Some Efficient Algorithms for Differential Equations

Some Efficient Algorithms for Differential Equations

By

Muhammad Usman 11-MS-MT-014

MS Thesis In Mathematics

HITEC University, Taxila Cantt. Taxila, Pakistan Spring, 2013 i

HITEC University, Taxila Cantt.

Some Efficient Algorithms for Differential Equations A Thesis Presented to

HITEC University, Taxila Cantt. In partial fulfillment of the requirement for the degree of

MS Mathematics

By

Muhammad Usman 11-MS-MT-014

Spring, 2013

ii

Some Efficient Algorithms for Differential Equations

A thesis submitted to the Department of Mathematics as partial fulfillment of the requirement for the award of Degree of MS Mathematics.

Name Muhammad Usman

Registration Number 11-MS-MT-014

Supervisor Prof. Dr. Syed Tauseef Mohyud-Din Department of Mathematics

HITEC University, Taxila Cantt.

iii

Final Approval

This thesis titled

Some Efficient Algorithms for Differential Equations By

Muhammad Usman 11-MS-MT-014 has been approved for the HITEC University Taxila Cantt.

Internal Examiner

: __________________________ Dr. Asif Waheed Department of Mathematics HITEC University Taxila Cantt.

External Examiner

: __________________________ Dr. Sarfraz Nawaz Malik Department of Mathematics GC University, Faisalabad

Supervisor

: __________________________ Prof. Dr. Syed Tauseef Mohyud-Din Department of Mathematics HITEC University, Taxila Cantt.

Chairperson

: __________________________ Prof. Dr. Syed Tauseef Mohyud-Din Department of Mathematics HITEC University, Taxila Cantt. iv

Declaration I, Muhammad Usman, 11-MS-MT-014 hereby declare that I have produced the work presented in this thesis, during the scheduled period of study. I also declare that I have not taken any material from any source except referred to and wherever due that amount of plagiarism is within acceptable range. If a violation of HEC rules on research has occurred in this thesis, I shall be liable to punishable action under the plagiarism rules of the HEC.

Date: ___________________ Signature of student:

Muhammad Usman

11-MS-MT-014

v

Certificate It is certified that Muhammad Usman 11-MS-MT-014 has carried out all the work related to this thesis under my supervision at the Department of Mathematics, HITEC University Taxila Cantt., and the work fulfills the requirement for award of MS degree.

Date: ____________________

Supervisor:

___________________________ Prof. Dr. Syed Tauseef Mohyud-Din Department of Mathematics

Chairperson:

__________________ Prof. Dr. Syed Tauseef Mohyud-Din Department of Mathematics HITEC University, Taxila Cantt.

vi

DEDICATION

Dedicated To

My Beloved Parents

vii

ACKNOWLEDGEMENT I bow my head before Him, Who is worth of all praise, The Creator of the universe and offer countless Darood and Salaams to my beloved Holy Prophet Hazrat Muhammad (PBUH), for Whom this universe has been manifested. I would like to thank Allah Almighty, Whose benediction bestowed upon me talented teachers, provided me sufficient opportunities and enabled me to undertake and execute this research work. My heartfelt appreciation goes to my affectionate, sincere, kind and most respected supervisor Prof. Dr. Syed Tauseef Mohyud-Din for so patiently bearing and guiding me with invaluably suggesting and continuously encouraging me with his precious contributions in completing this thesis. I am grateful for the constructive criticism as well as his encouraging comments. If not for his invaluable advice and guidance, this thesis would not have come to realization. He has been very kind in extending all possible help to make this work a success. His ideologies and concepts have a remarkable impact on my research contrivances. He genuinely facilitated me without which my objective would not have been obtained. I have learnt a lot from his abilities. He is my model as a teacher, a researcher, a supervisor and a mentor. I extend my deepest gratitude to my teachers Prof. Dr. Muhammad Tahir, Prof. Dr. Muhammad Rafiq, Dr. Saira Zainab and my internal examiner Dr. Asif Waheed for their moral support, kind comments and morale boosting attitude. I am grateful to all my teachers. They always guided me sincerely and honestly throughout my course work as well as research work. I would like to pay the most sincere gratitude to the Vice Chancellor HITEC University, Brig (R) Qamar Zaman for the provision of research oriented and conducive environment coupled with the appropriate facilities. His sincere efforts to streamline the system can be considered a role model. His personal interest in the subject proved to be the real driving force and the pivotal motivating factor behind the success of this work. It was his dynamic and proactive approach that enabled us to learn mathematics and do the research under the guidance of eminent mathematicians of national and international repute. viii

It is a little sad that I could not spare time for my old parents; though their long lasting prayers have opened new horizons for my success. I pay regards to my caring and loving parents, my sisters, my brother, whose sincere prayers and best wishes always make me courageous and daring throughout my life. I am also grateful to Sir Syed Adeel Akhtar Shah, Umer Iqbal, Muhammad Saleem, Nabila Afzal, Sidra Nazir and Aqsa Nazir for their guidance. Muhammad Usman 11-MS-MT-014

ix

ABSTRACT Some Efficient Algorithms for Differential Equations This work is devoted to the implementation and extension of some exiting schemes for differential equations. Moreover, in order to calculate various types of solutions including solitary wave solution, a wide range of new algorithms are also developed which proved extremely useful to cope with the complexity of physical problems. It is observed that proposed techniques are very efficient, reliable and accurate. Numerical results and graphical representations explicitly re-confirm the compatibility of suggested schemes.

x

TABLE OF CONTENTS

Introduction ___________________________________________

1

1. Preliminary Definitions__________________________________

5

1.1.

Introduction______________________________________

6

1.2.

Some Basic Definitions of Solitary Theory______________

6

1.2.1. Soliton_____________________________________

6

1.2.2. Solitary Wave_______________________________

6

1.2.3. Travelling Wave_____________________________

6

1.2.4. Types of Soliton Solutions_____________________

7

1.2.4.1. Solitary Wave and Solitons_____________

7

1.2.4.2. Kink Waves_________________________

8

1.2.4.3. Periodic Waves______________________

8

1.2.4.4. Peakons____________________________

9

1.2.4.5. Cuspons____________________________

9

1.2.4.6. Compactons_________________________

10

Preliminaries of Fractional Calculus___________________

11

1.3.1. History_____________________________________

11

1.3.2. Fractional Derivative of a Basic Power Function____

12

1.3.3. Laplace Transform __________________________

13

1.3.4. Fractional Integral____________________________

14

1.3.4.1. Riemann-Liouville Fractional Integral_____

14

1.3.4.2. Erdélyi-Kober Integral Operator__________

15

1.3.4.3. Hadamard Fractional Integral____________

15

1.3.5. Fractional Derivative__________________________

15

1.3.5.1. Riemann-Liouville Fractional Derivative___

15

1.3.

1.3.5.2. Modified Riemann-Liouville Fractional Derivative___________________________ xi

16

1.4.

1.3.5.3. Caputo Fractional Derivative_____________

16

1.3.5.4. Jumarie Fractional Derivative____________

16

Introductory Concepts of Integral Equations______________

17

1.4.1. Integral Equations_____________________________

17

1.4.2. Classification of Integral Equations_______________

18

1.4.2.1. Fredholm Integral Equations_____________

18

1.4.2.2. Volterra Integral Equations______________

18

1.4.2.3. First Kind of Fredholm Integral Equations__

18

1.4.2.4. First Kind of Volterra Integral Equations___

18

1.4.2.5. Second Kind of Fredholm Integral Equations_

18

1.4.2.6. Second Kind of Volterra Integral Equations__

19

1.4.3. Homogeneous Fredholm Integral Equations_________

19

1.4.4. Homogeneous Volterra Integral Equations__________

19

1.4.5. Integro-Differential Equations____________________

19

1.4.5.1. Fredholm Integro-Differential Equations____

19

1.4.5.2. Volterra Integro-Differential Equations_____

19

1.4.6. Linearity Concept_____________________________

19

1.4.7. Homogeneity Concept__________________________

20

1.4.8. Singular Integral Equations______________________

20

1.4.8.1. Weakly Singular Integral Equation________

20

1.4.8.2. Hypersingular Integral Equation__________

21

1.4.9. System of Integral Equations___________________

1.5.

21

1.4.9.1. System of Fredholm Integral Equations____

21

1.4.9.2. System of Volterra Integral Equations_____

21

1.4.9.3. System of Singular Integral Equations _____

22

Some Basic Definitions of Wavelets Theory______________

23

1.5.1. History______________________________________

23

1.5.2. Wavelets____________________________________

23

1.5.3. Types of Wavelets_____________________________

23

xii

1.5.4. Functions Approximation ______________________

24

2. Approximate Solutions of Differential Equations_____________

25

2.1.

Introduction_______________________________________

26

2.2.

Adomian’s Decomposition Method____________________

27

2.2.1. Methodology________________________________

27

2.2.2. Micro Polar Flow in a Porous Channel with Mass Injection____________________________________

28

2.2.3. Squeezing Flow and Heat Transfer between Two Parallel

2.3.

2.4.

2.5.

Disks with Velocity Slip and Temperature Jump _____

34

Modified Adomian’s Decomposition Method_____________

42

2.3.1. Methodology_________________________________

42

2.3.2. Goursat Equation______________________________

43

2.3.3. Fourth-order Parabolic Equation__________________

45

2.3.4. Wave Equation_______________________________

50

Optimal Adomian’s Decomposition Method______________

54

2.4.1. Methodology_________________________________

54

2.4.2. Fredholm Equation____________________________

54

Variational Iteration Method__________________________

71

2.5.1. Methodology_________________________________

72

2.5.2. MHD Flow over a Nonlinear Stretching Sheet_______

73

2.5.3. MHD Flow of an Incompressible Viscous Fluid through Convergent or Divergent Channels in Presence of a High

2.6.

Magnetic Field________________________________

75

Modified Variational Iteration Method__________________

83

2.6.1. Methodology_________________________________

84

2.6.2. Nonlinear Partial Differential Equation_____________

84

2.6.3. Inhomogeneous Burger’s Equation________________

85

2.6.4. Homogeneous Burger’s Equation_________________

86

2.6.5. Diffusion Equation____________________________

87

xiii

2.7.

2.6.6. Homogeneous Fourth-order Parabolic Equation_____

89

2.6.7. Inhomogeneous Fourth-order Parabolic Equation____

90

2.6.8. Inhomogeneous Diffusion Equation_______________

91

2.6.9. Parabolic Equation____________________________

92

2.6.10. Nonlinear Parabolic Equation____________________

94

2.6.11. Reaction-Diffusion Equation_____________________

95

Homotopy Perturbation Method _______________________

99

2.7.1. Methodology_________________________________

99

2.7.2. Flow of a Viscoelastic Fluid through a Porous Channel with Expanding or Contracting Walls__________________

100

2.7.3. MHD Flow of an Incompressible Viscous Fluid through Convergent or Divergent Channels in Presence of a High

2.8.

Magnetic Field_______________________________

105

Modified Homotopy Perturbation Method_______________

111

2.8.1. Methodology_________________________________

112

2.8.2. Nonlinear Dispersive K(2,2,1) Equation____________

112

2.8.3. Nonlinear Dispersive K(3,3,1) Equation____________

113

2.8.4. Kawahara Equation____________________________

115

2.8.5. Nonlinear Dispersive K(2,2) Equation_____________

124

2.8.6. Nonlinear Dispersive K(3,3) Equation_____________

127

2.8.7. (2+1)-Dimensional Nonlinear Dispersive K(2,2)

2.9.

Equation_____________________________________

130

2.8.8. Homogeneous Convection-Diffusion Problem_______

134

Homotopy Analysis Method___________________________

141

2.9.1. Methodology_________________________________

142

2.9.2. Goursat Problem______________________________

143

2.9.3. System of Partial Differential Equations____________

153

2.9.4. K(m,p,1) Equation_____________________________

163

2.9.5. K(2,2,1) Equation_____________________________

165

xiv

2.9.6. Nonlinear Schrodinger Equation__________________

167

2.9.7. Veriational Problem____________________________

170

2.10. Modified Homotopy Analysis Method___________________

176

2.10.1. Methodology_________________________________

176

2.10.2. Helmholtz Equation____________________________

176

2.10.3.Seventh-order Generalized Korteweg-deVries Equation

179

2.10.4.Cubic Nonlinear Schrodinger Equation_____________

182

2.10.5. Heat and Wave-like Equation____________________

188

2.10.6.System of Partial Differential Equations____________

195

2.10.7.Zakharov Kuznetsov (2,2,2) Equation______________

201

2.10.8.Zakharov Kuznetsov (3,3,3) Equation______________

205

2.11. Variation of Parameters Method________________________

209

2.11.1. Methodology_________________________________

209

2.11.2. MHD Flow of an Incompressible Viscous Fluid through Convergent or Divergent Channels in Presence of a High Magnetic Field________________________________

210

2.11.3. Flow of a Viscoelastic Fluid through a Porous Channel with Expanding or Contracting Wall___________________

216

3. Traveling Wave Solutions of Nonlinear Evolution Equations____

219

3.1.

Introduction_______________________________________

220

3.2.

Exp-function Method______________________________

220

3.2.1. Methodology_________________________________

221

3.2.2. Korteweg-deVries Equation_____________________

221

3.2.3. Burger’s Equation_____________________________

223

3.2.4. Burger’s Hierarchy in (2+1)-Dimensional Equation___

226

3.2.5. Shallow Water Wave Equation___________________

231

3.2.6. Modified Generalized Degasperis–Procesi Equation__

234

3.2.7. System of Nonlinear Volterra Integral Equations_____

238

Modified Exp-function Method________________________

240

3.3.

xv

3.4.

3.3.1. Methodology_________________________________

240

3.3.2. Pochhammer-Chree (PC) Equation________________

241

3.3.3. (2+1)-Dimensional Jimbo-Miwa Equation__________

244

3.3.4. Generalized Nonlinear Heat Conduction Equation____

246

3.3.5. Generalized Nonlinear Heat Equation______________

251

3.3.6. Konopelchenko-Dubrovsky Equation______________

256

3.3.7. Long Short Wave Interaction System______________

258

3.3.8. Nizhnik-Novikov-Veselov Equations______________

261

3.3.9. System of Zakhrov____________________________

263

3.3.10. Nonlinear Parabolic Equation____________________

265

(G´/G)-Expansion Method____________________________

269

3.4.1. Methodology_________________________________

270

3.4.2. Sine-Gordon Equation__________________________

271

3.4.3. Doubly Sine-Gordon Equation___________________

276

3.4.4. (2+1)-Dimensional Sine-Gordon Equation__________

278

3.4.5. Klein Gordon Equation_________________________

283

3.4.6. Potential Kadomtsev-Petviashvili Equation_________

285

3.4.7. Nonlinear Wave Equation of Longitudinal

3.5.

Oscillation___________________________________

288

3.4.8. Nonlinear Schrö dinger Equation__________________

290

3.4.9. Vankhnenko Equation__________________________

293

3.4.10. Complex Equation____________________________

295

3.4.11. Kadomtsev-Petviashvili Equation ________________

297

F-Expansion Method________________________________

300

3.5.1. Methodology_________________________________

300

3.5.2. (2+1)−Dimensional Davey-Stewartson System______

305

3.5.3. Higher Dimensional Schrö dinger Equation_________

308

3.5.4. (2+1)-Dimensional Broer–Kaup Kupershmidt System_

312

3.5.5. Drinfeld-Sokolov Equation______________________

315

xvi

3.6.

3.5.6. Hirota–Satsuma Coupled Korteweg-deVries System__

317

3.5.7. Long-Short-Wave Interaction System______________

322

Rational Sinh-Cosh Method___________________________

325

3.6.1. Methodology_________________________________

325

3.6.1.1. The Rational Sinh and Cosh Functions Methods_____________________________

326

3.6.1.2. The Rational Exponential Functions Methods_____________________________

326

3.6.2. (2+1)-Dimensional Breaking Soliton Equation_______

326

3.6.3. (2+1)-Zakharov-Kuznetsov Equation______________

332

3.6.4. Benjamin-Bona-Mahony Equation________________

335

3.6.5. Good Boussinesq Equation______________________

338

3.6.6. Regularized Long Width (RLW) Equation__________

340


343

3.6.8. Zakharov Kuznetsov-Benjamin Bona Mahony Equation

345

3.6.9. Kadomtsev Petviashvili-Benjamin Bona Mahony Equation_____________________________________

3.7.

348

3.6.10. Modified Korteweg-deVries Equation_____________

351

(U´/U)-Expansion Method____________________________

354

3.7.1. Methodology_________________________________

354

3.7.2. Lax Seventh-order Equation_____________________

355

3.7.3. Cahn-Hilliard Equation_________________________

357

3.7.4. The Kaup-Kuperschmidt Equation________________

358

3.7.5. Sawada-Kotera Equation________________________

359

3.7.6. Sawada-Kotera-Ito Seventh-order Equation_________

361

3.7.7. (3+1)-Dimensional Korteweg-deVries Equation_____

362

3.7.8. Kaup-Kuperschmidt Seventh-order Equation________

363

3.7.9. (2+1)-Dimensional Nonlinear Kadomtsev PetviashviliBenjamin Bona Mahony Equation________________ xvii

365

3.8.

Extended Tanh Method______________________________

366

3.8.1. Methodology_________________________________

366

3.8.2. Benjamin-Bona-Mahony Equation________________

368

3.8.3. Kadomtsov-Petviashivilli Equation_______________

371

3.8.4. Nonlinear Elastic Structural Element of a large Defection

3.9.

Equation____________________________________

375


381

3.8.6. (2+1)-Dimensional Burger’s Equation_____________

386

3.8.7. (3+1)-Dimensional Burger’s Equation_____________

389


394

3.8.9. Modified Zakharov-Kuznetsov Equation___________

399

3.8.10.Modified Equal-Width Equation__________________

402

New Approach of (G´/G)-Expansion Method_____________

405

3.9.1. Methodology_________________________________

405

3.9.1.1. New Approach of (G´/G) – Expansion Method______________________________

406

3.9.1.2. New Approach of generalized (G´/G) – Expansion Method______________________________

408


408

3.10. Combined Tanh-Coth (CTC) and Combined Sinh-Cosh (CSC) Method ___________________________________________

414

3.10.1. Methodology_________________________________

414

3.10.1.1. Combined Tanh-Coth (CTC) Method_____

415

3.10.1.2. Combined Sinh-Cosh (CSC) Method______

415

3.10.2. Generalized Korteweg-deVries Equation with Two Power Nonlinearities________________________________

416

3.10.3. Klein-Gordon-Zakharov System__________________

418

3.11. U-Expansion Method________________________________

422

3.11.1. Methodology_________________________________

422

xviii

3.11.2. Cubic Ginzburg-Landau Equation_________________

423

3.11.3. Camassa-Holm Equation________________________

425

3.11.4. Schrö dinger Equation of Cubic Nonlinearity________

431

3.11.5. Schrodinger Equation with Power Law Nonlinearity__

434

3.11.6. Sine-Gordon Equation_________________________

436

3.11.7. Third Variant of Korteweg-deVries Equation________

441

3.11.8. (2+1)-Dimensional Burgers Equation______________

442

3.11.9. Medium Equal Width Equation___________________

444

4. Traveling Wave Solutions Nonlinear Evolution Equation of Fractionalorder _________________________________________________

447

4.1.

Introduction_______________________________________

448

4.2.

F-Expansion Method________________________________

448

4.2.1. Methodology_________________________________

449

4.2.2. The Zakharov Kuznetsov-Modified Equal Width Equation_____________________________________

453

4.2.3. The Korteweg-deVries Equation__________________

456

4.2.4. Schamel Korteweg–deVries Equation_____________

458

4.2.5. (2+1) Dimensional Broer–Kaup Kupershmidt

4.3.

System______________________________________

462

Generalized Tanh Method____________________________

466

4.3.1. Methodology_________________________________

466

4.3.2. (2+1) Korteweg-deVries Equation________________

468

4.3.3. Combined Korteweg-deVries and Modified Korteweg-

4.4.

deVries Equation______________________________

470

4.3.4. Konopelchenko-Dubrovsky Equation______________

472

4.3.5. Newel Whiched Equation_______________________

474

4.3.6. Boiti-Leon-Pempinelli Equation__________________

476

(G’/G)-Expansion Method Based on Riccati Equation_______

478

4.4.1. Methodology_________________________________

479

xix

4.5.


481


483

4.4.4. Fisher’s Equation______________________________

484

4.4.5. Huxley Equation______________________________

486

4.4.6. Hirota Ramani Equation________________________

488

4.4.7. Breaking Soliton Equation______________________

489

4.4.8. Benjamin Bona Mahony Equation________________

491

4.4.9. Generalized Benjamin Bona Mahony Equation______

492

New Approach of (G’/G)-Expansion Method______________

494

4.5.1. Methodology_________________________________

494

4.5.2. Calogero–Bogoyavlenskii–Schiff Equation_________

496

4.5.3. (3 + 1) Dimensional Potential- Yu Toda Sasa Fukuyama Equation_____________________________________

4.6.

498

4.5.4. Symmetric Regularized Long Wave Equation_______

500

4.5.5. Modified Equal Width Equation__________________

506

4.5.6. Sharma–Tasso–Olver (STO) Equation_____________

511

4.5.7. Sine-Gordon Equation__________________________

516

4.5.8. Boussinesq Equation___________________________

521


526

4.5.10.Klein Gordon Equation_________________________

531

4.5.11.Burger Hierarchy Equation______________________

535

Rational Hyperbolic Function (RHF) Method_____________

538

4.6.1. Methodology_________________________________

539


539


541

4.6.4. Newel Whiched Equation__________________ _____

542

4.6.5. Zakharov Kuznetsov-Modified Equal Width Equation

543

4.6.6. Fisher’s Equation_____________________________

545

4.6.7. Gas Dynamical Equation_______________________

546

xx

4.7.

4.8.

4.9.


547

4.6.9. Nizhnik-Novikov-Vesselov System_______________

549

4.6.10. Konopelchenko-Dubrovsky Equation_____________

552

4.6.11. Nonlinear System_____________________________

554

U-expansion Method________________________________

556

4.7.1. Methodology_________________________________

558

4.7.2. Fifth-order Caudrey Dodd Gibbon Equation_________

559

4.7.3. Fifth-order Ito Equation________________________

566

4.7.4. Fifth-order Kaup Kuperschmidt Equation__________

570

4.7.5. Fifth-order Lax Equation________________________

573

4.7.6. Seventh-order Kaup Kuperschmidt Equation________

580

4.7.7. Seventh-order Lax Equation_____________________

583

4.7.8. Ninth-order Sawada Kotera Equation______________

589

4.7.9. Fifth-order Sawada Kotera Equation_______________

598

4.7.10. Seventh-order Sawada Kotera Equation____________

605

Modified U-expansion Method________________________

612

4.8.1. Methodology_________________________________

612

4.8.2. (2+1) Dimensional Equation_____________________

614

4.8.3. (3+1) Dimensional Burger’s Equation_____________

619

4.8.4. (2+1) Dimensional Burger’s Equation_____________

623


626

4.8.6. (2+1) Dimensional FitzHugh-Nagumo Equation_____

632

4.8.7. Klein Gordon Method__________________________

639

4.8.8. Modified Kawahara Equation____________________

644


647

4.8.10. Zakharov Kuznetsov Equation___________________

650

4.8.11. Regularized Long Width Equation________________

656

Generalized U-expansion Method______________________

658

4.9.1. Analysis of Generalized U-expansion Method_______

658

xxi

4.9.2. Negative Gardener Equation_____________________

659


663

4.9.4. (2+1) Dimension Burger’s Equation_______________

665

4.9.5. (3+1) Dimension Burger’s Equation_______________

668

4.9.6. Burger’s Equation_____________________________

672


675

4.9.8. FitzHugh-Nagumo Equation_____________________

677


683

4.9.10. Generalized Benjamin Bona Mahony Equation______

688

4.9.11. Kadomtsev Petviashvili-Benjamin Bona Mahony Equation_____________________________________

693

4.10. (U’/U)-Expansion Method____________________________

695

4.10.1. Methodology_________________________________

696

4.10.2. Caudrey Dodd Gibbon (CDG) Equation____________

698

4.10.3. Ito Equation__________________________________

701

4.10.4. Kaup Kuperschmidt Equation____________________

703

4.10.5. Lax Equation_________________________________

705

4.10.6. Ninth-order Sawada Kotera (SK) Equation__________

710

4.10.7. Fifth-order Kaup Kuperschmidt Equation___________

714


717

4.10.9. Sawada Kotera (SK) Equation____________________

720


723

4.11. (G𝛂/G)-Expansion method____________________________

727

4.11.1. Analysis of (G𝛂/G)-Expansion method_____________

728


730

4.11.3. Boussinesq Equation___________________________

732

4.11.4. Breaking Soliton Equation______________________

734

4.11.5. Generalized Shallow Water Wave Equation_________

736


738

xxii

4.11.7.Goursat Equation______________________________

740


742

4.11.9.Benjoman Bona Mahony Equation_______________

744

4.11.10.Korteweg-de-Vries Equation___________________

746

4.11.11.Modified Korteweg-deVries Equation____________

749

5. Numerical and Exact Solutions of Linear and Nonlinear Integral Equations______________________________________________

752

5.1.

Introduction_______________________________________

753

5.2.

New Scheme Using Riemann-Liouville Fractional Integral__

754

5.2.1. Methodology_________________________________

754

5.2.2. Abel’s Integral Equation________________________

755

5.3.

5.4.

5.5.

5.6.

5.7.

New Alternative Scheme Using Riemann-Liouville Fractional Integral and Derivative_______________________________

759

5.3.1. Methodology_________________________________

759

5.3.2. Abel’s Integral Equations_______________________

760

New Scheme Using Generalized Riemann-Liouville Fractional Integral and Derivative_______________________________

769

5.4.1. Methodology_________________________________

769


770

New Scheme Using Taylor’s Series_______________________

772

5.5.1. Methodology_________________________________

772


772

Adomian’s Decomposition Method Using Fractional Calculus_________________________________________

776

5.6.1. Methodology_________________________________

776

5.6.2. Weakly Singular Integral Equations_______________

777

Modified Adomian’s Decomposition Method Using Fractional Calculus__________________________________________

781

5.7.1. Methodology_________________________________

781

xxiii

5.8.


782

Kravchuk Polynomials Method (KPM)__________________

785

5.8.1. Methodology_________________________________

785

5.8.1.1. Integral Equation of the 1st Kind__________

785

5.8.1.2. Integral Equation of the 2nd Kind__________

786

5.8.2. Volterra Integral Equation_______________________

787

5.8.3. Fredholm Integral Equation______________________

791

5.8.4. Volterra Integro-Differential Equation_____________

793

5.8.5. Volterra-Fredholm Integral Equation______________

797

5.8.6. Nonlinear Abel’s Integral Equation________________

802

5.8.7. Weakly Singular Integral Equation________________

805

5.8.8. System of Fredholm Integral Equations____________

806

5.8.9. System of Weakly Singular Integral Equations of 1st Kind_____________________________________

807

Newton’s Polynomials Method (NPM)__________________

809

5.9.1. Methodology_________________________________

810

5.9.1.1. Integral Equation of the First Kind________

810

5.9.1.2. Integral Equation of the Second Kind______

811

5.9.2. Volterra Integral Equations______________________

811

5.9.3. Fredholm Integral Equations_____________________

815


819


827


837


839

5.9.8. System of Volterra Integral Equations of 1st Kind____

842

5.9.9. System of Volterra Integral Equations_____________

849

5.10. Touchard Polynomials Method (TPM)__________________

850

5.10.1. Methodology_________________________________

851


851

5.9.

xxiv

5.10.1.2. Integral Equation of the Second Kind_____

852


852


856


860


864


869


872


874


878

5.11. Zernike Polynomial Method (ZPM)_____________________

880

5.11.1. Methodology_________________________________

880


880


881


882


885


890


894

5.11.6. Nonlinear Abel’s Integral Equations_______________

900


903


905


908

5.12. Meixner Polynomial Method (MPM)____________________

910

5.12.1. Methodology_________________________________

910


910


911


912

5.12.3. Abel’ Integral Equation_________________________

915


916

5.12.5. System of Volterra Integral Equation______________

921

xxv

5.13. Rook Polynomials Method (RPM)______________________

926

5.13.1. Methodology_________________________________

927

5.13.1.1. Integral Equation of the First Kind_______

927


928


928


933


938


943


945


950

5.13.8. System of Volterra Integral Equation______________

955

5.14. Fibonicca Polynomials Method (FPM)_____________ _____

960

5.14.1. Methodology_________________________________

961

5.14.1.1.

Integral Equation of the First Kind_______

961


962

5.14.2.Volterra Integral Equations______________________

962


968

5.14.4. Fredholm Integral Integral Equation_______________

969


973

5.15. Gegenbauer Polynomials Method (GPM)________________

976

5.15.1. Methodology_________________________________

976


976


977


978


983

5.15.4. Fredholm Integral Integral Equation_______________

984


987

5.16. Charlier's Polynomials Method (CPM)__________________

989

5.16.1. Methodology_________________________________

990

xxvi


990


991


991

5.16.3. Volterra Integral Equation_______________________

993

5.16.4. Weakly-Singular Integral Equation________________

994


995

5.17. Bessel's Polynomials Method (BPM)____________________

996

5.17.1. Methodology_________________________________

997


997


998


998

5.17.3. Volterra Integral Equation_______________________ 1000 5.17.4. Weakly-Singular Integral Equation________________ 1001 5.17.5. Abel’s Integral Equation________________________

1002

5.18. Bell Polynomials Method (BPM)_______________________ 1003 5.18.1. Methodology_________________________________

1003

5.18.1.1. Integral Equation of the First Kind________ 1003 5.18.1.2. Integral Equation of the Second Kind_____

1004

5.18.2. Volterra Integral Equations______________________ 1005 5.18.3. Weakly Singular Integral Equation________________ 1009 5.18.4. Fredholm Integral Equation______________________ 1012 5.18.5. Nonlinear Abel’s Integral Equation________________ 1013 5.19. Appell Polynomials Method (APM)_____________________ 1014 5.19.1. Methodology_________________________________

1014

5.19.1.1. Integral Equation of the First Kind________ 1014 5.19.1.2. Integral Equation of the Second Kind_____

1015

5.19.2. Fredholm Integral Equation______________________ 1016 5.19.3. Volterra Integral Equation_______________________ 1019 5.19.4. Abel’s Integral Equation________________________ xxvii

1023

6. Wavelets Theory________________________________________

1025

6.1.

Introduction_______________________________________

1026

6.2.

Physicists Hermite Wavelet Method (PHWM)____________

1027

6.2.1. Methodology_________________________________

1027

6.2.1.1. Physicists Hermite Polynomials___________ 1027 6.2.1.2. Wavelets and Physicists Hermite Wavelets__ 1027 6.2.1.3. Functions Approximation_______________

1028

6.2.1.4. Analysis_____________________________

1028

6.2.2. Singular Differential Equations___________________ 1029 6.3.

Newton’s Wavelet Method (NWM)_____________________ 1038 6.3.1. Methodology_________________________________

1038

6.3.1.1. Newton’s Polynomials__________________

1038

6.3.1.2. Wavelets and Newton’s Wavelets_________

1039

6.3.1.3. Functions Approximation________________ 1039 6.3.1.4. Analysis_____________________________

1040

6.3.2. Fifth-order Boundary Value Problems (BVPs)_______ 1040

6.4.

6.5.

6.3.3. Sixth-order Boundary Value Problems (BVPs)______

1046

6.3.4. Ninth-order Boundary Value Problems (BVPs)______

1052

6.3.5. Tenth-order Boundary Value Problems (BVPs)______

1058

6.3.6. Twelfth-order Boundary Value Problems (BVPs)____

1064

Laurent Wavelet Method (LWM)______________________

1070

6.4.1. Methodology________________________________

1070

6.4.1.1. Laurent Polynomials___________________

1070

6.4.1.2. Wavelets and Laurent Wavelets__________

1071

6.4.1.3. Functions Approximation_______________

1071

6.4.1.4. Analysis_____________________________

1072

6.4.2. System of Differential Equation of 2×2____________

1072

6.4.3. System of Differential Equation of 3×3____________

1100

Kravchuk Wavelet Method (KWM)_____________________ 1128 xxviii

6.5.1. Methodology_________________________________

6.6.

1128

6.5.1.1. Kravchuk Polynomials_________________

1128

6.5.1.2. Wavelets and Kravchuk Wavelets________

1128


1129

6.5.1.4. Analysis_____________________________

1129

6.5.2. Volterra Integral Equation of 2nd Kind_____________

1130

6.5.3. Fredholm Integral Equation_____________________

1135


1146

Lommal Wavelet Method (LWM)______________________ 1158 6.6.1. Methodology_________________________________

1158

6.6.1.1. Lommal Polynomials___________________

1158

6.6.1.2. Wavelets and Lommal Wavelets__________

1159

6.6.1.3. Functions Approximation________________ 1159

6.7.

6.6.1.4. Analysis_____________________________

1160

6.6.2. Delay Differential Equations (DDEs)______________

1160

Bernoulli’s Wavelet Method (BWM)____________________ 1182 6.7.1. Methodology_________________________________

1182

6.7.1.1. Bernoulli’s Polynomials_________________ 1182 6.7.1.2. Wavelets and Bernoulli’s Wavelets________

1183

6.7.1.3. Functions Approximation________________ 1183 6.7.1.4. Analysis_____________________________

1184

6.7.2. Second-order Differential Equations with Robin

6.8.

Conditions___________________________________

1184

Probabilist Hermite Wavelet Method (PWM)_____________

1214

6.8.1. Methodology_________________________________

1215

6.8.1.1. Probabilist Hermite Polynomials_________

1215

6.8.1.2. Wavelets and Probabilist Hermite Wavelets_____________________________

1215

6.8.1.3. Functions Approximation________________ 1216 xxix

6.8.1.4. Analysis_____________________________

1216

6.8.2. Nonlinear Integral Equations of Second Kind________ 1217 6.8.3. Nonlinear Integral Equations of First Kind__________ 1228 6.9.

Zernike Wavelet Method (ZWM)_______________________ 1242 6.9.1. Methodology_________________________________

1242

6.9.1.1. Zernike Polynomials___________________

1242

6.9.1.2. Wavelets and Zernike Wavelets__________

1242


1243

6.9.1.4. Analysis_____________________________

1243

6.9.2. First-order Stiff Delay Differential Equation________

1244

6.9.3. Third-order Difference Differential Equation________ 1250 6.9.3. Delay Differential Equation_____________________

1256

6.9.3. Delay Integro-Differential Equation_______________

1262

6.9.5. System of Stiff Delay Differential Equation_________ 1268 6.10. Rook Wavelet Method (RWM)________________________

1278

6.10.1. Methodology_________________________________

1278

6.10.1.1. Rook Polynomials____________________

1278

6.10.1.2. Wavelets and Rook Wavelets____________ 1279 6.10.1.3. Operational Matrix for Integration and Derivative of Fractional-order____________________

1279

6.10.1.4. Functions Approximation______________

1281

6.10.1.4. Analysis____________________________

1281

6.10.2. Differential Equations of Fractional-order__________

1282

6.11. Touchard Wavelet Method (TWM)_____________________

1291

6.11.1. Methodology_________________________________

1291

6.11.1.1. Touchard Polynomials_________________

1291

6.11.1.2. Wavelets and Touchard Wavelets________

1291

6.11.1.3. Functions Approximation_______________ 1292 6.11.1.4. Analysis____________________________ xxx

1292

6.11.2. Homogenous Cauchy Problem___________________

1293

6.11.3. Non-homogenous Cauchy Problem_______________

1300

6.11.4. Transport Equation____________________________

1308

7. Conclusion_____________________________________________

1311

8. References_____________________________________________

1312

xxxi

LIST OF ABBRIVATION

PDEs

Partial Differential Equations

DEs

Differential Equations

NLEEs

Nonlinear Linear Evolution Equations

RL

Riemann-Liouville

ADM

Adomian’s Decomposition Method

MADM

Modified Adomian’s Decomposition Method

OADM

Optimal Adomian’s Decomposition Method

VIM

Variational Iteration Method

MVIM

Modified Variational Iteration Method

HPM

Homotopy Perturbation Method

MHPM

Modified Homotopy Perturbation Method

HAM

Homotopy Analysis Method

MHAM

Modified Homotopy Analysis Method

VPM

Variation of Parameter Method

BPM

Bernoulli’s Polynomial Method

NPM

Newton’s Polynomial Method

KPM

Kravchuk Polynomial Method

TPM

Touchard Polynomial Method

ZPM

Zernike Polynomial Method

MPM

Meixner Polynomial Method

RPM

Rook Polynomial Method

FPM

Fibonicca Polynomial Method

GPM

Gegenbauer Polynomial Method

CPM

Charlier’s Polynomial Method

BsPM

Bessel Polynomial Method

BPM

Bell Polynomial Method xxxii

APM

Appell Polynomial Method

PHWM

Physicists Hermite Wavelet Method

NWM

Newton’s Wavelet Method

LWM

Laurent Wavelet Method

KWM

Kravchuk Wavelet Method

LmWM

Lommal Polynomial Method

BWM

Bernoulli’s Polynomial Method

PWM

Probabilist Hermite Polynomial Method

ZWM

Zernike Polynomial Method

RWM

Rook Polynomial Method

TWM

Touchard Polynomial Method

DDE

Delay Differential Equation

BVPs

Boundary Value Problems

FDEs

Fractional Differential Equations

IVPs

Initial Value Problems

IE

Integral Equation

VIE

Volterra Integal Equation

FIV

Fredholm Integral Equation

2D

Two Dimensional

3D

Three Dimensional

Γ

Gamma Function

xxxiii

LIST OF FIGURES Fig 2.1 The Effect of Reynolds Number on Velocity Component in 𝑥 −direction...

34

Fig 2.2 The Effect of Reynolds Number on 𝑓(𝜂) when 𝑅𝑒 varies…………..……..

34

Fig 2.3 Effect of Prandtl Number 𝑃𝑟 on Temperature Profile……………………...

38

Fig 2.4 Effect of Magnetic Parameter 𝑀 on the Radial Velocity…………………..

39

Fig 2.5 Effect of Temperature Slip Parameter 𝛾 on Temperature Profile………….

39

Fig 2.6 Effect of Eckert number 𝐸𝑐 on temperature Profile………………………..

40

Fig 2.7 Effect of Squeeze Parameter on the Radial Velocity………………………

40

Fig 2.8 Comparison of Exact and Approximate Solutions of Eq. (2.36)…....……..

47

Fig 2.9 Comparison of Exact and Approximate Solutions of Eq. (2.37)…...………

48

Fig 2.10 Comparison of Exact and Approximate Solutions of Eq. (2.38)………….

50


53

Fig 2.12 (a)-(b) Comparison of Exact, ADM and OADM Solution of Eq. (2.40)…

56


58


60


62


64


66


69


71

Fig 2.20 VIM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝛼 = −2.50 …………………………………………………………………………..

78

Fig 2.21 VIM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝛼 = 2.50 …………………………………………………………………………….

78

Fig 2.22 VIM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝛼 = 50 ………………………………………………………………………………

79

Fig 2.23 VIM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝐻 = 1500…………………………………………………………………………..

79

Fig 2.24 VIM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝐻 = 1500………………………………………………………………………….. xxxiv

80

Fig 2.25 VIM Solution for Velocity is Convergent Channel for 𝛼 = 2.50 and 𝐻 = 1500…………………………………………………………………………..

80

Fig 2.26 VIM Solution for Velocity is Convergent Channel for 𝛼 = −2.50 and 𝐻 = 1500………………………………………………………………………….

81

Fig 2.27 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (2.73)….

85


86


87


88


90


91


92


94


95


96


98


99

Fig 2.39 Flow of a Viscoelastic Fluid through a Porous Channel with Expanding or Contracting Walls………………………………………………………………..

101

Fig 2.40 Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for Different Values of Re with 𝜔 = 0.02 and 𝛼 = 2………………………………………………………………..

103

Fig 2.41 Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for Different Values of Re with 𝜔 = 0.02 and 𝛼 = −2……………………………………………………………...

104

Fig 2.42 Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for Different Values of 𝛼 with 𝜔 = 0.02 and Re = 2.5…………………………………………………………….

104

Fig 2.43 Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for Different Values of 𝜔 with 𝛼 = 2 and Re = −5………………………………………………………………………..

104

Fig 2.44 HPM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝛼 = −2.50 …………………………………………………………………………..

106

Fig 2.45 HPM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝛼 = 2.50 ……………………………………………………………………………. Fig 2.46 HPM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and xxxv

107

𝛼 = 50 ………………………………………………………………………………

107

Fig 2.47 HPM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝐻 = 1500…………………………………………………………………………..

108

Fig 2.48 HPM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝐻 = 1500…………………………………………………………………………..

108

Fig 2.49 HPM Solution for Velocity is Convergent Channel for 𝛼 = 2.50 and 𝐻 = 1500…………………………………………………………………………..

109

Fig 2.50 HPM Solution for Velocity is Convergent Channel for 𝛼 = −2.50 and 𝐻 = 1500…………………………………………………………………………..

109

Fig 2.51 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (2.109)...

113


115

Fig 2.53 (a)-(f) Comparison of Exact and Approximate Solutions of Eq. (2.122) for different values of 𝑐 and 𝑡……………………………………………………...

127

Fig 2.54 (a)-(f) Comparison of Exact and Approximate Solutions of Eq. (2.125) for different values of 𝑐 and 𝑡………………………………………………………

130

Fig 2.55 (a)-(i) Comparison of Exact and Approximate Solutions of Eq. (2.128) for different values of 𝑦 and 𝑡………………………………………………………

133

Fig 2.56 (a)-(d) Comparison of Exact and Approximate Solutions of Eq. (2.131)...

135


137


139


141

Fig 2.60 Comparison of Exact and Approximate Solutions of Eq. (2.139)………...

145

Fig 2.61 Comparison of Exact and Approximate Solutions of Eq. (2.140) ………..

146


148


150


151


153

Fig 2.66 (a)-(d) Comparison of Exact and Approximate Solutions of Eq. (2.1452.146)……………………………………………………………………………….

156

Fig 2.67 (a)-(f) Comparison of Exact and Approximate Solutions of Eq. (2.1472.149) ……………………………………………………………………………… xxxvi

159

Fig 2.68 (a)-(f) Comparison of Exact and Approximate Solutions of Eq. (2.1502.152)……………………………………………………………………………….

163


165


167


171


173


175


177


179


182


188


190


192


194

Fig 2.81 (a)-(d) Comparison of Exact and Approximate Solutions of Eq. (2.1742.175)……………………………………………………………………………….

196


203


205


207


209

Fig 2.86 VPM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝛼 = −2.50 …………………………………………………………………………..

211

Fig 2.87 VPM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝛼 = 2.50 …………………………………………………………………………….

212

Fig 2.88 VPM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝛼 = 50 ………………………………………………………………………………

212

Fig 2.89 VPM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝐻 = 1500…………………………………………………………………………..

213

Fig 2.90 VPM Solution for Velocity is Convergent Channel for 𝑅𝑒 = 100 and 𝐻 = 1500………………………………………………………………………….. Fig 2.91 VPM Solution for Velocity is Convergent Channel for 𝛼 = 2.50 and xxxvii

213

𝐻 = 1500…………………………………………………………………………..

214

Fig 2.92 VPM Solution for Velocity is Convergent Channel for 𝛼 = −2.50 and 𝐻 = 1500…………………………………………………………………………..

214

Fig 2.93 Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for Different Values of 𝑅𝑒 with 𝜔 = 0.02 and 𝛼 = 2………………………………………………………………..

217

Fig 2.94 Characteristics of 𝑓(𝜂) and 𝑓 ′ (𝜂) for Different Values of 𝑅𝑒 with 𝜔 = 0.02 and 𝛼 = −2……………………………………………………………...

218

Fig 2.95 Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for Different Values of 𝑅𝑒 with 𝜔 = 0.02 and 𝑅𝑒 = 2.5…………………………………………………………….

218

Fig 2.96 Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for Different Values of 𝑅𝑒 with 𝜔 = 0.02 and 𝑅𝑒 = −5…………………………………………………………….

218

Fig 3.1 Graphical Representation of Eq. (3.6a-3.6b) for Different Values of Parameters……………………………………………………………......................

223

Fig 3.2 Graphical Representation of Eq. (3.7a-3.7b) for Different Values of Parameters……………………………………………………………......................

225

Fig 3.3 (a)-(b) Graphical Representation of Eq. (3.7a-3.7b) for Different Values of Parameters……………………………………………………………......................

226

Fig 3.4 (a)-(b) Graphical Representation of Eq. (3.8a) for Different Values of Parameters……………………………………………………………......................

228


229


229


230


231


231

Fig 3.10 (a)-(b) Graphical Representation of Eq. (3.9a) for Different Values of Parameters……………………………………………………………...................... xxxviii

234


234


236


237


238


238

Fig 3.16 Graphical Representation of Eq. (3.11a) for Different Values of Parameters……………………………………………………………......................

240

Fig 3.17 Graphical Representation of Eq. (3.11a) for Different Values of Parameters……………………………………………………………......................

240


243


243

Fig 3.20 (a)-(b) Graphical Representation of Eq. (3.17) for Different Values of Parameters……………………………………………………………......................

245


246


246


248

Fig 3.24 (a)-(b) Graphical Representation of Eq. (3.20a) for Different Values of Parameter…………………………………………………………….......................

249

Fig 3.25 (a)-(b) Graphical Representation of Eq. (3.20a) for Different Values of Parameters……………………………………………………………...................... Fig 3.26 (a)-(b) Graphical Representation of Eq. (3.20a) for Different Values of xxxix

249

Parameters……………………………………………………………......................

250


250


251


253


253


254


255


255

Fig 3.34 (a)-(b) Graphical Representation of Eq. (3.24a-3.25a) for Different Values of Parameters…………………………………………………………….....

257


258


258

Fig 3.37 (a)-(b) Graphical Representation of Eq. (3.29) for Different Values of Parameters…………………………………………………………..........................

260


261


262


263

Fig 3.41 (a)-(b) Graphical Representation of Eq. (3.45) for Different Values of Parameters………………………………………………………….......................... xl

267


267


268


269

Fig 3.45 (a)-(b) Graphical Representation of Eq. (3.52a) for Different Values of Parameters…………………………………………………………..........................

272


273


274


274


275


276


277


278

Fig 3.53(a)-(b) Graphical Representation of Eq. (3.54a) for Different Values of Parameters…………………………………………………………..........................

278


280


280

Fig 3.56 (a)-(b) Graphical Representation of Eq. (3.56a) for Different Values of Parameters………………………………………………………….......................... Fig 3.57 (a)-(b) Graphical Representation of Eq. (3.56a) for Different Values of xli

281

Parameters…………………………………………………………..........................

282


282


283


284


285


285


287


287


288


289


290


290


294


294


295

Fig 3.72 (a)-(b) Graphical Representation of Eq. (3.67a) for Different Values of Parameters………………………………………………………….......................... xlii

296


297


297


298


299


300

Fig 3.78 (a)-(b) Graphical Representation of Eq. (3.116a-3.116b) for Different Values of Parameters………………………………………………………….........

329

Fig 3.79 (a)-(c) Graphical Representation of Eq. (3.116a-3.116b) for Different Values of Parameters………………………………………………………….........

329


330

Fig 3.81 (a)-(f) Graphical Representation of Eq. (3.116a-3.116b) for Different Values of Parameters………………………………………………………….........

331


332

Fig 3.83 (a)-(d) Graphical Representation of Eq. (3.116a-3.116b) for Different Values of Parameters………………………………………………………….........

332


335


338


340

Fig 3.87 (a)-(b) Graphical Representation of Eq. (3.147) for Different Values of Parameter………………………………………………………….......................... Fig 3.88 (a)-(b) Graphical Representation of Eq. (3.154) for Different Values of xliii

343

Parameters…………………………………………………………..........................

345


351


354


357


358


359


361


362


363


365


366


371


375


381


386

Fig 3.103 (a)-(b) Graphical Representation of Eq. (3.237) for Different Values of Parameters………………………………………………………….......................... xliv

389


394


399


402


405

Fig 3.108 (a)-(e) Graphical Representation of Eq. (3.274) for Different Values of Parameters…………………………………………………………..........................

411


411


413


414


417


417


418


418

Fig 3.116 (a)-(b) Graphical Representation of Eq. (3.288-3.289) for Different Values of Parameters………………………………………………………….........

421

Fig 3.117 (a)-(b) Graphical Representation of Eq. (3.288-3.289) for Different Values of Parameters………………………………………………………….........

421

Fig 3.118 (a)-(b) Graphical Representation of Eq. (3.242) for Different Values of Parameters………………………………………………………….......................... Fig 3.119 (a)-(b) Graphical Representation of Eq. (3.242) for Different Values of xlv

426

Parameters…………………………………………………………..........................

427


428


428


429


429


430


430


431


432


433


433


434


434


436


438

Fig 3.134 (a)-(b) Graphical Representation of Eq. (3.251) for Different Values of Parameters………………………………………………………….......................... xlvi

438


439


439


440


441


442


443


444


444


446

Fig 4.1 (a)-(b) Graphical Representation of Eq. (4.23)……………………………..

470

Fig 4.2 (a)-(b) Graphical Representation of Eq. (4.28) …………………………….

472


476


478


783


484


486


487


489

Fig 4.10 (a)-(b) Graphical Representation of Eq. (4.64) …………………………...

491


492


494

Fig 4.13 (a) 2D Kink Wave Solution of Eq. (4.74) for Different Values of xlvii

Parameters…………………………………………………………..........................

498

Fig 4.13 (b) 3D Kink Wave Solution of Eq. (4.74) for Different Values of Parameters…………………………………………………………..........................

498

Fig 4.14 (a) 2D Kink Wave Solution of Eq. (4.80) for Different Values of Parameters…………………………………………………………..........................

500


500


506


506


511


511


516


516


521


521


526


526

Fig 4.20 (a) 2D Solitary wave sol of Eq. (4.116) for Different Values of Parameters…………………………………………………………..........................

531

Fig 4.20 (b) 3D Solitary wave sol of Eq. (4.116) for Different Values of Parameters………………………………………………………….......................... xlviii

531

Fig 4.21 (a) 2D Solitary wave sol of Eq. (4.122) for Different Values of Parameters…………………………………………………………..........................

535

Fig 4.21 (b) 3D Solitary wave sol of Eq. (4.122) for Different Values of Parameters…………………………………………………………..........................

535


538


538

Fig 4.23 (a)-(b) Graphical Representation of Eq. (4.137)…………………………..

541

Fig 4.24 (a)-(b) Graphical Representation of Eq. (4.140) ………………………….

542


543


545


546


547

Fig 4.29 (a)-(b) Travelling Wave Solution of MKdV Equation for Different Values of Parameters……………………………………………………………….

549

Fig 4.30 (a)-(b) Travelling Wave Solution of Nizhnik-Novikov-Vesselov Equation for Different Values of Parameters……………………………………….

550


551


552

Fig 4.33 (a)-(b) Travelling Wave Solution of Konopelchenko-Dubrovsky Equation for Different Values of Parameters……………………………………….

554

Fig 4.34 (a)-(b) Travelling Wave Solution of nonlinear system of the third order Equation for Different Values of Parameters……………………………………….

556

Fig 4.35 Solitary Wave Solution for 𝑡 = 0, 𝑡 = 0.5 and 𝑡 = 1…………………….

566

Fig 4.36 Combined Graph for Various of 𝑡………………………………………...

566


569


570

Fig 4.39 Solitary Wave Solution for 𝑡 = 1, 𝑡 = 5 and 𝑡 = 10……………………..

573

xlix


573


580


580


582


583


589


589


598


598


605


605


612


612

Fig 4.53 (a) 2D Travelling Wave Solution of Eq. (4.226) for Different Values of Parameters and 𝑡 = 1……………………………………………………………….

619

Fig 4.53 (b) 3D Travelling Wave Solution of Eq. (4.226) for Different Values of Parameters…………………………………………………………………………..

619

Fig 4.54 (a) 2D Kink Wave Solution of Eq. (4.230) for Different Values of Parameters…………………………………………………………………………..

623

Fig 4.54 (b) 3D Kink Wave Solution of Eq. (4.230) for Different Values of Parameters…………………………………………………………………………..

623

Fig 4.55 (a) 2D Kink Wave Solution of Eq. (4.234) Different Values of Parameters and 𝑡 = 1……………………………………………………………….

626

Fig 4.55 (b) 3D Kink Wave Solution of for Eq. (4.234) for Different Values of Parameters…………………………………………………………………………..

626


632


632

Fig 4.57 (a) 2D Travelling Wave Solution of Eq. (4.242) for Different Values of Parameters and 𝑡 = 1………………………………………………………………. l

639


639

Fig 4.58 (a) 2D Travelling Wave Solution of Eq. (4.246) for Different Values of Parameters…………………………………………………………………………..

644


644


647


647


650


650

Fig 4.61 (a) 2D Periodic Wave Solution of Eq. (4.258) for Different Values of Parameters and 𝑡 = 1……………………………………………………………….

655

Fig 4.61 (b) 3D Periodic Wave Solution of Eq. (4.258) for Different Values of Parameters…………………………………………………………………………..

655

Fig 4.62 (a) 2D Solitary Wave Solution of Eq. (4.262) for Different Values of Parameters…………………………………………………………………………..

658

Fig 4.62 (b) 3D Solitary Wave Solution of Eq. (4.262) for Different Values of Parameters…………………………………………………………………………..

658


663


663

Fig 4.64 (a) 2D Periodic Wave Solution of Eq. (4.273) for Different Values of Parameters…………………………………………………………………………..

665

Fig 4.64 (b) 3D Periodic Wave Solution of Eq. (4.273) for Different Values of Parameters………………………………………………………………………….. Fig 4.65 (a) 2D Periodic Wave Solution of Eq. (4.277) for Different Values of li

665

Parameters…………………………………………………………………………..

668


668


672


672


675


675


677


677


683


683


688


688


693


693


695

Fig 4.72 (b) 3D Periodic Wave Solution of Eq. (4.305) for Different Values of Parameters………………………………………………………………………….. lii

695


701


701

Fig 4.75 Solitary Wave Solution for 𝑡 = 0, 𝑡 = 1 and 𝑡 = 2………………………

703


703


705


705

Fig 4.79 Solitary Wave Solution for 𝑡 = 0………………………………………....

710


713


714


716


716


720

Fig 4.85 Travelling Wave Solution for 𝑡 = 0………………………………………

723


727

Fig 4.87 (a)-(b) 3D Graph of Eq. (4.339) for Different Values of Parameters……..

732

Fig 4.88 3D Graph of Eq. (4.343) for Different Values Parameters………………..

734

Fig 4.89 (a)-(b) 3D Soliton Solution of Eq. (4.357) for Different Values Parameters…………………………………………………………………………..

736

Fig 4.90 3D Graph of Eq. (4.361) for Different Values Parameters………………..

738

Fig 4.91 (a)-(b) 3D Soliton Solution of Eq. (4.365) for Different Values of Parameters…………………………………………………………………………..

740

Fig 4.92 3D Graph of Eq. (4.369) for Different Values of Parameters…………….

742

Fig 4.93 (a)-(b) 3D Soliton Solution of Eq. (4.373) for Different Values of Parameters…………………………………………………………………………..

744


746

Fig 4.95 (a)-(b) 3D Soliton Solution of Eq. (4.381) for Different Values Parameters…………………………………………………………………………..

749


751

Fig 5.1 Comparison of Exact and Approximate Solutions of Eq. (5.67)…………...

773

Fig 5.2 Comparison of Exact and Approximate Solutions of Eq. (5.68a)………….

774

Fig 5.3 Comparison of Exact and Approximate Solutions of Eq. (5.68b)………….

775

liii


776

Fig 5.5 Comparison of Exact and Approximate Solutions of Eq. (5.79) for n =3…

788

Fig 5.6 Comparison of Exact and Approximate Solutions of Eq. (5.79) for n =50..

789

Fig 5.7 Comparison of Exact and Approximate Solutions of Eq. (5.79) for n =100

790


792

Fig 5.9 Comparison of Exact and Approximate Solutions of Eq. (5.89) for n =3…

794


796


797


799


802


812

Fig 5.15 Comparison of Exact and Approximate Solutions of Eq. (5.142) for n =42………………………………………………………………………………….

813


815


816


817


819


821


822


823


825


826


827


830

Fig 5.27 Comparison of Exact and Approximate Solutions of Eq. (5.167) for n liv

=30………………………………………………………………………………….

831


832


834


835


836


841


841

Fig 5.34 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (5.193) for n =5……………………………………………………………………………..

846


847


848


853


854


855


857


858


858


859


861


862

Fig 5.46 Comparison of Exact and Approximate Solutions of Eq. (5.228) for n =80…………………………………………………………………………………. lv

863


865


866


868

Fig 5.50 (a)-(c) Comparison of Exact and Approximate Solutions of Eq. (5.262) for n =2……………………………………………………………………………...

876


883


884


885


887


888


889


892


892


893


898


898


899

Fig 5.63 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (5.322) for n =2……………………………………………………………………………..

907

Fig 5.64 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (5.322) for n =10…………………………………………………………………………… lvi

907

Fig 5.65 Comparison of Exact and Approximate Solutions of Eq. (5.340) for k =3

913

Fig 5.66 Comparison of Exact and Approximate Solutions of Eq. (5.340) for k =50………………………………………………………………………………….

914

Fig 5.67 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (5.351) for k =2……………………………………………………………………………..

917

Fig 5.68 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (5.351) for k =20……………………………………………………………………………

919


920


923


924


926

Fig 5.73 Comparison of Exact and Approximate Solutions of Eq. (5.369) for k =3.

929


931


932


934


936


937


939


941


942

Fig 5.82 Comparison of Exact and Approximate Solutions of Eq. (5.400) for k =2

947

lvii


948


950


952


953


955


957


959


960


964


966

Fig 5.93 Comparison of Exact and Approximate Solutions of Eq. (5.422) for n =100………………………………………………………………………………...

968


971


972


975


979


981

Fig 5.99 Comparison of Exact and Approximate Solutions of Eq. (5.469) for n =100………………………………………………………………………………... lviii

983

Fig 5.100 Comparison of Exact and Approximate Solutions of Eq. (5.488)……….

985


988

Fig 5.102 Comparison of Exact and Approximate Solutions of Eq. (5.514) for n =2…………………………………………………………………………………...

993


1000


1006


1008


1009


1011


1017


1020


1022

Fig 5.111 Comparison of Exact and Approximate Solutions of Eq. (5.641) for n = 50……………………………………………………………………………………

1023

Fig 6.1 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.62) for k =1 and M =10…………………………………………………………………….

1043


1045

Fig 6.3 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.62) for k =1 and M =100…………………………………………………………………...

1046


1049

Fig 6.5 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.71) for k =1 and M =25……………………………………………………………………. lix

1051


1052


1055


1057


1058


1061


1063


1064

Fig 6.13 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.103) for k =1 and M =10…………………………………………………………………

1067


1068

Fig 6.15 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.103) for k =1 and M =100………………………………………………………………..

1070


1076


1077


1079


1080

Fig 6.20 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.122) for k =1 and M =50………………………………………………………………… Fig 6.21 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.123) lx

1081

for k =1 and M =50…………………………………………………………………

1082


1085


1086


1088


1089


1090


1091


1094


1095


1097


1098


1099


1100


1104


1104

Fig 6.36 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.163) for k =1 and M =10………………………………………………………………… lxi

1105


1106


1107


1107


1108


1109


1109


1113


1113


1114


1115


1116


1116


1117


1118

Fig 6.51 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.183) for k =1 and M =50………………………………………………………………… Fig 6.52 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.183) k lxii

1118

= 1 and M = 100………………………………………………………………….....

1118


1122


1122


1123


1124


1125


1125


1126


1127


1127


1127


1132


1133


1135


1137

Fig 6.67 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.230) for k =1 and M =25………………………………………………………………… Fig 6.68 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.230) lxiii

1139

for k =1 and M =50…………………………………………………………………

1140


1143


1145


1146


1152


1153


1154


1155


1157


1158


1163


1165


1166


1169


1171

Fig 6.83 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.283) for k =1 and M =75………………………………………………………………… lxiv

1172


1175


1177


1178


1187


1189


1190


1193


1195


1196


1199


1201


1202


1205


1207

Fig 6.97 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.354) for k =1 and M =100……………………………………………………………….. Fig 6.98 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.363) lxv

1208

for k =1 and M =10…………………………………………………………………

1211


1213


1214


1219


1221


1222


1225


1227


1228


1232


1234


1235


1238


1240


1241

Fig 6.113 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.422) for k =1 and M =10………………………………………………………………… lxvi

1247


1249


1250


1253


1255


1256


1259


1261


1262


1265


1267


1268

Fig 6.125 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.458a6.458b) for k =1 and M =10………………………………………………………..

1272


1273


1274

Fig 6.128 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.458a6.458b) for k =1 and M =25……………………………………………………….. Fig 6.129 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.458alxvii

1275

6.458b) for k =1 and M =50………………………………………………………..

1277


1278


1295


1296

Fig 6.133 Comparison of Exact and Approximate Solutions of Eq. (6.536) for k =2 and M =10………………………………………………………………………

1296


1297


1298


1298


1299


1300


1300


1303


1303


1303


1305

Fig 6.144 (a)-(b) Comparison of Exact and Approximate Solutions of Eq. (6.545) for k =3 and M =20………………………………………………………………… lxviii

1305


1305


1307


1307

Fig 6.148 Comparison of Exact and Approximate Solutions of Eq. (6.545) for k =3 and M =4………………………………………………………………………..

lxix

1308

LIST OF TABLES Table 2.1 The Results of ADM and RK Numerical Method for 𝑓(𝜂) when Re = −10………………………………………………………………………………….

32

Table 2.2 The Results of ADM and Runge–Kutta Numerical Method for 𝑔(𝜂) when Re = −10……………………………………………………………………...

33

Table 2.3 Comparison of Numerical and Different Order ADM Solutions for 𝑓(𝜂) when 𝑆 = 𝑀 = 𝑃𝑟 = 𝐸𝑐 = 1.0, 𝛽 = 𝛾 = 𝛽 = 0.1………………………………...

41

Table 2.4 Comparison of Numerical and Different order ADM Solutions for 𝜃(𝜂) when 𝑆 = 𝑀 = 𝑃𝑟 = 𝐸𝑐 = 1.0, 𝛽 = 𝛾 = 𝛽 = 0.1………………………………...

42

Table 2.5 Comparison of Exact and Approximate Solutions of Eq. (2.39)………...

54

Table 2.6 Comparison of the Exact and Approximate Solutions of Eq. (2.40) Obtained from ADM and ODM…………………………………………………….

56


58


60


62


64


66


68


70

Table 2.14 Error Analysis of Eq. (2.63) for Different Values of 𝑀 and 𝛽…………

75

Table 2.15 Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = −2.50 and Different Values of 𝐻…….

77

Table 2.16 Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = 2.50 and Different Values of 𝐻………

78

Table 2.17 Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = 50 and Different Values of 𝐻………...

78

lxx

Table 2.18 Values of 𝑎 for 𝑅𝑒 = 100, 𝐻 = 1500 and Different Values of 𝛼……...

79


80

Table 2.20 Values of 𝑎 for 𝛼 = 2.50 , 𝐻 = 1500 and Different Values of 𝑅𝑒……..

80

Table 2.21 Values of 𝑎 for 𝛼 = −2.50 , 𝐻 = 1500 and Different Values of 𝑅𝑒…...

81

Table 2.22 (a) The Comparison Between the Numerical Results and VIM Solution of Eq. (2.71) for Velocity when 𝛼 = 2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500……………..

82

Table 2.22 (b) The Comparison Between the Numerical Results and VIM Solution of Eq. (2.71) for Velocity when 𝛼 = 2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500……………..

83


106


106


107


107


108


108

Table 2.29 Values of 𝑎 for 𝛼 = −2.50 , 𝐻 = 1500 and Different Values of 𝑅𝑒…...

109

Table 2.30 (a).The Comparison Between the Numerical Results and HPM Solution of Eq. (2.71) for Velocity when 𝛼 = 2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500……

110

Table 2.30 (b) The Comparison Between the Numerical Results and HPM Solution of Eq. (2.71) for Velocity when 𝛼 = −2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500….

111

Table 2.31 Numerical Solution of Eq. (2.113), where 𝑝 = 0.001, 𝜂 = 0.001, 𝑡 = 0.005………………………………………………………………………………..

117


117


117


119


119

Table 2.36 Numerical Solution of Eq. (2.113), where 𝑝 = 0.005, 𝜂 = 0.005, 𝑡 = 0.005……………………………………………………………………………….. lxxi

119

Table 2.37 Numerical Solution of Eq. (2.117), where 𝑝 = 0.001, 𝜂 = 0.001, 𝑥0 = 0.001, 𝑡 = 0.005……………………………………………………………………

121


121


121


123


124


124

Table 2.43 Comparison of Exact and Approximate Solutions of Eq. (2.166)……...

181

Table 2.44 Comparison of Exact and Approximate Solutions of Eq. (2.168-169)…

185


211


211


212


212


213


214


214

Table 2.52 (a) The Comparison Between the Numerical Results and VPM Solution for Velocity when 𝛼 = 2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500……………………………

215

Table 2.52 (b) The Comparison Between the Numerical Results and VPM Solution for Velocity when 𝛼 = 2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500……………………………

216

Table 3.1 Exact Solutions of Eq. (3.73) for Different Values of 𝑃, 𝑄 and 𝑅, in Terms of Jacobi Elliptic Functions………………………………………………….

301

Table 3.2 Solutions of Eq. (3.238) for Different Values of 𝛼1 and 𝛼2 ……………..

423

Table 4.1 (a) Solutions of Eq. (4.4) for Different Values of 𝑃, 𝑄 and 𝑅…………...

450

Table 4.1 (b) Solutions of Eq. (4.4) for Different Values of 𝑃, 𝑄 and 𝑅…………...

453


559

lxxii


616


659

Table 5.1 Comparison of Exact and Approximate Solutions of Eq. (5.50) for n =10, n =20 and n =30………………………………………………………………

767

Table 5.2 Comparison of Exact and Approximate Solutions of Eq. (5.51) for n =10, n =20 and n =30………………………………………………………………

768

Table 5.3 Comparison of Exact and Approximate Solutions of Eq. (5.52) for k =1, M =2 and k =1, M =3………………………………………………………………

769

Table 5.4 Comparison of Exact and Approximate Solutions of Eq. (5.79) Obtained from Kravchuk Polynomials Method (KPM) for n =3…………………..

788

Table 5.5 Comparison of Exact and Approximate Solutions of Eq. (5.79) Obtained from Kravchuk Polynomials Method (KPM) for n =50…………………

789

Table 5.6 Comparison of Exact and Approximate Solutions of Eq. (5.79) Obtained from Kravchuk Polynomials Method (KPM) for n =100………………..

790


792


795


796


797


799


801

Table 5.13 Comparison of Exact and Approximate Solutions of Eq. (5.142) Obtained from Newton’s Polynomials Method (NPM) for n =3…………………..

813

Table 5.14 Comparison of Exact and Approximate Solutions of Eq. (5.142) Obtained from Newton’s Polynomials Method (NPM) for n =42………………… Table 5.15 Comparison of Exact and Approximate Solutions of Eq. (5.142) lxxiii

814

Obtained from Newton’s Polynomials Method (NPM) for n =50…………………

815


817


818

Table 5.18 Comparison of Exact and Approximate Solutions of Eq. (5.147) Obtained from Newton’s Polynomials Method (NPM) for n =12…………………

819


821


822


823


825


826


827


830


831


832


834


835

Table 5.30 Comparison of Exact and Approximate Solutions of Eq. (5.172) Obtained from Newton’s Polynomials Method (NPM) for n =70………………… lxxiv

836


840


842


846


847


848

Table 5.36 Comparison of Exact and Approximate Solutions of Eq. (5.216) Obtained from Touchard Polynomials Method (TPM) for n =3…………………...

854

Table 5.37 Comparison of Exact and Approximate Solutions of Eq. (5.216) Obtained from Touchard Polynomials Method (TPM) for n =40………………….

855


856


858


859


860


862


863


864

Table 5.45 Comparison of Exact and Approximate Solutions of Eq. (5.234) Obtained from Touchard Polynomials Method (TPM) for n =3…………………... Table 5.46 Comparison of Exact and Approximate Solutions of Eq. (5.234) lxxv

866

Obtained from Touchard Polynomials Method (TPM) for n =20………………….

867


868


876


877


877

Table 5.51 Comparison of Exact and Approximate Solutions of Eq. (5.279) Obtained from Zernike Polynomials Method (ZPM) for n =3……………………..

883

Table 5.52 Comparison of Exact and Approximate Solutions of Eq. (5.279) Obtained from Zernike Polynomials Method (ZPM) for n =10……………………

884


885


887


888


889


891


893


894


897

Table 5.61 Comparison of Exact and Approximate Solutions of Eq. (5.300) Obtained from Zernike Polynomials Method (ZPM) for n =10…………………… lxxvi

899


900


906


908

Table 5.65 Comparison of Exact and Approximate Solutions of Eq. (5.340) Obtained from Meixner Polynomials Method (MPM) for k =3……………………

913

Table 5.66 Comparison of Exact and Approximate Solutions of Eq. (5.340) Obtained from Meixner Polynomials Method (MPM) for k =50…………………..

914


918

Table 5.68 Comparison of Exact and Approximate Solutions of Eq. (5 351) Obtained from Meixner Polynomials Method (MPM) for k =20…………………..

919


921


923


925


926

Table 5.73 Comparison of Exact and Approximate Solutions of Eq. (5.369) Obtained from Rook Polynomials Method (RPM) for k =3………………………..

930

Table 5.74 Comparison of Exact and Approximate Solutions of Eq. (5 369) Obtained from Rook Polynomials Method (RPM) for k =30………………………

931

Table 5.75 Comparison of Exact and Approximate Solutions of Eq. (5.369) Obtained from Rook Polynomials Method (RPM) for k =50………………………

933

Table 5.76 Comparison of Exact and Approximate Solutions of Eq. (5.376) Obtained from Rook Polynomials Method (RPM) for k =2……………………….. Table 5.77 Comparison of Exact and Approximate Solutions of Eq. (5 376) lxxvii

935

Obtained from Rook Polynomials Method (RPM) for k =30………………………

936


938


940


941


943


947


949

Table 5.84 Comparison of Exact and Approximate Solutions of Eq. (5.400) Obtained from Rook Polynomials Method (RPM) for k =30………………………

950

Table 5.85 Comparison of Exact and Approximate Solutions of Eq. (5 405) Obtained from Rook Polynomials Method (RPM) for k =2………………………..

952


954


955


957


959


960

Table 5.91 Comparison of Exact and Approximate Solutions of Eq. (5.422) Obtained from Fibonicca Polynomials Method (FPM) for n =25………………….

964

Table 5.92 Comparison of Exact and Approximate Solutions of Eq. (5.422) Obtained from Fibonicca Polynomials Method (FPM) for n =50…………………. lxxviii

966

Table 5.93 Comparison of Exact and Approximate Solutions of Eq. (5.422) Obtained from Fibonicca Polynomials Method (FPM) for n =100………………...

968

Table 5.94 Comparison of Exact and Approximate Solutions of Eq. (5.438) Obtained from Fibonicca Polynomials Method (FPM) for n =2…………………...

970


972


974

Table 5.97 Comparison of Exact and Approximate Solutions of Eq. (5.469) Obtained from Gegenbauer Polynomials Method (GPM) for n =25……………….

979

Table 5.98 Comparison of Exact and Approximate Solutions of Eq. (5.469) Obtained from Gegenbauer Polynomials Method (GPM) for n =50……………….

981

Table 5.99 Comparison of Exact and Approximate Solutions of Eq. (5.469) Obtained from Gegenbauer Polynomials Method (GPM) for n =100……………...

983

Table 5.100 Comparison of Exact and Approximate Solutions of Eq. (5.488) Obtained from Gegenbauer Polynomials Method (GPM) for n =2………………...

986

Table 5.101 Comparison of Exact and Approximate Solutions of Eq. (5.494) Obtained from Gegenbauer Polynomials Method (GPM) for n =1………………...

988

Table 5.102 Comparison of Exact and Approximate Solutions of Eq. (5.514) Obtained from Charlier's Polynomials Method (CPM) for n =2…………………...

992

Table 5.103 Comparison of Exact and Approximate Solutions of Eq. (5.554) Obtained from Bessel's Polynomials Method (BPM) for n =2…………………….

999

Table 5.104 Comparison of Exact and Approximate Solutions of Eq. (5.589) Obtained from Bell Polynomials Method (BPM) for n =5………………………...

1006

Table 5.105 Comparison of Exact and Approximate Solutions of Eq. (5.589) Obtained from Bell Polynomials Method (BPM) for n =25……………………….

1007

Table 5.106 Comparison of Exact and Approximate Solutions of Eq. (5.589) Obtained from Bell Polynomials Method (BPM) for n =50……………………….

1009

Table 5.107 Comparison of Exact and Approximate Solutions of Eq. (5.601) Obtained from Bell Polynomials Method (BPM) for n =1………………………... Table 5.108 Comparison of Exact and Approximate Solutions of Eq. (5.630) lxxix

1011

Obtained from Appell Polynomials Method (APM) for n =2……………………...

1017

Table 5.109 Comparison of Exact and Approximate Solutions of Eq. (5.641) Obtained from Appell Polynomials Method (APM) for n =5……………………...

1020

Table 5.110 Comparison of Exact and Approximate Solutions of Eq. (5.641) Obtained from Appell Polynomials Method (APM) for n =25…………………….

1021

Table 5.111 Comparison of Exact and Approximate Solutions of Eq. (5.641) Obtained from Appell Polynomials Method (APM) for n =50…………………….

1023

Table 6.1 Comparison of the Exact Solution and Approximate Solution of Eq. (6.62) Obtained from Newton’s Wavelet Method (NWM) for k =1 and M =10…..

1043


1044

Table 6.3 Comparison of the Exact Solution and Approximate Solution of Eq. (6.62) Obtained from Newton’s Wavelet Method (NWM) for k =1 and M =100…

1046


1049


1050


1052

Table 6.7 Comparison of the Exact Solution and Approximate Solutions Obtained from Newton’s Wavelet Method (NWM) for k =1 and M =10…………………….

1055

Table 6.8 Comparison of the Exact Solution and Approximate Solutions Obtained from Newton’s Wavelet Method (NWM) for k =1 and M =25…………………….

1056


1058


1061


1062

Table 6.12 Comparison of the Exact Solution and Approximate Solution of Eq. (6.92) Obtained from Newton’s Wavelet Method (NWM) for k =1 and M =100… lxxx

1064


1067


1068

Table 6.15 Comparison of the Exact Solution and Approximate Solution of Eq. (6.103) Obtained from Newton’s Wavelet Method (NWM) for k =1 and M =100..

1069

Table 6.16 (a) Comparison of the Exact Solution and Approximate Solutions of 𝑦1 (𝑥) Obtained from Laurent Wavelet Method (LWM) for k =1 and M =10……..

1076

Table 6.16 (b) Comparison of the Exact Solution and Approximate Solutions of 𝑦2 (𝑥) Obtained from Laurent Wavelet Method (LWM) for k =1 and M =10……..

1077


1078


1079


1081


1082


1085


1086


1087


1088


1090

Table 6.21 (b) Comparison of the Exact Solution and Approximate Solutions of 𝑦2 (𝑥) Obtained from Laurent Wavelet Method (LWM) for k =1 and M =50…….. Table 6.22 (a) Comparison of the Exact Solution and Approximate Solutions of lxxxi

1091

𝑦1 (𝑥) Obtained from Laurent Wavelet Method (LWM) for k =1 and M =10……..

1094


1095


1096


1097


1099


1100

Table 6.25 Comparison of the Exact Solution and Approximate Solution of Eq. (6.161-6.163) Obtained from Laurent Wavelet Method (LWM) for k =1 and M =10………………………………………………………………………………….

1103


1106


1108


1112


1115

Table 6.30 Comparison of the Exact Solution and Approximate Solution of Eq. (6.181-6.183) Obtained from Laurent Wavelet Method (LWM) for k =1 and M =50…………………………………………………………………………………. Table 6.31 Comparison of the Exact Solution and Approximate Solution of Eq. (6.198-6.200) Obtained from Laurent Wavelet Method (LWM) for k =1 and M lxxxii

1117

=10………………………………………………………………………………….

1121


1124


1126

Table 6.34 Comparison of the Exact Solution and Approximate Solution of Eq. (6.223) Obtained from Kravchuk Wavelet Method (KWM) for k =1 and M =10…

1132


1133


1134

Table 6.37 Comparison of the Exact Solution and Approximate Solution of Eq. (230) Obtained from Kravchuk Wavelet Method (KWM) for k =1 and M =10…...

1137

Table 6.38 Comparison of the Exact Solution and Approximate Solution of Eq. (230) Obtained from Kravchuk Wavelet Method (KWM) for k =1 and M =25…...

1138

Table 6.39 Comparison of the Exact Solution and Approximate Solutions of Eq. (6.230) Obtained from Kravchuk Wavelet Method (KWM) for k =1 and M =50…

1140


1143


1144


1146

Table 6.43 (a) Comparison of the Exact Solution and Approximate Solutions of 𝑦1 (𝑥) Obtained from Kravchuk Wavelet Method (KWM) for k =1 and M =10…..

1151

Table 6.43 (b) Comparison of the Exact Solution and Approximate Solutions of 𝑦2 (𝑥) Obtained from Kravchuk Wavelet Method (KWM) for k =1 and M =10…..

1152

Table 6.44 (a) Comparison of the Exact Solution and Approximate Solutions of 𝑦1 (𝑥) Obtained from Kravchuk Wavelet Method (KWM) for k =1 and M =25….. lxxxiii

1154


1155

Table 6.45 (a) Comparison of the Exact Solution and Approximate Solutions of 𝑦1 (𝑥) Obtained from Kravchuk Wavelet Method (KWM) for k =1 and M =50…..

1156


1157

Table 6.46 Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) Obtained from Bessel Wavelet Method (LWM) for k =1 and M =10……………..

1163


1164

Table 6.48 Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) Obtained from Bessel Wavelet Method (LWM) for k =1 and M =75………..

1166


1169


1170


1172


1175


1176


1178

Table 6.55 Comparison of the Exact Solution and Approximate Solution of Eq. (6.327) Obtained from Bernoulli’s Wavelet Method (BWM) for k =1 and M =10..

1187


1188

Table 6.57 Comparison of the Exact Solution and Approximate Solution of Eq. (6.327) Obtained from Bernoulli’s Wavelet Method (BWM) for k =1 and M =100 Table 6.58 Comparison of the Exact Solution and Approximate Solution of Eq. lxxxiv

1191

(6.336) Obtained from Bernoulli’s Wavelet Method (BWM) for k =1 and M =10..

1193


1194

Table 6.60 Comparison of the Exact Solution and Approximate Solution of Eq. (6.343) Obtained from Bernoulli’s Wavelet Method (BWM) for k =1 and M =100

1196


1199


1200


1202


1205


1206


1208


1211


1212


1214

Table 6.70 Comparison of the Exact Solution and Approximate Solution of Eq. (6.379) Obtained from Probabilist Hermite Wavelet Method (PWM) for k =1 and M =10………………………………………………………………………………

1219

Table 6.71 Comparison of the Exact Solution and Approximate Solution of Eq. (6.379) Obtained from Probabilist Hermite Wavelet Method (PWM) for k =1 and M =12……………………………………………………………………………… Table 6.72 Comparison of the Exact Solution and Approximate Solution of Eq. (6.379) Obtained from Probabilist Hermite Wavelet Method (PWM) for k =1 and lxxxv

1220

M =15………………………………………………………………………………

1222


1225


1226


1228


1232


1233


1235


1238


1239


1241

Table 6.82 Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) of Eq. (6.422) Obtained from Zernike Wavelet Method (ZWM) for k =1 and M =10…………………………………………………………………………………. lxxxvi

1247

Table 6.83 Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) of Eq. (6.422) Obtained from Zernike Wavelet Method (ZWM) for k =1 and M =25………………………………………………………………………………….

1248


1250


1253


1254


1256


1259


1260


1262


1265

Table 6.92 Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) of Eq. (6.449) Obtained from Zernike Wavelet Method (ZWM) for k =1 and M =25…………………………………………………………………………………. Table 6.93 Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) lxxxvii

1266

of Eq. (6.455) Obtained from Zernike Wavelet Method (ZWM) for k =1 and M =50………………………………………………………………………………….

1268

Table 6.94 Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) of Eq. (6.458a-6.458b) Obtained from Zernike Wavelet Method (ZWM) for k =1 and M =10…………………………………………………………………………..

1271


1272


1274


1275


1276


1277

Table 6.100 Comparison of the Exact Solution and Approximate Solutions of Eq. (6.536) Obtained from Touchard Wavelet Method (TWM) for k =2 and M =10….

1295


1297


1299


1302

Table 6.104 Comparison of the Exact Solution and Approximate Solutions of Eq. (6.545) Obtained from Touchard Wavelet Method (TWM) for k =3 and M =20…. Table 6.105 Comparison of the Exact Solution and Approximate Solutions of Eq. lxxxviii

1304

(6.545) Obtained from Touchard Wavelet Method (TWM) for k =3 and M =49….

lxxxix

1306

Introduction There has been an unprecedented development in nonlinear sciences [13, 24, 26, 35-45, 56, 88, 90, 92, 102-108, 114, 117, 120, 124-140, 155, 160, 164, 170-185, 199, 207, 209, 211, 215-255, 260, 270] in the recent past and in the similar context wide range of analytical [229-230], semi analytical [12-17, 79, 81-86, 115-120, 127-129, 186, 206, 232237, 266-268, 270], numerical [69, 91, 106, 139, 175, 188-190, 196] and semi numerical [7, 25, 29, 60, 64, 70, 97, 100, 177-178, 251] schemes have been developed to cope with the complexity of diversified physical problems. The thorough study of literature witnesses some useful modifications in the existing techniques. It is an established fact that most of the physical phenomenon are nonlinear in nature and hence appropriate solutions of the same are extremely important. Consequently, there is a dire need to search for new algorithms. The basic inspiration of this work is the development of some new schemes, modifications of some existing techniques and extension of few traditional methods for linear and nonlinear physical problems [5-105, 107, 115-199, 202, 210, 212, 215, 250-270]. In this thesis, we have used/introduced fifty eight different approaches [14, 18, 56, 75, 78-79, 115, 128, 184, 186, 202, 204, 213, 223-224, 262, 266, 270] which are applied on a wide range of problems. It is observed that proposed algorithms are highly efficient, accurate and user-friendly. Computational work and subsequent numerical results are fully supportive of the reliability of suggested schemes. The work is organized as follows: First Chapter is divided into four sections, in first section some basic definitions regarding solitary wave theory [230] are given, second section contains some basic preliminaries of fractional calculus [38, 104-105, 132, 192, 263], third section consists of some basic definitions of integral equations [229] and in the fourth section wavelet theory is discussed [155, 244, 250]. Second Chapter contains new work on series solutions of linear and nonlinear partial differential equations and integral equations. Adomian’s Decomposition Method (ADM) [5-6, 12-15, 125, 131, 215-218, 221, 224, 225] has been applied for solving Micro polar flow in a porous channel with mass injection and squeezing flow [187] and heat transfer between two parallel disks with velocity slip and temperature jump [99], Modified Adomian’s Decomposition Method (MADM) [224] is applied on Goursat [230], Fourth1

order parabolic [230], inhomogeneous wave [230] and inhomogeneous nonlinear wave [230] equations, Optimal Adomian’s Decomposition Method (OADM) has been applied to obtain the numerical solutions of Fredholm integral equations [229]. Variational Iteration Method (VIM) [8, 10-11, 78, 80, 82-87, 92-93, 159-160, 166, 197, 231-233, 235-238, 248, 266-268] has been applied to obtain the numerical solutions of MHD flow over a nonlinear stretching sheet [94] and MHD flow of an incompressible viscous fluid through convergent or divergent channels in the presence of a high magnetic field [99], Modified Variational Iteration Method (MVIM) [266-268] is applied to find the approximate solutions of nonlinear partial differential [230], Burger’s [230], fourth-order parabolic [230], Diffusion [230] and Reaction-diffusion [230] equations, Homotopy Perturbation Method (HPM) [79, 81] for solving flow of a viscoelastic fluid through a porous channel with expanding or contracting walls [40] and MHD flow of an incompressible viscous fluid through convergent or divergent channels in presence of a high magnetic field [99], Modified Homotopy Perturbation Method (MHPM) [16-17, 186] is applied to construct the numerical solutions of nonlinear Dispersive K(2,2,1) [230], nonlinear Dispersive K(3,3,1) [230], Kawahara [230], modified Kawahara [230], nonlinear Dispersive K(2,2) [230], nonlinear Dispersive K(3,3) [230], (2+1)-Dimensional nonlinear Dispersive K(2,2) [230] and homogeneous Convection-Diffusion [130] equations, Homotopy Analysis Method (HAM) [1-4, 73-74, 115-121, 141, 205, 247, 271] is used to solve homogenous linear Goursat [230], inhomogeneous linear Goursat [230], inhomogeneous nonlinear Goursat [230], system of initial value [230], K(m,p,1) [234], nonlinear Schrodinger [30] and veriational [200] equations, Modified Homotopy Analysis Method (MHAM) [206, 270] implemented to homogenous Helmholtz [270], SOG-KdV [134, 270], cubic nonlinear Schrodinger [30], Heat and wave-like [149], system of partial differential [230] and ZK(m,n,p) [26, 101] equations, and Variation of Parameters Method (VPM) [127-129, 147-148, 157, 168, 185] applied to construct the series solutions of MHD flow of an incompressible viscous fluid through convergent or divergent channels in the presence of a high magnetic field [99] and Flow of a viscoelastic fluid through a porous channel with expanding or contracting walls [40]. The third Chapter discusses the traveling wave solutions of nonlinear evolution equations [24, 33-34, 45, 58, 65, 110-114, 135-136, 143, 153, 176, 208, 220, 222, 230, 246, 254, 2

259] tackled by Exp-function Method [75-76, 103, 126, 142, 144-145, 156, 202, 242-243, 258, 264, 269], Modified Exp-function Method [198, 199, 202], (G´/G)-Expansion Method [23, 27, 39, 49, 59, 66, 152, 174, 213, 253, 255, 260, 261], F-Expansion Method [9, 61, 98, 123-124, 203, 211, 213-214, 249, 256, 262], Rational Sinh-Cosh Method [223], (U´/U)-Expansion Method, Extended Tanh Method [50-51, 54-56], New Approach of (G´/G)-Expansion Method, Combined Tanh-Coth (CTC) and Combined Sinh-Cosh (CSC) Method and U-Expansion Method [204]. The physical properties of several nonlinear traveling wave solutions [230] by plotting and analyzing their figures have also been studied. In Chapter four, we have applied F-Expansion Method [9, 61, 98, 123-124, 203, 211, 213-214, 249, 256, 262], Generalized Tanh Method [184, 195], (G’/G)-Expansion Method based on Riccati Equation [18], New Approach of (G’/G)-Expansion Method, Rational Hyperbolic Function (RHF) Method, U-expansion Method [204], Modified Uexpansion Method, Generalized U-expansion Method, (U’/U)-expansion Method and (G𝛂/G)-Expansion Method coupled with complex transformation [20-21, 32, 257] for fractional-order partial differential equations [37, 52, 90, 96, 122, 170, 179, 201, 265] (PDEs) including Boussinesq [240], Breaking Soliton [222], potential KadomtsevPetviashvili [230], Benjoman–Bona-Mahony [114], Korteweg-de-Vries [230] equations etc. Moreover, these methods are greatly capable of minimizing the size of computational work as compared to the other existing methods. In Chapter five, New scheme using Riemann-Liouville fractional integral [38, 104-105, 132, 192, 263], New alternative scheme using Riemann-Liouville fractional integral and derivative [38, 104-105, 132, 192, 263], New scheme using generalized RiemannLiouville Fractional integral and derivative [38, 104-105, 132, 192, 263] and New approach using Taylor’s series for solving linear and nonlinear Abel’s integral equations [229], Adomian’s Decomposition Method [5-6, 12-15, 125, 131, 215-218, 221, 224-225] and Modified Adomian’s Decomposition Method [228] using fractional calculus [38, 104-105, 132, 192, 263] were implemented to find the exact solutions of linear weakly singular integral equations [229]. Approximate solutions obtained using Newton Polynomials [154] (NPM), Bernoulli's Polynomials [100] (BPM), Touchard Polynomials [41] (TPM), Zernike Polynomials [182] (ZPM), Kravchuk Polynomials [183] (KPM), 3

Meixner Polynomials [183] (MPM), Rook Polynomials [62] (RPM), Fibonicca Polynomials [28] (FPM), Gegenbauer Polynomials [53] (GPM), Charlier's Polynomials [44] (CPM), Bessel's Polynomials [108] (BPM), Bell Polynomials [35] (BPM) and Appell Polynomials [107] Methods (APM) have been used to find out the exact and approximate solutions of linear and nonlinear integral equations [229]. In Chapter six, Physicists Hermite Wavelet Method (PHWM), has been used for solving singular ordinary differential equations [46, 72, 210] Newton’s Wavelet Method (NWM) has been implemented for higher order BVPs [67, 158, 219, 226-227], Laurent Wavelet Method (LWM) was used to solve the system of ordinary differential equations [19, 31, 43, 194, 201], Kravchuk Wavelet Method (KWM) for solving integral equations [229], Lommal Wavelet Method (LWM) applying to find the exact solutions of Delay differential equations [68], Bernoulli’s Wavelet Method (BWM) was applied to solve second-order differential equations with Robin conditions [207], Probabilist Hermite Wavelet Method (PWM) is used to construct the approximate solutions of nonlinear integral equations [229], Zernike Wavelet Method (ZWM) stiff Delay differential equation and difference Differential equation [22, 36, 181], Rook Wavelet Method (RWM) for solving Fractional differential equations [89] and Touchard Wavelet Method (TWM) is applied to find the exact and approximate solutions of partial differential equations [191]. It is to be observed that the proposed techniques have been applied on a wide range of nonlinear diversified physical problems including high-dimensional nonlinear evolution equations. It is also to be highlighted that the suggested algorithms are extremely simple but highly effective and may be extended to other singular problems of diversified physical nature.

4

Chapter 1 Preliminary Definitions

5

1.1 . Introduction This Chapter is divided into four sections, in first section some basic definitions of solitary wave theory are given, in second section some basic definitions of fractional calculus are given, in third section some basic definitions of integral equations are given and in four section some basic definitions of wavelets theory are given.

1.2. Some Basic Definitions of Solitary Wave Theory 1.2.1. Soliton A soliton [230] is a solitary wave which asymptotically preserves its shape and velocity upon nonlinear interaction with other solitary waves. Soliton has following properties 1. They are of permanent form. 2. They are localized within a region. 3. They can interact with other solitons and emerge from the collision unchanged, except for a phase shift. 4. Soliton is caused by a delicate balance between nonlinear and dispersive effects.

1.2.2. Solitary Wave These waves have soliton-like solutions of nonlinear evolution equations describing wave process in dispersive and dissipation media [230]. A sketch of these types of waves is shown below

Fig. 1.1: General representation of Solitary Waves

1.2.3. Travelling Wave 6

Travelling wave [230] can be defined as a wave in which the medium is travelling in the direction of propagation of wave. Travelling waves arises in the study of nonlinear differential equations where waves are represented by the form 𝑢(𝑥, 𝑡) = 𝑓(𝑥 − 𝑐𝑡),

Fig. 1.2: Travelling Waves

and 𝑐 is the speed of wave propagation. For 𝑐 > 0, the wave moves in the positive 𝑥 direction whereas the wave moves in the negative 𝑥 direction for 𝑐 < 0.

1.2.4. Types of Travelling Wave Solutions Travelling wave solutions are of many types and they are of particular importance in solitary wave theory. Here we are going to describe six different types of travelling wave solutions with their figures.

1.2.4.1. Solitary Waves and Solitons Solitary waves [230] are localized travelling waves with constant speeds and shapes, asymptotically zero at large distance solutions. Solitons are special kind of solitary waves.

The

soliton

solution

is

a

spatially

localized

𝑢′ (𝜉), 𝑢′′ (𝜉) and 𝑢′′′ (𝜉) → 0 as 𝜉 → ±∞ and as 𝜉 = 𝑥 − 𝑐𝑡.

7

solution,

hence

Fig. 1.3 (a): General representation of Solitons

Fig. 1.3 (b): 2D Animation

The KdV equation is a pioneer model for analytic bell-shaped 𝑠𝑒𝑐ℎ2 solitary wave solutions.

1.2.4.2. Kink Waves Kink waves [230] are travelling waves which rise or descend from one asymptotic state to another. The Kink solution approaches a constant at infinity. The standard dissipative burgers equation 𝑢𝑡 + 𝑢𝑢𝑥 = 𝑣𝑢𝑥𝑥 , is a well-known equation that gives kink solutions where v is the viscosity coefficient. The graphical representation of Kink wave is



1.2.4.3. Periodic Waves

8

Periodic waves [230] are travelling wave that are periodic such as cos(𝑥 − 𝑡). The standard wave equation 𝑢𝑡𝑡 = 𝑢𝑥𝑥 gives periodic solutions. Graphical representation is given below



1.2.4.4. Peakons Peakons [230] are the solitary wave solutions. In this case the travelling solutions are the smooth except for a peak at a corner of its crest. The Peakons are solutions retaining their shape and speed after interacting. Peakons were investigated and classified as a periodic peakons and peakons with exponential decay. Some graphical representations of peakons are given below



1.2.4.5. Cuspons

9

Cuspons [230] are other form of solitons where solution exhibit cusps at their crests. Unlike Peakons where the derivative at the peak differs only by a sign, the derivative at the jump of a Cuspons diverges. It is important to note that the Soliton solution 𝑢(𝑥, 𝑡), along with its derivatives tends to zero as |𝑥| → ∞. Cuspons were investigated and classified as periodic Cuspons and Cuspons with exponential decay.

Fig. 1.7 (a): General representation of Cuspons


1.2.4.6. Compactons It is new class of Soliton with compact spatial support such that each compacton is a Soliton confined to a finite core. Compactons [230] has some properties and we can describe it in one of the following ways: 1. Compactons are Solitons with finite wavelength. 2. Compactons are solitary waves with compact support. 3. Compactons are Solitons free of exponential tail. 4. Compactons are Solitons characterized by the absence of infinite wings. 5. Compactons are robust Soliton-like solutions. Graphical representation is given below

10



1.3. Preliminaries of Fractional Calculus 1.3.1. History Fractional calculus [263] has its origin in the question of the extension of meaning. A well known example is the extension of meaning of real numbers to complex numbers, and another is the extension of meaning of factorials of integers to factorials

of

complex

numbers. In generalized integration and differentiation the

question of the extension of meaning is: Can the meaning of derivatives of integral order

𝑑𝑛 𝑦

be extended to have meaning where n is any number irrational,

𝑑𝑥 𝑛

fractional or complex? Leibnitz invented the above notation. Perhaps, it was naive play with symbols that prompted L’Hôpital to ask Leibnitz about the possibility that n be a fraction. "What if 𝑛 be ½?” asked L’Hôpital. Leibnitz in 1695 replied, "It will lead to a paradox." But he added prophetically, "From this apparent paradox, one day useful consequences will be 𝜋

drawn." In 1697, Leibnitz [192], referring to Wallis's infinite product for 2 , used the 1

notation 𝑑 2 𝑦 and stated that differential calculus might have been used to achieve the same result. In 1819 the first mention of a derivative of arbitrary order appears in a text. The French mathematician S. F. Lacroix [132] published a 700 page text on differential and integral calculus in which he devoted less than two pages to this topic. Summary of above discussion

11

In the letters to J. Wallis and J. Bernulli (in 1697) Leibniz mentioned the possible approach to fractional-order differentiation in that sense, that for non-integer values of n the definition could be the following: 𝑑𝑛 𝑥 𝑚 𝑑𝑥 𝑛

= 𝑚(𝑚 − 1) … (𝑚 − 𝑛 + 1)𝑥 𝑚−𝑛 .

Γ(𝑚 + 1) = 𝑚(𝑚 − 1) … (𝑚 − 𝑛 + 1)Γ(𝑚 − 𝑛 + 1). 𝑑𝑛 𝑥 𝑚 𝑑𝑥 𝑛

Γ(𝑚+1)

= Γ(𝑚−𝑛+1) 𝑥 𝑚−𝑛 .

(1.1) (1.2) (1.3)

Euler suggested using this relationship also for negative or non-integer (rational) values of n. Taking m = 1 and n = ½ Euler obtained: 𝑑1/2 𝑥 𝑑𝑥 1/2

4𝑥

= √ 𝜋 𝑥 𝑚−𝑛 . 1

(1.4) 𝑥

𝑑

𝐷𝑥𝛼 𝑓(𝑥) = Γ(1−𝛼) 𝑑𝑥 ∫0

1 𝑓(𝑡)d𝑡, 0 (𝑥−𝑡)𝛼

≤ 𝛼 < 1.

(1.5)

1.3.2. Fractional Derivative of a Basic Power Function The half derivative (purple curve) of the function 𝑓(𝑥) = 𝑥 (blue curve) together with the first derivative (red curve). Let us assume that 𝑓(𝑥) is a monomial of the form 𝑓(𝑥) = 𝑥 𝑘 .

(1.6)

The first derivative is as usual 𝑑

𝑓 ′ (𝑥) = 𝑑𝑥 𝑓(𝑥) = 𝑘𝑥 𝑘−1 .

(1.7)

Repeating this gives the more general result that 𝑑𝛼 𝑑𝑥 𝛼

𝑘!

𝑥 𝑘 = (𝑘−𝑎)! 𝑥 𝑘−𝑎 ,

(1.8)

which, after replacing the factorials with the Gamma function, leads us to 𝑑𝛼 𝑑𝑥 𝛼

Γ(𝑘+1)

𝑥 𝑘 = Γ(𝑘−𝑎+1) 𝑥 𝑘−𝑎 .

(1.9)

1

For 𝑘 = 1 and 𝑎 = 2, we obtain the half-derivative of the function 𝑥 as 1

𝑑2

Γ(1+1)

𝑑𝑥 2

Γ(1− +1)

1𝑥 =

1 2

1

𝑥1−2 =

1! 3 2

Γ( )

1

𝑥2 =

1

2𝑥 2 √𝜋

,

Repeating this process yields 12

(1.10)

1

𝑑2 1 𝑑𝑥 2

−

1 2

1 2

2𝜋 𝑥 = 2𝜋

−

1 2

1 2

Γ(1+ ) 1 1 Γ( − +1) 2 2

𝑥

1 1 − 2 2

3

=

1 Γ( ) 2 2𝜋 −2 Γ(1) 𝑥0

=

2√𝜋𝑥 0 2√𝜋0!

= 1,

(1.11)

which is indeed the expected result of 1

(

1

𝑑2 𝑑2

1 𝑑𝑥 2

1 𝑑𝑥 2

𝑑

) 𝑥 = 𝑑𝑥 𝑥 = 1

(1.12)

This extension of the above differential operator need not be constrained only to real powers. For example, the (1 + 𝑖)th derivative of the (1 + 𝑖)th derivative yields the 2nd derivative. Also notice that setting negative values for yields integrals. For a general function 𝑓(𝑥) and 0 < 𝛼 < 1, the complete fractional derivative is 1

𝑑

𝑥 𝑓(𝑡) d𝑡 . (𝑥−𝑡)𝛼

𝐷𝛼 𝑓(𝑥) = Γ(1−𝛼) 𝑑𝑥 ∫0

(1.13)

For arbitrary 𝛼, since the gamma function is undefined for arguments whose real part is a negative integer, it is necessary to apply the fractional derivative after the integer derivative has been performed. For example, 3

1

1

𝑑

𝐷2 𝑓(𝑥) = 𝐷 2 𝐷1 𝑓(𝑥) = 𝐷2 𝑑𝑥 𝑓(𝑥).

(1.14)

1.3.3. Laplace Transform We can also come at the question via the Laplace transform. Noting that 𝑡

1

ℒ{𝐽𝑓}(𝑠) = ℒ {∫0 𝑓(𝑡)d𝑡} (𝑠) = 𝑠 (ℒ{𝑓})(𝑠),

(1.15)

and 1

1

ℒ{𝐽2 𝑓} = 𝑠 (ℒ{𝐽𝑓})(𝑠) = 𝑠2 (ℒ{𝑓})(𝑠).

(1.16)

etc., we assert 𝐽𝛼 𝑓 = ℒ −1 {𝑠 −𝛼 (ℒ{𝑓})(𝑠)}.

(1.17)

For example Γ(𝑘+1)

𝐽𝛼 (𝑡 𝑘 ) = ℒ −1 {𝑠𝑎+𝑘+1 },

(1.18)

Γ(𝑘+1)

= Γ(𝑘+𝑎+1) 𝑡 𝑎+𝑘 .

(1.19)

13

as expected. Indeed, given the convolution rule ℒ{𝑓 ∗ 𝑔} = (ℒ{𝑓})(ℒ{𝑔}) (and short handing 𝑝(𝑥) = 𝑥 𝛼−1 for clarity) we find that (𝐽𝛼 𝑓)(𝑡) =

1 Γ(𝛼)

ℒ −1 {(ℒ{𝑝})(ℒ{𝑓})},

1

= Γ(𝛼) (𝑝 ∗ 𝑓), 1

𝑡

1

𝑡

(1.20) (1.21)

= Γ(𝛼) ∫0 𝑝(𝑡 − 𝑥)𝑓(𝑥)d𝑥,

(1.22)

= Γ(𝛼) ∫0 (𝑡 − 𝑥)𝛼−1 𝑓(𝑥)d𝑥.

(1.23)

which is what Cauchy gave us above. Laplace transforms "work" on relatively few functions, but they are often useful for solving fractional differential equations.

1.3.4. Fractional Integrals 1.3.4.1. Riemann–Liouville Fractional Integral The classical form of fractional calculus is given by the Riemann–Liouville integral, essentially what has been described above. The theory for periodic functions, therefore including the 'boundary condition' of repeating after a period, is the Weyl integral. It is defined on Fourier series, and requires the constant Fourier coefficient to vanish (so, applies to functions on the unit circle integrating to 0). The Riemann-Liouville fractional integral operator [263] of order 𝛼 ≥ 0, of a function 𝑓 ∈ 𝐶𝜇 , 𝜇 ≥ −1, is defined as 1

𝑡

𝐽𝛼 𝑓(𝑡) = 𝛤(𝛼) ∫0 (𝑡 − 𝜏)𝛼−1 𝑓(𝜏)𝑑𝜏, 𝛼 > 0, 𝑡 > 0,

(1.24)

𝐽0 𝑓(𝑡) = 𝑓(𝑡).

(1.25)

properties of the operator 𝐽𝛼 , we mention only the following: For 𝑓 ∈ 𝐶𝜇 , 𝜇 ≥ −1, 𝛼, 𝛽 ≥ 0 and 𝛾 > −1: 1. 𝐽𝛼 𝐽𝛽 𝑓(𝑡) = 𝐽𝛼+𝛽 𝑓(𝑡).

(1.26)

2. 𝐽𝛼 𝐽𝛽 𝑓(𝑡) = 𝐽𝛽 𝐽𝛼 𝑓(𝑡).

(1.27)

𝛤(𝛾+1)

3. 𝐽𝛼 𝑡 𝛾 = 𝛤(𝛼+𝛾+1) 𝑡 𝛼+𝛾 .

(1.28)

The Riemann-Liouville derivative has certain disadvantages when trying to model realworld phenomena with fractional differential equations.

14

1.3.4.2. Erdélyi–Kober Integral Operator The Erdélyi–Kober operator is an integral operator introduced by Arthur Erdélyi (1940) and Hermann Kober (1940) and is given by 𝑥 −𝜈−𝛼+1 Γ(𝛼)

𝑥

∫0 (𝑡 − 𝑥)𝛼−1 𝑡 −𝛼−𝜈 𝑓(𝑡)d𝑡,

(1.29)

which generalizes the Riemann fractional integral and the Weyl integral.

1.3.4.3. Hadamard Fractional Integral The Hadamard fractional integral is introduced by J. Hadamard and is given by the following formula, 𝑥 𝛼−1

𝑥

1

𝐷𝑡−𝛼 𝑓(𝑥) = Γ(𝛼) ∫0 (log 𝑡 )

𝑓(𝑡)

d𝑡 𝑡

.

(1.30)

Note: A recent generalization is the following, which generalizes the Riemann-Liouville fractional integral and the Hadamard fractional integral. It is given by 𝜌

( 𝑎+𝐽𝑥𝛼 𝑓)(𝑥) =

𝜌1−𝛼

𝑥 𝑡 𝜌−1 𝑓(𝑡)

∫ Γ(𝛼) 0

(𝑥 𝜌 −𝑡 𝜌 )

d𝑡,

(1.31)

for 𝑥 > 𝑎.

1.3.5. Fractional Derivatives Not like classical Newtonian derivatives, a fractional derivative is defined via a fractional integral.

1.3.5.1. Riemann–Liouville Fractional Derivative The Riemann–Liouville fractional derivative [263] of order α for a function f is defined by 𝐷𝑥𝛼 𝑓(𝑥) =

1

𝑑

𝑛

𝑥

( ) ∫0 (𝑥 − 𝑠)𝑛−𝛼−1 𝑓(𝑠)𝑑𝑠, 𝛤(𝑛−𝛼) 𝑑𝑥

(1.32)

Note: the main disadvantage of Riemann–Liouville fractional derivative is that the fractional derivative of a constant is not zero. If 𝑓 = 𝐶, then 𝑥 −𝛼

𝐷𝑥𝛼 𝐶 = Γ(𝐶−𝛼).

(1.33)

15

1.3.5.2. Modified Riemann–Liouville Fractional Derivative The modified Riemann–Liouville [105] derivative is defined as 𝐷𝑥𝛼 𝑓(𝑥) =

1

𝑑

𝑛

𝑥

( ) ∫0 (𝑥 − 𝑠)𝑛−𝛼 (𝑓(𝑠) − 𝑓(0))𝑑𝑠, 𝛤(𝑛−𝛼) 𝑑𝑥

(1.34)

1.3.5.3. Caputo Fractional Derivative There is another option for computing fractional derivatives; the Caputo fractional derivative. It was introduced by M. Caputo [38] in 1967 in his celebrated paper. In contrast to the Riemann Liouville fractional derivative, when solving differential equations using Caputo's definition, it is not necessary to define the fractional order initial conditions. Caputo's definition is illustrated as follows; the fractional derivative of 𝑓(𝑡) in the Caputo sense is defined as 𝐶 𝛼 𝑎𝐷𝑡 𝑓(𝑡)

1

𝑡

= 𝐽𝑚−𝛼 𝐷𝑚 𝑓(𝑡) = 𝛤(𝑚−𝛼) ∫0 (𝑡 − 𝜏)𝑚−𝛼−1 𝑓 𝑚 (𝑡)𝑑𝑡,

(1.35)

𝑚 for 𝑚 − 1 < 𝛼 ≤ 1𝑚, 𝑚 ∈ 𝑁, 𝑡 > 0, 𝑓 ∈ 𝐶−1 .

1.3.5.4. Jumarie Fractional Derivative Let 𝑓 be a continuous function defined on 𝑅, then the Jumarie’s modified RiemannLiouville derivative [3, 7] of 𝑓 is defined as follows 𝐷𝑥𝛼 𝑓(𝑥) =

1

𝑑

𝑛

𝑥

( ) ∫𝑎 (𝑥 − 𝑡)𝑛−1−𝛼 (𝑓(𝑡) − 𝑓(0))d𝑡, 𝛤(𝑛−𝛼) 𝑑𝑥

(1.36)

where 0 ≤ 𝑛 − 1 ≤ 𝛼 < 𝑛 with 𝑛 ∈ 𝑁 ∗ . This fractional order derivative is in fact defined through the fractional difference ∆𝛼 [𝑓(𝑥)−𝑓(0)]

𝑓 𝛼 (𝑥) = lim (

ℎ𝛼

),

(1.37)

where ℎ > 0, FW 𝑓(𝑡) = 𝑓(𝑡 + ℎ) and ∆𝛼 𝑓(𝑥) = (FW − 1)𝛼 𝑓(𝑡),

(1.38)

𝛼!

𝑖 = ∑∞ 𝑖=0(−1) (𝛼−𝑖)!𝑖! 𝑓[𝑡 + (𝛼 − 𝑖)ℎ].

(1.39)

Let us recall that the Riemann-Liouville derivative is defined as follows 𝐷𝑥𝛼 𝑓(𝑥) =

1

𝑑

𝑛

𝑥

( ) ∫0 (𝑥 − 𝑠)𝑛−𝛼−1 𝑓(𝑠)𝑑𝑠, 𝛤(𝑛−𝛼) 𝑑𝑥

(1.40)

and 𝑑𝛼 𝑑𝑥 𝛼

Γ(𝑛+1)

𝑥 𝑛 = Γ(𝑛−𝑎+1) 𝑥 𝑛−𝑎 .

(1.41) 16

Consequently, the Jumarie’s modified Riemann-Liouville derivative can be expressed as follows 𝑡 −𝛼

𝐷𝑥𝛼 𝑓(𝑥) = 𝐷𝑥𝛼 𝑓(𝑡) − Γ(1−𝛼) 𝑓(0).

(1.42)

Note that the Jumarie and the Riemann–Liouville fractional derivatives are equal if 𝑓(0) = 0. The advantage of Jumarie’s definition with respect to the classical definition of Riemann–Liouville is that the fractional derivative of a constant is now zero, as desired. One of some useful properties of the Jumarie’s modified Riemann-Liouville derivative is the fractional Leibniz derivative rule Table 1.1. Fractional derivative of any order Lacroix

𝑑𝑛 𝑥 𝑚 Γ(𝑚 + 1) = 𝑥 𝑚−𝑛 𝑛 𝑑𝑥 Γ(𝑚 − 𝑛 + 1)

Liouville

𝑑 𝜈 𝑥 −𝛼 Γ(𝛼 + 𝜈) −𝛼−𝜈 = (−1)𝜈 𝑥 𝜈 𝑑𝑥 Γ(𝛼)

Laurent

𝑚−𝜌

𝐷𝑥𝜈 𝑓(𝑥) = 𝐷𝑥

𝐷∗𝛼 𝑓(𝑡) =

Caputo

Jumarie

𝑥 𝑑𝑚 1 (𝑥 − 𝑡)𝜌−1 𝑓(𝑡)d𝑡 ] [ ∫ 𝑑𝑥 𝑚 Γ(𝜌) 0

𝑓+𝛼 = ∫ 𝑓(𝑡)

Cauchy

Riemann-Liouville

𝑓(𝑥) =

𝐷𝑥𝛼 𝑓(𝑥) 𝐷𝑥𝛼 𝑓(𝑥)

(𝑥 − 𝑡)−𝛼−1 d𝑡 Γ(−𝛼)

𝑡 1 𝑓 𝑚 (𝑠) ∫ 𝑑𝑠 Γ(𝑚 − 𝛼) 0 (𝑡 − 𝑠)𝛼+1−𝑚

1 𝑑 𝑛 𝑥 = ( ) ∫ (𝑥 − 𝑠)𝑛−𝛼−1 𝑓(𝑠)𝑑𝑠 𝛤(𝑛 − 𝛼) 𝑑𝑥 0

1 𝑑 𝑛 𝑥 = ( ) ∫ (𝑥 − 𝑡)𝑛−1−𝛼 (𝑓(𝑡) − 𝑓(0))d𝑡 𝛤(𝑛 − 𝛼) 𝑑𝑥 𝑎

1.4. Introductory Concepts of Integral Equations In this section we discuss some basic definition regarding integral equation as below:

1.4.1. Integral Equations Any functional equation in which the unknown function appears under the sign of integration is called an integral equation. 𝛽(𝑥)

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡, where 17

(1.43)



𝛼(𝑥) and 𝛽(𝑥) are limits of integration,



𝜆 is a constant parameter



𝐾(𝑥, 𝑡) is a function of two variables 𝑥 and 𝑡



The function 𝑢(𝑥) is unknown and



The functions 𝑓(𝑥) real function.

1.4.2. Classifications of Integral Equations Integral equations appear in many forms. Two distinct ways that depend on the limits of integration are used to characterize integral equations, namely:

1.4.2.1. Fredholm Integral Equations If the limits of integration are fixed, the integral equation is called a Fredholm integral equation given in the form: 𝑏

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡.

(1.44)

where a and b are constants.

1.4.2.2. Volterra Integral Equations If at least one limit is a variable, the equation is called a Volterra integral equation given in the form: 𝛽(𝑥)

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫𝑎

𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡.

(1.45)

Moreover, two other distinct kinds, which depend on the appearance of the unknown function u(x), are defined as follows:

1.4.2.3. First Kind of Fredholm Integral Equation If the unknown function 𝑢(𝑥) appears only under the integral sign of Fredholm integral equation, the integral equation is called a first kind Fredholm integral equation. or

𝑏

𝑓(𝑥) = 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡.

(1.46)

1.4.2.4. First Kind of Volterra Integral Equations If the unknown function 𝑢(𝑥) appears only under the integral sign of Volterra integral equation, the integral equation is called a first kind Volterra integral equation. 𝛽(𝑥)

𝑓(𝑥) = 𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡,

(1.47)

1.4.2.5. Second Kind Fredholm Integral Equations 18

If the unknown function 𝑢(𝑥) appears both inside and outside the integral sign of Fredholm integral equation, the integral equation is called a second kind Fredholm integral equation 𝑏

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡.

(1.48)

1.4.2.6. Second Kind Volterra Integral Equations If the unknown function 𝑢(𝑥) appears both inside and outside the integral sign of Volterra integral equation, the integral equation is called a second kind Volterra integral equation 𝛽(𝑥)

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡,

(1.49)

1.4.3. Homogeneous Fredholm Integral Equations In all Fredholm integral equations presented above, if 𝑓(𝑥) is identically zero, the resulting equation: is called homogeneous Fredholm integral equation or

𝑏

𝑢(𝑥) = 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡.

(1.50)

1.4.4. Homogeneous Volterra Integral Equations In all Volterra integral equations presented above, if 𝑓(𝑥) is identically zero, the resulting equation: is called homogeneous Volterra integral equation 𝛽(𝑥)

𝑢(𝑥) = 𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡,

(1.51)

1.4.5. Integro-Differential Equation It is interesting to point out that any equation that includes both integrals and derivatives of the unknown function 𝑢(𝑥) is called integro-differential equation.

1.4.5.1. Fredholm Integro-Differential Equation The Fredholm integro-differential equation is of the form: 𝑑𝑘 𝑢

𝑏

𝑢𝑘 (𝑥) = 𝑓(𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡, where 𝑢𝑘 (𝑥) = 𝑑𝑥 𝑘 .

(1.52)

1.4.5.2. Volterra Integro-Differential Equation The Volterra integro-differential equation is of the form: 𝑑𝑘 𝑢

𝛽(𝑥)

𝑢𝑘 (𝑥) = 𝑓(𝑥) + 𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡, where 𝑢𝑘 (𝑥) = 𝑑𝑥 𝑘 .

(1.53)

The integro-differential equations will be defined and classified in this text.

1.4.6. Linearity Concept 19

If the exponent of the unknown function 𝑢(𝑥) inside the integral sign is one, the integral equation or the integro-differential equation is called linear. If the unknown function 𝑢(𝑥) has exponent other than one, or if the equation contains nonlinear functions of 𝑢(𝑥), such as sinh(𝑢), cos(𝑢), ln(1 + 𝑢), the integral equation or the integro-differential equation is called nonlinear. To explain this concept, we consider the equations: 𝑥

𝑢(𝑥) = 1 + 𝑥 + ∫0 (𝑥 − 𝑡) cos(𝑢) d𝑡, 1

𝑢′′ (𝑥) = 1 + ∫0 𝑥𝑡 exp(𝑢) 𝑢(𝑡)d𝑡.

(1.54) (1.55)

1.4.7. Homogeneity Concept Integral equations and integro-differential equations of the second kind are classified as homogeneous or inhomogeneous, if the function 𝑓(𝑥) in the second kind of Volterra or Fredholm integral equations or integro-differential equations is identically zero, the equation is called homogeneous. Otherwise it is called inhomogeneous. Notice that this property holds for equations of the second kind only. To clarify this concept we consider the following equations 𝑥

𝑢(𝑥) = ∫0 (𝑥 − 𝑡)𝑢(t)d𝑡,

(1.56)

1

𝑢′ (𝑥) = ∫0 𝑥𝑡 exp(𝑢) 𝑢(𝑡)d𝑡.

(1.57)

1.4.8. Singular Integral Equations Volterra integral equations of the first kind 𝛽(𝑥)

𝑓(𝑥) = 𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡,

(1.58)

or of the second kind 𝛽(𝑥)

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡,

(1.59)

are called singular if: 1.

One of the limits of integration 𝑔(𝑥), ℎ(𝑥) or both are infinite, or

2.

If the kernel 𝐾(𝑥, 𝑡) becomes infinite at one or more points at the range of

integration. 𝑥 𝑢(𝑡)

𝑥 + 1 = ∫0

1

d𝑡,

(1.60)

(𝑥−𝑡)4 𝑥 𝑢(𝑡)

𝑢(𝑥) = 𝑥 2 + ∫0

√𝑥−𝑡

d𝑡,

(1.61)

1.4.8.1. Abel’s Integral Equation 20

The Abel’s integral equations is given as 𝑥

𝑓(𝑥) = ∫0

1 (𝑥 𝜌 −𝑡 𝜌 )𝛼

𝑢(𝑡)d𝑡.

(1.62)

1.4.8.2. Weakly Singular Integral Equation If α < n the kernel (and the integral) is called weakly singular then Eq. (1.43) is called weakly singular integral equation.

1.4.9. System of Integral Equations 1.4.9.1. System of Fredholm Integral Equations Systems of Fredholm integral equation have attractive much concern in applied sciences the system of Fredholm integral equation appears in two kinds. The System of Fredholm integral equation of the first kind reads 𝑏 ̃1 (𝑥, 𝑡)𝑣(𝑡)) d𝑡, 𝑓1 (𝑥) = ∫𝑎 (𝑘1 (𝑥, 𝑡)𝑢(𝑡) + 𝑘

(1.63)

𝑏 ̃2 (𝑥, 𝑡)𝑣(𝑡)) d𝑡, 𝑓2 (𝑥) = ∫𝑎 (𝑘2 (𝑥, 𝑡)𝑢(𝑡) + 𝑘

(1.64)

where the unknown functions 𝑢(𝑥) and 𝑣(𝑥) appear only under the integral sign, and 𝑎 and 𝑏 are constants. However, for systems of Fredholm integral equations of the second kind, the unknown functions 𝑢(𝑥) and 𝑣(𝑥) appear inside and outside the integral sign. The second kind is represented by the form 𝑏 ̃1 (𝑥, 𝑡)𝑣(𝑡)) d𝑡 , 𝑢(𝑥) = 𝑓1 (𝑥) + ∫𝑎 (𝑘1 (𝑥, 𝑡)𝑢(𝑡) + 𝑘

(1.65)

𝑏 ̃2 (𝑥, 𝑡)𝑣(𝑡)) d𝑡. 𝑣(𝑥) = 𝑓2 (𝑥) + ∫𝑎 (𝑘2 (𝑥, 𝑡)𝑢(𝑡) + 𝑘

(1.66)

The systems of Fredholm integro-differential equations have also attracted a considerable size of interest. These systems are given by 𝑏 ̃1 (𝑥, 𝑡)𝑣(𝑡)) d𝑡, 𝑢𝑖 (𝑥) = 𝑓1 (𝑥) + ∫𝑎 (𝑘1 (𝑥, 𝑡)𝑢(𝑡) + 𝑘

(1.67)

𝑏 ̃2 (𝑥, 𝑡)𝑣(𝑡)) d𝑡. 𝑣 𝑖 (𝑥) = 𝑓2 (𝑥) + ∫𝑎 (𝑘2 (𝑥, 𝑡)𝑢(𝑡) + 𝑘

(1.68)

where the initial conditions for the last system should be prescribed.

1.4.9.2. System of Volterra Integral Equations Systems of integral equations, linear or nonlinear, appear in scientific applications in engineering, physics, chemistry and populations growth models. Studies of systems of integral equations have attracted much concern in applied sciences. The general ideas and the essential features of these systems are of wide applicability. The systems of Volterra 21

integral equations appear in two kinds. For systems of Volterra integral equations of the first kind, the unknown functions appear only under the integral sign in the form: 𝑥 ̃1 (𝑥, 𝑡)𝑣(𝑡) + ⋯ )d𝑡, 𝑓1 (𝑥) = ∫0 (𝑘1 (𝑥, 𝑡)𝑢(𝑡) + 𝑘 𝑥

̃2 (𝑥, 𝑡)𝑣(𝑡) + ⋯ )d𝑡. 𝑓2 (𝑥) = ∫0 (𝑘2 (𝑥, 𝑡)𝑢(𝑡) + 𝑘

(1.69) (1.70)

However, systems of Volterra integral equations of the second kind, the unknown functions appear inside and outside the integral sign of the form: 𝑥 ̃1 (𝑥, 𝑡)𝑣(𝑡) + ⋯ )d𝑡, 𝑢(𝑥) = 𝑓1 (𝑥) + ∫0 (𝑘1 (𝑥, 𝑡)𝑢(𝑡) + 𝑘 𝑥

̃2 (𝑥, 𝑡)𝑣(𝑡) + ⋯ )d𝑡. 𝑣(𝑥) = 𝑓2 (𝑥) + ∫0 (𝑘2 (𝑥, 𝑡)𝑢(𝑡) + 𝑘

(1.71) (1.72)

̃1 (𝑥, 𝑡), and the functions𝑓𝑖 (𝑥), 𝑖 = 1, 2, . . ., n are given realThe kernels 𝐾𝑖 (𝑥, 𝑡) and 𝑘 valued functions. A variety of analytical and numerical methods are used to handle systems of Volterra integral equations. The existing techniques encountered some difficulties in terms of the size of computational work, especially when the system involves several integral equations.

1.4.9.3. System of Singular Integral Equations Systems of singular integral equations appear in many branches of scientific fields, such as microscopy, seismology, radio astronomy, electron emission, atomic scattering, radar ranging, plasma diagnostics, X-ray radiography, and optical fiber evaluation. Studies of systems of singular integral equations have attracted much concern in applied sciences. The use of computer symbolic systems such as Maple and Mathematica facilitates the tedious work of computation. The general ideas and the essential features of these systems are of wide applicability. The well known systems of singular integral equations are given by 𝑥

𝑓1 (𝑥) = ∫0 (𝑘11 (𝑥, 𝑡)𝑢(𝑡) + 𝑘12 (𝑥, 𝑡)𝑣(𝑡))d𝑡, 𝑥

𝑓2 (𝑥) = ∫0 (𝑘21 (𝑥, 𝑡)𝑢(𝑡) + 𝑘22 (𝑥, 𝑡)𝑣(𝑡))d𝑡,

(1.73) (1.74)

and 𝑥

𝑢(𝑥) = 𝑓1 (𝑥) + ∫0 (𝑘11 (𝑥, 𝑡)𝑢(𝑡) + 𝑘12 (𝑥, 𝑡)𝑣(𝑡))d𝑡 , 𝑥

𝑣(𝑥) = 𝑓2 (𝑥) + ∫0 (𝑘21 (𝑥, 𝑡)𝑢(𝑡) + 𝑘22 (𝑥, 𝑡)𝑣(𝑡))d𝑡. Where the kernels 𝑘𝑖𝑗 are singular kernel given by 𝑘𝑖𝑗 =

1 𝛼 , (𝑥−𝑡) 𝑖𝑗

1 ≤ 𝑖 , 𝑗 ≤ 2. 22

(1.75) (1.76)

1.5. Some Basic Definitions of Wavelets Theory 1.5.1. History Wavelet analysis developed in the largely mathematical literature in the 1980's and began to be used commonly in geophysics in the 1990's. Wavelets can be used in signal analysis, image processing and data compression. They are useful for sorting out scale information, while still maintaining some degree of time or space locality.

1.5.2. Wavelets A wavelet [155, 244] is a wave-like oscillation that is localized in the sense that it grows from zero, reaches maximum amplitude, and then decreases back to zero amplitude again. It thus has a location where it maximizes, a characteristic oscillation period, and also a scale over which it amplifies and declines.

Wavelets constitute a family of function constructed from dilaition

1.5.3. Types of Wavelets According to Meyer (1993), two fundamental types of wavelets [244] can be considered, 1. The Grossmann-Morlet time-scale wavelets 2. The Gabor-Malvar time-frequency wavelets. Time-scale wavelets are defined in reference to a "mother function" 𝜓(𝑡) of some real variable 𝑡. The mother function can be used to generate a whole family of wavelets by translating and scaling the mother wavelet. 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|−2 𝜓 (

𝑎

) , 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

Here 𝑏 is the translation parameter and 𝑎 is the scaling parameter. Provided that 𝜓(𝑡) is real-valued, this collection of wavelets can be used as an orthonormal basis. If we restrict

23

the parameters 𝑎 and 𝑏 to discrete values as 𝑎 = 𝑎0−𝑘 , 𝑏 = 𝑛𝑏0 𝑎0−𝑘 , 𝑎0 > 1, 𝑏0 > 0 and 𝑛, 𝑘 are positive integers, we have the following family of discrete wavelets 1

𝜓𝑘,𝑛 (𝑡) = |𝑎|−2 𝜓((𝑎0 )𝑘 𝑡 − 𝑛𝑏0 ), where 𝜓𝑘,𝑛 (𝑡) forms an orthogonal basis.

1.5.4. Functions Approximation A function 𝑓(𝑡) define over [0, 1) may be expanded [250] as ∞ 𝑓(𝑡) = ∑∞ 𝑛=1 ∑𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡),

(1.77)

where 𝑐𝑛𝑚 = 〈𝑓(𝑡), 𝜓𝑛𝑚 (𝑡)〉, in which 〈∙,∙〉 denotes the inner product. If the infinite series in Eq. (7) is truncated, then Eq. (7) can be written as 𝑘−1

𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡).

(1.78)

It can be written as 𝑘−1

𝑇 𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡) = 𝐂 𝛙(𝑡).

(1.79)

where 𝐶 and 𝜓(𝑡) are 2𝑘−1 𝑀 × 1 matrices given as 𝑇

𝐂 = [𝑐10 , 𝑐11 , … , 𝑐1𝑀−1 , 𝑐20 , 𝑐21 , … , 𝑐2𝑀−1 , … , 𝑐2𝑘−1 0 , … , 𝑐2𝑘−1 𝑀−1 ] , and

𝑇

𝛙(𝑡) = [𝜓10 , 𝜓11 , … , 𝜓1𝑀−1 , 𝜓20 , 𝜓21 , … , 𝜓2𝑀−1 , … , 𝜓2𝑘−1 0 , … , 𝜓2𝑘−1 𝑀−1 ] .

24

Chapter 2 Approximate Solutions of Differential Equations

25

2.1. Introduction This Chapter, contains new work on series solutions of linear and nonlinear partial differential equations and integral equations. Adomian’s Decomposition Method (ADM) has been applied for solving Micro polar flow in a porous channel with mass injection and squeezing flow and heat transfer between two parallel disks with velocity slip and temperature jump, Modified Adomian’s Decomposition Method (MADM) is applied on Goursat equation, fourth-order Parabolic equation, inhomogeneous wave equation and inhomogeneous nonlinear wave equation, Optimal Adomian’s Decomposition Method (OADM) has been applied to obtain the numerical solutions of Fredholm integral equations. Variational Iteration Method (VIM) has been applied to obtain the numerical solutions of MHD flow over a nonlinear stretching sheet and MHD flow of an incompressible viscous fluid through convergent or divergent channels in presence of a high magnetic field, Modified Variational Iteration Method (MVIM) is applied to find the approximate solutions of nonlinear Partial Differential Equation, Burger’s equation, fourth-order Parabolic equation, Diffusion equation and Reaction-diffusion equation, Homotopy Perturbation Method (HPM) for solving Flow of a viscoelastic fluid through a porous channel with expanding or contracting walls and MHD flow of an incompressible viscous fluid through convergent or divergent channels in presence of a high magnetic field, Modified Homotopy Perturbation Method (MHPM) is applied to construct the numerical solutions of nonlinear Dispersive K(2,2,1) equation, nonlinear Dispersive K(3,3,1) equation, Kawahara equation, modified Kawahara equation, nonlinear Dispersive K(2,2) equation, nonlinear Dispersive K(3,3) equation, (2+1)-Dimensional nonlinear Dispersive K(2,2) equation and homogeneous Convection-Diffusion problem, Homotopy Analysis Method (HAM) is used to solve homogenous linear Goursat problem, inhomogeneous linear Goursat problem, inhomogeneous nonlinear Goursat problem, system of initial value problems, K(m,p,1) equation, nonlinear Schrodinger equation and veriational problem, Modified Homotopy Analysis Method (MHAM) implemented to homogenous Helmholtz equation, SOG-KdV equation, cubic nonlinear Schrodinger equation, Heat and wave-like equation, system of Partial Differential Equations (PDEs) and ZK(m,n,p) equation, MHD flow of an incompressible viscous fluid through convergent or divergent channels in presence of a high magnetic field and flow 26

of a viscoelastic fluid through a porous channel with expanding or contracting walls. A concrete relation between the basic ideas of the proposed technique and the existing literature is also presented. These methods appeared to be easier and more convenient by means of a symbolic computation system.

2.2. Adomian’s Decomposition Method 2.2.1. Methodology The Adomian’s decomposition method has been getting a lot of attention in recent years in applied mathematics in general, and in the area of series solutions in particular. The method proved to be powerful, effective, and can easily handle a wide class of linear or nonlinear, ordinary or partial differential equations, and linear and nonlinear integral equations. The decomposition method demonstrates fast convergence of the solution and therefore provides several significant advantages. The method attacks the problem in a direct way and in a straightforward fashion without using linearization, perturbation or any other restrictive assumption that may change the physical behavior of the model under discussion. The Adomian’s decomposition method was introduced and developed by George Adomian in [14, 15] and is well addressed in the literature. Adomian applied Adomian’s decomposition method for solving [13] and [12] equations, Wazwaz implemented Adomian’s decomposition method to obtain the exact solutions of fourth-order parabolic partial differential equations [221], third-order dispersive partial differential equations [216] and higher order boundary value problems [217]. Marasi and Nikbakht used Adomian decomposition method for boundary eigenvalue problems [131] and Abbasbandy applied Adomian’s decomposition method for solving Riccati differential equations of second kind [5], and nonlinear equations [6]. Many modifications have been made;

Wazwaz [215] introduced a new modification by decomposing initial guess into two parts, this modification was applied to obtain the exact solutions of Volterra-Fredholm integral equations [218], linear integral equations [224] and nonlinear differential equations [225]. Luo used two step Adomian’s decomposition method to construct analytical solution of nonlinear evolution equation [125]. Later on, Wazwaz [224] introduced a new modification by decomposing initial guess into Taylor’s series for solving integral equations. Consider the nonlinear differential equation 27

𝐿𝑢 + 𝑅𝑢 + 𝑁𝑢 = 𝑔,

(2.1)

where L is, mostly the lower order derivative which is assumed to be invertible, R is other linear differential operator, N is a non linear term and g is a source term. We next apply the inverse operator 𝐿−1 on both sides of equation (2.1) and using the given conditions to obtain, 𝑢 = 𝑓 − 𝐿−1 (𝑅𝑢 + 𝑁𝑢),

(2.2)

where the function f represents the terms arising from integrating the source term g and from using the given conditions that are assumed to be prescribed. As indicated before, Adomian’s method defines the solution u by an infinite series of components given by, 𝑢 = ∑∞ 𝑛=𝑥0 𝑢𝑛 ,

(2.3)

where the components 𝑢0 , 𝑢1 , 𝑢2 …are usually recurrently determined. Substituting (2.3) into both sides of (2.2) leads to, ∞ ∞ −1 ∑∞ 𝑛=0 𝑢𝑛 = 𝑓 − 𝐿 (𝑅(∑𝑛=0 𝑢𝑛 ) + 𝑁(∑𝑛=0 𝑢𝑛 )).

(2.4)

Accordingly, the formal recursive relation is defined by 𝑢0 = 𝑓, 𝑢𝑘+1= − 𝐿−1 (𝑅𝑢𝑘 + 𝐴𝑘 ).

2.2.2. Micro Polar Flow in a Porous Channel with Mass Injection We consider steady, incompressible, laminar flow of a micropolar fluid along a two dimensional channel with porous walls through which fluid is uniformly injected or removed with speed 𝑞 [187]. Using Cartesian coordinates, the channel walls are parallel to the 𝑥 −axis and located at 𝑦 = ±ℎ, where 2ℎ is the channel width. The relevant equations governing the flow are 𝜕𝑢

𝜕𝑣

+ 𝜕𝑦 = 0, 𝜕𝑥

(2.5)

𝜕𝑢

𝜕𝑢

1 𝜕𝑃

𝜅

𝜕2 𝑢

𝜕2 𝑢

𝜅 𝜕𝑁

𝜕𝑣

𝜕𝑣

1 𝜕𝑃

𝜅

𝜕2 𝑣

𝜕2 𝑣

𝜅 𝜕𝑁

𝜕𝑁

𝜕𝑁

𝜅

𝑢 𝜕𝑥 + 𝑣 𝜕𝑦 = − 𝜌 𝜕𝑥 + (𝜈 + 𝜌) (𝜕𝑥 2 + 𝜕𝑦 2 ) + 𝜌 𝜕𝑦 , 𝑢 𝜕𝑥 + 𝑣 𝜕𝑦 = − 𝜌 𝜕𝑦 + (𝜈 + 𝜌) (𝜕𝑥 2 + 𝜕𝑦 2 ) − 𝜌 𝜕𝑥 , 𝜕𝑢

𝜕𝑣

𝜈

𝜕2 𝑁

𝜕2 𝑁

𝑢 𝜕𝑥 + 𝑣 𝜕𝑦 = − 𝜌𝑗 (2𝑁 + 𝜕𝑦 − 𝜕𝑥) + 𝜌𝑗𝑠 ( 𝜕𝑥 2 + 𝜕𝑦 2 ).

(2.6) (2.7) (2.8)

Compared with Newtonian fluids, the governing equations include the micro rotation or angular velocity N whose direction of rotation is in the 𝑥𝑦 −plane, and the material parameters 𝑗, 𝜅 and 𝜈𝑠 . For consistency with other micropolar studies, all material 28

parameters are taken as independent and constant. When these constants are zero, the governing equations reduce to those given by [26]. The appropriate physical boundary conditions are 𝜕𝑢

𝑢(𝑥, ±ℎ) = 0, 𝑣(𝑥, ±ℎ) = ±𝑞, 𝑁(𝑥, ±ℎ) = −𝑠 𝜕𝑦|

,

(2.9)

(𝑥,±ℎ)

For symmetric flow about 𝜕𝑢 𝜕𝑦

(𝑥, 0) = 𝑣(𝑥, 0) = 0,

(2.10)

where 𝑞 > 0 corresponds to suction, 𝑞 < 0 to injection, and s is a boundary parameter that is used to model the extent to which microelements are free to rotate in the vicinity of the channel walls. For example, the value s = 0 corresponds to the case where 1

microelements close to a wall are unable to rotate, whereas the value 𝑠 = 2 corresponds to the case where the micro rotation is equal to the fluid vorticity at the boundary. To simplify the governing equations, we use following similarity transforms 𝑞𝑥

𝜓 = −𝑞𝑥𝑓(𝜂), 𝑦

where 𝜂 = ℎ , 𝑢 =

𝜕𝜓 𝜕𝑦

=−

𝑁 = ℎ2 𝑔(𝜂), 𝑞𝑥 ℎ

(2.11)

𝜕𝜓

𝑓 ′ (𝜂), 𝑣 = − 𝜕𝑥 = 𝑞𝑓(𝜂).

(2.12)

Substitute Eq. (2.11) into Eq. (2.5-2.8), we have (1 + 𝑁1 )𝑓 ′′′′ (𝜂) − 𝑁1 𝑔′′ (𝜂) − 𝑅𝑒(𝑓(𝜂)𝑓 ′′′ (𝜂) − 𝑓 ′ (𝜂)𝑓 ′′ (𝜂)) = 0,

(2.13)

𝑁2 𝑔′′ (𝜂) + 𝑁1 (𝑓 ′′ (𝜂) − 2𝑔(𝜂)) − 𝑁3 𝑅𝑒(𝑓(𝜂)𝑔′ (𝜂) − 𝑓 ′ (𝜂)𝑔(𝜂)) = 0, (2.14) where primes denote differentiation with respect to 𝜂. 𝑁1 , 𝑁2 , 𝑁3 and 𝑅𝑒 are dimensionless parameters and introduced as follow 𝜅

𝑁1 = 𝜌𝜈,

𝜈

𝑁2 = 𝜌𝜈ℎ𝑠 2 ,

𝑗

𝑁3 = ℎ2

and

𝑅𝑒 =

𝑞ℎ 𝜈

,

(2.15)

where 𝑅𝑒 > 0 corresponds to suction, and 𝑅𝑒 < 0 to injection. The boundary conditions are 𝑓(±) = 1,

𝑓 ′ (±1) = 0,

𝑔(±1) = 𝑠𝑓 ′′ (±1).

(2.16)

For the symmetric flow 𝑓(0) = 𝑓 ′′ (0) = 𝑓 ′ (1) = 0,

𝑓(1) = 1, 𝑔(1) = 𝑠𝑓 ′′ (1).

(2.17)

In this work, we set 𝑠 = 0, 𝑁1 = 𝑁2 = 1, 𝑁3 = 0.1, 2𝑓 ′′′′ (𝜂) − 𝑔′′ (𝜂) − 𝑅𝑒(𝑓(𝜂)𝑓 ′′′ (𝜂) − 𝑓 ′ (𝜂)𝑓 ′′ (𝜂)) = 0,

(2.18)

𝑔′′ (𝜂) + (𝑓 ′′ (𝜂) − 2𝑔(𝜂)) − 0.1 𝑅𝑒(𝑓(𝜂)𝑔′ (𝜂) − 𝑓 ′ (𝜂)𝑔(𝜂)) = 0, 29

(2.19)

with 𝑓(0) = 𝑓 ′′ (0) = 𝑓 ′ (1) = 0, 𝑓(1) = 1, 𝑔(0) = 0, 𝑔(1) = 0,

(2.20)

and investigate solution behavior of the symmetric flow with mass injection (𝑞 < 0) as 𝑅𝑒 is varied. According to Adomian’s Decomposition Method (ADM) Eq. (2.18)-Eq. (2.20) reduces to 𝐿1 𝑓(𝜂) = 𝑔′′ (𝜂) + 𝑅𝑒(𝑓(𝜂)𝑓 ′′′ (𝜂) − 𝑓 ′ (𝜂)𝑓 ′′ (𝜂)),

(2.21)

𝐿2 𝑔(𝜂) = −𝑓 ′′ (𝜂) + 2𝑔(𝜂) + 0.1 𝑅𝑒(𝑓(𝜂)𝑔′ (𝜂) − 𝑓 ′ (𝜂)𝑔(𝜂)), (2.22) 𝑑4

𝑑2

where 𝐿1 = 𝑑𝜂4 and 𝐿2 = 𝑑𝜂2 . Applying 𝐿1 −1 and 𝐿−1 2 on both sides of Eq. (2.212.-22) we get, 𝑓(𝜂) = 𝑓(0) + 𝜂𝑓 ′ (0) + 𝜂

𝜂

𝜂

𝜂2

𝑓 ′′ (0) + 2!

𝜂3 3!

𝑓 ′′′ (0)

𝜂

+ ∫0 ∫0 ∫0 ∫0 [𝑔′′ (𝜂) + 𝑅𝑒(𝑓(𝜂)𝑓 ′′′ (𝜂) − 𝑓 ′ (𝜂)𝑓 ′′ (𝜂))]d𝜂d𝜂d𝜂d𝜂,

(2.23)

𝑔(𝜂) = 𝑔(0) + 𝜂𝑔′ (0) 𝜂

𝜂

+ ∫0 ∫0 [2𝑔(𝜂) − 𝑓 ′′ (𝜂) + 0.1 𝑅𝑒(𝑓(𝜂)𝑔′ (𝜂) − 𝑓 ′ (𝜂)𝑔(𝜂))]d𝜂d𝜂 ,

(2.24)

Using BCs and letting 𝑓 ′ (0) = 𝑎1 , 𝑓 ′′′ (0) = 𝑎2 and 𝑔′ (0) = 𝑎3 in Eq. (2.23-2.24), we obtained 𝑓(𝜂) = 𝜂𝑎1 + 𝜂

𝜂

𝜂

𝜂3 3!

𝑎2

𝜂

+ ∫0 ∫0 ∫0 ∫0 [𝑔′′ (𝜂) + 𝑅𝑒(𝑓(𝜂)𝑓 ′′′ (𝜂) − 𝑓 ′ (𝜂)𝑓 ′′ (𝜂))]d𝜂d𝜂d𝜂d𝜂, 𝜂

(2.25)

𝜂

𝑔(𝜂) = 𝜂𝑎3 + ∫0 ∫0 [2𝑔(𝜂) − 𝑓 ′′ (𝜂) + 0.1 𝑅𝑒(𝑓(𝜂)𝑔′ (𝜂) − 𝑓 ′ (𝜂)𝑔(𝜂))]d𝜂d𝜂, (2.26) According to Adomian’s Decomposition Method (ADM), we have the following recurrence relation 𝑓0 (𝜂) = 𝜂𝑎1 +

𝜂3 3!

𝑎2 ,

𝜂

𝜂

𝜂

𝜂

𝜂

𝑔0 (𝜂) = 𝜂𝑎3 , 𝜂

𝑓𝑘+1 (𝜂) = ∫0 ∫0 ∫0 ∫0 [𝑔𝑘′′ (𝜂) + 𝑅𝑒(𝐴𝑘 − 𝐵𝑘 )]d𝜂d𝜂d𝜂d𝜂, 𝑘 ≥ 0, 𝑔𝑘+1 (𝜂) = ∫0 ∫0 [2𝑔(𝜂) − 𝑓 ′′ (𝜂) + 0.1 𝑅𝑒(𝐶𝑘 − 𝐷𝑘 )]d𝜂d𝜂 , 𝑘 ≥ 0, where 𝐴𝑘 , 𝐵𝑘 , 𝐶𝑘 𝑎𝑛𝑑 𝐷𝑘 are Adomian’s polynomials which are given by 𝐴0 = 𝑓0 (𝜂)𝑓0 ′′′ (𝜂), 𝐴1 = 𝑓1 (𝜂)𝑓0 ′′′ (𝜂) + 𝑓0 (𝜂)𝑓1 ′′′ (𝜂), 𝐴2 = 𝑓2 (𝜂)𝑓0 ′′′ (𝜂) + 𝑓1 (𝜂)𝑓1 ′′′ (𝜂) + 𝑓0 (𝜂)𝑓2 ′′′ (𝜂), 30

𝐴3 = 𝑓3 (𝜂)𝑓0 ′′′ (𝜂) + 𝑓2 (𝜂)𝑓1 ′′′ (𝜂) + 𝑓1 (𝜂)𝑓2 ′′′ (𝜂) + 𝑓0 (𝜂)𝑓3 ′′′ (𝜂), . . .. 𝐵0 = 𝑓0 ′ (𝜂)𝑓0 ′′ (𝜂), 𝐵1 = 𝑓1 ′ (𝜂)𝑓0 ′′ (𝜂) + 𝑓0 ′ (𝜂)𝑓1 ′′ (𝜂), 𝐵2 = 𝑓2 ′ (𝜂)𝑓0 ′′ (𝜂) + 𝑓1 ′ (𝜂)𝑓1 ′′ (𝜂) + 𝑓0 ′ (𝜂)𝑓2 ′′ (𝜂), 𝐵3 = 𝑓3 ′ (𝜂)𝑓0 ′′ (𝜂) + 𝑓2 ′ (𝜂)𝑓1 ′′ (𝜂) + 𝑓1 ′ (𝜂)𝑓2 ′′ (𝜂) + 𝑓0 ′ (𝜂)𝑓3 ′′ (𝜂), . . .. ′ 𝐶0 = 𝑓0 (𝜂)𝑔0 (𝜂), ′ ′ 𝐶1 = 𝑓1 (𝜂)𝑔0 (𝜂) + 𝑓0 (𝜂)𝑔1 (𝜂), ′ ′ ′ 𝐶2 = 𝑓2 (𝜂)𝑔0 (𝜂) + 𝑓1 (𝜂)𝑔1 (𝜂) + 𝑓0 (𝜂)𝑔2 (𝜂), ′ ′ ′ ′ 𝐶3 = 𝑓3 (𝜂)𝑔0 (𝜂) + 𝑓2 (𝜂)𝑔1 (𝜂) + 𝑓1 (𝜂)𝑔2 (𝜂) + 𝑓0 (𝜂)𝑔3 (𝜂),

. . .. 𝐷0 = 𝑓0′ (𝜂)𝑔0 (𝜂), 𝐷1 = 𝑓1′ (𝜂)𝑔0 (𝜂) + 𝑓0′ (𝜂)𝑔1 (𝜂), 𝐷2 = 𝑓2′ (𝜂)𝑔0 (𝜂) + 𝑓1′ (𝜂)𝑔1 (𝜂) + 𝑓0′ (𝜂)𝑔2 (𝜂), 𝐷3 = 𝑓3′ (𝜂)𝑔0 (𝜂) + 𝑓2′ (𝜂)𝑔1 (𝜂) + 𝑓1′ (𝜂)𝑔2 (𝜂) + 𝑓0′ (𝜂)𝑔3 (𝜂), . . .. Consequently, we have the following approximants 𝜂3

𝑓0 (𝜂) = 𝜂𝑎1 + 𝑎2 3! , 𝑔0 (𝜂) = 𝜂𝑎3 , 1

𝑓1 (𝜂) = − 5040 𝑅𝑒𝑎22 𝜂7 , 𝑔1 (𝜂) = −0.00166666667𝑅𝑒𝑎2 𝑎3 𝜂5 + 0.33333333𝜂3 𝑎3 − 0.16666666𝑎3 𝜂3 , 31

𝑓2 (𝜂) = −1.002084335 × 10−7 𝑅𝑒 2 𝑎23 𝜂11 − 0.0000055114638𝑅𝑒 2 𝑎1 𝑎22 𝜂9 + ⋯, 𝑔2 (𝜂) = 8.818342150 × 10−7 𝑅 2 𝑎22 𝑎3 𝜂9 + 0.0001984126984𝑅𝑎22 𝜂7 − ⋯, . . .. The series solution is given as 𝑓(𝜂) = 𝑓0 (𝜂) + 𝑓1 (𝜂) + 𝑓2 (𝜂) + ⋯, 𝜂3

1

𝑓(𝜂) = 𝜂𝑎1 + 𝑎2 3! − 5040 𝑅𝑒𝑎22 𝜂7 − 1.002084335 × 10−7 𝑅𝑒 2 𝑎23 𝜂11 −0.0000055114638𝑅𝑒 2 𝑎1 𝑎22 𝜂9 + ⋯, 𝑔(𝜂) = 𝑔0 (𝜂) + 𝑔1 (𝜂) + 𝑔2 (𝜂) + ⋯, 𝑔(𝜂) = 𝜂𝑎3 − 0.00166666667𝑅𝑒𝑎2 𝑎3 𝜂5 + 0.33333333𝜂3 𝑎3 −0.16666666𝑎3 𝜂3 8.818342150 × 10−7 𝑅 2 𝑎22 𝑎3 𝜂9 + 0.0001984126984𝑅𝑎22 𝜂7 − ⋯. Table 2.1 shows results obtained from ADM for the functions 𝑓(𝜂) at Re = −10 and Table 2.2 shows results obtained from ADM for the function 𝑔(𝜂). The results are compared fourth-order Runge–Kutta method. Excellent agreement can be seen. Table 2.1. The results of ADM and R–K numerical method for 𝑓(𝜂) when Re = −10. ___________________________________________________________________________ 𝜂

0th Order

1th Order

Approx.

Approx.

2nd Order Approx.

3rd Order Approx.

4th Order Approx.

Numerical Approx.

___________________________________________________________________________ 0.0 0.00000000

0.00000000

0.00000000

0.00000000

0.00000000

0.00000000

0.1 0.14950000

0.15406378

0.15227665

0.15263708

0.15265417

0.15261477

0.2 0.29600000

0.30471085

0.30124513

0.30195051

0.30197954

0.30190458

0.3 0.43650000

0.44852842

0.44361340

0.44463218

0.44466288

0.44456001

0.4 0.56800000

0.58212311

0.57613141

0.57741277

0.57743004

0.57731069

0.5 0.68750000

0.70215997

0.69563311

0.69709789

0.69708337

0.69696245

0.6

0.79200000

0.80543635

0.79909460

0.80061790

0.80055540

0.80045027

0.7 0.87850000

0.88900258

0.88369924

0.88509018

0.88497837

0.88490551

0.8 0.94400000

0.95034099

0.94688873

0.94789484

0.94776636

0.94773361

0.9 0.98550000

0.98761502

0.98636587

0.98677566

0.98670109

0.98669724

1.0 1.00000000

1.00000000

1.00000000

1.00000000

1.00000000

1.00000001

32

Table 2.2: The results of ADM and Runge–Kutta numerical method for 𝑔(𝜂) when Re = −10. ___________________________________________________________________________ 𝜂

0th Order

1th Order

2nd Order

3rd Order

4th Order

Numerical

Approx.

Approx.

Approx.

Approx.

Approx.

Approx.

___________________________________________________________________________ 0.0 0.00000000

0.00000000

0.00000000

0.00000000

0.00000000

0.00000000

0.1 0.00000000 -0.04419716 -0.03912644

-0.03946779 -0.03962257 -0.03953400

0.2 0.00000000 -0.08586207 -0.07573261

-0.07640988 -0.07671728 -0.07654220

0.3

0.00000000 -0.12243197 -0.10730274

-0.10831714 -0.10877194 -0.10851455

0.4 0.00000000 -0.15128315 -0.13134204

-0.13272345 -0.13331291 -0.13297987

0.5 0.00000000 -0.16970039 -0.14541767

-0.14724746 -0.14793938 -0.14754153

0.6 0.00000000 -0.17484651 -0.14723649

-0.14964796 -0.15036845 -0.14992643

0.7 0.00000000 -0.16373190 -0.13477262

-0.13787908 -0.13848843 -0.13804462

0.8 0.00000000 -0.13318395 -0.10645829

-0.11011710 -0.11042174 -0.11005419

0.9 0.00000000 -0.07981664 -0.06145212

-0.06471234 -0.06461445 -0.06442561

1.0 0.00000000 -0.00000000 -0.00000000

0.00000000

-0.00000000

-0.00000001

____________________________________________________________________________

Now, we examine the effect of Reynolds number on the dimensionless velocity component in 𝑥 −direction 𝑓′(𝜂) is considered in Figs. 2.1 and 2.2. These figures elucidate that increasing the magnitude of Reynolds number (i.e., increasing mass injection) increases the maximum value of this component.

33

Fig. 2.1. The effect of Reynolds number on velocity component in 𝑥 −direction.

Fig. 2.2. The effect of Reynolds number on 𝑓(𝜂) when 𝑅𝑒 varies.

2.2.3. Squeezing Flow and Heat Transfer between Two Parallel Disks with Velocity Slip and Temperature Jump Consider the axisymmetric flow of an incompressible electrically conducting viscous fluid squeezed between two parallel disks separated [99] by a variable distance ℎ(𝑡) = 𝐻√1 − 𝛼𝑡, where 𝛼 is a characteristic parameter having dimensions that of time inverse. The upper disk at 𝑧 = ℎ(𝑡) is approaching the lower stationary disk at 𝑧 = 0 with a velocity 𝑣(𝑡) =

𝑑ℎ 𝑑𝑡

. We consider the velocity profile in the form (𝑢, 0, 𝑤), where 𝑢 and 𝑤

are radial and axial velocities along 𝑟 and 𝑧 axes, respectively. A uniform magnetic field proportional to

𝐵0 √1−𝛼𝑡

is applied along the axial component and electric field is taken zero.

34

Under the stated conditions the continuity, momentum and energy equations can be written as: 𝜕𝑢 𝜕𝑟

𝑢

+𝑟+

𝜕𝑤 𝜕𝑧

= 0,

𝜕𝑢

𝜕𝑢

𝜕𝑤

𝜕𝑤

(2.27a) 𝜕𝑢

𝜕2 𝑢

𝜕𝑝

1 𝜕𝑢

𝜕2 𝑢

𝑢

𝜎𝐵2

0 𝜌 ( 𝜕𝑡 + 𝑢 𝜕𝑟 + 𝑤 𝜕𝑧 ) = − 𝜕𝑟 + 𝜇 ( 𝜕𝑟 2 + 𝑟 𝜕𝑟 + 𝜕𝑧 2 − 𝑟 2 ) − 1−𝛼𝑡 𝑢, (2.27b)

𝜌 ( 𝜕𝑡 + 𝑢 𝜕𝑟 + 𝑤 𝜕𝑇

𝜕𝑤

𝜕2 𝑤

𝜕𝑝

1 𝜕𝑤

) = − 𝜕𝑟 + 𝜇 ( 𝜕𝑟 2 + 𝑟 𝜕𝑟 + 𝜕𝑧

𝜕𝑇

𝜕2 𝑇

𝜕𝑇

1 𝜕𝑇

𝜕2 𝑇

𝜕2 𝑤 𝜕𝑧 2

),

(2.27c) 𝜕𝑢 2

𝑢 2

𝜕𝑤 2

𝜌𝐶𝑝 ( 𝜕𝑡 + 𝑢 𝜕𝑟 + 𝑤 𝜕𝑧 ) = 𝑘 ( 𝜕𝑟 2 + 𝑟 𝜕𝑟 + 𝜕𝑧 2 ) + 2𝜇 [( 𝜕𝑟 ) + ( 𝑟 ) + ( 𝜕𝑧 ) + 1 𝜕𝑢

( + 2 𝜕𝑧

𝜕𝑤 2 𝜕𝑟

) ],

(2.27d)

subject to the boundary conditions 𝜕𝑢

𝑢 = 𝛽1 𝜕𝑧 , 𝜕𝑢

𝑢 = −𝛽1 𝜕𝑧 ,

𝜕𝑇

𝑤 = 0, 𝑤=

𝑇 = 𝛾1 𝜕𝑧 + 𝑇0 ,

𝑑ℎ(𝑡) 𝑑𝑡

,

𝜕𝑇

𝑇 = −𝛾1 𝜕𝑧 + 𝑇1 ,

at 𝑧 = 0,

(2.28a)

at 𝑧 = ℎ(𝑡),

(2.28b)

Where 𝜌 is fluid density, 𝜇 is dynamic viscosity, p is the pressure, 𝜎 is electrical conductivity, 𝐶𝑝 is the specific heat at constant pressure, k is the thermal conductivity, 𝛽1 and 𝛾1 are the slip parameters with respect to velocity and temperature. Note that 𝑇, 𝑇0 and T1 ( T1  T0 ) are the temperature of the fluid, constant temperatures of the lower and upper disks, respectively. Introducing the dimensionless quantities: 𝛼𝑟

𝑢 = 2(1−𝛼𝑡) 𝑓 ′ (𝜂), 𝑤 = −

𝛼𝐻 √1−𝛼𝑡

𝑧

𝑓(𝜂), 𝑇 = 𝑇0 + (𝑇1 − 𝑇0 )𝜃(𝜂), 𝜂 = 𝐻√1−𝛼𝑡,(2.29)

Substituting 𝑢 and 𝑤 into Eqs. (2.27b-2.27c) and eliminating the pressure gradient, we obtain the following dimensionless differential equation: 𝑓 ′′′′ (𝜂) − 𝑆(𝜂𝑓 ′′′ (𝜂) + 3𝑓 ′′ (𝜂) − 2𝑓(𝜂)𝑓 ′′′ (𝜂)) − 𝑀2 𝑓 ′′ (𝜂) = 0, (2.30) By virtue of (3.29), the energy equation (3.27d) and the boundary conditions (3.28) take the following form 𝜃 ′′ (𝜂) + Pr 𝑆(2𝑓(𝜂)𝜃 ′ (𝜂) − 𝜂𝜃′(𝜂)) + Pr 𝐸𝑐 (𝑓 ′′ (𝜂)2 + 12𝛿 2 𝑓 ′ (𝜂)2 ) = 0.(2.31) 𝑓(0) = 0, 1

𝑓(1) = 2,

𝑓 ′ (0) − 𝛽𝑓 ′′ (0) = 0,

𝜃(0) − 𝛾𝜃 ′ (0) = 0,

(2.32a)

𝑓 ′ (1) − 𝛽𝑓 ′′ (1) = 0,

𝜃(1) − 𝛾𝜃 ′ (1) = 1.

(2.32b)

35

Where the respective values of squeeze number 𝑆, the Hartman number 𝑀 the Prandtl number Pr the Eckert number 𝐸𝑐, the dimensionless slip parameters 𝛽 and 𝛾 with respect to velocity and temperature, and the dimensionless number 𝛿 are 𝑆=

𝛼𝐻 2 2𝜈

, 𝑀2 =

𝜎𝐻 2 𝐵02 𝜇

𝛽

, Pr =

𝜇𝐶𝑝 𝑘

𝛾

1 1 𝛽 = 𝐻√1−𝛼𝑡 , 𝛾 = 𝐻√1−𝛼𝑡 ,𝛿 =

1

, 𝐸𝑐 = 𝐶

𝛼𝑟

𝑝 (𝑇1 −𝑇0 )

𝐻√1−𝛼𝑡 𝑟

(2(1−𝛼𝑡)),

.

The dimensionless physical quantities like skin friction coefficient and local Nusselt number can be calculated from the following: 𝜕𝑢

𝜕𝑇

𝜇𝜕𝑧 |

𝑧=ℎ(𝑡) 𝜌[𝑣(𝑡)]2

𝐶𝑓𝑟 =

, 𝑁𝑢 = −

𝐻𝑘 𝜕𝑧 |

𝑧=ℎ(𝑡)

𝑘(𝑇1 −𝑇0 )

,

which by virtue of (6) reduces to 𝐻2

′′

′ √1 − 𝛼𝑡 𝑟2 𝑅𝑒𝑟 𝐶𝑓𝑟 = 𝑓 (1), √1 − 𝛼𝑡 𝑁𝑢 = −𝜃 (1),

where 𝑅𝑒𝑟 =

𝛼𝐻𝑟 2𝜈

(2.33)

is the local squeeze Reynolds number. Applying ADM the equation

(2.30) and (2.31) in operator form will be 𝐿1 𝑓 = 𝑆(𝜂𝑓 ′′′ + 3𝑓′′ − 2𝑓𝑓 ′′′ ) + 𝑀2 𝑓′′, 𝐿2 𝜃 = −𝑃𝑟𝑆(2𝑓𝜃 ′ − 𝜂𝜃 ′ ) − 𝑃𝑟𝐸𝑐 (𝑓 ′′2 + 12𝛿 2 𝑓 ′2 ), 𝑑4

𝑑2

where 𝐿1 = 𝑑𝜂4 , and 𝐿2 = 𝑑𝜂2 , −1 Applying 𝐿−1 1 and 𝐿2 on both sides of above equations we get, 𝜂2

𝑓(𝜂) = 𝑓(0) + 𝜂 𝑓 ′ (0) + 2 𝑓 ′′ (0) +

𝜂3 3!

′′′ 𝑓 ′′′ (0) + 𝐿−1 + 3𝑓′′ − 1 {𝑆(𝜂𝑓

2𝑓𝑓 ′′′ ) + 𝑀2 𝑓′′}, ′ ′ ′′2 𝜃(𝜂) = 𝜃(0) + 𝜂𝜃 ′ (0) + 𝐿−1 + 12𝛿 2 𝑓 ′2 )}, 2 {−𝑃𝑟𝑆(2𝑓𝜃 − 𝜂𝜃 ) − 𝑃𝑟𝐸𝑐 (𝑓 𝜂

𝜂

𝜂

𝜂

𝜂

𝜂

−1 where 𝐿−1 1 = ∫0 ∫0 ∫0 ∫0 (. )𝑑𝜂𝑑𝜂𝑑𝜂𝑑𝜂 ,and 𝐿2 = ∫0 ∫0 (. )𝑑𝜂𝑑𝜂 ,

let

𝑓(0) = 𝐴, 𝑓 ′ (0) = 𝐵, 𝑓 ′′ (0) = 𝐶, 𝑓 ′′′ (0) = 𝐷, 𝜃(0) = 𝐸, 𝜃 ′ (0) = 𝐹

we get, 𝜂2

𝑓(𝜂) = 𝐴 + 𝜂 𝐵 + 2 𝐶 +

𝜂3 3!

′′′ 𝐷 + 𝐿−1 + 3𝑓′′ − 2𝑓𝑓 ′′′ ) + 𝑀2 𝑓′′}, 1 {𝑆(𝜂𝑓

′ ′ ′′2 𝜃(𝜂) = 𝐸 + 𝜂𝐹 + 𝐿−1 + 12𝛿 2 𝑓 ′2 )}, 2 {−𝑃𝑟𝑆(2𝑓𝜃 − 𝜂𝜃 ) − 𝑃𝑟𝐸𝑐 (𝑓

According to ADM we have recurrence relation as 𝜂2

𝑓0 (𝜂) = 𝐴 + 𝜂 𝐵 + 2 𝐶 +

𝜂3 3!

𝐷, 36

′′′ ′′ 2 ′′ 𝑓𝑘+1 (𝜂) = 𝐿−1 1 {𝑆(𝜂𝑓𝑘 + 3𝑓𝑘 − 2𝐴𝑘 ) + 𝑀 𝑓𝑘 },

𝑘≥0,

𝜃0 (𝜂) = 𝐸 + 𝜂𝐹, ′ 2 𝜃𝑘+1 (𝜂) = 𝐿−1 2 {−𝑃𝑟𝑆(2𝐵𝑘 − 𝜂𝜃𝑘 ) − 𝑃𝑟𝐸𝑐 (𝐶𝑘 + 12𝛿 𝐷𝑘 )}, 𝑘 ≥ 0,

where 𝐴𝑘 , 𝐵𝑘 , 𝐶𝑘 and 𝐷𝑘 are Adomian’s polynomials ′′′ ′′ 2 ′′ 𝑓1 (𝜂) = 𝐿−1 1 {𝑆(𝜂𝑓0 + 3𝑓0 − 2𝐴0 ) + 𝑀 𝑓0 }, ′ 2 𝜃1 (𝜂) = 𝐿−1 2 {−𝑃𝑟𝑆(2𝐵0 − 𝜂𝜃0 ) − 𝑃𝑟𝐸𝑐 (𝐶0 + 12𝛿 𝐷0 )}, ′′′ ′′ 2 ′′ 𝑓2 (𝜂) = 𝐿−1 1 {𝑆(𝜂𝑓1 + 3𝑓1 − 2𝐴1 ) + 𝑀 𝑓1 }, ′ 2 𝜃2 (𝜂) = 𝐿−1 2 {−𝑃𝑟𝑆(2𝐵1 − 𝜂𝜃1 ) − 𝑃𝑟𝐸𝑐 (𝐶1 + 12𝛿 𝐷1 )}, ′′′ ′′ 2 ′′ 𝑓3 (𝜂) = 𝐿−1 1 {𝑆(𝜂𝑓2 + 3𝑓2 − 2𝐴2 ) + 𝑀 𝑓2 }, ′ 2 𝜃3 (𝜂) = 𝐿−1 2 {−𝑃𝑟𝑆(2𝐵2 − 𝜂𝜃2 ) − 𝑃𝑟𝐸𝑐 (𝐶2 + 12𝛿 𝐷2 )},

. . ., where 𝐴0 = 𝑓0 𝑓0′′′ , 𝐴2 = 𝑓1 𝑓0′′′ + 𝑓0 𝑓1′′′ , , 𝑎𝑛𝑑 , 𝐴1 = 𝑓2 𝑓0′′′ + 𝑓1 𝑓1′′′ + 𝑓0 𝑓2′′′ , . . .. 𝐵0 = 𝑓0 𝜃0′ , 𝐵1 = 𝑓1 𝜃0′ + 𝑓0 𝜃1′ 𝐵2 = 𝑓2 𝜃0′ + 𝑓1 𝜃1′ + 𝑓0 𝜃2′ , . . .. 𝐶0 = (𝑓0′′ )2 , 𝐶1 = 2𝑓0′′ 𝑓1′′ , 2

𝐶2 = (𝑓1′′ ) + 2𝑓0′′ 𝑓2′′ , . . 37

.. 𝐷0 = (𝑓0′ )2 , 𝐷1 = 2𝑓0′ 𝑓1′ , 𝐷2 = (𝑓1′ )2 + 2𝑓0′ 𝑓2′ , . . .. The series solution is obtained as, 𝑓(𝜂) = 𝑓0 (𝜂) + 𝑓1 (𝜂) + 𝑓2 (𝜂) + 𝑓3 (𝜂) …, 𝜂2

𝑓(𝜂) = 𝐴 + 𝜂 𝐵 + 2 𝐶 +

𝜂3 3!

1

1

1

𝐷 − 2520 𝑆𝐷2 𝜂7 − 360 𝑆𝐶𝐷𝜂6 + 30 𝑆𝐷𝜂5 + ⋯,

𝜃(𝜂) = 𝜃0 (𝜂) + 𝜃1 (𝜂) + 𝜃2 (𝜂) + 𝜃3 (𝜂) + ⋯, 1

1

3

𝜃(𝜂) = 𝐸 + 𝜂𝐹 − 10 𝛿 2 𝑃𝑟𝐸𝑐𝐷2 𝜂6 − 60 𝜂5 𝑃𝑟𝑆𝐷𝐹 − 5 𝜂5 𝑃𝑟𝐸𝑐𝐶𝐷𝛿 2 + ⋯. Graph of 𝜃(𝜂) for 𝑀 = 1.0 = 𝑆 = 𝐸𝑐, 𝛽 = 𝛾 = 𝛿 = 0.1

Fig. 2.3. Effect of Prandtl number Pr on temperature profile

Graph of 𝑓 ′ (𝜂) for 𝑃𝑟 = 1.0 = 𝑆, 𝐸𝑐 = 0.5, 𝛽 = 𝛾 = 𝛿 = 0.1

38

Fig. 2.4. Effect of magnetic parameter M on the radial velocity

Graph of 𝜃(𝜂) for 𝑆 = 1.0 = 𝑀 = 𝑃𝑟, 𝐸𝑐 = 0, 𝛽 = 𝛿 = 0.1

Fig. 2.5. Effect of temperature slip parameter 𝛾 on temperature profile

Graph of 𝜃(𝜂) for 𝑆 = 1.0 = 𝑀 = 𝑃𝑟, 𝛽 = 𝛿 = 𝛾 = 0.1

39

Fig. 2.6. Effect of Eckert number Ec on temperature Profile

Graph of 𝑓 ′ (𝜂) for 𝑃𝑟 = 1.0 = 𝑀, 𝐸𝑐 = 0, 𝛽 = 𝛾 = 𝛿 = 0.1

Fig. 2.7. Effect of squeeze parameter on the radial velocity

40

Table 2.3. Comparison of Numerical and different order ADM solutions for 𝑓(𝜂) when 𝑆 = 𝑀 = 𝑃𝑟 = 𝐸𝑐 = 1.0, 𝛽 = 𝛾 = 𝛽 = 0.1. ________________________________________________________________________ 𝜂

0𝑡ℎ order 1𝑠𝑡 order 2𝑛𝑑 order 3𝑟𝑑 order 4𝑡ℎ order 5𝑡ℎ order 6𝑡ℎ order

(NM)

________________________________________________________________________ 0.0

0.00000

0.00000

0.00000

0.00000

0.00000

0.00000

0.00000

0.00000

0.1

0.02679

0.02682

0.02839

0.02839

0.02838

0.02838

0.02837

0.02837

0.2

0.06745

0.06788

0.07157

0.07157

0.07156

0.07154

0.07153

0.07153

0.3

0.11748

0.11955

0.12537

0.12537

0.12536

0.12532

0.12532

0.12532

0.4

0.17238

0.17852

0.18604

0.18604

0.18603

0.18598

0.18597

0.18597

0.5

0.22767

0.24167

0.25007

0.25008

0.25007

0.25001

0.25000

0.25000

0.6

0.27885

0.30587

0.31406

0.31411

0.31410

0.31403

0.31402

0.31402

0.7

0.32143

0.36778

0.37460

0.37476

0.37475

0.37469

0.37468

0.37468

0.8

0.35091

0.42372

0.42809

0.42853

0.42852

0.42847

0.42846

0.42846

0.9

0.36279

0.46946

0.47059

0.47165

0.47165

0.47163

0.47163

0.47163

1.0

0.35259

0.50000

0.49758

0.50000

0.50000

0.50000

0.50000

0.50000

________________________________________________________________________

41

Table 2.4. Comparison of Numerical and different order ADM solutions for 𝜃(𝜂) when 𝑆 = 𝑀 = 𝑃𝑟 = 𝐸𝑐 = 1.0, 𝛽 = 𝛾 = 𝛽 = 0.1. ________________________________________________________________________ η

0𝑡ℎ order 1𝑠𝑡 order 2𝑛𝑑 order 3𝑟𝑑 order 4𝑡ℎ order 5𝑡ℎ order 6𝑡ℎ order

(NM)

________________________________________________________________________ 0.0

0.14709

0.14709

0.13489

0.13489

0.13809

0.13986

0.14012

0.14006

0.1

0.29419

0.28001

0.25388

0.25388

0.26028

0.26382

0.26434

0.26423

0.2

0.44128

0.39387

0.35096

0.35092

0.36054

0.36588

0.36666

0.36650

0.3

0.58837

0.49929

0.43694

0.43664

0.44948

0.45663

0.45768

0.45747

0.4

0.73546

0.60261

0.51880

0.51748

0.53351

0.54247

0.54379

0.54352

0.5

0.88256

0.70602

0.60077

0.59701

0.61598

0.62675

0.62833

0.62801

0.6

1.02965

0.80772

0.68458

0.67659

0.69784

0.71034

0.71219

0.71182

0.7

1.17674

0.90206

0.76892

0.75587

0.77791

0.79191

0.79402

0.79360

0.8

1.32383

0.97961

0.84796

0.83279

0.85301

0.86780

0.87013

0.86967

0.9

1.47093

1.02731

0.90878

0.90294

0.91787

0.93160

0.93396

0.93351

1.0

1.61802

1.02849

0.92733

0.95770

0.96496

0.97346

0.97506

0.97477

________________________________________________________________________

2.3. Modified Adomian’s Decomposition Method Wazwaz [224] did a new modification by decomposing the initial guess into the Taylor’s series and set the recurrence relation accordingly. This modification has been applied many nonlinear PDEs and integral equation [215]. In this modified version we face less computational work as compare to traditional Adomian’s decomposition method. This modification is more reliable and more effective.

2.3.1. Methodology Applying the Taylor series on 𝑓 then the above equation (2.4) it becomes ∞ −1 ∑∞ 𝑛=0 𝑢𝑛 = 𝑓0 + 𝑓1 + 𝑓2 + 𝑓3 + ⋯ − 𝐿 (𝑅(∑𝑛=0 𝑢𝑛 )),

Accordingly, the formal recursive relation is defined by 𝑢𝑘 = 𝑓0 , 𝑢𝑘+1 = 𝑓𝑘+1 − 𝐿−1 (𝑅(∑∞ 𝑛=0 𝑢𝑛 )).

42

2.3.2. Goursat Equation Example 2.1. Consider the following Goursat equation [230] 𝑢𝑥𝑦 = −𝑥 + 𝑢, with initial conditions, 𝑢(𝑥, 0) = 𝑥 + 𝑒 𝑥 , 𝑢(0, 𝑦) = 𝑒 𝑦 , 𝑢(0,0) = 1. Apply Adomian’s decomposition method, we get 1

−1 𝑢(𝑥, 𝑦) = 𝑥 + 𝑒 𝑥 + 𝑒 𝑦 − 1 − 2 𝑥 2 𝑦 + 𝐿−1 𝑥 𝐿𝑦 (𝑢),

Applying Taylor’s series, we get 1

1

1

𝑥

1

𝑦

𝑢(𝑥, 𝑦) = 𝑒 𝑦 + 2𝑥 + (2 − 2 𝑦) 𝑥 2 + 6 𝑥 3 + 24 𝑥 4 + ⋯ ∫0 ∫0 𝑢𝑑𝑦𝑑𝑥 . According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑦) = 𝑒 𝑦 , 𝑥

𝑦

𝑢𝑘+1 (𝑥, 𝑦) = 𝑓𝑘+1 (𝑥) + ∫𝑥 ∫0 𝑢𝑘 (𝑥, 𝑦)𝑑𝑥𝑑𝑦. Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑦) = 𝑒 𝑦 , 𝑥

𝑦

𝑢1 (𝑥, 𝑦) = 2𝑥 + ∫𝑥 ∫0 𝑒 𝑦 𝑑𝑥𝑑𝑦, = 𝑥 + 𝑒 𝑦 𝑥. 1

𝑥

1

𝑦

𝑢2 (𝑥, 𝑦) = 2 𝑥 2 − 2 𝑥 2 𝑦 + ∫0 ∫0 (𝑥 + 𝑒 𝑦 𝑥)𝑑𝑥𝑑𝑦, 1

= 2 𝑒𝑦𝑥2. 𝑥

𝑦

𝑢3 (𝑥, 𝑦) = ∫𝑥 ∫0 𝑒 𝑦 𝑥 2 𝑑𝑥𝑑𝑦, 1

= 6 𝑒𝑦𝑥3. 1

𝑢4 (𝑥, 𝑦) = 24 𝑥 4 + =

1 24

1

𝑥

𝑦

∫ ∫ 𝑒 𝑦 𝑥 3 𝑑𝑥𝑑𝑦, 6 𝑥 0

𝑒 𝑦 𝑥4.

. . .. The series solution is given by 1

1

1

𝑢(𝑥, 𝑦) = 𝑒 𝑦 + 𝑒 𝑦 (1 + 𝑥 + 2 𝑥 2 + 3! 𝑥 3 + 4! 𝑥 4 + ⋯ ), 43

and the close form is given by 𝑢(𝑥, 𝑦) = 𝑥 + 𝑒 𝑥+𝑦 .

Example 2.2. Consider the following Goursat equation [230] 𝑢𝑥𝑦 = 𝑒 𝑥+𝑦 𝑢,

(2.34)

with initial conditions 𝑢(𝑥, 0) = 𝑙𝑛2 − 2 ln(1 + 𝑒 𝑥 ) , 𝑢(𝑥, 0) = 𝑙𝑛2 − 2 ln(1 + 𝑒 𝑦 ) , 𝑢(0,0) = −𝑙𝑛2. Applying Adomian’s decomposition method, we get −1 𝑥+𝑦 𝑦 ), 𝑢(𝑥, 𝑦) = 3𝑙𝑛2 − 2 ln(1 + 𝑒 𝑥 ) − 2 ln(1 + 𝑒 𝑦 ) + 𝐿−1 𝑒 𝑥 𝐿𝑦 (𝑒

Applying Taylor series, we get 1

1

1

𝑢(𝑥, 𝑦) = ln 2 + 2 ln(1 + 𝑒 𝑦 ) − 𝑥 − 4 𝑥 2 + 96 𝑥 4 − 1440 𝑥 6 + ⋯ −1 𝑥+𝑦 𝑦 +𝐿−1 𝑒 ), 𝑥 𝐿𝑦 (𝑒

According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑦) = 𝑙𝑛2 − 2ln(1 + 𝑒 𝑦 ), 𝑥

𝑦

𝑢𝑘+1 (𝑥, 𝑦) = 𝑓𝑘+1 (𝑥) + ∫0 ∫0 𝑒 𝑥+𝑦 𝐴𝑘 𝑑𝑥𝑑𝑦,

𝑘 ≥ 0.

Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑦) = 𝑙𝑛2 − 2ln(1 + 𝑒 𝑦 ), 𝑥

𝑦

𝑢1 (𝑥, 𝑦) = −𝑥 + ∫0 ∫0 𝑒 𝑥+𝑦 𝑒 𝑢0 𝑑𝑥𝑑𝑦 , = −𝑥 − (

−1+𝑒 𝑦 𝑒 𝑥 𝑒 𝑥+𝑦 1+𝑒 𝑦 𝑥

1

𝑦

)𝑒 𝑙𝑛2−2ln(1+𝑒 ) .

𝑦

𝑢2 (𝑥, 𝑦) = − 4 𝑥 4 + ∫0 ∫0 𝑒 𝑥+𝑦 𝑢1 𝑒 𝑢0 , 1

1

1

= − 4 𝑥 4 − 4 (1+𝑒 𝑦 )2 (4𝑒 𝑥+2𝑦 . 𝑥 − 2𝑒 𝑥+2𝑦 + 3𝑒 2𝑦 − 𝑒 2𝑥+2𝑦 + 2𝑒 2𝑥+𝑦 + 2𝑒 𝑦 − 4𝑒 𝑥+𝑦 − 𝑒 2𝑥 − 5 − 4𝑥𝑒 𝑥 + 6𝑒 𝑥 ). . . .. The series solution is given by 𝑢(𝑥) = 𝑙𝑛2 − 2 ln(1 + 𝑒 𝑦 ) − 𝑥 − ( 1

1

4 (1+𝑒 𝑦 )2

−1+𝑒 𝑦 𝑒 𝑥 𝑒 𝑥+𝑦 1+𝑒 𝑦

) 𝑒 𝑙𝑛2−2ln(1+𝑒

𝑦)

1

− 4 𝑥4 −

(4𝑒 𝑥+2𝑦 𝑥 − 2𝑒 𝑥+2𝑦 + 3𝑒 2𝑦 − 𝑒 2𝑥+2𝑦 + 2𝑒 2𝑥+𝑦 + 2𝑒 𝑦 − 4𝑒 𝑥+𝑦 − 𝑒 2𝑥 − 5 −

4𝑥𝑒 𝑥 + 6𝑒 𝑥 ) + ⋯ 44

Example 2.3. Consider the following partial differential equation [230] 𝑢𝑥𝑦 = 4𝑥𝑦 − 𝑥 2 𝑦 2 + 𝑢,

(2.35)

with initial conditions , 𝑢(𝑥, 0) = 𝑒 𝑥 , 𝑢(0, 𝑦) = 𝑒 𝑦 , 𝑢(0,0) = 1. Applying Adomian’s decomposition method, we get 1

−1 𝑢(𝑥, 𝑦) = 𝑥 2 𝑦 2 − 9 𝑥 3 𝑦 3 − 1 + 𝑒 𝑥 + 𝑒 𝑦 + 𝐿−1 𝑥 𝐿𝑦 (𝑢(𝑥, 𝑦)).

Applying Taylor series, we get 1

−1

1

−1 𝑢(𝑥, 𝑦) = 𝑒 𝑦 + 𝑥 (𝑦 2 + 2) 𝑥 2 + ( 9 𝑦 3 + 6) 𝑥 3 + ⋯ + 𝐿−1 𝑥 𝐿𝑦 (𝑢(𝑥, 𝑦)),

According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑦) = 𝑒 𝑦 , 𝑥

𝑦

𝑢𝑘+1 (𝑥, 𝑦) = 𝑓𝑘+1 (𝑥) + ∫0 ∫0 𝑢𝑘 𝑑𝑥𝑑𝑦

𝑘 ≥ 0.

Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑦) = 𝑒 𝑦 , 𝑥

𝑦

𝑢1 (𝑥, 𝑦) = 𝑥 + ∫0 ∫0 𝑒 𝑦 𝑑𝑥𝑑𝑦 = 𝑥𝑒 𝑦 . 𝑥

1

𝑦

1

𝑢2 (𝑥, 𝑦) = (𝑦 2 + 2)𝑥 2 + ∫0 ∫0 𝑥𝑒 𝑦 𝑑𝑥𝑑𝑦 = 𝑥 2 𝑦 2 + 2! 𝑥 2 𝑒 𝑦 . −1

1

𝑥

𝑦

1

1

𝑢3 (𝑥, 𝑦) = ( 9 𝑦 3 + 6)𝑥 3 + ∫0 ∫0 𝑥 2 𝑦 2 + 2! 𝑥 2 𝑒 𝑦 𝑑𝑥𝑑𝑦 = 3! 𝑥 3 𝑒 𝑦 . . . .. The series solution is given by 1

1

𝑢(𝑥, 𝑦) = 𝑒 𝑦 + 𝑥𝑒 𝑦 + 𝑥 2 𝑦 2 + 2! 𝑥 2 𝑒 𝑦 + 3! 𝑥 3 𝑒 𝑦 + ⋯ and the closed form is given by 𝑢(𝑥, 𝑦) = 𝑥 2 𝑦 2 + 𝑒 𝑥+𝑦

2.3.3. Fourth-order Parabolic Equation Example 2.4. Consider the following partial differential equation [230] 𝜕2 𝑢 𝜕𝑡 2

𝜕4 𝑢

+ 𝜕𝑥 4 = 0,

(2.36)

with initial conditions, 𝑢(𝑥, 0) = cos(𝑥) , 𝑢𝑡 (𝑥, 0) = − sin(𝑥). 45

Applying Adomian’s Decomposition Method, we get ∂4 u

𝑢(𝑥, 𝑡) = cos(𝑥) − 𝑡 sin(𝑥) − 𝐿−1 𝑡 (∂𝑥 4 ) Applying Taylor’s Series, we get 1

1

1

1

∂4 u

1

𝑢(𝑥, 𝑡) = 1 − 𝑡𝑥 − 2 𝑥 2 + 6 𝑡𝑥 3 + 24 𝑥 4 − 120 𝑡𝑥 5 − 720 𝑥 6 + ⋯ −𝐿−1 𝑡 (∂𝑥 4 ) According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑡) = 1, 𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 (𝑥) − 𝐿−1 𝑡 (

∂4 𝑢𝑘 ∂𝑥 4

), 𝑘 ≥ 0,

Consequently, following approximants are obtained 1

𝑢1 (𝑥, 𝑡) = −𝑡𝑥 − 2 𝑥 2 , 1

1

𝑢2 (𝑥, 𝑡) = 6 𝑡𝑥 3 + 24 𝑥 4 , 𝑢3 (𝑥, 𝑡) = −

1 120

𝑡𝑥 5 –

1 720

1

1

𝑥6 − 2 𝑡2 ,

1

1

1

𝑢4 (𝑥, 𝑡) = 5040 𝑡𝑥 7 + 40320 𝑥 8 + 6 𝑥𝑡 3 + 4 𝑥 2 𝑡 2 , . . .. The series solution is given by 𝑢(𝑥, 𝑡) =

1

1 − 𝑡𝑥 − 2 𝑥 2 + 1 40320

1

1

1

1

𝑡𝑥 3 + 24 𝑥 4 − 6

1

𝑡𝑥 5 – 120

𝑥 8 + 6 𝑥𝑡 3 + 4 𝑥 2 𝑡 2 + ⋯

and closed form solution is obtained as 𝑢(𝑥, 𝑡) = cos(𝑥 + 𝑡)

46

1

1

1

𝑥 6 − 2 𝑡 2 + 5040 𝑡𝑥 7 + 720

Fig. 2.8. Comparison of Exact and Approximate solution of Eq. (2.36)

Example 2.5. Consider the following partial differential equation [230] 𝜕2 𝑢 𝜕𝑡 2

𝜕4 𝑢

+ 𝜕𝑥 4 = (π4 − 1)sinπ𝑥𝑠𝑖𝑛𝑡

(2.37)

with initial conditions, 𝑢(𝑥, 0) = 0, 𝑢𝑡 (𝑥, 0) = sin(𝜋𝑥). Applying Adomian’s Decomposition Method, we get ∂4 𝑢

𝑢(𝑥, 𝑡) = 𝑡𝑠𝑖𝑛𝜋𝑥 + (𝜋 4 − 1) sin(𝜋𝑥) (𝑡 − sin(𝑡)) − 𝑡𝑠𝑖𝑛𝑥 − 𝐿−1 𝑡 (∂𝑥 4 ). Applying Taylor’s Series, we get 1

1

𝑢(𝑥, 𝑡) = (𝑡𝜋 5 − 𝜋 5 sin(𝑡) + 𝜋 sin(𝑡))𝑥 + (− 6 𝑡𝜋 7 + 6 𝜋 7 sin(𝑡) − 1 6

1

1

1

1

𝜋 3 sin(𝑡)) 𝑥 3 + (120 𝑡𝜋 9 − 120 𝜋 9 sin(𝑡) + 120 𝜋 5 sin(𝑡)) 𝑥 5 + (− 5040 𝑡𝜋11 + 1

∂4 u

1

𝜋11 sin(𝑡) − 5040 𝜋 7 sin(𝑡)) x 7 + ⋯ − 𝐿−1 𝑡 (∂𝑥 4 ). 5040 According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑡) = (𝑡𝜋 + (𝜋 4 − 1) 𝜋 (𝑡 − sin(𝑡))) 𝑥, ∂4 𝑢

𝑘 𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 (𝑥) + 𝐿−1 𝑡 [ ∂𝑥 4 ] , 𝑘 ≥ 0,


𝑢1 (𝑦, 𝑡) = (− 6 (𝜋 4 − 1)𝜋 3 (𝑡 − sin(𝑡)) − 1

𝑢2 (𝑥, 𝑡) = (120 (𝜋 4 − 1)𝜋 5 (𝑡 − sin(𝑡)) + 1

1 6

𝑡𝜋 3 ) 𝑥 3 ,

1 120

𝑢3 (𝑥, 𝑡) = (− 5040 (𝜋 4 − 1) 𝜋 7 (𝑡 − sin(𝑡)) − 1 6

𝑡𝜋 5 ) 𝜋 5 , 1 5040

𝑡𝜋 7 ) 𝑥 7 + 𝑥𝜋 9 𝑡 − 𝑥𝜋 5 𝑡 −

𝑥𝜋 9 𝑡 3 − 𝑥𝜋 9 sin(𝑡) + 𝑥𝜋 5 sin(𝑡),

47


𝑢(𝑥, 𝑡) = (𝑡𝜋 + (𝜋 4 − 1) 𝜋 (𝑡 − sin(𝑡))) 𝑥 + (− 6 (𝜋 4 − 1)𝜋 3 (𝑡 − 1

1

sin(𝑡)) − 6 𝑡𝜋 3 ) 𝑥 3 + (120 (𝜋 4 − 1)𝜋 5 (𝑡 − sin(𝑡)) + 1) 𝜋 7 (𝑡 − sin(𝑡)) −

1 5040

1 120

1

𝑡𝜋 5 )𝜋 5 + (− 5040 (𝜋 4 −

1

𝑡𝜋 7 ) 𝑥 7 + 𝑥𝜋 9 𝑡 − 𝑥𝜋 5 𝑡 − 6 𝑥𝜋 9 𝑡 3 − 𝑥𝜋 9 sin(𝑡) +

𝑥𝜋 5 sin(𝑡), and the closed form solution is obtained as 𝑢(𝑥, 𝑡) = sin(𝜋𝑥) sin(𝑡).


Example 2.6. Consider the following partial differential equation [230] 𝜕2 𝑢 𝜕𝑡 2

𝜕4 𝑢

6

+ (1 + 𝑥) 𝜕𝑥 4 = (𝑥 4 + 𝑥 3 − 7! 𝑥 7 ) cos(𝑡),

(2.38)

with initial conditions, 6

𝑢(𝑥, 0) = 7! 𝑥 7 , 𝑢𝑡 (𝑥, 0) = 0 Applying Adomian’s Decomposition Method, we get 6

𝑡

6

𝑡

𝜕4 𝑢

𝑢(𝑥, 𝑡) = 7! 𝑥 7 + (𝑥 4 + 𝑥 3 − 7!) (1 − 𝑐𝑜𝑠𝑡) − ∫0 ∫0 (1 + 𝑥) 𝜕𝑥 4 𝑑𝑡𝑑𝑡, Applying Taylor’s Series, we get

48

1

1

1

1

1

1

𝑢(𝑥, 𝑡) = 840 𝑥 7 + (2 𝑥 4 + 2 𝑥 3 − 1680 𝑥 7 ) 𝑡 2 + (− 24 𝑥 4 − 24 𝑥 3 + 1 20160 𝑡

1

𝑥7) 𝑡4 +

1

1

(720 𝑥 4 + 720 𝑥 3 − 604800 𝑥 7 ) 𝑡 6 − ⋯ −

𝜕4 𝑢

𝑡

∫0 ∫0 (1 + 𝑥) 𝜕𝑥 4 𝑑𝑡𝑑𝑡. According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑡) =

1 840

𝑥7, 𝑡

𝑡

𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 (𝑥) − ∫0 ∫0 (1 + 𝑥)

𝜕4 𝑢𝑘 𝜕𝑥 4

𝑑𝑡𝑦𝑡 , 𝑘 ≥ 0,


1

1

1

𝑢1 (𝑥, 𝑡) = (2 𝑥 4 + 2 𝑥 3 − 1680 𝑥 7 ) 𝑡 2 − 2 (1 + 𝑥)𝑥 3 𝑡 2 , 1

1

1

1

𝑢2 (𝑥, 𝑡) = (− 24 𝑥 4 − 24 𝑥 3 + 20160 𝑥 7 ) 𝑡 4 + 24 (1 + 𝑥)𝑥 3 𝑡 2 , 1

1

1

1

𝑢3 (𝑥, 𝑡) = (720 𝑥 4 + 720 𝑥 3 − 604800 𝑥 7 ) 𝑡 6 − 720 (1 + 𝑥)𝑥 3 𝑡 6 , . . .. The series solution is given by 1

1

1

1

1

1

𝑢(𝑥, 𝑡) = 840 𝑥 7 + (2 𝑥 4 + 2 𝑥 3 − 1680 𝑥 7 ) 𝑡 2 − 2 (1 + 𝑥)𝑥 3 𝑡 2 + (− 24 𝑥 4 − 1 24

1

1

1

1

1

𝑥 3 + 20160 𝑥 7 ) 𝑡 4 + 24 (1 + 𝑥)𝑥 3 𝑡 2 + (720 𝑥 4 + 720 𝑥 3 − 604800 𝑥 7 ) 𝑡 6 −

1 720

(1 + 𝑥)𝑥 3 𝑡 6 + ⋯,

and the closed form solution is obtained as 6

𝑢(𝑥, 𝑡) = 7! 𝑥 7 cos(𝑡).

49


2.3.4. Wave Equation Example 2.7. Consider the following partial differential equation [230] 𝑢𝑡𝑡 − 𝑢𝑥𝑥 + 𝑢 = 2 sin(𝑥), with initial conditions, 𝑢(𝑥, 0) = 𝑠𝑖𝑛𝑥 , 𝑢𝑡 (𝑥, 0) = 1. Applying Adomian’s decomposition method, we get 𝑢(𝑥, 𝑡) = 𝑠𝑖𝑛𝑥 + 𝑡 + 𝑡 2 𝑠𝑖𝑛𝑥 + 𝐿−1 𝑡𝑡 (𝑢𝑥𝑥 − 𝑢)). Applying Taylor series, we get −1

1

𝑢(𝑥, 𝑡) = 𝑡 + (1 + 𝑡 2 )𝑥 + ( 6 𝑡 2 − 6) 𝑥 3 + ⋯ + 𝐿−1 𝑡𝑡 (𝑢𝑥𝑥 − 𝑢)). According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑡) = 𝑡, 𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 (𝑥) + 𝐿−1 𝑡𝑡 (𝑢𝑥𝑥𝑘 − 𝑢𝑘 ) , 𝑘 ≥ 0. Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑡) = 𝑡, 𝑡

𝑡 𝜕2

1

𝑢1 (𝑥, 𝑡) = (1 + 𝑡 2 )𝑥 + ∫0 ∫0 (𝜕2 𝑥 (𝑡) − 𝑡)𝑑𝑡𝑑𝑡 = 𝑥 + 𝑥𝑡 2 − 3! 𝑡 3 , −1

𝑡

1

𝑡 𝜕2

1

𝑢2 (𝑥, 𝑡) = ( 6 𝑡 2 − 6) 𝑥 3 + ∫0 ∫0 (𝜕2 𝑥 (𝑥 + 𝑥𝑡 2 − 3! 𝑡 3 ) − (𝑥 + 𝑥𝑡 2 − 1 3 3!

𝑡 )) 𝑑𝑡𝑑, 𝑡 =

−1 3!

1

1

1

1

𝑥 3 − 6 𝑡 2 𝑥 3 + 5! 𝑡 5 − 12 𝑥𝑡 2 − 2 𝑥𝑡 2 , . 50

. .. The series solution is given by 1

1

1

1

1

1

𝑢(𝑥, 𝑡) = 𝑡 + 𝑥 + 𝑥𝑡 2 − 3! 𝑡 3 − 3! 𝑥 3 − 6 𝑡 2 𝑥 3 + 5! 𝑡 5 − 12 𝑥𝑡 2 − 2 𝑥𝑡 2 + ⋯ , and the close form is given by, 𝑢(𝑥, 𝑡) = sin(𝑥) + sin(𝑡).

Example 2.8. Consider the following partial differential equation [230] 𝑢𝑡𝑡 − 𝑢𝑥𝑥 − 𝑢 = 0, with initial conditions 𝑢(𝑥, 0) = 0, 𝑢𝑡 (𝑥, 0) = sin(𝑥). Applying Adomian’s decomposition method, we get 𝑢(𝑥, 𝑡) = 𝑡sin(𝑥) + 𝐿−1 𝑡𝑡 (𝑢𝑥𝑥 + 𝑢)). Applying Taylor series, we get 1

1

1

𝑢(𝑥, 𝑡) = 𝑡𝑥 − 6 𝑡𝑥 3 + 120 𝑡𝑥 5 − 5040 𝑡𝑥 7 + ⋯ + 𝐿−1 𝑡𝑡 (𝑢𝑥𝑥 + 𝑢)). According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑡) = 𝑡𝑥, 𝑡

𝑡 𝜕2

𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 (𝑥) + ∫0 ∫0 (𝜕2 𝑥 (𝑢𝑘 ) + 𝑢𝑘 )𝑑𝑡𝑑𝑡

𝑘 ≥ 0.

Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑡) = 𝑡𝑥, 𝑡

1

𝑡 𝜕2

𝑢1 (𝑥, 𝑡) = − 6 𝑡𝑥 3 + ∫0 ∫0 (𝜕2 𝑥 (𝑡𝑥) + 𝑡𝑥)𝑑𝑡𝑑𝑡, =

−1 6

1

𝑡𝑥 3 + 6 𝑥𝑡 3 . 𝑡

1

𝑡 𝜕2

−1

1

𝑢2 (𝑥, 𝑡) = 120 𝑡𝑥 5 + ∫0 ∫0 (𝜕2 𝑥 ( 6 𝑡𝑥 3 + 6 𝑥𝑡 3 ) + 1

1

1

1

−1 6

1

𝑡𝑥 3 + 6 𝑥𝑡 3 ) 𝑑𝑡𝑑𝑡,

1

= 120 𝑡𝑥 5 + 120 𝑥𝑡 5 + 3 (− 2 𝑥 − 12 𝑡 3 ) 𝑡 3 . 1

𝑡

1

𝑡 𝜕2

1

1

1

1

𝑢3 (𝑥, 𝑡) = − 5040 𝑡𝑥 7 + 5040 𝑥𝑡 7 + ∫0 ∫0 (𝜕2 𝑥 (120 𝑡𝑥 5 + 120 𝑥𝑡 5 + 3 (− 2 𝑥 − 1 12

𝑡 3 )𝑡 3 ) +

1 120

1

1

1

1

𝑡𝑥 5 + 120 𝑥𝑡 5 + 3 (− 2 𝑥 − 12 𝑡 3 )𝑡 3 ) 𝑑𝑡𝑑𝑡, 1

1

1

1

1

1

1

1

= − 5040 𝑡𝑥 7 + 5040 𝑥𝑡 7 + 5 (− 12 𝑥 − 144 𝑥 3 ) 𝑡 5 + 3 (12 𝑥 3 + 240 𝑥 5 ) 𝑡 3 . . 51

. .. The series solution is given by 1

1

1

1

𝑢(𝑥, 𝑡) = 𝑡 (𝑥 − 3! 𝑥 3 + 5! 𝑥 5 − 7! 𝑥 7 + 9! 𝑥 9 −, … ), And the close form solution is 𝑢(𝑥, 𝑡) = 𝑡 sin(𝑥).

Example 2.9. Consider the following partial differential equation [230] 𝑢𝑡𝑡 − 𝑢𝑥𝑥 + 𝑢2 = 2𝑥 2 − 2𝑡 2 + 𝑥 4 𝑡 4 ,

(2.39)

with initial conditions, 𝑢(𝑥, 0) = 0, 𝑢𝑡 (𝑥, 0) = 0. Applying Adomian’s decomposition method, we get 1

1

𝑢(𝑥, 𝑡) = 𝑥 2 𝑡 2 − 6 𝑡 4 + 30 𝑥 4 𝑡 6 + 𝐿−1 𝑡𝑡 (𝑢𝑥𝑥 − 𝐴)), Applying Taylor series, we get 1

1

𝑢(𝑥, 𝑡) = − 6 𝑡 4 +𝑥 2 𝑡 2 + 30 𝑥 4 𝑡 6 + 𝐿−1 𝑡𝑡 (𝑢𝑥𝑥 − 𝐴)), According to the proposed technique, we have the following recurrence relation 1

𝑢0 (𝑥, 𝑡) = − 6 𝑡 4 , 𝑡 𝜕2

𝑡

𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 (𝑥) + ∫0 ∫0 (𝜕2 𝑥 (𝑢𝑘 ) − 𝐴𝑘 )𝑑𝑡𝑑𝑡 , 𝑘 ≥ 0. Consequently, following approximants are obtained, 1

𝑢0 (𝑥, 𝑡) = − 6 𝑡 4 , 𝑡 𝜕2

𝑡

𝑢1 (𝑥, 𝑡) = 𝑥 2 𝑡 2 + ∫0 ∫0 (𝜕2 𝑥 (𝑢0 ) − 𝑢02 )𝑑𝑦𝑑𝑡, 1

= 𝑥 2 𝑡 2 − 3240 𝑡10 . 𝑢2 (𝑥, 𝑡) =

1 30

𝑡

𝑡 𝜕2

𝑥 4 𝑡 6 + ∫0 ∫0 (

1

𝜕2 𝑥

(𝑢1 ) − 2𝑢0 𝑢1 )𝑑𝑡𝑑𝑡,

1

1

1

= 30 𝑥 4 𝑡 6 − 2332800 𝑡16 + 168 𝑥 2 𝑡 8 + 6 𝑡 4 . 𝑡

𝑡 𝜕2

𝑢3 (𝑥, 𝑡) = 0 + ∫0 ∫0 (𝜕2𝑥 (𝑢2 ) − (2𝑢0 𝑢2 + 𝑢12 )𝑑𝑡𝑑𝑡, 1

59

1

17

= − 1939956480 𝑡 22 + 4127760 𝑥 2 𝑡14 + 11880 𝑥 4 𝑡12 + 22680 𝑡10 + 1 140

1

𝑥 2 𝑡 8 − 30 𝑥 4 𝑡 6 , 52

121

6907

13

𝑢4 (𝑥, 𝑡) = − 219991064832000 𝑡 28 + 282338784000 𝑥 2 𝑡 20 + 81793800 𝑥 4 𝑡18 + 787

47

29

1

1

1

𝑡16 + 2522520 𝑥 2 𝑡14 − 166320 𝑥 4 𝑡12 + 10 (630 − 135 𝑥 6 ) 𝑡10 − 495331200

11 840

𝑥2𝑡 8,


1

1

1

1

1

𝑢(𝑥, 𝑡) = − 6 𝑡 4 +𝑥 2 𝑡 2 − 3240 𝑡10 + 30 𝑥 4 𝑡 6 − 2332800 𝑡16 + 168 𝑥 2 𝑡 8 + 6 𝑡 4 − 1

59

1939956480 1 30

𝑥4𝑡6 − 787

495331200 11 840

1

17

1

𝑡 22 + 4127760 𝑥 2 𝑡14 + 11880 𝑥 4 𝑡12 + 22680 𝑡10 + 140 𝑥 2 𝑡 8 − 121

219991064832000 47

𝑡 28 +

6907 282338784000 29

𝑥 2 𝑡 20 + 1

1

13 81793800

𝑥 4 𝑡18 +

1

𝑡16 + 2522520 𝑥 2 𝑡14 − 166320 𝑥 4 𝑡12 + 10 (630 − 135 𝑥 6 ) 𝑡10 −

𝑥2𝑡8.

Since the exact solution is 𝑢(𝑥, 𝑡) = 𝑥 2 𝑡 2 .


53

Table 2.5. Comparison of Exact and Approximate solution of Eq. (2.39) 𝑥=𝑡

Exact solution

Series solution

*Error(absolute)

0.00,0,00 0.00000000000000 0.00000000000000 0.00000000000000 0.01,0.01 0.00000001000000 0.00000001000000 0.00000000000000 0.02,0.02 0.00000016000000 0.00000016000000 0.00000000000000 0.03,0.03 0.00000081000000 0.00000081000000 0.00000000000000 0.04,0.04 0.00000256000000 0.00000256000000 0.00000000000000 0.05,0.05 0.00000625000000 0.00000625000000 0.00000000000000 0.06,0.06 0.00001296000000 0.00001296000000 0.00000000000000 0.07,0.07 0.00002401000000 0.00002401000000 0.00000000000000 0.08,0.08 0.00004096000000 0.00004096000000 0.00000000000000 0.09,0.09 0.00006561000003 0.00006561000000 0.00000000000003 *Error=Exact solution−Series solutionn

2.4. Optimal Adomian’s Decomposition Method (OADM) We modified traditional Adomian’s decomposition method by applying optimality condition and name it Optimal Adomian’s Decomposition Method. This method is successively applied to Fredholm Integral equation and obtained better results as compare to Adomian’s decomposition method. This method is so simple and very reliable.

2.4.1. Methodology In Optimal decomposition method we have the following recurrence relation 𝑢0 = 𝑓, 𝑢𝑘+1 = −𝐴𝑘+1 𝐿−1 (𝑅𝑢𝑘 + 𝑁𝑢𝑘 ). The constants 𝐴𝑘+1 will determine as 𝜕𝑈 𝜕𝐶𝑘+1

= 0,

𝑘 = 0,1,2, …

𝑏

where 𝑈 = ∫𝑎 𝑅 2 (𝑥, 𝐶1 , 𝐶2 , 𝐶3 , … )d𝑥, and 𝑅(𝑥, 𝐶1 , 𝐶2 , 𝐶3 , … ) is residual.

2.4.2. Fredholm Integral Equation Example 2.10. Consider the following linear Fredholm integral equation [229] 𝜋

𝑢(𝑥) = −𝑥 + sin(𝑥) + ∫02 (1 + 𝑥 − 𝑡)𝑢(𝑡)d𝑡. 54

(2.40)

The exact solution of (2.40) is given as 𝑢(𝑥) = sin(𝑥). According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥) = −𝑥 + sin(𝑥), 𝜋

𝑢𝑘+1 (𝑥) = 𝐴𝑘+1 ∫02 (1 + 𝑥 − 𝑡)𝑢(𝑡)d𝑡. Consequently, we have the following approximants 𝑢0 (𝑥) = −𝑥 + sin(𝑥), 1

𝑢1 (𝑥) = 24 𝐴1 (24𝑥 − 3𝜋 2 + 𝜋 3 − 3𝑥𝜋 2 ), 1

𝑢2 (𝑥) = 192 𝐴1 𝐴2 𝜋 2 (−20𝜋 + 24 + 24𝑥 + 𝑥𝜋 2 + 4𝜋 2 − 12𝑥𝜋), . . .. After two iterations, we have the series solution as 2

− 3𝜋 ) + 𝑢𝑂𝐷𝑀 (𝑥) = −𝑥 + sin(𝑥) + 24 𝐴1 ( 24𝑥 𝐴 𝐴 𝜋 2 (−20𝜋 + 24 + 3 192 1 2 +𝜋 − 3𝑥𝜋 2 1

1

24𝑥 + 𝑥𝜋 2 + 4𝜋 2 − 12𝑥𝜋) + ⋯,

(2.41)

The residual vector can be constructing as 𝜋

𝑅 = 𝑢𝑂𝐷𝑀 (𝑥) + 𝑥 − sin(𝑥) − ∫02 (1 + 𝑥 − 𝑡)𝑢(𝑡)d𝑡. 𝜋

Also

𝐽 = ∫02 𝑅 2 d𝑥.

To determine 𝐴1 and 𝐴2 , setting 𝜕𝐽

𝜕𝐽

[𝜕𝐴 = 0, 𝜕𝐴 = 0]. 1

2

We obtained the solution set 𝐴1 = −

96(𝜋−2) 𝜋 4 −96192

, 𝐴2 = −

2

.

𝜋−2

After substituting the value of 𝐴1 and 𝐴2 into Eq. (2.41), we get the exact solution 𝑢𝑂𝐷𝑀 (𝑥) = sin(𝑥). If we take 𝐴1 = 𝐴2 = 1, we get the solution of Adomian’s decomposition method 1

1

1

1

𝑢𝐴𝐷𝑀 (𝑥) = sin(𝑥) − 16 𝜋 3 + 192 𝜋 4 𝑥 + 48 𝜋 4 − 16 𝜋 3 𝑥.

55

Table 2.6. Comparison of the Exact and Approximate Solutions of Eq. (2.40) obtained from ADM and ODM _______________________________________________________________________ 𝑥

Exact Solution ADM Solution

ODM Solution Error in ADM

Error in ODM

_______________________________________________________________________ 0

0.0000000000

0.0914637710

0.0000000000

9.1464E-02

0.0000E+00

1

0.8414709848

-0.4976185200

0.8414709848

1.3391E+00

0.0000E+00

2

0.9092974268

-1.8603453560

0.9092974268

2.7696E+00

0.0000E+00

3

0.1411200081

-4.0590760520

0.1411200081

4.2002E+00

0.0000E+00

4

-0.7568024953

-6.3875518310

-0.7568024953

5.6307E+00

0.0000E+00

5

-0.9589242747

-8.0202268860

-0.9589242747

7.0613E+00

0.0000E+00

6

-0.2794154982

-8.7712713900

-0.2794154982

8.4919E+00

0.0000E+00

7

0.6569865987

-9.2654225630

0.6569865987

9.9224E+00

0.0000E+00

8

0.9893582466 -10.3636042000

0.9893582466

1.1353E+01

0.0000E+00

9

0.4121184852 -12.3713972300

0.4121184852

1.2784E+01

0.0000E+00

10 -0.5440211109 -14.7580901000

-0.5440211109

1.4214E+01

0.0000E+00

______________________________________________________________________

(a)

(b)

Fig. 2.12. (a)-(b): Comparison of Exact, ADM and OADM solution of Eq. (2.40)

Example 2.11. Consider the following linear Fredholm integral equation [229] 𝜋

𝑢(𝑥) = cos(x) + 2x + ∫0 (𝑥𝑡)𝑢(𝑡)d𝑡. The exact solution of (2.42) is given as 56

(2.42)

𝑢(𝑥) = cos(𝑥). According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥) = cos(x) + 2𝑥, 𝜋

𝑢𝑘+1 (𝑥) = 𝐴𝑘+1 ∫0 (𝑥𝑡)𝑢(𝑡)d𝑡. Consequently, we have the following approximants 𝑢0 (𝑥) = cos(x) + 2𝑥, 2

𝑢1 (𝑥) = 3 𝐴1 𝑥(−3 + 𝜋 3 ), 2

𝑢2 (𝑥) = 9 𝐴1 𝐴2 𝑥(−3 + 𝜋 3 )𝜋 3 , . . .. After two iterations, we have the series solution as 2

2

𝑢𝑂𝐷𝑀 (𝑥) = cos(x) + 2𝑥 + 3 𝐴1 𝑥(−3 + 𝜋 3 ) + 9 𝐴1 𝐴2 𝑥(−3 + 𝜋 3 )𝜋 3 + ⋯, (2.43) The residual vector can be constructing as 𝜋

𝑅 = 𝑢𝑂𝐷𝑀 (𝑥) − cos(x) − 2x − ∫0 (𝑥𝑡)𝑢(𝑡)d𝑡. Also

𝜋

𝐽 = ∫0 𝑅 2 d𝑥.


𝜕𝐽

[𝜕𝐴 = 0, 𝜕𝐴 = 0]. 1

2

We obtained the solution set 𝐴1 = 𝐴1 ,

𝐴2 = −

3(𝐴1 𝜋 3 −3𝐴1 +3) 𝐴1 𝜋 3 (−3+𝜋3 )

.

After substituting the value of 𝐴1 and 𝐴2 into Eq. (2.43) we get the exact solution 𝑢𝑂𝐷𝑀 (𝑥) = cos(𝑥). If we take 𝐴1 = 𝐴2 = 1, we get the solution of Adomian’s decomposition method 2

𝑢𝐴𝐷𝑀 (𝑥) = cos(𝑥) + 9 𝑥𝜋 6 .

57

Table 2.7. Comparison of the Exact and Approximate Solutions of Eq. (2.42) obtained from ADM and ODM _______________________________________________________________________ x

Exact Solution

ADM Solution

ODM Solution

Error in ADM

Error in

ODM _______________________________________________________________________ 0.

1.0000000000

1.

0.5403023059 214.1823455000

0.5403023059

2.1364E+02

0.0000E+00

2.

-0.4161468365 426.8679395000

-0.4161468365

4.2728E+02

0.0000E+00

3.

-0.9899924966 639.9361371000

-0.9899924966

6.4093E+02

0.0000E+00

4.

-0.6536436209 853.9145291000

-0.6536436209

8.5457E+02

0.0000E+00

5.

0.2836621855 1068.4938780000

0.2836621855

1.0682E+03

0.0000E+00

6.

0.9601702867 1282.8124290000

0.9601702867

1.2819E+03

0.0000E+00

7.

0.7539022543 1496.2482050000

0.7539022543

1.4955E+03

0.0000E+00

8.

-0.1455000338 1708.9908460000

-0.1455000338

1.7091E+03

0.0000E+00

9.

-0.9111302619 1921.8672590000

-0.9111302619

1.9228E+03

0.0000E+00

-0.8390715291

2.1364E+03

0.0000E+00

1.0 -0.839071529

1.0000000000

12135.5813600000

1.0000000000

0.0000E+00

0.0000E+00

_______________________________________________________________________

(a)

(b)


Example 2.12. Consider the following linear Fredholm integral equation [229] 58

𝜋

𝑢(𝑥) = sin(𝑥) − 𝑥 + ∫02 (𝑥𝑡)𝑢(𝑡)d𝑡.

(2.44)

The exact solution of (2.44) is given as 𝑢(𝑥) = sin(𝑥). According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥) = sin(𝑥) − 𝑥, 𝜋

𝑢𝑘+1 (𝑥) = 𝐴𝑘+1 ∫02 (𝑥𝑡)𝑢(𝑡)d𝑡. Consequently, we have the following approximants 𝑢0 (𝑥) = sin(𝑥) − 𝑥, 1

𝑢1 (𝑥) = 24 𝐴1 𝑥(−24+𝜋 3 ), 1

𝑢2 (𝑥) = − 576 𝐴1 𝐴2 𝑥(−24+𝜋 3 )𝜋 3 , . . .. After two iterations, we have the series solution as 1

1

𝑢𝑂𝐷𝑀 (𝑥) = sin(𝑥) − 𝑥 + 24 𝐴1 𝑥(−24+𝜋 3 ) − 576 𝐴1 𝐴2 𝑥(−24+𝜋 3 )𝜋 3 + ⋯,

(2.45)


𝑅 = 𝑢𝑂𝐷𝑀 (𝑥) − sin(𝑥) + 𝑥 − ∫02 (𝑥𝑡)𝑢(𝑡)d𝑡. 𝜋

Also

𝐽 = ∫02 𝑅 2 d𝑥.


𝜕𝐽

[𝜕𝐴 = 0, 𝜕𝐴 = 0]. 1

2

We obtained the solution set 𝐴1 = 𝐴1 , 𝐴2 = −

24(𝐴1 𝜋 3 −24𝐴1 +24) 𝐴1 𝜋 3 (−24+𝜋3 )

.

After substituting the value of 𝐴1 and 𝐴2 into Eq. (2.45), we get the exact solution 𝑢𝑂𝐷𝑀 (𝑥) = sin(𝑥). If we take 𝐴1 = 𝐴2 = 1, we get the solution of Adomian’s decomposition method 1

𝑢𝐴𝐷𝑀 (𝑥) = sin(𝑥) − 576 𝜋 6 .

59


Exact Solution

ADM Solution ODM Solution Error in ADM Error in ODM

_______________________________________________________________________ 0.

0.0000000000

0.0000000000

0.0000000000

0.0000E+00

0.0000E+00

1.

0.0099998333

-0.0066909513

0.0099998333

1.6691E-02

0.0000E+00

2.

0.0199986667

-0.0133829026

0.0199986667

3.3382E-02

0.0000E+00

3.

0.0299955002

-0.0200768537

0.0299955002

5.0072E-02

0.0000E+00

4.

0.0399893342

-0.0267738043

0.0399893342

6.6763E-02

0.0000E+00

5.

0.0499791693

-0.0334747538

0.0499791693

8.3454E-02

0.0000E+00

6.

0.0599640065

-0.0401807013

0.0599640065

1.0014E-01

0.0000E+00

7.

0.0699428473

-0.0468926451

0.0699428473

1.1684E-01

0.0000E+00

8.

0.0799146940

-0.0536115830

0.0799146940

1.3353E-01

0.0000E+00

9.

0.0898785492

-0.0603385124

0.0898785492

1.5022E-01

0.0000E+00

1.0

0.0998334166

-0.0670744296

0.0998334166

1.6691E-01

0.0000E+00

_______________________________________________________________________

(a)

(b)


Example 2.13. Consider the following linear Fredholm integral equation [229] 1

𝜋

𝑢(𝑥) = 𝑥 cos(𝑥) + 1 + 2 ∫0 𝑢(𝑡)d𝑡. The exact solution of (2.46) is given as 60

(2.46)

𝑢(𝑥) = x cos(x). According to the proposed technique we have the following recurrence relation 𝑢0 (𝑥) = x cos(x) + 1, 𝜋

1

𝑢𝑘+1 (𝑥) = 𝐴𝑘+1 2 ∫0 𝑢(𝑡)d𝑡. Consequently, we have the following approximants 𝑢0 (𝑥) = x cos(x), 1

𝑢1 (𝑥) = 2 𝐴1 (−2 + 𝜋), 1

𝑢2 (𝑥) = 4 𝐴1 𝐴2 (−2 + 𝜋)𝜋, . . .. After two iterations, we have the series solution as 1

1

𝑢𝑂𝐷𝑀 (𝑥) = x cos(x) + 2 𝐴1 (−2 + 𝜋) + 4 𝐴1 𝐴2 (−2 + 𝜋)𝜋 + ⋯,

(2.47)

The residual vector can be constructing as 1

𝜋

𝑅 = 𝑢𝑂𝐷𝑀 (𝑥) − x cos(x) − 1 − 2 ∫0 𝑢(𝑡)d𝑡. Also

𝜋

𝐽 = ∫0 𝑅 2 d𝑥.


𝜕𝐽

[𝜕𝐴 = 0, 𝜕𝐴 = 0]. 1

2

We obtained the solution set 𝐴1 = 𝐴1 , 𝐴2 = −

2(𝐴1 𝜋−2𝐴1 +2) 𝐴1 𝜋(−2+𝜋)

.

After substituting the value of 𝐴1 and 𝐴2 into Eq. (2.47), we get the exact solution 𝑢𝑂𝐷𝑀 (𝑥) = x cos(x). If we take 𝐴1 = 𝐴2 = 1, we get the solution of Adomian’s decomposition method 1

𝑢𝐴𝐷𝑀 (𝑥) = x cos(x) + 4 π2 .

61

Table 2.9. Comparison of the Exact and Approximate Solutions of Eq. (2.46) obtained from ADM and ODM _____________________________________________________________________ x

Exact Solution

ADM Solution

ODM Solution Error in ADM Error in ODM

______________________________________________________________________ 0.

0.0000000000

0.0000000000

0.0000000000

0.0000E+00

0.0000E+00

1.

0.0099998333

-0.0052203371

0.0099998333

1.5220E-02

0.0000E+00

2.

0.0199986667

-0.0104416743

0.0199986667

3.0440E-02

0.0000E+00

3.

0.0299955002

-0.0156650112

0.0299955002

4.5661E-02

0.0000E+00

4.

0.0399893342

-0.0208913477

0.0399893342

6.0881E-02

0.0000E+00

5.

0.0499791693

-0.0261216831

0.0499791693

7.6101E-02

0.0000E+00

6.

0.0599640065

-0.0313570164

0.0599640065

9.1321E-02

0.0000E+00

7.

0.0699428473

-0.0365983461

0.0699428473

1.0654E-01

0.0000E+00

8.

0.0799146940

-0.0418466698

0.0799146940

1.2176E-01

0.0000E+00

9.

0.0898785492

-0.0471029851

0.0898785492

1.3698E-01

0.0000E+00

1.0

0.0998334166

-0.0523682882

0.0998334166

1.5220E-01

0.0000E+00

______________________________________________________________________

(a)

(b)


Example 2.14. Consider the following linear Fredholm integro differential equation [229] as 1

𝑢´ (𝑥) = 3 + 6𝑥 + ∫0 (𝑥𝑡)𝑢(𝑡)d𝑡.

(2.48) 62

The exact solution of (2.48) is given as 𝑢(𝑥) = 3x + 4x 2 . According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥) = 3𝑥 + 3𝑥 2 , 𝑥

1

𝑢𝑘+1 (𝑥) = 𝐴𝑘+1 ∫0 ∫0 (𝑥𝑡)𝑢(𝑡)d𝑡𝑑𝑥. Consequently, we have the following approximants 𝑢0 (𝑥) = 3𝑥 + 3𝑥 2 , 7

𝑢1 (𝑥) = 8 𝐴1 𝑥 2 , 7

𝑢2 (𝑥) = 64 𝐴1 𝐴2 𝑥 2 , . . .. After two iterations we have the series solution as 7

7

𝑢𝑂𝐷𝑀 (𝑥) = 3𝑥 + 3𝑥 2 + 8 𝐴1 𝑥 2 + 64 𝐴1 𝐴2 𝑥 2 + ⋯

(2.49)

The residual vector can be constructing as 1

𝑅 = 𝑢´ 𝑂𝐷𝑀 − 3 − 6𝑥 − 𝑥 ∫0 (𝑡)𝑢(𝑡)d𝑡. Also

1

𝐽 = ∫0 𝑅 2 d𝑥.


𝜕𝐽

[𝜕𝐴 = 0, 𝜕𝐴 = 0]. 1

2


𝐴2 = −

8(7𝐴1 −8) 7𝐴1

.

After substituting the value of 𝐴1 and 𝐴2 into Eq. (2.49), we get the exact solution 𝑢𝑂𝐷𝑀 (𝑥) = 3x + 4x 2 . If we take 𝐴1 = 𝐴2 = 1, we get the solution of Adomian’s decomposition method 𝑢𝐴𝐷𝑀 (𝑥) = 3x +

255 2 x . 64

63


Exact Solution

ADM Solution ODM Solution Error in ADM Error in ODM

_______________________________________________________________________ 0.

0.0000000000

0.0000000000

0.0000000000

0.0000E+00

0.0000E+00

1.

0.0304000000

0.0303984375

0.0304000000

1.5625E-06

0.0000E+00

2.

0.0616000000

0.0615937500

0.0616000000

6.2500E-06

0.0000E+00

3.

0.0936000000

0.0935859375

0.0936000000

1.4062E-05

0.0000E+00

4.

0.1264000000

0.1263750000

0.1264000000

2.5000E-05

0.0000E+00

5.

0.1600000000

0.1599609375

0.1600000000

3.9062E-05

0.0000E+00

6.

0.1944000000

0.1943437500

0.1944000000

5.6250E-05

0.0000E+00

7.

0.2296000000

0.2295234375

0.2296000000

7.6562E-05

0.0000E+00

8.

0.2656000000

0.2655000000

0.2656000000

1.0000E-04

0.0000E+00

9.

0.3024000000

0.3022734375

0.3024000000

1.2656E-04

0.0000E+00

10.

0.3400000000

0.3398437500

0.3400000000

1.5625E-04

0.0000E+00

_______________________________________________________________________

(a)

(b)


Example 2.15. Consider the following linear Fredholm integro differential equation [229] as 𝜋

𝑢´´´ (𝑥) = 2 + sin(x) − ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡. 64

(2.50)

With boundary condition 𝑢(0) = 1, 𝑢´ (0) = 0, 𝑢 ´´(0) = −1 The exact solution of (2.50) is given as 𝑢(𝑥) = cos(x) According to the proposed technique, we have the following recurrence relation 1

𝑢0 (𝑥) = 3 x 3 + cos(x), 𝑥

𝑥

𝑥

𝜋

𝑢𝑘+1 (𝑥) = −𝐴𝑘+1 ∫0 ∫0 ∫0 ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡𝑑𝑥𝑑𝑥𝑑𝑥. Consequently, we have the following approximants 1

𝑢0 (𝑥) = 3 x 3 + cos(x), 1

𝑢1 (𝑥) = 1440 𝐴1 𝑥 3 (−5𝑥𝜋 4 − 480 + 16𝜋 5 ), 1

𝑢2 (𝑥) = 518400 𝐴1 𝐴2 𝑥 3 𝜋 4 (−45𝜋 5 + 1800𝑥 + 142𝜋 6 − 5760), . . .. After two iterations, we have the series solution as 1 3 x 3

𝑢𝑂𝐷𝑀 (𝑥) =

1

+ cos(x) + 1440 𝐴1 𝑥 3 (−5𝑥𝜋 4 − 480 + 16𝜋 5 )

1

+ 518400 𝐴1 𝐴2 𝑥 3 𝜋 4 (−45𝜋 5 + 1800𝑥 + 142𝜋 6 − 5760) + ⋯,

(2.51)


𝑅 = 𝑢´´´ 𝑂𝐷𝑀 (𝑥) − 2 − sin(x) + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡. Also

𝜋

𝐽 = ∫0 𝑅 2 d𝑥.


𝜕𝐽

[𝜕𝐴 = 0, 𝜕𝐴 = 0]. 1

2

We obtained the solution set 2160(𝜋 5 −40)

𝐴1 = − 𝜋10−2160𝜋5 +86400 ,

40

𝐴2 = − 𝜋5 −40)

After substituting the value of 𝐴1 and 𝐴2 into Eq. (2.51), we get the exact solution 𝑢𝑂𝐷𝑀 (𝑥) = cos(x). If we take 𝐴1 = 𝐴2 = 1, we get the solution of Adomian’s decomposition method 65

1

71

𝑢𝐴𝐷𝑀 (𝑥) = cos(x) − 11520 𝑥 4 𝜋 9 + 259200 𝑥 3 𝜋10 . Table 2.11. Comparison of the Exact and Approximate Solutions of Eq. (2.50) obtained from ADM and ODM _______________________________________________________________________ x

Exact Solution ADM Solution ODM Solution

Error in ADM

Error in ODM

_______________________________________________________________________ 0.

1.0000000000

1.0000000000

1.0000000000

0.0000E+00

0.0000E+00

1.

0.9999500004

0.9999756266

0.9999500004

2.5626E-05

0.0000E+00

2.

0.9998000067

1.0000048090

0.9998000067

2.0480E-04

0.0000E+00

3.

0.9995500337

1.0002405430

0.9995500337

6.9051E-04

0.0000E+00

4.

0.9992001067

1.0008352140

0.9992001067

1.6351E-03

0.0000E+00

5.

0.9987502604

1.0019405940

0.9987502604

3.1903E-03

0.0000E+00

6.

0.9982005399

1.0037078480

0.9982005399

5.5073E-03

0.0000E+00

7.

0.9975510003

1.0062875250

0.9975510003

8.7365E-03

0.0000E+00

8.

0.9968017063

1.0098295680

0.9968017063

1.3028E-02

0.0000E+00

9.

0.9959527330

1.0144833050

0.9959527330

1.8531E-02

0.0000E+00

10.

0.9950041653

1.0203974560

0.9950041653

2.5393E-02

0.0000E+00

_______________________________________________________________________

(a)

(b)


66

Example 2.16. Consider the following linear Fredholm integro differential equation [229] as 𝜋

𝑢´´´´(𝑥) = (2𝑥 − 𝜋) + sin(x) + cos(x) − ∫02 (𝑥 − 2𝑡)𝑢(𝑡)d𝑡.

(2.52)

With boundary conditions 𝑢(0) = 𝑢´ (0) = 1,

𝑢´´ (0) = 𝑢´´´ (0) = −1

The exact solution of (2.52) is given as 𝑢(𝑥) = sin(x) + cos(x). According to the proposed technique, we have the following recurrence relation 1

1

𝑢0 (𝑥) = 60 x 5 − 24 πx 4 + sin(x) + cos(x), 𝑥

𝑥

𝑥

𝑥

𝜋

𝑢𝑘+1 (𝑥) = 𝐴𝑘+1 ∫0 ∫0 ∫0 ∫0 ∫0 (𝑥 − 2𝑡)𝑢(𝑡)d𝑡𝑑𝑥𝑑𝑥𝑑𝑥𝑑𝑥. Consequently, we have the following approximants 1

1

𝑢0 (𝑥) = 60 x 5 − 24 πx 4 + sin(x) + cos(x), 1

𝑢1 (𝑥) = − 3870720 𝐴1 𝑥 4 (64512𝑥 − 7𝑥𝜋 6 + 29𝜋 7 − 161280𝜋), 1

𝑢2 (𝑥) = − 891813888000 𝐴1 𝐴2 𝑥 4 𝜋 6 (1612800𝑥 − 313𝑥𝜋 6 − 661600𝜋 + 1300𝜋 7 ), . . .. After two iterations, we have the series solution as 1

1

1

𝑢𝑂𝐷𝑀 (𝑥) = 60 x 5 − 24 πx 4 + sin(x) + cos(x) + − 3870720 𝐴1 𝑥 4 (64512𝑥 − 1

7𝑥𝜋 6 + 29𝜋 7 − 161280𝜋) + − 891813888000 𝐴1 𝐴2 𝑥 4 𝜋 6 (1612800𝑥 − 313𝑥𝜋 6 − 661600𝜋 + 1300𝜋 7 ) + ⋯,

(2.53)


𝑅 = 𝑢´´´´ 𝑂𝐷𝑀 (𝑥) − (2𝑥 − 𝜋) − sin(x) − cos(x) + ∫02 (𝑥 − 2𝑡)𝑢(𝑡)d𝑡. 𝜋

Also

𝐽 = ∫02 𝑅 2 d𝑥.


𝜕𝐽

[𝜕𝐴 = 0, 𝜕𝐴 = 0]. 1

2

67

We obtained the solution set 1451520(𝜋 6 −5120)

5120

𝐴1 = − 𝜋12−1451520𝜋6 +7431782400 , 𝐴2 = − 𝜋6 −5210. After substituting of the 𝐴1 and 𝐴2 into Eq. (2.53), we get the exact solution 𝑢𝑂𝐷𝑀 (𝑥) = sin(x) + cos(x). If we take 𝐴1 = 𝐴2 = 1, we get the solution of Adomian’s decomposition method 313

13

𝑢𝐴𝐷𝑀 (𝑥) = sin(x) + cos(x) + 891813888000 x 5 π12 − 8918138880 x 4 π12 . Table 2.12. Comparison of the Exact and Approximate Solutions of Eq. (2.52) obtained from ADM and ODM _______________________________________________________________________ x

Exact Solution ADM Solution ODM Solution

Error in ADM

Error in ODM

_______________________________________________________________________ 0.

1.0000000000

1.0000000000

1.0000000000

0.0000E+00

0.0000E+00

1.

1.0099498340

1.0099498340

1.0099498340

0.0000E+00

0.0000E+00

2.

1.0197986730

1.0197986720

1.0197986730

1.0000E-09

0.0000E+00

3.

1.0295455340

1.0295455310

1.0295455340

3.0000E-09

0.0000E+00

4.

1.0391894410

1.0391894300

1.0391894410

1.1000E-08

0.0000E+00

5.

1.0487294300

1.0487294040

1.0487294300

2.6000E-08

0.0000E+00

6.

1.0581645460

1.0581644910

1.0581645460

5.5000E-08

0.0000E+00

7.

1.0674938480

1.0674937470

1.0674938480

1.0100E-07

0.0000E+00

8.

1.0767164000

1.0767162280

1.0767164000

1.7200E-07

0.0000E+00

9.

1.0858312820

1.0858310060

1.0858312820

2.7600E-07

0.0000E+00

10.

1.0948375820

1.0948371620

1.0948375820

4.2000E-07

0.0000E+00

_______________________________________________________________________

68

(a)

(b)


Example 2.17. Consider the following linear Fredholm integro differential equation [229] as 1

𝑢´ (𝑥) = 12𝑥 + ∫0 𝑢(𝑡)d𝑡.

(2.54)

With boundary condition 𝑢(0) = 0. The exact solution of (2.54) is given as 𝑢(𝑥) = 4x + 6x 2 . According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥) = 6x 2 , 𝑥

1

𝑢𝑘+1 (𝑥) = 𝐴𝑘+1 ∫0 ∫0 𝑢(𝑡)d𝑡𝑑𝑥. Consequently, we have the following approximants 𝑢0 (𝑥) = 6x 2 , 𝑢1 (𝑥) = 2𝐴1 𝑥, 𝑢2 (𝑥) = 𝐴1 𝐴2 𝑥, . . .. After two iterations, we have the series solution as 𝑢𝑂𝐷𝑀 (𝑥) = 6x 2 + 2𝐴1 𝑥 + 𝐴1 𝐴2 𝑥 + ⋯, The residual vector can be constructing as 69

(2.55)

1

𝑅 = 𝑢´ 𝑂𝐷𝑀 − 12𝑥 − ∫0 𝑢(𝑡)d𝑡. Also

1

𝐽 = ∫0 𝑅 2 d𝑥.


𝜕𝐽

[𝜕𝐴 = 0, 𝜕𝐴 = 0]. 1

2


𝐴2 = −

2(𝐴1 −2) 𝐴1

.

After substituting the value of 𝐴1 and 𝐴2 into Eq. (2.55), we get the exact solution 𝑢𝑂𝐷𝑀 (𝑥) = 4x + 6x 2 . If we take 𝐴1 = 𝐴2 = 1, we get the solution of Adomian’s decomposition method 𝑢𝐴𝐷𝑀 (𝑥) = 6x 2 + 3x. Table 2.13. Comparison of the Exact and Approximate Solutions of Eq. (2.54) obtained from ADM and ODM _______________________________________________________________________ x

Exact Solution

ADM Solution

ODM Solution

Error in ADM

Error in

ODM _______________________________________________________________________ 0.

0.0000000000

0.0000000000

0.0000000000

0.0000E+00

0.0000E+00

1.

0.0406000000

0.0306000000

0.0406000000

1.0000E-02

0.0000E+00

2.

0.0824000000

0.0624000000

0.0824000000

2.0000E-02

0.0000E+00

3.

0.1254000000

0.0954000000

0.1254000000

3.0000E-02

0.0000E+00

4.

0.1696000000

0.1296000000

0.1696000000

4.0000E-02

0.0000E+00

5.

0.2150000000

0.1650000000

0.2150000000

5.0000E-02

0.0000E+00

6.

0.2616000000

0.2016000000

0.2616000000

6.0000E-02

0.0000E+00

7.

0.3094000000

0.2394000000

0.3094000000

7.0000E-02

0.0000E+00

8.

0.3584000000

0.2784000000

0.3584000000

8.0000E-02

0.0000E+00

9.

0.4086000000

0.3186000000

0.4086000000

9.0000E-02

0.0000E+00

10.

0.4600000000

0.3600000000

0.4600000000

1.0000E-01

0.0000E+00

_______________________________________________________________________

70

(a)

(b)


2.5. Variational Iteration Method The variational iteration method (VIM) established by Ji-Huan He [78] is thoroughly used by mathematicians to handle a wide variety of scientific and engineering applications: linear and nonlinear, and homogeneous and inhomogeneous as well. It was shown that this method is effective and reliable for analytic and numerical purposes. The method gives rapidly convergent successive approximations of the exact solution if such a solution exists. The VIM does not require specific treatments for nonlinear problems as in Adomian method, perturbation techniques, etc. He constructed the approximate solutions of strongly nonlinear equations [82-83], autonomous ordinary differential systems [84], Construction of solitary solution and compaction-like solution [86], seepage flow with fractional derivatives in porous media [80], nonlinear partial differential equations with variable coefficient [85] and eighthorder initial boundary value problems [87] by variational iteration method. Noor and Mohyud-Din implemented Variational Iteration method to obtain the solutions of fourth order boundary value problems [160], twelfth-order boundary value problems [166] and higher order boundary value problems [159]. Abdou and Soliman used Variational iteration method for solving Burger’s and coupled Burger’s equations [11] and nonlinear evolution equations [10, 197]. Abbasbandy find the numerical solutions of nonlinear Klein-Gordon equation [8] by using He’s Variational iteration method. Later on, Wazwaz gave a comparison of Adomian’s decomposition method and Variational iteration method 71

[231] and also applied for finding the exact solutions of Laplace equation [233], linear and nonlinear systems of partial differential equations [235], wave equations [237], diffusion equations [236], Goursat problem [232], Schrodinger equations [238], ordinary differential equations [239] and rational solutions for KdV,𝐾(2,2), Burgers and cubic Boussinesq equations [240]. Many modifications have been made; Noor and MohyudDin [162] developed the elegant coupling of Adomian’s polynomials and the correctional functional of VIM calling it as modified variational iteration method (MVIM) and applied this reliable version for solving eighth-order boundary value problems [161], singular and nonsingular initial and boundary value problems [165], Thomas-Fermi equation [169], fifth-order boundary value problems [167], heat and wave-like equations [163], tarvelling wave solutions of seventh-order generalized KdV equations [146] and Goursat and Laplace problems [164]. Herisanu and Marinca’s modification is much more attractive, where the variational iteration method is coupled with the least squares technology, and one iteration leads to ideal results [88]. Yilmaz and Inc constructed a variational iteration algorithm, where an auxiliary parameter was introduced to adjust the convergence rate, but they did not give a general rule for best choice of the auxiliary parameter [248]. Recently, Hosseini et al. [92] proposed a new algorithm using an auxiliary parameter in Variational iteration method and applied for solving linear and nonlinear partial differential equation [93]. Tamour et al. developed a new modification using Taylor’s series; this modification is extremely effective for the selection of initial value. It is observed that coupling of VIM and Taylor’s series is very effective to tackle such problems and is more user-friendly as compare to the traditional variational iteration method (VIM) and applied to solve Advection problems [266], Burger’s equation [267] and fourth order parabolic equation [268].

2.5.1. Methodology We consider the following general differential equation 𝐿𝑢 + 𝑁𝑢 = 𝑔(𝑥),

(2.56)

where 𝐿 is a linear operator, 𝑁 a nonlinear operator and 𝑔(𝑥) is the forcing term. According to variational iteration method, we can construct a correction functional as follows 𝑥

𝑢𝑛+1 (𝑥) = 𝑢𝑛 (𝑥) + ∫0 𝜆(𝑠)[𝐿𝑢𝑛 (𝑠) + 𝑁𝑢 ̃(𝑠) − 𝑔(𝑠)]ds, 𝑛 72

(2.57)

where 𝜆 is a Lagrange multiplier, which can be identified optimally via variational iteration method (VIM). The subscripts n denotes the nth approximation,

𝑢 ̃𝑛 is

considered as restricted variation. i.e. 𝛿𝑢 ̃𝑛 = 0, (2.57) is called a correction functional. The solution of the linear problems can be found in a single iteration step by the exact identification of the Lagrange multiplier. The principles of variational iteration method and its applicability for various kinds of differential equations are given in. In this method, it is required first to determine the Lagrange multiplier 𝜆 optimally. The successive approximation 𝑢𝑛+1 , 𝑛 ≥ 0, of the solution 𝑢 will be readily obtained upon using the determined Lagrange multiplier and any selective function u 0 , consequently, the solution is given by 𝑢 = lim 𝑢𝑛 . 𝑛→∞

2.5.2. MHD Flow over a Nonlinear Stretching Sheet Let us consider the MHD flow of an incompressible viscous fluid over a stretching sheet at 𝑦 = 0.The fluid is electrically conducting under the influence of an applied magnetic field 𝐵(𝑥) normal to the stretching sheet .The induced magnetic field is ignored .The resulting boundary layer equations are 𝜕𝑢 𝜕𝑥

+

𝜕𝑣 𝜕𝑦

= 0,

𝜕𝑢

(2.58) 𝜕2 𝑢

𝜕𝑢

𝑢 𝜕𝑥 + 𝑣 𝜕𝑦 = 𝜈 𝜕𝑦 2 −

𝜎𝐵2 (𝑥) 𝜌

𝑢,

(2.59)

Here u and v are velocity components in x and y directions respectively, νis kinematic viscosity, ρ is the fluid density , 𝜎 is the electrical conductivity of the fluid .In equation (2.59), the external electrical field and polarization effects are negligible 𝐵(𝑥) = 𝐵0 𝑥

𝑛−1 2

.

(2.60)

The boundary conditions corresponding to the non linear stretching of a sheet are 𝑢(𝑥, 0) = 𝑐𝑥 𝑛 ,

𝑣(𝑥, 0) = 0, 𝑢(𝑥, 𝑦) → 0, 𝑦 → ∞.

(2.61)

Upon making use of following substitutions: 𝑐(𝑛+1)

𝜂=√

2𝜈

𝑥

𝑐(𝑛+1)

𝜈 = −√

2𝜈

(𝑛−1) 2

𝑥

𝑦,

(𝑛−1) 2

𝑢 = 𝑐𝑥 𝑛 𝑓 ′ (𝜂), 𝑛−1

[𝑓(𝜂) + 𝑛+1 𝜂𝑓 ′ (𝜂)],

(2.62)

The resulting non linear differential system is of the following form: 𝑓 ′′′ + 𝑓𝑓 ′′ − 𝛽(𝑓 ′ )2 − 𝑀𝑓 ′ = 0,

(2.63) 73

𝑓(0) = 0, 𝑓 ′ (0) = 1 𝑓 ′ (∞) = 0,

(2.64)

where 2𝜎𝐵2

2𝑛

0 𝑀 = 𝜎𝐶(1+𝑛)

𝛽 = 1+𝑛 ,

(2.65)

The Correction functional for Eq. (2.63) is given as ̃2 ̃ 𝜂 𝑑3 𝑢 𝑑2 𝑢 𝑑𝑢 𝑑𝑢 ̃ 𝑢𝑛+1 (𝜂) = 𝑢𝑛 (𝜂) + ∫0 𝜆(𝑠) ( 𝑑𝑠3𝑛 + 𝑢𝑛 𝑑𝑠2𝑛 − 𝛽 ( 𝑑𝑠𝑛 ) − 𝑀 𝑑𝑠𝑛 ) 𝑑𝑠, making the correction functional stationary, the Lagrange multiplier can be identified as 𝜆(𝑠) = −

(𝑠−𝜂)2 2

,

we get the following iterative formula 𝜂 (𝑠−𝜂)2

𝑢𝑛+1 (𝜂) = 𝑢𝑛 (𝜂) − ∫0

2

̃2 ̃ 𝑑3 𝑢 𝑑2 𝑢 𝑑𝑢 𝑑𝑢 ̃ ( 𝑑𝑠3𝑛 + 𝑢𝑛 𝑑𝑠2𝑛 − 𝛽 ( 𝑑𝑠𝑛) − 𝑀 𝑑𝑠𝑛 ) 𝑑𝑠.

Consequently, the following approximants are obtained 𝑢0 (𝜂) = 𝜂 + 𝑢1 (𝜂) = 𝜂 + 𝑢2 (𝜂) = 𝜂 +

𝐴𝜂 2 2 𝐴𝜂 2 2 𝐴𝜂 2 2

, + +

𝜂 5 𝐴2 𝛽 60 𝜂 5 𝐴2 𝛽 60

−

𝜂 5 𝐴2 120

−

𝜂4 𝐴 24

+

𝜂 4 𝐴𝜷 12

+

𝜂 4 𝐴𝑀 24

+

𝜂3 𝑀 6

+

𝜂3 𝛽 6

,

+ ⋯,

. . .. The series solution is given by 𝑢(𝜂) = lim 𝑢𝑛+1 (𝜂), 𝑛→∞

𝑢(𝜂) = 𝜂 +

𝐴𝜂 2 2

+

𝜂 5 𝐴2 𝛽 60

−

𝜂 5 𝐴2 120

−

𝜂4 𝐴 24

+

𝜂 4 𝐴𝜷 12

+

𝜂 4 𝐴𝑀 24

+

𝜂3 𝑀 6

+

The value of constant can be evaluated by using Pade approximation

74

𝜂3 𝛽 6

….

Table 2.14. Error Analysis of Eq. (2.63) for different values of 𝑀 and 𝛽. 𝛽

1.5

𝑀

Pade [2/2]

Pade [3/3]

Pade [4/4]

Shooting Method

1.0

-1.564019301 -1.507824385 -1.533045283 -1.5252

5.0

-2.597272802 -2.506647097 -2.512831092 -2.5161

10.0

-3.479473099 -3.355115756 -3.359942776 -3.3663

50.0

-7.414014013 -7.142737684 -7.149633379 -7.1647

100.0

-10.41823182 -10.03581039 -10.04508281 -10.0664

500.0

-23.17566344 -22.32376958 -22.34268094 -

1000.0 -32.75402011 -31.55486722 -31.57625735 -

-1

1.0

-0.942809042 -0.949073679 -0.0

5.0

-2.290482629 -2.140943930 -2.163519576 -2.1628

10.0

-3.253971680 -3.094087909 -3.109374075 -3.1100

50.0

-7.0

100.0

-10.34410312 -9.951908855 -9.962644851 -9.9833

500.0

-23.14239336 -22.28588228 -22.30567963 -

-7.0

-7.0

-0.8511

-

1000.0 -32.73048190 -31.52599789 -31.55008194 1.0

-0.584385448 -0.837814817 -0.643216839 -0.8511

5.0

-2.221255544 -2.047659774 -2.086163149 -2.1628

10.0

-3.206471884 -3.037379743 -3.055858076 -3.1100

-1.5 50.0

-7.0

-7.0

-7.0

-

100.0

-10.32920606 -9.935003747 -9.946092151 -9.9833

500.0

-23.13573288 -22.27797711 -22.29827311 -

1000.0 -32.72577349 -31.51898464 -31.54484464 -

2.5.3. MHD Flow of an Incompressible Viscous Fluid through Convergent or Divergent Channels in Presence of a High Magnetic Field Consider a system of cylindrical polar coordinates (𝑟, 𝜃, 𝑧), where the steady twodimensional flow of an incompressible conducting viscous fluid from a source or sink at 75

channel walls lie in planes and intersect in z-axis [94]. The schematic diagram of problem is illustrated. Now we assumed that 𝑢𝜃 = 0; it means that there are no changes with respect to 𝑧 direction; thus the motion is purely in radial direction and merely depends on 𝑟 and 𝜃 and there is no magnetic field along 𝑧 − axis. The polar form of equation of continuity, Navier-Stokes and Maxwell’s in reduce form is given as 𝜕

𝜌 𝑟𝜕𝑟 [𝑟𝑢(𝑟, 𝜃)] = 0,

(2.66) 𝜕2 𝑢(𝑟,𝜃)

𝑢(𝑟, 𝜃)

𝜕𝑢(𝑟,𝜃) 𝜕𝑟 1 𝜕𝑃 𝜌𝑟 𝜕𝜃

1 𝜕𝑃

= − 𝜌 𝜕𝑟 + 𝑣 [ 2𝑣 𝜕𝑢(𝑟,𝜃)

− 𝑟2

𝜕𝜃

+

1 𝜕𝑢(𝑟,𝜃)

𝜕𝑟 2 𝑟 1 𝜕2 𝑢(𝑟,𝜃)

+ 𝑟2

𝜕𝜃2

−

𝜕𝑟 ] 𝑢(𝑟,𝜃)

𝜎𝐵2

− 𝜌𝑟 02 𝑢(𝑟, 𝜃),

(2.67)

𝑟2

= 0,

(2.68)

where 𝐵0 = Electromagnetic induction strength, 𝜎 = Conductivity of the fluid, 𝑢 = Velocity along radial direction, 𝑃 = Fluid pressure, 𝜈 = Coefficient of kinematic viscosity, and

𝜌 = Fluid density.

Now from Eq. (2.66), we have 𝑓(𝜃) = 𝑟𝑢(𝑟, 𝜃).

(2.69)

𝜃

Using 𝜂 = 𝛼, i.e., the dimensionless parameters, where 𝛼 is the semiangle between the inclined walls 𝑓(𝜃)

𝑓(𝜂) = 𝑓

max

,

(2.70)

substituting Eq. (2.70) into Eq. (2.67) and Eq. (2.68), we have 𝑓 ′′′ (𝜂) + 2𝛼𝑅𝑒𝑓(𝜂)𝑓 ′ (𝜂) + (4 − 𝐻)𝛼 2 𝑓 ′ (𝜂) = 0,

(2.71)

𝜎𝐵2

where 𝐻 = √ 𝜌𝑣0 is Hertmann number and 𝑅𝑒 is the Reynolds number is 𝑅𝑒 =

𝑓max 𝛼 𝜈

{

diverdent − channel: 𝛼 > 0, 𝑓max > 0, convergent − channel: 𝛼 < 0, 𝑓max < 0.

So, we have the BCs 𝑓(0) = 1, 𝑓 ′ (0) = 0, 𝑓(1) = 0.

(2.72)

76

According to the Variational iteration method the correctional function of Eq. (2.71) is written as 𝜂 ′ ̃′ (𝑠) + (4 − 𝐻)𝛼 2 𝑓̃ 𝑓𝑘+1 (𝜂) = 𝑓𝑘 (𝜂) + ∫0 𝜆(𝑠)(𝑓𝑘′′′ (𝑠) + 2𝛼𝑅𝑒𝑓𝑘 (𝑠)𝑓 𝑘 𝑘 (𝑠))ds.

Substitute the obtained value of Lagrange multiplier

𝜆(𝑠) = −

(𝑠−𝑡)2 2!

into correction

functional we have 𝜂 (𝑠−𝑡)2

𝑓𝑘+1 (𝜂) = 𝑓𝑘 (𝜂) − ∫0

2!

(𝑓𝑘′′′ (𝑠) + 2𝛼𝑅𝑒𝑓𝑘 (𝑠)𝑓𝑘′ (𝑠) + (4 − 𝐻)𝛼 2 𝑓𝑘′ (𝑠))ds.

Consequently, we have the following approximants 𝑓0 (𝜂) = 1 −

𝑎𝜂 2 2

,

1

1

1

1

1

𝑓1 (𝜂) = 1 + 2 𝑎𝜂2 − 120 𝑅𝑒𝛼𝑎2 𝜂6 − 12 𝜂4 𝑅𝑒𝛼𝑎 − 6 𝜂4 𝛼 2 𝑎 + 24 𝜂4 𝛼 2 𝑎𝐻, 1

1

1

𝑓2 (𝜂) = 1 + 2 𝑎𝜂2 − 95040 𝜂12 𝑅𝑒 3 𝛼 3 𝑎3 − 47520 𝑅𝑒 2 𝛼 4 𝑎3 + ⋯ . . .. 𝑓(𝜂) = lim 𝑓𝑘 (𝜂). 𝑘→∞

1

1

1

𝑓(𝜂) = 1 + 2 𝑎𝜂2 − 95040 𝜂12 𝑅𝑒 3 𝛼 3 𝑦3 − 47520 𝑅𝑒 2 𝛼 4 𝑎3 + ⋯. The objective of the present study was to apply the Variational iteration method to obtain an explicit analytic solution of MHD flow through convergent or divergent channels in presence of a high magnetic field. The magnetic field plays its role in no dimensional parameter, namely, the Hartmann number. If we fix Re number reveals the fact that by increasing magnetic field the velocity profile becomes flat and thickness of boundary layer decreases. In fact magnetic field induces a force in opposite of the momentum’s direction that stabilizes the velocity profile. We have the following table Table 2.15. Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = −2.50 and different values of 𝐻. 𝐻 0 𝑎

1000

2000

4000

-1.119985350 -0.9672680211 -0.8391122178 -0.6389447538

77

Fig. 2.20. VIM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝛼 = −2.50 .

Now we take inverse case of Fig. 2.20 in which we see by decreasing Hartman number, the velocity profile becomes flat and thickness of boundary layer decreases, and we have the following table Table 2.16. Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = 2.50 and different values of 𝐻. 𝐻 0 𝑎

1000

2000

4000

-3.532651597 -3.016558576 -2.583688243 -1.912969976

Fig. 2.21. VIM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝛼 = 2.50 .

We take 𝛼 = 50 instead of 𝛼 = 2.50 as discuss in Fig. 2.20, notice that velocity profile is clearer. Table 2.17. Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = 50 and different values of 𝐻. 𝐻 0 𝑎

1000

2000

4000

-5.869196478 -3.272798354 -1.854115346 -0.6512075718 78

Fig. 2.22. VIM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝛼 = 50 .

We examine if we increases 𝛼 the effect of walls on fluid flow decreases when we move away from them which lead to an increase of velocity and the velocity profile in divergent channels, we have the following tables Table 2.18. Values of 𝑎 for 𝑅𝑒 = 100, 𝐻 = 1500 and different values of 𝛼. 𝛼 2.50

50

100

150

200

𝑎 -2.790676065 -2.455715650 -0.6698633971 -0.07459514756 -0.00667889575

Fig. 2.23. VIM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝐻 = 1500.

If we decrease 𝛼 the behavior in velocity profile is overturned as in Fig. 2.23. The velocity profile in convergent channels and we have

79

Table 2.19. Values of 𝑎 for 𝑅𝑒 = 100, 𝐻 = 1500 and different values of 𝛼. 𝛼 -2.50

-50

-100

-150

-200

𝑎 -0.89774945 -0.306202517 -0.039179482 -0.007564057565 -0.002014462370

Fig. 2.24. VIM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝐻 = 1500.

Now we examine the behavior of velocity profile if 𝑅𝑒 varies, and fixed Hartmann numbers we obtained Table 2.20. Values of 𝑎 for 𝛼 = 2.50 , 𝐻 = 1500 and different values of 𝑅𝑒. 𝑅𝑒 -200

-100

50

100

200

𝑎

-

-

-

-2.790676065

-4.704402105

0.5226120092

0.9004377158

2.108823833

Fig. 2.25. VIM solution for velocity is convergent channel for 𝛼 = 2.50 and 𝐻 = 1500.

The reverse behavior of velocity profile is notice when we increase 𝑅𝑒. We can infer from Figures 2.24 and 2.25 for inflow regime, back flow is prevented in the case of 80

convergent channels but is possible for large Reynolds numbers in the case of divergent channels. Table 2.21. Values of 𝑎 for 𝛼 = −2.50 , 𝐻 = 1500 and different values of 𝑅𝑒. 𝑅𝑒 -200 𝑎

-100

50

100

200

-4.704402105 -2.790676065 -1.192874651 -0.5226120092 -0.1925220645

Fig. 2.26. VIM solution for velocity is convergent channel for 𝛼 = −2.50 and 𝐻 = 1500.

The comparison between the numerical results and VIM solution (1st, 2nd, 3rd and 4th order approximate) for velocity when 𝑅𝑒 = 100 and 𝐻 = 1500 is shown in Table 7.

81

Table 2.22 (a). The comparison between the numerical results and VIM solution of Eq. (2.71) for velocity when 𝛼 = 2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500. __________________________________________________________________ 𝑥

0th Order 1st Order 2nd Order 3rd Order 4th Order 5th Order Numerical Approx.

Approx. Approx.

Approx.

Approx.

Approx. Solution

__________________________________________________________________ 0.00 1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

0.05 0.99750

0.99650

0.99652

0.99652

0.99652

0.99652

0.99652

0.10 0.99000

0.98606

0.98612

0.98611

0.98611

0.98611

0.98611

0.15 0.97750

0.96883

0.96896

0.96895

0.96895

0.96895

0.96895

0.20 0.96000

0.94505

0.94528

0.94525

0.94525

0.94525

0.94525

0.25 0.93750

0.91506

0.91541

0.91536

0.91536

0.91536

0.91536

0.30 0.91000

0.87928

0.87973

0.87966

0.87966

0.87966

0.87966

0.35 0.87750

0.83817

0.83869

0.83860

0.83860

0.83860

0.83860

0.40 0.84000

0.79227

0.79276

0.79265

0.79265

0.79265

0.79265

0.45 0.79750

0.74210

0.74243

0.74230

0.74230

0.74230

0.74230

0.50 0.75000

0.68822

0.68818

0.68804

0.68804

0.68804

0.68804

0.55 0.69750

0.63115

0.63047

0.63032

0.63032

0.63032

0.63032

0.60 0.64000

0.57132

0.56970

0.56956

0.56956

0.56956

0.56956

0.65 0.57750

0.50912

0.50624

0.50612

0.50612

0.50612

0.50612

0.70 0.51000

0.44478

0.44037

0.44030

0.44029

0.44029

0.44029

0.75 0.43750

0.37837

0.37232

0.37230

0.37229

0.37229

0.37229

0.80 0.36000

0.30974

0.30220

0.30224

0.30223

0.30224

0.30224

0.85 0.27750

0.23848

0.23005

0.23014

0.23013

0.23013

0.23013

0.90 0.19000

0.16389

0.15579

0.15590

0.15589

0.15589

0.15589

0.95 0.09750

0.08490

0.07922

0.07930

0.07930

0.07930

0.07930

1.00 0.00000

0.00000

0.00000

0.00000

0.00000

0.00000

0.00000

__________________________________________________________________

82

Table 2.22 (b). The comparison between the numerical results and VIM solution of Eq. (2.71) for velocity when 𝛼 = −2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500. __________________________________________________________________ 𝑥


Approx. Approx.

Approx.

Approx.

Approx. Solution

__________________________________________________________________ 0.00 1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

0.05 0.99750

0.99867

0.99885

0.99887

0.99887

0.99887

0.99887

0.10 0.99000

0.99465

0.99535

0.99545

0.99545

0.99545

0.99545

0.15 0.97750

0.98782

0.98941

0.98965

0.98965

0.98965

0.98965

0.20 0.96000

0.97799

0.98085

0.98128

0.98128

0.98129

0.98129

0.25 0.93750

0.96490

0.96944

0.97013

0.97013

0.97013

0.97013

0.30 0.91000

0.94821

0.95484

0.95586

0.95586

0.95586

0.95586

0.35 0.87750

0.92751

0.93665

0.93807

0.93807

0.93808

0.93808

0.40 0.84000

0.90232

0.91437

0.91627

0.91627

0.91629

0.91629

0.45 0.79750

0.87210

0.88739

0.88987

0.88987

0.88989

0.88989

0.50 0.75000

0.83625

0.85502

0.85815

0.85815

0.85817

0.85817

0.55 0.69750

0.79410

0.81644

0.82029

0.82028

0.82031

0.82031

0.60 0.64000

0.74495

0.77070

0.77533

0.77531

0.77534

0.77534

0.65 0.57750

0.68802

0.71677

0.72216

0.72211

0.72215

0.72215

0.70 0.51000

0.62250

0.65347

0.65955

0.65946

0.65950

0.65950

0.75 0.43750

0.54756

0.57953

0.58613

0.58596

0.58601

0.58601

0.80 0.36000

0.46231

0.49356

0.50039

0.50009

0.50014

0.50014

0.85 0.27750

0.36586

0.39412

0.40074

0.40022

0.40027

0.40027

0.90 0.19000

0.25728

0.27970

0.28552

0.28469

0.28473

0.28473

0.95 0.09750

0.13563

0.14881

0.15308

0.15180

0.15183

0.15183

1.00 0.00000

0.00000

0.00000

0.00187

0.00000

0.00000

0.00000

__________________________________________________________________

2.6. Modified Variational Iteration Method Usman et al. [266-268] did a new modification by decomposing the initial guess into the Taylor’s series and set the recurrence relation accordingly. This modification has been 83

applied many nonlinear PDEs. In this modified version we face less computational work as compare to traditional Variational iteration method. This modification is more reliable and more effective.

2.6.1. Methodology In this modified version we decompose initial condition into Taylor’s series i.e., 𝑢(𝑥) = 𝑓0 + 𝑓1 + 𝑓2 + ⋯, letting 𝑢0 (𝑥) = 𝑓0 , then we have the correctional function is 𝑥 𝑢𝑛+1 (𝑥) = 𝑢𝑛 (𝑥) + 𝑓𝑛+1 + ∫0 𝜆(𝑠) (𝐿(𝑢𝑛 + 𝑓𝑛+1 ) + 𝑁(𝑢𝑛 ̃ + 𝑓𝑛+1 ) − 𝑔(𝑠)) d𝑠,

2.6.2. Nonlinear Partial Differential Equation Consider the following partial differential equation [230] 1

1

𝑢𝑡 − 2 𝑢𝑥2 = − sin(𝑥 + 𝑡) − 2 sin 2(𝑥 + 𝑡),

(2.73)

with initial conditions, 𝑢(𝑥, 0) = cos(𝑥). Applying Taylor’s series on initial condition, we get 1

1

1

1

𝑢(𝑥, 𝑡) = 1 − 2 𝑥 2 + 24 𝑥 4 − 720 𝑥 6 + 40320 𝑥 8 − ⋯ The correction functional for the given problem is 𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 +

𝜕 [𝑢𝑛 (𝑥, 𝑠) 𝑡 𝜕𝑠 ∫0 λ(s) [

1 ̃ (𝑥, 𝑠)+𝑓 + 𝑓𝑘+1 ] + 2 [𝑢𝑛̃ 𝑘+1 ] 1

+ sin(𝑥 + 𝑠) + 2 sin 2(𝑥 + 𝑠)

2

] 𝑑𝑠.

