## Abstract

In this paper, new homotopy perturbation method (NHPM) (Biazar et al., 2007) is applied to obtain solutions of system of Burgers equations. In this method, the solution is considered as an infinite series expansion where it converges rapidly to the exact solution. In order to show the ability and reliability of the method some examples are provided. The results reveal that the method is very effective and simple. The modified method accelerates the rapid convergence of the series solution and reduces the size of work.

## Keywords

System of Burger equation; New homotopy perturbation method

## 1. Introduction

The investigation of the numerical and exact solutions for nonlinear partial differential equations (NLPDEs) plays an important role in the study of nonlinear physical phenomena. Nonlinear wave phenomena appear in various scientific and engineering fields, such as fluid mechanics, plasma physics, optical fibers, biology, solid-state physics, chemical kinematics, chemical physics and geochemistry. The dispersion, dissipation, diffusion, reaction and convection of nonlinear wave phenomena are very important in nonlinear wave equations. In the past several decades, there have been significant improvements in the study of exact solutions. In the recent years, many authors mainly had paid attention to obtain solutions of NLPDEs by using various methods.

The purpose of this paper was to apply new homotopy perturbation method (NHPM) [10] to coupled Burgers equations.

General form of system of Burgers equations can be considered as the following forms

 ${\displaystyle {\frac {\partial u}{\partial t}}-{\frac {{\partial }^{2}u}{\partial x^{2}}}+}$${\displaystyle \eta u{\frac {\partial u}{\partial x}}+\alpha {\frac {\partial }{\partial x}}(uv)=}$${\displaystyle f(x{\mbox{,}}t){\mbox{,}}}$ ${\displaystyle {\frac {\partial v}{\partial t}}-{\frac {\partial v}{\partial x^{2}}}+}$${\displaystyle \eta v{\frac {\partial v}{\partial x}}+\beta {\frac {\partial }{\partial x}}(uv)=}$${\displaystyle g(x{\mbox{,}}t){\mbox{,}}}$
(1)

with initial conditions

 ${\displaystyle u(x{\mbox{,}}0)={\varphi }_{1}(x){\mbox{,}}}$ ${\displaystyle v(x{\mbox{,}}0)={\varphi }_{2}(x){\mbox{,}}}$

where ${\textstyle \alpha {\mbox{,}}\beta {\mbox{,}}\eta }$ are constants, and ${\textstyle f{\mbox{,}}g}$ are inhomogeneous terms.

The Burgers model of turbulence is a very important fluid dynamic model and the study of this model and the theory of shock waves have been considered by many authors, both to obtain a conceptual understanding of a class of physical flows and for testing various numerical methods. The distinctive feature of Eq. (1) is that it is the simplest mathematical formulation of the competition between nonlinear advection and viscous diffusion. Homotopy perturbation method (HPM) was proposed by He [1] and [2] which is, in fact, a coupling of the traditional perturbation method and Homotopy in topology. This method has been applied to solve linear and nonlinear equations of heat transfer [3], sixth-order boundary value problems [4], quadratic Riccati differential equation of fractional order [5], conservative truly nonlinear oscillators [6], and linear and nonlinear weakly singular VIE of second kind [7], and many others [8], [9], [10], [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28] and [29].

This paper is arranged as follows. In Section 2, the new modification of HPM for solving systems of Burgers equations is presented. The efficiency of this new method is verified by the numerical results for two examples in Section 3. Conclusion will appear in Section 4.

## 2. NHPM for coupled Burgers equations

For solving system (1) by NHPM we construct the following homotopies:

 ${\displaystyle (1-p)\left({\frac {\partial U}{\partial t}}-u_{0}\right)+}$${\displaystyle p\left({\frac {\partial U}{\partial t}}-{\frac {{\partial }^{2}U}{\partial x^{2}}}+\right.}$${\displaystyle \left.\eta U{\frac {\partial U}{\partial x}}+\alpha {\frac {\partial }{\partial x}}(UV)-\right.}$${\displaystyle \left.f(x{\mbox{,}}t)\right)=0{\mbox{,}}}$ ${\displaystyle (1-p)\left({\frac {\partial V}{\partial t}}-v_{0}\right)+}$${\displaystyle p\left({\frac {\partial V}{\partial t}}-{\frac {\partial V}{\partial x^{2}}}+\right.}$${\displaystyle \left.\eta V{\frac {\partial V}{\partial x}}+\beta {\frac {\partial }{\partial x}}(UV)-\right.}$${\displaystyle \left.g(x{\mbox{,}}t)\right)=0{\mbox{,}}}$
(2)

or

 ${\displaystyle {\frac {\partial U}{\partial t}}=u_{0}-p\left(u_{0}-{\frac {{\partial }^{2}U}{\partial x^{2}}}+\right.}$${\displaystyle \left.\eta U{\frac {\partial U}{\partial x}}+\alpha {\frac {\partial }{\partial x}}(UV)-\right.}$${\displaystyle \left.f(x{\mbox{,}}t)\right){\mbox{,}}}$ ${\displaystyle {\frac {\partial V}{\partial t}}=v_{0}-p\left(v_{0}-{\frac {\partial V}{\partial x^{2}}}+\right.}$${\displaystyle \left.\eta V{\frac {\partial V}{\partial x}}+\beta {\frac {\partial }{\partial x}}(UV)-\right.}$${\displaystyle \left.g(x{\mbox{,}}t)\right){\mbox{,}}}$
(3)

