Analytic Study of Coupled Burgers’ Equation

Abstract

In this paper, we construct an analytical solution of the coupled Burgers’ equation, using the homotopy analysis method, which is a semi-analytical method, the approximate solution obtained by this method is convergent for different values of the convergence control parameter ℏ, the optimal value of ℏ corresponding with the minimum error to be determined by the residual. The results obtained by the present method are compared with other obtained solutions by different numerical methods.

1. Introduction

The coupled Burgers’ equation is one of the most famous differential equations used and interpreted for various phenomena in engineering and applied science, such as the status of quantum particles, shock waves, and acoustic transmission and traffic flow [1]; however, because most partial and ordinary differential equations are not easy to solve, many researchers have tried to find approximation solutions with the best value of error to guarantee the convergence of the series solutions, such as the Adomian decomposition method, variational iteration method, perturbation method, and so on.

The homotopy analysis method is one of the most important methods used for solving all types of differential equations. It was created by Shijun Liao in his Ph.D. in 1992 at Shanghai university. The philosophy of this method is based on homotopy theory, and it constructs a deformation between a family of equations, starting with an equation that has a known solution, and determined by a proposed equation that has an unknown solution.

In this paper, we investigate the effectiveness and accuracy of the homotopy analysis method [2] applied to the coupled Burgers’ equation:

$$\left\{\begin{array}{lll}\frac{\partial u}{\partial t}-{\alpha }_1\frac{{\partial }^{2}u}{\partial x^2}+{\beta }_{1,1}\frac{\partial u}{\partial x}+{\beta }_{1,2}\frac{\partial uv}{\partial x}=0& t\in [0,T],& x\in [a,b]\\ \frac{\partial v}{\partial t}-{\alpha }_2\frac{{\partial }^{2}v}{\partial x^2}+{\beta }_{2,2}\frac{\partial v}{\partial x}+{\beta }_{2,1}\frac{\partial uv}{\partial x}=0& t\in [0,T],& x\in [a,b]\end{array}\right.$$

(1)

with the initial conditions:

$$\left\{\begin{array}{ll}u(x,0)={\varphi }_1\left(x\right)& x\in [a,b]\\ v(x,0)={\varphi }_2\left(x\right)& x\in [a,b]\end{array}\right.$$

(2)

and the boundary conditions:

$$\left\{\begin{array}{lll}u(a,t)={\psi }_1\left(t\right)& u(b,t)={\psi }_2\left(t\right)& x\in [a,b]\\ v(a,t)={\psi }_3\left(t\right)& v(b,t)={\psi }_4\left(t\right)& x\in [a,b]\end{array}\right.$$

(3)

where ${\alpha }_1$ and ${\alpha }_2$ are the positive viscosity parameters, and ${\beta }_{1,1},{\beta }_{1,2},{\beta }_{2,1}$, and ${\beta }_{2,2}$ are the constants of the Stokes velocity.

In recent years, many methods have been used for solving the coupled Burgers’ equation, such as the modified variational iteration algorithm by C. Cesarano [3], the Chebyshev spectral collocation method by Hassan [4], the Chebyshev–Legendre Pseudo-Spectral method by Rashid [5]. Ahmed used Variational Iteration Algorithm-I with an Auxiliary Parameter for this Equation [6] and while the Pseudo-Spectral method by Darvishi’s preconditioning [7] have also been used.

2. The Homotopy Analysis Method

We briefly give some basic ideas of the homotopy analysis method. Let us consider:

$$\mathfrak{L}u(x,t)+\mathfrak{N}u(x,t)=f(x,t),$$

(4)

where $\mathfrak{L}$ and $\mathfrak{N}$ are the linear and non-linear operators, respectively, and $f(x,t)$ is the source term. The operator N is given as [8,9,10]

$$N\left[\varphi \right(x,t,q\left)\right]=\mathfrak{R}\varphi (x,t,q)+\mathfrak{N}\varphi (x,t,q)-f(x,t)$$

(5)

where $\varphi (x,t,q)$ is the real function of x and t, namely $q\in [0,1]$, and the zero-order deformation constructed by Liao [9,10] is

$$(1-q)\mathcal{L}[\varphi (x,t,q)-u_{x,0}\left(t\right)]=\hslash qH(x,t)N\left[\varphi (x,t,q)\right]$$

(6)

where ℏ is the non-zero auxiliary parameter, the auxiliary function $H(x,t)\ne 0$, $u_0(x,t)$ is the initial guess of $u(x,t)$, and $\varphi (x,t,q)$ is an unknown function, which is obviously $\varphi (x,t,0)=u_0(x,t)$ and $\varphi (x,t,1)=u(x,t)$. We expand $\varphi (x,t,q)$ in the Taylor series with respect to q and t,

$$\varphi (x,t,q)=\sum _{i=0}^n{u}_i(x,t)q^i$$

where

$$u_i(x,t)=\frac{1}{m!}\frac{{\partial }^m\varphi (x,t,q)}{\partial q^m}{|}_{q=0}$$

