doi:10.4236/am.2011.28134 Published Online August 2011 (http://www.SciRP.org/journal/am)
Adomian Decomposition Method for Solving
Goursat’s Problems
Mariam A. Al-Mazmumy
Mathematics Department, Science College, King Abdulaziz University, Jeddah, Saudi Arabia E-mail: mome0505@hotmail.com
Received May 18, 2011; revised June 22, 2011; accepted June 29, 2011
Abstract
In this paper, Goursat’s problems for: linear and nonlinear hyperbolic equations of second-order, systems of nonlinear hyperbolic equations and fourth-order linear hyperbolic equations in which the attached conditions are given on the characteristics curves are transformed in such a manner that the Adomian decomposition method (ADM) can be applied. Some examples with closed-form solutions are studied in detail to further illustrate the proposed technique, and the results obtained indicate this approach is indeed practical and effi-cient.
Keywords:Goursat’s Problem, Linear and Nonlinear Hyperbolic Equation of Second and Fourth-Orders, System of Linear Hyperbolic Equations of Second Order, Adomian Decomposition Method
1. Introduction
The simple Goursat’s problem concerns a class of linear hyperbolic equations of second-order in two independent variables with given values on two characteristics curves [1]. For example, for the linear hyperbolic equation
= ,
tt xx
u u u f t x
Goursat’s problem is posed as follows:
0 1
= , ,
, =
,
tt xx
u u u f t x
u t x x for x t
u t x x for x t
0 = 0
.
(1)
Several numerical methods such as Range-Kutta method, finite difference method and finite elements method have been used to approach the problem.
The general difficult which arises to us is the presence of the attached conditions on two characteristics curves and which complicates the application of numerical methods and Adomian decomposition me- thod [2-12].
= 0
x t x t = 0
The clue of this one consists in transforming this type of problems into classical problems where the conditions can be converted into initial conditions. In this technique, we use the variables and Hence we show that the linear Goursat models will be approached more effectively and rapidly by using the Adomian
decomposition method (ADM) to obtain the exact solu- tions to this type of problems. In a manner parallel to this problem, we study a class of Goursat’s problems for non- linear hyperbolic equations, systems of nonlinear hyper- bolic equations and fourth-order linear hyperbolic equa- tions. Our techniques are easily applicable and offer a very direct way to determine the solutions.
=
w x t z=x t
The present paper extends some results of [13].
2. The Goursat’s Problem for Linear
Hyperbolic Equations of Second Order
In this section consider Goursat’s problem (1).
Our first approach consists in converting problem (1) into a classical problem by introducing new variables
=
w x t and z=x t. The second one is that in order to find the solution for the given problem we consider this form as the most general form for ADM.
Make the substitutions w=x t and z=x t. into the first Equation of (1), and applying the chain rule to obtain
=
u u u
x w z
=
u u u
t w z
2 2 2 2
2 = 2 2
u u u
w z 2
u
x w z
2 2 2 2
2 = 2 2
u u u
w z
t w z
2
u
Then the first Equation of (1) becomes
2
4 = ,
2 2
u w z w z
f u
w z
(2) Also the given conditions of (1) can be converted into
, 0 =
02 w
u w
(3)
0, = 12 z
u z
(4)
Then,
Lemma 1. Goursat’s problem (1) is equivalent to (2)-(4).
Now we shall use the Adomian’s decomposition method [2-12] for solving (2)-(4).
Consider Equation (2) in an operator form as
1 =
4 4
wz
L u f u (5)
where
2= wz
u
L u
w z
Operating with the inverse operator we have
1
0 0
(.) = w z(.)d d ,
wz
L
w z
0 1
0 0 0 0
, = 0, 0
2 2
1
, d
4 2 2
, d d 4
w z
w z
w z
u w z u
w z w z d
f w z
u w z w z
(6)
where u
0, 0 =0
0 1
0
.
Following the Adomian decomposition method the unknown solution u is assumed to be given by a series of the form u w z
,
= n=0un
w z, where the com-
ponents un
w z, are going to be determined recurrently. Thus, we get the scheme
0 0 1
0 0
1 0 0
= 0
2 2
1
, d
4 2 2
= d d , 1
4 w z
w z
n n
w z
u u
w z w z
, 0
d
f w z
u u w z n
(7)
Now, by summing the first n1 terms of
,
= n=0 n
,u w z
u w z
we obtain the nth approxi- mation to the solution as=0
= ,
n
n i
i
S
u n0i
(8)
or
0 =1
= n n
i
S u
u (9)By substitution of the recursive scheme (7) into this sum, we conclude that the ADM for Goursat’s problem (1) can be converted to an equivalent problem, which we state as follows
Theorem 1. The ADM for Goursat’s problem (1) is equivalent to the following problem:
Find the sequence Sn such that
and satisfies
0 1
=
n n
S u u u
0 0 1
0 0
0 0 0 1
0, 0
2 2
1
, d
4 2 2
= d d ,
4 w z
w z
n n
w z
u u
w z w z
d
1
f w z
S u S w z n
(10)
Also, the result concerning the convergence analysis of the ADM for problem (1) can be stated as follows
Theorem 2. Let Sn be a sequence defined by (10). If 0 <
u and there exist a constant 0< 1, such that un1 un then Sn is convergent.
