Skip to main content


We’d like to understand how you use our websites in order to improve them. Register your interest.

Solitons for the modified \((2 + 1)\)-dimensional Konopelchenko–Dubrovsky equations


In the paper, we consider the modified \((2 + 1)\)-dimensional Konopelchenko–Dubrovsky equations which possess high order nonlinear terms. Under the aid of Maple, we derive the exact traveling wave solutions of the mKDs by the auxiliary equation approach. Under some special conditions, Jacobi elliptic function solutions, degenerated triangular function solutions, and solitons for the mKD equations are constructed.


Konopelchenko and Dubrovsky [1] ever presented a \((2 + 1)\)-dimensional Konopelchenko and Dubrovsky (KD) model as follows:

$$\begin{aligned} \textstyle\begin{cases} u_{t}-u_{xxx}-6b_{0}uu_{x}+\frac{3}{2}a_{0}^{2}u^{2}u_{x}-3v_{y}+3a _{0}u_{x}v=0, \\ u_{y}=v_{x}, \end{cases}\displaystyle \end{aligned}$$

where u and v are two analytic functions corresponding to the variables x, y, and t; \(a_{0}\) and \(b_{0}\) are the real parameters. If \(a_{0}=0\), Eq. (1) reshapes into the well-known Kadomtsev–Petviashvili (KP) equation; If \(b_{0}=0\), it turns into the modified KP equation; If \(u_{y}=0\), the first row of Eq. (1) recasts into the Gardner equation, the combination of KdV and modified KdV [2].

In the field of nonlinear partial differential equations (PDEs), one knows that all kinds of solutions for these equations are of particular interest to scientists [3,4,5,6,7,8,9,10,11,12,13,14,15]. Particularly, many researchers have made use of some effective methods to study the exact solutions of KD Eq. (1). Special and general N-soliton solutions of Eq. (1) have been obtained via Hirota’s bilinear approach [16]. Bilinear-form Bäcklund transformation and single-soliton solutions have been presented by [17]. Based on the Hirota bilinear form of Eq. (1), the lump waves, solitary waves, as well as interaction between lump waves and solitary waves, have been obtained [18]. Based on the Hirota direct method and linear superposition principle, Eq. (1) was investigated to have the complexiton and resonant multiple wave solutions [19]. By means of the variable separation method and improved mapping approach, new types of variable-separation solutions (including solitary wave solutions, periodic wave solutions, and rational function solutions) for Eq. (1) have been successfully obtained [20]. Based on the similarity transformations approach together with Lie group theory, the KD system has been reshaped into another system of ordinary differential equations, new closed form solutions of Eq. (1) have been obtained [21]. The tanh-sech method, cosh-sinh method, and the exponential method were employed to discuss Eq. (1) by Wazwaz [22], and some exact solutions of distinct physical structures, solitons, kinks, and periodic wave solutions were derived.

Motivated by these works [1, 2, 19,20,21,22], the presented paper will study the following modified Konopelchenko and Dubrovsky (mKD) equations:

$$\begin{aligned} \textstyle\begin{cases} u_{t}-u_{xxx}-bu^{n}u_{x}+au^{2n}u_{x}-3v_{y}+ku_{x}v^{n}=0, \\ u_{y}=v_{x}, \end{cases}\displaystyle \end{aligned}$$

where a, b, k, and n are nonzero real constants. Furthermore, when \(n=1\), \(b=6b_{0}\), \(a=\frac{3}{2}a_{0}^{2}\), and \(k=3a_{0}\) in Eq. (2), it turns into Eq. (1). The paper will make use of the auxiliary differential equation approach to solve Eq. (2) analytically and obtains abundant new exact solutions of Eq. (2) in general and special cases. Moreover, some Jacobi elliptic function solutions will degenerate solitons and triangular function solutions when the module approaches to 0 or 1. It shows that most of the solutions provided in this work are different to those presented in [1, 16,17,18,19,20,21,22,23,24].

Method description

The main idea of the auxiliary equation approach is described as follows. Firstly, we introduce a transform \(\xi =\mu (x-ct)\) to a given partial differential equation (PDE) with dependent variable u, independent variables x and t. The PDE will be changed into an ordinary differential equation (ODE). Of course, μ and c should be nonzero. Secondly, we seek for the solution of the ODE by the following:

$$ u(\xi )=\sum_{i=0}^{n} h_{i}\omega ^{i}(\xi ), $$

where the integer n and the constants \(h_{0}, h_{1},\ldots , h_{n}\) are unknown. For nonnegative integers q and p, the highest order of \(\frac{\partial ^{p} u}{\partial \xi ^{p}}\) and \(u^{q}\frac{\partial ^{p} u}{\partial \xi ^{p}}\) are \(n+p\) and \(qn+p\), respectively. Balancing the highest order in ODE, n will be determined. Thirdly, we introduce an expression \(\omega (\xi )\) defined as

$$ \biggl(\frac{d \omega }{d\xi }\biggr)^{2}=c_{1}+c_{2} \omega ^{2}+\frac{c_{3}}{2} \omega ^{4}, $$

where \(c_{i}\in R\) (\(i=1, 2, 3\)). Substituting Eqs. (3) and (4) into ODE and then forcing the coefficients of each power of \(\omega (\xi )\) to be zero, several algebraic expressions will be obtained. Solving these expressions with the aid of Maple, some exact solutions for PDE can be obtained.

Exact solutions to Eq. (2)

Introducing \(\xi =\mu (x+y-ct)\) (where \(c \neq 0\), \(\mu \in R\)) into (2), we obtain the following form:

$$\begin{aligned} \textstyle\begin{cases} -cu_{\xi }-\mu ^{2}u_{\xi \xi \xi }-b(u^{n})u_{\xi }+ a(u^{2n})u_{ \xi }-3v_{\xi }+ku_{\xi }v^{n}=0, \\ u_{\xi }=v_{\xi }. \end{cases}\displaystyle \end{aligned}$$

Substituting \(u=v\) into the first equation of (5) yields

$$ -(3+c)u'-\mu ^{2}u'''+ a\bigl(u^{2n}\bigr)u'+(k-b) u^{n} u'=0. $$

Using a transform \(u^{n}=W\), we know that

$$\begin{aligned} \textstyle\begin{cases} u'={\frac{1}{n-1}}^{{\frac{1}{n-1}}-1}W' , \\ u''={\frac{1}{n-1}}({\frac{1}{n-1}}-1)W^{{\frac{1}{n-1}}-2}(W')^{2}+ {\frac{1}{n-1}}W^{{\frac{1}{n-1}}-1}W'', \\ u'''=\frac{(2-n)(3-2n)}{(n-1)^{3}}W^{{\frac{1}{n-1}}-3}W^{\prime \,2}+ \frac{3(2-n)}{(n-1)^{2}}W^{{\frac{1}{n-1}}-2}W'W''+\frac{1}{n-1}W^{ {\frac{1}{n-1}}-1}W'''. \end{cases}\displaystyle \end{aligned}$$

Eq. (6) can be rewritten as

$$\begin{aligned}& -(c+3)W^{2}W'-\mu ^{2}\biggl( \frac{1}{n}-1\biggr) \biggl(\frac{1}{n}-2\biggr)W^{\prime \,3}-3 \mu ^{2}\biggl( \frac{1}{n}-1\biggr)WW'W''- \mu ^{2}W^{2}W''' \\& \quad {}+(k-b)W^{3}W'+aW^{4}W'=0, \end{aligned}$$

hence \(n=1\), Eq. (8) has the following solution:

$$ W(\xi )=h_{0}+h_{1}\omega , $$

where \(\omega (\xi )\) satisfies the sub-equation \((\frac{d \omega }{d \xi })^{2}=c_{1}+c_{2}\omega ^{2}+\frac{c_{3}}{2}\omega ^{4}\), some of the solutions can be expressed in Table 1 (also listed in [23] and [24]).

