Application of homotopy perturbation method for

Applied Mathematics and Computation 191 (2007) 334–346 www.elsevier.com/locate/amc

Application of homotopy perturbation method for solving eighth-order boundary value problems A. Golbabai, M. Javidi

*

Department of Mathematics, Iran University of Science and Technology, Narmak, Tehran 16844, Iran

Abstract Homotopy perturbation method is applied to the numerical solution for solving eighth-order boundary value problems. Comparison of the result obtained by the present method with that obtained by modified decomposition method [M. Mesrovic, The modified decomposition method for eighth-order boundary value problems, Appl. Math. Comput. (2006), doi:10.1016/j.amc.2006.11.015] reveals that the present method is very effective and convenient. Ó 2007 Elsevier Inc. All rights reserved. Keywords: Eighth-order boundary value problems; Homotopy perturbation method

1. Introduction A new perturbation method called homotopy perturbation method (HPM) was proposed by He in 1997 and systematical description in 2000 which is, in fact, a coupling of the traditional perturbation method and homotopy in topology [1]. This new method was further developed and improved by He and applied to nonlinear oscillators with discontinuities [2], nonlinear wave equations [3], asymptotology [4], boundary value problem [5], Limit cycle and bifurcation of nonlinear problems [6] and many other subjects. Thus He’s method is a universal one which can solve various kinds of nonlinear equations. For example, it was applied to the quadratic Ricatti differential equation by Abbasbandy [7]; to the axisymmetric flow over a stretching sheet by Ariel et al. [8]; to the nonlinear systems of reaction–diffusion equations by Ganji and Sadighi [9]; to the Helmholtz equation and fifth-order KdV equation by Rafei and Ganji [10]; for the thin film flow of a fourth grade fluid down a vertical cylinder by Siddiqui et al. [11]; to the nonlinear Voltra–Fredholm integral equations by Ghasemi et al. [12]. Recently various powerful mathematical methods such as variational iteration method [13–21], Exp-function method [22,23], F-expansion method [24], Adomian decomposition method [25] and others [26,27] have been proposed to obtain exact and approximate analytic solutions for linear and nonlinear problems.

*

Corresponding author. E-mail address: [email protected] (M. Javidi).

0096-3003/$ - see front matter Ó 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2007.02.091

A. Golbabai, M. Javidi / Applied Mathematics and Computation 191 (2007) 334–346

335

In this paper, we consider the general eighth-order boundary value problems of the type: y ðviiiÞ ðxÞ ¼ f ðx; y; y 0 ; y 00 ; y 000 ; y ðivÞ ; y ðvÞ ; y ðviÞ ; y ðviiÞ Þ;

a < x < b;

ð1Þ

with boundary conditions yðaÞ ¼ a0 ; y ðivÞ ðaÞ ¼ a4 ;

y 0 ðaÞ ¼ a1 ;

y 00 ðaÞ ¼ a2 ;

y ðvÞ ðaÞ ¼ a5 ;

y 000 ðaÞ ¼ a3 ;

yðbÞ ¼ b0 ;

y 0 ðbÞ ¼ b1 ;

ð2Þ

where f is continuous function on [a,b] and the parameters ai ; i ¼ 0; 1; . . . ; 6 and bi ; i ¼ 0; 1 are real constants. Such type of boundary value problems arise in the mathematical modeling of the viscoelastic flows and other branches of mathematical, physical and engineering sciences, see [28–31] and references therein. Several numerical methods including spectral Galerkin and collocation [29,30], sixth B-spline method [28], decomposition method [32], spline collocation approximation [33], Chow–Yorke algorithm [34] and others [35,36] have been developed for solving the problem of type (1). This paper, applies the homotopy perturbation method [1– 12] to the discussed problem. 2. Homotopy perturbation method Using the transformation dy d2 y d3 y ¼ y2; ¼ y ; ¼ y4; 3 dx dx2 dx3 ð3Þ d4 y d5 y d6 y d7 y ¼ y5; ¼ y6; ¼ y7; ¼ y8; dx4 dx5 dx6 dx7 we can rewrite the eighth-order boundary value problem (1), (2) as the system of ordinary differential equations: 8 dy 1 ¼ y2; > > dx > > > dy 2 > ¼ y3; > dx > > > dy 3 > > ¼ y4; > dx > > > < dy 4 ¼ y ; 5 dx ð4Þ dy 5 > ¼ y > 6; dx > > > dy 6 > > ¼ y7; > dx > > > dy 7 > > ¼ y8; > dx > > : dy 8 ¼ f ðx; y 1 ; y 2 ; y 3 ; y 4 ; y 5 ; y 6 ; y 7 Þ; dx y 1 ¼ y;

