# On classification of the solutions for general elliptic equation

Sirendaoreji
Mathematical Science College,Inner Mongolia Normal University,
Huhhot 010022,PR China
Email<EMAIL_ADDRESS>
(October,11,2018)

Abstract: The Bäcklund transformations and the superposition formulas for two
sub–equations of the general elliptic equation are constructed from the
Riccati equation by using an indirect mapping method. The thirty-six
previously known solutions of the general elliptic equation are proved to be
equivalent to the another ten solutions. The classification of solutions for
the general elliptic equation is obtained based on the equivalence relations.
The general elliptic equation expansion method with new classified solutions
are used to obtain abundant new exact traveling wave solutions of a modified
Camassa-Holm equation.

##  1 Introduction

Seeking exact solutions of nonlinear evolution equations (NLEEs) is a
practical and attractive problem in mathematical physics and nonlinear
science. People have made continuous effort on the problem and various direct
methods have been proposed to construct exact solutions for NLEEs. Among them
the Jacobi elliptic function expansion method [1],the F–expansion method
[2],the auxiliary equation method [3] and the sub-equation method [4, 5, 6]
are related to the general elliptic equation of the form [7, 8, 9, 10, 11, 12]

${F^{\prime}}^{2}(\xi)=c_{0}+c_{1}F(\xi)+c_{2}F^{2}(\xi)+c_{3}F^{3}(\xi)+c_{4}F^{4}(\xi),$
(1)

where $c_{i}~{}(i=0,1,\cdots,4)$ are constants.Except for the sub–equation
method ,the Eq.(1) was also chosen as the auxiliary equation in other direct
methods.It is the fact that the Eq.(1) with
$c_{0}=1,c_{1}=0,c_{2}=-(m^{2}+1),c_{3}=0,c_{4}=m^{2}$ and
$c_{0}=1-m^{2},c_{1}=0,c_{2}=2m^{2}-1,c_{3}=0,c_{4}=-m^{2}$ were employed in
the Jacobi elliptic function expansion method to seek the snoidal wave
solutions and the cnoidal wave solutions of NLEEs,respectively. The
F–expansion method for seeking twelve types of Jacobi elliptic function
solutions of NLEEs was proposed by using the following sub-equation

${F^{\prime}}^{2}(\xi)=c_{0}+c_{2}F^{2}(\xi)+c_{4}F^{4}(\xi).$ (2)

Another sub–equation [13, 14, 15, 16, 17, 18, 19, 20, 21]

${F^{\prime}}^{2}(\xi)=c_{2}F^{2}(\xi)+c_{3}F^{3}(\xi)+c_{4}F^{4}(\xi),$ (3)

or

${G^{\prime}}^{2}(\xi)=h_{2}G^{2}(\xi)+h_{4}G^{4}(\xi)+h_{6}G^{6}(\xi),$ (4)

were used in the auxiliary equation method to find different types of exact
traveling wave solutions of NLEEs simultaneously. Because of Eq.(3) and Eq.(4)
can be transformed into each other by the transformation

$G(\xi)=F^{\frac{1}{2}}(\xi),h_{2}={\frac{c_{2}}{4}},h_{4}={\frac{c_{3}}{4}},h_{6}={\frac{c_{4}}{4}},$
(5)

so we only need to study Eq.(3).In addition,the following sub–equations [22,
23, 24]

${F^{\prime}}^{2}(\xi)=c_{0}+c_{1}F(\xi)+c_{2}F^{2}(\xi)+c_{3}F^{3}(\xi),$ (6)
${F^{\prime}}^{2}(\xi)=c_{0}+c_{1}F(\xi)+c_{2}F^{2}(\xi),$ (7)

were also widely applied to seek exact traveling wave solutions of NLEEs. The
Eq.(1) generalized to the case of involving six degree nonlinear term was also
studied by many authors [25, 26, 27, 28, 29, 30, 31], but according to the
relation (5) their results can be converted into the case of Eq.(3). All these
facts show that the Eq.(1) has a strong application background and its
solutions are worth to further study. At present, in the area of studying
exact solutions of Eq.(1),there exist some problems such as

  * (1)

Many different solutions have been constructed and some of them are found to
be equivalent,but these equivalence relations have not been proved.

  * (2)

The classification of its solutions is not given.

  * (3)

Its theoretical system such as the Bäcklund transformation (BT) and the
superposition formula (SF) has not been established.

For this reason,the paper aims to give a systematic study of Eq.(1) and try to
solve the problems in a comprehensive way. This effort lead us to organize the
paper as follows. In Sec.2,we shall construct the BT and the SF for the sub-
equations of Eq.(1).This lead us to construct the BT and the SF of Eq.(3) and
Eq.(7) from the Riccati equation in terms of an indirect mapping method. In
Sec.3,we shall give the proof of the equivalence relations of the known
solutions for Eq.(1). As a result,the thirty six previously known solutions
are proved to be equivalent to the another ten solutions for Eq.(1). In
Sec.4,we shall give the classification of the solutions for Eq.(1).It is shown
that the solutions of Eq.(1) are classified into five classes which consist of
thirty eight independent solutions. In Sec.5,we shall construct abundant new
exact traveling wave solutions for the modified Camassa–Holm equation by means
of the general elliptic equation expansion method.Sec.6 is devoted to
concluding remarks.

##  2 Bäcklund transformations and superposition formulas

Although the BT and the SF of Eq.(1) is difficult to establish in a simple
way,but it usually can be constructed for its sub-equations through use of
some tedious integration procedure.For avoid the complexity,below we shall
construct the BT and the SF for Eq.(3) and Eq.(7) by using an indirect mapping
method.

###  2.1 The Bäcklund transformation and superposition formula of Eq.(3)

Introducing the transformation

$F(\xi)={\frac{4f^{2}(\xi)-c_{2}}{c_{3}-4\sqrt{c_{4}}f(\xi)}},$ (8)

which,substituted into the Eq.(3),implies that

$\displaystyle\left({\frac{dF}{d\xi}}\right)^{2}-c_{2}F^{2}-c_{3}F^{3}-c_{4}F^{4}={\frac{16\left({\frac{df}{d\xi}}-f^{2}+{\frac{c_{2}}{4}}\right)\left({\frac{df}{d\xi}}+f^{2}-{\frac{c_{2}}{4}}\right)H(f)}{\left(4\sqrt{c_{4}}f-c_{3}\right)^{4}}},$
$\displaystyle
H(f)=16c_{4}f^{4}-16c_{3}\sqrt{c_{4}}f^{3}+\left(4c_{3}^{2}+8c_{2}c_{4}\right)f^{2}-4c_{2}c_{3}\sqrt{c_{4}}f+c_{2}^{2}c_{4}.$

From this identity we deduce that $F(\xi)$ is a solution of Eq.(3) if and only
if $f(\xi)$ is a solutions of the Riccati equation

$f^{\prime}(\xi)=f^{2}(\xi)-{\frac{1}{4}}{c_{2}}.$ (9)

This shows that the BT and the SF of Eq.(3) can be induced from the BT and the
SF of the Riccati equation (9). Using the BT and SF for Riccati equation [32],
we obtain the BT and the SF of Eq.(9) as follows

$f_{n}={\frac{f_{n-1}+{\frac{c_{2}}{4}}\alpha}{1+\alpha{f_{n-1}}}}={\frac{4f_{n-1}+\alpha{c_{2}}}{4+4\alpha{f_{n-1}}}},$
(10)

and

$f_{3}={\frac{\alpha_{1}f_{2}-\alpha_{2}f_{1}}{\alpha_{1}f_{1}-\alpha_{2}f_{2}}}f_{0},$
(11)

where $f_{n},f_{n-1},f_{i}~{}(i=0,1,2,3)$ are solutions of Eq.(9) and
$\alpha,\alpha_{i}~{}(i=1,2)$ are arbitrary constants. Taking the mapping (8)
into consideration,we are confident that two solutions $F_{n}$ and $F_{n-1}$
of Eq.(3) can be written as

$\displaystyle F_{n}={\frac{4f_{n}^{2}-c_{2}}{c_{3}-4\sqrt{c_{4}}f_{n}}},$
(12) $\displaystyle
F_{n-1}={\frac{4f_{n-1}^{2}-c_{2}}{c_{3}-4\sqrt{c_{4}}f_{n-1}}}.$ (13)

It solves from (13) that

$f_{n-1}=-{\frac{1}{2}}\sqrt{c_{4}}F_{n-1}+{\frac{\varepsilon}{2}}T_{n-1},T_{n-1}=\sqrt{c_{2}+c_{3}F_{n-1}+c_{4}F_{n-1}^{2}},\varepsilon=\pm{1}.$
(14)

Therefore,substituting (10) with (14) into (12) leads the BT of Eq.(3) of the
form

$F_{n}={\frac{H_{n-1}^{2}-c_{2}}{c_{3}-2\sqrt{c_{4}}H_{n-1}}},H_{n-1}={\frac{2(\varepsilon{T_{n-1}}-\sqrt{c_{4}}F_{n-1})+\alpha{c_{2}}}{2+\alpha(\varepsilon{T_{n-1}}-\sqrt{c_{4}}F_{n-1})}},$
(15)

where $T_{n-1}$ is determined by (14).It is now easy to see from (11) and (12)
that the SF of Eq.(3) is given by

$\left\\{\begin{array}[]{l}F_{3a}={\frac{4f_{3}^{2}-c_{2}}{c_{3}-4\sqrt{c_{4}}f_{3}}},f_{3}={\frac{\alpha_{1}f_{2}-\alpha_{2}f_{1}}{\alpha_{1}f_{1}-\alpha_{2}f_{2}}}f_{0},\\\
f_{n}={\frac{1}{2}}\left(\varepsilon{T_{n}}-\sqrt{c_{4}}F_{n}\right),T_{n}=\sqrt{c_{2}+c_{3}F_{n}+c_{4}F_{n}^{2}},n=0,1,2.\end{array}\right.$
(16)

###  2.2 The Bäcklund transformation and superposition formula of Eq.(7)

By virtue of the indirect mapping method as used above,we define a mapping

$F(\xi)={\frac{c_{1}+4\sqrt{c_{0}}g(\xi)}{4g^{2}(\xi)-c_{2}}},$ (17)

and substitute it into Eq.(7) leads

$\displaystyle\left({\frac{dF}{d\xi}}\right)^{2}-c_{0}-c_{1}F-c_{2}F^{2}={\frac{16\left({\frac{dg}{d\xi}}-g^{2}+{\frac{c_{2}}{4}}\right)\left({\frac{dg}{d\xi}}+g^{2}-{\frac{c_{2}}{4}}\right)K(f)}{\left(4g^{2}-c_{2}\right)^{4}}},$
$\displaystyle
K(f)=16c_{0}g^{4}+16c_{1}\sqrt{c_{0}}g^{3}+\left(4c_{1}^{2}+8c_{0}c_{2}\right)g^{2}+4c_{1}c_{2}\sqrt{c_{0}}g+c_{0}c_{2}^{2}.$

This indicates that $F(\xi)$ is a solution of Eq.(7) if and only if $g(\xi)$
satisfies the Riccati equation

$g^{\prime}(\xi)=g^{2}(\xi)-{\frac{1}{4}}{c_{2}}.$ (18)

Following the same procedure as used to derive the BT and the SF for Eq.(9),
we can now obtain the BT and the SF of Eq.(18)

$g_{n}={\frac{4g_{n-1}+\beta{c_{2}}}{4+4\beta{g_{n-1}}}},$ (19)

and

$g_{3}={\frac{\beta_{1}{g_{2}}-\beta_{2}{g_{1}}}{\beta_{1}{g_{1}}-\beta_{2}{g_{2}}}}g_{0},$
(20)

respectively.It can be induced from the mapping (17) that

$\displaystyle F_{n}={\frac{c_{1}+4\sqrt{c_{0}}g_{n}}{4g_{n}^{2}-c_{2}}},$
(21) $\displaystyle
F_{n-1}={\frac{c_{1}+4\sqrt{c_{0}}g_{n-1}}{4g_{n-1}^{2}-c_{2}}}.$ (22)

At the same time,it solves from (22) that

$g_{n-1}={\frac{1}{2F_{n-1}}}\left(\sqrt{c_{0}}+\varepsilon{R_{n-1}}\right),R_{n-1}=\sqrt{c_{0}+c_{1}F_{n-1}+c_{2}F_{n-1}^{2}},\varepsilon=\pm{1}.$
(23)

The BT of Eq.(7) cab be obtained by taking (19) and (23) into Eq.(21),which is
now determined to be

$F_{n}={\frac{c_{1}+2\sqrt{c_{0}}K_{n-1}}{K_{n-1}^{2}-c_{2}}},K_{n-1}={\frac{2\varepsilon{R_{n-1}}+\beta{c_{2}}F_{n-1}+2\sqrt{c_{0}}}{\beta\varepsilon{R_{n-1}}+2F_{n-1}+\beta\sqrt{c_{0}}}}.$
(24)

Finally,the superposition formula of Eq.(7) is found to be

$\left\\{\begin{array}[]{l}F_{3a}={\frac{c_{1}+4\sqrt{c_{0}}g_{3}}{4g_{3}^{2}-c_{2}}},9_{3}={\frac{\beta_{1}{g_{2}}-\beta_{2}{g_{1}}}{\beta_{1}{g_{1}}-\beta_{2}{g_{2}}}}g_{0},\\\
g_{n}={\frac{\sqrt{c_{0}}+\varepsilon{R_{n}}}{2F_{n}}},R_{n}=\sqrt{c_{0}+c_{1}F_{n}+c_{2}F_{n}^{2}},n=0,1,2.\end{array}\right.$
(25)

##  3 Equivalence relations for solutions

###### Definition 1

Two waves are called equivalence in the sense of wave translation if they have
same waveform but their phase difference is equal to a constant.Here this
equivalence relation will be expressed by the notation “$\cong$”.

