Homotopy perturbation method for special nonlinear partial differential equations

A. Roozi; E. Alibeiki; S.S. Hosseini; S.M. Shafiof; M. Ebrahimi

doi:10.1016/j.jksus.2010.06.014

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Original article

23 (

); 99-103

doi:

10.1016/j.jksus.2010.06.014

Homotopy perturbation method for special nonlinear partial differential equations

A. Roozi^a,*, E. Alibeiki^b, S.S. Hosseini^b, S.M. Shafiof^c, M. Ebrahimi^d

a Payame Noor University of Gonbad Kavos, Gonbad, Iran

b Department of Engineering, Islamic Azad University of Gonbad Kavos Branch, Gonbad, Iran

c Payame Noor University of Khorasgan, Isfahan, Iran

d Industrial Management Institute, Gorgan, Iran

*Corresponding author. Tel.: +98 1723345324 roozi_afshin@yahoo.com (A. Roozi)

Received: 2010-6-11, Accepted: 2010-6-19, Published: 2010-01-01

Disclaimer:
This article was originally published by Elsevier and was migrated to Scientific Scholar after the change of Publisher.

Available online 25 June 2010

Abstract

In this article, homotopy perturbation method is applied to solve nonlinear parabolic–hyperbolic partial differential equations. Examples of one-dimensional and two-dimensional are presented to show the ability of the method for such equations.

Keywords

Homotopy perturbation method

Special nonlinear partial differential equations

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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 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)

A (u) - f (r) = 0, r \in Ω,

(2)

B (u, \partial u / \partial n) = 0, r \in Γ,

where

A

is a general differential operator,

f (r)

is a known analytic function,

B

is a boundary condition and

Γ

is the boundary of the domain

Ω

The operator $A$ can be generally divided into two operators, $L$ and $N$ , where $L$ is a linear, while $N$ is a nonlinear operator. Eq. (1) can be, therefore, written as follows

(3)

L (u) + N (u) - f (r) = 0 .

Using the homotopy technique, we construct a homotopy

U (r, p) : Ω \times [0, 1] \to R

which satisfies

(4)

H (U, p) = (1 - p) [L (U) - L (u_{0}))] + p [A (U) - f (r)] = 0, p \in [0, 1], r \in Ω,

(5)

H (U, p) = L (U) - L (u_{0}) + pL (u_{0}) + p [N (U) - f (r)] = 0,

where

p \in [0, 1]

, is called homotopy parameter, and

u_{0}

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)

H (U, 0) = L (U) - L (u_{0}) = 0,

(7)

H (U, 1) = A (U) - f (r) = 0,

we can assume that the solution of (4) or (5) can be expressed as a series in

p

, as follows

(8)

U = U_{0} + {pU}_{1} + p^{2} U_{2} + \dots .

Setting

p = 1

, results in the approximate solution of Eq. (1)

(9)

u = \lim_{p \to 1} U = U_{0} + U_{1} + U_{2} + \dots .

In this paper, we consider Cauchy problem for the nonlinear parabolic–hyperbolic equation of the following type

(\frac{\partial}{\partial t} - Δ) (\frac{\partial^{2}}{\partial t^{2}} - Δ) u = F (u),

with initial conditions

\frac{\partial^{k} u}{\partial t^{k}} (0, X) = φ_{k} (X), X = (x_{1}, x_{2}, \dots, x_{i}), k = 0, 1, 2 .

where the nonlinear term is represented by

F (u)

, and

Δ

is the Laplace operator in

R^{n}

3 Examples

Example 1

Consider the following equation

(10)

(\frac{\partial}{\partial t} - \frac{\partial^{2}}{\partial x^{2}}) (\frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}}) u = - {(\frac{1}{3} \frac{\partial^{2} u}{\partial x^{2}})}^{2} + {(\frac{1}{6} \frac{\partial^{2} u}{\partial t^{2}})}^{3} - 16 u,

subject to the following initial conditions

(11)

\begin{matrix} u (0, x) = - x^{4}, \\ \frac{\partial u}{\partial t} (0, x) = 0, \\ \frac{\partial^{2} u}{\partial t^{2}} (0, x) = 0 . \end{matrix}

With the exact solution

u (t, x) = - x^{4} + 4 t^{3} .

