Extended generalized $(Zakh\frac{G^{\prime }}{G})$-expansion method for solving the nonlinear quantum Zakharov–Kuznetsov equation

Abstract

In this article, we apply the extended generalized $(\frac{G^{\prime }}{G})$-expansion method combined with the Jacobi elliptic equation to find new exact solutions of the nonlinear quantum Zakharov–Kuznetsov (QZK) equation with the aid of computer algebraic system Maple. Soliton solutions, periodic solutions, rational functions solutions and Jacobi elliptic functions solutions are obtained. Based on reductive perturbation technique and a series of transformation, the nonlinear QZK had been derived by many authors which can be reduced to a nonlinear ordinary differential equation (ODE) using the wave transformation. The extended generalized $(\frac{G^{\prime }}{G})$-expansion method is straightforward and concise, and it can also be applied to other nonlinear PDEs in mathematical physics.

1 Introduction

Investigation of exact solutions to nonlinear partial differential equations (PDEs) plays an important role in the study of nonlinear physical phenomena in many fields, such that fluid mechanics, hydrodynamics, optics, plasma physics, solid state physics, biology and so on. Several methods for finding the exact solutions to nonlinear PDEs in mathematical physics have been presented, such as the inverse scattering method [1], the Hirota bilinear transform method [2], the truncated Painlevé expansion method [3, 4], the Bäcklund transform method [5, 6], the exp-function method [7–9], the tanh-function method [10, 11], the Jacobi elliptic function expansion method [12–14], the $(\frac{G^{\prime }}{G})-$expansion method [15–19], the modified $(\frac{G^{\prime }}{G})$-expansion method [20], the $(\frac{G^{\prime }}{G},\frac{1}{G})$-expansion method [21–24], the modified simple equation method [25–27], the multiple exp-function algorithm method [28, 29], the transformed rational function method [30], the Frobenius decomposition technique [31], the local fractional variation iteration method [32], the local fractional series expansion method [33], the first integral method [34–36] and so on.

The objective of this article is to apply two methods, namely, an extended generalized $(\frac{G^{\prime }}{G})$-expansion method and a direct method to construct the exact solutions of the following nonlinear quantum Zakharov–Kuznetsov equation [37–39]:

$$\begin{aligned} \frac{\partial \phi }{\partial T}+A\phi \frac{\partial \phi }{\partial Z}+B \frac{\partial ^{3}\phi }{\partial Z^{3}}+C\frac{\partial }{\partial Z} \left( \frac{\partial ^{2}}{\partial X^{2}}+\frac{\partial ^{2}}{\partial Y^{2}}\right) \phi =0, \end{aligned}$$

(1.1)

which arises in quantum magneto plasma, and A,  B,  C are well known constants which represent the nonlinear and dispersive coefficients. Equation (1.1) has been derived in [37] using the reductive perturbation technique and in [38] using a series of transformations. Here $\phi$ is the electrostatic potential, while X,  Y,  Z,  T are the stretched space-time coordinates which are defined in [37]. Moslem et al [37] have derived Eq. (1.1) for electron-ion quantum plasma and solitary explosive and periodic solutions are presented. In [39], the authors applied the auxiliary equation method and Hirota bilinear method to study Eq. (1.1) and some types of exact solutions are obtained. The investigation of ion-acoustic waves and structures in dense quantum plasma has attracted much attention (see [40]).

This paper is organized as follows: In Sect. 2, the description of the extended generalized $(\frac{G^{\prime }}{G})$-expansion method is given. In Sect. 3, we use the given method described in Sect. 2, to find new exact solutions of the QZK equation (1.1). In Sect. 4, we solve Eq. (1.1) using a direct method. In Sect. 5, some conclusions are obtained.

2 Description of the extended generalized $(\frac{G^{\prime }}{G})$-expansion method

Consider a nonlinear PDE in the form

$$\begin{aligned} F(\phi ,\phi _{T},\phi _{X},\phi _{Y},\phi _{Z},\phi _{XX},\ldots )=0, \end{aligned}$$

(2.1)

where $\phi =\phi (X,Y,Z,T)$ is an unknown function, F is a polynomial in $\phi$ and its partial derivatives in which the highest order derivatives and nonlinear terms are involved. Let us now give the main steps [41] of this method as follows:

Step 1 :

We look for the electrostatic potential $\phi$ in the traveling form:

$$\begin{aligned} \phi (X,Y,Z,T)=\phi (\xi ),\qquad \xi =\alpha X+\beta Y+\gamma Z-\upsilon T, \end{aligned}$$

(2.2)

where $\alpha ,$ $\beta ,$ $\gamma$ are the direction cosines, and $\upsilon$ is the quantum ion-acoustic wave speed, to reduce Eq. (2.1) to the following nonlinear ordinary differential equation (ODE):

$$\begin{aligned} H(\phi ,\phi ^{\prime },\phi ^{\prime \prime },\ldots )=0, \end{aligned}$$

(2.3)

where H is a polynomial of $\phi (\xi )$ and its total derivatives $\phi ^{\prime },\phi ^{\prime \prime },\ldots$ and $^{\prime }= \dfrac{d}{d\xi }$.

Step 2 :

We suppose that the solution of the ODE. (2.3) has the form:

$$\begin{aligned} \phi (\xi )=\sum \limits _{i=-N}^{N}a_{i}\left( \frac{G^{\prime }}{G}\right) ^{i}, \end{aligned}$$

(2.4)

where $G=G(\xi )$ satisfies the Jacobi elliptic equation:

$$\begin{aligned} G^{\prime 2}(\xi )=R+QG^{2}(\xi )+PG^{4}(\xi ), \end{aligned}$$

(2.5)

where $a_{i},$ R,  Q,  P ($i=1,2,\ldots ,N)$ are constants to be determined later, provided $a_{N}\ne 0$ or $a_{-N}\ne 0.$

Step 3 :

We determine the positive integer N in (2.4) by balancing the highest-order derivatives and the nonlinear terms in Eq. (2.3).

Step 4 :

Substituting (2.4) along with Eq. (2.5) into Eq. (2.3) and collecting all the coefficients of $\left( \frac{G^{\prime }}{G}\right) ^{i}$ ($i=0,1,2,\ldots )$, then setting these coefficients to zero, yield a set of algebraic equations, which can be solved by using the Maple or Mathematica to find the values of $a_{i},R,$ Q,  P,  $\alpha ,$ $\beta ,$ $\gamma$ and $\upsilon$.

