Traveling wave solutions of Zakharov–Kuznetsov-modified equal-width and Burger’s equations via ${\text{exp}}( - \varphi \left( \eta \right))$-expansion method

Abstract

In this article, a technique is proposed for obtaining better and accurate results for nonlinear PDEs. We constructed abundant exact solutions via exp$( - \varphi \left( \eta \right))$-expansion method for the Zakharov–Kuznetsov-modified equal-width (ZK-MEW) equation and the (2 + 1)-dimensional Burgers equation. The traveling wave solutions are found through the hyperbolic functions, the trigonometric functions and the rational functions. The specified idea is very pragmatic for PDEs, and could be extended to engineering problems.

Background

Over the past few decades, researchers have shown keen interest in the solutions of nonlinear partial differential equations (PDEs).In the study of nonlinear physical phenomena, the investigation of solitary wave solutions [1–44] of nonlinear wave equations shows an important role. Scientific problems arise nonlinearly in numerous fields of mathematical physics, such as fluid mechanics, plasma physics, solid-state physics and geochemistry. Due to exact interpretation of nonlinear phenomena, these problems have gained much importance. However, in recent years, a variety of effective analytical methods has been developed to study soliton solutions of nonlinear equations, such as Backlund transformation method [1], tanh method [2–6], extended tanh method [7–12], pseudo-spectral method [13], trial function [14], sine–cosine method [15], Hirota method [16], exp function method [17–25], $(G^{'} /G)$-expansion method [26–30], homogeneous balance method [31, 32], F-expansion method [33–35] and Jacobi elliptic function expansion method [36–38]. Ma et al. [39–44] established the complexiton solutions for Toda lattice equation. The theme of the method is that the exact solutions of nonlinear evolution equations can be articulated by exp$( - \varphi \left( \eta \right))$, where $\varphi \left( \eta \right)$ gratifies the ordinary differential equation (ODE):

$$ \left( {\varphi '\left( \eta \right)} \right) = { \exp }\left( { - \varphi \left( \eta \right)} \right) + \mu \,\, { \exp }\left( {\varphi \left( \eta \right)} \right) + \lambda $$

(1)

where $\eta = x - Vt.$

Explanation of exp$( - \varphi \left( \eta \right))$-expansion method

Now, the exp$( - \varphi \left( \eta \right))$-expansion method will be explained for constructing traveling wave solutions. Consider the general nonlinear partial differential equation for $u\left( {x,t} \right)$ is given by,

$$ \phi \left( {u,u_{t} ,u_{x} ,u_{tt} ,u_{xx} ,u_{xxx} , \ldots } \right) = 0, $$

(2)

where $u\left( \eta \right) = u\left( {x,t} \right),$ $\phi$ is a polynomial of $u$ and its derivatives. Solving (2), the following steps are as.

Step 1 We Combine the variables by $\eta ,$

$$ u = u\left( \eta \right),\quad \eta = x - Vt, $$

(3)

where V is the speed of wave. Using Eqs. (3, 2) reduced to the following ODE for $u = u\left( \eta \right)$

$$ G\left( {u, u^{\prime}, u^{\prime\prime}, u^{\prime\prime\prime}, u^{\prime\prime\prime\prime}, \ldots } \right) = 0, $$

(4)

Step 2 The solution of Eq. (4) can be articulated as

$$ u\left( \eta \right) = \mathop \sum \limits_{n = 0}^{M} \,\, a_{n} \,\, \left( {{ \exp }\left( { - \varphi \left( \eta \right)} \right)} \right)^{\text{n}} , $$

(5)

where $a_{n} 0 \le n \le M$ are constants such that $a_{n} \ne 0$ and $\varphi \left( \eta \right)$ satisfies Eq. (1). Our solutions now depend on the parameters involved in (1).