Substitute the obtained value of Lagrange multiplier λ(s) = −1 into correction functional 𝑡 𝜕

1

𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 − ∫0 (𝜕𝑠 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] + 2 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ]2 + 1

sin(𝑥 + 𝑠) + 2 𝑠𝑖𝑛2(𝑥 + 𝑠) ])𝑑𝑠. Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑡) = 1, 1

1

1

1

𝑢1 (𝑥, 𝑡) = 1 − 2 𝑥 2 − 𝑐𝑜𝑠𝑥 − 4 𝑐𝑜𝑠2𝑥 + 𝑥𝑡 − 2 𝑥 3 𝑡𝑐𝑜𝑠(𝑥 + 𝑡) + 4 cos(2𝑥 + 2𝑡), 1

1

1

1

1

𝑢2 (𝑥, 𝑡) = 1 − 2 𝑥 2 + 96 𝑥 4 𝑐𝑜𝑠2𝑥 + 24 𝑥 4 𝑐𝑜𝑠𝑥 + 8 𝑥 5 𝑡 − 12 𝑥 3 sin(2𝑥 + 2𝑡) + 2𝑥 2 cos(𝑥 + 𝑡). . . 84

.. The closed form solution is 𝑢(𝑥, 𝑡) = cos(𝑥 + 𝑡).

(a)

(b)

Fig. 2.27. (a)-(b): Comparison of Exact and Approximate solution of Eq. (2.73)

2.6.3. Inhomogeneous Burger’s Equation Consider the following partial differential equation [230] 𝑢𝑡 − 𝑢𝑢𝑥 = 1 − 𝑒 −𝑥 (𝑡 + 𝑒 −𝑥 ),

(2.74)

with initial conditions, 𝑢(𝑥, 0) = 𝑒 −𝑥 . Applying Taylor’s series on initial condition, we get 1

1

1

𝑢(𝑥, 𝑡) = 1 − 𝑥 + 2 𝑥 2 − 6 𝑥 3 + 24 𝑥 4 + ⋯. The correction functional for the given problem is 𝑡 𝜕 (𝑥, 𝑠) + 𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 + ∫0 λ(s) (𝜕𝑠 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] + ([𝑢𝑛̃

̃ 𝜕 𝑓𝑘+1 ])(𝜕𝑥 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ]) – 1 + 𝑒 −𝑥 (𝑠 + 𝑒 −𝑥 ])𝑑𝑠. Substitute the obtained value of Lagrange multiplier λ(s) = −1 into correction functional 𝑡 𝜕

𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 − ∫0 (𝜕𝑠 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] + ([𝑢𝑛 (𝑥, 𝑠) + 𝜕

𝑓𝑘+1 ])(𝜕𝑥 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ]) – 1 + 𝑒 −𝑥 (𝑠 + 𝑒 −𝑥 ])𝑑𝑠. Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑡) = 1, 85

1

𝑢1 (𝑥, 𝑡) = 1 − 𝑥 − 2 𝑒 −𝑥 𝑡 2 + 2𝑡 − 𝑥𝑡 − 𝑒 −2𝑥 𝑡, 5

3

2

3

1

𝑢2 (𝑥, 𝑡) = 1 − 3 𝑒 −2𝑥 𝑡 3 + 2 𝑡 2 𝑥𝑒 −2𝑥 3 𝑡 3 𝑥𝑒 −2𝑥 − 2 𝑡 2 𝑒 −2𝑥 − 2 𝑡 2 𝑥 2 𝑒 −2𝑥 2

+ 3 𝑡 3 𝑒 −4𝑥 + ⋯. . . .. The closed form solution is 𝑢(𝑥, 𝑡) = 𝑡 + 𝑒 −𝑥 .

(a)

(b)


2.6.4. Homogeneous Burger’s Equation Consider the following partial differential equation [230] 𝑢𝑡 − 𝑢𝑢𝑥 = 𝑢𝑥𝑥 ,

(2.75)

with initial conditions, 𝑢(𝑥, 0) = 2𝑡𝑎𝑛𝑥. Applying Taylor’s series on initial condition, we get 2

4

34

124

𝑢(𝑥, 𝑡) = 2𝑥 + 3 𝑥 3 + 15 𝑥 5 + 315 𝑥 7 + 2835 𝑥 9 + ⋯ The correction functional for the given problem is 𝑡 𝜕 (𝑥, 𝑠) + 𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 + ∫0 λ(s) (𝑦𝑠 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] − ([𝑢𝑛̃

̃ ̃ 𝜕 𝜕2 𝑓𝑘+1 ]) (𝜕𝑥 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ]) − 𝜕𝑥 2 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ])𝑑𝑠. 86

Substitute the obtained value of Lagrange multiplier λ(s) = −1 into correction functional 𝑡 𝜕

𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 − ∫0 (𝜕𝑠 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] − ([𝑢𝑛 (𝑥, 𝑠) + 𝜕2

𝜕

𝑓𝑘+1 ])(𝜕𝑠 (𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 )) − 𝜕𝑥 2 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ])𝑑𝑠. Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑡) = 2𝑥, 2

𝑢1 (𝑥, 𝑡) = 2𝑥 + 3 𝑥 3 − 2

16 3

4

𝑡𝑥 3 − 3 𝑡𝑥 5 ,

4

𝑢2 (𝑥, 𝑡) = 2𝑥 + 3 𝑥 3 + 15 𝑥 5 − 8𝑡 2 𝑥 3 +

56 2 5 𝑡 𝑥 3

+

416 2 7 𝑡 𝑥 45

+

16 2 9 𝑡 𝑥 9

256 3 5 𝑡 𝑥 9 68

−

512 3 7 𝑡 𝑥 27

64

80

− 27 𝑡 3 𝑥 9 − 16𝑡 2 𝑥 +

16

− 15 𝑡𝑥 5 − 45 𝑡𝑥 7 − 45 𝑡𝑥 9 .

. . .. The closed form solution is 𝑢(𝑥, 𝑡) = 2tan𝑥.

(a)

(b)


2.6.5. Diffusion Equation Consider the following partial differential equation [230] 𝑢𝑡 − 𝑢𝑢𝑥 = 0,

(2.76)

with initial conditions, 1

𝑢(𝑥, 0) = 1+𝑥. 87

Applying Taylor’s series on initial condition, we get 𝑢(𝑥, 𝑡) = 1 − 𝑥 + 𝑥 2 − 𝑥 3 + 𝑥 4 − 𝑥 5 + 𝑥 6 + ⋯. The correction functional for the given problem is 𝑡 𝜕 (𝑥, 𝑠) + 𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 + ∫0 λ(s) (𝜕𝑠 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] − ([𝑢𝑛̃

̃ 𝜕 𝑓𝑘+1 ]) (𝜕𝑥 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ]) 𝑑𝑠. Substitute the obtained value of Lagrange multiplier λ(s) = −1 into correction functional 𝑡 𝜕

𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 − ∫0 (𝜕𝑠 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] − ([𝑢𝑛 (𝑥, 𝑠) + 𝜕

𝑓𝑘+1 ])(𝜕𝑠 (𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 )))𝑑𝑠. Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑡) = 1, 𝑢1 (𝑥, 𝑡) = 1 − 𝑥 + 𝑡 − 𝑥𝑡, 1

1

3

𝑢2 (𝑥, 𝑡) = 1 − 𝑥 + 𝑥 2 + 3 𝑡 3 − 3 𝑡 3 𝑥 + 𝑡 2 − 2𝑡 2 𝑥 + 2 𝑡 2 𝑥 2 + 𝑡 − 3𝑥𝑡 +3𝑡𝑥 2 − 2𝑡𝑥 3 . . . .. The closed form solution is 1

1

1

𝑢(𝑥, 𝑡) = 1+𝑥 (1 + (1+𝑥)2 𝑡 + (1+𝑥)4 𝑡 2 ).

(a)

(b)


88

2.6.6. Homogeneous Fourth-order Parabolic Equation Consider the following partial differential equation [230] 𝑢𝑡𝑡 − 𝑢𝑥𝑥𝑥𝑥 = 0,

(2.77)

with initial conditions, 𝑢(𝑥, 0) = cos(𝑥), 𝑢𝑡 (𝑥, 0) = − sin(𝑥). Applying Taylor’s series on initial condition, we get 1

1

1

1

𝑢(𝑥, 𝑡) = 1 − 𝑡𝑥 − 2 𝑥 2 + 6 𝑡𝑥 3 + 24 𝑥 4 − 120 𝑡𝑥 5 + ⋯ The correction functional for the given problem is 𝜕2

𝑡

𝜕4

𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 + ∫0 λ(s) (𝜕𝑠2 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] − 𝜕𝑥 4 [𝑢 ̃(𝑥, 𝑡) + 𝑛 𝑓𝑘+1 ]𝑑𝑠. Substitute the obtained value of Lagrange multiplier λ(s) = (𝑠 − 𝑥) into correction functional 𝜕2

𝑡

𝜕4

𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 + ∫0 {(𝜕𝑠2 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] − 𝜕𝑥 4 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ]}𝑑𝑠. Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑡) = 1, 1

𝑢1 (𝑥, 𝑡) = 1 − 𝑡𝑥 + 𝑥 − 2 𝑥 2 , 1

1

1

1

1

1

1

1

𝑢2 (𝑥, 𝑡) = 1 − 𝑡𝑥 − 2 𝑥 2 + 6 𝑡𝑥 3 + 24 𝑥 4 − 2 𝑡 2 , 1

1

𝑢3 (𝑥, 𝑡) = 1 − 𝑥𝑡 − 2 𝑡 2 + 6 𝑡𝑥 3 + 24 𝑥 5 − 2 𝑡 2 − 120 𝑥 5 𝑡 − 720 𝑥 6 1

1

+ 2 𝑥𝑡 3 − 4 𝑡 2 𝑥 2 , . . .. The closed form solution is 𝑢(𝑥, 𝑡) = cos(𝑥 + 𝑡).

89

(a)

(b)


2.6.7. Inhomogeneous Fourth-order Parabolic Equation Consider the following partial differential equation [230] 𝑢𝑡𝑡 + 𝑢𝑥𝑥𝑥𝑥 = (𝜋 4 − 1) sin(𝜋𝑥) sin(𝑡),

(2.78)

with initial conditions, 𝑢(𝑥, 0) = 0, 𝑢𝑡 (𝑥, 0) = sin(𝜋𝑥). Applying Taylor’s series on initial condition, we get 1

1

1

𝑢(𝑥, 𝑡) = 𝜋𝑡𝑥 − 6 𝜋 3 𝑡𝑥 3 + 120 𝜋 4 𝑡𝑥 5 − 5040 𝜋 7 𝑡𝑥 7 + ⋯. The correction functional for the given problem is 𝜕2

𝑡

𝜕4

𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 + ∫0 {λ(s) (𝜕𝑠2 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] + 𝜕𝑥 4 [𝑢 ̃(𝑥, 𝑡) + 𝑛 𝑓𝑘+1 ] − (𝜋 4 − 1) sin(𝜋𝑥) sin(𝑡)}𝑑𝑠. Substitute the obtained value of Lagrange multiplier λ(s) = (𝑠 − 𝑡) into correction functional 𝜕2

𝑡

𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 + ∫0 {(𝑠 − 𝑡) (𝜕𝑠2 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] + 𝜕4 𝜕𝑥 4

[𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 ] − (𝜋 4 − 1) sin(𝜋𝑥) sin(𝑡)}𝑑𝑠.

Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑡) = 𝜋𝑡𝑥, 1

𝑢1 (𝑥, 𝑡) = 𝜋𝑡𝑥 − 6 𝑡𝜋 3 𝑥 3 + sin(𝜋𝑥) . 𝑡𝜋 4 − sin(𝜋𝑥) 𝑡𝑥 − sin(𝜋𝑥) sin(𝑡) 𝜋 4 90

+ sin(𝜋𝑥) sin(𝑡), 1

1

𝑢2 (𝑥, 𝑡) = 𝜋𝑡𝑥 − 6 𝑡𝜋 3 𝑥 3 + sin(𝜋𝑥) 𝑡 − sin(𝜋𝑥) sin(𝑡) + 120 𝑡𝜋 5 𝑥 5 + 1

1

1

𝑡 sin(𝜋𝑥) 𝜋 8 − 2 𝑡 3 𝜋 5 𝑥 − 6 sin(𝜋𝑥) 𝜋 8 𝑡 3 + 6 sin(𝜋𝑥) 𝜋 4 𝑡 3 − sin(𝜋𝑥)𝜋 8 sin(𝑡). . . .. The closed form solution is 𝑢(𝑥, 𝑡) = sin(𝜋𝑥) sin(𝑡).

(a)

(b)


2.6.8. Inhomogeneous Diffusion Equation Consider the following nonlinear partial differential equation [230] 𝑢𝑡 + 𝑢𝑢𝑥 = 1 + 𝑡cos(𝑥) + sin(𝑥) cos(𝑥), with initial condition, 𝑢(𝑥, 0) = sin(𝑥), Applying Taylor’s series on initial condition, we get 1

1

1

1

𝑓(𝑥) = 𝑥 − 6 𝑥 3 + 120 𝑥 5 − 5040 𝑥 7 + 362880 𝑥 9 − ⋯. The correction functional for the given problem is 𝑢0 (𝑥, 𝑡) = 𝑥,

91

(2.79)

𝑡

𝜕

𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 + ∫0 λ(s) (𝜕𝑠 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] + [𝑢 ̃(𝑥, 𝑠) + 𝑛 𝜕

𝑓𝑘+1 ] 𝜕𝑥 [𝑢 ̃(𝑥, 𝑡) + 𝑓𝑘+1 ] − 1 − 𝑠cos(𝑥) + sin(𝑥) cos(𝑥)) 𝑑𝑠. 𝑛 Substitute the obtained value of Lagrange multiplier λ(s) = −1 into correction functional 𝑡

𝜕

𝑢𝑛+1 (𝑥, 𝑡) = 𝑢𝑛 (𝑥, 𝑡) + 𝑓𝑘+1 − ∫0 (𝜕𝑠 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] +

[𝑢𝑛 (𝑥, 𝑠) +

𝜕

𝑓𝑘+1 ] 𝜕𝑥 [𝑢𝑛 (𝑥, 𝑠) + 𝑓𝑘+1 ] − 1 − 𝑠cos(𝑥) + sin(𝑥) cos(𝑥)) 𝑑𝑠 . Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑡) = 𝑥, 1

2

1

1

𝑢1 (𝑥, 𝑡) = 𝑥 − 6 𝑥 3 + 𝑡 − 𝑥𝑡 + 3 𝑥 3 𝑡 − 12 𝑥 5 𝑡 + 2 𝑐𝑜𝑠𝑥. 𝑡 2 + 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥. 𝑡, 3

1

1

1

2

𝑢2 (𝑥, 𝑡) = 2 𝑡 2 𝑥 + 𝑡 + 𝑥 − 36 𝑡 3 𝑠𝑖𝑛𝑥. 𝑥 3 + 120 𝑥 5 − 6 𝑥 3 − 𝑥𝑡 − 3 𝑡 3 𝑠𝑖𝑛𝑥. 𝑐𝑜𝑠𝑥 3 , . . .. The closed form solution is 𝑢(𝑥, 𝑡) = 𝑡 + sin(𝑥).

(a)

(b)


2.6.9. Parabolic Equation Consider the following nonlinear partial differential equation [230] 𝑢𝑥𝑥 − 𝑢𝑥 𝑢𝑦𝑦 = −𝑥 + 𝑢,

(2.80)

with initial condition, 92

𝑢(0, 𝑦) = sin(𝑦), 𝑢𝑥 (0, 𝑦) = 1, Applying Taylor’s series on initial condition, we get 1

1

1

𝑓(𝑦) = 𝑥 + 𝑦 − 6 𝑦 3 + 120 𝑦 5 − 5040 𝑦 7 + ⋯. The correction functional for the given problem is 𝑢0 (𝑥, 𝑦) = 𝑥, 𝜕2

𝑥

𝜕

𝑢𝑛+1 (𝑥, 𝑦) = 𝑢𝑛 (𝑥, 𝑦) + 𝑓𝑘+1 + ∫0 λ(s) (𝜕𝑠2 [𝑢𝑛 (𝑠, 𝑦) + 𝑓𝑘+1 ] − 𝜕𝑠 [𝑢 ̃(𝑠, 𝑦) + 𝑛 𝑓𝑠𝑘+1 ]

𝜕2 𝜕𝑦 2

[𝑢 ̃(𝑠, 𝑡) + 𝑓𝑘+1 ] + 𝑠 − [𝑢 ̃(𝑠, 𝑦) + 𝑓𝑘+1 ])𝑑𝑠. 𝑛 𝑛

Substitute the obtained value of Lagrange multiplier λ(s) = −1 into correction functional 𝑥 𝜕2

𝜕

𝑢𝑛+1 (𝑥, 𝑦) = 𝑢𝑛 (𝑥, 𝑦) + 𝑓𝑘+1 − ∫0 (𝜕𝑠2 [𝑢𝑛 (𝑠, 𝑦) + 𝑓𝑘+1 ] − 𝜕𝑠 [𝑢𝑛 (𝑠, 𝑦) + 𝜕2

𝑓𝑘+1 ] 𝜕𝑦 2 [𝑢𝑛 (𝑠, 𝑡) + 𝑓𝑘+1 ] + 𝑠 − [𝑢𝑛 (𝑠, 𝑦) + 𝑓𝑘+1 ])𝑑𝑠 . Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑦) = 𝑥, 1

𝑢1 (𝑥, 𝑦) = 𝑥 + 𝑦 + 2 𝑦𝑥 2 , 1

1

1

1

𝑢2 (𝑥, 𝑦) = 𝑥 + 𝑦 − 6 𝑦 3 + 24 𝑦𝑥 4 − 6 𝑥 3 𝑦 2 − 12 𝑥 2 𝑦 3, 1

1

1

1

1

11

𝑢3 (𝑥, 𝑦) = − 6 𝑦 3 + 120 𝑦 5 + 𝑥 + 𝑦 − 60 𝑥 5 − 1008 𝑦𝑥 8 + 504 𝑥 7 𝑦 2 + 1080 𝑥 6 𝑦 3 + 1 720 1

1

1

5

1

1

𝑥 6 𝑦 − 60 𝑥 5 𝑦 2 + 180 𝑥 5 𝑦 4 + 144 𝑥 4 𝑦 3 − 144 𝑥 4 𝑦 5 + 36 𝑥 3 𝑦 4 − 1

𝑥 2 𝑦 5 − 24 𝑦𝑥 4 , 240 . . .. The closed form solution is 𝑢(𝑥, 𝑦) = 𝑥 + sin(𝑦).

93

1 216

𝑥3𝑦6 +

(a)

(b)


2.6.10. Nonlinear Parabolic Equation Consider the following non linear partial differential equation [230] 𝑢𝑥𝑥 + 𝑢2 − 𝑢𝑦 2 = 0,

(2.81)

with initial condition, 𝑢(0, 𝑦) = 0, 𝑢𝑥 (0, 𝑦) = 𝑒 𝑦 , Applying Taylor’s series on initial condition, we get 1

1

1

𝑓(𝑦) = 𝑥 + 𝑥𝑦 + 2 𝑥𝑦 2 + 6 𝑥𝑦 3 + 24 𝑥𝑦 4 + ⋯. The correction functional for the given problem is 𝑢0 (𝑥, 𝑦) = 𝑥, 𝜕2

𝑥

𝑢𝑛+1 (𝑥, 𝑦) = 𝑢𝑛 (𝑥, 𝑦) + 𝑓𝑘+1 + ∫0 λ(s) (𝜕𝑠2 [𝑢𝑛 (𝑠, 𝑦) + 𝑓𝑘+1 ] + ([𝑢 ̃(𝑠, 𝑦) + 𝑛 𝜕

𝑓𝑘+1 ])2 − (𝜕𝑦 [𝑢 ̃(𝑠, 𝑡) + 𝑓𝑘+1 ])2 )𝑑𝑠. 𝑛 Substitute the obtained value of Lagrange multiplier λ(s) = (𝑠 − 𝑥) into correction functional 𝜕2

𝑥

𝑢𝑛+1 (𝑥, 𝑦) = 𝑢𝑛 (𝑥, 𝑦) + 𝑓𝑘+1 − ∫0 (𝑠 − 𝑥) (𝜕𝑠2 [𝑢𝑛 (𝑠, 𝑦) + 𝑓𝑘+1 ] + ([𝑢𝑛 (𝑠, 𝑦) + 𝑓𝑘+1 ])2 − (

2

𝜕

[𝑢𝑛 (𝑠, 𝑦) + 𝑓𝑘+1 ]) ) 𝑑𝑠 . 𝜕𝑦

Consequently, following approximants are obtained, 𝑢0 (𝑥, 𝑦) = 𝑥, 94

5

1

1

𝑢1 (𝑥, 𝑦) = 𝑥 + 𝑥𝑦 + 12 𝑥 4 − 3 𝑥 4 𝑦 − 2 𝑥 4 𝑦 2 , 1

1

17

47

1

𝑢2 (𝑥, 𝑦) = 𝑥 + 𝑥𝑦 + 2 𝑥𝑦 2 − 28 𝑥 7 + 1620 𝑥10 𝑦 + 3240 𝑥10 𝑦 2 − 270 𝑥10 𝑦 3 − 1 360 1 4

31

103

11

1

1

1

1

𝑥10 𝑦 4 − 420 𝑥 7 𝑦 − 2520 𝑥 7 𝑦 2 + 35 𝑥 7 𝑦 3 + 60 𝑥 7 𝑦 4 − 6 𝑥 4 𝑦 3 − 8 𝑥 4 𝑦 4 + 6 𝑥 4 𝑦 + 1

𝑥 4 𝑦 2 − 1440 𝑥10 , . . .. The closed form solution is 𝑢(𝑥, 𝑦) = 𝑥𝑒 𝑦 .

(a)

(b)


2.6.11. Reaction-diffusion Equation Example 2.18. Consider the reaction-diffusion equation [230] 𝑢𝑡 = 𝑢𝑥𝑥 + (cos(𝑥) − sin2 (𝑥) − 1) 𝑢,

(2.82)

with initial condition 1

𝑢(𝑥, 0) = 10 exp(cos(𝑥) − 11) , 𝑥 ∈ 𝑅, 1

𝑢(0, 𝑡) = 10 exp(−t − 10) , 𝜕 𝜕𝑥

𝑢(0, 𝑡) = 0,

𝑡 ∈ 𝑅,

𝑡 ∈ 𝑅,

Applying variational iteration method taking approximate langrage multiplier, we get

95

1

𝑢(𝑥, 𝑡) = 10 exp(cos(𝑥) − 11) −

𝜕 𝑢 𝑡 𝜕𝑠 ( ∫0

𝜕2

− 𝜕𝑥 2 𝑢 − (cos(𝑥)

− sin2 (𝑥) − 1)𝑢

) 𝑑𝑠.

Applying Taylor’s on the initial condition, we get 1

1

1

31

𝑓 = 10 𝑒 −10 − 20 𝑒 −10 𝑥 2 + 60 𝑒 −10 𝑥 4 − 7200 𝑒 −10 𝑥 6 + ⋯. According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑡) = 𝑓0 , 𝑡

𝜕2

𝜕

𝑢𝑘+1 (𝑥, 𝑡) = 𝑢𝑘 + 𝑓𝑘+1 − ∫0 (𝜕𝑠 (𝑢𝑘 + 𝑓𝑘+1 ) − 𝜕𝑥 2 (𝑢𝑘 + 𝑓𝑘+1 )(cos(𝑥) − sin2(𝑥) − 1)(𝑢𝑘 + 𝑓𝑘+1 )) 𝑑𝑠, 𝑘 ≥ 0 . Consequently, following approximants are obtained 1

𝑢0 = 10 𝑒 −10 , 1

𝑢1 = 20 𝑒 −10 (−2 − 2𝑡𝑐𝑜𝑠(𝑥) + 𝑡𝑐𝑜𝑠(𝑥)𝑥 2 + 𝑥 2 + 6𝑡 − 2𝑡𝑥 2 − 2cos2 (𝑥)𝑡 + co s2 (𝑥) 𝑡𝑥 2 ), 1

𝑢2 = − 120 𝑒 −10 (−12𝑡 2 sin(𝑥) 𝑥 − 24𝑡 2 𝑐𝑜𝑠(𝑥)𝑥𝑠𝑖𝑛(𝑥) − 12 + 36𝑡 + 6𝑡 2 𝑐𝑜𝑠 3 (𝑥)𝑥 2 + ⋯, . . .. The solution in closed form is given by 1

𝑢(𝑥, 𝑡) = 10 exp(cos(𝑥) − 𝑡 − 11).

96

(a)

(b)


Example 2.19. Consider the reaction-diffusion equation [230] 𝑢𝑡 = 𝑢𝑥𝑥 − 16𝑡 𝑢,

(2.83)

with initial condition 𝑢(𝑥, 0) = exp(−𝑥 − 4) , 𝑥 ∈ 𝑅, 𝑢(0, 𝑡) = exp(−𝑡(8𝑡 − 1) − 4) , 𝜕 𝜕𝑥

𝑡 ∈ 𝑅,

𝑢(0, 𝑡) = − exp(−𝑡(8𝑡 − 1) − 4) ,

𝑡 ∈ 𝑅,

Applying variational iteration method taking approximate langrage multiplier, we get 𝑡

𝜕2

𝜕

𝑢(𝑥, 𝑡) = exp(−𝑥 − 4) − ∫0 (𝜕𝑠 𝑢 − 𝜕𝑥 2 𝑢 + 16𝑠𝑢) 𝑑𝑠. Applying Taylor’s on the initial condition, we get 1

1

1

𝑓 = 𝑒 −4 − 𝑒 −4 𝑥 + 2 𝑒 −4 𝑥 2 − 6 𝑒 −4 𝑥 3 + 24 𝑒 −4 𝑥 4 + ⋯. According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑡) = 𝑓0 , 𝑢𝑘+1 (𝑥, 𝑡) = 𝑢𝑘 + 𝑓𝑘+1 −

𝜕 (𝑢𝑘 𝑡 𝜕𝑠 ∫0 (

+ 𝑓𝑘+1 ) −

𝜕2 𝜕𝑥 2

(𝑢𝑘 + 𝑓𝑘+1 )

+16𝑠(𝑢𝑘 + 𝑓𝑘+1 )

) 𝑑𝑠, 𝑘 ≥ 0 .

Consequently, following approximants are obtained 𝑢0 = 𝑒 −4 , 𝑢1 = 𝑒 −4 (1 − 𝑥 − 16𝑡 2 + 16𝑡 2 𝑥), 1

𝑢2 = − 6 𝑒 −4 (−6 + 6𝑥 + 96𝑡 2 − 16𝑡 2 𝑥 − 3𝑥 2 − 512𝑡 4 + 521𝑡 4 𝑥 − 6𝑡 + 48𝑡 2 𝑥 2 ), 97

1

𝑢3 = 30 𝑒 −4 (−5𝑥 3 + 15𝑥 2 − 400𝑡 3 − 30𝑥𝑡 − 8192𝑡 6 + 8192𝑡 6 𝑥 + 1280𝑡 4 𝑥 2 + 80𝑡 2 𝑥 3 − 30𝑥 − 480𝑡 2 + 30 + 480𝑡 2 𝑥 + 30𝑡 + 2560𝑡 4 − 2560𝑡 4 𝑥 − 240𝑡 2 𝑥 2 ), 1

𝑢4 = − 2520 𝑒 −4 (−2520 − 2520𝑡 + 2520𝑥 − 105𝑥 4 − 1260𝑥 2 𝑡 + ⋯, . . .. The solution in closed form is given by 𝑢(𝑥, 𝑡) = exp(−x − 𝑡(8𝑡 − 1) − 4).

(a)

(b)


Example 2.20. Consider equations reaction-diffusion equation [230] 1

𝑢𝑡 = 𝑢𝑥𝑥 − 4 𝑢,

(2.84)


𝑢(𝑥, 0) = 2 𝑥 + e−𝑥⁄2 , 𝑥 ∈ 𝑅, 𝑢(0, 𝑡) = 1, 𝜕 𝜕𝑥

𝑡 ∈ 𝑅, 1

1

𝑢(0, 𝑡) = 2 e−𝑡⁄4 − 2

𝑡 ∈ 𝑅,

Applying variational iteration method taking approximate langrage multiplier, we get 1

𝑡

𝜕

𝜕2

1

𝑢(𝑥, 𝑡) = 2 𝑥 + e−𝑥⁄2 − ∫0 (𝜕𝑠 𝑢 − 𝜕𝑥 2 𝑢 − 4 𝑢) 𝑑𝑠. Applying Taylor’s on the initial condition, we get 98

1

1

1

1

𝑓 = 1 + 8 𝑥 2 − 48 𝑥 3 + 384 𝑥 4 − 3840 𝑥 5 + ⋯. According to the proposed technique, we have the following recurrence relation 𝑢0 (𝑥, 𝑡) = 𝑓0 , 𝜕 𝑡

𝑢𝑘+1 (𝑥, 𝑡) = 𝑢𝑘 + 𝑓𝑘+1 − ∫0 (𝜕𝑠

(𝑢𝑘 + 𝑓𝑘+1 ) −

𝜕2 𝜕𝑥 2

(𝑢𝑘 + 𝑓𝑘+1 )

1

+ 4 (𝑢𝑘 + 𝑓𝑘+1 )

) 𝑑𝑠, 𝑘 ≥ 0 .

Consequently, following approximants are obtained 𝑢0 = 1, 1

1

8

32

1

1

1

1

𝑢1 = 1 + 𝑥 2 −

𝑥2𝑡 , 1

1

1

1

1

𝑢2 = 1 + 8 𝑥 2 − 32 𝑥 2 𝑡 − 48 𝑥 3 − 32 𝑡 2 + 256 𝑥 2 𝑡 2 − 8 𝑥𝑡 + 192 𝑥 3 𝑡 , 1

1

1

1

1

1

𝑢3 = 1 + 8 𝑥 2 − 48 𝑥 3 − 32 𝑡 2 + 256 𝑥 2 𝑡 2 − 8 𝑥𝑡 + 192 𝑥 3 𝑡 + 384 𝑥 4 − 3072 𝑥 2 𝑡 3 + 1 192 1

1

1

1

𝑡 3 + 32 𝑡 2 𝑥 − 1536 𝑥 3 𝑡 2 − 1536 𝑥 4 𝑡 , 1

1

1

1

1

𝑢4 = 1 + 8 𝑥 2 − 48 𝑥 3 − 256 𝑥 2 𝑡 2 − 8 𝑥𝑡 + 384 𝑥 4 − 3072 𝑥 2 𝑡 3 + + ⋯, The solution in closed form is given by 1

𝑢(𝑥, 𝑡) = 2 𝑥e−𝑡⁄4 + e−𝑥⁄2 .

(a)

(b)

Fig. 2.38.(a)-(b): Comparison of Exact and Approximate solution of Eq. (2.84)

2.7. Homotopy Perturbation Method

99

Homotopy Perturbation Method, which was first proposed by He in 1999, and was successfully applied to various engineering problems [79, 81], has been shown to effectively, easily, and accurately solve a large class of linear and non-linear problems with components converging rapidly to accurate solutions. It is an established fact that most of the physical phenomenon is nonlinear in nature and hence there is a dire need to find their appropriate solutions.

2.7.1. Methodology To illustrate the basic concept of Homotopy perturbation method, consider the following nonlinear functional equation 𝐴(𝑢) = 𝑓(𝑟), 𝑟𝜖 Ω,

(2.85)

with the boundary conditions 𝜕𝑢

𝐵 (𝑢, 𝜕𝑛) = 0, 𝑟𝜕Ω,

(2.86)

where 𝐴 = General functional operator, 𝐵 = Boundary operator, 𝑓(𝑟) = Known analytic function, and

∂Ω = Boundary of the domain Ω.

The operator A is decomposed as 𝐴 = 𝐿 + 𝑁, where 𝐿 is the linear and 𝑁 is the nonlinear operator. Hence Equation (2.85) can be written as 𝐿(𝑢) + 𝑁(𝑢) − 𝑓(𝑟) = 0, 𝑟 ∈ Ω.

(2.87)

We construct a Homotopy 𝑣(𝑟, 𝑝): Ω × [0,1] → 𝑅 satisfying 𝐻(𝑣, 𝑝) = (1 − 𝑝)[𝐿(𝑣) − 𝐿(𝑢0 )] + 𝑝[𝐴(𝑣) − 𝑓(𝑟)] = 0, 𝑝𝜖[0,1], 𝑟 ϵ Ω. (2.88) Hence 𝐻(𝑣, 𝑝) = 𝐿(𝑣) − 𝐿(𝑢0 ) + 𝑝𝐿(𝑢0 ) + 𝑝[𝑁(𝑣) − 𝑓(𝑟)] = 0,

(2.89)

where 𝑢0 is an initial approximation for the solution of (2.85). As 𝐻(𝑣, 0) = 𝐿(𝑣) − 𝐿(𝑢0 ) and 𝐻(𝑣, 1) = 𝐴(𝑣) − 𝑓(𝑟),

(2.90)

It shows that 𝐻(𝑣, 𝑝) continuously traces an implicitly defined curve from a starting point 𝐻(𝑢0 , 0) to a solution 𝐻(𝑣, 1). The embedding parameter p increases monotonously from zero to one as the trivial linear part 𝐿(𝑢) = 0 deforms continuously to the original problem 𝐴(𝑢) = 𝑓(𝑟). The embedding parameter 𝑝 ϵ [0,1] can be considered as an expanding parameter to obtain 100

𝑣 = 𝑣0 + 𝑝𝑣1 + 𝑝2 𝑣2 + ⋯.

(2.91)

The solution is obtained by taking the limit as 𝑝 tends 0 to 1 in equation (2.91). Hence 𝑢 = 𝑙𝑖𝑚 𝑣 = 𝑣0 + 𝑣1 + 𝑣2 + ⋯.

(2.92)

𝑝→1

2.7.2. Flow of a Viscoelastic Fluid through a Porous Channel with Expanding or Contracting Walls We consider an unsteady viscoelastic fluid in a porous semi-infinite channel with expanding or contracting walls [40]. The distance 2𝑎(𝑡) between the porous walls is much smaller than the width and the length of the channels. One end of the channel is closed by a complicated solid membrane. Both walls have equal permeability 𝑣𝜔 and expand or contract uniformly at a time-dependent rate 𝑎(𝑡), as show

Fig. 2.39. Flow of a Viscoelastic Fluid through a Porous Channel with Expanding or Contracting Walls

According to the above conditions, we have ∇ ∙ 𝑉 = 0,

(2.93)

𝐷𝑉

𝜌 𝐷𝑡 = ∇ ∙ (−𝑝𝐼 + 𝜇𝐴1 + 𝛼1 𝐴2 + 𝛼2 𝐴12 ),

(2.94)

Where 𝜌, 𝜇, 𝛼1 and 𝛼2 are the dimensional density, viscosity coefficient, viscoelastic coefficients and cross-viscosity coefficient of viscoelastic fluid. 𝐴1 and 𝐴2 are acceleration tensors is given by 𝐴1 = ∇ 𝑉 + (∇ 𝑉)𝑇 , 𝐴2 = where

𝑑 𝑑𝑡

𝑑𝐴1 𝑑𝑡

(2.95a)

+ 𝐴1 (∇ 𝑉) + (∇ 𝑉)𝑇 𝐴1 .

(2.95b)

denote the material time derivative and ∇ the gradient operator, the following

condition must be hold, 101

𝑎1 ≥ 0, 𝑎1 + 𝑎2 = 0,

(2.96)

The appropriate BCs are 𝑢(𝑥, 𝑎) = 0, 𝑣(𝑥, 𝑎) = −𝑣𝜔 = −𝐴𝑎̇ , 𝜕𝑢

|

𝜕𝑦 (𝑥,0)

where 𝐴 =

𝑣𝜔 𝑎̇

(2.97)

= 0, 𝑣(𝑥, 0), 𝑢(0, 𝑦) = 0,

(2.98)

is the injection coefficient, which is the measure of wall permeability.

Consider that 𝑦

𝑢 = 𝑣𝑎 −2 𝑥𝐹𝜂 (𝜂, 𝑡), 𝑣 = −𝑣𝑎−1 𝐹(𝜂, 𝑡), 𝜂 = 𝑎(𝑡).

(2.99)

Substituting Eq. (2.99) into Eq. (2.93) and Eq. (2.94), we get 𝐹𝜂𝜂𝜂𝜂 (𝜂) + 𝑎 (3𝐹𝜂𝜂 (𝜂) + 𝜂𝐹𝜂𝜂𝜂 (𝜂)) − 𝐹𝜂 (𝜂)𝐹𝜂𝜂 (𝜂) + 𝐹(𝜂)𝐹𝜂𝜂𝜂 (𝜂) − 𝜔𝑎 (5𝐹𝜂𝜂𝜂𝜂 (𝜂) + 𝜂𝐹𝜂𝜂𝜂𝜂 (𝜂)) + 𝜔 (𝐹𝜂 (𝜂)𝐹𝜂𝜂𝜂𝜂 (𝜂) − 𝐹(𝜂)𝐹𝜂𝜂𝑦𝜂𝜂 (𝜂)) − 𝑎2 𝑣 −1 𝐹𝜂𝜂𝑡 (𝜂) + 𝜔𝑎2 𝑣 1 𝐹𝜂𝜂𝜂𝜂𝑡 (𝜂) = 0.

(2.100)

where 𝛼 = 𝑎𝑣̇ is the wall expansion ratio, and 𝜔 is the viscoelastic parameter defined by 𝑎

𝜔 = 𝜌𝑎12 . Note that the expansion ratio is positive for expansion and negative for contractions. Accordingly the BCs are 𝐹𝜂𝜂 (0, 𝑡) = 0, 𝐹(0, 𝑡) = 0, 𝐹𝜂 (1, 𝑡) = 0, 𝐹(1, 𝑡) = 𝑅𝑒, where 𝑅𝑒 is Reynolds number defined by 𝑅𝑒 = 𝐹

𝑎𝑣𝜔 𝑣

(2.101)

. Consider that

𝑦

𝑓 = 𝑅𝑒 , 𝑙 = 𝑎 .

(2.102)

Using the transformation we get 𝑓 iv (𝜂) + 𝛼(3𝑓 ′′ (𝜂) + 𝜂𝑓 ′′′ (𝜂)) + 𝑅𝑒(𝑓(𝜂)𝑓 ′′′ (𝜂) − 𝑓 ′ (𝜂)𝑓 ′′ (𝜂)) − 𝜔𝛼 (5𝑓 iv (𝜂) + 𝜂𝑓 v (𝜂)) + 𝜔𝑅𝑒(𝑓 ′ (𝜂)𝑓 iv (𝜂) − 𝑓(𝜂)𝑓 v ) = 0.

(2.103)

The BCs are 𝑓 ′′ (0) = 0, 𝑓(0) = 0, 𝑓 ′ (1) = 0, 𝑓(1) = 1.

(2.104)

Using Homotopy perturbation method. Assuming 𝑓 = 𝑣 the Eq. (2.103-2.104) we have 𝑣 ′′′′ + 𝛼(3𝑣 ′′ + 𝜂𝑣′′′) + 𝑅𝑒(𝑣𝑣 ′′′ − 𝑣′𝑣′′) − 𝜔𝛼(5𝑣 (4) − 𝜂𝑣 (5) ) + 𝜔𝑅𝑒(𝑣 ′ 𝑣 (4) −

𝑣𝑣 (5) = 0)

(2.105) According to the Homotopy perturbation method we construct the homotopy in the form: 102

𝐻(𝑣, 𝑝) = (1 − 𝑝)[𝑣 ′′′′ (𝜂) − 𝑓0 ′′′′ (𝜂)] + 𝑝[𝑣 ′′′′ + 𝛼(3𝑣 ′′ + 𝜂𝑣′′′) + 𝑅𝑒(𝑣𝑣 ′′′ − 𝑣′𝑣′′) − 𝜔𝛼(5𝑣 (4) − 𝜂𝑣 (5) ) + 𝜔𝑅𝑒(𝑣 ′ 𝑣 (4) − 𝑣𝑣 (5) )]. (2.106) Substituting equation (2.91) into equation (2.106) and rearranging based on powers of pterms, it is reduced to 𝑣0 ′′′′ (𝜂) − 𝑓0 ′′′′ (𝜂) = 0, 𝑣1′′′′ (𝜂) + 𝛼 (3𝑣0′′ (𝜂) + 𝜂𝑣0′′′ (𝜂)) + 𝑅𝑒(𝑣0 (𝜂)𝑣0′′′ (𝜂) − 𝑣0′ (𝜂)𝑣0′′ (𝜂)) − (4) (5) (4) (5) 𝜔𝛼(5𝑣0 (𝜂) − 𝜂𝑣0 (𝜂)) + 𝜔𝑅𝑒(𝑣0′ (𝜂)𝑣0 (𝜂) − 𝑣0 (𝜂)𝑣0 (𝜂)) = 0

. . .. Solving the above equations 𝜂3

𝑣0 (𝜂) = 𝑏𝜂 + 𝑡 3! , 1

1

𝑣1 (𝜂) = − 30 𝜂5 𝛼𝑡 + 2520 𝑅𝑡 2 𝜂7 , 2

1

1

𝑣2 (𝜂) = − 15 𝜔𝜂5 𝛼 2 𝑡 + 420 𝑅𝜔𝑡 2 𝛼𝜂7 + 1260 𝜂7 𝜔𝑅 2 𝑏𝑡 2 + ⋯, . . .. The solution of the given equation when p → 1 is as follows 𝑓(𝜂) = 𝑣0 (𝜂) + 𝑣1 (𝜂) + 𝑣2 (𝜂) + ⋯. 𝜂3

1

1

2

1

𝑓(𝜂) = 𝑏𝜂 + 𝑡 3! − 30 𝜂5 𝛼𝑡 + 2520 𝑅𝑡 2 𝜂7 + − 15 𝜔𝜂5 𝛼 2 𝑡 + 420 𝑅𝜔𝑡 2 𝛼𝜂7 + 1 1260

𝜂7 𝜔𝑅 2 𝑏𝑡 2 + ⋯.

Our concern now is to determine the parameters b and t by using boundary conditions.

103

Fig. 2.40. Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for different values of Re with 𝜔 = 0.02 and 𝛼 = 2.

Fig. 2.41. Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for different values of Re with 𝜔 = 0.02 and 𝛼 = −2.

Fig. 2.42. Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for different values of 𝛼 with 𝜔 = 0.02 and Re = 2.5.

104

Fig. 2.43. Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for different values of 𝜔 with 𝛼 = 2 and Re = −5.

2.7.3. MHD Flow of an Incompressible Viscous Fluid through Convergent or Divergent Channels in Presence of a High Magnetic Field Applying Homotopy perturbation method on Eq. (2.71)-Eq. (2.72) Assuming 𝑓 = 𝑣 the Eq. (2.71) we have 𝑣 ′′′ (𝜂) + 2𝛼𝑅𝑒𝑣(𝜂)𝑣 ′ (𝜂) + (4 − 𝐻)𝛼 2 𝑣 ′ (𝜂) = 0,

(2.107)

According to the Homotopy perturbation method we construct the homotopy in the form: 𝐻(𝑣, 𝑝) = (1 − 𝑝)[𝑣 ′′′ (𝜂) − 𝑓0 ′′′ (𝜂)] − 𝑝[𝑣 ′′′ (𝜂) + 2𝛼𝑅𝑒𝑣(𝜂)𝑣 ′ (𝜂) + (4 − 𝐻)𝛼 2 𝑣 ′ (𝜂)].

(2.108)

Substituting equation (2.91) into equation (2.108) and rearranging based on powers of pterms, it is reduced to 𝑣0 ′′′ (𝜂) − 𝑓0 ′′′ (𝜂) = 0, 𝑣1 ′′′ (𝜂) + 2𝛼𝑅𝑒𝑣0 (𝜂)𝑣0 ′ (𝜂) + (4 − 𝐻)𝛼 2 𝑣0 ′ (𝜂) = 0, . . .. Solving the above equations 𝜂2

𝑣0 (𝜂) = 1 + 𝑎 2! , 1

1

1

1

𝑣1 = − 120 𝑅𝑒𝛼𝑎2 𝜂6 − 12 𝑅𝑒𝛼𝑎𝜂4 − 6 𝛼 2 𝑎𝜂4 − 120 𝐻𝛼 2 𝑎𝜂4 , . 105

. .. The solution of the given equation, when p → 1 is as follows 𝑓(𝜂) = 𝑣0 (𝜂) + 𝑣1 (𝜂) + 𝑣2 (𝜂) + ⋯. 𝜂2

1

1

1

1

𝑓(𝜂) = 1 + 𝑎 2! − 120 𝑅𝑒𝛼𝑎2 𝜂6 − 12 𝑅𝑒𝛼𝑎𝜂4 − 6 𝛼 2 𝑎𝜂4 + − 120 𝐻𝛼 2 𝑎𝜂4 + 1

1

𝑅𝑒 2 𝛼 2 𝑎3 𝜂10 + 560 𝑅𝑒 2 𝛼 2 𝑎2 𝜂8 + ⋯. 10800 The objective of the present study was to apply the Homotopy perturbation method to obtain an explicit analytic solution of MHD flow through convergent or divergent channels in presence of a high magnetic field. The magnetic field plays its role in no dimensional parameter, namely, the Hartmann number. If we fix Re number reveals the fact that by increasing magnetic field the velocity profile becomes flat and thickness of boundary layer decreases. In fact magnetic field induces a force in opposite of the momentum’s direction that stabilizes the velocity profile. We have the following table Table 2.23. Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = −2.50 and different values of 𝐻. 𝐻 0 𝑎

1000

2000

4000

-1.117418863 -0.9644432630 -0.8367326331 -0.6392026162

Fig. 2.44. HPM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝛼 = −2.50 .

Now we take inverse case of Fig. 2.44 in which we see by decreasing Hartman number, the velocity profile becomes flat and thickness of boundary layer decreases, and we have the following table Table 2.24. Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = 2.50 and different values of 𝐻. 106

𝐻 0 𝑎

1000

2000

4000

- 3.512069452 - 3.011764524 -2.583264460 -1.912965835

Fig. 2.45. HPM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝛼 = 2.50 .

We take 𝛼 = 50 instead of 𝛼 = 2.50 as discuss in Fig. 2.44, notice that velocity profile is more clear. Table 2.25. Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = 50 and different values of 𝐻. 𝐻 0 𝑎

1000

2000

4000

-5.458959462 -3.262595571 -1.854117794 -0.6557546191

Fig. 2.46. HPM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝛼 = 50 .

we examine if we increases 𝛼 the effect of walls on fluid flow decreases when we move away from them which lead to an increase of velocity and the velocity profile in divergent channels, we have the following tables Table 2.26. Values of 𝑎 for 𝑅𝑒 = 100, 𝐻 = 1500 and different values of 𝛼. 107

𝛼 2.50 𝑎

50

100

150

200

-2.788876071 -2.458506932 -0.6923394340 -0.1033788301 -0.01711080411

Fig. 2.47. HPM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝐻 = 1500.

If we decrease 𝛼 the behavior in velocity profile is overturned as in Fig. 2.47. The velocity profile in convergent channels and we have Table 2.27. Values of 𝑎 for 𝑅𝑒 = 100, 𝐻 = 1500 and different values of 𝛼. 𝛼 -2.50

-50

-100

-150

-200

𝑎 -

-

-

-

-0.00201446237

0.306202513

0.03917948224

0.007564057560

0.89774945

Fig. 2.48. HPM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝐻 = 1500.

Now we examine the behavior of velocity profile if 𝑅𝑒 varies, and fixed Hartmann numbers we obtained Table 2.28. Values of 𝑎 for 𝛼 = 2.50 , 𝐻 = 1500 and different values of 𝑅𝑒. 108

𝑅𝑒 -200 𝑎

-100

50

100

200

-0.5181038551 -0.8977494510 -2.108830241 -2.788876073 -4.704765922

Fig. 2.49. HPM solution for velocity is convergent channel for 𝛼 = 2.50 and 𝐻 = 1500.

The reverse behavior of velocity profile is notice when we increase 𝑅𝑒. We can infer from Figures 2.48 and 2.49 for inflow regime, back flow is prevented in the case of convergent channels but is possible for large Reynolds numbers in the case of divergent channels. Table 2.29. Values of 𝑎 for 𝛼 = −2.50 , 𝐻 = 1500 and different values of 𝑅𝑒. 𝑅𝑒 -200 𝑎

-100

50

100

200

-4.704765923 -2.788876073 -1.192205819 -0.5181038541 -0.1989016484

Fig. 2.50. HPM solution for velocity is convergent channel for 𝛼 = −2.50 and 𝐻 = 1500.

The comparison between the numerical results and HPM solution (1st, 2nd, 3rd and 4th order approximate) for velocity when 𝑅𝑒 = 100 and 𝐻 = 1500 is shown in Table 2.30. 109

Table 2.30 (a).The comparison between the numerical results and HPM solution of Eq. (2.71) for velocity when 𝛼 = 2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500. _____________________________________________________________________________ 𝑥

0th Order Approx.

1st Order Approx.

2nd Order

3rd Order

4th Order

Approx.

Approx.

Approx.

Numerical Solution

_____________________________________________________________________________ 0.00

1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

0.05

0.99650

0.99650

0.99648

0.99652

0.99652

0.99652

0.10

0.98599

0.98606

0.98598

0.98612

0.98612

0.98612

0.15

0.96848

0.96883

0.96865

0.96897

0.96895

0.96895

0.20

0.94397

0.94505

0.94473

0.94529

0.94525

0.94525

0.25

0.91245

0.91506

0.91456

0.91542

0.91536

0.91536

0.30

0.87393

0.87928

0.87852

0.87974

0.87966

0.87966

0.35

0.82840

0.83817

0.83706

0.83871

0.83860

0.83860

0.40

0.77587

0.79227

0.79067

0.79279

0.79269

0.79265

0.45

0.71633

0.74210

0.73983

0.74247

0.74232

0.74230

0.50

0.64980

0.68822

0.68503

0.68824

0.68809

0.68804

0.55

0.57625

0.63115

0.62674

0.63056

0.63038

0.63032

0.60

0.49571

0.57132

0.56538

0.56984

0.56956

0.56956

0.65

0.40815

0.50912

0.50135

0.50643

0.50620

0.50612

0.70

0.31360

0.44478

0.43498

0.44063

0.44029

0.44029

0.75

0.21204

0.37837

0.36655

0.37264

0.37240

0.37229

0.80

0.10348

0.30974

0.29631

0.30257

0.30225

0.30224

0.85

-0.01209

0.23848

0.22443

0.23042

0.23012

0.23013

0.90

-0.13466

0.16389

0.15105

0.15609

0.15602

0.15589

0.95

-0.26424

0.08490

0.07625

0.07939

0.07939

0.07930

1.00

-0.40082

0.00000

-0.00000

0.00000

-0.00000

-0.00000

____________________________________________________________________________

110

Table 2.30 (b). The comparison between the numerical results and HPM solution of Eq. (2.71) for velocity when 𝛼 = −2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500. ______________________________________________________________________________ 𝑥

0th Order Approx.

1st Order Approx.

2nd Order Approx.

3rd Order Approx.

4th Order Approx.

Numerical Solution

_____________________________________________________________________________ 0.00

1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

0.05

0.99868

0.99867

0.99886

0.99888

0.99887

0.99887

0.01

0.99470

0.99465

0.99539

0.99547

0.99546

0.99545

0.15

0.98808

0.98782

0.98951

0.98968

0.98966

0.98965

0.20

0.97881

0.97799

0.98103

0.98134

0.98131

0.98129

0.25

0.96689

0.96490

0.96972

0.97022

0.97016

0.97013

0.30

0.95232

0.94821

0.95526

0.95599

0.95591

0.95586

0.35

0.93510

0.92751

0.93723

0.93826

0.93808

0.93808

0.40

0.91523

0.90232

0.91515

0.91654

0.91638

0.91629

0.45

0.89272

0.87210

0.88843

0.89021

0.89001

0.88989

0.50

0.86755

0.83625

0.85634

0.85859

0.85832

0.85817

0.55

0.83974

0.79410

0.81809

0.82085

0.82030

0.82031

0.60

0.80927

0.74495

0.77274

0.77601

0.77558

0.77534

0.65

0.77616

0.68802

0.71923

0.72299

0.72213

0.72215

0.70

0.74040

0.62250

0.65636

0.66052

0.65986

0.65951

0.75

0.70199

0.54756

0.58281

0.58723

0.58600

0.58601

0.80

0.66093

0.46231

0.49714

0.50155

0.50062

0.50014

0.85

0.61722

0.36586

0.39776

0.40182

0.40028

0.40028

0.90

0.57086

0.25728

0.28299

0.28624

0.28523

0.28473

0.95

0.52186

0.13563

0.15103

0.15294

0.15219

0.15283

1.00

0.47020

0.00000

-0.00000

-0.00000

0.00000

0.00000

____________________________________________________________________________

2.8. Modified Homotopy Perturbation Method In 2012, Afzal et al. introduce a new modification by decomposition the initial guess into the Taylor’s series. Modified Homotopy Perturbation Method, successively applied to construct the numerical solutions of Kawahara equations [186] nonlinear Dispersive equations [16] and nonlinear Dispersive K(2,2,1) and K(3,3,1) equations [17]. This 111

modified version has less computational work as compare to traditional Homotopy Perturbation method. It is an established fact that most of the physical phenomenon is nonlinear in nature and hence there is a dire need to find their appropriate solutions.

2.8.1. Methodology Applying the Taylor’s series on the initial condition and source term then set the recurrence relation accordingly.

2.8.2. Nonlinear Dispersive K(2,2,1) Equation Consider the following nonlinear dispersive 𝐾(2,2,1) equation [230] 𝑢𝑡 + (𝑢2 )𝑥 − (𝑢2 )𝑥𝑥𝑥 + 𝑢5𝑥 = 0,

(2.109)

with initial condition 𝑢(𝑥, 0) =

16𝑐−1 12

1

cosh2 (4 𝑥).

To apply HPM Taylor’s on given equation, firstly we have 𝑢(𝑥, 𝑡) =

16𝑐−1 12

𝑡

1

𝜕3

𝜕

𝜕5

cosh2 (4 𝑥) − ∫0 (𝜕𝑥 (𝑢2 ) − 𝜕𝑥 3 (𝑢2 ) + 𝜕𝑥 5 (𝑢)) 𝑑𝑡,(1)

Suppose that 𝑢 = 𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯,

(2.110)

Use Eq. (2.110) in Eq. (2.109), we get 𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ =

16𝑐−1 12

𝑡

1

𝜕

cosh2 (4 𝑥) − 𝑝 ∫0 [𝜕𝑥 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 +

𝜕3

𝜕5

⋯ )2 − 𝜕𝑥 3 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )2 + 𝜕𝑥 5 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )], Compare the like power of 𝑝, we get 𝑝𝑜 : 𝑢0 (𝑥, 𝑡) =

16𝑐−1 12

1

cosh2 (4 𝑥),

Apply Taylor series, we get 4

1

3

12

𝑢0 (𝑥, 𝑡) = 𝑐 − 1

1

1

12

192

+( 𝑐−

1

) 𝑥2 + (

1

576

𝑐−

1

1

) 𝑥4 + (

9216

69120

𝑐−

1

) 𝑥 6 + (15482880 𝑐 − 247726080) 𝑥 8 + ⋯, 1105920 = 𝑓0 + 𝑓1 + ⋯, 4

1

1

1

Take 𝑢0 (𝑥, 𝑡) = 𝑓0 = 3 𝑐 − 12 + (12 𝑐 − 192) 𝑥 2 , we have following recurrence relation 𝑡

𝜕3

𝜕

𝜕5

𝑝𝑘+1 : 𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 − ∫0 (𝜕𝑥 (𝐴𝑘 ) − 𝜕𝑥 3 (𝐴𝑘 ) + 𝜕𝑥 5 (𝑢𝑘 )) 𝑑𝑡, 112

where 𝐴𝑘 are He’s polynomials. Consequently, we have following components of solution 1

1

1

1

4

1

𝑢1 (𝑥, 𝑡) = (576 𝑐 − 9216) 𝑥 4 + (69120 𝑐 − 1105920) 𝑥 6 − 4 (3 𝑐 − 12 + 1

1

1

1

1

2

1

(12 𝑐 − 192) 𝑥 2 ) (12 𝑐 − 192) 𝑥𝑡 + 24 (12 𝑐 − 192) 𝑥𝑡, . . .. The series solution is 4

1

1

1

4

1

1

1

1

1

1

𝑢(𝑥, 𝑡) = 3 𝑐 − 12 + (12 𝑐 − 192) 𝑥 2 + (576 𝑐 − 9216) 𝑥 4 + (69120 𝑐 − 1

1

1

1

1

2

) 𝑥 6 − 4 (3 𝑐 − 12 + (12 𝑐 − 192) 𝑥 2 ) (12 𝑐 − 192) 𝑥𝑡 + 24 (12 𝑐 − 192) 𝑥𝑡 + ⋯, 1105920 The closed form solution is 𝑢(𝑥, 𝑡) =

16𝑐−1 12

𝑐𝑡−𝑥

cosh2 (

4

).

(a)

(b)


2.8.3. Nonlinear Dispersive K(3,3,1) Equation Consider the following nonlinear dispersive 𝐾(3,3,1) 𝑢𝑡 + (𝑢3 )𝑥 − (𝑢3 )𝑥𝑥𝑥 + 𝑢5𝑥 = 0,

(2.111)

with initial condition 18𝑐−1

𝑢(𝑥, 0) = √

54

1

cosh (3 𝑥). 113

To apply HPM Taylor’s on given equation, firstly we have 18𝑐−1

𝑢(𝑥, 𝑡) = √

54

𝑡

1

𝜕3

𝜕

𝜕5

cosh (3 𝑥) − ∫0 (𝜕𝑥 (𝑢3 ) − 𝜕𝑥 3 (𝑢3 ) + 𝜕𝑥 5 (𝑢)) 𝑑𝑡,


(2.112)

Use Eq. (2.112) in Eq. (2.111), we get 𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ = √ 𝜕3

⋯ )3 −

𝜕𝑥 3

18𝑐−1 54

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )3 +

𝑡

1

𝜕

cosh (3 𝑥) − 𝑝 ∫0 [𝜕𝑥 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + 𝜕5

𝜕𝑥 5

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )],

Compare the like power of 𝑝, we get 18𝑐−1

𝑝𝑜 : 𝑢0 (𝑥, 𝑡) = √

54

1

cosh (3 𝑥),

Apply Taylor series, we get 1

1

1

𝑢0 (𝑥, 𝑡) = 18 √486𝑐 − 6 + 324 √486𝑐 − 6 𝑥 2 + 34992 √486𝑐 − 6 𝑥 4 + 1 9447840

1

1

√486𝑐 − 6 𝑥 6 + 4761711360 √486𝑐 − 6 𝑥 8 + 3856986201600 √486𝑐 − 6 𝑥10 + ⋯, = 𝑓0 + 𝑓1 + ⋯, 1

1

Take 𝑢0 (𝑥, 𝑡) = 𝑓0 = 18 √486𝑐 − 6 + 324 √486𝑐 − 6 𝑥 2 , we have following recurrence relation 𝑡

𝜕3

𝜕

𝜕5

𝑝𝑘+1 : 𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 − ∫0 (𝜕𝑥 (𝐴𝑘 ) − 𝜕𝑥 3 (𝐴𝑘 ) + 𝜕𝑥 5 (𝑢𝑘 )) 𝑑𝑡, where 𝐴𝑘 are He’s polynomials. Consequently, we have following components of solution 𝑢1 (𝑥, 𝑡) =

1

1

1

1

√486𝑐 − 6 𝑥 4 + 9447840 √486𝑐 − 6 𝑥 6 − 54 (18 √486𝑐 − 6 + 34992 2

1

1

√486𝑐 − 6 𝑥 2 ) √486𝑐 − 6 𝑥𝑡 + 708588 (486𝑐 − 6)3⁄2 𝑥 3 𝑡 + 324 1

1

1

( √486𝑐 − 6 + 324 √486𝑐 − 6 𝑥 2 ) (486𝑐 − 6)𝑥𝑡, 1458 18 . . .. The series solution is 114

1

1

1

𝑢(𝑥, 𝑡) = 18 √486𝑐 − 6 + 324 √486𝑐 − 6 𝑥 2 + 34992 √486𝑐 − 6 𝑥 4 + 1

1

1

2

1

√486𝑐 − 6 𝑥 6 − 54 (18 √486𝑐 − 6 + 324 √486𝑐 − 6 𝑥 2 ) √486𝑐 − 6 𝑥𝑡 + 9447840 1 708588

(486𝑐 − 6)3⁄2 𝑥 3 𝑡 +

1

1

1

( √486𝑐 − 6 + 324 √486𝑐 − 6 𝑥 2 ) (486𝑐 − 6)𝑥𝑡 +

1458 18

⋯, The closed form solution is 𝑢(𝑥, 𝑡) = √

18𝑐−1 54

𝑐𝑡−𝑥

cosh (

3

).

(a)

(b)


2.8.4. Kawahara Equation Example 2.21. Consider the following Kawahara equation [230] 𝑢𝑡 + 𝑢𝑢𝑥 + 𝑢3𝑥 − 𝜂𝑢5𝑥 = 0,

(2.113)

with the initial condition 72

105

𝑢(𝑥, 0) = − 169 + 169 sec ℎ4 (𝑝𝑥). Homotopy Perturbation Method: To apply HPM on given system of equations, firstly we have 72

105

𝑡

𝜕

𝜕3 𝑢

𝜕5 𝑢

𝑢(𝑥, 𝑡) = − 169 + 169 sec ℎ4 (𝑝𝑥) − ∫0 (𝑢 𝜕𝑥 𝑢 + 𝜕𝑥 3 − 𝜂 𝜕𝑥 5 ) 𝑑𝑡, Suppose that 𝑢 = 𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯,

(2.114)

Use Eq. (2.114) in Eq. (2.113), we get 115

72

𝑡

105

𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ = − 169 + 169 sec ℎ4 (𝑝𝑥) − 𝑝 ∫0 ((𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + 𝜕3

𝜕

𝜕5

⋯ ) 𝜕𝑥 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) + 𝜕𝑥 3 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) − 𝜂 𝜕𝑥 5 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )) 𝑑𝑡, Compare the like power of 𝑝, we get 72

105

𝑝𝑜 : 𝑢0 = − 169 + 169 sec ℎ(𝑝𝑥)4 , 420

72

105

𝑝1 : 𝑢1 = 169 (− 169 + 169 sec ℎ(𝑝𝑥)4 ) sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥) 𝑝𝑡 + 6720 169

sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥)3 𝑝3 𝑡 −

𝜂 (−

107520 169

5880 169

sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥) 𝑝3 (1 − tan ℎ(𝑝𝑥)2 )𝑡 + 440160

sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥)5 𝑝5 +

tan ℎ(𝑝𝑥)2 ) −

157920 169

169

sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥)3 𝑝5 (1 −

sec ℎ(𝑝𝑥)4 tan ℎ(𝑦𝑥) 𝑝5 (1 − tan ℎ(𝑝𝑥)2 )2 ) 𝑡,

. . .. The series solutions are given by 𝑢(𝑥, 𝑡) = − 105 169

+

72 169

+

105 169

sec ℎ(𝑝𝑥)4 +

420 169

(−

72 169

+

sec ℎ(𝑝𝑥)4 ) sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥) 𝑝𝑡

6720 169

𝜂 (−

sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥)3 𝑝3 𝑡 −

107520 169

5880 169

sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥)5 𝑝5 +

tan ℎ(𝑝𝑥)2 ) −

157920 169

sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥) 𝑝3 (1 − tan ℎ(𝑝𝑥)2 )𝑡 + 440160 169

sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥)3 𝑝5 (1 −

sec ℎ(𝑝𝑥)4 tan ℎ(𝑝𝑥) 𝑝5 (1 − tan ℎ(𝑝𝑥)2 )2 ) 𝑡 + ⋯,

The close form solution is 𝑢(𝑥, 𝑡) = −

72 169

+

105 169

sec ℎ (𝑝 (𝑥 +

18 13

4

𝑡)) .

Table 2.31. Numerical solution of Eq. (2.113), where 𝑝 = 0.001, 𝜂 = 0.001, 𝑡 = 0.005. 116

𝑥

Exact Solution

Approximate Solution

Error

10

0.195140300

0.195142268

1.97x10−6

20

0.194766018

0.194769946

3.96x10−6

30

0.194143945

0.194149824

5.88x10−6

40

0.193274950

0.193282769

7.82x10−6

50

0.192160245

0.192169990

9.75x10−6

Table 2.32. Numerical solution of Eq. (2.113), where 𝑝 = 0.003, 𝜂 = 0.003, 𝑡 = 0.005. 𝑥

Exact Solution


Error

10

0.194133596

0.194151266

1.77x10−5

20

0.190780864

0.190815855

3.50x10−5

30

0.185249976

0.185301708

5.17x10−5

40

0.177609612

0.177677238

6.76x10−5

50

0.167953547

0.168035982

8.24x10−5


Exact Solution


Error

10

0.192125894

0.192174725

4.88x10−5

20

0.182899734

0.182995003

9.53x10−5

30

0.167904480

0.168041903

1.37x10−4

40

0.147644128

0.147817766

1.74x10−4

50

0.122777822

0.122980584

2.03x10−4

Homotopy Perturbation Method Coupled with Taylors Series To apply HPM Taylor’s on given system of equations, firstly we have 72

105

𝑡

𝜕

𝜕3 𝑢

𝜕5 𝑢

𝑢(𝑥, 𝑡) = − 169 + 169 sec ℎ4 (𝑝𝑥) − ∫0 (𝑢 𝜕𝑥 𝑢 + 𝜕𝑥 3 − 𝜂 𝜕𝑥 5 ) 𝑑𝑡, (2.115) Suppose that 𝑢 = 𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯,

(2.116)

Use Eq. (2.116) in Eq. (2.115), we get

117

72

𝑡

105

𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ = − 169 + 169 sec ℎ4 (𝑝𝑥) − 𝑝 ∫0 ((𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + 𝜕3

𝜕

𝜕5

⋯ ) 𝜕𝑥 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) + 𝜕𝑥 3 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) − 𝜂 𝜕𝑥 5 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )) 𝑑𝑡, Compare the like power of 𝑝,we get 72

105

𝑝𝑜 : 𝑢0 = − 169 + 169 sec ℎ(𝑝𝑥)4 , Apply Taylor’s series, we get 33

210

245

658

502

15484

𝑢0 (𝑥, 𝑡) = 169 − 169 𝑝2 𝑥 2 + 169 𝑝4 𝑥 4 − 507 𝑝6 𝑥 6 + 507 𝑝8 𝑥 8 − 22815 𝑝10 𝑥10 + 324419 752895

17698598

21746206

12888218164

𝑝12 𝑥12 − 68513445 𝑝14 𝑥14 + 146814525 𝑝16 𝑥16 − 157238356275 𝑝18 𝑥18 + ⋯, = 𝑓0 + 𝑓1 + ⋯, 33

210

245

Take 𝑢0 (𝑥, 𝑡) = 𝑓0 = 169 − 169 𝑝2 𝑥 2 + 169 𝑝4 𝑥 4 , we have the following recurrence relation 𝜕3 𝑢

𝑡

𝜕5 𝑢

𝑝𝑘+1 : 𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 − ∫0 (∑∞ 𝑘=0 𝐴𝑘 + 𝜕𝑥 3 − 𝜂 𝜕𝑥 5 ) 𝑑𝑡 , where 𝐴𝑘 He’s polynomial. Consequently, we have following components of solution 658

502

15484

𝑢1 = − 507 𝑝6 𝑥 6 + 507 𝑝8 𝑥 8 − 22815 𝑝10 𝑥10 − 33

210

245

420

980

(169 − 169 𝑝2 𝑥 2 + 169 𝑝4 𝑥 4 ) (− 169 𝑝2 𝑥 169 𝑝4 𝑥 3 ) 𝑡 −

5880 169

𝑝4 𝑥𝑡,

. . .. The series solutions are given by 33

210

245

658

502

15484

𝑢(𝑥, 𝑡) = 169 − 169 𝑝2 𝑥 2 + 169 𝑝4 𝑥 4 − 507 𝑝6 𝑥 6 + 507 𝑝8 𝑥 8 − 22815 𝑝10 𝑥10 − 33

210

245

420

980

(169 − 169 𝑝2 𝑥 2 + 169 𝑝4 𝑥 4 ) (− 169 𝑝2 𝑥 + 169 𝑝4 𝑥 3 ) 𝑡 − The close form solution is 72

105

18

4

𝑢(𝑥, 𝑡) = − 169 + 169 sec ℎ (𝑝 (𝑥 + 13 𝑡)) .

118

5880 169

𝑝4 𝑥𝑡 + ⋯,


Exact Solution


Error

10

0.195140300

0.195142268

1.97x10−6

20

0.194766018

0.194769946

3.93x10−6

30

0.194143945

0.194149824

5.88x10−6

40

0.193274950

0.193282769

7.82x10−6

50

0.192160245

0.192169990

9.75x10−6


Exact Solution


Error

10

0.194133596

0.194151266

1.77x10−5

20

0.190780864

0.190815855

3.50x10−5

30

0.185249976

0.185301708

5.17x10−5

40

0.177609612

0.177677238

6.76x10−5

50

0.167953547

0.168035982

8.24x10−5


Exact Solution


Error

10

0.192125894

0.192174725

4.88x10−5

20

0.182899734

0.182995003

9.53x10−5

30

0.167904480

0.168041903

1.37x10−4

40

0.147644128

0.147817767

1.74x10−4

50

0.122777822

0.122980584

2.03x10−4

Example 2.22. Consider the following Kawahara equation [230] as 𝑢𝑡 + 𝑢𝑥 + 𝑢𝑢𝑥 + 𝑢3𝑥 − 𝜂𝑢5𝑥 = 0,

(2.117)

with the initial condition 105

𝑢(𝑥, 0) = 169 sec ℎ4 (𝑝𝑥 − 𝑥0 ). Homotopy Perturbation Method: To apply HPM on given system of equations, firstly we have

119

𝑡 𝜕𝑢

105

𝜕𝑢

𝜕3 𝑢

𝜕5 𝑢

𝑢(𝑥, 𝑡) = 169 sec ℎ4 (𝑝𝑥 − 𝑥0 ) − ∫0 (𝜕𝑥 + 𝑢 𝜕𝑥 + 𝜕𝑥 3 − 𝜂 𝜕𝑥 5 ) 𝑑𝑡, (2.118) Suppose that 𝑢 = 𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯,

(2.119)

Use Eq. (2.119) in Eq. (2.118), we get 𝑡

105

𝜕

𝑢(𝑥, 𝑡) = 169 sec ℎ4 (𝑝𝑥 − 𝑥0 ) − 𝑝 ∫0 (𝜕𝑥 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) + (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + 𝑥0 ⋯ )

𝜕3

𝜕

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) + 𝜕𝑥 3 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + 𝜕𝑥

𝜕5

⋯ ) − 𝜂 𝜕𝑥 5 (𝑢0 + 𝑥0 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )) 𝑑𝑡, Compare the like power of 𝑝, we get 105

𝑝𝑜 : 𝑢0 = 169 sec ℎ(𝑝𝑥 − 𝑥0 )4 , 44100

420

𝑝1 : 𝑢1 = 28561 sec ℎ (𝑝𝑥 − 𝑥0 )8 tanh(𝑝𝑥 − 𝑥0 ) 𝑝𝑡 + 169 sech(𝑝𝑥 − 𝑥0 )4 tanh(𝑝𝑥 − 𝑥0 ) 𝑝𝑡 +

6720 169

sech(𝑝𝑥 − 𝑥0 )4 tanh(𝑝𝑥 − 𝑥0 )3 𝑝3 𝑡 −

𝑥0 )4 tanh(𝑝𝑥 − 𝑥0 ) 𝑝3 (1 − tanh(𝑝𝑥 − 𝑥0 )2 )𝑡 + 𝜂 (− 𝑥0 )5 𝑝5 + 157920 169

440160 169

107520 169

5880 169

sech(𝑝𝑥 −

sech(𝑝𝑥 − 𝑥0 )4 tanh(𝑝𝑥 −

sech(𝑝𝑥 − 𝑥0 )4 tanh(𝑝𝑥 − 𝑥0 )3 𝑝5 (1 − tanh(𝑝𝑥 − 𝑥0 )2 ) −

sech(𝑝𝑥 − 𝑥0 )4 tanh(𝑝𝑥 − 𝑥0 ) 𝑝5 (1 − tanh(𝑝𝑥 − 𝑥0 )2 )2 ) 𝑡, . . ..

The series solutions are given by 105

44100

𝑢(𝑥, 𝑡) = 169 sec ℎ(𝑝𝑥 − 𝑥0 )4 + 28561 sec ℎ (𝑝𝑥 − 𝑥0 )8 tanh(𝑝𝑥 − 𝑥0 ) 𝑝𝑡 + 420 169

sech(𝑝𝑥 − 𝑥0 )4 tanh(𝑝𝑥 − 𝑥0 ) 𝑝𝑡 + ⋯,

The close form solution is 105

205

4

𝑢(𝑥, 𝑡) = 169 sec ℎ (𝑦𝑥 − 169 𝑥0 ) .

120

Table 2.37. Numerical solution of Eq. (2.117), where where 𝑝 = 0.001, 𝜂 = 0.001, 𝑥0 = 0.001, 𝑡 = 0.005. 𝑥

Exact Solution


Error

10

0.621282535

0.621201315

8.12x10−5

20

0.621060538

0.620853766

2.07x10−4

30

0.620590388

0.620258352

3.32x10−4

40

0.619872740

0.619415906

4.57x10−4

50

0.618908599

0.618327603

5.81x10−4

Table 2.38. Numerical solution of Eq. (2.117), where 𝑝 = 0.003, 𝜂 = 0.003, 𝑥0 = 0.003, 𝑡 = 0.005. 𝑥

Exact Solution


Error

10

0.620590388

0.620398315

1.92x10−4

20

0.617699308

0.617283225

4.16x10−4

30

0.612619048

0.611984138

6.35x10−4

40

0.605412814

0.604566917

8.46x10−4

50

0.596169295

0.595122722

1.05x10−3


Exact Solution


Error

10

0.618908599

0.618795940

1.13x10−4

20

0.610449316

0.610213714

2.36x10−4

30

0.596169295

0.595818694

3.51x10−4

40

0.576550112

0.576096045

4.54x10−4

50

0.552233126

0.551690002

5.43x10−4

Homotopy Perturbation Method Coupled with Taylors Series To apply HPM Taylor’s on given system of equations, firstly we have 105

𝑡 𝜕𝑢

𝜕𝑢

𝜕3 𝑢

𝜕5 𝑢

𝑢(𝑥, 𝑡) = 169 sec ℎ4 (𝑝𝑥 − 𝑥0 ) − ∫0 (𝜕𝑥 + 𝑢 𝜕𝑥 + 𝜕𝑥 3 − 𝜂 𝜕𝑥 5 ) 𝑑𝑡, (2.120) 121


(2.121)


105

𝜕

𝑢(𝑥, 𝑡) = 169 sec ℎ4 (𝑝𝑥 − 𝑥0 ) − 𝑝 ∫0 (𝜕𝑥 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) + 𝜕3

𝜕

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + 𝑥0 ⋯ )

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) + 𝜕𝑥 3 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + 𝜕𝑥

𝜕5

⋯ ) − 𝜂 𝜕𝑥 5 (𝑢0 + 𝑥0 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )) 𝑑𝑡, Compare the like power of 𝑝, we get 105

𝑝𝑜 : 𝑢0 = 169 sec ℎ(𝑝𝑥 − 𝑥0 )4 , Apply Taylors series, we get 𝑢0 =

1680 1

4 −

169(𝑒 −𝑥0 + −𝑥0 ) 𝑒

𝑝

𝑒 −𝑥0 (𝑒 −𝑥0 𝑝− −𝑥0 ) 𝑒

6720

169 (𝑒 −𝑥0 + 1 )4 ((𝑒 −𝑥0 )2 +1) −𝑥0

𝑥+⋯

𝑒

= 𝑓0 + 𝑓1 + ⋯, According to the purposed technique, take 𝑢0 = 𝑓0 =

1680 1

169(𝑒 −𝑥0 + −𝑥0 ) 𝑒

𝑝

𝑒 −𝑥0 (𝑒 −𝑥0 𝑝− −𝑥0 ) 𝑒

6720

4 −

169 (𝑒 −𝑥0 + 1 )4 ((𝑒 −𝑥0 )2 +1) −𝑥0

𝑥,

𝑒

we get the following recurrence relation 𝑡 𝜕𝑢

𝑝𝑘+1 : 𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 − ∫0 ( 𝜕𝑥𝑘 + 𝑢𝑘

𝜕𝑢𝑘 𝜕𝑥

+

𝜕3 𝑢𝑘 𝜕𝑥 3

−𝜂

𝜕5 𝑢𝑘 𝜕𝑥 5

) 𝑑𝑡,

Consequently, we get the following components

6720

𝑢1 =

𝑝

(𝑒 −𝑥0 )2 (𝑒 −𝑥0 𝑝− −𝑥0 ) 𝑒

2

169 (𝑒 −𝑥0 𝑝+ 𝑝 )4 ((𝑒 −𝑥0 )2 +1)2 −𝑥0

840

+ 169

𝑒

2

𝑝

1 1

(𝑒 −𝑥0 + −𝑥0 ) 𝑒

2

(

4(𝑒 −𝑥0 )2 (𝑒 −𝑥0 𝑝− −𝑥0 ) 𝑒 𝑝

2

((𝑒 −𝑥0 )2 +1)2 (𝑒 −𝑥0 𝑝+ −𝑥0 ) 𝑒

( 1

1 1

(𝑒 −𝑥0 + −𝑥0 ) 𝑒

2

(8 (−

1 𝑝2

𝑒 −𝑥0 ( 𝑒 −𝑥0 𝑝2 + −𝑥0 ) 2 2𝑒 (𝑒 −𝑥0 )2 +1

𝑝

+

𝑒 −𝑥0 𝑝((𝑒 −𝑥0 )2 +1)(𝑒 −𝑥0 𝑝− −𝑥0 ) 𝑒 ))) ((𝑒 −𝑥0 )2 +1)2

)

122

𝑥2 +

+

6720

1 1680 (( 4 1 169 (𝑒 −𝑥0 + 1 )4 ((𝑒 −𝑥0 )2 +1) −𝑥 169(𝑒 0 + −𝑥0 ) −𝑥0 𝑒

6720

−

𝑒

𝑝

𝑒 −𝑥0 (𝑒 −𝑥0 𝑝− −𝑥0 )𝑥 𝑒

)𝑒

169 (𝑒 −𝑥0 + 1 )4 ((𝑒 −𝑥0 )2 +1) −𝑥0

−𝑥0

(𝑒

−𝑥0

𝑝

𝑝 − 𝑒 −𝑥0 ) 𝑡) +

𝑒

6720

𝑝

𝑒 −𝑥0 (𝑒 −𝑥0 𝑝− −𝑥0 )𝑡 𝑒

,

169 (𝑒 −𝑥0 + 1 )4 ((𝑒 −𝑥0 )2 +1) −𝑥0 𝑒

. . .. The series solutions are given by 𝑢(𝑥, 𝑡) =

1680 1

169(𝑒 −𝑥0 + −𝑥0 ) 𝑒

4

−

6720

𝑝

𝑒 −𝑥0 (𝑒 −𝑥0 𝑝− −𝑥0 ) 𝑒

169 (𝑒 −𝑥0 + 1 )4 ((𝑒 −𝑥0 )2 +1) −𝑥0

𝑥 + ⋯,

𝑒


205

4

𝑢(𝑥, 𝑡) = 169 sech (𝑝𝑥 − 169 𝑥0 ) . Table 2.40. Numerical solution of Eq. (2.117), where 𝑝 = 0.001, 𝜂 = 0.001, 𝑥0 = 0.001, 𝑡 = 0.005. 𝑥

Exact Solution


Error

10

0.621282535

0.621201117

8.14x10−5

20

0.621060538

0.620853190

2.07x10−4

30

0.620590388

0.620256745

3.34x10−4

40

0.619872740

0.619411781

4.61x10−4

50

0.618908599

0.618318297

5.90x10−4

123


Exact Solution


Error

10

0.620590388

0.620395913

1.94x10−4

20

0.617699308

0.617264759

4.35x10−4

30

0.612619048

0.611897059

7.22x10−4

40

0.605412814

0.604292814

1.12x10−3

50

0.596169295

0.594452023

1.72x10−3


Exact Solution


Error

10

0.618908599

0.618785818

1.23x10−4

20

0.610449316

0.610089187

3.60x10−4

30

0.596169295

0.595180626

9.89x10−4

40

0.576550112

0.574060135

2.49x10−3

50

0.552233126

0.546727713

5.51x10−3

2.8.5. Nonlinear Dispersive K(2,2) Equation Consider the following initial value problem 𝐾(2, 2) [230] as 𝑢𝑡 + (𝑢2 )𝑥 + (𝑢2 )𝑥𝑥𝑥 = 0,

(2.122)


1

𝑢(𝑥, 0) = 3 𝑐 cos2 (4 𝑥). To apply HPM Taylor’s on given equation, firstly we have 4

1

𝑡

𝜕3

𝜕

𝑢(𝑥, 𝑡) = 3 𝑐 cos2 (4 𝑥) − ∫0 (𝜕𝑥 (𝑢2 ) + 𝜕𝑥 3 (𝑢2 )) 𝑑𝑡,

(2.123)


(2.124)

Use Eq. (2.124) in Eq. (2.123), we get

124

4

𝑡

1

𝜕

𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ = 3 𝑐 cos2 (4 𝑥) − 𝑝 ∫0 [𝜕𝑥 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )2 + 𝜕3 𝜕𝑥 3

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )2 ] 𝑑𝑡,

Compare the like power of 𝑝, we get 4

1

𝑝𝑜 : 𝑢0 (𝑥, 𝑡) = 3 𝑐 cos2 (4 𝑥), Apply Taylor series, we get 4

1

1

1

1

1

𝑢0 (𝑥, 𝑡) = 3 𝑐 − 12 𝑐𝑥 2 + 576 𝑐𝑥 4 − 69120 𝑐𝑥 6 + 15482880 𝑐𝑥 8 − 5573836800 𝑐𝑥10 + 1 2942985830400

𝑐𝑥12 −

1

1

2142493684531200

1

𝑐𝑥14 + 2056793937149952000 𝑐𝑥16 − 1

2517515779071541248000

𝑐𝑥18 + 3826623984188742696960000 𝑐𝑥 20 −

1 7071601122780796503982080000

𝑐𝑥 22 +

1 15614095279099998680792432640000

𝑐𝑥 24 + ⋯,

= 𝑓0 + 𝑓1 + ⋯, Take 4

1

1

1

1

𝑢0 (𝑥, 𝑡) = 𝑓0 = 3 𝑐 − 12 𝑐𝑥 2 + 576 𝑐𝑥 4 − 69120 𝑐𝑥 6 + 15482880 𝑐𝑥 8 − 1 5573836800

𝑐𝑥10 +

1

1

2942985830400

1 2056793937149952000

𝑐𝑥16 −

𝑐𝑥12 − 2142493684531200 𝑐𝑥14 + 1

1

2517515779071541248000

𝑐𝑥18 + 3826623984188742696960000 𝑐𝑥 20 ,

we have following recurrence relation 𝑡

𝜕3

𝜕

𝑝𝑘+1 : 𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 − ∫0 (𝜕𝑥 (𝑢𝑘2 ) + 𝜕𝑥 3 (𝑢𝑘2 )) 𝑑𝑡, Consequently, we have following components of solution 4

1

1

1

1

1

𝑢0 (𝑥, 𝑡) = 3 𝑐 − 12 𝑐𝑥 2 + 576 𝑐𝑥 4 − 69120 𝑐𝑥 6 + 15482880 𝑐𝑥 8 − 5573836800 𝑐𝑥10 + 1 2942985830400

𝑐𝑥12 −

1

1

2142493684531200

1 2517515779071541248000

𝑐𝑥18 +

𝑐𝑥14 + 2056793937149952000 𝑐𝑥16 − 1

3826623984188742696960000

1

𝑐𝑥 20 , 1

𝑢1 (𝑥, 𝑡) = − 7071601122780796503982080000 𝑦𝑥 22 + 15614095279099998680792432640 𝑐𝑥 24 + ⋯, . . ..

125

The series solution is 4

1

1

1

1

1

𝑢(𝑥, 𝑡) = 3 𝑐 − 12 𝑐𝑥 2 + 576 𝑐𝑥 4 − 69120 𝑐𝑥 6 + 15482880 𝑐𝑥 8 − 5573836800 𝑐𝑥10 + 1

1

2942985830400

1

𝑐𝑥12 − 2142493684531200 𝑐𝑥14 + 2056793937149952000 𝑐𝑥16 −

1 2517515779071541248000

𝑐𝑥18 +

1

1 3826623984188742696960000

𝑐𝑥 20 −

1

7071601122780796503982080000

𝑐𝑥 22 + 15614095279099998680792432640000 𝑐𝑥 24 + ⋯,


𝑥−𝑐𝑡

𝑢(𝑥, 𝑡) = 3 𝑐 cos2 (

4

).

(a)

(b)

(c)

(d)

126

(e)

(f)

Fig. 2.53.(a)-(f): Comparison of Exact and Approximate solutions of Eq. (2.122) for different values of 𝑐 and 𝑡

2.8.6. Nonlinear Dispersive K(3,3) Equation Consider the following initial value problem 𝐾(3, 3) [230] as 𝑢𝑡 + (𝑢3 )𝑥 + (𝑢3 )𝑥𝑥𝑥 = 0,

(2.125)

with initial condition 3𝑐

1

2

3

𝑢(𝑥, 0) = √ cos ( 𝑥). To apply HPM Taylor’s on given equation, firstly we have 3𝑐

𝑡

1

𝜕3

𝜕

𝑢(𝑥, 𝑡) = √ 2 cos (3 𝑥) − ∫0 (𝜕𝑥 (𝑢3 ) + 𝜕𝑥 3 (𝑢3 )) 𝑑𝑡,

(2.126)


(2.127)

Use Eq. (2.127) in Eq. (2.126), we get 3𝑐

𝑡

1

𝜕

𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ = √ 2 cos (3 𝑥) − 𝑝 ∫0 [𝜕𝑥 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )3 + 𝜕3 𝜕𝑥 3

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )3 ] 𝑑𝑡,

Compare the like power of 𝑝, we get 3𝑐

1

𝑝𝑜 : 𝑢0 (𝑥, 𝑡) = √ 2 cos (3 𝑥), Apply Taylor series, we get

127

1

1

1

𝑢0 (𝑥, 𝑡) = − 36 √6√𝑐 𝑥 2 + 3888 √6√𝑐 𝑥 4 − 1049760 √6√𝑐 𝑥 6 + 1 529079040

√6√𝑐 𝑥 8

1

1

1

− 428554022400 √6√𝑐 𝑥10 + 509122178611200 √6√𝑐 𝑥12 − 833942128565145600 √6√𝑐 𝑥14 + 1

1

1801314997700714496000

√6√𝑐 𝑥16 − 4960821503667767721984000 √6√𝑐 𝑥18 +

1

1

16966009542543765609185280000

√6√𝑐 𝑥 20 − 70544667677896977402992394240000 √6√𝑐 𝑥 24 +

1 350465909023792183738066214584320000

√6√𝑐 𝑥 24 + ⋯,

= 𝑓0 + 𝑓1 + ⋯, Take 1

1

1

1

𝑢0 (𝑥, 𝑡) = 𝑓0 = 2 √6√𝑐 − 36 √6√𝑐 𝑥 2 + 3888 √6√𝑐 𝑥 4 − 1049760 √6√𝑐 𝑥 6 + 1

1

1

√6√𝑐 𝑥 8 − 428554022400 √6√𝑐 𝑥10 + 509122178611200 √6√𝑐 𝑥 12 − 529079040 1 833942128565145600

√6√𝑐 𝑥14 + ⋯,

we have following recurrence relation 𝑡

𝜕3

𝜕

𝑝𝑘+1 : 𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 − ∫0 (𝜕𝑥 (𝑢𝑘3 ) + 𝜕𝑥 3 (𝑢𝑘3 )) 𝑑𝑡, Consequently, we have following components of solution 1

1

1

1

𝑢0 (𝑥, 𝑡) = 𝑓0 = 2 √6√𝑐 − 36 √6√𝑐 𝑥 2 + 3888 √6√𝑐 𝑥 4 − 1049760 √6√𝑐 𝑥 6 + 1

1

529079040

1

√6√𝑐 𝑥 8 − 428554022400 √6√𝑐 𝑥10 + 509122178611200 √6√𝑐 𝑥 12 −

1 833942128565145600

√6√𝑐 𝑥14 + ⋯, 1

𝑢1 (𝑥, 𝑡) = − 70544667677896977402992394240000 √6√𝑐 𝑥 24 + 1 350465909023792183738066214584320000

√6√𝑐 𝑥 24 + ⋯,

. . .. The series solution is 1

1

1

1

𝑢(𝑥, 𝑡) = 2 √6√𝑐 − 36 √6√𝑐 𝑥 2 + 3888 √6√𝑐 𝑥 4 − 1049760 √6√𝑐 𝑥 6 + 1 529079040

√6√𝑐 𝑥 8 −

1 428554022400

1

√6√𝑐 𝑥10 + 509122178611200 √6√𝑐 𝑥 12 − 128

1 833942128565145600

√6√𝑐 𝑥14 +

1 4960821503667767721984000

1 1801314997700714496000

√6√𝑐 𝑥16 −

√6√𝑐 𝑥18 + ⋯,

The close form solution is 𝑢(𝑥, 𝑡) =

𝑥−𝑐𝑡 √6𝑐 cos ( 3 ). 2

(a)

(b)

(c)

(d)

129

(e)

(f)

Fig. 2.54.(a)-(f): Comparison of Exact and Approximate solutions of Eq. (2.125) for different values of 𝑐 and 𝑡

2.8.7. (2+1)-Dimensional Nonlinear Dispersive K(2,2) Equation Consider the following initial value problem 𝐾(3, 3) [230] as 𝑢𝑡 + (𝑢3 )𝑥 + (𝑢3 )𝑥𝑥𝑥 + (𝑢3 )𝑦𝑦𝑦 = 0,

(2.128)

with initial condition 𝑢(𝑥, 𝑦, 0) = 3 cos (

1

3√2

(𝑥 + 𝑦)).

To apply HPM Taylor’s on given equation, firstly we have 𝑡

1

𝜕3

𝜕

𝜕3

𝑢(𝑥, 𝑦, 𝑡) = 3 cos (3√2 (𝑥 + 𝑦)) − ∫0 (𝜕𝑥 (𝑢3 ) + 𝜕𝑥 3 (𝑢3 ) + 𝜕𝑦 3 (𝑢3 )) 𝑑𝑡, (2.129) Suppose that 𝑢 = 𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯,

(2.130)


1

𝜕

𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ = 3 cos (3√2 (𝑥 + 𝑦)) − 𝑝 ∫0 [𝜕𝑥 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + 𝜕3

𝜕3

⋯ )3 + 𝜕𝑥 3 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )3 + 𝜕𝑥 3 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )3 ] 𝑑𝑡, Compare the like power of 𝑝, we get 1

𝑝𝑜 : 𝑢0 (𝑥, 𝑦, 𝑡) = 3 cos (3√2 (𝑥 + 𝑦)), Apply Taylor series, we get

130

1

1

1

1

1

𝑢0 (𝑥, 𝑦, 𝑡) = 3 cos (6 𝑦√2) − 2 sin (6 𝑦√2) √2 𝑥 − 12 cos (6 𝑦√2) 𝑥 2 + 1

1

216

1

1

1

1

sin (6 𝑦√2) √2 𝑥 3 + 2592 cos (6 𝑦√2) 𝑥 4 − 77760 sin (6 𝑦√2) √2 𝑥 5 −

1

1

1399680

1

1

1

1

cos (6 𝑦√2) 𝑥 6 + 58786560 sin (6 𝑦√2) √2 𝑥 7 + 1410877440 cos (6 𝑦√2) 𝑥 8 −

1

1

76187381760

1

1

sin (6 𝑦√2) √2 𝑥 9 − 2285621452800 cos (6 𝑦√2) 𝑥10 +

1

1

150851015884800

1

1

sin (6 𝑦√2) √2 𝑥11 + 5430636571852800 cos (6 𝑦√2) 𝑥12 + ⋯,

= 𝑓0 + 𝑓1 + ⋯, Take 1

1

1

1

1


1

216

1

1

1

1


1

1

1399680

1

1

1

1


1

1

76187381760

1

1

sin (6 𝑦√2) √2 𝑥 9 − 2285621452800 cos (6 𝑦√2) 𝑥10 +

1

1

1

1

sin (6 𝑦√2) √2 𝑥11 + 5430636571852800 cos (6 𝑦√2) 𝑥12 + ⋯, 150851015884800 we have following recurrence relation 𝑡

𝜕3

𝜕

𝜕3

𝑝𝑘+1 : 𝑢𝑘+1 (𝑥, 𝑦, 𝑡) = 𝑓𝑘+1 − ∫0 (𝜕𝑥 (𝑢𝑘3 ) + 𝜕𝑥 3 (𝑢𝑘3 ) + 𝜕𝑦 3 (𝑢𝑘3 )) 𝑑𝑡, Consequently, we have following components of solution 1

1

1

1

1


1

216

1

1

1

1


1

1

1399680

1

1

1

1

cos (6 𝑦√2) 𝑥 6 + 58786560 sin (6 𝑦√2) √2 𝑥 7 + 1410877440 cos (6 𝑦√2) 𝑦8 −

1

1

76187381760

1

1

sin (6 𝑦√2) √2 𝑥 9 − 2285621452800 cos (6 𝑦√2) 𝑥10 +

1

1

150851015884800

1

1

1

𝑢1 (𝑥, 𝑦, 𝑡) = 150851015884800 sin (6 𝑦√2) √2 𝑥11 + 1

1

sin (6 𝑦√2) √2 𝑥11 + 5430636571852800 cos (6 𝑦√2) 𝑥12 + ⋯,

1

cos (6 𝑦√2) 𝑥12 + ⋯, 5430636571852800 . . 131

.. The series solution is 1

1

1

1

1

𝑢(𝑥, 𝑦, 𝑡) = 3 cos (6 𝑦√2) − 2 sin (6 𝑦√2) √2 𝑥 − 12 cos (6 𝑦√2) 𝑥 2 + 1

1

216

1

1

1

1


1

1

1399680

1

1

1

1


1

1

1

1

sin (6 𝑦√2) √2 𝑥 9 − 2285621452800 cos (6 𝑦√2) 𝑥10 + 76187381760 1

1

1

1

sin (6 𝑦√2) √2 𝑥11 + 5430636571852800 cos (6 𝑦√2) 𝑥12 + ⋯, 150851015884800 The close form solution is 1

𝑢(𝑥, 𝑦, 𝑡) = 3 cos (3√2 (𝑥 + 𝑦 − 6𝑡)).

(a)

(b)

(c)

(d)

132

(e)

(f)

(g)

(h)

(i)

133

Fig. 2.55.(a)-(i): Comparison of Exact and Approximate solutions of Eq. (2.128) for different values of 𝑦 and 𝑡

2.8.8. Convection-Diffusion Problem Example 2.23. Consider the following homogeneous Convection-diffusion problem [130] as 𝜕𝑢 𝜕𝑡

𝜕𝑢

𝜕𝑢

𝜕2 𝑢

𝜕2 𝑢

+ 𝜕𝑥 + 𝜕𝑦 − 𝜕𝑥 2 − 𝜕𝑦 2 = 0,

(2.131)

with the initial conditions 2 −(𝑦−0.5)2

𝑢(𝑥, 𝑦, 0) = 𝑒 −(𝑥−0.5)

.

Homotopy Perturbation Method To apply HPM on given system of equations, firstly we have 𝜕 𝜕𝑡

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) = 𝑝 [−

𝜕2

𝜕2

𝜕𝑥

𝜕𝑦 2

(𝑦0 + 𝑝𝑢1 + ⋯ ) + 2

𝜕 𝜕𝑥

(𝑢0 + 𝑝𝑢1 + ⋯ ) −

𝜕 𝜕𝑦

(𝑢0 + 𝑝𝑢1 + ⋯ ) +

(𝑢0 + 𝑝𝑢1 + ⋯ )],

Equating the terms with identical powers of ‘𝑝’ leads to 𝑝0 ∶ 𝑝1 ∶ 𝑝2 ∶ 𝑝3 ∶

𝜕 𝜕𝑡 𝜕 𝜕𝑡 𝜕 𝜕𝑡 𝜕 𝜕𝑡

2 −(𝑦−0.5)2

𝑢0 (𝑥, 𝑦, 0) = 𝑒 −(𝑥−0.5)

(𝑢0 ) = 0, (𝑢1 ) = − (𝑢2 ) = − (𝑢3 ) = −

𝜕 𝜕𝑥 𝜕 𝜕𝑥 𝜕 𝜕𝑥

(𝑢0 ) − (𝑢1 ) − (𝑢2 ) −

𝜕 𝜕𝑦 𝜕 𝜕𝑦 𝜕 𝜕𝑦

(𝑢0 ) + (𝑢1 ) + (𝑢2 ) +

𝜕2

𝜕2

𝜕𝑥

𝜕𝑦 2

(𝑢0 ) + 2

𝜕2

𝜕2

𝜕𝑥

𝜕𝑦 2

(𝑢1 ) + 2

𝜕2 𝜕𝑥 2

(𝑢2 ) +

𝜕2 𝜕𝑦 2

,

(𝑢0 ),

𝑢1 (𝑥, 𝑦, 0) = 0,

(𝑢1 ),

𝑢1 (𝑥, 𝑦, 0) = 0,

(𝑢2 ),

𝑢2 (𝑥, 𝑦, 0) = 0,

. . .. Consequently, we have the following component of solutions 2 −(𝑦−0.5)2

𝑢0 (𝑥, 𝑦, 𝑡) = 𝑒 −(𝑥−0.5)

, 2 −1.(𝑦−0.5000000000)2

𝑢1 (𝑥, 𝑦, 𝑡) = −(−2𝑥 + 1)𝑒 −1.(𝑥−0.5000000000) . . .. The series solution is

134

𝑡 + ⋯,

𝑢(𝑥, 𝑦, 𝑡) = 𝑒

−(𝑥−0.5)2 −(𝑦−0.5)2

− (−2𝑥 +

−1.(𝑥−0.5000000000)2 − ( 2 ) 1)𝑒 1.(𝑦−0.5000000000) 𝑡

+ ⋯,

The closed form solution is 1

𝑢(𝑥, 𝑦, 𝑡) = 4𝑡+1 𝑒

(−

(𝑥−𝑡−0.5)2 (𝑦−𝑡−0.5)2 ) − 4𝑡+1 4𝑡+1

.

(a)

(b)

(c)

(d)

Fig. 2.56.(a)-(d): Comparison of Exact and Approximate solution of Eq. (2.131)

Homotopy Perturbation Method Coupled with Taylor’s Series To apply HPM Taylor’s on given system of equations, firstly we have 2 −(𝑦−0.5)2

𝑢(𝑥, 𝑦, 𝑡) = 𝑒 −(𝑥−0.5)

𝑡 𝜕𝑢

𝜕𝑢

𝜕2 𝑢

𝜕2 𝑢

− ∫0 (𝜕𝑥 + 𝜕𝑦 − 𝜕𝑥 2 − 𝜕𝑦 2 ) 𝑑𝑡,

(2.132)


(2.133)

Use Eq. (2.133) into Eq. (2.132), we get 135

2 −(𝑦−0.5)2

𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ = 𝑒 −(𝑥−0.5)

𝑡

𝜕

− ∫0 (𝜕𝑥 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) +

𝜕2

𝜕

𝜕2

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) − 𝜕𝑥 2 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) − 𝜕𝑦 2 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + 𝜕𝑦 ⋯ )) 𝑑𝑡, Comparing the like power of 𝑝, we get 2 −(𝑦−0.5)2

𝑝0 ∶ 𝑢0 (𝑥, 𝑦, 𝑡) = 𝑒 −(𝑥−0.5)

,

Applying Taylor’s series 2

2

𝑢0 (𝑥, 𝑦, 𝑡) = e−0.25−1.(𝑦−0.5) + 1.0e−0.25−1.(𝑦−0.5) 𝑥 2

−0.5000000000e−0.25−1.(𝑦−0.5) 𝑥 2 + ⋯, = 𝑓0 + 𝑓1 + 𝑓2 + ⋯, 2

2

Take 𝑢0 (𝑥, 𝑦, 𝑡) = 𝑓0 = e−0.25−1.(𝑦−0.5) + 1.0e−0.25−1.(𝑦−0.5) 𝑥 2

−0.5000000000e−0.25−1.(𝑦−0.5) 𝑥 2 , we have the following recurrence relation 𝑡 𝜕𝑢

𝑝𝑘+1 ∶ 𝑢𝑘+1 (𝑥, 𝑦, 𝑡) = 𝑓𝑘+1 − ∫0 ( 𝜕𝑥𝑘 +

𝜕𝑢𝑘 𝜕𝑦

−

𝜕2 𝑢𝑘 𝜕𝑥 2

−

𝜕2 𝑢𝑘 𝜕𝑦 2

) 𝑑𝑡,

Consequently, we have the following component of solutions 2

2

𝑢0 (𝑥, 𝑦, 𝑡) = e−0.25−1.(𝑦−0.5) + 1.0e−0.25−1.(𝑦−0.5) 𝑥 2

−0.5000000000e−0.25−1.(𝑦−0.5) 𝑥 2 , 𝑢1 (𝑥, 𝑦, 𝑡) = 2

−0.8333333333𝑦−0.25−1.(𝑦−0.5) 𝑥 3 + 2

0.0416666666𝑒 −0.25−1.(𝑦−0.5) 𝑥 4 + ⋯, . . .. The series solution is 2

2

𝑢(𝑥, 𝑦, 𝑡) = e−0.25−1.(𝑦−0.5) + 1.0e−0.25−1.(𝑦−0.5) 𝑥 − 2

2

0.5000000000e−0.25−1.(𝑦−0.5) 𝑥 2 − 0.8333333333𝑒 −0.25−1.(𝑦−0.5) 𝑥 3 + ⋯, The closed form solution is 1

𝑢(𝑥, 𝑦, 𝑡) = 4𝑡+1 exp (−

(𝑥−𝑡−0.5)2 4𝑡+1

−

(𝑦−𝑡−0.5)2

136

4𝑡+1

).

(a)

(b)

(c)

(d)


Example 2.24. Consider the following homogeneous Convection-diffusion problem [130] as 𝜕𝑢 𝜕𝑡

𝜕2 𝑢

𝜕2 𝑢

− 𝜕𝑥 2 − 𝜕𝑦 2 = 0,

(2.134)

with the initial conditions 𝑢(𝑥, 𝑦, 0) = sin(𝜋𝑥)sin(𝜋𝑦). Homotopy Perturbation Method To apply HPM on given system of equations, firstly we have 𝜕 𝜕𝑡

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) = 𝑝 [

𝜕2 𝜕𝑥 2

(𝑢0 + 𝑝𝑢1 + ⋯ ) +

Equating the terms with identical powers of ‘𝑝’ leads to

137

𝜕2 𝜕𝑦 2

(𝑢0 + 𝑝𝑢1 + ⋯ )],

𝑝0 ∶ 𝑝1 ∶ 𝑝2 ∶ 𝑝3 ∶

𝜕 𝜕𝑡 𝜕 𝜕𝑡 𝜕 𝜕𝑡 𝜕 𝜕𝑡

(𝑢0 ) = 0, (𝑢1 ) = (𝑢2 ) = (𝑢3 ) =

𝑢0 (𝑥, 𝑦, 0) = sin(𝜋𝑥) sin(𝜋𝑦),

𝜕2 𝜕𝑥 2 𝜕2 𝜕𝑥 2

(𝑢0 ) + (𝑢1 ) +

𝜕2 𝜕𝑦 2 𝜕2 𝜕𝑦 2

(𝑢0 ),

𝑢1 (𝑥, 𝑦, 0) = 0,

(𝑢1 ),

𝑢1 (𝑥, 𝑦, 0) = 0,

(𝑢2 ),

𝑢2 (𝑥, 𝑦, 0) = 0,

𝜕2

𝜕2

𝜕𝑥

𝜕𝑦 2

(𝑢2 ) + 2

. . .. Consequently, we have the following component of solutions 𝑢0 (𝑥, 𝑦, 𝑡) = sin(𝜋𝑥) sin(𝜋𝑦), 𝑢1 (𝑥, 𝑦, 𝑡) = −2sin(𝜋𝑥)𝜋 2 sin(𝜋𝑦)𝑡, 𝑢2 (𝑥, 𝑦, 𝑡) = 2sin(𝜋𝑥)𝜋 4 sin(𝜋𝑦)𝑡 2 , 4

𝑢3 (𝑥, 𝑦, 𝑡) = − 3 sin(𝜋𝑥)𝜋 6 sin(𝜋𝑦)𝑡 3 , . . ..

The series solution is 𝑢(𝑥, 𝑦, 𝑡) = sin(𝜋𝑥) sin(𝜋𝑦) −2sin(𝜋𝑥)𝜋 2 sin(𝜋𝑦)𝑡 + 2sin(𝜋𝑥)𝜋 4 sin(𝜋𝑦)𝑡 2 + ⋯, The closed form solution is 2

𝑢(𝑥, 𝑦, 𝑡) = 𝑒 −2𝜋 𝑡 sin(𝜋𝑦) sin(𝜋𝑦).

138

(a)

(b)

(c)

(d)


Homotopy Perturbation Method coupled with Taylors series To apply HPM Taylor’s on given system of equations, firstly we have 𝑡 𝜕2 𝑢

𝜕2 𝑢

𝑢(𝑥, 𝑦, 𝑡) = sin(𝜋𝑥) sin(𝜋𝑦) + ∫0 (𝜕𝑥 2 + 𝜕𝑦 2 ) 𝑑𝑡,

(2.135)


(2.136)

Use Eq. (2.136) into Eq. (2.135), we get 𝑡

𝜕2

𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ = sin(𝜋𝑥) sin(𝜋𝑦) + ∫0 (𝜕𝑥 2 (𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ ) + 𝜕2 𝜕𝑦 2

(𝑢0 + 𝑝𝑢1 + 𝑝2 𝑢2 + ⋯ )) 𝑑𝑡,

139

Comparing the like power of 𝑝, we get 𝑝0 ∶ 𝑢0 (𝑥, 𝑦, 𝑡) = sin(𝜋𝑥) sin(𝜋𝑦) , Applying Taylor’s series 𝑢0 (𝑥, 𝑦, 𝑡) = 𝜋 sin(𝜋𝑦)𝑥 − 1

1 6

1

𝜋 3 sin(𝜋𝑦)𝑥 3 + 120 𝜋 5 sin(𝜋𝑦) 𝑥 5 −

1

5040

1

𝜋 7 sin(𝜋𝑦) 𝑥 7 − 5040 𝜋 7 sin(𝜋𝑦) 𝑥 7 + 362880 𝜋 9 sin(𝜋𝑦) 𝑥 9 ⋯,

= 𝑓0 + 𝑓1 + 𝑓2 + ⋯, Take 𝑢0 (𝑥, 𝑡) = 𝑓0 = 𝜋 sin(𝜋𝑦)𝑥 −

1 6

1

𝜋 3 sin(𝜋𝑦)𝑥 3 + 120 𝜋 5 sin(𝜋𝑦) 𝑥 5 ,

we have the following recurrence relation 𝑡 𝜕2 𝑢

𝑦𝑘+1 ∶ 𝑢𝑘+1 (𝑥, 𝑡) = 𝑓𝑘+1 + ∫0 ( 𝜕𝑥 2𝑘 +

𝜕2 𝑢𝑘 𝜕𝑦 2

) 𝑑𝑡,

Consequently, we have the following component of solutions 𝑢0 (𝑥, 𝑦, 𝑡) = 𝜋 sin(𝜋𝑦)𝑥 −

1 6

1

𝜋 3 sin(𝜋𝑦)𝑥 3 + 120 𝜋 5 sin(𝜋𝑦) 𝑥 5 ,

1

1

𝑢1 (𝑥, 𝑦, 𝑡) = − 5040 𝜋 7 sin(𝜋𝑦) 𝑥 7 + 362880 𝜋 9 sin(𝜋𝑦) 𝑥 9 − 1

1

39916800

𝜋11 sin(𝜋𝑦) 𝑥11 − 2𝜋 3 sin(𝜋𝑦) 𝑥𝑡 + 3 𝜋 5 sin(𝜋𝑦) 𝜋 3 𝑡

1

− 120 𝜋 7 sin(𝜋𝑦) 𝜋 5 𝑡, 1

1

𝑢2 (𝑥, 𝑦, 𝑡) = 6227020800 𝜋13 sin(𝜋𝑦) 𝑥13 − 1307674368000 𝜋15 sin(𝜋𝑦) 𝑥15 + ⋯, . . .. The series solution is 𝑢(𝑥, 𝑦, 𝑡) = 𝜋sin(𝜋𝑦)𝑥 − 1 5040

1 6

1

𝜋 3 sin(𝜋𝑦)𝑥 3 + 120 𝜋 5 sin(𝜋𝑦) 𝑥 5 −

1

1

𝜋 7 sin(𝜋𝑦) 𝑥 7 + 362880 𝜋 9 sin(𝜋𝑦) 𝑥 9 − 39916800 𝜋11 sin(𝜋𝑦) 𝑥11 − 1

1

2𝜋 3 sin(𝜋𝑦) 𝑥𝑡 + 3 𝜋 5 sin(𝜋𝑦) 𝜋 3 𝑡 − 120 𝜋 7 sin(𝜋𝑦) 𝜋 5 𝑡 + ⋯, The closed form solution is 2

𝑢(𝑥, 𝑦, 𝑡) = 𝑒 −2𝜋 𝑡 sin(𝜋𝑥) sin(𝜋𝑦).

140

(a)

(b)

(c)

(d)


2.9. Homotopy Analysis Method Liao [115-121] was developed Homotopy Analysis Method (HAM) which is being applied on a wide range of nonlinear problems of physical nature. Abbasbandy applied Homotopy Analysis Method for solving nonlinear equations arising in heat transfer [1], generalized Hirota Satsuma coupled KdV equations [2], Benjamin-Bona-Mahony equation [3] and Kawahara equation [4], Mohyud-Din and Yildirim implement Homotopy Analysis Method for three-dimensional Helmholtz [141] and fractional order Burger’s equations [247], Hayat et al. using Homotopy Analysis Method to obtained the numerical solution of generalized second-grade fluid past a porous Plate [73] and MHD flows of an Oldroyd 8– constant fluid [74] and Usman et al. used Homotopy Analysis

141

Method to obtained the numerical solutions of system of PDEs [271] and Goursat problems [205]. Moreover, this method (HAM) is more user-friendly and it overcomes the complexities of selection of initial value. Several examples are given which reveal the efficiency and reliability of the proposed algorithm.

2.9.1. Methodology We consider the following differential equation 𝑁[𝑢(𝜏)] = 0,

(2.137)

where N is a nonlinear operator, τ denotes independent variables, 𝑢(𝜏) is an unknown function, respectively. For simplicity, we ignore all boundary or initial conditions, which can be treated in the similar way. By means of generalizing the traditional Homotopy method, Liao constructs the so called zero - order deformation equation. (1 − 𝑝)𝐿[𝜑(𝜏; 𝑝) − 𝑢0 (𝜏)] = 𝑝ℎ𝐻(𝜏)𝑁[𝜑(𝜏; 𝑝)]

(2.138)

where 𝑝 ∈ [0,1] is the embedding parameter, ℎ ≠ 0 is a nonzero parameter, 𝐻(𝜏) ≠ 0, is an auxiliary function, 𝐿 is an auxiliary linear operator, 𝑢0 (𝜏) is an initial guess of 𝑢(𝜏), 𝑢(𝜏; 𝑝) is a unknown function, respectively. It is important, that one has great freedom to choose auxiliary things in HAM. Obviously, when, 𝑝 = 0 and 𝑝 = 1 it holds 𝜑(𝜏; 0) = 𝑢0 (𝜏), 𝜑(𝜏; 1) = 𝑢(𝜏). respectively. Thus, as 𝑝 increases from 0 to 1, the solution 𝜑(𝜏; 𝑝) varies from the initial guesses 𝑢0 (𝜏) to the solution 𝑢(𝜏). Expanding in Taylor series with respect to p, we have 1 𝜕𝑚 𝜑(𝜏;𝑝)

𝜑(𝜏; 𝑝) = 𝑢0 (𝜏) + ∑∞ 𝑚=1 (𝑚! 1 𝜕𝑚 𝜑(𝜏;𝑝)

where 𝑢𝑚 (𝜏) = 𝑚!

𝜕𝑝𝑚

) 𝑝𝑚 |

𝜕𝑝𝑚

, 𝑝=0

at 𝑝 = 0.

If the auxiliary linear operator, the initial guess, the auxiliary h, and the auxiliary function are so properly chosen, the above series converges at 𝑝 = 1, then we have 𝑢(𝜏) = 𝑢0 (𝜏) + ∑∞ 𝑚=1 𝑢𝑚 (𝜏), define the vector 𝑢 ⃗ = {𝑢0 (𝜏), 𝑢1 (𝜏), 𝑢2 (𝜏), … , 𝑢0 (𝜏)}. Differentiating equation (2.138) 𝑚𝑡ℎ times with respect to the embedding parameter 𝑝 and then setting 𝑝 = 0 and finally dividing them by 𝑚!, we obtain the 𝑚𝑡ℎ-order deformation equation 𝐿[𝑢𝑚 (𝜏) − 𝜒𝑚 𝑢𝑚−1 (𝜏)] = ℎ𝐻(𝜏)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ), 142

1

𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = (𝑚−1)!

𝜕𝑚−1 𝜑(𝜏;𝑝) 𝜕𝑝𝑚−1

|

,

𝑝=0

𝜒𝑚 = 0, 𝑚 ≤ 1, = 1, 𝑚 > 1. Applying 𝐿−1 both sides of (2.58), we get 𝑢𝑚 (𝜏: ℎ) = 𝜒𝑚 𝑢𝑚−1 (𝜏) + ℎ𝐿−1 [𝐻(𝜏)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )]. In this way, it is easily to obtain 𝑢𝑚 for m ≥ 1, at 𝑚𝑡ℎ- order, we have 𝑢(𝜏) = ∑𝑀 𝑚=0 𝑢𝑚 (𝜏). when 𝑀 → ∞, we get an accurate approximation of the original equation (2.137). For the convergence of the above method we refer the reader Liao’s work. If equation (2.137) admits unique solution, then this method will produce the unique solution. If equation (2.137) does not possess unique solution, the HAM will give a solution among many other (possible) solutions.

2.9.2. Goursat Problem Example 2.25. We first consider the following homogenous linear Goursat problem [230] 𝑢𝑥𝑡 = 𝑢,

(2.139)

with initial conditions 𝑢(𝑥, 0) = 𝑒 𝑥 ,

𝑢(0, 𝑡) = 𝑒 𝑡 ,

𝑢(0,0) = 1

To solve the given Equation by HAM we choose the linear operator 𝜕2

𝐿[𝜑(𝑥, 𝑡; 𝑞)] = 𝜕𝑥𝜕𝑡 (𝜑(𝑥, 𝑡; 𝑞)), with the property 𝐿[𝑐1 + 𝑥𝑐2 ] = 0, 𝑦[𝑐1 + 𝑡𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑥

𝑡

𝐿−1 = ∫0 ∫0 (. )𝑑𝑥𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑥𝑡 − 𝜑(𝑥, 𝑡; 𝑞), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡; 𝑞)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. 143

for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑢𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 0) = 0, 𝑢𝑚 (0,0) = 1. 𝜕2

where 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = [𝜕𝑥𝜕𝑡 𝑢𝑚−1 − 𝑢𝑚−1 ]. Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑢0 (𝑥, 𝑡) = 𝑒 𝑥 + 𝑒 𝑡 − 1, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢0 (𝑥, 𝑡) = 𝑒 𝑥 + 𝑒 𝑡 − 1, 1

𝑢1 (𝑥, 𝑡) = 2 𝑡(−2 − 𝑡 + 2𝑒 𝑡 ), 1

1

1

𝑢2 (𝑥, 𝑡) = − 2 𝑡 2 + 2𝑒 𝑡 𝑡 + 2 − 6 𝑡 3 − 24 𝑡 4 − 2𝑒 𝑡 , . . .. the series solution is given by 1

1

1

1

𝑢(𝑥, 𝑡) = 𝑒 𝑥 + 𝑒 𝑡 − 1 + 2 𝑡(−2 − 𝑡 + 2𝑒 𝑡 ) − 2 𝑡 2 + 2𝑒 𝑡 𝑡 + 2 − 6 𝑡 3 − 24 𝑡 4 − 2𝑒 𝑡 + ⋯. and the close form solution is 𝑢(𝑥, 𝑡) = 𝑒 𝑥+𝑡 .

144


Example 2.26. We first consider the following homogenous linear Goursat problem [230] 𝑢𝑥𝑡 = −2𝑢,

(2.140)


𝑢(0, 𝑡) = 𝑒 −2𝑡 ,

𝑢(0,0) = 1


𝐿[𝜑(𝑥, 𝑡; 𝑞)] = 𝜕𝑥𝜕𝑡 (𝜑(𝑥, 𝑡; 𝑞)), with the property 𝐿[𝑐1 + 𝑥𝑐2 ] = 0, 𝐿[𝑐1 + 𝑡𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑥

𝑡

𝐿−1 = ∫0 ∫0 (. )𝑑𝑥𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑥𝑡 + 2𝜑(𝑥, 𝑡; 𝑞), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡; 𝑞)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑢𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). 145

with initial condition 𝑢𝑚 (𝑥, 0) = 0, 𝑢𝑚 (0,0) = 1. 𝜕2

where 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = [𝜕𝑥𝜕𝑡 𝑢𝑚−1 + 2𝑢𝑚−1 ]. Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑢0 (𝑥, 𝑡) = 𝑒 𝑥 + 𝑒 −2𝑡 − 1, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢0 (𝑥, 𝑡) = 𝑒 𝑥 + 𝑒 −2𝑡 − 1, 3

1

𝑢1 (𝑥, 𝑡) = − 2 − 𝑡 − 𝑡 2 − 2𝑡𝑒 𝑡 + 2𝑒 𝑡 − 2 𝑒 −2𝑡 , 2

𝑢2 (𝑥, 𝑡) = 7 + 2𝑡 + 2𝑡 2 + 2𝑒 𝑡 𝑡 − 6𝑒 𝑡 − 𝑒 −2𝑡 − 3 𝑡 3 . . .. the series solution is given by 3

1

𝑢(𝑥, 𝑡) = 𝑒 𝑥 + 𝑒 −2𝑡 − 1 − 2 − 𝑡 − 𝑡 2 − 2𝑡𝑒 𝑡 + 2𝑒 𝑡 − 2 𝑒 −2𝑡 + 7 + 2𝑡 + 2𝑡 2 + 2

2𝑒 𝑡 𝑡 − 6𝑒 𝑡 − 𝑒 −2𝑡 − 3 𝑡 3 + ⋯. and the close form solution is 𝑢(𝑥, 𝑡) = 𝑒 𝑥−2𝑡 .


146

Example 2.27. We first consider the following inhomogeneous linear Goursat problem [230] 𝑢𝑥𝑡 = 𝑢 − 𝑡,

(2.141)


𝑢(0, 𝑡) = 𝑡 + 𝑒 𝑡 ,

𝑢(0,0) = 1.



𝑡

𝐿−1 = ∫0 ∫0 (. )𝑑𝑥𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑥𝑡 − 𝜑(𝑥, 𝑡; 𝑞) + 𝑡, using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡; 𝑞)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).


where 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = [𝜕𝑥𝜕𝑡 𝑢𝑚−1 − 𝑢𝑚−1 + 𝑡]. Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑢0 (𝑥, 𝑡) = 𝑒 𝑥 + 𝑡 + 𝑒 𝑡 − 1, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢0 (𝑥, 𝑡) = 𝑒 𝑥 + 𝑡 + 𝑒 𝑡 − 1,

147

1

𝑢1 (𝑥, 𝑡) = 2 𝑡(−2 − 𝑡 + 2𝑒 𝑡 ), 1

1

1

𝑢2 (𝑥, 𝑡) = − 2 𝑡 2 + 2𝑡𝑒 𝑡 + 2 − 6 𝑡 3 − 24 𝑡 4 − 2𝑒 𝑡 , . . .. the series solution is given by 1

1

1

𝑢(𝑥, 𝑡) = 𝑒 𝑥 + 𝑡 + 𝑒 𝑡 − 1 + 2 𝑡(−2 − 𝑡 + 2𝑒 𝑡 ) − 2 𝑡 2 + 2𝑡𝑒 𝑡 + 2 − 6 𝑡 3 − 1 24

𝑡 4 − 2𝑒 𝑡 + ⋯.

and the close form solution is 𝑢(𝑥, 𝑡) = 𝑡 + 𝑒 𝑥+𝑡 .


Example 2.28. We first consider the following inhomogeneous linear Goursat problem [230] as 𝑢𝑥𝑡 = 𝑢 + 4𝑥𝑡 − 𝑡 2 𝑥 2 ,

(2.142)


𝑢(0, 𝑡) = 𝑒 𝑡 ,

𝑢(0,0) = 1.


𝐿[𝜑(𝑥, 𝑡; 𝑞)] = 𝜕𝑥𝜕𝑡 (𝜑(𝑥, 𝑡; 𝑞)), with the property 𝐿[𝑐1 + 𝑥𝑐2 ] = 0, 148

𝐿[𝑐1 + 𝑡𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑥

𝑡

𝐿−1 = ∫0 ∫0 (. )𝑑𝑥𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑥𝑡 − 𝜑(𝑥, 𝑡; 𝑞) − 4𝑥𝑡 + 𝑥 2 𝑡 2 , using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡; 𝑞)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).


where 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = [𝜕𝑥𝜕𝑡 𝑢𝑚−1 + 𝑢𝑚−1 − 4𝑥𝑡 + 𝑥 2 𝑡 2 ]. Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑢0 (𝑥, 𝑡) = 𝑒 𝑥 + 𝑒 𝑡 − 1, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢0 (𝑥, 𝑡) = 𝑒 𝑥 + 𝑒 𝑡 − 1, 1

𝑢1 (𝑥, 𝑡) = 18 𝑡(−18 − 9𝑡 + 18𝑒 𝑡 + 9𝑡 3 − 𝑡 5 ), 1

11

7

1

𝑢2 (𝑥, 𝑡) = − 2 𝑡 2 + 2𝑡𝑒 𝑡 + 24 𝑡 4 − 180 𝑡 6 + 2 − 6 𝑡 3 − 2𝑒 𝑡 , . . .. the series solution is given by 1

1

𝑢(𝑥, 𝑡) = 𝑒 𝑥 + 𝑒 𝑡 − 1 + 18 𝑡(−18 − 9𝑡 + 18𝑒 𝑡 + 9𝑡 3 − 𝑡 5 ) − 2 𝑡 2 + 2𝑡𝑒 𝑡 + 11 4 𝑡 24

7

1

− 180 𝑡 6 + 2 − 6 𝑡 3 − 2𝑒 𝑡 + ⋯.

149

and the close form solution is 𝑢(𝑥, 𝑡) = 𝑡 2 𝑥 2 + 𝑒 𝑥+𝑡 .


Example 2.29. We first consider the following inhomogeneous linear Goursat problem [230] as 𝑢𝑥𝑡 = −𝑢3 + 𝑥 3 + 3𝑥 2 𝑡 + 3𝑥𝑡 2 + 𝑡 3 ,

(2.143)

with initial conditions 𝑢(𝑥, 0) = 𝑥,

𝑢(0, 𝑡) = 𝑡,

𝑢(0,0) = 0.



𝑡

𝐿−1 = ∫0 ∫0 (. )𝑑𝑥𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑥𝑡 + 𝜑 3 (𝑥, 𝑡; 𝑞) − 𝑥 3 − 3𝑥 2 𝑡 − 3𝑥𝑡 2 − 𝑡 3 , using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡; 𝑞)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡). 150


𝑠 3 2 2 where 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = [𝜕𝑥𝜕𝑡 𝑢𝑚−1 + ∑𝑚−1 𝑟=0 ∑𝑘=0 𝑢𝑠 𝑢𝑠−𝑘 𝑢𝑚−1−𝑟 − 𝑥 − 3𝑥 𝑡 − 3𝑥𝑡 −

𝑡 3 ]. Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑢0 (𝑥, 𝑡) = 𝑥 + 𝑡, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢0 (𝑥, 𝑡) = 𝑥 + 𝑡, 𝑢1 (𝑥, 𝑡) = 0, 𝑢2 (𝑥, 𝑡) = 0, . . .. and the close form solution is 𝑢(𝑥, 𝑡) = 𝑥 + 𝑡.


151

Example 2.30. We first consider the following inhomogeneous linear Goursat problem [230] as 𝑢𝑥𝑡 = −𝑢2 + 𝑒 2𝑥 + 𝑒 2𝑡 + 2𝑒 𝑥+𝑡 ,

(2.144)

with initial conditions 𝑢(𝑥, 0) = 1 + 𝑒 𝑥 ,

𝑢(0, 𝑡) = 1 + 𝑒 𝑡 ,

𝑢(0,0) = 2.



𝑡

𝐿−1 = ∫0 ∫0 (. )𝑑𝑥𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑥𝑡 + 𝜑 2 (𝑥, 𝑡; 𝑞) − (𝑒 2𝑥 + 𝑒 2𝑡 + 2𝑒 𝑥+𝑡 ), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡; 𝑞)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).


𝑛 2𝑥 where 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = [𝜕𝑥𝜕𝑡 𝑢𝑚−1 + ∑𝑚−1 + 𝑒 2𝑡 + 2𝑒 𝑥+𝑡 )]. 𝑟=0 ∑𝑗=0 𝑢𝑗 𝑢𝑛−𝑗 − (𝑒

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑢0 (𝑥, 𝑡) = 𝑒 𝑡 + 𝑒 𝑥 , by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢0 (𝑥, 𝑡) = 𝑒 𝑡 + 𝑒 𝑥 ,

152

𝑢1 (𝑥, 𝑡) = 0, 𝑢2 (𝑥, 𝑡) = 0, . . .. the series solution is given by 𝑢(𝑥, 𝑡) = 𝑒 𝑡 + 𝑒 𝑥 . and the close form solution is 𝑢(𝑥, 𝑡) = 𝑒 𝑡 + 𝑒 𝑥 .


2.9.3. System of Initial Value Problems Example 2.31. We first consider the following system of initial value problems [230] 𝑢𝑡 + 𝑣𝑥 − (𝑢 − 𝑣) = 0,

(2.145)

𝑣𝑡 + 𝑢𝑥 − (𝑢 + 𝑣) = 0,

(2.146)

with initial conditions 𝑢(𝑥, 0) = 𝑠𝑖𝑛ℎ𝑥, 𝑣(𝑥, 0) = 𝑐𝑜𝑠ℎ𝑥. To solve the given Equation by HAM we choose the linear operator 𝜕

𝐿[𝜑1 (𝑥, 𝑡; 𝑞1 )] = 𝜕𝑡 (𝜑1 (𝑥, 𝑡; 𝑞1 )), 153

𝜕

𝐿[𝜑2 (𝑥, 𝑡; 𝑞2 )] = 𝜕𝑡 (𝜑2 (𝑥, 𝑡; 𝑞2 )), with the property 𝐿[𝑐1 + 𝑡𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑡

𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑1 (𝑥, 𝑡; 𝑞1 )] = 𝜑1 (𝑥, 𝑡; 𝑞1 )𝑡 + 𝜑2 (𝑥, 𝑦; 𝑞1 )𝑥 − (𝜑1 (𝑥, 𝑡; 𝑞1 ) + 𝜑2 (𝑥, 𝑡; 𝑞1 )), 𝑁[𝜑2 (𝑥, 𝑡; 𝑞2 )] = 𝜑2 (𝑥, 𝑡; 𝑞2 )𝑡 + 𝜑1 (𝑥, 𝑡; 𝑞2 )𝑥 − (𝜑1 (𝑥, 𝑡; 𝑞2 ) + 𝜑2 (𝑥, 𝑡; 𝑞2 )), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞1 )𝐿[𝜑1 (𝑥, 𝑡; 𝑞1 ) − 𝑢0 (𝑥, 𝑡)] = 𝑞1 ℎ𝐻(𝑥, 𝑡; 𝑞1 )𝑁[𝜑1 (𝑥, 𝑡; 𝑞1 )]. for 𝑞1 = 0 and 𝑞1 = 1, we can write 𝜑1 (𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑1 (𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).

(1 − 𝑞2 )𝐿[𝜑2 (𝑥, 𝑡; 𝑞2 ) − 𝑢0 (𝑥, 𝑡)] = 𝑞2 ℎ𝐻(𝑥, 𝑡; 𝑞2 )𝑁[𝜑2 (𝑥, 𝑡; 𝑞2 )]. for 𝑞2 = 0 and 𝑞2 = 1, we can write 𝜑2 (𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑2 (𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equations 𝐿[𝑢𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). 𝐿[𝑣𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑣𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 0) = 0, 𝑣𝑚 (𝑥, 0) = 0. where 𝜕

𝜕

𝜕

𝜕

𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ [𝜕𝑡 𝑢𝑚−1 + 𝜕𝑥 𝑣𝑚−1 − (𝑢𝑚−1 + 𝑣𝑚−1 )]. 𝑅𝑚 (𝑣𝑚−1 ) = ℎ [𝜕𝑡 𝑣𝑚−1 + 𝜕𝑥 𝑢𝑚−1 − (𝑢𝑚−1 + 𝑣𝑚−1 )]. Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, 𝑣𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑣𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximations 𝑢0 (𝑥, 𝑡) = 𝑠𝑖𝑛ℎ𝑥, 𝑣0 (𝑥, 𝑡) = 𝑐𝑜𝑠ℎ𝑥,

154

by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢1 (𝑥, 𝑡) = 𝑡𝑐𝑜𝑠ℎ𝑥, 𝑣1 (𝑥, 𝑡) = 𝑡𝑠𝑖𝑛ℎ𝑥, 𝑢2 (𝑥, 𝑡) = 𝑣2 (𝑥, 𝑡) =

𝑡2 2! 𝑡2 2!

𝑠𝑖𝑛ℎ𝑥, 𝑐𝑜𝑠ℎ𝑥, . . ..

The series solutions are 𝑡2

𝑢(𝑥, 𝑡) = 𝑠𝑦𝑛ℎ𝑥 + 𝑡𝑐𝑜𝑠ℎ𝑥 + 2! 𝑠𝑖𝑛ℎ𝑥 + ⋯. 𝑡2

𝑣(𝑥, 𝑡) = 𝑐𝑜𝑠ℎ𝑥 + 𝑡𝑠𝑖𝑛ℎ𝑥 + 2! 𝑐𝑜𝑠ℎ𝑥 + ⋯. and the close form solution is 𝑢(𝑥, 𝑡) = sinh(𝑥 + 𝑡). 𝑣(𝑥, 𝑡) = cosh(𝑥 + 𝑡).

(a)

(b)

155

(c)

(d)

Fig. 2.66.(a)-(d): Comparison of Exact and Approximate solution of Eq. (2.145-2.146)

Example 2.32. We first consider the following system of initial value problems [230] 𝑢𝑡 + 𝑢𝑥 + 2𝑤 = 0,

(2.147)

𝑣𝑡 + 𝑣𝑥 + 2𝑢 = 0,

(2.148)

𝑤𝑡 + 𝑤𝑥 − 2𝑢 = 0,

(2.149)

with initial conditions 𝑢(𝑥, 𝑦, 0) = sin(𝑥 + 𝑦), 𝑣(𝑥, 𝑦, 0) = cos(𝑥 + 𝑦). 𝑤(𝑥, 𝑦, 0) = −cos(𝑥 + 𝑦). To solve the given Equation by HAM we choose the linear operator 𝜕

𝐿[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )] = 𝜕𝑡 (𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )), 𝜕

𝐿[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )] = 𝜕𝑡 (𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )), 𝜕

𝐿[𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )] = 𝜕𝑡 (𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )), with the property 𝐿[𝑐1 + 𝑡𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑡

𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )] = 𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )𝑡 + 𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )𝑥 + 2𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 ), 𝑁[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )] = 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑡 + 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑥 + 2𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ), 156

𝑁[𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )] = 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )𝑡 + 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )𝑥 − 2𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞1 )𝐿[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ) − 𝑢0 (𝑥, 𝑦, 𝑡)] = 𝑞1 ℎ𝐻(𝑥, 𝑦, 𝑡; 𝑞1 )𝑁[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )]. for 𝑞1 = 0 and 𝑞1 = 1, we can write 𝜑1 (𝑥, 𝑦, 𝑡; 0) = 𝑢0 (𝑥, 𝑦, 𝑡),

𝜑1 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).

(1 − 𝑞2 )𝐿[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 ) − 𝑢0 (𝑥, 𝑦, 𝑡)] = 𝑞1 ℎ𝐻(𝑥, 𝑦, 𝑡; 𝑞2 )𝑁[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )]. for 𝑞2 = 0 and 𝑞2 = 1, we can write 𝜑2 (𝑥, 𝑦, 𝑡; 0) = 𝑢0 (𝑥, 𝑦, 𝑡),

𝜑2 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).


𝜑3 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equations 𝐿[𝑢𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). 𝐿[𝑣𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑣𝑚−1 ). 𝐿[𝑤𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑤𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 𝑦, 0) = 0, 𝑣𝑚 (𝑥, 𝑦, 0) = 0, 𝑤𝑚 (𝑥, 𝑦, 0) = 0. where 𝜕

𝜕

𝜕

𝜕

𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ [𝜕𝑡 𝑢𝑚−1 + 𝜕𝑥 𝑢𝑚−1 + 2𝑤𝑚−1 ]. 𝑅𝑚 (𝑣𝑚−1 ) = ℎ [𝜕𝑡 𝑣𝑚−1 + 𝜕𝑥 𝑣𝑚−1 + 2𝑢𝑚−1 ]. 𝜕

𝜕

𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ) = ℎ [𝜕𝑡 𝑤𝑚−1 + 𝜕𝑥 𝑤𝑚−1 − 2𝑢𝑚−1 ]. Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, 𝑣𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑣𝑚−1 )], 𝑚 ≥ 1, 𝑤𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑤𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximations 𝑢0 (𝑥, 𝑦, 𝑡) = 𝑠𝑖𝑛(𝑥 + 𝑦), 𝑣0 (𝑥, 𝑦, 𝑡) = 𝑐𝑜𝑠(𝑥 + 𝑦), 𝑤0 (𝑥, 𝑦, 𝑡) = −𝑐𝑜𝑠(𝑥 + 𝑦),

157

by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components are 𝑢1 (𝑥, 𝑦, 𝑡) = 𝑡𝑐𝑜𝑠(𝑥 + 𝑦), 𝑣1 (𝑥, 𝑦, 𝑡) = −𝑡𝑠𝑖𝑛(𝑥 + 𝑦), 𝑤1 (𝑥, 𝑦, 𝑡) = 𝑠𝑖𝑛(𝑥 + 𝑦), . . .. The series solutions are 𝑢(𝑥, 𝑦, 𝑡) = 𝑠𝑖𝑛(𝑥 + 𝑦) + 𝑡𝑐𝑜𝑠(𝑥 + 𝑦) + ⋯. 𝑣(𝑥, 𝑦, 𝑡) = 𝑐𝑜𝑠(𝑥 + 𝑦) − 𝑡𝑠𝑖𝑛(𝑥 + 𝑦) + ⋯. 𝑤(𝑥, 𝑦, 𝑡) = 𝑐𝑜𝑠(𝑥 + 𝑦) + 𝑡𝑠𝑖𝑛(𝑥 + 𝑦) + ⋯. and the close form solution is 𝑢(𝑥, 𝑦, 𝑡) = 𝑠𝑖𝑛(𝑥 + 𝑦 + 𝑡). 𝑣(𝑥, 𝑦, 𝑡) = 𝑐𝑜𝑠(𝑥 + 𝑦 + 𝑡). 𝑤(𝑥, 𝑦, 𝑡) = − 𝑐𝑜𝑠(𝑥 + 𝑦 + 𝑡).

(a)

(b)

158

(c)

(d)

(e)

(f)

Fig. 2.67.(a)-(f): Comparison of Exact and Approximate solution of Eq. (2.147-2.149)

Example 2.33. We first consider the following system of initial value problems [230] 𝑢𝑡 + 𝑣𝑥 − 𝑤𝑦 = 𝑤,

(2.150)

𝑣𝑡 + 𝑤𝑥 + 𝑢𝑦 = 𝑢,

(2.151)

𝑤𝑡 + 𝑣𝑥 − 𝑣𝑦 = 𝑣,

(2.152)

with initial conditions 𝑢(𝑥, 𝑦, 0) = sin(𝑥 + 𝑦). 𝑣(𝑥, 𝑦, 0) = cos(𝑥 + 𝑦). 𝑤(𝑥, 𝑦, 0) = −sin(𝑥 + 𝑦). To solve the given Equation by HAM we choose the linear operator 159

𝜕

𝐿[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )] = 𝜕𝑡 (𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )), 𝜕

𝐿[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )] = 𝜕𝑡 (𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )), 𝜕


𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )] = 𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )𝑡 + 𝑦2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑥 − 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )𝑦 − 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 ), 𝑁[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )] = 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑡 + 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )𝑥 + 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑦 − 𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ), 𝑁[𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )] = 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )𝑡 + 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑥 − 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑦 − 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 ), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞1 )𝐿[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ) − 𝑢0 (𝑥, 𝑦, 𝑡)] = 𝑞1 ℎ𝐻(𝑥, 𝑦, 𝑡; 𝑞1 )𝑁[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )]. for 𝑞1 = 0 and 𝑞1 = 1, we can write 𝜑1 (𝑥, 𝑦, 𝑡; 0) = 𝑢0 (𝑥, 𝑦, 𝑡),

𝜑1 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).


𝜑2 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).


𝜑3 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equations 𝐿[𝑢𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). 𝐿[𝑣𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑣𝑚−1 ). 𝐿[𝑤𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑤𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ). 160

with initial condition 𝑢𝑚 (𝑥, 𝑦, 0) = 0, 𝑣𝑚 (𝑥, 𝑦, 0) = 0, 𝑤𝑚 (𝑥, 𝑦, 0) = 0. where 𝜕 𝜕 𝜕 ⃗⃗ 𝑅𝑚 (𝑦 ) = ℎ [𝜕𝑡 𝑢𝑚−1 + 𝜕𝑥 𝑣𝑚−1 − 𝜕𝑦 𝑤𝑚−1 − 𝑤𝑚−1 ]. 𝑚−1 𝜕

𝜕

𝜕

𝑅𝑚 (𝑣𝑚−1 ) = ℎ [𝜕𝑡 𝑣𝑚−1 + 𝜕𝑥 𝑤𝑚−1 + 𝜕𝑦 𝑢𝑚−1 − 𝑢𝑚−1 ]. 𝜕

𝜕

𝜕

𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ) = ℎ [𝜕𝑡 𝑤𝑚−1 + 𝜕𝑥 𝑣𝑚−1 − 𝜕𝑦 𝑣𝑚−1 − 𝑣𝑚−1 ]. Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, 𝑣𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑣𝑚−1 )], 𝑚 ≥ 1, 𝑤𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑤𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximations 𝑢0 (𝑥, 𝑦, 𝑡) = 𝑠𝑖𝑛(𝑥 + 𝑦), 𝑣0 (𝑥, 𝑦, 𝑡) = 𝑐𝑜𝑠(𝑥 + 𝑦), 𝑤0 (𝑥, 𝑦, 𝑡) = −𝑠𝑖𝑛(𝑥 + 𝑦), by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components are 𝑢1 (𝑥, 𝑦, 𝑡) = −𝑡𝑐𝑜𝑠(𝑥 + 𝑦), 𝑣1 (𝑥, 𝑦, 𝑡) = 𝑡𝑠𝑖𝑛(𝑥 + 𝑦), 𝑤1 (𝑥, 𝑦, 𝑡) = 𝑡𝑐𝑜𝑠(𝑥 + 𝑦), 𝑡2

𝑢2 (𝑥, 𝑦, 𝑡) = − 2! 𝑠𝑖𝑛(𝑥 + 𝑦), 𝑡2

𝑣2 (𝑥, 𝑦, 𝑡) = − 2! 𝑐𝑜𝑠(𝑥 + 𝑦), 𝑤2 (𝑥, 𝑦, 𝑡) =

𝑡2 2!

𝑐𝑜𝑠(𝑥 + 𝑦), . . ..

The series solutions are 𝑢(𝑥, 𝑦, 𝑡) = 𝑠𝑖𝑛(𝑥 + 𝑦) + 𝑡𝑐𝑜𝑠(𝑥 + 𝑦) −

1 2 2! 1

𝑡 𝑠𝑖𝑛(𝑥 + 𝑦) + ⋯ .

𝑣(𝑥, 𝑦, 𝑡) = 𝑠𝑖𝑛(𝑥 + 𝑦) − 𝑡𝑐𝑜𝑠(𝑥 + 𝑦) − 2! 𝑡 2 𝑠𝑖𝑛(𝑥 + 𝑦) + ⋯ .

161

1

𝑤(𝑥, 𝑦, 𝑡) = − 𝑠𝑖𝑛(𝑥 + 𝑦) + 𝑡𝑐𝑜𝑠(𝑥 + 𝑦) + 2! 𝑡 2 𝑠𝑖𝑛(𝑥 + 𝑦) + ⋯ . and the close form solution is 𝑢(𝑥, 𝑦, 𝑡) = − 𝑠𝑖𝑛(𝑥 − 𝑦 − 𝑡). 𝑣(𝑥, 𝑦, 𝑡) = − 𝑠𝑖𝑛(−𝑥 − 𝑦 + 𝑡). 𝑤(𝑥, 𝑦, 𝑡) = − 𝑠𝑖𝑛(𝑥 − 𝑦 + 𝑡).

(a)

(b)

(c)

(d)

162

(e)

(f)

Fig. 2.68.(a)-(f): Comparison of Exact and Approximate solution of Eq. (2.150-2.152)

2.9.4. K(m,p,1) Equation We first consider the following initial value problem of the 𝐾(𝑚, 𝑝, 1) equation [234] 𝑢𝑡 + (𝑢2 )𝑥 − (𝑢2 )𝑥𝑥𝑥 + (𝑢)𝑥𝑥𝑥𝑥𝑥 = 0, 𝑢(𝑥, 0) =

16𝑐−1 12

(2.153)

1

cosh2 (4 𝑥).

To solve the given Equation by HAM we choose the linear operator 𝜕

𝐿[𝜑(𝑥, 𝑡; 𝑞)] = 𝜕𝑡 (𝜑(𝑥, 𝑡; 𝑞)), with the property 𝐿[𝑐1 + 𝑡𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑡

𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑡 + (𝜑(𝑥, 𝑡; 𝑞)2 )𝑥 − (𝜑(𝑥, 𝑡; 𝑞)2 )𝑥𝑥𝑥 + (𝜑(𝑥, 𝑡; 𝑞))𝑥𝑥𝑥𝑥𝑥 using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).


with initial condition 𝑢𝑚 (𝑥, 0) = 0, 𝑣𝑚 (𝑥, 0) = 0. where 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ [

(𝑢𝑚−1 (𝑥, 𝑡))𝑡 + (∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑡))𝑥 −(∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑡))𝑥𝑥𝑥 + (𝑢𝑚−1 (𝑥, 𝑡))𝑥𝑥𝑥𝑥𝑥

].

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 16𝑐−1

𝑢0 (𝑥, 𝑡) =

12

1

cosh2 (4 𝑥),

by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 1

1

1

𝑢1 = − 24 𝑐(16𝑐 − 1)𝑡 cosh (4 𝑥) sinh (4 𝑥), 1

1

𝑢2 = 192 𝑐 2 (16𝑐 − 1)𝑡 2 (2cosh2 (4 𝑥) − 1), 1

1

1

𝑢3 = − 576 𝑐 3 (16𝑐 − 1)𝑡 3 cosh (4 𝑥) sinh (4 𝑥), . . .. 𝑢(𝑥, 𝑡) = 1

1

16𝑐−1 12

1

1

cosh2 (4 𝑥) − 24 𝑐(16𝑐 −

1

1

1

1)𝑡 cosh (4 𝑥) sinh (4 𝑥) + 192 𝑐 2 (16𝑐 − 1)𝑡 2 (2cosh2 (4 𝑥) − 1) − 576 𝑐 3 (16𝑐 − 1

1

1)𝑡 3 cosh (4 𝑥) sinh (4 𝑥) + ⋯ Rest of the components of the iteration formulae can be obtained using the Maple.

164

(a)

(b)

Fig. 2.69.(a)-(b):Comparison of Exact and Approximate solution of Eq. (2.153)

2.9.5. K(2,2,1) Equation We first consider the following initial value problem of the 𝐾(2,2,1) equation [234] 𝑢𝑡 + (𝑢2 )𝑥 − (𝑢2 )𝑥𝑥𝑥 + (𝑢)𝑥𝑥𝑥𝑥𝑥 = 0, 81𝑐−1

𝑢(𝑥, 0) = √

54

(2.154)

1

cosh (3 𝑥).

To solve the given Equation by HAM we choose the linear operator 𝜕


𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑡 + (𝜑(𝑥, 𝑡; 𝑞)2 )𝑥 − (𝜑(𝑥, 𝑡; 𝑞)2 )𝑥𝑥𝑥 + (𝜑(𝑥, 𝑡; 𝑞))𝑥𝑥𝑥𝑥𝑥 using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).


with initial condition 𝑢𝑚 (𝑥, 0) = 0, 𝑣𝑚 (𝑥, 0) = 0. where 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ [

(𝑢𝑚−1 (𝑥, 𝑡))𝑡 + (∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑡))𝑥 − (∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑡))𝑥𝑥𝑥 + (𝑢𝑚−1 (𝑥, 𝑡))𝑥𝑥𝑥𝑥𝑥

].

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 81𝑐−1

𝑢0 (𝑥, 𝑡) = √

54

1

cosh (3 𝑥),


1

𝑢1 = − 54 √486𝑐 − 6 𝑦sinh (3 𝑥) 𝑡, 1

1

𝑢2 = 324 √486𝑐 − 6 𝑐 2 cosh (3 𝑥) 𝑡 2 , 1

1

𝑢3 = − 2916 √486𝑐 − 6 𝑐 3 sinh (3 𝑥) 𝑡 3 , . . .. 81𝑐−1

𝑢(𝑥, 𝑡) = √ 1

54

1

1

1

cosh (3 𝑥) − 54 √486𝑐 − 6 𝑐sinh (3 𝑥) 𝑡 + 1

1

1

√486𝑐 − 6 𝑐 2 cosh (3 𝑥) 𝑡 2 − 2916 √486𝑐 − 6 𝑐 3 sinh (3 𝑥) 𝑡 3 + ⋯ 324 Rest of the components of the iteration formulae can be obtained using the Maple.

166

(a)

(b)


2.9.6. Nonlinear Schrodinger Equation Consider the following Schrodinger equation [30] with the following initial condition 𝜕𝜓(𝑥,𝑡) 𝜕𝑡

1

= − 2 ∇2 𝜓 + 𝑉𝑑 (𝑥)𝜓 + 𝛽𝑑 |𝜓|2 𝜓,

𝜓(𝑥, 0) = 𝜓0 (𝑥),

𝑥 ∈ 𝑅, 𝑡 ≥ 0,

(2.155)

𝑥 ∈ 𝑅𝑑 ,

where 𝑉𝑑 (𝑥) is the trapping potential and 𝛽𝑑 is the real constant. To solve the given Equation by HAM we choose the linear operator 𝜕


𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 1

𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝑖𝜑(𝑥, 𝑡; 𝑞)𝑡 + 2 ∇2 𝜑(𝑥, 𝑡; 𝑞) − 𝑉𝑑 (𝑥)𝑦(𝑥, 𝑡; 𝑞) + 𝛽𝑑 |𝜑(𝑥, 𝑡; 𝑞)|2 𝜑(𝑥, 𝑡; 𝑞), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝜓0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡; 𝑞)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝜓0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝜓(𝑥, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 167

𝐿[𝜓𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝜓𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝜓⃗𝑚−1 ).

(2.156)

with initial condition 𝜓𝑚 (𝑥, 0) = 0. 𝜕

1

𝜓𝑚−1 + 2 ∇2 𝜓𝑚−1 − 𝑉𝑑 (𝑥)𝜓𝑚−1 − 𝜕𝑡 ⃗ 𝑅𝑚 (𝜓𝑚−1 ) = [ ]. 𝑗−1 𝑗−𝑖−1 𝛽𝑑 ∑𝑖=0 ∑𝑘=0 |𝜓𝑖 ||𝜓𝑘 ||𝜓𝑗−𝑘−𝑖−1 |

where

(2.157)

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝜓𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝜓𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝜓⃗𝑚−1 )], 𝑚 ≥ 1 , 𝜕2

Case 1: Consider we have ∇2 = 𝜕𝑥 2 , 𝑉𝑑 (𝑥) = 0 and 𝛽𝑑 = −1,with initial condition 𝜓(𝑥, 0) = 𝑒 𝑖𝑥 . Starting with

𝜓0 (𝑥, 𝑡) = 𝑒 𝑖𝑥 ,by using (2.157) and (2.156), we have the following

components of solution 𝜓0 (𝑥, 𝑡) = 𝑒 𝑖𝑥 , 1

1 1

𝜓1 (𝑥, 𝑡) = 2 𝑖𝑡𝑒 𝑖𝑥 = 1! (2 𝑖𝑡)𝑒 𝑖𝑥 , 1

1 1

𝜓2 (𝑥, 𝑡) = − 8 𝑡 2 𝑒 𝑖𝑥 = 2! (2 𝑖𝑡)2 𝑒 𝑖𝑥 , . . .. General form can be recognized as 1

1

𝜓𝑛 (𝑥, 𝑡) = 𝑛! (2 𝑖𝑡)𝑛 𝑒 𝑖𝑥 , and the exact solution will be as 1

1

1

1

𝑖(𝑥+ 𝑡) ∞ 𝑛 𝑖𝑥 𝑖𝑥 𝑖𝑡 2 . 𝜓(𝑥, 𝑡) = ∑∞ 𝑛=0 𝜓𝑛 (𝑥, 𝑡) = ∑𝑛=0 𝑛! (2 𝑖𝑡) 𝑒 = 𝑒 𝑒 2 = 𝑒 𝜕2

Case 2: Consider we have ∇2 = 𝜕𝑥 2 , 𝑉𝑑 (𝑥) = 𝑐𝑜𝑠 2 (𝑥) and 𝛽𝑑 = 1,with initial condition 𝜓(𝑥, 0) = 𝑠𝑖𝑛𝑥. Starting with 𝜓0 (𝑥, 𝑡) = 𝑠𝑖𝑛𝑥,by using (2.157) and (2.156), we have the following components of solution 𝜓0 (𝑥, 𝑡) = 𝑠𝑖𝑛𝑥, 𝜓1 (𝑥, 𝑡) =

−3

𝜓2 (𝑥, 𝑡) =

−9

2 8

1 −3

𝑖𝑡𝑠𝑖𝑛𝑥 = 1! ( 2 𝑖𝑡)𝑠𝑖𝑛𝑥, 1 −3

𝑖𝑡 2 𝑠𝑖𝑛𝑥 = 1! ( 2 𝑖𝑡)2 𝑠𝑖𝑛𝑥, 168

. . .. Thus, 1

−3

−3

𝑖𝑡 ∞ 𝑛 𝜓(𝑥, 𝑡) = ∑∞ 𝑛=0 𝜓𝑛 (𝑥, 𝑡) = ∑𝑛=0 𝑛! ( 2 𝑖𝑡) 𝑠𝑖𝑛𝑥 = 𝑠𝑖𝑛𝑥𝑒 2 .

which is exact solution. 𝜕2

𝜕2

Case 3: Consider we have ∇2 = 𝜕𝑥 2 + 𝜕𝑦 2 , 𝑉𝑑 (𝑥) = 𝑉(𝑥, 𝑦) and 𝛽𝑑 = 1,with initial condition 𝜓(𝑥, 𝑦, 0) = 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦. Starting with

𝜓0 (𝑥, 𝑦, 𝑡) = 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦,by using (2.157) and (2.156), we have the

following components of solution 𝜓0 (𝑥, 𝑦, 𝑡) = 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦, 𝜓1 (𝑥, 𝑦, 𝑡) = −2𝑖𝑡𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦 = (

−2𝑖𝑡

𝜓2 (𝑥, 𝑦, 𝑡) = −2𝑡 2 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦 = (

1!

)𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦,

(−2𝑖𝑡)2 2!

) 𝑠𝑖𝑛𝑥𝑠𝑦𝑛𝑦,

. . .. The solution in the series form is ∞ 𝜓(𝑥, 𝑦, 𝑡) = ∑∞ 𝑛=0 𝜓𝑛 (𝑥, 𝑦, 𝑡) = ∑𝑛=0

(−2𝑖𝑡)𝑛 𝑛!

𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦 = 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦𝑒 −2𝑖𝑡 .

which is exact solution. 𝜕2

𝜕2

𝜕2

Case 4: Consider we have ∇2 = 𝜕𝑥 2 + 𝜕𝑦 2 + 𝜕𝑧 2 , 𝑉𝑑 (𝑥) = 𝑉(𝑥, 𝑦, 𝑧) and 𝛽𝑑 = 1,with initial condition 𝜓(𝑥, 𝑦, 𝑧, 0) = 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦𝑠𝑖𝑛𝑧. Starting with 𝜓0 (𝑥, 𝑦, 𝑧, 𝑡) = 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦, we have the following components of solution 𝜓0 (𝑥, 𝑦, 𝑧, 𝑡) = 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦𝑠𝑖𝑛𝑧, 5

1

−5𝑖𝑡

𝜓1 (𝑥, 𝑦, 𝑧, 𝑡) = − 2 𝑖𝑡𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦𝑠𝑖𝑛𝑧 = 1! ( 𝜓2 (𝑥, 𝑦, 𝑧, 𝑡) = −

25 2 𝑡 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦𝑠𝑖𝑛𝑧 8

1

2

= 2! (

. 169

) 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦𝑠𝑖𝑛𝑧,

−5𝑖𝑡 2 2

) 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦𝑠𝑖𝑛𝑧

. .. The solution in the series form is 1

∞ 𝜓(𝑥, 𝑦, 𝑧, 𝑡) = ∑∞ 𝑛=0 𝜓𝑛 (𝑥, 𝑦, 𝑧, 𝑡) = ∑𝑛=0 𝑛! (

−5𝑖𝑡 𝑛 ) 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦𝑠𝑖𝑛𝑧 2

= 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦𝑠𝑖𝑛𝑧𝑒

−5𝑖𝑡 2

.

which is exact solution.

2.9.7. Veriational Problem Example 2.34. We consider the following veriational problem [200] 1

min 𝑣 = ∫0 (𝑦(𝑥) + 𝑦 ′ (𝑥) − 4 exp(3𝑥))2 ,

(2.158)

with the boundary conditions 𝑦(1) = 𝑒 3 .

𝑦(0) = 1,

(2.159)

The corresponding Euler-Langrage equation is 𝑦 ′′ − 𝑦 − 8 exp(3𝑥) = 0, with the boundary conditions (2.159).The exact solution of this problem is 𝑦(𝑥) = exp(3𝑥). To solve the given Equation by HAM we choose the linear operator 𝑑2

𝐿[𝜑(𝑥; 𝑞)] = 𝑑𝑥 2 (𝜑(𝑥; 𝑞)), with the property 𝐿[𝑐1 + 𝑥𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑥

𝑥

𝐿−1 = ∫0 ∫0 (. )𝑑𝑥𝑑𝑥, we now define a nonlinear operator as 𝑁[𝜑(𝑥; 𝑞)] = 𝜑 ′′ (𝑥; 𝑞) − 8𝑒 3𝑥 − 𝜑(𝑥; 𝑞), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥; 𝑞) − 𝑦0 (𝑥)] = 𝑞ℎ𝐻(𝑥; 𝑞)𝑁[𝜑(𝑥; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥; 0) = 𝑦0 (𝑥),

𝜑(𝑥; 1) = 𝑦(𝑥).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑦𝑚 (𝑥) − 𝜒𝑚 𝑦𝑚−1 (𝑥)] = ℎ𝐻(𝑥)𝑅𝑚 (𝑦𝑚−1 ). with initial condition 𝑦𝑚 (0) = 0. 𝑑2

Where 𝑅𝑚 (𝑦𝑚−1 ) = [𝑑𝑥 2 𝑦𝑚−1 − 8𝑒 3𝑥 − 𝑦𝑚−1 ]. 170

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑦𝑚 (𝑥) = 𝜒𝑚 𝑦𝑚−1 (𝑥) + 𝐿−1 [ℎ𝐻(𝑥)𝑅𝑚 (𝑦𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑦0 (𝑥) = 1 + 𝐴𝑥, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 8

8

1

8

4

1

8

𝑦1 (𝑥) = − 9 − 3 𝑥 + 2 𝑥 2 + 6 𝐴𝑥 3 + 9 𝑒 3𝑥 , 𝑦2 (𝑥) =

−8 81

8

4

1

1

− 27 𝑥 − 9 𝑥 2 + 81 𝑒 3𝑥 − 9 𝑥 3 − 24 𝑥 4 + 120 𝐴𝑥 5 ,

. . .. the series solution is given by 8

8

1

1

8

−8

8

4

8

𝑦(𝑥) = 1 + 𝐴𝑥, − 9 − 3 𝑥 + 2 𝑥 2 + 6 𝐴𝑥 3 + 9 𝑒 3𝑥 + 81 − 27 𝑥 − 9 𝑥 2 + 81 𝑒 3𝑥 − 4 9

1

1

𝑥 3 − 24 𝑥 4 + 120 𝐴𝑥 5 + ⋯.

Use the boundary conditions, we get 560

𝐴 = 479763 𝑒 3 +

1428175 479763

,

Fig. 2.71:Comparison of Exact and Approximate solution of Eq. (2.158)

Example 2.35. We consider the following veriational problem [200] 1 1+𝑦 2 (𝑥)

min 𝑣 = ∫0

𝑦 ′ 2 (𝑥)

𝑑𝑥,

(2.160)

Let the boundary conditions be 171

𝑦(0) = 0,

𝑦(1) = 0.5.

(2.161)

The corresponding Euler-Langrage equation is 𝑦 ′′ + 𝑦 ′′ 𝑦 2 − 𝑦𝑦 ′

2

= 0,

with the boundary conditions (2.161).The exact solution of this problem is 𝑦(𝑥) = sinh(0.4812118250𝑥). To solve the given Equation by HAM we choose the linear operator 𝑑2

𝐿[𝜑(𝑥; 𝑞)] = 𝑑𝑥 2 (𝜑(𝑥; 𝑞)), with the property 𝐿[𝑐1 + 𝑥𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑥

𝑥

𝐿−1 = ∫0 ∫0 (. )𝑑𝑥𝑑𝑥, we now define a nonlinear operator as 𝑁[𝜑(𝑥; 𝑞)] = 𝜑 ′′ (𝑥; 𝑞) + (𝑥; 𝑞)𝜑 2 (𝑥; 𝑞) − 𝜑(𝑥; 𝑞)𝜑 ′ 2 (𝑥; 𝑞), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥; 𝑞) − 𝑦0 (𝑥)] = 𝑞ℎ𝐻(𝑥; 𝑞)𝑁[𝜑(𝑥; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥; 0) = 𝑦0 (𝑥),

𝜑(𝑥; 1) = 𝑦(𝑥).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑦𝑚 (𝑥) − 𝜒𝑚 𝑦𝑚−1 (𝑥)] = ℎ𝐻(𝑥)𝑅𝑚 (𝑦𝑚−1 ). with initial condition 𝑦𝑚 (0) = 0. 𝑑2

where 𝑅𝑚 (𝑦𝑚−1 ) = [

𝑑𝑥 2

𝑑

𝑠 𝑦𝑚−1 + (∑𝑚−1 𝑟=0 ∑𝑘=0 (𝑑𝑥 𝑦𝑠 ) 𝑦𝑠−𝑘 𝑦𝑚−1−𝑟 ) 𝑑 𝑠 −(∑𝑚−1 𝑟=0 ∑𝑘=0(𝑦𝑠 ) 𝑑𝑥 {𝑦𝑠−𝑘 𝑦𝑚−1−𝑟 })

].

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑦𝑚 (𝑥) = 𝜒𝑚 𝑦𝑚−1 (𝑥) + 𝐿−1 [ℎ𝐻(𝑥)𝑅𝑚 (𝑦𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑦0 (𝑥) = 𝐴 + 𝐵𝑥, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 1

1

𝑦1 (𝑥) = 6 𝐵 3 𝑥 3 + 2 𝐴𝐵 2 𝑥 2 , 172

1

1

1

𝑦2 (𝑥) = 20 𝐵 5 𝑥 5 + 4 𝐴𝐵 4 𝑥 4 + 3 𝑥 3 𝐴2 𝐵 3 , . . .. the series solution is given by 1

1

1

1

1

𝑦(𝑥) = 𝐴 + 𝐵𝑥 + 6 𝐵 3 𝑥 3 + 2 𝐴𝐵 2 𝑥 2 + 20 𝐵 5 𝑥 5 + 4 𝐴𝐵 4 𝑥 4 + 3 𝑥 3 𝐴2 𝐵 3 + ⋯. Use the boundary conditions, we get 𝐴 = 0, 𝐵 = 0.480260494.

Fig. 2.72: Comparison of Exact and Approximate solution of Eq. (2.160)

Example 2.36. We consider the following problem [200] 𝜋

𝑣[𝑦(𝑥), 𝑧(𝑥)] = ∫02 [𝑦 ′ 2 (𝑥) + 𝑧 ′ 2 (𝑥) + 2𝑦(𝑥)𝑧(𝑥)] 𝑑𝑥,

(2.162)

with the boundary conditions as follows: 𝜋

𝑦(0) = 0,

𝑦 (2 ) = 1,

𝑧(0) = 0,

𝑧 ( 2 ) = −1.

𝜋

(2.163)

The system of Euler differential equations is 𝑦 ′′ − 𝑧 = 0, 𝑧 ′′ − 𝑦 = 0, with the boundary conditions (2.163). To solve the given Equation by HAM we choose the linear operator 𝑑2

𝐿[𝜑1 (𝑥; 𝑞1 )] = 𝑑𝑥 2 (𝜑1 (𝑥; 𝑞1 )), 173

𝑑2

𝐿[𝜑2 (𝑥; 𝑞2 )] = 𝑑𝑥 2 (𝜑2 (𝑥; 𝑞2 )), with the property 𝐿[𝑐1 + 𝑥𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑥

𝑥

𝐿−1 = ∫0 ∫0 (. )𝑑𝑥𝑑𝑥, we now define a nonlinear operator as 𝑁[𝜑1 (𝑥; 𝑞1 )] = 𝜑 ′′1 (𝑥; 𝑞1 ) − 𝜑2 (𝑥; 𝑞2 ), 𝑁[𝜑2 (𝑥; 𝑞2 )] = 𝜑 ′′ 2 (𝑥; 𝑞2 ) − 𝜑1 (𝑥; 𝑞1 ), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞1 )𝐿[𝜑1 (𝑥; 𝑞2 ) − 𝑦0 (𝑥)] = 𝑞1 ℎ𝐻(𝑥; 𝑞1 )𝑁[𝜑1 (𝑥; 𝑞1 )]. for 𝑞1 = 0 and 𝑞1 = 1, we can write 𝜑1 (𝑥; 0) = 𝑦0 (𝑥),

𝜑1 (𝑥; 1) = 𝑦(𝑥).

(1 − 𝑞2 )𝐿[𝜑2 (𝑥; 𝑞2 ) − 𝑧0 (𝑥)] = 𝑞2 ℎ𝐻(𝑥; 𝑞2 )𝑁[𝜑2 (𝑥; 𝑞2 )]. for 𝑞2 = 0 and 𝑞2 = 1, we can write 𝜑2 (𝑥; 0) = 𝑧0 (𝑥),

𝜑2 (𝑥; 1) = 𝑧(𝑥).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑦𝑚 (𝑥) − 𝜒𝑚 𝑦𝑚−1 (𝑥)] = ℎ𝐻(𝑥)𝑅𝑚 (𝑦𝑚−1 ). with initial condition 𝑦𝑚 (0) = 0. 𝐿[𝑧𝑚 (𝑥) − 𝜒𝑚 𝑧𝑚−1 (𝑥)] = ℎ𝐻(𝑥)𝑅𝑚 (𝑧𝑚−1 ). with initial condition 𝑧𝑚 (0) = 0. 𝑑2

where 𝑅𝑚 (𝑦𝑚−1 ) = [𝑑𝑥 2 𝑦𝑚−1 − 𝑧𝑚−1 )]. 𝑑2

𝑅𝑚 (𝑧𝑚−1 ) = [𝑑𝑥 2 𝑧𝑚−1 − 𝑦𝑚−1 )] Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑦𝑚 (𝑥) = 𝜒𝑚 𝑦𝑚−1 (𝑥) + 𝐿−1 [ℎ𝐻(𝑥)𝑅𝑚 (𝑦𝑚−1 )], 𝑚 ≥ 1, 𝑧𝑚 (𝑥) = 𝜒𝑚 𝑧𝑚−1 (𝑥) + 𝐿−1 [ℎ𝐻(𝑥)𝑅𝑚 (𝑧𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑦0 (𝑥) = 𝐴1 + 𝐵1 𝑥, 𝑧0 (𝑥) = 𝐴2 + 𝐵2 𝑥,

174


1

1

1

𝑦1 (𝑥) = 2 𝐴2 𝑥 2 + 6 𝐵2 𝑥 3 , 𝑧1 (𝑥) = 2 𝐴1 𝑥 2 + 6 𝐵1 𝑥 3 , 1

1

1

1

𝑦2 (𝑥) = 24 𝐴2 𝑥 4 + 120 𝐵1 𝑥 5 , 𝑧2 (𝑥) = 24 𝐴2 𝑥 4 + 120 𝐵2 𝑥 5 , . . .. the series solution is given by 1

1

1

1

1

1

1

1

𝑦(𝑥) = 𝐴1 + 𝐵1 𝑥 + 2 𝐴2 𝑥 2 + 6 𝐵2 𝑥 3 + 24 𝐴2 𝑥 4 + 120 𝐵1 𝑥 5 + ⋯. 𝑧(𝑥) = 𝐴2 + 𝐵2 𝑥 + 2 𝐴1 𝑥 2 + 6 𝐵1 𝑥 3 + 24 𝐴2 𝑥 4 + 120 𝐵2 𝑥 5 + ⋯. Use the boundary conditions, we get 3840

𝐴1 = 0, 𝐵1 = (𝜋4 −80𝜋2+1920)𝜋. −3840

𝐴2 = 0, 𝐵2 = (𝜋4 −80𝜋2 +1920)𝜋 and the close form solution is 𝑦(𝑥) = sin(𝑥) and 𝑧(𝑥) = −sin(𝑥).

(a)

(b)


175

2.10. Modified Homotopy Analysis Method Recently Usman et al. proposed a new modification and successively applied to solve Heat and Wave-like equations [206] and Helmholtz equation [270]. It is observed that proposed technique is highly effective. This modified version has less computational work as compare to traditional Homotopy Perturbation method. It is an established fact that most of the physical phenomenon is nonlinear in nature and hence there is a dire need to find their appropriate solutions. Several examples are given which reveal the efficiency and reliability of the proposed algorithm.

2.10.1. Methodology Applying the Taylor’s series on the initial condition and source term then set the recurrence relation accordingly.

2.10.2. Helmholtz Equation Example 2.37. Consider the following Helmholtz equation [134] 𝑢𝑦𝑦 + 𝑢𝑥𝑥 − 𝑢 = 0, 𝑢(0, 𝑦) = 𝑦,

(2.164)

𝑢𝑥 (0, 𝑦) = 𝑦 + cosh(𝑦).

To solve the given Equation by HAM, we choose the linear operator 𝜕2

𝐿[𝜑(𝑥, 𝑦; 𝑞)] = 𝜕𝑥 2 (𝜑(𝑥, 𝑦; 𝑞)), with the property 𝐿[𝑐1 + 𝑥𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑥

𝑥

𝐿−1 = ∫0 ∫0 ( . )𝑑𝑥𝑑𝑥, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑦; 𝑞)] = 𝜑(𝑦; 𝑞)𝑦𝑦 + 𝜑(𝑦; 𝑞)𝑥𝑥 − 𝜑(𝑦; 𝑞) using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑦; 𝑞) − 𝑢0 (𝑦)] = 𝑞ℎ𝐻(𝑦)𝑁[𝜑(𝑦; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(0, 𝑦) = 𝑢0 (𝑦),

𝜑(1, 𝑦) = 𝑢(𝑦).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑢𝑚 (𝑦) − 𝜒𝑚 𝑢𝑚−1 (𝑦)] = ℎ𝐻(𝑦)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ).

176

with initial condition 𝑢𝑚 (𝑥, 0) = 0. After applying the Taylor’s series in initial condition, we introduce 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑦) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑢0 (𝑥, 𝑡) = 𝑥 + (1 + 𝑥)𝑦, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢0 (𝑥, 𝑡) = 𝑥 + (1 + 𝑥)𝑦, 1

1

6

144

𝑢1 (𝑥, 𝑡) = 𝑥 3 𝑦 +

1

𝑥 3 𝑦 4 + 𝑥 2 𝑦, 2

1

1

1

1

1

1

𝑢2 (𝑥, 𝑡) = − 144 𝑥 3 𝑦 4 + 120 𝑥 5 𝑦 + 2880 𝑥 5 𝑦 4 − 240 𝑥 5 𝑦 2 + 24 𝑦𝑥 4 + 241920 𝑥 3 𝑦 8 , .

. .. The series form solution is given by 1

1

1

1

1

𝑢(𝑥, 𝑡) = 𝑥 + (1 + 𝑥)𝑦 + 6 𝑥 3 𝑦 + 144 𝑥 3 𝑦 4 + 2 𝑥 2 𝑦 − 144 𝑥 3 𝑦 4 + 120 𝑥 5 𝑦 + 1

1

2880

1

1

𝑥 5 𝑦 4 − 240 𝑥 5 𝑦 2 + 24 𝑦𝑥 4 + 241920 𝑥 3 𝑦 8 + ⋯,

Rest of the components of the iteration formulae can be obtained using the Maple.

(a)

(b)


177

Example 2.38. Consider the following Helmholtz equation [134] 𝑢𝑦𝑦 + 𝑢𝑥𝑥 + 8𝑢 = 0, 𝑢(0, 𝑦) = 𝑠𝑖𝑛2𝑦,

(2.165)

𝑢𝑥 (0, 𝑦) = 0.

To solve the given Equation by HAM, we choose the linear operator 𝜕2

𝐿[𝜑(𝑥, 𝑦; 𝑞)] = 𝜕𝑥 2 (𝜑(𝑥, 𝑦; 𝑞)), with the property 𝐿[𝑐1 + 𝑥𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑥

𝑥

𝐿−1 = ∫0 ∫0 ( . )𝑑𝑥𝑑𝑥, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑦; 𝑞)] = 𝜑(𝑦; 𝑞)𝑦𝑦 + 𝜑(𝑦; 𝑞)𝑥𝑥 + 8𝜑(𝑦; 𝑞) using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑦; 𝑞) − 𝑢0 (𝑦)] = 𝑞ℎ𝐻(𝑦)𝑁[𝜑(𝑦; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(0, 𝑦) = 𝑢0 (𝑦),

𝜑(1, 𝑦) = 𝑢(𝑦).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑢𝑚 (𝑦) − 𝜒𝑚 𝑢𝑚−1 (𝑦)] = ℎ𝐻(𝑦)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 0) = 0. After applying the Taylor’s series in initial condition, we introduce 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑦) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 4

𝑢0 (𝑥, 𝑡) = 2𝑦 − 3 𝑦 3 , by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 4

𝑢0 (𝑥, 𝑡) = 2𝑦 − 3 𝑦 3 , 4

𝑢1 (𝑥, 𝑡) = 315 𝑦(−315 + 210𝑦 2 − 42𝑦 4 + 8𝑦 6 )𝑥 2 16

8

64

4

32

8

𝑢2 (𝑥, 𝑡) = − 315 𝑥 2 𝑦 7 − 9 𝑥 4 𝑦 3 − 945 𝑥 4 𝑦 7 + 3 𝑥 4 𝑦 + 155925 𝑥 2 𝑦11 − 2835 𝑥 2 𝑦 9 , . 178

. .. The series form solution is given by 4

4

16

𝑢(𝑥, 𝑡) = 2𝑦 − 3 𝑦 3 + 315 𝑦(−315 + 210𝑦 2 − 42𝑦 4 + 8𝑦 6 )𝑥 2 − 315 𝑥 2 𝑦 7 − 8 9

𝑥4𝑦3 −

64 945

4

32

8

𝑥 4 𝑦 7 + 3 𝑥 4 𝑦 + 155925 𝑥 2 𝑦11 − 2835 𝑥 2 𝑦 9 + ⋯

Rest of the components of the iteration formulae can be obtained using the Maple.

(a)

(b)


2.10.3. SOG-KdV Equation Example 2.39. Consider the following SOG-KdV equation [270] 𝑢𝑡 + 𝑢𝑢𝑥 + 𝑢𝑥𝑥𝑥 − 𝑢5𝑥 + 𝑢7𝑥 = 0,

(2.166)

𝑢(𝑥, 0) = 𝑎0 + 𝑎6 𝑠𝑒𝑐ℎ4 (𝑘𝑥) + 𝑎6 𝑠𝑒𝑐ℎ6 (𝑘𝑥), To solve the given Equation by HAM, we choose the linear operator 𝜕

𝐿[𝜑(𝑥; 𝑞)] = 𝜕𝑡 (𝜑(𝑥; 𝑞)), with the property 𝐿[𝑦1 + 𝑡𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑡

𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as

179

𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥; 𝑞)𝑡 + 𝜑(𝑥; 𝑞)𝜑(𝑥; 𝑞)𝑥 + 𝜑(𝑥; 𝑞)𝑥𝑥𝑥 − 𝜑(𝑥; 𝑞)5𝑥 + 𝜑(𝑥; 𝑞)7𝑥 using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥; 𝑞) − 𝑢0 (𝑥)] = 𝑞ℎ𝐻(𝑥)𝑁[𝜑(𝑥; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥; 0) = 𝑢0 (𝑥),

𝜑(𝑥; 1) = 𝑢(𝑥).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)] = ℎ𝐻(𝑥)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 0) = 0. Appling the Taylor’s series on the initial condition and introduce the expression of 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation. 𝑢0 (𝑥, 𝑡) = 𝑎0 + 2𝑎6 , by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢0 (𝑥, 𝑡) = 𝑎0 + 2𝑎6 , 𝑢1 (𝑥, 𝑡) = −10(−𝑎0 − 2𝑎6 + 5𝑎6 𝑘 2 𝑥 2 )𝑎6 𝑘 2 𝑥𝑡, 𝑢2 (𝑥, 𝑡) = 150𝑎62 𝑘 4 𝑡 2 − 176𝑎6 𝑘 4 𝑥𝑡, . . .. The series form solution is given by 𝑢(𝑥, 𝑡) = 𝑎0 + 2𝑎6 − 10(−𝑎0 − 2𝑎6 + 5𝑎6 𝑘 2 𝑥 2 )𝑎6 𝑘 2 𝑥𝑡 + 150𝑎62 𝑘 4 𝑡 2 − 176𝑎6 𝑘 4 𝑥𝑡 + ⋯ Rest of the components of the iteration formulae can be obtained using the Maple.

180

Table 2.43. Comparison of Exact and Approximate Solutions of Eq. (2.166) Exact solution

Approximate solution

0.01

0.9307996773

0.9308017722

2.094 × 10−6

0.02

0.9307970955

0.9308034443

6.348 × 10−6

0.03

0.9307919320

0.9308051166

1.318 × 10−5

0.04

0.9307841866

0.9308067888

2.260 × 10−5

0.05

0.9307738597

0.9308084610

3.460 × 10−5

x

Error

Example 2.40. Consider the following SOG-KdV equation [270] 𝑢𝑡 + 𝑢𝑢𝑥 + 𝑢𝑥𝑥𝑥 − 𝑢5𝑥 + 𝑢7𝑥 = 0,

(2.167)

𝑢(𝑥, 0) = 𝑎0 + 𝑎6 𝑠𝑒𝑐ℎ6 (𝑘𝑥), To solve the given Equation by HAM, we choose the linear operator 𝜕

𝐿[𝜑(𝑥; 𝑞)] = 𝜕𝑡 (𝜑(𝑥; 𝑞)), with the property 𝐿[𝑐1 + 𝑡𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑦

𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥; 𝑞)𝑡 + 𝜑(𝑥; 𝑞)𝜑(𝑥; 𝑞)𝑥 + 𝜑(𝑥; 𝑞)𝑥𝑥𝑥 − 𝜑(𝑥; 𝑞)5𝑥 + 𝜑(𝑥; 𝑞)7𝑥 using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥; 𝑞) − 𝑢0 (𝑥)] = 𝑞ℎ𝐻(𝑥)𝑁[𝜑(𝑥; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥; 0) = 𝑢0 (𝑥),

𝜑(𝑥; 1) = 𝑢(𝑥).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)] = ℎ𝐻(𝑥)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 0) = 0. Appling the Taylor’s series on the initial condition and introduce the expression of 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 181

𝑢𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation. 𝑢0 (𝑥, 𝑡) = 𝑎0 + 𝑎6 , by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢0 (𝑥, 𝑡) = 𝑎0 + 𝑎6 , 𝑢1 (𝑥, 𝑡) = −6(−𝑎0 − 𝑎6 + 6𝑎6 𝑘 2 𝑥 2 )𝑎6 𝑘 2 𝑥𝑡, 𝑢2 (𝑥, 𝑡) = 54𝑎62 𝑘 4 𝑡 2 − 120𝑎6 𝑘 4 𝑥𝑡, . . .. The series form solution is given by 𝑢(𝑥, 𝑡) = 𝑎0 + 𝑎6 − 6(−𝑎0 − 𝑎6 + 6𝑎6 𝑘 2 𝑥 2 )𝑎6 𝑘 2 𝑥𝑡 + 54𝑎62 𝑘 4 𝑦2 − 120𝑎6 𝑘 4 𝑥𝑡 + ⋯ Rest of the components of the iteration formulae can be obtained using the Maple.

(a)

(b)


2.10.4. Cubic Nonlinear Schrodinger Equation Example 2.41. Consider the following Cubic Nonlinear Schrodinger Equation [30] 𝑣𝑡 + 𝑤𝑥𝑥 + 𝑞(𝑣 2 + 𝑤 2 )𝑤 = 0,

(2.168)

𝑤𝑡 − 𝑣𝑥𝑥 − 𝑞(𝑣 2 + 𝑤 2 )𝑣 = 0,

(2.169)

Subject to the following initial conditions 𝑣(𝑥, 0) = √2α⁄q cos(0.5𝑥) sech(√α𝑥), 182

𝑤(𝑥, 0) = √2α⁄q cos(0.5𝑥) sech(√α𝑥). To solve the given Equation by HAM we choose the linear operator 𝜕

𝐿[𝜑1 (𝑥, 𝑡; 𝑞1 )] = 𝜕𝑡 (𝜑1 (𝑥, 𝑡; 𝑞1 )), 𝜕


𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑1 (𝑥, 𝑡; 𝑞1 )] = 𝜑1 (𝑥, 𝑡; 𝑞1 )𝑡 + 𝜑2 (𝑥, 𝑡; 𝑞2 )𝑥𝑥 + 𝑞((𝜑1 2 (𝑥, 𝑡; 𝑞1 ) + 𝜑2 2 (𝑥, 𝑡; 𝑞2 ))𝜑2 (𝑥, 𝑡; 𝑞2 ), 𝑁[𝜑2 (𝑥, 𝑡; 𝑞2 )] = 𝜑2 (𝑥, 𝑡; 𝑞2 )𝑡 − 𝜑1 (𝑥, 𝑡; 𝑞2 )𝑥𝑥 − 𝑞((𝜑1 2 (𝑥, 𝑡; 𝑞1 ) + 𝜑2 2 (𝑥, 𝑡; 𝑞2 ))𝜑1 (𝑥, 𝑡; 𝑞1 ), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞1 )𝐿[𝜑1 (𝑥, 𝑡; 𝑞1 ) − 𝑣0 (𝑥, 𝑡)] = 𝑞1 ℎ𝐻(𝑥, 𝑡; 𝑞1 )𝑁[𝜑1 (𝑥, 𝑡; 𝑞1 )]. for 𝑦1 = 0 and 𝑞1 = 1, we can write 𝜑1 (𝑥, 𝑡; 0) = 𝑣0 (𝑥, 𝑡),

𝜑1 (𝑥, 𝑡; 1) = 𝑣(𝑥, 𝑡).

(1 − 𝑞2 )𝐿[𝜑2 (𝑥, 𝑡; 𝑞2 ) − 𝑤0 (𝑥, 𝑡)] = 𝑞2 ℎ𝐻(𝑥, 𝑡; 𝑞2 )𝑁[𝜑2 (𝑥, 𝑡; 𝑞2 )]. for 𝑞2 = 0 and 𝑞2 = 1, we can write 𝜑2 (𝑥, 𝑡; 0) = 𝑤0 (𝑥, 𝑡),

𝜑2 (𝑥, 𝑡; 1) = 𝑤(𝑥, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equations 𝐿[𝑣𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑣𝑚−1 ). 𝐿[𝑤𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑤𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ). with initial condition 𝑤𝑚 (𝑥, 0) = 0, 𝑤𝑚 (𝑥, 0) = 0. 𝜕2

𝜕

where 𝑅𝑚 (𝑣𝑚−1 ) = [𝜕𝑡 𝑣𝑚−1 + 𝜕𝑥 2 𝑤𝑚−1 + 𝑞 ( 𝜕

𝜕2

𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ) = [𝜕𝑡 𝑤𝑚−1 − 𝜕𝑥 2 𝑣𝑚−1 − 𝑞 (

𝑠 ∑𝑚−1 𝑟−0 ∑𝑘=0 𝑣𝑠 𝑣𝑠−𝑘 𝑤𝑚−1−𝑟 + )]. 𝑠 ∑𝑚−1 𝑟−0 ∑𝑘=0 𝑤𝑠 𝑤𝑠−𝑘 𝑤𝑚−1−𝑟

𝑠 ∑𝑚−1 𝑟−0 ∑𝑘=0 𝑣𝑠 𝑣𝑠−𝑘 𝑣𝑚−1−𝑟 + )]. 𝑠 ∑𝑚−1 𝑟−0 ∑𝑘=0 𝑤𝑠 𝑤𝑠−𝑘 𝑣𝑚−1−𝑟

Appling the Taylor’s series on the initial condition and introduce the expression of 𝑅𝑚 (𝑣𝑚−1 ) and 𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ) accordingly. 183

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑣𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑣𝑚−1 )], 𝑚 ≥ 1, 𝑤𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑤𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximations 𝛼

𝑣0 (𝑥, 𝑡) = 1. √2√𝑞 , 𝑤0 = 0, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝛼

𝑣0 (𝑥, 𝑡) = 1. √2√𝑞 , 𝑤0 = 0, 𝛼

𝑣1 = −0.353553390𝛼(4. +𝑥 2 𝑐 2 )𝑥√𝑞 𝑐𝑡,

𝑤1 =

𝛼

0.7071067810𝛼(4. +𝑥 2 𝑐 2 )𝑥√𝑞 , 𝛼

𝛼

𝑣2 = −0.353553390𝛼𝑥 3 √𝑞 𝑐 3 𝑡 − .7071067810𝛼 √𝑞 𝑐 2 𝑡 2 − 𝛼

𝛼

1.41421356𝛼 2 √𝑞 𝑡 2 𝑐 2 𝑥 2 − 0.1767766955𝛼 2 √𝑞 𝑡 2 𝑐 4 𝑥 4 𝛼

−2.828427125𝛼 2 √𝑞 𝑡 2 + ⋯, 𝛼

𝛼

𝑤2 = −1.060660172𝛼 √𝑞 𝑡𝑐 2 𝑥 2 − 0.08838834760𝛼 2 √𝑞 𝑡 2 𝑐 5 𝑥 5 + 𝛼

𝛼

0.7071067810𝛼 √𝑞 𝑡 2 𝑐 2 − 1.414213562𝛼 2 √𝑞 𝑡 2 𝑥𝑐 − 𝛼

0.7071067810𝛼 2 √ 𝑡 2 𝑐 3 𝑥 3 + ⋯, 𝑞

. . ..

184

𝛼

The series form solution is given by 𝑣(𝑥, 𝑡) = 1. √2√𝑞 − 𝛼

0.353553390𝛼(4. +𝑥 2 𝑐 2 )𝑥√𝑞 𝑐𝑡 − 𝛼

𝛼

0.353553390𝛼𝑥 3 √𝑞 𝑐 3 𝑡 − 0.7071067810𝛼√𝑦 𝑐 2 𝑡 2 + ⋯, 𝛼

𝛼

𝑤(𝑥, 𝑡) = 0.7071067810𝛼(4. +𝑥 2 𝑐 2 )𝑥√𝑞 − 1.060660172𝛼 √𝑞 𝑡𝑐 2 𝑥 2 𝛼

𝛼

−0.08838834760𝛼 2 √𝑞 𝑡 2 𝑐 5 𝑥 5 + 0.7071067810𝛼 √𝑞 𝑡 2 𝑐 2 + ⋯. Rest of the components of the iteration formulae can be obtained using the Maple. Table 2.44. Comparison of Exact and Approximate Solutions of Eq. (2.168-169) |𝑣(𝑥, 𝑡) − 𝑣2 (𝑥, 𝑡)|

|𝑤(𝑥, 𝑡) − 𝑤2 (𝑥, 𝑡)|

0.01

6.300 × 10−8

5.303 × 10−5

0.02

3.052 × 10−7

1.237 × 10−4

0.03

7.241 × 10−7

1.944 × 10−4

0.04

1.319 × 10−6

2.651 × 10−4

0.05

2.094 × 10−5

3.358 × 10−4

X

Example 2.42. Consider the following Cubic Nonlinear Schrodinger Equation [30] 𝑖𝑢𝑡 + ∇2 𝑢 + 𝑞|𝑢|2 𝑢 = 0,

𝐿0 < 𝑥, 𝑦 < 𝐿1 ,

(2.170)

with initial condition 𝑢(𝑥, 𝑦, 0) = √2𝑎 ⁄𝑞 𝑒 (0.5𝑖𝑐(𝑥+𝑦)) sech(√𝑎(𝑥 + 𝑦)), and boundary conditions 𝜕𝑢(𝐿0 ,𝑡) 𝜕𝑥 ∂2

=

𝜕𝑢(𝐿1 ,𝑡) 𝜕𝑥

= 0,

𝜕𝑢(𝐿0 ,𝑡) 𝜕𝑦

=

𝜕𝑢(𝐿1 ,𝑡) 𝜕𝑦

= 0,

∂2

where ∇2 = ∂x2 + ∂y2 is the Laplace operator,𝑖 = √−1; 𝑞 ≥ 0 is the real parameter. The exact solution of this equation is 𝑥+𝑦 )). −𝑐𝑡

𝑢(𝑥, 𝑦, 𝑡) = √2𝑎 ⁄𝑞 exp (𝑖(0.5𝑐(𝑥 + 𝑦))) − (0.25𝑐 2 + 𝑎)𝑡𝑠𝑒𝑐ℎ (√𝑎 ( Equation (1) can be reduced to the following system of equations 𝑣𝑡 + ∇2 𝑤 + 𝑞(𝑣 2 + 𝑤 2 )𝑤 = 0, 185

𝑤𝑡 − ∇2 𝑣 − 𝑞(𝑣 2 + 𝑤 2 )𝑣 = 0, where 𝑢 = 𝑢(𝑥, 𝑦, 𝑡) = 𝑣(𝑥, 𝑦, 𝑡) + 𝑖𝑤(𝑥, 𝑦, 𝑡), then the initial data can be written as 𝑣(𝑥, 𝑦, 0) = √2𝑎 ⁄𝑞 cos(0.5𝑐(𝑥 + 𝑦))sec(√𝑎(𝑥 + 𝑦)), 𝑤(𝑥, 𝑦, 0) = √2𝑎 ⁄𝑞 sin(0.5𝑐(𝑥 + 𝑦))sec(√𝑎(𝑥 + 𝑦)), and the exact solution of this system can be written as 𝑣(𝑥, 𝑦, 𝑡) = √2𝑎 ⁄𝑞 𝑐𝑜𝑠(0.5𝑐(𝑥 + 𝑦) − (0.25𝑐 2 − 𝑎)𝑡) 𝑠𝑒𝑐ℎ (√𝑎(𝑥 + 𝑦 − 𝑐𝑡)), 𝑤(𝑥, 𝑦, 𝑡) = √2𝑎 ⁄𝑞 𝑠𝑖𝑛(0.5𝑐(𝑥 + 𝑦) − (0.25𝑐 2 − 𝑎)𝑡) 𝑠𝑒𝑐ℎ (√𝑎(𝑥 + 𝑦 − 𝑐𝑡)), To solve the given Equation by HAM we choose the linear operator 𝐿[𝜑1 (𝑥, 𝑦; 𝑞1 )] =

𝜕 𝜕𝑡

(𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )),

𝜕

𝐿[𝜑2 (𝑥, 𝑦; 𝑞2 )] = 𝜕𝑡 (𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )), with the property 𝐿[𝑐1 + 𝑡𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑡

𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )] = 𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )𝑡 + ∇2 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 ) + 𝑞((𝜑1 2 (𝑥, 𝑦, 𝑡; 𝑞1 ) + 𝜑2 2 (𝑥, 𝑦, 𝑡; 𝑞2 ))𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 ), 𝑁[𝜑2 (𝑥, 𝑡; 𝑞2 )] = 𝜑2 (𝑥, 𝑡; 𝑞2 )𝑡 − ∇2 𝜑1 (𝑥, 𝑦, 𝑡; 𝑞2 ) − 𝑞((𝜑1 2 (𝑥, 𝑦, 𝑡; 𝑞1 ) + 𝜑2 2 (𝑥, 𝑦, 𝑡; 𝑞2 ))𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞1 )𝐿[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ) − 𝑣0 (𝑥, 𝑦, 𝑡)] = 𝑞1 ℎ𝐻(𝑥, 𝑦, 𝑡; 𝑞1 )𝑁[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )]. for 𝑞1 = 0 and 𝑞1 = 1, we can write 𝜑1 (𝑥, 𝑦, 𝑡; 0) = 𝑣0 (𝑥, 𝑦, 𝑡),

𝜑1 (𝑥, 𝑦, 𝑡; 1) = 𝑣(𝑥, 𝑦, 𝑡).

(1 − 𝑞2 )𝐿[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 ) − 𝑤0 (𝑥, 𝑦, 𝑡)] = 𝑞2 ℎ𝐻(𝑥, 𝑦, 𝑡; 𝑞2 )𝑁[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )]. for 𝑞2 = 0 and 𝑞2 = 1, we can write 𝜑2 (𝑥, 𝑦, 𝑡; 0) = 𝑤0 (𝑥, 𝑦, 𝑡),

𝜑2 (𝑥, 𝑦, 𝑡; 1) = 𝑤(𝑥, 𝑦, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equations 𝐿[𝑣𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑣𝑚−1 ). 𝐿[𝑤𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑤𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ). 186

with initial condition 𝑤𝑚 (𝑥, 𝑦, 0) = 0, 𝑤𝑚 (𝑥, 𝑦, 0) = 0. ∑𝑚−1 ∑𝑠𝑘=0 𝑣𝑠 𝑣𝑠−𝑘 𝑤𝑚−1−𝑟 + 𝜕 where 𝑅𝑚 (𝑣𝑚−1 ) [𝜕𝑡 𝑣𝑚−1 + ∇2 𝑤𝑚−1 + 𝑞 ( 𝑟−0 )]. 𝑠 ∑𝑚−1 𝑟−0 ∑𝑘=0 𝑤𝑠 𝑤𝑠−𝑘 𝑤𝑚−1−𝑟 𝜕

2

𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ) = [𝜕𝑡 𝑤𝑚−1 − ∇ 𝑣𝑚−1

𝑠 ∑𝑚−1 𝑟−0 ∑𝑘=0 𝑣𝑠 𝑣𝑠−𝑘 𝑣𝑚−1−𝑟 + − 𝑞 ( 𝑚−1 𝑠 )]. ∑𝑟−0 ∑𝑘=0 𝑤𝑠 𝑤𝑠−𝑘 𝑣𝑚−1−𝑟

Appling the Taylor’s series on the initial condition and introduce the expression of 𝑅𝑚 (𝑣𝑚−1 ) and 𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ) accordingly. Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑣𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑣𝑚−1 )], 𝑚 ≥ 1, 𝑤𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑤𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximations 𝑣0 = √2α⁄q cos(0.5c𝑥) sech(√α𝑥), 𝑤0 = √2α⁄q sin(0.5c𝑥) sech√α𝑥. by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑣0 = √2α⁄q cos(0.5c𝑥) sech(√α𝑥), 𝑤0 = √2α⁄q sin(0.5c𝑥) sech√α𝑥. 𝑣1 =

1 6

cosh(√α𝑥)

(2.000000000 × 10−10 √α⁄q (8.83883476 5

× 108 𝑦𝑐𝑜𝑠ℎ(√α𝑥) cos(0.5000000000𝑐𝑥) 𝑐 3 + 7.071067810 × 109 cos(0.5000000000𝑐𝑥)𝑐𝑠𝑖𝑛ℎ(√α𝑥)√α𝑐𝑜𝑠ℎ(√α𝑥)4 + ⋯, 𝑤1 = −

1 cosh(√α𝑥)

6

(2.000000000 × 10−10 √α⁄q (8.83883476 5

× 108 𝑦𝑐𝑜𝑠ℎ(√α𝑥) cos(0.5000000000𝑐𝑥) 𝑐 3 + ⋯, . . .. The series form solution is given by 𝑣(𝑥, 𝑡) = √2α⁄q cos(0.5c𝑥) sech(√α𝑥) + 5

1 6

cosh(√α𝑥)

(2.000000000 ×

10−10 √α⁄q (8.83883476 × 108 𝑦𝑐𝑜𝑠ℎ(√α𝑥) cos(0.5000000000𝑐𝑥) 𝑐 3 + ⋯, 187

𝑤(𝑥, 𝑡) = √2α⁄q sin(0.5c𝑥) sech√α𝑥 −

1 6

cosh(√α𝑥)

(2.000000000 ×

5

10−10 √α⁄q (8.83883476 × 108 𝑦𝑐𝑜𝑠ℎ(√α𝑥) cos(0.5000000000𝑐𝑥) 𝑐 3 + ⋯, Rest of the components of the iteration formulae can be obtained using the Maple.

(a)

(b)


2.10.5. Heat and Wave-like Equation Example 2.43. Consider the heat and wave like equation [149] 2

𝑦𝑥𝑥 + 𝑥 𝑦𝑥 − (6 + 4𝑥 2 − cos(𝑡))𝑦 = 𝑦𝑡 ,

(2.171)

𝑦(0, 𝑡) = esin(𝑡) , 𝑦𝑥 (0, 𝑡) = 0. To solve the given Equation by HAM, we choose the linear operator 𝜕2

𝐿[𝜑(𝑥, 𝑦; 𝑞)] = 𝜕𝑥 2 (𝜑(𝑥, 𝑦; 𝑞)), with the property 𝐿[𝑐1 + 𝑥𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑡

𝑡

𝐿−1 = ∫0 ∫0 ( . )𝑑𝑥𝑑𝑥, we now define a nonlinear operator as 2 𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑥𝑥 + 𝜑(𝑥, 𝑡; 𝑞)𝑥 − (6 + 4𝑥 2 − 𝑐𝑜𝑠𝑡)𝜑(𝑥, 𝑡; 𝑞) − 𝜑(𝑥, 𝑡; 𝑞)𝑡 𝑥 using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑦0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 188

𝜑(𝑥, 𝑡; 0) = 𝑦0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑦(𝑥, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑦𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑦𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑦𝑚−1 ). with initial condition 𝑦𝑚 (𝑥, 0) = 0. 𝜕2

2 𝜕

𝜕

where 𝑅𝑚 (𝑦𝑚−1 ) = ℎ [𝜕𝑥 2 𝑦𝑚−1 + 𝑥 𝜕𝑥 𝑦𝑚−1 + (6 + 4𝑥 2 − 𝑐𝑜𝑠𝑡)𝑦𝑚−1 − 𝜕𝑡 𝑦𝑚−1 ]. after applying the Taylor’s series we have 𝜕2

𝑅𝑚 (𝑦𝑚−1 ) = ℎ [

𝜕𝑥 2

2 𝜕

(𝑦𝑚−1 + 𝑓𝑚 ) + 𝑥 𝜕𝑥 (𝑦𝑚−1 + 𝑓𝑚 )(6 + 4𝑥 2 − 𝑐𝑜𝑠𝑡) (𝑦𝑚−1 + 𝑓𝑚 ) −

𝜕 𝜕𝑡

].

(𝑦𝑚−1 + 𝑓𝑚 )

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑦𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑦𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑦0 (𝑥, 𝑡) = 1 + 𝑡, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑦0 (𝑥, 𝑡) = 1 + 𝑡, 1

2

4

4

1

1

𝑦1 (𝑥, 𝑡) = − 4 (− 3 t 2 − 3 − 3 t) 𝑥 4 − 2 (−7 − 7t − 3t 2 + cos(t) t + 2 cos(t) t 2 + cos(t)) 𝑥 2 , 1

𝑦2 (𝑥, 𝑡) = − 5040 𝑥 2 (35280 + 392𝑥 4 𝑐𝑜𝑠(𝑡) 𝑡 + 196𝑥 4 cos(𝑡)𝑡 2 − 210𝑥 2 cos 2 (t)t − 105𝑥 2 cos 2 (t)t 2 + ⋯, . . .. The series form solution is given by 1

2

4

4

1

𝑦(𝑥, 𝑡) = 1 + 𝑡 − 4 (− 3 t 2 − 3 − 3 t) 𝑥 4 − 2 (−7 − 7t − 3t 2 + cos(t) t + 1

1

cos(t) t 2 + cos(t)) 𝑥 2 − 5040 𝑥 2 (35280 + 392𝑥 4 𝑐𝑜𝑠(𝑡) 𝑡 + 196𝑥 4 cos(𝑡)𝑡 2 − 2 210𝑥 2 cos 2 (t)t − 105𝑥 2 cos2 (t)t 2 + ⋯, Rest of the components of the iteration formulae can be obtained using the MAPLE.

189

(a)

(b)


Example 2.44. Consider the heat and wave like equation [149] 2

𝑦𝑥𝑥 + 𝑥 𝑦𝑥 − (5 + 4𝑥 2 )𝑦 = 𝑦𝑡 + (6 − 5𝑥 2 − 4𝑥 4 ),

(2.172)

𝑦(0, 𝑡) = e(t) , 𝑦𝑥 (0, 𝑡) = 0. To solve the given Equation by HAM we choose the linear operator 𝜕2


𝑡

𝐿−1 = ∫0 ∫0 ( . )𝑑𝑥𝑑𝑥, we now define a nonlinear operator as 2

𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑥𝑥 + 𝑥 𝜑(𝑥, 𝑡; 𝑞)𝑥 − (5 − 4𝑥 2 )𝜑(𝑥, 𝑡; 𝑞) − (6 − 5𝑥 2 − 4𝑥 4 ) − 𝜑(𝑥, 𝑡; 𝑞)𝑡 using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑦0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝑦0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑦(𝑥, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑦𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑦𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑦𝑚−1 ). with initial condition 𝑦𝑚 (𝑥, 0) = 0. 190

𝜕2

where 𝑅𝑚 (𝑦𝑚−1 ) = ℎ [

𝜕𝑥 2

2 𝜕

𝑦𝑚−1 + 𝑥 𝜕𝑥 𝑦𝑚−1 − (5 − 4𝑥 2 )𝑦𝑚−1 𝜕

−(6 − 5𝑥 2 − 4𝑥 4 ) − 𝜕𝑡 𝑦𝑚−1

].

after applying the Taylor’s series we get 𝜕2

𝑅𝑚 (𝑦𝑚−1 ) = ℎ [

𝜕𝑥 2

2 𝜕

(𝑦𝑚−1 + 𝑓𝑚 ) + 𝑥 𝜕𝑥 (𝑦𝑚−1 + 𝑓𝑚 ) − 𝜕

(5 − 4𝑥 2 )(𝑦𝑚−1 + 𝑓𝑚 ) − (6 − 5𝑥 2 − 4𝑥 4 ) − 𝜕𝑡 (𝑦𝑚−1 + 𝑓𝑚 )

].

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑦𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑦𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑦0 (𝑥, 𝑡) = 1 + 𝑡, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑦0 (𝑥, 𝑡) = 1 + 𝑡, 2

1 1

4

2

1

5

𝑦1 (𝑥, 𝑡) = − 15 𝑥 6 − 4 (3 − 3 t − 3 t 2 ) 𝑥 4 − 2 (−12 − 6t − 2 t 2 ) 𝑥 2 , 1291

43

89

59

𝑦2 (𝑥, 𝑡) = 1800 𝑥 6 + 81 𝑥 4 + 72 𝑥 4 t + 144 𝑥 4 𝑡 2 − 9𝑥 2 − 6𝑥 2 t + ⋯, . . .. The series form solution is given by 2

1 1

4

2

1

5

𝑦(𝑥, 𝑡) = 1 + 𝑡 − 15 𝑥 6 − 4 (3 − 3 t − 3 t 2 ) 𝑥 4 − 2 (−12 − 6t − 2 t 2 ) 𝑥 2 + 1291

43

89

59

𝑥 6 + 81 𝑥 4 + 72 𝑥 4 t + 144 𝑥 4 𝑡 2 − 9𝑥 2 − 6𝑥 2 t + ⋯ 1800 Rest of the components of the iteration formulae can be obtained using the MAPLE.

191

(a)

(b)


Example 2.45. Consider the heat and wave like equation [149] 2

𝑦𝑥𝑥 + 𝑥 𝑦𝑥 − (5 + 4𝑥 2 )𝑦 = 𝑦𝑡𝑡 + (12𝑥 − 5𝑥 3 − 4𝑥 5 ),

(2.173)

𝑦(0, 𝑡) = e(−t) , 𝑦𝑥 (0, 𝑡) = 0. To solve the given Equation by HAM we choose the linear operator 𝜕2


𝑡

𝐿−1 = ∫0 ∫0 ( . )𝑑𝑥𝑑𝑥, we now define a nonlinear operator as 2

𝑁[𝜑(𝑥, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑡; 𝑞)𝑥𝑥 + 𝑥 𝜑(𝑥, 𝑡; 𝑞)𝑥 − (5 − 4𝑥 2 )𝜑(𝑥, 𝑡; 𝑞) − (12𝑥 − 5𝑥 3 − 4𝑥 5 ) − 𝜑(𝑥, 𝑡; 𝑞)𝑡𝑡 using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑡; 𝑞) − 𝑦0 (𝑥, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑡)𝑁[𝜑(𝑥, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑡; 0) = 𝑦0 (𝑥, 𝑡),

𝜑(𝑥, 𝑡; 1) = 𝑦(𝑥, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑦𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑦𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑦𝑚−1 ). with initial condition 𝑦𝑚 (𝑥, 0) = 0. 192

𝜕2

𝑅𝑚 (𝑦𝑚−1 ) = ℎ [

where

2 𝜕

𝑦𝑚−1 + 𝑥 𝜕𝑥 𝑦𝑚−1 − (5 − 4𝑥 2 )𝑦𝑚−1 − ]. 𝜕2 (12𝑥 − 5𝑥 3 − 4𝑥 5 ) − 2 𝑦𝑚−1 𝜕𝑡

𝜕𝑥 2

After applying the Taylor’s series we have 𝜕2 𝜕𝑥 2

2 𝜕

(𝑦𝑚−1 + 𝑓𝑚 ) + 𝑥 𝜕𝑥 (𝑦𝑚−1 + 𝑓𝑚 ) −

𝑅𝑚 (𝑦𝑚−1 ) = ℎ (5 − 4𝑥 2 )(𝑦𝑚−1 + 𝑓𝑚 ) − (12𝑥 − 5𝑥 3 − 4𝑥 5 ) . 𝜕2

− 𝜕𝑡 2 (𝑦𝑚−1 + 𝑓𝑚 )

[

]

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑦𝑚 (𝑥, 𝑡) = 𝜒𝑚 𝑦𝑚−1 (𝑥, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑦0 (𝑥, 𝑡) = 1 − 𝑡, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑦0 (𝑥, 𝑡) = 1 − 𝑡, 2

1

1

4

4

2

1

𝑦1 (𝑥, 𝑡) = − 21 𝑥 7 − 4 𝑥 5 − 4 (− 3 + 3 𝑡 − 3 𝑡 2 ) 𝑥 4 + 2𝑥 3 − 2 (

−6 + 5𝑡 5 ) 𝑥2, − 2 𝑡2

7 3 1 4 4 2 1 −6 + 5𝑡 𝑦2 (𝑥, 𝑡) = 72 𝑥 7 + 8 𝑥 5 − 4 (− 3 + 3 𝑡 − 3 𝑡 2 ) 𝑥 4 − 2 ( 5 2 ) 𝑥 2 + ⋯, −2𝑡

. . .. The series form solution is given by 2

1

1

4

4

2

1

𝑦(𝑥, 𝑡) = 1 − 𝑡 − 21 𝑥 7 − 4 𝑥 5 − 4 (− 3 + 3 𝑡 − 3 𝑡 2 ) 𝑥 4 + 2𝑥 3 − 2 (−6 + 5𝑡 − 5 2 𝑡 ) 𝑥2 2

7

3

+ 72 𝑥 7 + 8 𝑦5 − ⋯,

Rest of the components of the iteration formulae can be obtained using the MAPLE.

193

(a)

(b)


3.10.6. System of Partial Differential Equations Example 2.46. Consider the following linear PDEs [230] 𝑢𝑡 + 𝑣𝑥 − (𝑢 − 𝑣) = 0,

(2.174)

𝑣𝑡 + 𝑢𝑥 − (𝑢 + 𝑣) = 0,

(2.175)

with initial conditions 𝑢(𝑥, 0) = 𝑠𝑖𝑛ℎ𝑥, 𝑣(𝑥, 0) = 𝑐𝑜𝑠ℎ𝑥. To solve the given Equation by MHAM we choose the linear operator 𝜕

𝐿[𝜑1 (𝑥, 𝑡; 𝑞1 )] = 𝜕𝑡 (𝜑1 (𝑥, 𝑡; 𝑞1 )), 𝜕


𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑1 (𝑥, 𝑡; 𝑞1 )] = 𝜑1 (𝑥, 𝑡; 𝑞1 )𝑡 + 𝜑2 (𝑥, 𝑦; 𝑞1 )𝑥 − (𝜑1 (𝑥, 𝑡; 𝑞1 ) + 𝜑2 (𝑥, 𝑡; 𝑞1 )), 𝑁[𝜑2 (𝑥, 𝑡; 𝑞2 )] = 𝜑2 (𝑥, 𝑡; 𝑞2 )𝑡 + 𝜑1 (𝑥, 𝑡; 𝑞2 )𝑥 − (𝜑1 (𝑥, 𝑡; 𝑞2 ) + 𝜑2 (𝑥, 𝑡; 𝑞2 )), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞1 )𝐿[𝜑1 (𝑥, 𝑡; 𝑞1 ) − 𝑢0 (𝑥, 𝑡)] = 𝑞1 ℎ𝐻(𝑥, 𝑡; 𝑞1 )𝑁[𝜑1 (𝑥, 𝑡; 𝑞1 )]. 194

for 𝑞1 = 0 and 𝑞1 = 1, we can write 𝜑1 (𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑1 (𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).

(1 − 𝑞2 )𝐿[𝜑2 (𝑥, 𝑡; 𝑞2 ) − 𝑢0 (𝑥, 𝑡)] = 𝑞1 ℎ𝐻(𝑥, 𝑡; 𝑞2 )𝑁[𝑦2 (𝑥, 𝑡; 𝑞2 )]. for 𝑞2 = 0 and 𝑞2 = 1, we can write 𝜑2 (𝑥, 𝑡; 0) = 𝑢0 (𝑥, 𝑡),

𝜑2 (𝑥, 𝑡; 1) = 𝑢(𝑥, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equations 𝐿[𝑢𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). 𝐿[𝑣𝑚 (𝑥, 𝑡) − 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑡)] = ℎ𝐻(𝑥, 𝑡)𝑅𝑚 (𝑣𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 0) = 0, 𝑣𝑚 (𝑥, 0) = 0. Apply the Taylor series on initial condition and choose 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) and 𝑅𝑚 (𝑣𝑚−1 ) accordingly. we start with an initial approximation 𝑢0 (𝑥, 𝑡) = 𝑥, 𝑣0 (𝑥, 𝑡) = 1, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢0 (𝑥, 𝑡) = 𝑥, 𝑣0 (𝑥, 𝑡) = 1, 1

1

𝑢1 (𝑥, 𝑡) = 𝑡 + 6 𝑥 3 𝑡 + 2 𝑥 2 𝑡, 1

𝑣1 (𝑥, 𝑡) = 𝑥𝑡 + 6 𝑥 3 𝑡, 1

1

1

1

1

𝑢2 (𝑥, 𝑡) = − 6 𝑥 3 𝑡 + 6 𝑥 3 𝑡 2 + 2 𝑥𝑡 2 + 120 𝑥 5 𝑡 + 24 𝑥 4 𝑡, 1

1

1

𝑣2 (𝑥, 𝑡) = 2 𝑡 2 + 6 𝑥 3 𝑡 2 + 120 𝑥 5 𝑡, . . .. The series form solution is given by 1

1

1

1

1

1

1

𝑢(𝑥, 𝑡) = 𝑥 + 𝑡 + 6 𝑥 3 𝑡 + 2 𝑥 2 𝑡 − 6 𝑥 3 𝑡 + 6 𝑥 3 𝑡 2 + 2 𝑥𝑡 2 + 120 𝑥 5 𝑡 + 24 𝑥 4 𝑡 + ⋯ 1

1

1

1

𝑣(𝑥, 𝑡) = 1 + 𝑥𝑡 + 6 𝑥 3 𝑡 + 2 𝑡 2 + 6 𝑥 3 𝑡 2 + 120 𝑥 5 𝑡 + ⋯ and the close form solution is 195

𝑢(𝑥, 𝑡) = sinh(𝑥 + 𝑡). 𝑣(𝑥, 𝑡) = cosh(𝑥 + 𝑡). Rest of the components of the iteration formulae can be obtained using the Maple.

(a)

(b)

(c)

(d)

Fig. 2.81.(a)-(d):Comparison of Exact and Approximate solution of Eq. (2.174-2.175)

Example 2.47. We first consider the following system of initial value problems [230] 𝑢𝑡 + 𝑢𝑥 + 2𝑤 = 0,

(2.176)

𝑣𝑡 + 𝑣𝑥 + 2𝑢 = 0,

(2.177)

𝑤𝑡 + 𝑤𝑥 − 2𝑢 = 0,

(2.178)

with initial conditions 𝑢(𝑥, 𝑦, 0) = sin(𝑥 + 𝑦), 196

𝑣(𝑥, 𝑦, 0) = cos(𝑥 + 𝑦). 𝑤(𝑥, 𝑦, 0) = −cos(𝑥 + 𝑦). To solve the given Equation by HAM we choose the linear operator 𝜕

𝐿[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )] = 𝜕𝑡 (𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )), 𝜕

𝐿[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )] = 𝜕𝑡 (𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )), 𝜕


𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )] = 𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )𝑡 + 𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )𝑥 + 2𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 ), 𝑁[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )] = 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑡 + 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑥 + 2𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ), 𝑁[𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )] = 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )𝑡 + 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )𝑥 − 2𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞1 )𝐿[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ) − 𝑢0 (𝑥, 𝑦, 𝑡)] = 𝑞1 ℎ𝐻(𝑥, 𝑦, 𝑡; 𝑞1 )𝑁[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )]. for 𝑞1 = 0 and 𝑞1 = 1, we can write 𝜑1 (𝑥, 𝑦, 𝑡; 0) = 𝑢0 (𝑥, 𝑦, 𝑡),

𝜑1 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).


𝜑2 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).


𝜑3 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equations 𝐿[𝑢𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). 𝐿[𝑣𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑣𝑚−1 ). 𝐿[𝑤𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑤𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ). 197

with initial condition 𝑢𝑚 (𝑥, 𝑦, 0) = 0, 𝑣𝑚 (𝑥, 𝑦, 0) = 0, 𝑤𝑚 (𝑥, 𝑦, 0) = 0. Apply the Taylor series on initial condition and choose𝑅𝑚 (𝑢 ⃗ 𝑚−1 ), 𝑅𝑚 (𝑣𝑚−1 ) and 𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ) accordingly. we start with an initial approximation 𝑢0 (𝑥, 𝑦, 𝑡) = sin(𝑥) + cos(𝑥)𝑦, 𝑣0 (𝑥, 𝑦, 𝑡) = cos(𝑥) − sin(𝑥) 𝑦, 𝑤0 (𝑥, 𝑦, 𝑡) = −cos(𝑥) + sin(𝑥)𝑦, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢1 (𝑥, 𝑦, 𝑡) = 𝑣1 (𝑥, 𝑦, 𝑡) =

1 6 1 6

𝑡(6 cos(𝑥) − 6 sin(𝑥) 𝑦 − 3 cos(𝑥) 𝑦 2 + sin(𝑥)𝑦 3 ), 𝑡(−6 sin(𝑥) − 6 cos(𝑥) 𝑦 + 3 sin(𝑥) 𝑦 2 + 𝑐𝑜s(𝑥)𝑦 3 ), 1

𝑤1 (𝑥, 𝑦, 𝑡) = − 6 𝑡(−6 sin(𝑥) − 6 cos(𝑥) 𝑦 + 3 sin(𝑥) 𝑦 2 + 𝑐𝑜s(𝑥)𝑦 3 ), 𝑢2 (𝑥, 𝑦, 𝑡) =

1

𝑡(30𝑡𝑠𝑖𝑛(𝑥)𝑦 2 + 10𝑡 cos(𝑥) 𝑦 3 − 60𝑡𝑠𝑖𝑛(𝑥) + ⋯,

120

1

𝑣2 (𝑥, 𝑦, 𝑡) = − 120 𝑡(60𝑡𝑐𝑜𝑠(𝑥) − 60𝑡 sin(𝑥) 𝑦 − 30𝑡 cos(𝑥) 𝑦 2 + ⋯, 1

𝑤2 (𝑥, 𝑦, 𝑡) = − 120 𝑡(60𝑡𝑐𝑜𝑠(𝑥) − 60𝑡 sin(𝑥) 𝑦 − 30𝑡 cos(𝑥) 𝑦 2 + ⋯, . . .. The series form solution is given by 𝑢(𝑥, 𝑦, 𝑡) = sin(𝑥) + cos(𝑥) 𝑦 + sin(𝑥) 𝑦 3 ) +

1 120

1 6

𝑡(6 cos(𝑥) − 6 sin(𝑥) 𝑦 − 3 cos(𝑥) 𝑦 2 +

𝑡(30𝑡𝑠𝑖𝑛(𝑥)𝑦 2 + 10𝑡 cos(𝑥) 𝑦 3 − 60𝑡𝑠𝑖𝑛(𝑥) + ⋯, 1

𝑣(𝑥, 𝑦, 𝑡) = cos(𝑥) − sin(𝑥) 𝑦 + 6 𝑡(−6 sin(𝑥) − 6 cos(𝑥) 𝑦 + 3 sin(𝑥) 𝑦 2 + cos(𝑥)𝑦 3 ) −

1 120

𝑡(60𝑡𝑐𝑜𝑠(𝑥) − 60𝑡 sin(𝑥) 𝑦 − 30𝑡 cos(𝑥) 𝑦 2 + ⋯, 1

𝑤(𝑥, 𝑦, 𝑡) = −cos(𝑥) + sin(𝑥)𝑦 − 6 𝑡(−6 sin(𝑥) − 6 cos(𝑥) 𝑦 + 3 sin(𝑥) 𝑦 2 + cos(𝑥)𝑦 3 ) −

1 120

𝑡(60𝑡𝑐𝑜𝑠(𝑥) − 60𝑡 sin(𝑥) 𝑦 − 30𝑡 cos(𝑥) 𝑦 2 + ⋯,

and the close form solution is 𝑢(𝑥, 𝑦, 𝑡) = 𝑠𝑖𝑛(𝑥 + 𝑦 + 𝑡). 198

𝑣(𝑥, 𝑦, 𝑡) = 𝑐𝑜𝑠(𝑥 + 𝑦 + 𝑡). 𝑤(𝑥, 𝑦, 𝑡) = − 𝑐𝑜𝑠(𝑥 + 𝑦 + 𝑡). Rest of the components of the iteration formulae can be obtained using the Maple.

Example 2.48. We first consider the following system of initial value problems [230] 𝑢𝑡 + 𝑣𝑥 − 𝑤𝑦 = 𝑤,

(2.179)

𝑣𝑡 + 𝑤𝑥 + 𝑢𝑦 = 𝑢,

(2.180)

𝑤𝑡 + 𝑣𝑥 − 𝑣𝑦 = 𝑣,

(2.181)

with initial conditions 𝑢(𝑥, 𝑦, 0) = sin(𝑥 + 𝑦). 𝑣(𝑥, 𝑦, 0) = cos(𝑥 + 𝑦). 𝑤(𝑥, 𝑦, 0) = −sin(𝑥 + 𝑦). To solve the given Equation by HAM we choose the linear operator 𝜕

𝐿[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )] = 𝜕𝑡 (𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )), 𝜕

𝐿[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )] = 𝜕𝑡 (𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )), 𝜕


𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )] = 𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )𝑡 + 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑥 −𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )𝑦 − 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 ), 𝑁[𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )] = 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑡 + 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )𝑥 +𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑦 − 𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ), 𝑁[𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )] = 𝜑3 (𝑥, 𝑦, 𝑡; 𝑞3 )𝑡 + 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑥 −𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 )𝑦 − 𝜑2 (𝑥, 𝑦, 𝑡; 𝑞2 ), using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞1 )𝐿[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 ) − 𝑢0 (𝑥, 𝑦, 𝑡)] = 𝑞1 ℎ𝑦(𝑥, 𝑦, 𝑡; 𝑞1 )𝑁[𝜑1 (𝑥, 𝑦, 𝑡; 𝑞1 )]. for 𝑞1 = 0 and 𝑞1 = 1, we can write 199

𝜑1 (𝑥, 𝑦, 𝑡; 0) = 𝑢0 (𝑥, 𝑦, 𝑡),

𝜑1 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).


𝜑2 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).


𝜑3 (𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equations 𝐿[𝑢𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻 (𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). 𝐿[𝑣𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑣𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑣𝑚−1 ). 𝐿[𝑤𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑤𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 𝑦, 0) = 0, 𝑣𝑚 (𝑥, 𝑦, 0) = 0, 𝑤𝑚 (𝑥, 𝑦, 0) = 0. Apply the Taylor series on initial condition and choose𝑅𝑚 (𝑢 ⃗ 𝑚−1 ), 𝑅𝑚 (𝑣𝑚−1 ) and 𝑅𝑚 (𝑤 ⃗⃗ 𝑚−1 ) accordingly. we start with an initial approximation 𝑢0 (𝑥, 𝑦, 𝑡) = sin(𝑥) + cos(𝑥)𝑦, 𝑣0 (𝑥, 𝑦, 𝑡) = cos(𝑥) − sin(𝑥) 𝑦, 𝑤0 (𝑥, 𝑦, 𝑡) = −sin(𝑥) − 𝑐𝑜s(𝑥)𝑦, by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢1 (𝑥, 𝑦, 𝑡) =

1 2

𝑡(−2 cos(𝑥) + 2 sin(𝑥) 𝑦 + cos(𝑥) 𝑦 2 ), 1

𝑣1 (𝑥, 𝑦, 𝑡) = − 6 𝑡(− sin(𝑥)𝑦 3 − 6 cos(𝑥) − 6cos(𝑥)𝑦 + 3 sin(𝑥) 𝑦 2 + 𝑐𝑜s(𝑥)𝑦 3 ), cos(𝑥) 𝑦 3 − 6 cos(𝑥) + 6 sin(𝑥) 𝑦 𝑤1 (𝑥, 𝑦, 𝑡) = − 6 𝑡 ( ), +3𝑐𝑜s(𝑥)𝑦 2 − sin(𝑥) 𝑦 3 1

1

𝑢2 (𝑥, 𝑦, 𝑡) = − 24 𝑡(12𝑡 cos(𝑥)𝑦 − 6 tsin(𝑥)𝑦 2 + 6𝑡 cos(𝑥) 𝑦 2 +12𝑡𝑠𝑖𝑛(𝑥) + ⋯, 1

𝑣2 (𝑥, 𝑦, 𝑡) = − 120 𝑡(20 sin(𝑥)𝑦 3 + 10𝑡 sin(𝑥)𝑦 3 + 30𝑡 cos(𝑥) 𝑦 2 −60𝑡 sin(𝑥) 𝑦 + ⋯,

200

1

𝑤2 (𝑥, 𝑦, 𝑡) = − 120 𝑡(−20 cos(𝑥) 𝑦 3 − 60𝑡 s in(𝑥) − 60𝑡 cos(𝑥) 𝑦 +20𝑡𝑐𝑜s(𝑥)𝑦 3 + ⋯, . . .. The series form solution is given by 1

𝑢(𝑥, 𝑦, 𝑡) = sin(𝑥) + cos(𝑥) 𝑦 + 2 𝑡(−2 cos(𝑥) + 2 sin(𝑥) 𝑦 + cos(𝑥) 𝑦 2 ) − 1 24

𝑡(12𝑡 cos(𝑥)𝑦 − 6 tsin(𝑥)𝑦 2 + 6𝑡 cos(𝑥) 𝑦 2 + ⋯, 1

𝑣(𝑥, 𝑦, 𝑡) = cos(𝑥) − sin(𝑥) 𝑦 + − 6 𝑡(− sin(𝑥)𝑦 3 − 6 cos(𝑥) − 6cos(𝑥)𝑦 + 1

3 sin(𝑥) 𝑦 2 + 𝑐𝑜s(𝑥)𝑦 3 ) − 120 𝑡(20 sin(𝑥)𝑦 3 + ⋯, 1

𝑤(𝑥, 𝑦, 𝑡) = −sin(𝑥) − 𝑐𝑜s(𝑥)𝑦 − − 6 𝑡(cos(𝑥) 𝑦 3 − 6 cos(𝑥) + 6 sin(𝑥) 𝑦 + 1

3𝑐𝑜s(𝑥)𝑦 2 − sin(𝑥)𝑦 3 ) − 120 𝑡(−20 cos(𝑥) 𝑦 3 + ⋯, and the close form solution is 𝑢(𝑥, 𝑦, 𝑡) = − 𝑠𝑖𝑛(𝑥 − 𝑦 − 𝑡). 𝑣(𝑥, 𝑦, 𝑡) = − 𝑠𝑖𝑛(−𝑥 − 𝑦 + 𝑡). 𝑤(𝑥, 𝑦, 𝑡) = − 𝑠𝑖𝑛(𝑥 − 𝑦 + 𝑡).

2.10.7. ZK(2,2,2) Equation Example 2.49. Consider the ZK(2,2,2) equation [26] 1

1

𝑢𝑡 + (𝑢2 )𝑥 + 8 (𝑢2 )𝑥𝑥𝑥 + 8 (𝑢2 )𝑦𝑦𝑥 = 0,

(2.182)

4

𝑢(𝑥, 𝑦, 0) = 3 𝜆 sinh2(𝑥 + 𝑦), To solve the given Equation by HAM we choose the linear operator 𝜕

𝐿[𝜑(𝑥, 𝑦, 𝑡; 𝑞)] = 𝜕𝑡 (𝜑(𝑥, 𝑦, 𝑡; 𝑞)), with the property 𝐿[𝑐1 + 𝑡𝑐2 ] = 0, where 𝑐1 and 𝑐2 are the integral constants. The inverse operator 𝐿−1 is given by 𝑡


1

𝑁[𝜑(𝑥, 𝑦, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑦, 𝑡; 𝑞)𝑡 + (𝜑(𝑦, 𝑦, 𝑡; 𝑞)2 )𝑥 + 8 (𝜑(𝑥, 𝑦, 𝑡; 𝑞)2 )𝑥𝑥𝑥 + 1 8

(𝜑(𝑥, 𝑦, 𝑡; 𝑞)2 )𝑦𝑦𝑥

using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑦, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑦, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑦, 𝑡)𝑁[𝜑(𝑥, 𝑦, 𝑡; 𝑞)]. for 𝑦 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑦, 𝑡; 0) = 𝑢0 (𝑥, 𝑦, 𝑡),

𝜑(𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑢𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 𝑦, 0) = 0. 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ [(𝑢𝑚−1 (𝑥, 𝑦, 𝑡))𝑡 + (∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))𝑥 +

Where 1 8

(∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))

1

𝑥𝑥𝑥

+ 8 (∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))𝑦𝑦𝑥 ].

after applying the Taylor’s series we get (𝑢𝑚−1 (𝑥, 𝑦, 𝑡) + 𝑓𝑚 )𝑡 + (∑𝑚−1 𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 ) (𝑥, × (𝑢𝑚−1−𝑟 𝑦, 𝑡) + 𝑓𝑚−𝑟 ))𝑥 + 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ

1

. (∑𝑚−1 𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 )(𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑚−𝑟 ))𝑥𝑥𝑥

8 1

𝑚−1

[+ 8 (∑𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 )(𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑚−𝑟 ))𝑦𝑦𝑥 ] Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑢0 (𝑥, 𝑡) =

4 3

𝑠𝑖𝑛ℎ2 (𝑦),

by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢1 = −

32 2 𝜆 (−14𝑥𝑐𝑜𝑠ℎ2 (𝑦) 9

+ 2𝑥 + 10𝑥𝑠𝑖𝑛ℎ(𝑦) 𝑐𝑜𝑠ℎ(𝑦) 𝑠𝑖𝑛ℎ(𝑥 +

𝑦) 𝑐𝑜𝑠ℎ(𝑥 + 𝑦) − 8 𝑠𝑖𝑛ℎ(𝑥 + 𝑦) 𝑐𝑜𝑠ℎ(𝑥 + 𝑦) − 5𝑠𝑖𝑛ℎ(𝑦) 𝑐𝑜𝑠ℎ(𝑦) − 2𝑥𝑐𝑜𝑠ℎ2 (𝑥 + 𝑦) + 10 𝑠𝑖𝑛ℎ(𝑥 + 𝑦) 𝑐𝑜𝑠ℎ3 (𝑥 + 𝑦) + 12𝑥𝑐𝑜𝑠ℎ4 (𝑥 + 𝑦) + 7 𝑠𝑖𝑛ℎ(𝑦) 𝑐𝑜𝑠ℎ(𝑦) 𝑐𝑜𝑠ℎ2 (𝑥 + 𝑠𝑦) + 2 𝑠𝑖𝑛ℎ(𝑥 + 𝑦) 𝑐𝑜𝑠ℎ(𝑥 + 𝑦) 𝑐𝑜𝑠ℎ2 (𝑦) + 4𝑥𝑐𝑜𝑠ℎ2 (𝑦)𝑐𝑜𝑠ℎ2 (𝑥 + 𝑦)𝑡),

202

16

𝑢2 = 27 𝜆2 𝑡(−84𝑥𝑐𝑜𝑠ℎ2 (𝑦)𝑐𝑜𝑠ℎ2 (𝑥 + 𝑦) − 18𝑠𝑖𝑛ℎ(𝑥 + 𝑦) 𝑐𝑜𝑠ℎ(𝑥 + 𝑦) 𝑐𝑜𝑠ℎ2 (𝑦) − 24𝑥𝑠𝑖𝑛ℎ(𝑦) 𝑐𝑜𝑠ℎ(𝑦) 𝑠𝑖𝑛ℎ(𝑥 + 𝑦) 𝑐𝑜ℎ(𝑥 + 𝑦) + ⋯ . . .. The series form solution is given by 4

𝑢(𝑥, 𝑡) = 3 𝑠𝑖𝑛ℎ2 (𝑦) + −

32 2 𝜆 (−14𝑥𝑐𝑜𝑠ℎ2 (𝑦) 9

+ 2𝑥 + 10𝑥𝑠𝑖𝑛ℎ(𝑦) 𝑐𝑜𝑠ℎ(𝑦) 𝑠𝑖𝑛ℎ(𝑥 +

𝑦) 𝑐𝑜𝑠ℎ(𝑥 + 𝑦) − 8 𝑠𝑖𝑛ℎ(𝑥 + 𝑦) 𝑐𝑜𝑠ℎ(𝑥 + 𝑦) − 5𝑠𝑖𝑛ℎ(𝑦) 𝑐𝑜𝑠ℎ(𝑦) − 2𝑥𝑐𝑜𝑠ℎ2 (𝑥 + 𝑦) + 10 𝑠𝑖𝑛ℎ(𝑥 + 𝑦) 𝑐𝑜𝑠ℎ3 (𝑥 + 𝑦) + ⋯ Rest of the components of the iteration formulae can be obtained using the MAPLE.

Fig. 2.82:Comparison of Exact and Approximate solution of Eq. (2.182)

Example 2.50. Consider the following ZK nonlinear [26] 1

1

𝑢𝑡 + (𝑢2 )𝑥 + 8 (𝑢2 )𝑥𝑥𝑥 + 8 (𝑢2 )𝑦𝑦𝑥 = 0,

(2.183)

4

𝑢(𝑥, 𝑦, 0) = − 3 𝜆 cosh2 (𝑥 + 𝑦), To solve the given Equation by HAM we choose the linear operator 𝜕


𝐿−1 = ∫0 (. )𝑑𝑡, 203

we now define a nonlinear operator as 1

𝑁[𝜑(𝑥, 𝑦, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑦, 𝑡; 𝑞)𝑡 + (𝜑(𝑥, 𝑦, 𝑡; 𝑞)2 )𝑥 + 8 (𝜑(𝑥, 𝑦, 𝑡; 𝑞)2 )𝑥𝑥𝑥 + 1 8


using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑦, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑦, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑦, 𝑦)𝑁[𝜑(𝑥, 𝑦, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑦, 𝑡; 0) = 𝑢0 (𝑥, 𝑦, 𝑡),

𝜑(𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑢𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑦−1 ). with initial condition 𝑢𝑚 (𝑥, 𝑦, 0) = 0. 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ [(𝑢𝑚−1 (𝑥, 𝑦, 𝑡))𝑡 + (∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))𝑥 +

Where 1 8

(∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))

1

𝑥𝑥𝑥

+ 8 (∑𝑚−1 𝑟=0 𝑢𝑦 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))𝑦𝑦𝑥 ].

after applying the Taylor’s series we get (𝑢𝑚−1 (𝑥, 𝑦, 𝑡) + 𝑓𝑚 )𝑡 + (∑𝑚−1 𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 ) 𝑚−1 1 ∑𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 ) (𝑢 (𝑥, )) 𝑦, 𝑡) + 𝑓 + ( ) 𝑚−1−𝑟 𝑚−𝑟 𝑥 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ . 8 (𝑢 𝑚−1−𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑚−𝑟 ) 𝑥𝑥𝑥 1

𝑚−1 [ + 8 (∑𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 )(𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑚−𝑟 ))𝑦𝑦𝑥 ]

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 4

𝑢0 (𝑥, 𝑡) = − 3 𝜆𝑐𝑜𝑠ℎ2 (𝑦), by means of the iteration formula as discuss above, if ℎ = −1, 𝐻 = 1, we can obtain directly the others components as 𝑢1 = −

16 2 𝜆 (−3 cosh(𝑦) sinh(𝑦) + 9

12cosℎ3 (𝑦) sinh(𝑦) + 2𝑥 + 24𝑐𝑜𝑠ℎ4 (𝑦)𝑥 −

24𝑐𝑜𝑠ℎ2 (𝑦)𝑥)𝑡), 16

𝑢2 = − 27 𝜆2 𝑡(−9 cosh(𝑦) sinh(𝑦) + 18cosℎ3 (𝑦) sinh(𝑦) + 3𝑥 + 72𝑐𝑜𝑠ℎ4 (𝑦)𝑥 − 54𝑐𝑜𝑠ℎ2 (𝑦)𝑦 + 336𝜆 cosh(𝑦) sinh(𝑦) 𝑡𝑥 + ⋯, . . 204

.. The series form solution is given by 4

𝑢(𝑥, 𝑡) = − 3 𝑐𝑜𝑠ℎ2 (𝑦) −

16 2 𝜆 (−3 cosh(𝑦) sinh(𝑦) 9

+ 12cosℎ3 (𝑦) sinh(𝑦) +

16

2𝑥 + 24𝑐𝑜𝑠ℎ4 (𝑦)𝑥 − 24𝑐𝑜𝑠ℎ2 (𝑦)𝑥)𝑡) + − 27 𝜆2 𝑡(−9 cosh(𝑦) sinh(𝑦) − 16 2 𝜆 𝑡(−9 cosh(𝑦) sinh(𝑦) + 27

⋯


Fig. 2.83.:Comparison of Exact and Approximate solution of Eq. (2.183)

2.10.8. ZK(3,3,3) Equation Example 2.51. Consider the following nonlinear PDE [101] 𝑢𝑡 + (𝑢3 )𝑥 + 2(𝑢3 )𝑥𝑥𝑥 + 2(𝑢3 )𝑦𝑦𝑥 = 0, 3

(2.184)

1

𝑢(𝑥, 𝑦, 0) = 2 𝜆 sinh (6 (𝑥 + 𝑦)), To solve the given Equation by HAM we choose the linear operator 𝜕


𝐿−1 = ∫0 (. )𝑑𝑡, we now define a nonlinear operator as 𝑁[𝜑(𝑥, 𝑦, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑦, 𝑡; 𝑞)𝑡 + (𝜑(𝑥, 𝑦, 𝑡; 𝑞)2 )𝑥 + 2(𝜑(𝑥, 𝑦, 𝑡; 𝑞)2 )𝑥𝑥𝑥 + 2(𝜑(𝑥, 𝑦, 𝑡; 𝑞)2 )𝑦𝑦𝑥 205

using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑦, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑦, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑦, 𝑡)𝑁[𝜑(𝑥, 𝑦, 𝑡; 𝑞)]. for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑦, 𝑡; 0) = 𝑢0 (𝑥, 𝑦, 𝑡),

𝜑(𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑢𝑚 (𝑥, 𝑦, 𝑦) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 𝑦, 0) = 0. 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ [(𝑢𝑚−1 (𝑥, 𝑦, 𝑡))𝑡 + (∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))𝑥 +

Where

𝑚−1 2(∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))𝑥𝑥𝑥 + 2(∑𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))𝑦𝑦𝑥 ].

after applying the Taylor’s series we get (𝑢𝑚−1 (𝑥, 𝑦, 𝑡) + 𝑓𝑚 )𝑡 + (∑𝑚−1 𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 ) ∑𝑚−1 𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 ) ) 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ (𝑢𝑚−1−𝑦 (𝑥, 𝑦, 𝑡) + 𝑓𝑚−𝑟 ))𝑥 + 2 ( . (𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑚−𝑟 ) 𝑥𝑥𝑥 𝑚−1 [ +2(∑𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 )(𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑚−𝑟 ))𝑦𝑦𝑥 ] Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑢0 (𝑥, 𝑡) =

3 2

1

𝜆𝑠𝑖𝑛ℎ (6 𝑦),


1

1

1

1

𝑢1 = − 128 𝜆3 (−8 𝑠𝑖𝑛ℎ (6 𝑦) 𝑥 + 108 𝑠𝑖𝑛ℎ (6 𝑦) 𝑥𝑐𝑜𝑠ℎ2 (6 𝑦) − 300 𝑐𝑜𝑠ℎ (6 𝑦) + 1

1

1

9𝑥 2 𝑐𝑜𝑠 ℎ3 (6 𝑦) + 348𝑐𝑜𝑠ℎ3 (6 𝑦) − 2𝑐𝑜𝑠ℎ (6 𝑦) 𝑥 2 ), 1

1

1

1

𝑢2 = 18432 𝜆3 𝑡(6048 𝑠𝑖𝑛ℎ (6 𝑦) 𝑥 − 11232 𝑠𝑖𝑛ℎ (6 𝑦) 𝑥𝑐𝑜𝑠ℎ2 (6 𝑦) 1

1

1

+10368 𝑐𝑜𝑠ℎ (6 𝑦) − 3888𝑥 2 − 10368𝑐𝑜𝑠ℎ3 (6 𝑦) + 3600𝑐𝑜𝑠ℎ (6 𝑦) 𝑥 2 + ⋯, . . .. The series form solution is given by

206

𝑢(𝑥, 𝑡) =

3

1

1

1

1

1

𝜆𝑠𝑖𝑛ℎ (6 𝑦) − 128 𝜆3 (−8 𝑠𝑖𝑛ℎ (6 𝑦) 𝑥 + 108 𝑠𝑖𝑛ℎ (6 𝑦) 𝑥𝑐𝑜𝑠ℎ2 (6 𝑦) − 2 1

1

1

1

300 𝑐𝑜𝑠ℎ (6 𝑦) + 9𝑥 2 𝑐𝑜𝑠 ℎ3 (6 𝑦) + 348𝑐𝑜𝑠ℎ3 (6 𝑦) − 2𝑐𝑜𝑠ℎ (6 𝑦) 𝑥 2 ) + 1 18432

1

𝜆3 𝑡(6048 𝑠𝑖𝑛ℎ (6 𝑦) 𝑥 − ⋯.



Example 2.52. Consider the following nonlinear PDE [101] 1

1

𝑢𝑡 + (𝑢3 )𝑥 + 8 (𝑢3 )𝑥𝑥𝑥 + 8 (𝑢3 )𝑦𝑦𝑥 = 0, 3

(2.185)

1

𝑢(𝑥, 𝑦, 0) = 2 𝜆 cosh (6 (𝑥 + 𝑦)), To solve the given Equation by HAM we choose the linear operator 𝜕



𝑁[𝜑(𝑥, 𝑦, 𝑡; 𝑞)] = 𝜑(𝑥, 𝑦, 𝑡; 𝑞)𝑡 + (𝜑(𝑥, 𝑦, 𝑡; 𝑞)2 )𝑥 + 8 (𝜑(𝑥, 𝑦, 𝑡; 𝑞)2 )𝑦𝑥𝑥 + 1 8


using the above definition, we construct the zeroth-order deformation equation (1 − 𝑞)𝐿[𝜑(𝑥, 𝑦, 𝑡; 𝑞) − 𝑢0 (𝑥, 𝑦, 𝑡)] = 𝑞ℎ𝐻(𝑥, 𝑦, 𝑡)𝑁[𝜑(𝑥, 𝑦, 𝑡; 𝑞)]. 207

for 𝑞 = 0 and 𝑞 = 1, we can write 𝜑(𝑥, 𝑦, 𝑡; 0) = 𝑢0 (𝑥, 𝑦, 𝑡),

𝜑(𝑥, 𝑦, 𝑡; 1) = 𝑢(𝑥, 𝑦, 𝑡).

Thus, we obtain the 𝑚𝑡ℎ −order deformation equation 𝐿[𝑢𝑚 (𝑥, 𝑦, 𝑡) − 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡)] = ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 ). with initial condition 𝑢𝑚 (𝑥, 𝑦, 0) = 0. 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ [(𝑢𝑚−1 (𝑥, 𝑦, 𝑡))𝑡 + (∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))𝑥 +

Where 1 8

(∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))

1

𝑥𝑥𝑥

+ 8 (∑𝑚−1 𝑟=0 𝑢𝑟 (𝑥, 𝑦, 𝑡) 𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡))𝑦𝑦𝑥 ].

after applying the Taylor’s series we get (𝑢𝑚−1 (𝑥, 𝑦, 𝑡) + 𝑓𝑚 )𝑡 + (∑𝑚−1 𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 ) 𝑚−1 1 ∑𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 ) (𝑢 (𝑥, )) 𝑦, 𝑡) + 𝑓 + ( ) 𝑚−1−𝑟 𝑚−𝑟 𝑥 𝑅𝑚 (𝑢 ⃗ 𝑚−1 ) = ℎ . 8 (𝑢 𝑚−1−𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑚−𝑟 ) 𝑥𝑥𝑥 1

𝑚−1 [ + 8 (∑𝑟=0 (𝑢𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑟+1 )(𝑢𝑚−1−𝑟 (𝑥, 𝑦, 𝑡) + 𝑓𝑚−𝑟 ))𝑦𝑦𝑥 ]

Now the solutions of the 𝑚𝑡ℎ −order deformation equation are 𝑢𝑚 (𝑥, 𝑦, 𝑡) = 𝜒𝑚 𝑢𝑚−1 (𝑥, 𝑦, 𝑡) + 𝐿−1 [ℎ𝐻(𝑥, 𝑦, 𝑡)𝑅𝑚 (𝑢 ⃗ 𝑚−1 )], 𝑚 ≥ 1, we start with an initial approximation 𝑢0 (𝑥, 𝑡) =

3 2

1

𝜆𝑐𝑜𝑠ℎ (6 𝑦),


1

1

1

𝑢1 = − 2048 𝜆3 (3588 𝑠𝑖𝑛ℎ (6 𝑦) 𝑐𝑜𝑠ℎ2 (6 𝑦) − 97 𝑠𝑖𝑛ℎ (6 𝑦) 𝑥 2 − 1

1

1

1

48 𝑠𝑖𝑛ℎ (6 𝑦) + 99𝑥 2 𝑠𝑖𝑛ℎ (6 𝑦) 𝑐𝑜𝑠 ℎ2 (6 𝑦) − 1180𝑐𝑜𝑠 ℎ (6 𝑦) 𝑥 + 1

1188𝑐𝑜𝑠ℎ3 (6 𝑦) 𝑥), 𝑢2 =

1 4718592

1

1

𝜆3 𝑡(−165888 𝑠𝑖𝑛ℎ (6 𝑦) 𝑐𝑜𝑠ℎ2 (6 𝑦) + ⋯, . . ..

The series form solution is given by

208

𝑢(𝑥, 𝑡) =

3

1

1

1

1

𝜆𝑐𝑜𝑠ℎ (6 𝑦) − 2048 𝜆3 (3588 𝑠𝑖𝑛ℎ (6 𝑦) 𝑐𝑜𝑠ℎ2 (6 𝑦) − 2

1

1

1

1

1

97 𝑠𝑖𝑛ℎ (6 𝑦) 𝑥 2 − 48 𝑠𝑖𝑛ℎ (6 𝑦) + 99𝑥 2 𝑠𝑖𝑛ℎ (6 𝑦) 𝑐𝑜𝑠 ℎ2 (6 𝑦) − 1180𝑐𝑜𝑠 ℎ (6 𝑦) 𝑥 + 1

1188𝑐𝑜𝑠ℎ3 (6 𝑦) 𝑥) +

1 4718592

1

1

𝜆3 𝑡(−165888 𝑠𝑖𝑛ℎ (6 𝑦) 𝑐𝑜𝑠ℎ2 (6 𝑦) + ⋯



2.11. Variation of Parameter Method (VPM) Recently Variation of parameters method has been developed and applied to solving [127-129, 185]. Mohyud-Din et al. applied variation of parameter method for solving nonlinear boundary value problems [147], fifth-order boundary values problems [168], sixth-order boundary value problems [157] and initial and boundary value problems [148]. This suggested algorithm so simple and more effective.

2.11.1. Methodology Consider the 2nd order partial differential equation: 𝑦𝑡𝑡 = 𝑓(𝑡, 𝑥, 𝑦, 𝑧, 𝑦, 𝑦𝑥 , 𝑦𝑦 , 𝑦𝑧 , 𝑦𝑥𝑥 , 𝑦𝑦𝑦 , 𝑦𝑧𝑧 ),

(2.186)

where t such that (−∞ < 𝑡 < ∞) is the time and f is linear or non linear function of 𝑦, 𝑦𝑥 , 𝑦𝑦 , 𝑦𝑧 , 𝑦𝑥𝑥 , 𝑦𝑦𝑦 , 𝑦𝑧𝑧 . The homogenous solution of (2.186) is 𝑦(𝑡, 𝑥, 𝑦, 𝑧) = 𝐴 + 𝐵𝑡, where 𝐴 and 𝐵 are the function of 𝑥, 𝑦, 𝑧 and 𝑡. Using Variational of parameters method we have the following system of equations 𝜕𝐴 𝜕𝑡

+

𝜕𝐵 𝜕𝑡

= 0, 209

𝜕𝐵 𝜕𝑡

= 𝑓.

and hence 𝑡

𝐴(𝑥, 𝑦, 𝑧, 𝑡) = 𝐷 − (𝑥, 𝑦, 𝑧) ∫0 𝑠𝑓ds, 𝑡

𝐵(𝑥, 𝑦, 𝑧, 𝑡) = 𝐶(𝑥, 𝑦, 𝑧) − ∫0 𝑓ds. Therefore, we have 𝑦(𝑥, 𝑦, 𝑧, 𝑡) = 𝑦(𝑥, 𝑦, 𝑧, 0) + 𝑡𝑦𝑡 (𝑥, 𝑦, 𝑧, 0) 𝑡

+ ∫0 (𝑡 − 𝑠) 𝑓(𝑡, 𝑥, 𝑦, 𝑧, 𝑦, 𝑦𝑥 , 𝑦𝑦 , 𝑦𝑧 , 𝑦𝑥𝑥 , 𝑦𝑦𝑦 , 𝑦𝑧𝑧 )ds, which can be solved iteratively as 𝑦𝑘+1 (𝑥, 𝑦, 𝑧, 𝑡) = 𝑦(𝑥, 𝑦, 𝑧, 0) + 𝑡𝑦𝑡 (𝑥, 𝑦, 𝑧, 0) 𝑡

+ ∫0 (𝑡 − 𝑠) 𝑓(𝑡, 𝑥, 𝑦, 𝑧, 𝑦𝑘 , 𝑦𝑘,𝑥 , 𝑦𝑘,𝑦 , 𝑦𝑘,𝑧 , 𝑦𝑘,𝑥𝑥 , 𝑦𝑘,𝑦𝑦 , 𝑦𝑘,𝑧𝑧 )ds, 𝑘 = 0,1,2, …

2.11.2. MHD Flow of an Incompressible Viscous Fluid through Convergent or Divergent Channels in Presence of a High Magnetic Field According to the Variation of parameter method Eq. (2.71)-Eq. (2.72) 𝑓𝑘+1 (𝜂) = 1 −

𝑎𝜂 2 2

𝜂 𝑠2

𝜂2

2

2

− ∫0 ( − 𝑠𝜂 +

) (2𝛼𝑅𝑒𝑓𝑘 (𝑠)𝑓𝑘′ (𝑠) + (4 − 𝐻)𝛼 2 𝑓𝑘′ (𝑠))ds.

Consequently, we have the following approximants 𝑓0 (𝜂) = 1 −

𝑎𝜂 2 2

,

1

1

1

1

1

𝑓1 (𝜂) = 1 + 2 𝑎𝜂2 − 120 𝑅𝑒𝛼𝑎2 𝜂6 − 12 𝜂4 𝑅𝑒𝛼𝑎 − 6 𝜂4 𝛼 2 𝑎 + 24 𝜂4 𝛼 2 𝑎𝐻, 1

1

1

𝑓2 (𝜂) = 1 + 2 𝑎𝜂2 − 95040 𝜂12 𝑅𝑒 3 𝛼 3 𝑎3 − 47520 𝑅𝑒 2 𝛼 4 𝑎3 + ⋯ . . .. 𝑓(𝜂) = lim 𝑓𝑘 (𝜂). 𝑘→∞

1

1

1

𝑓(𝜂) = 1 + 2 𝑎𝜂2 − 95040 𝜂12 𝑅𝑒 3 𝛼 3 𝑎3 − 47520 𝑅𝑒 2 𝛼 4 𝑎3 + ⋯. The objective of the present study was to apply the Variation of parameter method to obtain an explicit analytic solution of MHD flow through convergent or divergent channels in presence of a high magnetic field. The magnetic field plays its role in no 210

dimensional parameter, namely, the Hartmann number. If we fix Re number reveals the fact that by increasing magnetic field the velocity profile becomes flat and thickness of boundary layer decreases. In fact magnetic field induces a force in opposite of the momentum’s direction that stabilizes the velocity profile. We have the following table Table 2.45. Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = −2.50 and different values of 𝐻. 𝐻 0 𝑎

1000

2000

4000

-1.119985350 -0.9672680211 -0.8391122178 -0.6389447538

Fig.2.86: VPM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝛼 = −2.50 .

Now we take inverse case of Fig 2.86 in which we see by decreasing Hartman number, the velocity profile becomes flat and thickness of boundary layer decreases, and we have the following table Table 2.46. Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = 2.50 and different values of 𝐻. 𝐻 0 𝑎

1000

2000

4000

-3.532651597 -3.016558576 -2.583688243 -1.912969976

211

Fig.2.87: VPM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝛼 = 2.50 .

We take 𝛼 = 50 instead of 𝛼 = 2.50 as discuss in Fig 2.86, notice that velocity profile is more clear. Table 2.47. Values of 𝑎 for 𝑅𝑒 = 100, 𝛼 = 50 and different values of 𝐻. 𝐻 0 𝑎

1000

2000

4000

-5.869196481 -3.272798353 -1.854115346 -0.6512075716

Fig.2.88: VPM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝛼 = 50 .

We examine if we increases 𝛼 the effect of walls on fluid flow decreases when we move away from them which lead to an increase of velocity and the velocity profile in divergent channels, we have the following tables Table 2.48. Values of 𝑎 for 𝑅𝑒 = 100, 𝐻 = 1500 and different values of 𝛼. 𝛼 2.50

50

100

150 212

200

𝑎 2.790676065 2.455715650 0.6698633972 0.07459514752 0.006678895736

Fig.2.89: VPM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝐻 = 1500.

If we decrease 𝛼 the behavior in velocity profile is overturned as in Fig 2.89. The velocity profile in convergent channels and we have Table 2.49. Values of 𝑎 for 𝑅𝑒 = 100, 𝐻 = 1500 and different values of 𝛼. 𝛼 -2.50

-50

-100

-150

-200

𝑎 -0.900437715 -0.2991101463 -0.024018603 -0.0021689878 -0.0002728671

Fig.2.90: VPM solution for velocity is convergent channel for 𝑅𝑒 = 100 and 𝐻 = 1500.

Now we examine the behavior of velocity profile if 𝑅𝑒 varies, and fixed Hartmann numbers we obtained

213

Table 2.50. Values of 𝑎 for 𝛼 = 2.50 , 𝐻 = 1500 and different values of 𝑅𝑒. 𝑅𝑒 -200

-100

50

100

200

𝑎

-

-

-

-2.790676065

-4.704402105

0.5226120092

0.9004377158

2.108823833

Fig.2.91: VPM solution for velocity is convergent channel for 𝛼 = 2.50 and 𝐻 = 1500.

The reverse behavior of velocity profile is notice when we increase 𝑅𝑒. We can infer from Figures 2.90 and 2.91 for inflow regime, back flow is prevented in the case of convergent channels but is possible for large Reynolds numbers in the case of divergent channels. Table 2.51. Values of 𝑎 for 𝛼 = 2.50 , 𝐻 = 1500 and different values of 𝑅𝑒. 𝑅𝑒 -200 𝑎

-100

50

100

200

-4.704402105 -2.790676065 -1.192874651 -0.5226120092 -0.1925220645

Fig.2.92: VPM solution for velocity is convergent channel for 𝛼 = −2.50 and 𝐻 = 1500.

214

The comparison between the numerical results and VPM solution (1st, 2nd, 3rd and 4th order approximate) for velocity when 𝑅𝑒 = 100 and 𝐻 = 1500 is shown in Table 2.52. Table 2.52 (a). The comparison between the numerical results and VPM solution for velocity when 𝛼 = 2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500. _______________________________________________________________ 𝑥


Approx. Approx.

Approx. Approx.

Approx. Solution

_______________________________________________________________ 0.00 1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

0.05 0.99750

0.99650

0.99652

0.99652

0.99652

0.99652

0.99652

0.10 0.99000

0.98606

0.98612

0.98611

0.98611

0.98611

0.98611

0.15 0.97750

0.96883

0.96896

0.96895

0.96895

0.96895

0.96895

0.20 0.96000

0.94505

0.94528

0.94525

0.94525

0.94525

0.94525

0.25 0.93750

0.91506

0.91541

0.91536

0.91536

0.91536

0.91536

0.30 0.91000

0.87928

0.87973

0.87966

0.87966

0.87966

0.87966

0.35 0.87750

0.83817

0.83869

0.83860

0.83860

0.83860

0.83860

0.40 0.84000

0.79227

0.79276

0.79265

0.79265

0.79265

0.79265

0.45 0.79750

0.74210

0.74243

0.74230

0.74230

0.74230

0.74230

0.50 0.75000

0.68822

0.68818

0.68804

0.68804

0.68804

0.68804

0.55 0.69750

0.63115

0.63047

0.63032

0.63032

0.63032

0.63032

0.60 0.64000

0.57132

0.56970

0.56956

0.56956

0.56956

0.56956

0.65 0.57750

0.50912

0.50624

0.50612

0.50612

0.50612

0.50612

0.70 0.51000

0.44478

0.44037

0.44030

0.44029

0.44029

0.44029

0.75 0.43750

0.37837

0.37232

0.37230

0.37229

0.37229

0.37229

0.80 0.36000

0.30974

0.30220

0.30224

0.30223

0.30224

0.30224

0.85 0.27750

0.23848

0.23005

0.23014

0.23013

0.23013

0.23013

0.90 0.19000

0.16389

0.15579

0.15590

0.15589

0.15589

0.15589

0.95 0.09750

0.08490

0.07922

0.07930

0.07930

0.07930

0.07930

1.00 0.00000

0.00000

0.00000

0.00000

0.00000

0.00000

0.00000

________________________________________________________________

215

Table 2.52 (b): The comparison between the numerical results and VPM solution for velocity when 𝛼 = −2.50 , 𝑅𝑒 = 100 and 𝐻 = 1500. _______________________________________________________________ 𝑥


Approx. Approx.

Approx.

Approx.

Approx. Solution

_______________________________________________________________ 0.00 1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

0.05 0.99750

0.99867

0.99885

0.99887

0.99887

0.99887

0.99887

0.01 0.99000

0.99465

0.99535

0.99544

0.99545

0.99545

0.99545

0.15 0.97750

0.98782

0.98941

0.98963

0.98965

0.98965

0.98965

0.20 0.96000

0.97799

0.98085

0.98125

0.98128

0.98129

0.98129

0.25 0.93750

0.96490

0.96944

0.97007

0.97013

0.97013

0.97013

0.30 0.91000

0.94821

0.95484

0.95577

0.95586

0.95586

0.95586

0.35 0.87750

0.92751

0.93665

0.93794

0.93807

0.93808

0.93808

0.40 0.84000

0.90232

0.91437

0.91610

0.91627

0.91629

0.91629

0.45 0.79750

0.87210

0.88739

0.88965

0.88987

0.88989

0.88989

0.50 0.75000

0.83625

0.85502

0.85787

0.85815

0.85817

0.85817

0.55 0.69750

0.79410

0.81644

0.81993

0.82028

0.82031

0.82031

0.60 0.64000

0.74495

0.77070

0.77487

0.77531

0.77534

0.77534

0.65 0.57750

0.68802

0.71677

0.72160

0.72211

0.72215

0.72215

0.70 0.51000

0.62250

0.65347

0.65887

0.65946

0.65950

0.65950

0.75 0.43750

0.54756

0.57953

0.58530

0.58596

0.58601

0.58601

0.80 0.36000

0.46231

0.49356

0.49940

0.50009

0.50014

0.50014

0.85 0.27750

0.36586

0.39412

0.39957

0.40022

0.40027

0.40027

0.90 0.19000

0.25728

0.27970

0.28414

0.28469

0.28473

0.28473

0.95 0.09750

0.13563

0.14881

0.15146

0.15180

0.15183

0.15183

1.00 0.00000

0.00000

0.00000

0.00000

0.00000

0.00000

0.00000

________________________________________________________________

2.11.3. Flow of a Viscoelastic Fluid through a Porous Channel with Expanding or Contracting Wall According to the Variation of parameter method Eq. (2.103)-Eq. (2.104), we have 216

𝜂3

𝑠3

𝜂

𝑓𝑘+1 (𝜂) = 𝑏𝜂 + 𝑡 3! − ∫0 (− 3! +

𝑠2 𝜂 2

−

𝑠𝜂 2 2

𝜂3

+

3!

) (𝛼 (3𝑓𝑘′′ (𝑠) + 𝜂𝑓𝑘′′′ (𝑠) +

(4) (5) 𝑅𝑒(𝑓𝑘 (𝑠)𝑓𝑘′′′ (𝑠) − 𝑓𝑘′ (𝑠)𝑓𝑘′′ (𝑠)) − 𝜔𝛼 (5𝑓𝑘 (𝑠) − 𝜂𝑓𝑘 (𝑠)) + (4)

(5)

𝜔𝑅𝑒(𝑦′𝑘 (𝑠)𝑓𝑘 (𝑠) – 𝑓𝑘 (𝑠)𝑓𝑘 (𝑠))) ds. Consequently, we have the following approximants 𝜂3

𝑓0 (𝜂) = 𝑏𝜂 + 𝑡 3! , 𝜂3

1

1

𝜂3

1

1

𝜂3

1

2

𝑓1 (𝜂) = 𝑏𝜂 + 𝑡 3! − 30 𝜂5 𝛼𝑡 + 2520 𝑅𝑡 2 𝜂7 , 2

𝑓2 (𝜂) = 𝑏𝜂 + 𝑡 3! − 30 𝜂5 𝛼𝑡 + 210 𝑡𝛼 2 𝜂7 − 15 𝜂5 𝜔𝛼 2 𝑡 …, 1

𝑓3 (𝜂) = 𝑏𝜂 + 𝑡 3! − 30 𝜂5 𝛼𝑡 − 15 𝜂5 𝜔𝛼 2 𝑡 + 210 𝑡𝛼 2 𝜂7 …, . . ., 𝑓(𝜂) = lim 𝑓𝑘 (𝜂). 𝑘→∞

𝜂3

1

2

1

𝑓(𝜂) = 𝑏𝜂 + 𝑡 3! − 30 𝜂5 𝛼𝑡 − 15 𝜂5 𝜔𝛼 2 𝑡 + 210 𝑡𝛼 2 𝜂7 + ⋯. Our concern now is to determine the parameters b and t by using boundary conditions.

Fig.2.93: Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for different values of Re with 𝜔 = 0.02 and 𝛼 = 2.

217

Fig.2.94: Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for different values of Re with 𝜔 = 0.02 and 𝛼 = −2.

Fig.2.95: Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for different values of Re with 𝜔 = 0.02 and 𝑅𝑒 = 2.5.

Fig.2.96: Characteristics of 𝑓(𝜂) and 𝑓′(𝜂) for different values of Re with 𝜔 = 0.02 and 𝑅𝑒 = −5.