Step 5 :

It is well-known [41] that Eq. (2.5) has families of Jacobi elliptic function solutions as follows:

No	P	Q	R	$G(\xi )$
1	$m^{2}$	$-(1+m^{2})$	1	$\text{ sn }(\xi )$
2	$-m^{2}$	$2m^{2}-1$	$1-m^{2}$	cn$(\xi )$
3	$-1$	$2-m^{2}$	$m^{2}-1$	$\text{ dn }(\xi )$
4	$m^{2}$	$-(1+m^{2})$	1	cd$(\xi )$
5	$1-m^{2}$	$2-m^{2}$	1	sc$(\xi )$
6	$\dfrac{1}{4}$	$\dfrac{1-2m^{2}}{2}$	$\tfrac{1}{4}$	ns$(\xi )\pm$ cs$(\xi )$
7	$\tfrac{m^{2}}{4}$	$\dfrac{m^{2}-2}{2}$	$\tfrac{m^{2}}{4}$	$\dfrac{ \text{ sn }(\xi )}{1\pm \text{ dn }(\xi )}$
8	$\tfrac{1}{4}$	$\tfrac{m^{2}+1}{2}$	$\tfrac{( 1-m^{2}) ^{2}}{4}$	$m\text{ cn }(\xi )\pm \text{ dn }(\xi )$
9	1	$2-4m^{2}$	1	sc$(\xi )\text{ dn }(\xi )$
10	$m^{4}$	$2m^{2}-4$	1	sd$(\xi )\text{ cn }(\xi )$

In this table, $\text{ sn }\xi =\text{ sn }(\xi ,m)$, $\text{ cn }\xi =\text{ cn } (\xi ,m)$, $\text{ dn }\xi =\text{ dn }(\xi ,m)$, ns$\xi =$ ns$(\xi ,m)$, cs$\xi =$ cs$(\xi ,m)$, ds$\xi =$ ds$(\xi ,m)$, sc$\xi =$ sc$(\xi ,m)$, sd$\xi =$ sd$(\xi ,m)$ are the Jacobi elliptic function with modulus m, where $0<m<1$. These functions degenerate into hyperbolic functions when $m\rightarrow 1$ as follows:

$\text{ sn }\xi \rightarrow \tanh \xi ,$ $\text{ cn }\xi \rightarrow \sec$h$\xi$, $\text{ dn }\xi \rightarrow \sec$h$\xi$, ns$\xi =\coth \xi$, cs$\xi = \text{ csch }\xi ,$ ds$\xi =\text{ csch }\xi ,$ sc$\xi =\sinh \xi ,$ sd$\xi =\sinh \xi ,$nc$\xi =\cosh \xi$ and into trigonometric functions when $m\rightarrow 0$ as follows:

$\text{ sn }\xi \rightarrow \sin \xi ,$ $\text{ cn }\xi \rightarrow \cos \xi ,$ $\text{ dn }\xi \rightarrow 1,$ ns$\xi \rightarrow \csc \xi ,$ cs$\xi \rightarrow \cot \xi ,$ ds$\xi \rightarrow \csc \xi ,$ sc$\xi \rightarrow \tan \xi ,$ sd$\xi \rightarrow \sin \xi$, nc$\xi \rightarrow \sec \xi .$

Also, these functions satisfy the following formulas:

$\text{ sn }^{2}\xi +\text{ cn }^{2}\xi =1,$ $\text{ dn }^{2}\xi +m^{2}\text{ sn } ^{2}\xi =1,$

and $\text{ sn }^{\prime }\xi =\text{ cn }\xi \text{ dn }\xi ,$ $\text{ cn }^{\prime }\xi =-\text{ sn }\xi \text{ dn }\xi ,$ $\text{ dn }^{\prime }\xi =-m^{2}\text{ sn } \xi \text{ cn }\xi ,$ cd$^{\prime }\xi =-(1-m^{2})$sd$\xi$nd$\xi ,$ ns$^{\prime }\xi =-$cs$\xi$ds$\xi ,$ dc$^{\prime }\xi =(1-m^{2})$nc$\xi$sc$\xi ,$ $\text{ cn }^{\prime }\xi =$ sc$\xi$dc$\xi ,$ nd$^{\prime }\xi =m^{2}$cd $\xi$sd$\xi ,$ sc$^{\prime }\xi =$ dc$\xi$nc$\xi ,$ cs$^{\prime }\xi =-$ns$\xi$ds$\xi ,$ ds$^{\prime }\xi =-$cs$\xi$ns$\xi ,$ sd$^{\prime }\xi =$ nd$\xi$cd$\xi ,$

where $^{\prime }=\frac{d}{d\xi }.$

Step 6 :

Substituting the values of $a_{i},R,$ Q,  P,  $\alpha ,$ $\beta ,$ $\gamma$ and $\upsilon$ as well as the solutions of Eq. (2.5) listed in Step 5, into (2.4) we have the exact solutions of Eq. (2.1).

3 Exact solutions of the QZK equation (1.1) using the extended generalized $(\frac{G^{\prime }}{G})$-expansion method

In this section, we apply the extended generalized $(\frac{ G^{\prime }}{G})$-expansion method to find families of new Jacobi elliptic function solutions of Eq. (1.1). To this aim, we use the wave transformation (2.2) to reduce Eq. (1.1) into the following nonlinear ODE:

$$\begin{aligned} V\phi ^{\prime }-A\phi \phi ^{\prime }-\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) \phi ^{\prime \prime \prime }=0, \end{aligned}$$

(3.1)

where $V=\dfrac{\upsilon }{\gamma }.$

Integrating Eq. (3.1) once with respect to $\xi$ and vanishing the constant of integration, we find the following ODE:

$$\begin{aligned} V\phi -\frac{1}{2}A\phi ^{2}-\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) \phi ^{\prime \prime }=0. \end{aligned}$$

(3.2)

Balancing $\phi ^{\prime \prime }$ with $\phi ^{2}$ gives $N=2$. Therefore, (2.4) reduces to

$$\begin{aligned} \phi (\xi )=a_{0}+a_{1}\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) +a_{2}\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{2}+a_{-1}\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{-1}+a_{-2}\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{-2}, \end{aligned}$$

(3.3)

where $a_{0}$, $a_{1}$, $a_{2}$, $a_{-1}$ and $a_{-2}$ are constants to be determined such that $a_{2}\ne 0$ or $a_{-2}\ne 0$.