Applying the inverse operator, ${\textstyle L^{-1}={\int }_{t_{0}}^{t}(.)dt}$ to both sides of Eq. (3), we obtain

 ${\displaystyle U(x{\mbox{,}}t)=U(x{\mbox{,}}t_{0})+{\int }_{t_{0}}^{t}u_{0}dt-}$${\displaystyle p{\int }_{t_{0}}^{t}\left(u_{0}-{\frac {{\partial }^{2}U}{\partial x^{2}}}+\right.}$${\displaystyle \left.\eta U{\frac {\partial U}{\partial x}}+\alpha {\frac {\partial }{\partial x}}(UV)-\right.}$${\displaystyle \left.f(x{\mbox{,}}t)\right)dt{\mbox{,}}}$ ${\displaystyle V(x{\mbox{,}}t)=V(x{\mbox{,}}t_{0})+{\int }_{t_{0}}^{t}v_{0}dt-}$${\displaystyle p{\int }_{t_{0}}^{t}\left(v_{0}-{\frac {\partial V}{\partial x^{2}}}+\right.}$${\displaystyle \left.\eta V{\frac {\partial V}{\partial x}}+\beta {\frac {\partial }{\partial x}}(UV)-\right.}$${\displaystyle \left.g(x{\mbox{,}}t)\right)dt{\mbox{,}}}$
(4)

where

 ${\displaystyle U(x{\mbox{,}}t_{0})=u(x{\mbox{,}}t_{0}){\mbox{,}}}$ ${\displaystyle V(x{\mbox{,}}t_{0})=v(x{\mbox{,}}t_{0}){\mbox{.}}}$

Let’s present the solution of the system (4) as the following

 ${\displaystyle U=U_{0}+{pU}_{1}+p^{2}U_{2}+\cdots {\mbox{,}}}$ ${\displaystyle V=V_{0}+{pV}_{1}+p^{2}V_{2}+\cdots {\mbox{,}}}$
(5)

where ${\textstyle U_{i}{\mbox{,}}V_{i}{\mbox{,}}\quad i=1\ldots n}$, are functions which should be determined.

Suppose that the initial approximation of Eq. (1) is in the following form

 ${\displaystyle u_{0}(x{\mbox{,}}t)=\sum _{j=0}^{\infty }a_{j}(x)P_{j}(t){\mbox{,}}}$ ${\displaystyle v_{0}(x{\mbox{,}}t)=\sum _{j=0}^{\infty }b_{j}(x)P_{j}(t){\mbox{,}}}$
(6)

where ${\textstyle a_{i{\mbox{,}}j}(x_{1}{\mbox{,}}x_{2}{\mbox{,}}\ldots {\mbox{,}}x_{n-1}){\mbox{,}}i=}$${\displaystyle 1\ldots n{\mbox{,}}j=0\ldots n}$, are unknown coefficients and ${\textstyle P_{0}(t){\mbox{,}}P_{1}(t){\mbox{,}}P_{2}(t){\mbox{,}}\ldots }$ are specific functions.

Substituting (5) and (6) into (4) and equating the coefficients of p with the same powers lead to