Proof. We have
1 1 2 2 1
=
n m n n n n n m m
S S S S S S S S S for nm, then
1 1 2 ... 1
n m n n n n m m
S S S S S S S S
Thus
1 1
0 0
n n m
n m
S S u u u0
Therefore
1 1
0
1
m n m
n m
S S u
As
i=0i is a geometric series with the common ratio , 0< 1. By induction we see that 1
0 1
m
n m
S S u
and SnSm 0, as thus converges for
,
m Sn
977
0
3. The Goursat’s Problem for Nonlinear
Hyperbolic Equations of the Second
Order
The same procedure can be adopted to resolve the following nonlinear hyperbolic equation of the second order
0 1
= ,
, = = 0
, =
tt xx
u u u f t x F u
u t x x for x t
u t x x for x t
(11)
where the nonlinear term is represented by F u
. As before, we haveLemma 2. The nonlinear Goursat’s problem (11) is equivalent to
20
1
4 ,
2 2
, 0
2 0,
2
u w z w z
f u F u
w z w u w
z
u z
(12)
Substituting the decomposition
,
= n=0 n
,u w z
u w z
into (12) and define F u
by
= n=0 n
0, 1, , n
F u
A u u u where theAnare the Ado-mian polynomials [2, 3], which depend only on the components . Thus we get the recursive scheme
0, 1, , n
u u u
0 0 1
1 1
0 0 0 0
0, 0
2 2
1
d d d d , 1
4 4
w z w z
n n n
w z
u u
u u w z A w z n
(13)
As before, we conclude that the ADM for Goursat’s problem (11) can be converted to an equivalent problem.
Theorem 3. The ADM for Goursat’s problem (11) is equivalent to the following problem:
Find the sequence n such that
and satisfies
S u
0 1
=
n n
S u u
0 0 1
0 0
0 0 0 1
0, 0
2 2
1
, d d
4 2 2
d d , 1
4 w z
w z
n n
w z
u u
w z w z
f w z
S u F S w z n
(14)
4. The Goursat’s Problem for System of
Nonlinear Hyperbolic Equations of the
Second Order
We shall extend this method to the following system of nonlinear hyperbolic equations of the second order
1 1
2 2
0 1 0 1
, ,
, ,
, =
, =
, =
, = = 0
tt xx tt xx
u u F u v f t x
v v F u v f t x
u t x x for x t
u t x x for x t
v t x x for x t
v t x x for x t
0
0
0
(15)
where the nonlinear terms are represented by F u vi
, ,. = 1, 2 i
By the same way as given in the nonlinear hyperbolic Equation (11), the scheme for system (15) can be expressed as
0 0 1
1 0 0
0 0 1
2 0 0
1 0 0
1 0 0
0, 0
2 2
1
( , )d d
4 2 2
0, 0
2 2
1
( , )d d
4 2 2
1
= d d , 1
4 1
= d d , 1
4 w z
w z
w z
n n
w z
n n
w z
u u
w z w z
f w z
w z
v v
w z w z
f w z
u A w z n
v B w z n
(16)
where F u v1
, = n=0A u un
0, 1, ,u v vn; 0, ,1 ,vn
2 , = n=0 n 0, ,1 , n; 0, ,1 , n
and F u v
B u u u v v v and An andn
B are the Adomian polynomials [2,3].
5. The Goursat’s Problem for Linear
Hyperbolic Equation of Fourth Order
22 2
2 2 u u= f t x,
t x
(17)
To Equation (17), we attach the following conditions
, = 0
= 0u t x x for x t (18)
1
= ( ) = 0
u u
x for x t
t x
(19)
, = 2
= 0u t x x for xt (20)
and
3
= =
u u
x for x t
t x
0
,
(21)
We mention that the attached conditions are given on the multiple characteristics curves and
of Equation (17).