218

Chapter 3 Traveling Wave Solutions of Nonlinear Evolution Equations

219

3.1. Introduction John Scott Russell was the first one who observed the solitary waves in 1834. He observed a large protrusion of water slowly travelling on the Edinburgh-Glasgow canal without changing its shape. He observed the bulge of water and called it ‘‘great wave of translation” [193] was travelling along the channel of water for a long period of time while still retaining its shape. The single humped wave of bulge of water is now called solitary wave or soliton. In 1895, Diederik Korteweg and Gustav de Vries provided what is now known as the Korteweg de Vries equation (KdV), they also gave its solitary wave and periodic conidial wave solutions. In 1965, Norman Zabusky and Martin Kruskal investigated numerically the nonlinear interaction of a large solitary-wave with a smaller one, and the recurrence of initial states [252]. They discovered that solitary waves undergo nonlinear interaction following with KdV equation. The remarkable discovery, of Russell that solitary waves retain their identities and their character resembles particle like behavior, motivated Zabusky and Kruskal to call these solitary waves to solitons. In this Chapter, we will discusses the traveling wave solutions of nonlinear evolution equations tackled by Exp-function Method, Modified Exp-function Method, (G´/G)Expansion Method, F-Expansion Method, Modified Exp-function Method, Rational Sinh-cosh Method, (U´/U)-Expansion Method, Extended Tanh Method, New Approach of (G´/G)-Expansion Method, Combined Tanh-Coth (CTC) and Combined Sinh-Cosh (CSC) Method and U-Expansion Method on nonlinear evolution equations. The physical properties of several nonlinear traveling wave solutions by plotting and analyzing their figures have also been studied.

3.2. Exp-function Method In 2006 He and Wu [75] devolved Exp-function method to find the Solitary as well as periodic solutions of nonlinear PDEs, Exp function method applied to construct the travelling wave solutions of nonlinear evolution equations [76, 242-243, 258, 264]. This method is discussed by Mohyud-Din [142, 144-145, 156] for solving KuramotoSivashinsky, Boussinesq equations and Calogero-Degasperis-Fokas equations and Usman et al. for solving Burger’s and KdV equations [202], simplified modified Camassa Holm equation [103] and modified generalized Degasperis equation [269]. Later on, Ma’s et al. proposed a new method named as “Multiple Exp-function Method” and successively 220

applied to construct the multiple (3+1)-dimensional potential-Yu–Toda–Sasa–Fukuyama equation (potential-YTSF equation) [126]. This method is very reliable and straight forward.

3.2.1. Methodology We consider the general nonlinear PDE of the type 𝑃(𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑦 , 𝑢𝑡𝑡 , 𝑢𝑥𝑥 , 𝑢𝑦𝑦 , 𝑢𝑥𝑡 , 𝑢𝑥𝑦 , 𝑢𝑡𝑦 , … ) = 0.

(3.1)

using a transformation 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝜔𝑡.

(3.2)

where 𝑘, 𝑙 and 𝜔 are constants, we can rewrite equation (3.1) in the following nonlinear ODE; 𝑄(𝑢, 𝑢′ , 𝑢′′ , 𝑢′′′ , … ) = 0.

(3.3)

According to Exp-function method, we assume that the wave solution can be expressed in the following form ∑𝑐

𝑏 exp[𝑛𝜉]

𝑢(𝜉) = ∑𝑝𝑚=−𝑑𝑏 𝑚exp[𝑚𝜉]

(3.4)

𝑚=−𝑞 𝑚

where 𝑝, 𝑞, 𝑐 and 𝑑 are positive integers which are known to be further determined , 𝑎𝑛 and 𝑏𝑚 are unknown constants. We can rewrite equation (3.4) in the following equivalent form. 𝑎 exp[𝑐𝜉]+⋯+𝑎

exp[−𝑑𝜉]

𝑢(𝜉) = 𝑏 𝑐 exp[𝑝𝜉]+⋯+𝑏−𝑑 exp[−𝑞𝜉]. 𝑝

(3.5)

−𝑞

This equivalent formulation plays an important and fundamental part for finding the analytic solution of problems. To determine the value of 𝑐 and 𝑝, we balance the linear term of highest order of equation (3.4) with the highest order non-linear term. Similarly, to determine the value of 𝑑 and 𝑞, we balance the linear term of lowest order of equation (3.4) with lowest order non linear term.

3.2.2. KdV Equation Consider the following KdV equation [230] 𝑢𝑡 + 6𝑢𝑢𝑥 − 𝑢𝑥𝑥𝑥 = 0,

(3.6a)

Introducing the transformation as 𝜂 = 𝑘𝑥 + 𝜔𝑡, we can convert the given equation into ordinary differential equations 𝑤𝑢′ + 6𝑘𝑢𝑢′ − 𝑘 3 𝑢′′′ = 0.

(3.6b) 221

Consider the solution expressed in this form 𝑎 exp[𝑐𝜂]+⋯+𝑎

exp[−𝑑𝜂]

𝑢(𝜂) = 𝑏𝑐 exp[𝑝𝜂]+⋯+𝑏−𝑑 exp[−𝑞𝜂]. 𝑝

−𝑞

To determine the value of 𝑐 and 𝑝, we balance the linear term of the highest order of (3.5) with the highest order nonlinear term 𝑢′′′ (𝜂) =

𝑐1 exp[(7𝑝+𝑐)𝜂]+⋯

𝑢𝑢′ (𝜂) =

𝑐3 exp[(𝑝+2𝑐)𝜂]+⋯

𝑐2 exp[8𝑝𝜂]+⋯

and 𝑐4 exp[3𝑝𝜂]+⋯

=

𝑐3 exp[(6𝑝+2𝑐)𝜂]+⋯ 𝑐4 exp[8𝑝𝜂]+⋯

,

where 𝑐𝑖 are determine coefficient only for simplicity; balancing the highest order of Exp-function in above expressions, we have 7𝑝 + 𝑐 = 6𝑝 + 2𝑐, This in turn gives 𝑝 = 𝑐. To determine the value of 𝑑 and 𝑞, we balance the linear term of the lowest order of the above expression with the lowest order nonlinear term 𝑢′′′ (𝜂) =

…+𝑑1 exp[(−7𝑞−𝑑)𝜂]

𝑢𝑢′ (𝜂) =

…+𝑑3 exp[(−6𝑞−2𝑑)𝜂]

…+𝑑2 exp[−8𝑞𝜂] …+𝑑4 exp[−8𝑞𝜂]

=

…+𝑑3 exp[(−6𝑞−2𝑑)𝜂] …+𝑑4 exp[−8𝑞𝜂]

,

where 𝑑𝑖 are determine coefficient only for simplicity. Now, balancing the lowest order of the Exp-function in above expressions, we have −7𝑞 − 𝑑 = −6𝑞 − 2𝑑, which in turn gives 𝑞 = 𝑑. Case 3.1.1 We can freely choose the value of 𝑐 and 𝑑, but we will illustrate that the final solution do not strongly depend upon the choice of value of 𝑐 and 𝑑. For simplicity, we set 𝑝 = 𝑐 = 1 and 𝑞 = 𝑑 = 1, then the trial solution reduce to 𝑢(𝜂) =

𝑎1 exp[𝜂]+𝑎0 +𝑎−1 exp[−𝜂] 𝑏1 exp[𝜂]+𝑏0 +𝑏−1 exp[−𝜂]

.

Substituting this expression into the (1), we have 1 (𝑏1 exp[𝜂]+𝑏0 +𝑏−1

exp[−𝜂])4

𝑐3 exp[3𝜂] + 𝑐2 exp[2𝜂] + 𝑐1 exp[𝜂] + 𝑐0 ( )=0 +𝑐−1 exp[−𝜂] + 𝑐−2 exp[−2𝜂] + 𝑐−3 exp[−3𝜂] 222

The constant 𝑐𝑖 (𝑖 = −3, −2, −1,0,1,2,3)can obtain by Maple 11, given as 3 3 2 2 2 𝑐−3 = −𝜔𝑎−1 𝑏0 𝑏−1 + 𝑘 3 𝑎0 𝑏−1 + 𝜔𝑎0 𝑏−1 + 6𝑘𝑎−1 𝑎0 𝑏−1 − 6𝑘𝑎−1 𝑏0 𝑏−1 2 −𝑘 3 𝑎−1 𝑏0 𝑏−1

. . . 𝑐3 = −𝑘 3 𝑎0 𝑏13 − 𝜔𝑎0 𝑏13 − 6𝑘𝑎1 𝑎0 𝑏12 + 𝜔𝑎1 𝑏0 𝑏12 + 𝑘 3 𝑎1 𝑏0 𝑏12 + 6𝑘𝑎12 𝑏0 𝑏1 . Equating the coefficient of exp[𝑛𝜂] to be zero, we obtain [𝑐3 = 0, 𝑐1 = 0, 𝑐2 = 0, 𝑐0 = 0, 𝑐−3 = 0, 𝑐−1 = 0𝑐−2 = 0]. Solutions of the above system will yield 𝑎1 =

−𝑏02 (𝜔+𝑘 3 ) 24(𝑘𝑏−1 )

, 𝑎0 = −

𝑏0 (5𝑘 3 −𝜔) 6𝑘

, 𝑎−1 = −

𝑏−1 (5𝑘 3 +𝜔) 6𝑘

𝑏02

𝑏1 = 4𝑏 , 𝑏0 = 𝑏0 , 𝑏−1 = 𝑏−1 , 𝜔 = 𝜔. −1

Fig. 3.1: Graphical representation of Eq. (3.6a-3.6b) for different values of parameters

We, therefore, obtained the following generalized solitary solution 𝑢(𝑥, 𝑡) of the given equation 𝑢(𝑥, 𝑡) =

2 −1(𝜔+𝑘3 )𝑏0 𝑒𝑘𝑥+𝑤𝑡 1𝑏0 (−𝜔+5𝑘3 ) 1𝑏−1 (𝜔+𝑘3 )𝑒−𝑘𝑥−𝜔𝑡 + − 24 𝑏−1 6 𝑘 6 𝑘 2 𝑘𝑥+𝑤𝑡 𝑏0 𝑒 −𝑘𝑥−𝜔𝑡 +𝑏0 +𝑏−1 𝑒 4𝑏−1

.

where 𝑏0 , 𝑏−1 , 𝜔 and 𝑘 are the real numbers. Figure 1 depict the soliton solution of the given problem, when 𝑏0 = 𝑏−1 = 𝜔 = 𝑘 = 1.

3.2.3. Burger’s Equation Consider the following Burger’s equation [230]

223

𝑢𝑡 + 𝑢𝑢𝑥 = 𝑢𝑥𝑥 ,

(3.7a)

Introducing the transformation as 𝜂 = 𝑘𝑥 + 𝑤𝑡, we can convert the given equation into ordinary differential equations 𝑤𝑢′ + 𝑘𝑢𝑢′ = 𝑘 2 𝑢′′ .

(3.7b)


exp[−𝑑𝜂]


−𝑞

To determine the value of 𝑐 and 𝑝, we balance the linear term of the highest order of (3.7) with the highest order nonlinear term 𝑢′′ (𝜂) =


𝑢𝑢′ (𝜂) =




=


,

where 𝑐𝑖 are determine coefficient only for simplicity; balancing the highest order of Exp-function in above expressions, we have 3𝑝 + 𝑐 = 2𝑝 + 2𝑐, This in turn gives 𝑝 = 𝑐. To determine the value of 𝑑 and 𝑞, we balance the linear term of the lowest order of the above expression with the lowest order nonlinear term 𝑢′′ (𝜂) =

…+𝑑1 exp[(−𝑑−3𝑞)𝜂]

𝑢𝑢′ (𝜂) =

…+𝑑2 exp[−4𝑞𝜂] …+𝑑3 exp[(−2𝑑−𝑞)𝜂] …+𝑑4 exp[−3𝑞𝜂]

=

…+𝑑3 exp[(−2𝑑−2𝑞)𝜂] …+𝑑4 exp[−4𝑞𝜂]

,

where 𝑑𝑖 are determine coefficient only for simplicity. Now, balancing the lowest order of the Exp-function in above expressions, we have −2𝑑 − 2𝑞 = −𝑑 − 3𝑞, which in turn gives 𝑞 = 𝑑. Case 3.1.1 We can freely choose the value of 𝑐 and 𝑑, but we will illustrate that the final solution do not strongly depend upon the choice of value of 𝑐 and 𝑑. For simplicity, we set 𝑝 = 𝑐 = 1 and 𝑞 = 𝑑 = 1, then the trial solution reduce to

224

𝑢(𝜂) =


.

Substituting this expression into the (1), we have 𝑐 exp[2𝜂] + 𝑐1 exp[𝜂] + 𝑐0 1 ( 2 ) (𝑏1 exp[𝜂]+𝑏0 +𝑏−1 exp[−𝜂])3 +𝑐 −1 exp[−𝜂] + 𝑐−2 exp[−2𝜂]

=0

The constant 𝑐𝑖 (𝑖 = −2, −1,0,1,2)can obtain by Maple 11, given as 2 𝑐−2 = −𝜔𝑎−1 𝑏0 𝑏−1 + 𝑘𝑎−1 𝑎0 𝑏−1 + ⋯ + 𝜔𝑎0 𝑏−1 =0

. . . 𝑐2 = 𝑘 2 𝑎1 𝑏1 𝑏0 − 𝑘𝑎1 𝑎0 𝑏1 + ⋯ + 𝜔𝑎1 𝑏1 𝑏0 = 0. Equating the coefficient of exp[𝑛𝜂] to be zero, we obtain [𝑐1 = 0, 𝑐2 = 0, 𝑐0 = 0, 𝑐−1 = 0, 𝑐−2 = 0]. Solutions of the above system will yield 𝑎1 = 0, 𝑎0 = −

𝑏0 (𝑘 2 +𝜔) 𝑘

, 𝑎−1 =

𝑏−1 (𝑘 2 −𝜔) 𝑘

𝑏1 = 0, 𝑏0 = 𝑏0 , 𝑏−1 = 𝑏−1 , 𝜔 = 𝜔.

Fig. 3.2: Graphical representation of Eq. (3.7a-3.7b) for different values of parameters

We, therefore, obtained the following generalized solitary solution 𝑢(𝑥, 𝑡) of the given equation we get 𝜔

𝑢(𝑥, 𝑡) = − 𝑘 −

(𝑏0 −𝑏−1 𝑒 −(𝜔𝑡+𝑘𝑥) )𝑘 (𝑏0 +𝑏−1 𝑒 −(𝜔𝑡+𝑘𝑥) )

.

where 𝑏0 , 𝑏−1 , 𝜔 and 𝑘 are the real numbers. Figure 1 depict the soliton solution of the given problem, when 𝑏0 = 𝑏−1 = 𝜔 = 𝑘 = 1. In case 𝑘 is the imaginary number, the obtained soliton solution can converted into periodic or compact-like solutions. Therefore, we write 𝑘 = 𝑖𝐾, consequently the above expression becomes 225

𝑢(𝑥, 𝑡) = 𝑢(𝑥, 𝑡) =

𝑖𝜔 𝐾 𝑖𝜔 𝐾

− −

(𝑏0 −𝑏−1 𝑒 −(𝜔𝑡+𝑖𝐾𝑥) )𝑖𝐾 (𝑏0 +𝑏−1 𝑒 −(𝜔𝑡+𝑖𝐾𝑥) )

.

(𝑏0 −𝑏−1 𝑒 −𝜔𝑡 (cos(𝐾𝑥)−𝑖𝑠𝑖𝑛(𝐾𝑥)))𝑖𝐾 (𝑏0 +𝑏−1 𝑒 −𝜔𝑡 (cos(𝐾𝑥)−𝑖𝑠𝑖𝑛(𝐾𝑥)))

.

Now, to obtained periodic solutions from soliton solutions, we put the imaginary part of the above expression equal to zero we have 𝑢(𝑥, 𝑡) = − (𝑏

0 +𝑏−1

𝑏0 −𝜔𝑡 𝑒 cos(𝐾𝑥)

.

(a)

(b)

Fig. 3.3.(a)-(b): Graphical representation of Eq. (3.7a-3.7b) for different values of parameters

when 𝑏0 = 𝑏−1 = 𝜔 = 𝐾 = 1. Case 3.1.2 If 𝑝 = 𝑐 = 2 and 𝑞 = 𝑑 = 2, then the trial solution reduce to 𝑎 exp[2𝜂]+𝑎 exp[𝜂]+𝑎 +𝑎−1 exp[−𝜂]+𝑎−2 exp[−2𝜂]

𝑢(𝜂) = 𝑏 2exp[2𝜂]+𝑏 1exp[𝜂]+𝑏 0+𝑏 2

1

0

.

−1 exp[−𝜂]++𝑏−2 exp[−2𝜂]

Proceeding as before, we get 𝑎2 =

𝑏2 (3𝑘 2 +𝜔) 𝑘

, 𝑎1 = 0, 𝑎0 = 0, 𝑎−1 =

𝑏−1 (3𝑘 2 −𝜔) 𝑘

, 𝑏2 = 𝑏2 , 𝑏1 = 0,𝑏0 = 0,

𝑏−1 = 𝑏−1 , 𝜔 = 𝜔. After simplification we get 𝜔

𝑢(𝑥, 𝑡) = − 𝑘 −

(𝑏0 −𝑏−1 𝑒 −(𝜔𝑡+𝑘𝑥) )𝑘 (𝑏0 +𝑏−1 𝑒 −(𝜔𝑡+𝑘𝑥) )

,

where 𝑏0 , 𝑏1 , 𝜔 and 𝑘 are real numbers.

3.2.4. Burger’s Hierarchy in (2+1)-Dimensional Equation Consider the following Burger hierarchy in (2 + 1) − dimension equation [220] 𝑢𝑡 + 𝛼𝑢𝑥𝑥 + 2𝛼𝑢𝑢𝑥 + 𝛽(𝑢𝑥 + 𝑢𝑦 ) = 0,

(3.8a)

Introducing the transformation as 𝜂 = 𝑘𝑥 + 𝑚𝑦 + 𝑤𝑡, we can convert the given equation into ordinary differential equations 226

𝑤𝑢′ + 𝛼𝑘 2 𝑢′′ + 2𝛼𝑘𝑢𝑢′ + 𝛽(𝑘 + 𝑚)𝑢′ = 0.

(3.8)


exp[−𝑑𝜂]


−𝑞

To determine the value of 𝑐 and 𝑝, we balance the linear term of the highest order of (3.8) with the highest order nonlinear term 𝑢′′ (𝜂) =


𝑢𝑢′ (𝜂) =



,


=


,


…+𝑑1 exp[(−𝑑−3𝑞)𝜂]

𝑢𝑢′ (𝜂) =

…+𝑑2 exp[−4𝑞𝜂] …+𝑑3 exp[(−2𝑑−𝑞)𝜂] …+𝑑4 exp[−3𝑞𝜂]

…+𝑑3 exp[(−2𝑑−2𝑞)𝜂]

=

…+𝑑4 exp[−4𝑞𝜂]

,

where 𝑑𝑖 are determine coefficient only for simplicity. Now, balancing the lowest order of the Exp-function in above expressions, we have −2𝑑 − 2𝑞 = −𝑑 − 3𝑞, which in turn gives 𝑞 = 𝑑. Case 3.1.1 We can freely choose the value of 𝑐 and 𝑑, but we will illustrate that the final solution do not strongly depend upon the choice of value of 𝑐 and 𝑑. For simplicity, we set 𝑝 = 𝑐 = 1 and 𝑞 = 𝑑 = 1, then the trial solution reduce to 𝑢(𝜂) =


.

Substituting this expression into the (1), we have

227

1 (𝑏1 exp[𝜂]+𝑏0 +𝑏−1

exp[−𝜂])3

𝑐 exp[2𝜂] + 𝑐1 exp[𝜂] + 𝑐0 ( 2 )=0 +𝑐−1 exp[−𝜂] + 𝑐−2 exp[−2𝜂]

The constant 𝑐𝑖 (𝑖 = −2, −1,0,1,2)can obtain by Maple 13, given as 𝑐−2 = 𝑤𝑏12 𝑎0 + 𝛼𝑘 2 𝑏0 𝑏1 𝑎1 + ⋯ + 2𝛼𝑘𝑎0 𝑎1 𝑏1 = 0, . . . 𝑐2 = −𝑤𝑎0 𝑏12 − 𝛼𝑘 2 𝑎1 𝑏1 𝑏0 − 2𝛼𝑘𝑎1 𝑎0 𝑏1 = 0. Equating the coefficient of exp[𝑛𝜂] to be zero, we obtain [𝑐1 = 0, 𝑐2 = 0, 𝑐0 = 0, 𝑐−1 = 0, 𝑐−2 = 0], We have four solution set satisfy the given equation 1st Solution Set: 1 𝑏0 (𝛼𝑘 2 +𝛽𝑘+𝛽𝑚+𝑤)

𝑤 = 𝑤, 𝑎−1 = 0, 𝑎0 = − 2

𝛼𝑘

1 𝑏1 (−𝑤+𝛼𝑘 2 −𝛽𝑘−𝛽𝑚)

, 𝑎1 = − 2

𝛼𝑘

,

𝑏−1 = 0, 𝑏0 = 𝑏0 , 𝑏1 = 𝑏1 . The above solution yields 𝑢(𝜂) =

(−

1𝑏0 (𝛼𝑘2 +𝛽𝑘+𝛽𝑚+𝑤) 1𝑏 (−𝑤+𝛼𝑘2 −𝛽𝑘−𝛽𝑚) ) exp[𝜂]+𝑎0 +(− 1 ) exp[−𝜂] 2 𝛼𝑘 2 𝛼𝑘

.

𝑏1 exp[𝜂]+𝑏0

(a)

(b)

Fig. 3.4.(a)-(b): Graphical representation of Eq. (3.8a) for different values of parameters

2nd Solution Set: 𝑤 = 𝑤, 𝑎−1 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑏−1 = 0, 𝑏0 = 𝑏0 , 𝑏1 = The above solution yields 228

𝑏0 𝑎1 𝑎0

,

𝑢(𝜂) =

𝑎0 +𝑎1 exp[−𝜂] 𝑏0 𝑎1 ( ) exp[𝜂]+𝑏0 𝑎0

.

(a)

(b)


3rd Solution Set: 𝑏−1 𝑎1

𝑤 = 𝑤, 𝑎−1 =

𝑏1

, 𝑎0 =

𝑏0 𝑎1 𝑏1

, 𝑎1 = 𝑎1 , 𝑏−1 = 𝑏−1 , 𝑏0 = 𝑏0 , 𝑏1 = 𝑏1 .

The above solution yields 𝑢(𝜂) =

𝑏 𝑎 𝑏 𝑎 ( −1 1 ) exp[𝜂]+ 0 1 +𝑎1 exp[−𝜂] 𝑏1

𝑏1

𝑏1 exp[𝜂]+𝑏0 +𝑏−1 exp[−𝜂]

.

(a)

(b)


Case 3.1.2 If 𝑝 = 𝑐 = 2 and 𝑞 = 𝑑 = 1, then the trial solution reduce to 𝑢(𝜂) =

𝑎2 exp[2𝜂]+𝑎1 exp[𝜂]+𝑎0 +𝑎−1 exp[−𝜂]+𝑎−2 exp[−2𝜂] 𝑏1 exp[𝜂]+⋯+𝑏−1 exp[−𝜂]

229

.

Proceeding as before, we get four solution sets 1st Solution Set: 𝑤 = 𝑤, 𝑎−1 = 0, 𝑎−2 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏−1 = 0, 𝑏0 = 𝑏0 , 𝑏1 = The above solution yields 𝑎 exp[𝜂]+𝑎0

𝑢(𝜂) = 𝑎11𝑏0 𝑎0

exp[𝜂]+𝑏0

.

(a)

(b)


2nd Solution Set: 𝑤 = 𝑤, 𝑎−1 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑏−1 = 0, 𝑏0 = 𝑏0 , 𝑏1 = The above solution yields 𝑢(𝜂) =

𝑎0 +𝑎1 exp[−𝜂] 𝑏0 𝑎1 ( ) exp[𝜂]+𝑏0 𝑎0

.

230

𝑏0 𝑎1 𝑎0

,

𝑎1 𝑏0 𝑎0

.

(a)

(b)


3rd Solution Set: 𝑤 = 𝑤, 𝑎−1 =

𝑏−1 𝑎1 𝑏1

, 𝑎0 =

𝑏0 𝑎1 𝑏1

, 𝑎1 = 𝑎1 , 𝑏−1 = 𝑏−1 , 𝑏0 = 𝑏0 , 𝑏1 = 𝑏1 .


𝑏 𝑎 𝑏 𝑎 ( −1 1 ) exp[𝜂]+ 0 1 +𝑎1 exp[−𝜂] 𝑏1

𝑏1


.

(a)

(b)


3.2.5. Shallow Water Wave Equation Consider the following Shallow water equation [230] 𝑣𝑡 − 𝑣𝑥𝑥𝑡 − 4𝑣𝑣𝑡 − 2𝑣𝑥 ∫ 𝑣𝑡 𝑑𝑥 + 𝑣𝑥 + 𝑣𝑥𝑥𝑥 + 6𝑣𝑣𝑥 = 0, Using the potential 𝑣 = 𝑢𝑥 , then equation become 231

(3.9a)

𝑢𝑥𝑡 − 𝑢𝑥𝑥𝑥𝑡 − 4𝑢𝑥 𝑢𝑥𝑡 − 2𝑢𝑥𝑥 𝑢𝑡 + 𝑢𝑥𝑥 + 𝑢𝑥𝑥𝑥𝑥 + 6𝑢𝑥 𝑢𝑥𝑥 = 0, Introducing the transformation as 𝜂 = 𝑘𝑥 + 𝜔𝑡, we can convert the given equation into ordinary differential equations 𝑤𝑘𝑢′′ + 𝑘 3 𝑤𝑢′′′′ − 6𝑘 2 𝑤𝑢′ 𝑢′′ + 𝑘 2 𝑢′′ + 𝑘 4 𝑢′′′′ + 6𝑘 3 𝑢′′ = 0.

(3.9)


exp[−𝑑𝜂]


−𝑞

To determine the value of 𝑐 and 𝑝, we balance the linear term of the highest order of (3.9) with the highest order nonlinear term 𝑢′′′′ (𝜂) =

𝑐1 exp[(15𝑝+𝑐)𝜂]+⋯

𝑢′ 𝑢′′ (𝜂) =

𝑐3 exp[(4𝑝+2𝑐)𝜂]+⋯


,


=

𝑐3 exp[(14𝑝+2𝑐)𝜂]+⋯ 𝑐4 exp[16𝑝𝜂]+⋯

,

where 𝑐𝑖 are determine coefficient only for simplicity; balancing the highest order of Exp-function in above expressions, we have 15𝑝 + 𝑐 = 14𝑝 + 2𝑐, This in turn gives 𝑝 = 𝑐. To determine the value of 𝑑 and 𝑞, we balance the linear term of the lowest order of the above expression with the lowest order nonlinear term 𝑢′′′′ (𝜂) =

…+𝑑1 exp[(−𝑑−15𝑞)𝜂]

𝑢′ 𝑢′′ (𝜂) =

…+𝑑2 exp[−16𝑞𝜂] …+𝑑3 exp[(−2𝑑−4𝑞)𝜂] …+𝑑4 exp[−6𝑞𝜂]

=

…+𝑑3 exp[(−2𝑑−14𝑞)𝜂] …+𝑑4 exp[−16𝑞𝜂]

Where 𝑑𝑖 are determine coefficient only for simplicity. Now, balancing the lowest order of the Exp-function in above expressions, we have −𝑑 − 15𝑞 = −2𝑑 − 14𝑞, which in turn gives 𝑞 = 𝑑. Case 3.1.1 We can freely choose the value of 𝑐 and 𝑑, but we will illustrate that the final solution do not strongly depend upon the choice of value of 𝑐 and 𝑑. For simplicity, we set 𝑝 = 𝑐 = 1 and 𝑞 = 𝑑 = 1, then the trial solution reduce to

232

𝑢(𝜂) =


.

Substituting this expression into the (1), we have

𝑐5 exp[5𝜂] + 𝑐4 exp[4𝜂] + 𝑐3 exp[3𝜂] + 𝑐2 exp[2𝜂] ( +𝑐1 exp[𝜂] + 𝑐0 + 𝑐−1 exp[−𝜂] + 𝑐−2 exp[−2𝜂] + ) = 0, (𝑏1 exp[𝜂]+𝑏0 +𝑏−1 exp[−𝜂])5 𝑐−3 exp[−3𝜂] + 𝑐−4 exp[−4𝜂] + 𝑐−5 exp[−5𝜂] 1

The constant 𝑐𝑖 (𝑖 = −5, −4, −3, −2, −1,0,1,2,3,4,5)can obtain by Maple 11, given as 𝑐−5 = 𝑘 2 𝜔𝑎1 𝑏0 𝑏13 − 𝑘 3 𝑎1 𝑏0 𝑏13 + ⋯ + 𝑘 3 𝑎0 𝑏14 = 0, . . . 3 4 𝑐5 = 𝑘 2 𝜔𝑎1 𝑏−1 𝑏0 + 𝜔𝑎0 𝑏−1 + ⋯ + 𝑘 3 𝑎0 𝑏14 = 0.

Equating the coefficient of exp[𝑛𝜂] to be zero, we obtain [𝑐1 = 0, 𝑐2 = 0, 𝑐3 = 0, 𝑐4 = 0, 𝑐5 = 0, 𝑐0 = 0, 𝑐−1 = 0, 𝑐−2 = 0, 𝑐−3 = 0, 𝑐−4 = 0, 𝑐−5 = 0], We have four solution set satisfy the given equation 1st Solution Set: 𝜔=

𝑘(𝑘 2 +1) 𝑘 2 −1

, 𝑎−1 = −𝑏−1, 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑏−1 = 𝑏−1 , 𝑏0 = 𝑏0 , 𝑏1 = 0.


𝑎0 −𝑏−1 exp[−𝜂] 𝑏0 +𝑏−1 exp[−𝜂]

.

(a)

(b)

233


2nd Solution Set: 𝜔=

𝑘(𝑘 2 +1) 𝑘 2 −1

−6𝑎1 𝑏0 𝑎0 𝑏12 𝑘 + 4𝑘𝑏1 𝑏02 𝑎12 − 𝑎13 𝑏02 − 2𝑎02 𝑏13 𝑘 1 1 , 𝑎−1 = 4 𝑘 2 𝑏4 ( 2 2 2 2 ) , 𝑎0 = 2 3 2 2 1 𝑎1 𝑎0 𝑏1 + 2𝑎1 𝑏0 𝑎0 𝑏1 + 4𝑘 𝑏1 𝑎0 𝑏0 − 4𝑘 𝑏1 𝑎1 𝑏0

2𝑎 𝑏 𝑏 2 𝑘 − 2𝑘𝑏1 𝑏02 𝑎1 + 𝑎12 𝑏02 1 1 𝑎0 , 𝑎1 = 𝑎1 , 𝑏−1 = 4 𝑘 2 𝑏3 ( 0 0 12 2 ) , 𝑏0 = 𝑏0 , 𝑏1 = 0. 1 𝑎0 𝑏1 − 2𝑎0 𝑎1 𝑏0 𝑏1 The above solution set yields

(a)

(b)


3.2.6. Modified Generalized Degasperis–Procesi Equation Consider the following modified generalized Degasperis–Procesi equation [269] 𝑢𝑡 − 𝑢𝑥𝑥𝑡 + (𝑏 + 1)𝑢2 𝑢𝑥 − 𝑏𝑢𝑥 𝑢𝑥𝑥 − 𝑢𝑥𝑥𝑥 = 0,

(3.10a)

Introducing the transformation as 𝜂 = 𝑘𝑥 + 𝜔𝑡, we can convert the given equation into ordinary differential equations 𝜔𝑢′ − 𝜔𝑘 2 𝑢′′′ + (𝑏 + 1)𝑘𝑢2 𝑢′ − 𝑏𝑘 3 𝑢′ 𝑢′′ − 𝑘 3 𝑢′′′ = 0.

(3.10)


exp[−𝑑𝜂]


−𝑞

To determine the value of 𝑐 and 𝑝, we balance the linear term of the highest order of (3.10) with the highest order nonlinear term 𝑢′′′ (𝜂) =

𝑐1 exp[(7𝑝+𝑐)𝜂]+⋯ 𝑐2 exp[8𝑝𝜂]+⋯

,

and

234

𝑢2 𝑢′ (𝜂) =

𝑐3 exp[(𝑝+3𝑐)𝜂]+⋯ 𝑐4 exp[4𝑝𝜂]+⋯

𝑐3 exp[(6𝑝+2𝑐)𝜂]+⋯

=


,


…+𝑑1 exp[(−𝑑−7𝑞)𝜂]

𝑢𝑢′ (𝜂) =


,

…+𝑑3 exp[(−3𝑑−𝑞)𝜂] …+𝑑4 exp[−4𝑞𝜂]

…+𝑑3 exp[(−2𝑑−6𝑞)𝜂]

=


,

where 𝑑𝑖 are determine coefficient only for simplicity. Now, balancing the lowest order of the Exp-function in above expressions, we have −𝑑 − 7𝑞 = −2𝑑 − 6𝑞, which in turn gives 𝑞 = 𝑑. Case 3.1.1 We can freely choose the value of 𝑐 and 𝑑, but we will illustrate that the final solution do not strongly depend upon the choice of value of 𝑐 and 𝑑. For simplicity, we set 𝑝 = 𝑐 = 1 and 𝑞 = 𝑑 = 1, then the trial solution reduce to 𝑢(𝜂) =


.

Substituting this expression into the (1), we have 𝑐4 exp[4𝜂] + 𝑐3 exp[3𝜂] + 𝑐2 exp[2𝜂] (+𝑐1 exp[𝜂] + 𝑐0 + 𝑐−1 exp[−𝜂] + 𝑐−2 exp[−2𝜂] +) = 0 (𝑏1 exp[𝜂]+𝑏0 +𝑏−1 exp[−𝜂])5 𝑐−3 exp[−3𝜂] + 𝑐−4 exp[−4𝜂] 1

The constant 𝑐𝑖 (𝑖 = −4, −3, −2, −1,0,1,2,3,4)can obtain by Maple 11, given as 𝑐−4 = 𝜔𝑏14 𝑎0 + 𝑘 2 𝜔𝑏0 𝑏13 𝑎1 − ⋯ + 𝑏14 𝑎0 = 0, . . . 3 4 2 2 𝑐4 = −𝜔𝑏−1 𝑎0 + 𝑘 3 𝑎−1 𝑎0 𝑏−1 + ⋯ + 𝑘𝑏𝑎−1 𝑎0 𝑏−1 = 0.

235

Equating the coefficient of exp[𝑛𝜂] to be zero, we obtain [𝑐1 = 0, 𝑐2 = 0, 𝑐3 = 0, 𝑐4 = 0, 𝑐0 = 0, 𝑐−1 = 0, 𝑐−2 = 0, 𝑐−3 = 0, 𝑐−4 = 0] Finally, we have two solutions 1st Solution Set 𝜔 = 𝜔, 𝑎−1 =

𝑎0 𝑏−1 𝑏0

, 𝑎−2 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = 0, 𝑏−1 = 𝑏−1 , 𝑏0 = 𝑏0 , 𝑏1 = 0.

Solutions of the above system will yields 𝑢(𝜂) =

𝑎 𝑏 𝑎0 +( 0 −1 ) exp[−𝜂] 𝑏0

𝑏0 +𝑏−1 exp[−𝜂]

.

(a)

(b)


2nd Solution Set 𝜔 = 𝜔, 𝑎−1 =

𝑎1 𝑏−1 𝑏1

, 𝑎−2 = 0, 𝑎0 =

𝑎1 𝑏0 𝑏1

, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏−1 = 𝑏−1 , 𝑏0 = 𝑏0 , 𝑏1 = 𝑏1 .


𝑎 𝑏 𝑎1 exp[𝜂]+𝑎0 +( 1 −1 ) exp[−𝜂] 𝑏1


.

236

(a)

(b)


Case 3.1.2 We can freely choose the value of 𝑐 and 𝑑, but we will illustrate that the final solution do not strongly depend upon the choice of value of 𝑐 and 𝑑. For simplicity, we set 𝑝 = 𝑐 = 1 and 𝑞 = 𝑑 = 2, then the trial solution reduce to 𝑎1 exp[𝜂]+𝑎0 +𝑎−1 exp[−𝜂]

𝑢(𝜂) = 𝑏

.

2 exp[2𝜂]+𝑏0 +𝑏−2 exp[−2𝜂]

Proceeding as before, we get two solution sets 1st Solution Set 𝜔 = 𝜔, 𝑎−1 =

𝑎0 𝑏−1 𝑏0

, 𝑎−2 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = 0, 𝑏−1 = 𝑏−1 , 𝑏0 = 𝑏0 , 𝑏1 = 0.


𝑎 𝑏 𝑎0 +( 0 −1 ) exp[−𝜂] 𝑏0

𝑏0 +𝑏−1 exp[−𝜂]

.

(a)

(b)

237


2nd Solution Set 𝜔 = 𝜔, 𝑎−1 =

𝑎1 𝑏−1 𝑏1

, 𝑎−2 = 0, 𝑎0 =

𝑎1 𝑏0 𝑏1

, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏−1 = 𝑏−1 , 𝑏0 = 𝑏0 , 𝑏1 = 𝑏1 .


𝑎 𝑏 𝑎1 exp[𝜂]+𝑎0 +( 1 −1 ) exp[−𝜂] 𝑏1


.

(a)

(b)


3.2.7. System of Nonlinear Volterra Integral Equations Consider the following system of Volterra integral equations [229] 𝑥

𝑢(𝑥) = 1 + ∫0 𝑣(𝑥)𝑑𝑥, 𝑥 𝑣2

𝑣(𝑥) = 1 + 𝛼 ∫0

𝑢

𝑥

𝑑𝑥 + 𝛽 ∫0 𝑢(𝑥)𝑑𝑥.

(3.11a)

Taking derivative of the above system, convert the system into one equation finally, we get 𝑢𝑢′′ = 𝛼(𝑢′ )2 + 𝛽𝑢2 .

(3.11)

Consider the solution expressed in this form 𝑎 exp[𝑐𝑥]+⋯+𝑎

exp[−𝑑𝑥]

𝑢(𝜂) = 𝑏𝑐 exp[𝑝𝑥]+⋯+𝑏−𝑑 exp[−𝑞𝑥]. 𝑝

−𝑞

To determine the value of 𝑐, 𝑝, 𝑞 and 𝑝, we balance the linear term of the highest /lowest order of (3.11) with the highest /lowest order nonlinear term, we get 𝑐 = 𝑝,

and

𝑞 = 𝑑.

238

Case 3.1.1 We can freely choose the value of 𝑐 and 𝑑, but we will illustrate that the final solution do not strongly depend upon the choice of value of 𝑐 and 𝑑. For simplicity, we set 𝑝 = 𝑐 = 1 and 𝑞 = 𝑑 = 1, then the trial solution reduce to 𝑢(𝜂) =

𝑎1 exp[𝑥]+𝑎0 +𝑎−1 exp[−𝑥] 𝑏1 exp[𝑥]+𝑏0 +𝑏−1 exp[−𝑥]

.

Substituting this expression into the (31), we have 1 (𝑐 exp[4𝑥] (𝑏−1 exp[𝑥]+𝑏0 +𝑏1 exp[−𝑥])4 4

+ ⋯ + 𝑐−4 exp[−4𝑥]) = 0

The constant 𝑐𝑖 (𝑖 = −2, −1,0,1,2)can obtain by Maple 13, given as 𝑐4 : 𝛽𝑎12 𝑏12 = 0, . . . 2 2 𝑐−4 : 𝛽𝑎−1 𝑏−1 = 0.

Solve the following system, we get [𝐶−4 = 0 … 𝐶−4 = 0], We have two solution set satisfy the given equation 1st Solution Set: 1

𝑎−1 = 0, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑏−1 = 𝑏−1 , 𝑏0 = 0, 𝑏1 = 0, 𝛼 = − 4 𝛽 + 1. The above solution yields 𝑎1 e𝑥

𝑢(𝑥) = 𝑏

−𝑥 −1 e

.

Solution of 𝑣(𝑥) can be obtained by using retransformation.

239

Fig. 3.16: Graphical representation of Eq. (3.11a) for different values of parameters

2nd Solution set: 𝑎−1 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑏−1 = 𝑏−1 , 𝑏0 = 0, 𝑏1 = 0, 𝛼 = −𝛽 + 1. The above solution yields 𝑢(𝑥) = 𝑏

𝑎0

−𝑥 −1 e

.

Solution of 𝑣(𝑥) can be obtained by using retransformation.

Fig. 3.17: Graphical representation of Eq. (3.11a) for different values of parameters

Case 3.2.1 We can freely choose the value of 𝑐 and 𝑑, but we will illustrate that the final solution do not strongly depend upon the choice of value of 𝑐 and 𝑑. For simplicity, we set 𝑝 = 𝑐 = 2 and 𝑞 = 𝑑 = 2, then the trial solution reduce to 𝑢(𝜂) =

𝑎2 exp[2𝜂]+𝑎1 exp[𝜂]+𝑎0 +𝑎−1 exp[−𝜂]++𝑎−2 exp[−2𝜂] 𝑏2 exp[2𝜂]+𝑏1 exp[𝜂]+𝑏0 +𝑏−1 exp[−𝜂]+𝑏−2 exp[−2𝜂]

.

the solitary wave solution in this case are same as discuss in case 3.1.1.

3.3. Modified Exp-function Method Recently Usman et al. [202] introduce a modification in Exp-function method. The proposed modification is the elegant coupling of Exp-function and homogenous balancing. The main advantages of modified Exp-function method over tanh method [198-199] is it has less computational work, easy to apply and it finish the case which we discuss Exp-function method. It is observed that the proposed modification is highly compatible to find solitary wave solutions of nonlinear problems of diversified physical nature and the same can be extended to other problems even with very strong nonlinearity.

3.3.1. Methodology We consider the general nonlinear PDE of the type 240

𝑃(𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑦 , 𝑢𝑧 , 𝑢𝑡𝑡 , 𝑢𝑥𝑥 , 𝑢𝑦𝑦 , 𝑢𝑥𝑥 , 𝑢𝑥𝑡 , 𝑢𝑦𝑡 , 𝑢𝑧𝑡 , 𝑢𝑥𝑦 , 𝑢𝑥𝑧 , 𝑢𝑦𝑧 , … ) = 0. (3.12) where 𝑃 is a polynomial in its arguments. The essence of the Modified Exp-function Method can be presented in the following steps: Step 1: Seek Solitary wave solutions of Eq. (3.12) by taking 𝑢(𝑥, 𝑡) = 𝑈(𝓏), 𝓏 = 𝛼𝑥 + 𝛽𝑦 + 𝛾𝑧 + 𝜔𝑡. and transform Eq. (3.12) to the ordinary differential equation. 𝑄(𝑢, 𝜔𝑢′ , 𝑘𝑢′ , 𝑙𝑢′ , … ) = 0.

(3.13)

where 𝛼, 𝛽, 𝛾 and 𝜔 are constants and where prime denotes the derivative with respect to 𝓏. Step 2: If possible, integrate Eq. (3.13) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero. Step 3: According to Modified Exp-function method, we assume that the wave solution can be expressed in the following form 𝑢(𝓏) =

∑2𝑀 𝑛=0 𝑎𝑛 exp[𝑛𝓏] ∑2𝑀 𝑛=0 𝑏𝑛 exp[𝑛𝓏]

,

(3.14)

where 𝑎𝑖 and 𝑏𝑖 are real constants to be determined, 𝑀 is a positive integer to be determined. Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term(s) of highest order with the highest order nonlinear term(s) in Eq. (3.13). Step 5: Substituting (3.14) into Eq. (3.13) yields an algebraic equation involving powers of exp[𝓏]. Equating the coefficients of each power of exp[𝓏] to zero gives a system of algebraic equations for 𝑎𝑖 , 𝑏𝑖 and 𝑐. Then, we solve the system with the aid of a computer algebra system (CAS), such as MAPLE 11, to determine these constants. Step 6: Use the results obtained in above steps into trial solution and then we can obtain exact solutions of Eq. (3.12).

3.3.2. Pochhammer-Chree (PC) Equation Consider the Pochhammer-Chree (PC) equation [111] 𝑢𝑡𝑡 − 𝑢𝑡𝑡𝑥𝑥 − (𝛼𝑢 − 𝛽𝑢3 − 𝛾𝑢5 )𝑥𝑥 = 0, Let we consider 𝛾 = 0, we get the above equation 𝑢𝑡𝑡 − 𝑢𝑡𝑡𝑥𝑥 − (𝛼𝑢 − 𝛽𝑢3 )𝑥𝑥 = 0,

241

(3.15a)

introducing the transformation as 𝜉 = 𝑘𝑥 + 𝜔𝑡, we can convert the given equation into ordinary differential equations 𝜔2 𝑢′′ − 𝑘 2 𝜔2 𝑢′′′′ − (𝛼 − 𝛽𝑢3 )′′ = 0.

(3.15)

Integrating two times we get 𝜔2 𝑢 − 𝑘 2 𝜔2 𝑢′′ − (𝛼 − 𝛽𝑢3 ) = 0. Balance the 𝑢3 and 𝑢′′ by using homogenous principal, we have 𝑀 + 2 = 3𝑀, 𝑀 = 1. Then we have the trail solution is 𝑢(𝜉) =

𝑎0 +𝑎1 𝑒 𝜉 +𝑎2 𝑒 2𝜉 𝑏0 +𝑏1 𝑒 𝜉 +𝑏2 𝑒 2𝜉

,

(3.16)

where 𝜉 = 𝑘𝑥 + 𝜔𝑡. put (3.16) in (3.15), we have ODE as 1 (𝑐 (𝑏0 +𝑏1 exp[𝜉]+𝑏2 exp[2𝜉])3 0

+ 𝑐1 exp[𝜉] + 𝑐2 exp[2𝜉] + 𝑐3 exp[3𝜉] +

𝑐4 exp[4𝜉] + 𝑐5 exp[5𝜉] + 𝑐6 exp[6𝜉]) = 0, the constant 𝑐𝑖 (𝑖 = 0,1,2,3,4,5,6) can obtain by Maple 13, given as 𝑐0 = −𝜔2 𝑎0 𝑏02 − 𝛽𝑎03 + 𝛼𝑎0 𝑏02 , . . .. 𝑐6 = −𝜔2 𝑎2 𝑏22 − 𝛽𝑎23 + 𝛼𝑎2 𝑏22 . Equating the coefficient of exp[𝑛𝜉] to be zero, we obtain [𝑐0 = 0, 𝑐1 = 0, 𝑐2 = 0, 𝑐3 = 0, 𝑐4 = 0, 𝑐5 = 0, 𝑐6 = 0], we have two solution sets satisfying the given equation 1st Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑎2 = 1 𝛼𝑘 2 𝑏12 −𝑘2 𝛽𝑎12 −2𝛽𝑎12 4

𝛼𝑘√−

−2𝛽−𝑘2 𝛽 𝑎0 𝛼

1 𝛼𝑘 2 𝑏12 −𝑘2 𝛽𝑎12 −2𝛽𝑎12 −4 , 𝑏0 𝑎0 𝛽(2+𝑘 2 )

.

The above solution yields

242

=

2 √−−2𝛽−𝑘 𝛽𝑎0 𝛼

𝑘

, 𝑏1 = 𝑏1 , 𝑏2 =

𝑎0 +𝑎1 e𝜉 +(−

𝑢(𝜉) =

2 2 2 1𝛼𝑘2 𝑏1 −𝑘2 𝛽𝑎1 −2𝛽𝑎1 2𝜉 )e 4 𝑎0 𝛽(2+𝑘2 )

2 √−−2𝛽−𝑘 𝛽𝑎0 2 2 2 1𝛼𝑘2 𝑏1 −𝑘2 𝛽𝑎1 −2𝛽𝑎1 𝛼 )e2𝜉 +𝑏1 e𝜉 +( 2𝛽 𝑘 4 −2𝛽−𝑘 𝛼𝑘√− 𝑎0 𝛼

.

where 𝜉 = 𝑘𝑥 + 𝜔𝑡.

(a)

(b)


2nd Solution Set: 1 𝛽𝑎12 (𝑘 2 −1)

𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏0 = 𝑏0 , 𝑏1 = 0, 𝑏2 = 8

𝛼𝑘 2 𝑏0

, 𝑘 = 𝑘, 𝜔 = 𝜔.

The above solution yields 𝑎1 e𝜉

𝑢(𝜉) = 𝑏0 +(

2 1𝛽𝑎1 (𝑘2 −1) 2𝜉 )e 8 𝛼𝑘2 𝑏0

.

where 𝜉 = 𝑘𝑥 + 𝜔𝑡.

(a)

(b)


243

3.3.3. (2+1)-Dimensional Jimbo-Miwa Equation We consider the (3 + 1)-dimensional Jimbo–Miwa equation [202] 𝑢𝑥𝑥𝑥𝑦 + 3𝑢𝑦 𝑢𝑥𝑥 + 3𝑢𝑥 𝑢𝑥𝑦 + 2𝑢𝑦𝑡 − 3𝑢𝑥𝑧 = 0.

(3.17)

introducing the transformation as 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡, we can convert the given equation into ordinary differential equations 𝑙𝑘 3 𝑢′′′′ + 6𝑙𝑘 2 𝑢′ 𝑢′′ + 2𝑙𝜔𝑢′′ + 3𝑘𝑚𝑢′′ = 0. Integrating once we have 𝑙𝑘 3 𝑢′′′ + 3𝑙𝑘 2 (𝑢′ )2 + 2𝑙𝜔𝑢′ + 3𝑘𝑚𝑢′ = 0.

(3.18)

Balancing the (𝑢′ )2 and 𝑢′′′ by using homogenous principal, we have 2𝑀 + 2 = 𝑀 + 3, 𝑀 = 1. Therefore, we have trial solution 𝑢(𝜉) =

𝑎0 +𝑎1 𝑒 𝜉 +𝑎2 𝑒 2𝜉 𝑏0 +𝑏1 𝑒 𝜉 +𝑏2 𝑒 2𝜉

,

(3.19)

put (3.19) in (3.18), we have − exp[𝜉] (𝑐 (𝑏0 +𝑏1 exp[𝜉]+𝑏2 exp[2𝜉])4 0

+ 𝑐1 exp[𝜉] + 𝑐2 exp[2𝜉] + 𝑐3 exp[3𝜉] +

𝑐4 exp[4𝜉] + 𝑐5 exp[5𝜉] + 𝑐6 exp[6𝜉]) = 0, the constant 𝑐𝑖 (𝑖 = 0,1,2,3,4,5,6)can obtain by Maple 13, given as 𝑐0 = 2𝑙𝜔𝑎0 𝑏1 𝑏02 − 2𝑙𝜔𝑎1 𝑏03 − 3𝑘𝑚𝑎0 𝑏1 𝑏02 + 3𝑘𝑚𝑎1 𝑏03 + 𝑘 3 𝑙𝑎0 𝑏1 𝑏02 − −𝑘 3 𝑙𝑎1 𝑏03 . . .. 𝑐6 = 2𝑙𝜔𝑎1 𝑏23 + 3𝑘𝑚𝑎2 𝑏1 𝑏22 + 𝑘 3 𝑙𝑎1 𝑏23 − 3𝑘𝑚𝑎1 𝑏23 − 2𝑙𝜔𝑎2 𝑏1 𝑏22 − 𝑘 3 𝑙𝑎2 𝑏1 𝑏22 Equating the coefficient of exp[𝑛𝜉] to be zero, we obtain [𝑐0 = 0, 𝑐1 = 0, 𝑐2 = 0, 𝑐3 = 0, 𝑐4 = 0, 𝑐5 = 0, 𝑐6 = 0], we have three solutions set satisfying the given equation 1st Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 =

𝑏2 (𝑎0 +4𝑘𝑏0 ) 𝑏0

, 𝑏0 = 𝑏0 , 𝑏1 = 0, 𝑏2 = 𝑏2 , 𝑘 = 𝑘, 𝜔 = 𝜔. 244

The above solution yields 𝑢(𝜉) =

𝑏 (𝑎 +4𝑘𝑏0 ) 2𝜉 𝑎0 +( 2 0 )e 𝑏0

𝑏0 +𝑏2 e2𝜉

.

where 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡.

(a)

(b)

Fig. 3.20.(a)-(b): Graphical representation of Eq. (3.17) for different values of parameters

2nd Solution Set: 1

𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑎2 = − 4𝑘 2 𝑏4 (4𝑎02 𝑏12 𝑘𝑏0 − 6𝑎1 𝑏02 𝑎0 𝑏1 𝑘 + 𝑎03 𝑏12 − 0

2𝑎1 𝑏0 𝑎02 𝑏1 + 𝑎0 𝑎12 𝑏02 − 4𝑘 2 𝑎1 𝑏03 𝑏1 + 4𝑘 2 𝑎0 𝑏12 𝑏02 + 2𝑎13 𝑏03 𝑘), 𝑏0 = 1

𝑏0 , 𝑏1 = 𝑏1 , 𝑏2 = − 4𝑘 2 𝑏3 (𝑎02 𝑏12 − 2𝑎1 𝑏0 𝑎0 𝑏1 − 2𝑘𝑎1 𝑏02 𝑏1 + 0

2𝑘𝑎0 𝑏12 𝑏0

+

𝑎12 𝑏02 ), 𝑘

= 𝑘, 𝜔 = 𝜔.


𝑎0 +𝑎1 e𝜉 +𝑎2 e2𝜉 𝑏0 +𝑏1 e𝜉 +𝑏2 e2𝜉

.


245

(a)

(b)


3rd Solution Set: 𝑎0 = 0, 𝑎1 = −

𝑏1 (2𝑏2 𝑘−𝑎2 ) 𝑏2

, 𝑎2 = 𝑎2 , 𝑏0 = 0, 𝑏1 = 𝑏1 , 𝑏2 = 𝑏2 .


𝑏 (2𝑏 𝑘−𝑎2 ) 𝜉 (− 1 2 )e +𝑎2 e2𝜉 𝑏2

𝑏1 e𝜉 +𝑏2 e2𝜉

,


(a)

(b)


3.3.4. Generalized Nonlinear Heat Conduction Equation Consider the generalized nonlinear Heat Conduction equation [153] 𝑢𝑡 − 𝛼(𝑢𝑛 )𝑥𝑥 − 𝑢 + 𝑢𝑛 = 0,

𝛼 > 0, 𝑛 > 1,

246

(3.20a)

introducing the transformation as 𝜉 = 𝑘𝑥 + 𝜔𝑡, we can convert the given equation into ordinary differential equation, 𝜔𝑢′ − 𝛼𝑘 2 (𝑢𝑛 )′′ − 𝑢 + 𝑢𝑛 = 0. Balancing 𝑢′ and (𝑢𝑛 )′′ by using homogenous principal, we have 𝑀 + 1 = 𝑛𝑀 + 2, −1

𝑀 = 𝑛−1. 1

By making the transformation 𝑢 = 𝑣 −𝑛−1 ODE (4) reduces to 𝜔(𝑛−1) 3

(𝑣 3 )′ + (𝑛 − 1)2 𝑣 2 (𝑣 − 1) + 𝛼𝑘 2 𝑛(2𝑛 − 1)(𝑣 ′ )2

−𝛼𝑘 2 𝑛(𝑛 − 1)𝑣𝑣 ′′ = 0,

(3.20b)

balancing 𝑣𝑣 ′′ with (𝑣 3 )′ by using Homogeneous principle, 𝑀 + 𝑀 + 2 = 3𝑀 + 1, 𝑀 = 1, Then we have the trail solution is 𝑣(𝜉) =

𝑎0 +𝑎1 𝑒 1𝜉 +𝑎2 𝑒 2𝜉 𝑏0 +𝑏1 𝑒 𝜉 +𝑏2 𝑒 2𝜉

,

(3.21)

put (3.21) in (3.20), we have ODE is 1

− (𝑏

0 +𝑏1 exp[𝜉]+𝑏2 exp[2𝜉])

4

(𝐶0 + 𝐶1 exp[𝜉] + 𝐶2 exp[2𝜉] + 𝐶3 exp[3𝜉] +

𝐶4 exp[4𝜉] + 𝐶5 exp[5𝜉] + 𝐶6 exp[6𝜉] + 𝐶7 exp[7𝜉] + 𝐶8 exp[8𝜉]) = 0, the constants 𝐶𝑖 (𝑖 = 0,1,2,3,4,5,6,7,8) can obtain by Maple 13, given as 𝐶0 = −2𝑛𝑎03 𝑏0 + 𝑛2 𝑎03 𝑏0 − 𝑎02 𝑏02 + 𝑎03 𝑏0 − 𝑛2 𝑎02 𝑏02 + 2𝑛𝑎02 𝑏02 , . . .. 𝐶8 = 2𝑛𝑎22 𝑏22 − 2𝑛𝑎23 𝑏2 − 𝑛2 𝑎22 𝑏22 + 𝑛2 𝑎23 𝑏2 + 𝑎23 𝑏2 − 𝑎22 𝑏22 . Equating the coefficients of exp[𝑛𝜉] to be zero, we obtain [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0, 𝐶7 = 0, 𝐶8 = 0], we have six solution sets satisfying the given equation. 1st solution Set: 𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝑏0 = 0, 𝑏1 = 𝑏1 , 𝑏2 = 0, 𝑘 = 𝑘, 𝜔 = 1 − 𝑛. The above solution yields 247

𝑣(𝜉) =

𝑎2 e2𝜉 𝑏1 e𝜉

, −

𝑎2 e2𝜉

𝑢(𝜉) = ( 𝑏

)

𝜉 1e

1 𝑛−1

.

where 𝜉 = 𝑘𝑥 + (1 − 𝑛)𝑡.

(a)

(b)


2nd Solution Set: 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏0 = 0, 𝑏1 = 𝑎1 , 𝑏2 = 𝑏2 , 𝑘 = 𝑘, 𝜔 = − The above solution yields 𝑎1 e𝜉

𝑣(𝜉) = 𝑎

𝜉 2𝜉 1 e +𝑏2 e

𝑢(𝜉) = (𝑎

,

𝑎1 e𝜉

𝜉 2𝜉 1 e +𝑏2 e

where 𝜉 = 𝑘𝑥 − (

−

)

1 𝑛−1

.

𝑛−1 𝑛

) 𝑡.

248

𝑛−1 𝑛

.

(a)

(b)

Fig. 3.24.(a)-(b): Graphical representation of Eq. (3.20a) for different values of parameter

3rd Solution Set: 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏0 = 0, 𝑏1 = 0, 𝑏2 = 𝑏2 , 𝑘 = 𝑘, 𝜔 = 𝑛 − 1. The above solution yields 𝑎 e𝜉

𝑣(𝜉) = 𝑏 1e2𝜉 , 2

1 𝑛−1

−

𝑎1 e𝜉

𝑢(𝜉) = (𝑏

)

2𝜉 2e

.

where 𝜉 = 𝑘𝑥 + (𝑛 − 1)𝑡.

(a)

(b)


4th Solution Set: 𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝑏0 = 0, 𝑏1 = 𝑏1 , 𝑏2 = 𝑎2 , 𝑘 = 𝑘, 𝜔 = The above solution yields 𝑎2 e2𝜉

𝑣(𝜉) = 𝑏

𝜉 2𝜉 1 e +𝑎2 e

𝑢(𝜉) = (𝑏

,

𝑎2 e2𝜉

𝜉 2𝜉 1 e +𝑎2 e

where 𝜉 = 𝑘𝑥 + (

−

)

1 𝑛−1

.

𝑛−1 𝑛

) 𝑡.

249

𝑛−1 𝑛

.

(a)

(b)


5th Solution Set: 1

𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝑏0 = 𝑏0 , 𝑏1 = 0, 𝑏2 = 0, 𝑘 = 𝑘, 𝜔 = − 2 𝑛. The above solution yields 𝑣(𝜉) =

𝑎2 e2𝜉 𝑏0

,

𝑎2 e2𝜉

𝑢(𝜉) = (

𝑏0

−

)

1 𝑛−1

.

1

where 𝜉 = 𝑘𝑥 − 2 𝑛𝑡.

(a)

(b)


6th Solution Set: 1 𝑛−1

𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝑏0 = 𝑏0 , 𝑏1 = 0, 𝑏2 = 𝑎2 , 𝑘 = 𝑘, 𝜔 = 2 250

𝑛

.

The above solution yields 𝑣(𝜉) = 𝑏

𝑎2 e2𝜉

2𝜉 0 +𝑏2 e

𝑢(𝜉) = (𝑏

, 1 𝑛−1

−

𝑎2 e2𝜉

2𝜉 0 +𝑏2 e

)

.

1 𝑛−1

where 𝜉 = 𝑘𝑥 + 2 (

𝑛

) 𝑡.

(a)

(b)


3.3.5. Generalized Nonlinear Heat Equation Consider the generalized nonlinear heat conduction equation [153] 𝑢𝑡 − 𝑎(𝑢𝑛 )𝑥𝑥 − 𝑎(𝑢𝑛 )𝑦𝑦 − 𝑢 + 𝑢𝑛 = 0,

𝑎 > 1, 𝑛 > 1. (3.22a)

introducing the transformation as 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝜔𝑡, we can convert the given equation into ordinary differential equation, 𝜔𝑢′ − 𝑎(𝑘 2 + 𝑙 2 )(𝑢𝑛 )′′ − 𝑢 + 𝑢𝑛 = 0, balancing (𝑢𝑛 )′′ and 𝑢′ by using homogenous principal, we have 𝑛𝑀 + 2 = 𝑀 + 1, −1

𝑀 = 𝑛−1. 1

By making the transformation 𝑢 = 𝑣 −𝑛−1 ODE (4) reduces to 𝜔(𝑛−1) 3

(𝑣 3 )′ + (𝑛 − 1)2 𝑣 2 (𝑣 − 1) + 𝛼(𝑘 2 + 𝑙 2 )(𝑛(2𝑛 − 1)(𝑣 ′ )2 −

𝑛(𝑛 − 1)𝑣𝑣 ′′ ) = 0,

(3.22b)

balancing 𝑣𝑣 ′′ with (𝑣 3 )′ by using Homogeneous principle, 𝑀 + 𝑀 + 2 = 3𝑀 + 1, 251

𝑀 = 1. Then we have the trail solution is 𝑣(𝜉) =

𝑎0 +𝑎1 𝑒 𝜉 +𝑎2 𝑒 2𝜉 𝑏0 +𝑏1 𝑒 𝜉 +𝑏2 𝑒 2𝜉

,

(3.23)

put (3.23) in (3.22), we have ODE is 1 (𝐶0 (𝑏0 +𝑏1 exp[𝜉]+𝑏2 exp[2𝜉])4

+ 𝐶1 exp[𝜉] + 𝐶2 exp[2𝜉] + 𝐶3 exp[3𝜉] +

𝐶4 exp[4𝜉] + 𝐶5 exp[5𝜉] + 𝐶6 exp[6𝜉] + 𝐶7 exp[7𝜉] + 𝐶8 exp[8𝜉]) = 0, the constants 𝐶𝑖 (𝑖 = 0,1,2,3,4,5,6,7,8) can obtain by Maple 13, given as 𝐶0 = −2𝑛𝑎03 𝑏0 − 𝑎02 𝑏02 + 𝑎03 𝑏0 + 2𝑛𝑎02 𝑏02 − 𝑛2 𝑎02 𝑏02 + 𝑛2 𝑎03 𝑏0 , . . .. 𝐶8 = −𝑎22 𝑏22 + 𝑎23 𝑏2 − 2𝑛𝑎23 𝑏2 + 2𝑛𝑎22 𝑏22 − 𝑛2 𝑎22 𝑏22 + 𝑛2 𝑎23 𝑏2 . Equating the coefficients of exp[𝑛𝜉] to be zero, we obtain [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0, 𝐶7 = 0, 𝐶8 = 0], we have five solution sets satisfying the given equation. 1st Solution Set: 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏0 = 0, 𝑏1 = 0, 𝑏2 = 𝑏2 , 𝑘 = 𝑘, 𝜔 = 𝑛 − 1. The above solution yields 𝑎 e𝜉

𝑣(𝜉) = 𝑏 1e2𝜉 , 2

𝑎1 e𝜉

𝑢(𝜉) = (𝑏

2𝜉 2e

1 𝑛−1

−

)

.

where 𝜉 = 𝑘𝑥 + (𝑛 − 1)𝑡.

252

(a)

(b)


2nd Solution Set: 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏0 = 0, 𝑏1 = 𝑎1 , 𝑏2 = 𝑏2 , 𝑘 = 𝑘, 𝜔 = −

𝑛−1 𝑛

.


𝑣(𝜉) = 𝑎

𝜉 2𝜉 1 e +𝑏2 e

𝑢(𝜉) = (𝑎

,

𝑎1 e𝜉

𝜉 2𝜉 1 e +𝑏2 e

where 𝜉 = 𝑘𝑥 − (

−

)

1 𝑛−1

.

𝑛−1 𝑛

) 𝑡.

(a)

(b)


3rd Solution Set: 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏0 = 𝑏0 , 𝑏1 = 0, 𝑏2 = 0, 𝑘 = 𝑘, 𝜔 = 1 − 𝑛. The above solution yields 253

𝑣(𝜉) =

𝑎1 e𝜉 𝑏0

,

𝑎1 e𝜉

𝑢(𝜉) = (

𝑏0

−

)

1 𝑛−1

.

where 𝜉 = 𝑘𝑥 + (1 − 𝑛)𝑡.

(a)

(b)


4th Solution Set: 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏0 = 𝑏0 , 𝑏1 = 𝑎1 , 𝑏2 = 0, 𝑘 = 𝑘, 𝜔 = The above solution yields 𝑣(𝜉) = 𝑏

𝑎1 e𝜉

𝜉 0 +𝑎1 e

𝑢(𝜉) = (𝑏

,

𝑎1 e𝜉

𝜉 0 +𝑎1 e

where 𝜉 = 𝑘𝑥 + (

1 𝑛−1

−

)

.

𝑛−1 𝑛

) 𝑡.

254

𝑛−1 𝑛

.

(a)

(b)


5th Solution Set: 𝑛𝑎

1

𝑛2 𝑎12 2,𝑘 0 (𝑛+1)

𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏0 = 𝑏0 , 𝑏1 = 𝑛+11 , 𝑏2 = 4 𝑏

= 𝑘, 𝜔 = 0.


𝑣(𝜉) = 𝑏0 +(

2 𝑛𝑎1 𝜉 1 𝑛2 𝑎1 )e +( )e2𝜉 𝑛+1 4𝑏0 (𝑛+1)2

, −

𝑢(𝜉) = ( 𝑏0 +(

𝑎1 e𝜉 2 𝑛𝑎1 𝜉 1 𝑛2 𝑎1 )e +( )e2𝜉 𝑛+1 4𝑏0 (𝑛+1)2

1 𝑛−1

)

.

where 𝜉 = 𝑘𝑥.

(a)

(b)


255

3.3.6. KD Equation Consider the KD equation [254] 3

𝑢𝑡 − 𝑢𝑥𝑥𝑥 − 6𝑏 2 𝑢𝑢𝑥 + 2 𝑎2 𝑢2 𝑢𝑥 − 3𝑣𝑦 + 3𝑎𝑢𝑥 𝑣 = 0,

(3.24a)

𝑢𝑦 = 𝑣𝑥 .

(3.25a)

Using the transformation 𝜉 = 𝑥 + 𝑦 + 𝜔𝑡, the above equation we have 3

𝜔𝑢′ − 𝑢′′′ − 6𝑏 2 𝑢𝑢′ + 2 𝑎2 𝑢2 𝑢′ − 3𝑣 ′ + 3𝑎𝑢′ 𝑣 = 0,

(3.24b)

𝑙𝑢′ = 𝑘𝑣 ′ .

(3.25b)

Integrating the second equation in the system and neglecting constant of integration we find 𝑙𝑢 = 𝑘𝑣,

(3.26)

using expression (3.26) in (3.24) and then integrating once, we get 1

3

𝜔𝑢(𝜉) − 𝑢′′ (𝜉) − 3𝑏 2 𝑢(𝜉)2 + 2 𝑎2 𝑢(𝜉)3 − 3𝑢(𝜉) + 2 𝑎𝑢(𝜉)2 = 0,(3.27) balance 𝑢3 and 𝑢′′ by using homogenous principal, we have 3𝑀 = 𝑀 + 2, 𝑀 = 1. So, the trial solution is given as 𝑢(𝜉) =

𝑎0 +𝑎1 exp[𝜉]+𝑎2 exp[2𝜉] 𝑏0 +𝑏1 exp[𝜉]+𝑏2 exp[2𝜉]

,

(3.28)

putting (3.28) into (3.27), we have 1 3 (𝑏0 +𝑏1 𝑒 𝜉 +𝑏2 𝑒 2𝜉 )

(𝐶0 + 𝐶1 exp[𝜉] + 𝐶2 exp[2𝜉] + 𝐶3 exp[3𝜉] + 𝐶4 exp[4𝜉] +

𝐶5 exp[5𝜉] + 𝐶6 exp[6𝜉]) = 0. The constants 𝐶𝑖 = (𝑖 = 0,1,2,3,4,5,6) can be obtained by MAPLE 11. 𝐶0 = 2𝜔𝑎0 𝑏02 − 6𝑎0 𝑏02 + 𝑎2 𝑎03 − 6𝑏 2 𝑎02 𝑏0 + 3𝑎𝑎02 𝑏0 , . . .. 𝐶6 = 2𝜔𝑎2 𝑏22 + 𝑎2 𝑎23 − 6𝑏 2 𝑎22 𝑏2 + 3𝑎𝑎22 𝑏2 − 6𝑎2 𝑏22 . After solving we have three solution sets which satisfy the given equation 1st Solution Set: 256

1

𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏0 = 𝑏0 , 𝑏1 = − 2 𝑎𝑎1 + 𝑏 2 𝑎1 , 𝑏2 = 1 𝑏 2 𝑎12 (−𝑏2 +𝑎)

−4

𝑏0

, 𝑘 = 𝑘, 𝜔 = 4.

Their corresponding solution is 𝑎1 exp[𝜉]

𝑢(𝜉) =

1 2

𝑏0 +(− 𝑎𝑎1 +𝑏 2 𝑎1 ) exp[𝜉]+(−

2 1𝑏2 𝑎1 (−𝑏2 +𝑎) )exp[2𝜉] 4 𝑏0

,

where 𝜉 = 𝑥 + 𝑦 + 4𝑡.

(a)

(b)

Fig. 3.34.(a)-(b): Graphical representation of Eq. (3.24a-3.25a) for different values of parameters

2nd Solution Set: 1

1

𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑎2 = − 4 𝑘 8 𝑏4 (−6𝑎1 𝑏02 𝑎0 𝑏1 𝑘 4 + 4𝑘 8 𝑎0 𝑏12 𝑏02 − 2𝑎1 𝑏0 𝑎02 𝑏1 + 0

4𝑏0 𝑎02 𝑏12 𝑘 4 1

+

𝑎03 𝑏12

+

𝑎0 𝑎12 𝑏02

− 4𝑘 8 𝑎1 𝑏03 𝑏1 + 2𝑎12 𝑏03 𝑘 4 ), 𝑏0 = 𝑏0 , 𝑏1 = 𝑏1 , 𝑏2 =

1

− 4 𝑘 8 𝑏3 (−2𝑎1 𝑏0 𝑎0 𝑏1 + 2𝑘 4 𝑎0 𝑏12 𝑏0 − 2𝑘 4 𝑎1 𝑏02 𝑏1 + 𝑎02 𝑏12 + 𝑎12 𝑏02 ), 𝑘 = 𝑘, 𝜔 = 0

1 𝑘 4 +3𝑙2

−4

𝑘

.

Their corresponding solution is 𝑢(𝜉) =

1 1 (−6𝑎1 𝑏02 𝑎0 𝑏1 𝑘 4 +⋯+2𝑎12 𝑏03 𝑘 4 )exp[2𝜉] 4𝑘8 𝑏4 0 1 1 𝑏0 +𝑏1 exp[𝜉]− 8 3 (−2𝑎1 𝑏0 𝑎0 𝑏1 +⋯+𝑎12 𝑏02 )exp[2𝜉] 4𝑘 𝑏 0

𝑎0 +𝑎1 exp[𝜉]−

1 𝑘 4 +3𝑙2

where 𝜉 = 𝑥 + 𝑦 − 4

𝑘

𝑡.

257

,

(a)

(b)


3rd Solution Set: 𝑎0 = 0, 𝑎1 = − 1 𝑘 4 +3𝑙2

−4

𝑘

𝑏1 (2𝑏2 𝑘 4 −𝑎2 ) 𝑏2

, 𝑎2 = 𝑎2 , 𝑏0 = 0, 𝑏1 = 𝑏1 , 𝑏2 = 𝑏2 , 𝑘 = 𝑘, 𝜔 =

.

Their corresponding solution is 𝑢(𝜉) =

𝑏 (2𝑏 𝑘4 −𝑎2 ) (− 1 2 ) exp[𝜉]+𝑎2 exp[2𝜉] 𝑏2

𝑏1 exp[𝜉]+𝑏2 exp[2𝜉] 1 𝑘 4 +3𝑙2

where 𝜉 = 𝑥 + 𝑦 − 4

𝑘

,

𝑡.

(a)

(b)


3.3.7. Long Short Wave Interaction System Consider the long-short-wave interaction system [246] 258

𝑖𝜓𝑡 + 𝜓𝑥𝑥 − 𝜓𝑣 = 0, 𝑣𝑡 + 𝑣𝑥 + (|𝜓|2 )𝑥 = 0.

(3.29)

Where 𝜓(𝑥, 𝑡) is a complex function and 𝑣(𝑥, 𝑡)is a real function. The sonic-Langmuir soliton of system were obtained by using the inverse scattering method. Let 𝜓(𝑥, 𝑡) = 𝑒 𝑖𝜂 𝑢(𝑥, 𝑡),

𝜂 = 𝛼𝑥 + 𝛽𝑡,

𝑣(𝑥, 𝑡) = 𝑉(𝜂),

𝑢(𝑥, 𝑡) = 𝑈(𝜂),

(3.30)

where 𝜂 = 𝑥 − 𝜔𝑡, 𝛼 and 𝛽 are constants and 𝑢(𝑥, 𝑡) is a real function. Substituting (3.30) into (3.29), we find 𝜔 = 2𝛼and 𝑈, 𝑉 satisfy the following coupled ODEs: 𝑈 ′′ − 𝑈𝑉 − (𝛼 2 + 𝛽)𝑈 = 0,

(3.31)

(1 − 2𝛼)𝑉 ′ + 2𝑈𝑈 ′ = 0.

(3.32)

Integrating(3.32), we get 𝑈2

𝑉 = − 1−2𝛼 , using the above transformation in (3.31), we get 𝑈3

𝑈 ′′ + 1−2𝛼 − (𝛼 2 + 𝛽)𝑈 = 0, applying the homogeneous balancing principle on 𝑈 3 and 𝑈 ′′ 3𝑀 = 𝑀 + 2, 𝑀 = 1. So, the trial solution is given as 𝑈(𝜉) =

𝑎0 +𝑎1 exp[𝜉]+𝑎2 exp[2𝜉] 𝑏0 +𝑏1 exp[𝜉]+𝑏2 exp[2𝜉]

,

(3.33)

putting (3.33) into (3.32), we have 1 3 (𝑏0 +𝑏1 𝑒 𝜉 +𝑏2 𝑒 2𝜉 ) (−1+2𝛼)


𝐶4 exp[4𝜉] + 𝐶5 exp[5𝜉] + 𝐶6 exp[6𝜉]) = 0. The constants 𝐶𝑖 = (𝑖 = 0,1,2,3,4,5,6) can be obtained by MAPLE 11. 𝐶0 = −𝛽𝑎0 𝑏02 + 2𝛼 3 𝑎0 𝑏02 + 𝑎03 − 𝛼 2 𝑎0 𝑏02 + 2𝛽𝑎0 𝑏02 𝛼, . . .. 𝐶6 = −𝛼 2 𝑎2 𝑏22 − 𝛽𝑎2 𝑏22 + 2𝛽𝑎2 𝑏22 𝛼 + 2𝛼 3 𝑎2 𝑏22 + 𝑎23 . After solving we have two solution sets which satisfy the given system 259

1st Solution Set: 1 2𝑎12 +𝑏12

𝛼=2

𝑏12

𝑏1 , 𝑏2 = −

1 3𝑏14 +4𝑎12 𝑏12 +4𝑎14

,𝛽 = −4

𝑎2 𝑏1 𝑎1

𝑏14

, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 𝑎2 , 𝑏0 = 0, 𝑏1 =

, 𝜔 = 𝜔.

Their corresponding solution is 𝑢(𝜂) =

𝑎1 exp[𝜉]+𝑎2 exp[2𝜉]

, 𝑎 𝑏 𝑏1 exp[𝜉]−( 2 1 )exp[2𝜉] 𝑎1

where 𝜂 = 𝑥 − 𝜔𝑡.

(a)

(b)

Fig. 3.37.(a)-(b): Graphical representation of Eq. (3.29) for different values of parameter

2nd Solution Set: 1 𝑏02 +2𝑎02

𝛼=2

𝑏02

1 3𝑏04 +4𝑎02 𝑏02 +4𝑎04

,𝛽 = −4

𝑏04

1 𝑎12 𝑏02 −𝑎02 𝑏12

𝑏0 , 𝑏1 = 𝑏1 , 𝑏2 = − 4

𝑎02 𝑏0

1 𝑎12 𝑏02 −𝑎02 𝑏12

, 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑎2 = 4

, 𝜔 = 𝜔.

Their corresponding solution is 2 2 2 2 1𝑎1 𝑏0 −𝑎0 𝑏1 )exp[2𝜉] 4 𝑎0 𝑏2 0 2 2 2 2 1𝑎 𝑏 −𝑎 𝑏 𝑏0 +𝑏1 exp[𝜉]+(− 1 02 0 1 )exp[2𝜉] 4 𝑎0 𝑏0

𝑎0 +𝑎1 exp[𝜉]+(

𝑢(𝜂) =

where 𝜂 = 𝑥 − 𝜔𝑡.

260

,

𝑎0 𝑏02

, 𝑏0 =

(a)

(b)


3.3.8. Nizhnik-Novikov-Veselov Equations Consider the Nizhnik-Novikov-Veselov equation [65] 𝑢𝑡 + 𝑎𝑢𝑥𝑥𝑥 + 𝑏𝑢𝑦𝑦𝑦 − 3𝑎(𝑢𝑣)𝑥 − 3𝑏(𝑢𝑤)𝑦 = 0, 𝑢𝑥 = 𝑣𝑦 , 𝑢𝑦 = 𝑤𝑥, Introducing the transformation as 𝜉 = 𝑥 + 𝑦 + 𝑐𝑡, we can convert the given equation into ordinary differential equations, 𝑐𝑢′ + 𝑎𝑢′′′ + 𝑏𝑢′′′ − 3𝑎(𝑢𝑣)′ − 3𝑏(𝑢𝑤)′ = 0, 𝑢′ = 𝑣 ′ , 𝑢′ = 𝑤 ′ . Integrating once the above equation 𝑐𝑢 + 𝑎𝑢′′ + 𝑏𝑢′′ − 3𝑎(𝑢𝑣) − 3𝑏(𝑢𝑤) = 0.

(3.34)

𝑢 = 𝑣,

(3.35)

𝑢 = 𝑤.

(3.36)

Using (3.35) and (3.36) in (3.34) we have, 𝑐𝑢 + (𝑎 + 𝑏)𝑢′′ − 3(𝑎 + 𝑏)𝑢2 .

(3.37)

Balance the 𝑢2 and 𝑢′′ by using homogenous principal, we have 𝑀 + 2 = 2𝑀, 𝑀 = 2. Then we have the trail solution is 261

𝑎0 +𝑎2 𝑒 2𝜉 +𝑎4 𝑒 4𝜉

𝑢(𝜉) =

𝑏0 +𝑏2 𝑒 2𝜉 +𝑏4 𝑒 4𝜉

,

(3.38)

put (3.38) in (3.37), we have ODE is 1 (𝑏0 +𝑏2 exp[2𝜉]+𝑏4 exp[4𝜉])3


𝐶4 exp[4𝜉] + 𝐶5 exp[5𝜉] + 𝐶6 exp[6𝜉]) = 0, the constant 𝐶𝑖 (𝑖 = 0,1,2,3,4,5,6) can obtain by Maple 13, given as 𝐶0 = 𝑐𝑎0 𝑏02 − 3𝑎02 𝑎𝑏0 − 3𝑎02 𝑏𝑏0 , . . . 𝐶6 = −3𝑎42 𝑎𝑏4 − 3𝑎42 𝑏𝑏4 + 𝑐𝑎4 𝑏42 . Equating the coefficients of exp[𝑛𝜉] to be zero, we obtain [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0], we have two solution sets satisfying the given equation. 1st Solution Set: 3 𝑎2

4

3

9 𝑎2

𝑐 = 4𝑏 + 4𝑎, 𝑎0 = 3 𝑏0 , 𝑎2 = 𝑎2 , 𝑎4 = 64 𝑏2 , 𝑏0 = 𝑏0 , 𝑏2 = − 8 𝑎2 , 𝑏4 = 256 𝑏2 . 0

0


2 4 3 𝑎2 4𝜉 𝑏 +𝑎2 𝑒 2𝜉 +( )𝑒 3 0 64𝑏0 3 8

𝑏0 −( 𝑎2 )𝑒 2𝜉 +(

2 9 𝑎2 4𝜉 )𝑒 256𝑏0

,

where 𝜉 = 𝑥 + 𝑦 + (4𝑏 + 4𝑎)𝑡.

(a)

(b)


262

2nd Solution Set: 1 𝑏2

𝑐 = −4𝑏 − 4𝑎, 𝑎0 = 0, 𝑎2 = −4𝑏2 , 𝑎4 = 0, 𝑏0 = 𝑏0 , 𝑏2 = 𝑏2 , 𝑏4 = 4 𝑏2 . 0


−4𝑏2 𝑒 2𝜉 𝑏0 +𝑏2 𝑒 2𝜉 +(

2 1𝑏2 4𝜉 )𝑒 4𝑏0

,

where 𝜉 = 𝑥 + 𝑦 + (−4𝑏 − 4𝑎)𝑡.

(a)

(b)


3.3.9. System of Zakhrov We consider the following system of Zakharov [110] 𝑢𝑡𝑡 − 𝑐𝑠2 𝑢𝑥𝑥 − 𝛽(|𝑣|2 )𝑥𝑥 = 0,

(3.39a)

𝑖𝑣𝑡 + 𝛼𝑢𝑥𝑥 − 𝛿𝑢𝑣 = 0.

(3.39b)

where 𝑢 = 𝑢(𝑥, 𝑡) is the perturbed number density of the ion (in the low-frequency response), 𝑣 = 𝑣(𝑥, 𝑡) is the slow variation amplitude of the electric field intensity, 𝑐𝑠 is the thermal transportation velocity of the electron-ion,𝛼 ≠ 0, 𝛽 ≠ 0, 𝛿 ≠ 0, 𝑐𝑠 are constants. The above system is one of the fundamental models governing dynamics of nonlinear waves, and describes the interactions between high- and low- frequency waves. The physically most important example involves the interaction between Langmuir and ion-acoustic waves in plasma. Since 𝑣(𝑥, 𝑡) is a complex function thus we introduce a guage transformation: 𝑢 = 𝑢(𝑥, 𝑡) = 𝑢(𝜉),

(3.40a)

𝑣 = 𝑣(𝑥, 𝑡) = 𝑣(𝜉) = 𝜙(𝜉) exp[𝑖(𝑠𝑥 − 𝑤𝑡)],

(3.40b)

263

𝜉 = 𝑘(𝑥 − 𝑐𝑡) + 𝜉0 .

(3.40c)

Where 𝜙(𝜉) a real-valued is function,𝑠, 𝑤, 𝑘, 𝑐 is four real constants to be determined, and 𝜉0 is an arbitrary constant. Substituting (3.40) into (3.39) we have: 𝑘 2 (𝑐 2 − 𝑐𝑠2 )𝑢′′ − 𝛽𝑘 2 (𝜙 2 )′′ ) = 0,

(3.41a)

𝛼𝑘 2 𝜙 ′′ + (𝑤 − 𝛼𝑠 2 )𝜙 − 𝛿𝑢𝜙 + 𝑖(2𝛼𝑠𝑘 − 𝑘𝑐)𝜙 ′ = 0.

(3.41b)

Integrating equation (3.41a) twice with respect to 𝑢 and put the integration constants to zero, we obtain 𝛽

𝑢 = 𝑐 2 −𝑐 2 𝜙 2 ,

(3.42)

𝑠

substituting (3.42) into (3.41b), we obtain: 𝛿𝛽

𝛼𝑘 2 𝜙 ′′ + (𝑤 − 𝛼𝑠 2 )𝜙 − 𝑐 2 −𝑐 2 𝜙 3 = 0,

(3.43)

𝑠

balance the 𝜙 3 and 𝜙 ′′ by using homogenous principal, we have 3𝑀 = 𝑀 + 2, 𝑀 = 1. Then the trail solution is 𝑎0 +𝑎1 𝑒 𝜉 +𝑎2 𝑒 2𝜉

𝑢(𝜉) =

𝑏0 +𝑏1 𝑒 𝜉 +𝑏2 𝑒 2𝜉

.

(3.44)

put (3.44) in (3.43), we have −1 (𝑐 2 −𝑐𝑠2 )(𝑏0 +𝑏1 exp[𝜉]+𝑏2 exp[2𝜉])3


𝐶4 exp[4𝜉] + 𝐶5 exp[5𝜉] + 𝐶6 exp[6𝜉]) = 0, the constant 𝐶𝑖 (𝑖 = 0,1,2,3,4,5,6) can obtain by Maple 13, given as 𝐶0 = −𝜔𝑎0 𝑐 2 𝑏02 + 𝜔𝑎0 𝑐𝑠2 𝑏02 + 𝛼𝑠 2 𝑎0 𝑐 2 𝑏02 − 𝛼𝑠 2 𝑎0 𝑐𝑠2 𝑏02 + 𝛿𝛽𝑎03 , . . .. 𝐶6 = 𝛿𝛽𝑎23 − 𝛼𝑠 2 𝑎2 𝑐𝑠2 𝑏22 − 𝜔𝑎2 𝑐 2 𝑏22 + 𝜔𝑎2 𝑐𝑠2 𝑏22 + 𝛼𝑠 2 𝑎2 𝑐 2 𝑏22 Equating the coefficient of exp[𝑛𝜉] to be zero, we obtain [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0], we have three solutions set satisfying the given equation. 1st Solution Set: 264

𝛿𝛽

𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 𝑎2 , 𝑏0 = 0, 𝑏1 = √−𝜔𝑐 2 −𝛼𝑠2 𝑐 2 +𝛼𝑠2 𝑐 2 +𝜔𝑐 2 𝑎1 , 𝑏2 = 𝑠

−

𝑎2 𝛿𝛽 𝛿𝛽 2 2 2 𝑐2 +𝛼𝑠2 𝑐2 +𝜔𝑐2 (𝑐 −𝑐𝑠 ) −𝜔𝑐2 −𝛼𝑠 𝑠 𝑠

(𝛼𝑠2 −𝜔)√

𝑠

, 𝑘 = 𝑘, 𝑐 = 𝑐.

2nd Solution Set: 𝛿𝛽

𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑏0 = √−𝜔𝑐 2 −𝛼𝑠2 𝑐 2 +𝛼𝑠2 𝑐 2 +𝜔𝑐 2 𝑎0 , 𝑏1 = 𝑠

−

𝑎1 𝛿𝛽 𝛿𝛽 2 2 2 𝑐2 +𝛼𝑠2 𝑐2 +𝜔𝑐2 (𝑐 −𝑐𝑠 ) −𝜔𝑐2 −𝛼𝑠 𝑠 𝑠

(𝛼𝑠2 −𝜔)√

𝑠

, 𝑏2 = 0, 𝑘 = 𝑘, 𝑐 = 𝑐.

3rd Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝑏0 = −

𝑎0 𝛿𝛽 𝛿𝛽 2 2 2 (𝛼𝑠 −𝜔)√ 2 𝑐2 +𝛼𝑠2 𝑐2 +𝜔𝑐2 (𝑐 −𝑐𝑠 ) −𝜔𝑐2 −𝛼𝑠 𝑠 𝑠

, 𝑏1 =

𝛿𝛽

0, 𝑏2 = √−𝜔𝑐 2 −𝛼𝑠2 𝑐 2 +𝛼𝑠2 𝑐 2 +𝜔𝑐 2 𝑎2 , 𝑘 = 𝑘, 𝑐 = 𝑐. 𝑠

𝑠

4th Solution Set: 𝛿𝛽

𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑎2 = 𝑎2 , 𝑏0 = √−𝜔𝑐 2 −𝛼𝑠2 𝑐 2 +𝛼𝑠2 𝑐 2 +𝜔𝑐 2 𝑎0 , 𝑏1 = 𝑠

−

𝑎1 𝛿𝛽 𝛿𝛽 2 2 2 2 2 2 2 (𝑐 −𝑐𝑠 ) −𝜔𝑐2 𝑠 −𝛼𝑠 𝑐 +𝛼𝑠 𝑐𝑠 +𝜔𝑐

(𝛼𝑠2 −𝜔)√

𝑎2 𝛿𝛽 𝛿𝛽 2 2 2 2 2 2 2 (𝑐 −𝑐𝑠 ) −𝜔𝑐2 𝑠 −𝛼𝑠 𝑐 +𝛼𝑠 𝑐𝑠 +𝜔𝑐

(𝛼𝑠2 −𝜔)√

𝑠

, 𝑏2 =

, 𝑘 = 𝑘, 𝑐 = 𝑐.

Using these solution sets we get the four solutions of 𝜑(𝑥, 𝑡) and the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡) obtained by using the transformation 𝛽

𝑢(𝑥, 𝑡) = 𝑐 2 −𝑐 2 𝜙 2 (𝑥, 𝑡) 𝑠

𝑣(𝑥, 𝑡) = 𝜙(𝑥, 𝑡) exp[𝑖(𝑠𝑥 − 𝑤𝑡)].

3.3.10. Nonlinear Parabolic Equation Consider the following nonlinear parabolic equation [230] 1

𝑢𝑡 − 𝑢2 +𝑢2 (𝑢𝑦2 𝑢𝑥𝑥 + 𝑢𝑥2 𝑢𝑦𝑦 − 2𝑢𝑥 𝑢𝑦 𝑢𝑦𝑥 ) = 0 𝑥

𝑦

Introducing the transformation as 𝜂 = 𝑘𝑥 + 𝑚𝑦 + 𝑤𝑡, we can convert the given equation into ordinary differential equations 1

𝑤𝑢′ − 𝑘 2 𝑢′′ +𝑚2 𝑢′′ (2𝑚2 𝑘 2 𝑢′ 𝑢′ 𝑢′′ − 2𝑚2 𝑘 2 𝑢′𝑢′𝑢′′) = 0. 265

(3.45)

For simplification,we take 𝑀 = 1. Therefore, we have trial solution 𝑢(𝜂) =

𝑎0 +𝑎1 𝑒 𝜂 +𝑎2 𝑒 2𝜂 𝑏0 +𝑏1 𝑒 𝜂 +𝑏2 𝑒 2𝜂

,

(3.46)

put (3.46) in(3.45), we have ODE is 1 (𝑐 (𝑏0 +𝑏1 exp[𝜂]+𝑏2 exp[2𝜂])4 0

+ 𝑐1 exp[𝜂] + 𝑐2 exp[2𝜂] + 𝑐3 exp[3𝜂] +

𝑐4 exp[4𝜂] + 𝑐5 exp[5𝜂] + 𝑐6 exp[6𝜂] + 𝑐7 exp[7𝜂] + 𝑐8 exp[8𝜂] + 𝑐9 exp[9𝜂]) = 0, The constant 𝑐𝑖 (𝑖 = 0,1,2,3,4,5,6,7,8,9)can obtain by Maple 13, given as 𝑐0 = 0, . . .. 𝑐9 = 𝑤𝑚𝑎22 𝑏1 𝑏22 − 𝑤𝑚𝑎1 𝑏23 𝑎2 + 𝑤𝑘𝑎22 𝑏1 𝑏22 − 𝑤𝑘𝑎1 𝑏23 𝑎2 + 𝑘𝑚2 𝑎22 𝑏1 𝑏22 − 𝑘𝑚2 𝑎1 𝑏23 𝑎2 + 𝑚𝑘 2 𝑎22 𝑏1 𝑏22 − 𝑚𝑘 2 𝑎1 𝑏23 𝑎2 . Equating the coefficient of exp[𝑛𝜂] to be zero, we obtain [𝑐0 = 0, 𝑐1 = 0, 𝑐2 = 0, 𝑐3 = 0, 𝑐4 = 0, 𝑐5 = 0, 𝑐6 = 0, 𝑐7 = 0, 𝑐8 = 0, 𝑐9 = 0], We have four solution set satisfy the given equation 1st Solution Set: 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 =

𝑎1 𝑏1 𝑏0

, 𝑏0 = 𝑏0 , 𝑏1 = 𝑏1 , 𝑏2 = 0, 𝑘 = 𝑘, 𝑤 = −


𝑎 𝑏 𝑎1 𝑒 𝜂 +( 1 1 )𝑒 2𝜂 𝑏0

𝑏0 +𝑏1 𝑒 𝜂

where 𝜂 = 𝑘𝑥 + 𝑚𝑦 −

,

𝑚𝑘(𝑚𝑘−𝑘−𝑚) 𝑚+𝑘

𝑡.

266


.

(a)

(b)


2nd Solution Set: 𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝑏0 = 0, 𝑏1 = 𝑏1 , 𝑏2 = 0, 𝑘 = 𝑘, 𝑤 = −


.


𝑎2 𝑒 2𝜂 𝑏1 𝑒 𝜂

,



𝑡.

(a)

(b)


3rd Solution Set: 267

𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝑏0 = 𝑏0 , 𝑏1 = 0, 𝑏2 = 0, 𝑘 = 𝑘, 𝑤 = −

2𝑚𝑘(2𝑚𝑘−𝑘−𝑚) 𝑚+𝑘

.


𝑎2 𝑒 2𝜂 𝑏0

,


2𝑚𝑘(2𝑚𝑘−𝑘−𝑚) 𝑚+𝑘

𝑡.

(a)

(b)


4th Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = 0, 𝑏0 = 0, 𝑏1 = 0, 𝑏2 = 𝑏2 , 𝑘 = 𝑘, 𝑤 = − The above solution yields 𝑢(𝜂) = 𝑏

𝑎0

2𝑒

2𝜂

,


2𝑚𝑘(2𝑚𝑘+𝑘+𝑚) 𝑚+𝑘

𝑡.

268

2𝑚𝑘(2𝑚𝑘+𝑘+𝑚) 𝑚+𝑘

.

(a)

(b)


3.4. (G´/G)-Expansion Method Recently, Wang et al. [213] presented a reliable technique which is called the  G / G  expansion method and obtained exact traveling wave solutions for the nonlinear evolution equations (NLEEs). In this method, second order linear ordinary differential equation with constant coefficients 𝐺 ′′ (𝜂) + 𝜆𝐺 ′ (𝜂) + 𝜇𝐺(𝜂) is used, as an auxiliary equation. (G´/G)-expansion [23, 27, 59, 66, 152, 174, 255, 260] applied to solve the nonlinear evolution equations. Zhao et al. [261] introduced a new modification in (G´/G)expansion method. Letar on, Zayad and Abdelaziz [49] introduce extended (G´/G)expansion method for solving (2+1)-dimensional nonlinear evolution equations. In the recent year, Chen and Li [39] applied multiple (G´/G)-expansion method for nonlinear evolution equations arising in mathematical physics. Recently, Zareen and Usman applied generalized improved rational (G´/G)-Expansion method applied to solve modified Zakharov-Kuznetsov and Equal-Width equations [253]. The purposed technique expresses the solutions in term of rational hyperbolic, the rational trigonometric and rational functions. The proposed scheme is fully compatible with the complexity of such problems and is very user-friendly. Numerical results are very encouraging.

269

3.4.1. Methodology We briefly show what (G´/G)-expansion method is and how to use it to obtain various periodic wave solutions to nonlinear equations. Suppose a nonlinear equation for 𝜓(𝑥, 𝑡) is given by 𝜓(𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑥𝑥 , 𝑢𝑡𝑡 , 𝑢𝑥𝑡 , … ) = 0,

(3.47)

in which both nonlinear term(s) and higher order derivatives of 𝜓(𝑥, 𝑡) are all involved. In general, the left-hand side of Eq. (3.47) is a polynomial in 𝜓 and its various derivatives. The proposed technique for solving Eq. (3.47) proceeds in the following five steps: Step 1: Look for travelling wave solution of Eq. (3.47) by taking 𝑢 = 𝑢(𝜉),

𝜉 = 𝑘𝑥 ± 𝜔𝑡,

(3.48)

where 𝑉 is nonzero constant, 𝑢(𝜉) the function of 𝜉. Substituting (3.48) into Eq. (3.47) yields an ordinary differential equation (ODE) for 𝑢(𝜉). 𝜑(𝑢, ±𝜔𝑢′ , 𝑘𝑢′ , 𝜔2 𝑢′′ , 𝑘 2 𝑢′′ , … ) = 0.

(3.48)

Step 2: If possible, integrate Eq. (3.48) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) may be set to zero. Step 3: According (G´/G)-expansion method the proposed technique, we assume that the wave solution can be expressed in the following form ′ 𝑛 𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 (𝐺 /𝐺) .

(3.49)

where 𝐺 is the solution of first order nonlinear equation in the form 𝐺 ′′ + 𝜆𝐺 ′ + 𝜇𝐺 = 0,

(3.50)

where 𝜆 and 𝜇 are arbitrary constant .Using the general solution of Eq. (3.50), we have 𝐺 ′ (𝜉) 𝐺(𝜉)

=

=

√𝜆2 −4𝜇 2

(

√−𝜆2 +4𝜇 2

=𝑐

2𝑐2

1 +𝑐2 𝜉

1 1 2 2 1 1 𝑐1 cosh( √𝜆2 −4𝜇)+𝑐2 sinh( √𝜆2 −4𝜇𝜉) 2 2

𝑐1 sinh( √𝜆2 −4𝜇𝜉)+𝑐2 cosh( √𝜆2 −4𝜇𝜉)

1 1 2 2 1 1 𝑐1 cos( √−𝜆2 +4𝜇𝜉)+𝑐2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−𝑐1 sin( √−𝜆2 +4𝜇𝜉)+𝑐2 cos( √−𝜆2 +4𝜇𝜉)

(

, 𝜆2 − 4𝜇 = 0.

where 𝑐1 , 𝑐2 are arbitrary constants and it follows, 𝐺 ′ (𝜉)

′

𝐺′

2

𝐺′

( 𝐺(𝜉) ) = − [( 𝐺 ) + 𝜆 ( 𝐺 ) + 𝜇], 270

𝜆

) − 2 , 𝜆2 − 4𝜇 > 0, (3.51) 𝜆

) − 2 , 𝜆2 − 4𝜇 < 0,

𝐺 ′ (𝜉)

′′

𝐺′

3

𝐺′

2

𝐺′

( 𝐺(𝜉) ) = [2 ( 𝐺 ) + 3𝜆 ( 𝐺 ) + (𝜆2 + 2𝜇) ( 𝐺 ) + 𝜇𝜆], . . .. and so on, here the primes denotes the derivative with respective to 𝜉. To determine 𝑢 explicitly, we take the following four steps. Step 4: Substitute Eq. (3.49) along with Eq. (3.50) into Eq. (3.51) and collect all terms with the same order of (𝐺 ′ /𝐺) together, the left-hand side of Eq. (3.48) is converted into a polynomial in (𝐺 ′ /𝐺) .Then set each coefficient of this polynomial to zero to derive a set of algebraic equations for 𝑘, 𝜔, 𝜀 and 𝑎𝑛 , 𝑛 = 0,1, . . . , 𝑀. Step 5: Solve the system of algebraic equations obtained, for 𝑘, 𝑙, 𝑚, 𝜔, 𝜀 and 𝑎𝑛 , 𝑛 =0,1, . . . , 𝑀 by use of MAPLE 13. Step 6: Use the results obtained in above steps to derive a series of fundamental solutions 𝑢(𝜉) of Eq. (3.48) depending on (𝐺 ′ /𝐺) since the solutions of Eq. (3.50) have been well known for us, and then we can obtain exact solutions of Eq. (3.45).

3.4.2. Sine-Gordon Equation We consider Sine-Gordon equation [230] 𝑣𝑡𝑡 − 𝛼 2 𝑣𝑥𝑥 + 𝛽sin(𝑣) = 0, using the transformation sin(𝑣) =

(3.52a)

𝑢−𝑢−1 2𝑖

, 𝑢 = exp(𝑖𝑣). we have the above equation is

2𝑢𝑢𝑡𝑡 − 2(𝑢𝑡 )2 − 2𝛼 2 𝑢𝑢𝑥𝑥 + 2𝛼 2 (𝑢𝑥 )2 + 𝛽𝑢3 − 𝛽𝑢 = 0. Introducing the transformation as 𝜂 = 𝑥 + 𝑐𝑡, we can convert the given equation into ordinary differential equations, 2𝑐 2 𝑢𝑢′′ − 2(𝑢′ )2 𝑐 2 − 2𝛼 2 𝑢𝑢′′ + 2𝛼 2 (𝑢′ )2 + 𝛽𝑢3 − 𝛽𝑢 = 0.

(3.52b)

Balance the 𝑢3 and 𝑢𝑢′′ by using homogenous principal, we have 𝑀 + 𝑀 + 2 = 3𝑀, 𝑀 = 2. Then we have the trail solution is 𝐺′

𝐺′

2

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ) + 𝑎2 ( 𝐺 )

(3.53)

271

By equation (3.50) and (3.53), substitute into (3.52), we obtain the system of algebraic equations for 𝑎0 , 𝑎1 , 𝑐, 𝜆 and 𝜇 as follow: 𝐺′

0

𝐺′

1

2

𝐺′

𝐺′

3

𝐺′

4

𝐺′

5

𝐺′

6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) + 𝐶5 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0, To find the values of unknown constant, we set [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0]. We have two solution sets 1st Solution Set: 𝑐 = √−

−4𝜇𝛼2 −𝛽+𝜆2 𝛼2 −𝜆2 +4𝜇

, 𝑎0 = −

𝜆2 −𝜆2 +4𝜇

, 𝑎1 = −

4𝜆 −𝜆2 +4𝜇

, 𝑎2 = −

4 −𝜆2 +4𝜇

.

Case 1: When λ2 − 4μ > 0, we obtain the travelling wave solution 𝜆2

√𝜆2 −4𝜇

4𝜆

𝑢(𝜉) = − −𝜆2 +4𝜇 − −𝜆2 +4𝜇 ( 4 −𝜆2 +4𝜇

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2

√𝜆2 −4𝜇

(

2

2

1 1 2 2 1 1 c1 cosh( √𝜆2 −4𝜇𝜉)+c2 sinh( √𝜆2 −4𝜇𝜉) 2 2

c1 sinh( √𝜆2 −4𝜇𝜉)+c2 cosh( √𝜆2 −4𝜇𝜉)

(



(



(

λ

) − 2) −

2 λ

) − 2) ,

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 2: When λ2 − 4μ < 0, we obtain the travelling wave solution

272

𝜆2

√−𝜆2 +4𝜇

4𝜆

𝑢(𝜉) = − −𝜆2 +4𝜇 − −𝜆2 +4𝜇 ( λ

4

√−𝜆2 +4𝜇

) − −𝜆2 +4𝜇 (

2

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√−𝜆2 +4𝜇 2

(

2

2

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

)− 2

λ

) − 2) ,

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 3: When λ2 − 4μ = 0, we obtain the travelling wave solution 𝜆2

4𝜆

𝑢(𝜉) = − −𝜆2 +4𝜇 − −𝜆2 +4𝜇 (c where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2

2c2 1 +c2

𝜆

4

− 2) − −𝜆2 +4𝜇 (c 𝜉

𝜆

− 2, 𝜉

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

273

2c2 1 +c2

𝜆 2

− 2) , 𝜉

(a)

(b)


2nd Solution Set: 𝑐 = √−

−4𝜇+𝛽+𝜆2 𝛼2 −𝜆2 +4𝜇

𝜆2

4𝜆

4

, 𝑎0 = −𝜆2 +4𝜇 , 𝑎1 = −𝜆2 +4𝜇 , 𝑎2 = −𝜆2 +4𝜇.


√𝜆2 −4𝜇

4𝜆

𝑢(𝜉) = −𝜆2 +4𝜇 + −𝜆2 +4𝜇 ( 4 −𝜆2 +4𝜇

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2

√𝜆2 −4𝜇

(

2

2



(



(



(

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)

274

2 λ

) − 2) ,

λ

) − 2) +


Case 2: When λ2 − 4μ < 0, we obtain the travelling wave solution 𝜆2

√−𝜆2 +4𝜇

4𝜆

𝑢(𝜉) = −𝜆2 +4𝜇 + −𝜆2 +4𝜇 ( √−𝜆2 +4𝜇

4 −𝜆2 +4𝜇

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

(

√−𝜆2 +4𝜇 2

(

2

2

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

λ

) − 2) +

2 λ

) − 2) ,

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)



4𝜆

𝑢(𝜉) = −𝜆2 +4𝜇 + −𝜆2 +4𝜇 (c where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2 𝜉

2c2 1 +c2

𝜆

4

− 2) + −𝜆2 +4𝜇 (c 𝜉

𝜆

− 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

275

2c2 1 +c2

𝜆 2

− 2) , 𝜉

(a)

(b)


3.4.3. Doubly Sine Gordon Equation Consider the Sine-Gordon equation [230] 𝑣𝑥𝑡 = sin(𝑣) + sin(2𝑣),

(3.54a)

using the transformation sin(𝑣) =

𝑢−𝑢−1 2𝑖

, sin(2𝑣) =

𝑢2 −𝑢−2 2𝑖

, 𝑢 = exp(𝑖𝑣). we have the

above equation is 2𝑢𝑢𝑥𝑡 − 2𝑢𝑥 𝑢𝑡 − 𝑢4 − 𝑢3 + 𝑢 + 1 = 0. Introducing the transformation as 𝜂 = 𝑥 + 𝑐𝑡, we can convert the given equation into ordinary differential equations, after integration we get 2𝑐𝑢𝑢′′ − 2(𝑢′ )2 − 𝑢4 − 𝑢3 + 𝑢 + 1 = 0.

(3.54b)


𝑢(𝜉) = 𝑎0 + 𝑎1 ( )

(3.55)

𝐺


0

𝐺′

1

𝐺′

2

𝐺′

3

𝐺′

4

𝐺′

5

𝐺′

6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) + 𝐶5 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0. To find the values of unknown constant, we set [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0]. 276

We get the solution set 3

1

1

1

1

𝑐 = 2(−𝜆2 +4𝜇) , 𝑎0 = 2 √3√−𝜆2 +4𝜇 𝜆 − 2 , 𝑎1 = √3√−𝜆2 +4𝜇. Case 1: When λ2 − 4μ > 0, we obtain the travelling wave solution 1

1

1

𝑢(𝜉) = 2 √3√−𝜆2 +4𝜇 𝜆 − 2 + √𝜆2 −4𝜇

1

√3√−𝜆2 +4𝜇 ( where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2

2



(



(

λ

) − 2),

λ

)− , 2

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 2: When λ2 − 4μ < 0, we obtain the travelling wave solution 1

1

1

√−𝜆2 +4𝜇

1

𝑢(𝜉) = 2 √3√−𝜆2 +4𝜇 𝜆 − 2 + √3√−𝜆2 +4𝜇 ( where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√−𝜆2 +4𝜇 2

2

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

277

λ

) − 2,

λ

) − 2),

(a)

(b)


Case 3: When λ2 − 4μ = 0, we obtain the travelling wave solution 1

1

1

1

𝑢(𝜉) = 2 √3√−𝜆2 +4𝜇 𝜆 − 2 + √3√−𝜆2 +4𝜇 (c where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2 𝜉

2c2 1 +c2 𝜉

𝜆

− 2),

𝜆

− 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)

Fig. 3.53(a)-(b): Graphical representation of Eq. (3.54a) for different values of parameters

3.4.4. (2+1)-Dimensional Sine Gordon Equation Consider the following Sine-Gordon equation [230]

278

𝑣𝑡𝑡 − 𝑣𝑥𝑥 + 𝑣𝑦𝑦 + 𝑚2 sin(𝑣) = 0, using the transformation sin(𝑣) =

𝑢−𝑢−1 2𝑖

(3.56a)

, 𝑢 = exp(𝑖𝑣). we have the above equation is

2(𝑢𝑢𝑡𝑡 − 𝑢𝑡 2 ) − 2(𝑢𝑢𝑥𝑥 − 𝑢𝑥 2 ) + 2(𝑢𝑢𝑦𝑦 − 𝑢𝑦 2 ) + 𝑚2 (𝑢3 − 𝑢) = 0. Introducing the transformation as 𝜂 = 𝑥 + 𝑐𝑡, we can convert the given equation into ordinary differential equations, after integration we get 2𝑐 2 𝑢𝑢′′ − 2𝑐 2 (𝑢′ )2 + 𝑚2 𝑢3 − 𝑚2 𝑢 = 0.

(3.56b)


2

𝐺′

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ) + 𝑎2 ( 𝐺 )

(3.57)


0

𝐺′

1

𝐺′

2

𝐺′

3

𝐺′

4

𝐺′

5

𝐺′

6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) + 𝐶5 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0, To find the values of unknown constant, we set [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0]. We have two solution sets 1st Solution Set: 𝜆2

1

4𝜆

4

𝑐 = √4𝜇−𝜆2 𝑚, 𝑎0 = − 4𝜇−𝜆2 , 𝑎1 = − 4𝜇−𝜆2 , 𝑎2 = − 4𝜇−𝜆2 . Case 1: When λ2 − 4μ > 0, we obtain the travelling wave solution 𝜆2

√𝜆2 −4𝜇

4𝜆

𝑢(𝜉) = − 4𝜇−𝜆2 − 4𝜇−𝜆2 ( 4 4𝜇−𝜆2

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2

√𝜆2 −4𝜇

(

2

(

2



(





(

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

279

λ

) − 2,

2 λ

) − 2) ,

λ

) − 2) −

(a)

(b)



√−𝜆2 +4𝜇

4𝜆

𝑢(𝜉) = − 4𝜇−𝜆2 − 4𝜇−𝜆2 ( 4 4𝜇−𝜆2

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√−𝜆2 +4𝜇

(

√−𝜆2 +4𝜇 2

2

(

2

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

λ

) − 2) −

2 λ

) − 2) ,

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 3: When λ2 − 4μ = 0, we obtain the travelling wave solution

280

𝜆2

4𝜆

𝑢(𝜉) = − 4𝜇−𝜆2 − 4𝜇−𝜆2 (c where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2 𝜉

2c2 1 +c2

𝜆

4

− 2) − 4𝜇−𝜆2 (c 𝜉

2c2 1 +c2

𝜆 2

− 2) , 𝜉

𝜆

− 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


2nd Solution Set: 𝜆2

1

4𝜆

4

𝑐 = √− 4𝜇−𝜆2 𝑚, 𝑎0 = 4𝜇−𝜆2 , 𝑎1 = 4𝜇−𝜆2 , 𝑎2 = 4𝜇−𝜆2 . Case 1: When λ2 − 4μ > 0, we obtain the travelling wave solution 𝜆2

√𝜆2 −4𝜇

4𝜆

𝑢(𝜉) = 4𝜇−𝜆2 + 4𝜇−𝜆2 ( 4 4𝜇−𝜆2

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2

√𝜆2 −4𝜇

(

2

(

2



(





(

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

281

λ

) − 2,

2 λ

) − 2) ,

λ

) − 2) +

(a)

(b)



4𝜆

√−𝜆2 +4𝜇

𝑢(𝜉) = 4𝜇−𝜆2 + 4𝜇−𝜆2 ( 4 4𝜇−𝜆2

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√−𝜆2 +4𝜇

(

√−𝜆2 +4𝜇 2

2

(

2

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

λ

) − 2) +

2 λ

) − 2) ,

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


282


4𝜆

𝑢(𝜉) = 4𝜇−𝜆2 + 4𝜇−𝜆2 (c where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2 𝜉

2c2 1 +c2

𝜆

4

− 2) + 4𝜇−𝜆2 (c 𝜉

𝜆 2

2c2 1 +c2

− 2) , 𝜉

𝜆

− 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


3.4.5. Klein Gordon Equation Consider the Klein Gordon equation [230] 𝑢𝑡𝑡 + 𝛼𝑢𝑥𝑥 − 𝛽𝑢 − 𝛾𝑢3 = 0,

(3.58a)

Introducing the transformation as 𝜂 = 𝑥 + 𝑐𝑡, we can convert the given equation into ordinary differential equations, 𝑐 2 𝑢′′ + 𝛼𝑢′′ − 𝛽𝑢 − 𝛾𝑢3 = 0.

(3.58b)

Balance the 𝑢3 and 𝑢′′ by using homogenous principal, we have 𝑀 + 2 = 3𝑀, 𝑀 = 1. Then we have the trail solution is 𝐺′

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 )

(3.59)

By equation and (3.50), substitute into (3.58), we obtain the system of algebraic equations for 𝑎0 , 𝑎1 , 𝑐, 𝜆 and 𝜇 as follow: 𝐺′

0

𝐺′

1

𝐺′

2

𝐺′

3

𝐺′

4

𝐺′

5

𝐺′

6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) + 𝐶5 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0, 283

To find the values of unknown constant, we set [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0]. Solve the system of equation, we have the solution set 𝑐 = √−

−4𝛼𝜇+𝛼𝜆2 +2𝛽

𝛽

, 𝑎0 = √− −4𝜇𝛾+𝛾𝜆2 𝜆, 𝑎1 =

−4𝜇+𝜆2

2𝛽 (−4𝜇+𝜆2 )𝛾√−

𝛽 −4𝜇𝛾+𝛾𝜆2

.

Case 1: When λ2 − 4μ > 0, we obtain the travelling wave solution 𝛽

𝑢(𝜉) = √− −4𝜇𝛾+𝛾𝜆2 𝜆 + √𝜆2 −4𝜇

2𝛽 𝛽 (−4𝜇+𝜆2 )𝛾√− −4𝜇𝛾+𝛾𝜆2

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2

(

2



(



(

λ

) − 2),

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 2: When λ2 − 4μ < 0, we obtain the travelling wave solution 𝛽

𝑢(𝜉) = √− −4𝜇𝛾+𝛾𝜆2 𝜆 + √𝜆2 −4𝜇

2𝛽 𝛽 (−4𝜇+𝜆2 )𝛾√− −4𝜇𝛾+𝛾𝜆2

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√−𝜆2 +4𝜇 2

(

(

2



(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

284

λ

) − 2,

λ

) − 2),

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 3: When λ2 − 4μ = 0, we obtain the travelling wave solution 𝛽

𝑢(𝜉) = √− −4𝜇𝛾+𝛾𝜆2 𝜆 + where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2 𝜉

2𝛽 𝛽 (−4𝜇+𝜆2 )𝛾√− −4𝜇𝛾+𝛾𝜆2

(c

2c2

𝜆

1 +c2 𝜉

− 2),

𝜆

− 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


3.4.6. Potential Kadomtsev-Petviashvili Equation Consider the Potential Kadomtsev-Petviashvili equation [230] 285

4𝑢𝑥𝑡 + 6𝑢𝑥 𝑢𝑥𝑥 + 4𝑢𝑥𝑥𝑥𝑥 + 3𝑢𝑦𝑦𝑦 = 0,

(3.60a)

Introducing the transformation as 𝜂 = 𝑥 + 𝑐𝑡, we can convert the given equation into ordinary differential equations, after integration we get 4𝑐𝑢′ + 3(𝑢′ )2 + 𝑢′′′ + 3𝑢′ = 0.

(3.60b)

Balance the (𝑢′ )2 and 𝑢′′′ by using homogenous principal, we have 2𝑀 + 2 = 𝑀 + 3, 𝑀 = 1. Then we have the trail solution is 𝐺′

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 )

(3.61)


0

𝐺′

1

𝐺′

2

𝐺′

3

𝐺′

4

𝐺′

5

𝐺′

6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) + 𝐶5 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0, To find the values of unknown constant, we set [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0]. Solve the system of equation, we have 1

3

𝑐 = 𝜇 − 4 𝜆2 − 4 , 𝑎0 = 𝑎0 , 𝑎1 = 2. Case 1: When λ2 − 4μ > 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + 2 ( where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2

√𝜆2 −4𝜇 2

(





(

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

286

λ

) − 2,

λ

) − 2),

(a)

(b)


Case 2: When λ2 − 4μ < 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + 2 ( where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√−𝜆2 +4𝜇 2

√−𝜆2 +4𝜇

(

2

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

λ

) − 2),

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 3: When λ2 − 4μ = 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + 2 (c where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2

2c2 1 +c2 𝜉

𝜆

− 2),

𝜆

− 2, 𝜉 287

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


3.4.7. Nonlinear Wave Equation of Longitudinal Oscillation Consider the nonlinear wave equation of longitudinal oscillation of a nonlinear elastic rod with lateral inertia [113] with 𝑛 = 2. 𝜕2 𝑢 𝜕𝑡 2

𝜕𝑢 𝜕2 𝑢

− 𝑐02 [1 + 2𝑎2 𝜕𝑥 ] 𝜕𝑥 2 −

𝑣 2 𝐽𝑝

𝜕4 𝑢

𝑠

𝜕𝑡 2 𝜕𝑥 2

= 0.

(3.62a)

Introducing the transformation as 𝜂 = 𝑥 + 𝑐𝑡, we can convert the given equation into ordinary differential equations, after integration we get 𝑐 2 𝑢′ − 𝑐02 𝑢′ − 𝑐02 𝑎2 (𝑢′ )2 − 𝑐 2

𝑣 2 𝐽𝑝 𝑠

𝑢′′′ = 0.

(3.62b)

Balance the (𝑢′ )2 and 𝑢′′′ by using homogenous principal, we have 2𝑀 + 2 = 𝑀 + 3, 𝑀 = 1. Then we have the trail solution is 𝐺′

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ).

(3.63)


0

𝐺′

1

𝐺′

2

𝐺′

3

𝐺′

4

𝐺′

5

𝐺′

6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) + 𝐶5 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0, To find the values of unknown constant, we set [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0]. 288

Solve the system of equation, we have the solution set is 6𝑣 2 𝐽𝑝

𝑠

𝑐 = √𝑠+4𝑣 2 𝐽 𝜇−𝑣2 𝐽 𝜆2 𝑐0 , 𝑎0 = 𝑎0 , 𝑎1 = (𝑠+4𝑣2 𝐽 𝜇−𝑣2 𝐽 𝜆2 )𝑎 . 𝑝 𝑝 𝑝 𝑝 2 Case 1: When λ2 − 4μ > 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + (𝑠+4𝑣2 𝐽

6𝑣 2 𝐽𝑝

𝑝 𝜇−𝑣

λ

√𝜆2 −4𝜇

2 𝐽 𝜆2 )𝑎 𝑝 2

(

2



(

)−

),

2

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2



(

λ

)− , 2

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 2: When λ2 − 4μ < 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + (𝑠+4𝑣2 𝐽

6𝑣 2 𝐽𝑝

2 2 𝑝 𝜇−𝑣 𝐽𝑝 𝜆 )𝑎2

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√−𝜆2 +4𝜇 2

(

√−𝜆2 +4𝜇

(

2

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

289

λ

) − 2,

λ

) − 2),

(a)

(b)


Case 3: When λ2 − 4μ = 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + (𝑠+4𝑣2 𝐽

6𝑣 2 𝐽𝑝

(

2c2

2 2 𝑝 𝜇−𝑣 𝐽𝑝 𝜆 )𝑎2 c1 +c2 𝜉

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2

𝜆

− 2),

𝜆

− 2, 𝜉

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


3.4.8. Nonlinear Schr𝐨̈ dinger Equation We consider higher dimensional Schrö dinger equation [136] 𝜕𝜓 𝜕𝑥

= 𝑖𝛼1

𝜕2 𝜓 𝜕𝑡 2

+ 𝑖𝛼2 𝜓|𝜓|2 + 𝛼3

𝜕3 𝜓 𝜕𝑡 3

+ 𝛼4 290

𝜕𝜓|𝜓|2 𝜕𝑡

+ 𝛼5 𝜓

𝜕|𝜓|2 𝜕𝑡

.

using the gauge transformation 𝜓(𝑥, 𝑡) = 𝑢(𝑥, 𝑡) exp[𝑖(𝑘𝑥 − 𝜔𝑡)]. where 𝑘 and 𝜔 are constants |𝜓|2 = 𝑢2 exp[𝑖(𝑘𝑥 − 𝜔𝑡)], 𝜓|𝜓|2 = 𝑢3 exp[𝑖(𝑘𝑥 − 𝜔𝑡)], system of equations are, (𝑎1 − 3𝑎3 𝜔)𝑢𝑡𝑡 + (𝑎3 𝜔3 − 𝑎1 𝜔2 − 𝑘)𝑢 + (𝛼2 − 𝛼4 𝜔)𝑢3 = 0, 𝛼3 𝑢𝑡𝑡𝑡 + 2𝛽𝛼1 𝜔𝑢𝑡 − 3𝛽𝛼3 𝜔2 𝑢𝑡 + 𝜆1 𝑘𝑢𝑥 + (3𝛼4 + 2𝛼5 )𝑢2 𝑢𝑡 = 0. Introducing the transformation 𝜉 = 𝛽𝑡 + 𝜆1 𝑥, the above system of PDEs reduces to system of ODEs as, (𝛽 2 𝑎1 − 3𝛽 2 𝑎3 𝜔)𝑢′′ + (𝑎3 𝜔3 − 𝑎1 𝜔2 − 𝑘)𝑢 + (𝛼2 − 𝛼4 𝜔)𝑢3 = 0, 𝛽 3 𝛼3 𝑢′′′ + (2𝛽𝛼1 𝜔 − 3𝛽𝛼3 𝜔2 + 𝜆1 𝑘)𝑢′ + (3𝛽𝛼4 + 2𝛽𝛼5 )𝑢2 𝑢′ = 0. It is easy to see that the above system reduce to an equation as, 𝑢′′ +

2𝛽𝛼1 𝜔−3𝛽𝛼2 𝜔 2 +𝜆1 𝛽 3 𝛼3

𝑢+

3𝛼4 +2𝛼5 3𝛽 2 𝛼3

𝑢3 = 0.

(3.64)

Under the constraint conditions 𝜔=

3𝛼1 𝛼4 +2𝛼1 𝛼5 −3𝛼2 𝛼3 6𝛼3 (𝛼4 +𝛼5 ) 1

,

1

𝑘 = 𝛼 [𝛽 (3𝛼3 𝜔 − 𝛼1 )𝜆1 − 2𝜔(𝛼1 − 3𝛼3 𝜔)2 ], 3

Balance 𝑢3 and 𝑢′′ by using homogenous principal, we have 𝑀 + 2 = 3𝑀, 𝑀 = 1. Then we have the trail solution as, 𝐺′

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 )

(3.65)


0

𝐺′

1

𝐺′

2

𝐺′

3

𝐺′

4

𝐺′

5

𝐺′

6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) + 𝐶5 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0, To find the values of unknown constant, we set [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0]. After solving we have 𝑐 = √−

−4𝜇𝛼2 −𝛽+𝜆2 𝛼2 −𝜆2 +4𝜇

𝜆2

4𝜆

4

, 𝑎0 = − −𝜆2 +4𝜇 , 𝑎1 = − −𝜆2 +4𝜇 , 𝑎2 = − −𝜆2 +4𝜇 291


√𝜆2 −4𝜇

4𝜆

𝑢(𝜉) = − −𝜆2 +4𝜇 − −𝜆2 +4𝜇 ( √𝜆2 −4𝜇

4 −𝜆2 +4𝜇

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2

(

2

2



(



(



(

λ

) − 2) −

2 λ

) − 2) ,

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡. Case 2: When λ2 − 4μ < 0, we obtain the travelling wave solution 𝜆2

√−𝜆2 +4𝜇

4𝜆

𝑢(𝜉) = − −𝜆2 +4𝜇 − −𝜆2 +4𝜇 ( λ 2

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√−𝜆2 +4𝜇

4

) − −𝜆2 +4𝜇 (

√−𝜆2 +4𝜇 2

(

2

2

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

2

λ

) − 2,


4𝜆

where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2

2c2 1 +c2

𝜆

4

− 2) − −𝜆2 +4𝜇 (c 𝜉

2c2 1 +c2

𝜆 2

− 2) , 𝜉

𝜆

− 2, 𝜉

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡. where 𝜉 = 𝛽𝑡 + 𝜆1 𝑥, 𝛽 =

16𝛼4 −4𝛼5 𝑎0 𝛼3

√−3

𝜆

and also

1 1

𝜆1 = 9 𝜆3 ((−6𝛼1 𝜔𝜆2 + 9𝜔2 𝛼3 𝜆2 − 3𝛼4 𝑎02 𝜆2 + 2𝛼5 𝑎02 𝜆2 + 12𝛼4 𝑎02 𝜇 − 8𝛼5 𝑎02 𝜇)𝑎0 √−

18𝛼4 −12𝛼5 𝛼3

),

in all three cases the solution is given as 𝜓(𝑥, 𝑡) = 𝑢(𝑥, 𝑡) exp[𝑖(𝑘𝑥 − 𝜔𝑡)]. 292

λ

) − 2) ,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

𝑢(𝜉) = − −𝜆2 +4𝜇 − −𝜆2 +4𝜇 (c

)−

3.4.9. Vankhnenko Equation We consider Vakhnenko equation [208] 𝑢𝑥𝑡 − 𝑢𝑥 2 + 𝑢𝑢𝑥 + 𝑢 = 0.

(3.66a)

we introduce new independent variables 𝑋, 𝑇 defined by 𝑥 = 𝑇 + 𝑊(𝑋, 𝑇) + 𝑥0 , 𝑡 = 𝑋, where 𝑢(𝑥, 𝑡) = 𝑊𝑥 (𝑋, 𝑇), 𝑥0 is an arbitrary constant, we have 𝜕 𝜕𝑋

𝜕

𝜕

𝜕

= 𝜕𝑡 + 𝑢 𝜕𝑥,

𝜕𝑇

𝜕

= (1 − 𝑊𝑇 ) 𝜕𝑥.

The Vakhnenko equation is given as 𝑊𝑋𝑋𝑇 + 𝑊𝑋 𝑊𝑇 + 𝑊𝑋 = 0. Introducing the transformation 𝜉 = 𝑥 + 𝑐𝑡, the above PDE reduces to ODE as, 𝑐𝑊 ′′′ + 𝑐(𝑊 ′ )2 + 𝑊 ′ = 0. Balance the (𝑊 ′ )2 and 𝑊 ′′′ by using homogenous principal, we have 2𝑀 + 2 = 𝑀 + 3, 𝑀 = 1. Then we have the trail solution is 𝐺′

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 )

(3.66)


0

𝐺′

1

𝐺′

2

𝐺′

3

𝐺′

4

𝐺′

5

𝐺′

6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) + 𝐶5 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0, To find the values of unknown constant, we set [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0]. After solving we have 1

𝑐 = − −4𝜇+𝜆2 , 𝑎0 = 𝑎0 , 𝑎1 = 6. Case 1: When λ2 − 4μ > 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + 6 ( where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2

√𝜆2 −4𝜇 2

(





(

293

λ

) − 2,

λ

) − 2),

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 2: When λ2 − 4μ < 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + 6 ( where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√−𝜆2 +4𝜇 2

√−𝜆2 +4𝜇

(

2

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

λ

) − 2),

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡. 𝑢(𝑥, 𝑡) = 𝑊𝑋 (𝑋, 𝑇).

(a)

(b)


Case 3: When λ2 − 4μ = 0, we obtain the travelling wave solution 294

𝑢(𝜉) = 𝑎0 + 6 (c where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2 𝜉

2c2 1 +c2

𝜆

− 2), 𝜉

𝜆

− 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡. 𝑢(𝑥, 𝑡) = 𝑊𝑋 (𝑋, 𝑇).

(a)

(b)


3.4.10. Complex Equation Consider the Complex equation [33] 𝑢𝑦𝑡 − 𝛼𝑢𝑥𝑥𝑥𝑦 − 𝛽(𝑢𝑥 𝑢𝑦 )𝑥 + 𝛾𝑢𝑥𝑥 = 0.

(3.67a)

Introducing the transformation 𝜉 = 𝑥 + 𝑦 + 𝑐𝑡, the above PDE reduces to ODE as, 𝑐𝑢′′ − 𝛼𝑢′′′′ − 2𝛽𝑢′ 𝑢′′ + 𝛾𝑢′′ = 0. Integrate the above equation we have 𝑐𝑢′ − 𝛼𝑢′′′ − 𝛽(𝑢′ )2 + 𝛾𝑢′ = 0. Balance the (𝑢′ )2 and 𝑢′′′ by using homogenous principal, we have 2𝑀 + 2 = 𝑀 + 3, 𝑀 = 1. Then we have the trail solution is 𝐺′

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 )

(3.67)

By equation (3.50) and (3.67), substitute into (3.66), we obtain the system of algebraic equations for 𝑎0 , 𝑎1 , 𝑐, 𝜆 and 𝜇 as follow: 295

𝐺′

0

1

𝐺′

𝐺′

2

3

𝐺′

𝐺′

4

𝐺′

5

𝐺′

6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) + 𝐶5 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0, To find the values of unknown constant, we set [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0]. After solving we have 𝑐 = −4𝛼𝜇 + 𝛼𝜆2 − 𝛾, 𝑎0 = 𝑎0 , 𝑎1 =

6𝛼 𝛽

.

Case 1: When λ2 − 4μ > 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

6𝛼 𝛽

√𝜆2 −4𝜇 2

(

√𝜆2 −4𝜇 2

(





(

λ

) − 2),

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 2: When λ2 − 4μ < 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

6𝛼 𝛽

√−𝜆2 +4𝜇 2

(

(

√−𝜆2 +4𝜇 2

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

296

λ

) − 2,

λ

) − ), 2

(a)

(b)


Case 3: When λ2 − 4μ = 0, we obtain the travelling wave solution 𝑢(𝜉) = 𝑎0 + where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2

6𝛼 𝛽

(c

2c2 1 +c2 𝜉

𝜆

− 2),

𝜆

− 2, 𝜉

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


3.4.11. Kadomtsev-Petviashvili Equation Consider the Kadomtsev-Petviashvili equation [230] 𝑢𝑥𝑡 + 𝛼𝑢𝑥 𝑢𝑥𝑧 + 𝛽𝑢𝑥𝑥 𝑢𝑧 + 𝛾𝑢𝑥𝑥𝑥𝑧 = 0.

(3.69a)

Introducing the transformation 𝜉 = 𝑥 + 𝑧 + 𝑐𝑡, the above PDE reduces to ODE as, 𝑐𝑢′′ + 𝛼𝑢′ 𝑢′′ + 𝛽𝑢′ 𝑢′′ + 𝛾𝑢′′′′ = 0. 297

Integrate the above equation we have 𝛼

𝛽

𝑐𝑢′ + 2 (𝑢′ )2 + 2 (𝑢′ )2 + 𝛾𝑢′′′ = 0. Balance the (𝑢′ )2 and 𝑢′′′ by using homogenous principal, we have 2𝑀 + 2 = 𝑀 + 3, 𝑀 = 1. Then we have the trail solution is 𝐺′

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 )

(3.69)

By equation (3.50) and (3.69), we obtain the system of algebraic equations for 𝑎0 , 𝑎1 , 𝑐, 𝜆 and 𝜇 as follow: 𝐺′

0

𝐺′

1

𝐺′

2

𝐺′

3

𝐺′

4

𝐺′

5

𝐺′

6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) + 𝐶5 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0, To find the values of unknown constant, we set [𝐶0 = 0, 𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0]. After solving we have 12𝛾

𝑐 = 4𝛾𝜇 − 𝛾𝜆2 , 𝑎0 = 𝑎0 , 𝑎1 = 𝛼+𝛽. Case 1: When λ2 − 4μ > 0, we obtain the travelling wave solution 12𝛾

√𝜆2 −4𝜇

𝑢(𝜉) = 𝑎0 + 𝛼+𝛽 ( where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√𝜆2 −4𝜇 2

2

(





(

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

298

λ

) − 2,

λ

) − 2),

(a)

(b)


Case 2: When λ2 − 4μ < 0, we obtain the travelling wave solution √−𝜆2 +4𝜇

12𝛾

𝑢(𝜉) = 𝑎0 + 𝛼+𝛽 ( where

𝐺 ′ (𝜉) 𝐺(𝜉)

=

√−𝜆2 +4𝜇 2

(

2

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

(

1 1 2 2 1 1 c1 cos( √−𝜆2 +4𝜇𝜉)+c2 sin( √−𝜆2 +4𝜇𝜉) 2 2

−c1 sin( √−𝜆2 +4𝜇𝜉)+c2 cos( √−𝜆2 +4𝜇𝜉)

λ

) − 2),

λ

) − 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

(a)

(b)


Case 3: When λ2 − 4μ = 0, we obtain the travelling wave solution 12𝛾

𝑢(𝜉) = 𝑎0 + 𝛼+𝛽 (c where

𝐺 ′ (𝜉) 𝐺(𝜉)

=c

2c2 1 +c2 𝜉

2c2 1 +c2

𝜆

− 2), 𝜉

𝜆

− 2,

and 𝜉 = 𝑥 + √1 + √−4𝜇 + 𝜆2 𝑡.

299

(a)

(b)


3.5. F-Expansion Method In 2003, Zhou et al. [262] proposed new algorithm named as F-Expansion method. Fexpansion method applied to seek the travelling wave solution of nonlinear evolution equation [9, 98, 211, 214, 256] which can be regarded as generalization of the Jacobi method [61, 124]. Liu and Yang [123] proposed another extended F-expansion method to get not only single non-degenerate Jacobi elliptic function solutions, but also combined non-degenerate Jacobi elliptic function solutions, and their corresponding degenerate solutions. Yomba [249] and Wang [213] proposed the Homogenous balance method. Usman et. al. [203] applied F-Expansion method on (2+1)-Dimensional DaveyStewartson equations to obtain the solution terms of Weierstrass-Elliptic and JacobianElliptic functions. In F-expansion method, elliptic equations have been applied as a mapping to obtain many kinds of travelling wave solutions to a large variety of NLPDEs whose odd- and even-order derivative terms do not coexist. The main advantages of the F-expansion approach are that with the aid of elliptic equations one only needs to calculate the function which is solution of the elliptic equations, instead of calculating the Jacobi elliptic functions one by one the coefficients of the elliptic equations can be selected so that the corresponding solution is a Jacobi elliptic function. The proposed scheme is fully compatible with the complexity of such problems and is very userfriendly. Numerical results are very encouraging.

3.5.1. Methodology We consider the general nonlinear PDE of the type 300

𝑃(𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑦 , 𝑢𝑧 , 𝑢𝑡𝑡 , 𝑢𝑥𝑥 , 𝑢𝑦𝑦 , 𝑢𝑥𝑥 , 𝑢𝑥𝑡 , 𝑢𝑦𝑡 , 𝑢𝑧𝑡 , 𝑢𝑥𝑦 , 𝑢𝑥𝑧 , 𝑢𝑦𝑧 , … ) = 0.

(3.70)

where 𝑃 is a polynomial in its arguments. The essence of the F-expansion method can be presented in the following steps:

Step 1: Seek Solitary wave solutions of Eq. (3.70) by taking 𝑢(𝑥, 𝑡) = 𝑈(𝜉), 𝜉 = 𝛼𝑥 + 𝛽𝑦 + 𝛾𝑧 + 𝜔𝑡. and transform Eq. (3.70) to the ordinary differential equation. 𝑄(𝑢, 𝜔𝑢′ , 𝛼𝑢′ , 𝛽𝑢′ , 𝛾𝑢′ , … ) = 0.

(3.71)

where 𝛼, 𝛽, 𝛾 and 𝜔 are constants and where prime denotes the derivative with respect to 𝜉.

Step 2: If possible, integrate Eq. (3.71) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero.

Step 3: According to F-expansion method, we assume that the wave solution can be expressed in the following form 𝑛 𝑢(𝜉) = ∑𝑀 𝑛=0 𝑎𝑛 𝐹 (𝜉).

(3.72)

where 𝑎𝑖 real constants to be determined are, 𝑀 is a positive integer to be determined. 𝐹(𝜉) satisfies the following auxiliary equation [𝐹 ′ (𝜉)]2 = 𝑃𝐹 4 (𝜉) + 𝑄𝐹 2 (𝜉) + 𝑅.

(3.73)

𝑃, 𝑄, and 𝑅 are constants, The last equation hence holds for 𝐹 ′′ (𝜉) = 2𝑃𝐹 3 (𝜉) + 𝑄𝐹(𝜉), 𝐹 ′′′ (𝜉) = 6𝑃𝐹 2 𝐹 ′ (𝜉) + 𝑄𝐹 ′ (𝜉), 𝐹 (iv) (𝜉) = 24𝑃2 𝐹 5 (𝜉) + 20𝑃𝑄𝐹 3 (𝜉) + 12𝑃𝑅𝐹(𝜉) + 𝑄 2 𝐹(𝜉), 𝐹 (v) (𝜉) = 120𝑃2 𝐹 4 𝐹 ′ (𝜉) + 60𝑃𝑄𝐹 3 𝐹 ′ (𝜉) + 12𝑃𝑅𝐹 ′ (𝜉) + 𝑄 2 𝐹 ′ (𝜉), . . . In Table 3.1 and B, we present 52 types of exact solution for Eq. (3.73). In fact, these exact solutions can be used to construct more exact solutions for Eq. (3.70). Table 3.1. (a): Exact solutions of Eq. (3.73) for different values of 𝑃, 𝑄 and 𝑅, in terms of Jacobi elliptic functions. Case

𝑷

𝑸

𝑹 301

𝑭(𝝃)

1

𝑚2

−(1 + 𝑚2 )

1

Sn𝜉

2

𝑚2

−(1 + 𝑚2 )

1

cd𝜉 = cn𝜉/dn𝜉

3

−𝑚2

2𝑚2 − 1

1 − 𝑚2

Cn𝜉

4

−1

2 − 𝑚2

𝑚2 − 1

Dn𝜉

2

2

ns𝜉 = (sn𝜉)−1

5

1

−(1 + 𝑚 )

𝑚

6

1

−(1 + 𝑚2 )

𝑚2

dc𝜉 = dn𝜉/cn𝜉

7

1 − 𝑚2

2𝑚2 − 1

−𝑚2

nc𝜉 = (cn𝜉)−1

8

𝑚2 − 1

2 − 𝑚2

−1

nd𝜉 = (dn𝜉)−1

9

1 − 𝑚2

2 − 𝑚2

1

sc𝜉 = sn𝜉/cn𝜉

10

−𝑚2 (1 − 𝑚2 )

2𝑚2 − 1

1

sd𝜉 = sn𝜉/dn𝜉

11

1

2 − 𝑚2

1 − 𝑚2

cs𝜉 = cn𝜉/sn𝜉

12

1

2𝑚2 − 1

−𝑚2 (1 − 𝑚2 )

ds𝜉 = dn𝜉/sn𝜉

13

¼

¼

ns𝜉 ± cn𝜉

14

(1 − 𝑚2 )/4

(1 − 𝑚2 )/4

nc𝜉 ± sc𝜉

15

¼

𝑚2 /4

ns𝜉 ± ds𝜉

16

𝑚2 /4

𝑚2 /4

sn𝜉 ± 𝑖cn𝜉

17

2

𝑚 /4

𝑚2 /4

√1 − 𝑚2 sd𝜉 ± cd𝜉

18

¼

¼

𝑚cd𝜉 ± 𝑖 √1 − 𝑚2 nd𝜉

19

¼

¼

𝑚sn𝜉 ± 𝑖dc𝜉

20

¼

¼

√1 − 𝑚2 sc𝜉 ± dc𝜉

21

(𝑚2 − 1)/4

(m2 − 1)/4

𝑚sd𝜉 ± nd𝜉

22

𝑚2 /4

¼

sn𝜉 1 ± dn𝜉

23

−¼

(1 − 𝑚2 )2 /4

𝑚cn𝜉 ± dn𝜉

(1 − 2𝑚2 )/2 (1 + 𝑚2 )/2 (𝑚2 − 2)/2 (𝑚2 − 2)/2 (𝑚2 − 2)/2 (1 − 𝑚2 ) /2 (1 − 2𝑚2 )/2 (1 − 𝑚2 ) /2 (𝑚2 + 1)/2 (𝑚2 − 2)/2 (𝑚2 + 1)/2

302

24

(1 − 𝑚2 )2 /4

25

𝑚4 (1 − 𝑚2 ) 2(2 − 𝑚2 )

26

(𝑚2

¼

ds𝜉 ± cs𝜉

2(1 − 𝑚2 ) 𝑚2 − 2

1 − 𝑚2 2(2 − 𝑚2 )

dc𝜉 ± √1 − 𝑚2 nc𝜉

𝑃>0

𝑄 0, 𝐺 ′ (𝜉)

√Ω

𝐵

√Ω

√Ω 𝐶1 sinh( 2𝐴 𝜉)+𝐶2 cosh( 2𝐴 𝜉)

𝑈(𝜉) = ( 𝐺(𝜉) ) = 2Ψ + 2Ψ

.

√Ω √Ω 𝜉)+𝐶2 sinh( 𝜉) 2𝐴 2𝐴

𝐶1 cosh(

Family 2: When 𝐵 ≠ 0, Ψ = 𝐴 − 𝐶 and Ω = 𝐵 2 + 4𝐸(𝐴 − 𝐶) < 0, 𝐺 ′ (𝜉)

√−Ω

𝐵

𝑈(𝜉) = ( 𝐺(𝜉) ) = 2Ψ +

√−Ω

√−Ω −𝐶1 sin( 2𝐴 𝜉)+𝐶2 cos( 2𝐴 𝜉) . 2Ψ 𝐶 cos(√−Ω𝜉)+𝐶 sin(√−Ω𝜉) 1

2

2𝐴

2𝐴

Family 3: When 𝐵 ≠ 0, Ψ = 𝐴 − 𝐶 and Ω = 𝐵 2 + 4𝐸(𝐴 − 𝐶) = 0, 𝐺 ′ (𝜉)

𝐵

𝑈(𝜉) = ( 𝐺(𝜉) ) = 2Ψ + 𝐶

𝐶2

1 +𝐶2 𝜉

.

Family 4: When 𝐵 = 0, Ψ = 𝐴 − 𝐶 and ∆= Ψ𝐸 > 0, 𝐺 ′ (𝜉)

𝑈(𝜉) = ( 𝐺(𝜉) ) =

√Δ

√Δ

√Δ 𝐶1 sinh( 𝐴 𝜉)+𝐶2 cosh( 𝐴 𝜉) Ψ 𝐶 cosh(√Δ𝜉)+𝐶 sinh(√Δ𝜉) 1 2 𝐴

.

𝐴

Family 5: When 𝐵 = 0, Ψ = 𝐴 − 𝐶 and ∆= Ψ𝐸 < 0, 𝐺 ′ (𝜉)

𝑈(𝜉) = ( 𝐺(𝜉) ) =

√−Δ

√−Δ

√−Δ 𝐶1 sin( 𝐴 𝜉)+𝐶2 cos( 𝐴 𝜉) . Ψ 𝐶 cos(√−Δ𝜉)+𝐶 sin(√−Δ𝜉) 1

𝐴

2

𝐴

Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term of highest order with the highest order nonlinear term which obtained in Step 2. Step 5: Substituting (3.270) into Eq. (3.269) with Eq. (3.271) will yield a polynomial in 𝑈 𝑀 . Compare the like powers of 𝑈 equal to zero, yields a set of algebraic equations for 𝑎𝑛 (𝑛 = 0,1,2, … , 𝑀) and 𝑉. 407

Step 6: Solve the system which obtained in step 5 for 𝑎𝑛 and 𝑉 with the help of Maple 13, to determine these constants. Putting these constant into Eq. (3.270), coupled with the well known solutions of Eq. (3.271), we can obtained the more general type and new exact traveling wave solution of the nonlinear partial differential equation (3.267).

3.9.1.2. New Approach of generalized (G´/G)–Expansion Method Step 7: According to the new approach of generalized (G´/G) – expansion method suppose that 𝑢(𝜉) can be expressed by a finite power series of 𝑛 𝑢(𝜉) = ∑𝑀 𝑛=0 𝑎𝑛 (𝑠 + 𝑈) ,

(3.272)

where 𝑠 and 𝑎𝑛 (𝑛 = 0,1,2, … , 𝑀) are constants to be determine later and 𝑈(𝜉) is 𝐺 ′ (𝜉)

𝑈(𝜉) = ( 𝐺(𝜉) ), where 𝐺(𝜉) satisfied the following auxiliary equation 𝐴𝐺𝐺 ′′ − 𝐵𝐺𝐺 ′ − 𝐶(𝐺 ′ )2 − 𝐸𝐺 = 0, 𝐺 ′ (𝜉)

′

𝐺 ′ (𝜉)

′′

𝐸

𝐵 𝐺 ′ (𝜉)

𝐶 𝐺 ′ (𝜉)

2

(3.273) 𝐺 ′ (𝜉)

2

( 𝐺(𝜉) ) = 𝐴 + 𝐴 ( 𝐺(𝜉) ) + 𝐴 ( 𝐺(𝜉) ) − ( 𝐺(𝜉) ) , ( 𝐺(𝜉) ) = −

3𝐵 𝐺 ′ (𝜉) 𝐴

𝐵𝐸 𝐴2

𝐵2

𝐺 ′ (𝜉)

+ 𝑦2 ( 𝐺(𝜉) ) +

2

( 𝐺(𝜉) ) +

2𝐶 2 𝐺 ′ (𝜉) 𝐴2

2𝐶𝐸 𝐺 ′ (𝜉) 𝐴2

3

( 𝐺(𝜉) ) −

( 𝐺(𝜉) ) −

4𝐶 𝐺 ′ (𝜉) 𝐴

3

2𝐸 𝐺 ′ (𝜉) 𝐴

( 𝐺(𝜉) ) + 𝐺 ′ (𝜉)

3𝐵𝐶 𝐺 ′ (𝜉) 𝐴2

2

( 𝐺(𝜉) )

3

( 𝐺(𝜉) ) + 2 ( 𝐺(𝜉) ) ,

. . .. where ′ denote the derivative w.r.t 𝜉. 𝐴, 𝐵, 𝐶 and 𝐸 are real constant. Step 8: Determine 𝑀. This, usually, can be accomplished by balancing the linear term of highest order with the highest order nonlinear term which obtained in Step 2. Step 9: Substituting (3.270) into Eq. (3.269) with Eq. (3.271) will yield a polynomial in 𝑈 𝑀 . Compare the like powers of 𝑈 equal to zero, yields a set of algebraic equations for 𝑠, 𝑎𝑛 (𝑛 = 0,1,2, … , 𝑀) and 𝑉. Step 10: Solve the system which obtained in step 5 for 𝑠, 𝑎𝑛 and 𝑉 with the help of Maple 13, to determine these constants. Putting these constant into Eq. (3.270), coupled 408

with the well known solutions of Eq. (3.271), we can obtained the more general type and new exact traveling wave solution of the nonlinear partial differential equation (3.268).

3.9.2. Modified Korteweg-deVries (mKdV) Equation The Korteweg-deVries (KdV) equation in its simplest form [230] is given by 𝑢𝑡 + 𝑎𝑢2 𝑢𝑥 + 𝛿𝑢𝑥𝑥𝑥 = 0.

(3.274)

The KdV equation arises in the study of shallow water waves. In particular, the KdV equation is used to describe long waves traveling in canals. It is formally proved that this equation has solitary waves as solutions; hence it can have any number of solitons. The KdV equation has received a lot of attention and has been extensively studied. Several numerical and analytical techniques were employed to study the solitary waves that result from this equation. Consider the transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), where 𝜉 = 𝑥 − 𝑉𝑡 we haves −𝑉𝑢′ + 𝑎𝑢2 𝑢′ + 𝛿𝑢′′′ = 0, Integrating once we have 𝑎

−𝑉𝑢 + 3 𝑢3 + 𝛿𝑢′′ + 𝐾 = 0,

(3.275)

by applying the balancing principle we have 3𝑀 = 𝑀 + 2, 𝑀 = 1. Application of new Approach of (G´/G) – Expansion Method: According to (G´/G) – expansion method we consider the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈(𝜉),

(3.276) 𝐺 ′ (𝜉)

where 𝑎0 , 𝑎1 arbitrary constant to be determine, 𝑈(𝜉) = ( 𝐺(𝜉) ), and 𝐺(𝜉) satisfied the Eq. (3.275). Putting (3.276) in to (3.275) with (3.273) we have 1 1

𝐺 ′ (𝜉)

0

𝐺 ′ (𝜉)

1

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

3

[𝑆 ( ) + 𝑆1 ( 𝐺(𝜉) ) + 𝑆2 ( 𝐺(𝜉) ) + 𝑆3 ( 𝐺(𝜉) ) ] = 0. 3 𝐴2 0 𝐺(𝜉) Compare the like powers of U we have system of equation 𝐺 ′ (𝜉)

0

𝐺 ′ (𝜉)

1

𝐺 ′ (𝜉)

2

( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) :

3𝑉𝐴2 𝑎0 − 𝑎𝐴2 𝑎03 − 3𝛿𝑎1 𝐵𝐸 = 0, 3𝑉𝐴2 𝑎1 − 3𝑎𝐴2 𝑎02 𝑎1 − 3𝛿𝑎1 𝐵 2 − 6𝛿𝑎1 𝐶𝐸 + 6𝛿𝑎1 𝐴𝐸 = 0, −3𝑎𝐴2 𝑎0 𝑎12 − 9𝛿𝑎1 𝐵𝐶 + 9𝛿𝑎1 𝐴𝐵 = 0, 409

𝐺 ′ (𝜉)

3

( 𝐺(𝜉) ) :

−𝑎𝐴2 𝑎13 − 6𝛿𝑎1 𝐶 2 + 12𝛿𝑎1 𝐴𝐶 − 6𝛿𝑎1 𝐴2 = 0.

Solving these algebraic equations with the help of any mathematical software such as Maple 13, we obtain following Solution Set: 1 𝛿Ω

𝐵

3𝛿

𝑉 = − 2 𝐴2 , 𝑎0 = 𝐴 √− 2𝑎 , 𝑎1 = −

2Ψ 𝐴

√−

3𝛿 2𝑎

, 𝐾 = 𝐾.

Family 1: When 𝐵 ≠ 0, Ψ = 𝐴 − 𝐶 and Ω = 𝐵 2 + 4𝐸(𝐴 − 𝐶) > 0, 𝑈1 (𝜉) = (

𝐺 ′ (𝜉) 𝐺(𝜉)

√Ω

𝐵

3𝛿

𝐴

2𝑎

) = √−

−

2Ψ 𝐴

√−

3𝛿 2𝑎

(

𝐵 2Ψ

+

√Ω

√Ω 𝐶1 sinh( 2𝐴 𝑦)+𝐶2 cosh( 2𝐴 𝜉) 2Ψ 𝐶 cosh(√Ω𝜉)+𝐶 sinh(√Ω𝜉) 1 2 2𝐴

).

2𝐴

Family 2: When 𝐵 ≠ 0, Ψ = 𝐴 − 𝐶 and Ω = 𝐵 2 + 4𝐸(𝐴 − 𝐶) < 0, 𝐺 ′ (𝜉)

√−Ω

𝐵

3𝛿

𝑈2 (𝜉) = ( 𝐺(𝜉) ) = 𝐴 √− 2𝑎 −

2Ψ 𝐴

√−

3𝛿 2𝑎

𝐵

(2Ψ +

√−Ω

√−Ω −𝐶1 sin( 2𝐴 𝜉)+𝐶2 cos( 2𝐴 𝜉) ). 2Ψ 𝐶 cos(√−Ω𝜉)+𝐶 sin(√−Ω𝜉) 1

2

2𝐴

2𝐴

Family 3: When 𝐵 ≠ 0, Ψ = 𝐴 − 𝐶 and Ω = 𝐵 2 + 4𝐸(𝐴 − 𝐶) = 0, 𝐺 ′ (𝜉)

𝐵

3𝛿

𝑈3 (𝜉) = ( 𝐺(𝜉) ) = 𝐴 √− 2𝑎 −

2Ψ 𝐴

√−

3𝛿

(

𝐵

2𝑎 2Ψ

+𝐶

𝐶2

1 +𝐶2 𝜉

).

Family 4: When 𝐵 = 0, Ψ = 𝐴 − 𝐶 and ∆= Ψ𝐸 > 0, 𝐺 ′ (𝜉)

√Δ

𝐵

3𝛿

𝑈4 (𝜉) = ( 𝐺(𝜉) ) = 𝐴 √− 2𝑎 −

2Ψ 𝐴

√−

3𝛿 2𝑎

√Δ

√Δ 𝐶1 sinh( 𝐴 𝜉)+𝐶2 cosh( 𝐴 𝜉)

(Ψ

).

√Δ √Δ 𝜉)+𝐶2 sinh( 𝜉) 𝐴 𝐴

𝐶1 cosh(


√−Δ

𝐵

3𝛿

𝑈5 (𝜉) = ( 𝐺(𝜉) ) = 𝐴 √− 2𝑎 −

√−Δ

√−Δ 𝐶1 sin( 𝐴 𝜉)+𝐶2 cos( 𝐴 𝜉) √− ( ). 𝐴 2𝑎 Ψ 𝐶 cos(√−Δ𝜉)+𝐶 sin(√−Δ𝜉)

2Ψ

3𝛿

1

1 𝛿Ω

In all families we have 𝜉 = 𝑥 − 2 𝐴2 .

410

𝐴

2

𝐴

(a)

(c)

(b)

(d)

(e)

Fig. 3.108.(a)-(e): Graphical representation of Eq. (3.274) for different values of parameters

(a)

(b)

411

(c)

(d)

(e)


Application of new Approach of generalized (G´/G) – Expansion Method: According to (G´/G) – expansion method we consider the trail solution 𝑢 = 𝑎0 + 𝑎1 (𝑠 + 𝑈(𝜉)),

(3.277) 𝐺 ′ (𝜉)

where 𝑎0 , 𝑎1 , 𝑑 arbitrary constant to be determine, 𝑈(𝜉) = ( 𝐺(𝜉) ), and 𝐺(𝜉) satisfied the Eq. (3.275). Putting (3.277) in to (3.277) with (3.275) we have 1 1

𝐺 ′ (𝜉)

0

𝐺 ′ (𝜉)

1

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

3

[𝑆 ( ) + 𝑆1 ( 𝐺(𝜉) ) + 𝑆2 ( 𝐺(𝜉) ) + 𝑆3 ( 𝐺(𝜉) ) ] = 0. 3 𝐴2 0 𝐺(𝜉) Compare the like powers of U we have system of equation 𝐺 ′ (𝜉)

0

𝐺 ′ (𝜉)

1

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

3

( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) :

3𝑉𝐴2 𝑎0 + 3𝑉𝐴2 𝑎1 𝑠 + ⋯ + 𝑎𝐴2 𝑎13 𝑠 3 − 3𝛿𝑎1 𝐵𝐸 = 0, 3𝑉𝐴2 𝑎1 − 3𝑎𝐴2 𝑎02 𝑎1 + ⋯ + 6𝛿𝑎1 𝐴𝐸 = 0, −3𝑎𝐴2 𝑎0 𝑎12 − 3𝑎𝐴2 𝑎13 𝑠 + ⋯ + 9𝛿𝑎1 𝐴𝐵 = 0, −𝑎𝐴2 𝑎13 − 6𝛿𝑎1 𝐶 2 + 12𝛿𝑎1 𝐴𝐶 − 6𝛿𝑎1 𝐴2 = 0.

Solving these algebraic equations with the help of any mathematical software such as Maple 13, we obtain following Solution Set: 1 𝛿Ω

𝑉 = − 2 𝐴2 , 𝑎0 = 𝑎0 , 𝑎1 =

Ψ 𝐴

√−

6𝛿 𝑎

1

𝑎𝐴𝑎0

, 𝐾 = 𝐾, 𝑠 = 6 (

𝛿Ψ

√−

Family 1: When 𝐵 ≠ 0, Ψ = 𝐴 − 𝐶 and Ω = 𝐵 2 + 4𝐸(𝐴 − 𝐶) > 0,

412

6𝛿 𝑎

−

3𝐵 Ψ

).

√Ω

Ψ

𝑈1 (𝜉) = 𝑎0 + 𝐴 √−

6𝛿 𝑎

1

(6 (

𝑎𝐴𝑎0 𝛿Ψ

√−

6𝛿

−

𝑎

3𝐵

√Ω

√Ω 𝐶1 sinh( 2𝐴 𝜉)+𝐶2 cosh( 2𝐴 𝜉)

𝐵

) + 2Ψ + 2Ψ Ψ

).

√Ω √Ω 𝜉)+𝐶2 sinh( 𝜉) 2𝐴 2𝐴

𝐶1 cosh(

Family 2: When 𝐵 ≠ 0, Ψ = 𝐴 − 𝐶 and Ω = 𝐵 2 + 4𝐸(𝐴 − 𝐶) < 0, Ψ

𝑈2 (𝜉) = 𝑎0 + 𝐴 √− √−Ω

6𝛿 𝑎

1

(6 (


√−

6𝛿

−

𝑎

3𝐵

𝐵

) + 2Ψ + Ψ

√−Ω

√−Ω −𝐶1 sin( 2𝐴 𝜉)+𝐶2 cos( 2𝐴 𝑦) ). 2Ψ 𝐶 cos(√−Ω𝜉)+𝐶 sin(√−Ω𝜉) 1

2

2𝐴

2𝐴

Family 3: When 𝐵 ≠ 0, Ψ = 𝐴 − 𝐶 and Ω = 𝐵 2 + 4𝐸(𝐴 − 𝐶) = 0, Ψ

𝑈3 (𝜉) = 𝑎0 + 𝐴 √−

6𝛿 𝑎

1

𝑎𝐴𝑎0

(6 (

𝛿Ψ

√−

6𝛿

−

𝑎

3𝐵 Ψ

𝐵

) + 2Ψ + 𝐶

𝐶2

1 +𝐶2 𝜉

).

Family 4: When 𝐵 = 0, Ψ = 𝐴 − 𝐶 and ∆= Ψ𝐸 > 0, √Δ

Ψ

𝑈4 (𝜉) = 𝑎0 + 𝐴 √−

6𝛿 𝑎

1

(6 (


√−

6𝛿

−

𝑎

3𝐵

)+ Ψ

Ψ 𝐶 cosh(√Δ𝜉)+𝐶 sinh(√Δ𝜉) 1 2 𝐴


Ψ

𝑈5 (𝜉) = ( 𝐺(𝜉) ) = 𝑎0 + 𝐴 √− √−Δ

6𝛿 𝑎

1

(6 (


√−

6𝛿 𝑎

−

√−Δ

√−Δ 𝐶1 sin( 𝐴 𝜉)+𝐶2 cos( 𝐴 𝜉) ). Ψ 𝐶 cos(√−Δ𝜉)+𝐶 sin(√−Δ𝜉) 1

𝐴

2

𝐴

1 𝛿Ω

In all families we have 𝜉 = 𝑥 − 2 𝐴2 .

(a)

(b)

413

√Δ

√Δ 𝐶1 sinh( 𝐴 𝜉)+𝐶2 cosh( 𝐴 𝜉)

3𝐵 Ψ

)+

𝐴

).

(c)

(d)

(e)


(a)

(c)

(b)

(d)

(e)


414

3.10. Combined Tanh-Coth (CTC) and Combined Sinh-Cosh (CSC) Method We developed Combined Tanh-Coth and Combined Sinh-Cosh Method to fine the traveling wave solutions of NLEEs. These methods free of auxiliary equation, simple and straight forward.

3.10.1. Methodology Suppose a nonlinear wave equation for 𝜓(𝑥, 𝑡) is given by 𝜓(𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑥𝑥 , 𝑢𝑡𝑡 , 𝑢𝑥𝑡 , … ) = 0.

(3.278)

in which both nonlinear term(s) and higher order derivatives of 𝜓(𝑥, 𝑡) are all involved. In general, the left-hand side of Eq. (3.278) is a polynomial in 𝜓 and its various derivatives. The CTC method for solving Eq. (3.278) proceeds in the following five steps: Step 1: Look for traveling wave solution of Eq. (3.278) by taking 𝜓 = 𝜓(𝜉),

𝜉 = 𝑘𝑥 + 𝜔𝑡.

(3.279)

where 𝑘 and 𝜔 is nonzero constants, 𝜓(𝜉) the function of 𝜉. Substituting (24) into Eq. (3.278) yields an ordinary differential equation (ode) for 𝜓(𝜉). 𝜑(𝑢, 𝑘𝜔𝑢′ , 𝑘𝑢′ , 𝑘 2 𝜔2 𝑢′′ , 𝑘 2 𝑢′′ , 𝑘 2 𝜔2 𝑢′′ , … ) = 0.

(3.280)


3.10.1.1. Combined Tanh-Coth (CTC) Method Step 3: According to the CTC Method suppose that 𝜓(𝜉) can be expressed by a finite power series of 𝜓(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎1 [tanh(𝜇𝑛𝜉) + coth(𝜇𝑛𝜉)].

(3.281)

Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term of highest order with the highest order nonlinear term which obtained in Step 1. Step 5: Substituting (3.281) into Eq. (3.280) will yields an algebraic equation. Equating the coefficients of 𝑈 𝑛 = [tanh(𝜇𝑛𝜉) + coth(𝜇𝑛𝜉)] to zero gives a system of algebraic equations for 𝑎𝑖 , 𝑏𝑖 , 𝑘, 𝜇 and 𝜔. Then, we solve the system with the help of MAPLE 11, to determine these constants.

3.10.1.2. Combined Sinh-Cosh (CSC) Method 415

Step 6: According to the CSC Method suppose that 𝜓(𝜉) can be expressed by a finite power series of 𝑛 𝜓(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎1 [sinh(𝜇𝜉) + cosh(𝜇𝜉)] −𝑛 + ∑−𝑀 𝑛=1 𝑎1 [sinh(𝜇𝜉) + cosh(𝜇𝜉)] .

(3.282)

Step 7: Determine 𝑀. This, usually, can be accomplished by balancing the linear term of highest order with the highest order nonlinear term which obtained in Step 1. Step 8: Substituting (3.282) into Eq. (3.281) will yields an algebraic equation. Equating the coefficients of [sinh(𝜇𝜉) + cosh(𝜇𝜉)]𝑛 to zero gives a system of algebraic equations for 𝑎𝑖 , 𝑏𝑖 , 𝑘, 𝜇 and 𝜔. Then, we solve the system with the help of MAPLE 11, to determine these constants.

3.10.2. Generalized KdV Equation With Two Power Nonlinearities Generalized KdV equation with two power nonlinearities [230] 𝑢𝑡 + (𝑎𝑢𝑛 − 𝑏𝑢2𝑛 )𝑢𝑥 + 𝑢𝑥𝑥𝑥 = 0.

(3.283)

This equation describes the propagation of non linear long acoustic-type waves it is to be noted that for 𝑛 = 1, Eq. (3.283) is the well-known gardner that is also called the combined KdV-MKdV equation. using the transform 𝜉 = 𝑥 + 𝜔𝑡, in (35) we have 𝜔𝑢′ + (𝑎𝑢𝑛 − 𝑏𝑢2𝑛 )𝑢′ + 𝑢′′′ = 0, integrating once, we get 𝜔𝑢(𝜉) + (𝑎

𝑢𝑛+1 𝑛+1

−𝑏

𝑢2𝑛+1 2𝑛+1

) + 𝑢′′ (𝜉) = 0,

(3.284)

balance the 𝑢2𝑛+1 and 𝑢’’ by using homogenous principle, we have (2𝑛 + 1)𝑀 = 𝑀 + 2, 1

𝑀=𝑛 to achieve our goal, we use the transformation 1

𝑢(𝑥, 𝑡) = 𝑣 𝑛 ,

(3.285)

the substitution given in (3.285) converts (3.284) into 𝜔𝑛2 (2𝑛 + 1)(𝑛 + 1)𝑣 2 + 𝑎𝑛2 (2𝑛 + 1)𝑣 3 − 𝑏𝑛2 (𝑛 + 1)𝑣 4 + 𝑛(2𝑛 + 1)(𝑛 + 1)𝑣𝑣 ′′ + (1 − 𝑛2 )(2𝑛 + 1)(𝑣’)2 = 0,

(3.286)

balance the 𝑣 4 and 𝑣𝑣’’ by using homogenous principle, we have 4𝑀 = 𝑀 + 𝑀 + 2, 416

𝑀 = 1. so the trial solution is given as 𝑣(𝜉) = 𝑎0 + 𝑎1 [tanh(𝜇𝜉) + coth(𝜇𝜉)].

(3.287)

according to the purposed technique we get two solution sets. 1st solution set: 𝑎2 (2𝑛+1)

𝜔 = − 𝑏(𝑛3 +5𝑛2 +8𝑛+4) , 𝜇 =

√−

−2𝑛−1 𝑛𝑎 16𝑏𝑛+16𝑏

𝑛+2

1 𝑎(2𝑛+1)

, 𝑎0 = 2

𝑏(𝑛+2)

1 𝑎(2𝑛+1)

, 𝑎1 = − 4

𝑏(𝑛+2)

,

substituting in (39) we get 1 𝑎(2𝑛+1)

𝑣(𝜉) = 4

𝑏(𝑛+2)

[2 − [tanh(𝜇𝜉) + coth(𝜇𝜉)]].

(a)

(b)

Fig. 3.112.(a)-(b): Graphical representation of Eq. (3.283) for different values of parameters 1 𝑎(2𝑛+1)

𝑢(𝜉) = (4

𝑏(𝑛+2)

𝟏

[2 − [tanh(𝜇𝜉) + coth(𝜇𝜉)]])𝒏 .

(a)

(b)

417


2nd solution set: 𝑎2 (2𝑛+1)

𝜔 = − 𝑏(𝑛3 +5𝑛2 +8𝑛+4) , 𝜇 =

√−

−2𝑛−1 𝑛𝑎 16𝑏𝑛+16𝑏

𝑛+2

1 𝑎(2𝑛+1)

, 𝑎0 = 2

𝑏(𝑛+2)

1 𝑎(2𝑛+1)

, 𝑎1 = − 4

𝑏(𝑛+2)

,

substituting in (39) we get 1 𝑎(2𝑛+1)

𝑣(𝜉) = 4

𝑏(𝑛+2)

[2 + [tanh(𝜇𝜉) + coth(𝜇𝜉)]].

(a)

(b)

Fig. 3.114.(a)-(b): Graphical representation of Eq. (3.283) for different values of parameters 1 𝑎(2𝑛+1)

𝑢(𝜉) = (4

𝑏(𝑛+2)

𝟏

[2 + [tanh(𝜇𝜉) + coth(𝜇𝜉)]])𝒏 .

(a)

(b)


3.10.3. Klein-Gordon-Zakharov System

418

We consider the exact wave solution of nonlinear Klein-Gordon-Zakharov system [34] which describes the interaction of Longmuir wave and ion sound wave in plasma 𝐸𝑡𝑡 − 𝐸𝑥𝑥 + 𝐸 − 𝑎𝑁𝐸 − 𝛽|𝐸| 𝐸 = 0, 𝑁𝑡𝑡− 𝑁𝑥𝑥 = 𝛾(|𝐸 2 |)𝑥𝑥 . Where the complex valued unknown function 𝐸 = 𝐸(𝑥, 𝑡) denotes the fst time scale component of electric field raised by electrons,and the real valued unknown function 𝑁 = 𝑁(𝑥, 𝑡)represents the deviation of ion density.𝛼, 𝛽 and 𝛾 are some real parameters.if 𝛽 = 0 and 𝛾 = 1, it is a conversed system .moreover if 𝛼 = 1, 𝛽 = 0 and 𝛾 = 1,it will reduce to the classical Klein-Gordon-Zakharov system regime 𝐸𝑡𝑡 − 𝐸𝑥𝑥 + 𝐸 − 𝑁𝐸 = 0,

(3.288)

𝑁𝑡𝑡− 𝑁𝑥𝑥 = (|𝐸 2 |)𝑥𝑥 .

(3.289)

In the theoretical investigation of the dynamics of strong Langmuir turbulence in plasma physics, various types of Zakharov equation play a very important role. In modern physics since the discovery ok Kruskal and Zabusky for the solitons, searching for exact solution of non linear problems has become more attractive and motivated a considerable research. Because of the complexity of the nonlinear wave equation, there is no unified method to find all solution of these equations. Consider the transformation 𝜂 = 𝑘𝑥 + 𝜔𝑡 and the following gauge transformation 𝐸(𝑥, 𝑡) = 𝑒𝑥𝑝[𝑖𝜂]𝑒(𝑥, 𝑡), Using transformation in (3.288) and (3.289) we have 2i(𝜔𝑒𝑡 − 𝑘𝑒𝑥 ) + 𝑒𝑡𝑡 − 𝑒𝑥𝑥 + (𝑘 2 − 𝑤 2 + 1)𝑒 − 𝛼𝑁𝑒 − 𝛽𝑒 3 = 0, (3.290) 𝑁𝑡𝑡− 𝑁𝑥𝑥 = 𝛾(𝑒 2 )𝑥𝑥 .

(3.291)

Where 𝑒(𝑥, 𝑡)a real-valued function, k is,𝜔 are two real constant to b determined by (3.290) suppose that 𝑒(𝑥, 𝑡) = 𝑒(𝜉) = 𝑒(𝜔𝑥 + 𝑘𝑡) Substituting it in into (43), we infer that 𝑁(𝑥, 𝑡) =

𝑘 2 −𝑤 2 𝑒′′ 𝑎

𝑒

+

𝑘 2 −𝑤 2 +1 𝑎

𝛽

− 𝛼 𝑒 2,

The above equation implies 𝑁(𝑥, 𝑡) has the following form as 𝑁(𝑥, 𝑡) = 𝑛(𝜉) = 𝑛(𝜔𝑥 + 𝑘𝑡),

419

using the relation of 𝑒(𝑥, 𝑡) and 𝑁(𝑥, 𝑡) into (3.290) and integrating the resultant equation twice w.r.t 𝜉, we have 𝜔 2 𝛾𝑒 2

𝑛(𝜉) = 𝑘 2 −𝑤2 .

(3.292)

Here we take integration constant 𝐶 = 0, using the relation (3.291) we have (𝑘 2 − 𝑤 2 )𝑒 ′ − (𝑘 2 − 𝑤 2 + 1)𝑒 − (𝛽 +

𝜔 2 𝛾𝑒 2 𝑘 2 −𝑤 2

) 𝑒 3 = 0,

(3.293)

𝜔 2 𝛾𝑒 2

denote 𝑝 = 𝑘 2 − 𝑤 2 + 1,𝑞 = (𝛽 + 𝑘 2 −𝑤2 ). Hence (3.293) becomes the following equation (𝑝 − 1)𝑒 ′′ − 𝑝𝑒 − 𝑞𝑒 3 = 0. balancing the 𝑒 3 and 𝑒 ′′′ by using the homogenous principle, we have 3𝑀 = 𝑀 + 2, 𝑀 = 1. So the trail solution is 𝑒(𝜉) = 𝑎0 + 𝑎1 [sinh(𝜇𝜉) + cosh(𝜇𝜉)]1 +𝑏1 [sinh(𝜇𝜉) + cosh(𝜇𝜉)]−1 .

(3.294)

Putting (3.294) into (3.293) compare the like power of [𝑠𝑖𝑛ℎ(𝜇𝜉) + 𝑐𝑜𝑠ℎ(𝜇𝜉)], we get [sinh(𝜇𝜉) + cosh(𝜇𝜉)]−3 :

𝛼𝛾𝜔𝑏13 + 𝛽𝑘 2 𝑏13 − 𝛽𝑤 2 𝑏13 = 0,

[sinh(𝜇𝜉) + cosh(𝜇𝜉)]−2 :

3𝛽𝑘 2 𝑏12 + 3𝛼𝛾𝜔2 𝑏12 − 3𝛽𝑤 2 𝑏12 = 0,

. . . [sinh(𝜇𝜉) + cosh(𝜇𝜉)]2 :

3𝛼𝛾𝜔𝑎0 𝑎12 + 3𝛽𝑘 2 𝑎0 𝑎12 − 3𝛽𝑤 2 𝑎0 𝑎12 = 0,

[sinh(𝜇𝜉) + cosh(𝜇𝜉)]3 :

−𝛽𝜔2 𝑎13 + 𝛽𝑘 2 𝑎13 − 𝛼𝛾𝜔𝑎13 = 0.

Solving the above system by using MAPLE, we have 𝑎0 = 𝑜, 𝑏1 = 𝑏1 , 𝑎1 = 𝑎1 , 𝑘 = 𝑘, 𝜔 = − −1−𝑘 2 +𝑤2

sinh (√

𝑘 2 −𝑤 2

(−

−1−𝑘 2 +𝑤 2

[

𝑘 2 −𝑤 2

𝛼𝛾

𝛽(𝑘 2 −𝑤 2 )

𝑒(𝑥, 𝑡) = 𝑎0 + 𝑎1 cosh (√

𝛽(𝑘 2 −𝑤 2)

(−

𝛼𝛾

420

𝑘 2 −𝑤 2

𝑥 + 𝑘𝑡)) +

𝛽(𝑘 2 −𝑤 2 ) 𝛼𝛾

−1−𝑘 2 +𝑤 2

,𝜇 = √

𝑥 + 𝑘𝑡))

]

.

−1

sinh (√

−1−𝑘2 +𝑤 2 𝑘 2 −𝑤 2

(−

𝛽(𝑘 2 −𝑤 2 )

+𝑏1

𝛼𝛾

𝑥 + 𝑘𝑡)) .

−1−𝑘 2 +𝑤 2

+cosh (√

𝑘 2 −𝑤 2

[

(−

𝛽(𝑘 2 −𝑤 2 ) 𝛼𝛾

(a)

𝑥 + 𝑘𝑡))

]

(b)

Fig. 3.116.(a)-(b): Graphical representation of Eq. (3.288-3.289) for different values of parameters −1−𝑘 2 +𝑤2

sinh (√

𝑘 2 −𝑤 2

(−

𝛽(𝑘 2 −𝑤 2 ) 𝛼𝛾

𝐸(𝑥, 𝑡) = 𝑎0 + 𝑎1 −1−𝑘2 +𝑤 2

+ cosh (√

𝑘 2 −𝑤 2

[

(−

𝑥 + 𝑘𝑡))

𝛽(𝑘 2 −𝑤 2 ) 𝛼𝛾

𝑥 + 𝑘𝑡))

] −1

−1−𝑘2 +𝑤 2

sinh (√

𝑘 2 −𝑤 2

(−

𝛽(𝑘 2 −𝑤 2 )

+𝑏1

𝛼𝛾

𝑥 + 𝑘𝑡)) exp(𝑖𝜂)

−1−𝑘 2 +𝑤 2

[

+cosh (√

𝑘 2 −𝑤 2

421

(−

𝛽(𝑘 2 −𝑤 2 ) 𝛼𝛾

𝑥 + 𝑘𝑡))

]

(a)

(b)

Fig. 3.117.(a)-(b): Graphical representation of Eq. (3.288-3.289) for different values of parameters

The exact solution of 𝑁(𝑥, 𝑡) is given as 𝑁(𝑥, 𝑡) =

𝑘 2 −𝑤 2 𝑒 ′′ (𝑥,𝑡) 𝛼

𝑒(𝑥,𝑡)

+

𝑘 2 −𝑤 2 +1 𝛼

𝛽

− 𝛼 𝑒 2 (𝑥, 𝑡).

3.11. U-expansion method Recently Usman et al. [204] developed U-expansion method to seek the soliton solutions of nonlinear evolution equations. This method is so simple and straight forward. This method have less computational work as compare to Exp-function method, (G`/G)expansion method, F-expansion method etc. This method is very effective and more reliable.

3.11.1. Methodology We consider the general nonlinear PDE of the type 𝑃(𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑦 , 𝑢𝑧 , 𝑢𝑡𝑡 , 𝑢𝑥𝑥 , 𝑢𝑦𝑦 , 𝑢𝑥𝑥 , 𝑢𝑥𝑡 , 𝑢𝑦𝑡 , 𝑢𝑧𝑡 , 𝑢𝑥𝑦 , 𝑢𝑥𝑧 , 𝑢𝑦𝑧 , … ) = 0, (3.235) where 𝑃 is a polynomial in its arguments. The essence of the U-expansion method can be presented in the following steps: Step 1: Seek Solitary wave solutions of Eq. (3.235) by taking 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡, and transform Eq. (3.235) to the ordinary differential equation. 𝑄(𝑢, 𝜔𝑢′ , 𝑘𝑢′ , 𝑙𝑢′ , 𝑚𝑢′ , 𝜔2 𝑢′′ , 𝑘 2 𝑢′′ , 𝑙 2 𝑢′′ , … ) = 0,

(3.236)

where 𝜔 is constant and where prime denotes the derivative with respect to 𝜉.

422

Step 2: If possible, integrate Eq. (3.236) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero. Step 3: According to U-Expansion method, we assume that the wave solution can be expressed in the following form 𝑛 𝑢(𝜉) = ∑𝑀 𝑛=0 𝑎𝑛 𝑈 ,

(3.237)

where 𝑈 is the solution of first order nonlinear equation in the form 𝑈 ′ (𝜉) = 𝛼1 𝑈 + 𝛼2 𝑈 2 .

(3.238)

where 𝛼1 and 𝛼2 are real constants, 𝑀 is a positive integer to be determined. The Table 1 shows the solution of Eq. (3.238) for different value of 𝛼1 and 𝛼2 .

Table 3.2. Solutions of Eq. (3.238) for different values of 𝛼1 and 𝛼2 . 𝑈(𝜉)

𝛼1

𝛼2

Cases

−2i

2i

I

1 i − tan(𝜉) 2 2

−i

I

II

1 (1 + i sec(𝜉) − i tan(𝜉)) 2

4

2

III

1 1 −1 − tanh(𝜉) − coth(𝜉) 2 2

−√3i √3i IV

√3 √3 i 𝐶1 sin ( 2 𝜉) + 𝐶2 cos ( 2 𝜉) 1

1 − 2 2

√3 √3 𝐶1 cos ( 2 𝜉) + i𝐶2 sin ( 2 𝜉)

Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term(s) of highest order with the highest order nonlinear term(s) in Eq. (3.236). Step 5: Substituting (3.237) into Eq. (3.236) with (3.238) will yields an algebraic equation involving power of U. Equating the coefficients of like power of U to zero gives a system of algebraic equations for 𝑎𝑖 , 𝑘, 𝑙, 𝑚 and 𝜔. Then, we solve the system with the

423

aid of a computer algebra system (CAS), such as MAPLE 13, to determine these constants. Step 6: Putting these constant into Eq. (3.237), coupled with the well known solutions of Eq. (3.238), we can obtained the more general type and new exact traveling wave solution of the nonlinear partial differential equation (3.235).

3.11.2. Cubic Ginzburg-Landau Equation Consider Cubic Ginzburg-Landau equation [230] 𝑤𝑡 = (1 + 𝑖𝑎)𝑤𝑥𝑥 + 𝑅𝑤 − (1 + 𝑖𝑏)|𝑤|2 𝑤 = 0,

(3.239)

We assume the complex field 𝑤(𝑥, 𝑡) can be expressed as 𝑤(𝑥, 𝑡) = 𝑢(𝑡)𝑒 𝑖𝛼𝑥 , 𝑖 2 = −1, This transformation converts it into following ODE 𝑢′ (𝑡) = 𝛽𝑢(𝑡) + 𝛾𝑢3 (𝑡), where 𝛽 = (1 − 𝛼 2 ) − 𝑖𝑎𝛼 2 , 𝛾 = −1 − 𝑖𝑏 To get the analytical solution, we introduce transformation 1

𝑢 = 𝑣2, This transformation converts the equation into following 1 2

𝑣 ′ = 𝛽𝑣 + 𝛾𝑣 2 .

(3.240)

Balance the 𝑣 2 and 𝑣 ′ by using homogenous principal, we have 𝑀 + 1 = 2𝑀, 𝑀 = 1. Then we have the trail solution is 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑉

(3.241)

By equation (3.241) and (3.235), substitute into (3.240), we obtain the system of algebraic equations as follow: 𝐶0 = −𝑎0 + 𝛼 2 𝑎0 + 𝑙𝑎𝛼 2 𝑎0 + 𝑎02 + 𝑙𝑏𝑎02 , 1

𝐶1 = 2 𝑎1 𝛼1 − 𝑎1 + 𝛼 2 𝑎1 + 𝑙𝑎𝛼 2 𝑎1 + 2𝑎0 𝑎1 + 2𝑙𝑏𝑎0 𝑎1 , 1

𝐶2 = 2 𝑎1 𝛼2 + 𝑎12 + 𝑙𝑏𝑎12 . Equating the coefficient of like powers of 𝑉 to be zero, we obtain [𝐶1 = 0, 𝐶2 = 0, 𝐶0 = 0], 424

We have six solution set satisfy the given equation 1st Solution set: 𝜔 = 𝜔, 𝑎0 = −

−1+𝛼2 +i𝑎𝛼2 1+i𝑏

, 𝑎1 = 𝑎1 , 𝛼1 = −2 + 2𝛼 2 + 2i𝑎𝛼 2 , 𝛼2 = 2𝑎1 −

2i𝑏𝑎1 . Their corresponding solution is 𝑤(𝑥, 𝑡) = ((−

−1+𝛼2 +i𝑎𝛼2 1+i𝑏

1 2

+ 𝑎1 𝑉) ) ei𝛼𝑥 .

2nd Solution set: 𝜔 = 𝜔, 𝑎0 = −

−1+𝛼2 +i𝑎𝛼2 1+i𝑏

, 𝑎1 = 0, 𝛼1 = 𝛼1 , 𝛼2 = 𝛼2 .

Their corresponding solution is 𝑤(𝑥, 𝑡) = √−

−1+𝛼2 +i𝑎𝛼2 i𝛼𝑥 e . 1+i𝑏

3rd Solution set: 𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝛼1 = 2 − 2𝛼 2 − 2i𝑎𝛼 2 , 𝛼2 = −2𝑎1 − 2i𝑏𝑎1 . Their corresponding solution is 𝑤(𝑥, 𝑡) = √𝑎1 𝑉ei𝛼𝑥 . 4th Solution set: 𝜔 = 𝜔, 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝛼1 = −2 + 2𝛼 2 + 2𝑙𝑎𝛼 2 , 𝛼2 = − −

−1+𝛼2 +𝑙𝑎𝛼2 +𝑎0 𝑙𝑎0

2𝑎1 (−1+𝛼2 +𝑙𝑎𝛼2 ) 𝑎0

.

Their corresponding solution is 𝑤(𝑥, 𝑡) = √𝑎0 + 𝑎1 𝑉ei𝛼𝑥 . 5th Solution set: 𝜔 = 𝜔, 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝛼1 = 𝛼1 , 𝛼2 = 𝛼2 , 𝑏 = − Their corresponding solution is 𝑤(𝑥, 𝑡) = √𝑎0 ei𝛼𝑥 . 6th Solution set:

425

−1+𝛼2 +𝑙𝑎𝛼2 +𝑎0 𝑙𝑎0

.

,𝑏 =

𝜔 = 𝜔, 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝛼1 = −2 + 2𝛼 2 + 2𝑙𝑎𝛼 2 , 𝛼2 = −

−1+𝛼2 +𝑙𝑎𝛼2 +𝑎0 𝑙𝑎0

2𝑎1 (−1+𝛼2 +𝑙𝑎𝛼2 ) 𝑦0

,𝑏 =

, 𝛼 = 𝛼.

Their corresponding solution is 𝑤(𝑥, 𝑡) = √𝑎0 + 𝑎1 𝑉ei𝛼𝑥 .

3.11.3. Camassa-Holm Equation Consider the Family of Camassa-Holm equation [230] 𝑢𝑡 − 𝑢𝑥𝑥𝑡 + 𝛼𝑢𝑥 + 𝑏𝑢𝑢𝑥 = 𝑘𝑢𝑥 𝑢𝑥𝑥 + 𝑢𝑢𝑥𝑥𝑥 ,

(3.242)

Introducing the transformation as 𝜉 = 𝑥 + 𝜔𝑡, we can convert the given equation into ordinary differential equations, after integration we get 𝑑3

𝑑

𝑑

𝑑

(𝑑𝜉 𝑢(𝜉)) 𝜔 − (𝑑𝜉3 𝑢(𝜉)) 𝜔 + 𝑎 (𝑑𝜉 𝑢(𝜉)) + 𝑏𝑢(𝜉) (𝑑𝜉 𝑢(𝜉)) − 𝑑2

𝑑

𝑑3

𝑘 (𝑑𝜉 𝑢(𝜉)) (𝑑𝜉2 𝑢(𝜉)) − 𝑢(𝜉) (𝑑𝜉3 𝑢(𝜉)) = 0,

(3.243)

Balance the 𝑢𝑢′ and 𝑢′′′ by using homogenous principal, we have 𝑀 + 3 = 𝑀 + 𝑀 + 1, 𝑀 = 2. Then we have the trail solution is 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈 + 𝑎2 𝑈 2 .

(3.244)

By equation (3.244) and (3.235), substitute into (3.243), we obtain the system of algebraic equations as follow: 𝐶1 = 𝜔𝑎1 𝛼1 − 𝜔𝑎1 𝛼13 + 𝑎𝑎1 𝛼1 − 𝑎0 𝑎1 𝛼13 + 𝑏𝑎0 𝑎1 𝛼1 , . . . 𝐶7 = −8𝑘𝑎22 𝛼23 − 16𝑎22 𝛼23 . Equating the coefficient of like powers of 𝑈 to be zero, we obtain [𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0, 𝐶7 = 0], We have nine solution set satisfy the given equation 1st Solution set:

426

𝜔 = 𝜔, 𝑎0 = −

𝜔+𝑎+𝑘𝜔+𝑘𝑎−𝑏𝜔 𝑘𝑏

𝑏

, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝛼1 = √1+𝑘 , 𝛼2 = 0.

Their corresponding solution is 𝑢(𝜉) = −


+ 𝑎1 𝑈,

where 𝜉 = 𝑥 + 𝜔𝑡.

(a)

(b)


2nd

Solution set: 𝜔 = 𝜔, 𝑎0 = −


𝑏

, 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝛼1 = √3+3𝑘 , 𝛼2 = 0.

Their corresponding solution is 𝑢(𝜉) = −


+ 𝑎2 𝑈 2 ,

where 𝜉 = 𝑥 + 𝜔𝑡.

(a)

(b)

427


3rd Solution set: 1 𝑏𝑎12 +12𝑎𝑎2 , 𝑎0 𝑎2 (𝑏−1)

𝜔 = 12 1 3

1 3𝑎12 +12𝑎𝑎2 −2𝑏𝑎12 , 𝑎1 𝑎2 (𝑏−1)

= − 12

= 𝑎1 , 𝑎2 = 𝑎2 , 𝛼1 =

1

√2√3√𝑏, 𝛼2 = 0, 𝑘 = − 2.

Their corresponding solution is 1 3𝑎12 +12𝑎𝑎2 −2𝑏𝑎12 𝑎2 (𝑏−1)

𝑢(𝜉) = − 12

+ 𝑎1 𝑈 + 𝑎2 𝑈 2 ,

1 𝑏𝑎12 +12𝑎𝑎2 𝑡. 𝑎2 (𝑏−1)

where 𝜉 = 𝑥 + 12

(a)

(b)


4th Solution set: 1 𝑏 3⁄2 𝑎1 −√𝑏𝑎1 −2𝑎𝛼2 , 𝑎1 𝛼2 (𝑏−1)

𝑎

𝜔 = 𝑏−1 , 𝑎0 = 2

= 𝑎1 , 𝑎2 = 0, 𝛼1 = √𝑏, 𝛼2 = 𝛼2 , 𝑘 = −3.

Their corresponding solution is 1 𝑏 3⁄2 𝑎1 −√𝑏𝑎1 −2𝑎𝛼2 𝛼2 (𝑏−1)

𝑢(𝜉) = 2

+ 𝑎1 𝑈,

𝑎

where 𝜉 = 𝑥 + 𝑏−1 𝑡.

428

(a)

(b)


5th Solution set: 𝑎−𝑎0 𝛼12 +𝑏𝑎0

𝜔 = (𝛼

1 −1)(𝛼1

, 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝛼2 = 0, 𝑘 = +1)

−𝛼12 +𝑏 𝛼12

.

Their corresponding solution is 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈, 𝑎−𝑎0 𝛼12 +𝑏𝑎0

where 𝜉 = 𝑥 + (𝛼

1 −1)(𝛼1 +1)

𝑡.

(a)

(b)


6th Solution set: 𝑎−𝑎0 𝛼12 +𝑏𝑎0

𝜔 = (𝛼

1 −1)(𝛼1

, 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝛼1 = 𝛼1 , 𝛼2 = 0, 𝑘 = +1)

Their corresponding solution is 429

−𝛼12 +𝑏 𝛼12

.

𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈, 𝑎−𝑎0 𝛼12 +𝑏𝑎0

where 𝜉 = 𝑥 + (𝛼

1 −1)(𝛼1 +1)

𝑡.

(a)

(b)


7thSolution set: 𝜔=

𝑎−3𝑎0 𝛼12 +𝑏𝑎0 −1+3𝛼12

1 −3𝛼12 +𝑏

, 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝛼1 = 𝛼1 , 𝛼2 = 0, 𝑘 = 3

𝛼12

.

Their corresponding solution is 𝑢(𝜉) = 𝑎0 + 𝑎2 𝑈 2 , where 𝜉 = 𝑥 +

𝑎−3𝑎0 𝛼12 +𝑏𝑎0 −1+3𝛼12

𝑡.

(a)

(b)


430

8th Solution set: 𝑎

1

𝜔 = 𝜔, 𝑎0 = − 𝑏−1 , 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝛼1 = 3 √3, 𝛼2 = 0, 𝑘 = 𝑏 − 1. Their corresponding solution is 𝑎

𝑢(𝜉) = − 𝑏−1 + 𝑎2 𝑈 2 , where 𝜉 = 𝑥 + 𝜔𝑡.

(a)

(b)


9thSolution set: 𝑎

𝜔 = 𝜔, 𝑎0 = − 𝑏−1 , 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝛼1 = −1, 𝛼2 = 0, 𝑘 = 𝑏 − 1. Their corresponding solution is 𝑎

𝑢(𝜉) = − 𝑏−1 + 𝑎1 𝑈, where 𝜉 = 𝑥 + 𝜔𝑡.

(a)

(b)

431


3.11.4. Schr𝐨̈ dinger Equation of Cubic Nonlinearity Consider Schr𝑜̈ dinger equation of Cubic nonlinearity [230] 𝑖𝑤𝑡 + 𝛼𝑤𝑥𝑥 − 𝑏𝑢𝑥𝑥𝑥𝑥 + 𝑐|𝑤|2 𝑤 = 0,

(3.245)

We first the transformation 𝑤(𝑥, 𝑡) = 𝑢(𝑥)𝑒 𝑖𝛼𝑡 , 𝑖 2 = −1, where 𝛼 is real parameter. The assumption the equation into ODE 𝑑4

𝑑2

𝑏 (𝑑ξ4 𝑢(𝜉)) − 𝑎 (𝑑ξ2 𝑢(𝜉)) + 𝑎𝑢(𝜉) + 𝑐𝑢(𝜉)3 = 0.

(3.246)

Balance the 𝑢′′′′ and 𝑢3 by using homogenous principal, we have 𝑀 + 4 = 3𝑀, 𝑀 = 2. Then we have the trail solution is 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈+𝑎2 𝑈 2

(3.247)

By equation (3.247) and (3.235), substitute into (3.246), we obtain the system of algebraic equations as follow: 𝐶0 = 𝛼𝑎0 + 𝑐𝑎03 = 0, 𝐶1 = 𝛼𝑎1 + 𝑏𝑎1 𝛼14 − 𝑎𝑎1 𝛼12 + 3𝑐𝑎02 𝑎1 = 0, 𝐶2 = 𝛼𝑎2 + 11𝑏𝑎2 𝛼14 − 3𝑎𝑎2 𝛼12 + 3𝑐𝑎02 𝑎2 + 3𝑐𝑎0 𝑎12 + 15𝑏𝑎1 𝛼13 𝛼2 − 3𝑎𝑎1 𝛼1 𝛼2 = 0, 𝐶3 = 𝑐𝑎13 − 2𝑎𝑎1 𝛼22 + 50𝑏𝑎1 𝛼12 𝛼22 + 85𝑏𝑎2 𝛼13 𝛼2 − 7𝑎𝑎2 𝛼1 𝛼2 + 6𝑐𝑎0 𝑎1 𝑎2 = 0, 𝐶4 = −4𝑎𝑎2 𝛼22 + 3𝑐𝑎0 𝑎22 + 3𝑐𝑎12 𝑎2 + 60𝑏𝑎1 𝛼1 𝛼23 + 210𝑏𝑎2 𝛼12 𝛼22 = 0, 𝐶5 = 24𝑏𝑎1 𝛼24 + 3𝑐𝑎1 𝑎22 + 210𝑏𝑎2 𝛼1 𝛼23 = 0, 𝐶6 = 𝑐𝑎23 + 74𝑏𝑎2 𝛼24 = 0. Solve the system of six equations. We have five solution set satisfy the given equation We have four solution set satisfy the given equation 1st Solution set: 𝜔 = 𝜔, 𝑎0 = −

𝛼 𝛼 𝑐√− 𝑐

, 𝑎1 =

𝛼 𝑐

2√− 𝛼2 𝛼1

2𝛼

, 𝑎2 = 0, 𝛼1 = 𝛼1 , 𝑎 = − 𝛼2 , 𝑏 = 0, 𝛼2 = 𝛼2 . 1

432

Their corresponding solution is

𝑤(𝑥, 𝑡) = √−

𝛼 𝛼 𝑐√− 𝑐

+

𝛼 𝑐

2√− 𝛼2 𝛼1

𝑈ei𝛼𝑡 .

(a)

(b)


2nd Solution set: 5 𝛼

1 𝛼

1

1

𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 𝑎2 , 𝛼1 = 𝛼1 , 𝑎 = 4 𝛼2 , 𝑏 = 4 𝛼4 , 𝑦 = 0, 𝛼2 = 0. Their corresponding solution is 𝑤(𝑥, 𝑡) = √𝑎1 𝑈 + 𝑎2 𝑈 2 ei𝛼𝑡 .

(a)

(b)


433

3rd Solution set: 𝜔 = 𝜔, 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝛼1 =

2𝛼2 𝑎0 𝑎1

1 𝛼𝑎2

𝛼

2 0

0

, 𝑎 = − 2 𝛼2 𝑎12 , 𝑏 = 0, 𝑐 = − 𝑎2 .

Their corresponding solution is 𝑤(𝑥, 𝑡) = √𝑎0 + 𝑎1 𝑈ei𝛼𝑡 .

(a)

(b)


4th Solution set: 𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝛼1 = 𝛼1 , 𝑎 =

𝑏𝛼14 +𝛼 𝛼12

, 𝑏 = 𝑏, 𝑐 = 0, 𝛼2 = 0.

Their corresponding solution is 𝑤(𝑥, 𝑡) = √𝑎1 𝑈ei𝛼𝑡 .

(a)

(b)

434


3.11.5. Schrodinger Equation with Power Law Nonlinearity Consider Schrodinger equation with power law nonlinearity [230] 𝑖𝑤𝑡 + 𝛼𝑤𝑥𝑥 − 𝑏𝑢𝑥𝑥𝑥𝑥 + 𝑐|𝑤|2𝑛 𝑤 = 0,

(3.248)

We first the transformation 𝑤(𝑥, 𝑡) = 𝑢(𝑥)𝑒 𝑖𝛼𝑡 , 𝑖 2 = −1, where 𝛼 is real parameter. The assumption the equation into ODE 𝑏𝑢′′′′ − 𝑎𝑢′′ + 𝛼𝑢 − 𝑐𝑢2𝑛+1 = 0, Balance the 𝑢4 and 𝑢2𝑛+1 by using homogenous principal, we have 8 + 𝑀 − 4 = (2𝑛 + 1)𝑀, 2

𝑀 = 𝑛. To obtain an analytical solution, we should use the transformation 1

𝑢(𝑥) = 𝑣 𝛾 (𝑥) , 𝛾 = 𝑛, so that equation become 3

𝜕4

2

𝜕3

𝜕

2

𝜕2

2

𝑏𝛾𝜈 (𝜕𝑥 4 𝜈(𝜉)) + 4𝑏𝛾𝜈 (𝜕𝑥 𝜈(𝜉)) (𝜕𝑥 3 𝜈(𝜉)) + 3𝑏𝛾(𝛾 − 1)𝜈 (𝜕𝑥 2 𝜈(𝜉)) + 𝜕

2

𝜕2

6𝑏𝛾(𝛾 − 1)(𝛾 − 2)𝜈 (𝜕𝑥 𝜈(𝜉)) (𝜕𝑥 2 𝜈(𝜉)) + 𝑏𝛾(𝛾 − 1)(𝛾 − 2)(𝛾 − 𝜕

4

3

𝜕2

2

𝜕

2

3) (𝜕𝑥 𝜈(𝜉)) − 𝑎𝛾𝜈 (𝜕𝑥 2 𝜈(𝜉)) − 𝑎𝛾(𝛾 − 1)𝜈 (𝜕𝑥 𝜈(𝜉)) + 𝛼𝜈 4 − 𝑐𝜈 6 = 0. (3.249) Balance the 𝑣 6 and 𝑣 3 𝑣 4 by using homogenous principal, we have 6𝑀 = 3𝑀 + 4𝑀, 𝑀 = 2. Then we have the trail solution is 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈 + 𝑎2 𝑈 2 .

(3.250)

By equation (3.250) and (3.235), substitute into (3.249), we obtain the system of algebraic equations as follow: 𝐶1 = −𝑎𝛾𝑎03 𝑎1 𝛼12 + 𝑏𝛾𝑎03 𝑎1 𝛼14 + 4𝛼𝑎03 𝑎1 − 6𝑐𝑎05 𝑎1 = 0, 𝐶2 = 15𝑏𝛾𝑎03 𝑎1 𝛼13 𝛼2 − 3𝑎𝛾𝑎03 𝑎2 𝛼12 + ⋯ − 15𝑐𝑎04 𝑎12 − 3𝑎𝛾𝑎03 𝑎1 𝛼1 𝛼2 = 0, . 435

. .. Solve the system of twelve equations. We have five solution set satisfy the given equation 1st Solution set: 𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝛼1 = 𝛼1 , 𝛼2 = 0, 𝑎 = 𝛼+32𝑏𝛾𝛼14 −13𝑏𝛾2 𝛼14 −24𝑏𝛾3 𝛼14 +16𝑏𝛾4 𝛼14 ,𝑏 𝛾𝛼12 (4𝛾−1)

= 𝑏, 𝑐 = 0.


𝑤(𝑥, 𝑡) = (𝑎2 𝑈 2 )𝑛 ei𝛼𝑡 .

(a)

(b)


2nd Solution set: 𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝛼1 = 𝛼1 , 𝛼2 = 0, 𝑎 = 𝛼+𝑏𝛾4 𝛼14 +4𝑏𝛾𝛼14 −4𝑏𝛾2 𝛼14 𝛾 2 𝛼12

, 𝑏 = 𝑏, 𝑐 = 0.


𝑤(𝑥, 𝑡) = (𝑎1 𝑈)𝑛 ei𝛼𝑡 .

436

(a)

(b)


3.11.6. Sine-Gordon Equation We will consider Sine-Gordon equation [230] 𝑢𝑡𝑡 − 𝑢𝑥𝑥 + sin(𝑢) = 0,

(3.251)

Firstly,we introduce a transformation sin(𝑢) =

𝑣−𝑣 −1 2𝑖

, cos(𝑢) =

𝑣+𝑣 −1 2𝑖

,

This transformation converts the equation of the following 𝜕2

𝜕2

𝜕

𝜕

2𝜈 𝜕𝑡 2 𝑢 − 2𝜈 𝜕𝑥 2 𝜈 − 2 𝜕𝑡 𝜈 2 + 2 𝜕𝑥 𝜈 2 + 𝜈 3 − 𝜈 = 0. Introducing the transformation as 𝜉 = 𝑥 + 𝜔𝑡, we can convert the given equation into ordinary differential equations, after integration we get 𝑑2

𝑑2

𝑑

2𝜈(𝜉) (𝑑𝜉2 𝑣(𝜉)) 𝜔2 − 2𝜈(𝜉) (𝑑𝜉2 𝑣(𝜉)) − 4𝜈(𝜉) (𝑑𝜉 𝑣(𝜉)) 𝜔 + 𝑑

4𝜈(𝜉) (𝑑𝜉 𝑣(𝜉)) + 𝜈(𝜉)3 − 𝜈(𝜉) = 0.

(3.252)

Balance the 𝑣 3 and 𝑣𝑣 ′ by using homogenous principal, we have 𝑀 + 𝑀 + 1 = 3𝑀, 𝑀 = 2. Then we have the trail solution is 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑉 + 𝑎2 𝑉 2 .

(3.253)

By equation (3.253) and (3.235), substitute into (3.252), we obtain the system of algebraic equations as follow: 437

𝐶1 = −𝑎1 + 3𝑎02 𝑎1 + 2𝜔2 𝑎0 𝑎1 𝛼12 − 4𝜔𝑎0 𝑎1 𝛼1 − 2𝑎0 𝑎1 𝛼12 + 4𝑎0 𝑎1 𝛼1 , . . . 𝐶0 = −𝑎0 + 𝑎03 . Equating the coefficient of like powers of 𝑉 to be zero, we obtain [𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0, 𝐶0 = 0]. We have five solution set satisfy the given equation 1st Solution set: 7

64

9

9

𝜔 = , 𝑎0 = 1, 𝑎1 =

𝛼2 , 𝑎2 =

256 81

9

𝛼22 , 𝛼1 = . 4

Their corresponding solution is 𝑢(𝜉) = 1 +

64 9

𝛼2 𝑈 +

256 81

𝛼22 𝑈 2 ,

7

where 𝜉 = 𝑥 + 9 𝑡.

(a)

(b)


2nd Solution set: 7

𝜔 = 9 , 𝑎0 = 1, 𝑎1 = −

16 3

𝛼2 , 𝑎2 =

256 81

Their corresponding solution is 𝑢(𝜉) = 1 −

16 3

𝛼2 𝑈 +

256 81

9

𝛼22 , 𝛼1 = − 8.

𝛼22 𝑈 2 ,

7

where 𝜉 = 𝑥 + 9 𝑡. 438

(a)

(b)


3rd Solution set: 9

𝜔 = 7 , 𝑎0 = −1, 𝑎1 = −

64 7

𝛼2 , 𝑎2 = −

256 49

7

𝛼22 , 𝛼1 = 4.

Their corresponding solution is 𝑢(𝜉) = −1 −

64 7

𝛼2 𝑈 −

256 49

𝛼22 𝑈 2 ,

9


(a)

(b)


4th Solution set: 15

𝜔 = 113 , 𝑎0 = −1, 𝑎1 = 0, 𝑎2 =

100352 12769

439

113

𝛼22 , 𝛼1 = 224 , 𝛼2 = 𝛼2 .

Their corresponding solution is 𝑢(𝜉) = −1 +

100352 12769

𝛼22 𝑈 2 ,

15


(a)

(b)


5th Solution set: 𝜔=

113 15

, 𝑎0 = 1, 𝑎1 = 0, 𝑎2 = −

100352 225

Their corresponding solution is 𝑢(𝜉) = 1 − where 𝜉 = 𝑥 +

113 15

100352 225

𝛼22 𝑈 2 ,

𝑡.

440

15

𝛼22 , 𝛼1 = 224 , 𝛼2 = 𝛼2 .

(a)

(b)


6th Solution set: 1

1

𝜔 = −1, 𝑎0 = 1, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝛼1 = − 4 , 𝛼2 = − 8 𝑎1 . Their corresponding solution is 𝑢(𝜉) = 1 + 𝑎1 𝑈, where 𝜉 = 𝑥 − 𝑡.

(a)

(b)


3.11.7. Third Variant of KdV Consider the Third Variant of KdV [230] as 𝑢𝑡 + (𝑎𝑢 + 𝑏𝑢𝑛 )𝑥 + 𝑘(𝑢𝑛 )𝑥𝑥𝑥 = 0,

(3.254)

Introducing the transformation as 𝜉 = 𝑥 + 𝑤𝑡, we can convert the given equation into ordinary differential equations, after integration we get (𝜔 + 𝑐)𝑢 + 𝑏𝑢𝑛 + 𝑘𝑛𝑢𝑛−1 𝑢′′ + 𝑘𝑛(𝑛 − 1)𝑢𝑛−2 (𝑢′ )2 = 0. Balance the 𝑢(𝑥, 𝑡) and 𝑢𝑛−1 𝑢′′ by using homogenous principal, we have −2

𝑀 = 𝑛−1, We introduced another transformation −1

𝑢 = 𝑣 𝑛−1 , This transformation converts it into following

441

(3.255)

𝜕2

(𝑤 + 𝑎)(𝑛 − 1)2 𝜈 3 + 𝑏(𝑛 − 1)𝜈 2 − 𝑘𝑛(𝑛 − 1)𝜈 (

𝜕𝜉 2

𝜈) + 2𝑛(2𝑛 −

2

𝜕

1) (𝜕𝜉 𝜈) = 0, Balance the 𝑣𝑣 ′′ and 𝑣𝑣 ′′ gives 𝑀 = 2. Then we have the trail solution is 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑉 + 𝑎2 𝑉 2 .

(3.256)

By equation (3.256) and (3.235), substitute into (3.255), we obtain the system of algebraic equations as follow 𝐶1 = 𝑘𝑛𝑎0 𝑎1 𝛼12 − 𝑘𝑛2 𝑎0 𝑎1 𝛼12 + ⋯ − 6𝑎𝑛𝑎02 𝑎1 + 2𝑏𝑛𝑎0 𝑎1 , . . . 𝐶0 = 𝑎𝑛2 𝑎03 − 2𝑎𝑛𝑎03 + ⋯ + 𝑎𝑎03 − 𝑏𝑎02 . Equating the coefficient of like powers of 𝑉 to be zero, we obtain [𝐶1 = 0, 𝐶2 = 0, 𝐶3 = 0, 𝐶4 = 0, 𝐶5 = 0, 𝐶6 = 0, 𝐶0 = 0], We have one solution set satisfy the given equation 1st Solution set: 𝑤=−

𝛼2 √−𝑎 𝑎1

, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝛼1 = √−𝑎.


𝑢(𝜉) = (𝑎1 𝑉)−𝑛−1 , where 𝜉 = 𝑥 −

𝛼2 √−𝑎 𝑎1

𝑡.

(a)

(b)

442


3.11.8. (2+1)-Dimensional Burgers Equation We will consider (2+1)-dimensional Burgers equation [230] 𝑢𝑡 + 𝑢𝑥𝑥 − (𝑘 − 𝑢)(𝑢 − 1) = 0, 𝑘 ≠ 0.

(3.257)

Introducing the transformation as 𝜉 = 𝑥 + 𝑤𝑡, we can convert the given equation into ordinary differential equations, after integration we get 𝑤𝑢′ − 𝑢′′ − 𝑘𝑢2 + 𝑘𝑢 + 𝑢3 − 𝑢2 = 0.

(3.258)

Balance the 𝑢3 and 𝑢′′ by using homogenous principal, we have 𝑀 + 2 = 3𝑀, 𝑀 = 1. Then we have the trail solution is 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈

(3.259)

By equation (3.259) and (3.235), substitute into (3.258), we obtain the system of algebraic equations as follow: 𝑐0 = −𝑘𝑎02 + 𝑘𝑎0 + 𝑎03 − 𝑎02 = 0, 𝑐1 = 𝑎1 𝜔𝛼1 − 𝑎1 𝛼12 − 2𝑘𝑎0 𝑎1 + 𝑘𝑎1 + 3𝑎02 𝑎1 − 2𝑎0 𝑎1 = 0, 𝑐2 = 𝑎1 𝜔𝛼2 − 3𝑎1 𝛼1 𝛼2 − 𝑘𝑎12 + 3𝑎0 𝑎12 − 𝑎12 = 0, 𝑐3 = −2𝑎1 𝛼22 + 𝑎13 . Equating the coefficient of like powers of 𝑈 to be zero, we obtain [𝑐1 = 0, 𝑐2 = 0, 𝑐3 = 0, 𝑐0 = 0], We have three solution set satisfy the given equation 1st Solution set: 1

1

𝜔 = − 2 𝑘√2 + √2, 𝑎0 = 𝑘, 𝑎1 = √2𝛼2 , 𝛼1 = 2 𝑘√2. Their corresponding solution is 𝑢(𝜉) = 𝑘 + √2𝛼2 𝑈. 1

where 𝜉 = 𝑥 − 2 𝑘√2𝑡. Fig 3.1(a)

443

(a)

(b)


2nd Solution set: 1

1

𝜔 = − 2 √2 + 𝑘√2, 𝑎0 = 1, 𝑎1 = √2𝛼2 , 𝛼1 = 2 √2. Their corresponding solution is 𝑢(𝜉) = 1 + √2𝛼2 𝑈. 1

where 𝜉 = 𝑥 − 2 √2𝑡.

(a)

(b)


3rd Solution set: 1

1

𝜔 = − 2 √2 + 𝑘√2, 𝑎0 = 0, 𝑎1 = √2𝛼2 , 𝛼1 = − 2 √2. Their corresponding solution is 𝑢(𝜉) = √2𝛼2 𝑈. 1

where 𝜉 = 𝑥 − 2 √2𝑡.

444

(a)

(b)


3.11.9. Medium Equal Width (MEW) Equation Consider the Medium Equal Width (MEW) equation [230] 𝑢𝑡 + 3𝑢2 𝑢𝑥 − 𝑎𝑢𝑥𝑥𝑡 = 0,

(3.260)

Introducing the transformation as 𝜉 = 𝑥 + 𝑤𝑡, we can convert the given equation into ordinary differential equations, after integration we get 𝑤𝑢 + 𝑢3 − 𝑎𝑢′′ 𝑤 = 0.

(3.261)

Balance the 𝑢3 and 𝑢′′ by using homogenous principal, we have 𝑀 + 2 = 3𝑀, 𝑀 = 1. Then we have the trail solution is 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈.

(3.262)

By equation (3.262) and (3.235), substitute into (3.261), we obtain the system of algebraic equations as follow 𝑐0 = 𝜔𝑎0 + 𝑎03 = 0, 𝑐1 = 𝜔𝑎1 + 3𝑎02 𝑎1 − 𝑎𝑎1 𝜔𝛼12 , 𝑐2 = 3𝑎0 𝑎12 − 3𝑎𝑎1 𝜔𝛼1 𝛼2 = 0, 𝑐3 = 𝑎13 − 2𝑎𝑎1 𝜔𝛼22 = 0. Equating the coefficient of like powers of 𝑈 to be zero, we obtain [𝑐1 = 0, 𝑐2 = 0, 𝑐3 , 𝑐0 = 0], We have one solution set satisfy the given equation 1st Solution set:

445

1

9

8

1

𝑤 = − 2 , 𝛽 = − 5 , 𝑎0 = 9 , 𝑎1 = − 2 , 𝑘 = 2, 𝐶 = −10, 𝑎 = −6. Their corresponding solution is 8

1

𝑢(𝜉) = 9 − 2 𝑈. 1

where 𝜉 = 𝑥 − 2 𝑡.

(a)

(b)


446

Chapter 4 Traveling Wave Solutions Nonlinear Evolution Equation of Fractional-order

447

4.1. Introduction It is to be noted that the fractional partial differential equations [112, 173, 180] (FDEs) are applied in many fields of sciences. The exact solutions of nonlinear fractional partial differential equations have a great importance in nonlinear sciences. In the past, many analytical and numerical methods have been proposed to obtain solutions of nonlinear FDEs, such as finite difference method [63], finite element method [42], differential transform method [150, 171], Adomian decomposition method [47-48, 95], variational iteration method [102, 172, 241], homotopy perturbation method [71, 77, 157] and so on. Many authors find the soliton solutions [20-21, 257] of fractional order PDEs. In this Chapter, I am going to propose some new algorithms which are very helpful for finding the soliton solutions of fractional-order partial differential equations. we have applied F-Expansion Method , Generalized Tanh Method, (G’/G)-Expansion Method Based on Riccati Equation, New Approach of (G’/G)-Expansion Method, Rational Hyperbolic Function (RHF) Method, U-expansion Method, Modified U-expansion Method, Generalized U-expansion Method, (U’/U)-expansion Method and (G𝛂/G)expansion method for fractional-order Partial Differential Equations (PDEs) including Boussinesq equation, Breaking Soliton equation, potential Kadomtsev-Petviashvili equation, Benjoman–Bona-Mahony (BBM) equation, Korteweg-de-Vries (KdV) equation etc. Moreover, these methods are greatly capable of minimizing the size of computational work as compared to the other existing methods. It is to be observed that the proposed technique has been applied on a wide range of nonlinear diversified physical problems including, high-dimensional nonlinear evolution equation. The proposed scheme is fully compatible with the complexity of such problems and is very user-friendly. Numerical results are very encouraging.

4.2. F-Expansion Method 448

As discuss in Chapter 3 the Zhou et al. proposed new algorithm named as F-Expansion method. This method has used to obtained the exact solutions of nonlinear evolution equations so far. Here we are applied F-expansion method for fractional-order PDEs using complex transformation [32]. The proposed scheme is fully compatible with the complexity of such problems and is very user-friendly. Numerical results are very encouraging.

4.2.1. Methodology We consider the general nonlinear PDE of the type 𝑃(𝑢, 𝐷𝑡𝛼 𝑢, 𝐷𝑥𝛼 𝑢, 𝐷𝑦𝛼 𝑢, 𝐷𝑧𝛼 𝑢, 𝐷𝑡2𝛼 𝑢, 𝐷𝑥2𝛼 𝑢, 𝐷𝑦2𝛼 𝑢, 𝐷𝑧2𝛼 𝑢, … ) = 0,

(4.1)

where 𝑃 is a polynomial in its arguments. The essence of the F-expansion method can be presented in the following steps:

Step 1: Seek Solitary wave solutions of Eq. (4.1) by taking 𝑥𝛽

𝑦𝛾

𝑧𝛿

𝑡𝛼

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘 Γ(𝛽+1) + 𝑙 Γ(𝛾+1) + 𝑚 Γ(𝛿+1) + 𝜔 Γ(𝛼+1), and transform Eq. (4.1) to the ordinary differential equation. 𝑄(𝑢, 𝜔𝑢′ , 𝛼𝑢′ , 𝛽𝑢′ , 𝛾𝑢′ , … ) = 0.

(4.2)

where 𝛼, 𝛽, 𝛾 and 𝜔 are constants and where prime denotes the derivative with respect to 𝜉.


Step 3: According to F-expansion method, we assume that the wave solution can be expressed in the following form 𝑛 𝑢(𝜉) = ∑𝑀 𝑛=0 𝑎𝑛 𝐹 (𝜉).

(4.3)

where 𝑎𝑖 real constants to be determined are, 𝑀 is a positive integer to be determined. 𝐹(𝜉) satisfies the following auxiliary equation [𝐹 ′ (𝜉)]2 = 𝑃𝐹 4 (𝜉) + 𝑄𝐹 2 (𝜉) + 𝑅. 𝑃, 𝑄, and 𝑅 are constants, The last equation hence holds for 𝐹 ′′ (𝜉) = 2𝑃𝐹 3 (𝜉) + 𝑄𝐹(𝜉), 𝐹 ′′′ (𝜉) = 6𝑃𝐹 2 𝐹 ′ (𝜉) + 𝑄𝐹 ′ (𝜉), 𝐹 (iv) (𝜉) = 24𝑃2 𝐹 5 (𝜉) + 20𝑃𝑄𝐹 3 (𝜉) + 12𝑃𝑅𝐹(𝜉) + 𝑄 2 𝐹(𝜉), 𝐹 (v) (𝜉) = 120𝑃2 𝐹 4 𝐹 ′ (𝜉) + 60𝑃𝑄𝐹 3 𝐹 ′ (𝜉) + 12𝑃𝑅𝐹 ′ (𝜉) + 𝑄 2 𝐹 ′ (𝜉), 449

(4.4)

. . .. In Table 3.1 and B, we present 52 types of exact solution for Eq. (4.4). In fact, these exact solutions can be used to construct more exact solutions for Eq. (4.1). Table 4.1 (a): Solutions of Eq. (4.4) for different values of 𝑃, 𝑄 and 𝑅. Case

𝑷

𝑸

𝑹

𝑭(𝝃)

1

𝑚2

−(1 + 𝑚2 )

1

Sn𝜉

2

𝑚2

−(1 + 𝑚2 )

1

cd𝜉 = cn𝜉/dn𝜉

3

−𝑚2

2𝑚2 − 1

1 − 𝑚2

Cn𝜉

4

−1

2 − 𝑚2

𝑚2 − 1

Dn𝜉

2

2

ns𝜉 = (sn𝜉)−1

5

1

−(1 + 𝑚 )

𝑚

6

1

−(1 + 𝑚2 )

𝑚2

dc𝜉 = dn𝜉/cn𝜉

7

1 − 𝑚2

2𝑚2 − 1

−𝑚2

nc𝜉 = (cn𝜉)−1

8

𝑚2 − 1

2 − 𝑚2

−1

nd𝜉 = (dn𝜉)−1

9

1 − 𝑚2

2 − 𝑚2

1

sc𝜉 = sn𝜉/cn𝜉

10

−𝑚2 (1 − 𝑚2 )

2𝑚2 − 1

1

sd𝜉 = sn𝜉/dn𝜉

11

1

2 − 𝑚2

1 − 𝑚2

cs𝜉 = cn𝜉/sn𝜉

12

1

2𝑚2 − 1

−𝑚2 (1 − 𝑚2 )

ds𝜉 = dn𝜉/sn𝜉

13

¼

¼

ns𝜉 ± cn𝜉

14

(1 − 𝑚2 )/4

(1 − 𝑚2 )/4

nc𝜉 ± sc𝜉

15

¼

𝑚2 /4

ns𝜉 ± ds𝜉

16

𝑚2 /4

𝑚2 /4

sn𝜉 ± 𝑖cn𝜉

17

2

𝑚 /4

𝑚2 /4

√1 − 𝑚2 sd𝜉 ± cd𝜉

18

¼

¼

𝑚cd𝜉 ± 𝑖 √1 − 𝑚2 nd𝜉

(1 − 2𝑚2 )/2 (1 + 𝑚2 )/2 (𝑚2 − 2)/2 (𝑚2 − 2)/2 (𝑚2 − 2)/2 (1 − 𝑚2 ) /2

450

19

¼

(1

¼

𝑚sn𝜉 ± 𝑖dc𝜉

¼

√1 − 𝑚2 sc𝜉 ± dc𝜉

(m2 − 1)/4

𝑚sd𝜉 ± nd𝜉

¼

sn𝜉 1 ± dn𝜉

(1 − 𝑚2 )2 /4

𝑚cn𝜉 ± dn𝜉

¼

ds𝜉 ± cs𝜉

− 2𝑚2 )/2 (1 − 𝑚2 )

20

¼

21

(𝑚2 − 1)/4

22

𝑚2 /4

23

−¼

24

(1 − 𝑚2 )2 /4

25

𝑚4 (1 − 𝑚2 ) 2(2 − 𝑚2 )

2(1 − 𝑚2 ) 𝑚2 − 2

1 − 𝑚2 2(2 − 𝑚2 )

dc𝜉 ± √1 − 𝑚2 nc𝜉

26

𝑃>0

𝑄 0,

𝐵 2 − 4𝐴𝐶 > 0,

,

𝐴 > 0,

𝐵 2 − 4𝐴𝐶 < 0,

,

𝐴 < 0,

𝐵 2 − 4𝐴𝐶 > 0,

,

𝐴 < 0,

𝐵 2 − 4𝐴𝐶 < 0,

𝑢1 (𝜉) = −

𝐵 sech(√𝐴𝜉)−√𝐵2 −4𝐶𝐴

𝑢2 (𝜉) = −

𝐵 csch(√𝐴𝜉)+√4𝐶𝐴−𝐵2

𝑢3 (𝜉) = −

𝐵 sech(√−𝐴𝜉)−√𝐵2 −4𝐶𝐴

𝑢4 (𝜉) = −

2𝐴 csch(√𝐴𝜉)

2𝐴 sech(√−𝐴𝜉)

2𝐴 csch(√−𝐴𝜉) 𝐵 csch(√−𝐴𝜉)−√𝐵2 −4𝐶𝐴 2𝐴 sech(√𝐴𝜉)

, 𝐴 > 0,

𝐵 2 − 4𝐴2 − 4𝐴𝐶 < 0,

, 𝐴 > 0,

𝐵 2 + 4𝐴2 − 4𝐴𝐶 < 0,

, 𝐴 < 0,

𝐵 2 − 4𝐴2 − 4𝐴𝐶 > 0,

, 𝐴 < 0,

𝐵 2 − 4𝐴2 − 4𝐴𝐶 > 0,

𝑢5 (𝜉) = −

𝐵 sech(√𝐴𝜉)+√4𝐴2 +4𝐶𝐴−𝐵2 tanh(√𝐴𝜉)−2𝐴

𝑢6 (𝜉) = −

𝐵 csch(√𝐴𝜉)−√𝐵2 −4𝐶𝐴+4𝐴2 coth(√𝐴𝜉)+2𝐴

𝑢7 (𝜉) = −

𝐵 sec(√−𝐴𝜉)+√𝐵2 −4𝐴2 −4𝐶𝐴 tan(√−𝐴𝜉)−2𝐴

𝑢8 (𝜉) = −

2𝐴 csch(√𝐴𝜉)

2𝐴 sec(√−𝐴𝜉)

2𝐴 csc(√−𝐴𝜉) 𝐵 csc(√−𝐴𝜉)−√𝐵2 −4𝐴2 −4𝐶𝐴 cot(√−𝐴𝜉)+2𝐴

467

Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term of highest order with the highest order nonlinear term which obtained in Step 2. Step 5: Substituting (4.21) into Eq. (4.20) with Eq. (4.22) will yield a polynomial in 𝑗

𝑈 𝑖 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) (𝑖 = 0,1,2, … , 𝑗 = 0,1).

Compare

the

like

powers

of

𝑗

𝑈 𝑖 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) equal to zero, yields a set of algebraic equations for unknown parameters. Step 6: Solve the system which obtained in step 5 for unknown parameters with the help of Maple 13. Putting these constant into Eq. (4.21), coupled with the well known solutions of Eq. (4.22), we can obtained the more general type and new exact traveling wave solution of the nonlinear partial differential equation (4.20).

4.3.2. (2+1) Korteweg-deVries (KdV) Equation Let us consider the (2 + 1)-dimensional KdV equations [230] 𝐷𝑡𝛼 𝑢 + 𝑢𝑥𝑥𝑥 − 3𝑣𝑥 𝑢 − 3𝑣𝑢𝑥 = 0,

(4.23a)

𝑢𝑥 − 𝐷𝑦𝛼 𝑣 = 0,

(4.23b)

0 < 𝛼 < 1.

Using the following transformation 𝑢(𝜉) = 𝑢(𝑥, 𝑦, 𝑡), 𝜉 = 𝑘𝑥 + 𝑙

𝑦𝛼 Γ(𝛼+1)

+𝜆

𝑡𝛼

.

Γ(𝛼+1)

We have Eq. (4.23) as 𝜆𝑢′ + 𝑘 3 𝑢′′′ − 3𝑘𝑣 ′ 𝑢 − 3𝑘𝑣𝑢′ = 0,

(4.24a)

𝑘𝑢′ − 𝑙𝑣 ′ = 0.

(4.24b)

Substituting Eq. (4.24b) into Eq. (4.24a), we obtained 𝜆𝑢′ + 𝑘 3 𝑢′′′ −

3𝑘 2 𝑙

𝑢′ 𝑢 −

3𝑘 2 𝑙

𝑢′ 𝑢 = 0.

(4.25)

Integrating Eq. (4.25) once we have 𝜆𝑢 + 𝑘 3 𝑢′′ −

3𝑘 2 𝑙

𝑢2 = 0.

(4.26)

According to the proposed technique we obtained 𝑏

√𝐴𝑈 2 +𝐵𝑈 3 +𝐶𝑈 4

1 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈 + (𝜇𝑈+1 ) + 𝑐1 (

𝜇𝑈+1

),

(4.27)

where 𝑎0 , 𝑎1 , 𝑏1 , 𝑐1 arbitrary constant to be determine, 𝑈(𝜉) satisfied the Eq. (4.22). Putting Eq. (4.27) in to Eq. (4.26) with Eq. (4.22) we have

468

1

1

− 2 (𝜇𝑈+1)3 𝑙 [𝐻0 𝑈 0 + 𝐻1 𝑈1 + 𝐻2 𝑈 2 + 𝐻3 𝑈 3 + 𝐻4 𝑈 4 + 𝐻5 𝑈 5 + 𝐻6 𝑈 6 + 0

1

2

3

𝐹0 𝑈 0 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) + 𝐹1 𝑈1 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) + 𝐹2 𝑈 2 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) + 𝐹3 𝑈 3 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) + 4

𝐹4 𝑈 4 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) ] = 0. Compare the like powers of U we have system of equation 𝑈0:

−2𝜆𝑙𝑏1 + 6𝑘𝑎02 + ⋯ − 2𝜆𝑙𝑎0 = 0, . . .

𝑈6:

−4𝑘 3 𝑙𝑎1 𝐶𝜇 3 = 0, 0

𝑈 0 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) : −2𝜆𝑙𝑐0 + 12𝑘𝑎0 𝑐1 + ⋯ − 2𝑘3 𝑙𝑐1 𝐴 = 0, . . . 4

𝑈 4 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) : −4𝑘 3 𝑙𝑐1 𝜇 2 𝐶 = 0. Solving these algebraic equations with the help of any mathematical software such as Maple 13, we obtain following 1st Solution Set: 1

𝑎0 =

2 2 1 𝑙𝑘 (−2𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴 −2 , 𝑎1 1 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵

1

= 0, 𝑏1 =

2𝐴

2 2 1 𝑙𝑘 (−2𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴 , 1 2 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵 2𝐴

1

𝑐1 = 0, 𝜇 = 2𝐴 (𝐵 + √𝐵 2 − 4𝐴𝐶)𝐵, 𝜆 = −𝐴𝑘 3 . Using these solution set into the trail solution we have 1

2 2 1 𝑙𝑘 (−2𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴 1 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵

𝑢(𝜉) = − 2

2𝐴

+( 1 2𝐴

𝑘

𝑣(𝜉) = − 2𝑙

1

1

2 2 1 𝑙𝑘 (−2𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴 , 1 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵

)2

(𝐵+√𝐵2 −4𝐴𝐶)𝐵𝑈+1

2𝐴

1 𝑙𝑘 2 (−2𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵)𝐴 2𝐴 1 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵 2𝐴

469

+(

1

2 2 𝑘 𝑙𝑘 (−2𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴 ) , 1 1 (𝐵+√𝐵2 −4𝐴𝐶)𝐵𝑈+1 𝑙 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵

1

2𝐴

2𝐴

𝑦𝛼

𝑡𝛼

where 𝜉 = 𝑘𝑥 + 𝑙 Γ(𝛼+1) − 𝐴𝑘 3 Γ(𝛼+1), and 𝑈 = 𝑈(𝜉) is the solution of Eq. (4.22). 2nd Solution Set: 1

2 2 1 𝑙𝑘 (−4𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴 , 𝑎1 1 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵

𝑎0 = − 6

2𝐴

1


= 0, 𝑏1 = 2

2𝐴

1

𝑐1 = 0, 𝜇 = 2𝐴 (𝐵 + √𝐵 2 − 4𝐴𝐶)𝐵, 𝜆 = 𝐴𝑘 3 . Using these solution set into the trail solution we have 1

𝑢(𝜉) =

2 2 1 𝑙𝑘 (−4𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴 −6 1 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵 2𝐴

+( 1 2𝐴

1

1


)2

(𝐵+√𝐵2 −4𝐴𝐶)𝐵𝑈+1

,

2𝐴

1

2 2 𝑘 𝑙𝑘 (−4𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴

𝑣(𝜉) = − 6𝑙

+(

−𝐶+

1 (𝐵+√𝐵2 −4𝐴𝐶)𝐵 2𝐴

1

2 2 𝑘 𝑙𝑘 (−2𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴 ) 1 1 (𝐵+√𝐵2 −4𝐴𝐶)𝐵𝑈+1 𝑙 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵

1

2𝐴

,

2𝐴

𝑦𝛼

𝑡𝛼

where 𝜉 = 𝑘𝑥 + 𝑙 Γ(𝛼+1) + 𝐴𝑘 3 Γ(𝛼+1), and 𝑈 = 𝑈(𝜉) is the solution of Eq. (4.22).

(a)

(b)

Fig. 4.1. (a)-(b): Graphical representation of Eq. (4.23)

4.3.3. Combined KdV and MKdV Equation Let us consider the Combined KdV and MKdV Equation [230] 𝐷𝑡𝛼 𝑢 + 𝑎𝑢𝑢𝑥 + 𝑏𝑢2 𝑢𝑥 + 𝑢𝑥𝑥𝑥 = 0, 0 < 𝛼 < 1. 470

(4.28)

Using the following transformation 𝑡𝛼

𝑢(𝜉) = 𝑢(𝑥, 𝑡), 𝜉 = 𝑘𝑥 + 𝜆 Γ(𝛼+1). We have Eq. (4.28) as 𝜆𝑢′ + 𝑎𝑘𝑢𝑢′ + 𝑏𝑘𝑢2 𝑢′ + 𝑘 3 𝑢′′′ = 0.

(4.29)

Integrating Eq. (4.29) once we have 𝜆𝑢 +

𝑎𝑘 2

𝑢2 +

𝑏𝑘 3

𝑢3 + 𝑘 3 𝑢′′ = 0.

(4.30)

According to the proposed technique we obtained 𝑏1

𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈 + (

√𝐴𝑈 2 +𝐵𝑈 3 +𝐶𝑈 4

) + 𝑐1 (

𝜇𝑈+1

𝜇𝑈+1

),

(4.31)

where 𝑎0 , 𝑎1 , 𝑏1 , 𝑐1 arbitrary constant to be determine, 𝑈(𝜉) satisfied the Eq. (4.22). Putting Eq. (4.31) in to Eq. (4.30) with Eq. (4.22) we have 1

1


1

2

3



−6𝑎𝑘𝑎0 𝑏1 − 6𝑏𝑘𝑎02 𝑏1 − ⋯ − 2𝑏𝑘𝑏13 = 0, . . ..

𝑈6:

−6𝑏𝑘𝑎1 𝜇𝑐12 𝐶 − 2𝑏𝑘𝑎13 𝜇 3 − 12𝑘 3 𝑎1 𝐶𝜇 3 = 0, 0

𝑈 0 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) : −6𝑘 3 𝑐1 𝐴 − 6𝑎𝑘𝑎0 𝑐1 − ⋯ − 12𝑏𝑘𝑎0 𝑏1 𝑐1 = 0, . . .. 4

𝑈 4 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) : −6𝑏𝑘𝑎12 𝜇 2 𝑐1 − 2𝑘𝑏𝑐13 𝐶 − 12𝑘 3 𝑐1 𝜇 2 𝐶 = 0. Solving these algebraic equations with the help of any mathematical software such as Maple 13, we obtain following 1st Solution Set: 471

3𝑎

𝑎0 = − 4 𝑏 , 𝑎1 = 0, 𝑏1 = 1 5𝐵+Δ

𝜇=6

𝐴

31 𝑏

5 (5𝐵+Δ)𝐵3 −40𝐴𝐶 2 +16𝐶𝐵2 ) 3𝐴 5 𝑏(4𝐴𝐶−𝐵2 )( (5𝐵+Δ)𝐵−8𝐶) 6𝐴

𝑎( (5𝐵+Δ)𝐵𝐶−

4

1

1

, 𝑐1 = 0, 𝜆 = costant,

7

, 𝑘 = 4𝐴𝐶−𝐵2 √− 96𝑏 (𝐴 (5𝐵 + Δ)𝐵 3 − 2 (5𝐵 + Δ)𝐵𝐶 + 40𝐴𝐶 2 − 12𝐶𝐵 2 ) 𝑎.

Using these solution set into the trail solution we have 3𝑎

𝑢(𝜉) = − 4 𝑏 + 𝑎1 𝑈 + (

(

where

4

1

1

)

31 𝑏

5 (5𝐵+Δ)𝐵3 −40𝐴𝐶 2 +16𝐶𝐵2 ) 3𝐴 5 𝑏(4𝐴𝐶−𝐵2 )( (5𝐵+Δ)𝐵−8𝐶) 6𝐴

𝑎( (5𝐵+Δ)𝐵𝐶−

15𝐵+Δ )𝑈+1 6 𝐴

1

,

7

𝜉 = 4𝐴𝐶−𝐵2 √− 96𝑏 (𝐴 (5𝐵 + Δ)𝐵 3 − 2 (5𝐵 + Δ)𝐵𝐶 + 40𝐴𝐶 2 − 12𝐶𝐵 2 ) 𝑎𝑥 +

𝑡𝛼

𝜆 Γ(𝛼+1), and ∆= √25𝐵 2 − 96𝐴𝐶 and 𝑈 = 𝑈(𝜉) is the solution of Eq. (4.22). 2nd Solution Set: 𝑎0 = −

2𝑎(𝜇 2 𝐴−𝜇𝐵+𝐶) (2𝐴𝜇−𝐵)𝑏𝜇

2

, 𝑎1 = 0, 𝑏1 =

1 𝜇 2 𝐴−𝜇𝐵+𝐶

𝑘 = 2𝐴𝜇−𝐵 √− 6

𝑏


, 𝑐1 = 0, 𝜇 = 𝜇, 1𝜇2 𝐴−𝜇𝐵+𝐶

𝑎, 𝜆 =

3 2 4 𝑎 (𝜇 𝐴−𝜇𝐵+𝐶)𝐴√−6 𝑏 3 𝑏(4𝐴2 𝜇2 −4𝐴𝜇𝐵+𝐵2 )(2𝐴𝜇−𝐵)

.

Using these solution set into the trail solution we have 𝑢(𝜉) = − 2


1 𝜇 2 𝐴−𝜇𝐵+𝐶

where 𝜉 = 2𝐴𝜇−𝐵 √− 6

𝑏

1

+ (𝜇𝑈+1)


, 1𝜇2 𝐴−𝜇𝐵+𝐶

𝑎𝑥 +

3 2 4 𝑎 (𝜇 𝐴−𝜇𝐵+𝐶)𝐴√−6 𝑡𝛼 𝑏 , 3 𝑏(4𝐴2 𝜇 2 −4𝐴𝜇𝐵+𝐵2 )(2𝐴𝜇−𝐵) Γ(𝛼+1)

is the solution of Eq. (4.22).

(a)

(b)


4.3.4. Konopelchenko-Dubrovsky Equation 472

and 𝑈 = 𝑈(𝜉)

Let us consider the Konopelchenko–Dubrovsky equation [122] 3

𝐷𝑡𝛼 𝑢 − 𝑢𝑥𝑥𝑥 − 6𝑏𝑢𝑢𝑥 + 2 𝑎2 𝑢2 𝑢𝑥 − 𝐷𝑦𝛼 𝑣 + 3𝑎𝑢𝑥 𝑣 = 0,

(4.32a)

𝐷𝑦𝛼 𝑢 = 𝑣𝑥 .

(4.32b)

Using the following transformation 𝑦𝛽

𝑡𝛼

𝑢(𝜉) = 𝑢(𝑥, 𝑦, 𝑡), 𝜉 = 𝑘𝑥 + 𝑙 Γ(𝛽+1) + 𝑤 Γ(𝛼+1). We have Eq. (4.32) as 3

𝑤𝑢′ − 𝑘 3 𝑢′′′ − 6𝑏𝑘𝑢𝑢′ + 2 𝑎2 𝑘𝑢2 𝑢′ − 𝑙𝑣 ′ + 3𝑎𝑘𝑢′ 𝑣 = 0,

(4.33)

𝑙𝑢′ = 𝑘𝑣 ′ . Integrating the second equation we find 𝑙𝑢 = 𝑘𝑣.

(4.44)

Substituting Eq. (4.44) into Eq. (4.43), we obtained 𝑙2

3

(𝑤 − 𝑘 ) 𝑢′ − 𝑘 3 𝑢′′′ − 6𝑏𝑘𝑢𝑢′ + 2 𝑎2 𝑘𝑢2 𝑢′ + 3𝑎𝑙𝑢𝑢′ = 0.

(4.35)

Integrating Eq. (4.35) once we have 𝑙2

1

3

(𝑤 − 𝑘 ) 𝑢 − 𝑘 3 𝑢′′ − 3𝑏𝑘𝑢2 + 2 𝑎2 𝑘𝑢3 + 2 𝑎𝑙𝑢2 = 0.

(4.36)

Applying the Homogenous balancing principle we obtained 𝑀 + 2 = 3𝑀, 𝑀 = 1. According to the proposed technique we obtained √𝐴𝑈 2 +𝐵𝑈 3 +𝐶𝑈 4

𝑏

1 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈 + (𝜇𝑈+1 ) + 𝑐1 (

𝜇𝑈+1

),

(4.37)


1


1

2

3



2𝑤𝑘𝑎0 + 2𝑤𝑘𝑏1 − ⋯ + 3𝑎2 𝑘 2 𝑎0 𝑏12 = 0, 473

. . .. 𝑈6:

3𝑎2 𝑘 2 𝑎1 𝜇𝑐12 𝐶 + 𝑎2 𝑘 2 𝑎13 𝜇 3 − 4𝑘 3 𝑎1 𝐶𝜇 3 = 0, 0

𝑈 0 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) : −6𝑙 2 𝑐1 + 6𝑘𝑎𝑙𝑎0 𝑐1 + ⋯ − 2𝑘 3 𝑐1 𝐴 = 0, . . .. 4

𝑈 4 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) : 3𝑎2 𝑘 2 𝑎12 𝜇 2 𝑐1 + 𝑎2 𝑘 2 𝑐13 𝐶 − 4𝑘 2 𝑐1 𝜇 2 𝐶 = 0. Solving these algebraic equations with the help of any mathematical software such as Maple 13, we obtain following Solution Set: 1

𝑎 2 𝑎 2 𝜇2

0 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑏1 = −𝑎0 , 𝑐1 = 0, 𝜇 = 𝜇, 𝑘 = 4 𝐴𝜇2 −𝜇𝐵+𝐶 , 𝑤 = constant,

1 𝑎𝑎02 𝜇 2 (2𝑎2 𝑎0 𝜇 2 𝐴−4𝑏𝐴𝜇 2 +4𝑏𝜇𝐵+4𝑏𝐶−𝑎2 𝑎0 𝜇𝐵) . (𝐴𝜇 2 −𝜇𝐵+𝐶)2

𝑙 = − 16

Using these solution set into the trail solution we have 𝑎

0 𝑢(𝜉) = 𝑎0 − 𝜇𝑈+1 ,

where 𝑈 = 𝑈(𝜉) is the solution of Eq. (4.22).

4.3.5. Newel Whiched Equation Let us consider the Newel Whiched equation [170] as 𝑢𝑡 − 𝑢𝑥𝑥 − 𝛽𝑢(1 − 𝑢2 ) = 0,

(4.38)

Using the following transformation 𝑢(𝜉) = 𝑢(𝑥, 𝑡), 𝜉 = 𝑘𝑥 + 𝜆𝑡. We have Eq. (4.38) as 𝜆𝑢′ − 𝑘 2 𝑢′′ − 𝛽𝑢 + 𝛽𝑢3 = 0.

(4.39)

Applying the Homogenous balancing principle we obtained 𝑀 + 2 = 3𝑀, 𝑀 = 1. According to the proposed technique we obtained

474

√𝐴𝑈 2 +𝐵𝑈 3 +𝐶𝑈 4

𝑏

1 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈 + (𝜇𝑈+1 ) + 𝑐1 (

𝜇𝑈+1

),

(4.40)


1


1

2

3



6𝛽𝑎02 𝑏1 + 6𝛽𝑎0 𝑏12 − ⋯ + 2𝛽𝑏13 = 0, . . .

𝑈6:

−4𝑘 2 𝑎1 𝐶𝜇 3 + 2𝛽𝑎13 𝜇 3 + 6𝛽𝑎1 𝜇𝑐12 𝐶 = 0, 0

𝑈 0 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) : −2𝑘 2 𝑐1 𝐴 − 2𝜆𝑏1 𝜇 + ⋯ + 12𝛽𝑎0 𝑏1 𝑐1 = 0, . . . 4

𝑈 4 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) : 6𝛽𝑎12 𝜇 2 𝑐1 − 4𝑘 2 𝑐1 𝜇 2 𝐶 + 2𝛽𝑐13 𝐶 = 0. Solving these algebraic equations with the help of any mathematical software such as Maple 13, we obtain following 1st Solution Set: 𝑎0 = −

√2𝐵2 −8𝐶𝐴 𝐵

, 𝑎1 = 0, 𝑏1 =

√2𝐵2 −8𝐶𝐴 𝐵

Using these solution set into the trail solution we have 𝑢(𝜉) = −

√2𝐵2 −8𝐶𝐴 𝐵

1

+ (𝜇𝑈+1)

√2𝐵2 −8𝐶𝐴

,

𝐵

where 𝑈 = 𝑈(𝜉) is the solution of Eq. (4.22).

475

𝛽

1𝐵

𝐴

2𝐴

, 𝑐1 = 0, 𝜆 = 0, 𝑘 = 𝜆√− , 𝜇 =

.

(a)

(b)


4.3.6. Boiti-Leon-Pempinelli Equation Let us consider the Boiti-Leon-Pempinelli equation [37] 𝐷𝑡𝛼 𝑢𝑦 = (𝑢2 − 𝑢𝑥 )𝑥𝑦 + 2𝑣𝑥𝑥𝑥 ,

(4.41a)

𝐷𝑡𝛼 𝑣 = 𝑣𝑥𝑥 + 2𝑢𝑣𝑥 .

(4.41b)

Using the following transformation 𝑥𝛼

𝑢(𝜉) = 𝑢(𝑥, 𝑦, 𝑡), 𝜉 = 𝜇 (

Γ(𝛼+1)

+ 𝑦 − 𝑐𝑡).

We have Eq. (3.104) as −𝑐𝑢′′ = (𝑢2 )′′ − 𝜇𝑢′′′ + 2𝜇𝑣 ′′′ ,

(4.42a)

−𝑐𝑣 ′ = 𝜇𝑣 ′′ + 2𝜇𝑣 ′ .

(4.42b)

Integrating (4.42a) twice w.r.t 𝜉 gives 1

𝑣 ′ = 2 𝑢′ −

𝑢2 +𝑐𝑢 2𝜇

,

integrating above gives 1

1

2

2𝜇

𝑣(𝜉) = 𝑢 +

∫(𝑢2 + 𝑐𝑢) 𝑑𝜉.

(4.43)

Substituting Eq. (4.43) into Eq. (4.42a), we obtained 𝜇 2 𝑢′′ − 2𝑢3 − 3𝑐𝑢2 − 𝑐 2 𝑢 = 0.

(4.44)

By homogeneous balancing principle 3𝑀 = 𝑀 + 2, 𝑀 = 1. According to the proposed technique we obtained 476

√𝐴𝑈 2 +𝐵𝑈 3 +𝐶𝑈 4

𝑏

1 𝑢(𝜉) = 𝑎0 + 𝑎1 𝑈 + (𝜇𝑈+1 ) + 𝑐1 (

𝜇𝑈+1

),

(4.45)


1


1

2

3



12𝑎02 𝑏1 + 12𝑎0 𝑏12 + ⋯ + 2𝑐 2 𝑏1 = 0, . . .

𝑈6:

4𝑎13 𝜇 3 + 12𝑎1 𝜇𝑐12 𝐶 − 4𝑘 2 𝑎1 𝐶𝜇 3 = 0, 0

𝑈 0 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) : 24𝑎0 𝑏1 𝑐1 + 12𝑐𝑎0 𝑐1 + ⋯ − 2𝑘 2 𝑐1 𝐴 = 0, . . . 4

𝑈 4 (√𝐴𝑈 2 + 𝐵𝑈 3 + 𝐶𝑈 3 ) : 4𝑐13 𝐶 + 12𝑎12 𝜇 2 𝑐1 − 4𝑘 2 𝑐1 𝜇 2 𝐶 = 0. Solving these algebraic equations with the help of any mathematical software such as Maple 13, we obtain following 1st Solution Set: 1

𝑎0 = −

2 2 1 𝑙𝑘 (−2𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴 , 𝑎1 1 2 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵

1

= 0, 𝑏1 =

2𝐴


1

𝑐1 = 0, 𝜇 = 2𝐴 (𝐵 + √𝐵 2 − 4𝐴𝐶)𝐵, 𝜆 = −𝐴𝑘 3 . Using these solution set into the trail solution we have 1


𝑢(𝜉) = − 2

2𝐴

1

477

1


+ (𝜇𝑈+1) 2

2𝐴

where ∆= 𝐵 2 − 4𝐴𝐶 and 𝑈 = 𝑈(𝜉) is the solution of Eq. (4.22). The exact solution of 𝑣(𝜉) given as 1

1

𝑣(𝜉) = 2 𝑢 + 2𝜇 ∫(𝑢2 + 𝑐𝑢) 𝑑𝜉. 2nd Solution Set: 1

𝑎0 =

2 2 1 𝑙𝑘 (−4𝐶+2𝐴 (𝐵+√𝐵 −4𝐴𝐶)𝐵)𝐴 −6 , 𝑎1 1 −𝐶+ (𝐵+√𝐵2 −4𝐴𝐶)𝐵

1

= 0, 𝑏1 =

2𝐴


1

𝑐1 = 0, 𝜇 = 2𝐴 (𝐵 + √𝐵 2 − 4𝐴𝐶)𝐵, 𝜆 = 𝐴𝑘 3 . Using these solution set into the trail solution we have 1


𝑢(𝜉) = − 6

2𝐴

1

1


+ (𝜇𝑈+1) 2

2𝐴

2

where ∆= 𝐵 − 4𝐴𝐶 and 𝑈 = 𝑈(𝜉) is the solution of Eq. (4.22). The exact solution of 𝑣(𝜉) given as 1

1

𝑣(𝜉) = 2 𝑢 + 2𝜇 ∫(𝑢2 + 𝑐𝑢) 𝑑𝜉.

(a)

(b)


4.4. (G ’/G)-Expansion Method Based on Riccati Equation Mohyud-Din et al. [18] developed a new algorithm (G’/G)-Expansion method using Riccati equation. I this method they used 1st order nonlinear equations 𝐺 ′ (𝜉) = ℎ1 + ℎ2 𝐺 2 (𝜉) as a Riccati equation. In this technique we obtained the different travelling wave solutions of nonlinear evolution equations. The proposed scheme is fully compatible with

478

the complexity of such problems and is very user-friendly. Numerical results are very encouraging.

4.4.1. Methodology Suppose the general nonlinear partial differential Equation 𝑃(𝑢, 𝐷𝑡𝛼 𝑢, 𝐷𝑥𝛼 𝑢, 𝐷𝑦𝛼 𝑢, 𝐷𝑧𝛼 𝑢, 𝐷𝑡2𝛼 𝑢, 𝐷𝑥2𝛼 𝑢, 𝐷𝑦2𝛼 𝑢, 𝐷𝑧2𝛼 𝑢, … ) = 0,

(4.1)

where 𝑢 = 𝑢 (𝑥, 𝑦, 𝑧, 𝑡) is an unknown function, ∆ is a polynomial in 𝑢 (𝑥, 𝑦, 𝑧, 𝑡) and its partial derivatives in which the highest order partial derivatives and the nonlinear terms are involved. The main steps of the (G`/G)-expansion method combined with the Riccati equation is as follows: Step 1: To seek Solitary wave solutions of Eq. (4.1) we assume 𝑥𝛽

𝑦𝛾

𝑧𝛿

𝑡𝛼

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘 Γ(𝛽+1) + 𝑙 Γ(𝛾+1) + 𝑚 Γ(𝛿+1) + 𝜔 Γ(𝛼+1), and transform Eq. (4.1) to the ordinary differential equation. ∇(𝑢, 𝜔𝑢′ , 𝛼𝑢′ , 𝛽𝑢′ , 𝛾𝑢′ , 𝜔2 𝑢′′ , 𝛼 2 𝑢′′ , 𝛽 2 𝑢′′ , 𝛾 2 𝑢′′ , 𝜔𝛼𝑢′′ , … ) = 0,

(4.46)

where 𝛼, 𝛽, 𝛾 and 𝜔 are constants and where prime denotes the derivative with respect to 𝜉. Step 2: If possible, integrate Eq. (4.46) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero. Step 3: According to the (G´/G)-expansion method combine with the Riccati equation, we assume that the wave solution can be expressed by polynomial in (G´/G), as 𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 (

𝐺′ 𝐺

𝑛

) .

(4.47)

The variable 𝐺(𝜉) satisfied the following equation 𝐺 ′ (𝜉) = ℎ1 + ℎ2 𝐺 2 (𝜉),

(4.48)

where 𝑎𝑛 (𝑛 = 0,1,2,3, … ) ℎ1 and ℎ2 are arbitrary constants to be determined later. The Riccati Eq. (4.48) plays important role in manipulating nonlinear equations to get exact solutions by the (G`/G)-expansion method. It has the following twenty one exact solutions [210]. Family 1: When ℎ1 and ℎ2 have same sign and ℎ1 ℎ2 ≠ 0, the solutions of Eq. (4.48) are: 1

𝐺1 = ℎ [√ℎ1 ℎ2 tan(√ℎ1 ℎ2 𝜉)]. 2

1

𝐺2 = − ℎ [√ℎ1 ℎ2 cot(√ℎ1 ℎ2 𝜉)]. 2

479

1

𝐺3 = ℎ [√ℎ1 ℎ2 (tan(2√ℎ1 ℎ2 𝜉) ± sec(2√ℎ1 ℎ2 𝜉))]. 2

1

𝐺4 = − ℎ [√ℎ1 ℎ2 (cot(2√ℎ1 ℎ2 𝜉) ± csc(2√ℎ1 ℎ2 𝜉))]. 2

1

1

1

𝐺5 = 2ℎ [√ℎ1 ℎ2 (tan (2 √ℎ1 ℎ2 𝜉) − cot (2 √ℎ1 ℎ2 𝜉))]. 2

𝐺6 = 𝐺7 =

√ℎ1 ℎ2 √(𝑀2 −𝑁2 )−𝑀 cos(2√ℎ1 ℎ2 𝜉)

[

ℎ2

𝑀 sin(2√ℎ1 ℎ2 𝜉)+𝑁

√ℎ1 ℎ2 √(𝑀2 −𝑁2 )+𝑀 sin(2√ℎ1 ℎ2 𝜉)

[

ℎ2

𝑀 cos(2√ℎ1 ℎ2 𝜉)+𝑁

].

].

Where 𝑀 and 𝑁 are two non-zero real constants and satisfies the condition 𝑀2 − 𝑁 2 > 0. 𝐺8 = 𝐺9 =

−ℎ1 cos(2√ℎ1 ℎ2 𝜉) √ℎ1 ℎ2 sin(2√ℎ1 ℎ2 𝜉)±√ℎ1 ℎ2

.

ℎ1 sin(2√ℎ1 ℎ2 𝜉) √ℎ1 ℎ2 cos(2√ℎ1 ℎ2 𝜉)±√ℎ1 ℎ2 1

𝐺10 =

.

1

2ℎ1 sin( √ℎ1 ℎ2 𝜉) cos( √ℎ1 ℎ2 𝜉) 2 2 1

2√ℎ1 ℎ2 cos2 ( √ℎ1 ℎ2 𝜉)−√ℎ1 ℎ2 2

.

Family 2: When ℎ1 and ℎ2 possess opposite sign and ℎ1 ℎ2 ≠ 0, the solutions of Eq. (4.48) are: 1

𝐺11 = − ℎ [√−ℎ1 ℎ2 tanh(√−ℎ1 ℎ2 𝜉)]. 2

1

𝐺12 = − ℎ [√−ℎ1 ℎ2 coth(√−ℎ1 ℎ2 𝜉)]. 2

1

𝐺13 = − ℎ [√−ℎ1 ℎ2 (tanh(2√−ℎ1 ℎ2 𝜉) ± isech(2√−ℎ1 ℎ2 𝜉))]. 2

1

𝐺14 = − ℎ [√−ℎ1 ℎ2 (coth(2√−ℎ1 ℎ2 𝜉) ± csch(2√−ℎ1 ℎ2 𝜉))]. 2

1

1

1

𝐺15 = − 2ℎ [√−ℎ1 ℎ2 (tanh (2 √−ℎ1 ℎ2 𝜉) + coth (2 √−ℎ1 ℎ2 𝜉))]. 2

𝐺16 = 𝐺17 =

√−ℎ1 ℎ2 √(𝑀2 +𝑁 2 )−𝑀 cosh(2√−ℎ1 ℎ2 𝜉) ℎ2

[

𝑀 sinh(2√−ℎ1 ℎ2 𝜉)+𝑁

√−ℎ1 ℎ2 √(𝑁 2 −𝑀2 )+𝑀 sinh(2√−ℎ1 ℎ2 𝜉) ℎ2

[

𝑀 cosh(2√−ℎ1 ℎ2 𝜉)+𝑁

].

].

Where 𝑀 and 𝑁 are two non-zero real constants and satisfies the condition 𝑀2 − 𝑁 2 > 0. 𝐺18 = 𝐺19 =

ℎ1 cosh(2√−ℎ1 ℎ2 𝜉) √−ℎ1 ℎ2 sinh(2√−ℎ1 ℎ2 𝜉)±𝑖√−ℎ1 ℎ2 ℎ1 sinh(2√−ℎ1 ℎ2 𝜉) √−ℎ1 ℎ2 cosh(2√−ℎ1 ℎ2 𝜉)±√−ℎ1 ℎ2

.

.

480

1

𝐺20 =

1

2ℎ1 sinh( √−ℎ1 ℎ2 𝜉) cosh( √−ℎ1 ℎ2 𝜉) 2 2 1

2√−ℎ1 ℎ2 cosh2 ( √−ℎ1 ℎ2 𝜉)−√−ℎ1 ℎ2 2

.

Family 3: When ℎ2 ≠ 0 but ℎ1 = 0, the solution of Eq. (4.48) is 𝐺21 = − ℎ

1 2 𝜉+𝑑

,

where 𝑑 is an arbitrary constant. The above solutions help to generate various traveling wave solutions, including solitary, periodic and rational solutions, in elementary functions. Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term(s) of highest order with the highest order nonlinear term(s) in Eq. (4.46). Step 5: Substituting (4.47) into Eq. (4.46) with (4.48) yields an algebraic equation involving powers of 𝐺 𝑛 . Equating the coefficients of each power of 𝐺 𝑛 to zero gives a system of algebraic equations for 𝑎𝑛 ’s. Then, we solve the system with the aid of a computer algebra system (CAS), such as MAPLE 13, to determine required constants.

4.4.2. KdV Equation Consider the KdV equation [230] 𝐷𝑡𝛼 𝑢 + 𝑎𝑢𝑢𝑥 + 𝑢𝑥𝑥𝑥 = 0,

0 < 𝛼 < 1.

(4.49) 𝜔𝑡 𝛼

Consider the transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), where 𝜉 = 𝑘𝑥 + Γ(𝛼+1), we have 𝜔𝑢′ + 𝑘𝑎𝑢𝑢′ + 𝑘 3 𝑢′′′ = 0. Integrating once we have 𝜔𝑢 +

𝑘𝑎 2

𝑢2 + 𝑘 3 𝑢′′ = 0,

(4.50)

by applying the balancing principle we have 2𝑀 = 𝑀 + 2, 𝑀 = 2. Therefore, the trail solution is 𝑢(𝜉) = 𝑎0 + 𝑎1 (

𝐺′ 𝐺

) +𝑎2 (

𝐺′ 𝐺

2

) .

(4.51)

Putting Eq. (4.51) into Eq. (4.50) with Eq. (4.48) we have 1 1

[ 2 𝐺4

𝐶0 𝐺 0 + 𝐶1 𝐺 1 + 𝐶2 𝐺 2 + 𝐶3 𝐺 3 + 𝐶4 𝐺 4 ] = 0. +𝐶5 𝐺 5 + 𝐶6 𝐺 6 + 𝐶7 𝐺 7 + 𝐶8 𝐺 8

Compare the like powers of 𝐺 we have system of equation 481

𝐺 0:

12𝑘 3 𝑎2 ℎ14 + 𝑎𝑘𝑎22 ℎ14 = 0, . . .

𝐺 8:

12𝑘 3 𝑎2 ℎ24 + 𝑎𝑘𝑎22 ℎ24 = 0.

Solving the above system for 𝑎0 , 𝑎1 , 𝑎2 , 𝑘 and ω we have 1st Solution Set: 𝑎0 =

32𝑘 2 ℎ1 ℎ2 𝑎

, 𝑎1 = 0, 𝑎2 = −

12𝑘 2 𝑎

, 𝜔 = −16𝑘 3 ℎ1 ℎ2 , 𝑘 = 𝑘.

Using the above information in Eq. (4.51), we have 𝑢(𝜉) =

32𝑘 2 ℎ1 ℎ2 𝑎

−

12𝑘 2 𝑎

(

𝐺′ 𝐺

2

) ,

𝑡𝛼

where 𝜉 = 𝑘𝑥 − 16𝑘 3 ℎ2 ℎ1 Γ(𝛼+1). With the aid of Step 3 there are 21 solutions of Eq. (4.49). 2nd Solution Set: 𝑎0 = 0, 𝑎1 = 0, 𝑎2 = −

12𝑘 2 𝑎

, 𝜔 = 16𝑘 3 ℎ1 ℎ2 , 𝑘 = 𝑘.

Using the above information in Eq. (4.51), we have 𝑢(𝜉) = −

12𝑘 2 𝑎

(

𝐺′ 𝐺

2

) , 𝑡𝛼

where 𝜉 = 𝑘𝑥 + 16𝑘 3 ℎ2 ℎ1 Γ(𝛼+1). With the aid of Step 3 there are 21 solutions of Eq. (4.49).

