Modified method of simplest equation for obtaining exact solutions of the Zakharov–Kuznetsov equation, the modified Zakharov–Kuznetsov equation, and their generalized forms

Abstract

In this paper, we study the application of a version of the method of simplest equation for obtaining exact traveling wave solutions of the Zakharov–Kuznetsov equation, the modified Zakharov–Kuznetsov equation, and their generalized forms. The Duffing-type equation is used as simplest auxiliary equation. In the meantime, the proposed method is proved to be a powerful mathematical tool for obtaining exact solutions of nonlinear partial differential equations in mathematical physics.

1 Introduction

In the last years, constructing exact solutions of nonlinear partial differential equations (PDEs) has been of great significance, because a lot of nonlinear complex physical phenomena arise in many research fields, for example physics, biology, ecology, and engineering. The study of exact solutions can help us understand the mechanism that governs nonlinear complex physical phenomena. Moreover, the exact analytical solutions of the nonlinear PDEs play a key role in several research directions, such as descriptions of different kinds of waves, as initial condition for simulation process. People have paid lots of attention to this significant research field by constructing exact solutions of various nonlinear PDEs [1–8]. A lot of excellent methods have been derived to find exact solutions of nonlinear PDEs [9–12]. Some important methods among them are the transformed rational function method [13], the homogeneous balance method [14–16], the hyperbolic function method [17], the F-expansion method [18], the variable-detached method [19–22], the Bäcklund transformation method [23–26], the Jacobi elliptic function method [27], the extended $\tanh$-function method [28], the multiple $\exp$-function approach [29], and so on [30–33]. One of the direct methods is the method of simplest equation, which was presented by Kudryashov in 1988 [34]. This method and modified method of simplest equation have been used to find exact solutions of nonlinear PDEs [35–42], such as Fisher equation and Fisher-like equations [43], generalized Kuramoto–Sivashinsky equation [44]. The aim of this research is to develop a modified method of simplest equation based on the Duffing-type equation and then apply this method to solve the Zakharov–Kuznetsov equation, the modified Zakharov–Kuznetsov equation, and their generalized forms. The aforementioned equations have been studied by researchers, and the exact and numerical simulation papers [45–54] can help us deeply understand our theoretical results.

This paper is organized as follows. In Sect. 2, we introduce the Duffing-type equation and its special case with exact solutions. In Sect. 3, we give a brief description of the modified method of simplest equation. In Sect. 4, we obtain exact solutions of the Zakharov–Kuznetsov equation. In Sect. 5, using the similar procedure, we obtain exact solutions of the generalized Zakharov–Kuznetsov equation. In Sect. 6, exact solutions of modified Zakharov–Kuznetsov equation are obtained. Exact solutions of the generalized modified Zakharov–Kuznetsov equation are determined in Sect. 7. Some comparisons will be given in Sect. 8. Finally, we conclude this article with some conclusions and discussions in Sect. 9.

2 Exact solutions of a Duffing-type equation

In this section, we briefly present the descriptions of some new exact solutions of a Duffing-type equation. It is well known that many practically important applied problems, which arise from physics, engineering, and so on, require to solve a Duffing-type ordinary differential equation (ODE) [55–57], which is an autonomous ODE written in the following form

$$\begin{aligned} x^{\prime \prime }(t)=\sum _{j\in K}a_jx^j(t), \end{aligned}$$

(1)

where $a_j^{\prime }s$ are all real constants and K is a finite set of integers.

One example is that a very simple model of a spring, which is suspended from a ceiling, subjects to resistant offered by medium proportional to the square displacement. Then, it can be described by an ODE as

$$\begin{aligned} x^{\prime \prime }(t)=-\omega ^2 x-\beta x^2 \end{aligned}$$

which is a direct application of Newton’s second law.

The other example is the cubic–quintic Duffing oscillatory problem presented by the following ODE

$$\begin{aligned} x^{\prime \prime }(t)+ax+bx^3+cx^5=0, \end{aligned}$$

which is the mathematical models for free vibrations of a restrained uniform beam with intermediate lumped mass, a nonlinear generalized compound KdV equation, the nonlinear dynamics of slender elastica, and so on.

Actually, Eq. (1) can be integrated after multiply it through by $x^{\prime }(t)$, the result equation is

$$\begin{aligned} (x^{\prime }(t))^2=\sum _{j\in K}b_jx^{j+1}(t), \end{aligned}$$

(2)

where $b_j=\frac{2a_j}{(j+1)}$.

In this research, we will use the particular case of Eq. (2) as simplest equation, which is described as follows:

$$\begin{aligned} (h^{\prime }(\xi ))^2=c_2h^2(\xi )+c_3h^3(\xi )+c_4h^4(\xi ), \end{aligned}$$

(3)

where $c_2,c_3,c_4$ are all real parameters to be determined.

Theorem 2.1

Suppose that $h(\xi )$ is a solution of Eq. (3), $\varDelta =c_3^2-4c_2c_4$ and $\epsilon =\pm 1$. Then, we have the following results:

• (1) If $c_2>0$, then Eq. (3) possesses the following solutions:

  $$\begin{aligned}&h_1(\xi )=\frac{-c_2c_3sech ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }{c_3^2-c_2c_4\left( 1+\epsilon \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2}, \end{aligned}$$

  (4)
  $$\begin{aligned}&h_2(\xi )=\frac{c_2c_3csch ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }{c_3^2-c_2c_4\left( 1+\epsilon \coth \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2}, \end{aligned}$$

  (5)
  $$\begin{aligned}&h_3(\xi )=\frac{4c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }}{(\mathrm{e}^{\epsilon \sqrt{c_2}\xi }-c_3)^2-4c_2c_4}. \end{aligned}$$

  (6)

• (2) If $c_2>0, \varDelta >0$, then Eq. (3) has the following solution:

  $$\begin{aligned}&h_4(\xi )=\frac{2c_{2} sech (\sqrt{c_2}\xi )}{\epsilon \sqrt{\varDelta }-c_3 sech (\sqrt{c_2}\xi )}. \end{aligned}$$

  (7)

• (3) If $c_2>0, \varDelta <0$, then a solution of Eq. (3) is

  $$\begin{aligned}&h_5(\xi )=\frac{2c_2 csch (\sqrt{c_2}\xi )}{\epsilon \sqrt{-\varDelta }-c_3 csch (\sqrt{c_2}\xi )}. \end{aligned}$$

  (8)

• (4) If $c_2<0, \varDelta >0$, then Eq. (3) has the following solutions:

  $$\begin{aligned}&h_6(\xi )=\frac{2c_2 \sec (\sqrt{-c_2}\xi )}{\epsilon \sqrt{\varDelta }-c_3 \sec (\sqrt{-c_2}\xi )}, \end{aligned}$$

  (9)
  $$\begin{aligned}&h_7(\xi )=\frac{2c_2 \csc (\sqrt{-c_2}\xi )}{\epsilon \sqrt{\varDelta }-c_3 \csc (\sqrt{-c_2}\xi )}. \end{aligned}$$

  (10)

• (5) If $c_2>0, \varDelta =0$, then the solutions of Eq. (3) are as follows:

  $$\begin{aligned}&h_8(\xi )=-\frac{c_2}{c_3}\left( 1+\epsilon \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) , \end{aligned}$$

  (11)
  $$\begin{aligned}&h_9(\xi )=-\frac{c_2}{c_3}\left( 1+\epsilon \coth \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) . \end{aligned}$$

  (12)

• (6) If $c_2>0, c_4>0$, then Eq. (3) possesses the following solutions:

  $$\begin{aligned}&h_{10}(\xi )=\frac{-c_2sech ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }{c_3+2\epsilon \sqrt{c_2c_4} \tanh (\frac{\sqrt{c_2}}{2}\xi )}, \end{aligned}$$

  (13)
  $$\begin{aligned}&h_{11}(\xi )=\frac{c_2csch ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }{c_3+2\epsilon \sqrt{c_2c_4} \coth (\frac{\sqrt{c_2}}{2}\xi )}. \end{aligned}$$

  (14)

• (7) If $c_2<0, c_4>0$, then Eq. (3) possesses the following solutions:

  $$\begin{aligned}&h_{12}(\xi )=\frac{-c_2\sec ^2\left( \frac{\sqrt{-c_2}}{2}\xi \right) }{c_3+2\epsilon \sqrt{-c_2c_4} \tan \left( \frac{\sqrt{c_2}}{2}\xi \right) }, \end{aligned}$$

  (15)
  $$\begin{aligned}&h_{13}(\xi )=\frac{-c_2\csc ^2\left( \frac{\sqrt{-c_2}}{2}\xi \right) }{c_3+2\epsilon \sqrt{-c_2c_4} \cot \left( \frac{\sqrt{c_2}}{2}\xi \right) }. \end{aligned}$$

  (16)

• (8) If $c_2>0, c_3=0$, then we have the solution of Eq. (3) as

  $$\begin{aligned}&h_{14}(\xi )=\frac{\pm 4c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }}{1-4c_2c_4\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }}. \end{aligned}$$

  (17)

• (9) If $c_2=c_3=0$ and $c_4>0$, then Eq. (3) has a rational solution

  $$\begin{aligned}&h_{15}(\xi )=-\frac{\epsilon }{\sqrt{c_4}\xi }. \end{aligned}$$

  (18)

• (10) If $c_2=c_4=0$, then Eq. (3) has a rational solution

  $$\begin{aligned}&h_{16}(\xi )=\frac{1}{c_3\xi ^2}. \end{aligned}$$

  (19)

In the present paper, we will apply a modified method of simplest equation, which is Eq. (3), to the Zakharov–Kuznetsov equation, modified Zakharov–Kuznetsov equation, and their generalized forms, and construct more exact solutions of them.

