A note on the use of modified Adomian decomposition method for solving singular boundary value problems of higher-order ordinary differential equations

Commun Nonlinear Sci Numer Simulat 14 (2009) 3261–3265

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A note on the use of modified Adomian decomposition method for solving singular boundary value problems of higher-order ordinary differential equations q Yahya Qaid Hasan, Liu Ming Zhu * Department of Mathematics, Harbin Institute of Technology Harbin, 150001, PR China

a r t i c l e

i n f o

Article history: Received 9 November 2008 Accepted 5 December 2008 Available online 25 December 2008 PACS: 02.60.Lj Keywords: Adomian decomposition method Singular boundary value problems High-order ordinary differential equation Linear and non-linear ordinary differential equation

a b s t r a c t This paper extend the work [Yahya Qaid Hasan, Liu Ming Zhu. Solving singular boundary value problems of higher-order ordinary differential equations by modified Adomian decomposition method. Commun Nonlinear Sci Numer Simul. doi :10.1016/j.cnsns.2008.09.027] to high order of singular boundary value problems. Solution of these problems is considered by proposed modification of Adomian decomposition method. The proposed method can be applied to linear and nonlinear problems. Some examples are presented to show the ability of the method for linear and non-linear ordinary differential equation. Ó 2008 Elsevier B.V. All rights reserved.

1. Introduction Adomian’s decomposition method (ADM) [2–5,14] is a new and ingenious method for solving nonlinear equations of various kinds. In recent years, the ADM has been successfully applied to solve many nonlinear equations in applied sciences, see for example [6,7,11]. The purpose of this paper is to introduce a new reliable modification of ADM. For this reason, a new differential operator is defined which can be used for high-order singular boundary value problems. The theoretical treatment of the convergence of Adomian’s method has been considered in [1,8–10]. 2. Modified Adomian decomposition method Consider the singular boundary value problem of n þ 1 order-ordinary differential equation in the form

yðnþ1Þ þ mx yðnÞ þ Ny ¼ gðxÞ; yð0Þ ¼ a0 ; y0 ð0Þ ¼ a1 ; . . . ; yðn1Þ ð0Þ ¼ an1 ; y0 ðbÞ ¼ c;

ð1Þ

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 q

This paper is supported by the National Natural Science Foundation of China (10471067). * Corresponding author. 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.2008.12.015

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d

Lð:Þ ¼ x1

n1

dx

xnm

d mnþ1 d x ð:Þ; dx dx

ð2Þ

where m 6 n  1; 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

Z

x

xnm1

0

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

Z

x

xmn b

Z

Z x ... xð:Þdx . . . dx: 0 |fflfflfflfflfflffl ffl{zfflfflfflfflfflffl0ffl} x

ð4Þ

n1

1

By applying L

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 [12,13] to formulate Adomian polynomials. The following algorithm:

A0 ¼ FðuÞ; A1 ¼ F 0 ðu0 Þu1 ; A2 ¼ F 0 ðu0 Þu2 þ 12 F 00 ðu0 Þu21 ;

ð8Þ

A3 ¼ F 0 ðu0 Þu3 þ F 00 ðu0 Þu1 u2 þ 3!1 F 000 ðu0 Þu31 ; .. . 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

y3 ¼ L1 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 approximant

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.

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3. Numerical examples Example 1. We consider the linear boundary value problem:

y00 þ qx y0 ¼ x1q cos x  ð2  qÞxq sin x; q

yð0Þ ¼ 0; y0 ðbÞ ¼ b ð1  qÞ cos b  b

1q

ð13Þ

sin b:

We put

Lð:Þ ¼ xq

d q d x ð:Þ; dx dx

so

L1 ð:Þ ¼

Z

x

x q

Z

0

x

xq ð:Þdxdx:

b

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 q

b y0 ðbÞx1q þ L1 ðx1q cos x  ð2  qÞxq sin xÞ 1q Z x Z x ðð1  qÞ cos b  b sin bÞx1q þ x q ðx1q cos x  ð2  qÞxq sin xÞdxdx;

y ¼ yð0Þ þ ¼

1 1q

0

b

and it implies,

yðxÞ ¼

1 1 ðð1  qÞ cos b  b sin bÞx1q þ ðð1  qÞ cos b þ b sin bÞx1q þ x1q cos x ¼ x1q cos x: 1q 1q

so, the exact solution is easily obtained by this method. Example 2. We consider the non-linear boundary value problem: 2

y00  1x y0 ¼ x3 y5 ; yð0Þ ¼ 1; y0 ð1Þ ¼ 0:216506:

ð15Þ

We put

Lð:Þ ¼ x

d 1 d x ð:Þ; dx dx

so

L1 ð:Þ ¼

Z

x

x 0

Z

x

x1 ð:Þdxdx:

1

In an operator form, Eq. (15) becomes

Ly ¼

x2 5 y : 3

ð16Þ

Applying L1 on both sides of (16) we find

y ¼ 1  0:108253x2 þ

x2 1 5 L y : 3

Proceeding as before we obtained the recursive relationship

y0 ¼ 1  0:108253x2 ; 2

ynþ1 ¼ x3 L1 An ;

n P 0:

ð17Þ

Where An ’s are Adomian polynomials of nonlinear term y5 , as below,

A0 ¼ y50 ; A1 ¼ 5y40 y1 ; A2 ¼ 5y2 y40 þ 10y30 y21 ; .. .

