Solving singular boundary value problems of higher-order ordinary differential equations by modified Adomian decomposition method

Commun Nonlinear Sci Numer Simulat 14 (2009) 2592–2596

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Solving singular boundary value problems of higher-order ordinary differential equations by modified Adomian decomposition method q Yahya Qaid Hasan, Liu Ming Zhu * Department of Mathematics, Harbin Institute of Technology, 11, Siling Street, Harbin, Helongjiang 150001, PR China

a r t i c l e

i n f o

Article history: Received 28 June 2008 Received in revised form 30 September 2008 Accepted 30 September 2008 Available online 10 October 2008

a b s t r a c t In this paper, we use modified Adomian decomposition method to solving singular boundary value problems of higher-order ordinary differential equations. The proposed method can be applied to linear and nonlinear problems. The scheme is tested for some examples and the obtained results demonstrate efficiency of the proposed method. Ó 2008 Elsevier B.V. All rights reserved.

PACS: 02.60.Lj Keywords: Adomian decomposition method Singular boundary value problems Higher-order ordinary differential equation

1. Introduction The decomposition method has been shown [2–4,8–14] to solve effectively, easily and accurately a large class of linear and nonlinear, ordinary, partial, deterministic or stochastic differential equations with approximate solutions which converge rapidly to accurate solutions. The goal of this paper is to introduce a new reliable modification of Adomian decomposition method. For this reason, a new differential operator is proposed which can be used for singular boundary value problem. In this paper we find solutions without massive computations and restrictive assumptions which change the physics problem into a mathematically tractable problem. This method is easy than the traditional methods. The theoretical treatment of the Convergence of Adomian’s method has been considered in[1,5–7]. 2. Modified Adomian decomposition method Consider the singular boundary value problem of n þ 1 order ordinary differential equation in the form

m ðnÞ y þ Ny ¼ gðxÞ; x yð0Þ ¼ a0 ; y0 ð0Þ ¼ a1 ; . . . ; yðn1Þ ð0Þ ¼ an1 ; yðbÞ ¼ c; yðnþ1Þ þ

q

This paper is supported by National Natural Science Foundation of China (10471067). * Corresponding author. Tel.: +68 451 86 403055; fax: +86 451 86403055. E-mail addresses: [email protected] (Y.Q. Hasan), [email protected] (L.M. Zhu).

1007-5704/$ - see front matter Ó 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.cnsns.2008.09.027

ð1Þ

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where N is a nonlinear differential operator of order less than n, gðxÞ is given function and a0 ; a1 ; . . . ; an1 ; c; b are given constants. We propose the new differential operator, as below n

Lð:Þ ¼ x1

d 1þnm d mn x ð:Þ; nx dx dx

ð2Þ

where m 6 n; n P 1, so, the problem (1) can be written as

Ly ¼ gðxÞ  Ny:

ð3Þ 1

The inverse operator L

L1 ð:Þ ¼ xnm

Z

is therefore considered a n þ 1-fold integral operator, as below

x

xmn1

b

Z

x

Z

Z

x

x

xð:Þdx . . . dx:

... 0

0

ð4Þ

0

By applying L1 on (3), we have

yðxÞ ¼ UðxÞ þ L1 gðxÞ  L1 Ny:

ð5Þ

such that

LUðxÞ ¼ 0: The Adomian decomposition method introduce the solution yðxÞ and the nonlinear function Ny by infinite series

yðxÞ ¼

1 X

yn ðxÞ;

ð6Þ

n¼0

and

Ny ¼

1 X

An ;

ð7Þ

n¼0

where the components yn ðxÞ of the solution yðxÞ will be determined recurrently. Specific algorithms were seen in [8,12] to formulate Adomian polynomials. The following algorithm:

A0 ¼ FðuÞ; A1 ¼ F 0 ðu0 Þu1 ; 1 A2 ¼ F 0 ðu0 Þu2 þ F 00 ðu0 Þu21 ; 2 A3 ¼ F 0 ðu0 Þu3 þ F 00 ðu0 Þu1 u2 þ

ð8Þ 1 000 F ðu0 Þu31 ; 3!