3 Description of modified method of simplest equation

We will introduce briefly the method of simplest equation as follows. First, we will reduce the given nonlinear PDE to a nonlinear ODE by an appropriate ansatz, such as the traveling wave ansatz, described by

$$\begin{aligned} H(u,u_{\xi },u_{\xi \xi },\ldots )=0. \end{aligned}$$

(20)

Suppose that

$$\begin{aligned} u(\xi )=\sum _{i=0}^n a_ih^i(\xi ), \end{aligned}$$

(21)

where $h(\xi )$ is a solution of simpler ODE called simplest equation and n is positive integer determined by the homogeneous method. Plugging (21) into (20), we obtain a polynomial of $h(\xi )$. Let all the coefficients of the obtained polynomial be zero, then a system of nonlinear algebraic equations with respect to the coefficients $a_i^{\prime }s$ is obtained. The solutions of this system lead to the solutions of the studied nonlinear PDE.

Taking full advantage of the previous work, especially the Duffing-type equation, we present the following algorithm for the modified method of simplest equation.

Algorithm 3.1.

Step 1: For any given nonlinear PDE in three independent variables x, y, t and dependent variable $u=u(x,y,t)$,

$$\begin{aligned} H(u,u_t,u_x,u_y,u_{tt},u_{tx},u_{xx},u_{yy},u_{ty},u_{xy},\ldots )=0, \end{aligned}$$

(22)

where H is a polynomial of function u and its various derivatives $u_t,u_x,u_y,u_{tt},u_{tx},$ $u_{xx},u_{yy},u_{ty},u_{xy},\ldots$. By the traveling wave transform

$$\begin{aligned} u(x,y,t)=u(\xi ), \xi =\alpha x+\beta y-ct, \end{aligned}$$

(23)

Equation (22) is reduced to an ODE (ODE)

$$\begin{aligned} L(u,u^{\prime }, u^{\prime \prime }, u^{'''}, \ldots )=0, \end{aligned}$$

(24)

where $\alpha , \beta$, and $\lambda$ are constants to be determined later and $u^{\prime }=\frac{\mathrm{d}u}{\mathrm{d}\xi }$.

Step 2: Using modified method of simplest equation, we seek the solution of Eq. (24) by using the following ansatz

$$\begin{aligned} u(\xi )=\sum _{i=0}^{N}a_ih^i(\xi ),\ a_N\ne 0, \end{aligned}$$

(25)

where N is a positive integer, which can be determined by balancing the highest order derivative terms with the highest power nonlinear terms in Eq. (24), $a_i^{\prime }$s are all real constants to be determined, and $h(\xi )$ is a solution of Eq. (3) given in Sect. 2.

Step 3: Inserting ansatz (23) and (25) into Eq. (24) with computerized symbolic computation, collecting the coefficients of all powers of $h^i(\xi )$, and then equating all the obtained coefficients to zero lead to a system of algebraic equations for the parameters $\alpha ,\beta , c, c_2, c_3, c_4, a_i (i=0, 1, 2, \ldots , N).$

Step 4: Applying computer algebraic technique to solve this system of algebraic equations obtained in Step 3, we can obain the values of $\alpha ,\beta ,c, c_2, c_3, c_4, a_i\ (i=0, 1, 2, \ldots , N).$

Step 5: Plugging these obtained coefficients and solutions of Eq. (3) into (25), and setting $\xi =\alpha x+\beta y-ct$, finally, we obtain the new exact traveling wave solutions of Eq. (24).

In the following sections, we will apply the aforementioned modified method of simplest equation to find exact traveling wave solutions of the Zakharov–Kuznetsov equation, the modified Zakharov–Kuznetsov equation, and their generalized forms.

4 The Zakharov–Kuznetsov equation (ZK)

In this section, we will apply the proposed modified method of simplest equation to the Zakharov–Kuznetsov equation in the (2+1) dimensions. The ZK equation is written in the following form:

$$\begin{aligned} u_t+auu_x+b(u_{xxx}+u_{yyx})=0, \end{aligned}$$

(26)

where a, b are arbitrary constants. It is well known that the ZK equation [58] governs the behavior of weakly nonlinear ion-acoustic waves in a plasma comprising cold ions and hot isothermal electrons in the presence of a uniform magnetic field [60]. Moreover, the ZK equation cannot be integrated by using the inverse scattering transform method, and the solitary wave solutions of the ZK equation are inelastic.

To seek exact solutions of Eq. (26), substituting (23) into Eq. (26), integrating once with respect to $\xi$, and setting the integration constant equal to zero, we have

$$\begin{aligned} -cu+\frac{a\alpha u^2}{2}+b(\alpha ^3+\alpha \beta ^2)u_{\xi \xi }=0. \end{aligned}$$

(27)

Balancing $u^2$ with $u_{\xi \xi }$ gives $N+2=2N$, which implies $N=2$. So that we can take the ansatz

$$\begin{aligned} u(\xi )=a_0+a_1h(\xi )+a_2h^2(\xi ), \end{aligned}$$

(28)

where $a_0, a_1, a_2$ are constants to be determined and $h(\xi )$ is a solution of Eq. (3) presented in Sect. 2. Substituting (28) into (27), collecting the coefficients of $h^{i}(\xi )$, and setting them to zero, we obtain a system of algebraic equations:

$$\begin{aligned} \left\{ \begin{array}{l} -ca_0+\frac{1}{2}a\alpha a_0^2=0,\\ \frac{1}{2}a\alpha a_2^2+6b\alpha (\alpha ^3+\alpha \beta ^2)a_2c_4=0,\\ a\alpha a_1a_2+b\alpha (\alpha ^3+\alpha \beta ^2)(2a_1c_4+5a_2c_3)=0,\\ -ca_1+a\alpha a_0a_1+b\alpha (\alpha ^3+\alpha \beta ^2)a_1c_2=0,\\ -ca_2+(1/2)\frac{1}{2}a\alpha (2a_0a_2+a_1^2)\\ \quad +\,b\alpha (\alpha ^3+\alpha \beta ^2)(\frac{3}{2}a_1c_3+4a_2c_2)=0. \end{array} \right. \end{aligned}$$

Solving this resulting system of algebraic equations by the use of Maple, we obtain

$$\begin{aligned} c= & {} \alpha ^4bc_2+\alpha ^2b\beta ^2c_2, a_0 = 0,\nonumber \\&a_1 = -\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}, a_2 = 0, c_4 = 0, \end{aligned}$$

(29)

$$\begin{aligned} c= & {} -(\alpha ^4bc_2+\alpha ^2b\beta ^2c_2), a_0=- \frac{2b\alpha c_2(\alpha ^2+\beta ^2)}{a},\nonumber \\&a_1 = -\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}, a_2 = 0, c_4 = 0, \end{aligned}$$

(30)

$$\begin{aligned} c= & {} 4\alpha ^2bc_2(\alpha ^2+\beta ^2), a_0 = 0, a_1 = 0,\nonumber \\&a_2= -\frac{12\alpha bc_4(\alpha ^2+\beta ^2)}{a}, c_3 = 0, \end{aligned}$$

(31)

$$\begin{aligned} c= & {} -4\alpha ^2bc_2(\alpha ^2+\beta ^2), a_0=-\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a},\nonumber \\&a_1=0, a_2= -\frac{12\alpha bc_4(\alpha ^2+\beta ^2)}{a}, c_3 = 0, \end{aligned}$$

(32)

$$\begin{aligned} \displaystyle c= & {} \frac{1}{4}\frac{\alpha ^2bc_3^2(\alpha ^2+\beta ^2)}{c_4},a_0=0, \nonumber \\&a_1=-\frac{6\alpha b c_3(\alpha ^2+\beta ^2)}{a},\nonumber \\&a_2= -\frac{12\alpha bc_4(\alpha ^2+\beta ^2)}{a}, c_2 = \frac{1}{4}\frac{c_3^2}{c_4}, \end{aligned}$$

(33)

$$\begin{aligned} \displaystyle c= & {} -\frac{1}{4}\frac{\alpha ^2bc_3^2(\alpha ^2+\beta ^2)}{c_4},a_0=-\frac{1}{2}\frac{\alpha b c_3^2(\alpha ^2+\beta ^2)}{ac_4},\nonumber \\&a_1=-\frac{6\alpha b c_3(\alpha ^2+\beta ^2)}{a},\nonumber \\&a_2= -\frac{12\alpha bc_4(\alpha ^2+\beta ^2)}{a}, c_2 = \frac{1}{4}\frac{c_3^2}{c_4}. \end{aligned}$$