To solve Eq. (10) by homotopy perturbation method, we construct the following homotopy

(1 - p) (\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}}) + p (\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} U}{\partial t \partial x^{2}} - \frac{\partial^{4} U}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U}{\partial x^{4}} + {(\frac{1}{3} \frac{\partial^{2} U}{\partial x^{2}})}^{2} - {(\frac{1}{6} \frac{\partial^{2} U}{\partial t^{2}})}^{3} + 16 U) = 0,

(12)

\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}} + p (\frac{\partial^{3} u_{0}}{\partial t^{3}} - \frac{\partial^{3} U}{\partial t \partial x^{2}} - \frac{\partial^{4} U}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U}{\partial x^{4}} + {(\frac{1}{3} \frac{\partial^{2} U}{\partial x^{2}})}^{2} - {(\frac{1}{6} \frac{\partial^{2} U}{\partial t^{2}})}^{3} + 16 U) = 0 .

Suppose the solution of Eq. (12) has the following form

(13)

U = U_{0} + {pU}_{1} + p^{2} U_{2} + \dots .

Substituting (13) into (12) and equating the coefficients of the terms with the identical powers of

p

leads to

\begin{matrix} p^{0} : \frac{\partial^{3} U_{0}}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}} = 0, \\ p^{1} : \frac{\partial^{3} U_{1}}{\partial t^{3}} + \frac{\partial^{3} u_{0}}{\partial t^{3}} - \frac{\partial^{3} U_{0}}{\partial t \partial x^{2}} - \frac{\partial^{4} U_{0}}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U_{0}}{\partial x^{4}} \\ + {(\frac{1}{3} \frac{\partial^{2} U_{0}}{\partial x^{2}})}^{2} - {(\frac{1}{6} \frac{\partial^{2} U_{0}}{\partial t^{2}})}^{3} + 16 U_{0} = 0, \\ p^{2} : \frac{\partial^{3} U_{2}}{\partial t^{3}} - \frac{\partial^{3} U_{1}}{\partial t \partial x^{2}} - \frac{\partial^{4} U_{1}}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U_{1}}{\partial x^{4}} + \frac{2}{9} \frac{\partial^{2} U_{0}}{\partial x^{2}} \frac{\partial^{2} U_{1}}{\partial x^{2}} \\ - \frac{1}{72} \frac{\partial^{2} U_{1}}{\partial t^{2}} \frac{\partial^{2} U_{0}}{\partial t^{2}} \frac{\partial^{2} U_{0}}{\partial t^{2}} + 16 U_{1} = 0, \\ ⋮ \\ p^{j} : \frac{\partial^{3} U_{j}}{\partial t^{3}} - \frac{\partial^{3} U_{j - 1}}{\partial t \partial x^{2}} - \frac{\partial^{4} U_{j - 1}}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U_{j - 1}}{\partial x^{4}} + \frac{1}{9} \sum_{k = 0}^{j - 1} \frac{\partial^{2} U_{k}}{\partial x^{2}} \frac{\partial^{2} U_{j - 1 - k}}{\partial x^{2}} \\ - \frac{1}{216} \sum_{i = 0}^{j - 1} \sum_{k = 0}^{j - i - 1} \frac{\partial U_{i}}{\partial t} \frac{\partial U_{k}}{\partial t} \frac{\partial U_{j - k - i - 1}}{\partial t} + 16 U_{j - 1} = 0, \\ ⋮ \end{matrix}

For simplicity we take

U_{0} = u_{0} = - x^{4}

. So we derive the following recurrent relation for

j = 1, 2, 3, \dots

(14)

U_{j} = - \int_{0}^{t} \int_{0}^{t} \int_{0}^{t} (\begin{matrix} - \frac{\partial^{3} U_{j - 1}}{\partial x^{2} \partial ξ_{1}} - \frac{\partial^{4} U_{j - 1}}{\partial x^{2} \partial ξ_{1}^{2}} + \frac{\partial^{4} U_{j - 1}}{\partial x^{4}} + \frac{1}{9} \sum_{k = 0}^{j - 1} \frac{\partial^{2} U_{k}}{\partial x^{2}} \frac{\partial^{2} U_{j - 1 - k}}{\partial x^{2}} \\ - \frac{1}{216} \sum_{i = 0}^{j - 1} \sum_{k = 0}^{j - i - 1} \frac{\partial U_{i}}{\partial t} \frac{\partial U_{k}}{\partial t} \frac{\partial U_{j - k - i - 1}}{\partial t} + 16 U_{j - 1} \end{matrix}) d ξ_{1} d ξ_{2} dt .