with the boundary conditions y 1 ðaÞ ¼ a0 ;

y 2 ðaÞ ¼ a1 ;

y 3 ðaÞ ¼ a2 ;

y 4 ðaÞ ¼ a3 ;

y 5 ðaÞ ¼ a4 ;

y 6 ðaÞ ¼ a5 ;

y 1 ðbÞ ¼ b0 ;

y 2 ðbÞ ¼ b1 ;

which can be written as a system of integral equations: 8 Rx y 1 ¼ a0 þ 0 y 2 ðtÞ dt; > > > Rx > > > > y 2 ¼ a1 þ R0 y 3 ðtÞ dt; > > > > y 3 ¼ a2 þ 0x y 4 ðtÞ dt; > > > < y ¼ a þ R x y ðtÞ dt; 3 4 R0x 5 > y ¼ a þ y ðtÞ dt; 4 > 5 > R0x 6 > > > y ¼ a5 þ 0 y 7 ðtÞ dt; > > Rx > 6 > > > > y 7 ¼ A þ R0 y 8 ðtÞ dt; > x : y 8 ¼ B þ 0 f ðt; y 1 ðtÞ; y 2 ðtÞ; y 3 ðtÞ; y 4 ðtÞ; y 5 ðtÞ; y 6 ðtÞ; y 7 ðtÞÞ dt;

ð5Þ

ð6Þ

336


where y 7 ðaÞ ¼ A;

y 8 ðaÞ ¼ B:

To explain HPM, we consider (6) as 2 3 L1 ðy 1 ; y 2 ; . . . ; y 8 Þ 6 7 .. 7 ¼ 0; Lðy 1 ; y 2 ; . . . ; y 8 Þ ¼ 6 . 4 5 L8 ðy 1 ; y 2 ; . . . ; y 8 Þ

ð7Þ

with solution ðf1 ; f2 ; . . . ; f8 Þ where Rx 8 L1 ðy 1 ; y 2 ; . . . ; y 8 Þ ¼ y 1 a0 0 y 2 ðtÞ dt; > > < .. . > > Rx : L8 ðy 1 ; y 2 ; . . . ; y 8 Þ ¼ y 8 B 0 f ðt; y 1 ðtÞ; y 2 ðtÞ; y 3 ðtÞ; y 4 ðtÞ; y 5 ðtÞ; y 6 ðtÞ; y 7 ðtÞÞ dt:

ð8Þ

We can define homotopy Hðy 1 ; y 2 ; . . . ; y 8 ; pÞ by Hðy 1 ; y 2 ; . . . ; y 8 ; 0Þ ¼ Fðy 1 ; y 2 ; . . . ; y 8 Þ;

Hðy 1 ; y 2 ; . . . ; y 8 ; 1Þ ¼ Lðy 1 ; y 2 ; . . . ; y 8 Þ;

where T

Fðy 1 ; y 2 ; . . . ; y 8 Þ ¼ ½F 1 ðy 1 ; y 2 ; . . . ; y 8 Þ; . . . ; F 8 ðy 1 ; y 2 ; . . . ; y 8 Þ T

ð9Þ

¼ ½y 1 a0 ; . . . ; y 7 A; y 8 B ; T

Hðy 1 ; y 2 ; . . . ; y 8 ; pÞ ¼ ½H 1 ðy 1 ; y 2 ; . . . ; y 8 ; pÞ; . . . ; H 8 ðy 1 ; y 2 ; . . . ; y 8 ; pÞ : Typically we may choose a convex homotopy by Hðy 1 ; y 2 ; . . . ; y 8 ; pÞ ¼ ð1 pÞFðy 1 ; y 2 ; . . . ; y 8 Þ þ pLðy 1 ; y 2 ; . . . ; y 8 Þ ¼ 0:

ð10Þ

The convex homotopy (10) continuously trace an implicitly defined curve from a starting point Hðy 1 a0 ; . . . ; y 7 A; y 8 B; 0Þ ¼ 0 to a solution function Hðy 1 ; . . . ; y 7 ; y 8 ; 1Þ ¼ 0. The embedding parameter p monotonically increases from zero to unit as trivial problem Fðy 1 ; y 2 ; . . . ; y 8 Þ ¼ 0 is continuously deformed to original problem Lðy 1 ; y 2 ; . . . ; y 8 Þ ¼ 0. The HPM uses the homotopy parameter p as an expanding parameter to obtain [16]: y 1 ¼ y 10 þ py 11 þ p2 y 12 þ ; .. .

