Numerical implementation of Adomian decomposition method

Applied Mathematics and Computation 153 (2004) 301–305 www.elsevier.com/locate/amc

Numerical implementation of Adomian decomposition method E. Babolian, A. Davari

*

Faculty of Mathematical Sciences and Computer Engineering, University for Teacher Education, 599 Taleghani Avenue, Tehran 15618, Iran

Abstract In this paper, we propose new ideas to implement Adomian decomposition method to solve integral equations. In this paper we verify use of numericalPintegration in AdomianÕs method for cases that evaluation of terms of the series u ¼ ui analytically is impossible. Ó 2003 Published by Elsevier Inc.

1. Introduction There are many papers [3,4] in which Adomian decomposition method has been used for solving linear and nonlinear equations, the solution is obtained as an infinite series which converges to exact solution [1], under mild conditions. Consider the following Fredholm integral equation Z b uðxÞ ¼ f ðxÞ þ k kðx; tÞuðtÞ dt ð1Þ a

for x 2 ½a; b
where f , k are known functions, u is to be determined. The AdomianÕs technique consists of representing u as a series u¼

1 X

un :

n¼0

*

Corresponding author. E-mail address: [email protected] (A. Davari).

0096-3003/$ - see front matter Ó 2003 Published by Elsevier Inc. doi:10.1016/S0096-3003(03)00646-5

ð2Þ

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E. Babolian, A. Davari / Appl. Math. Comput. 153 (2004) 301–305

Now if we set (2) in (1), we will have Z b 1 1 X X un ðxÞ ¼ f ðxÞ þ k kðx; tÞ un ðtÞ dt a

n¼0

ð3Þ

n¼0

Note that the Adomian method uses the recursive relations u0 ðxÞ ¼ f ðxÞ Z unþ1 ðxÞ ¼ k

b

kðx; tÞun ðtÞ dt

ð4Þ

a

Example 1 9 uðxÞ ¼ x3 þ 10

Z

1 0

1 3 x tuðtÞ dt 2

with the exact solution is uðxÞ ¼ x3 . Applying (4) we obtain 9 3 9 x ; un ðxÞ ¼ nþ1 x3 10 10 1 1 X X 9 3 uðxÞ ¼ un ðxÞ ¼ x ¼ x3 n 10 n¼0 n¼1 u0 ðxÞ ¼

Obviously the series is convergent to the exact solution. Now we consider the following example. Example 2 uðxÞ ¼ cos x þ

Z

p=6

uðtÞ cos x2 dt 0

If we apply the decomposition method, we have u0 ðxÞ ¼ cos x 1 cos x2 2 Z p=6 cos t2 dt u2 ðxÞ ¼ cos x2 2 0 u1 ðxÞ ¼

or cos x2 u2 ðxÞ ¼ 2

Z

p=6

cos t2 dt 0

R p=6 to evaluate 0 cos t2 dt, also We know that evaluation of u2 ðxÞ is conditioned R we know that the indefinite integral cos t2 dt is not available and practically

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303

the decomposition method fails. So, for such problem we have to use numerical integration [2]. Now we apply Simpson rule with n ¼ 10, h ¼ 0:1. u0 ðxÞ ¼ cos x u1 ðxÞ ¼ 0:5 cos x2 u2 ðxÞ ’ 0:239838 cos x2 u3 ðxÞ ’ 0:135032 cos x2 u4 ðxÞ ’ 0:070173 cos x2 u5 ðxÞ ’ 0:0364673 cos x2 u6 ðxÞ ’ 0:0189512 cos x2 u7 ðxÞ ’ 0:0098485 cos x2 u8 ðxÞ ’ 0:00511804 cos x2 u9 ðxÞ ’ 0:00265973 cos x2 u10 ðxÞ ’ 0:0013822 cos x2 u11 ðxÞ ’ 0:000718297 cos x2 u12 ðxÞ ’ 0:000373282 cos x2 u13 ðxÞ ’ 0:000193986 cos x2 u14 ðxÞ ’ 0:0001008 cos x2 u15 ðxÞ ’ 0:0000523886 cos x2  u15 ¼