The solution reads

\begin{matrix} U_{1} = 0, \\ U_{2} = 0, \\ U_{3} = 4 t^{3}, \\ U_{4} = 0, \end{matrix}

and by repeating this approach we obtain,

U_{5} = U_{6} = \dots = 0

. Therefore, the approximate solution of Example 1 can be readily obtained by

u = \sum_{i = 0}^{\infty} U_{i} = - x^{4} + t^{3},

and hence,

u = - x^{4} + 4 t^{3}

, which is an exact solution.

Example 2

Consider the following equation

(15)

(\frac{\partial}{\partial t} - \frac{\partial^{2}}{\partial x^{2}}) (\frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}}) u = {(\frac{\partial^{2} u}{\partial t^{2}})}^{2} - {(\frac{\partial^{2} u}{\partial x^{2}})}^{2} - 2 u^{2},

subject to the initial conditions

(16)

\begin{matrix} u (0, x) = e^{x}, \\ \frac{\partial u}{\partial t} (0, x) = e^{x}, \\ \frac{\partial^{2} u}{\partial t^{2}} (0, x) = e^{x} . \end{matrix}

With the exact solution

u (t, x) = e^{x + t} .

To solve Eq. (15) by homotopy perturbation method, we construct the following homotopy

(1 - p) (\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}}) + p (\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} U}{\partial x \partial t^{2}} - \frac{\partial^{4} U}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U}{\partial x^{4}} - {(\frac{\partial^{2} U}{\partial t^{2}})}^{2} + {(\frac{\partial^{2} U}{\partial x^{2}})}^{2} + 2 U^{2}) = 0,

(17)

\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}} + p (\frac{\partial^{3} u_{0}}{\partial t^{3}} - \frac{\partial^{3} U}{\partial x \partial t^{2}} - \frac{\partial^{4} U}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U}{\partial x^{4}} - {(\frac{\partial^{2} U}{\partial t^{2}})}^{2} + {(\frac{\partial^{2} U}{\partial x^{2}})}^{2} + 2 U^{2}) = 0 .

Suppose the solution of Eq. (17) has the following form

(18)

U = U_{0} + {pU}_{1} + p^{2} U_{2} + \dots .

Substituting (18) into (12) and equating the coefficients of the terms with the identical powers of

p

leads to

\begin{matrix} p^{0} : \frac{\partial^{3} U_{0}}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}} = 0, \\ p^{1} : \frac{\partial^{3} U_{1}}{\partial t^{3}} + \frac{\partial^{3} u_{0}}{\partial t^{3}} - \frac{\partial^{3} U_{0}}{\partial t \partial x^{2}} - \frac{\partial^{4} U_{0}}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U_{0}}{\partial x^{4}} - {(\frac{\partial^{2} U_{0}}{\partial t^{2}})}^{2} \\ + {(\frac{\partial^{2} U_{0}}{\partial x^{2}})}^{2} + 2 U_{0}^{2} = 0, \\ p^{2} : \frac{\partial^{3} U_{2}}{\partial t^{3}} - \frac{\partial^{3} U_{1}}{\partial t \partial x^{2}} - \frac{\partial^{4} U_{1}}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U_{1}}{\partial x^{4}} - 2 \frac{\partial^{2} U_{1}}{\partial t^{2}} \frac{\partial^{2} U_{0}}{\partial t^{2}} \\ + 2 \frac{\partial^{2} U_{0}}{\partial x^{2}} \frac{\partial^{2} U_{1}}{\partial x^{2}} + 4 U_{0} U_{1} = 0, \\ ⋮ \\ p^{j} : \frac{\partial^{3} U_{j}}{\partial t^{3}} - \frac{\partial^{3} U_{j - 1}}{\partial t \partial x^{2}} - \frac{\partial^{4} U_{j - 1}}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U_{j - 1}}{\partial x^{4}} - \sum_{k = 0}^{j - 1} \frac{\partial^{2} U_{k}}{\partial t^{2}} \frac{\partial^{2} U_{j - 1 - k}}{\partial t^{2}} \\ + \sum_{k = 0}^{j - 1} \frac{\partial^{2} U_{k}}{\partial x^{2}} \frac{\partial^{2} U_{j - 1 - k}}{\partial x^{2}} + 2 \sum_{k = 0}^{j - 1} U_{k} U_{j - 1 - k} = 0, \\ ⋮ \end{matrix}