(7)

by differentiating (6) m times with respect q, and taking $q=0$, we obtain the m-th order deformation equation

$$L[u_m(x,t)-{\chi }_m{u}_{m-1}(x,t)]=\hslash H(x,t)R_m\left(u_{m-1}(x,t)\right)$$

(8)

Applying $L^{-1}$ in (8), we obtain:

$$u_m(x,t)={\chi }_m{u}_{m-1}(x,t)+\hslash L^{-1}\left[R_m\left(u_{m-1}(x,t)\right)\right]$$

(9)

we mention that we choose here $L^{-1}$ as the Riemann integral operator $L^{-1}$, where

$${\chi }_m=\left\{\begin{array}{lll}0,& & m\le 1\\ 1,& & m>1\end{array}\right.$$

We mention that $u(x,t)$ is written as a series

$$u(x,t)=\sum _{i=0}^{\infty }{u}_i(x,t)$$

(10)

3. Convergence Analysis

Theorem 1 

([11,12,13,14,15,16]). If the series solutions ${\sum }_{i=0}^n{u}_i(x,t)$ are convergent, where $u_m$ is governed by Equation (7), then they must be solutions of (1).

Proof. 

Suppose that ${\sum }_{i=0}^n{u}_i(x,t)$ is convergent, i.e., $\underset{n⟶\infty }{lim}{u}_n(x,t)=0$. From Equation (8), we obtain:

$$\begin{array}{lllll}\hslash H\sum _{m=1}^{\infty }{R}_m& =& \underset{k⟶\infty }{lim}\sum _{m=0}^{k}L[u_m-{\chi }_m{u}_{m-1}]& & \\ & =& L\left[\underset{k⟶\infty }{lim}\sum _{m=0}^k[u_m-{\chi }_m{u}_{m-1}]\right]& =& L\left[\underset{k⟶\infty }{lim}{u}_k\right]\end{array}$$

where L is a Laplace operator. Since $\underset{k⟶\infty }{lim}{u}_k=0$, $H\ne 0$ and $\hslash \ne 0$, this implies ${\sum }_{m=1}^{\infty }{K}_m=0$. Now, expand $N[\Phi (x,t,q\left)\right]$ about $q=0$, then set $q=1$,

$$N[\Phi (x,t,1\left)\right]=0,$$

we can see that $u(x,t)=\Phi (x,t,1)={\sum }_{i=0}^{\infty }{u}_i(x,t)$ solves Equation (1). □

Theorem 2 

([11]). Let the solution component $u_0(x,t),u_1(x,t),u_2(x,t),...$ be defined as (9). The series solution ${\sum }_{m=0}^{\infty }{u}_m(x,t)$ defined in (10) converges if there exist $0<\gamma <1$, such that $\parallel u_{m+1}(x,t)\parallel \le \gamma \parallel u_m(x,t)\parallel ,\forall m>m_0$ for some $m_0\in \mathbb{N}$.

Proof. 

Let ${\left(T_n\right)}_{n\in \mathbb{N}}$ be a sequence, and define as $T_n=u_0+u_1+...+u_n$. For every $p,q\in \mathbb{N},p>q\ge m_0$, we have

$$\begin{array}{lll}\Vert T_p-T_q\Vert & \le & \Vert (T_p-T_{p-1})\Vert +\Vert (T_{p-1}-T_{p-2})\Vert +...+\Vert (T_{q+1}-T_q)\Vert \\ & \le & {\gamma }^{p-m_0}\Vert u_{m_0}\Vert +{\gamma }^{p-m_0-1}\Vert u_{m_0}\Vert +...+{\gamma }^{q-m_0+1}\Vert u_{m_0}\Vert \\ & \le & ({\gamma }^{p-m_0}+{\gamma }^{p-m_0-1}+...+{\gamma }^{q-m_0+1})\Vert u_{m_0}\Vert \\ & \le & {\gamma }^{q-m_0+1}\frac{{\gamma }^{p-q}}{1-\gamma }\Vert u_{m_0}\Vert \end{array}$$

(11)

Because of $0<\gamma <1$, we obtain

$$\underset{p,q\to \infty }{lim}\Vert T_p-T_q\Vert =0$$

So, $\left(T_n\right)$ is a Cauchy sequence in $\mathbb{R}$, which implies the convergence of $\left(T_n\right)$, so the series solution $\sum u_i(x,t)$ converges. □