482

(a)

(b)


4.4.3. Modified KdV Equation Consider the modified KdV equation [230] 𝑢𝑡 + 𝑎𝑢2 𝐷𝛼𝑥 𝑢 + 𝐷3𝛼 𝑥 𝑢 = 0,

0 < 𝛼 < 1.

(4.52) 𝑘𝑥 𝛼

Consider the transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), where 𝜉 = Γ(𝛼+1) + 𝜔𝑡, we have 𝜔𝑢′ + 𝑘𝑎𝑢2 𝑢′ + 𝑘 3 𝑢′′′ = 0. Integrating once we have 𝜔𝑢 +

𝑘𝑎 3

𝑢3 + 𝑘 3 𝑢′′ = 0,

(4.53)


𝐺′ 𝐺

).

(4.54)

Putting Eq. (4.54) into Eq. (4.53) with Eq. (4.48) we have 1 1 3 𝐺3

[𝐶0 𝐺 0 + 𝐶1 𝐺 1 + 𝐶2 𝐺 2 + 𝐶3 𝐺 3 + 𝐶4 𝐺 4 + 𝐶5 𝐺 5 + 𝐶6 𝐺 6 ] = 0.

Compare the like powers of 𝐺 we have system of equation 𝐺 0:

𝑎𝑘𝑎13 ℎ13 + 6𝑎1 𝑘 3 ℎ13 = 0, . . . 483

𝐺 6:

𝑎𝑘𝑎13 ℎ23 + 6𝑎1 𝑘 3 ℎ23 = 0.

Solving the above system for 𝑎0 , 𝑎1 , 𝑘 and ω we have Solution Set: 2

1

1

𝑎0 = 0, 𝑎1 = 𝑎1 , 𝜔 = − 3 𝑎√− 6 𝑎𝑎13 ℎ1 ℎ2 , 𝑘 = √− 6 𝑎𝑎1 . Using the above information in Eq. (4.54), we have 𝑢(𝜉) = 𝑎1 (

𝐺′ 𝐺

), 𝑥𝛼

1

2

1

where 𝜉 = √− 6 𝑎𝑎1 Γ(𝛼+1) − 3 𝑎√− 6 𝑎𝑎13 ℎ1 ℎ2 𝑡. With the aid of Step 3 there are 21 solutions of Eq. (4.52).

(a)

(b)


4.4.4. Fisher’s Equation Consider the Fisher’s equation [230] 𝐷𝑡𝛼 𝑢 − 𝑢𝑥𝑥 = 𝑢(1 − 𝑢), 0 < 𝛼 < 1.

(4.55) 𝜔𝑡 𝛼

Consider the transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), where 𝜉 = 𝑘𝑥 + Γ(𝛼+1), we have 𝜔𝑢′ − 𝑘 2 𝑢′′ − 𝑢 + 𝑢2 = 0.

(4.56)

By applying the balancing principle we have 2𝑀 = 𝑀 + 2, 𝑀 = 2. Therefore, the trail solution is

484

𝑢(𝜉) = 𝑎0 + 𝑎1 (

𝐺′

) + 𝑎2 (

𝐺

𝐺′ 𝐺

2

) .

(4.57)

Putting Eq. (4.57) into Eq. (4.56) with Eq. (4.48) we have 1

− 𝐺4 [

𝐶0 𝐺 0 + 𝐶1 𝐺 1 + 𝐶2 𝐺 2 + 𝐶3 𝐺 3 + 𝐶4 𝐺 4 + 𝐶5 𝐺 5 ] = 0. +𝐶6 𝐺 6 + 𝐶7 𝐺 7 + 𝐶8 𝐺 8


−𝑎22 ℎ14 + 6𝑘 2 𝑎2 ℎ14 = 0, . . .

𝐺 8:

−𝑎22 ℎ24 + 6𝑘 2 𝑎2 ℎ24 = 0.

Solving the above system for 𝑎0 , 𝑎1 , 𝑎2 , 𝑘 and ω we have 1st Solution Set: 3

𝑎0 = 1, 𝑎1 = 0, 𝑎2 = − 8ℎ

1

1 ℎ2

, 𝜔 = 0, 𝑘 = √− 16ℎ

1 ℎ2

.

Using the above information in Eq. (3.36), we have 3

𝑢(𝜉) = 1 − 8ℎ

2 ℎ1

1

where 𝜉 = √− 16ℎ

2 ℎ1

(

𝐺′ 𝐺

2

) ,

𝑥. With the aid of Step 3 there are 21 solutions of Eq. (4.55).

2nd Solution Set: 3

𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 8ℎ

1 ℎ2

1

, 𝜔 = 0, 𝑘 = √16ℎ

1 ℎ2

.


𝑢(𝜉) = 8ℎ where 𝜉 =

1 16

2 ℎ1

(

√16√ℎ

𝐺′ 𝐺

1

2 ℎ1

2

) ,

𝑥. With the aid of Step 3 there are 21 solutions of Eq. (4.55).

485

(a)

(b)


4.4.5. Huxley Equation Consider the Huxley equation [230] 𝐷𝑡𝛼 𝑢 − 𝐷𝑥2𝛼 𝑢 = 𝑢(𝜇 − 𝑢)(𝑢 − 1), 0 < 𝛼 < 1.

(4.58) 𝑘𝑥 𝛼

𝜔𝑡 𝛼

Consider the transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), where 𝜉 = Γ(𝛼+1) + Γ(𝛼+1), we have 𝜔𝑢′ − 𝑘 2 𝑢′′ − 𝜇𝑢2 + 𝜇𝑢 + 𝑢3 − 𝑢2 = 0,

(4.59)


𝐺′ 𝐺

).

(4.60)


− 𝐺3 [𝐶0 𝐺 0 + 𝐶1 𝐺 1 + 𝐶2 𝐺 2 + 𝐶3 𝐺 3 + 𝐶4 𝐺 4 + 𝐶5 𝐺 5 + 𝐶6 𝐺 6 ] = 0. Compare the like powers of 𝐺 we have system of equation 𝐺 0:

2𝑎1 𝑘 2 ℎ13 − 𝑎13 ℎ13 = 0, . . .

𝐺 6:

−𝑎13 ℎ23 + 2𝑎1 𝑘 2 ℎ23 = 0.0

Solving the above system for 𝑎0 , 𝑎1 , 𝑘 and ω we have 486

1st Solution Set: 1

𝑎0 = 1, 𝑎1 =

1

√2 √ℎ1 ℎ2

, 𝜔 = 0, 𝑘 =

1

1

√4 √ℎ1 ℎ2

, 𝜇 = 2.


1

𝑢(𝜉) = 1 + 2 √2√ℎ

1 ℎ2

1

where 𝜉 = 4 √4√ℎ

𝑥𝛼

1

1 ℎ2

𝐺′

(

𝐺

),

. With the aid of Step 3 there are 21 solutions of Eq. (4.58).

Γ(𝛼+1)

2nd Solution Set: 1

1

𝑎0 = , 𝑎1 = 8

1

, 𝜔 = 0, 𝑘 =

√8 √ℎ1 ℎ2

1

1

1

√16 √ℎ1 ℎ2

,𝜇 = . 2


1

𝑢(𝜉) = 2 + 8 √8√ℎ

1

1 ℎ2

1

where 𝜉 = 16 √16√ℎ

𝑥𝛼

1

1 ℎ2

(

𝐺′ 𝐺

),


Γ(𝛼+1)

3rd Solution Set: 𝑎0 = 0, 𝑎1 =

1

1

√2 √ℎ1 ℎ2

, 𝜔 = 0, 𝑘 =

1

1

√4 √ℎ1 ℎ2

, 𝜇 = −1.


𝑢(𝜉) = 2 √2√ℎ

1

1 ℎ2

1

where 𝜉 = 4 √4√ℎ

1

1 ℎ2

𝑥𝛼

𝐺′

(

𝐺

),


Γ(𝛼+1)

(a)

(b)


487

4.4.6. Hirota-Ramani Equation Consider the Hirota-Ramani equation [230] 𝐷𝑡𝛼 𝑢 + 𝑢𝑥𝑥𝑡 + 𝛼𝑢𝑥 (1 − 𝐷𝑡𝛼 𝑢) = 0, 0 < 𝛼 < 1.

(4.61) 𝜔𝑡 𝛼

Consider the transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), where 𝜉 = 𝑘𝑥 + Γ(𝛼+1), we have 𝜔𝑢′ + 𝜔𝑘 2 𝑢′′′ + 𝛼𝑘𝑢′ − 𝛼𝑘𝜔(𝑢′ )2 = 0. Integrating once we have 𝜔𝑢 + 𝜔𝑘 2 𝑢′′ + 𝛼𝑘𝑢 − 𝛼𝑘𝜔(𝑢)2 = 0,

(4.62)


𝐺′ 𝐺

) +𝑎2 (

𝐺′ 𝐺

2

) .

(4.63)


[ 𝐺4



6𝑘 2 𝜔𝑎2 ℎ14 − 𝛼𝑘𝜔𝑎22 ℎ14 = 0, . . .

𝐺 8:

−𝛼𝑘𝜔𝑎22 ℎ24 + 6𝑘 2 𝜔𝑎2 ℎ24 = 0.

Solving the above system for 𝑎0 , 𝑎1 , 𝑎2 , 𝑘, 𝛼 and ω we have 1st Solution Set: 𝛼2 𝑎

3

𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝜔 = 2 4𝛼2 𝑎2 ℎ 2ℎ

2 1 2 −9

1

, 𝑘 = 6 𝛼𝑎2 .

Using the above information in Eq. (4.63), we have 𝑢(𝜉) = 𝑎2 ( 1

𝐺′ 𝐺 3

2

) , 𝛼2 𝑎

where 𝜉 = 6 𝛼𝑎2 𝑥 + 2 4𝛼2 𝑎2 ℎ 2ℎ

𝑡𝛼

. With the aid of Step 3 there are 21 solutions of

2 1 2 −9 Γ(𝛼+1)

Eq. (4.61). 488

2nd Solution Set: 8

3

𝛼2 𝑎

𝑎0 = − 3 ℎ2 ℎ1 𝑎2 , 𝑎1 = 0, 𝑎2 = 𝑎2 , 𝜔 = − 2 4𝛼2 𝑎2 ℎ 2ℎ

2 1 2 +9

1

, 𝑘 = 6 𝛼𝑎2 .


𝑢(𝜉) = − 3 ℎ2 𝑎2 ℎ1 + 𝑎2 ( 1

where 𝜉 = 6 𝛼𝑎2 𝑥 −

𝐺′ 𝐺

2

) ,

3 𝛼2 𝑎2 𝑡𝛼 2 2 4𝛼2 𝑎2 ℎ1 ℎ2 +9 Γ(𝛼+1)

. With the aid of Step 3 there are 21 solutions of

Eq. (4.61).

(a)

(b)


4.4.7. Breaking Soliton Equation Consider the Breaking Soliton equation

[222]

𝑢𝑥𝑡 − 4𝑢𝑥 𝐷𝛼𝑦 𝑢𝑥 − 2𝑢𝑥𝑥 𝐷𝛼𝑦 𝑢 + 𝐷𝛼𝑦 𝑢𝑥𝑥𝑥 = 0, 0 < 𝛼 < 1.

(4.64) 𝑦𝛼

Consider the transformation 𝑢(𝑥, 𝑦, 𝑡) = 𝑢(𝜉), where 𝜉 = 𝑘𝑥 + 𝑙 Γ(𝛼+1) + 𝜔𝑡 we have 𝜔𝑘𝑢′′ − 6𝑘 2 𝑙𝑢′ 𝑢′′ + 𝑘 3 𝑙𝑢′′′′ = 0, Integrating twice we have 𝜔𝑘𝑢 − 3𝑘 2 𝑙𝑢2 + 𝑘 3 𝑙𝑢′′ = 0,

(4.65)

by applying the balancing principle we have 2𝑀 = 𝑀 + 2, 𝑀 = 2. Therefore, the trail solution is

489

𝑢(𝜉) = 𝑎0 + 𝑎1 (

𝐺′ 𝐺

) + 𝑎2 (

𝐺′ 𝐺

2

) .

(4.66)

Putting Eq. (4.66) into Eq. (4.65) with Eq. (4.48) we have 𝐶0 𝐺 0 + 𝐶1 𝐺 1 + 𝐶2 𝐺 2 + 𝐶3 𝐺 3 + 𝐶4 𝐺 4 [ ] = 0. 𝐺4 +𝐶5 𝐺 5 + 𝐶6 𝐺 6 + 𝐶7 𝐺 7 + 𝐶8 𝐺 8 1


−3𝑙𝑘𝑎22 ℎ14 + 6𝑘 2 𝑙𝑎2 ℎ14 = 0, . . .

𝐺 8:

−3𝑙𝑘𝑎22 ℎ24 + 6𝑘 2 𝑙𝑎2 ℎ24 = 0.

Solving the above system for 𝑎0 , 𝑎1 , 𝑎2 , 𝑘, 𝑙 and ω we have 1st Solution Set: 𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 2𝑘, 𝜔 = 16𝑘 2 𝑙ℎ1 ℎ2 , 𝑘 = 𝑘, 𝑙 = 𝑙. Using the above information in Eq. (4.66), we have 𝑢(𝜉) = 2 (

𝐺′ 𝐺

2

) ,

𝑦𝛼

where 𝜉 = 𝑘𝑥 + 𝑙 Γ(𝛼+1) + 16𝑘 2 𝑙ℎ2 ℎ1 𝑡. With the aid of Step 3 there are 21 solutions of Eq. (4.64). 2nd Solution Set: 𝑎0 = −

16 3

𝑘ℎ1 ℎ2 , 𝑎1 = 0, 𝑎2 = 2𝑘, 𝜔 = −16𝑘 2 𝑙ℎ1 ℎ2 , 𝑘 = 𝑘, 𝑙 = 𝑙.


16 3

𝑘ℎ1 ℎ2 + 2 (

𝐺′ 𝐺

2

) ,

𝑦𝛼

where 𝜉 = 𝑘𝑥 + 𝑙 Γ(𝛼+1) − 16𝑘 2 𝑙ℎ2 ℎ1 𝑡. With the aid of Step 3 there are 21 solutions of Eq. (4.64).

490

(a)

(b)


4.4.8. BBM Equation Consider the BBM equation [114] 𝐷𝑡𝛼 𝑢 + 𝐷𝑥𝛼 𝑢 + 𝑎𝑢𝐷𝑥𝛼 𝑢 + 𝐷𝑥3𝛼 𝑢 = 0, 0 < 𝛼 < 1.

(4.67) 𝑘𝑥 𝛼

𝜔𝑡 𝛼

Consider the transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), where 𝜉 = Γ(𝛼+1) + Γ(𝛼+1), we have 𝜔𝑢′ + 𝑘𝑢′ + 𝑘𝑎𝑢𝑢′ + 𝑘 3 𝑢′′′ = 0. Integrating once we have 𝜔𝑢 + 𝑘𝑢 +

𝑘𝑎 2

𝑢2 + 𝑘 3 𝑢′′ = 0,

(4.68)


𝐺′ 𝐺

) +𝑎2 (

𝑦′ 𝐺

2

) .

(4.69)

Putting Eq. (4.69) into Eq. (4.68) with Eq. (4.48) we have 1 1

[ 2 𝐺4



12𝑘 3 𝑎2 ℎ14 + 𝑎𝑘𝑎22 ℎ14 = 0, . . 491

. 𝐺 8:

12𝑘 3 𝑎2 ℎ24 + 𝑎𝑘𝑎22 ℎ24 = 0.

Solving the above system for 𝑎0 , 𝑎1 , 𝑎2 , 𝑘 and ω we have 1st Solution Set: 𝑎0 = 0, 𝑎1 = 0, 𝑎2 = −

12𝑘 2 𝑎

, 𝜔 = 𝑘(−1 + 16𝑘 2 ℎ1 ℎ2 ), 𝑘 = 𝑘.


12𝑘 2 𝑎

(

𝐺′ 𝐺

2

) ,

𝑥𝛼

𝑡𝛼

where 𝜉 = 𝑘 Γ(𝛼+1) + (−1 + 16𝑘 2 ℎ2 ℎ1 ) Γ(𝛼+1). With the aid of Step 3 there are 21 solutions of Eq. (4.67). 2nd Solution Set: 𝑎0 =

32𝑘 2 ℎ1 ℎ2 𝑎

, 𝑎1 = 0, 𝑎2 = −

12𝑘 2 𝑎

, 𝜔 = −𝑘(1 + 16𝑘 2 ℎ1 ℎ2 ), 𝑘 = 𝑘.

Using the above information in Eq. (4.69), we have 𝑢(𝜉) =

32𝑘 2 ℎ2 ℎ1 𝑎

−

12𝑘 2 𝑎

(

𝐺′ 𝐺

𝑥𝛼

2

) , 𝑡𝛼

where 𝜉 = 𝑘 Γ(𝛼+1) − 𝑘(1 + 16𝑘 2 ℎ2 ℎ1 ) Γ(𝛼+1). With the aid of Step 3 there are 21 solutions of Eq. (4.67)

(a)

(b)


4.4.9. gBBM Equation Consider the gBBM equation [114] 492

𝛽

𝛽

3𝛽

𝐷𝑡𝛼 𝑢 + 𝐷𝑥 𝑢 + 𝑎𝑢2 𝐷𝑥 𝑢 + 𝐷𝑥 𝑢 = 0, 0 < 𝛼, 𝛽 < 1.

(4.70)

𝑘𝑥 𝛽

𝜔𝑡 𝛼

Consider the transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), where 𝜉 = Γ(𝛽+1) + Γ(𝛼+1), we have 𝜔𝑢′ + 𝑘𝑢′ + 𝑘𝑎𝑢2 𝑢′ + 𝑘 3 𝑢′′′ = 0, Integrating once we have 𝜔𝑢 + 𝑘𝑢 +

𝑘𝑎 3

𝑢3 + 𝑘 3 𝑢′′ = 0,

(4.71)


𝐺′ 𝐺

).

(4.72)

Putting Eq. (4.72) into Eq. (4.71) with Eq. (4.48) we have 1 1 3 𝐺3

[𝐶0 𝐺 0 + 𝐶1 𝐺 1 + 𝐶2 𝐺 2 + 𝐶3 𝐺 3 + 𝐶4 𝐺 4 + 𝐶5 𝐺 5 + 𝐶6 𝐺 6 ] = 0.


𝑎𝑘𝑎13 ℎ13 + 6𝑎1 𝑘 3 ℎ13 = 0, . . .

𝐺 6:

𝑎𝑘𝑎13 ℎ23 + 6𝑎1 𝑘 3 ℎ23 = 0.

Solving the above system for 𝑎0 , 𝑎1 , 𝑘 and ω we have Solution Set: 1

1

1

𝑎0 = 0, 𝑎1 = 𝑎1 , 𝜔 = − 3 (3 + 2𝑎𝑎12 ℎ1 ℎ2 )√− 6 𝑎𝑎1 , 𝑘 = √− 6 𝑎𝑎1 . Using the above information in Eq. (4.72), we have 𝑢(𝜉) = 𝑎1 ( 1

𝐺′ 𝐺

), 𝑘𝑥 𝛽

1

1

𝜔𝑡 𝛼

where 𝜉 = √− 6 𝑎𝑎1 Γ(𝛽+1) − 3 (3 + 2𝑎𝑎12 ℎ2 ℎ1 )√− 6 𝑎𝑎1 Γ(𝛼+1). With the aid of Step 3 there are 21 solutions of Eq. (4.70).

493

(a)

(b)


4.5. New Approach of (G`/G)-Expansion Method As discuss in Chapter 3 the (G`/G)-expansion method introduced by Wang et al. in this method they used second order linear differential equation as an auxiliary equation. In this method we used second order nonlinear equations 𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 2

𝛿2 (𝐺 ′ (𝜉)) = 0, as a Riccati equation. In this technique we obtained the different

travelling wave solutions of nonlinear evolution equations. The proposed scheme is fully compatible with the complex.

4.5.1. Methodology We consider the general nonlinear PDE of the type 𝑃(𝑢, 𝐷𝑡𝛼 𝑢, 𝐷𝑥𝛼 𝑢, 𝐷𝑦𝛼 𝑢, 𝐷𝑧𝛼 𝑢, 𝐷𝑡2𝛼 𝑢, 𝐷𝑥2𝛼 𝑢, 𝐷𝑦2𝛼 𝑢, 𝐷𝑧2𝛼 𝑢, … ) = 0,

(4.1)

where 𝑃 is a polynomial in its arguments. The essence of the (G`/G)-expansion method can be presented in the following steps: Step 1: Seek Solitary wave solutions of Eq. (4.1) by taking 𝑥𝛽

𝑦𝛾

𝑧𝛿

𝑡𝛼

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘 Γ(𝛽+1) + 𝑙 Γ(𝛾+1) + 𝑚 Γ(𝛿+1) + 𝜔 Γ(𝛼+1), and transform Eq. (4.1) to the ordinary differential equation. 𝑃(𝑢, 𝜔𝑢′ , 𝑘𝑢′ , 𝑙𝑢′ , 𝑚𝑢′ , 𝜔2 𝑢′′ , 𝑘 2 𝑢′′ , 𝑙 2 𝑢′′ , … ) = 0,

(4.71)

where 𝜔, 𝑘, 𝑙 and 𝑚 is constant and where prime denotes the derivative with respect to 𝜉. Step 2: If possible, integrate Eq. (4.71) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero. 494

Step 3: According to (G`/G)-expansion method, we assume that the wave solution can be expressed in the following form 𝐺(𝜉)′

𝑀

𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 ( 𝐺(𝜉) ) ,

(4.72)

where 𝐺(𝜉) is the solution of first order nonlinear equation in the form 2

𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0, 𝐺(𝜉)′

′

𝐺(𝜉)′

2

(4.73)

2

𝐺(𝜉)′

( 𝐺(𝜉) ) = 𝛿1 − 𝛿2 ( 𝐺(𝜉) ) − ( 𝐺(𝜉) ) , 𝐺(𝜉)′

(

𝐺(𝜉)

′′

𝐺(𝜉)′

) = −2𝛿2 𝛿1 ( 𝐺(𝜉)′

𝐺(𝜉)

′

𝐺(𝜉)′

) + 2𝛿22 (

𝐺(𝜉)

3

) + 4𝛿2 (

𝐺(𝜉)′ 𝐺(𝜉)

3

) − 2𝛿1 (

𝐺(𝜉)′ 𝐺(𝜉)

)

3

+2 ( 𝐺(𝜉) ) , . . .. Where 𝛿1 and 𝛿2 are real constants. The Riccati Eq. (4.73) plays important role in manipulating nonlinear equations to get exact solutions by the (G method. It has the following type of exact solutions. Family 1: When 𝛿1 , 𝛿2 ≠ 0, [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)

( 𝐺(𝜉) ) =

[cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 𝐺 ′ (𝜉)

( 𝐺(𝜉) ) =

[cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1 [cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 𝐺 ′ (𝜉)

( 𝐺(𝜉) ) =

[cosh(2√𝛿1 𝜉)+sinh(2√𝛿1 𝜉)]√𝛿1 +√𝛿1 [cosh(2√𝛿1 𝜉)+sinh(2√𝛿1 𝜉)]−1

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

1

1

( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝐺 ′ (𝜉)

1

( 𝐺(𝜉) ) = 1+𝜉.

495

.

.

G)-expansion

Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term(s) of highest order with the highest order nonlinear term(s) in Eq. (4.71). Step 5: Substituting (4.72) into Eq. (4.71) with (4.73) will yields an algebraic equation involving power of (G

G). Equating the coefficients of like power of (G

G) to zero

gives a system of algebraic equations for 𝑎𝑖 , 𝑘, 𝑙, 𝑚 and 𝜔. Then, we solve the system with the aid of a computer algebra system (CAS), 5such as MAPLE 13, to determine these constants. Step 6: Putting these constant into Eq. (4.72), coupled with the well known solutions of Eq. (4.73), we can obtained the more general type and new exact traveling wave solution of the nonlinear partial differential equation (4.1).

4.5.2. Calogero–Bogoyavlenskii–Schiff Equation Consider the Calogero–Bogoyavlenskii–Schiff equation [230] 𝐷𝑡𝛼 𝑢𝑥 + 𝑢𝑥𝑥𝑥𝑧 + 4𝑢𝑥 𝑢𝑥𝑧 + 2𝑢𝑥𝑥 𝑢𝑧 = 0.

(4.74)

To convert Eq. (4.74) into ODE we use following transformation 𝑡𝛼

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔 Γ(𝛼+1) + 𝑚𝑧,

(4.75)

where 𝑘,𝜔 and 𝑚 are arbitrary constant. Substituting Eq. (4.75) into Eq. (4.74) and using the chain rule and 𝜉𝑥 = 𝑘, 𝜉𝑡 = 𝜔, 𝜉𝑧 = 𝑚 we obtained 𝜔𝑘𝑢′′ − 𝑘 4 𝑢′′′′ + 6𝑢′ 𝑘 2 𝑢′′ 𝑚.

Integrating the above equation once, ignoring the constant of integration equal to zero we have the following equation 𝑢′ + 𝑘 4 𝑢′′′ + 3𝑘 2 𝑚(𝑢′ )2 .

(4.76)

Consider the trial solution for Eq. (4.74) as follow 𝐺(𝜉)′

𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 (

𝐺(𝜉)

𝑀

) ,

(4.77)

where 𝑀 can be obtained by applying the homogenous balancing principle we have 𝑀 + 3 = 2𝑀 + 2, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.77), we obtained the trail solution 𝐺(𝜉)′

𝑢 = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ).

(4.78)

496

where 𝐺(𝜉) satisfying the following Riccati equation 2

𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0.

(4.79)

Putting Eq. (4.79) into Eq. (4.76) coupled with Eq. (4.79); the Eq. (4.76) yields an 𝐺(𝜉)′

algebraic equation involving power of ( 𝐺(𝜉) ) as 𝐺(𝜉)′

0

𝐺(𝜉)′

2

𝐺(𝜉)′

4

𝐶0 ( 𝐺(𝜉) ) + 𝑦2 ( 𝐺(𝜉) ) + 𝐶4 ( 𝐺(𝜉) ) = 0. 𝐺(𝜉)′

Compare the like powers of ( 𝐺(𝜉) ) we have system of equations 𝐺(𝜉)′

0

𝐺(𝜉)′

2

𝐺(𝜉)′

4

−𝑎1 𝑘𝜔𝛿1 − 2𝑎1 𝑘 2 𝜔𝛿12 𝛿2 − ⋯ + 3𝑎12 𝑘 2 𝑚𝛿12 = 0,

( 𝐺(𝜉) ) :

-𝑎1 𝑘𝜔𝛿2 − 𝑎1 𝑘𝜔+⋯ − 6m𝑎12 𝑘 2 𝛿1 = 0,

( 𝐺(𝜉) ) :

−6𝑎1 𝛿23 𝑘 4 − 18𝑎1 𝛿22 𝑘 4 + ⋯ + 3𝑎12 𝑘 2 𝑚 = 0.

( 𝐺(𝜉) ) :

Solving the above system for unknown parameters, we have the following solution sets Solution Set: 𝑘 = 𝑘, 𝜔 = −4𝑘 3 𝛿2 𝛿1 − 4𝑘 3 𝛿1 , 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑚 =

2𝑘 2 (𝛿2 +1) 𝑎1

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

where ( 𝐺(𝜉) ) =

[cosh(2√𝛿1 (1+𝛿2 )𝑦)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1 [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.


𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)




𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

where (

𝐺(𝜉)

)=


Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0,

497

.

.

.

𝐺 ′ (𝜉)

𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

where ( 𝐺(𝜉) ) = 1+𝜉. 𝑡𝛼

In all cases 𝜉 = 𝑘𝑥 + (−4𝑘 3 𝛿2 𝛿1 − 4𝑘 3 𝛿1 ) Γ(𝛼+1) +

2𝑘 2 (𝛿2 +1) 𝑎1

𝑧.

Fig 1: Kink wave solution for the given PDE

Fig 4.13 (a): 2D Kink wave solution of Eq. (4.74)

Fig 4.13 (b): 3D Kink wave solution of Eq. (4.74)

for different values of parameters


4.5.3. (3 + 1) Dimensional Potential-YTSF Equation Consider the following PDE [230] as −4𝑢𝑥𝑡 + 𝑢𝑥𝑥𝑥𝑧 + 4𝑢𝑥 𝑢𝑥𝑧 + 2𝑢𝑥𝑥 𝑢𝑧 + 𝐷2𝛼 𝑦 𝑢 = 0,

0 < 𝛼 < 1.

(4.80)

To convert Eq. (13) into ODE we use following transformation 𝑦𝛼

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡 + 𝑙 Γ(𝛼+1) + 𝑚𝑧,

(4.81)

where 𝑘, 𝑚, 𝑙 and 𝜔 are arbitrary constant. Substituting Eq. (4.81) into Eq. (4.80) and using the chain rule and 𝜉𝑥 = 𝑘, 𝜉𝑡 = 𝜔, 𝜉𝑦 = 𝑙, 𝜉𝑧 = 𝑚, we obtained −4𝜔𝑢′′ 𝑘 + 𝑘 3 𝑚𝑢′′′′ + 6𝑢′ 𝑘 2 𝑚𝑢′′ + 3𝑙 2 𝑢′′ = 0. Integrating the above equation once, ignoring the constant of integration equal to zero we have the following equation 498

−4𝜔𝑘𝑢′ + 𝑚𝑘 3 𝑢′′′ + 3𝑘 2 𝑚(𝑢′ )2 + 3𝑙 2 𝑢 = 0.

(4.82)


𝑀

𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 ( 𝐺(𝜉) ) ,

(4.83)

where 𝑀 can be obtained by applying the homogenous balancing principle we have 𝑀 + 3 = 2𝑀 + 2, 𝑀 = 1.

(4.84)

Using the value of 𝑀 into Eq. (4.84), we obtained the trail solution 𝐺(𝜉)′

𝑢 = 𝑎0 + 𝑎1 (

𝐺(𝜉)

).

(4.85)


𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0.

(4.86)



0

𝐺(𝜉)′

2

𝐺(𝜉)′

4

𝐶0 ( 𝐺(𝜉) ) + 𝐶2 ( 𝐺(𝜉) ) + 𝐶4 ( 𝐺(𝜉) ) = 0. 𝐺(𝜉)′


0

𝐺(𝜉)′

2

𝐺(𝜉)′

4

( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) :

−4𝑘𝑎1 𝜔𝛿1 − 2𝑚𝑘 3 𝑎1 𝛿2 𝛿12 − ⋯ + 3𝑙 2 𝑎1 𝛿1 = 0, 4𝑘𝑎1 𝜔𝛿2 + 4𝑤𝑘𝑎1 + ⋯ − 3𝑙 2 𝑎1 = 0, −6𝑚𝑘 3 𝑎1 𝛿23 − 18𝑚𝑘 3 𝑎1 𝛿22 − ⋯ + 3𝑘 2 𝑚𝑎12 = 0,

Solving the above system for unknown parameters, we have the following solution sets Solution Set: 𝑘 = 𝑘, 𝜔 =

1 4𝑚𝑘 3 𝛿2 𝛿1 +4𝑚𝑘 3 𝛿1 +3𝑙2 4

𝑘

, 𝑎0 = 𝑎0 , 𝑎1 = 2𝑘𝛿2 + 2𝑘.

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

𝑢(𝜉) = 𝑎0 + 2𝑘𝛿2 + 2𝑘 ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)


[cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1 [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 499

𝐺 ′ (𝜉)

𝑢(𝜉) = 𝑎0 + 2𝑘𝛿2 + 2𝑘 ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)



.


𝑢(𝜉) = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)



.


𝑢(𝜉) = 𝑎0 + 2𝑘𝛿2 + 2𝑘 ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


𝑢(𝜉) = 𝑎0 + 2𝑘𝛿2 + 2𝑘 ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

where ( 𝐺(𝜉) ) = 1+𝜉. 𝑦𝛼

1 4𝑚𝑘 3 𝛿2 𝛿1 +4𝑚𝑘 3 𝛿1 +3𝑙2

In all cases 𝜉 = 𝑘𝑥 + 𝑙 Γ(𝛼+1) + (4

𝑘

) 𝑡.

Fig 1: Solitary wave solution for the given PDE of fractional order (12)





4.5.4. Symmetric Regularized Long Wave Equation Consider the Symmetric Regularized Long Wave equation [230] 𝐷𝑡2𝛼 𝑢 + 𝑢𝑥𝑥 + 𝑢𝐷𝑡𝛼 𝑢𝑥 + 𝑢𝑥 𝑢𝑡 + 𝐷𝑡2𝛼 𝑢𝑥𝑥 = 0, 0 < 𝛼 < 1.

500

(4.84)


𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔 Γ(𝛼+1),

(4.85)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.85) into Eq. (4.84) and using the chain rule and 𝜉𝑥 = 𝑘, 𝜉𝑡 = 𝜔, we obtained 𝜔2 𝑢′′ + 𝑘𝑢′′ (𝑘 + 𝑢𝑤) + (𝑢′ )2 𝑘𝑤 + 𝑤 2 𝑘 2 𝑢′′′′ . Integrating the above equation twice, ignoring the constant of integration equal to zero we have the following equation 𝑢

𝑢

1 𝑢2

𝜔2 𝑘 2 (𝑘 2 + 𝜔2 + 2 𝜔𝑘) + 𝜔2 𝑘 2 𝑢′′ = 0. Consider the trial solution for Eq. (4.84) as follow 𝐺(𝜉)′

𝑛

𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 ( 𝐺(𝜉) ) ,

(4.86)

where 𝑀 can be obtained by applying the homogenous balancing principle we have 𝑀 + 2 = 2𝑀, 𝑀 = 2.

(4.87)


𝐺(𝜉)′

2

𝑢 = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ) + 𝑎2 ( 𝐺(𝜉) ) .

(4.88)


𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0.

(4.89)



0

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐺(𝜉)′

3

𝐺(𝜉)′

4

𝐶0 ( 𝐺(𝜉) ) + 𝐶1 ( 𝐺(𝜉) ) + 𝐶2 ( 𝐺(𝜉) ) + 𝐶3 ( 𝐺(𝜉) ) + 𝐶4 ( 𝐺(𝜉) ) = 0. 𝐺(𝜉)′


0

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐺(𝜉)′

3

( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) :

𝑎0 𝜔2 + 𝑎0 𝑘 2 + ⋯ + 2𝑘 2 𝜔2 𝑎2 𝛿12 = 0, 𝑎0 𝜔𝑘𝑎1 + 𝑎1 𝜔2 +⋯ − 2k 2 𝑤 2 𝑎1 𝛿1 = 0, 1

𝜔𝑘𝑎0 𝑎2 + 2 𝜔𝑎12 𝑘 + ⋯ − 8𝑘 2 𝜔2 𝑎1 𝛿1 = 0. 𝑎1 𝜔𝑘𝑎2 + 2𝑘 2 𝑎1 𝜔2 𝛿22+⋯ + 2𝑘 2 𝜔2 𝑎1 = 0, 501

𝐺(𝜉)′

4

1

( 𝐺(𝜉) ) :

2

𝜔𝑎22 𝑘 + 6𝜔2 𝑎2 𝛿22 𝑘 2 … + 6𝑘 2 𝜔2 𝑎2 = 0.

Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: 𝑘=

𝜔 √4𝜔 2 𝛿2 𝛿1 −1+4𝜔2 𝛿1 12𝜔 2 𝛿22

−

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

−

, 𝜔 = 𝜔, 𝑎0 =

4𝜔 2 𝛿1 (𝛿2 +1) √4𝜔2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

−

, 𝑎1 = 0, 𝑎2 =

12𝜔 2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

.

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 4𝜔 2 𝛿1 (𝛿2 +1)

𝑢(𝜉) =

√4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

12𝜔 2 𝛿22

−

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

[

−

−

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

𝐺 ′ (𝜉)

2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1 [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)



.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 12𝜔 2 𝛿22

𝑢(𝜉) =

4𝜔 2 𝛿1 (𝛿2 +1) √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔

𝐺 ′ (𝜉)


√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

−[ 2

−

−

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

](

𝐺 ′ (𝜉) 𝐺(𝜉)

2

) ,

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1


.

Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 12𝜔 2 𝛿22

𝑢(𝜉) = 𝐺 ′ (𝜉)

where (

𝐺(𝜉)

4𝜔 2 𝛿1 (𝛿2 +1)

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

√4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

)=

−[

−


−

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 12𝜔 2 𝛿22

𝑢(𝜉) = 𝐺 ′ (𝜉)

4𝜔 2 𝛿1 (𝛿2 +1) √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔

1

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

−[ 2

−

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

502

−

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2 √4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

] ( 𝐺(𝜉) ) ,

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 12𝜔 2 𝛿22

𝑢(𝜉) =

4𝜔 2 𝛿1 (𝛿2 +1) √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔

𝐺 ′ (𝜉)

−

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

−[ 2

−

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

𝐺 ′ (𝜉)

2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

1


𝜔

In all cases 𝜉 =

√4𝜔2 𝛿2 𝛿1 −1+4𝜔2 𝛿1

𝑥 + 𝜔 Γ(𝛼+1).

2nd Solution Set: 𝑘=−

𝜔 √4𝜔2 𝛿2 𝛿1 −1+4𝜔2 𝛿1

12𝜔 2 𝛿22 √4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

+

4𝜔 2 𝛿1 (𝛿2 +1)

, 𝜔 = 𝜔, 𝑎0 = − 24𝜔 2 𝛿2

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

+

√4𝜔2 𝑦1 𝛿2 −1+4𝛿1 𝜔2 12𝜔 2

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

, 𝑎1 = 0, 𝑎2 =

.

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 12𝜔 2 𝛿22

𝑢(𝜉) =

4𝜔 2 𝛿1 (𝛿2 +1) √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

𝐺 ′ (𝜉)


+[

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

+

+

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1


.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 12𝜔 2 𝛿22

𝑢(𝜉) =

4𝜔 2 𝛿1 (𝛿2 +1) √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

𝐺 ′ (𝜉)


+[

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

+

+

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1


.

Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 12𝜔 2 𝛿22

𝑢(𝜉) = 𝐺 ′ (𝜉)

4𝜔 2 𝛿1 (𝛿2 +1) √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2


+[

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

+



503

+

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2 √4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

.

] ( 𝐺(𝜉) ) ,

12𝜔 2 𝛿22

𝑢(𝜉) =

4𝜔 2 𝛿1 (𝛿2 +1) √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔

𝐺 ′ (𝜉)

1

+[ 2

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

+

24𝜔 2 𝛿2

+

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 12𝜔 2 𝛿22

𝑢(𝜉) =

4𝜔 2 𝛿1 (𝛿2 +1) √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔

𝐺 ′ (𝜉)

+[ 2

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

+

24𝜔 2 𝛿2

+

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1

1

where ( 𝐺(𝜉) ) = 1+𝜉. In all cases 𝜉 = −

𝑡𝛼

𝜔 √4𝜔 2 𝛿2 𝛿1 −1+4𝜔2 𝛿1

𝑥 + 𝜔 Γ(𝛼+1).

3rd Solution Set: 𝑘=−

𝜔 √−4𝜔2 𝛿2 𝛿1 −1−4𝜔2 𝛿1 12𝜔 2 𝛿22

0, 𝑎2 =

√−4𝛿1 𝛿2 𝜔2 −1−4𝜔2 𝛿1

, 𝜔 = 𝜔, 𝑎0 = +

12𝜔 2 𝛿1 (𝛿2 +1)√−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔2 √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

24𝜔 2 𝛿2 √−4𝛿1 𝛿2 𝜔2 −1−4𝜔2 𝛿1

+

12𝜔 2 √−4𝛿1 𝛿2 𝜔2 −1−4𝜔2 𝛿1

, 𝑎1 =

.

Family 1: When 𝛿1 , 𝛿2 ≠ 0, (

𝑢(𝜉) =

𝐺 ′ (𝜉)


12𝜔 2 𝛿1 (𝛿2 +1)× √−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔2 √4𝜔2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

12𝜔 2 𝛿22

)

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

+[

+

+

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1


.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 (

𝑢(𝜉) =

𝐺 ′ (𝜉)


12𝜔 2 𝛿1 (𝛿2 +1)× √−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔2 √4𝜔2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

12𝜔 2 𝛿22

)

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

+[

+

+

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1


.

Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, (

𝑢(𝜉) =

12𝜔 2 𝛿1 (𝛿2 +1)× √−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔

) 2

√4𝜔2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

12𝜔 2 𝛿22 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

+[

+ 504

+

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

] ( 𝐺(𝜉) ) ,

𝐺 ′ (𝜉)



.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, (

𝑢(𝜉) =

12𝜔 2 𝛿1 (𝑦2 +1)× √−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔2 √4𝜔2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

𝐺 ′ (𝜉)

1

12𝜔 2 𝛿22

)

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

+[

+

+

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. (

𝑢(𝜉) =

12𝜔 2 𝛿1 (𝛿2 +1)× √−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔

) 2

√4𝜔2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

𝐺 ′ (𝜉)

12𝜔 2 𝛿22 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

+[

+

+

24𝜔 2 𝛿2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

12𝜔 2

] ( 𝐺(𝜉) ) ,

√4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1

1


𝑡𝛼

𝜔 √4𝜔 2 𝛿2 𝛿1 −1+4𝜔2 𝛿1

𝑥 + 𝜔 Γ(𝛼+1).

4th Solution Set: 𝑘=− 0, 𝑎2 = −

𝜔 √−4𝜔2 𝛿2 𝛿1 −1−4𝜔2 𝛿1 12𝜔 2 𝛿22

√−4𝛿1 𝛿2 𝜔2 −1−4𝜔2 𝛿1

−

, 𝜔 = 𝜔, 𝑎0 = −

12𝜔 2 𝛿1 (𝛿2 +1)√−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔2

24𝜔 2 𝛿2 √−4𝛿1 𝛿2 𝜔2 −1−4𝜔2 𝛿1

√4𝜔2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

−

12𝜔 2 √−4𝛿1 𝛿2 𝜔2 −1−4𝜔2 𝛿1

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = − 12𝜔 2 𝛿 2

12𝜔 2 𝛿1 (𝛿2 +1) ( ) √−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔2 √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

+

24𝜔 2 𝛿

2 2 − − 2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1 √4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1 𝐺 ′ (𝜉) [ ] ( ) , 2 12𝜔 𝐺(𝜉) − 2 2

√4𝛿1 𝛿2 𝜔 −1+4𝜔 𝛿1

𝐺 ′ (𝜉)



.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0

505

.

, 𝑎1 =

𝑢(𝜉) = −

12𝜔 2 𝛿1 (𝛿2 +1) ( ) √−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔2 √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

12𝜔 2 𝛿 2

+

24𝜔 2 𝛿

2 2 − − 2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1 √4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1 𝐺 ′ (𝜉) [ ] ( ) , 12𝜔 2 𝐺(𝜉) − 2 2

√4𝛿1 𝛿2 𝜔 −1+4𝜔 𝛿1

[cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 𝑢(𝜉) = −

12𝜔 2 𝛿1 (𝛿2 +1) ( ) √−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔2 √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

12𝜔 2 𝛿 2

+

24𝜔 2 𝛿


√4𝛿1 𝛿2 𝜔 −1+4𝜔 𝛿1

𝐺 ′ (𝜉)



.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑢(𝜉) = −

12𝜔 2 𝛿1 (𝛿2 +1) ( ) √−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔2 √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

12𝜔 2 𝛿 2

+

24𝜔 2 𝛿

2 2 − − 2 √4𝛿1 𝛿2 𝜔2 −1+4𝜔2 𝛿1 √4𝛿1 𝛿2 𝜔 2 −1+4𝜔2 𝛿1 𝐺 ′ (𝜉) [ ] ( ) , 2 12𝜔 𝐺(𝜉) − 2 2

√4𝛿1 𝛿2 𝜔 −1+4𝜔 𝛿1

𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑢(𝜉) = −

12𝜔 2 𝛿1 (𝛿2 +1) ( ) √−4𝜔2 𝛿2 𝛿1 −1−4𝛿1 𝜔2 √4𝜔 2 𝛿1 𝛿2 −1+4𝛿1 𝜔2

12𝜔 2 𝛿 2

+

24𝜔 2 𝛿


√4𝛿1 𝛿2 𝜔 −1+4𝜔 𝛿1

𝐺 ′ (𝜉)

1


𝜔 √−4𝜔2 𝛿2 𝛿1 −1−4𝜔2 𝛿1

𝑡𝛼

𝑥 + 𝜔 Γ(𝛼+1). 506

.





4.5.5. Modified Equal Width Equation Consider the Modified Equal Width a equation [230] 𝛽

2𝛽

𝐷𝑡𝛼 𝑢 + 3𝑢2 𝐷𝑥 𝑢 − 𝑎𝐷𝑥 𝐷𝑡𝛼 𝑢 = 0, 0 < 𝛼, 𝛽 < 1.

(4.90)

To convert Eq. (4.90) into ODE we use following transformation 𝑘𝑥 𝛽

𝜔𝑡 𝛼

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = Γ(𝛽+1) + Γ(𝛼+1),

(4.91)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.91) into Eq. (4.90) and using the chain rule and 𝜉𝑥 = 𝑘, 𝜉𝑡 = 𝜔, we obtained 𝜔𝑢′ + 3𝑘𝑢2 𝑢′ − 𝑘 2 𝜔𝑎𝑢′′′ = 0. Integrating the above equation once, ignoring the constant of integration equal to zero we have the following equation 𝜔𝑢 + 𝑘𝑢3 + 𝑘 2 𝜔𝑎𝑢′′ = 0.

(4.92)

Consider the trial solution for Eq. (13) as follow 𝐺(𝜉)′

𝑛

𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 ( 𝐺(𝜉) ) ,

(4.93)

where 𝑀 can be obtained by applying the homogenous balancing principle we have 𝑀 + 3 = 2𝑀 + 2, 𝑀 = 1.

(4.94)


𝑢 = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ).

(4.95)


2

𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0.

(4.96)



0

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐺(𝜉)′

3

𝐶0 ( 𝐺(𝜉) ) + 𝐶1 ( 𝐺(𝜉) ) + 𝐶2 ( 𝐺(𝜉) ) + 𝐶3 ( 𝐺(𝜉) ) = 0. 𝐺(𝜉)′


0

1

𝑎0 + 6 𝑎03 = 0,

( 𝐺(𝜉) ) : 𝐺(𝜉)′

1

𝐺(𝜉)′

2

1

−2𝑎1 𝜔2 𝛿1 𝛿2 − 2𝑎1 𝜔2 𝛿1 + ⋯ + 2 𝑎02 𝑎1 = 0,

( 𝐺(𝜉) ) :

1

( 𝐺(𝜉) ) : 𝐺(𝜉)′

2

3

𝑎0 𝑎12 = 0, 1

2𝑎1 𝜔2 𝛿22 + 4𝑎1 𝜔2 𝛿2 + ⋯ + 6 𝑎13 = 0.

( 𝐺(𝜉) ) :

Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: 1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1)

𝑘=2

𝛿1 (𝛿2 +1)

√−6𝛿1 (𝛿2 +1)

, 𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 =

𝛿1

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) =

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1

𝐺 ′ (𝜉)


( 𝐺(𝜉) ),


.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 𝑢(𝜉) =

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1

𝐺 ′ (𝜉)


( 𝐺(𝜉) ),


Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 𝑢(𝜉) = 𝐺 ′ (𝜉)

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉)


𝛿1

( 𝐺(𝜉) ),


Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 508

.

.

.

𝑢(𝜉) =

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝛿1

𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑢(𝜉) =

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝛿1

𝐺 ′ (𝜉)

1

where ( 𝐺(𝜉) ) = 1+𝜉. 1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1)

In all cases 𝜉 = 2

𝛿1 (𝛿2 +1)

𝑥𝛽

𝑡𝛼

+ 𝜔 Γ(𝛼+1). Γ(𝛽+1)

2nd Solution Set: 1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1)

𝑘 = −2

𝛿1 (𝛿2 +1)

, 𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 =

√−6𝛿1 (𝛿2 +1) 𝛿1


√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1

[cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)


2

( 𝐺(𝜉) ) ,


.


√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1

𝐺 ′ (𝜉)


2

( 𝐺(𝜉) ) ,



√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1


2

( 𝐺(𝜉) ) ,


Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑢(𝜉) = 𝐺 ′ (𝜉)

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1 1

2

( 𝐺(𝜉) ) ,

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1

2

( 𝐺(𝜉) ) , 509

.

.

.

𝐺 ′ (𝜉)

1


In all cases 𝜉 = − 2

𝛿1 (𝛿2 +1)

𝑥𝛽

𝑡𝛼

+ 𝜔 Γ(𝛼+1). Γ(𝛽+1)

3rd Solution Set: 1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1) ,𝜔 𝛿1 (𝛿2 +1)

𝑘=2

= 𝜔, 𝑎0 = 0, 𝑎1 = −

√−6𝛿1 (𝛿2 +1)

.

𝛿1

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = −

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1


𝐺 ′ (𝜉)


2

( 𝐺(𝜉) ) ,


.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 𝑢(𝜉) = −

𝛿1

2

( 𝐺(𝜉) ) ,

[cos(2√−𝛿1 (1+𝛿2 )𝑦)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉)

[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.

Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 𝑢(𝜉) = − 𝐺 ′ (𝜉)


√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1

(

𝐺(𝜉)

2

) ,


.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑢(𝜉) = − 𝐺 ′ (𝜉)

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1 1

2

( 𝐺(𝜉) ) ,

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑢(𝜉) = − 𝐺 ′ (𝜉)

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1

2

( 𝐺(𝜉) ) ,

1

where ( 𝐺(𝜉) ) = 1+𝜉. 1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1) 𝑥 𝛽 𝛿1 (𝛿2 +1) Γ(𝛽+1)


𝑡𝛼

+ 𝜔 Γ(𝛼+1).

4th Solution Set: 1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1) ,𝜔 𝛿1 (𝛿2 +1)

𝑘 = −2

510

= 𝜔, 𝑎0 = 0, 𝑎1 = −

√−6𝛿1 (𝛿2 +1) 𝛿1

.

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = −

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1


𝐺 ′ (𝜉)


2

( 𝐺(𝜉) ) ,


.


2

( 𝐺(𝜉) ) ,

𝛿1

[cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉)

[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.



√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉)

2

( 𝐺(𝜉) ) ,

𝛿1


.


where (

𝐺(𝜉)

)=

√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1 1

1

1+𝜉 1+𝛿2

2

( 𝐺(𝜉) ) ,

.


√−6𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) 𝛿1

2

( 𝐺(𝜉) ) ,

1

where ( 𝐺(𝜉) ) = 1+𝜉. 1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1) 𝑥 𝛽 𝛿1 (𝛿2 +1) Γ(𝛽+1)


511

𝑡𝛼

+ 𝜔 Γ(𝛼+1).





4.5.6. Sharma–Tasso–Olver (STO) Equation Consider the Sharma–Tasso–Olver (STO) equation [52] 3 2

𝐷𝑡𝛼 𝑢 + 3𝛼𝑢2 𝑢𝑥 + 𝛼𝑢2𝑥𝑥 + 𝛼𝑢𝑥𝑥𝑥 = 0.

(4.97)

To convert Eq. (7) into ODE we use following transformation 𝑡𝛼

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔 Γ(𝛼+1),

(4.98)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.98) into Eq. (4.97) and using the chain rule and 𝜉𝑥 = 𝑘, 𝜉𝑡 = 𝜔, we obtained 𝜔𝑢′ + 3𝑎𝑘(𝑢2 𝑢′ + ( 𝑢′ )2 𝑘 + 𝑢𝑢′′ 𝑘) + 𝑎𝑘 3 𝑢′′′ . Integrating the above equation once, ignoring the constant of integration equal to zero we have the following equation 𝜔𝑢 + 𝑎𝑘𝑢3 + 3𝑢𝑘 2 𝑢′ 𝑎 + 𝑢′′ 𝑎𝑘 3 = 0.

(4.99)


𝑛

𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 ( 𝐺(𝜉) ) ,

(4.100)

where 𝑀 can be obtained by applying the homogenous balancing principle we have 𝑀 + 2 = 3𝑀, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.100), we obtained the trail solution 𝐺(𝜉)′

𝑢 = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ).

(4.101)


𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0.

(4.102)



0

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐺(𝜉)′

3

𝐶0 ( 𝐺(𝜉) ) + 𝐶1 ( 𝐺(𝜉) ) + 𝐶2 ( 𝐺(𝜉) ) + 𝐶3 ( 𝐺(𝜉) ) = 0. 𝐺(𝜉)′


0

( 𝐺(𝜉) ) :

𝑎0 𝜔 + 𝑎𝑘𝑎03 + 3𝑘 2 𝑎1 𝑎𝑎0 𝛿1 = 0, 512

𝐺(𝜉)′

1

( 𝐺(𝜉) ) : 𝐺(𝜉)′

2

𝑦(𝜉)′

3

( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) :

𝑎1 𝜔 + 3𝑎𝑘𝑎1 𝑎02 + ⋯ − 2𝑎𝑎1 𝑘 3 𝛿1 = 0, 3𝑎𝑘𝑎0 𝑎12 − 3𝑘 2 𝑎1 𝑎0 𝛿1 − 3𝑘 2 𝑎1 𝑎𝑎0 = 0, 𝑎𝑘𝑎13 − 3𝑘 2 𝑎12 𝑎𝛿2 − ⋯ + 2𝑎1 𝑎𝑘 3 = 0.

Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: 𝑘 = 𝑘, 𝜔 = −𝑎𝑘 3 𝛿2 𝛿1 − 𝑎𝑘 3 𝛿1 , 𝑚 = 𝑚, 𝑎0 = 0, 𝑎1 = 𝑘𝛿2 + 𝑘. Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

𝑢(𝜉) = [𝑘𝛿2 + 𝑘] ( 𝐺(𝜉) ), [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)



.


𝑢(𝜉) = [𝑘𝛿2 + 𝑘] ( 𝐺(𝜉) ), [cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2


𝑢(𝜉) = [𝑘𝛿2 + 𝑘] ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)



.


𝑢(𝜉) = [𝑘𝛿2 + 𝑘] ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


𝑢(𝜉) = [𝑘𝛿2 + 𝑘] ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1


In all cases 𝜉 = 𝑘𝑥 + [−𝑎𝑘 3 𝛿2 𝛿1 − 𝑎𝑘 3 𝛿1 ] Γ(𝛼+1). 2nd Solution Set: 513

.

𝑘 = 𝑘, 𝜔 = −4𝑎𝑘 3 𝛿1 − 4𝑎𝑘 3 𝛿2 𝛿1 , 𝑚 = 𝑚, 𝑎0 = 0, 𝑎1 = 2𝑘 + 2𝑘𝛿2 . Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

𝑢(𝜉) = [2𝑘 + 2𝑘𝛿2 ] ( 𝐺(𝜉) ), [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)



.


𝑢(𝜉) = [2𝑘 + 2𝑘𝛿2 ] ( 𝐺(𝜉) ), [cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.


𝑢(𝜉) = [2𝑘 + 2𝑘𝛿2 ] ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)



.


𝑢(𝜉) = [2𝑘 + 2𝑘𝛿2 ] ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


𝑢(𝜉) = [2𝑘 + 2𝑘𝛿2 ] ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1


In all cases 𝜉 = 𝑘𝑥 + [−4𝑎𝑘 3 𝛿1 − 4𝑎𝑘 3 𝛿2 𝛿1 ] Γ(𝛼+1). 3rd Solution Set: 𝑘=

𝑎0 √𝛿2 𝛿1 +𝛿1

, 𝑚 = 𝑚, 𝜔 = −

4𝑎𝑎03 √𝛿2 𝛿1 +𝛿1

, 𝑎0 = 𝑎0 , 𝑎1 =

𝑎0 𝛿2 √𝛿2 𝛿1 +𝛿1

+

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑎0 𝛿2

𝑢(𝜉) = 𝑎0 + [

√𝛿2 𝛿1 +𝛿1

𝐺 ′ (𝜉)


+

𝑎0 √𝛿2 𝛿1 +𝛿1

𝐺 ′ (𝜉)

] ( 𝐺(𝜉) ),


.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 514

𝑎0 √𝛿2 𝛿1 +𝛿1

.

𝑎0 𝛿2

𝑢(𝜉) = 𝑎0 + [

√𝛿2 𝛿1 +𝛿1

𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1

] ( 𝐺(𝜉) ),

[cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


+

[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.

Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 𝑎0 𝛿2

𝑢(𝜉) = 𝑎0 + [

√𝛿2 𝛿1 +𝛿1

𝐺 ′ (𝜉)


+

𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1

] ( 𝐺(𝜉) ),


.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑎0 𝛿2

𝑢(𝜉) = 𝑎0 + [

√𝛿2 𝛿1 +𝛿1

𝐺 ′ (𝜉)

1

+

𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1

] ( 𝐺(𝜉) ),

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑎0 𝛿2

𝑢(𝜉) = 𝑎0 + [

√𝛿2 𝛿1 +𝛿1

𝐺 ′ (𝜉)

+

𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1

] ( 𝐺(𝜉) ),

1

where ( 𝐺(𝜉) ) = 1+𝜉. In all cases 𝜉 =

𝑎0 √𝛿2 𝛿1 +𝛿1

𝑥−

4𝑎𝑎03

𝑡𝛼

.

√𝛿2 𝛿1 +𝛿1 Γ(𝛼+1)

4th Solution Set: 𝑘=−

𝑎0 √𝛿2 𝛿1 +𝛿1

, 𝑚 = 𝑚, 𝜔 =

4𝑎𝑎03 √𝛿2 𝛿1 +𝛿1

, 𝑎0 = 𝑎0 , 𝑎1 = −

𝑎0 𝛿2 √𝛿2 𝛿1 +𝛿1

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = 𝑎0 + [− 𝐺 ′ (𝜉)


𝑎0 𝛿2 √𝛿2 𝛿1 +𝛿1

−

𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1

] ( 𝐺(𝜉) ),


.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 𝑢(𝜉) = 𝑎0 + [− 𝐺 ′ (𝜉)


𝑎0 𝛿2 √𝛿2 𝛿1 +𝛿1

−

𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1

] ( 𝐺(𝜉) ),


Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 𝑢(𝜉) = 𝑎0 + [−

𝑎0 𝛿2 √𝛿2 𝛿1 +𝛿1

−

𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1

] ( 𝐺(𝜉) ),

515

.

−

𝑎0 √𝛿2 𝛿1 +𝛿1

.

𝐺 ′ (𝜉)



.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑢(𝜉) = 𝑎0 + [− 𝐺 ′ (𝜉)

1

𝑎0 𝛿2 √𝛿2 𝛿1 +𝛿1

−

𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1

] ( 𝐺(𝜉) ),

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑢(𝜉) = 𝑎0 + [− 𝐺 ′ (𝜉)

𝑎0 𝛿2 √𝛿2 𝛿1 +𝛿1

−

𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1

] ( 𝐺(𝜉) ),

1


𝑎0 √𝛿2 𝛿1 +𝛿1

𝑥+

4𝑎𝑎03

𝑡𝛼

.

√𝛿2 𝛿1 +𝛿1 Γ(𝛼+1)





4.5.7. Sine-Gordon Equation Consider the Sine-Gordon equation [230] 2𝛽

1 6

𝐷𝑡2𝛼 𝑢 − 𝐷𝑥 𝑢 + 𝑢 − 𝑢3 = 0, 0 < 𝛼, 𝛽 < 1.

(4.103)


𝜔𝑡 𝛼

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = Γ(𝛽+1) + Γ(𝛼+1),

(4.104)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.104) into Eq. (4.103) and using the chain rule and 𝜉𝑥 = 𝑘, 𝜉𝑡 = 𝜔, we obtained 1

𝑢′′ (𝜔2 − 𝑘 2 ) + 𝑢 − 6 𝑢3 = 0. (4.105) 516


𝑀

𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 ( 𝐺(𝜉) ) ,

(4.106)

where 𝑀 can be obtained by applying the homogenous balancing principle we have 𝑀 + 2 = 3𝑀, 𝑀 = 1.

(4.107)


𝑢 = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ).

(4.108)


𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0.

(4.109)


algebraic equation involving power of ( 𝐺(𝜉) ) as 0

𝐺(𝜉)′

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐺(𝜉)′

3

𝐶0 ( 𝐺(𝜉) ) + 𝐶1 ( 𝐺(𝜉) ) + 𝐶2 ( 𝐺(𝜉) ) + 𝐶3 ( 𝐺(𝜉) ) = 0. 𝐺(𝜉)′


0

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐺(𝜉)′

3

1

𝑎0 − 6 𝑎03 = 0,

( 𝐺(𝜉) ) :

1

−2𝑎1 𝜔2 𝛿1 𝛿2 − 2𝑎1 𝜔2 𝛿1 + ⋯ − 2 𝑎02 𝑎1 = 0,

( 𝐺(𝜉) ) :

1

− 2 𝑎0 𝑎12 = 0,

( 𝐺(𝜉) ) :

1

2𝑎1 𝑦2 𝛿22 + 4𝑎1 𝜔2 𝛿2 + ⋯ − 6 𝑎13 = 0.

( 𝐺(𝜉) ) :

Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: 1

𝑘=2

√2 √𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1)

, 𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 =

√6√𝛿1 (𝛿2 +1) . 𝛿1

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = 𝐺 ′ (𝜉)

√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1



517

.


√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1 [cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.

Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 𝑢(𝜉) =

√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

𝐺 ′ (𝜉)



.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑢(𝜉) =

√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

𝐺 ′ (𝜉)

1

where ( 𝐺(𝜉) ) = 1+𝜉. 1


𝑥𝛽

√2 √𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1)

𝑡𝛼

+ 𝜔 Γ(𝛼+1). Γ(𝛽+1)

2nd Solution Set: 1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1)

𝑘 = −2

𝛿1 (𝛿2 +1)

, 𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 =

√6√𝛿1 (𝛿2 +1) . 𝛿1


√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

𝐺 ′ (𝜉)



.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 𝑢(𝜉) = 𝐺 ′ (𝜉)

√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1



Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 518

.

𝑢(𝜉) =

√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

𝐺 ′ (𝜉)



.


√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

𝐺 ′ (𝜉)

1



𝛿1 (𝛿2 +1)

𝑥𝛽

𝑡𝛼

+ 𝜔 Γ(𝛼+1). Γ(𝛽+1)

3rd Solution Set: 1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1)

𝑘=2

𝛿1 (𝛿2 +1)

, 𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 = −

√6√𝛿1 (𝛿2 +1) 𝛿1

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = −


𝐺 ′ (𝜉)


√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1


.


√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

𝐺 ′ (𝜉)

[cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1


[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2



√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1



√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

519

.

.

.

𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑢(𝜉) = −

√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

𝐺 ′ (𝜉)

1



𝛿1 (𝛿2 +1)

𝑥𝛽

𝑡𝛼

+ 𝜔 Γ(𝛼+1). Γ(𝛽+1)

4th Solution Set: 1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1)

𝑘 = −2

𝛿1 (𝛿2 +1)

, 𝜔 = 𝜔, 𝑎0 = 0, 𝑎1 = −

√6√𝛿1 (𝛿2 +1) 𝛿1

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = −

√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1 [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)



.


√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1

𝐺 ′ (𝜉)

[cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1


[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2



√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1



√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( ), 𝛿1 𝐺(𝜉)

𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


√6√𝛿1 (𝛿2 +1) 𝐺 ′ (𝜉) ( 𝐺(𝜉) ), 𝛿1 1

where ( 𝐺(𝜉) ) = 1+𝜉. 520

.

.

.

1 √2√𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 −1)


𝛿1 (𝛿2 +1)


𝑥𝛽

𝑡𝛼

+ 𝜔 Γ(𝛼+1). Γ(𝛽+1)




4.5.8. Boussinesq Equation Consider the Boussinesq equation [230] 𝐷𝑡𝛼 𝑢 − 𝑢𝑥𝑥𝑡 + 𝑎𝑢𝑥 (1 − 𝐷𝑡𝛼 𝑢) = 0, 0 < 𝛼 < 1.

(4.110)


𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔 Γ(𝛼+1),

(4.111)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.111) into Eq. (4.110) and using the chain rule and 𝜉𝑥 = 𝑘, 𝜉𝑡 = 𝜔, we obtained 𝑢′′ (𝜔2 − 𝑘 2 − 2𝑢𝑘 2 ) − 𝑘 3 𝑢′′′ − 2(𝑢′ )2 𝑘 2 = 0.

(4.112)

Integrating the above equation once, ignoring the constant of integration equal to zero we have the following equation 𝜔2 𝑢 − 𝑘 2 𝑢 − 𝑘 2 𝑢2 − 𝑘 3 𝑢′ = 0. Consider the trial solution for Eq. (7) as follow 𝐺(𝜉)′

𝑛

𝑢 ( 𝜉 ) = 𝑎 0 + ∑𝑀 𝑛=1 𝑎𝑛 ( 𝐺(𝜉) ) ,

(4.113)


𝑢 = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ).

(4.114) 521


𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0.

(4.115)



𝐺(𝜉)′

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐶0 ( 𝐺(𝜉) ) + 𝐶1 ( 𝐺(𝜉) ) + 𝐶2 ( 𝐺(𝜉) ) = 0. 𝐺(𝜉)′


0

𝐺(𝜉)′

1

𝐺(𝜉)′

2

( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) :

𝑎0 𝜔2 − 𝑘 2 𝑎0 − 𝑘 2 𝑎02 − 𝑘 3 𝑎1 𝛿1 = 0, -2𝑎0 𝑘 2 𝑎1 − 𝑎1 𝜔2 − 𝑘 2 𝑎1 = 0, −𝑎12 𝑘 2 + 𝑘 3 𝑎1 𝛿2 + 𝑘 3 𝑎1 = 0.

Solving the above system for unknown parameters, we have the following solution sets 𝟏𝐬𝐭 Solution Set: 𝑘 = 𝑘, 𝜔 = √1 + 2√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 𝑘, 𝑎0 = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 , 𝑎1 = 𝑘(𝛿2 + 1). Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

𝑢(𝜉) = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)



.


𝑢(𝜉) = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)




𝑢(𝜉) = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)




522

.

.

𝐺 ′ (𝜉)

𝑢(𝜉) = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


𝑢(𝜉) = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

where ( 𝐺(𝜉) ) = 1+𝜉. In all cases 𝜉 = 𝑘𝑥 + √1 + 2√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1

𝑡𝛼

.

Γ(𝛼+1)

𝟐𝐧𝐝 Solution Set: 𝑘 = 𝑘, 𝜔 = −√1 + 2√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 𝑘, 𝑎0 = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 , 𝑎1 = 𝑘(𝛿2 + 1). Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

𝑢(𝜉) = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)



.


𝑢(𝜉) = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)




𝑢(𝜉) = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

where (

𝐺(𝜉)

)=


.


𝑢(𝜉) = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0.

523

.

𝐺 ′ (𝜉)

𝑢(𝜉) = √𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

where ( 𝐺(𝜉) ) = 1+𝜉. In all cases 𝜉 = 𝑘𝑥 + √1 + 2√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1

𝑡𝛼

.

Γ(𝛼+1)

𝟑𝐫𝐝 Solution Set: 𝑘 = 𝑘, 𝜔 = √1 − 2√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 𝑘, 𝑎0 = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 , 𝑎1 = 𝑘(𝛿2 + 1). Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

𝑢(𝜉) = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)



.


𝑢(𝜉) = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), [cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


[cos(2√−𝑦1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2


𝑢(𝜉) = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)



.


𝑢(𝜉) = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


𝑢(𝜉) = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1


In all cases 𝜉 = 𝑘𝑥 + √1 − 2√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 𝑘 Γ(𝛼+1). 𝟒𝐭𝐡 Solution Set: 524

.

𝑘 = 𝑘, 𝜔 = −√1 − 2√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 𝑘, 𝑎0 = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 , 𝑎1 = 𝑘(𝛿2 + 1). Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

𝑢(𝜉) = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)

where ( 𝑦(𝜉) ) =


.


𝑢(𝜉) = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)




𝑢(𝜉) = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)



.


𝑢(𝜉) = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


𝑢(𝜉) = −√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 + 𝑘(𝛿2 + 1) ( 𝐺(𝜉) ), 𝐺 ′ (𝜉)

1

where ( 𝐺(𝜉) ) = 1+𝜉. In all cases 𝜉 = 𝑘𝑥 + −√1 − 2√𝑘 2 𝛿2 𝛿1 + 𝑘 2 𝛿1 𝑘

525

𝑡𝛼

.

Γ(𝛼+1)

.





4.5.9. Good Boussinesq Equation Consider the Good Boussinesq equation [230] 𝐷𝑡2𝛼 𝑢 − 𝑢𝑥𝑥 + 𝑢𝑥𝑥𝑥𝑥 − (𝑢2 )𝑥𝑥 = 0, 0 < 𝛼 < 1.

(4.116)


𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔 Γ(𝛼+1),

(4.117)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.117) into Eq. (4.116) and using the chain rule and 𝜉𝑥 = 𝑘, 𝜉𝑡 = 𝜔, we obtained 𝑢′′ (𝜔2 − 𝑘 2 − 2𝑢𝑘 2 ) − 𝑘 4 𝑢′′′′ − 2(𝑢′ )2 𝑘 2 = 0. Integrating the above equation once, ignoring the constant of integration equal to zero we have the following equation 𝜔2 𝑢 − 𝑘 2 𝑢 − 𝑘 2 𝑢2 + 𝑘 4 𝑢′′ = 0.

(4.118)


𝑛

𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 ( 𝐺(𝜉) ) ,

(4.119)


𝐺(𝜉)′

2

𝑢 = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ) + 𝑎2 ( 𝐺(𝜉) ) .

(4.120)


2

𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0.

(4.121)



𝐺(𝜉)′

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐺(𝜉)′

3

𝐺(𝜉)′

4

𝐶0 ( 𝐺(𝜉) ) + 𝐶1 ( 𝐺(𝜉) ) + 𝐶2 ( 𝐺(𝜉) ) + 𝐶3 ( 𝐺(𝜉) ) + +𝐶4 ( 𝐺(𝜉) ) = 0. 𝐺(𝜉)′


0

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐺(𝜉)′

3

𝐺(𝜉)′

4

( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) :

𝑎0 𝜔2 − 𝑘 2 𝑎0 − ⋯ + 2𝑘 4 𝑎2 𝛿12 = 0, −2𝑎0 𝑘 2 𝑎1 + 𝑎1 𝜔2 − ⋯ − 2𝑎1 𝑘 4 𝛿1 = 0, −2𝑎0 𝑎2 𝑘 2 − 𝑎12 𝑘 2 + ⋯ − 8𝑘 4 𝑎2 𝛿1 = 0, −2𝑘 2 𝑎1 𝑎2 + 2𝑘 4 𝑎1 𝛿22 + ⋯ + 2𝑎1 𝑘 4 = 0, −𝑎22 𝑘 2 + 6𝑘 4 𝑎2 𝛿22 + ⋯ + 6𝑎2 𝑘 4 = 0.

Solving the above system for unknown parameters, we have the following solution sets 𝟏𝐬𝐭 Solution Set: 𝑘 = 𝑦, 𝜔 = √4𝑘 2 𝛿2 𝛿1 + 4𝑘 2 𝛿1 𝑘, 𝑎0 = −2𝑘 2 𝛿1 (𝛿2 + 1), 𝑎1 = 0, 𝑎2 = 6𝑘 2 (𝛿2 + 1)2 . Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

2

𝑢(𝜉) = −2𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)



.


2

𝑢(𝜉) = −2𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)




2

𝑢(𝜉) = −2𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 527

.

𝐺 ′ (𝜉)



.


2

𝐺 ′ (𝜉)

2

𝑢(𝜉) = −2𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑢(𝜉) = −2𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)

1


In all cases 𝜉 = 𝑘𝑥 + √4𝑘 2 𝛿2 𝛿1 + 4𝑘 2 𝛿1 𝑘 Γ(𝛼+1). 𝟐𝐧𝐝 Solution Set: 𝑘 = 𝑘, 𝜔 = −√4𝑘 2 𝛿2 𝛿1 + 4𝑘 2 𝛿1 + 1 𝑘, 𝑎0 = −2𝑘 2 𝛿1 (𝛿2 + 1), 𝑎1 = 0, 𝑎2 = 6𝑘 2 (𝛿2 + 1)2 . Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = −2𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺 ′ (𝜉)


𝐺 ′ (𝜉) 𝐺(𝜉)

2

) ,


.


2

𝑢(𝜉) = −2𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)




2

𝐺 ′ (𝜉)

2

𝑢(𝜉) = −2𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)



.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑢(𝜉) = −2𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) ,

528

.

𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


2

𝑢(𝜉) = −2𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)

1


In all cases 𝜉 = 𝑘𝑥 + −√4𝑘 2 𝛿2 𝛿1 + 4𝑘 2 𝛿1 + 1 𝑘 Γ(𝛼+1). 𝟑𝐫𝐝 Solution Set: 𝑘 = 𝑘, 𝜔 = √−4𝑘 2 𝛿2 𝛿1 − 4𝑘 2 𝛿1 + 1 𝑘, 𝑎0 = −6𝑘 2 𝛿1 (𝛿2 + 1), 𝑎1 = 0, 𝑎2 = 6𝑘 2 (𝛿2 + 1)2 . Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

2

𝑢(𝜉) = −6𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)



.


2

𝑢(𝜉) = −6𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)




2

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

2

𝑢(𝜉) = −6𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)



.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑢(𝜉) = −6𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑢(𝜉) = −6𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) ,

529

.

𝐺 ′ (𝜉)

1


In all cases 𝜉 = 𝑘𝑥 + √−4𝑘 2 𝛿2 𝛿1 − 4𝑘 2 𝛿1 + 1 𝑘 Γ(𝛼+1). 𝟒𝐭𝐡 Solution Set: 𝑘 = 𝑘, 𝜔 = −√−4𝑘 2 𝛿2 𝛿1 − 4𝑘 2 𝛿1 + 1 𝑘, 𝑎0 = −6𝑘 2 𝛿1 (𝛿2 + 1), 𝑎1 = 0, 𝑎2 = 6𝑘 2 (𝛿2 + 1)2 . Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝐺 ′ (𝜉)

2

𝑢(𝜉) = −6𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , [cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝛿2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)



.


2

𝑢(𝜉) = −6𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)




2

𝐺 ′ (𝜉)

2

𝐺 ′ (𝜉)

2

𝑢(𝜉) = −6𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)



.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑢(𝜉) = −6𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑢(𝜉) = −6𝑘 2 𝛿1 (𝛿2 + 1) + 6𝑘 2 (𝛿2 + 1)2 ( 𝐺(𝜉) ) , 𝐺 ′ (𝜉)

1

where ( 𝐺(𝜉) ) = 1+𝜉. In all cases 𝜉 = 𝑘𝑥 + −√−4𝑘 2 𝛿2 𝛿1 − 4𝑘 2 𝛿1 + 1 𝑘

530

𝑡𝛼

.

Γ(𝛼+1)

.

Fig 4.20 (a): 2D Solitary wave sol of Eq. (4.116)

Fig 4.20 (b): 3D Solitary wave sol of Eq. (4.116)



4.5.10. Klein Gordon Equation Consider the Klein Gordon equation [230] 2𝛽

𝐷𝑡2𝛼 𝑢 + 𝑎𝐷𝑥 𝑢 − 𝑏𝑢 − 𝑐𝑢3 = 0, 0 < 𝛼, 𝛽 < 1.

(4.122)


𝜔𝑡 𝛼

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = Γ(𝛽+1) + Γ(𝛼+1),

(4.123)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.123) into Eq. (4.122) and using the chain rule and 𝜉𝑥 = 𝑘, 𝜉𝑡 = 𝜔, we obtained 𝑢′′ (𝜔2 + 𝑎𝑘 2 ) − 𝑏𝑢 − 𝑐𝑢3 = 0.

(4.124)


𝑛

𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 ( 𝐺(𝜉) ) ,

(4.125)


𝑢 = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ).

(4.126)


𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0.

(4.127)



0

𝐺(𝜉)′

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐺(𝜉)′

3

𝐶0 ( 𝐺(𝜉) ) + 𝐶1 ( 𝐺(𝜉) ) + 𝐶2 ( 𝐺(𝜉) ) + 𝐶3 ( 𝐺(𝜉) ) = 0. 𝐺(𝜉)′


0

𝐺(𝜉)′

1

𝐺(𝜉)′

2

−𝑏𝑎0 − 𝑐𝑎03 = 0,

( 𝐺(𝜉) ) :

−2𝑎1 𝜔2 𝛿2 𝛿1 − 2𝑎1 𝜔2 𝛿1 − ⋯ − 3𝑐𝑎02 𝑎1 = 0,

( 𝐺(𝜉) ) :

−3𝑐𝑎12 𝑎0 = 0,

( 𝐺(𝜉) ) : 𝐺(𝜉)′

3

2𝑎1 𝜔2 𝛿22 + 4𝑎1 𝜔2 𝛿2 + ⋯ − 𝑐𝑎13 = 0,

( 𝐺(𝜉) ) :

Solving the above system for unknown parameters, we have the following solution sets 𝟏𝐬𝐭 Solution Set: 1 √−2𝑎𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 +𝑏)

𝑘=2

𝑎𝛿1 (𝛿2 +1)

, 𝜔 = 𝜔, 𝑎0 = 0 , 𝑎1 = −

√𝑐𝛿1 𝑏(𝛿2 +1)

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = − 𝐺 ′ (𝜉)

where (

𝐺(𝜉)

)=

√𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉) 𝑐𝛿1

( 𝐺(𝜉) ),


.


𝑐𝛿1

( 𝐺(𝜉) ),

[cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


√𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2



√𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉) 𝑐𝛿1

( 𝐺(𝜉) ),



√𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉) 𝑐𝛿1 1

( 𝐺(𝜉) ),

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 532

.

.

𝑐𝛿1

.

𝑢(𝜉) = − 𝐺 ′ (𝜉)

√𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝑐𝛿1 1

where ( 𝐺(𝜉) ) = 1+𝜉. 1 √−2𝑎𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 +𝑏)


𝑎𝛿1 (𝛿2 +1)

𝑥𝛽

𝑡𝛼

+ 𝜔 Γ(𝛼+1). Γ(𝛽+1)

𝟐𝐧𝐝 Solution Set: 1 √−2𝑎𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 +𝑏)

𝑘 = −2

𝑎𝛿1 (𝛿2 +1)

, 𝜔 = 𝜔, 𝑎0 = 0 , 𝑎1 =

√−𝑐𝛿1 𝑏(𝛿2 +1)


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝑐𝛿1


𝐺 ′ (𝜉)



.


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝑐𝛿1

[cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝑐𝛿1



.


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝑐𝛿1

𝐺 ′ (𝜉)

1

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

𝐺 ′ (𝜉)

𝑐𝛿1

( 𝐺(𝜉) ),

1



𝑎𝛿1 (𝛿2 +1)

𝟑𝐫𝐝 Solution Set: 533

𝑥𝛽

𝑡𝛼

+ 𝜔 Γ(𝛼+1). Γ(𝛽+1)

𝑐𝛿1

.

1 √−2𝑎𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 +𝑏)

𝑘=2

𝑎𝛿1 (𝛿2 +1)

, 𝜔 = 𝜔, 𝑎0 = 0 , 𝑎1 = −

√−𝑐𝛿1 𝑏(𝛿2 +1) 𝑐𝛿1

.

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = −

( 𝐺(𝜉) ),

𝑐𝛿1

[cosh(2√𝛿1 (1+𝛿2 )𝜉)+sinh(2√𝛿1 (1+𝑦2 )𝜉)]√𝛿1 +√𝛿1

𝐺 ′ (𝜉)


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)


.


( 𝐺(𝑦) ),

𝑐𝛿1

[cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.



√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝑐𝛿1


.


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

(

𝑐𝛿1 1

𝐺(𝜉)

),

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝑐𝛿1 1



𝑎𝛿1 (𝛿2 +1)

𝑥𝛽

𝑡𝛼

+ 𝜔 Γ(𝛼+1). Γ(𝛽+1)

𝟒𝐭𝐡 Solution Set: 1 √−2𝑎𝛿1 (𝛿2 +1)(2𝜔 2 𝛿2 𝛿1 +2𝜔2 𝛿1 +𝑏)

𝑘 = −2

𝑎𝛿1 (𝛿2 +1)

, 𝑦 = 𝜔, 𝑎0 = 0 , 𝑎1 = −

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = − 𝐺 ′ (𝜉)


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉) 𝑐𝛿1

( 𝐺(𝜉) ),


534

.

√−𝑐𝛿1 𝑏(𝛿2 +1) 𝑐𝛿1

.


𝑐𝛿1

( 𝐺(𝜉) ),

[cos(2√−𝛿1 (1+𝛿2 )𝜉)−sin(2√−𝛿1 (1+𝛿2 )𝜉)]√−𝛿1 +√−𝛿1

𝐺 ′ (𝜉)


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉)

[cos(2√−𝛿1 (1+𝛿2 )𝜉)+sin(2√−𝛿1 (1+𝛿2 )𝜉)]√1+𝛿2 −√1+𝛿2

.



√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉) 𝑐𝛿1

( 𝐺(𝜉) ),


.


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉) 𝑐𝛿1 1

( 𝐺(𝜉) ),

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2


√−𝑐𝛿1 𝑏(𝛿2 +1) 𝐺 ′ (𝜉) 𝑐𝛿1

( 𝐺(𝜉) ),

1



𝑎𝛿1 (𝛿2 +1)

Fig 4.21 (a): 2D Solitary wave sol of Eq. (4.122)

𝑡𝛼

𝑥 + 𝜔 Γ(𝛼+1).

Fig 4.21 (b): 3D Solitary wave sol of Eq. (4.122)



4.5.11. Burger Hierarchy Equation Consider the Burger Hierarchy equation [220] as 𝐷𝑡𝛼 𝑢 + 𝑎𝑢𝑥𝑥 + 2𝑎𝑢𝑢𝑥 + 𝑏𝑢𝑥 + 𝑏𝑢𝑦 = 0, 0 < 𝛼 < 1. 535

(4.128)

To convert Eq. (4.128) into ODE we use following transformation 𝜔𝑡 𝛼

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + Γ(𝛼+1),

(4.129)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.129) into Eq. (4.128) and using the chain rule and 𝜉𝑥 = 𝑘, 𝜉𝑡 = 𝜔, we obtained 𝑢′ (𝜔 + 2𝑎𝑢𝑘 + 𝑏𝑘 + 𝑏𝑙) + 𝑎𝑢′′ 𝑘 2 = 0. Integrating the above equation once, ignoring the constant of integration equal to zero we have the following equation 𝜔𝑢 + 𝑎𝑘 2 𝑢′ + 𝑎𝑘𝑢2 + 𝑏𝑘𝑢 + 𝑏𝑙𝑢 = 0.

(4.130)


𝑛

𝑢(𝜉) = 𝑎0 + ∑𝑀 𝑛=1 𝑎𝑛 ( 𝐺(𝜉) ) ,

(4.131)


𝑢 = 𝑎0 + 𝑎1 ( 𝐺(𝜉) ).

(4.132)


𝐺(𝜉)𝐺 ′′ (𝜉) − 𝛿1 𝐺 2 (𝜉) + 𝛿2 (𝐺 ′ (𝜉)) = 0.

(4.133)



0

𝐺(𝜉)′

1

𝐺(𝜉)′

2

𝐶0 ( 𝐺(𝜉) ) + 𝐶1 ( 𝐺(𝜉) ) + 𝐶2 ( 𝐺(𝜉) ) = 0. 𝐺(𝜉)′


0

𝐺(𝜉)′

1

𝐺(𝜉)′

2

( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) : ( 𝐺(𝜉) ) :

𝜔𝑎0 𝛿1 + 𝑎𝑘 2 𝑎1 𝛿1 + ⋯ + 𝑏𝑙𝑎0 = 0, 𝜔𝑎1 + 2𝑎𝑘𝑎0 𝑎1 − ⋯ + 𝑏𝑙𝑎1 = 0, −𝑎𝑘 2 𝑎1 𝛿2 − 𝑎𝑘 2 𝑎1 + 𝑎𝑘𝑎12 = 0,

Solving the above system for unknown parameters, we have the following solution sets 𝟏𝐬𝐭 Solution Set: 536

𝑘=

𝑎0 √𝛿2 𝛿1 +𝛿1

,𝜔 = −

2𝑎𝑎02 −𝑏𝑎0 √𝛿2 𝛿1 +𝛿1

− 𝑏𝑙, 𝑎0 = 𝑎0 , 𝑎1 =

𝑎0 √𝛿2 𝛿1 +𝛿1 𝛿1

.

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = 𝑎0 +

( 𝐺(𝜉) ),

𝛿1


𝐺 ′ (𝜉)


𝑎0 √𝛿2 𝛿1 +𝛿1 𝐺 ′ (𝜉)


.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 𝑢(𝜉) = 𝑎0 + 𝐺 ′ (𝜉)


𝑎0 √𝛿2 𝛿1 +𝛿1 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝛿1


.

Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 𝑢(𝜉) = 𝑎0 + 𝐺 ′ (𝜉)


𝑎0 √𝛿2 𝛿1 +𝛿1 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝛿1


.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑢(𝜉) = 𝑎0 + 𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1 𝐺 ′ (𝜉)

(

𝛿1

1

𝐺(𝜉)

),

1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑢(𝜉) = 𝑎0 + 𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝛿1

1

where ( 𝐺(𝜉) ) = 1+𝜉. 𝑎0

In all cases 𝜉 =

√𝛿2 𝛿1 +𝛿1

𝑥−

2𝑎𝑎02 −𝑏𝑎0

𝑡𝛼

.

√𝛿2 𝛿1 +𝛿1 Γ(𝛼+1)

𝟐𝐧𝐝 Solution Set: 𝑘=−

𝑎0 √𝛿2 𝛿1 +𝛿1

,𝜔 =

2𝑎𝑎02 +𝑏𝑎0 √𝛿2 𝛿1 +𝛿1

− 𝑏𝑙, 𝑎0 = 𝑎0 , 𝑎1 = −

𝑎0 √𝛿2 𝛿1 +𝛿1 𝛿1

.

Family 1: When 𝛿1 , 𝛿2 ≠ 0, 𝑢(𝜉) = 𝑎0 − 𝐺 ′ (𝜉)


𝑎0 √𝛿2 𝛿1 +𝛿1 𝐺 ′ (𝜉) 𝛿1

( 𝐺(𝜉) ),


537

.

Family 2: When 𝛿1 < 0, and (1 + 𝛿2 ) > 0, or 𝛿1 > 0, and (1 + 𝛿2 ) < 0 𝑢(𝜉) = 𝑎0 − 𝐺 ′ (𝜉)


𝑎0 √𝛿2 𝛿1 +𝛿1 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝛿1


.

Family 3: When 𝛿1 ≠ 0, and 𝛿2 = 0, 𝑢(𝜉) = 𝑎0 − 𝐺 ′ (𝜉)


𝑎0 √𝛿2 𝛿1 +𝛿1 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝛿1


.

Family 4: When 𝛿1 = 0, and 𝛿2 ≠ 0, 𝑢(𝜉) = 𝑎0 − 𝐺 ′ (𝜉)

1

𝑎0 √𝛿2 𝛿1 +𝛿1 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝛿1 1

where ( 𝐺(𝜉) ) = 1+𝜉 1+𝛿 . 2

Family 5: When 𝛿1 = 0, and 𝛿2 = 0. 𝑢(𝜉) = 𝑎0 − 𝐺 ′ (𝜉)

𝑎0 √𝛿2 𝛿1 +𝛿1 𝐺 ′ (𝜉)

( 𝐺(𝜉) ),

𝛿1

1


𝑎0 √𝛿2 𝛿1 +𝛿1

𝑥+

2𝑎𝑎02 −𝑏𝑎0

𝑡𝛼

.

√𝛿2 𝛿1 +𝛿1 Γ(𝛼+1)





4.6. Rational Hyperbolic Function (RHF) Method We proposed a new algorithm “Rational Hyperbolic Function (RHF) Method” to find the soliton solutions of nonlinear evolution equations. In this the trial solution in rational 538

form. The computational work is less as compare to Exp-function method. This method is more effevtive and reliable.

4.6.1. Mehtodolgy Suppose a nonlinear equation for 𝑢(𝑥, 𝑡) is given by 𝑃(𝑢, 𝐷𝑡𝛼 𝑢, 𝐷𝑥𝛼 𝑢, 𝐷𝑦𝛼 𝑢, 𝐷𝑧𝛼 𝑢, 𝐷𝑡2𝛼 𝑢, 𝐷𝑥2𝛼 𝑢, 𝐷𝑦2𝛼 𝑢, 𝐷𝑧2𝛼 𝑢, … ) = 0,

(4.1)

in which both nonlinear term(s) and higher order derivatives of 𝑢(𝑥, 𝑦, 𝑧, 𝑡) are all involved. In general, the left-hand side of Eq. (4.1) is a polynomial in 𝜓 and its various derivatives. The rational hyperbolic function method for solving Eq. (4.1) proceeds in the following five steps: Step 1: Look for traveling wave solution of Eq. (4.1) by taking 𝑥𝛽

𝑦𝛾

𝑧𝛿

𝑡𝛼

𝑢(𝑥, 𝑦, 𝑧, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘 Γ(𝛽+1) + 𝑙 Γ(𝛾+1) + 𝑚 Γ(𝛿+1) + 𝜔 Γ(𝛼+1),

(4.134)

where 𝑘 and 𝜔 is nonzero constants, 𝑢(𝜉) the function of 𝜉. Substituting (4.134) into Eq. (4.1) yields an ordinary differential equation (ode) for 𝑢(𝜉). 𝜑(𝑢, 𝑘𝜔𝑢′ , 𝑘𝑢′ , 𝑘 2 𝜔2 𝑢′′ , 𝑘 2 𝑢′′ , 𝑘 2 𝜔2 𝑢′′ , … ) = 0.

(4.135)

Step 2: If possible, integrate Eq. (4.135) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero. Step 3: According to the rational hyperbolic function method suppose that 𝑢(𝜉) can be expressed by a finite power series of 𝑢(𝜉) = ∑𝑀 𝑛=1

𝑎𝑛+1 sinh(𝑛𝜇𝜉)+𝑎𝑛+2 cosh(𝑛𝜇𝜉)+𝑎𝑛+3 tanh(𝑛𝜇𝜉) 𝑎𝑛−1 +𝑎𝑛 sinh(𝑛𝜇𝜉) cosh(𝑛𝜇𝜉)

.

(4.136)

Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term of highest order with the highest order nonlinear term which obtained in Step 1. Step 5: Substituting (4.136) into Eq. (4.135) will yields an algebraic equation. Solve these system for 𝑎𝑛 , 𝑘, 𝜇 and 𝜔. Then, we solve the system with the help of MAPLE 11, to determine these constants.

4.6.2. Klein Gordon Equation Consider the Klein Gordon equation [230] 𝐷𝑡2𝛼 𝑢 + 𝛼𝑢𝑥𝑥 − 𝛽𝑢 − 𝛾𝑢3 = 0,

0 < 𝛼 < 1.

(4.137)

𝜔𝑡 𝛼

Consider the transformation 𝜉 = 𝑘𝑥 + Γ(𝛼+1), we have Eq. (4.137) is 𝜔2 𝑈 ′′ + 𝑎𝑘 2 𝑈 ′′ − 𝛽𝑈 − 𝛾𝑈 3 = 0.

(4.138) 539

Applying the balancing term phenomena on Eq. (4.138) we obtained 𝑀 + 2 = 3𝑀, 𝑀 = 1. Then we have the trail solution is 𝑎2 sinh(𝜇𝑥)+𝑎3 cosh(𝜇𝑥)+𝑎4 tanh(𝜇𝑥)

𝑈(𝜉) =

.

𝑎0 +𝑎1 sinh(𝜇𝑥) cosh(𝜇𝑥)

(4.139)

Putting (4.139) in (4.138) and according to the purposed technique we obtained two solution set 1st Solution Set: 𝛾

𝑎0 = √− 𝛽 𝑎4 , 𝑎1 = 0, 𝑎2 = 0, 𝑎3 = 0, 𝑎4 = 𝑎4 , 𝜇 = 𝜇, 𝑘=

2 2 √−12𝜇 𝜔 +𝛽 2

𝛼

𝜇

, 𝜔 = 𝜔.

Substituting in (4.139) we obtained, 2 2 √−12𝜇 𝜔 +𝑦 𝑡𝛼 2 𝛼 ) 𝜇( 𝑥+𝜔 𝜇 Γ(𝛼+1)

𝑎4 tanh (

𝑢(𝑥, 𝑡) =

) 𝛾

,

√−𝛽

2nd Solution Set: 1

𝑎0 = − 2

1 𝛾𝑎2 2 −𝛾𝑎3 2

𝛾𝑎3 𝑎2 1 𝛾𝑎2 2 −𝛾𝑎3 2 𝛽 √− 2 𝛽

𝜇 = 𝜇, 𝑘 =

2 2 √−2𝜇 𝜔 −𝛽 𝛼

𝜇

, 𝑎1 = √− 2

𝛽

, 𝑎2 = 𝑎2 , 𝑎3 = 𝑎3 , 𝑎4 = 0,

, 𝜔 = 𝜔.

Substituting in (4.139) we obtained 𝑎2 sinh 𝜇(

2 2 √−2𝜇 𝜔 −𝛽 𝑡𝛼 𝛼 ) 𝑥+𝜔 𝜇 Γ(𝛼+1)

(

𝑢(𝑥, 𝑡) =

2 2 √−2𝜇 𝜔 −𝛽 𝑡𝛼 𝛼 ) 𝑥+𝜔 𝜇 Γ(𝛼+1)

+𝑎3 cosh 𝜇(

)

1 𝛾𝑎2 2 −𝛾𝑎3 2 √− sinh 2 𝛽

(

2 2 √−2𝜇 𝜔 −𝛽 𝑡𝛼 𝛼 ) 𝜇( 𝑥+𝜔 𝜇 Γ(𝛼+1)

( − [

540

1 2

𝛾𝑎3 𝑎2

)

cosh

)

1 𝛾𝑎2 2 −𝛾𝑎3 2 𝛽√− 2 𝛽

(

2 2 √−2𝜇 𝜔 −𝛽 𝑡𝛼 𝛼 ) 𝜇( 𝑥+𝜔 𝜇 Γ(𝛼+1)

.

)

]

(a)

(b)


4.6.3. Potential Kadomtsev-Petviashvili Equation Consider the Potential Kadomtsev-Petviashvili equation [230] 4𝑢𝑥𝑡 + 6𝑢𝑥 𝑢𝑥𝑥 + 4𝑢𝑥𝑥𝑥𝑥 + 3𝐷𝑦3𝛼 𝑢 = 0,

0 < 𝛼 < 1.

(4.140)

𝑦𝛼

Consider the transformation 𝜉 = 𝑘𝑥 + 𝑙 Γ(𝛼+1) + 𝜔𝑡, we have Eq. (4.140) is 4𝑘𝜔𝑈 ′′ + 6𝑘 3 𝑈 ′ 𝑈 ′′ + 4𝑘 3 𝑈 ′′′′ + 3𝑙 3 𝑈 ′′ = 0.

(4.141)

Integrate Eq. (4.141) once we have 4𝑘𝜔𝑈 ′ + 3𝑘 3 (𝑈 ′ )2 + 4𝑘 3 𝑈 ′′′ + 3𝑙 3 𝑈 ′ = 0.

(4.142)

Applying the balancing term phenomena on Eq. (4.142) we obtained 2𝑀 + 2 = 𝑀 + 3, 𝑀 = 1. Then we have the trail solution is 𝑈(𝜉) =

𝑎2 sinh(𝜇𝜉)+𝑎3 cosh(𝜇𝜉)+𝑎4 tanh(𝜇𝜉) 𝑎0 +𝑎1 sinh(𝜇𝜉) cosh(𝜇𝜉)

.

(4.143)

Putting (4.143) in (4.142) and according to the purposed technique we obtained one solution set, 1 𝑎4

𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = 0, 𝑎3 = 0, 𝑎4 = 𝑎4 , 𝜇 = 𝜇, 𝑘 = 8 Substituting in (4.143) we have 𝑎4 tanh(𝜇(

𝑢(𝑥, 𝑡) =

3 1 𝑎4 𝑦𝛼 1 𝑎4 𝑥+𝑙 − 𝑡)) 8 𝜇𝑎0 Γ(𝛼+1) 128 𝜇𝑎3 0

.

𝑎0

541

𝜇𝑎0

1

, 𝜔 = − 128

𝑎43 𝜇𝑎03

.

(a)

(b)


4.6.4. Newel whiched Equation Consider the Newel whiched equation [170] 2𝛽

𝐷𝑡𝛼 𝑢 − 𝐷𝑥 𝑢 − 𝛽𝑢(1 − 𝑢2 ) = 0,

0 < 𝛼, 𝛽 < 1.

(4.144)

Consider the transformation 𝜉 = 𝑘𝑥 + 𝜔𝑡, we have Eq. (4.144) is 𝜔𝑈 ′ − 𝑘 2 𝑈 ′′ − 𝛽𝑈(1 − 𝑈 2 ) = 0.

(4.145)

Applying the balancing term phenomena on Eq. (4.145) we obtained 𝑀 + 2 = 3𝑀, 𝑀 = 1. Then we have the trail solution is 𝑈(𝜉) =


.

(4.146)

Putting (4.146) in (4.145) and according to the purposed technique we obtained two solution set, 1st Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = 0, 𝑎3 = 0, 𝑎4 = −𝑎0 , 𝑘 = 𝑘, 𝜔 = 𝜔. Substituting in (4.146) we obtained, 𝑥𝛽

𝑡𝛼

𝑈(𝑦) = − tanh (𝜇 (𝑘 Γ(𝛽+1) + 𝜔 Γ(𝛼+1))). 2nd Solution Set: 1

𝑎0 = 2

𝑎3 𝑎2 1 𝛽√ (𝑎2 2 −𝑎3 2 ) 2

1

, 𝑎1 = √2 (𝑎2 2 − 𝑎3 2 ), 𝑎2 = 𝑎2 , 𝑎3 = 𝑎3 , 𝑎4 = 0. 542

Substituting in (4.146) we obtained 𝑎2 sinh(𝜇(𝑘

𝑦(𝜉) =

1 2

𝑥𝛽 𝑡𝛼 𝑥𝛽 𝑡𝛼 +𝜔 ))+𝑎3 cosh(𝜇(𝑘 +𝜔 )) Γ(𝛽+1) Γ(𝛼+1) Γ(𝛽+1) Γ(𝛼+1)

.

1 𝑥𝛽 𝑡𝛼 𝑥𝛽 𝑡𝛼 )) cosh(𝜇(𝑘 )) +√ (𝑎2 2 −𝑎3 2 ) sinh(𝜇(𝑘 +𝜔 +𝜔 2 Γ(𝛽+1) Γ(𝛼+1) Γ(𝛽+1) Γ(𝛼+1) 1 𝛽√ (𝑎2 2 −𝑎3 2 ) 2 𝑎3 𝑎2

(a)

(b)


4.6.5. ZK-MEW Equation Consider the ZK-MEW equation [201] 𝐷𝑡𝛼 𝑢 + 𝑎(𝑢3 )𝑥 + 𝑏𝑢𝑥𝑥𝑡 + 𝑟𝑢𝑥𝑦𝑦 = 0, 0 < 𝛼 < 1.

(4.147)

𝑡𝛼

Consider the transformation 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝜔 Γ(𝛼+1), we have Eq. (4.147) is 𝜔𝑈 ′ + 𝑎𝑘(𝑈 3 )′ + 𝑏𝜔𝑘 2 𝑈 ′′′ + 𝑟𝑘𝑙 2 𝑈 ′′′ = 0,

(4.148)

Integrate Eq. (4.148) once we have 𝜔𝑈 + 𝑎𝑘𝑈 3 + 𝑏𝜔𝑘 2 𝑈 ′′ + 𝑟𝑘𝑙 2 𝑈 ′′ = 0,

(4.149)

Applying the balancing term phenomena on Eq. (4.149) we obtained 𝑀 + 2 = 3𝑀, 𝑀 = 1. Then we have the trail solution is 𝑈(𝜉) =


,

(4.150)

Putting (4.150) in (4.149) and according to the purposed technique we obtained one solution set,

543

𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 𝑎2 , 𝑎3 = 0, 𝑎4 = 0, 𝑘 = 2

2 2 𝜇2

1 𝑎𝑎 −2𝑟𝑎 𝑙 𝜔 = − 2 √− 2 𝑏𝑎 1

2 2 2 2 √−𝑎𝑎2 −2𝑟𝑎1 𝑙 𝜇 𝑏𝑎

𝜇𝑎2

, 𝜇 = 𝜇,

.

Substituting in (4.150) we have 𝑎2

𝑢(𝑥, 𝑡) = 2 2 2 2 √−𝑎𝑎2 −2𝑟𝑎1 𝑙 𝜇 𝑏𝑎

𝑎1 cosh 𝜇

𝜇𝑎2

.

2 2 𝑡𝛼 𝑎𝑎2 −2𝑟𝑎2 1 1𝑙 𝜇 𝑥+𝑙𝑦− √− 2 2

𝑏𝑎

Γ(𝛼+1)

( (

))

2nd Solution Set: 𝑎0 =

20 9

5

1

𝑎1 , 𝑎1 = 𝑎1 , 𝑎2 = 4 𝑎3 , 𝑎3 = 𝑎3 , 𝑎4 = 0, 𝑘 = 𝑎

3𝜇

1 9𝑎𝑎32 −32𝑎12 𝑙2 𝜇 2 𝑟

9

𝜔 = − 32𝜇𝑎2 √− 9

𝑏𝑎

1

2

2 2 𝜇2𝑟

√− 1 9𝑎𝑎3 −32𝑎1 𝑙 9

𝑏𝑎

𝑎3 𝑎, 𝑙 = 𝑙, 𝜇 = 𝜇.

Substitute in (4.150) we have 2 2 2 2 5 1 √ 19𝑎𝑎3 −32𝑎1 𝑙2 𝜇2 𝑟 9 √ 19𝑎𝑎3 −32𝑎1 𝑙2 𝜇2 𝑟 𝑡𝛼 𝑎 sinh(𝜇( − 𝑥+𝑙𝑦− − ))+ 4 3 𝑎3 𝜇 9 𝑏𝑎 9 𝑏𝑎 Γ(𝛼+1) 32𝜇𝑎2 1

𝑎3 cosh(𝜇(

𝑈(𝜉) =

2 2 2 2 1 √ 19𝑎𝑎3 −32𝑎1 𝑙2 𝜇2 𝑟 9 √ 19𝑎𝑎3 −32𝑎1 𝑙2 𝜇2 𝑟 𝑡𝛼 − 𝑥+𝑙𝑦− − )) 2 𝑎3 𝜇 9 𝑏𝑎 9 𝑏𝑎 Γ(𝛼+1) 32𝜇𝑎1

.

2 2 2 2 20 1 √ 19𝑎𝑎3 −32𝑎1 𝑙2 𝜇2 𝑟 9 √ 19𝑎𝑎3 −32𝑎1 𝑙2 𝜇2 𝑟 𝑡𝛼 +𝑎1 sinh(𝜇( − 𝑥+𝑙𝑦− − ))+ 2 9 𝑎3 𝜇 9 𝑏𝑎 9 𝑏𝑎 Γ(𝛼+1) 32𝜇𝑎1

cosh(𝜇(

2 2 2 2 1 √ 19𝑎𝑎3 −32𝑎1 𝑙2 𝜇2 𝑟 9 √ 19𝑎𝑎3 −32𝑎1 𝑙2 𝜇2 𝑟 𝑡𝛼 − 𝑥+𝑙𝑦− − )) 𝑎3 𝜇 9 𝑏𝑎 9 𝑏𝑎 Γ(𝛼+1) 32𝜇𝑎2 1

3rd Solution set: 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑎3 = 𝑎3 , 𝑎4 = 0, 𝑘 = 1

𝜔 = 2 √−

𝑎𝑎32 +2𝑟𝑎12 𝑙2 𝜇 2 𝑏𝑎12

2 2 2 2 √−𝑎𝑎3 +2𝑟𝑎1 𝑙 𝜇 𝑏𝑎

𝜇𝑎3

,

𝑎𝑎3 .

Substituting in (4.150) we have 𝑎3

𝑢(𝑥, 𝑡) = 𝑎1 cosh 𝜇

2 2 2 2 √−𝑎𝑎3 +2𝑟𝑎1 𝑙 𝜇 𝑏𝑎 𝜇𝑎3

. 1 2

𝑥+𝑙𝑦+ √−

( (

2 2 2 𝑎𝑎2 𝑡𝛼 3 +2𝑟𝑎1 𝑙 𝜇 𝑎𝑎 3 Γ(𝛼+1) 2 𝑏𝑎1

))

4th Solution Set:

544

,

√−

𝑎0 =

𝑎1 𝑎2 𝑎3 𝑎22 −𝑎32 √−

𝜔=

, 𝑎1 = 𝑎1 , 𝑎2 = 𝑎2 , 𝑎3 = 𝑎3 , 𝑎4 = 0, 𝑘 =

2 2 2 2 𝑎𝑎2 2 −𝑎𝑎3 −2𝑟𝑎1 𝑙 𝜇 2𝑏 𝑎𝑎2 𝑏−𝑎𝑎 2 3

𝜇𝑎12

2 2 2 2 𝑎𝑎2 2 −𝑎𝑎3 −2𝑟𝑎1 𝑙 𝜇 2 2 𝑎𝑎2 𝑏−𝑎𝑎3 𝑏

𝜇

,

, 𝜇 = 𝜇, 𝑙 = 𝑙.

Substituting in (4.150) we have 2 2 2 2 𝑎𝑎2 2 −𝑎𝑎3 −2𝑟𝑎1 𝑙 𝜇 2 2 𝑎𝑎2 𝑏−𝑎𝑎3 𝑏

√–

𝑎2 sinh 𝜇

𝑈(𝜉) =

(

𝜇

√–

𝑥+𝑙𝑦

2 2 2 𝑎𝑎2 −𝑎𝑎2 3 −2𝑟𝑎1 𝑙 𝜇 √− 2 2 2 𝑎𝑎2 𝑏−𝑎𝑦3 𝑏 𝑡𝛼 + Γ(𝛼+1)) 𝜇𝑎2 ( 1 ) 2 2 2 𝑎𝑎2 −𝑎𝑎2 3 −2𝑟𝑎1 𝑙 𝜇 √– 2 2𝑏 𝑎𝑎2 𝑏−𝑎𝑎 2 3

𝑎1 𝑎2 𝑎3 2 +𝑎1 sinh 𝜇 𝑎2 2 −𝑎3

𝜇

+𝑎3 cosh 𝜇

(

𝜇

(a)

𝑥+𝑙𝑦

2 2 2 𝑎𝑎2 −𝑎𝑎2 3 −2𝑟𝑎1 𝑙 𝜇 √− 2 2 2 𝑎𝑎2 𝑏−𝑎𝑎3 𝑏 𝑡𝛼 + Γ(𝛼+1)) 𝜇𝑎2 ( 1 ) 2 2 2 𝑎𝑎2 −𝑎𝑎2 3 −2𝑟𝑎1 𝑙 𝜇 √– 2 2𝑏 𝑎𝑎2 𝑏−𝑎𝑎 2 3 𝑥+𝑙𝑦 𝜇

𝑥+𝑙𝑦

2 2 2 𝑎𝑎2 −𝑎𝑎2 3 −2𝑟𝑎1 𝑙 𝜇 √− 2 2 2 𝑎𝑎2 𝑏−𝑎𝑎3 𝑏 𝑡𝛼 + Γ(𝛼+1)) 𝜇𝑎2 ( 1

2 2 2 2 𝑎𝑎2 2 −𝑎𝑎3 −2𝑟𝑎1 𝑙 𝜇 2 2 𝑎𝑎2 𝑏−𝑎𝑎3 𝑏

cosh 𝜇

√−

( (

+

.

2 2 2 2 𝑎𝑎2 2 −𝑎𝑎3 −2𝑟𝑎1 𝑙 𝜇 2 2 𝑎𝑎2 𝑏−𝑎𝑎3 𝑏 𝑡𝛼 Γ(𝛼+1)) 𝜇𝑎2 1 )

(b)


4.6.6. Fisher’s Equations Consider the Fisher’s equation [230] 𝛽

2𝛽

𝐷𝑡𝛼 𝑢 − 𝑢𝐷𝑥 𝑢 − 𝐷𝑥 𝑢 − 𝑢(1 − 𝑢) = 0.

(4.151)

Consider the transformation 𝜉 = 𝑘𝑥 + 𝜔𝑡, we have Eq. (4.151) is 𝜔𝑈 ′ − 𝑘𝑈𝑈 ′ − 𝑘 2 𝑈 ′′ − 𝑈(1 − 𝑈) = 0. Then we have the trail solution is

545

(4.152)

𝑈(𝜉) =


,

(4.153)

Putting (4.153) in (4.152) and according to the purposed technique we obtained two solution set, 1st Solution Set: 2

1

𝜔 = 𝜇 , 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = 𝑎3 , 𝑎3 = 𝑎3 , 𝑎4 = 0, 𝑘 = 𝜇. Substituting in (4.153) we obtained, 𝑎3 cosh(

𝑢(𝑥, 𝑡) =

𝑥𝛽 𝑡𝛼 +2 ) Γ(𝛽+1) Γ(𝛼+1)

.

𝑎0

(a)

(b)


4.6.7. Gas Dynamical Equation Consider the Gas Dynamical equation [90] 1

𝐷𝑡𝛼 𝑢 + 2 𝐷𝑥𝛼 𝑢2 − 𝑢(1 − 𝑢) = 0, 0 < 𝛼 < 1. 𝑘𝑥 𝛼

(4.154)

𝜔𝑡 𝛼

Consider the transformation 𝜉 = Γ(𝛼+1) + Γ(𝛼+1), we have Eq. (4.154) is 𝑘

𝜔𝑈 ′ + 2 (𝑈 2 )′ − 𝑈(1 − 𝑈) = 0.

(4.155)

Then we have the trail solution is 𝑈(𝜉) =


,

(4.156)

Putting (4.156) in (4.155) and according to the purposed technique we obtained the following solution set,

546

1

1

𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = −𝑎3 , 𝑎3 = 𝑎3 . 𝑎4 = 0, 𝜔 = − 𝜇 , 𝑘 = 𝜇. Substituting in (4.156) we have 𝑢(𝑥, 𝑡) =

𝑥𝛼 𝑡𝛼 𝑥𝛼 𝑡𝛼 )+𝑎3 cosh( ) − − Γ(𝛼+1) Γ(𝛼+1) Γ(𝛼+1) Γ(𝛼+1)

−𝑎3 sinh(

.

𝑎0

(a)

(b)


4.6.8. MKdV Equation Consider the MKdV equation [230] 𝐷𝑡𝛼 𝑢 + 𝛼𝑢2 𝑢𝑥 + 𝑢𝑥𝑥𝑥 = 0,

0 < 𝛼 < 1.

(4.157)

𝜔𝑡 𝛼

Consider the transformation 𝜉 = 𝑘𝑥 + Γ(𝛼+1), we have 𝜔𝑈 ′ + 𝑎𝑘𝑈 2 𝑈 ′ + 𝑘 3 𝑈 ′′′ = 0, integrate once we have, 𝑎

𝜔𝑈 + 3 𝑘𝑈 3 + 𝑘 3 𝑈 ′′ = 0.

(4.158)

Then we have the trail solution is, 𝑈(𝜉) =


,

(4.159)

put(4.159) in (4.158) and according to the purposed technique we obtained the following solution set, 1st Solution Set: 1 √6√𝑎𝑎2

𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎1 = 𝑎2 , 𝑎2 = 𝑎3 = 0, 𝑎4 = 0, 𝜇 = 𝜇, 𝑘 = 6 3

1 √6𝑎2 𝑎3 − 36 𝑎3 𝜇 2 , 1

547

𝑎1 𝜇

,𝜔 =

substituting in (4.159) we have 𝑎2

𝑢(𝑥, 𝑡) = 𝑎1 cosh(𝜇(

.

3 1 √6𝑎𝑎𝑎2 𝑡𝛼 √6𝑎 𝑎 𝑥− )) 𝑎1 𝜇 2 36 𝑎3 𝜇 Γ(𝛼+1) 1

2nd Solution Set: 𝑎0 = 1 6 3 𝑎1 𝜇

𝑎2 𝑎3 𝑎1 𝑎2 −𝑎3 1 6

𝑎(𝑎22 −𝑎32 )√ 𝑎𝑎22 − 𝑎𝑎32

1

1

, 𝑎1 = 𝑎1 , 𝑎2 = 𝑎2 , 𝑎3 = 𝑎3 , 𝑎4 = 0, 𝜇 = 𝜇, 𝑘 = √6 𝑎𝑎22 − 6 𝑎𝑎32 , 𝜑 = .

substitute in (4.159) we have 𝑎2 sinh 𝜇(

1 1 1 2 2 2 1 √ 𝑎𝑎2 − 𝑎𝑎2 1𝑎(𝑎2 −𝑎3 )√6𝑎𝑎2 −6𝑎𝑎3 𝑡𝛼 6 6 3 𝑥− ) + 3 𝑎1 𝜇 6 Γ(𝛼+1) 𝑎1 𝜇

(

)

1 1 1 2 2 2 1 √ 𝑎𝑎2 − 𝑎𝑎2 1𝑎(𝑦2 −𝑎3 )√6𝑎𝑎2 −6𝑎𝑎3 𝑡𝛼 6 6 3 𝑎3 cosh 𝜇( 𝑥− ) 𝑎1 𝜇 6 Γ(𝛼+1) 𝑎3 1𝜇

[

𝑢(𝑥, 𝑡) =

)]

(

.

1 1 1 2 2 2 1 √ 𝑎𝑎2− 𝑎𝑎2 𝑎1 𝑎2 𝑎3 1𝑎(𝑎2 −𝑎3 )√6𝑎𝑎2 −6𝑎𝑎3 𝑡𝛼 6 6 3 +𝑎1 sinh 𝜇( 𝑥− ) 2 2 3 𝑎 𝜇 6 Γ(𝛼+1) 𝑎2 −𝑎3 𝑎1 𝜇 1

(

)

1 1 1 2 2 2 1 √ 𝑎𝑎2 − 𝑎𝑎2 1𝑎(𝑎2 −𝑎3 )√6𝑎𝑎2−6𝑎𝑎3 𝑡𝛼 6 6 3 cosh 𝜇( 𝑥− ) 3 𝑎1 𝜇 6 Γ(𝛼+1) 𝑎1 𝜇

[

3rd

(

)

]

Solution Set: 1 6

− 𝑎𝑎3

𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑎3 = 𝑎3 , 𝑎4 = 0, 𝜇 = 𝜇, 𝑘 = √ substitute in (4.159) we obtained 𝑢(𝑥, 𝑡) =

𝑎3 3 1 √𝑎𝑎3 𝑎 𝑡𝛼 √𝑎𝑎3 𝑎0 +𝑎1 sinh(𝜇( 𝑥+ )) 6√−6𝑎3 𝜇Γ(𝛼+1) √−6𝑎1 𝜇 1

,

548

𝑎1 𝜇

,𝜑 =

1 1 √− 𝑎 𝑎33 𝑎 6 6 𝑎13 𝜇

,

(a)

(b)

Fig.4.29. (a)-(b). Travelling wave solution of MKdV equation for different values of parameters.

4.6.9. Nizhnik-Novikov-Vesselov System Consider the Nizhnik-Novikov-Vesselov equation [65] 𝐷𝑡𝛼 𝑢 + 𝑎𝑢𝑥𝑥𝑥 + 𝑏𝑢𝑦𝑦𝑦 − 3𝑎(𝑢𝑣)𝑥 − 3𝑏(𝑢𝑤)𝑦 = 0,

(4.160)

𝑢𝑥 = 𝑣𝑦 ,

(4.161)

𝑢𝑦 = 𝑤𝑥 .

(4.162) 𝑡𝛼

Introducing the transformation as 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝜔 Γ(𝛼+1), we can convert the above equations into ordinary differential equations, 𝜔𝑢′ + 𝑎𝑘 3 𝑢′′′ + 𝑏𝑙 3 𝑢′′′ − 3𝑎𝑘(𝑢𝑣)′ − 3𝑏𝑙(𝑢𝑤)′ = 0, 𝑘𝑢′ = 𝑙𝑣 ′ , 𝑙𝑢′ = 𝑘𝑤 ′ . Integrating once the above equation 𝜔𝑢 + 𝑎𝑘 3 𝑢′′ + 𝑏𝑙 3 𝑢′′ − 3𝑎𝑘(𝑢𝑣) − 3𝑏𝑙(𝑢𝑤) = 0.

(4.163)

𝑘𝑢 = 𝑙𝑣,

(4.164)

𝑙𝑢 = 𝑘𝑤.

(4.165)

Using (4.165) and (4.164) in (4.163) we have, 𝑎𝑘 2

𝑐𝑢 + (𝑘 3 𝑎 + 𝑙 3 𝑏)𝑢′′ − 3 (

𝑙

+

𝑏𝑙2 𝑘

) 𝑢2 = 0.

(4.166)

Consider the trial solution 𝑢(𝜉) =


549

,

(4.167)

put (4.167) in (4.166) and according to the purposed technique we obtained the following solution sets 1st Solution Set: 𝜔 = −4𝜇 2 𝑘 3 𝑎 − 4𝜇 2 𝑙 3 𝑏, 𝑘 = 𝑘, 𝑙 = 𝑙, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎3 = 0, 𝑎4 = −2𝜇 2 𝑙𝑘𝑎1 . Substitute in (4.167), we obtained, −2𝜇 2 𝑙𝑘𝑎1 tanh(𝜇𝜉)

𝑢(𝜉) = 𝑎

1 sinh(𝜇𝜉) cosh(𝜇𝜉)

, 𝑡𝛼

where 𝜉 = 𝑘𝑥 + 𝑙𝑦 + (−4𝜇 2 𝑘 3 𝑎 − 4𝜇 2 𝑙 3 𝑏) Γ(𝛼+1). The solution of 𝑣(𝑥, 𝑦, 𝑡) and 𝑤(𝑥, 𝑦, 𝑡) can be obtained by using the Eq. (4.164) and Eq. (4.165)

(a)

(b)

Fig.4.30. (a)-(b). Travelling wave solution of Nizhnik-Novikov-Vesselov equation for different values of parameters

2nd Solution Set: 1 1 2 3 𝑎 3 1 𝑏𝑙27 (𝑏2 )

𝜔 = 0, 𝑘 = − 3 1 𝑎3 √27𝑎

−9

𝑏 𝑎2 √ 2 𝑏

𝑎

, 𝑙 = 𝑙. 𝑎0 = 0, 𝑎1 =

1 1 2 6 𝑎 276 ( 2 ) 𝑎3 𝑏

𝜇2 𝑙2

, 𝑎2 = 0, 𝑎3 = 𝑎3 , 𝑎4 =

,

proceeding as before we get the solution of 𝑢(𝑥, 𝑦, 𝑡), 𝑣(𝑥, 𝑦, 𝑡) and 𝑤(𝑥, 𝑦, 𝑡).

550

(a)

(b)


3rd Solution Set: 𝑎

𝑎3 𝑎 3 √3√8√ 𝑑 3

𝜔 = − 64

𝑎13 𝜇

𝑎

𝑎 1 √3√8√𝑑 3

,𝑘 = 8

𝑎1 𝜇

20

, 𝑎0 = −

9

5

𝑎1 , 𝑎1 = 𝑎1 , 𝑎2 = − 4 𝑎3 , 𝑎3 = 𝑎3 , 𝑎4 =

0, substitute in (4.167) we obtained, 𝑢(𝜉) =

5 4 20 − 𝑎1 +𝑎1 sinh(𝜇𝜉) cosh(𝜇𝜉) 9

− 𝑎3 sinh(𝜇𝜉)+𝑎3 cosh(𝜇𝜉)

𝑎

𝑎 1 √3√8√ 𝑑 3

where 𝜉 = 8

𝑎1 𝜇

𝑎

𝑎3 𝑎 3 √3√8√𝑑 3

𝑥 − 64

𝑎13 𝜇

, 𝑡𝛼

,

Γ(𝛼+1)

the solution of 𝑣(𝑥, 𝑦, 𝑡) and 𝑤(𝑥, 𝑦, 𝑡) can be obtained by using the Eq. (4.164) and Eq. (4.165)

551

(a)

(b)


4.6.10. Konopelchenko-Dubrovsky Equations Consider the KD equation [254] as 3

𝛽

𝐷𝑡𝛼 𝑢 − 𝑢𝑥𝑥𝑥 − 6𝑏 2 𝑢𝑢𝑥 + 2 𝑎2 𝑢2 𝑢𝑥 − 3𝐷𝑦 𝑣 + 3𝑎𝑢𝑥 𝑣 = 0, 𝛽

𝐷𝑦 𝑢 = 𝑣𝑥 ,

0 < 𝛼, 𝛽 < 1.

(4.168) (4.169)

𝑦𝛽

𝑡𝛼

Using the transformation 𝜉 = 𝑥 + Γ(𝛽+1) + 𝜔 Γ(𝛼+1), the above equation we have 3

𝜔𝑢′ − 𝑢′′′ − 6𝑏 2 𝑢𝑢′ + 2 𝑎2 𝑢2 𝑢′ − 3𝑣 ′ + 3𝑎𝑢′ 𝑣 = 0,

(4.170)

𝑢′ = 𝑣 ′ .

(4.171)

Integrating the second equation in the system and neglecting constant of integration we find 𝑢 = 𝑣.

(4.172)

Using expression (4.172) in (4.170) and then integrating once, we get 1

3

2

2

𝜔𝑢(𝜉) − 𝑢′′ (𝜉) − 3𝑏 2 𝑢(𝜉)2 + 𝑎2 𝑢(𝜉)3 − 3𝑢(𝜉) + 𝑎𝑢(𝜉)2 = 0,

(4.173)

consider the trial solution 𝑢(𝜉) =


,

(4.174)

put (4.174) in (4.173) and according to the purposed technique we obtained the following solution sets 1st Solution Set: 552

𝜔 = 2𝜇 2 + 3, 𝑎0 =

𝑙𝑏 2 𝑎4 𝜇

, 𝑎1 = 0, 𝑎2 = 0, 𝑎3 = 0, 𝑎4 = 𝑎4 , 𝑎 = 2𝑏 2 .

Substitute in (4.174) we obtained, 𝑢(𝜉) =

𝜇 2𝑏 2 tanh(𝜇𝜉) i 𝑏 2 𝑎4 𝑎0

,

𝑦𝛽

𝑡𝛼

where 𝜉 = 𝑥 + Γ(𝛽+1) + (2𝜇 2 + 3) Γ(𝛼+1). The solution of 𝑣(𝑥, 𝑡) obtained using the Eq. (4.172) 2nd Solution Set: 2

𝜔 = 3 − 𝜇 , 𝑎0 =

𝑏 2 𝑎3 𝑎2 𝜇√𝑎22 −𝑎32

, 𝑎1 =

√𝑎22 −𝑎32 𝑏2 𝜇

, 𝑎2 = 𝑎2 , 𝑎3 = 𝑎3 , 𝑎4 = 0, 𝑎 = 2𝑏 2 .

Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡). 3rd Solution Set: 5 𝑏 2 𝑎3

𝜔 = −𝜇 2 + 3, 𝑎0 = 3

𝜇

3 𝑏 2 𝑎3

, 𝑎1 = − 4

𝜇

5

, 𝑎2 = − 4 𝑎3 , 𝑎3 = 𝑎3 , 𝑎4 = 0, 𝑎 = 2𝑏 2 .

Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡). 4th Solution Set: 5 𝑏 2 𝑎3

𝜔 = 3 − 𝜇 2 , 𝑎0 = − 3

𝜇

3 𝑏 2 𝑎3

, 𝑎1 = 4

𝜇

5

, 𝑎2 = − 4 𝑎3 , 𝑎3 = 𝑎3 , 𝑎4 = 0, 𝑎 = 2𝑏 2 .

Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡). 5th Solution Set: 𝜔 = −𝜇 2 + 3, 𝑎0 = 0, 𝑎1 = −

𝑏 2 𝑎2 𝜇

, 𝑎2 = 𝑎2 , 𝑎3 = 0, 𝑎4 = 0, 𝑎 = 2𝑏 2 .

Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡). 6th Solution Set: 𝜔 = −𝜇 2 + 3, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2

𝜇𝑎1 𝑏2

, 𝑎3 = 0, 𝑎4 = 0, 𝑎 = 2𝑏 2 .

Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡). 7th Solution Set: 𝜔 = −𝜇 2 + 3, 𝑎0 = 0, 𝑎1 =

𝑙𝑎3 𝑏 2 𝜇

, 𝑎2 = 0, 𝑎3 , 𝑎3 = 𝑎3 , 𝑎4 = 0, 𝑎 = 2𝑏 2 .

Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡).

553

(a)

(b)

Fig.4.33. (a)-(b). Travelling wave solution of Konopelchenko_Dubrovsky equation for different values of parameters

4.6.11. Nonlinear System Consider the nonlinear system [254] of the third order 𝐷𝑡𝛼 𝑢𝑥 + 𝑣𝑥 𝐷𝑡𝛼 𝑣 = 0, 𝑣𝑡 + 𝑣𝑥𝑥𝑥 + (𝑣𝑥 )3 + 3𝑣𝑥 𝑢𝑥𝑥 = 0,

(4.175) 0 < 𝛼 < 1.

(4.176)

𝑡𝛼

Introducing the transformation as 𝜉 = 𝑘𝑥 + 𝜔 Γ(𝛼+1), we can convert the given equation into ordinary differential equations 𝑘𝜔𝑢′′ + 𝑘𝜔(𝑣 ′ )2 = 0, 𝜔𝑣 ′ + 𝑘 3 𝑣 ′′′ + (𝑘𝑣 ′ )3 + 3𝑘 3 𝑣 ′ 𝑢′′ = 0. From the first equation of the system

let

𝑢′′ = −(𝑣 ′ )2 ,

(4.177)

ℎ = 𝑣′.

(4.178)

Using this equation in the second equation of the system we get 𝜔ℎ + 𝑘 3 ℎ′′ − 2𝑘 3 ℎ3 = 0.

(4.179)

Consider the trial solution ℎ(𝜉) =


,

(4.180)

put (4.180) in (4.179) and according to the purposed technique we obtained the following solution sets 1st Solution Set: 554

𝜔 = −𝜇 2 𝑘 3 , 𝑘 = 𝑘, 𝑎0 = 0, 𝑎1 =

𝑙𝑎2 𝜇

, 𝑎2 = 𝑎2 , 𝑎3 = 0, 𝑎4 = 0.

Substitute in (4.178) we obtained, 𝜇

ℎ(𝜉) = i cosh(𝜇𝜉), where

𝑡𝛼

𝜉 = 𝑘𝑥 − 𝜇 2 𝑘 3 Γ(𝛼+1).

The solution of 𝑣(𝑥, 𝑡) and 𝑢(𝑥, 𝑡) obtained using the Eq. (4.177) and Eq. (4.178) 2nd Solution Set: 𝜔 = −𝜇 2 𝑘 3 , 𝑘 = 𝑘, 𝑎0 = −

𝑎2 𝑎3 𝜇√−𝑎22 +𝑎32

, 𝑎1 =

√−𝑎22 +𝑎32 𝜇

, 𝑎2 = 𝑎2 , 𝑎3 = 𝑎3 , 𝑎4 = 0.

Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡). 3rd Solution Set: 2 3

𝜔 = −𝜇 𝑘 , 𝑘 = 𝑘, 𝑎0 = −

5 𝑙𝑎 3 3

𝜇

, 𝑎1 =

3 𝑙𝑎 4 3

𝜇

5

, 𝑎2 = − 4 𝑎3 , 𝑎3 = 𝑎3 , 𝑎4 = 0.

Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡). 4th Solution Set: 𝜔 = −𝜇 2 𝑘 3 , 𝑘 = 𝑘, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑎3 = 𝜇𝑎1 , 𝑎4 = 0. Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡). 5th Solution Set: 𝜔 = −𝜇 2 𝑘 3 , 𝑘 = 𝑘, 𝑎0 = 0, 𝑎1 = 𝑎1 , 𝑎2 = 0, 𝑎3 = −𝜇𝑎1 , 𝑎4 = 0. Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡). 6th Solution Set: 𝜔 = 2𝜇 2 𝑘 3 , 𝑘 = 𝑘, 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = 0, 𝑎3 = 0, 𝑎4 = 𝜇𝑎0 . Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡). 7th Solution Set: 𝜔 = 2𝜇 2 𝑘 3 , 𝑘 = 𝑘, 𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑎2 = 0, 𝑎3 = 0, 𝑎4 = −𝜇𝑎0 . Proceeding as before we get the solution of 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡).

555

(a)

(b)

Fig.4.34. (a)-(b). Travelling wave solution of nonlinear system of the third order equation for different values of parameters

4.7. U-expansion Method As discussed in Chapter 3, U-Expansion method has been developed by Usman et. al. in 2013 to find the solitary wave solution of nonlinear partial differential equations. In this method we use first order nonlinear ordinary differential equation 𝑈𝜉 = 𝛼1 𝑈 + 𝛼2 𝑈 2 , as a Riccati equation. In this thesis I suggested a new approach for finding the solitary wave solutions of nonlinear partial differential equations of fractional-order. In the proposed algorithm we use the 𝑈𝜉 𝛼 = 𝛼1 𝑈 + 𝛼2 𝑈 2 , where 0 < 𝛼 ≤ 1, as a Riccati equation. The proposed technique is highly compatible for solving nonlinear PDEs of fractional order. As we know that in U-Expansion method the solution of 𝑈𝜉 = 𝛼1 𝑈 + 𝛼2 𝑈 2 , plays an important role, so before analysis of U-Expansion method we first find the exact solution of 𝑈𝜉 𝛼 = 𝛼1 𝑈 + 𝛼2 𝑈 2 , therefore we consider 𝐷𝜉𝛼 𝑈 = 𝛼1 𝑈 + 𝛼2 𝑈 2 ,

0 < 𝛼 ≤ 1,

(4.181)

subject to the initial condition 𝑈(0) = 𝑓. Applying 𝐽𝛼 on both sides of Eq. (4.110) and using the given initial condition we have 𝑈(𝜉) = 𝑈(0) + 𝐽𝛼 [𝛼1 𝑈 + 𝛼2 𝑈 2 ]. Following the discussion presented in the decomposition method section, we can obtain the recurrence relation 556

𝑈0 (𝜉) = 𝑈(0) = 𝑓, 𝑈𝑘+1 (𝜉) = 𝐽𝛼 [𝛼1 𝑈𝑘 + 𝛼2 𝑈𝑘2 ]. Case 1: Consider that 𝑓 =

1 2

, 𝛼1 = −2i and 𝛼2 = 2i we have the above recurrence

relation 1

𝑈0 (𝜉) = 𝑈(0) = 2, 1

i

𝑈1 (𝜉) = 𝐽𝛼 [−2i𝑈0 + 2i𝑈02 ] = − 2 Γ(𝛼+1) 𝑥 𝛼 , 𝑈2 (𝜉) = 𝐽𝛼 [−2i𝑈1 + 4i𝑈0 𝑈1 ] = 0, 1

i

Γ(2𝛼+1)

𝑈3 (𝜉) = 𝐽𝛼 [−2i𝑈2 + 2i𝑈12 + 4i𝑈0 𝑈2 ] = 2 Γ(𝛼+1)2 Γ(3𝛼+1) 𝑥 3𝛼 , 𝑈4 (𝜉) = 𝐽𝛼 [−2i𝑈3 + 4i𝑈0 𝑈3 + 4i𝑈1 𝑈2 ] = 0, 1

i

Γ(2𝛼+1) Γ(4𝛼+1)

𝑈5 (𝜉) = 𝐽𝛼 [−2i𝑈4 + 2i𝑈22 + 4i𝑈0 𝑈4 + 4i𝑈1 𝑈3 ] = 2 Γ(𝛼+1)3 Γ(3𝛼+1) Γ(5𝛼+1) 𝑥 3𝛼 , . . .. The series solution is given as 1

1

1

𝑈(𝜉) = − [ 2

2 Γ(𝛼+1)

𝑥𝛼 −

1

Γ(2𝛼+1)

Γ(𝛼+1)2

Γ(3𝛼+1)

𝑥 3𝛼 +

1

Γ(2𝛼+1) Γ(4𝛼+1)

Γ(𝛼+1)3

Γ(3𝛼+1) Γ(5𝛼+1)

𝑥 3𝛼 − ⋯ ].

The closed form solution is given as i 𝐸 (i𝜉 𝛼 )−𝐸 (−i𝜉 𝛼 )

1

1

i

𝑈(𝜉) = 2 − 2 𝐸𝛼 (i𝜉𝛼 )+𝐸𝛼(−i𝜉𝛼) = 2 − 2 tan𝛼 (𝜉). 𝛼

𝛼

Case 2: Consider that 𝑓 =

1

i

+ 2 , 𝛼1 = −i and 𝛼2 = i proceeding as before we get 2

1

𝐸 (i𝜉 𝛼 )−𝐸 (−i𝜉 𝛼 )

2

𝑈(𝜉) = 2 [1 + i 𝐸

𝛼 𝛼 𝛼 (i𝜉 )+𝐸𝛼 (−i𝜉 )

1

− i 𝐸𝛼 (i𝜉𝛼 )+𝐸𝛼 (−i𝜉𝛼 )] = 2 (1 + i sec(𝜉) − 𝛼

𝛼

i tan𝛼 (𝜉)). Case 3: Consider that 𝑓 = −1, 𝛼1 = 4 and 𝛼2 = 2 proceeding as before we get 𝐸 (𝜉 𝛼 )−𝐸 (−𝜉 𝛼 )

1

𝐸 (𝜉 𝛼 )+𝐸 (−𝜉 𝛼 )

1

𝑈(𝜉) = − 2 [2 + 𝐸𝛼 (𝜉𝛼 )+𝐸𝛼 (−𝜉𝛼 ) + 𝐸𝛼 (𝜉𝛼 )−𝐸𝛼(−𝜉𝛼)] = −1 − 2 tanh𝛼 (𝜉) − 𝛼

1 2

𝛼

𝛼

𝛼

coth𝛼 (𝜉).

Case 4: Consider that 𝑓 = −1, 𝛼1 = −√3i and 𝛼2 = √3i proceeding as before we get 1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

𝑈(𝜉) = 2 − 2

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

,

557

where 𝐸𝛼 denote the Mittage-Leffler function [199], given as 𝑧𝑘

𝐸𝛼 (𝑧) = ∑∞ 𝑘=0 Γ(𝛼𝑘+1) .

4.7.1. Methodology In order to simultaneously obtain more periodic wave solutions expressed in rational hyperbolic function and rational trigonometry function to nonlinear equations, we introduce U-expansion Method. We briefly show what U-expansion method is and how to use it to obtain various periodic wave solutions to nonlinear equations. Suppose a nonlinear equation for 𝑢(𝑥, 𝑡) is given by We consider the general nonlinear PDE of the type 𝑃(𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑦 , 𝑢𝑧 , 𝑢𝑡𝑡 , 𝑢𝑥𝑥 , 𝑢𝑦𝑦 , 𝑢𝑥𝑥 , 𝑢𝑥𝑡 , 𝑢𝑦𝑡 , 𝑢𝑧𝑡 , 𝑢𝑥𝑦 , 𝑢𝑥𝑧 , 𝑢𝑦𝑧 , … ) = 0,

(4.182)

where 𝑃 is a polynomial in its arguments. The essence of the U-expansion method can be presented in the following steps: Step 1: Seek Solitary wave solutions of Eq. (4.182) by taking 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡, and transform Eq. (4.182) to the ordinary differential equation. 𝑄(𝑢, 𝜔𝑢′ , 𝑘𝑢′ , 𝑙𝑢′ , 𝑚𝑢′ , 𝜔2 𝑢′′ , 𝑘 2 𝑢′′ , 𝑙 2 𝑢′′ , … ) = 0,

(4.183)

where 𝜔 is constant and where prime denotes the derivative with respect to 𝜉. Step 2: If possible, integrate Eq. (4.183) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero. Step 3: According to U-Expansion method, we assume that the wave solution can be expressed in the following form 𝑛 𝑢(𝜉) = ∑𝑀 𝑛=0 𝑎𝑛 𝑈 ,

(4.184)

where 𝑈 is the solution of first order nonlinear equation in the form 𝑈𝜉 𝛼 = 𝛼1 𝑈 + 𝛼2 𝑈 2 .

(4.185)


558

Table 4.2: Solutions of Eq. (4.185) for different values of 𝛼1 and 𝛼2 . 𝛼1

𝛼2

Cases

𝑈(𝜉)

−2i

2i

I

1 i − tan𝛼 (𝜉) 2 2

−i

I

II

4

2

III

−√3i √3i

IV

1 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)) 2 1 1 −1 − tanh𝛼 (𝜉) − coth𝛼 (𝜉) 2 2 √3 √3 i 𝐶1 sin𝛼 ( 2 𝜉) + 𝐶2 cos𝛼 ( 2 𝜉) 1

1 − 2 2

√3 √3 𝐶1 cos𝛼 ( 2 𝜉) + i𝐶2 sin𝛼 ( 2 𝜉)

Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term(s) of highest order with the highest order nonlinear term(s) in Eq. (4.182). Step 5: Substituting (4.184) into Eq. (4.183) with (4.185) will yields an algebraic equation involving power of U. Equating the coefficients of like power of U to zero gives a system of algebraic equations for 𝑎𝑖 , 𝑘, 𝑙, 𝑚 and 𝜔. Then, we solve the system with the aid of a computer algebra system (CAS), such as MAPLE 13, to determine these constants. Step 6: Putting these constant into Eq. (4.184), coupled with the well known solutions of Eq. (4.185), we can obtained the more general type and new exact traveling wave solution of the nonlinear partial differential equation (4.182).

4.7.2. 5th order Caudrey Dodd Gibbon Equation Consider the CDG equation of fraction order [230] 𝐷𝑡𝛼 𝑢 + 180𝑢2 𝐷𝑥𝛼 𝑢 + 30𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 30𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0,0 < 𝛼 ≤ 1.

(4.186)

To convert Eq. (12) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡,

(4.187)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.187) into Eq. (4.186) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝑘 𝛼 180𝑢2 𝐷𝜉𝛼 𝑢 + 30𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 30𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0. By applying the homogenous balancing principle for 𝛼 = 1, we have 559

(4.188)

𝑀 + 5 = 2𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.184), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑎2 𝑈 2 .

(4.189)

Putting Eq. (4.189) into Eq. (4.188) coupled with Eq. (4.187); the Eq. (4.189) yields an algebraic equation involving power of U as 𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 = 0. Compare the like powers of U we have system of equations 𝑈0:

𝜔𝛼 𝑎1 𝛼1 + 𝑘 5𝛼 𝑎1 𝛼15 + 180𝑘 𝛼 𝑎02 𝑎1 𝛼1 + 30𝑘 3𝛼 𝑎0 𝑎1 𝛼13 = 0,

𝑈1 :

𝜔𝛼 𝑎1 𝛼2 + 2𝜔𝛼 𝑎2 𝛼1 + 32𝑘 5𝛼 𝑎2 𝛼15 + ⋯ + 31𝑘 5𝛼 𝑎1 𝛼14 𝛼2 + 60𝑘 5𝛼 𝑎12 𝛼13 = 0,

𝑈2:

2𝜔𝛼 𝑎2 𝛼2 + 360𝑘 𝛼 𝑎02 𝑎2 𝛼2 + 360𝑘 𝛼 𝑎0 𝑎12 𝛼2 + ⋯ + 180𝑘 𝛼 𝑎13 𝛼1 = 0,

𝑈3:

720𝑘 𝛼 𝑎0 𝑎22 𝛼1 + 720𝑘 𝛼 𝑎12 𝑎2 𝛼1 + 510𝑘 3𝛼 𝑎12 𝛼1 𝛼22 + ⋯ + 480𝑘 3𝛼 𝑎22 𝛼13 = 0,

𝑈4:

720𝑘 𝛼 𝑎0 𝑎22 𝛼2 + 720𝑦𝛼 𝑎12 𝑎2 𝛼2 + 900𝑘 𝛼 𝑎1 𝑎22 𝛼1 + ⋯ + 240𝑘 3𝛼 𝑎12 𝛼23 = 0,

𝑈5:

120𝑘 5𝛼 𝑎1 𝛼25 + 900𝑘 𝛼 𝑎1 𝑎22 𝛼2 + 1200𝑘 3𝛼 𝑎1 𝛼23 𝑎2 + ⋯ + 360𝑘 𝛼 𝑎23 𝛼1 = 0,

𝑈6:

720𝑘 5𝛼 𝑎2 𝛼25 + 360𝑘 𝛼 𝑎23 𝛼2 + 1080𝑘 3𝛼 𝑎22 𝛼23 = 0.

Solving the above system for unknown parameters, we have the six solution sets 1st Solution Set: 2

1

2

2

𝑎0 = − 6 [e𝛼 ln(𝑘) ] 𝛼12 , 𝑎1 = −2[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 , 𝑎2 = −2[e𝛼 ln(𝑘) ] 𝛼22 , 𝑘 = 𝑘, 𝜔=e

ln(−𝛼4 1 )+5𝛼 ln(𝑘) 𝛼

.

Case I: Substituting the values of unknown into Eq. (4.189) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1

2

2

2

𝑢1 (𝜉) = − 6 [e𝛼 ln(𝑘) ] 𝛼12 − 2[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 2[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , 1

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.189) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 1

2

2

2


where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)).

560

Case III: Substituting the values of unknown into Eq. (4.189) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 2

1

2

2


1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.189) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 2

1

2

2

𝑢4 (𝜉) = − 6 [e𝛼 ln(𝑘) ] 𝛼12 − 2[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 2[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , √3

where 𝑈 = 2 − 2 e

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1

ln(−𝛼4 1 )+5𝛼 ln(𝑘) 𝛼

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1. In all cases 𝜉 = 𝑘𝑥 +

𝑡.

2nd Solution Set: 2

2

𝑎0 = 𝑎0 , 𝑎1 = −[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 , 𝑎2 = −[e𝛼 ln(𝑘) ] 𝛼22 , 𝑘 = 𝑘, 𝜔=e

2 𝛼 2 3𝛼 ln(−𝑘5𝛼 𝛼4 1 −180𝑘 𝑎0 −30𝑘 𝑎0 𝑎1 )+5𝛼 ln(𝑘) 𝛼

.


2

𝑢5 (𝜉) = 𝑎0 − [e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − [e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , 1

i


2


where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.189) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 2

2


1


2

2


√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

where 𝑈 = 2 − 2

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

and i = √−1. In all cases 𝜉 = 𝑘𝑥 + e

,

2 𝛼 2 3𝛼 ln(−𝑘5𝛼 𝛼4 1 −180𝑘 𝑎0 −30𝑘 𝑎0 𝑎1 )+5𝛼 ln(𝑘) 𝛼

𝑡.

3rd Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = − √−15𝑎0 +3√5𝑎0

(

𝛼1

(−15𝑎0 +3√5𝑎0 )𝛼2 𝛼1

, 𝑎2 = −

(−15𝑎0 +3√5𝑎0 )𝛼22 𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.

Case I: Substituting the values of unknown into Eq. (4.189) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢9 (𝜉) = 𝑎0 − 1

(−15𝑎0 +3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 +3√5𝑎0 )𝛼22 𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.189) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢10 (𝜉) = 𝑎0 −

(−15𝑎0 +3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 +3√5𝑎0 )𝛼22 𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.189) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢11 (𝜉) = 𝑎0 −

(−15𝑎0 +3√5𝑎0 )𝛼2 𝛼1

1

𝑈−

(−15𝑎0 +3√5𝑎0 )𝛼22 𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.189) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢12 (𝜉) = 𝑎0 −

(−15𝑎0 +3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 +3√5𝑎0 )𝛼22

562

𝛼12

𝑈2,

1

√3

√−15𝑎0 +3√5𝑎0

(

𝛼1

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1. In all cases 𝜉 =

1 𝛼

) 𝑥.

4th Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = −

(−

√−15𝑎0 +3√5𝑎0 𝛼1

(−15𝑎0 +3√5𝑎0 )𝛼2 𝛼1

, 𝑎2 = −

(−15𝑎0 +3√5𝑎0 )𝛼22 𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.


(−15𝑎0 +3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 +3√5𝑎0 )𝛼22 𝛼12

𝑈2,

i


(−15𝑎0 +3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 +3√5𝑎0 )𝛼22 𝛼12

𝑈2,

1


(−15𝑎0 +3√5𝑎0 )𝛼2 𝛼1

1

1

2

2

𝑈−

(−15𝑎0 +3√5𝑎0 )𝛼22 𝛼12

𝑈2,

where 𝑈 = −1 − tanh𝛼 (𝜉) − coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.189) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢16 (𝜉) = 𝑎0 −

(−15𝑎0 +3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 +3√5𝑎0 )𝛼22

563

𝛼12

𝑈2,

1

√3

(−

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√−15𝑎0 +3√5𝑎0 𝛼1

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


1 𝛼

) 𝑥.

5th Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = − √−15𝑎0 −3√5𝑎0

(

𝛼1

(−15𝑎0 −3√5𝑎0 )𝛼2 𝛼1

, 𝑎2 = −

(−15𝑎0 −3√5𝑎0 )𝛼22 𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.


(−15𝑎0 −3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 −3√5𝑎0 )𝛼22 𝛼12

𝑈2,

i


(−15𝑎0 −3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 −3√5𝑎0 )𝛼22 𝛼12

𝑈2,

1


(−15𝑎0 −3√5𝑎0 )𝛼2 𝛼1

1

1

2

2

𝑈−

(−15𝑎0 −3√5𝑎0 )𝛼22 𝛼12

𝑈2,


(−15𝑎0 −3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 −3√5𝑎0 )𝛼22

564

𝛼12

𝑈2,

1

√3

(−

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√−15𝑎0 −3√5𝑎0 𝛼1

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


1 𝛼

) 𝑥.

6th Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = −

(−

√−15𝑎0 −3√5𝑎0 𝛼1

(−15𝑎0 −3√5𝑎0 )𝛼2 𝛼1

, 𝑎2 = −

(−15𝑎0 −3√5𝑎0 )𝛼22 𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.


(−15𝑎0 −3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 −3√5𝑎0 )𝛼22 𝛼12

𝑈2,

i


(−15𝑎0 −3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 −3√5𝑎0 )𝛼22 𝛼12

𝑈2,

1


(−15𝑎0 −3√5𝑎0 )𝛼2 𝛼1

1

1

2

2

𝑈−

(−15𝑎0 −3√5𝑎0 )𝛼22 𝛼12

𝑈2,


(−15𝑎0 −3√5𝑎0 )𝛼2 𝛼1

𝑈−

(−15𝑎0 −3√5𝑎0 )𝛼22

565

𝛼12

𝑈2,

1

(−

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√−15𝑎0 −3√5𝑎0 𝛼1

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


1 𝛼

) 𝑥.

𝑡=0

𝑡 = 0.5

𝑡=1

Fig 4.35: Solitary wave solution for 𝑡 = 0, 𝑡 = 0.5 and 𝑡 = 1.

Fig 4.36: Combined graph for various of 𝑡.

4.7.3. 5th order Ito Equation Consider the Ito equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 2𝑢2 𝐷𝑥𝛼 𝑢 + 6𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 3𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0,

0 < 𝛼 ≤ 1.

(4.190)

To convert Eq. (4.190) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡,

(4.191)

566

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.191) into Eq. (4.190) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 2𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 6𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 3𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0. (4.192) By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 5 = 2𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.184), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑎2 𝑈 2 .

(4.193)



𝑈1 :

21𝑘 3𝛼 𝑎0 𝑎1 𝛼12 𝛼2 + 2𝑘 𝛼 𝑎02 𝑎1 𝛼2 + 4𝑘 𝛼 𝑎02 𝑎2 𝛼1 + ⋯ + 32𝑘 5𝛼 𝑎2 𝛼15 + 9𝑘 3𝛼 𝑎12 𝛼13 = 0,

𝑈2:

12𝑘 𝛼 𝑎0 𝑎1 𝑎2 𝛼1 + 36𝑘 3𝛼 𝑎0 𝑎1 𝛼1 𝛼22 + 114𝑘 3𝛼 𝑎0 𝑎2 𝛼12 𝛼2 + ⋯ 2𝑘 𝛼 𝑎13 𝛼2 = 0

𝑈3:

12𝑘 𝛼 𝑎0 𝑎1 𝑎2 𝛼2 + 267𝑘 3𝛼 𝑎1 𝛼12 𝑎2 𝛼2 + 162𝑘 3𝛼 𝑎0 𝑎2 𝛼1 𝛼22 + ⋯ + 72𝑘 3𝛼 𝑎22 𝛼13 = 0,

𝑈4:

354𝑘 3𝛼 𝑎1 𝛼2 𝑎2 𝛼22 + 8𝑘 𝛼 𝑎0 𝑎22 𝛼2 + 8𝑘 𝛼 𝑎12 𝑎2 𝛼2 + ⋯ + 3000𝑘 5𝛼 𝑎2 𝛼12 𝛼13 + 30𝑘 3𝛼 𝑎12 𝛼23 = 0,

𝑈5:

10𝑘 𝛼 𝑎1 𝑎22 𝛼2 + 150𝑘 3𝛼 𝑎1 𝛼23 𝑎2 + 354𝑘 3𝛼 𝑎22 𝛼1 𝛼22 + ⋯ + 120𝑘 5𝛼 𝑎1 𝛼25 + 4𝑘 𝛼 𝑎23 𝛼1 = 0,

𝑈6:

720𝑘 5𝛼 𝑎2 𝛼25 + 4𝑘 𝛼 𝑎23 𝛼2 + 144𝑘 3𝛼 𝑎22 𝛼23 = 0.

Solving the above system for unknown parameters, we have the two solution sets 1st Solution Set: 5

2

2

2


ln(−6𝛼4 1 )+5𝑦 ln(𝑘) 𝛼

.

Case I: Substituting the values of unknown into Eq. (4.193) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have

567

2

5

2

2


i


5

2

2



5

2

2


1


5

2

2


√3

1

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

2

2 𝐶 cos (√3𝜉)+i𝐶 sin (√3𝜉) 1 𝛼 2 𝛼

where 𝑈 = −

2

,

2


ln(−𝛼4 1 )+5𝛼 ln(𝑘) 𝛼

𝑡.

2nd Solution Set: 2

1

2

2

𝑎0 = − 2 [e𝛼 ln(𝑘) ] 𝛼12 , 𝑎1 = −6[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 , 𝑎2 = −6[e𝛼 ln(𝑘) ] 𝛼22 , 𝑘 = 𝑘, 𝜔 = 0. Case I: Substituting the values of unknown into Eq. (4.193) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1

2

2

2

𝑢5 (𝜉) = − 2 [e𝛼 ln(𝑘) ] 𝛼12 − 6[e𝛼 ln(𝑦) ] 𝛼1 𝛼2 𝑈 − 6[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , 1

i


2

2

2


1


1

2

2


1


1

2

2


√3

√3

1 i 𝑦1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


𝑡=0

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1. In all cases 𝜉 = 𝑘𝑥.

𝑡 = 0.5

𝑡=1


569


4.7.4. 5th order Kaup Kuperschmidt Equation Consider the Kaup Kuperschmidt Equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 20𝑢2 𝐷𝑥𝛼 𝑢 + 25𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 10𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1. (4.194) To convert Eq. (12) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡,

(4.195)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.195) into Eq. (4.194) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 20𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 25𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 10𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0.

(4.196)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 5 = 2𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.184), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑎2 𝑈 2 .

(4.197)


20𝑘 𝛼 𝑎02 𝑎1 𝛼1 + 10𝑘 3𝛼 𝑎0 𝑎1 𝛼13 + 𝜔𝛼 𝑎1 𝛼1 + 𝑘 5𝛼 𝑎1 𝛼15 = 0,

570

𝑈1 :

20𝑘 𝛼 𝑎02 𝑎1 𝛼2 + 40𝑦𝛼 𝑎02 𝑎2 𝛼1 + 40𝑘 𝛼 𝑎0 𝑎12 𝛼1 + ⋯ + 70𝑘 3𝛼 𝑎0 𝑎1 𝛼12 𝛼2 + 35𝑘 3𝛼 𝑎12 𝛼13 = 0,

𝑈2:

170𝑘 3𝛼 𝑎12 𝛼12 𝛼2 + 240𝑘 3𝛼 𝑎1 𝛼13 𝑎2 + 40𝑘 𝛼 𝑎02 𝑎2 𝛼2 + ⋯ + 380𝑘 3𝛼 𝑎0 𝑎2 𝛼12 𝛼2 = 0,

𝑈3:

245𝑘 3𝛼 𝑎12 𝛼1 𝛼22 + 80𝑘 𝛼 𝑎0 𝑎22 𝛼1 + 80𝑘 𝛼 𝑎12 𝑎2 𝛼1 + ⋯ + 280𝑘 3𝛼 𝑎22 𝛼13 = 0,

𝑈4:

1080𝑘 3𝛼 𝑎22 𝛼12 𝛼2 + 80𝑘 𝛼 𝑎0 𝑎22 𝛼2 + 80𝑘 𝛼 𝑎12 𝑎2 𝛼2 + ⋯ + 110𝑘 3𝛼 𝑎12 𝛼23 = 0,

𝑈5:

550𝑘 3𝛼 𝑎1 𝛼23 𝑎2 + 1340𝑘 3𝛼 𝑎22 𝛼1 𝛼22 + 100𝑘 𝛼 𝑎1 𝑎22 𝛼2 + ⋯ + 40𝑘 𝛼 𝑎23 𝛼1 = 0,

𝑈6:

720𝑘 5𝛼 𝑎2 𝛼25 + 40𝑘 𝛼 𝑎23 𝛼2 + 540𝑘 3𝛼 𝑎22 𝛼23 = 0.


1

2

3

2

3

𝑎0 = − 8 [e𝛼 ln(𝑘) ] 𝛼12 , 𝑎1 = − 2 [e𝛼 ln(𝑘) ] 𝛼1 𝛼2 , 𝑎2 = − 2 [e𝛼 ln(𝑘) ] 𝛼22 , 𝑘 = 𝑘, 𝜔=e

1 ln(− 𝛼4 )+5𝛼 ln(𝑘) 16 1 𝛼

.


2

2

3

3

2

𝑢1 (𝜉) = − 8 [e𝛼 ln(𝑘) ] 𝛼12 − 2 [e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 2 [e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , 1

i


2

2

3

3

2



2

2

3

3

2


1


2

3

2

3

2

𝑢4 (𝜉) = − 8 [e𝛼 ln(𝑘) ] 𝛼12 − 2 [e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 2 [e𝛼 ln(𝑘) ] 𝑦22 𝑈 2 , 571

1

√3

e

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

where 𝑈 = 2 − 2 1 ln(− 𝛼4 )+5𝛼 ln(𝑘) 16 1 𝛼

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


𝑡.

2nd Solution Set: 2

2

2

𝑎0 = −[e𝛼 ln(𝑘) ] 𝛼12 , 𝑎1 = −12[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 , 𝑎2 = −12[e𝛼 ln(𝑘) ] 𝛼22 , 𝑘 = 𝑘, 𝜔=e

ln(−11𝛼4 1 )+5𝛼 ln(𝑘) 𝛼

.


2

2

𝑢5 (𝜉) = −[e𝛼 ln(𝑘) ] 𝛼12 − 12[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 12[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , 1

i


2

2



2

2


1


2

2


√3

e

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

where 𝑈 = 2 − 2 ln(−11𝛼4 1 )+5𝛼 ln(𝑘) 𝛼

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


𝑡.

572

𝑡=1

𝑡=5

𝑡 = 10

Fig 4.39: Solitary wave solution for 𝑡 = 1, 𝑡 = 5 and 𝑡 = 10.


4.7.5. 5th order Lax Equation Consider the Lax equation [230] of fraction order 𝑦𝛼𝑡 𝑢 + 30𝑢2 𝐷𝑥𝛼 𝑢 + 20𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 10𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0,

0 < 𝛼 ≤ 1. (4.198)

To convert Eq. (17) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡,

(4.199)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (18) into Eq. (17) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 30𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 20𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 10𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0. By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 5 = 2𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.184), we obtained the trail solution 573

(4.200)

𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑎2 𝑈 2 .

(4.201)


𝜔𝛼 𝛼1 𝑎1 + 𝑘 5𝛼 𝛼15 𝑎1 + 30𝑘 𝛼 𝑎02 𝑎1 𝛼1 + 10𝑘 3𝛼 𝑎0 𝑎1 𝛼13 = 0,

𝑈1 :

30𝑘 3𝛼 𝑎12 𝛼13 + 31𝑘 5𝛼 𝑎1 𝛼14 𝛼2 + ⋯ + 60𝑘 𝛼 𝑎02 𝑎2 𝛼1 + 60𝑘 𝛼 𝑎0 𝑎12 𝛼1 = 0,

𝑈2:

30𝑘 𝛼 𝑎13 𝛼1 + 180𝑘 5𝛼 𝑎1 𝛼13 𝛼22 + 422𝑘 5𝛼 𝑎2 𝛼14 𝛼2 + ⋯ 60𝑘 𝛼 𝑎02 𝑎2 𝛼2 + 60𝑘 𝛼 𝑎0 𝑎12 𝛼2 = 0,

𝑈3:

30𝑘 𝛼 𝑎13 𝛼2 + 240𝑘 3𝛼 𝛼13 𝑎22 + 390𝑘 5𝛼 𝑎1 𝛼12 𝛼23 + ⋯ + 120𝑘 𝛼 𝑎2 𝑎12 𝛼1 = 0,

𝑈4:

100𝑘 3𝛼 𝑎12 𝛼23 + 360𝑘 5𝛼 𝑎1 𝛼24 𝛼1 + 3000𝑘 5𝛼 𝑎2 𝛼12 𝛼23 + ⋯ + 150𝑘 𝛼 𝑎1 𝑎22 𝛼1 = 0,

𝑈5:

60𝑘 𝛼 𝑎23 𝛼1 + 2400𝑘 5𝛼 𝑎2 𝛼1 𝛼24 + 120𝑘 5𝛼 𝑎1 𝛼15 + ⋯ + 150𝑘 𝛼 𝑎1 𝑎22 𝛼2 = 0,

𝑈6:

60𝑘 𝛼 𝑎23 𝛼2 + 480𝑘 3𝛼 𝑎22 𝛼23 + 720𝑘 5𝛼 𝑎2 𝛼25 = 0.

Solving the above system for unknown parameters, we have the two solution sets 1st Solution Set: 2

1

2

2


7 ln(− 𝛼4 )+5𝛼 ln(𝑘) 2 1 𝛼

.


2

2

2


i


2

2

2



2

2

2


1

1


1

2

2


√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


,

7 ln(− 𝛼4 )+5𝛼 ln(𝑘) 2 1 𝛼

𝑡.

2nd Solution Set: 2

2

𝑎0 = 𝑎0 , 𝑎1 = −2[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 , 𝑎2 = −2[e𝛼 ln(𝑘) ] 𝛼22 , 𝑘 = 𝑘, 𝜔=e

4 𝛼 2 3𝛼 ln(−𝑘5𝛼 𝛼4 1 −30𝑘 𝑎0 −10𝑘 𝑎0 𝛼1 )+5𝛼 ln(𝑘) 𝛼

.


2

𝑢5 (𝜉) = 𝑎0 − 2[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 2[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , 1

i


2



2


1


2

𝑢8 (𝜉) = 𝑎0 − 2[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 2[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 ,

575

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


,

4 𝛼 2 3𝛼 ln(−𝑘5𝛼 𝛼4 1 −30𝑘 𝑎0 −10𝑘 𝑎0 𝛼1 )+5𝛼 ln(𝑘) 𝛼

𝑡.

3rd Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = − √−5𝑎0 +i√5𝑎0

(

𝛼1

2(−5𝑎0 +i√5𝑎0 )𝛼2 𝛼1

, 𝑎2 = −

2(−5𝑎0 +i√5𝑎0 )𝛼22 𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.


2(−5𝑎0 +i√5𝑎0 )𝛼2 𝛼1

𝑈−

2(−5𝑎0 +i√5𝑎0 )𝛼22 𝛼12

𝑈2,

i


2(−5𝑎0 +i√5𝑎0 )𝛼2 𝛼1

2(−5𝑎0 +i√5𝑎0 )𝛼22

𝑈−

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (21) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢11 (𝜉) = 𝑎0 −

2(−5𝑎0 +i√5𝑎0 )𝛼2 𝛼1

1

2(−5𝑎0 +i√5𝑎0 )𝛼22

𝑈−

𝛼12

𝑈2,

1


2(−5𝑎0 +i√5𝑎0 )𝛼2 𝛼1

𝑈−

2(−5𝑎0 +i√5𝑎0 )𝛼22 𝛼12

576

𝑈2,

1

√3

√−5𝑎0 +i√5𝑎0

(

𝛼1

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


1 𝛼

) 𝑥.

4th Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = − −√−5𝑎0 +i√5𝑎0

(

𝛼1

2(−5𝑎0 +i√5𝑎0 )𝛼2 𝛼1

, 𝑎2 = −

2(−5𝑎0 +i√5𝑎0 )𝛼22 𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.


2(−5𝑎0 +i√5𝑎0 )𝛼2 𝛼1

2(−5𝑎0 +i√5𝑎0 )𝛼22

𝑈−

𝛼12

𝑈2,

i


2(−5𝑎0 +i√5𝑎0 )𝛼2 𝛼1

2(−5𝑎0 +i√5𝑎0 )𝛼22

𝑈−

𝛼12

𝑈2,

1


2(−5𝑎0 +i√5𝑎0 )𝛼2 𝛼1

1

1

2

2

2(−5𝑎0 +i√5𝑎0 )𝛼22

𝑈−

𝛼12

𝑈2,


2(−5𝑎0 +i√5𝑎0 )𝛼2 𝛼1

𝑈−

2(−5𝑎0 +i√5𝑎0 )𝛼22 𝛼12

577

𝑈2,

√3

(−

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1


√−5𝑎0 +i√5𝑎0 𝛼1

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


1 𝛼

) 𝑥.

5th Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = − √−5𝑎0 −i√5𝑎0

(

𝛼1

2(−5𝑎0 −i√5𝑎0 )𝛼2 𝛼1

, 𝑎2 = −

2(−5𝑎0 −i√5𝑎0 )𝛼22 𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.


2(−5𝑎0 −i√5𝑎0 )𝛼2 𝛼1

2(−5𝑎0 −i√5𝑎0 )𝛼22

𝑈−

𝛼12

𝑈2,

i


2(−5𝑎0 −i√5𝑎0 )𝛼2 𝛼1

2(−5𝑎0 −i√5𝑎0 )𝛼22

𝑈−

𝛼12

𝑈2,

1


2(−5𝑎0 −i√5𝑦0 )𝛼2 𝛼1

1

1

2

2

𝑈−

2(−5𝑎0 −i√5𝑎0 )𝛼22 𝛼12

𝑈2,

where 𝑈 = −1 − tanh𝛼 (𝜉) − coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.201) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and α2 = √3i, we have 𝑢20 (𝜉) = 𝑎0 −

2(−5𝑎0 −i√5𝑎0 )𝛼2 𝛼1

𝑈−

2(−5𝑎0 −i√5𝑎0 )𝛼22 𝛼12

578

𝑈2,

1

√3

√−5𝑎0 −i√5𝑎0

(

𝛼1

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


1 𝛼

) 𝑥.

6th Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = − −√−5𝑎0 −i√5𝑎0

(

𝛼1

2(−5𝑎0 −i√5𝑎0 )𝛼2 𝛼1

, 𝑎2 = −

2(−5𝑎0 −i√5𝑎0 )𝛼22 𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.


2(−5𝑎0 −i√5𝑎0 )𝛼2 𝛼1

2(−5𝑎0 −i√5𝑎0 )𝛼22

𝑈−

𝛼12

𝑈2,

i


2(−5𝑎0 −i√5𝑎0 )𝛼2 𝛼1

2(−5𝑎0 −i√5𝑎0 )𝛼22

𝑈−

𝛼12

𝑈2,

1


2(−5𝑎0 −i√5𝑎0 )𝛼2 𝛼1

1

1

2

2

2(−5𝑎0 −i√5𝑎0 )𝛼22

𝑈−

𝛼12

𝑈2,


2(−5𝑎0 −i√5𝑎0 )𝛼2 𝛼1

𝑈−

2(−5𝑎0 −i√5𝑎0 )𝛼22 𝛼12

579

𝑈2,

1

√3

(−

√−5𝑎0 −i√5𝑎0 𝛼1

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


1 𝛼

) 𝑥.

𝑡=0

𝑡 = 0.5

𝑡=1



4.7.6. 7th order Kaup Kuperschmidt Equation Consider the 7th order Kaup Kuperschmidt [230] of fraction order 𝐷𝑡𝛼 𝑢 + 2016𝑢3 𝐷𝑥𝛼 𝑢 + 630(𝐷𝑥𝛼 𝑢)3 + 2268𝑢𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 504𝑢2 𝐷𝑥3𝛼 𝑢 + 252𝐷𝑥2𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 147𝐷𝑥𝛼 𝑢𝐷𝑥4𝛼 𝑢 + 42𝑢𝐷𝑥5𝛼 𝑢 + 𝐷𝑥7𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1, (4.202) To convert Eq. (4.202) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡,

(4.203)

580

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.203) into Eq. (4.202) and using the chain rule we obtained 3

𝜔𝛼 𝐷𝜉𝛼 𝑢 + 2016𝑘 𝛼 𝑢3 𝐷𝜉𝛼 𝑢 + 630𝑘 3𝛼 (𝐷𝜉𝛼 𝑢) + 2268𝑘 3𝛼 𝑢𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 504𝑘 3𝛼 𝑢2 𝐷𝜉3𝛼 𝑢 + 252𝑘 5𝛼 𝐷𝜉2𝑦 𝑢𝐷𝜉3𝛼 𝑢 + 147𝑘 5𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉4𝛼 𝑢 + 42𝑘 5𝛼 𝑢𝐷𝜉5𝛼 𝑢 + 𝑘 7𝛼 𝐷𝜉7𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1,

(4.204)


(4.205)

Putting Eq. (4.205) into Eq. (4.204) coupled with Eq. (4.184); the Eq. (4.205) yields an algebraic equation involving power of U as 𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 + 𝐶7 𝑈 7 + 𝐶8 𝑈 8 = 0. Compare the like powers of U we have system of equations 𝑈0:

2016𝑘 𝛼 𝑎03 𝑎1 𝛼1 + 504𝑘 3𝛼 𝑎02 𝑎1 𝛼13 + ⋯ + 𝜔𝛼 𝑎1 𝛼1 = 0,

𝑈1 :

441𝑘 5𝛼 𝑎12 𝛼15 + 2016𝑘 𝛼 𝑎03 𝑎1 𝛼2 + ⋯ + 1302𝑘 5𝛼 𝑎0 𝑎1 𝛼14 𝛼2 = 0,

𝑈2:

4032𝑘 𝛼 𝑎03 𝑎2 𝛼2 + 6048𝑘 𝛼 𝑎02 𝑎12 𝛼2 + ⋯ + 17724𝑘 5𝛼 𝑎0 𝑎2 𝛼14 𝛼22 = 0,

𝑈3:


𝑈4:


𝑈5:


𝑈6:

11592𝑘 5𝛼 𝑎12 𝛼25 + 12096𝑘 𝛼 𝑎0 𝑎23 𝛼2 + ⋯ + 286272𝑘 5𝛼 𝑎1 𝑎2 𝛼1 𝛼24 = 0,

𝑈7:

14112𝑘 𝛼 𝑎1 𝑎23 𝛼2 + 84672𝑘 3𝛼 𝑎1 𝛼23 𝑎22 + ⋯ + 4032𝑘 𝛼 𝑎24 𝛼1 = 0,

𝑈8:

101808𝑘 5𝛼 𝑎22 𝛼25 + 40320𝑘 7𝛼 𝑎2 𝛼27 + 4032𝑘 𝛼 𝑎24 𝛼2 + 44352𝑘 3𝛼 𝛼23 𝑎23 = 0.

Solving the above system for unknown parameters, we have the one solution sets Solution Set: 1

2

2

1

1

2

𝑎0 = − 24 [e𝛼 ln(𝑘) ] 𝛼12 , 𝑎1 = − 2 [e𝛼 ln(𝑘) ] 𝛼1 𝛼2 , 𝑎2 = − 2 [e𝛼 ln(𝑘) ] 𝛼22 , 𝑘 = 𝑘, ln(

𝜔=e

1 6 𝛼 )+7𝛼 ln(𝑘) 48 1 𝛼

.


2

1

2

1

1

2


i

where 𝑈 = 2 − 2 tan𝛼 (𝑦). Case II: Substituting the values of unknown into Eq. (4.205) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 2

1

2

1

1

2



1

2

1

1

2


1


1

2

1

1

2

𝑢4 (𝜉) = − 24 [e𝛼 ln(𝑘) ] 𝛼12 − 2 [e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 2 [e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , √3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

2

2 𝐶 cos (√3𝜉)+i𝐶 sin (√3𝜉) 1 𝑦 2 𝛼

where 𝑈 = −

2

ln(

e

√3

1

1 6 𝛼 )+7𝛼 ln(𝑘) 48 1 𝛼


2

𝑡.

𝑡=1

𝑡=5 Fig 4.43: Solitary wave solution for 𝑡 = 1, 𝑡 = 5 and 𝑡 = 10.

582

𝑡 = 10


4.7.7. 7th Order Lax Equation Consider the 7th order Lax equations [230] of fraction order 𝐷𝑡𝛼 𝑢 + 140𝑢3 𝐷𝑥𝛼 𝑢 + 70(𝐷𝑥𝛼 𝑢)3 + 280𝑢𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 70𝑢2 𝐷𝑥3𝛼 𝑢 + 70𝐷𝑥2𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 42𝐷𝑥𝛼 𝑢𝐷𝑥4𝛼 𝑢 + 14𝑢𝐷𝑥5𝛼 𝑢 + 𝐷𝑥7𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.206)


(4.207)


𝜔𝛼 𝐷𝜉𝛼 𝑢 + 140𝑘 𝛼 𝑢3 𝐷𝜉𝛼 𝑢 + 70𝑘 3𝛼 (𝐷𝜉𝛼 𝑢) + 280𝑘 3𝛼 𝑢𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 70𝑘 3𝛼 𝑢2 𝐷𝜉3𝛼 𝑢 + 70𝑘 5𝛼 𝐷𝜉2𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 42𝑘 5𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉4𝛼 𝑢 + 14𝑘 5𝛼 𝑢𝐷𝜉5𝛼 𝑢 + 𝑘 7𝛼 𝐷𝜉7𝛼 𝑢 = 0,

(4.208)


(4.209)

Putting Eq. (4.209) into Eq. (4.208) coupled with Eq. (4.185); the Eq. (4.209) yields an algebraic equation involving power of U as 𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 + 𝐶7 𝑈 7 + 𝐶8 𝑈 8 = 0. Compare the like powers of U we have system of equations 583

𝑈0:

140𝑘 𝛼 𝑎03 𝑎1 𝛼1 + 70𝑘 3𝛼 𝑎02 𝑎1 𝛼13 + ⋯ + 𝜔𝛼 𝑎1 𝛼1 = 0,

𝑈1 :

126𝑘 5𝛼 𝑎12 𝛼15 + 128𝑘 7𝛼 𝑎2 𝛼17 + 𝜔𝛼 𝑎1 𝛼2 + ⋯ + 127𝑘 7𝛼 𝑎1 𝛼16 𝛼2 = 0,

𝑈2:

2𝜔𝛼 𝑎2 𝛼2 + 420𝑘 3𝛼 𝑎13 𝛼13 + 1260𝑘 𝛼 𝑎02 𝑎1 𝑎2 𝛼1 + ⋯ + 4118𝑘 7𝛼 𝑎2 𝛼16 𝛼2 = 0,

𝑈3:

4032𝑘 5𝛼 𝑎22 𝛼15 + 140𝑘 𝛼 𝑎14 𝛼1 + ⋯ + 36414𝑘 7𝛼 𝑎2 𝛼15 𝛼22 = 0,

𝑈4:

140𝑘 𝛼 𝑎14 𝛼2 + 12180𝑘 3𝛼 𝑎12 𝛼12 𝑎2 𝛼2 + ⋯ + 141624𝑘 7𝛼 𝑎2 𝛼14 𝛼23 = 0,

𝑈5:

1050𝑘 3𝛼 𝑎13 𝛼23 + 3360𝑘 3𝛼 𝑎23 𝛼13 + ⋯ + 285600𝑘 7𝛼 𝑎2 𝛼13 𝛼24 = 0,

𝑈6:

3528𝑘 5𝛼 𝑎12 𝛼25 + 27160𝑘 3𝛼 𝑎1 𝛼1 𝑎22 𝛼22 + ⋯ + 312480𝑘 7𝛼 𝑎2 𝛼12 𝛼25 = 0,

𝑈7:

5040𝑘 7𝛼 𝑎1 𝛼27 + 280𝑘 𝛼 𝛼1 𝑎24 + ⋯ + 176400𝑘 7𝛼 𝑎2 𝛼1 𝛼26 = 0,

𝑈8:

30240𝑘 5𝛼 𝑎22 𝛼25 + 40320𝑘 7𝛼 𝑎2 𝛼27 + 280𝑘 𝛼 𝑎24 𝛼2 + 5600𝑘 3𝛼 𝛼23 𝑎23 = 0.

Solving the above system for unknown parameters, we have the five solution sets 1st Solution Set: 2

2


3𝛼 2 2 5𝛼 2 4 7𝛼 6 ln(−140𝑘𝛼 𝑎3 0 −70𝑘 𝑎0 𝛼1 −14𝑘 𝑎0 𝛼1 −𝑘 𝛼1 ) 𝛼

.


2


i


2



2


1


2


√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


,

3𝛼 2 2 5𝛼 2 4 7𝛼 6 ln(−140𝑘𝛼 𝑎3 0 −70𝑘 𝑎0 𝛼1 −14𝑘 𝑎0 𝛼1 −𝑘 𝛼1 ) 𝛼

𝑡.

2nd Solution Set: 2 (−15𝑎0 +3i𝑎0 √5)𝛼22

2 (−15𝑎0 +3i𝑎0 √5)𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 3 1 𝛼

√−15𝑎0 +3i𝑎0 √5

(

ln(−

𝛼1

𝛼12

,𝑘 =

20 2 71 14 𝑎 √−15𝑎0 +3i𝑎0 √5( 𝑎0 + i𝑎0 √5)) 27𝛼1 0 3 3

) ,𝜔 = e

3𝛼1

, 𝑎2 = − 3

.

𝛼

Case I: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 2 (−15𝑎0 +3i𝑎0 √5)𝛼22

2 (−15𝑎0 +3i𝑎0 √5)𝛼2

𝑢5 (𝜉) = 𝑎0 − 3 1

𝑈−3

𝛼1

𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 2 (−15𝑎0 +3i𝑎0 √5)𝛼22

2 (−15𝑎0 +3i𝑎0 √5)𝛼2

𝑢6 (𝜉) = 𝑎0 − 3

𝑈−3

𝛼1

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 2 (−15𝑎0 +3i𝑎0 √5)𝛼22

2 (−15𝑎0 +3i𝑎0 √5)𝛼2

𝑢7 (𝜉) = 𝑎0 − 3

𝑈−3

𝛼1

1

𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 2 (−15𝑎0 +3i𝑎0 √5)𝛼22

2 (−15𝑎0 +3i𝑎0 √5)𝛼2

𝑢8 (𝜉) = 𝑎0 − 3 1

𝑈−3

𝛼1 √3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

,

585

𝛼12

𝑈2,

√−15𝑎0 +3i𝑎0 √5

and i = √−1. In all cases 𝜉 = (

1 𝛼

) 𝑥+

3𝛼1

20 2 71 14 ln(− 𝑎 √−15𝑎0 +3i𝑎0 √5( 𝑎0 + i𝑎0 √5)) 27𝛼1 0 3 3

e

𝑡.

𝛼

3rd Solution Set: 2 (−15𝑎0 +3i𝑎0 √5)𝛼22

2 (−15𝑎0 +3i𝑎0 √5)𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 3

(−

𝛼1

1 𝛼

√−15𝑎0 +3i𝑎0 √5

, 𝑎2 = − 3

ln(

,𝑘 =

20 2 71 14 𝑎 √−15𝑎0 +3i𝑎0 √5( 𝑎0 + i𝑎0 √5)) 27𝛼1 0 3 3

) ,𝜔 = e

3𝛼1

𝛼12

.

𝛼

Case I: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 2 (−15𝑎0 +3i𝑎0 √5)𝛼22

2 (−15𝑎0 +3i𝑦0 √5)𝛼2

𝑢9 (𝜉) = 𝑎0 − 3 1

𝑈−3

𝛼1

𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 2 (−15𝑎0 +3i𝑎0 √5)𝛼2

𝑢10 (𝜉) = 𝑎0 − 3

𝛼1

2 (−15𝑎0 +3i𝑎0 √5)𝛼22

𝑈−3

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 2 (−15𝑎0 +3i𝑎0 √5)𝛼2

𝑢11 (𝜉) = 𝑎0 − 3

𝛼1

1

2 (−15𝑎0 +3i𝑎0 √5)𝛼22

𝑈−3

𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 2 (−15𝑎0 +3i𝑎0 √5)𝛼2

𝑢12 (𝜉) = 𝑎0 − 3 1

𝛼1 √3

2 (−15𝑎0 +3i𝑎0 √5)𝛼22

𝑈−3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

,

586

𝛼12

𝑈2,

and i = √−1. In all cases 𝜉 = (−

ln(

1 𝛼

√−15𝑎0 +3i𝑎0 √5

) 𝑥+

3𝛼1

20 2 71 14 𝑎 √−15𝑎0 +3i𝑦0 √5( 𝑎0 + i𝑎0 √5)) 27𝛼1 0 3 3

e

𝑡.

𝛼

4th Solution Set: 2 (−15𝑎0 −3i𝑎0 √5)𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 3 1 𝛼

√−15𝑎0 −3i𝑎0 √5

(

𝛼1 ln(−

2 (−15𝑎0 −3i𝑎0 √5)𝛼22

, 𝑎2 = − 3

,𝑘 =

20 2 71 14 𝑎 √−15𝑎0 −3i𝑎0 √5( 𝑎0 − i𝑎0 √5)) 27𝛼1 0 3 3

) ,𝜔 = e

3𝛼1

𝛼12

.

𝛼

Case I: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 2 (−15𝑎0 −3i𝑎0 √5)𝛼2

𝑢13 (𝜉) = 𝑎0 − 3 1

𝛼1

2 (−15𝑎0 −3i𝑎0 √5)𝛼22

𝑈−3

𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 2 (−15𝑎0 −3i𝑎0 √5)𝛼2

𝑢14 (𝜉) = 𝑎0 − 3

𝛼1

2 (−15𝑎0 −3i𝑎0 √5)𝛼22

𝑈−3

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 2 (−15𝑎0 −3i𝑎0 √5)𝛼2

𝑢15 (𝜉) = 𝑎0 − 3

𝛼1

1

2 (−15𝑎0 −3i𝑎0 √5)𝛼22

𝑈−3

𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝑦 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 2 (−15𝑎0 −3i𝑎0 √5)𝛼2

𝑢16 (𝜉) = 𝑎0 − 3 1

𝛼1 √3

2 (−15𝑎0 −3i𝑎0 √5)𝑦22

𝑈−3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

,

587

𝛼12

𝑈2,

√−15𝑎0 −3i𝑎0 √5

and i = √−1. In all cases 𝜉 = (

1 𝛼

) 𝑥+

3𝛼1

20 2 71 14 ln(− 𝑎 √−15𝑎0 −3i𝑎0 √5( 𝑎0 − i𝑎0 √5)) 27𝛼1 0 3 3

e

𝑡.

𝛼

5th Solution Set: 2 (−15𝑎0 −3i𝑎0 √5)𝛼22

2 (−15𝑎0 −3i𝑎0 √5)𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 3

(−

√−15𝑎0 −3i𝑎0 √5

, 𝑎2 = − 3

𝛼1 1 𝛼

ln(

,𝑘 =

20 2 71 14 𝑎 √−15𝑎0 −3i𝑎0 √5( 𝑎0 − i𝑎0 √5)) 27𝛼1 0 3 3

) , 𝜔=e

3𝛼1

𝛼12

.

𝛼

Case I: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 2 (−15𝑎0 −3i𝑎0 √5)𝛼2

𝑢17 (𝜉) = 𝑎0 − 3 1

𝛼1

2 (−15𝑎0 −3i𝑎0 √5)𝛼22

𝑈−3

𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 2 (−15𝑎0 −3i𝑎0 √5)𝛼2

𝑢18 (𝜉) = 𝑎0 − 3

𝛼1

2 (−15𝑎0 −3i𝑎0 √5)𝛼22

𝑈−3

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 2 (−15𝑎0 −3i𝑎0 √5)𝛼2

𝑢19 (𝜉) = 𝑎0 − 3

𝛼1

1

2 (−15𝑎0 −3i𝑎0 √5)𝛼22

𝑈−3

𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.209) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 2 (−15𝑎0 −3i𝑎0 √5)𝛼2

𝑢20 (𝜉) = 𝑎0 − 3 1

𝛼1 √3

2 (−15𝑎0 −3i𝑎0 √5)𝛼22

𝑈−3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

,

588

𝛼12

𝑈2,

and i = √−1. In all cases 𝜉 = (−

ln(

√−15𝑎0 −3i𝑎0 √5

1 𝛼

) 𝑥+

3𝛼1

20 2 71 14 𝑎 √−15𝑎0 −3i𝑎0 √5( 𝑎0 − i𝑎0 √5)) 27𝛼1 0 3 3

e

𝛼

𝑡=0

𝑡.

𝑡 = 0.5

𝑡=1



4.7.8. 9th order Sawada Kotera Equation Consider the 9th order Sawada Kotera equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 45𝐷𝑥𝛼 𝑢𝐷𝑥6𝛼 𝑢 + 45𝑢𝐷𝑥7𝛼 𝑢 + 210𝐷𝑥4𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 210𝐷𝑥2𝛼 𝑢𝐷𝑥5𝛼 𝑢 + 1575𝐷𝑥𝑦 𝑢(𝐷𝑥2𝛼 𝑢)2 + 3150𝑢𝐷𝑥2𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 1260𝑢𝐷𝑥𝛼 𝑢𝐷𝑥4𝛼 𝑢 + 630𝑢2 𝐷𝑥5𝛼 𝑢 + 9450𝑢2 𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 3150𝑢3 𝐷𝑥3𝛼 𝑢 + 4725𝑢4 𝐷𝑥𝛼 𝑢 + 𝐷𝑥9𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1,

(4.210)

To convert Eq. (4.210) into ODE we use following transformation 589

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡,

(4.211)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.211) into Eq. (4.210) and using the chain rule we obtained 𝜔𝛼 𝐷𝑡𝛼 𝑢 + 45𝑘 7𝛼 𝐷𝑥𝛼 𝑢𝐷𝑥6𝛼 𝑢 + 45𝑘 7𝛼 𝑢𝐷𝑥7𝛼 𝑢 + 210𝑘 7𝛼 𝐷𝑥4𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 210𝑘 7𝛼 𝐷𝑥2𝛼 𝑢𝐷𝑥5𝛼 𝑢 + 1575𝑘 5𝛼 𝐷𝑥𝛼 𝑢(𝐷𝑥2𝛼 𝑢)2 + 3150𝑘 5𝛼 𝑢𝐷𝑥2𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 1260𝑘 5𝛼 𝑢𝐷𝑥𝛼 𝑢𝐷𝑥4𝛼 𝑢 + 630𝑘 5𝛼 𝑢2 𝐷𝑥5𝛼 𝑢 + 9450𝑘 3𝛼 𝑢2 𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 3150𝑘 3𝛼 𝑢3 𝐷𝑥3𝛼 𝑢 + 4725𝑘 𝛼 𝑢4 𝐷𝑥𝛼 𝑢 + 𝑘 9𝛼 𝐷𝑥9𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1,

(4.212)


(4.213)

Putting Eq. (4.213) into Eq. (4.212) coupled with Eq. (4.185); the Eq. (4.213) yields an algebraic equation involving power of U as 𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 + 𝐶7 𝑈 7 + 𝐶8 𝑈 8 +𝐶9 𝑈 9 + 𝐶10 𝑈10 = 0. Compare the like powers of U we have system of equations 𝑈0:

𝜔𝛼 𝑎1 𝛼1 + 𝑘 9𝛼 𝑎1 𝛼19 + 4725𝑘 𝛼 𝑎04 𝑎1 𝛼1 + ⋯ + 630𝑘 5𝛼 𝑎02 𝑎1 𝛼15 = 0,

𝑈1 :

512𝑘 9𝛼 𝑎2 𝛼19 + 𝜔𝛼 𝑎1 𝛼2 + 2𝜔𝛼 𝑎2 𝛼1 + ⋯ + 18900𝑘 𝛼 𝑎03 𝑎12 𝛼1 = 0,

𝑈2:

6615𝑘 5𝛼 𝑎13 𝛼15 + 2𝜔𝛼 𝑎2 𝛼2 + ⋯ + 38342𝑘 9𝛼 𝑎2 𝛼18 𝛼2 = 0,

𝑈3:

113400𝑘 𝛼 𝑎02 𝑎12 𝑎2 𝛼1 + 65280𝑘 7𝛼 𝑎22 𝛼17 + ⋯ + 18900𝑘 3𝛼 𝛼03 𝑎1 𝛼23 = 0,

𝑈4:

113400𝑘 𝛼 𝑎02 𝑎12 𝑎2 𝛼2 + 141750𝑘 𝛼 𝑎02 𝑎22 𝑎1 𝛼1 + ⋯ + 75600𝑘 3𝛼 𝛼03 𝑎2 𝛼23 = 0,

𝑈5:

141750𝑘 𝛼 𝑎02 𝑎22 𝑎1 𝛼2 + 94500𝑘 𝛼 𝑎0 𝑎13 𝑎2 𝛼2 + ⋯ + 18799200𝑘 9𝛼 𝑎2 𝛼15 𝛼24 = 0,

𝑈6:

132300𝑘 𝛼 𝑎0 𝑎23 𝑎1 𝛼1 + ⋯ + 418370400𝑘 9𝛼 𝑎2 𝛼14 𝛼25 = 0,

𝑈7:

149940𝑘 5𝛼 𝑎13 𝛼25 + 100800𝑘 3𝛼 𝑎24 𝛼13 + ⋯ + 56095200𝑘 9𝛼 𝑎2 𝛼13 𝛼26 = 0,

𝑈8:

339840𝑘 7𝛼 𝑎12 𝛼27 + 13705920𝑘 7𝛼 𝑎1 𝛼1 𝑎2 𝛼26 + ⋯ + 44755200𝑘 9𝛼 𝑎2 𝛼12 𝛼27 = 0,

𝑈9:

9450𝑘 𝛼 𝑎25 𝛼1 + 362880𝑘 9𝛼 𝑎1 𝛼29 + ⋯ + 19595520𝑘 9𝛼 𝑎2 𝛼1 𝛼28 = 0,

𝑈10 : 9450𝑘 𝛼 𝑎25 𝛼2 + 362880𝑘 9𝛼 𝑎2 𝛼29 + ⋯ + 189000𝑘 3𝛼 𝑎24 𝛼23 = 0. Solving the above system for unknown parameters, we have the six solution sets 1st Solution Set: 590

2

2


6 𝛼 4 3𝛼 3 2 7𝛼 5𝛼 2 4 ln(−𝑘9𝛼 𝛼8 1 −4725𝑘 𝛼0 −3150𝑘 𝑎0 𝛼1 −45𝑘 𝑎0 𝛼1 −630𝑘 𝑎0 𝛼1 ) 𝛼

.


2


i


2



2


1


2


√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

In all cases 𝜉 = 𝑘𝑥 + e

, and i = √−1.

6 𝛼 4 3𝛼 3 2 7𝛼 5𝛼 2 4 ln(−𝑘9𝛼 𝛼8 1 −4725𝑘 𝛼0 −3150𝑘 𝑎0 𝛼1 −45𝑘 𝑎0 𝛼1 −630𝑘 𝑎0 𝛼1 ) 𝛼

𝑡.

2nd Solution Set: 1

4 ∆𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 289 𝜔=e

𝛼1

4 ∆𝛼22

, 𝑎2 = − 289

𝛼12

1 √∆ 𝛼

, 𝑘 = (17 𝛼 ) ,

63 992 2 8520 ln(− 𝑎2 ∆ + 𝑎 ∆+15915𝑎2 0 )) 4913𝛼1 0 √∆(83521 289 0 𝛼

1

.


591

2

2


i


2



2


1


2

𝑢8 (𝜉) = 𝑎0 − 6[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 6[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , √3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1

where 𝑈 = 2 − 2 6783𝑎02

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

(−3717𝑎03 +357i𝑎03 √390)3

1

, ∆=

17(−3717𝑎03

+

357i𝑎03 √390)3

+

− 1071𝑎0 , and i = √−1. 1

ln(−

1 √∆ 𝛼

In all cases 𝜉 = (17 𝛼 ) 𝑥 + e

63 992 2 8520 𝑎2 ∆ + 𝑎 ∆+15915𝑎2 0 )) 4913𝛼1 0 √∆(83521 289 0 𝛼

1

𝑡.

3rd Solution Set: 1

4 ∆𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 289 ln(

𝜔=e

𝛼1

4 ∆𝛼22

, 𝑎2 = − 289

𝛼12

1 √∆ 𝛼

, 𝑘 = (− 17 𝛼 ) ,

63 992 2 8520 𝑎2 ∆ + 𝑎 ∆+15915𝑎2 0 )) 4913𝛼1 0 √∆(83521 289 0 𝛼

1

.


2


i

where 𝑈 = 2 − 2 tan𝛼 (𝜉).

592

Case II: Substituting the values of unknown into Eq. (4.213) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 2

2



2


1

2

2

where 𝑈 = −1 − tanh𝛼 (𝜉) − coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.213) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 2

2

𝑢12 (𝜉) = 𝑎0 − 6[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 6[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , √3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1

where 𝑈 = 2 − 2 6783𝑎02

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

(−3717𝑎03 +357i𝑎03 √390)3

1

, ∆= 17(−3717𝑎03 + 357i𝑎03 √390)3 +

− 1071𝑎0 , and i = √−1.

1 √∆

1 𝛼

ln(

In all cases 𝜉 = (− 17 𝛼 ) 𝑥 + e

63 992 2 8520 𝑎2 ∆ + 𝑎 ∆+15915𝑎2 0 )) 4913𝛼1 0 √∆(83521 289 0 𝛼

1

𝑡.

4th Solution Set: 1

1 ∆𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 289 𝜔=e

𝛼1

1 ∆𝛼22

, 𝑎2 = − 289

𝛼12

1 √∆ 𝛼

, 𝑘 = (34 𝛼 ) ,

63 62 2130 ln(− 𝑦2 ∆2 + 𝑎 ∆+15915𝑎2 0 )) 9826𝛼1 0 √∆(83521 289 0 𝛼

1

.

Case I: Substituting the values of unknown into Eq. (4.213) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1 ∆𝛼2

𝑢13 (𝜉) = 𝑎0 − 289 1

𝛼1

1 ∆𝛼22

𝑈 − 289

𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.213) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have

593

1 ∆𝛼2

𝑢14 (𝜉) = 𝑎0 − 289

𝛼1

1 ∆𝛼22

𝑈 − 289

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.213) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 1 ∆𝛼2

𝑢15 (𝜉) = 𝑎0 − 289

𝛼1

1 ∆𝛼22

𝑈 − 289

1

𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.213) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 1 ∆𝛼2

𝑢16 (𝜉) = 𝑎0 − 289

𝛼1

1 ∆𝛼22

𝑈 − 289

√3

𝛼12

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1

where 𝑈 = 2 − 2 13566𝑎02

𝑈2,

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

, ∆= −34(−3717𝑎03 + 357i𝑎03 √390)3 −

1

(−3717𝑎03 +357i𝑎03 √390)3 1

1

−4284𝑎0 + 578i√3 ( (−3717𝑎03 + 357i𝑎03 √390)3 − 17

399

𝑎02

7

(−3717𝑎03 +357i𝑎03 √390)3

1

),i =

√−1. 1 √∆

1 𝛼

ln(−


63 62 2130 𝑎2 ∆2 + 𝑎 ∆+15915𝑎2 0 )) 9826𝛼1 0 √∆(83521 289 0 𝛼

1

𝑡.

5th Solution Set: 1

1 ∆𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 289 ln(

𝜔=e

𝛼1

1 ∆𝛼22

, 𝑎2 = − 289

𝛼12

1 √∆ 𝛼

, 𝑘 = (− 34 𝛼 ) ,

63 62 2130 𝑎2 ∆2 + 𝑎 ∆+15915𝑎2 0 )) 9826𝛼1 0 √∆(83521 289 0 𝛼

1

.


𝑢17 (𝜉) = 𝑎0 − 289 1

𝛼1

1 ∆𝛼22

𝑈 − 289

𝛼12

𝑈2,

i


594

Case II: Substituting the values of unknown into Eq. (4.213) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 1 ∆𝛼2

𝑢18 (𝜉) = 𝑎0 − 289

𝛼1

1 ∆𝛼22

𝑈 − 289

𝛼12

𝑈2,

1


𝑢19 (𝜉) = 𝑎0 − 289

𝛼1

1 ∆𝛼22

𝑈 − 289

1

𝛼12

𝑈2,

1


𝑢20 (𝜉) = 𝑎0 − 289

𝛼1

1 ∆𝛼22

𝑈 − 289

√3

𝛼12

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1

where 𝑈 = 2 − 2 13566𝑎02

𝑈2,

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

, ∆=

−34(−3717𝑎03

+

357i𝑎03 √390)3

−

1

(−3717𝑎03 +357i𝑎03 √390)3 1

1

−4284𝑎0 + 578i√3 (17 (−3717𝑎03 + 357i𝑎03 √390)3 −

399

𝑎02

7

(−3717𝑎03 +357i𝑎03 √390)3

1

),i =

√−1. 1

ln(

1 √∆ 𝛼


63 62 2130 𝑎2 ∆2 + 𝑎 ∆+15915𝑎2 0 )) 9826𝛼1 0 √∆(83521 289 0 𝛼

1

𝑡.

6th Solution Set: 1

1 ∆𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 289 𝜔=e

𝛼1

1 ∆𝛼22

, 𝑎2 = − 289

𝛼12

1 √∆ 𝛼

, 𝑘 = (34 𝛼 ) ,

63 62 2130 ln(− 𝑎2 ∆2 + 𝑎 ∆+15915𝑎2 0 )) 9826𝛼1 0 √∆(83521 289 0 𝛼

1

.


𝑢21 (𝜉) = 𝑎0 − 289

𝑦1

1 ∆𝛼22

𝑈 − 289

𝛼12

𝑈2, 595

1

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.213) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 1 ∆𝛼2

𝑢22 (𝜉) = 𝑎0 − 289

𝛼1

1 ∆𝛼22

𝑈 − 289

𝛼12

𝑈2,

1


𝑢23 (𝜉) = 𝑎0 − 289

𝛼1

1 ∆𝛼22

𝑈 − 289

1

𝛼12

𝑈2,

1


𝑢24 (𝜉) = 𝑎0 − 289

𝛼1

1 ∆𝛼22

𝑈 − 289

√3

𝛼12

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1

where 𝑈 = 2 − 2 13566𝑎02

𝑈2,

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

, ∆=

−34(−3717𝑎03

+

357i𝑎03 √390)3

−

1

(−3717𝑎03 +357i𝑎03 √390)3 1

1

−4284𝑎0 − 578i√3 (17 (−3717𝑎03 + 357i𝑎03 √390)3 −

399

𝑎02

7

(−3717𝑎03 +357i𝑎03 √390)3

1

),i =

√−1. 1

ln(−

1 √∆ 𝛼


63 62 2130 𝑎2 ∆2 + 𝑎 ∆+15915𝑎2 0 )) 9826𝛼1 0 √∆(83521 289 0 𝛼

1

𝑡.

7th Solution Set: 1

1 ∆𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 289 ln(

𝜔=e

𝛼1

1 ∆𝛼22

, 𝑎2 = − 289

𝛼12

1 √∆ 𝛼

, 𝑘 = (− 34 𝛼 ) ,

63 62 2130 𝑎2 ∆2 + 𝑎 ∆+15915𝑎2 0 )) 9826𝛼1 0 √∆(83521 289 0 𝛼

1

.


1 ∆𝛼2

𝑢25 (𝜉) = 𝑎0 − 289 1

𝛼1

1 ∆𝛼22

𝑈 − 289

𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.213) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 1 ∆𝛼2

𝑢26 (𝜉) = 𝑎0 − 289

𝛼1

1 ∆𝛼22

𝑈 − 289

𝛼12

𝑈2,

1


𝑢27 (𝜉) = 𝑎0 − 289

𝛼1

1

1 ∆𝛼22

𝑈 − 289

𝛼12

𝑈2,

1


𝑢28 (𝜉) = 𝑎0 − 289 1

√3

𝛼12

𝑈2,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

where 𝑈 = 2 − 2 13566𝑎02

𝛼1

1 ∆𝛼22

𝑈 − 289

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

, ∆=

−34(−3717𝑎03

+

357i𝑎03 √390)3

−

1

(−3717𝑎03 +357i𝑎03 √390)3 1

1

−4284𝑎0 − 578i√3 (17 (−3717𝑎03 + 357i𝑎03 √390)3 −

399

𝑎02

7

(−3717𝑎03 +357i𝑎03 √390)3

√−1. 1

1 √∆ 𝛼


ln(

63 62 2130 𝑎2 ∆2 + 𝑎 ∆+15915𝑎2 0 )) 9826𝛼1 0 √∆(83521 289 0 𝛼

1

597

𝑡.

1

),i =

𝑡=1

𝑡=5

𝑡 = 10



4.7.9. 5th order Sawada Kotera Equation Consider the 5th order Sawada Kotera equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 5𝑢2 𝐷𝑥𝛼 𝑢 + 5𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 5𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.214)


(4.215)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.215) into Eq. (4.214) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 5𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 5𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 5𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0. By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 5 = 2𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.184), we obtained the trail solution 598

(4.216)

𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑎2 𝑈 2 .

(4.217)



𝑈1 :

𝜔𝛼 𝑎1 𝛼2 + 2𝜔𝛼 𝑎2 𝛼1 + 32𝑘 5𝛼 𝑎2 𝛼15 + ⋯ + 10𝑦𝛼 𝑎02 𝑎2 𝛼1 + 10𝑘 𝛼 𝑎0 𝑎12 𝛼1 = 0,

𝑈2:

2𝜔𝛼 𝑎2 𝛼2 + 30𝑘 𝛼 𝑎0 𝑎1 𝑎2 𝛼1 + 60𝑘 3𝛼 𝑎0 𝑎1 𝛼1 𝛼12 + ⋯ + 55𝑘 3𝛼 𝑎12 𝛼12 𝛼2 = 0,

𝑈3:

30𝑘 𝛼 𝑎0 𝑎1 𝑎2 𝛼2 + 335𝑘 3𝛼 𝑎1 𝛼12 𝑎2 𝛼2 + ⋯ + 85𝑘 3𝛼 𝑎12 𝛼1 𝛼22 = 0,

𝑈4:

460𝑘 3𝛼 𝑎1 𝑎2 𝛼1 𝛼22 + 330𝑘 3𝛼 𝑎22 𝛼12 𝛼2 + 120𝑘 3𝛼 𝑎0 𝑎2 𝛼23 + ⋯ + 25𝑘 𝛼 𝑎1 𝑎22 𝛼1 = 0,

𝑈5:

120𝑘 5𝛼 𝑎1 𝛼25 + 200𝑘 3𝛼 𝑎1 𝛼23 𝑎2 + 430𝑘 3𝛼 𝑎22 𝛼1 𝛼22 + ⋯ + 25𝑘 𝛼 𝑎1 𝑎22 𝛼2 = 0,

𝑈6:

720𝑘 5𝛼 𝑎2 𝛼25 + 10𝑘 𝛼 𝑎23 𝛼2 + 180𝑘 3𝛼 𝑎22 𝛼23 = 0.


2

2

𝑎0 = −[e𝛼 ln(𝑘) ] 𝛼12 , 𝑎1 = −12[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 , 𝑎2 = −12[e𝛼 ln(𝑘) ] 𝛼22 , 𝑘 = 𝑘, 𝜔=e

ln(−𝛼4 1 )+5𝛼 ln(𝑘) 𝛼

.


2

2


i


2

2



2

2


1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). 599

Case IV: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 2

2

2

𝑢4 (𝜉) = −[e𝑦 ln(𝑘) ] 𝛼12 − 12[e𝛼 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 12[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , √3


√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1

ln(−𝛼4 1 )+5𝛼 ln(𝑘) 𝛼

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


𝑡.

2nd Solution Set: 2

2


2 𝛼 2 3𝛼 ln(−𝑘5𝛼 𝛼4 1 −5𝑘 𝑎0 −5𝑘 𝑎0 𝛼1 ) 𝛼

.


2


i


2



2


1


2

𝑢8 (𝜉) = 𝑎0 − 6[e𝛼 ln(𝑦) ] 𝛼1 𝛼2 𝑈 − 6[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , 1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

,

600


2 𝛼 2 3𝛼 ln(−𝑘5𝛼 𝛼4 1 −5𝑘 𝑎0 −5𝑘 𝑎0 𝛼1 ) 𝛼

𝑡.

3rd Solution Set: 3 2(−5𝑎0 +√5𝑎0 )𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 2 √ 1 −10𝑎0 +2√5𝑎0

(2

𝛼1

𝛼1

3 2(−5𝑎0 +√5𝑎0 )𝛼22

, 𝑎2 = − 2

𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.

Case I: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 3 2(−5𝑎0 +√5𝑎0 )𝛼2

𝑢9 (𝜉) = 𝑎0 − 2 1

𝛼1

3 2(−5𝑎0 +√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 3 2(−5𝑎0 +√5𝑎0 )𝛼2

𝑢10 (𝜉) = 𝑎0 − 2

𝛼1

3 2(−5𝑎0 +√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 3 2(−5𝑎0 +√5𝑎0 )𝛼2

𝑢11 (𝜉) = 𝑎0 − 2

𝛼1

1

3 2(−5𝑎0 +√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢12 (𝜉) = 𝑎0 − 1

√ 1 −10𝑎0 +2√5𝑎0 𝛼1

2

𝛼1 √3

𝑈−

3 2(−5𝑎0 +√5𝑎0 )𝛼22 2

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

𝛼12

𝑈2,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


(2

3 2(−5𝑎0 +√5𝑎0 )𝛼2


1 𝛼

) 𝑥.

4th Solution Set: 601

3 2(−5𝑎0 +√5𝑎0 )𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 2 √ 1 −10𝑎0 +2√5𝑎0

(− 2

𝛼1

𝛼1

3 2(−5𝑎0 +√5𝑎0 )𝛼22

, 𝑎2 = − 2

𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.

Case I: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 3 2(−5𝑎0 +√5𝑎0 )𝛼2

𝑢13 (𝜉) = 𝑎0 − 2 1

𝛼1

3 2(−5𝑎0 +√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 3 2(−5𝑎0 +√5𝑎0 )𝛼2

𝑢14 (𝜉) = 𝑎0 − 2

𝛼1

3 2(−5𝑎0 +√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 3 2(−5𝑎0 +√5𝑎0 )𝛼2

𝑢15 (𝜉) = 𝑎0 − 2

𝛼1

1

3 2(−5𝑎0 +√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 3 2(−5𝑎0 +√5𝑎0 )𝛼2

𝑢16 (𝜉) = 𝑎0 − 2 1

√3

√ 1 −10𝑎0 +2√5𝑎0 𝛼1

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

𝛼12

𝑈2,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


(− 2

𝛼1

3 2(−5𝑎0 +√5𝑎0 )𝛼22

𝑈−2


1 𝛼

) 𝑥.

5th Solution Set:

602

3 2(−5𝑎0 −√5𝑎0 )𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 2 √ 1 −10𝑎0 −2√5𝑎0

(2

𝛼1

𝛼1

3 2(−5𝑎0 −√5𝑎0 )𝛼22

, 𝑎2 = − 2

𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.

Case I: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 3 2(−5𝑎0 −√5𝑎0 )𝛼2

𝑢17 (𝜉) = 𝑎0 − 2 1

𝛼1

3 2(−5𝑎0 −√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 3 2(−5𝑎0 −√5𝑎0 )𝛼2

𝑢18 (𝜉) = 𝑎0 − 2

𝛼1

3 2(−5𝑎0 −√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 3 2(−5𝑎0 −√5𝑎0 )𝛼2

𝑢19 (𝜉) = 𝑎0 − 2

𝛼1

1

3 2(−5𝑎0 −√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 3 2(−5𝑎0 −√5𝑎0 )𝛼2

𝑢20 (𝜉) = 𝑎0 − 2 1

√3

√ 1 −10𝑎0 −2√5𝑎0 𝛼1

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

𝛼12

𝑈2,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


(2

𝛼1

3 2(−5𝑎0 −√5𝑎0 )𝛼22

𝑈−2


1 𝛼

) 𝑥.

6th Solution Set:

603

3 2(−5𝑎0 −√5𝑎0 )𝛼2

𝑎0 = 𝑎0 , 𝑎1 = − 2 √ 1 −10𝑎0 −2√5𝑎0

(− 2

𝛼1

𝛼1

3 2(−5𝑎0 −√5𝑎0 )𝛼22

, 𝑎2 = − 2

𝛼12

,𝑘 =

1 𝛼

) , 𝜔 = 0.

Case I: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 3 2(−5𝑎0 −√5𝑎0 )𝛼2

𝑢21 (𝜉) = 𝑎0 − 2 1

𝛼1

3 2(−5𝑎0 −√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −i and 𝛼2 = i, we have 3 2(−5𝑎0 −√5𝑎0 )𝛼2

𝑢22 (𝜉) = 𝑎0 − 2

𝛼1

3 2(−5𝑎0 −√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 3 2(−5𝑎0 −√5𝑎0 )𝛼2

𝑢23 (𝜉) = 𝑎0 − 2

𝛼1

1

3 2(−5𝑎0 −√5𝑎0 )𝛼22

𝑈−2

𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.217) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and α2 = √3i, we have 3 2(−5𝑎0 −√5𝑎0 )𝛼2

𝑢24 (𝜉) = 𝑎0 − 2 1

√3

√ 1 −10𝑎0 −2√5𝑎0 𝛼1

√3 √3 𝜉)+i𝐶2 sin𝑦 ( 𝜉) 2 2

𝐶1 cos𝛼 (

𝛼12

𝑈2,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


(− 2

𝛼1

3 2(−5𝑎0 −√5𝑎0 )𝛼22

𝑈−2


1 𝛼

) 𝑥.

604

𝑡=1

𝑡=5

𝑡 = 10



4.7.10. 7th order Sawada Kotera Equation Consider the 7th order Sawada Kotera equations [230] of fraction order 𝐷𝑡𝛼 𝑢 + 252𝑢3 𝐷𝑥𝛼 𝑢 + 63(𝐷𝑥𝛼 𝑢)3 + 378𝑢𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 126𝑢2 𝐷𝑥3𝛼 𝑢 + 63𝐷𝑥2𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 42𝐷𝑥𝛼 𝑢𝐷𝑥4𝛼 𝑢 + 21𝑢𝐷𝑥5𝛼 𝑢 + 𝐷𝑥7𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.218)


(4.219)


𝜔𝛼 𝐷𝜉𝛼 𝑢 + 252𝑘 𝛼 𝑢3 𝐷𝜉𝛼 𝑢 + 63𝑘 3𝛼 (𝐷𝜉𝛼 𝑢) + 378𝑘 3𝛼 𝑢𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 126𝑘 3𝛼 𝑢2 𝐷𝜉3𝛼 𝑢 + 63𝑘 5𝛼 𝐷𝜉2𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 42𝑘 5𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉4𝛼 𝑢 + 21𝑘 5𝛼 𝑢𝐷𝜉5𝛼 𝑢 + 𝑘 7𝛼 𝐷𝜉7𝛼 𝑢 = 0,

(4.220)

605


(4.221)


5 21𝑘 5𝛼 𝑎0 𝑎1 𝛼1` + 252𝑘 𝛼 𝑎03 𝑎1 𝛼1 + 126𝑘 3𝛼 𝑎02 𝑎1 𝛼13 + 𝜔𝛼 𝑎1 𝛼1 + 𝑘 7𝛼 𝑎1 𝛼17 = 0,

𝑈1 :

4 𝜔𝛼 𝑎1 𝛼2 + 2𝜔𝛼 𝑎2 𝛼1 + 128𝑘 7𝛼 𝑎2 𝛼17 + ⋯ + 651𝑘 5𝛼 𝑎0 𝑎1 𝛼1` 𝛼2 = 0,

𝑈2:

4 2𝜔𝛼 𝑎2 𝛼2 + 567𝑘 3𝛼 𝑎13 𝛼13 + 2205𝑘 5𝛼 𝑎1 𝑎2 𝛼15 + ⋯ + 8862𝑘 5𝛼 𝑎0 𝑎2 𝑦1` 𝛼2 = 0,

𝑈3:

252𝑘 𝛼 𝑎14 𝛼1 + 4032𝑘 5𝛼 𝑎22 𝛼15 + ⋯ + 35910𝑘 5𝛼 𝑎0 𝑎2 𝛼13 𝛼22 = 0,

𝑈4:

252𝑘 𝛼 𝑎14 𝛼2 + 16338𝑘 5𝛼 𝑎12 𝛼12 𝛼23 + ⋯ + 63000𝑘 5𝛼 𝑎0 𝑎2 𝛼12 𝛼23 = 0,

𝑈5:

1575𝑘 3𝛼 𝑎13 𝛼23 + 4536𝑘 3𝛼 𝑎23 𝛼13 + ⋯ + 50400𝑘 5𝛼 𝑎0 𝑎2 𝛼1 𝛼24 = 0,

𝑈6:

4284𝑘 5𝛼 𝑎12 𝛼25 + 173376𝑘 5𝛼 𝑎22 𝛼12 𝛼23 + ⋯ + 103824𝑘 5𝛼 𝑎1 𝛼1 𝛼24 𝑎2 = 0,

𝑈7:

504𝑘 𝛼 𝑎24 𝑦1 + 5040𝑘 7𝛼 𝑎1 𝛼27 + ⋯ + 20412𝑘 3𝛼 𝑎23 𝛼1 𝛼23 = 0,

𝑈8:

504𝑘 𝛼 𝑎24 𝛼2 + 8064𝑘 3𝛼 𝑎23 𝛼23 + 34272𝑘 5𝛼 𝑎22 𝛼25 + 40320𝑘 7𝛼 𝑎2 𝛼27 = 0.


1

2

2


4 ln(− 𝛼4 )+7𝛼 ln(𝑘) 3 1 𝛼

.


2

2

2


i


2

2

2


1


1

2

2

𝑢3 (𝜉) = − 3 [e𝛼 ln(𝑘) ] 𝛼12 − 4[e𝑦 ln(𝑘) ] 𝛼1 𝛼2 𝑈 − 4[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , 1

1


1

2

2


e

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1

where 𝑈 = 2 − 2 4 ln(− 𝛼4 )+7𝛼 ln(𝑘) 3 1 𝛼

√3 √3 𝜉)+i𝐶2 sin𝑦 ( 𝜉) 2 2

𝐶1 cos𝛼 (


𝑡.

2nd Solution Set: 2

2


𝛼 3 3𝛼 2 2 7𝛼 6 ln(−21𝑘5𝛼 𝑎0 𝛼4 1 −252𝑘 𝑎0 −126𝑘 𝑎0 𝛼1 −𝑘 𝛼1 ) 𝛼

.


2

𝑢5 (𝜉) = 𝑎0 − 2[e𝛼 ln(𝑘) ] 𝑦1 𝛼2 𝑈 − 2[e𝛼 ln(𝑘) ] 𝛼22 𝑈 2 , 1

i


2



2


1



2


√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


,

𝛼 3 3𝛼 2 2 7𝛼 6 ln(−21𝑘5𝛼 𝑎0 𝛼4 1 −252𝑘 𝑎0 −126𝑘 𝑎0 𝛼1 −𝑘 𝛼1 ) 𝛼

𝑡.

3rd Solution Set: 1

2

1

𝑎0 = 𝑎0 , 𝑎1 = −

2

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼22

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼2

, 𝑎2 = −

𝛼1

, 𝜔=

𝛼12

1/𝛼 1 2 √((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )

0, 𝑘 =

.

𝛼1

( ) Case I: Substituting the values of unknown into Eq. (4.221) coupled with the solution of Eq. (4.185) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1

𝑢9 (𝜉) = 𝑎0 − 1

2

1

2

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼22

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼2

𝑈−

𝛼1

𝛼12

𝑈2,

i


𝑢10 (𝜉) = 𝑎0 −

2

1

𝑈−

𝛼1

2

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼22

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼2

𝛼12

𝑈2,

1


𝑢11 (𝜉) = 𝑎0 − 1

2

1

𝑈−

𝛼1 1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉).

608

2

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼22

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼2

𝛼12

𝑈2,


𝑢12 (𝜉) = 𝑎0 −

2

1

𝑈−

𝛼1 √3

2

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼22

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼2

𝛼12

𝑈2,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, 1/𝛼

1 2 √((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )

and i = √−1. In all cases 𝜉 =

𝑥.

𝛼1

( 4th

)

Solution Set: 1

2

1

𝑎0 = 𝑎0 , 𝑎1 = −

2

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼22

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼2

, 𝑎2 = −

𝛼1

, 𝜔=

𝛼12

1/𝛼 1 2 √((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )

0, 𝑘 =

−

.

𝛼1

(

)


𝑢13 (𝜉) = 𝑎0 − 1

2

1

𝑈−

𝛼1

2

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼22

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼2

𝛼12

𝑈2,

i


𝑢14 (𝜉) = 𝑎0 −

2

1

𝑈−

𝛼1

2

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼22

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼2

𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.221) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have

609

1

𝑢15 (𝜉) = 𝑎0 −

2

1

𝑈−

𝛼1

1

2

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼22

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼2

𝛼12

𝑈2,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝑦 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.221) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 1

𝑢16 (𝜉) = 𝑎0 − 1

2

1

𝑈−

𝛼1 √3

2

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼22

2((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )𝛼2

𝛼12

𝑈2,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, 1/𝛼

1 2 √((−7)3 𝑎0 −(−7)3 𝑎0 −7𝑎0 )

and i = √−1. In all cases 𝜉 =

−

𝑥.

𝛼1

(

)

5th Solution Set: 1

𝑎0 = 𝑎0 , 𝑎1 = −

2𝛽𝛼2 𝛼1

, 𝑎2 = −

2𝛽𝛼22

√𝛽 𝛼

, 𝑘 = (2𝛼 ) , 𝜔 = 0.

𝛼12

1


2𝛽𝛼2 𝛼1

𝑈−

2𝛽𝛼22 𝛼12

𝑈2,

i


2𝛽𝛼2 𝛼1

𝑈−

2𝛽𝛼22 𝛼12

𝑈2,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.221) coupled with the solution of Eq. (4.185) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢19 (𝜉) = 𝑎0 − 1

2𝛽𝛼2 𝛼1

𝑈−

2𝛽𝛼22 𝛼12

𝑈2,

1


Case IV: Substituting the values of unknown into Eq. (4.221) coupled with the solution of Eq. (4.185) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢20 (𝜉) = 𝑎0 − 1

2𝛽𝛼2 𝛼1

𝑈−

2𝛽𝛼22 𝛼12

√3

𝑈2, 1

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝑦 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

√𝛽 𝛼

, and i = √−1. In all cases 𝜉 = (2𝛼 ) 𝑥, and 𝛽 = 1

2

1

2

−2𝑎0 (−7)3 + 2𝑎0 (−7)3 − 28𝑎0 + 2 i√3 (𝑎0 (−7)3 + 𝑎0 (−7)3 ). 6th Solution Set: 1

𝑎0 = 𝑎0 , 𝑎1 = −

2𝛽𝛼2 𝛼1

, 𝑎2 = −

2𝛽𝛼22

√𝛽 𝛼

, 𝑘 = (− 2𝛼 ) , 𝜔 = 0.

𝛼12

1


2𝛽𝛼2 𝛼1

𝑈−

2𝛽𝛼22 𝛼12

𝑈2,

i


2𝛽𝛼2 𝛼1

𝑈−

2𝛽𝛼22 𝛼12

𝑈2,

1


2𝛽𝛼2 𝛼1

1

𝑈−

2𝛽𝛼22 𝛼12

𝑈2,

1


2𝛽𝛼2 𝛼1

𝑈−

2𝛽𝛼22 𝛼12

𝑈2,

611

1

√3

1

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

√𝛽 𝛼

, and i = √−1. In all cases 𝜉 = (− 2𝛼 ) 𝑥, and 1

2

1

2

𝛽 = −2𝑎0 (−7)3 + 2𝑎0 (−7)3 − 28𝑎0 + 2 i√3 (𝑎0 (−7)3 + 𝑎0 (−7)3 ),

𝑡=0

𝑡 = 0.5

𝑡=1



4.8. Modified U-expansion Method I modified the U-Expansion Method and applied it for different problems. This modification gives more solution set as compare to U-Expansion method. This is reliable and effective.

4.8.1. Methodology In order to simultaneously obtain more periodic wave solutions expressed in rational hyperbolic function and rational trigonometry function to nonlinear equations, we introduce modified U-expansion method. We briefly show what modified U-expansion 612

method is and how to use it to obtain various periodic wave solutions to nonlinear equations. Suppose a nonlinear equation for 𝑈(𝑥, 𝑡) is given by We consider the general nonlinear PDE of the type 𝑃(𝑢, 𝐷𝑡𝛼 𝑢, 𝐷𝑥𝛼 𝑢, 𝐷𝑦𝛼 𝑢, 𝐷𝑧𝛼 𝑢, 𝐷𝑡2𝛼 𝑢, 𝐷𝑥2𝛼 𝑢, 𝐷𝑦2𝛼 𝑢, 𝐷𝑧2𝛼 𝑢, 𝐷𝑡𝛼 𝐷𝑥𝛼 𝑢, … ) = 0,

(4.222)

where 𝑃 is a polynomial in its arguments. The essence of the modified U-expansion method can be presented in the following steps: Step 1: Seek Solitary wave solutions of Eq. (4.222) by taking 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡, and transform Eq. (4.222) to the ordinary differential equation. 𝑃(𝑢, 𝜔𝛼 𝐷𝜉𝛼 𝑢, 𝑘 𝛼 𝐷𝜉𝛼 𝑢, 𝑙 𝛼 𝐷𝜉𝛼 𝑢, 𝑚𝛼 𝐷𝜉𝛼 𝑢, 𝜔2𝛼 𝐷𝜉2𝛼 𝑢, 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢, 𝑙 2𝛼 𝐷𝜉2𝛼 𝑢, … ) = 0,

(4.223)

where 𝜔, 𝑘, 𝑙 and 𝑚 is constant and where prime denotes the derivative with respect to 𝜉. Step 2: If possible, integrate Eq. (4.223) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero. Step 3: According to modified U-expansion method, we assume that the wave solution can be expressed in the following form 𝑀 𝑛 −𝑛 𝑢(𝜉) = ∑𝑀 , 𝑛=0 𝑎𝑛 𝑈 + ∑𝑛=1 𝑏𝑛 𝑈

(4.224)


(4.225)


613

Table 4.3: Solutions of Eq. (4.225) for different values of 𝛼1 and 𝛼2 . 𝛼1

𝛼2

Cases

𝑈(𝜉)

−2i

2i

I

1 i − tan𝛼 (𝜉) 2 2

−i

I

II

4

2

III

−√3i √3i

IV

1 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)) 2 1 1 −1 − tanh𝛼 (𝜉) − coth𝛼 (𝜉) 2 2 √3 √3 i 𝐶1 sin𝛼 ( 2 𝜉) + 𝐶2 cos𝛼 ( 2 𝜉) 1

1 − 2 2

√3 √3 𝐶1 cos𝛼 ( 2 𝜉) + i𝐶2 sin𝛼 ( 2 𝜉)

Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term(s) of highest order with the highest order nonlinear term(s) in Eq. (4.223). Step 5: Substituting (4.224) into Eq. (4.223) with (4.225) will yields an algebraic equation involving power of U. Equating the coefficients of like power of U to zero gives a system of algebraic equations for 𝑎𝑖 , 𝑘, 𝑙, 𝑚 and 𝜔. Then, we solve the system with the aid of a computer algebra system (CAS), such as MAPLE 13, to determine these constants. Step 6: Putting these constant into Eq. (4.224), coupled with the well known solutions of Eq. (4.225), we can obtained the more general type and new exact traveling wave solution of the nonlinear partial differential equation (4.222).

4.8.2. (2+1) Burger’s Dimensional Equation Consider the (2+1)-dimensional Burger’s equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝛽(𝑢𝐷𝑥𝛼 𝑢 + 𝑢𝐷𝑦𝑦 𝑢) + 𝛾(𝐷𝑥2𝛼 𝑢 + 𝐷𝑦2𝛼 𝑢) = 0, 0 < 𝛼 ≤ 1.

(4.226)

To convert Eq. (4.226) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝜔𝑡,

(4.227)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.227) into Eq. (4.226) and using the chain rule we obtained 𝐷𝜉𝛼 𝑢 + 𝛽(𝑘 𝛼 𝑢𝐷𝜉𝛼 𝑢 + 𝑙 𝛼 𝑢𝐷𝜉𝛼 𝑢) + 𝛾(𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 + 𝑙 2𝛼 𝐷𝜉2𝛼 𝑢) = 0, By applying the homogenous balancing principle for 𝛼 = 1, we have 614

(4.228)

𝑀 + 2 = 𝑀 + 𝑀 + 1, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.224), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑏1 𝑈 −1 .

(4.229)

Putting Eq. (4.229) into Eq. (4.228) coupled with Eq. (4.225); the Eq. (4.229) yields an algebraic equation involving power of U as 𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 = 0. Compare the like powers of U we have system of equations 𝑈0:

−𝛼𝑘 𝛼 𝛼1 𝑏12 − 𝛼𝑙 𝛼 𝛼1 𝑏12 = 0,

𝑈1 :

−𝛼𝑘 𝛼 𝛼2 𝑏12 − 𝛼𝑙 𝛼 𝛼2 𝑏12 − 𝛽𝑘 2𝑦 𝑏1 𝛼12 − ⋯ − 𝛼𝑙 𝛼 𝑎0 𝛼1 𝑏1 = 0,

𝑈2:

−𝜔𝛼 𝛼2 𝑏1 − 𝛼𝑘 𝛼 𝑎0 𝛼2 𝑏1 − 𝛼𝑙 𝛼 𝑎0 𝛼2 𝑏1 − 𝛽𝑘 2𝛼 𝑏1 𝛼1 𝛼2 − 𝛽𝑙 2𝛼 𝑏1 𝛼1 𝛼2 = 0,

𝑈3:

−𝛽𝑘 2𝛼 𝑎1 𝛼12 − 𝛽𝑙 2𝛼 𝑎1 𝛼12 + 𝜔𝛼 𝛼1 𝑎1 + 𝛼𝑘 𝛼 𝑎0 𝛼1 𝑎1 + 𝛼𝑙 𝛼 𝑎0 𝛼1 𝑎1 = 0,

𝑈4:

𝛼𝑘 𝛼 𝛼1 𝑎12 + 𝛼𝑙 𝛼 𝛼1 𝑎12 + 𝜔𝛼 𝛼2 𝑎1 + ⋯ − 3𝛽𝑙 2𝛼 𝑎1 𝛼1 𝛼2 = 0,

𝑈5:

𝛼𝑘 𝛼 𝛼2 𝑎12 + 𝛼𝑙 𝛼 𝛼2 𝑎12 − 2𝛽𝑘 2𝛼 𝑎1 𝛼22 − 2𝛽𝑙 2𝛼 𝑎1 𝛼22 = 0.

Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: 𝑘 = 𝑘, 𝜔 = e

ln(𝛽𝑘2𝛼 𝛼1 +𝛽𝑙2𝛼𝛼1 −𝛼𝑘𝛼 𝑎0 −𝛼𝑙𝛼 𝑎0 ) ) 𝛼

(

, 𝑎0 = 𝑎0 , 𝑎1 =

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 )

, 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢1 (𝜉) = 𝑎0 + 1

i

2

2

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝑦 ) 𝛼(𝑘 𝛼 +𝑙𝛼 )

𝑈,

where 𝑈 = − tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢2 (𝜉) = 𝑎0 +

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑦𝛼 +𝑙𝛼 )

𝑈,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢3 (𝜉) = 𝑎0 +

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 )

𝑈, 615

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢4 (𝜉) = 𝑎0 +

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 ) √3

𝑈, √3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


, and i = √−1.

ln(𝛽𝑘2𝛼 𝛼1 +𝛽𝑙2𝛼 𝛼1 −𝛼𝑘𝛼 𝑎0 −𝛼𝑙𝛼 𝑎0 ) ) 𝛼

(

𝑡.

2nd Solution Set: 𝑘 = 0, 𝜔 = e

ln(𝛽𝑙2𝛼 𝛼1 −𝛼𝑙𝛼 𝑎0 ) ) 𝛼

(

, 𝑎0 = 𝑎0 , 𝑎1 =

2𝛽𝑙𝛼 𝛼2 𝛼

, 𝑏1 = 0.



𝑈,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝑦1 = −i and 𝛼2 = i, we have 𝑢6 (𝜉) = 𝑎0 +


𝑈,

1



𝑈,

1

1

2

2

where 𝑈 = −1 − tanh𝛼 (𝜉) − coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢8 (𝜉) = 𝑎0 + 1

2𝛽𝑙𝛼 𝛼2 𝛼 √3

𝑈, √3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

616

ln(−

In all cases 𝜉 = e

1 1 ln( 2 ) 2𝛼 2𝛼1

(

𝑥+e

1−1+2𝑎 ) 2 𝛼1 ) 𝛼

𝑡.

3rd Solution Set: 𝑎0 =

𝛽𝛼1 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 )

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 )

, 𝑎1 =

𝛼(𝑘 𝛼 +𝑙𝛼 )

, 𝑏1 = 0, 𝑘 = 𝑘, 𝜔 = 0.

Case I: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝛽𝛼1 (𝑘 2𝛼 +𝑙2𝛼 )

𝑢9 (𝜉) = 1

𝛼(𝑘 𝛼 +𝑙𝛼 )

+

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 )

𝑈,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢10 (𝜉) =

𝛽𝛼1 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 )

+

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 )

𝑈,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢11 (𝜉) =

𝛽𝛼1 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 ) 1

+

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝑦 +𝑙𝛼 )

𝑈,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢12 (𝜉) = 1

𝛽𝛼1 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 ) √3

+

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 )

𝑈,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

In all cases 𝜉 = 𝑘𝑥. 4th Solution Set: 1

𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑏1 = 𝑏1 , 𝜔 = e𝛼ln(−2𝛽𝛼1

)+2𝛼ln(𝑙)

1

, 𝑘 = (−𝑙 𝛼 )𝛼 .

Case I: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢13 (𝜉) = 𝑎0 + 𝑏1 𝑈 −1 , 617

1

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢14 (𝜉) = 𝑎0 + 𝑏1 𝑈 −1 , 1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢15 (𝜉) = 𝑎0 + 𝑏1 𝑈 −1 , 1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢16 (𝜉) = 𝑎0 + 𝑏1 𝑈 −1 , 1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 ( 1

1

In all cases 𝜉 = (−𝑙𝛼 )𝛼 𝑥 + e𝛼ln(−2𝛽𝛼1

, and i = √−1.

)+2𝛼ln(𝑙)

𝑡.

5th Solution Set: 1

𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝜔 = e𝛼ln(−𝛼𝑘

𝛼 𝑎 −𝛼𝑙 𝛼 𝑎 −𝛽𝑘 2𝛼 𝛼 −𝛽𝑙 2𝛼 𝛼 ) 0 0 1 1

, 𝑘 = 𝑘.

Case I: Substituting the values of unknown into Eq. (16) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢17 (𝜉) = 𝑎0 + 𝑎1 𝑈, 1

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢18 (𝜉) = 𝑎0 + 𝑎1 𝑈, 1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢19 (𝜉) = 𝑎0 + 𝑎1 𝑈, 618

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.229) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢20 (𝜉) = 𝑎0 + 𝑎1 𝑈, 1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

In all cases 𝜉 = 𝑘𝑥 + e𝛼ln(−𝛼𝑘

, and i = √−1.

𝛼 𝑎 −𝛼𝑙 𝛼 𝑎 −𝛽𝑘 2𝛼 𝛼 −𝛽𝑙 2𝛼 𝛼 ) 0 0 1 1

𝑡.

Fig 4.53: Solitary wave solution for the given PDE of fractional order (12)

Fig 4.53 (a): 2D travelling wave solution of Eq. (4.226) for different values of parameters and 𝑡 = 1.

Fig 4.53 (b): 3D travelling wave solution of Eq. (4.226) for different values of parameters

4.8.3. (3+1) Dimensional Burger’s Equation Consider the (3+1)-Dimension Burger’s equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝛽(𝑢𝐷𝑥𝛼 𝑢 + 𝑢𝐷𝑦𝛼 𝑢 + 𝑢𝐷𝑧𝛼 𝑢) + 𝛾(𝐷𝑥2𝛼 𝑢 + 𝐷𝑦2𝛼 𝑢 + 𝐷𝑧2𝛼 𝑢) = 0,0 < 𝛼 ≤ 1. (4.230) To convert Eq. (4.230) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡,

(4.231)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.231) into Eq. (4.230) and using the chain rule we obtained 𝐷𝜉𝛼 𝑢 + 𝛽(𝑘 𝛼 + 𝑙 𝛼 + 𝑚𝛼 )𝑢𝐷𝜉𝛼 𝑢 + 𝛾(𝑘 2𝛼 + 𝑙 2𝛼 + 𝑚2𝛼 )𝐷𝜉2𝛼 𝑢 = 0. By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 2 = 𝑀 + 𝑀 + 1, 𝑀 = 1. 619

(4.232)

Using the value of 𝑀 into Eq. (4.224), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑏1 𝑈 −1 .

(4.233)

Putting Eq. (4.233) into Eq. (4.232) coupled with Eq. (4.225); the Eq. (4.233) yields an algebraic equation involving power of U as 1

− (1+𝜀𝑈)3 [𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 ] = 0. Compare the like powers of U we have system of equations 𝑈0:

−𝛼𝑘 𝛼 𝛼1 𝑏12 − 𝛼𝑙 𝛼 𝛼1 𝑏12 − 𝛼𝑚𝛼 𝛼1 𝑏12 = 0,

𝑈1 :

−𝛼𝑘 𝛼 𝛼2 𝑏12 − 𝛼𝑙 𝛼 𝛼2 𝑏12 − ⋯ − 𝛼𝑙 𝛼 𝑎0 𝛼1 𝑏1 − 𝛼𝑚𝛼 𝑎0 𝛼1 𝑏1 = 0,

𝑈2:

−𝜔𝛼 𝛼2 𝑏1 − 𝛼𝑘 𝛼 𝑎0 𝛼2 𝑏1 − ⋯ − 𝛽𝑙 2𝛼 𝑏1 𝛼1 𝛼2 − 𝛽𝑚2𝛼 𝑏1 𝛼1 𝛼2 = 0,

𝑈3:

−𝛽𝑘 2𝛼 𝑎1 𝛼12 − 𝛽𝑙 2𝛼 𝑎1 𝛼12 − ⋯ + 𝛼𝑙 𝛼 𝑎0 𝛼1 𝑎1 + 𝛼𝑚𝛼 𝑎0 𝛼1 𝑎1 = 0,

𝑈4:

𝛼𝑘 𝛼 𝛼1 𝑎12 + 𝛼𝑙 𝛼 𝛼1 𝑎12 + 𝛼𝑚𝛼 𝛼1 𝑎12 + ⋯ − 3𝛽𝑚2𝛼 𝑎1 𝛼1 𝛼2 = 0,

𝑈5:

𝛼𝑘 𝛼 𝛼2 𝑎12 + 𝛼𝑙 𝛼 𝛼2 𝑎12 + 𝛼𝑚𝛼 𝛼2 𝑎12 − ⋯ − 2𝛽𝑙 2𝛼 𝑎1 𝛼22 − 2𝛽𝑚2𝛼 𝑎1 𝛼22 = 0.

Solving the above system for unknown parameters, we have the following solution set 1st Solution Set: 𝛽𝑘2𝛼 𝛼1 +𝛽𝑙2𝛼 𝛼1 +𝛽𝑚2𝛼 𝛼1 ln( ) −𝛼𝑘𝛼 𝑎0 −𝛼𝑙𝛼 𝑎0 −𝛼𝑚𝛼 𝑎0 ( ) 𝛼

𝑘 = 𝑘, 𝜔 = e

, 𝑎0 = 𝑎0 , 𝑎1 =

2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 )

,

𝑏1 = 0. Case I: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢1 (𝜉) = 𝑎0 + 1


𝑈,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢2 (𝜉) = 𝑎0 +


𝑈,

1



𝑈, 620

1

1


2𝛽𝛼2 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 ) √3

𝑈,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


, and i = √−1.

ln(𝛽𝑘2𝛼 𝛼1 +𝛽𝑙2𝛼 𝛼1 +𝛽𝑚2𝛼 𝛼1 −𝛼𝑘𝛼 𝑎0 −𝛼𝑙𝛼 𝑎0 −𝛼𝑚𝛼 𝑎0 ) ) 𝛼

(

𝑡.

2nd Solution Set: 1

𝑘 = 𝑘, 𝜔 = e𝛼ln(−𝛼𝑘

𝛼 𝑎 −𝛼𝑙 𝛼 𝑎 −𝛼𝑚𝛼 𝑎 −𝛽𝑘 2𝛼 𝛼 −𝛽𝑙 2𝛼 𝛼 −𝛽𝑚2𝛼 𝛼 ) 0 0 0 1 1 1

, 𝑎0 = 𝑎0 , 𝑎1 = 𝑎1 , 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢5 (𝜉) = 𝑎0 + 𝑎1 𝑈, 1

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢6 (𝜉) = 𝑎0 + 𝑎1 𝑈, 1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢7 (𝜉) = 𝑎0 + 𝑎1 𝑈 , 1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢8 (𝜉) = 𝑎0 + 𝑎1 𝑈, 1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

In all cases 𝜉 = 𝑘𝑥 + e𝛼ln(−𝛼𝑘

, and i = √−1.

𝛼 𝑎 −𝛼𝑙 𝛼 𝑎 −𝛼𝑚𝛼 𝑎 −𝑦𝑘 2𝛼 𝛼 −𝛽𝑙 2𝛼 𝛼 −𝛽𝑚2𝛼 𝛼 ) 0 0 0 1 1 1


𝑡.

𝑎0 =

𝛽𝛼1 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 )

, 𝑎1 =


, 𝑏1 = 0, 𝑘 = 𝑘, 𝜔 = 0.

Case I: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝛽𝛼1 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 )

𝑢9 (𝜉) = 1

𝛼(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 )

+


𝑈,

i


𝛽𝛼1 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 )

+


𝑈,

1


𝛽𝛼1 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 ) 1

+


𝑈,

1


𝛽𝛼1 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 ) 𝛼(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 ) √3

+


𝑈,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

In all cases 𝜉 = 𝑘𝑥. 4th Solution Set: 1

𝑎0 = 𝑎0 , 𝑎1 = 0, 𝑏1 = 𝑏1 , 𝜔 = e𝛼ln(−𝛽𝑙

2𝛼 𝛼 −2𝛽𝛼 𝑙 𝛼 𝑚𝛼 −𝛽𝑚2𝛼 𝛼 −𝛽𝑙 2𝛼 𝛼 −𝛽𝑚2𝛼 𝛼 ) 1 1 1 1 1

,

1

𝑘 = (−𝑙 𝛼 − 𝑚𝛼 )𝛼 . Case I: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢13 (𝜉) = 𝑎0 + 𝑏1 𝑈 −1 , 1

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). 622

Case II: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢14 (𝜉) = 𝑎0 + 𝑏1 𝑈 −1 , 1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢15 (𝜉) = 𝑎0 + 𝑏1 𝑈 −1 , 1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.233) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢16 (𝜉) = 𝑎0 + 𝑏1 𝑈 −1 , 1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

, and i = √−1.

1

In all cases 𝜉 = (−𝑙𝛼 − 𝑚𝛼 )𝛼 𝑥 + e𝛼ln(−𝛽𝑙

2𝛼 𝛼 −2𝛽𝛼 𝑙 𝛼 𝑚𝛼 −𝛽𝑚2𝛼 𝛼 −𝛽𝑙 2𝛼 𝛼 −𝛽𝑚2𝛼 𝛼 ) 1 1 1 1 1

Fig 4.54 (a): 2D Kink wave solution of

𝑡.

Fig 4.54 (b): 3D Kink wave solution of

Eq. (4.230) for different values of parameters


4.8.4. (2+1) Dimensional Burger’s Equation Consider the (2+1)-dimensional Burger’s equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝑎𝑢𝐷𝑥𝛼 𝑢 + 𝜈𝐷𝑥2𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.234)


(4.235) 623

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.235) into Eq. (4.234) and using the chain rule we obtained 𝐷𝜉𝛼 𝑢 + 𝑘 𝛼 𝑢𝐷𝜉𝛼 𝑢 + 𝜈𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.236)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 2 = 𝑀 + 𝑀 + 1, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.224), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑏1 𝑈 −1 .

(4.237)

Putting Eq. (4.237) into Eq. (4.236) coupled with Eq. (4.225); the Eq. (4.237) yields an algebraic equation involving power of U as 𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 = 0. Compare the like powers of U we have system of equations 𝑈0:

−𝑎𝑘 𝛼 𝛼1 𝑏12 = 0,

𝑈1 :

−𝜔𝛼 𝛼1 𝑏1 − 𝑎𝑘 𝛼 𝑎0 𝛼1 𝑏1 − 𝑎𝑘 𝛼 𝛼2 𝑏12 − 𝑣𝑘 2𝛼 𝑏1 𝛼12 = 0,

𝑈2:

−𝜔𝛼 𝛼2 𝑏1 − 𝑎𝑘 𝛼 𝑎0 𝛼2 𝑏1 − 𝑣𝑘 2𝛼 𝑏1 𝛼1 𝛼2 = 0,

𝑈3:

𝜔𝛼 𝛼1 𝑎1 + 𝑎𝑘 𝛼 𝑎0 𝛼1 𝑎1 − 𝑣𝑘 2𝛼 𝑎1 𝛼12 = 0,

𝑈4:

𝜔𝛼 𝛼2 𝑎1 + 𝑎𝑘 𝛼 𝑎0 𝛼2 𝑎1 + 𝑎𝑘 𝛼 𝛼1 𝑎12 − 3𝑣𝑘 2𝛼 𝑎1 𝛼1 𝛼2 = 0,

𝑈5:

𝑎𝑘 𝛼 𝛼2 𝑎12 − 2𝑣𝑘 2𝛼 𝑎1 𝛼22 = 0.

Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: 𝑘 = 𝑘, 𝜔 = e

ln(𝜈𝑘2𝛼 𝛼1 −𝛼𝑘𝛼 𝑎0 ) ) 𝛼

(

, 𝑎0 = 𝑎0 , 𝑎1 =

2𝜈e𝛼ln(𝑘) 𝛼2 𝛼(𝑘 𝛼 +𝑙𝛼 )

, 𝑏1 = 0.



𝑈,

i



𝑈,

624

1



1

𝑈,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.237) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢4 (𝜉) = 𝑎0 + 1


𝑈,

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


ln(𝜈𝑘2𝛼 𝛼1 −𝛼𝑘𝛼 𝑎0 ) ) 𝛼

(

, and i = √−1.

𝑡.

2nd Solution Set: 𝑘 = 𝑘, 𝜔 = 0, 𝑎0 =

𝜈e𝛼ln(𝑘) 𝛼1 𝑎

, 𝑎1 = −

2𝜈e𝛼ln(𝑘) 𝛼2 𝑎

, 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.237) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝜈e𝛼ln(𝑘) 𝛼1

𝑢5 (𝜉) = 1

𝑎

−


𝑈,

i



−


𝑈,

1


𝜈e𝛼ln(𝑘) 𝛼1 𝑎 1

−


𝑈,

1


625

Case IV: Substituting the values of unknown into Eq. (4.237) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢8 (𝜉) = 1


−

2𝜈e𝛼ln(𝑘) 𝛼2

√3

𝑎

𝑈,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


, and i = √−1.

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝑦) 2 2

𝐶1 cos𝛼 (

In all cases 𝜉 = 𝑘𝑥.

Fig 4.55 (a): 2D Kink wave solution of Eq. (4.234) different values of parameters and 𝑡 = 1.

Fig 4.55 (b): 3D Kink wave solution of for Eq. (4.234) for different values of parameters

4.8.5. Fisher’s Equation Consider the Fisher’s equation [230] of fraction order 𝐷𝑡𝛼 𝑢 − 𝐷𝑥2𝛼 𝑢 − 𝑢(1 − 𝑢) = 0, 0 < 𝛼 ≤ 1.

(4.238)


(4.239)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.239) into Eq. (4.238) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 − 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 − 𝑢 + 𝑢2 = 0.

(4.240)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 2 = 2𝑀, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.224), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑎2 𝑈 2 + 𝑏1 𝑈 −1 + 𝑏2 𝑈 −2 . 626

(4.241)


𝑏22 = 0,

𝑈1 :

2𝑏1 𝑏2 = 0,

𝑈2:

−2𝜔𝛼 𝛼1 𝑏2 − 4𝑘 2𝛼 𝑏2 𝛼12 − 𝑏2 + 𝑏12 + 2𝑎0 𝑏2 = 0,

𝑈3:

−6𝑘 2𝛼 𝑏2 𝛼1 𝛼2 − 𝜔𝛼 𝛼1 𝑏1 − 2𝜔𝛼 𝛼2 𝑏2 − 𝑘 2𝛼 𝑏1 𝛼12 − 𝑏1 + 2𝑎0 𝑏1 + 2𝑎1 𝑏2 = 0,

𝑈4:

−𝑘 2𝛼 𝑏1 𝛼1 𝛼2 − 𝜔𝛼 𝛼2 𝑏1 − 2𝑘 2𝛼 𝑏2 𝛼22 − 𝑎0 + 𝑎02 + 2𝑎1 𝑏1 + 2𝑎2 𝑏2 = 0,

𝑈5:

𝜔𝛼 𝛼1 𝑎1 − 𝑘 2𝛼 𝑎1 𝛼12 − 𝑎1 + 2𝑎0 𝑎1 + 2𝑎2 𝑏1 = 0,

𝑈6:

−3𝑘 2𝛼 𝑎1 𝛼1 𝛼2 + 2𝜔𝛼 𝛼1 𝑎2 + 𝜔𝛼 𝛼2 𝑎1 − 4𝑘 2𝛼 𝑎2 𝛼12 − 𝑎2 + 𝑎12 + 2𝑎0 𝑎2 = 0,

𝑈7:

−10𝑘 2𝛼 𝑎2 𝛼1 𝛼2 + 2𝜔𝛼 𝛼2 𝑎2 − 2𝑘 2𝛼 𝑎1 𝛼22 + 2𝑎1 𝑎2 = 0,

𝑈8:

−6𝑘 2𝛼 𝑎2 𝛼22 + 𝑎22 = 0.

Solving the above system for unknown parameters, we have the following solution set 1st Solution Set: ln(

1 ( 2

1 ) 6𝛼2 1 ) 𝛼

𝑘=e

5 ln( 2 ) 6𝛼1 ( ) 𝛼

,𝜔 = e

2𝛼

1 ) 6𝛼2 1 ) 𝛼

ln(−

1 ( 2

𝛼22 , 𝑏1 =

, 𝑎0 = 0, 𝑎1 = 0, 𝑎2 = 6 e [

]

0, 𝑏2 = 0. Case I: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 2𝛼

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

𝛼22 𝑈 2 ,

𝑢1 (𝜉) = 6 e [ 1

]

i


627

2𝛼

1 ) 6𝛼2 1 ) 𝛼

ln(−

1 ( 2

𝛼22 𝑈 2 ,

𝑢2 (𝜉) = 6 e [

]

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 2𝛼

1 ) 6𝛼2 1 ) 𝛼

ln(−

1 ( 2

𝛼22 𝑈 2 ,

𝑢3 (𝜉) = 6 e [

]

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 2𝛼

1 ) 6𝛼2 1 ) 𝛼

ln(−

1 ( 2

𝛼22 𝑈 2 ,

𝑢4 (𝜉) = 6 e [ 1

] √3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1 ln( 2 ) 6𝛼1 1 ( ) 2 𝛼


ln(

(

𝑥+e

, and i = √−1.

5 ) 6𝛼2 1 ) 𝛼

𝑡.

2nd Solution Set: 1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

𝑘=e

5 ) 6𝛼2 1 ) 𝛼

(

,𝜔 = e

2𝛼

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

ln(−

𝛼22 ,

, 𝑎0 = 1, 𝑎1 = 0, 𝑎2 = 6 e [

]

𝑏1 = 0, 𝑏2 = 0, Case I: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 628

2𝛼

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

𝛼22 𝑈 2 ,

𝑢5 (𝜉) = 1 + 6 e [ 1

]

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 2𝛼

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

𝛼22 𝑈 2 ,

𝑢6 (𝜉) = 1 + 6 e [

]

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 2𝛼

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

𝛼22 𝑈 2 ,

𝑢7 (𝜉) = 1 + 6 e [

]

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 2𝛼

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

𝛼22 𝑈 2 ,

𝑢8 (𝜉) = 1 + 6 e [ 1

]

√3 √3 1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼


, and i = √−1.

5 ) 6𝛼2 1 ) 𝛼

ln(−

(

𝑥+e

𝑡.


1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

𝑘=e

ln(

(

5 ) 6𝛼2 1 ) 𝛼

,𝜔 = e

2𝛼

1 ) 6𝛼2 1 ) 𝛼

ln(−

, 𝑎0 = 0, 𝑎1 = −

2𝛼2 𝛼1

1 ( 2

𝛼22 ,

, 𝑎2 = 6 e [

]

𝑏1 = 𝑏2 = 0. Case I: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 2𝛼

1 ) 6𝛼2 1 ) 𝛼

ln(−

𝑢9 (𝜉) = −

2𝛼2 𝛼1

1 ( 2

𝛼22 𝑈 2 ,

𝑈+6 e [

1

]

i


𝑢10 (𝜉) = −

2𝛼2 𝛼1

2𝛼

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

𝛼22 𝑈 2 ,

𝑈+6 e [

]

1


𝑢11 (𝜉) = −

2𝛼2 𝛼1

𝛼22 𝑈 2 ,

𝑈+6 e [

1

2𝛼

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

] 1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have

630

𝑢12 (𝜉) = −

2𝛼2 𝛼1

𝛼22 𝑈 2 ,

𝑈+6 e [

1

2𝛼

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼

]

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1 ln(− 2 ) 6𝛼1 1 ( ) 2 𝛼


, and i = √−1.

5 ln( 2 ) 6𝛼1 ( ) 𝛼

𝑥+e

𝑡.

4th Solution Set: 1

𝛼

)

𝑎0 = 0, 𝑎1 = − 𝛼2 , 𝑎2 = 0, 𝑏1 = 0, 𝑏2 = 0, 𝜔 = e−𝛼ln(𝛼1 , 𝑘 = 0. 1

Case I: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝛼

𝑢13 (𝜉) = − 𝛼2 𝑈, 1

1

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝛼

𝑢14 (𝜉) = − 𝛼2 𝑈, 1

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝛼

𝑢15 (𝜉) = − 𝛼2 𝑈, 1

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.241) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝛼

𝑢16 (𝜉) = − 𝛼2 𝑈, 1

1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

631

1

)

In all cases 𝜉 = e−𝛼ln(𝛼1 𝑡.





4.8.6. (2+1) Dimensional FitzHugh-Nagumo Equation Consider the (2+1)-dimensional FitzHugh-Nagumo equation [230] of fraction order 𝐷𝑡𝛼 𝑢 − 𝑎𝐷𝑥2𝛼 𝑢 + 𝑢(1 − 𝑢)(𝑎 − 𝑢) = 0, 0 < 𝛼 ≤ 1.

(4.242)


(4.243)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.243) into Eq. (4.242) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 − 𝑎𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 + 𝑢(1 − 𝑢)(𝑎 − 𝑢) = 0, 0 < 𝛼 ≤ 1.

(4.244)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 2 = 3𝑀, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.224), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑏1 𝑈 −1 .

(4.245)


𝑏13 = 0,

𝑈1 :

−𝑏12 − 𝑎𝑏12 + 3𝑎0 𝑏12 = 0, 632

𝑈2:

−2𝑎𝑎0 𝑏1 − 𝜔𝛼 𝛼1 𝑏1 − 𝑘 2𝛼 𝑏1 𝛼12 + 𝑎𝑏1 − 2𝑎0 𝑏1 + 3𝑎02 𝑏1 + 3𝑎1 𝑏12 = 0,

𝑈3:

−2𝑎𝑎1 𝑏1 + 6𝑎0 𝑎1 𝑏1 − 𝜔𝛼 𝛼2 𝑏1 + 𝑎03 − ⋯ − 𝑎02 + 𝑎𝑎0 − 2𝑎1 𝑏1 − 𝑎𝑎02 = 0,

𝑈4:

−2𝑎𝑎0 𝑎1 + 𝜔𝛼 𝛼1 𝑎1 − 𝑘 2𝛼 𝑎1 𝛼12 + 𝑎𝑎1 − 2𝑎0 𝑎1 + 3𝑎02 𝑎1 + 3𝑎12 𝑏1 = 0,

𝑈5:

𝜔𝛼 𝛼2 𝑎1 − 3𝑘 2𝛼 𝑎1 𝛼1 𝛼2 − 𝑎12 − 𝑎𝑎12 + 3𝑎0 𝑎12 = 0,

𝑈6:

−2𝑘 2𝛼 𝑎1 𝛼22 + 𝑎13 = 0.

Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: 1 ln( 2 ) 2𝛼1 1 ( ) 2 𝛼

𝑘=e

2𝑎−1 ln(− ) 2𝛼1 ) 𝛼

(

,𝜔 = e

𝛼

, 𝑎0 = 0, 𝑎1 = − 𝛼2 , 𝑏1 = 0. 1


𝑢1 (𝜉) = − 𝛼2 𝑈, 1

1

i


𝑢2 (𝜉) = − 𝛼2 𝑈, 1

1


𝑢3 (𝜉) = − 𝛼2 𝑈, 1

1

1


𝑢4 (𝜉) = − 𝛼2 𝑈, 1

1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

633

1 ln( 2 ) 2𝛼1 1 ( ) 2 𝛼


2𝑎−1 ln(− ) 2𝛼1 ) 𝛼

(

𝑥+e

𝑡.

2nd Solution Set: 𝑎2 ln( 2 ) 2𝛼1 1 ( ) 2 𝛼

𝑘=e

ln(

(

,𝜔 = e

𝑎(𝑎−2) ) 2𝛼1 ) 𝛼

, 𝑎0 = 0, 𝑎1 = −

𝑎𝛼2 𝛼1

, 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢5 (𝜉) = − 1

𝑎𝛼2 𝛼1

𝑈,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢6 (𝜉) = −

𝑎𝛼2 𝛼1

𝑈,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢7 (𝜉) = −

𝑎𝛼2 𝛼1

𝑈,

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢8 (𝜉) = −

𝑎𝛼2 𝛼1

𝑈, √3

√3

1

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

2

2 𝐶 cos (√3𝜉)+i𝐶 sin (√3𝜉) 1 𝛼 2 𝛼

where 𝑈 = −

2

𝑎2 ln( 2 ) 2𝛼1 1 ( ) 2 𝛼


, and i = √−1.

2

ln(

(

𝑥+e

𝑎(𝑎−2) ) 2𝛼1 ) 𝛼

𝑡.

3rd Solution Set:

634

1 ln( 2 ) 2𝛼1 1 ( ) 2 𝛼

𝑘=e

2𝑎−1 ln(− ) 2𝛼1 ) 𝛼

(

,𝜔 = e

𝛼

, 𝑎0 = 1, 𝑎1 = 𝛼2 , 𝑏1 = 0. 1


𝑢9 (𝜉) = 1 + 𝛼2 𝑈, 1

1

i


𝑢10 (𝜉) = 1 + 𝛼2 𝑈, 1

1


𝑢11 (𝜉) = 1 + 𝛼2 𝑈, 1

1

1


𝑢12 (𝜉) = 1 + 𝛼2 𝑈, 1

1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1 ln( 2 ) 2𝛼1 1 ( ) 2 𝛼


, and i = √−1.

2𝑎−1 ln(− ) 2𝛼1 ) 𝛼

(

𝑥+e

𝑡.

4th Solution Set: 𝑎2 ln( 2 ) 2𝛼1 1 ( ) 2 𝛼

𝑘=e

𝑎(𝑎−2) ln(− ) 2𝛼1 ) 𝛼

(

,𝜔 = e

, 𝑎0 = 𝑎, 𝑎1 =

𝑎𝛼2 𝛼1

, 𝑏1 = 0.


635

𝑎𝛼2

𝑢13 (𝜉) = 𝑎 + 1

𝛼1

𝑈,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑎𝛼2

𝑢14 (𝜉) = 𝑎 +

𝛼1

𝑈,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢15 (𝜉) = 𝑎 +

𝑎𝛼2 𝛼1

𝑈,

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢16 (𝜉) = 𝑎 +

𝑎𝛼2 𝛼1

𝑈,

√3

√3

1

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

2

2 𝐶 cos (√3𝜉)+i𝐶 sin (√3𝜉) 1 𝛼 2 𝛼

where 𝑈 = −

2

ln(

1 ( 2

, and i = √−1.

2

𝑎2 ) 2𝛼2 1 ) 𝛼


𝑎(𝑎−2) ln(− ) 2𝛼1 ) 𝛼

(

𝑥+e

𝑡.

5th Solution Set: (𝑎−1)2 ln( ) 2𝛼2 1 1 ( ) 2 𝛼

𝑘=e

ln(

(

,𝜔 = e

(𝑎+1)(𝑎−1) ) 2𝛼1 ) 𝛼

, 𝑎0 = 1, 𝑎1 = −

𝛼2 (𝑎−1) 𝛼1

, 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢17 (𝜉) = 1 − 1

𝛼2 (𝑎−1) 𝛼1

𝑈,

i


𝑢18 (𝜉) = 1 −

𝛼2 (𝑎−1) 𝛼1

𝑈,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝛼2 (𝑎−1)

𝑢19 (𝜉) = 1 −

𝑈,

𝛼1

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢20 (𝜉) = 1 − 1

𝛼2 (𝑎−1) 𝛼1

𝑈,

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 ( ln(

1 ( 2

(𝑎−1)2 ) 2𝛼2 1 ) 𝛼


ln(

(

𝑥+e

, and i = √−1.

(𝑎+1)(𝑎−1) ) 𝛼1 ) 𝛼

𝑡.

6th Solution Set: ln(

1 ( 2

𝑎2 ) 2𝛼2 1 ) 𝛼

𝑘=e

𝑎(𝑎−2) ln(− ) 2𝛼1 ) 𝛼

(

,𝜔 = e

, 𝑎0 = 0, 𝑎1 =

𝑎𝛼2 𝛼1

, 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢21 (𝜉) = 1

𝑎𝛼2 𝛼1

𝑈,

i


𝑎𝛼2 𝛼1

𝑈,

1


𝑎𝛼2

𝑢23 (𝜉) =

𝛼1

𝑈,

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑎𝛼2

𝑢24 (𝜉) = 1

𝛼1

𝑈, √3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

𝑎2 ln( 2 ) 2𝛼1 1 ( ) 2 𝛼


, and i = √−1.

𝑎(𝑎−2) ln(− ) 2𝛼1 ) 𝛼

(

𝑥+e

𝑡.

7th Solution Set: (𝑎−1)2 ln( ) 2𝛼2 1 1 ( ) 2 𝛼

𝑘=e

(

,𝜔 = e

(𝑎+1)(𝑎−1) ln(− ) 2𝛼1 ) 𝛼

, 𝑎0 = 𝑎, 𝑎1 =

𝛼2 (𝑎−1) 𝛼1

, 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢25 (𝜉) = 𝑎 − 1

𝛼2 (𝑎−1) 𝛼1

𝑈,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢26 (𝜉) = 𝑎 −

𝛼2 (𝑎−1) 𝛼1

𝑈,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.245) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢27 (𝜉) = 𝑎 − 1

𝛼2 (𝑎−1) 𝛼1

𝑈, 1


𝑢28 (𝜉) = 𝑎 − 1

𝛼2 (𝑎−1) 𝛼1

𝑈,

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

(𝑎−1)2 ln( ) 2𝛼2 1 1 ( ) 2 𝛼


ln(

(

𝑥+e

, and i = √−1.

(𝑎+1)(𝑎−1) ) 𝛼1 ) 𝛼

𝑡.



4.8.7. Klein Gordon Equation Consider the Klein Gordon equation [230] of fraction order 𝐷𝑡2𝛼 𝑢 + 𝑎𝐷𝑥2𝛼 𝑢 − 𝑢 − 𝑢3 = 0, 0 < 𝛼 ≤ 1.

(4.246)


(4.247)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.247) into Eq. (4.246) and using the chain rule we obtained 𝜔2𝛼 𝐷𝜉2𝛼 𝑢 + 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 − 𝑢 − 𝑢3 = 0.

(4.248)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 2 = 3𝑀, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.224), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑏1 𝑈 −1 .

(4.249)

639


−𝑏13 = 0,

𝑈1 :

−3𝑎0 𝑏12 = 0,

𝑈2:

𝜔2𝛼 𝑏1 𝛼12 + 𝑎𝑘 2𝛼 𝑏1 𝛼12 − 𝑏1 − 3𝑎02 𝑏1 − 3𝑎1 𝑏12 = 0,

𝑈3:

𝜔2𝛼 𝑏1 𝛼1 𝛼2 + 𝑎𝑘 2𝛼 𝑏1 𝛼1 𝛼2 − 𝑎0 − 𝑎03 − 6𝑎0 𝑎1 𝑏1 = 0,

𝑈4:

𝜔2𝛼 𝑎1 𝛼12 + 𝑎𝑘 2𝛼 𝑎1 𝛼12 − 𝑎1 − 3𝑎02 𝑎1 − 3𝑎12 𝑏1 = 0,

𝑈5:

3𝜔2𝛼 𝑎1 𝛼1 𝛼2 + 3𝑎𝑘 2𝛼 𝑎1 𝛼1 𝛼2 − 3𝑎0 𝑎12 = 0,

𝑈6:

2𝜔2𝛼 𝑎1 𝛼22 + 2𝑎𝑘 2𝛼 𝑎1 𝛼22 − 𝑎13 = 0.

