Solving Second-order Singular Problems Using He's ... - CiteSeerX

World Applied Sciences Journal 6 (6): 769-775, 2009 ISSN 1818-4952 © IDOSI Publications, 2009

Solving Second-order Singular Problems Using He’s Polynomials Syed Tauseef Mohyud-Din, Muhammad Aslam Noor and Khalida Inayat Noor Department of Mathematics, COMSATS Institute of Information Technology, Islamabad, Pakistan Abstract: In this paper, we use He’s polynomials for solving singular second-order differential equations. The approach also introduces a transformation which is helpful to cope with the singularity behavior at x = 0. The proposed algorithm is applied to the reformulated problems which gives the solution in terms of transformed variable. The desired series solution is obtained by making use of the inverse transformation. The suggested algorithm is tested on Emden-Fowler, white-dwarf, generalized Lane-Emden and some other second-order singular differential equations. Key words: Homotopy perturbation method He’s polynomials Emden-Fowler equation white-dwarf equations Lane-Emden equation second order nonlinear singular ODES •

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PACS: 02.30 Jr 02.00.00 •

INTRODUCTION

Emden-Fowler equation, white-dwarf equations, generalized Lane-Emden equation and second order nonlinear singular ODES.

The singular differential equations [1-6, 16, 17, 22, 25-29] arise in several physical phenomena in mathematical physics, astrophysics, theory of stellar structure, thermal behavior of a spherical cloud of gas, isothermal gas spheres and theory of thermionic currents. Several techniques including decomposition, spline, finite difference, multi-integral, modified variational iteration and variational iteration have been applied for solving such singular equations, see [1-6, 16, 17, 22, 25-29] and the references therein. He [9-15] developed and formulated homotopy perturbation method (HPM) by merging the standard homotopy and perturbation. The homotopy perturbation method (HPM) has been applied to a wide class of diversified physical problems, see [7-15, 18-21, 23, 24] and the references therein. It is worth mentioning that He’s polynomials which were introduced by Ghorbani et al. [7, 8] are calculated from He’s homotopy perturbation method (HPM). Inspired and motivated by the ongoing research in this area, we use He’s polynomials for solving second-order singular problems. It is also to be highlighted, that we introduce a transformation u(x) = xy(x) which is helpful to cope with the singularity behavior at x = 0 which is a difficult element in such problems. The proposed algorithm is applied to the reformulated problems which leads the solution in terms of transformed variable. The desired series solution is obtained by implementing the inverse transformation. The suggested algorithm is tested on

HOMOTOPY PERTURBATION METHOD (HPM) AND HE’S POLYNOMIALS To explain the He’s homotopy perturbation method, we consider a general equation of the type, L(u) = 0

(1)

where L is any integral or differential operator. We define a convex homotopy H(u,p) by: H(u,p) = (1 − p)F(u) + pL(u)

(2)

where F(u) is a functional operator with known solutions ν0 , which can be obtained easily. It is clear that, for H(u,p) = 0

(3)

we have

= H(u,0) F(u), = H(u,1) L(u)

This shows that H(u,p) continuously traces an implicitly defined curve from a starting point H(ν0 ,0) to a solution function H(f,l). The embedding parameter monotonically increases from zero to unit as the trivial problem F(u) = 0, is continuously deforms the original problem L(u) = 0. The embedding parameter p∈(0,1]

Corresponding Author: Dr. Syed Tauseef Mohyud-Din, HITECH University Taxila Cantt, Pakistan

769

World Appl. Sci. J., 6 (6): 769-775, 2009

can be considered as an expanding parameter [7-15, 18-21, 23, 24]. The homotopy perturbation method uses the homotopy parameter p as an expanding parameter [9-15] to obtain ∞

u =∑ p i u i =u 0 + p u1 + p 2 u 2 + p3 u3 + ,

x x

= u 0 (x) − p∫ u 0 + pu1 + p 2u 2 +

0 0

(4)

p(0 ) : u 0 (x) = x3 + x 4 +

if p→1, then (4) corresponds to (2) and becomes the approximate solution of the form,

p →1

∞

∑u i= 0

1 5 1 6 x + x 20 30

1 5 1 6 1 5 x + x − x 20 30 20 1 1 7 1 8 − x6 − x − x 30 840 1680

p(1) : u1( x ) = x 3 + x 4 +

(5)

i

  dxdx  

Comparing the coefficients of like powers of p

i =0

= f limu =

∫

 ( u 0 + pu 1+ p 2 u 2 + )   − ( 6x + 12x 2 + x 3 + x 4 ) 