4. Application

4.1. Test Example 1

Let us consider the coupled Burgers’ Equation (1) with ${\alpha }_1={\alpha }_2=1$ and ${\beta }_{1,1}={\beta }_{2,2}=-2$

$$\left\{\begin{array}{lll}\frac{\partial u}{\partial t}-{\alpha }_1\frac{{\partial }^{2}u}{\partial x^2}-2\frac{\partial u}{\partial x}+{\beta }_{1,2}\frac{\partial uv}{\partial x}=0& t\in [0,T],& x\in [-10,10]\\ \frac{\partial v}{\partial t}-{\alpha }_2\frac{{\partial }^{2}v}{\partial x^2}-2\frac{\partial v}{\partial x}+{\beta }_{2,1}\frac{\partial uv}{\partial x}=0& t\in [0,T],& x\in [-10,10]\end{array}\right.$$

(12)

with:

$$\left\{\begin{array}{l}u(x,0)=K(1-tanh(B\left(x\right)\left)\right)\\ v(x,0)=K\left(\frac{2{\beta }_{2,1}-1}{2{\beta }_{1,2}-1}-tanh\left(B\left(x\right)\right)\right),\end{array}\right.$$

(13)

the exact solution of (12) is

$$\left\{\begin{array}{l}u(x,0)=K(1-tanh(B(x-2Bt)\left)\right)\\ v(x,0)=K\left(\frac{2{\beta }_{2,1}-1}{2{\beta }_{1,2}-1}-tanh\left(B(x-2Bt)\right)\right),\end{array}\right.$$

(14)

The zero-order deformation equation is:

$$(1-q)L_i[{\varphi }_i(x,t,q)-u_{i,0}(x,t)]=q\hslash N_i\left[{\varphi }_i(x,t,q)\right],i\in \{1,2\}$$

where $L_i=\frac{\partial {\varphi }_i}{\partial t}$, and

$$\left\{\begin{array}{l}{N}_1[{\varphi }_1(x,t,q),{\varphi }_2(x,t,q)]=\frac{\partial {\varphi }_1(x,t,q)}{\partial t}-\frac{{\partial }^2{\varphi }_1(x,t,q)}{\partial x^2}-2\frac{\partial {\varphi }_1(x,t,q)}{\partial x}+{\beta }_{1,2}\frac{\partial {\varphi }_1(x,t,q){\varphi }_2(x,t,q)}{\partial x}\\ N_2[{\varphi }_1(x,t,q),{\varphi }_2(x,t,q)]=\frac{\partial {\varphi }_2(x,t,q)}{\partial t}-\frac{{\partial }^2{\varphi }_2(x,t,q)}{\partial x^2}-2\frac{\partial {\varphi }_2(x,t,q)}{\partial x}+{\beta }_{2,1}\frac{\partial {\varphi }_1(x,t,q){\varphi }_2(x,t,q)}{\partial x}\end{array}\right.$$

(15)

The corresponding m-th order deformation equation for (15) becomes:

$$\left\{\begin{array}{l}{u}_m(x,t)={\chi }_m{u}_{m-1}(x,t)+\hslash L^{-1}\left[R_{1,m}(u_{m-1}(x,t),v_{m-1}(x,t))\right]\\ v_m(x,t)={\chi }_m{v}_{m-1}(x,t)+\hslash L^{-1}\left[R_{2,m}(u_{m-1}(x,t),v_{m-1}(x,t))\right]\end{array}\right.$$

(16)

where

$$\left\{\begin{array}{ll}{R}_{1,m}& =\frac{\partial u_{m-1}(x,t)}{\partial t}-\frac{{\partial }^2{u}_{m-1}(x,t)}{\partial x^2}-2\frac{\partial u_{m-1}(x,t,q)}{\partial x}\\ & +{\beta }_{1,2}\sum _{i=0}^{m-1}\left(\frac{\partial u_i(x,t)}{\partial x}{v}_{m-1-i}(x,t)+\frac{\partial v_{m-1-i}(x,t)}{\partial x}{u}_i(x,t)\right)\\ R_{2,m}& =\frac{\partial v_{m-1}(x,t)}{\partial t}-\frac{{\partial }^2{v}_{m-1}(x,t)}{\partial x^2}-2\frac{\partial v_{m-1}(x,t)}{\partial x}\\ & +{\beta }_{2,1}\sum _{i=0}^{m-1}\left(\frac{\partial u_i(x,t)}{\partial x}{v}_{m-1-i}(x,t)+\frac{\partial v_{m-1-i}(x,t)}{\partial x}{u}_i(x,t)\right)\end{array}\right.$$