###### Lemma 1

For any real numbers $\alpha$ and $\beta$,if $\alpha>{0},\beta>{0}$,then we
hold that

$\displaystyle\left\\{\begin{array}[]{l}\alpha{e^{\xi}}+\beta{e^{-\xi}}=2\sqrt{\alpha\beta}\cosh(\xi+{\frac{1}{2}}\ln{\frac{\alpha}{\beta}}),\\\
\alpha{e^{\xi}}-\beta{e^{-\xi}}=2\sqrt{\alpha\beta}\sinh(\xi+{\frac{1}{2}}\ln{\frac{\alpha}{\beta}}).\end{array}\right.$

###### Lemma 2

For any real numbers $A$ and $B$,if $A^{2}>B^{2}$,then we have

$\displaystyle\left\\{\begin{array}[]{l}A\sinh\eta\pm{B}\cosh\eta=\sqrt{A^{2}-B^{2}}\sinh\left(\eta\pm{\frac{1}{2}}\ln{\frac{A+B}{A-B}}\right),A>B,\\\
A\sinh\eta\pm{B}\cosh\eta=-\sqrt{A^{2}-B^{2}}\sinh\left(\eta\pm{\frac{1}{2}}\ln{\frac{A+B}{A-B}}\right),A<B.\end{array}\right.$

###### Lemma 3

For any real numbers $A$ and $B$,if $A^{2}<B^{2}$,then we have

$\displaystyle\left\\{\begin{array}[]{l}A\sinh\eta\pm{B}\cosh\eta=\mp\sqrt{B^{2}-A^{2}}\cosh\left(\eta\pm{\frac{1}{2}}\ln{\frac{B+A}{B-A}}\right),A>B,\\\
A\sinh\eta\pm{B}\cosh\eta=\pm\sqrt{B^{2}-A^{2}}\cosh\left(\eta\pm{\frac{1}{2}}\ln{\frac{B+A}{B-A}}\right),A<B.\end{array}\right.$

###### Lemma 4

For any real numbers $A$ and $B$,we have

$\displaystyle\left\\{\begin{array}[]{l}A\sin\eta\pm{B}\cos\eta=\sqrt{A^{2}+B^{2}}\sin\left(\eta\pm\theta\right),\\\
A\cos\eta\pm{B}\sin\eta=\sqrt{A^{2}+B^{2}}\cos\left(\eta\mp\theta\right),\end{array}\right.$

where $\theta=\arctan(B/A)$.

The above four lemmas are very important in classifying the traveling wave
solutions of NLEEs.The proof of the Lemma 1 and the Lemma 4 are quite
evidently.The Lemma 2 and the Lemma 3 can be proved by using the Lemma 1.So
the proofs are omitted here.

###  3.1 The equivalence relations for solutions of Eq.(3)

We first prove the solutions of Eq.(3) given by Table 1 are equivalent to the
solutions given by Table 2.The fourteen solutions were used in our previous
studies [14, 18].

Table 1: Solutions of Eq.(3) with $\Delta=c_{3}^{2}-4c_{2}c_{4},\varepsilon=\pm{1}$. No | $z(\xi)$ | Condition |  No | $z(\xi)$ | Condition
---|---|---|---|---|---
1 | ${\frac{-c_{2}c_{3}\mathrm{sech}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}{c_{3}^{2}-c_{2}c_{4}\left(1+\varepsilon\tanh\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right)^{2}}}$ | $c_{2}>0$ |  5 | ${\frac{-c_{2}\sec^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}{c_{3}+2\varepsilon\sqrt{-c_{2}c_{4}}\tan\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}}$ | $c_{2}<0,c_{4}>0$
2 | ${\frac{c_{2}c_{3}\mathrm{csch}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}{c_{3}^{2}-c_{2}c_{4}\left(1+\varepsilon\coth\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right)^{2}}}$ | $c_{2}>0$ |  6 | ${\frac{-c_{2}\csc^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}{c_{3}+2\varepsilon\sqrt{-c_{2}c_{4}}\cot\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}}$ | $c_{2}<0,c_{4}>0$
3 | ${\frac{-c_{2}\mathrm{sech}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}{c_{3}+2\varepsilon\sqrt{c_{2}c_{4}}\tanh\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}}$ | $c_{2}>0,c_{4}>0$ |  7 | ${\frac{4c_{2}{e^{\varepsilon\sqrt{c_{2}}\xi}}}{(e^{\varepsilon\sqrt{c_{2}}\xi}-c_{3})^{2}-4c_{2}c_{4}}}$ | $c_{2}>0$
4 | ${\frac{c_{2}\mathrm{csch}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}{c_{3}+2\varepsilon\sqrt{c_{2}c_{4}}\coth\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}}$ | $c_{2}>0,c_{4}>0$ |  8 | ${\frac{4\varepsilon{e^{\varepsilon\sqrt{c_{2}}\xi}}}{1-4c_{2}c_{4}e^{2\varepsilon\sqrt{c_{2}}\xi}}}$ | $c_{2}>0,c_{3}=0$
Table 2: Solutions of Eq.(3) with $\Delta=c_{3}^{2}-4c_{2}c_{4},\varepsilon=\pm{1}$. No | $F(\xi)$ | Condition |  No | $F(\xi)$ | Condition
---|---|---|---|---|---
1 | ${\frac{2c_{2}}{\varepsilon\sqrt{\Delta}\cosh\left(\sqrt{c_{2}}\xi\right)-c_{3}}}$ | $\Delta>0,c_{2}>0$ |  3b | ${\frac{2c_{2}}{\varepsilon\sqrt{\Delta}\sin\left(\sqrt{-c_{2}}\xi\right)-c_{3}}}$ | $\Delta>0,c_{2}<0$
2 | ${\frac{2c_{2}}{\varepsilon\sqrt{-\Delta}\sinh\left(\sqrt{c_{2}}\xi\right)-c_{3}}}$ | $\Delta<0,c_{2}>0$ |  4 | $-{\frac{c_{2}}{c_{3}}}\left[1+\varepsilon\tanh\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right]$ | $\Delta=0,c_{2}>0$
3a | ${\frac{2c_{2}}{\varepsilon\sqrt{\Delta}\cos\left(\sqrt{-c_{2}}\xi\right)-c_{3}}}$ | $\Delta>0,c_{2}<0$ |  5 | $-{\frac{c_{2}}{c_{3}}}\left[1+\varepsilon\coth\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right]$ | $\Delta=0,c_{2}>0$

Let $A=2c_{2}c_{4},B=c_{3}^{2}-2c_{2}c_{4}$,then we find
$A^{2}-B^{2}=-c_{3}^{2}\Delta$.Hence the condition $\Delta>0$ implies that
$A^{2}<B^{2}$ and $A<B$.But for condition $\Delta<0$ we obtain that
$A^{2}>B^{2}$ and $A>B$.Therefore,noting the definition of hyperbolic
functions,using the formulas

$\sinh{x}=2\sinh{\frac{x}{2}}\cosh{\frac{x}{2}},\cosh^{2}{\frac{x}{2}}={\frac{\cosh{x}+1}{2}},\sinh^{2}{\frac{x}{2}}={\frac{\cosh{x}-1}{2}},$
(30)

the Lemma 3 and the Lemma 2 we can derive that

$\displaystyle\begin{array}[]{l}z_{1}(\xi)={\frac{-2c_{2}c_{3}}{c_{3}^{2}\mp\left[2c_{2}c_{4}\sinh\left(\sqrt{c_{2}}\xi\right)\right.\left.\mp\left(c_{3}^{2}-2c_{2}c_{4}\right)\cosh\left(\sqrt{c_{2}}\xi\right)\right]}},\\\
\quad=\left\\{\begin{array}[]{l}{\frac{2c_{2}}{\varepsilon\sqrt{\Delta}\cosh\left(\sqrt{c_{2}}\xi\mp{\frac{1}{2}}\ln({\frac{c_{3}^{2}}{\Delta}})\right)-c_{3}}}\cong{F_{1}(\xi)},\Delta>0,c_{2}>0,\\\
{\frac{2c_{2}}{\varepsilon\sqrt{-\Delta}\sinh\left(\sqrt{c_{2}}\xi\mp{\frac{1}{2}}\ln({\frac{c_{3}^{2}}{-\Delta}})\right)-c_{3}}}\cong{F_{2}(\xi)},\Delta<0,c_{2}>0.\end{array}\right.\end{array}$

Following the same procedure we can derive that
$z_{2}(\xi)\cong{F_{1}(\xi)}$,for $\Delta>0,c_{2}>0$ and
$z_{2}(\xi)\cong{F_{2}(\xi)}$,for $\Delta<0,c_{2}>0$ and the details of the
proof is omitted here.
Set $A=2\sqrt{c_{2}c_{4}},B=c_{3}$,then $A^{2}-B^{2}=-\Delta$.Thus for
$\Delta>0$ we hold $A^{2}<B^{2}$.But for $\Delta<0$,we have
$A^{2}>B^{2}$.Hence for $c_{2}>0,c_{4}>0$, the formulas (30),the Lemma 2 and
the Lemma 3 are used to obtain that

$\displaystyle\begin{array}[]{l}z_{3}(\xi)={\frac{-2c_{2}}{c_{3}\pm\left[2\sqrt{c_{2}c_{4}}\sinh\left(\sqrt{c_{2}}\xi\right)\pm{c_{3}}\cosh\left(\sqrt{c_{2}}\xi\right)\right]}},\\\
\quad=\left\\{\begin{array}[]{l}{\frac{2c_{2}}{\varepsilon\sqrt{\Delta}\cosh\left(\sqrt{c_{2}}\xi\pm{\frac{1}{2}}\ln({\frac{c_{3}+2\sqrt{c_{2}c_{4}}}{c_{3}-2\sqrt{c_{2}c_{4}}}})\right)-c_{3}}}\cong{F_{1}(\xi)},\Delta>0,\\\
{\frac{2c_{2}}{\varepsilon\sqrt{-\Delta}\sinh\left(\sqrt{c_{2}}\xi\pm{\frac{1}{2}}\ln({\frac{2\sqrt{c_{2}c_{4}}+c_{3}}{2\sqrt{c_{2}c_{4}}-c_{3}}})\right)-c_{3}}}\cong{F_{2}(\xi)},\Delta<0.\end{array}\right.\end{array}$

Applying the same procedure as used to prove $z_{3}(\xi)$,we obtain the
results $z_{4}(\xi)\cong{F_{1}(\xi)},\Delta>0$ and
$z_{4}(\xi)\cong{F_{2}(\xi)},\Delta<0$.