(34)

Substituting (23) and Eq. (29) with $h(\xi )$ in Sect. 2 into Eq. (28) gives the solitons solutions, periodic solutions of Eq. (27) as follows:

• (1) For $c_2>0$, then

  $$\begin{aligned}&\displaystyle u_1(x,y,t)=\frac{3\alpha b(\alpha ^2+\beta ^2)c_2sech ^2(\frac{\sqrt{c_2}}{2}\xi )}{a},\\&u_2(x,y,t)=-\frac{3\alpha b(\alpha ^2+\beta ^2)c_2csch ^2(\frac{\sqrt{c_2}}{2}\xi )}{a},\\&\displaystyle u_3(x,y,t)=-\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}\frac{4c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }}{(\mathrm{e}^{\epsilon \sqrt{c_2}\xi }-c_3)^2}. \end{aligned}$$

• (2) For $c_2>0, \varDelta >0$, we have

  $$\begin{aligned}&\displaystyle u_4(x,y,t)\\&\quad =-\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}\frac{2c_2 sech (\sqrt{c_2}\xi )}{\epsilon \sqrt{\varDelta }-c_3 sech (\sqrt{c_2}\xi )}. \end{aligned}$$

• (3) However, for $c_2<0, \varDelta >0$, then

  $$\begin{aligned}&\displaystyle u_5(x,y,t)\\&\quad =-\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}\frac{2c_2 \sec (\sqrt{-c_2}\xi )}{\epsilon \sqrt{\varDelta }-c_3 \sec (\sqrt{-c_2}\xi )},\\&\displaystyle u_6(x,y,t)\\&\quad =-\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}\frac{2c_2 \csc (\sqrt{-c_2}\xi )}{\epsilon \sqrt{\varDelta }-c_3 \csc (\sqrt{-c_2}\xi )}. \end{aligned}$$

• (4) For $c_2>0, c_3=0$, then

  $$\begin{aligned} u_7(x,y,t)=\mp 4\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }, \end{aligned}$$

  where $\xi =\alpha x+\beta y-(\alpha ^4bc_2+\alpha ^2b\beta ^2c_2)t$.

Substituting (23) and Eq. (30) with $h(\xi )$ in Sect. 2 into Eq. (28), we obtain the solutions of Eq. (27)

• (1) For $c_2>0$

  $$\begin{aligned} \displaystyle u_1(x,y,t)= & {} -\frac{2b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&+\frac{3\alpha b(\alpha ^2+\beta ^2)c_2sech ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }{a},\\ \displaystyle u_2(x,y,t)= & {} -\frac{2b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&-\frac{3\alpha b(\alpha ^2+\beta ^2)c_2csch ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }{a},\\ \displaystyle u_3(x,y,t)= & {} -\frac{2b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&-\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}\frac{4c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }}{(\mathrm{e}^{\epsilon \sqrt{c_2}\xi }-c_3)^2}. \end{aligned}$$

• (2) For $c_2>0, \varDelta >0$, then

  $$\begin{aligned}&\displaystyle u_4(x,y,t)\\&\quad =-\frac{2b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&\qquad -\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}\frac{2c_2 sech (\sqrt{c_2}\xi )}{\epsilon \sqrt{\varDelta }-c_3 sech (\sqrt{c_2}\xi )}. \end{aligned}$$

• (3) For $c_2<0, \varDelta >0$,

  $$\begin{aligned}&\displaystyle u_5(x,y,t)\\&\quad =-\frac{2b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&\qquad -\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}\frac{2c_2 \sec (\sqrt{-c_2}\xi )}{\epsilon \sqrt{\varDelta }-c_3 \sec (\sqrt{-c_2}\xi )},\\&u_6(x,y,t)\\&\quad =-\frac{2b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&\qquad -\frac{3\alpha bc_3(\alpha ^2+\beta ^2)}{a}\frac{2c_2 \csc (\sqrt{-c_2}\xi )}{\epsilon \sqrt{\varDelta }-c_3 \csc (\sqrt{-c_2}\xi )}. \end{aligned}$$

• (4) For $c_2>0, c_3=0$,

  $$\begin{aligned} \displaystyle u_7(x,y,t)= & {} -\frac{2b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&\mp \frac{12\alpha bc_3(\alpha ^2+\beta ^2)}{a} c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }, \end{aligned}$$

  where $\xi =\alpha x+\beta y+(\alpha ^4bc_2+\alpha ^2b\beta ^2c_2)t$.

Putting (23) and Eq. (31) with $h(\xi )$ in Sect. 2 into Eq. (28), we get the solutions of Eq. (27)

• (1) For $c_2>0$, then

  $$\begin{aligned} \displaystyle u_1(x,y,t)= & {} -\frac{48\alpha b c_4(\alpha ^2+\beta ^2)c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }}{a(\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }-4c_2c_4)^2}. \end{aligned}$$

• (2) For $\varDelta >0$, if $c_2>0$ then

  $$\begin{aligned} \displaystyle u_2(x,y,t)=\frac{12\alpha b(\alpha ^2+\beta ^2)c_2 sech ^2(\sqrt{c_2}\xi )}{a}, \end{aligned}$$

  else if $c_2<0$ then

  $$\begin{aligned} \displaystyle u_3(x,y,t)= & {} \frac{12\alpha b(\alpha ^2+\beta ^2)c_2 \sec ^2(\sqrt{-c_2}\xi )}{a},\\ \displaystyle u_4(x,y,t)= & {} \frac{12\alpha b(\alpha ^2+\beta ^2)c_2 \csc ^2(\sqrt{-c_2}\xi )}{a}. \end{aligned}$$

• (3) For $c_2>0, \varDelta <0$, we have

  $$\begin{aligned} \displaystyle u_5(x,y,t)=-\frac{12\alpha b(\alpha ^2+\beta ^2)c_2 csch ^2(\sqrt{c_2}\xi )}{a}. \end{aligned}$$

• (4) For $c_2>0, c_4>0$, then

  $$\begin{aligned} \displaystyle u_6(x,y,t)= & {} -\frac{3\alpha b c_2(\alpha ^2+\beta ^2)sech ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) }{a \tanh ^2(\frac{\sqrt{c_2}}{2}\xi )},\\ \displaystyle u_7(x,y,t)= & {} -\frac{3\alpha b c_2(\alpha ^2+\beta ^2)csch ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) }{a \coth ^2(\frac{\sqrt{c_2}}{2}\xi )}. \end{aligned}$$

• (4) For $c_2<0, c_4>0$, then

  $$\begin{aligned} \displaystyle u_8(x,y,t)= & {} \frac{3\alpha b c_2(\alpha ^2+\beta ^2)\sec ^4\left( \frac{\sqrt{-c_2}}{2}\xi \right) }{a tan^2(\frac{\sqrt{-c_2}}{2}\xi )},\\ \displaystyle u_9(x,y,t)= & {} \frac{3\alpha b c_2(\alpha ^2+\beta ^2)\csc ^4\left( \frac{\sqrt{-c_2}}{2}\xi \right) }{a \cot ^2(\frac{\sqrt{-c_2}}{2}\xi )}, \end{aligned}$$

• (5) For $c_2>0, c_3=0$, then

  $$\begin{aligned}&u_{10}(x,y,t)\\&\quad =-\frac{12\alpha b c_4(\alpha ^2+\beta ^2)}{a}\left( \frac{4c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }}{1-4c_2c_4\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }}\right) ^2, \end{aligned}$$

  where $\xi =\alpha x+\beta y-4\alpha ^2bc_2(\alpha ^2+\beta ^2)t$.

