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A SIMPLE PROGRAM FOR SOLVING NONLINEAR INITIAL VALUE PROBLEM USING ADOMIAN DECOMPOSITION METHOD

F.A. Hendi, H.O.Bakodah, M. Almazmumy & H. Alzumi

Department of Mathematics, Science Faculty for Girls, King Abdulaziz University, Saudi Arabia [email protected], [email protected], [email protected], [email protected]

ABSTRACT

The Adomian method is widely used in approximate calculation, its main demerit is that it is very difficult and complex to calculate Adomian's polynomials. Many researchers have suggested different methods and algorithm for computing these polynomials. In this paper, a mathematica program is prepared to solve the initial value problem in ordinary differential equation of the first order. Simplicity and efficiency of the algorithm presented in this paper are illustrated briefly in the examples.

Keywords: Adomian decomposition method, Adomian polynomials, Initial value problem.

2000 Methematics Subject Classification : 49M27,34A12 1. INTRODUCTION

Adomian decomposition method (ADM) is an approximate approach for solving nonlinear differential equations by substitution of nonlinear parts of equation with Adomian polynomials and use a step by step method for finding the solutions [1]. This method is a powerful approach in nonlinear differential equations and an accuracy of it depends on number of used partial solutions. Also, the solution of this method has a fast convergence to the exact solution generally. In recent years, some modifications on this method have been presented [2,3], these modifications affect on convergence of the method. In some papers [4-7], authers modified the method of computing Adomian's polynomials, which are rapidly convergence by least number of components as compared than standard (ADM). (ADM) has been successfully applied to solve many nonlinear equations, ordinary or partial, determinstic or stochastic, with approximants which converges rapidly to accurate solutions. We shall deal here only with the simple first order differential equation. ADM is one of the new method for solving initial value problem in ordinary differential equation of various kind arising not only in the field of medicine, physical and biological science but also in the area of engineering [8]. In this paper, we present many cases of nonlinearities of I.V.P. and applying (ADM) to get the solution. Adomian and Rach considered nonlinearities for polynomial [9], Negative power [10], decimal power [11], Redicals [12] and composite nonlinearities [13]. They are also used transform techniques with different equations containing convolution product nonlinearities to yield an algebraic equations. ADM has been used to solve nonlinear initial value problems, but applying this method needs some computations which is sometimes boring, having a program to do all computations would be interesting and helpful. In this artical a mathematica program is prepared to solve I.V.P. with different nonlinearity.

2. ADOMIAN DECOMPOSITION METHOD

We use (ADM) for solving first order ordinary differential equations of the form

  , ,   =

0

= f x y y a y

y

' (2.1)

The basic concepts of (ADM) [14]:

we assume that

y   x

is sufficiently differentiable and that the solution of (2.1) exists and satisfy the lipschitz condition. (ADM) usaually defines an equation in an operator form by considering the highest -order derivative in the problem. In an operator form, equation (2.1) can be written as

  x y f

Ly = ,

(2.2)

where the differential operator

L

is given as

dx

L = d

(2.3)

The inverse operator

L

1 is considered a one fold integral operator defined by

 dx

L =

x

.

0

1

(2.4)

(2)

398

If we operate

L

1 on both side of (2.2) and use the initial condition

y   a = y

0 , we have

  x y L f   x y

y =

0

1

,

(2.5)

The (ADM) introduces the solution

y   x

in an infinite series form as

  x y   x

y

n

n

0

=

=

(2.6)

where the components

y

n

  x

will be determined recursively. Moreover, the method defines the nonlinear function

  x y

f ,

by the infinite series of the form

 

n

n

A y

x

f

0

=

=

,

(2.7)

If we now use equation (2.6) and (2.7) in (2.5), we have

 

n

n n

n

A L y x

y

0

= 1 0 0

=

=

(2.8)

The next step is to seek a way to determine the component

y

n

  x

. We first identify the zeroth component

y

0

  x

by the term that arises from the initial condition. The remaining component is determind by using the preceding component.

Each term of the series (2.6) is given by the recurrence relation.