Solving the above system for unknown parameters, we have the following solution set 1st Solution Set: ln(−

𝑎e2𝛼ln(𝑘) 𝛼2 1 +2) 𝛼2 1 𝛼

𝑘 = 𝑘, 𝜔 = e(

) , 𝑎0

= i, 𝑎1 =

2i𝛼2 𝛼1

, 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢1 (𝜉) = i + 1

2i𝛼2 𝛼1

𝑈,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢2 (𝜉) = i +

2i𝛼2 𝛼1

𝑈,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢3 (𝜉) = i + 1

2i𝛼2 𝛼1

𝑈, 1


Case IV: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢4 (𝜉) = i + 1

2i𝛼2

𝑈,

𝛼1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

𝑎e2𝛼ln(𝑘) 𝛼2 1 +2) ln(− 𝛼2 1 𝛼

In all cases 𝜉 = 𝑘𝑥 + e(

) 𝑡.

2nd Solution Set: ln(−

𝑎e2𝛼ln(𝑘) 𝛼2 1 +2) 𝛼2 1 𝛼

𝑘 = 𝑘, 𝜔 = e(

) , 𝑎0

= −i, 𝑎1 = −

2i𝛼2 𝛼1

, 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢5 (𝜉) = − (i + 1

2i𝛼2 𝛼1

𝑈),

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢6 (𝜉) = − (i +

2i𝛼2 𝛼1

𝑈),

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢7 (𝜉) = − (i +

2i𝛼2 𝛼1

𝑈),

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢8 (𝜉) = − (i +

2i𝛼2 𝛼1

𝑈),

641

1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

𝑎e2𝛼ln(𝑘) 𝛼2 1 +2) ln(− 𝛼2 1 𝛼

In all cases 𝜉 = 𝑘𝑥 + e(

) 𝑡.

3rd Solution Set: 2 ln(− 2 ) 𝛼1 1 ( ) 2 𝛼

𝑘 = 0, 𝜔 = e

, 𝑎0 = −i, 𝑎1 = −

2i𝛼2 𝛼1

, 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢9 (𝜉) = −i − 1

2i𝛼2 𝛼1

𝑈,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢10 (𝜉) = −i −

2i𝛼2 𝛼1

𝑈,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢11 (𝜉) = −i −

2i𝛼2 𝛼1

𝑈,

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢12 (𝜉) = −i − 1

2i𝛼2 𝛼1

𝑈,

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

2 ln(− 2 ) 𝛼1 1 ( ) 2 𝛼


𝑡. 642

4th Solution Set: 2 ln(− 2 ) 𝛼1 1 ( ) 2 𝛼

𝑘 = 0, 𝜔 = e

, 𝑎0 = i, 𝑎1 =

2i𝛼2 𝛼1

, 𝑏1 = 0.


2i𝛼2 𝛼1

𝑈,

i


2i𝛼2 𝛼1

𝑈,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢15 (𝜉) = i +

2i𝛼2 𝛼1

𝑈,

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.249) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢16 (𝜉) = i + 1

2i𝛼2 𝛼1

𝑈,

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

2 ln(− 2 ) 𝛼1 1 ( ) 2 𝛼


𝑡.

643

Fig 4.58 (a): 2D travelling wave solution of

Fig 4.58 (b): 3D travelling wave solution of



4.8.8. Modified Kawahara Equation Consider the (2+1)-dimensional modified Kawahara equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 6𝑢2 𝐷𝑥𝛼 𝑢 + 𝐷𝑥3𝛼 𝑢 − 𝐷𝑥5𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.250)


(4.251)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.251) into Eq. (4.250) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 6𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 𝐷𝜉3𝛼 𝑢 − 𝐷𝜉5𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.252)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 5 = 2𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.224), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑎2 𝑈 2 + 𝑏1 𝑈 −1 + 𝑏2 𝑈 −2 .

(4.253)

Putting Eq. (4.253) into Eq. (4.252) coupled with Eq. (4.225); the Eq. (4.253) yields an algebraic equation involving power of U as 𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 + 𝐶7 𝑈 7 + 𝐶8 𝑈 8 + 𝐶9 𝑈 9 +

𝐶10 𝑈10 + 𝐶11 𝑈11 + 𝐶12 𝑈12 + 𝐶13 𝑈13 = 0.

Compare the like powers of U we have system of equations 𝑈0:

−12𝑘 𝛼 𝑏23 𝛼1 = 0,

𝑈1 :

−12𝑘 𝛼 𝑏23 𝛼2 − 30𝑘 𝛼 𝑏22 𝛼1 𝑏1 = 0, 644

𝑈2:

−24𝑘 𝛼 𝑏12 𝛼1 𝑏2 − 30𝑘 𝛼 𝑏22 𝛼2 𝑏1 − 24𝑘 𝛼 𝑎0 𝑏22 𝛼1 = 0,

𝑈3:

−36𝑘 𝛼 𝑎0 𝑏1 𝛼1 𝑏2 − 6𝑘 𝛼 𝑏13 𝛼1 − 18𝑘 𝛼 𝑏22 𝛼1 𝑎1 − ⋯ − 24𝑘 𝛼 𝑎0 𝑏22 𝛼2 = 0,

𝑈4:

−36𝑘 𝛼 𝑎0 𝑏1 𝛼2 𝑏2 − 24𝑘 𝛼 𝑎1 𝑏1 𝛼1 𝑏2 − ⋯ − 18𝑘 𝛼 𝑏22 𝛼2 𝑎1 − 12𝑘 𝛼 𝑎0 𝑏12 𝛼1 = 0,

𝑈5:

−12𝑘 𝛼 𝑎0 𝑎1 𝛼1 𝑏2 − 24𝑘 𝛼 𝑎1 𝑏1 𝛼2 𝑏2 − 12𝑘 𝛼 𝑎2 𝑏1 𝛼1 𝑏2 − ⋯ − 12𝑘 𝛼 𝑎0 𝑏12 𝛼2 = 0,

𝑈6:

−12𝑘 𝛼 𝑎0 𝑎1 𝛼2 𝑏2 − 12𝑘 𝛼 𝑎2 𝑏1 𝛼2 𝑏2 − ⋯ + 𝑘 5𝛼 𝑏1 𝛼14 𝛼2 + 30𝑘 5𝛼 𝑏2 𝛼13 𝛼22 = 0,

𝑈7:

−12𝑘 𝛼 𝑎0 𝑎2 𝛼1 𝑏1 + 12𝑘 𝛼 𝑎1 𝑎2 𝛼1 𝑏2 − ⋯ + 6𝑘 𝛼 𝑎12 𝛼1 𝑏1 + 6𝑘 𝛼 𝑎02 𝛼1 𝑎1 = 0,

𝑈8:

−12𝑘 𝛼 𝑎0 𝑎2 𝛼2 𝑏1 + 24𝑘 𝛼 𝑎1 𝑎2 𝛼1 𝑏1 − ⋯ − 31𝑘 5𝛼 𝑎1 𝛼14 𝛼2 + 12𝑘 𝛼 𝑎0 𝑎12 𝛼1 = 0,

𝑈9:

36𝑘 𝛼 𝑎0 𝑎1 𝛼1 𝑎2 + 24𝑘 𝛼 𝑎1 𝑎2 𝛼2 𝑏1 + ⋯ − 180𝑘 5𝛼 𝑎1 𝛼13 𝛼22 + 12𝑘 𝛼 𝑎0 𝑎12 𝛼2 = 0,

𝑈10 : 36𝑘 𝛼 𝑎0 𝑎1 𝛼2 𝑎2 + 6𝑘 3𝛼 𝑎1 𝛼23 + ⋯ + 24𝑘 𝛼 𝑎0 𝑎22 𝛼1 + 24𝑘 𝛼 𝑎12 𝑎2 𝛼1 = 0, 𝑈11 : 24𝑘 3𝛼 𝑎2 𝛼23 + 24𝑘 𝛼 𝑎12 𝑎2 𝛼2 − ⋯ + 24𝑘 𝛼 𝑎0 𝑎22 𝛼2 + 30𝑘 𝛼 𝑎1 𝑎22 𝛼1 = 0, 𝑈12 : −120𝑘 5𝛼 𝑎1 𝛼25 + 12𝑘 𝛼 𝑎23 𝛼1 + 30𝑘 𝛼 𝑎1 𝑎22 𝛼2 − 2400𝑘 5𝛼 𝑎2 𝛼1 𝛼24 = 0, 𝑈13 : −720𝑘 5𝛼 𝑎2 𝛼25 + 12𝑘 𝛼 𝑎23 𝛼2 = 0. Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: 3 1 𝛼 ln(− 𝑘5𝛼𝛼4 1 −10𝑘 ) 2 ) 𝛼

(

𝜔=e

1

1

, 𝑎0 = − 30𝑘 𝛼 √15𝑘 6𝛼 (−5𝛼12 + 𝑘 −2𝛼 ), 𝑎1 = 𝑘 𝛼 2√15𝑘 6𝛼 𝛼1 𝛼2 ,

1

𝑎2 = 𝑘 𝛼 2√15𝑘 6𝛼 𝛼22 , 𝑏1 = 𝑏2 = 0. Case I: Substituting the values of unknown into Eq. (4.253) coupled with the solution of Eq. (2.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1

1

1

𝑢1 (𝜉) = − 30𝑘 𝛼 √15𝑘 6𝛼 (−5𝛼12 + 𝑘 −2𝛼 ) + 𝑘 𝛼 2√15𝑘 6𝛼 𝛼1 𝛼2 𝑈 + 𝑘 𝛼 2√15𝑘 6𝛼 𝛼22 𝑈 2 , 1

i


1

1



1

1

𝑢3 (𝜉) = − 30𝑘 𝛼 √15𝑘 6𝛼 (−5𝛼12 + 𝑘 −2𝛼 ) + 𝑘 𝛼 2√15𝑘 6𝛼 𝛼1 𝛼2 𝑈 + 𝑘 𝛼 2√15𝑘 6𝛼 𝛼22 𝑈 2 ,

645

1

1


1

1


√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

3 1 𝛼 ln(− 𝑘5𝛼 𝛼4 1 −10𝑘 ) 2 ) 𝛼

(


𝑡.

2nd Solution Set: 3 1 𝛼 ln(− 𝑘5𝛼𝛼4 1 −10𝑘 ) 2 ) 𝛼

(

𝜔=e

1

1

, 𝑎0 = 30𝑘 𝛼 √15𝑘 6𝛼 (−5𝛼12 + 𝑘 −2𝛼 ), 𝑎1 = − 𝑘 𝛼 2√15𝑘 6𝛼 𝛼1 𝛼2 ,

1

𝑎2 = − 𝑘 𝛼 2√15𝑘 6𝛼 𝛼22 , 𝑏1 = 𝑏2 = 0. Case I: Substituting the values of unknown into Eq. (4.253) coupled with the solution of Eq. (2.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1

1

1

𝑢5 (𝜉) = 30𝑘 𝛼 √15𝑘 6𝛼 (−5𝛼12 + 𝑘 −2𝛼 ) − 𝑘 𝛼 2√15𝑘 6𝛼 𝛼1 𝛼2 𝑈 − 𝑘 𝛼 2√15𝑘 6𝛼 𝛼22 𝑈 2 , 1

i


1

1



1

1


1


1

1

𝑢8 (𝜉) = 30𝑘 𝛼 √15𝑘 6𝛼 (−5𝛼12 + 𝑘 −2𝛼 ) − 𝑘 𝛼 2√15𝑘 6𝛼 𝛼1 𝛼2 𝑈 − 𝑘 𝛼 2√15𝑘 6𝛼 𝛼22 𝑈 2 ,

646

1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

3 1 𝛼 ln(− 𝑘5𝛼 𝛼4 1 −10𝑘 ) 2 ) 𝛼

(


𝑡.

Fig 4.59 (a): 2D travelling wave solution of Eq. (4.250)

Fig 4.59 (b): 3D travelling wave solution of

for different values of parameters and 𝑡 = 1.


4.8.9. Modified Equal Width Equation Consider the modified equal width equation [230] of fraction order 𝐷𝑡𝛼 𝑢 − 3𝑢2 𝐷𝑥𝛼 𝑢 − 𝑎𝐷𝑡𝛼 𝐷𝑥2𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.254)


(4.255)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.255) into Eq. (4.254) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 − 3𝑢2 𝑘 𝛼 𝐷𝜉𝛼 𝑢 − 𝑎𝜔𝛼 𝑘 2𝛼 𝐷𝜉3𝛼 𝑢 = 0.

(4.256)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 3 = 2𝑀 + 𝑀 + 1, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.224), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑏1 𝑈 −1 .

(4.257)

Putting Eq. (4.257) into Eq. (4.256) coupled with Eq. (4.225); the Eq. (4.257) yields an algebraic equation involving power of U as 𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 = 0. 647


𝑘 𝛼 𝑏13 = 0,

𝑈1 :

3𝑘 𝛼 𝑎0 𝑏12 = 0,

𝑈2:

−𝜔𝛼 𝑏1 + 3𝑘 𝛼 𝑎02 𝑏1 + 3𝑘 𝛼 𝑎1 𝑏12 + 𝑎𝜔𝑎 𝑘 2𝛼 𝑏1 𝛼12 = 0,

𝑈3:

−𝜔𝛼 𝑎0 + 𝑘 𝛼 𝑎03 + 6𝑘 𝛼 𝑎0 𝑎1 𝑏1 + 𝑎𝜔𝑎 𝑘 2𝛼 𝑏1 𝛼1 𝛼2 = 0,

𝑈4:

−𝜔𝛼 𝑎1 + 3𝑘 𝛼 𝑎02 𝑎1 + 3𝑘 𝛼 𝑎12 𝑏1 + 𝑎𝜔𝑎 𝑘 2𝛼 𝑎1 𝛼12 = 0,

𝑈5:

3𝑘 𝛼 𝑎0 𝑎12 + 3𝑎𝜔𝑎 𝑘 2𝛼 𝑎1 𝛼1 𝛼2 = 0,

𝑈6:

𝑘 𝛼 𝑎13 + 2𝑎𝜔𝑎 𝑘 2𝛼 𝑎1 𝛼22 = 0.

Solving the above system for unknown parameters, we have the following solution set 1st Solution Set: 𝑘 = (−

2 √−2𝑎𝛼1

1 𝛼

1 𝛼

2

) , 𝜔 = [(𝑎0 ) ] (−

1 𝛼

2 √−2𝑎𝛼1

) , 𝑎0 = 𝑎0 , 𝑎1 =

2𝑎0 𝛼2 𝛼1

, 𝑏1 = 0.


2𝑎0 𝛼2 𝛼1

𝑈,

i


2𝑎0 𝛼2 𝛼1

𝑈,

1


2𝑎0 𝛼2 𝛼1

𝑈,

1

1


2𝑎0 𝛼2 𝛼1

𝑈,

648

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

In all cases 𝜉 = (−

1 𝛼

2 √−2𝑎𝛼1

, and i = √−1.

1 𝛼

2

) 𝑥 + [(𝑎0 ) ] (−

2 √−2𝑎𝛼1

1 𝛼

) 𝑡.

2nd Solution Set: 1 𝛼

𝑘=(

2 √−2𝑎𝛼1

1

2

) , 𝜔 = [(𝑎0 )𝛼 ] (

1 𝛼

2

√−2𝑎𝛼1

) , 𝑎0 = 𝑎0 , 𝑎1 =

2𝑎0 𝛼2 𝛼1

, 𝑏1 = 0.

Case I: Substituting the values of unknown into Eq. (4.257) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢5 (𝜉) = − (i + 1

2i𝛼2 𝛼1

𝑈),

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.257) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢6 (𝜉) = − (i +

2i𝛼2 𝛼1

𝑈),

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.257) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢7 (𝜉) = − (i +

2i𝛼2 𝛼1

𝑈),

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.257) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢8 (𝜉) = − (i + 1

2i𝛼2 𝛼1 √3

𝑈), √3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

In all cases 𝜉 = (

2 √−2𝑎𝛼1

1 𝛼

1 𝛼

, and i = √−1.

2

) 𝑥 + [(𝑎0 ) ] (

2 √−2𝑎𝛼1

649

1 𝛼

) 𝑡.





4.8.10. ZK Equation Consider the ZK equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝑎𝑢𝐷𝑥𝛼 𝑢 + 𝑏𝑢2 𝐷𝑥𝛼 𝑢 + 𝐷𝑥3𝛼 𝑢 + 𝑐𝐷𝑥𝛼 𝐷𝑦2𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.258)


(4.259)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.259) into Eq. (4.258) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝑎𝑘 𝛼 𝑢𝐷𝜉𝛼 𝑢 + 𝑏𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 𝑘 3𝛼 𝐷𝜉3𝛼 𝑢 + 𝑐𝜔𝛼 𝑙 2𝛼 𝐷𝜉3𝛼 𝑢 = 0.

(4.260)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 3 = 2𝑀 + 𝑀 + 1, 𝑀 = 1. Using the value of 𝑀 into Eq. (2.224), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑏1 𝑈 −1 .

(4.261)

Putting Eq. (4.261) into Eq. (4.260) coupled with Eq. (4.225); the Eq. (4.261) yields an algebraic equation involving power of U as 𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 + 𝐶7 𝑈 7 = 0. Compare the like powers of U we have system of equations 𝑈0:

−𝑏𝑘 𝛼 𝛼1 𝑏13 = 0,

𝑈1 :

−2𝑏𝑘 𝛼 𝑎0 𝛼1 𝑏12 − 𝑎𝑘 𝛼 𝛼1 𝑏12 − 𝑏𝑘 𝛼 𝛼2 𝑏13 = 0,

𝑈2:

−2𝑏𝑘 𝛼 𝑎0 𝛼2 𝑏12 − 𝑏𝑘 𝛼 𝑎02 𝛼1 𝑏1 − ⋯ − 𝑎𝑘 𝛼 𝛼2 𝑏12 − 𝜔𝛼 𝛼1 𝑏1 − 𝑘 3𝛼 𝑏1 𝛼13 = 0, 650

𝑈3:

−𝑐𝑘 𝛼 𝑙 2𝛼 𝑏1 𝛼12 𝛼2 − 𝑏𝑘 𝛼 𝑎02 𝛼2 𝑏1 − ⋯ − 𝑘 3𝛼 𝑏1 𝛼12 𝛼2 − 𝜔𝛼 𝛼2 𝑏1 = 0,

𝑈4:

𝑏𝑘 𝛼 𝛼1 𝑎12 𝑏1 + 𝑏𝑘 𝛼 𝑎02 𝛼1 𝑎1 + ⋯ + 𝑐𝑘 𝛼 𝑙 2𝛼 𝑎1 𝛼13 + 𝜔𝛼 𝛼1 𝑎1 + 𝑘 3𝛼 𝑎1 𝛼13 = 0,

𝑈5:

7𝑐𝑘 𝛼 𝑙 2𝛼 𝑎1 𝛼12 𝛼2 + 2𝑏𝑘 𝛼 𝑎0 𝛼1 𝑎12 + 𝑎𝑘 𝛼 𝛼2 𝑎12 + 𝑏𝑘 𝛼 𝛼1 𝑎13 + 12𝑘 3𝛼 𝑎1 𝛼1 𝛼22 = 0,

𝑈6:

12𝑐𝑘 𝛼 𝑙 2𝛼 𝑎1 𝛼1 𝛼22 + 2𝑏𝑘 𝛼 𝑎0 𝛼2 𝑎12 + ⋯ + 𝑘 5𝛼 𝑏1 𝛼14 𝛼2 + 30𝑘 5𝛼 𝑏2 𝛼13 𝛼22 = 0,

𝑈7:

6𝑐𝑘 𝛼 𝑙 2𝛼 𝑎1 𝛼23 + 𝑏𝑘 𝛼 𝛼2 𝑎13 + 6𝑘 3𝛼 𝑎1 𝛼23 = 0.

Solving the above system for unknown parameters, we have the following solution sets

1st Solution Set:

𝑘=(

(3𝑎2 𝛼22 −𝛼12 𝑏2 𝑎12 )√−36𝑐𝑙2𝛼 𝛼22 −6𝑏𝑎12

(

72𝛼23 𝑏

√−36𝑐𝑙2𝛼 𝛼22 −6𝑏𝑎12 𝛼2

1 𝛼

) ,𝜔 =

1 𝛼

) , 𝑎0 =

−

𝛼2 𝑎−𝑎1 𝛼1 𝑏 2𝛼2 𝑏

, 𝑎1 = 𝑎1 , 𝑏1 = 0.


𝛼2 𝑎−𝑎1 𝛼1 𝑏 2𝛼2 𝑏

+ 𝑎1 𝑈,

i


𝛼2 𝑎−𝑎1 𝛼1 𝑏 2𝛼2 𝑏

+ 𝑎1 𝑈,

1


𝛼2 𝑎−𝑎1 𝛼1 𝑏 2𝛼2 𝑏

1

+ 𝑎1 𝑈, 1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢4 (𝜉) = −

𝛼2 𝑎−𝑎1 𝛼1 𝑏 2𝛼2 𝑏

+ 𝑎1 𝑈,

651

1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


√−36𝑐𝑙2𝛼 𝛼22 −6𝑏𝑎12 𝛼2

, and i = √−1.

1 𝛼

) 𝑥+(

(3𝑎2 𝛼22 −𝛼12 𝑏 2 𝑎12 )√−36𝑐𝑙2𝛼 𝛼22 −6𝑏𝑎12 72𝛼23 𝑏

1 𝛼

) 𝑡.

2nd Solution Set:

𝑘 = (− −

√−36𝑐𝑙2𝛼 𝛼22 −6𝑏𝑎12 𝛼2

𝛼2 𝑎−𝑎1 𝛼1 𝑏 2𝛼2 𝑏

1 𝛼

) , 𝜔 = (−

(3𝑎2 𝛼22 −𝛼12 𝑏2 𝑎12 )√−36𝑐𝑙2𝛼 𝛼22 −6𝑏𝑎12 72𝛼23 𝑏

1 𝛼

) , 𝑎0 =

, 𝑎1 = 𝑎1 , 𝑏1 = 0.


𝛼2 𝑎−𝑎1 𝛼1 𝑏 2𝛼2 𝑏

+ 𝑎1 𝑈,

i


𝛼2 𝑎−𝑎1 𝛼1 𝑏 2𝛼2 𝑏

+ 𝑎1 𝑈,

1


𝛼2 𝑎−𝑎1 𝛼1 𝑏 2𝛼2 𝑏

1

+ 𝑎1 𝑈, 1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢8 (𝜉) = − 1

𝛼2 𝑎−𝑎1 𝛼1 𝑏 2𝛼2 𝑏 √3

+ 𝑎1 𝑈, √3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

652

√−36𝑐𝑙2𝛼 𝛼22 −6𝑏𝑎12

In all cases 𝜉 = (−

𝛼2

1 𝛼

) 𝑥 + (−

(3𝑎2 𝛼22 −𝛼12 𝑏2 𝑎12 )√−36𝑐𝑙2𝛼 𝛼22 −6𝑏𝑎12 72𝛼23 𝑏

1 𝛼

) 𝑡.

3rd Solution Set: 2 2𝛼 2 1 √−2𝑏(2𝛼1 𝑏𝑐𝑙 +𝑎 )

𝑘 = (2

𝑏𝛼1

1 𝛼

1 𝑎(√3+1)

) , 𝜔 = 0, 𝑎0 = − 2

2𝑏

, 𝑎1 = −

√3𝛼2 𝑎 , 𝑏1 𝛼1 𝑏

= 0.

Case I: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1 𝑎(√3+1)

𝑢9 (𝜉) = − 2 1

2𝑏

−

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 1 𝑎(√3+1)

𝑢10 (𝜉) = − 2

2𝑏

−

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 1 𝑎(√3+1)

𝑢11 (𝜉) = − 2

2𝑏

−

1

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏 1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 1 𝑎(√3+1)

𝑢12 (𝜉) = − 2 1

2𝑏

−

√3

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏 √3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

2 2𝛼 2 1 √−2𝑏(2𝛼1 𝑏𝑐𝑙 +𝑎 )

In all cases 𝜉 = (2

𝑏𝛼1

, and i = √−1.

1 𝛼

) 𝑥.


2 2𝛼 2 1 √−2𝑏(2𝛼1 𝑏𝑐𝑙 +𝑎 )

𝑘 = (− 2

𝑏𝛼1

1 𝛼

1 𝑎(√3−1)

) , 𝜔 = 0, 𝑎0 = 2

2𝑏

, 𝑎1 =

√3𝛼2 𝑎 , 𝑏1 𝛼1 𝑏

= 0.

Case I: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1 𝑎(√3−1)

𝑢13 (𝜉) = 2 1

2𝑏

+

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 1 𝑎(√3−1)

𝑢14 (𝜉) = 2

2𝑏

+

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 1 𝑎(√3−1)

𝑢15 (𝜉) = 2

2𝑏

+

1

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏 1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 1 𝑎(√3−1)

𝑢16 (𝜉) = 2

2𝑏

+

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏

√3

√3

1

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

2

2 𝐶 cos (√3𝜉)+i𝐶 sin (√3𝜉) 1 𝛼 2 𝛼

where 𝑈 = −

2

, and i = √−1.

2

2 2𝛼 2 1 √−2𝑏(2𝛼1 𝑏𝑐𝑙 +𝑎 )

In all cases 𝜉 = (− 2

𝑏𝛼1

1 𝛼

) 𝑥.

5th Solution Set: 2 2𝛼 2 1 √−2𝑏(2𝛼1 𝑏𝑐𝑙 +𝑎 )

𝑘 = (2

𝑏𝛼1

1 𝛼

1 𝑎(√3−1)

) , 𝜔 = 0, 𝑎0 = 2

654

2𝑏

, 𝑎1 =

√3𝛼2 𝑎 , 𝑏1 𝛼1 𝑏

= 0.

Case I: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1 𝑎(√3−1)

𝑢17 (𝜉) = 2 1

2𝑏

+

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −i and 𝛼2 = i, we have 1 𝑎(√3−1)

𝑢18 (𝜉) = 2

2𝑏

+

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = 4 and 𝛼2 = 2, we have 1 𝑎(√3−1)

𝑢19 (𝜉) = 2

2𝑏

+

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.261) coupled with the solution of Eq. (4.225) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 1 𝑎(√3−1)

𝑢20 (𝜉) = 2 1

2𝑏

+

√3𝛼2 𝑎 𝑈, 𝛼1 𝑏

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

2 2𝛼 2 1 √−2𝑏(2𝛼1 𝑏𝑐𝑙 +𝑎 )


𝑏𝛼1

, and i = √−1.

1 𝛼

) 𝑥.

655

Fig 4.61 (a): 2D Periodic wave solution of Eq. (4.258) for different values of parameters and 𝑡 = 1.

Fig 4.61 (b): 3D Periodic wave solution of Eq. (4.258) for different values of parameters

4.8.11. RLW Equation Consider the RLW equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝑎𝐷𝑥𝛼 𝑢 + 2𝑢𝐷𝑥𝛼 𝑢 + 𝑏𝐷𝑥2𝛼 𝐷𝑡𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.262)


(4.263)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.263) into Eq. (4.262) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝑘 𝛼 𝐷𝜉𝛼 𝑢 + 2𝑢𝑘 𝛼 𝐷𝜉𝛼 𝑢 + 𝑏𝜔𝛼 𝑘 2𝛼 𝐷𝜉3𝛼 𝑢 = 0.

(4.264)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 3 = 𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.224), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1 𝑈 + 𝑎2 𝑈 2 + 𝑏1 𝑈 −1 + 𝑏2 𝑈 −2 .

(4.265)

Putting Eq. (4.265) into Eq. (4.264) coupled with Eq. (4.225); the Eq. (4.265) yields an algebraic equation involving power of U as 𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 + 𝑐7 𝑈 7 +𝐶8 𝑈 8 + 𝐶9 𝑈 9 = 0. Compare the like powers of U we have system of equations 𝑈0:

−4𝑘 𝛼 𝛼1 𝑏22 = 0,

𝑈1 :

−6𝑘 𝛼 𝛼1 𝑏1 𝑏2 − 4𝑘 𝛼 𝛼2 𝑏22 = 0, 656

𝑈2:

−8𝑏𝜔𝛼 𝑘 2𝛼 𝑏2 𝛼13 − 2𝑎𝑘 𝛼 𝛼1 𝑏2 − ⋯ − 2𝜔𝛼 𝛼1 𝑏2 − 2𝑘 𝛼 𝛼1 𝑏12 = 0,

𝑈3:

−𝑏𝜔𝛼 𝑘 2𝛼 𝑏1 𝛼13 − 14𝑏𝜔𝛼 𝑘 2𝛼 𝑏2 𝛼12 𝛼2 − ⋯ − 2𝜔𝛼 𝛼2 𝑏2 − 2𝑘 𝛼 𝛼2 𝑏12 = 0,

𝑈4:

−𝑏𝜔𝛼 𝑘 2𝛼 𝑏1 𝛼12 𝛼2 − 6𝑏𝜔𝛼 𝑘 2𝛼 𝑏2 𝛼22 𝛼1 − ⋯ − 2𝑘 𝛼 𝛼2 𝑎1 𝑏2 − 𝜔𝛼 𝛼2 𝑏1 = 0,

𝑈5:

𝑏𝜔𝛼 𝑘 2𝛼 𝑎1 𝛼13 + 𝑎𝑘 𝛼 𝛼1 𝑎1 + 2𝑘 𝛼 𝑎0 𝛼1 𝑎1 + 2𝑘 𝛼 𝛼1 𝑎2 𝑏1 + 𝜔𝛼 𝛼1 𝑎1 = 0,

𝑈6:

7𝑏𝜔𝛼 𝑘 2𝛼 𝑎1 𝛼12 𝛼2 + 8𝑏𝜔𝛼 𝑘 2𝛼 𝑎2 𝛼13 + ⋯ + 𝜔𝛼 𝛼2 𝑎1 + 2𝑘 𝛼 𝛼1 𝑎12 = 0,

𝑈7:

𝑏𝜔𝛼 𝑘 2𝛼 𝑎1 𝛼13 + 𝑎𝑘 𝛼 𝛼1 𝑎1 + 2𝑘 𝛼 𝑎0 𝛼1 𝑎1 + 2𝑘 𝛼 𝛼1 𝑎2 𝑏1 + 𝜔𝛼 𝛼1 𝑎1 = 0,

𝑈8:

6𝑏𝜔𝛼 𝑘 2𝛼 𝑎1 𝛼23 + 54𝑏𝜔𝛼 𝑘 2𝛼 𝑎2 𝛼1 𝛼22 + 6𝑘 𝛼 𝛼2 𝑎1 𝑎2 + 4𝑘 𝛼 𝛼1 𝑎22 = 0,

𝑈9:

24𝑏𝜔𝛼 𝑘 2𝛼 𝑎2 𝛼23 + 4𝑘 𝛼 𝛼2 𝑎22

Solving the above system for unknown parameters, we have the following solution set 1st Solution Set: 2

1 e𝛼ln(𝜔) [e𝛼ln(𝑘) ] 𝛼12 +𝑎e𝛼ln(𝑘) +e𝛼ln(𝜔)

𝑘 = 𝑘, 𝜔 = 𝜔, 𝑎0 = − 2

e𝛼ln(𝑘)

, 𝑎1 = −6𝑏e𝛼ln(𝑘) e𝛼ln(𝜔) 𝛼1 𝛼2 ,

𝑎2 = −6𝑏e𝛼ln(𝑘) e𝛼ln(𝜔) 𝛼22 , 𝑏1 = 𝑏2 = 0. Case I: Substituting the values of unknown into Eq. (4.265) coupled with the solution of Eq. (4.225) when 𝛼1 = −2i and 𝛼2 = 2i, we have 2


𝑢1 (𝜉) = − 2

e𝛼ln(𝑘)

− 6𝑏e𝛼ln(𝑘) e𝛼ln(𝜔) 𝛼1 𝛼2 𝑈

−6𝑏e𝛼ln(𝑘) e𝛼ln(𝜔) 𝛼22 𝑈 2 , 1

i



𝑢2 (𝜉) = − 2

e𝛼ln(𝑘)





𝑢3 (𝜉) = − 2

e𝛼ln(𝑘)


1





𝑢4 (𝜉) = − 2

e𝛼ln(𝑘)



√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

In all cases 𝜉 = 𝑘𝑥 + 𝜔𝑡.

Fig 4.62 (a): 2D Solitary wave solution of

Fig 4.62 (b): 3D Solitary wave solution of



4.9. Generalized U-expansion Method 4.9.1. Analysis of Generalized U-expansion method In order to simultaneously obtain more periodic wave solutions expressed in rational hyperbolic function and rational trigonometry function to nonlinear equations, we introduce generalized U-expansion Method. We briefly show what generalized Uexpansion method is and how to use it to obtain various periodic wave solutions to nonlinear equations. Suppose a nonlinear equation for 𝑈(𝑥, 𝑡) is given by We consider the general nonlinear PDE of the type 𝑃(𝑢, 𝐷𝑡𝛼 𝑢, 𝐷𝑥𝛼 𝑢, 𝐷𝑦𝛼 𝑢, 𝐷𝑧𝛼 𝑢, 𝐷𝑡2𝛼 𝑢, 𝐷𝑥2𝛼 𝑢, 𝐷𝑦2𝛼 𝑢, 𝐷𝑧2𝛼 𝑢, 𝐷𝑡𝛼 𝐷𝑥𝛼 𝑢, … ) = 0,

(4.1)

where 𝑃 is a polynomial in its arguments. The essence of the generalized U-expansion method can be presented in the following steps: Step 1: Seek Solitary wave solutions of Eq. (4.1) by taking 658

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡, and transform Eq. (4.1) to the ordinary differential equation. 𝑃(𝑢, 𝜔𝛼 𝐷𝜉𝛼 𝑢, 𝑘 𝛼 𝐷𝜉𝛼 𝑢, 𝑙 𝛼 𝐷𝜉𝛼 𝑢, 𝑚𝛼 𝐷𝜉𝛼 𝑢, 𝜔2𝛼 𝐷𝜉2𝛼 𝑢, 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢, 𝑙 2𝛼 𝐷𝜉2𝛼 𝑢, … ) = 0,

(4.266)

where 𝜔, 𝑘, 𝑙 and 𝑚 is constant and where prime denotes the derivative with respect to 𝜉. Step 2: If possible, integrate Eq. (4.266) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero. Step 3: According to generalized U-Expansion method, we assume that the wave solution can be expressed in the following form 𝑈𝑛

𝑢(𝜉) = ∑𝑀 𝑛=0 𝑎𝑛 (1+𝜀𝑈 𝑛 ),

(4.267)


(4.268)

where 𝛼1 and 𝛼2 are real constants, 𝑀 is a positive integer to be determined. The Table 1 shows the solution of Eq. (4.268) for different value of 𝛼1 and 𝛼2 . Table 4.4: Solutions of Eq. (4.268) for different values of 𝛼1 and 𝛼2 . 𝛼1

𝛼2

Cases

𝑈(𝜉)

−2i

2i

I

1 i − tan𝛼 (𝜉) 2 2

−i

I

II

1 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)) 2

4

2

III

1 1 −1 − tanh𝛼 (𝜉) − coth𝛼 (𝜉) 2 2

−√3i √3i

IV

√3 √3 i 𝐶1 sin𝛼 ( 2 𝜉) + 𝐶2 cos𝛼 ( 2 𝜉) 1

1 − 2 2

√3 √3 𝐶1 cos𝛼 ( 2 𝜉) + i𝐶2 sin𝛼 ( 2 𝜉)

Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term(s) of highest order with the highest order nonlinear term(s) in Eq. (4.266). Step 5: Substituting (4.267) into Eq. (4.266) with (4.268) will yields an algebraic equation involving power of U. Equating the coefficients of like power of U to zero gives a system of algebraic equations for 𝑎𝑖 , 𝑘, 𝑙, 𝑚 and 𝜔. Then, we solve the system with the

659

aid of a computer algebra system (CAS), such as MAPLE 13, to determine these constants. Step 6: Putting these constant into Eq. (4.267), coupled with the well known solutions of Eq. (4.268), we can obtained the more general type and new exact traveling wave solution of the nonlinear partial differential equation (4.1).

4.9.2. Negative Gardener Equation Consider the negative gardener equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 2𝑎𝑢𝐷𝑥𝛼 𝑢 − 3𝑏𝑢2 𝐷𝑥𝛼 𝑢 + 𝐷𝑥3𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.269)


(4.270)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.270) into Eq. (4.269) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 2𝑎𝑢𝑘 𝛼 𝐷𝜉𝛼 𝑢 − 3𝑏𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 𝑘 3𝛼 𝐷𝜉3𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.271)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 3 = 2𝑀 + 𝑀 + 1, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.267), we obtained the trail solution 𝑈

𝑢 = 𝑎0 + 𝑎1 1+𝜀𝑈.

(4.272)

Putting Eq. (4.272) into Eq. (4.271) coupled with Eq. (4.268); the Eq. (4.272) yields an algebraic equation involving power of U as 1 [𝑎1 𝑈(𝐶0 𝑈 0 (1+𝜀𝑈)4

+ 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 )] = 0.


𝜔𝛼 𝛼1 + 𝑘 3𝛼 𝛼13 + 2𝑎𝑘 𝛼 𝑎0 𝛼1 − 3𝑏𝑘 𝛼 𝑎02 𝛼1 = 0,

𝑈1 :

−6𝑏𝑎1 𝑘 𝛼 𝑎0 𝛼1 − 6𝑏𝑘 𝛼 𝑎02 𝜀𝛼1 + ⋯ + 7𝑘 3𝛼 𝛼2 𝛼12 − 4𝑘 3𝛼 𝛼13 𝜀 = 0,

𝑈2:

−6𝑏𝑎1 𝑘 𝛼 𝑎0 𝛼2 + 4𝑎𝑘 𝛼 𝑎0 𝜀𝛼2 + ⋯ + 12𝑘 3𝛼 𝛼22 𝛼1 + 𝑘 3𝛼 𝛼13 𝜀 2 = 0,

𝑈3:

−3𝑏𝑘 𝛼 𝑎02 𝜀 2 𝛼2 + 2𝑎𝑘 𝛼 𝑎0 𝜀 2 𝛼2 + ⋯ + 6𝑘 3𝛼 𝛼23 + 𝜔𝛼 𝛼2 𝜀 2 = 0.

Solving the above system for unknown parameters, we have the following solution sets

660

1

1st

1 2𝑎𝛼2 +3𝑏𝑎1 𝛼1 −2𝑎𝜀𝛼1

𝑎0 = 6

Solution Set:

e

−𝜀𝛼1 +𝛼2

1 2 2 2 2 2 √2𝑎1 (4𝑎2 𝛼2 −8𝑎2 𝛼2 𝜀𝛼1 +4𝑎1 𝜀2 𝛼1 −3𝑏2 𝑎1 𝛼1 ) 𝛼 ( ) 12√𝑏(−𝜀𝛼1 +𝛼2 )3

,𝑘 =

𝛼 √𝑏𝑎 ( (−𝜀𝛼 1+𝛼 )) , 𝜔 √2 1 2

=

, 𝑎1 = 𝑎1 .

Case I: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1 2𝑎𝛼2 +3𝑏𝑎1 𝛼1 −2𝑎𝜀𝛼1

𝑢1 (𝜉) = 6 1

𝑈

+ 1+𝜀𝑈 𝑎1 ,

−𝜀𝛼1 +𝛼2

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 1 2𝑎𝛼2 +3𝑏𝑎1 𝛼1 −2𝑎𝜀𝛼1

𝑢2 (𝜉) = 6

−𝜀𝛼1 +𝛼2

𝑈

+ 1+𝜀𝑈 𝑎1 ,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 1 2𝑎𝛼2 +3𝑏𝑎1 𝛼1 −2𝑎𝜀𝛼1

𝑢3 (𝜉) =

6

𝑈

+ 1+𝜀𝑈 𝑎1 ,

−𝜀𝛼1 +𝛼2

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 1 2𝑎𝛼2 +3𝑏𝑎1 𝛼1 −2𝑎𝜀𝛼1

𝑢4 (𝜉) = 6

−𝜀𝛼1 +𝛼2 √3

𝑈

+ 1+𝜀𝑈 𝑎1 ,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


√𝑏𝑎1 ) √2(−𝜀𝛼1 +𝛼2 )

1 𝛼

𝑥+e

, and i = √−1.

1 2 2 2 2 2 √2𝑎1 (4𝑎2 𝛼2 −8𝑎2 𝛼2 𝜀𝛼1 +4𝑎1 𝜀2 𝛼1 −3𝑏2 𝑎1 𝛼1 ) 𝛼 ( ) 12√𝑏(−𝜀𝛼1 +𝛼2 )3

2nd Solution Set: 1 (1+√3)𝑎

𝑎0 = 3

𝑏

, 𝑎1 =

2(−𝜀𝛼1 +𝛼2 )√3 3𝑏𝛼1

1 √2√3√𝑏𝑎

,𝑘 = (

661

3𝛼1

1 𝛼

) , 𝜔 = 0.

𝑡.

Case I: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1 (1+√3)𝑎

𝑢5 (𝜉) = 3 1

𝑏

+

2(−𝜀𝛼1 +𝛼2 )√3

𝑈

3𝑏𝛼1

1+𝜀𝑈

,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 1 (1+√3)𝑎

𝑢6 (𝜉) = 3

𝑏

+

2(−𝜀𝛼1 +𝛼2 )√3

𝑈

3𝑏𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 1 (1+√3)𝑎

𝑢7 (𝜉) = 3

𝑏

+

2(−𝜀𝛼1 +𝛼2 )√3

𝑈

3𝑏𝛼1

1+𝜀𝑈

1

,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 1 (1+√3)𝑎

𝑢8 (𝜉) = 3

𝑏

+

2(−𝜀𝛼1 +𝛼2 )√3

𝑈

3𝑏𝛼1

1+𝜀𝑈

, 1 √2√3√𝑏𝑎

√3 √3 1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1. In all cases 𝜉 = (

3𝛼1

1 𝛼

) 𝑥.

3rd Solution Set: 1 (1−√3)𝑎

𝑎0 = 3

𝑏

, 𝑎1 = −

2(−𝜀𝛼1 +𝛼2 )√3 3𝑏𝛼1

1 √2√3√𝑏𝑎

,𝑘 = (

3𝛼1

1 𝛼

) , 𝜔 = 0.

Case I: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 1 (1−√3)𝑎

𝑢9 (𝜉) = 3 1

𝑏

−

2(−𝜀𝛼1 +𝛼2 )√3

𝑈

3𝑏𝛼1

1+𝜀𝑈

,

i


662

Case II: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 1 (1−√3)𝑎

𝑢10 (𝜉) = 3

𝑏

−

2(−𝜀𝛼1 +𝛼2 )√3

𝑈

3𝑏𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 1 (1−√3)𝑎

𝑢11 (𝜉) = 3

𝑏

−

1

2(−𝜀𝛼1 +𝛼2 )√3

𝑈

3𝑏𝛼1

1+𝜀𝑈

,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.272) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 1 (1−√3)𝑎

𝑢12 (𝜉) = 3 1

𝑏

−

2(−𝜀𝛼1 +𝛼2 )√3

𝑈

3𝑏𝛼1

1+𝜀𝑈

, 1 √2√3√𝑏𝑎

√3 √3 1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1. In all cases 𝜉 = (

3𝛼1

1 𝛼

) 𝑥.




4.9.3. Modified Equal Width Equation Consider the modified equal width equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 3𝑢2 𝐷𝑥𝛼 𝑢 + 𝐷𝑥2𝛼 𝐷𝑡𝛼 𝑢 = 0,

0 < 𝛼 ≤ 1. 663

(4.273)


(4.274)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.274) into Eq. (4.273) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 3𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 𝑘 2𝛼 𝜔𝛼 𝐷𝜉3𝛼 𝑢 = 0.

(4.275)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 3 = 2𝑀 + 𝑀 + 1, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.267), we obtained the trail solution 𝑈

𝑢 = 𝑎0 + 𝑎1 1+𝜀𝑈.

(4.276)


− (1+𝜀𝑈)4 [𝑎1 𝑈(𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 )] = 0. Compare the like powers of U we have system of equations 𝑈0:

−3𝑘 𝛼 𝑎02 𝛼1 + 𝑎𝑘 2𝛼 𝜔𝛼 𝛼13 − 𝜔𝛼 𝛼1 = 0,

𝑈1 :

−6𝑎1 𝑘 𝛼 𝑎0 𝛼1 − 6𝑘 𝛼 𝑎02 𝜀𝛼1 − 𝜔𝛼 𝑎2 − ⋯ + 7𝑎𝑘 2𝛼 𝜔𝛼 𝛼2 𝑎12 = 0,

𝑈2:

−6𝑎1 𝑘 𝛼 𝑎0 𝛼2 − 6𝑘 𝛼 𝑎02 𝜀𝛼2 − 3𝑘 𝛼 𝑎02 𝜀 2 𝛼1 − ⋯ − 6𝑎1 𝑘 𝛼 𝑎0 𝜀𝛼1 = 0,

𝑈2:

6𝑎𝑘 2𝛼 𝜔𝛼 𝛼23 − 3𝑘 𝛼 𝑎02 𝜀 2 𝛼2 − 𝜔𝛼 𝛼2 𝜀 2 − ⋯ − 6𝑎1 𝑘 𝛼 𝑎0 𝜀𝛼2 = 0.

Solving the above system for unknown parameters, we have the following solution set Solution Set:

𝑎0 = 𝑎0 , 𝑎1 =

2𝑎0 (−𝜀𝛼1 +𝛼2 ) 𝛼1

2 √− 𝑎

1 𝛼

, 𝑘 = ( 𝛼 ) , 𝜔 = (− 1

2 √− 𝑎02 𝑎

𝛼1

1 𝛼

) .


2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i


𝑢2 (𝜉) = 𝑎0 +

2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1


2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

1

,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.276) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢4 (𝜉) = 𝑎0 + 1

2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

√3

,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

2 √− 𝑎

1 𝛼

In all cases 𝜉 = ( 𝛼 ) 𝑥 + (− 1

2 √− 𝑎02 𝑎

𝛼1

, and i = √−1. 1 𝛼

) 𝑡.


Fig 4.64 (a): 2D Periodic wave solution of

Fig 4.64 (b): 3D Periodic wave solution of



4.9.4. (2+1) Dimension Burger’s Equation Consider the (2+1)-Dimension Burger’s equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝛽(𝑢𝐷𝑥𝛼 𝑢 + 𝑢𝐷𝑦𝛼 𝑢) + 𝛾(𝐷𝑥2𝛼 𝑢 + 𝐷𝑦2𝛼 𝑢) = 0, 0 < 𝛼 ≤ 1. 665

(4.277)


(4.278)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.278) into Eq. (4.277) and using the chain rule we obtained 𝐷𝜉𝛼 𝑢 + 𝛽(𝑘 𝛼 𝑢𝐷𝜉𝛼 𝑢 + 𝑙 𝛼 𝑢𝐷𝜉𝛼 𝑢) + 𝛾(𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 + 𝑙 2𝛼 𝐷𝜉2𝛼 𝑢) = 0, 0 < 𝛼 ≤ 1.

(4.279)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 2 = 𝑀 + 𝑀 + 1, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.267), we obtained the trail solution 𝑢 = 𝑎0 + 𝑎1

𝑈 1+𝜀𝑈

.

(4.280)


+ 𝐶1 𝑈1 + 𝐶2 𝑈 2 )] = 0.


𝛽𝑘 𝛼 𝑎0 𝛼1 + 𝛽𝑙 𝛼 𝑎0 𝛼1 + 𝛾𝑘 2𝛼 𝛼12 + 𝛾𝑙 2𝛼 𝛼12 + 𝜔𝛼 𝛼1 = 0,

𝑈1 :

𝜔𝛼 𝜀𝛼1 + 𝛽𝑘 𝛼 𝑎0 𝛼2 + 𝛽𝑙 𝛼 𝑎0 𝛼2 − ⋯ + 𝛽𝑙 𝛼 𝑎0 𝜀𝛼1 + 𝜔𝛼 𝛼2 = 0,

𝑈2:

𝜔𝛼 𝛼2 𝜀 + 2𝛾𝑘 2𝛼 𝛼22 + 2𝛾𝑙 2𝛼 𝛼22 + ⋯ − 𝛾𝑘 2𝛼 𝛼2 𝜀𝛼1 − 𝛾𝑙 2𝛼 𝛼2 𝜀𝛼1 = 0.

Solving the above system for unknown parameters, we have the six solution sets 1st Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = − e

2𝛾(−𝑘 2𝛼 𝜀𝛼1 −𝑙2𝛼 𝜀𝛼1 +𝑘 2𝛼 𝛼2 +𝑙2𝛼 𝛼2 ) 𝛽(𝑙𝛼 +𝑘 𝛼 )

ln(−𝛽𝑘𝛼 𝑎0 −𝛽𝑙𝛼 𝑎0 −𝛾𝑘2𝛼 𝛼1 −𝛾𝑙2𝛼 𝛼1 ) 𝛼

, 𝑘 = 𝑘, 𝜔 =

.


2𝛾(−𝑘 2𝛼 𝜀𝛼1 −𝑙2𝛼 𝜀𝛼1 +𝑘 2𝛼 𝛼2 +𝑙2𝛼 𝛼2 )

𝑈

𝛽(𝑙𝛼 +𝑘 𝛼 )

1+𝜀𝑈

,

i



𝑈


1+𝜀𝑈

666

,

1



𝑈


1+𝜀𝑈

1

,

1



𝑈


1+𝜀𝑈

√3

,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


, and i = √−1.

ln(−𝛽𝑘𝛼 𝑎0 −𝛽𝑙𝛼 𝑎0 −𝛾𝑘2𝛼 𝛼1 −𝛾𝑙2𝛼 𝛼1 ) 𝛼

𝑡.

2nd Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = −

2𝛾𝑙𝛼 (−𝜀𝛼1 +𝛼2 ) 𝛽

, 𝑘 = 0, 𝜔 = e

ln(−𝛽𝑙𝛼 𝑎0 −𝛾𝑙2𝛼𝛼1 ) 𝛼

.


2𝛾𝑙𝛼 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛽

1+𝜀𝑈

,

i


2𝛾𝑙𝛼 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛽

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.280) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢7 (𝜉) = 𝑎0 − 1

2𝛾𝑙𝛼 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛽

1+𝜀𝑈

,

1


667

Case IV: Substituting the values of unknown into Eq. (4.280) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢8 (𝜉) = 𝑎0 − 1

2𝛾𝑙𝛼 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛽

1+𝜀𝑈

√3

,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1. In all cases 𝜉 = e

ln(−𝛽𝑙𝛼 𝑎0 −𝛾𝑙2𝛼𝛼1 ) 𝛼

𝑡.


𝛾𝛼1 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛽(𝑘 𝛼 +𝑙𝛼 )

, 𝑎1 = −

2𝛾(−𝑘 2𝛼 𝜀𝛼1 −𝑙2𝛼 𝜀𝛼1 +𝑘 2𝛼 𝛼2 +𝑙2𝛼 𝛼2 ) 𝛽(𝑙𝛼 +𝑘 𝛼 )

, 𝑘 = 𝑘, 𝜔 = 0.

Case I: Substituting the values of unknown into Eq. (4.280) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝛾𝛼1 (𝑘 2𝛼 +𝑙2𝛼 )

𝑢9 (𝜉) = 1

𝛽(𝑘 𝛼 +𝑙𝛼 )

−


𝑈


1+𝜀𝑈

,

i


𝛾𝛼1 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛽(𝑘 𝛼 +𝑙𝛼 )

−


𝑈


1+𝜀𝑈

,

1


𝛾𝛼1 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛽(𝑘 𝛼 +𝑙𝛼 ) 1

−


𝑈


1+𝜀𝑈

,

1


𝛾𝛼1 (𝑘 2𝛼 +𝑙2𝛼 ) 𝛽(𝑘 𝛼 +𝑙𝛼 ) √3

−


𝑈


1+𝜀𝑈

,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (



668

Fig 4.65 (a): 2D Periodic wave solution of Eq. (4.277) for different values of parameters


4.9.5. (3+1) Dimension Burger’s Equation Consider the (3+1)-Dimension Burger’s equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝛽(𝑢𝐷𝑥𝛼 𝑢 + 𝑢𝐷𝑦𝛼 𝑢 + 𝑢𝐷𝑧𝛼 𝑢) + 𝛾(𝐷𝑥2𝛼 𝑢 + 𝐷𝑦2𝛼 𝑢 + 𝐷𝑧2𝛼 𝑢) = 0,0 < 𝛼 ≤ 1. (4.281) To convert Eq. (4.281) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡,

(4.282)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.282) into Eq. (4.281) and using the chain rule we obtained 𝐷𝜉𝛼 𝑢 + 𝛽(𝑘 𝛼 + 𝑙 𝛼 + 𝑚𝛼 )𝑢𝐷𝜉𝛼 𝑢 + 𝛾(𝑘 2𝛼 + 𝑙 2𝛼 + 𝑚2𝛼 )𝐷𝜉2𝛼 𝑢 = 0.

(4.283)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 2 = 𝑀 + 𝑀 + 1, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.267), we obtained the trail solution 𝑈

𝑢 = 𝑎0 + 𝑎1 1+𝜀𝑈.

(4.284)


+ 𝐶1 𝑈1 + 𝐶2 𝑈 2 )] = 0.


𝛾𝑘 2𝛼 𝛼12 + 𝛾𝑙 2𝛼 𝛼12 + 𝛾𝑚2𝛼 𝛼12 + ⋯ + 𝛽𝑚𝛼 𝑎0 𝛼1 + 𝜔𝛼 𝛼1 = 0,

𝑈1 :

𝜔𝛼 𝜀𝛼1 + 𝛽𝑎1 𝑘 𝛼 𝛼1 + 𝛽𝑎1 𝑙 𝛼 𝛼1 + ⋯ + 𝛽𝑙 𝛼 𝑎0 𝜀𝛼1 + 𝛽𝑚𝛼 𝑎0 𝜀𝛼1 = 0, 669

𝑈2:

𝜔𝛼 𝜀𝛼2 + 𝛽𝑎1 𝑘 𝛼 𝛼2 + 𝛽𝑎1 𝑙 𝛼 𝛼2 + ⋯ − 𝛾𝑙 2𝛼 𝛼2 𝜀𝛼1 − 𝛾𝑚2𝛼 𝛼2 𝜀𝛼1 = 0.

Solving the above system for unknown parameters, we have the six solution sets 1st Solution Set: 𝑎0 = 𝑎0 , 𝑎1 = − 𝜔=e

2𝛾(−𝑘 2𝛼 𝜀𝛼1 −𝑙2𝛼 𝜀𝛼1 −𝑚2𝛼 𝜀𝛼1 +𝑘 2𝛼 𝛼2 +𝑙2𝛼 𝛼2 +𝑚2𝛼 𝛼2 ) 𝛽(𝑙𝛼 +𝑘 𝛼 +𝑚𝛼 )

ln(−𝛽𝑘𝛼 𝑎0 −𝛽𝑙𝛼 𝑎0 −−𝛽𝑚𝛼 𝑎0 −𝛾𝑘2𝛼 𝛼1 −𝛾𝑙2𝛼 𝛼1 −𝛾𝑚2𝛼 𝛼1 ) 𝛼

, 𝑘 = 𝑘,

.


2𝛾(−𝑘 2𝛼 𝜀𝛼1 −𝑙2𝛼 𝜀𝛼1 −𝑚2𝛼 𝜀𝛼1 +𝑘 2𝛼 𝛼2 +𝑙2𝛼 𝛼2 +𝑚2𝛼 𝛼2 )

𝑈

𝛽(𝑙𝛼 +𝑘 𝛼 +𝑚𝛼 )

1+𝜀𝑈

,

i



𝑈


1+𝜀𝑈

,

1



𝑈


1+𝜀𝑈

1

,

1



𝑈


1+𝜀𝑈

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


, and i = √−1.

ln(−𝛽𝑘𝛼 𝑎0 −𝛽𝑙𝛼 𝑎0 −−𝛽𝑚𝛼 𝑎0 −𝛾𝑘2𝛼 𝛼1 −𝛾𝑙2𝛼𝛼1 −𝛾𝑚2𝛼 𝛼1 ) 𝛼

2nd Solution Set:

670

𝑡.

,

𝑎0 = 𝑎0 , 𝑎1 = − e

2𝛾(−𝑙2𝛼 𝜀𝛼1 +𝑙2𝛼 𝛼2 −𝑚2𝛼 𝜀𝛼1 +𝑚2𝛼 𝛼2 ) 𝛽(𝑙𝛼 +𝑚𝛼 )

ln(−𝛽𝑙𝛼 𝑎0 −𝛽𝑚𝛼 𝑎0 −𝛾𝑙2𝛼 𝛼1 −𝛾𝑚2𝛼 𝛼1 ) 𝛼

, 𝑘 = 0, 𝜔 =

.

Case I: Substituting the values of unknown into Eq. (4.284) coupled with the solution of Eq. (11) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢5 (𝜉) = 𝑎0 − 1

2𝛾(−𝑙2𝛼 𝜀𝛼1 +𝑙2𝛼 𝛼2 −𝑚2𝛼 𝜀𝛼1 +𝑚2𝛼 𝛼2 )

𝑈

𝛽(𝑙𝛼 +𝑚𝛼 )

1+𝜀𝑈

,

i



𝑈


1+𝜀𝑈

,

1



𝑈


1+𝜀𝑈

1

,

1



𝑈


1+𝜀𝑈

√3

,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


, and i = √−1.

ln(−𝛽𝑙𝛼 𝑎0 −𝛽𝑚𝛼 𝑎0 −𝛾𝑙2𝛼 𝛼1 −𝛾𝑚2𝛼 𝛼1 ) 𝛼

𝑡.


𝛾𝛼1 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 ) 𝛽(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 )

, 𝑎1 = −

2𝛾(−𝑘 2𝛼 𝜀𝛼1 −𝑙2𝛼 𝜀𝛼1 −𝑚2𝛼 𝜀𝛼1 +𝑘 2𝛼 𝛼2 +𝑙2𝛼 𝛼2 +𝑚2𝛼 𝛼2 ) 𝛽(𝑙𝛼 +𝑘 𝛼 +𝑚𝛼 )

,𝑘 =

𝑘, 𝜔 = 0. Case I: Substituting the values of unknown into Eq. (4.284) coupled with the solution of Eq. (4.678) when 𝛼1 = −2i and 𝛼2 = 2i, we have

671

𝛾𝛼1 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 )

𝑢9 (𝜉) = 1

𝛽(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 )

−


𝑈


1+𝜀𝑈

,

i


𝛾𝛼1 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 ) 𝛽(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 )

−


𝑈


1+𝜀𝑈

,

1


𝛾𝛼1 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 ) 𝛽(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 ) 1

−


𝑈


1+𝜀𝑈

,

1


𝛾𝛼1 (𝑘 2𝛼 +𝑙2𝛼 +𝑚2𝛼 ) 𝛽(𝑘 𝛼 +𝑙𝛼 +𝑚𝛼 ) √3

−


𝑈


1+𝜀𝑈

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (






4.9.6. Burger’s Equation Consider the Burger’s equation [230] of fraction order 672

,

𝐷𝑡𝛼 𝑢 + 𝑢𝐷𝑥𝛼 𝑢 + 𝐷𝑥2𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.285)


(4.286)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.286) into Eq. (4.285) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝑘 𝛼 𝑢𝐷𝜉𝛼 𝑢 + 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.287)


𝑢 = 𝑎0 + 𝑎1 1+𝜀𝑈.

(4.288)


+ 𝐶1 𝑈1 + 𝐶2 𝑈 2 )] = 0.


2𝜔𝛼 𝑎0 + 𝑘 𝛼 𝑎02 = 0,

𝑈1 :

4𝜔𝛼 𝑎0 𝜀 + 2𝜔𝛼 𝑎1 + 2𝑘 𝛼 𝑎02 𝜀 + 2𝑘 𝛼 𝑎0 𝑎1 + 2𝑎1 𝑘 2𝛼 𝛼1 = 0,

𝑈2:

2𝜔𝛼 𝑎0 𝜀 2 + 2𝜔𝛼 𝑎1 𝜀 + 𝑘 𝛼 𝑎02 𝜀 2 + 2𝑘 𝛼 𝑎0 𝑎1 𝜀 + 𝑘 𝛼 𝑎12 + 2𝑎1 𝑘 2𝛼 𝛼2 = 0.

Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: 𝑎0 = 0, 𝑎1 = −2(−𝜀𝛼1 + 𝛼2 )e𝛼 ln(𝑘) , 𝑘 = 𝑘, 𝜔 = e

ln(−𝛼1 )+2𝛼ln(𝑘) 𝛼

.

Case I: Substituting the values of unknown into Eq. (4.288) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑈

𝑢1 (𝜉) = −2(−𝜀𝛼1 + 𝛼2 )e𝛼 ln(𝑘) 1+𝜀𝑈, 1

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.288) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 𝑈

𝑢2 (𝜉) = −2(−𝜀𝛼1 + 𝛼2 )e𝛼 ln(𝑘) 1+𝜀𝑈,

673

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.288) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑈

𝑢3 (𝜉) = −2(−𝜀𝛼1 + 𝛼2 )e𝛼 ln(𝑘) 1+𝜀𝑈, 1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.288) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑈

𝑢4 (𝜉) = −2(−𝜀𝛼1 + 𝛼2 )e𝛼 ln(𝑘) 1+𝜀𝑈, √3


ln(−𝛼1 )+2𝛼ln(𝑘) 𝛼

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)

1

√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


𝑡.

2nd Solution Set: 𝑎0 = −2e𝛼 ln(𝑘) , 𝑎1 = −2(−𝜀𝛼1 + 𝛼2 )e𝛼 ln(𝑘) , 𝑘 = 𝑘, 𝜔 = e

ln(𝛼1 )+2𝛼ln(𝑘) 𝛼

.

Case I: Substituting the values of unknown into Eq. (4.288) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑈

𝑢1 (𝜉) = −2e𝛼 ln(𝑘) − 2(−𝜀𝛼1 + 𝛼2 )e𝛼 ln(𝑘) 1+𝜀𝑈, 1

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.288) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 𝑈


where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.288) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑈


1


𝑈



ln(𝛼1 )+2𝛼ln(𝑘) 𝛼

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉) √3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


𝑡.




4.9.7. Modified KdV Equation Consider the Modified KdV equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝑢2 𝐷𝑥𝛼 𝑢 + 𝐷𝑥3𝛼 𝑢 = 0,

0 < 𝛼 ≤ 1.

(4.289)

To convert Eq. (4.289) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡,

(4.290)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.290) into Eq. (4.289) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 𝑘 3𝛼 𝐷𝜉3𝛼 𝑢 = 0.

(4.291)


𝑢 = 𝑎0 + 𝑎1 1+𝜀𝑈.

(4.292)

675


1

3 (1+𝜀𝑈)3

[𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 ] = 0.


𝑘 𝛼 𝑎03 + 3𝜔𝛼 𝑎0 = 0,

𝑈1 :

3𝑎1 𝑘 3𝛼 𝛼12 + 9𝜔𝛼 𝑎0 𝜀 + 3𝑘 𝛼 𝑎03 𝜀 + 3𝑘 𝛼 𝑎02 𝑎1 + 3𝜔𝛼 𝑎1 = 0,

𝑈2:

9𝜔𝛼 𝑎0 𝜀 2 + 6𝜔𝛼 𝑎1 𝜀 + 3𝑘 𝛼 𝑎03 𝜀 2 + ⋯ + 9𝑎1 𝑘 3𝛼 𝛼2 𝛼1 = 0,

𝑈3:

𝑘 𝛼 𝑎03 𝜀 3 + 6𝑎1 𝑘 3𝛼 𝛼22 + ⋯ + 𝑘 𝛼 𝑎13 − 3𝑎1 𝑘 3𝛼 𝛼2 𝜀𝛼1 = 0.

Solving the above system for unknown parameters, we have the following solution sets Solution Set: 1

𝑎0 = 𝑎0 , 𝑎1 =

2𝑎0 (−𝜀𝛼1 +𝛼2 ) 𝛼1

1 √−6 𝛼

, 𝑘 = (3

𝛼1

) ,𝜔 = e

1 1√−6 ln(− 𝑎0 )+ln( ) 3 3 𝛼1 𝛼

.


2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i


2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1


2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

1

,

1


2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

676

1

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1

1 √−6 𝛼


𝛼1

) 𝑥+e

, and i = √−1.

1 1√−6 ln(− 𝑎0 )+ln( ) 3 3 𝛼1

𝑡.

𝛼





4.9.8. FitzHugh–Nagumo Equation Consider the FitzHugh-Nagumo equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝐷𝑥2𝛼 𝑢 + 𝑢(1 − 𝑢)(𝑎 − 𝑢) = 0, 0 < 𝛼 ≤ 1.

(4.293)


(4.294)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.294) into Eq. (4.293) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 + 𝑢(1 − 𝑢)(𝑎 − 𝑢) = 0, 0 < 𝛼 ≤ 1.

(4.295)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 2 = 3𝑀, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.267), we obtained the trail solution 𝑈

𝑢 = 𝑎0 + 𝑎1 1+𝜀𝑈.

(4.296)


1 [𝐶0 𝑈 0 (1+𝜀𝑈)3

+ 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 ] = 0.


𝑎𝑎0 − 𝑎02 + 𝑎03 − 𝑎𝑎02 = 0,

𝑈1 :

𝑎𝑎1 − 3𝑎02 𝜀 − 2𝑎0 𝑎1 + 3𝑎03 𝜀 + ⋯ + 𝑎1 𝜔𝛼 𝛼1 − 𝑎1 𝑘 2𝛼 𝛼12 = 0,

𝑈2:

−𝑎12 − 3𝑎02 𝜀 2 − 𝑎𝑎12 + 3𝑎03 𝜀 2 + ⋯ + 𝑎1 𝑘 2𝛼 𝛼12 𝜀 − 3𝑎1 𝑘 2𝛼 𝛼2 𝛼1 = 0,

𝑈3:

𝑎13 + 𝑎1 𝑘 2𝛼 𝛼2 𝜀𝛼1 − 𝑎02 𝜀 3 − ⋯ − 2𝑎𝑎0 𝜀 2 𝑎1 + 𝑎1 𝜔𝛼 𝛼2 𝜀 = 0.

Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: ln(− 1

ln(

𝑘 = e2𝛼

1 ) 2𝛼2 1

, 𝑎1 = −

−𝜀𝛼1 +𝛼2 𝛼1

(

, 𝑎0 = 0, 𝜔 = e

1−1+2𝑎 ) 2 𝛼1 ) 𝛼

.


−𝜀𝛼1 +𝛼2

𝑈

𝛼1

1+𝜀𝑈

,

i


−𝜀𝛼1 +𝛼2

𝑈

𝛼1

1+𝜀𝑈

,

1


−𝜀𝛼1 +𝛼2

𝑈

𝛼1

1+𝜀𝑈

1

,

1


−𝜀𝛼1 +𝛼2

𝑈

𝛼1

1+𝜀𝑈 √3

, √3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

, and i = √−1.

678

ln(−


1 1 ln( 2 ) 2𝛼 2𝛼1

(

𝑥+e

1−1+2𝑎 ) 2 𝛼1 ) 𝛼

𝑡.

2nd Solution Set:

𝑘=e

1 𝑎2 ln( 2 ) 2𝛼 𝛼1

ln(−

𝑎(−𝜀𝛼1 +𝛼2 )

, 𝑎1 = −

𝛼1

(

, 𝑎0 = 0, 𝜔 = e

1𝑎(𝑎−2) ) 2 𝛼1 ) 𝛼

.

Case I: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢5 (𝜉) = 1

𝑎(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i


𝑎(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1


𝑎(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

1

,

1


𝑎(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

ln(−


1 1 ln( 2 ) 2𝛼 2𝛼1

(

𝑥+e

, and i = √−1.

1−1+2𝑎 ) 2 𝛼1 ) 𝛼

𝑡.

3rd Solution Set:

𝑎0 = 1, 𝑎1 =

(−𝜀𝛼1 +𝛼2 ) 𝛼1

1

ln(

, 𝑘 = e2𝛼

1 ) 2𝛼2 1

,𝜔 = e

679

1−1+2𝑎 ln(− ) 2 𝛼1 ( ) 𝛼

.

Case I: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢9 (𝜉) = 1 + 1

(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢10 (𝜉) = 1 +

(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢11 (𝜉) = 1 +

(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

1

,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢12 (𝜉) = 1 + 1

(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

√3

, √3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

ln(− 1

1 ) 2𝛼2 1

ln(

In all cases 𝜉 = e2𝛼

(

𝑥+e

, and i = √−1.

1−1+2𝑎 ) 2 𝛼1 ) 𝛼

𝑡.

4th Solution Set: ln(

𝑎0 = 1, 𝑎1 = −

𝛼2 𝑎−𝛼2 −𝛼1 𝑎𝜀+𝜀𝛼1 𝛼1

,𝜔 = e

1 𝑎2 −1 ) ln( 2𝛼 2𝛼1

(

,𝑘 = e

1 𝑎2 +1−2𝑎 ) 2 𝛼2 1 ) 𝛼

.

Case I: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢13 (𝜉) = 1 − 1

𝛼2 𝑎−𝛼2 −𝛼1 𝑎𝜀+𝜀𝛼1

𝑈

𝛼1

1+𝜀𝑈

,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). 680

Case II: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢14 (𝜉) = 1 −


𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢15 (𝜉) = 1 −


𝑈

𝛼1

1+𝜀𝑈

1

,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢16 (𝜉) = 1 − 1


𝑈

𝛼1

1+𝜀𝑈

√3

,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

ln(


1 𝑎2 −1 ) ln( 2𝛼 2𝛼1

(

, and i = √−1.

1 𝑎2 +1−2𝑎 ) 2 𝛼2 1 ) 𝛼

𝑥+e

𝑡.

5th Solution Set:

𝑎0 = 𝑎, 𝑎1 =

𝛼2 𝑎−𝛼2 −𝛼1 𝑎𝜀+𝜀𝛼1 𝛼1

1

𝑎2 −1 ) 2𝛼1

ln(

, 𝜔 = e2𝛼

1 𝑎2 +1−2𝑎 ln( ) 2 𝛼2 1 ( ) 𝛼

,𝑘 = e

.

Case I: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢17 (𝜉) = 𝑎 + 1


𝑈

𝛼1

1+𝜀𝑈

,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢18 (𝜉) = 𝑎 +


𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). 681

Case III: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢19 (𝜉) = 𝑎 +


𝑈

𝛼1

1+𝜀𝑈

1

,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢20 (𝜉) = 𝑎 + 1


𝑈

𝛼1

1+𝜀𝑈

√3

,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

ln(


1 𝑎2 −1 ) ln( 2𝛼 2𝛼1

(

, and i = √−1.

1 𝑎2 +1−2𝑎 ) 2 𝛼2 1 ) 𝛼

𝑥+e

𝑡.

6th Solution Set:

𝑎0 = 𝑎, 𝑎1 =

𝑎(−𝜀𝛼1 +𝛼2 ) 𝛼1

,𝑘 = e

1 𝑎2 ln( 2 ) 2𝛼 2𝛼1

ln(−

(

,𝜔 = e

1 𝑎(𝑎−2) ) 2 𝛼1 ) 𝛼

.

Case I: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢21 (𝜉) = 𝑎 + 1

(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢22 (𝜉) = 𝑎 +

(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢23 (𝜉) = 𝑎 + 1

(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1


Case IV: Substituting the values of unknown into Eq. (4.296) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢24 (𝜉) = 𝑎 + 1

(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

√3

, √3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


1 𝑎2 ln( 2 ) 2𝛼 2𝛼1

ln(−

(

𝑥+e

, and i = √−1.

1 𝑎(𝑎−2) ) 2 𝛼1 ) 𝛼

𝑡.



4.9.9. Klein Gordon Equation Consider the Klein Gordon equation [230] of fraction order 𝐷𝑡2𝛼 𝑢 + 𝐷𝑥2𝛼 𝑢 + 𝑢 + 𝑢3 = 0,

0 < 𝛼 ≤ 1.

(4.297)


(4.298)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.298) into Eq. (4.297) and using the chain rule we obtained 𝜔2𝛼 𝐷𝜉2𝛼 𝑢 + 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 + 𝑢 + 𝑢3 = 0. By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 2 = 3𝑀, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.267), we obtained the trail solution 683

(4.299)

𝑈

𝑢 = 𝑎0 + 𝑎1 1+𝜀𝑈.

(4.300)


− (1+𝜀𝑈)3 [𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 ] = 0. Compare the like powers of U we have system of equations 𝑈0:

−𝑎0 − 𝑎03 = 0,

𝑈1 :

−𝑎1 𝜔2𝛼 𝛼12 − 𝑎1 𝑘 2𝛼 𝛼12 − 3𝑎0 𝜀 − 𝑎1 − 3𝑎03 𝜀 − 3𝑎02 𝑎1 = 0,

𝑈2:

−6𝑎02 𝜀𝑎1 + 𝑎1 𝜔2𝛼 𝛼12 𝜀 − 3𝑎1 𝜔2𝛼 𝛼2 𝛼1 + ⋯ − 2𝑎1 𝜀 − 3𝑎03 𝜀 2 − 3𝑎0 𝑎12 = 0,

𝑈2:

𝑎1 𝜔2𝛼 𝛼2 𝜀𝛼1 + 𝑎1 𝑘 2𝛼 𝛼2 𝜀𝛼1 − 3𝑎02 𝜀 2 𝑎1 − ⋯ − 𝑎1 𝜀 2 − 𝑎03 𝜀 3 − 𝑎13 = 0.

Solving the above system for unknown parameters, we have the following solution set 1st Solution Set: ln(

𝑎0 = i, 𝑎1 =

2i(−𝜀𝛼1 +𝛼2 ) 𝛼1

2 e2𝛼 ln(𝑘)𝛼1 −2 ) 𝛼2 1

, 𝑘 = 𝑘, 𝜔 = e

𝛼

.


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.300) coupled with the solution of Eq. (4.678) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢3 (𝜉) = i + 1

2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1


𝑢4 (𝜉) = i + 1

2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

√3

, √3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

ln(

, and i = √−1.

2 e2𝛼 ln(𝑘)𝛼1 −2 ) 2 𝛼1


𝑡.

𝛼

2nd Solution Set: 𝑎0 = i, 𝑎1 =

2i(−𝜀𝛼1 +𝛼2 ) 𝛼1

, 𝑘 = 0, 𝜔 = e

2 ln( 2 ) 𝛼1 2𝛼

.


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.300) coupled with the solution of Eq. (4.678) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢7 (𝜉) = i +

2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

1

,

1


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

√3 √3 𝜉)+𝐶2 cos𝛼 ( 𝜉) 2 2

1 i 𝐶1 sin𝛼 (


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


3rd Solution Set:

685

2 ln( 2 ) 𝛼1 2𝛼

𝑡.

2i(−𝜀𝛼1 +𝛼2 )

𝑎0 = −i, 𝑎1 = −

𝛼1

, 𝑘 = 0, 𝜔 = e

2 ln( 2 ) 𝛼1 2𝛼

.


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.300) coupled with the solution of Eq. (4.678) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢10 (𝜉) = −i −

2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

1

,

1


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

√3 √3 𝜉)+𝐶2 cos𝛼 ( 𝜉) 2 2

1 i 𝐶1 sin𝛼 (


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


2 ln( 2 ) 𝛼1 2𝛼

𝑡.

4th Solution Set: 𝑎0 = −i, 𝑎1 = −

2i(−𝜀𝛼1 +𝛼2 ) 𝛼1

,𝑘 = e

2 ln( 2 ) 𝛼1 2𝛼

, 𝜔 = 0.


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i


686

Case II: Substituting the values of unknown into Eq. (4.300) coupled with the solution of Eq. (4.678) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢14 (𝜉) = −i −

2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

1

,

1


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

√3 √3 𝜉)+𝐶2 cos𝛼 ( 𝜉) 2 2

1 i 𝐶1 sin𝛼 (


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


2 ln( 2 ) 𝛼1 2𝛼

𝑥.

6th Solution Set: 𝑎0 = i, 𝑎1 =

2i(−𝜀𝛼1 +𝛼2 ) 𝛼1

,𝑘 = e

2 ln( 2 ) 𝛼1 2𝛼

, 𝜔 = 0.


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1


687

𝑢7 (𝜉) = i +

2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

1

,

1


2i(−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

√3 √3 𝜉)+𝐶2 cos𝛼 ( 𝜉) 2 2

1 i 𝐶1 sin𝛼 (


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


2 ln( 2 ) 𝛼1 2𝛼

𝑥.





4.9.10. gBBM Equation Consider the gBBM equation [114] of fraction order 𝐷𝑡𝛼 𝑢 + 𝐷𝑥𝛼 𝑢 + 𝑢2 𝐷𝑥𝛼 𝑢 + 𝐷𝑥3𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.301)


(4.302)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.302) into Eq. (4.301) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝑘 𝛼 𝐷𝜉𝛼 𝑢 + 𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 𝑘 3𝛼 𝐷𝜉3𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1. By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 3 = 2𝑀 + 𝑀 + 1, 𝑀 = 1. 688

(4.303)

Using the value of 𝑀 into Eq. (4.267), we obtained the trail solution 𝑈

𝑢 = 𝑎0 + 𝑎1 1+𝜀𝑈.

(4.304)


+ 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 )] = 0.


𝑘 𝛼 𝑎02 𝛼1 + 𝑘 𝛼 𝛼1 + 𝜔𝛼 𝛼1 + 𝑘 3𝛼 𝛼13 = 0,

𝑈1 :

2𝑎1 𝑘 𝛼 𝑎0 𝛼1 + 2𝑘 𝛼 𝑎02 𝜀𝛼1 + 𝜔𝛼 𝛼2 + ⋯ + 7𝑘 3𝛼 𝛼2 𝛼12 − 4𝑘 3𝛼 𝛼13 𝜀 = 0,

𝑈2:

2𝑎1 𝑘 𝛼 𝑎0 𝛼2 + 2𝑘 𝛼 𝑎02 𝜀𝛼2 + 𝑘 𝛼 𝑎02 𝜀 2 𝛼1 − ⋯ + 𝑎12 𝑘 𝛼 𝛼1 + 2𝑎1 𝑘 𝛼 𝑎0 𝜀𝛼1 = 0,

𝑈3:

𝑘 𝛼 𝑎02 𝜀 2 𝛼2 + 𝑘 3𝛼 𝛼2 𝜀 2 𝛼12 − 6𝑘 3𝛼 𝛼22 𝜀𝛼1 + ⋯ + 𝑎12 𝑘 𝛼 𝛼2 + 2𝑎1 𝑘 𝛼 𝑎0 𝜀𝛼2 = 0.

Solving the above system for unknown parameters, we have the following solution sets 1st Solution Set: 1

𝑎 𝛼

1 𝑎0 = 2 −𝜀𝛼1 +𝛼 , 𝑎1 = 𝑎1 , 𝑘 = ( 1

2

i𝑎1 ) √6(−𝜀𝛼1 +𝛼2 )

1 𝛼

,𝜔 = e

1 2 2 2 2 𝛼 i𝑎1 ln(12𝛼2 2 −24𝛼2 𝜀𝛼1 +𝑎1 𝛼1 −12𝜀 𝛼1 ) ( ) 12√6(−𝜀𝛼1 +𝛼2 )3

.


𝑎 𝛼

𝑈

1 𝑢1 (𝜉) = 2 −𝜀𝛼1 +𝛼 + 1+𝜀𝑈 𝑎1 , 1

1

2

i


𝑎 𝛼

𝑈

1 𝑢2 (𝜉) = 2 −𝜀𝛼1 +𝛼 + 1+𝜀𝑈 𝑎1 , 1

2

1


1

𝑎1 𝛼1

2 −𝜀𝛼1 +𝛼2 1

𝑈

+ 1+𝜀𝑈 𝑎1 , 1


1

𝑎 𝛼

𝑈

1 𝑢4 (𝜉) = 2 −𝜀𝛼1 +𝛼 + 1+𝜀𝑈 𝑎1 , 1

1

2

√3

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


i𝑎1 ) √6(−𝜀𝛼1 +𝛼2 )

1 𝛼

𝑥+e

, and i = √−1.

1 2 2 2 2 𝛼 i𝑎1 ln(12𝛼2 2 −24𝛼2 𝜀𝛼1 +𝑎1 𝛼1 −12𝜀 𝛼1 ) ( ) 12√6(−𝜀𝛼1 +𝛼2 )3

𝑡.

2nd Solution Set: 1

𝑎0 = √3i, 𝑎1 =

2(−𝜀𝛼1 +𝛼2 )√3i 𝛼1

,𝑘 =

√2 𝛼 (𝛼 ) , 𝜔 1

= 0.

Case I: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢5 (𝜉) = √3i + 1

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢6 (𝜉) = √3i +

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢7 (𝜉) = √3i +

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

1

,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢8 (𝜉) = √3i + 1

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

√3

, 1

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


3rd Solution Set:

690

√2 𝛼 (𝛼 ) 𝑥. 1

1

𝑎0 = √3i, 𝑎1 =

2(−𝜀𝛼1 +𝛼2 )√3i 𝛼1

√2 𝛼 (− 𝛼 ) , 𝜔 1

,𝑘 =

= 0.

Case I: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢9 (𝜉) = √3i + 1

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢10 (𝜉) = √3i +

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢11 (𝜉) = √3i +

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

1

,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢12 (𝜉) = √3i + 1

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

√3

, 1

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


√2 𝛼 (− 𝛼 ) 𝑥. 1

4th Solution Set: 1

𝑎0 = −√3i, 𝑎1 = −

2(−𝜀𝛼1 +𝛼2 )√3i 𝛼1

,𝑘 =

√2 𝛼 (𝛼 ) , 𝜔 1

= 0.

Case I: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢13 (𝜉) = −√3i − 1

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

,

i


691

Case II: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢14 (𝜉) = −√3i −

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢15 (𝜉) = −√3i −

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

1

,

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢16 (𝜉) = −√3i − 1

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

√3

, 1

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


√2 𝛼 (𝛼 ) 𝑥. 1

5th Solution Set: 1

𝑎0 = −√3i, 𝑎1 = −

2(−𝜀𝛼1 +𝛼2 )√3i 𝛼1

,𝑘 =

√2 𝛼 (− 𝛼 ) , 𝜔 1

= 0.

Case I: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −2i and 𝛼2 = 2i, we have 𝑢17 (𝜉) = −√3i − 1

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

,

i

where 𝑈 = 2 − 2 tan𝛼 (𝜉). Case II: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −i and 𝛼2 = i, we have 𝑢18 (𝜉) = −√3i −

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢19 (𝜉) = −√3i −

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

,

692

1

1

where 𝑈 = −1 − 2 tanh𝛼 (𝜉) − 2 coth𝛼 (𝜉). Case IV: Substituting the values of unknown into Eq. (4.304) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢20 (𝜉) = −√3i − 1

2(−𝜀𝛼1 +𝛼2 )√3i

𝑈

𝛼1

1+𝜀𝑈

√3

, 1

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (


√2 𝛼 (− 𝛼 ) 𝑥. 1




4.9.11. KP-BBM Equation Consider the KP-BBM equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝑢2 𝐷𝑥𝛼 𝑢 + 𝐷𝑥3𝛼 𝑢 + 𝐷𝑥3 𝐷𝑦2𝛼 𝑢 = 0,

0 < 𝛼 ≤ 1.

(4.305)


(4.306)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.306) into Eq. (4.305) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 𝑘 3𝛼 𝐷𝜉3𝛼 𝑢 + 𝑘 𝛼 𝑙 2𝛼 𝐷𝜉3𝛼 𝑢 = 0. By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 3 = 2𝑀 + 𝑀 + 1, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.267), we obtained the trail solution 693

(4.307)

𝑈

𝑢 = 𝑎0 + 𝑎1 1+𝜀𝑈.

(4.308)


1 [𝐶0 𝑈 0 3 (1+𝜀𝑈)3

+ 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 ] = 0.


3𝜔𝛼 𝑎0 + 𝑘 𝛼 𝑎03 = 0,

𝑈1 :

9𝜔𝛼 𝑎0 𝜀 + 3𝑘 𝛼 𝑎03 𝜀 + 3𝑘 𝛼 𝑎02 𝑎1 + ⋯ + 3𝜔𝛼 𝑎1 = 0,

𝑈2:

9𝜔𝛼 𝑎0 𝜀 2 + 6𝜔𝛼 𝑎1 𝜀 + 3𝑘 𝛼 𝑎03 𝜀 2 + ⋯ − 3𝑎1 𝑘 𝛼 𝑙 2𝛼 𝛼12 𝜀 + 9𝑎1 𝑘 𝛼 𝑙 2𝛼 𝛼2 𝛼1 = 0,

𝑈2:

3𝜔𝛼 𝑎0 𝜀 3 + 3𝜔𝛼 𝑎1 𝜀 2 + 𝑘 𝛼 𝑎03 𝜀 3 + ⋯ − 3𝑎1 𝑘 𝛼 𝑙 2𝛼 𝛼2 𝜀𝛼1 = 0.

Solving the above system for unknown parameters, we have the following solution set Solution Set:

𝑎0 = 𝑎0 , 𝑎1 =

2𝑎0 (−𝜀𝛼1 +𝛼2 ) 𝛼1

1

, 𝑘 = (3

2 3

√−𝑙2𝛼 𝛼12 − 𝑎02 𝛼1

1 𝛼

) ,𝜔 = e

2 2 √−𝑙2𝛼𝛼2 1 −3𝑎0 1 1 ln(− 𝑎0 )+ln 3 3 𝛼1 ( ) 𝛼

.


2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

i


2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1

where 𝑈 = 2 (1 + i sec𝛼 (𝜉) − i tan𝛼 (𝜉)). Case III: Substituting the values of unknown into Eq. (4.308) coupled with the solution of Eq. (4.268) when 𝛼1 = 4 and 𝛼2 = 2, we have 𝑢3 (𝜉) = 𝑎0 + 1

2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

,

1


694

Case IV: Substituting the values of unknown into Eq. (4.308) coupled with the solution of Eq. (4.268) when 𝛼1 = −√3i and 𝛼2 = √3i, we have 𝑢4 (𝜉) = 𝑎0 + 1

2𝑎0 (−𝜀𝛼1 +𝛼2 )

𝑈

𝛼1

1+𝜀𝑈

√3

,

√3

1 i 𝐶1 sin𝛼 ( 2 𝜉)+𝐶2 cos𝛼 ( 2 𝜉)


√3 √3 𝜉)+i𝐶2 sin𝛼 ( 𝜉) 2 2

𝐶1 cos𝛼 (

1


2 3

√−𝑙2𝛼 𝛼12 − 𝑎02 𝛼1

1 𝛼

) 𝑥+e

, and i = √−1.

2 2 √−𝑙2𝛼 𝛼2 1 −3𝑎0 1 1 ln(− 𝑎0 )+ln 3 3 𝛼1 ( ) 𝛼

𝑡.





4.10. (U`/U)-expansion Method As discussed in Chapter 3, (U`/U)-Expansion method has been developed by Usman et. al. in 2013 to find the solitary wave solution of nonlinear partial differential equations. In this method we use first order nonlinear ordinary differential equation 𝑈𝜉 = 𝐴 + 𝐵𝑈, as a Riccati equation. In this thesis I suggested a new approach for finding the solitary wave solutions of nonlinear partial differential equations of fractional-order. In the proposed algorithm we use the 𝑈𝜉 𝛼 = 𝐴 + 𝐵𝑈, where 0 < 𝛼 ≤ 1, as a Riccati equation. The proposed technique is highly compatible for solving nonlinear PDEs of fractional order. As we know that in (U`/U)-Expansion method the solution of 𝑈𝜉 𝛼 = 𝐴 + 𝐵𝑈, plays an important role, so before analysis of (U`/U)-Expansion method we first find the exact solution of 𝑈𝜉 𝛼 = 𝐴 + 𝐵𝑈, therefore we consider 695

𝐷𝜉𝛼 𝑈 = 𝐴𝑈 + 𝐵,

0 < 𝛼 ≤ 1,

(4.309)

subject to the initial condition 𝑈(0) = 𝑓. Appling both sides of 𝐽𝛼 both sides of Eq. (4.309) using the given initial condition we have 𝐵

𝑈(𝜉) = Γ(𝛼+1) 𝑥 𝛼 + 𝑈(0) + 𝐽𝛼 [𝐴𝑈]. Following the discussion presented in the decomposition method section, we can obtain the recurrence relation 𝐵

𝐵

𝑈0 (𝜉) = Γ(𝛼+1) 𝑥 𝛼 + 𝑈(0) = Γ(𝛼+1) 𝑥 𝛼 + 𝑓, 𝑈𝑘+1 (𝜉) = 𝐽𝛼 [𝐴𝑈𝑘 ]. Case 1: Consider that 𝑓 = 0, 𝐴 = 1 and 𝐵 = 1 we have the above recurrence relation 1

𝑈0 (𝜉) = 𝑈(0) = Γ(𝛼+1) 𝑥 𝛼 , 1

𝑈1 (𝜉) = 𝐽𝛼 [𝑈0 ] = Γ(2𝛼+1) 𝑥 2𝛼 , 1

𝑈2 (𝜉) = 𝐽𝛼 [𝑈1 ] = Γ(3𝛼+1) 𝑥 3𝛼 , 1

𝑈3 (𝜉) = 𝐽𝛼 [𝑈2 ] = Γ(4𝛼+1) 𝑥 4𝛼 , . . .. The series solution is given as 1

1

1

𝑈(𝜉) = Γ(𝛼+1) 𝑥 𝛼 + Γ(2𝛼+1) 𝑥 2𝛼 + Γ(3𝛼+1) 𝑥 3𝛼 + ⋯ . = −1 + 𝐸𝛼 (𝜉 𝛼 ). where 𝐸𝛼 denote the Mittage-Leffler function [11], given as 𝐸𝛼 (𝑧) = ∑∞ 𝑘=0

𝑧𝑘 Γ(𝛼𝑘+1)

.

Generally we have 𝐵

𝑈(𝜉) = − 𝐴 + 𝐸𝛼 [(𝐴𝜉)𝛼 ].

4.10.1. Methodology In order to simultaneously obtain more periodic wave solutions expressed in rational hyperbolic function and rational trigonometry function to nonlinear equations, we 696

introduce a (Uα/U)-expansion method. We briefly show what (Uα/U)-expansion method is and how to use it to obtain various periodic wave solutions to nonlinear equations. Suppose a nonlinear equation for 𝑈(𝑥, 𝑡) is given by We consider the general nonlinear PDE of the type 𝑃(𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑦 , 𝑢𝑧 , 𝑢𝑡𝑡 , 𝑢𝑥𝑥 , 𝑢𝑦𝑦 , 𝑢𝑥𝑥 , 𝑢𝑥𝑡 , 𝑢𝑦𝑡 , 𝑢𝑧𝑡 , 𝑢𝑥𝑦 , 𝑢𝑥𝑧 , 𝑢𝑦𝑧 , … ) = 0,

(4.310)

where 𝑃 is a polynomial in its arguments. The essence of the (Uα/U)-expansion method can be presented in the following steps: Step 1: Seek Solitary wave solutions of Eq. (4.310) by taking 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡, and transform Eq. (4.310) to the ordinary differential equation. 𝑄(𝑢, 𝜔𝑢′ , 𝑘𝑢′ , 𝑙𝑢′ , 𝑚𝑢′ , 𝜔2 𝑢′′ , 𝑘 2 𝑢′′ , 𝑙 2 𝑢′′ , … ) = 0,

(4.311)

where 𝜔 is constant and where prime denotes the derivative with respect to 𝜉. Step 2: If possible, integrate Eq. (4.311) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero. Step 3: According to (Uα/U)-expansion method, we assume that the wave solution can be expressed in the following form 𝑈𝛼 𝑛

𝑢(𝜉) = ∑𝑀 𝑛=0 𝑎𝑛 ( 𝑈 ) ,

(4.312)

where 𝑈 is the solution of first order nonlinear equation in the form 𝐷𝜉 𝛼 𝑈 = 𝐴𝑈 + 𝐵,

(4.313)

where 𝐴 and 𝐵 are real constants, 𝑀 is a positive integer to be determined and the Eq. (4.313) has solution 𝐵

𝑈(𝜉) = − 𝐴 + 𝐸𝛼 [(𝐴𝜉)𝛼 ]. Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term(s) of highest order with the highest order nonlinear term(s) in Eq. (4.310). Step 5: Substituting (4.312) into Eq. (4.311) with (4.313) will yields an algebraic equation involving power of U. Equating the coefficients of like power of U to zero gives a system of algebraic equations for 𝑎𝑖 , 𝑘, 𝑙, 𝑚 and 𝜔. Then, we solve the system with the aid of a computer algebra system (CAS), such as MAPLE 13, to determine these constants.

697

Step 6: Putting these constant into Eq. (4.312), coupled with the well known solutions of Eq. (4.313), we can obtained the more general type and new exact traveling wave solution of the nonlinear partial differential equation (4.310).

4.10.2. Caudrey Dodd Gibbon (CDG) Equation Consider the 5th order Caudrey Dodd Gibbon (CDG) equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 180𝑢2 𝐷𝑥𝛼 𝑢 + 30𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 30𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1. (4.314) To convert Eq. (4.314) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡,

(4.315)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.315) into Eq. (4.314) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝑘 𝛼 180𝑢2 𝐷𝜉𝛼 𝑢 + 30𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 30𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0.

(4.316)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 5 = 2𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.312), we obtained the trail solution 𝑈𝛼

𝑈𝛼 2

𝑎0 + 𝑎1 ( 𝑈 ) + 𝑎2 ( 𝑈 ) .

(4.317)


− 𝑈 7 [(𝐴𝑈 + 𝐵)𝐵(𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 )] = 0. Compare the like powers of U we have system of equations 𝑈0:

1080𝐵 5 𝑘 3𝛼 𝑎22 + 360𝑘 𝛼 𝑎23 𝐵 5 + 720𝑘 5𝛼 𝑎2 𝐵 5 = 0,

𝑈1 :

900𝑘 𝛼 𝑎1 𝐵 4 𝑎22 + 1800𝑘 𝛼 𝑎23 𝐴𝐵 4 + ⋯ + 1920𝑘 5𝛼 𝑎2 𝐴𝐵 4 + 120𝑘 5𝛼 𝑎1 𝐵 4 = 0,

𝑈2:

720𝑘 𝛼 𝑎0 𝑎22 𝐵 3 + 720𝑘 𝛼 𝑎12 𝐵 3 𝑎2 + ⋯ + 240𝐵 3 𝑘 3𝛼 𝑎12 + 3600𝑘 𝛼 𝑎1 𝐴𝑎22 𝐵 3 = 0,

𝑈3:

3600𝑘 𝛼 𝑎23 𝐴3 𝐵 2 + 450𝐵 2 𝑘 3𝛼 𝑎12 𝐴 + ⋯ + 5400𝑘 𝛼 𝑎1 𝐴2 𝑎22 𝐵 2 + 2160𝑘 𝛼 𝑎12 𝐴𝑎2 𝐵 2 = 0,

𝑈4:

2160𝑘 𝛼 𝑎0 𝑎2 𝐴𝐵𝑎1 + 360𝑘 𝛼 𝑎0 𝑎12 𝐵 + ⋯ + 3600𝑘 𝛼 𝑎1 𝐴3 𝑎22 𝐵 + 2𝜔𝛼 𝑎2 𝐵 = 0,

𝑈5:

360𝑘 𝛼 𝑎0 𝑎12 𝐴 + 720𝑘 𝛼 𝑎0 𝑎22 𝐴3 + 720𝑘 𝛼 𝑎12 𝐴3 𝑎2 + ⋯ + 2𝑘 5𝛼 𝑎2 𝐴5 + 2𝜔𝛼 𝑎2 𝐴 = 0.

Solving the above system for unknown parameters, we have 1st Solution Set: 698

𝑘 = 𝑘, 𝜔 = e

ln(−𝐴4 )+5𝛼 ln(𝑘) 𝛼

2

1

2

, 𝑎0 = − 6 [e𝛼 ln(𝑘) ] 𝐴2 , 𝑎1 = 2[e𝛼 ln(𝑘) ] 𝐴, 𝑎2 =

2

−2[e𝛼 ln(𝑘) ] . Substituting the values of unknown into Eq. (4.317) coupled with the solution of Eq. (4.313) 2

1

2 𝑈𝛼 2

𝑈𝛼

2

𝑢1 (𝜉) = − 6 [e𝛼 ln(𝑘) ] 𝐴2 − 2[e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − 2[e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈𝛼

where ( 𝑈 ) =

𝐴𝛼 𝐸𝛼 [(𝐴𝜉)𝛼 ]

. and 𝜉 = 𝑘𝑥 + e 𝛼

𝐵 𝐴

ln(−𝐴4 )+5𝛼 ln(𝑘) 𝛼

− +𝐸𝛼 [(𝐴𝜉) ]

𝑡.

2nd Solution Set: 5 3 ln(− [e𝛼 ln(𝑘) ] 𝐴4 −30[e𝛼 ln(𝑘) ] 𝐴2 𝑎0 −180e𝛼 ln(𝑘) 𝑎2 0)

𝑘 = 𝑘, 𝜔 = e

, 𝑎0 = 𝑎0 , 𝑎1 =

𝛼

2

[e𝛼 ln(𝑘) ] 𝐴, 2

𝑎2 = −[e𝛼 ln(𝑘) ] . Substituting the values of unknown into Eq. (4.317) coupled with the solution of Eq. (4.313) 2

2 𝑈𝛼 2

𝑈𝛼

𝑢2 (𝜉) = 𝑎0 + [e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − [e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈𝛼

where ( 𝑈 ) =

𝐴𝛼 𝐸𝛼 [(𝐴𝜉)𝛼 ] 𝐵 − +𝐸𝛼 [(𝐴𝜉)𝛼 ] 𝐴

5 3 ln(− [e𝛼 ln(𝑘) ] 𝐴4 −30[e𝛼 ln(𝑘) ] 𝐴2 𝑎0 −180e𝛼 ln(𝑘) 𝑎2 0)

. and 𝜉 = 𝑘𝑥 + e

𝑡.

𝛼

3rd Solution Set:

𝑘=(

√−15𝑎0 +3√5𝑎0 𝐴

1 𝛼

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

−15𝑎0 +3√5𝑎0 𝐴

, 𝑎2 = −

−15𝑎0 +3√5𝑎0 𝐴2

.

Substituting the values of unknown into Eq. (4.317) coupled with the solution of Eq. (4.313) 𝑢3 (𝜉) = 𝑎0 + 𝑈𝛼

where ( 𝑈 ) =

−15𝑎0 +3√5𝑎0 𝑈 𝛼


𝐴

(𝑈)−

−15𝑎0 +3√5𝑎0 𝑈 𝛼 2

√−15𝑎0 +3√5𝑎0

. and 𝜉 = (

(𝑈) ,

𝐴

𝐴

4th Solution Set:

699

1 𝛼

) 𝑥.

𝑘 = (−

√−15𝑎0 +3√5𝑎0 𝐴

1 𝛼

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

−15𝑎0 +3√5𝑎0 𝐴

, 𝑎2 = −

−15𝑎0 +3√5𝑎0 𝐴2

.


where ( 𝑈 ) =

−15𝑎0 +3√5𝑎0 𝑈 𝛼

(𝑈)−

𝐴


. and 𝜉 = (− 𝛼

𝐵 𝐴

−15𝑎0 +3√5𝑎0 𝑈 𝛼 2

(𝑈) ,

𝐴

√−15𝑎0 +3√5𝑎0 𝐴

− +𝐸𝛼 [(𝐴𝜉) ]

1 𝛼

) 𝑥.

5th Solution Set:

𝑘=(

√−15𝑎0 −3√5𝑎0 𝐴

1 𝛼

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

−15𝑎0 −3√5𝑎0 𝐴

, 𝑎2 = −

−15𝑎0 −3√5𝑎0 𝐴2

.


where ( 𝑈 ) =

−15𝑎0 −3√5𝑎0 𝑈 𝛼

(𝑈) −

𝐴


−15𝑎0 −3√5𝑎0 𝑈 𝛼 2

√−15𝑎0 −3√5𝑎0

. and 𝜉 = (

(𝑈) ,

𝐴

𝐴

1 𝛼

) 𝑥.

6th Solution Set:

𝑘 = (−

√−15𝑎0 −3√5𝑎0 𝐴

1 𝛼

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

−15𝑎0 −3√5𝑎0 𝐴

, 𝑎2 = −

−15𝑎0 −3√5𝑎0 𝐴2

.


where ( 𝑈 ) =

−15𝑎0 −3√5𝑎0 𝑈 𝛼 𝐴

𝐴𝛼 𝐸𝛼 [(𝐴𝜉)𝛼 ] 𝐵 𝐴

(𝑈) −

. and 𝜉 = (− 𝛼

− +𝐸𝛼 [(𝐴𝜉) ]

−15𝑎0 −3√5𝑎0 𝑈 𝛼 2 𝐴

√−15𝑎0 −3√5𝑎0 𝐴

700

(𝑈) , 1 𝛼

) 𝑥.

𝑡=0

𝑡 = 0.5

𝑡=1



4.10.3. Ito Equation Consider the 5th order Ito equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 2𝑢2 𝐷𝑥𝛼 𝑢 + 6𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 3𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.318)


(4.319)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.319) into Eq. (4.318) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 2𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 6𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 3𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0. (4.320) By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 5 = 2𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.312), we obtained the trail solution 701

𝑈𝛼 2

𝑈𝛼

𝑎0 + 𝑎1 ( 𝑈 ) + 𝑎2 ( 𝑈 ) .

(4.321)


− 𝑈 7 [(𝐴𝑈 + 𝐵)𝐵(𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 )] = 0. Compare the like powers of U we have system of equations 𝑈1 :

720𝑘 5𝛼 𝑎2 𝐵 5 + 4𝑘 𝛼 𝑎23 𝐵 5 + 144𝐵 5 𝑘 3𝛼 𝑎22 = 0,

𝑈2:

120𝑘 5𝛼 𝑎1 𝐵 4 + 1920𝑘 5𝛼 𝑎2 𝐴𝐵 4 + ⋯ + 510𝐵 4 𝑘 3𝛼 𝑎22 𝐴 = 0,

𝑈3:

240𝑘 5𝛼 𝑎1 𝐴𝐵 3 + 1800𝑘 5𝛼 𝑎2 𝐴2 𝐵 3 + ⋯ + 72𝑘 3𝛼 𝑎0 𝑎2 𝐵 3 = 0,

𝑈4:

690𝑘 5𝛼 𝑎2 𝐴3 𝐵 2 + 150𝑘 5𝛼 𝑎1 𝐴2 𝐵 2 + ⋯ + 18𝑘 3𝛼 𝑎0 𝑎1 𝐵 2 = 0,

𝑈5:

2𝜔𝛼 𝑎2 𝐵 + 30𝑘 5𝛼 𝑎1 𝐴3 𝐵 + ⋯ + 27𝐵𝑘 3𝛼 𝑎12 𝐴2 + 96𝐵𝑘 3𝛼 𝑎22 𝐴4 = 0,

𝑈6:

𝜔𝛼 𝑎1 + 2𝜔𝛼 𝑎2 𝐴 + 𝑘 5𝛼 𝑎1 𝐴4 + ⋯ + 6𝑘 3𝛼 𝑎0 𝑎2 𝐴3 + 9𝑘 3𝛼 𝑎1 𝐴4 𝑎2 = 0.

Solving the above system for unknown parameters, we have 1st Solution Set: 𝑘 = 𝑘, 𝜔 = e

ln(−6𝐴4 )+5𝛼 ln(𝑘) 𝛼

2

5

2

, 𝑎0 = − 2 [e𝛼 ln(𝑘) ] 𝐴2 , 𝑎1 = 30[e𝛼 ln(𝑘) ] 𝐴, 𝑎2 =

2

−30[e𝛼 ln(𝑘) ] . Substituting the values of unknown into Eq. (4.321) coupled with the solution of Eq. (4.313) 5

2

2

2 𝑈𝛼 2

𝑈𝛼

𝑢1 (𝜉) = − 6 [e𝛼 ln(𝑘) ] 𝐴2 + 30[e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − 30[e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈𝛼

where ( 𝑈 ) =



ln(−6𝐴4 )+5𝛼 ln(𝑘) 𝛼

𝑡.

2nd Solution Set: 1

2

2

2

𝑘 = 𝑘, 𝜔 = 0, 𝑎0 = − 2 [e𝛼 ln(𝑘) ] 𝐴2 , 𝑎1 = 6[e𝛼 ln(𝑘) ] 𝐴, 𝑎2 = −6[e𝛼 ln(𝑘) ] . Substituting the values of unknown into Eq. (4.321) coupled with the solution of Eq. (4.313) 1

2

2

𝑈𝛼

2 𝑈𝛼 2

𝑢2 (𝜉) = − 2 [e𝛼 ln(𝑘) ] 𝐴2 + 6[e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − 6[e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈𝛼

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

. and 𝜉 = 𝑘𝑥.

702

𝑡=0

𝑡 = 0.5

𝑡=1



4.10.4. Kaup Kuperschmidt Equation Consider the 7th order Kaup Kuperschmidt equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 2016𝑢3 𝐷𝑥𝛼 𝑢 + 630(𝐷𝑥𝛼 𝑢)3 + 2268𝑢𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 504𝑢2 𝐷𝑥3𝛼 𝑢 + 252𝐷𝑥2𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 147𝐷𝑥𝛼 𝑢𝐷𝑥4𝛼 𝑢 + 42𝑢𝐷𝑥5𝛼 𝑢 + 𝐷𝑥7𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1. (4.322) To convert Eq. (4.322) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡,

(4.323)


𝜔𝛼 𝐷𝜉𝛼 𝑢 + 2016𝑘 𝛼 𝑢3 𝐷𝜉𝛼 𝑢 + 630𝑘 3𝛼 (𝐷𝜉𝛼 𝑢) + 2268𝑘 3𝛼 𝑢𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 504𝑘 3𝛼 𝑢2 𝐷𝜉3𝛼 𝑢 + 252𝑘 5𝛼 𝐷𝜉2𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 147𝑘 5𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉4𝛼 𝑢 + 42𝑘 5𝛼 𝑢𝐷𝜉5𝛼 𝑢 + 𝑘 7𝛼 𝐷𝜉7𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1,

(4.324)

703

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 7 = 3𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.312), we obtained the trail solution 𝑈′

𝑈′

2

𝑢 = 𝑎0 + 𝑎1 ( 𝑈 ) + 𝑎2 ( 𝑈 ) .

(4.325)


− 𝑈 9 [(𝐴𝑈 + 𝐵)𝐵(𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 + 𝐶7 𝑈 7 )] = 0. Compare the like powers of U we have system of equations 𝑈0:

4032𝑘 𝛼 𝑎24 𝐵 7 + 44352𝐵 7 𝑘 3𝛼 𝑎23 + 101808𝐵 7 𝑘 5𝛼 𝑎22 + 40320𝑘 7𝛼 𝑎2 𝐵 7 = 0,

𝑈1 :

5040𝑘 7𝛼 𝑎1 𝐵 6 + 239904𝐵 6 𝑘 3𝛼 𝐴𝑎23 + ⋯ + 146160𝑘 7𝛼 𝑎2 𝐴𝐵 6 = 0,

𝑈2:

84672𝑘 𝛼 𝑎1 𝐵 5 𝑎23 𝐴 + 378000𝐵 5 𝑘 3𝛼 𝐴𝑎1 𝑎22 + ⋯ + 206640𝑘 7𝛼 𝑎2 𝐴2 𝐵 5 = 0,

𝑈3:

52920𝐵 4 𝑘 3𝛼 𝑎0 𝑎1 𝑎2 + 60480𝑘 𝛼 𝑎0 𝑎23 𝐵 4 𝐴 + ⋯ + 16800𝑘 7𝛼 𝑎1 𝐴2 𝐵 4 = 0,

𝑈4:

120960𝑘 𝛼 𝑎0 𝑎1 𝐴𝑎22 𝐵 3 + 139104𝐵 3 𝑘 3𝛼 𝑎0 𝑎1 𝐴𝑎2 + ⋯ + 8400𝑘 7𝛼 𝑎1 𝐴3 𝐵 3 = 0,

𝑈5:

7602𝑘 7𝛼 𝑎2 𝐴5 𝐵 2 + 145152𝐵 2 𝑘 3𝛼 𝐴5 𝑎23 + ⋯ + 6300𝑘 5𝛼 𝑎0 𝑎1 𝐴2 𝐵 2 = 0,

𝑈6:

4032𝑘 𝛼 𝑎03 𝑎2 𝐵 + 6048𝑘 𝛼 𝑎02 𝑎12 𝐵 + ⋯ + 1260𝑘 5𝛼 𝑎0 𝑎1 𝐴3 𝐵 = 0,

𝑈7:

4032𝑘 𝛼 𝑎03 𝑎2 𝐴 + 6048𝑘 𝛼 𝑎02 𝑎12 𝐴 + ⋯ + 3024𝑘 3𝛼 𝑎0 𝑎1 𝐴4 𝑎2 + 𝜔𝛼 𝑎1 = 0.

Solving the above system for unknown parameters, we have 1st Solution Set: ln(−

𝑘 = 𝑘, 𝜔 = e

1 6 𝐴 )+7𝛼 ln(𝑘) 48 𝛼

2

1

2

1

, 𝑎0 = − 24 [e𝛼 ln(𝑘) ] 𝐴2 , 𝑎1 = 2 [e𝛼 ln(𝑘) ] 𝐴,

2

1

𝑎2 = − 2 [e𝛼 ln(𝑘) ] . Substituting the values of unknown into Eq. (4.325) coupled with the solution of Eq. (4.313) 1

2

2

3

𝑈′

3

2 𝑈′ 2

𝑢1 (𝜉) = − 8 [e𝛼 ln(𝑘) ] 𝐴2 + 2 [e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − 2 [e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]


1 ln(− 𝐴4 )+5𝛼 ln(𝑘) 16 𝛼

704

𝑡.

𝑡=0

𝑡 = 0.5

𝑡=1



4.10.5. Lax Equation Consider the 7th order Lax equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 140𝑢3 𝐷𝑥𝛼 𝑢 + 70(𝐷𝑥𝛼 𝑢)3 + 280𝑢𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 70𝑢2 𝐷𝑥3𝛼 𝑢 + 70𝐷𝑥2𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 42𝐷𝑥𝛼 𝑢𝐷𝑥4𝛼 𝑢 + 14𝑢𝐷𝑥5𝛼 𝑢 + 𝐷𝑥7𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.326)


(4.327)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.327) into Eq. (4.326) and using the chain rule we obtained

705

3

𝜔𝛼 𝐷𝜉𝛼 𝑢 + 140𝑘 𝛼 𝑢3 𝐷𝜉𝛼 𝑢 + 70𝑘 3𝛼 (𝐷𝜉𝛼 𝑢) + 280𝑘 3𝛼 𝑢𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 70𝑘 3𝛼 𝑢2 𝐷𝜉3𝛼 𝑢 + 70𝑘 5𝛼 𝐷𝜉2𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 42𝑘 5𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉4𝛼 𝑢 + 14𝑘 5𝛼 𝑢𝐷𝜉5𝛼 𝑢 + 𝑘 7𝛼 𝐷𝜉7𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1,

(4.328)


𝑈′

𝑈

𝑈

2

𝑢 = 𝑎0 + 𝑎1 ( ) + 𝑎2 ( ) .

(4.329)


− 𝑈 9 (𝐴𝑈 + 𝐵)𝐵[𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 + 𝐶7 𝑈 7 ] = 0. Compare the like powers of U we have system of equations 𝑈0:

280𝑘 𝛼 𝑎24 𝐵 7 + 5600𝐵 7 𝑘 3𝛼 𝑎23 + 30240𝐵 7 𝑘 5𝛼 𝑎22 + 40320𝑘 7𝛼 𝑎2 𝐵 7 = 0,

𝑈1 :

980𝑘 𝛼 𝑎1 𝐵 6 𝑎23 + 1960𝑘 𝛼 𝑎24 𝐴𝐵 6 + ⋯ + 146160𝑘 7𝛼 𝑎2 𝐴𝐵 6 + 5040𝑘 7𝛼 𝑎1 𝐵 6 = 0,

𝑈2:

840𝑘 𝛼 𝑎0 𝑎23 𝐵 5 + 1260𝑘 𝛼 𝑎12 𝐵 5 𝑎22 + ⋯ + 86184𝐵 5 𝑘 5𝛼 𝑎1 𝐴𝑎2 = 0,

𝑈3:

700𝑘 𝛼 𝑎!3 𝐵 4 𝑎2 + 9800𝑘 𝛼 𝑎24 𝐴3 𝐵 4 + ⋯ + 6300𝑘 𝛼 𝑎12 𝐴𝐵 4 𝑎22 = 0,

𝑈4:

840𝑘 𝛼 𝑎02 𝑎22 𝐵 3 + 9800𝑘 𝛼 𝑎24 𝐴4 𝐵 3 + ⋯ 18480𝐵 3 𝑘 3𝛼 𝑎0 𝑎1 𝐴𝑎2 = 0,

𝑈5:

2520𝐵 2 𝑘 3𝛼 𝐴2 𝑎13 + 18900𝐵 2 𝑘 3𝛼 𝐴5 𝑎23 + ⋯ 5880𝑘 𝛼 𝑎24 𝐴5 𝐵 2 = 0,

𝑈6:

546𝐵𝑘 5𝛼 𝑎12 𝐴4 + 840𝐵𝑘 3𝛼 𝑎13 𝐴3 + ⋯ + 280𝑘 𝛼 𝑎03 𝑎2 𝐵 = 0,

𝑈7:

700𝑘 𝛼 𝑎13 𝐴4 𝑎2 + 980𝑘 𝛼 𝑎23 𝐴6 𝑎1 + ⋯ + 𝑘 7𝛼 𝑎1 𝐴6 + 2𝑘 7𝛼 𝑎2 𝐴7 + 𝜔𝛼 𝑎1 = 0.

Solving the above system for unknown parameters, we have 1st Solution Set: 20 2 71 14 ln(− 𝑎 𝑎 + i 𝑎0 √5)) 27𝐴 0 √Ω( 3 0 3

1

1 √Ω 𝛼

𝑘 = (3

𝐴

) ,𝜔 = e

2Ω

2 Ω

, 𝑎0 = 𝑎0 , 𝑎1 = 3 𝐴 , 𝑎2 = − 3 A2 .

𝛼

Substituting the values of unknown into Eq. (4.329) coupled with the solution of Eq. (4.313) 2 Ω 𝑈′

2 Ω

𝑈′

2

𝑢1 (𝜉) = 𝑎0 + 3 𝐴 ( 𝑈 ) − 3 𝐴2 ( 𝑈 ) , 706

1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

ln(−

1 √Ω 𝛼

, 𝜉 = (3

𝐴

20 2 71 14 𝑎 𝑎 + i 𝑎0 √5)) 27𝐴 0 √Ω( 3 0 3

) 𝑥+e

𝑡.

𝛼

and Ω = −15𝑎0 + 3i 𝑎0 √5. 2nd Solution Set: 1 √Ω

𝑘 = (− 3

𝐴

1 𝛼

ln(

20 2 71 14 𝑎 𝑎 + i 𝑎0 √5)) 27𝐴 0 √Ω( 3 0 3

) ,𝜔 = e

2Ω

2 Ω

, 𝑎0 = 𝑎0 , 𝑎1 = 3 𝐴 , 𝑎2 = − 3 A2 .

𝛼


2 Ω

2

𝑈′

𝑢2 (𝜉) = 𝑎0 + 3 𝐴 ( 𝑈 ) − 3 𝐴2 ( 𝑈 ) , 𝑈′

where ( 𝑈 ) =


1 √Ω

, 𝜉 = (− 3

𝐴

20 2 71 14 ln( 𝑎 𝑎 + i 𝑎0 √5)) 27𝐴 0 √Ω( 3 0 3

1 𝛼

) 𝑥+e

𝑡.

𝛼

and Ω = −15𝑎0 + 3i 𝑎0 √5. 3rd Solution Set: 20 2 71 14 ln(− 𝑎 𝑎 − i 𝑎0 √5)) 27𝐴 0 √Ω( 3 0 3

1

1 √Ω 𝛼

𝑘 = (3

) ,𝜔 = e 𝐴

2Ω

2 Ω

, 𝑎0 = 𝑎0 , 𝑎1 = 3 𝐴 , 𝑎2 = − 3 A2 .

𝛼


2 Ω

𝑈′

2

𝑢3 (𝜉) = 𝑎0 + 3 𝐴 ( 𝑈 ) − 3 𝐴2 ( 𝑈 ) , 1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

ln(−

1 √Ω 𝛼

, 𝜉 = (3

𝐴

20 2 71 14 𝑎 𝑎 − i 𝑎0 √5)) 27𝐴 0 √Ω( 3 0 3

) 𝑥+e

𝛼

𝑡.

and Ω = −15𝑎0 − 3i 𝑎0 √5. 4th Solution Set: 1

ln(

1 √Ω 𝛼

𝑘 = (− 3

𝐴

20 2 71 14 𝑎 𝑎 − i 𝑎0 √5)) 27𝐴 0 √Ω( 3 0 3

) ,𝜔 = e

2Ω

2 Ω

, 𝑎0 = 𝑎0 , 𝑎1 = 3 𝐴 , 𝑎2 = − 3 A2 .

𝛼


2 Ω

𝑈′

2

𝑢4 (𝜉) = 𝑎0 + 3 𝐴 ( 𝑈 ) − 3 𝐴2 ( 𝑈 ) ,

707

20 2 71 14 ln( 𝑎 𝑎 − i 𝑎0 √5)) 27𝐴 0 √Ω( 3 0 3

1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

1 √Ω 𝛼

, 𝜉 = (− 3

𝐴

) 𝑥+e

𝑡.

𝛼

and Ω = −15𝑎0 − 3i 𝑎0 √5. 5th Solution Set: 2 6 4 2 𝛼 ln(𝑘) ] 𝐴6 −14[e𝛼 ln(𝑘) ] 𝑎 𝐴4 −140𝑎3 )+𝛼 ln(𝑘) ln(−70[e𝛼 ln(𝑘) ] 𝑎2 0 0 𝐴 −[e 0

𝑘 = 𝑘, 𝜔 = e

, 𝑎0 = 𝑎0 ,

𝛼

2

2

𝑎1 = 2[e𝛼 ln(𝑘) ] 𝐴, 𝑎2 = −2[e𝛼 ln(𝑘) ] . Substituting the values of unknown into Eq. (4.329) coupled with the solution of Eq. (4.313) 2 𝑈′ 2

𝑈′

2

𝑢5 (𝜉) = 𝑎0 + 2[e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − 2[e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈′

(𝑈) =

where


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

, 𝜉 = 𝑘𝑥 +

2 6 4 2 𝛼 ln(𝑘) ] 𝐴6 −14[e𝛼 ln(𝑘) ] 𝑎 𝐴4 −140𝑎3 )+𝛼 ln(𝑘) ln(−70[e𝛼 ln(𝑘) ] 𝑎2 0 0 𝐴 −[e 0

e

𝑡.

𝛼

6th Solution Set: 1

1 √Ω 𝛼

2Ω

3 𝐴

9𝐴

𝑘=(

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

, 𝑎2 = −

2 Ω 9 A2

.


𝑈′

2 Ω

2

𝑢6 (𝜉) = 𝑎0 + 3 𝐴 ( 𝑈 ) − 9 𝐴2 ( 𝑈 ) , 1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

1 √Ω 𝛼

, 𝜉 = (3

𝐴

) 𝑥.

1

and Ω = 3(−224𝑎03 + 42√30𝑎03 )3 −

42𝑎02

1

(−224𝑎03 +42√30𝑎03 )3

− 42𝑎0 .

8th Solution Set: 1

1 √Ω 𝛼

𝑘 = (6

𝐴

1 Ω

1 Ω

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 18 𝐴 , 𝑎2 = − 18 A2 .


1 Ω

𝑈′

2

𝑢8 (𝜉) = 𝑎0 + 18 𝐴 ( 𝑈 ) − 18 𝐴2 ( 𝑈 ) , 708

1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

1 √Ω 𝛼

, 𝜉 = (6

𝐴

) 𝑥, Ω = 𝜏1 + i 𝜏2 , 1

𝜏1 =

and

−6(−224𝑎03

+

42√30𝑎03 )3

84𝑎02

+

1

(−224𝑎03 +42√30𝑎03 )3 1

1

𝜏2 = 18i√3 [3 (−224𝑎03 + 42√30𝑎03 )3 +

− 168𝑎0 ,

14𝑎02

1

].

3(−224𝑎03 +42√30𝑎03 )3

9th Solution Set: 1

1 √Ω 𝛼

𝑘 = (− 6

𝐴

1 Ω

1 Ω

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 18 𝐴 , 𝑎2 = − 18 A2 .


2

𝑈′

1 Ω

𝑢9 (𝜉) = 𝑎0 + 18 𝐴 ( 𝑈 ) − 18 𝐴2 ( 𝑈 ) , 1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

1 √Ω 𝛼

, 𝜉 = (− 6

𝐴

) 𝑥, Ω = 𝜏1 + i 𝜏2 , 1

𝜏1 =

and

𝜏2 =

−6(−224𝑎03

+

42√30𝑎03 )3

84𝑎02

+ 1

1

18i√3 [3 (−224𝑎03

42√30𝑎03 )3

+

1

(−224𝑎03 +42√30𝑎03 )3

− 168𝑎0 ,

14𝑎02

+

1

].

3(−224𝑎03 +42√30𝑎03 )3


1 √Ω 𝛼

𝑘 = (6

𝐴

1 Ω

1 Ω

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 18 𝐴 , 𝑎2 = − 18 A2 .


𝑈′

1 Ω

2

𝑢10 (𝜉) = 𝑎0 + 18 𝐴 ( 𝑈 ) − 18 𝐴2 ( 𝑈 ) , 1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

1 √Ω 𝛼

, 𝜉 = (6

𝐴

) 𝑥, Ω = 𝜏1 + i 𝜏2 , 1

𝜏1 =

and

𝜏2 =

−6(−224𝑎03 1

+

42√30𝑎03 )3

−18i√3 [3 (−224𝑎03

84𝑎02

+ 1

+

1

(−224𝑎03 +42√30𝑎03 )3

42√30𝑎03 )3 709

+

− 168𝑎0 ,

14𝑎02

1

3(−224𝑎03 +42√30𝑎03 )3

].


1 √Ω 𝛼

𝑘 = (− 6

𝐴

1 Ω

1 Ω

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 18 𝐴 , 𝑎2 = − 18 A2 .


𝑈′

1 Ω

2

𝑢11 (𝜉) = 𝑎0 + 18 𝐴 ( 𝑈 ) − 18 𝐴2 ( 𝑈 ) , 1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

1 √Ω 𝛼

, 𝜉 = (− 6

𝐴

) 𝑥, Ω = 𝜏1 + i 𝜏2 , 1

𝜏1 =

and

𝜏2 =

−6(−224𝑎03 1

+

42√30𝑎03 )3

84𝑎02

+ 1

−18i√3 [3 (−224𝑎03

+

1

(−224𝑎03 +42√30𝑎03 )3

42√30𝑎03 )3

− 168𝑎0 ,

14𝑎02

+

1

].

3(−224𝑎03 +42√30𝑎03 )3

Fig 4.79: Solitary wave solution for 𝑡 = 0.

4.10.6. 9th order Sawada Kotera (SK) Equation Consider the 9th order Sawada Kotera equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 45𝐷𝑥𝛼 𝑢𝐷𝑥6𝛼 𝑢 + 45𝑢𝐷𝑥7𝛼 𝑢 + 210𝐷𝑥4𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 210𝐷𝑥2𝛼 𝑢𝐷𝑥5𝛼 𝑢 + 1575𝐷𝑥𝛼 𝑢(𝐷𝑥2𝛼 𝑢)2 + 3150𝑢𝐷𝑥2𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 1260𝑢𝐷𝑥𝛼 𝑢𝐷𝑥4𝛼 𝑢 + 630𝑢2 𝐷𝑥5𝛼 𝑢 + 9450𝑢2 𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 3150𝑢3 𝐷𝑥3𝛼 𝑢 + 4725𝑢4 𝐷𝑥𝛼 𝑢 + 𝐷𝑥9𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1,

(4.330)


(4.331)

710

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.331) into Eq. (4.330) and using the chain rule we obtained 𝜔𝛼 𝐷𝑡𝛼 𝑢 + 45𝑘 7𝛼 𝐷𝑥𝛼 𝑢𝐷𝑥6𝛼 𝑢 + 45𝑘 7𝛼 𝑢𝐷𝑥7𝛼 𝑢 + 210𝑘 7𝛼 𝐷𝑥4𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 210𝑘 7𝛼 𝐷𝑥2𝛼 𝑢𝐷𝑥5𝛼 𝑢 + 1575𝑘 5𝛼 𝐷𝑥𝛼 𝑢(𝐷𝑥2𝛼 𝑢)2 + 3150𝑘 5𝛼 𝑢𝐷𝑥2𝛼 𝑢𝐷𝑥3𝛼 𝑢 + 1260𝑘 5𝛼 𝑢𝐷𝑥𝛼 𝑢𝐷𝑥4𝛼 𝑢 + 630𝑘 5𝛼 𝑢2 𝐷𝑥5𝛼 𝑢 + 9450𝑘 3𝛼 𝑢2 𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 3150𝑘 3𝛼 𝑢3 𝐷𝑥3𝛼 𝑢 + 4725𝑘 𝛼 𝑢4 𝐷𝑥𝛼 𝑢 + 𝑘 9𝛼 𝐷𝑥9𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1,

(4.332)


𝑈′

2

𝑢 = 𝑎0 + 𝑎1 ( 𝑈 ) + 𝑎2 ( 𝑈 ) .

(4.333)


− 𝑈 11 [(𝐴𝑈 + 𝐵)𝐵 (

𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 )] = +𝐶7 𝑈 7 + 𝐶8 𝑈 8 + 𝐶9 𝑈 9

0. Compare the like powers of U we have system of equations 𝑈0:

3780000𝐵 9 𝑘 7𝛼 𝑎22 + 1323000𝐵 9 𝑘 5𝛼 𝑎23 + ⋯ + 3628800𝑘 9𝛼 𝑎2 𝐵 9 = 0,

𝑈1 :

362880𝑘 9𝛼 𝑎1 𝐵 8 + 20068560𝐵 8 𝑘 7𝛼 𝑎22 𝐴 + ⋯ + 567000𝐵 8 𝑘 3𝛼 𝑎1 𝑎23 = 0,

𝑈2:

339840𝐵 7 𝑘 7𝛼 𝑎12 + 44830080𝐵 7 𝑘 7𝛼 𝑎22 𝐴2 + ⋯ + 4636800𝐵 7 𝑘 3𝛼 𝑎24 𝐴2 = 0,

𝑈3:

1302840𝐵 6 𝑘 7𝛼 𝑎12 𝐴 + 54558000𝐵 6 𝑘 7𝛼 𝑎22 𝐴3 + ⋯ + 264600𝐵 6 𝑘 3𝛼 𝑎13 𝑎2 = 0,

𝑈4:

37800𝐵 5 𝑘 3𝛼 𝑎14 + 1970640𝐵 5 𝑘 7𝛼 𝑎12 𝐴2 + ⋯ + 170100𝑘 𝛼 𝑎0 𝑎12 𝐵 5 𝑎22 = 0,

𝑈5:

1480500𝐵 4 𝑘 7𝛼 𝑎12 𝐴3 + 16582650𝐵 4 𝑘 7𝛼 𝑎22 𝐴5 + ⋯ + 141750𝐵 4 𝑘 3𝛼 𝑎14 𝐴 = 0,

𝑈6:

567840𝐵 3 𝑘 7𝛼 𝑎12 𝐴4 + 3957840𝐵 3 𝑘 7𝛼 𝑎22 𝐴6 + ⋯ + 1323000𝑘 𝛼 𝑎13 𝐴3 𝐵 3 𝑎22 = 0,

𝑈7:

101115𝐵 2 𝑘 7𝛼 𝑎12 𝐴5 + 464220𝐵 2 𝑘 7𝛼 𝑎22 𝐴7 + ⋯ + 56700𝑘 𝛼 𝑎0 𝑎14 𝐴𝐵 2 = 0,

𝑈8:

6180𝐵𝑘 7𝛼 𝑎12 𝐴6 + 19140𝐵𝑘 7𝛼 𝑎22 𝐴8 + ⋯ + 56700𝑘 𝛼 𝑎0 𝑎14 𝐴2 𝐵 = 0,

𝑈9:

90𝑘 7𝛼 𝑎22 𝐴9 + 4725𝑘 𝛼 𝑎15 𝐴4 + ⋯ + 170100𝑘 𝛼 𝑎0 𝑎12 𝐴5 𝑎22 = 0,

Solving the above system for unknown parameters, we have 1st Solution Set:

711

63 992 2 8520 ln(− ( 𝜓 + 𝜓𝑎0 +15915𝑎2 0 )) 4913𝐴 √𝜓𝑎0 83521 289

1

1 √𝜓 𝛼

𝑘 = (17

𝐴

) ,𝜔 = e

4 𝜓

, 𝑎0 = 𝑎0 , 𝑎1 = 289 𝐴 ,

𝛼

4 𝜓

𝑎2 = − 289 𝐴2 . Substituting the values of unknown into Eq. (4.333) coupled with the solution of Eq. (4.313) 4 𝜓 𝑈′

4 𝜓

𝑈′

2

𝑢1 (𝜉) = 𝑎0 + 289 𝐴 ( 𝑈 ) − 289 𝐴2 ( 𝑈 ) , 1

𝑈′


𝑈

− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

where ( ) =

𝐵 𝐴

,𝜉 = (

ln(−

1 √𝜓 𝛼

17 𝐴

) 𝑥+e

1

𝜓=

17(−3717𝑎03

+ 357i

𝑎03 √390)3

63 992 2 8520 ( 𝜓 + 𝜓𝑎0 +15915𝑎2 0 )) 4913𝐴 √𝜓𝑎0 83521 289

6783𝑎02

+

𝑡,

𝛼

1

(−3717𝑎03 +357i 𝑎03 √390)3

− 1071𝑎0 .

2nd Solution Set: 1 √𝜓

𝑘 = (− 17

𝐴

63 992 2 8520 ln( ( 𝜓 + 𝜓𝑎0 +15915𝑎2 0 )) 4913𝐴 √𝜓𝑎0 83521 289

1 𝛼

) ,𝜔 = e

4 𝜓

, 𝑎0 = 𝑎0 , 𝑎1 = 289 𝐴 ,

𝛼

4 𝜓


4 𝜓

𝑈′

2

𝑢2 (𝜉) = 𝑎0 + 289 𝐴 ( 𝑈 ) − 289 𝐴2 ( 𝑈 ) , 1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

ln(

1 √𝜓 𝛼

, 𝜉 = (− 17

𝐴

) 𝑥+e

1

𝜓 = 17(−3717𝑎03 + 357i 𝑎03 √390)3 +

63 992 2 8520 ( 𝜓 + 𝜓𝑎0 +15915𝑎2 0 )) 4913𝐴 √𝜓𝑎0 83521 289

6783𝑎02

𝛼

1

(−3717𝑎03 +357i 𝑎03 √390)3

𝑡,

− 1071𝑎0 .

3rd Solution Set: 1

1 √𝜓 𝛼

𝑘 = (34

𝐴

) ,𝜔 = e

63 62 2130 ln(− ( 𝜓2 + 𝜓𝑎0 +15915𝑎2 0 )) 9826𝐴 √𝜓𝑎0 83521 289 𝛼

4 𝜓

, 𝑎0 = 𝑎0 , 𝑎1 = 289 𝐴 ,

4 𝜓

𝑎2 = − 289 𝐴2 . Substituting the values of unknown into Eq. (4.333) coupled with the solution of Eq. (4.313)

712

4 𝜓 𝑈′

𝑈′

4 𝜓

2

𝑢3 (𝜉) = 𝑎0 + 289 𝐴 ( 𝑈 ) − 289 𝐴2 ( 𝑈 ) , 1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

ln(−

1 √𝜓 𝛼

, 𝜉 = (34

𝐴

63 62 2130 ( 𝜓2 + 𝜓𝑎0 +15915𝑎2 0 )) 9826𝐴 √𝜓𝑎0 83521 289

) 𝑥+e 1

13566𝑎02

𝜓 = −34(−3717𝑎03 + 357i 𝑎03 √390)3 +

1

(−3717𝑎03 +357i 𝑎03 √390)3 1

1

𝑡,

𝛼

578i √3 [17 (−3717𝑎03 + 357i 𝑎03 √390)3 −

− 4284𝑎0 +

399𝑎02

1

]

17(−3717𝑎03 +357i 𝑎03 √390)3

4th Solution Set: 63 62 2130 ln( ( 𝜓2 + 𝜓𝑎0 +15915𝑎2 0 )) 9826𝐴 √𝜓𝑎0 83521 289

1

1 √𝜓 𝛼

𝑘 = (− 34

𝐴

) ,𝜔 = e

4 𝜓

, 𝑎0 = 𝑎0 , 𝑎1 = 289 𝐴 ,

𝛼

4 𝜓


𝑈′

4 𝜓

2

𝑢4 (𝜉) = 𝑎0 + 289 𝐴 ( 𝑈 ) − 289 𝐴2 ( 𝑈 ) , 1

𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]

ln(

1 √𝜓 𝛼

, 𝜉 = (− 34

𝐴

63 62 2130 ( 𝜓2 + 𝜓𝑎0 +15915𝑎2 0 )) 9826𝐴 √𝜓𝑎0 83521 289

) 𝑥+e

1

𝜓 = −34(−3717𝑎03 + 357i 𝑎03 √390)3 +

13566𝑎02

1

578i √3 [17 (−3717𝑎03 + 357i 𝑎03 √390)3 −

𝑡=0

1

(−3717𝑎03 +357i 𝑎03 √390)3

1

𝑡 = 0.5

713

𝑡,

𝛼

− 4284𝑎0 +

399𝑎02

1

17(−3717𝑎03 +357i 𝑎03 √390)3

𝑡=1

]



4.10.7. 5th order Kaup Kuperschmidt Equation Consider the 5th order Kaup Kuperschmidt equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 20𝑢2 𝐷𝑥𝛼 𝑢 + 25𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 10𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0,0 < 𝛼 ≤ 1.

(4.333a)

We use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡, where 𝑘 and 𝜔 are arbitrary constant. Using the above transformation we get 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 20𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 25𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 10𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0. By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 5 = 2𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (10), we obtained the trail solution 𝑈′

𝑈′

2

𝑢 = 𝑎0 + 𝑎1 ( 𝑈 ) + 𝑎2 ( 𝑈 ) . Putting the trial solution into the reduced ODE coupled with auxiliary equation yields an algebraic equation involving power of U as 1

− 𝑈 7 [(𝐴𝑈 + 𝐵)𝐵(𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 )] = 0. Compare the like powers of U we have system of equations 𝑈0:

720𝑘 5𝛼 𝑎2 𝐵 5 + 540𝐵 5 𝑘 3𝛼 𝑎22 + 40𝑘 𝛼 𝑎23 𝐵 5 = 0,

𝑈1 :

120𝑘 5𝛼 𝑎1 𝐵 4 + 1920𝑘 5𝛼 𝑎2 𝐴𝐵 4 + ⋯ + 1900𝐵 4 𝑘 3𝛼 𝑎22 𝐴 = 0,

𝑈2:

240𝑘 5𝛼 𝑎1 𝐴𝐵 3 + 1800𝑘 5𝛼 𝑎2 𝐴2 𝐵 3 + ⋯ + 240𝑘 3𝛼 𝑎0 𝑎2 𝐵 3 = 0, 714

𝑈3:


𝑈4:

2𝜔𝛼 𝑎2 𝐵 + 30𝑘 5𝛼 𝑎1 𝐴3 𝐵 + ⋯ + 340𝐵𝑘 3𝛼 𝑎22 𝐴4 = 0,

𝑈5:

𝜔𝛼 𝑎1 + 2𝜔𝛼 𝑎2 𝐴 + 𝑘 5𝛼 𝑎1 𝐴4 + ⋯ + 30𝐵𝑘 3𝛼 𝑎1 𝐴4 𝑎2 = 0.

Solving the above system for unknown parameters, we have 1st Solution Set: ln(−

𝑘 = 𝑘, 𝜔 = e

1 4 𝐴 )+5𝛼 ln(𝑘) 16 𝛼

2

1

2

3

, 𝑎0 = − 8 [e𝛼 ln(𝑘) ] 𝐴2 , 𝑎1 = 2 [e𝛼 ln(𝑘) ] 𝐴,

2

3

𝑎2 = − 2 [e𝛼 ln(𝑘) ] . Substituting the values of unknown we get 2

1

2

3

𝑈′

3

2 𝑈′ 2

𝑢1 (𝜉) = − 8 [e𝛼 ln(𝑘) ] 𝐴2 + 2 [e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − 2 [e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]


1 ln(− 𝐴4 )+5𝛼 ln(𝑘) 16 𝛼

𝑡.

2nd Solution Set: 𝑘 = 𝑘, 𝜔 = e

ln(− 11𝐴4 )+5𝛼 ln(𝑘) 𝛼

2

2

, 𝑎0 = −[e𝛼 ln(𝑘) ] 𝐴2 , 𝑎1 = 12[e𝛼 ln(𝑘) ] 𝐴,

2

𝑎2 = −12[e𝛼 ln(𝑘) ] . Substituting the values of unknown we get 2

2

2 𝑈′ 2

𝑈′

𝑢2 (𝜉) = −[e𝛼 ln(𝑘) ] 𝐴2 + 12[e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − 12[e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈′

where ( 𝑈 ) =



ln(− 11𝐴4 )+5𝛼 ln(𝑘) 𝛼

− +𝐸𝛼 [(𝐴𝜉) ]

𝑡.

Fig 1: Solitary wave solution for 𝑡 = 0, 𝑡 = 0.5 and 𝑡 = 1.

715

𝑡=0

𝑡 = 0.5

𝑡=1 Fig 4.82: Solitary wave solution for 𝑡 = 0, 𝑡 = 0.5 and 𝑡 = 1.


716

4.10.8. 5th order Lax Equation Consider the 5th order Lax equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 30𝑢2 𝐷𝑥𝛼 𝑢 + 20𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 10𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.334)


(4.335)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.335) into Eq. (4.334) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 30𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 20𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 10𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0.

(4.336)


𝑈′

2

𝑢 = 𝑎0 + 𝑎1 ( 𝑈 ) + 𝑎2 ( 𝑈 ) .

(4.337)

Putting Eq. (4.337) into Eq. (4.336) coupled with Eq. (4.313); the Eq. (4.337) yields an algebraic equation involving power of U as −

1 𝑈7

(𝐴𝑈 + 𝐵)𝐵[𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 ] = 0.


720𝑘 5𝛼 𝑎2 𝐵 5 + 60𝑘 𝛼 𝑎23 𝐵 5 + 480𝐵 5 𝑘 3𝛼 𝑎22 = 0,

𝑈1 :

120𝑘 5𝛼 𝑎1 𝐵 4 + 1920𝑘 5𝛼 𝑎2 𝐴𝐵 4 + ⋯ + 1700𝐵 4 𝑘 3𝛼 𝑎22 𝐴 = 0,

𝑈2:


𝑈3:


𝑈4:

2𝜔𝛼 𝑎2 𝐵 + 30𝑘 5𝛼 𝑎1 𝐴3 𝐵 + ⋯ + 320𝐵𝑘 3𝛼 𝑎22 𝐴4 = 0,

𝑈5:

𝜔𝛼 𝑎1 + 2𝜔𝛼 𝑎2 𝐴 + 𝑘 5𝛼 𝑎1 𝐴4 + ⋯ + 30𝑘 3𝛼 𝑎1 𝐴4 𝑎2 = 0.


7 ln(− 𝐴4 )+5𝛼 ln(𝑘) 2 𝛼

1

2

2

, 𝑎0 = − 2 [e𝛼 ln(𝑘) ] 𝐴2 , 𝑎1 = 6[e𝛼 ln(𝑘) ] 𝐴,

2

𝑎2 = −6[e𝛼 ln(𝑘) ] . Substituting the values of unknown into Eq. (4.337) coupled with the solution of Eq. (4.313) 717

2

1

2 𝑈′ 2

𝑈′

2

𝑢1 (𝜉) = − 2 [e𝛼 ln(𝑘) ] 𝐴2 + 6[e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − 6[e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]


7 ln(− 𝐴4 )+5𝛼 ln(𝑘) 2 𝛼

𝑡.

2nd Solution Set: 2 3 1 ln(−[e𝛼 ln(𝑘) ] 𝐴4 −10[e𝛼 ln(𝑘) ] 𝑎0 𝐴2 −30[e𝛼 ln(𝑘) ] 𝑎2 0)

𝑘 = 𝑘, 𝜔 = e

, 𝑎0 = 𝑎0 , 𝑎1 =

𝛼

2

2[e𝛼 ln(𝑘) ] 𝐴, 2


2 𝑈′ 2

𝑈′


where ( 𝑈 ) =


2 3 1 ln(−[e𝛼 ln(𝑘) ] 𝐴4 −10[e𝛼 ln(𝑘) ] 𝑎0 𝐴2 −30[e𝛼 ln(𝑘) ] 𝑎2 0)


𝑡.

𝛼

3rd Solution Set:

𝑘=(

1 𝛼

√−5𝑎0 +i𝑎0 √5

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

𝐴

2(−5𝑎0 +i𝑎0 √5) 𝐴

, 𝑎2 = −

2(−5𝑎0 +i𝑎0 √5) 𝐴2

,

2

𝑎2 = −2[e𝛼 ln(𝑘) ] . Substituting the values of unknown into Eq. (4.337) coupled with the solution of Eq. (4.313) 𝑢3 (𝜉) = 𝑎0 + 𝑈′

where ( 𝑈 ) =

2(−5𝑎0 +i𝑎0 √5) 𝑈 ′


𝐴

(𝑈) −

2(−5𝑎0 +i𝑎0 √5) 𝑈 ′ 𝐴2

√−5𝑎0 +i𝑎0 √5

. and 𝜉 = (

𝐴

4th Solution Set:

718

2

(𝑈) ,

1 𝛼

) 𝑥.

𝑘 = (−

−

1 𝛼

√−5𝑎0 +i𝑎0 √5

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

𝐴

2(−5𝑎0 +i𝑎0 √5) 𝐴

, 𝑎2 =

2(−5𝑎0 +i𝑎0 √5) 𝐴2

, 2


where ( 𝑈 ) =

2(−5𝑎0 +i𝑎0 √5) 𝑈 ′

( ) −

𝐴

. and 𝜉 = (− 𝛼

𝑈

1 𝛼

√−5𝑎0 +i𝑎0 √5

) 𝑥.

𝐴

− +𝐸𝛼 [(𝐴𝜉) ]

2

( ) ,

𝐴2

𝑈


2(−5𝑎0 +i𝑎0 √5) 𝑈 ′

5th Solution Set:

𝑘=(

√−5𝑎0 −i𝑎0 √5

1 𝛼

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

𝐴

2(−5𝑎0 −i𝑎0 √5) 𝐴

, 𝑎2 = −

2(−5𝑎0 −i𝑎0 √5) 𝐴2

,

2


where ( 𝑈 ) =

2(−5𝑎0 −i𝑎0 √5) 𝑈 ′

(𝑈) −

𝐴

. and 𝜉 = ( 𝛼

− +𝐸𝛼 [(𝐴𝜉) ]

𝐴

2

(𝑈) ,

𝐴2

√−5𝑎0 −i𝑎0 √5


2(−5𝑎0 −i𝑎0 √5) 𝑈 ′ 1 𝛼

) 𝑥.

6th Solution Set:

𝑘 = (−

−

√−5𝑎0 −i𝑎0 √5

1 𝛼

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

𝐴

2(−5𝑎0 −i𝑎0 √5) 𝐴2

, 2

𝑎2 = −2[e𝛼 ln(𝑘) ] .

719

2(−5𝑎0 −i𝑎0 √5) 𝐴

, 𝑎2 =

Substituting the values of unknown into Eq. (4.337) coupled with the solution of Eq. (4.313) 𝑢6 (𝜉) = 𝑎0 + 𝑈′

where ( 𝑈 ) =

2(−5𝑎0 −i𝑎0 √5) 𝑈 ′

(𝑈) −

𝐴


. and 𝜉 = (−

2(−5𝑎0 −i𝑎0 √5) 𝑈 ′

√−5𝑎0 −i𝑎0 √5 𝐴

2

(𝑈) ,

𝐴2 1 𝛼

) 𝑥.


4.10.9. Sawada Kotera (SK) Equation Consider the Sawada Kotera equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 5𝑢2 𝐷𝑥𝛼 𝑢 + 5𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 5𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.338)


(4.339)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.339) into Eq. (4.338) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 5𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 5𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 5𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0. (4.340) By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 5 = 2𝑀 + 𝑀 + 1, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.312), we obtained the trail solution 𝑈′

𝑈′

2

𝑢 = 𝑎0 + 𝑎1 ( 𝑈 ) + 𝑎2 ( 𝑈 ) .

(4.341)

720


− 𝑈 7 [(𝐴𝑈 + 𝐵)𝐵(𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 )] = 0. Compare the like powers of U we have system of equations 𝑈0:

180𝐵 5 𝑘 3𝛼 𝑎22 + 10𝑘 𝛼 𝑎23 𝐵 5 + 720𝑘 5𝛼 𝑎2 𝐵 5 = 0,

𝑈1 :

120𝑘 5𝛼 𝑎1 𝐵 4 + 1920𝑘 5𝛼 𝑎2 𝐴𝐵 4 + ⋯ + 650𝐵 4 𝑘 3𝛼 𝑎22 𝐴 = 0,

𝑈2:


𝑈3:


𝑈4:

2𝜔𝛼 𝑎2 𝐵 + 30𝑘 5𝛼 𝑎1 𝐴3 𝐵 + ⋯ + 140𝐵𝑘 3𝛼 𝑎22 𝐴4 = 0,

𝑈5:

2𝜔𝛼 𝑎2 𝐴 + 𝑘 5𝛼 𝑎1 𝐴4 + ⋯ + 15𝑘 3𝛼 𝑎1 𝐴4 𝑎2 + 𝜔𝛼 𝑎1 = 0.

Solving the above system for unknown parameters, we have 1st Solution Set: 5 3 1 ln(−[e𝛼 ln(𝑘) ] 𝐴4 −5[e𝛼 ln(𝑘) ] 𝑎0 𝐴2 −5[e𝛼 ln(𝑘) ] 𝑎2 0)

𝑘 = 𝑘, 𝜔 = e

, 𝑎0 = 𝑎0 , 𝑎1 =

𝛼

2

2

6[e𝛼 ln(𝑘) ] 𝐴, 𝑎2 = −6[e𝛼 ln(𝑘) ] . Substituting the values of unknown into Eq. (4.341) coupled with the solution of Eq. (4.313) 2

2 𝑈′ 2

𝑈′


where ( 𝑈 ) =




𝑡.

𝛼

2nd Solution Set: 𝑘 = 𝑘, 𝜔 = e

ln(− 𝐴4 )+5𝛼 ln(𝑘) 𝛼

2

2

, 𝑎0 = −[e𝛼 ln(𝑘) ] 𝐴2 , 𝑎1 = 12[e𝛼 ln(𝑘) ] 𝐴,

2


2

2 𝑈′ 2

𝑈′

𝑢2 (𝜉) = −[e𝛼 ln(𝑘) ] 𝐴2 + 12[e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − 12[e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈′

where ( 𝑈 ) =



ln(− 𝐴4 )+5𝛼 ln(𝑘) 𝛼

− +𝐸𝛼 [(𝐴𝜉) ]


𝑡.

√ 1 −10𝑎0 +2√5𝑎0

𝑘 = (2

𝐴

3 −10𝑎0 +2√5𝑎0

−2

𝐴2

1 𝛼

3 −10𝑎0 +2√5𝑎0

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 2

𝐴

, 𝑎2 =

.

Substituting the values of unknown into Eq. (4.341) coupled with the solution of Eq. (4.313) 3 −10𝑎0 +2√5𝑎0 𝑈 ′

𝑢3 (𝜉) = 𝑎0 + 2 𝑈′

where ( 𝑈 ) =

𝐴


3 −10𝑎0 +2√5𝑎0 𝑈 ′

(𝑈) − 2

𝐴2

√ 1 −10𝑎0 +2√5𝑎0

. and 𝜉 = (2

𝐴

2

(𝑈) ,

1 𝛼

) 𝑥.

4th Solution Set: √ 1 −10𝑎0 +2√5𝑎0

𝑘 = (− 2

𝐴

3 −10𝑎0 +2√5𝑎0

−2

𝐴2

1 𝛼

3 −10𝑎0 +2√5𝑎0

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 2

𝐴

, 𝑎2 =

.

Substituting the values of unknown into Eq. (4.341) coupled with the solution of Eq. (4.313) 3 −10𝑎0 +2√5𝑎0 𝑈 ′

𝑢4 (𝜉) = 𝑎0 + 2 𝑈′

where ( 𝑈 ) =


3 −10𝑎0 +2√5𝑎0 𝑈 ′

(𝑈) − 2

𝐴

𝐴2

√ 1 −10𝑎0 +2√5𝑎0

. and 𝜉 = (− 2

𝐴

2

(𝑈) ,

1 𝛼

) 𝑥.

5th Solution Set: √ 1 −10𝑎0 −2√5𝑎0

𝑘 = (2

3 −10𝑎0 −2√5𝑎0

−2

𝐴2

𝐴

1 𝛼

3 −10𝑎0 −2√5𝑎0

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 2

𝐴

, 𝑎2 =

.

Substituting the values of unknown into Eq. (4.341) coupled with the solution of Eq. (4.313) 3 −10𝑎0 −2√5𝑎0 𝑈 ′

𝑢5 (𝜉) = 𝑎0 + 2

𝐴

3 −10𝑎0 −2√5𝑎0 𝑈 ′

(𝑈) − 2

722

𝐴2

2

(𝑈) ,

𝑈′

where ( 𝑈 ) =

√ 1 −10𝑎0 −2√5𝑎0


. and 𝜉 = (2 𝛼

𝐵 𝐴

𝐴

− +𝐸𝛼 [(𝐴𝜉) ]

1 𝛼

) 𝑥.

6th Solution Set: √ 1 −10𝑎0 −2√5𝑎0

𝑘 = (− 2

𝐴

3 −10𝑎0 −2√5𝑎0

−2

𝐴2

1 𝛼

3 −10𝑎0 −2√5𝑎0

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 2

𝐴

, 𝑎2 =

.

Substituting the values of unknown into Eq. (4.341) coupled with the solution of Eq. (4.313) 3 −10𝑎0 −2√5𝑎0 𝑈 ′

𝑢6 (𝜉) = 𝑎0 + 2 𝑈′

where ( 𝑈 ) =

𝐴


3 −10𝑎0 −2√5𝑎0 𝑈 ′

(𝑈) − 2

𝐴2

√ 1 −10𝑎0 −2√5𝑎0

. and 𝜉 = (− 2 𝛼

𝐴

− +𝐸𝛼 [(𝐴𝜉) ]

2

(𝑈) ,

1 𝛼

) 𝑥.

Fig 4.85: Travelling wave solution the given PDE for 𝑡 = 0.

4.10.10. 5th order Lax Equation Consider the 5th order Lax equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 30𝑢2 𝐷𝑥𝛼 𝑢 + 20𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 10𝑢𝐷𝑥3𝛼 𝑢 + 𝐷𝑥5𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.342)

To convert Eq. (4.342) into ODE we use following transformation 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝜔𝑡, where 𝑘 and 𝜔 are arbitrary constant. Substituting the above transformation into Eq. (4.331) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 30𝑘 𝛼 𝑢2 𝐷𝜉𝛼 𝑢 + 20𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 10𝑘 3𝛼 𝑢𝐷𝜉3𝛼 𝑢 + 𝑘 5𝛼 𝐷𝜉5𝛼 𝑢 = 0. 723

(4.343)


𝑈′

2

𝑢 = 𝑎0 + 𝑎1 ( 𝑈 ) + 𝑎2 ( 𝑈 ) .

(4.344)


− 𝑈 7 (𝐴𝑈 + 𝐵)𝐵[𝐶0 𝑈 0 + 𝐶1 𝑈1 + 𝐶2 𝑈 2 + 𝐶3 𝑈 3 + 𝐶4 𝑈 4 + 𝐶5 𝑈 5 + 𝐶6 𝑈 6 ] = 0. Compare the like powers of U we have system of equations 𝑈0:

720𝑘 5𝛼 𝑎2 𝐵 5 + 60𝑘 𝛼 𝑎23 𝐵 5 + 480𝐵 5 𝑘 3𝛼 𝑎22 = 0,

𝑈1 :

120𝑘 5𝛼 𝑎1 𝐵 4 + 1920𝑘 5𝛼 𝑎2 𝐴𝐵 4 + ⋯ + 170𝐵 4 𝑘 3𝛼 𝑎22 𝐴 = 0,

𝑈2:


𝑈3:


𝑈4:

2𝜔𝛼 𝑎2 𝐵 + 30𝑘 5𝛼 𝑎1 𝐴3 𝐵 + ⋯ + 320𝐵𝑘 3𝛼 𝑎22 𝐴4 = 0,

𝑈5:

𝜔𝛼 𝑎1 + 2𝜔𝛼 𝑎2 𝐴 + 𝑘 5𝛼 𝑎1 𝐴4 + ⋯ + 30𝑘 3𝛼 𝑎1 𝐴4 𝑎2 = 0.


7 ln(− 𝐴4 )+5𝛼 ln(𝑘) 2 𝛼

2

1

2

, 𝑎0 = − 2 [e𝛼 ln(𝑘) ] 𝐴2 , 𝑎1 = 6[e𝛼 ln(𝑘) ] 𝐴,

2


1

2

2 𝑈′ 2

𝑈′

𝑢1 (𝜉) = − 2 [e𝛼 ln(𝑘) ] 𝐴2 + 6[e𝛼 ln(𝑘) ] 𝐴 ( 𝑈 ) − 6[e𝛼 ln(𝑘) ] ( 𝑈 ) , 𝑈′

where ( 𝑈 ) =


− +𝐸𝛼 [(𝐴𝜉)𝛼 ]


7 ln(− 𝐴4 )+5𝛼 ln(𝑘) 2 𝛼

𝑡.

2nd Solution Set: 2 3 1 ln(−[e𝛼 ln(𝑘) ] 𝐴4 −10[e𝛼 ln(𝑘) ] 𝑎0 𝐴2 −30[e𝛼 ln(𝑘) ] 𝑎2 0)

𝑘 = 𝑘, 𝜔 = e

, 𝑎0 = 𝑎0 , 𝑎1 =

𝛼

2

2[e𝛼 ln(𝑘) ] 𝐴, 724

2


2 𝑈′ 2

𝑈′


where ( 𝑈 ) =




𝑡.

𝛼

3rd Solution Set:

𝑘=(

1 𝛼

√−5𝑎0 +i𝑎0 √5

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

𝐴

2(−5𝑎0 +i𝑎0 √5) 𝐴

, 𝑎2 = −

2(−5𝑎0 +i𝑎0 √5) 𝐴2

,

2


where ( 𝑈 ) =

2(−5𝑎0 +i𝑎0 √5) 𝑈 ′

(𝑈) −

𝐴


2(−5𝑎0 +i𝑎0 √5) 𝑈 ′ 𝐴2

√−5𝑎0 +i𝑎0 √5

. and 𝜉 = (

𝐴

2

(𝑈) ,

1 𝛼

) 𝑥.

4th Solution Set:

𝑘 = (−

−

1 𝛼

√−5𝑎0 +i𝑎0 √5

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

𝐴

2(−5𝑎0 +i𝑎0 √5) 𝐴

, 𝑎2 =

2(−5𝑎0 +i𝑎0 √5) 𝐴2

, 2

𝑎2 = −2[e𝛼 ln(𝑘) ] . Substituting the values of unknown into Eq. (4.344) coupled with the solution of Eq. (4.313) 𝑢4 (𝜉) = 𝑎0 +

2(−5𝑎0 +i𝑎0 √5) 𝑈 ′ 𝐴

(𝑈) −

2(−5𝑎0 +i𝑎0 √5) 𝑈 ′ 𝐴2

725

2

(𝑈) ,

𝑈′

where ( 𝑈 ) =


. and 𝜉 = (− 𝛼

𝐵 𝐴

1 𝛼

√−5𝑎0 +i𝑎0 √5

) 𝑥.

𝐴

− +𝐸𝛼 [(𝐴𝜉) ]

5th Solution Set:

𝑘=(

1 𝛼

√−5𝑎0 −i𝑎0 √5

2(−5𝑎0 −i𝑎0 √5)

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

𝐴

𝐴

, 𝑎2 = −

2(−5𝑎0 −i𝑎0 √5) 𝐴2

,

2


where ( 𝑈 ) =

2(−5𝑎0 −i𝑎0 √5) 𝑈 ′

(𝑈) −

𝐴

. and 𝜉 = ( 𝛼

1 𝛼

) 𝑥.

𝐴

− +𝐸𝛼 [(𝐴𝜉) ]

2

(𝑈) ,

𝐴2

√−5𝑎0 −i𝑎0 √5


2(−5𝑎0 −i𝑎0 √5) 𝑈 ′

6th Solution Set:

𝑘 = (−

−

1 𝛼

√−5𝑎0 −i𝑎0 √5

) , 𝜔 = 0, 𝑎0 = 𝑎0 , 𝑎1 =

𝐴

2(−5𝑎0 −i𝑎0 √5) 𝐴

, 𝑎2 =

2(−5𝑎0 −i𝑎0 √5) 𝐴2

, 2


where ( 𝑈 ) =

2(−5𝑎0 −i𝑎0 √5) 𝑈 ′


𝐴

(𝑈) −

. and 𝜉 = (−

2(−5𝑎0 −i𝑎0 √5) 𝑈 ′ 𝐴2

√−5𝑎0 −i𝑎0 √5 𝐴

726

2

(𝑈) ,

1 𝛼

) 𝑥.


4.11. (G𝛂/G)-expansion method As already discussed in Chapter 3 (G`/G)-Expansion method developed by Wang et. al. in 2008 to find the solitary wave solutions of nonlinear partial differential equations. In this method we used second order linear ordinary differential equation 𝐺 ′′ (𝜉) + 𝜆𝐺 ′ (𝜉) + 𝜇𝐺(𝜉) = 0, as a Riccati equation. In this thesis, I proposed a new algorithm for finding the solitary wave solutions of nonlinear partial differential equations of fractional-order. In the proposed algorithm we used the 𝐺 2𝛼 (𝜉) + 𝜆𝐺 𝛼 (𝜉) + 𝜇𝐺(𝜉) = 0, where 0 < 𝛼 ≤ 1, as a Riccati equation. The proposed technique is highly compatible for solving nonlinear PDEs of fractional order. As we know that in (G`/G)-Expansion method the solution of 𝐺 ′′ (𝜉) + 𝜆𝐺 ′ (𝜉) + 𝜇𝐺(𝜉) = 0, plays an important role, so before analysis of (Gα/G)-Expansion method we first find the exact solution of 𝐺 2𝛼 (𝜉) + 𝜆𝐺 𝛼 (𝜉) + 𝜇𝐺(𝜉) = 0, therefore consider the equation 𝐺 2𝛼 (𝜉) + 𝜆𝐺 𝛼 (𝜉) + 𝜇𝐺(𝜉) = 0,

0 < 𝛼 ≤ 1,

(4.345)

subject to the initial conditions 𝐺(0) = 𝑓, 𝐺 ′ (0) = 𝑔. Appling 𝐽𝛼 on both sides of Eq. (4.335) using the given initial condition we have 𝐺(𝜉) = 𝐺(0) + 𝜉𝐺 ′ (0) − 𝐽2𝛼 [𝜆𝐺 𝛼 + 𝜇𝐺], 𝐺(𝜉) = 𝑓 + 𝜉𝑔 − 𝐽2𝛼 [𝜆𝐺 𝛼 + 𝜇𝐺]. Following the discussion presented in the decomposition method section, we can obtain the recurrence relation 727

𝐺0 (𝜉) = 𝑓 + 𝜉𝑔, 𝐺𝑘+1 (𝜉) = −𝐽2𝛼 [𝜆𝐺𝑘𝛼 + 𝜇𝐺𝑘 ]. 1 √𝜆2 −4𝜇𝐴2

Case I: Consider that 𝑓 = 2

𝐴1

1

1 (𝜆2 −4𝜇)𝐴22

1

− 2 𝜆 and 𝑔 = 2 𝜆2 − 𝜇 − 4

𝐴21

, we have the

above recurrence relation 1 √𝜆2 −4𝜇𝐴2

𝐺0 (𝜉) = 𝐺(0) + 𝜉𝐺 ′ (0) = 2

𝐴1

1

1 (𝜆2 −4𝜇)𝐴22

1

− 2 𝜆 + (2 𝜆2 − 𝜇 − 4

𝐴21

) 𝜉,

𝐺1 (𝜉) = −𝐽2𝛼 [𝜆𝐺0𝛼 + 𝜇𝐺0 ], 𝐺2 (𝜉) = −𝐽2𝛼 [𝜆𝐺1𝛼 + 𝜇𝐺1 ], . . .. The closed form solution is given as 𝐺𝛼

(𝐺)=

1

1

2 2 √𝜆2 −4𝜇 𝐴1 sinh𝛼(2√𝜆 −4𝜇𝜉)+𝐴2 cosh𝛼 (2√𝜆 −4𝜇𝜉)

2

where sinh𝛼 (𝑥) =

[

1 1 𝐴1 cosh𝛼 ( √𝜆2 −4𝜇𝜉)+𝐴2 sinh𝛼 ( √𝜆2 −4𝜇𝜉) 2 2

𝐸𝛼 (𝑥 𝛼 )−𝐸𝛼 (−𝑥 𝛼 ) 2

, cosh𝛼 (𝑥) =

𝐸𝛼 (𝑥 𝛼 )+𝐸𝛼 (−𝑥 𝛼 ) 2

1

] − 2 𝜆, and 𝐸𝛼 denotes the

Mittage-Leffler function [11], given as 𝑧𝑘

𝐸𝛼 (𝑧) = ∑∞ 𝑘=0 Γ(𝛼𝑘+1) . 1 √−𝜆2 +4𝜇𝐴2

Case II: Consider that 𝑓 = 2

𝐴1

1

1 (−𝜆2 +4𝜇)𝐴22

1

− 2 𝜆 and 𝑔 = 4 𝜆2 − 𝜇 − 4

𝐴21

, we have

the solution of the form 𝐺𝛼

(𝐺)=

1

2

where sin𝛼 (𝑥) =

[

1 1 𝐴1 cos𝛼( √𝜆2 −4𝜇𝜉)+𝐴2 sin𝛼( √𝜆2 −4𝜇𝜉) 2 2

𝐸𝛼 (i𝑥 𝛼 )−𝐸𝛼 (−i𝑥 𝛼 ) 2i

Case III: Consider that 𝑓 = 𝐺𝛼

1

2 2 √𝜆2 −4𝜇 −𝐴1 sin𝛼 (2√𝜆 −4𝜇𝜉)+𝐴2 cos𝛼(2√𝜆 −4𝜇𝜉)

(𝐺)=𝐴

𝐴2

1 +𝐴2 𝜉

𝐴2 𝐴1

and cos𝛼 (𝑥) =

𝐸𝛼 (i𝑥 𝛼 )+𝐸𝛼 (−i𝑥 𝛼 )

1

𝐴22

2

𝐴21

− 𝜆 and 𝑔 = −

1

] − 2 𝜆,

2

.

, we have the solution of the form

1

− 2 𝜆.

4.11.1. Analysis of (G𝛂/G)-expansion method In order to obtain simultaneously more periodic wave solutions expressed in terms of rational hyperbolic function and rational trigonometric function to nonlinear equations, we introduce (G𝛂/G)-expansion method. We will briefly show what (G𝛂/G)-expansion 728

method is, and how we use it to obtain various periodic wave solutions to nonlinear equations. We consider the general nonlinear PDE of fractional order 𝑃(𝑢, 𝐷𝑡𝛼 𝑢, 𝐷𝑥𝛼 𝑢, 𝐷𝑦𝛼 𝑢, 𝐷𝑧𝛼 𝑢, 𝐷𝑡2𝛼 𝑢, 𝐷𝑥2𝛼 𝑢, 𝐷𝑦2𝛼 𝑢, 𝐷𝑧2𝛼 𝑢, … ) = 0, where 𝑃 is a polynomial in its arguments. The essence of the (G𝛂/G)-expansion method can be presented in the following steps: Step 1: Seek Solitary wave solutions of nonlinear PDE by taking 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝑚𝑧 + 𝜔𝑡, and transform Eq. (4.1) to into ordinary differential equation. 𝑃(𝑢, 𝜔𝛼 𝐷𝜉𝛼 𝑢, 𝑘 𝛼 𝐷𝜉𝛼 𝑢, 𝑙 𝛼 𝐷𝜉𝛼 𝑢, 𝑚𝛼 𝐷𝜉𝛼 𝑢, 𝜔2𝛼 𝐷𝜉2𝛼 𝑢, 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢, … ) = 0,

(4.346)

where 𝜔, 𝑘, 𝑙 and 𝑚 are constants and prime denotes the derivative with respect to 𝜉. Step 2: If possible, integrate Eq. (4.346) term by term one or more times. This yields constant(s) of integration. For simplicity, the integration constant(s) can be set to zero. Step 3: According to (G𝛂/G)-expansion method, we assume that the wave solution can be expressed in the following form 𝐺𝛼 𝑛

𝑢(𝜉) = ∑𝑀 𝑛=0 𝑎𝑛 ( 𝐺 ) ,

(4.347)

where 𝐺 is the solution of first order nonlinear equation in the form 𝐺 2𝛼 (𝜉) + 𝜆𝐺 𝛼 (𝜉) + 𝜇𝐺(𝜉) = 0,

0 < 𝛼 ≤ 1,

(4.348)

where 𝜆 and 𝜇 are real constants, the solution of Eq. (4.348) is Case I: When 𝜆2 − 4𝜇 > 0, then 𝐺𝛼

(𝐺)=

1

1


2

[


1

] − 2 𝜆.

Case II: When 𝜆2 − 4𝜇 < 0, then 𝐺𝛼

(𝐺)=

1

1


2

[

1 1 𝐴1 cos𝛼( √𝜆2 −4𝜇𝜉)+𝐴2 sin𝛼( √𝜆2 −4𝜇𝜉) 2 2

1

] − 2 𝜆.

Case III: When 𝜆2 − 4𝜇 = 0, then 𝐺𝛼

(𝐺)=𝐴

𝐴2

1 +𝐴2 𝜉

1

− 2 𝜆.

Step 4: Determine 𝑀. This, usually, can be accomplished by balancing the linear term(s) of highest order with the highest order nonlinear term(s) in Eq. (4.346).

729

Step 5: Substituting (4.347) into Eq. (4.346) with (4.348) will yields an algebraic 𝐺𝛼

𝐺𝛼

equation involving power of ( 𝐺 ).Equating the coefficients of like powers of ( 𝐺 ) to zero gives a system of algebraic equations for 𝑎𝑖 , 𝑘, 𝑙, 𝑚 and 𝜔. Then, we solve the system with the aid of a computer algebra system (CAS), such as MAPLE 13, to determine these constants. Step 6: Putting these constants into Eq. (4.347), coupled with the well known solutions of Eq. (4.348), we can obtained the more general type and new exact traveling wave solution of the nonlinear partial differential equation.

4.11.2. Good Boussinesq Equation Consider the good Boussinesq equation of [230] fraction order 𝐷𝑡2𝛼 𝑢 − 𝐷𝑥2𝛼 𝑢 + 𝐷𝑥4𝛼 𝑢 − 𝐷𝑥2𝛼 𝑢2 = 0,

0 < 𝛼 ≤ 1.

(4.349)


(4.350)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.350) into Eq. (4.349) and using the chain rule we obtained 𝜔2𝛼 𝐷𝜉2𝛼 𝑢 − 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 + 𝐷𝜉4𝛼 𝑢𝑘 4𝛼 − 2(𝐷𝜉𝛼 )2 𝑘 2𝛼 − 𝑘 2𝛼 2𝑢𝐷𝜉2𝛼 𝑢 = 0.

(4.351)

By applying the homogenous balancing principle for 𝛼 = 1, we have 2𝑀 + 2 = 𝑀 + 4, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.337), we obtained the trial solution 𝐺𝛼 2

𝐺𝛼

𝑈(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ) + 𝑎2 ( 𝐺 ) .

(4.352)

Putting Eq. (4.352) into Eq. (4.351) coupled with Eq. (4.348); the Eq. (4.352) yields an 𝐺𝛼

algebraic equation involving power of ( 𝐺 ) as 𝐺𝛼 0

𝐺𝛼 1

𝐺𝛼 2

𝐺𝛼 3

𝐺𝛼 6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + ⋯ + 𝐶6 ( 𝐺 ) = 0. 𝐺𝛼

Compare the like powers of ( 𝐺 ) we have system of equations 𝐺𝛼 0

( 𝐺 ) : 𝜔2𝛼 𝑎1 𝜆𝜇 − 𝑘 2𝛼 𝑎1 𝜆𝜇 + 8𝑘 4𝛼 𝑎1 𝜆𝜇 2 + 𝑘 4𝛼 𝑎1 𝜆3 𝜇 ⋯ − 2𝑘 2𝛼 𝑎0 𝑎1 𝜆𝜇 = 0, 𝐺𝛼 1

( 𝐺 ) : 2𝜔2𝛼 𝑎1 𝜇 + 𝜔2𝛼 𝑎1 𝜆2 − 2𝑘 2𝛼 𝑎1 𝜇 − 𝑘 2𝛼 𝑎1 𝜆2 ⋯ − 2𝑘 2𝛼 𝑎0 𝑎1 𝜆2 = 0, 730

𝐺𝛼 2

( 𝐺 ) : 3𝜔2𝛼 𝑎1 𝜆 + 8𝜔2𝛼 𝑎2 𝜇 + 4𝜔2𝛼 𝑎2 𝜆2 − 3𝑘 2𝛼 𝑎1 𝜆 ⋯ − 8𝑘 2𝛼 𝑎0 𝑎2 𝜆2 = 0, 𝐺𝛼 3

( 𝐺 ) : 10𝜔2𝛼 𝑎2 𝜆 − 10𝑘 2𝛼 𝑎2 𝜆 + 40𝑘 4𝛼 𝑎1 𝜇 + 50𝑘 4𝛼 𝑎1 𝜆2 ⋯ − 20𝑘 2𝛼 𝑎0 𝑎2 𝜆 = 0, 𝐺𝛼 4

( 𝐺 ) : 𝜔2𝛼 𝑎1 𝜆𝜇 − 𝑘 2𝛼 𝑎1 𝜆𝜇 + 8𝑘 4𝛼 𝑎1 𝜆𝜇 2 + 𝑘 4𝛼 𝑎1 𝜆3 𝜇 ⋯ − 2𝑘 2𝛼 𝑎0 𝑎1 𝜆𝜇 = 0, 𝐺𝛼 5

( 𝐺 ) : 336𝑘 4𝛼 𝑎2 𝜆 − 24𝑘 2𝛼 𝑎1 𝑎2 − 36𝑘 2𝛼 𝜆𝑎22 + 24𝑘 4𝛼 𝑎1 = 0, 𝐺𝛼 6

( 𝐺 ) : 120𝑘 4𝛼 𝑎2 − 20𝑘 2𝛼 𝑎22 = 0. Solving the above system for unknown parameters, we have 1st Solution Set 2 2 ln(−(e2𝛼ln(𝑘) ) 𝜆2 −8(e2𝛼ln(𝑘) ) +2e2𝛼ln(𝑘) 𝑎0 +e2𝛼ln(𝑘) )

𝑘 = 𝑘, 𝜔 = e

, 𝑎0 = 𝑎0 , 𝑎1 =

2𝛼

6e2𝛼ln(𝑘) 𝜆, 𝑎2 = e2𝛼ln(𝑘) Case I: Substituting the values of unknown into Eq. (4.352) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 𝐺𝛼 2

𝐺𝛼

𝑢(𝜉) = 𝑎0 + 6e2𝛼ln(𝑘) 𝜆 ( 𝐺 ) + e2𝛼ln(𝑘) ( 𝐺 ) , 𝐺𝛼

1

where ( 𝐺 ) =

1


2

[


1

] − 2 𝜆.

Case II: Substituting the values of unknown into Eq. (4.352) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 < 0, we have 𝐺𝛼 2

𝐺𝛼


where ( 𝐺 ) =

1

1


2

[

1 2

1 2

𝐴1 cos𝛼 ( √𝜆2 −4𝜇𝜉)+𝐴2 sin𝛼 ( √𝜆2 −4𝜇𝜉)

1

]− 2𝜆

Case III: Substituting the values of unknown into Eq. (4.352) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 = 0, we have 𝐺𝛼 2

𝐺𝛼


where ( 𝐺 ) = 𝐴

𝐴2

1 +𝐴2

1

− 2 𝜆. 𝜉 2 2 ln(−(e2𝛼ln(𝑘) ) 𝜆2 −8(e2𝛼ln(𝑘) ) +2e2𝛼ln(𝑘) 𝑎0 +e2𝛼ln(𝑘) )


2𝛼

2nd Solution Set 731

𝑡.

1

1

𝑘 = 𝑘, 𝜔 = 0, 𝑎0 = 2 e2𝛼ln(𝑘) 𝜆2 + 4e2𝛼ln(𝑘) 𝜇 − 2 , 𝑎1 = 6e2𝛼ln(𝑘) 𝜆, 𝑎2 = e2𝛼ln(𝑘) Case I: Substituting the values of unknown into Eq. (4.352) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 1

𝐺𝛼 2

𝐺𝛼

1

𝑢(𝜉) = 2 e2𝛼ln(𝑘) 𝜆2 + 4e2𝛼ln(𝑘) 𝜇 − 2 + 6e2𝛼ln(𝑘) 𝜆 ( 𝐺 ) + e2𝛼ln(𝑘) ( 𝐺 ) , 1

𝐺𝛼

where ( 𝐺 ) =

1


2

[


1

] − 2 𝜆.

Case II: Substituting the values of unknown into Eq. (4.342) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 < 0, we have 1

𝐺𝛼 2

𝐺𝛼

1

𝑢(𝜉) = 2 e2𝛼ln(𝑘) 𝜆2 + 4e2𝛼ln(𝑘) 𝜇 − 2 + 6e2𝛼ln(𝑘) 𝜆 ( 𝐺 ) + e2𝛼ln(𝑘) ( 𝐺 ) , 𝐺𝛼

1

where ( 𝐺 ) =

1


2

[

1 1 𝐴1 cos𝛼 ( √𝜆2 −4𝜇𝜉)+𝐴2 sin𝛼 ( √𝜆2 −4𝜇𝜉) 2 2

1

]− 2𝜆

Case III: Substituting the values of unknown into Eq. (4.352) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 = 0, we have 1

1

𝐺𝛼

𝐺𝛼 2

𝑢(𝜉) = 2 e2𝛼ln(𝑘) 𝜆2 + 4e2𝛼ln(𝑘) 𝜇 − 2 + 6e2𝛼ln(𝑘) 𝜆 ( 𝐺 ) + e2𝛼ln(𝑘) ( 𝐺 ) , 𝐺𝛼


𝐴2

1 +𝐴2

1

− 2 𝜆. 𝜉

In all cases 𝜉 = 𝑘𝑥.

Fig 4.87(a): 3D graph of Eq. (4.339) for different values of parameters.

Fig 4.87(b): 3D graph of Eq. (4.339) for different values of parameters

4.11.3. Boussinesq Equation 732

Consider the boussinesq equation [230] of fraction order 𝐷𝑡2𝛼 𝑢 − 𝐷𝑥2𝛼 𝑢 − 𝐷𝑥3𝛼 𝑢 − 𝐷𝑥2𝛼 𝑢2 = 0,

0 < 𝛼 ≤ 1.

(4.353)


(4.354)

Where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.354) into Eq. (4.353) and using the chain rule we obtained 𝜔2𝛼 𝐷𝜉2𝛼 𝑢 − 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 − 𝐷𝜉3𝛼 𝑢𝑘 3𝛼 − 2(𝐷𝜉𝛼 )2 𝑘 2𝛼 − 𝑘 2𝛼 2𝑢𝐷𝜉2𝛼 𝑢 = 0.

(4.355)

By applying the homogenous balancing principle for 𝛼 = 1, we have 2𝑀 + 2 = 𝑀 + 3 𝑀 = 1. Using the value of 𝑀 into Eq. (4.337), we obtained the trail solution 𝐺𝛼

𝑈(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ).

(4.356)



𝐺𝛼 1

𝐺𝛼 2

𝐺𝛼 3

𝐺𝛼 4

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) = 0. 𝐺𝛼


( 𝐺 ) : −2𝑎1 𝑘 2𝛼 𝜇 2 + 𝜔2𝛼 𝜆𝜇 − 𝑘 2𝛼 𝜆𝜇 + 𝑘 3𝛼 𝜆2 𝜇 ⋯ − 2𝑘 3𝛼 𝜇 2 = 0, 𝐺𝛼 1

( 𝐺 ) : 2𝜔2𝛼 𝜇 + 𝜔2𝛼 𝜆2 − 2𝑘 2𝛼 𝜇 − 𝑘 2𝛼 𝜆2 + 𝑘 3𝛼 𝜆3 ⋯ − 2𝑘 2𝛼 𝑎0 𝜆2 = 0, 𝐺𝛼 2

( 𝐺 ) : 3𝜔2𝛼 𝜆 − 3𝑘 2𝛼 𝜆 + 8𝑘 3𝛼 𝜇 + 7𝑘 3𝛼 𝜆2 ⋯ − 4𝑎1 𝑘 2𝛼 𝜆2 = 0, 𝐺𝛼 3

( 𝐺 ) : 12𝑘 3𝛼 𝜆 − 4𝑘 2𝛼 𝑎0 − 10𝑎1 𝑘 2𝛼 𝜆 + 2𝜔2𝛼 − 2𝑘 2𝛼 = 0, 𝐺𝛼 4

( 𝐺 ) : − 6𝑎1 𝑘 2𝛼 + 6𝑘 3𝛼 = 0. Solving the above system for unknown parameters, we have 1st Solution Set 3 2 2 ln(−(e𝛼ln(𝑘) ) 𝜆+2(e𝛼ln(𝑘) ) 𝑎0 +(e𝛼ln(𝑘) ) )

𝑘 = 𝑘, 𝜔 = e

, 𝑎0 = 𝑎0 . 𝑎1 = e𝛼ln(𝑘) .

2𝛼

Case I: Substituting the values of unknown into Eq. (4.356) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 733

𝐺𝛼

𝑢(𝜉) = 𝑎0 − e𝛼ln(𝑘) ( 𝐺 ), 𝐺𝛼

1

where ( 𝐺 ) =

1


[


2

1

] − 2 𝜆.

Case II: Substituting the values of unknown into Eq. (4.356) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 < 0, we have 𝐺𝛼

𝑢(𝜉) = 𝑎0 − e𝛼ln(𝑘) ( 𝐺 ), 𝐺𝛼

where ( 𝐺 ) =

1

1


2

[

1 2

1 2


1

]− 2𝜆

Case III: Substituting the values of unknown into Eq. (4.356) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 = 0, we have 𝐺𝛼

𝑢(𝜉) = 𝑎0 − e𝛼ln(𝑘) ( 𝐺 ), 𝐺𝛼


𝐴2

1 +𝐴2 𝜉

1

− 2 𝜆. 3 2 2 ln(−(e𝛼ln(𝑘) ) 𝜆+2(e𝛼ln(𝑘) ) 𝑎0 +(e𝛼ln(𝑘) ) )

In all cases 𝜉 = 𝑘𝑥 + 𝑒

𝑡

2𝛼

Graphical Representation

Fig 4.88:3D graph of Eq. (4.343) for different values parameters.

4.11.4. Breaking Soliton Equation Consider the breaking soliton [230] equation of fraction order 2𝛼 2𝛼 4𝛼 𝐷𝑥𝑡 𝑢 − 4𝐷𝑥𝛼 𝑢𝐷𝑥𝑦 𝑢 − 2𝐷𝑥2𝛼 𝑢𝐷𝑦𝛼 𝑢 + 𝐷𝑥𝑥𝑥𝑦 𝑢 = 0,

734

0 < 𝛼 ≤ 1.

(4.357)


(4.358)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.358) into Eq. (4.357) and using the chain rule we obtained 𝑘 𝛼 𝜔𝛼 𝐷𝜉2𝛼 𝑢 − 6𝑘 2𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢𝑙 𝛼 + 𝐷𝜉4𝛼 𝑢𝑘 3𝛼 𝑙 𝛼 = 0.

(4.359)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 1 + 𝑀 + 2 = 𝑀 + 4, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.337), we obtained the trial solution 𝐺𝛼

𝑈(𝜉) = 𝑎0 + 𝑎1 ( ).

(4.360)

𝐺



𝐺𝛼 1

𝐺𝛼 2

𝐺𝛼 3

𝐺𝛼 5

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + ⋯ + 𝐶5 ( 𝐺 ) = 0. 𝐺𝛼


𝛼

( 𝐺 ) : 𝑘 𝛼 𝜔𝛼 𝜆𝜇 + 8𝑘 3𝛼 𝑙 𝛼 𝜆𝜇 2 + ⋯ + 6𝑎1 𝑘 2 𝑙 𝛼 𝜆𝜇 2 = 0, 𝐺𝛼 1

𝛼

( 𝐺 ) : 22𝑘 3𝛼 𝑙 𝛼 𝜆2 𝜇 + 2𝑘 𝛼 𝜔𝛼 𝜇 + 𝑘 𝛼 𝜔𝛼 𝜆2 ⋯ + 12𝑎1 𝑘 2 𝑙 𝛼 𝜆2 𝜇 = 0, 𝐺𝛼 2

𝛼

( 𝐺 ) : 60𝑘 3𝛼 𝑙 𝛼 𝜆𝜇 + 3𝑘 𝛼 𝜔𝛼 𝜆 + ⋯ + 36𝑎1 𝑘 2 𝑙 𝛼 𝜆𝜇 = 0, 𝐺𝛼 3

𝛼

( 𝐺 ) : 2𝑘 𝛼 𝜔𝛼 + 40𝑘 3𝛼 𝑙 𝛼 𝜇 + 50𝑘 3𝛼 𝑙 𝛼 𝜆2 ⋯ + 24𝑎1 𝑘 2 𝑙 𝛼 𝜆2 = 0, 𝐺𝛼 4

𝛼

( 𝐺 ) : 60𝑘 3𝛼 𝑙 𝛼 𝜆 + 30𝑎1 𝑘 2 𝑙 𝛼 𝜆 = 0, 𝐺𝛼 5

𝛼

( 𝐺 ) : 24𝑘 3𝛼 𝑙 𝛼 + 12𝑎1 𝑘 2 𝑙 𝛼 = 0. Solving the above system for unknown parameters, we have 1st Solution Set 𝑘 = 𝑘, 𝜔 = 𝑒

ln(4𝜇−𝜆2 )+𝛼 ln(𝑘2 𝑙) 𝛼

𝛼

, 𝑎0 = 𝑎0 , 𝑎1 = −2𝑘 3𝛼−2 .

Case I: Substituting the values of unknown into Eq. (4.360) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have

735

𝛼

𝐺𝛼

𝑢(𝜉) = 𝑎0 − 2𝑘 3𝛼−2 ( 𝐺 ), 1

𝐺𝛼

where ( 𝐺 ) =

1


[


2

1

] − 2 𝜆.

Case II: Substituting the values of unknown into Eq. (4.360) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 < 0, we have 𝛼

𝐺𝛼

𝑢(𝜉) = 𝑎0 − 2𝑘 3𝛼−2 ( 𝐺 ), 𝐺𝛼

1

where ( 𝐺 ) =

1


2

[

1 2

1 2


1

]− 2𝜆

Case III: Substituting the values of unknown into Eq. (4.360) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 = 0, we have 𝛼

𝐺𝛼

𝑢(𝜉) = 𝑎0 − 2𝑘 3𝛼−2 ( 𝐺 ), 𝐺𝛼


𝐴2

1 +𝐴2 𝜉

1

− 2 𝜆.


ln(4𝜇−𝜆2 )+𝛼 ln(𝑘2 𝑙) 𝛼

𝑡.

Fig 4.89(a): 3D Soliton solution of Eq. (4.357) for different values parameters.

Fig 4.89(b): 3D graph of Eq. (4.357) for different values parameters

4.11.5. Generalized Shallow Water Wave Equation Consider the generalized shallow water wave equation [230] of fraction order 2𝛼 2𝛼 4𝛼 𝐷𝑥𝑡 𝑢 − 𝛽𝐷𝑡𝛼 𝑢𝐷𝑥2𝛼 𝑢 − 𝛿𝐷𝑥𝛼 𝑢𝐷𝑥𝑡 𝑢 − 𝐷𝑥𝑥𝑥𝑡 𝑢 + 𝐷𝑥2𝛼 𝑢 = 0, 0 < 𝛼 ≤ 1.

(4.361)


(4.362) 736

Where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.362) into Eq. (4.361) and using the chain rule we obtained 𝑘 𝛼 𝜔𝛼 𝐷𝜉2𝛼 𝑢 − 𝛽𝑘 2𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢𝜔𝛼 − 𝛿𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢𝑘 2𝛼 𝜔𝛼 − 𝐷𝜉4𝛼 𝑢𝑘 3𝛼 𝜔𝛼 + 𝐷𝜉2𝛼 𝑢𝑘 2𝛼 = 0.

(4.363)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀+1+𝑀+2= 𝑀+4 𝑀 = 1. Using the value of 𝑀 into Eq. (4.347), we obtained the trail solution 𝐺𝛼

𝑈(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ).

(4.364)



𝐺𝛼 1

𝐺𝛼 2

𝐺𝛼 3

𝐺𝛼 5

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶5 ( 𝐺 ) = 0. 𝐺𝛼


( 𝐺 ) : 𝑘 2𝛼 𝜆𝜇 + 𝛿𝑎1 𝑘 2𝛼 𝜔𝛼 𝜆𝜇 2 + 𝑘 𝛼 𝜔𝛼 𝜆𝜇 ⋯ + 𝛽𝑎1 𝜔𝛼 𝑘 2𝛼 𝜆𝜇 2 = 0, 𝐺𝛼 1

( 𝐺 ) : − 22𝑘 3𝛼 𝜔𝛼 𝜆2 𝜇 + 2𝛽𝑎1 𝑘 2𝛼 𝜔𝛼 𝜆2 𝜇 ⋯ + 2𝛽𝑎1 𝜔𝛼 𝑘 2𝛼 𝜇 2 = 0, 𝐺𝛼 2

( 𝐺 ) : − 60𝑘 3𝛼 𝜔𝛼 𝜆𝜇 + 6𝛽𝑎1 𝑘 2𝛼 𝜔𝛼 𝜆𝜇 ⋯ + 𝛿𝑎1 𝜔𝛼 𝑘 2𝛼 𝜆3 = 0, 𝐺𝛼 3

( 𝐺 ) : 2𝑘 2𝛼 + 2𝑘 𝛼 𝜔𝛼 − 40𝑘 3𝛼 𝜔𝛼 𝜇 ⋯ + 4𝛿𝑎1 𝜔𝛼 𝑘 2𝛼 𝜆2 = 0, 𝐺𝛼 4

( 𝐺 ) : −60𝑘 3𝛼 𝜔𝛼 𝜆 + 5𝛽𝑎1 𝜔𝛼 𝑘 2𝛼 𝜆 + 5𝛿𝑎1 𝑘 2𝛼 𝜔𝛼 𝜆 = 0, Gα 5

( G ) : 2𝛽𝑎1 𝜔𝛼 𝑘 2𝛼 + 2𝛿𝑎1 𝑘 2𝛼 𝜔𝛼 − 24𝑘 3𝛼 𝜔𝛼 = 0. Solving the above system for unknown parameters, we have 1st Solution Set 𝑘=𝑒

1 1 𝑙𝑛( 𝛽𝑎1 + 𝛿𝑎1 ) 12 12 𝛼

ln(−

,𝜔 = e

12a1 (β+δ) ) 2 2 2 2 2 2 2 2 2 2 2 2 8𝛿𝑎2 1 𝜇𝛽+4𝛿 𝑎1 𝜇+11+4𝜇𝛽 𝑎1 −𝜆 𝛽 𝑎1 −2𝜆 𝛽𝑎1 𝛿−𝜆 𝛿 𝑎1 𝛼

, 𝑎0 =

𝑎0 , 𝑎1 = 𝑎1 . Case I: Substituting the values of unknown into Eq. (4.354) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 737

𝐺𝛼

𝑢(𝜉) = 𝑎0 − 𝑎1 ( 𝐺 ), 𝐺𝛼

1

where ( 𝐺 ) =

1


[


2

1

] − 2 𝜆.


𝑢(𝜉) = 𝑎0 − 𝑎1 ( 𝐺 ), 𝐺𝛼

where ( 𝐺 ) =

1

1


2

[

1 2

1 2


1

]− 2𝜆


𝑢(𝜉) = 𝑎0 − 𝑎1 ( 𝐺 ), 𝐺𝛼


𝐴2

1

1 +𝐴2 𝜉 𝑙𝑛(

In all cases 𝜉 = 𝑒

− 2 𝜆.

1 1 𝛽𝑎 + 𝛿𝑎 ) 12 1 12 1 𝛼

ln(−

𝑥+e

12a1 (β+δ) ) 2 2 2 2 2 2 2 2 2 2 2 2 8𝛿𝑎2 1 𝜇𝛽+4𝛿 𝑎1 𝜇+11+4𝜇𝛽 𝑎1 −𝜆 𝛽 𝑎1 −2𝜆 𝛽𝑎1 𝛿−𝜆 𝛿 𝑎1 𝛼

𝑡

Graphical Representation:


4.11.6. Potential Kadomtsev-Petviashvili Equation Consider the potential KP equation [230] of fraction order 3𝛼 2𝛼 4𝛼 4𝐷𝑥𝑡 𝑢 + 6𝐷𝑥𝛼 𝑢𝐷𝑥2𝛼 𝑢 + 4𝐷4𝑥 𝑢 + 3𝐷3𝑦 𝑢 = 0,

0 < 𝛼 ≤ 1.

To convert Eq. (4.365) into ODE we use following transformation

738

(4.365)

𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝑘𝑥 + 𝑙𝑦 + 𝜔𝑡,

(4.366)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.366) into Eq. (4.365) and using the chain rule we obtained 4𝑘 𝛼 𝜔𝛼 𝐷𝜉2𝛼 𝑢 + 6𝑘 3𝛼 𝐷𝜉𝛼 𝑢𝐷𝜉2𝛼 𝑢 + 4𝐷𝜉4𝛼 𝑢𝑘 4𝛼 + 3𝐷𝜉3𝛼 𝑢𝑙 3𝛼 = 0.

(4.367)

By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀 + 𝑀 + 2 = 𝑀 + 3, 𝑀 = 1. Using the value of 𝑀 into Eq. (4.347), we obtained the trial solution 𝐺𝛼

𝑈(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ).

(4.368)



𝐺𝛼 1

𝐺𝛼 2

𝐺𝛼 3

𝐺𝛼 5

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + ⋯ + 𝐶5 ( 𝐺 ) = 0. 𝐺𝛼


( 𝐺 ) : 4𝑘 4𝛼 𝜆3 𝜇 + 32𝑘 4𝛼 𝜆𝜇 2 ⋯ − 6𝑙 3𝛼 𝜇 2 = 0, 𝐺𝛼 1

( 𝐺 ) : −12𝑎1 𝑘 3𝛼 𝜆2 𝜇 + 64𝑘 4𝛼 𝜇 2 ⋯ + 88𝑘 4𝛼 𝜆2 𝜇 = 0, 𝐺𝛼 2

( 𝐺 ) : −36𝑎1 𝑘 3𝛼 𝜆𝜇 − 21𝑙 3𝛼 𝜆2 ⋯ − 240𝑘 4𝛼 𝜆𝜇 = 0, 𝐺𝛼 3

( 𝐺 ) : 200𝑘 4𝛼 𝜆2 + 160𝑘 4𝛼 𝜇 ⋯ − 24𝑎1 𝑘 3𝛼 𝜇 = 0, 𝐺𝛼 4

( 𝐺 ) : 240𝑘 4𝛼 𝜆 − 30𝑎1 𝑘 3𝛼 𝜆 − 18𝑙 3𝛼 = 0, 𝐺𝛼 5

( 𝐺 ) : − 12𝑎1 𝑘 3𝛼 + 96𝑘 4𝛼 = 0. Solving the above system for unknown parameters, we have 1st Solution Set 3 ln(−(e𝛼 ln(𝑘) ) 𝜆2 )+4(e𝛼ln(𝑘) )3 𝜇)

𝑘 = 𝑘, 𝜔 = 𝑒

, 𝑙 = 0, 𝑎0 = 𝑎0 , 𝑎1 = 8𝑒 𝛼ln(𝑘) .

𝛼

Case I: Substituting the values of unknown into Eq. (4.368) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 𝐺𝛼

𝑢(𝜉) = 𝑎0 − 8𝑒 𝛼ln(𝑘) ( 𝐺 ), 739

𝐺𝛼

where ( 𝐺 ) =

1

1

1 2

1 2


[

𝐴1 cosh𝛼 ( √𝜆2 −4𝜇𝜉)+𝐴2 sinh𝛼 ( √𝜆2 −4𝜇𝜉)

2

1

] − 2 𝜆.


𝑢(𝜉) = 𝑎0 − 8𝑒 𝛼ln(𝑘) ( 𝐺 ), 𝐺𝛼

1

where ( 𝐺 ) =

1


2

[


1

]− 2𝜆


𝑢(𝜉) = 𝑎0 − 8𝑒 𝛼ln(𝑘) ( 𝐺 ), 𝐺𝛼


𝐴2

1 +𝐴2 𝜉

1

− 2 𝜆. 3 ln(−(e𝛼 ln(𝑘) ) 𝜆2 )+4(e𝛼ln(𝑘) )3 𝜇)


𝑡.

𝛼

Fig 4.91(a): 3D Soliton solution of Eq. (4.365) for different values parameters.

Fig 4.91(b): 3D graph of Eq. (4.365) for different values parameters

4.11.7. Goursat Equation Consider the Goursat equation [230] of fraction order 2𝛼 𝑢𝐷𝑥𝑡 𝑢 − 𝐷𝑡𝛼 𝑢𝐷𝑥𝛼 𝑢 − 𝑢3 = 0,

0 < 𝛼 ≤ 1.

(4.369)


(4.370) 740

Where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.370) into Eq. (4.369) and using the chain rule we obtained 2

𝑘 𝛼 𝜔𝛼 𝑢𝐷𝜉2𝛼 𝑢 − (𝐷𝜉𝛼 𝑢) 𝑘 𝛼 𝜔𝛼 − 𝑢3 = 0.

(4.371)

By applying the homogenous balancing principle for 𝛼 = 1, we have 3𝑀 = 𝑀 + 𝑀 + 2 𝑀 = 2. Using the value of 𝑀 into Eq. (4.347), we obtained the trail solution 𝐺𝛼 2

𝐺𝛼

𝑈(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ) + 𝑎2 ( 𝐺 ) .

(4.372)



𝐺𝛼 1

𝐺𝛼 2

𝐺𝛼 3

𝐺𝛼 6

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶6 ( 𝐺 ) = 0. 𝐺𝛼


( 𝐺 ) : 2𝜔𝛼 𝑘 𝛼 𝑎0 𝑎2 𝜇 2 ⋯ − 𝜔𝛼 𝑘 𝛼 𝑎0 𝑎1 𝜆𝜇 = 0, 𝐺𝛼 1

( ) : 6𝜔𝛼 𝑘 𝛼 𝑎0 𝑎2 𝜆𝜇 − 3𝑎02 𝑎1 ⋯ − 2𝜔𝛼 𝑘 𝛼 𝑎1 𝑎2 𝜇2 = 0, 𝐺

𝐺𝛼 2

( 𝐺 ) : − 2𝜔𝛼 𝑘 𝛼 𝑎22 𝜇 2 − 𝜔𝛼 𝑘 𝛼 𝑎1 𝑎2 𝜆𝜇 − 3𝑎02 𝑎2 ⋯ + 4𝜔𝛼 𝑘 𝛼 𝑎0 𝑎2 𝜆2 = 0, 𝐺𝛼 3

( 𝐺 ) : 2𝜔𝛼 𝑘 𝛼 𝑎0 𝑎1 + 𝜔𝛼 𝑘 𝛼 𝑎12 𝜆 − 6𝑎0 𝑎1 𝑎2 ⋯ − 𝜔𝛼 𝑘 𝛼 𝑎1 𝑎2 𝜆2 = 0, 𝐺𝛼 4

( 𝐺 ) : 6𝜔𝛼 𝑘 𝛼 𝑎0 𝑎2 + 𝜔𝛼 𝑘 𝛼 𝑎12 ⋯ − 5𝜔𝛼 𝑘 𝛼 𝑎1 𝑎2 𝜆 = 0, Gα 5

( G ) : 4𝜔𝛼 𝑘 𝛼 𝑎1 𝑎2 + 2𝜔𝛼 𝑘 𝛼 𝑎22 𝜆 − 3𝑎1 𝑎22 = 0, Gα 6

( G ) : 2𝜔𝛼 𝑘 𝛼 𝑎22 ⋯ − 𝑎23 = 0. Solving the above system for unknown parameters, we have 1st Solution Set 𝑘 = 𝑘, 𝜔 = 𝜔, 𝑎0 = 2𝑒 𝛼ln(𝜔) 𝑒 𝛼ln(𝑘) 𝜇, 𝑎1 = 2𝑒 𝛼ln(𝜔) 𝑒 𝛼ln(𝑘) 𝜆, 𝑎2 = 2𝜔𝛼 𝑘 𝛼 Case I: Substituting the values of unknown into Eq. (4.372) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 𝐺𝛼

𝐺𝛼 2

𝑢(𝜉) = 2𝑒 𝛼ln(𝜔) 𝑒 𝛼ln(𝑘) 𝜇 − 2𝑒 𝛼ln(𝜔) 𝑒 𝛼ln(𝑘) 𝜆 ( 𝐺 ) + 2𝜔𝛼 𝑘 𝛼 ( 𝐺 ) , 741

𝐺𝛼

where ( 𝐺 ) =

1

1

1 2

1 2


[


2

1

] − 2 𝜆.

Case II: Substituting the values of unknown into Eq. (4.372) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 < 0, we have 𝐺𝛼 2

𝐺𝛼

𝑢(𝜉) = 2𝑒 𝛼ln(𝜔) 𝑒 𝛼ln(𝑘) 𝜇 − 2𝑒 𝛼ln(𝜔) 𝑒 𝛼ln(𝑘) 𝜆 ( 𝐺 ) + 2𝜔𝛼 𝑘 𝛼 ( 𝐺 ) , 𝐺𝛼

where ( 𝐺 ) =

1

1


2

[


1

] − 2 𝜆.


𝐺𝛼 2

𝑢(𝜉) = 2𝑒 𝛼ln(𝜔) 𝑒 𝛼ln(𝑘) 𝜇 − 2𝑒 𝛼ln(𝜔) 𝑒 𝛼ln(𝑘) 𝜆 ( 𝐺 ) + 2𝜔𝛼 𝑘 𝛼 ( 𝐺 ) ,, 𝐺𝛼


𝐴2

1 +𝐴2 𝜉

1

− 2 𝜆.

In all cases 𝜉 = 𝑘𝑥 + ω𝑡.


4.11.8. Fisher’s Equation Consider the Fisher equation [230] of fraction order 𝐷𝑡𝛼 𝑢 − 𝐷𝑥2𝛼 𝑢 − 𝑢 + 𝑢2 = 0,

0 < 𝛼 ≤ 1.

(4.373)


(4.374)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.374) into Eq. (4.373) and using the chain rule we obtained

742

𝜔𝛼 𝐷𝜉𝛼 𝑢 − 𝑘 2𝛼 𝐷𝜉2𝛼 𝑢 − 𝑢 + 𝑢2 = 0.

(4.375)

By applying the homogenous balancing principle for 𝛼 = 1, we have 2𝑀 = 𝑀 + 2, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.347), we obtained the trial solution 𝐺𝛼 2

𝐺𝛼

𝑈(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ) + 𝑎2 ( 𝐺 ) .

(4.376)


algebraic equation involving power of ( ) as 𝐺

𝐺𝛼 0

𝐺𝛼 1

𝐺𝛼 2

𝐺𝛼 3

𝐺𝛼 4

𝐶0 ( ) + 𝐶1 ( ) + 𝐶2 ( ) + 𝐶3 ( ) + 𝐶4 ( ) = 0. 𝐺

𝐺

𝐺

𝐺

𝐺

𝐺𝛼


( 𝐺 ) : 𝑎02 − 𝑘 2𝛼 𝑎1 𝜆𝜇 − 𝜔𝛼 𝑎1 𝜇 − 2𝑘 2𝛼 𝑎2 𝜇 2 − 𝑎0 = 0, 𝐺𝛼 1

( 𝐺 ) : −6𝑘 2𝛼 𝑎2 𝜆𝜇 − 𝑎1 − 𝜔𝛼 𝑎1 𝜆 ⋯ + 2𝑎0 𝑎1 = 0, 𝐺𝛼 2

( 𝐺 ) : −𝑎2 + 𝑎12 − 2𝜔𝛼 𝑎2 𝜆 − 3𝑘 2𝛼 𝑎1 𝜆 ⋯ − 𝜔𝛼 𝑎1 = 0, 𝐺𝛼 3

( 𝐺 ) : −10𝑘 2𝛼 𝑎2 𝜆 + 2𝑎1 𝑎2 − 2𝜔𝛼 𝑎2 − 2𝑘 2𝛼 𝑎1 = 0, 𝐺𝛼 4

( 𝐺 ) : 𝑎22 − 6𝑘 2𝛼 𝑎2 = 0. Solving the above system for unknown parameters, we have 1st Solution Set 𝑘 = 0, 𝜔 = (

1 √−4𝜇+𝜆

1

1

1

) 𝛼 , 𝑎0 = 2 + 2 2

𝜆 √−4𝜇+𝜆2

, 𝑎1 = −

√−4𝜇+𝜆2 4𝜇−𝜆2

, 𝑎2 = 0.

Case I: Substituting the values of unknown into Eq. (4.376) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 1

1

𝑢(𝜉) = 2 + 2 𝐺𝛼

where ( 𝐺 ) =

𝜆 √−4𝜇+𝜆2

−

√−4𝜇+𝜆2 𝐺 𝛼 4𝜇−𝜆2

( 𝐺 ),

1

1

1 2

1 2


2

[


1

] − 2 𝜆.


1

1

𝑢(𝜉) = 2 + 2 𝐺𝛼

𝜆 √−4𝜇+𝜆2

−

√−4𝜇+𝜆2 𝐺 𝛼 4𝜇−𝜆2

( 𝐺 ),

1

where ( 𝐺 ) =

1


[

2


1

]− 2𝜆

Case III: Substituting the values of unknown into Eq. (4.376) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 = 0, we have 1

1

𝑢(𝜉) = 2 + 2 𝐺𝛼


𝜆 √−4𝜇+𝜆2

𝐴2

√−4𝜇+𝜆2 𝐺 𝛼 4𝜇−𝜆2

( 𝐺 ),

1

1 +𝐴2 𝜉


−

− 2 𝜆.

1 √−4𝜇+𝜆2

1

)𝛼 𝑡.

Fig 4.93(a): 3D Soliton solution of Eq. (4.373)

Fig 4.93(b): 3D Soliton solution of Eq. (4.373)

for different values parameters.

for different values parameters

4.11.9. Benjoman –Bona- Mahony (BBM) Equation Consider the BBM equation [114] of fraction order 𝐷𝑡𝛼 𝑢 + 𝐷𝑥𝛼 𝑢 + 𝛽𝑢𝐷𝑥𝛼 𝑢 + 𝐷𝑥3𝛼 𝑢 = 0,

0 < 𝛼 ≤ 1.

(4.377)


(4.378)

Where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.378) into Eq. (4.377) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝑘 𝛼 𝐷𝜉𝛼 𝑢 + 𝛽𝑢𝐷𝜉𝛼 𝑢𝑘 𝛼 + 𝐷𝜉3𝛼 𝑢𝑘 3𝛼 = 0. By applying the homogenous balancing principle for 𝛼 = 1, we have 𝑀+𝑀+1=𝑀+3 744

(4.379)

𝑀 = 2. Using the value of 𝑀 into Eq. (4.347), we obtained the trail solution 𝐺𝛼 2

𝐺𝛼

𝑈(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ) + 𝑎2 ( 𝐺 ) .

(4.380)



𝐺𝛼 1

𝐺𝛼 2

𝐺𝛼 3

𝐺𝛼 5

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶5 ( 𝐺 ) = 0. 𝐺𝛼


( 𝐺 ) : −𝛽𝑘 𝛼 𝑎0 𝑎1 𝜇 − 𝑘 3𝛼 𝑎1 𝜆2 𝜇 ⋯ − 2𝑘 3𝛼 𝑎1 𝜇 2 = 0, 𝐺𝛼 1

( 𝐺 ) : −𝛽𝑘 𝛼 𝑎0 𝑎1 𝜆 − 2𝛽𝑘 𝛼 𝑎0 𝑎2 𝜇 ⋯ − 16𝑘 3𝛼 𝑎2 𝜇 2 = 0, 𝐺𝛼 2

( 𝐺 ) : − 2𝛽𝑘 𝛼 𝑎0 𝑎2 𝜆 − 3𝛽𝑘 𝛼 𝑎1 𝑎2 𝜇 ⋯ − 𝑘 𝛼 𝑎1 = 0, 𝐺𝛼 3

( 𝐺 ) : −3𝛽𝑘 𝛼 𝑎1 𝑎2 𝜆 − 2𝛽𝑘 𝛼 𝑎0 𝑎2 ⋯ − 2𝑘 𝛼 𝑎2 = 0, 𝐺𝛼 4

( 𝐺 ) : −3𝛽𝑘 𝛼 𝑎1 𝑎2 − 2𝛽𝑘 𝛼 𝑎22 𝜆 − 54𝑘 3𝛼 𝑎2 𝜆 − 6𝑘 3𝛼 𝑎1 = 0, Gα 5

( G ) : −2𝛽𝑘 𝛼 𝑎22 − 24𝑘 3𝛼 𝑎2 = 0. Solving the above system for unknown parameters, we have 1st Solution Set 2 2 ln(−(e𝛼ln(𝑘) ) 𝜆2 −8𝜇(e𝛼ln(𝑘) ) 𝜇−𝛽𝑎0 −1)+𝛼ln(𝑘)

𝑘 = 𝑘, 𝜔 = 𝑒 −

12(e𝛼ln(𝑘) )

2

𝛽

, 𝑎0 = 𝑎0 , 𝑎1 =

𝛼

𝜆, 𝑎2 = −

12(e𝛼ln(𝑘) )

2

𝛽

.

Case I: Substituting the values of unknown into Eq. (4.380) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 𝑢(𝜉) = 𝑎0 − 𝐺𝛼

where ( 𝐺 ) =

12(e𝛼ln(𝑘) )

√𝜆2 −4𝜇 2

𝛽

2

𝐺𝛼

𝜆(𝐺 )−

12(e𝛼ln(𝑘) ) 𝛽

2

𝐺𝛼 2

(𝐺) ,

1 1 𝐴1 sinh𝛼( √𝜆2 −4𝜇𝜉)+𝐴2 cosh𝛼 ( √𝜆2 −4𝜇𝜉) 2 2 1 1 𝐴1 cosh𝛼 ( √𝜆2 −4𝜇𝜉)+𝐴2 sinh𝛼 ( √𝜆2 −4𝜇𝜉) 2 2

[

1

] − 2 𝜆.


𝑢(𝜉) = 𝑎0 − 𝐺𝛼

where ( 𝐺 ) =

12(e𝛼ln(𝑘) )

2

𝐺𝛼

𝜆(𝐺 )−

𝛽

12(e𝛼ln(𝑘) )

2

𝛽

1

𝐺𝛼 2

(𝐺) ,

1


2

[


1

]− 2𝜆

Case III: Substituting the values of unknown into Eq. (4.380) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 = 0, we have 𝑢(𝜉) = 𝑎0 − 𝐺𝛼


12(e𝛼ln(𝑘) )

𝐴2

1 +𝐴2 𝜉

𝛽

2

𝐺𝛼

𝜆(𝐺 )−


2

𝐺𝛼 2

(𝐺) ,

1

− 2 𝜆. 2 2 ln(−(e𝛼ln(𝑘) ) 𝜆2 −8𝜇(e𝛼ln(𝑘) ) 𝜇−𝛽𝑎0 −1)+𝛼ln(𝑘)


𝑡.

𝛼


4.11.10. Korteweg-de-Vries (KdV) Equation Consider the KdV equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝛽𝑢𝐷𝑥𝛼 𝑢 + 𝐷𝑥3𝛼 𝑢 = 0,

0 < 𝛼 ≤ 1.

(4.381)


(4.382)

where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.382) into Eq. (4.381) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝛽𝑢𝑘 𝛼 𝐷𝜉𝛼 𝑢 + 𝐷𝜉3𝛼 𝑢𝑘 3𝛼 = 0. By applying the homogenous balancing principle for 𝛼 = 1, we have 746

(4.383)

𝑀 + 𝑀 + 1 = 𝑀 + 3, 𝑀 = 2. Using the value of 𝑀 into Eq. (4.347), we obtained the trial solution 𝐺𝛼 2

𝐺𝛼

𝑈(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ) + 𝑎2 ( 𝐺 ) .

(4.384)



𝐺𝛼 1

𝐺𝛼 2

𝐺𝛼 3

𝐺𝛼 5

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + ⋯ + 𝐶5 ( 𝐺 ) = 0. 𝐺𝛼


( 𝐺 ) : −𝑘 3𝛼 𝑎1 𝜆2 𝜇 − 6𝑘 3𝛼 𝑎2 𝜆𝜇 2 − 𝜔𝛼 𝑎1 𝜇 ⋯ − 𝛽𝑘 𝛼 𝑎0 𝑎1 𝜇 = 0, 𝐺𝛼 1

( 𝐺 ) : −𝜔𝛼 𝑎1 𝜆 − 2𝜔𝛼 𝑎2 𝜇 − 𝑘 3𝛼 𝑎1 𝜆3 − 16𝑘 3𝛼 𝑎2 𝜇 2 ⋯ − 14𝑘 3𝛼 𝑎2 𝜆2 𝜇 = 0, 𝐺𝛼 2

( 𝐺 ) : −2𝜔𝛼 𝑎2 𝜆 − 8𝑘 3𝛼 𝑎1 𝜇 − 7𝑘 3𝛼 𝑎1 𝜆2 ⋯ − 52𝑘 3𝛼 𝑎2 𝜆𝜇 = 0, 𝐺𝛼 3

( 𝐺 ) : −𝛽𝑘 𝛼 𝑎12 − 12𝑘 3𝛼 𝑎1 𝜆 − 40𝑘 3𝛼 𝑎2 𝜇 ⋯ − 2𝛽𝑘 𝛼 𝑎22 𝜇 = 0, 𝐺𝛼 4

( 𝐺 ) : −54𝑘 3𝛼 𝑎2 𝜆 − 6𝑘 3𝛼 𝑎1 − 3𝛽𝑘 𝛼 𝑎1 𝑎2 − 2𝛽𝑘 𝛼 𝑎22 𝜆 = 0, 𝐺𝛼 5

( 𝐺 ) : −2𝛽𝑘 𝛼 𝑎22 − 24𝑘 3𝛼 𝑎2 = 0. Solving the above system for unknown parameters, we have 1st Solution Set 2 2 ln(−(e𝛼ln(𝑘) ) 𝜆2 −8(e𝛼ln(𝑘) ) 𝜇−𝛽𝑎0 )+𝛼ln(𝑘)

𝑘 = 𝑘, 𝜔 = 𝑒 −

12(e𝛼ln(𝑘) )

2

𝛽

, 𝑎0 = 𝑎0 , 𝑎1 =

𝛼

𝜆, 𝑎2 = −

12(e𝛼ln(𝑘) )

2

𝛽

.

Case I: Substituting the values of unknown into Eq. (4.384) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 𝑢(𝜉) = 𝑎0 − 𝐺𝛼

where ( 𝐺 ) =


2

𝐺𝛼

𝜆(𝐺 )−

12(e𝛼ln(𝑘) )

2

𝛽

1

𝐺𝛼 2

(𝐺) , 1


2

[


747

1

] − 2 𝜆.

Case II: Substituting the values of unknown into Eq. (4.384) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 < 0, we have 𝑢(𝜉) = 𝑎0 − 𝐺𝛼

where ( 𝐺 ) =

12(e𝛼ln(𝑘) )

2

𝐺𝛼

𝜆(𝐺 )−

𝛽

12(e𝛼ln(𝑘) )

2

𝛽

1

𝐺𝛼 2

(𝐺) ,

1


[

2

1 2

1 2


1

]− 2𝜆

Case III: Substituting the values of unknown into Eq. (4.384) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 = 0, we have 𝑢(𝜉) = 𝑎0 −


𝐺𝛼

𝐴2

𝐺

𝐴1 +𝐴2 𝜉

where ( ) =

2

𝐺𝛼

12(e𝛼ln(𝑘) )

𝐺

𝛽

𝜆( )−

2

𝐺𝛼 2

( ) , 𝐺

1

− 𝜆. 2

2 2 ln(−(e𝛼ln(𝑘) ) 𝜆2 −8(e𝛼ln(𝑘) ) 𝜇−𝛽𝑎0 )+𝛼 ln(𝑘)


𝑡.

𝛼

2nd Solution Set 𝑘 = 𝑘, 𝜔 = 0, 𝑎0 = −

(e𝛼ln(𝑘) )2 (𝜆2 +8𝜇) 𝛽

, 𝑎1 = −


2

𝜆, 𝑎2 = −

12(e𝛼ln(𝑘) )

2

𝛽

Case I: Substituting the values of unknown into Eq. (4.384) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 𝑢(𝜉) == − 𝐺𝛼

where ( 𝐺 ) =

(e𝛼ln(𝑘) )2 (𝜆2 +8𝜇) 𝛽

√𝜆2 −4𝜇 2

−−

12(e𝛼ln(𝑘) )

2

𝐺𝛼

𝜆(𝐺 )−

𝛽

12(e𝛼ln(𝑘) )

𝐺𝛼 2

(𝐺) ,

𝛽

1 1 𝐴1 sinh𝛼( √𝜆2 −4𝜇𝜉)+𝐴2 cosh𝛼 ( √𝜆2 −4𝜇𝜉) 2 2 1 1 𝐴1 cosh𝛼 ( √𝜆2 −4𝜇𝜉)+𝐴2 sinh𝛼 ( √𝜆2 −4𝜇𝜉) 2 2

[

2

1

] − 2 𝜆.

Case II: Substituting the values of unknown into Eq. (4.384) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 < 0, we have 𝑢(𝜉) == − 𝐺𝛼

where ( 𝐺 ) =

(e𝛼ln(𝑘) )2 (𝜆2 +8𝜇) 𝛽

√𝜆2 −4𝜇 2

−−

12(e𝛼ln(𝑘) )

2

𝐺𝛼

12(e𝛼ln(𝑘) )

𝐺

𝛽

𝜆( )−

𝛽

1 1 −𝐴1 sin𝛼 ( √𝜆2 −4𝜇𝜉)+𝐴2 cos𝛼( √𝜆2 −4𝜇𝜉) 2 2 1 1 𝐴1 cos𝛼 ( √𝜆2 −4𝜇𝜉)+𝐴2 sin𝛼 ( √𝜆2 −4𝜇𝜉) 2 2

[

2

𝐺𝛼 2

( ) , 𝐺

1

]− 2𝜆

Case III: Substituting the values of unknown into Eq. (4.384) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 = 0, we have 𝑢(𝜉) == −

(e𝛼ln(𝑘) )2 (𝜆2 +8𝜇) 𝛽

−−


748

2

𝐺𝛼

𝜆(𝐺 )−


2

𝐺𝛼 2

(𝐺) ,

𝐺𝛼


𝐴2

1 +𝐴2 𝜉

1

− 2 𝜆.

In all cases 𝜉 = 𝑘𝑥. Graphical Representation:

Fig 4.95(a): 3D Soliton solution of Eq. (4.381)

Fig 4.95(b): 3D Soliton solution of Eq. (4.381)

for different values parameters.

for different values parameters

4.11.11. Modified KdV Equation Consider the modified KdV equation [230] of fraction order 𝐷𝑡𝛼 𝑢 + 𝛽𝑢2 𝐷𝑥𝛼 𝑢 + 𝐷𝑥3𝛼 𝑢 = 0,

0 < 𝛼 ≤ 1.

(4.385)


(4.386)

Where 𝑘 and 𝜔 are arbitrary constant. Substituting Eq. (4.386) into Eq. (4.385) and using the chain rule we obtained 𝜔𝛼 𝐷𝜉𝛼 𝑢 + 𝛽𝑢2 𝑘 𝛼 𝐷𝜉𝛼 𝑢 + 𝐷𝜉3𝛼 𝑢𝑘 3𝛼 = 0.

(4.387)

By applying the homogenous balancing principle for 𝛼 = 1, we have 2𝑀 + 𝑀 + 1 = 𝑀 + 3 𝑀 = 1. Using the value of 𝑀 into Eq. (4.347), we obtained the trail solution 𝐺𝛼

𝑈(𝜉) = 𝑎0 + 𝑎1 ( 𝐺 ).

(4.388)


algebraic equation involving power of ( 𝐺 ) as 749

𝐺𝛼 0

𝐺𝛼 1

𝐺𝛼 2

𝐺𝛼 3

𝐺𝛼 4

𝐶0 ( 𝐺 ) + 𝐶1 ( 𝐺 ) + 𝐶2 ( 𝐺 ) + 𝐶3 ( 𝐺 ) + 𝐶4 ( 𝐺 ) = 0. 𝐺𝛼


( 𝐺 ) : 𝜔𝛼 𝜇 + 𝛽𝑘 𝛼 𝑎02 𝜇 + 2𝑘 3𝛼 𝜇 2 + 𝑘 3𝛼 𝜆2 𝜇 = 0, 𝐺𝛼 1

( 𝐺 ) : 𝜔𝛼 𝜆 + 𝛽𝑘 𝛼 𝑎02 𝜆 + 2𝛽𝑎1 𝑘 𝛼 𝑎0 𝜇 + 8𝑘 3𝛼 𝜆𝜇 + 𝑘 3𝛼 𝜆3 = 0, 𝐺𝛼 2

( 𝐺 ) : 𝜔𝛼 + 𝛽𝑘 𝛼 𝑎02 + 2𝛽𝑎1 𝑘 𝛼 𝑎0 𝜆 + 𝛽𝑎12 𝑘 𝛼 𝜇 + 8𝑘 3𝛼 𝜇 + 7𝑘 3𝛼 𝜆2 = 0, 𝐺𝛼 3

( 𝐺 ) : 2𝛽𝑎1 𝑘 𝛼 𝑎0 + 𝛽𝑎12 𝑘 𝛼 𝜆 + 12𝑘 3𝛼 𝜆 = 0, 𝐺𝛼 4

( 𝐺 ) : 𝛽𝑎12 𝑘 𝛼 + 6𝑘 3𝛼 = 0. Solving the above system for unknown parameters, we have 1st Solution Set 1 𝛼

1

𝑘 = (6 √−6𝛽𝑎1 ) , 𝜔 = (

1 1 𝛽(−𝜆2 +4𝜇)𝑎13 √−6𝛽)𝛼 6 1 12𝛼

1

) , 𝑎0 = 2 𝑎1 𝜆, 𝑎1 = 𝑎1 .

Case I: Substituting the values of unknown into Eq. (4.388) coupled with the solution of Eq. (4.338) when 𝜆2 − 4𝜇 > 0, we have 𝐺𝛼

1

𝑢(𝜉) = 2 𝑎1 𝜆 − 𝑎1 ( 𝐺 ), 𝐺𝛼

where ( 𝐺 ) =

1

1

1 2

1 2


[


2

1

] − 2 𝜆.


1

𝑢(𝜉) = 2 𝑎1 𝜆 − 𝑎1 ( 𝐺 ), 𝐺𝛼

where ( 𝐺 ) =

1

1


2

[


1

]− 2𝜆


1

𝑢(𝜉) = 2 𝑎1 𝜆 − 𝑎1 ( 𝐺 ), 𝐺𝛼


𝐴2

1 +𝐴2 𝜉

1

− 2 𝜆.

750

In all cases 𝜉 = 𝑘𝑥 + (

1 1 𝛽(−𝜆2 +4𝜇)𝑎13 √−6𝛽)𝛼 6 1 12𝛼

)𝑡

Graphical Representation:


751

Chapter 5 Numerical and Exact Solutions of Linear and Nonlinear Integral Equations

752

5.1. Introduction Many physical phenomena when mathematical modeled yield an integral equation. Many algorithms have been developed to find the exact and approximate solution of linear and nonlinear integral equations. Fettis use the Gauss–Jacobi quadrature rule [60] to determine the a numerical form of the solution of Abel equation, Huang et al. [97] used the Taylor expansion of the unknown function and obtained an approximate solution, latter on Piessens and Verbaeten [178] and Piessens [177] used Chebyshev polynomials to develop an approximate solution to Abel equation, Yousefi used Legender wavelets [251] to present a numerical method for the solution of Abel integral equation. When input signal is with noisy error, Murio et al. [151] suggested a stable numerical solution. Furthermore, Garza et al. [64] and Hall et al. [70] used the wavelet method to invert the inversion of noisy Abel equation, Sunil Kumar et al. [109] used modified Homotopy perturbation to solve the generalized Abel’s equation, Avazzadeh et al. [25] used Chebyshev polynomials to solve Abel’s Integral equations, Wazwaz [228] used efficient method to solve nonlinear integral equations and Abbasbandy [7], and Bhattacharya [29] implemented Bernstein Polynomials in numerical solutions of Volterra integral equations and Hwang and Shih [100] find the solution of integral equations via Laguerre polynomials. In this Chapter, I am going to develop some new algorithms which are very helpful for finding the exact and approximate solutions of linear and nonlinear integral equations. New scheme using Riemann-Liouville fractional integral, New Alternative scheme using Riemann-Liouville fractional integral and derivative and new scheme using generalized Riemann-Liouville Fractional integral and derivative for solving linear and nonlinear Abel’s integral equations, Adomian’s Decomposition Method and Modified Adomian’s Decomposition method using fractional calculus were implemented to find the exact solutions of linear weakly singular integral equations. Approximate solution, obtained for Taylor’s series method of linear and nonlinear Abel’s integral equations, Newton Polynomials Method (NPM), Bernoulli's Polynomials Method (BPM), Touchard Polynomial Method (TPM), Zernike Polynomial Method (ZPM), Kravchuk Polynomials Method (KPM), Meixner Polynomial Method (MPM), Rook Polynomials Method (RPM), Fibonicca Polynomials Method (FPM), Gegenbauer Polynomials Method 753

(GPM), Charlier's Polynomials Method (CPM), Bessel's Polynomials Method (BPM), Bell Polynomials Method (BPM) and Appell Polynomials Method (APM) have been used to find out the exact and approximate solutions of linear and nonlinear integral equations. It is to be observe that the proposed technique has been applied on a wide range of nonlinear diversified physical problems including, high-dimensional nonlinear evolution equation. It is to be highlighted that suggested algorithm is extremely simple but highly effective and may be extended to other singular problems of diversified physical nature. Moreover, this new scheme is capable of reducing the computational work to a tangible level while still maintaining a very high level of accuracy.

5.2. New Scheme Using Riemann-Liouville Fractional Integral I proposed a new algorithm for finding the exact solutions of linear and nonlinear Abel’s integral equations. In this technique we used Riemann-Liouville fractional integral and Riemann-Liouville fractional derivative. The proposed technique is very simple and highly compatible for solving linear and nonlinear Abel’s equations.

5.2.1. Methodology Consider the Abel’s integral equation 𝑥

𝑓(𝑥) = 𝜆 ∫0

1 𝑢(𝑡)d𝑡. (𝑥−𝑡)𝛼

(5.1)

By using the definition of Riemann-Liouville fractional integral we obtained 𝑓(𝑥) = 𝜆Γ(1 − 𝛼)𝐽𝑥1−𝛼 𝑢(𝑥), where

is gamma function, simplify Eq. (5.1), we have 1

𝑢(𝑥) = 𝜆Γ(1−𝛼) 𝐷𝑥1−𝛼 𝑓(𝑥), Using the definition of fractional derivative 1

𝑑

𝑥

𝐷𝑥𝛼 𝑓(𝑥) = Γ(1−𝛼) 𝑑𝑥 ∫0 and

1 𝑓(𝑡)d𝑡, (𝑥−𝑡)𝛼

Γ(𝑘+1)

𝐷𝑥𝛼 𝑥 𝑘 = Γ(𝑘−𝛼+1) 𝑥 𝑘−𝛼 .

(5.2) (5.3)

Decomposing 𝑓(𝑥) into Taylor’s series 𝑓(𝑥) = 𝑐1 𝑥 𝑘1 + 𝑐2 𝑥 𝑘2 + 𝑐3 𝑥 𝑘3 + ⋯ 𝑐𝑛 𝑥 𝑘𝑛 + ⋯, and then using Eq. (5.2) and Eq. (5.3) into Eq. (5.1), we have 1

Γ(𝑘 +1)

𝑛 𝑘𝑛 +𝛼−1 𝑢(𝑥) = 𝜆Γ(1−𝛼) ∑∞ . 𝑛 Γ(𝑘 +𝛼) 𝑐𝑛 𝑥 𝑛

754

(5.4)

5.2.2. Abel’s Integral Equations Example 5.1. Consider the following generalized linear Abel’s integral equation [229] 11

128

𝑥

𝑥 4 = ∫0 231

1 1

𝑢(𝑡)d𝑡,

(5.5)

(𝑥−𝑡)4

Taking 1

3

11

𝛼 = 4 , 𝛽 = 4 , 𝑘1 =

4

128

, 𝑐1 = 231.

Using the above information in Eq. (5.5), we obtained 𝑢(𝑥) =

11 4 11 1 Γ( + ) 4 4

11 1

Γ( +1) 128

1 1 Γ(1− ) 4

[

𝑥 4 +4−1 ], 231

𝑢(𝑥) = 𝑥 2 , which is the exact solution.

Example 5.2. Consider the following generalized linear Abel’s integral equation [229] 1

7

27

𝑥

2𝜋𝑥 3 − 14 𝑥 3 = ∫0

1 2

𝑢(𝑡)d𝑡,

(5.6)

(𝑥−𝑡)3

Taking 2

1

1

7

27

𝛼 = 3 , 𝛽 = 3 , 𝑘1 = 3 , 𝑘2 = 3 , 𝑐1 = 2𝜋, 𝑐2 = − 14. Using the above information in Eq. (5.6), we obtained 𝑢(𝑥) =

1 3 1 2 Γ( + ) 3 3

Γ( +1)

1 2 Γ(1− ) 3

[

7

1 2

Γ( +1) 27 3

2𝜋𝑥 3+3−1 −

7 2 Γ( + ) 3 3

7 2

𝑥 3+3−1 ], 14

𝑢(𝑥) = 𝜋 − 𝑥 2 , which is the exact solution.


7

128

11

𝑥

𝑥 4 + 231 𝑥 4 = ∫0 21

1 1

𝑢(𝑡)d𝑡,

(5.7)

(𝑥−𝑡)4

Taking 1

3

7

𝛼 = 4 , 𝛽 = 4 , 𝑘1 = 4 , 𝑘2 =

11 4

16

128

, 𝑐1 = 21 , 𝑐2 = 231.


1 1 Γ(1− ) 4

7 4 7 1 Γ( + ) 4 4

Γ( +1) 16

[

𝑥 21

7 1 + −1 4 4

+

11 4 11 1 Γ( + ) 4 4

Γ( +1) 128

𝑢(𝑥) = 𝑥 + 𝑥 2 , which is the exact solution. 755

11 1

𝑥 4 +4−1 ], 231

Example 5.4. Consider the following generalized nonlinear Abel’s integral equation [229] 2 3

5 6

9Γ( )Γ( ) 7√π

7

x

𝑥 6 = ∫0

1 1

ln(𝑢(𝑡)) d𝑡.

(5.8)

𝑣(𝑡)d𝑡,

(5.9)

(𝑥−𝑡)3

Using the transformation 𝑣(𝑥) = ln(𝑢(𝑥)), we obtained 2 3

5 6

9Γ( )Γ( ) 7√π

7

𝑥

𝑥 6 = ∫0

1 1

(𝑥−𝑡)3

Taking 1

2

7

𝛼 = 3 , 𝛽 = 3 , 𝑘1 = 6 , 𝑐1 =

2 3

5 6

9Γ( )Γ( ) 7√π

.

Using the above information in Eq. (5.9), we obtained 𝑣(𝑥) =

7 6 7 1 Γ( + ) 6 3

2 3

5 6

Γ( +1) 9Γ( )Γ( )

1 1 Γ(1− ) 3

[

7√π

7 1

𝑥 6+3−1 ],

𝑣(𝑥) = √𝑥, The inverse transformation yields 𝑢(𝑥) = exp(𝑣(𝑥)) = exp(√𝑥), which is the exact solution.

Example 5.5. Consider the following generalized nonlinear Abel’s integral equation [229] 1

1

𝑥

𝑥 2 (30 + 40𝑥 + 16𝑥 2 ) = ∫0 15

1 √𝑥−𝑡

𝑢2 (𝑡)d𝑡.

(5.10)

𝑣(𝑡)d𝑡.

(5.11)

Using the transformation 𝑣(𝑥) = 𝑢2 (𝑥), we obtained 1

1

𝑥

𝑥 2 (30 + 40𝑥 + 16𝑥 2 ) = ∫0 15

1 √𝑥−𝑡

Taking 1

1

1

3

5

8

16

𝛼 = 2 , 𝛽 = 2 , 𝑘1 = 2 , 𝑘2 = 2 , 𝑘3 = 2 , 𝑐1 = 2, 𝑐2 = 3 , 𝑐3 = 15. we obtained

756

𝑣(𝑥) =

1 2 1 1 Γ( + ) 2 2

Γ( +1)

1 1 Γ(1− ) 2

[

2𝑥

1 1 + −1 2 2

3

+

Γ( +1) 8 2 3 1 Γ( + ) 2 2

𝑥 3

3 1 + −1 2 2

5

+

Γ( +1) 16 2 5 1 Γ( + ) 2 2

5 1

𝑥 2+2−1 ], 15

𝑣(𝑥) = 1 + 2𝑥 + 𝑥 2 , The inverse transformation yields 𝑢(𝑥) = ±√𝑣(𝑥) = ±√1 + 2𝑥 + 𝑥 2 , which is the exact solution.


17

𝑥

𝑥 6 = ∫0 935

1 1

𝑢(𝑡)d𝑡,

(5.12)

(𝑥−𝑡)6

Taking 1

5

𝛼 = 6 , 𝛽 = 6 , 𝑘1 =

17 6

432

, 𝑐1 = 935 .


17

1 1 Γ(1− ) 6

Γ( +1) 432 6

[

17 1 Γ( + ) 6 6

17 1

𝑥 6 +6−1 ], 935



2

5

9

𝑥

𝑥 3 + 10 𝑥 3 = ∫0 2

1 1

𝑢(𝑡)d𝑡,

(5.13)

(𝑥−𝑡)3

Taking 1

2

2

5

3

9

𝛼 = 3 , 𝛽 = 3 , 𝑘1 = 3 , 𝑘2 = 3 , 𝑐1 = 2 , 𝑐2 = 10. Using the above information in Eq. (5.13), we obtained 𝑢(𝑥) =

2 3 2 1 Γ( + ) 3 3

Γ( +1) 3

1 1 3

Γ(1− )

[

2 1

𝑥 3+3−1 + 2

5

Γ( +1) 9 3 5 1 3 3

Γ( + )

5 1

𝑥 3+3−1 ], 10

𝑢(𝑥) = 1 + 𝑥, which is the exact solution.


8

3

𝑥

2𝑥 2 + 3 𝑥 2 = ∫0

1 1

𝑢(𝑡)d𝑡,

(5.14)

(𝑥−𝑡)2

Taking

757

1

1

1

3

8

𝛼 = 2 , 𝛽 = 2 , 𝑘1 = 2 , 𝑘2 = 2 , 𝑐1 = 2, 𝑐2 = 3. Using the above information in Eq. (5.14), we obtained 𝑢(𝑥) =

1

1 1 Γ(1− ) 2

Γ( +1) 1 2

[

1 1 Γ( + ) 2 2

𝑥 2

1 1 + −1 2 2

3

+

Γ( +1) 8 2 3 1 Γ( + ) 2 2

3 1

𝑥 2+2−1 ], 3

𝑢(𝑥) = 1 + 2𝑥, which is the exact solution.


243

𝑥

𝑥 3 = ∫0 440

1 1 (𝑥−𝑡)3

𝑢3 (𝑡)d𝑡.

(5.15)

𝑣(𝑡)d𝑡.

(5.16)


243

𝑥

𝑥 3 = ∫0 440

1 1

(𝑥−𝑡)3

Taking 1

2

11

3

3

3

𝛼 = , 𝛽 = , 𝑘1 =

, 𝑐1 =

243

.

440


1 1 Γ(1− ) 3

11 3 11 1 Γ( + ) 3 3

Γ( +1) 243

[

11 1

𝑥 3 +3−1 ], 440

𝑣(𝑥) = 𝑥 3 . The inverse transformation yields 1

𝑢(𝑥) = (𝑣(𝑥))3 = 𝑥, which is the exact solution.


5

𝑥

𝑥 6 (11 + 6𝑥) = ∫0 55

1 1 (𝑥−𝑡)6

sin−1 (𝑢(𝑡)) d𝑡.

Using the transformation 𝑣(𝑥) = sin−1(𝑢(𝑥)),

758

(5.17)

we obtained 6

5

𝑥

𝑥 6 (11 + 6𝑥) = ∫0 55

1 1

𝑣(𝑡)𝑑𝑡,

(5.18)

(𝑥−𝑡)6

Taking 1

5

5

𝛼 = 6 , 𝛽 = 6 , 𝑘1 = 6 , 𝑘2 =

11 6

6

36

, 𝑐1 = 5 , 𝑐2 = 55.

we obtained 𝑣(𝑥) =

5 6 5 1 Γ( + ) 6 6

Γ( +1) 6

1 1 Γ(1− ) 6

[

𝑥 5

5 1 + −1 6 6

+

11 6 11 1 Γ( + ) 6 6

Γ( +1) 36

11 1

𝑥 6 +6−1 ], 55

𝑣(𝑥) = 1 + 𝑥, The inverse transformation yields 𝑢(𝑥) = sin(𝑣(𝑥)) = sin(1 + 𝑥), which is the exact solution.

5.3. New Alternative Scheme Using Riemann-Liouville Fractional Integral and Derivative We proposed a new algorithm for finding the exact solutions of linear and nonlinear Abel’s integral equations. In this technique we used Riemann-Liouville fractional integral and Riemann-Liouville fractional derivative. The proposed technique is very simple and highly compatible for solving linear and nonlinear Abel’s equations.

5.3.1. Methodology Taking Laplace transformation of both sides of Eq. (5.1), we have ℒ[𝑓(𝑥)] = ℒ[𝑢(𝑥)]ℒ[𝑥 −𝛼 ].

(5.19)

Eq. (5.19) can be written as 𝐹(𝑠) = 𝑈(𝑠) where

Γ(1−𝛼) 𝑠1−𝛼

,

(5.20)

is gamma function, ℒ[𝑓(𝑥)] = 𝐹(𝑠), and ℒ[𝑢(𝑥)] = 𝑈(𝑠).

Eq. (5.20) may be re-written as 𝑠1−𝛼

𝑈(𝑠) = Γ(1−𝛼) 𝐹(𝑠).

(5.21)

Eq. (5.21) can be written as 𝑠

ℒ[𝑢(𝑥)] = Γ(𝛼)Γ(1−𝛼) ℒ[𝑝(𝑥)], 𝑥

where 𝑝(𝑥) = ∫0

(5.22)

1 𝑓(𝑡)d𝑡. (𝑥−𝑡)𝛼−1

(5.23) 759

As we know that ℒ[𝑝(𝑥)′ ] = 𝑠ℒ[𝑝(𝑥)] − 𝑝(0),

(5.24)

𝜋

Γ(𝛼)Γ(1 − 𝛼) = sin(𝛼𝜋).

and

(5.25)

Using Eq. (5.24) and Eq. (5.25) into Eq. (5.22) we have ℒ[𝑢(𝑥)] =

sin(𝛼𝜋) 𝜋

ℒ[𝑝(𝑥)′ ].

(5.26)

Now, applying inverse Laplace transformation and using Eq. (5.23) we obtained 𝑢(𝑥) =

𝑥 1 𝑓(𝑡)d𝑡, ∫ 𝑑𝑥 0 (𝑥−𝑡)𝛽

sin(𝛼𝜋) 𝑑 𝜋

where 𝛽 = 1 − 𝛼.

(5.27)

Using the definition of fractional derivative 1

𝛽

𝑑

𝑥

𝐷𝑥 𝑓(𝑥) = Γ(1−𝛽) 𝑑𝑥 ∫0 and

1 𝑓(𝑡)d𝑡, (𝑥−𝑡)𝛽

(5.28)

Γ(𝑘+1)

𝛽

𝐷𝑥 𝑥 𝑘 = Γ(𝑘−𝛽+1) 𝑥 𝑘−𝛽 .

(5.29)

Decomposing 𝑓(𝑥) into Taylor’s series 𝑓(𝑥) = 𝑐1 𝑥 𝑘1 + 𝑐2 𝑥 𝑘2 + 𝑐3 𝑥 𝑘3 + ⋯ 𝑐𝑛 𝑥 𝑘𝑛 + ⋯, and then using Eq. (5.28) and Eq. (5.29) into Eq. (5.27), we have Γ(𝑘1 +1)

𝑢(𝑥) =

sin(𝛼𝜋) 𝜋

[

Γ(𝑘2 +1)

𝑐 𝑥 𝑘1 −𝛽 + Γ(𝑘 −𝛽+1) 1

Γ(𝑘1 Γ(𝑘3 +1)

+ Γ(𝑘

3

2 −𝛽+1)

𝑐2 𝑥 𝑘2 −𝛽

Γ(𝑘𝑛 +1)

𝑐 𝑥 𝑘3 −𝛽 + ⋯ + Γ(𝑘 −𝛽+1) 3

𝑛

𝑐 𝑥 𝑘𝑛−𝛽 + ⋯ −𝛽+1) 𝑛

] Γ(1 − 𝛽).

(5.30)

Remark: If we have a nonlinear Abel’s Integral Equation i.e. 𝑥 𝐹(𝑢(𝑡))

𝑓(𝑥) = ∫0

(𝑥−𝑡)𝛼

d𝑡, 𝑥 > 0, 0 < 𝛼 ≤ 1.

(5.31)

First we using the transformation 𝑣(𝑥) = 𝐹(𝑢(𝑥)). Converting Eq. (5.31) into linear then solve proceeding as discuss in analysis, at the end we use the retransformation 𝑢(𝑥) = 𝐹 −1 (𝑣(𝑥)).

5.3.2. Abel’s Integral Equations Example 5.11. Consider the following generalized linear Abel’s integral equation [229] as 128

11

𝑥

𝑥 4 = ∫0 231

1 1

𝑢(𝑡)d𝑡,

(5.32)

(𝑥−𝑡)4

Taking 760

1

3

11

𝛼 = 4 , 𝛽 = 4 , 𝑘1 =

4

128

, 𝑐1 = 231.


1 4

sin( 𝜋)

11

[

Γ( +1) 128 4

11 3 Γ( − +1) 231 4 4

𝜋

11 3

3

𝑥 4 −4 ] Γ (1 − 4),


Example 5.12. Consider the following generalized linear Abel’s integral equation [229] as 1

7

27

𝑥

2𝜋𝑥 3 − 14 𝑥 3 = ∫0

1 2

𝑢(𝑡)d𝑡,

(5.33)

(𝑥−𝑡)3

Taking 2

1

1

7

27

𝛼 = 3 , 𝛽 = 3 , 𝑘1 = 3 , 𝑘2 = 3 , 𝑐1 = 2𝜋, 𝑐2 = − 14. Using the above information in Eq. (5.30), we obtained 𝑢(𝑥) =

2 3

sin( 𝜋)

[

1 3

1 1 Γ( − +1) 3 3

𝜋

7

1 1

Γ( +1)

2𝜋𝑥 3−3 −

Γ( +1) 27 3

7 1 Γ( − +1) 14 3 3

7 1

1

𝑥 3−3 ] Γ (1 − 3),

𝑢(𝑥) = 𝜋 − 𝑥 2 , which is the exact solution.


7

11

128

𝑥

𝑥 4 + 231 𝑥 4 = ∫0 21

1 1

𝑢(𝑡)d𝑡,

(5.34)

(𝑥−𝑡)4

Taking 1

3

7

𝛼 = 4 , 𝛽 = 4 , 𝑘1 = 4 , 𝑘2 =

11 4

16

128

, 𝑐1 = 21 , 𝑐2 = 231.


1 4

sin( 𝜋) 𝜋

7

[

Γ( +1) 16 4

7 3 Γ( − +1) 4 4

𝑥 21

7 3 − 4 4

11

+

Γ( +1) 128 4

11 3 Γ( − +1) 4 4

11 3

3

𝑥 4 −4 ] Γ (1 − 4), 231

𝑢(𝑥) = 𝑥 + 𝑥 2 , which is the exact solution.

Example 5.14. Consider the following generalized nonlinear Abel’s integral equation [229] as

761

11

243

𝑥

𝑥 3 = ∫0 440

1 1

𝑢3 (𝑡)d𝑡.

(5.35)

𝑣(𝑡)d𝑡.

(5.36)

(𝑥−𝑡)3


243

𝑥

𝑥 3 = ∫0 440

1 1

(𝑥−𝑡)3

Taking 1

2

𝛼 = 3 , 𝛽 = 3 , 𝑘1 =

11 3

243

, 𝑐1 = 440.


1 3

sin( 𝜋) 𝜋

11

[

Γ( +1) 243 3

11 2 Γ( − +1) 440 3 3

11 2

2

𝑥 3 −3 ] Γ (1 − 3),

𝑣(𝑥) = 𝑥 3 . The inverse transformation yields 1

𝑢(𝑥) = (𝑣(𝑥))3 = 𝑥, which is the exact solution.

Example 5.15. Consider the following generalized nonlinear Abel’s integral equation [229] as 6

5

𝑥

𝑥 6 (11 + 6𝑥) = ∫0 55

1 1

sin−1 (𝑢(𝑡)) d𝑡.

(5.37)

𝑣(𝑡)d𝑡,

(5.38)

(𝑥−𝑡)6

Using the transformation 𝑣(𝑥) = sin−1(𝑢(𝑥)), We obtained 6

5

𝑥

𝑥 6 (11 + 6𝑥) = ∫0 55

1 1

(𝑥−𝑡)6

Taking 1

5

5

𝛼 = 6 , 𝛽 = 6 , 𝑘1 = 6 , 𝑘2 =

11 6

6

36

, 𝑐1 = 5 , 𝑐2 = 55.

we obtained 𝑣(𝑥) =

1 6

sin( 𝜋) 𝜋

5

[

Γ( +1) 6 6

5 5 Γ( − +1) 6 6

5 5

𝑥 6 −6 + 5

11

Γ( +1) 36 6

11 5 Γ( − +1) 6 6

𝑣(𝑥) = 1 + 𝑥, 762

11 5

5

𝑥 6 −6 ] Γ (1 − 6), 55

The inverse transformation yields 𝑢(𝑥) = sin(𝑣(𝑥)) = sin(1 + 𝑥), which is the exact solution.

Example 5.16. Consider the following generalized nonlinear Abel’s integral equation [229] as 2 3

5 6

9Γ( )Γ( ) 7 √𝜋

7

𝑥

𝑥 6 = ∫0

1 1


(5.39)

𝑣(𝑡)d𝑡,

(5.40)

(𝑥−𝑡)3

Using the transformation 𝑣(𝑥) = ln(𝑢(𝑥)), We obtained 2 3

5 6

9Γ( )Γ( ) 7 √𝜋

7

𝑥

𝑥 6 = ∫0

1 1

(𝑥−𝑡)3

Taking 1

2

7

𝛼 = 3 , 𝛽 = 3 , 𝑘1 = 6 , 𝑐1 =

2 3

5 6

9Γ( )Γ( ) 7√𝜋

.


1 3

sin( 𝜋)

7 6

2 3

[

7 2 Γ( − +1) 6 3

𝜋

5 6

Γ( +1) 9Γ( )Γ( ) 7√𝜋

7 2

2

𝑥 6−3 ] Γ (1 − 3),

𝑣(𝑥) = √𝑥, The inverse transformation yields 𝑢(𝑥) = exp(𝑣(𝑥)) = exp(√𝑥), which is the exact solution.

Example 5.17. Consider the following linear Abel’s integral equation [229] as 𝑥

2𝜋√𝑥 = ∫0

1 √𝑥−𝑡

𝑢(𝑡)d𝑡,

(5.41)

Taking 1

1

1

𝛼 = 2 , 𝛽 = 2 , 𝑘1 = 2 , 𝑐1 = 2𝜋. Using the above information in Eq. (5.30), we obtained 𝑢(𝑥) =

1 2

sin( 𝜋) 𝜋

[

1 2

Γ( +1)

1 1 Γ( − +1) 2 2

1 1

1

2𝜋𝑥 2−2 ] Γ (1 − 2),

𝑢(𝑥) = 𝜋. 763

which is the exact solution.

Example 5.18. Consider the following linear Abel’s integral equation [229] as 8

3

𝑥2 + 3

16 5

5

𝑥

𝑥 2 = ∫0

1 √𝑥−𝑡

𝑢(𝑡)d𝑡,

(5.42)

Taking 1

1

3

5

8

𝛼 = 2 , 𝛽 = 2 , 𝑘1 = 2 , 𝑘2 = 2 , 𝑐1 = 3 , 𝑐2 =

16 5

.


1 2

sin( 𝜋)

3

Γ( +1) 8 2

[

3 1 Γ( − +1) 3 2 2

𝜋

𝑥

3 1 − 2 2

5

+

Γ( +1) 16 2

5 1 Γ( − +1) 2 2

5

5 1

1

𝑥 2−2 ] Γ (1 − 2),

𝑢(𝑥) = 2𝑥 + 3𝑥 2 . which is the exact solution.


sin(𝑥) = ∫0 1

1 √𝑥−𝑡

𝑢(𝑡)d𝑡, 𝑥

1

𝑥 − 3! 𝑥 3 + 5! 𝑥 5 − ⋯ = ∫0

1 √𝑥−𝑡

𝑢(𝑡)d𝑡,

(5.43)

Taking 1

1

1

1

𝛼 = 2 , 𝛽 = 2 , 𝑘1 = 1, 𝑘2 = 3, 𝑘3 = 5, … , 𝑐1 = 1, 𝑐2 = − 6 , 𝑐3 = 120 , … Using the above information in Eq. (5.30), we obtained 1

Γ(1+1)

𝑢(𝑥) =

1 Γ(1− +1) 2

1 sin( 𝜋) 2

𝜋

+

[

Γ(3+1) 1

𝑥1−2 −

Γ(5+1) 1 2

1 Γ(3− +1) 6 2 1 1 5− 2

Γ(5− +1) 120

5

5

8𝑥 2

32𝑥 2

𝑥

1

𝑥 3−2

−⋯

1

Γ (1 − 2), ]

13

𝑢(𝑥) =

2√𝑥 𝜋

128𝑥 2

𝑢(𝑥) =

2√𝑥 22𝑛 ∞ (−1)𝑛 [∑ 𝑥 2𝑛 ]. 𝑛=0 𝜋 1∙3∙5∙∙∙(4𝑛−1)

− 15𝜋 + 945𝜋 − 135135𝜋 + ⋯


Example 5.20. Consider the following nonlinear Abel’s integral equation [229] as 𝑥

2𝜋√𝑥 = ∫0

1 √𝑥−𝑡

𝑢2 (𝑡)d𝑡,

(5.44)

using the transformation 𝑣(𝑥) = 𝑢2 (𝑥), we have the Eq. (5.44), we obtained 764

𝑥

2𝜋√𝑥 = ∫0

1 √𝑥−𝑡

𝑣(𝑡)d𝑡,

(5.45)

Taking 1

1

1

𝛼 = 2 , 𝛽 = 2 , 𝑘1 = 2 , 𝑐1 = 2𝜋. Using the above information in Eq. (5.30), we obtained 𝑣(𝑥) =

1 2

sin( 𝜋) 𝜋

[

1 2

Γ( +1)

1 1 Γ( − +1) 2 2

1 1

1

2𝜋𝑥 2−2 ] Γ (1 − 2),

𝑣(𝑥) = 𝜋, the solution of Eq. (5.44), given as 𝑢(𝑥) = ±√𝑣(𝑥) = ±√𝜋. which is the exact solution.

Example 5.21. Consider the following nonlinear Abel’s integral equation [229] as 1

𝑥

2

6𝑥 2 (1 + 9 𝑥) = ∫0

1 √𝑥−𝑡

sin−1(𝑢(𝑡)) d𝑡,

(5.46)

using the transformation 𝑣(𝑥) = sin−1(𝑢(𝑥)), we have the Eq. (5.46), we obtained 1

𝑥

2

6𝑥 2 (1 + 9 𝑥) = ∫0

1 √𝑥−𝑡

𝑣(𝑡)d𝑡,

(5.47)

Taking 1

1

1

3

4

𝛼 = 2 , 𝛽 = 2 , 𝑘1 = 2 , 𝑘2 = 2 , 𝑐1 = 6, 𝑐2 = 3. Using the above information in Eq. (5.30), we obtained 𝑣(𝑥) =

1 2

sin( 𝜋) 𝜋

[

1 2

Γ( +1)

1 1 Γ( − +1) 2 2

6𝑥

1 1 − 2 2

3

+

Γ( +1) 4 2

3 1 Γ( − +1) 2 2

3 1

1

𝑥 2−2 ] Γ (1 − 2), 3

𝑣(𝑥) = 3 + 𝑥, the solution of Eq. (5.46), given as 𝑢(𝑥) = sin(𝑣(𝑥)) = sin(𝑥 + 3). which is the exact solution.

Example 5.22. Consider the following nonlinear Abel’s integral equation [229] as 2

1

𝑥

𝑒𝑥 2 (3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡

exp(𝑢(𝑡)) d𝑡,

using the transformation 765

(5.48)

𝑣(𝑥) = exp(𝑢(𝑥)), we have the Eq. (5.48), we obtained 2

1

𝑥

𝑒𝑥 2 (3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡

𝑣(𝑡)d𝑡,

(5.49)

Taking 1

1

1

3

4

𝛼 = 2 , 𝛽 = 2 , 𝑘1 = 2 , 𝑘2 = 2 , 𝑐1 = 2𝑒, 𝑐2 = 3 𝑒. Using the above information in Eq. (5.30), we obtained 𝑣(𝑥) =

1 2

sin( 𝜋)

1 2

Γ( +1)

[

1 1 Γ( − +1) 2 2

𝜋

3

1 1

2𝑒𝑥 2−2 +

Γ( +1) 4 2

3 1 Γ( − +1) 3 2 2

3 1

1

𝑒𝑥 2−2 ] Γ (1 − 2),

𝑣(𝑥) = 𝑒 + 𝑒𝑥, the solution of Eq. (5.48), given as 𝑢(𝑥) = ln(𝑣(𝑥)) = 1 + ln(1 + 𝑥). which is the exact solution.

Example 5.23. Consider the following generalized linear Abel’s integral equation [229] as 𝑥

1

1 + 𝑥 = ∫0

4

𝑢(𝑡)d𝑡,

(5.50)

(𝑥−𝑡)5

Taking 4

1

𝛼 = 5 , 𝛽 = 5 , 𝑘1 = 0, 𝑘2 = 1, 𝑐1 = 1, 𝑐2 = 1. Using the above information in Eq. (5.30), we obtained 𝑢(𝑥) = 𝑢(𝑥) =

4 5

sin( 𝜋) 𝜋

[

(1+1.25) 1 𝜋𝑥 5

Γ(0+1)

1 Γ(0− +1) 5

1

𝑥 0−5 +

Γ(1+1) 1 Γ(1− +1) 5

1

sin (5 𝜋),


766

1

1

𝑥1−5 ] Γ (1 − 5),

Table 5.1. Comparison of Exact and Approximate solutions of Eq. (5.50) x

n = 10

n = 20

n = 30

Exact

0.1 0.32882 0.33219 0.33317 0.33359 0.2 0.32529 0.32117 0.32371 0.32268 0.3 0.32346 0.32567 0.32715 0.32730 0.4 0.34031 0.33805 0.33679 0.33709 0.5 0.34828 0.34960 0.34902 0.34924 0.6 0.36120 0.36162 0.36247 0.36264 0.7 0.37894 0.37715 0.37648 0.37674 0.8 0.38938 0.39207 0.39078 0.39127 0.9 0.40782 0.40670 0.40638 0.40604 1.0 0.42415 0.42245 0.42190 0.42097


𝑥

𝑥 6 = ∫0

1 1

𝑢(𝑡)d𝑡,

(5.51)

(𝑥−𝑡)3

Taking 1

2

7

𝛼 = 3 , 𝛽 = 3 , 𝑘1 = 6 , 𝑐1 = 1. Using the above information in Eq. (5.30), we obtained 𝑢(𝑥) =

1 3

sin( 𝜋) 𝜋 7

𝑢(𝑥) = 9

[

7 2 Γ( − +1) 6 3

√𝑥𝜋 2 3

7 6

Γ( +1)

5 6

Γ( )Γ( )

7 2

2

𝑥 6−3 ] Γ (1 − 3),

.

This is the exact solution.

767


n = 10

n = 20

n = 30

Exact

0.1 0.28424 0.28484 0.28501 0.28520 0.2 0.40334 0.40327 0.40333 0.40334 0.3 0.49462 0.49444 0.49413 0.49399 0.4 0.56879 0.57040 0.57056 0.57041 0.5 0.63918 0.63742 0.63787 0.63774 0.6 0.69839 0.69891 0.69873 0.69861 0.7 0.75358 0.75468 0.75472 0.75458 0.8 0.80828 0.80635 0.80681 0.80669 0.9 0.85375 0.85519 0.85545 0.85562 1.0 0.89394 0.89918 0.90048 0.90190

Example 5.25. Consider the following generalized linear Abel’s integral equation [229] as 𝑥

2

√𝑥(105 − 56𝑥 2 + 48𝑥 3 ) = ∫0 105

1 √𝑥−𝑡

𝑢(𝑡)d𝑡,

(5.52)

Taking 1

1

1

3

7

112

96

𝛼 = 2 , 𝛽 = 2 , 𝑘1 = 2 , 𝑘2 = 2 , 𝑘3 = 2 , 𝑐1 = 2, 𝑐2 = − 105 , 𝑐3 = 105. Using the above information in Eq. (5.30), we obtained 𝑢(𝑥) =

1 2

sin( 𝜋) 𝜋

[2

1 2

Γ( +1) 1 1 Γ( − +1) 2 2

1 1

3

3 1 112 Γ(2+1) − 2 2 𝑥 3 1 Γ( − +1)

𝑥 2−2 − 105

2 2

𝑢(𝑥) = 1 − 𝑥 2 + 𝑥 3 .

768

96

+ 105

7 2

Γ( +1) 7 1 Γ( − +1) 2 2

7 1

1

𝑥 2−2 ] Γ (1 − 2),


Exact solution Approximate solution k = 1, m = 2

k = 1, m = 3

0.0 1.000

1.013060

1.003960

0.1 0.991

0.986939

0.988503

0.2 0.968

0.960816

0.967075

0.3 0.937

0.934694

0.939672

0.4 0.904

0.908571

0.906296

0.5 0.875

0.908164

0.845819

0.6 0.856

0.907899

0.835359

0.7 0.853

0.907635

0.843653

0.8 0.872

0.907370

0.870702

0.9 0.919

0.907106

0.916505

5.4. New Scheme Using Riemann-Liouville Fractional Integral and Derivative We proposed a new algorithm for finding the exact solutions of linear and nonlinear generalized Abel’s integral equations. In this technique we used Riemann-Liouville fractional integral and Riemann-Liouville fractional derivative. The proposed technique is very simple and highly compatible for solving linear and nonlinear Abel’s equations.

5.4.1 Methodology Consider the generalized Abel’s integral equation, 𝑥

𝑓(𝑥) = ∫0

𝑢(𝑡) (𝑥 𝜌 −𝑡 𝜌 )𝛼

d𝑡,

𝑥 > 0, 0 < 𝛼 ≤ 1.

(5.53)

The solution of Eq. (5.53) is given as 𝑢(𝑥) =


𝑥

∫ 𝑑𝑥 0

𝜌𝑡 𝜌−1 𝑓(𝑡)d𝑡. (𝑥 𝜌 −𝑡 𝜌 )𝛼

(5.54)

Decomposing 𝑓(𝑥) into Taylor’s series 𝑘𝑛 𝑓(𝑡) = 𝑐1 𝑡 𝑘1 + 𝑐2 𝑡 𝑘2 + 𝑐3 𝑡 𝑘3 + ⋯ 𝑐𝑛 𝑡 𝑘𝑛 + ⋯ = ∑∞ 𝑛=0 𝑐𝑛 𝑥 .