 ${\displaystyle p^{0}:{\begin{array}{l}U_{0}(x{\mbox{,}}t)={\varphi }_{1}(x)+\sum _{j=0}^{\infty }a_{j}{\int }_{t_{0}}^{t}P_{j}(t)dt{\mbox{,}}\\V_{0}(x{\mbox{,}}t)={\varphi }_{2}(x)+\sum _{j=0}^{\infty }b_{j}{\int }_{t_{0}}^{t}P_{j}(t)dt\end{array}}}$ ${\displaystyle p^{1}:{\begin{array}{l}U_{1}(x{\mbox{,}}t)=-\sum _{j=0}^{\infty }a_{j}{\int }_{t_{0}}^{t}P_{j}(t)dt-{\int }_{t_{0}}^{t}\left(-{\frac {{\partial }^{2}U_{0}}{\partial x^{2}}}+\eta U_{0}{\frac {\partial U_{0}}{\partial x}}+\alpha {\frac {\partial }{\partial x}}(U_{0}V_{0})-f(x{\mbox{,}}t)\right)dt\\V_{1}(x{\mbox{,}}t)=-\sum _{j=0}^{\infty }b_{j}{\int }_{t_{0}}^{t}P_{j}(t)dt-{\int }_{t_{0}}^{t}\left(-{\frac {\partial V_{0}}{\partial x^{2}}}+\eta V_{0}{\frac {\partial V_{0}}{\partial x}}+\beta {\frac {\partial }{\partial x}}(U_{0}V_{0})-g(x{\mbox{,}}t)\right)dt\end{array}}}$ ${\displaystyle p^{2}:{\begin{array}{l}U_{2}(x{\mbox{,}}t)=-{\int }_{t_{0}}^{t}(-{\frac {{\partial }^{2}U_{1}}{\partial x^{2}}}+\eta U_{1}{\frac {\partial U_{0}}{\partial x}}+\eta U_{0}{\frac {\partial U_{1}}{\partial x}}+\alpha {\frac {\partial U_{0}}{\partial x}}V_{1}+\alpha {\frac {\partial V_{0}}{\partial x}}U_{1}+\alpha {\frac {\partial V_{1}}{\partial x}}U_{0}+\alpha {\frac {\partial U_{1}}{\partial x}}V_{0})dt\\V_{2}(x{\mbox{,}}t)=-{\int }_{t_{0}}^{t}(-{\frac {{\partial }^{2}V_{1}}{\partial x^{2}}}+\eta V_{1}{\frac {\partial V_{0}}{\partial x}}+\eta V_{0}{\frac {\partial V_{1}}{\partial x}}+\beta {\frac {\partial U_{0}}{\partial x}}V_{1}+\beta {\frac {\partial V_{0}}{\partial x}}U_{1}+\beta {\frac {\partial V_{1}}{\partial x}}U_{0}+\beta {\frac {\partial U_{1}}{\partial x}}V_{0})dt\end{array}}}$ ${\displaystyle \vdots }$ ${\displaystyle p^{j+1}:{\begin{array}{l}U_{j+1}(x{\mbox{,}}t)=-{\int }_{t_{0}}^{t}\left(-{\frac {{\partial }^{2}U_{j}}{\partial x^{2}}}+\eta \sum _{k=0}^{j}U_{j}{\frac {\partial U_{j-k}}{\partial x}}+\alpha \sum _{k=0}^{j}{\frac {\partial U_{j}}{\partial x}}V_{j-k}+\alpha \sum _{k=0}^{j}{\frac {\partial V_{j}}{\partial x}}U_{j-k}\right)dt\\V_{j+1}(x{\mbox{,}}t)=-{\int }_{t_{0}}^{t}\left(-{\frac {{\partial }^{2}V_{j}}{\partial x^{2}}}+\eta \sum _{k=0}^{j}V_{j}{\frac {\partial V_{j-k}}{\partial x}}+\beta \sum _{k=0}^{j}{\frac {\partial U_{j}}{\partial x}}V_{j-k}+\beta \sum _{k=0}^{j}{\frac {\partial V_{j}}{\partial x}}U_{j-k}\right)dt\end{array}}}$ ${\displaystyle \vdots }$
(7)

Now if we solve these equations in such a way that ${\textstyle U_{i{\mbox{,}}1}(x_{1}x_{2}{\mbox{,}}\ldots {\mbox{,}}x_{n-1}{\mbox{,}}t)=}$${\displaystyle 0}$, then Eq. (7) results in ${\textstyle U_{i{\mbox{,}}1}(x_{1}x_{2}{\mbox{,}}\ldots {\mbox{,}}x_{n}{\mbox{,}}t)=}$${\displaystyle U_{i{\mbox{,}}2}(x_{1}x_{2}{\mbox{,}}\ldots {\mbox{,}}x_{n}{\mbox{,}}t)=}$${\displaystyle \cdots =0}$.

Therefore the exact solution may be obtained as the following.

 ${\displaystyle u_{i}(x_{1}{\mbox{,}}x_{2}{\mbox{,}}\ldots {\mbox{,}}x_{n-1}{\mbox{,}}t)=}$${\displaystyle U_{i{\mbox{,}}0}(x_{1}{\mbox{,}}x_{2}{\mbox{,}}\ldots {\mbox{,}}x_{n-1}{\mbox{,}}t)=}$${\displaystyle f_{i}(x_{1}{\mbox{,}}x_{2}{\mbox{,}}\ldots {\mbox{,}}x_{n-1})+}$${\displaystyle \sum _{j=0}^{\infty }a_{i{\mbox{,}}j}{\int }_{t_{0}}^{t}P_{j}(t)dt{\mbox{,}}\quad i=}$${\displaystyle 1{\mbox{,}}\ldots {\mbox{,}}n{\mbox{.}}}$
(8)

It is worthwhile to mention that if ${\textstyle g(x_{1}{\mbox{,}}x_{2}{\mbox{,}}\ldots {\mbox{,}}x_{n-1}{\mbox{,}}t)}$, and ${\textstyle u_{i{\mbox{,}}0}(x_{1}{\mbox{,}}x_{2}{\mbox{,}}\ldots {\mbox{,}}x_{n-1}{\mbox{,}}t)}$, are analytic around ${\textstyle t=t_{0}}$, then their Taylor series can be defined as