starting with

(19)

U_{0} = u_{0} = (1 + t + \frac{t^{2}}{2}) e^{x} .

We have the following recurrent equations for

j = 1, 2, 3, \dots

(20)

U_{j} = - \int_{0}^{t} \int_{0}^{t} \int_{0}^{t} (\begin{matrix} - \frac{\partial^{3} U_{j - 1}}{\partial x^{2} \partial ξ_{1}} - \frac{\partial^{4} U_{j - 1}}{\partial x^{2} \partial ξ_{1}^{2}} + \frac{\partial^{4} U_{j - 1}}{\partial x^{4}} - \sum_{k = 0}^{j - 1} \frac{\partial U_{k}}{\partial ξ_{1}} \frac{\partial^{2} U_{j - 1 - k}}{\partial ξ_{1}^{2}} \\ + \sum_{k = 0}^{j - 1} \frac{\partial^{2} U_{k}}{\partial x^{2}} \frac{\partial^{2} U_{j - 1 - k}}{\partial x^{2}} + 2 \sum_{k = 0}^{j - 1} U_{k} U_{j - 1 - k} \end{matrix}) d ξ_{1} d ξ_{2} dt .

We obtain the following results

\begin{matrix} U_{1} = \frac{t^{3}}{6} e^{x}, \\ U_{2} = \frac{t^{4}}{24} e^{x}, \\ U_{3} = \frac{t^{5}}{120} e^{x}, vdots \end{matrix}

Solution of Eq. (15) will be derived by adding these terms, so

u = \sum_{i = 0}^{\infty} U_{i} = e^{x + t} .

Example 3

Consider the following equation

(21)

(\frac{\partial}{\partial t} - \frac{\partial^{2}}{\partial x^{2}}) (\frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}}) u = u \frac{\partial u}{\partial t} + \frac{\partial^{2} u}{\partial t^{2}} \frac{\partial u}{\partial x},

with initial condition,

\begin{matrix} u (0, x) = \cos x, \\ \frac{\partial u}{\partial t} (0, x) = - \sin x, \\ \frac{\partial^{2} u}{\partial t^{2}} (0, x) = - \cos x . \end{matrix}

To solve Eq. (21) by homotopy perturbation method, we construct the following homotopy

(1 - p) (\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}}) + p (\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} U}{\partial x \partial t^{2}} - \frac{\partial^{4} U}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U}{\partial x^{4}} - U \frac{\partial U}{\partial t} - \frac{\partial^{2} U}{\partial t^{2}} \frac{\partial U}{\partial x}) = 0,

(22)

\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}} + p (\frac{\partial^{3} u_{0}}{\partial t^{3}} - \frac{\partial^{3} U}{\partial x \partial t^{2}} - \frac{\partial^{4} U}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U}{\partial x^{4}} - U \frac{\partial U}{\partial t} - \frac{\partial^{2} U}{\partial t^{2}} \frac{\partial U}{\partial x}) = 0 .