Using Eq. (5.55) into Eq. (5.54), we have

769

(5.55)

𝑢(𝑥) =

sin(𝛼𝜋) 𝑑

𝑥

∫ 𝑑𝑥 0

𝜋

𝑑

𝑢(𝑥) =

sin(𝛼𝜋) 𝜋

[

𝜌𝑡 𝜌−1 [𝑐 𝑡 𝑘1 (𝑥 𝜌 −𝑡 𝜌 )𝛼 1

+ 𝑐2 𝑡 𝑘2 + 𝑐3 𝑡 𝑘3 + ⋯ 𝑐𝑛 𝑡 𝑘𝑛 + ⋯ ]d𝑡,

𝜌𝑡 𝜌−1 𝑐 𝑡 𝑘1 (𝑥 𝜌 −𝑡 𝜌 )𝛼 1

+ 𝑑𝑥 ∫0

𝑥

∫ 𝑑𝑥 0

𝑥 𝜌𝑡 𝜌−1 + 𝑑𝑥 ∫0 (𝑥 𝜌 −𝑡 𝜌 )𝛼 𝑐3 𝑡 𝑘3 𝑑

𝑥

𝑑

+ ⋯+

𝜌𝑡 𝜌−1 𝑐 𝑡 𝑘2 (𝑥 𝜌 −𝑡 𝜌 )𝛼 2

𝑥 𝜌𝑡 𝜌−1 𝑐 𝑡 𝑘𝑛 ∫ 𝑑𝑥 0 (𝑥 𝜌 −𝑡 𝜌 )𝛼 𝑛 𝑑

] d𝑡.(5.56) +⋯

Using the definition of generalized fractional derivative 𝑛

𝑑

𝜌

𝜌

( 𝑎+𝐷𝑥𝛼 𝑓)(𝑥) = (𝑥1−𝜌 𝑑𝑥) ( 𝑎+𝐼𝑥𝑛−𝛼 𝑓)(𝑥), 𝜌𝛼−𝑛+1

𝑛

𝑑

𝑥

= Γ(𝑛−𝛼) (𝑥1−𝜌 𝑑𝑥) ∫𝑎 and

(5.57)

𝑡 𝜌−1 𝑓(𝑡)d𝑡, (𝑥 𝜌 −𝑡 𝜌 )𝛼

Γ(𝑘+1)

𝐷𝑥𝛼 𝑥 𝑘 = Γ(𝑘−𝛼+1) 𝑥 𝑘−𝛼 .

(5.58)

And then using Eq. (5.57) and Eq. (5.58) into Eq. (5.56), we have (𝑘1 + 𝜌𝛼) 𝑢(𝑥) =

sin(𝛼𝜋)

𝑘 Γ(𝛼)Γ( 1 +1) 𝜌

𝑘 Γ( 1 +𝛼+1) 𝜌

𝑐1 𝑥

𝑘1 +𝛼𝜌−1

𝜋

+ ⋯ + (𝑘𝑛 + 𝜌𝛼)

[ 𝑢(𝑥) =

+ (𝑘2 + 𝜌𝛼)

𝑘 Γ(𝛼)Γ( 𝑛 +1) 𝜌

𝑘 Γ( 𝑛 +𝛼+1)

𝑐𝑛 𝑥

𝑘 Γ(𝛼)Γ( 2 +1) 𝜌 𝑘2 Γ( +𝛼+1) 𝜌

𝑘𝑛 +𝛼𝜌−1

𝑐2 𝑥 𝑘2 +𝛼𝜌−1 .

+⋯

𝜌

sin(𝛼𝜋) 𝜋

∑∞ 𝑛=1(𝑘𝑛

+ 𝜌𝛼)

𝑘 Γ(𝛼)Γ( 𝑛 +1) 𝜌 𝑘𝑛 Γ( +𝛼+1) 𝜌

𝑐𝑛 𝑥 𝑘𝑛+𝛼𝜌−1 .

]

(5.59)

5.4.2. Abel’s Integral Equations Example 5.26. Consider the following generalized linear Abel’s integral equation [229] as 6

25

𝑥

1

𝑥 6 = ∫0 25

1 (𝑥 5 −𝑡 5 )6

𝑢(𝑡)d𝑡,

(5.60)

Taking 1

𝛼 = 6 , 𝑘1 =

25 6

6

, 𝑐1 = 25 , 𝜌 = 5.

Using the above information in Eq. (5.60), we obtained 1

𝑢(𝑥) =

sin( 𝜋) 25 6 𝜋

1

( 6 + 5 (6)) [

1 25 +1) 6 6 6∗5 1 25 Γ( + +1) 25 6 6∗5

Γ( )Γ(

25

1

𝑥 6 −5(6)−1 ],

𝑢(𝑥) = 𝑥 4 , This is the exact solution.


𝑥

𝑥 2 = ∫0

1 √𝑥 2 −𝑡 2

𝑢(𝑡)d𝑡,

(5.61)

Taking 1

𝛼 = 2 , 𝑘1 = 2, 𝑐1 = 1, 𝜌 = 2. Using the above information in Eq. (5.61), we obtained 𝑢(𝑥) = 𝑢(𝑥) =

1 2

sin( 𝜋)

1

(2 + 2 (2)) [

𝜋 4𝑥 2 𝜋

1 2 2 2 1 2 Γ( + +1) 2 2

Γ( )Γ( +1)

1

𝑥 2−2(2)−1 ],

.

This is the exact solution.

Example 5.28. Consider the following generalized linear Abel’s integral equation [229] as 2 3

𝑥

1

𝜋𝑥 3 = ∫0

√𝑥 2 −𝑡 2

𝑢(𝑡)d𝑡,

(5.62)

Taking 1

2

𝛼 = 2 , 𝑘1 = 3, 𝑐1 = 3 𝜋, 𝜌 = 2. Using the above information in Eq. (5.62), we obtained 𝑢(𝑥) =

1 2

sin( 𝜋) 𝜋

1

1 3 2 2 1 3 Γ( + +1) 2 2

Γ( )Γ( +1)

(3 + 2 ( )) [ 2

1

𝑥 3−2(2)−1 ],

𝑢(𝑥) = 𝜋𝑥 3 , This is the exact solution.

Example 5.29. Consider the following generalized nonlinear Abel’s integral equation [229] as 64

11

𝑥

𝑥 2 = ∫0 231

1 1

(𝑥 2 −𝑡 2 )4

𝑢(𝑡)d𝑡.

(5.63)

Taking 1

𝛼 = 4 , 𝑘1 =

11 2

64

, 𝑐1 = 231 , 𝜌 = 2.

Using the above information in Eq. (5.63), we obtained 1

𝑢(𝑥) =

sin( 𝜋) 11 4 𝜋

1

( 2 + 2 (4)) [

1 11 4 4 1 11 Γ( + +1) 4 4

Γ( )Γ( +1)

𝑢(𝑥) = 𝑥 5 . This is the exact solution. 771

11

1

𝑥 2 −2(4)−1 ],

5.5. New scheme Using Taylor’s Series Using Taylor’s series we, proposed a new algorithm for finding the exact solutions of linear nonlinear Abel’s integral equations. It is observed that the proposed scheme is extremely simple, highly effective, utmost accurate and can be extended to other singular problems.

5.5.1. Methodology Consider the Abel’s equations 𝑥 𝑢(𝑡)

𝑓(𝑥) = ∫0

(𝑥−𝑡)𝛼

d𝑡, 𝑥 > 0, 0 < 𝛼 ≤ 1.

Proceeding as before we have the exact solution 𝑢(𝑥) =

𝑥 1 𝑓(𝑡)d𝑡, ∫ 𝑑𝑥 0 (𝑥−𝑡)1−𝛼


(5.64)

1

Decompose 𝐾(𝑥, 𝑡) = (𝑥−𝑡)1−𝛼 , into Taylor’s series Eq. (5.64) becomes 𝑢(𝑥) =

𝑥 𝑥 [ ∫ 0 𝑑𝑥


𝛼−1

1

− (𝛼 − 1)𝑥 𝛼−2 𝑡 + 2 (𝛼 − 1) (𝛼 − 2)𝑥 𝛼−3 𝑡 2 − ⋯

] 𝑓(𝑡)d𝑡,

(5.65)

Remark: If we have a nonlinear Abel’s Integral Equation i. e., 𝑥 𝐹(𝑢(𝑡))

𝑓(𝑥) = ∫0

(𝑥−𝑡)𝛼

d𝑡, 𝑥 > 0, 0 < 𝛼 ≤ 1.

(5.66)

First we using the transformation 𝑣(𝑥) = 𝐹(𝑢(𝑥)), convert Eq. (5.31) into linear then solve proceeding as discuss in analysis, at the end we use the retransformation 𝑢(𝑥) = 𝐹 −1 (𝑣(𝑥)).

5.5.2. Abel’s Integral Equations Example 5.30. Consider the following linear Abel’s integral equation [229] as 𝑥

𝜋

𝑥 = ∫0 2 1

𝑢(𝑥) =

1 √𝑥−𝑡

sin( 𝜋) 𝑑 2 𝜋

𝑥

𝑢(𝑡)d𝑡, 1

(5.67) 1

1

1 1

1

1

𝜋

−1 −2 −3 ∫ [𝑥 2 − (2 − 1) 𝑥 2 𝑡 + 2 (2 − 1) (2 − 2) 𝑥 2 𝑡 2 − ⋯ ] ( 2 𝑡) d𝑡, 𝑑𝑥 0

If we take the first 100 term of Taylor’s series then 𝑢(𝑥) = 0.91575623√𝑥. If we take the first 500 term of Taylor’s series then 772

𝑢(𝑥) = 0.96218765√𝑥. If we take the first 1000 term of Taylor’s series then 𝑢(𝑥) = 0.97325039√𝑥. If we take the first 5000 term of Taylor’s series then 𝑢(𝑥) = 0.98803283√𝑥. If we take the first 10000 term of Taylor’s series then 𝑢(𝑥) = 0.99153754√𝑥. If we take the first 30000 term of Taylor’s series then 𝑢(𝑥) = 0.99511405√𝑥. The exact solution is given as 𝑢(𝑥) = √𝑥.

Fig 5.1. Comparison of Exact and Approximate solutions of Eq. (5.67)


𝜋𝑥 = ∫0 2

𝑢(𝑥) =

1 2


𝑢(𝑡)d𝑡,

(5.68a)

(𝑥−𝑡)3 𝑥

2

2

2

1 2

2

2

−1 −2 −3 ∫ [𝑥 3 − (3 − 1) 𝑥 3 𝑡 + 2 (3 − 1) (3 − 2) 𝑥 3 𝑡 2 − ⋯ ] (𝜋𝑡)d𝑡, 𝑑𝑥 0

If we take the first 100 term of Taylor’s series then 2

𝑢(𝑥) = 1.26171581𝑥 3 . If we take the first 500 term of Taylor’s series then 2


2




𝑢(𝑥) = 1.29820105𝑥 3 . The exact solution is given by 𝑢(𝑥) =

3√3 4

2

𝑥3.

Fig 5.2. Comparison of Exact and Approximate solutions of Eq. (5.68a)


1 + 𝑥 = ∫0 1

𝑢(𝑥) =


1 1

𝑢(𝑡)d𝑡,

(5.68b)

(𝑥−𝑡)4 𝑥

1

1

1

1 1

1

1

−1 −2 −3 ∫ [𝑥 4 − (4 − 1) 𝑥 4 𝑡 + 2 (4 − 1) (4 − 2) 𝑥 4 𝑡 2 − ⋯ ] (1 + 𝑡)d𝑡, 𝑑𝑥 0


1

𝑢(𝑥) = 0.16705016𝑥 − 4 + 0.6174624𝑥 4 . If we take the first 500 term of Taylor’s series then 3

1

𝑢(𝑥) = 0.18624378𝑥 − 4 + 0.70621699𝑥 4 . If we take the first 1000 term of Taylor’s series then

774

3

1


1


1


1

𝑢(𝑥) = 0.2150790790𝑥 − 4 + 0.8503163158𝑥 4 . The exact solution is given by 𝑢(𝑥) =

1 3 √2 𝜋𝑥 4

2√2

+

𝜋

1

𝑥4.

Fig 5.3. Comparison of Exact and Approximate solutions of Eq. (5.68b)

Example 5.33. Consider the following nonlinear Abel’s integral equation [229] as 𝑥

1 + 𝑥 = ∫0 1

𝑢(𝑥) =


1 √𝑥−𝑡 𝑥

𝑢(𝑡)d𝑡,

1

(5.69)

1

1

1 1

1

1

−1 −2 −3 ∫ [𝑥 2 − (2 − 1) 𝑥 2 𝑡 + 2 (2 − 1) (2 − 2) 𝑥 2 𝑡 2 − ⋯ ] (1 + 𝑡)d𝑡, 𝑑𝑥 0


1


1


1

𝑢(𝑥) = 0.31263155𝑥 − 2 + 0.61959045𝑥 2 . 775


1


1


1

𝑢(𝑥) = 0.31851401𝑥 − 2 + 0.63323239𝑥 2 . The exact solution is given by 𝑢(𝑥) =

1

2

1

𝑥2. 1 +

𝜋𝑥 2

𝜋


5.6. Adomian’s Decomposition Method Using Fractional Calculus In this session, we develop a new algorithm which is mainly based upon RiemannLiouville fractional derivatives and fractional integral to obtain solutions of linear and nonlinear generalized weakly singular equations. It is observed that the proposed scheme is extremely simple, highly effective, utmost accurate and can be extended to other singular problems

5.6.1. Methodology Consider the Weakly Singular equations, we have 𝑥

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫0


(5.70)

Eq. (5.70) can be converted into initial value problem 𝑢′ (𝑥) = 𝑓 ′ (𝑥) + 𝜆𝐷𝑥𝛼 𝑢(𝑡),

776

𝑢(0) = 𝑓(0).

According to the Adomian decomposition method 𝑥

𝑢(𝑥) = 𝑓(0) + ∫0 [𝑓(𝑥) + 𝜆𝐷𝑥𝛼 𝑢(𝑡) ]d𝑥, 𝑥

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫0 [𝐷𝑥𝛼 𝑢(𝑥) ]d𝑥 .

The recurrence relation read as 𝑢0 (𝑥) = 𝑓(𝑥), 𝑥

𝑢𝑘+1 (𝑥) = 𝜆 ∫0 [𝐷𝑥𝛼 𝑢𝑘 (𝑥) ]d𝑥, 𝑘 ≥ 0.

Using the Riemann-Liouville fractional and derivative and fractional integral 1

𝑥 𝑢(𝑡) d𝑡, (𝑥−𝑡)𝛼

𝑑

𝐷𝑥𝛼 𝑢(𝑥) = Γ(1−𝛼) 𝑑𝑥 ∫0 Γ(𝑘+1)

𝐷𝑥𝛼 𝑥 𝑘 = Γ(𝑘−𝛼+1) 𝑥 𝑘−𝛼 , Γ(𝑘+1)

and 𝐽𝑥𝛼 𝑥 𝑘 = Γ(𝑘+𝛼+1) 𝑥 𝑘+𝛼 . We have the recurrence relation 1

1 𝑘𝑛 𝑢0 (𝑥) = 𝑇(𝑓(𝑥)) = ∑∞ 𝑛=0 𝑐𝑛 𝑥 , 𝑚 𝑢𝑚 (𝑥) = 𝜆 ∑∞ 𝑛=0 𝑐𝑛

𝑚 +1)Γ(1−𝛼) 𝑚 Γ(𝑘𝑛 𝑥 𝑘𝑛 −𝛼+1 , 𝑚 𝑚 Γ(𝑘𝑛 −𝛼+2)

≥ 1.

5.6.2. Weakly Singular Integral Equation Example 5.34. Consider the following weakly singular integral equation [229] as 16

5

𝑥

𝑢(𝑥) = 𝑥 2 + 15 𝑥 2 − ∫0

1 𝑢(𝑡)d𝑡, √𝑥−𝑡

which is equivalent to initial value problem as 80

3

𝑢′ (𝑥) = 2𝑥 + 30 𝑥 2 − 𝐷𝑥𝛼 𝑢(𝑡), 𝑢(0) = 𝑓(0).

According to Adomian Decomposition method the recurrence relation read as 16

5

𝑢0 (𝑥) = 𝑥 2 + 15 𝑥 2 , 𝑥

𝑢𝑘+1 (𝑥) = − ∫0 𝐷𝑥𝛼 𝑢𝑘 (𝑡)d𝑥 , 𝑘 ≥ 0.

Now according to the proposed algorithm, 1

1 𝑘𝑛 𝑢0 (𝑥) = ∑∞ 𝑛=0 𝑐𝑛 𝑥 , 𝑚 𝑢𝑚 (𝑥) = − ∑∞ 𝑛=0 𝑐𝑛

𝑚 +1)Γ(1−𝛼) 𝑚 Γ(𝑘𝑛 𝑥 𝑘𝑛 −𝛼+1 , 𝑚 𝑚 −𝛼+2) Γ(𝑘𝑛

We have following approximations

777

≥ 1.

𝑢0 (𝑥) = 𝑥 2 +

16 5 𝑥 2, 15 5

16

1

𝑢1 (𝑥) = − 15 𝑥 2 − 3 𝜋𝑥 3 , 1 3

𝑢2 (𝑥) = 𝜋𝑥 3 − 𝑢3 (𝑥) =

7 32 𝜋𝑥 2 105

7 32 𝜋𝑥 2 , 105

1 𝜋𝑥 4 , 12

−

. . .. 𝑢(𝑥) = 𝑥 2 +

16 5 𝑥2 15

16 5 𝑥2 15

−

1 3

1 3

− 𝜋𝑥 3 𝜋𝑥 3 −

7 32 𝜋𝑥 2 105

+

7 32 𝜋𝑥 2 105

−

1 𝜋𝑥 4 , 12

𝑢(𝑥) = 𝑥 2 .

Example 5.35. Consider the following weakly singular integral equation [229] as 7

32

3

4

𝑥

1

𝑢(𝑥) = 1 − 2𝑥 − 21 𝑥 4 + 3 𝑥 4 − ∫0

1

𝑢(𝑡)d𝑡.

(𝑥−𝑡)4

According to Adomian Decomposition method the recurrence relation read as 𝑢0 (𝑥) = 1 − 2𝑥 −

32 7 𝑥4 21

3

4 3

+ 𝑥 4,

𝑥

𝑢𝑘+1 (𝑥) = − ∫0 𝐷𝑥𝛼 𝑢𝑘 (𝑡)d𝑥, 𝑘 ≥ 0.


1 𝑘𝑛 𝑢0 (𝑥) = ∑∞ 𝑛=0 𝑐𝑛 𝑥 , 𝑚 𝑢𝑚 (𝑥) = −𝜆 ∑∞ 𝑛=0 𝑐𝑛


≥ 1.

We have following approximations 7

32

3

4

𝑢0 (𝑥) = 1 − 2𝑥 − 21 𝑥 4 + 3 𝑥 4 , 𝑢1 (𝑥) =

32 7 𝑥4 21

𝑢2 (𝑥) = − 𝑢3 (𝑥) =

4 3 − 𝑥4 3

3 2 16Γ( ) 4

15√𝜋

5 2

𝑥 +

3 4 256Γ( ) √2 4

585𝜋

+

3 2 4

16Γ( ) 15√𝜋 3 2 4Γ( ) 4

3√𝜋

13

𝑥4 −

5 2

𝑥 − 3 2

𝑥 −

3 4 32Γ( ) √2 4

45𝜋

3 2 4

4Γ( ) 3 √𝜋

3

𝑥 2,

3 4

256Γ( ) √2 4 585𝜋 9

1

𝑥

13 4

3 4

+

32Γ( ) √2 4

3 4

45𝜋 1

9

𝑥 4, 3 4

𝑥 4 + 24 √2Γ (4) 𝑥 4 − 12 √2Γ (4) 𝑥 3 ,

. . ..

778

𝑢(𝑥) = 1 − 2𝑥 − −

3 2 16Γ( ) 4

15√𝜋

5 2

𝑥 +

3 4 32Γ( ) √2 4

45𝜋

32 7 − 𝑥4 21

3 2 4Γ( ) 4

3 2

𝑥 −

3√𝜋

9

+

4 3 𝑥4 3

+

32 7 𝑥4 21

3 4 256Γ( ) √2 4

585𝜋

3 4

1

𝑥

− 13 4

4 3 𝑥4 3

+

+

3 2 4

16Γ( ) 15√𝜋

3 4 32Γ( ) √2 4

45𝜋

9 4

5 2

𝑥 −

𝑥 +

3 2 4

4Γ( ) 3√𝜋

3

𝑥2

3 4

256Γ( ) √2 4 585𝜋

13

𝑥4

3 4

1

𝑥 4 + 24 √2Γ (4) 𝑥 4 − 12 √2Γ (4) 𝑥 3 ,

𝑢(𝑥) = 1 − 2𝑥.


𝑥

5

𝑢(𝑥) = 𝑥 2 − 16 𝜋𝑥 3 + ∫0

1 𝑢(𝑡)d𝑡. √𝑥−𝑡


𝑢0 (𝑥) = 𝑥 2 −

5 𝜋𝑥 3 , 16

𝑥

𝑢𝑘+1 (𝑥) = ∫0 𝐷𝑥𝛼 𝑢𝑘 (𝑡)d𝑥, 𝑘 ≥ 0.


1 𝑘𝑛 𝑢0 (𝑥) = ∑∞ 𝑛=0 𝑐𝑛 𝑥 , 𝑚 𝑢𝑚 (𝑥) = 𝜆 ∑∞ 𝑛=0 𝑐𝑛

𝑚 +1)Γ(1−𝛼) 𝑚 −𝛼+1 Γ(𝑘𝑛 𝑘𝑛 ,𝑚 𝑚 −𝛼+2) 𝑥 Γ(𝑘𝑛

≥ 1.


5

𝑢0 (𝑥) = 𝑥 2 − 16 𝜋𝑥 3 , 𝑢1 (𝑥) =

5 𝜋𝑥 3 16 2 7

7

7

2 7

− 𝜋𝑥 2 ,

𝑢2 (𝑥) = 𝜋𝑥 2 −

5 𝜋𝑥 4 , 64

5

9

4

𝑢3 (𝑥) = 64 𝜋𝑥 4 − 63 𝜋𝑥 2 ,

. . .. 5

5

5

2

7

2

7

5

5

4

9

𝑢(𝑥) = 𝑥 2 − 16 𝜋𝑥 3 + 16 𝜋𝑥 3 − 7 𝜋𝑥 2 + 7 𝜋𝑥 2 − 64 𝜋𝑥 4 +64 𝜋𝑥 4 − 63 𝜋𝑥 2 , 5

𝑢(𝑥) = 𝑥 2 .

Example 5.37. Consider the following weakly singular integral equation [229] as 𝑥

𝑢(𝑥) = 3 − 6√𝑥 + ∫0


According to Adomian Decomposition method the recurrence relation read as 𝑢0 (𝑥) = 3 − 6√𝑥,

779

𝑥

𝑢𝑘+1 (𝑥) = ∫0 𝐷𝑥𝛼 𝑢𝑘 (𝑡)d𝑥, 𝑘 ≥ 0.




≥ 1.

We have following approximations 𝑢0 (𝑥) = 3 − 6√𝑥, 𝑢1 (𝑥) = 6√𝑥 − 3𝜋𝑥, 3

𝑢2 (𝑥) = 3𝜋𝑥 − 4𝜋𝑥 2 , 3

3

𝑢3 (𝑥) = 4𝜋𝑥 2 − 2 𝜋𝑥 2 ,

. . .. 3

3

3

𝑢(𝑥) = 3 − 6√𝑥 + 6√𝑥 − 3𝜋𝑥 +3𝜋𝑥 − 4𝜋𝑥 2 + 4𝜋𝑥 2 − 2 𝜋𝑥 2 , 𝑢(𝑥) = 3.


16

𝑥

𝑢(𝑥) = 1 + 𝑥 2 − 2√𝑥 − 15 𝑥 2 + ∫0

1 1

𝑢(𝑡)d𝑡.

(𝑥−𝑡)2


5

𝑢0 (𝑥) = 1 + 𝑥 2 − 2√𝑥 − 15 𝑥 2 , 𝑥

𝑢𝑘+1 (𝑥) = ∫0 𝐷𝑥𝛼 𝑢𝑘 (𝑡)d𝑥, 𝑘 ≥ 0.



𝑚 +1)Γ(1−𝛼) 𝑚 −𝛼+1 Γ(𝑘𝑛 𝑘𝑛 𝑥 ,𝑚 𝑚 Γ(𝑘𝑛 −𝛼+2)

We have following approximations 𝑢0 (𝑥) = 1 + 𝑥 2 − 2√𝑥 − 16

16 5 𝑥 2, 15

5

1

𝑢1 (𝑥) = 2√𝑥 + 15 𝑥 2 − 𝜋𝑥 − 3 𝜋𝑥 3 , 1

4

3

7

1

7

32

𝑢2 (𝑥) = 𝜋𝑥 + 3 𝜋𝑥 3 − 3 𝜋𝑥 2 − 105 𝜋𝑥 2 , 4

3

32

1

𝑢3 (𝑥) = 3 𝜋𝑥 2 + 105 𝜋𝑥 2 − 2 𝜋𝑥 2 − 12 𝜋𝑥 4 ,

780

≥ 1.

. . .. 16

5

16

5

1

1

𝑢(𝑥) = 1 + 𝑥 2 − 2√𝑥 − 15 𝑥 2 + 2√𝑥 + 15 𝑥 2 − 𝜋𝑥 − 3 𝜋𝑥 3 + 𝜋𝑥 + 3 𝜋𝑥 3 4

3

7

32

3

4

32

7

1

1

− 3 𝜋𝑥 2 − 105 𝜋𝑥 2 +3 𝜋𝑥 2 + 105 𝜋𝑥 2 − 2 𝜋𝑥 2 − 12 𝜋𝑥 4 + ⋯,

𝑢(𝑥) = 1 + 𝑥 2 .

5.7. Modified Adomian’s Decomposition Method Using Fractional Calculus In this session, we develop a new algorithm modified Adomian’s decomposition method using fractional calculus which is mainly based upon Riemann-Liouville fractional derivatives and fractional integral to obtain solutions of linear and nonlinear generalized weakly singular equations. It is observed that the proposed scheme is extremely simple, highly effective, utmost accurate and can be extended to other singular problems

5.7.1. Methodology Consider the Weakly Singular equations, we have 𝑥

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫0


(5.71)

Eq. (5.71) can be converted into initial value problem 𝑢′ (𝑥) = 𝑓 ′ (𝑥) + 𝜆𝐷𝑥𝛼 𝑢(𝑡), 𝑢(0) = 𝑓(0).

According to the Adomian decomposition method 𝑥

𝑢(𝑥) = 𝑓(0) + ∫0 [𝑓(𝑥) + 𝜆𝐷𝑥𝛼 𝑢(𝑡) ]d𝑥, 𝑥

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫0 [𝐷𝑥𝛼 𝑢(𝑥) ]d𝑥 .

The recurrence relation read as 𝑢0 (𝑥) = 𝑓(𝑥), 𝑥

𝑢1 (𝑥) = 𝑓1 (𝑥) + 𝜆 ∫0 [𝐷𝑥𝛼 𝑢0 (𝑥) ]d𝑥, 𝑥

𝑢𝑘+1 (𝑥) = 𝜆 ∫0 [𝐷𝑥𝛼 𝑢𝑘 (𝑥) ]d𝑥, 𝑘 ≥ 1.

Using the Riemann-Liouville fractional and derivative and fractional integral 𝐷𝑥𝛼 𝑢(𝑥) =

1 𝑑 𝑥 𝑢(𝑡) d𝑡, ∫ Γ(1−𝛼) 𝑑𝑥 0 (𝑥−𝑡)𝛼

Γ(𝑘+1)

𝐷𝑥𝛼 𝑥 𝑘 = Γ(𝑘−𝛼+1) 𝑥 𝑘−𝛼 ,

781

Γ(𝑘+1)

and 𝐽𝑥𝛼 𝑥 𝑘 = Γ(𝑘+𝛼+1) 𝑥 𝑘+𝛼 . We have the recurrence relation 1

1 𝑘𝑛 𝑢0 (𝑥) = 𝑇(𝑓0 (𝑥)) = ∑∞ 𝑛=0 𝑐𝑛 𝑥 , 2

2 𝑘𝑛 𝑢1 (𝑥) = 𝑇(𝑓1 (𝑥)) + ∑∞ 𝑛=0 𝑐𝑛 𝑥 , 𝑚 𝑢𝑚 (𝑥) = 𝜆 ∑∞ 𝑛=0 𝑐𝑛


≥ 2.

5.7.2 Weakly Singular Integral Equation Example 5.39. Consider the following weakly singular integral equation [229] as 16

5

𝑥

𝑢(𝑥) = 𝑥 2 + 15 𝑥 2 − ∫0

1 𝑢(𝑡)d𝑡, 𝑥−𝑡 √

which is equivalent to initial value problem as 𝑢′ (𝑥) = 2𝑥 +

80 3 𝑥2 30

− 𝐷𝑥𝛼 𝑢(𝑡),

𝑢(0) = 𝑓(0).

According to Adomian Decomposition method the recurrence relation read as 𝑢0 (𝑥) = 𝑥 2 , 16

5

𝑥

𝑢0 (𝑥) = 15 𝑥 2 − ∫0 𝐷𝑥𝛼 𝑢0 (𝑡)d𝑥 , 𝑥

𝑢𝑘+1 (𝑥) = ∫0 𝐷𝑥𝛼 𝑢𝑘 (𝑡)d𝑥, 𝑘 ≥ 1.

Now according to the proposed algorithm, 𝑢0 (𝑥) = 𝑥 2 , 16

5

𝑚 𝑢1 (𝑥) = 15 𝑥 2 + ∑∞ 𝑛=0 𝑐𝑛 𝑚 𝑢𝑚 (𝑥) = ∑∞ 𝑛=0 𝑐𝑛

1 +1)Γ(1−𝛼) Γ(𝑘𝑛 1 −𝛼+2) Γ(𝑘𝑛

1

𝑥 𝑘𝑛 −𝛼+1 ,

𝑚 +1)Γ(1−𝛼) 𝑚 −𝛼+1 Γ(𝑘𝑛 𝑘𝑛 𝑥 ,𝑚 𝑚 Γ(𝑘𝑛 −𝛼+2)

≥ 2.

We have following approximations 𝑢0 (𝑥) = 𝑥 2 , 𝑢1 (𝑥) =

16 5 𝑥2 15

−

16 5 𝑥2 15

= 0,

𝑢𝑛 (𝑥) = 0, ∀ 𝑛 ≥ 1.

Therefore we have 𝑢(𝑥) = 𝑥 2 .


7

4

3

𝑥

𝑢(𝑥) = 1 − 2𝑥 − 21 𝑥 4 + 3 𝑥 4 − ∫0

1 1

𝑢(𝑡)d𝑡.

(𝑥−𝑡)4

782

According to Adomian Decomposition method the recurrence relation read as 𝑢0 (𝑥) = 1 − 2𝑥, 7

32

4

3

𝑥

𝑢0 (𝑥) = − 21 𝑥 4 + 3 𝑥 4 − ∫0 𝐷𝑥𝛼 𝑢0 (𝑡)d𝑥 , 𝑥

𝑢𝑘+1 (𝑥) = − ∫0 𝐷𝑥𝛼 𝑢𝑘 (𝑡)d𝑥, 𝑘 ≥ 1.

Now according to the proposed algorithm, 𝑢0 (𝑥) = 1 − 2𝑥, 7

32

4

3

𝑚 𝑢1 (𝑥) = − 21 𝑥 4 + 3 𝑥 4 − ∑∞ 𝑛=0 𝑐𝑛 𝑚 𝑢𝑚 (𝑥) = − ∑∞ 𝑛=0 𝑐𝑛

1 +1)Γ(1−𝛼) Γ(𝑘𝑛 1 −𝛼+2) Γ(𝑘𝑛


1

𝑥 𝑘𝑛 −𝛼+1 , ≥ 2.

We have following approximations 𝑢0 (𝑥) = 1 − 2𝑥, 𝑢1 (𝑥) =

32 7 𝑥4 21

4 3

3

− 𝑥4 −

32 7 𝑥4 21

4 3

3

+ 𝑥 4 = 0,

𝑢𝑛 (𝑥) = 0, ∀ 𝑛 ≥ 1.

Therefore we have 𝑢(𝑥) = 1 − 2𝑥.


𝑥

5

𝑢(𝑥) = 𝑥 2 − 16 𝜋𝑥 3 + ∫0



𝑢0 (𝑥) = 𝑥 2 , 𝑥

5

𝑢0 (𝑥) = − 16 𝜋𝑥 3 + ∫0 𝐷𝑥𝛼 𝑢0 (𝑡)d𝑥 , 𝑥

𝑢𝑘+1 (𝑥) = ∫0 𝐷𝑥𝛼 𝑢𝑘 (𝑡)d𝑥 , 𝑘 ≥ 1.


𝑢0 (𝑥) = 𝑥 2 , 5

𝑚 𝑢1 (𝑥) = − 16 𝜋𝑥 3 − ∑∞ 𝑛=0 𝑐𝑛 𝑚 𝑢𝑚 (𝑥) = ∑∞ 𝑛=0 𝑐𝑛

1 +1)Γ(1−𝛼) Γ(𝑘𝑛 1 −𝛼+2) Γ(𝑘𝑛



𝑢0 (𝑥) = 𝑥 2 , 5

5

𝑢1 (𝑥) = 16 𝜋𝑥 3 − 16 𝜋𝑥 3 = 0,

783

1

𝑥 𝑘𝑛 −𝛼+1 , ≥ 2.

𝑢𝑛 (𝑥) = 0, ∀ 𝑛 ≥ 1.

Therefore we have 5

𝑢(𝑥) = 𝑥 2 .

Example 5.42. Consider the following weakly singular integral equation [229] as 𝑥

𝑢(𝑥) = 3 − 6√𝑥 + ∫0

1 𝑢(𝑡)d𝑡. 𝑥−𝑡 √

According to Adomian Decomposition method the recurrence relation read as 𝑢0 (𝑥) = 3, 𝑥

𝑢0 (𝑥) = −6√𝑥 + ∫0 𝐷𝑥𝛼 𝑢0 (𝑡)d𝑥 , 𝑥

𝑢𝑘+1 (𝑥) = ∫0 𝐷𝑥𝛼 𝑢𝑘 (𝑡)d𝑥 , 𝑘 ≥ 1.

Now according to the proposed algorithm, 𝑢0 (𝑥) = 3, 𝑚 𝑢1 (𝑥) = −6√𝑥 − ∑∞ 𝑛=0 𝑐𝑛 𝑚 𝑢𝑚 (𝑥) = ∑∞ 𝑛=0 𝑐𝑛

1 +1)Γ(1−𝛼) Γ(𝑘𝑛

1

𝑥 𝑘𝑛 −𝛼+1 ,

1 −𝛼+2) Γ(𝑘𝑛


≥ 2.

We have following approximations 𝑢0 (𝑥) = 3, 𝑢1 (𝑥) = −6√𝑥 + 6√𝑥 = 0, 𝑢𝑛 (𝑥) = 0, ∀ 𝑛 ≥ 1.

Therefore we have 𝑢(𝑥) = 3.


16

𝑥

𝑢(𝑥) = 1 + 𝑥 2 − 2√𝑥 − 15 𝑥 2 + ∫0

1 1

𝑢(𝑡)d𝑡.

(𝑥−𝑡)2

According to Adomian Decomposition method the recurrence relation read as 𝑢0 (𝑥) = 1 + 𝑥 2 , 16

5

𝑥

𝑢0 (𝑥) = −2√𝑥 − 15 𝑥 2 + ∫0 𝐷𝑥𝛼 𝑢0 (𝑡)d𝑥 , 𝑥

𝑢𝑘+1 (𝑥) = ∫0 𝐷𝑥𝛼 𝑢𝑘 (𝑡)d𝑥 .

Now according to the proposed algorithm, 𝑢0 (𝑥) = 1 + 𝑥 2 , 16

5

𝑚 𝑢1 (𝑥) = −2√𝑥 − 15 𝑥 2 + ∑∞ 𝑛=0 𝑐𝑛

1 +1)Γ(1−𝛼) Γ(𝑘𝑛 1 −𝛼+2) Γ(𝑘𝑛

784

1

𝑥 𝑘𝑛 −𝛼+1 ,

𝑚 𝑢𝑚 (𝑥) = ∑∞ 𝑛=0 𝑐𝑛


≥ 2.

We have following approximations 𝑢0 (𝑥) = 1 + 𝑥 2 , 16

5

16

5

𝑢1 (𝑥) = −2√𝑥 − 15 𝑥 2 + 2√𝑥 + 15 𝑥 2 = 0, 𝑢𝑛 (𝑥) = 0, ∀ 𝑛 ≥ 1.

Therefore we have 𝑢(𝑥) = 1 + 𝑥 2 .

5.8. Kravchuk Polynomials Method (KPM) I develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Kravchuk polynomials [183]. Kravchuk polynomials is given as (−1)𝑚 𝑥!

(𝑎−𝑥)!

𝐾(𝑥) = ∑𝑛𝑚=0 [𝑚!(𝑥−𝑚)! (𝑛−𝑚)!(𝑎−𝑥−𝑛+𝑚)!]. First four Kravchuk polynomials is given as 𝐾0 (𝑥) = 1, 𝐾1 (𝑥) = 𝑎 − 2𝑥, 1

1

1

1

𝐾2 (𝑥) = 2 𝑎2 − 2𝑎𝑥 − 2 𝑎 + 2𝑥 2 , 1

4

2

𝐾3 (𝑥) = 6 𝑎3 − 𝑎2 𝑥 − 2 𝑎2 + 2𝑎𝑥 2 + 𝑎𝑥 + 3 𝑎 − 3 𝑥 3 − 3 𝑥,

5.8.1. Methodology 5.8.1.1. Integral Equation of the 1st Kind Consider the integral equation of the 1st kind is given as 𝛽(𝑥)

𝑓(𝑥) = 𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡 , 𝑎 ≤ 𝑥 ≤ 𝑏.

(5.72)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, to find an approximate solution 𝑢̃(𝑥) of Eq. (5.72). For this, we assume that 𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘 (𝑥),

(5.73)

where 𝐾𝑘 (𝑥) are Kravchuk polynomials of degree 𝑘 defined in equation (1) and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.73) into (5.72), we get 785

𝑥

𝑓(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝐾𝑘 (𝑡)d𝑡,

(5.74)

Then the Galerkin equations are obtained by multiplying both sides of (4) by 𝐾𝑗 (𝑥) and then integrating with respect to 𝑥 from 𝑎 to 𝑏, we have 𝑏

𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝐾𝑗 (𝑥)d𝑥 = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐾𝑘 (𝑡)d𝑡)𝐾𝑗 (𝑡)d𝑥, 𝑗 = 0,1,2, … 𝑛,(5.75) Since in each equation, there are two integrals. The inner integrand of the left side is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.75) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.76)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐾𝑘 (𝑡)d𝑡 )𝐾𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛 𝑏

𝐹𝑗 = ∫𝑎 𝑓(𝑥) 𝐾𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛 Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.76) and substituting these values of parameters in (5.73), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.72).

5.8.1.2. Integral Equation of the 2nd Kind Consider the integral equation of the 2nd kind is 𝛽(𝑥)

𝑓(𝑥) = 𝑐(𝑥)𝑢(𝑥) + 𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡, 𝑎 ≤ 𝑥 ≤ 𝑏.

(5.77)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) and 𝑢(𝑥) being the known function and 𝜆 is a constant. Proceeding as before ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.78)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝐾𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝐾𝑘 (𝑡)d𝑡)𝐾𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛.

786

Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.78) and substituting these values of parameters in (5.73), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.77).

5.8.2. Volterra Integral Equation Consider the Volterra integral equation of the 1st Kind [229] as 𝑥

1

−𝑥 + 2 𝑥 2 + (1 + 𝑥) ln(1 + 𝑥) = ∫0 𝑢(𝑡)d𝑡 .

0≤𝑥≤1

(5.79)

The exact solution of Eq. (5.79) is 𝑢(𝑥) = 𝑥 + ln(1 + 𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥).

(5.80)

To solve Eq. (5.79) we take 𝑛 = 3. Eq. (5.80) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑3𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥),

(5.81)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 , 𝜓3 ]𝑇 .

and

Putting Eq. (5.81) into Eq. (5.79), we obtained 𝑥

1

−𝑥 + 2 𝑥 2 + (1 + 𝑥) ln(1 + 𝑥) = ∫0 ∑𝑛𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡.

(5.82)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.82) is 1

1

∫0 [−𝑥 + 2 𝑥 2 + (1 + 𝑥) ln(1 + 𝑥)] 𝛙𝑗 (𝑥)d𝑥 = 1

𝑥

∫0 [∫0 ∑𝑛𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡 ]𝛙𝑗 (𝑥)d𝑥.

(5.83)

for 𝑗 = 0,1,2,3. The matrix form of Eq. (5.83) is given as

=

.

After solving the above we get 𝛼0 = 0.00057703609, 𝛼1 = −0.96276944, 𝛼2 = −0.20460353, 𝛼3 = −0.08915597. 𝑢(𝑥) = 0.00057703609 + 1.984976199𝑥 − 0.409207057𝑥 2 + 0.1188746285𝑥 3 . 787


Table 5.4. Comparison of Exact and Approximate solutions of Eq. (5.79) obtained from Kravchuk Polynomials Method (KPM) _____________________________________________________________________________ 𝑥

Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00057700242314413165

5.77002E-04

0.1

0.19531017980432486004

0.19510145889907311785

2.08721E-04

0.2

0.38232155679395462621

0.38215500347237734931

1.66553E-04

0.3

0.56236426446749105204

0.56245088906340211607

8.66246E-05

0.4

0.73647223662121293050

0.73670236859249270815

2.30132E-04

0.5

0.90546510810816438198

0.90562269497999441557

1.57587E-04

0.6

1.07000362924573555365

1.06992512114625252836

7.85081E-05

0.7

1.23062825106217039623

1.23032290001161233655

3.05351E-04

0.8

1.38778666490211900819

1.38752928449641913017

2.57380E-04

0.9

1.54185388617239477599

1.54225752752101819923

4.03641E-04

1.0

1.69314718055994530942

1.69522088200575483378

2.07370E-03

_____________________________________________________________________________

For 𝑛 = 50. Eq. (5.80) is 50 𝑇 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑𝑘=0 𝐂 𝛙𝑘 (𝑥),

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼50 ]𝑇 , and

𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 , … , 𝜓50 ]𝑇 .

Proceeding as before we have 788

0.5000 [ ⋮ 0.2994

⋯ ⋱ ⋯

0.30296 0.68954 𝛼0 ⋮ ] [ ] = [ ]. ⋮ ⋮ 𝛼 50 0.23138 0.06421

After solving we obtained the approximate solution 𝑢(𝑥) = 1.0000 + ⋯ − 2.4590 × 10−39 .



Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00000000000000000000

2.76755E-28

0.1

0.19531017980432486004

0.19531017980432486004

4.04837E-29

0.2

0.38232155679395462621

0.38232155679395462621

2.80269E-30

0.3

0.56236426446749105204

0.56236426446749105204

4.28755E-29

0.4

0.73647223662121293050

0.73647223662121293050

9.50490E-30

0.5

0.90546510810816438198

0.90546510810816438198

5.06701E-29

0.6

1.07000362924573555365

1.07000362924573555365

5.50943E-29

0.7

1.23062825106217039623

1.23062825106217039623

1.46400E-29

0.8

1.38778666490211900819

1.38778666490211900819

8.07536E-29

0.9

1.54185388617239477599

1.54185388617239477599

3.55607E-29

1.0

1.69314718055994530942

1.69314718055994530942

1.08737E-26

_____________________________________________________________________________

Finally we take 𝑛 = 100. Eq. (5.80) is 789

100 𝑇 𝑢(𝑥) = ∑100 𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑𝑘=0 𝐂 𝛙𝑘 (𝑥),

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼100 ]𝑇 , and

𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 , … , 𝜓100 ]𝑇 .

Proceeding as before we have 𝐀101×101 𝒙101×1 = 𝐛101×1 .



Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00000000000000000000

3.24048E-49

0.1

0.19531017980432486004

0.19531017980432486004

3.20756E-49

0.2

0.38232155679395462621

0.38232155679395462621

2.39513E-50

0.3

0.56236426446749105204

0.56236426446749105204

4.78396E-49

0.4

0.73647223662121293050

0.73647223662121293050

5.10307E-49

0.5

0.90546510810816438198

0.90546510810816438198

6.38330E-49

0.6

1.07000362924573555365

1.07000362924573555365

7.84334E-49

0.7

1.23062825106217039623

1.23062825106217039623

9.22096E-49

0.8

1.38778666490211900819

1.38778666490211900819

5.83654E-49

0.9

1.54185388617239477599

1.54185388617239477599

1.89200E-48

1.0

1.69314718055994530942

1.69314718055994530942

3.42869E-46

_____________________________________________________________________________

790

5.8.3. Fredholm Integral Equation: Consider the Fredholm integral equation [229] of the form 1

𝑢(𝑥) = 4 + 45𝑥 + 26𝑥 2 − ∫0 (1 + 30𝑥𝑡 2 + 12𝑥 2 𝑡)𝑢(𝑡)d𝑡 , 0 ≤ 𝑥 ≤ 1. (5.84) The exact solution of Eq. (5.84) is 𝑢(𝑥) = 1 + 2𝑥 + 3𝑥 2 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥).

(5.85)


(5.86)

where 𝐂 = [𝛼0 , 𝛼1 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 ]𝑇 . Putting Eq. (5.86) into Eq. (5.84), we obtained 1

∑𝑛𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥) = 4 + 45𝑥 + 26𝑥 2 − ∫0 (1 + 30𝑥𝑡 2 + 12𝑥 2 𝑡) ∑𝑛𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡. (5.87) Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.87) is 1

1

∫0 ∑𝑛𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥)𝛙𝑗 (𝑥)d𝑥 = ∫0 [4 + 45𝑥 + 26𝑥 2 ]𝛙𝑗 (𝑥)d𝑥 1

1

+ ∫0 [∫0 (1 + 30𝑥𝑡 2 + 12𝑥 2 𝑡) ∑𝑛𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡] 𝛙𝑗 (𝑥)d𝑥 .(5.88) for 𝑗 = 0,1,2,3. The matrix form of Eq. (5.88) is given as =

.

After solving the above we get 46

𝛼0 = 91 , 𝛼1 = − 𝑢(𝑥) =

46 91

+

458 91

229 91

.

𝑥.

791


Table 5.7. Comparison of Exact and Approximate solutions of Eq. (5.84) obtained from Kravchuk Polynomials Method (KPM) _____________________________________________________________________________ x

Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

0.50549450549450549451

4.94505E-01

0.1

1.23000000000000000000

1.00879120879120879121

2.21209E-01

0.2

1.52000000000000000000

1.51208791208791208791

7.91209E-03

0.3

1.87000000000000000000

2.01538461538461538462

1.45385E-01

0.4

2.28000000000000000000

2.51868131868131868132

2.38681E-01

0.5

2.75000000000000000000

3.02197802197802197802

2.71978E-01

0.6

3.28000000000000000000

3.52527472527472527473

2.45275E-01

0.7

3.87000000000000000000

4.02857142857142857143

1.58571E-01

0.8

4.52000000000000000000

4.53186813186813186813

1.18681E-02

0.9

5.23000000000000000000

5.03516483516483516484

1.94835E-01

1.0

6.00000000000000000000

5.53846153846153846154

4.61538E-01

_____________________________________________________________________________

For 𝑛 = 2. Eq. (5.85) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥), where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 ]𝑇 . Proceeding as before we have 792

=

.

After solving we obtained the approximate solution. 3

𝛼0 = 1, 𝛼1 = −1, 𝛼2 = 2. 𝑢(𝑥) = 1 + 2𝑥 + 3𝑥 2 , which is exact solution.

5.8.4. Volterra Integro-Differential Equation: Consider the Volterra integrodifferential equation [229] as 𝑥

1

𝑢′′′ (𝑥) = 1 + 3! 𝑥 3 − e−𝑥 + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 , 0 ≤ 𝑥 ≤ 1

(5.89)

𝑢(0) = 1 = 𝑢′′ (0), 𝑢′ (0) = 0. The exact solution of Eq. (5.89) is 𝑢(𝑥) = cos h(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥).

(5.90)


(5.91)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and

𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 , 𝜓3 ]𝑇 .


1

[𝐂 𝑇 𝛙𝑘 (𝑥)]′′′ = 1 + 𝑥 3 − e−𝑥 + ∫0 (𝑥 − 𝑡) ∑𝑛𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡. 3!

(5.92)


1

1

∫0 ∑𝑛𝑘=0[𝐂 𝑇 𝛙𝑘 (𝑥)]′′′ 𝛙𝑗 (𝑥)d𝑥 = ∫0 [1 + 3! 𝑥 3 − e−𝑥 ] 𝛙𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 (𝑥 − 𝑡) ∑𝑛𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡]𝛙𝑗 (𝑥)d𝑥. for 𝑗 = 0,1,2,3. The matrix form of Eq. (5.93) is given as

793

(5.93)

=

.

After solving the above we get 𝛼0 = 0.99927069, 𝛼1 = 0.016014918, 𝛼2 = 0.21869639, 𝛼3 = −0.06951932520. 𝑢(𝑥) = 0.99927069 + 0.0143163809𝑥 + 0.4373927742𝑥 2 + 0.092692433𝑥 3 .


794


Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

0.99927070435414601068

7.29296E-04

0.1

1.00500416805580359899

1.00516895464404488571

1.64787E-04

0.2

1.02006675561907584630

1.02037121839593191992

3.04463E-04

0.3

1.04533851412886048503

1.04543364949281453014

9.51354E-05

0.4

1.08107237183845480928

1.08091240181770013323

1.59970E-04

0.5

1.12762596520638078523

1.12736362925359614605

2.62336E-04

0.6

1.18546521824226770375

1.18534348568350998542

1.21733E-04

0.7

1.25516900563094301816

1.25540812499044906822

2.39119E-04

0.8

1.33743494630484459800

1.33811370105742081128

6.78755E-04

0.9

1.43308638544877438784

1.43401636776743263145

9.29982E-04

1.0

1.54308063481524377848

1.54367227900349194559

5.91644E-04

_____________________________________________________________________________

For 𝑛 = 50. Eq. (5.90) is 50 𝑇 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑𝑘=0 𝐂 𝛙𝑘 (𝑥),

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼50 ]𝑇 , and

𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 , … , 𝜓50 ]𝑇 .

Proceeding as before we have 0.9986 [ ⋮ −0.5983

⋯ ⋱ ⋯

0.38294 0.3583 𝛼0 ⋮ ]. ⋮ ⋮ ][ ] = [ 0.4948 0.3733 𝛼50

After solving we obtained the approximate solution.

795



Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

4.69050E-51

0.1

1.00500416805580359899

1.00500416944469248813

1.38889E-09

0.2

1.02006675561907584630

1.02006684450796524825

8.88889E-08

0.3

1.04533851412886048503

1.04533952662890486408

1.01250E-06

0.4

1.08107237183845480928

1.08107806072839445978

5.68889E-06

0.5

1.12762596520638078523

1.12764766660750214964

2.17014E-05

0.6

1.18546521824226770375

1.18553001833315602892

6.48001E-05

0.7

1.25516900563094301816

1.25533240751519410468

1.63402E-04

0.8

1.33743494630484459800

1.33779903734569455642

3.64091E-04

0.9

1.43308638544877438784

1.43382450581040031006

7.38120E-04

1.0

1.54308063481524377848

1.54446954875628884760

1.38891E-03

_____________________________________________________________________________

Finally we take 𝑛 = 80. Eq. (5.90) is 100 𝑇 𝑢(𝑥) = ∑80 𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑𝑘=0 𝐂 𝛙𝑘 (𝑥),

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼80 ]𝑇 , and

𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 , … , 𝜓80 ]𝑇 . 796

Proceeding as before we have 𝐀 81×81 𝒙81×1 = 𝐛81×1 .


Table 5.10. Comparison of Exact and Approximate solutions of Eq. (5.89) obtained from Kravchuk Polynomials Method (KPM) ____________________________________________________________________________ 𝑥

Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

1.25144E-100

0.1

1.00500416805580359899

1.00500416944469248813

1.38889E-09

0.2

1.02006675561907584630

1.02006684450796524825

8.88889E-08

0.3

1.04533851412886048503

1.04533952662890486408

1.01250E-06

0.4

1.08107237183845480928

1.08107806072839445978

5.68889E-06

0.5

1.12762596520638078523

1.12764766660750214964

2.17014E-05

0.6

1.18546521824226770375

1.18553001833315602892

6.48001E-05

0.7

1.25516900563094301816

1.25533240751519410468

1.63402E-04

0.8

1.33743494630484459800

1.33779903734569455642

3.64091E-04

0.9

1.43308638544877438784

1.43382450581040031006

7.38120E-04

1.0

1.54308063481524377848

1.54446954875628884760

1.38891E-03

_____________________________________________________________________________

5.8.5. Volterra-Fredholm Integral Equation Example 5.44. Consider the Volterra-Fredholm integral equation [229] as 𝑥

1

𝑢(𝑥) = 2 − 𝑥 − 𝑥 2 − 6𝑥 3 + 𝑥 5 + ∫0 𝑡𝑢(𝑡)d𝑡 + ∫−1(𝑥 + 𝑡)𝑢(𝑡)d𝑡, 797

(5.94)

The exact solution of Eq. (5.94) is 𝑢(𝑥) = 2 + 3𝑥 − 5𝑥 3 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥).

(5.95)


(5.96)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 ]𝑇 . Putting Eq. (5.96) into Eq. (5.94), we obtained ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥) = 2 − 𝑥 − 𝑥 2 − 6𝑥 3 + 𝑥 5 + 𝑥

1

∫0 𝑡 ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡 + ∫−1(𝑥 + 𝑡) ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡.

(5.97)


1

∫−1 ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥)𝛙𝑗 (𝑥)d𝑥 = ∫−1[2 − 𝑥 − 𝑥 2 − 6𝑥 3 + 𝑥 5 ]𝛙𝑗 (𝑥)d𝑥 1

𝑥

1

𝑥

+ ∫−1[∫0 𝑡 ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡 ]𝛙𝑗 (𝑥)d𝑥 + ∫−1[∫0 (𝑥 + 𝑡) ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡]𝛙𝑗 (𝑥)d𝑥.(5.98) for 𝑗 = 0,1,2,3. The matrix form of Eq. (5.98) is given as

=

.

After solving the above we get 𝛼0 =

113507 55804

𝑢(𝑥) =

3501

90

, 𝛼1 = − 55804 , 𝛼2 = 1993.

29882 13951

+

6021 27902

𝑥+

180 1993

𝑥2.

798



Exact Solution


Error

_____________________________________________________________________________ 0.0

2.00000000000000000000

2.14192531001361909541

1.41925E-01

0.1

2.29500000000000000000

2.16440756934986739302

1.30592E-01

0.2

2.56000000000000000000

2.18869615081356175185

3.71304E-01

0.3

2.76500000000000000000

2.21479105440470217189

5.50209E-01

0.4

2.88000000000000000000

2.24269228012328865314

6.37308E-01

0.5

2.87500000000000000000

2.27239982796932119561

6.02600E-01

0.6

2.72000000000000000000

2.30391369794279979930

4.16086E-01

0.7

2.38500000000000000000

2.33723389004372446420

4.77661E-02

0.8

1.84000000000000000000

2.37236040427209519031

5.32360E-01

0.9

1.05500000000000000000

2.40929324062791197764

1.35429E+00

1.0

0.00000000000000000000

2.44803239911117482618

2.44803E+00

_____________________________________________________________________________

For 𝑛 = 3. Eq. (5.95) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑3𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥), where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 , 𝜓3 ]𝑇 . Proceeding as before we have

799

=

.


𝛼0 = − 4 , 𝛼1 = −

11 4

, 𝛼2 =

15 4

, 𝛼3 =

15 4

.

𝑢(𝑥) = 2 + 3𝑥 − 5𝑥 3 , which is exact solution.

Example 5.45. Consider the Volterra-Fredholm integral equation [229] as 𝑥

9

1

𝑢(𝑥) = 2 + 4𝑥 − 8 𝑥 2 − 5𝑥 3 + ∫0 ∫0 (𝑟 − 𝑡)𝑢(𝑡)d𝑡d𝑟 .

(5.99)

The exact solution of Eq. (5.99) is 𝑢(𝑥) = 2 + 3𝑥 − 5𝑥 3 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥).

(5.100)


(5.101)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 , 𝜓3 ]𝑇 . Putting Eq. (5.101) into Eq. (5.99), we obtained 𝑥

9

1

∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥) = 2 + 4𝑥 − 𝑥 2 − 5𝑥 3 + ∫0 ∫0 (𝑟 − 𝑡) ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡d𝑟.(5.102) 8 Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.102) is 1

1

9

∫0 ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥)𝛙𝑗 (𝑥)d𝑥 = ∫0 [2 + 4𝑥 − 8 𝑥 2 − 5𝑥 3 ] 𝛙𝑗 (𝑥)d𝑥 1

𝑥

1

+ ∫0 [∫0 ∫0 (𝑟 − 𝑡) ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡d𝑟] 𝛙𝑗 (𝑥)d𝑥. (5.103) for 𝑗 = 0,1,2. The matrix form of Eq. (5.103) is given as

800

=

.


15

𝛼0 = 4 , 𝛼1 = −3, 𝛼2 = − 4 . 7

𝑢(𝑥) = 4 + 6𝑥 −

15 2

𝑥2.


Exact Solution

Approx. Solutions

Error

_____________________________________________________________________________ 0.0

2.00000000000000000000

1.75000000000000000000

2.50000E-01

0.1

2.29500000000000000000

2.27500000000000000000

2.00000E-02

0.2

2.56000000000000000000

2.65000000000000000000

9.00000E-02

0.3

2.76500000000000000000

2.87500000000000000000

1.10000E-01

0.4

2.88000000000000000000

2.95000000000000000000

7.00000E-02

0.5

2.87500000000000000000

2.87500000000000000000

0.00000E+00

0.6

2.72000000000000000000

2.65000000000000000000

7.00000E-02

0.7

2.38500000000000000000

2.27500000000000000000

1.10000E-01

0.8

1.84000000000000000000

1.75000000000000000000

9.00000E-02

0.9

1.05500000000000000000

1.07500000000000000000

2.00000E-02

1.0

0.00000000000000000000

0.25000000000000000000

2.50000E-01

801


For 𝑛 = 3. Eq. (5.100) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑3𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥), where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 , 𝜓3 ]𝑇 . Proceeding as before we have

=

.

After solving we obtained 𝛼0 = 2, 𝛼1 = −

11 4

, 𝛼2 = 0, 𝛼3 =

15 4

.

𝑢(𝑥) = 2 + 3𝑥 − 5𝑥 3 , which is the exact solution.

5.8.6. Nonlinear Abel’s Integral Equation Example 5.46. Consider the nonlinear Abel’s integral equation [229] as 5

4

𝑥

𝑥 5 (9 − 5𝑥) = ∫0 36

1 1

ln(𝑢(𝑡)) d𝑡 .

(𝑥−𝑡)5

The exact solution of Eq. (5.104) is 𝑢(𝑥) = e1−𝑥 . 802

(5.104)

Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation 𝑣(𝑡) = ln(𝑢(𝑡)) , 𝑢(𝑡) = e𝑣(𝑡) . Using the above transformation into the given integral equation we get 5

4

𝑥

𝑥 5 (9 − 5𝑥) = ∫0 36

1 1

𝑣(𝑡)d𝑡.

(5.105)

(𝑥−𝑡)5

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥).

(5.106)

To solve Eq. (5.104) we take 𝑛 = 2. Eq. (5.106) is 𝑣(𝑥) = ∑2𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥),

(5.107)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 ]𝑇 . Putting Eq. (5.107) into Eq. (5.104), we obtained 5

4

𝑥

𝑥 5 (9 − 5𝑥) = ∫0 36

1 1 (𝑥−𝑡)5

∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡.

(5.108)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2. Then integrating over [0,1] we have Eq. (5.108) is 1 5

∫0

4

1

𝑥

𝑥 5 (9 − 5𝑥)𝛙𝑗 (𝑥)d𝑥 = ∫0 [∫0 36

1 1

(𝑥−𝑡)5

∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡 ] 𝛙𝑗 (𝑥)d𝑥.

for 𝑗 = 0,1,2. The matrix form of Eq. (5.109) is given as

=

.


𝛼0 = 1, 𝛼1 = 2 , 𝛼2 = 0. 𝑣(𝑥) = 1 − 𝑥. After applying the reverse transformation we get 𝑢(𝑥) = exp(1 − 𝑥). which is the exact solution.

803

(5.109)

Example 5.47. Consider the nonlinear Abel’s integral equation [229] as 𝜋2 2

𝑥

+ 𝑥 = ∫0

1 1 (𝑥 2 −𝑡 2 )2

sinh−1(𝑢(𝑡)) d𝑡.

(5.110)

The exact solution of Eq. (5.110) is 𝑢(𝑥) = sinh(𝜋 + 𝑥). Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation 𝑣(𝑡) = sinh−1 (𝑢(𝑡)) , 𝑢(𝑡) = sinh(𝑢(𝑡)). Using the above transformation into the given integral equation we get 𝜋2 2

𝑥

+ 𝑥 = ∫0

1 1

(𝑥 2 −𝑡 2 )2

𝑣(𝑡)d𝑡.

(5.111)

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥).

(5.112)

To solve Eq. (5.111) we take 𝑛 = 2. Eq. (5.112) is 𝑣(𝑥) = ∑2𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥),

(5.113)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 ]𝑇 . Putting Eq. (5.113) into Eq. (5.111), we obtained 𝜋2 2

𝑥

+ 𝑥 = ∫0

1 1 (𝑥 2 −𝑡 2 )2


(5.114)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2. Then integrating over [0,1] we have Eq. (5.114) is 1 𝜋2

1

𝑥

∫0 ( 2 + 𝑥) 𝛙𝑗 (𝑥)d𝑥 = ∫0 [∫0

1 1 (𝑥 2 −𝑡 2 )2

∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡] 𝛙𝑗 (𝑥)d𝑥 .


=

.

After solving the above we get

804

(5.115)

1

𝛼0 = 𝜋, 𝛼1 = − 2 , 𝛼2 = 0. 𝑣(𝑥) = 𝜋 + 𝑥. After applying the reverse transformation we get 𝑢(𝑥) = sinh(𝜋 + 𝑥). which is the exact solution.

5.8.7. Weakly Singular Integral Equation Consider the weakly singular integral equation [229] as 5

16

𝑥

𝑢(𝑥) = 1 + 𝑥 2 − 2√𝑥 − 15 𝑥 2 + ∫0

1 1

𝑢(𝑡)d𝑡.

(5.116)

(𝑥−𝑡)2

The exact solution of Eq. (5.116) is 𝑢(𝑥) = 1 + 𝑥 2 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥).

(5.117)


(5.118)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 ]𝑇 . Putting Eq. (5.118) into Eq. (5.116), we obtained ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥) = 1 + 𝑥 2 − 2√𝑥 −

16

5

𝑥

𝑥 2 + ∫0 15

1 1

(𝑥−𝑡)2


(5.119)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2. Then integrating over [0,1] we have Eq. (5.119) is 1

1

16

5

∫0 ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥) 𝛙𝑗 (𝑥)d𝑥 = ∫0 (1 + 𝑥 2 − 2√𝑥 − 15 𝑥 2 ) 𝛙𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0

1 1

(𝑥−𝑡)2

∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡 ] 𝛙𝑗 (𝑥)d𝑥. (5.120)


=

.


1

𝛼0 = 1, 𝛼1 = 0, 𝛼2 = 2. 𝑢(𝑥) = 1 + 𝑥 2 , which is the exact solution.

5.8.8. System of Fredholm Integral Equations Consider the system of Fredholm integral equation [229] as 3

2

3

8

𝑢(𝑥) = 1 + 𝑥 − 𝑥 2 + 40 𝑥 3 (9𝑥 2 − 12𝑥 − 20) − 231 𝑥 4 (32𝑥 2 − 44𝑥 + 77) 𝑥

+ ∫0

𝑥

1 1 (𝑥−𝑡)3

𝑢(𝑡)d𝑡 − ∫0 5

2 1

𝑣(𝑡)d𝑡.

(5.121a)

(𝑥−𝑡)4

4

6

5

𝑣(𝑥) = 1 − 𝑥 + 𝑥 2 + 126 𝑥 5 (25𝑥 2 − 35𝑥 − 63) + 935 𝑥 6 (72𝑥 2 − 102 + 187) 𝑥

+ ∫0

𝑥

2 1 (𝑥−𝑡)5

𝑢(𝑡)d𝑡 − ∫0

1 1

𝑣(𝑡)d𝑡.

(5.121b)

(𝑥−𝑡)6

The exact solution of Eq. (5.121) is 𝑢(𝑥) = 1 + 𝑥 − 𝑥 2 , 𝑣(𝑥) = 1 − 𝑥 + 𝑥 2 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥),

(5.122a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝜓𝑘 (𝑥).

(5.122b)


(5.123a)

𝑣(𝑥) = ∑2𝑘=0 𝛽𝑘 𝜓𝑘 (𝑥) = ∑2𝑘=0 𝐃𝑇 𝛙𝑘 (𝑥),

(5.123b)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1 , 𝛽2 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 ]𝑇 . Putting Eq. (5.123) into Eq. (5.121), we obtained ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥) = 1 + 𝑥 − 𝑥 2 + 𝑥

77) + ∫0

1 1 (𝑥−𝑡)3

𝑥

∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡 − ∫0

𝑥

2 1 (𝑥−𝑡)5

2

3

8

𝑥 3 (9𝑥 2 − 12𝑥 − 20) − 231 𝑥 4 (32𝑥 2 − 44𝑥 + 40 2 1

(𝑥−𝑡)4

∑2𝑘=0 𝐃𝑇 𝛙𝑘 (𝑥) = 1 − 𝑥 + 𝑥 2 + 102 + 187) + ∫0

3

5

∑2𝑘=0 𝐃𝑇 𝛙𝑘 (𝑡) d𝑡. 4

(5.124a) 6

5

𝑥 5 (25𝑥 2 − 35𝑥 − 63) + 935 𝑥 6 (72𝑥 2 − 126 𝑥

∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡 − ∫0

1 1 (𝑥−𝑡)6

∑2𝑘=0 𝐃𝑇 𝛙𝑘 (𝑡) d𝑡 .

(5.124b)


1

1

2

3

∫0 ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥)𝛙𝑗 (𝑥)d𝑥 = ∫0 (1 + 𝑥 − 𝑥 2 + 40 𝑥 3 (9𝑥 2 − 12𝑥 − 20) − 3

8

1

𝑥

𝑥 4 (32𝑥 2 − 44𝑥 + 77)) 𝛙𝑗 (𝑥)d𝑥 + ∫0 [∫− 231 1

𝑥

− ∫0 [∫0

2 1 (𝑥−𝑡)4

1 1 (𝑥−𝑡)3

∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡 ] 𝛙𝑗 (𝑥)d𝑥

∑2𝑘=0 𝐃𝑇 𝛙𝑘 (𝑡) d𝑡 ] 𝛙𝑗 (𝑥)d𝑥,

1

(5.125a)

1

2

3

∫0 ∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑥)𝛙𝑗 (𝑥)d𝑥 = ∫0 (1 + 𝑥 − 𝑥 2 + 40 𝑥 3 (9𝑥 2 − 12𝑥 − 20) − 3

8

1

𝑥

𝑥 4 (32𝑥 2 − 44𝑥 + 77)) 𝛙𝑗 (𝑥)d𝑥 + ∫0 [∫− 231 1

𝑥

− ∫0 [∫0

2 1

(𝑥−𝑡)4

1 1 (𝑥−𝑡)3

∑2𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) d𝑡 ] 𝛙𝑗 (𝑥)d𝑥

∑2𝑘=0 𝐃𝑇 𝛙𝑘 (𝑡) d𝑡 ] 𝛙𝑗 (𝑥)d𝑥,

(5.125b)


=

.


1

1

1

𝛼0 = 1, 𝛼1 = − 2 , 𝛼2 = − 2 , 𝛽0 = 1, 𝛽1 = 2 , 𝛽2 = 2. 𝑢(𝑥) = 1 + 𝑥 − 𝑥 2 , 𝑣(𝑥) = 1 − 𝑥 + 𝑥 2 .

5.8.9. System of Weakly Singular Integral Equations of 1st Kind Consider the system of weakly singular integral equation of 1st Kind [229] as 27

5

𝑥

𝑥 3 (45𝑥 2 + 33𝑥 + 44) = ∫0 440 25

8

𝑥

𝑥 5 (25𝑥 2 + 30𝑥 + 39) = ∫0 468

1 1

[𝑢(𝑡) + 𝑣(𝑡) + 𝑤(𝑡)]d𝑡.

(5.126a)

[𝑣(𝑡) + 𝑤(𝑡)]d𝑡.

(5.126b)

(𝑥−𝑡)3 1 2

(𝑥−𝑡)5

807

32

5

𝑥

𝑥 4 (96𝑥 2 + 208𝑥 + 117) = ∫0 585

1 1

[𝑢(𝑡) + 𝑣(𝑡)]d𝑡.

(5.126c)

(𝑥−𝑡)3

The exact solution of Eq. (5.126) is 𝑢(𝑥) = 𝑥 + 𝑥 2 + 3𝑥 3 , 𝑣(𝑥) = 𝑥 + 3𝑥 2 − 𝑥 3 , 𝑤(𝑥) = 𝑥 − 𝑥 2 + 3𝑥 3 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥),

(5.127a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝜓𝑘 (𝑥),

(5.127b)

𝑤(𝑥) = ∑𝑛𝑘=0 𝜇𝑘 𝜓𝑘 (𝑥).

(5.127c)


(5.128a)

𝑣(𝑥) = ∑3𝑘=0 𝛽𝑘 𝜓𝑘 (𝑥) = ∑3𝑘=0 𝐃𝑇 𝛙𝑘 (𝑥),

(5.128b)

𝑤(𝑥) = ∑3𝑘=0 𝜇𝑘 𝜓𝑘 (𝑥) = ∑3𝑘=0 𝐄𝑇 𝛙𝑘 (𝑥),

(5.128c)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1 , 𝛽2 , 𝛽3 ]𝑇 , 𝐄 = [𝜇0 , 𝜇1 , 𝜇2 , 𝜇3 ]𝑇 , and 𝛙𝑘 (𝑥) = [𝜓0 , 𝜓1 , 𝜓2 ]𝑇 . Putting Eq. (5.128) into Eq. (5.126), we obtained 27

5

𝑥

𝑥 3 (45𝑥 2 + 33𝑥 + 44) = ∫0 440 25

8

1 (𝑥−𝑡)3

𝑥

𝑥 5 (25𝑥 2 + 30𝑥 + 39) = ∫0 468 32

1

5

1 2 (𝑥−𝑡)5

𝑥

𝑥 4 (96𝑥 2 + 208𝑥 + 117) = ∫0 585

∑3𝑘=0[𝐂 𝑇 + 𝐃𝑇 + 𝐄𝑇 ]𝛙𝑘 (𝑡) d𝑡.

(5.129a)

[∑3𝑘=0 𝐃𝑇 𝛙𝑘 (𝑥) + ∑3𝑘=0 𝐄𝑇 𝛙𝑘 (𝑡)]d𝑡. (5.129b)

1 1 (𝑥−𝑡)3

[∑3𝑘=0 𝐂 𝑇 𝛙𝑘 (𝑡) + ∑3𝑘=0 𝐃𝑇 𝛙𝑘 (𝑡)]d𝑡.( 5.129c)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.129) is 1 27

∫0

5

𝑥 3 (45𝑥 2 + 33𝑥 + 44)𝛙𝑗 (𝑥)d𝑥 = 440 1

𝑥

∫0 [∫0 1 25

∫0

468

1 1 (𝑥−𝑡)3

∑3𝑘=0[𝐂 𝑇 + 𝐃𝑇 + 𝐄𝑇 ]𝛙𝑘 (𝑡) d𝑡] 𝛙𝑗 (𝑥)d𝑥.

(5.130a)

8

𝑥 5 (25𝑥 2 + 30𝑥 + 39) 𝛙𝑗 (𝑥)d𝑥 = 1

𝑥

∫0 [∫0

1 2

(𝑥−𝑡)5

∑3𝑘=0[𝐃𝑇 + 𝐄𝑇 ]𝛙𝑘 (𝑡) d𝑡] 𝛙𝑗 (𝑥)d𝑥. 808

(5.130b)

1 32

∫0

5

𝑥 4 (96𝑥 2 + 208𝑥 + 117)𝛙𝑗 (𝑥)d𝑥 = 585 1

𝑥

∫0 [∫0

1 1 (𝑥−𝑡)3

∑3𝑘=0[𝐂 𝑇 + 𝐃𝑇 ]𝛙𝑘 (𝑡) d𝑡 ] 𝛙𝑗 (𝑥)d𝑥.

(5.130c)

for 𝑗 = 0,1,2. The matrix form of Eq. (5.130) is given as 𝐀 𝟏𝟏 [𝐁𝟐𝟑 𝐂𝟑𝟏

𝐀 𝟏𝟐 𝐁𝟐𝟐 𝐂𝟑𝟐

𝐛𝟏 𝐀 𝟏𝟑 𝒙𝟏 𝐁𝟐𝟑 ] [𝒙𝟐 ] = [𝐛𝟐 ]. 𝐂𝟑𝟑 𝒙𝟑 𝐛𝟑

where 𝐀 𝟏𝟏 , 𝐀 𝟏𝟐 , 𝐀 𝟏𝟑 , 𝐁𝟐𝟏 , 𝐁𝟐𝟐 , 𝐁𝟐𝟑 , 𝐂𝟑𝟏 , 𝐂𝟑𝟐 and 𝐂𝟑𝟑 are matrices of order 3×3, 𝒙𝟏 , 𝒙𝟐 , 𝒙𝟑 , 𝐛𝟏 , 𝐛𝟐 , and 𝐛𝟑 are matrices of order 3×1. After solving the above we get 1

1

9

1

3

3

𝛼0 = 0, 𝛼1 = 4 , 𝛼2 = 2 , 𝛼3 = − 4 , 𝛽0 = 0, 𝛽1 = − 4 , 𝛽2 = 2 , 𝛽3 = 4, 1

1

9

𝜇0 = 0, 𝜇1 = 4 , 𝜇2 = − 2 , 𝜇3 = − 4. 𝑢(𝑥) = 𝑥 + 𝑥 2 + 3𝑥 3 , 𝑣(𝑥) = 𝑥 + 3𝑥 2 − 𝑥 3 , 𝑤(𝑥) = 𝑥 − 𝑥 2 + 3𝑥 3 .

5.9. Newton’s Polynomials Method I develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Newton’s polynomials [154]. Assume that 𝑓 ∈ 𝐶 𝑛+1 [𝑎, 𝑏] and 𝑥𝑘 [𝑎, 𝑏] for 𝑘 = 0,1, … , 𝑛 are distinct. Then 𝑓(𝑥) = 𝑃𝑛 (𝑥) + 𝑅𝑛 (𝑥),

(5.131)

where 𝑃𝑛 (𝑥) is a polynomial that can be used to approximate 𝑓(𝑥), 𝑃𝑛 (𝑥) = 𝑎0 + 𝑎1 (𝑥 − 𝑥0 ) + 𝑎2 (𝑥 − 𝑥0 )(𝑥 − 𝑥1 ) + 𝑎3 (𝑥 − 𝑥0 )(𝑥 − 𝑥1 )(𝑥 − 𝑥2 ) + ⋯ + 𝑎𝑛 (𝑥 − 𝑥0 )(𝑥 − 𝑥1 )(𝑥 − 𝑥2 ) … (𝑥 − 𝑥𝑛−1 ). 𝑗−1

𝑃𝑛 (𝑥) = ∑𝑛𝑗=0(∏𝑖=0 𝑎𝑖 (𝑥 − 𝑥𝑖 )).

(5.132) (5.133)

It can be written as 𝑓(𝑥) ≈ 𝑃𝑛 (𝑥). The Newton polynomial goes through the 𝑛 + 1 points {(𝑥𝑘 , 𝑦𝑘 )}𝑛𝑘=0 , i.e. 𝑃𝑛 (𝑥𝑘 ) = 𝑓(𝑥𝑘 ),

for 𝑘 = 0,1,2, … , 𝑛

(5.134)

and 𝑅𝑛 (𝑥) is the remainder. The coefficients 𝑎𝑖 are constructed using divided differences.

809



(5.135)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, to find an approximate solution 𝑢̃(𝑥) of Eq. (5.135). For this, we assume that 𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝑁𝑘 (𝑥),

(5.136)

where 𝑁𝑘 (𝑥) are Newton’s polynomials of degree 𝑘 defined earlier and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.136) into (5.135), we get 𝑥

𝑓(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝑁𝑘 (𝑡)d𝑡,

(5.137)

Then the Galerkin equations are obtained by multiplying both sides of (5.137) by 𝑁𝑗 (𝑥) and then integrating with respect to 𝑥 from 𝑎 to 𝑏, we have 𝑏

𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝑁𝑗 (𝑥)d𝑥 = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝑁𝑘 (𝑡)d𝑡 )𝑁𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛,( 5.138) Since in each equation, there are two integrals. The inner integrand of the left side is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.138) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.139)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝑁𝑘 (𝑡)d𝑡 )𝑁𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛 𝑏

𝐹𝑗 = ∫𝑎 𝑓(𝑥) 𝑁𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛 Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.139) and substituting these values of parameters in (5.137), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.136). 810



(5.140)


(5.141)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝑁𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑁𝑘 (𝑡)d𝑡 )𝑁𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛. Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.141) and substituting these values of parameters in (5.137), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.140).

5.9.2. Volterra Integral Equation Consider the Volterra integral equation of the 1st Kind [229] as 𝑥

𝑢(𝑥) = 2 cosh(𝑥) − 2 + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 ,

0 ≤ 𝑥 ≤ 1.

(5.142)

The exact solution of Eq. (11) is 𝑢(𝑥) = 𝑥 sinh(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥).

(5.143)

To solve Eq. (5.142) we take 𝑛 = 3. Eq. (5.143) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑3𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥),

(5.144)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , 𝑁3 ]𝑇 .


∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥) = 2 cosh(𝑥) − 2 + ∫0 (𝑥 − 𝑡) ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡.

(5.145)


1

∫0 ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [2 cosh(𝑥) − 2]𝐍𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 (𝑥 − 𝑡) ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡]𝐍𝑗 (𝑥)d𝑥. 811

(5.146)


=

.

After solving the above we get 𝛼0 = −0.350755945, 𝛼1 = −0.77196520, 𝛼2 = 0.7580156,

𝛼3 = 0.36190250.

𝑢(𝑥) = 0.002803072 + 0.05506962𝑥 + 0.7580156035𝑥 2 + 0.3619024665𝑥 3 .


812

Table 5.13. Comparison of Exact and Approximate solutions of Eq. (5.142) obtained from Newton’s polynomials Method (NPM) _____________________________________________________________________________ 𝑥

Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00280315270657469879

2.80315E-03

0.1

0.01001667500198440258

0.01064592101008597807

6.29246E-04

0.2

0.04026720050821879753

0.04142670138579126722

1.15950E-03

0.3

0.09135608803414278569

0.09171060699168307847

3.54519E-04

0.4

0.16430093032112620342

0.16366905639890332166

6.31874E-04

0.5

0.26054765274687368081

0.25947346817859390661

1.07418E-03

0.6

0.38199214928894476267

0.38129526090189674317

6.96888E-04

0.7

0.53100859128767345242

0.53130585313995374115

2.97262E-04

0.8

0.71048478575009840526

0.71167666346390681039

1.19188E-03

0.9

0.92386505313735774836

0.92457911044489786072

7.14057E-04

1.0

1.17520119364380145688

1.17218461265406880196

3.01658E-03

_____________________________________________________________________________

For 𝑛 = 42. Eq. (5.142) is 42 𝑇 𝑢(𝑥) = ∑42 𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑𝑘=0 𝐂 𝐍𝑘 (𝑥),

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼42 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁42 ]𝑇 . Proceeding as before we have we obtained the approximate solution of given integral equation.


813


Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00000000000000000000

1.47196E-73

0.1

0.01001667500198440258

0.01001667500198440258

2.37452E-74

0.2

0.04026720050821879753

0.04026720050821879753

2.19810E-74

0.3

0.09135608803414278569

0.09135608803414278569

2.31767E-74

0.4

0.16430093032112620342

0.16430093032112620342

1.36706E-74

0.5

0.26054765274687368081

0.26054765274687368081

1.11511E-74

0.6

0.38199214928894476267

0.38199214928894476267

2.59831E-74

0.7

0.53100859128767345242

0.53100859128767345242

2.26905E-74

0.8

0.71048478575009840526

0.71048478575009840526

2.81106E-74

0.9

0.92386505313735774836

0.92386505313735774836

1.76670E-74

1.0

1.17520119364380145688

1.17520119364380145688

2.18150E-73

_____________________________________________________________________________

Finally we take 𝑛 = 50. Eq. (5.142) is 50 𝑇 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑𝑘=0 𝐂 𝐍𝑘 (𝑥),

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼50 ]𝑇 , and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁50 ]𝑇 .

Proceeding as before we have 𝐀 51×51 𝒙51×1 = 𝐛51×1 . Proceeding as before we have we obtained the approximate solution of given integral equation.

814


Table 5.15. Comparison of Exact and Approximate solutions of Eq. (5.142) obtained from Newton’s polynomials Method (NPM) _____________________________________________________________________________ x

Exact Solution

Approx. Solutions

Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00000000000000000000

4.33793E-95

0.1

0.01001667500198440258

0.01001667500198440258

3.73318E-96

0.2

0.04026720050821879753

0.04026720050821879753

5.34277E-96

0.3

0.09135608803414278569

0.09135608803414278569

3.59207E-96

0.4

0.16430093032112620342

0.16430093032112620342

4.01372E-96

0.5

0.26054765274687368081

0.26054765274687368081

5.08515E-98

0.6

0.38199214928894476267

0.38199214928894476267

3.97147E-96

0.7

0.53100859128767345242

0.53100859128767345242

3.69051E-96

0.8

0.71048478575009840526

0.71048478575009840526

5.42425E-96

0.9

0.92386505313735774836

0.92386505313735774836

3.73206E-96

1.0

1.17520119364380145688

1.17520119364380145688

4.43030E-95

_____________________________________________________________________________

5.9.3. Fredholm Integral Equation Consider the Fredholm integral equation [229] of the form 𝑢(𝑥) = 𝑥 2 + 𝑥 cos(𝑥) +

𝜋3 3

𝜋

𝑥 − 2𝑥 − ∫0 𝑥𝑢(𝑡)d𝑡 .

The exact solution of Eq. (5.147) is 𝑢(𝑥) = 𝑥 2 + 𝑥 cos(𝑥). 815

0 ≤ 𝑥 ≤ 1. (5.147)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥).

(5.148)


(5.149)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , 𝑁3 ]𝑇 . Putting Eq. (5.149) into Eq. (5.147), we obtained ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥) = 𝑥 2 + 𝑥 cos(𝑥) +

𝜋3 3

𝜋

𝑥 − 2𝑥 − ∫0 𝑥 ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡 .

(5.150)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0, 𝜋] we have Eq. (5.150) is 𝜋

𝜋

∫0 ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [𝑥 2 + 𝑥 cos(𝑥) + 𝜋

𝜋3 3

𝑥 − 2𝑥] 𝐍𝑗 (𝑥)d𝑥

𝜋

− ∫0 [∫0 𝑥 ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡 ]𝐍𝑗 (𝑥)d𝑥 .

(5.151)


=

After solving the above we get 𝛼0 = −0.100750996, 𝛼1 = 10.998555, 𝛼2 = −34.5077972, 𝛼3 = −73.5060631. 𝑢(𝑥) = −0.100750996 + 27.00693120𝑥 − 69.01559538𝑥 2 + 98.00808408𝑥 3 .


816

.


Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.20186199111380257471

2.01862E-01

0.1

0.10950041652780257661

0.01841373477479369844

9.10867E-02

0.2

0.23601331556824832622

0.22494111947929046536

1.10722E-02

0.3

0.37660094673768180589

0.41908733838069153878

4.24864E-02

0.4

0.52842439760115403312

0.60221956686000073143

7.37952E-02

0.5

0.68879128094518635806

0.77570498029822185607

8.69137E-02

0.6

0.85520136894580697834

0.94091075407635872543

8.57094E-02

0.7

1.02538953109914189838

1.09920406357541515224

7.38145E-02

0.8

1.19736536747773233674

1.25195208417639494925

5.45867E-02

0.9

1.36944897144359801084

1.40052199126030192919

3.10730E-02

1.0

1.54030230586813971740

1.54628096020813990481

5.97865E-03

_____________________________________________________________________________

For 𝑛 = 5. Eq. (5.147) is 𝑢(𝑥) = ∑5𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑5𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥), where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼5 ]𝑇 , and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁5 ]𝑇 .

Proceeding as before we obtained the approximate solution.


817


Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00788073538687907252

7.88074E-03

0.1

0.10950041652780257661

0.10979457812625839934

2.94162E-04

0.2

0.23601331556824832622

0.23319318620657095364

2.82013E-03

0.3

0.37660094673768180589

0.37341107127460963580

3.18988E-03

0.4

0.52842439760115403312

0.52631527707424000382

2.10912E-03

0.5

0.68879128094518635806

0.68829110232660795888

5.00179E-04

0.6

0.85520136894580697834

0.85622782361034743095

1.02645E-03

0.7

1.02538953109914189838

1.02750441824178806441

2.11489E-03

0.8

1.19736536747773233674

1.19997528715516290358

2.60992E-03

0.9

1.36944897144359801084

1.37195597778281607837

2.50701E-03

1.0

1.54030230586813971740

1.54220890693541048986

1.90660E-03

_____________________________________________________________________________


where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼12 ]𝑇 , and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁12 ]𝑇 .

Proceeding as before we have 𝐀13×13 𝒙13×1 = 𝐛13×1 , we obtained the approximate solution after solving the above matrix equation.

818



Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00000000003249437317

3.24944E-11

0.1

0.10950041652780257661

0.10950041653942645009

1.16239E-11

0.2

0.23601331556824832622

0.23601331555769001432

1.05583E-11

0.3

0.37660094673768180589

0.37660094673020961916

7.47219E-12

0.4

0.52842439760115403312

0.52842439760948094405

8.32691E-12

0.5

0.68879128094518635806

0.68879128095731116725

1.21248E-11

0.6

0.85520136894580697834

0.85520136894705858822

1.25161E-12

0.7

1.02538953109914189838

1.02538953108813774695

1.10042E-11

0.8

1.19736536747773233674

1.19736536746511457748

1.26178E-11

0.9

1.36944897144359801084

1.36944897144081077082

2.78724E-12

1.0

1.54030230586813971740

1.54030230587780452492

9.66481E-12

_____________________________________________________________________________

5.9.4.Volterra Integro-Differential Equation Example 5.48. Consider the Volterra integro-differential equations [229] as 𝑥

𝑢′′ (𝑥) = 1 − ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 ,

0 ≤ 𝑥 ≤ 1.

𝑢(0) = 1, 𝑢′ (0) = 0. The exact solution of Eq. (5.152) is 819

(5.152)

𝑢(𝑥) = cosh(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥).

(5.153)


(5.154)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , 𝑁3 ]𝑇 .


[𝐂 𝑇 𝐍𝑘 (𝑥)]′′ = 1 + ∫0 (𝑥 − 𝑡) ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡.

(5.155)


1

∫0 ∑𝑛𝑘=0[𝐂 𝑇 𝐍𝑘 (𝑥)]′′ 𝐍𝑗 (𝑥)d𝑥 = ∫0 [1]𝐍𝑗 (𝑥)d𝑥 1

𝑥

− ∫0 [∫0 (𝑥 − 𝑡) ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡]𝐍𝑗 (𝑥)d𝑥. for 𝑗 = 0,1,2,3. The matrix form of Eq. (5.156) is given as

=

.


2192848

34328

42392

𝛼0 = 5839047 , 𝛼1 = − 5839047 , 𝛼2 = 72087 , 𝛼3 = 648783. 34328

42392

𝑢(𝑥) = 1 + 72087 𝑥 2 + 648783 𝑥 3 .

820

(5.156)



Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

0.00000E+00

0.1

1.00500416805580359899

1.00482736446546842319

1.76804E-04

0.2

1.02006675561907584630

1.01957082106035454073

4.95935E-04

0.3

1.04533851412886048503

1.04462241458237962462

7.16100E-04

0.4

1.08107237183845480928

1.08037418982926494683

6.98182E-04

0.5

1.12762596520638078523

1.12721819159873177935

4.07774E-04

0.6

1.18546521824226770375

1.18554646468850139415

8.12464E-05

0.7

1.25516900563094301816

1.25575105389629506322

5.82048E-04

0.8

1.33743494630484459800

1.33822400401983405854

7.89058E-04

0.9

1.43308638544877438784

1.43335735985683965209

2.70974E-04

1.0

1.54308063481524377848

1.54154316620503311585

1.53747E-03

_____________________________________________________________________________


where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼30 ]𝑇 , and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁30 ]𝑇 . 821




Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

0.00000E+00

0.1

1.00500416805580359899

1.00500416805580359899

4.65245E-53

0.2

1.02006675561907584630

1.02006675561907584630

2.43023E-53

0.3

1.04533851412886048503

1.04533851412886048503

1.24403E-53

0.4

1.08107237183845480928

1.08107237183845480928

4.71918E-54

0.5

1.12762596520638078523

1.12762596520638078523

2.78457E-54

0.6

1.18546521824226770375

1.18546521824226770375

1.11301E-54

0.7

1.25516900563094301816

1.25516900563094301816

5.75213E-54

0.8

1.33743494630484459800

1.33743494630484459800

2.01461E-53

0.9

1.43308638544877438784

1.43308638544877438784

1.79391E-53

1.0

1.54308063481524377848

1.54308063481524377848

1.27747E-52

_____________________________________________________________________________


where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼50 ]𝑇 , 822

and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁50 ]𝑇 .




Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

0.00000E+00

0.1

1.00500416805580359899

1.00500416805580359899

7.20000E-98

0.2

1.02006675561907584630

1.02006675561907584630

9.60000E-98

0.3

1.04533851412886048503

1.04533851412886048503

6.50000E-98

0.4

1.08107237183845480928

1.08107237183845480928

7.80000E-98

0.5

1.12762596520638078523

1.12762596520638078523

4.00000E-99

0.6

1.18546521824226770375

1.18546521824226770375

7.20000E-98

0.7

1.25516900563094301816

1.25516900563094301816

7.20000E-98

0.8

1.33743494630484459800

1.33743494630484459800

1.02000E-97

0.9

1.43308638544877438784

1.43308638544877438784

7.00000E-98

1.0

1.54308063481524377848

1.54308063481524377848

8.31000E-97

_____________________________________________________________________________

Example 5.49. Consider the Volterra integro-differential equations [229] as 823

𝑥

𝑢′′ (𝑥) = 1 + 𝑥 + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 ,

0 ≤ 𝑥 ≤ 1.

(5.157)

𝑢(0) = 1, 𝑢′ (0) = 1. The exact solution of Eq. (5.157) is 𝑢(𝑥) = exp(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥).

(5.158)


(5.159)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , 𝑁3 ]𝑇 .


[𝐂 𝑇 𝐍𝑘 (𝑥)]′′ = 1 + 𝑥 + ∫0 (𝑥 − 𝑡) ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡 .

(5.160)


1

∫0 ∑𝑛𝑘=0[𝐂 𝑇 𝐍𝑘 (𝑥)]′′ 𝐍𝑗 (𝑥)d𝑥 = ∫0 [1 + 𝑥]𝐍𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 (𝑥 − 𝑡) ∑𝑛𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡]𝐍𝑗 (𝑥)d𝑥.

(5.161)


=

.


8193085

101428

480868

𝛼0 = − 17517141 , 𝛼1 = 17517141 , 𝛼2 = 216261 , 𝛼3 = 1946349. 101428

480868

𝑢(𝑥) = 1 + 𝑥 + 216261 𝑥 2 + 1946349 𝑥 3 .

824



Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

0.00000E+00

0.1

1.10517091807564762481

1.10493713511811088351

2.33783E-04

0.2

1.22140275816016983392

1.22073678667083858034

6.65971E-04

0.3

1.34985880757600310398

1.34888132395577565997

9.77484E-04

0.4

1.49182469764127031782

1.49085311627051469187

9.71581E-04

0.5

1.64872127070012814685

1.64813453291264824551

5.86738E-04

0.6

1.82211880039050897488

1.82220794317976889037

8.91428E-05

0.7

2.01375270747047652162

2.01455571636946919592

8.03009E-04

0.8

2.22554092849246760458

2.22666022177934173162

1.11929E-03

0.9

2.45960311115694966380

2.46000382870697906696

4.00718E-04

1.0

2.71828182845904523536

2.71606890644997377140

2.21292E-03

_____________________________________________________________________________


where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼30 ]𝑇 , and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁30 ]𝑇 . 825




Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

1.00000E-99

0.1

1.10517091807564762481

1.10517091807564762481

1.48304E-52

0.2

1.22140275816016983392

1.22140275816016983392

7.72880E-53

0.3

1.34985880757600310398

1.34985880757600310398

3.91077E-53

0.4

1.49182469764127031782

1.49182469764127031782

1.44981E-53

0.5

1.64872127070012814685

1.64872127070012814685

8.33753E-54

0.6

1.82211880039050897488

1.82211880039050897488

3.04576E-54

0.7

2.01375270747047652162

2.01375270747047652162

1.85361E-53

0.8

2.22554092849246760458

2.22554092849246760458

6.33916E-53

0.9

2.45960311115694966380

2.45960311115694966380

5.64630E-53

1.0

2.71828182845904523536

2.71828182845904523536

3.99808E-52

_____________________________________________________________________________


where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼60 ]𝑇 , 826

and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁60 ]𝑇 .




Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

0.00000E+00

0.1

1.10517091807564762481

1.10517091807564762481

2.35181E-62

0.2

1.22140275816016983392

1.22140275816016983392

1.01335E-62

0.3

1.34985880757600310398

1.34985880757600310398

2.52358E-62

0.4

1.49182469764127031782

1.49182469764127031782

2.33896E-62

0.5

1.64872127070012814685

1.64872127070012814685

2.33838E-62

0.6

1.82211880039050897488

1.82211880039050897488

1.96907E-62

0.7

2.01375270747047652162

2.01375270747047652162

2.07806E-62

0.8

2.22554092849246760458

2.22554092849246760458

1.07479E-62

0.9

2.45960311115694966380

2.45960311115694966380

2.07824E-62

1.0

2.71828182845904523536

2.71828182845904523536

1.77888E-61

_____________________________________________________________________________

5.9.5. Volterra-Fredholm Integral Equation 827


1

𝑢(𝑥) = −5 − 𝑥 + 12𝑥 2 − 𝑥 3 − 𝑥 4 + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 + ∫0 (𝑥 + 𝑡)𝑢(𝑡)d𝑡, (5.162) The exact solution of Eq. (5.162) is 𝑢(𝑥) = 6𝑥 + 12𝑥 2 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥).

(5.163)


(5.164)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 ]𝑇 . Putting Eq. (5.164) into Eq. (5.162), we obtained ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥) = −5 − 𝑥 + 12𝑥 2 − 𝑥 3 − 𝑥 4 + 𝑥

1

∫0 (𝑥 − 𝑡) ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡 + ∫0 (𝑥 + 𝑡) ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡. (5.165) Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.165) is 1 1 −5 − 𝑥 + 12𝑥 2 ] 𝐍𝑗 (𝑥)d𝑥 + ∫0 ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [ −𝑥 3 − 𝑥 4 1

𝑥

∫0 [∫0 (𝑥 − 𝑡) ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡 ]𝐍𝑗 (𝑥)d𝑥 1

1

+ ∫0 [∫0 (𝑥 + 𝑡) ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡] 𝐍𝑗 (𝑥)d𝑥 .

(5.166)


=

.

After solving the above we get 𝛼0 = 3, 𝛼1 = 22, 𝛼2 = 12. 𝑢(𝑥) = 6𝑥 + 12𝑥 2 . which is exact solution.


1

𝑢(𝑥) = −2 − 2𝑥 + 2e𝑥 + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 + ∫0 𝑥𝑢(𝑡)d𝑡 , 828

(5.167)

The exact solution of Eq. (5.167) is 𝑢(𝑥) = 𝑥e𝑥 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥).

(5.168)


(5.169)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 ]𝑇 . Putting Eq. (5.169) into Eq. (5.167), we obtained ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥) = −2 − 2𝑥 + 2e𝑥 + 𝑥

1

∫0 (𝑥 − 𝑡) ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡 + ∫0 𝑥 ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡.

(5.170)


1

∫0 ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [−2 − 2𝑥 + 2e𝑥 ]𝐍𝑗 (𝑥)d𝑥 1

𝑥

1

1

+ ∫0 [∫0 (𝑥 − 𝑡) ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡]𝐍𝑗 (𝑥)d𝑥 + ∫0 [∫0 𝑥 ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡 ] 𝐍𝑗 (𝑥)d𝑥.( 5.171) for 𝑗 = 0,1,2,3. The matrix form of Eq. (5.171) is given as

=

.


215140 613

𝑢(𝑥) = −

−

79080 613

129145 613

+

e, 𝛼1 = −

47520 613

e+

2080370 5517

688570 613

+

𝑥−

829

257360 1839

253200 613

e, 𝛼2 = −

e𝑥 −

2069375 1839

2069375 1839

𝑥2 +

+

254240 613

254240 613

e.

e𝑥 2 .



Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.04527322736350666284

4.52732E-02

0.1

0.11051709180756476248

0.11566010733835915940

5.14302E-03

0.2

0.24428055163203396678

0.22860181278665354736

1.56787E-02

0.3

0.40495764227280093120

0.38409834370838982671

2.08593E-02

0.4

0.59672987905650812713

0.58214970010356799746

1.45802E-02

0.5

0.82436063535006407342

0.82275588197218805961

1.60475E-03

0.6

1.09327128023430538493

1.10591688931425001316

1.26456E-02

0.7

1.40962689522933356514

1.43163272212975385810

2.20058E-02

0.8

1.78043274279397408366

1.79990338041869959444

1.94706E-02

0.9

2.21364280004125469742

2.21072886418108722218

2.91394E-03

1.0

2.71828182845904523536

2.66410917341691674132

5.41727E-02

_____________________________________________________________________________


where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼30 ]𝑇 , and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁30 ]𝑇 . 830




Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00000000000000000000

3.13082E-27

0.1

0.11051709180756476248

0.11051709180756476248

1.06432E-27

0.2

0.24428055163203396678

0.24428055163203396678

7.06158E-28

0.3

0.40495764227280093120

0.40495764227280093120

1.18924E-27

0.4

0.59672987905650812713

0.59672987905650812713

3.11242E-28

0.5

0.82436063535006407342

0.82436063535006407342

7.74861E-28

0.6

1.09327128023430538493

1.09327128023430538493

1.62672E-27

0.7

1.40962689522933356514

1.40962689522933356514

1.59992E-27

0.8

1.78043274279397408366

1.78043274279397408366

6.65914E-28

0.9

2.21364280004125469742

2.21364280004125469742

1.34374E-27

1.0

2.71828182845904523536

2.71828182845904523536

1.18841E-26

_____________________________________________________________________________


where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼60 ]𝑇 , 831

and

𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁60 ]𝑇 .




Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00000000000000000000

1.64341E-56

0.1

0.11051709180756476248

0.11051709180756476248

5.28349E-58

0.2

0.24428055163203396678

0.24428055163203396678

3.91868E-57

0.3

0.40495764227280093120

0.40495764227280093120

2.96163E-58

0.4

0.59672987905650812713

0.59672987905650812713

2.74395E-57

0.5

0.82436063535006407342

0.82436063535006407342

4.40174E-57

0.6

1.09327128023430538493

1.09327128023430538493

4.68122E-57

0.7

1.40962689522933356514

1.40962689522933356514

2.49196E-57

0.8

1.78043274279397408366

1.78043274279397408366

3.87858E-57

0.9

2.21364280004125469742

2.21364280004125469742

1.87883E-57

1.0

2.71828182845904523536

2.71828182845904523536

3.14447E-56

_____________________________________________________________________________

Example 5.52. Consider the Volterra-Fredholm integral equation [229] as 1

𝑥

1

𝑢(𝑥) = 𝑥e𝑥 − 2 𝑥 2 + ∫0 ∫0 𝑟𝑢(𝑡)d𝑡d𝑟. 832

(5.172)

The exact solution of Eq. (5.172) is 𝑢(𝑥) = 𝑥e𝑥 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥).

(5.173)


(5.174)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 ]𝑇 . Putting Eq. (5.174) into Eq. (5.172), we obtained 𝑥

1

1

∑3𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥) = 𝑥e𝑥 − 𝑥 2 + ∫0 ∫0 𝑟 ∑3𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡d𝑟. 2

(5.175)


1

1

∫0 ∑3𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [𝑥e𝑥 − 2 𝑥 2 ] 𝐍𝑗 (𝑥)d𝑥 1

𝑥

1

+ ∫0 [∫0 ∫0 𝑟 ∑3𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡d𝑟 ] 𝐍𝑗 (𝑥)d𝑥 . (5.176) for 𝑗 = 0,1,2. The matrix form of Eq. (5.176) is given as

=

.

After solving the above we get 𝛼0 = −489 + 180e, 𝛼1 = 460 − 168e, 𝛼2 = 1470 − 540e. 𝑢(𝑥) = 261 − 96e − 1500𝑥 + 552𝑥e + 1470𝑥 2 − 540𝑥 2 e.

833



Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.04494446793165740541

4.49445E-02

0.1

0.11051709180756476248

0.11537952519211012635

4.86243E-03

0.2

0.24428055163203396678

0.22837083509487430541

1.59097E-02

0.3

0.40495764227280093120

0.38391839763994994257

2.10392E-02

0.4

0.59672987905650812713

0.58202221282733703784

1.47077E-02

0.5

0.82436063535006407342

0.82268228065703559121

1.67835E-03

0.6

1.09327128023430538493

1.10589860112904560270

1.26273E-02

0.7

1.40962689522933356514

1.43167117424336707230

2.20443E-02

0.8

1.78043274279397408366

1.80000000000000000000

1.95673E-02

0.9

2.21364280004125469742

2.21088507839894438581

2.75772E-03

1.0

2.71828182845904523536

2.66432640944020022974

5.39554E-02

_____________________________________________________________________________


where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼30 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁30 ]𝑇 . Proceeding as before we obtained the approximate solutions.

834



Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00000000000000000000

4.53348E-24

0.1

0.11051709180756476248

0.11051709180756476248

1.25257E-25

0.2

0.24428055163203396678

0.24428055163203396678

7.43404E-25

0.3

0.40495764227280093120

0.40495764227280093120

2.08026E-25

0.4

0.59672987905650812713

0.59672987905650812713

7.08063E-25

0.5

0.82436063535006407342

0.82436063535006407342

8.39535E-25

0.6

1.09327128023430538493

1.09327128023430538493

7.06225E-25

0.7

1.40962689522933356514

1.40962689522933356514

2.05384E-25

0.8

1.78043274279397408366

1.78043274279397408366

7.43431E-25

0.9

2.21364280004125469742

2.21364280004125469742

1.22975E-25

1.0

2.71828182845904523536

2.71828182845904523536

4.51915E-24

_____________________________________________________________________________

Finally 𝑛 = 70. Eq. (5.173) is 70 𝑇 𝑢(𝑥) = ∑70 𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑𝑘=0 𝐂 𝐍𝑘 (𝑥),

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼70 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁70 ]𝑇 . 835

Proceeding as before we have 𝐀 71×71 𝒙71×1 = 𝐛71×1 . After solving we obtained the approximate solutions.



Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00000000000000000000

1.87740E-67

0.1

0.11051709180756476248

0.11051709180756476248

9.26880E-68

0.2

0.24428055163203396678

0.24428055163203396678

1.52881E-67

0.3

0.40495764227280093120

0.40495764227280093120

9.60501E-68

0.4

0.59672987905650812713

0.59672987905650812713

4.71280E-68

0.5

0.82436063535006407342

0.82436063535006407342

1.70393E-67

0.6

1.09327128023430538493

1.09327128023430538493

2.40938E-67

0.7

1.40962689522933356514

1.40962689522933356514

1.82893E-67

0.8

1.78043274279397408366

1.78043274279397408366

1.48825E-67

0.9

2.21364280004125469742

2.21364280004125469742

1.48140E-67

1.0

2.71828182845904523536

2.71828182845904523536

1.68910E-66

_____________________________________________________________________________

836


2

𝑥

𝑥 2 (3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡


(5.177)

The exact solution of Eq. (5.177) is 𝑢(𝑥) = ln(𝑥 + 1). Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation ln(𝑢(𝑡)) = 𝑣(𝑡), 𝑢(𝑡) = exp(𝑣(𝑡)). Using the above transformation into the given integral equation we get 1

2

𝑥

𝑥 2 (3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡

𝑣(𝑡)d𝑡.

(5.178)

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥).

(5.179)

To solve Eq. (5.177) we take 𝑛 = 2. Eq. (5.179) is 𝑣(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥),

(5.180)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 ]𝑇 . Putting Eq. (5.180) into Eq. (5.177), we obtained 1

2

𝑥

𝑥 2 (3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡

∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡.

(5.181)


1

1

𝑥

∫0 3 𝑥 2 (3 + 2𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [∫0

1 √𝑥−𝑡

∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡] 𝐍𝑗 (𝑥)d𝑥 .(5.182)


=

.

After solving the above we get 𝛼0 = 0, 𝛼1 = 1, 𝛼2 = 0. 837

𝑣(𝑥) = 𝑥 + 1. After applying the reverse transformation we get 𝑢(𝑥) = exp(𝑥 + 1). which is the exact solution.

Example 5.54. Consider the nonlinear Abel’s integral equation [229] as 2

1

𝑥

𝑥 2 e(3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡

exp(𝑢(𝑡)) d𝑡.

(5.183)

The exact solution of Eq. (5.183) is 𝑢(𝑥) = 1 + ln(𝑥 + 1). Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation exp(𝑢(𝑡)) = 𝑣(𝑡), 𝑢(𝑡) = ln(𝑣(𝑡)). Using the above transformation into the given integral equation we get 2

1

𝑥

𝑥 2 e(3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡

𝑣(𝑡)d𝑡.

(5.184)

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥).

(5.185)

To solve Eq. (5.183) we take 𝑛 = 2. Eq. (12) is 𝑣(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥),

(5.186)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 ]𝑇 . Putting Eq. (5.186) into Eq. (5.183), we obtained 2

1

𝑥

𝑥 2 e(3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡

∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡.

(5.187)


1

1

𝑥

∫0 3 𝑥 2 e(3 + 2𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [∫0

1 √𝑥−𝑡

∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡 ] 𝐍𝑗 (𝑥)d𝑥.


838

(5.188)

=

.

After solving the above we get 𝛼0 = 0, 𝛼1 = e, 𝛼2 = 0. 𝑣(𝑥) = (𝑥 + 1)e. After applying the reverse transformation we get 𝑢(𝑥) = 1 + ln(𝑥 + 1). which is the exact solution.


1

𝑥

𝑢(𝑥) = 𝑥 7 (1 − 6435 𝑥 2 ) + ∫0

1 1

𝑢(𝑡)d𝑡 .

(5.188)

(𝑥−𝑡)2

The exact solution of Eq. (5.188) is 𝑢(𝑥) = 𝑥 7 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥).

(5.189)

To solve Eq. (5.188) we take 𝑛 = 2. Eq. (5.189) is 𝑣(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥),

(5.190)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 ]𝑇 . Putting Eq. (5.190) into Eq. (5.188), we obtained ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥) = 𝑥 7 (1 −

4096

1

𝑥

𝑥 2 ) + ∫0 6435

1 1 (𝑥−𝑡)2

∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡 .

(5.191)


1

4096

1

∫0 ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥) 𝐍𝑗 (𝑥)d𝑥 = ∫0 𝑥 7 (1 − 6435 𝑥 2 ) 𝐍𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0

1 1 (𝑥−𝑡)2

∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑡) d𝑡 ] 𝐍𝑗 (𝑥)d𝑥. (5.192)

for 𝑗 = 0,1,2. The matrix form of Eq. (5.192) is given as 839

=

.


9076159141

343420539

𝛼0 = 4500101892 , 𝛼1 = − 3000067928 , 𝛼2 = 136366724. 𝑢(𝑥) =

2179192373 900203784

−

1324633303 750016982

343420539

+ 136366724 𝑥 2 .


Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.24212700348785791448

2.42127E-01

0.1

0.00000010000000000000

0.09069683139743450058

9.06967E-02

0.2

0.00001280000000000000

-0.01036613542082882242

1.03789E-02

0.3

0.00021870000000000000

-0.06106189696693205452

6.12806E-02

0.4

0.00163840000000000000

-0.06139045324087519572

6.30289E-02

0.5

0.00781250000000000000

-0.01135180424265824601

1.91643E-02

0.6

0.02799360000000000000

0.08905405002771879459

6.10605E-02

0.7

0.08235430000000000000

0.23982710957025592611

1.57473E-01

0.8

0.20971520000000000000

0.44096737438495314852

2.31252E-01

0.9

0.47829690000000000000

0.69247484447181046184

2.14178E-01

1.0

1.00000000000000000000

0.99434951983082786606

5.65048E-03

_____________________________________________________________________________

840


For 𝑛 = 5. Eq. (5.190) is 𝑢(𝑥) = ∑5𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑5𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥), where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼5 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁5 ]𝑇 . Proceeding as before we obtained the approximate solutions.


841


Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00454443004771315095

4.54443E-03

0.1

0.00000010000000000000

0.00180977563133575316

1.80968E-03

0.2

0.00001280000000000000

-0.00037511574451315215

3.87916E-04

0.3

0.00021870000000000000

-0.00114470464918431265

1.36340E-03

0.4

0.00163840000000000000

0.00155967753617760509

7.87225E-05

0.5

0.00781250000000000000

0.00898773356865777490

1.17523E-03

0.6

0.02799360000000000000

0.02866696538902127204

6.73365E-04

0.7

0.08235430000000000000

0.08148943916716313440

8.64861E-04

0.8

0.20971520000000000000

0.20879855034755842356

9.16650E-04

0.9

0.47829690000000000000

0.47947578869471228596

1.17889E-03

1.0

1.00000000000000000000

0.99702750333861001397

2.97250E-03

_____________________________________________________________________________

Finally 𝑛 = 7. Eq. (5.190) is 𝑢(𝑥) = ∑7𝑘=0 𝛼𝑘 𝜓𝑘 (𝑥) = ∑7𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥), where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼7 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁7 ]𝑇 . Proceeding as before we have 𝐀 7×76 𝒙7×1 = 𝐛7×1 . After solving we obtained the exact solution. 𝑢(𝑥) = 𝑥 7 .

5.9.8. System of Volterra Integral Equations of 1st Kind Example 5.55. Consider the system of Volterra integral equation of 1st Kind [229] as 27

5

𝑥

𝑥 3 (45𝑥 2 + 33𝑥 + 44) = ∫0 440 25

8

5

1

[𝑢(𝑡) + 𝑣(𝑡) + 𝑤(𝑡)]d𝑡.

(5.193a)

[𝑣(𝑡) + 𝑤(𝑡)]d𝑡.

(5.193b)

(𝑥−𝑡)3

𝑥

𝑥 5 (25𝑥 2 + 30𝑥 + 39) = ∫0 468 32

1

1 2

(𝑥−𝑡)5 𝑥

𝑥 4 (96𝑥 2 + 208𝑥 + 117) = ∫0 585

1 3

(𝑥−𝑡)4

842

[𝑢(𝑡) − 𝑣(𝑡)]d𝑡.

(5.193c)

The exact solution of Eq. (5.193) is 𝑢(𝑥) = 𝑥 + 𝑥 2 + 3𝑥 3 , 𝑣(𝑥) = 𝑥 + 3𝑥 2 − 𝑥 3 , 𝑤(𝑥) = 𝑥 − 𝑥 2 + 3𝑥 3 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥),

(5.194a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝑁𝑘 (𝑥),

(5.194b)

𝑤(𝑥) = ∑𝑛𝑘=0 𝜇𝑘 𝑁𝑘 (𝑥).

(5.194c)

To solve Eq. (11) we take 𝑛 = 3. Eq. (5.194) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑3𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥),

(5.195a)

𝑣(𝑥) = ∑3𝑘=0 𝛽𝑘 𝑁𝑘 (𝑥) = ∑3𝑘=0 𝐃𝑇 𝐍𝑘 (𝑥),

(5.195b)

𝑤(𝑥) = ∑3𝑘=0 𝜇𝑘 𝑁𝑘 (𝑥) = ∑3𝑘=0 𝐄𝑇 𝐍𝑘 (𝑥),

(5.195c)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1 , 𝛽2 , 𝛽3 ]𝑇 , 𝐄 = [𝜇0 , 𝜇1 , 𝜇2 , 𝜇3 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , 𝑁3 ]𝑇 . Putting Eq. (5.195) into Eq. (5.193), we obtained 27

5

𝑥

𝑥 3 (45𝑥 2 + 33𝑥 + 44) = ∫0 440 25

8

1 (𝑥−𝑡)3

𝑥

𝑥 5 (25𝑥 2 + 30𝑥 + 39) = ∫0 468 32

1

5

1 2 (𝑥−𝑡)5

𝑥

𝑥 4 (96𝑥 2 + 208𝑥 + 117) = ∫0 585

∑3𝑘=0[𝐂 𝑇 + 𝐃𝑇 + 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡 . (5.196a) ∑3𝑘=0[𝐃𝑇 + 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡 .

1 3 (𝑥−𝑡)4

(5.196b)

∑3𝑘=0[𝐂 𝑇 + 𝐃𝑇 ]𝐍𝑘 (𝑥) d𝑡. (5.196c)


27

5

∫0 [440 𝑥 3 (45𝑥 2 + 33𝑥 + 44)] 𝐍𝑗 (𝑥)d𝑥 = 1

𝑥

∫0 [∫0 1

25

1 1 (𝑥−𝑡)3

∑3𝑘=0[𝐂 𝑇 + 𝐃𝑇 + 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡] 𝐍𝑗 (𝑡)d𝑥 ,

(5.197a)

8

∫0 [468 𝑥 5 (25𝑥 2 + 30𝑥 + 39)] 𝐍𝑗 (𝑥)d𝑥 = 1

𝑥

∫0 [∫0 1

32

1 2 (𝑥−𝑡)5

∑3𝑘=0[𝐃𝑇 + 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡] 𝐍𝑗 (𝑡)d𝑥 ,

5

∫0 [585 𝑥 4 (96𝑥 2 + 208𝑥 + 117)] 𝐍𝑗 (𝑥)d𝑥 = 843

(5.197b)

1

𝑥

∫0 [∫0

1 3 (𝑥−𝑡)4

∑3𝑘=0[𝐂 𝑇 + 𝐃𝑇 ]𝐍𝑘 (𝑥) d𝑡] 𝐍𝑗 (𝑡)d𝑥,

(5.197c)


𝐀 𝟏𝟐 𝐁𝟐𝟐 𝐂𝟑𝟐

𝐛𝟏 𝐀 𝟏𝟑 𝒙𝟏 𝐁𝟐𝟑 ] [𝒙𝟐 ] = [𝐛𝟐 ]. 𝐂𝟑𝟑 𝒙𝟑 𝐛𝟑

where 𝐀 𝟏𝟏 , 𝐀 𝟏𝟐 , 𝐀 𝟏𝟑 , 𝐁𝟐𝟏 , 𝐁𝟐𝟐 , 𝐁𝟐𝟑 , 𝐂𝟑𝟏 , 𝐂𝟑𝟐 and 𝐂𝟑𝟑 are matrices of order 4×4, 𝒙𝟏 , 𝒙𝟐 , 𝒙𝟑 , 𝐛𝟏 , 𝐛𝟐 , and 𝐛𝟑 are matrices of order 4×1. After solving the above 5

7

𝛼0 = − 3 , 𝛼1 = − 3 , 𝛼2 = 1, 𝛼3 = 3, 𝛽0 = −

17 9

1

, 𝛽1 = − 9 , 𝛽2 = −1, 𝛽3 = −1,

𝜇0 = −1, 𝜇1 = −1, 𝜇2 = −1, 𝜇3 = 3. Substituting the values we have 𝑢(𝑥) = 𝑥 + 𝑥 2 + 3𝑥 3 , 𝑣(𝑥) = 𝑥 + 3𝑥 2 − 𝑥 3 , 𝑤(𝑥) = 𝑥 − 𝑥 2 + 3𝑥 3 .

Example 5.56. Consider the system of Volterra integral equation of 1st Kind [229] as 𝑥

𝑢(𝑥) = 1 − 2𝑥 + sin(𝑥) + ∫0 (𝑢(𝑡) + 𝑣(𝑡))d𝑡, 𝑥

𝑣(𝑥) = 1 − 𝑥 2 − sin(𝑥) + ∫0 (𝑡𝑢(𝑡) + 𝑡𝑣(𝑡))d𝑡.

(5.198a) (5.198b)

The exact solution of Eq. (5.198) is 𝑢(𝑥) = 1 + sin(𝑥), 𝑣(𝑥) = 1 − sin(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥),

(5.199a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝑁𝑘 (𝑥),

(5.199b)


(5.200a)

𝑣(𝑥) = ∑2𝑘=0 𝛽𝑘 𝑁𝑘 (𝑥) = ∑2𝑘=0 𝐃𝑇 𝐍𝑘 (𝑥),

(5.200b)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1 , 𝛽2 ]𝑇 and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 ]𝑇 . Putting Eq. (5.200) into Eq. (5.198), we obtained 𝑥

∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥) = 1 − 2𝑥 + sin(𝑥) + ∫0 ∑2𝑘=0(𝐂 𝑇 + 𝐃𝑇 )𝐍𝑘 (𝑥) d𝑡 , (5.201a) 844

𝑥

∑2𝑘=0 𝐃𝑇 𝐍𝑘 (𝑥) = 1 − 𝑥 2 − sin(𝑥) + ∫0 ∑2𝑘=0(𝑡𝐂 𝑇 + 𝑡𝐃𝑇 )𝐍𝑘 (𝑥) d𝑡 . (5.201b) Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.201) is 1

1

∫0 ∑2𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [1 − 2𝑥 + sin(𝑥)]𝐍𝑗 (𝑥)d𝑥 + 1

𝑥

∫0 [∫0 ∑2𝑘=0(𝐂 𝑇 + 𝐃𝑇 )𝐍𝑘 (𝑥) d𝑡]𝐍𝑗 (𝑥)d𝑥 . 1

(5.202a)

1

∫0 ∑2𝑘=0 𝐃𝑇 𝐍𝑘 (𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [1 − 𝑥 2 − sin(𝑥)] 𝐍𝑗 (𝑥)d𝑥 + 1

𝑥

∫0 [∫0 ∑2𝑘=0(𝑡𝐂 𝑇 + 𝑡𝐃𝑇 )𝐍𝑘 (𝑥) d𝑡 ] 𝐍𝑗 (𝑥)d𝑥. (5.202b) for 𝑗 = 0,1,2. The matrix form of Eq. (5.202) is given as

=

.

After solving the above we get 𝛼0 = 0.013657, 𝛼1 = 1.2214, 𝛼2 = −0.24385, 𝛽0 = 2.0001, 𝛽1 = −1.2503, 𝛽2 = 0.26246. 𝑢(𝑥) = 0.99121 + 1.0995𝑥 − 0.24385𝑥 2 , 𝑣(𝑥) = 1.0123 − 1.1191𝑥 + 0.26246𝑥 2 .

845

(a) (b) Fig 5.34. (a)-(b). Comparison of Exact and Approximate solutions of Eq. (5.193)

Table 5.33. Comparison of Exact and Approximate solutions of Eq. (5.193) obtained from Newton’s polynomials Method (NPM) ______________________________________________________________________________ 𝑥

𝑢(𝑥)

𝑣(𝑥)

𝑢Approx. (𝑥)

𝑣Approx. (𝑥) Error in 𝑢(𝑥) Error in 𝑣(𝑥)

______________________________________________________________________________ 0.0 1.0000000000 1.0000000000 0.9912100000 1.0123000000 8.7900E-03 1.2300E-02 0.1 1.0998000000 0.9001700000 1.0988000000 0.9030100000 1.0000E-03 2.8400E-03 0.2 1.1987000000 0.8013300000 1.2013000000 0.7989800000 2.6000E-03 2.3500E-03 0.3 1.2955000000 0.7044800000 1.2992000000 0.7001900000 3.7000E-03 4.2900E-03 0.4

1.3894000000 0.6105800000 1.3920000000 0.6066500000 2.6000E-03 3.9300E-03

0.5 1.4794000000 0.5205700000 1.4800000000 0.5183600000 6.0000E-04 2.2100E-03 0.6 1.5646000000 0.4353600000 1.5631000000 0.4353300000 1.5000E-03 3.0000E-05 0.7 1.6442000000 0.3557800000 1.6414000000 0.3575400000 2.8000E-03 1.7600E-03 0.8 1.7174000000 0.2826400000 1.7147000000 0.2849900000 2.7000E-03 2.3500E-03 0.9 1.7833000000 0.2166700000 1.7833000000 0.2176900000 0.0000E+00 1.0200E-03 1.0 1.8415000000 0.1585300000 1.8468000000 0.1556600000 5.3000E-03 2.8700E-03 ______________________________________________________________________________

For 𝑛 = 15. Eq. (5.199) is 15 𝑇 𝑢(𝑥) = ∑15 𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑𝑘=0 𝐂 𝐍𝑘 (𝑥), 15 𝑇 𝑣(𝑥) = ∑15 𝑘=0 𝛽𝑘 𝑁𝑘 (𝑥) = ∑𝑘=0 𝐃 𝐍𝑘 (𝑥),

846

where

𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼15 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1 , 𝛽2 … , 𝛽15 ]𝑇

and

𝐍𝑘 (𝑥) =

[𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁15 ]𝑇 . Proceeding as before we have the following error Table 3.1nd graphical representation



𝑢(𝑥)

𝑣(𝑥)

𝑢Approx. (𝑥)


______________________________________________________________________________ 0.0 1.0000000000 1.0000000000 1.0000000000 1.0000000000 3.6924E-23 3.6924E-23 0.1 1.0998334166 0.9001665834 1.0998334166 0.9001665834 8.7450E-24 8.7450E-24 0.2 1.1986693308 0.8013306692 1.1986693308 0.8013306692 2.8144E-24 2.8144E-24 0.3 1.2955202067 0.7044797933 1.2955202067 0.7044797933 6.6327E-24 6.6327E-24 0.4 1.3894183423 0.6105816577 1.3894183423 0.6105816577 7.3364E-24 7.3364E-24 0.5 1.4794255386 0.5205744614 1.4794255386 0.5205744614 7.4611E-24 7.4611E-24 0.6 1.5646424734 0.4353575266 1.5646424734 0.4353575266 7.4884E-24 7.4884E-24 0.7 1.6442176872 0.3557823128 1.6442176872 0.3557823128 6.9671E-24 6.9671E-24 0.8 1.7173560909 0.2826439091 1.7173560909 0.2826439091 3.2472E-24 3.2472E-24 0.9 1.7833269096 0.2166730904 1.7833269096 0.2166730904 9.0069E-24 9.0069E-24 1.0 1.8414709848 0.1585290152 1.8414709848 0.1585290152 3.9031E-23 3.9031E-23 ______________________________________________________________________________

For 𝑛 = 35. Eq. (5.199) is 847

35 𝑇 𝑢(𝑥) = ∑35 𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥) = ∑𝑘=0 𝐂 𝐍𝑘 (𝑥), 35 𝑇 𝑣(𝑥) = ∑35 𝑘=0 𝛽𝑘 𝑁𝑘 (𝑥) = ∑𝑘=0 𝐃 𝐍𝑘 (𝑥),

where

𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼35 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1 , 𝛽2 … , 𝛽35 ]𝑇

and

𝐍𝑘 (𝑥) =

[𝑁0 , 𝑁1 , 𝑁2 , … , 𝑁35 ]𝑇 . Proceeding as before we have the following error Table 3.1nd graphical representation



𝑢(𝑥)

𝑣(𝑥)

𝑢Approx. (𝑥)


______________________________________________________________________________ 0.0 1.0000000000 1.0000000000 1.0000000000 1.0000000000 3.0617E-28 2.4832E-27 0.1 1.0998334166 0.9001665834 1.0998334166 0.9001665834 9.9177E-29 4.1996E-28 0.2 1.1986693308 0.8013306692 1.1986693308 0.8013306692 1.2768E-29 2.8866E-28 0.3 1.2955202067 0.7044797933 1.2955202067 0.7044797933 2.2701E-28 2.7800E-28 0.4 1.3894183423 0.6105816577 1.3894183423 0.6105816577 7.6827E-29 2.6211E-28 0.5 1.4794255386 0.5205744614 1.4794255386 0.5205744614 1.1076E-28 4.0552E-29 0.6 1.5646424734 0.4353575266 1.5646424734 0.4353575266 2.5264E-28 1.9815E-28 0.7 1.6442176872 0.3557823128 1.6442176872 0.3557823128 2.1066E-28 2.8546E-28 0.8

1.7173560909 0.2826439091 1.7173560909 0.2826439091 2.5847E-28 1.9881E-28

0.9 1.7833269096 0.2166730904 1.7833269096 0.2166730904 3.4261E-28 3.3061E-28 1.0 1.8414709848 0.1585290152 1.8414709848 0.1585290152 1.9395E-27 1.8418E-27 ______________________________________________________________________________

848

5.9.9. System of Volterra Integral Equations Consider the system of Volterra integral equation [229] as 𝑥

1

𝑢(𝑥) = 1 − 𝑥 2 + 4 𝑥 4 + ∫0 [𝑢(𝑡) + 𝑣(𝑡) − 𝑤(𝑡)]d𝑡, 2

1

𝑥

1

𝑥

𝑣(𝑥) = 2𝑥 + 𝑥 2 − 3 𝑥 3 − 4 𝑥 4 + ∫0 [𝑣(𝑡) + 𝑤(𝑡) − 𝑢(𝑡)]d𝑡, 𝑤(𝑥) = −𝑥 + 𝑥 2 + 𝑥 3 − 4 𝑥 4 + ∫0 [𝑢(𝑡) − 𝑣(𝑡) + 𝑤(𝑡)]d𝑡.

(5.203a) (5.203b) (5.203c)

The exact solution of Eq. (5.203) is 𝑢(𝑥) = 1 + 𝑥, 𝑣(𝑥) = 𝑥 + 𝑥 2 , 𝑤(𝑥) = 𝑥 2 + 𝑥 3 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥),

(5.204a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝑁𝑘 (𝑥),

(5.204b)

𝑤(𝑥) = ∑𝑛𝑘=0 𝜇𝑘 𝑁𝑘 (𝑥).

(5.204c)


(5.205a)

𝑣(𝑥) = ∑3𝑘=0 𝛽𝑘 𝑁𝑘 (𝑥) = ∑3𝑘=0 𝐃𝑇 𝐍𝑘 (𝑥),

(5.205b)

𝑤(𝑥) = ∑3𝑘=0 𝜇𝑘 𝑁𝑘 (𝑥) = ∑3𝑘=0 𝐄𝑇 𝐍𝑘 (𝑥),

(5.205c)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1 , 𝛽2 , 𝛽3 ]𝑇 , 𝐄 = [𝜇0 , 𝜇1 , 𝜇2 , 𝜇3 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑁0 , 𝑁1 , 𝑁2 , 𝑁3 ]𝑇 . Putting Eq. (5.205) into Eq. (5.203), we obtained 𝑥

1

∑3𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥) = 1 − 𝑥 2 + 𝑥 4 + ∫0 ∑3𝑘=0[𝐂 𝑇 + 𝐃𝑇 − 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡, 4 2

(5.206a)

𝑥

1

∑3𝑘=0 𝐃𝑇 𝐍𝑘 (𝑥) = 2𝑥 + 𝑥 2 − 𝑥 3 − 𝑥 4 + ∫0 ∑3𝑘=0[−𝐂 𝑇 + 𝐃𝑇 + 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡 ,(5.206b) 3 4 𝑥

1

∑3𝑘=0 𝐄𝑇 𝐍𝑘 (𝑥) = −𝑥 + 𝑥 2 + 𝑥 3 − 𝑥 4 + ∫0 ∑3𝑘=0[𝐂 𝑇 − 𝐃𝑇 + 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡. (5.206c) 4 Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.206) is 𝑥

1

∑3𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥) = 1 − 𝑥 2 + 𝑥 4 + ∫0 ∑3𝑘=0[𝐂 𝑇 + 𝐃𝑇 − 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡, 4 2

1

𝑥

(5.207a)

∑3𝑘=0 𝐃𝑇 𝐍𝑘 (𝑥) = 2𝑥 + 𝑥 2 − 𝑥 3 − 𝑥 4 + ∫0 ∑3𝑘=0[−𝐂 𝑇 + 𝐃𝑇 + 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡 ,(5.207b) 3 4 849

𝑥

1

∑3𝑘=0 𝐄𝑇 𝐍𝑘 (𝑥) = −𝑥 + 𝑥 2 + 𝑥 3 − 𝑥 4 + ∫0 ∑3𝑘=0[𝐂 𝑇 − 𝐃𝑇 + 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡. (5.207c) 4 1

1

1

∫0 ∑3𝑘=0 𝐂 𝑇 𝐍𝑘 (𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [1 − 𝑥 2 + 4 𝑥 4 ] 𝐍𝑗 (𝑥)d𝑥 + 1

𝑥

∫0 [∫0 ∑3𝑘=0[𝐂 𝑇 + 𝐃𝑇 − 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡]𝐍𝑗 (𝑡)d𝑥, 1

1

2

(5.208a)

1

∫0 ∑3𝑘=0 𝐃𝑇 𝐍𝑘 (𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [2𝑥 + 𝑥 2 − 3 𝑥 3 − 4 𝑥 4 ] 𝐍𝑗 (𝑥)d𝑥 + 1

𝑥

∫0 [∫0 ∑3𝑘=0[−𝐂 𝑇 + 𝐃𝑇 + 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡]𝐍𝑗 (𝑡)d𝑥, 1

1

(5.208b)

1

∫0 ∑3𝑘=0 𝐃𝑇 𝐍𝑘 (𝑥)𝐍𝑗 (𝑥)d𝑥 = ∫0 [−𝑥 + 𝑥 2 + 𝑥 3 − 4 𝑥 4 ] 𝐍𝑗 (𝑥)d𝑥 + 1

𝑥

∫0 [∫0 ∑3𝑘=0[𝐂 𝑇 − 𝐃𝑇 + 𝐄𝑇 ]𝐍𝑘 (𝑥) d𝑡]𝐍𝑗 (𝑡)d𝑥,

(5.208c)


𝐀 𝟏𝟐 𝐁𝟐𝟐 𝐂𝟑𝟐

𝐛𝟏 𝐀 𝟏𝟑 𝒙𝟏 𝐁𝟐𝟑 ] [𝒙𝟐 ] = [𝐛𝟐 ]. 𝐂𝟑𝟑 𝒙𝟑 𝐛𝟑

where 𝐀 𝟏𝟏 , 𝐀 𝟏𝟐 , 𝐀 𝟏𝟑 , 𝐁𝟐𝟏 , 𝐁𝟐𝟐 , 𝐁𝟐𝟑 , 𝐂𝟑𝟏 , 𝐂𝟑𝟐 and 𝐂𝟑𝟑 are matrices of order 4×4, 𝒙𝟏 , 𝒙𝟐 , 𝒙𝟑 , 𝐛𝟏 , 𝐛𝟐 , and 𝐛𝟑 are matrices of order 4×1. After solving the above 1

4

𝛼0 = −1, 𝛼1 = 1, 𝛼2 = 0, 𝛼3 = 0, 𝛽0 = 3 , 𝛽1 = − 3 , 𝛽2 = 1, 𝛽3 = 0, 1

1

𝜇0 = − 9 , 𝜇1 = 9 , 𝜇2 = −1, 𝜇3 = 1. Substituting the values we have 𝑢(𝑥) = 1 + 𝑥, 𝑣(𝑥) = 𝑥 + 𝑥 2 , 𝑤(𝑥) = 𝑥 2 + 𝑥 3 .

5.10. Touchard Polynomials Method (TPM) We develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Touchard polynomials [41]. Touchard polynomials is given as 𝑛!

𝑇(𝑥) = ∑𝑛𝑘=0 [𝑘!(𝑛−𝑘)! . (𝑥)𝑘 ]. First four Touchard polynomials is given as 𝑇0 (𝑥) = 1, 𝑇1 (𝑥) = 1 + 𝑥, 850

𝑇2 (𝑥) = 1 + 2𝑥 + 𝑥 2 , 𝑇3 (𝑥) = 1 + 3𝑥 + 3𝑥 2 + 𝑥 3 ,



(5.209)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, to find an approximate solution 𝑢̃(𝑥) of Eq. (5.209). For this, we assume that 𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝑇𝑘 (𝑥),

(5.210)

where 𝑇𝑘 (𝑥) are Touchard polynomials of degree 𝑘 defined earlier and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.209) into (5.210), we get 𝑥

𝑓(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝑇𝑘 (𝑡)d𝑡,

(5.211)

Then the Galerkin equations are obtained by multiplying both sides of (5.211) by 𝑇𝑗 (𝑥) and then integrating with respect to 𝑥 from 𝑎 to 𝑏, we have 𝑏

𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝑇𝑗 (𝑥)d𝑥 = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝑇𝑘 (𝑡)d𝑡)𝑇𝑗 (𝑡)d𝑥, 𝑗 = 0,1,2, … 𝑛,( 5.212) Since in each equation, there are two integrals. The inner integrand of the left side is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.212) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.213)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝑇𝑘 (𝑡)d𝑡 )𝑇𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛 𝑏

𝐹𝑗 = ∫𝑎 𝑓(𝑥) 𝑇𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛

851




(5.214)


(5.215)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝑇𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑇𝑘 (𝑡)d𝑡)𝑇𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛. Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.215) and substituting these values of parameters in (5.209), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.214).

5.10.2. Volterra Integral Equation Consider the Volterra integral equation of the form [229] as 1

1

𝑥

𝑢(𝑥) = 1 − 𝑥 − 2 𝑥 2 + 2 ∫0 (𝑡 − 𝑥)2 𝑢(𝑡)d𝑡 .

0≤𝑥≤1

(5.216)

The exact solution of Eq. (5.216) is 𝑢(𝑥) = exp(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.217)

To solve Eq. (5.216) we take 𝑛 = 3. Eq. (5.217) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.218)

Putting Eq. (5.218) into Eq. (5.216), we obtained 1

𝑥

∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥) = 1 − 𝑥 − 𝑥 2 − ∫0 (𝑡 − 𝑥) ∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑡) d𝑡. 2

(5.219)

1

[4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ] = 1 − 𝑥 − 𝑥 2 + 2 𝑥 ∫ (𝑡 2 0 1

− 𝑥)2 [4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ]d𝑡.

852

(5.220)

Multiplying both sides by 𝑇𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.220) is 1

1

1

∫0 [4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ]𝑇𝑗 (𝑥)d𝑥 = ∫0 [1 − 𝑥 − 2 𝑥 2 ] 𝑇𝑗 (𝑥)d𝑥 1 1

𝑥

+ ∫0 [2 ∫0 (𝑡 − 𝑥)2 [4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ]d𝑡 ] 𝑇𝑗 (𝑥)d𝑥.(5.221) for 𝑗 = 0,1,2,3. The matrix form of Eq. (5.221) is given as

=

.


25901388072

10614750048

1426749576

𝛼0 = 25604772985 , 𝛼1 = 25604772985 , 𝛼2 = − 25604772985 , 𝛼3 = 5120954597. 25580715272

26073131616

10786493592

1426749576

𝑢(𝑥) = 25604772985 + 25604772985 𝑥 + 25604772985 𝑥 2 + 5120954597 𝑥 3 .


853

Table 5.36. Comparison of Exact and Approximate solutions of Eq. (5.216) obtained from Touchard polynomials Method (TPM) _____________________________________________________________________________ Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

0.99906042076553095438

9.39579E-04

0.1

1.10517091807564762481

1.10538090433298172825

2.09986E-04

0.2

1.22140275816016983392

1.22179842563911722180

3.95667E-04

0.3

1.34985880757600310398

1.34998464517103001372

1.25838E-04

0.4

1.49182469764127031782

1.49161122341581268271

2.13474E-04

0.5

1.64872127070012814685

1.64834982086055780744

3.71450E-04

0.6

1.82211880039050897488

1.82187209799235796661

2.46702E-04

0.7

2.01375270747047652162

2.01384971529830573891

9.70078E-05

0.8

2.22554092849246760458

2.22595433326549370303

4.13405E-04

0.9

2.45960311115694966380

2.45985761238101443765

2.54501E-04

1.0

2.71828182845904523536

2.71723121313196052146

1.05062E-03

_____________________________________________________________________________

We take 𝑛 = 40. Eq. (5.217) is 𝑢(𝑥) = ∑25 𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥). Proceeding as before we have we obtained the approximate solution of given integral equation.


854



Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

1.15499E-73

0.1

1.10517091807564762481

1.10517091807564762481

1.31496E-74

0.2

1.22140275816016983392

1.22140275816016983392

1.60311E-74

0.3

1.34985880757600310398

1.34985880757600310398

1.47146E-74

0.4

1.49182469764127031782

1.49182469764127031782

1.26815E-74

0.5

1.64872127070012814685

1.64872127070012814685

8.56474E-77

0.6

1.82211880039050897488

1.82211880039050897488

1.27936E-74

0.7

2.01375270747047652162

2.01375270747047652162

1.47522E-74

0.8

2.22554092849246760458

2.22554092849246760458

1.61467E-74

0.9

2.45960311115694966380

2.45960311115694966380

1.31814E-74

1.0

2.71828182845904523536

2.71828182845904523536

1.16899E-73

_____________________________________________________________________________

For 𝑛 = 75. Eq. (5.217) is 𝑢(𝑥) = ∑75 𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥). Proceeding as before we have we obtained the approximate solution of given integral equation.


855



Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

2.36479E-156

0.1

1.10517091807564762481

1.10517091807564762481

7.06648E-158

0.2

1.22140275816016983392

1.22140275816016983392

1.23536E-157

0.3

1.34985880757600310398

1.34985880757600310398

2.25242E-157

0.4

1.49182469764127031782

1.49182469764127031782

2.08709E-157

0.5

1.64872127070012814685

1.64872127070012814685

2.16426E-157

0.6

1.82211880039050897488

1.82211880039050897488

2.08564E-157

0.7

2.01375270747047652162

2.01375270747047652162

2.25916E-157

0.8

2.22554092849246760458

2.22554092849246760458

1.22934E-157

0.9

2.45960311115694966380

2.45960311115694966380

6.99765E-158

1.0

2.71828182845904523536

2.71828182845904523536

2.38029E-156

_____________________________________________________________________________

5.10.3. Fredholm Integral Equation Consider the Fredholm integral equation of the form [229] as 𝜋

𝑢(𝑥) = −2𝑥 + sin(𝑥) + cos(𝑥) − ∫0 𝑥𝑢(𝑡)d𝑡 .

0 ≤𝑥≤1

(5.222)

The exact solution of Eq. (5.222) is 𝑢(𝑥) = sin(𝑥) + cos(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.223)


(5.224)

Putting Eq. (5.224) into Eq. (5.222), we obtained 𝜋

∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥) = −2𝑥 + sin(𝑥) + cos(𝑥) + ∫0 𝑥 ∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥) d𝑡. (5.225) [4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ] = −2𝑥 + sin(𝑥) + cos(𝑥) + 𝜋

∫0 𝑥[4𝛼0 + 6𝛼1 𝑡 + 4𝛼2 𝑡 2 + 𝛼3 𝑡 3 ]d𝑡.

856

(5.226)

Multiplying both sides by 𝑇𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0, 𝜋] we have Eq. (5.226) is 𝜋

𝜋

∫0 [4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ]𝑇𝑗 (𝑥)d𝑥 = ∫0 [−2𝑥 + sin(𝑥) + cos(𝑥)]𝑇𝑗 (𝑥)d𝑥 − 𝜋

𝜋

∫0 [∫0 𝑥[4𝛼0 + 6𝛼1 𝑡 + 4𝛼2 𝑡 2 + 𝛼3 𝑡 3 ]d𝑡 ]𝑇𝑗 (𝑥)d𝑥.

(5.227)


=

.

After solving the above we get 𝛼0 = 0.30460, 𝛼1 = 1.2609, 𝛼2 = −0.37859, 𝛼3 = −0.0041117. 𝑢(𝑥) = 1.1828 + 0.49142𝑥 − 0.39092𝑥 2 − 0.0041117.


857



Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

0.94049736997266509844

5.95026E-02

0.1

1.09483758192485391840

1.06935167781706304510

2.54859E-02

0.2

1.17873590863630284658

1.17705064759113470914

1.68526E-03

0.3

1.25085669578694559475

1.26446465017472129913

1.36080E-02

0.4

1.31047933631153557446

1.33246405644766402361

2.19847E-02

0.5

1.35700810049457571639

1.38191923728980409116

2.49111E-02

0.6

1.38997808830471365444

1.41370056358098271034

2.37225E-02

0.7

1.40905987452217947993

1.42867840620104108969

1.96185E-02

0.8

1.41406280024668818255

1.42772313602982043779

1.36603E-02

0.9

1.40493687789814784495

1.41170512394716196320

6.76825E-03

1.0

1.38177329067603622405

1.38149474083290687446

2.78550E-04

_____________________________________________________________________________

We take 𝑛 = 40. Eq. (5.223) is 𝑢(𝑥) = ∑25 𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥). Proceeding as before we have we obtained the approximate solution of given integral equation.


858



Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

1.64358E-53

0.1

1.09483758192485391840

1.09483758192485391840

1.70391E-55

0.2

1.17873590863630284658

1.17873590863630284658

1.45664E-55

0.3

1.25085669578694559475

1.25085669578694559475

2.63930E-54

0.4

1.31047933631153557446

1.31047933631153557446

8.89913E-55

0.5

1.35700810049457571639

1.35700810049457571639

6.41365E-55

0.6

1.38997808830471365444

1.38997808830471365444

1.32221E-54

0.7

1.40905987452217947993

1.40905987452217947993

1.50058E-54

0.8

1.41406280024668818255

1.41406280024668818255

1.39849E-54

0.9

1.40493687789814784495

1.40493687789814784495

1.07670E-54

1.0

1.38177329067603622405

1.38177329067603622405

5.39839E-55

_____________________________________________________________________________

For 𝑛 = 80. Eq. (5.223) is 𝑢(𝑥) = ∑80 𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥). Proceeding as before we have we obtained the approximate solution of given integral equation.


859



Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

9.40261E-126

0.1

1.09483758192485391840

1.09483758192485391840

1.36054E-126

0.2

1.17873590863630284658

1.17873590863630284658

9.49415E-127

0.3

1.25085669578694559475

1.25085669578694559475

9.02520E-127

0.4

1.31047933631153557446

1.31047933631153557446

2.46605E-127

0.5

1.35700810049457571639

1.35700810049457571639

6.57210E-127

0.6

1.38997808830471365444

1.38997808830471365444

8.44101E-127

0.7

1.40905987452217947993

1.40905987452217947993

8.24062E-127

0.8

1.41406280024668818255

1.41406280024668818255

7.04832E-127

0.9

1.40493687789814784495

1.40493687789814784495

3.83547E-127

1.0

1.38177329067603622405

1.38177329067603622405

1.66619E-127

_____________________________________________________________________________

5.10.4. Volterra Integro-Differential Equation Consider the Volterra integro-differential equations [229] as 𝑥

𝑢′ (𝑥) = 1 + 𝑥 + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 ,

0 ≤ 𝑥 ≤ 1.

(5.228)

𝑢(0) = 1. The exact solution of Eq. (5.228) is 𝑢(𝑥) = exp(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.229)


(5.230)


[∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥)]′ = 1 + 𝑥 + ∫0 (𝑥 − 𝑡) ∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥) d𝑡.

(5.231)

[4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ]′ = 1 + 𝑥 + 𝑥

∫0 (𝑥 − 𝑡)[4𝛼0 + 6𝛼1 𝑡 + 4𝛼2 𝑡 2 + 𝛼3 𝑡 3 ]d𝑡. 860

(5.232)


1

∫0 [4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ]′ 𝑇𝑗 (𝑥)d𝑥 = ∫0 [1 + 𝑥]𝑇𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 (𝑥 − 𝑡)[4𝛼0 + 6𝛼1 𝑡 + 4𝛼2 𝑡 2 + 𝛼3 𝑡 3 ]d𝑡 ]𝑇𝑗 (𝑥)d𝑥. (5.233) for 𝑗 = 0,1,2,3. The matrix form of Eq. (5.233) is given as

=

.


19047651

7203672

5534172

𝛼0 = 21074971 , 𝛼1 = 21074971 , 𝛼2 = − 21074971 , 𝛼3 = 21074971. 21242823

494676

5534172

𝑢(𝑥) = 1 + 21074971 𝑥 + 1109209 𝑥 2 + 21074971 𝑥 3 .


861



Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

0.00000E+00

0.1

1.10517091807564762481

1.10551876498430294400

3.47847E-04

0.2

1.22140275816016983392

1.22153253430336867367

1.29776E-04

0.3

1.34985880757600310398

1.34961687510744380146

2.41932E-04

0.4

1.49182469764127031782

1.49134735454677493981

4.77343E-04

0.5

1.64872127070012814685

1.64829953977160870115

4.21731E-04

0.6

1.82211880039050897488

1.82204899793219169792

6.98025E-05

0.7

2.01375270747047652162

2.01417129617877054255

4.18589E-04

0.8

2.22554092849246760458

2.22624200166159184750

7.01073E-04

0.9

2.45960311115694966380

2.45983668153090222520

2.33570E-04

1.0

2.71828182845904523536

2.71653090293694828809

1.75093E-03

_____________________________________________________________________________

For 𝑛 = 40. Eq. (5.229) is 𝑢(𝑥) = ∑40 𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥). Proceeding as before we obtained the approximate solution.


862



Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

0.00000E+00

0.1

1.10517091807564762481

1.10517091807564762481

4.71457E-75

0.2

1.22140275816016983392

1.22140275816016983392

9.09025E-75

0.3

1.34985880757600310398

1.34985880757600310398

9.00533E-75

0.4

1.49182469764127031782

1.49182469764127031782

1.83389E-74

0.5

1.64872127070012814685

1.64872127070012814685

1.15965E-74

0.6

1.82211880039050897488

1.82211880039050897488

1.11691E-74

0.7

2.01375270747047652162

2.01375270747047652162

2.27393E-74

0.8

2.22554092849246760458

2.22554092849246760458

2.45089E-74

0.9

2.45960311115694966380

2.45960311115694966380

2.64206E-74

1.0

2.71828182845904523536

2.71828182845904523536

1.98935E-73

_____________________________________________________________________________

Finally we take 𝑛 = 80. Eq. (5.229) is 𝑢(𝑥) = ∑80 𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥). Proceeding as before we have 𝐀 81×81 𝒙81×1 = 𝐛81×1 . After solving the above equation we obtained approximate solution.


863



Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

1.00000E-200

0.1

1.10517091807564762481

1.10517091807564762481

6.86587E-113

0.2

1.22140275816016983392

1.22140275816016983392

2.37782E-112

0.3

1.34985880757600310398

1.34985880757600310398

1.85546E-112

0.4

1.49182469764127031782

1.49182469764127031782

3.35905E-112

0.5

1.64872127070012814685

1.64872127070012814685

3.05711E-112

0.6

1.82211880039050897488

1.82211880039050897488

2.17635E-112

0.7

2.01375270747047652162

2.01375270747047652162

1.97002E-112

0.8

2.22554092849246760458

2.22554092849246760458

1.09863E-112

0.9

2.45960311115694966380

2.45960311115694966380

4.40935E-112

1.0

2.71828182845904523536

2.71828182845904523536

4.58409E-111

_____________________________________________________________________________


1

𝑢(𝑥) = 1 + 𝑥 + e𝑥 + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 + ∫0 e𝑥−𝑡 𝑢(𝑡)d𝑡 ,

(5.234)

The exact solution of Eq. (5.234) is 𝑢(𝑥) = e𝑥 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.235)

To solve Eq. (5.234) we take 𝑛 = 3. Eq. (5.235) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑁𝑘 (𝑥).

(5.236)

Putting Eq. (5.236) into Eq. (5.234), we obtained ∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥) = 1 + 𝑥 + e𝑥 + 𝑥

1

∫0 (𝑥 − 𝑡) ∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑡) d𝑡 + ∫0 e𝑥−𝑡 ∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑡) d𝑡 . [4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ] = 1 + 𝑥 + e𝑥 +

864

(5.237)

𝑥

∫0 (𝑥 − 𝑡) [

4𝛼0 + 6𝛼1 𝑡 4𝛼0 + 6𝛼1 𝑡 1 𝑥−𝑡 [ ] d𝑡. 2 3 ] d𝑡 + ∫0 e +4𝛼2 𝑡 + 𝛼3 𝑡 +4𝛼2 𝑡 2 + 𝛼3 𝑡 3

(5.238)


1

∫0 [4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ]𝑇𝑗 (𝑥)d𝑥 = ∫0 [1 + 𝑥 + e𝑥 ]𝑇𝑗 (𝑥)d𝑥 1

𝑥

1

1

+ ∫0 [∫0 (𝑥 − 𝑡)[4𝛼0 + 6𝛼1 𝑡 + 4𝛼2 𝑡 2 + 𝛼3 𝑡 3 ]d𝑡 ]𝑇𝑗 (𝑥)d𝑥 + ∫0 [∫0 e𝑥−𝑡 [4𝛼0 + 6𝛼1 𝑡 + 4𝛼2 𝑡 2 + 𝛼3 𝑡 3 ]d𝑡] 𝑇𝑗 (𝑥)d𝑥 .

(5.239)


=

.

After solving the above we get 𝛼0 = 0.010935, 𝛼1 = 1.1597, 𝛼2 = −0.45866, 𝛼3 = 0.27682. 𝑢(𝑥) = 0.98880 + 1.0729𝑥 + 0.37180𝑥 2 + 0.27682𝑥 3 .


865



Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

0.98880000000000000000

1.12000E-02

0.1

1.10520000000000000000

1.10010000000000000000

5.10000E-03

0.2

1.22140000000000000000

1.22050000000000000000

9.00000E-04

0.3

1.34990000000000000000

1.35170000000000000000

1.80000E-03

0.4

1.49180000000000000000

1.49520000000000000000

3.40000E-03

0.5

1.64870000000000000000

1.65280000000000000000

4.10000E-03

0.6

1.82210000000000000000

1.82620000000000000000

4.10000E-03

0.7

2.01380000000000000000

2.01690000000000000000

3.10000E-03

0.8

2.22550000000000000000

2.22670000000000000000

1.20000E-03

0.9

2.45960000000000000000

2.45740000000000000000

2.20000E-03

1.0

2.71830000000000000000

2.71030000000000000000

8.00000E-03

_____________________________________________________________________________

For 𝑛 = 20. Eq. (5.235) is 𝑢(𝑥) = ∑20 𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥). Proceeding as before we obtained the approximate solution.


866



Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

5.94372E-32

0.1

1.10517091807564762481

1.10517091807564762481

1.18028E-32

0.2

1.22140275816016983392

1.22140275816016983392

1.10105E-32

0.3

1.34985880757600310398

1.34985880757600310398

5.75007E-33

0.4

1.49182469764127031782

1.49182469764127031782

9.71557E-33

0.5

1.64872127070012814685

1.64872127070012814685

1.17532E-34

0.6

1.82211880039050897488

1.82211880039050897488

9.67290E-33

0.7

2.01375270747047652162

2.01375270747047652162

5.99236E-33

0.8

2.22554092849246760458

2.22554092849246760458

1.10995E-32

0.9

2.45960311115694966380

2.45960311115694966380

1.19351E-32

1.0

2.71828182845904523536

2.71828182845904523536

6.08355E-32

_____________________________________________________________________________

Finally we take 𝑛 = 50. Eq. (5.235) is 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥). Proceeding as before we have 𝐀 51×51 𝒙51×1 = 𝐛51×1 . After solving we obtained the approximate solution.

867




Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000000000000

2.40034E-50

0.1

1.10517091807564762481

1.10517091807564762481

4.45252E-51

0.2

1.22140275816016983392

1.22140275816016983392

2.74740E-51

0.3

1.34985880757600310398

1.34985880757600310398

3.56524E-51

0.4

1.49182469764127031782

1.49182469764127031782

3.39233E-51

0.5

1.64872127070012814685

1.64872127070012814685

3.37850E-51

0.6

1.82211880039050897488

1.82211880039050897488

3.36640E-51

0.7

2.01375270747047652162

2.01375270747047652162

3.49213E-51

0.8

2.22554092849246760458

2.22554092849246760458

2.62000E-51

0.9

2.45960311115694966380

2.45960311115694966380

4.26891E-51

1.0

2.71828182845904523536

2.71828182845904523536

2.27694E-50

_____________________________________________________________________________


𝑥

1

𝑢(𝑥) = 𝑥 2 − 3 𝑥 + ∫0 ∫−1(1 + 𝑟𝑡)𝑢(𝑡)d𝑡d𝑟. The exact solution of Eq. (5.240) is 𝑢(𝑥) = 𝑥 2 . According to the proposed technique, consider the trail solution 868

(5.240)

𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.241)


(5.242)


2

1

∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥) = 𝑥 2 − 𝑥 + ∫0 ∫−1(1 + 𝑟𝑡) ∑3𝑘=0 𝛼𝑘 𝑇𝑘 (𝑡) d𝑡d𝑟 . 3

(5.243)

2

[4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ] = 𝑥 2 − 𝑥 + 3 𝑥

1

∫0 ∫−1(1 + 𝑟𝑡)[4𝛼0 + 6𝛼1 𝑡 + 4𝛼2 𝑡 2 + 𝛼3 𝑡 3 ]d𝑡d𝑟 . (5.244) Multiplying both sides by 𝑇𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [−1,1] we have Eq. (5.244) is 1

1

2

∫−1[4𝛼0 + 6𝛼1 𝑥 + 4𝛼2 𝑥 2 + 𝛼3 𝑥 3 ]𝑇𝑗 (𝑥)d𝑥 = ∫−1 [𝑥 2 − 3 𝑥] 𝑇𝑗 (𝑥)d𝑥 1

𝑥

1

+ ∫−1 [∫0 ∫−1(1 + 𝑟𝑡)[4𝛼0 + 6𝛼1 𝑡 + 4𝛼2 𝑡 2 + 𝛼3 𝑡 3 ]d𝑡d𝑟 ] 𝑇𝑗 (𝑥)d𝑥.(5.245) for 𝑗 = 0,1,2,3. The matrix form of Eq. (5.245) is given as

=

.

After solving the above we get 𝛼0 = 1, 𝛼1 = −2, 𝛼2 = 1, 𝛼3 = 0. Substituting the values we get 𝑢(𝑥) = 𝑥 2 , which is the exact solution.


5

𝑥

𝑥 6 (11 + 6𝑥) = ∫0 55

1 1

e𝑢(𝑡) d𝑡.

(5.246)

(𝑥−𝑡)6

The exact solution of Eq. (5.246) is 𝑢(𝑥) = ln(1 + 𝑥). 869

Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation e𝑢(𝑡) = 𝑣(𝑡), 𝑢(𝑡) = ln(𝑣(𝑡)). Using the above transformation into the given integral equation we get 6

5

𝑥

𝑥 6 (11 + 6𝑥) = ∫0 55

1 1

𝑣(𝑡)d𝑡.

(5.247)

(𝑥−𝑡)6

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.248)

To solve Eq. (5.246) we take 𝑛 = 2. Eq. (5.248) is 𝑣(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.249)


5

𝑥

𝑥 6 (11 + 6𝑥) = ∫0 55 6

5

𝑥

𝑥 6 (11 + 6𝑥) = ∫0 55

1 1

(𝑥−𝑡)6 1 1 (𝑥−𝑡)6

∑2𝑘=0 𝛼𝑘 𝑇𝑘 (𝑡) d𝑡.

(5.250)

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡.

Multiplying both sides by 𝑇𝑗 (𝑥), 𝑗 = 0,1,2. Then integrating over [0,1] we have Eq. (5.250) is 1 6

∫0

5

1

𝑥

𝑥 6 (11 + 6𝑥)𝑇𝑗 (𝑥)d𝑥 = ∫0 [∫0 55 1 6

for 𝑗 = 0, ∫0

1

𝑥

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡] 𝑇𝑗 (𝑥)d𝑥.(5.251) 1

1 (𝑥−𝑡)6

5

1

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡] d𝑥, 𝑥

𝑥 6 (11 + 6𝑥)(1 + 𝑥)d𝑥 = ∫0 [∫0 55

1 6

for 𝑗 = 2, ∫0

1

(𝑥−𝑡)6

𝑥 6 (11 + 6𝑥)d𝑥 = ∫0 [∫0 55

1 6

for 𝑗 = 1, ∫0

5

1

1 1 (𝑥−𝑡)6

[

3𝛼0 + 3𝛼1 𝑡 ] (1 + 𝑥)d𝑡] d𝑥, +𝛼2 𝑡 2

5

𝑥 6 (11 + 6𝑥)(1 + 2𝑥 + 𝑥 2 )d𝑥 = 55 1

𝑥

∫0 [∫0

1 1

(𝑥−𝑡)6

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ](1 + 2𝑥 + 𝑥 2 )d𝑡] d𝑥,

The matrix form is given as

870

=

.

After solving the above we get 𝛼0 = 0, 𝛼1 = 1, 𝛼2 = 0. 𝑣(𝑥) = 𝑥 + 1. After applying the reverse transformation we get 𝑢(𝑥) = ln(𝑥 + 1). which is the exact solution.


2

𝑥

𝑥 3 (5𝜋 + 3𝑥) = ∫0 10

1 1

sinh−1(𝑢(𝑡)) d𝑡 .

(5.252)

(𝑥−𝑡)3

The exact solution of Eq. (5.252) is 𝑢(𝑥) = sinh(𝜋 + 𝑥). Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation sinh−1(𝑢(𝑡)) = 𝑣(𝑡), 𝑢(𝑡) = sinh(𝑣(𝑡)). Using the above transformation into the given integral equation we get 3

2

𝑥

𝑥 3 (5𝜋 + 3𝑥) = ∫0 10

1 1

𝑣(𝑡)d𝑡.

(5.253)

(𝑥−𝑡)3

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.254)


(5.255)


2

𝑥

𝑥 3 (5𝜋 + 3𝑥) = ∫0 10 3

2

𝑥

𝑥 3 (5𝜋 + 3𝑥) = ∫0 10

1 1 (𝑥−𝑡)3

1 1 (𝑥−𝑡)3

∑2𝑘=0 𝛼𝑘 𝑇𝑘 (𝑡) d𝑡. [3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡.

871

(5.256)

Multiplying both sides by 𝑇𝑗 (𝑥), 𝑗 = 0,1,2. Then integrating over [0,1] we have Eq. (5.256) is 1 3

∫0

2

1

𝑥

𝑥 3 (5𝜋 + 3𝑥)𝑇𝑗 (𝑥)d𝑥 = ∫0 [∫0 10 1 3

for 𝑗 = 0, ∫0

1

𝑥

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡] 𝑇𝑗 (𝑥)d𝑥.(5.256) 1

1 (𝑥−𝑡)3

2

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡] d𝑥,

1

𝑥

𝑥 3 (5𝜋 + 3𝑥)(1 + 𝑥)d𝑥 = ∫0 [∫0 10

1 3

for 𝑗 = 2, ∫0

1 (𝑥−𝑡)3

𝑥 3 (5𝜋 + 3𝑥)d𝑥 = ∫0 [∫0 10

1 3

for 𝑗 = 1, ∫0

2

1

10

1 1 (𝑥−𝑡)3

[

3𝛼0 + 3𝛼1 𝑡 ] (1 + 𝑥)d𝑡] d𝑥, +𝛼2 𝑡 2

2

𝑥 3 (5𝜋 + 3𝑥)(1 + 2𝑥 + 𝑥 2 )d𝑥 = 1

𝑥

∫0 [∫0

1 1 (𝑥−𝑡)3

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ](1 + 2𝑥 + 𝑥 2 )d𝑡] d𝑥,

The matrix form is given as

=

.

After solving the above we get 𝛼0 = −1 + 𝜋, 𝛼1 = 1, 𝛼2 = 0. 𝑣(𝑥) = 𝜋 + 𝑥. After applying the reverse transformation we get 𝑢(𝑥) = sinh(𝑥 + 𝜋). which is the exact solution.

5.10.7. Weakly Singular Integral Equation Consider the weakly singular integral equation [229] as 𝑢(𝑥) = 𝑥 2 −

128 45

9

𝑥

𝑥 4 + ∫0

1 3

𝑢(𝑡)d𝑡.

(5.257)

(𝑥−𝑡)4

The exact solution of Eq. (5.257) is 𝑢(𝑥) = 𝑥 2 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.258) 872


(5.259)

Putting Eq. (5.259) into Eq. (5.257), we obtained ∑2𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥) = 𝑥 2 −

128 45

9

𝑥

1

𝑥 4 + ∫0

[3𝛼0 + 3𝛼1 𝑥 + 𝛼2 𝑥 2 ] = 𝑥 2 −

128 45

3 (𝑥−𝑡)4 9

∑2𝑘=0 𝛼𝑘 𝑇𝑘 (𝑡) d𝑡. 𝑥

𝑥 4 + ∫0

1 3 (𝑥−𝑡)4

(5.260)

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡 .


1

∫0 [3𝛼0 + 3𝛼1 𝑥 + 𝛼2 𝑥 2 ]𝑇𝑗 (𝑥)d𝑥 = ∫0 [𝑥 2 − 1

𝑥

+ ∫0 [∫0 for 𝑗 = 0,

1 3

(𝑥−𝑡)4

1

1

1

𝑥

1 3 (𝑥−𝑡)4

45

9

𝑥 4 ] 𝑇𝑗 (𝑥)d𝑥

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡] 𝑇𝑗 (𝑥)d𝑥.

∫0 [3𝛼0 + 3𝛼1 𝑥 + 𝛼2 𝑥 2 ]d𝑥 = ∫0 [𝑥 2 − + ∫0 [∫0

128

128 45

9

𝑥 4 ] d𝑥

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡] d𝑥,

1

1

for 𝑗 = 1, ∫0 [3𝛼0 + 3𝛼1 𝑥 + 𝛼2 𝑥 2 ](1 + 𝑥)d𝑥 = ∫0 [𝑥 2 − 1

𝑥

+ ∫0 [∫0 for 𝑗 = 2,

1

∫0 [

1 3 (𝑥−𝑡)4

(5.261)

128 45

9

𝑥 4 ] (1 + 𝑥)d𝑥

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ](1 + 𝑥)d𝑡] d𝑥,

3𝛼0 + 3𝛼1 𝑥 1 128 9 2 )d𝑥 2 (1 ] + 2𝑥 + 𝑥 = [𝑥 − 𝑥 4 ] (1 + 2𝑥 + 𝑥 2 )d𝑥 ∫ 0 45 +𝛼2 𝑥 2 1

𝑥

+ ∫0 [∫0

1 3 (𝑥−𝑡)4

[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ](1 + 2𝑥 + 𝑥 2 )d𝑡] d𝑥,

The matrix form of Eq. (5.261) is given as

=

.

After solving the above we get 𝛼0 = 1, 𝛼1 = −2, 𝛼2 = 1. Substituting the values we have 873


5.10.8 System of Volterra Integral Equations Consider the system of Volterra integral equation [229] as 𝑥

1

𝑢(𝑥) = 1 − 𝑥 2 + 4 𝑥 4 + ∫0 [𝑢(𝑡) + 𝑣(𝑡) − 𝑤(𝑡)]d𝑡, 2

1

𝑥

1

𝑥

𝑣(𝑥) = 2𝑥 + 𝑥 2 − 3 𝑥 3 − 4 𝑥 4 + ∫0 [𝑣(𝑡) + 𝑤(𝑡) − 𝑢(𝑡)]d𝑡, 𝑤(𝑥) = −𝑥 + 𝑥 2 + 𝑥 3 − 4 𝑥 4 + ∫0 [𝑢(𝑡) − 𝑣(𝑡) + 𝑤(𝑡)]d𝑡.

(5.262a) (5.262b) (5.262c)

The exact solution of Eq. (5.262) is 𝑢(𝑥) = 1 + 𝑥, 𝑣(𝑥) = 𝑥 + 𝑥 2 , 𝑤(𝑥) = 𝑥 2 + 𝑥 3 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥),

(5.263a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝑇𝑘 (𝑥),

(5.263b)

𝑤(𝑥) = ∑𝑛𝑘=0 𝜇𝑘 𝑇𝑘 (𝑥).

(5.263c)

To solve Eq. (5.262) we take 𝑛 = 2. Eq. (5.263) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥),

(5.264a)

𝑣(𝑥) = ∑2𝑘=0 𝛽𝑘 𝑇𝑘 (𝑥),

(5.264b)

𝑤(𝑥) = ∑2𝑘=0 𝜇𝑘 𝑇𝑘 (𝑥).

(5.264c)


1

∑2𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥) = 1 − 𝑥 2 + 𝑥 4 + ∫0 ∑2𝑘=0[𝛼𝑘 + 𝛽𝑘 − 𝜇𝑘 ] 𝑇𝑘 (𝑥)d𝑡, 4 2

(5.265a)

𝑥

1

∑2𝑘=0 𝛽𝑘 𝑇𝑘 (𝑥) = 2𝑥 + 𝑥 2 − 𝑥 3 − 𝑥 4 + ∫0 ∑2𝑘=0[−𝛼𝑘 + 𝛽𝑘 + 𝜇𝑘 ] 𝑇𝑘 (𝑥)d𝑡,(5.265b) 3 4 1

𝑥

∑2𝑘=0 𝜇𝑘 𝑇𝑘 (𝑥) = −𝑥 + 𝑥 2 + 𝑥 3 − 𝑥 4 + ∫0 ∑2𝑘=0[𝛼𝑘 − 𝛽𝑘 + 𝜇𝑘 ] 𝑇𝑘 (𝑥)d𝑡. (5.265c) 4

𝑥

1

[3𝛼0 + 3𝛼1 𝑥 + 𝛼2 𝑥 2 ] = 1 − 𝑥 2 + 𝑥 4 + ∫0 [3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡 4 𝑥

𝑥

+ ∫0 [3𝛽0 + 3𝛽1 𝑡 + 𝛽2 𝑡 2 ]d𝑡 − ∫0 [3𝜇0 + 3𝜇1 𝑡 + 𝜇2 𝑡 2 ]d𝑡, (5.266a) 2

1

𝑥

[3𝛽0 + 3𝛽1 𝑥 + 𝛽2 𝑥 2 ] = 2𝑥 + 𝑥 2 − 𝑥 3 − 𝑥 4 − ∫0 [3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡 3 4 𝑥

𝑥

+ ∫0 [3𝛽0 + 3𝛽1 𝑡 + 𝛽2 𝑡 2 ]d𝑡 + ∫0 [3𝜇0 + 3𝜇1 𝑡 + 𝜇2 𝑡 2 ]d𝑡, (5.266b)

874

1

𝑥

[3𝜇0 + 3𝜇1 𝑥 + 𝜇2 𝑥 2 ] = −𝑥 + 𝑥 2 + 𝑥 3 − 𝑥 4 + ∫0 [3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡 4 𝑥

𝑥

− ∫0 [3𝛽0 + 3𝛽1 𝑡 + 𝛽2 𝑡 2 ]d𝑡 + ∫0 [3𝜇0 + 3𝜇1 𝑡 + 𝜇2 𝑡 2 ]d𝑡. (5.266c) Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.266) is 1

1

1

1

𝑥

∫0 [3𝛼0 + 3𝛼1 𝑥 + 𝛼2 𝑥 2 ] 𝑇𝑗 (𝑥)d𝑥 = ∫0 [1 − 𝑥 2 + 4 𝑥 4 ] 𝑇𝑗 (𝑥) + ∫0 [∫0 [3𝛼0 + 3𝛼1 𝑡 + 𝑥

𝑥

𝛼2 𝑡 2 ]d𝑡 + ∫0 [3𝛽0 + 3𝛽1𝑡 + 𝛽2 𝑡 2 ]d𝑡 − ∫0 [3𝜇0 + 3𝜇1 𝑡 + 𝜇2 𝑡 2 ]d𝑡] 𝑇𝑗 (𝑥)d𝑥, (5.267a) 1

∫0 [

2𝑥 + 𝑥 2 3𝛽0 + 3𝛽1 𝑥 1 1 𝑥 (𝑥)d𝑥 2 1 4 ] 𝑇𝑗 (𝑥) + ∫ [− ∫ [3𝛼0 + 3𝛼1 𝑡 + ] 𝑇 = [ ∫ 3 𝑗 2 0 0 0 − 𝑥 − 𝑥 +𝛽2𝑥 3 4 𝑥

𝛼2 𝑡 2 ]d𝑡 + ∫0 [

3𝛽0 + 3𝛽1 𝑡 𝑥 3𝜇0 + 3𝜇1 𝑡 ] d𝑡 ] 𝑇𝑗 (𝑥)d𝑥, 2 ] d𝑡 + ∫0 [ +𝜇2 𝑡 2 +𝛽2 𝑡

1 3𝜇0 + 3𝜇1 𝑥 ] 𝑇𝑗 (𝑥)d𝑥 ∫0 [ +𝜇2 𝑥 2 𝑥

(5.267b)

−𝑥 + 𝑥 2 1 1 𝑥 = ∫0 [ 3 1 4 ] 𝑇𝑗 (𝑥) + ∫0 [∫0 [3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 ]d𝑡 − +𝑥 − 4 𝑥 𝑥

∫0 [3𝛽0 + 3𝛽1 𝑡 + 𝛽2 𝑡 2 ]d𝑡 + ∫0 [3𝜇0 + 3𝜇1 𝑡 + 𝜇2 𝑡 2 ]d𝑡 ] 𝑇𝑗 (𝑥)d𝑥.

(5.267c)


=

After solving we obtained the approximate solution.

875

.

(a) (b) (c) Fig 5.50. (a)-(b)-(c). Comparison of Exact and Approximate solutions of Eq. (5.262)


Approximate Solution of u(x)

Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

0.99597032745926106087

4.02967E-03

0.1

1.10000000000000000000

1.09817354677515643310

1.82645E-03

0.2

1.20000000000000000000

1.19992106485064231543

7.89351E-05

0.3

1.30000000000000000000

1.30121288168571870785

1.21288E-03

0.4

1.40000000000000000000

1.40204899728038561037

2.04900E-03

0.5

1.50000000000000000000

1.50242941163464302297

2.42941E-03

0.6

1.60000000000000000000

1.60235412474849094567

2.35412E-03

0.7

1.70000000000000000000

1.70182313662192937847

1.82314E-03

0.8

1.80000000000000000000

1.80083644725495832136

8.36447E-04

0.9

1.90000000000000000000

1.89939405664757777434

6.05943E-04

1.0

2.00000000000000000000

1.99749596479978773741

2.50404E-03

_____________________________________________________________________________

876


Approximate Solution of v(x)

Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00468304331483405930

4.68304E-03

0.1

0.11000000000000000000

0.11260054834501514582

2.60055E-03

0.2

0.24000000000000000000

0.24091007583965330444

9.10076E-04

0.3

0.39000000000000000000

0.38961162579874853517

3.88374E-04

0.4

0.56000000000000000000

0.55870519822230083800

1.29480E-03

0.5

0.75000000000000000000

0.74819079311031021293

1.80921E-03

0.6

0.96000000000000000000

0.95806841046277665996

1.93159E-03

0.7

1.19000000000000000000

1.18833805027970017910

1.66195E-03

0.8

1.44000000000000000000

1.43899971256108077034

1.00029E-03

0.9

1.71000000000000000000

1.71005339730691843368

5.33973E-05

1.0

2.00000000000000000000

2.00149910451721316912

1.49910E-03

_____________________________________________________________________________


Approximate Solution of w(x)

Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.05327017047338978929

5.32702E-02

0.1

0.01100000000000000000

0.01618767550357087580

5.18768E-03

0.2

0.04800000000000000000

0.02959139452097198576

1.84086E-02

0.3

0.11700000000000000000

0.09348132752559311915

2.35187E-02

0.4

0.22400000000000000000

0.20785747451743427599

1.61425E-02

0.5

0.37500000000000000000

0.37271983549649545625

2.28016E-03

0.6

0.57600000000000000000

0.58806841046277665996

1.20684E-02

0.7

0.83300000000000000000

0.85390319941627788710

2.09032E-02

0.8

1.15200000000000000000

1.17022420235699913768

1.82242E-02

0.9

1.53900000000000000000

1.53703141928494041170

1.96858E-03

1.0

2.00000000000000000000

1.95432485020010170916

4.56751E-02

877


(5.264a)

𝑣(𝑥) = ∑3𝑘=0 𝛽𝑘 𝑇𝑘 (𝑥),

(5.264b)

𝑤(𝑥) = ∑3𝑘=0 𝜇𝑘 𝑇𝑘 (𝑥).

(5.264c)

Proceeding as before we obtained 𝛼0 = 0, 𝛼1 = 1, 𝛼2 = 0, 𝛼3 = 0, 𝛽0 = 0, 𝛽1 = −1, 𝛽2 = 1, 𝛽3 = 0, 𝜇0 = 0, 𝜇1 = 1, 𝜇2 = −2, 𝜇3 = 1. Substituting the values we have 𝑢(𝑥) = 1 + 𝑥, 𝑣(𝑥) = 𝑥 + 𝑥 2 , 𝑤(𝑥) = 𝑥 2 + 𝑥 3 .


1

𝑢(𝑥) = 𝑥 − 3 + ∫−1[𝑣(𝑡) + 𝑤(𝑡)]d𝑡, 1

2

𝑣(𝑥) = 𝑥 + 𝑥 2 − 3 + ∫−1[𝑤(𝑡) + 𝑢(𝑡)]d𝑡, 2

𝑥

𝑤(𝑥) = 𝑥 2 + 𝑥 3 − + ∫0 [𝑢(𝑡) + 𝑣(𝑡)]d𝑡. 3

(5.265a) (5.265b) (5.265c)

The exact solution of Eq. (5.265) is 𝑢(𝑥) = 𝑥, 𝑣(𝑥) = 𝑥 + 𝑥 2 , 𝑤(𝑥) = 𝑥 2 + 𝑥 3 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥),

(5.266a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝑇𝑘 (𝑥),

(5.266b)

𝑤(𝑥) = ∑𝑛𝑘=0 𝜇𝑘 𝑇𝑘 (𝑥).

(5.266c)


(5.267a)

𝑣(𝑥) = ∑3𝑘=0 𝛽𝑘 𝑇𝑘 (𝑥),

(5.267b)

𝑤(𝑥) = ∑3𝑘=0 𝜇𝑘 𝑇𝑘 (𝑥).

(5.267c)

Putting Eq. (13) into Eq. (5.267), we obtained 878

1

1

∑2𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥) = 1 − 𝑥 2 + 𝑥 4 + ∫−1 ∑2𝑘=0[𝛼𝑘 + 𝛽𝑘 − 𝜇𝑘 ] 𝑇𝑘 (𝑥)d𝑡, 4 2

(5.268a)

1

1

∑2𝑘=0 𝛽𝑘 𝑇𝑘 (𝑥) = 2𝑥 + 𝑥 2 − 𝑥 3 − 𝑥 4 + ∫−1 ∑2𝑘=0[−𝛼𝑘 + 𝛽𝑘 + 𝜇𝑘 ] 𝑇𝑘 (𝑥)d𝑡,(5.268b) 3 4 1

1

∑2𝑘=0 𝜇𝑘 𝑇𝑘 (𝑥) = −𝑥 + 𝑥 2 + 𝑥 3 − 𝑥 4 + ∫−1 ∑2𝑘=0[𝛼𝑘 − 𝛽𝑘 + 𝜇𝑘 ] 𝑇𝑘 (𝑥)d𝑡. (5.268c) 4 1

4

[3𝛼0 + 3𝛼1 𝑥 + 𝛼2 𝑥 2 + ⋯ ] = 𝑥 − + ∫−1[3𝛽0 + 3𝛽1 𝑡 + 𝛽2 𝑡 2 + ⋯ ]d𝑡 3 1

+ ∫−1[3𝜇0 + 3𝜇1 𝑡 + 𝜇2 𝑡 2 + ⋯ ]d𝑡,

(5.269a)

1

2

[3𝛽0 + 3𝛽1 𝑥 + 𝛽2 𝑥 2 + ⋯ ] = 𝑥 + 𝑥 2 − + ∫−1[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 + ⋯ ]d𝑡 3 1

+ ∫−1[3𝜇0 + 3𝜇1 𝑡 + 𝜇2 𝑡 2 + ⋯ ]d𝑡,

(5.269b)

1

2

[3𝜇0 + 3𝜇1 𝑥 + 𝜇2 𝑥 2 + ⋯ ] = 𝑥 2 + 𝑥 3 − + ∫−1[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 + ⋯ ]d𝑡 3 1

+ ∫−1[3𝛽0 + 3𝛽1 𝑡 + 𝛽2 𝑡 2 + ⋯ ]d𝑡.

(5.269c)


1

1

4

1

∫−1[3𝛼0 + 3𝛼1 𝑥 + 𝛼2 𝑥 2 + ⋯ ] 𝑇𝑗 (𝑥)d𝑥 = ∫−1 [𝑥 − 3] 𝑇𝑗 (𝑥) + ∫−1 [∫−1[3𝛽0 + 1

3𝛽1 𝑡 + 𝛽2 𝑡 2 + ⋯ ]d𝑡 + ∫−1[3𝜇0 + 3𝜇1 𝑡 + 𝜇2 𝑡 2 + ⋯ ]d𝑡] 𝑇𝑗 (𝑥)d𝑥, 1

1

(5.270a)

2

∫−1[3𝛽0 + 3𝛽1 𝑥 + 𝛽2 𝑥 2 + ⋯ ]𝑇𝑗 (𝑥)d𝑥 = ∫−1 [𝑥 − 𝑥 2 − 3] 𝑇𝑗 (𝑥) + 1

1

1

∫−1 [∫−1[3𝛼0 + 3𝛼1 𝑡 + 𝛼2 𝑡 2 + ⋯ ]d𝑡 + + ∫−1[3𝜇0 + 3𝜇1 𝑡 + 𝜇2 𝑡 2 + ⋯ ]d𝑡] 𝑇𝑗 (𝑥)d𝑥, (5.270b) 1

1

2

∫−1[3𝜇0 + 3𝜇1 𝑥 + 𝜇2 𝑥 2 + ⋯ ]𝑇𝑗 (𝑥)d𝑥 = ∫−1 [𝑥 2 + 𝑥 3 − 3] 𝑇𝑗 (𝑥) + 1

1

∫−1 [∫−1 [

3𝛼0 + 3𝛼1 𝑡 1 3𝛽0 + 3𝛽1 𝑡 ] d𝑡 + ∫−1 [ ] d𝑡] 𝑇𝑗 (𝑥)d𝑥. +𝛼2 𝑡 2 + ⋯ +𝛽2 𝑡 2 + ⋯

(5.270c)

for 𝑗 = 0,1,2,3. The matrix form of Eq. (5.270) is given as 𝐀 𝟏𝟏 [𝐁𝟐𝟑 𝐂𝟑𝟏

𝐀 𝟏𝟐 𝐁𝟐𝟐 𝐂𝟑𝟐

𝐛𝟏 𝐀 𝟏𝟑 𝒙𝟏 𝐁𝟐𝟑 ] [𝒙𝟐 ] = [𝐛𝟐 ]. 𝐂𝟑𝟑 𝒙𝟑 𝐛𝟑

where 𝐀 𝟏𝟏 , 𝐀 𝟏𝟐 , 𝐀 𝟏𝟑 , 𝐁𝟐𝟏 , 𝐁𝟐𝟐 , 𝐁𝟐𝟑 , 𝐂𝟑𝟏 , 𝐂𝟑𝟐 and 𝐂𝟑𝟑 are matrices of order 4×4, 𝒙𝟏 , 𝒙𝟐 , 𝒙𝟑 , 𝐛𝟏 , 𝐛𝟐 , and 𝐛𝟑 are matrices of order 4×1. After solving the above 𝛼0 = −1, 𝛼1 = 1, 𝛼2 = 0, 𝛼3 = 0, 𝛽0 = 0, 𝛽1 = −1, 𝛽2 = 1, 𝛽3 = 0, 879

𝜇0 = 0, 𝜇1 = 1, 𝜇2 = −2, 𝜇3 = 1. Substituting the values we have 𝑢(𝑥) = 𝑥, 𝑣(𝑥) = 𝑥 + 𝑥 2 , 𝑤(𝑥) = 𝑥 2 + 𝑥 3 .

5.11. Zernike Polynomials Method I develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Zernike polynomials [182]. Zernike polynomials is given as 𝑛−𝑚 2 𝑍(𝑥) = ∑𝑘=0 [

(−1)𝑘 (𝑛−𝑘)! 𝑘!

1 Γ(

𝑛+𝑚 𝑛−𝑚 +1)Γ( +1) 2+𝑘 2+𝑘

𝑥 𝑛−2𝑘 ].

(5.271)

For 𝑚 = 0, first four Zernike polynomials is given as 𝑍0 (𝑥) = 1, 𝑍1 (𝑥) =

4𝑥 𝜋

, 2 2

𝑍2 (𝑥) =

2 1 8𝑥 Γ(3) −9 , 2 2 4 Γ( ) 3

2 𝑥(−16𝑥 2 +3𝜋)

𝑍3 (𝑥) = − 3

𝜋

,



(5.272)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, to find an approximate solution 𝑢̃(𝑥) of Eq. (5.272). For this, we assume that 𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝑍𝑘 (𝑥),

(5.273)

where 𝑍𝑘 (𝑥) are Zernike polynomials of degree 𝑘 defined in equation (5.271) and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.273) into (5.272), we get 𝑥

𝑓(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝑍𝑘 (𝑡)d𝑡, 880

(5.274)

Then the Galerkin equations are obtained by multiplying both sides of (4) by 𝑍𝑗 (𝑥) and then integrating with respect to 𝑥 from 𝑎 to 𝑏, we have 𝑏

𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝑍𝑗 (𝑥)d𝑥 = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝑍𝑘 (𝑡)d𝑡)𝑍𝑗 (𝑡)d𝑥, 𝑗 = 0,1,2, … 𝑛,(5.275) Since in each equation, there are two integrals. The inner integrand of the left side is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.275) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.276)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝑍𝑘 (𝑡)d𝑡)𝑍𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛 𝑏

𝐹𝑗 = ∫𝑎 𝑓(𝑥) 𝑍𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛 Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.276) and substituting these values of parameters in (5.273), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.272).



(5.277)


(5.278)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝑍𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑍𝑘 (𝑡)d𝑡)𝑍𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛. Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.278) and substituting these values of parameters in (5.273), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.277). 881

5.11.2. Volterra Integral Equation Consider the Volterra integral equation [229] of the form 𝑥

e𝑥 − 1 − 𝑥 = ∫0 (𝑥 − 𝑡 + 1)𝑢(𝑡)d𝑡 .

0≤𝑥≤1

(5.279)

The exact solution of Eq. (5.279) is 𝑢(𝑥) = sinh(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥).

(5.280)

To solve Eq. (5.279) we take 𝑛 = 3. Eq. (5.280) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥) = ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥),

(5.281)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝐙𝑘 (𝑥) = [𝑍0 , 𝑍1 , 𝑍2 , 𝑍3 ]𝑇 . Putting Eq. (5.281) into Eq. (5.279), we obtained 𝑥

e𝑥 − 1 − 𝑥 = ∫0 (𝑥 − 𝑡 + 1) ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 .

(5.282)

Multiplying both sides by 𝑍𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.282) is 1

1

𝑥

∫0 [e𝑥 − 1 − 𝑥]𝑇𝑗 (𝑥)d𝑥 − ∫0 [∫0 (𝑥 − 𝑡 + 1) ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 ]𝑇𝑗 (𝑥)d𝑥 = 0. 1

1

(5.283)

𝑥

for 𝑗 = 0, ∫0 [e𝑥 − 1 − 𝑥]d𝑥 = ∫0 [∫0 (𝑥 − 𝑡 + 1) ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 ]d𝑥 . 1

for 𝑗 = 1, ∫0 [e𝑥 − 1 − 𝑥] for 𝑗 = 2,

1 ∫0 [e𝑥

4𝑥 𝜋

1

𝑥

4𝑡

d𝑥 = ∫0 [∫0 (𝑥 − 𝑡 + 1) ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡] 𝜋 d𝑥 . 2 2

−1−

2 1 8𝑥 Γ(3) −9 𝑥] [4 ] d𝑥 2 2 Γ( )

2 2

=

3

1

2 𝑥(−16𝑥 2 +3𝜋)

1

for 𝑗 = 3, ∫0 [1 − 𝑥 − 2 𝑥 2 ] [− 3

1

𝜋 𝑥

∫0 [∫0

2 (𝑥 − 𝑡 + 1) 1 𝑥 1 8𝑡 Γ(3) −9 ] [ ] d𝑥 . ∫0 [∫0 3 ∑𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 4 Γ(2)2 3

] d𝑥 =

(𝑥 − 𝑡 + 1) 2 𝑡(−16𝑡 2 +3𝜋) ] [− 3 ] d𝑥 . 3 𝑇 𝜋 ∑𝑘=0 𝐂 𝐙𝑘 (𝑡) d𝑡


=

After solving the above we get

882

.

𝛼0 = −0.28918, 𝛼1 = 1.1474, 𝛼2 = −0.22709, 𝛼3 = 0.15363. 𝑢(𝑥) = −0.010520 + 1.1533𝑥 − 0.45419𝑥 + 0.52229𝑥 2 . Table 5.51. Comparison of Exact and Approximate solutions of Eq. (5.279) obtained from Zernike polynomials Method (ZPM) _____________________________________________________________________________ 𝑥

Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00013085303588964035

1.30853E-04

0.1

0.10016675001984402582

0.10021257290225667775

4.58229E-05

0.2

0.20133600254109398763

0.20138800433862002910

5.20018E-05

0.3

0.30452029344714261896

0.30450810059899568932

1.21928E-05

0.4

0.41075232580281550854

0.41068552100917893403

6.68048E-05

0.5

0.52109530549374736162

0.52103292489496503884

6.23806E-05

0.6

0.63665358214824127112

0.63666297158214927938

9.38943E-06

0.7

0.75858370183953350346

0.75868832039652693126

1.04619E-04

0.8

0.88810598218762300657

0.88822163066389327011

1.15648E-04

0.9

1.02651672570817527596

1.02637556171004357154

1.41164E-04

1.0

1.17520119364380145688

1.17426277286077311117

9.38421E-04

_____________________________________________________________________________


If we take Eq. (5.279) we take 𝑛 = 10. Eq. (5.280) is 10 𝑇 𝑢(𝑥) = ∑10 𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥) = ∑𝑘=0 𝐂 𝐙𝑘 (𝑥),

883

(5.281)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼10 ]𝑇 , and 𝐙𝑘 (𝑥) = [𝑍0 , 𝑍1 , 𝑍2 , … , 𝑍10 ]𝑇 . Proceeding as before we have we obtained the approximate solution of given integral equation.


Table 5.52. Comparison of Exact and Approximate solutions of Eq. (5.279) obtained from Zernike polynomials Method (ZPM) _____________________________________________________________________________ 𝑥

Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00000000000002020645

2.02064E-14

0.1

0.10016675001984402582

0.10016675001984912937

5.10355E-15

0.2

0.20133600254109398763

0.20133600254109037166

3.61596E-15

0.3

0.30452029344714261896

0.30452029344714764965

5.03069E-15

0.4

0.41075232580281550854

0.41075232580280931342

6.19512E-15

0.5

0.52109530549374736162

0.52109530549375210685

4.74522E-15

0.6

0.63665358214824127112

0.63665358214824090228

3.68847E-16

0.7

0.75858370183953350346

0.75858370183952846087

5.04259E-15

0.8

0.88810598218762300657

0.88810598218763288328

9.87670E-15

0.9

1.02651672570817527596

1.02651672570816240383

1.28721E-14

1.0

1.17520119364380145688

1.17520119364354441917

2.57038E-13

_____________________________________________________________________________

Finally we take Eq. (5.279) we take 𝑛 = 20. Eq. (5.280) is 20 𝑇 𝑢(𝑥) = ∑20 𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥) = ∑𝑘=0 𝐂 𝐙𝑘 (𝑥),

884

(5.281)




Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00000000000000000000

2.05923E-32

0.1

0.10016675001984402582

0.10016675001984402582

3.33846E-33

0.2

0.20133600254109398763

0.20133600254109398763

3.19170E-33

0.3

0.30452029344714261896

0.30452029344714261896

4.03458E-33

0.4

0.41075232580281550854

0.41075232580281550854

2.24785E-33

0.5

0.52109530549374736162

0.52109530549374736162

3.54150E-33

0.6

0.63665358214824127112

0.63665358214824127112

5.01761E-33

0.7

0.75858370183953350346

0.75858370183953350346

2.69187E-33

0.8

0.88810598218762300657

0.88810598218762300657

6.31821E-33

0.9

1.02651672570817527596

1.02651672570817527596

1.08588E-32

1.0

1.17520119364380145688

1.17520119364380145688

4.67978E-31

_____________________________________________________________________________

5.11.3. Fredholm Integral Equation: Consider the Fredholm integral equation of the form [229] as 885

𝜋

𝑢(𝑥) = 𝑥 + sec 2 (𝑥) − ∫04 𝑥𝑢(𝑡)d𝑡 .

0 ≤𝑥≤

𝜋

(5.284)

4

The exact solution of Eq. (5.284) is 𝑢(𝑥) = sec 2 (𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥).

(5.285)


(5.286)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝐙𝑘 (𝑥) = [𝑍0 , 𝑍1 , 𝑍2 , 𝑍3 ]𝑇 . Putting Eq. (5.286) into Eq. (5.284), we obtained 𝜋

∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥) = 𝑥 + sec 2 (𝑥) − ∫04 𝑥 ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 .

(5.287) 𝜋

Multiplying both sides by 𝑍𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0, 4 ] we have Eq. (5.287) is 1 ∫0 ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥) 𝑇𝑗 (𝑥)d𝑥

𝜋

= ∫04 [𝑥 + sec 2 (𝑥)]𝑇𝑗 (𝑥)d𝑥 𝜋

𝜋

− ∫04 [∫04 𝑥 ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 ] 𝑇𝑗 (𝑥)d𝑥. 𝜋

𝜋

𝜋

(5.288)

𝜋

for 𝑗 = 0, ∫04 ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥) d𝑥 = ∫04 [𝑥 + sec 2(𝑥)]d𝑥 − ∫04 [∫04 ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡] d𝑥. 𝜋

for 𝑗 = 1, ∫04 ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥)

4𝑥 𝜋

𝜋

d𝑥 = ∫04 [𝑥 + sec 2 (𝑥)] 𝜋

4𝑥 𝜋

𝜋

d𝑥 4𝑡

− ∫04 [∫04 𝑥 ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 ] 𝜋 d𝑥 . for 𝑗 = 2,

2 2

𝜋 4

2 1 8𝑥 Γ(3) −9 3 𝑇 ∑ 𝐂 𝐙 (𝑥) [ ] d𝑥 ∫0 𝑘=0 𝑘 2 2 4 Γ( )

2 2

𝜋 4

= ∫0 [𝑥 +

3

𝜋 4

2 1 8𝑥 Γ(3) −9 sec 2(𝑥)] [4 ] d𝑥 2 2 Γ( ) 3

2 2

𝜋 4

2 1 8𝑡 Γ(3) −9 − ∫0 [∫0 𝑥 ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 ] [4 ] d𝑥 . 2 2 Γ( ) 3

𝜋 4

2 𝑥(−16𝑥 2 +3𝜋)

for 𝑗 = 3, ∫0 ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥) [− 3

𝜋 𝜋

𝜋 4

2 𝑥(−16𝑥 2 +3𝜋)

] d𝑥 = ∫0 [𝑥 + sec 2 (𝑥)] [− 3 𝜋

2 𝑡(−16𝑡 2 +3𝜋)

− ∫04 [∫04 𝑥 ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 ] [− 3 The matrix form of Eq. (5.288) is given as

886

𝜋

𝜋

] d𝑥

] d𝑥 .

=

.

After solving the above we get 𝛼0 = 0.62749, 𝛼1 = 1.3139, 𝛼2 = −0.28909, 𝛼3 = 0.66025. 𝑢(𝑥) = 0.98225 + 0.35243𝑥 − 0.57817𝑥 2 + 2.2417𝑥 3 .



Exact Solution


Error

_____________________________________________________________________________ 0.1

1.01006704642249488883

1.01379298542554050688

3.72594E-03

0.2

1.04109135849592725905

1.04484927359426130257

3.75792E-03

0.3

1.09568891532254712979

1.09422820590098030696

1.46071E-03

0.4

1.17875410581097509805

1.17391078718823504462

4.84332E-03

0.5

1.29844641040952483688

1.29587802229856304011

2.56839E-03

0.6

1.46804317252795744635

1.47211091607450181799

4.06774E-03

0.7

1.70944971586311727656

1.71459047335858890282

5.14076E-03

0.8

2.06015555816475570782

2.03529769899336181917

2.48579E-02

0.9

2.58799873325964786090

2.44621359782135809160

1.41785E-01

1.0

3.42551882081475976094

2.95931917468511524466

4.66200E-01

887


(5.286)




Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000018102767837298

1.81028E-07

0.1

1.01006704642249488883

1.01006706599697085016

1.95745E-08

0.2

1.04109135849592725905

1.04109136979243769901

1.12965E-08

0.3

1.09568891532254712979

1.09568890438316583595

1.09394E-08

0.4

1.17875410581097509805

1.17875410727401212214

1.46304E-09

0.5

1.29844641040952483688

1.29844641802777754953

7.61825E-09

0.6

1.46804317252795744635

1.46804316923135868775

3.29660E-09

0.7

1.70944971586311727656

1.70944966283086541814

5.30323E-08

0.8

2.06015555816475570782

2.06015402136227382045

1.53680E-06

0.9

2.58799873325964786090

2.58774714938760950978

2.51584E-04

1.0

3.42551882081475976094

3.42062700875587655807

4.89181E-03

_____________________________________________________________________________

888


(5.286)




Exact Solution


Error

_____________________________________________________________________________ 0.0

1.00000000000000000000

1.00000000000001017745

1.01775E-14

0.1

1.01006704642249488883

1.01006704642249353263

1.35620E-15

0.2

1.04109135849592725905

1.04109135849592527095

1.98811E-15

0.3

1.09568891532254712979

1.09568891532254546383

1.66596E-15

0.4

1.17875410581097509805

1.17875410581097464784

4.50211E-16

0.5

1.29844641040952483688

1.29844641040952590568

1.06880E-15

0.6

1.46804317252795744635

1.46804317252795915710

1.71075E-15

0.7

1.70944971586311727656

1.70944971586311623441

1.04215E-15

0.8

2.06015555816475570782

2.06015555816384149031

9.14218E-13

0.9

2.58799873325964786090

2.58799871588253649539

1.73771E-08

1.0

3.42551882081475976094

3.42551447440630274481

4.34641E-06

889

5.11.4. Volterra Integro-Differential Equation Consider the Volterra integro-differential equations [229] as 𝑥

𝑢′′ (𝑥) = 𝑥 + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 ,

0 ≤ 𝑥 ≤ 1.

(5.289)

𝑢(0) = 0, 𝑢′ (0) = 1. The exact solution of Eq. (5.289) is 𝑢(𝑥) = sinh(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥).

(5.290)


(5.291)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝐙𝑘 (𝑥) = [𝑍0 , 𝑍1 , 𝑍2 , 𝑍3 ]𝑇 . Putting Eq. (5.291) into Eq. (5.289), we obtained 𝑥

[∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥)]′′ = 𝑥 + ∫0 (𝑥 − 𝑡) ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 .

(5.292)


1

∫0 [∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥)]′′ d𝑥 = ∫0 𝑥𝑇𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 (𝑥 − 𝑡) ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 ]𝑇𝑗 (𝑥)d𝑥. (5.293) 1

1

for 𝑗 = 0, ∫0 [∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥)]′′ d𝑥 = ∫0 𝑥d𝑥 1

𝑥

1

4𝑥

+ ∫0 [∫0 (𝑥 − 𝑡) ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 ]d𝑥. 1

for 𝑗 = 1, ∫0 [∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥)]′′

4𝑥 𝜋

d𝑥 = ∫0 𝑥 1

𝜋

d𝑥

𝑥

4𝑡

+ ∫0 [∫0 (𝑥 − 𝑡) ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 ] 𝜋 d𝑥 . 2 2

for 𝑗 = 2,

8𝑥 2 Γ( ) −9 1 3 3 𝑇 ′′ 1 [∑ (𝑥) ] 𝐂 𝐙 [ ] d𝑥 ∫0 𝑘=0 𝑘 2 2 4 Γ( )

2 2

=

3

1 𝑥 + ∫0 [∫0 (𝑥

2 1 1 8𝑥 Γ(3) −9 𝑥 [ ] d𝑥 ∫0 4 2 2 Γ( ) 3

2 2

−

2 1 8𝑡 Γ(3) −9 𝑡) ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡 ] [4 ] d𝑥 . 2 2 Γ( ) 3

1

2 𝑥(−16𝑥 2 +3𝜋)

for 𝑗 = 3, ∫0 [∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑥)]′′ [− 3

𝜋

890

1

2 𝑥(−16𝑥 2 +3𝜋)

] d𝑥 = ∫0 𝑥 [− 3

𝜋

] d𝑥

1

𝑥

+ ∫0 [∫0 (𝑥 − 2 𝑡(−16𝑡 2 +3𝜋)

𝑡) ∑3𝑘=0 𝐂 𝑇 𝐙𝑘 (𝑡) d𝑡] [− 3

𝜋

] d𝑥.


=

.

After solving the above we get 𝛼0 = −0.013717, 𝛼1 = 0.87839, 𝛼2 = −0.010915, 𝛼3 = 0.056355. 𝑢(𝑥) = −0.00032290 + 1.0057𝑥 − 0.021830𝑥 2 + 0.19134𝑥 3 . 9)


Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00027793483850218315

2.77935E-04

0.1

0.10016675001984402582

0.10022452665238661305

5.77766E-05

0.2

0.20133600254109398763

0.20145803478533292347

1.22032E-04

0.3

0.30452029344714261896

0.30456595497333715020

4.56615E-05

0.4

0.41075232580281550854

0.41069165262939969529

6.06732E-05

0.5

0.52109530549374736162

0.52097849316652096082

1.16812E-04

0.6

0.63665358214824127112

0.63656984199770134886

8.37402E-05

0.7

0.75858370183953350346

0.75860906453594126148

2.53627E-05

0.8

0.88810598218762300657

0.88823952619424110075

1.33544E-04

0.9

1.02651672570817527596

1.02660459238560126875

8.78667E-05

1.0

1.17520119364380145688

1.17484762852302216754

3.53565E-04

_____________________________________________________________________________

891

Fig 5.57. Comparison of Exact and Approximate solutions of Eq. (5.28

Now we take 𝑛 = 10. Eq. (5.290) is 𝑇 𝑢(𝑥) = ∑10 𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥) = 𝐂 𝐙(𝑥),

(5.291)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼10 ]𝑇 , and 𝐙(𝑥) = [𝑍0 , 𝑍1 , 𝑍2 , … , 𝑍10 ]𝑇 . Proceeding as before we obtained the approximate solution. Fig 2: Comparison of Approximate and exact solution


892


Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00000000000000000000

4.02385E-23

0.1

0.10016675001984402582

0.10016675001984402582

9.54843E-24

0.2

0.20133600254109398763

0.20133600254109398763

3.03851E-24

0.3

0.30452029344714261896

0.30452029344714261896

7.22480E-24

0.4

0.41075232580281550854

0.41075232580281550854

8.01124E-24

0.5

0.52109530549374736162

0.52109530549374736162

8.16230E-24

0.6

0.63665358214824127112

0.63665358214824127112

8.20775E-24

0.7

0.75858370183953350346

0.75858370183953350346

7.65720E-24

0.8

0.88810598218762300657

0.88810598218762300657

3.59839E-24

0.9

1.02651672570817527596

1.02651672570817527596

9.88719E-24

1.0

1.17520119364380145688

1.17520119364380145688

4.29659E-23

_____________________________________________________________________________

Finally we take 𝑛 = 25. Eq. (5.290) is 𝑇 𝑢(𝑥) = ∑25 𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥) = 𝐂 𝐙(𝑥),

(5.291)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , … , 𝛼25 ]𝑇 , and 𝐙(𝑥) = [𝑍0 , 𝑍1 , 𝑍2 , … , 𝑍25 ]𝑇 . Proceeding as before we obtained the approximate solution.


893


Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00000000000000000000

5.48478E-42

0.1

0.10016675001984402582

0.10016675001984402582

1.19280E-42

0.2

0.20133600254109398763

0.20133600254109398763

1.01203E-42

0.3

0.30452029344714261896

0.30452029344714261896

1.19414E-42

0.4

0.41075232580281550854

0.41075232580281550854

9.26035E-43

0.5

0.52109530549374736162

0.52109530549374736162

4.04321E-43

0.6

0.63665358214824127112

0.63665358214824127112

1.40398E-42

0.7

0.75858370183953350346

0.75858370183953350346

1.12012E-42

0.8

0.88810598218762300657

0.88810598218762300657

8.13132E-43

0.9

1.02651672570817527596

1.02651672570817527596

2.08249E-42

1.0

1.17520119364380145688

1.17520119364380145688

1.07126E-41

_____________________________________________________________________________


1

𝑢(𝑥) = −5 − 𝑥 + 12𝑥 2 − 𝑥 3 − 𝑥 4 + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 + ∫0 (𝑥 + 𝑡)𝑢(𝑡)d𝑡, (5.294) The exact solution of Eq. (5.294) is 𝑢(𝑥) = 6𝑥 + 12𝑥 2 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥).

(5.295)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥).

(5.296)

To solve Eq. (5.294) we take 𝑛 = 3. Eq. (5.296) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥) = 𝐂 𝑇 𝐙(𝑥),

(5.297)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝐙𝑘 (𝑥) = [𝑍0 , 𝑍1 , 𝑍2 , 𝑍3 ]𝑇 . Putting Eq. (5.297) into Eq. (5.294), we obtained

894

𝑥

𝐂 𝑇 𝐙(𝑥) = −5 − 𝑥 + 12𝑥 2 − 𝑥 3 − 𝑥 4 + ∫0 (𝑥 − 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡 1

+ ∫0 (𝑥 + 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡.

(5.298)


1

∫0 𝐂 𝑇 𝐙(𝑥)d𝑥 = ∫0 [−5 − 𝑥 + 12𝑥 2 − 𝑥 3 − 𝑥 4 ]𝑇𝑗 (𝑥)d𝑥 1

𝑥

1

+ ∫0 [∫0 (𝑥 − 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡 + ∫0 (𝑥 + 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡] 𝑇𝑗 (𝑥)d𝑥. (5.299) 1

1

for 𝑗 = 0, ∫0 𝐂 𝑇 𝐙(𝑥)d𝑥 = ∫0 [−5 − 𝑥 + 12𝑥 2 − 𝑥 3 − 𝑥 4 ]d𝑥 1

𝑥

+ ∫0 [∫0 (𝑥 − 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡]d𝑥 . 1

for 𝑗 = 1, ∫0 𝐂 𝑇 𝐙(𝑥)

4𝑥 𝜋

1

d𝑥 = ∫0 [−5 − 𝑥 + 12𝑥 2 − 𝑥 3 − 𝑥 4 ] 1

𝑥

4𝑥 𝜋

d𝑥

1

4𝑡

+ ∫0 [∫0 (𝑥 − 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡 + ∫0 (𝑥 + 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡] 𝜋 d𝑥 . 2 2

for 𝑗 = 2,

2 1 1 8𝑥 Γ(3) −9 ] d𝑥 ∫0 𝐂 𝑇 𝐙(𝑥) [4 2 2 Γ( )

=

1 ∫0 [−5

2 2

2

3

− 𝑥 + 12𝑥 − 𝑥 −

3

1 𝑥 + ∫0 [∫0 (𝑥

𝑇

2 1 8𝑥 Γ(3) −9 𝑥 4 ] [4 ] d𝑥 2 2 Γ( ) 3

− 𝑡)𝐂 𝐙(𝑡)d𝑡 +

1 ∫0 (𝑥

2 2

+ 𝑡)𝐂

𝑇

2 1 8𝑡 Γ(3) −9 𝐙(𝑡)d𝑡] [4 ] d𝑥. 2 2 Γ( ) 3

1 + 12𝑥 2 ] [− 2 𝑥(−16𝑥 2 +3𝜋)] d𝑥 ] d𝑥 = ∫0 [−5 − 𝑥 3 3 𝜋 −𝑥 − 𝑥 4

2 𝑥(−16𝑥 2 +3𝜋)

1

for 𝑗 = 3, ∫0 𝐂 𝑇 𝐙(𝑥) [− 3 1

𝜋

𝑥

2 𝑡(−16𝑡 2 +3𝜋)

1

+ ∫0 [∫0 (𝑥 − 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡 + ∫0 (𝑥 + 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡] [− 3

𝜋


=

.


27 2 2 2Γ( ) 3

3

, 𝛼1 = 2 𝜋, 𝛼2 = 6, 𝛼3 = 0.

𝑢(𝑥) = 6𝑥 + 12𝑥 2 . which is exact solution.


] d𝑥 .

𝑥

1

𝜋

𝑢(𝑥) = 𝑥 sin(𝑥) − 2 𝑥 2 + (𝜋 − 2)𝑥 + ∫0 ∫02 (𝑟 − 𝑡)𝑢(𝑡)d𝑡d𝑟.

(5.300)

The exact solution of Eq. (5.300) is 𝑢(𝑥) = 𝑥 sin(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥).

(5.301)


(5.302)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝐙𝑘 (𝑥) = [𝑍0 , 𝑍1 , 𝑍2 , 𝑍3 ]𝑇 . Putting Eq. (5.302) into Eq. (5.300), we obtained 𝜋

𝑥

1

𝐂 𝑇 𝐙(𝑥) = 𝑥 sin(𝑥) − 2 𝑥 2 + (𝜋 − 2)𝑥 + ∫0 ∫02 (𝑟 − 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡d𝑟.( 5.303) 𝜋

Multiplying both sides by 𝑍𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0, 2 ] we have Eq. (5.303) is 𝜋 2

𝜋 2

𝑇

1

∫0 𝐂 𝐙(𝑥)d𝑥 = ∫0 [𝑥 sin(𝑥) − 2 𝑥 2 + (𝜋 − 2)𝑥] 𝑇𝑗 (𝑥)d𝑥 𝜋

𝜋

𝑥

+ ∫02 [∫0 ∫02 (𝑟 − 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡d𝑟 ] 𝑇𝑗 (𝑥)d𝑥. 𝜋 2

𝜋 2

𝑇

(5.304)

1

for 𝑗 = 0, ∫0 𝐂 𝐙(𝑥)d𝑥 = ∫0 [𝑥 sin(𝑥) − 2 𝑥 2 + (𝜋 − 2)𝑥] d𝑥 𝜋

𝑥

𝜋

+ ∫02 [∫0 ∫02 (𝑟 − 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡d𝑟 ] d𝑥. 𝜋 2

𝑇

for 𝑗 = 1, ∫0 𝐂 𝐙(𝑥)

4𝑥 𝜋

𝜋 2

1

d𝑥 = ∫0 [ln(1 + 𝑥) − 2 𝑥 2 ] 𝜋

𝑥

4𝑥 𝜋

d𝑥

𝜋

+ ∫02 [∫0 ∫02 (𝑟 − 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡d𝑟 ] 𝜋 2

for 𝑗 = 2, ∫0 𝐂

2 2

𝑇

2 1 8𝑥 Γ(3) −9 𝐙(𝑥) [4 ] d𝑥 2 2 Γ( )

𝜋 2

1

= ∫0 [ln(1 + 𝑥) − 2 𝑥

3

𝜋 2

4𝑡 𝜋

d𝑥 . 2 2

2

2 1 8𝑥 Γ(3) −9 ] [4 ] d𝑥 2 2 Γ( ) 3

2 2

𝜋

𝑥 + ∫0 [∫0 ∫02 (𝑟

− 𝑡)𝐂

𝑇

2 1 8𝑡 Γ(3) −9 𝐙(𝑡)d𝑡d𝑟 ] [4 ] d𝑥. 2 2 Γ( ) 3

e−1

for 𝑗 = 3, ∫0

2 𝑥(−16𝑥 2 +3𝜋)

𝐂 𝑇 𝐙(𝑥) [− 3

𝜋 𝜋

𝑥

𝜋 2

2 𝑥(−16𝑥 2 +3𝜋)

1

] d𝑥 = ∫0 [ln(1 + 𝑥) − 2 𝑥 2 ] [− 3 𝜋

2 𝑡(−16𝑡 2 +3𝜋)

+ ∫02 [∫0 ∫02 (𝑟 − 𝑡)𝐂 𝑇 𝐙(𝑡)d𝑡d𝑟 ] [− 3 The matrix form of Eq. (5.304) is given as 896

𝜋

𝜋

] d𝑥 .

] d𝑥

=

.

After solving the above we get 𝛼0 = 0.86613, 𝛼1 = −0.30635, 𝛼2 = 0.69897, 𝛼3 = −0.12693. Substituting the values we get 𝑢(𝑥) = 0.011406 − 0.13621𝑥 + 1.3979𝑥 2 − 0.43096, Table 5.60. Comparison of Exact and Approximate solutions of Eq. (5.300) obtained from Zernike polynomials Method (ZPM) _____________________________________________________________________________ 𝑥

Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

0.00903374723334873316

9.03375E-03

0.1

0.00998334166468281523

0.01021796079471232964

2.34619E-04

0.2

0.03973386615901224309

0.03641964760301951778

3.31422E-03

0.3

0.08865606199840187253

0.08509970096676115455

3.55636E-03

0.4

0.15576733692346019667

0.15371901419442809691

2.04832E-03

0.5

0.23971276930210150014

0.23973848059451120183

2.57113E-05

0.6

0.33878548403702121432

0.34061899347550132627

1.83351E-03

0.7

0.45095238106638373757

0.45382144614588932721

2.86907E-03

0.8

0.57388487271961820930

0.57680673191416606160

2.92186E-03

0.9

0.70499421866473504962

0.70703574408882238642

2.04153E-03

1.0

0.84147098480789650665

0.84196937597834915863

4.98391E-04

_____________________________________________________________________________

897


Now we take 𝑛 = 10. Eq. (5.301) is 𝑇 𝑢(𝑥) = ∑10 𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥) = 𝐂 𝐙(𝑥),

(5.302)



898


Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00000000004049081697

4.04908E-11

0.1

0.00998334166468281523

0.00998334166020687481

4.47594E-12

0.2

0.03973386615901224309

0.03973386615573351400

3.27873E-12

0.3

0.08865606199840187253

0.08865606200917053055

1.07687E-11

0.4

0.15576733692346019667

0.15576733691989073665

3.56946E-12

0.5

0.23971276930210150014

0.23971276929279736843

9.30413E-12

0.6

0.33878548403702121432

0.33878548404059082806

3.56961E-12

0.7

0.45095238106638373757

0.45095238107585139961

9.46766E-12

0.8

0.57388487271961820930

0.57388487271782163714

1.79657E-12

0.9

0.70499421866473504962

0.70499421865480179641

9.93325E-12

1.0

0.84147098480789650665

0.84147098480804944540

1.52939E-13

_____________________________________________________________________________

Finally we take 𝑛 = 25. Eq. (5.301) is 𝑇 𝑢(𝑥) = ∑25 𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥) = 𝐂 𝐙(𝑥),

(5.302)



899


Exact Solution


Error

_____________________________________________________________________________ 0.0

0.00000000000000000000

-0.00000000000000000000

1.13118E-35

0.1

0.00998334166468281523

0.00998334166468281523

2.46824E-36

0.2

0.03973386615901224309

0.03973386615901224309

2.04770E-36

0.3

0.08865606199840187253

0.08865606199840187253

7.06700E-37

0.4

0.15576733692346019667

0.15576733692346019667

9.84026E-37

0.5

0.23971276930210150014

0.23971276930210150014

1.64389E-36

0.6

0.33878548403702121432

0.33878548403702121432

1.75714E-36

0.7

0.45095238106638373757

0.45095238106638373757

1.67134E-36

0.8

0.57388487271961820930

0.57388487271961820930

1.50696E-36

0.9

0.70499421866473504962

0.70499421866473504962

1.25995E-36

1.0

0.84147098480789650665

0.84147098480789650665

8.44416E-37

_____________________________________________________________________________


2

𝑥

𝑥 3 (20 − 24𝑥 + 9𝑥 2 ) = ∫0 40

1 1 (𝑥−𝑡)3

𝑢2 (𝑡)d𝑡.

(5.305)

The exact solution of Eq. (5.305) is 𝑢(𝑥) = ±(1 − 𝑥). Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation 𝑢2 (𝑡) = 𝑣(𝑡), 𝑢(𝑡) = ±√𝑣(𝑡). Using the above transformation into the given integral equation we get 3

2

𝑥

𝑥 3 (20 − 24𝑥 + 9𝑥 2 ) = ∫0 40

1 1

𝑣(𝑡)d𝑡.

(𝑥−𝑡)3

According to the proposed technique, consider the trail solution

900

(5.306)

𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥).

(5.307)


(5.308)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝐙𝑘 (𝑥) = [𝑍0 , 𝑍1 , 𝑍2 , 𝑍3 ]𝑇 . Putting Eq. (5.308) into Eq. (5.305), we obtained 3

2

𝑥

𝑥 3 (20 − 24𝑥 + 9𝑥 2 ) = ∫0 40

1 1 (𝑥−𝑡)3

𝐂 𝑇 𝐙(𝑡)d𝑡.

(5.309)


3

2

1

𝑥

∫0 [40 𝑥 3 (20 − 24𝑥 + 9𝑥 2 )] 𝑇𝑗 (𝑥)d𝑥 = ∫0 [∫0 1

3

2

1

1

𝑥

for 𝑗 = 0, ∫0 [40 𝑥 3 (20 − 24𝑥 + 9𝑥 2 )] d𝑥 = ∫0 [∫0 1

3

2

for 𝑗 = 1, ∫0 [40 𝑥 3 (20 − 24𝑥 + 9𝑥 2 )] for 𝑗 = 2,

1 3 2 ∫0 [40 𝑥 3 (20

4𝑥 𝜋

𝐂 𝑇 𝐙(𝑡)d𝑡] 𝑇𝑗 (𝑥)d𝑥 . (5.310)

1

(𝑥−𝑡)3

1 1

𝐂 𝑇 𝐙(𝑡)d𝑡] d𝑥 .

(𝑥−𝑡)3

1

𝑥

d𝑥 = ∫0 [∫0

1 1 (𝑥−𝑡)3

4𝑡

𝐂 𝑇 𝐙(𝑡)d𝑡] 𝜋 d𝑥 .

2 2

− 24𝑥 +

2 1 8𝑥 Γ(3) −9 9𝑥 2 )] [4 ] d𝑥 2 2 Γ( ) 3

= 1

3

1 𝑥 ∫0 [∫0

2

2 2

1 1

𝐂

𝑇

(𝑥−𝑡)3

3

2 𝑥(−16𝑥 2 +3𝜋)

for 𝑗 = 3, ∫0 [40 𝑥 3 (20 − 24𝑥 + 9𝑥 2 )] [− 3 1

𝜋 𝑥

= ∫0 [∫0

1 1 (𝑥−𝑡)3

2 1 8𝑡 Γ(3) −9 𝐙(𝑡)d𝑡] [4 ] d𝑥. 2 2 Γ( )

] d𝑥 2 𝑡(−16𝑡 2 +3𝜋)

𝐂 𝑇 𝐙(𝑡)d𝑡] [− 3


=

After solving the above we get 2 2

1 9+8Γ(3)

𝛼0 = 8

2 2 Γ( ) 3

1

1

, 𝛼1 = − 2 𝜋, 𝛼2 = 2 , 𝛼3 = 0.

901

.

𝜋

] d𝑥 .

Substituting the values we have 𝑣(𝑥) = 𝑥 2 − 2𝑥 + 1. After applying the reverse transformation we get 𝑢(𝑥) = ±(1 − 𝑥). which is the exact solution.


4

𝑥

𝑥 4 (77 − 88𝑥 + 32𝑥 2 ) = ∫0 231

1 1

𝑢2 (𝑡)d𝑡.

(5.311)

(𝑥−𝑡)4

The exact solution of Eq. (5.311) is 𝑢(𝑥) = ±(1 − 𝑥). Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation 𝑢2 (𝑡) = 𝑣(𝑡), 𝑢(𝑡) = ±√𝑣(𝑡). Using the above transformation into the given integral equation we get 3

4

𝑥

𝑥 4 (77 − 88𝑥 + 32𝑥 2 ) = ∫0 231

1 1

𝑣(𝑡)d𝑡.

(5.312)

(𝑥−𝑡)4

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥).

(5.313)


(5.314)


2

𝑥

𝑥 3 (20 − 24𝑥 + 9𝑥 2 ) = ∫0 40

1 1


(5.315)

(𝑥−𝑡)3


4

3

1

𝑥

∫0 [231 𝑥 4 (77 − 88𝑥 + 32𝑥 2 )] 𝑇𝑗 (𝑥)d𝑥 = ∫0 [∫0 1

4

3

1 1 (𝑥−𝑡)4

1

𝑥

for 𝑗 = 0, ∫0 [231 𝑥 4 (77 − 88𝑥 + 32𝑥 2 )] d𝑥 = ∫0 [∫0

902

𝐂 𝑇 𝐙(𝑡)d𝑡] 𝑇𝑗 (𝑥)d𝑥.(5.316) 1

1 (𝑥−𝑡)4

𝐂 𝑇 𝐙(𝑡)d𝑡] d𝑥 .

1

3

4

for 𝑗 = 1, ∫0 [231 𝑥 4 (77 − 88𝑥 + 32𝑥 2 )] for 𝑗 = 2,

3 1 4 ∫0 [231 𝑥 4 (77

4𝑥 𝜋

1

𝑥

d𝑥 = ∫0 [∫0

1 1 (𝑥−𝑡)4

4𝑡


2 2

− 88𝑥 +

2 1 8𝑥 Γ(3) −9 32𝑥 2 )] [4 ] d𝑥 2 2 Γ( ) 3

= 1

1 𝑥 ∫0 [∫0

3

4

2 2

1 1

𝐂

𝑇

(𝑥−𝑡)3

2 1 8𝑡 Γ(3) −9 𝐙(𝑡)d𝑡] [4 ] d𝑥. 2 2 Γ( ) 3

2 𝑥(−16𝑥 2 +3𝜋)

for 𝑗 = 3, ∫0 [231 𝑥 4 (77 − 88𝑥 + 32𝑥 2 )] [− 3 1

𝜋

𝑥

= ∫0 [∫0

1 1 (𝑥−𝑡)4

] d𝑥

𝐂 𝑇 𝐙(𝑡)d𝑡] [−

2 𝑡(−16𝑡 2 +3𝜋) 3

𝜋

] d𝑥 .


=

.

After solving the above we get 2 2

1 9+8Γ(3)

𝛼0 = 8

2 2 Γ( ) 3

1

1

, 𝛼1 = − 2 𝜋, 𝛼2 = 2 , 𝛼3 = 0.

Substituting the values we have 𝑣(𝑥) = 𝑥 2 − 2𝑥 + 1. After applying the reverse transformation we get 𝑢(𝑥) = ±(1 − 𝑥). which is the exact solution.


21

𝑥

𝑢(𝑥) = 𝑥 5 − 63 𝑥 4 + ∫0

1 1

(𝑥 3 −𝑡 3 )4

𝑢(𝑡)d𝑡.

(5.317)

The exact solution of Eq. (5.317) is 𝑢(𝑥) = 𝑥 5 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥).

(5.318) 903


(5.319)


16

𝑥

𝐂 𝑇 𝐙(𝑥) = 𝑥 5 − 63 𝑥 4 + ∫0

1 1

(𝑥 3 −𝑡 3 )4


(5.320)


1

21

16

∫0 [𝐂 𝑇 𝐙(𝑥 )]𝑇𝑗 (𝑥)d𝑥 = ∫0 [𝑥 5 − 63 𝑥 4 ] 𝑇𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 1

1 1 (𝑥 3 −𝑡 3 )4

1

𝐂 𝑇 𝐙(𝑡)d𝑡] 𝑇𝑗 (𝑥)d𝑥 .

16

(5.321)

21

for 𝑗 = 0, ∫0 [𝐂 𝑇 𝐙(𝑥)]d𝑥 = ∫0 [𝑥 5 − 63 𝑥 4 ] d𝑥 1

𝑥

+ ∫0 [∫0 1

for 𝑗 = 1, ∫0 [𝐂 𝑇 𝐙(𝑥)]

4𝑥 𝜋

1

16

21

d𝑥 = ∫0 [𝑥 5 − 63 𝑥 4 ] 1

2 2

2 1 1 8𝑡 Γ(3) −9 ] d𝑥 ∫0 [𝐂 𝑇 𝐙(𝑥)] [4 2 2 Γ( )

=

4𝑥

1

4𝑡

1 (𝑥 3 −𝑡 3 )4

1 ∫0 [𝑥 5

𝐂 𝑇 𝐙(𝑡)d𝑡] d𝑥.

d𝑥

𝜋

𝑥

+ ∫0 [∫0 for 𝑗 = 2,

1 1 (𝑥 3 −𝑡 3 )4


16

− 63 𝑥

21 4

3

2 2

2 1 8𝑡 Γ(3) −9 ] [4 ] d𝑥 2 2 Γ( ) 3

1 𝑥 + ∫0 [∫0

2 2

1 1

(𝑥 3 −𝑡 3 )4

𝐂

𝑇

2 1 8𝑡 Γ(3) −9 𝐙(𝑡)d𝑡] [4 ] d𝑥 . 2 2 Γ( ) 3

16

for 𝑗 =

1 2 𝑥(−16𝑥 2 +3𝜋) 3, ∫0 [𝐂 𝑇 𝐙(𝑥)] [− 3 ] d𝑥 𝜋

1

=

1 ∫0 (

[𝑥 5 − 63 𝑥 ] 2 𝑥(−16𝑥 2 +3𝜋)

[− 3

𝑥

+ ∫0 [∫0

1 1 (𝑥 3 −𝑡 3 )4

. . .. Proceeding as before we get 904

21 4

𝜋

) d𝑥 ] 2 𝑡(−16𝑡 2 +3𝜋)

𝐂 𝑇 𝐙(𝑡)d𝑡] [− 3

𝜋

] d𝑥 .

9

𝛼0 = 0, 𝛼1 = 81920

3 2 4

2 2 3

243𝜋 3 −1024Γ( ) Γ( ) 2 2 Γ( ) 3

2187

, 𝛼2 = 0, 𝛼3 = 40960

𝜋2 2 2 Γ( ) 3

15

, 𝛼4 = 0, 𝛼5 = 512 𝜋 .

Substituting the values we have 𝑢(𝑥) = 𝑥 5 , which is the exact solution.

5.11.8. System of Volterra Integral Equation Consider the system of Volterra integral equation of 1st Kind [229] as 𝑥

𝑢(𝑥) = cos(𝑥) − 𝑥 sin(𝑥) + ∫0 (sin(𝑥 − 𝑡) 𝑢(𝑡) + cos(𝑥 − 𝑡) 𝑣(𝑡))d𝑡, 𝑥

𝑣(𝑥) = sin(𝑥) − 𝑥 cos(𝑥) + ∫0 (cos(𝑥 − 𝑡) 𝑢(𝑡) − sin(𝑥 − 𝑡) 𝑣(𝑡))d𝑡.

(5.322a) (5.322b)

The exact solution of Eq. (5.322) is 𝑢(𝑥) = cos(𝑥), 𝑣(𝑥) = sin(𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥),

(5.323a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝑍𝑘 (𝑥),

(5.323b)

To solve Eq. (5.322) we take 𝑛 = 2. Eq. (5.323) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥) = 𝐂 𝐓 𝐙(𝑥),

(5.324a)

𝑣(𝑥) = ∑2𝑘=0 𝛽𝑘 𝑍𝑘 (𝑥) = 𝐃𝐓 𝐙(𝑥).

(5.324b)


1

7

1

𝐂 𝐓 𝐙(𝑥) = 6 + 𝑥 2 − 280 𝑥 3 (21𝑥 3 + 280𝑥 3 − 60𝑥 2 + 560) 𝑥

+ ∫0

1 1 (𝑥−𝑡)3

𝑥

𝐂 𝐓 𝐙(𝑡)d𝑡 + ∫0 5

1

1 2

𝐃𝐓 𝐙(𝑡)d𝑡.

(5.325a)

(𝑥−𝑡)3

11

1

𝐃𝐓 𝐙(𝑥) = 6 − 𝑥 2 − 924 𝑥 5 (21𝑥 5 + 2772𝑥 5 − 700𝑥 2 + 5544) 𝑥

+ ∫0

1 3 (𝑥−𝑡)5

𝑥

𝐂 𝐓 𝐙(𝑡)d𝑡 + ∫0

1 4

𝐃𝐓 𝐙(𝑡)d𝑡.

(5.325b)

(𝑥−𝑡)5

Multiplying both sides by 𝑍𝑗 (𝑥), 𝑗 = 0,1,2. Then integrating over [0,1]. The matrix form of Eq. (5.325) is given as 1

7

1

1 1 9 280𝑥 3 )) 𝑇𝑗 (𝑥)d𝑥 ∫0 𝐂 𝐓 𝐙(𝑥)𝑍𝑗 (𝑥) d𝑥 = ∫0 (6 + 𝑥 2 − 280 𝑥 3 (21𝑥 3 + 2 −60𝑥 + 560

905

1

𝑥

+ ∫0 (∫0

1 1 (𝑥−𝑡)3

1

𝑥

𝐂 𝐓 𝐙(𝑡)d𝑡 + ∫0 1

1 2 (𝑥−𝑡)3

11

1

5

𝐃𝐓 𝐙(𝑡)d𝑡) 𝑇𝑗 (𝑥)d𝑥, (5.326a) 1

∫0 𝐃𝐓 𝐙(𝑥)𝑍𝑗 (𝑥) d𝑥 = ∫0 (6 − 𝑥 2 − 924 𝑥 5 (21𝑥 5 +2 2772𝑥 5 )) 𝑇𝑗 (𝑥)d𝑥 −700𝑥 + 5544 1

𝑥

+ ∫0 (∫0

1 3 (𝑥−𝑡)5

𝑥

𝐂 𝐓 𝐙(𝑡)d𝑡 + ∫0

1 4 (𝑥−𝑡)5

𝐃𝐓 𝐙(𝑡)d𝑡) 𝑇𝑗 (𝑥)d𝑥, (5.326b)

for 𝑗 = 0,1,2. After solving the above we get 𝑢(𝑥) = 0.99912 + 0.014957𝑥 − 0.55934𝑥 2 + 0.085394𝑥 3 , 𝑣(𝑥) = −0.000036804 + 1.0023𝑥 − 0.012963𝑥 2 − 0.14827𝑥 3 . Table 5.63. Comparison of Exact and Approximate solutions of Eq. (5.322) obtained from Zernike polynomials Method (ZPM) ______________________________________________________________________________ 𝑥

𝑢(𝑥)

𝑣(𝑥)

𝑢Approx. (𝑥)


______________________________________________________________________________ 0.0 1.0000000000 0.0000000000 0.9994492616 -0.0002818084 5.5074E-04 2.8181E-04 0.1 0.9950041653 0.0998334166 0.9951270848 0.0998850797 1.2292E-04 5.1663E-05 0.2 0.9800665778 0.1986693308 0.9802854306 0.1987906302 2.1885E-04 1.2130E-04 0.3 0.9553364891 0.2955202067 0.9553989943 0.2955728686 6.2505E-05 5.2662E-05 0.4 0.9210609940 0.3894183423 0.9209424709 0.3893698203 1.1852E-04 4.8522E-05 0.5 0.8775825619 0.4794255386 0.8773905554 0.4793195106 1.9201E-04 1.0603E-04 0.6 0.8253356149 0.5646424734 0.8252179430 0.5645599650 1.1767E-04 8.2508E-05 0.7 0.7648421873 0.6442176872 0.7648993287 0.6442292088 5.7141E-05 1.1522E-05 0.8 0.6967067093 0.7173560909 0.6969094077 0.7174652674 2.0270E-04 1.0918E-04 0.9 0.6216099683 0.7833269096 0.6217228750 0.7834061662 1.1291E-04 7.9257E-05 1.0 0.5403023059 0.8414709848 0.5398144257 0.8411899306 4.8788E-04 2.8105E-04 ______________________________________________________________________________

906


For 𝑛 = 10. Eq. (5.323) is 𝐓 𝑢(𝑥) = ∑10 𝑘=0 𝛼𝑘 𝑇𝑘 (𝑥) = 𝐂 𝐙(𝑥), 𝐓 𝑣(𝑥) = ∑10 𝑘=0 𝛽𝑘 𝑇𝑘 (𝑥) = 𝐃 𝐙(𝑥).

Proceeding as before we have the following error Table 3.1nd graphical representation


907

Table 5.64. Comparison of Exact and Approximate solutions of Eq. (5.322) obtained from Zernike polynomials Method (ZPM) ______________________________________________________________________________ 𝑥

𝑢(𝑥)

𝑣(𝑥)

𝑢Approx. (𝑥)


______________________________________________________________________________ 0.0 1.0000000000 0.0000000000 1.0000000000 -0.0000000000 1.5602E-14 3.0996E-14 0.1 0.9950041653 0.0998334166 0.9950041653 0.0998334166 4.5709E-15 8.8882E-15 0.2 0.9800665778 0.1986693308 0.9800665778 0.1986693308 3.8225E-15 7.2859E-15 0.3 0.9553364891 0.2955202067 0.9553364891 0.2955202067 4.0252E-15 7.6176E-15 0.4 0.9210609940 0.3894183423 0.9210609940 0.3894183423 2.9116E-15 5.5226E-15 0.5 0.8775825619 0.4794255386 0.8775825619 0.4794255386 1.3861E-17 1.7394E-16 0.6 0.8253356149 0.5646424734 0.8253356149 0.5646424734 2.9872E-15 5.2909E-15 0.7 0.7648421873 0.6442176872 0.7648421873 0.6442176872 4.2902E-15 7.6157E-15 0.8 0.6967067093 0.7173560909 0.6967067093 0.7173560909 4.2181E-15 7.4293E-15 0.9 0.6216099683 0.7833269096 0.6216099683 0.7833269096 5.2011E-15 8.9873E-15 1.0 0.5403023059 0.8414709848 0.5403023059 0.8414709848 1.8300E-14 3.0997E-14 ______________________________________________________________________________


5

𝑥

𝑥 3 (45𝑥 2 + 33𝑥 + 44) = ∫0 440 25

8

5

1

[𝑢(𝑡) + 𝑣(𝑡) + 𝑤(𝑡)]d𝑡.

(5.327a)

[𝑣(𝑡) + 𝑤(𝑡)]d𝑡.

(5.327b)

(𝑥−𝑡)3

𝑥

𝑥 5 (25𝑥 2 + 30𝑥 + 39) = ∫0 468 32

1

1 2

(𝑥−𝑡)5 𝑥

𝑥 4 (96𝑥 2 + 208𝑥 + 117) = ∫0 585

1 3

[𝑢(𝑡) − 𝑣(𝑡)]d𝑡.

(5.327c)

(𝑥−𝑡)4

The exact solution of Eq. (5.327) is 𝑢(𝑥) = 𝑥 + 𝑥 2 + 3𝑥 3 , 𝑣(𝑥) = 𝑥 + 3𝑥 2 − 𝑥 3 , 𝑤(𝑥) = 𝑥 − 𝑥 2 + 3𝑥 3 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑍𝑘 (𝑥),

(5.328a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝑍𝑘 (𝑥),

(5.328b) 908

𝑤(𝑥) = ∑𝑛𝑘=0 𝜇𝑘 𝑍𝑘 (𝑥).

(5.328c)


(5.329a)

𝑣(𝑥) = ∑3𝑘=0 𝛽𝑘 𝑁𝑘 (𝑥) = 𝐃𝑇 𝐙(𝑥),

(5.329b)

𝑤(𝑥) = ∑3𝑘=0 𝜇𝑘 𝑁𝑘 (𝑥) = 𝐄𝑇 𝐙(𝑥),

(5.329c)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1 , 𝛽2 , 𝛽3 ]𝑇 , 𝐄 = [𝜇0 , 𝜇1 , 𝜇2 , 𝜇3 ]𝑇 , and 𝐍𝑘 (𝑥) = [𝑍0 , 𝑍1 , 𝑍2 , 𝑍3 ]𝑇 . Putting Eq. (5.329) into Eq. (5.327), we obtained 5

27

𝑥

𝑥 3 (45𝑥 2 + 33𝑥 + 44) = ∫0 440 8

25

1 (𝑥−𝑡)3

𝑥

𝑥 5 (25𝑥 2 + 30𝑥 + 39) = ∫0 468 5

32

1

1 2 (𝑥−𝑡)5

𝑥

𝑥 4 (96𝑥 2 + 208𝑥 + 117) = ∫0 585

[𝐂 𝑇 + 𝐃𝑇 + 𝐄𝑇 ]𝐙(𝑡)d𝑡.

(5.330a)

[𝐃𝑇 + 𝐄𝑇 ]𝐙(𝑡)d𝑡.

(5.330b)

1 3 (𝑥−𝑡)4

[𝐂 𝑇 + 𝐃𝑇 ]𝐙(𝑡)d𝑡.

(5.330c)


27

5

∫0 [440 𝑥 3 (45𝑥 2 + 33𝑥 + 44)] 𝐙𝑗 (𝑥)d𝑥 = 1

𝑥

∫0 [∫0 1

25

1 1 (𝑥−𝑡)3

[𝐂 𝑇 + 𝐃𝑇 + 𝐄𝑇 ]𝐙(𝑡)d𝑡 ] 𝐙𝑗 (𝑡)d𝑥 ,

(5.331a)

8

∫0 [468 𝑥 5 (25𝑥 2 + 30𝑥 + 39)] 𝐙𝑗 (𝑥)d𝑥 = 1

𝑥

∫0 [∫0 1

32

1 2 (𝑥−𝑡)5

[𝐃𝑇 + 𝐄𝑇 ]𝐙(𝑡)d𝑡 ] 𝐙𝑗 (𝑡)d𝑥 ,

(5.331b)

5

∫0 [585 𝑥 4 (96𝑥 2 + 208𝑥 + 117)] 𝐙𝑗 (𝑥)d𝑥 = 1

𝑥

∫0 [∫0

1 3 (𝑥−𝑡)4

[𝐂 𝑇 + 𝐃𝑇 ]𝐙(𝑡)d𝑡] 𝐙𝑗 (𝑡)d𝑥,

(5.331c)


𝐀 𝟏𝟐 𝐁𝟐𝟐 𝐂𝟑𝟐

𝐛𝟏 𝐀 𝟏𝟑 𝒙𝟏 𝒙 𝐁𝟐𝟑 ] [ 𝟐 ] = [𝐛𝟐 ]. 𝐂𝟑𝟑 𝒙𝟑 𝐛𝟑

where 𝐀 𝟏𝟏 , 𝐀 𝟏𝟐 , 𝐀 𝟏𝟑 , 𝐁𝟐𝟏 , 𝐁𝟐𝟐 , 𝐁𝟐𝟑 , 𝐂𝟑𝟏 , 𝐂𝟑𝟐 and 𝐂𝟑𝟑 are matrices of order 4×4, 𝒙𝟏 , 𝒙𝟐 , 𝒙𝟑 , 𝐛𝟏 , 𝐛𝟐 , and 𝐛𝟑 are matrices of order 4×1. 909

After solving the above 𝛼0 =

9

1

2 2 8Γ( ) 3

1

9

, 𝛼1 = 64 𝜋(16 + 9𝜋), 𝛼2 = 2 , 𝛼3 = 32 𝜋, 𝛽0 =

1

3

3

𝛽1 = − 64 𝜋(16 + 9𝜋), 𝛽2 = 2 , 𝛽3 = − 32 𝜋, 𝜇0 = − 1

9 2 2 8Γ( ) 3

27 2 2 3

,

8Γ( )

1

, 𝜇1 = 64 𝜋(16 + 9𝜋),

9

𝜇2 = − 2 , 𝜇3 = 32 𝜋. Substituting the values we have 𝑢(𝑥) = 𝑥 + 𝑥 2 + 3𝑥 3 , 𝑣(𝑥) = 𝑥 + 3𝑥 2 − 𝑥 3 , 𝑤(𝑥) = 𝑥 − 𝑥 2 + 3𝑥 3 .

5.12. Meixner Polynomials Method We develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Meixner polynomials [133]. Meixner polynomials is given as 𝑛−𝑚 2

(−1)𝑘 (𝑛)!

𝑀(𝑥) = ∑𝑘=0 [ 𝑘!(𝑛−𝑘)!

𝑘! Γ(x−β+1)

𝑥!

𝑛−𝑚 Γ(𝑥−𝛽−𝑛+𝑘+1)Γ( +1) 𝑘!(𝑥−𝑘)! 2+𝑘

𝑔−𝑘 ].

(5.332)

For 𝑚 = 0, first four Meixner polynomials is given as 𝑀0 (𝑥) = 1, 1

𝑀1 (𝑥) = 2 𝑥 − 1, 1

9

1

21

𝑀2 (𝑥) = 4 𝑥 2 − 4 𝑥 + 2, 𝑀3 (𝑥) = 8 𝑥 3 −

8

𝑥2 +

17 2

𝑥 − 6,



(5.333)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, to find an approximate solution 𝑢̃(𝑥) of Eq. (5.333). For this, we assume that

910

𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝑀𝑘 (𝑥),

(5.334)

where 𝑍𝑘 (𝑥) are Zernike polynomials of degree 𝑘 defined in equation (5.332) and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.334) into (5.333), we get 𝑥

𝑓(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝑀𝑘 (𝑡)d𝑡,

(5.335)

Then the Galerkin equations are obtained by multiplying both sides of (4) by 𝑀𝑗 (𝑥) and then integrating with respect to 𝑥 from 𝑎 to 𝑏, we have 𝑏

𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝑀𝑗 (𝑥)d𝑥 = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝑀𝑘 (𝑡)d𝑡)𝑀𝑗 (𝑡)d𝑥, 𝑗 = 0,1,2, … 𝑛,(5.336) Since in each equation, there are two integrals. The inner integrand of the left side is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.336) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.337)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝑀𝑘 (𝑡)d𝑡)𝑀𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛 𝑏

𝐹𝑗 = ∫𝑎 𝑓(𝑥) 𝑀𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛 Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.337) and substituting these values of parameters in (5.334), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.333).



(5.338)


(5.339)

where 911

𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝑀𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑀𝑘 (𝑡)d𝑡)𝑀𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛. Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.339) and substituting these values of parameters in (5.334), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.338).

5.12.2. Volterra-Fredholm Integral Equation Consider the Volterra-Fredholm integral equation [229] as 𝑥

1

𝑢(𝑥) = −6 − 2𝑥 + 19𝑥 3 − 𝑥 5 + ∫0 (𝑥 − 𝑡)𝑢(𝑡)d𝑡 + ∫0 (𝑥 + 𝑡)𝑢(𝑡)d𝑡,

(5.340)

The exact solution of Eq. (5.340) is 𝑢(𝑥) = 6𝑥 + 20𝑥 3 . According to the proposed technique, consider the path solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑀𝑘 (𝑥).

(5.341)

To solve Eq. (5.340) we take 𝒌 = 𝟑. Eq. (5.341) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝑀𝑘 (𝑥) = ∑3𝑘=0 𝐂 𝑇 𝑴𝑘 (𝑥).

(5.342)

Where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and 𝑴𝑘 (𝑥) = [𝑀0 , 𝑀1 , 𝑀2 , 𝑀3 ]𝑇 . Putting Eq. (5.342) into Eq. (5.340), we obtained ∑3𝑘=0 𝐂 𝑇 𝑅𝑘 (𝑥) = −6 − 2𝑥 + 19𝑥 3 − 𝑥 5 + 𝑥

1

∫0 (𝑥 − 𝑡) ∑3𝑘=0 𝐂 𝑇 𝑴𝑘 (𝑡) d𝑡 + ∫0 (𝑥 + 𝑡) ∑3𝑘=0 𝐂 𝑇 𝑴𝑘 (𝑡) d𝑡.(5.343) Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.343) is 1

1

∫0 ∑3𝑘=0 𝐂 𝑇 𝑴𝑘 (𝑥)𝑅𝑗 (𝑥)d𝑥 = ∫0 [ 1

= −6 − 2𝑥 ] 𝑅 (𝑥)d𝑥 +19𝑥 3 − 𝑥 5 𝑗

𝑥

+ ∫0 [∫0 (𝑥 − 𝑡) ∑3𝑘=0 𝐂 𝑇 𝑴𝑘 (𝑡) d𝑡]𝑅𝑗 (𝑥)d𝑥 1

1

+ ∫0 [∫0 (𝑥 + 𝑡) ∑3𝑘=0 𝐂 𝑇 𝑴𝑘 (𝑡) d𝑡 ] 𝑅𝑗 (𝑥)d𝑥. For 𝑗 = 0,1,2,3. The matrix form of Eq. (5.344) is given as

After solving the above we get 𝛼0 = 6.37, 𝛼1 = 5.12, 𝛼2 = −0.801, 𝛼3 = 0.753. 912

(5.344)

𝑢(𝑥) = −1.3030 + 8.7050𝑥 + 12.332𝑥 2 + 4.522472𝑥 3 . Table 5.65. Comparison of Exact and Approximate solutions of Eq. (5.340) obtained from Meixner polynomials Method (MPM) 𝑥

Exact Solution


Error

0.0

0.000000000000000

-0.000000000000001

1.33800E-15

0.1

0.620000000000000

0.620000000000000

2.57470E-16

0.2

1.360000000000000

1.360000000000000

3.17900E-16

0.3

2.340000000000000

2.340000000000001

5.04700E-16

0.4

3.680000000000000

3.680000000000000

4.19600E-16

0.5

5.500000000000000

5.500000000000000

1.79300E-16

0.6

7.920000000000000

7.920000000000000

9.95000E-17

0.7

11.060000000000000

11.060000000000000

3.00000E-16

0.8

15.040000000000000

15.040000000000000

3.06000E-16

0.9

19.980000000000000

19.980000000000000

2.00000E-18

1.0

26.000000000000000

26.000000000000001

7.32000E-16


Now we consider 50th order Meixner Polynomials, i.e. for 𝒌 = 𝟓𝟎, and we apply the proposed technique to solve Eq. (5.340) with 𝑘 = 50. We have Eq. (5.341) is 50 𝑇 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝑀𝑘 (𝑥) = ∑𝑘=0 𝑪 𝑴𝑘 (𝑥),

where 𝑪 = [𝛼0 , 𝛼1 , … , 𝛼50 ]𝑇 , 𝑴(𝑥) = [𝑀0 , 𝑀1 , … , 𝑀50 ]𝑇 . For 𝑗 = 0,1,2,3. The matrix form of Eq. (5.341) is given as

913

(5.342)

−0.16666666 [ ⋮ −6.35727 ∗ 1067

⋯ ⋱ ⋯

−2.41666666 −3.180212 ∗ 1067 𝑎0 ⋮ ][ ] = [ ] ⋮ ⋮ 𝑎50 5.945049 ∗ 1068 9.0773 ∗ 10136

We have the approximate values of 𝛼. 𝛼0 = 0.1704074455, 𝛼1 = −4.521590, … , 𝛼50 = 1.19033 ∗ 10−71 , We obtained the approximate solution of 𝑢(𝑥), their error estimate and graphical representation as shown below 𝑢(𝑥) = 8.72300 ∗ 10−24 + 5.999999𝑥 + ⋯ + 3.62029 ∗ 10−7 𝑥 50 , Graphical representation


Table 5.66. Comparison of Exact and Approximate solutions of Eq. (5.340) obtained from Meixner polynomials Method (MPM) 𝑥

Exact Solution


0.0

0.000000000000000

0.000000000000000

8.72300E-24

0.1

0.620000000000000

0.620000000000000

2.23218E-24

0.2

1.360000000000000

1.360000000000000

1.70219E-24

0.3

2.340000000000000

2.340000000000000

3.52167E-24

0.4

3.680000000000000

3.680000000000000

1.73955E-24

0.5

5.500000000000000

5.500000000000000

2.26779E-24

0.6

7.920000000000000

7.920000000000000

3.64022E-24

0.7

11.060000000000000

11.060000000000000

2.01627E-24

0.8

15.040000000000000

15.040000000000000

4.16472E-25

0.9

19.980000000000000

19.980000000000000

4.76919E-24

914

Error

5.12.3. Abel’ Integral Equation Consider the nonlinear Abel’s integral equation [229] as 2

1

𝑥

𝑒𝑥 2 (3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡

𝑒 𝑢(𝑡) d𝑡 .

(5.345)

The exact solution of Eq. (5.345) is 𝑢(𝑥) = 1 + ln(1 + 𝑥). Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation 𝑒 𝑢(𝑡) = 𝑣(𝑡), 𝑢(𝑡) = ln(𝑣(𝑡)). Using the above transformation into the given integral equation we get 2

1

𝑥

𝑒𝑥 2 (3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡

𝑣(𝑡)d𝑡.

(5.346)

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑀𝑘 (𝑥).

(5.347)

To solve Eq. (5.345) we take 𝑘 = 2. Eq. (5.347) is 𝑣(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑀𝑘 (𝑥) = ∑2𝑘=0 𝐂 𝑇 𝑴𝑘 (𝑥),

(5.348)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝑀𝑘 (𝑥) = [𝑀0 , 𝑀1 , 𝑀2 ]𝑇 . Putting Eq. (5.348) into Eq. (5.345), we obtained 2

1

𝑥

𝑒𝑥 2 (3 + 2𝑥) = ∫0 3

1 √𝑥−𝑡

∑2𝑘=0 𝐂 𝑇 𝑴𝑘 (𝑡) d𝑡.

(5.349)


1

1

𝑥

∫0 3 𝑒𝑥 2 (3 + 2𝑥)𝑅𝑗 (𝑥)d𝑥 = ∫0 [∫0

1 √𝑥−𝑡

∑2𝑘=0 𝐂 𝑇 𝑀𝑘 (𝑡) d𝑡] 𝑅𝑗 (𝑥)d𝑥 .


=

.

After solving the above, we get 𝛼0 = 0.02654, 𝛼1 = 2.744, 𝛼2 = −0.003572. 𝑣(𝑥) = 𝑒(𝑥 + 1). 915

(5.350)

After applying the reverse transformation we get 𝑢(𝑥) = 1 + ln(1 + 𝑥). This is the exact solution.


1

2

𝑢(𝑥) = 𝑥 − 3 𝑥 2 + 5 + ∫−1((𝑥 2 − 𝑡 2 )𝑢(𝑡) + (𝑥 2 − 𝑡 2 )𝑣(𝑡)d𝑡)d𝑡. (5.351a) 1

2

𝑣(𝑥) = 𝑥 2 + 𝑥 3 + 5 𝑥 + ∫−1((𝑥 2 − 𝑡 2 )𝑢(𝑡) − (𝑥𝑡)𝑣(𝑡))d𝑡.

(5.351b)

The exact solution of Eq. (5.351) is 𝑢(𝑥) = 𝑥,

𝑣(𝑥) = 𝑥 2 + 𝑥 3 .

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑴𝒌 (𝑥),

(5.352a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝑴𝒌 (𝑥).

(5.352b)

To solve Eq. (5.351) we take 𝑲 = 𝟐. Eq. (5.352) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑴𝒌 (𝑥) = ∑2𝑘=0 𝐂 𝑇 𝑴𝒌 (𝑥),

(5.353a)

𝑣(𝑥) = ∑2𝑘=0 𝛽𝑘 𝑴𝒌 (𝑥) = ∑2𝑘=0 𝐃𝑇 𝑴𝒌 (𝑥),

(5.353b)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1 , 𝛽2 ]𝑇 , and 𝑴𝒌 (𝑥) = [𝑀0 , 𝑀1 , 𝑀2 ]𝑇 . Putting Eq. (5.353) into Eq. (5.351), we obtained 2

2

1

∑2𝑘=0 𝐂 𝑇 𝑴𝑘 (𝑥) = 𝑥 − 𝑥 2 + + ∫−1 ∑2𝑘=0(𝑥 2 − 𝑡 2 )[𝐂 𝑇 + 𝐃𝑇 ]𝑴𝑘 (𝑡) d𝑡 , (5.354a) 3 5 2

1

∑2𝑘=0 𝐃𝑇 𝑴𝑘 (𝑥) = 𝑥 2 + 𝑥 3 + 𝑥 + ∫−1 ∑2𝑘=0[(𝑥 2 − 𝑡 2 )𝐂 𝑇 − (𝑥 𝑡)𝐃𝑇 ]𝑴𝑘 (𝑡) d𝑡 , (5.354b) 5 Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2. Then integrating over [0,1] we have Eq. (5.354) is 1

1

2

2

∫0 ∑2𝑘=0 𝐂 𝑇 𝑴𝑘 (𝑥)𝑅𝑗 (𝑥)d𝑥 = ∫0 [(𝑥 − 3 𝑥 2 + 5) 𝑴𝑗 (𝑥)d𝑥] 𝐑𝑗 (𝑥)𝑑𝑥 1

1

+ ∫0 ([∫−1 ∑2𝑘=0(𝑥 2 − 𝑡 2 )[𝐂 𝑇 + 𝐃𝑇 ]𝑴𝑘 (𝑡) d𝑡 ] 𝐑𝑗 (𝑥)) d𝑥 . 1

1

2

(5.355a)

2

∫0 ∑2𝑘=0 𝐃𝑇 𝑴𝑘 (𝑥)𝐑𝑗 (𝑥)d𝑥 = ∫0 [(𝑥 − 3 𝑥 2 + 5) 𝑴𝑗 (𝑥)d𝑥] 𝐑𝑗 (𝑥)𝑑𝑥 1

1

+ ∫0 ([∫−1 ∑2𝑘=0[(𝑥 2 − 𝑡 2 )𝐂 𝑇 − (𝑥 𝑡)𝐃𝑇 ]𝑴𝑘 (𝑡) d𝑡 ] 𝐑𝑗 (𝑥)) d𝑥 . for 𝑗 = 0,1,2. The matrix form of Eq. (5.355) is given as

916

(5.355b)

=

After solving the above we get 𝛼0 = −1.01, 𝛼1 = 0.981, 𝛼2 = 0.00243, 𝛽0 = 1.05, 𝛽1 = −1.65, 𝛽2 = 0.561. 𝑢(𝑥) = −0.027 + 0.991𝑥 + 0.00486𝑥 2 , 𝑣(𝑥) = −0.04 + 0.59𝑥 + 1.12𝑥 2 .


917


𝑢(𝑥)

𝑣(𝑥)

𝑢(𝑥)𝑎𝑝𝑝𝑟𝑜𝑥 .

𝑣(𝑥)𝑎𝑝𝑝𝑟𝑜𝑥 .

Error in 𝑢(𝑥)

Error in 𝑣(𝑥)

0.0

0.00000000

0.00000000 -0.027000000

0.0400000000 2.7000E-02

4.0000E-02

0.1

0.10000000

0.01100000 0.0721000000 0.0302000000 2.7900E-02

1.9200E-02

0.2

0.20000000

0.04800000 0.1710000000 0.1230000000 2.9000E-02

7.5000E-02

0.3

0.30000000

0.11700000 0.2700000000 0.2380000000 3.0000E-02

1.2100E-01

0.4

0.40000000

0.22400000 0.3700000000 0.3750000000 3.0000E-02

1.5100E-01

0.5

0.50000000

0.37500000 0.4700000000 0.5350000000 3.0000E-02

1.6000E-01

0.6

0.60000000

0.57600000 0.5700000000 0.7170000000 3.0000E-02

1.4100E-01

0.7

0.70000000

0.83300000 0.6690000000 0.9220000000 3.1000E-02

8.9000E-02

0.8

0.80000000

1.15000000 0.7690000000 1.1500000000 3.1000E-02

0.0000E+00

0.9

0.90000000

1.54000000 0.8690000000 1.4000000000 3.1000E-02

1.4000E-01

1.0

1.00000000

2.00000000 0.9690000000 1.6700000000 3.1000E-02

3.3000E-01

To solve Eq. (5.351) we take 𝑲 = 𝟐𝟎. Eq. (5.352) is 20 𝑇 𝑢(𝑥) = ∑20 𝑘=0 𝛼𝑘 𝑴𝒌 (𝑥) = ∑𝑘=0 𝐂 𝑴𝒌 (𝑥),

(5.353a)

20 𝑇 𝑣(𝑥) = ∑20 𝑘=0 𝛽𝑘 𝑴𝒌 (𝑥) = ∑𝑘=0 𝐃 𝑴𝒌 (𝑥),

(5.353b)

where 𝐂 = [𝛼0 , 𝛼1 , … , 𝛼20 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1, … , 𝛽20 ]𝑇 , and 𝑴𝒌 (𝑥) = [𝑀0 , 𝑀1 , … , 𝑀20 ]𝑇 . Proceeding as before, we obtained the approximate values of 𝛼 and 𝛽. 𝛼0 = −1.714582, 𝛼1 = 2.033141, … , 𝛼20 = −1.19069 ∗ 10−34 , 𝛽0 = −0.11003544, 𝛽1 = 0.2494245, … , 𝛽20 = 2.759441 ∗ 10−34 . Consequently we get the approximate solution of 𝑢(𝑥), 𝑣(𝑥) and their error estimate and graphical representation as shown below 𝑢(𝑥) = −3.2778 ∗ 10−22 + 0.99999𝑥 + ⋯ − 2.89683 ∗ 10−16 𝑥 20 , 𝑣(𝑥) = 1.05272 ∗ 10−21 − 1.2363 ∗ 10−20 𝑥 + ⋯ + 6.7131 ∗ 10−16 𝑥 20 .

918

Graphical representation


Table 5.68. Comparison of Exact and Approximate solutions of Eq. (5. 351) obtained from Meixner polynomials Method (MPM) 𝑥

𝑢(𝑥)

𝑣(𝑥)


𝑣(𝑥)𝑎𝑝𝑝𝑟𝑜𝑥 . 0.000000000

Error

Error

in 𝑢(𝑥)

in 𝑣(𝑥)

0.0

0.00000000

0.00000000

0.0000000000

3.2779E-22

1.0527E-21

0.1

0.10000000

0.01100000

0.1000000000 0.0110000000 1.7528E-22

1.0482E-21

0.2

0.20000000

0.04800000

0.2000000000 0.0480000000 4.3035E-22

1.6879E-22

0.3

0.30000000

0.11700000

0.3000000000 0.1170000000 2.2671E-22

1.2501E-21

0.4

0.40000000

0.22400000

0.4000000000 0.2240000000 1.4162E-22

1.1382E-21

0.5

0.50000000

0.37500000

0.5000000000 0.3750000000 3.3287E-22

1.0237E-22

0.6

0.60000000

0.57600000

0.6000000000 0.5760000000 2.9921E-22

9.3373E-22

0.7

0.70000000

0.83300000

0.7000000000 0.8330000000 1.9768E-22

1.4885E-21

0.8

0.80000000

1.15200000

0.8000000000 1.1520000000 1.7282E-22

1.6925E-21

0.9

0.90000000

1.53900000

0.9000000000 1.5390000000 2.2667E-22

9.5223E-22

1.0

1.00000000

2.00000000

1.0000000000 2.0000000000 4.1294E-22

8.0334E-21

To solve Eq. (5.351) we take 𝑲 = 𝟓𝟎. Eq. (5.352) is 50 𝑇 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝑴𝒌 (𝑥) = ∑𝑘=0 𝐂 𝑴𝒌 (𝑥),

(5.353a)

50 𝑇 𝑣(𝑥) = ∑50 𝑘=0 𝛽𝑘 𝑴𝒌 (𝑥) = ∑𝑘=0 𝐃 𝑴𝒌 (𝑥),

(5.353b)

where 𝐂 = [𝛼0 , 𝛼1 , … , 𝛼50 ]𝑇 , 𝐃 = [𝛽0 , 𝛽1, … , 𝛽50 ]𝑇 , and 𝑴𝒌 (𝑥) = [𝑀0 , 𝑀1 , … , 𝑀50 ]𝑇 . Proceeding as before, we obtained the approximate values of 𝛼 and 𝛽. 919

𝛼0 = −2.54114, 𝛼1 = 3.229481, … , 𝛼50 = −4.383255 ∗ 10−80 , 𝛽0 = −4.563667, 𝛽1 = 6.721260, … , 𝛽50 = −3.971876 ∗ 10−79 . Consequently we get the approximate solution of 𝑢(𝑥), 𝑣(𝑥) and their error estimate and graphical representation as shown below 𝑢(𝑥) = −1.367778 ∗ 10−27 + 0.99999𝑥 + ⋯ − 1.3331274 ∗ 10−15 𝑥 50 , 𝑣(𝑥) = −1.23939 ∗ 10−26 − 2.317133 ∗ 10−25 𝑥 + ⋯ − 1.208010 ∗ 10−14 𝑥 50 . Graphical representation


920


𝑢(𝑥)

𝑣(𝑥)



Error

Error in

in 𝑢(𝑥)

𝑣(𝑥)

0.0

0.00000000

0.00000000

0.0000000000 0.0000000000

1.3678E-27

1.2394E-26

0.1

0.10000000

0.01100000

0.1000000000 0.0110000000 3.1289E-28

2.8353E-27

0.2

0.20000000

0.04800000

0.2000000000 0.0480000000 1.4313E-27

1.2970E-26

0.3

0.30000000

0.11700000

0.3000000000 0.1170000000 4.2242E-28

3.8276E-27

0.4

0.40000000

0.22400000

0.4000000000 0.2240000000 1.0116E-27

9.1663E-27

0.5

0.50000000

0.37500000

0.5000000000 0.3750000000 1.4497E-27

1.3137E-26

0.6

0.60000000

0.57600000

0.6000000000 0.5760000000 1.3188E-27

1.1950E-26

0.7

0.70000000

0.83300000

0.7000000000 0.8330000000 1.4655E-27

1.3280E-26

0.8

0.80000000

1.15200000

0.8000000000 1.1520000000 1.1713E-27

1.0614E-26

0.9

0.90000000

1.53900000

0.9000000000 1.5390000000 1.3973E-27

1.2661E-26

1.0

1.00000000

2.00000000

1.0000000000 2.0000000000 1.1014E-26

9.9807E-26

5.12.5. System of Volterra Integral Equation Consider the system of Volterra integral equation of 2nd kind [229] as 𝑥

𝑢(𝑥) = 1 + 𝑥 + 𝑥 2 − 𝑥 3 + ∫0 ((𝑥 𝑡)𝑢(𝑡) + (𝑥 𝑡)𝑣(𝑡)), 1

1

(5.356a)

𝑥

𝑣(𝑥) = 1 − 𝑥 − 𝑥 2 − 3 𝑥 3 − 6 𝑥 4 − ∫0 ((𝑥 − 𝑡)𝑢(𝑡) + (𝑥 − 𝑡)𝑣(𝑡)),(5.356b) The exact solution of Eq. (5.356) is 𝑢(𝑥) = 1 + 𝑥 + 𝑥 2 . 𝑣(𝑥) = 1 − 𝑥 − 𝑥 2 . According to the proposed technique, consider the trial solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑀𝑘 (𝑥).