ð11Þ

y 8 ¼ y 80 þ py 81 þ p2 y 82 þ ; when p ! 1, (8) corresponds to f1 ¼ lim y 1 ¼ y 10 þ y 11 þ y 12 þ ; p!1

.. .

ð12Þ

f8 ¼ lim y 8 ¼ y 80 þ y 81 þ y 82 þ : p!1

For the application of HPM to (6) we can write (10) as follows: 8 H 1 ðy 1 ; y 2 ; . . . ; y 8 ; pÞ ¼ ð1 pÞF 1 ðy 1 ; y 2 ; . . . ; y 8 Þ þ pL1 ðy 1 ; y 2 ; . . . ; y 8 Þ ¼ 0; > > < .. . > > : H 8 ðy 1 ; y 2 ; . . . ; y 8 ; pÞ ¼ ð1 pÞF 8 ðy 1 ; y 2 ; . . . ; y 8 Þ þ pL8 ðy 1 ; y 2 ; . . . ; y 8 Þ ¼ 0:

ð13Þ


Substitution of (8), (9) and (11) into (13) yields Rx 8 ð1 pÞðy 1 a0 Þ þ p y 1 a0 0 y 2 ðtÞ dt ¼ 0; > > < .. . > > Rx : ð1 pÞðy 8 BÞ þ p y 8 B 0 f ðt; y 1 ðtÞ; y 2 ðtÞ; y 3 ðtÞ; y 4 ðtÞ; y 5 ðtÞ; y 6 ðtÞ; y 7 ðtÞÞ dt ¼ 0:

337

ð14Þ

By equating the terms with identical powers of p, we have 8 y 10 > > > > > . > > y 70 > > : y 70 8 y 11 > > > > > > < .. . p1 : > > > > y 71 > > : y 81 8 > y > > 12 > > > . > < .. p2 : > > y 72 > > > > > :y 81

¼ a0 ;

ð15Þ ¼ A; ¼ B ! y 80 ¼ B Rx ¼ 0 y 20 ðtÞ dt;

¼ ¼

ð16Þ

Rx

y ðtÞ dt; 0 80 Rx d f ðt; y 1 ðtÞ; y 2 ðtÞ; y 3 ðtÞ; y 4 ðtÞ; y 5 ðtÞ; y 6 ðtÞ; y 7 ðtÞÞjp¼0 dt; 0 dp

¼

Rx

¼

Rx

0

0

¼ 2!1

y 21 ðtÞ dt;

ð17Þ

y 81 ðtÞ dt; R x d2 Rx f ðt; y 1 ðtÞ; y 2 ðtÞ; y 3 ðtÞ; y 4 ðtÞ; y 5 ðtÞ; y 6 ðtÞ; y 7 ðtÞÞjp¼0 dt ! y 82 ¼ 2!1 0 0 dp2

d2 dp2

f ðt; y 1 ðtÞ; y 2 ðtÞ; y 3 ðtÞ; y 4 ðtÞ; y 5 ðtÞ; y 6 ðtÞ; y 7 ðtÞÞjp¼0 dt;

.. . Rx 8 y 1;n ¼ 0 y 2;n1 ðtÞ dt; > > > > > > < .. . n p : Rx > > y 7;n ¼ 0 y 8;n1 ðtÞ dt; > > > > R x dn : y 8;n ¼ n!1 0 dp n f ðt; y 1 ðtÞ; y 2 ðtÞ; y 3 ðtÞ; y 4 ðtÞ; y 5 ðtÞ; y 6 ðtÞ; y 7 ðtÞÞjp¼0 dt:

ð18Þ

Combining all the terms: (15)–(18) gives the solution of the problem. Using the boundary conditions y 1 ðbÞ ¼ b0 and y 2 ðbÞ ¼ b1 we can obtained A and B. 3. Applications In this section, in order to verify numerically whether the proposed methodology leads to higher accuracy, we evaluate the numerical solution of the problem (1). To show the efficiency of the present method for our problem in comparison with the exact solution we report absolute error which is defined by Ey N ðxÞ ¼ jy Exact ðxÞ y N ðxÞj; P where y N ðxÞ ¼ Nm¼0 y 1m for N = 0, 1, 2, . . .. Example 1. Consider the following linear eighth-order problem [32] y ðviiiÞ ðxÞ ¼ 8ex þ yðxÞ;

