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1. Introduction

The well known Adomian decomposition method is a powerful device forsolving functional equations in the following canonical form in a functional

space, say F,

u f Gu; 1wherefis a known function, uis the function we are looking for in F, andGis

an analytical functional operator. Adomian decomposition method[24]con-

siders the solution, u and G(u) as the sum of two series, say

u X1

n0 un; 2

Gu X1n0

Anu0;. . . ; un; 3

where An(u0, . . . , un) are called Adomian polynomials.

To define these polynomials by using a parameter k Adomian rewrites (2)

and (3)as

uk X

1

n0 unk

n; 4

Guk X1n0

Anu0;. . . ; unkn 5

and defines these polynomials as the following

Anu0;. . . ; un 1n!

dn

dknG

X1n0

unkn

!" #k0

.

An alternate algorithm for computing Adomian polynomials in special casesuses the same presentations (4) and (5) and compute Guk GP1

n0unkn,

by using algebraic manipulation and trigonometric identities. Then by compar-

ing the terms with identity degrees in(5), Adomian polynomials can be recog-

nized easily[1].

The restriction of the method refers to the cases in which computation of

GP1n0unkn is very complicated or maybe impossible.In these cases we can consider the Taylor expansion ofG(u) and consider a

few first terms of the expansion and then apply the main idea of the algorithm.

2. Examples

To illustrate the modification some examples, are presented in the following.

J. Biazar et al. / Appl. Math. Comput. 173 (2006) 582592 583
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Example 1. Lets start with a simple, but illustrative, following integral

equation

u Z

x

0

cosudt.

With initial condition u(0) = 0, and the exact solution x ln j secu tanuj.The solution, by Adomian decomposition method, consists of the following

scheme[2]

u0 0;u

n1 Z x

0

An

u0; u

1;. . . ; u

ndt n

0; 1; 2.

6

To find An

let ukP1

n0unkn, by considering Taylor expansion of G(u) =

cos(u), and taking just two terms approximation, Gu 1 12

u2, we would have:

Gku 1 12

u20

u0u1k 1

22u0u2 u21k2 u0u3 u1u2k3

122u0u4 2u1u3 u22k4 7

And in comparison with(5)we have the following approximations for Ado-mian polynomials,

A0u0 1 12

u20;

A1u0; u1 u0u1;A2u0; u1; u2 u0u2 1

2u21;

A3u0; u1; u2; u3 u0u3 u1u2;A4u0; u1; u2; u3; u4 u0u4 u1u3 1

2u22;

.

.

.

From(6)we have:

u0 0;u1

Z x0

A0u0dt x;

u2Z x

0

A1u0; u1dt 0;

u3Z x

0

A2u0; u1; u2 16

x3;

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u4Z x

0

A3u0; u1; u2; u3 0;

u5Z

x

0

A4u0; u1; u2; u3; u4 130

x5;

.

.

.

A six terms approximation to the solution is u x 16x3 1

30x5 (Table 1).

Example 2. Consider the differential equation u0 sinu 0, with initialconditionu(0) = 0, and the exact solution

x ln j cscu cotuj.The canonical form of the equation is

u p2

Z x

0

sinudt.

Adomian decomposition method consists of the following scheme:

u0p

2;

un1 Z x

0

Anu0; u1;. . . ; undt; n 0; 1; 2.8

To find An

, let ukP1

n0unkn; considering the following approximation

Gu sinu u 16

u3.

Using the idea presented in [1],

Guk u0 16

u30

u1 1

2u20u1

k u2 1

2u20u2

1

2u0u

21

k

2

Table 1

Some numerical results of Example 1

x u (Adomian) u (Exact) E(u(x))

0.1 0.099834 0.099834 0.000000

0.2 0.198677 0.198679 0.000002

0.3 0.295581 0.295597 0.000016

0.4 0.389675 0.389741 0.000066

0.5 0.480208 0.480381 0.000073


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From(5)we have

A0u0 u0 1

6 u2

0;

A1u0; u1 u1 12

u20u1;

A2u0; u1; u2 u2 12

u20u2 1

2u0u

21;

.