Choosing from (14),

$$\left\{\begin{array}{l}u(x,0)=K(1-tanh(B(x-2Bt)\left)\right)\\ v(x,0)=K\left(\frac{2{\beta }_{2,1}-1}{2{\beta }_{1,2}-1}-tanh\left(B(x-2Bt)\right)\right),\end{array}\right.$$

we obtain for ${\beta }_{1,2}=0.1$, and ${\beta }_{2,1}=0.3$:

$$\begin{array}{ll}{u}_1(x,t)& =\hslash .t{\sec \mathrm{h}}^2\left(0.00625x\right)(7.03125\times {10}^{-6}tanh\left(0.00625x\right))\\ u_2(x,t)& =\hslash .t.{\sec \mathrm{h}}^2\left(0.00625x\right)\left(t\hslash (1.733\times {10}^{-9}tanh\left(0.00625x\right)-8.739\times {10}^{-8}){\sec \mathrm{h}}^2\left(0.00625x\right)-0.000622\hslash \right.\\ & -0.00062-1.18408\times {10}^{-9}t.\hslash {tanh}^3\left(0.0062x\right)+1.7478\times {10}^{-7}t\hslash {tanh}^2\left(0.00625x\right)+\\ & \left.(-7.754162\times {10}^{-6}t\hslash +7.031249\times {10}^{-6}\hslash +7.031249\times {10}^{-6})tanh\left(0.00625x\right)\right)\\ u_3(x,t)& =\hslash .t{\sec \mathrm{h}}^2\left(0.00625x\right)\\ & \left(t^2{\hslash }^2(5.6642\times {10}^{-13}tanh\left(0.00625x\right)-2.1527\times {10}^{-11}){\sec \mathrm{h}}^4\left(0.00625x\right)\right.\\ & -0.000622656{(1.\hslash +1.)}^{2}tanh\left(0.00625x\right)+\left(\hslash (0.0000140625-0.0000155083t)+\right)\\ & (7.03125\times {10}^{-6}-0.0000155083t){\hslash }^2+7.03125\times {10}^{-6}+....\end{array}$$

and

$$\begin{array}{ll}{v}_1(x,t)& =\hslash t(0.0000132813tanh\left(0.00625x\right)-0.000617969){\sec \mathrm{h}}^2\left(0.00625x\right)\\ v_2(x,t)& =t\hslash {\sec \mathrm{h}}^2\left(0.00625x\right)\left(\hslash t(3.9795\times {10}^{-9}tanh\left(0.00625x\right)-1.6451\times {10}^{-7}){\sec \mathrm{h}}^2\left(0.0062x\right)\right.\\ & -0.000617969\hslash +(-7.6375\times {10}^{-6}\hslash t+0.0000132813\hslash +0.0000132813)\\ & \left.-2.942\times {10}^{-9}\hslash t{tanh}^3\left(0.00625x\right)+3.29\times {10}^{-7}\hslash t{tanh}^2\left(0.00625x\right)+tanh\left(0.0062x\right)\right)\\ v_3(x,t)& =\hslash t{\sec \mathrm{h}}^2\left(0.00625x\right)\left(t^2{\hslash }^2{\sec \mathrm{h}}^4\left(0.00625x\right)-0.000617969{(1.\hslash +1.)}^2\right.\\ & 1.4290\times {10}^{-12}tanh\left(0.00625x\right)-4.9323\times {10}^{-11}+(\hslash (0.0000265625-0.000015275t)\\ & +0.0000132813\left.{\hslash }^2+(0.0000132813-0.000015275t)\right)tanh\left(0.00625x\right)\hslash \\ & +\hslash t((6.5806\times {10}^{-7}-6.2926\times {10}^{-8}t)+6.5806\times {10}^{-7}){tanh}^2\left(0.00625x\right)+...\end{array}$$

4.2. Test Example 2

Let us consider the coupled Burgers’ Equation (1) with ${\alpha }_1={\alpha }_2=1$ and ${\beta }_{1,1}={\beta }_{2,2}=-2,{\beta }_{1,2}={\beta }_{2,1}=2.5$

$$\left\{\begin{array}{lll}\frac{\partial u}{\partial t}-\frac{{\partial }^{2}u}{\partial x^2}-2\frac{\partial u}{\partial x}+2.5\frac{\partial uv}{\partial x}=0& t>0,& x\in [-20,20]\\ \frac{\partial v}{\partial t}-\frac{{\partial }^{2}v}{\partial x^2}-2\frac{\partial v}{\partial x}+2.5\frac{\partial uv}{\partial x}=0& t>0,& x\in [-20,20]\end{array}\right.$$

(17)

with:

$$\left\{\begin{array}{l}u(x,0)=K\left(1-tanh\left(\frac{3}{2}K\left(x\right)\right)\right)\\ v(x,0)=K\left(1-tanh\left(\frac{3}{2}K\left(x\right)\right)\right),\end{array}\right.$$