If $\Delta=0$,then $4c_{2}c_{4}=c_{3}^{2}$.In this case,for $c_{2}>0,c_{4}>0$
we find

$\displaystyle\begin{array}[]{l}z_{3}(\xi)={\frac{-c_{2}\left(1-\tanh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right)}{c_{3}+\varepsilon{c_{3}}\tanh\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}}=-{\frac{c_{2}}{c_{3}}}\left[1+\varepsilon\tanh\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right]=F_{4}(\xi),\\\
z_{4}(\xi)={\frac{-c_{2}\left(1-\coth^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right)}{c_{3}+\varepsilon{c_{3}}\coth\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}}=-{\frac{c_{2}}{c_{3}}}\left[1+\varepsilon\coth\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right]=F_{5}(\xi).\end{array}$

When $\Delta>0,c_{2}<0,c_{4}>0$,using the expressions
$\sec{x}={\frac{1}{\cos{x}}},\csc{x}={\frac{1}{\sin{x}}}$,the identities

$\sin{x}=2\sin{\frac{x}{2}}\cos{\frac{x}{2}},\sin^{2}{\frac{x}{2}}={\frac{1-\cos{x}}{2}},\cos^{2}{\frac{x}{2}}={\frac{1+\cos{x}}{2}},$
(36)

and the Lemma 4,the solutions $z_{5}(\xi)$ can be simplified as

$\displaystyle\begin{array}[]{l}z_{5}(\xi)={\frac{-2c_{2}}{2\varepsilon\sqrt{-c_{2}c_{4}}\sin\left(\sqrt{-c_{2}}\xi\right)+{c_{3}}\cos\left(\sqrt{-c_{2}}\xi\right)+c_{3}}},\\\
\quad=\left\\{\begin{array}[]{l}{\frac{2c_{2}}{-\sqrt{\Delta}\cos\left(\sqrt{-c_{2}}\xi\mp\theta_{1}\right)-c_{3}}}{\cong}F_{3a}(\xi)|_{\varepsilon=-1},\\\
{\frac{2c_{2}}{\varepsilon\sqrt{\Delta}\sin\left(\sqrt{-c_{2}}\xi\pm\theta_{2}\right)-c_{3}}}{\cong}F_{3b}(\xi),\end{array}\right.\end{array}$

where
$\theta_{1}=\arctan\left({\frac{2\sqrt{-c_{2}c_{4}}}{c_{3}}}\right),\theta_{2}=\arctan\left({\frac{c_{3}}{2\sqrt{-c_{2}c_{4}}}}\right)$.
Similarly,when $\Delta>0,c_{2}<0,c_{4}>0$,we can prove that
$z_{6}(\xi){\cong}F_{3a}(\xi)|_{\varepsilon=-1}$ and
$z_{6}(\xi){\cong}F_{3b}(\xi)$.

Converting $z_{7}(\xi)$ into the exponential form and then using the Lemma
1,we obtain

$\displaystyle\begin{array}[]{l}z_{7}(\xi)={\frac{4c_{2}}{e^{\varepsilon\sqrt{c_{2}}\xi}+(c_{3}^{2}-4c_{2}c_{4})e^{-\varepsilon\sqrt{c_{2}}\xi}-2c_{3}}},\\\
\quad=\left\\{\begin{array}[]{l}{\frac{2c_{2}}{\sqrt{\Delta}\cosh\left(\varepsilon\sqrt{c_{2}}\xi+{\frac{1}{2}}\ln({\frac{1}{\Delta}})\right)-c_{3}}},\\\
{\frac{2c_{2}}{\sqrt{-\Delta}\sinh\left(\varepsilon\sqrt{c_{2}}\xi+{\frac{1}{2}}\ln({\frac{1}{-\Delta}})\right)-c_{3}}},\end{array}\right.\\\
\quad=\left\\{\begin{array}[]{l}{\frac{2c_{2}}{\sqrt{\Delta}\cosh\left(\sqrt{c_{2}}\xi\pm{\frac{1}{2}}\ln({\frac{1}{\Delta}})\right)-c_{3}}}\cong{F_{1}(\xi)|_{\varepsilon=1},\Delta>0,c_{2}>0},\\\
{\frac{2c_{2}}{\varepsilon\sqrt{-\Delta}\sinh\left(\sqrt{c_{2}}\xi\pm{\frac{1}{2}}\ln({\frac{1}{-\Delta}})\right)-c_{3}}}\cong{F_{2}(\xi)},\Delta<0,c_{2}>0.\end{array}\right.\end{array}$

When $\Delta=0,c_{2}>0$,it is obtained from the identities
${\frac{1}{1+e^{x}}}={\frac{1}{2}}\left(1-\tanh{\frac{x}{2}}\right)$ and
${\frac{1}{1-e^{x}}}={\frac{1}{2}}\left(1-\coth{\frac{x}{2}}\right)$ that

$\displaystyle\begin{array}[]{l}z_{7}(\xi)={\frac{4c_{2}}{e^{\varepsilon\sqrt{c_{2}}\xi}-2c_{3}}}=\left\\{\begin{array}[]{l}-{\frac{2c_{2}}{c_{3}}}{\frac{1}{1-e^{\varepsilon\sqrt{c_{2}}\xi+\ln({\frac{1}{2c_{3}}})}}},\\\
-{\frac{2c_{2}}{c_{3}}}{\frac{1}{1+e^{\varepsilon\sqrt{c_{2}}\xi+\ln({\frac{-1}{2c_{3}}})}}},\\\
\end{array}\right.\\\
\quad=\left\\{\begin{array}[]{l}-{\frac{c_{2}}{c_{3}}}\left[1-\coth\left({\frac{\varepsilon\sqrt{c_{2}}}{2}\xi+\ln({\frac{1}{2c_{3}}})}\right)\right],\\\
-{\frac{c_{2}}{c_{3}}}\left[1-\tanh\left({\frac{\varepsilon\sqrt{c_{2}}}{2}\xi+\ln({\frac{-1}{2c_{3}}})}\right)\right],\\\
\end{array}\right.\\\
\quad=\left\\{\begin{array}[]{l}-{\frac{c_{2}}{c_{3}}}\left[1+\varepsilon\coth\left({\frac{\sqrt{c_{2}}}{2}\xi\pm\ln({\frac{1}{2c_{3}}})}\right)\right]\cong{F_{5}(\xi)},c_{3}>0,\\\
-{\frac{c_{2}}{c_{3}}}\left[1+\varepsilon\tanh\left({\frac{\sqrt{c_{2}}}{2}\xi\pm\ln({\frac{-1}{2c_{3}}})}\right)\right]\cong{F_{4}(\xi)},c_{3}<0.\end{array}\right.\end{array}$

For $c_{2}>0,c_{3}=0$,it yields from the Lemma 1 that

$\displaystyle\begin{array}[]{l}z_{8}(\xi)={\frac{4\varepsilon{c_{2}}}{e^{-\varepsilon\sqrt{c_{2}}\xi}-4c_{2}c_{4}e^{\varepsilon\sqrt{c_{2}}\xi}}}=\left\\{\begin{array}[]{l}{\frac{4\varepsilon{c_{2}}}{2\sqrt{-4c_{2}c_{4}}\cosh\left(\varepsilon\sqrt{c_{2}}\xi+{\frac{1}{2}}\ln(-4c_{2}c_{4})\right)}},\\\
{\frac{4\varepsilon{c_{2}}}{-2\sqrt{4c_{2}c_{4}}\sinh\left(\varepsilon\sqrt{c_{2}}\xi+{\frac{1}{2}}\ln(4c_{2}c_{4})\right)}},\end{array}\right.\\\
\quad=\left\\{\begin{array}[]{l}\varepsilon\sqrt{-\frac{c_{2}}{c_{4}}}\mathrm{sech}\left(\sqrt{c_{2}}\xi\pm{\frac{1}{2}}\ln(-4c_{2}c_{4})\right)\cong{F_{1}(\xi),c_{4}<0},\\\
\varepsilon\sqrt{\frac{c_{2}}{c_{4}}}\mathrm{csch}\left(\sqrt{c_{2}}\xi\pm{\frac{1}{2}}\ln(4c_{2}c_{4})\right)\cong{F_{2}(\xi),c_{4}>0.}\end{array}\right.\end{array}$

The above obtained results on the equivalence relations for solutions of
Eq.(3) are consistent with the Liu’s results [33, 34], but our proofs are more
simple and clear than that of Liu’s proofs because the proving processes have
been simplified by the above four Lemmas.

We next consider the following four solutions [35]

$\displaystyle\begin{array}[]{l}\phi_{5}(\xi)={\frac{-2c_{2}\mathrm{sech}\left(\sqrt{c_{2}}\xi\right)}{c_{3}\mathrm{sech}\left(\sqrt{c_{2}}\xi\right)+\sqrt{4c_{2}^{2}-c_{3}^{2}+4c_{2}c_{4}}\tanh\left(\sqrt{c_{2}}\xi\right)-2c_{2}}},c_{2}>0,\\\
\phi_{6}(\xi)={\frac{-2c_{2}\mathrm{csch}\left(\sqrt{c_{2}}\xi\right)}{c_{3}\mathrm{csch}\left(\sqrt{c_{2}}\xi\right)-\sqrt{4c_{2}^{2}+c_{3}^{2}-4c_{2}c_{4}}\coth\left(\sqrt{c_{2}}\xi\right)+2c_{2}}},c_{2}>0,\\\
\phi_{7}(\xi)={\frac{-2c_{2}\sec\left(\sqrt{-c_{2}}\xi\right)}{c_{3}\sec\left(\sqrt{-c_{2}}\xi\right)+\sqrt{c_{3}^{2}-4c_{2}^{2}-4c_{2}c_{4}}\tan\left(\sqrt{-c_{2}}\xi\right)-2c_{2}}},c_{2}<0,\\\
\phi_{8}(\xi)={\frac{-2c_{2}\csc\left(\sqrt{-c_{2}}\xi\right)}{c_{3}\csc\left(\sqrt{-c_{2}}\xi\right)-\sqrt{c_{3}^{2}-4c_{2}^{2}-4c_{2}c_{4}}\cot\left(\sqrt{-c_{2}}\xi\right)+2c_{2}}},c_{2}<0.\end{array}$

Let $A=\sqrt{4c_{2}^{2}-c_{3}^{2}+4c_{2}c_{4}},B=2c_{2}$,then we get
$A^{2}-B^{2}=-\Delta$. However,for $c_{2}>0$,the condition $\Delta>0$ yields
that $A^{2}<B^{2}$ and $A<B$. And the condition $\Delta<0$ gives that
$A^{2}>B^{2}$ and $A>B$. Therefore,when $c_{2}>0$,it follows from the Lemma 3
and the Lemma 2 that

$\displaystyle\begin{array}[]{l}\phi_{5}(\xi)={\frac{-2c_{2}}{c_{3}-2c_{2}\cosh\left(\sqrt{c_{2}}\xi\right)+\sqrt{4c_{2}^{2}-c_{3}^{2}+4c_{2}c_{4}}\sinh\left(\sqrt{c_{2}}\xi\right)}},\\\
\quad=\left\\{\begin{array}[]{l}{\frac{2c_{2}}{\sqrt{\Delta}\cosh\left(\sqrt{c_{2}}\xi-{\frac{1}{2}}\ln(\theta_{1})\right)-c_{3}}}\cong{F_{1}(\xi)}|_{\varepsilon=1},\Delta>0,\\\
{\frac{2c_{2}}{-\sqrt{-\Delta}\sinh\left(\sqrt{c_{2}}\xi-{\frac{1}{2}}\ln(\theta_{2})\right)-c_{3}}}\cong{F_{2}(\xi)}|_{\varepsilon=-1},\Delta<0.\end{array}\right.\end{array}$

where

$\displaystyle\theta_{1}={\frac{2c_{2}+\sqrt{4c_{2}^{2}-c_{3}^{2}+4c_{2}c_{4}}}{2c_{2}-\sqrt{4c_{2}^{2}-c_{3}^{2}+4c_{2}c_{4}}}},\theta_{2}={\frac{\sqrt{4c_{2}^{2}-c_{3}^{2}+4c_{2}c_{4}}+2c_{2}}{\sqrt{4c_{2}^{2}-c_{3}^{2}+4c_{2}c_{4}}-2c_{2}}}.$

By introducing $A=2c_{2},B=\sqrt{4c_{2}^{2}+c_{3}^{2}-4c_{2}c_{4}}$,we find
that $A^{2}-B^{2}=-\Delta$. When $c_{2}>0$,the condition $\Delta>0$ implies
that $A^{2}<B^{2}$ and $A<B$.On the other hand,from the condition $\Delta<0$
we can deduce that $A^{2}>B^{2}$ and $A>B$.Hence for $c_{2}>0$,it is obtained
from the Lemma 3 and the Lemma 2 that