ð19Þ

0 < x < 1;

with the following boundary conditions: yð0Þ ¼ 1;

y 0 ð0Þ ¼ 0;

y ðivÞ ð0Þ ¼ 3;

y 00 ð0Þ ¼ 1;

y ðvÞ ð0Þ ¼ 4;

y 000 ð0Þ ¼ 2;

y ðviÞ ð0Þ ¼ 5;

y ðviiÞ ð0Þ ¼ 6:

ð20Þ

338


The analytic solution is yðxÞ ¼ ð1 xÞex : Using the transformation (3) we can rewrite the eighth-order boundary value problem (19) as the system integral equations Rx 8 y 1 ¼ 1 þ 0 y 2 ðtÞ dt; > > > Rx > > > y 2 ¼ 0 y 3 ðtÞ dt; > > > Rx > > > y 3 ¼ 1 þ 0 y 4 ðtÞ dt; > > > > R > < y 4 ¼ 2 þ x y 5 ðtÞ dt; 0 ð21Þ Rx > y ¼ 3 þ y 6 ðtÞ dt; > 5 > 0 > > Rx > > > y 6 ¼ 4 þ 0 y 7 ðtÞ dt; > > > Rx > > > y 7 ¼ 5 þ 0 y 8 ðtÞ dt; > > > Rx : y 8 ¼ 6 þ 0 ð8et þ y 1 ðtÞÞ dt: Using (14) for (21) we have Rx 8 y 10 þ py 11 þ p2 y 12 þ ¼ 1 þ p 0 ðy 20 þ py 21 þ p2 y 22 þ Þ dt; > > > Rx > > > < y 20 þ py 21 þ p2 y 22 þ ¼ p 0 ðy 30 þ py 31 þ p2 y 32 þ Þ dt; .. > > > . > > > Rx : y 80 þ py 81 þ p2 y 82 þ ¼ 6 þ p 0 ð8et þ y 10 þ py 11 þ p2 y 12 þ Þ dt: Comparing the coefficient of like powers of p, we have: 8 y 10 ¼ 1; > > > > > > > > y 20 ¼ 0; > > > > y 30 ¼ 1; > > > > > < y 40 ¼ 2; p0 : > y 50 ¼ 3; > > > > > > > y 60 ¼ 4; > > > > > > y 70 ¼ 5; > > > : y 80 ¼ 6; 8 y 11 ¼ 0; > > > > > > y 21 ¼ x; > > > > > > y 31 ¼ 2x; > > > > > < y 41 ¼ 3x; p1 : > > > y 51 ¼ 4x; > > > > > y 61 ¼ 5x; > > > > > > y 71 ¼ 6x; > > > : y 81 ¼ 8ex þ x þ 8;

ð22Þ

ð23Þ

ð24Þ


8 > y > > 12 > > > > y 22 > > > > > > y 32 > > > > > y 52 > > > > > y 62 > > > > > > y 72 > > > > : y 82

339

¼ 12 x2 ; ¼ x2 ; ¼ 32 x2 ; ¼ 2x2 ;

ð25Þ

¼ 52 x2 ; ¼ 3x2 ; ¼ 8ex þ 12 x2 þ 8x þ 8; ¼ 0;

.. .

p3 :

8 y 13 > > > > > > y 23 > > > > > > y 33 > > > > > < y 43 > y 53 > > > > > > y 63 > > > > > > y 73 > > > > : y 83

¼ 13 x3 ; ¼ 12 x3 ; ¼ 23 x3 ; ¼ 56 x3 ;

ð26Þ

¼ x3 ; ¼ 8ex þ 16 x3 þ 4x2 þ 8x þ 8; ¼ 0; ¼ 16 x3 :

Combining all the terms: (23)–(26) gives 7 5 1 1 6 1 7 1 y 8 ðxÞ ¼ 9 þ x2 þ x3 þ x4 þ x5 þ x þ x 8ex þ x8 þ 8x: 2 24 30 240 2520 40320 The numerical results obtained in Table 1. In Table 1, we list the results obtained by homotopy perturbation method and compared with modified decomposition method (MDM) results given in [32] at x = 0.25(0.25)1. As we see from this Table, it is clear that the result obtained by the present method is very superior to that obtained by MDM method. As can be seen from Table 1, the error decreased when the integer N is increased until N = 8. Example 2. Consider the following nonlinear eighth-order problem [32]: y ðviiiÞ ðxÞ ¼ ex y 2 ðxÞ;

ð27Þ

0 < x < 1;

with the following boundary conditions: y ðiÞ ð0Þ ¼ 1;

i ¼ 0; 1; . . . ; 7:

ð28Þ

Table 1 Numerical results for Example 1 x

0.25 0.50 0.75 1.00

Error HPM, N = 2

HPM, N = 4

HPM, N = 8

MDM

0.0057 0.0506 0.1895 0.5000

3.4323e5 0.0012 0.0093 0.0417

2.7611e13 2.9519e10 1.7849e8 3.3265e7

1.0e7 3.1e6 5.5e5 4.2e4

340


The analytic solution is yðxÞ ¼ ex : Using the transformation (3) we can rewrite the eighth-order boundary value problem (27) as the system integral equations 8 Rx y 1 ¼ 1 þ 0 y 2 ðtÞ dt; > > > Rx > > > y 2 ¼ 1 þ 0 y 3 ðtÞ dt; > > Rx > > > y 3 ¼ 1 þ 0 y 4 ðtÞ dt; > > Rx > < y 4 ¼ 1 þ 0 y 5 ðtÞ dt; Rx ð29Þ > y 5 ¼ 1 þ 0 y 6 ðtÞ dt; > > Rx > > > y 6 ¼ 1 þ 0 y 7 ðtÞ dt; > > Rx > > > y 7 ¼ 1 þ 0 y 8 ðtÞ dt; > > > Rx : y 8 ¼ 1 þ 0 ðet y 21 ðtÞÞ dt: Using (14) for (29) we have 8 Rx y 10 þ py 11 þ p2 y 12 þ ¼ 1 þ p 0 ðy 20 þ py 21 þ p2 y 22 þ Þ dt; > > > Rx > > < y 20 þ py 21 þ p2 y 22 þ ¼ 1 þ p 0 ðy 30 þ py 31 þ p2 y 32 þ Þ dt; .. > > . > > > Rx : y 80 þ py 81 þ p2 y 82 þ ¼ 1 þ p 0 et ðy 10 þ py 11 þ p2 y 12 þ Þ2 dt:

ð30Þ

Comparing the coefficient of like powers of p, we have: p0 : y i;0 ¼ 1; i ¼ 1; 2; . . . ; 8; y i;1 ¼ x; i ¼ 1; 2; . . . ; 7; 1 p : y ¼ ex þ 1; 8 81 1 2 > < y i;2 ¼ 2 x ; i ¼ 1; 2; . . . ; 6; p2 : y 72 ¼ x þ ex 1; > : y 82 ¼ 2ex x 2ex þ 2; .. . 8 y i;3 ¼ 13 x3 ; i ¼ 1; 2; . . . ; 5; > > > < y ¼ 1 x2 ex x þ 1; 63 2 p3 : > y ¼ 2ex x þ 4ex þ 2x 4; > 73 > : y 83 ¼ 2ex x2 4ex x 4ex þ 4:

ð31Þ ð32Þ

ð33Þ

ð34Þ

Combining all the terms: (31)–(34) gives 1 1 1 1 5 1 6 1 7 y 7 ðxÞ ¼ 1 þ x þ x2 þ x3 þ x4 þ x þ x þ x: 2 6 24 120 720 5040 The numerical results obtained in Table 2. In Table 2, we list the results obtained by homotopy perturbation method and compared with modified decomposition method (MDM) results given in [32] at Table 2 Numerical results for Example 2 x

Error HPM, N = 3

HPM, N = 5

HPM, N = 7

MDM

0.25 0.50 0.75

0.0314 0.1279 0.2967

1.6311e4 0.0026 0.0135

3.8922e10 1.0255e7 2.7065e6

4.91e5 7.04e5 4.98e5


341

x = 0.25(0.25)0.75. As we see from this Table, it is clear that the result obtained by the present method is very superior to that obtained by MDM method. As can be seen from Table 2, the error decreased when the integer N is increased until N = 7. Example 3. Consider the following linear eighth-order problem [32]: y ðviiiÞ ðxÞ ¼ 8ex þ yðxÞ;

ð35Þ

0 < x < 1;

with the following boundary conditions: yð0Þ ¼ 1;

y 0 ð0Þ ¼ 0;

y ðivÞ ð0Þ ¼ 3;

y 00 ð0Þ ¼ 1;

y ðvÞ ð0Þ ¼ 4;

y 000 ð0Þ ¼ 2;

y 0 ð1Þ ¼ e;

y 00 ð1Þ ¼ 2e:

ð36Þ

The analytic solution is yðxÞ ¼ ð1 xÞex : Using the transformation (3) we can rewrite the eighth-order boundary value problem (19) as the system integral equations Rx 8 y 1 ¼ 1 þ 0 y 2 ðtÞ dt; > > > Rx > > > y 2 ¼ 0 y 3 ðtÞ dt; > > > Rx > > > y 3 ¼ 1 þ 0 y 4 ðtÞ dt; > > > R > < y ¼ 2 þ x y ðtÞ dt; 4 0 5 ð37Þ Rx > y 5 ¼ 3 þ 0 y 6 ðtÞ dt; > > > Rx > > > y 6 ¼ 4 þ 0 y 7 ðtÞ dt; > > > Rx > > > ¼ A þ y ðtÞ dt; y > 7 0 8 > > Rx : y 8 ¼ B þ 0 ð8et þ y 1 ðtÞÞ dt: Using (14) for (37) we have Rx 8 y 10 þ py 11 þ p2 y 12 þ ¼ 1 þ p 0 ðy 20 þ py 21 þ p2 y 22 þ Þ dt; > > R > x > > y þ py 21 þ p2 y 22 þ ¼ p 0 ðy 30 þ py 31 þ p2 y 32 þ Þ dt; > > < 20 .. . > > Rx > > > y 70 þ py 71 þ p2 y 72 þ ¼ A þ p 0 ðy 80 þ py 81 þ p2 y 82 þ Þ dt; > > Rx : y 80 þ py 81 þ p2 y 82 þ ¼ B þ p 0 ð8et þ y 10 þ py 11 þ p2 y 12 þ Þ dt: Comparing the coefficient of like powers of p, we have: 8 y 10 ¼ 1; > > > > > y 20 ¼ 0; > > > > > > > > y 30 ¼ 1; > > < y ¼ 2; 40 0 p : > y 50 ¼ 3; > > > > > > y 60 ¼ 4; > > > > > y 70 ¼ A; > > > : y 80 ¼ B;

ð38Þ

ð39Þ

342


8 y 11 > > > > > y 21 > > > > > y 31 > > > > y 51 > > > > > > y 61 > > > > > y 71 > > : y 81 8 y 12 > > > > > y 22 > > > > > y > > > 32 > y > 52 > > > > > y 62 > > > > > y 72 > > : y 82 .. . 8 y 13 > > > > > y 23 > > > > > > > > y 33 > > y 53 > > > > > > y 63 > > > >y > > 73 > > : y 83

¼ 0; ¼ x; ¼ 2x; ¼ 3x;

ð40Þ

¼ 4x; ¼ Ax; ¼ Bx; ¼ 8ex þ x þ 8; ¼ 12 x2 ; ¼ x2 ; ¼ 32 x2 ; ¼ 2x2 ;

ð41Þ

¼ 12 Ax2 ; ¼ 12 Bx2 ; ¼ 8ex þ 12 x2 þ 8x þ 8; ¼ 0;

¼ 13 x3 ; ¼ 12 x3 ; ¼ 23 x3 ; ¼ 16 Ax3 ;

ð42Þ

¼ 16 Bx3 ; ¼ 8ex þ 16 x3 þ 4x2 þ 8x þ 8; ¼ 0; ¼ 16 x3 :

Combining all the terms: (39)–(42) gives 7 5 1 1 1 1 1 7 1 y 8 ðxÞ ¼ 9 þ x2 þ x3 þ x4 þ x5 þ Ax6 þ Bx7 8ex þ x8 þ x þ x6 þ 8x: 2 24 30 720 5040 40320 630 90

ð43Þ

Using the boundary conditions y 0 ð1Þ ¼ e and y 00 ð1Þ ¼ 2e leads to the system for unknown coefficients DX ¼ b; where

" D¼

and

" b¼

1 120 1 24

1 720 1 120

# ð44Þ

# þ 8e 2:7183 32051 1680 ; 5:4366 11929 þ 8e 720

X ¼ ½A; BT :

ð45Þ

The solution of this system is A ¼ 5:0074;

B ¼ 5:9710:

ð46Þ


343


Error

0.25 0.50 0.75 1.00

HPM, N = 6

HPM, N = 7

HPM, N = 8

7.2883e8 9.8635e6 1.7626e4 0.0014

5.7238e9 6.0777e7 1.8121e5 1.9386e4

2.1630e9 1.1571e7 1.0479e6 4.2188e6

The numerical results were obtained in Table 3. In Table 3, we list the results obtained by homotopy perturbation method at x = 0.25(0.25)1. As can be seen from Table 3, the error decreased when the integer N is increased until N = 8. Example 4. One of the problems considered by Scott and Watts [37] is the linear fourth-order problem 1 y ð4Þ ðxÞ ðc þ 1Þy 00 ðxÞ þ cyðxÞ ¼ cx2 1 ¼ gðxÞ; 2