.. . can be used to construct Adomian polynomials, when FðuÞ is a nonlinear function. By substituting (6) and (7) into (5), 1 X

yn ¼ UðxÞ þ L1 gðxÞ  L1

n¼0

1 X

An :

ð9Þ

n¼0

Through using Adomian decomposition method, the components yn ðxÞ can be determined as

y0 ¼ UðxÞ þ L1 gðxÞ; ynþ1 ¼ L1 An ; n P 0;

ð10Þ

which gives

y0 ¼ UðxÞ þ L1 gðxÞ; y1 ¼ L1 A0 ; y2 ¼ L1 A1 ;

ð11Þ

1

y3 ¼ L A3 ; ... From (8) and (11), we can determine the components yn ðxÞ, and hence the series solution of yðxÞ in (6) can be immediately obtained. For numerical purposes, the n-term approximate

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Wn ¼

n1 X

yk ;

ð12Þ

n¼0

can be used to approximate the exact solution. The approach presented above can be validated by testing it on a variety of several linear and nonlinear initial value problems. 3. Numerical examples Example 1. We consider the linear boundary value problem:

q

y0 ¼ x1q cos x  ð2  qÞxq sin x; x yð0Þ ¼ 0; yð1Þ ¼ cos 1:

y00 þ

ð13Þ

We put

Lð:Þ ¼ x1

d 2q d 1þq ð:Þ; x x dx dx

so

L1 ð:Þ ¼ x1q

Z

x

x2þq

1

Z

x

xð:Þdxdx:

0

In an operator form, Eq. (13) becomes

Ly ¼ x1q cos x  ð2  qÞxq sin x: 1

By applying L

ð14Þ

to both sides of (14) we have

y ¼ yð0Þ þ ðyð1Þ  yð0ÞÞx1q þ L1 ðx1q cos x  ð2  qÞxq sin xÞ Z x Z x ¼ x1q cos 1 þ x1q x2þq ðx1q cos x  ð2  qÞxq sin xÞdxdx; 1

0

and it implies,

yðxÞ ¼ x1q cos 1  x1q cos 1 þ x1q cos x ¼ x1q cos x: so, the exact solution is easily obtained by this method. Example 2. We consider the non-linear boundary value problem:

1 4x2 y y00  y0 ¼ e; x 4 þ x2     1 1 ; yð1Þ ¼ ln : yð0Þ ¼ ln 4 5

ð15Þ

We put

Lð:Þ ¼ x1

d 3 d 2 x x ð:Þ; dx dx

so

L1 ð:Þ ¼ x2

Z

x

1

x3

Z

x

xð:Þdxdx: 0

In an operator form, Eq. (15) becomes

Ly ¼

4x2 y e: 4 þ x2

ð16Þ

Applying L1 on both sides of (16) we find

y ¼ yð0Þ þ ðyð1Þ  yð0ÞÞx2 þ 4L1

x2 ey : 4 þ x2

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By modified Adomian decomposition method [9] we obtain

  1 ; 4   4 2 x2 x þ 4L1 A0 ; y1 ¼ ln 5 4 þ x2 2 x Ak ; k P 1: ykþ1 ¼ 4L1 4 þ x2 y0 ¼ ln

ð17Þ

The Adomian polynomials for the nonlinear term FðyÞ ¼ ey are computed as follows:

A0 ¼ ey0 ; A1 ¼ y1 ey0 ;   1 A2 ¼ y2 þ y21 ey0 ; 2   1 A3 ¼ y3 þ y1 y2 þ y31 ey0 ; 3!

