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A new algorithm for calculatingAdomian polynomials

Yonggui Zhu * , Qianshun Chang, Shengchang WuInstitute of Applied Mathematics, Academy of Mathematics & Systems Science,

Chinese Academy of Sciences, Beijing 100080, People s Republic of China

Abstract

In this paper, a new algorithm for calculating Adomian polynomials for nonlinearoperators will be established by parametrization. The algorithm requires less formulathan the previous method developed by Adomian [Nonlinear Stochastic Operator Equa-tions, Academic Press, 1986, G. Adomian, R. Rach, On composite nonlinearities anddecomposition method. J. Math. Anal. Appl. 113 (1986) 504509, G. Adomian, Appli-cations of Nonlinear Stochastic Systems Theory to Physics, Kluwer, 1988]. Many formsof nonlinearity will be studied to illustrate the new algorithm. The new algorithm will beextended to calculate Adomian polynomials for nonlinearity of several variables.

2004 Elsevier Inc. All rights reserved.

Keywords: Adomian decomposition method; Adomian polynomials; Nonlinear operators; Non-linearity of several variables

1. Introduction

The Adomian decomposition method [13] has been applied to a largeclass of linear and nonlinear problems in mathematics, physics, biology and

0096-3003/$ - see front matter 2004 Elsevier Inc. All rights reserved.doi:10.1016/j.amc.2004.09.082

* Corresponding author.E-mail address: ygzhu@amss.ac.cn (Y. Zhu).

Applied Mathematics and Computation 169 (2005) 402416www.elsevier.com/locate/amc

mailto:ygzhu@amss.ac.cnmailto:ygzhu@amss.ac.cn

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chemistry. For many problems, the decomposition method has shown reliableresults in providing analytical approximation that converges rapidly. It is well

known that the Adomian decomposition method decomposes the term u(x , t)into an innite sum of components un (x , t) dened by

u x; t X1

n0un x; t : 1

And the decomposition method decomposes the nonlinear term F (u(x , t))into A0 (u0 ) + A 1 (u0 , u1 ) + A 2 (u0 , u1 , u2 ) + , that is

F u x; t

X1

n0

Anu0 ; u1 ; . . . ; un; 2

where An are the so-called Adomian polynomials. Adomian [13] introducedmany formulas to generate Adomian polynomials for all forms of nonlinearoperators.

Recently, some feasible methods for the calculation of Adomian polynomi-als in a simple way without any need for the formulas introduced by Adomian[13] have been studied by many authors [48].We believe that a new simpleand reliable algorithm can be established to calculate Adomian polynomialsless dependable on so many formulas as before.

The main goal of this paper is to provide a promising algorithm that can beused to calculate Adomian polynomials for nonlinear operators in an easy way.The newly developed algorithm has been established mainly by parametriza-tion. The new method can also be used to generate Adomian polynomialsfor non-linear terms of several variables.

2. The new algorithm for generating Adomian polynomials

At rst a theorem will be given.

Theorem 2.1. Suppose nonlinear function Nu = F(u), and the parameterized representation of u is uk P

1k 0k

k uk , where k is a parameter, then wehave

o n F uk o k n

k 0

o n F P1k 0

k k uk o k n k 0 o n F P

nk 0

k k uk o k n k 0 : 3Proof 1. Since

uk X1

k 0

k k uk Xn

k 0

k k uk X1

k n 1

k k uk ;

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we have such result as following:

o n F u

k

o k nk 0

o n F

P1

k 0k k uk

o k n k 0 o n F

Pn

k 0k k uk

P1

k n 1k k uk

o k n k 0

o n F Pnk 0

k k uk o k n

k 0:

Therefore, we obtain

o n F uk o k n

k 0

o n F

P

1k 0

k k uk

o k n

k0

o n F

P

nk 0

k k uk

o k n

k0

:

As the representation introduced by Adomian [1], we assume the followingform again:

Nuk F uk X1

k 0

k k Ak : 4

So we have

F uk F X1

k 0

k k uk ! X1

k 0

k k Ak : 5

In order to obtain An , we give n-order derivative of both sides of (5) with re-spect to k and let k = 0, that is

o n F uk o k n

k 0

o n P1k 0

k k Ak o k n k 0 : 6According to Theorem 2.1

o n F uk o k n

k 0

o n F Pnk 0

k k uk o k n

k 07

and

o n P1k 0

k k Ak o k n k 0 o n P

nk 0

k k Ak o k n k 0 : 8So

o n F Pnk 0

k k uk o k n k 0 o n P

nk 0

k k Ak o k n k 0 : 9

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For (9) when n = 0, we can get A0 ; when n = 1, we can obtain A1 ; go on thiscourse, we will get A2 , A3 , . . ., An 1 , An .