Table 1 Different solutions of the sub-equation

Substituting Eq. (9) and the expression \((\frac{d \omega }{d\xi })^{2}=c_{1}+c_{2}\omega ^{2}+\frac{c _{3}}{2}\omega ^{4}\) into Eq. (8) and equating to zero the coefficients of all powers of \(\omega (\xi )\) yields

$$\begin{aligned}& -a h_{1}^{5}+\frac{\mu ^{2}}{2}\biggl( \frac{1}{n}-1\biggr) \biggl(\frac{1}{n}-2\biggr)h_{1} ^{3}c_{3}+3\biggl(\frac{1}{n}-1\biggr)\mu ^{2}h_{1}^{3}c_{3}+3\mu ^{2}h_{1}^{3}c _{3}=0, \end{aligned}$$
$$\begin{aligned}& \bigl[(b-k) h_{1}-2a h_{0}h_{1} \bigr]h_{1}^{3}-2a h_{0}h_{1}^{4}+3 \biggl( \frac{1}{p}-1\biggr)\mu ^{2}h_{0}h_{1}^{2}c_{3}+6 \mu ^{2}h_{0}h_{1}^{2}c_{3}=0, \end{aligned}$$
$$\begin{aligned}& ch_{1}^{3}+\bigl[3 +(b-k) h_{0}- a h_{0}^{2}\bigr]h_{1}^{3}+2 \bigl[(b-k) h_{1}-2 a h_{0}h_{1} \bigr]h_{0}h_{1}^{2}-a h_{0}^{2}h_{1}^{3} \\& \quad {}+\mu ^{2}\biggl(\frac{1}{n}-1\biggr) \biggl( \frac{1}{n}-2\biggr)h_{1}^{3}c_{2}+3 \biggl(\frac{1}{n}-1\biggr) \mu ^{2}h_{1}^{3}c_{2}+3c_{3} \mu ^{2}h_{0}^{2}+\mu ^{2}c_{2}h_{1}^{3}=0, \end{aligned}$$
$$\begin{aligned}& 2ch_{0}h_{1}^{2}+2\bigl[3 +(b-k) h_{0}-a h_{0}^{2}\bigr]h_{1}^{2}h_{0}+ \bigl[(b-k) h_{1}-2a h_{0}h_{1} \bigr]h_{0}^{2}h_{1} \\& \quad {}+3\biggl(\frac{1}{n}-1\biggr)\mu ^{2}h_{1}^{2}h_{0}c_{2}+2 \mu ^{2}h_{1}^{2}h_{0}c _{2}=0, \end{aligned}$$
$$\begin{aligned}& ch_{0}^{2}h_{1}+\bigl[3 +(b-k) h_{0}-a h_{0}^{2}\bigr]h_{0}^{2}h_{1}+ \biggl( \frac{1}{n}-1\biggr) \biggl(\frac{1}{n}-2\biggr)\mu ^{2}h_{1}^{3}c_{1} \\& \quad {}+\mu ^{2}h_{0}^{2}h_{1}c_{2}=0. \end{aligned}$$

Under the help of Maple, the above algebraic equation system can be solved as follows:

$$\begin{aligned}& \textstyle\begin{cases} n=1, \qquad h_{0}= \frac{(b-k)}{2a}, \qquad h_{1}=\pm \frac{(b-k)}{2a }\sqrt{-\frac{c_{3}}{c_{2}}}, \\ \mu =\pm \frac{(b-k) }{2}\sqrt{-\frac{1}{3a c_{2}}}, \qquad c=-3-\frac{(b-k)^{2}}{6a}, \end{cases}\displaystyle \end{aligned}$$
$$\begin{aligned}& \textstyle\begin{cases} n=\frac{1}{2}, \qquad h_{0}=\frac{2(b-k)}{5a}, \qquad h_{1}=\pm \frac{2(b-k)}{5a }\sqrt{-\frac{c_{3}}{c_{2}}}, \\ \mu =\pm \frac{(b-k) }{5}\sqrt{-\frac{2}{3a c_{2}}}, \qquad c=-3-\frac{16(b-k)^{2}}{75a}, \end{cases}\displaystyle \end{aligned}$$
$$\begin{aligned}& \textstyle\begin{cases} n=2, \qquad \mu =\pm (\frac{-5c_{2}(b-k)^{2}+5\epsilon (b-k)^{2}\sqrt{c_{2}^{2}-2c _{1}c_{3}}}{48c_{1}c_{3}a})^{\frac{1}{2}}, \\ h_{0}=\frac{5(b-k)}{8a }, \qquad h_{1}=\pm \frac{\mu }{4 }\sqrt{\frac{30c_{3}}{a}}, \qquad c=-3-\frac{5(b-k)^{2}}{32a }-\frac{1}{4}\mu ^{2}c_{2}, \end{cases}\displaystyle \end{aligned}$$


$$\begin{aligned} \textstyle\begin{cases} h_{0}=\frac{(b-k)(2n+1)}{2a (n+2)}, \qquad h_{1}=\pm \frac{(b-k)(2n+1)}{2a (n+2)}\sqrt{-\frac{c_{3}}{c_{2}}}, \qquad 2c_{1}c_{3}=c_{2}^{2}, \\ \mu =\pm \frac{(b-k) n}{n+2}\sqrt{-\frac{2n+1}{2a (n+1)c_{2}}}, \qquad c=-3-\frac{(b-k)^{2}(2n+1)}{a (n+2)^{2}(n+1)}, \end{cases}\displaystyle \end{aligned}$$

where \(\epsilon =\pm 1\).

Solutions in the case \(n=1\)

Consider Eq. (15). From Table 1, some exact solutions for Eq. (2) can be obtained.