Suppose the solution of Eq. (22) has the form (8), substituting (8) into (22), and comparing the terms with identical powers of

p

, leads to

\begin{matrix} p^{0} : \frac{\partial^{3} U_{0}}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}} = 0, \\ p^{1} : \frac{\partial^{3} U_{1}}{\partial t^{3}} + \frac{\partial^{3} u_{0}}{\partial t^{3}} - \frac{\partial^{3} U_{0}}{\partial t \partial x^{2}} - \frac{\partial^{4} U_{0}}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U_{0}}{\partial x^{4}} \\ - U_{0} \frac{\partial U_{0}}{\partial t} - \frac{\partial^{2} U_{0}}{\partial t^{2}} \frac{\partial U_{0}}{\partial x} = 0, \\ p^{2} : \frac{\partial^{3} U_{2}}{\partial t^{3}} - \frac{\partial^{3} U_{1}}{\partial t \partial x^{2}} - \frac{\partial^{4} U_{1}}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U_{1}}{\partial x^{4}} - U_{0} \frac{\partial U_{1}}{\partial t} - U_{1} \frac{\partial U_{0}}{\partial t} \\ - \frac{\partial^{2} U_{1}}{\partial t^{2}} \frac{\partial U_{0}}{\partial x} - \frac{\partial^{2} U_{0}}{\partial t^{2}} \frac{\partial U_{1}}{\partial x} = 0, \\ ⋮ \\ p^{j} : \frac{\partial^{3} U_{j}}{\partial t^{3}} - \frac{\partial^{3} U_{j - 1}}{\partial t \partial x^{2}} - \frac{\partial^{4} U_{j - 1}}{\partial x^{2} \partial t^{2}} + \frac{\partial^{4} U_{j - 1}}{\partial x^{4}} - \sum_{k = 0}^{j - 1} U_{k} \frac{\partial U_{j - 1 - k}}{\partial t} \\ - \sum_{k = 0}^{j - 1} \frac{\partial^{2} U_{k}}{\partial t^{2}} \frac{\partial U_{j - 1 - k}}{\partial x} = 0, \\ ⋮ \end{matrix}

We take

(23)

U_{0} = u_{0} = \cos x - t \sin x - \frac{t^{2}}{2} \cos x .

We have the following recurrent equations for

j = 1, 2, 3, \dots

(24)

U_{j} = - \int_{0}^{t} \int_{0}^{t} \int_{0}^{t} (\begin{matrix} - \frac{\partial^{3} U_{j - 1}}{\partial x^{2} \partial ξ_{1}} - \frac{\partial^{4} U_{j - 1}}{\partial x^{2} \partial ξ_{1}^{2}} + \frac{\partial^{4} U_{j - 1}}{\partial x^{4}} - \sum_{k = 0}^{j - 1} \frac{\partial U_{k}}{\partial ξ_{1}} \frac{\partial^{2} U_{j - 1 - k}}{\partial ξ_{1}^{2}} \\ + \sum_{k = 0}^{j - 1} \frac{\partial^{2} U_{k}}{\partial x^{2}} \frac{\partial^{2} U_{j - 1 - k}}{\partial x^{2}} + 2 \sum_{k = 0}^{j - 1} U_{k} U_{j - 1 - k} \end{matrix}) d ξ_{1} d ξ_{2} dt .

With the aid of the initial approximation given by Eq. (23) and the iteration formula (24) we get the other of component as follows

\begin{matrix} U_{1} = \frac{t^{3}}{3!} \sin x + \frac{t^{4}}{4!} \cos x, \\ U_{2} = - \frac{t^{5}}{5!} \sin x - \frac{t^{6}}{6!} \cos x, \\ U_{3} = \frac{t^{7}}{7!} \sin x + \frac{t^{8}}{8!} \cos x, \\ ⋮ \end{matrix}

The solution in a series form is

u = U_{0} + U_{1} + U_{2} + \dots = \cos x (1 - \frac{t^{2}}{2} + \frac{t^{4}}{4!} + \dots) - \sin x (t - \frac{t^{3}}{3!} + \frac{t^{5}}{5!} + \dots) = \cos (x + t),

which is an exact solution.