 

0

0

x = y

y

(2.9)

  =

1

  , 0

1

x L A n

y

n n (2.10)

We must state here that in practice all term of the series in (2.6) cannot be determined and the solution will be approximated by the series of the form

  x y   x y   x

n

  x

n n N

n

n

 = , = lim

1

0

=

(2.11)

with (2.11), we obtain the series of the solutions for our system (2.1).

3. SOME CASES OF NONLINEARITIES

In the 1980's, G. Adomian introduced a new powerful method for solving nonlinear functional equations. The technique is based on a decomposition of a solution of nonlinear operator equation in a series of functions. Each term of the series is obtained from a polynomial generated from an expansion of analytic function into a power series. The Adomian technique is very simple in an abstract formulation but the difficulty arises in calculating the polynomials.

The purpose of this section is to study many cases of nonlinearities of this method in application to the initial value problems for nonlinear ordinary differential equations.

3.1 polynomial nonlinearities

Adomian and Rach [9] solved equations of the form

  t x Ny

Ly  =

(3.1)

where the nonlinear term

Ny = y

m for positive integer

m

. With the application of the decomposition method, the solutions to nonlinear differential equations can be made without linearization procedures. Adomian introduced these polynomials, for functions with one vriable, by the following formula [15]

  , = 0,1,2,...

!

= 1 ,..., , ,

0 0 =

= 2

1

0

f y n

d d y n

y y y

A

i i

n

i n n n

n

 

 

 

 

 

(3.2)

Hence, the solution for any equation involving a polynomial nonlinearity is quick and easy.

3.2 Negative power nonlinearities

Differential equations involving the term

y

m, where

m

is a positive integer, are solved by decomposition method

(3)

399 [10], Then the

A

n are given by

y

m

A

0

=

0

 

1 1 0

1

= my y

A

m

 

2

1 0 2 1 2 0

2

1

2

= 1 m m y y my y

A

m

m

  

 

3

1 0 2 1 2 0 3

1 3 0

3

1 2 1

6

= 1 m m m y y m m y y y my y

A   

m

 

m

m

. . .

Consequently,

   

1 ! 1

=

1

1

0

=

n

k t m

y

n n nm n

n n

 

so that

     

1 ! 1

=

1

1

0

= 1

0

=

n

k t m

t y

n n nm nn

n

 

a convergent series ( whose sum is m1

m 1t k

m1 ) .

The higher derivative and higher powers in

Ny

or

x   t

are simple generalization. For example :

r n m m

t dt y

y

d

=

is immediately calculable from expressions for

A

n for negative powers of

y

remembering

L

1represents

m

-fold integration from

0

to

t

.

3.3 Decimal power nonlinearities

If

Ny

is a nonlinear term of the form

y

with

a decimal number [11], Let

y

'

y

= 0

and

y   0 = k

(

extension to cases

y x   t dt

y d

m m

=

with appropriate given conditions is a simple generalization ) . The Adomian polynomials are

0 0

= y A

1 1 0

1

= y y

A

 

12

2 0 2

1 0

2

1

2

= y y 1 y y

A

   

    

13

3 0 2

1 2 0 3

1 0

3

1 2

5 1 1

= y y y y y y y

A

   

     

. . .

Consequently,

       

 

1 ! 1

=

1

1

0

= 0

=

mn

k t t

y

mn m m m m

m

 

 

  

 



3.4 Equations containing radicals Suppose we are considering equation with

(4)

400

2

21

=  y   Ny

We treat the nonlinear or radical term

y

2

21 as the composite nonlinearity [12] with

 

0

0

= 0 0

0

= =

n

n

A u

u

N

where

u

0

=   N

1

u

1

and 1

0

= 1

1

=

n

n

A u

N

with

u =

1

y

.

Calculating the

A

n0 polynomials

  

2 10

21

1 0 0 0

0

= u = A

A   

       

2 11 1 1 0 0

2 1 1 0 0 0

1

2

= 1 2

= 1 u u A A

A

  

       

2 10 2 3 0 0 0

2 2 1 0 0 0

2

8

1 2

= 1 u u u u

A

. . .

Then, the solution for

y y   k

dt

dy  

2

  = 0, 0 =

is

    ...