From (2.5) and (3.3) we deduce that

$$\begin{aligned} \phi ^{\prime \prime }(\xi )&=2a_{1}\left( \frac{G^{\prime }(\xi )}{G(\xi ) }\right) \left[ \left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{2}-Q\right] \nonumber \\&\quad +2a_{2}\left[ 3\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{4}-4Q\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{2}+Q^{2}-4RP\right] \nonumber \\&\quad +2a_{-1}\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{-1}\left[ \left( Q^{2}-4RP\right) \left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{-2}-Q \right] \nonumber \\&\quad +2a_{-2}\left[ \left( 3Q^{2}-12RP\right) \left( \frac{G^{\prime }(\xi )}{ G(\xi )}\right) ^{-4}-4Q\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{-2}+1\right] , \end{aligned}$$

(3.4)

Substituting (3.3) and (3.4) into Eq. (3.2) and collecting all the coefficients of $( \tfrac{G^{\prime }}{G}) ^{i}$ ,  $(i=0,\pm 1,\pm 2,\pm 3,\pm 4)$ and setting them to zero, we have the following algebraic equations:

• $\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{4}:Aa_{2}^{2}+12\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) a_{2}=0,$

• $\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{3}:4\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) a_{1}+2Aa_{1}a_{2}=0,$

• $\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{2}:Aa_{1}^{2}-2Va_{2}-16\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) Qa_{2}+2Aa_{0}a_{2}=0,$

• $\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) :2Va_{1}+4\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) Qa_{1}-2Aa_{0}a_{1}-2Aa_{2}a_{-1}=0,$

• $\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{0}:4\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) a_{2}Q^{2}+Aa_{0}^{2}-2Va_{0}+4\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) a_{-2}$

•       $+\,2Aa_{1}a_{-1}+2Aa_{2}a_{-2}-16\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) PRa_{2}=0,$

• $\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{-1}:2Va_{-1}+4\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) Qa_{-1}-2Aa_{0}a_{-1}-2Aa_{1}a_{-2}=0,$

• $\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{-2}:Aa_{-1}^{2}-2Va_{-2}-16\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) Qa_{-2}+2Aa_{0}a_{-2}=0,$

• $\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{-3}:4\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) a_{-1}Q^{2}+2Aa_{-1}a_{-2}-16\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) PRa_{-1}=0,$

• $\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{-4}:Aa_{-2}^{2}+12\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) Q^{2}a_{-2}-48\left( B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2})\right) PRa_{-2}=0.$

On solving the above algebraic equations with the aid of Maple or Mathematica, we have the following results:

Result 1

$$\begin{aligned} a_{0}= & {} \frac{4\left( 2Q\pm \sqrt{Q^{2}+12RP}\right) \left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A},\quad a_{1}=0,a_{-1}=0, \nonumber \\ a_{2}= & {} -\frac{12\left[ B\gamma ^{2}+C\left( \alpha ^{2}+\beta ^{2}\right) \right] }{A},\quad a_{-2}=0,\alpha =\alpha ,\beta =\beta ,\gamma =\gamma , \nonumber \\ V= & {} \pm 4\left[ B\gamma ^{2}+C\left( \alpha ^{2}+\beta ^{2}\right) \right] \sqrt{Q^{2}+12RP}. \end{aligned}$$

(3.5)

Substituting (3.5) into (3.3) yields

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ \left( 2Q\pm \sqrt{Q^{2}+12RP}\right) -3\left( \frac{G^{\prime }(\xi ) }{G(\xi )}\right) ^{2}\right] , \end{aligned}$$

(3.6)

where $\xi =\alpha X+\beta Y+\gamma Z\pm \left[ 4\gamma \left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{Q^{2}+12RP}\right] T.$

According to the table of Step 5, we deduce that the exact wave solutions of Eq. (1.1) as follows:

Case 1 Choosing $P=m^{2},Q=-(1+m^{2}),R=1$ and $G(\xi )= \text{ sn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ \left( -2(1+m^{2})\pm \sqrt{1+14m^{2}+m^{4}}\right) -3\text {cs} ^{2}(\xi )\text{ dn }^{2}(\xi )\right] , \end{aligned}$$

(3.7)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1+14m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then Eq. (1.1) has the trigonometric solutions

$$\begin{aligned} \phi (\xi )=-\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 1+3\cot ^{2}(\xi )\right] , \end{aligned}$$

(3.8)

where $\xi =\alpha X+\beta Y+\gamma Z+4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 1+\cot ^{2}(\xi )\right] , \end{aligned}$$

(3.9)

where $\xi =\alpha X+\beta Y+\gamma Z-4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

If $m\rightarrow 1,$ then Eq. (1.1) has the hyperbolic solutions

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \text{ sech }^{2}\left( \xi \right) \text{ csch }^{2}\left( \xi \right) , \end{aligned}$$

(3.10)

where $\xi =\alpha X+\beta Y+\gamma Z+16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 8+3\text{ sech }^{2}\left( \xi \right) \text{ csch }^{2}\left( \xi \right) \right] , \end{aligned}$$

(3.11)

where $\xi =\alpha X+\beta Y+\gamma Z-16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

Case 2 Choosing $P=-m^{2},Q=2m^{2}-1,R=1-m^{2}$ and $G(\xi )=\text{ cn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A } \nonumber \\&\times \left[ \left( 2(2m^{2}-1)\pm \sqrt{16m^{4}-16m^{2}+1}\right) -3 \text {sc}^{2}(\xi )\text{ dn }^{2}(\xi )\right] , \end{aligned}$$

(3.12)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{16m^{4}-16m^{2}+1}.$

If $m\rightarrow 0,$ then Eq. (1.1) has the trigonometric solutions

$$\begin{aligned} \phi (\xi )=-\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 1+3\tan ^{2}(\xi )\right] , \end{aligned}$$

(3.13)

where $\xi =\alpha X+\beta Y+\gamma Z+4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 1+\tan ^{2}(\xi )\right] , \end{aligned}$$

(3.14)

where $\xi =\alpha X+\beta Y+\gamma Z-4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

while if $m\rightarrow 1,$ then Eq. (1.1) has the hyperbolic solution (the bright 1-soliton solution):

$$\begin{aligned} \phi (\xi )=\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \text{ sech }^{2}\left( \xi \right) , \end{aligned}$$

(3.15)

where $\xi =\alpha X+\beta Y+\gamma Z+4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or the hyperbolic solution (the dark 1-soliton solution):

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 1-3\tanh ^{2}\left( \xi \right) \right] , \end{aligned}$$

(3.16)

where $\xi =\alpha X+\beta Y+\gamma Z-4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

Case 3 Choosing $P=-1,Q=2-m^{2},R=m^{2}-1$ and $G(\xi )= \text{ dn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A } \nonumber \\&\times \left[ \left( 2(2-m^{2})\pm \sqrt{16-16m^{2}+m^{4}}\right) -3m^{2} \text {sd}^{2}(\xi )\text{ cn }^{2}(\xi )\right] , \qquad \end{aligned}$$

(3.17)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{16-16m^{2}+m^{4}}.$

If $m\rightarrow 1,$ then we have the same hyperbolic function solutions (3.15) and (3.16).