The following is the algorithm for calculating A0 , A1 , A2 , . . ., An 1 , An :

Step 1: Input nonlinear term Nu = F (u) and n that is the order of Adomianpolynomials

Step 2: Set u = u0 + ku1 + k2 u2 + + k

n unStep 3: Let P

nk 0

k k Ak F u0 k u1 k 2u2 k nunStep 4: For i = 0, 1, . . ., n do(a) i th-order derivative of both sides of the above equality with respect to k :

o i

Pn

k 0k k Ak

o k i o F u0 k u1 k 2u2 k nun

o k i : 10

(b) For (10), let k = 0 and determine A i by solving the equation with respect toA i .

End do.Step 5: Output A0 , A1 , . . ., An .

3. The calculation of Adomian polynomials for all forms of nonlinearity

by using the new algorithm

3.1. The case of nonlinear polynomials

Example 1. F (u) = u2 , solve A0 , A1 , A2 , A3 , A4 , A5 , . . .At rst, let

uk u0 k u1 k 2u2 k 3u3 k 4u4 k 5u5 11

By (5) we have

X1

k 0

k k Ak u0 k u1 k 2u2 k 3u3 k 4u4 k 5u5 2 12

let k = 0, we can obtain

A0 u20 :

If we give one-order derivative of the two sides of (12) with respect to k and letk = 0, then using (9) we have

o A0 A1

ko k

k 0

ou0

ku1

2

o kk 0

: 13

Solve this equation with respect to A1 , we obtain A1 2u0u1 :

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If we give two-order derivative of the two sides of (12) with respect to k and letk = 0, by (9) we have

o 2 A0 A1k A2k 2o k 2 k 0

o 2u0 k u1 k 2u2

2

o k 2k 0

; 14

that is

2 A2 2u21 4u0u2

so

A2 u21 2u0u2 :

If we give three-order derivative of the two sides of (12) with respect to k andlet k = 0, by (9) we have

o 3 A0 A1k A2k 2 A3k 3o k 3

k 0

o 3u0 k u1 k 2u2 k 3u3

2

o k 3k 0

; 15

6 A3 12u0u3 12u1u2 ;

A3 2u0u3 2u1u2 :

If four-order derivative of the two sides of (12) with respect to k is given and letk = 0, by (9) we have

o 4 A0 A1k A2k 2 A3k 3 A4k 4o k 4 k 0

o 4u0 k u1 k 2u2 k 3u3 k 4u42

o k4

k 0

; 16

24 A4 48u0u4 24u22 48u1u3 ;

A4 2u0u4 u22 2u1u3 :

When ve-order derivative of the two sides of (13) with respect to k is given andlet k = 0, by (9) we obtain

o 5 A0 A1k A2k 2 A3k 3 A4k 4 A5k 5o k 5 k 0

o 5u0 k u1 k 2u2 k 3u3 k 4u4 k 5u5

2

o k 5k 0

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simplify the above equity, we have

120 A5 240u

0u

5 240u

1u

4 240u

2u

3;

that is

A5 2u0u5 2u1u4 2u2u3 :

Continue this course, we can get A6 , A7 , . . .