6

= 2

3 2 1 2 2

2 1

2

    

t

t k k t k

k

y      

4. COMPOSITE NONLINEARITIES

The composite nonlinearities of the form

N   x = N

0

N

1

N

2

...   x ...   

where

N   x

a nonlinear term in an equation to be solved by decomposition. Terms such as

x

2

, e

x

, sin x

etc are viewed as

N

0

u

0 where

u =

0

x

, and expanded in Adomian's polynomial [13]. Now identified as

A

n0 to correspond to the

N

0 nonlinear operator.

Thus, 0

0

= 0

0

=

n

n

A u

N

.

A first order composite nonlinearity is defined as

N

1

x = N

0

  N

1

u

1

or as

N

0

N

1

u

1 where

u =

1

x

and

1 1 0

= N u

u

with 0

0

= 0

0

=

n

n

A u

N

and 1

0

= 1

1

=

n

n

A u

N

.

In general, 1

0

=

=

=

u A u

N

n

n

for

1    m

with

u

m

= x

and n

n

u u

0

=

=

when the results apply to differential equations in the form

  x g Ny Ly  =

where

Ny

is a composite nonlinearity, we get

  x L Ny

g L Ly

L

1

=

1

1

 

n

n

A L x g

L

0

= 1

=

1

If

y   k

dx

Ly = dy , 0 =

, Then

(5)

401

n n n

n

A L g L k y

y

0

= 1 1

0

=

=

=

where

g L k y

0

= 

1

n

n

L A

y

1

= 

1 for

n  0

and the

A

n are calculated by the methods discussed.

5. A SIMPLE ALGORITHM FOR SOLVING I.V.P.

Many researchers have suggested different methods and algorithm for computing Adomian polynomials. Wazwaz suggestion [4] is to substitute the series solution (2.6) into (2.2) and then by manipulation terms, based on algebraic operations, trigonometric identities and Taylor series as appropriate, all terms can be collceted such that the subscripts of the components of in each terms is the same. With this step preformed, the calculation of the Adomian polynomials is thus completed. In [5], Rach introduced a new definition of Adomian polynomial to provide a new proof of convergence of the Adomian decomposition series for solving nonlinear ordinary and partial differential equations.

Adomian polynomials are expressed in terms of new objects called reduced polynomials. In [6], these new objects, which carry two subscripts, are independent of the form of the nonlinear operator. In [7], the authers modified the method of computing Adomian's polynomial to find the numerical solution for nonlinear systems of partial differential equations with less number of components, more accuracy and faster convergence when compared with the standard Adomian decomposition method.

The main goal of this work is to present a mathematica program to compute easily these polynomials for nonlinear terms in solving I.V.P. in ordinary differential equations and to compute

A

n in general for any kind of nonlinearities.

6. MATHEMATICA PROGRAM

Consider the following first order initial value problem of the form

  , ,   =

0

= f x y y a y y

'

In Adomian decomposition method, we can consider the solution as summation of a series

j j j

y Y

0

=

=

and

f   x , y

, as a series, say:

 

j j

j

A y

x

f

0

=

= ,

where

A

j

y

0

, y

1

,..., y

n

, which depends on

y

0

, y

1

,..., y

j called Adomian polynomials and are defined as

0,1,2,...

= ,

! ,

= 1

0 0 =

=

n y

x d f

d

A n

j j

n

j n

n

n

 

 

 

 

 

The inputs of the algorithm are the following:

n

, the number of terms of the approximations of the solutions.

  x

g

, the first term in the right hand side of the canonical form

  x g Ny Ly  = 5

= n

 

i

i n

i

y S   

0

=

:=

Ad=Table

      ,   , /. 0,  ,0,5  ;

!

1  

  D f S nn

n   

(6)

402 Table Form

[

ad,Table Aligmments

Left, TabeHeadings

 

''

A

0

" ,

''

A

1

" ,

''

A

2

" ,

''

A

3

" ,

''

A

5

" , None].

