1. Introduction

During the past decades, the exact solutions of nonlinear partial differential equations have been investigated by many authors. Meanwhile, many powerful methods have been proposed by them, such as Backlund transformation method ^[1], multiple exp-function method ^[2], homogeneous balance principle ^[3], tanh-sech method ^[4], $G'/G$-expansion method ^[5,6,7], the first integral method ^[8,9] and so on.

The $G'/G$-expansion method was first presented by Wang ^[5] which can be used to deal with all types of nonlinear evolution equations. The first integral method was first proposed by Feng ^[8] for obtaining the exact solutions of Burgers-KdV equation which is based on the ring theory of commutative algebra. The basic idea of the first integral method is to construct a first integral with polynomial coefficients of an explicit form to an equivalent autonomous planer system by using the division theorem. Both the $G'/G$-expansion method and the first integral method are powerful methods for computing the exact solutions of nonlinear partial differential equations. They are direct, elementary and effective algebraic methods.

In this paper, we consider the following generalized (2+1)-dimensional BKP equation ^[10]

$\begin{cases} (w^n)_{t}+(w^m)_{xxx}+(w^m)_{yyy}+\alpha(uw)_{x}+\beta(vw)_{y}=0, \\ u_{y}=w_{x}, \\ v_{x}=w_{y}, \end{cases}$

(1.1)

where $\alpha, \beta$ are arbitrary constants and $\alpha+\beta\neq0, m, n$ are integers and $m, n\geq2$. In ^[10], authors studied traveling wave solutions in the parameter space of this system by bifurcation theory of dynamical systems and they obtained some exact explicit parametric representations of periodic cusp wave solutions, solitary wave solutions and compacton solutions. In this paper, we continue to consider the problem of solving system (1.1) by using the $G'/G$-expansion method and the first integral method and we obtain the rational function solutions, periodic function solutions and the hyperbolic function solutions of (1.1) under some parametric conditions and the values of $m$, $n$ in several cases.

Specially, when $m=1, n=1, \alpha=\beta=6$, (1.1) becomes

$\begin{cases} w_{t}+w_{xxx}+w_{yyy}+6(uw)_{x}+6(vw)_{y}=0, \\ u_{y}=w_{x}, \\ v_{x}=w_{y}. \end{cases}$

It is the famous (2+1)-dimensional BKP equation which was introduced by Date et al. ^[11] and describes the processes of interaction of exponentially localized structures. It is one of a hierarchy of integrable systems emerging from a bilinear identity related to a Clifford algebra which is generated by two neutral fermion fields ^[12]. This equation has been studied by using many methods, such as the sine-cosine method ^[13], the $G'/G$-expansion method ^[6], the improved $G'/G$-expansion method ^[14] and so on.

The aim of this paper is to extract the exact solutions of the generalized (2+1)-dimensional BKP equation by using the $G'/G$-expansion method and the first integral method. The paper is arranged as follows: In section 2, we apply the $G'/G$-expansion method to this equation. In section 3, we apply the first integral method to solve this equation. In section 4, we give the conclusion of the paper.

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2. Application of the $G^{\prime}/ G$-expansion method to the generalized (2+1)-dimensional BKP equation

We suppose the wave transformations

$w(x, y, t)=w(\xi), u(x, y, t)=u(\xi), v(x, y, t)=v(\xi), ~~~~~~\xi=k_{1}x+l_{1}y+\lambda_{1}t$

(2.1)

where $k_{1}, l_{1}, \lambda_{1}$ are constants. By using the wave transformations (2.1), (1.1) can be converted into ODEs

$\begin{cases} \lambda_{1}(w^n)^{\prime}+(k^{3}_{1}+l^{3}_{1})(w^{m})^{\prime\prime\prime}+\alpha k_{1}(u^{\prime}w+uw^{\prime})+\beta l_{1}(v^{\prime}w+vw^{\prime})=0, \\ l_{1}u^{\prime}=k_{1}w^{\prime}, \\ k_{1}v^{\prime}=l_{1}w^{\prime}, \end{cases}$

(2.2)

where $"'"$ is the derivative with respect to $\xi$. Integrating the second and third equation of (2.2) and neglecting integral constants, we obtain

$\begin{cases} l_{1}u=k_{1}w, \\ k_{1}v=l_{1}w. \end{cases}$

Substituting the above equations into the first equation of (2.2) and integrating it, then it becomes

$\lambda_{1}w^n+(k^{3}_{1}+l^{3}_{1})(w^{m})^{\prime\prime}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})w^{2}=g,$

(2.3)

where $g$ is an integral constant. We assume that (2.3) has the following formal solutions ^[7,15]:

$w(\xi)=D{\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{N}, ~~~~D\neq0,$

(2.4)

where $D$ is a constant to be determined later. $N$ is determined by balancing the linear term of the highest order derivatives with the highest order nonlinear term in (2.3) and $G$ satisfies a second order constant coefficient ODE which is

$G''(\xi)+\lambda G'(\xi)+\mu G(\xi)=0,$

(2.5)

where $\lambda, ~ \mu$ are constants and will be determined later. Next, we will obtain the exact solutions of (1.1) by considering the values of $m$ and $n$ in several cases.