0 < x < 1;

ð47Þ

subject to the boundary conditions yð0Þ ¼ 1; y 0 ð0Þ ¼ 1;

3 þ sinhð1Þ; 2 y 0 ð1Þ ¼ 1 þ coshð1Þ:

yð1Þ ¼

ð48Þ

The analytic solution is 1 yðxÞ ¼ 1 þ x2 þ sinhðxÞ: 2 This is an interesting problem and has some unusual behavior. If we regroup (47) in terms of c we get 1 2 ð4Þ 00 00 ½y ðxÞ y ðxÞ þ 1 c y ðxÞ yðxÞ þ x ¼ 0: ð49Þ 2 We can rewrite (49) as follows: 00 1 1 y 00 ðxÞ yðxÞ þ x2 c y 00 ðxÞ yðxÞ þ x2 ¼ 0: 2 2

ð50Þ

The solution of fourth-order problem is also a solution of 1 y 00 ðxÞ yðxÞ þ x2 ¼ 0: 2

ð51Þ

This results in the solution of the original problem being independent of constant c. However, for procedures which solve initial value problems, the solutions of the intermediate problems are not independent of c. Hence for large values of c, any procedure solving initial value problems will have difficulty. Using the transformation (3) we can rewrite the fourth-order boundary value problem (47) and (48) as the system integral equations Rx 8 y 1 ¼ 1 þ 0 y 2 ðtÞ dt; > > > < y ¼ 1 þ R x y ðtÞ dt; 2 R0x 3 ð52Þ > y ¼ A þ y ðtÞ dt; > 3 0 4 > Rx : y 4 ¼ B þ 0 ððc þ 1Þy 3 ðtÞ cy 1 ðtÞ þ gðtÞÞ dt:

344


Using (14) for (49) we have 8 Rx y 10 þ py 11 þ p2 y 12 þ ¼ 1 þ p 0 ðy 20 þ py 21 þ p2 y 22 þ Þ dt; > > > Rx > 2 2 > > < y 20 þ py 21 þ p y 22 þ ¼ 1 þ p R0 ðy 30 þ py 31 þ p y 32 þ Þ dt; x y 30 þ py 31 þ p2 y 32 þ ¼ A þ p 0 ðy 40 þ py 41 þ p2 y 42 þ Þ dt; > Rx > > > y 40 þ py 41 þ p2 y 42 þ ¼ B þ p 0 ððc þ 1Þðy 30 þ py 31 þ p2 y 32 þ Þ > > : cðy 10 þ py 11 þ p2 y 12 þ Þ þ gðtÞÞ dt: Comparing the coefficient of like powers of p, we have: 8 y 10 ¼ 1; > > > < y ¼ 1; 20 p0 : > y > 30 ¼ A; > : y 40 ¼ B; 8 y 11 ¼ x; > > > < y ¼ Ax; 21 p1 : > y > 31 ¼ Bx; > : y 41 ¼ Acx þ Ax cx x þ 16 cx3 ; 8 y 12 ¼ 12 12 Ax2 ; > > > < y 22 ¼ 12 12 Bx2 ; p2 : > y 32 ¼ 121 241 þ 12 ðAc þ A c 1Þx2 ; > > : y 42 ¼ 12 ððc þ 1Þb cÞx2 ; .. . 8 y 13 ¼ 16 Bx3 ; > > > < 1 y 23 ¼ 120 cx5 þ 13 12 Ac þ 12 A 12 c 12 x3 ; 3 p : > y 33 ¼ 13 12 ðc þ 1ÞB 12 c x3 ; > > : 1 y 43 ¼ 120 ðc þ 1Þcx5 þ 13 ðc þ 1Þ 12 Ac þ 12 A 12 c 12 12 Ac x3 :

ð53Þ

ð54Þ

ð55Þ

ð56Þ

ð57Þ

Combining all the terms: (51)–(54) gives

1 1 1 1 1 1 1 1 1 1 1 y 6 ðxÞ ¼ 1 þ x þ Ax2 þ Bx3 þ cx6 þ Ac þ A c x4 þ ðc þ 1ÞB c x5 2 6 720 4 6 6 6 6 5 24 24 1 1 1 1 1 1 1 1 ðc þ 1Þcx8 þ ðc þ 1Þ Ac þ A c Ac x6 : þ 40320 6 60 2 2 2 2 120