(18)

the exact solution of (17) is

$$\left\{\begin{array}{l}u(x,t)=K\left(1-tanh\left(\frac{3}{2}K(x-3Kt)\right)\right)\\ v(x,t)=K\left(1-tanh\left(\frac{3}{2}K(x-3Kt)\right)\right),\end{array}\right.$$

(19)

The zero-order deformation equation is:

$$(1-q)L_i[{\varphi }_i(x,t,q)-u_{i,0}(x,t)]=q\hslash N_i\left[{\varphi }_i(x,t,q)\right],i\in \{1,2\}$$

where $L_i=\frac{\partial {\varphi }_i}{\partial t}$, and

$$\left\{\begin{array}{l}{N}_1[{\varphi }_1(x,t,q),{\varphi }_2(x,t,q)]=\frac{\partial {\varphi }_1(x,t,q)}{\partial t}-\frac{{\partial }^2{\varphi }_1(x,t,q)}{\partial x^2}-2\frac{\partial {\varphi }_1(x,t,q)}{\partial x}+2.5\frac{\partial {\varphi }_1(x,t,q){\varphi }_2(x,t,q)}{\partial x}\\ N_2[{\varphi }_1(x,t,q),{\varphi }_2(x,t,q)]=\frac{\partial {\varphi }_2(x,t,q)}{\partial t}-\frac{{\partial }^2{\varphi }_2(x,t,q)}{\partial x^2}-2\frac{\partial {\varphi }_2(x,t,q)}{\partial x}+2.5\frac{\partial {\varphi }_1(x,t,q){\varphi }_2(x,t,q)}{\partial x}\end{array}\right.$$

(20)

The m-th order deformation equation for (20) becomes:

$$\left\{\begin{array}{l}{u}_m(x,t)={\chi }_m{u}_{m-1}(x,t)+\hslash L^{-1}\left[R_{1,m}(u_{m-1}(x,t),v_{m-1}(x,t))\right]\\ v_m(x,t)={\chi }_m{v}_{m-1}(x,t)+\hslash L^{-1}\left[R_{2,m}(u_{m-1}(x,t),v_{m-1}(x,t))\right]\end{array}\right.$$

(21)

where

$$\left\{\begin{array}{ll}{R}_{1,m}& =\frac{\partial u_{m-1}(x,t)}{\partial t}-\frac{{\partial }^2{u}_{m-1}(x,t)}{\partial x^2}-2\frac{\partial u_{m-1}(x,t,q)}{\partial x}\\ & +2.5\sum _{i=0}^{m-1}\left(\frac{\partial u_i(x,t)}{\partial x}{v}_{m-1-i}(x,t)+\frac{\partial v_{m-1-i}(x,t)}{\partial x}{u}_i(x,t)\right)\\ R_{2,m}& =\frac{\partial v_{m-1}(x,t)}{\partial t}-\frac{{\partial }^2{v}_{m-1}(x,t)}{\partial x^2}-2\frac{\partial v_{m-1}(x,t)}{\partial x}\\ & +2.5\sum _{i=0}^{m-1}\left(\frac{\partial u_i(x,t)}{\partial x}{v}_{m-1-i}(x,t)+\frac{\partial v_{m-1-i}(x,t)}{\partial x}{u}_i(x,t)\right).\end{array}\right.$$

Choosing from (18),

$$\left\{\begin{array}{l}{u}_0(x,t)=K\left(1-tanh\left(\frac{3}{2}K\left(x\right)\right)\right)\\ v_0(x,t)=K\left(1-tanh\left(\frac{3}{2}K\left(x\right)\right)\right),\end{array}\right.$$

we successfully obtain:

$$\begin{array}{ll}{u}_1(x,t)& =\hslash t(0.0000944531tanh\left(0.001875x\right)-0.00028125){\sec \mathrm{h}}^2\left(0.001875x\right)\\ u_2(x,t)& =\hslash t{\sec \mathrm{h}}^2\left(0.001875x\right)\left(\hslash t(8.9878\times {10}^{-8}tanh\left(0.001875x\right)-2.6565\times {10}^{-7})\right.\\ & {\sec \mathrm{h}}^2\left(0.001875x\right)-0.0003\hslash -0.00028125-8.9214\times {10}^{-8}\hslash t{tanh}^3\left(0.001875x\right)\\ & \left.+5.313\times {10}^{-7}\hslash t{tanh}^2\left(0.001875x\right)+(-7.9102\times {10}^{-7}\hslash t+0.0000944\hslash +0.0000944)\right)\\ & tanh\left(0.001875x\right).\end{array}$$