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2.1. $m\neq n, m>2, n>2$

Balancing $(w^{m})^{\prime\prime}$ with $w^{n}$ of (2.3), we have $mN+2=nN$, i.e., $N=2/(n-m)$. Thus, we assume

$w(\xi)=D_{1}{\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2}{n-m}}, ~~~~D_{1}\neq0$

(2.6)

where $D_{1}$ is a constant to be determined later. Then, we have

$\begin{eqnarray*} w^{n}&=&D^{n}_{1}{\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2n}{n-m}}, ~~~~~~~~~~~~~~~~~~w^{2}=D^{2}_{1}{\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{4}{n-m}}, \\ (w^{m})''&=&\frac{2m}{n-m}D^{m}_{1}{\bigg[}(\frac{2m}{n-m}+1){\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}+2}+(\frac{4m}{n-m}+1)\lambda {\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}+1} \nonumber\\ &+&\frac{2m}{n-m}(2\mu+\lambda^{2}){\bigg(}\frac{G'}{G}{\bigg)}^{\frac{2m}{n-m}}+(\frac{4m}{n-m}-1)\lambda\mu {\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}-1}\nonumber\\ &+&(\frac{2m}{n-m}-1)\mu^{2}{\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}-2}{\bigg]}. \end{eqnarray*}$

Substituting the above formulas into (2.3) and collecting all terms with the same order of $G^{\prime}/G$ together, we can convert the left-hand side of (2.3) into a polynomial in $G^{\prime}/G$. Then, setting each coefficient of each polynomial to zero, we can derive a set of algebraic equation for $\lambda, \mu$ and $D_{1}$:

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}+2}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}+1)\frac{2m}{n-m}D^{m}_{1}+\lambda_{1}D^{n}_{1}=0,$

(2.7)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}+1}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{4m}{n-m}+1)\frac{2m}{n-m}\lambda D^{m}_{1}=0.$

(2.8)

Here, we need to consider the value of $4/(n-m)$ in the following cases:

Case 1. $\frac{4}{n-m}=\frac{2m}{n-m}-1$

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m})^{2}(2\mu+\lambda^{2})D^{m}_{1}=0,$

(2.9)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}-1}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{4m}{n-m}-1)\frac{2m}{n-m}\lambda\mu D^{m}_{1}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})D^{2}_{1}=0,$

(2.10)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}-2}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}-1)\frac{2m}{n-m}\mu^{2} D^{m}_{1}=0.$

(2.11)

Solving the set of (2.7)-(2.11), we obtain

$\lambda=\mu=0, ~~~~ g=0, ~~~~~ \frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}=0, ~~~~~ D_{1}={\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}+1)\frac{2m}{n-m}}{\lambda_{1}} {\bigg)}^{1/(n-m)}.$

(2.12)

Case 2. $\frac{4}{n-m}=\frac{2m}{n-m}-2$

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m})^{2}(2\mu+\lambda^{2})D^{m}_{1}=0,$

(2.13)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}-1}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{4m}{n-m}-1)\frac{2m}{n-m}\lambda\mu D^{m}_{1}=0,$

(2.14)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}-2}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}-1)\frac{2m}{n-m}\mu^{2} D^{m}_{1}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})D^{2}_{1}=0.$

(2.15)

Solving the set of (2.7)-(2.8) and (2.13)-(2.15), we get the same results as those of Case 1.

Case 3. $\frac{4}{n-m}\neq\frac{2m}{n-m}-1$ } and $\frac{4}{n-m}\neq\frac{2m}{n-m}-2$

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m})^{2}(2\mu+\lambda^{2})D^{m}_{1}=0,$

(2.16)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}-1}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{4m}{n-m}-1)\frac{2m}{n-m}\lambda\mu D^{m}_{1}=0,$

(2.17)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{n-m}-2}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}-1)\frac{2m}{n-m}\mu^{2} D^{m}_{1}=0,$

(2.18)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{4}{n-m}}$ coeff:

$(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})D^{2}_{1}=0.$

(2.19)

Solving the set of (2.7)-(2.8) and (2.16)-(2.19), we obtain the same results as those of former cases. Substituting (2.12) into (2.5) and (2.6), then, we can get the rational function solutions

$\begin{eqnarray*} w(x, y, t)&=&{\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}+1)\frac{2m}{n-m}}{\lambda_{1}} {\bigg)}^{\frac{1}{n-m}}{\bigg(}\frac{C_{1}}{C_{1}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{2}}{\bigg)}^{\frac{2}{n-m}}, \\ v(x, y, t)&=&\frac{l_{1}}{k_{1}}{\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}+1)\frac{2m}{n-m}}{\lambda_{1}} {\bigg)}^{\frac{1}{n-m}}{\bigg(}\frac{C_{1}}{C_{1}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{2}}{\bigg)}^{\frac{2}{n-m}}, \\ u(x, y, t)&=&\frac{k_{1}}{l_{1}}{\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}+1)\frac{2m}{n-m}}{\lambda_{1}} {\bigg)}^{\frac{1}{n-m}}{\bigg(}\frac{C_{1}}{C_{1}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{2}}{\bigg)}^{\frac{2}{n-m}}, \end{eqnarray*}$

where $C_{1}, C_{2}$ are arbitrary constants and $\alpha k^{3}_{1}+\beta l^{3}_{1}=0$.

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2.2. $m=2, n>2$

(2.3) becomes

$\lambda_{1}w^n+(k^{3}_{1}+l^{3}_{1})(w^{2})^{\prime\prime}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})w^{2}=g.$

(2.20)

Balancing $(w^{2})^{\prime\prime}$ with $w^{n}$, we have $N=2/(n-2)$. Thus, (2.20) has the following formal solutions

$w(\xi)=D_{2}{\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2}{n-2}}, ~~~~D_{2}\neq0,$

(2.21)

where $D_{2}$ is a constant to be determined later and $G$ satisfies (2.5). Similarly, we can get a set of algebraic equations:

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{4}{n-2}+2}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{4}{n-2}+1)\frac{4}{n-2}D^{2}_{2}+\lambda_{1}D^{n}_{2}=0,$

(2.22)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{4}{n-2}+1}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{8}{n-2}+1)\frac{4}{n-2}\lambda D^{2}_{2}=0,$

(2.23)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{4}{n-2}}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{4}{n-2})^{2}(2\mu+\lambda^{2})D^{2}_{2}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})D^{2}_{2}=0,$

(2.24)

Ⅰ. The case $g=0$

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{4}{n-2}-1}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{8}{n-2}-1)\frac{4}{n-2}\lambda\mu D^{2}_{2}=0,$