$$\begin{aligned}& u_{1.1}=v_{1.1}=\frac{b-k}{2a}+ \frac{\epsilon (b-k)r}{2a }\sqrt{\frac{2}{r ^{2}+1}}sn\biggl[\frac{(b-k) }{2} \sqrt{\frac{1}{3a (r^{2}+1)}}(x+y-ct )\biggr], \end{aligned}$$
$$\begin{aligned}& u_{1.2}=v_{1.2}= \frac{b-k}{2a}+ \frac{\epsilon (b-k) r}{2a }\sqrt{\frac{2 }{r^{2}+1}}cd\biggl[\frac{(b-k) }{2} \sqrt{\frac{1}{3a (r ^{2}+1)}}(x+y-ct )\biggr] , \end{aligned}$$
$$\begin{aligned}& u_{1.3}=v_{1.3}= \frac{ b-k }{2a}+ \frac{\epsilon (b-k) r}{2a }\sqrt{\frac{2 }{2r^{2}-1}}cn\biggl[\frac{ b-k }{2} \sqrt{\frac{1}{3a (1-2r ^{2})}}(x+y-ct )\biggr] , \end{aligned}$$
$$\begin{aligned}& u_{1.4}=v_{1.4}=\frac{ b-k }{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{2}{2-r ^{2}}}dn\biggl[\frac{ b-k }{2} \sqrt{\frac{1}{3 a (r^{2}-2)}}(x+y-ct )\biggr] , \end{aligned}$$
$$\begin{aligned}& u_{1.5}=v_{1.5}=\frac{ b-k }{2a}+ \frac{\epsilon (b-k)}{2 a }\sqrt{\frac{2(r ^{2}-1)}{2r^{2}-1}}nc\biggl[\frac{ b-k }{2} \sqrt{\frac{1}{3 a (1-2r^{2})}}(x+y-ct )\biggr], \end{aligned}$$
$$\begin{aligned}& u_{1.6}=v_{1.6}=\frac{b-k}{2a}+ \frac{\epsilon (b-k) }{2 a }\sqrt{\frac{2}{r ^{2}+1}}ns\biggl[\frac{(b-k) }{2} \sqrt{ \frac{1}{3 a (r^{2}+1)}}(x+y-ct )\biggr], \end{aligned}$$
$$\begin{aligned}& u_{1.7}=v_{1.7}=\frac{b-k}{2a}+ \frac{\epsilon (b-k)}{2 a }\sqrt{\frac{2}{r ^{2}+1}}dc\biggl[\frac{(b-k) }{2} \sqrt{\frac{1}{3a (r^{2}+1)}}(x+y-ct )\biggr] , \end{aligned}$$
$$\begin{aligned}& u_{1.8}=v_{1.8}=\frac{ b-k }{2a}+ \frac{\epsilon (b-k) }{2 a }\sqrt{\frac{2(r ^{2}-1)}{r^{2}-2}}nd\biggl[\frac{ b-k }{2} \sqrt{\frac{1}{3 a (r^{2}-2)}}(x+y-ct )\biggr] , \end{aligned}$$
$$\begin{aligned}& u_{1.9}=v_{1.9}=\frac{(b-k)}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{2}{r ^{2}-2}}cs\biggl[\frac{(b-k) }{2} \sqrt{\frac{1}{3a (r^{2}-2)}}(x+y-ct )\biggr] , \end{aligned}$$
$$\begin{aligned}& u_{1.10}=v_{1.10}=\frac{b-k}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{2(1-r ^{2})}{r^{2}-2}}sc\biggl[\frac{b-k }{2} \sqrt{\frac{1}{3a (r^{2}-2)}}(x+y-ct )\biggr], \end{aligned}$$
$$\begin{aligned}& u_{1.11}=v_{1.11}=\frac{ b-k }{2a}+ \frac{\epsilon (b-k) r}{2a }\sqrt{\frac{2 (r^{2}-1)}{1-2r^{2}}}sd\biggl[\frac{ b-k }{2} \sqrt{ \frac{1}{3a (1-2r^{2})}}(x+y-ct )\biggr], \end{aligned}$$
$$\begin{aligned}& u_{1.12}=v_{1.12}=\frac{ b-k}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{2}{1-2r ^{2}}}ds\biggl[\frac{ b-k }{2} \sqrt{\frac{1}{3 a (1-2r^{2})}}(x+y-ct )\biggr] , \end{aligned}$$
$$\begin{aligned}& u_{1.13}=v_{1.13} \\& \hphantom{u_{1.13}}=\frac{b-k}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{1}{1+r ^{2}}} \biggl\{ rcn\biggl[ \frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r^{2}+1)}}(x+y-ct )\biggr] \\ & \hphantom{u_{1.13}=}{}\pm nd\biggl[\frac{(b-k) }{2}\sqrt{- \frac{2}{3a (r^{2}+1)}}(x+y-ct )\biggr] \biggr\} , \end{aligned}$$
$$\begin{aligned}& u_{1.14}=v_{1.14} \\ & \hphantom{u_{1.13}}=\frac{b-k}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{1}{2r ^{2}-1}} \biggl\{ ns\biggl[ \frac{(b-k) }{2}\sqrt{\frac{2}{3a (2r^{2}-1)}}(x+y-ct )\biggr] \\ & \hphantom{u_{1.14}=}{} \pm cs\biggl[\frac{(b-k) }{2}\sqrt{- \frac{2}{3a (1-2r^{2})}}(x+y-ct )\biggr] \biggr\} , \end{aligned}$$
$$\begin{aligned}& \begin{gathered}[b] u_{1.15}=v_{1.15}\\ \hphantom{u_{1.13}}=\frac{b-k}{2a}+ \frac{\epsilon (b-k)}{2da }\sqrt{\frac{r ^{2}-1}{1+r^{2}}} \biggl\{ nc\biggl[ \frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r ^{2}+1)}}(x+y-ct )\biggr] \\ \hphantom{u_{1.15}=}{}\pm sc\biggl[\frac{(b-k)}{2}\sqrt{- \frac{2}{3a (r^{2}+1)}}(x+y-ct )\biggr] \biggr\} , \end{gathered} \end{aligned}$$
$$\begin{aligned}& u_{1.16}=v_{1.16} \\ & \hphantom{u_{1.13}}=\frac{b-k}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{1}{2-r ^{2}}} \biggl\{ ns\biggl[ \frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r^{2}-2)}}(x+y-ct )\biggr] \\ & \hphantom{u_{1.16}=}{} \pm ds\biggl[\frac{(b-k) }{2}\sqrt{- \frac{2}{3a (r^{2}-2)}}(x+y-ct )\biggr] \biggr\} , \end{aligned}$$
$$\begin{aligned}& u_{1.17}=v_{1.17} \\ & \hphantom{u_{1.13}}=\frac{(b-k)}{2a}+ \frac{\epsilon (b-k) r}{2a }\sqrt{\frac{1 }{2-r^{2}}} \biggl\{ sn\biggl[ \frac{\beta }{2}\sqrt{-\frac{2}{3a (r^{2}-2)}}(x+y-ct )\biggr] \\ & \hphantom{u_{1.17}=}{}\pm icn\biggl[\frac{(b-k) }{2}\sqrt{- \frac{2}{3a (r^{2}-2)}}(x+y-ct )\biggr] \biggr\} , \end{aligned}$$
$$\begin{aligned}& u_{1.18}=v_{1.18} \\ & \hphantom{u_{1.13}}=\frac{(b-k)}{2a}+ \frac{\epsilon (b-k) r}{2a }\sqrt{\frac{ 1}{2-r^{2}}} \\ & \hphantom{u_{1.15}=}{}\times\frac{dn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r^{2}-2)}}(x+y-ct )]}{\sqrt{1-r ^{2}}sn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r^{2}-2)}}(x+y-ct )] \pm cn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r^{2}-2)}}(x+y-ct )]} , \end{aligned}$$
$$\begin{aligned}& u_{1.19}=v_{1.19} \\ & \hphantom{u_{1.13}}=\frac{(b-k)}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{1}{2r ^{2}-1}} \biggl\{ rsn\biggl[ \frac{(b-k) }{2}\sqrt{-\frac{2}{3a (1-2r^{2})}}(x+y-ct )\biggr] \\ & \hphantom{u_{1.19}=}{}\pm idn\biggl[\frac{(b-k) }{2}\sqrt{- \frac{2}{3a (1-2r^{2})}}(x+y-ct )\biggr] \biggr\} , \end{aligned}$$
$$\begin{aligned}& u_{1.20}=v_{1.20} \\ & \hphantom{u_{1.13}}=\frac{(b-k)}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{1}{2r ^{2}-1}}\cdot \frac{sn[\frac{(b-k) }{2}\sqrt{- \frac{1}{3a (1-2r^{2})}}(x+y-ct )]}{1\pm cn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (1-2r^{2})}}(x+y-ct )]} , \end{aligned}$$
$$\begin{aligned}& u_{1.21}=v_{1.21} \\ & \hphantom{u_{1.13}}=\frac{(b-k)}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{r ^{2}}{2-r^{2}}}\cdot \frac{sn[\frac{(b-k) }{2}\sqrt{-\frac{1}{3a (r ^{2}-2)}}(x+y-ct )]}{1\pm dn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r ^{2}-2)}}(x+y-ct )]} , \end{aligned}$$
$$\begin{aligned}& u_{1.22}=v_{1.22} \\& \hphantom{u_{1.13}}=\frac{(b-k)}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{1-r ^{2}}{r^{2}+1}}\cdot \frac{dn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r ^{2}+1)}}(x+y-ct )]}{1\pm rsn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r ^{2}+1)}}(x+y-ct )]} , \end{aligned}$$
$$\begin{aligned}& u_{1.23}=v_{1.23} \\& \hphantom{u_{1.13}}=\frac{(b-k)}{2a}+ \frac{\epsilon (b-k)}{2a }\sqrt{\frac{r ^{2}-1}{1+r^{2}}}\cdot \frac{cn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r ^{2}+1)}}(x+y-ct )]}{1\pm sn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r ^{2}+1)}}(x+y-ct )]} , \end{aligned}$$
$$\begin{aligned}& u_{1.24}=v_{1.24} \\& \hphantom{u_{1.13}}=\frac{(b-k)}{2a}+ \frac{\epsilon (b-k)(1-r^{2})}{2a }\sqrt{-\frac{1}{r^{2}+1}} \\& \hphantom{u_{1.24}=}{}\times \frac{sn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r^{2}+1)}}(x+y-ct )]}{dn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3 a (r^{2}+1)}}(x+y-ct )] \pm cn[\frac{(b-k) }{2}\sqrt{-\frac{2}{3a (r^{2}+1)}}(x+y-ct )]}, \end{aligned}$$
$$\begin{aligned}& u_{1.25}=v_{1.25} \\& \hphantom{u_{1.13}}=\frac{b-k}{2a}+ \frac{\epsilon (b-k) r^{2}}{2a }\sqrt{\frac{1}{2-r ^{2}}}\frac{cn[\frac{b-k }{2}\sqrt{-\frac{2}{3a (r^{2}-2)}}(x+y-ct )]}{\sqrt{1-r ^{2}}\pm dn[\frac{b-k }{2}\sqrt{-\frac{2}{3a (r^{2}-2)}}(x+y-ct )]}, \end{aligned}$$

where \(c=-3-\frac{(b-k)^{2}}{6a}\).

When r approaches 0 or 1, sn, cn, and dn degenerate into triangular or hyperbolic functions. Therefore, in the case \(n=1\), we can deduce Jacobi elliptic function solutions, solitons, and triangular function solutions for Eq. (2).

When \(k=l=\frac{2b-a}{2a}\sqrt{\frac{2}{1+r^{2}}}\) and \(w=k(3+ \frac{(b-k)^{2}}{6a})\), the solutions \(u_{1}\), \(v_{1}\), \(u_{2}\), \(v_{2}\), \(u_{3}\), \(v_{3}\), \(u_{4}\), \(v_{4}\) in [25] are the solutions \(u_{1.1}\), \(u_{1.2}\), \(u_{1.6}\), \(u_{1.7}\), respectively. When \(k=l= \frac{2b-a}{2a}\sqrt{\frac{2}{1-2r^{2}}}\) and \(w=k(3+ \frac{(b-k)^{2}}{6a})\), the solutions \(u_{1.5}\), \(u_{1.12}\) are the same as \(u_{5}\), \(v_{5}\), \(u_{8}\), \(v_{8}\) in [25]. When \(k=l=\frac{2b-a}{2a}\sqrt{\frac{2}{r^{2}-2}}\) and \(w=k(3+ \frac{(b-k)^{2}}{6a})\), the expressions \(u_{6}\), \(v_{6}\), \(u_{7}\), \(v_{7}\) are also obtained by Zhang [25].