ð18Þ

1 Substituting (18) into (17) and it must noted that, to compute y1 we use the Taylor series of 4þx 2 with order 10. In this case we obtain

y0 ¼ ln

  1 ; 4

y1 ¼ 0:2520728x2 þ 0:03125x4  0:0026042x6 þ 0:0003255x8  0:0000488x10 þ 8:1380208  106 x12  1:4532180  106 x14 ; y2 ¼ 0:0098820x2  0:0105030x6 þ 0:0006510x8  0:0000326x10 þ 2:7126736  106 x12  2:9064360  107 x14 þ 3:6330450  108 x16  5:0458959  109 x18 ; .. . In Fig. 1 we have plotted

P4

i¼0 yi ðxÞ,

which is almost equal to the exact solution yðxÞ ¼ ln



1 4þx2

 .

Example 3. Consider the non-linear boundary value problems:

2 y000  y00  y  y2 ¼ gðxÞ; x yð0Þ ¼ 0;

y0 ð0Þ ¼ 0;

ð19Þ

yð1Þ ¼ 2:71828182:

Where

gðxÞ ¼ 7x2 ex þ 6xex  6ex  x6 e2x : Here, we use Taylor series of g(x) with order 10,

gðxÞ  6 þ 10x2 þ 9x3 þ

17 4 41 5 2 6 325 7 5729 8 20137 9 67181 10 x þ x  x  x  x  x  x : 4 30 3 168 2880 15120 100800

y

-1.3865

-1.387

-1.3875

x -0.1

-0.05

0.05

0.1

-1.3885

1 Fig. 1. The exact solution y ¼ ln 4þx 2 and the Adomian decomposition solution y ¼

P4

i¼10 yiðxÞ .

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We put 2

Lð:Þ ¼ x1

d

2

dx

So

L1 ð:Þ ¼ x4

Z

x5

d 4 x ð:Þ: dx

x

x5

Z

1

x 0

Z

x

xð:Þdxdxdx:

0

In an operator form, Eq. (19) becomes

Ly ¼ gðxÞ þ y þ y2 : 1

Applying L

ð20Þ

on both sides of (20) we find

y ¼ 2:71828182x4 þ L1 gðxÞ þ L1 y þ L1 y2 : Proceeding as before we obtained the recursive relationship

y0 ¼ 2:71828182x4 þ L1 gðxÞ; ynþ1 ¼ L1 yn þ L1 An ;

n P 0:

ð21Þ

Where An ’s are Adomian polynomials of nonlinear term y2 , as below

A0 ¼ y20 ; A1 ¼ 2y0 y1 ; A2 ¼ 2y2 y0 þ y21 ; .. .

ð22Þ

So, by substituting (22) into (21), we have

y0 ¼ x3 þ 1:0382041x4 þ 0:5x5 þ 0:15x6 þ 0:0337302x7 þ 0:0061012x8 þ 0:001851852x9  0:003582451x10  0:00258342x11  0:001261185x12  0:0004746x13 þ . . . ; y1 ¼ 0:0388054x4 þ 0:016666x6 þ 0:00823972x7 þ 0:00223214x8 þ 0:00319444x9 þ 0:00390766x10 þ 0:00270645x11 þ 0:00126549x12 þ 0:000445398x13 þ . . . ; y2 ¼ 0:000413381x4  0:000307979x7 þ 0:0000462963x9  0:000128465x10  0:000101745x11  2:15681  106 x12 þ 0:0000308777x13 þ . . . ; .. . This means that the solution in a series form is given by

y ¼ y0 þ y1 þ y2 ¼ x3 þ 0:999812x4 þ 0:5x5 þ 0:1666666x6 þ 0:0416619x7 þ 0:0083333x8 þ 0:00138889x9 þ 0:000196747x10 þ . . . : Note that, the Taylor series of exact solution yðxÞ ¼ xex with order 10 is as below

yðxÞ ¼ x3 þ x4 þ 0:5x5 þ 0:166666x6 þ 0:04166666x7 þ 0:008333x8 þ 0:0013888x9 þ 0:000198x10 þ . . . : References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]

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