(5.357a)

𝑣(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝑀𝑘 (𝑥).

(5.357b)

Consider 2nd order Meixner Bessel Polynomials, i.e. for 𝒌 = 𝟐, and we apply the proposed technique to solve Eq. (5.356) with 𝑘 = 2. We have Eq. (5.357) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑀𝑘 (𝑥) = ∑2𝑘=0 𝑪𝑇 𝑴𝑘 (𝑥),

(5.358a)

𝑣(𝑥) = ∑2𝑘=0 𝛽𝑘 𝑀𝑘 (𝑥) = ∑2𝑘=0 𝑫𝑇 𝑴𝑘 (𝑥),

(5.358b)

where 𝑪 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , 𝑫 = [𝛽0 , 𝛽1 , 𝛽2 ]𝑇 , 𝑴(𝑥) = [𝑀0 , 𝑀1 , 𝑀2 ]𝑇 . 921

Putting Eq. (5.358) into Eq. (5.356), we obtained ∑2𝑘=0 𝑪𝑇 𝑴𝑘 (𝑥) = 1 + 𝑥 + 𝑥 2 − 𝑥 3 𝑥

+ ∫0 (𝑥 𝑡)(𝑢(𝑡) + 𝑣(t)) ∑2𝑘=0 𝑪𝑇 𝑴𝑘 (𝑡) d𝑡. 1

(5.359a)

1

∑2𝑘=0 𝑫𝑇 𝑹𝑘 (𝑥) = 1 − 𝑥 − 𝑥 2 − 𝑥 3 − 𝑥 4 3 6 𝑥

− ∫0 (𝑥 − 𝑡)(𝑢(𝑡) + 𝑣(t)) ∑2𝑘=0 𝑫𝑇 𝑹𝑘 (𝑡) d𝑡.

(5.359b)

Multiplying both sides by 𝑹𝒋 (𝑥), 𝑗 = 0,1,2. Then integrating over [0,1] we have Eq. (5.359) is 1

1

∫0 ∑2𝑘=0 𝑪𝑇 𝑴𝑘 (𝑥)𝑹𝒋 (𝑥) d𝑥 = ∫0 [1 + 𝑥 + 𝑥 2 − 𝑥 3 ] 𝑹𝒋 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 (𝑥 𝑡)(𝑢(𝑡) + 𝑣(t)) ∑2𝑘=0 𝑪𝑇 𝑴𝑘 (𝑡) d𝑡 ]𝑹𝒋 (𝑥)d𝑥, 1

1

1

(5.360a)

1

∫0 ∑2𝑘=0 𝐷 𝑇 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [1 − 𝑥 − 𝑥 2 − 3 𝑥 3 − 6 𝑥 4 ] 𝑹𝒋 (𝑥)d𝑥 1

𝑥

− ∫0 [∫0 ((𝑥 − 𝑡)𝑢(𝑡) + 𝑣(t)) ∑2𝑘=0 𝑫𝑇 𝑴𝑘 (𝑡) d𝑡]𝑹𝒋 (𝑥)d𝑥,

(5.360b)


After solving we get the values of 𝛼 , 𝛽. 𝛼0 = −0.035605, 𝛼1 = 0.83918, 𝛼2 = 0.15036 . 𝛽0 = −1.7508, 𝛽1 = 3.9800 , 𝛽2 = −1.2224 . Consequently, we get the approximate solution and its graphical solution and error table. 𝑢(𝑥) = 0.95394 + 1.4406𝑥 + 0.30072𝑥 2 . 𝑣(𝑥) = 1.0068 − 0.9096𝑥 − 2.4448𝑥 2 .

922



𝑢(𝑥)

𝑣(𝑥)


𝑣(𝑥)𝑎𝑝𝑝𝑟𝑜𝑥

Error in 𝑣(𝑥)

Error in 𝑢(𝑥)

0.0

1.00000000

1.00000000

0.95394000

1.00680000

4.6060E-02

6.8000E-03

0.1

1.11000000

0.89000000

1.10100000

0.89139000

9.0000E-03

1.3900E-03

0.2

1.24000000

0.76000000

1.25410000

0.72709000

1.4100E-02

3.2910E-02

0.3

1.39000000

0.61000000

1.41320000

0.51389000

2.3200E-02

9.6110E-02

0.4

1.56000000

0.44000000

1.57830000

0.25179000

1.8300E-02

1.8821E-01

0.5

1.75000000

0.25000000

1.74940000

-0.0592000

6.000E-04

3.0920E-01

0.6

1.96000000

0.04000000

1.92660000

-0.4190900

3.340E-02

4.5909E-01

0.7

2.19000000

-0.1900000

2.10960000

-0.8279200

8.040E-02

6.3792E-01

0.8

2.44000000

-0.4400000

2.29890000

-1.2856000

1.411E-01

8.4560E-01

0.9

2.71000000

-0.7100000

2.49400000

-1.7921000

2.160E-01

1.0821E+00

1.0

3.00000000

-1.0000000

2.69520000

-2.3476000

3.048E-01

1.3476E+00

Now we consider 20th order Meixner Polynomials, i.e. for 𝒌 = 𝟐𝟎, and we apply the proposed technique to solve Eq. (5.356) with 𝑘 = 20. We have Eq. (5.357) is 20 𝑇 𝑢(𝑥) = ∑20 𝑘=0 𝛼𝑘 𝑀𝑘 (𝑥) = ∑𝑘=0 𝑪 𝑴𝑘 (𝑥),

(5.358a)

20 𝑇 𝑣(𝑥) = ∑20 𝑘=0 𝛽𝑘 𝑀𝑘 (𝑥) = ∑𝑘=0 𝑫 𝑴𝑘 (𝑥),

(5.358b)

where 𝑪 = [𝛼0 , 𝛼1 , … , 𝛼20 ]𝑇 , 𝑫 = [𝛽0 , 𝛽1 , … , 𝛽20 ]𝑇 , 𝑴(𝑥) = [𝑀0 , 𝑀1 , … , 𝑀20 ]𝑇 . We have the approximate values of 𝛼, and 𝛽. 923

𝛼0 = −3791.1639, 𝛼1 = −7586.010, … , 𝛼20 = 4.2532 ∗ 10−17 , 𝛽0 = 1495.0293, 𝛽1 = 2989.74155, … , 𝛽20 = 5.615093 ∗ 10−17 . We obtained the approximate solution of 𝑢(𝑥), 𝑣(𝑥) and their error estimate and graphical representation as shown below 𝑢(𝑥) = 0.9999999 + 1.0000𝑥 + ⋯ + 4.0562459 ∗ 10−23 𝑥 20 , 𝑣(𝑥) = 1.000 + 1.000𝑥 + ⋯ + 5.354970 ∗ 10−23 𝑥 20 . Graphical representation


924


𝑢(𝑥)

𝑣(𝑥)



Error

Error in

in 𝑢(𝑥)

𝑣(𝑥)

0.0

1.00000000

1.000000000

1.0000000000 1.0000000000 7.4752E-23

9.1892E-21

0.1

1.09983341

0.9001665834 1.0998334166 0.9001665834 3.1510E-23

2.3426E-21

0.2

1.19866933

0.8013306692 1.1986693308 0.8013306692 2.7038E-24

1.0677E-21

0.3

1.29552020

0.7044797933 1.2955202067 0.7044797933 2.0801E-23

1.5343E-22

0.4

1.38941834

0.6105816577 1.3894183423 0.6105816577 2.6255E-23

7.3522E-23

0.5

1.47942553

0.5205744614 1.4794255386 0.5205744614 2.5710E-23

3.5118E-23

0.6

1.56464247

0.4353575266 1.5646424734 0.4353575266 2.4325E-23

3.9200E-24

0.7

1.64421768

0.3557823128 1.6442176872 0.3557823128 3.0840E-23

2.2004E-22

0.8

1.71735609

0.2826439091 1.7173560909 0.2826439091 6.0265E-23

1.1173E-21

0.9

1.78332690

0.2166730904 1.7833269096 0.2166730904 9.0111E-23

2.3182E-21

1.0

1.84147098

0.1585290152 1.8414709848 0.1585290152 4.0095E-22

9.1256E-21

Now we consider 50th order Meixner Polynomials, i.e. for 𝒌 = 𝟓𝟎, and we apply the proposed technique to solve Eq. (5.356) with 𝑘 = 20. We have Eq. (5.357) is 50 𝑇 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝑀𝑘 (𝑥) = ∑𝑘=0 𝑪 𝑴𝑘 (𝑥),

(5.358a)

50 𝑇 𝑣(𝑥) = ∑50 𝑘=0 𝛽𝑘 𝑀𝑘 (𝑥) = ∑𝑘=0 𝑫 𝑴𝑘 (𝑥),

(5.358b)

where 𝑪 = [𝛼0 , 𝛼1 , … , 𝛼50 ]𝑇 , 𝑫 = [𝛽0 , 𝛽1 , … , 𝛽50 ]𝑇 , 𝑴(𝑥) = [𝑀0 , 𝑀1 , … , 𝑀50 ]𝑇 . We have the approximate values of 𝛼, and 𝛽. 𝛼0 = 3617.7533, 𝛼1 = 7231.8236, … , 𝛼50 = −3.44068 ∗ 10−63 , 𝛽0 = 188.83308, 𝛽1 = 377.349114, … , 𝛽20 = −3.427173 ∗ 10−64 . We obtained the approximate solution of 𝑢(𝑥), 𝑣(𝑥) and their error estimate and graphical representation as shown below 𝑢(𝑥) = 1.000 + 0.99999𝑥 + ⋯ − 3.0559 ∗ 10−78 𝑥 50 , 𝑣(𝑥) = 1.000 − 1.000𝑥 + ⋯ − 3.043941 ∗ 10−79 𝑥 50 .

925



𝑢(𝑥)

𝑣(𝑥)



Error

Error in

in 𝑢(𝑥)

𝑣(𝑥)

0.0

1.00000000

1.00000000

1.00000000

1.0000000000

2.9524E-46

2.6971E-48

0.1

1.09983341

0.90016658

1.09983341

0.9001665834

1.6024E-47

7.0444E-50

0.2

1.19866933

0.80133066

1.19866933

0.8013306692

4.4196E-47

5.8193E-49

0.3

1.29552020

0.70447979

1.29552020

0.7044797933

3.1770E-47

5.3425E-49

0.4

1.38941834

0.61058165

1.38941834

0.6105816577

4.0162E-47

6.1982E-49

0.5

1.47942553

0.52057446

1.47942553

0.5205744614

4.7749E-47

5.6567E-49

0.6

1.56464247

0.43535752

1.56464247

0.4353575266

4.2471E-47

2.9415E-49

0.7

1.64421768

0.35578231

1.64421768

0.3557823128

3.5261E-47

7.1455E-50

0.8

1.71735609

0.28264390

1.71735609

0.2826439091

4.9288E-47

2.2560E-49

0.9

1.78332690

0.21667309

1.78332690

0.2166730904

1.6410E-47

4.7603E-49

1.0

1.84147098

0.15852901

1.84147098

0.1585290152

3.4279E-46

1.5025E-48

5.13. Rook Polynomials Method (RPM) I develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Rook polynomials [62]. Rook polynomials is given as 𝑛−𝑚

(−1)𝑚 (𝑛)!

2 𝑅(𝑥) = 𝑥 𝑛 𝑛! ∑𝑘=0 [(𝑚!)2 ((𝑛−𝑚)!) (−𝑥) − 𝑚].

926

(5.361)

For 𝑚 = 0, first four Rook polynomials is given as 𝑅0 (𝑥) = 1, 𝑅1 (𝑥) = 𝑥 + 1, 𝑅2 (𝑥) = 2𝑥 2 + 4𝑥 + 1, 𝑅3 (𝑥) = 6𝑥 3 + 18𝑥 2 + 9𝑥 + 1,



(5.362)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, to find an approximate solution 𝑢̃(𝑥) of Eq. (5.362). For this, we assume that 𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝑅𝑘 (𝑥),

(5.363)

where 𝑅𝑘 (𝑥) are Rook polynomials of degree 𝑘 defined in equation (5.361) and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.363) into (5.362), we get 𝑥

𝑓(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝑅𝑘 (𝑡)d𝑡,

(5.364)

Then the Galerkin equations are obtained by multiplying both sides of (4) by 𝑍𝑗 (𝑥) and then integrating with respect to 𝑥 from 𝑎 to 𝑏, we have 𝑏

𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝑅𝑗 (𝑥)d𝑥 = ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝑅𝑘 (𝑡)d𝑡 )𝑅𝑗 (𝑡)d𝑥, 𝑗 = 0,1,2, … 𝑛,(5.365) Since in each equation, there are two integrals. The inner integrand of the left side is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.365) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.366)

where 927

𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝑅𝑘 (𝑡)d𝑡)𝑅𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛 𝑏

𝐹𝑗 = ∫𝑎 𝑓(𝑥) 𝑅𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛 Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.366) and substituting these values of parameters in (5.362), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.363).



(5.367)


(5.368)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝑅𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑅𝑘 (𝑡)d𝑡)𝑅𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛. Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.361) and substituting these values of parameters in (5.363), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.367).

5.13.2. Volterra Integral Equation Consider the linear Volterra Integral equation of 2nd kind [229] as 1

𝑥

1

𝑢(𝑥) = − 2 𝑥 − 4 sin(2𝑥) + cos 2 (𝑥) + ∫0 𝑢(𝑡)d𝑡.

(5.369)

The exact solution of Eq. (5.369) is 𝑢(𝑥) = cos 2 (𝑥). According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥).

(5.370)

Consider 3rd order Rook Polynomials, i.e. for 𝒌 = 𝟑, and we apply the proposed technique to solve Eq. (5.369) with 𝑘 = 3. We have Eq. (14) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑3𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥), where 𝐶 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and

(5.371) (5.372)

𝑅(𝑥) = [𝑅0 , 𝑅1 , 𝑅2 , 𝑅3 ]𝑇 .

(5.373) 928


𝑥

1

∑3𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥) = − 𝑥 − sin(2𝑥) + cos 2 (𝑥) + ∫0 ∑3𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑡) d𝑡. 2 4

(5.374)


1

1

1

∫0 ∑3𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [− 2 𝑥 − 4 sin(2𝑥) + cos2 (𝑥)] 𝑅𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 ∑3𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑡) d𝑡]𝑅𝑗 (𝑥)d𝑥,

(5.375)


=

.

After solving we get 𝛼0 = −0.15247, 𝛼1 = 1.6660, 𝛼2 = −0.51246, 𝛼3 = 0.019997. Consequently, we have the approximate solution is 𝑢(𝑥) = 1.0210 − 0.20380𝑥 − 0.66495𝑥 2 + 0.11998𝑥 3 After solving we obtained the approximate solution.


929

Table 5.73. Comparison of Exact and Approximate solutions of Eq. (5.369) obtained from Rook polynomials Method (RPM) x

Exact Solution


Error

0.0

1.000000000000000

1.021000000000000

2.10000E-02

0.1

0.990020000000000

0.994070000000000

4.05000E-03

0.2

0.960540000000000

0.954600000000000

5.94000E-03

0.3

0.912670000000000

0.903250000000000

9.42000E-03

0.4

0.848350000000000

0.840770000000000

7.58000E-03

0.5

0.770150000000000

0.767860000000000

2.29000E-03

0.6

0.681190000000000

0.685260000000000

4.07000E-03

0.7

0.584980000000000

0.593660000000000

8.68000E-03

0.8

0.485400000000000

0.493820000000000

8.42000E-03

0.9

0.386400000000000

0.386440000000000

4.00000E-05

1.0

0.291920000000000

0.272230000000000

1.96900E-02

Now we consider 30th order Rook Polynomials, i.e. for 𝒌 = 𝟑𝟎, and we apply the proposed technique to solve Eq. (5.369) with 𝑘 = 30. We have Eq. (5.370) is 30 𝑇 𝑢(𝑥) = ∑30 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐶 𝑅𝑘 (𝑥),

(5.371)

where 𝐶 = [𝛼0 , 𝛼1 , … , 𝛼30 ]𝑇 , and

(5.372)

𝑅(𝑥) = [𝑅0 , 𝑅1 , … , 𝑅30 ]𝑇 .

(5.373)

Putting Eq. (5.371) into Eq. (13), we obtained 1

𝑥

1

30 𝑇 2 𝑇 ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑥) = − 2 𝑥 − 4 sin(2𝑥) + cos (𝑥) + ∫0 ∑𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡.

(5.374)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2, … ,30. Then integrating over [0,1] we have Eq. (5.374) is 1

1

1

1

𝑇 2 ∫0 ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [− 2 𝑥 − 4 sin(2𝑥) + cos (𝑥)] 𝑅𝑗 (𝑥)d𝑥 1

𝑥

𝑇 + ∫0 [∫0 ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡 ]𝑅𝑗 (𝑥)d𝑥 ,

(5.375)

for 𝑗 = 0,1,2, … ,30. The matrix form of Eq. (5.375) is given as 0.50000 [ ⋮ 1.75683 ∗ 1033

0.30030354 ⋯ 4.57394 ∗ 1034 𝑎0 ⋮ ][ ] = [ ] ⋮ ⋱ ⋮ 34 70 𝑎30 1.8466 ∗ 10 ⋯ 2.886072 ∗ 10

After solving we get 930

𝛼0 = 0.0905269, 𝛼1 = 1.15962, … , 𝛼30 = −8.298184 ∗ 10−47 . Consequently, we have the approximate solution is 𝑢(𝑥) = 0.9999 − 4.29860 ∗ 10−24 𝑥 + ⋯ − 2.20111 ∗ 10−14 𝑥 30


Table 5.74. Comparison of Exact and Approximate solutions of Eq. (5. 369) obtained from Rook polynomials Method (RPM) 𝑥

Exact Solution


Error

0.0

1.000000000000000

1.000000000000000

5.45675E-27

0.1

0.990033288920621

0.990033288920621

9.85473E-30

0.2

0.960530497001443

0.960530497001443

7.58615E-28

0.3

0.912667807454839

0.912667807454839

5.94132E-28

0.4

0.848353354673583

0.848353354673583

5.87327E-28

0.5

0.770151152934070

0.770151152934070

5.05774E-28

0.6

0.681178877238337

0.681178877238337

3.13449E-28

0.7

0.584983571450120

0.584983571450120

1.90935E-28

0.8

0.485400238849356

0.485400238849356

3.50036E-28

0.9

0.386398952653456

0.386398952653456

2.54223E-28

1.0

0.291926581726429

0.291926581726429

2.68651E-27

Now we again consider 50th order Rook Polynomials, i.e. for 𝑘 = 50, and we apply the proposed technique to solve Eq. (5.369) with 𝑘 = 50. We have Eq. (5.370) is 50 𝑇 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐶 𝑅𝑘 (𝑥),

where 𝐶 = [𝛼0 , 𝛼1 , … , 𝛼50 ]𝑇 ,

(5.371) (5.372)

931

and

𝑅(𝑥) = [𝑅0 , 𝑅1 , … , 𝑅50 ]𝑇 .

(5.373)


𝑥

1

50 𝑇 2 𝑇 ∑50 𝑘=0 𝐶 𝑅𝑘 (𝑥) = − 2 𝑥 − 4 sin(2𝑥) + cos (𝑥) + ∫0 ∑𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡.

(5.374)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2, … ,50. Then integrating over [0,1] we have Eq. (5.374) is 1

1

1

1

𝑇 2 ∫0 ∑50 𝑘=0 𝐶 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [− 2 𝑥 − 4 sin(2𝑥) + cos (𝑥)] 𝑅𝑗 (𝑥)d𝑥 1

𝑥

𝑇 + ∫0 [∫0 ∑50 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡 ]𝑅𝑗 (𝑥)d𝑥 ,

(5.375)

for 𝑗 = 0,1,2, … ,50. The matrix form of Eq. (5.375) is given as 0.50000 [ ⋮ 2.713015 ∗ 1067

⋯ ⋱ ⋯

𝑎0 0.30030600 6.500226 ∗ 1067 ⋮ ][ ] = [ ] ⋮ ⋮ 1.8466 ∗ 1034 9.7253119 ∗ 10136 𝑎50

After solving we get 𝛼0 = 0.59087, 𝛼1 = 0.43612, … , 𝛼50 = 5.1983192 ∗ 10−92 . Consequently, we have the approximate solution is 𝑢(𝑥) = 0.9999 + 5.4818731 ∗ 10−51 𝑥 + ⋯ + 1.58102 ∗ 10−27 𝑥 50 .


932

Table 5.75. Comparison of Exact and Approximate solutions of Eq. (5.369) obtained from Rook polynomials Method (RPM) 𝑥

Exact Solution


Error

0.0

1.000000000000000

1.000000000000000

2.24285E-54

0.1

0.990033288920621

0.990033288920621

3.01384E-55

0.2

0.960530497001443

0.960530497001443

1.30641E-55

0.3

0.912667807454839

0.912667807454839

2.58576E-55

0.4

0.848353354673583

0.848353354673583

1.40037E-55

0.5

0.770151152934070

0.770151152934070

1.41704E-56

0.6

0.681178877238337

0.681178877238337

1.13797E-55

0.7

0.584983571450120

0.584983571450120

2.48375E-55

0.8

0.485400238849356

0.485400238849356

1.01580E-55

0.9

0.386398952653456

0.386398952653456

2.88291E-55

1.0

0.291926581726429

0.291926581726429

2.05682E-54

5.13.3. Fredholm Integral equation Consider the linear Fredholm Integral equation [229] as 2

1

7

𝑢(𝑥) = 15 + 12 𝑥 + 𝑥 2 + 𝑥 3 − ∫0 (1 + 𝑥 − 𝑡)𝑢(𝑡)d𝑡.

(5.376)

The exact solution of Eq. (5.376) is 𝑢(𝑥) = 𝑥 2 + 𝑥 3 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥).

(5.377)

Consider 2nd order Rook Polynomials, i.e. for 𝒌 = 𝟐, and we apply the proposed technique to solve Eq. (5.376) with 𝑘 = 2. We have Eq. (5.377) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥). Where 𝐶 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and

(5.378) (5.379)

𝑅(𝑥) = [𝑅0 , 𝑅1 , 𝑅2 ]𝑇 .

(5.380)

Putting Eq. (5.378) into Eq. (5.376), we obtained ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥) =

2 15

7

1

+ 12 𝑥 + 𝑥 2 + 𝑥 3 − ∫0 (1 + 𝑥 − 𝑡) ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑡) d𝑡. (5.381)


1

1

2

7

∫0 ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [15 + 12 𝑥 + 𝑥 2 + 𝑥 3 ] 𝑅𝑗 (𝑥)d𝑥 1

1

− ∫0 [∫0 (1 + 𝑥 − 𝑡) ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑡) d𝑡 ] 𝑅𝑗 (𝑥)d𝑥,

(5.382)


=

After solving we get 𝛼0 =

22 5

, 𝛼1 = −

28 5

5

, 𝛼2 = 4.

Consequently, we have the approximate solution is 𝑢(𝑥) = 0.079000 − 0.75690𝑥 + 2.6480𝑥 2 .


934


Exact Solution


Error

0.0

0.000000000000000

0.079000000000000

7.90000E-02

0.1

0.011000000000000

0.029790000000000

1.87900E-02

0.2

0.048000000000000

0.033540000000000

1.44600E-02

0.3

0.117000000000000

0.090250000000000

2.67500E-02

0.4

0.224000000000000

0.199920000000000

2.40800E-02

0.5

0.375000000000000

0.362550000000000

1.24500E-02

0.6

0.576000000000000

0.578140000000000

2.14000E-03

0.7

0.833000000000000

0.846670000000000

1.36700E-02

0.8

1.152000000000000

1.168200000000000

1.62000E-02

0.9

1.539000000000000

1.542700000000000

3.70000E-03

1.0

2.000000000000000

1.970100000000000

2.99000E-02

Consider 30th order Rook Polynomials, i.e. for 𝒌 = 𝟑𝟎, and we apply the proposed technique to solve Eq. (5.376) with 𝑘 = 30. We have Eq. (5.377) is 30 𝑇 𝑢(𝑥) = ∑30 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐶 𝑅𝑘 (𝑥).

where 𝐶 = [𝛼0 , 𝛼1 , 𝛼2 , … 𝛼30 ]𝑇 , and

(5.378) (5.379)

𝑅(𝑥) = [𝑅0 , 𝑅1 , 𝑅2 , … 𝑅30 ]𝑇 .

(5.380)

Putting Eq. (5.378) into Eq. (5.376), we obtained 𝑇 ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑥) =

2 15

1

7

𝑇 + 12 𝑥 + 𝑥 2 + 𝑥 3 + ∫0 (1 + 𝑥 − 𝑡) ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡. (5.381)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2 … 30. Then integrating over [0,1] we have Eq. (5.381) is 1

1

2

7

𝑇 2 3 ∫0 ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [15 + 12 𝑥 + 𝑥 + 𝑥 ] 𝑅𝑗 (𝑥)d𝑥 1

1

𝑇 + ∫0 [∫0 (1 + 𝑥 − 𝑡) ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡 ] 𝑅𝑗 (𝑥)d𝑥 ,

(5.382)

for 𝑗 = 0,1,2 … 30. The matrix form of Eq. (5.382) is given as 2 [ ⋮ 1.169838 ∗ 1035

1.00833333 ⋯ 7.30012 ∗ 1034 𝑎0 ⋮ ][ ] = [ ] ⋮ ⋱ ⋮ 35 70 𝑎30 1.197123 ∗ 10 ⋯ 3.22445 ∗ 10

After solving we get the values of 𝛼. 935

𝛼0 = 0.232644, 𝛼1 = −0.2465190, … , 𝛼30 = 2.557596881 ∗ 10−44 . Consequently, we have the approximate solution is 𝑢(𝑥) = 7.5062 ∗ 10−26 + 5.9568340 ∗ 10−23 𝑥 + ⋯ + 6.78409 ∗ 10−12 𝑥 30 ,


Table 5.77. Comparison of Exact and Approximate solutions of Eq. (5. 376) obtained from Rook polynomials Method (RPM) x

Exact Solution


Error

0.0

0.000000000000000

0.000000000000000

7.50625E-26

0.1

0.011000000000000

0.011000000000000

1.06124E-27

0.2

0.048000000000000

0.048000000000000

1.14244E-26

0.3

0.117000000000000

0.117000000000000

9.10760E-27

0.4

0.224000000000000

0.224000000000000

9.75159E-27

0.5

0.375000000000000

0.375000000000000

9.25931E-27

0.6

0.576000000000000

0.576000000000000

6.43535E-27

0.7

0.833000000000000

0.833000000000000

4.36917E-27

0.8

1.152000000000000

1.152000000000000

7.48556E-27

0.9

1.539000000000000

1.539000000000000

4.80759E-27

1.0

2.000000000000000

2.000000000000000

5.79936E-26

Consider 75th order Rook Polynomials, i.e. for 𝑘 = 75, and we apply the proposed technique to solve Eq. (5.376) with 𝑘 = 75. We have Eq. (5.377) is 75 𝑇 𝑢(𝑥) = ∑75 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐶 𝑅𝑘 (𝑥),

where 𝐶 = [𝛼0 , 𝛼1 , 𝛼2 , … 𝛼75 ]𝑇 ,

(5.378) (5.379)

936

and

𝑅(𝑥) = [𝑅0 , 𝑅1 , 𝑅2 , … 𝑅75 ]𝑇 .

(5.380)

Putting Eq. (5.378) into Eq. (5.376), we obtained 𝑇 ∑75 𝑘=0 𝐶 𝑅𝑘 (𝑥) =

2 15

1

7

𝑇 + 12 𝑥 + 𝑥 2 + 𝑥 3 + ∫0 (1 + 𝑥 − 𝑡) ∑75 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡. (5.381)


1

2

7

𝑇 2 3 ∫0 ∑75 𝑘=0 𝐶 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [15 + 12 𝑥 + 𝑥 + 𝑥 ] 𝑅𝑗 (𝑥)d𝑥 1

1

𝑇 + ∫0 [∫0 (1 + 𝑥 − 𝑡) ∑75 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡 ] 𝑅𝑗 (𝑥)d𝑥 ,

(5.382)

for 𝑗 = 0,1,2 … 75. The matrix form of Eq. (5.382) is given as 2 [ ⋮ 1.8862439 ∗ 10114

⋯ ⋱ ⋯

1.00833 31.14936 ∗ 10114 𝑎0 ⋮ ] [ ] = [ ] ⋮ ⋮ 114 229 𝑎75 2.0014 ∗ 10 2.029252 ∗ 10

After solving we get 𝛼0 = −0.205135, 𝛼1 = 0.388433240, … , 𝛼75 = 1.2445406 ∗ 10−124 . Consequently, we have the approximate solution is 𝑢(𝑥) = 2.476701 ∗ 10−49 + 1.010126 ∗ 10−45 𝑥 + ⋯ + 3.087598 ∗ 10−15 𝑥 75 ,


937


Exact


Error

0.0

0.000000000000000

0.000000000000000

2.47670E-49

0.1

0.011000000000000

0.011000000000000

2.84733E-50

0.2

0.048000000000000

0.048000000000000

1.79062E-50

0.3

0.117000000000000

0.117000000000000

6.45166E-51

0.4

0.224000000000000

0.224000000000000

2.13475E-50

0.5

0.375000000000000

0.375000000000000

1.29382E-50

0.6

0.576000000000000

0.576000000000000

4.95874E-51

0.7

0.833000000000000

0.833000000000000

1.95127E-50

0.8

1.152000000000000

1.152000000000000

2.03154E-50

0.9

1.539000000000000

1.539000000000000

1.73308E-50

1.0

2.000000000000000

2.000000000000000

1.88281E-49

5.13.4. Volterra-Fredholm Integral Equation Consider the Volterra-Fredholm integral equation [229] as 𝑥

1

𝑢(𝑥) = −𝑥 2 − 𝑥 4 − 𝑥 6 + ∫0 (3𝑥𝑡 + 4𝑥 2 𝑡 2 )𝑢(𝑡)d𝑡 + ∫0 (4𝑥𝑡 2 + 3𝑥 2 𝑡)𝑢(𝑡)d𝑡 ,

(5.383)

The exact solution of Eq. (5.383) is 𝑢(𝑥) = 𝑥. According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥).

(5.384)

To solve Eq. (5.383) we take 𝑛 = 1. Eq. (5.384) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑2𝑘=0 𝐂 𝑇 𝑅𝑘 (𝑥),

(5.385)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and 𝑅𝑘 (𝑥) = [𝑅0 , 𝑅1 , 𝑅2 ]𝑇 . Putting Eq. (5.385) into Eq. (5.383), we obtained ∑2𝑘=0 𝐂 𝑇 𝑅𝑘 (𝑥) = −𝑥 2 − 𝑥 4 − 𝑥 6 + 𝑥

1

∫0 (3𝑥𝑡 + 4𝑥 2 𝑡 2 ) ∑2𝑘=0 𝐂 𝑇 𝑅𝑘 (𝑡) d𝑡 + ∫0 (4𝑥𝑡 2 + 3𝑥 2 𝑡) ∑1𝑘=0 𝐂 𝑇 𝑅𝑘 (𝑡) d𝑡. (5.386) Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2. Then integrating over [0,1] we have Eq. (5.386) is

938

1

1

∫0 ∑2𝑘=0 𝐂 𝑇 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥)d𝑥 = ∫0 [−𝑥 2 − 𝑥 4 − 𝑥 6 ]𝑅𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 (3𝑥𝑡 + 4𝑥 2 𝑡 2 ) ∑2𝑘=0 𝐂 𝑇 𝑅𝑘 (𝑡) d𝑡]𝑅𝑗 (𝑥)d𝑥 1

1

+ ∫0 [∫0 (4𝑥𝑡 2 + 3𝑥 2 𝑡) ∑2𝑘=0 𝐂 𝑇 𝑅𝑘 (𝑡) d𝑡] 𝑅𝑗 (𝑥)d𝑥 .

(5.387)

for 𝑗 = 0,1,2. The matrix form of Eq. (5.387) is given as =

After solving the above we get 𝛼0 = −1.621, 𝛼1 = 1.690, 𝛼2 = −0.1138. 𝑢(𝑥) = 0.04500 + 1.235𝑥 − 0.2276𝑥 2 .


939


Exact Solution


Error

0.0

0.000000000000000

-0.045000000000000

4.50000E-02

0.1

0.100000000000000

0.076220000000000

2.37800E-02

0.2

0.200000000000000

0.192900000000000

7.10000E-03

0.3

0.300000000000000

0.305000000000000

5.00000E-03

0.4

0.400000000000000

0.412600000000000

1.26000E-02

0.5

0.500000000000000

0.515600000000000

1.56000E-02

0.6

0.600000000000000

0.614100000000000

1.41000E-02

0.7

0.700000000000000

0.708000000000000

8.00000E-03

0.8

0.800000000000000

0.797300000000000

2.70000E-03

0.9

0.900000000000000

0.882600000000000

1.74000E-02

1.0

1.000000000000000

0.962400000000000

3.76000E-02

To solve Eq. (5.383) we take 𝒌 = 𝟐𝟎. Eq. (5.384) is 20 𝑇 𝑢(𝑥) = ∑20 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐂 𝑅𝑘 (𝑥),

(5.385)

where 𝐂 = [𝛼0 , 𝛼1 , … , 𝛼20 ]𝑇 , and 𝑅𝑘 (𝑥) = [𝑅0 , 𝑅1 , … , 𝑅20 ]𝑇 . Putting Eq. (5.385) into Eq. (5.383), we obtained 𝑇 2 4 6 ∑20 𝑘=0 𝐂 𝑅𝑘 (𝑥) = −𝑥 − 𝑥 − 𝑥 + 𝑥

1

20 𝑇 2 2 𝑇 ∫0 (3𝑥𝑡 + 4𝑥 2 𝑡 2 ) ∑20 𝑘=0 𝐂 𝑅𝑘 (𝑡) d𝑡 + ∫0 (4𝑥𝑡 + 3𝑥 𝑡) ∑𝑘=0 𝐂 𝑅𝑘 (𝑡) d𝑡 . (5.386)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1, … ,20. Then integrating over [0,1] we have Eq. (5.386) is 1

1

𝑇 2 4 6 ∫0 ∑20 𝑘=0 𝐂 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥)d𝑥 = ∫0 [−𝑥 − 𝑥 − 𝑥 ]𝑅𝑗 (𝑥)d𝑥 1

𝑥

𝑇 + ∫0 [∫0 (3𝑥𝑡 + 4𝑥 2 𝑡 2 ) ∑20 𝑘=0 𝐂 𝑅𝑘 (𝑡) d𝑡 ]𝑅𝑗 (𝑥)d𝑥 1

1

𝑇 + ∫0 [∫0 (4𝑥𝑡 2 + 3𝑥 2 𝑡) ∑20 𝑘=0 𝐂 𝑅𝑘 (𝑡) d𝑡 ] 𝑅𝑗 (𝑥)d𝑥 .

(5.387)

for 𝑗 = 0,1, … ,20. The matrix form of Eq. (5.387) is given as −0.76388888 [ ⋮ −3.9468144 ∗ 1020

𝑎0 −0.6761904761 ⋯ 2.05886 ∗ 1020 ⋮ ] [ ] = [ ] ⋮ ⋱ ⋮ 20 38 𝑎20 2.46864 ∗ 10 ⋯ −7.793781 ∗ 10


𝛼0 = −0.878779, 𝛼1 = 0.8247392, … , 𝛼20 = −3.201650 ∗ 10−32 . 𝑢(𝑥) = 3.5394 ∗ 10−23 + 0.999999999𝑥 + ⋯ − 7.7893 ∗ 10−14 𝑥 20 .



Exact Solution


Error

0.0

0.000000000000000

0.000000000000000

3.53942E-23

0.1

0.100000000000000

0.100000000000000

4.81676E-24

0.2

0.200000000000000

0.200000000000000

9.05415E-25

0.3

0.300000000000000

0.300000000000000

5.76265E-24

0.4

0.400000000000000

0.400000000000000

4.73069E-24

0.5

0.500000000000000

0.500000000000000

1.00961E-24

0.6

0.600000000000000

0.600000000000000

3.10078E-24

0.7

0.700000000000000

0.700000000000000

5.49098E-24

0.8

0.800000000000000

0.800000000000000

1.00691E-24

0.9

0.900000000000000

0.900000000000000

4.89380E-24

1.0

1.000000000000000

1.000000000000000

2.57593E-23

To solve Eq. (5.383) we take 𝒌 = 𝟓𝟎. Eq. (5.384) is 50 𝑇 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐂 𝑅𝑘 (𝑥),

where 𝐂 = [𝛼0 , 𝛼1 , … , 𝛼50 ]𝑇 , and 𝑅𝑘 (𝑥) = [𝑅0 , 𝑅1 , … , 𝑅50 ]𝑇 . Putting Eq. (5.385) into Eq. (5.383), we obtained 𝑇 2 4 6 ∑50 𝑘=0 𝐂 𝑅𝑘 (𝑥) = −𝑥 − 𝑥 − 𝑥 +

941

(5.385)

𝑥

1

50 𝑇 2 2 𝑇 ∫0 (3𝑥𝑡 + 4𝑥 2 𝑡 2 ) ∑50 𝑘=0 𝐂 𝑅𝑘 (𝑡) d𝑡 + ∫0 (4𝑥𝑡 + 3𝑥 𝑡) ∑𝑘=0 𝐂 𝑅𝑘 (𝑡) d𝑡 . (5.386)


1

𝑇 2 4 6 ∫0 ∑50 𝑘=0 𝐂 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥)d𝑥 = ∫0 [−𝑥 − 𝑥 − 𝑥 ]𝑅𝑗 (𝑥)d𝑥 1

𝑥

𝑇 + ∫0 [∫0 (3𝑥𝑡 + 4𝑥 2 𝑡 2 ) ∑50 𝑘=0 𝐂 𝑅𝑘 (𝑡) d𝑡 ]𝑅𝑗 (𝑥)d𝑥 1

1

𝑇 + ∫0 [∫0 (4𝑥𝑡 2 + 3𝑥 2 𝑡) ∑50 𝑘=0 𝐂 𝑅𝑘 (𝑡) d𝑡 ] 𝑅𝑗 (𝑥)d𝑥 .

(5.387)


⋯ ⋱ ⋯

−0.67619047 1.34992 ∗ 1068 𝑎0 ⋮ ][ ] = [ ] ⋮ ⋮ 1.8315 ∗ 1068 5.60919 ∗ 10136 𝑎50

After solving the above we get 𝛼0 = −0.89268, 𝛼1 = 0.84476822, … , 𝛼50 = 8.054678 ∗ 10−93 . 𝑢(𝑥) = 3.457895 ∗ 10−55 + 1.0000𝑥 + ⋯ − 2.44975 ∗ 10−28 𝑥 50 .


942


Exact Solution


Error

0.0

0.000000000000000

-0.000000000000000

3.45790E-55

0.1

0.100000000000000

0.100000000000000

4.66714E-56

0.2

0.200000000000000

0.200000000000000

2.00131E-56

0.3

0.300000000000000

0.300000000000000

4.01783E-56

0.4

0.400000000000000

0.400000000000000

2.16600E-56

0.5

0.500000000000000

0.500000000000000

2.11161E-57

0.6

0.600000000000000

0.600000000000000

1.77657E-56

0.7

0.700000000000000

0.700000000000000

3.87840E-56

0.8

0.800000000000000

0.800000000000000

1.54924E-56

0.9

0.900000000000000

0.900000000000000

4.47895E-56

1.0

1.000000000000000

1.000000000000000

3.14957E-55

5.13.5. Abel’s Integral Equation Example 5.65. Consider the nonlinear Abel’s integral equation [229] as 343

11

𝑥

𝑥 3 = ∫0 440

1 3

√(𝑥−𝑡)

𝑢3 (𝑡)d𝑡.

(5.388)

The exact solution of Eq. (5.388) is 𝑢(𝑥) = 𝑥. Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation 𝑢3 (𝑡) = 𝑣(𝑡), 1

𝑢(𝑡) = 𝑣(𝑡)3 . Using the above transformation into the given integral equation we get 343

11

𝑥

𝑥 3 = ∫0 440

1 3

√(𝑥−𝑡)

𝑣(𝑡)d𝑡.

(5.389)

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 ∅𝑘 (𝑥).

(5.390)

To solve Eq. (5.388) we take 𝒌 = 𝟑. Eq. (5.390) is 𝑣(𝑥) = ∑3𝑘=0 𝛼𝑘 ∅𝑘 (𝑥) = ∑3𝑘=0 𝐂 𝑇 ∅𝑘 (𝑥), 943

(5.391)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 , 𝛼3 ]𝑇 , and ∅𝑘 (𝑥) = [∅0 , ∅1 , ∅2 , ∅3 ]𝑇 . Putting Eq. (5.391) into Eq. (5.388), we obtained 343

11

𝑥

𝑥 3 = ∫0 440

1 3

√(𝑥−𝑡)

∑3𝑘=0 𝐂 𝑇 ∅𝑘 (𝑡) d𝑡.

(5.392)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2,3. Then integrating over [0,1] we have Eq. (5.392) is 1 343

∫0

11

1

𝑥

𝑥 3 ∅𝑗 (𝑥)d𝑥 = ∫0 [∫0 440

1 3

√(𝑥−𝑡)

∑2𝑘=0 𝐂 𝑇 ∅𝑘 (𝑡) d𝑡] ∅𝑗 (𝑥)d𝑥 .

(5.393)


=

After solving the above, we get 𝛼0 = −

19 6

9

3

1

, 𝛼1 = 2 , 𝛼2 = − 2 , 𝛼3 = 6.

𝑣(𝑥) = 𝑥 3 . After applying the reverse transformation we get 𝑢(𝑥) = 𝑥. This is the exact solution.


3

𝑥

𝑥 4 (77 − 88𝑥 + 32𝑥 2 ) = ∫0 231

1 1 (𝑥−𝑡)4

𝑢2 (𝑡)d𝑡.

(5.394)

The exact solution of Eq. (5.394) is 𝑢(𝑥) = 1 − 𝑥. Consider the following transformation to convert given nonlinear Abel’s integral equation to linear integral equation 𝑢2 (𝑡) = 𝑣(𝑡), 𝑢(𝑡) = √𝑣(𝑡). Using the above transformation into the given integral equation we get

944

4

3

𝑥

𝑥 4 (77 − 88𝑥 + 32𝑥 2 ) = ∫0 231

1 1

𝑣(𝑡)d𝑡.

(5.395)

(𝑥−𝑡)4

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 ∅𝑘 (𝑥).

(5.396)

To solve Eq. (5.394) we take 𝒌 = 𝟐. Eq. (5.396) is 𝑣(𝑥) = ∑2𝑘=0 𝛼𝑘 ∅𝑘 (𝑥) = ∑2𝑘=0 𝐂 𝑇 ∅𝑘 (𝑥),

(5.397)

where 𝐂 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , and ∅𝑘 (𝑥) = [∅0 , ∅1 , ∅2 ]𝑇 . Putting Eq. (5.397) into Eq. (5.394), we obtained 4 231

3

𝑥

𝑥 4 (77 − 88𝑥 + 32𝑥 2 ) = ∫0

1 1 (𝑥−𝑡)4

∑2𝑘=0 𝐂 𝑇 ∅𝑘 (𝑡) d𝑡.

(5.398)

Multiplying both sides by 𝑅𝑗 (𝑥), 𝑗 = 0,1,2. Then integrating over [0,1] we have Eq. (5.398) is 1 4

∫0

3

1

𝑥

𝑥 4 (77 − 88𝑥 + 32𝑥 2 )∅𝑗 (𝑥)d𝑥 = ∫0 [∫0 231

1 1 (𝑥−𝑡)4

∑2𝑘=0 𝐂 𝑇 ∅𝑘 (𝑡) d𝑡 ] ∅𝑗 (𝑥)d𝑥. (5.399)


=

.

After solving the above, we get 9

1

𝛼0 = 4 , 𝛼1 = −4, 𝛼2 = 2. 𝑣(𝑥) = 1 − 2𝑥 + 𝑥 2 . After applying the reverse transformation we get 𝑢(𝑥) = (−1 + 𝑥). This is the exact solution.


3

𝑥

𝑢(𝑥) = 𝑥 − 3 𝑥 2 + ∫0

1 √(𝑥−𝑡)

𝑢(𝑡)d𝑡.

The exact solution of Eq. (5.400) is 𝑢(𝑥) = 𝑥. 945

(5.400)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥).

(5.401)

Consider 2nd order Rook Polynomials, i.e. for 𝑘 = 2, and we apply the proposed technique to solve Eq. (5.400) with 𝑘 = 2. We have Eq. (5.401) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥),

(5.402)

where 𝐶 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 and 𝑅(𝑥) = [𝑅0 , 𝑅1 , 𝑅2 ]𝑇 . Putting Eq. (5.402) into Eq. (5.400), we obtained 4

3

𝑥

∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥) = 𝑥 − 𝑥 2 + ∫0 3

1 √(𝑥−𝑡)

∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑡) d𝑡.

(5.403)


1

9

4

∫0 ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [𝑥 − 4 𝑥 3 ] 𝑅𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0

1 √(𝑥−𝑡)

∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑡) d𝑡] 𝑅𝑗 (𝑥)d𝑥,


=

After solving we get 𝛼0 = −0.94099, 𝛼1 = 0.93124, 𝛼2 = 0.013386. Consequently, we have the approximate solution is 𝑢(𝑥) = 0.0036400 + 0.98478𝑥 + 0.026772𝑥 2 .

946

(5.404)



Exact Solution


Error

0.0

0.000000000000000

0.003640000000000

3.64000E-03

0.1

0.100000000000000

0.102390000000000

2.39000E-03

0.2

0.200000000000000

0.201670000000000

1.67000E-03

0.3

0.300000000000000

0.301480000000000

1.48000E-03

0.4

0.400000000000000

0.401830000000000

1.83000E-03

0.5

0.500000000000000

0.502720000000000

2.72000E-03

0.6

0.600000000000000

0.604150000000000

4.15000E-03

0.7

0.700000000000000

0.706110000000000

6.11000E-03

0.8

0.800000000000000

0.808590000000000

8.59000E-03

0.9

0.900000000000000

0.911620000000000

1.16200E-02

1.0

1.000000000000000

1.015200000000000

1.52000E-02

Consider 25th order Rook Polynomials, i.e. for 𝒌 = 𝟐𝟓, and we apply the proposed technique to solve Eq. (5.400) with 𝑘 = 25. We have Eq. (5.401) is 25 𝑇 𝑢(𝑥) = ∑25 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐶 𝑅𝑘 (𝑥),

(5.402)

where 𝐶 = [𝛼0 , 𝛼1 , … , 𝛼25 ]𝑇 and 𝑅(𝑥) = [𝑅0 , 𝑅1 , … , 𝑅25 ]𝑇 . Putting Eq. (5.402) into Eq. (5.400), we obtained 4

3

𝑥

𝑇 ∑25 𝑘=0 𝐶 𝑅𝑘 (𝑥) = 𝑥 − 3 𝑥 2 + ∫0

1 √(𝑥−𝑡)

947

𝑇 ∑25 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡.

(5.403)


1

9

4

𝑇 ∫0 ∑25 𝑘=0 𝐶 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [𝑥 − 4 𝑥 3 ] 𝑅𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0

1 √(𝑥−𝑡)

𝑇 ∑25 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡 ] 𝑅𝑗 (𝑥)d𝑥 ,

(5.404)


𝑎0 −0.033333 ⋯ 8.5411 ∗ 1026 ⋮ ] [ ] = [ ] ⋮ ⋱ ⋮ 26 55 𝑎25 −3.9786 ∗ 10 ⋯ 1.280477 ∗ 10

After solving we get 𝛼0 = −0.8424376, 𝛼1 = 0.772196944, … , 𝛼25 = 3.488432 ∗ 10−40 . Consequently, we have the approximate solution is 𝑢(𝑥) = 3.4735087 ∗ 10−27 + 0.9999𝑥 + ⋯ + 5.41098 ∗ 10−15 𝑥 25 .


948


Exact Solution


Error

0.0

0.000000000000000

0.000000000000000

3.47351E-27

0.1

0.100000000000000

0.100000000000000

4.58848E-28

0.2

0.200000000000000

0.200000000000000

5.72808E-28

0.3

0.300000000000000

0.300000000000000

3.50104E-28

0.4

0.400000000000000

0.400000000000000

2.07170E-28

0.5

0.500000000000000

0.500000000000000

4.64497E-28

0.6

0.600000000000000

0.600000000000000

1.86665E-30

0.7

0.700000000000000

0.700000000000000

4.16711E-28

0.8

0.800000000000000

0.800000000000000

4.70549E-28

0.9

0.900000000000000

0.900000000000000

4.40397E-28

1.0

1.000000000000000

1.000000000000000

2.22023E-27

Consider 30th order Rook Polynomials, i.e. for 𝒌 = 𝟑𝟎, and we apply the proposed technique to solve Eq. (5.400) with 𝑘 = 30. We have Eq. (5.401) is 30 𝑇 𝑢(𝑥) = ∑30 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐶 𝑅𝑘 (𝑥),

(5.402)

where 𝐶 = [𝛼0 , 𝛼1 , … , 𝛼30 ]𝑇 and 𝑅(𝑥) = [𝑅0 , 𝑅1 , … , 𝑅30 ]𝑇 . Putting Eq. (5.402) into Eq. (5.400), we obtained 4

3

𝑥

𝑇 ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑥) = 𝑥 − 3 𝑥 2 + ∫0

1 √(𝑥−𝑡)

𝑇 ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡.

(5.403)


1

9

4

𝑇 ∫0 ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [𝑥 − 4 𝑥 3 ] 𝑅𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0

1 √(𝑥−𝑡)

𝑇 ∑30 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡 ] 𝑅𝑗 (𝑥)d𝑥 ,

(5.404)


−0.033333 ⋯ 3.123693 ∗ 1034 𝑎0 ][ ⋮ ] = [ ] ⋮ ⋱ ⋮ 34 70 𝑎 50 −1.4139 ∗ 10 ⋯ 1.96354 ∗ 10

After solving we get 𝛼0 = −1.00000, 𝛼1 = 1.000000150, … , 𝛼30 = −2.858377 ∗ 10−68 . 949

Consequently, we have the approximate solution is 𝑢(𝑥) = −7.079844 ∗ 10−53 + 1.0000𝑥 + ⋯ − 7.58192 ∗ 10−36 𝑥 30 .



Exact Solution


Error

0.0

0.000000000000000

-0.000000000000000

7.07984E-53

0.1

0.100000000000000

0.100000000000000

1.29920E-53

0.2

0.200000000000000

0.200000000000000

7.27562E-54

0.3

0.300000000000000

0.300000000000000

1.01766E-53

0.4

0.400000000000000

0.400000000000000

9.69633E-54

0.5

0.500000000000000

0.500000000000000

9.37847E-54

0.6

0.600000000000000

0.600000000000000

9.04687E-54

0.7

0.700000000000000

0.700000000000000

9.37577E-54

0.8

0.800000000000000

0.800000000000000

7.73297E-54

0.9

0.900000000000000

0.900000000000000

1.09609E-53

1.0

1.000000000000000

1.000000000000000

5.31214E-53

5.13.7. System of Fredholm Integral Equations Consider the system of Fredholm Integral equation [229] as 16

1

𝑢(𝑥) = 𝑥 + 𝑥 2 − 15 + ∫−1( 𝑢(𝑡) + 𝑣(t))d𝑡. 16

2

1

𝑣(𝑥) = 𝑥 3 + 𝑥 4 − 15 𝑥 + 3 + ∫−1((𝑥 − 𝑡)𝑢(𝑡) + (𝑥𝑡)𝑣(t))d𝑡. 950

(5.405a) (5.405b)

The exact solution of Eq. (5.405) is 𝑢(𝑥) = 𝑥 + 𝑥 2 , 𝑣(𝑥) = 𝑥 3 + 𝑥 4 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥).

(5.406a)

𝑢(𝑥) = ∑𝑛𝑘=0 𝛽𝑘 𝑅𝑘 (𝑥).

(5.406b)

Consider 2nd order Rook Polynomials, i.e. for 𝒌 = 𝟐, and we apply the proposed technique to solve Eq. (5.405) with 𝑘 = 2. We have Eq. (5.406) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥),

(5.407a)

𝑣(𝑥) = ∑2𝑘=0 𝑏𝑘 𝑅𝑘 (𝑥) = ∑2𝑘=0 𝐷𝑇 𝑅𝑘 (𝑥),

(5.407b)

where 𝐶 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , 𝐷 = [𝑏0 , 𝑏 1 , 𝑏2 ]𝑇 , 𝑅(𝑥) = [𝑅0 , 𝑅1 , 𝑅2 ]𝑇 . Putting Eq. (5.407) into Eq. (5.405), we obtained ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥) = 𝑥 + 𝑥 2 −

16 15

∑2𝑘=0 𝐷𝑇 𝑅𝑘 (𝑥) = 𝑥 3 + 𝑥 4 −

1

+ ∫−1(𝑢(𝑡) + 𝑣(t)) ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑡) d𝑡.

16

1

2

𝑥 + 3 + ∫−1 ( 15

(5.408a)

(𝑥 − 𝑡)𝑢(𝑡) 2 ) ∑𝑘=0 𝐷𝑇 𝑅𝑘 (𝑡) d𝑡.(5.408b) +(𝑥𝑡)𝑣(t)


1

16

∫0 ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [𝑥 + 𝑥 2 − 15] 𝑅𝑗 (𝑥)d𝑥 1

1

+ ∫0 [∫−1(𝑢(𝑡) + 𝑣(t)) ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑡) d𝑡] 𝑅𝑗 (𝑥)d𝑥 , 1

1

16

(5.409a)

2

∫0 ∑2𝑘=0 𝐷 𝑇 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [𝑥 3 + 𝑥 4 − 15 𝑥 + 3] 𝑅𝑗 (𝑥)d𝑥 1

1

+ ∫0 [∫−1((𝑥 − 𝑡)𝑢(𝑡) + (𝑥𝑡)𝑣(t)) ∑2𝑘=0 𝐷𝑇 𝑅𝑘 (𝑡) d𝑡] 𝑅𝑗 (𝑥)d𝑥 , (5.409b) for 𝑗 = 0,1,2. The matrix form of Eq. (5.409) is given as

=

After solving we get 𝛼0 = 0.4948, 𝛼1 = −0.9895, 𝛼2 = 0.4976, 𝛽0 = 0.5800 , 𝛽1 = −1.090 , 𝛽2 = 0.4255 . 951

Consequently, we have the approximate solution is 𝑢(𝑥) = 0.0029 + 1.00𝑥0.995𝑥 2 , 𝑣(𝑥) = −0.0845 + 0.612𝑥 + 0.8510𝑥 2 .


Table 5.85. Comparison of Exact and Approximate solutions of Eq. (5. 405) obtained from Rook polynomials Method (RPM) 𝑢(𝑥)𝑎𝑝𝑝𝑟𝑜𝑥 .


Error in 𝑢(𝑥)

Error in 𝑣(𝑥)

0.00000000

0.00290000

-0.08450000

2.9000E-03

8.4500E-02

0.1 0.11000000

0.00110000

0.11290000

-0.01479000

2.9000E-03

1.5890E-02

0.2 0.24000000

0.00960000

0.24270000

0.07194000

2.7000E-03

6.2340E-02

0.3 0.39000000

0.03510000

0.39250000

0.17570000

2.5000E-03

1.4060E-01

0.4 0.56000000

0.08960000

0.56210000

0.29650000

2.1000E-03

2.0690E-01

0.5 0.75000000

0.18750000

0.75170000

0.43430000

1.7000E-03

2.4680E-01

0.6 0.96000000

0.34560000

0.96120000

0.58910000

1.2000E-03

2.4350E-01

0.7 1.19000000

0.58310000

1.19000000

0.76090000

0.0000E+00

1.7780E-01

0.8 1.44000000

0.92160000

1.44000000

0.94970000

0.0000E+00

2.8100E-02

0.9 1.71000000

1.38500000

1.70900000

1.15600000

1.0000E-03

2.2900E-01

1.0 2.00000000

2.00000000

1.99800000

1.37800000

2.0000E-03

6.2200E-01

𝑢(𝑥)

𝑣(𝑥)

0.0 0.00000000

𝑥

Consider 15th order Rook Polynomials, i.e. for 𝑘 = 15, and we apply the proposed technique to solve Eq. (5.406). 15 𝑇 𝑢(𝑥) = ∑15 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐶 𝑅𝑘 (𝑥),

952

(5.407a)

15 𝑇 𝑣(𝑥) = ∑15 𝑘=0 𝑏𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐷 𝑅𝑘 (𝑥),

(5.407b)

where 𝐶 = [𝛼0 , 𝛼1 , … , 𝛼15 ]𝑇 , 𝐷 = [𝑏0 , 𝑏1 , … , 𝑏15 ]𝑇 , 𝑅(𝑥) = [𝑅0 , 𝑅1 , … , 𝑅15 ]𝑇 . Putting Eq. (5.407) into Eq. (5.405), we obtained 𝑇 2 ∑15 𝑘=0 𝐶 𝑅𝑘 (𝑥) = 𝑥 + 𝑥 −

16

1

𝑇 + ∫−1(𝑢(𝑡) + 𝑣(t)) ∑15 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡. 15

𝑇 3 4 ∑15 𝑘=0 𝐷 𝑅𝑘 (𝑥) = 𝑥 + 𝑥 −

16 15

1

2

𝑥 + 3 + ∫−1 (

(5.408a)

(𝑥 − 𝑡)𝑢(𝑡) 15 𝑇 ) ∑𝑘=0 𝐷 𝑅𝑘 (𝑡) d𝑡.(5.408b) +(𝑥𝑡)𝑣(t)


1

16

𝑇 2 ∫0 ∑15 𝑘=0 𝐶 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [𝑥 + 𝑥 − 15] 𝑅𝑗 (𝑥)d𝑥 1

1

𝑇 + ∫0 [∫−1(𝑢(𝑡) + 𝑣(t)) ∑15 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡 ] 𝑅𝑗 (𝑥)d𝑥 , 1

1

16

(5.409a)

2

𝑇 3 4 ∫0 ∑15 𝑘=0 𝐷 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [𝑥 + 𝑥 − 15 𝑥 + 3] 𝑅𝑗 (𝑥)d𝑥 1

1

𝑇 + ∫0 [∫−1((𝑥 − 𝑡)𝑢(𝑡) + (𝑥𝑡)𝑣(t)) ∑15 𝑘=0 𝐷 𝑅𝑘 (𝑡) d𝑡 ] 𝑅𝑗 (𝑥)d𝑥 ,

(5.409b)

Proceeding as before, we obtained the approximate solution of 𝑢(𝑥), 𝑣(𝑥) their error estimate and graphical representation as shown below


953


𝑢(𝑥)

𝑣(𝑥)



Error in 𝑢(𝑥) Error in 𝑣(𝑥)

0.0

0.00000000

0.00000000

0.00000000

0.00000000

1.5241E-26

2.2911E-26

0.1

0.11000000

0.00110000

0.11000000

0.00110000

3.1602E-25

4.7506E-25

0.2

0.24000000

0.00960000

0.24000000

0.00960000

2.0778E-26

3.1235E-26

0.3

0.39000000

0.03510000

0.39000000

0.03510000

3.2024E-25

4.8140E-25

0.4

0.56000000

0.08960000

0.56000000

0.08960000

1.8126E-26

2.7249E-26

0.5

0.75000000

0.18750000

0.75000000

0.18750000

3.2471E-25

4.8813E-25

0.6

0.96000000

0.34560000

0.96000000

0.34560000

1.7094E-25

2.5697E-25

0.7

1.19000000

0.58310000

1.19000000

0.58310000

1.9708E-25

2.9626E-25

0.8

1.44000000

0.92160000

1.44000000

0.92160000

3.8538E-25

5.7932E-25

0.9

1.71000000

1.38510000

1.71000000

1.38510000

4.5962E-25

6.9093E-25

1.0

2.00000000

2.00000000

2.00000000

2.00000000

1.4991E-24

2.2536E-24

Consider 50th order Rook Polynomials, i.e. for 𝑘 = 50, and we apply the proposed technique to solve Eq. (5.406). 50 𝑇 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐶 𝑅𝑘 (𝑥), 50 𝑇 𝑣(𝑥) = ∑50 𝑘=0 𝑏𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐷 𝑅𝑘 (𝑥),

(5.407a) (5.407b)

where 𝐶 = [𝛼0 , 𝛼1 , … , 𝛼50 ]𝑇 , 𝐷 = [𝑏0 , 𝑏1 , … , 𝑏50 ]𝑇 , 𝑅(𝑥) = [𝑅0 , 𝑅1 , … , 𝑅50 ]𝑇 . Proceeding as before, we obtained the approximate solution of 𝑢(𝑥), 𝑣(𝑥) their error estimate and graphical representation as shown below

954


Table 5.87. Comparison of Exact and Approximate solutions of Eq. (5. 405) obtained from Rook polynomials Method (RPM). 𝑥

𝑢(𝑥)

𝑣(𝑥)


𝑣(𝑥)𝑎𝑝𝑝𝑟𝑜𝑥 . Error in 𝑢(𝑥)

Error in 𝑣(𝑥)

0.0

0.00000000

0.00000000

0.00000000

0.00000000

8.8660E-62

2.1506E-63

0.1

0.11000000

0.00110000

0.11000000

0.00110000

8.2325E-62

1.9970E-62

0.2

0.24000000

0.00960000

0.24000000

0.00960000

3.5431E-62

8.5944E-63

0.3

0.39000000

0.03510000

0.39000000

0.03510000

5.5444E-62

1.3449E-62

0.4

0.56000000

0.08960000

0.56000000

0.08960000

8.3854E-62

2.0340E-62

0.5

0.75000000

0.18750000

0.75000000

0.18750000

5.5591E-62

1.3485E-62

0.6

0.96000000

0.34560000

0.96000000

0.34560000

3.1266E-62

7.5843E-63

0.7

1.19000000

0.58310000

1.19000000

0.58310000

5.4044E-62

1.3110E-62

0.8

1.44000000

0.92160000

1.44000000

0.92160000

9.7550E-62

2.3663E-62

0.9

1.71000000

1.38510000

1.71000000

1.38510000

1.0872E-61

2.6371E-62

1.0

2.00000000

2.00000000

2.00000000

2.00000000

6.8688E-61

1.6662E-61

5.13.8. System of Volterra integral equation Consider the system of Volterra Integral equation [229] as 𝑥

𝑢(𝑥) = 2 − 𝑒 −𝑥 + ∫0 ((𝑥 − 𝑡)𝑢(𝑡) + (𝑥 − 𝑡)𝑣(t))d𝑡. 𝑥

𝑣(𝑥) = 2𝑥 − 𝑒 𝑥 + 2𝑒 −𝑥 + ∫0 ((𝑥 − 𝑡)𝑢(𝑡) − (𝑥 − 𝑡)𝑣(t))d𝑡. The exact solution of Eq. (5.410) is 955

(5.410a) (5.410b)

𝑢(𝑥) = 𝑒 𝑥 . 𝑣(𝑥) = 𝑒 −𝑥 . According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥).

(5.411a)

𝑢(𝑥) = ∑𝑛𝑘=0 𝑏𝑘 𝑅𝑘 (𝑥).

(5.411b)

Consider 2nd order Rook Polynomials, i.e. for 𝑘 = 2, and we apply the proposed technique to solve Eq. (5.410) with 𝑘 = 2. We have Eq. (5.411) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥),

(5.412a)

𝑣(𝑥) = ∑2𝑘=0 𝑏𝑘 𝑅𝑘 (𝑥) = ∑2𝑘=0 𝐷𝑇 𝑅𝑘 (𝑥),

(5.412b)

where 𝐶 = [𝛼0 , 𝛼1 , 𝛼2 ]𝑇 , 𝐷 = [𝑏0 , 𝑏 1 , 𝑏2 ]𝑇 , 𝑅(𝑥) = [𝑅0 , 𝑅1 , 𝑅2 ]𝑇 . Putting Eq. (5.412) into Eq. (5.410), we obtained ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥) = 2 − 𝑒 −𝑥 𝑥

+ ∫0 ((𝑥 − 𝑡)𝑢(𝑡) + (𝑥 − 𝑡)𝑣(t)) ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑡) d𝑡.

(5.413a)

∑2𝑘=0 𝐷𝑇 𝑅𝑘 (𝑥) = 2𝑥 − 𝑒 𝑥 + 2𝑒 −𝑥 𝑥

+ ∫0 ((𝑥 − 𝑡)𝑢(𝑡) − (𝑥 − 𝑡)𝑣(t)) ∑2𝑘=0 𝐷𝑇 𝑅𝑘 (𝑡) d𝑡.

(5.413b)


1

∫0 ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [2 − 𝑒 −𝑥 ] 𝑅𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 ((𝑥 − 𝑡)𝑢(𝑡) + (𝑥 − 𝑡)𝑣(t)) ∑2𝑘=0 𝐶 𝑇 𝑅𝑘 (𝑡) d𝑡 ]𝑅𝑗 (𝑥)d𝑥, (5.414a) 1

1

∫0 ∑2𝑘=0 𝐷 𝑇 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [2𝑥 − 𝑒 𝑥 + 2𝑒 −𝑥 ] 𝑅𝑗 (𝑥)d𝑥 1

𝑥

+ ∫0 [∫0 ((𝑥 − 𝑡)𝑢(𝑡) − (𝑥 − 𝑡)𝑣(t)) ∑2𝑘=0 𝐷𝑇 𝑅𝑘 (𝑡) d𝑡]𝑅𝑗 (𝑥)d𝑥, for 𝑗 = 0,1,2. The matrix form of Eq. (5.414) is given as

=

After solving we get the values of 𝛼 and 𝛽. 𝛼0 = −0.359, 𝛼1 = 1.18, 𝛼2 = 0.0820. 956

.

(5.414b)

𝛽0 = 0.517, 𝛽1 = 0.527, 𝛽2 = −0.184. Consequently, we have the approximate solution is 𝑢(𝑥) = 0.901 + 1.15𝑥 + 0.164𝑥 2 , 𝑣(𝑥) = 0.860 − 0.209𝑥 − 0.368𝑥 2 . Graphical Representation


Table 5.88. Comparison of Exact and Approximate solutions of Eq. (5.410) obtained from Rook polynomials Method (RPM) 𝑥

𝑢(𝑥)

𝑣(𝑥)

𝑢(𝑥)𝑎𝑝𝑝𝑟 .

0.0

1.00000000

1.0000000000 0.9010000

0.1

1.11000000

0.2

𝑣(𝑥)𝑎𝑝𝑝𝑟 .

Error in 𝑢(𝑥)

Error in 𝑣(𝑥)

0.86000000

9.9000E-02

1.4000E-01

0.9050000000 1.0500000

0.83500000

6.0000E-02

7.0000E-02

1.22000000

0.81900000

1.210000

0.80300000

1.0000E-02

1.6000E-02

0.3

1.35000000

0.74100000

1.360000

0.76400000

1.0000E-02

2.3000E-02

0.4

1.49000000

0.67000000

1.530000

0.71700000

4.0000E-02

4.7000E-02

0.5

1.65000000

0.60700000

1.700000

0.66400000

5.0000E-02

5.7000E-02

0.6

1.82000000

0.54900000

1.870000

0.60300000

5.0000E-02

5.4000E-02

0.7

2.01000000

0.49700000

2.040000

0.53400000

3.0000E-02

3.7000E-02

0.8

2.23000000

0.44900000

2.220000

0.45700000

1.0000E-02

8.0000E-03

0.9

2.46000000

0.40700000

2.390000

0.37400000

7.0000E-02

3.3000E-02

1.0

2.72000000

0.36800000

2.570000

0.28300000

1.5000E-01

8.5000E-02

Consider 20th order Rook Polynomials, i.e. for 𝑘 = 20, and we apply the proposed technique to solve Eq. (5.410) with 𝑘 = 20. We have Eq. (5.411) is 957

20 𝑇 𝑢(𝑥) = ∑20 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐶 𝑅𝑘 (𝑥),

(5.412a)

20 𝑇 𝑣(𝑥) = ∑20 𝑘=0 𝑏𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐷 𝑅𝑘 (𝑥),

(5.412b)

where 𝐶 = [𝛼0 , 𝛼1 , … , 𝛼20 ]𝑇 , 𝐷 = [𝑏0 , 𝑏 1 , … , 𝑏20 ]𝑇 , 𝑅(𝑥) = [𝑅0 , 𝑅1 , … , 𝑅20 ]𝑇 . Putting Eq. (5.412) into Eq. (5.410), we obtained 𝑇 −𝑥 ∑20 𝑘=0 𝐶 𝑅𝑘 (𝑥) = 2 − 𝑒 𝑥

𝑇 + ∫0 ((𝑥 − 𝑡)𝑢(𝑡) + (𝑥 − 𝑡)𝑣(t)) ∑20 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡.

(5.413a)

𝑇 𝑥 −𝑥 ∑20 𝑘=0 𝐷 𝑅𝑘 (𝑥) = 2𝑥 − 𝑒 + 2𝑒 𝑥

𝑇 + ∫0 ((𝑥 − 𝑡)𝑢(𝑡) − (𝑥 − 𝑡)𝑣(t)) ∑20 𝑘=0 𝐷 𝑅𝑘 (𝑡) d𝑡.

(5.413b)


1

𝑇 −𝑥 ∫0 ∑20 𝑘=0 𝐶 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [2 − 𝑒 ] 𝑅𝑗 (𝑥)d𝑥 1

𝑥

𝑇 + ∫0 [∫0 ((𝑥 − 𝑡)𝑢(𝑡) + (𝑥 − 𝑡)𝑣(t)) ∑20 𝑘=0 𝐶 𝑅𝑘 (𝑡) d𝑡 ]𝑅𝑗 (𝑥)d𝑥 , 1

(5.414a)

1

𝑇 𝑥 −𝑥 ∫0 ∑20 𝑘=0 𝐷 𝑅𝑘 (𝑥)𝑅𝑗 (𝑥) d𝑥 = ∫0 [2𝑥 − 𝑒 + 2𝑒 ] 𝑅𝑗 (𝑥)d𝑥 1

𝑥

𝑇 + ∫0 [∫0 ((𝑥 − 𝑡)𝑢(𝑡) − (𝑥 − 𝑡)𝑣(t)) ∑20 𝑘=0 𝐷 𝑅𝑘 (𝑡) d𝑡 ]𝑅𝑗 (𝑥)d𝑥 ,

After solving we get the values of 𝛼 and 𝛽. 𝛼0 = 0.01824579, 𝛼1 = 1.0458443, … , 𝛼20 = 1.6435 ∗ 10−25 . 𝛽0 = 4.73743, 𝛽1 = −4.8622564, … , 𝛽20 = −4.77131 ∗ 10−24 . Consequently, we have the approximate solution is 𝑢(𝑥) = 1.000 + 0.9999𝑥 + ⋯ + 3.9984 ∗ 10−7 𝑥 20 , 𝑣(𝑥) = 0.99999 − 0.999999𝑥 + ⋯ − 0.000011608𝑥 2 .

958

(5.414b)



𝑢(𝑥)

𝑣(𝑥)



Error in 𝑢(𝑥)

Error in 𝑣(𝑥)

0.0

1.00000000

1.00000000

1.00000000

1.0000000000 1.8606E-15

5.5916E-14

0.1

1.10517091

0.90483741

1.10517091

0.9048374180 1.6348E-16

1.8750E-15

0.2

1.22140275

0.81873075

1.22140275

0.8187307531 3.2553E-16

1.0217E-14

0.3

1.34985880

0.74081822

1.34985880

0.7408182207 1.4480E-16

1.3412E-15

0.4

1.49182469

0.67032004

1.49182469

0.6703200460 3.3279E-17

5.0362E-15

0.5

1.64872127

0.60653065

1.64872127

0.6065306597 1.3857E-16

7.9742E-15

0.6

1.82211880

0.54881163

1.82211880

0.5488116361 1.8730E-16

8.2614E-15

0.7

2.01375270

0.49658530

2.01375270

0.4965853038 1.5648E-16

4.6605E-15

0.8

2.22554092

0.44932896

2.22554092

0.4493289641 2.9056E-17

5.1165E-15

0.9

2.45960311

0.40656965

2.45960311

0.4065696597 1.0695E-16

3.2240E-15

1.0

2.71828182

0.36787944

2.71828182

0.3678794412 5.9854E-16

4.0693E-14

Consider 50th order Rook Polynomials, i.e. for 𝒌 = 𝟓𝟎, and we apply the proposed technique to solve Eq. (5.410) with 𝑘 = 50. We have Eq. (5.411) is 50 𝑇 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐶 𝑅𝑘 (𝑥),

(5.412a)

50 𝑇 𝑣(𝑥) = ∑50 𝑘=0 𝑏𝑘 𝑅𝑘 (𝑥) = ∑𝑘=0 𝐷 𝑅𝑘 (𝑥).

(5.412b)

Where 𝐶 = [𝛼0 , 𝛼1 , … , 𝛼50 ]𝑇 , 𝐷 = [𝑏0 , 𝑏 1 , … , 𝑏50 ]𝑇 , 𝑅(𝑥) = [𝑅0 , 𝑅1 , … , 𝑅50 ]𝑇 . Proceeding as before, we obtained the approximate solution of 𝑢(𝑥), 𝑣(𝑥) their error estimate and graphical representation as shown below 959

Graphical representation



𝑢(𝑥)

0.0

1.00000000

0.1

𝑣(𝑥)



Error in 𝑢(𝑥)

Error in 𝑣(𝑥)

1.00000000

1.00000000

1.00000000

3.2060E-53

8.4653E-54

1.10517091

0.90483741

1.10517091

0.90483741

4.3197E-54

1.1403E-54

0.2

1.22140275

0.81873075

1.22140275

0.81873075

1.8548E-54

4.8999E-55

0.3

1.34985880

0.74081822

1.34985880

0.74081822

3.7098E-54

9.7918E-55

0.4

1.49182469

0.67032004

1.49182469

0.67032004

1.9942E-54

5.2664E-55

0.5

1.64872127

0.60653065

1.64872127

0.60653065

1.8298E-55

4.8642E-56

0.6

1.82211880

0.54881163

1.82211880

0.54881163

1.6565E-54

4.3683E-55

0.7

2.01375270

0.49658530

2.01375270

0.49658530

3.5823E-54

9.4522E-55

0.8

2.22554092

0.44932896

2.22554092

0.44932896

1.4826E-54

3.9097E-55

0.9

2.45960311

0.40656965

2.45960311

0.40656965

4.1606E-54

1.0978E-54

1.0

2.7182812

0.36787944

2.71828182

0.36787944

2.9751E-53

7.8494E-54

5.14. Fibonicca Polynomials Method (FPM) We develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Fibonicca polynomials [28]. Fibonicca polynomials is given as 𝐹𝑚 (𝑥) = 𝑥(𝑇𝐼𝑛−1 + 𝑇𝐼𝑛−2 ).

(5.415) 960

𝐹0 = 1, 𝐹1 = 1, 𝐹2 = 𝑥, 𝐹3 = 𝑥(𝑥 + 1), 𝐹4 = 𝑥 2 (𝑥 + 2), . . ..

5.14.1 Methodology 5.14.1.1 Integral Equation of the 1st Kind Consider the integral equation of the 1st kind is given as 𝛽(𝑥)

𝑓(𝑥) = 𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡)𝑢(𝑡)d𝑡 , 𝑎 ≤ 𝑥 ≤ 𝑏,

(5.416)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, [Lewis, 11], to find an approximate solution 𝑢̃(𝑥) of Eq. (5.416). For this, we assume that 𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝐴𝑘 (𝑥),

(5.417)

where 𝐴𝑘 (𝑥) are Fibonicca polynomials of degree 𝑘 defined in equation (5.415) and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.417) into (5.416), we get 𝑥

𝑓(𝑥) = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝐴𝑘 (𝑡)d𝑡.

(5.418)

Then the Galerkin equations are obtained by multiplying both sides of (5.418) by 𝐴𝑗 (𝑥) and then integrating with respect to 𝑥 from 𝑎 to 𝑏, we have 𝑏

𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝐴𝑗 (𝑥)d𝑥 = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐴𝑘 (𝑡)d𝑡 )𝐴𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛.

(5.418)

Since in each equation, there are two integrals. The inner integrand of the left side is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏

961

yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.418) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by 𝜆 ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.419)

Where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐴𝑘 (𝑡)d𝑡)𝐴𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏

𝐹𝑗 = ∫𝑎 𝑓(𝑥) 𝐴𝑗 (𝑥)d𝑥,

𝑗 = 0,1,2, … 𝑛.




(5.420)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) and 𝑢(𝑥) being the known function and 𝜆 is a constant. Proceeding as before ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ;

𝑘, 𝑗 = 0,1,2, … 𝑛

(5.421)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝑇𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝑇𝑘 (𝑡)d𝑡)𝑇𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏


𝑗 = 0,1,2, … 𝑛.


5.14.2. Volterra Integral Equation Consider the Volterra Integral equation [229] as 1

1

𝑥

𝑢(𝑥) = 1 − 2 𝑥 2 + 6 ∫0 (𝑥 − 𝑡)3 𝑢(𝑡)d𝑡.

(5.422)

The exact solution of Eq. (5.422) is 𝑢(𝑥) = cos(𝑥).

(5.423) 962

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐹𝑘 (𝑥).

(5.424)

Consider 25th order Fibonicca Polynomials, i.e. for 𝑛 = 25, we have Eq. (5.424) is 𝑢(𝑥) = ∑25 𝑘=0 𝛼𝑘 𝐹𝑘 (𝑥) = 𝑎0 𝐹0 (𝑥) + 𝑎1 𝐹1 (𝑥) + ⋯ + 𝑎25 𝐹0 (𝑥), 𝑢(𝑥) = 𝑎1 + 𝑥𝑎2 + 𝑥(𝑥 + 1)𝑎3 + ⋯ + 𝑥12 (𝑥12 + 23𝑥11 + 231𝑥10 + 1330𝑥 9 + 4845𝑥 8 + 11628𝑥 7 + 18564𝑥 6 + 19448𝑥 5 + 12870𝑥 4 + 5005𝑥 3 + 1001𝑥 2 + 78𝑥 + 1)𝑎25 .

(5.425)

Substituting Eq. (5.425) into Eq. (5.422) 𝑢(𝑥) = 𝑎1 + 𝑥𝑎2 + 𝑥(𝑥 + 1)𝑎3 + ⋯ + 𝑥12 (𝑥12 + 23𝑥11 + 231𝑥10 + 1330𝑥 9 + 4845𝑥 8 + 11628𝑥 7 + 18564𝑥 6 + 19448𝑥 5 + 12870𝑥 4 + 5005𝑥 3 + 1001𝑥 2 + 1

1

𝑥

78𝑥 + 1)𝑎25 = 1 − 2 𝑥 2 + 6 ∫0 (𝑥 − 𝑡)3 [𝑎1 + 𝑡𝑎2 + 𝑡(𝑡 + 1)𝑎3 + ⋯ 𝑡12 (𝑡12 + 23𝑡11 + 231𝑡10 + 1330𝑡 9 + 4845𝑡 8 + 11628𝑡 7 + 18564𝑡 6 + 19448𝑡 5 + 12870𝑡 4 + 5005𝑡 3 + 1001𝑡 2 + 78𝑡 + 1)𝑎25 ]d𝑡.

(5.426)

Now multiply Eq. (5.426) by 𝐹𝑗 (𝑥), 𝑗 = 0,1. . .25 and integrating both side from -1 to 1, we have following system of equations, 1.98333 0 0.65476

[3943.8455

0 0.66428 0.664285

3844.5826

0.66587 0.66428 . . . 1.06367 . . . 7516.5354

𝑎1 1.6667 𝑎2 0 𝑎3 . = 0.4667 . . . . . . 1.596219108 ] [𝑎25 ] [2243.19]

4083.4162 1598.0142 7551.7528

After solving we get, 𝑎1 = 1.000033958, 𝑎2 = −1.295553300, 𝑎3 = 1.295556509, … 𝑎25 = 0.00005495847733 Putting the value back we get the approximate solution 𝑢(𝑥) 1.000033958 + 0.00000320900000𝑥 − 0.503268874𝑥 2 + … 0.00005295847733𝑥 24 .

(5.427)

963

Table 5.91. Comparison of Exact and Approximate solutions of Eq. (5.422) obtained from Fibonicca polynomials Method (FPM) __________________________________________________________________ 𝑥

Exact Solution


Error Analysis

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

5.15494e-35

0.1

0.995004165300000

0.995004165000000

3.00000e-10

0.2

0.980066577800000

0.980066578000000

2.00000e-10

0.3

0.955336489100000

0.955336489000000

1.00000e-10

0.4

0.921060994000000

0.921060994000000

0.00000e+00

0.5

0.877582561900000

0.877582562000000

1.00000e-10

0.6

0.825335614900000

0.825335615000000

1.00000e-10

0.7

0.764842187300000

0.764842187000000

3.00000e-10

0.8

0.696706709300000

0.696706709000000

3.00000e-10

0.9

0.621609968300000

0.621609969000000

7.00000e-10

1.0

0.540302305900000

0.540302306000000

1.00000e-10

__________________________________________________________________

Fig 5.91. Comparison of Exact and Approximate solutions of Eq. ( 5.422)

Consider 50th order Fibonicca Polynomials, i.e. for 𝑛 = 50, we have Eq. (5.425) is 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝐹𝑘 (𝑥) = 𝑎0 𝐹0 (𝑥) + 𝑎1 𝐹1 (𝑥) + ⋯ 𝑎50 𝐹50 (𝑥)

964

𝑢(𝑥) = 𝑎1 + 𝑥𝑎2 + 𝑥(𝑥 + 1)𝑎3 + ⋯ 𝑥 25 (25 + 2600𝑥 + 80730𝑥 2 + ⋯ 48𝑥 23 + 𝑥 24 )𝑎50

(5.428)

Substituting Eq. (5.425) into Eq. (5.422) 𝑢(𝑥) = 𝑎1 + 𝑥𝑎2 + 𝑥(𝑥 + 1)𝑎3 + ⋯ 𝑥 25 (25 + 2600𝑥 + 80730𝑥 2 + ⋯ 48𝑥 23 + 1

𝑥

1

𝑥 24 )𝑎50 = 1 − 2 𝑥 2 + 6 ∫0 (𝑥 − 𝑡)3 [𝑎1 + 𝑡𝑎2 + 𝑡(𝑡 + 1)𝑎3 + ⋯ 𝑥 25 (25 + 2600𝑡 + 80730𝑡 2 + ⋯ 48𝑡 23 + 𝑡 24 )𝑎50 ]d𝑡.

(5.429)

Now multiply Eq. (5.429) by 𝐹𝑗 (𝑥), 𝑗 = 0,1 … 50 and integrating both side from -1 to 1, we have the following system of equations, 1.98333 0 0.65476

[3.324 × 108

0 0.66428 0.664285

0.66587 0.66428 . . . 1.06367.

3.343 × 108

. . 6.598 × 108

3.45206134 3.35961949 × 108 12.17462957

2.19 × 1018

𝑎1 1.6667 𝑎2 0 𝑎3 0.4667 . = . . . . . . ] [𝑎10 ] [1.81605]

After solving we get, 𝑎1 = 1.00000131081, 𝑎2 = −0.61836284995, 𝑎3 = 0.6181003491, … 𝑎50 = −5.268867421 × 10−10 Putting the values back we get the approximate solution as, 𝑢(𝑥) = 1.00001281 − 0.0002624958𝑥 − 0.5011059247𝑥 2 + ⋯ −5.268867421 × 10−10 𝑥 49 .

965


Exact Solution


Error Analysis

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

4.10003e-80

0.1

0.995004165300000

0.995004165000000

3.00000e-10

0.2

0.980066577800000

0.980066578000000

2.00000e-10

0.3

0.955336489100000

0.955336489000000

1.00000e-10

0.4

0.921060994000000

0.921060994000000

0.00000e+00

0.5

0.877582561900000

0.877582562000000

1.00000e-10

0.6

0.825335614900000

0.825335615000000

1.00000e-10

0.7

0.764842187300000

0.764842187000000

3.00000e-10

0.8

0.696706709300000

0.696706709000000

3.00000e-10

0.9

0.621609968300000

0.621609968000000

3.00000e-10

1.0

0.540302305900000

0.540302306000000

1.00000e-10

_________________________________________________________________


Consider 100th order Fibonicca Polynomials, i.e. for 𝑛 = 100, we have Eq. (5.424) is 𝑢(𝑥) = ∑100 𝑘=0 𝛼𝑘 𝐹𝑘 (𝑥) = 𝑎0 𝐹0 (𝑥) + 𝑎1 𝐹1 (𝑥) + ⋯ … 𝑎100 𝐹100 (𝑥) 966

𝑢(𝑥) = 𝑎1 + 𝑥𝑎2 + 𝑥(𝑥 + 1)𝑎3 + ⋯ 𝑥 50 (50 + 20825𝑥 + 2598960𝑥 2 + ⋯ 98𝑥 48 + 𝑥 49 )𝑎100

(5.430)

Substituting Eq. (5.430) into Eq. (5.422) 𝑢(𝑥) = 𝑎1 + 𝑥𝑎2 + 𝑥(𝑥 + 1)𝑎3 + ⋯ 𝑥 50 (50 + 20825𝑥 + 2598960𝑥 2 + 1

1

𝑥

⋯ 98𝑥 48 + 𝑥 49 )𝑎100 = 1 − 2 𝑥 2 + 6 ∫0 (𝑥 − 𝑡)3 [𝑎1 + 𝑡𝑎2 + 𝑡(𝑡 + 1)𝑎3 + ⋯ 𝑥 50 (50 + 20825𝑡 + 2598960𝑡 2 + ⋯ 98𝑡 48 + 𝑡 49 )𝑎100 ]d𝑡.

(5.431)

Now multiply Eq. (5.431) by 𝐹𝑗 (𝑥), 𝑗 = 0,1 … 100 and integrating both side from -1 to 1, we have the following system of equations, 1.98333 0 0.65476

0 0.66428 0.664285

0.66587 0.66428 . . . 1.06367 . . .

[4.684 × 1018

4.772 × 1018

9.505 × 1018

𝑎1 1.667 𝑎2 0 𝑎3 0.467 . = . . . . . . 8.690 × 1018 ] [𝑎100 ] [−2.5 × 1038 ]

4.876 × 1018 4.810 × 1018 12.17462957

After solving we get, 𝑎1 = 1.0000010815, 𝑎2 = 0.6224979323, 𝑎3 = −0.6223555723, … 𝑎100 = 4.619397432 × 10−21 Putting the values back we get the approximate solution as, 𝑢(𝑥) = 1.000010815 + 0.0001423600𝑥 − 0.5010424578𝑥 2 + … + 4.61939748310−21 𝑥 99 .

967


Exact Solution


Error Analysis

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.19099e-188

0.1

0.995004165278026

0.995004165278026

7.00000e-100

0.2

0.980066577841242

0.980066577841242

3.00000e-100

0.3

0.955336489125606

0.955336489125606

3.00000e-100

0.4

0.921060994002885

0.921060994002885

1.60000e-99

0.5

0.877582561890373

0.877582561890373

1.00000e-100

0.6

0.825335614909678

0.825335614909678

1.40000e-99

0.7

0.764842187284488

0.764842187284488

2.20000e-99

0.8

0.696706709347165

0.696706709347165

3.00000e-100

0.9

0.621609968270664

0.621609968270664

5.00000e-100

1.0

0.540302305868140

0.540302305868140

3.00000e-100

__________________________________________________________________


5.14.3. Weakly Singular Integral Equation Consider the Weakly singular Integral equation [229] as

968

𝑥

𝑢(𝑥) = 3 − 6√𝑥 + ∫0

1 √𝑥−𝑡

𝑢(𝑡)d𝑡.

(5.432)

The exact solution of Eq. (5.432) is 𝑢(𝑥) = 3.

(5.433)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐹𝑘 (𝑥)

(5.434)

Consider 1st order Fibonicca Polynomials, i.e. for 𝑛 = 1, we have Eq. (24) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐹𝑘 (𝑥) = 𝑎0 𝐹0 (𝑥) + 𝑎1 𝐹1 (𝑥). or

𝑢(𝑥) = 𝑎1 .

(5.435)

Substituting Eq. (5.435) into Eq. (5.432) 𝑥

𝑢(𝑥) = 𝑎1 = 3 − 6√𝑥 + ∫0

1 √𝑥−𝑡

𝑢(𝑡)d𝑡.

(5.436)

Now multiply Eq. (5.436) by 𝐹𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from 0 to 1, we have 1

1

1

𝑥

∫0 [𝑎1 ]𝐹𝑗 (𝑥)d𝑥 = ∫0 (3 − 6√𝑥)𝐹𝑗 (𝑥)d𝑥 + ∫0 [∫0

1 √𝑥−𝑡

[𝑎1 ]d𝑡] 𝐹𝑗 (𝑥)d𝑥, 𝑗 = 0,1 (5.437)

Now by solving Eq. (5.437) we get, 1 3

𝑎1 = 1.00

After solving we get 𝛼1 = 3. consequently we have the exact solution is 𝑢(𝑥) = 3.

5.14.4. Fredholm Integral Integral Equation Consider the Fredholm Integral equation [229] as 1

𝑢(𝑥) = 3𝑥 2 − 5𝑥 3 + ∫−1 𝑥 4 𝑡𝑢(𝑡)d𝑡.

(5.438)

The exact solution of Eq. (5.438) is 𝑢(𝑥) = 3𝑥 2 − 5𝑥 3 − 2𝑥 4 .

(5.439)

Consider 2nd order Fibonicca Polynomials, i.e. for 𝑛 = 2, we have Eq. (29) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝐹𝑘 (𝑥) = 𝑎0 𝐹0 (𝑥) + 𝑎1 𝐹1 (𝑥) + 𝑎2 𝐹2 (𝑥). Or

𝑢(𝑥) = 𝑎1 + 𝑎2 𝑥.

(5.440)


𝑢(𝑥) = 𝑎1 + 𝑎2 (𝑥) = 3𝑥 2 − 5𝑥 3 − 2𝑥 4 + ∫0 𝑥 4 𝑡[𝑎1 + 𝑎2 (𝑡)]d𝑡. (5.441) Now multiply Eq. (31) by 𝐹𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from -1 to 1, we have

969

1

1

1

1

∫−1[𝑎1 + 𝑎2 (𝑥)]𝐹𝑗 (𝑥)d𝑥 = ∫−1(3𝑥 2 − 5𝑥 3 − 2𝑥 4 )𝐹𝑗 (𝑥)d𝑥 + ∫−1 [∫−1 𝑥 4 𝑡[𝑎1 + 𝑎2 (𝑥)]d𝑡] 𝐹𝑗 (𝑥)d𝑥, 𝑗 = 0,1

(5.442)

the matrix form of Eq. (5.442) is [

−2 0

4 𝛼1 15 ] [ 2 𝛼2 ]

−3

=[

−2 ], 2 3

After solving we get 𝛼1 = 5 and 𝛼2 = −3.

consequently we have the approximate

solution is 3

𝑢(𝑥) = 5 − 3𝑥. Table 5.94. Comparison of Exact and Approximate solutions of Eq. (5.438) obtained from Fibonicca polynomials Method (FPM) __________________________________________________________________ 𝑥

Exact Solution


Error Analysis

__________________________________________________________________ 0.0

0.000000000000000

0.600000000000000

6.00000e-01

0.1

0.024800000000000

0.300000000000000

2.75200e-01

0.2

0.076800000000000

0.000000000000000

7.68000e-02

0.3

0.118800000000000

-0.300000000000000

4.18800e-01

0.4

0.108800000000000

-0.600000000000000

7.08800e-01

0.5

0.000000000000000

-0.900000000000000

9.00000e-01

0.6

-0.259200000000000

-1.200000000000000

9.40800e-01

0.7

-0.725200000000000

-1.500000000000000

7.74800e-01

0.8

-1.459200000000000

-1.800000000000000

3.40800e-01

0.9

-2.527200000000000

-2.100000000000000

4.27200e-01

1.0

-4.000000000000000

-2.400000000000000

1.60000e+00

__________________________________________________________________

970


Consider 4th order Fibonicca Polynomials, i.e. for 𝑛 = 4, we have Eq. (5.440) is 𝑢(𝑥) = ∑4𝑘=0 𝛼𝑘 𝐹𝑘 (𝑥) = 𝑎0 𝐹0 (𝑥) + 𝑎1 𝐹1 (𝑥) + 𝑎2 𝐹2 (𝑥) + 𝑎3 𝐹3 (𝑥) + 𝑎4 𝐹4 (𝑥). or

𝑢(𝑥) = 𝑎1 + 𝑎2 𝑥 + 𝑥(1 + 𝑥)𝑎3 + 𝑥 2 (𝑥 + 2)𝑎4 .

(5.443)

Substituting Eq. (5.443) into Eq. (5.438) 1

𝑢(𝑥) = 𝑎1 + 𝑎2 𝑥 + 𝑥(1 + 𝑥)𝑎3 + 𝑥 2 (𝑥 + 2)𝑎4 = 3𝑥 2 − 5𝑥 3 + ∫−1 𝑥 4 𝑡[𝑎1 + 𝑎2 𝑡 + 𝑡(1 + 𝑡)𝑎3 + 𝑡 2 (𝑡 + 2)𝑎4 ]d𝑡.

(5.444)

Now multiply Eq. (5.444) by 𝐹𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from -1 to 1, we have 1

1

∫−1[𝑎1 + 𝑎2 𝑥 + 𝑥(1 + 𝑥)𝑎3 + 𝑥 2 (𝑥 + 2)𝑎4 ]𝐹𝑗 (𝑥)d𝑥 = ∫−1(3𝑥 2 − 1

1

5𝑥 3 )𝐹𝑗 (𝑥)d𝑥 + ∫−1 [∫−1

𝑥 4 𝑡[𝑎1 + 𝑎2 𝑡 + 𝑡(1 + 𝑡)𝑎3 ] 𝐹𝑗 (𝑥)d𝑥, 𝑗 = 0,1. .4 (5.445) +𝑡 2 (𝑡 + 2)𝑎4 d𝑡

the matrix form of Eq. (5.445) is 𝛼1 −2.0000 0.2667 − 0.4000 − 1.1733 −2.00 𝛼 0.0000 − 0.6667 − 0.667 − 0.400 2 2.000 [ ][ ] = [ ], −0.6667 − 0.4761 − 0.87619 − 1.08571 𝛼3 0.800 −1.3334 − 0.01904 − 0.81904 − 1.65714 𝛼4 −0.971 6

After solving we get 𝛼1 = 35 ,𝛼2 = −

79 7

, 𝑎3 =

the approximate solution is 6

9

𝑢(𝑥) = 35 + 7 𝑥 2 − 5𝑥 3

971

79 7

𝑎𝑛𝑑 𝑎4 = −5. consequently we have


Exact Solution


Error Analysis

__________________________________________________________________ 0.0

0.000000000000000

0.171428571400000

1.71429e-01

0.1

0.024800000000000

0.179285714300000

1.54486e-01

0.2

0.076800000000000

0.182857142800000

1.06057e-01

0.3

0.118800000000000

0.152142857100000

3.33429e-02

0.4

0.108800000000000

0.057142857100000

5.16571e-02

0.5

0.000000000000000

-0.132142857200000

1.32143e-01

0.6

-0.259200000000000

-0.445714285700000

1.86514e-01

0.7

-0.725200000000000

-0.913571428600000

1.88371e-01

0.8

-1.459200000000000

-1.565714286000000

1.06514e-01

0.9

-2.527200000000000

-2.432142858000000

9.50571e-02

1.0

-4.000000000000000

-3.542857143000000

4.57143e-01

__________________________________________________________________


Consider 5th order Fibonicca Polynomials, i.e. for 𝑛 = 5, we have Eq. (5.440) is 𝑢(𝑥) = ∑5𝑘=0 𝛼𝑘 𝐹𝑘 (𝑥) = 𝑎0 𝐹0 (𝑥) + 𝑎1 𝐹1 (𝑥) + 𝑎2 𝐹2 (𝑥) + 𝑎3 𝐹3 (𝑥) + 𝑎4 𝐹4 (𝑥) + 𝑎5 𝐹5 (𝑥). 𝑢(𝑥) = 𝑎1 + 𝑎2 𝑥 + 𝑥(1 + 𝑥)𝑎3 + 𝑥 2 (𝑥 + 2)𝑎4 + 𝑥 2 (𝑥 2 + 3𝑥 + 1)𝑎5 . 972

(5.446)

Substituting Eq. (5.446) into Eq. (5.438) 𝑢(𝑥) = 𝑎1 + 𝑎2 𝑥 + 𝑥(1 + 𝑥)𝑎3 + 𝑥 2 (𝑥 + 2)𝑎4 + 𝑥 2 (𝑥 2 + 3𝑥 + 1)𝑎5 = 3𝑥 2 1

−5𝑥 3 + ∫−1 𝑥 4 𝑡 [

𝑡

2 (𝑡

𝑎1 + 𝑎2 𝑡 + 𝑡(1 + 𝑡)𝑎3 + ] d𝑡. + 2)𝑎4 + 𝑡 2 (𝑡 2 + 3𝑡 + 1)𝑎5

(5.447)

Now multiply Eq. (5.447) by 𝐹𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from -1 to 1, we have the matrix form as, −2.0000 0.2667 − 0.4000 − 1.1733 0.0000 − 0.6667 − 0.6667 − 0.4000 −0.6667 − 0.4762 − 0.87620 − 1.0857 −1.334 − 0.0190 − 0.81904 − 1.6714 [ −1.0667 − 0.8614 − 1.54708 − 2.0253

𝑎1 −2.000 − 0.5867 𝑎 2 2.0000 − 1.2000 𝑎 = 3 0.8000 − 1.5428 𝑎 4 −0.9714 − 1.5428 [ 𝑎 ] [ 5 2.22857 ] ] − 3.1555

After solving we get 𝛼1 = 0 ,𝛼2 = −3, 𝑎3 = 3, 𝑎4 = 1 𝑎𝑛𝑑 𝑎5 = −2. Consequently we have exact solution as 𝑢(𝑥) = 3𝑥 2 − 5𝑥 3 − 2𝑥 4 .

5.14.5. Abel’s Integral Equations Consider the Abel’s Integral equation [229] as 16

5

𝑥

𝑥 2 = ∫0 15

1 √𝑥−𝑡

( 𝑢3 (𝑡))d𝑡.

(5.448)

The exact solution of Eq. (5.448) is 2

𝑢(𝑥) = 𝑥 3 .

(5.449)

The transformation 1

𝑣(𝑥) = ( 𝑢3 (𝑥)), 𝑢(𝑥) = (𝑣(𝑥))3 ,

(5.450)

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥)

(5.451)

Consider 1st order Fibonicca Polynomials, i.e. for 𝑛 = 1, we have Eq. (5.451) is 𝑣(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥) = 𝑎0 𝐴0 (𝑥) + 𝑎1 𝐴1 (𝑥). or

𝑣(𝑥) = 𝑎1 .

(5.452)

Substituting Eq. (5.452) into Eq. (5.449)

973

16

5

𝑥

𝑥 2 = ∫0 15

1 √𝑥−𝑡

[𝑎1 ]d𝑡.

(5.453)

Now multiply Eq. (5.453) by 𝐴𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from 0 to 1, we have 1 16

5

1

𝑥

∫0 (15 𝑥 2 ) 𝐴𝑗 (𝑥)d𝑥 = ∫0 [∫0

1 √𝑥−𝑡

[𝑎1 ]d𝑡] 𝐴𝑗 (𝑥)d𝑥, 𝑗 = 0,1

(5.454)

Eq. (5.454) can be written as, 4

𝑎 3 1

32

= 105, 8

After solving we get 𝑎1 = 35. consequently we have the approximate solution is 8

𝑣(𝑥) = 35,

(5.455)

Now by putting back into Eq(5.450) we get 1

𝑢(𝑥) =

8 3 (35) .

(5.456)


Exact Solution


Error Analysis

__________________________________________________________________ 0.0

0.000000000000000

0.611421417700000

6.11421e-01

0.1

0.215443469000000

0.611421417700000

3.95978e-01

0.2

0.341995189300000

0.611421417700000

2.69426e-01

0.3

0.448140474700000

0.611421417700000

1.63281e-01

0.4

0.542883523300000

0.611421417700000

6.85379e-02

0.5

0.629960524900000

0.611421417700000

1.85391e-02

0.6

0.711378660900000

0.611421417700000

9.99572e-02

0.7

0.788373516300000

0.611421417700000

1.76952e-01

0.8

0.861773876000000

0.611421417700000

2.50352e-01

0.9

0.932169751800000

0.611421417700000

3.20748e-01

1.0

1.000000000000000

0.611421417700000

3.88579e-01

__________________________________________________________________

974


Consider 3rd order Fibonicca Polynomials, i.e. for 𝑛 = 3, we have Eq. (5.452) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝐹𝑘 (𝑥) = 𝑎0 𝐹0 (𝑥) + 𝑎1 𝐹1 (𝑥) + 𝑎2 𝐹2 (𝑥) + 𝑎3 𝐹3 (𝑥). Or

𝑢(𝑥) = 𝑎1 + 𝑎2 𝑥 + 𝑥(1 + 𝑥)𝑎3 .

(5.457)


5

𝑥

𝑥 2 = ∫0 15

1 √𝑥−𝑡

[𝑎1 + 𝑎2 𝑡 + 𝑡(1 + 𝑡)𝑎3 ]d𝑡.

(5.458)

Now multiply Eq. (5.458) by 𝐹𝑗 (𝑥), 𝑗 = 0,1,2,3 and integrating both side from 0 to 1, we have the matrix form of Eq. (5.459) is 4

8

88

3 4

15 8

105 584

5 21 945 48 128 256 [35 189 231]

32

𝑎1 [𝑎 2 ] = 𝑎3

105 32 135 128

,

(5.459)

[297]

After solving we get 𝛼1 = 0, 𝑎2 = −1 𝑎𝑛𝑑 𝛼3 = 1. Consequently we have 𝑣 (𝑥 ) = 𝑥 2 .

(5.460)

This gives the exact solution by 2

𝑢(𝑥) = 𝑥 3 .

975

5.15. Gegenbauer Polynomials Method (GPM) We develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Gegenbauer polynomials. Gegenbauer polynomials [53] is given as 1

𝐺𝑚 (𝑥) = (𝑛 (2𝑥 (𝑛 + 𝛼 − 1)) 𝑇𝐼𝑛−1 − (𝑛 + 2 𝛼 − 2)𝑇𝐼𝑛−2 )).

(5.461)

𝐺0 = 1, 1

𝐺1 = 𝑥 − 2, 4

2

𝐺2 = 𝑥 2 − 3 𝑥 + 3, 𝐺3 = 𝑥 3 −

11

𝐺4 = 𝑥 4 −

26

4

5

𝑥2 +

11

𝑥3 +

143

3

𝑥−

10

11 6

𝑥2 −

, 286 15

𝑥+

143 15

,

. . ..



(5.462)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, [Lewis, 11], to find an approximate solution 𝑢̃(𝑥) of Eq. (5.462). For this, we assume that 𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝐺𝑘 (𝑥),

(5.463)

where 𝐴𝑘 (𝑥) are Gegenbauer polynomials of degree 𝑘 defined in equation (5.461) and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.463) into (5.462), we get 𝑥

𝑓(𝑥) = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝐺𝑘 (𝑡)d𝑡. 976

(5.464)

Then the Galerkin equations are obtained by multiplying both sides of (5.464) by 𝐺𝑗 (𝑥) and then integrating with respect to 𝑥 from 𝑎 to 𝑏, we have 𝑏

𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝐺𝑗 (𝑥)d𝑥 = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐺𝑘 (𝑡)d𝑡)𝐺𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛.

(5.465)

Since in each equation, there are two integrals. The inner integrand of the left side is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.465) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by 𝜆 ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝑌𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.466)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐺𝑘 (𝑡)d𝑡)𝐺𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏

𝑌𝑗 = ∫𝑎 𝑓(𝑥) 𝐺𝑗 (𝑥)d𝑥,

𝑗 = 0,1,2, … 𝑛.




(5.467)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) and 𝑢(𝑥) being the known function and 𝜆 is a constant. Proceeding as before ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝑌𝑗 ;

𝑘, 𝑗 = 0,1,2, … 𝑛

(5.468)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝐺𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝐺𝑘 (𝑡)d𝑡)𝐺𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏

𝑌𝑗 = ∫𝑎 𝑓(𝑥) 𝐺𝑗 (𝑥)d𝑥,

𝑗 = 0,1,2, … 𝑛.

977


5.15.2. Volterra Integral Equation Consider the Volterra Integral equation [229] as 1

1

𝑥

𝑢(𝑥) = 1 − 2 𝑥 2 + 6 ∫0 (𝑥 − 𝑡)3 𝑢(𝑡)d𝑡.

(5.469)


(5.470)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥).

(5.471)

Cosider 25th order Gegenbauer Polynomials, i.e. for 𝑛 = 25, we have Eq. (5.471) is 𝑢(𝑥) = ∑25 𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥) = 𝑎0 𝐺0 (𝑥) + 𝑎1 𝐺1 (𝑥) + ⋯ + 𝑎25 𝐺0 (𝑥), 1

𝑢(𝑥) = 𝑎0 + 2𝛽𝑥𝑎1 + ⋯ (10 𝛽 + ⋯ + 𝑥 25 𝛽 25 ) 𝑎25.

(5.472)

Subsituting Eq. (5.472) into Eq. (5.469) 1

1

1

𝑥

𝑢(𝑥) = 𝑎0 + 2𝛽𝑥𝑎1 + ⋯ (10 𝛽 + ⋯ + 𝑥 25 𝛽 25 ) 𝑎25 = 1 − 2 𝑥 2 + 6 ∫0 (𝑥 − 1

𝑡)3 [𝑎0 + 2𝛽𝑡𝑎1 + ⋯ (10 𝛽 + ⋯ + 𝑡 25 𝛽 25 ) 𝑎25 ]d𝑡.

(5.473)

Now multiply Eq. (5.473) by 𝐺𝑗 (𝑥), 𝑗 = 0,1. . .25 and integrating both side from -1 to 1, we have 𝑎0 + 2𝛽𝑥𝑎1 + ⋯ + 1 1 1 ] 𝐺𝑗 (𝑥)d𝑥 = ∫−1 (1 − 2 𝑥 2 ) 𝐺𝑗 (𝑥)d𝑥 + ∫−1 [ 1 25 25 (10 𝛽 + ⋯ + 𝑥 𝛽 ) 𝑎25 𝑎0 + 2𝛽𝑡𝑎1 + ⋯ 1 1 𝑥 ] d𝑡] 𝐺𝑗 (𝑥)d𝑥, 𝑗 = 0,1 , … 25 1 ∫−1 [6 ∫0 (𝑥 − 𝑡)3 [ + (10 𝛽 + ⋯ + 𝑡 25 𝛽 25 ) 𝑎25 Applying the prescribed technique on eq.(5.474), we get the following system of equations, a0 ⋯ 1.6667 2.0167 0 0 a ⋯ 2.6762 0.1592] [ 1 ] = [ 0 ]. [ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0.1554 ⋯ 5.2222 a25 0 After solving we get, 𝑎0 = 0.86990, 𝑎1 = −4.3682 × 10−55 , … 𝑎25 = 1.0146 × 10−53 , 978

(5.474)

Putting the value back we get the approximate solution 𝑢(𝑥) = 0.9999 + 4.5964 × 10−51 𝑥 − ⋯ + 3.4044 × 10−46 .

(5.475)

Table 5.97. Comparison of Exact and Approximate solutions of Eq. (5.469) obtained from Gegenbauer polynomials Method (GPM) __________________________________________________________________ 𝑥

Exact Solution


Error

__________________________________________________________________ 0.0e+00

1.000000000000000

1.000000000000000

5.24568e-35

1.0e-01

0.995004165278026

0.994995834722470

8.33056e-06

2.0e-01

0.980066577841242

0.979933422285686

1.33156e-04

3.0e-01

0.955336489125606

0.954663514125604

6.72975e-04

4.0e-01

0.921060994002885

0.918939038447260

2.12196e-03

5.0e-01

0.877582561890373

0.872417631333799

5.16493e-03

6.0e-01

0.825335614909678

0.814665214900607

1.06704e-02

7.0e-01

0.764842187284488

0.745160656671296

1.96815e-02

8.0e-01

0.696706709347165

0.663301553505691

3.34052e-02

9.0e-01

0.621609968270664

0.568411192096670

5.31988e-02

1.0e+00

0.540302305868140

0.459746746160174

8.05556e-02

__________________________________________________________________


Consider 50th order Gegenbauer Polynomials, i.e. for 𝑛 = 50, we have Eq. (5.472) is 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥) = 𝑎0 𝐺0 (𝑥) + 𝑎1 𝐺1 (𝑥) + ⋯ 𝑎50 𝐺50 (𝑥) 979

1

𝑢(𝑥) = 𝑎0 + 2𝛽𝑥𝑎1 + ⋯ + (− 25 𝛽 + ⋯ 𝑥 50 𝛽 50 ) 𝑎50

(5.476)


1

𝑢(𝑥) = 𝑎0 + 2𝛽𝑥𝑎1 + ⋯ + (− 25 𝛽 + ⋯ 𝑥 50 𝛽 50 ) 𝑎50 = 1 − 2 𝑥 2 + 𝑥 ∫ (𝑥 6 0 1

1

− 𝑡)3 [𝑎0 + 2𝛽𝑡𝑎1 + ⋯ + (− 25 𝛽 + ⋯ 𝑡 50 𝛽 50 ) 𝑎50 ]d𝑡.

(5.477)

Now multiply Eq. (5.477) by 𝐺𝑗 (𝑥), 𝑗 = 0,1 … 50 and integrating both side from -1 to 1, we have 𝑎0 + 2𝛽𝑥𝑎1 + ⋯ + 1 [ ]𝐺 1 ∫−1 (− 𝛽 + ⋯ 𝑥 50 𝛽 50 ) 𝑎50 𝑗 25 1 1 𝑥 ∫−1 [6 ∫0 (𝑥

1

1

(𝑥)d𝑥 = ∫−1 (1 − 𝑥 2 ) 𝐺𝑗 (𝑥)d𝑥 + 2

𝑎0 + 2𝛽𝑥𝑎1 + ⋯ − 𝑡)3 [ ] d𝑡 ] 𝐺𝑗 (𝑥)d𝑥, 1 + (− 25 𝛽 + … 𝑡 50 𝛽 50 ) 𝑎50

(5.478)

Applying the prescribed technique on eq. (5.478), we get the following system of equations, ⋯ 0.03908 a0 1.6667 2.0167 0 a1 ⋯ 0 0 2.6762 0 [ ][ ⋮ ] = [ ]. ⋮ ⋮ ⋮ ⋮ ⋮ a50 0 0.0102 ⋯ 0 0 After solving we get, 𝑎1 = 1.00000131081, 𝑎2 = −0.61836284995, 𝑎3 = 0.6181003491, … 𝑎50 = −5.268867421 × 10−10 Putting the values back we get the approximate solution as, 𝑢(𝑥) = 1.00001281 − 0.0002624958𝑥 − 0.5011059247𝑥 2 + ⋯ − 5.268867421 ×

10−10 𝑥 49 .

980

Table 5.98. Comparison of Exact and Approximate solutions of Eq. (5.469) obtained from Gegenbauer polynomials Method (GPM) __________________________________________________________________ x

Exact Solution


Error in OADM

0.0

1.000000000000000

1.000000000000000

4.10003e-80

0.1

0.995004165300000

0.995004165000000

3.00000e-10

0.2

0.980066577800000

0.980066578000000

2.00000e-10

0.3

0.955336489100000

0.955336489000000

1.00000e-10

0.4

0.921060994000000

0.921060994000000

0.00000e+00

0.5

0.877582561900000

0.877582562000000

1.00000e-10

0.6

0.825335614900000

0.825335615000000

1.00000e-10

0.7

0.764842187300000

0.764842187000000

3.00000e-10

0.8

0.696706709300000

0.696706709000000

3.00000e-10

0.9

0.621609968300000

0.621609968000000

3.00000e-10

1.0

0.540302305900000

0.540302306000000

1.00000e-10

__________________________________________________________________


Consider 100th order Gegenbauer Polynomials, i.e. for 𝑛 = 100, we have Eq. (5.471) is 𝑢(𝑥) = ∑100 𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥) = 𝑎0 𝐺0 (𝑥) + 𝑎1 𝐺1 (𝑥) + ⋯ … 𝑎100 𝐺100 (𝑥) 𝑢(𝑥) = 𝑎1 + 𝑥𝑎2 + 𝑥(𝑥 + 1)𝑎3 + ⋯ 𝑥 50 (50 + 20825𝑥 + 2598960𝑥 2 + ⋯ 98𝑥 48 + 𝑥 49 )𝑎100

(5.479)


𝑢(𝑥) = 𝑎1 + 𝑥𝑎2 + 𝑥(𝑥 + 1)𝑎3 + ⋯ 𝑥 50 (50 + 20825𝑥 + 2598960𝑥 2 + 1

1

𝑥

⋯ 98𝑥 48 + 𝑥 49 )𝑎100 = 1 − 2 𝑥 2 + 6 ∫0 (𝑥 − 𝑡)3 [𝑎1 + 𝑡𝑎2 + 𝑡(𝑡 + 1)𝑎3 + ⋯ 𝑥 50 (50 + 20825𝑡 + 2598960𝑡 2 + ⋯ 98𝑡 48 + 𝑡 49 )𝑎100 ]d𝑡.

(5.480)

Now multiply Eq. (5.480) by 𝐺𝑗 (𝑥), 𝑗 = 0,1 … 100 and integrating both side from -1 to 1, we have the following system of equations, 1.98333 0 0.65476

0 0.66428 0.664285

0.66587 0.66428 . . . 1.06367 . . .

[4.684 × 1018

4.772 × 1018

9.505 × 1018

𝑎1 1.667 𝑎2 0 𝑎3 0.467 . = . . . . . . 18 ] [𝑎 [ ] −2.5 × 1038 ] 8.690 × 10 100

4.876 × 1018 4.810 × 1018 12.17462957

After solving we get, 𝑎1 = 1.0000010815, 𝑎2 = 0.6224979323, 𝑎3 = −0.6223555723, … 𝑎100 = 4.619397432 × 10−21 Putting the values back we get the approximate solution as, 𝑢(𝑥) = 1.000010815 + 0.0001423600𝑥 − 0.5010424578𝑥 2 + ⋯ + 4.61939748310−21 𝑥 99 .

(5.481)

982


Exact Solution


Error in OADM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.19099e-188

0.1

0.995004165278026

0.995004165278026

7.00000e-100

0.2

0.980066577841242

0.980066577841242

3.00000e-100

0.3

0.955336489125606

0.955336489125606

3.00000e-100

0.4

0.921060994002885

0.921060994002885

1.60000e-99

0.5

0.877582561890373

0.877582561890373

1.00000e-100

0.6

0.825335614909678

0.825335614909678

1.40000e-99

0.7

0.764842187284488

0.764842187284488

2.20000e-99

0.8

0.696706709347165

0.696706709347165

3.00000e-100

0.9

0.621609968270664

0.621609968270664

5.00000e-100

1.0

0.540302305868140

0.540302305868140

3.00000e-100

__________________________________________________________________


5.15.3. Weakly Singular Integral Equation Consider the Weakly singular Integral equation [229] as 𝑥

𝑢(𝑥) = 3 − 6√𝑥 + ∫0

1 √𝑥−𝑡

𝑢(𝑡)d𝑡.

(5.482)

983

The exact solution of Eq. (5.482) is 𝑢(𝑥) = 3.

(5.483)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥)

(5.484)

Consider 1st order Gegenbauer Polynomials, i.e. for 𝑛 = 1, we have Eq. (5.484) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥) = 𝑎0 𝐺0 (𝑥) + 𝑎1 𝐺1 (𝑥). or

𝑢(𝑥) = 𝑎0 + 2𝛽𝑥𝑎1 .

(5.485)


𝑢(𝑥) = 𝑎0 + 2𝛽𝑥𝑎1 = 3 − 6√𝑥 + ∫0

1 √𝑥−𝑡

[𝑎0 + 2𝛽𝑡𝑎1 ]d𝑡.

(5.486)

Now multiply Eq. (5.486) by 𝐺𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from 0 to 1, we have 1

1

1

𝑥

∫0 [𝑎0 + 2𝛽𝑥𝑎1 ]𝐺𝑗 (𝑥)d𝑥 = ∫0 (3 − 6√𝑥)𝐺𝑗 (𝑥)d𝑥 + ∫0 [∫0 2𝛽𝑡𝑎1 ]d𝑡] 𝐺𝑗 (𝑥)d𝑥, 𝑗 = 0,1

1 √𝑥−𝑡

[𝑎0 + (5.487)

Now for 𝛽 = 1 by solving Eq. (5.487) we get, 1

1

5

21

− − −1 15 𝑎0 9 [ 33 ] [ ] = [ 4 𝑎1 − ] − − 5

After solving we get 𝛼0 = 3 and 𝑎1 = 0. Consequently we have the exact solution is 𝑢(𝑥) = 3.

5.15.4. Fredholm Integral Equation Consider the Fredholm Integral equation [229] as 1

𝑢(𝑥) = 3𝑥 2 − 5𝑥 3 + ∫−1 𝑥 4 𝑡𝑢(𝑡)d𝑡.

(5.488)

The exact solution of Eq. (5.488) is 𝑢(𝑥) = 3𝑥 2 − 5𝑥 3 − 2𝑥 4 .

(5.489)

Consider 2nd order Gegenbauer Polynomials, i.e. for 𝑛 = 2, we have Eq. (5.489) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥) = 𝑎0 𝐺0 (𝑥) + 𝑎1 𝐺1 (𝑥) + 𝑎2 𝐺2 (𝑥). Or

𝑢(𝑥) = 𝑎0 + 2𝛽𝑥𝑎1 + (2𝑥 2 𝛽 + 2𝑥 2 𝛽 2 − 𝛽)𝑎2 .


(5.490)

𝑢(𝑥) = 𝑎0 + 2𝛽𝑥𝑎1 + (2𝑥 2 𝛽 + 2𝑥 2 𝛽 2 − 𝛽)𝑎2 = 3𝑥 2 − 5𝑥 3 − 2𝑥 4 + 𝑥

∫0 𝑥 4 𝑡[𝑎0 + 2𝛽𝑡𝑎1 + (2𝑡 2 𝛽 + 2𝑡 2 𝛽 2 − 𝛽)𝑎2 ]d𝑡.

(5.491)

Now multiply Eq. (5.491) by 𝐺𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from -1 to 1, we have 1

1

∫−1[𝑎0 + 2𝛽𝑥𝑎1 + (2𝑥 2 𝛽 + 2𝑥 2 𝛽 2 − 𝛽)𝑎2 ]𝐺𝑗 (𝑥)d𝑥 = ∫−1(3𝑥 2 − 5𝑥 3 − 1

𝑥

2𝑥 4 )𝐺𝑗 (𝑥)d𝑥 + ∫−1[∫0 𝑥 4 𝑡[𝑎0 + 2𝛽𝑥𝑎1 + (2𝑥 2 𝛽 + 2𝑥 2 𝛽 2 − 𝛽)𝑎2 ]d𝑡]𝐺𝑗 (𝑥)d𝑥, (5.492) Now for 𝛽 = 1, the matrix form of Eq. (5.492) is 8

2 0 2

[3

− 15 8 3 104

− 105

2

2 a0 ], 0 [a1 ] = [−4 14 46 a 2 5 15] 3

After solving we get 69

3

9

𝑎0 = 140 , 𝑎1 = − 2 and 𝛼2 = 28. consequently we have the approximate solution is 𝑢(𝑥) =

6 35

9

− 3𝑥 + 𝑥 2 . 7


985


Exact Solution


Error in OADM

__________________________________________________________________ 0.0e+00

0.000000000000000

0.171428571400000

1.71429e-01

1.0e-01

0.024800000000000

-0.115714285700000

1.40514e-01

2.0e-01

0.076800000000000

-0.377142857200000

4.53943e-01

3.0e-01

0.118800000000000

-0.612857142900000

7.31657e-01

4.0e-01

0.108800000000000

-0.822857143300000

9.31657e-01

5.0e-01

0.000000000000000

-1.007142858000000

1.00714e+00

6.0e-01

-0.259200000000000

-1.165714286000000

9.06514e-01

7.0e-01

-0.725200000000000

-1.298571429000000

5.73371e-01

8.0e-01

-1.459200000000000

-1.405714286000000

5.34857e-02

9.0e-01

-2.527200000000000

-1.487142858000000

1.04006e+00

1.0e+00

-4.000000000000000

-1.542857143000000

2.45714e+00

__________________________________________________________________ Consider 4th order Gegenbauer Polynomials, i.e. for 𝑛 = 4, we have Eq. (5.490) is 𝑢(𝑥) = ∑4𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥) = 𝑎0 𝐺0 (𝑥) + 𝑎1 𝐺1 (𝑥) + 𝑎2 𝐺2 (𝑥) + 𝑎3 𝐺3 (𝑥) + 𝑎4 𝐺4 (𝑥). 1

1

2

𝑢(𝑥) = 𝑎0 + 2𝛽𝑥𝑎1 + ⋯ + (2 𝛽 + 2 𝛽 2 + ⋯ 3 𝑥 4 𝛽 4 ) 𝑎4 . Proceeding as before we have 16

8

2 0 2 3

0 2 5

[

2

− 75

2

8

3

16

5

3 104

0 46

15 208

− 15 − 105 16

15

− 525

0

352

142 15 664 105 − 945

0

142 105

0

1126 105 1328 315 − 4725 ]

2 𝑎0 −4 14 𝑎1 𝑎2 = 5 24 𝑎3 −7 [𝑎4 ] 46 [ 35 ]

After solving we get, 1

5

3

5

1

𝛼0 = 2 ,𝛼1 = − 4 , 𝑎2 = 8 , 𝑎3 = − 8 𝑎𝑛𝑑 𝑎4 = − 8. 986

(5.493)

Consequently we have exact solution as 𝑢(𝑥) = 3𝑥 2 − 5𝑥 3 − 2𝑥 4 .

5.15.5. Abel’s Integral Equation Consider the Abel’s Integral equation [229] as 16

5

𝑥

𝑥 2 = ∫0 15

1 √𝑥−𝑡

( 𝑢3 (𝑡))d𝑡.

(5.494)


𝑢(𝑥) = 𝑥 3 .

(5.495)

The transformation 𝑣(𝑥) = ( 𝑢

3 (𝑥)),

1 3

𝑢(𝑥) = (𝑣(𝑥)) ,

(5.496)

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥)

(5.497)

Consider 1st order Gegenbauer Polynomials, i.e. for 𝑛 = 1, we have Eq. (5.497) is 𝑣(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥) = 𝑎0 𝐺0 (𝑥) + 𝑎1 𝐺1 (𝑥). or

𝑣(𝑥) = 𝑎1 .

(5.498)


5

𝑥

𝑥 2 = ∫0 15

1 √𝑥−𝑡

[𝑎1 ]d𝑡.

(5.499)

Now multiply Eq. (5.499) by 𝐺𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from 0 to 1, we have 1 16

5

1

𝑥

∫0 (15 𝑥 2 ) 𝐺𝑗 (𝑥)d𝑥 = ∫0 [∫0

1 √𝑥−𝑡

[𝑎1 ]d𝑡] 𝐺𝑗 (𝑥)d𝑥, 𝑗 = 0,1

(5.500)

Eq. (5.500) can be written as, 4

𝑎 3 1

32

= 105, 8

After solving we get 𝑎1 = 35. consequently we have the approximate solution is 8

𝑣(𝑥) = 35,

(5.501)

Now by putting back into Eq(5.499) we get 1

𝑢(𝑥) =

8 3 ( ). 35

(5.502)

987


Exact Solution


Error in OADM

__________________________________________________________________ 0.0

0.000000000000000

0.611421417700000

6.11421e-01

0.1

0.215443469000000

0.611421417700000

3.95978e-01

0.2

0.341995189300000

0.611421417700000

2.69426e-01

0.3

0.448140474700000

0.611421417700000

1.63281e-01

0.4

0.542883523300000

0.611421417700000

6.85379e-02

0.5

0.629960524900000

0.611421417700000

1.85391e-02

0.6

0.711378660900000

0.611421417700000

9.99572e-02

0.7

0.788373516300000

0.611421417700000

1.76952e-01

0.8

0.861773876000000

0.611421417700000

2.50352e-01

0.9

0.932169751800000

0.611421417700000

3.20748e-01

1.0

1.000000000000000

0.611421417700000

3.88579e-01

__________________________________________________________________


Consider 3rd order Gegenbauer Polynomials, i.e. for 𝑛 = 3, we have Eq. (5.500) is 𝑢(𝑥) = ∑3𝑘=0 𝛼𝑘 𝐺𝑘 (𝑥) = 𝑎0 𝐺0 (𝑥) + 𝑎1 𝐺1 (𝑥) + 𝑎2 𝐺2 (𝑥) + 𝑎3 𝐺3 (𝑥). Or

𝑢(𝑥) = 𝑎1 + 𝑎2 𝑥 + 𝑥(1 + 𝑥)𝑎3 .

(5.503)


16

5

𝑥

𝑥 2 = ∫0 15

1 √𝑥−𝑡

[𝑎1 + 𝑎2 𝑡 + 𝑡(1 + 𝑡)𝑎3 ]d𝑡.

(5.504)

Now multiply Eq. (5.504) by 𝐺𝑗 (𝑥), 𝑗 = 0,1,2,3 and integrating both side from 0 to 1, we have the matrix form of Eq. (50) is 4

8

88

3 4

15 8

105 584

5 21 945 48 128 256 [35 189 231]

32

𝑎1 [𝑎 2 ] = 𝑎3

105 32 135 128

,

[297]

After solving we get 𝛼1 = 0, 𝑎2 = −1 𝑎𝑛𝑑 𝛼3 = 1. Consequently we have 𝑣(𝑥) = 𝑥 2 .

(5.505)

This gives the exact solution by 2

𝑢(𝑥) = 𝑥 3 .

5.16. Charlier's r Polynomials Method (CPM) We develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Charlier's polynomials [44] Charlier's polynomials is given as 𝑛!

𝐶𝑚 (𝑥) = ∑𝑚 𝑛=0 (𝑛−𝑘)!

𝑥! 𝑘!(𝑥−𝑘)!

(−𝛼)2 − 𝑘

𝐶0 = 1, 𝐶1 = 𝑥 − 𝛼, 𝐶2 = 𝛼 2 − 2𝑥𝛼 + 𝑥 2 − 𝑥, 𝐶3 = −𝛼 3 + 3𝑥𝛼 − 3𝑥 2 𝛼 + 3𝑥𝛼 + 𝑥 3 − 3𝑥 2 + 2𝑥, 𝐶4 = 𝛼 4 − 4𝛼 3 𝑥 + 6𝑥 2 𝛼 2 − 6𝑥𝛼 2 − 4𝑥 3 𝛼 + 12𝑥 2 𝛼, . . ..

989

(5.506)



(5.507)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, [Lewis, 11], to find an approximate solution 𝑢̃(𝑥) of Eq. (2). For this, we assume that 𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝐶𝑘 (𝑥),

(5.508)

where 𝐴𝑘 (𝑥) are Charlier's polynomials of degree 𝑘 defined in equation (5.506) and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.508) into (5.507), we get 𝑥

𝑓(𝑥) = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝐶𝑘 (𝑡)d𝑡.

(5.509)

Then the Galerkin equations are obtained by multiplying both sides of (5.509) by 𝐶𝑗 (𝑥) and then integrating with respect to 𝑥 from 𝑎 to 𝑏, we have 𝑏

𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝐶𝑗 (𝑥)d𝑥 = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐶𝑘 (𝑡)d𝑡)𝐶𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛.

(5.510)

Since in each equation, there are two integrals. The inner integrand of the left side is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.510) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by 𝜆 ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.511)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐶𝑘 (𝑡)d𝑡 )𝐶𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏

𝐹𝑗 = ∫𝑎 𝑓(𝑥) 𝐶𝑗 (𝑥)d𝑥,

𝑗 = 0,1,2, … 𝑛.

Now the unknown parameters 𝑎𝑘 are determined by solving the system of equations (5.511) and substituting these values of parameters in (5.508), we get the approximate solution 𝑢̃(𝑥) of the integral equation (5.507). 990



(5.512)


𝑘, 𝑗 = 0,1,2, … 𝑛

(5.513)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝐶𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝐶𝑘 (𝑡)d𝑡)𝐶𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏

𝐹𝑗 = ∫𝑎 𝑓(𝑥) 𝐶𝑗 (𝑥)d𝑥,

𝑗 = 0,1,2, … 𝑛.


5.16.2. Fredholm Integral Equation Consider the linear Fredholm Integral equation [229] as 1

𝑢(𝑥) = 2 − 2𝑥 + 5𝑥 4 + 7𝑥 5 + ∫−1(𝑥 − 𝑡)𝑢(𝑡)d𝑡.

(5.514)

The exact solution of Eq. (5.514) is 𝑢(𝑥) = 5𝑥 4 + 7𝑥 5 .

(5.515)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐶𝑘 (𝑥).

(5.516)

Consider 2nd order Charlier's Polynomials, i.e. for 𝑛 = 2, we have Eq. (5.516) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐶𝑘 (𝑥) = 𝑎0 𝐶0 (𝑥) + 𝑎1 𝐶1 (𝑥) + 𝑎2 𝐶2 (𝑥), or 𝑢(𝑥) = 𝑎0 + 𝑎1 (−𝛽 + 𝑥) + 𝑎2 (𝛽 2 − 2𝑥𝛽 + 𝑥 2 − 𝑥).

(5.517)

Substituting Eq. (5.517) into Eq. (5.514) 𝑎0 + 𝑎1 (−𝛽 + 𝑥) + 𝑎2 (𝛽 2 − 2𝑥𝛽 + 𝑥 2 − 𝑥) = 2 − 2𝑥 + 5𝑥 4 + 7𝑥 5 + 1

∫−1(𝑥 − 𝑡)[𝑎0 + 𝑎1 (−𝛽 + 𝑥) + 𝑎2 (𝛽 2 − 2𝑥𝛽 + 𝑥 2 − 𝑥)]d𝑡.

(5.518)

Now multiply Eq. (5.518) by 𝐵𝑗 (𝑥), 𝑗 = 0,1,2 and integrating both side from -1 to 1,we have 991

1

1

∫−1 [𝑎0 + 𝑎1 (−𝛽 + 𝑥) + 𝑎2 (𝛽 2 − 2𝑥𝛽 + 𝑥 2 − 𝑥)] 𝐵𝑗 (𝑥)𝑑𝑥=∫−1[2 − 2𝑥 + 5𝑥 4 + 1

7𝑥 5 + ∫−1(𝑥 − 𝑡)[𝑎0 + 𝑎1 (−𝛽 + 𝑥) + 𝑎2 (𝛽 2 − 2𝑥𝛽 + 𝑥 2 − 𝑥)]d𝑡] 𝐵𝑗 (𝑥)𝑑𝑥. (5.519) If 𝛽 = 1.Then the matrix form of Eq. (5.519) is 2 − [

10

3 20 3

2

−3 8 3 62

−

9

4

−3 −

22

9 146 15

]

after solving we get, 𝛼0 =

6 a0 16 [a1 ] = [− 3 ], 142 a2 21

78 7

, 𝑎1 =

111 7

, 𝑎2 =

30 7

.Consequently we have the approximate

solution is 3

𝑢(𝑥) = − 7 + 3𝑥 +

30 7

𝑥2.

(5.520)

Table 5.102. Comparison of Exact and Approximate solutions of Eq. (5.514) obtained from Charlier's polynomials Method (CPM) _________________________________________________________________ x

Exact Solution


Error

__________________________________________________________________ 0.0e+00

0.000000000000000

-0.428571428571429

4.28571e-01

1.0e-01

0.000570000000000

-0.085714285714286

8.62843e-02

2.0e-01

0.010240000000000

0.342857142857143

3.32617e-01

3.0e-01

0.057510000000000

0.857142857142857

7.99633e-01

4.0e-01

0.199680000000000

1.457142857142857

1.25746e+00

5.0e-01

0.531250000000000

2.142857142857143

1.61161e+00

6.0e-01

1.192320000000000

2.914285714285714

1.72197e+00

7.0e-01

2.376990000000000

3.771428571428571

1.39444e+00

8.0e-01

4.341760000000000

4.714285714285714

3.72526e-01

9.0e-01

7.413930000000000

5.742857142857143

1.67107e+00

1.0e+00

12.000000000000000

6.857142857142857

5.14286e+00

__________________________________________________________________

992

Fig 5.102. Comparison of Exact and Approximate solutions of Eq. ( 5.514) for n=2

Consider 5th order Charlier's Polynomials, i.e. for 𝑛 = 5, we have Eq. (5.516) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐶𝑘 (𝑥) = 𝑎0 𝐶0 (𝑥) + 𝑎1 𝐶1 (𝑥) + 𝑎2 𝐶2 (𝑥) + ⋯ + 𝑎5 𝐶5 (𝑥), or 𝑢(𝑥) = 𝑎0 + 𝑎1 (−𝛽 + 𝑥) + 𝑎2 (𝛽 2 − 2𝑥𝛽 + 𝑥 2 − 𝑥) + 𝑎3 (−𝛽 3 + 3𝑥𝛽 2 − 3𝑥 2 𝛽 + 3𝑥𝛽 + 𝑥 3 − 3𝑥 2 + 2𝑥) + ⋯

(5.521)

substituting Eq. (5.521) into Eq. (5.514) 𝑎0 + 𝑎1 (−𝛽 + 𝑥) + 𝑎2 (𝛽 2 − 2𝑥𝛽 + 𝑥 2 − 𝑥) + 𝑎3 (−𝛽 3 + 3𝑥𝛽 2 − 3𝑥 2 𝛽 + 1

3𝑥𝛽 + 𝑥 3 − 3𝑥 2 + 2𝑥) + ⋯ = 2 − 2𝑥 + 5𝑥 4 + 7𝑥 5 + ∫−1(𝑥 − 𝑡)[𝑎0 + 𝑎1 (−𝛽 + 𝑡) + 𝑎2 (𝛽 2 − 2𝑡𝛽 + 𝑡 2 − 𝑡) + 𝑎3 (

−𝛽 3 + 3𝑡𝛽 2 − 3𝑡 2 𝛽 + ) + ⋯ ]𝑑𝑡. 3𝑡𝛽 + 𝑡 3 − 3𝑡 2 + 2𝑡

(5.522)

Now multiply Eq. (5.522) by 𝐵𝑗 (𝑥), 𝑗 = 0,1,2,3,4,5 and integrating both side from -1 to 1,we have the matrix form of Eq. (5.523) is 2 10

−3 ⋮ ⋮ 5962 [− 21

2

−3 8 3

⋮ ⋮

19136 63

⋯ ⋯ ⋮ ⋮ ⋯

522

⋯ ⋯ ⋮ ⋮ ⋯

6 a0 16 −3 a1 63 ⋮ = ⋮ , ⋮ ⋮ ⋮ ⋮ 24520 5133368 [a 5 ] [− 77 ] ] 7 5344

(5.523)

231

after solving we get, 𝛼0 = 439, 𝑎1 = 1242, ⋯ 𝑎5 = 7.Consequently we have the approximate solution is 𝑢(𝑥) = 5𝑥 4 + 7𝑥 5 .

(5.524)

5.16.3. Volterra Integral Equation 993

Consider the Volterra Integral equation of the first kind [229] as 𝑥

9𝑥 2 + 5𝑥 3 = ∫0 (10𝑥 − 10𝑡 + 6)𝑢(𝑡) dt.

(5.525)

Differentiating both side above equation and using Leibnitz rule we obtain the Volterra equation of second kind 5

𝑥

5

𝑢(𝑥) = 3𝑥 + 2 𝑥 2 − 3 ∫0 𝑢(𝑡)d𝑡.

(5.526)

The exact solution of Eq. (5.526) 𝑢(𝑥) = 3𝑥.

(5.527)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐶𝑘 (𝑥).

(5.528)

Consider 1st order Charlier's Polynomials, i.e. for 𝑛 = 1, we have Eq. (5.528) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐶𝑘 (𝑥) = 𝑎0 𝐶0 (𝑥) + 𝑎1 𝐶1 (𝑥), or

𝑢(𝑥) = 𝑎0 + 𝑎1 (−𝛽 + 𝑥).

(5.529)


5

𝑥

𝑎0 + 𝑎1 (−𝛽 + 𝑥) = 3𝑥 + 2 𝑥 2 − 3 ∫0 [𝑎0 + 𝑎1 (−𝛽 + 𝑡)]d𝑡.

(5.530)

Now multiply Eq. (5.530) by 𝐵𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from 0 to 1, we have 1

∫0 (𝑎0 + 𝑎1 (−𝛽 + 𝑥))𝐵𝑗 (𝑥)d𝑥 = 1

5

𝑥

5

∫0 [[3𝑥 + 2 𝑥 2 − 3 ∫0 [𝑎0 + 𝑎1 (−𝛽 + 𝑡)]d𝑡 ]] 𝐵𝑗 (𝑥)d𝑥,

(5.531)

if 𝛽 = 1.Then the matrix form of Eq. (5.531) is 4.6667 3.2500 𝛼0 4.2500 [ ][ ] = [ ], −1.4167 1.0833 𝛼1 1.0000 after solving we get a0 = 3 and 𝛼1 = 3. Consequently we have the exact solution is 𝑢(𝑥) = 3𝑥.

5.16.4. Weakly-Singular Integral Equation Consider the weakly-singular integral equation [229] as 16

5

𝑥

𝑢(𝑥) = 𝑥 2 − 15 𝑥 2 + ∫0

1 √𝑥−𝑡

𝑢(𝑡)d𝑡.

(5.532)

The exact solution of Eq. (5.532) 𝑢(𝑥) = 𝑥 2 .

(5.533)

According to the proposed technique, consider the trail solution 994

𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐶𝑘 (𝑥).

(5.534)

Consider 2nd order Charlier's Polynomials, i.e. for 𝑛 = 2, we have Eq. (5.534) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐶𝑘 (𝑥) = 𝑎0 𝐶0 (𝑥) + 𝑎1 𝐶1 (𝑥) + 𝑎2 𝐶2 (𝑥), or

𝑢(𝑥) = 𝑎0 + 𝑎1 (−𝛽 + 𝑥) + 𝑎2 (𝛽 2 − 2𝑥𝛽 + 𝑥 2 − 𝑥).

(5.535)


𝑥

𝑎0 + 𝑎1 (−𝛽 + 𝑥) + 𝑎2 (𝛽 2 − 2𝑥𝛽 + 𝑥 2 − 𝑥) = 𝑥 2 − 𝑥 2 + ∫0

1 √𝑥−𝑡

𝑎1 (−𝛽 + 𝑡) + 𝑎2 (𝛽 2 − 2𝑡𝛽 + 𝑡 2 − 𝑡]d𝑡.

[𝑎0 + (5.536)

Now multiply Eq. (5.536) by 𝐵𝑗 (𝑥), 𝑗 = 0,1,2 and integrating both side from 0 to 1, we have 1

1

16

5

∫0 (𝑎0 + 𝑎1 (−𝛽 + 𝑥) + 𝑎2 (𝛽 2 − 2𝑥𝛽 + 𝑥 2 − 𝑥))𝐵𝑗 (𝑥)d𝑥 = ∫0 [𝑥 2 − 15 𝑥 2 + 𝑥

∫0

1 √𝑥−𝑡

(𝑎0 + 𝑎1 (−𝛽 + 𝑡) + 𝑎2 (𝛽 2 − 2𝑡𝛽 + 𝑡 2 − 𝑡]) d𝑡 ] 𝐵𝑗 (𝑥)d𝑥, (5.537)

if 𝛽 = 1.The matrix form of Eq. (5.537) is −0.3333 [ 0.3333 0.3285

0.3000 −0.0476 −0.2653

−0.20476 a0 0.0285 a 0.0605 ] [ 1 ] = [0.0156], a2 0.1346 0.0042

after solving we get 𝛼0 = 2, 𝛼1 = 3and a2 = 1. Consequently we have the exact solution is 𝑢(𝑥) = 𝑥 2 .

5.16.5. Abel Integral Equation Consider the generalized nonlinear Abel integral equation [229] as 𝑥

𝑥 = ∫0

1 1

(x2 −t2 )2

√𝑢(𝑡)d𝑡,

(5.538)

The exact solution of Eq. (5.538) is 𝑢(𝑥) = 𝑥 2 .

(5.539)

Consider √𝑢(𝑥) = 𝑣(𝑥),

(5.540)

so we have Eq.( 5.538) of the form 𝑥

𝑥 = ∫0

1 1

(x2 −t2 )2

𝑣(𝑡)d𝑡.

(5.541)

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐶𝑘 (𝑥).

(5.542) 995

Consider 1st order Charlier's Polynomials, i.e. for 𝑛 = 1, we have Eq. (5.542) is 𝑣(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐶𝑘 (𝑥) = 𝑎0 𝐶0 (𝑥) + 𝑎1 𝐶1 (𝑥), or

𝑣(𝑥) = 𝑎0 + 𝑎1 (−𝛽 + 𝑥).

(5.543)


𝑥 = ∫0

1 1

(x2 −t2 )2

[𝑎0 + 𝑎1 (−𝛽 + 𝑡)]d𝑡.

(5.544)


1

𝑥

∫0 (𝑥)𝐵𝑗 (𝑥)d𝑥 = ∫0 [∫0

1 1

(x2 −t2 )2

[𝑎1 + 𝑎1 (−𝛽 + 𝑡)]d𝑡] 𝐵𝑗 (𝑥)d𝑥, (5.545)

if 𝛽 = 1.Then the matrix form of Eq. (5.545) is 1.5707 −1.0707 𝛼0 0.5000 [ ] [𝛼 ] = [ ], 1 −0.7853 0.6187 −0.1667 after solving we get 𝛼0 = −1 and 𝛼1 = 1. Consequently we have the approximate solution is 𝑣(𝑥) = 𝑥. Using Eq. (5.540) we have solution 𝑢(𝑥) = 𝑥 2 , which is exact solution.

5.17. Bessel's Polynomials Method (BPM) We develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Bessel's polynomials [108]. Bessel’s polynomials is given as 1

𝑛 𝑘 𝑛 𝑚 𝐵𝑚 (𝑥) = ∑𝑚 𝑛=0 𝑛+1 ∑𝑘=0(−1) (𝑘 )(𝑥 + 𝑘) .

𝐵0 = 1, 1

𝐵1 = 𝑥 − 2, 4

2

𝐵2 = 𝑥 2 − 3 𝑥 + 3, 𝐵3 = 𝑥 3 −

11 4

𝑥2 +

11 3

𝑥−

11 6

,

996

(5.546)

𝐵4 = 𝑥 4 −

26 5

𝑥3 +

143 10

𝑥2 −

286 15

𝑥+

143 15

,

. . ..



(5.547)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, [Lewis, 11], to find an approximate solution 𝑢̃(𝑥) of Eq. (2). For this, we assume that 𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝐵𝑘 (𝑥),

(5.548)

where 𝐴𝑘 (𝑥) are Bessel's polynomials of degree 𝑘 defined in equation (5.546) and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.548) into (5.547), we get 𝑥

𝑓(𝑥) = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝐵𝑘 (𝑡)d𝑡.

(5.549)

Then the Galerkin equations are obtained by multiplying both sides of (5.549) by 𝐵𝑗 (𝑥) and then integrating with respect to 𝑥 from 𝑎 to 𝑏, we have 𝑏

𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝐵𝑗 (𝑥)d𝑥 = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐵𝑘 (𝑡)d𝑡)𝐵𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛.

(5.550)

Since in each equation, there are two integrals. The inner integrand of the left side is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.550) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by 𝜆 ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛 where

997

(5.551)

𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐵𝑘 (𝑡)d𝑡)𝐵𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏


𝑗 = 0,1,2, … 𝑛.




(5.552)


𝑘, 𝑗 = 0,1,2, … 𝑛

(5.553)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝐵𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝐵𝑘 (𝑡)d𝑡)𝐵𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏

𝐹𝑗 = ∫𝑎 𝑓(𝑥) 𝐵𝑗 (𝑥)d𝑥,

𝑗 = 0,1,2, … 𝑛.


5.17.2. Fredholm Integral Equation Consider the linear Fredholm integral equation [229] as 1

𝑢(𝑥) = −15 + 10𝑥 3 + 𝑥 4 − ∫0 (20𝑥 3 − 56𝑡 3 )𝑢(𝑡)𝑑𝑡.

(5.554)

The exact solution of Eq. (5.554) is 𝑢(𝑥) = 𝑥 3 + 𝑥 4 .

(5.555)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥)

(5.556)

Consider 2nd order Bessel's Polynomials, i.e. for 𝑛 = 2, we have Eq. (5.556) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥) + 𝑎2 𝐵2 (𝑥). or

𝑢(𝑥) = 𝑎1 + 𝑎1 (1 + 𝑥) + 𝑎2 (3𝑥 2 + 3𝑥 + 1),


(5.557)

1

𝑎0 + 𝑎1 (1 + 𝑥) + 𝑎2 (3𝑥 2 + 3𝑥 + 1)=−15 + 10𝑥 3 + 𝑥 4 − ∫0 (20𝑥 3 − 56𝑡 3 )[𝑎0 + 𝑎1 (1 + 𝑡) + 𝑎2 (3𝑡 2 + 3𝑡 + 1)]𝑑𝑡.

(5.558)

Now multiply Eq. (5.558) by 𝐵𝑗 (𝑥), 𝑗 = 0,1,2 and integrating both side from 0 to 1,we have 1

1

∫0 [𝑎0 + 𝑎1 (1 + 𝑥) + 𝑎2 (3𝑥 2 + 3𝑥 + 1)] 𝐵𝑗 (𝑥)𝑑𝑥=∫0 [−15 + 10𝑥 3 + 𝑥 4 − 1

∫0 (20𝑥 3 − 56𝑡 3 )[𝑎0 + 𝑎1 (1 + 𝑡) + 𝑎2 (3𝑡 2 + 3𝑡 + 1)]𝑑𝑡]𝐵𝑗 (𝑥)𝑑𝑥. (5.559) The matrix form of Eq. (5.559) is −8.0000 [−10.5000 −18.5000

16.2000 −21.9666 −41.9500

−12.3000 −54.6000 a0 a ] [ ] = [ −17.6333], −76.1500 1 −37.8714 −154.800 a2

after solving we get, 𝛼0 = 3.8127, 𝑎1 = −47594, 𝑎2 = 1.0787.Consequently we have the approximate solution is 𝑢(𝑥) = 0.1320 − 1.5231𝑥 + 1.0787𝑥 2 . Table 5.103. Comparison of Exact and Approximate solutions of Eq. (5.554) obtained from Bessel's polynomials Method (BPM) __________________________________________________________________ x

Exact Solution


Error

__________________________________________________________________ 0.0e+00

0.000000000000000

0.132038749000000

1.32039e-01

1.0e-01

0.001100000000000

0.012090857770000

1.09909e-02

2.0e-01

0.009600000000000

-0.043130061900000

5.27301e-02

3.0e-01

0.035100000000000

-0.033624010100000

6.87240e-02

4.0e-01

0.089600000000000

0.040609013300000

4.89910e-02

5.0e-01

0.187500000000000

0.179569008200000

7.93099e-03

6.0e-01

0.345600000000000

0.383255975000000

3.76560e-02

7.0e-01

0.583100000000000

0.651669913000000

6.85699e-02

8.0e-01

0.921600000000000

0.984810822000000

6.32108e-02

9.0e-01

1.385100000000000

1.382678703000000

2.42130e-03

1.0e+00

2.000000000000000

1.845273556000000

1.54726e-01

__________________________________________________________________

999


Consider 4th order Bessel's Polynomials, i.e. for 𝑛 = 4, we have Eq. (5.556) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥) + 𝑎2 𝐵2 (𝑥) + 𝑎3 𝐵3 (𝑥) + 𝑎4 𝐵4 (𝑥), or

𝑢(𝑥) = 𝑎1 + 𝑎1 (1 + 𝑥) + 𝑎2 (3𝑥 2 + 3𝑥 + 1) + 𝑎3 (15𝑥 3 + 15𝑥 2 + 6𝑥 + 1) + 𝑎4 (105𝑥 4 + 105𝑥 3 + 45𝑥 2 + 10𝑥 + 1).

(5.560)

Substituting Eq. (5.560) into Eq. (5.554) an proceeding as before we have the matrix form of Eq. (5.560) is −8.0000 −10.5000 [ ⋮ −129.75

−16.2000 −21.9666 ⋮ −461.816

⋯ −1711.5 a0 −12.3000 ⋯ −2445.4 ] [a1 ] = [−17.6333]. ⋯ ⋮ ⋮ ⋮ ⋯ −83008.7 a4 −611.913

After solving we get 𝛼0 = −0.2023, 𝛼1 = 0.3369, 𝑎2 = −0.1443, 𝑎3 = 0.0002, 𝑎4 = 0.0095. consequently we have the exact solution is 𝑢(𝑥) = 𝑥 3 + 𝑥 4 .

5.17.3. Volterra Integral Equation Consider the linear Volterra Integral equation [229] as 𝑥

𝑢(𝑥) = 6𝑥 + 3𝑥 2 − ∫0 𝑢(𝑡)d𝑡.

(5.561)

The exact solution of Eq. (5.561) 𝑢(𝑥) = 6𝑥.

(5.562)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥).

(5.563)

1000

Consider 1st order Bessel's Polynomials, i.e. for 𝑛 = 1, we have Eq. (5.563) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥), or

𝑢(𝑥) = 𝑎0 + 𝑎1 (1 + 𝑥).

(5.564)


𝑎0 + 𝑎1 (1 + 𝑥) = 6𝑥 + 3𝑥 2 − ∫0 [𝑎0 + 𝑎1 (1 + 𝑡)]d𝑡 .

(5.565)


1

𝑥

∫0 (𝑎0 + 𝑎1 (1 + 𝑥))𝐵𝑗 (𝑥)d𝑥 = ∫0 [6𝑥 + 3𝑥 2 − ∫0 [𝑎0 + 𝑎1 (1 + 𝑡)]d𝑡]𝐵𝑗 (𝑥)d𝑥, 𝑗 =

0,1.

(5.566)

The matrix form of Eq. (5.566) is 4.0000 1.5000 2.1666 𝛼0 [ ][ ] = [ ], 6.7500 2.3333 3.4583 𝛼1 after solving we geta0 = −6 and 𝛼1 = 6. Consequently we have the exact solution is 𝑢(𝑥) = 6𝑥.


3

𝑥

𝑢(𝑥) = 1 − 𝑥 − 2√𝑥 + 3 𝑥 2 + ∫0

1 √𝑥−𝑡

𝑢(𝑡)d𝑡.

(5.567)

The exact solution of Eq. (5.567) 𝑢(𝑥) = 1 − 𝑥.

(5.568)


(5.568)

Consider 1st order Bessel's Polynomials, i.e. for 𝑛 = 1, we have Eq. (5.568) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥), or

𝑢(𝑥) = 𝑎0 + 𝑎1 (1 + 𝑥).

(5.569)


3

𝑥

𝑎0 + 𝑎1 (1 + 𝑥) = 1 − 𝑥 − 2√𝑥 + 3 𝑥 2 + ∫0

1 √𝑥−𝑡

[𝑎0 + 𝑎1 (1 + 𝑡)]d𝑡.

(5.570)

Now multiply Eq. (5.570) by 𝐵𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from 0 to 1, we have

1001

1

1

4

3

𝑥

∫0 (𝑎0 + 𝑎1 (1 + 𝑥))𝐵𝑗 (𝑥)d𝑥 = ∫0 [1 − 𝑥 − 2√𝑥 + 3 𝑥 2 + ∫0 𝑎1 (1 + 𝑡)]) d𝑡] 𝐵𝑗 (𝑥)d𝑥, 𝑗 = 0,1

1 √𝑥−𝑡

([𝑎0 + (5.571)

The matrix form of Eq. (5.571) is −0.3333 −0.3667 𝛼0 −0.3000 [ ][ ] = [ ], −0.6333 −0.7143 𝛼1 −0.5523 after solving we get 𝛼0 = 2 and 𝛼1 = −1. Consequently we have the exact solution is 𝑢(𝑥) = 1 − 𝑥.

5.17.5. Abel Integral Equation Consider the nonlinear Abel integral equation [229] as 9

5

𝑥

𝑥 3 = ∫0 10

1 1 (𝑥−𝑡)3

𝑢2 (t)d𝑡.

(5.572)

The exact solution of Eq. (5.572) is 𝑢(𝑥) = √𝑥 .

(5.573)

Consider 𝑢2 (𝑥) = 𝑣(𝑥).

(5.574)

So we have Eq.( 5.572) of the form 9

5

𝑥

𝑥 3 = ∫0 10

1 1

𝑣(𝑡)d𝑡.

(5.575)

(𝑥−𝑡)3

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥).

(5.576)

Consider 1st order Bessel's Polynomials, i.e. for 𝑛 = 1, we have Eq. (5.576) is 𝑣(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥). 𝑣(𝑥) = 𝑎0 + 𝑎1 (1 + 𝑥).

or

(5.577)


5

𝑥

𝑥 3 = ∫0 10

1 1

(x−t)3

[𝑎0 + 𝑎1 (1 + 𝑡)]d𝑡.

(5.578)

Now multiply Eq. (5.578) by 𝐵𝑗 (𝑥), 𝑗 = 0,1 and integrating both side from 0 to 1, we have 1 9

∫0

5

1

𝑥

𝑥 3 𝐵𝑗 (𝑥)d𝑥 = ∫0 [∫0 10

1 1

(x−t)3

([𝑎0 + 𝑎1 (1 + 𝑡)])d𝑡] 𝐵𝑗 (𝑥)d𝑥,

The matrix form of Eq. (5.579) is 1002

(5.579)

[

0.9000 1.4625

1.2375 𝛼0 0.3375 ] [𝛼 ] = [ ], 2.0454 1 0.5829

after solving we get 𝛼0 = −1 and 𝛼1 = 1. Consequently we have the approximate solution is 𝑣(𝑥) = 𝑥.

(5.580)

Using Eq.( 5.580) we have solution 𝑢(𝑥) = ±√𝑥.

5.18. Bell Polynomials Method (BPM) We develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Bell polynomials [35]. Bell polynomials is given as 𝑑

𝐵𝑚 (𝑥) = (𝑥(𝑇𝐼𝑛 + 𝑑𝑥 𝑇𝐼𝑛 )

(5.581)

𝐵0 = 1, 𝐵1 = 𝑥, 𝐵2 = 𝑥(𝑥 + 1), 𝐵3 = 𝑥(𝑥 2 + 3𝑥 + 1), 𝐵4 = 𝑥(𝑥 3 + 6𝑥 2 + 7𝑥 + 1), . . ..



(5.582)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, [Lewis, 11], to find an approximate solution 𝑢̃(𝑥) of Eq. (5.582). For this, we assume that 1003

𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝐵𝑘 (𝑥),

(5.583)

where 𝐵𝑘 (𝑥) are Bell polynomials of degree 𝑘 defined in equation (5.581) and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.583) into (5.582), we get 𝑥

𝑓(𝑥) = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝐵𝑘 (𝑡)d𝑡.

(5.584)


𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝐵𝑗 (𝑥)d𝑥 = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐵𝑘 (𝑡)d𝑡)𝐵𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛.

(5.585)

Since in each equation, there are two integrals. The inner integrand of the left sides is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.585) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by 𝜆 ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.586)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐵𝑘 (𝑡)d𝑡)𝐵𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏


𝑗 = 0,1,2, … 𝑛.




(5.587)


𝑘, 𝑗 = 0,1,2, … 𝑛

where 1004

(5.588)

𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝐵𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝐵𝑘 (𝑡)d𝑡)𝐵𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏


𝑗 = 0,1,2, … 𝑛.


5.18.2. Volterra integral equation Consider the Volterra Integral equation [229] as 𝑥

𝑢(𝑥) = 1 − 𝑥 − ∫0 (𝑥 − 𝑡) 𝑢(𝑡)𝑑𝑡.

(5.589)

The exact solution of Eq. (5.589) is 𝑢(𝑥) = cos(𝑥) − sin(𝑥).

(5.590)


(5.591)

Consider 5th order Bell Polynomials, i.e. for 𝑛 = 5, we have Eq. (5.591) is 𝑢(𝑥) = ∑5𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥) + 𝑎2 𝐵2 (𝑥) + 𝑎3 𝐵3 (𝑥) +𝑎4 𝐵4 (𝑥) + 𝑎5 𝐵5 (𝑥). 4 3 𝑢(𝑥) = 𝑎0 + 𝑎1 (𝑥) + 𝑎2 (𝑥 2 + 𝑥) + 𝑎3 (𝑥 3 + 3𝑥 2 + 𝑥) + 𝑎4 (𝑥 +26𝑥 +) 7𝑥 + 𝑥

+𝑎5 (𝑥 5 + 10𝑥 4 + 25𝑥 3 + 15𝑥 2 + 𝑥).

(5.592)

Substituting Eq. (5.592) into Eq. (5.589) 𝑎0 + 𝑎1 (𝑥) + 𝑎2 (𝑥 2 + 𝑥) + 𝑎3 (𝑥 3 + 3𝑥 2 + 𝑥) + 𝑎4 (𝑥 4 + 6𝑥 3 + 7𝑥 2 + 𝑥) + 𝑥

𝑎5 (𝑥 5 + 10𝑥 4 + 25𝑥 3 + 15𝑥 2 + 𝑥) = 1 − 𝑥 − ∫0 (𝑥 − 𝑡) [𝑎0 + 𝑎1 (𝑡) + 𝑎2 (𝑡 2 + 𝑡) + 𝑎3 (𝑡 3 + 3𝑡 2 + 𝑡) + 𝑎4 (𝑡 4 + 6𝑡 3 + 7𝑡 2 + 𝑡) + 𝑎5 (𝑡 5 + 10𝑡 4 + 25𝑡 3 + 15𝑡 2 + 𝑡)]𝑑𝑡.

(5.593)

Now multiply Eq. (5.593) by 𝐵𝑗 (𝑥), 𝑗 = 0,1,2 and integrating both sides from 0 to 1, we have the matrix form of Eq. (5.594) is 1.166666667 [ ⋮ 18.40178572

⋯ ⋱ ⋯

𝛼0 0.5000000000 14.46726191 𝛼 1 0.1666666667 ][ ⋮ ] = [ ]. ⋮ ⋮ 427.4156363 𝛼 5 3.023809524

(5.594)

After solving we get, 𝛼0 = 0.9999607287, 𝛼1 = −0.268914403, …, 𝛼5 = −0.001402873352 . 1005

Consequently we have the approximate solution is 𝑢(𝑥) = 0.9999607287 − 0.9988406770𝑥 + ⋯ − 0.001402873352𝑥 5 . (5.595) Table 5.104. Comparison of Exact and Approximate solutions of Eq. (5.589) obtained from Bell polynomials Method (BPM) __________________________________________________________________ X

Exact Solution


Error

__________________________________________________________________ 0.0e+00

1.000000000000000

1.000000669000000

6.69410e-07

1.0e-01

0.895170748600000

0.895170497300000

2.51300e-07

2.0e-01

0.781397247000000

0.781397374300000

1.27300e-07

3.0e-01

0.659816282400000

0.659816455000000

1.72600e-07

4.0e-01

0.531642651700000

0.531642580300000

7.14000e-08

5.0e-01

0.398157023300000

0.398156836800000

1.86500e-07

6.0e-01

0.260693141500000

0.260693116000000

2.55000e-08

7.0e-01

0.120624500100000

0.120624674800000

1.74700e-07

8.0e-01

-0.020649381600000

-0.020649304670000

7.69300e-08

9.0e-01

-0.161716941300000

-0.161717155500000

2.14200e-07

1.0e+00

-0.301168678900000

-0.301168165300000

5.13600e-07


Now Consider 25th order Bell Polynomials, i.e. for 𝑛 = 25, we have Eq. (5.591) is 𝑢(𝑥) = ∑25 𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥) + ⋯ + 𝑎25 𝐵25 (𝑥). (5.596) 1006

Now multiply Eq. (5.596) by 𝐵𝑗 (𝑥), 𝑗 = 0,1, … ,25 and integrating both sides from 0 to 1, we have the matrix form of Eq. (5.597) is 1.166666667 [ ⋮ 6.271098472 1017

⋯ ⋱ ⋯

𝛼0 0.50000000000 4.452081399 1017 𝛼 1 0.16666666667 ][ ⋮ ] = [ ] ⋮ ⋮ 36 1.074254826 10 𝛼25 3.909136696 1016

After solving we get, 𝛼0 = 0.9999647312, 𝛼1 = −0.1029473149,…, 𝛼24 = 1.153886859 10−18 , 𝛼25 = −3.166048280 10−19 . Consequently we have the approximate solution is 𝑢(𝑥) = −0.008748652963𝑥 2 − 0.1184838938𝑥 3 + ⋯ + 07.003307393 10−12 𝑥 21 + 1.000648076𝑥.

(5.598)

Table 5.105. Comparison of Exact and Approximate solutions of Eq. (5.589) obtained from Bell polynomials Method (BPM) __________________________________________________________________ X

Exact Solution


Error

__________________________________________________________________ 0.0e+00

1.000000000000000

1.000000000000000

2.04977e-42

1.0e-01

0.895170748600000

0.895170748700000

1.00000e-10

2.0e-01

0.781397247000000

0.781397247000000

0.00000e+00

3.0e-01

0.659816282400000

0.659816282400000

0.00000e+00

4.0e-01

0.531642651700000

0.531642651900000

2.00000e-10

5.0e-01

0.398157023300000

0.398157023200000

1.00000e-10

6.0e-01

0.260693141500000

0.260693141500000

0.00000e+00

7.0e-01

0.120624500100000

0.120624500100000

0.00000e+00

8.0e-01

-0.020649381600000

-0.020649383000000

1.40000e-09

9.0e-01

-0.161716941300000

-0.161716942000000

7.00000e-10

1.0e+00

-0.301168678900000

-0.301168678000000

9.00000e-10

1007


Now Consider 50th order Bell Polynomials, i.e. for 𝑛 = 50, we have Eq. (5.591) is 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥) + 𝑎2 𝐵2 (𝑥) … . +𝑎50 𝐵50 (𝑥). Proceeding as before we have the matrix form is 1.16666666667 [ ⋮ 1.547960211 1046

⋯ ⋱ ⋯

𝛼0 0.50000000000 1.072211729 1046 𝛼 1 0.16666666667 ][ ⋮ ] = [ ](5.599) ⋮ ⋮ 1.019358344 1093 𝛼 50 5.8809464271044

After solving we get, 𝛼0 = 1.000004238, 𝛼1 = −0.9368270280, … , 𝛼49 = −4.172945015 10−46 , 𝛼50 = 3.0111826666 10−47. Consequently we have the approximate solution is 𝑢(𝑥) = 0.000105𝑥 25 − 3.0111 10−47 𝑥 50 + ⋯ + 2.3714 10−8 𝑥 29 +1.1.887 10−9 𝑥 45 .

(5.600)

1008


Exact Solution


Error

__________________________________________________________________ 0.0e+00

1.000000000000000

1.000000000000000

2.18346e-96

1.0e-01

0.895170748600000

0.895170748600000

0.00000e+00

2.0e-01

0.781397247000000

0.781397247000000

0.00000e+00

3.0e-01

0.659816282400000

0.659816282500000

1.00000e-10

4.0e-01

0.531642651700000

0.531642651700000

0.00000e+00

5.0e-01

0.398157023300000

0.398157023200000

1.00000e-10

6.0e-01

0.260693141500000

0.260693141500000

0.00000e+00

7.0e-01

0.120624500100000

0.120624500100000

0.00000e+00

8.0e-01

-0.020649381600000

-0.020649382000000

4.00000e-10

9.0e-01

-0.161716941300000

-0.161716941000000

3.00000e-10

1.0e+00

-0.301168678900000

-0.301168679000000

1.00000e-10


5.18.3. Weakly Singular Integral Equation Consider the weakly singular Integral equation [229] as 𝑢(𝑥) = 𝑥 2 −

128 45

9

𝑥

𝑥 4 + ∫0

1 3

𝑢(𝑡)d𝑡.

(𝑥−𝑡)4


(5.601)

𝑢(𝑥) = 𝑥 2 .

(5.602)


(5.603)

Consider 1st order Bell’s Polynomials, i.e. for 𝑛 = 1, we have Eq. (5.603) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥). Or

𝑢(𝑥) = 𝑎0 + 𝑎1 (1 + 𝑥).

(5.604)

Substituting Eq. (5.604) into Eq. (5.601) 𝑎0 + 𝑎1 (𝑥) = 𝑥 2 −

128 45

9

𝑥

𝑥 4 + ∫0

1 3

(𝑥−𝑡)4

[𝑎1 + 𝑎1 (𝑡)]d𝑡.

(5.605)

Now multiply Eq. (5.605) by 𝐵𝑗 (𝑥), 𝑗 = 0,1 and integrating both sides from 0 to 1, we have 1

1

16

5

1

𝑥

∫0 [𝑎0 + 𝑎1 (𝑥)]𝐵𝑗 (𝑥)d𝑥 = ∫0 (𝑥 2 − 15 𝑥 2 ) 𝐵𝑗 (𝑥)d𝑥 + ∫0 [∫0 𝑎1 (𝑡)]d𝑡] 𝐵𝑗 (𝑥)d𝑥, 𝑗 = 0,1.

1 3

(𝑥−𝑡)4

[𝑎0 +

(5.606)

The matrix form of Eq. (5.606) is −0.3333 −0.3667 𝛼0 0.02857 [ ][ ] = [ ]. −0.6333 −0.7142 𝛼1 0.04153 After solving we get, 𝛼0 = −0.88181039 and 𝛼1 = 0.72372. Consequently we have the approximate solution is 𝑢(𝑥) = −0.1865251865 + 1.06654628020𝑥.

1010


Exact Solution


Error

__________________________________________________________________ 0.0

0.000000000000000

-0.158086660900000

1.58087e-01

0.1

0.010000000000000

-0.085714287350000

9.57143e-02

0.2

0.040000000000000

-0.013341913800000

5.33419e-02

0.3

0.090000000000000

0.059030459700000

3.09695e-02

0.4

0.160000000000000

0.131402833300000

2.85972e-02

0.5

0.250000000000000

0.203775206900000

4.62248e-02

0.6

0.360000000000000

0.276147580400000

8.38524e-02

0.7

0.490000000000000

0.348519953900000

1.41480e-01

0.8

0.640000000000000

0.420892327500000

2.19108e-01

0.9

0.810000000000000

0.493264701100000

3.16735e-01

1.0

1.000000000000000

0.565637074600000

4.34363e-01


Consider 2nd order Bell’s Polynomials, i.e. for 𝑛 = 2, we have Eq. (5.603) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥) + 𝑎2 𝐵2 (𝑥). Or

𝑢(𝑥) = 𝑎0 + 𝑎1 (1 + 𝑥) + 𝑎2 (1 + 𝑥 2 + 2𝑥).

Substituting Eq. (5.607) into Eq. (5.601)

1011

(5.607)

𝑎0 + 𝑎1 (𝑥) + 𝑎2 (𝑥 2 + 𝑥) = 𝑥 2 −

128 45

9

𝑥

𝑥 4 + ∫0

1 3 (𝑥−𝑡)4

[

𝑎1 + 𝑎1 (𝑡) + ] d𝑡. 𝑎2 (𝑡 2 + 𝑡)

(5.608)

Now multiply Eq. (5.608) by 𝐵𝑗 (𝑥), 𝑗 = 0,1,2 and integrating both sides from 0 to 1, we have the matrix form of Eq. (5.609) is 0.0285714285 −0.37142857 𝛼0 𝛼 −0.75370370 ] [ 1 ] = [ 0.0415343915], − 1.46060606 𝛼2 0.0605579605

−0.33333333 −0.36666667 [ −0.63333333 −0.71428571 −1.17142857 − 1.34629629 After solving we get, 𝛼0 = 1 and 𝛼1 = −2, 𝛼2 = 1 .

Consequently we have the approximate solution is 𝑢(𝑥) = 𝑥 2 .

5.18.4. Fredholm Integral Equation Consider the Fredholm Integral equation [229] as 1

𝑢(𝑥) = 𝑥 + 𝑥 4 + 9𝑒 𝑥+1 −23𝑒 𝑥 − ∫0 𝑒 𝑥+𝑡 𝑢(𝑡)𝑑𝑡.

(5.610)

The exact solution of Eq. (5.610) is 𝑢(𝑥) = 𝑥 + 𝑥 4 .

(5.611)


(5.612)

Consider 5th order Bell Polynomials, i.e. for 𝑛 = 5, we have Eq. (5.612) is 𝑢(𝑥) = ∑5𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥) + 𝑎2 𝐵2 (𝑥) + 𝑎3 𝐵3 (𝑥) +𝑎4 𝐵4 (𝑥) + 𝑎5 𝐵5 (𝑥). 5 4 𝑢(𝑥) = 𝑎0 + 𝑎1 (𝑥) + 𝑎2 (𝑥 2 + 𝑥) + ⋯ + 𝑎5 ( 𝑥 3 + 10𝑥 2 + ). (5.613) 25𝑥 + 15𝑥 + 𝑥

Substituting Eq. (5.613) into Eq. (5.610) 𝑎0 + 𝑎1 (𝑥) + 𝑎2 (𝑥 2 + 𝑥) + ⋯ + 𝑎5 (𝑥 5 + 10𝑥 4 + 25𝑥 3 + 15𝑥 2 + 𝑥) = 2 − 1

2𝑥 + 5𝑥 4 + 7𝑥 5 + ∫−1(𝑥 − 𝑡) [𝑎0 + 𝑎1 (𝑡) + 𝑎2 (𝑡 2 + 𝑡) + ⋯ + 𝑎5 (𝑡 5 + 10𝑡 4 + 25𝑡 3 + 15𝑡 2 + 𝑡)]𝑑𝑡.

(5.614)

Now multiply Eq. (5.614) by 𝐵𝑗 (𝑥), 𝑗 = 0,1,2,3,4,5 and integrating both sides from 0 to 1, we have the matrix form of Eq. (5.615) is

1012

3.952492442 [ ⋮ 67.01346547

𝛼0 3.216486380 67.01346547 𝛼 1 1.964536456 ][ ⋮ ] = [ ]. ⋮ ⋮ 1363.480802 𝛼 5 62.79427971

⋯ ⋱ ⋯

After solving we get, 𝛼0 = 0, 𝛼1 = −5, 𝛼2 = 11, 𝛼3 = −6, 𝛼4 = 1, 𝛼5 = 0. Consequently we have the approximate solution is 𝑢(𝑥) = 𝑥 + 𝑥 4 .

5.18.5. Nonlinear Abel Integral Equation Consider the non-linear Abel Integral equation [229] as 11

36 55

𝑥

𝑥 6 = ∫0

1 1 (𝑥−𝑡)6

cos−1 (𝑢(𝑡)) 𝑑𝑡.

(5.616)

The exact solution of Eq. (5.616) is 𝑢(𝑥) = 𝑥.

(5.617)


(5.618)

Consider 1st order Bell Polynomials, i.e. for 𝑛 = 1, we have Eq. (5.618) is 𝑢(𝑥) = ∑1𝑘=0 𝛼𝑘 𝐵𝑘 (𝑥) = 𝑎0 𝐵0 (𝑥) + 𝑎1 𝐵1 (𝑥). Or

𝑢(𝑥) = 𝑎0 + 𝑎1 (𝑥).

(5.619)


11

𝑥

𝑥 6 = ∫0 55

1 1 (𝑥−𝑡)6

cos −1 (𝑎0 + 𝑎1 (𝑡)) 𝑑𝑡.

(5.620)

Now multiply Eq. (44) by 𝐵𝑗 (𝑥), 𝑗 = 0,1 and integrating both sides from 0 to 1, we have 1 36

11

1

𝑥

∫0 [55 𝑥 6 ] 𝐵𝑗 (𝑥)d𝑥 = ∫0 [∫0

1 1

(𝑥−𝑡)6

cos −1(𝑎0 + 𝑎1 (𝑡)) 𝑑𝑡] 𝐵𝑗 (𝑥)d𝑥, 𝑗 = 0,1.

The matrix form of Eq. (5.621) is 36

[55 36 85

216

216 𝛼0 935 935 216 ] [𝛼1 ]=[ 216 ]. 1265

1265

After solving we get, 𝛼0 = 0, 𝛼1 = 1. Consequently we have the approximate solution is 1013

(5.621)

𝑢(𝑥) = 𝑥.

5.19. Appell Polynomials Method (APM) We develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Appell polynomials [107]. Appell polynomials is given as 𝑛!

𝑛−𝑘 𝐴𝑚 (𝑥) = ∑𝑚 . 𝑘=0 (𝑛−𝑘)!𝑘! 𝑥

(5.622)

𝐴0 = 1, 𝐴1 = 1 + 𝑥, 𝐴2 = 1 + 2𝑥 + 𝑥 2 , 𝐴3 = 1 + 3𝑥 2 + 3𝑥 + 𝑥 3 , 𝐴4 = 1 + 4𝑥 3 + 6𝑥 2 + 4𝑥 + 𝑥 4 , . . ..



(5.623)

where 𝑢(𝑡) is the unknown function, to be determined, 𝑘(𝑥, 𝑡), the kernel, is a continuous or discontinuous and square integrable function, 𝑓(𝑥) being the known function. Now we use the technique of Galerkin method, [Lewis, 11], to find an approximate solution 𝑢̃(𝑥) of Eq. (2). For this, we assume that 𝑢̃(𝑥) = ∑𝑛𝑘=0 𝑎𝑘 𝐴𝑘 (𝑥),

(5.624)

where 𝐴𝑘 (𝑥) are Appell’s polynomials of degree 𝑘 defined in equation (5.622) and 𝑎𝑘 are unknown parameters, to be determined. Substituting (5.624) into (5.623), we get 𝑥

𝑓(𝑥) = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 𝐾(𝑥, 𝑡)𝐴𝑘 (𝑡)d𝑡.

1014

(5.625)


𝑏

𝑥

∫𝑎 𝑓(𝑥) 𝐴𝑗 (𝑥)d𝑥 = 𝜆 ∑𝑛𝑘=0 𝑎𝑘 ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐴𝑘 (𝑡)d𝑡 )𝐴𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2, … 𝑛.

(5.626)

Since in each equation, there are two integrals. The inner integrand of the left sides is a function of 𝑥, and 𝑡, and is integrated with respect to 𝑡 from 𝑎 to 𝑥. As a result the outer integrand becomes a function of 𝑥 only and integration with respect to 𝑥 from 𝑎 to 𝑏 yields a constant. Thus for each 𝑗 = 0,1,2, … , 𝑛 we have a linear equation with 𝑛 + 1 unknowns 𝑎𝑘 , 𝑘 = 0,1,2, … , 𝑛. Finally (5.626) represents the system of 𝑛 + 1 linear equations in 𝑛 + 1 unknowns, are given by 𝜆 ∑𝑛𝑘=0 𝑎𝑘 𝐾𝑘,𝑗 = 𝐹𝑗 ; 𝑘, 𝑗 = 0,1,2, … 𝑛

(5.627)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (∫𝑎 𝐾(𝑥, 𝑡)𝐴𝑘 (𝑡)d𝑡)𝐴𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏


𝑗 = 0,1,2, … 𝑛.




(5.628)


𝑘, 𝑗 = 0,1,2, … 𝑛

(5.629)

where 𝑏

𝑥

𝐾𝑘,𝑗 = ∫𝑎 (𝑐(𝑥)𝐴𝑘 (𝑥) + 𝜆 ∫𝑎 𝐾(𝑥, 𝑡)𝐴𝑘 (𝑡)d𝑡 )𝐴𝑗 (𝑥)d𝑥, 𝑘, 𝑗 = 0,1,2, … 𝑛, and

𝑏


𝑗 = 0,1,2, … 𝑛.

1015


5.19.2. Fredholm Integral Equation Example 5.67. Consider the Fredholm Integral equation [229] as 1

𝑢(𝑥) = −15 + 10𝑥 3 + 𝑥 4 − ∫0 (20𝑥 3 − 56𝑡 3 ) 𝑢(𝑡)𝑑𝑡.

(5.630)

The exact solution of Eq. (5.630) is 𝑢(𝑥) = 𝑥 3 + 𝑥 4 .

(5.631)

According to the proposed technique, consider the trail solution 𝑢(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥).

(5.632)

Consider 2nd order Appell’s Polynomials, i.e. for 𝑛 = 2, we have Eq. (5.632) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥) = 𝑎0 𝐴0 (𝑥) + 𝑎1 𝐴1 (𝑥) + 𝑎2 𝐴2 (𝑥) . 𝑢(𝑥) = 𝑎0 + 𝑎1 (1 + 𝑥) + 𝑎2 (1 + 𝑥 2 + 2𝑥).

(5.633)


𝑎0 + 𝑎1 (1 + 𝑥) + 𝑎2 (1 + 𝑥 2 + 2𝑥) = −15 + 10𝑥 3 + 𝑥 4 − ∫0 (20𝑥 3 − 56𝑡 3 ) [𝑎0 + 𝑎1 (1 + 𝑡) + 𝑎2 (1 + 𝑡 2 + 2𝑡)]𝑑𝑡.

(5.634)

Now multiply Eq. (5.634) by 𝐴𝑗 (𝑥), 𝑗 = 0,1,2 and integrating both sides from 0 to 1, we have the matrix form of Eq. (5.634) is −8.000000111 −16.2000000 [ −10.50000000 21.966666670 −14.00000000 − 30.55000000

12.30000000 −31.73333334 𝛼0 𝛼 −43.85000000 ] [ 1 ] = [ 17.63333333]. − 62.40000000 𝛼2 26.15714286

After solving we get, 𝛼0 = 4.891533926, 𝛼1 = −7.995854485, 𝛼2 = 3.236362539. Consequently we have the approximate solution is 𝑢(𝑥) = 0.1320419800 − 1.523129407𝑥 + 3.236362539𝑥 2 .

1016

Table 5.108. Comparison of Exact and Approximate solutions of Eq. (5.630) obtained from Appell polynomials Method (APM) __________________________________________________________________ 𝑥

Exact Solution


Error

__________________________________________________________________ 0.0e+00

0.000000000000000

0.132037815100000

1.32038e-01

1.0e-01

0.001100000000000

0.012090336080000

1.09903e-02

2.0e-01

0.009600000000000

-0.043130252100000

5.27303e-02

3.0e-01

0.035100000000000

-0.033623949600000

6.87239e-02

4.0e-01

0.089600000000000

0.040609243600000

4.89908e-02

5.0e-01

0.187500000000000

0.179569327700000

7.93067e-03

6.0e-01

0.345600000000000

0.383256302900000

3.76563e-02

7.0e-01

0.583100000000000

0.651670168100000

6.85702e-02

8.0e-01

0.921600000000000

0.984810924000000

6.32109e-02

9.0e-01

1.385100000000000

1.382678572000000

2.42143e-03

1.0e+00

2.000000000000000

1.845273109000000

1.54727e-01


Consider 4th order Appell’s Polynomials, i.e. for 𝑛 = 4, we have Eq. (5.632) is 𝑢(𝑥) = ∑4𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥) = 𝑎0 𝐴0 (𝑥) + 𝑎1 𝐴1 (𝑥) + 𝑎2 𝐴2 (𝑥) + 𝑎3 𝐴3 (𝑥) + +𝑎4 𝐴4 (𝑥).

1017

Substituting Eq. (14) into Eq. (5.630) and proceeding as before we get the matrix form of Eq. (17) is −8.00000 −16.2000 −31.733333 −61.100000 −10.5000 21.96666 −43.850000 −85.450000 −14.0000 −30.55000 −62.400000 −123.31666 −18.8928 −43.51425 −91.333333 −183.39285 [ −25.6714 −63.34714 −137.23714 −280.46285

−116.600 𝛼0 12.300 𝛼 −164.400 1 17.633 −239.414 𝛼2 = 26.157 . 40.067 −359.771 𝛼3 [ ] [ 𝛼 ] 4 63.200 ] −556.207

After solving we get, 𝛼0 = 0, 𝛼1 = −1, 𝛼2 = 3, 𝛼3 = −3, 𝛼4 = 1. Consequently we have the approximate solution is 𝑢(𝑥) = 𝑥 3 + 𝑥 4 .

Example 5.68. Consider the Fredholm Integral equation 1

𝑢(𝑥) = 𝑥 + 𝑥 4 + 9𝑒 𝑥+1 −23𝑒 𝑥 − ∫0 𝑒 𝑥+𝑡 𝑢(𝑡)𝑑𝑡.

(5.635)

The exact solution of Eq. (5.635) is 𝑢(𝑥) = 𝑥 + 𝑥 4 .

(5.636)


(5.637)

Consider 5th order Appell’s Polynomials, i.e. for 𝑛 = 5, we have Eq. (5.637) is 𝑢(𝑥) = ∑5𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥) = 𝑎0 𝐴0 (𝑥) + 𝑎1 𝐴1 (𝑥) + 𝑎2 𝐴2 (𝑥) + 𝑎3 𝐴3 (𝑥) + 𝑎4 𝐴4 (𝑥) + 𝑎5 𝐴5 (𝑥) + 𝑎6 𝐴6 (𝑥). 5 4 + ). 𝑢(𝑥) = 𝑎0 + 𝑎1 (1 + 𝑥) + 𝑎2 (1 + 𝑥 2 + 2𝑥) + ⋯ + 𝑎5 ( 1 +3 𝑥 + 5𝑥 2 10𝑥 + 10𝑥 + 5𝑥

(5.638)


2 − 2𝑥 + 5𝑥 4 + 7𝑥 5 + ∫−1(𝑥 − 𝑡) [𝑎0 + 𝑎1 (𝑥) + 𝑎2 (𝑥 2 + 𝑥) + ⋯ + 𝑎5 (1 + 𝑥 5 + 5𝑥 4 + 10𝑥 3 + 10𝑥 2 + 5𝑥)]𝑑𝑡.

(5.639)

Now multiply Eq. (5.639) by 𝐴𝑗 (𝑥), 𝑗 = 0,1 and integrating both sides from 0 to 1, we have 1 𝑎0 ∫0 [ +𝑎 1

1

+ 𝑎1 (1 + 𝑥) + 𝑎2 (1 + 𝑥 2 + 2𝑥) + ⋯ 1 2 − 2𝑥 + ] 𝐴𝑗 (𝑥)d𝑥 = ∫0 ( 4 ) 𝐴 (𝑥)d𝑥 − 5 4 3 2 5𝑥 + 7𝑥 5 𝑗 (1 + 𝑥 + 5𝑥 + 10𝑥 + 10𝑥 + 5𝑥) 5

∫0 [∫−1(𝑥 − 𝑡) [

𝑎0 + 𝑎1 (1 + 𝑡) + 𝑎2 (1 + 𝑡 2 + 2𝑡) + ⋯ ] 𝑑𝑡] 𝐴𝑗 (𝑥)d𝑥, (5.640) +𝑎5 (1 + 𝑡 5 + 5𝑡 4 + 10𝑡 3 + 10𝑡 2 + 5𝑡)

The matrix form of Eq. (5.640) is

1018

3.952492442 [ ⋮ 48.73820629

𝛼0 3.216486380 48.73820629 𝛼 1 5.181022836 ][ ⋮ ] = [ ]. ⋮ ⋮ 681.3200922 𝛼 5 44.60173884

⋯ ⋱ ⋯

After solving we get, 𝛼0 = 0, 𝛼1 = −3, 𝛼2 = 6, 𝛼3 = −4, 𝛼4 = 1, 𝛼5 = 0. Consequently we have the approximate solution is, 𝑢(𝑥) = 𝑥 + 𝑥 4 .

5.19.3. Volterra Integral Equation Consider the Volterra Integral equation [229] as 𝑥

𝑢(𝑥) = 1 − ∫0 (𝑥 − 𝑡) 𝑢(𝑡)𝑑𝑡.

(5.641)


(5.642)


(5.643)

Consider 5th order Appell’s Polynomials, i.e. for 𝑛 = 5, we have Eq. (5.643) is 𝑢(𝑥) = ∑5𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥) = 𝑎0 𝐴0 (𝑥) + 𝑎1 𝐴1 (𝑥) + 𝑎2 𝐴2 (𝑥) + 𝑎3 𝐴3 (𝑥) + +𝑎4 𝐴4 (𝑥) + 𝑎5 𝐴5 (𝑥). Substituting Eq. (5.643) into Eq. (5.641) 𝑎0 + 𝑎1 (1 + 𝑥) + 𝑎2 (1 + 𝑥 2 + 2𝑥) + ⋯ + 𝑎5 (1 + 𝑥 5 + 5𝑥 4 + 10𝑥 3 + 10𝑥 2 + 𝑥

5𝑥) = 1 − ∫0 (𝑥 − 𝑡) [𝑎0 + 𝑎1 (1 + 𝑡) + 𝑎2 (1 + 𝑡 2 + 2𝑡) + ⋯ + 𝑎5 (1 + 𝑡 5 + 5𝑡 4 + 10𝑡 3 + 10𝑡 2 + 5𝑡)]𝑑𝑡.

(5.644)

Now multiply Eq. (5.644) by 𝐴𝑗 (𝑥), 𝑗 = 0,1,2,3,4,5 and integrating both sides from 0 to 1, we have 1

∫0 [𝑎0 + 𝑎1 (1 + 𝑥) + 𝑎2 (1 + 𝑥 2 + 2𝑥) + ⋯ + 𝑎5 (1 + 𝑥 5 + 5𝑥 4 + 10𝑥 3 + 1

1

𝑥

10𝑥 2 + 5𝑥)]𝐴𝑗 (𝑥)d𝑥 = ∫0 𝐴𝑗 (𝑥) d𝑥 − ∫0 [∫0 (𝑥 − 𝑡) [𝑎0 + 𝑎1 (1 + 𝑡) + 𝑎2 (1 + 𝑡 2 + 2𝑡) + ⋯ + 𝑎5 (1 + 𝑡 5 + 5𝑡 4 + 10𝑡 3 + 10𝑡 2 + 5𝑡)]𝑑𝑡]𝐴𝑗 (𝑥)d𝑥, The matrix form of Eq. (5.645) is 1.166666667 [ ⋮ 13.54464286

⋯ ⋱ ⋯

𝛼0 1.000000000 11.15178571 𝛼 1 1.500000000 ][ ⋮ ] = [ ]. ⋮ ⋮ 199.5689311 𝛼 5 10.50000000 1019

(5.645)

After solving we get, 𝛼0 = 0.492648276, 𝛼1 = 0.999515950 …, 𝛼5 = 0.00513664243. Consequently we have the approximate solution is, 𝑢(𝑥) = 0.9999668187 + 0.0009972422000𝑥 + ⋯ + 0.005131664243𝑥 5 . Table 5.109. Comparison of Exact and Approximate solutions of Eq. (5.641) obtained from Appell polynomials Method (APM) __________________________________________________________________ x

Exact Solution


Error

__________________________________________________________________ 0.0e+00

1.000000000000000

1.000001332000000

1.33165e-06

1.0e-01

0.995004165300000

0.995003647100000

5.18200e-07

2.0e-01

0.980066577800000

0.980066812800000

2.35000e-07

3.0e-01

0.955336489100000

0.955336872600000

3.83500e-07

4.0e-01

0.921060994000000

0.921060880400000

1.13600e-07

5.0e-01

0.877582561900000

0.877582152200000

4.09700e-07

6.0e-01

0.825335614900000

0.825335517500000

9.74000e-08

7.0e-01

0.764842187300000

0.764842571600000

3.84300e-07

8.0e-01

0.696706709300000

0.696706926600000

2.17300e-07

9.0e-01

0.621609968300000

0.621609463200000

5.05100e-07

1.0e+00

0.540302305900000

0.540303583000000

1.27710e-06


Now Consider 25th order Appell’s Polynomials, i.e. for 𝑛 = 25, we have Eq. (5.643) is 1020

𝑢(𝑥) = ∑25 𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥) = 𝑎0 𝐴0 (𝑥) + 𝑎1 𝐴1 (𝑥) + 𝑎2 𝐴2 (𝑥) … . +𝑎25 𝐴25 (𝑥). Proceeding as before we have 1.166666667 [ ⋮ 3.694128515 106

⋯ ⋱ ⋯

𝛼0 1.00000000000 2.594766762 106 𝛼 1 1.50000000000 ][ ⋮ ] = [ ]. ⋮ ⋮ 13 4.439502739 10 𝛼25 2.581110115 106

After solving we get, 𝛼0 = 0.7678382600, 𝛼1 = 0.3261059426 ,…, 𝛼24 = −4.117159982 10−9 , 𝛼25 = 1.297664685 10−8 . Consequently we have the approximate solution is, 𝑢(𝑥) = −0.005149426944𝑥 − 0.4498178557𝑥 2 + ⋯ + 1.297664685 10−8 𝑥 25 + 1.000120809. Table 5.110. Comparison of Exact and Approximate solutions of Eq. (5.641) obtained from Appell polynomials Method (APM) __________________________________________________________________ X

Exact Solution


Error

__________________________________________________________________ 0.0e+00

1.000000000000000

1.000000000000000

4.39974e-42

1.0e-01

0.995004165300000

0.995004165300000

0.00000e+00

2.0e-01

0.980066577800000

0.980066577800000

0.00000e+00

3.0e-01

0.955336489100000

0.955336489100000

0.00000e+00

4.0e-01

0.921060994000000

0.921060994000000

0.00000e+00

5.0e-01

0.877582561900000

0.877582561900000

0.00000e+00

6.0e-01

0.825335614900000

0.825335614900000

0.00000e+00

7.0e-01

0.764842187300000

0.764842187300000

0.00000e+00

8.0e-01

0.696706709300000

0.696706709300000

0.00000e+00

9.0e-01

0.621609968300000

0.621609968300000

0.00000e+00

1.0e+00

0.540302305900000

0.540302305900000

0.00000e+00

1021


Now Consider 50th order Appell’s Polynomials, i.e. for 𝑛 = 50, we have Eq. (5.643) is 𝑢(𝑥) = ∑50 𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥) = 𝑎0 𝐴0 (𝑥) + 𝑎1 𝐴1 (𝑥) + 𝑎2 𝐴2 (𝑥) … . +𝑎50 𝐴50 (𝑥). The matrix form for 𝑛 = 50, is 1.16666666667 [ ⋮ 6.459529901 1013

⋯ ⋱ ⋯

𝛼0 1.000000000000 4.421702016 1013 𝛼 1 1.500000000000 ][ ⋮ ] = [ ]. ⋮ ⋮ 28 2.513911814 10 𝛼50 4.415293752 1013

After solving we get, 𝛼0 = 0.6194221220, 𝛼1 = 0.6285767987 , … , 𝛼49 = −6.534147506 10−16 , 𝛼50 = −1.100609211 10−15 . Consequently we have the approximate solution is, 𝑢(𝑥) = −0.00000866 𝑥 40 − 0.00000228 𝑥 41 + ⋯ + 0.0250623𝑥19 −0.0092102𝑥 20 .

1022

Table 5.111. Comparison of Exact and Approximate solutions of Eq. (5.641) obtained from Appell polynomials Method (APM) __________________________________________________________________ X

Exact Solution


Error

__________________________________________________________________ 0.0e+00

1.000000000000000

1.000000000000000

7.65644e-97

1.0e-01

0.995004165300000

0.995004165300000

0.00000e+00

2.0e-01

0.980066577800000

0.980066577900000

1.00000e-10

3.0e-01

0.955336489100000

0.955336489100000

0.00000e+00

4.0e-01

0.921060994000000

0.921060994100000

1.00000e+00

5.0e-01

0.877582561900000

0.877582561900000

0.00000e+00

6.0e-01

0.825335614900000

0.825335614900000

0.00000e+00

7.0e-01

0.764842187300000

0.764842187300000

0.00000e+00

8.0e-01

0.696706709300000

0.696706709300000

0.00000e+00

9.0e-01

0.621609968300000

0.621609968300000

0.00000e+00

1.0e+00

0.540302305900000

0.540302305900000

0.00000e+00

Fig 5.111. Comparison of Exact and Approximate solutions of Eq. ( 5.641) for n= 50

5.19.4. Abel Integral Equation Consider the generalized nonlinear Abel Integral equation [229] as 3 40

2

𝑥

𝑥 3 (20 + 24𝑥 + 9𝑥 2 ) = ∫0

1 1

𝑢2 (𝑡)𝑑𝑡.

(𝑥−𝑡)3

1023

(5.646)

The exact solution of Eq. (5.646) is 𝑢(𝑥) = 𝑥.

(5.647)

The transformation 𝑣(𝑥) = 𝑢2 (𝑥), 𝑢(𝑥) = ± √𝑣(𝑥).

(5.648)

According to the proposed technique, consider the trail solution 𝑣(𝑥) = ∑𝑛𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥).

(5.649)

Consider 2nd order Appell’s Polynomials, i.e. for 𝑛 = 2, we have Eq. (5.649) is 𝑢(𝑥) = ∑2𝑘=0 𝛼𝑘 𝐴𝑘 (𝑥) = 𝑎0 𝐴0 (𝑥) + 𝑎1 𝐴1 (𝑥) + 𝑎2 𝐴2 (𝑥) . Or

𝑢(𝑥) = 𝑎0 + 𝑎1 (1 + 𝑥) + 𝑎2 (1 + 𝑥 2 + 2𝑥).

(5.650)

Substituting Eq. (5.650) into Eq. (5.646) 3 40

2

𝑥

𝑥 3 (20 + 24𝑥 + 9𝑥 2 ) = ∫0

1 1

(𝑥−𝑡)3

2 𝑎 + 𝑎1 (1 + 𝑡) [ 0 ] 𝑑𝑡. +𝑎2 (1 + 𝑡 2 + 2𝑡)

(5.651)

Now multiply Eq. (5.651) by 𝐴𝑗 (𝑥), 𝑗 = 0,1,2 and integrating both sides from 0 to 1, we have 1

3

2

1

𝑥

∫0 [40 𝑥 3 (20 + 24𝑥 + 9𝑥 2 )] 𝐴𝑗 (𝑥)d𝑥 = ∫0 [∫0 𝑎2 (1 + 𝑡 2 + 2𝑡)]2 𝑑𝑡] 𝐴𝑗 (𝑥)d𝑥, 𝑗 = 0,1,2.

1 1

(𝑥−𝑡)3

[𝑎0 + 𝑎1 (1 + 𝑡) + (5.652)

The matrix form of Eq. (5.652) is 0.900000000 1.237500000 [1.462500000 2.045454545 2.434090909 3.455357143

1.759090909 1.759090909 𝛼0 𝛼 2.957142857] [ 1 ] = [ 2.957142857]. 5.069117647 𝛼2 5.069117647

After solving we get, 𝛼0 = 0, 𝛼1 = 0, 𝛼2 = 1. Consequently we have the approximate solution is, 𝑢(𝑥) = 1 + 𝑥 2 + 2𝑥.

1024

Chapter 6 Wavelets Theory

1025

6.1. Introduction Wavelets theory is a relatively new and emerging area in mathematical research. As a powerful tool, wavelets have been extensively used in signal processing, numerical analysis and many other areas. Wavelets theory is a relatively new and an emerging area in mathematical research. Wavelets allow the accurate representation of a variety of functions and operators. Moreover, wavelets establish a connection with fast numerical algorithms. Hafshejani et al. [69] applied the Legendre wavelet methods to obtain the numerical solution of Delay differential equations. Soleymanivaraki et al. [196] implement the Lengendre wavelets method to solve fractional order differential equations. Razzaghi and Yousefi used Legendre wavelets method [188-190]. In addition, Chebyshev, Legender and Shannon wavelet method have been used for obtaining the numerical solutions of the Fredholm integral equation in [175]. Kajani and Mahdavi [106] solved the nonlinear integral equations by using Galerkin method with hybrid Block-Pulse function. Mohammadi et al. [139] used Legendre wavelet Galerkin method for solving the ordinary differential equations. Hosseini et al. [91] solved the telegraph equations with a new Rothe-wavelet method. Venkatesh et al. [209] used Legendre Wavelets Based Approximation Method for Cauchy Problems. In this Chapter, I modified original Legendre wavelets method by inserting different polynomial which gives us very reliable solutions. Physicists Hermite Wavelet Method (PHWM), is used for solving singular ordinary differential equations Newton’s Wavelet Method (NWM) is implemented for higher order BVPs, Laurent Wavelet Method (LWM) was used to solve the system of ordinary differential equations, Kravchuk Wavelet Method (KWM) for solving integral equations, Lommal Wavelet Method (LWM) applying to find the exact solutions of Delay differential equations, Bernoulli’s Wavelet Method (BWM) was applied to solve second-order differential equations with Robin conditions, Probabilist Hermite Wavelet Method (PWM) is used to construct the approximate solution of nonlinear integral equations, Zernike Wavelet Method (ZWM) stiff Delay differential equation and difference Differential equation, Rook Wavelet Method (RWM) for solving Fractional differential equation and Touchard Wavelet Method (TWM) is applied to find the exact and approximate solution of Partial differential equation. It is to be observe that the proposed technique has been applied on a 1026

wide range of nonlinear diversified physical problems including, high-dimensional nonlinear evolution equation. It is to be highlighted that suggested algorithm is extremely simple but highly effective and may be extended to other singular problems of diversified physical nature. Moreover, this new scheme is capable of reducing the computational work to a tangible level while still maintaining a very high level of accuracy.

6.2. Physicists Hermite Wavelet Method (PHWM) We proposed a new algorithm by inserting Physicists Hermite polynomials in traditional Legendre wavelets method. This technique is successively applied to find the exact solution singular ordinary differential equations. The proposed technique is very simple and highly compatible for solving such kind of problems.

6.2.1. Methodology 6.2.1.1. Physicists Hermite Polynomials Physicists Hermite polynomial, named after Charles Hermite (1864) although they were studied earlier by Laplace and Chebyshev polynomials 𝑥2

𝑑𝑛

𝑥2

𝑃𝐻𝑛 (𝑥) = (−1)𝑛 e 2 . 𝑑𝑥 𝑛 e− 2

(6.1)

and in which 𝑝0 (𝑥) is a non-zero constant, few Physicists Hermite polynomials given below 𝑃𝐻0 = 1, 𝑃𝐻1 = 2𝑥, 𝑃𝐻2 = −2 + 4𝑥 2 , 𝑃𝐻3 = 4𝑥(−3 + 2𝑥 2 ), 𝑃𝐻4 = 12 − 48𝑥 2 + 16𝑥 4 , . . ..

6.2.1.2. Wavelets and Physicists Hermite Wavelets In recent years, wavelets have found their way into many different fields of science and engineering [137-140]. Wavelets constitute a family of functions constructed from the dilation and translation of a single function called the mother wavelet. When the dilation 1027

parameter a and the translation parameter 𝑏 vary continuously, we have the following family of continuous wavelets: 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|− 2 𝜓 (

𝑎

) 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

(6.2)

Physicists Hermite wavelets 𝜓𝑛𝑚 (𝑡) = 𝜓(𝑘, 𝑛, 𝑚, 𝑡) have four arguments; 𝑛, 𝑘 can assume any positive integer, 𝑚 is the order for Physicists Hermite polynomials and t is the normalized time. They are defined on the interval [0, 1) by; 1

𝜓𝑛𝑚

𝑘

𝑘 2 (𝑡) = {√𝑚 + 2 2 𝑃𝐻𝑚 (2 𝑡 − (2𝑛 + 1))

0

𝑛 2𝑘

≤𝑡≤

𝑛+1 2𝑘

(6.3)

otherwise 1

where 𝑚 = 0,1,2, … , 𝑀 − 1 and 𝑛 = 0,1, … , 2𝑘−1 The coefficient (𝑚 + 2) is for orthonormality. Here 𝑃𝐻𝑚 (𝑡) are the well-known Physicists Hermite polynomials of order 𝑚, which have been previously described.

6.2.1.3. Functions Approximation A function 𝑓(𝑡) define over [0,1) may be expanded as ∞ 𝑓(𝑡) = ∑∞ 𝑛=1 ∑𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡),

(6.4)

where 𝑐𝑛𝑚 = 〈𝑓(𝑡), 𝜓𝑛𝑚 (𝑡)〉, in which 〈∙,∙〉 denotes the inner product. If the infinite series in Eq. (6.4) is truncated, then Eq. (6.4) can be written as 𝑘−1

𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡).

(6.5)


𝑇 𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡) = 𝐶 𝜓(𝑡).

(6.6)

where 𝐶 and 𝜓(𝑡) are 2𝑘−1 𝑀 × 1 matrices given as

and

𝑐10 , 𝑐11 , … , 𝑐1𝑀−1 , 𝑐20 , 𝑐21 , … , 𝑇 𝐶 = [𝑐 ] , 2𝑀−1 , … , 𝑐2𝑘−1 0 , … , 𝑐2𝑘−1 𝑀−1

(6.7)

𝜓10 , 𝜓11 , … , 𝜓1𝑀−1 , 𝜓20 , 𝜓21 , … , 𝑇 𝜓(𝑡) = [ ] . 𝜓2𝑀−1 , … , 𝜓2𝑘−1 0 , … , 𝜓2𝑘−1 𝑀−1

(6.8)

6.2.1.4. Analysis Consider the singular differential equation 𝑟

𝑦 ′′ (𝑥) + 𝑥 𝑦 ′ (𝑥) + 𝑝(𝑥)𝑦(𝑥) = 𝑞(𝑥), 𝑦(0) = 𝛼, 𝑦(𝐴) = 𝛽,

1028

𝑎 ≤ 𝑥 ≤ 𝑏,

where 𝑟 is nonzero constant and 𝑝(𝑥), 𝑞(𝑥) are real valued functions, 𝛼, 𝛽, 𝑎, 𝑏 and 𝐴 are constants. In order to solve singular differential equation, we suppose that [𝑎, 𝑏] = [0, 1], otherwise the problem can be mapped from [𝑎, 𝑏] to [0, 1] easily. For solving singular differential equation, it is sufficient to suppose that the approximate solution is as 𝑦(𝑥) = 𝐂 𝑇 𝛙(𝑥), where 𝐂 and 𝛙(𝑥) are given in above. Therefore, we obtain 𝑟

𝐂 𝑇 𝛙′′ (𝑥) + 𝑥 𝐂 𝑇 𝛙′ (𝑥) + 𝑝(𝑥)𝐂 𝑇 𝛙(𝑥) = 𝑞(𝑥),

𝑎 ≤ 𝑥 ≤ 𝑏,

𝐂 𝑇 𝛙(0) = 𝛼, 𝐂 𝑇 𝛙(𝐴) = 𝛽. Therefore, in order to apply the Physicists Hermite Wavelet Method (PHWM); we require 2𝑘−1 𝑀 − 1 collocating points. Suitable collocating points are as 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2, … , 2𝑘−1 𝑀 − 1.

Implementing the collocating points and imposing the initial value to the singular differential equation, we obtain 𝑟

𝐂 𝑇 𝛙′′ (𝑥𝑖 ) + 𝑥 𝐂 𝑇 𝛙′ (𝑥𝑖 ) + 𝑝(𝑥𝑖 )𝐂 𝑇 𝛙(𝑥𝑖 ) = 𝑞(𝑥𝑖 ), 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑖

𝐂 𝑇 𝛙(0) = 𝛼, 𝐂 𝑇 𝛙(𝐴) = 𝛽. The differential equation yields 2𝑘−1 𝑀 − 3 equations and initial condition produces 2 number of equations. Therefore, the obtained system has 2𝑘−1 𝑀 − 1 equations and 2𝑘−1 𝑀 − 1 unknowns. Solving this system gives the unknown coefficients 𝐂.

6.2.2. Singular Equations Example 6.1 [46]. Consider the following singular BVP as 2

−𝑦 ′′ (𝑥) − 𝑥 𝑦 ′ (𝑥) + (1 − 𝑥 2 )𝑦(𝑥) = 𝑥 4 − 2𝑥 2 + 7,

(6.9)

subject to the boundary conditions 𝑦(0) = 1, 𝑦(1) = 0.

(6.10)

The exact solution of Eq. (6.9-6.10) is 𝑦(𝑥) = 1 − 𝑥 2 .

(6.11)

According to the Physicists Hermite Wavelets Method (PHWM), we assume the trial solution 𝑘−1

𝐓 𝑦(𝑥) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥),

where 𝐶 and 𝜓(𝑡) are 2𝑘−1 𝑀 × 1 matrices given as 1029

(6.12)

𝑇


𝑇

𝛙(𝑥) = [𝜓10 , 𝜓11 , … , 𝜓1𝑀−1 , 𝜓20 , 𝜓21 , … , 𝜓2𝑀−1 , … , 𝜓2𝑘−1 0 , … , 𝜓2𝑘−1 𝑀−1 ] .

We apply the proposed technique to solve Eq. (6.9) with 𝐾 = 1, and 𝑀 = 3. We have Eq. (6.12) is 𝑦(𝑥) = ∑1𝑛=1 ∑2𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) = 𝐂 𝐓 𝛙(𝑥),

(6.13) 𝑘

1

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑃𝐻𝑚 (2𝑘 𝑡 − (2𝑛 + 1)).

(6.14)

In the above expression, 𝑃𝐻𝑚 is the Physicists Hermite polynomials given as 𝑃𝐻0 = 1, 𝑃𝐻1 = 2𝑥, 𝑃𝐻2 = −2 + 4𝑥 2 , 𝑃𝐻3 = 4𝑥(−3 + 2𝑥 2 ), 𝑃𝐻4 = 12 − 48𝑥 2 + 16𝑥 4 , . . .. Therefore, the Physicists Hermite wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = 4√5𝑥(𝑥 − 1), 𝜓1,3 (𝑥) = 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + 4√5𝑥(𝑥 − 1)𝑐1,2 = 𝐂 𝐓 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 ] and 𝛙(𝑥) = [1, √3(2𝑥 − 1), 4√5𝑥(𝑥 − 1)] . Substituting into the given problem we get

1030

2

−𝐂 𝐓 𝛙′′ (𝑥) − 𝑥 𝐂 𝐓 𝛙′ (𝑥) + (1 + 𝑥 2 )𝐂 𝐓 𝛙(𝑥) = 𝑥 4 − 2𝑥 2 + 7,

(6.15)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 1, 𝐂 𝐓 𝛙(1) = 0.

(6.16)

Substitute the collocating points are in Eq. (6.15), we have the system of equations 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 3.

(6.17)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 3×3 𝐂3×1 = 𝐛3×1 . After solving we get the following exact solution 𝑦(𝑥) = 1 − 𝑥 2 .

Example 6.2 [72].Consider the following singular IVP as 4

2

𝑦 ′′ (𝑥) + 𝑥 𝑦 ′ (𝑥) + 𝑥 2 𝑦(𝑥) = 12,

(6.18)

subject to the boundary conditions 𝑦(0) = 0, 𝑦 ′ (0) = 0.

(6.19)

The exact solution of Eq. (6.18-6.19) is 𝑦(𝑥) = 𝑥 2 .

(6.20)


𝐓 𝑦(𝑥) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥).

(6.21)



𝑇


We apply the proposed technique to solve Eq. (6.18) with 𝐾 = 1, and 𝑀 = 3. We have Eq. (6.21) is 𝑦(𝑥) = ∑1𝑛=1 ∑2𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) = 𝐂 𝐓 𝛙(𝑥),

(6.22) 1

𝑘


(6.23)

In the above expression, 𝑃𝐻𝑚 is the Physicists Hermite polynomials given as 1031

𝑃𝐻0 = 1, 𝑃𝐻1 = 2𝑥, 𝑃𝐻2 = −2 + 4𝑥 2 , 𝑃𝐻3 = 4𝑥(−3 + 2𝑥 2 ), 𝑃𝐻4 = 12 − 48𝑥 2 + 16𝑥 4 , . . .. Therefore, the Physicists Hermite wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = 4√5𝑥(𝑥 − 1), 𝜓1,3 (𝑥) = 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + 4√5𝑥(𝑥 − 1)𝑐1,2 = 𝐂 𝐓 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 ] and 𝛙(𝑥) = [1, √3(2𝑥 − 1), 4√5𝑥(𝑥 − 1)] . Substituting into the given problem we get 4

2

𝐂 𝐓 𝛙′′ (𝑥) + 𝑥 𝐂 𝐓 𝛙′ (𝑥) + 𝑥 2 𝐂 𝐓 𝛙(𝑥) = 12,

(6.24)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 1, 𝐂 𝐓 𝛙′ (0) = 0.

(6.25)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2.

(6.26)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 3×3 𝐂3×1 = 𝐛3×1 . After solving we get the following exact solution 1032

𝑦(𝑥) = 𝑥 2 .

Example 6.3 [72]. Consider the following singular IVP as 2

𝑦 ′′ (𝑥) + 𝑥 𝑦 ′ (𝑥) + 𝑦(𝑥) = 6 + 12𝑥 + 𝑥 2 + 𝑥 3 ,

(6.27)


(6.28)

The exact solution of Eq. (6.27-6.28) is 𝑦(𝑥) = 𝑥 2 + 𝑥 3 .

(6.29)



(6.30)



𝑇


We apply the proposed technique to solve Eq. (6.27) with 𝐾 = 1, and 𝑀 = 4. We have Eq. (6.30) is 𝑦(𝑥) = ∑1𝑛=1 ∑4𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) +𝑐13 𝜓13 (𝑥) = 𝐂 𝐓 𝛙(𝑥), 1

(6.31)

𝑘


(6.32)

In the above expression, 𝑃𝐻𝑚 is the Physicists Hermite polynomials given as 𝑃𝐻0 = 1, 𝑃𝐻1 = 2𝑥, 𝑃𝐻2 = −2 + 4𝑥 2 , 𝑃𝐻3 = 4𝑥(−3 + 2𝑥 2 ), 𝑃𝐻4 = 12 − 48𝑥 2 + 16𝑥 4 , . . .. Therefore, the Physicists Hermite wavelets given as 𝜓1,0 (𝑥) = 1, 1033

𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = 4√5𝑥(𝑥 − 1), 𝜓1,3 (𝑥) = 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + 4√5𝑥(𝑥 − 1)𝑐1,2 + 2√7(2𝑥 − 1) (2𝑥 2 − 2𝑥 − 1)𝑐1,3 = 𝐂 𝐓 𝛙(𝑥), 𝑇

1, √3(2𝑥 − 1), 4√5𝑥(𝑥 − 1), where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 ] and 𝛙(𝑥) = [ ] . 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1) 𝑇

Substituting into the given problem we get 2

𝐂 𝐓 𝛙′′ (𝑥) + 𝑥 𝐂 𝐓 𝛙′ (𝑥) + 𝐂 𝐓 𝛙(𝑥) = 6 + 12𝑥 + 𝑥 2 + 𝑥 3 ,

(6.33)


(6.34)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2,3.

(6.35)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 4×4 𝐂4×1 = 𝐛4×1 . After solving we get the following exact solution 𝑦(𝑥) = 𝑥 2 + 𝑥 3 .

Example 6.4 [210]. Consider the following singular IVP as 8

𝑦 ′′ (𝑥) + 𝑥 𝑦 ′ (𝑥) + 𝑥𝑦(𝑥) = −30𝑥 + 44𝑥 2 − 𝑥 4 + 𝑥 5 ,

(6.36)


(6.37)

The exact solution of Eq. (6.36-6.37) is 𝑦(𝑥) = 𝑥 4 − 𝑥 3 .

(6.38)

1034



(6.39)



𝑇


We apply the proposed technique to solve Eq. (6.36) with 𝐾 = 1, and 𝑀 = 5. We have Eq. (6.39) is 𝑦(𝑥) = ∑1𝑛=1 ∑4𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) +𝑐13 𝜓13 (𝑥) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 1

(6.40)

𝑘


(6.41)

In the above expression, 𝑃𝐻𝑚 is the Physicists Hermite polynomials given as 𝑃𝐻0 = 1, 𝑃𝐻1 = 2𝑥, 𝑃𝐻2 = −2 + 4𝑥 2 , 𝑃𝐻3 = 4𝑥(−3 + 2𝑥 2 ), 𝑃𝐻4 = 12 − 48𝑥 2 + 16𝑥 4 , . . .. Therefore, the Physicists Hermite wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = 4√5𝑥(𝑥 − 1), 𝜓1,3 (𝑥) = 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), . . .. Therefore, we have the trial solution is 1035

𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + 4√5𝑥(𝑥 − 1)𝑐1,2 + 2√7(2𝑥 − 1) (2𝑥 2 − 2𝑥 − 1)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

1, √3(2𝑥 − 1), 4√5𝑥(𝑥 − 1), where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] and 𝛙(𝑥) = [ ] . 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), … 𝑇


𝐂 𝐓 𝛙′′ (𝑥) + 𝑥 𝐂 𝐓 𝛙′ (𝑥) + 𝑥𝐂 𝐓 𝛙(𝑥) = −30𝑥 + 44𝑥 2 − 𝑥 4 + 𝑥 5 , (6.42) subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 0, 𝐂 𝐓 𝛙′ (0) = 0.

(6.43)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2,3,4.

(6.44)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 5×4 𝐂5×1 = 𝐛5×1 . After solving we get the following exact solution 𝑦(𝑥) = 𝑥 4 − 𝑥 3 .

Example 6.5 [72]. Consider the following singular BVP as 2

𝑦 ′′ (𝑥) + 𝑥 𝑦 ′ (𝑥) − 10𝑦(𝑥) = 12 − 10𝑥 4 ,

(6.45)


(6.46)

The exact solution of Eq. (6.45-6.46) is 𝑦(𝑥) = 2𝑥 2 + 𝑥 4 .

(6.47)



(6.48)



𝑇


We apply the proposed technique to solve Eq. (6.45) with 𝐾 = 1, and 𝑀 = 5. We have Eq. (6.48) is 1036

𝑦(𝑥) = ∑1𝑛=1 ∑4𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) +𝑐13 𝜓13 (𝑥) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 1

(6.49)

𝑘


(6.50)

In the above expression, 𝑃𝐻𝑚 is the Physicists Hermite polynomials given as 𝑃𝐻0 = 1, 𝑃𝐻1 = 2𝑥, 𝑃𝐻2 = −2 + 4𝑥 2 , 𝑃𝐻3 = 4𝑥(−3 + 2𝑥 2 ), 𝑃𝐻4 = 12 − 48𝑥 2 + 16𝑥 4 , . . .. Therefore, the Physicists Hermite wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = 4√5𝑥(𝑥 − 1), 𝜓1,3 (𝑥) = 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + 4√5𝑥(𝑥 − 1)𝑐1,2 + 2√7(2𝑥 − 1) (2𝑥 2 − 2𝑥 − 1)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇



𝐂 𝐓 𝛙′′ (𝑥) + 𝐂 𝐓 𝛙′ (𝑥) − 10𝐂 𝐓 𝛙(𝑥) = 12 + 10𝑥 4 , 𝑥

(6.51)


(6.52) 1037


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2,3,4.

(6.53)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 5×4 𝐂5×1 = 𝐛5×1 . After solving we get the following exact solution 𝑦(𝑥) = 2𝑥 2 + 𝑥 4 .

6.3. Newton’s Wavelet Method (PHWM) We proposed a new algorithm by inserting Newton’s polynomials in traditional Legendre wavelets method. This technique is successively applied to find the numerical solutions of higher order boundary values problems. The proposed technique is very simple and highly compatible for solving such kind of problems.

6.3.1. Methodology 6.3.1.1. Newton’s Polynomials Sir Isaac Newton (1643 – 1727) introduced Newton’s polynomials, these polynomials have a center 𝑥0 . Polynomial interpolation can be used to construct the polynomial of degree ≤ 𝑛 that passes through the 𝑛 + 1 points (𝑥𝑘 , 𝑦𝑘 ) = (𝑥𝑘 , 𝑓(𝑥𝑘 )), for 𝑘 = 0,1, … , 𝑛. If multiple "centers" 𝑥0 , 𝑥1 , … , 𝑥𝑛 are used, then the result is the so called Newton polynomials.

Theorem 6.1. Assume that 𝑓 ∈ 𝐶 𝑛+1 [𝑎, 𝑏] and 𝑥𝑘 [𝑎, 𝑏] for 𝑘 = 0,1, … , 𝑛 are distinct. Then 𝑓(𝑥) = 𝑃𝑛 (𝑥) + 𝑅𝑛 (𝑥), where 𝑃𝑛 (𝑥) is a polynomial that can be used to approximate 𝑓(𝑥), 𝑃𝑛 (𝑥) = 𝑎0 + 𝑎1 (𝑥 − 𝑥0 ) + 𝑎2 (𝑥 − 𝑥0 )(𝑥 − 𝑥1 ) + 𝑎3 (𝑥 − 𝑥0 )(𝑥 − 𝑥1 )(𝑥 − 𝑥2 ) + ⋯ + 𝑎𝑛 (𝑥 − 𝑥0 )(𝑥 − 𝑥1 )(𝑥 − 𝑥2 ) … (𝑥 − 𝑥𝑛−1 ). 𝑗−1

𝑃𝑛 (𝑥) = ∑𝑛𝑗=0(∏𝑖=0 𝑎𝑖 (𝑥 − 𝑥𝑖 )).

(6.54)

It can be written as 𝑓(𝑥) ≈ 𝑃𝑛 (𝑥). The Newton polynomial goes through the 𝑛 + 1 points {(𝑥𝑘 , 𝑦𝑘 )}𝑛𝑘=0 , i.e. 𝑃𝑛 (𝑥𝑘 ) = 𝑓(𝑥𝑘 ),

for 𝑘 = 0,1,2, … , 𝑛 1038

6.3.1.2. Wavelets and Newton’s Wavelets In recent years, wavelets have found their way into many different fields of science and engineering. Wavelets constitute a family of functions constructed from the dilation and translation of a single function called the mother wavelet. When the dilation parameter a and the translation parameter 𝑏 vary continuously, we have the following family of continuous wavelets: 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|− 2 𝜓 (

𝑎

) 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

(6.55)

Newton’s wavelets 𝜓𝑛𝑚 (𝑡) = 𝜓(𝑘, 𝑛, 𝑚, 𝑡) have four arguments; 𝑛, 𝑘 can assume any positive integer, 𝑚 is the order for Newton’s polynomials and t is the normalized time. They are defined on the interval [0, 1) by; 1

𝜓𝑛𝑚

𝑘

𝑘 2 (𝑡) = {√𝑚 + 2 2 𝑁𝑚 (2 𝑡 − (2𝑛 + 1))

0

𝑛 2𝑘

≤𝑡≤

𝑛+1 2𝑘

(6.56)

otherwise 1

where 𝑚 = 0,1,2, … , 𝑀 − 1 and 𝑛 = 0,1, … , 2𝑘−1 The coefficient (𝑚 + 2) is for orthonormality. Here 𝑁𝑚 (𝑡) are the well-known Newton’s polynomials of order 𝑚, which have been previously described.


(6.57)


𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡).

(6.58)



(6.59)

where 𝐶 and 𝜓(𝑡) are 2𝑘−1 𝑀 × 1 matrices given as 𝑐10 , 𝑐11 , … , 𝑐1𝑀−1 , 𝑐20 , 𝑐21 , … , 𝑇 𝐂 = [𝑐 ] , , … , 𝑐 𝑘−1 , … , 𝑐 𝑘−1 2𝑀−1

and

𝛙(𝑡) = [

2

0

2

𝑀−1

𝜓10 , 𝜓11 , … , 𝜓1𝑀−1 , 𝜓20 , 𝜓21 , … , 𝑇 ] . 𝜓2𝑀−1 , … , 𝜓2𝑘−1 0 , … , 𝜓2𝑘−1 𝑀−1

6.3.1.4. Analysis 1039

(6.60) (6.61)

Consider the nth order boundary values problem 𝑦 𝑛 (𝑥) = 𝑓 (𝑥, 𝑦(𝑥), 𝑦 ′ (𝑥), 𝑦 ′′ (𝑥), … , 𝑦 𝑛−1 (𝑥)), 𝑥0 ≤ 𝑥 ≤ 𝑥𝑓 𝑦 2𝑗 (𝑥0 ) = 𝐴𝑗 , 𝑦 2𝑗 (𝑥𝑓 ) = 𝐵𝑗 , 𝑗 = 0,1,2, … , (𝑛 − 1). It is interesting to point out that 𝑦(𝑥) and 𝑓(𝑥, 𝑦) are assumed real and as many times differentiable functions and 𝐴𝑗 and 𝐵𝑗 are the real constant. In order to solve nth order boundary values problem, we suppose that [𝑥0 , 𝑥𝑓 ] = [0, 1], otherwise the problem can be mapped from [𝑥0 , 𝑥𝑓 ] to [0, 1] easily. For solving nth order boundary values problem, it is sufficient to suppose that the approximate solution is as 𝑦(𝑥) = 𝐂 𝑇 𝛙(𝑥), where 𝐂 and 𝛙(𝑥) are given in above. Therefore, we obtain ′

𝑛

′′

𝐂 𝑇 (𝛙(𝑥)) = 𝑓 (𝑥, 𝐂 𝑇 (𝛙(𝑥)), 𝐂 𝑇 (𝛙(𝑥)) , 𝐂 𝑇 (𝛙(𝑥)) , … , 𝐂 𝑇 (𝛙(𝑥))

𝑛−1

),

𝐂 𝑇 (𝛙)2𝑗 (𝑥0 ) = 𝐴𝑗 , 𝐂 𝑇 (𝛙)2𝑗 (𝑥𝑓 ) = 𝐵𝑗 , Therefore, in order to apply the Newton’s Wavelet Method (NWM); we require 2𝑘−1 𝑀 − 1 collocating points. Suitable collocating points are as 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 𝑛, … , 2𝑘−1 𝑀 − 1.

Implementing the collocating points and imposing the initial value to the nth order boundary values problem, we obtain ′

𝑛

′′

𝐂 𝑇 (𝛙(𝑥𝑖 )) = 𝑓 (𝑥, 𝐂 𝑇 (𝛙(𝑥𝑖 )), 𝐂 𝑇 (𝛙(𝑥𝑖 )) , 𝐂 𝑇 (𝛙(𝑥𝑖 )) , … , 𝐂 𝑇 (𝛙(𝑥𝑖 ))

𝑛−1

),

𝐂 𝑇 (𝛙)2𝑗 (𝑥0 ) = 𝐴𝑗 , 𝐂 𝑇 (𝛙)2𝑗 (𝑥𝑓 ) = 𝐵𝑗 , The differential equation yields 2𝑘−1 𝑀 − 𝑛 + 1 equations and initial condition produces 𝑛 number of equations. Therefore, the obtained system has 2𝑘−1 𝑀 − 1 equations and 2𝑘−1 𝑀 − 1 unknowns. Solving this system gives the unknown coefficients 𝐂.

6.3.2. Fifth Order BVP Consider the 5th order ODE [226] as follow 𝑦 (v) (𝑥) = 𝑦(𝑥) − 15e𝑥 − 10𝑥e𝑥 ,

0 < 𝑥 < 1,

(6.62)

subject to the boundary conditions 𝑦(0) = 0, 𝑦 ′ (0) = 1, 𝑦 ′′ (0) = 0, 𝑦(1) = 0, 𝑦 ′ (1) = −e. The exact solution of Eq. (6.62-6.63) is 1040

(6.63)

𝑦(𝑥) = 𝑥(1 − 𝑥)e𝑥 .

(6.64)

According to the Newton’s Wavelets Method (NWM), we assume the trial solution 𝑘−1


(6.65)



𝑇


We apply the proposed technique to solve Eq. (6.62) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.65) is 𝑦(𝑥) = ∑1𝑛=1 ∑10 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) , = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) +𝑐13 𝜓13 (𝑥) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 1

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑁𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝑁𝑚 is the Newton’s polynomials given as 𝑁0 = 1, 𝑁1 = 1 + 𝑥, 4

𝑁2 = 1 + 5 𝑥 + 𝑥 2 , 22

2

𝑁3 = 1 + 25 𝑥 + 5 𝑥 2 + 𝑥 3 , 104

21

(6.66)

1

𝑁4 = 1 + 125 𝑥 + 25 𝑥 2 − 5 𝑥 3 + 𝑥 4 , . . .. Therefore, the Newton’s wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 2

𝜓1,2 (𝑥) = 5 √5(3 − 6𝑥 + 10𝑥 2 ), 2

𝜓1,3 (𝑥) = 25 √7(−6 + 77𝑥 − 130𝑥 2 + 100𝑥 3 ), . . .. 1041

(6.67)

Therefore, we have the trial solution is 2

𝑦(𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + 5 √5(3 − 6𝑥 + 10𝑥 2 )𝑐1,2 + 2 25

√7(−6 + 77𝑥 − 130𝑥 2 + 100𝑥 3 ) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

2

1, 2√3𝑥, 5 √5(3 − 6𝑥 + 10𝑥 2 ), 𝑇 where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] and 𝛙(𝑥) = [ 2 ] . −6 + 77𝑥 − ( ) , … √7 2 3 25 130𝑥 + 100𝑥 Substituting into the given problem we get 𝐂 𝐓 𝛙(𝑥)(v) = 𝐂 𝐓 𝛙(𝑥) − 15e𝑥 − 10𝑥e𝑥 ,

0 < 𝑥 < 1,

(6.68)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 1, 𝐂 𝐓 𝛙′ (0) = 1, 𝐂 𝐓 𝛙′′ (0) = 0, 𝐂 𝐓 𝛙(1) = 0, 𝐂 𝐓 𝛙′ (1) = −e.

(6.69)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 5,6,7,8,9.

(6.70)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following approximate solution 𝑦(𝑥) = 0.99999𝑥 + 0.00000𝑥 2 − 0.49686𝑥 3 + ⋯.

Table 6.1: Comparison of the Exact Solution and Approximate Solution of Eq. (6.62) obtained from Newton’s Wavelet Method (NWM) 1042

__________________________________________________________________ 𝑥

Exact Solution


Error in NWM

__________________________________________________________________ 0.0

0.000000000000000

0.000000000000000

1.00000E-200

0.1

0.099465382626808

0.099468107407824

2.72478E-06

0.2

0.195424441305627

0.195442930390957

1.84891E-05

0.3

0.283470349590961

0.283521792062770

5.14425E-05

0.4

0.358037927433905

0.358134682191875

9.67548E-05

0.5

0.412180317675032

0.412322471445376

1.42154E-04

0.6

0.437308512093722

0.437479299709762

1.70788E-04

0.7

0.422888068568800

0.423054257147050

1.66189E-04

0.8

0.356086548558795

0.356206975557239

1.20427E-04

0.9

0.221364280004125

0.221411218352115

4.69383E-05

1.0

0.000000000000000

0.000000000000000

4.00000E-200

_________________________________________________________________

Fig. 6.1 (a)-(b): Comparison of Exact and Approximate Solution of Eq. (6.62)

Now to solve Eq. (6.62) with 𝐾 = 1, and 𝑀 = 25. We have Eq. (6.65) is 𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑24 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] .

Proceeding as before we have the matrix form 1043

𝐀 𝟐𝟓×𝟐𝟓 𝐂𝟐𝟓×𝟏 = 𝐛𝟐𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.2: Comparison of the Exact Solution and Approximate Solution of Eq. (6.62) obtained from Newton’s Wavelet Method (NWM) __________________________________________________________________ 𝑥

Exact Solution


Error in NWM

__________________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

1.00000E-200

0.1

0.099465382626808

0.099465382626808

4.30907E-25

0.2

0.195424441305627

0.195424441305627

3.05172E-24

0.3

0.283470349590961

0.283470349590961

8.96460E-24

0.4

0.358037927433905

0.358037927433905

1.80852E-23

0.5

0.412180317675032

0.412180317675032

2.91425E-23

0.6

0.437308512093722

0.437308512093722

3.96794E-23

0.7

0.422888068568800

0.422888068568800

4.60521E-23

0.8

0.356086548558795

0.356086548558795

4.34325E-23

0.9

0.221364280004125

0.221364280004125

2.59101E-23

1.0

0.000000000000000

-0.000000000000000

3.02632E-199

____________________________________________________________

(a)

(b)

1044

Fig. 6.2 (a)-(b): Comparison of Exact and Approximate Solution of Eq. (6.62)

Finally we solve Eq. (6.62) with 𝐾 = 1, and 𝑀 = 100. We have Eq. (6.65) is 𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑99 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,99 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,99 ] .

Proceeding as before we have the matrix form 𝐀 𝟏𝟎𝟎×𝟏𝟎𝟎 𝐂𝟏𝟎𝟎×𝟏 = 𝐛𝟏𝟎𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table.

Table 6.3: Comparison of the Exact Solution and Approximate Solution of Eq. (6.62) obtained from Newton’s Wavelet Method (NWM) __________________________________________________________________ 𝑥

Exact Solution


Error in NWM

__________________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

5.00000E-200

0.1

0.099465382626808

0.099465382626808

2.19370E-142

0.2

0.195424441305627

0.195424441305627

1.55958E-141

0.3

0.283470349590961

0.283470349590961

4.60420E-141

0.4

0.358037927433905

0.358037927433905

9.35065E-141

0.5

0.412180317675032

0.412180317675032

1.52103E-140

1045

0.6

0.437308512093722

0.437308512093722

2.10084E-140

0.7

0.422888068568800

0.422888068568800

2.49845E-140

0.8

0.356086548558795

0.356086548558795

2.47923E-140

0.9

0.221364280004125

0.221364280004125

1.75000E-140

1.0

0.000000000000000

-0.000000000000000

9.88236E-201

____________________________________________________________

(a)

(b)

Fig. 6.3. (a)-(b): Comparison of Exact and Approximate Solution of Eq. (6.62)

6.3.3. Sixth-order BVP We first consider the 6th order linear BVP [227] as follow 𝑦 (vi) (𝑥) = 𝑦(𝑥) − 6e𝑥 ,

0 < 𝑥 < 1,

(6.71)

subject to the boundary conditions 𝑦(0) = 1, 𝑦 ′′ (0) = −1, 𝑦 (iv) (0) = −3,

(6.72)

𝑦(1) = 0, 𝑦 ′′ (1) = −2e, 𝑦 (iv) (1) = −4e

(6.73)

The exact solution of Eq. (6.71-6.73) is 𝑦(𝑥) = (1 − 𝑥)e𝑥 .

(6.74)



(6.75)


𝐂 = [𝑐10 , 𝑐11 , … , 𝑐1𝑀−1 , 𝑐20 , 𝑐21 , … , 𝑐2𝑀−1 , … , 𝑐2𝑘−1 0 , … , 𝑐2𝑘−1 𝑀−1 ] ,

1046

and

𝑇



(6.76)

𝑘


𝑁2 = 1 + 5 𝑥 + 𝑥 2 , 22

2

𝑁3 = 1 + 25 𝑥 + 5 𝑥 2 + 𝑥 3 , 104

21

1


𝜓1,2 (𝑥) = 5 √5(3 − 6𝑥 + 10𝑥 2 ), 2

𝜓1,3 (𝑥) = 25 √7(−6 + 77𝑥 − 130𝑥 2 + 100𝑥 3 ), . . .. Therefore, we have the trial solution is 2

𝑦(𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + 5 √5(3 − 6𝑥 + 10𝑥 2 )𝑐1,2 + 2 25

√7(−6 + 77𝑥 − 130𝑥 2 + 100𝑥 3 ) + ⋯ = 𝐂 𝐓 𝛙(𝑥),

1047

(6.77)

𝑇

2

1, 2√3𝑥, 5 √5(3 − 6𝑥 + 10𝑥 2 ), 𝑇 where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] and 𝛙(𝑥) = [ 2 ] . −6 + 77𝑥 − ( ) , … √7 25 130𝑥 2 + 100𝑥 3 Substituting into the given problem we get 𝐂 𝐓 𝛙(𝑥)(vi) = 𝐂 𝐓 𝛙(𝑥) − 6e𝑥 ,

0 < 𝑥 < 1,

(6.78)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 0, 𝐂 𝐓 𝛙′′ (0) = −1, 𝐂 𝐓 𝛙(iv) (0) = −3,

(6.79)

𝐂 𝐓 𝛙(1) = 0, 𝐂 𝐓 𝛙′′ (1) = −2e, 𝐂 𝐓 𝛙(iv) (1) = −4e

(6.80)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 6,7,8,9.

(6.81)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following approximate solution 𝑦(𝑥) = 1 − 0.00259𝑥 − 0.50000𝑥 2 − 0.32933𝑥 3 + ⋯.

Table 6.4: Comparison of the Exact Solution and Approximate Solution of Eq. (6.71) obtained from Newton’s Wavelet Method (NWM) ___________________________________________________________ 𝑥

Exact Solution


Error in NWM

___________________________________________________________ 0.0

1.000000000000000

1.000000000000000

3.00000E-200

0.1

0.994653826268083

0.994398983745573

2.54843E-04

0.2

0.977122206528136

0.976636056782961

4.86150E-04

0.3

0.944901165303202

0.944228905723056

6.72260E-04

0.4

0.895094818584762

0.894299651139807

7.95167E-04

0.5

0.824360635350064

0.823518482992040

8.42152E-04

1048

0.6

0.728847520156204

0.728040361994211

8.07158E-04

0.7

0.604125812241143

0.603434001927407

6.91810E-04

0.8

0.445108185698494

0.444602278519964

5.05907E-04

0.9

0.245960311115695

0.245693137726984

2.67173E-04

1.0

0.000000000000000

-0.000000000000000

1.73610E-199

____________________________________________________________

(a)

(b)



where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] .

Proceeding as before we have the matrix form 𝐀 𝟐𝟓×𝟐𝟓 𝐂𝟐𝟓×𝟏 = 𝐛𝟐𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.5: Comparison of the Exact Solution and Approximate Solution of Eq. (6.71) obtained from Newton’s Wavelet Method (NWM) ___________________________________________________________ 𝑥

Exact Solution


Error in NWM

___________________________________________________________ 0.0

1.000000000000000

1.000000000000000 1049

1.70000E-199

0.1

0.994653826268083

0.994653826268083

5.98382E-21

0.2

0.977122206528136

0.977122206528136

1.14534E-20

0.3

0.944901165303202

0.944901165303202

1.59260E-20

0.4

0.895094818584762

0.895094818584762

1.89822E-20

0.5

0.824360635350064

0.824360635350064

2.02971E-20

0.6

0.728847520156204

0.728847520156204

1.96725E-20

0.7

0.604125812241143

0.604125812241143

1.70678E-20

0.8

0.445108185698494

0.445108185698494

1.26325E-20

0.9

0.245960311115695

0.245960311115695

6.73679E-21

1.0

0.000000000000000

0.000000000000000

1.20000E-199

____________________________________________________________

(a)

(b)



where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,99 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,99 ] .


1050

Table 6.6: Comparison of the Exact Solution and Approximate Solution of Eq. (6.71) obtained from Newton’s Wavelet Method (NWM) ____________________________________________________________ 𝑥

Exact Solution


Error in NWM

____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

6.92175E-191

0.1

0.994653826268083

0.994653826268083

1.76510E-134

0.2

0.977122206528136

0.977122206528136

3.37904E-134

0.3

0.944901165303202

0.944901165303202

4.69987E-134

0.4

0.895094818584762

0.895094818584762

5.60403E-134

0.5

0.824360635350064

0.824360635350064

5.99559E-134

0.6

0.728847520156204

0.728847520156204

5.81541E-134

0.7

0.604125812241143

0.604125812241143

5.05043E-134

0.8

0.445108185698494

0.445108185698494

3.74280E-134

0.9

0.245960311115695

0.245960311115695

1.99920E-134

1.0

0.000000000000000

0.000000000000000

5.30033E-200

____________________________________________________________

1051

(a)

(b)


6.3.4. Ninth-order BVP We consider the following 9th order BVP [219] as follow 𝑦 (ix) (𝑥) = 𝑦(𝑥) − 9e𝑥 ,

0 < 𝑥 < 1,

(6.82)

subject to the boundary conditions 𝑦(0) = 1, 𝑦 ′ (0) = 0, 𝑦 ′′ (0) = −1, 𝑦 ′′′ (0) = −2, 𝑦 (iv) (0) = −3, (6.83) 𝑦(1) = 0, 𝑦 ′ (1) = −e, 𝑦 ′′ (1) = −2e, 𝑦 ′′′ (1) = −3e.

(6.84)


(6.85)



(6.86)



𝑇



(6.87)

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑁𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝑁𝑚 is the Newton’s polynomials given as 1052

(6.88)

𝑁0 = 1, 𝑁1 = 1 + 𝑥, 4

𝑁2 = 1 + 5 𝑥 + 𝑥 2 , 22

2

𝑁3 = 1 + 25 𝑥 + 5 𝑥 2 + 𝑥 3 , 104

21

1


𝜓1,2 (𝑥) = 5 √5(3 − 6𝑥 + 10𝑥 2 ), 2


𝑦(𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + 5 √5(3 − 6𝑥 + 10𝑥 2 )𝑐1,2 + 2 25

√7(−6 + 77𝑥 − 130𝑥 2 + 100𝑥 3 ) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

2

1, 2√3𝑥, 5 √5(3 − 6𝑥 + 10𝑥 2 ), 𝑇 where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] and 𝛙(𝑥) = [ 2 ] . −6 + 77𝑥 − ( ) , … √7 25 130𝑥 2 + 100𝑥 3 Substituting into the given problem we get 𝐂 𝐓 𝛙(ix) (𝑥) = 𝐂 𝐓 𝛙(𝑥) − 9e𝑥 ,

0 < 𝑥 < 1,

(6.89)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 1, 𝐂 𝐓 𝛙′ (0) = 0, 𝐂 𝐓 𝛙′′ (0) = −1, 𝐂 𝐓 𝛙′′′ (0) = −2, 𝐂 𝐓 𝛙(iv) (0) = −3,

(6.90)

𝐂 𝐓 𝛙(1) = 0, 𝐂 𝐓 𝛙′ (1) = −e, 𝐂 𝐓 𝛙′′ (1) = −2e, 𝐂 𝐓 𝛙′′′ (1) = −3e. (6c) Substitute the collocating points are in Eq. (6.89), we have the system of equations 1053

𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 9.

(6.91)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following approximate solution 𝑦(𝑥) = 1 − 0.0000𝑥 − 0.50000𝑥 2 − 0.333333𝑥 3 + ⋯.

Table 6.7: Comparison of the Exact Solution and Approximate Solutions obtained from Newton’s Wavelet Method (NWM) ___________________________________________________________ 𝑥

Exact Solution


Error in NWM

___________________________________________________________ 0.0

1.000000000000000

1.000000000000000

0.00000E+00

0.1

0.994653826268083

0.994653826453032

1.84949E-10

0.2

0.977122206528136

0.977122210273641

3.74551E-09

0.3

0.944901165303202

0.944901182205811

1.69026E-08

0.4

0.895094818584762

0.895094857565638

3.89809E-08

0.5

0.824360635350064

0.824360693520868

5.81708E-08

0.6

0.728847520156204

0.728847580278743

6.01225E-08

0.7

0.604125812241143

0.604125853939665

4.16985E-08

0.8

0.445108185698494

0.445108201986150

1.62877E-08

0.9

0.245960311115695

0.245960312976428

1.86073E-09

1.0

0.000000000000000

0.000000000000000

1.31820E-199

___________________________________________________________

1054

(a)

(b)



where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] .

Proceeding as before we have the matrix form 𝐀 𝟐𝟓×𝟐𝟓 𝐂𝟐𝟓×𝟏 = 𝐛𝟐𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.8: Comparison of the Exact Solution and Approximate Solutions obtained from Newton’s Wavelet Method (NWM) ___________________________________________________________ 𝑥

Exact Solution


Error in NWM

___________________________________________________________ 0.0

1.000000000000000

1.000000000000000

3.00000E-200

0.1

0.994653826268083

0.994653826268083

3.13431E-24

0.2

0.977122206528136

0.977122206528136

6.86119E-23

0.3

0.944901165303202

0.944901165303202

3.37581E-22

0.4

0.895094818584762

0.895094818584762

8.57551E-22

0.5

0.824360635350064

0.824360635350064

1.42706E-21

0.6

0.728847520156204

0.728847520156204

1.66940E-21

0.7

0.604125812241143

0.604125812241143

1.33428E-21

1055

0.8

0.445108185698494

0.445108185698494

6.13757E-22

0.9

0.245960311115695

0.245960311115695

8.47189E-23

1.0

0.000000000000000

0.000000000000000

8.00000E-200

____________________________________________________________

(a)

(b)



where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,99 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,99 ] .


1056


Exact Solution


Error in NWM

____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

8.47477E-192

0.1

0.994653826268083

0.994653826268083

1.84757E-136

0.2

0.977122206528136

0.977122206528136

4.14651E-135

0.3

0.944901165303202

0.944901165303202

2.10562E-134

0.4

0.895094818584762

0.895094818584762

5.57427E-134

0.5

0.824360635350064

0.824360635350064

9.81145E-134

0.6

0.728847520156204

0.728847520156204

1.24371E-133

0.7

0.604125812241143

0.604125812241143

1.12460E-133

0.8

0.445108185698494

0.445108185698494

6.38893E-134

0.9

0.245960311115695

0.245960311115695

1.36952E-134

1.0

0.000000000000000

-0.000000000000000

3.00000E-199

____________________________________________________________

1057

(a)

(b)


6.3.5. Tenth-order BVP We consider the following 10th order BVP [158] as follow 𝑦 (x) (𝑥) = 𝑦 ′′ (𝑥) − 8e𝑥 ,

0 < 𝑥 < 1,

(6.92)

subject to the boundary conditions 𝑦(0) = 1, 𝑦 ′ (0) = 0, 𝑦 ′′ (0) = −1, 𝑦 ′′′ (0) = −2, 𝑦 (iv) (0) = −3,

(6.93)

𝑦(1) = 0, 𝑦 ′ (1) = −e, 𝑦 ′′ (1) = −2e, 𝑦 ′′′ (1) = −3e, 𝑦 (iv) (1) = −4e.

(6.94)


(6.95)



(6.96)



𝑇



𝑘


𝑁2 = 1 + 5 𝑥 + 𝑥 2 , 22

2

𝑁3 = 1 + 25 𝑥 + 5 𝑥 2 + 𝑥 3 , 104

21

(6.97)

1

𝑁4 = 1 + 125 𝑥 + 25 𝑥 2 − 5 𝑥 3 + 𝑥 4 , . .

1058

(6.98)

.. Therefore, the Newton’s wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 2

𝜓1,2 (𝑥) = 5 √5(3 − 6𝑥 + 10𝑥 2 ), 2


𝑦(𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + 5 √5(3 − 6𝑥 + 10𝑥 2 )𝑐1,2 + 2 25

√7(−6 + 77𝑥 − 130𝑥 2 + 100𝑥 3 ) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

2

1, 2√3𝑥, 5 √5(3 − 6𝑥 + 10𝑥 2 ), 𝑇 where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] and 𝛙(𝑥) = [ 2 ] . −6 + 77𝑥 − ( ) , … √7 2 3 25 130𝑥 + 100𝑥 Substituting into the given problem we get 𝐂 𝐓 𝛙(x) (𝑥) = 𝐂 𝐓 𝛙′′ (𝑥) − 8e𝑥 ,

0 < 𝑥 < 1,

(6.99)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 1, 𝐂 𝐓 𝛙′ (0) = 0, 𝐂 𝐓 𝛙′′ (0) = −1, 𝐂 𝐓 𝛙′′′ (0) = −2, 𝐂 𝐓 𝛙(iv) (0) = −3,

(6.100)

𝐂 𝐓 𝛙(1) = 0, 𝐂 𝐓 𝛙′ (1) = −e, 𝐂 𝐓 𝛙′′ (1) = −2e, 𝐂 𝐓 𝛙′′′ (1) = −3e, 𝐂 𝐓 𝛙(iv) (1) = −4e,

(6.101)


(2𝑖+1)𝜋 2𝑘 𝑀

),

(6.102)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following approximate solution 𝑦(𝑥) = 1 − 0.0000𝑥 − 0.50000𝑥 2 − 0.333333𝑥 3 + ⋯. 1059


Exact Solution


Error in NWM

___________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.00000E-199

0.1

0.994653826268083

0.994653826243300

2.47830E-11

0.2

0.977122206528136

0.977122206083736

4.44400E-10

0.3

0.944901165303202

0.944901163555214

1.74799E-09

0.4

0.895094818584762

0.895094815142845

3.44192E-09

0.5

0.824360635350064

0.824360631086411

4.26365E-09

0.6

0.728847520156204

0.728847516644553

3.51165E-09

0.7

0.604125812241143

0.604125810421605

1.81954E-09

0.8

0.445108185698494

0.445108185226528

4.71966E-10

0.9

0.245960311115695

0.245960311088841

2.68539E-11

1.0

0.000000000000000

0.000000000000000

1.10000E-199

___________________________________________________________

(a)

(b)

1060



where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,49 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] .

Proceeding as before we have the matrix form 𝐀 𝟓𝟎×𝟓𝟎 𝐂𝟓𝟎×𝟏 = 𝐛𝟓𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.11: Comparison of the Exact Solution and Approximate Solution of Eq. (6.92) obtained from Newton’s Wavelet Method (NWM) ___________________________________________________________ 𝑥

Exact Solution


Error in NWM

___________________________________________________________ 0.0

1.000000000000000

1.000000000000000

6.00000E-199

0.1

0.994653826268083

0.994653826268083

1.44891E-63

0.2

0.977122206528136

0.977122206528136

2.86914E-62

0.3

0.944901165303202

0.944901165303202

1.26282E-61

0.4

0.895094818584762

0.895094818584762

2.82981E-61

0.5

0.824360635350064

0.824360635350064

4.07922E-61

0.6

0.728847520156204

0.728847520156204

4.03197E-61

0.7

0.604125812241143

0.604125812241143

2.62286E-61

0.8

0.445108185698494

0.445108185698494

9.17870E-62

0.9

0.245960311115695

0.245960311115695

8.01866E-63

1.0

0.000000000000000

-0.000000000000000

2.50000E-200

____________________________________________________________

1061

(a)

(b)



where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,99 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,99 ] .


1062


Exact Solution


Error in NWM

____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.99222E-191

0.1

0.994653826268083

0.994653826268083

4.07328E-134

0.2

0.977122206528136

0.977122206528136

8.11994E-133

0.3

0.944901165303202

0.944901165303202

3.60450E-132

0.4

0.895094818584762

0.895094818584762

8.16880E-132

0.5

0.824360635350064

0.824360635350064

1.19607E-131

0.6

0.728847520156204

0.728847520156204

1.20975E-131

0.7

0.604125812241143

0.604125812241143

8.16872E-132

0.8

0.445108185698494

0.445108185698494

3.06765E-132

0.9

0.245960311115695

0.245960311115695

3.20949E-133

1.0

0.000000000000000

0.000000000000000

4.58220E-199

____________________________________________________________

(a)

(b)


6.3.6. Twelfth-order BVP We consider the following 12th order BVP [67] as follow 𝑦 (xii) + 𝑥𝑦(𝑥) + 𝑥 3 e𝑥 + 23𝑥e𝑥 + 120e𝑥 = 0,0 < 𝑥 < 1, 1063

(6.103)

subject to the boundary conditions 𝑦(0) = 0, 𝑦 ′ (0) = 1, 𝑦 ′′ (0) = 0, 𝑦 ′′′ (0) = −3, 𝑦 (iv) (0) = −8, 𝑦 (v) (0) = −15,

(6.104)

𝑦(1) = 0, 𝑦 ′ (1) = −e, 𝑦 ′′ (1) = −4e, 𝑦 ′′′ (1) = −9e, 𝑦 (iv) (1) = −16e, 𝑦 (v) (1) = −25e.

(6.105)

The exact solution of Eq. (6.103-6.105) is 𝑦(𝑥) = 𝑥(1 − 𝑥)e𝑥 .

(6.106)



(6.107)



𝑇



𝑘


𝑁2 = 1 + 5 𝑥 + 𝑥 2 , 22

2

𝑁3 = 1 + 25 𝑥 + 5 𝑥 2 + 𝑥 3 , 104

21

(6.108)

1

𝑁4 = 1 + 125 𝑥 + 25 𝑥 2 − 5 𝑥 3 + 𝑥 4 , . . .. Therefore, the Newton’s wavelets given as 𝜓1,0 (𝑥) = 1, 1064

(6.109)

𝜓1,1 (𝑥) = 2√3𝑥, 2

𝜓1,2 (𝑥) = 5 √5(3 − 6𝑥 + 10𝑥 2 ), 2


𝑦(𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(3 − 6𝑥 + 10𝑥 2 )𝑐1,2 + 5 2 25

√7(−6 + 77𝑥 − 130𝑥 2 + 100𝑥 3 ) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

2

1, 2√3𝑥, 5 √5(3 − 6𝑥 + 10𝑥 2 ), 𝑇 where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] and 𝛙(𝑥) = [ 2 ] . −6 + 77𝑥 − ( ) , … √7 25 130𝑥 2 + 100𝑥 3 Substituting into the given problem we get 𝐂 𝐓 𝛙(xii) (𝑥) + 𝑥𝐂 𝐓 𝛙(𝑥) + 𝑥 3 e𝑥 + 23𝑥e𝑥 + 120e𝑥 = 0,

(6.110)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 0, 𝐂 𝐓 𝛙′ (0) = 1, 𝐂 𝐓 𝛙′′ (0) = 0, 𝐂 𝐓 𝛙′′′ (0) = −3, 𝐂 𝐓 𝛙(iv) (0) = −8, 𝐂 𝐓 𝛙(v) (0) = −15,

(6.111)

𝐂 𝐓 𝛙(1) = 0, 𝐂 𝐓 𝛙′ (1) = −e, 𝐂 𝐓 𝛙′′ (1) = −4e, 𝐂 𝐓 𝛙′′′ (1) = −9e, 𝐂 𝐓 𝛙(iv) (1) = −16e, 𝐂 𝐓 𝛙(v) (1) = −25e.

(6.112)


(2𝑖+1)𝜋 2𝑘 𝑀

),

(6.113)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following approximate solution 𝑦(𝑥) = 𝑥 + 0.00000𝑥 2 − 0.4444449𝑥 3 − 0.33333𝑥 4 + ⋯.

1065


Exact Solution


Error in NWM

___________________________________________________________ 0.0

0.000000000000000

0.000000000000000

1.00000E-200

0.1

0.099465382626808

0.099465382393068

2.33740E-10

0.2

0.195424441305627

0.195424437110061

4.19557E-09

0.3

0.283470349590961

0.283470333071422

1.65195E-08

0.4

0.358037927433905

0.358037894871987

3.25619E-08

0.5

0.412180317675032

0.412180277296707

4.03783E-08

0.6

0.437308512093722

0.437308478801581

3.32921E-08

0.7

0.422888068568800

0.422888051299981

1.72688E-08

0.8

0.356086548558795

0.356086544074536

4.48426E-09

0.9

0.221364280004125

0.221364279748695

2.55430E-10

1.0

0.000000000000000

0.000000000000000

4.31600E-200

(a)

(b)



where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,49 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] . 1066

Proceeding as before we have the matrix form 𝐀 𝟓𝟎×𝟓𝟎 𝐂𝟓𝟎×𝟏 = 𝐛𝟓𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.14: Comparison of the Exact Solution and Approximate Solution of Eq. (6.103) obtained from Newton’s Wavelet Method (NWM) ___________________________________________________________ 𝑥

Exact Solution


Error in NWM

___________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

1.02223E-199

0.1

0.099465382626808

0.099465382626808

1.82132E-60

0.2

0.195424441305627

0.195424441305627

6.38188E-59

0.3

0.283470349590961

0.283470349590961

3.66505E-58

0.4

0.358037927433905

0.358037927433905

9.31448E-58

0.5

0.412180317675032

0.412180317675032

1.38420E-57

0.6

0.437308512093722

0.437308512093722

1.29411E-57

0.7

0.422888068568800

0.422888068568800

7.20009E-58

0.8

0.356086548558795

0.356086548558795

1.84626E-58

0.9

0.221364280004125

0.221364280004125

8.38937E-60

1.0

0.000000000000000

0.000000000000000

1.43004E-199

____________________________________________________________

(a)

(b)


1067


where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,99 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,99 ] .

Proceeding as before we have the matrix form 𝐀 𝟏𝟎𝟎×𝟏𝟎𝟎 𝐂𝟏𝟎𝟎×𝟏 = 𝐛𝟏𝟎𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.15: Comparison of the Exact Solution and Approximate Solution of Eq. (6.103) obtained from Newton’s Wavelet Method (NWM) ____________________________________________________________ 𝑥

Exact Solution


Error in NWM

____________________________________________________________ 0.0

0.000000000000000

0.000000000000000

7.83189E-192

0.1

0.099465382626808

0.099465382626808

1.02004E-129

0.2

0.195424441305627

0.195424441305627

3.61089E-128

0.3

0.283470349590961

0.283470349590961

2.10077E-127

0.4

0.358037927433905

0.358037927433905

5.43068E-127

0.5

0.412180317675032

0.412180317675032

8.26090E-127

0.6

0.437308512093722

0.437308512093722

7.98956E-127

0.7

0.422888068568800

0.422888068568800

4.69009E-127

0.8

0.356086548558795

0.356086548558795

1.32526E-127

0.9

0.221364280004125

0.221364280004125

7.53637E-129

1.0

0.000000000000000

0.000000000000000

6.06314E-200

____________________________________________________________

1068

(a)

(b)


6.4. Laurent Wavelet Method (BWM) We proposed a new algorithm by inserting Bessel polynomials in traditional Legendre wavelets method. This technique is successively applied to find the approximate solution higher order boundary value problems. The proposed technique is very simple and highly compatible for solving such kind of problems.

6.4.1. Methodology 6.4.1.1. Laurent Polynomials The Laurent polynomials are orthogonal sequence of polynomials, the recurrence relation is given by 𝐿𝑛 = ∑𝑛𝑘=0 𝑥 𝑘 .

(6.114)

First five Laurent polynomials given below 𝐿0 = 1, 𝐿1 = 1 + 𝑥, 𝐿2 = 1 + 𝑥 + 𝑥 2 , 𝐿3 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 , 𝐿4 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 , . . ..

1069

6.4.1.2. Wavelets and Laurent Wavelets In recent years, wavelets have found their way into many different fields of science and engineering. Wavelets constitute a family of functions constructed from the dilation and translation of a single function called the mother wavelet. When the dilation parameter a and the translation parameter 𝑏 vary continuously, we have the following family of continuous wavelets: 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|− 2 𝜓 (

𝑎

) 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

(6.114)

Laurent wavelets 𝜓𝑛𝑚 (𝑡) = 𝜓(𝑘, 𝑛, 𝑚, 𝑡) have four arguments; 𝑛, 𝑘 can assume any positive integer, 𝑚 is the order for Laurent polynomials and t is the normalized time. They are defined on the interval [0, 1) by; 1

𝜓𝑛𝑚

𝑘

𝑘 2 (𝑡) = {√𝑚 + 2 2 𝐿𝑚 (2 𝑡 − (2𝑛 + 1))

0

𝑛 2𝑘

≤𝑡≤

𝑛+1 2𝑘

(6.115)

otherwise 1

where 𝑚 = 0,1,2, … , 𝑀 − 1 and 𝑛 = 0,1, … , 2𝑘−1 . The coefficient (𝑚 + 2) is for orthonormality. Here 𝐿𝑚 (𝑡) are the well-known Laurent polynomials of order 𝑚, which have been previously described.


(6.116)


𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡).

(6.117)



(6.118)


and

𝛙(𝑡) = [

2

0

2

𝑀−1

𝜓10 , 𝜓11 , … , 𝜓1𝑀−1 , 𝜓20 , 𝜓21 , … , 𝑇 ] . 𝜓2𝑀−1 , … , 𝜓2𝑘−1 0 , … , 𝜓2𝑘−1 𝑀−1

1070

(6.119) (6.120)

6.4.1.4. Analysis System of Ordinary Differential Equations of 2×2 Consider the system of differential equations 𝑦1′′ (𝑥) + 𝛼1 𝑦1 (𝑥) + 𝛽1 𝑦2 (𝑥) = 𝑓1 (𝑥) + 𝛾1 , 𝑦2′′ (𝑥) + 𝛼2 𝑦1 (𝑥) + 𝛽2 𝑦2 (𝑥) = 𝑓2 (𝑥) + 𝛾2 , 0 ≤ 𝑥 ≤ 𝐿, 𝑦1 (0) = 𝛼3 , 𝑦2 (0) = 𝛽3 , 𝑦1 (𝐴1 ) = 𝛼4 , 𝑦2 (𝐴2 ) = 𝛽4 , where 𝛼1 , 𝛼2 , 𝛼3 , 𝛼4 , 𝛽1 , 𝛽2 , 𝛽3 , 𝛽4 , 𝛾1 , 𝛾2 , 𝐴1 , 𝐴2 are constants and 𝑓1 (𝑥), 𝑓2 (𝑥) are real valued functions. In order to solve system of differential equations, we suppose that [0, 𝐿] = [0, 1], otherwise the problem can be mapped from [0, 𝐿] to [0, 1] easily. For solving system of differential equations, it is sufficient to suppose that the approximate solution is as 𝑦1 (𝑥) = 𝐂 𝑇 𝛙(𝑥), 𝑦2 (𝑥) = 𝐃𝑇 𝛙(𝑥), where 𝐂, 𝐃 and 𝛙(𝑥) are given in above. Therefore, we obtain 𝐂 𝑇 𝛙′′ (𝑥) + 𝛼1 𝐂 𝑇 𝛙(𝑥) + 𝛽1 𝐃𝑇 𝛙(𝑥) = 𝑓1 (𝑥) + 𝛾1 , 𝐃𝑇 𝛙′′ (𝑥) + 𝛼2 𝐂 𝑇 𝛙(𝑥) + 𝛽2 𝐃𝑇 𝛙(𝑥) = 𝑓2 (𝑥) + 𝛾2 ,

0 ≤ 𝑥 ≤ 𝐿,

𝐂 𝑇 𝛙(0) = 𝛼3 , 𝐃𝑇 𝛙(0) = 𝛽3 . 𝐂 𝑇 𝛙(𝐴1 ) = 𝛼4 , 𝐃𝑇 𝛙(𝐴2 ) = 𝛽4 . Therefore, in order to apply the Laurent Wavelet Method (LWM); we require 2𝑘−1 𝑀 − 1 collocating points. Suitable collocating points are as 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2, … , 2𝑘−1 𝑀 − 1.