When the modulus \(r\rightarrow 1\), choosing \(b=6b_{0}\), \(a=\frac{3}{2}a _{0}^{2}\), and \(k=3a_{0}\) in the solutions \(u_{1.1}\) and \(u_{2.3}\), we obtain

$$ u_{1.1.1} =v_{1.1.1} =\frac{2b_{0}-a_{0}}{a^{2}_{0}} \biggl(1\pm \tanh \biggl[\frac{2b_{0}-a_{0}}{2a_{0}} \biggl(x+y+\frac{4(a^{2}_{0}+b^{2} _{0}-a_{0}b_{0})}{a^{2}_{0}}t \biggr) \biggr] \biggr) $$


$$ u_{1.6.1} =v_{1.6.1} =\frac{2b_{0}-a_{0}}{a^{2}_{0}} \biggl(1\pm \coth \biggl[\frac{2b_{0}-a_{0}}{2a_{0}} \biggl(x+y+\frac{4(a^{2}_{0}+b^{2} _{0}-a_{0}b_{0})}{a^{2}_{0}}t \biggr) \biggr] \biggr), $$

which have also been established in Wazwaz [22].

Solutions in the case \(n=\frac{1}{2}\)

From Eq. (16) and Table 1, we obtain solutions of Eq. (2) when \(n=\frac{1}{2}\):

$$\begin{aligned}& u_{2.1}=v_{2.1} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2(b-k) r}{5a }\sqrt{\frac{2 }{r^{2}+1}}sn\biggl[\frac{(b-k) }{5} \sqrt{\frac{2}{3a (r ^{2}+1)}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.2}=v_{2.2} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2(b-k) r}{5a }\sqrt{\frac{2 }{r^{2}+1}}cd\biggl[\frac{(b-k) }{5} \sqrt{\frac{2}{3a (r ^{2}+1)}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.3}=v_{2.3} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2(b-k) r}{5a }\sqrt{\frac{2 }{2r^{2}-1}}cn\biggl[\frac{(b-k) }{5} \sqrt{ \frac{2}{3a (1-2r ^{2})}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.4}=v_{2.4} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5 a}\pm \frac{2(b-k)}{5 a }\sqrt{\frac{2}{2-r ^{2}}}dn\biggl[\frac{(b-k) }{5} \sqrt{-\frac{2}{3a (2-r^{2})}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.5}=v_{2.5} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2(b-k)}{5a }\sqrt{ \frac{2r^{2}-2}{2r^{2}-1}} nc\biggl[\frac{(b-k) }{5} \sqrt{ \frac{2}{3a (1-2r^{2})}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.6}=v_{2.6} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2(b-k)}{5a }\sqrt{\frac{2}{r ^{2}+1}}ns\biggl[\frac{(b-k) }{5} \sqrt{\frac{2}{3a (r^{2}+1)}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.7}=v_{2.7} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2(b-k)}{5a }\sqrt{\frac{2}{r ^{2}+1}}dc\biggl[\frac{\beta }{5} \sqrt{\frac{2}{3a (r^{2}+1)}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.8}=v_{2.8} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{\frac{2-2r ^{2}}{2-r^{2}}}nd\biggl[\frac{(b-k) }{5} \sqrt{ \frac{2}{3a (r^{2}-2)}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.9}=v_{2.9} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{ \frac{2}{r^{2}-2}}cs\biggl[\frac{(b-k) }{5} \sqrt{\frac{2}{3a (r ^{2}-2)}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.10}=v_{2.10} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{\frac{2r ^{2}-2}{2-r^{2}}} sc\biggl[\frac{(b-k) }{5} \sqrt{\frac{2}{3a(r^{2}-2)}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& \begin{gathered}[b] u_{2.11}=v_{2.11} \\ \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{\frac{2r ^{2}(1-r^{2})}{2r^{2}-1}} sd\biggl[\frac{b-k }{5} \sqrt{\frac{2}{3a -6ar ^{2}}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{gathered} \end{aligned}$$
$$\begin{aligned}& u_{2.12}=v_{2.12} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{\frac{2}{1-2r ^{2}}}ds\biggl[\frac{b-k }{5} \sqrt{\frac{2}{3a (1-2r^{2})}}(x+y-ct )\biggr] \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.13}=v_{2.13} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{\frac{1}{r ^{2}+1}} \biggl(rcn\biggl[ \frac{(b-k) }{5}\sqrt{-\frac{4}{3a (r^{2}+1)}}(x+y-ct )\biggr] \\ & \hphantom{u_{2.13}=}{}\pm dn\biggl[\frac{\beta }{5}\sqrt{- \frac{4}{3a (r^{2}+1)}}(x+y-ct )\biggr] \biggr) \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.14}=v_{2.14} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{-\frac{1}{1-2r ^{2}}} \biggl(ns\biggl[ \frac{(b-k) }{5}\sqrt{-\frac{4}{3a (-2r^{2}+1)}}(x+y-ct )\biggr] \\ & \hphantom{u_{2.14}=}{} \pm cs\biggl[\frac{(b-k) }{5}\sqrt{- \frac{4}{3a (-2r^{2}+1)}}(x+y-ct )\biggr] \biggr) \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.15}=v_{2.15} \\ & \hphantom{u_{2.15}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{- \frac{1-r ^{2}}{r^{2}+1}} \biggl(nc\biggl[\frac{(b-k) }{5}\sqrt{- \frac{4}{3a (r^{2}+1)}}(x+y-ct )\biggr] \\ & \hphantom{u_{2.15}=}{} \pm sc\biggl[\frac{(b-k) }{5}\sqrt{- \frac{4}{3a (r^{2}+1)}}(x+y-ct )\biggr] \biggr) \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.16}=v_{2.16} \\ & \hphantom{u_{2.16}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{- \frac{1}{r ^{2}-2}} \biggl(ns\biggl[\frac{(b-k) }{5}\sqrt{- \frac{4}{3a (r^{2}-2)}}(x+y-ct )\biggr] \\ & \hphantom{u_{2.16}=}{} \pm ds\biggl[\frac{(b-k) }{5}\sqrt{- \frac{4}{3a (r^{2}-2)}}(x+y-ct )\biggr] \biggr) \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.17}=v_{2.17} \\ & \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k) r}{5a }\sqrt{- \frac{1}{r ^{2}-2}} \biggl(sn\biggl[\frac{(b-k) }{5}\sqrt{- \frac{4}{3a (r^{2}-2)}}(x+y-ct )\biggr] \\ & \hphantom{u_{2.17}=}{} \pm icn\biggl[\frac{(b-k) }{5}\sqrt{- \frac{4}{3a (r^{2}-2)}}(x+y-ct )\biggr] \biggr) \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& \begin{gathered}[b] u_{2.18}=v_{2.18} \\ \hphantom{u_{2.18}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k) r}{5a }\sqrt{- \frac{1}{r ^{2}-2}} \\ \hphantom{u_{2.18}=}{}\times \frac{dn[\frac{(b-k) }{5}\sqrt{-\frac{4}{3a (r^{2}-2)}}(x+y-ct )]}{\sqrt{1-r ^{2}}sn[\frac{(b-k) }{5}\sqrt{-\frac{4}{3a (r^{2}-2)}}(x+y-ct )] \pm cn[\frac{(b-k) }{5}\sqrt{-\frac{4}{3a(r^{2}-2)}}(x+y-ct )]} \biggr\} ^{2}, \end{gathered} \end{aligned}$$
$$\begin{aligned}& u_{2.19}=v_{2.19} \\& \hphantom{u_{2.17}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{\frac{1 }{2r^{2}-1}}\frac{sn[\frac{(b-k) }{5}\sqrt{\frac{4}{3a (2r^{2}-1)}}(x+y-ct )]}{1\pm cn[ \frac{(b-k) }{5}\sqrt{\frac{4}{3a (2r^{2}-1)}}(x+y-ct )]} \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.20}=v_{2.20} \\& \hphantom{u_{2.20}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)}{5a }\sqrt{ \frac{1}{2r ^{2}-1}} \biggl(rsn\biggl[\frac{(b-k) }{5}\sqrt{- \frac{4}{3a (1-2r^{2})}}(x+y-ct )\biggr] \\& \hphantom{u_{2.20}=}{}\pm i dn\biggl[\frac{(b-k) }{5}\sqrt{- \frac{4}{3a (1-2r^{2})}}(x+y-ct )\biggr] \biggr) \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.21}=v_{2.21} \\& \hphantom{u_{2.17}}= \biggl\{ \frac{ 2(b-k) }{5a}\pm \frac{2 (b-k) r}{5a }\sqrt{ \frac{1}{2-r^{2}}}\frac{sn[\frac{ b-k }{5}\sqrt{ \frac{4}{3a (r^{2}-2)}}(x+y-ct )]}{1\pm dn[\frac{(b-k) }{5}\sqrt{ \frac{4}{3a (r ^{2}-2)}}(x+y-ct )]} \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.22}=v_{2.22}= \biggl\{ \frac{2(b-k) }{5a}\pm \frac{2 (b-k)}{5a }\sqrt{\frac{1-r ^{2}}{r^{2}+1}}\frac{dn[\frac{ b-k }{5}\sqrt{ \frac{-4}{3a (r^{2}+1)}}(x+y-ct )]}{1\pm r sn[\frac{ b-k }{5}\sqrt{\frac{-4}{3a(r ^{2}+1)}}(x+y-ct )]} \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.23}=v_{2.23}= \biggl\{ \frac{ 2(b-k) }{5a}\pm \frac{2 (b-k)}{5a }\sqrt{ \frac{r^{2}-1}{r^{2}+1}}\frac{cn[\frac{ b-k }{5}\sqrt{\frac{-4}{3a (r^{2}+1)}}(x+y-ct )]}{1\pm sn[\frac{ b-k }{5}\sqrt{\frac{-4}{3a (r^{2}+1)}}(x+y-ct )]} \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.24}=v_{2.24} \\& \hphantom{u_{2.24}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k)(1-r ^{2})}{5a }\sqrt{- \frac{1}{r^{2}+1}} \\& \hphantom{u_{2.24}=}{}\times\frac{sn[\frac{(b-k) }{5}\sqrt{-\frac{4}{3a (r^{2}+1)}}(x+y-ct )]}{dn[\frac{(b-k) }{5}\sqrt{-\frac{4}{3a (r^{2}+1)}}(x+y-ct )] \pm cn[\frac{(b-k) }{5}\sqrt{-\frac{4}{3a (r^{2}+1)}}(x+y-ct )]} \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned}& u_{2.25}=v_{2.25} \\& \hphantom{u_{2.25}}= \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k) r^{2}}{5a }\sqrt{ \frac{1}{2-r^{2}}} \\& \hphantom{u_{2.25}=}{}\times \frac{cn[\frac{(b-k) }{5}\sqrt{-\frac{4}{3a (r^{2}-2)}}(x+y-ct )]}{\sqrt{1-r^{2}}\pm dn[\frac{(b-k) }{5}\sqrt{-\frac{4}{3a (r ^{2}-2)}}(x+y-ct )]} \biggr\} ^{2}, \end{aligned}$$