Plugging (23) and Eq. (32) with $h(\xi )$ in Sect. 2 into Eq. (28), we get the solutions of Eq. (27)

• (1) For $c_2>0$, we have

  $$\begin{aligned}&\displaystyle u_1(x,y,t)\\&\quad =-\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&\qquad -\frac{12\alpha b c_4(\alpha ^2+\beta ^2)}{a}\left( \frac{4c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }}{\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }-4c_2c_4}\right) ^2. \end{aligned}$$

• (2) For $\varDelta >0$, if $c_2>0$, then

  $$\begin{aligned} \displaystyle u_2(x,y,t)= & {} -\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&+\frac{12\alpha b(\alpha ^2+\beta ^2)c_2 sech ^2(\sqrt{c_2}\xi )}{a}, \end{aligned}$$

  else if $c_2<0$ then we obtain

  $$\begin{aligned} u_3(x,y,t)= & {} -\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&+\frac{12\alpha b(\alpha ^2+\beta ^2)c_2 \sec ^2(\sqrt{-c_2}\xi )}{a}. \end{aligned}$$

• (3) For $c_2>0, \varDelta <0$,

  $$\begin{aligned} u_4(x,y,t)= & {} -\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&-\frac{12\alpha b(\alpha ^2+\beta ^2)c_2 csch ^2(\sqrt{c_2}\xi )}{a}, \end{aligned}$$

  However, for $c_2<0, \varDelta >0$, we get

  $$\begin{aligned} u_5(x,y,t)= & {} -\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&+\frac{12\alpha b(\alpha ^2+\beta ^2)c_2 \csc ^2(\sqrt{-c_2}\xi )}{a}. \end{aligned}$$

• (4) For $c_4>0$, when $c_2>0$ then

  $$\begin{aligned} \displaystyle u_6(x,y,t)= & {} -\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&-\frac{3\alpha b c_2(\alpha ^2+\beta ^2)sech ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) }{a \tanh ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) },\\ \displaystyle u_7(x,y,t)= & {} -\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&+\frac{3\alpha b c_2(\alpha ^2+\beta ^2)\sec ^4\left( \frac{\sqrt{-c_2}}{2}\xi \right) }{a \tan ^2\left( \frac{\sqrt{-c_2}}{2}\xi \right) }, \\ \end{aligned}$$

  but when $c_2<0$, the solutions are

  $$\begin{aligned} \displaystyle u_8(x,y,t)= & {} -\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&-\frac{3\alpha b c_2(\alpha ^2+\beta ^2)csch ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) }{a \coth ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) },\\ \displaystyle u_9(x,y,t)= & {} -\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&+\frac{3\alpha b c_2(\alpha ^2+\beta ^2)\csc ^4\left( \frac{\sqrt{-c_2}}{2}\xi \right) }{a \cot ^2\left( \frac{\sqrt{-c_2}}{2}\xi \right) }. \end{aligned}$$

• (4) For $c_2>0, c_3=0$, then

  $$\begin{aligned}&u_{10}(x,y,t)\\&\quad =-\frac{8b\alpha c_2(\alpha ^2+\beta ^2)}{a}\\&\quad -\frac{12\alpha b c_4(\alpha ^2+\beta ^2)}{a}\left( \frac{4c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }}{1-4c_2c_4\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }}\right) ^2, \end{aligned}$$

  where $\xi =\alpha x+\beta y+4\alpha ^2bc_2(\alpha ^2+\beta ^2)t$.

Taking (23) and Eq. (33) with $h(\xi )$ in Sect. 2 into Eq. (28), we get the solutions of Eq. (27). For $c_2>0, \varDelta =0$, we get the following two solutions as

$$\begin{aligned}&\displaystyle u_1(x,y,t)\\&\quad =\frac{6\alpha b c_2(\alpha ^2+\beta ^2)}{a}\left( 1+\epsilon \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) \\&\qquad -\frac{12\alpha b c_4c_2^2(\alpha ^2+\beta ^2)}{ac_3^2}\left( 1+\epsilon \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2,\\&\displaystyle u_2(x,y,t)\\&\quad =\frac{6\alpha b c_2(\alpha ^2+\beta ^2)}{a}\left( 1+\epsilon \coth \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) \\&\qquad -\frac{12\alpha b c_4c_2^2(\alpha ^2+\beta ^2)}{ac_3^2}\left( 1+\epsilon \coth \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2, \end{aligned}$$

where $\xi =\alpha x+\beta y-\frac{\alpha ^2bc_3^2(\alpha ^2+\beta ^2)}{4c_4}t$.

Substituting (23) and Eq. (34) with $h(\xi )$ in Sect. 2 into Eq. (28), we get the solutions of Eq. (27). Similarly, when $c_2>0, \varDelta =0$, we get the solutions as follows

$$\begin{aligned}&\displaystyle u_1(x,y,t)\\&\quad =-\dfrac{\alpha bc_3^2(\alpha ^2+\beta ^2)}{2ac_4}\\&\qquad +\dfrac{6\alpha b c_2(\alpha ^2+\beta ^2)}{a}\left( 1+\epsilon \tanh \left( \dfrac{\sqrt{c_2}}{2}\xi \right) \right) \\&\qquad \displaystyle -\dfrac{12\alpha b c_4c_2^2(\alpha ^2+\beta ^2)}{ac_3^2}\left( 1+\epsilon \tanh \left( \dfrac{\sqrt{c_2}}{2}\xi \right) \right) ^2,\\&\displaystyle u_2(x,y,t)\\&\quad =-\dfrac{\alpha bc_3^2(\alpha ^2+\beta ^2)}{2ac_4}\dfrac{6\alpha b c_2(\alpha ^2+\beta ^2)}{a}\\&\qquad \times \left( 1+\epsilon \coth \left( \dfrac{\sqrt{c_2}}{2}\xi \right) \right) \\&\qquad \displaystyle -\dfrac{12\alpha b c_4c_2^2(\alpha ^2+\beta ^2)}{ac_3^2}\left( 1+\epsilon \coth \left( \dfrac{\sqrt{c_2}}{2}\xi \right) \right) ^2, \end{aligned}$$

where $\xi =\alpha x+\beta y+\frac{\alpha ^2bc_3^2(\alpha ^2+\beta ^2)}{4c_4}t$.

5 The generalized ZK equation (gZK)

The generalized ZK equation will be studied in this section, which is given as

$$\begin{aligned} u_t+au^nu_x+b(u_{xx}+u_{yy})_x=0,\ (n\ge 1). \end{aligned}$$

(35)

Substituting (23) into Eq. (35), integrating once with respect to $\xi$, and setting the integration constant equal to zero, we have:

$$\begin{aligned} -cu+\frac{a\alpha u^{n+1}}{n+1}+b(\alpha ^3+\alpha \beta ^2)u_{\xi \xi }=0. \end{aligned}$$

(36)

Balancing $u^{n+1}$ with $u_{\xi \xi }$, we found $(n+1)N=N+2$, which implies $N=\frac{2}{n}$. In order to get a closed-form solution, N must be an integer, and then, we have to make use of the transformation $u(\xi )=v^{\frac{1}{n}}(\xi )$. Thus, Eq. (36) is transformed into

$$\begin{aligned}&-cn^2(n+1)v^2+an^2\alpha v^3+(\alpha ^3+\alpha \beta ^2)vv_{\xi \xi }\nonumber \\&\quad -\,(\alpha ^3+\alpha \beta ^2)(n^2-1)(v_{\xi })^2=0. \end{aligned}$$

(37)

Balancing $v^3$ and $vv_{\xi \xi }$ in this equation, we get $3N=2N+2$, which implies $N=2$. So that we can take the ansatz

$$\begin{aligned} v(\xi )=a_0+a_1h(\xi )+a_2h^2(\xi ), \end{aligned}$$

(38)

where $a_0, a_1, a_2$ are constants to be determined and $h(\xi )$ is a solution of Eq. (3) in Sect. 2. Substituting (38) into (37), collecting the coefficients of $h(\xi )$, and setting them to zero, we obtain a system of algebraic equations, and proceeding as before, we find the following set of solutions:

$$\begin{aligned}&\displaystyle c = \frac{\alpha b c_2(\alpha ^2+\beta ^2)}{n^2},\nonumber \\&\quad a_0= 0,\nonumber \\&\quad a_1= -\frac{bc_3(\alpha ^2n^2 +\beta ^2n^2+3\alpha ^2n +3\beta ^2n+2\alpha ^2+2\beta ^2)}{2an^2},\nonumber \\&\quad a_2 = 0, c_4 = 0, \end{aligned}$$

(39)

$$\begin{aligned}&\displaystyle c = \frac{4\alpha b c_2(\alpha ^2+\beta ^2)}{n^2},a_0= 0,\nonumber \\&\quad a_1=0,\nonumber \\&\quad a_2=-\frac{2bc_4(\alpha ^2n^2 +\beta ^2n^2+3\alpha ^2n +3\beta ^2n+2\alpha ^2 +2\beta ^2)}{an^2},\nonumber \\&\quad c_3 = 0, \end{aligned}$$

(40)

$$\begin{aligned}&\displaystyle c = \frac{\alpha b c_3^2(\alpha ^2+\beta ^2)}{4c_4n^2}, a_0= 0,\nonumber \\&\quad a_1= -\frac{bc_3(\alpha ^2n^2 +\beta ^2n^2+3\alpha ^2n +3\beta ^2n+2\alpha ^2 +2\beta ^2)}{an^2}, \nonumber \\&\quad \displaystyle a_2 = -\frac{2bc_4(\alpha ^2n^2 +\beta ^2n^2+3\alpha ^2n+3\beta ^2n +2\alpha ^2+2\beta ^2)}{an^2},\nonumber \\&\quad c_2=\frac{c_3^2}{4c_4}. \end{aligned}$$

(41)

Substituting (23) and Eq. (39) with $h(\xi )$ in Sect. 2 into Eq. (38) and recalling that $u(\xi )=v^{\frac{1}{n}}(\xi )$, we get the solutions of Eq. (37) as follows:

• (1) When $c_2>0$, then

  $$\begin{aligned} \displaystyle&u_1(x,y,t)=\left\{ \frac{bc_2(\alpha ^2n^2 +\beta ^2n^2+3\alpha ^2n +3\beta ^2n+2\alpha ^2 +2\beta ^2)sech ^2(\frac{\sqrt{c_2}}{2}\xi )}{2an^2}\right\} ^{\frac{1}{n}},\\&\displaystyle u_2(x,y,t)=\left\{ -\frac{bc_2(\alpha ^2n^2 +\beta ^2n^2+3\alpha ^2n+3\beta ^2n +2\alpha ^2+2\beta ^2)csch ^2(\frac{\sqrt{c_2}}{2}\xi )}{2an^2}\right\} ^{\frac{1}{n}},\\&\displaystyle u_3(x,y,t)=\left\{ -\frac{4bc_2c_3(\alpha ^2n^2 +\beta ^2n^2+3\alpha ^2n +3\beta ^2n+2\alpha ^2 +2\beta ^2)\mathrm{e}^{\epsilon \sqrt{c_2}\xi }}{2an^2(\mathrm{e}^{\epsilon \sqrt{c_2}\xi }-c_3)^2}\right\} ^{\frac{1}{n}}\\ \end{aligned}$$

• (2) For $\varDelta >0$, if $c_2>0$, then

  $$\begin{aligned} \displaystyle&u_4(x,y,t)=\left\{ -\frac{bc_2c_3(\alpha ^2n^2 +\beta ^2n^2 +3\alpha ^2n+3\beta ^2n +2\alpha ^2+2\beta ^2)sech (\sqrt{c_2}\xi )}{an^2(\epsilon \sqrt{\varDelta }-c_3 sech (\sqrt{c_2}\xi ))}\right\} ^{\frac{1}{n}}, \end{aligned}$$

  However when $c_2<0$, we get

  $$\begin{aligned} \displaystyle&u_5(x,y,t)=\left\{ -\frac{bc_2c_3(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\sec (\sqrt{-c_2}\xi )}{an^2(\epsilon \sqrt{\varDelta }-c_3 \sec (\sqrt{-c_2}\xi ))}\right\} ^{\frac{1}{n}},\\ \displaystyle&u_6(x,y,t)=\left\{ -\frac{bc_2c_3(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\csc (\sqrt{-c_2}\xi )}{an^2(\epsilon \sqrt{\varDelta }-c_3 \csc (\sqrt{-c_2}\xi ))}\right\} ^{\frac{1}{n}}. \end{aligned}$$

  where $\xi =\alpha x+\beta y-\dfrac{4\alpha b c_2(\alpha ^2+\beta ^2)}{n^2}t$.

Putting (23) and Eq. (40) with $h(\xi )$ in Sect. 2 into Eq. (38), we get the solutions of Eq. (37) as follows:

• (1) For $c_2>0$, the solution is

  $$\begin{aligned}&\displaystyle u_1(x,y,t)=\left\{ -\frac{32bc_2^2c_4(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }}{an^2(\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }-4c_2c_4)^2}\right\} ^{\frac{1}{n}},\\&\displaystyle u_2(x,y,t)=\left\{ -\frac{32bc_2c_4(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }}{an^2(1-4c_2c_4\mathrm{e}^{2\epsilon \sqrt{c_2}\xi })^2}\right\} ^{\frac{1}{n}}. \end{aligned}$$

• (2) When $c_2>0, \varDelta <0$

  $$\begin{aligned} \displaystyle&u_3(x,y,t)=\left\{ -\frac{b(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)csch ^2(\sqrt{c_2}\xi )}{an^2}\right\} ^{\frac{1}{n}}. \end{aligned}$$

• (3) However, when $c_2<0, \varDelta >0$

  $$\begin{aligned} \displaystyle&u_4(x,y,t)=\left\{ \frac{b(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\sec ^2(\sqrt{-c_2}\xi )}{an^2}\right\} ^{\frac{1}{n}},\\ \displaystyle&u_5(x,y,t)=\left\{ \frac{b(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\csc ^2(\sqrt{-c_2}\xi )}{an^2}\right\} ^{\frac{1}{n}}. \end{aligned}$$

• (4) For $c_2>0$ and $c_4>0$

  $$\begin{aligned} \displaystyle&u_6(x,y,t)=\left\{ -\frac{b(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)sech ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) }{2an^2\tanh ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }\right\} ^{\frac{1}{n}},\\ \displaystyle&u_7(x,y,t)=\left\{ -\frac{b(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)csch ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) }{2an^2\coth ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }\right\} ^{\frac{1}{n}}. \end{aligned}$$

• (5) But when $c_2<0$ and $c_4>0$

  $$\begin{aligned}&\displaystyle u_8(x,y,t)=\left\{ \frac{b(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\sec ^4\left( \frac{\sqrt{-c_2}}{2}\xi \right) }{2an^2\tan ^2\left( \frac{\sqrt{-c_2}}{2}\xi \right) }\right\} ^{\frac{1}{n}},\\&\displaystyle u_9(x,y,t)=\left\{ \frac{b(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\csc ^4\left( \frac{\sqrt{-c_2}}{2}\xi \right) }{2an^2\cot ^2\left( \frac{\sqrt{-c_2}}{2}\xi \right) }\right\} ^{\frac{1}{n}}, \end{aligned}$$

  where $\xi =\alpha x+\beta y+\dfrac{4\alpha b c_2(\alpha ^2+\beta ^2)}{n^2}t$.

Plugging (23) and Eq. (41) with $h(\xi )$ in Sect. 2 into Eq. (38), we get the solutions of Eq. (37) as follows:

• (1) When $c_2>0$, then

  $$\begin{aligned} \displaystyle u_1(x,y,t)= & {} \left\{ \frac{bc_2c_3^2(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)sech ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }{an^2\left( c_3^2-c_2c_4\left( 1+\epsilon \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2\right) }\right. \\&\displaystyle -\left. \frac{2bc_2^2c_3^2c_4(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)sech ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) }{an^2\left( c_3^2-c_2c_4\left( 1+\epsilon \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2\right) ^2}\right\} ^{\frac{1}{n}},\\ \displaystyle u_2(x,y,t)= & {} \left\{ \frac{bc_2c_3^2(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)csch ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }{an^2\left( c_3^2-c_2c_4\left( 1+\epsilon \coth \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2\right) }\right. \\&\displaystyle -\left. \frac{2bc_2^2c_3^2c_4(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)csch ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) }{an^2\left( c_3^2-c_2c_4\left( 1+\epsilon \coth \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2\right) ^2} \right\} ^{\frac{1}{n}},\\ \displaystyle u_3(x,y,t)= & {} \left\{ -\frac{bc_3(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)}{an^2}\right. \\&\displaystyle -\left. \frac{32bc_2^2c_4(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }}{an^2(\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }-4c_2c_4)^2}\right\} ^{\frac{1}{n}}. \end{aligned}$$

• (2) For $c_2>0, \varDelta =0$

  $$\begin{aligned} \displaystyle u_4(x,y,t)= & {} \left\{ \frac{bc_2(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\left( 1+\epsilon \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) }{an^2}\right. \\&\displaystyle -\left. \frac{2bc_2^2c_4(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\left( 1+\epsilon \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2}{an^2c_3^2}\right\} ^{\frac{1}{n}},\\ \displaystyle u_5(x,y,t)= & {} \left\{ \frac{bc_2(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\left( 1+\epsilon \coth \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) }{an^2}\right. \\&\displaystyle -\left. \frac{2bc_2^2c_4(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)\left( 1+\epsilon \coth \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2}{an^2c_3^2}\right\} ^{\frac{1}{n}}. \end{aligned}$$

• (3) For $c_2>0, c_4>0$

  $$\begin{aligned} \displaystyle u_6(x,y,t)= & {} \left\{ \frac{bc_2c_3(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)}{an^2\left( c_3+2\epsilon \sqrt{c_2c_4} \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) }\right. \\&\displaystyle -\left. \frac{2bc_2^2c_4(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)sech ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) }{an^2\left( c_3+2\epsilon \sqrt{c_2c_4} \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2}\right\} ^{\frac{1}{n}},\\ \displaystyle u_7(x,y,t)= & {} \left\{ -\frac{bc_2c_3(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)csch ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }{an^2\left( c_3+2\epsilon \sqrt{c_2c_4} \coth \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) }\right. \\&\displaystyle -\left. \frac{2bc_2^2c_4(\alpha ^2n^2+\beta ^2n^2+3\alpha ^2n+3\beta ^2n+2\alpha ^2+2\beta ^2)csch ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) }{an^2\left( c_3+2\epsilon \sqrt{c_2c_4} \tanh \left( \frac{\sqrt{c_2}}{2}\xi \right) \right) ^2}\right\} ^{\frac{1}{n}}, \end{aligned}$$

where $\xi =\alpha x+\beta y+\frac{\alpha bc_3^2(\alpha ^2+\beta ^2)}{4c_4n^2}t$.