and

$$\begin{array}{ll}{v}_1(x,t)& =\hslash t(0.0000944531tanh\left(0.001875x\right)-0.00028125){\sec \mathrm{h}}^2\left(0.001875x\right)\\ v_2(x,t)& =\hslash t{\sec \mathrm{h}}^2\left(0.001875x\right)\left(\hslash t(8.9878\times {10}^{-8}tanh\left(0.001875x\right)-2.6565\times {10}^{-7})\right.\\ & {\sec \mathrm{h}}^2\left(0.001875x\right)-0.00028125\hslash -0.00028125-8.92139\times {10}^{-8}t\hslash {tanh}^3\left(0.001875x\right)\\ & \left.+5.312\times {10}^{-7}\hslash t{tanh}^2\left(0.001875x\right)-7.9102\times {10}^{-7}\hslash t+0.00009\hslash +0.000094)\right)\\ & \times tanh\left(0.0018x\right).\end{array}$$

4.3. Test Example 3

Let us consider the coupled Burgers’ Equation (1) with ${\alpha }_1={\alpha }_2=1$ and ${\beta }_{1,1}={\beta }_{2,2}=-2,{\beta }_{1,2}={\beta }_{2,1}=1$

$$\left\{\begin{array}{lll}\frac{\partial u}{\partial t}-\frac{{\partial }^{2}u}{\partial x^2}-2u\frac{\partial u}{\partial x}+\frac{\partial uv}{\partial x}=0& t>0,& x\in [-\pi ,\pi ]\\ \frac{\partial v}{\partial t}-\frac{{\partial }^{2}v}{\partial x^2}-2v\frac{\partial v}{\partial x}+\frac{\partial uv}{\partial x}=0& t>0,& x\in [-\pi ,\pi ]\end{array}\right.$$

(22)

with:

$$u(x,0)=v(x,0)=sin\left(x\right)$$

(23)

the exact solution of (22) is

$$u(x,t)=v(x,t)=e^{-t}sin\left(x\right)$$

(24)

The m-th order deformation equation for (22) is:

$$\left\{\begin{array}{l}{u}_m(x,t)={\chi }_m{u}_{m-1}(x,t)+\hslash L^{-1}\left[R_{1,m}(u_{m-1}(x,t),v_{m-1}(x,t))\right]\\ v_m(x,t)={\chi }_m{v}_{m-1}(x,t)+\hslash L^{-1}\left[R_{2,m}(u_{m-1}(x,t),v_{m-1}(x,t))\right]\end{array}\right.$$

(25)

where

$$\left\{\begin{array}{l}{R}_{1,m}=\frac{\partial u_{m-1}}{\partial t}-\frac{{\partial }^2{u}_{m-1}}{\partial x^2}-2\sum _{i=0}^{m-1}\left(u_{m-1-i}\frac{\partial u_i}{\partial x}\right)+\sum _{i=0}^{m-1}\left(\frac{\partial u_i}{\partial x}{v}_{m-1-i}+\frac{\partial v_{m-1-i}}{\partial x}{u}_i\right)\\ R_{2,m}=\frac{\partial v_{m-1}}{\partial t}-\frac{{\partial }^2{v}_{m-1}}{\partial x^2}-2\sum _{i=0}^{m-1}\left(v_{m-1-i}\frac{\partial v_i}{\partial x}\right)+\sum _{i=0}^{m-1}\left(\frac{\partial u_i}{\partial x}{v}_{m-1-i}+\frac{\partial v_{m-1-i}}{\partial x}{u}_i(x,t)\right)\end{array}\right.$$

Choosing from (23),

$$\left\{\begin{array}{l}{u}_0(x,t)=sin\left(x\right)\\ v_0(x,t)=sin\left(x\right),\end{array}\right.$$

we successfully obtain the first terms of the homotopy series solution:

$$\begin{array}{ll}{u}_1(x,t)& =\hslash tsin\left(x\right)\\ u_2(x,t)& =\hslash \left(\frac{1}{2}\hslash t^{2}sin\left(x\right)+\hslash tsin\left(x\right)\right)+\hslash tsin\left(x\right)\\ u_3(x,t)& =\frac{1}{6}\hslash t\left(\left(t^2+6t+6\right){\hslash }^2+6(t+2)\hslash +6\right)sin\left(x\right)\end{array}$$

and

$$\begin{array}{ll}{v}_1(x,t)& =\hslash tsin\left(x\right)\\ v_2(x,t)& =\frac{1}{2}{\hslash }^2{t}^{2}sin\left(x\right)+{\hslash }^{2}tsin\left(x\right)+\hslash tsin\left(x\right)\\ v_3(x,t)& =\frac{1}{6}\hslash t\left(\left(t^2+6t+6\right){\hslash }^2+6\hslash (t+2)+6\right)sin\left(x\right)\end{array}$$