 ${\displaystyle u_{0}(x{\mbox{,}}t)=\sum _{j=0}^{\infty }a_{j}(x){\left(t-t_{0}\right)}^{n}{\mbox{,}}}$ ${\displaystyle v_{0}(x{\mbox{,}}t)=\sum _{j=0}^{\infty }b_{j}(x){\left(t-t_{0}\right)}^{n}{\mbox{,}}}$ ${\displaystyle f(x{\mbox{,}}t)=\sum _{j=0}^{\infty }a_{j}^{_{\ast }}(x){\left(t-t_{0}\right)}^{n}{\mbox{,}}}$ ${\displaystyle g(x{\mbox{,}}t)=\sum _{j=0}^{\infty }b_{j}^{_{\ast }}(x){\left(t-t_{0}\right)}^{n}{\mbox{,}}}$
(9)

can be used in Eq. (7), where ${\textstyle a_{j}(x){\mbox{,}}\quad j=0{\mbox{,}}\ldots {\mbox{,}}n{\mbox{,}}b_{j}(x){\mbox{,}}\quad j=}$${\displaystyle 0{\mbox{,}}\ldots {\mbox{,}}n}$, are unknown coefficients which must be computed, and ${\textstyle a_{j}^{_{\ast }}(x){\mbox{,}}\quad j=0{\mbox{,}}\ldots {\mbox{,}}n.b_{j}^{_{\ast }}(x){\mbox{,}}\quad j=}$${\displaystyle 0{\mbox{,}}\ldots {\mbox{,}}n}$. are known ones.

The two most important steps in application of the new Homotopy perturbation method are to construct a suitable Homotopy equation and to choose a suitable initial guess.

To show the capability of the method, NHPM has been applied to some examples in the next section.

## 3. Numerical results

To demonstrate the effectiveness of the method two examples of system of Burgers equations are presented.

## Example 1.

Consider the following system of Burgers equations

 ${\displaystyle {\begin{array}{l}{\frac {\partial u}{\partial t}}-{\frac {{\partial }^{2}u}{\partial x^{2}}}-2u{\frac {\partial u}{\partial x}}+{\frac {\partial }{\partial x}}(uv)=0{\mbox{,}}\\{\frac {\partial v}{\partial t}}-{\frac {{\partial }^{2}v}{\partial x^{2}}}-2v{\frac {\partial v}{\partial x}}-{\frac {\partial }{\partial x}}(uv)=0{\mbox{,}}\end{array}}}$
(10)

with initial conditions

 ${\displaystyle u(x{\mbox{,}}0)=sinx{\mbox{,}}}$ ${\displaystyle v(x{\mbox{,}}0)=sinx{\mbox{.}}}$

The exact solutions are

 ${\displaystyle u(x{\mbox{,}}t)=e^{-t}sinx{\mbox{,}}}$ ${\displaystyle v(x{\mbox{,}}t)=e^{-t}sinx{\mbox{.}}}$

To solve Eq. (10), by the NHPM, we construct the following homotopies

 ${\displaystyle {\frac {\partial U}{\partial t}}=u_{0}-p\left(u_{0}-{\frac {{\partial }^{2}u}{\partial x^{2}}}-\right.}$${\displaystyle \left.2u{\frac {\partial u}{\partial x}}+{\frac {\partial }{\partial x}}(uv)\right){\mbox{,}}}$ ${\displaystyle {\frac {\partial V}{\partial t}}=v_{0}-p\left(v_{0}-{\frac {{\partial }^{2}v}{\partial x^{2}}}-\right.}$${\displaystyle \left.2v{\frac {\partial v}{\partial x}}-{\frac {\partial }{\partial x}}(uv)\right){\mbox{.}}}$
(11)

Applying the inverse operator, ${\textstyle L^{-1}={\int }_{0}^{t}(.)dt}$ to both sides of these equations, we obtain

 ${\displaystyle U(x{\mbox{,}}t)=U(x{\mbox{,}}0)+{\int }_{0}^{t}u_{0}(x{\mbox{,}}t)dt-}$${\displaystyle p{\int }_{0}^{t}\left(u_{0}(x{\mbox{,}}t)-{\frac {{\partial }^{2}U}{\partial x^{2}}}-\right.}$${\displaystyle \left.2U{\frac {\partial U}{\partial x}}+{\frac {\partial }{\partial x}}(UV)\right)dt{\mbox{,}}}$ ${\displaystyle V(x{\mbox{,}}t)=V(x{\mbox{,}}0)+{\int }_{0}^{t}v_{0}(x{\mbox{,}}t)dt-}$${\displaystyle p{\int }_{0}^{t}\left(v_{0}(x{\mbox{,}}t)-{\frac {{\partial }^{2}V}{\partial x^{2}}}-\right.}$${\displaystyle \left.2V{\frac {\partial V}{\partial x}}-{\frac {\partial }{\partial x}}(UV)\right)dt{\mbox{.}}}$
(12)

Suppose the solutions of system (12) are, as assumed in (5), substituting Eq. (5) into Eq. (12), collecting the same powers of p, and equating each coefficient of p to zero, results in