Case 4 Choosing $P=m^{2},Q=-(1+m^{2}),R=1$ and $G(\xi )=$ cd$(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A } \nonumber \\&\times \left[ \left( -2(1+m^{2})\pm \sqrt{1+14m^{2}+m^{4}}\right) -3(1-m^{2})\text {sd}^{2}(\xi )\text {nc}^{2}(\xi )\right] , \nonumber \\ \end{aligned}$$

(3.18)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1+14m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then we have the same trigonometric solutions (3.13) and (3.14).

Case 5 Choosing $P=1-m^{2},Q=2-m^{2},R=1$ and $G(\xi )= \text{ sc }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A } \nonumber \\&\times \left[ \left( 2(2-m^{2})\pm \sqrt{16-16m^{2}+m^{4}}\right) -3\text { ds}^{2}(\xi )\text {nc}^{2}(\xi )\right] , \end{aligned}$$

(3.19)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{16-16m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then Eq. (1.1) has the trigonometric solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 8-3\sec ^{2}(\xi )\csc ^{2}(\xi )\right] , \end{aligned}$$

(3.20)

where $\xi =\alpha X+\beta Y+\gamma Z+16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \sec ^{2}(\xi )\csc ^{2}(\xi ), \end{aligned}$$

(3.21)

where $\xi =\alpha X+\beta Y+\gamma Z-16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

If $m\rightarrow 1,$ then Eq. (1.1) has the hyperbolic solution (the singular bright 1-soliton solution):

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \text{ csch }^{2}(\xi ), \end{aligned}$$

(3.22)

where $\xi =\alpha X+\beta Y+\gamma Z+4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or the hyperbolic solution (the singular dark 1-soliton solution):

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 1-3\coth ^{2}(\xi )\right] , \end{aligned}$$

(3.23)

where $\xi =\alpha X+\beta Y+\gamma Z-4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

Case 6 Choosing $P=\tfrac{1}{4},Q=\tfrac{1-2m^{2}}{2},R= \tfrac{1}{4}$ and $G(\xi )=$ ns$(\xi )~\pm ~$cs$(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ \left( (1-2m^{2})\pm \sqrt{1-m^{2}+m^{4}}\right) -3\text {ds}^{2}(\xi ) \right] , \qquad \end{aligned}$$

(3.24)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then Eq. (1.1) has the trigonometric solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 2-3\sec ^{2}(\xi )\right] , \end{aligned}$$

(3.25)

where $\xi =\alpha X+\beta Y+\gamma Z+4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \sec ^{2}(\xi ), \end{aligned}$$

(3.26)

where $\xi =\alpha X+\beta Y+\gamma Z-4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

If $m\rightarrow 1,$ then Eq. (1.1) has the hyperbolic solution (the singular bright 1-soliton solution):

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \text{ csch }^{2}(\xi ), \end{aligned}$$

(3.27)

where $\xi =\alpha X+\beta Y+\gamma Z+4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 2+3\text{ csch }^{2}(\xi )\right] , \end{aligned}$$

(3.28)

where $\xi =\alpha X+\beta Y+\gamma Z-4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

Case 7 Choosing $P=\tfrac{m^{2}}{4},Q=\tfrac{m^{2}-2}{2} ,R=\tfrac{m^{2}}{4}$ and $G(\xi )=\tfrac{\text{ sn }(\xi )}{1\pm \text{ dn }(\xi )}$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ \left( (m^{2}-2)\pm \sqrt{1-m^{2}+m^{4}}\right) -3\text {cs}^{2}(\xi ) \right] , \end{aligned}$$

(3.29)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 0$; then we have the same trigonometric function solutions (3.8) and (3.9), while if $m\rightarrow 1$; then we have the same hyperbolic function solutions (3.22) and (3.23).

Case 8 Choosing $P=\tfrac{1}{4},Q=\tfrac{m^{2}+1}{2},R= \tfrac{\left( 1-m^{2}\right) ^{2}}{4}$ and $G(\xi )=m\text{ cn }(\xi )\pm \text{ dn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ \left( m^{2}+1\pm \sqrt{1-m^{2}+m^{4}}\right) -3m^{2}\text{ sn } ^{2}(\xi )\right] , \end{aligned}$$

(3.30)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 1,$ then we have the same hyperbolic solutions (3.15) and (3.16).

Case 9 Choosing $P=1,Q=2-4m^{2},R=1$ and $G(\xi )=$ sc$(\xi )\text{ dn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 4\left( 1-2m^{2}\pm \sqrt{1-m^{2}+m^{4}}\right) \right. \nonumber \\&\left. -3\left( \frac{m^{2}\text{ cn }^{2}(\xi )\text{ sn }^{2}(\xi )-\text{ dn } ^{2}(\xi )}{\text{ cn }(\xi )\text{ sn }(\xi )\text{ dn }(\xi )}\right) ^{2}\right] \end{aligned}$$

(3.31)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 0$; then we have the same trigonometric function solutions (3.20) and (3.21), while if $m\rightarrow 1$; then we have the same hyperbolic function solutions (3.10) and (3.11).