(2.25)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{4}{n-2}-2}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{4}{n-2}-1)\frac{4}{n-2}\mu^{2} D^{2}_{2}=0.$

(2.26)

Solving that set of (2.22)-(2.26), we obtain

$\lambda=\mu=0, ~~~~~ \frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}=0, ~~~~~ D_{2}={\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{4}{n-2}+1)\frac{4}{n-2}}{\lambda_{1}} {\bigg)}^{1/(n-2)}.$

(2.27)

Specially, when $\frac{4}{n-2}-1=0$, i.e., $n=6$, we obtain

$\lambda=0, ~~~~ ~ \mu=-\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}, ~~~~~ D_{2}={\bigg(}\frac{-2(k^{3}_{1}+l^{3}_{1})}{\lambda_{1}} {\bigg)}^{1/4}.$

(2.28)

Substituting (2.27) into (2.5) and (2.21), then, we can get the rational function solutions

$\begin{eqnarray*} w(x, y, t)&=&{\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{4}{n-2}+1)\frac{4}{n-2}}{\lambda_{1}} {\bigg)}^{\frac{1}{n-2}}{\bigg(}\frac{C_{3}}{C_{3}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{4}}{\bigg)}^{\frac{2}{n-2}}, \\ v(x, y, t)&=&\frac{l_{1}}{k_{1}}{\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{4}{n-2}+1)\frac{4}{n-2}}{\lambda_{1}} {\bigg)}^{\frac{1}{n-2}}{\bigg(}\frac{C_{3}}{C_{3}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{4}}{\bigg)}^{\frac{2}{n-2}}, \\ u(x, y, t)&=&\frac{k_{1}}{l_{1}}{\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{4}{n-2}+1)\frac{4}{n-2}}{\lambda_{1}} {\bigg)}^{\frac{1}{n-2}}{\bigg(}\frac{C_{3}}{C_{3}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{4}}{\bigg)}^{\frac{2}{n-2}}, \end{eqnarray*}$

where $C_{3}, C_{4}$ are arbitrary constants and $\alpha k^{3}_{1}+\beta l^{3}_{1}=0$. Substituting (2.28) into (2.5) and (2.21), then, we have

$G''+\bigg{(}-\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}G=0.$

Case 1. $\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}>0$

We obtain the hyperbolic function solutions

$\begin{eqnarray*} w(x, y, t)=&&\bigg{(} \frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})^{2}}{2\lambda_{1}(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/4}\\ &&\left\{ \frac{C_{5}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{6}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{5}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{6}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}^{1/2}, \\ v(x, y, t)=&&\frac{l_{1}}{k_{1}}\bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})^{2}}{2\lambda_{1}(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/4}\\ &&\left\{ \frac{C_{5}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{6}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{5}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{6}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}^{1/2}, \\ u(x, y, t)=&&\frac{k_{1}}{l_{1}}\bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})^{2}}{2\lambda_{1}(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/4}\\ &&\left\{ \frac{C_{5}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{6}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{5}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{6}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}^{1/2}, \end{eqnarray*}$

where $C_{5}, C_{6}$ are arbitrary constants.

Case 2. $\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})} < 0$

We obtain the hyperbolic function solutions

$\begin{eqnarray*} w(x, y, t)=&&\bigg{(} \frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})^{2}}{2\lambda_{1}(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/4}\\ &&\left\{ \frac{-C_{7}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{8}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{7}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{8}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}^{1/2}, \\ v(x, y, t)=&&\frac{l_{1}}{k_{1}}\bigg{(} \frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})^{2}}{2\lambda_{1}(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/4}\\ &&\left\{ \frac{-C_{7}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{8}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{7}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{8}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}^{1/2}, \\ u(x, y, t)=&&\frac{k_{1}}{l_{1}}\bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})^{2}}{2\lambda_{1}(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/4}\\ &&\left\{ \frac{-C_{7}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{8}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{7}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{8}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}^{1/2}, \end{eqnarray*}$

where $C_{7}, C_{8}$ are arbitrary constants.

Case 3. $\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}=0$

We obtain the rational function solutions

$\begin{eqnarray*} w(x, y, t)=&&{\bigg(}\frac{-2(k^{3}_{1}+l^{3}_{1})}{\lambda_{1}} {\bigg)}^{1/4}\left\{ \frac{C_{9}}{C_{9}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{10}} \right\}^{1/2}, \\ v(x, y, t)=&&\frac{l_{1}}{k_{1}}{\bigg(}\frac{-2(k^{3}_{1}+l^{3}_{1})}{\lambda_{1}} {\bigg)}^{1/4}\left\{ \frac{C_{9}}{C_{9}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{10}} \right\}^{1/2}, \\ u(x, y, t)=&&\frac{k_{1}}{l_{1}}{\bigg(}\frac{-2(k^{3}_{1}+l^{3}_{1})}{\lambda_{1}} {\bigg)}^{1/4}\left\{ \frac{C_{9}}{C_{9}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{10}} \right\}^{1/2}, \end{eqnarray*}$

where $C_{9}, C_{10}$ are arbitrary constants.

Ⅱ. The case $g\neq0$

When $\frac{4}{n-2}-2=0$, i.e, $n=4$.

