1
1 Introduction
Homotopy perturbation method has been used by many mathematicians and engineers to solve various functional equations (Yıldırım and Ozis, 2007; Biazar et al., 2007; Noor and Mohyud-Din, 2008; Ozis and Yıldırım, 2007; Odibat and Momani, 2008; Siddiqui et al., 2008; Ghori et al., 2007). This method was further developed and improved by He and applied to nonlinear oscillators with discontinuities (He, 2004), nonlinear wave equations (He, 2005a), boundary value problem (He, 2006), limit cycle and bifurcation of nonlinear problems (He, 2005b), and many other subjects (He, 1999, 2000, 2003, 2004). It can be said that homotopy perturbation method is a universal one, is able to solve various kinds of nonlinear functional equations.
2
2 Basic idea of method
For the purpose of applications illustration of the methodology of the proposed method, using homotopy perturbation method, we consider the following nonlinear differential equation
(1)
(2)
where
is a general differential operator,
is a known analytic function,
is a boundary condition and
is the boundary of the domain
.
The operator
can be generally divided into two operators,
and
, where
is a linear, while
is a nonlinear operator. Eq. (1) can be, therefore, written as follows
(3)
Using the homotopy technique, we construct a homotopy
which satisfies
(4)
or
(5)
where
, is called homotopy parameter, and
is an initial approximation for the solution of Eq. (1), which satisfies the boundary conditions. Obviously from Eqs. (4) and (5) we will have
(6)
(7)
we can assume that the solution of (4) or (5) can be expressed as a series in
, as follows
(8)
Setting
, results in the approximate solution of Eq. (1)
(9)
In this paper, we consider Cauchy problem for the nonlinear parabolic–hyperbolic equation of the following type
with initial conditions
where the nonlinear term is represented by
, and
is the Laplace operator in
.
3
3 Examples
Example 1
Consider the following equation
(10)
subject to the following initial conditions
(11)
With the exact solution
To solve Eq. (10) by homotopy perturbation method, we construct the following homotopy
or
(12)
Suppose the solution of Eq. (12) has the following form
(13)
Substituting (13) into (12) and equating the coefficients of the terms with the identical powers of
leads to
For simplicity we take
. So we derive the following recurrent relation for
.
(14)
The solution reads
and by repeating this approach we obtain,
. Therefore, the approximate solution of Example 1 can be readily obtained by
and hence,
, which is an exact solution.
Example 2
Consider the following equation
(15)
subject to the initial conditions
(16)
With the exact solution
To solve Eq. (15) by homotopy perturbation method, we construct the following homotopy
or
(17)
Suppose the solution of Eq. (17) has the following form
(18)
Substituting (18) into (12) and equating the coefficients of the terms with the identical powers of
leads to
starting with
(19)
We have the following recurrent equations for
(20)
We obtain the following results
Solution of Eq. (15) will be derived by adding these terms, so
Example 3
Consider the following equation
(21)
with initial condition,
To solve Eq. (21) by homotopy perturbation method, we construct the following homotopy
or
(22)
Suppose the solution of Eq. (22) has the form (8), substituting (8) into (22), and comparing the terms with identical powers of
, leads to
We take
(23)
We have the following recurrent equations for
.
(24)
With the aid of the initial approximation given by Eq. (23) and the iteration formula (24) we get the other of component as follows
The solution in a series form is
which is an exact solution.
Example 4
Consider the following equation
(25)
subject to the initial condition,
(26)
With the exact solution
To solve Eq. (25) by homotopy perturbation method, we construct the following homotopy
or
(27)
Suppose the solution of Eq. (27) has the following form (8), substituting (8) into (27) and equating the coefficients of the terms with the identical powers of
leads to
We take
(28)
So we have
(29)
With the aid of the initial approximation given by Eq. (28) and the iteration formula (29) we get the other of component as follows
Therefore, the approximate solution can be readily obtained by
hence,
, which is an exact solution of Example 4.
4
4 Conclusion
In this work, we used homotopy perturbation method for solving nonlinear partial differential parabolic–hyperbolic equations. The results have been approved the efficiency of this method for solving these problems. The solution obtained by homotopy perturbation method is valid for not only weakly nonlinear equations but also for strong ones. Furthermore, accurate solutions were derived from first-order approximations in the examples presented in this paper.