$\displaystyle\begin{array}[]{l}\phi_{6}(\xi)={\frac{-2c_{2}}{c_{3}+{2c_{2}}\sinh\left(\sqrt{c_{2}}\xi\right)-\sqrt{4c_{2}^{2}+c_{3}^{2}-4c_{2}c_{4}}\cosh\left(\sqrt{c_{2}}\xi\right)}},\\\
\quad=\left\\{\begin{array}[]{l}{\frac{2c_{2}}{\sqrt{\Delta}\cosh\left(\sqrt{c_{2}}\xi-{\frac{1}{2}}\ln(\theta_{1})\right)-c_{3}}}\cong{F_{1}(\xi)}|_{\varepsilon=1},\Delta>0,\\\
{\frac{2c_{2}}{-\sqrt{-\Delta}\sinh\left(\sqrt{c_{2}}\xi-{\frac{1}{2}}\ln(\theta_{2})\right)-c_{3}}}\cong{F_{2}(\xi)}|_{\varepsilon=-1},\Delta<0.\end{array}\right.\end{array}$

where

$\displaystyle\theta_{1}={\frac{\sqrt{4c_{2}^{2}+c_{3}^{2}-4c_{2}c_{4}}+2c_{2}}{\sqrt{4c_{2}^{2}+c_{3}^{2}-4c_{2}c_{4}}-2c_{2}}},\theta_{2}={\frac{2c_{2}+\sqrt{4c_{2}^{2}+c_{3}^{2}-4c_{2}c_{4}}}{2c_{2}-\sqrt{4c_{2}^{2}+c_{3}^{2}-4c_{2}c_{4}}}}.$

For $\Delta>0,c_{2}<0$ we obtain from the Lemma 4 that

$\displaystyle\begin{array}[]{l}\phi_{7}(\xi)={\frac{-2c_{2}}{c_{3}-2c_{2}\cos\left(\sqrt{-c_{2}}\xi\right)+\sqrt{c_{3}^{2}-4c_{2}^{2}-4c_{2}c_{4}}\sin\left(\sqrt{-c_{2}}\xi\right)}},\\\
\quad=\left\\{\begin{array}[]{l}{\frac{2c_{2}}{\sqrt{\Delta}\cos\left(\sqrt{-c_{2}}\xi+\theta_{1}\right)-c_{3}}}{\cong}F_{3a}(\xi)|_{\varepsilon=1},\\\
{\frac{2c_{2}}{-\sqrt{\Delta}\sin\left(\sqrt{-c_{2}}\xi-\theta_{2}\right)-c_{3}}}{\cong}F_{3b}(\xi)|_{\varepsilon=-1},\end{array}\right.\\\
\phi_{8}(\xi)={\frac{-2c_{2}}{c_{3}+2c_{2}\sin\left(\sqrt{-c_{2}}\xi\right)-\sqrt{c_{3}^{2}-4c_{2}^{2}-4c_{2}c_{4}}\cos\left(\sqrt{-c_{2}}\xi\right)}},\\\
\quad=\left\\{\begin{array}[]{l}{\frac{2c_{2}}{\sqrt{\Delta}\cos\left(\sqrt{-c_{2}}\xi+\theta_{1}\right)-c_{3}}}{\cong}F_{3a}(\xi)|_{\varepsilon=1},\\\
{\frac{2c_{2}}{-\sqrt{\Delta}\sin\left(\sqrt{-c_{2}}\xi-\theta_{2}\right)-c_{3}}}{\cong}F_{3b}(\xi)|_{\varepsilon=-1},\end{array}\right.\end{array}$

where

$\displaystyle\theta_{1}=\arctan\left({\frac{\sqrt{c_{3}^{2}-4c_{2}^{2}-4c_{2}c_{3}}}{2c_{2}}}\right),\theta_{2}=\arctan\left({\frac{2c_{2}}{\sqrt{c_{3}^{2}-4c_{2}^{2}-4c_{2}c_{3}}}}\right).$

Third we turn to consider the eight solutions listed in Table 3 which were
obtained by Yang et al [36].

Table 3: Solutions of Eq.(3) with $\Delta=c_{3}^{2}-4c_{2}c_{4}>0$. No | $\varphi(\xi)$ | Condition |  No | $\varphi(\xi)$ | Condition
---|---|---|---|---|---
1 | ${\frac{2c_{2}\mathrm{sech}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}{2\sqrt{\Delta}-(\Delta+c_{3})\mathrm{sech}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}}$ | $c_{2}>0$ |  5 | ${\frac{2c_{2}\sec^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}{2\sqrt{\Delta}-(\Delta+c_{3})\sec^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}}$ | $c_{2}<0$
2 | ${\frac{2c_{2}\mathrm{sech}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}{-2\sqrt{\Delta}+(\Delta-c_{3})\mathrm{sech}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}}$ | $c_{2}>0$ |  6 | ${\frac{2c_{2}\sec^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}{-2\sqrt{\Delta}+(\Delta-c_{3})\sec^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}}$ | $c_{2}<0$
3 | ${\frac{2c_{2}\mathrm{csch}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}{2\sqrt{\Delta}+(\Delta-c_{3})\mathrm{csch}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}}$ | $c_{2}>0$ |  7 | ${\frac{2c_{2}\csc^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}{2\sqrt{\Delta}-(\Delta+c_{3})\csc^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}}$ | $c_{2}<0$
4 | ${\frac{2c_{2}\mathrm{csch}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}{-2\sqrt{\Delta}-(\Delta+c_{3})\mathrm{csch}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)}}$ | $c_{2}>0$ |  8 | ${\frac{2c_{2}\csc^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}{-2\sqrt{\Delta}+(\Delta-c_{3})\csc^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}}$ | $c_{2}<0$

Using the definition of the hyperbolic functions and trigonometric functions,
the formulas (30) and (36),we find that

$\displaystyle\begin{array}[]{l}\varphi_{1}(\xi)={\frac{2c_{2}}{2\sqrt{\Delta}\cosh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)-\sqrt{\Delta}-c_{3}}}={\frac{2c_{2}}{\sqrt{\Delta}\cosh\left(\sqrt{c_{2}}\xi\right)-c_{3}}}=F_{1}(\xi)|_{\varepsilon=1},\\\
\varphi_{3}(\xi)={\frac{2c_{2}}{2\sqrt{\Delta}\sinh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)+\sqrt{\Delta}-c_{3}}}={\frac{2c_{2}}{\sqrt{\Delta}\cosh\left(\sqrt{c_{2}}\xi\right)-c_{3}}}=F_{1}(\xi)|_{\varepsilon=1},\\\
\varphi_{5}(\xi)={\frac{2c_{2}}{2\sqrt{\Delta}\cos^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)-\sqrt{\Delta}-c_{3}}}={\frac{2c_{2}}{\sqrt{\Delta}\cos\left(\sqrt{-c_{2}}\xi\right)-c_{3}}}=F_{3a}(\xi)|_{\varepsilon=1},\\\
\varphi_{8}(\xi)={\frac{2c_{2}}{-2\sqrt{\Delta}\sin^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)+\sqrt{\Delta}-c_{3}}}={\frac{2c_{2}}{\sqrt{\Delta}\cos\left(\sqrt{-c_{2}}\xi\right)-c_{3}}}=F_{3a}(\xi)|_{\varepsilon=1}.\end{array}$

Proceeding as before, we can easily prove that
$\varphi_{2}(\xi)={F_{1}(\xi)}|_{\varepsilon=-1},\varphi_{4}(\xi)={F_{1}(\xi)}|_{\varepsilon=-1},\varphi_{6}(\xi)={F_{3a}(\xi)}|_{\varepsilon=-1},\varphi_{7}(\xi)={F_{3a}(\xi)}|_{\varepsilon=-1}$.

Corresponding to the case $c_{4}=0$, we shall now prove the following two
solutions [37] of Eq.(3)

$\displaystyle\begin{array}[]{l}f_{1}(\xi)=-{\frac{4c_{2}\left(\cosh\left(\sqrt{c_{2}}\xi\right)+\sinh\left(\sqrt{c_{2}}\xi\right)\right)}{\left(b+\cosh\left(\sqrt{c_{2}}\xi\right)+\sinh\left(\sqrt{c_{2}}\xi\right)\right)^{2}}},c_{2}>0,\\\
f_{2}(\xi)={\frac{4c_{2}\left(\cosh\left(\sqrt{c_{2}}\xi\right)+\sinh\left(\sqrt{c_{2}}\xi\right)\right)}{\left(-b+\cosh\left(\sqrt{c_{2}}\xi\right)+\sinh\left(\sqrt{c_{2}}\xi\right)\right)^{2}}},c_{2}>0,\end{array}$

satisfy the equivalent relation
$f_{1}(\xi)\cong{F_{1}(\xi)}|_{\varepsilon=-1}$ and
$f_{2}(\xi)\cong{F_{1}(\xi)}|_{\varepsilon=1}$.

For $c_{4}=0$,using the identity $\sinh{x}+\cosh{x}=e^{x}$ and the Lemma 1
yields

$\displaystyle\begin{array}[]{l}f_{1}(\xi)=-{\frac{4c_{2}}{\left(e^{{\frac{\sqrt{c_{2}}}{2}}\xi}+c_{3}e^{-{\frac{\sqrt{c_{2}}}{2}}\xi}\right)^{2}}}=\left\\{\begin{array}[]{l}{\frac{-4c_{2}}{\left[2\sqrt{c_{3}}\cosh\left({\frac{\sqrt{c_{2}}}{2}}\xi-\ln(\sqrt{c_{3}})\right)\right]^{2}}},\\\
{\frac{-4c_{2}}{\left[2\sqrt{-c_{3}}\sinh\left({\frac{\sqrt{c_{2}}}{2}}\xi-\ln(\sqrt{-c_{3}})\right)\right]^{2}}},\end{array}\right.\\\
\quad\quad\;=\left\\{\begin{array}[]{l}-{\frac{c_{2}}{c_{3}}}\mathrm{sech}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi-\ln(\sqrt{c_{3}})\right),c_{3}>0,\\\
{\frac{c_{2}}{c_{3}}}\mathrm{csch}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi-\ln(\sqrt{-c_{3}})\right),c_{3}<0,\end{array}\right.\end{array}$
$\displaystyle\begin{array}[]{l}f_{2}(\xi)={\frac{4c_{2}}{\left(e^{{\frac{\sqrt{c_{2}}}{2}}\xi}-c_{3}e^{-{\frac{\sqrt{c_{2}}}{2}}\xi}\right)^{2}}},=\left\\{\begin{array}[]{l}{\frac{4c_{2}}{\left[2\sqrt{c_{3}}\sinh\left({\frac{\sqrt{c_{2}}}{2}}\xi-\ln(\sqrt{c_{3}})\right)\right]^{2}}},\\\
{\frac{4c_{2}}{\left[2\sqrt{-c_{3}}\cosh\left({\frac{\sqrt{c_{2}}}{2}}\xi-\ln(\sqrt{-c_{3}})\right)\right]^{2}}},\end{array}\right.\\\
\quad\quad\;=\left\\{\begin{array}[]{l}{\frac{c_{2}}{c_{3}}}\mathrm{csch}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi-\ln(\sqrt{c_{3}})\right),c_{3}>0,\\\
-{\frac{c_{2}}{c_{3}}}\mathrm{sech}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi-\ln(\sqrt{-c_{3}})\right),c_{3}<0.\end{array}\right.\end{array}$

These are just the results we want to prove.

