Explicit Solution of Hyperbolic Partial Differential Equations by an Iterative Decomposition Method
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(IJCSIS) International Journal of Computer Science and Information Security,
Vol. 9, No. 6, June 2011
Explicit Solution of Hyperbolic Partial
Differential Equations by an Iterative
Decomposition Method
Adekunle, Y.A.1*, Kadiri, K.O.2, Odetunde, O.S.3
1
Department of Computer Science and Mathematics, Babcock University, Ilisan-Remo Ogun State, Nigeria
adekunleya@gmail.com
2
. Department Electrical/Electronics Engineering, Federal Polytechnic, Offa, Kwara State, Nigeria
kadiritoyin2007@yahoo.com
3
. Department of Mathematical Science, Olabisi Onabanjo University, Ago-Iwoye, Ogun State,Nigeria
todetunde@yahoo.co.uk
Abstract- In this paper, an iterative decomposition method such as the wave equations, much attention is given
is applied to solve partial differential equations. The to the study of this class of partial differential
solution of a partial differential equation of the equation by mathematicians.
hyperbolic form is obtained by the stated method in the Various numerical methods have been applied to
form of an infinite series of easily computable terms.
Some examples are given and the solutions obtained by
solve hyperbolic partial differential equations. See
the method are found to compare favourably with the Ref[4 ]. However, many of these methods, commonly
known exact solutions. used as characteristic methods require large size of
computational work, and round-off error causes loss
Keywords: Hyperbolic partial differential equation, of accuracy [2]. The present paper attempts to present
iterative decomposition method, analytical solution. a method for the expilict solution of hyperbolic
partial differential equation. The results obtained are
compared with known analytical solution. For
I. INTRODUCTION problems whose analytical solutions are unknown,
An equation involving two or more partial the method will be appropriate.
derivatives of an unknown function of two or more The paper is organised as follows: section 2
independent variables is called a Partial Differential presents the iterative decomposition method, section
equation. 3 considers the method of section 2 for hyperbolic
The most general second order linear partial equations. In section 4, we present examples to
differential equation in two independent variables x illustrate the simplicity, efficiency and accuracy of
and y in this paper can be expressed generally in the the method. A conclusion is drawn in section 5
form
+ + + +e (1) II THE ITERATIVE DECOMPOSITION
METHOD
The idea of the iterative decomposition method
Or can be conveyed considering equation (1) as an
equation in the form
+ + = H (x, y, u, ) (2)
(4)
Where k is a constant and N(u) is the nonlinear term.
where a, b and c are function of x and y and H is a We may find the solution of (4) in a series form such
linear function of u, , . as
Thus equation (2) is called linear, and is a form of (5)
(1). The second order derivatives occur only to the We can decompose the nonlinear operator N as
first degree .If in equation (2), b2_ 4ac >0, the
equation is called Hyperbolic.
Since the governing equation in many
phenomena of engineering, science, and (6)
mathematical physics lead to hyperbolic equations,
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ISSN 1947-5500
(IJCSIS) International Journal of Computer Science and Information Security,
Vol. 9, No. 6, June 2011
From equations (5) and (6), equation (4) is then
equivalent to = {- } (15)
(7)
From equation (7), we define the following
Then, the solution of equation two will be given as
iterative scheme
(x, y) = N ( (8)
We can determine the component un as far as our
desired accuracy of the approximation. The n-terms
can be used to approximate the
m=1, 2, 3... solution.
Thus,
(9)
III Application of the iterative Decomposition
Method. III Numerical Experiments.
Consider the hyperbolic equation (2) with the
following initial conditions; We shall now consider the application of the iterative
decomposition method.
(10)
(x,0) =g(x) (11) Example 4. [ ] Consider the hyperbolic partial
differential equation
Using , we can put
(16)
, with the initial condition
(12)
(17)
in operation form as
u = H(x,y,u, , )-
The exact solution of this problem is
(13)
The inverse operator of denoted is
Putting equation (16) in operator form, we have
defined as
(.) = - (18)
Then L=
u(x,y) = u(x,0) + (x,0) + H Applying the inverse operator L1 to both sides of
(18)
(x,y,u, , )- } (14) where
and
308 http://sites.google.com/site/ijcsis/
ISSN 1947-5500
(IJCSIS) International Journal of Computer Science and Information Security,
Vol. 9, No. 6, June 2011
Then .
Then we have the exact solution
-( } (19)
By the iterative decomposition method,
after only two terms of the solution series.
-( IV Conclusion
} dy dy =0 (20) Our aim in this work has been to derive an
approximation algorithm for the explicit solution of
hyperbolic partial differential equations. This we
have been able to do via a method which does not
require elaborate mathematical computation and
Then the solution is which is assumptions. This method is a significant
the same as the given exact solution. improvement over some other decomposition
methods, notably the well-known Adomian
Example 4.2 [ ] consider the equation Decomposition Method.
REFERENCES
(21)
[1] G. Adomian, Nonlinear Stochastic Systems and Application to
Physics, Kluwer Academic Press. 1989
With the associated initial condition
[2] G. Adomian, Solving Frontier Problems of Physics: The
Decomposition Method, Kluwer Academic Press. 1994.
u(x,0) =x (22)
[3] J. Biazar, and H. Ebrahimi, An approximation to the solution of
hyperbolic equations by Adomian decomposition method and
comparison with characteristic method, Appl. Math. Comput. 163,
633-638. 2005.
The exact solution of the problem is [4] G. Evens, J. Blackledge, and P. Yardley, Numerical Method for
Partial Differential Equations, Springer. 2000.
The equation may be written in operator form as
[5] O. Taiwo, S. Odetunde, and .Y Adekunle, Numerical
Approximation of One Dimensional Biharmonic Equations by an
LU= Iterative Decomposition Method, J. Math. Sc. 20 (1) 37-44. 2008
Applying the inverse operator ,we have
From (14) and (15)
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