Example 2. F (u) = u3 + u2 , determine A0 , A1 , A2 , A3 , . . .Proceeding as Example 1, we set

uk u0 k u1 k 2u2 k 3u3 17By (5) we have

X1

k 0

k k Ak u0 k u1 k 2u2 k 3u3 3

u0 k u1 k 2u2 k 3u3 2 18

let k = 0, we can obtain

A0 u30 u20 :

If we give one-order derivative of the two sides of (18) with respect to k and letk = 0, then using (9) we have

o A0 A1k o k

k 0

o u0 k u13 u0 k u12o k

k 0

: 19

Solve Eq. (19) with respect to A1 , we obtain

A1 3u20u1 2u0u1 :

If we give two-order derivative of the two sides of (18) with respect to k and letk = 0, by (9) we have

o 2 A0 A1k A2k 2o k 2 k 0

o 2u0 k u1 k 2u2

3 u0 k u1 k 2u22

o k 2k 0

;

20

that is

2 A2 6u0u21 6u20u2 2u

21 4u0u2

so

A2 3u0u21 3u20u2 u

21 2u0u2 :

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If we give three-order derivative of both sides of (18) with respect to k andlet k = 0, by (9) we have

o 3 A0 A1k A2k 2 A3k 3o k 3 k 0

o 3u0 k u1 k 2u2 k 3u33 u0 k u1 k 2u2 k 3u32

o k 3k 0

; 21

6 A3 6u31 36u0u1u2 18u20u3 12u0u3 12u1u2 ;

A3 u3

1 6u0u1u2 3u2

0u3 2u0u3 2u1u2:

Go on as the above, we will get A4 , A5 , . . .

3.2. Case of nonlinear derivatives

Example 1. For F (u) = uu x uxx , determine A0 , A1 , A2 , A3 , . . .We rst set

u

k X

1

k 0

k k uk

:

22

By (5), we have

X1

k 0

k k Ak X1

k 0

k k uk ! X1

k 0

k k uk ! x X

1

k 0

k k uk ! xx

23

let k = 0, we can obtain

A0 u0u0 xu0 xx :

If we give one-order derivative of the two sides of (23) with respect to k and letk = 0, then using (9) we have

o A0 A1k o k

k 0

o u0 k u1u0 x k u1 xu0 xx k u1 xxo k

k 0: 24

Solve the above equation with respect to A1 , we obtain

A1 u1u0 xu0 xx u0u1 xu0 xx u0u0 xu1 xx :

Similarly, if we give two-order derivative of the two sides of (23) with respectto k and let k = 0, and using (9) and by solving an equation with respect to A2we can get

A2 u2u0 xu0 xx u1u1 xu0 xx u1u0 xu1 xx u0u2 xu0 xx u0u1 xu1 xx u0u0 xu2 xx :

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Continue as this course, we can obtain

A3 u3u0 xu0 xx u2u1 xu0 xx u2u0 xu1 xx u1u2 xu0 xx u1u1 xu1 xx u1u0 xu2 xx u0u3 xu0 xx u0u2 xu1 xx u0u1 xu2 xx u0u0 xu3 xx ; . . .

Example 2. F u uu2 x uu xx2, determine A0 , A1 , A2 . . .

Set uk P1k 0k

k uk , using (5) we have

X1

k 0

k k Ak X1

k 0

k k uk ! X1

k 0

k k uk !2

x

X1

k 0

k k uk ! X1

k 0

k k uk ! xx

!2

:

25

Using the new algorithm, the Adomian polynomials for F u uu2 x uu xx

2

are thus given by

A0 u0u20 x u0u0 xx2

;

A1 u1u20 x 2u0u0 xu1 x 2u0u20 xxu1 2u

20u0 xxu1 xx ;

A2 u2u20 x 2u1u0 xu1 x u0u21 x 2u0u0 xu2 x u

21u

20 xx 4u0u0 xxu1u1 xx

2u0u20 xxu2 u20u

21 xx 2u

20u0 xxu2 xx ; . . .