###  3.2 The equivalence relations for solutions of Eq.(7)

Finally,we shall prove the twelve solutions [38] of Eq.(7) listed in Table 4

Table 4: Solutions of Eq.(7). No | $Z(\xi)$ | Condition
---|---|---
1 | ${\frac{1}{c_{2}}}\left[c_{1}-2\sqrt{c_{0}c_{2}}\coth\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right]\sinh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)$ | $c_{0}>0,c_{2}>0$
2 | $-{\frac{1}{c_{2}}}\left[c_{1}-2\sqrt{c_{0}c_{2}}\tanh\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right]\cosh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)$ | $c_{0}>0,c_{2}>0$
3 | $-{\frac{1}{c_{2}}}\left[c_{1}+2\sqrt{-c_{0}c_{2}}\tan\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)\right]\cos^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)$ | $c_{0}>0,c_{2}<0$
4 | $-{\frac{1}{c_{2}}}\left[c_{1}-2\sqrt{-c_{0}c_{2}}\cot\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)\right]\sin^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)$ | $c_{0}>0,c_{2}<0$
5 | ${\frac{1}{4c_{2}}}\left[-2c_{1}-\left(1+c_{1}^{2}-4c_{0}c_{2}\right)\cosh\left(\sqrt{c_{2}}\xi\right)\right.$ |
|  $+\left.\left(1-c_{1}^{2}+4c_{0}c_{2}\right)\sinh\left(\sqrt{c_{2}}\xi\right)\right]$ | $c_{2}>0$
6 | $-{\frac{1}{2c_{2}}}+{\frac{c_{1}^{2}-4c_{0}c_{2}+4c_{2}^{2}\left(\cosh\left(\sqrt{c_{2}}\xi\right)+\sinh\left(\sqrt{c_{2}}\xi\right)\right)^{2}}{8c_{2}^{2}\left(\cosh\left(\sqrt{c_{2}}\xi\right)+\sinh\left(\sqrt{c_{2}}\xi\right)\right)}}$ | $c_{2}>0$
7 | $-{\frac{c_{1}+2\sqrt{-c_{0}c_{2}}\left(\tan\left(\sqrt{-c_{2}}\xi\right)+\sec\left(\sqrt{-c_{2}}\xi\right)\right)}{c_{2}\left[1+\left(\tan\left(\sqrt{-c_{2}}\xi\right)+\sec\left(\sqrt{-c_{2}}\xi\right)\right)^{2}\right]}}$ | $c_{2}<0,c_{0}>0$
8 | $-{\frac{c_{1}+2\sqrt{-c_{0}c_{2}}\left(\tan\left(\sqrt{-c_{2}}\xi\right)-\sec\left(\sqrt{-c_{2}}\xi\right)\right)}{c_{2}\left[1+\left(\tan\left(\sqrt{-c_{2}}\xi\right)-\sec\left(\sqrt{-c_{2}}\xi\right)\right)^{2}\right]}}$ | $c_{2}<0,c_{0}>0$
9 | ${\frac{-c_{1}+2\sqrt{-c_{0}c_{2}}\left(\cot\left(\sqrt{-c_{2}}\xi\right)+\csc\left(\sqrt{-c_{2}}\xi\right)\right)}{c_{2}\left[1+\left(\cot\left(\sqrt{-c_{2}}\xi\right)+\csc\left(\sqrt{-c_{2}}\xi\right)\right)^{2}\right]}}$ | $c_{2}<0,c_{0}>0$
10 | ${\frac{-c_{1}+2\sqrt{-c_{0}c_{2}}\left(\cot\left(\sqrt{-c_{2}}\xi\right)-\csc\left(\sqrt{-c_{2}}\xi\right)\right)}{c_{2}\left[1+\left(\cot\left(\sqrt{-c_{2}}\xi\right)-\csc\left(\sqrt{-c_{2}}\xi\right)\right)^{2}\right]}}$ | $c_{2}<0,c_{0}>0$
11 | ${\frac{c_{1}-2\sqrt{c_{0}c_{2}}\left(\tanh\left(\sqrt{c_{2}}\xi\right)+i\mathrm{sech}\left(\sqrt{c_{2}}\xi\right)\right)}{-c_{2}\left[1-\left(\tanh\left(\sqrt{c_{2}}\xi\right)+i\mathrm{sech}\left(\sqrt{c_{2}}\xi\right)\right)^{2}\right]}}$ | $c_{2}>0,c_{0}>0$
12 | ${\frac{c_{1}-2\sqrt{c_{0}c_{2}}\left(\tanh\left(\sqrt{c_{2}}\xi\right)-\mathrm{sech}\left(\sqrt{c_{2}}\xi\right)\right)}{-c_{2}\left[1-\left(\tanh\left(\sqrt{c_{2}}\xi\right)-\mathrm{sech}\left(\sqrt{c_{2}}\xi\right)\right)^{2}\right]}}$ | $c_{2}>0,c_{0}>0$

are equivalent to the following solutions

$\displaystyle
F_{8}(\xi)=-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{\delta}}{2c_{2}}}\cosh\left(\sqrt{c_{2}}\xi\right),\delta>0,c_{2}>0,$
$\displaystyle
F_{9}(\xi)=-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{-\delta}}{2c_{2}}}\sinh\left(\sqrt{c_{2}}\xi\right),\delta<0,c_{2}>0,$
$\displaystyle
F_{10a}(\xi)=-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{\delta}}{2c_{2}}}\cos\left(\sqrt{-c_{2}}\xi\right),\delta>0,c_{2}<0,$
$\displaystyle
F_{10b}(\xi)=-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{\delta}}{2c_{2}}}\sin\left(\sqrt{-c_{2}}\xi\right),\delta>0,c_{2}<0.$

where $\delta=c_{1}^{2}-4c_{0}c_{2}$.

Let $A=\sqrt{\frac{c_{0}}{c_{2}}},B={\frac{c_{1}}{2c_{2}}}$,then
$A^{2}-B^{2}=-{\frac{\delta}{4c_{2}^{2}}}$. Thus,the condition $\delta>0$
gives that $A^{2}<B^{2}$,and the condition $\delta<0$ implies that
$A^{2}>B^{2}$.Hence,for $c_{0}>0,c_{2}>0$ we obtain from the formulas (30),the
Lemma 3 and the Lemma 2 that

$\displaystyle
Z_{1}(\xi)=-{\frac{c_{1}}{2c_{2}}}-\left(\sqrt{\frac{c_{0}}{c_{2}}}\sinh\left(\sqrt{c_{2}}\xi\right)-{\frac{c_{1}}{2c_{2}}}\cosh\left(\sqrt{c_{2}}\xi\right)\right),$
$\displaystyle=\left\\{\begin{array}[]{l}-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{\delta}}{2c_{2}}}\cosh\left(\sqrt{c_{2}}\xi-{\frac{1}{2}}\ln({\frac{c_{1}+2\sqrt{c_{0}c_{2}}}{c_{1}-2\sqrt{c_{0}c_{2}}}})\right)\cong{F_{8}(\xi)},\delta>0,\\\
-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{-\delta}}{2c_{2}}}\sinh\left(\sqrt{c_{2}}\xi-{\frac{1}{2}}\ln({\frac{2\sqrt{c_{0}c_{2}}+c_{1}}{2\sqrt{c_{0}c_{2}}-c_{1}}})\right)\cong{F_{9}(\xi)},\delta<0.\end{array}\right.$
$\displaystyle
Z_{2}(\xi)=-{\frac{c_{1}}{2c_{2}}}+\sqrt{\frac{c_{0}}{c_{2}}}\sinh\left(\sqrt{c_{2}}\xi\right)-{\frac{c_{1}}{2c_{2}}}\cosh\left(\sqrt{c_{2}}\xi\right),$
$\displaystyle=\left\\{\begin{array}[]{l}-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{\delta}}{2c_{2}}}\cosh\left(\sqrt{c_{2}}\xi-{\frac{1}{2}}\ln({\frac{c_{1}+2\sqrt{c_{0}c_{2}}}{c_{1}-2\sqrt{c_{0}c_{2}}}})\right)\cong{F_{8}(\xi)},\delta>0,\\\
-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{-\delta}}{2c_{2}}}\sinh\left(\sqrt{c_{2}}\xi-{\frac{1}{2}}\ln({\frac{2\sqrt{c_{0}c_{2}}+c_{1}}{2\sqrt{c_{0}c_{2}}-c_{1}}})\right)\cong{F_{9}(\xi)},\delta<0.\end{array}\right.$

Using the identities (36) and the Lemma 4 yields

$\displaystyle
Z_{3}(\xi)=-{\frac{c_{1}}{2c_{2}}}-\left({\frac{\sqrt{-c_{0}c_{2}}}{c_{2}}}\sin\left(\sqrt{-c_{2}}\xi\right)+{\frac{c_{1}}{2c_{2}}}\cos\left(\sqrt{-c_{2}}\xi\right)\right),$
$\displaystyle\quad=\left\\{\begin{array}[]{l}-{\frac{c_{1}}{2c_{2}}}-{\frac{\sqrt{\delta}}{2c_{2}}}\cos(\sqrt{-c_{2}}\xi-\theta_{1})\cong{F_{10a}(\xi)|_{\varepsilon=-1}},\delta>0,c_{0}>0,c_{2}<0,\\\
-{\frac{c_{1}}{2c_{2}}}-{\frac{\sqrt{\delta}}{2c_{2}}}\sin(\sqrt{-c_{2}}\xi+\theta_{2})\cong{F_{10b}(\xi)|_{\varepsilon=-1}},\delta>0,c_{0}>0,c_{2}<0.\end{array}\right.$
$\displaystyle
Z_{4}(\xi)=-{\frac{c_{1}}{2c_{2}}}+\left({\frac{\sqrt{-c_{0}c_{2}}}{c_{2}}}\sin\left(\sqrt{-c_{2}}\xi\right)+{\frac{c_{1}}{2c_{2}}}\cos\left(\sqrt{-c_{2}}\xi\right)\right),$
$\displaystyle\quad=\left\\{\begin{array}[]{l}-{\frac{c_{1}}{2c_{2}}}-{\frac{\sqrt{\delta}}{2c_{2}}}\cos(\sqrt{-c_{2}}\xi-\theta_{1})\cong{F_{10a}(\xi)|_{\varepsilon=-1}},\delta>0,c_{0}>0,c_{2}<0,\\\
-{\frac{c_{1}}{2c_{2}}}-{\frac{\sqrt{\delta}}{2c_{2}}}\sin(\sqrt{-c_{2}}\xi+\theta_{2})\cong{F_{10b}(\xi)|_{\varepsilon=-1}},\delta>0,c_{0}>0,c_{2}<0,\\\
\end{array}\right.$

where
$\theta_{1}=\arctan\left({\frac{2\sqrt{-c_{0}c_{2}}}{c_{1}}}\right),\theta_{2}=\arctan\left({\frac{c_{1}}{2\sqrt{-c_{0}c_{2}}}}\right).$
Set $A=1-c_{1}^{2}+4c_{0}c_{4},B=1+c_{1}^{2}-4c_{0}c_{4}$ then we find
$A^{2}-B^{2}=-4\delta,A-B=-2\delta$.Therefore,from the condition $\delta>0$ we
obtain that $A^{2}<B^{2}$ and $A<B$,and the condition $\delta<0$ implies that
$A^{2}>B^{2}$ and $A>B$.Hence,it follows from the Lemma 3 and the Lemma 2 that

$\displaystyle
Z_{5}(\xi)=-{\frac{c_{1}}{2c_{2}}}+{\frac{1}{4c_{2}}}\left[(1-c_{1}^{2}+4c_{0}c_{4})\sinh\left(\sqrt{c_{2}}\xi\right)-(1+c_{1}^{2}-4c_{0}c_{4})\cosh\left(\sqrt{c_{2}}\xi\right)\right],$
$\displaystyle\quad=\left\\{\begin{array}[]{l}-{\frac{c_{1}}{2c_{2}}}-{\frac{\sqrt{\delta}}{2c_{2}}}\cosh\left(\sqrt{c_{2}}\xi-{\frac{1}{2}}\ln({\frac{1}{\delta}})\right)\cong{F_{8}(\xi)|_{\varepsilon=-1}},\delta>0,c_{2}>0,\\\
-{\frac{c_{1}}{2c_{2}}}+{\frac{\sqrt{-\delta}}{2c_{2}}}\sinh\left(\sqrt{c_{2}}\xi-{\frac{1}{2}}\ln({\frac{-1}{\delta}})\right)\cong{F_{9}(\xi)|_{\varepsilon=1}},\delta<0,c_{2}>0.\end{array}\right.$

Using $\sinh{x}+\cosh{x}=e^{x}$ and the Lemma 1,we obtain

$\displaystyle
Z_{6}(\xi)=-{\frac{c_{1}}{2c_{2}}}+{\frac{1}{8c_{2}^{2}}}\left[4c_{2}^{2}e^{\sqrt{c_{2}}\xi}+(c_{1}^{2}-4c_{0}c_{4})e^{-\sqrt{c_{2}}\xi}\right],$
$\displaystyle\quad=\left\\{\begin{array}[]{l}-{\frac{c_{1}}{2c_{2}}}+{\frac{\sqrt{\delta}}{2c_{2}}}\cosh\left(\sqrt{c_{2}}\xi+{\frac{1}{2}}\ln({\frac{4c_{2}^{2}}{\delta}})\right)\cong{F_{8}(\xi)|_{\varepsilon=1}},\delta>0,c_{2}>0,\\\
-{\frac{c_{1}}{2c_{2}}}+{\frac{\sqrt{-\delta}}{2c_{2}}}\sinh\left(\sqrt{c_{2}}\xi+{\frac{1}{2}}\ln({\frac{4c_{2}^{2}}{-\delta}})\right)\cong{F_{9}(\xi)|_{\varepsilon=1}},\delta<0,c_{2}>0.\end{array}\right.$

The following identities are useful to prove the equivalence relations for
solutions of the auxiliary equations and we can prove these identities by
means of the trigonometric and hyperbolic function identities with the aid of
some direct calculations.