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}$ coeff:

$6(k^{3}_{1}+l^{3}_{1})\lambda\mu D^{2}_{2}=0,$

(2.29)

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{0}$ coeff:

$2(k^{3}_{1}+l^{3}_{1})\mu^{2} D^{2}_{2}=g.$

(2.30)

Solving the set of (2.22)-(2.24) and (2.29)-(2.30), we obtain

$\lambda=0, ~~~ \mu=-\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}, ~~~D_{2}={\bigg(}\frac{-6(k^{3}_{1}+l^{3}_{1})}{\lambda_{1}} {\bigg)}^{1/2}, ~~~ \frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}\neq0, ~~~ \frac{-3(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})^{2}}{\lambda_{1}}=g.$

(2.31)

Similarly, we can obtain the hyperbolic function solutions and trigonometric function solutions

Case 1. $\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}>0$

$\begin{eqnarray} w(x, y, t)=&&\bigg{(} \frac{-3(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{\lambda_{1}}\bigg{)}^{1/2}\nonumber \\ &&\left\{ \frac{C_{11}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{12}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{11}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{12}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}, \\ v(x, y, t)=&&\frac{l_{1}}{k_{1}}\bigg{(} \frac{-3(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{\lambda_{1}}\bigg{)}^{1/2}\nonumber \\ &&\left\{ \frac{C_{11}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{12}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{11}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{12}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}, \nonumber\\ u(x, y, t)=&&\frac{k_{1}}{l_{1}}\bigg{(} \frac{-3(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{\lambda_{1}}\bigg{)}^{1/2}\nonumber \\ &&\left\{ \frac{C_{11}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{12}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{11}\cosh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{12}\sinh \bigg{(}\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}\nonumber, \end{eqnarray}$

(2.32)

where $C_{11}, C_{12}$ are arbitrary constants and $\lambda_{1}(k_{1}^{3}+l_{1}^{3}) < 0$.

Case 2. $\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})} < 0$

$\begin{eqnarray} w(x, y, t)=&&\bigg{(} \frac{-3(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{\lambda_{1}}\bigg{)}^{1/2}\nonumber\\ &&\left\{ \frac{-C_{13}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{14}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{13}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{14}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}, \\ v(x, y, t)=&&\frac{l_{1}}{k_{1}}\bigg{(} \frac{-3(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{\lambda_{1}}\bigg{)}^{1/2}\nonumber\\ &&\left\{ \frac{-C_{13}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{14}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{13}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{14}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}, \nonumber\\ u(x, y, t)=&&\frac{k_{1}}{l_{1}}\bigg{(} \frac{-3(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{\lambda_{1}}\bigg{)}^{1/2}\nonumber\\ &&\left\{ \frac{-C_{13}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{14}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}{C_{13}\cos \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{14}\sin \bigg{(}\frac{-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})}{2(k_{1}^{3}+l_{1}^{3})}\bigg{)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}\right\}\nonumber, \end{eqnarray}$

(2.33)

where $C_{13}, C_{14}$ are arbitrary constants and $\lambda_{1}(k_{1}^{3}+l_{1}^{3}) < 0$.

---

2.3. $m>2, n=2$

(2.3) becomes

${\bigg(}\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}{\bigg)}w^{2}+(k^{3}_{1}+l^{3}_{1})(w^{m})^{\prime\prime}=g.$

(2.34)

Balancing $(w^{m})^{\prime\prime}$ with $w^{2}$, we have $N=2/(2-m)$. Thus, (2.34) has the following formal solution

$w(\xi)=D_{3}{\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2}{2-m}}, ~~~~D_{3}\neq0,$

(2.35)

where $D_{3}$ is a constant to be determined later and $G$ satisfies (2.5). Similarly, we can get a set of algebraic equations:

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{2-m}+2}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)(\frac{2m}{2-m})D^{m}_{3}+{\bigg(}\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}{\bigg)}D^{2}_{3}=0,$

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{2-m}+1}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{4m}{2-m}+1)\frac{2m}{2-m}\lambda D^{m}_{3}=0,$

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{2-m}}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m})^{2}(2\mu+\lambda^{2})D^{m}_{3}=0,$

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{2-m}-1}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{4m}{2-m}-1)\frac{2m}{2-m}\lambda\mu D^{m}_{3}=0,$

${\bigg(}\frac{G^{\prime}}{G}{\bigg)}^{\frac{2m}{2-m}-2}$ coeff:

$(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}-1)\frac{2m}{2-m}\mu^{2} D^{m}_{3}=0.$

Solving the above algebraic equations, we obtain

$\lambda=\mu=0, ~~~~ g=0, ~~~~D_{3}={\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}}{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}} {\bigg)}^{1/(2-m)}.$

(2.36)

Substituting (2.36) into (2.5) and (2.35), then, when $m\neq n=2$, we have the rational function solutions

$\begin{eqnarray*} w(x, y, t)&=&{\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}}{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}} {\bigg)}^{\frac{1}{2-m}}{\bigg(}\frac{C_{15}}{C_{15}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{16}}{\bigg)}^{\frac{2}{2-m}}, \\ v(x, y, t)&=&\frac{l_{1}}{k_{1}}{\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}}{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}} {\bigg)}^{\frac{1}{2-m}}{\bigg(}\frac{C_{15}}{C_{15}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{16}}{\bigg)}^{\frac{2}{2-m}}, \\ u(x, y, t)&=&\frac{k_{1}}{l_{1}}{\bigg(}-\frac{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}}{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}} {\bigg)}^{\frac{1}{2-m}}{\bigg(}\frac{C_{15}}{C_{15}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{16}}{\bigg)}^{\frac{2}{2-m}}, \end{eqnarray*}$

where $C_{15}, C_{16}$ are arbitrary constants and $\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}\neq0$.

---

2.4. $m=n=2$

Now, (2.3) can be converted into a second order ODE

${\bigg(}\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}{\bigg)}w^{2}+(k^{3}_{1}+l^{3}_{1})(w^{2})^{\prime\prime}=g.$

(2.37)

Obviously, the characteristic equation of (2.37) is $r^{2}+{\bigg(}\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}=0$, $r$ is the characteristic value.