1 5 1 6 1 5 x + x − x 20 30 20 1 6 1 7 1 1 7 − x − x − x8 + x 30 840 1680 840 1 8 1 1 + x + x8 + x9 1680 60480 151200

p(2 ) : u2( x ) = x3 + x 4 +

It is well known that series (5) is convergent for most of the cases and also the rate of convergence is dependent on L (u); see [9-15]. We assume that (5) has a unique solution. The comparisons of like powers of p give solutions of various orders. In sum, according to [7, 8], He’s HPM considers the nonlinear term N(u) as:



where pi s are He’s polynomials. The series solution is given by

∞

N(u) =∑ pi H i =H 0 + pH1 + p 2H 2 + ... i =0

3

u(x) = x +x

where Hn ’s are the so-called He’s polynomials [7, 8], which can be calculated by using the formula = H n (u 0 , ,u n )

and using the inverse transformation, the required series solution is given by

n 1 ∂n   = N( ∑ pi ui )  , n 0,1,2,  . n  n! ∂p  i= 0  p =0

y(x) =

In this section, we use He’s polynomials for solving various singular differential equations. Numerical results are very encouraging. Example 3.1 [29] Consider the following secondorder singular initial value problem

y′ +

2 y′= 2(2x 2 + 3)y x

with initial conditions y(0) = 1, y′(0) = 0.

2 y′ + y = 6 + 12x + x 2 + x 3 x

Using the transformation, u(x) = xy(x) the above problem can be re-formulated as

with initial conditions

u′( x ) − 2(2x 2 + 3)u(x) = 0

y(0) = 0, y′(0) = 0.

Applying the convex homotopy

Using the transformation, u(x) = xy(x), the above problem can be re-formulated as u′( x ) + u(x) = 6x + 12x + x + x 2

x2 + x3

Example 3.2 [29] Consider the following second-order singular initial value problem

NUMERICAL APPLICATIONS

y′ +

4

3

x x  u 0 + pu1  u0 + pu 1 + p 2 u 2 +  =− 1 p ∫ ∫ 2(2x2 + 3)  2  dxdx  + p u2 +   0 0

4


Comparing the coefficients of like powers of p 770

World Appl. Sci. J., 6 (6): 769-775, 2009


p(0 ) : u 0 (x) = x

1 8 1 7 x − x + x5 − x4 98 78

1 p(1) : u1( x ) =x + x3 + x 5 5

p(0) : u 0 (x)=

1 3 13 7 1 9 p(2 ) : u 2( x ) =x + x 3 + x 5 + x 5 + x + x 5 10 105 90

p(1) : u1 (x) = x 5 − x 4 +

1 8 1 7 1 11 x − x − x 98 78 1660 1 1 8 1 7 + x9 − x + x 11232 98 78

1 5 3 5 13 7 1 9 x + x + x + x 5 10 105 90 3 7 17 9 59 11 1 13 x + x + x + x + 70 630 11550 3510

p(3) : u 3( x ) =x + x 3 +






1 1 1 1 1 u(x) = x + x + x5 + x7 + x9 + x11 + x12 + 2! 3! 4! 5! 6! 3


and using the inverse transformation, the required series solution is given by y(x) = 1+ x2 +

u(x) = x5 − x 4

1 4 1 6 1 8 1 10 x + x + x + x + 2! 3! 4! 5!

and using the inverse transformation, the required solution is given by

and in the closed form

y(x) = y(x) = e

x2

x4 − x3

Example 3.4 Consider the following Emden-Fowler equation

Example 3.3 [29] Consider the linear initial value problem y′ (x) +

1 8 1 7 1 11 1 x − x − x + x9 98 78 1660 11232 1 1 1 − x8 + x7 + x14 98 78 4331600 1 1 1 − x13 + x 11 − x 11 2560896 1660 11232

p(2 ) : u 2 (x) = x5 − x 4 +

y′ +

8 y′ (x) + x y (x ) = x 5 − x 4 + 44x 2 − 30x x

2 y′ + ax m yn = 0 x

with boundary conditions y(0) = 1, y′(0) = 0.