A few cases of nonlinear operators are given . 6.1 Nonlinear polynomials

f   y = y

m

Substituting

S    942

instead of

f   S   

in the above program we get

  =

0

 

2

;

0

x y x

A

1

  x = 2 y

0

    x y

1

x ; A

   2   4      ;

2

= 1

1 2 0 2

2

x y x y x y x

A

   12     12      ; 6

= 1

1 2 0 3

3

x y x y x y x y x

A

   24   48     48      ;

24

= 1

2 2 1 3 0 4

4

x y x y x y x y x y x

A  

   240     240     240      ;

120

= 1

2 3 1 4 0 5

5

x y x y x y x y x y x y x

A  

6.2 Negative power nonlinearities

  y y

m

f =

  x y

m

A

0

=

0

 

1

1 0

1

x = my y

A

m

    

2

1 0 2

1 2 0

2

1 2

2

= 1 m m y y my y

x

A   

m

m

  x m m m

A ( ( 2 )( 1 )

6

= 1

3

     y

03m

y

13

 6(  1  m ) my

02m

y

1

y

2

 6 my

01m

y

3

  x m m m my y m m m

A ( ( 3 )( 2 )( 1 )

m

12( 2 )( 1 )

24

= 1

04 14

4

      

    

4 1 0 3

1 2 0 2

2 2 0 2

2 1 3

0

y y 12( 1 m ) my y 24( 1 m ) my y y 24 my y

y

m

  

m

  

m

m

  x m m m m m

A ( ( 4 )( 3 )( 2 )( 1 )

120

= 1

5

         y

05m

y

15

m m m

m )( 2 )( 1 ) 3

20(      

y

04m

y

13

y

2

 60(  2  m )(  1  m ) ) 1 120(

) 1 )(

2

60(

03 12 3

2 2 1 3

0

y y m m my y y m

my

m

    

m

   m

y y my m y

y

my

02m 2 3

 120(  1  )

02m 1 4

 120 y

01m

y

5

6.3 Decimal power nonlinearities

  y = y

31

f

 

031 0

x = y

A

(7)

403

 

1 03 1

x = y /3 y A

    

 

2 3

0 2

3 5

0 2 1 2

2 2 9 2 2

= 1

y y y x y

A

    

 

3 2

0 3

3 5

0 2 1

3 8

0 3 1 3

2 3

4 27 10 6

= 1

y y y

y y y x y

A

    

 

3 2

0 4

3 5

0 3 1

3 5

0 2 2

3 8

0 2 2 1

3 11

0 4 1 4

8 3

16 3

8 9

40 81

80 24

= 1

y y y

y y y

y y

y y y

x y A

    

 

3 2

0 5

3 5

0 4 1

3 5

0 3 2

3 8

0 3 2 1

3 8

0 2 2 1

3 11

0 2 3 1

3 14

0 5 1 5

40 3

80 3

80 9

200 9

200 81

1600 243

880 120

= 1

y y y

y y y

y y y

y y y

y y y

y y y

x y A

6.4 Equations containing radicals

2

21

= y

Ny   

 

02

0

x = y

A   

 

2

0 1 0

1

=

y y x y

A  

    )

( 2 2

= 1

2 0 2 0 2

0 2 1

2 3 2 0 2 1 2 0 2

2

y

y y y

y y

y x y

A  

 

 

        )

6 6

6 3

( 3 6

= 1

2 0 3 0 2

0 2 0

2 3 2 0

2 1 2 0 2

2 3 2 0 3 1 0 2

2 5 2 0 3 1 3 0 3

3

y

y y y

y y y

y y y y

y y y

y x y

A  

 

 

          

02

23

2 2 1 0 2

2 5 2 0

2 2 1 3 0 2

2 3 2 0 4 1 2

2 5 2 0 4 1 2 0 3

2 7 2 0

4 1 4 0 4

4

36 36

3 18

( 15 24

= 1

y y y y y

y y y y

y y

y y y

y x y

A

    )

24 24

24 12

12

2 0 4 0 2

0 3 1

2 3 2 0

3 1 2 0 2

2 0 2 2

2 3 2 0

2 2 2 0 2

y y y y

y y y

y y y y

y y

y y

 

 

 

          

02

25

2 3 1 2 0 3

2 7 2 0

2 3 1 4 0 4

2 5 2 0 5 1 0 3

2 7 2 0

5 1 3 0 4

2 9 2 0

5 1 5 0 5

5

360 300

45 150

( 105 120

= 1

y y y y y

y y y y

y y y

y y y

y x y

A

        

02

23

3 2 1 0 2

2 5 2 0

3 2 1 3 0 2

2 3 2 0

2 2 1 0 2

2 5 2 0

2 2 1 3 0 3

2 3 2 0 2 3 1

3

180 180 180 180

60

y y y y y

y y y y

y y y y

y y y y

y y

    )