So, the HAM series solution becomes:

$$\left\{\begin{array}{ll}u(x,t)& =\frac{1}{120}\left(t^5{\hslash }^5+5t^4(4\hslash +5){\hslash }^4+20t^3\left(6{\hslash }^2+15\hslash +10\right){\hslash }^3\right.\\ & +\left.60t^2\left(4{\hslash }^3+15{\hslash }^2+20\hslash +10\right){\hslash }^2+...\right)sin\left(x\right)\\ v(x,t)& =\frac{1}{120}\left(t^5{\hslash }^5+5t^4(4\hslash +5){\hslash }^4+20t^3\left(6{\hslash }^2+15\hslash +10\right){\hslash }^3\right.\\ & \left.+60t^2\left(4{\hslash }^3+15{\hslash }^2+20\hslash +10\right){\hslash }^2+...\right)sin\left(x\right)\end{array}\right.$$

and for $\hslash =-1$ we obtain

$$u(x,t)=v(x,t)=-\frac{1}{120}\left(t^5-5t^4+20t^3-60t^2+120t-120....\right)sin\left(x\right)=e^{-t}sin\left(x\right)$$

5. Numerical Result and Discussion

In this part, we first plot the HAM approximate solution of u and v and the exact solution for examples 1, 2 and 3 in Figure 1, Figure 2, Figure 3 and Figure 4, and the absolute error for example 1 is presented in Figure 5 for t = 1. In Figure 6, we display the error with different values of t for example 2. Figure 7 is the error of the example, and to show the efficiency and accuracy of the present algorithm, we compare the results with others obtained by different methods, as shown in the following Table 1, Table 2 and Table 3:

Figure 1. The HAM solution (left) and exact solution (right) for $u(x,t)$ example 1.

Figure 2. The HAM solution (left) and exact solution (right) for $v(x,t)$ example 1.

Figure 3. The HAM solution of u,v and the exact solution (from left to right) of example 2.

Figure 4. The HAM solution and exact solution (from left to right) for $u,v$ of example 3.

Figure 5. The absolute error for $t=1$ of example 1.

Figure 6. The absolute error for different values of t of example 2.

Figure 7. The absolute error for example 3.

Table 1. Comparison of HAM solution with the exact solution for $u(x,t)$ example 1.

x	t	Exact	HAM	Absolute Error
−10	$0.5$	$0.046881$	$0.047189$	$3.0848\times {10}^{-4}$
	1	$0.046883$	$0.047500$	$6.1714\times {10}^{-4}$
	5	$0.046898$	$0.049988$	$3.0897\times {10}^{-3}$
	10	$0.04691$	$0.05309$	$6.1753\times {10}^{-3}$
$-5$	$0.5$	$0.04844$	$0.04874$	$3.0921\times {10}^{-4}$
	1	$0.04844$	$0.04906$	$6.1848\times {10}^{-4}$
	5	$0.04845$	$0.05154$	$3.0916\times {10}^{-3}$
	10	$0.04847$	$0.05464$	$6.1671\times {10}^{-3}$
0	$0.5$	$0.05000$	$0.050311$	$3.0934\times {10}^{-4}$
	1	$0.05000$	$0.050622$	$6.1863\times {10}^{-4}$
	5	$0.05002$	$0.053107$	$3.0875\times {10}^{-3}$
	10	$0.05004$	$0.056186$	$6.1469\times {10}^{-3}$
5	$0.5$	$0.05156$	$0.05162$	$6.1789\times {10}^{-5}$
	1	$0.05156$	$0.05218$	$6.1756\times {10}^{-4}$
	5	$0.05158$	$0.05465$	$3.0775\times {10}^{-3}$
	10	$0.05160$	$0.05771$	$6.1151\times {10}^{-3}$
10	$0.5$	$0.05312$	$0.05343$	$3.07805\times {10}^{-4}$
	1	$0.05312$	$0.05374$	$6.15303\times {10}^{-4}$
	5	$0.05314$	$0.05620$	$3.06143\times {10}^{-3}$
	10	$0.05315$	$0.05923$	$6.07167\times {10}^{-3}$

Table 2. Comparison of HAM solution with the exact solution for $v(x,t)$ example 1.