6 The modified ZK equation (mZK)

Schame [59] derived the (2+1)-dimensional equations as

$$\begin{aligned}&u_t+u^{\frac{1}{2}}u_x+b(u_{xx}+u_{yy})_x=0, \end{aligned}$$

(42)

$$\begin{aligned}&u_t+(1+bu^{\frac{1}{2}})u_x+\frac{1}{2}u_{xxx}=0, \end{aligned}$$

(43)

which describe ion-acoustic waves in a cold-ion plasma, and where the electrons do not behave isothermally during their passage of the wave. Monro and Parkes [60, 61] showed that if the electrons are non-isothermal, the governing equation of the ZK equation is a modified form named the mZK equation. They also showed that with an appropriate modified form of the electron number density proposed by Schamel [59], a reductive procedure leads to a modified form of the ZK equation, that is as follows:

$$\begin{aligned} 16u_t+20u^{\frac{1}{2}}u_x+(u_{xx}+u_{yy})_x=0. \end{aligned}$$

(44)

Substituting Eq. (23) into (44), then integrating the result equation, and neglecting the integration constant, we obtain

$$\begin{aligned} -16cu+20\alpha u^{\frac{3}{2}}+(\alpha ^3+\alpha \beta ^2)u_{\xi \xi }=0. \end{aligned}$$

(45)

In order to achieve our goal, we must use the transformation $u(\xi )=v(\xi )^2$, which converts Eq. (45) into

$$\begin{aligned} -16cv^2+20\alpha v^3+2(\alpha ^3+\alpha \beta ^2)(vv_{\xi \xi }+v_{\xi }^2)=0. \end{aligned}$$

(46)

Balancing $v^3$ with $vv_{\xi \xi }$ gives $N=2$. The method of simplest equation assumes that the solutions be written in the following form

$$\begin{aligned} v(\xi )=a_0+a_1h(\xi )+a_2h^2(\xi ). \end{aligned}$$

(47)

where $a_0, a_1, a_2$ are constants to be determined and $h(\xi )$ is a solution of Eq. (3) in Sect. 2. Substituting (47) into (46), collecting the coefficients of $h(\xi )$, and setting them to zero, we obtain a system of algebraic equations, and proceeding as before, we find the following set of solutions:

$$\begin{aligned}&a_0=0,a_1=\gamma ,c_2=\frac{4c}{\alpha (\beta ^2+\alpha ^2)},\nonumber \\&\quad c_3=-\frac{2\gamma }{\beta ^2+\alpha ^2},c_4=-\frac{a_2}{\beta ^2+\alpha ^2}, \end{aligned}$$

(48)

$$\begin{aligned}&a_0=0, a_1=0, c_2 =\frac{c}{\alpha (\beta ^2+\alpha ^2)},\nonumber \\&\quad c_3 = 0, c_4 = -\frac{a_2}{\beta ^2+\alpha ^2}, \end{aligned}$$

(49)

$$\begin{aligned}&a_0=0, a_2=0, c_2=\frac{4c}{\alpha (\beta ^2+\alpha ^2)},\nonumber \\&\quad c_3=-\frac{4a_1}{\beta ^2+\alpha ^2}, c_4=0, \end{aligned}$$

(50)

where $\gamma =2RootOf(\_Z^2\alpha +ca_2)$.

Substituting (23) and Eq. (48) with $h(\xi )$ in Sect. 2 into Eq. (47) and recalling that $u(\xi )=v^2(\xi )$, we get the solutions of Eq. (45) when $c_2>0$

$$\begin{aligned}&\displaystyle u_1(x,y,t)\\&\quad =\frac{36\gamma ^4 sech ^4\left( \sqrt{\frac{c}{\alpha (\beta ^2+\alpha ^2)}}\xi \right) }{a_2^2\left( 4-\left( 1+\epsilon \tanh \left( \sqrt{\frac{c}{\alpha (\beta ^2+\alpha ^2)}}\xi \right) \right) ^2\right) ^2},\\&\displaystyle u_2(x,y,t)\\&\quad =\frac{4\gamma ^2 csch ^4(\sqrt{\frac{c}{\alpha (\beta ^2+\alpha ^2)}}\xi )}{a_2^2\left( 4-\left( 1+\epsilon \coth \left( \sqrt{\frac{c}{\alpha (\beta ^2+\alpha ^2)}}\xi \right) \right) ^2\right) ^2}, \end{aligned}$$

$$\begin{aligned}&\displaystyle u_3(x,y,t)\\&\quad =\left\{ \frac{16\gamma ^2 c(\beta ^2+\alpha ^2)\mathrm{e}^{\epsilon \sqrt{\frac{4c}{\alpha (\beta ^2+\alpha ^2)}}\xi }}{\alpha \left( (\beta ^2+\alpha ^2)\mathrm{e}^{\epsilon \sqrt{\frac{4c}{\alpha (\beta ^2+\alpha ^2)}}\xi }+2\gamma \right) ^2+4ca_2}\right. \\&\qquad \displaystyle +\left. a_2\left( \frac{16\gamma c(\beta ^2+\alpha ^2)\mathrm{e}^{\epsilon \sqrt{\frac{4c}{\alpha (\beta ^2+\alpha ^2)}}\xi }}{\alpha \left( (\beta ^2+\alpha ^2)\mathrm{e}^{\epsilon \sqrt{\frac{4c}{\alpha (\beta ^2+\alpha ^2)}}\xi }+2\gamma \right) ^2+4ca_2}\right) ^2\right\} ^2. \end{aligned}$$

Substituting (23) and Eq. (49) with $h(\xi )$ in Sect. 2 into Eq. (47) gives rise to the solutions of Eq. (45) as follows:

• (1) For $c_2>0$

  $$\begin{aligned}&\displaystyle u_1(x,y,t)\\&\quad =256a_2^2\left( \frac{c(\alpha ^2+\beta ^2)\mathrm{e}^{\epsilon \sqrt{\frac{c}{\alpha (\beta ^2+\alpha ^2)}}\xi }}{\alpha (\alpha ^2+\beta ^2)^2\mathrm{e}^{2\epsilon \sqrt{\frac{c}{\alpha (\beta ^2+\alpha ^2)}}\xi }+4ca_2}\right) ^4. \end{aligned}$$

• (2) When $c_2>0$ and $\varDelta >0$, so

  $$\begin{aligned}&\displaystyle u_2(x,y,t)=\frac{c^2sech ^4\left( \sqrt{\frac{c}{\alpha (\alpha ^2+\beta ^2)}}\xi \right) }{\alpha ^2}. \end{aligned}$$

• (3) However, while $c_2>0$ and $\varDelta >0$, thus

  $$\begin{aligned}&\displaystyle u_3(x,y,t)=\frac{c^2csch ^4\left( \sqrt{\frac{c}{\alpha (\alpha ^2+\beta ^2)}}\xi \right) }{\alpha ^2}. \end{aligned}$$

• (4) For $c_2>0$ and $c_4>0$, we have

  $$\begin{aligned}&\displaystyle u_4(x,y,t)\\&\quad =\frac{c^2sech ^8\left( \sqrt{\frac{c}{4\alpha (\alpha ^2+\beta ^2)}}\xi \right) }{16\alpha ^2\tanh ^4\left( \sqrt{\frac{c}{4\alpha (\alpha ^2+\beta ^2)}}\xi \right) },\\&\displaystyle u_5(x,y,t)\\&\quad =\frac{c^2csch ^8\left( \sqrt{\frac{c}{2\alpha (\alpha ^2+\beta ^2)}}\xi \right) }{16\alpha ^2\coth ^4\left( \sqrt{\frac{c}{2\alpha (\alpha ^2+\beta ^2)}}\xi \right) }. \end{aligned}$$

• (5) Take $c_2>0$ and $c_3=0$, so that

  $$\begin{aligned} \displaystyle u_6(x,y,t)=256a_2^2\frac{c^4\mathrm{e}^{4\epsilon \sqrt{\frac{c}{\alpha (\beta ^2+\alpha ^2)}}\xi }}{\left( \alpha (\alpha ^2 +\beta ^2)^2+4c\mathrm{e}^{2\epsilon \sqrt{2}}\xi \right) ^4}. \end{aligned}$$

Plugging (23) and Eq. (50) with $h(\xi )$ in Sect. 2 into Eq. (47), we obtain the solutions of Eq. (45)

• (1) For $c_2>0$, we get

  $$\begin{aligned}&\displaystyle u_1(x,y,t)=\frac{c^2sech ^4\left( \sqrt{\frac{c}{\alpha (\alpha ^2+\beta ^2)}}\xi \right) }{16\alpha ^2},\\&\displaystyle u_2(x,y,t)=\frac{c^2csch ^4\left( \sqrt{\frac{c}{\alpha (\alpha ^2+\beta ^2)}}\xi \right) }{16\alpha ^2},\\&\displaystyle u_3(x,y,t)=\frac{16a_1^2c^2(\alpha ^2+\beta ^2)\mathrm{e}^{4\epsilon \sqrt{\frac{c}{\alpha (\alpha ^2+\beta ^2)}}\xi }}{\alpha ^2\left( \mathrm{e}^{2\epsilon \sqrt{\frac{c}{\alpha (\alpha ^2+\beta ^2)}}\xi }+4a_1\right) ^4}. \end{aligned}$$

• (2) For $c_2>0$ and $\varDelta >0$, then

  $$\begin{aligned}&\displaystyle u_3(x,y,t)\\&\quad =\frac{c^2sech ^2\left( \sqrt{\frac{4c}{\alpha (\alpha ^2+\beta ^2)}}\xi \right) }{4\alpha ^2\left( \epsilon +sign(a_1)sech \left( \sqrt{\frac{4c}{\alpha (\alpha ^2+\beta ^2)}}\xi \right) \right) ^2}, \end{aligned}$$

  where $\xi =\alpha x+\beta y-ct$.