$\displaystyle\tan{\eta}+\sec{\eta}=\tan\left({\frac{\eta}{2}}+{\frac{\pi}{4}}\right),\tan{\eta}-\sec{\eta}=-\cot\left({\frac{\eta}{2}}+{\frac{\pi}{4}}\right),$
(71)
$\displaystyle\cot{\eta}+\csc{\eta}=\cot\left({\frac{\eta}{2}}\right),\cot{\eta}-\csc{\eta}=-\tan\left({\frac{\eta}{2}}\right)$
(72)
$\displaystyle\tanh{\eta}+i\mathrm{sech}{\eta}=\tanh\left({\frac{\eta}{2}}+{\frac{\pi{i}}{4}}\right),\tanh{\eta}-i\mathrm{sech}{\eta}=\coth\left({\frac{\eta}{2}}+{\frac{\pi{i}}{4}}\right).$
(73)

Now the identities (71),(72),(73) and the trigonometric identities are
employed to obtain

$\displaystyle\begin{array}[]{l}Z_{7}(\xi)=-{\frac{c_{1}+2\sqrt{-c_{0}c_{2}}\tan\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)}{c_{2}\left(1+\tan^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)\right)}}=-{\frac{c_{1}+2\sqrt{-c_{0}c_{2}}\tan\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)}{c_{2}\sec^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)}},\\\
\quad=-{\frac{1}{c_{2}}}\left[c_{1}+2\sqrt{-c_{0}c_{2}}\tan\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)\right]\cos^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)\cong{Z_{3}(\xi)},\\\
Z_{8}(\xi)=-{\frac{c_{1}+2\sqrt{-c_{0}c_{2}}\cot\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)}{c_{2}\left(1+\cot^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)\right)}}=-{\frac{c_{1}-2\sqrt{-c_{0}c_{2}}\cot\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)}{c_{2}\csc^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)}},\\\
\quad=-{\frac{1}{c_{2}}}\left[c_{1}-2\sqrt{-c_{0}c_{2}}\cot\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)\right]sin^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi+{\frac{\pi}{4}}\right)\cong{Z_{4}(\xi)},\\\
Z_{9}(\xi)={\frac{-c_{1}+2\sqrt{-c_{0}c_{2}}\cot\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}{c_{2}\left(1+\cot^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)\right)}}={\frac{-c_{1}+2\sqrt{-c_{0}c_{2}}\cot\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}{c_{2}\csc^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}},\\\
\quad=-{\frac{1}{c_{2}}}\left[c_{1}-2\sqrt{-c_{0}c_{2}}\cot\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)\right]\sin^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)={Z_{4}(\xi)},\\\
Z_{10}(\xi)={\frac{-c_{1}-2\sqrt{-c_{0}c_{2}}\tan\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}{c_{2}\left(1+\tan^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)\right)}}={\frac{-c_{1}-2\sqrt{-c_{0}c_{2}}\tan\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}{c_{2}\sec^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)}},\\\
\quad=-{\frac{1}{c_{2}}}\left[c_{1}+2\sqrt{-c_{0}c_{2}}\tan\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)\right]\cos^{2}\left({\frac{\sqrt{-c_{2}}}{2}}\xi\right)={Z_{3}(\xi)},\\\
Z_{11}(\xi)=-{\frac{c_{1}-2\sqrt{c_{0}c_{2}}\tanh\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)}{c_{2}\left(1-\tanh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)\right)}}=-{\frac{c_{1}-2\sqrt{c_{0}c_{2}}\tanh\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)}{c_{2}\mathrm{sech}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)}},\\\
\quad=-{\frac{1}{c_{2}}}\left[c_{1}-2\sqrt{c_{0}c_{2}}\tanh\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)\right]\cosh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)\cong{Z_{2}(\xi)},\\\
Z_{12}(\xi)=-{\frac{c_{1}-2\sqrt{c_{0}c_{2}}\coth\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)}{c_{2}\left(1-\coth^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)\right)}}=-{\frac{c_{1}-2\sqrt{c_{0}c_{2}}\coth\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)}{c_{2}\mathrm{csch}^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)}},\\\
\quad=-{\frac{1}{c_{2}}}\left[c_{1}-2\sqrt{c_{0}c_{2}}\coth\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)\right]\sinh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi+{\frac{\pi{i}}{4}}\right)\cong{Z_{1}(\xi)}.\end{array}$

In accordance with the transitivity,we have proved the equivalence relations
$Z_{i}(\xi)\cong{F_{10a}(\xi)|_{\varepsilon=-1}}\\\
(c_{0}>0,c_{2}<0,\delta>0,i=7,8,9,10)$,
$Z_{i}(\xi)\cong{F_{8}(\xi)}~{}(c_{0}>0,c_{2}>0,\delta>0,i=11,12)$ and
$Z_{i}(\xi)\cong{F_{9}(\xi)}~{}(c_{0}>0,c_{2}>0,\delta<0,i=11,12)$.

For the following two solutions [7] of Eq.(7)

$\displaystyle\begin{array}[]{l}y_{1}(\xi)={\frac{-c_{1}+\sqrt{\delta}}{2c_{2}}}+{\frac{\sqrt{\delta}}{c_{2}}}\sinh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right),\delta=c_{1}^{2}-4c_{0}c_{2}>0,c_{2}>0,\\\
y_{2}(\xi)={\frac{-c_{1}-\sqrt{\delta}}{2c_{2}}}-{\frac{\sqrt{\delta}}{c_{2}}}\sinh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right),\delta=c_{1}^{2}-4c_{0}c_{2}>0,c_{2}>0,\end{array}$

we have $y_{1}(\xi)=F_{8}(\xi)|_{\varepsilon=1}$ and
$y_{2}(\xi)=F_{8}(\xi)|_{\varepsilon=-1}$ which can be obtained by
substituting the identity
$\sinh^{2}\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)={\frac{1}{2}}\left(\cosh\left(\sqrt{c_{2}}\xi\right)-1\right)$
into $y_{1}(\xi)$ and $y_{2}(\xi)$.

##  4 The classification of the solutions

The classification of the solutions of Eq.(1) is very important to clarify
whether the solutions given in literatures are new or equivalent in the sense
of wave translation.If this problem is clear then we can choose the new
solutions and avoid the repeated solutions effectively. Based on the
equivalence relations of the solutions for Eq.(1) as proved in Sec.3,by
removing all those repeated solutions and collecting the simple independent
solutions presented in the literatures, we now conclude that the solutions of
Eq.(1) can be classified into six classes such as the hyperbolic function
solutions,the trigonometric function solutions,the elliptic function
solutions,the exponential function solutions,the polynomial solutions and the
rational solutions.In other words, the six classes of solutions containing
thirty eight independent solutions can be divided into the five cases as
below.

###### Case 1

$c_{0}=c_{1}=0$.

$\displaystyle\begin{array}[]{l}F_{1}(\xi)={\frac{2c_{2}}{\varepsilon\sqrt{\Delta}\cosh\left(\sqrt{c_{2}}\xi\right)-c_{3}}},\Delta>0,c_{2}>0,\\\
F_{2}(\xi)={\frac{2c_{2}}{\varepsilon\sqrt{-\Delta}\sinh\left(\sqrt{c_{2}}\xi\right)-c_{3}}},\Delta<0,c_{2}>0,\\\
F_{3a}(\xi)={\frac{2c_{2}}{\varepsilon\sqrt{\Delta}\cos\left(\sqrt{-c_{2}}\xi\right)-c_{3}}},\\\
\quad\,F_{3b}(\xi)={\frac{2c_{2}}{\varepsilon\sqrt{\Delta}\sin\left(\sqrt{-c_{2}}\xi\right)-c_{3}}},\Delta>0,c_{2}<0,\\\
F_{4}(\xi)=-{\frac{c_{2}}{c_{3}}}\left[1+\varepsilon\tanh\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right],\Delta=0,c_{2}>0,\\\
F_{5}(\xi)=-{\frac{c_{2}}{c_{3}}}\left[1+\varepsilon\coth\left({\frac{\sqrt{c_{2}}}{2}}\xi\right)\right],\Delta=0,c_{2}>0,\\\
F_{6}(\xi)={\frac{\varepsilon}{\sqrt{c_{4}}\xi}},c_{2}=c_{3}=0,c_{4}>0,\\\
F_{7}(\xi)={\frac{4c_{3}}{c_{3}^{2}\xi^{2}-4c_{4}}},c_{2}=0,\end{array}$

where $\Delta=c_{3}^{2}-4c_{2}c_{4},\varepsilon=\pm{1}$.

###### Case 2

$c_{3}=c_{4}=0$.

$\displaystyle\begin{array}[]{l}F_{8}(\xi)=-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{\delta}}{2c_{2}}}\cosh\left(\sqrt{c_{2}}\xi\right),\delta>0,c_{2}>0,\\\
F_{9}(\xi)=-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{-\delta}}{2c_{2}}}\sinh\left(\sqrt{c_{2}}\xi\right),\delta<0,c_{2}>0,\\\
F_{10a}(\xi)=-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{\delta}}{2c_{2}}}\cos\left(\sqrt{-c_{2}}\xi\right),\\\
\quad\,F_{10b}(\xi)=-{\frac{c_{1}}{2c_{2}}}+{\frac{\varepsilon\sqrt{\delta}}{2c_{2}}}\sin\left(\sqrt{-c_{2}}\xi\right),\delta>0,c_{2}<0,\\\
F_{11}(\xi)=-{\frac{c_{1}}{2c_{2}}}+e^{\varepsilon\sqrt{c_{2}}\xi},\delta=0,c_{2}>0,\\\
F_{12}(\xi)=\varepsilon\sqrt{c_{0}}\xi,c_{1}=c_{2}=0,\\\
F_{13}(\xi)=-{\frac{c_{0}}{c_{1}}}+{\frac{c_{1}}{4}}\xi^{2},c_{2}=0,\end{array}$

where $\delta=c_{1}^{2}-4c_{0}c_{2},\varepsilon=\pm{1}$.

###### Case 3

$c_{1}=c_{3}=0$.

$\displaystyle\begin{array}[]{l}F_{14}(\xi)=\varepsilon\sqrt{-\frac{c_{2}}{2c_{4}}}\tanh\left(\sqrt{-\frac{c_{2}}{2}}\xi\right),\Delta_{1}=0,c_{2}<0,c_{4}>0,\\\
F_{15}(\xi)=\varepsilon\sqrt{-\frac{c_{2}}{2c_{4}}}\coth\left(\sqrt{-\frac{c_{2}}{2}}\xi\right),\Delta_{1}=0,c_{2}<0,c_{4}>0,\\\
F_{16a}(\xi)=\varepsilon\sqrt{\frac{c_{2}}{2c_{4}}}\tan\left(\sqrt{\frac{c_{2}}{2}}\xi\right),\\\
\quad\,F_{16b}(\xi)=\varepsilon\sqrt{\frac{c_{2}}{2c_{4}}}\cot\left(\sqrt{\frac{c_{2}}{2}}\xi\right),\Delta_{1}=0,c_{2}>0,c_{4}>0,\\\
F_{17}(\xi)=\sqrt{\frac{-c_{2}m^{2}}{c_{4}(m^{2}+1)}}\mathrm{sn}\left(\sqrt{\frac{-c_{2}}{m^{2}+1}}\xi\right),c_{0}={\frac{c_{2}^{2}m^{2}}{c_{4}(m^{2}+1)^{2}}},c_{2}<0,c_{4}>0,\\\
F_{18}(\xi)=\sqrt{\frac{-c_{2}m^{2}}{c_{4}(2m^{2}-1)}}\mathrm{cn}\left(\sqrt{\frac{c_{2}}{2m^{2}-1}}\xi\right),c_{0}={\frac{c_{2}^{2}m^{2}(m^{2}-1)}{c_{4}(2m^{2}-1)^{2}}},c_{2}>0,c_{4}<0,\\\
F_{19}(\xi)=\sqrt{\frac{-c_{2}}{c_{4}(2-m^{2})}}\mathrm{dn}\left(\sqrt{\frac{c_{2}}{2-m^{2}}}\xi\right),c_{0}={\frac{c_{2}^{2}(1-m^{2})}{c_{4}(2-m^{2})^{2}}},c_{2}>0,c_{4}<0,\\\
F_{20}(\xi)=\varepsilon\left(-{\frac{4c_{0}}{c_{4}}}\right)^{\frac{1}{4}}\mathrm{ds}\left((-4c_{0}c_{4})^{\frac{1}{4}}\xi,{\frac{\sqrt{2}}{2}}\right),c_{2}=0,c_{0}c_{4}<0,\\\
F_{21}(\xi)=\varepsilon\left({\frac{c_{0}}{c_{4}}}\right)^{\frac{1}{4}}\left[\mathrm{ns}\left(2(c_{0}c_{4})^{\frac{1}{4}}\xi,{\frac{\sqrt{2}}{2}}\right)+\mathrm{cs}\left(2(c_{0}c_{4})^{\frac{1}{4}}\xi,{\frac{\sqrt{2}}{2}}\right)\right],c_{2}=0,c_{0}c_{4}>0,\end{array}$

where $\Delta_{1}=c_{2}^{2}-4c_{0}c_{4},\varepsilon=\pm{1}$.