120 120

120 120

120

2 0 5 0 2

0 4 1

2 3 2 0

4 1 2 0 2

2 0 3 2

2 3 2 0

3 2 2 0 2

y y y y

y y y

y y y y

y y y

y y y

 

 

 

6.5 Composite nonlinearities

(8)

404

  y Expy

f = sin

 

sin

 

0

0

x = e

y

A

     

0 1 sin 0

1

x = e cos y y

A

y

              

0 2

sin 0 2 1 0 sin 0

2 1 2 0 sin 0

2

cos sin 2 cos

2

= 1 e y y e y y e y y

x

A

y

y

y

               

0 0 13 sin 0

3 1 3 0 sin 0

3 1 0 sin 0

3

( cos cos 3 cos sin

6

= 1 e y y e y y e y y y

x

A

y

y

y

  cos   6   sin   6   cos   ) 6 e

siny0

y

0 2

y

1

y

2

e

siny0

y

0

y

1

y

2

e

siny0

y

0

y

3

               

0 0 14 sin 0

4 1 4 0 sin 0

4 1 2 0 sin 0

4

( 4 cos cos cos sin

24

= 1 e y y e y y y y y

x

A

y

y

y

             

2

2 1 0 sin 0

4 1 2 0 sin 0

4 1 0 2 0 sin 0

cos 12

sin 3

sin cos

6 e

y

y y ye

y

y ye

y

y y y

              

22

2 0 sin 0

2 1 0 0 sin 0

2 2 1 3 0 sin 0

cos 12

sin cos 36

cos

12 e

y

y y ye

y

y y y ye

y

y y

12   sin   24   cos   24   sin   24   cos  

0 4

)

sin 0

3 1 0 sin 0

3 1 2 0 sin 0

2 2 0 sin 0

y y e

y y y e

y y y e

y y

e

y

y

y

y

             

15

5 0 sin 0

5 1 3 0 sin 0

5 1 0 sin 0

5

( cos 10 cos cos

120

= 1 e y y e y y e y y

x

A

y

y

y

                 

15

2 0 0 sin 0

5 1 0 3 0 sin 0

5 1 0 0 sin 0

sin cos 15

sin cos

10 sin

cos

15 e

y

y y ye

y

y y ye

y

y y y

            

2

3 1 0 sin 0

2 3 1 4 0 sin 0

2 3 1 2 0 sin 0

sin 20

cos 20

cos

80 e

y

y y ye

y

y y ye

y

y y y

              

0 1 22

sin 0 2

3 1 2 0 sin 0

2 3 1 0 2 0 sin 0

cos 60

sin 60

sin cos

120 e

y

y y y ye

y

y y ye

y

y y y

              

3

2 1 0 sin 0

2 2 1 0 0 sin 0

2 2 1 3 0 sin 0

cos 60

sin cos 180

cos

60 e

y

y y ye

y

y y y ye

y

y y y

              

2 3

2 0 sin 0

3 2 1 0 0 sin 0

3 2 1 3 0 sin 0

cos 120

sin cos 180

cos

60 e

y

y y ye

y

y y y ye

y

y y y

 120 e

sin

 

y0

sin   y

0

y

2

y

3

 120 e

sin

 

y0

cos   y

0 2

y

1

y

4

 120 e

sin

 

y0

sin   y

0

y

1

y

4

 120 e

sin

 

y0

cos   y

0

y

5

)

7. NUMERICAL EXAMPLES

Adomian decomposition method as a possible approach can be applied to the analysis of equation modeling water flows [8] from an inverted conical tank with circular orifice at the rate

    A y g y dt r

dy =  0.6 

2

 2

where

r

is the radius of the orifice,

x

is the height of the liquid level from the vertex of the cone, and

A   y

is the

area of the cross section of the tank

y

unit above the orific. Suppose

r = 0.1

ft,

g =  32.17

ft/s2, and the tank has an initial water level of

8

ft and initial volume of

 

 512  3

ft3.

We need to compute the water level after

10

min with

h = 205

, and determine, to within

1

min, when the tank will be empty.

Solution of the problem:

References

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