Family 1: When $\lambda^{2} - 4\mu > 0,$ we have

$$ \varphi \left( \eta \right) = { \ln }\left\{ {\frac{1}{2\mu }\left( { - \sqrt {\left( {\lambda^{2} - 4\mu } \right)} { \tanh }\left( {\frac{{\sqrt {\left( {\lambda^{2} - 4\mu } \right)} }}{2}(\eta + c_{1} )} \right) - \lambda } \right)} \right\}. $$

(6)

Family 2: When $\lambda^{2} - 4\mu < 0,$ we have

$$ \varphi \left( \eta \right) = { \ln }\left\{ {\frac{1}{2\mu }\left( {\sqrt {\left( {\lambda^{2} - 4\mu } \right)} { \tan }\left( {\frac{{\sqrt {\left( {\lambda^{2} - 4\mu } \right)} }}{2}(\eta + c_{1} )} \right) - \lambda } \right)} \right\}. $$

(7)

Family 3: When $\lambda^{2} - 4{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \mu > 0{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \mu = 0$ and $\lambda \ne 0,$

$$ \varphi \left( \eta \right) = - { \ln }\left\{ {\frac{\lambda }{{\exp \left( {\lambda \left( {\eta + k} \right)} \right) - 1}}} \right\}. $$

(8)

Family 4: When $\lambda^{2} - 4{\kern 1pt} {\kern 1pt} {\kern 1pt} \mu = 0,{\kern 1pt} {\kern 1pt} {\kern 1pt} \lambda \ne 0,$ and $\mu \ne 0,$

$$ \varphi \left( \eta \right) = { \ln }\left\{ {\frac{{2\left( {\lambda \left( {\eta + k} \right) + 2} \right)}}{{\left( {\lambda^{2} \left( {\eta + k} \right)} \right)}}} \right\}. $$

(9)

Family 5: When $\lambda^{2} - 4{\kern 1pt} {\kern 1pt} {\kern 1pt} \mu = 0,{\kern 1pt} {\kern 1pt} {\kern 1pt} \lambda = 0,$ and $\mu = 0,$

$$ \varphi \left( \eta \right) = { \ln }(\eta + k) $$

(10)

Step 3 By considering the homogenous principal, in Eq. (4). Considering Eqs. (1, 4, 5), we have ${\text{e}}^{M\varphi \left( \eta \right)}$. We get algebraic equations with $a_{n} ,V,\lambda , \mu ,$ after comparing the same powers of ${\text{e}}^{\varphi \left( \eta \right)}$ to zero. We put the above values in Eq. (5) and with Eq. (1), we get some valuable traveling wave solutions of Eq. (2).

Solution procedure

Zakharov–Kuznetsov-modified equal-width equation

Consider the equation,

$$ u_{t} + \alpha \left( {u^{n} } \right)_{x} + \left( {\beta u_{xt} + \delta u_{yy} } \right)_{x} = 0, $$

(11)

where $\alpha ,{\kern 1pt} \beta$ and $\delta$ are some nonzero parameters. We use $u = u\left( \eta \right),\eta = x + y - Vt,$ we can convert Eq. (11) into an ODE.

$$ - Vu^{\prime} - \beta Vu^{\prime\prime\prime} + \delta u^{\prime\prime\prime} + 2\alpha uu^{\prime} = 0, $$

(12)

where the dash denotes the derivative w. r. t. $\eta$. Now integrating Eq. (12), we have,

$$ - Vu - \beta Vu^{\prime\prime} + \delta u^{\prime\prime} + \alpha u^{2} + C = 0, $$

(13)

Using homogenous principle, balancing $u^{\prime\prime}$ and $u^{2}$, we have

$$ \begin{aligned} 2M &= M + 2, \hfill \\ M &= 2. \hfill \\ \end{aligned} $$

The trial solution of Eq. (12) can be stated as,

$$ u\left( \eta \right) = a_{2} \left( {{ \exp }\left( { - \varphi \left( \eta \right)} \right)} \right)^{2} + a_{1} \left( {{ \exp }\left( { - \varphi \left( \eta \right)} \right)} \right) + a_{0} , $$

(14)

where $a_{2} \ne 0,{\kern 1pt} {\kern 1pt} a_{1}$ and $a_{0}$ are constants, while $\lambda ,\mu$ are any constants.