Case 10 Choosing $P=m^{4},Q=2m^{2}-4,R=1$ and $G(\xi )=$ sd$(\xi )\text{ cn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 4\left( m^{2}-2\pm \sqrt{1-m^{2}+m^{4}}\right) \right. \nonumber \\&\left. -3\left( \text{ cn }(\xi )\text {ns}(\xi )\text {nd}(\xi )-\text{ sn } (\xi )\text{ dn }(\xi )\text {nc}(\xi )\right) ^{2}\right] \end{aligned}$$

(3.32)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then Eq. (1.1) has the trigonometric solutions

$$\begin{aligned} \phi (\xi )=-\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 4+3\left( \frac{\cos ^{2}(\xi )-\sin ^{2}(\xi )}{\sin (\xi )\cos (\xi )}\right) ^{2}\right] , \end{aligned}$$

(3.33)

where $\xi =\alpha X+\beta Y+\gamma Z+16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 4+\left( \frac{\cos ^{2}(\xi )-\sin ^{2}(\xi )}{\sin (\xi )\cos (\xi ) }\right) ^{2}\right] , \end{aligned}$$

(3.34)

where $\xi =\alpha X+\beta Y+\gamma Z-16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) ,$ while if $m\rightarrow 1,$ then we have the same hyperbolic function solutions (3.10) and (3.11).

Result 2

$$\begin{aligned} a_{0}= & {} \frac{4\left( 2Q\pm \sqrt{Q^{2}+12RP}\right) \left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A},\quad a_{1}=0,a_{-1}=0, \nonumber \\ a_{2}= & {} 0,a_{-2}=\frac{12\left( -Q^{2}+4RP\right) \left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A},\quad \alpha =\alpha ,\beta =\beta ,\gamma =\gamma , \nonumber \\ V= & {} \pm 4\left( B\gamma ^{2}+C\left( \alpha ^{2}+\beta ^{2}\right) \right) \sqrt{Q^{2}+12RP}. \end{aligned}$$

(3.35)

Substituting (3.35) into (3.3) yields

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A}\times \left[ \left( 2Q\pm \sqrt{Q^{2}+12RP}\right) \right. \nonumber \\&\left. +3\left( -Q^{2}+4RP\right) \left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{-2}\right] , \end{aligned}$$

(3.36)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{Q^{2}+12RP}.$

According to the table of Step 5, we deduce that the exact wave solutions of equation (1.1) as follows:

Case 1 Choosing $P=m^{2},Q=-(1+m^{2}),R=1$ and $G(\xi )= \text{ sn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ \left( -2(1+m^{2})\pm \sqrt{1+14m^{2}+m^{4}}\right) \right. \nonumber \\&\left. -\,3\left( m-1\right) ^{2}\left( m+1\right) ^{2}\text {sc}^{2}(\xi )\text {nd}^{2}(\xi ) \right] , \end{aligned}$$

(3.37)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1+14m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then we have the same trigonometric solutions (3.13) and (3.14).

Case 2 Choosing $P=-m^{2},Q=2m^{2}-1,R=1-m^{2}$ and $G(\xi )=\text{ cn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \nonumber \\&\times \left[ \left( 2(2m^{2}-1)\pm \sqrt{16m^{4}-16m^{2}+1}\right) -3 \text {cs}^{2}(\xi )nd^{2}(\xi )\right] , \quad \end{aligned}$$

(3.38)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{16m^{4}-16m^{2}+1}.$

If $m\rightarrow 0,$ then we have the same trigonometric solutions (3.8) and (3.9), while if $m\rightarrow 1$; then we have the same hyperbolic function solutions (3.22) and (3.23).

Case 3 Choosing $P=-1,Q=2-m^{2},R=m^{2}-1$ and $G(\xi )= \text{ dn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \nonumber \\&\times \left[ \left( 2(2-m^{2})\pm \sqrt{16-16m^{2}+m^{4}}\right) -3\text { ds}^{2}(\xi )\text {nc}^{2}(\xi )\right] , \end{aligned}$$

(3.39)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{16-16m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then we have the same trigonometric function solutions (3.20) and (3.21), while if $m\rightarrow 1$; then we have the same hyperbolic function solutions (3.22) and (3.23).

Case 4 Choosing $P=m^{2},Q=-(1+m^{2}),R=1$ and $G(\xi )=$ cd$(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \nonumber \\&\times \left[ \left( -2(1+m^{2})\pm \sqrt{1+14m^{2}+m^{4}}\right) -3\text { ds}^{2}(\xi )\text{ cn }^{2}(\xi )\right] , \end{aligned}$$

(3.40)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1+14m^{2}+m^{4}}.$

If $m\rightarrow 0$; then we have the same trigonometric function solutions (3.8) and (3.9), while if $m\rightarrow 1$; then we have the same hyperbolic function solutions (3.10) and (3.11).

Case 5 Choosing $P=1-m^{2},Q=2-m^{2},R=1$ and $G(\xi )= \text{ sc }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \nonumber \\&\times \left[ \left( 2(2-m^{2})\pm \sqrt{16-16m^{2}+m^{4}}\right) -3m^{4} \text {sd}^{2}(\xi )\text{ cn }^{2}(\xi )\right] , \end{aligned}$$

(3.41)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{16-16m^{2}+m^{4}}.$

If $m\rightarrow 1$; then we have the same hyperbolic function solutions (3.15) and (3.16).

Case 6 Choosing $P=\tfrac{1}{4},Q=\tfrac{1-2m^{2}}{2},R= \tfrac{1}{4}$ and $G(\xi )=$ ns$(\xi )~\pm ~$cs$(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ \left( (1-2m^{2}){\pm } \sqrt{1-m^{2}+m^{4}}\right) -3m^{2}(m^{2}-1) \text {sd}^{2}(\xi )\right] , \end{aligned}$$

(3.42)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

Case 7 Choosing $P=\tfrac{m^{2}}{4},Q=\tfrac{m^{2}-2}{2} ,R=\tfrac{m^{2}}{4}$ and $G(\xi )=\tfrac{\text{ sn }(\xi )}{1\pm \text{ dn }(\xi )}$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ \left( (m^{2}-2){\pm } \sqrt{1-m^{2}+m^{4}}\right) +3(m^{2}-1)\text {sc} ^{2}(\xi )\right] , \end{aligned}$$

(3.43)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 0$; then we have the same trigonometric function solutions (3.13) and (3.14).

Case 8 Choosing $P=\tfrac{1}{4},Q=\tfrac{m^{2}+1}{2},R= \tfrac{\left( 1-m^{2}\right) ^{2}}{4}$ and $G(\xi )=m\text{ cn }(\xi )\pm \text{ dn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ \left( m^{2}+1\pm \sqrt{1-m^{2}+m^{4}}\right) -3\text {ns}^{2}(\xi ) \right] , \end{aligned}$$

(3.44)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then Eq. (1.1) has the trigonometric solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 2-3\csc ^{2}(\xi )\right] , \end{aligned}$$

(3.45)

where $\xi =\alpha X+\beta Y+\gamma Z+4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \csc ^{2}(\xi ), \end{aligned}$$

(3.46)

where $\xi =\alpha X+\beta Y+\gamma Z-4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

If $m\rightarrow 1,$ then we have the same hyperbolic solutions (3.22) and (3.23).