7 Generalized form of modified ZK equation (gmZK)

Motivated by the rich treasure of the ZK equation and its modified forms in the nonlinear development of ion-acoustic waves in a magnetized plasma, we will carry this research with the analytic study on the generalized form of modified ZK equation described by

$$\begin{aligned} u_t+au^{\frac{n}{2}}u_x+b(u_{xx}+u_{yy})_x=0, n\ge 1. \end{aligned}$$

(51)

Substituting Eq. (23) into (51), then integrating the result equation, and neglecting the integration constant, we obtain the following ODE

$$\begin{aligned} -cu+\frac{2a\alpha }{n+2}u^{\frac{n+2}{2}}+b(\alpha ^3+\alpha \beta ^2)u_{\xi \xi }=0. \end{aligned}$$

(52)

Now, if we balance $u_{\xi \xi }$ and $u^{\frac{n+2}{2}}$, then $N=\frac{4}{n}$. In order to get a closed-form solution, N must be an integer. Thus, we have to make use of the transformation $u(\xi )=v^{\frac{2}{n}}(\xi )$, which converts Eq. (52) into the ODE as follows:

$$\begin{aligned}&-\,cn^2(n+2)v^2+2an^2\alpha v^3\nonumber \\&\quad +\,2b(\alpha ^3+\alpha \beta ^2)n(n+2)vv_{\xi \xi }\nonumber \\&\quad -\,2b(\alpha ^3+\alpha \beta ^2)(n^2-4)v_{\xi }^2=0. \end{aligned}$$

(53)

Balancing $vv_{\xi \xi }$ with $v^3$, we can get $N=2$. Accordingly, the method of simplest equation assumes that the solution can be described by the expression

$$\begin{aligned} v=a_0+a_1h(\xi )+a_2h^2(\xi ), \end{aligned}$$

(54)

where $a_0, a_1, a_2$ are arbitrary constants to be determined and $h(\xi )$ is a solution of Eq. (3) in Sect. 2.

Substituting (54) into (53) and proceeding as in the previous sections, we find the following sets of solutions:

$$\begin{aligned}&\displaystyle c = \frac{4\alpha bc_2(\alpha ^2+\beta ^2)}{n^2}, a_0 = 0,\nonumber \\&\quad a_1 = -\frac{bc_3(\alpha ^2n^2 +\beta ^2n^2+6\alpha ^2n+6\beta ^2n +8\alpha ^2+8\beta ^2)}{2an^2},\nonumber \\&\quad a_2 = 0,c_4= 0. \end{aligned}$$

(55)

$$\begin{aligned}&\displaystyle c = \frac{16\alpha bc_2(\alpha ^2+\beta ^2)}{n^2}, a_0 = 0,\nonumber \\&\quad a_1 =0,\nonumber \\&\quad a_2=-\frac{2bc_4(\alpha ^2n^2 +\beta ^2n^2+6\alpha ^2n+6\beta ^2n +8\alpha ^2+8\beta ^2)}{an^2},\nonumber \\&\quad c_3= 0. \end{aligned}$$

(56)

$$\begin{aligned}&\displaystyle c = \frac{\alpha bc_3^2(\alpha ^2+\beta ^2)}{c_4n^2}, a_0 = 0,\nonumber \\&\quad a_1=-\frac{bc_3(\alpha ^2n^2 +\beta ^2n^2+6\alpha ^2n+6\beta ^2n +8\alpha ^2+8\beta ^2)}{2an^2},\nonumber \\&\quad \displaystyle a_2=-\frac{2bc_4(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n +6\beta ^2n+8\alpha ^2+8\beta ^2)}{an^2},\nonumber \\&\quad c_2=\frac{c_3^2}{4c_4}. \end{aligned}$$

(57)

Substituting (23) and Eq. (29) with $h(\xi )$ in Sect. 2 into Eq. (54) and recalling that $u(\xi )=v^{\frac{2}{n}}(\xi )$, we obtain the solutions of Eq. (52) as

• (1) While $c_2>0$, we can obtain

  $$\begin{aligned}&\displaystyle u_1(x,y,t)=\left\{ \frac{c_2sech ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) b(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{2an^2}\right\} ^{\frac{2}{n}},\\&\displaystyle u_2(x,y,t)=\left\{ \frac{c_2csch ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) b(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{2an^2}\right\} ^{\frac{2}{n}},\\&\displaystyle u_3(x,y,t)=\left\{ \frac{2c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }bc_3(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{(\mathrm{e}^{\epsilon \sqrt{c_2}\xi }-c_3)^2an^2}\right\} ^{\frac{2}{n}}. \end{aligned}$$

• (2) When $c_2>0$ and $\varDelta >0$, so

  $$\begin{aligned}&\displaystyle u_4(x,y,t)=\left\{ \frac{c_2sech (\sqrt{c_2}\xi )bc_3(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{\left( \epsilon \sqrt{c_3^2}-c_3sech (\sqrt{c_2}\xi )\right) an^2}\right\} ^{\frac{2}{n}},\\ \end{aligned}$$

• (3) For $c_2<0, \varDelta >0$, then

  $$\begin{aligned}&\displaystyle u_5(x,y,t)=\left\{ \frac{c_2\sec (\sqrt{-c_2}\xi )bc_3(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{\left( \epsilon \sqrt{c_3^2}-c_3\sec (\sqrt{-c_2}\xi )\right) an^2}\right\} ^{\frac{2}{n}},\\&\displaystyle u_6(x,y,t)=\left\{ \frac{c_2\csc (\sqrt{-c_2}\xi )bc_3(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{\left( \epsilon \sqrt{c_3^2}-c_3\csc (\sqrt{-c_2}\xi )\right) an^2}\right\} ^{\frac{2}{n}}. \end{aligned}$$

• (4) For $c_2>0$ and $c_3=0$, the solution is

  $$\begin{aligned}&\displaystyle u_7(x,y,t)=\left\{ \frac{2c_2\mathrm{e}^{\epsilon \sqrt{c_2}\xi }bc_3(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{an^2}\right\} ^{\frac{2}{n}},\\ \end{aligned}$$

where $\xi =\alpha x+\beta y-\frac{4\alpha bc_2(\alpha ^2+\beta ^2)}{n^2}t$.

Plugging (23) and Eq. (56) with $h(\xi )$ in Sect. 2 into Eq. (54) gives rise to the solutions of Eq. (52) as follows:

• (1) Take $c_2>0$,

  $$\begin{aligned}&\displaystyle u_1(x,y,t)=\left\{ \frac{32c_2^2\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }bc_4(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{(\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }-4c_2c_4)^2an^2}\right\} ^{\frac{2}{n}}. \end{aligned}$$

• (2) For $c_2>0$ and $\varDelta <0$, then

  $$\begin{aligned}&\displaystyle u_2(x,y,t)=\left\{ \frac{8bc_2csch ^2(\sqrt{c_2}\xi )(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{an^2}\right\} ^{\frac{2}{n}}. \end{aligned}$$

• (3) When $c_2>0$ and $c_4>0$, we obtain

  $$\begin{aligned}&\displaystyle u_3(x,y,t)=\left\{ \frac{bc_2sech ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) (\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{2an^2\tanh ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }\right\} ^{\frac{2}{n}},\\&\displaystyle u_4(x,y,t)=\left\{ \frac{bc_2csch ^4\left( \frac{\sqrt{c_2}}{2}\xi \right) (\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{2an^2\coth ^2\left( \frac{\sqrt{c_2}}{2}\xi \right) }\right\} ^{\frac{2}{n}}. \end{aligned}$$

• (4) However, when $c_2<0$ and $c_4>0$, we obtain the solutions as

  $$\begin{aligned}&\displaystyle u_5(x,y,t)=\left\{ \frac{bc_2\sec ^4\left( \frac{\sqrt{-c_2}}{2}\xi \right) (\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{2an^2\tan ^2\left( \frac{\sqrt{-c_2}}{2}\xi \right) }\right\} ^{\frac{2}{n}},\\ \displaystyle&u_6(x,y,t)=\left\{ \frac{bc_2\csc ^4\left( \frac{\sqrt{-c_2}}{2}\xi \right) (\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{2an^2\cot ^2\left( \frac{\sqrt{-c_2}}{2}\xi \right) }\right\} ^{\frac{2}{n}}. \end{aligned}$$

• (5) For $c_2>0, c_3=0$, the solution is

  $$\begin{aligned} \displaystyle&u_7(x,y,t)=\left\{ \frac{-32bc_4c_2^2\mathrm{e}^{2\epsilon \sqrt{c_2}\xi }(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)}{an^2(1-4c_2c_4\mathrm{e}^{2\epsilon \sqrt{c_2}\xi })}\right\} ^{\frac{2}{n}}, \end{aligned}$$

  where $\xi =\alpha x+\beta y-\frac{16\alpha bc_2(\alpha ^2+\beta ^2)}{n^2}t$.