Case 1. $\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}} < 0$

We can obtain the exact solution

$\begin{eqnarray} w(x, y, t)&=&{\bigg(}C_{17}e^{{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}+C_{18}e^{-{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}+\frac{g}{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{\bigg)}^{1/2}, \\ v(x, y, t)&=&\frac{l_{1}}{k_{1}}{\bigg(}C_{17}e^{{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}+C_{18}e^{-{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}+\frac{g}{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{\bigg)}^{1/2}, \nonumber\\ u(x, y, t)&=&\frac{k_{1}}{l_{1}}{\bigg(}C_{17}e^{{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}+C_{18}e^{-{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)}+\frac{g}{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{\bigg)}^{1/2}\nonumber, \end{eqnarray}$

(2.38)

where $C_{17}, C_{18}$ are arbitrary constants.

Case 2. $\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}>0$

We can obtain the periodic function solutions

$\begin{eqnarray} w(x, y, t)&=&\bigg{(}C_{19}\cos\bigg{[}{\bigg(}\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)\bigg{]}\nonumber\\ &&+C_{20}\sin\bigg{[}{\bigg(}\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)\bigg{]}+\frac{g}{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}\bigg{)}^{1/2}, \\ v(x, y, t)&=&\frac{l_{1}}{k_{1}}\bigg{(}C_{19}\cos\bigg{[}{\bigg(}\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)\bigg{]}\nonumber\\ &&+C_{20}\sin\bigg{[}{\bigg(}\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)\bigg{]}+\frac{g}{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}\bigg{)}^{1/2}, \nonumber\\ u(x, y, t)&=&\frac{k_{1}}{l_{1}}\bigg{(}C_{19}\cos\bigg{[}{\bigg(}\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)\bigg{]}\nonumber\\ &&+C_{20}\sin\bigg{[}{\bigg(}\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}{\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)\bigg{]}+\frac{g}{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}\bigg{)}^{1/2}\nonumber, \end{eqnarray}$

(2.39)

where $C_{19}, C_{20}$ are arbitrary constants.

Case 3. $\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}=0$

We can obtain the rational function solutions

$\begin{eqnarray} w(x, y, t)&=&\bigg{(}\frac{g}{2(k_{1}^{3}+l_{1}^{3})}(k_{1}x+l_{1}y+\lambda_{1}t)^{2}+C_{21}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{22}\bigg{)}^{1/2}, \\ v(x, y, t)&=&\frac{l_{1}}{k_{1}}\bigg{(}\frac{g}{2(k_{1}^{3}+l_{1}^{3})}(k_{1}x+l_{1}y+\lambda_{1}t)^{2}+C_{21}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{22}\bigg{)}^{1/2}, \nonumber\\ u(x, y, t)&=&\frac{k_{1}}{l_{1}}\bigg{(}\frac{g}{2(k_{1}^{3}+l_{1}^{3})}(k_{1}x+l_{1}y+\lambda_{1}t)^{2}+C_{21}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{22}\bigg{)}^{1/2}, \nonumber \end{eqnarray}$

(2.40)

where $C_{21}, C_{22}$ are arbitrary constants.

---

3. Application of the first integral method to the generalized (2+1)-dimensional BKP equation

---

3.1. $m\neq n$

For simplicity, we let $g=0$ and propose a transformation $w=\varphi^{\frac{2}{n-m}}$. Then, (2.3) is converted to

$\lambda_{1}\varphi^{4}+(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}-1)\frac{2m}{n-m}(\varphi')^{2}+(k^{3}_{1}+l^{3}_{1})\frac{2m}{n-m}\varphi\varphi''+{\bigg(}\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}{\bigg)}\varphi^{2-\frac{2m-4}{n-m}}=0.$

(3.1)

Let $x=\varphi, y=\frac{\mathrm {d}\varphi}{\mathrm {d}\xi}$, thus (3.1) is equivalent to the two dimensional autonomous system

$\begin{cases} x'=y, \\ y'=-{\bigg(}\frac{ \lambda_{1}x^{4}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})x^{2-\frac{2m-4}{n-m}}+(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}-1)\frac{2m}{n-m}y^{2}}{(k^{3}_{1}+l^{3}_{1})\frac{2m}{n-m}x} {\bigg)}. \end{cases}$

(3.2)

Making the transformation $d\eta=\frac{d\xi}{(k^{3}_{1}+l^{3}_{1})\frac{2m}{n-m}x}$, then, (3.2) becomes

$\begin{cases} \frac{dx}{d\eta}=(k^{3}_{1}+l^{3}_{1})\frac{2m}{n-m}xy, \\ \frac{dy}{d\eta}=-\left(\lambda_{1}x^{4}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})x^{2-\frac{2m-4}{n-m}}+(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}-1)\frac{2m}{n-m}y^{2}\right). \end{cases}$

(3.3)

Then, we will apply the Division Theorem to seek the first integral of system (3.3). Suppose that $x=x(\eta), y=y(\eta)$ are the nontrivial solutions to (3.3), and $p(x, y)=\sum^{M}_{i=0}a_{i}(x)y^{i}$ is an irreducible polynomial in $C[x, y],$ where $a_{i}(x) (i=0, 1..., M)$ are polynomials of $x$ and $a_{i}(x)\neq0$. Let $p(x(\eta), y(\eta))=0$ be the first integral to system (3.3). $\frac{dp}{d\eta}$ is a polynomial in $x$, $y$ and $\frac{dp}{d\eta}\big|_{(3.3)}=0$. According to the Division Theorem, there exists a polynomial $g(x)+h(x)y$ in $C[x, y]$, such that

$\begin{eqnarray} \frac{dp}{d\eta}\bigg|_{(3.3)}&=&\left( \frac{\partial p}{\partial x}\frac{dx}{d\eta}+\frac{\partial p}{\partial y}\frac{dy}{d\eta} \right)\bigg|_{(3.3)} \nonumber \\ &=&\sum^{M}_{i=0}[a^{\prime}_{i}(x)y^{i}\cdot (k^{3}_{1}+l^{3}_{1})\frac{2m}{n-m}xy]\nonumber\\ &&-\sum^{2}_{i=0}\left[ia_{i}(x)y^{i-1}(\lambda_{1}x^{4}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})x^{2-\frac{2m-4}{n-m}}+(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}-1)\frac{2m}{n-m}y^{2})\right] \nonumber \\ &=&[g(x)+h(x)y] \bigg{[}\sum^{M}_{i=0}a_{i}(x)y^{i}\bigg{]}. \end{eqnarray}$