Case 9 Choosing $P=1,Q=2-4m^{2},R=1$ and $G(\xi )=$ sc$(\xi )\text{ dn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 4\left( 1-2m^{2}\pm \sqrt{1-m^{2}+m^{4}}\right) \right. \nonumber \\&\left. -\,48m^{2}(m^{2}-1)\left[ m^{2}\text{ cn }(\xi )\text{ sn }(\xi )\text {nd} (\xi )-\text{ dn }(\xi )\text {nc}(\xi )\text {ns}(\xi )\right] ^{-2}\right] \qquad \end{aligned}$$

(3.47)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

Case 10 Choosing $P=m^{4},Q=2m^{2}-4,R=1$ and $G(\xi )=$ sd$(\xi )\text{ cn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 4\left( m^{2}-2\pm \sqrt{1-m^{2}+m^{4}}\right) \right. \nonumber \\&\left. +\,48(m^{2}-1)\left[ \text{ cn }(\xi )\text {ns}(\xi )\text {nd}(\xi )- \text{ sn }(\xi )\text{ dn }(\xi )\text {nc}(\xi )\right] ^{-2}\right] \end{aligned}$$

(3.48)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then Eq. (1.1) has the trigonometric solutions

$$\begin{aligned} \phi (\xi )=-\frac{16\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 1+12\left( \frac{\sin (\xi )\cos (\xi )}{\cos ^{2}(\xi )-\sin ^{2}(\xi )}\right) ^{2}\right] , \end{aligned}$$

(3.49)

where $\xi =\alpha X+\beta Y+\gamma Z+16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{48\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 1+4\left( \frac{\sin (\xi )\cos (\xi )}{\cos ^{2}(\xi )-\sin ^{2}(\xi )}\right) ^{2}\right] , \end{aligned}$$

(3.50)

where $\xi =\alpha X+\beta Y+\gamma Z-16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right)$.

Result 3

$$\begin{aligned} a_{0}= & {} \frac{8\left( Q\pm 2\sqrt{Q^{2}-3RP}\right) \left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A},\quad a_{1}=0,a_{-1}=0, \nonumber \\ a_{2}= & {} -\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} ,\quad a_{-2}=\frac{12\left( -Q^{2}+4RP\right) \left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A}, \nonumber \\ \alpha= & {} \alpha ,\beta =\beta ,\gamma =\gamma ,V=\pm 16\left( B\gamma ^{2}+C\left( \alpha ^{2}+\beta ^{2}\right) \right) \sqrt{Q^{2}-3RP}. \end{aligned}$$

(3.51)

Substituting (3.51) into (3.3) yields

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 2\left( Q\pm 2\sqrt{Q^{2}-3RP}\right) -3\left( \frac{G^{\prime }(\xi )}{G(\xi )}\right) ^{2}\right. \nonumber \\&\quad \times \left. 3\left( -Q^{2}+4RP\right) \left( \frac{G^{\prime }(\xi )}{G(\xi )} \right) ^{-2}\right] , \end{aligned}$$

(3.52)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{Q^{2}+12RP}.$

According to the table of Step 5, we deduce that the exact wave solutions of equation (1.1) as follows:

Case 1 Choosing $P=m^{2},Q=-(1+m^{2}),R=1$ and $G(\xi )= \text{ sn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ \left( -2(1+m^{2}){\pm } 4\sqrt{1-m^{2}+m^{4}}\right) -3\text {cs} ^{2}(\xi )\text{ dn }^{2}(\xi )\right. \nonumber \\&\left. -\,3(m-1)^{2}(m+1)^{2}\text {sc}^{2}(\xi )\text {nd}^{2}(\xi )\right] \end{aligned}$$

(3.53)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then Eq. (1.1) has the trigonometric solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 2-3\left( \tan ^{2}(\xi )+\cot ^{2}(\xi )\right) \right] , \end{aligned}$$

(3.54)

where $\xi =\alpha X+\beta Y+\gamma Z+16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 2+\left( \tan ^{2}(\xi )+\cot ^{2}(\xi )\right) \right] , \end{aligned}$$

(3.55)

where $\xi =\alpha X+\beta Y+\gamma Z-16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

Case 2 Choosing $P=-m^{2},Q=2m^{2}-1,R=1-m^{2}$ and $G(\xi )=\text{ cn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 2\left( (2m^{2}-1)\pm 2\sqrt{m^{4}-m^{2}+1}\right) \right. \nonumber \\&\left. -\,3\left( \text {sc}^{2}(\xi )\text{ dn }^{2}(\xi )+\text {cs}^{2}(\xi ) \text {nd}^{2}(\xi )\right) \right] , \end{aligned}$$

(3.56)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{m^{4}-m^{2}+1}.$

If $m\rightarrow 0$; then we have the same trigonometric function solutions (3.54) and (3.55).

while if $m\rightarrow 1,$ then Eq. (1.1) has the hyperbolic solutions

$$\begin{aligned} \phi (\xi )=\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 2-3\left( \tanh ^{2}\left( \xi \right) +\coth ^{2}\left( \xi \right) \right) \right] , \end{aligned}$$

(3.57)

where $\xi =\alpha X+\beta Y+\gamma Z+16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 2+\tanh ^{2}\left( \xi \right) +\coth ^{2}\left( \xi \right) \right] , \end{aligned}$$

(3.58)

where $\xi =\alpha X+\beta Y+\gamma Z-16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

Case 3 Choosing $P=-1,Q=2-m^{2},R=m^{2}-1$ and $G(\xi )= \text{ dn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 2\left( (2-m^{2})\pm 2\sqrt{1-m^{2}+m^{4}}\right) \right. \nonumber \\&\left. -\,3m^{4}\text {sd}^{2}(\xi )\text{ cn }^{2}(\xi )-3m^{-4}\text {ds} ^{2}(\xi )\text {nc}^{2}(\xi )\right] , \end{aligned}$$