= 0 x t = 0
x t
Let be a solution of problem (18)-(21) , make the transformations and applying the chain rule to obtain
u
= , =
z x t w x t
4 2 2
1
= ,
16 2 2 16
u w z w z
f w z
u
(22)
and the given conditions (18)-(21) can be converted into
, 0 =
02 w
u w
(23)
01 , 0 =
2 2
u w z
w
(24)
0, = 22 z
u z
(25)
31
0, =
2 2
u z w
z
(26)
Then,
Lemma 3. Goursat’s problem (17)-(21) is now equiva- lent to (22)-(26).
As before, rewrite Equation (22) in an operator form as
1 =
16 16
wwzz
L u f u (27)
where
42 2
= wwzz
u
L u
w z
Operating with the inverse operator
1
0 0 0 0
. = w w z z . d d d d
wwzz
L
w
2
0 0 0 0
0 0 0 0
, = , 0 0, 0, 0
, 0 0, 0, 0
0, 0 0, 0
1
, d d d
16 2 2
, d d d d
16
w z w z
w z w z
u w z u w u z u
u u u
z w w z z
z w z
u u
w wz
w w z
w z w z
d
f w w z z
u w z w w z z
where
0, 0 = 0
0 2
0u
2
1
1
0, 0 = 0
4 u
w z
1
1
0, 0 = 0
2 u
z
and
3
1
0, 0 = 0
2 u
w
So that
0 2
1 3 1
3 1
0 0 0 0
0 0 0 0
, = 0, 0
2 2
0
2 2 2 2 2
0 0
2 4
1
, d d d
16 2 2
, d d d d
16
w z w z
w z w z
w z
u x t u
z w w z z
w wz
w z w z
d
f w w z z
u w z w w z z
Following the Adomian decomposition method we get the scheme
0
1 0 0 0 0
,
d d d d , 1
16
w z w z
n n
u w z
u u w w z z n
(28)
where
0 2
1 3
3 1
, = 0, 0
2 2
0
2 2 2 2 2
0 0
2 4
w z
w z u
z w w z z
w wz
w z
1
979 and
1 0 0 0 0
, = , d d d d
16
w z w z
w z
f t x w w z z 6. Applications
In order to illustrate a possible practical use of the ADM, we shall give in this section some examples.
Example 1. Consider problem (1) with
, = 0
0 = 0 = 0u t x for x t
, = 1
0 = 0 = 0u t x for xt
, = 4 = 0f t x and
Then by the above recurrent scheme, we obtain
0 0 0d d =
0, 1
w z
n
u w z
u n
wz,z
2
(29)
So
can be determined asWe return to the original variables
and we get which
is the exact solution.
=0
= n
n
u
u w
,z
=wz. t z=x t u w
=
w x u t x
, =x2tExample 2. In problem (1) we choose
, = 0
0 = 2 = 0u t x x for x t
, = 1
0 = 2 = 0u t x x for x t
, = 2 = 1f t x x and
Then by the above recurrent scheme, we obtain
2 2
0
2 2 3 2 3 2
1 2
3 2 3 2 4 3 4 3
2 2 3
1 1
2 1 2 1
1
1
4 2 2
1 1
4 2 2 4 6 2 6 2
1 1
6 2 6 2 24 6 24 6
4 4
1
1 ! ! 1 ! ! 4
1
2 ! 1 ! 2 ! 4
n n n n n n n
n n n n n
n
w z
u w z z w
w z w z z w
u z w
w z z w w z z w
u
w z z w
u
n n n n
w z z w
n n n n
, 1 1 ! n
(30) So u= n=0un
w can be determined as
,
=u w z wz. Thus u t x
, = 2x which is the exact solution.Example 3. Consider now the Goursat’s problem (17)-(21) with
40
, = = = 0
u t x x x for x t
21
= = =
u u
x x for x t
t x
0
2
21
, = = = 0
4
u t x x x for x t
3
= = 0 =
u u
x for x t
t x
0
, = 0 = 0f t x and
The recurrence relation successively yields
4 2 2
4 2 0
1 1
2 4 2 2 2 16 8
= 0, 1
n
w z z w
u w
u n
2
1
z zw
(31) Then
1
4 2
1, =
16 8
u w z w z zw2
Hence, direct calculation produces the exact solution
1
2
2, = 16
u t x x t x t
Example 4. Consider Goursat’s problem (15) with
0 x = sin 2x for x t = 0
1 x = 0 for xt= 0
0 x = cos 2x for x t = 0
1 x = 1 for x t = 0
2 2, = 0 , = 1, = 1, 2
i i
f t x and F u v u v i Then by the above recurrent scheme, we obtain
0 0
sin cos 0, 1 0, 1
n n
u w
v w
u n
v n
(32)
where the first few Adomian’s polynomial of the non- linear terms F u vi
, =u2v21, = 1,i 21
1
can be ex- pressed as
2 2 0 = 0 = 0 0
A B u v
1= 1= 2 0 1 2 0
A B u u v
2 = 2 = 2u u0 2 u1 2v v0 v 2
2 2
1,
A B v
z ,z
t
t
7. References
Thus u= n=0un
w, and can be
v= n=0vn
w
determined as and So
that and .