(3.4)

Here, let $M=1$, thus, $p(x, y)=a_{0}(x)+a_{1}(x)y$. By comparing with the coefficients of $y^{i}$ of both sides of (3.4), we have

$(k^{3}_{1}+l^{3}_{1})\frac{2m}{n-m}xa'_{1}(x)=h(x)a_{1}(x)+(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}-1)\frac{2m}{n-m}a_{1}(x),$

(3.5)

$(k^{3}_{1}+l^{3}_{1})\frac{2m}{n-m}xa_{0}'(x)=g(x)a_{1}(x)+h(x)a_{0}(x),$

(3.6)

$g(x)a_{0}(x)=-\left(\lambda_{1}x^{4}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})x^{2-\frac{2m-4}{n-m}}\right)a_{1}(x).$

(3.7)

Since $a_{i}(x)(i=0, 1)$ are polynomials, then from (3.5), we deduce that $h(x)=-(k^{3}_{1}+l^{3}_{1})(\frac{2m}{n-m}-1)\frac{2m}{n-m}$ and $a_{1}(x)$ is a constant. For simplicity, take $a_{1}(x)=1$. Balancing the degrees of $g(x)$ and $a_{0}(x)$, we conclude that $\deg(g(x))=\deg(a_{0}(x))$. Then, we derive $\deg(g(x))=\deg(a_{0}(x))=j, (j\in Z^{+}, j\geq2).$

When $2-\frac{2m-4}{n-m}=4$ ($n=2$) and $\deg(g(x))=\deg(a_{0}(x))=2$, we suppose that

$\begin{eqnarray} g(x)&=&A_{0}+A_{1}x+A_{2}x^{2}, \nonumber\\ a_{0}(x)&=&B_{0}+B_{1}x+B_{2}x^{2}, ~~(A_{2}\neq0, B_{2}\neq0), \end{eqnarray}$

(3.8)

where $A_{i}, B_{i}, (i=0, 1, 2)$ are all constants to be determined. Substituting (3.8) into (3.6), we obtain

$g(x)=(k^{3}_{1}+l^{3}_{1})\frac{2m}{2-m}\left[(\frac{2m}{2-m}-1)B_{0}+\frac{2m}{2-m}B_{1}x+(\frac{2m}{2-m}+1)B_{2}x^{2}\right].$

Substituting $a_{0}(x), a_{1}(x)$ and $g(x)$ into (3.7), and setting all the coefficients of powers $x$ to be zero, we can get a system of nonlinear algebraic equations. After solving it, we can get the following solutions

$B_{0}=B_{1}=0, ~~~~~~B_{2}=\pm{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}} {\bigg)}^{1/2}.$

(3.9)

Using the conditions (3.9) in $p(x, y)=a_{0}(x)+a_{1}(x)y=0$, we obtain

$y\pm{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}} {\bigg)}^{1/2}x^{2}=0.$

(3.10)

Combining (3.3) with (3.10), we find

$\frac{dx}{d\eta}=\pm(k_{1}^{3}+l_{1}^{3})\frac{2m}{n-m}{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}} {\bigg)}^{1/2}x^{3}.$

Thus, (3.10) can be reduced to

$\frac{d\varphi}{d\xi}=\pm{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}} {\bigg)}^{1/2}\varphi^{2}.$

Then, we have

$\varphi(\xi)=\bigg{[}\pm{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}} {\bigg)}^{1/2}\xi+C_{23}\bigg{]}^{-1}.$

Thus, we can have the rational function solutions

$\begin{eqnarray*} w(x, y, t)&=&\bigg{[}\pm{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}} {\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{23}\bigg{]}^{2/(m-2)}, \\ v(x, y, t)&=&\frac{l_{1}}{k_{1}}\bigg{[}\pm{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}} {\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{23}\bigg{]}^{2/(m-2)}, \\ u(x, y, t)&=&\frac{k_{1}}{l_{1}}\bigg{[}\pm{\bigg(}-\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}+1)\frac{2m}{2-m}} {\bigg)}^{1/2}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{23}\bigg{]}^{2/(m-2)}, \end{eqnarray*}$

where $C_{23}$ is an arbitrary constant and $k^{3}_{1}+l^{3}_{1}\neq0$.

Remark 1: When $\deg(g(x))=\deg(a_{0}(x))=2$ and $2-\frac{2m-4}{n-m}=i, (i\in Z, i < 4)$, there is no solution for them by using the method as that of $2-\frac{2m-4}{n-m}=4.$

Remark 2: When $\deg(g(x))=\deg(a_{0}(x))=j, (j\in Z, j>2)$, there is no exact solution of (1.1) by using the method as that of $\deg(g(x))=\deg(a_{0}(x))=2$.