Using the transformation, u(x) = xy(x), the above problem can be re-formulated as

y(0) = 0, y′(0) = 0. Using the transformation, u(x) = xy (x), the above problem can be re-formulated as

u′( x ) + a x m − n+ 1u n(x) = 0

Applying the convex homotopy  6  6 5 3 2 u′( x ) +  0  u′( x ) + xu(x) − ( x − x + 44x − 30x ) = 1 + x  n x x  u + pu1  2 m−n +1  0 u0 + pu 1 + p u 2 +  =− 1 p∫ ∫  a x  2  Applying the convex homotopy  0 0   +p u 2 +   x x

− p∫ u 0 + pu1 + =

∫

0 0

  6   du 0 du du  + p 1 + p 2 2 +     dx dx    1 + x  dx  u +  dxdx    x6 − x5 0  +x   −     pu1 +    +44x 3 − 30x 2    

Comparing the coefficients of like powers of p p(0 ) : u 0 (x) = x

771

  dxdx  

World Appl. Sci. J., 6 (6): 769-775, 2009 p(1) : u1( x )= x −

p(2 ) : u 2( x )= x −

a (m + 3)(m + 2)

x m +3

y(x) =

a x m +3 (m + 3)(m + 2)

y′ +

2

(3)

: u 3( x )= x −

a2n x 2m +5 2(2m + 5)(m + 3)(m + 2) 2

−

a 3n( 8n + 3mn − 2m − 5) x 3m + 7 6(3m + 7)(5 + 2m)(m + 3) 2( m + 2)3

2  x  y(x)=  1 + a  3  

y(x) = 1 −


−

3

a n( 8n + 3mn − 2m − 5) x 3m + 7 +  6(3m + 7)(5 + 2m)(m + 3)2 (m + 2) 3

a n(8n + 3mn − 2m − 5) x3 m + 6 +  6(3m + 7)(5 + 2m)(m + 3)2 (m + 2) 3

y′ +

4a 3 x 15

5

2 y′+ (y 2 − C) 3 2 = 0 x

with boundary conditions

3

y(0) = 1, y′(0) = 0

It is quite obvious that m can not take the values

Using the transformation, u(x) = xy(x), the above problem can be re-formulated as

5 5 −3, −2, − , − ,  2 3

 u2  u′( x ) + x  2 − C  x  

Substitution of m = 0, n = 0 gives the solution as: y(x)= 1 −

y(x) = 1 −

Example 3.5 [28] Consider the white-dwarf equation

2

−

4a 2 x 35

1 2

a x m +2 (m + 3)(m + 2) a n x 2 m +4 2 2(2m + 5)(m + 3)(m + 2)

5

y(x) = 1 −

are obtained for the values of m = ± .


+

a x m+ 2 (m + 3)(m + 2)

and

a 2n x 2m + 5 2(2m + 5)(m + 3)(m + 2)2

y(x)= x −

1 2

is obtained. Hence for n = 0, the exact solutions

a m +3 x (m + 3)(m + 2)

+

−

is obtained. Moreover, for n = 0, the exact solution



u(x)= x −

2 y′ + ay = 0 x

Similarly, for m = 0, n = 5, the exact solution

a m +3 x (m + 3)(m + 2)

+

ax

for the equation

a n + x 2m + 5 2 2(2m + 5)(m + 3)(m + 2) p

sin( ax)

a 2 x 3!

3

2

0 =


for the equation

x x

u 0 + pu1 + p u 2 + =− 1 p∫ 2

2 y′ + y′ + a = 0 x

∫

0 0

  2   1  u 0 + pu 1   x  x 2  +p 2u +  − C  2    


For m = 0, n = 1, we obtain the exact solution 772

3

   

2

   dxdx  

World Appl. Sci. J., 6 (6): 769-775, 2009


p(0) : u 0 (x) = x, 1 p(1) : u1( x ) =− x (C − 1)3 2 x3 , 6

1 1 y(x) = 1 − (C − 1) 3 2 x2 + (C − 1)2 x 4 6 40 1 52 6 − (5(C − 1) + 14)(C − 1) x 7! 1 + (339(C − 1) + 280)(C − 1)3 x8 3 ⋅ 9! 2  1  1425(C − 1) 7 2 10 +   (C − 1) x +  5 ⋅11! + 11436(C − 1) + 4256 

1 1 p(2 ) : u 2( x ) =− x (C − 1)3 2 x 3 + (C − 1)2 x 4 , 6 40

1 1 p(3) : u 3( x ) = x − (C − 1) 3 2 x3 + (C − 1) 2 x 4 6 40 1 52 7 − (5(C − 1) +14)(C − 1) x 7!