###### Case 4

$c_{2}=c_{4}=0$.

$\displaystyle
F_{22}(\xi)=\wp\left({\frac{\sqrt{c_{3}}}{2}}\xi,g_{2},g_{3}\right),g_{2}=-{\frac{4c_{1}}{c_{3}}},g_{3}=-{\frac{4c_{0}}{c_{3}}},c_{3}>0.$

###### Case 5

$c_{0}=0$.

$\displaystyle\begin{array}[]{l}F_{23}(\xi)=-{\frac{8c_{2}\tanh^{2}\left(\sqrt{-\frac{c_{2}}{12}}\xi\right)}{3c_{3}\left(3+\tanh^{2}\left(\sqrt{-\frac{c_{2}}{12}}\xi\right)\right)}},c_{2}<0,c_{1}={\frac{8c_{2}^{2}}{27c_{3}}},c_{4}={\frac{c_{3}^{2}}{4c_{2}}},\\\
F_{24}(\xi)=-{\frac{8c_{2}\coth^{2}\left(\sqrt{-\frac{c_{2}}{12}}\xi\right)}{3c_{3}\left(3+\coth^{2}\left(\sqrt{-\frac{c_{2}}{12}}\xi\right)\right)}},c_{2}<0,c_{1}={\frac{8c_{2}^{2}}{27c_{3}}},c_{4}={\frac{c_{3}^{2}}{4c_{2}}},\\\
F_{25}(\xi)={\frac{8c_{2}\tan^{2}\left(\sqrt{\frac{c_{2}}{12}}\xi\right)}{3c_{3}\left(3-\tan^{2}\left(\sqrt{\frac{c_{2}}{12}}\xi\right)\right)}},c_{2}>0,c_{1}={\frac{8c_{2}^{2}}{27c_{3}}},c_{4}={\frac{c_{3}^{2}}{4c_{2}}},\\\
F_{26}(\xi)={\frac{8c_{2}\cot^{2}\left(\sqrt{\frac{c_{2}}{12}}\xi\right)}{3c_{3}\left(3-\cot^{2}\left(\sqrt{\frac{c_{2}}{12}}\xi\right)\right)}},c_{2}>0,c_{1}={\frac{8c_{2}^{2}}{27c_{3}}},c_{4}={\frac{c_{3}^{2}}{4c_{2}}},\\\
F_{27}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+\varepsilon{\mathrm{sn}}\left({\frac{c_{3}}{4m\sqrt{c_{4}}}}\xi\right)\right],c_{4}>0,c_{1}={\frac{c_{3}^{3}(m^{2}-1)}{32m^{2}c_{4}^{2}}},c_{2}={\frac{c_{3}^{2}(5m^{2}-1)}{16m^{2}c_{4}}},\\\
F_{28}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+{\frac{\varepsilon}{m{\mathrm{sn}}\left({\frac{c_{3}}{4m\sqrt{c_{4}}}}\xi\right)}}\right],c_{4}>0,c_{1}={\frac{c_{3}^{3}(m^{2}-1)}{32m^{2}c_{4}^{2}}},c_{2}={\frac{c_{3}^{2}(5m^{2}-1)}{16m^{2}c_{4}}},\\\
F_{29}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+\varepsilon{m\mathrm{sn}}\left({\frac{c_{3}}{4\sqrt{c_{4}}}}\xi\right)\right],c_{4}>0,c_{1}={\frac{c_{3}^{3}(1-m^{2})}{32c_{4}^{2}}},c_{2}={\frac{c_{3}^{2}(5-m^{2})}{16c_{4}}},\\\
F_{30}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+{\frac{\varepsilon}{{\mathrm{sn}}\left({\frac{c_{3}}{4\sqrt{c_{4}}}}\xi\right)}}\right],c_{4}>0,c_{1}={\frac{c_{3}^{3}(1-m^{2})}{32c_{4}^{2}}},c_{2}={\frac{c_{3}^{2}(5-m^{2})}{16c_{4}}},\\\
F_{31}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+\varepsilon{\mathrm{cn}}\left({-\frac{c_{3}}{4m\sqrt{-c_{4}}}}\xi\right)\right],c_{4}<0,c_{1}={\frac{c_{3}^{3}}{32m^{2}c_{4}^{2}}},c_{2}={\frac{c_{3}^{2}(4m^{2}+1)}{16m^{2}c_{4}}},\\\
F_{32}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+{\frac{\varepsilon\sqrt{1-m^{2}}{\mathrm{sn}}\left({-\frac{c_{3}}{4m\sqrt{-c_{4}}}}\xi\right)}{{\mathrm{dn}}\left({-\frac{c_{3}}{4m\sqrt{-c_{4}}}}\xi\right)}}\right],c_{4}<0,c_{1}={\frac{c_{3}^{3}}{32m^{2}c_{4}^{2}}},c_{2}={\frac{c_{3}^{2}(4m^{2}+1)}{16m^{2}c_{4}}},\\\
F_{33}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+{\frac{\varepsilon}{\sqrt{1-m^{2}}}}{\mathrm{dn}}\left({\frac{c_{3}}{4\sqrt{c_{4}(m^{2}-1)}}}\xi\right)\right],\\\
\qquad\,c_{4}<0,c_{1}={\frac{c_{3}^{3}m^{2}}{32c_{4}^{2}(m^{2}-1)}},c_{2}={\frac{c_{3}^{2}(5m^{2}-4)}{16c_{4}(m^{2}-1)}},\\\
F_{34}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+{\frac{\varepsilon}{{\mathrm{dn}}\left({\frac{c_{3}}{4\sqrt{c_{4}(m^{2}-1)}}}\xi\right)}}\right],c_{4}<0,c_{1}={\frac{c_{3}^{3}m^{2}}{32c_{4}^{2}(m^{2}-1)}},c_{2}={\frac{c_{3}^{2}(5m^{2}-4)}{16c_{4}(m^{2}-1)}},\\\
F_{35}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+{\frac{\varepsilon}{{\mathrm{cn}}\left({\frac{c_{3}}{4\sqrt{c_{4}(1-m^{2})}}}\xi\right)}}\right],c_{4}>0,c_{1}={\frac{c_{3}^{3}}{32c_{4}^{2}(1-m^{2})}},c_{2}={\frac{c_{3}^{2}(4m^{2}-5)}{16c_{4}(m^{2}-1)}},\\\
F_{36}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+{\frac{\varepsilon{\mathrm{dn}}\left({\frac{c_{3}}{4\sqrt{c_{4}(1-m^{2})}}}\xi\right)}{\sqrt{1-m^{2}}{\mathrm{cn}}\left({\frac{c_{3}}{4\sqrt{c_{4}(1-m^{2})}}}\xi\right)}}\right],\\\
\qquad\,c_{4}>0,c_{1}={\frac{c_{3}^{3}}{32c_{4}^{2}(1-m^{2})}},c_{2}={\frac{c_{3}^{2}(4m^{2}-5)}{16c_{4}(m^{2}-1)}},\\\
F_{37}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+\varepsilon{\mathrm{dn}}\left({-\frac{c_{3}}{4\sqrt{-c_{4}}}}\xi\right)\right],c_{4}<0,c_{1}={\frac{c_{3}^{3}m^{2}}{32c_{4}^{2}}},c_{2}={\frac{c_{3}^{2}(m^{2}+4)}{16c_{4}}},\\\
F_{38}(\xi)=-{\frac{c_{3}}{4c_{4}}}\left[1+{\frac{\varepsilon\sqrt{1-m^{2}}}{{\mathrm{dn}}\left({-\frac{c_{3}}{4\sqrt{-c_{4}}}}\xi\right)}}\right],c_{4}<0,c_{1}={\frac{c_{3}^{3}m^{2}}{32c_{4}^{2}}},c_{2}={\frac{c_{3}^{2}(m^{2}+4)}{16c_{4}}}\end{array}$

where $\varepsilon=\pm{1}$.

##  5 New traveling wave solutions for mCH equation

As an illustrative example below we consider the modified Camassa–Holm (mCH)
equation in the form

$u_{t}-u_{xxt}+3u^{2}u_{x}=2u_{x}u_{xx}+uu_{xxx},$ (80)

was due to Wazwaz [39] obtained by using the term $u^{2}u_{x}$ instead of the
nonlinear convection term $uu_{x}$ in Camassa-Holm equation. Wazwaz [39, 40]
also found some exact traveling wave solutions of wave speed $c=1/2,1,2$ for
Eq.(80).Some traveling wave solutions of wave speed $c=1/3$ and some peakon
solutions of wave speed $c=3$ for Eq.(80) were obtained by Wang and Tang in
[41]. A rational solution and more exact traveling wave solutions of Eq.(80)
were also constructed by Liang and Jeffrey in [42].

In the present paper we are interested in finding more new exact traveling
wave solutions for Eq.(80) through use of the general elliptic equation
expansion method equipped with our new classified solutions in Sec.4. In order
that we make the wave transformation $u(x,t)=u(\xi),\xi=x-{\omega}t$ and
change the Eq.(80) into the form

$-\omega{u^{\prime}}+\omega{u^{\prime\prime\prime}}+3u^{2}u^{\prime}-2{u^{\prime}}u^{\prime\prime}-uu^{\prime\prime\prime}=0.$
(81)

According to the balancing principle to balance $uu^{\prime\prime\prime}$ with
$u^{2}{u^{\prime}}$ yields $m=2$, therefore the Eq.(81) admits the solution in
the form

$u(\xi)=a_{0}+a_{1}F(\xi)+a_{2}F^{2}(\xi),$ (82)

where $F(\xi)$ expresses the solution of Eq.(1) and $a_{k}~{}(k=0,1,2)$ are
constants to be determined.