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3.2. $m=n$

Similarly, we propose a transformation denoted by $w=\phi^{\frac{2}{2-m}}$. Then, (2.3) can be converted to

$\lambda_{1}\phi^{2}+{\bigg(}\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}{\bigg)}\phi^{4}+(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}-1)\frac{2m}{2-m}(\phi')^{2}+(k^{3}_{1}+l^{3}_{1})\frac{2m}{2-m}\phi\phi''-g\phi^{2-\frac{2m}{2-m}}=0.$

(3.11)

Let $x=\phi, y=\frac{\mathrm {d}\phi}{\mathrm {d}\xi}$, thus (3.11) is equivalent to the two dimensional autonomous system

$\begin{cases} x'=y, \\ y'=-\bigg{(}\frac{\lambda_{1}x^{2}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})x^{4}-gx^{2-\frac{2m}{2-m}}+(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}-1)\frac{2m}{2-m}y^{2}}{(k^{3}_{1}+l^{3}_{1})\frac{2m}{2-m}x} {\bigg)}. \end{cases}$

(3.12)

Making the transformation $d\eta=\frac{d\xi}{(k^{3}_{1}+l^{3}_{1})\frac{2m}{n-m}x}$, then, (3.12) becomes

$\begin{cases} \frac{dx}{d\eta}=(k^{3}_{1}+l^{3}_{1})\frac{2m}{2-m}xy, \\ \frac{dy}{d\eta}=-\left(\lambda_{1}x^{2}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})x^{4}-gx^{2-\frac{2m}{2-m}}+(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}-1)\frac{2m}{2-m}y^{2}\right). \end{cases}$

(3.13)

Similarly, let $M=1$, we have

$(k^{3}_{1}+l^{3}_{1})\frac{2m}{2-m}xa'_{1}(x)=h(x)a_{1}(x)+(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}-1)\frac{2m}{2-m}a_{1}(x),$

(3.14)

$(k^{3}_{1}+l^{3}_{1})\frac{2m}{2-m}xa_{0}'(x)=g(x)a_{1}(x)+h(x)a_{0}(x),$

(3.15)

$g(x)a_{0}(x)=-\left(\lambda_{1}x^{2}+(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})x^{4}-gx^{2-\frac{2m}{2-m}}\right)a_{1}(x).$

(3.16)

According to $m\neq n$, we have $h(x)=-(k^{3}_{1}+l^{3}_{1})(\frac{2m}{2-m}-1)\frac{2m}{2-m}$, $a_{1}(x)=1$ and $\deg(g(x))=\deg(a_{0}(x))=j, (j\in Z^{+}, j\geq2)$. Considering all cases, only when $\deg(g(x))=\deg(a_{0}(x))=3$, i.e., $2-\frac{2m}{2-m}=6$ $(m=n=4)$, there exist solutions of (1). We suppose that

$\begin{eqnarray} g(x)&=&a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}, \nonumber\\ a_{0}(x)&=&b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}, ~~(a_{3}\neq0, b_{3}\neq0), \end{eqnarray}$

(3.17)

where $a_{i}, b_{i}, (i=0, 1, 2, 3)$ are all constants to be determined. Substituting (3.17) into (3.15), we obtain

$g(x)=4(k^{3}_{1}+l^{3}_{1})\left(5b_{0}+4b_{1}x+3b_{2}x^{2}+2b_{3}x^{3}\right).$

Substituting $a_{0}(x), a_{1}(x)$ and $g(x)$ into (3.16), and setting all the coefficients of powers $x$ to be zero, we have

$b_{0}=b_{2}=0, ~~ 16(k_{1}^{3}+l_{1}^{3})b_{1}^{2}=-\lambda_{1}, ~~ 8(k_{1}^{3}+l_{1}^{3})b_{3}^{2}=g, ~~ 24(k_{1}^{3}+l_{1}^{3})b_{1}b_{3}=-(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}).$

Solving it, we find

$b_{0}=b_{2}=0, ~~~ b_{1}=\pm\sqrt{\frac{-\lambda_{1}}{16(k_{1}^{3}+l_{1}^{3})}}, ~~~ b_{3}=\pm\sqrt{\frac{g}{8(k_{1}^{3}+l_{1}^{3})}}, ~~~-9\lambda_{1}g=2(\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}})^{2}.$

(3.18)

Using the conditions (3.18) in $p(x, y)=a_{0}(x)+a_{1}(x)y=0$, we obtain

$y=\pm\sqrt{\frac{-\lambda_{1}}{16(k_{1}^{3}+l_{1}^{3})}}x\pm\sqrt{\frac{g}{8(k_{1}^{3}+l_{1}^{3})}}x^{3}.$

(3.19)

Then, (3.19) can be reduced to

$\frac{d\phi}{d\xi}=\pm\sqrt{\frac{-\lambda_{1}}{16(k_{1}^{3}+l_{1}^{3})}}\xi\pm\sqrt{\frac{g}{8(k_{1}^{3}+l_{1}^{3})}}\xi^{3}.$

(3.20)

Solving (3.20), we obtain

$\phi(\xi)=\pm\left( \pm\sqrt{\frac{-2g}{\lambda_{1}}}+C_{24}e^{\pm\frac{1}{2}\sqrt{\frac{-\lambda_{1}}{k_{1}^{3}+l_{1}^{3}}}\xi} \right)^{-1/2}.$

Thus, we can have the exact solution

$\begin{eqnarray*} w(x, y, t)=&&\pm\left(\pm\sqrt{\frac{-2g}{\lambda_{1}}}+C_{24}e^{\pm\frac{1}{2}\sqrt{\frac{-\lambda_{1}}{k_{1}^{3}+l_{1}^{3}}}(k_{1}x+l_{1}y+\lambda_{1}t)} \right)^{1/2}, \\ v(x, y, t)=&&\pm\frac{l_{1}}{k_{1}}\left( \pm\sqrt{\frac{-2g}{\lambda_{1}}}+C_{24}e^{\pm\frac{1}{2}\sqrt{\frac{-\lambda_{1}}{k_{1}^{3}+l_{1}^{3}}}(k_{1}x+l_{1}y+\lambda_{1}t)} \right)^{1/2}, \\ u(x, y, t)=&&\pm\frac{k_{1}}{l_{1}}\left( \pm\sqrt{\frac{-2g}{\lambda_{1}}}+C_{24}e^{\pm\frac{1}{2}\sqrt{\frac{-\lambda_{1}}{k_{1}^{3}+l_{1}^{3}}}(k_{1}x+l_{1}y+\lambda_{1}t)} \right)^{1/2}, \end{eqnarray*}$

where $C_{24}$ is an arbitrary constant and $\alpha k_{1}^{3}+\beta l_{1}^{3}\neq0$, $\lambda_{1}(k_{1}^{3}+l_{1}^{3}) < 0$.