.

.

And from(8)we derive (Table 2)

u0 p2

;

u1 Z x

0

A0u0dt p2

1 p2

24

x;

u2 Z x

0

A1u0; u1dt p4

1 p2

24

1 p

2

8

x2;

.

.

.

u p2

1 1 p2

24

x 1 p

2

24

1 p

2

8

x2

2

1 p2

24

1 p

2

8

2x3

6 p

80 1 p

2

24

1 p

2

8

2x5

!.

Example 3. Consider the following nonlinear homogeneous P.D.E.

ut expu tux 0;ux; 0 lnx.

With the solution, u(x, t) = ln(x) +t.

Table 2

Some numerical results for Example 2

x u (Adomian) u (Exact) E(u(x))

0.01 1.5607 1.5607 0.0000

0.02 1.5515 1.5508 0.0007

0.03 1.5422 1.5408 0.0014

0.04 1.5328 1.5308 0.0020

0.05 1.5235 1.5208 0.0027

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The equivalent canonical form of the equation is as follows:

ux; t lnx Z t

0 expu tux dt.Following Adomian method

u0 lnx;

un1Z t

0

Anu0; u1;. . . ; undt; n 0; 1; 2;. . .9

To findAn

letuk

P1n0unk

n, and by consideringGu 1 u u22 exptux,

we would have:

Guk u0x u0u0x 12u20u0x

expt u1x u0xu1 u1xu0 u0xu0u1 12u20u1x

exptk

From which Adomian polynomials can be approximated as

A0u0 u0x u0u0x 12 u20u0x expt;A1u0; u1 u1x u1u0x u0u1x u1u0u0x 12 u20u1x expt;

..

.

And from(9)we have:

u0 lnx;

u1Z t

0

A0u0dt 12x

1 expt2 2 lnx lnx2;

.

.

.

Just a two terms approximation to the solution would derive the following

nice results (Table 3)

u lnx 12x

1 expt2 2 lnx lnx2.

Example 4. Considered the following P.D.E.

t

1

ut

2x2uxx

x

t

1

exp

u

0;

ux; 0 lnx;utx; 0 1.

8>>>:

With the solution u(x, t) = ln(x(t+ 1)).


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The canonical form of the equation is

ux; t lnx Z t

0

2x2uxx

t 1 x expu

dt.

Following Adomian:

u0 lnx;

un

1

Z t

0

An

u0; u1;. . . ; un

dt; n

0; 1; 2.

10

To find An

let ukP1

n0unkn, considering the following approximation

Gu 2x2uxx

t 1 x 1 u 1

2u2

.

Using an alternate algorithm[1] we derive

Gku 2x2u0uu

t 1 x 1 u0 1

2 u2

0

2x2u1xx

t 1 x1 u1 u0u1

k

And Adomian polynomials can be approximated as the following

A0u0 2x2u0xx

t

1x

1 u0 1

2u20

;

A1u0; u1 2x2u1xx

t 1 x1 u1 u0u1;

.

.

.

Table 3

Some numerical results of Example 3 for different values oft

x u (Adomian) u (Exact) E(u(x, t)) x u (Adomian) u (Exact) E(u(x, t))t = 0.01 t = 0.02

1 0.00100 0.001 0.00000 1 0.00200 0.002 0.00000

1.1 0.09632 0.09631 0.00001 1.1 0.09731 0.09731 0.00000

1.2 0.18331 0.18332 0.00001 1.2 0.18432 0.18432 0.00000

1.3 0.26335 0.26336 0.00001 1.3 0.26435 0.26436 0.00001

1.4 0.33746 0.33747 0.00001 1.4 0.33845 0.33847 0.00002

1.5 0.40646 0.40647 0.00001 1.5 0.40745 0.40747 0.00002


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f2;0t 0;f2;n1

t

Rt

0Bn

f1;0

s

;. . . ;f1;n

s

ds;

n 0; 1; 2; 3;. . . 13

To find Bns let f1;kP1

n0f1;nkn, consider the Taylor expansion of

G2f1;ks ts expf21 s, and taking two terms 1 f21 s we would have:G2f2;ks ts 1 f220s

2tsf20sf21sk ts2f22sf20s f221sk2 2tsf23sf20s f22sf21sk3

From(5)we have:

B0

f10

s

ts

1

f210

s

;

B1f10s;f11s 2tsf10sf11s;B2f10s;f11s;f12s ts2f12sf10s f211s;B3f10s;f11s;f12s;f13s 2stf13sf10s f12sf11s;...