Putting $u, u^{\prime} , u^{\prime\prime} , u^{2}$ in Eq. (13) and comparing, we get,

$$ \begin{aligned} &\alpha a_{0}^{2} + \delta a_{1} \mu \lambda + C - 2\beta Va_{2} \mu^{2} - \beta Va_{1} \mu \lambda + 2\delta a_{2} \mu^{2} - Va_{0} = 0, \hfill \\ &2\alpha a_{0} a_{1} + \delta a_{1} \lambda^{2} + 2\delta a_{1} \mu + - 2\beta Va_{1} \mu - 6\beta V\mu \lambda - \beta Va_{1} \lambda^{2} + 6\delta a_{2} \mu \lambda - Va_{1} = 0, \hfill \\ &2\alpha a_{2} a_{1} + 10\delta a_{2} \lambda + 2\delta a_{1} + - 2\beta Va_{1} - 10\beta Va_{2} \lambda = 0, \hfill \\ &2\alpha a_{2} a_{1} + 10a_{2} \lambda + 2a_{1} + - 2\beta Va_{1} - 10\beta Va_{2} \lambda = 0, \hfill \\ &\alpha a_{2}^{2} + 6\delta a_{2} - 6\beta Va_{2} = 0, \hfill \\ \end{aligned} $$

(15)

By solving the algebraic equations, the required solution is given below.

$$ \left\{ {V = \frac{1}{6}\frac{{\alpha a_{2} + 6\delta }}{\beta }, {\kern 1pt} \lambda = 0, {\kern 1pt} a_{0} = a_{0} , a_{1} = 0, \mu = \frac{1}{2}\frac{1}{{\beta \alpha a_{2} }}\left(\sqrt 2 \sqrt {\beta \alpha \left( {6C\beta + 6\alpha \beta a_{0}^{2} - \alpha a_{0} a_{2} - 6a_{0} \delta } \right)} ,\right)} \}\right. $$

where $\lambda$ and $\mu$ are any constants.

Now putting the values in Eq. (14), we obtain

$$ u = a_{0} + a_{2} e^{ - 2\varphi \left( \eta \right)} , $$

(16)

where $\eta = x - Vt.$ By putting (6–10) in (16), we obtain the solutions which are given below.

Case 1 When $\lambda^{2} - 4\mu > 0$ and $\mu \ne 0,$ we have,

$$ u_{1} \left( \eta \right) = a_{0} + \frac{{4a_{2} \mu^{2} }}{{\left( { - \sqrt {\lambda^{2} - 4\mu } { \tanh }\left( {\frac{{\sqrt {\lambda^{2} - 4\mu } }}{2}\left( {\eta + c_{1} } \right)} \right) - \lambda } \right)^{2} }}, $$

where $\eta = x - \frac{1}{6}\frac{{\alpha a_{2} + 6\delta }}{\beta }t$ and where $c_{1}$ is any constant.

Case 2 When $\lambda^{2} - 4\mu < 0$ and $\mu \ne 0,$ we have,

$$ u_{2} \left( \eta \right) = a_{0} + \frac{{4a_{2} \mu^{2} }}{{\left( {\sqrt { - \lambda^{2} + 4\mu } { \tan }\left( {\frac{{\sqrt { - \lambda^{2} + 4\mu } }}{2}\left( {\eta + c_{1} } \right)} \right) - \lambda } \right)^{2} }}, $$

where $\eta = x - \frac{1}{6}\frac{{\alpha a_{2} + 6\delta }}{\beta }t$ and where $c_{1}$ is any constant.

Case 3 When $\mu = 0$ and $\lambda \ne 0,$ we have,

$$ u_{3} \left( \eta \right) = a_{0} + \frac{{a_{2} \lambda^{2} }}{{\left( {\exp \left( {\eta + c_{1} } \right)^{\lambda } - 1} \right)^{2} }}, $$

where $\eta = x - \frac{1}{6}\frac{{\alpha a_{2} + 6\delta }}{\beta }t$ and where $c_{1}$ is any constant.

Case 4 When $\lambda^{2} - 4\mu = 0,{\kern 1pt} \lambda \ne 0,$ and $\mu \ne 0,$ we obtain,

$$ u_{4} \left( \eta \right) = a_{0} + \frac{{a_{2} \left( {\eta + c_{1} } \right)^{2} \lambda^{4} }}{{\left( {2\left( {\eta + c_{1} } \right)^{\lambda } + 2} \right)^{2} }} , $$

where $\eta = x - \frac{1}{6}\frac{{\alpha a_{2} + 6\delta }}{\beta }t$ and where $c_{1}$ is any constant.