x	t	Exact	HAM	Absolute Error
−10	$0.5$	$0.02188$	$0.02218$	$3.0632\times {10}^{-4}$
	1	$0.02188$	$0.02249$	$6.1278\times {10}^{-4}$
	5	$0.02189$	$0.02496$	$3.0663\times {10}^{-3}$
	10	$0.02191$	$0.02804$	$6.1248\times {10}^{-3}$
$-5$	$0.5$	$0.02344$	$0.02374$	$3.0695\times {10}^{-4}$
	1	$0.02344$	$0.02405$	$6.1392\times {10}^{-4}$
	5	$0.02345$	$0.02652$	$3.0673\times {10}^{-3}$
	10	$0.02347$	$0.02959$	$6.1149\times {10}^{-3}$
0	$0.5$	$0.02500$	$0.02530$	$3.0698\times {10}^{-4}$
	1	$0.05000$	$0.050622$	$6.1863\times {10}^{-4}$
	5	$0.02501$	$0.02808$	$3.0623\times {10}^{-3}$
	10	$0.02503$	$0.03113$	$6.0932\times {10}^{-3}$
5	$0.5$	$0.02656$	$0.02687$	$3.0642\times {10}^{-4}$
	1	$0.02656$	$0.02717$	$6.1261\times {10}^{-4}$
	5	$0.02658$	$0.02963$	$3.0513\times {10}^{-3}$
	10	$0.02660$	$0.03266$	$6.0599\times {10}^{-3}$
10	$0.5$	$0.02812$	$0.02842$	$3.0525\times {10}^{-4}$
	1	$0.02812$	$0.02873$	$6.1017\times {10}^{-4}$
	5	$0.02814$	$0.03117$	$3.0345\times {10}^{-3}$
	10	$0.02815$	$0.03417$	$6.0153\times {10}^{-3}$

Table 3. Comparison of HAM solution with the exact solution for example 2.

x	t	Exact	HAM u	Absolute Error u	HAM v	Absolute Error v
−10	$0.5$	$0.09809$	$0.09826$	$1.6951\times {10}^{-4}$	$0.09826$	$1.6951\times {10}^{-4}$
	1	$0.09806$	$0.09840$	$3.3889\times {10}^{-4}$	$0.09840$	$3.3889\times {10}^{-4}$
	5	$0.09784$	$0.09953$	$1.6893\times {10}^{-3}$	$0.09953$	$1.6893\times {10}^{-3}$
	10	$0.09756$	$0.10092$	$3.3657\times {10}^{-3}$	$0.10092$	$3.3657\times {10}^{-3}$
−5	$0.5$	$0.09903$	$0.09920$	$1.6911\times {10}^{-4}$	$0.09920$	$1.6911\times {10}^{-4}$
	1	$0.09900$	$0.09934$	$3.3809\times {10}^{-4}$	$0.09934$	$3.3809\times {10}^{-4}$
	5	$0.09878$	$0.10046$	$1.6852\times {10}^{-3}$	$0.10046$	$1.6852\times {10}^{-3}$
	10	$0.09850$	$0.10185$	$3.3571\times {10}^{-3}$	$0.10185$	$3.3571\times {10}^{-3}$
0	$0.5$	$0.09997$	$0.10014$	$1.6868\times {10}^{-4}$	$0.10014$	$1.6868\times {10}^{-4}$
	1	$0.09994$	$0.10028$	$3.3723\times {10}^{-4}$	$0.10028$	$3.3723\times {10}^{-4}$
	5	$0.09971$	$0.1014$	$1.6808\times {10}^{-3}$	$0.1014$	$1.6808\times {10}^{-3}$
	10	$0.09943$	$0.10278$	$3.3479\times {10}^{-3}$	$0.10278$	$3.3479\times {10}^{-3}$
5	$0.5$	$0.10090$	$0.10107$	$1.6822\times {10}^{-4}$	$0.10107$	$1.6822\times {10}^{-4}$
	1	$0.10088$	$0.10121$	$3.3631\times {10}^{-4}$	$0.10121$	$3.3631\times {10}^{-4}$
	5	$0.10065$	$0.10233$	$1.6760\times {10}^{-3}$	$0.10233$	$1.6760\times {10}^{-3}$
	10	$0.10037$	$0.10371$	$3.3382\times {10}^{-3}$	$0.10371$	$3.3382\times {10}^{-3}$
10	$0.5$	$0.10184$	$0.10201$	$1.6773\times {10}^{-4}$	$0.10201$	$1.6773\times {10}^{-4}$
	1	$0.10181$	$0.10215$	$3.3533\times {10}^{-3}$	$0.10215$	$3.3533\times {10}^{-3}$
	5	$0.10159$	$0.10326$	$1.6710\times {10}^{-3}$	$0.10326$	$1.6710\times {10}^{-3}$
	10	$0.10131$	$0.10464$	$3.3279\times {10}^{-3}$	$0.10464$	$3.3279\times {10}^{-3}$

6. Conclusions

In this work, we successfully applied the HAM for the coupled Burgers’ equation, and we show that this method is efficient and applicable for this type of partial differential equation. We compared our results with the exact solution and NIM method, and we present graphically the HAM solution and the error.