Plugging (23) and Eq. (57) with $h(\xi )$ in Sect. 2 into Eq. (54), we get the solutions of Eq. (52) as follows:

• (1) When $c_2>0$,

  $$\begin{aligned}&\displaystyle u_1(x,y,t)\\&\quad =\left\{ bc_3^2c_2(\alpha ^2n^2+\beta ^2n^2\right. \\&\qquad +\left. 6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)\right\} ^{\frac{2}{n}}\\&\qquad \displaystyle \times \left\{ \left( \frac{sech ^2\left( \frac{\sqrt{2}}{2}\xi \right) }{an^2\left( c_3^2-c_2c_4\left( 1+\epsilon \tanh \left( \frac{\sqrt{2}}{2}\xi \right) \right) ^2\right) }\right. \right. \\&\qquad \left. \left. -\frac{2c_3c_4sech ^4\left( \frac{\sqrt{2}}{2}\xi \right) }{an^2\left( c_3^2 -c_2c_4\left( 1+\epsilon \tanh \left( \frac{\sqrt{2}}{2}\xi \right) \right) ^2\right) ^2}\right) \right\} ^{\frac{2}{n}},\\&\displaystyle u_2(x,y,t)\nonumber \\&\quad =\left\{ bc_3^2c_2(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n\right. \\&\qquad \left. +\,6\beta ^2n+8\alpha ^2+8\beta ^2)\right\} ^{\frac{2}{n}}\\&\qquad \displaystyle \times \left\{ \left( \frac{csch ^2\left( \frac{\sqrt{2}}{2}\xi \right) }{an^2\left( c_3^2-c_2c_4\left( 1+\epsilon \coth \left( \frac{\sqrt{2}}{2}\xi \right) \right) ^2\right) }\right. \right. \\&\qquad \left. \left. +\frac{2c_3c_4csch ^4\left( \frac{\sqrt{2}}{2}\xi \right) }{an^2\left( c_3^2-c_2c_4\left( 1+\epsilon \coth \left( \frac{\sqrt{2}}{2}\xi \right) \right) ^2\right) }\right) \right\} ^{\frac{2}{n}},\\&\displaystyle u_3(x,y,t)\\&\quad =\left\{ 4bc_2(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n+6\beta ^2n +8\alpha ^2+8\beta ^2)\right. \\&\qquad \displaystyle \times \left. \left( \frac{c_3\mathrm{e}^2(\epsilon \sqrt{2}\xi )}{an^2((\mathrm{e}^{\epsilon \sqrt{2}\xi }-c_3)^2-4c_2c_4)}\right. \right. \\&\qquad \left. \left. +\frac{8c_2c_4\mathrm{e}^{2\epsilon \sqrt{2}\xi }}{an^2((\mathrm{e}^{\epsilon \sqrt{2}\xi }-c_3)^2-4c_2c_4)^2}\right) \right\} ^{\frac{2}{n}},\\ \end{aligned}$$

• (2) For $c_2>0$ and $\varDelta =0$,

  $$\begin{aligned}&\displaystyle u_4(x,y,t)\\&\quad =\left\{ bc_2(\alpha ^2n^2+\beta ^2n^2 +6\alpha ^2n+6\beta ^2n+8\alpha ^2 +8\beta ^2)\right\} ^{\frac{2}{n}}\\&\qquad \displaystyle \times \left\{ \left( \frac{1+\epsilon \tanh \left( \frac{\sqrt{2}}{2}\xi \right) }{an^2} -\frac{2c_2c_4\tanh ^2\left( \frac{\sqrt{2}}{2}\xi \right) }{an^2c_3^2}\right) \right\} ^{\frac{2}{n}},\\&u_5(x,y,t)=\left\{ bc_2(\alpha ^2n^2+\beta ^2n^2\right. \\&\quad \left. +\,6\alpha ^2n+6\beta ^2n+8\alpha ^2+8\beta ^2)\right\} ^{\frac{2}{n}}\\&\quad \displaystyle \times \left\{ \left( \frac{1+\epsilon \coth \left( \frac{\sqrt{2}}{2}\xi \right) }{an^2} -\frac{2c_2c_4\coth ^2\left( \frac{\sqrt{2}}{2}\xi \right) }{an^2c_3^2}\right) \right\} ^{\frac{2}{n}},\\ \end{aligned}$$

• (3) Take $c_2>0, c_4>0$, then

  $$\begin{aligned}&\displaystyle u_6(x,y,t) =\left\{ bc_2(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n\right. \\&\qquad \left. +\,6\beta ^2n+8\alpha ^2+8\beta ^2)\right\} ^{\frac{2}{n}}\\&\qquad \displaystyle \times \left\{ \left( \frac{c_3 sech ^2\left( \frac{\sqrt{2}}{2}\xi \right) }{an^2\left( c_3+2\epsilon \sqrt{c_2c_4\tanh \left( \frac{\sqrt{2}}{2}\xi \right) }\right) }\right. \right. \\&\qquad \left. \left. -\frac{2c_2c_4sech ^4(\frac{\sqrt{2}}{2}\xi )}{an^2\left( c_3 +2\epsilon \sqrt{c_2c_4\tanh \left( \frac{\sqrt{2}}{2}\xi \right) }\right) ^2}\right) \right\} ^{\frac{2}{n}},\\&\displaystyle u_7(x,y,t)\\&\quad =\left\{ bc_2(\alpha ^2n^2+\beta ^2n^2+6\alpha ^2n\right. \\&\qquad \left. +\,6\beta ^2n+8\alpha ^2+8\beta ^2)\right\} ^{\frac{2}{n}}\\&\qquad \displaystyle \times \left\{ \left( \frac{c_3 csch ^2(\frac{\sqrt{2}}{2}\xi )}{an^2\left( c_3+2\epsilon \sqrt{c_2c_4\coth \left( \frac{\sqrt{2}}{2}\xi \right) }\right) }\right. \right. \\&\qquad \left. \left. -\frac{2c_2c_4csch ^4\left( \frac{\sqrt{2}}{2}\xi \right) }{an^2\left( c_3 +2\epsilon \sqrt{c_2c_4\coth \left( \frac{\sqrt{2}}{2}\xi \right) }\right) ^2}\right) \right\} ^{\frac{2}{n}}, \end{aligned}$$

where $\xi =\alpha x+\beta y-\frac{\alpha bc_3^2(\alpha ^2+\beta ^2)}{c_4n^2}t$.

Remark

In fact, some types of exact solutions appear in pairs with other corresponding solutions. For example, $\sec (\xi ), \tan (\xi ), sech (\xi )$, and $\tanh (\xi )$ appear in pairs with the corresponding $\csc (\xi ), \cot (\xi ), csch (\xi )$ and $\coth (\xi ).$

8 Comparisons

We carefully compare the method of simplest equation with the expansion method around integrable ODEs [2], and it is found out that this method of simplest equation is just one particular application of the expansion method, since the proposed method is based on the different simplest equation: Duffing-type equation (3), which can help us find a lot new exact solutions. In the meantime, it is worth pointing out that the proposed method can not only construct more new exact traveling wave solutions to the studied ZK equation according to new exact solutions of Eq. (3) (see [62]), but also be used to derive some new solitons and periodic solutions to the gZK equation, the mZK equation, and the gmZK equation. All the obtained results have clearly demonstrated that the reliability and stability of the used method. Kudryashov determined the value of N in formula (25) by using the pole order of general solution of Eq. (24). In this study, we use the homogeneous balance method to determine the value of N in formula (25).

9 Conclusion

In this paper, we have successfully applied the method of simplest equation, which is Duffing-type equation, to study the Zakharov–Kuznetsov equation, the generalized ZK equation, the modified ZK equation, and a generalized form of the modified ZK equation. Some new solitons and periodic solutions have been formally obtained. These solutions may be helpful to describe features of nonlinear waves in plasma physics and other research fields. Moreover, all the obtained results in this work can clearly illustrate the validity and reliability of the method of simplest equation. Furthermore, this method is valid for a large number of nonlinear PDEs with variable coefficients.

During the study, it is noted that different types of exact solutions of Eq. (3) can help us construct different types of exact traveling wave solutions of nonlinear PDEs. Thus, it is one open problem for us to find more new exact solutions of Eq. (3), and the other open problem is how to find and apply new exact solutions of other Duffing-type equation of higher degree, such as even 10th and even 8th power of Duffing-type ODEs, to constructing new exact solutions of given nonlinear PDEs.