where \(c=-3-\frac{16(b-k)^{2}}{75a}\).

Therefore, when \(n=\frac{1}{2}\), we can obtain exact solutions for Eq. (2), some of them will degenerate solitons and triangular function solutions when r approaches 0 or 1.

Solutions in the case \(n=2\)

From the expressions of Eq. (17), we get

$$\begin{aligned}& u_{3.1} =v_{3.1} = \biggl\{ \frac{5(b-k)}{8a }+ \frac{\epsilon \mu r}{2 }\sqrt{ \frac{15 }{ a}}sn\biggl[\mu \biggl(x+y+3 t+ \frac{5(b-k)^{2}}{32a } t- \frac{t}{4}\bigl(r^{2}+1\bigr)\mu ^{2}\biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4r}\biggl(\frac{-5(r^{2}+1)+5\epsilon (r^{2}-1)}{6a} \biggr)^{ \frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.2}=v_{3.2}= \biggl\{ \frac{5(b-k)}{8a }+ \frac{\epsilon \mu r}{2 }\sqrt{\frac{15 }{ a}}cd \biggl[\mu \biggl(x+y+3 t+ \frac{5(b-k)^{2}}{32a } t- \frac{t}{4}\bigl(r^{2}+1\bigr)\mu ^{2}\biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4r}\biggl(\frac{-5(r^{2}+1)+5\epsilon (r^{2}-1)}{6a} \biggr)^{ \frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.3}=v_{3.3} \\& \hphantom{u_{2.3}}= \biggl\{ \frac{5(b-k)}{8a }+ \frac{\epsilon \mu r}{2 }\sqrt{-\frac{15 }{ a}}cn\biggl[\mu \biggl(x+y+3 t+ \frac{5(b-k)^{2}}{32a } t+ \frac{t}{4}\bigl(2r^{2}-1\bigr)\mu ^{2}\biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4r}\biggl(- \frac{5(2r^{2}-1) +5\epsilon }{6(r^{2}-1)a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.4}=v_{3.4}= \biggl\{ \frac{5(b-k)}{8a }+ \frac{\epsilon \mu }{2 }\sqrt{- \frac{15}{a}}dn\biggl[\mu \biggl(x+y+3 t+ \frac{5(b-k)^{2}}{32a } t+\frac{t}{4}\bigl(2-r ^{2}\bigr)\mu ^{2} \biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4}\biggl(\frac{-5(2-r^{2})+5\epsilon \ r^{2}}{6(1-r ^{2})a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.5}=v_{3.5} \\& \hphantom{u_{2.1}}= \biggl\{ \frac{5(b-k)}{8a } \\& \hphantom{u_{2.3}=}{}+ \frac{\epsilon \mu }{2 }\sqrt{ \frac{15(1-r^{2})}{ a}}nc\biggl[\mu \biggl(x+y+3 t+ \frac{5(b-k)^{2}}{32 a } t+\frac{t}{4}\bigl(2r^{2}-1\bigr)\mu ^{2} \biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4r}\biggl(\frac{-5(2r^{2}-1)+5\epsilon }{6(r^{2}-1)a} \biggr)^{ \frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.6}=v_{3.6}= \biggl\{ \frac{5(b-k)}{8 a }+ \frac{\epsilon \mu }{2 }\sqrt{\frac{15}{ a}}ns\biggl[\mu \biggl(x+y+3t+ \frac{5(b-k)^{2}}{32 a } t- \frac{t}{4}\bigl(r^{2}+1\bigr)\mu ^{2}\biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4r}\biggl(\frac{-5(r^{2}+1)+5\epsilon (r^{2}-1)}{6a} \biggr)^{ \frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.7}=v_{3.7}= \biggl\{ \frac{5(b-k)}{8 a }+ \frac{\epsilon \mu }{2 }\sqrt{ \frac{15}{ a}}dc\biggl[\mu \biggl(x+y+3 t+ \frac{5(b-k)^{2}}{32a } t- \frac{t}{4}\bigl(r^{2}+1\bigr)\mu ^{2}\biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4r}\biggl(\frac{-5(r^{2}+1)+5\epsilon (r^{2}-1)}{6a} \biggr)^{ \frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.8}=v_{3.8} \\& \hphantom{u_{2.8}}= \biggl\{ \frac{5(b-k)}{8a }+ \frac{\epsilon \mu }{2 }\sqrt{\frac{15(r ^{2}-1)}{a}}nd\biggl[\mu \biggl(x+y+3 t+ \frac{5(b-k)^{2}}{32a } t+\frac{t}{4}\bigl(2-r ^{2}\bigr)\mu ^{2}\biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4}\biggl(- \frac{5(2-r^{2})+5\epsilon r^{2}}{6(1-r^{2})a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.9}=v_{3.9}= \biggl\{ \frac{5(b-k)}{8a }+ \frac{\epsilon \mu }{2 }\sqrt{ \frac{15}{ a}}cs\biggl[\mu \biggl(x+y+3 t+ \frac{5(b-k)^{2}}{32a } t+ \frac{t}{4}\bigl(2-r^{2}\bigr)\mu ^{2}\biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4}\biggl(-\frac{5(2-r^{2})+5\epsilon r^{2}}{6\lambda (1-r^{2})} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.10}=v_{3.10} \\& \hphantom{u_{2.17}}= \biggl\{ \frac{5(b-k)}{8a }+ \frac{\epsilon \mu }{2 }\sqrt{\frac{15(1-r^{2})}{a}}sc\biggl[\mu \biggl(x+y+3 t+ \frac{5(b-k)^{2}}{32a } t+\frac{t}{4}\bigl(2-r^{2}\bigr)\mu ^{2}\biggr)\biggr] \biggr\} ^{ \frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4}\biggl(-\frac{5(2-r^{2})+5\epsilon r^{2}}{6a(1-r ^{2})} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.11}=v_{3.11} \\& \hphantom{u_{2.17}}= \biggl\{ \frac{5(b-k)}{8a } \\& \hphantom{u_{2.17}=}{}+ \frac{\epsilon \mu r}{2 }\sqrt{\frac{15(r^{2}-1)}{a}}sd\biggl[\mu \biggl(x+y+3t+ \frac{5(b-k)^{2}}{32a } t+\frac{t}{4}\bigl(2r^{2}-1\bigr)\mu ^{2}\biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4r}\biggl(- \frac{5(2r^{2}-1) +5\epsilon }{6(r^{2}-1)a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.12}=v_{3.12} \\& \hphantom{u_{2.17}}= \biggl\{ \frac{5(b-k)}{8a }+ \frac{\epsilon \mu }{2 }\sqrt{\frac{15}{ a}}ds\biggl[\mu \biggl(x+y+3 t+ \frac{5(b-k)^{2}}{32a } t+ \frac{t}{4}\bigl(2r^{2}-1\bigr)\mu ^{2}\biggr)\biggr] \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{4r}\biggl(- \frac{5(2r^{2}-1)+5\epsilon }{6(r^{2}-1) a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.13}=v_{3.13} \\& \hphantom{u_{3.13}}= \biggl\{ \frac{5(b-k)}{8 a }+\frac{\epsilon \mu }{4 }\sqrt{- \frac{15}{a}} \biggl(rcn\biggl[\mu \biggl(x+y+3t+\frac{5(b-k)^{2}}{32a } t+ \frac{t}{8}\bigl(r^{2}+1\bigr)\mu ^{2}\biggr) \biggr] \\& \hphantom{u_{3.13}=}{}\pm dn\biggl[\mu \biggl(x+y+3t+\frac{5(b-k)^{2}}{32a } t+ \frac{t}{8}\bigl(r^{2}+1\bigr)\mu ^{2}\biggr) \biggr] \biggr) \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{2(1-r^{2})}\biggl(-\frac{5(r^{2}+1)+10\epsilon r}{3a} \biggr)^{ \frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.14}=v_{3.14} \\& \hphantom{u_{3.14}}= \biggl\{ \frac{5(b-k)}{8a }+\frac{\epsilon \mu }{4 }\sqrt{ \frac{15}{ a}} \biggl(ns\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32a } t+ \frac{t}{8}\bigl(1-2r^{2}\bigr)\biggr)\biggr] \\& \hphantom{u_{3.14}=}{}\pm cs\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32a } t+ \frac{t}{8}\bigl(1-2r^{2}\bigr)\mu ^{2}\biggr) \biggr] \biggr) \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{2}\biggl(-\frac{5(1-2r^{2})+10\epsilon \ r\sqrt{r ^{2}-1}}{3a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.15}=v_{3.15} \\& \hphantom{u_{3.15}}= \biggl\{ \frac{5(b-k)}{8a }+\frac{\epsilon \mu }{4 }\sqrt{ \frac{15(1-r^{2})}{ a}} \biggl(nc\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32a } t+ \frac{t}{8}\bigl(r^{2}+1\bigr)\mu ^{2}\biggr) \biggr] \\ & \hphantom{u_{3.15}=}{}\pm dc\biggl[\mu \biggl(x+y+3t+\frac{5(b-k)^{2}}{32a } t+ \frac{t}{8}\bigl(r^{2}+1\bigr)\mu ^{2}\biggr) \biggr] \biggr) \biggr\} ^{\frac{1}{2}}, \\ & \quad \mu =\pm \frac{(b-k)}{2(1-r^{2})}\biggl(- \frac{5(r^{2}+1)+10\epsilon \ r}{3a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.16}=v_{3.16} \\ & \hphantom{u_{3.16}}= \biggl\{ \frac{5(b-k)}{8 a }+\frac{\epsilon \mu }{4 }\sqrt{ \frac{15}{ a}} \biggl( ns\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}\bigl(r^{2}-2\bigr)\mu ^{2}\biggr) \biggr] \\ & \hphantom{u_{3.16}=}{}\pm ds\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32a } t+ \frac{t}{8}\bigl(r^{2}-2\bigr)\mu ^{2}\biggr) \biggr] \biggr) \biggr\} ^{\frac{1}{2}}, \\ & \quad \mu =\pm \frac{(b-k)}{2r^{2}}\biggl(-\frac{5(r^{2}-2)+10\epsilon \sqrt{1-r ^{2}}}{3a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.17}=v_{3.17} \\ & \hphantom{u_{3.17}}= \biggl\{ \frac{5(b-k)}{8a }+\frac{\epsilon \mu r}{4 }\sqrt{ \frac{15}{a}} \biggl(sn\biggl[\mu \biggl(x+y+3t+\frac{5(b-k)^{2}}{32a } t+ \frac{t}{8}\bigl(r^{2}-2\bigr)\mu ^{2}\biggr) \biggr] \\ & \hphantom{u_{3.17}=}{}\pm icn\biggl[\mu \biggl(x+y+3t+\frac{5(b-k)^{2}}{32 a } t+ \frac{t}{8}\bigl(r^{2}-2\bigr)\mu ^{2}\biggr) \biggr] \biggr) \biggr\} ^{\frac{1}{2}}, \\ & \quad \mu =\pm \frac{(b-k)}{2r^{2}}\biggl(-\frac{5(r^{2}-2)+10\epsilon \sqrt{1-r ^{2}}}{3a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.18}=v_{3.18} \\ & \hphantom{u_{3.18}}= \biggl\{ \frac{5(b-k)}{8a }+\frac{\epsilon \mu r}{4 }\sqrt{ \frac{15}{a}} \\ & \hphantom{u_{3.13}=}{}\times \biggl(sn\biggl[\mu \biggl(x+y+3t+\frac{5(b-k)^{2}}{32a } t+\frac{t}{8}\bigl(r^{2}-2\bigr) \mu ^{2}\biggr)\biggr]\biggr) \\ & \hphantom{u_{3.13}=}{}\times \biggl(\sqrt{1-r^{2}}sn\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32a } t+ \frac{t}{8}\bigl(r^{2}-2\bigr)\mu ^{2}\biggr)\biggr] \\ & \hphantom{u_{3.13}=}{}\pm cn\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}\bigl(r^{2}-2\bigr)\mu ^{2}\biggr)\biggr]\biggr)^{-1} \biggr\} ^{\frac{1}{2}}, \\ & \quad \mu =\pm \frac{(b-k)}{2r^{2}}\biggl(\frac{-5(r^{2}-2)+10\epsilon \sqrt{1-r ^{2}}}{3a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.19}=v_{3.19} \\& \hphantom{u_{3.19}}= \biggl\{ \frac{5(b-k)}{8a }+\frac{\epsilon \mu }{4 }\sqrt{ \frac{15}{a}} \biggl(rsn\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32a } t+\frac{t}{8}\bigl(1-2r^{2}\bigr)\mu ^{2} \biggr)\biggr] \\& \hphantom{u_{3.19}=}{}\pm idn\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32 a } t+ \frac{t}{8}\bigl(1-2r^{2}\bigr) \mu ^{2}\biggr) \biggr] \biggr) \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{2}\biggl(-\frac{5(1-2r^{2})+10\epsilon \ r\sqrt{r ^{2}-1}}{3 a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.20}=v_{3.20}= \biggl\{ \frac{5(b-k)}{8 a }+ \frac{\epsilon \mu }{4 }\sqrt{ \frac{15}{ a}}\frac{sn[\mu (x+y+3 t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}(1-2r^{2})\mu ^{2})]}{1\pm cn[ \mu (x+y+3t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}(1-2r^{2})\mu ^{2})]} \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{2}\biggl(-\frac{5(1-2r^{2})+10\epsilon \ r\sqrt{r ^{2}-1}}{3 a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.21}=v_{3.21}= \biggl\{ \frac{5(b-k)}{8 a }+ \frac{\epsilon \mu r}{4 }\sqrt{ \frac{15}{ a}}\frac{sn[\mu (x+y+3 t+\frac{5(b-k)^{2}}{32a } t+\frac{t}{8}(r^{2}-2)\mu ^{2}]}{1\pm dn[ \mu (x+y+3t+\frac{5(b-k)^{2}}{32a } t+\frac{t}{8}(r^{2}-2)\mu ^{2}]} \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{2r}\biggl(-\frac{5(r^{2}-2)+5\epsilon \sqrt{r^{4}-5r ^{2}+4}}{3 a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.22}=v_{3.22} \\& \hphantom{u_{2.17}}= \biggl\{ \frac{5(b-k)}{8 a }+ \frac{\epsilon \mu }{4 }\sqrt{\frac{15(r^{2}-1)}{ a}}\frac{dn[\mu (x+y+3 t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}(r^{2}+1)\mu ^{2})]}{1\pm rsn[ \mu (x+y+3 t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}(r^{2}+1)\mu ^{2})]} \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{\beta }{2(1-r^{2})}\biggl(- \frac{5(r^{2}+1)+10\epsilon r}{3 a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.23}=v_{3.23} \\& \hphantom{u_{3.23}}= \biggl\{ \frac{5(b-k)}{8 a}+\frac{\epsilon \mu }{4 }\sqrt{- \frac{15(r^{2}-1)}{ a}} \\& \hphantom{u_{3.23}=}{}\times \frac{cn[\mu (x+y+3t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}(r ^{2}+1)\mu ^{2})]}{1\pm sn[\mu (x+y+3 t+\frac{5(b-k)^{2}}{32a } t+ \frac{t}{8}(r^{2}+1)\mu ^{2})]} \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{2(1-r^{2})}\biggl(-\frac{5(r^{2}+1)+10\epsilon r}{3 a} \biggr)^{ \frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.24}=v_{3.24} \\& \hphantom{u_{3.24}}= \biggl\{ \frac{5(b-k)}{8 a }+\frac{\epsilon \mu (r^{2}-1)}{4 }\sqrt{ \frac{15}{ a}} \\& \hphantom{u_{3.24}=}{}\times\biggl(sn\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}\bigl(r^{2}+1\bigr) \mu ^{2}\biggr)\biggr]\biggr) \\& \hphantom{u_{3.24}=}{}\times\biggl(dn\biggl[\mu \biggl(x+y+3 t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}\bigl(r ^{2}+1\bigr)\mu ^{2}\biggr)\biggr] \\& \hphantom{u_{3.24}=}{}\pm cn\biggl[\mu \biggl(x+y+3t+\frac{5(b-k)^{2}}{32 a } t+ \frac{t}{8}\bigl(r^{2}+1\bigr)\mu ^{2}\biggr)\biggr]\biggr)^{-1} \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{2(1-r^{2})}\biggl(- \frac{5(r^{2}+1)+10\epsilon r}{3 a} \biggr)^{\frac{1}{2}}, \end{aligned}$$
$$\begin{aligned}& u_{3.25}=v_{3.25} \\& \hphantom{u_{2.17}}= \biggl\{ \frac{5(b-k)}{8 a }+ \frac{\epsilon \mu r^{2}}{4 }\sqrt{ \frac{15}{ a}}\frac{cn[\mu (x+y+3 t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}(r^{2}-2)\mu ^{2}])]}{\sqrt{1-r ^{2}}\pm dn[\mu (x+y+3 t+\frac{5(b-k)^{2}}{32 a } t+\frac{t}{8}(r^{2}-2) \mu ^{2}])]} \biggr\} ^{\frac{1}{2}}, \\& \quad \mu =\pm \frac{(b-k)}{2r^{2}}\biggl(\frac{-5(r^{2}-2)+10\epsilon \sqrt{1-r ^{2}}}{3 a} \biggr)^{\frac{1}{2}}. \end{aligned}$$