Example 3.6 [28] Consider the following generalized Lane-Emden

1 1 p(4 ) : u 4 (x) =− x (C − 1) 3 2 x3 + (C − 1) 2 x 4 6 40 1 − (5(C − 1) + 14)(C − 1)5 2 x 7 7! 1 + (339(C − 1) + 280)(C − 1)3 x 9 3 ⋅ 9! 

y′ +


where pi s are He’s polynomials. The series solution is given as

y(0) = 1, y′(0) = 0. Using the transformation, u(x) = xy(x), the above problem can be re-formulated as

1 1 32 3 2 4 u(x) =− x (C −1) x + (C − 1) x 6 40 1 − (5(C −1) + 14)(C − 1)5 2 x 7 7! 1 (339(C − 1) + 280)(C − 1)3 x 9 +  + 3 ⋅ 9! x

u0 + pu 1 + p 2 u 2 +  =− 1 p∫ 0

x

∫ 0

n y′ + y m= 0, n ≥ 0 x

m

 n−2  u u′( x ) +  0 u ′( x ) +  x  = 1+x   

Applying the convex homotopy m  n −2 1    ( u ′0 + pu′′2 + ) +  ( u0 + pu 1+ p 2u 2 + )   x   1 + x 

 dxdx 

Comparing the coefficients of like powers of p p(0 ) : u 0 (x) = x

p(2 ) : u 2( x )= x −

p

(3)

p(1) : u1( x )= x −

1 2(n + 1)

x3

1 m x3 x5 2(n + 1) 8 ( n + 1)( n + 3)

( 2n + 4) m2 − (n + 3)m 7 1 m 3 5 : u 3( x ) = x− x x + x 2 2(n + 1) 8 ( n + 1)( n + 3) 48 ( n + 1) (n + 3)(n + 5)

1 m ( 2n + 4) m 2 − (n + 3)m 7 p(4) : u 4( x ) = x− x3 x5 + x 2 2(n + 1) 8 ( n+ 1)( n+ 3) 48 ( n+ 1) (n + 3)(n + 5) (6n 2 + 32n + 34)m 2 − ( 97n 2 + 46n + 630m ) + (2n2 + 16n + 30) 2

+

3

384(n + 1) (n + 3)(n + 5)(n + 7)



773

x8

World Appl. Sci. J., 6 (6): 769-775, 2009

where p i s are He’s polynomials. The series solution is given by

( 2n + 4 ) m 2 − (n + 3)m 7 1 m u(x) = x− x3 x5 + x 2 2(n + 1) 8 ( n+ 1)( n+ 3) 48( n + 1 ) (n + 3)(n + 5) (6n 2 + 32n + 34)m 2 − ( 97n 2 + 46n + 630m ) + (2n 2 + 16n + 30) 2

+

384(n + 1)3 (n+ 3)(n + 5)(n + 7)

x8 + 

and the inverse transformation will yield

( 2n + 4 ) m 2 − (n + 3)m 6 1 m y(x) = 1− x2 + x4 + x 2 2(n + 1) 8( n + 1)( n + 3) 48( n + 1 ) (n + 3)(n + 5) (6n 2 + 32n + 34)m 2 − ( 97n 2 + 46n + 630m ) + (2n 2 + 16n + 30) 2

+

384(n + 1)3 (n+ 3)(n + 5)(n + 7)

For a fixed n = 0 and m = 0,1,2, we obtain the following solutions respectively:

x8 +

obtained by implementing the inverse transformation. The suggested algorithm is tested on Emden-Fowler equation, white-dwarf equation, generalized LaneEmden equation, second order nonlinear singular ODES. Acknowledgment The authors are highly grateful to Dr S. M. Junaid Zaidi Rector CIIT for the provision of excellent research facilities and environment.

1 2  y(x)= 1 − 2 x ,  y(x) = cosx,  1 2 1 4 1 6 1 8 y(x) = 1− x + x − x + x + 2 12 72 504 

REFERENCES

where x = 0 is just an ordinary point. Similarly, for fixed n = 0.5 and m = 0,1,2, we get the following solutions respectively:

1.

1 2  y(x)= 1 − 3 x ,  1 1 4 1 6 1  x 8 + , 1 − x2 + x − x + y(x) = 3 42 1386 83160  1 2 1 4 13 6 23 8  x + . 1− x + x − x + y(x) = 3 21 2079 31185 

2. 3. 4.

Finally for n = 1 and m = 0,1,2, we obtain the following solutions respectively:

5.

1 2  y(x)= 1 − 4 x ,  1 1 1 1  x 8 + , 1 − x2 + x4 − x6 + y(x) = 4 64 2304 147456  1 2 1 4 1 6 13  8 1− x + x − x + x + . y(x) = 4 32 288 36864 

6.

7.

CONCLUSION In this paper, we use He’s polynomials coupled with a transformation which is helpful to cope with the singularity behavior for solving various singular differential equations. The desired series solution is

8.

774

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