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Several examples are given to illustrate the efficiency and its comparison with other problems. *Corresponding author. E-mail address: noormaslam@hotmail.
ISSN 1749-3889 (print), 1749-3897 (online) International Journal of Nonlinear Science Vol.7(2009) No.2,pp.143-148

A New Approach to Fifth-Order Boundary Value Problems Muhammad Aslam Noor ∗ , Syed Tauseef Mohyud-Din Mathematics Department, COMSATS Institute of Information Technology, Islamabad, Pakistan (Received 26 March 2008, accepted 23 July 2008)

Abstract: In this paper, we use a new iterative method for solving the fifth order boundary value problems. The analytical results of the equations have been obtained in terms of convergent series with easily computable components. The algorithm is quite efficient and is practically well suited for nonlinear problems. Several examples are given to illustrate the reliability and implementation of the iterative method. This new decomposition method can be viewed as an alternative to Adomian decomposition and homotopy perturbation methods. Key words: Noor decomposition method; Iterative method; nonlinear problems; boundary value problems; error estimates

1

Introduction

In this paper, we consider the general fifth-order boundary value problems of the type: y v (x) = f (x, y, y 0 , y 00 , . . .) with boundary conditions y(a) = A1 ,

y 0 (a) = A2 ,

y 00 (a) = A3 ,

y(b) = A4 ,

y 0 (b) = A5 ,

where f (., ., ., . . .) is a continuous function on a given domain and Ai , i = 1, 2, 3, 4, 5, are constants. This type of boundary value problems arises in the mathematical modeling of the viscoelastic flows and other branches of mathematical, physical and engineering sciences, see [1, 2, 3, 4] and the references therein. Several numerical methods including spectral Galerkin and collocation [2, 4], decomposition [12] and sixth order B-spline have been developed for solving fifth order boundary value problems. The use of spline function in the context of fifth order boundary value problems was studied by Fyfe[3] who used the quintic polynomial spline functions to develop consistency relation connecting the values of solution with fifth order derivative at the respective nodal points. Variational iteration methods and homotopy perturbation methods are employed by Noor and Mohyud-Din [8, 9, 11] to find the series solution of such problems. It is worth mentioning that to implement the Adomian’s method, one has to find the Adomian’s polynomials which is itself a difficult task, This fact has motivated to develop other analytical and numerical techniques for solving boundary value problems of fifth order. To overcome these drawbacks and deficiencies, Noor and Noor [5,6] have suggested and analyzed a new iterative method for solving nonlinear boundary and initial value problems. The main motivation of this paper is to use this new decomposition method for solving fifthorder boundary value problems. It is shown that this new iterative method [5, 6] solves effectively, easily and accurately large classes of linear and nonlinear, ordinary, partial, deterministic or stochastic differential equations with approximate solutions which converge rapidly to accurate solutions. It is shown that this method provides the solution in a rapid convergent series. It is observed that the new iterative method may be considered as an alternative method for solving linear and nonlinear problems to Adomian decomposition, variational iteration method and homotopy perturbation method. . Several examples are given to illustrate the efficiency and its comparison with other problems. ∗

Corresponding author.

E-mail address: [email protected] c Copyright°World Academic Press, World Academic Union IJNS.2009.04.15/211

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International Journal of Nonlinear Science,Vol.7(2009),No.2,pp. 143-148

Iterative Method

To convey an idea of the technique of the method of Noor and Noor [5, 6], we consider the following general functional equations: f (y) = 0, which can be rewritten in the following equivalent form as: y = f + N (y)

(1)

where N is a nonlinear operator and f is a known function. We are looking for a solution of equation (1) having the series form: y=

∞ X

yi

(2)

i=0

The nonlinear operator N can be decomposed as Ã∞ ! ( à ∞ !) ∞ X X X N yi = N (y0 ) + N yi . i=0

i=0

(3)

i=0

From (1), (2) and (3), we have y = f + N (y0 ) +

∞ X

(

à ∞ !) X N yi .

i=0

(4)

i=0

From (2) and (4), we have the following iterative scheme for solving the nonlinear equation (1) as: y0 = f y1 = N (y0 ), .. . ym+1 = N (y0 , y1 , . . . , ym ),

(5)

m = 1, 2, . . .

Then y1 + y2 + . . . , ym+1 = N (y0 ) + N (y0 + y1 ) + . . . + N (y0 + y1 + . . . + ym ),

m = 1, 2, . . . ,

(6)

and y=f+

∞ X

y1 .

(7)

i=0

For the convergence analysis of this decomposition method, see Noor and Noor [5,6]. This decomposition is quite different than the Adomian decomposition. It is also called the Noor and Noor decomposition method. In the implementation of this method, one does not have to evaluate the so-called Adomian polynomials, which is advantage of this method. It is worth to point out that no differentiability criteria is needed.