Case 5 When $\lambda = 0,$ and $\mu = 0,$ we have, $u_{5} \left( \eta \right) = a_{0} + \frac{{a_{2} }}{{\left( {\eta + c_{1} } \right)^{2} }},$ where $\eta = x - \frac{1}{6}\frac{{\alpha a_{2} + 6\delta }}{\beta }t$ and where $c_{1}$ is any constant.

Graphical demonstration

The graphs are given in Figs. 1, 2, 3, 4 and 5.

Fig. 1

Kink wave solution of $u_{1}$ when $a_{2} = 1, {\kern 1pt} {\kern 1pt} {\kern 1pt} a_{0} = 2,{\kern 1pt} {\kern 1pt} {\kern 1pt} y = 0, {\kern 1pt} {\kern 1pt} {\kern 1pt} \lambda = 3,{\kern 1pt} {\kern 1pt} {\kern 1pt} \mu = 2, {\kern 1pt} {\kern 1pt} {\kern 1pt} c_{1} = 1$

Fig. 2

Singular kink wave solution $u_{2}$ when ${\kern 1pt} a_{2} = 10, {\kern 1pt} {\kern 1pt} a_{0} = 8,{\kern 1pt} {\kern 1pt} {\kern 1pt} y = 0, {\kern 1pt} {\kern 1pt} {\kern 1pt} \lambda = 7,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \mu = 5,{\kern 1pt} {\kern 1pt} {\kern 1pt} c_{1} = - 10$

Fig. 3

Singular kink wave solution $u_{3}$ when $a_{2} = 1, a_{0} = 2, y = 0, \lambda = 1, c_{1} = - 1$

Fig. 4

Singular kink wave solution $u_{4}$ when ${\kern 1pt} a_{2} = 3, a_{0} = 2, y = 0, \lambda = 5, \mu = 4, c_{1} = - 2$

Fig. 5

Singular kink wave solution $u_{5}$ when $a_{2} = 0.5, a_{0} = 0.2, y = 0, \lambda = 0.1, c_{1} = - 0.1$

(2 + 1)-dimensional Burger’s equation

Consider the equation,

$$ u_{t} - uu_{x} - u_{xx} - u_{yy} = 0, $$

(17)

where $\alpha ,\beta$ and $\delta$ are some nonzero parameters. We have, $u = u\left( \eta \right)$, $\eta = x + y - Vt,$ we can convert Eq. (17) into an ODE.

$$ - Vu^{\prime} - 2u^{\prime\prime} - uu^{\prime} = 0, $$

(18)

where dash denotes the derivative w. r. t.$\eta.$

Integrating Eq. (18), we have,

$$ - Vu - 2u^{\prime} - \frac{1}{2}u^{2} + C = 0, $$

(19)

Using homogenous principle, balancing the $u'$ and $u^{2} ,$ we have, $M = 1$.

The trial solution of Eq. (18) can be stated as,

$$ u\left( \eta \right) = a_{1} \,\, \left( {{ \exp }\left( { - \varphi \left( \eta \right)} \right)} \right) + a_{0} , $$

(20)

where $a_{1} \ne 0,{\kern 1pt} {\kern 1pt} {\kern 1pt} a_{0}$ is a constant, while $\lambda ,\mu$ are any constants. By putting $u, u^{\prime} , u^{\prime\prime} , u^{2}$ in Eq. (19) and comparing, we get

$$ \begin{aligned} &- \frac{1}{2}a_{0}^{2} + 2a_{1} \mu + C - Va_{0} = 0, \hfill \\ &- a_{0} a_{1} + 2a_{1} \lambda - Va_{1} = 0, \hfill \\ &- \frac{1}{2}a_{1}^{2} + 2a_{1} = 0, \hfill \\ \end{aligned} $$

(21)

By solving the algebraic equations, the required solution is given below.

$$ \left\{ {\begin{array}{*{20}c} {\lambda = \frac{1}{2}\sqrt {V^{2} + 2C + 16\mu } , a_{0} = - V + \sqrt {V^{2} + 2C + 16\mu } , \quad a_{1} = 4 ,} \\ {} \\ \end{array} } \right\}. $$

where $\lambda$ and $\mu$ are any constants. Now putting the values in Eq. (20), we obtain,

$$ u = - {\text{V}} + \sqrt {V^{2} + 2C + 16\mu } + 4e^{ - \varphi \left( \eta \right)} , $$

(22)

where $\eta = x - Vt.$

Now putting (6–10) in (22), we obtain the solutions as.