Implementing the collocating points and imposing the initial value to the system of differential equations, we obtain 𝐂 𝑇 𝛙′ (𝑥𝑖 ) + 𝛼1 𝐂 𝑇 𝛙(𝑥𝑖 ) + 𝛽1 𝐃𝑇 𝛙(𝑥𝑖 ) = 𝑓1 (𝑥𝑖 ) + 𝛾1 , 𝐃𝑇 𝛙′ (𝑥𝑖 ) + 𝛼2 𝐂 𝑇 𝛙(𝑥𝑖 ) + 𝛽2 𝐃𝑇 𝛙(𝑥𝑖 ) = 𝑓2 (𝑥𝑖 ) + 𝛾2 , 𝐂 𝑇 𝛙(0) = 𝛼3 , 𝐃𝑇 𝛙(0) = 𝛽3 . 𝐂 𝑇 𝛙(𝐴1 ) = 𝛼4 , 𝐃𝑇 𝛙(𝐴2 ) = 𝛽4 .

1071

0 ≤ 𝑥 ≤ 𝐿,

The differential equation yields 2𝑘−1 𝑀 − 3 equations and initial condition produces 2 number of equations. Therefore, the obtained system has 2𝑘−1 𝑀 − 1 equations and 2𝑘−1 𝑀 − 1 unknowns. Solving this system gives the unknown coefficients 𝐂.

System of Ordinary Differential Equations of 3×3 Consider the system of differential equations 𝑦1′ (𝑥) + 𝛼1 𝑦1 (𝑥) + 𝛽1 𝑦2 (𝑥) + 𝛿1 𝑦3 (𝑥) = 𝑓1 (𝑥) + 𝛾1 , 𝑦2′ (𝑥) + 𝛼2 𝑦1 (𝑥) + 𝛽2 𝑦2 (𝑥) + 𝛿2 𝑦3 (𝑥) = 𝑓2 (𝑥) + 𝛾2 , 𝑦3′ (𝑥) + 𝛼3 𝑦1 (𝑥) + 𝛽3 𝑦2 (𝑥) + 𝛿3 𝑦3 (𝑥) = 𝑓3 (𝑥) + 𝛾3 ,

0 ≤ 𝑥 ≤ 𝐿,

𝑦1 (0) = 𝛼, 𝑦2 (0) = 𝛽, 𝑦3 (0) = 𝛾. where 𝛼1 , 𝛼2 , 𝛼3 , 𝛽1 , 𝛽2 , 𝛽3 , 𝛾1 , 𝛾2 , 𝛾3 , 𝛿1 , 𝛿2 , 𝛿3 , 𝛼, 𝛽, 𝛾 are constants and 𝑓1 (𝑥), 𝑓2 (𝑥), 𝑓3 (𝑥) are real valued functions 𝐂 𝑇 𝛙′ (𝑥𝑖 ) + 𝛼1 𝐂 𝑇 𝛙(𝑥𝑖 ) + 𝛽1 𝐃𝑇 𝛙(𝑥𝑖 ) + 𝛿1 𝐄𝑇 𝛙(𝑥𝑖 ) = 𝑓1 (𝑥𝑖 ) + 𝛾1 , 𝐃𝑇 𝛙′ (𝑥𝑖 ) + 𝛼2 𝐂 𝑇 𝛙(𝑥𝑖 ) + 𝛽2 𝐃𝑇 𝛙(𝑥𝑖 ) + 𝛿2 𝐄𝑇 𝛙(𝑥𝑖 ) = 𝑓2 (𝑥𝑖 ) + 𝛾2 , 𝐄𝑇 𝛙′ (𝑥𝑖 ) + 𝛼3 𝐂 𝑇 𝛙(𝑥𝑖 ) + 𝛽3 𝐃𝑇 𝛙(𝑥𝑖 ) + 𝛿3 𝐄𝑇 𝛙(𝑥𝑖 ) = 𝑓3 (𝑥𝑖 ) + 𝛾3 ,

0≤

𝑥 ≤ 𝐿, 𝛙(0) = 𝛼, 𝛙(0) = 𝛽, 𝛙(0) = 𝛾. The differential equation yields 2𝑘−1 𝑀 − 1 equations and initial condition produces 𝑛 number of equations. Therefore, the obtained system has 2𝑘−1 𝑀 − 1 equations and 2𝑘−1 𝑀 − 1 unknowns. Solving this system gives the unknown coefficients 𝐂.

6.4.2. System of Ordinary Differential Equations of 2×2 Example 6.6 [194]. Consider the following system of linear ODEs as 𝑦1′ (𝑥) = 𝑦1 (𝑥) + 𝑦2 (𝑥),

(6.121)

𝑦2′ (𝑥) = −𝑦1 (𝑥) + 𝑦2 (𝑥),

(6.122)

subject to the boundary conditions 𝑦1 (0) = 0, 𝑦2 (0) = 1.

(6.123)

The exact solution of the above system is 𝑦1 (𝑥) = e𝑥 sin(𝑥) , 𝑦2 (𝑥) = e𝑥 cos(𝑥).

(6.124)

According to the Laurent Wavelets Method (LWM), we assume the trial solution 𝑘−1

𝐓 𝑦1 (𝑥) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥). 𝑘−1

𝐓 𝑦2 (𝑥) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑑𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐃 𝛙(𝑥).

1072

(6.125) (6.126)


𝐂 = [𝑐10 , 𝑐11 , … , 𝑐1𝑀−1 , 𝑐20 , 𝑐21 , … , 𝑐2𝑀−1 , … , 𝑐2𝑘−1 0 , … , 𝑐2𝑘−1 𝑀−1 ] , 𝑇

𝐃 = [𝑑10 , 𝑑11 , … , 𝑑1𝑀−1 , 𝑑20 , 𝑑21 , … , 𝑑2𝑀−1 , … , 𝑑2𝑘−1 0 , … , 𝑑2𝑘−1 𝑀−1 ] , and

𝑇


We apply the proposed technique to solve Eq. (6.122-6.123) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.126-6.127) is 𝑦1 (𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) + ⋯ + 𝑐1,9 𝜓1,9 (𝑥) = 𝐂 𝐓 𝛙(𝑥),

(6.127)

𝑦2 (𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑑𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑑1,0 𝜓1,0 (𝑥) + 𝑑1,1 𝜓1,1 (𝑥) + 𝑑1,2 𝜓1,2 (𝑥) + ⋯ + 𝑑1,9 𝜓1,9 (𝑥) = 𝐃𝐓 𝛙(𝑥), 1

(6.128)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐿𝑚 is the Laurent polynomials given as 𝐿0 = 1, 𝐿1 = 1 + 𝑥, 𝐿2 = 1 + 𝑥 + 𝑥 2 , 𝐿3 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 , 𝐿4 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 , . . .. Therefore, the Laurent wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(1 − 2𝑥 + 4𝑥 2 ), 𝜓1,3 (𝑥) = 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), . . .. Therefore, we have the trial solution is 1073

(6.129)

𝑦1 (𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑐1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑦2 (𝑥) = 𝑑1,0 + 2√3𝑥𝑑1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑑1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑑1,3 + ⋯ = 𝐃𝐓 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , 𝐃 = [𝑑1,0 , 𝑑1,1 , 𝑑1,2 , 𝑑1,3 , … ] , 𝑇

1, 2√3𝑥, √5(1 − 2𝑥 + 4𝑥 2 ), and 𝛙(𝑥) = [ ] . 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), … Substituting into the given problem we get 𝐂 𝐓 𝛙′ (𝑥) = 𝐂 𝐓 𝛙(𝑥) + 𝐃𝐓 𝛙(𝑥),

(6.130)

𝐃𝐓 𝛙′ (𝑥) = −𝐂 𝐓 𝛙(𝑥) + 𝐃𝐓 𝛙(𝑥),

(6.131)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 1, 𝐃𝐓 𝛙(0) = 1.

(6.132)

Substitute the collocating points are in Eq. (6.131-6.132), we have the system of equations 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 1,2,3, … 9.

(6.133)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 20×20 𝐂20×1 = 𝐛20×1 . After solving we get the following approximate solution 𝑦1 (𝑥) = 0.0000 + 0.99999𝑥 + 0.99999𝑥 2 + 0.33333𝑥 3 , 𝑦2 (𝑥) = 1 + 0.99999𝑥 + 0.00000𝑥 2 − 0.33333𝑥 3 .

1074

Table 6.16 (a): Comparison of the Exact Solution and Approximate Solutions of 𝑦1 (𝑥) obtained from Laurent Wavelet Method (LWM) __________________________________________________________________ 𝑥

Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

0.000000000000000

2.72353E-200

0.1

0.110332988730204

0.110332974183055

1.45471E-08

0.2

0.242655268594923

0.242655251432294

1.71626E-08

0.3

0.398910553778490

0.398910552329761

1.44873E-09

0.4

0.580943900770567

0.580943921561324

2.07908E-08

0.5

0.790439083213615

0.790439107558266

2.43447E-08

0.6

1.028845666272092

1.028845661724243

4.54785E-09

0.7

1.297295111875269

1.297295076277757

3.55975E-08

0.8

1.596505340600251

1.596505340818293

2.18042E-10

0.9

1.926673303972717

1.926673372327307

6.83546E-08

1.0

2.287355287178842

2.287354861443428

4.25735E-07

__________________________________________________________________

(a)

(b)


1075

Table 6.16 (b): Comparison of the Exact Solution and Approximate Solutions of 𝑦2 (𝑥) obtained from Laurent Wavelet Method (LWM) __________________________________________________________________ 𝑥

Exact Solution


Error in LWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

0.00000E+00

0.1

1.099649666829409

1.099649666824914

4.49520E-12

0.2

1.197056021355891

1.197056023109000

1.75311E-09

0.3

1.289569374044936

1.289569378061872

4.01694E-09

0.4

1.374061538887522

1.374061543488860

4.60134E-09

0.5

1.446889036584169

1.446889039082203

2.49803E-09

0.6

1.503859540558786

1.503859540275820

2.82966E-10

0.7

1.540203025431780

1.540203025791200

3.59420E-10

0.8

1.550549296807422

1.550549302080235

5.27281E-09

0.9

1.528913811884699

1.528913817047290

5.16259E-09

1.0

1.468693939915885

1.468693925707422

1.42085E-08

__________________________________________________________________

(a)

(b)


Now to solve Eq. (6.122-6.123) with 𝐾 = 1, and 𝑀 = 25. We have Eq. (6.126-6.127) is 𝑇 𝑦1 (𝑥) = ∑1𝑛=1 ∑24 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥),

1076

𝑇 𝑦2 (𝑥) = ∑1𝑛=1 ∑24 𝑚=0 𝑑𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐃 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , 𝐃 = [𝑑1,0 , 𝑑1,1 , 𝑑1,2 , … , 𝑑1,24 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] .

Proceeding as before we have the matrix form 𝐀 𝟓𝟎×𝟓𝟎 𝐂𝟓𝟎×𝟏 = 𝐛𝟓𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.17 (a): Comparison of the Exact Solution and Approximate Solutions of 𝑦1 (𝑥) obtained from Laurent Wavelet Method (LWM) __________________________________________________________________ 𝑥

Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

4.23575E-201

0.1

0.110332988730204

0.110332988730204

3.07924E-29

0.2

0.242655268594923

0.242655268594923

1.64472E-29

0.3

0.398910553778490

0.398910553778490

1.98532E-29

0.4

0.580943900770567

0.580943900770567

4.46259E-29

0.5

0.790439083213615

0.790439083213615

1.11307E-29

0.6

1.028845666272092

1.028845666272092

6.59369E-29

0.7

1.297295111875269

1.297295111875269

1.24931E-29

0.8

1.596505340600251

1.596505340600251

6.34520E-29

0.9

1.926673303972717

1.926673303972717

1.15831E-28

1.0

2.287355287178842

2.287355287178842

9.61105E-28

__________________________________________________________________

1077

(a)

(b)



Exact Solution


Error in LWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

9.00000E-200

0.1

1.099649666829409

1.099649666829409

2.95007E-29

0.2

1.197056021355891

1.197056021355891

8.22263E-30

0.3

1.289569374044936

1.289569374044936

9.91050E-30

0.4

1.374061538887522

1.374061538887522

2.72277E-29

0.5

1.446889036584169

1.446889036584169

1.19377E-29

0.6

1.503859540558786

1.503859540558786

3.61476E-29

0.7

1.540203025431780

1.540203025431780

2.60354E-29

0.8

1.550549296807422

1.550549296807422

1.46411E-29

0.9

1.528913811884699

1.528913811884699

6.16073E-29

1.0

1.468693939915885

1.468693939915885

1.06733E-27

__________________________________________________________________

1078

(a)

(b)


Now to solve Eq. (6.122-6.123) with 𝐾 = 1, and 𝑀 = 50. We have Eq. (6.126-6.127) is 𝑇 𝑦1 (𝑥) = ∑1𝑛=1 ∑49 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇 𝑦2 (𝑥) = ∑1𝑛=1 ∑49 𝑚=0 𝑑𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐃 𝛙(𝑥), 𝑇

𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] .


1079


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

0.000000000000000

3.64471E-201

0.1

0.110332988730204

0.110332988730204

2.03600E-72

0.2

0.242655268594923

0.242655268594923

1.42143E-72

0.3

0.398910553778490

0.398910553778490

1.17503E-72

0.4

0.580943900770567

0.580943900770567

2.87663E-72

0.5

0.790439083213615

0.790439083213615

2.86192E-72

0.6

1.028845666272092

1.028845666272092

2.63096E-72

0.7

1.297295111875269

1.297295111875269

3.99501E-72

0.8

1.596505340600251

1.596505340600251

3.81893E-72

0.9

1.926673303972717

1.926673303972717

4.26291E-72

1.0

2.287355287178842

2.287355287178842

2.63276E-70

__________________________________________________________________

(a)

(b)


1080


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.00000E-200

0.1

1.099649666829409

1.099649666829409

5.07866E-74

0.2

1.197056021355891

1.197056021355891

9.75179E-74

0.3

1.289569374044936

1.289569374044936

8.20239E-74

0.4

1.374061538887522

1.374061538887522

3.17431E-74

0.5

1.446889036584169

1.446889036584169

2.35744E-75

0.6

1.503859540558786

1.503859540558786

1.34519E-74

0.7

1.540203025431780

1.540203025431780

1.23583E-74

0.8

1.550549296807422

1.550549296807422

7.15917E-74

0.9

1.528913811884699

1.528913811884699

3.92543E-74

1.0

1.468693939915885

1.468693939915885

2.46051E-72

__________________________________________________________________

(a)

(b)


Example 6.7 [210]. Consider the following system of linear ODEs as 𝑦1′ (𝑥) + 𝑦2′ (𝑥) + 𝑦1 (𝑥) + 𝑦2 (𝑥) = 1, 1081

(6.134)

𝑦2′ (𝑥) = 2𝑦1 (𝑥) + 𝑦2 (𝑥),

(6.135)

subject to the boundary conditions 𝑦1 (0) = 0, 𝑦2 (0) = 1.

(6.136)

The exact solution of the above system is 𝑦1 (𝑥) = e−𝑥 − 1, 𝑦2 (𝑥) = 2 − e−𝑥 .

(6.137)




(6.138) (6.139)




𝑇



(6.140)


(6.141)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐿𝑚 is the Laurent polynomials given as 𝐿0 = 1, 𝐿1 = 1 + 𝑥, 𝐿2 = 1 + 𝑥 + 𝑥 2 , 𝐿3 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 , 𝐿4 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 , . . .. Therefore, the Laurent wavelets given as 1082

(6.142)

𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(1 − 2𝑥 + 4𝑥 2 ), 𝜓1,3 (𝑥) = 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), . . .. Therefore, we have the trial solution is 𝑦1 (𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑐1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑦2 (𝑥) = 𝑑1,0 + 2√3𝑥𝑑1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑑1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑑1,3 + ⋯ = 𝐃𝐓 𝛙(𝑥), 𝑇

𝑇


1, 2√3𝑥, √5(1 − 2𝑥 + 4𝑥 2 ), and 𝛙(𝑥) = [ ] . 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), … Substituting into the given problem we get 𝐂 𝐓 𝛙′ (𝑥) + 𝐃𝐓 𝛙′ (𝑥) + 𝐂 𝐓 𝛙(𝑥) + 𝐃𝐓 𝛙(𝑥) = 1,

(6.143)

𝐃𝐓 𝛙′ (𝑥) = 2𝐂 𝐓 𝛙(𝑥) + 𝐃𝐓 𝛙(𝑥),

(6.144)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 0, 𝐃𝐓 𝛙(0) = 1.

(6.145)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 1,2,3, … 9.

(6.146)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 20×20 𝐂20×1 = 𝐛20×1 . After solving we get the following approximate solution 𝑦1 (𝑥) = 0.0000 − 1.000000𝑥 + 0.499999𝑥 2 + 0.166666𝑥 3 + ⋯, 𝑦2 (𝑥) = 1 + 1.00000𝑥 − 0.4999999𝑥 2 + 0.1666666𝑥 3 + ⋯.

1083


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

0.000000000000000

4.20000E-200

0.1

-0.095162581964040

-0.095162582486670

5.22630E-10

0.2

-0.181269246922018

-0.181269247425636

5.03618E-10

0.3

-0.259181779318282

-0.259181779141116

1.77167E-10

0.4

-0.329679953964361

-0.329679953014541

9.49820E-10

0.5

-0.393469340287367

-0.393469339412430

8.74937E-10

0.6

-0.451188363905974

-0.451188364273623

3.67649E-10

0.7

-0.503414696208590

-0.503414697626302

1.41771E-09

0.8

-0.550671035882778

-0.550671035684646

1.98133E-10

0.9

-0.593430340259401

-0.593430337700866

2.55853E-09

1.0

-0.632120558828558

-0.632120575420709

1.65922E-08

__________________________________________________________________

(a)

(b)


1084


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.00000E-199

0.1

1.095162581964040

1.095162582486670

5.22630E-10

0.2

1.181269246922018

1.181269247425636

5.03618E-10

0.3

1.259181779318282

1.259181779141116

1.77167E-10

0.4

1.329679953964361

1.329679953014541

9.49820E-10

0.5

1.393469340287367

1.393469339412430

8.74937E-10

0.6

1.451188363905974

1.451188364273623

3.67649E-10

0.7

1.503414696208590

1.503414697626302

1.41771E-09

0.8

1.550671035882778

1.550671035684646

1.98133E-10

0.9

1.593430340259401

1.593430337700866

2.55853E-09

1.0

1.632120558828558

1.632120575420709

1.65922E-08

__________________________________________________________________

(a)

(b)



1085


𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] .


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

1.27311E-200

0.1

-0.095162581964040

-0.095162581964040

7.43968E-33

0.2

-0.181269246922018

-0.181269246922018

1.36335E-33

0.3

-0.259181779318282

-0.259181779318282

2.12529E-33

0.4

-0.329679953964361

-0.329679953964361

6.33046E-33

0.5

-0.393469340287367

-0.393469340287367

3.32683E-33

0.6

-0.451188363905974

-0.451188363905974

9.46713E-33

0.7

-0.503414696208590

-0.503414696208590

5.85603E-33

0.8

-0.550671035882778

-0.550671035882778

5.28134E-33

0.9

-0.593430340259401

-0.593430340259401

1.91598E-32

1.0

-0.632120558828558

-0.632120558828558

2.75196E-31

__________________________________________________________________

1086

(a)

(b)



Exact Solution


Error in LWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.00000E-199

0.1

1.095162581964040

1.095162581964040

7.43968E-33

0.2

1.181269246922018

1.181269246922018

1.36335E-33

0.3

1.259181779318282

1.259181779318282

2.12529E-33

0.4

1.329679953964361

1.329679953964361

6.33046E-33

0.5

1.393469340287367

1.393469340287367

3.32683E-33

0.6

1.451188363905974

1.451188363905974

9.46713E-33

0.7

1.503414696208590

1.503414696208590

5.85603E-33

0.8

1.550671035882778

1.550671035882778

5.28134E-33

0.9

1.593430340259401

1.593430340259401

1.91598E-32

1.0

1.632120558828558

1.632120558828558

2.75196E-31

__________________________________________________________________

1087

(a)

(b)



𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] .


1088


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

9.53408E-200

0.1

-0.095162581964040

-0.095162581964040

6.46848E-80

0.2

-0.181269246922018

-0.181269246922018

4.82699E-80

0.3

-0.259181779318282

-0.259181779318282

3.25035E-80

0.4

-0.329679953964361

-0.329679953964361

8.81677E-80

0.5

-0.393469340287367

-0.393469340287367

9.04509E-80

0.6

-0.451188363905974

-0.451188363905974

8.46780E-80

0.7

-0.503414696208590

-0.503414696208590

1.26813E-79

0.8

-0.550671035882778

-0.550671035882778

1.16704E-79

0.9

-0.593430340259401

-0.593430340259401

1.27693E-79

1.0

-0.632120558828558

-0.632120558828558

8.20644E-78

__________________________________________________________________

(a)

(b)


1089


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.80000E-199

0.1

1.095162581964040

1.095162581964040

6.46848E-80

0.2

1.181269246922018

1.181269246922018

4.82699E-80

0.3

1.259181779318282

1.259181779318282

3.25035E-80

0.4

1.329679953964361

1.329679953964361

8.81677E-80

0.5

1.393469340287367

1.393469340287367

9.04509E-80

0.6

1.451188363905974

1.451188363905974

8.46780E-80

0.7

1.503414696208590

1.503414696208590

1.26813E-79

0.8

1.550671035882778

1.550671035882778

1.16704E-79

0.9

1.593430340259401

1.593430340259401

1.27693E-79

1.0

1.632120558828558

1.632120558828558

8.20644E-78

__________________________________________________________________

(a)

(b)


Example 6.8 [43]. Consider the following system of linear ODEs as 𝑦1′′ (𝑥) + 𝑦2 (𝑥) = 1,

(6.147) 1090

𝑦2′′ (𝑥) + 𝑦1 (𝑥) = 0,

(6.148)

subject to the boundary conditions 𝑦1 (0) = 𝑦1′ (0) = 𝑦2 (0) = 𝑦2′ (0) = 0.

(6.149)

The exact solution of the above system is 𝑦1 (𝑥) =

e𝑥 4

+

e−𝑥 4

−

cos(𝑥) 2

, 𝑦2 (𝑥) = 1 −

e𝑥 4

−

e−𝑥 4

−

cos(2) 2

.

(6.150)




(6.151) (6.152)




𝑇



(6.153)


(6.154)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐿𝑚 is the Laurent polynomials given as 𝐿0 = 1, 𝐿1 = 1 + 𝑥, 𝐿2 = 1 + 𝑥 + 𝑥 2 , 𝐿3 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 , 𝐿4 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 , . . .. 1091

(6.155)

Therefore, the Laurent wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(1 − 2𝑥 + 4𝑥 2 ), 𝜓1,3 (𝑥) = 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), . . .. Therefore, we have the trial solution is 𝑦1 (𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑐1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑦2 (𝑥) = 𝑑1,0 + 2√3𝑥𝑑1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑑1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑑1,3 + ⋯ = 𝐃𝐓 𝛙(𝑥), 𝑇

𝑇


1, 2√3𝑥, √5(1 − 2𝑥 + 4𝑥 2 ), and 𝛙(𝑥) = [ ] . 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), … Substituting into the given problem we get 𝐂 𝐓 𝛙′′ (𝑥) + 𝐃𝐓 𝛙(𝑥) = 1,

(6.156)

𝐃𝐓 𝛙′′ (𝑥) + 𝐂 𝐓 𝛙(𝑥) = 0,

(6.157)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 𝐂 𝐓 𝛙′ (0) = 𝐃𝐓 𝛙(0) = 𝐃𝐓 𝛙′ (0) = 0.

(6.158)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 1,2,3, … 9.

(6.159)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 20×20 𝐂20×1 = 𝐛20×1 . After solving we get the following approximate solution 𝑦1 (𝑥) = 0.000000𝑥 + 0.500000𝑥 2 + 1.95950𝑥 3 + ⋯, 1092

𝑦2 (𝑥) = 0.00000𝑥 − 1.27594𝑥 2 − 6.685018 × 10−10 𝑥 3 + ⋯. Table 6.22 (a): Comparison of the Exact Solution and Approximate Solutions of 𝑦1 (𝑥) obtained from Laurent Wavelet Method (LWM) __________________________________________________________________ 𝑥

Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

0.000000000000000

0.00000E+00

0.1

0.005000001388889

0.005000001660275

2.71386E-10

0.2

0.020000088888917

0.020000089753285

8.64368E-10

0.3

0.045001012501627

0.045001013570701

1.06907E-09

0.4

0.080005688917785

0.080005689092379

1.74595E-10

0.5

0.125021701658004

0.125021700166457

1.49155E-09

0.6

0.180064801666295

0.180064799866212

1.80008E-09

0.7

0.245163409173227

0.245163410831329

1.65810E-09

0.8

0.320364118478840

0.320364122434806

3.95597E-09

0.9

0.405738208589055

0.405738181415412

2.71736E-08

1.0

0.501389164473552

0.501388971179394

1.93294E-07

__________________________________________________________________

(a)

(b)


1093


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

0.000000000000000

4.00000E-202

0.1

-0.000004166666915

-0.000004166679489

1.25746E-11

0.2

-0.000066666730159

-0.000066666772578

4.24190E-11

0.3

-0.000337501627233

-0.000337501692217

6.49835E-11

0.4

-0.001066682920670

-0.001066682975777

5.51070E-11

0.5

-0.002604263548377

-0.002604263566385

1.80087E-11

0.6

-0.005400416575973

-0.005400416586981

1.10081E-11

0.7

-0.010005596457716

-0.010005596562262

1.04546E-10

0.8

-0.017070827826005

-0.017070828013924

1.87919E-10

0.9

-0.027348176859719

-0.027348176318507

5.41212E-10

1.0

-0.041691470341692

-0.041691465674752

4.66694E-09

__________________________________________________________________

(a)

(b)



1094


𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] .


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

1.14961E-200

0.1

0.005000001388889

0.005000001388889

6.33174E-34

0.2

0.020000088888917

0.020000088888917

1.95783E-33

0.3

0.045001012501627

0.045001012501627

1.86931E-33

0.4

0.080005688917785

0.080005688917785

3.61809E-33

0.5

0.125021701658004

0.125021701658004

3.46078E-33

0.6

0.180064801666295

0.180064801666295

5.14709E-33

0.7

0.245163409173227

0.245163409173227

4.66750E-33

0.8

0.320364118478840

0.320364118478840

7.94963E-33

0.9

0.405738208589055

0.405738208589055

7.57894E-33

1.0

0.501389164473552

0.501389164473552

3.87242E-31

__________________________________________________________________

1095

(a)

(b)



Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

0.000000000000000

8.90345E-203

0.1

-0.000004166666915

-0.000004166666915

7.40419E-36

0.2

-0.000066666730159

-0.000066666730159

2.97222E-35

0.3

-0.000337501627233

-0.000337501627233

5.43935E-35

0.4

-0.001066682920670

-0.001066682920670

1.19807E-34

0.5

-0.002604263548377

-0.002604263548377

1.98909E-34

0.6

-0.005400416575973

-0.005400416575973

3.33825E-34

0.7

-0.010005596457716

-0.010005596457716

4.95840E-34

0.8

-0.017070827826005

-0.017070827826005

7.47014E-34

0.9

-0.027348176859719

-0.027348176859719

1.02628E-33

1.0

-0.041691470341692

-0.041691470341692

4.90875E-33

__________________________________________________________________

1096

(a)

(b)



𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] .


1097


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

0.000000000000000

1.89998E-200

0.1

0.005000001388889

0.005000001388889

1.97882E-80

0.2

0.020000088888917

0.020000088888917

9.84515E-80

0.3

0.045001012501627

0.045001012501627

1.21666E-79

0.4

0.080005688917785

0.080005688917785

4.46650E-80

0.5

0.125021701658004

0.125021701658004

7.03903E-80

0.6

0.180064801666295

0.180064801666295

1.59184E-79

0.7

0.245163409173227

0.245163409173227

1.40353E-79

0.8

0.320364118478840

0.320364118478840

3.96117E-79

0.9

0.405738208589055

0.405738208589055

1.09469E-78

1.0

0.501389164473552

0.501389164473552

5.67023E-76

__________________________________________________________________

(a)

(b)


1098


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

1.40872E-201

0.1

-0.000004166666915

-0.000004166666915

3.96447E-82

0.2

-0.000066666730159

-0.000066666730159

1.55531E-81

0.3

-0.000337501627233

-0.000337501627233

2.74588E-81

0.4

-0.001066682920670

-0.001066682920670

3.63669E-81

0.5

-0.002604263548377

-0.002604263548377

4.45379E-81

0.6

-0.005400416575973

-0.005400416575973

5.41402E-81

0.7

-0.010005596457716

-0.010005596457716

6.81048E-81

0.8

-0.017070827826005

-0.017070827826005

1.06164E-80

0.9

-0.027348176859719

-0.027348176859719

3.79024E-81

1.0

-0.041691470341692

-0.041691470341692

2.86399E-78

__________________________________________________________________

(a)

(b)


6.4.3. System of Ordinary Differential Equation of 3×3 Example 6.9 [43]. Consider the following system of linear ODEs as 1099

𝑦1′ (𝑥) + cos(𝑥) = 𝑦3 (𝑥),

(6.160)

𝑦2′ (𝑥) + exp(𝑥) = 𝑦3 (𝑥),

(6.161)

𝑦3′ (𝑥) = 𝑦1 (𝑥) − 𝑦2 (𝑥),

(6.162)

subject to the boundary conditions 𝑦1 (0) = 1, 𝑦2 (0) = 0, 𝑦3 (0) = 2.

(6.163)

The exact solution of the above system is 𝑦1 (𝑥) = e𝑥 , 𝑦2 (𝑥) = sin(𝑥) , 𝑦3 (𝑥) = e𝑥 + cos(𝑥).

(6.164)



𝐓 𝑦2 (𝑥) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑑𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐃 𝛙(𝑥). 𝑘−1

𝐓 𝑦3 (𝑥) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑒𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐄 𝛙(𝑥).

(6.165) (6.166) (6.167)



𝐃 = [𝑑10 , 𝑑11 , … , 𝑑1𝑀−1 , 𝑑20 , 𝑑21 , … , 𝑑2𝑀−1 , … , 𝑑2𝑘−1 0 , … , 𝑑2𝑘−1 𝑀−1 ] , 𝑇

𝐄 = [𝑒10 , 𝑒11 , … , 𝑒1𝑀−1 , 𝑒20 , 𝑒21 , … , 𝑒2𝑀−1 , … , 𝑒2𝑘−1 0 , … , 𝑒2𝑘−1 𝑀−1 ] , and

𝑇



(6.168)

𝑦2 (𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑑𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑑1,0 𝜓1,0 (𝑥) + 𝑑1,1 𝜓1,1 (𝑥) + 𝑑1,2 𝜓1,2 (𝑥) + ⋯ + 𝑑1,9 𝜓1,9 (𝑥) = 𝐃𝐓 𝛙(𝑥),

(6.169)

𝑦3 (𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑑𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑒1,0 𝜓1,0 (𝑥) + 𝑒1,1 𝜓1,1 (𝑥) + 𝑒1,2 𝜓1,2 (𝑥) + ⋯ + 𝑒1,9 𝜓1,9 (𝑥) = 𝐄𝐓 𝛙(𝑥), 1

(6.170)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐿𝑚 is the Laurent polynomials given as 𝐿0 = 1, 𝐿1 = 1 + 𝑥, 1100

(6.171)

𝐿2 = 1 + 𝑥 + 𝑥 2 , 𝐿3 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 , 𝐿4 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 , . . .. Therefore, the Laurent wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(1 − 2𝑥 + 4𝑥 2 ), 𝜓1,3 (𝑥) = 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), . . .. Therefore, we have the trial solution is 𝑦1 (𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑐1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑦2 (𝑥) = 𝑑1,0 + 2√3𝑥𝑑1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑑1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑑1,3 + ⋯ = 𝐃𝐓 𝛙(𝑥), 𝑦3 (𝑥) = 𝑒1,0 + 2√3𝑥𝑒1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑒1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑒1,3 + ⋯ = 𝐄𝐓 𝛙(𝑥), 𝑇

𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … ] , 𝐃 = [𝑑1,0 , 𝑑1,1 , 𝑑1,2 , … ] , 𝐄 = [𝑒1,0 , 𝑒1,1 , 𝑒1,2 , … ] , 𝑇

1, 2√3𝑥, √5(1 − 2𝑥 + 4𝑥 2 ), and 𝛙(𝑥) = [ ] . 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), … Substituting into the given problem we get 𝐂 𝐓 𝛙′ (𝑥) + cos(𝑥) = 𝐄𝐓 𝛙(𝑥),

(6.172)

𝐃𝐓 𝛙′ (𝑥) + exp(𝑥) = 𝐄𝐓 𝛙(𝑥),

(6.173)

𝐄𝐓 𝛙′ (𝑥) = 𝐂 𝐓 𝛙(𝑥) − 𝐃𝐓 𝛙(𝑥),

(6.174)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 1, 𝐃𝐓 𝛙(0) = 0, 𝐄𝐓 𝛙(0) = 3. 1101

(6.175)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 1,2,3, … 9.

(6.176)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 30×30 𝐂30×1 = 𝐛30×1 . After solving we get the following approximate solution 𝑦1 (𝑥) = 0.999999 + 0.999999𝑥 + 0.499999𝑥 2 + 0.166666𝑥 3 + ⋯, 𝑦2 (𝑥) = 2.0000 × 10−200 + 0.99999𝑥 + 1.6847𝑥 2 − 0.16666𝑥 3 + ⋯. 𝑦3 (𝑥) = 2 + 1.00000𝑥 − 0.000000𝑥 2 + 0.166667𝑥 3 + ⋯. Table 6.25: Comparison of the Exact Solution and Approximate Solution of Eq. (6.1616.163) obtained from Laurent Wavelet Method (LWM) __________________________________________________________________ 𝑥

Error in 𝑦1 (𝑥)



__________________________________________________________________ 0.0

0.0000000000E+00

2.0000000000E-200

0.0000000000E+00

0.1

4.4899552332E-10

4.7827736644E-11

5.7541040532E-12

0.2

4.8136724030E-10

5.2678793886E-11

3.8318462125E-11

0.3

6.2478754010E-11

1.2840218419E-11

9.2360730327E-11

0.4

7.5290290716E-10

2.9568051239E-11

9.1749310743E-11

0.5

7.6898772643E-10

2.6217188746E-11

1.1509511459E-11

0.6

2.6728721941E-10

2.3616093432E-11

8.0109401240E-11

0.7

1.2570874952E-09

5.7830200791E-11

5.5719202497E-11

0.8

8.0677619668E-11

2.1715615543E-11

1.0036895522E-10

0.9

2.3074079569E-09

1.2774681928E-11

7.6451489657E-11

1.0

1.5097839516E-08

7.8668811422E-11

6.0279354989E-10

__________________________________________________________________

1102

(a)

(b)


(a)

(b)


1103

(a)

(b)


Now to solve Eq. (6.161-6.163) with 𝐾 = 1, and 𝑀 = 20. We have Eq. (6.166-6.168) is 𝑇 𝑦1 (𝑥) = ∑1𝑛=1 ∑19 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇 𝑦2 (𝑥) = ∑1𝑛=1 ∑19 𝑚=0 𝑑𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐃 𝛙(𝑥), 𝑇 𝑦3 (𝑥) = ∑1𝑛=1 ∑19 𝑚=0 𝑒𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐄 𝛙(𝑥), 𝑇

𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , … , 𝑐1,19 ] , 𝐃 = [𝑑1,0 , 𝑑1,1 , … , 𝑑1,19 ] , 𝐄 = [𝑒1,0 , 𝑒1,1 , … , 𝑒1,19 ] , and

𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,19 ] .

Proceeding as before we have the matrix form 𝐀 𝟔𝟎×𝟔𝟎 𝐱 𝟔𝟎×𝟏 = 𝐛𝟔𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table.

1104

Table 6.26: Comparison of the Exact Solution and Approximate Solution of Eq. (6.1616.163) obtained from Laurent Wavelet Method (LWM) __________________________________________________________________ 𝑥




__________________________________________________________________ 0.0

0.0000000000E+00

2.0000000000E-200

1.0000000000E-199

0.1

8.1228000715E-25

7.4468041757E-26

1.4787329874E-24

0.2

6.4827829507E-25

1.7467115224E-25

1.5271842024E-24

0.3

2.1133283219E-25

2.4015087376E-26

3.8466785580E-25

0.4

1.0940678607E-24

3.1252973774E-26

2.0959268285E-24

0.5

1.0670373854E-24

2.0241133544E-25

2.4003855106E-24

0.6

4.1807410555E-25

1.1194321104E-25

9.7757385048E-25

0.7

3.6188344260E-25

1.5616315204E-25

4.7523909899E-25

0.8

6.1416184770E-25

5.2119447645E-26

1.1447187237E-24

0.9

1.4508637065E-24

3.2764470041E-25

3.3538256539E-24

1.0

3.9447395352E-23

4.0861754579E-24

8.3934299515E-23

__________________________________________________________________

(a)

(b)


1105

(a)

(b)


(a)

(b)



𝑇

𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] .

Proceeding as before we have the matrix form 𝐀 𝟏𝟓𝟎×𝟏𝟓𝟎 𝐱 𝟏𝟓𝟎×𝟏 = 𝐛𝟏𝟓𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. 1106





__________________________________________________________________ 0.0

1.1000000000E-199

1.3030132872E-200 1.0000000000E-199

0.1

6.1645440183E-80

1.0079912061E-81

1.2428908336E-81

0.2

4.4591452206E-80

4.9902901296E-82

2.4172508430E-81

0.3

3.3546017193E-80

7.7800452479E-82

1.7734649150E-81

0.4

8.6391187298E-80

1.4085359086E-81

7.0953481556E-83

0.5

8.7600851373E-80

1.2945002379E-81

1.1708374189E-81

0.6

8.1609872415E-80

1.1135241221E-81

1.7092540945E-81

0.7

1.2313869951E-79

1.2450256009E-81

1.8384074311E-81

0.8

1.1583271551E-79

1.0202531357E-81

5.8035525315E-82

0.9

1.2643705869E-79

2.4843704114E-82

5.6628432104E-82

1.0

8.0460539476E-78

2.0962930286E-81

7.7304425498E-80

__________________________________________________________________

(a)

(b)


1107

(a)

(b)


(a)

(b)


Example 6.10 [19]. Consider the following of linear ODEs as 1

𝑦 ′′′ (𝑥) = 𝑥 𝑦(𝑥) + 𝑦 ′ (𝑥),

(6.177)

subject to the boundary conditions 𝑦(0) = 0, 𝑦 ′ (0) = 1, 𝑦 ′′ (0) = 2,

(6.178)

The exact solution of the above system is 𝑦(𝑥) = 𝑥e𝑥 .

(6.179)

Considering three functions, 𝑦1 (𝑥) = 𝑦(𝑥), 𝑦2 (𝑥) = 𝑦 ′ (𝑥), and 𝑦3 (𝑥) = 𝑦 ′′ (𝑥), we can convert (6.178) into the following non-linear system of three differential equation of order one 1108

𝑦1′ (𝑥) = 𝑦2 (𝑥),

(6.180)

𝑦2′ (𝑥) = 𝑦3 (𝑥),

(6.181)

1

𝑦3′ (𝑥) = 𝑥 𝑦1 (𝑥) + 𝑦2 (𝑥).

(6.182)

subject to the boundary conditions 𝑦1 (0) = 0, 𝑦2 (0) = 1, 𝑦2 (0) = 2,

(6.183)

The exact solution of the above system is 𝑦1 (𝑥) = 𝑥e𝑥 , 𝑦2 (𝑥) = e𝑥 + 𝑥e𝑥 , 𝑦3 (𝑥) = 2e𝑥 + 𝑥e𝑥 .

(6.184)





(6.185) (6.186) (6.187)





𝑇


We apply the proposed technique to solve Eq. (6.181-6.183) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.186-6.188) is 𝑦1 (𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) , = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) + ⋯ + 𝑐1,9 𝜓1,9 (𝑥) = 𝐂 𝐓 𝛙(𝑥),

(6.188)


(6.189)


(6.190)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐿𝑚 is the Laurent polynomials given as 𝐿0 = 1, 𝐿1 = 1 + 𝑥, 1109

(6.191)

𝐿2 = 1 + 𝑥 + 𝑥 2 , 𝐿3 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 , 𝐿4 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 , . . .. Therefore, the Laurent wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(1 − 2𝑥 + 4𝑥 2 ), 𝜓1,3 (𝑥) = 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), . . .. Therefore, we have the trial solution is 𝑦1 (𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑐1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑦2 (𝑥) = 𝑑1,0 + 2√3𝑥𝑑1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑑1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑑1,3 + ⋯ = 𝐃𝐓 𝛙(𝑥), 𝑦3 (𝑥) = 𝑒1,0 + 2√3𝑥𝑒1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑒1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑒1,3 + ⋯ = 𝐄𝐓 𝛙(𝑥), 𝑇

𝑇

𝑇


1, 2√3𝑥, √5(1 − 2𝑥 + 4𝑥 2 ), and 𝛙(𝑥) = [ ] . 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), … Substituting into the given problem we get 𝐂 𝐓 𝛙′ (𝑥) = 𝐃𝐓 𝛙(𝑥),

(6.192)

𝐃𝐓 𝛙′ (𝑥) = 𝐄𝐓 𝛙(𝑥),

(6.193)

1

𝐄𝐓 𝛙′ (𝑥) = 𝑥 𝐂 𝐓 𝛙(𝑥) + 𝐃𝐓 𝛙(𝑥),

(6.194)

subject to the boundary conditions 1110

𝐂 𝐓 𝛙(0) = 0, 𝐃𝐓 𝛙(0) = 1, 𝐄𝐓 𝛙(0) = 2.

(6.195)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 1,2,3, … 9.

(6.196)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 30×30 𝐂30×1 = 𝐛30×1 . After solving we get the following approximate solution 𝑦1 (𝑥) = 0.000000 + 0.999999𝑥 + 0.999999𝑥 2 + 0.500000𝑥 3 + ⋯, 𝑦2 (𝑥) = 0.99999 + 1.999999𝑥 + 1.49999𝑥 2 + ⋯. 𝑦3 (𝑥) = 2 + 2.99999𝑥 + 1.999999𝑥 2 + 0.8333333𝑥 3 + ⋯. Table 6.28: Comparison of the Exact Solution and Approximate Solution of Eq. (6.1816.183) obtained from Laurent Wavelet Method (LWM) __________________________________________________________________ 𝑥




__________________________________________________________________ 0.0

2.9290369600E-200

0.0000000000E+00

1.0000000000E-199

0.1

4.6859185600E-09

5.3777226999E-09

1.0359294565E-08

0.2

5.6637579745E-09

7.0930858888E-09

1.5938917538E-08

0.3

9.2460259151E-10

2.8110518721E-09

1.3226155144E-08

0.4

5.7802848803E-09

3.5365972885E-09

6.1853837071E-09

0.5

6.4256363150E-09

3.3095670288E-09

4.8688702150E-09

0.6

3.4157154812E-09

8.2780932855E-09

1.6017834242E-08

0.7

1.4160668291E-08

2.0776543285E-08

2.9970664554E-08

0.8

3.7800379720E-09

1.0187174940E-08

2.0926895784E-08

0.9

1.6823874449E-08

1.1396628506E-08

1.6967621712E-09

1.0

1.4292111625E-07

1.6489089779E-07

1.7763249844E-07

__________________________________________________________________

1111

(a)

(b)


(a)

(b)


1112

(a)

(b)



𝑇

𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,19 ] .


1113





__________________________________________________________________ 0.0

6.1397917793E-202

4.0000000000E-200

0.0000000000E+00

0.1

1.5260852014E-23

1.7360753300E-23

4.2761910654E-23

0.2

1.3926591117E-23

1.0451400062E-23

1.7466048941E-23

0.3

2.8218944906E-24

3.6924209889E-24

2.5595563252E-23

0.4

2.3322668114E-23

3.3366159795E-23

5.9497220926E-23

0.5

1.9600448558E-23

9.0212949645E-24

1.8090400185E-23

0.6

1.4553789684E-23

2.9375784090E-23

5.4101413464E-23

0.7

1.3071048225E-23

3.0292317574E-23

5.9632437241E-23

0.8

1.5489507733E-24

1.7787620177E-23

4.4831749557E-23

0.9

1.8677887321E-23

2.4310530719E-24

3.3833237213E-23

1.0

8.1191968820E-22

8.2736250783E-22

7.9909698108E-22

__________________________________________________________________

(a)

(b)


1114

(a)

(b)


(a)

(b)



𝑇

𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] .






__________________________________________________________________ 0.0

5.3601960290E-201

2.0000000000E-199

1.0000000000E-199

0.1

3.0235049603E-78

2.6733168673E-78

1.8305185692E-78

0.2

2.0276641515E-78

1.2088276311E-78

3.6956587698E-78

0.3

1.9916435613E-78

3.3467728134E-78

8.2348474872E-78

0.4

4.7345682851E-78

6.6120793180E-78

1.1394434709E-77

0.5

4.9793317572E-78

7.3400841016E-78

1.2122719038E-77

0.6

4.9804151199E-78

7.8300864073E-78

1.2736294813E-77

0.7

7.4595469363E-78

1.0860470522E-77

1.5993986494E-77

0.8

7.7346047494E-78

1.1659476075E-77

1.7175024265E-77

0.9

3.7176048068E-78

5.2556197417E-79

6.1033938605E-78

1.0

3.9770738018E-76

4.1053943539E-76

4.1678154746E-76

__________________________________________________________________

(a)

(b)


1116

(a)

(b)


(a)

(b)



1117

Example 6.11 [31]. Consider the following system of linear ODEs as 𝑦1′′ (𝑥) = (𝑥 + 1)𝑦2′ (𝑥) + 𝑥𝑦1 (𝑥),

(6.197)

𝑦2′ (𝑥) = (𝑥 − 1)e𝑥 − (𝑥 − 1)𝑦3 (𝑥) − 𝑦1 (𝑥),

(6.198)

𝑦3′ (𝑥) = 𝑥𝑦1 (𝑥) + e𝑥 − 𝑥 sin(𝑥),

(6.199)

subject to the boundary conditions 𝑦1 (0) = 0, 𝑦1′ (0) = 1, 𝑦2 (0) = 2, 𝑦3 (0) = 1.

(6.200)

The exact solution of the above system is 𝑦1 (𝑥) = sin(𝑥) , 𝑦2 (𝑥) = cos(𝑥) + 1, 𝑦3 (𝑥) = e𝑥 .

(6.201)





(6.202) (6.203) (6.204)





𝑇



(6.205)


(6.206)


(6.207)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐿𝑚 is the Laurent polynomials given as 𝐿0 = 1, 1118

(6.208)

𝐿1 = 1 + 𝑥, 𝐿2 = 1 + 𝑥 + 𝑥 2 , 𝐿3 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 , 𝐿4 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 , . . .. Therefore, the Laurent wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(1 − 2𝑥 + 4𝑥 2 ), 𝜓1,3 (𝑥) = 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), . . .. Therefore, we have the trial solution is 𝑦1 (𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑐1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑦2 (𝑥) = 𝑑1,0 + 2√3𝑥𝑑1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑑1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑑1,3 + ⋯ = 𝐃𝐓 𝛙(𝑥), 𝑦3 (𝑥) = 𝑒1,0 + 2√3𝑥𝑒1,1 + √5(1 − 2𝑥 + 4𝑥 2 )𝑒1,2 + 4√7𝑥(1 − 2𝑥 + 2𝑥 2 )𝑒1,3 + ⋯ = 𝐄𝐓 𝛙(𝑥), 𝑇

𝑇

𝑇


1, 2√3𝑥, √5(1 − 2𝑥 + 4𝑥 2 ), and 𝛙(𝑥) = [ ] . 4√7𝑥(1 − 2𝑥 + 2𝑥 2 ), … Substituting into the given problem we get 𝐂 𝐓 𝛙′′ (𝑥) = (𝑥 + 1)𝐃𝐓 𝛙′ (𝑥) + 𝑥𝐂 𝐓 𝛙(𝑥),

(6.209)

𝐃𝐓 𝛙′ (𝑥) = (𝑥 − 1)e𝑥 − (𝑥 − 1)𝐄𝐓 𝛙(𝑥) − 𝐂 𝐓 𝛙(𝑥),

(6.210)

𝐄𝐓 𝛙′ (𝑥) = 𝑥𝐂 𝐓 𝛙(𝑥) + e𝑥 − 𝑥 sin(𝑥),

(6.211)


𝐂 𝐓 𝛙(0) = 0, 𝐂 𝐓 𝛙′ (0) = 1, 𝐃𝐓 𝛙(0) = 2, 𝐄𝐓 𝛙(0) = 1.

(6.212)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 1,2,3, … 9.

(6.213)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 30×30 𝐂30×1 = 𝐛30×1 . After solving we get the following approximate solution 𝑦1 (𝑥) = 0.000000 + 0.999999𝑥 + 6.6917 × 10−9 𝑥 2 − 0.166666𝑥 3 + ⋯, 𝑦2 (𝑥) = 2 + 0.47585𝑥 − 0.49999𝑥 2 + ⋯. 𝑦3 (𝑥) = 1 + 0.99999𝑥 + 0.499999𝑥 2 + 0.166667𝑥 3 + ⋯. Table 6.31: Comparison of the Exact Solution and Approximate Solution of Eq. (6.1986.200) obtained from Laurent Wavelet Method (LWM) __________________________________________________________________ 𝑥




__________________________________________________________________ 0.0

2.8388164400E-200

0.0000000000E+00

1.0000000000E-199

0.1

5.4398250496E-11

4.6094875056E-10

4.4200553413E-10

0.2

1.6834403244E-10

4.4052665338E-10

4.7083052446E-10

0.3

2.0524084290E-10

1.8594970938E-10

7.2166058719E-11

0.4

4.6790689046E-11

9.1591799007E-10

7.5970939747E-10

0.5

2.3701318146E-10

8.8320835725E-10

7.7656152739E-10

0.6

3.0727296168E-10

2.4558107481E-10

2.4964968954E-10

0.7

1.6684753117E-10

1.2564309584E-09

1.2235929503E-09

0.8

5.0451623081E-10

1.6298158059E-10

1.0516723713E-10

0.9

3.1306283738E-09

2.3145552614E-09

2.2094977434E-09

1.0

2.1368709475E-08

1.5721796006E-08

1.5626051199E-08

__________________________________________________________________

1120

(a)

(b)


(a)

(b)


1121

(a)

(b)



𝑇

𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,19 ] .


1122





__________________________________________________________________ 0.0

3.2404428729E-200

2.0000000000E-199

1.0000000000E-200

0.1

1.1591797657E-25

7.4513420842E-25

7.0349808088E-25

0.2

1.1314712020E-25

7.3669874800E-25

7.9301550941E-25

0.3

7.1717063487E-26

2.0140497628E-25

1.7750527239E-25

0.4

1.1186562302E-25

1.0155628242E-24

1.0347315782E-24

0.5

8.1987745884E-26

1.1823456771E-24

1.2362365103E-24

0.6

1.7204321855E-25

4.7156492361E-25

5.2810795967E-25

0.7

2.8704672298E-25

2.0213776028E-25

2.0813500684E-25

0.8

4.4076971907E-25

5.4571146915E-25

5.5096206811E-25

0.9

4.8356126477E-25

1.6946124890E-24

1.7641286834E-24

1.0

5.9964163813E-23

4.1988282587E-23

4.3048742974E-23

__________________________________________________________________

(a)

(b)


1123

(a)

(b)


(a)

(b)



𝑇

𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] .






__________________________________________________________________ 0.0

3.0000000000E-200

3.0000000000E-199

0.0000000000E+00

0.1

8.7919504133E-82

6.3905928767E-80

6.1751728354E-80

0.2

3.7610894985E-81

4.7843974342E-80

4.4804116731E-80

0.3

4.3051124484E-81

3.1637001595E-80

3.3214912921E-80

0.4

1.4269915513E-81

8.6050250951E-80

8.5947429496E-80

0.5

2.3147114263E-81

8.7668195738E-80

8.7077804514E-80

0.6

4.8630732806E-81

8.1524596707E-80

8.1051074417E-80

0.7

4.2739888886E-81

1.2309923278E-79

1.2258576611E-79

0.8

8.9777300818E-81

1.1496854908E-79

1.1535078659E-79

0.9

2.5256749771E-80

1.2806286943E-79

1.2672022598E-79

1.0

1.1591366820E-77

8.0603354444E-78

8.0604516092E-78

__________________________________________________________________

(a)

(b)


1125

(a)

(b)


(a)

(b)


Fig. 6.62. Comparison of Exact and Approximate Solution of Eq. (6.200)

1126

6.5. Kravchuk Wavelet Method (KWM) We proposed a new algorithm by inserting Kravchuk polynomials in traditional Legendre wavelets method. This technique is successively applied to find the approximate solutions of integral equations. The proposed technique is very simple and highly compatible for solving such kind of problems.

6.5.1. Methodology 6.5.1.1. Kravchuk Polynomials Kravchuk polynomials have the following recurrence relation (−1)𝑚 (𝑥!)

(𝑎−𝑥)!

𝐾𝑛 (𝑥) = ∑𝑛𝑚=0 (𝑚!)((𝑥−𝑚)!) (𝑛−𝑚)!(𝑎−𝑥−𝑛+𝑚)!.

(6.214)

First five Kravchuk polynomials given below 𝐾0 = 1, 𝐾1 = −2𝑥, 𝐾2 = 2𝑥 2 , 2

4

𝐾3 = − 3 𝑥 − 3 𝑥 3 , 2

4

𝐾4 = 3 𝑥 4 + 3 𝑥 2 , . . ..

6.5.1.2. Wavelets and Kravchuk Wavelets In recent years, wavelets have found their way into many different fields of science and engineering. Wavelets constitute a family of functions constructed from the dilation and translation of a single function called the mother wavelet. When the dilation parameter a and the translation parameter 𝑏 vary continuously, we have the following family of continuous wavelets: 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|− 2 𝜓 (

𝑎

) 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

(6.215)

Kravchuk wavelets 𝜓𝑛𝑚 (𝑡) = 𝜓(𝑘, 𝑛, 𝑚, 𝑡) have four arguments; 𝑛, 𝑘 can assume any positive integer, 𝑚 is the order for Kravchuk polynomials and t is the normalized time. They are defined on the interval [0, 1) by;

1127

1

𝜓𝑛𝑚

𝑘

𝑘 2 (𝑡) = {√𝑚 + 2 2 𝐾𝑚 (2 𝑡 − (2𝑛 + 1))

0

𝑛 2𝑘

≤𝑡≤

𝑛+1 2𝑘

(6.216)

otherwise 1

where 𝑚 = 0,1,2, … , 𝑀 − 1 and 𝑛 = 0,1, … , 2𝑘−1 . The coefficient (𝑚 + 2) is for orthonormality. Here 𝐾𝑚 (𝑡) are the well-known Kravchuk polynomials of order 𝑚, which have been previously described.


(6.217)


𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡).

(6.218)



(6.219)

where 𝐶 and 𝜓(𝑡) are 2𝑘−1 𝑀 × 1 matrices given as 𝑐10 , 𝑐11 , … , 𝑐1𝑀−1 , 𝑐20 , 𝑐21 , … , 𝑇 𝐂 = [𝑐 ] , , … , 𝑐 𝑘−1 , … , 𝑐 𝑘−1

(6.220)

𝜓10 , 𝜓11 , … , 𝜓1𝑀−1 , 𝜓20 , 𝜓21 , … , 𝑇 𝛙(𝑡) = [ ] . 𝜓2𝑀−1 , … , 𝜓2𝑘−1 0 , … , 𝜓2𝑘−1 𝑀−1

(6.221)

2𝑀−1

and

2

0

2

𝑀−1

6.5.1.4. Analysis Consider the linear integral equation 𝛽(𝑥)

𝑦(𝑥) = 𝑓(𝑥) +𝜆 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡) 𝑦(𝑡) d𝑡, 𝑙 ≤ 𝑥 ≤ 𝐿. It is interesting to point out that 𝑦(𝑥) and 𝐾(𝑥, 𝑡) are assumed real. In order to solve linear integral equation, we suppose that [𝑙, 𝐿] = [0, 1], otherwise the problem can be mapped from [𝑙, 𝐿] to [0, 1] easily. For solving linear integral equation, it is sufficient to suppose that the approximate solution is as 𝑦(𝑥) = 𝐂 𝑇 𝛙(𝑥), where 𝐂 and 𝛙(𝑥) are given in above. Therefore, we obtain 𝛽(𝑥)

𝐂 𝑇 𝛙(𝑥) = 𝑓(𝑥) +𝜆𝐂 𝑇 ∫𝛼(𝑥) 𝐾(𝑥, 𝑡) 𝛙(𝑡) d𝑡, 𝑙 ≤ 𝑥 ≤ 𝐿.

1128

Therefore, in order to apply the Kravchuk Wavelet Method (KWM); we require 2𝑘−1 𝑀 − 1 collocating points. Suitable collocating points are as 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2, … , 2𝑘−1 𝑀 − 1.

Implementing the collocating points and imposing the initial value to the linear integral equation, we obtain 𝛽(𝑥 )

𝐂 𝑇 𝛙(𝑥𝑖 ) = 𝑓(𝑥𝑖 ) +𝜆𝐂 𝑇 ∫𝛼(𝑥 𝑖) 𝐾(𝑥𝑖 , 𝑡) 𝛙(𝑡) d𝑡. 𝑖

The differential equation yields 2𝑘−1 𝑀 − 1 equations. Therefore, the obtained system has 2𝑘−1 𝑀 − 1 equations and 2𝑘−1 𝑀 − 1 unknowns. Solving this system gives the unknown coefficients 𝐂.

6.5.2. Volterra Integral Equation Consider the following Volterra integral equation [229] as follow 𝑥

𝑦(𝑥) = sin(𝑥) + cos(𝑥) + 2 ∫0 sin(𝑥 − 𝑡)𝑦(𝑡) d𝑡,

(6.222)

The exact solution of Eq. (6.223) is 𝑦(𝑥) = exp(𝑥).

(6.223)

According to the Kravchuk Wavelets Method (KWM), we assume the trial solution 𝑘−1


(6.224)



𝑇


We apply the proposed technique to solve Eq. (6.223) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.225) is 𝑦(𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) , = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) + ⋯ = 𝐂 𝐓 𝛙(𝑥),

(6.225) 1

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝐾𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐾𝑚 is the Kravchuk polynomials given as 𝐾0 = 1, 𝐾1 = −2𝑥, 𝐾2 = 2𝑥 2 , 1129

(6.226)

2

4

𝐾3 = − 3 𝑥 − 3 𝑥 3 , 2

4

𝐾4 = 3 𝑥 4 + 3 𝑥 2 , . . .. Therefore, the Kravchuk wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = −2√3(2𝑥 − 1), 𝜓1,2 (𝑥) = 2√5(2𝑥 − 1)2 , 2

𝜓1,3 (𝑥) = − 3 √7(−3 + 14𝑥 − 24𝑥 2 + 16𝑥 3 ), . . .. Therefore, we have the trial solution is 2 −3 + 14𝑥 − 𝑦(𝑥) = 𝑐1,0 − 2√3(2𝑥 − 1)𝑐1,1 + 2√5(2𝑥 − 1)2 𝑐1,2 − 3 √7 ( ) 24𝑥 2 + 16𝑥 3

+ ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

1, −2√3(2𝑥 − 1), 2√5(2𝑥 − 1)2 , where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] and 𝛙(𝑥) = [ ] . 2 −3 + 14𝑥 − − 3 √7 ( 2 3) , … 24𝑥 + 16𝑥 𝑇

Substituting into the given problem we get 𝑥

𝐂 𝐓 𝛙(𝑥) = sin(𝑥) + cos(𝑥) + 2 ∫0 sin(𝑥 − 𝑡)𝐂 𝐓 𝛙(𝑡) d𝑡 ,

(6.227)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2, … ,9.

(6.228)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following exact solution 𝑦(𝑥) = 1.0000 + +1.00000𝑥 + 0.499999𝑥 2 + 0.166667𝑥 3 + ⋯.

1130

Table 6.34: Comparison of the Exact Solution and Approximate Solution of Eq. (6.223) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000558292

5.58292E-10

0.1

1.105170918075648

1.105170918384351

3.08703E-10

0.2

1.221402758160170

1.221402757932638

2.27532E-10

0.3

1.349858807576003

1.349858807027897

5.48106E-10

0.4

1.491824697641270

1.491824697334964

3.06306E-10

0.5

1.648721270700128

1.648721271000500

3.00372E-10

0.6

1.822118800390509

1.822118800975819

5.85310E-10

0.7

2.013752707470477

2.013752707540811

7.03344E-11

0.8

2.225540928492468

2.225540927922016

5.70452E-10

0.9

2.459603111156950

2.459603111293860

1.36911E-10

1.0

2.718281828459045

2.718281827871083

5.87963E-10

__________________________________________________________________

(a)

(b)



𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] . 1131

Proceeding as before we have the matrix form 𝐀 𝟐𝟓×𝟐𝟓 𝐂𝟐𝟓×𝟏 = 𝐛𝟐𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.35: Comparison of the Exact Solution and Approximate Solution of Eq. (6.223) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

4.00000E-199

0.1

1.105170918075648

1.105170918075648

2.34685E-33

0.2

1.221402758160170

1.221402758160170

3.64338E-33

0.3

1.349858807576003

1.349858807576003

3.90717E-33

0.4

1.491824697641270

1.491824697641270

2.86421E-33

0.5

1.648721270700128

1.648721270700128

2.13898E-33

0.6

1.822118800390509

1.822118800390509

1.27286E-33

0.7

2.013752707470477

2.013752707470477

2.26925E-33

0.8

2.225540928492468

2.225540928492468

3.43931E-33

0.9

2.459603111156950

2.459603111156950

1.42910E-33

1.0

2.718281828459045

2.718281828459045

4.39821E-33

__________________________________________________________________

(a)

(b)


1132

Finally to solve Eq. (6.223) with 𝐾 = 1, and 𝑀 = 50. We have Eq. (6.225) is 𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑49 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,49 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] . Proceeding as before we have the matrix form 𝐀 𝟓𝟎×𝟓𝟎 𝐂𝟓𝟎×𝟏 = 𝐛𝟓𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.36: Comparison of the Exact Solution and Approximate Solution of Eq. (6.223) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

7.09045E-42

0.1

1.105170918075648

1.105170918075648

7.01563E-42

0.2

1.221402758160170

1.221402758160170

6.92844E-42

0.3

1.349858807576003

1.349858807576003

6.84144E-42

0.4

1.491824697641270

1.491824697641270

7.01018E-42

0.5

1.648721270700128

1.648721270700128

7.17333E-42

0.6

1.822118800390509

1.822118800390509

6.49724E-42

0.7

2.013752707470477

2.013752707470477

2.11909E-42

0.8

2.225540928492468

2.225540928492468

6.50203E-42

0.9

2.459603111156950

2.459603111156950

4.13122E-42

1.0

2.718281828459045

2.718281828459045

7.28792E-42

__________________________________________________________________

1133

(a)

(b)


6.5.3. Fredholm Integral Equations Example 6.12 [229]. Consider the following integral equation as follow 1

1

𝑦(𝑥) = 3 𝑥 + exp(4𝑥) + 16 (17 + 3 exp(4)) + ∫0 𝑡𝑢(𝑡)d𝑡,

(6.229)

The exact solution of Eq. (6.230) is 𝑦(𝑥) = 3𝑥 + exp(4𝑥).

(6.230)



(6.231)



𝑇



(6.232) 1

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝐾𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐾𝑚 is the Kravchuk polynomials given as 𝐾0 = 1, 𝐾1 = −2𝑥, 1134

(6.233)

𝐾2 = 2𝑥 2 , 2

4

𝐾3 = − 3 𝑥 − 3 𝑥 3 , 2

4


𝜓1,3 (𝑥) = − 3 √7(−3 + 14𝑥 − 24𝑥 2 + 16𝑥 3 ), . . .. Therefore, we have the trial solution is 2 −3 + 14𝑥 − 𝑦(𝑥) = 𝑐1,0 − 2√3(2𝑥 − 1)𝑐1,1 + 2√5(2𝑥 − 1)2 𝑐1,2 − 3 √7 ( ) 24𝑥 2 + 16𝑥 3

+ ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇



1

𝐂 𝐓 𝛙(𝑥) = 3 𝑥 + exp(4𝑥) + 16 (17 + 3 exp(4)) + ∫0 𝑡𝐂 𝐓 𝛙(𝑡)d𝑡, (6.234) Substitute the collocating points are in Eq. (6a), we have the system of equations 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2, … ,9.

(6.235)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 9×9 𝐂9×1 = 𝐛9×1 . After solving we get the following exact solution 𝑦(𝑥) = 1.00077 + 7.00027𝑥 + 7.96197𝑥 2 + ⋯. 1135

(6.236)

Table 6.37: Comparison of the Exact Solution and Approximate Solution of Eq. (230) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

1.000000000000000

1.000770724702654

7.70725E-04

0.1

1.791824697641270

1.792258730770402

4.34033E-04

0.2

2.825540928492468

2.825198678721783

3.42250E-04

0.3

4.220116922736547

4.219279229248058

8.37693E-04

0.4

6.153032424395115

6.152544095926620

4.88328E-04

0.5

8.889056098930650

8.889524463934968

4.68365E-04

0.6

12.823176380641602

12.824132287976461

9.55907E-04

0.7

18.544646771097050

18.544755185709043

1.08415E-04

0.8

26.932530197109349

26.931505455003592

1.02474E-03

0.9

39.298234443677988

39.298462664355989

2.28221E-04

1.0

57.598150033144239

57.597011295736455

1.13874E-03

__________________________________________________________________

(a)

(b)


Now to solve Eq. (6.230) with 𝐾 = 1, and 𝑀 = 25. We have Eq. (6.232) is 𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑24 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥),

1136

𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] . Proceeding as before we have the matrix form 𝐀 𝟐𝟓×𝟐𝟓 𝐂𝟐𝟓×𝟏 = 𝐛𝟐𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.38: Comparison of the Exact Solution and Approximate Solution of Eq. (230) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.91402E-20

0.1

1.791824697641270

1.791824697641270

2.99471E-18

0.2

2.825540928492468

2.825540928492468

4.89924E-18

0.3

4.220116922736547

4.220116922736547

5.06079E-18

0.4

6.153032424395115

6.153032424395115

4.05414E-18

0.5

8.889056098930650

8.889056098930650

2.67937E-18

0.6

12.823176380641602

12.823176380641602

2.05210E-18

0.7

18.544646771097050

18.544646771097050

2.82739E-18

0.8

26.932530197109349

26.932530197109349

5.29055E-18

0.9

39.298234443677988

39.298234443677988

1.57716E-18

1.0

57.598150033144239

57.598150033144239

5.85748E-18

__________________________________________________________________

1137

(a)

(b)


Finally to solve Eq. (6.230) with 𝐾 = 1, and 𝑀 = 50. We have Eq. (6.232) is 𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑49 𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑪 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,49 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] . Proceeding as before we have the matrix form 𝐀 𝟓𝟎×𝟓𝟎 𝐂𝟓𝟎×𝟏 = 𝐛𝟓𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table.

1138

Table 6.39: Comparison of the Exact Solution and Approximate Solutions of Eq. (6.230) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

7.98389E-50

0.1

1.791824697641270

1.791824697641270

2.34747E-50

0.2

2.825540928492468

2.825540928492468

6.48946E-50

0.3

4.220116922736547

4.220116922736547

7.27572E-50

0.4

6.153032424395115

6.153032424395115

1.27569E-50

0.5

8.889056098930650

8.889056098930650

4.15453E-50

0.6

12.823176380641602

12.823176380641602

6.07682E-50

0.7

18.544646771097050

18.544646771097050

4.04963E-50

0.8

26.932530197109349

26.932530197109349

6.17688E-50

0.9

39.298234443677988

39.298234443677988

7.27496E-50

1.0

57.598150033144239

57.598150033144239

8.65897E-50

__________________________________________________________________

(a)

(b)


Example 6.13 [229]. Consider the following integral equation as follow 𝑦(𝑥) = 𝑥 + sin−1 (

𝑥+1

2−𝜋

2

2

)+

1

1

𝑥 2 + 2 ∫−1 𝑥 2 𝑢(𝑡)d𝑡 ,


(6.237)

𝑦(𝑥) = 𝑥 + sin−1 (

𝑥+1 2

).

(6.238)



(6.239)



𝑇



(6.240) 1

𝑘


4

𝐾3 = − 3 𝑥 − 3 𝑥 3 , 2

4


𝜓1,3 (𝑥) = − 3 √7(−3 + 14𝑥 − 24𝑥 2 + 16𝑥 3 ), . . .. 1140

(6.241)

Therefore, we have the trial solution is 2 −3 + 14𝑥 − 𝑦(𝑥) = 𝑐1,0 − 2√3(2𝑥 − 1)𝑐1,1 + 2√5(2𝑥 − 1)2 𝑐1,2 − 3 √7 ( ) 24𝑥 2 + 16𝑥 3

+ ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇


Substituting into the given problem we get 𝑥+1

2−𝜋

2

2

𝐂 𝐓 𝛙(𝑥) = 𝑥 + sin−1 (

)+

1

1

𝑥 2 − 2 ∫−1 𝑥 2 𝐂 𝐓 𝛙(𝑡)d𝑡,

(6.242)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2, … ,9.

(6.243)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following exact solution 𝑦(𝑥) = 0.52444 + 1.57815𝑥 + 0.0549978𝑥 2 + ⋯.

1141


Exact Solution


Error in KWM

__________________________________________________________________ 0.0

0.523598775598299

0.524440870663751

8.42095E-04

0.1

0.682364237868743

0.682866719602896

5.02482E-04

0.2

0.843501108793284

0.843060729886135

4.40379E-04

0.3

1.007584436725356

1.006423659770028

1.16078E-03

0.4

1.175397496610753

1.174653505097773

7.43992E-04

0.5

1.348062078981481

1.348893691742286

8.31613E-04

0.6

1.527295218001612

1.529275839392213

1.98062E-03

0.7

1.715985293814825

1.716296129005252

3.10835E-04

0.8

1.919769514998634

1.916324286218898

3.44523E-03

0.9

2.153235897503375

2.154570068363502

1.33417E-03

1.0

2.570796326794897

2.500023913466260

7.07724E-02

__________________________________________________________________

(a)

(b)



1142

𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] . Proceeding as before we have the matrix form 𝐀 𝟐𝟓×𝟐𝟓 𝐂𝟐𝟓×𝟏 = 𝐛𝟐𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.41: Comparison of the Exact Solution and Approximate Solution of Eq. (6.237) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

0.523598775598299

0.523598775598299

9.07339E-177

0.1

0.682364237868743

0.682327656707781

3.65812E-05

0.2

0.843501108793284

0.843567057800906

6.59490E-05

0.3

1.007584436725356

1.007508349256442

7.60875E-05

0.4

1.175397496610753

1.175468515532612

7.10189E-05

0.5

1.348062078981481

1.348008952629007

5.31264E-05

0.6

1.527295218001612

1.527348561847816

5.33438E-05

0.7

1.715985293814825

1.715896552933404

8.87409E-05

0.8

1.919769514998634

1.920022839717528

2.53325E-04

0.9

2.153235897503375

2.153099678354837

1.36219E-04

1.0

2.570796326794897

2.542507807622168

2.82885E-02

__________________________________________________________________

1143

(a)

(b)


Finally to solve Eq. (6.237) with 𝐾 = 1, and 𝑀 = 50. We have Eq. (6.239) is 𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑49 𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇

𝑇


1144


Exact Solution


Error in KWM

__________________________________________________________________ 0.0

0.523598775598299

0.523605743751780

6.96815E-06

0.1

0.682364237868743

0.682366504821847

2.26695E-06

0.2

0.843501108793284

0.843494179985914

6.92881E-06

0.3

1.007584436725356

1.007575669388234

8.76734E-06

0.4

1.175397496610753

1.175395849346004

1.64726E-06

0.5

1.348062078981481

1.348069261944755

7.18296E-06

0.6

1.527295218001612

1.527308143103836

1.29251E-05

0.7

1.715985293814825

1.715996821686958

1.15279E-05

0.8

1.919769514998634

1.919745075656893

2.44393E-05

0.9

2.153235897503375

2.153294441326245

5.85438E-05

1.0

2.570796326794897

2.556653664400843

1.41427E-02

__________________________________________________________________

(a)

(b)


6.5.4. System of Integral Equations Example 6.14 [229]. Consider the following system of Volterra integral equations as follow 1145

1

𝑥

1

𝑦1 (𝑥) = 1 + 𝑥 2 − 3 𝑥 3 − 3 𝑥 4 + ∫0 [ 1

(𝑥 − 𝑡)3 𝑦1 (𝑡) ] d𝑡, +(𝑥 − 𝑡)2 𝑦2 (𝑡) 𝑥

1

𝑦2 (𝑥) = 1 + 𝑥 − 𝑥 3 − 4 𝑥 4 − 4 𝑥 5 + ∫0 [

(𝑥 − 𝑡)4 𝑦1 (𝑡) ] d𝑡 . +(𝑥 − 𝑡)3 𝑦2 (𝑡)

(6.244) (6.245)

The exact solution of Eq. (6.244-6.245) is 𝑦1 (𝑥) = 1 + 𝑥 2 , 𝑦2 (𝑥) = 1 + 𝑥 − 𝑥 3 .

(6.246)




(6.247) (6.248)




𝑇


We apply the proposed technique to solve Eq. (6.244-6.245) with 𝐾 = 1, and 𝑀 = 4. We have Eq. (6.247-6.248) is 𝑦1 (𝑥) = ∑1𝑛=1 ∑3𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) + 𝑐1,3 𝜓1,3 (𝑥) = 𝐂 𝐓 𝛙(𝑥),

(6.249)

𝑦2 (𝑥) = ∑1𝑛=1 ∑3𝑚=0 𝑑𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑑1,0 𝜓1,0 (𝑥) + 𝑑1,1 𝜓1,1 (𝑥) + 𝑑1,2 𝜓1,2 (𝑥) + 𝑑1,3 𝜓1,3(𝑥) = 𝐃𝐓 𝛙(𝑥), 1

𝑘


4

𝐾3 = − 3 𝑥 − 3 𝑥 3 , 2

(6.250)

4

𝐾4 = 3 𝑥 4 + 3 𝑥 2 , . . .. 1146

(6.251)

Therefore, the Kravchuk wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = −2√3(2𝑥 − 1), 𝜓1,2 (𝑥) = 2√5(2𝑥 − 1)2 , 2

𝜓1,3 (𝑥) = − 3 √7(−3 + 14𝑥 − 24𝑥 2 + 16𝑥 3 ), . . .. Therefore, we have the trial solution is −3 + 14𝑥 − )𝑐 24𝑥 2 + 16𝑥 3 1,3

2

𝑦1 (𝑥) = 𝑐1,0 − 2√3(2𝑥 − 1)𝑐1,1 + 2√5(2𝑥 − 1)2 𝑐1,2 − 3 √7 ( = 𝐂 𝐓 𝛙(𝑥), 2

−3 + 14𝑥 − )𝑑 24𝑥 2 + 16𝑥 3 1,3

𝑦2 (𝑥) = 𝑑1,0 − 2√3(2𝑥 − 1)𝑑1,1 + 2√5(2𝑥 − 1)2 𝑑1,2 − 3 √7 ( = 𝐃𝐓 𝛙(𝑥), 𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 ] , 𝐃 = [𝑑1,0 , 𝑑1,1 , 𝑑1,2 , 𝑑1,3 ]

𝑇

𝑇

1, −2√3(2𝑥 − 1), 2√5(2𝑥 − 1)2 , and 𝛙(𝑥) = [ ] . 2 −3 + 14𝑥 − − 3 √7 ( 2 3) 24𝑥 + 16𝑥 Substituting into the given problem we get 1

1

𝑥

𝐂 𝐓 𝛙(𝑥) = 1 + 𝑥 2 − 3 𝑥 3 − 3 𝑥 4 + ∫0 [

(𝑥 − 𝑡)3 𝐂 𝐓 𝛙(𝑡) ] d𝑡, +(𝑥 − 𝑡)2 𝐃𝐓 𝛙(𝑡)

4 𝐓 𝑥 (𝑥 − 𝑡) 𝐂 𝛙(𝑡) 1 1 𝐃𝐓 𝛙(𝑥) = 1 + 𝑥 − 𝑥 3 − 4 𝑥 4 − 4 𝑥 5 + ∫0 [ ] d𝑡. +(𝑥 − 𝑡)3 𝐃𝐓 𝛙(𝑡)

(6.252) (6.253)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2,3.

(6.254)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 8×8 𝐱 8×1 = 𝐛8×1 . After solving we get the following exact solution

1147

𝑦1 (𝑥) = 1 + 𝑥 2 , 𝑦2 (𝑥) = 1 − 𝑥 + 𝑥 3 .

Example 6.15 [229]. Consider the following system of Fredholm integral equations as follow 𝜋3

𝜋

𝜋3

𝜋

𝑦1 (𝑥) = 𝑥 + sec 2 (𝑥) − 96 + ∫04 𝑡[𝑦1 (𝑡) + 𝑦2 (𝑡)]d𝑡, 𝑦2 (𝑥) = 𝑥 − sec 2 (𝑥) − 16 + ∫04 [𝑦1 (𝑡) + 𝑦2 (𝑡)]d𝑡 .

(6.255) (6.256)

The exact solution of Eq. (6.255-6.256) is 𝑦1 (𝑥) = 𝑥 + sec 2 (𝑥) , 𝑦2 (𝑥) = 𝑥 − sec 2 (𝑥).

(6.257)




(6.258) (6.259)




𝑇



(6.260)


𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 22 𝐾𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). 2

In the above expression, 𝐾𝑚 is the Kravchuk polynomials given as 𝐾0 = 1, 𝐾1 = −2𝑥, 𝐾2 = 2𝑥 2 , 2

(6.261)

4

𝐾3 = − 3 𝑥 − 3 𝑥 3 , 1148

(6.262)

2

4


𝜓1,3 (𝑥) = − 3 √7(−3 + 14𝑥 − 24𝑥 2 + 16𝑥 3 ), . . .. Therefore, we have the trial solution is −3 + 14𝑥 − )𝑐 24𝑥 2 + 16𝑥 3 1,3

2

𝑦1 (𝑥) = 𝑐1,0 − 2√3(2𝑥 − 1)𝑐1,1 + 2√5(2𝑥 − 1)2 𝑐1,2 − 3 √7 ( + ⋯ = 𝐂 𝐓 𝛙(𝑥), 2

−3 + 14𝑥 − )𝑑 24𝑥 2 + 16𝑥 3 1,3

𝑦2 (𝑥) = 𝑑1,0 − 2√3(2𝑥 − 1)𝑑1,1 + 2√5(2𝑥 − 1)2 𝑑1,2 − 3 √7 ( + ⋯ = 𝐃𝐓 𝛙(𝑥), 𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , 𝐃 = [𝑑1,0 , 𝑑1,1 , 𝑑1,2 , 𝑑1,3 , … ]

𝑇

𝑇

1, −2√3(2𝑥 − 1), 2√5(2𝑥 − 1)2 , and 𝛙(𝑥) = [ ] . 2 −3 + 14𝑥 − − 3 √7 ( 2 3) , … 24𝑥 + 16𝑥 Substituting into the given problem we get 𝐂 𝐓 𝛙(𝑥) = 𝑥 + sec 2 (𝑥) −

𝜋3 96 𝜋3

𝜋

+ ∫04 𝑡[𝐂 𝐓 + 𝐃𝐓 ]𝛙(𝑡)d𝑡,

(6.263)

𝜋

𝐃𝐓 𝛙(𝑥) = 𝑥 − sec 2 (𝑥) − 16 + ∫04 [𝐂 𝐓 + 𝐃𝐓 ]𝛙(𝑡)d𝑡.

(6.264)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2, … ,9.

1149

(6.265)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 20×10 𝐱 20×1 = 𝐛20×1 . After solving we get the following approximate solution 𝑦1 (𝑥) = 1.000815 + 1.000000𝑥 + 0.959632𝑥 2 + ⋯. 𝑦2 (𝑥) = −1.000815 + 1.000000𝑥 − 0.959632𝑥 2 + ⋯. Table 6.43 (a): Comparison of the Exact Solution and Approximate Solutions of 𝑦1 (𝑥) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

1.000000000000000

1.000814908094249

8.14908E-04

0.1

1.110067046422495

1.110508238460232

4.41192E-04

0.2

1.241091358495927

1.240735615590510

3.55743E-04

0.3

1.395688915322547

1.394842380443099

8.46535E-04

0.4

1.578754105810975

1.578259202819314

4.94903E-04

0.5

1.798446410409525

1.798907122341410

4.60712E-04

0.6

2.068043172527957

2.069009383162706

9.66211E-04

0.7

2.409449715863117

2.409554839751628

1.05124E-04

0.8

2.860155558164756

2.859013361760076

1.14220E-03

0.9

3.487998733259648

3.488259316655499

2.60583E-04

1.0

4.425518820814760

4.424014859465046

1.50396E-03

__________________________________________________________________

1150

(a)

(b)


Table 6.43 (b): Comparison of the Exact Solution and Approximate Solutions of 𝑦2 (𝑥) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

-1.000000000000000

-1.000814908094249

8.14908E-04

0.1

-0.910067046422495

-0.910508238460232

4.41192E-04

0.2

-0.841091358495927

-0.840735615590510

3.55743E-04

0.3

-0.795688915322547

-0.794842380443099

8.46535E-04

0.4

-0.778754105810975

-0.778259202819314

4.94903E-04

0.5

-0.798446410409525

-0.798907122341410

4.60712E-04

0.6

-0.868043172527957

-0.869009383162706

9.66211E-04

0.7

-1.009449715863117

-1.009554839751628

1.05124E-04

0.8

-1.260155558164756

-1.259013361760076

1.14220E-03

0.9

-1.687998733259648

-1.688259316655499

2.60583E-04

1.0

-2.425518820814760

-2.424014859465046

1.50396E-03

__________________________________________________________________

1151

(a)

(b)



𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] .

Proceeding as before we have the matrix form 𝐀 𝟓𝟎×𝟓𝟎 𝐂𝟓𝟎×𝟏 = 𝐛𝟓𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table.

1152

Table 6.44 (a): Comparison of the Exact Solution and Approximate Solutions of 𝑦1 (𝑥) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

5.32322E-176

0.1

1.110067046422495

1.110067046405998

1.64968E-11

0.2

1.241091358495927

1.241091358549215

5.32877E-11

0.3

1.395688915322547

1.395688915238819

8.37277E-11

0.4

1.578754105810975

1.578754105901051

9.00757E-11

0.5

1.798446410409525

1.798446410332238

7.72868E-11

0.6

2.068043172527957

2.068043172600514

7.25562E-11

0.7

2.409449715863117

2.409449715738350

1.24767E-10

0.8

2.860155558164756

2.860155558447318

2.82562E-10

0.9

3.487998733259648

3.487998733154509

1.05139E-10

1.0

4.425518820814760

4.425518820332266

4.82493E-10

__________________________________________________________________

(a)

(b)


1153


Exact Solution


Error in KWM

__________________________________________________________________ 0.0

-1.000000000000000

-1.000000000000000

9.09633E-176

0.1

-0.910067046422495

-0.910067046405998

1.64968E-11

0.2

-0.841091358495927

-0.841091358549215

5.32877E-11

0.3

-0.795688915322547

-0.795688915238819

8.37277E-11

0.4

-0.778754105810975

-0.778754105901051

9.00757E-11

0.5

-0.798446410409525

-0.798446410332238

7.72868E-11

0.6

-0.868043172527957

-0.868043172600514

7.25562E-11

0.7

-1.009449715863117

-1.009449715738350

1.24767E-10

0.8

-1.260155558164756

-1.260155558447318

2.82562E-10

0.9

-1.687998733259648

-1.687998733154509

1.05139E-10

1.0

-2.425518820814760

-2.425518820332266

4.82493E-10

__________________________________________________________________

(a)

(b)


Finally to solve Eq. (6.255-6.256) with 𝐾 = 1, and 𝑀 = 50. We have Eq. (6.258-6.259) is 1154

𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑49 𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,49 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] . Proceeding as before we have the matrix form 𝐀 𝟏𝟎𝟎×𝟏𝟎𝟎 𝐂𝟏𝟎𝟎×𝟏 = 𝐛𝟏𝟎𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.45 (a): Comparison of the Exact Solution and Approximate Solutions of 𝑦1 (𝑥) obtained from Kravchuk Wavelet Method (KWM) __________________________________________________________________ 𝑥

Exact Solution


Error in KWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

6.44437E-21

0.1

1.110067046422495

1.110067046422495

1.88757E-21

0.2

1.241091358495927

1.241091358495927

5.24458E-21

0.3

1.395688915322547

1.395688915322547

5.95983E-21

0.4

1.578754105810975

1.578754105810975

1.06945E-21

0.5

1.798446410409525

1.798446410409525

3.59775E-21

0.6

2.068043172527957

2.068043172527957

5.50199E-21

0.7

2.409449715863117

2.409449715863117

3.88446E-21

0.8

2.860155558164756

2.860155558164756

6.37943E-21

0.9

3.487998733259648

3.487998733259648

8.24812E-21

1.0

4.425518820814760

4.425518820814760

1.10604E-20

__________________________________________________________________

1155

(a)

(b)



Exact Solution


Error in KWM

__________________________________________________________________ 0.0

-1.000000000000000

-1.000000000000000

6.44437E-21

0.1

-0.910067046422495

-0.910067046422495

1.88757E-21

0.2

-0.841091358495927

-0.841091358495927

5.24458E-21

0.3

-0.795688915322547

-0.795688915322547

5.95983E-21

0.4

-0.778754105810975

-0.778754105810975

1.06945E-21

0.5

-0.798446410409525

-0.798446410409525

3.59775E-21

0.6

-0.868043172527957

-0.868043172527957

5.50199E-21

0.7

-1.009449715863117

-1.009449715863117

3.88446E-21

0.8

-1.260155558164756

-1.260155558164756

6.37943E-21

0.9

-1.687998733259648

-1.687998733259648

8.24812E-21

1.0

-2.425518820814760

-2.425518820814760

1.10604E-20

__________________________________________________________________

1156

(a)

(b)


6.6. Lommal Wavelet Method (LWM) We proposed a new algorithm by inserting Lommal polynomials in traditional Legendre wavelets method. This technique is successively applied to find the approximate solutions of Delay Differential Equations (DDEs). The proposed technique is very simple and highly compatible for solving such kind of problems.

6.6.1. Methodology 6.6.1.2. Lommal Polynomials The Lommal polynomials are used in physics and engineering. The recurrence relation of Lommal polynomial is given by 𝑛

𝑥

k (𝑛−𝑘)! (−1) Γ(− 𝜀 +𝑛−𝑘)

2 𝐿𝑛 (𝑥) = ∑𝑘=0 𝑘!(𝑛−2𝑘)!

𝑥 𝜀

Γ(− +𝑘)

(−𝜀)𝑛−2𝑘 .

First three Lommel polynomials given below 𝐿0 = 1, 𝐿1 = 𝑥, 𝐿2 = −1 + 𝑥 2 − 𝑥, 𝐿3 = 𝑥 3 − 3𝑥 2 + 2, . . .. In all problems, I take 𝜀 = 1. 1157

(6.266)

6.6.2.2. Wavelets and Lommal Wavelets In recent years, wavelets have found their way into many different fields of science and engineering. Wavelets constitute a family of functions constructed from the dilation and translation of a single function called the mother wavelet. When the dilation parameter a and the translation parameter 𝑏 vary continuously, we have the following family of continuous wavelets: 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|− 2 𝜓 (

𝑎

) 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

(6.267)

Lommal wavelets 𝜓𝑛𝑚 (𝑡) = 𝜓(𝑘, 𝑛, 𝑚, 𝑡) have four arguments; 𝑛, 𝑘 can assume any positive integer, 𝑚 is the order for Lommal polynomials and t is the normalized time. They are defined on the interval [0, 1) by; 1

𝜓𝑛𝑚

𝑘

𝑘 2 (𝑡) = {√𝑚 + 2 2 𝐿𝑚 (2 𝑡 − (2𝑛 + 1))

0

𝑛 2𝑘

≤𝑡≤

𝑛+1 2𝑘

(6.268)

otherwise 1

where 𝑚 = 0,1,2, … , 𝑀 − 1 and 𝑛 = 0,1, … , 2𝑘−1 . The coefficient (𝑚 + 2) is for orthonormality. Here 𝐿𝑚 (𝑡) are the well-known Lommal polynomials of order 𝑚, which have been previously described.


(6.269)


𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡).

(6.270)



(6.271)


and

𝛙(𝑡) = [

2

0

2

𝑀−1

𝜓10 , 𝜓11 , … , 𝜓1𝑀−1 , 𝜓20 , 𝜓21 , … , 𝑇 ] . 𝜓2𝑀−1 , … , 𝜓2𝑘−1 0 , … , 𝜓2𝑘−1 𝑀−1

1158

(6.272) (6.273)

6.6.1.4. Analysis Consider the nth order Delay differential equation 𝑦 𝑛 (𝑥) = 𝑓(𝑥, 𝑦(𝑥), 𝑦(𝛼1 𝑥), 𝑦(𝛼2 𝑥), 𝑦(𝛼3 𝑥), … , 𝑦(𝛼𝑟 𝑥)), 0 ≤ 𝑥 ≤ 1. with the physical initial conditions 𝑦 𝑛 (0) = 𝐴𝑗 , 𝑛 = 0,1,2, … , (𝑛 − 1). For solving equation nth order Delay differential equation, it is sufficient to suppose that the approximate solution is as 𝑦(𝑥) = 𝐂 𝑇 𝛙(𝑥), where 𝐂 and 𝛙(𝑥) are given in above. Therefore, we obtain 𝛙𝑛 (𝑥) = 𝑓(𝑥, 𝛙(𝑥), 𝛙(𝛼1 𝑥), 𝛙(𝛼2 𝑥), 𝛙(𝛼3 𝑥), … , 𝛙(𝛼𝑟 𝑥)), with the physical initial conditions 𝛙𝑛 (0) = 𝐴𝑗 , 𝑛 = 0,1,2, … , (𝑛 − 1). Therefore, in order to apply the Lommel Wavelet Method (LWM); we require 2𝑘−1 𝑀 − 1 collocating points. Suitable collocating points are as 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 𝑛, … , 2𝑘−1 𝑀 − 1.

Implementing the collocating points and imposing the initial value to the nth order Delay differential equation, we obtain 𝛙𝑛 (𝑥𝑖 ) = 𝑓(𝑥, 𝛙(𝑥𝑖 ), 𝛙(𝛼1 𝑥𝑖 ), 𝛙(𝛼2 𝑥𝑖 ), 𝛙(𝛼3 𝑥𝑖 ), … , 𝛙(𝛼𝑟 𝑥𝑖 )), 𝑖 = 𝑛, … ,2𝑘−1 𝑀 − 1. The differential equation yields 2𝑘−1 𝑀 − 𝑛 − 1 equations and initial condition produces 𝑛 number of equations. Therefore, the obtained system has 2𝑘−1 𝑀 − 1 equations and 2𝑘−1 𝑀 − 1 unknowns. Solving this system gives the unknown coefficients 𝐂.

6.6.2. Delay Differential Equations Example 6.16 [68]. Consider the following linear DDE [68] as follow 1 𝑥

𝑥

1

𝑦 ′ (𝑥) = 2 e2 𝑦 (2) + 2 𝑦(𝑥),

(6.274)

subject to the boundary conditions 𝑦(0) = 1.

(6.275)

The exact solution of the above system is 𝑦(𝑥) = e𝑥 .

(6.276) 1159

According to the Lommal Wavelets Method (LWM), we assume the trial solution 𝑘−1


(6.277)



𝑇


We apply the proposed technique to solve Eq. (1) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (5) is 𝑦(𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) , = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) + ⋯ + 𝑐1,9 𝜓1,9 (𝑥) = 𝐂 𝐓 𝛙(𝑥), 1

(6.278)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐿𝑚 is the Lommel polynomials given as 𝐿0 = 1, 𝐿1 = 𝑥, 𝐿2 = −1 + 𝑥 2 − 𝑥, 𝐿3 = 𝑥 3 − 3𝑥 2 + 2, 𝐿4 = −5 + 3𝑥 + 8𝑥 2 − 6𝑥 3 + 𝑥 4 , . . .. Therefore, the Lommel wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = √5(4𝑥 2 − 6𝑥 + 1), 𝜓1,3 (𝑥) = 2√7(−1 + 9𝑥 − 12𝑥 2 + 4𝑥 3 ), . . .. Therefore, we have the trial solution is

1160

(6.279)

−1 + 9𝑥 − 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + √5(4𝑥 2 − 6𝑥 + 1)𝑐1,2 + 2√7 ( ) 12𝑥 2 + 4𝑥 3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

1, √3(2𝑥 − 1), √5(4𝑥 2 − 6𝑥 + 1), where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] and 𝛙(𝑥) = [ ] . 2√7(−1 + 9𝑥 − 12𝑥 2 + 4𝑥 3 ), … 𝑇

Substituting into the given problem we get 1 𝑥

𝑥

1

𝐂 𝐓 𝛙′ (𝑥) = 2 e2 𝐂 𝐓 𝛙 (2) + 2 𝐂 𝐓 𝛙(𝑥),

(6.280)

Subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 1.

(6.281)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 1,2,3, … 9.

(6.282)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following approximate solution 𝑦(𝑥) = 1.0000 + 0.99999𝑥 + 0.499999𝑥 2 + 0.000189𝑥 3 + ⋯,

1161

Table 6.46: Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) obtained from Bessel Wavelet Method (LWM) __________________________________________________________________ 𝑥

Exact Solution


Error in LWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

5.00000E-200

0.1

1.105170918075648

1.105170917613166

4.62482E-10

0.2

1.221402758160170

1.221402757619899

5.40271E-10

0.3

1.349858807576003

1.349858807526211

4.97921E-11

0.4

1.491824697641270

1.491824698252717

6.11446E-10

0.5

1.648721270700128

1.648721271339750

6.39622E-10

0.6

1.822118800390509

1.822118800026426

3.64083E-10

0.7

2.013752707470477

2.013752706111536

1.35894E-09

0.8

2.225540928492468

2.225540928394773

9.76945E-11

0.9

2.459603111156950

2.459603113305860

2.14891E-09

1.0

2.718281828459045

2.718281814056567

1.44025E-08

__________________________________________________________________

(a)

(b)



1162

𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] . Proceeding as before we have the matrix form 𝐀 𝟐𝟓×𝟐𝟓 𝐂𝟐𝟓×𝟏 = 𝐛𝟐𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.47: Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) obtained from Bessel Wavelet Method (LWM) _____________________________________________________________ 𝑥

Exact Solution


Error in LWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

5.00000E-200

0.1

1.105170918075648

1.105170918075648

7.34208E-33

0.2

1.221402758160170

1.221402758160170

2.91140E-33

0.3

1.349858807576003

1.349858807576003

4.07036E-33

0.4

1.491824697641270

1.491824697641270

9.37116E-33

0.5

1.648721270700128

1.648721270700128

5.51365E-34

0.6

1.822118800390509

1.822118800390509

1.34374E-32

0.7

2.013752707470477

2.013752707470477

1.72728E-34

0.8

2.225540928492468

2.225540928492468

1.21897E-32

0.9

2.459603111156950

2.459603111156950

2.57981E-32

1.0

2.718281828459045

2.718281828459045

2.50892E-31

_____________________________________________________________

1163

(a)

(b)



𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,74 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,74 ] . Proceeding as before we have the matrix form 𝐀 𝟕𝟓×𝟕𝟓 𝐂𝟕𝟓×𝟏 = 𝐛𝟕𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table.

1164

Table 6.48: Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) obtained from Bessel Wavelet Method (LWM) _____________________________________________________________ 𝑥

Exact Solution


Error in LWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.60000E-199

0.1

1.105170918075648

1.105170918075648

1.58527E-132

0.2

1.221402758160170

1.221402758160170

4.68700E-132

0.3

1.349858807576003

1.349858807576003

4.75183E-132

0.4

1.491824697641270

1.491824697641270

3.59419E-133

0.5

1.648721270700128

1.648721270700128

3.51187E-132

0.6

1.822118800390509

1.822118800390509

5.62050E-132

0.7

2.013752707470477

2.013752707470477

4.45079E-132

0.8

2.225540928492468

2.225540928492468

1.00819E-132

0.9

2.459603111156950

2.459603111156950

1.16001E-131

1.0

2.718281828459045

2.718281828459045

4.42988E-130

_____________________________________________________________

(a)

(b)


Example 6.17 [68]. Consider the following linear DDE as follow 𝑥

𝑦 ′ (𝑥) − 𝑦 (2) = 0,

(6.283) 1165


(6.284)

The exact solution of the above system is 1

𝑦(𝑥) = ∑∞ 𝑘=0

1 𝑘(𝑘−1) ( )2 2

𝑘!

𝑥𝑘 .

(6.285)



(6.286)



𝑇


We apply the proposed technique to solve Eq. (6.283) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.285) is 𝑦(𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) , = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) + ⋯ + 𝑐1,9 𝜓1,9 (𝑥) = 𝐂 𝐓 𝛙(𝑥), 1

(6.287)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐿𝑚 is the Lommel polynomials given as 𝐿0 = 1, 𝐿1 = 𝑥, 𝐿2 = −1 + 𝑥 2 − 𝑥, 𝐿3 = 𝑥 3 − 3𝑥 2 + 2, 𝐿4 = −5 + 3𝑥 + 8𝑥 2 − 6𝑥 3 + 𝑥 4 , . . .. Therefore, the Lommel wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = √5(4𝑥 2 − 6𝑥 + 1), 𝜓1,3 (𝑥) = 2√7(−1 + 9𝑥 − 12𝑥 2 + 4𝑥 3 ), 1166

(6.288)

. . .. Therefore, we have the trial solution is −1 + 9𝑥 − 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + √5(4𝑥 2 − 6𝑥 + 1)𝑐1,2 + 2√7 ( ) 12𝑥 2 + 4𝑥 3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇



𝐂 𝐓 𝛙′ (𝑥) − 𝐂 𝐓 𝛙 (2) = 0,

(6.289)


(6.290)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 1,2,3, … 9.

(6.291)


1167


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

0.00000E+00

0.1

1.102520898518923

1.102520898518923

1.40533E-23

0.2

1.210167710940214

1.210167710940214

1.61101E-23

0.3

1.323067793243810

1.323067793243810

1.34194E-24

0.4

1.441350083507100

1.441350083507100

1.75775E-23

0.5

1.565145111746998

1.565145111746998

1.68744E-23

0.6

1.694585009792689

1.694585009792689

1.37309E-23

0.7

1.829803521189073

1.829803521189073

4.18853E-23

0.8

1.970936011130968

1.970936011130968

2.69883E-24

0.9

2.118119476428119

2.118119476428119

6.21111E-23

1.0

2.271492555501061

2.271492555501061

4.18581E-22

__________________________________________________________________

(a)

(b)



1168

𝑇

𝑇


Exact Solution


Error in LWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.00000E-199

0.1

1.102520898518923

1.102520898518923

3.66056E-123

0.2

1.210167710940214

1.210167710940214

1.37042E-123

0.3

1.323067793243810

1.323067793243810

2.09032E-123

0.4

1.441350083507100

1.441350083507100

4.63354E-123

0.5

1.565145111746998

1.565145111746998

1.40081E-124

0.6

1.694585009792689

1.694585009792689

6.40887E-123

0.7

1.829803521189073

1.829803521189073

3.87870E-124

0.8

1.970936011130968

1.970936011130968

5.63215E-123

0.9

2.118119476428119

2.118119476428119

1.23059E-122

1.0

2.271492555501061

2.271492555501061

1.25114E-121

_____________________________________________________________

1169

(a)

(b)



𝑇


1170


Exact Solution


Error in LWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.00000E-399

0.1

1.102520898518923

1.102520898518923

2.00000E-399

0.2

1.210167710940214

1.210167710940214

3.00000E-399

0.3

1.323067793243810

1.323067793243810

3.00000E-399

0.4

1.441350083507100

1.441350083507100

0.00000E+00

0.5

1.565145111746998

1.565145111746998

6.00000E-399

0.6

1.694585009792689

1.694585009792689

1.00000E-399

0.7

1.829803521189073

1.829803521189073

6.00000E-399

0.8

1.970936011130968

1.970936011130968

1.00000E-399

0.9

2.118119476428119

2.118119476428119

5.00000E-399

1.0

2.271492555501061

2.271492555501061

4.00000E-399

_____________________________________________________________

(a)

(b)


Example 6.18 [68]. Consider the following linear DDE as follow 𝑦 ′′′ (𝑥) + 𝑦(𝑥) + 𝑦(𝑥 − 0.3) = e−𝑥+0.3 , 1171

(6.292)

subject to the boundary conditions 𝑦(0) = 1, 𝑦 ′ (0) = −1, 𝑦(0) = 1.

(6.293)

The exact solution of the above system is 𝑦(𝑥) = exp(−𝑥).

(6.294)



(6.295)



𝑇


We apply the proposed technique to solve Eq. (6.292) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.295) is 𝑦(𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) , = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) + ⋯ + 𝑐1,9 𝜓1,9 (𝑥) = 𝐂 𝐓 𝛙(𝑥), 1

(6.296)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐿𝑚 is the Lommel polynomials given as 𝐿0 = 1, 𝐿1 = 𝑥, 𝐿2 = −1 + 𝑥 2 − 𝑥, 𝐿3 = 𝑥 3 − 3𝑥 2 + 2, 𝐿4 = −5 + 3𝑥 + 8𝑥 2 − 6𝑥 3 + 𝑥 4 , . . .. Therefore, the Lommel wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = √5(4𝑥 2 − 6𝑥 + 1), 𝜓1,3 (𝑥) = 2√7(−1 + 9𝑥 − 12𝑥 2 + 4𝑥 3 ), . 1172

(6.297)

. .. Therefore, we have the trial solution is −1 + 9𝑥 − 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + √5(4𝑥 2 − 6𝑥 + 1)𝑐1,2 + 2√7 ( ) 12𝑥 2 + 4𝑥 3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇


Substituting into the given problem we get 𝐂 𝐓 𝛙′′′ (𝑥) + 𝐂 𝐓 𝛙(𝑥) + 𝐂 𝐓 𝛙(𝑥 − 0.3) = e−𝑥+0.3 ,

(6.298)

Subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 1, 𝐂 𝐓 𝛙′ (0) = −1, 𝐂 𝐓 𝛙′′ (0) = 1.

(6.299)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 3,4, … 9.

(6.300)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following approximate solution 𝑦(𝑥) = 1.0000 − 0.99999𝑥 + 0.499999𝑥 2 + 0.16666𝑥 3 + ⋯,

1173


Exact Solution


Error in LWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.20000E-199

0.1

0.904837418035960

0.904837417885435

1.50525E-10

0.2

0.818730753077982

0.818730751978017

1.09996E-09

0.3

0.740818220681718

0.740818217898862

2.78286E-09

0.4

0.670320046035639

0.670320042806103

3.22954E-09

0.5

0.606530659712633

0.606530660461651

7.49018E-10

0.6

0.548811636094026

0.548811643616776

7.52275E-09

0.7

0.496585303791410

0.496585296939500

6.85191E-09

0.8

0.449328964117222

0.449328837174443

1.26943E-07

0.9

0.406569659740599

0.406569089224090

5.70517E-07

1.0

0.367879441171442

0.367877627369487

1.81380E-06

__________________________________________________________________

(a)

(b)



1174

𝑇

𝑇


Exact Solution


Error in LWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.00000E-200

0.1

0.904837418035960

0.904837418035960

6.14928E-33

0.2

0.818730753077982

0.818730753077982

5.67354E-32

0.3

0.740818220681718

0.740818220681718

1.16066E-31

0.4

0.670320046035639

0.670320046035639

2.05364E-31

0.5

0.606530659712633

0.606530659712633

3.36539E-31

0.6

0.548811636094026

0.548811636094026

4.46925E-31

0.7

0.496585303791410

0.496585303791410

6.53873E-31

0.8

0.449328964117222

0.449328964117222

8.68874E-31

0.9

0.406569659740599

0.406569659740599

7.71498E-31

1.0

0.367879441171442

0.367879441171442

2.06580E-28

_____________________________________________________________

1175

(a)

(b)



𝑇


1176


Exact Solution


Error in LWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

3.00000E-199

0.1

0.904837418035960

0.904837418035960

5.78906E-120

0.2

0.818730753077982

0.818730753077982

2.96376E-119

0.3

0.740818220681718

0.740818220681718

7.28990E-119

0.4

0.670320046035639

0.670320046035639

1.35552E-118

0.5

0.606530659712633

0.606530659712633

2.17490E-118

0.6

0.548811636094026

0.548811636094026

3.18519E-118

0.7

0.496585303791410

0.496585303791410

4.38320E-118

0.8

0.449328964117222

0.449328964117222

5.76409E-118

0.9

0.406569659740599

0.406569659740599

7.32104E-118

1.0

0.367879441171442

0.367879441171442

9.04481E-118

_____________________________________________________________

(a)

(b)


Example 6.19 [68]. Consider the following linear DDE as follow 3

𝑥

𝑦 ′′ (𝑥) = 4 𝑦(𝑥) + 𝑦 (2) + 2 − 𝑥 2 ,

(6.301) 1177


(6.302)

The exact solution of the above system is 𝑦(𝑥) = 𝑥 2 .

(6.303)



(6.304)



𝑇


We apply the proposed technique to solve Eq. (6.301) with 𝐾 = 1, and 𝑀 = 3. We have Eq. (6.304) is 𝑦(𝑥) = ∑1𝑛=1 ∑2𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) , = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) = 𝐂 𝐓 𝛙(𝑥), 1

(6.305)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐿𝑚 is the Lommel polynomials given as 𝐿0 = 1, 𝐿1 = 𝑥, 𝐿2 = −1 + 𝑥 2 − 𝑥, 𝐿3 = 𝑥 3 − 3𝑥 2 + 2, 𝐿4 = −5 + 3𝑥 + 8𝑥 2 − 6𝑥 3 + 𝑥 4 , . . .. Therefore, the Lommel wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = √5(4𝑥 2 − 6𝑥 + 1), 𝜓1,3 (𝑥) = 2√7(−1 + 9𝑥 − 12𝑥 2 + 4𝑥 3 ), . 1178

(6.306)

. .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + √5(4𝑥 2 − 6𝑥 + 1)𝑐1,2 = 𝐂 𝐓 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 ] and 𝛙(𝑥) = [1, √3(2𝑥 − 1), √5(4𝑥 2 − 6𝑥 + 1)] . Substituting into the given problem we get 3

𝑥

𝐂 𝐓 𝛙′′ (𝑥) = 4 𝐂 𝐓 𝛙(𝑥) + 𝐂 𝐓 𝛙 (2) + 2 − 𝑥 2 ,

(6.307)

Subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 0, 𝐂 𝐓 𝛙′ (0) = 0.

(6.308)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2.

(6.309)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 3×3 𝐂3×1 = 𝐛3×1 . After solving we get the following exact solution 𝑦(𝑥) = 𝑥 2 .

Example 6.20. Consider the following linear DDE [68] as follow 𝑥

𝑥

1

𝑥

𝑦 ′′′ (𝑥) = 𝑦(𝑥) + 𝑦 ′ (2) + 𝑦 ′′ (3) + 2 𝑦 ′′′ (4) − 𝑥 4 −

𝑥3 3

4

− 3 𝑥 2 + 21𝑥,

(6.310)

subject to the boundary conditions 𝑦(0) = 0, 𝑦 ′ (0) = 0, 𝑦 ′′ (0) = 0.

(6.311)

The exact solution of the above system is 𝑦(𝑥) = 𝑥 4 .

(6.312)



(6.313)



𝑇


1179

We apply the proposed technique to solve Eq. (6.310) with 𝐾 = 1, and 𝑀 = 5. We have Eq. (6.313) is 𝑦(𝑥) = ∑1𝑛=1 ∑4𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) , = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) + 𝑐1,3 𝜓1,3 (𝑥) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 1

(6.314)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝐿𝑚 (2𝑘 𝑡 − (2𝑛 + 1)).

(6.315)

In the above expression, 𝐿𝑚 is the Lommel polynomials given as 𝐿0 = 1, 𝐿1 = 𝑥, 𝐿2 = −1 + 𝑥 2 − 𝑥, 𝐿3 = 𝑥 3 − 3𝑥 2 + 2, 𝐿4 = −5 + 3𝑥 + 8𝑥 2 − 6𝑥 3 + 𝑥 4 , . . .. Therefore, the Lommel wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = √5(4𝑥 2 − 6𝑥 + 1), 𝜓1,3 (𝑥) = 2√7(−1 + 9𝑥 − 12𝑥 2 + 4𝑥 3 ), . . .. Therefore, we have the trial solution is −1 + 9𝑥 − 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + √5(4𝑥 2 − 6𝑥 + 1)𝑐1,2 + 2√7 ( ) 12𝑥 2 + 4𝑥 3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇



𝑥

1

𝑥

𝐂 𝐓 𝛙′′′ (𝑥) = 𝐂 𝐓 𝛙(𝑥) + 𝐂 𝐓 𝛙′ (2) + 𝐂 𝐓 𝛙′′ (3) + 2 𝐂 𝐓 𝛙′′′ (4) − 1180

𝑥4 −

𝑥3 3

4

− 3 𝑥 2 + 21𝑥,

(6.316)

Subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 0, 𝐂 𝐓 𝛙′ (0) = 0, 𝐂 𝐓 𝛙′′ (0) = 0.

(6.317)

Substitute the collocating points are in Eq. (7a), we have the system of equations 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 3,4.

(6.318)


6.7. Bernoulli’s Wavelet Method (BWM) We proposed a new algorithm by inserting Bernoulli’s polynomials in traditional Legendre wavelets method. This technique is successively applied to find the numerical solutions of second order differential equations with Robin conditions. The proposed technique is very simple and highly compatible for solving such kind of problems.

6.7.1. Methodology 6.7.1.1. Bernoulli’s Polynomials The Bernoulli’s polynomials are orthogonal sequence of polynomials; the recurrence relation is given by (𝑛)!

(𝐵

)

𝑛−1 𝐵𝑛 (𝑥) = 𝑥 𝑛 − ∑𝑛−1 𝑘=0 𝑘!(𝑛−𝑘)! 𝑛−𝑘+1.

(6.319)

First four Bernoulli’s polynomials given below 𝐵0 = 1, 1

𝐵1 = 𝑥 − 2, 4

2

𝐵2 = 𝑥 2 − 3 𝑥 + 3, 𝐵3 = 𝑥 3 −

11 4

𝑥2 +

11 3

𝑥−

11 6

,

. . .. 1181

6.7.1.2. Wavelets and Bernoulli’s Wavelets In recent years, wavelets have found their way into many different fields of science and engineering. Wavelets constitute a family of functions constructed from the dilation and translation of a single function called the mother wavelet. When the dilation parameter a and the translation parameter 𝑏 vary continuously, we have the following family of continuous wavelets: 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|− 2 𝜓 (

𝑎

) 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

(6.320)

Bernoulli’s wavelets 𝜓𝑛𝑚 (𝑡) = 𝜓(𝑘, 𝑛, 𝑚, 𝑡) have four arguments; 𝑛, 𝑘 can assume any positive integer, 𝑚 is the order for Bernoulli’s polynomials and t is the normalized time. They are defined on the interval [0, 1) by; 1

𝜓𝑛𝑚

𝑘

𝑘 2 (𝑡) = {√𝑚 + 2 2 𝐵𝑚 (2 𝑡 − (2𝑛 + 1))

0

𝑛 2𝑘

≤𝑡≤

𝑛+1 2𝑘

(6.321)

otherwise 1

where 𝑚 = 0,1,2, … , 𝑀 − 1 and 𝑛 = 0,1, … , 2𝑘−1 . The coefficient (𝑚 + 2) is for orthonormality. Here 𝐵𝑚 (𝑡) are the well-known Bernoulli’s polynomials of order 𝑚, which have been previously described.


(6.322)


𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡).

(6.323)



(6.324)


and

𝛙(𝑡) = [

2

0

2

𝑀−1

𝜓10 , 𝜓11 , … , 𝜓1𝑀−1 , 𝜓20 , 𝜓21 , … , 𝑇 ] . 𝜓2𝑀−1 , … , 𝜓2𝑘−1 0 , … , 𝜓2𝑘−1 𝑀−1

1182

(6.325) (6.326)

6.7.1.4. Analysis Consider the second order BVPs with Robin conditions 𝑎(𝑥)𝑦 ′′ (𝑥) + 𝑏(𝑥)𝑦 ′ (𝑥) + 𝑐(𝑥)𝑦(𝑥) = 𝑓(𝑥), 𝑟1 (𝑥)𝑦(𝑥) + 𝑟2 (𝑥)𝑦 ′ (𝑥) = 𝑔1 (𝑥), at 𝑥 = 𝑎, ℎ1 (𝑥)𝑦(𝑥) + ℎ2 (𝑥)𝑦 ′ (𝑥) = 𝑔2 (𝑥), at 𝑥 = 𝑏. For solving second order BVPs with Robin conditions, it is sufficient to suppose that the approximate solution is as 𝑦(𝑥) = 𝐂 𝑇 𝛙(𝑥), where 𝐂 and 𝛙(𝑥) are given in above. Therefore, we obtain 𝑎(𝑥)𝐂 𝑇 𝛙′′ (𝑥) + 𝑏(𝑥)𝐂 𝑇 𝛙′ (𝑥) + 𝑐(𝑥)𝐂 𝑇 𝛙(𝑥) = 𝑓(𝑥), 𝑟1 (𝑥)𝐂 𝑇 𝛙(𝑥) + 𝑟2 (𝑥)𝐂 𝑇 𝛙′ (𝑥) = 𝑔1 (𝑥), at 𝑥 = 𝑎, ℎ1 (𝑥)𝐂 𝑇 𝛙(𝑥) + ℎ2 (𝑥)𝐂 𝑇 𝛙′ (𝑥) = 𝑔2 (𝑥), at 𝑥 = 𝑏. Therefore, in order to apply the Bernoulli’s Wavelet Method (BWM); we require 2𝑘−1 𝑀 − 1 collocating points. Suitable collocating points are as 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2,3, … , 2𝑘−1 𝑀 − 1.

Implementing the collocating points and imposing the initial value to the second order BVPs with Robin conditions, we obtain 𝑎(𝑥𝑖 )𝐂 𝑇 𝛙′′ (𝑥𝑖 ) + 𝑏(𝑥𝑖 )𝐂 𝑇 𝛙′ (𝑥𝑖 ) + 𝑐(𝑥𝑖 )𝐂 𝑇 𝛙(𝑥𝑖 ) = 𝑓(𝑥𝑖 ), 𝑖 = 2, … ,2𝑘−1 𝑀 − 1. The differential equation yields 2𝑘−1 𝑀 − 3 equations and initial condition produces 2 number of equations. Therefore, the obtained system has 2𝑘−1 𝑀 − 1 equations and 2𝑘−1 𝑀 − 1 unknowns. Solving this system gives the unknown coefficients 𝐂.

6.7.2. Second-order Differential Equations with Robin Conditions Example 6.21 [207]. Consider the following second order boundary value problems with Robin conditions as follow 𝑦 ′′ (𝑥) − sin(𝑥) 𝑦 ′ (𝑥) + 𝑦(𝑥) = sin2 (𝑥),

(6.327)

subject to the boundary conditions 𝑦(0) + 𝑦 ′ (0) = 1, sin(1) 𝑦(1) + cos(1) 𝑦 ′ (1) = 0.

(6.328)

The exact solution of Eq. (6.334-6.335) is 𝑦(𝑥) = cos(𝑥).

(6.329)

According to the Bernoulli’s Wavelets Method (BWM), we assume the trial solution 1183

𝑘−1


(6.330)



𝑇


We apply the proposed technique to solve Eq. (6.327-6.328) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.330) is 𝑦(𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) + ⋯ = 𝐂 𝐓 𝛙(𝑥),

(6.331) 1

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 22 𝐵𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). 2

In the above expression, 𝐵𝑚 is the Bernoulli’s polynomials given as 𝐵0 = 1, 1

𝐵1 = 𝑥 − 2, 4

2

𝐵2 = 𝑥 2 − 3 𝑥 + 3, 11

𝐵3 = 𝑥 3 − 𝐵4 =

143 15

4

−

𝑥2 +

286 15

𝑥+

11 3

𝑥−

143 10

11 6

𝑥2 −

, 26 5

𝑥3 + 𝑥4,

. . .. Therefore, the Bernoulli’s wavelets given as 𝜓1,0 (𝑥) = 1, 1

𝜓1,1 (𝑥) = 2 √3(4𝑥 − 3), 1

𝜓1,2 (𝑥) = 3 √5(12𝑥 2 − 20𝑥 + 9), 1


(6.332)

1

1

𝑦(𝑥) = 𝑐1,0 + 2 √3(4𝑥 − 3)𝑐1,1 + 3 √5(12𝑥 2 − 20𝑥 + 9)𝑐1,2 + 1 12

√7(−111 + 292𝑥 − 276𝑥 2 + 96𝑥 3 ) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , 1

1

1, 2 √3(4𝑥 − 3), 3 √5(12𝑥 2 − 20𝑥 + 9),

𝑇

and 𝛙(𝑥) = [ 1

] . (−111 + 292𝑥 − 276𝑥 2 + 96𝑥 3 ), … √7 12

Substituting into the given problem we get 𝐂 𝐓 𝛙′′ (𝑥) − sin(𝑥) 𝐂 𝐓 𝛙′ (𝑥) + 𝐂 𝐓 𝛙(𝑥) = sin2(𝑥),

(6.333)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) + 𝐂 𝐓 𝛙′ (0) = 1, sin(1) 𝐂 𝐓 𝛙(1) + cos(1) 𝐂 𝐓 𝛙′ (1) = 0. (6.334) Substitute the collocating points are in Eq. (6.333), we have the system of equations 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2,3,4, … 9.

(6.335)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following exact solution 𝑦(𝑥) = 1.00000 − 0.00000𝑥 − 0.500000𝑥 2 + ⋯.

1185

Table 6.55: Comparison of the Exact Solution and Approximate Solution of Eq. (6.327) obtained from Bernoulli’s Wavelet Method (BWM) _____________________________________________________________ 𝑥

Exact Solution


Error in BWM

_____________________________________________________________ 0.00

1.000000000000000

1.000001074346365

1.07435E-06

0.01

0.999950000416665

0.999951063963089

1.06355E-06

0.02

0.999800006666578

0.999801059300048

1.05263E-06

0.03

0.999550033748988

0.999551075356439

1.04161E-06

0.04

0.999200106660978

0.999201137129339

1.03047E-06

0.05

0.998750260394966

0.998751279611205

1.01922E-06

0.06

0.998200539935204

0.998201547786382

1.00785E-06

0.07

0.997551000253280

0.997551996626603

9.96373E-07

0.08

0.996801706302619

0.996802691085494

9.84783E-07

0.09

0.995952733011994

0.995953706092077

9.73080E-07

0.10

0.995004165278026

0.995005126543282

9.61265E-07

_____________________________________________________________

(a)

(b)


Now to solve Eq. (6.327-6.328) with 𝐾 = 1, and 𝑀 = 30. We have Eq. (6.330) is 𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑29 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥),

1186

𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,29 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,29 ] . Proceeding as before we have the matrix form 𝐀 𝟑𝟎×𝟑𝟎 𝐂𝟑𝟎×𝟏 = 𝐛𝟑𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.56: Comparison of the Exact Solution and Approximate Solution of Eq. (6.327) obtained from Bernoulli’s Wavelet Method (BWM) _____________________________________________________________ 𝑥

Exact Solution


Error in BWM

_____________________________________________________________ 0.00

1.000000000000000

1.000000000000000

1.12345E-36

0.10

0.995004165278026

0.995004165278026

1.00547E-36

0.20

0.980066577841242

0.980066577841242

8.76228E-37

0.30

0.955336489125606

0.955336489125606

7.35492E-37

0.40

0.921060994002885

0.921060994002885

5.83116E-37

0.50

0.877582561890373

0.877582561890373

4.18674E-37

0.60

0.825335614909678

0.825335614909678

2.41946E-37

0.70

0.764842187284488

0.764842187284488

5.23923E-38

0.80

0.696706709347165

0.696706709347165

1.50343E-37

0.90

0.621609968270664

0.621609968270664

3.66583E-37

1.00

0.540302305868140

0.540302305868140

5.72751E-37

_____________________________________________________________

1187

(a)

(b)


Finally we take 𝐾 = 1, and 𝑀 = 100. We have Eq. (6.330) is 𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑99 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,99 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,99 ] . Proceeding as before we have the matrix form 𝐀 𝟏𝟎𝟎×𝟏𝟎𝟎 𝐂𝟏𝟎𝟎×𝟏 = 𝐛𝟏𝟎𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table.

1188


Exact Solution


Error in BWM

_____________________________________________________________ 0.00

1.000000000000000

1.000000000000000

3.42279E-181

0.10

0.995004165278026

0.995004165278026

3.06338E-181

0.20

0.980066577841242

0.980066577841242

2.66955E-181

0.30

0.955336489125606

0.955336489125606

2.24084E-181

0.40

0.921060994002885

0.921060994002885

1.77651E-181

0.50

0.877582561890373

0.877582561890373

1.27562E-181

0.60

0.825335614909678

0.825335614909678

7.37052E-182

0.70

0.764842187284488

0.764842187284488

1.59620E-182

0.80

0.696706709347165

0.696706709347165

4.57880E-182

0.90

0.621609968270664

0.621609968270664

1.11654E-181

1.00

0.540302305868140

0.540302305868140

1.81065E-181

_____________________________________________________________

(a)

(b)


Example 6.22 [207]. Consider the following second order boundary value problems with Robin conditions as follow −𝑦 ′′ (𝑥) + 𝑦(𝑥) = 2cos(𝑥),

(6.336) 1189

subject to the boundary conditions 𝜋

𝜋

3𝑦 ( 2 ) + 𝑦 ′ ( 2 ) = −1, 4𝑦(𝜋) + 𝑦 ′ (𝜋) = −4.

(6.337)

The exact solution of Eq. (6.336-6.337) is 𝑦(𝑥) = cos(𝑥).

(6.338)

According to the Bernoulli’s Wavelets Method (BWM), we assume the trial solution 𝑘−1


(6.339)



𝑇



(6.340) 1

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝐵𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝐵𝑚 is the Bernoulli’s polynomials given as 𝐵0 = 1, 1

𝐵1 = 𝑥 − 2, 4

2

𝐵2 = 𝑥 2 − 3 𝑥 + 3, 11

𝐵3 = 𝑥 3 − 𝐵4 =

143 15

4

−

𝑥2 +

286 15

𝑥+

11 3

𝑥−

143 10

11 6

𝑥2 −

, 26 5

𝑥3 + 𝑥4,


𝜓1,1 (𝑥) = 2 √3(4𝑥 − 3), 1

𝜓1,2 (𝑥) = 3 √5(12𝑥 2 − 20𝑥 + 9), 1190

(6.341)

1


1

𝑦(𝑥) = 𝑐1,0 + 2 √3(4𝑥 − 3)𝑐1,1 + 3 √5(12𝑥 2 − 20𝑥 + 9)𝑐1,2 + 1 12

√7(−111 + 292𝑥 − 276𝑥 2 + 96𝑥 3 ) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , 1

1

1, 2 √3(4𝑥 − 3), 3 √5(12𝑥 2 − 20𝑥 + 9),

𝑇

and 𝛙(𝑥) = [ 1

] . (−111 + 292𝑥 − 276𝑥 2 + 96𝑥 3 ), … √7 12

Substituting into the given problem we get −𝐂 𝐓 𝛙′′ (𝑥) + 𝐂 𝐓 𝛙(𝑥) = 2 cos(𝑥),

(6.342)

subject to the boundary conditions 𝜋

𝜋

3𝐂 𝐓 𝛙 ( 2 ) + 𝐂 𝐓 𝛙′ ( 2 ) = −1,4𝐂 𝐓 𝛙(𝜋) + 𝐂 𝐓 𝛙′ (𝜋) = −4.

(6.343)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2,3,4, … 9.

(6.344)


1191


Exact Solution


Error in BWM

_____________________________________________________________ 0.00

1.000000000000000

1.104674953495537

1.04675E-01

0.10

0.995004165278026

1.089251046323935

9.42469E-02

0.20

0.980066577841242

1.064828640630297

8.47621E-02

0.30

0.955336489125606

1.031462061199986

7.61256E-02

0.40

0.921060994002885

0.989311966604971

6.82510E-02

0.50

0.877582561890373

0.938642014348089

6.10595E-02

0.60

0.825335614909678

0.879814649593607

5.44790E-02

0.70

0.764842187284488

0.813286045485081

4.84439E-02

0.80

0.696706709347165

0.739600234466021

4.28935E-02

0.90

0.621609968270664

0.659382481521784

3.77725E-02

1.00

0.540302305868140

0.573331961852427

3.30297E-02

_____________________________________________________________

(a)

(b)



1192

𝑇

𝑇


Exact Solution


Error in BWM

_____________________________________________________________ 0.00

1.000000000000000

1.000000000000000

1.59900E-17

0.10

0.995004165278026

0.995004165278026

1.43976E-17

0.20

0.980066577841242

0.980066577841242

1.29493E-17

0.30

0.955336489125606

0.955336489125606

1.16306E-17

0.40

0.921060994002885

0.921060994002885

1.04283E-17

0.50

0.877582561890373

0.877582561890373

9.33037E-18

0.60

0.825335614909678

0.825335614909678

8.32584E-18

0.70

0.764842187284488

0.764842187284488

7.40463E-18

0.80

0.696706709347165

0.696706709347165

6.55753E-18

0.90

0.621609968270664

0.621609968270664

5.77605E-18

1.00

0.540302305868140

0.540302305868140

5.05239E-18

_____________________________________________________________

1193

(a)

(b)



𝑇


1194


Exact Solution


Error in BWM

_____________________________________________________________ 0.00

1.000000000000000

1.000000000000000

1.20130E-108

0.10

0.995004165278026

0.995004165278026

1.08166E-108

0.20

0.980066577841242

0.980066577841242

9.72855E-109

0.30

0.955336489125606

0.955336489125606

8.73784E-109

0.40

0.921060994002885

0.921060994002885

7.83459E-109

0.50

0.877582561890373

0.877582561890373

7.00974E-109

0.60

0.825335614909678

0.825335614909678

6.25505E-109

0.70

0.764842187284488

0.764842187284488

5.56296E-109

0.80

0.696706709347165

0.696706709347165

4.92655E-109

0.90

0.621609968270664

0.621609968270664

4.33944E-109

1.00

0.540302305868140

0.540302305868140

3.79577E-109

_____________________________________________________________

(a)

(b)


Example 6.23 [207]. Consider the following second order boundary value problems with Robin conditions as follow 1195

𝑦 ′′ (𝑥) + (1 − 𝑥)𝑦 ′ (𝑥) + 2𝑦(𝑥) = (1 + 2𝑥 − 𝑥 2 )sin(𝑥),

(6.345)

subject to the boundary conditions 5𝑦(0) + 4𝑦 ′ (0) = 1, 𝑦(1) + cos(1)𝑦 ′ (1) = − cos 2 (1).

(6.346)

The exact solution of Eq. (6.345-6.346) is 𝑦(𝑥) = (1 − 𝑥) cos(𝑥).

(6.347)



(6.348)



𝑇



(6.349) 1

𝑘


𝐵1 = 𝑥 − 2, 4

2

𝐵2 = 𝑥 2 − 3 𝑥 + 3, 11

𝐵3 = 𝑥 3 − 𝐵4 =

143 15

4

−

𝑥2 +

286 15

𝑥+

11 3

𝑥−

143 10

11 6

𝑥2 −

, 26 5

𝑥3 + 𝑥4,


𝜓1,1 (𝑥) = 2 √3(4𝑥 − 3), 1

𝜓1,2 (𝑥) = 3 √5(12𝑥 2 − 20𝑥 + 9), 1196

(6.350)

1


1

𝑦(𝑥) = 𝑐1,0 + 2 √3(4𝑥 − 3)𝑐1,1 + 3 √5(12𝑥 2 − 20𝑥 + 9)𝑐1,2 + 1 12

√7(−111 + 292𝑥 − 276𝑥 2 + 96𝑥 3 ) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , 1

1

1, 2 √3(4𝑥 − 3), 3 √5(12𝑥 2 − 20𝑥 + 9),

𝑇

and 𝛙(𝑥) = [ 1

] . (−111 + 292𝑥 − 276𝑥 2 + 96𝑥 3 ), … √7 12

Substituting into the given problem we get 𝐂 𝐓 𝛙′′ (𝑥) + (1 − 𝑥)𝐂 𝐓 𝛙′ (𝑥) + 2𝐂 𝐓 𝛙(𝑥) = (1 + 2𝑥 − 𝑥 2 ) sin(𝑥),

(6.351)

subject to the boundary conditions 5𝐂 𝐓 𝛙(0) + 4𝐂 𝐓 𝛙′(0) = 1, 𝐂 𝐓 𝛙(1) + cos(1)𝐂 𝐓 𝛙′(1) = − cos2 (1).

(6.352)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2,3,4, … 9.

(6.353)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following exact solution 𝑦(𝑥) = 1.000000 − 1.000000𝑥 − .50000𝑥 2 . . ..

1197


Exact Solution


Error in BWM

_____________________________________________________________ 0.00

1.000000000000000

1.000004107868391

4.10787E-06

0.10

0.895503748750223

0.895507328172219

3.57942E-06

0.20

0.784053262272993

0.784056289284957

3.02701E-06

0.30

0.668735542387924

0.668738002528434

2.46014E-06

0.40

0.552636596401731

0.552638483829952

1.88743E-06

0.50

0.438791280945186

0.438792594665640

1.31372E-06

0.60

0.330134245963871

0.330134984552217

7.38588E-07

0.70

0.229452656185347

0.229452817551377

1.61366E-07

0.80

0.139341341869433

0.139340942706241

3.99163E-07

0.90

0.062160996827066

0.062160137542977

8.59284E-07

1.00

0.000000000000000

-0.000000985257330

9.85257E-07

_____________________________________________________________

(a)

(b)



1198

𝑇

𝑇


Exact Solution


Error in BWM

_____________________________________________________________ 0.00

1.000000000000000

1.000000000000000

3.80142E-36

0.10

0.895503748750223

0.895503748750223

3.31318E-36

0.20

0.784053262272993

0.784053262272993

2.80373E-36

0.30

0.668735542387924

0.668735542387924

2.27947E-36

0.40

0.552636596401731

0.552636596401731

1.74680E-36

0.50

0.438791280945186

0.438791280945186

1.21109E-36

0.60

0.330134245963871

0.330134245963871

6.78112E-37

0.70

0.229452656185347

0.229452656185347

1.52580E-37

0.80

0.139341341869433

0.139341341869433

3.60484E-37

0.90

0.062160996827066

0.062160996827066

8.56221E-37

1.00

0.000000000000000

-0.000000000000000

1.27874E-36

_____________________________________________________________

1199

(a)

(b)



𝑇


1200


Exact Solution


Error in BWM

_____________________________________________________________ 0.00

1.000000000000000

1.000000000000000

1.10389E-180

0.10

0.895503748750223

0.895503748750223

9.62126E-181

0.20

0.784053262272993

0.784053262272993

8.14169E-181

0.30

0.668735542387924

0.668735542387924

6.61948E-181

0.40

0.552636596401731

0.552636596401731

5.07240E-181

0.50

0.438791280945186

0.438791280945186

3.51704E-181

0.60

0.330134245963871

0.330134245963871

1.96898E-181

0.70

0.229452656185347

0.229452656185347

4.43065E-182

0.80

0.139341341869433

0.139341341869433

1.04644E-181

0.90

0.062160996827066

0.062160996827066

2.48571E-181

1.00

0.000000000000000

0.000000000000000

3.84772E-181

______________________________________________________________

(a)

(b)


Example 6.24 [207]. Consider the following second order boundary value problems with Robin conditions as follow 1201

𝑦 ′′ (𝑥) + 𝑥𝑦 ′ (𝑥) − (3 − 𝑥 − 𝑥 2 + 𝑥 3 ) sin(𝑥) = 4𝑥cos(𝑥),

(6.354)

subject to the boundary conditions 𝑦(0) + 𝑦 ′ (0) = −1, cos(1) 𝑦(1) − 𝑦 ′ (1) = −2 sin(1).

(6.355)

The exact solution of Eq. (6.354-6.355) is 𝑦(𝑥) = (𝑥 2 − 1) sin(𝑥).

(6.356)



(6.357)



𝑇



(6.358) 1

𝑘


𝐵1 = 𝑥 − 2, 4

2

𝐵2 = 𝑥 2 − 3 𝑥 + 3, 11

𝐵3 = 𝑥 3 − 𝐵4 =

143 15

4

−

𝑥2 +

286 15

𝑥+

11 3

𝑥−

143 10

11 6

𝑥2 −

, 26 5

𝑥3 + 𝑥4,


𝜓1,1 (𝑥) = 2 √3(4𝑥 − 3), 1

𝜓1,2 (𝑥) = 3 √5(12𝑥 2 − 20𝑥 + 9), 1202

(6.359)

1


1

𝑦(𝑥) = 𝑐1,0 + 2 √3(4𝑥 − 3)𝑐1,1 + 3 √5(12𝑥 2 − 20𝑥 + 9)𝑐1,2 + 1 12

√7(−111 + 292𝑥 − 276𝑥 2 + 96𝑥 3 ) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , 1

1

1, 2 √3(4𝑥 − 3), 3 √5(12𝑥 2 − 20𝑥 + 9),

𝑇

and 𝛙(𝑥) = [ 1

] . (−111 + 292𝑥 − 276𝑥 2 + 96𝑥 3 ), … √7 12

Substituting into the given problem we get 𝐂 𝐓 𝛙′′ (𝑥) + 𝑥𝐂 𝐓 𝛙′ (𝑥) − (3 − 𝑥 − 𝑥 2 + 𝑥 3 ) sin(𝑥) = 4𝑥cos(𝑥), (6.360) subject to the boundary conditions 𝐂 𝐓 𝛙(0) + 𝐂 𝐓 𝛙′ (0) = −1, cos(1) 𝐂 𝐓 𝛙(1) + 𝐂 𝐓 𝛙′ (1) = −2 sin(1).

(6.361)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2,3,4, … 9.

(6.362)


1203


Exact Solution


Error in BWM

_____________________________________________________________ 0.0

0.000000000000000

0.000029369597764

2.93696E-05

0.1

-0.098835082480360

-0.098808660461496

2.64220E-05

0.2

-0.190722557563259

-0.190699115955223

2.34416E-05

0.3

-0.268923388061819

-0.268902965645561

2.04224E-05

0.4

-0.327111407539266

-0.327094044480659

1.73631E-05

0.5

-0.359569153953152

-0.359554905347149

1.42486E-05

0.6

-0.361371182972823

-0.361360141193874

1.10418E-05

0.7

-0.328551020491222

-0.328543308336096

7.71216E-06

0.8

-0.258248192723828

-0.258243849878857

4.34284E-06

0.9

-0.148832112829222

-0.148830754241387

1.35859E-06

1.0

0.000000000000000

-0.000000087721404

8.77214E-08

_____________________________________________________________

(a)

(b)


Now to solve Eq. (6.354-6.355) with 𝐾 = 1, and 𝑀 = 30. We have Eq. (6.357) is 𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑29 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇

𝑇


Proceeding as before we have the matrix form 𝐀 𝟑𝟎×𝟑𝟎 𝐂𝟑𝟎×𝟏 = 𝐛𝟑𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.65: Comparison of the Exact Solution and Approximate Solution of Eq. (6.354) obtained from Bernoulli’s Wavelet Method (BWM) _____________________________________________________________ 𝑥

Exact Solution


Error in BWM

_____________________________________________________________ 0.0

0.000000000000000

0.000000000000000

9.48310E-35

0.1

-0.098835082480360

-0.098835082480360

8.53320E-35

0.2

-0.190722557563259

-0.190722557563259

7.57513E-35

0.3

-0.268923388061819

-0.268923388061819

6.60184E-35

0.4

-0.327111407539266

-0.327111407539266

5.60918E-35

0.5

-0.359569153953152

-0.359569153953152

4.59394E-35

0.6

-0.361371182972823

-0.361371182972823

3.55629E-35

0.7

-0.328551020491222

-0.328551020491222

2.49716E-35

0.8

-0.258248192723828

-0.258248192723828

1.42065E-35

0.9

-0.148832112829222

-0.148832112829222

3.33018E-36

1.0

0.000000000000000

-0.000000000000000

6.80057E-36

_____________________________________________________________

1205

(a)

(b)



𝑇


1206


Exact Solution


Error in BWM

_____________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

9.30223E-179

0.1

-0.098835082480360

-0.098835082480360

8.37053E-179

0.2

-0.190722557563259

-0.190722557563259

7.43062E-179

0.3

-0.268923388061819

-0.268923388061819

6.47601E-179

0.4

-0.327111407539266

-0.327111407539266

5.50214E-179

0.5

-0.359569153953152

-0.359569153953152

4.50643E-179

0.6

-0.361371182972823

-0.361371182972823

3.48836E-179

0.7

-0.328551020491222

-0.328551020491222

2.44955E-179

0.8

-0.258248192723828

-0.258248192723828

1.39377E-179

0.9

-0.148832112829222

-0.148832112829222

3.27031E-180

1.0

0.000000000000000

0.000000000000000

7.35726E-180

_____________________________________________________________

(a)

(b)


Example 6.25 [207]. Consider the following second order boundary value problems with Robin conditions as follow 𝑦 ′′ (𝑥) − 𝑦(𝑥) = cos(𝑥),

(6.363) 1207

subject to the boundary conditions 5𝑦(0) = 5, 1 (−3 cosh(1)−3 sinh(1)+cos(1)+2+cos(1) sinh(1)e+sin(1) sinh(1)e)e−1

𝑦(1) − 𝑦 ′ (1) = − 2

sinh(1)

. (6.364)

The exact solution of Eq. (6.363-6.364) is 1 (−3 cosh(1)+3 sinh(1)+cos(1)+2)e𝑥

𝑦(𝑥) = 4

sinh(1)

1 (3 cosh(1)+3 sinh(1)−cos(1)−2)e−𝑥

+4

sinh(1)

1

− 2 cos(𝑥).



(6.365)



𝑇



(6.366) 1

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 22 𝐵𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). 2

In the above expression, 𝐵𝑚 is the Bernoulli’s polynomials given as 𝐵0 = 1, 1

𝐵1 = 𝑥 − 2, 4

2

𝐵2 = 𝑥 2 − 3 𝑥 + 3, 11

𝐵3 = 𝑥 3 − 𝐵4 =

143 15

4

−

𝑥2 +

286 15

𝑥+

11 3

𝑥−

143 10

11 6

𝑥2 −

, 26 5

𝑥3 + 𝑥4,


𝜓1,1 (𝑥) = 2 √3(4𝑥 − 3), 1208

(6.367)

1

𝜓1,2 (𝑥) = 3 √5(12𝑥 2 − 20𝑥 + 9), 1


1

𝑦(𝑥) = 𝑐1,0 + 2 √3(4𝑥 − 3)𝑐1,1 + 3 √5(12𝑥 2 − 20𝑥 + 9)𝑐1,2 + 1 12

√7(−111 + 292𝑥 − 276𝑥 2 + 96𝑥 3 ) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , 1

1

1, 2 √3(4𝑥 − 3), 3 √5(12𝑥 2 − 20𝑥 + 9),

𝑇

and 𝛙(𝑥) = [ 1

] . (−111 + 292𝑥 − 276𝑥 2 + 96𝑥 3 ), … √7 12

Substituting into the given problem we get 𝐂 𝐓 𝛙′′ (𝑥) + 𝑥𝐂 𝐓 𝛙′ (𝑥) − (3 − 𝑥 − 𝑥 2 + 𝑥 3 ) sin(𝑥) = 4𝑥cos(𝑥), (6.368) subject to the boundary conditions 𝐂 𝐓 𝛙(0) + 𝐂 𝐓 𝛙′ (0) = −1, cos(1) 𝐂 𝐓 𝛙(1) + 𝐂 𝐓 𝛙′ (1) = −2 sin(1).

(6.369)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2,3,4, … 9.

(6.370)


1209


Exact Solution


Error in BWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

0.00000E+00

0.1

0.920980140321431

0.920981751643364

1.61132E-06

0.2

0.861127803855995

0.861131043345910

3.23949E-06

0.3

0.819694592387419

0.819699491606103

4.89922E-06

0.4

0.796018527453292

0.796025133007789

6.60555E-06

0.5

0.789519895991516

0.789528272296454

8.37630E-06

0.6

0.799698873072258

0.799709106909750

1.02338E-05

0.7

0.826134863704475

0.826147065644636

1.22019E-05

0.8

0.868487513723571

0.868501803427862

1.42897E-05

0.9

0.926499347717514

0.926515797033995

1.64493E-05

1.0

1.000000000000000

1.000018488065478

1.84881E-05

_____________________________________________________________

(a)

(b)


Now to solve Eq. (6.363-6.364) with 𝐾 = 1, and 𝑀 = 30. We have Eq. (6.365) is 𝑇 𝑦(𝑥) = ∑1𝑛=1 ∑29 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇

𝑇


Proceeding as before we have the matrix form 𝐀 𝟑𝟎×𝟑𝟎 𝐂𝟑𝟎×𝟏 = 𝐛𝟑𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.68: Comparison of the Exact Solution and Approximate Solution of Eq. (6.363) obtained from Bernoulli’s Wavelet Method (BWM) _____________________________________________________________ 𝑥

Exact Solution


Error in BWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.00000E-199

0.1

0.920980140321431

0.920980140321431

1.83373E-36

0.2

0.861127803855995

0.861127803855995

3.68575E-36

0.3

0.819694592387419

0.819694592387419

5.57474E-36

0.4

0.796018527453292

0.796018527453292

7.51941E-36

0.5

0.789519895991516

0.789519895991516

9.53948E-36

0.6

0.799698873072258

0.799698873072258

1.16549E-35

0.7

0.826134863704475

0.826134863704475

1.38870E-35

0.8

0.868487513723571

0.868487513723571

1.62582E-35

0.9

0.926499347717514

0.926499347717514

1.87922E-35

1.0

1.000000000000000

1.000000000000000

2.14644E-35

_____________________________________________________________

1211

(a)

(b)



𝑇


1212


Exact Solution


Error in BWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.00000E-199

0.1

0.920980140321431

0.920980140321431

2.78844E-181

0.2

0.861127803855995

0.861127803855995

5.60479E-181

0.3

0.819694592387419

0.819694592387419

8.47723E-181

0.4

0.796018527453292

0.796018527453292

1.14345E-180

0.5

0.789519895991516

0.789519895991516

1.45062E-180

0.6

0.799698873072258

0.799698873072258

1.77232E-180

0.7

0.826134863704475

0.826134863704475

2.11174E-180

0.8

0.868487513723571

0.868487513723571

2.47231E-180

0.9

0.926499347717514

0.926499347717514

2.85762E-180

1.0

1.000000000000000

1.000000000000000

3.27085E-180

_____________________________________________________________

(a)

(b)


6.8. Probabilist Hermite Wavelet Method (PWM) We proposed a new algorithm by inserting Probabilist Hermite polynomials in traditional Legendre wavelets method. This technique is successively applied to find the numerical 1213

solutions of nonlinear Integral equations. The proposed technique is very simple and highly compatible for solving such kind of problems

6.8.1. Methodology 6.8.1.1. Probabilist Hermite Polynomials The Probabilist Hermite polynomials are orthogonal sequence of polynomials, the recurrence relation is given by 𝑥2

𝑃𝑛 = (−1)𝑛 e 2

𝑥2

𝑑𝑛

e− 2 𝑑𝑥 𝑛

(6.371)

First five Probabilist Hermite given below 𝑃0 = 1, 𝑃1 = 𝑥, 𝑃2 = −1 + 𝑥 2 𝑃3 = −3𝑥 + 𝑥 3 , 𝑃4 = 3 − 6𝑥 2 + 𝑥 4 , . . ..

6.8.1.2. Wavelets and Probabilist Hermite Wavelets In recent years, wavelets have found their way into many different fields of science and engineering. Wavelets constitute a family of functions constructed from the dilation and translation of a single function called the mother wavelet. When the dilation parameter a and the translation parameter 𝑏 vary continuously, we have the following family of continuous wavelets: 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|− 2 𝜓 (

𝑎

) 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

(6.372)

Probabilist Hermite wavelets 𝜓𝑛𝑚 (𝑡) = 𝜓(𝑘, 𝑛, 𝑚, 𝑡) have four arguments; 𝑛, 𝑘 can assume any positive integer, 𝑚 is the order for Probabilist Hermite polynomials and t is the normalized time. They are defined on the interval [0, 1) by; 1

𝜓𝑛𝑚

𝑘

𝑘 2 (𝑡) = {√𝑚 + 2 2 𝑃𝑚 (2 𝑡 − (2𝑛 + 1))

0

𝑛 2𝑘

≤𝑡≤

otherwise

1214

𝑛+1 2𝑘

(6.373)

1

where 𝑚 = 0,1,2, … , 𝑀 − 1 and 𝑛 = 0,1, … , 2𝑘−1 . The coefficient (𝑚 + 2) is for orthonormality. Here 𝑃𝑚 (𝑡) are the well-known Probabilist Hermite polynomials of order 𝑚, which have been previously described.


(6.374)


𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡).

(6.375)



(6.376)


(6.377)


(6.378)

2𝑀−1

and

2

0

2

𝑀−1

6.8.1.4. Analysis Consider the nonlinear Volterra integral equation 𝑥

𝑐(𝑥)𝑦(𝑥) = 𝑓(𝑥) + ∫0 𝐾(𝑥, 𝑡)𝐹(𝑢(𝑡))d𝑡. For solving nonlinear Volterra integral equation, it is sufficient to suppose that the approximate solution is as 𝑦(𝑥) = 𝐂 𝑇 𝛙(𝑥), where 𝐂 and 𝛙(𝑥) are given in above. Therefore, we obtain 𝑥

𝑐(𝑥)𝐂 𝑇 𝛙(𝑥) = 𝑓(𝑥) + ∫0 𝐾(𝑥, 𝑡)𝐹(𝐂 𝑇 𝛙(𝑡))d𝑡. Therefore, in order to apply the Probabilist Hermite Wavelet Method (PWM); we require 2𝑘−1 𝑀 − 1 collocating points. Suitable collocating points are as 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 1,2,3, … , 2𝑘−1 𝑀.

Implementing the collocating points and imposing the initial value to the nonlinear Volterra integral equation, we obtain

1215

𝑥

𝑐(𝑥𝑖 )𝐂 𝑇 𝛙(𝑥𝑖 ) = 𝑓(𝑥𝑖 ) + (∫0 𝐾(𝑥, 𝑡)𝐹(𝐂 𝑇 𝛙(𝑡))d𝑡)

𝑥=𝑥𝑖

, 𝑖 = 0,2, … ,2𝑘−1 𝑀 − 1.

The differential equation yields 2𝑘−1 𝑀 − 1 equations. Therefore, the obtained system has 2𝑘−1 𝑀 − 1 equations and 2𝑘−1 𝑀 − 1 unknowns. Solving this system gives the unknown coefficients 𝐂.

6.8.2. Nonlinear Integral Equation of 2nd Kind Example 6.26 [229]. Consider the following nonlinear integral equations as follow 1

1

1

𝑥

𝑦(𝑥) = sin(𝑥) + cos(𝑥) + 4 sin(2𝑥) − 2 𝑥 − 2 𝑥 2 + ∫0 (𝑥 − 𝑡)𝑦 2 (𝑡),

(6.379)

The exact solution of Eq. (6.379) is 𝑦(𝑥) = sin(𝑥) + cos(𝑥).

(6.380)

According to the Probabilist Hermite Wavelets Method (PWM), we assume the trial solution 𝑘−1


(6.381)

where 𝐂 and 𝛙(𝑥) are 2𝑘−1 𝑀 × 1 matrices given as 𝑇


𝑇


We apply the proposed technique to solve Eq. (6.379) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.381) is 𝑦(𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) + ⋯ = 𝐂 𝐓 𝛙(𝑥),

(6.382) 1

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑃𝑚 (2𝑘 𝑡 − (2𝑛 + 1)).

(6.383)

In the above expression, 𝑃𝑚 is the Probabilist Hermite polynomials given as 𝑃0 = 1, 𝑃1 = 𝑥, 𝑃2 = −1 + 𝑥 2 𝑃3 = −3𝑥 + 𝑥 3 , 𝑃4 = 3 − 6𝑥 2 + 𝑥 4 , . . 1216

.. Therefore, the Probabilist Hermite wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = 4√5𝑥(𝑥 − 1), 𝜓1,3 (𝑥) = 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + 4√5𝑥(𝑥 − 1)𝑐1,2 + 2√7(2𝑥 − 1) (2𝑥 2 − 2𝑥 − 1)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇



1

1

𝐂 𝐓 𝛙(𝑥) = sin(𝑥) + cos(𝑥) + 4 sin(2𝑥) − 2 𝑥 − 2 𝑥 2 𝑥

2

+ ∫0 (𝑥 − 𝑡)(𝐂 𝐓 𝛙(𝑡)) .

(6.384)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2,3,4, … 7.

(6.385)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 8×8 𝐂8×1 = 𝐛8×1 . After solving we get the following exact solution 𝑦(𝑥) = 1.00000 − 0.00000𝑥 − 0.500000𝑥 2 + ⋯.

1217

Table 6.70: Comparison of the Exact Solution and Approximate Solution of Eq. (6.379) obtained from Probabilist Hermite Wavelet Method (PWM) _____________________________________________________________ 𝑥

Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

0.999999805599298

1.94401E-07

0.1

1.094837581924854

1.094837443070901

1.38854E-07

0.2

1.178735908636303

1.178735909941607

1.30530E-09

0.3

1.250856695786946

1.250856837750599

1.41964E-07

0.4

1.310479336311536

1.310479524338895

1.88027E-07

0.5

1.357008100494576

1.357008193415372

9.29208E-08

0.6

1.389978088304714

1.389977997553646

9.07511E-08

0.7

1.409059874522179

1.409059667381954

2.07140E-07

0.8

1.414062800246688

1.414062709728205

9.05185E-08

0.9

1.404936877898148

1.404937057482351

1.79584E-07

1.0

1.381773290676036

1.381773073938238

2.16738E-07

_____________________________________________________________

(a)

(b)



1218

𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,11 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,11 ] . Proceeding as before we have the matrix form 𝐀 𝟏𝟐×𝟏𝟐 𝐂𝟏𝟐×𝟏 = 𝐛𝟏𝟐×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.71: Comparison of the Exact Solution and Approximate Solution of Eq. (6.379) obtained from Probabilist Hermite Wavelet Method (PWM) _____________________________________________________________ 𝑥

Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

0.999999999998985

1.01483E-12

0.1

1.094837581924854

1.094837581924476

3.78210E-13

0.2

1.178735908636303

1.178735908637049

7.46289E-13

0.3

1.250856695786946

1.250856695787824

8.78258E-13

0.4

1.310479336311536

1.310479336311294

2.42019E-13

0.5

1.357008100494576

1.357008100493528

1.04778E-12

0.6

1.389978088304714

1.389978088304567

1.46174E-13

0.7

1.409059874522179

1.409059874523214

1.03427E-12

0.8

1.414062800246688

1.414062800246534

1.54034E-13

0.9

1.404936877898148

1.404936877897448

6.99590E-13

1.0

1.381773290676036

1.381773290674946

1.09054E-12

_____________________________________________________________

1219

(a)

(b)



𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,14 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,14 ] . Proceeding as before we have the matrix form 𝐀 𝟏𝟓×𝟏𝟓 𝐂𝟏𝟓×𝟏 = 𝐛𝟏𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table.

1220


Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.00000E-200

0.1

1.094837581924854

1.094837581924854

4.62404E-17

0.2

1.178735908636303

1.178735908636303

6.81557E-18

0.3

1.250856695786946

1.250856695786946

4.26478E-17

0.4

1.310479336311536

1.310479336311536

2.21799E-18

0.5

1.357008100494576

1.357008100494576

4.79800E-17

0.6

1.389978088304714

1.389978088304714

5.66165E-18

0.7

1.409059874522179

1.409059874522180

2.99397E-17

0.8

1.414062800246688

1.414062800246688

4.90422E-17

0.9

1.404936877898148

1.404936877898148

3.05433E-17

1.0

1.381773290676036

1.381773290676036

3.36048E-17

_____________________________________________________________

(a)

(b)


Example 6.27 [229]. Consider the following nonlinear integral equations as follow 1

1

1

𝑥

𝑦(𝑥) = − sin(𝑥) + cos(𝑥) − 4 sin(2𝑥) + 2 𝑥 − 2 𝑥 2 + ∫0 (𝑥 − 𝑡)𝑦 2 (𝑡), The exact solution of Eq. (6.386) is 1221

(6.386)

𝑦(𝑥) = − sin(𝑥) + cos(𝑥).

(6.387)



(6.388)



𝑇



(6.389) 1

𝑘


(6.390)

In the above expression, 𝑃𝑚 is the Probabilist Hermite polynomials given as 𝑃0 = 1, 𝑃1 = 𝑥, 𝑃2 = −1 + 𝑥 2 𝑃3 = −3𝑥 + 𝑥 3 , 𝑃4 = 3 − 6𝑥 2 + 𝑥 4 , . . .. Therefore, the Probabilist Hermite wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = 4√5𝑥(𝑥 − 1), 𝜓1,3 (𝑥) = 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), . . .. 1222

Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + 4√5𝑥(𝑥 − 1)𝑐1,2 + 2√7(2𝑥 − 1) (2𝑥 2 − 2𝑥 − 1)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇



1

1

𝐂 𝐓 𝛙(𝑥) = − sin(𝑥) + cos(𝑥) + 4 sin(2𝑥) + 2 𝑥 − 2 𝑥 2 𝑥

2

+ ∫0 (𝑥 − 𝑡)(𝐂 𝐓 𝛙(𝑡)) .

(6.391)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2,3,4, … 7.

(6.392)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 8×8 𝐂8×1 = 𝐛8×1 . After solving we get the following exact solution 𝑦(𝑥) = 1.00000 − 0.00000𝑥 − 0.500000𝑥 2 + ⋯.

1223


Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

0.999999805599298

1.94401E-07

0.1

0.895170748631198

0.895170613392118

1.35239E-07

0.2

0.781397247046180

0.781397248728388

1.68221E-09

0.3

0.659816282464266

0.659816414454431

1.31990E-07

0.4

0.531642651694235

0.531642820519673

1.68825E-07

0.5

0.398157023286170

0.398157100132523

7.68464E-08

0.6

0.260693141514643

0.260693052774153

8.87405E-08

0.7

0.120624500046797

0.120624311308047

1.88739E-07

0.8

-0.020649381552357

-0.020649469576872

8.80245E-08

0.9

-0.161716941356819

-0.161716806351661

1.35005E-07

1.0

-0.301168678939757

-0.301168863823500

1.84884E-07

_____________________________________________________________

(a)

(b)



1224

𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,11 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,11 ] . Proceeding as before we have the matrix form 𝐀 𝟏𝟐×𝟏𝟐 𝐂𝟏𝟐×𝟏 = 𝐛𝟏𝟐×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.74: Comparison of the Exact Solution and Approximate Solution of Eq. (6.386) obtained from Probabilist Hermite Wavelet Method (PWM) _____________________________________________________________ 𝑥

Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

0.999999999998985

1.01483E-12

0.1

0.895170748631198

0.895170748630826

3.71304E-13

0.2

0.781397247046180

0.781397247046902

7.22053E-13

0.3

0.659816282464266

0.659816282465097

8.30106E-13

0.4

0.531642651694235

0.531642651693991

2.43502E-13

0.5

0.398157023286170

0.398157023285180

9.89734E-13

0.6

0.260693141514643

0.260693141514491

1.52274E-13

0.7

0.120624500046797

0.120624500047707

9.09968E-13

0.8

-0.020649381552357

-0.020649381552516

1.58850E-13

0.9

-0.161716941356819

-0.161716941357456

6.37303E-13

1.0

-0.301168678939757

-0.301168678940719

9.61906E-13

_____________________________________________________________

1225

(a)

(b)

Fig. 6.105.(a)-(b): Comparison of Exact and Approximate Solution of Eq. (6.386)


𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,14 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,14 ] . Proceeding as before we have the matrix form 𝐀 𝟏𝟓×𝟏𝟓 𝐂𝟏𝟓×𝟏 = 𝐛𝟏𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table.

1226


Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.00000E-200

0.1

0.895170748631198

0.895170748631198

4.67399E-17

0.2

0.781397247046180

0.781397247046180

6.75563E-18

0.3

0.659816282464266

0.659816282464266

4.45688E-17

0.4

0.531642651694235

0.531642651694235

2.60555E-18

0.5

0.398157023286170

0.398157023286170

5.06790E-17

0.6

0.260693141514643

0.260693141514643

6.88653E-18

0.7

0.120624500046797

0.120624500046797

3.39364E-17

0.8

-0.020649381552357

-0.020649381552357

5.24761E-17

0.9

-0.161716941356819

-0.161716941356819

3.67933E-17

1.0

-0.301168678939757

-0.301168678939757

4.18279E-17

_____________________________________________________________

(a)

(b)


6.8.3. Nonlinear Integral Equation of 1st Kind Example 6.28 [229]. Consider the following nonlinear integral equations as follow 1 3

1

𝑥

𝑥 3 + 12 𝑥 4 = ∫0 (𝑥 − 𝑡 + 1)𝑦 2 (𝑡), 1227

(6.393)

The exact solution of Eq. (6.393) is 𝑦(𝑥) = 𝑥.

(6.394)



(6.395)



𝑇


We apply the proposed technique to solve Eq. (6.393) with 𝐾 = 1, and 𝑀 = 3. We have Eq. (6.395) is 𝑦(𝑥) = ∑1𝑛=1 ∑2𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥), = 𝐂 𝐓 𝛙(𝑥),

(6.396) 1

𝑘


(6.397)

In the above expression, 𝑃𝑚 is the Probabilist Hermite polynomials given as 𝑃0 = 1, 𝑃1 = 𝑥, 𝑃2 = −1 + 𝑥 2 𝑃3 = −3𝑥 + 𝑥 3 , 𝑃4 = 3 − 6𝑥 2 + 𝑥 4 , . . .. Therefore, the Probabilist Hermite wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = 4√5𝑥(𝑥 − 1), 𝜓1,3 (𝑥) = 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), . . 1228

.. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + 4√5𝑥(𝑥 − 1)𝑐1,2 = 𝐂 𝐓 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 ] and 𝛙(𝑥) = [1, √3(2𝑥 − 1), 4√5𝑥(𝑥 − 1)] . Substituting into the given problem we get 1

2

𝑥

1

𝑥 3 + 12 𝑥 4 = ∫0 (𝑥 − 𝑡 + 1) (𝐂 𝐓 𝛙(𝑡)) , 3

(6.398)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2,3,4, … 7.

(6.399)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 3×3 𝐂3×1 = 𝐛3×1 . After solving we get the following exact solution 𝑦(𝑥) = 𝑥.

Example 6.29 [229]. Consider the following nonlinear integral equations as follow 3 4

1

𝑥

3

exp(2𝑥) − 2 𝑥 − 4 = ∫0 (𝑥 − 𝑡 + 1)𝑦 2 (𝑡),

(6.400)

The exact solution of Eq. (6.400) is 𝑦(𝑥) = exp(𝑥).

(6.401)



(6.402)



𝑇



(6.403) 1

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑃𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). 1229

(6.404)

In the above expression, 𝑃𝑚 is the Probabilist Hermite polynomials given as 𝑃0 = 1, 𝑃1 = 𝑥, 𝑃2 = −1 + 𝑥 2 𝑃3 = −3𝑥 + 𝑥 3 , 𝑃4 = 3 − 6𝑥 2 + 𝑥 4 , . . .. Therefore, the Probabilist Hermite wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = 4√5𝑥(𝑥 − 1), 𝜓1,3 (𝑥) = 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + 4√5𝑥(𝑥 − 1)𝑐1,2 + 2√7(2𝑥 − 1) (2𝑥 2 − 2𝑥 − 1)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇


Substituting into the given problem we get 3 4

1

3

2

4

2

𝑥

exp(2𝑥) − 𝑥 − = ∫0 (𝑥 − 𝑡 + 1)(𝐂 𝐓 𝛙(𝑡)) ,

(6.405)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2,3,4, … 9.

(6.406)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . 1230

After solving we get the following exact solution 𝑦(𝑥) = 1.00000 + 2.00000𝑥 + 2.00000𝑥 2 + ⋯. Table 6.76: Comparison of the Exact Solution and Approximate Solution of Eq. (6.400) obtained from Probabilist Hermite Wavelet Method (PWM) _____________________________________________________________ 𝑥

Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000296469980

2.96470E-07

0.1

1.105170918075648

1.105171044967146

1.26891E-07

0.2

1.221402758160170

1.221402629394380

1.28766E-07

0.3

1.349858807576003

1.349858572392356

2.35184E-07

0.4

1.491824697641270

1.491824592571722

1.05070E-07

0.5

1.648721270700128

1.648721390456807

1.19757E-07

0.6

1.822118800390509

1.822118988425615

1.88035E-07

0.7

2.013752707470477

2.013752713188883

5.71841E-09

0.8

2.225540928492468

2.225540762744431

1.65748E-07

0.9

2.459603111156950

2.459603150009770

3.88528E-08

1.0

2.718281828459045

2.718281683183438

1.45276E-07

_____________________________________________________________

(a)

(b)


1231


𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] . Proceeding as before we have the matrix form 𝐀 𝟐𝟓×𝟐𝟓 𝐂𝟐𝟓×𝟏 = 𝐛𝟐𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.77: Comparison of the Exact Solution and Approximate Solution of Eq. (6.400) obtained from Probabilist Hermite Wavelet Method (PWM) _____________________________________________________________ 𝑥

Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

6.00000E-200

0.1

1.105170918075648

1.105170918075648

3.21690E-26

0.2

1.221402758160170

1.221402758160170

5.40938E-26

0.3

1.349858807576003

1.349858807576003

4.87162E-26

0.4

1.491824697641270

1.491824697641270

3.79096E-26

0.5

1.648721270700128

1.648721270700128

2.17549E-26

0.6

1.822118800390509

1.822118800390509

1.65307E-26

0.7

2.013752707470477

2.013752707470477

1.85850E-26

0.8

2.225540928492468

2.225540928492468

3.14446E-26

0.9

2.459603111156950

2.459603111156950

7.39652E-27

1.0

2.718281828459045

2.718281828459045

2.72979E-26

_____________________________________________________________

1232

(a)

(b)



𝑇


1233


Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

3.34992E-65

0.1

1.105170918075648

1.105170918075648

9.48204E-66

0.2

1.221402758160170

1.221402758160170

2.18649E-65

0.3

1.349858807576003

1.349858807576003

2.26885E-65

0.4

1.491824697641270

1.491824697641270

3.97006E-66

0.5

1.648721270700128

1.648721270700128

1.01535E-65

0.6

1.822118800390509

1.822118800390509

1.36124E-65

0.7

2.013752707470477

2.013752707470477

8.09963E-66

0.8

2.225540928492468

2.225540928492468

1.15788E-65

0.9

2.459603111156950

2.459603111156950

1.22428E-65

1.0

2.718281828459045

2.718281828459045

1.30805E-65

____________________________________________________________

(a)

(b)


Example 6.30 [229]. Consider the following nonlinear integral equations as follow 2

2

𝑥

sin(𝑥) + 3 cos(𝑥) − 3 cos(2𝑥) = ∫0 cos(𝑥 − 𝑡) 𝑦 2 (𝑡), 1234

(6.407)

The exact solution of Eq. (6.407) is 𝑦(𝑥) = sin(𝑥) + cos(𝑥).

(6.408)



(6.409)



𝑇



(6.410) 1

𝑘


(6.411)

In the above expression, 𝑃𝑚 is the Probabilist Hermite polynomials given as 𝑃0 = 1, 𝑃1 = 𝑥, 𝑃2 = −1 + 𝑥 2 𝑃3 = −3𝑥 + 𝑥 3 , 𝑃4 = 3 − 6𝑥 2 + 𝑥 4 , . . .. Therefore, the Probabilist Hermite wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = √3(2𝑥 − 1), 𝜓1,2 (𝑥) = 4√5𝑥(𝑥 − 1), 𝜓1,3 (𝑥) = 2√7(2𝑥 − 1)(2𝑥 2 − 2𝑥 − 1), . . 1235

.. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + √3(2𝑥 − 1)𝑐1,1 + 4√5𝑥(𝑥 − 1)𝑐1,2 + 2√7(2𝑥 − 1) (2𝑥 2 − 2𝑥 − 1)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇



2

𝑥

2

sin(𝑥) + 3 cos(𝑥) − 3 cos(2𝑥) = ∫0 cos(𝑥 − 𝑡) (𝐂 𝐓 𝛙(𝑥)) ,

(6.412)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2,3,4, … 9.

(6.413)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following exact solution 𝑦(𝑥) = 0.999999 + 2.00000𝑥 + 0.00000𝑥 2 + ⋯.

1236


Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.00000E-198

0.1

1.094837581924854

1.094837579591005

2.33385E-09

0.2

1.178735908636303

1.178735912044863

3.40856E-09

0.3

1.250856695786946

1.250856707040365

1.12534E-08

0.4

1.310479336311536

1.310479344495108

8.18357E-09

0.5

1.357008100494576

1.357008092127235

8.36734E-09

0.6

1.389978088304714

1.389978068714430

1.95903E-08

0.7

1.409059874522179

1.409059872486462

2.03572E-09

0.8

1.414062800246688

1.414062826339752

2.60931E-08

0.9

1.404936877898148

1.404936872265843

5.63230E-09

1.0

1.381773290676036

1.381773324191990

3.35160E-08

_____________________________________________________________

(a)

(b)



1237

𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] . Proceeding as before we have the matrix form 𝐀 𝟐𝟓×𝟐𝟓 𝐂𝟐𝟓×𝟏 = 𝐛𝟐𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.80: Comparison of the Exact Solution and Approximate Solution of Eq. (6.407) obtained from Probabilist Hermite Wavelet Method (PWM) _____________________________________________________________ 𝑥

Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

0.00000E+00

0.1

1.094837581924854

1.094837581924854

3.40385E-26

0.2

1.178735908636303

1.178735908636303

4.96338E-26

0.3

1.250856695786946

1.250856695786946

4.89203E-26

0.4

1.310479336311536

1.310479336311536

3.52793E-26

0.5

1.357008100494576

1.357008100494576

2.38120E-26

0.6

1.389978088304714

1.389978088304714

1.54753E-26

0.7

1.409059874522179

1.409059874522179

2.37057E-26

0.8

1.414062800246688

1.414062800246688

3.93055E-26

0.9

1.404936877898148

1.404936877898148

1.37816E-26

1.0

1.381773290676036

1.381773290676036

4.65783E-26

_____________________________________________________________

1238

(a)

(b)



𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,34 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,34 ] . Proceeding as before we have the matrix form 𝐀 𝟑𝟓×𝟑𝟓 𝐂𝟑𝟓×𝟏 = 𝐛𝟑𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table.

1239


Exact Solution


Error in PWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

0.00000E+00

0.1

1.094837581924854

1.094837581924854

3.04265E-41

0.2

1.178735908636303

1.178735908636303

5.57953E-41

0.3

1.250856695786946

1.250856695786946

7.05939E-41

0.4

1.310479336311536

1.310479336311536

7.01914E-41

0.5

1.357008100494576

1.357008100494576

3.37103E-41

0.6

1.389978088304714

1.389978088304714

3.31655E-41

0.7

1.409059874522179

1.409059874522179

6.19083E-41

0.8

1.414062800246688

1.414062800246688

5.88875E-41

0.9

1.404936877898148

1.404936877898148

6.84588E-41

1.0

1.381773290676036

1.381773290676036

6.60658E-41

____________________________________________________________

(a)

(b)


1240

6.9. Zernike Wavelet Method (ZWM) We proposed a new algorithm by inserting Zernike polynomials in traditional Legendre wavelets method. This technique is successively applied to find the approximate solutions of Stiff-Delay Differential Equation, Difference Differential Equation and Delay Integrodifferential Equations. The proposed technique is very simple and highly compatible for solving such kind of problems.

6.9.1. Methodology 6.9.1.1. Zernike Polynomials The recurrence relation of Zernike polynomials is given by 𝑛−𝑚 2 𝑍𝑛 (𝑥) = ∑𝑘=0

(−1)𝑘 (𝑛−𝑘)! k!

1 Γ((𝑛+𝑚)

1 1 +1)Γ((𝑛−𝑚) +1) 2+𝑘 2+𝑘

𝑥 𝑛−2𝑘 .

(6.414)

First four Zernike polynomials given below 𝑍0 = 1, 𝑍1 =

4𝑥 𝜋

, 2 2

𝑍2 =

2 1 8𝑥 Γ(3) −9 , 2 2 4 Γ( ) 3

2 𝑥(−16𝑥 2 +3𝜋)

𝑍3 = − 3

𝜋

,

. . ..

6.9.1.2. Wavelets and Zernike Wavelets In recent years, wavelets have found their way into many different fields of science and engineering. Wavelets constitute a family of functions constructed from the dilation and translation of a single function called the mother wavelet. When the dilation parameter a and the translation parameter 𝑏 vary continuously, we have the following family of continuous wavelets: 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|− 2 𝜓 (

𝑎

) 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

1241

(6.415)

Zernike wavelets 𝜓𝑛𝑚 (𝑡) = 𝜓(𝑘, 𝑛, 𝑚, 𝑡) have four arguments; 𝑛, 𝑘 can assume any positive integer, 𝑚 is the order for Zernike polynomials and t is the normalized time. They are defined on the interval [0, 1) by; 1

𝜓𝑛𝑚

𝑘

𝑘 2 (𝑡) = {√𝑚 + 2 2 𝑍𝑚 (2 𝑡 − (2𝑛 + 1))

0

𝑛 2𝑘

≤𝑡≤

𝑛+1 2𝑘

(6.416)

otherwise 1

where 𝑚 = 0,1,2, … , 𝑀 − 1 and 𝑛 = 0,1, … , 2𝑘−1 . The coefficient (𝑚 + 2) is for orthonormality. Here 𝑍𝑚 (𝑡) are the well-known Zernike polynomials of order 𝑚, which have been previously described.


(6.417)


𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡).

(6.418)



(6.419)


(6.420)


(6.421)

2𝑀−1

and

2

0

2

𝑀−1

6.9.1.4. Analysis Consider the stiff Delay differential equation 𝑓0 (𝑥)𝑦 𝑛 (𝑥) + 𝑓1 (𝑥)𝑦 𝑛−1 (𝑥) + 𝑓2 (𝑥)𝑦 𝑛−2 (𝑥) + ⋯ + 𝑔0 (𝑥)𝑦(𝑥) + 𝑔1 (𝑥)𝑦(𝛼1 𝑥) + 𝑔2 (𝑥)𝑦(𝛼2 𝑥) + ⋯ + ℎ(𝑥) = 0, 0 ≤ 𝑥 ≤ 1. with the physical initial conditions 𝑦(0) = 𝐴0 , 𝑦 ′ (0) = 𝐴1 , 𝑦 ′′ (0) = 𝐴2 , … , 𝑦 𝑛−1 = 𝐴𝑛−1 . For solving equation stiff Delay differential equation, it is sufficient to suppose that the approximate solution is as 𝑦(𝑥) = 𝐂 𝑇 𝛙(𝑥), 1242

where 𝐂 and 𝛙(𝑥) are given in above. Therefore, we obtain 𝑓0 (𝑥)𝐂 𝑇 𝛙𝑛 (𝑥) + 𝑓1 (𝑥)𝐂 𝑇 𝛙𝑛−1 (𝑥) + 𝑓2 (𝑥)𝐂 𝑇 𝛙𝑛−2 (𝑥) + ⋯ + 𝑔0 (𝑥)𝐂 𝑇 𝛙(𝑥) + 𝑔1 (𝑥)𝐂 𝑇 𝛙(𝛼1 𝑥) + 𝑔2 (𝑥)𝐂 𝑇 𝛙(𝛼2 𝑥) + ⋯ + ℎ(𝑥) = 0, 0 ≤ 𝑥 ≤ 1. with the physical initial conditions 𝐂 𝑇 𝛙(0) = 𝐴0 , 𝐂 𝑇 𝛙′(0) = 𝐴1 , 𝐂 𝑇 𝛙′′ (0) = 𝐴2 , … , 𝐂 𝑇 𝛙𝑛−1 = 𝐴𝑛−1 . Therefore, in order to apply the Zernike Wavelet Method (ZWM); we require 2𝑘−1 𝑀 − 1 collocating points. Suitable collocating points are as 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 𝑛, … , 2𝑘−1 𝑀 − 1.

Implementing the collocating points and imposing the initial value to the stiff Delay differential equation, we obtain 𝑓0 (𝑥𝑖 )𝐂 𝑇 𝛙𝑛 (𝑥𝑖 ) + 𝑓1 (𝑥𝑖 )𝐂 𝑇 𝛙𝑛−1 (𝑥𝑖 ) + 𝑓2 (𝑥𝑖 )𝐂 𝑇 𝛙𝑛−2 (𝑥𝑖 ) + ⋯ + 𝑔0 (𝑥𝑖 )𝐂 𝑇 𝛙(𝑥𝑖 ) + 𝑔1 (𝑥𝑖 )𝐂 𝑇 𝛙(𝛼1 𝑥𝑖 ) + 𝑔2 (𝑥𝑖 )𝐂 𝑇 𝛙(𝛼2 𝑥𝑖 ) + ⋯ + ℎ(𝑥𝑖 ) = 0. for 𝑖 = 𝑛, … ,2𝑘−1 𝑀 − 1. The differential equation yields 2𝑘−1 𝑀 − 𝑛 − 1 equations and initial condition produces 𝑛 number of equations. Therefore, the obtained system has 2𝑘−1 𝑀 − 1 equations and 2𝑘−1 𝑀 − 1 unknowns. Solving this system gives the unknown coefficients 𝐂.

6.9.2. First-order Stiff Delay Differential Equation Consider the following linear DDE [22] as follow 𝑦 ′ (𝑥) = 𝐴𝑦(𝑥) + 𝑦 (𝑥 −

3𝜋 2

) − 𝐴 sin(𝑥),

(6.422)

where 𝐴 = 𝑝 − e−𝑥𝑝 , 𝑝 = −2, subject to the boundary conditions 𝑦(0) = 1.

(6.423)

The exact solution of the above system is 𝑦(𝑥) = e𝑝𝑥 + sin(𝑥).

(6.424)

According to the Zernike Wavelets Method (ZWM), we assume the trial solution 𝑘−1


(6.425)



𝑇


1243

We apply the proposed technique to solve Eq. (6.422) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.425) is 𝑦(𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) + ⋯ + 𝑐1,9 𝜓1,9 (𝑥) = 𝐂 𝐓 𝛙(𝑥), 1

(6.426)

𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝑍𝑚 (2𝑘 𝑡 − (2𝑛 + 1)).

(6.427)

In the above expression, 𝑍𝑚 is the Zernike polynomials given as 𝑍0 = 1, 𝑍1 =

4𝑥 𝜋

, 2 2

𝑍2 =

2 1 8𝑥 Γ(3) −9 , 2 2 4 Γ( ) 3

2 𝑥(−16𝑥 2 +3𝜋)

𝑍3 = − 3

𝜋

, 2 2

2 4 2 2 1 32𝜋 +192𝑥 𝜋 −729𝑥 Γ(3)

𝑍4 = 32

𝜋2

,

. . .. Therefore, the Zernike wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 𝜓1,2 (𝑥) =

4√3 𝜋

(2𝑥 − 1), 2 2

√5 2 2 4Γ( ) 3

2 2

2 2

(32𝑥 2 Γ (3) − 32Γ (3) 𝑥 + 8Γ (3) − 9),

2

𝜓1,3 (𝑥) = − 3𝜋 √7(2𝑥 − 1)(−64𝑥 2 + 64𝑥 − 16 + 3𝜋), . . .. Therefore, we have the trial solution is 2 2

𝑦(𝑥) = 𝑐1,0 +

4√3 𝜋

2 2

32𝑥 2 Γ (3) − 32Γ (3) 𝑥 √5 (2𝑥 − 1)𝑐1,1 + 2 2 ( ) 𝑐1,2 − 2 2 4Γ( ) 3 +8Γ (3) − 9 1244

2 3𝜋

√7(2𝑥 − 1)(−64𝑥 2 + 64𝑥 − 16 + 3𝜋)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , 1, and 𝛙(𝑥) = [

4√3 𝜋

(2𝑥 − 1),

√5 2 2 4Γ( ) 3

2 2

2 2

2 2

𝑇

(32𝑥 2 Γ (3) − 32Γ (3) 𝑥 + 8Γ (3) − 9) ,

2

] .

2

− 3𝜋 √7(2𝑥 − 1)(−64𝑥 + 64𝑥 − 16 + 3𝜋), … Substituting into the given problem we get 𝐂 𝐓 𝛙′ (𝑥) = 𝐴𝐂 𝐓 𝛙(𝑥) + 𝐂 𝐓 𝛙 (𝑥 −

3𝜋 2

) − 𝐴 sin(𝑥),

(6.428)


(6.429)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 1,2,3, … 9.

(6.430)


1245

Table 6.82: Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) of Eq. (6.422) obtained from Zernike Wavelet Method (ZWM) __________________________________________________________________ 𝑥

Exact Solution


Error in ZWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.20402E-15

0.1

0.918564169724810

0.918564169724810

4.12202E-15

0.2

0.868989376830701

0.868989376830701

4.24002E-15

0.3

0.844331842755366

0.844331842755366

8.45202E-15

0.4

0.838747306425872

0.838747306425872

9.45602E-15

0.5

0.847304979775645

0.847304979775645

2.20158E-15

0.6

0.865836685307237

0.865836685307237

8.12478E-15

0.7

0.890814651179298

0.890814651179298

3.98514E-15

0.8

0.919252608894178

0.919252608894178

2.36457E-15

0.9

0.948625797849070

0.948625797849070

7.12547E-15

1.0

0.976806268044509

0.976806268044509

2.10330E-15

__________________________________________________________________

(a)

(b)



1246

𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,24 ] , and 𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] . Proceeding as before we have the matrix form 𝐀 𝟐𝟓×𝟐𝟓 𝐂𝟐𝟓×𝟏 = 𝐛𝟐𝟓×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.83: Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) of Eq. (6.422) obtained from Zernike Wavelet Method (ZWM) _____________________________________________________________ 𝑥

Exact Solution


Error in ZWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.00000E-00

0.1

0.918564169724810

0.918564169724810

1.12202E-35

0.2

0.868989376830701

0.868989376830701

4.24002E-35

0.3

0.844331842755366

0.844331842755366

8.45202E-35

0.4

0.838747306425872

0.838747306425872

2.45602E-35

0.5

0.847304979775645

0.847304979775645

2.20158E-35

0.6

0.865836685307237

0.865836685307237

7.12478E-35

0.7

0.890814651179298

0.890814651179298

5.98514E-35

0.8

0.919252608894178

0.919252608894178

3.36457E-35

0.9

0.948625797849070

0.948625797849070

1.12547E-35

1.0

0.976806268044509

0.976806268044509

2.10330E-36

_____________________________________________________________

1247

(a)

(b)



𝑇


1248

Table 6.84: Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) of Eq. (6.422) obtained from Zernike Wavelet Method (ZWM) _____________________________________________________________ 𝑥

Exact Solution


Error in ZWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

1.20402E-200

0.1

0.918564169724810

0.918564169724810

1.99202E-132

0.2

0.868989376830701

0.868989376830701

1.24002E-132

0.3

0.844331842755366

0.844331842755366

2.45202E-132

0.4

0.838747306425872

0.838747306425872

7.45602E-132

0.5

0.847304979775645

0.847304979775645

1.20158E-132

0.6

0.865836685307237

0.865836685307237

2.12478E-132

0.7

0.890814651179298

0.890814651179298

3.98514E-132

0.8

0.919252608894178

0.919252608894178

2.36457E-132

0.9

0.948625797849070

0.948625797849070

2.12547E-132

1.0

0.976806268044509

0.976806268044509

2.10330E-131

_____________________________________________________________

(a)

(b)


6.9.3. Third-order Difference Differential Equation Consider the following linear DDE [22] as follow 1249

𝑦 ′′′ (𝑥) − 𝑦 ′′ (𝑥) − 𝑦(𝑥) − (e − 2)𝑦 ′′ (𝑥 − 1) +𝑦 ′ (𝑥 − 1) + 𝑦(𝑥 − 1) = 2e − 7,

(6.431)

subject to the boundary conditions 𝑦(0) = 1, 𝑦 ′ (0) = 0, 𝑦 ′′ (0) = 1.

(6.432)

The exact solution of the above system is 𝑦(𝑥) = 2 + 𝑥 + 𝑥 2 + e−𝑥 .

(6.433)



(6.434)



𝑇



𝑘

where 𝜓𝑛,𝑚 (𝑥) = √𝑚 + 2 22 𝑍𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝑍𝑚 is the Zernike polynomials given as 𝑍0 = 1, 𝑍1 =

4𝑥 𝜋

, 2 2

𝑍2 =

2 1 8𝑥 Γ(3) −9 , 2 2 4 Γ( ) 3

2 𝑥(−16𝑥 2 +3𝜋)

𝑍3 = − 3

𝜋

, 2 2

2 4 2 2 1 32𝜋 +192𝑥 𝜋 −729𝑥 Γ(3)

𝑍4 = 32

(6.435)

𝜋2

,

. . .. Therefore, the Zernike wavelets given as 𝜓1,0 (𝑥) = 1, 1250

(6.436)

𝜓1,1 (𝑥) = 𝜓1,2 (𝑥) =

4√3 𝜋

(2𝑥 − 1), 2 2

√5 2 2 4Γ( ) 3

2 2

2 2

(32𝑥 2 Γ (3) − 32Γ (3) 𝑥 + 8Γ (3) − 9),

2


2 2

32𝑥 2 Γ (3) − 32Γ (3) 𝑥 4√3 √5 (2𝑥 𝑦(𝑥) = 𝑐1,0 + 𝜋 − 1)𝑐1,1 + 2 2 ( ) 𝑐1,2 − 2 2 4Γ( ) 3 +8Γ (3) − 9 2 3𝜋

√7(2𝑥 − 1)(−64𝑥 2 + 64𝑥 − 16 + 3𝜋)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇


4√3 𝜋

(2𝑥 − 1),

√5 2 2 4Γ( ) 3

2 2

2 2

2 2

𝑇

(32𝑥 2 Γ (3) − 32Γ (3) 𝑥 + 8Γ (3) − 9) ,

2

] .

2

− 3𝜋 √7(2𝑥 − 1)(−64𝑥 + 64𝑥 − 16 + 3𝜋), … Substituting into the given problem we get 𝐂 𝐓 𝛙′′′ (𝑥) − 𝐂 𝐓 𝛙′′ (𝑥) − 𝐂 𝐓 𝛙(𝑥) − (e − 2)𝐂 𝐓 𝛙′′ (𝑥 − 1) +𝐂 𝐓 𝛙′ (𝑥 − 1) + 𝐂 𝐓 𝛙(𝑥 − 1) = 2e − 7,

(6.437)


(6.438)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 3, … 9.

(6.439)


1251


Exact Solution


Error in ZWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

0.00000E+00

0.1

1.004829081924352

1.004828767095580

3.14829E-07

0.2

1.018597241839830

1.018594772207149

2.46963E-06

0.3

1.040141192423997

1.040133001793637

8.19063E-06

0.4

1.068175302358730

1.068156183304942

1.91191E-05

0.5

1.101278729299872

1.101241880698865

3.68486E-05

0.6

1.137881199609491

1.137818243628951

6.29560E-05

0.7

1.176247292529523

1.176148270508661

9.90220E-05

0.8

1.214459071507532

1.214312433046835

1.46638E-04

0.9

1.250396888843050

1.250189496428593

2.07392E-04

1.0

1.281718171540955

1.281435355084688

2.82816E-04

__________________________________________________________________

(a)

(b)



1252

𝑇

𝑇


Exact Solution


Error in ZWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

5.59669E-98

0.1

1.004829081924352

1.004829081924352

1.59643E-17

0.2

1.018597241839830

1.018597241839830

1.25230E-16

0.3

1.040141192423997

1.040141192423996

4.15315E-16

0.4

1.068175302358730

1.068175302358729

9.69381E-16

0.5

1.101278729299872

1.101278729299870

1.86815E-15

0.6

1.137881199609491

1.137881199609488

3.19157E-15

0.7

1.176247292529523

1.176247292529518

5.02041E-15

0.8

1.214459071507532

1.214459071507525

7.43760E-15

0.9

1.250396888843050

1.250396888843040

1.05296E-14

1.0

1.281718171540955

1.281718171540940

1.43877E-14

_____________________________________________________________

1253

(a)

(b)



𝑇


1254


Exact Solution


Error in ZWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

5.00000E-173

0.1

1.004829081924352

1.004829081924352

1.51940E-30

0.2

1.018597241839830

1.018597241839830

1.19187E-29

0.3

1.040141192423997

1.040141192423997

3.95275E-29

0.4

1.068175302358730

1.068175302358730

9.22606E-29

0.5

1.101278729299872

1.101278729299872

1.77800E-28

0.6

1.137881199609491

1.137881199609491

3.03757E-28

0.7

1.176247292529523

1.176247292529523

4.77817E-28

0.8

1.214459071507532

1.214459071507532

7.07873E-28

0.9

1.250396888843050

1.250396888843050

1.00215E-27

1.0

1.281718171540955

1.281718171540955

1.36935E-27

_____________________________________________________________

(a)

(b)


6.9.4. Third-order Delay Differential Equation Consider the following linear DDE [181] as follow 1255

𝜋

𝜋

𝑦 ′′′ (𝑥) + cos(𝑥) 𝑦 ′ (𝑥) + sin(𝑥) 𝑦 ′ (𝑥 − 3 ) + √2𝑦 (𝑥 − 4 ) = sin(𝑥) − 2 cos(𝑥) − 1,

(6.440)

subject to the boundary conditions 𝑦(0) = 0, 𝑦 ′ (0) = 1, 𝑦(0) = 0.

(6.441)

The exact solution of the above system is 𝑦(𝑥) = sin(𝑥).

(6.442)



(6.443)



𝑇



𝑘


4𝑥 𝜋

, 2 2

𝑍2 =

2 1 8𝑥 Γ(3) −9 , 2 2 4 Γ( ) 3

2 𝑥(−16𝑥 2 +3𝜋)

𝑍3 = − 3

𝜋

, 2 2

2 4 2 2 1 32𝜋 +192𝑥 𝜋 −729𝑥 Γ(3)

𝑍4 = 32

(6.444)

𝜋2

,

. . .. Therefore, the Zernike wavelets given as

1256

(6.445)

𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 𝜓1,2 (𝑥) =

4√3 𝜋

(2𝑥 − 1), 2 2

√5 2 2 4Γ( ) 3

2 2

2 2

(32𝑥 2 Γ (3) − 32Γ (3) 𝑥 + 8Γ (3) − 9),

2


2 2

32𝑥 2 Γ (3) − 32Γ (3) 𝑥 4√3 √5 𝑦(𝑥) = 𝑐1,0 + 𝜋 (2𝑥 − 1)𝑐1,1 + 2 2 ( ) 𝑐1,2 − 2 2 4Γ( ) 3 +8Γ (3) − 9 2 3𝜋

√7(2𝑥 − 1)(−64𝑥 2 + 64𝑥 − 16 + 3𝜋)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇


4√3 𝜋

√5

(2𝑥 − 1),

2 3

2 2

2 2

𝑇

2 2

2 2 (32𝑥 Γ ( ) − 32Γ ( ) 𝑥 + 8Γ ( ) − 9) , 3 3 3

4Γ( ) 2

] .

2

− 3𝜋 √7(2𝑥 − 1)(−64𝑥 + 64𝑥 − 16 + 3𝜋), … Substituting into the given problem we get 𝜋

𝜋

𝐂 𝐓 𝛙′′′ (𝑥) + cos(𝑥) 𝐂 𝐓 𝛙′ (𝑥) + sin(𝑥) 𝐂 𝐓 𝛙′ (𝑥 − 3 ) + √2𝐂 𝐓 𝛙 (𝑥 − 4 ) = sin(𝑥) − 2 cos(𝑥) − 1,

(6.446)


(6.447)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 3,4, … 9.

(6.448)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following approximate solution 𝑦(𝑥) = 1.0000 − 0.99999𝑥 + 0.499999𝑥 2 + 0.16666𝑥 3 + ⋯, 1257


Exact Solution


Error in ZWM

__________________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

2.62134E-200

0.1

0.099833416646828

0.099833416373610

2.73218E-10

0.2

0.198669330795061

0.198669328559334

2.23573E-09

0.3

0.295520206661340

0.295520199213886

7.44745E-09

0.4

0.389418342308650

0.389418325590926

1.67177E-08

0.5

0.479425538604203

0.479425507981767

3.06224E-08

0.6

0.564642473395035

0.564642419106538

5.42885E-08

0.7

0.644217687237691

0.644217576035982

1.11202E-07

0.8

0.717356090899523

0.717355818245878

2.72654E-07

0.9

0.783326909627483

0.783326198139568

7.11488E-07

1.0

0.841470984807897

0.841469193819305

1.79099E-06

__________________________________________________________________

(a)

(b)



1258

𝑇

𝑇


Exact Solution


Error in ZWM

_____________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

4.63010E-102

0.1

0.099833416646828

0.099833416646828

4.42116E-37

0.2

0.198669330795061

0.198669330795061

7.64424E-36

0.3

0.295520206661340

0.295520206661340

3.76820E-35

0.4

0.389418342308650

0.389418342308650

1.12262E-34

0.5

0.479425538604203

0.479425538604203

2.53879E-34

0.6

0.564642473395035

0.564642473395035

4.82234E-34

0.7

0.644217687237691

0.644217687237691

8.12421E-34

0.8

0.717356090899523

0.717356090899523

1.25473E-33

0.9

0.783326909627483

0.783326909627483

1.81556E-33

1.0

0.841470984807897

0.841470984807897

2.49882E-33

_____________________________________________________________

1259

(a)

(b)



𝑇


1260


Exact Solution


Error in ZWM

_____________________________________________________________ 0.0

0.000000000000000

-0.000000000000000

1.93010E-102

0.1

0.099833416646828

0.099833416646828

3.02116E-77

0.2

0.198669330795061

0.198669330795061

1.74424E-76

0.3

0.295520206661340

0.295520206661340

2.66820E-75

0.4

0.389418342308650

0.389418342308650

3.52262E-74

0.5

0.479425538604203

0.479425538604203

4.43879E-74

0.6

0.564642473395035

0.564642473395035

2.92234E-74

0.7

0.644217687237691

0.644217687237691

3.02421E-74

0.8

0.717356090899523

0.717356090899523

2.00473E-73

0.9

0.783326909627483

0.783326909627483

2.11556E-73

1.0

0.841470984807897

0.841470984807897

1.69882E-73

_____________________________________________________________

(a)

(b)


6.9.5. Delay Integro-Differential Equation Consider the following linear DDE [36] as follow 1261

𝑦 ′′ (𝑥) + 𝑥𝑦 ′ (𝑥) + 𝑥𝑦(𝑥) + 𝑦 ′ (𝑥 − 1) + 𝑦(𝑥 − 1) = e−𝑥 + 0

e + ∫−1 𝑡𝑦(𝑡 − 1) d𝑡,

(6.449)

subject to the boundary conditions 𝑦(0) = 1, 𝑦 ′ (0) = −1.

(6.450)

The exact solution of the above system is 𝑦(𝑥) = e−𝑥 .

(6.451)



(6.452)



𝑇


We apply the proposed technique to solve Eq. (6.449) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.452) is 𝑦(𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) + 𝑐1,3 𝜓1,3 (𝑥) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 1

𝑘


4𝑥 𝜋

, 2 2

𝑍2 =

2 1 8𝑥 Γ(3) −9 , 2 2 4 Γ( ) 3

2 𝑥(−16𝑥 2 +3𝜋)

𝑍3 = − 3

𝜋

, 2 2

2 4 2 2 1 32𝜋 +192𝑥 𝜋 −729𝑥 Γ(3)

𝑍4 = 32

(6.453)

𝜋2

,

. . .. Therefore, the Zernike wavelets given as

1262

(6.454)

𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 𝜓1,2 (𝑥) =

4√3 𝜋

(2𝑥 − 1), 2 2

√5 2 2 4Γ( ) 3

2 2

2 2

(32𝑥 2 Γ (3) − 32Γ (3) 𝑥 + 8Γ (3) − 9),

2


2 2

32𝑥 2 Γ (3) − 32Γ (3) 𝑥 4√3 √5 𝑦(𝑥) = 𝑐1,0 + 𝜋 (2𝑥 − 1)𝑐1,1 + 2 2 ( ) 𝑐1,2 − 2 2 4Γ( ) 3 +8Γ (3) − 9 2 3𝜋

√7(2𝑥 − 1)(−64𝑥 2 + 64𝑥 − 16 + 3𝜋)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇


4√3 𝜋

√5

(2𝑥 − 1),

2 3

2 2

2 2

2 2

𝑇

2 2 (32𝑥 Γ ( ) − 32Γ ( ) 𝑥 + 8Γ ( ) − 9) , 3 3 3

4Γ( ) 2

] .

2

− 3𝜋 √7(2𝑥 − 1)(−64𝑥 + 64𝑥 − 16 + 3𝜋), … Substituting into the given problem we get 𝐂 𝐓 𝛙′′ (𝑥) + 𝑥𝐂 𝐓 𝛙′ (𝑥) + 𝑥𝐂 𝐓 𝛙(𝑥) + 𝐂 𝐓 𝛙′ (𝑥 − 1) + 𝐂 𝐓 𝛙(𝑥 − 1) = e−𝑥 + 0

e + ∫−1 𝑡𝐂 𝐓 𝛙(𝑡 − 1) d𝑡,

(6.455)

Subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 1, 𝐂 𝐓 𝛙′ (0) = −1.

(6.456)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2 … 9.

(6.457)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get the following approximate solution 𝑦(𝑥) = 1.0000 − 𝑥 + 0.5005𝑥 2 − 0.16566𝑥 3 + ⋯, 1263


Exact Solution


Error in ZWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

0.00000E+00

0.1

0.904837418035960

0.904843746977722

6.32894E-06

0.2

0.818730753077982

0.818759752783918

2.89997E-05

0.3

0.740818220681718

0.740890861372353

7.26407E-05

0.4

0.670320046035639

0.670460610598714

1.40565E-04

0.5

0.606530659712633

0.606765296492157

2.34637E-04

0.6

0.548811636094026

0.549166819225108

3.55183E-04

0.7

0.496585303791410

0.497086262392618

5.00959E-04

0.8

0.449328964117222

0.449998164725152

6.69201E-04

0.9

0.406569659740599

0.407425449742134

8.55790E-04

1.0

0.367879441171442

0.368934984107957

1.05554E-03

__________________________________________________________________

(a)

(b)



1264

𝑇

𝑇


Exact Solution


Error in ZWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

6.00000E-199

0.1

0.904837418035960

0.904837418035925

3.42272E-14

0.2

0.818730753077982

0.818730753077825

1.56827E-13

0.3

0.740818220681718

0.740818220681325

3.92818E-13

0.4

0.670320046035639

0.670320046034879

7.60099E-13

0.5

0.606530659712633

0.606530659711365

1.26876E-12

0.6

0.548811636094026

0.548811636092106

1.92060E-12

0.7

0.496585303791410

0.496585303788701

2.70890E-12

0.8

0.449328964117222

0.449328964113603

3.61867E-12

0.9

0.406569659740599

0.406569659735972

4.62716E-12

1.0

0.367879441171442

0.367879441165737

5.70493E-12

_____________________________________________________________

1265

(a)

(b)



𝑇


1266


Exact Solution


Error in ZWM

_____________________________________________________________ 0.0

1.000000000000000

1.000000000000000

3.65450E-98

0.1

0.904837418035960

0.904837418035960

1.75198E-32

0.2

0.818730753077982

0.818730753077982

8.02747E-32

0.3

0.740818220681718

0.740818220681718

2.01070E-31

0.4

0.670320046035639

0.670320046035639

3.89069E-31

0.5

0.606530659712633

0.606530659712633

6.49437E-31

0.6

0.548811636094026

0.548811636094026

9.83089E-31

0.7

0.496585303791410

0.496585303791410

1.38660E-30

0.8

0.449328964117222

0.449328964117222

1.85228E-30

0.9

0.406569659740599

0.406569659740599

2.36849E-30

1.0

0.367879441171442

0.367879441171442

2.92016E-30

_____________________________________________________________

(a)

(b)


6.9.6. System of Stiff Delay Differential Equation Consider the following linear DDE [242] as follow 1267

𝑦1′ (𝑥) = −2e2 𝑦1 (𝑥) + 4𝑦2 (𝑥),

(6.458a)

𝑦2′ (𝑥) = 𝑦1 (𝑥) − 𝑦1 (𝑥 − 1),

(6.458b)


𝑦1 (0) = 1, 𝑦2 (0) = 2 (e2 − 1).

(6.459)

The exact solution of the above system is 1

𝑦1 (𝑥) = e−2𝑥 , 𝑦2 (𝑥) = 2 (e−2(𝑥−1) − e−2𝑥 ).

(6.460)



(6.461)



𝑇


We apply the proposed technique to solve Eq. (6.458a-6.458b) with 𝐾 = 1, and 𝑀 = 10. We have Eq. (6.461) is 𝑦1 (𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑐𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑐1,0 𝜓1,0 (𝑥) + 𝑐1,1 𝜓1,1 (𝑥) + 𝑐1,2 𝜓1,2 (𝑥) + 𝑐1,3 𝜓1,3 (𝑥) + ⋯ = 𝐂 𝐓 𝛙(𝑥),

(6.462a)

𝑦2 (𝑥) = ∑1𝑛=1 ∑9𝑚=0 𝑑𝑛,𝑚 𝜓𝑛,𝑚 (𝑥) = 𝑑1,0 𝜓1,0 (𝑥) + 𝑑1,1 𝜓1,1 (𝑥) + 𝑑1,2 𝜓1,2 (𝑥) + 𝑑1,3 𝜓1,3(𝑥) + ⋯ = 𝐃𝐓 𝛙(𝑥), 1

𝑘


4𝑥 𝜋

, 2 2

𝑍2 =

2 1 8𝑥 Γ(3) −9 , 2 2 4 Γ( ) 3

2 𝑥(−16𝑥 2 +3𝜋)

𝑍3 = − 3

𝜋

, 2 2

2 4 2 2 1 32𝜋 +192𝑥 𝜋 −729𝑥 Γ(3)

𝑍4 = 32

(6.462b)

𝜋2

,

. . 1268

(6.463)

.. Therefore, the Zernike wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 𝜓1,2 (𝑥) =

4√3 𝜋

(2𝑥 − 1), 2 2

√5 2 2 4Γ( ) 3

2 2

2 2

(32𝑥 2 Γ (3) − 32Γ (3) 𝑥 + 8Γ (3) − 9),

2


2 2

32𝑥 2 Γ (3) − 32Γ (3) 𝑥 4√3 √5 𝑦1 (𝑥) = 𝑐1,0 + 𝜋 (2𝑥 − 1)𝑐1,1 + 2 2 ( ) 𝑐1,2 − 2 2 4Γ( ) 3 +8Γ (3) − 9 2 3𝜋

√7(2𝑥 − 1)(−64𝑥 2 + 64𝑥 − 16 + 3𝜋)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 2 2

2 2

32𝑥 2 Γ ( ) − 32Γ ( ) 𝑥 4√3 √5 3 3 (𝑥) (2𝑥 𝑦2 = 𝑑1,0 + 𝜋 − 1)𝑑1,1 + 2 2 ( ) 𝑑1,2 − 2 2 4Γ( ) 3 +8Γ (3) − 9 2 3𝜋

√7(2𝑥 − 1)(−64𝑥 2 + 64𝑥 − 16 + 3𝜋)𝑑1,3 + ⋯ = 𝐃𝐓 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , 𝐃 = [𝑑1,0 , 𝑑1,1 , 𝑑1,2 , 𝑑1,3 , … ] , 1, and 𝛙(𝑥) = [

4√3 𝜋

(2𝑥 − 1),

√5 2 3

2 2

2 2

2 2

𝑇

2 2 (32𝑥 Γ ( ) − 32Γ ( ) 𝑥 + 8Γ ( ) − 9) , 3 3 3

4Γ( )

] .

2

− 3𝜋 √7(2𝑥 − 1)(−64𝑥 2 + 64𝑥 − 16 + 3𝜋), … Substituting into the given problem we get 𝐂 𝐓 𝛙′ (𝑥) = −2e2 𝐂 𝐓 𝛙(𝑥) + 4𝐃𝐓 𝛙(𝑥),

(6.464a)

𝐃𝐓 𝛙′ (𝑥) = 𝐂 𝐓 𝛙(𝑥) − 𝐂 𝐓 𝛙(𝑥 − 1),

(6.464b)

Subject to the boundary conditions 1

𝐂 𝐓 𝛙(0) = 1, 𝐃𝐓 𝛙(0) = 2 (e2 − 1).

(6.465)

Substitute the collocating points are in Eq. (6.464a-6.464b), we have the system of equations 1269

𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2,3, … 9.

(6.466)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 20×20 𝐂20×1 = 𝐛20×1 . After solving we have the approximate solutions. Table 6.94: Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) of Eq. (6.458a-6.458b) obtained from Zernike Wavelet Method (ZWM) __________________________________________________________________ 𝑥

Exact Solution


Error in ZWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

0.00000E+00

0.1

0.818730753077982

0.818725333476347

5.41960E-06

0.2

0.670320046035639

0.670314044595418

6.00144E-06

0.3

0.548811636094026

0.548808129676242

3.50642E-06

0.4

0.449328964117222

0.449325966589869

2.99753E-06

0.5

0.367879441171442

0.367870998116938

8.44305E-06

0.6

0.301194211912202

0.301177163720640

1.70482E-05

0.7

0.246596963941606

0.246579617237683

1.73467E-05

0.8

0.201896517994655

0.201896408705071

1.09290E-07

0.9

0.165298888221587

0.165302713616463

3.82539E-06

1.0

0.135335283236613

0.135173862338665

1.61421E-04

__________________________________________________________________

1270

(a)

(b)

Fig. 6.125. (a)-(b): Comparison of Exact and Approximate Solution of Eq. (6.458a-6.458b)

Table 6.95: Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) of Eq. (6.458a-6.458b) obtained from Zernike Wavelet Method (ZWM) __________________________________________________________________ 𝑥

Exact Solution


Error in ZWM

__________________________________________________________________ 0.0

3.194528049465325

3.194528049465325

4.00000E-199

0.1

2.615458355667482

2.615441466123929

1.68895E-05

0.2

2.141356189179738

2.141328325980977

2.78632E-05

0.3

1.753194165375324

1.753161383629379

3.27817E-05

0.4

1.435393979309663

1.435360268648785

3.37107E-05

0.5

1.175201193643801

1.175167666354246

3.35273E-05

0.6

0.962173358290133

0.962138894840555

3.44634E-05

0.7

0.787760918224451

0.787724029164206

3.68891E-05

0.8

0.644964089823307

0.644925042416084

3.90474E-05

0.9

0.528051934969292

0.528012516710029

3.94183E-05

1.0

0.432332358381694

0.432287324745214

4.50336E-05

__________________________________________________________________

1271

(a)

(b)


Now to solve Eq. (6.458a-6.458b) with 𝐾 = 1, and 𝑀 = 25. We have Eq. (6.461) is 𝑇 𝑦1 (𝑥) = ∑1𝑛=1 ∑24 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥), 𝑇 𝑦2 (𝑥) = ∑1𝑛=1 ∑24 𝑚=0 𝑑𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐃 𝛙(𝑥), 𝑇

𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,24 ] .

Proceeding as before we have the matrix form 𝐀 𝟓𝟎×𝟓𝟎 𝐂𝟓𝟎×𝟏 = 𝐛𝟓𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table.

1272


Exact Solution


Error in ZWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.59702E-192

0.1

0.818730753077982

0.818730753077983

6.69759E-16

0.2

0.670320046035639

0.670320046035641

1.35329E-15

0.3

0.548811636094026

0.548811636094027

1.02849E-15

0.4

0.449328964117222

0.449328964117221

2.95667E-16

0.5

0.367879441171442

0.367879441171440

2.10177E-15

0.6

0.301194211912202

0.301194211912198

3.83646E-15

0.7

0.246596963941606

0.246596963941601

5.16707E-15

0.8

0.201896517994655

0.201896517994649

6.02320E-15

0.9

0.165298888221587

0.165298888221580

6.51019E-15

1.0

0.135335283236613

0.135335283236606

6.79128E-15

__________________________________________________________________

(a)

(b)


1273


Exact Solution


Error in ZWM

__________________________________________________________________ 0.0

3.194528049465325

3.194528049465325

3.26012E-192

0.1

2.615458355667482

2.615458355667487

4.85736E-15

0.2

2.141356189179738

2.141356189179743

5.62059E-15

0.3

1.753194165375324

1.753194165375326

1.58271E-15

0.4

1.435393979309663

1.435393979309658

5.25984E-15

0.5

1.175201193643801

1.175201193643789

1.23869E-14

0.6

0.962173358290133

0.962173358290115

1.80857E-14

0.7

0.787760918224451

0.787760918224429

2.18060E-14

0.8

0.644964089823307

0.644964089823284

2.38666E-14

0.9

0.528051934969292

0.528051934969267

2.49450E-14

1.0

0.432332358381694

0.432332358381668

2.56606E-14

__________________________________________________________________

(a)

(b)


Now to solve Eq. (6.458a-6.458b) with 𝐾 = 1, and 𝑀 = 50. We have Eq. (6.461) is 𝑇 𝑦1 (𝑥) = ∑1𝑛=1 ∑49 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝐂 𝛙(𝑥),

1274


𝑇


𝑇

𝛙(𝑡) = [𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 ] .

Proceeding as before we have the matrix form 𝐀 𝟏𝟎𝟎×𝟏𝟎𝟎 𝐂𝟏𝟎𝟎×𝟏 = 𝐛𝟏𝟎𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.98: Comparison of the Exact Solution and Approximate Solutions of 𝑦(𝑥) of Eq. (6.458a-6.458b) obtained from Zernike Wavelet Method (ZWM) __________________________________________________________________ 𝑥

Exact Solution


Error in ZWM

__________________________________________________________________ 0.0

1.000000000000000

1.000000000000000

2.09852E-86

0.1

0.818730753077982

0.818730753077982

2.32635E-29

0.2

0.670320046035639

0.670320046035639

3.97566E-29

0.3

0.548811636094026

0.548811636094026

1.14585E-29

0.4

0.449328964117222

0.449328964117222

5.70739E-29

0.5

0.367879441171442

0.367879441171442

1.42111E-28

0.6

0.301194211912202

0.301194211912202

2.20447E-28

0.7

0.246596963941606

0.246596963941606

2.79270E-28

0.8

0.201896517994655

0.201896517994655

3.16961E-28

0.9

0.165298888221587

0.165298888221587

3.38891E-28

1.0

0.135335283236613

0.135335283236613

3.52317E-28

__________________________________________________________________

1275

(a)

(b)



Exact Solution


Error in ZWM

__________________________________________________________________ 0.0

3.194528049465325

3.194528049465325

4.61686E-77

0.1

2.615458355667482

2.615458355667482

1.69426E-28

0.2

2.141356189179738

2.141356189179738

1.52518E-28

0.3

1.753194165375324

1.753194165375324

6.36333E-29

0.4

1.435393979309663

1.435393979309663

3.85629E-28

0.5

1.175201193643801

1.175201193643801

7.06014E-28

0.6

0.962173358290133

0.962173358290133

9.56280E-28

0.7

0.787760918224451

0.787760918224451

1.11820E-27

0.8

0.644964089823307

0.644964089823307

1.20904E-27

0.9

0.528051934969292

0.528051934969292

1.25914E-27

1.0

0.432332358381694

0.432332358381694

1.29476E-27

__________________________________________________________________

1276

(a)

(b)


6.10. Rook Wavelet Method (RWM) We proposed a new algorithm by inserting Rook polynomials in traditional Legendre wavelets method. This technique is successively applied to find the exact solutions of differential equations of Fractional-order. The proposed technique is very simple and highly compatible for solving such kind of problems.

6.10.1. Methodology 6.10.1.1. Rook Polynomials Rook sequence, named after Paul Emile Rook, is any sequence {𝑝𝑛 (𝑥)}, 𝑛 = 0,1,2, … satisfying the identity (−1)𝑚 (𝑛)!

𝑅𝑛 = 𝑥 𝑛 𝑛! ∑𝑛𝑚=0 (𝑚!)2 (𝑛−𝑚)! (−𝑥)−𝑚 . few Rook polynomials given below 𝑅0 = 1, 𝑅1 = 1 + 𝑥, 𝑅2 = 1 + 2𝑥 2 + 4𝑥, 𝑅3 = 1 + 9𝑥 + 18𝑥 2 + 6𝑥 3 , . . ..

1277

(6.467)

6.10.1.2. Wavelets and Rook Wavelets In recent years, wavelets have found their way into many different fields of science and engineering. Wavelets constitute a family of functions constructed from the dilation and translation of a single function called the mother wavelet. When the dilation parameter a and the translation parameter 𝑏 vary continuously, we have the following family of continuous wavelets: 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|− 2 𝜓 (

𝑎

) 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

(6.468)

Rook wavelets 𝜓𝑛𝑚 (𝑡) = 𝜓(𝑘, 𝑛, 𝑚, 𝑡) have four arguments; 𝑛, 𝑘 can assume any positive integer, 𝑚 is the order for Rook polynomials and t is the normalized time. They are defined on the interval [0, 1) by; 1

𝜓𝑛𝑚

𝑘

𝑘 2 (𝑡) = {√𝑚 + 2 2 𝑅𝑚 (2 𝑡 − (2𝑛 + 1))

0

𝑛 2𝑘

≤𝑡≤

𝑛+1 2𝑘

(6.469)

otherwise 1

where 𝑚 = 0,1,2, … , 𝑀 − 1 and 𝑛 = 0,1, … , 2𝑘−1 . The coefficient (𝑚 + 2) is for orthonormality. Here 𝑅𝑚 (𝑡) are the well-known Rook polynomials of order 𝑚, which have been previously described.

6.10.1.3. Operational Matrix for Integration and Derivative of Fractional-order The integration of the vector 𝜓(𝑡) defined can be obtained by 𝑡

∫0 𝜓(𝑥)𝑑𝑥 ≈ 𝑝𝜓(𝑡), where P is the 𝑚 × 𝑚 operational matrix for integration. Now, we derive the wavelet operational matrix of fractional integration. For this purpose, we rewrite the RiemannLiouville fractional integration as (𝐼 𝛼 𝑓)(𝑡) =

𝑡 1 ∫ (𝑡 Γ(𝛼) 0

− 𝑥)𝛼−1 𝑓(𝑥)𝑑𝑥 =

1 Γ(𝛼)

𝑡 𝛼−1 ∗ 𝑓(𝑡),

where 𝛼 ∈ 𝑅 is the order of the integration and 𝑡 𝛼−1 ∗ 𝑓(𝑡) denotes the convolution product of 𝑡 𝛼−1 and 𝑓(𝑡). Now if 𝑓(𝑡) is expanded by the Legendre and Chebyshev wavelets, the Riemann-Liouville fractional integration becomes 1 Γ(𝛼)

1

𝑡 𝛼−1 ∗ 𝑓(𝑡) ≈ 𝐶 𝑇 Γ(𝛼) {𝑡 𝛼−1 ∗ 𝜓(𝑡)},

So that, if 𝑡 𝛼−1 and 𝑓(𝑡) can be integrated, then by expanding 𝑓(𝑡) in the Legendre and 1278

Chebyshev wavelets, the Riemann-Liouville fractional integration can be solved via the Legendre and Chebyshev wavelets. Also, we define an m-set of block pulse functions as 𝑏𝑖 (𝑡) = {

1, 0,

𝑖 𝑚

≤𝑡≤

𝑖+1 𝑚

,

otherwise,

where 𝑖 = 0, 1, 2, . . . , (𝑚 – 1). The functions 𝑏𝑖 (𝑡) are disjoint and orthogonal, that is, 𝑏𝑖 (𝑡)𝑏𝑙 (𝑡) = {

𝑖, 𝑏𝑖 (𝑡) ,

𝑖≠𝑙, 𝑖 = 𝑙,

𝑖, 𝑖≠𝑙, 1 ∫0 𝑏𝑖 (𝑥)𝑏𝑙 (𝑥) 𝑑𝑥 = { 1 , 𝑖 = 𝑙, 𝑚

(6.470)

Similarly, the Legendre and Chebyshev wavelets may be expanded into m-term block pulse functions as follows: 𝜓𝑚 (𝑡) = 𝛷𝑚×𝑚 𝐵𝑚 (𝑡),

(6.471)

where 𝐵𝑚 (𝑡) = [𝐵0 (𝑡), 𝐵1 (𝑡), 𝐵2 (𝑡), … , 𝐵𝑚−1 (𝑡)]𝑇 . In, Kilicman and Al Zhour have given the block pulse operational matrix of the fractional integration 𝐹 𝛼 as follows: (𝐼 𝛼 𝜓𝑚 )(𝑡) ≈ 𝐹 𝛼 𝐵𝑚 (𝑡),

(6.472)

where 𝜉2 . . . 𝜉𝑚−1 𝜉1 . . . 𝜉𝑚−2 1 . . . 𝜉 1 1 . . . . . . 𝑚−3 . 𝐹 𝛼 = 𝑚𝛼 Γ(𝛼+2) . (6.473) . . . . . . . . . . . . . . [0 0 0 0 0 1] and 𝜉𝑖 = (𝑖 + 1)𝛼+1 − 2𝑖 𝛼+1 + (𝑖 − 1)𝛼+1 . Next, we derive the Legendre and 1 0 0

𝜉1 1 0

Chebyshev wavelet operational matrices of the fractional integration. Let 𝛼 (𝐼 𝛼 𝜓𝑚 )(𝑡) ≈ 𝑃𝑚×𝑚 𝜓𝑚 (𝑡),

(6.474)

𝛼 where the matrix 𝑃𝑚×𝑚 is called the wavelet operational matrix of the fractional

integration. Using (6.472) and (6.473), we have (𝐼 𝛼 𝜓𝑚 )(𝑡) ≈ (𝐼 𝛼 𝛷𝑚×𝑚 𝐵𝑚×𝑚 )(𝑡) = 𝛷𝑚×𝑚 (𝐼 𝛼 𝐵𝑚 )(𝑡) ≈ 𝛷𝑚×𝑚 𝐹 𝛼 𝐵𝑚 (𝑡), (6.475) and from (6.474) and (6.475), we get 𝛼 𝛼 𝑃𝑚×𝑚 𝜓𝑚 (𝑡) ≈ 𝑃𝑚×𝑚 𝛷𝑚×𝑚 𝐵𝑚 (𝑡) ≈ 𝛷𝑚×𝑚 𝐹 𝛼 𝐵𝑚 (𝑡),

1279

(6.476)

𝛼 Thus, the operational matrices of the fractional integration 𝑃𝑚×𝑚 can be approximately

expressed by 𝛼 𝑃𝑚×𝑚 = 𝛷𝑚×𝑚 𝐹 𝛼 𝛷−1 𝑚×𝑚 .

(6.477)


(6.478)


𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡).

(6.479)


𝑇 𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡) = 𝐶 𝜓(𝑡).

(6.480)

where 𝐶 and 𝜓(𝑡) are 2𝑘−1 𝑀 × 1 matrices given as

and

𝑐10 , 𝑐11 , … , 𝑐1𝑀−1 , 𝑐20 , 𝑐21 , … , 𝑇 𝐶 = [𝑐 ] , 2𝑀−1 , … , 𝑐2𝑘−1 0 , … , 𝑐2𝑘−1 𝑀−1

(6.481)

𝜓10 , 𝜓11 , … , 𝜓1𝑀−1 , 𝜓20 , 𝜓21 , … , 𝑇 𝜓(𝑡) = [ ] . 𝜓2𝑀−1 , … , 𝜓2𝑘−1 0 , … , 𝜓2𝑘−1 𝑀−1

(6.482)

6.10.1.4. Analysis Consider the differential equation of fractional-order 𝑦 𝛼 (𝑥) = 𝑓 (𝑥, 𝑦(𝑥), 𝑦 ′ (𝑥), 𝑦 ′′ (𝑥), … , 𝑦 𝑛−1 (𝑥)) , 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑛 − 1 ≤ 𝛼 ≤ 𝑛. 𝑦 2𝑗 (𝑎) = 𝐴𝑗 , 𝑦 2𝑗 (𝑏) = 𝐵𝑗 , 𝑗 = 0,1,2, … , (𝑛 − 2). It is interesting to point out that 𝑦(𝑥) and 𝑓(𝑥, 𝑦(𝑥)) are assumed real and as many times differentiable functions and 𝐴𝑗 and 𝐵𝑗 are the real constant. In order to solve differential equation of fractional-order, we suppose that [𝑎, 𝑏] = [0, 1], otherwise the problem can be mapped from [𝑎, 𝑏] to [0, 1] easily. For solving differential equation of fractional-order, it is sufficient to suppose that the approximate solution is as 𝑦(𝑥) = 𝐂 𝑇 𝛙(𝑥), where 𝐂 and 𝛙(𝑥) are given in above. Therefore, we obtain 𝛼 𝐂 𝑇 𝐏𝑀×𝑀 𝛙(𝑥) = 𝑓 (𝑥, 𝐂 𝑇 𝛙(𝑥), 𝐂 𝑇 𝛙′ (𝑥), 𝐂 𝑇 𝛙′′ (𝑥), … , 𝐂 𝑇 𝛙𝑛−1 (𝑥)),

𝐂 𝑇 (𝛙)2𝑗 (𝑎) = 𝐴𝑗 , 𝐂 𝑇 (𝛙)2𝑗 (𝑏) = 𝐵𝑗 ,

1280

Therefore, in order to apply the Rook Wavelet Method (RWM); we require 2𝑘−1 𝑀 − 1 collocating points. Suitable collocating points are as 𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 𝑛, … , 2𝑘−1 𝑀 − 1.

Implementing the collocating points and imposing the initial value to the differential equation of fractional-order, we obtain 𝛼 𝐂 𝑇 𝐏𝑀×𝑀 𝛙(𝑥𝑖 ) = 𝑓 (𝑥, 𝐂 𝑇 𝛙(𝑥𝑖 ), 𝐂 𝑇 𝛙′ (𝑥𝑖 ), 𝐂 𝑇 𝛙′′ (𝑥𝑖 ), … , 𝐂 𝑇 𝛙𝑛−1 (𝑥𝑖 )),

Solving this system gives the unknown coefficients 𝐂.

6.10.2. Fractional-order Differential Equations Example 6.31 [196]. Consider the following differential equation of fractional-order as given below 1

𝑑2 1 𝑑𝑥 2

2

𝑦(𝑥) + 𝑦(𝑥) − 𝑥 2 = Γ(2.5) 𝑥1.5 ,

(6.483)


(6.484)

The exact solution of Eq. (6.483-6.484) is 𝑦(𝑥) = 𝑥.

(6.485)

According to the Rook Wavelets Method (RWM), we assume the trial solution 𝑘−1


(6.486)



𝑇


We apply the proposed technique to solve Eq. (6.483-6.484) with 𝐾 = 2, and 𝑀 = 3. We have Eq. (6.486) is 𝑦(𝑥) = ∑1𝑛=1 ∑2𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) +𝑐20 𝜓20 (𝑥) + 𝑐21 𝜓21 (𝑥) + 𝑐22 𝜓22 (𝑥) = 𝐂 𝐓 𝛙(𝑥), 1

(6.487)

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑅𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝑅𝑚 is the Rook polynomials given as 𝑅0 = 1, 1281

(6.488)

𝑅1 = 1 + 𝑥, 𝑅2 = 1 + 2𝑥 2 + 4𝑥, 𝑅3 = 1 + 9𝑥 + 18𝑥 2 + 6𝑥 3 , 𝑅4 = 1 + 16𝑥 + 72𝑥 2 + 96𝑥 3 + 24𝑥 4 , . . .. Therefore, the Rook wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(8𝑥 2 − 1), 𝜓1,3 (𝑥) = 2√7(24𝑥 3 − 9𝑥 + 2), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(8𝑥 2 − 1)𝑐1,2 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … ] , and 𝛙(𝑥) = [1, 2√3𝑥, √5(8𝑥 2 − 1), … ] . Substituting into the given problem we get 1

2

2 𝐂 𝑇 𝐏𝑀×𝑀 𝛙(𝑥) + 𝐂 𝐓 𝛙(𝑥) − 𝑥 2 = Γ(2.5) 𝑥1.5 ,

(6.489)

subject to the boundary conditions 𝐂 𝐓 𝛙(0) = 0.

(6.490)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 3.

(6.491)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 6×6 𝐂6×1 = 𝐛6×1 . After solving we get the following exact solution 𝑦(𝑥) = 𝑥. 1282

Example 6.32 [196]. Consider the following differential equation of fractional-order as given below 3

𝑦

′′ (𝑥)

+

𝑑2 3

𝑦(𝑥) + 𝑦(𝑥) = 1 + 𝑥,

(6.492)

𝑑𝑥 2


(6.493)

The exact solution of Eq. (6.492-6.493) is 𝑦(𝑥) = 1 + 𝑥.

(6.494)



(6.495)



𝑇


We apply the proposed technique to solve Eq. (6.492-6.493) with 𝐾 = 2, and 𝑀 = 3. We have Eq. (6.495) is 𝑦(𝑥) = ∑2𝑛=1 ∑2𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) 𝑐20 𝜓20 (𝑥) + 𝑐21 𝜓21 (𝑥) + 𝑐22 𝜓22 (𝑥) = 𝐂 𝐓 𝛙(𝑥), 1

(6.496)

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑅𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝑅𝑚 is the Rook polynomials given as 𝑅0 = 1, 𝑅1 = 1 + 𝑥, 𝑅2 = 1 + 2𝑥 2 + 4𝑥, 𝑅3 = 1 + 9𝑥 + 18𝑥 2 + 6𝑥 3 , 𝑅4 = 1 + 16𝑥 + 72𝑥 2 + 96𝑥 3 + 24𝑥 4 , . . .. Therefore, the Rook wavelets given as 𝜓1,0 (𝑥) = 1, 1283

(6.497)

𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(8𝑥 2 − 1), 𝜓1,3 (𝑥) = 2√7(24𝑥 3 − 9𝑥 + 2), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(8𝑥 2 − 1)𝑐1,2 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … ] , and 𝛙(𝑥) = [1, 2√3𝑥, √5(8𝑥 2 − 1), … ] . Substituting into the given problem we get 𝐓

𝐂 𝛙

′′ (𝑥)

𝑇

3 2

+ 𝐂 𝐏𝑀×𝑀 𝛙(𝑥) + 𝐂 𝐓 𝛙(𝑥) = 1 + 𝑡,

(6.498)


(6.499)


(2𝑖+1)𝜋 2𝑘 𝑀

),

(6.500)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 6×6 𝐂6×1 = 𝐛6×1 . After solving we get the following exact solution 𝑦(𝑥) = 1 + 𝑥.


𝑦

′′ (𝑥)

+

𝑑2 1

𝑦(𝑥) + 𝑦(𝑥) = 0,

(6.501)

𝑑𝑥 2


(6.502)

The exact solution of Eq. (6.501-6.502) is 𝑦(𝑥) = 1 + 𝑥.

(6.503)

According to the Rook Wavelets Method (RWM), we assume the trial solution 1284

𝑘−1


(6.504)



𝑇


We apply the proposed technique to solve Eq. (6.501-6.502) with 𝐾 = 2, and 𝑀 = 4. We have Eq. (6.504) is 𝑦(𝑥) = ∑2𝑛=1 ∑3𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) 𝑐20 𝜓20 (𝑥) + 𝑐21 𝜓21 (𝑥) + 𝑐22 𝜓22 (𝑥) = 𝐂 𝐓 𝛙(𝑥), 1

(6.505)

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 22 𝑅𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). 2

(6.506)

From Eq. (6.505) we obtained In the above expression, 𝑅𝑚 is the Rook polynomials given as 𝑅0 = 1, 𝑅1 = 1 + 𝑥, 𝑅2 = 1 + 2𝑥 2 + 4𝑥, 𝑅3 = 1 + 9𝑥 + 18𝑥 2 + 6𝑥 3 , 𝑅4 = 1 + 16𝑥 + 72𝑥 2 + 96𝑥 3 + 24𝑥 4 , . . .. Therefore, the Rook wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(8𝑥 2 − 1), 𝜓1,3 (𝑥) = 2√7(24𝑥 3 − 9𝑥 + 2), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(8𝑥 2 − 1)𝑐1,2 + 2√7(24𝑥 3 − 9𝑥 + 2)𝑐1,3 + ⋯ 1285

= 𝐂 𝐓 𝛙(𝑥), 𝑇

1, 2√3𝑥, √5(8𝑥 2 − 1), where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , and 𝛙(𝑥) = [ ] . 2√7(24𝑥 3 − 9𝑥 + 2), … 𝑇

Substituting into the given problem we get 𝟏

𝐓

𝐂 𝛙

′′ (𝑥)

+

𝟐 𝐂 𝐓 𝐏𝑀×𝑀 𝛙′ (𝑥)

+ 𝐂 𝐓 𝛙(𝑥) = 0,

(6.507)


(6.508)


(2𝑖+1)𝜋 2𝑘 𝑀

).

(6.509)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀 6×6 𝐂6×1 = 𝐛6×1 . After solving we get the following exact solution 𝑦(𝑥) = 1 + 𝑥.


𝑡

𝑦 ′′ (𝑥) + 𝑦 2 (𝑥) + 𝑦(𝑥) = 𝑥 2 + 4√𝜋 + 2,

(6.510)


(6.511)

The exact solution of Eq. (6.510-6.511) is 𝑦(𝑥) = 𝑥 2 .

(6.512)



(6.513)



𝑇


We apply the proposed technique to solve Eq. (6.510-6.511) with 𝐾 = 2, and 𝑀 = 4. We have Eq. (6.513) is 1286

𝑦(𝑥) = ∑2𝑛=1 ∑3𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) +𝑐20 𝜓20 (𝑥) + 𝑐21 𝜓21 (𝑥) + 𝑐22 𝜓22 (𝑥) = 𝐂 𝐓 𝛙(𝑥), 1

(6.514)

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑅𝑚 (2𝑘 𝑡 − (2𝑛 + 1)).

(6.515)

In the above expression, 𝑅𝑚 is the Rook polynomials given as 𝑅0 = 1, 𝑅1 = 1 + 𝑥, 𝑅2 = 1 + 2𝑥 2 + 4𝑥, 𝑅3 = 1 + 9𝑥 + 18𝑥 2 + 6𝑥 3 , 𝑅4 = 1 + 16𝑥 + 72𝑥 2 + 96𝑥 3 + 24𝑥 4 , . . .. Therefore, the Rook wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(8𝑥 2 − 1), 𝜓1,3 (𝑥) = 2√7(24𝑥 3 − 9𝑥 + 2), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(8𝑥 2 − 1)𝑐1,2 + 2√7(24𝑥 3 − 9𝑥 + 2)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇


Substituting into the given problem we get 𝟑

𝐓

𝐂 𝛙

′′ (𝑥)

+

𝟐 𝐂 𝐓 𝐏𝑀×𝑀 𝛙(𝑥)

𝑡

+ 𝐂 𝐓 𝛙(𝑥) = 𝑥 2 + 4√𝜋 + 2,

(6.516)


(6.517) 1287


(2𝑖+1)𝜋 2𝑘 𝑀

).

(6.518)



128

𝑦 2 (𝑥) + 𝑦(𝑥) = 𝑥 5 + 𝑥 4 + 7

√𝜋

64

𝑥 3.5 − 5

√𝜋

𝑥 2.5 ,

(6.519)


(6.520)

The exact solution of Eq. (6.519-6.520) is 𝑦(𝑥) = 𝑥 4 (1 − 𝑥).

(6.521)



(6.522)



𝑇


We apply the proposed technique to solve Eq. (6.519-6.520) with 𝐾 = 2, and 𝑀 = 6. We have Eq. (6.522) is 𝑦(𝑥) = ∑2𝑛=1 ∑5𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) +𝑐13 𝜓13 (𝑥) + ⋯ = 𝐂 𝐓 𝛙(𝑥), 1

(6.523)

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑅𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). In the above expression, 𝑅𝑚 is the Rook polynomials given as 𝑅0 = 1, 𝑅1 = 1 + 𝑥, 𝑅2 = 1 + 2𝑥 2 + 4𝑥, 1288

(6.524)

𝑅3 = 1 + 9𝑥 + 18𝑥 2 + 6𝑥 3 , 𝑅4 = 1 + 16𝑥 + 72𝑥 2 + 96𝑥 3 + 24𝑥 4 , . . .. Therefore, the Rook wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = √5(8𝑥 2 − 1), 𝜓1,3 (𝑥) = 2√7(24𝑥 3 − 9𝑥 + 2), . . .. Therefore, we have the trial solution is 𝑦(𝑥) = 𝑐1,0 + 2√3𝑥𝑐1,1 + √5(8𝑥 2 − 1)𝑐1,2 + 2√7(24𝑥 3 − 9𝑥 + 2)𝑐1,3 + ⋯ = 𝐂 𝐓 𝛙(𝑥), 𝑇


Substituting into the given problem we get 𝟑

128

𝟐 𝐂 𝐓 𝐏𝑀×𝑀 𝛙(𝑥) + 𝐂 𝐓 𝛙(𝑥) = 𝑥 5 + 𝑥 4 + 7

√𝜋

64

𝑥 3.5 − 5

√𝜋

𝑥 2.5 ,

(6.525)


(6.526)


(2𝑖+1)𝜋 2𝑘 𝑀

).

(6.527)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀12×12 𝐂12×1 = 𝐛12×1 . After solving we get the following exact solution 𝑦(𝑥) = 𝑥 4 (𝑥 − 1). 1289

6.11. Touchard Wavelet Method (TWM) We proposed a new algorithm by inserting Touchard polynomials in traditional Touchard wavelets method. This technique is successively applied to find the exact/numerical solutions of partial differential equations. The proposed technique is very simple and highly compatible for solving such kind of problems.

6.11.1. Methodology 6.11.1.1. Touchard Polynomials The Recurrence relation of Touchard polynomial is 𝑛!

𝑇𝑛 = ∑𝑛𝑘=0 𝑘!(𝑛−𝑘)! . (𝑥)𝑘

(6.528)

few Touchard polynomials given below 𝑇0 = 1, 𝑇1 = 1 + 𝑥, 𝑇2 = 1 + 𝑥 2 + 2𝑥, 𝑇3 = 1 + 3𝑥 + 3𝑥 2 + 𝑥 3 , . . ..

6.11.1.2. Wavelets and Touchard Wavelets In recent years, wavelets have found their way into many different fields of science and engineering [222-224]. Wavelets constitute a family of functions constructed from the dilation and translation of a single function called the mother wavelet. When the dilation parameter a and the translation parameter 𝑏 vary continuously, we have the following family of continuous wavelets: 1

𝑡−𝑏

𝜓𝑎,𝑏 (𝑡) = |𝑎|− 2 𝜓 (

𝑎

) 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≠ 0.

(6.529)

Touchard wavelets 𝜓𝑛𝑚 (𝑡) = 𝜓(𝑘, 𝑛, 𝑚, 𝑡) have four arguments; 𝑛, 𝑘 can assume any positive integer, 𝑚 is the order for Touchard polynomials and t is the normalized time. They are defined on the interval [0, 1) by; 1

𝜓𝑛𝑚

𝑘

𝑘 2 (𝑡) = {√𝑚 + 2 2 𝑇𝑚 (2 𝑡 − (2𝑛 + 1))

0

𝑛 2𝑘

≤𝑡≤

otherwise 1290

𝑛+1 2𝑘

(6.530)

1

where 𝑚 = 0,1,2, … , 𝑀 − 1 and 𝑛 = 0,1, … , 2𝑘−1 . The coefficient (𝑚 + 2) is for orthonormality. Here 𝑇𝑚 (𝑡) are the well-known Touchard polynomials of order 𝑚, which have been previously described.


(6.531)


𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡)𝜙𝑛 (𝑡).

(6.532)

where 𝜙𝑛 (𝑡) = 𝑡 𝑛−1 . It can be written as 𝑘−1

𝑇 𝑓(𝑡) = ∑2𝑛=1 ∑𝑀−1 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑡) = 𝐂 𝛙(𝑡)𝜙(𝑡).

(6.533)

where 𝐂 and 𝛙(𝑡) are 2𝑘−1 𝑀 × 1 matrices given as 𝑐10 , 𝑐11 , … , 𝑐1𝑀−1 , 𝑐20 , 𝑐21 , … , 𝑇 𝐂 = [𝑐 ] , , … , 𝑐 𝑘−1 , … , 𝑐 𝑘−1

(6.534)


(6.535)

2𝑀−1

and

2

0

2

𝑀−1

6.11.1.4. Analysis Consider the partial differential equation 𝑢𝑡 (𝑥, 𝑡) + 𝑝(𝑥)𝑢𝑥 (𝑥) = 𝑞(𝑥),

𝑎 ≤ 𝑥 ≤ 𝑏,

𝑢(𝑥, 0) = 𝛼, where 𝑝(𝑥), 𝑞(𝑥) are real valued functions, 𝛼, 𝑎, and 𝑏 are constants. In order to solve singular differential equation the given equation can be rewrite as 𝑡

𝑡

𝑢(𝑥, 𝑡) = 𝑢(𝑥, 0) − ∫0 𝑝(𝑥)𝑢𝑥 (𝑥)d𝑡 + ∫0 𝑞(𝑥)d𝑡. For solving singular differential equation, it is sufficient to suppose that the approximate solution is as 𝑢(𝑥, 𝑡) = 𝐂 𝑇 𝛙(𝑥)𝜙𝑛 (𝑡), where 𝜙𝑛 (𝑡) = 𝑡 𝑛−1 , 𝐂 and 𝛙(𝑥) are given in above. Therefore, we obtain 𝑡

𝑡

𝐂 𝑇 𝛙(𝑥)𝜙𝑛 (𝑡) = 𝛼 − ∫0 𝑝(𝑥)(𝐂 𝑇 𝛙(𝑥)𝜙𝑛 (𝑡))𝑥 d𝑡 + ∫0 𝑞(𝑥)d𝑡. Therefore, in order to apply the Touchard Wavelet Method (TWM); we require 2𝑘−1 𝑀 − 1 collocating points. Suitable collocating points are as 1291

𝑥𝑖 = cos (

(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 2, … , 2𝑘−1 𝑀 − 1.

Implementing the collocating points we get system of algebraic equations. Solving this system gives the unknown coefficients 𝐂.

6.11.2. Homogenous Cauchy Problem Consider the following partial differential equation [209] as given below 𝑢𝑡 (𝑥, 𝑡) + 𝑥𝑢𝑥 (𝑥, 𝑡) = 0, 𝑥 ∈ ℝ, 𝑡 > 0,

(6.536)

subject to the boundary conditions 𝑢(𝑥, 0) = 𝑥 2 .

(6.537)

The exact solution of Eq. (6.536) is 𝑢(𝑥, 𝑡) = 𝑥 2 e−2𝑡 .

(6.538)

Integrate w.r.t. 𝑡 of Eq. (6.536) both sides and applying the initial conditions we obtained 𝑡

𝑢(𝑥, 𝑡) = 𝑢(𝑥, 0) − ∫0 𝑥𝑢𝑥 (𝑥, 𝑡)d𝑡,

(6.539)

According to the Touchard Wavelets Method (TWM), we assume the trial solution 𝑘−1

𝑛−1 𝑢(𝑥, 𝑡) = ∑2𝑛=1 ∑𝑀−1 = 𝐂 𝐓 𝛙(𝑥)𝜙(𝑡), 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥)𝑡

(6.540)

where 𝐂 and 𝛙(𝑡) are 2𝑘−1 𝑀 × 1 matrices given as 𝑇


𝑇


We apply the proposed technique to solve Eq. (6.536) with 𝐾 = 2, and 𝑀 = 10. We have Eq. (6.540) is 𝑢(𝑥, 𝑡) = ∑1𝑛=1 ∑2𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥)𝜙𝑛 (𝑡) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) +𝑐20 𝜓20 (𝑥)𝑡 + 𝑐21 𝜓21 (𝑥)𝑡 + 𝑐22 𝜓22 (𝑥)𝑡 + ⋯ = 𝐂 𝐓 𝛙(𝑥)𝜙(𝑡), 1

(6.541)

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 22 𝑇𝑚 (2𝑘 𝑡 − (2𝑛 + 1)). 2

In the above expression, 𝑇𝑚 is the Touchard polynomials given as 𝑇0 = 1, 𝑇1 = 1 + 𝑥, 𝑇2 = 1 + 2𝑥 2 + 2𝑥, 𝑇3 = 1 + 3𝑥 + 3𝑥 2 + 𝑥 3 , 𝑇4 = 1 + 4𝑥 + 6𝑥 2 + 4𝑥 3 + 𝑥 4 , 1292

(6.542)

. . .. Therefore, the Touchard wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = 4√5𝑥 2 , 𝜓1,3 (𝑥) = 8√7𝑥 3 , . . .. Therefore, we have the trial solution is 𝑢(𝑥, 𝑡) = 𝑐1,0 + 2√3𝑥𝑐1,1 + 4√5𝑥 2 𝑐1,2 + 8√7𝑥 3 𝑐1,3 + 𝑐2,0 𝑡 + 2√3𝑥𝑡𝑐2,1 +4√5𝑥 2 𝑡𝑐2,2 + 8√7𝑥 3 𝑡𝑐3,3 + ⋯ = 𝐂 𝑇 𝛙(𝑥)𝜙(𝑡), 𝑇

𝑇

where 𝐂 = [𝑐1,0 , 𝑐1,1 , 𝑐1,2 , 𝑐1,3 , … ] , and 𝛙(𝑥) = [1, 2√3𝑥, 4√5𝑥 2 , 8√7𝑥 3 , … ] . Substituting into the Eq. (6.542) we get 𝑡

𝐂 𝑇 𝛙(𝑥)𝜙(𝑡) = 𝑢(𝑥, 0) − ∫0 𝑥𝐂 𝑇 𝛙𝑥 (𝑥)𝜙(𝑡)d𝑡,

(6.543)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2, … 9.

(6.544)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get series solution

1293

Table 6.100: Comparison of the Exact Solution and Approximate Solutions of Eq. (6.536) obtained from Touchard Wavelet Method (TWM) _____________________________________________________________________ 𝑡

Approx. Solution

Approx. Solution

Approx. Solution

Approx. Solution

at x = 0.05

at x = 0.10

at x = 0.50

at x = 1.00

_____________________________________________________________________ 0.0

0.00000E+00

0.00000E+00

0.00000E+00

0.00000E+00

0.1

6.92851E-17

2.77140E-16

6.92851E-15

2.77140E-14

0.2

6.96979E-14

2.78792E-13

6.96979E-12

2.78792E-11

0.3

3.94935E-12

1.57974E-11

3.94935E-10

1.57974E-09

0.4

6.89315E-11

2.75726E-10

6.89315E-09

2.75726E-08

0.5

6.31147E-10

2.52459E-09

6.31147E-08

2.52459E-07

0.6

3.84292E-09

1.53717E-08

3.84292E-07

1.53717E-06

0.7

1.76584E-08

7.06336E-08

1.76584E-06

7.06336E-06

0.8

6.60379E-08

2.64151E-07

6.60379E-06

2.64151E-05

0.9

2.11027E-07

8.44110E-07

2.11027E-05

8.44110E-05

1.0

5.95704E-07

2.38281E-06

5.95704E-05

2.38281E-04

_____________________________________________________________________

(a)

(b)


1294

(a)

(b)


Fig. 6.133: Comparison of Exact and Approximate Solution of Eq. (6.536)

Now to solve Eq. (6.536) with 𝐾 = 2, and 𝑀 = 20. We have Eq. (6.540) is 𝑛−1 𝑢(𝑥, 𝑡) = ∑2𝑛=1 ∑19 = 𝐂 𝑇 𝛙(𝑥)𝜙(𝑡), 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥)𝑡

𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,19 𝑇 𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,19 𝑇 where 𝐂 = [𝑐 , 𝑐 , 𝑐 , … , 𝑐 ] , and 𝛙(𝑥) = [ ] . 𝜓2,0 , 𝜓2,1 , 𝜓2,2 , … , 𝜓2,19 2,0 2,1 2,2 2,19 Proceeding as before we have the matrix form 𝐀 𝟒𝟎×𝟒𝟎 𝐂𝟒𝟎×𝟏 = 𝐛𝟒𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table.

1295


Approx. Solution

Approx. Solution

Approx. Solution

Approx. Solution

at x = 0.05

at x = 0.10

at x = 0.50

at x = 1.00

_____________________________________________________________________ 0.0

0.00000E+00

0.00000E+00

0.00000E+00

0.00000E+00

0.1

1.06733E-35

4.26930E-35

1.06733E-33

4.26930E-33

0.2

1.10870E-29

4.43480E-29

1.10870E-27

4.43480E-27

0.3

3.65250E-26

1.46100E-25

3.65250E-24

1.46100E-23

0.4

1.14117E-23

4.56469E-23

1.14117E-21

4.56469E-21

0.5

9.80778E-22

3.92311E-21

9.80778E-20

3.92311E-19

0.6

3.72605E-20

1.49042E-19

3.72605E-18

1.49042E-17

0.7

8.05878E-19

3.22351E-18

8.05878E-17

3.22351E-16

0.8

1.15405E-17

4.61621E-17

1.15405E-15

4.61621E-15

0.9

1.20622E-16

4.82487E-16

1.20622E-14

4.82487E-14

1.0

9.83457E-16

3.93383E-15

9.83457E-14

3.93383E-13

_____________________________________________________________________

(a)

(b)


1296

(a)

(b)



Finally we assume 𝐾 = 2, and 𝑀 = 49. We have Eq. (6.540) is 𝑛−1 𝑢(𝑥, 𝑡) = ∑2𝑛=1 ∑49 = 𝐂 𝑇 𝛙(𝑥)𝜙(𝑡), 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥)𝑡

𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,49 𝑇 𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 𝑇 where 𝐂 = [𝑐 , 𝑐 , 𝑐 , … , 𝑐 ] , and 𝛙(𝑥) = [ ] . 𝜓2,0 , 𝜓2,1 , 𝜓2,2 , … , 𝜓2,49 2,0 2,1 2,2 2,49 Proceeding as before we have the matrix form 𝐀 𝟏𝟎𝟎×𝟏𝟎𝟎 𝐂𝟏𝟎𝟎×𝟏 = 𝐛𝟏𝟎𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table.

1297


Approx. Solution

Approx. Solution

Approx. Solution

Approx. Solution

at x = 0.05

at x = 0.10

at x = 0.50

at x = 1.00

_____________________________________________________________________ 0.0

0.00000E+00

0.00000E+00

0.00000E+00

0.00000E+00

0.1

3.6154E-105

1.4461E-104

3.6154E-103

1.44616E-102

0.2

8.11010E-90

3.24404E-89

8.11010E-88

3.24404E-87

0.3

7.72726E-81

3.09090E-80

7.72726E-79

3.09090E-78

0.4

1.81239E-74

7.24956E-74

1.81239E-72

7.24956E-72

0.5

1.58132E-69

6.32528E-69

1.58132E-67

6.32528E-67

0.6

1.72039E-65

6.88155E-65

1.72039E-63

6.88155E-63

0.7

4.44919E-62

1.77968E-61

4.44919E-60

1.77968E-59

0.8

4.02018E-59

1.60807E-58

4.02018E-57

1.60807E-56

0.9

1.62706E-56

6.50826E-56

1.62706E-54

6.50826E-54

1.0

3.49481E-54

1.39792E-53

3.49481E-52

1.39792E-51

_____________________________________________________________________

(a)

(b)


1298

(a)

(b)



6.11.3. Non-homogenous Cauchy Problem Consider the following partial differential equation [209] as given below 𝑢𝑡 (𝑥, 𝑡) + 𝑢𝑥 (𝑥, 𝑡) = 𝑥, 𝑥 ∈ ℝ, 𝑡 > 0,

(6.545)

subject to the boundary conditions 𝑢(𝑥, 0) = e𝑥 .

(6.546)

The exact solution of Eq. (6.545) is 𝑡

𝑢(𝑥, 𝑡) = 𝑡 (𝑥 − 2) + e𝑥−𝑡 .

(6.547)

Integrate w.r.t. 𝑡 of Eq. (6.545) both sides and applying the initial conditions we obtained 𝑡

𝑢(𝑥, 𝑡) = 𝑥𝑡 + 𝑢(𝑥, 0) − ∫0 𝑢𝑥 (𝑥, 𝑡)d𝑡,

(6.548)

According to the Touchard Wavelets Method (TWM), we assume the trial solution 1299

𝑘−1


(6.549)



𝑇


We apply the proposed technique to solve Eq. (6.552) with 𝐾 = 3, and 𝑀 = 10. We have Eq. (6.549) is 𝑢(𝑥, 𝑡) = ∑1𝑛=1 ∑2𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥)𝜙𝑛 (𝑡) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) +𝑐20 𝜓20 (𝑥)𝑡 + 𝑐21 𝜓21 (𝑥)𝑡 + 𝑐22 𝜓22 (𝑥)𝑡 + ⋯ = 𝐂 𝐓 𝛙(𝑥)𝜙(𝑡), 1

(6.550)

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑇𝑚 (2𝑘 𝑡 − (2𝑛 + 1)).

(6.551)

In the above expression, 𝑇𝑚 is the Touchard polynomials given as 𝑇0 = 1, 𝑇1 = 1 + 𝑥, 𝑇2 = 1 + 2𝑥 2 + 2𝑥, 𝑇3 = 1 + 3𝑥 + 3𝑥 2 + 𝑥 3 , 𝑇4 = 1 + 4𝑥 + 6𝑥 2 + 4𝑥 3 + 𝑥 4 , . . .. Therefore, the Touchard wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = 4√5𝑥 2 , 𝜓1,3 (𝑥) = 8√7𝑥 3 , . . .. Therefore, we have the trial solution is 𝑢(𝑥, 𝑡) = 𝑐1,0 + 2√3𝑥𝑐1,1 + 4√5𝑥 2 𝑐1,2 + 8√7𝑥 3 𝑐1,3 + 𝑐2,0 𝑡 + 2√3𝑥𝑡𝑐2,1 1300

+4√5𝑥 2 𝑡𝑐2,2 + 8√7𝑥 3 𝑡𝑐3,3 + ⋯ = 𝐂 𝑇 𝛙(𝑥)𝜙(𝑡), 𝑇

𝑇


𝐂 𝑇 𝛙(𝑥)𝜙(𝑡) = 𝑥𝑡 + e𝑥 − ∫0 𝑥𝐂 𝑇 𝛙𝑥 (𝑥)𝜙(𝑡)d𝑡,

(6.552)


(2𝑖+1)𝜋 2𝑘 𝑀

) , 𝑖 = 0,1,2, … 9.

(6.553)

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀10×10 𝐂10×1 = 𝐛10×1 . After solving we get series solution Table 6.103: Comparison of the Exact Solution and Approximate Solutions of Eq. (6.545) obtained from Touchard Wavelet Method (TWM) _____________________________________________________________________ 𝑡

Approx. Solution

Approx. Solution

Approx. Solution

Approx. Solution

at x = 0.05

at x = 0.10

at x = 0.50

at x = 1.00

_____________________________________________________________________ 0.0

0.00000E+00

0.00000E+00

0.00000E+00

0.00000E+00

0.1

2.61188E-19

2.74579E-19

4.09624E-19

6.75355E-19

0.2

5.30520E-16

5.57720E-16

8.32020E-16

1.37177E-15

0.3

4.55144E-14

4.78480E-14

7.13808E-14

1.17687E-13

0.4

1.06892E-12

1.12372E-12

1.67639E-12

2.76391E-12

0.5

1.23437E-11

1.29766E-11

1.93588E-11

3.19173E-11

0.6

9.09823E-11

9.56470E-11

1.42689E-10

2.35254E-10

0.7

4.91940E-10

5.17162E-10

7.71515E-10

1.27201E-09

0.8

2.12025E-09

2.22896E-09

3.32522E-09

5.48235E-09

0.9

7.68515E-09

8.07918E-09

1.20527E-08

1.98716E-08

1.0

2.42994E-08

2.55452E-08

3.81090E-08

6.28311E-08

_____________________________________________________________________

1301

(a)

(b)


(a)

(b)



1302

Now to solve Eq. (6.545) with 𝐾 = 3, and 𝑀 = 20. We have Eq. (6.549) is 𝑛−1 𝑢(𝑥, 𝑡) = ∑2𝑛=1 ∑19 = 𝐂 𝑇 𝛙(𝑥)𝜙(𝑡), 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥)𝑡

𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,19 , 𝑇 𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,19 𝑇 where 𝐂 = [𝑐 , 𝑐 , 𝑐 , … , 𝑐 , …] , and 𝛙(𝑥) = [ ] . 𝜓2,0 , 𝜓2,1 , 𝜓2,2 , … , 𝜓2,19 , … 2,0 2,1 2,2 2,19 Proceeding as before we have the matrix form 𝐀 𝟒𝟎×𝟒𝟎 𝐂𝟒𝟎×𝟏 = 𝐛𝟒𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.104: Comparison of the Exact Solution and Approximate Solutions of Eq. (6.545) obtained from Touchard Wavelet Method (TWM) _____________________________________________________________________ 𝑡

Approx. Solution

Approx. Solution

Approx. Solution

Approx. Solution

at x = 0.05

at x = 0.10

at x = 0.50

at x = 1.00

_____________________________________________________________________ 0.0

0.00000E+00

0.00000E+00

0.00000E+00

0.00000E+00

0.1

2.04833E-41

2.15335E-41

3.21243E-41

5.29640E-41

0.2

4.27631E-35

4.49556E-35

6.70658E-35

1.10573E-34

0.3

2.12340E-31

2.23227E-31

3.33015E-31

5.49049E-31

0.4

8.88790E-29

9.34359E-29

1.39390E-28

2.29815E-28

0.5

9.59338E-27

1.00852E-26

1.50454E-26

2.48057E-26

0.6

4.39388E-25

4.61916E-25

6.89097E-25

1.13613E-24

0.7

1.11380E-23

1.17091E-23

1.74679E-23

2.87997E-23

0.8

1.83115E-22

1.92504E-22

2.87182E-22

4.73483E-22

0.9

2.16282E-21

2.27372E-21

3.39198E-21

5.59244E-21

1.0

1.96802E-20

2.06892E-20

3.08647E-20

5.08873E-20

_____________________________________________________________________

1303

(a)

(b)


(a)

(b)


1304


Finally we assume 𝐾 = 3, and 𝑀 = 49. We have Eq. (6.549) is 𝑛−1 𝑢(𝑥, 𝑡) = ∑2𝑛=1 ∑49 = 𝐂 𝑇 𝛙(𝑥)𝜙(𝑡), 𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥)𝑡

𝑐1,0 , 𝑐1,1 , 𝑐1,2 , … , 𝑐1,49 , 𝑇 𝜓1,0 , 𝜓1,1 , 𝜓1,2 , … , 𝜓1,49 , 𝑇 where 𝐂 = [𝑐 , 𝑐 , 𝑐 , … , 𝑐 , …] , and 𝛙(𝑥) = [ ] . 𝜓2,0 , 𝜓2,1 , 𝜓2,2 , … , 𝜓2,49 , … 2,0 2,1 2,2 2,49 Proceeding as before we have the matrix form 𝐀 𝟏𝟎𝟎×𝟏𝟎𝟎 𝐂𝟏𝟎𝟎×𝟏 = 𝐛𝟏𝟎𝟎×𝟏 . Proceeding as before we have the following graphical representation and error table. Table 6.105: Comparison of the Exact Solution and Approximate Solutions of Eq. (6.545) obtained from Touchard Wavelet Method (TWM) _____________________________________________________________________ 𝑡

Approx. Solution

Approx. Solution

Approx. Solution

Approx. Solution

at x = 0.05

at x = 0.10

at x = 0.50

at x = 1.00

_____________________________________________________________________ 0.0

0.00000E+00

0.00000E+00

0.00000E+00

0.00000E+00

0.1

6.7644E-118

7.1112E-118

1.0608E-117

1.74910E-117

0.2

1.5203E-102

1.5982E-102

2.3843E-102

3.93108E-102

0.3

1.45129E-93

1.52570E-93

2.27608E-93

3.75262E-93

0.4

3.41036E-87

3.58522E-87

5.34852E-87

8.81821E-87

0.5

2.98115E-82

3.13399E-82

4.67537E-82

7.70838E-82

0.6

3.24938E-78

3.41598E-78

5.09604E-78

8.40194E-78

0.7

8.41901E-75

8.85066E-75

1.32036E-74

2.17691E-74

0.8

7.62126E-72

8.01201E-72

1.19525E-71

1.97064E-71

0.9

3.09017E-69

3.24861E-69

4.84636E-69

7.99029E-69

1.0

6.64958E-67

6.99051E-67

1.04286E-66

1.71939E-66

_____________________________________________________________________

1305

(a)

(b)


(a)

(b)


1306

Fig. 6.148 : Comparison of Exact and Approximate Solution of Eq. (6.545)

6.11.4. Transport Equation Consider the following partial differential equation [209] as given below 𝑢𝑡 (𝑥, 𝑡) + 2𝑢𝑥 (𝑥, 𝑡) = 0, 𝑥 ∈ ℝ, 𝑡 > 0,

(6.554)

subject to the boundary conditions 𝑢(𝑥, 0) = 𝑥 2 .

(6.555)

The exact solution of Eq. (6.554-6.555) is 𝑢(𝑥, 𝑡) = 𝑥 2 − 4𝑥𝑡 + 𝑡 2 .

(6.556)

Integrate w.r.t. 𝑡 of Eq. (6.554-6.555) both sides and applying the initial conditions we obtained 𝑡

𝑢(𝑥, 𝑡) = 𝑥𝑡 + 𝑢(𝑥, 0) − 2 ∫0 𝑢𝑥 (𝑥, 𝑡)d𝑡,

(6.557)

According to the Touchard Wavelets Method (TWM), we assume the trial solution 𝑘−1


(6.558)



𝑇


We apply the proposed technique to solve Eq. (6.554-6.555) with 𝐾 = 3, and 𝑀 = 4. We have Eq. (6.558) is 𝑢(𝑥, 𝑡) = ∑1𝑛=1 ∑2𝑚=0 𝑐𝑛𝑚 𝜓𝑛𝑚 (𝑥)𝜙𝑛 (𝑡) = 𝑐10 𝜓10 (𝑥) + 𝑐11 𝜓11 (𝑥) + 𝑐12 𝜓12 (𝑥) +𝑐20 𝜓20 (𝑥)𝑡 + 𝑐21 𝜓21 (𝑥)𝑡 + 𝑐22 𝜓22 (𝑥)𝑡 + ⋯ 1307

= 𝐂 𝐓 𝛙(𝑥)𝜙(𝑡), 1

(6.559)

𝑘

where 𝜓𝑛𝑚 (𝑥) = √𝑚 + 2 22 𝑇𝑚 (2𝑘 𝑡 − (2𝑛 + 1)).

(6.560)

In the above expression, 𝑇𝑚 is the Touchard polynomials given as 𝑇0 = 1, 𝑇1 = 1 + 𝑥, 𝑇2 = 1 + 2𝑥 2 + 2𝑥, 𝑇3 = 1 + 3𝑥 + 3𝑥 2 + 𝑥 3 , 𝑇4 = 1 + 4𝑥 + 6𝑥 2 + 4𝑥 3 + 𝑥 4 , . . .. Therefore, the Touchard wavelets given as 𝜓1,0 (𝑥) = 1, 𝜓1,1 (𝑥) = 2√3𝑥, 𝜓1,2 (𝑥) = 4√5𝑥 2 , 𝜓1,3 (𝑥) = 8√7𝑥 3 , . . .. Therefore, we have the trial solution is 𝑢(𝑥, 𝑡) = 𝑐1,0 + 2√3𝑥𝑐1,1 + 4√5𝑥 2 𝑐1,2 + 8√7𝑥 3 𝑐1,3 + 𝑐2,0 𝑡 + 2√3𝑥𝑡𝑐2,1 +4√5𝑥 2 𝑡𝑐2,2 + 8√7𝑥 3 𝑡𝑐3,3 + ⋯ = 𝐂 𝑇 𝛙(𝑥)𝜙(𝑡), 𝑇

𝑇


𝐂 𝑇 𝛙(𝑥)𝜙(𝑡) = 𝑥 2 − 2 ∫0 𝑥𝐂 𝑇 𝛙𝑥 (𝑥)𝜙(𝑡)d𝑡,

(6.561)


(2𝑖+1)𝜋 2𝑘 𝑀

).

(6.562)

1308

Implementing the collocating points and imposing the initial value of the system, the matrix form is given as 𝐀12×12 𝐂12×1 = 𝐛12×1 . After solving we get the exact solution 𝑢(𝑥, 𝑡) = 𝑥 2 − 4𝑥𝑡 + 𝑡 2 .

1309

7. Conclusion Various types of solutions including traveling wave solutions of wide range of physical problems of different orders have been determined by making use of fifty-eight different approaches. It has been observed that the proposed techniques are quite useful, reliable, and efficient and can be extended for other linear and nonlinear physical problems. Computational work and subsequent numerical results re-confirm the accuracy of proposed algorithms.

1310

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Some Efficient Algorithms for Differential Equations

Some Efficient Algorithms for Differential Equations

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