ð18Þ

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So, by substituting (18) into (17), we have y0 ¼ 1  0:108253x2 ; y1 ¼ 0:063782x2 þ 0:0416666x4  0:0075176x6 þ 0:0008138x8  0:000053x10 þ 1:9073486  106 x12  2:94966494  108 x14 ; y2 ¼ 0:0062577x2  0:0044294x6 þ 0:00240574x8  0:0006259x10 þ 0:000101699x12  0:0000115x14 þ . .. ; y3 ¼ 0:0010668x2 þ 0:00043456x6 þ 0:00018843x8  0:0003596x10 þ 0:0001864x12  0:000055x14 þ . .. ;

This means that the solution in a series form is given by

y ¼ y0 þ y1 þ y2 þ y3 ¼ 1  0:1668451x2 þ 0:04166666x4  0:0115124x6 þ 0:00340798x8  0:0010384x10 þ 0:0002900x12  0:00006677x14 þ ...: 1 ffi Note that, the Taylor series of exact solution yðxÞ ¼ pffiffiffiffiffiffiffi with order 14 is as below, x2 1þ 3

yðxÞ ¼ 1  0:166667x2 þ 0:0416667x4  0:0115741x6 þ 0:00337577x8  0:00101273x10 þ 0:000309446x12  0:0000957808x14 þ . . . : Example 3. Consider the non-linear boundary value problems:

y000  2x y00  y3 ¼ gðxÞ; yð0Þ ¼ 0; y0 ð0Þ ¼ 0; y0 ð1Þ ¼ 10:8731:

ð19Þ

Where

gðxÞ ¼ 7x2 ex þ 6xex  6ex  x9 e3x þ x3 ex : Here, we use Taylor series of g(x)with order 10,

gðxÞ  6 þ 10x2 þ 10x3 þ

21 4 28 5 1 6 3 7 11 8 3769 9 100787 10 x þ x þ x þ x þ x  x  x : 4 15 2 28 576 3780 33600

We put

d 4 d 3 d x x ð:Þ: dx dx dx

Lð:Þ ¼ x1 So

L1 ð:Þ ¼

Z

x

x3

Z

0

x

x4

1

Z

x

xð:Þdxdxdx:

0

In an operator form, Eq. (19) becomes

Ly ¼ gðxÞ þ y3 : 1

Applying L

ð20Þ

on both sides of (20) we find

y ¼ 2:71828x4 þ L1 gðxÞ þ L1 y3 : Proceeding as before we obtained the recursive relationship

y0 ¼ 2:71828x4 þ L1 gðxÞ; ynþ1 ¼ L1 An ;

ð21Þ

n P 0:

Where An ’s are Adomian polynomials of nonlinear term y3 , as below,

A0 ¼ y30 ; A1 ¼ 3y20 y1 ; A2 ¼ 3y2 y30 þ 3y0 y21 ; .. . So, by substituting (22) into (21), we have

ð22Þ

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3265

y0 ¼ x3 þ 1:00978x4 þ 0:5x5 þ 0:166667x6 þ 0:0416667x7 þ 0:00833333x8 þ 0:00138889x9 þ 0:000198413x10 þ 0:0000248016x11  0:000944214x12  0:00213648x13 þ . . . ; y1 ¼ 0:0348465x4 þ 0:00094697x12 þ 0:00215765x13 þ 0:00250492x14 þ 0:00197358x15 þ . . . ; y2 ¼ 0:000667923x4  0:000129061x10  0:0000913954x11  0:0000329986x12  8:27314  106 x13  1:59553  106 x14 þ . . . ; y3 ¼ 0:0000229991x4 þ 4:72245  106 x10 þ 3:34423  106 x11 þ 1:20744  106 x12 þ 3:02721  107 x13 þ 5:83819  108 x14 þ . . . ; .. . This means that the solution in a series form is given by y ¼ y0 þ y1 þ y2 þ y3 ¼ x3 þ 0:975577x4 þ 0:5x5 þ 0:1666666x6 þ 0:0416667x7 þ 0:0083333x8 þ 0:00138889x9 þ 0:0000740741x10 þ . ..:

Note that, the Taylor series of exact solution yðxÞ ¼ x3 ex 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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