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4. Conclusion

This paper considered the generalized (2+1)-dimensional BKP equation, by the aid of the $G'/G$-expansion method and the first integral method. Rational function solutions, periodic function solutions and hyperbolic function solutions are obtained under some parametric conditions and the values of $m$ and $n$ in several cases. In ^[10], authors gave some exact solutions of system (1.1) under some parametric conditions by using the bifurcation theory of dynamical systems. Here, we make a simple comparison:

1. When $m=2, n=3, g=0$, in ^[10], authors gave the exact solution (3.20) in P2443 under the parametric conditions $\alpha+\beta < 0, c < 0$ and in this paper, we get $w(x, y, t)=\frac{-20(k^{3}_{1}+l^{3}_{1})}{\lambda_{1}}{\bigg(}\frac{C_{1}}{C_{1}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{2}}{\bigg)}^{2}$ under the parametric condition $\alpha k^{3}_{1}+\beta l^{3}_{1}=0$.

2. When $m=2, n=2(k+1), (k\in Z^{+}), g=0$, in ^[10], authors gave the solitary wave solutions (3.9) in P2441 under the parametric conditions $\alpha+\beta < 0, c < 0$ and in this paper, we get $w(x, y, t)=\bigg{(}\sqrt{\frac{-2(k+2)(k_{1}^{3}+l_{1}^{3})}{\lambda_{1}k^{2}}}\frac{C_{1}}{C_{1}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{2}}\bigg{)}^{1/k}$ under the parametric condition $\alpha k^{3}_{1}+\beta l^{3}_{1}=0$.

3. When $m=3, n=4, g=0$, in ^[10], authors gave the compacton solution (3.23) in P2443 under the parametric conditions $\alpha+\beta < 0, c < 0$ and in this paper, we get $w(x, y, t)=\frac{-42(k_{1}^{3}+l_{1}^{3})}{\lambda_{1}}\bigg{(}\frac{C_{1}}{C_{1}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{2}}\bigg{)}^{2}$ under the parametric condition $\alpha k^{3}_{1}+\beta l^{3}_{1}=0$.

4. When $m=3, n=5, g=0$, in ^[10], authors gave the exact solution (3.27) in P2443 under the parametric conditions $c < 0$ and in this paper, we get $w(x, y, t)=\sqrt{\frac{-12(k_{1}^{3}+l_{1}^{3})}{\lambda_{1}}}\frac{C_{1}}{C_{1}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{2}}$ under the parametric condition $\alpha k^{3}_{1}+\beta l^{3}_{1}=0$.

5. When $m=4, n=6, g=0$, in ^[10], authors gave the periodic cusp wave solutions (3.6) in P2441 under the parametric conditions $\alpha+\beta < 0, c < 0$ and in this paper, we get $w(x, y, t)=\sqrt{\frac{-20(k_{1}^{3}+l_{1}^{3})}{\lambda_{1}}}\frac{C_{1}}{C_{1}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{2}}$ under the parametric condition $\alpha k^{3}_{1}+\beta l^{3}_{1}=0$.

6. When $m=4, n=2k+1, (k\in Z^{+}), g=0$, in ^[10], authors gave the exact solutions (3.17) and (3.18) in P2442 under the parametric conditions $\alpha+\beta < 0, c < 0$ and in this paper, we get $w(x, y, t)=\bigg{(}\frac{-8(2k+5)(k_{1}^{3}+l_{1}^{3})}{\lambda_{1}(2k-3)^{2}}\bigg{)}^{\frac{1}{2k-3}}\bigg{(}\frac{C_{1}}{C_{1}(k_{1}x+l_{1}y+\lambda_{1}t)+C_{2}}\bigg{)}^{\frac{2}{2k-3}}$ under the parametric condition $\alpha k^{3}_{1}+\beta l^{3}_{1}=0$.

7. When $m=2, n=4, g\neq0$, in ^[10], authors gave the exact solutions (3.30) in P2443 under the parametric conditions $g < \frac{(\alpha+\beta)^{2}}{4c}, c>0, g>0, \alpha+\beta>0$, (3.33) in P2444 under the parametric conditions $g < \frac{(\alpha+\beta)^{2}}{4c}, c < 0, g>0, \alpha+\beta>0$ or $g>\frac{(\alpha+\beta)^{2}}{4c}, c < 0, g>0, \alpha+\beta < 0$ and (3.41), (3.43) in P2445 under the parametric conditions $g>\frac{(\alpha+\beta)^{2}}{4c}, c < 0, g < 0, \alpha+\beta < 0$ and in this paper, we get (2.32) under the parametric conditions $\lambda_{1}(k_{1}^{3}+l_{1}^{3}) < 0, ~ \frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})}>0$ and (2.33) under the parametric conditions $\lambda_{1}(k_{1}^{3}+l_{1}^{3}) < 0, ~\frac{\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{2(k_{1}^{3}+l_{1}^{3})} < 0$.

8. When $m=2, n=2, g\neq 0$, in ^[10], authors gave the exact solutions (3.36) and (3.38) in P2444 under the parametric conditions $\alpha+\beta-c>0, g>0$ and in this paper, we get the exact solutions (2.38) under the parametric conditions $\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}} < 0$, (2.39) under the parametric conditions $\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}>0$ and (2.40) under the parametric conditions $\frac{\lambda_{1}+\frac{\alpha k^{2}_{1}}{l_{1}}+\frac{\beta l^{2}_{1}}{k_{1}}}{k^{3}_{1}+l^{3}_{1}}=0$.

In addition, when let $m, n$ be other values, we have got other exact solutions of (1.1) under some parametric conditions that haven't been given in ^[10]. Certainly, system (1.1) should be studied further, which will be left to a further discussion.

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Conflict of Interest

All authors declare no conflicts of interest in this paper.

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