3.3. Case of trigonometric and hyperbolic nonlinearity

Example 1. F u sin u sin 2ucos 2u.Let uk P

1k 0k

k uk , by (5) we have

X1

k 0

k k Ak sin X1

k 0

k k uk ! sin2 X1

k 0

k k uk !cos 2 X1

k 0

k k uk !: 26According to the new algorithm, the Adomian polynomials for F u sin u sin 2ucos 2u are thus given by

A0 sin u0 sin 2u0cos 2u0 ; A1 u1 cos u0 2u1 sin u0cos 3u0 2u1sin 3u0 cos u0 ;

A2 12

u21 sin u0 u2 cos u0 u21cos

4u0 6u21sin2u0cos 2u0

2u2 sin u0cos 3u0 u21sin4u0 2u2sin

3u0 cos u0 ;

A3 16 u

31 cos u0 u1u2 sin u0 u3 cos u0

163 u

31cos

3u0 sin u0

2u1u2cos 4u0 163

u31sin3u0 cos u0 12u1u2sin

2u0cos 2u0

2u3 sin u0cos 3u0 2u1u2sin 4u0 2u3sin 3u0 cos u0 ; . . .

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Example 2. F u sinh 2u sinh u cosh u.Set uk

P1k 0k

k uk , by (5) we have

X1

k 0

k k Ak sin h2 X1

k 0

k k uk ! sin h X1

k 0

k k uk !cosh X1

k 0

k k uk !: 27Using the new algorithm, the Adomian polynomials for F u sin h2u sinh u cosh u are thus given by

A0 sinh 2u0 sinh u0 cosh u0 ;

A1 2u1 sinh u0 cosh u0 u1cosh 2u0 u1sinh 2u0 ;

A2 u21cosh 2u0 u21sinh 2u0 2u2 sinh u0 cosh u0 2u21 cosh u0 sinh u0

u2cosh 2u0 u2sinh 2u0 ;

A3 43

u31 cosh u0 sinh u0 2u1u2cosh2u0 2u1u2 sin h2u0

2u3 sinh u0 cosh u0 23

u31sinh2u0 4u1u2 cos hu 0 sinh u0

2

3u31cosh

2u0 u3cosh 2u0 u3sinh 2u0 ; . . .

3.4. Case of exponential and logarithmic nonlinearity

Example 1. F (u) = e u .If set uk P

1k 0k

k uk , by (5) we have

X

1

k 0

k k Ak eP1k 0

k k uk : 28

Using the new algorithm, we can obtain the Adomian polynomials forF (u) = e u as followings:

A0 eu0 ; A1 u1eu0 ;

A2 u2eu0 12

u21eu0

;

A3 u3eu0 u1u2eu0 1

6u31e

u0;

A4 u4eu0 u1u3eu0 12

u22eu0

12

u2u21eu0

124

u41eu0

;

. . .

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Example 2. F (u) = ln u + e u .Let uk

P1k 0k

k uk , by (5) we have

X1

k 0

k k Ak ln X1

k 0

k k uk ! eP1

k 0k k uk

: 29

Using the new algorithm, the Adomian polynomials for F (u) = ln u + e u arethus given by the followings:

A0 ln u0 eu0 ;

A1 u1u0

u1 eu0 ;

A2 u2

u0 u212u20

u2 eu0 12u21 e

u0;

A3 u3u0

u1u2u20

u313u30

u3 eu0 u1u2 eu0 16

u31 eu0

;

. . .

3.5. Case of composite nonlinearity

Example 1. F u e sin 2 u2 .Set u = u0 + ku1 + k

2 u2 + k3 u3 + , by (5) we have

X1

k 0

k k Ak e sin2 u0 k u1 k

2 u2 k3 u3

2; 30

let k = 0, then we have

A0 e sin2 u0

2:

If we give one-order derivative of both sides of (30) with respect to k and letk = 0, using (9) we have

o A0 A1k o k

k 0

o e sin2 u0 k u1

2

o kk 0

;

A1 u1 e sin2 u0

2 sinu02

cosu02 :

If two-order derivative of two sides of (30) with respect to k is given and let

k = 0, by (9) we have

o 2 A0 A1k A2k 2o k 2 k 0

o 2 e sin

2 u0 k u1 k2 u2

2

o k 2k 0

;

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2 A2 u21 esin 2

u02 sin 2

u02

cos2 u02

u212

e sin2 u0

2 cos 2 u02

u212

e sin2 u0

2 sin 2 u02

2u2 e sin 2 u02 sin u02 cos u0

2 ;

A2 u212

e sin2 u0

2 sin 2 u02

cos2 u02

u214

e sin2 u0

2 cos 2 u02

u214

e sin2 u0

2 sin 2 u02

u2 e sin2 u0

2 sinu02

cosu02 :