Of course, these are Jacobi elliptic function solutions.

Solutions in general case

From the properties of the Jacobi elliptic functions, we know that when \(r\rightarrow 1\), for these solutions listed from (19) to (40), some of them will become zero or constants, the others will degenerate soliton-like solutions, the degenerated solitons for Eq. (2) are expressed by

$$\begin{aligned} u_{4.1} =&v_{4.1} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2a (n+2)}-\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ & {}\times \tanh \biggl[\frac{(b-k) n}{n+2}\sqrt{ \frac{2n+1}{4 a (n+1)}} \biggl(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}\biggr) \biggr] \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.2} =&v_{4.2} \\ =& \biggl\{ \frac{(b-k )(2n+1)}{2 a (n+2)}-\epsilon \frac{\sqrt{2}(b-k)(2n+1)}{2 a (n+2)} \\ & {}\times \operatorname{sech} \biggl[\frac{(b-k) n}{n+2}\sqrt{- \frac{2n+1}{2 a (n+1)}}\biggl(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}\biggr) \biggr] \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.3} =&v_{4.3} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \coth \biggl[\frac{(b-k) n}{n+2}\sqrt{ \frac{2n+1}{4 a (n+1)}} \biggl(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}\biggr) \biggr] \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.4} =&v_{4.4} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-i\epsilon \frac{\sqrt{2}(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times csch \biggl[\frac{(b-k) n}{n+2}\sqrt{-\frac{2n+1}{2 a (n+1)}} \biggl(x+y+3 t+\frac{(b-k)^{2}(2 +1)t}{ a (n+2)^{2}(n+1)}\biggr) \biggr] \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.5} =&v_{4.5} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \frac{ \cosh (\frac{(b-k)n}{n+2}\sqrt{ \frac{2n+1}{ a (p+1)}}(x+y+3 t+ \frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}) )\pm 1}{\sinh (\frac{(b-k) n}{n+2}\sqrt{ \frac{2n+1}{ a (n+1)}}(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}) )} \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.6} =&v_{4.6} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (p+2)}-\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \frac{\sinh (\frac{(b-k) n}{n+2}\sqrt{ \frac{2n+1}{ a (n+1) }}[x+y+3t+\frac{(b-k)^{2}(2 n+1)t}{ a (n+2)^{2}(n+1)}] ) \pm i}{\cosh (\frac{(b-k) n}{n+2}\sqrt{ \frac{2n+1}{ a (n+1)}}[x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}] )} \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.7} =&v_{4.7} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2a (n+2)}-\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \frac{\sinh (\frac{(b-k) n}{n+2}\sqrt{ \frac{2n+1}{ a (n+1)}}(x+y+3t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}) )}{1\pm \cosh (\frac{(b-k) n}{n+2}\sqrt{ \frac{2n+1}{a (n+1)}}[x+y+3 t+\frac{(b-k)^{2}(2p+1)t}{ a (n+2)^{2}(n+1)}] )} \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.8} =&v_{4.8} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-\epsilon \frac{ \sqrt{2}(b-k )(2n+1)}{2 a (n+2)} \\ &{} \times \sec \biggl[\frac{(b-k) p}{p+2}\sqrt{ \frac{2n+1}{2 a (n+1)}} \biggl(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}\biggr) \biggr] \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.9} =&v_{4.9} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-\epsilon \frac{ \sqrt{2}(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \csc \biggl[\frac{(b-k) n}{n+2}\sqrt{ \frac{2n+1}{2 a (n+1)}} \biggl(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}\biggr) \biggr] \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.10} =&v_{4.10} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-i\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \cot \biggl[\frac{(b-k) n}{n+2}\sqrt{-\frac{2n+1}{4 a (n+1)}} \biggl(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}\biggr) \biggr] \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.11} =&v_{4.11} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-i\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \tan \biggl[\frac{(b-k) n}{n+2}\sqrt{-\frac{2n+1}{4 a (n+1)}} \biggl(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}\biggr) \biggr] \biggr\} ^{\frac{1}{p}}, \end{aligned}$$
$$\begin{aligned} u_{4.12} =&v_{4.12} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-i\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \frac{1\pm \cos (\frac{(b-k) n}{n+2}\sqrt{-\frac{2n+1}{ a (n+1)}}(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}) )}{ \sin (\frac{(b-k) n}{n+2}\sqrt{-\frac{2n+1}{ a (n+1)}}(x+y+3 t+\frac{(b-k)^{2}(2p+1)t}{a (n+2)^{2}(n+1)}) )} ] \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.13} =&v_{4.13} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-i\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \frac{1\pm \sin (\frac{(b-k) n}{n+2}\sqrt{-\frac{2n+1}{ a (n+1)}}[x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}] )}{ \cos (\frac{(b-k) n}{n+2}\sqrt{-\frac{2n+1}{ a (n+1)}}[x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}] )} ] \biggr\} ^{\frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.14} =&v_{4.14} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-i\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \frac{\sin (\frac{(b-k) n}{n+2}\sqrt{- \frac{2n+1}{ a (n+1)}}(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}) )}{1\pm \cos (\frac{(b-k)n}{n+2}\sqrt{- \frac{2n+1}{ a (n+1)}}[x+y+3 t+ \frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}] )} \biggr\} ^{ \frac{1}{n}}, \end{aligned}$$
$$\begin{aligned} u_{4.15} =&v_{4.15} \\ =& \biggl\{ \frac{(b-k)(2n+1)}{2 a (n+2)}-i\epsilon \frac{(b-k)(2n+1)}{2 a (n+2)} \\ &{} \times \frac{\cos (\frac{(b-k) n}{n+2}\sqrt{-\frac{2n+1}{ a (n+1) }}[x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (p+2)^{2}(n+1)}] )}{1\pm \sin (\frac{(b-k) n}{n+2}\sqrt{-\frac{2n+1}{2 a (n+1) }}(x+y+3 t+\frac{(b-k)^{2}(2n+1)t}{ a (n+2)^{2}(n+1)}) )} \biggr\} ^{ \frac{1}{n}}. \end{aligned}$$