(3.59)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 1,$ then Eq. (1.1) has the hyperbolic solutions

$$\begin{aligned} \phi (\xi )=\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 2-\left( \tanh ^{2}\left( \xi \right) +\coth ^{2}\left( \xi \right) \right) \right] , \end{aligned}$$

(3.60)

where $\xi =\alpha X+\beta Y+\gamma Z+16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )=-\frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A} \left[ 2+3\left( \tanh ^{2}\left( \xi \right) +\coth ^{2}\left( \xi \right) \right) \right] , \end{aligned}$$

(3.61)

where $\xi =\alpha X+\beta Y+\gamma Z-16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

Case 4 Choosing $P=m^{2},Q=-(1+m^{2}),R=1$ and $G(\xi )=$ cd$(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 2\left( -(1+m^{2})\pm 2\sqrt{1-m^{2}+m^{4}}\right) \right. \nonumber \\&\left. -\,3(1-m^{2})^{2}\text {sd}^{2}(\xi )\text {nc}^{2}(\xi )-3\text {ds} ^{2}(\xi )\text {cn}^{2}(\xi )\right] , \end{aligned}$$

(3.62)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 0,$ then we have the same trigonometric solutions (3.54) and (3.55).

Case 5 Choosing $P=1-m^{2},Q=2-m^{2},R=1$ and $G(\xi )= \text{ sc }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 2\left( (2-m^{2})\pm 2\sqrt{1-m^{2}+m^{4}}\right) \right. \nonumber \\&\left. -\,3\text {ds}^{2}(\xi )\text {nc}^{2}(\xi )-3m^{4}\text {sd}^{2}(\xi ) \text{ cn }^{2}(\xi )\right] , \end{aligned}$$

(3.63)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-m^{2}+m^{4}}.$

If $m\rightarrow 1,$ then we have the same hyperbolic solutions (3.60) and (3.61).

Case 6 Choosing $P=\tfrac{1}{4},Q=\tfrac{1-2m^{2}}{2},R= \tfrac{1}{4}$ and $G(\xi )=$ns$(\xi )\pm$cs$(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ \left( (1-2m^{2})\pm 4\sqrt{\frac{1}{16}-m^{2}+m^{4}}\right) \right. \nonumber \\&\left. -\,3\text {ds}^{2}(\xi )-3m^{2}(m^{2}-1)\text {sd}^{2}(\xi )\right] , \end{aligned}$$

(3.64)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{\frac{1}{16}-m^{2}+m^{4}}.$

Case 7 Choosing $P=\tfrac{m^{2}}{4},Q=\tfrac{m^{2}-2}{2} ,R=\tfrac{m^{2}}{4}$ and $G(\xi )=\tfrac{\text{ sn }(\xi )}{1\pm \text{ dn }(\xi )}$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ \left( (m^{2}-2)\pm 4\sqrt{\frac{1}{16}-m^{2}+m^{4}}\right) \right. \nonumber \\&\left. -\,3\text {cs}^{2}(\xi )-3(m^{2}-1)\text {sc}^{2}(\xi )\right] , \end{aligned}$$

(3.65)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{\frac{1}{16}-m^{2}+m^{4}}.$

If $m\rightarrow 0$; then we have the same trigonometric function solutions (3.54) and (3.55).

Case 8 Choosing $P=\tfrac{1}{4},Q=\tfrac{m^{2}+1}{2},R= \tfrac{\left( 1-m^{2}\right) ^{2}}{4}$ and $G(\xi )=m\text{ cn }(\xi )\pm \text{ dn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ \left( m^{2}+1\pm \sqrt{1+14m^{2}+m^{4}}\right) \right. \nonumber \\&\left. -\,3\left( m^{2}\text{ sn }^{2}(\xi )+\text {ns}^{2}(\xi )\right) \right] , \end{aligned}$$

(3.66)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 4\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1+14m^{2}+m^{4}}.$

If $m\rightarrow 1,$ then we have the same hyperbolic solutions (3.60) and (3.61).

Case 9 Choosing $P=1,Q=2-4m^{2},R=1$ and $G(\xi )=$sc$(\xi )\text{ dn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 4\left( 1-2m^{2}\pm 4\sqrt{1-16m^{2}+16m^{4}}\right) \right. \nonumber \\&\left. -\,3\left( m^{2}\text{ cn }(\xi )\text{ sn }(\xi )\text {nd}(\xi )-\text{ dn } (\xi )\text {nc}(\xi )\text {ns}(\xi )\right) ^{2}\right. \nonumber \\&\left. -\,48m^{2}(m^{2}-1)\left( m^{2}\text{ cn }(\xi )\text{ sn }(\xi )\text {nd} (\xi )-\text{ dn }(\xi )\text {nc}(\xi )\text {ns}(\xi )\right) ^{-2}\right] , \qquad \quad \end{aligned}$$

(3.67)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-16m^{2}+16m^{4}}.$

Case 10 Choosing $P=m^{4},Q=2m^{2}-4,R=1$ and $G(\xi )=$ sd$(\xi )\text{ cn }(\xi )$, we obtain the Jacobi elliptic function solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 4\left( m^{2}-2\pm 4\sqrt{1-16m^{2}+16m^{4}}\right) \right. \nonumber \\&\left. -\,3\left( \text{ cn }(\xi )\text {ns}(\xi )\text {nd}(\xi )-\text{ sn } (\xi )\text{ dn }(\xi )\text {nc}(\xi )\right) ^{2}\right. \nonumber \\&\left. -\,48(m^{2}-1)\left( \text{ cn }(\xi )\text {ns}(\xi )\text {nd}(\xi )- \text{ sn }(\xi )\text{ dn }(\xi )\text {nc}(\xi )\right) ^{-2}\right] \end{aligned}$$

(3.68)

where $\xi =\alpha X+\beta Y+\gamma Z\pm 16\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) \sqrt{1-16m^{2}+16m^{4}}.$

If $m\rightarrow 0,$ then Eq. (1.1) has the trigonometric solutions

$$\begin{aligned} \phi (\xi )= & {} \frac{4\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A }\left[ 8-3\left( \frac{\cos ^{2}(\xi )-\sin ^{2}(\xi )}{\sin (\xi )\cos (\xi )}\right) ^{2}\right. \nonumber \\&\left. -\,48\left( \frac{\sin (\xi )\cos (\xi )}{\cos ^{2}(\xi )-\sin ^{2}(\xi )}\right) ^{2}\right] , \end{aligned}$$