If give three-order derivative of two sides of (30) with respect to k and let k = 0,by (9) we have

o 3 A0 A1k A2k 2 A3k 3o k 3 k 0

o 3 e sin

2 u0 k u1 k2 u2 k

3 u32

o k 3k 0

:

Simplify the above equality

6 A3 u31 cosu02

sinu02

e sin2 u0

2 3u1u2cos 2 u02

e sin2 u0

2 32

u31 sinu02

cos 3 u02 e sin

2 u02 3u1u2sin 2

u02 e sin

2 u02

32 u31sin 3

u02 cos

u02 e sin

2 u02

6u3 sinu02

cosu02

e sin2u0

2 6u1u2sin 2 u02

cos2 u02

e sin2 u0

2 u31sin3 u0

2

cos 3 u02

e sin2 u0

2:

So

A3 16 u

31 cos

u02 sin

u02 e

sin 2u02

12 u1u2cos

2 u02 e

sin 2u02

14 u

31 sin

u02

cos 3 u02

e sin2 u0

2 12

u1u2sin2 u0

2 e sin

2 u02

14

u31sin3 u0

2 cos

u02

e sin2 u0

2

u3 sinu02

cosu02

e sin2 u0

2 u1u2sin 2 u02

cos2 u02

e sin2 u0

216

u31sin3 u0

2

cos 3 u02

e sin2 u0

2:

If continue this course, we can obtain A4 , A4 , A5 , . . .

Example 2. F u u2 esin u cosh u ln u.As the above-mentioned examples, using the new algorithm Adomian

polynomials for F u u2 esin u cosh u ln u can be given by

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A0 u20 esin u0 cosh u0 ln u0 ;

A1 2u1u0 esin u0 u1u20 cos u0 esin u0 12 u1 eu

0 ln u012 u1 ln u0 e u

0

u12u0

eu0 u12u0

e u0 ;

A2 u21 esin u0 2u0u21 cos u0 e

sin u0 2u0u2 esin u0

12

u20u21 sin u0 e

sin u0 u20u2 cos u0 esin u0

1

2u2

0u2

1esin u0

1

2u2

0u2

1sin 2u0 esin u0

1

4 eu0 u2

1ln u0

14

u21 ln u0 eu0

12

eu0 u2 ln u012

u2 ln u0 e u0 12

eu0u21u0

12

u21u0

e u0

12

u2u0

eu0 12

u2u0

e u014

u21u20

eu014

u21u20

e u0 ;

. . .

4. The extension of new algorithm to the case of nonlinearity of several variables

In the following, by using some examples we will show how the new algo-rithm is extended to multi-variable case.

Example 1. F u o2u

o xo y 2 , determine A0 , A1 , A2 , A3 , . . .

At rst, set u = u0 + k u1 + k2 u2 + k

3 u3 + , from (5) we have

X1

k 0

k k Ak o2u0

o xo y k o 2u1

o xo y k 2 o 2u2

o xo y k 3 o 2u3

o xo y

2

: 31

If let k = 0, we can obtain

A0 o 2u0

o xo y 2

:

If give one-order derivative of both sides of (31) with respect to k and letk = 0, by (9), we have

o A0 A1k o k

k 0

o o2u0

o xo y ko 2u1o xo y

2

o k

k 0

;

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A1 2 o 2u0o xo y

o 2u1o xo y

:

If give two-order derivative of both sides of (31) with respect to k and letk = 0, by (9) we have

o 2 A0 A1k A2k 2o k 2 k 0

o 2 o

2u0o xo y k

o 2u1o xo y k

2 o 2u2o xo y

2

o k 2

k 0

;

2 A2 2 o 2u1

o xo y 2

4 o 2u0

o xo y o 2u2

o xo y ;

that is

A2 o 2u1

o xo y 2

2 o 2u0o xo y

o 2u2o xo y

:

If give three-order derivative of both sides of (31) with respect to k and letk = 0, by (9) we can get

o3

A0 A1k A2k2

A3k3

o k 3 k 0

o 3 o

2 u0

o xo y k o

2 u1

o xo y k 2 o

2 u2

o xo y k 3 o

2u3

o xo y

2

o k 3

k 0

;

6 A3 12 o 2u1

o xo y o2u2

o xo y 12 o2u0

o xo y o2u3

o xo y ;that is

A3 2 o 2u1

o xo y

o 2u2

o xo y 2

o 2u0

o xo y

o 2u3

o xo y :

If we go on this course, we can obtain A4 , A5 , A6 ,. . .