 ${\displaystyle p^{0}:{\begin{array}{l}U_{0}(x{\mbox{,}}t)=U(x{\mbox{,}}0)+{\int }_{0}^{t}u_{0}(x{\mbox{,}}t)dt{\mbox{,}}\\V_{0}(x{\mbox{,}}t)=V(x{\mbox{,}}0)+{\int }_{0}^{t}v_{0}(x{\mbox{,}}t)dt{\mbox{,}}\end{array}}}$ ${\displaystyle p^{1}:{\begin{array}{l}U_{1}(x{\mbox{,}}t)=-{\int }_{0}^{t}\left(u_{0}(x{\mbox{,}}t)-{\frac {{\partial }^{2}U_{0}}{\partial x^{2}}}-2U_{0}{\frac {\partial U_{0}}{\partial x}}+{\frac {\partial }{\partial x}}(U_{0}V_{0})\right)dt{\mbox{,}}\\V_{1}(x{\mbox{,}}t)=-{\int }_{0}^{t}\left(v_{0}(x{\mbox{,}}t)-{\frac {{\partial }^{2}V_{0}}{\partial x^{2}}}-2V_{0}{\frac {\partial V_{0}}{\partial x}}-{\frac {\partial }{\partial x}}(U_{0}V_{0})\right)dt{\mbox{,}}\end{array}}}$ ${\displaystyle p^{2}:{\begin{array}{l}U_{2}(x{\mbox{,}}t)=-{\int }_{0}^{t}\left(-{\frac {{\partial }^{2}U_{1}}{\partial x^{2}}}-2U_{0}{\frac {\partial U_{1}}{\partial x}}-2U_{1}{\frac {\partial U_{0}}{\partial x}}+U_{1}{\frac {\partial V_{0}}{\partial x}}+U_{0}{\frac {\partial V_{1}}{\partial x}}+V_{1}{\frac {\partial U_{0}}{\partial x}}+V_{0}{\frac {\partial U_{1}}{\partial x}}\right)dt{\mbox{,}}\\V_{2}(x{\mbox{,}}t)=-{\int }_{0}^{t}\left(-{\frac {{\partial }^{2}V_{1}}{\partial x^{2}}}-2V_{1}{\frac {\partial V_{0}}{\partial x}}-2V_{0}{\frac {\partial V_{1}}{\partial x}}-U_{1}{\frac {\partial V_{0}}{\partial x}}-U_{0}{\frac {\partial V_{1}}{\partial x}}-V_{1}{\frac {\partial U_{0}}{\partial x}}-V_{0}{\frac {\partial U_{1}}{\partial x}}\right)dt{\mbox{,}}\end{array}}}$ ${\displaystyle \vdots }$

Assuming ${\textstyle {\begin{array}{l}u_{0}(x{\mbox{,}}t)={\sum }_{n=0}^{\infty }a_{n}(x)P_{n}(t){\mbox{,}}\quad P_{k}(t)=t^{k}{\mbox{,}}\quad U(x{\mbox{,}}0)=u(x{\mbox{,}}0){\mbox{,}}\\v_{0}(x{\mbox{,}}t)={\sum }_{n=0}^{\infty }b_{n}(x)P_{n}(t){\mbox{,}}\quad P_{k}(t)=t^{k}{\mbox{,}}\quad V(x{\mbox{,}}0)=v(x{\mbox{,}}0){\mbox{.}}\end{array}}}$

And solving the above equations for ${\textstyle U_{1}(x{\mbox{,}}y{\mbox{,}}t){\mbox{,}}V_{1}(x{\mbox{,}}y{\mbox{,}}t)}$ leads to the result

 ${\displaystyle U_{1}(x{\mbox{,}}t)=(-a_{0}(x)-sinx)t+\left(-{\frac {1}{2}}a_{1}(x)+\right.}$${\displaystyle \left.{\frac {1}{2}}a_{0xx}(x)+{\frac {1}{2}}a_{0x}(x)sinx+\right.}$${\displaystyle \left.a_{0}(x)cosx-{\frac {1}{2}}b_{0x}sinx-{\frac {1}{2}}cos{xb}_{0x}(x)-\right.}$${\displaystyle \left.{\frac {1}{2}}cos{xa}_{0x}(x)\right)t^{2}+\left({\begin{array}{l}-{\frac {1}{3}}a_{2}(x)+a_{1xx}(x)+{\frac {1}{3}}sin{xa}_{1x}(x)+{\frac {2}{3}}a_{0}(x)a_{0x}(x)+{\frac {2}{3}}a_{1}(x)cosx-{\frac {1}{6}}sin{xb}_{1x}(x)-{\frac {1}{3}}a_{0}(x)b_{0x}(x)-{\frac {1}{6}}a_{1x}(x)cosx\\-{\frac {1}{6}}sin{xa}_{1x}(x)-{\frac {1}{3}}b_{0}(x)a_{0x}(x)-{\frac {1}{6}}b_{1}(x)cosx\end{array}}\right)t^{3}+\cdots {\mbox{,}}}$
 ${\displaystyle V_{1}(x{\mbox{,}}t)=(-b_{0}(x)-sinx)t\left(-{\frac {1}{2}}b_{1}(x)+\right.}$${\displaystyle \left.{\frac {1}{2}}b_{0xx}(x)+{\frac {1}{2}}b_{0x}(x)sinx-\right.}$${\displaystyle \left.b_{0}(x)cosx+{\frac {1}{2}}a_{0x}sinx+{\frac {1}{2}}cos{xa}_{0x}(x)+\right.}$${\displaystyle \left.{\frac {1}{2}}cos{xb}_{0x}(x)\right)t^{2}+\left({\begin{array}{l}-{\frac {1}{3}}b_{2}(x)+{\frac {1}{3}}sin{xb}_{1x}(x)+{\frac {2}{3}}b_{0}(x)b_{0x}(x)+{\frac {2}{3}}b_{1}(x)cosx+{\frac {1}{6}}sin{xb}_{1x}(x)+{\frac {1}{3}}a_{0}(x)b_{0x}(x)+{\frac {1}{6}}a_{1x}(x)cosx+{\frac {1}{6}}sin{xa}_{1x}(x)+{\frac {1}{3}}b_{0}(x)a_{0x}(x)\\+{\frac {1}{6}}b_{1}(x)cosx\end{array}}\right)t^{3}+\cdots {\mbox{.}}}$