In the solutions from \(u_{1.1 }\), \(v_{1.1 }\) to \(u_{3.25 }\), \(v_{3.25 }\), if \(r\rightarrow 0\) or \(r\rightarrow 1\), these solutions can also be found from \(u_{4.1 }\), \(v_{4.1}\) to \(u_{4.15 }\), \(v_{4.15}\) while \(n=1, 2\) and \(\frac{1}{2}\) respectively. Specially, when \(r\rightarrow 1\), \(u_{2.17 }\) and \(u_{3.17 }\) can be expressed by

$$\begin{aligned} u_{2.17.1}=v_{2.17} =& \biggl\{ \frac{2(b-k)}{5a}\pm \frac{2 (b-k) }{5a } \biggl(\tanh \biggl[\frac{(b-k) }{5}\sqrt{ \frac{4}{3a }}(x+y-ct )\biggr] \\ & {}\pm i\operatorname{sech}\biggl[\frac{(b-k) }{5}\sqrt{\frac{4}{3a }} \biggl(x+y+3t+ \frac{16(b-k)^{2}}{75a} \biggr)\biggr] \biggr) \biggr\} ^{2}, \end{aligned}$$
$$\begin{aligned} u_{3.17.1}=v_{3.17} =& \biggl\{ \frac{5(b-k)}{8a }+ \frac{5\epsilon (b-k)}{8a }+ \biggl(\tanh \biggl[\frac{b-k}{2}\sqrt{ \frac{5}{3a}}\biggl(x+y+3t+\frac{5(b-k)^{2}}{48 a } t\biggr)\biggr] \\ & {}\pm i\operatorname{sech}\biggl[\frac{b-k}{2}\sqrt{\frac{5}{3a}} \biggl(x+y+3t+\frac{5(b-k)^{2}}{48 a } t\biggr)\biggr] \biggr) \biggr\} ^{\frac{1}{2}}, \end{aligned}$$

which are the same as \(u_{4.6 }\), \(v_{4.6 }\) in the case of \(n= \frac{1}{2}, 2\), respectively.


The auxiliary differential equation approach has been successfully employed to seek the exact solutions of the modified \((2 + 1)\)-dimensional Konopelchenko–Dubrovsky equations (2) which possess high order nonlinear terms. After handling the mKD equations, it shows some new solutions compared with these obtained in [22, 25] by this approach. On the one hand, for special exponent n such as 1, 2, and 1/2, we derive many Jacobi function solutions, some of which can degenerate triangular function solutions and solitons. On the other hand, the paper presents abundant new exact traveling wave solutions for general exponent n.


  1. 1.

    Konopelchenko, B.G., Dubrovsky, V.G.: Some new integral nonlinear evolution equations in \((2 + 1)\) dimensions. Phys. Lett. A 102, 15–17 (1984)

  2. 2.

    Wang, M., Li, D., Zhang, C., Tang, Y., Long, Y.: Long time behavior of solutions of gKdV equations. J. Math. Anal. Appl. 390, 136–150 (2012)

  3. 3.

    Triki, H., Wazwaz, A.M.: Trial equation method for solving the generalized Fisher equation with variable coefficients. Phys. Lett. A 380, 1260–1262 (2016)

  4. 4.

    Gordoa, P.R., Pickering, A.: Auto-backlund transformations for a matrix partial differential equation. Phys. Lett. A 382, 1908–1915 (2018)

  5. 5.

    Yuan, Y.Q., Tian, B., Liu, L., et al.: Solitons for the \((2 +1)\) dimensional Konopelchenko-Dubrovsky equations. J. Math. Anal. Appl. 460, 476–486 (2018)

  6. 6.

    Cheng, X., Duan, J., Li, D.: A novel compact ADI scheme for two-dimensional Riesz space fractional nonlinear reaction-diffusion equations. Appl. Math. Comput. 346, 452–464 (2019)

  7. 7.

    Wu, F., Cheng, X., Li, D.: A two-level linearized compact ADI scheme for two-dimensional nonlinear reaction-diffusion equations. Comput. Math. Appl. 75, 2835–2850 (2018)

  8. 8.

    Wen, J., Chang, X.: On the KZ reduction. IEEE Trans. Inf. Theory 65, 1921–1935 (2019)

  9. 9.

    Zhang, Q., Zhang, C., Wang, L.: The compact and Crank-Nicolson ADI schemes for two-dimensional semilinear multidelay parabolic equations. J. Comput. Appl. Math. 306, 217–230 (2016)

  10. 10.

    Li, D., Wu, C., Zhang, Z.: Linearized Galerkin FEMs for nonlinear time fractional parabolic problems with non-smooth solutions in time direction. J. Sci. Comput. 80, 403–419 (2019)

  11. 11.

    Christov, I.: Internal solitary waves in the ocean: analysis using the periodic, inverse scattering transform. Math. Comput. Simul. 80, 192–201 (2009)

  12. 12.

    Ji, J.L., Zhu, Z.N.: Soliton solutions of an integrable nonlocal modified Korteweg-de Vries equation through inverse scattering transform. J. Math. Anal. Appl. 453, 973–984 (2017)

  13. 13.

    Bekir, A., Boz, A.: Application of exp-function method for \((2 + 1)\)-dimensional nonlinear evolution equations. Chaos Solitons Fractals 40, 458–465 (2009)

  14. 14.

    Ji, J.L., Zhu, Z.N.: Soliton solutions of an integrable nonlocal modified Korteweg-de Vries equation through inverse scattering transform. J. Math. Anal. Appl. 453, 973–984 (2017)

  15. 15.

    Zhao, Z.L.: Backlund transformations, rational solutions and soliton-cnoidal wave solutions of the modified Kadomtsev-Petviashvili equation. Appl. Math. Lett. 89, 103–110 (2019)

  16. 16.

    Alvaro, H.S. (FIZMAKO Research Group): Construction of N-soliton solutions to \((2 + 1)\)-dimensional Konopelchenko-Dubrovsky (KD) equations. Appl. Math. Comput. 217, 7391–7399 (2011)

  17. 17.

    Xu, P.B., Gao, Y.T., Gai, X.L., et al.: Soliton solutions, Backlund transformation and Wronskian solutions for the extended \((2 + 1)\)-dimensional Konopelchenko-Dubrovsky equations in fluid mechanics. Appl. Math. Comput. 218, 2489–2496 (2011)

  18. 18.

    Liu, W.H., Zhang, Y.F., Shi, D.D.: Lump waves, solitary waves and interaction phenomena to the \((2 + 1)\)-dimensional Konopelchenko-Dubrovsky equation. Phys. Lett. A 383, 97–102 (2019)

  19. 19.

    Wu, P.X., Zhang, Y.F., Muhammad, I., et al.: Complexiton and resonant multiple wave solutions to the \((2 + 1)\)-dimensional Konopelchenko-Dubrovsky equation. Comput. Math. Appl. 76, 845–853 (2018)

  20. 20.

    Feng, W.G., Lin, C.: Explicit exact solutions for the \((2 + 1)\)-dimensional Konopelchenko-Dubrovsky equation. Appl. Math. Comput. 210, 298–302 (2019)

  21. 21.

    Kumar, M., Kumar, A., Kumar, R.: Similarity solutions of the Konopelchenko-Dubrovsky system using Lie group theory. Comput. Math. Appl. 71, 2051–2059 (2016)

  22. 22.

    Wazwaz, A.M.: New kinks and solitons solutions to the \((2 + 1)\)-dimensional Konopelchenko-Dubrovsky equation. Math. Comput. Model. 45, 473–479 (2017)

  23. 23.

    Lai, S.Y., Lv, X.M., Shuai, M.Y.: The Jacobi elliptic function solutions to a generalized Benjamin–Bona–Mahony equation. Math. Comput. Model. 49, 369–378 (2009)

  24. 24.

    Lv, X.M., Shao, T.W., Chen, J.C.: The Study of the Solution to a Generalized KdV-mKdV Equation. Abstract and Applied Analysis (2013)

  25. 25.

    Zhang, S.: The periodic wave solutions for the \((2 + 1)\)-dimensional Konopelchenko-Dubrovsky equations. Chaos Solitons Fractals 30, 1213–1220 (2006)

Download references


This research was supported in part by the National Social Science Fund of China 19XYJ022, NSFC grants 11401591, and in part by the Chongqing Municipal Education Commission Fund under Grant YJG183097.

Author information




All authors contributed equally and significantly in writing this paper. All authors have read and approved the final paper.

Corresponding author

Correspondence to Wei Gu.

Ethics declarations

Competing interests

The authors declare that they have no competing interests.

Additional information

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (, which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Lyu, X., Gu, W. Solitons for the modified \((2 + 1)\)-dimensional Konopelchenko–Dubrovsky equations. Adv Differ Equ 2019, 443 (2019).

Download citation


  • 35Q53
  • 35B


  • Modified Konopelchenko–Dubrovsky equations
  • Auxiliary equation approach
  • Jacobi elliptic function solutions
  • Degenerated triangle function solutions
  • Solitons