Example 2. For F u u2 o3u

o xo y o z , determine A0 , A1 , A2 , A3 ,. . .,At rst, we let

u u0 k u1 k 2u2 k 3u3

from (5), we have

X1

k 0

k k Ak u0 k u1 k 2u2 k 3u3 2

o 3u0 k u1 k 2u2 k 3u3 o xo y o z

: 32

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Let k = 0, we have

A0 u20o 3

u0o xo y o z :

If one-order derivative of both sides of (32) with respect to k is given and letk = 0, by (9) we have

o A0 A1k o k

k 0

o u0 k u12 o 3u0 k u1

o xo y o z o kk 0

;

A1 2u0u1o 3u0

o xo y o z u20

o 3u1o xo y o z

:

If give two-order derivative of the two sides of (32) with respect to k and letk = 0, by (9) we get

o 2 A0 A1k A2k 2o k 2 k 0

o 2 u0 k u1 k 2u22

o 3 u0 k u1 k 2u2 o xo y o z o k 2

k 0

;

2 A2 2u21o 3u0

o xo y o z 4u0u1

o 3u1o xo y o z

4u0u2o 3u0

o xo y o z 2u20

o 3u2o xo y o z

;

A2 u21o 3u0

o xo y o z 2u0u1

o 3u1o xo y o z

2u0u2o 3u0

o xo y o z u20

o 3u2o xo y o z

:

If give three-order derivative of both sides of (32) with respect to k and letk = 0, by (9) we can obtain

o 3 A0 A1k A2k 2 A3k 3o k 3 k 0

o 3 u0 k u1 k 2u2 k 3u32

o 3u0 k u1 k 2 u2 k 3 u3 o xo y o z o k 3

k 0

:

Simplify the above equity, we have

6 A3 12u1u2o 3

u0o xo y o z 6u21

o 3

u1o xo y o z 12u0u1o 3

u2o xo y o z 12u0u2o 3

u1o xo y o z

12u0u3o 3u0

o xo y o z 6u20

o 3u3o xo y o z

:

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Thus

A3 2u1u2o 3u0

o xo y o z u21o 3u1

o xo y o z 2u0u1o 3u2

o xo y o z 2u0u2o 3u1

o xo y o z

2u0u3o 3u0

o xo y o z u20

o 3u3o xo y o z

:

In the same way, we can calculate A4 , A5 , A6 , . . . , etc.

5. Discussion and conclusion

The Adomian decomposition method is a powerful method for solution of alarge class of nonlinear problems in physics, biology, chemistry and etc. How-ever, the implementation of the decomposition method mainly depends uponthe calculation of Adomian polynomials for nonlinear operators. So develop-ing some practical methods for the calculation of Adomian polynomials for allforms of nonlinearity is vital to solve nonlinear problems in many appliedsciences.

In this paper, we introduced a new reliable algorithm to generate Adomianpolynomials for all forms of nonlinearity. The new algorithm does not sufferfrom the numerous formulas required in [13]. Several cases of nonlinearitywere discussed for calculating Adomian polynomials and results are verypromising.

References

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102 (1984) 415419.[5] V. Seng, K. Abbaoui, Y. Cherruault, Adomian s polynomials for nonlinear operators, Math.

Comput. Model. 24 (1996) 5965.[6] K. Abbaooui, Y. Cherruault, Convergence of Adomian s method applied to differential

equations, Comput. Math. Appl. 102 (1999) 7786.[7] A.M. Wazwaz, The decomposition method for approximate solution of the Goursat problem,

Appl. Math. Comput. 69 (1995) 299311.[8] A.M. Wazwaz, A new algorithm for calculating Adomian polynomials for nonlinear operators,

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