ð58Þ

Using the boundary conditions yð1Þ ¼ 2 þ sinhð1Þ and y 0 ð1Þ ¼ 2 þ coshð1Þ leads to the system for unknown coefficients DX ¼ b; where

" D¼

13 24

29 1 þ 720 c þ 720 ðc þ 1Þ2

1 6

1 þ 120 ðc þ 1Þ

7 6

19 1 þ 120 c þ 120 ðc þ 1Þ2

1 2

þ 241 ðc þ 1Þ

# ð59Þ

and 2

3 62

b¼4

þ sinhð1Þ

n

23 24

n

2

o3

2

7 o5

7 1 1 144 c þ 40320 ðc þ 1Þc 720 ðc þ 1Þ

1 1 2 þ coshð1Þ 16 15 c þ 5040 ðc þ 1Þc 120 ðc þ 1Þ

ð60Þ


345


Error HPM, N = 5

HPM, N = 10

HPM, N = 14

c=1 0.25 0.5 0.75

N=5 3.2696e7 2.0146e5 2.2050e4

N = 10 2.4034e13 1.0804e11 7.9793e10

N = 14 2.3484e13 5.9036e13 5.5106e13

c = 100 0.25 0.50 0.75

1.4046e4 0.0034 0.0397

2.0768e4 5.0700e4 0.0017

2.0763e4 4.4177e4 6.3589e5

and X ¼ ½A; BT :

ð61Þ

We can solve this system to obtain A and B. The numerical results obtained in Table 4. In Table 4, we list the results obtained by homotopy perturbation method at x = 0.25(0.25)0.75 with c = 1 and c = 100. This results in the solution of the original problem being independent of constant c. However, As can be seen from Table 4, for procedures which solve initial value problems, the solutions of the intermediate problems are not independent of c. Hence for large values of c, any procedure solving initial value problems will have difficulty. 4. Conclusions Homotopy perturbation method is applied to the numerical solution for solving eighth-order boundary value problems. Comparison of the result obtained by the present method with that obtained by modified decomposition method [32] reveals that the present method is very effective and convenient. References [1] J.H. He, A coupling method of a homotopy technique and a perturbation technique for non-linear problems, Int. J. Non-Linear Mech. 35 (1) (2000) 37–43. [2] J.H. He, The homotopy perturbation method for non-linear oscillators with discontinuities, Appl. Math. Comput. 151 (1) (2004) 287– 292. [3] J.H. He, Application of homotopy perturbation method to nonlinear wave equations, Chaos, Solitons, Fractals 26 (3) (2005) 695–700. [4] J.H. He, Asymptotology by homotopy perturbation method, Appl. Math. Comput. 156 (3) (2004) 591–596. [5] J.H. He, Homotopy perturbation method for solving boundary problems, Phys. Lett. A 350 (1–2) (2006) 87–88. [6] J.H. He, Limit cycle and bifurcation of nonlinear problems, Chaos, Solitons, Fractals 26 (3) (2005) 827–833. [7] S. Abbasbandy, Iterated He’s homotopy perturbation method for quadratic Riccati differential equation, Appl. Math. Comput. 175 (1) (2006) 581–589. [8] P.D. Ariel, T. Hayat, S. Asghar, Homotopy perturbation method and axisymmetric flow over a stretching sheet, Int. J. Nonlinear Sci. Numer. Simul. 7 (4) (2006) 399–406. [9] D.D. Ganji, A. Sadighi, Application of He’s homotopy-perturbation method to nonlinear coupled systems of reaction–diffusion equations, Int. J. Nonlinear Sci. Numer. Simul. 7 (4) (2006) 411–418. [10] M. Rafei, D.D. Ganji, Explicit solutions of Helmholtz equation and fifth-order KdV equation using homotopy perturbation method, Int. J. Nonlinear Sci. Numer. Simul. 7 (3) (2006) 321–328. [11] A.M. Siddiqui, R. Mahmood, Q.K. Ghori, Homotopy perturbation method for thin film flow of a fourth grade fluid down a vertical cylinder, Phys. Lett. A 352 (4–5) (2006) 404–410. [12] M. Ghasemi et al., Numerical solution of the nonlinear Voltra–Fredholm integral equations by using homotopy perturbation method, Appl. Math. Comput. (2006), doi:10.1016/j.amc.2006.10.015. [13] J.H. He, Approximate solution of nonlinear differential equations with convolution product nonlinearities, Comput. Meth. Appl. Mech. Eng. 167 (12) (1998) 6973.

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