,
= sinu w z w
sin xt
,
= cosv w z w
, = cos xt
, =u t x v t x
[1] E. Goursat, “A Course in Mathematical Analysis, Vol. 3: Variation of Solutions and Partial Differential Equations of the Second Order and Integral Equations and Calculus of Variations,” Gauthier-Villars, Paris, 1923.
Example 5. Finally, consider Goursat’s problem (17)-
(21) with [2] G. Adomian, “Nonlinear Stochastic Operator Equations,” Academic Press, Orlando, 1986.
20 = e 1 = 0
x
x for x
[3] G. Adomian, “Solving Frontier Problems of Physics: The Decomposition Method,” Kluwer Academic Publishers, Boston, 1994.
1 x = 2 for x t = 0
22 = 1 e = 0
x
x for x
[4] G. Adomian and R. Rach, “Transformation of Series,”
Applied Mathematics Letters, Vol. 4, No. 4, 1991, pp. 69- 71. doi:10.1016/0893-9659(91)90058-4
3 x = 2 for x t = 0
[5] G. Adomian, R. Rach and R. E. Meyers, “A Modified Decomposition,” Computers & Mathematics with Appli-cations, Vol. 23, No. 1, January 1992, pp. 17-23.
doi:10.1016/0898-1221(92)90076-T
, = ex t ex t = 1 f t x and Then by the above recurrent scheme, we obtain
2 2
0 1 1
2 2
1 1 1
4 4
3 3
2
4 4
2 2 3 3
6 6
5 5
3
2 2
1 1
1
1
e e ,
16 2 2
1
16 2 2
1
( )
12 12
16 1
( )
12 12
16 1
( )
360 360
16
1 2 =
2 ! 2 !
16 1 2 16
w z
n n n
n n n
n n
z w
u w
z w
u w z
z w
w z
z w
u w z
z w
w z
z w
u w
n n
n
z
z
2 2 2 1 2 2
2 1
2 2 !
, 1 2 2 !
n n
n n
z
w n
w
z n
n
[6] G. Adomian and R. Rach, “Inhomogeneous Nonlinear Partial Differential Equations with Variable Coeffi-cients,” Applied Mathematics Letters, Vol. 5, No. 2, March 1992, pp. 11-12.
doi:10.1016/0893-9659(92)90101-E
[7] G. Adomian and R. Rach, “Nonlinear Transformation of Series Part II,” Computers & Mathematics with Applica-tions, Vol. 23, No. 10, May 1992, pp. 79-83.
doi:10.1016/0898-1221(92)90058-P
[8] G. Adomian and R. Rach, “Modified Decomposition Solution of Nonlinear Partial Differential Equations,” Applied Mathematics Letters, Vol. 5, No. 6, November 1992, pp. 29-30. doi:10.1016/0893-9659(92)90008-W
[9] G. Adomian and R. Rach, “Solution of Nonlinear Partial Differential Equations in One, Two, Three, and four Di-mensions,” World Scientific Series in Applicable Analysis, Vol. 2, 1993, pp. 1-13.
[10] G. Adomian and R. Rach, “Modified Decomposition Solution of Linear and Nonlinear Boundary-Value Prob-lems,” Nonlinear Analysis, Vol. 23, No. 5, September 1994, pp. 615-619.
doi:10.1016/0362-546X(94)90240-2
[11] G. Adomian and R. Rach, “Analytic Solution of Nonlin-ear Boundary-Value Problems in Several Dimensions by Decomposition,” Journal of Mathematical Analysis and Applications, Vol. 174, No. 1, 15 March 1993, pp. 118- 137. doi:10.1006/jmaa.1993.1105
[12] G. Adomian and R. Rach, “A New Algorithm for Match-ing Boundary Conditions in Decomposition Solutions,” Applied Mathematics and Computation, Vol. 58, No. 1, September 1993, pp. 61-68.
doi:10.1016/0096-3003(93)90012-4
where
=0
= e
!
k n w n
k
w w
k
[13] A. Wazwaz, “The Decomposition Method for Approxi-mate Solution of the Goursat Problem,” Applied Mathe-matics and Computation, Vol. 69, No. 2-3, May 1995, pp. 299-311. doi:10.1016/0096-3003(94)00137-S
So u= n=0un
w can be determined as