(3.69)

where $\xi =\alpha X+\beta Y+\gamma Z+64\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

or

$$\begin{aligned} \phi (\xi )= & {} -\frac{12\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) }{A}\left[ 8+3\left( \frac{\cos ^{2}(\xi )-\sin ^{2}(\xi )}{\sin (\xi )\cos (\xi )}\right) ^{2}\right. \nonumber \\&\left. +16\left( \frac{\sin (\xi )\cos (\xi )}{\cos ^{2}(\xi )-\sin ^{2}(\xi )}\right) ^{2}\right] , \end{aligned}$$

(3.70)

where $\xi =\alpha X+\beta Y+\gamma Z-64\gamma T\left( B\gamma ^{2}+C\alpha ^{2}+C\beta ^{2}\right) .$

4 Further results

In this section, we obtain new solutions for Eq. (1.1) using a direct method as follows:

Equation (3.1) can be written in the form

$$\begin{aligned} \phi ^{\prime \prime \prime }+\frac{A}{E}\phi \phi ^{\prime }-\frac{V}{E} \phi ^{\prime }=0, \end{aligned}$$

(4.1)

where $E=B\gamma ^{2}+C(\alpha ^{2}+\beta ^{2}).$

Integrating Eq. (4.1) once with respect to $\xi ,$ we have

$$\begin{aligned} \phi ^{\prime \prime }+\frac{A}{2E}\phi ^{2}-\frac{V}{E}\phi =\frac{1}{2} K_{1}, \end{aligned}$$

(4.2)

where $K_{1}$ is a constant of integration.

Multiplying both sides of Eq. (4.2) by $\phi ^{\prime }(\xi )$ and integrating, we get

$$\begin{aligned} (\phi ^{\prime })^{2}=-\frac{A}{3E}\phi ^{3}+\frac{V}{E}\phi ^{2}+K_{1}\phi +K_{2}, \end{aligned}$$

(4.3)

where $K_{2}$ is a constant of integration.

Let us now discuss the two cases:

Case 1 If $K_{1}=K_{2}=0,$ then Eq. (1.1) has the bright 1-soliton solution

$$\begin{aligned} \phi (\xi )=\frac{3V}{A}\text{ sech }{}^{2}\left[ \frac{1}{2}\sqrt{\frac{V}{E}} (\xi +\xi _{0})\right] , \end{aligned}$$

(4.4)

where $\xi _{0}$ is a constant and $\frac{V}{E}>0.$

Case 2 If $K_{1}\ne 0$ and $K_{2}\ne 0,$ we choose

$$\begin{aligned} \phi (\xi )=\frac{V}{A}-\frac{12E}{A}P_{1}(\xi ), \end{aligned}$$

(4.5)

where $P_{1}(\xi )$ satisfies the Weierstrass equation [23, 42]:

$$\begin{aligned} P_{1}^{\prime 2}=4P_{1}^{3}-g_{2}P_{1}-g_{3}, \end{aligned}$$

(4.6)

where $g_{2}=\frac{1}{12}(\frac{A}{E}K_{1}+\frac{V^{2}}{E^{2}}) ,g_{3}=\frac{V^{3}}{216E^{3}}+\frac{1}{144}(\frac{AV}{E^{2}}K_{1}- \frac{A^{2}}{E^{2}}K_{2})$.

If $g_{2}=g_{3}=0,$ then Eq. (4.6) reduces to the equation $P_{1}^{\prime 2}=4P_{1}^{3},$which admits the elementary solution

$$\begin{aligned} P_{1}(\xi )=\frac{1}{(\xi +\xi _{1})^{2}}, \end{aligned}$$

(4.7)

where $\xi _{1}$ is a constant of integration. From (4.5) and (4.7) we have the rational solution of Eq. (1.1) in the form:

$$\begin{aligned} \phi (\xi )=\frac{V}{A}-\frac{12E}{A(\xi +\xi _{1})^{2}}, \end{aligned}$$

(4.8)

From the well-known solutions of Eq. (4.6), we write the solution of Eq. (1.1) in terms of the Jacobi elliptic functions as follows:

$$\begin{aligned} \phi (\xi )=\frac{V}{A}-\frac{12E}{A}\left[ e_{2}-(e_{2}-e_{3})\text {cn} ^{2}(R\xi ,m)\right] , \end{aligned}$$

(4.9)

or

$$\begin{aligned} \phi (\xi )=\frac{V}{A}-\frac{12E}{A}\left[ e_{3}+(e_{1}-e_{3})\text {ns} ^{2}(R\xi ,m)\right] , \end{aligned}$$

(4.10)

where $R=\sqrt{e_{1}-e_{3}},$ while $m=\sqrt{\frac{e_{2}-e_{3}}{e_{1}-e_{3}} }$ is the modulus of the Jacobi elliptic function $e_{i}$ ,$\left( i=1,2,3\right)$ are three roots of the cubic equation $4z^{3}-g_{2}z-g_{3}=0$ such that $e_{1}>e_{2}>e_{3}$ and $0<m<1.$

When $m\rightarrow 1$, we get cn$(R\xi ,m)\rightarrow \sec$h$(R\xi )$ and ns $(R\xi ,m)\rightarrow \coth (R\xi ).$

Consequently, we have the bright 1-soliton solution

$$\begin{aligned} \phi (\xi )=\frac{V}{A}-\frac{12E}{A}\left[ e_{2}-(e_{2}-e_{3})\text {sech} ^{2}(R\xi )\right] , \end{aligned}$$

(4.11)

or the singular dark 1-soliton solution

$$\begin{aligned} \phi (\xi )=\frac{V}{A}-\frac{12E}{A}\left[ e_{3}+(e_{1}-e_{3})\coth ^{2}(R\xi )\right] , \end{aligned}$$

(4.12)

respectively

Note that our solutions (4.9)–(4.12) are new and not reported elsewhere.

5 Conclusions

In this article, we have solved the nonlinear quantum Zakharov–Kuznetsov equation (1.1) using the extended generalized $(\frac{G^{\prime }}{ G})$-expansion method combined with the Jacobi elliptic equation described in Sect. 2. By the aid of Maple or Mathematica, we have found many solutions of Eq. (1.1) which are new and not found elsewhere. In Sect. 4, we have solved Eq. (1.1) using a direct method which gives many other solutions. On comparing our results with the results obtain in [37–39] using other different methods, we deduce that our results are different and new. Finally, all solutions obtained in this article have been checked with the Maple by putting them back into the original equations.