By vanishing ${\textstyle U_{1}(x{\mbox{,}}t){\mbox{,}}V_{1}(x{\mbox{,}}t)}$ coefficients ${\textstyle a_{n}(x){\mbox{,}}b_{n}(x){\mbox{,}}(n=1{\mbox{,}}2{\mbox{,}}3{\mbox{,}}\ldots )}$ are determined as

 ${\displaystyle a_{0}(x)=-sinx{\mbox{,}}a_{1}(x)=sinx{\mbox{,}}a_{2}(x)=-sinx{\mbox{,}}a_{3}(x)=}$${\displaystyle sinx{\mbox{,}}a_{4}(x)=-sinx{\mbox{,}}\ldots }$ ${\displaystyle b_{0}(x)=-sinx{\mbox{,}}b_{1}(x)=sinx{\mbox{,}}b_{2}(x)=-sinx{\mbox{,}}b_{3}(x)=}$${\displaystyle sinx{\mbox{,}}b_{4}(x)=-sinx{\mbox{,}}\ldots }$

Therefore we gain the solution of Eq. (14) as

 ${\displaystyle u(x{\mbox{,}}t)=U_{0}(x{\mbox{,}}t)=sinx+a_{0}(x)t+{\frac {1}{2}}a_{1}(x)t^{2}+}$${\displaystyle {\frac {1}{3}}a_{2}(x)t^{3}+{\frac {1}{4}}a_{3}(x)t^{4}+\cdots =}$${\displaystyle sin{xe}^{-t}{\mbox{,}}}$
 ${\displaystyle v(x{\mbox{,}}t)=V_{0}(x{\mbox{,}}t)=sinx+b_{0}(x)t+{\frac {1}{2}}b_{1}(x)t^{2}+}$${\displaystyle {\frac {1}{3}}b_{2}(x)t^{3}+{\frac {1}{4}}b_{3}(x)t^{4}+\cdots =}$${\displaystyle sin{xe}^{-t}{\mbox{,}}}$

which is an exact solution (see Fig. 1).

 Figure 1. Numerical result of Example 1.

## Example 2.

Consider the following system of coupled Burgers equations

 ${\displaystyle {\begin{array}{l}{\frac {\partial u}{\partial t}}-{\frac {{\partial }^{2}u}{\partial x^{2}}}+u{\frac {\partial u}{\partial x}}+{\frac {\partial }{\partial x}}(uv)=x^{2}-2t+2x^{3}t^{2}+t^{2}{\mbox{,}}\\{\frac {\partial v}{\partial t}}-{\frac {{\partial }^{2}v}{\partial x^{2}}}+v{\frac {\partial v}{\partial x}}-{\frac {\partial }{\partial x}}(uv)={\frac {1}{x}}-2{\frac {t}{x^{3}}}+{\frac {t^{2}}{x^{3}}}+t^{2}{\mbox{,}}\end{array}}}$
(13)

with initial conditions

 ${\displaystyle u(x{\mbox{,}}0)=0{\mbox{,}}}$ ${\displaystyle v(x{\mbox{,}}0)=0{\mbox{,}}}$

The exact solution of Eq. (13) is

 ${\displaystyle u(x{\mbox{,}}t)=x^{2}t{\mbox{,}}}$ ${\displaystyle v(x{\mbox{,}}t)={\frac {t}{x}}{\mbox{.}}}$

To solve Eq. (13), by the NHPM, we construct the following homotopy

 ${\displaystyle {\frac {\partial U}{\partial t}}=u_{0}-p\left(u_{0}-{\frac {{\partial }^{2}u}{\partial x^{2}}}+\right.}$${\displaystyle \left.u{\frac {\partial u}{\partial x}}+{\frac {\partial }{\partial x}}(uv)-\right.}$${\displaystyle \left.(x^{2}-2t+2x^{3}t^{2}+t^{2})\right){\mbox{,}}}$ ${\displaystyle {\frac {\partial V}{\partial t}}=v_{0}-p\left(v_{0}-{\frac {{\partial }^{2}v}{\partial x^{2}}}+\right.}$${\displaystyle \left.v{\frac {\partial v}{\partial x}}-{\frac {\partial }{\partial x}}(uv)-\right.}$${\displaystyle \left.({\frac {1}{x}}-2{\frac {t}{x^{3}}}+{\frac {t^{2}}{x^{3}}}+\right.}$${\displaystyle \left.t^{2})\right){\mbox{.}}}$
(14)