Example 4

Consider the following equation

(25)

(\frac{\partial}{\partial t} - \frac{\partial^{2}}{\partial x_{1}^{2}} - \frac{\partial^{2}}{\partial x_{2}^{2}}) (\frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x_{1}^{2}} - \frac{\partial^{2}}{\partial x_{2}^{2}}) u = \frac{\partial u}{\partial t} - 2 u,

subject to the initial condition,

(26)

\begin{matrix} u (0, x_{1}, x_{2}) = \sinh (x_{1} + x_{2}), \\ \frac{\partial u}{\partial t} (0, x_{1}, x_{2}) = 2 \sinh (x_{1} + x_{2}), \\ \frac{\partial^{2} u}{\partial t^{2}} (0, x_{1}, x_{2}) = 4 \sinh (x_{1} + x_{2}) . \end{matrix}

With the exact solution

u (t, x_{1}, x_{2}) = \sinh (x_{1} + x_{2}) e^{2 t} .

To solve Eq. (25) by homotopy perturbation method, we construct the following homotopy

(1 - p) (\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}}) + p (\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} U}{\partial t \partial x_{1}^{2}} - \frac{\partial^{3} U}{\partial t \partial x_{2}^{2}} - \frac{\partial^{4} U}{\partial x_{1}^{2} \partial t^{2}} + \frac{\partial^{4} U}{\partial x_{1}^{4}} + \frac{\partial^{4} U}{\partial x_{1}^{2} \partial x_{2}^{2}} - \frac{\partial^{4} U}{\partial x_{2}^{2} \partial t^{2}} + \frac{\partial^{4} U}{\partial x_{2}^{2} \partial x_{1}^{2}} + \frac{\partial^{4} U}{\partial x_{2}^{4}} - \frac{\partial U}{\partial t} + 2 U) = 0,

(27)

\frac{\partial^{3} U}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}} + p (\frac{\partial^{3} u_{0}}{\partial t^{3}} - \frac{\partial^{3} U}{\partial t \partial x_{1}^{2}} - \frac{\partial^{3} U}{\partial t \partial x_{2}^{2}} - \frac{\partial^{4} U}{\partial x_{1}^{2} \partial t^{2}} + \frac{\partial^{4} U}{\partial x_{1}^{4}} + \frac{\partial^{4} U}{\partial x_{1}^{2} \partial x_{2}^{2}} - \frac{\partial^{4} U}{\partial x_{2}^{2} \partial t^{2}} + \frac{\partial^{4} U}{\partial x_{2}^{2} \partial x_{1}^{2}} + \frac{\partial^{4} U}{\partial x_{2}^{4}} - \frac{\partial U}{\partial t} + 2 U) = 0 .

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

p

leads to

\begin{matrix} p^{0} : \frac{\partial^{3} U_{0}}{\partial t^{3}} - \frac{\partial^{3} u_{0}}{\partial t^{3}} = 0, \\ p^{1} : \frac{\partial^{3} U_{1}}{\partial t^{3}} + \frac{\partial^{3} u_{0}}{\partial t^{3}} - \frac{\partial^{3} U_{0}}{\partial t \partial x_{1}^{2}} - \frac{\partial^{3} U_{0}}{\partial t \partial x_{2}^{2}} - \frac{\partial^{4} U_{0}}{\partial x_{1}^{2} \partial t^{2}} \\ + \frac{\partial^{4} U_{0}}{\partial x_{1}^{4}} + \frac{\partial^{4} U_{0}}{\partial x_{1}^{2} \partial x_{2}^{2}} - \frac{\partial^{4} U_{0}}{\partial x_{2}^{2} \partial t^{2}} \\ + \frac{\partial^{4} U_{0}}{\partial x_{2}^{2} \partial x_{1}^{2}} + \frac{\partial^{4} U_{0}}{\partial x_{2}^{4}} - \frac{\partial U_{0}}{\partial t} + 2 U_{0} = 0, \\ \frac{\partial U_{1}}{\partial t} (0, x_{1}, x_{2}) = 2 \sinh (x_{1} + x_{2}), \\ p^{2} : \frac{\partial^{3} U_{2}}{\partial t^{3}} - \frac{\partial^{3} U_{1}}{\partial t \partial x_{1}^{2}} - \frac{\partial^{3} U_{1}}{\partial t \partial x_{2}^{2}} - \frac{\partial^{4} U_{1}}{\partial x_{1}^{2} \partial t^{2}} + \frac{\partial^{4} U_{1}}{\partial x_{1}^{4}} \\ + \frac{\partial^{4} U_{1}}{\partial x_{1}^{2} \partial x_{2}^{2}} - \frac{\partial^{4} U_{1}}{\partial x_{2}^{2} \partial t^{2}} + \frac{\partial^{4} U_{1}}{\partial x_{2}^{2} \partial x_{1}^{2}} \\ + \frac{\partial^{4} U_{1}}{\partial x_{2}^{4}} - \frac{\partial U_{1}}{\partial t} + 2 U_{1} = 0, \\ \frac{\partial^{2} U_{2}}{\partial t^{2}} (0, x_{1}, x_{2}) = 4 \sinh (x_{1} + x_{2}), \\ ⋮ \\ p^{j} : \frac{\partial^{3} U_{j}}{\partial t^{3}} - \frac{\partial^{3} U_{j - 1}}{\partial t \partial x_{1}^{2}} - \frac{\partial^{3} U_{j - 1}}{\partial t \partial x_{2}^{2}} - \frac{\partial^{4} U_{j - 1}}{\partial x_{1}^{2} \partial t^{2}} + \frac{\partial^{4} U_{j - 1}}{\partial x_{1}^{4}} \\ + \frac{\partial^{4} U_{j - 1}}{\partial x_{1}^{2} \partial x_{2}^{2}} - \frac{\partial^{4} U_{j - 1}}{\partial x_{2}^{2} \partial t^{2}} + \frac{\partial^{4} U_{j - 1}}{\partial x_{2}^{2} \partial x_{1}^{2}} \\ + \frac{\partial^{4} U_{j - 1}}{\partial x_{2}^{4}} - \frac{\partial U_{j - 1}}{\partial t} + 2 U_{j - 1} = 0, \\ ⋮ \end{matrix}