15 X

ui ¼ cos x þ ð1:04091Þ cos x2

i¼0

E15 ¼  u15  cos x 

Z

p=6

 u15 ðtÞ cos x2 dt 0

E15 ¼

15 X

ui  cos x 

Z

i¼0

p=6

 u15 cos x2 dt 0

¼ cos x þ 1:04091 cos x2  cos x 

Z

p=6

ðcos t þ 1:04091Þ cos x2 dt 0

¼ 1:04091 cos x2  1:04094 cos x2 ¼ 3  105 cos x2

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E. Babolian, A. Davari / Appl. Math. Comput. 153 (2004) 301–305

Therefore jE15 j < 3  105

Example 3 1 uðxÞ ¼ sin 3x  cos 6x þ 3 3

Z

p=6

t cos 6x uðtÞ dt 0

If we use Simpson rule with n ¼ 10 and h ¼ p=60, we have 1 u0 ðxÞ ¼ sin 3x  cos 6x 3 Z p=6 1 u1 ðxÞ ¼ 3tðsin 3t  cos 6tÞ cos 6xdt ’ 0:388876 cos 6x 3 0 u2 ðxÞ ’ 0:0648019 cos 6x u3 ðxÞ ’ 0:0107985 cos 6x u4 ðxÞ ’ 0:00179945 cos 6x u5 ðxÞ ’ 0:000299859 cos 6x u6 ðxÞ ’ 0:0000499682 cos 6x u7 ðxÞ ’ 8:32665  106 cos 6x 7 X  un ðxÞ ¼ sin 3x þ 1:36745  106 cos 6x u7 ðxÞ ¼ n¼0

We notice that uðxÞ ¼ sin 3x is the exact solution of the integral equation. Therefore, we will have E7 ðxÞ ¼ uðxÞ   u7 ðxÞ ¼ 1:36745  106 cos 6x and jE7 ðxÞj 6 1:36745  106

Example 4 2

uðxÞ ¼ cos x þ

Z

p=6 2

uðtÞ sin x2 dt 0

If we use Adomian method [5] and Simpson rule with h ¼ p=120, n ¼ 20, we will have

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305

u0 ðxÞ ¼ cos x2 A0 ðxÞ ¼ u20 ¼ ðcos x2 Þ2 Z p=6 u1 ðxÞ ¼ sin x2 A0 ðtÞ dt ’ 0:515837 sin x2 0

A1 ðxÞ ¼ 2u0 u1 ¼ 2 cos x2 ð0:515837 sin x2 Þ Z p=2 u2 ðxÞ ¼ sin x2 A1 ðtÞ dt ’ 0:0483147 sin x2 0 2

A2 ðxÞ ¼ 2u0 u2 þ u21 ¼ 0:0966294 cos x2 sin x2 þ 0:266088ðsin x2 Þ Z p=6 u3 ðxÞ ¼ sin x2 A2 ðtÞ dt ’ 0:00659069 sin x2 0

A3 ðxÞ ¼ 2u0 u3 þ 2u1 u2 ¼ 0:0131814 cos x2 sin x2 þ 0:049845ðsin x2 Þ2 u4 ðxÞ ’ 0:0010042 sin x2 2

A4 ðxÞ ¼ 2u0 u4 þ 2u1 u3 þ u22 ¼ 0:0020084 cos x2 sin x2 þ 0:00913375ðsin x2 Þ u5 ðxÞ ’ 0:000164953 sin x2 5 X  ui ¼ cos x2 þ 0:571912 sin x2 u5 ¼ i¼0 2

E5 ¼ cos x þ

Z

p=6

 u5 ðxÞ ¼ 0:00003 sin x2 ; u25 sin x2 d dt  

jE5 j < 0:00003

0

References [1] Y. Cherruault, G. Adomian, K. Abbaoui, R. Rach, Further remarks on convergence of decomposition method, International Journal of Biomedical Computing 38 (1995) 89–93. [2] J. Stoer, R. Bulirsch, Introduction to Numerical Analysis, Springer-Verlag, 2002. [3] A.M. Wazwaz, S.A. Khuri, Two methods for solving integral equations, Applied Mathematics and Computation 77 (1996) 79–89. [4] A.M. Wazwaz, A first course in integral equations, WSPC, New Jersey, 1997. [5] A.M. Wazwaz, A new algorithm for calculating Adomian polynomials for nonlinear operators, Applied Mathematics and Computation III (2000) 53–69.