3

Numerical Application

In this section, we apply the iterative method developed in Section 2 for solving the fifth order boundary value problems. For the sake of comparison, we consider the same examples as that of [8, 9, 12]. IJNS email for contribution: [email protected]

M. A. Noor, S. T. Mohyud-Din: A New Approach to Fifth-Order Boundary Value Problems

145

Example 3.1 [8, 9, 12]. Consider the following linear boundary value problem of fifth-order of the form: y v (x) = y − 15ex − 10xex , with boundary conditions y(0) = 0,

y 0 (0) = 1,

y 00 (0) = 0,

y(1) = 0,

y 0 (1) = e.

The exact solution of this problem is y(x) = x(1 − x)ex . Using the iterative scheme developed in Section 2, we have following approximants: µ ¶ µ ¶ A1 5 A5 5 15x2 + − + y0 (x) = −35 − 24x − x3 + x4 + 35ex − 10xex 2 6 6 24 24 y1 (x) = N (y0 ) 65 55 15 7 1 1 7 = −85 − 75x − x2 − x3 − x4 − x5 − x6 − x 2 24 30 ¶ 336 ¶6 µ8 µ 1 1 1 1 A4 − x8 + A5 + x9 + 85ex − 10xex + 40320 8064 362880 72576 y2 (x) = N (y0 + y1 ) = 240ex − 20xex − 240 − 220x − 100x2 − 30x3 7 1 11 7 1 8 20 x − x − x4 − x5 − x6 − 3 6 6 µ ¶560 µ 504 ¶ 1 1 1 1 1 9 + A4 − x + A4 − x10 − x11 362880 6048 3628800 90720 1140480 µ ¶ 1 1 1 1 12 13 − x − x + A5 − x14 19958400 415134720 87178291200 17435658240 µ ¶ 1 1 + + x15 . 251534873600 1307674368000 The series solution is given by y(x) = y0 + y1 + y2 = y0 + N (y0 ) + N (y0 + y1 )

µ ¶ 1 = −360 − 319x − 140x2 + 360ex − 40xex + − 40 x3 6 µ ¶ µ ¶ 1 25 35 5 1 6 19 7 1 17 4 + A5 − x − x − x − x + A4 − x8 24 3 24 5 840 40320 8064 µ ¶ µ ¶ 1 1 11 1 1 A4 + A5 − x9 + A5 − x10 362880 362800 72576 3628800 90720 1 1 1 − x11 − x12 − x13 1140480 119958400 415134720 ¶ µ 1 1 x14 + A4 − 87178291200 17435658240 µ ¶ 1 1 + x15 . A5 + 1307674368000 261534873600

Now using the boundary conditions at x = 1, we have the following values of the unknown parameters: A4 = −2.996774777,

A5 = −8.017400823.

Consequently, we obtain the series solution of the boundary value problem as: y(x) = −360 − 319x − 140x2 − 40.49946246x3 − 8.667391701x4 −

35 4 1 6 x − x 24 5

19 7 x − 0.002182459692x8 − 0.0001819173710x9 − 0.00001323230848x10 840 1 1 1 − x11 − x12 − x13 1140480 19958400 415134720 −9.1172896907x1o−11 x14 − 2.307455814x1011 x15 . −

IJNS homepage:http://www.nonlinearscience.org.uk/

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International Journal of Nonlinear Science,Vol.7(2009),No.2,pp. 143-148

Table 3.1(Error Estimates)

x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Exact Solution 0.00000000 0.09946538 0.19542444 0.28347035 0.35803893 0.41218032 0.43730951 0.42288807 0.35608755 0.22136428 0.00000000

Approx.Solution 0.00000000 0.09946580 0.19542750 0.28347920 0.35805430 0.41218032 0.43730380 0.42291420 0.35610480 0.22137120 0.00000000

Errors 0.00000 4.0E -7 3.1E -6 8.8E - 6 1.6E - 5 2.3E - 5 2.7E - 5 2.6E - 5 1.8E - 5 6.9E - 6 0.00000

Table 3.1 shows the approximate solution obtained by new iterative method (NIM) after two iterations and errors obtained by comparing it with the exact solution. Higher accuracy can be obtained by evaluating more iterations. Example 3.2 [8, 9, 12]. Consider the following non-linear boundary value problem of fifth-order of the form y (v) (x) = e−x y 2 (x),

0 < x < 1,

with boundary conditions y(0) = y 0 (0) = y 00 (0) = 1;

y(1) = y 0 (1) = e.