We take

(28)

U_{0} = u_{0} = \sinh (x_{1} + x_{2}) .

So we have

(29)

U_{j} = - \int_{0}^{t} \int_{0}^{t} \int_{0}^{t} (\begin{matrix} - \frac{\partial^{3} U_{j - 1}}{\partial ξ_{1} \partial x_{1}^{2}} - \frac{\partial^{3} U_{j - 1}}{\partial ξ_{1} \partial x_{2}^{2}} - \frac{\partial^{4} U_{j - 1}}{\partial x_{1}^{2} \partial ξ_{1}^{2}} + \frac{\partial^{4} U_{j - 1}}{\partial x_{1}^{4}} + \frac{\partial^{4} U_{j - 1}}{\partial x_{1}^{2} \partial x_{2}^{2}} - \frac{\partial^{4} U_{j - 1}}{\partial x_{2}^{2} \partial ξ_{1}^{2}} \\ + \frac{\partial^{4} U_{j - 1}}{\partial x_{2}^{2} \partial x_{1}^{2}} + \frac{\partial^{4} U_{j - 1}}{\partial x_{2}^{4}} - \sum_{k = 0}^{j - 1} \frac{\partial U_{k}}{\partial ξ_{1}} \frac{\partial U_{j - 1 - k}}{\partial ξ_{1}} + 4 \sum_{k = 0}^{j - 1} U_{k} U_{j - 1 - k} \end{matrix}) d ξ_{1} d ξ_{2} dt .

With the aid of the initial approximation given by Eq. (28) and the iteration formula (29) we get the other of component as follows

\begin{matrix} U_{1} = \frac{4}{3} t^{3} \sinh (x_{1} + x_{2}), \\ U_{2} = \frac{2}{3} t^{4} \sinh (x_{1} + x_{2}), \\ U_{3} = \frac{4}{15} t^{4} \sinh (x_{1} + x_{2}), \\ ⋮ \end{matrix}

Therefore, the approximate solution can be readily obtained by

u = U_{0} + U_{1} + U_{2} + \dots = e^{2 t} \sinh (x_{1} + x_{2}),

hence,

u = e^{2 t} \sinh (x_{1} + x_{2})

, which is an exact solution of Example 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.

Acknowledgement

With special thankful from Azad University of Gonbad Kavos Branch that this article is resulting from research project published in this University.