Applying the inverse operator, ${\textstyle L^{-1}={\int }_{0}^{t}(.)dt}$ to the both sides of the above equations, we obtain

 ${\displaystyle U(x{\mbox{,}}t)=U(x{\mbox{,}}0)+{\int }_{0}^{t}u_{0}(x{\mbox{,}}t)dt-}$${\displaystyle p{\int }_{0}^{t}\left(u_{0}(x{\mbox{,}}t)-{\frac {{\partial }^{2}U}{\partial x^{2}}}+\right.}$${\displaystyle \left.U{\frac {\partial U}{\partial x}}+{\frac {\partial }{\partial x}}(UV)-\right.}$${\displaystyle \left.(x^{2}-2t+2x^{3}t^{2}+t^{2})\right)dt{\mbox{,}}}$ ${\displaystyle V(x{\mbox{,}}t)=V(x{\mbox{,}}0)+{\int }_{0}^{t}v_{0}(x{\mbox{,}}t)dt-}$${\displaystyle p{\int }_{0}^{t}\left(v_{0}(x{\mbox{,}}t)-{\frac {{\partial }^{2}V}{\partial x^{2}}}+\right.}$${\displaystyle \left.V{\frac {\partial V}{\partial x}}-{\frac {\partial }{\partial x}}(UV)-\right.}$${\displaystyle \left.({\frac {1}{x}}-2{\frac {t}{x^{3}}}+{\frac {t^{2}}{x^{3}}}+\right.}$${\displaystyle \left.t^{2})\right)dt{\mbox{.}}}$
(15)

Suppose the solutions of system (15) have the form (5), substituting Eq. (5) into Eq. (15), collecting the same powers of p, and equating each coefficient of p to zero, results in

 ${\displaystyle p^{0}:{\begin{array}{l}U_{0}(x{\mbox{,}}t)=U(x{\mbox{,}}0)+{\int }_{0}^{t}u_{0}(x{\mbox{,}}t)dt{\mbox{,}}\\V_{0}(x{\mbox{,}}t)=V(x{\mbox{,}}0)+{\int }_{0}^{t}v_{0}(x{\mbox{,}}t)dt{\mbox{,}}\end{array}}}$ ${\displaystyle p^{1}:{\begin{array}{l}U_{1}(x{\mbox{,}}t)=-{\int }_{0}^{t}\left(u_{0}(x{\mbox{,}}t)-{\frac {{\partial }^{2}U_{0}}{\partial x^{2}}}+U_{0}{\frac {\partial U_{0}}{\partial x}}+{\frac {\partial }{\partial x}}(U_{0}V_{0})-(x^{2}-2t+2x^{3}t^{2}+t^{2})\right)dt{\mbox{,}}\\V_{1}(x{\mbox{,}}t)=-{\int }_{0}^{t}\left(v_{0}(x{\mbox{,}}t)-{\frac {{\partial }^{2}V_{0}}{\partial x^{2}}}+V_{0}{\frac {\partial V_{0}}{\partial x}}-{\frac {\partial }{\partial x}}(U_{0}V_{0})-({\frac {1}{x}}-2{\frac {t}{x^{3}}}+{\frac {t^{2}}{x^{3}}}+t^{2})\right)dt{\mbox{,}}\end{array}}}$ ${\displaystyle p^{2}:{\begin{array}{l}U_{2}(x{\mbox{,}}t)=-{\int }_{0}^{t}\left(-{\frac {{\partial }^{2}U_{1}}{\partial x^{2}}}+U_{0}{\frac {\partial U_{1}}{\partial x}}+U_{1}{\frac {\partial U_{0}}{\partial x}}+U_{1}{\frac {\partial V_{0}}{\partial x}}+U_{0}{\frac {\partial V_{1}}{\partial x}}+V_{1}{\frac {\partial U_{0}}{\partial x}}+V_{0}{\frac {\partial U_{1}}{\partial x}}\right)dt{\mbox{,}}\\V_{2}(x{\mbox{,}}t)=-{\int }_{0}^{t}\left(-{\frac {{\partial }^{2}V_{1}}{\partial x^{2}}}+V_{1}{\frac {\partial V_{0}}{\partial x}}+V_{0}{\frac {\partial V_{1}}{\partial x}}-U_{1}{\frac {\partial V_{0}}{\partial x}}-U_{0}{\frac {\partial V_{1}}{\partial x}}-V_{1}{\frac {\partial U_{0}}{\partial x}}-V_{0}{\frac {\partial U_{1}}{\partial x}}\right)dt{\mbox{,}}\end{array}}}$ ${\displaystyle \vdots }$

By assuming ${\textstyle {\begin{array}{ll}u_{0}(x{\mbox{,}}t)={\sum }_{n=0}^{\infty }a_{n}(x)t^{k}{\mbox{,}}&U(x{\mbox{,}}0)=u(x{\mbox{,}}0){\mbox{,}}\\v_{0}(x{\mbox{,}}t)={\sum }_{n=0}^{\infty }b_{n}(x)t^{k}{\mbox{,}}&V(x{\mbox{,}}0)=v(x{\mbox{,}}0){\mbox{,}}\end{array}}}$and solving equations ${\textstyle U_{1}(x{\mbox{,}}t){\mbox{,}}V_{1}(x{\mbox{,}}t)}$ leads to the following results