The exact solution of the problem is y(x) = ex . In order to avoid lengthy expression, we use fact that e−x = 1−x+ 21 x2 . Now using the iterative scheme developed in Section 2, we have following approximants: 1 1 1 y0 (x) = 1 + x + x2 + + A4 x3 + A5 x4 2 6 24 y1 (x) = N (y0 ) µ ¶ 1 1 5 1 6 1 1 1 8 x9 = x + x + A4 x + A5 + 120 720 20160 9 20160 6720 µ ¶ 1 1 1 1 + x11 x10 + A24 + 120960 11 191440 40320 µ ¶ 1 1 1 1 2 + A4 A5 − A + A4 x12 12 570240 285120 4 95040 µ ¶ 1 1 1 1 1 + A25 − A4 A5 + A5 + A24 x13 13 6824880 855360 570240 855360 µ ¶ 1 −1 1 1 + A25 + A4 A5 x14 + A2 x15 . 14 9884160 2471040 415134720 5 IJNS email for contribution: [email protected]

M. A. Noor, S. T. Mohyud-Din: A New Approach to Fifth-Order Boundary Value Problems

147

In a similar way, we can find y2 , y3 , . . . . The series solution is given by y(x) = y0 + y1 + . . . = y0 + N (y0 ) + n(y0 + y1 ) + . . . 1 1 1 = 1 + x + x2 + + A4 x3 + A5 x4 2 6 24 µ ¶ 1 5 1 6 1 1 1 1 8 = x + x + A4 x + A5 + x9 120 720 20160 9 20160 6720 µ ¶ 1 1 1 1 10 2 + x + A + x11 120960 11 191440 4 40320 µ ¶ 1 1 1 1 2 A4 A5 − A + A4 x12 + 12 570240 285120 4 95040 µ ¶ 1 1 1 1 1 2 2 + A − A4 A5 + A5 + A x13 13 6824880 5 855360 570240 855360 4 µ ¶ 1 −1 1 1 2 + A5 + A4 A5 x14 + A2 x15 + . . . . 14 9884160 2471040 415134720 5 Using the boundary conditions at x = 1, we have A4 = 1.001621302,

A5 = 0.9921767676.

Thus, we have the following series solution of the fifth-order boundary value as: 1 1 5 1 6 y(x) = 1 + x + x2 + 0.1669368837x3 + 0.04134069865x4 + x + x 2 120 720 1 7 x + 4.968359633x10−5 x8 + 2.20027379x10−5 x9 + O(x10 ). + 720 Table 3.2(Error Estimates) x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Exact Solution 1.000000000 1.105170918 1.221402758 1.349858808 1.491824698 1.648721271 1.822118800 2.013752707 2.225540928 2.459603111 2.718281828

Approx.Solution 1.000000000 1.1051711550 1.2214043990 1.3498634640 1.4918336700 1.6487348180 1.8221355940 2.0137696290 2.2255536730 2.4596082900 2.7182818290

Errors 0.00000 2.3E - 7 1.6E -6 4.6E - 6 8.9E - 6 1.3E - 5 1.6E - 5 1.6E - 5 1.2E - 5 5.1E - 6 0.00000

Table 3.2 shows the approximate solution obtained by new iterative method (NIM) after two iterations and errors obtained by comparing it with the exact solution. Higher accuracy can be obtained by evaluating more iterations.

4

Conclusion

In this paper, we have used a new iterative method, which is called the Noor and Noor decomposition method, for finding the solution of linear and nonlinear boundary value problems for fifth-order. The method is used in a direct way without using linearization, perturbation or restrictive assumptions. It may be concluded that the new decomposition method is very powerful and efficient in finding the analytical solutions for a wide class of boundary value problems. The method gives more realistic series solutions that converge very rapidly in physical problems. Thus we conclude that iterative technique can be considered as an efficient method for solving linear and nonlinear problems. We believe that this new decompostion method will be used effectively for solving nonlinear problems and will open new research directions. IJNS homepage:http://www.nonlinearscience.org.uk/

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International Journal of Nonlinear Science,Vol.7(2009),No.2,pp. 143-148

Acknowledgement Authors would like to thank Dr. S. M. Junaid Zaidi, Rector, CIIT, for providing excellent research facilities.

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An Iterative method for solving Helmholtz equation. A. J.

[11] M. Aslam Noor , S. Tauseef Mohyud-Din: Variational iteration method for fifth-order boundary value problems using He’s polynomials, Math. Problems Eng. Article 954794, 12 pages(2008) [12] A.M. Wazwaz: The numerical solution of fifth-order boundary-value problems by Adomian decomposition method. J. Comput. Appl. Math. 136, 259-270(2001)

IJNS email for contribution: [email protected]