From(13)we have:

f20

t

0;

f21t Z

t

0

B0f10sds 12

t3;

f22t Z t

0

B1f10s;f11sds 0;

f23t Z t

0

B2f10s;f11s;f12sds 132

t9;

Adomian decomposition approach for the solution of(11)yields to the follow-

ing scheme:

f1;0t 0;

f1;n1t Z t

0

Anf1;0s;. . . ;f1;ns;f2;0s;. . . ;f2;nsds; n 0; 1; 2; 3;. . .

To find An

let us consider f1;ks P1

n0f1;nskn, and f2ks P1

n0f2nskn,from(11): We can recognize G1(f1,k(s), f2,k(s)) = 1 +f1(s)s exp(s2)2f2(s).

By substituting f1,k(s)and f2,k(s) into G1(f1,k(s), f2,k(s)) we would have:

G1f1;ks;f2;ks 1 s exps2 f10s 2f20s f11s 2f21sk f12s 2f22sk2 f13s 2f23sk3 f14s 2f24sk4

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From which Adomian polynomials could be recognized as [1]:

A0

f10

s

;f20

s

1

exp

s2

f10

s

2f20

s

;

A2f10s;f11s;f20s;f21s f11s 2f21s;A3f10s;f11s;f12s;f20s;f21s;f22s f12s 2f22s;

.

.

.

And f1;ns would be computed as

f10

t

0;

f11t Z t

0

A0f10s;f20sds t 12

expt2 12

;

f12t Z t

0

A1f10s;. . . ;f21sds 12

t2 14

ffiffiffip

p erft 1

2t 1

4t4;

f13t Z t

0

A2f10s;. . . ;f22sds 16

t3 14

14

expt2

14 tffiffiffipp erft 14 t2 120 t5;

A four terms approximation to the solution would be as follows ( Table 5):

ft 34

12

t 34

expt2 14

t2 14

ffiffiffip

p erft

14

t

ffiffiffip

p erft 1

6t3 1

4t4 1

20t5:

Table 5

Some numerical results for Example 5

t f(Ado) f(exa) E(f(t))

0.01 0.0099897 0.01 0.0000103

0.02 0.019999 0.02 0.000001

0.03 0.029986 0.03 0.000014

0.04 0.039999 0.04 0.000001

0.05 0.049989 0.05 0.000011


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3. Discussion

Many researchers are working on deriving more and easer computationalmethods for computing Adomian polynomial. In this paper one of the existed

algorithms has been improved to work on more comprehensive situations.

The only necessary and sufficient assumption for applying this improvement

is that the existence of the Taylor expansion of G(u) but this assumption is

guaranteed already. Since in the canonical form of the equation there is men-

tioned that G(u) is analytical.

Examples illustrate the variety of applications, simplicity and reliability, of

the modified algorithm.

Computations are performed using Maple 9 package.

References

[1] J. Biazar, E. Babolian, A. Nouri, R. Islam, An alternate algorithm for computing Adomian

decomposition method in special cases, App. Math. Comput. 138 (23) (2003) 523529.

[2] G. Adomian, G.E. Adomian, A global method for solution of complex systems, Math. Model.

5 (1984) 521568.

[3] G. Adomian, Nonlinear Stochastic Systems and Applications to Physics, Kluwer, 1989.

[4] G. Adomian, Solving Frontier Problems of Physics: the Decomposition Method, KluwerAcademic Publishers, Dordecht, 1994.


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