 ${\displaystyle U_{1}(x{\mbox{,}}t)=(-a_{0}(x)+x^{2})t+\left(-{\frac {1}{2}}a_{1}(x)+\right.}$${\displaystyle \left.{\frac {1}{2}}a_{0xx}(x)+1\right)t^{2}+\left({\begin{array}{l}-{\frac {1}{3}}a_{2}(x)+{\frac {1}{3}}a_{2xx}-{\frac {1}{3}}a_{0}(x)a_{2x}(x)+{\frac {1}{3}}a_{1}(x)a_{1x}(x)+{\frac {1}{6}}a_{0}(x)b_{2}(x)+{\frac {1}{6}}a_{2}(x)b_{0x}(x)+{\frac {1}{3}}a_{1}(x)b_{1x}(x)\\+{\frac {1}{6}}b_{0}(x)a_{2}(x)+{\frac {1}{6}}b_{2}(x)a_{0x}(x)+{\frac {1}{3}}b_{1}(x)a_{1x}(x)+{\frac {1}{3}}+{\frac {2}{3}}x^{3}\end{array}}\right)t^{3}+\cdots {\mbox{,}}}$
 ${\displaystyle V_{1}(x{\mbox{,}}t)=\left(-b_{0}(x)+{\frac {1}{x}}\right)t\left(-\right.}$${\displaystyle \left.{\frac {1}{2}}b_{1}(x)+{\frac {1}{2}}b_{0xx}(x)+{\frac {1}{x^{3}}}\right)t^{2}+}$${\displaystyle \left({\begin{array}{l}-{\frac {1}{3}}b_{2}(x)+{\frac {1}{3}}b_{2xx}-{\frac {1}{3}}b_{0}(x)b_{2x}(x)+{\frac {1}{3}}b_{1}(x)b_{1x}(x)-{\frac {1}{6}}a_{0}(x)b_{2}(x)-{\frac {1}{6}}a_{2}(x)b_{0x}(x)-{\frac {1}{3}}a_{1}(x)b_{1x}(x)\\-{\frac {1}{6}}b_{0}(x)a_{2}(x)-{\frac {1}{6}}b_{2}(x)a_{0x}(x)-{\frac {1}{3}}b_{1}(x)a_{1x}(x)+{\frac {1}{3}}+{\frac {2}{3}}x^{3}\end{array}}\right)t^{3}+\cdots {\mbox{.}}}$

Vanishing ${\textstyle U_{1}(x{\mbox{,}}t){\mbox{,}}V_{1}(x{\mbox{,}}t)}$, lets the coefficients ${\textstyle a_{n}(x){\mbox{,}}b_{n}(x){\mbox{,}}(n=1{\mbox{,}}2{\mbox{,}}3{\mbox{,}}\ldots )}$ to take the following values

 ${\displaystyle a_{0}(x)=x^{2}{\mbox{,}}a_{1}(x)=0{\mbox{,}}a_{2}(x)=0{\mbox{,}}a_{3}(x)=}$${\displaystyle 0{\mbox{,}}a_{4}(x)=0{\mbox{,}}\ldots }$ ${\displaystyle b_{0}(x)={\frac {1}{x^{2}}}{\mbox{,}}b_{1}(x)=0{\mbox{,}}b_{2}(x)=}$${\displaystyle 0{\mbox{,}}b_{3}(x)=0{\mbox{,}}b_{4}(x)=0{\mbox{,}}\ldots }$

This implies that

 ${\displaystyle u(x{\mbox{,}}t)=U_{0}(x{\mbox{,}}t)=a_{0}(x)t+{\frac {1}{2}}a_{1}(x)t^{2}+}$${\displaystyle {\frac {1}{3}}a_{2}(x)t^{3}+{\frac {1}{4}}a_{3}(x)t^{4}+\cdots =}$${\displaystyle x^{2}t{\mbox{,}}}$ ${\displaystyle v(x{\mbox{,}}t)=V_{0}(x{\mbox{,}}t)=b_{0}(x)t+{\frac {1}{2}}b_{1}(x)t^{2}+}$${\displaystyle {\frac {1}{3}}b_{2}(x)t^{3}+{\frac {1}{4}}b_{3}(x)t^{4}+\cdots =}$${\displaystyle {\frac {t}{x}}{\mbox{.}}}$

In this example we have also derived exact solutions (see Fig. 2).

 Figure 2. Numerical result of Example 2.

## 4. Conclusions

In this paper, the modified HPM is applied to approximate solutions of Burgers equations. The present method reduces the computational difficulties of the other methods and all the calculations can be made with simple manipulations. The solutions introduced in this study can be used to obtain the closed form of the solutions if they are required. Thus, it can be concluded that the MHPM is very powerful and efficient in finding analytical as well as numerical solutions for wide classes of partial differential equations. The computations associated with the examples were performed using Maple 15.

## Acknowledgments

The authors would like to thank the referees for their useful comments which lead to some improvements of the current paper.

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