References

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He J.H., . Homotopy perturbation method: a new nonlinear analytical technique. Applied Mathematics and Computation. 2003;135:73-79.
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He J.H., . The homotopy perturbation method for nonlinear oscillators with discontinuities. Applied Mathematics and Computation. 2004;151:287-292.
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He J.H., . Comparison of homotopy perturbation method and homotopy analysis method. Applied Mathematics and Computation. 2004;156:527-539.
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He J.H., . Limit cycle and bifurcation of nonlinear problems. Chaos, Solitons and Fractals. 2005;26(3):827-833.
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Odibat Z., Momani S., . Modified homotopy perturbation method: application to quadratic Riccati differential equation of fractional order. Chaos, Solitons and Fractals. 2008;36(1):167-174.
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Ozis T., Yıldırım A., . Traveling wave solution of Korteweg–de Vries equation using He’s homotopy perturbation method. International Journal of Nonlinear Sciences and Numerical Simulation. 2007;8(2):239-242.
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Siddiqui A.M., Mahmood R., Ghori Q.K., . Homotopy perturbation method for thin film flow of a third grade fluid down an inclined plane. Chaos, Solitons and Fractals. 2008;35(1):140-147.
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Introduction

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Examples

Conclusion

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[1] Biazar J., Eslami M., Ghazvini H., . Homotopy perturbation method for systems of partial differential equations. International Journal of Nonlinear Science and Numerical Simulation. 2007;8(3):411-416.
[Google Scholar]

[2] Ghori O.K., Ahmed M., Siddiqui A.M., . Application of homotopy perturbation method to squeezing flow of a Newtonian fluid. International Journal of Nonlinear Science and Numerical Simulation. 2007;8(2):179-184.
[Google Scholar]

[3] He J.H., . Homotopy perturbation technique. Computer Methods in Applied Mechanics and Engineering. 1999;178:257-262.
[Google Scholar]

[4] He J.H., . A coupling method of homotopy technique and perturbation technique for nonlinear problems. International Journal of Non-Linear Mechanics. 2000;35(1):37-43.
[Google Scholar]

[5] He J.H., . Homotopy perturbation method: a new nonlinear analytical technique. Applied Mathematics and Computation. 2003;135:73-79.
[Google Scholar]

[6] He J.H., . The homotopy perturbation method for nonlinear oscillators with discontinuities. Applied Mathematics and Computation. 2004;151:287-292.
[Google Scholar]

[7] He J.H., . Comparison of homotopy perturbation method and homotopy analysis method. Applied Mathematics and Computation. 2004;156:527-539.
[Google Scholar]

[8] He J.H., . Application of homotopy perturbation method to nonlinear wave equations. Chaos, Solitons and Fractals. 2005;26:695-700.
[Google Scholar]

[9] He J.H., . Limit cycle and bifurcation of nonlinear problems. Chaos, Solitons and Fractals. 2005;26(3):827-833.
[Google Scholar]

[10] He J.H., . Homotopy perturbation method for solving boundary value problems. Physics Letters A. 2006;350:87-88.
[Google Scholar]

[11] Noor Muhammad Aslam, Mohyud-Din Syed Tauseef, . Homotopy perturbation method for solving sixth-order boundary value problems. Computers & Mathematics with Applications. 2008;55(12):2953-2972.
[Google Scholar]

[12] Odibat Z., Momani S., . Modified homotopy perturbation method: application to quadratic Riccati differential equation of fractional order. Chaos, Solitons and Fractals. 2008;36(1):167-174.
[Google Scholar]

[13] Ozis T., Yıldırım A., . Traveling wave solution of Korteweg–de Vries equation using He’s homotopy perturbation method. International Journal of Nonlinear Sciences and Numerical Simulation. 2007;8(2):239-242.
[Google Scholar]

[14] Siddiqui A.M., Mahmood R., Ghori Q.K., . Homotopy perturbation method for thin film flow of a third grade fluid down an inclined plane. Chaos, Solitons and Fractals. 2008;35(1):140-147.
[Google Scholar]

[15] Yıldırım A., Ozis T., . Solutions of singular IVPs of Lane–Emden type by homotopy perturbation method. Physics Letters A. 2007;369:70-76.
[Google Scholar]