Case 1 When $\lambda^{2} - 4\mu > 0$ and $\mu \ne 0,$ we have,

$$ u_{6} \left( \eta \right) = - 1 + \sqrt {1 + 2C + 16\mu } + \frac{8\mu }{{\left( { - \sqrt {\lambda^{2} - 4\mu } { \tanh }\left( {\frac{{\sqrt {\lambda^{2} - 4\mu } }}{2}\left( {\eta + c_{1} } \right)} \right) - \lambda } \right)}}, $$

where $\eta = x - {\text{V}}t$ and where $c_{1}$ is any constant.

Case 2 When $\lambda^{2} - 4\mu < 0$ and $\mu \ne 0,$ we obtain,

$$ u_{7} \left( \eta \right) = = - 1 + \sqrt {1 + 2C + 16\mu } + \frac{8\mu }{{\left( { + \sqrt { - \lambda^{2} + 4\mu } { \tanh }\left( {\frac{{\sqrt { - \lambda^{2} + 4\mu } }}{2}\left( {\eta + c_{1} } \right)} \right) - \lambda } \right)}}, $$

where $\eta = x - {\text{V}}t$ and where $c_{1}$ is any constant.

Case 3 When $\mu = 0$ and $\lambda \ne 0,$ we have,

$$ u_{8} \left( \eta \right) = = = - 1 + \sqrt {1 + 2C + 16\mu } + \frac{4\lambda }{{\left( {\left( {\eta + c_{1} } \right)^{\lambda } - 1} \right)}}, $$

where $\eta = x - Vt$ and where $c_{1}$ is any constant.

Case 4 When $\lambda^{2} - 4\mu = 0, \lambda \ne 0,$ and $\mu \ne 0,$ we obtain,

$$ u_{9} \left( \eta \right) = - 1 + \sqrt {1 + 2C + 16\mu } + \frac{{4\left( {\eta + c_{1} } \right)\lambda^{2} }}{{\left( {2\left( {\eta + c_{1} } \right)^{\lambda } + 2} \right)}}, $$

where $\eta = x - Vt$ and where $c_{1}$ is any constant.

Case 5 When $\lambda = 0,$ and $\mu = 0,$ we have,

$$ u_{10} \left( \eta \right) = - 1 + \sqrt {1 + 2C + 16\mu } + \frac{4}{{\left( {\eta + c_{1} } \right)}}, $$

where $\eta = x - Vt$ and where $c_{1}$ is any constant.

Graphical illustration

The graphs are given in Figs. 6, 7, 8, 9 and 10.

Fig. 6

Kink wave solution $u_{6}$ when $C = 1, a_{0} = 1, y = 0, \lambda = 3, \mu = 1, c_{1} = 1$

Fig. 7

Periodic solution $u_{7} \left( \eta \right)$ when $a_{2} = 2, C = 1, y = 0, \lambda = 1,\mu = 2, c_{1} = - 1$

Fig. 8

Singular kink wave solution $u_{8}$ when $\mu = 1, C = 1, y = 0, \lambda = 3, c_{1} = - 1$

Fig. 9

Singular kink wave solution $u_{9}$ when $a_{2} = 1, C = 1, y = 0, \lambda = 13,\mu = 1, c_{1} = - 1$

Fig. 10

Singular kink wave solution $u_{10}$ when C = 15, y = 0, μ = 12, c ₁ = −1

Conclusions

The exp$( - \varphi \left( \eta \right))$-expansion method has been successfully applied to find the exact solutions of (ZK-MEW) equation and the Burger’s equation. The attained results show that the proposed technique is effective and capable for solving nonlinear partial differential equations. In this study, some exact solitary wave solutions, mostly solitons and kink solutions, are obtained through the hyperbolic and rational functions. This study shows that the proposed method is quite proficient and practically well organized in finding exact solutions of other physical problems.