Substituting (82) into (81) along with Eq.(1) and setting the coefficients of
$F^{j}F^{\prime}~{}(j=0,1,2,3,4,5)$ to zero,we obtain the following set of
algebraic equations

$\left\\{\begin{array}[]{l}6a_{2}^{3}-48a_{2}^{2}c_{4}=0,\\\
15a_{1}a_{2}^{2}-50a_{1}a_{2}c_{4}-35a_{2}^{2}c_{3}=0,\\\
12a_{0}a_{2}^{2}-24a_{0}a_{2}c_{4}+12a_{1}^{2}a_{2}-10a_{1}^{2}c_{4}-34a_{1}a_{2}c_{3}-24a_{2}^{2}c_{2}+24a_{2}c_{4}\omega=0,\\\
3a_{0}^{2}a_{1}-a_{0}a_{1}c_{2}-3a_{0}a_{2}c_{1}-a_{1}^{2}c_{1}-4a_{1}a_{2}c_{0}+a_{1}c_{2}\omega+3a_{2}c_{1}\omega-
a_{1}\omega=0,\\\
18a_{0}a_{1}a_{2}-6a_{0}a_{1}c_{4}-15a_{0}a_{2}c_{3}+3a_{1}^{3}-6a_{1}^{2}c_{3}-21a_{1}a_{2}c_{2}+6a_{1}c_{4}\omega\\\
\quad-15a_{2}^{2}c_{1}+15a_{2}c_{3}\omega=0,\\\
6a_{0}^{2}a_{2}+6a_{0}a_{1}^{2}-3a_{0}a_{1}c_{3}-8a_{0}a_{2}c_{2}-3a_{1}^{2}c_{2}-2a_{2}\omega-11a_{1}a_{2}c_{1}+3a_{1}c_{3}\omega\\\
\quad-8a_{2}^{2}c_{0}+8a_{2}c_{2}\omega=0.\end{array}\right.$ (83)

A. When $c_{0}=c_{1}=0$,the Eq.(83) solves that

$\displaystyle
a_{0}={\frac{1}{3}}\left(-\omega+\sqrt{2\omega(3-\omega)}\right),a_{1}=0,a_{2}=8c_{4},c_{2}={\frac{1}{8}}\sqrt{2\omega(3-\omega)},c_{3}=0,$
(84) $\displaystyle
a_{0}={\frac{1}{3}}\left(-\omega-\sqrt{2\omega(3-\omega)}\right),a_{1}=0,a_{2}=8c_{4},c_{2}=-{\frac{1}{8}}\sqrt{2\omega(3-\omega)},c_{3}=0,$
(85) $\displaystyle
a_{0}={\frac{1}{3}}\left(-\omega+\sqrt{2\omega(3-\omega)}\right),a_{1}=2c_{3},a_{2}=0,c_{2}={\frac{1}{2}}\sqrt{2\omega(3-\omega)},c_{4}=0,$
(86) $\displaystyle
a_{0}={\frac{1}{3}}\left(-\omega-\sqrt{2\omega(3-\omega)}\right),a_{1}=2c_{3},a_{2}=0,c_{2}=-{\frac{1}{2}}\sqrt{2\omega(3-\omega)},c_{4}=0,$
$\displaystyle
a_{0}={\frac{1}{3}}\left(-\omega+\sqrt{2\omega(3-\omega)}\right),a_{1}=\pm\left(8c_{4}\right)^{\frac{1}{2}}\left(2\omega(3-\omega)\right)^{\frac{1}{4}},a_{2}=8c_{4},$
$\displaystyle{\quad}c_{2}={\frac{1}{2}}\sqrt{2\omega(3-\omega)},c_{4}=\pm{\frac{1}{2}}\left(8c_{4}\right)^{\frac{1}{2}}\left(2\omega(3-\omega)\right)^{\frac{1}{4}},$
(88) $\displaystyle
a_{0}={\frac{1}{3}}\left(-\omega+\sqrt{2\omega(3-\omega)}\right),a_{1}=2c_{3},c_{2}={\frac{1}{2}}\sqrt{2\omega(3-\omega)},c_{4}=0,$
(89) $\displaystyle
a_{0}={\frac{1}{3}}\left(-\omega-\sqrt{2\omega(3-\omega)}\right),a_{1}=2c_{3},c_{2}=-{\frac{1}{2}}\sqrt{2\omega(3-\omega)},c_{4}=0.$
(90)

Inserting (84)–(90) together with the solutions given by Case 1 into (82),we
obtain the exact traveling wave solutions of Eq.(80) as follows

$\displaystyle\begin{array}[]{l}u_{1a}(x,t)=-{\frac{\omega}{3}}+{\frac{1}{3}}\sqrt{2\omega(3-\omega)}-\sqrt{2\omega(3-\omega)}\mathrm{sech}^{2}\eta,\\\
u_{1b}(x,t)=-{\frac{\omega}{3}}+{\frac{1}{3}}\sqrt{2\omega(3-\omega)}+\sqrt{2\omega(3-\omega)}\mathrm{csch}^{2}\eta,\\\
u_{2a}(x,t)=-{\frac{\omega}{3}}-{\frac{1}{3}}\sqrt{2\omega(3-\omega)}+\sqrt{2\omega(3-\omega)}\sec^{2}\eta,\\\
u_{2b}(x,t)=-{\frac{\omega}{3}}-{\frac{1}{3}}\sqrt{2\omega(3-\omega)}+\sqrt{2\omega(3-\omega)}\csc^{2}\eta,\\\
u_{3}(x,t)=-{\frac{\omega}{3}}+{\frac{1}{3}}\sqrt{2\omega(3-\omega)}+{\frac{2\sqrt{2\omega(3-\omega)}}{\varepsilon\cosh\eta-1}},\\\
u_{4a}(x,t)=-{\frac{\omega}{3}}-{\frac{1}{3}}\sqrt{2\omega(3-\omega)}-{\frac{2\sqrt{2\omega(3-\omega)}}{\varepsilon\cos\eta-1}},\\\
u_{4b}(x,t)=-{\frac{\omega}{3}}-{\frac{1}{3}}\sqrt{2\omega(3-\omega)}-{\frac{2\sqrt{2\omega(3-\omega)}}{\varepsilon\sin\eta-1}},\\\
u_{5a}(x,t)=-{\frac{\omega}{3}}-{\frac{2}{3}}\sqrt{2\omega(3-\omega)}+\sqrt{2\omega(3-\omega)}\tanh^{2}\eta,\\\
u_{5b}(x,t)=-{\frac{\omega}{3}}-{\frac{2}{3}}\sqrt{2\omega(3-\omega)}+\sqrt{2\omega(3-\omega)}\coth^{2}\eta,\\\
u_{6a}(x,t)=u_{1a}(x,t),u_{6b}(x,t)=u_{1b}(x,t),c_{4}=0,\\\
u_{7a}(x,t)=u_{2a}(x,t),u_{7b}(x,t)=u_{2b}(x,t),c_{4}=0,\\\
\qquad\eta={\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2}}\right)^{\frac{1}{4}}(x-{\omega}t),0<\omega<3.\end{array}$

B. When $c_{3}=c_{4}=0$,we cannot find the solutions of Eq.(83) so we can not
obtain the exact traveling wave solutions for Eq.(80).
C. When $c_{1}=c_{3}=0$,we have
(1) For $c_{0}={\frac{c_{2}^{2}m^{2}}{c_{4}(m^{2}+1)^{2}}},c_{2}<0,c_{4}>0$,
the Eq.(83) solves that

$\displaystyle
a_{0}=-{\frac{\omega}{3}}+{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=0,$
$\displaystyle{\quad}a_{2}=8c_{4},c_{2}=-{\frac{m^{2}+1}{8}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},$

which leads the Jacobian snoidal wave solution of Eq.(80) as

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}+m^{2}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{sn}^{2}(\eta,m),$
(92)
$\displaystyle\quad\eta={\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}}(x-{\omega}t),0<\omega<3.$

(2) For
$c_{0}={\frac{c_{2}^{2}m^{2}(m^{2}-1)}{c_{4}(2m^{2}-1)^{2}}},c_{2}>0,c_{4}<0$,
the solution of Eq.(83) is found to be

$\displaystyle
a_{0}=-{\frac{\omega}{3}}+{\frac{2m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=0,$
$\displaystyle{\quad}a_{2}=8c_{4},c_{2}={\frac{2m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},$

which gives the Jacobian cnoidal wave solution of Eq.(80) as following

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}+{\frac{2m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}-m^{2}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{cn}^{2}(\eta,m),$
$\displaystyle\quad\eta={\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}}(x-{\omega}t),0<\omega<3.$
(93)

(3) For
$c_{0}={\frac{c_{2}^{2}(1-m^{2})}{c_{4}(2-m^{2})^{2}}},c_{2}>0,c_{4}<0$, the
solution of Eq.(83) is obtained that

$\displaystyle
a_{0}=-{\frac{\omega}{3}}+{\frac{m^{2}-2}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=0,$
$\displaystyle{\quad}a_{2}=8c_{4},c_{2}={\frac{m^{2}-2}{8}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},$

which implies the Jacobian elliptic function solution of Eq.(80) as

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}-{\frac{m^{2}-2}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}-\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{dn}^{2}(\eta,m),$
$\displaystyle\quad\eta={\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}}(x-{\omega}t),0<\omega<3.$
(94)

(4) For $c_{2}=0$,the Eq.(83) solves

$\displaystyle
a_{0}=-{\frac{\omega}{3}},a_{1}=0,a_{2}=8c_{4},c_{0}={\frac{\omega(\omega-3)}{96c_{4}}},$

which gives the Jacobian elliptic function solution of Eq.(80) as following

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}+{\frac{2}{3}}\sqrt{6\omega(3-\omega)}\mathrm{ds}^{2}\left(\eta,{\frac{\sqrt{2}}{2}}\right),$
$\displaystyle{\quad}\eta={\frac{1}{6}}\left(54\omega(3-\omega)\right)^{\frac{1}{4}}(x-{\omega}t),0<\omega<3,$
(95) $\displaystyle
u(x,t)=-{\frac{\omega}{3}}+{\frac{\varepsilon}{3}}\sqrt{6\omega(\omega-3)}\left[\mathrm{ns}(\eta,{\frac{\sqrt{2}}{2}})+\mathrm{cs}(\eta,{\frac{\sqrt{2}}{2}})\right]^{2},$
$\displaystyle\quad\eta=384\left(6\omega(\omega-3)\right)^{\frac{1}{4}}(x-{\omega}t),\omega<0\;\mbox{or}\;\omega>3.$
(96)

D. When $c_{2}=c_{4}=0$,the solution of Eq.(83) is found to be

$\displaystyle
a_{0}=-{\frac{\omega}{3}},a_{1}=2c_{3},a_{2}=0,c_{1}={\frac{\omega(3-\omega)}{6c_{3}}},$

which gives the Weierstrass elliptic function solution of Eq.(80) as

$\displaystyle
u(x,t)=2c_{3}\wp\left({\frac{\sqrt{c_{3}}}{2}}(x-{\omega}t),{\frac{2\omega(3-\omega)}{3c_{3}^{2}}},-{\frac{4c_{0}}{c_{3}}}\right).$
(97)

E. When $c_{0}=0$,we have
(1) From the condition of $F_{27}$ we obtain the solution of Eq.(83) as
following

$\displaystyle
a_{0}=-{\frac{\omega}{3}}+{\frac{2m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}={\frac{2c_{3}^{2}}{m^{2}}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4m^{2}}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

from which we obtain the Jacobian snoidal wave solution of Eq.(80) as

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}+m^{2}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{sn}^{2}\left(\rho(x-{\omega}t,m\right),$
$\displaystyle{\quad}\rho=-{\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(98)

(2) Under condition of $F_{28}$,the Eq.(83) solves that

$\displaystyle
a_{0}=-{\frac{\omega}{3}}+{\frac{m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}={\frac{2c_{3}^{2}}{m^{2}}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4m^{2}}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

from which we obtain the Jacobian elliptic function solution of Eq.(80) in the
form

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}+\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{ns}^{2}\left(\rho(x-{\omega}t,m\right),$
$\displaystyle\quad\rho=-{\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(99)

(3) The solution of Eq.(83) induced by the condition of $F_{29}$ is that

$\displaystyle
a_{0}=-{\frac{\omega}{3}}-{\frac{m^{2}-2}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}=2c_{3}^{2}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

which leads the Jacobian snoidal wave solution of Eq.(80) as

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}+m^{2}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{sn}^{2}\left(\rho(x-{\omega}t,m\right),$
$\displaystyle\quad\rho=-{\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(100)

(4) Using the condition of $F_{30}$ we find the solution of Eq.(83) as
following

$\displaystyle
a_{0}=-{\frac{\omega}{3}}-{\frac{m^{2}-2}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}=2c_{3}^{2}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

from which we obtain the Jacobian elliptic function solution of Eq.(80) in the
form

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}+\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{ns}^{2}\left(\rho(x-{\omega}t,m\right),$
$\displaystyle\quad\rho=-{\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(101)

(5) From the condition of $F_{31}$ we yields the solution of Eq.(83) as
following

$\displaystyle
a_{0}=-{\frac{\omega}{3}}+{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}={\frac{2c_{3}^{2}}{m^{2}}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4m^{2}}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

which gives the Jacobian snoidal wave solution of Eq.(80) in the form

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}+m^{2}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{sn}^{2}\left(\rho(x-{\omega}t,m\right),$
$\displaystyle\quad\rho={\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(102)

(6) By the condition of $F_{32}$ we get the solution of Eq.(83) as following

$\displaystyle
a_{0}=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}={\frac{2c_{3}^{2}}{m^{2}}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4m^{2}}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

which leads the Jacobian elliptic function solution of Eq.(80) in the form

$\displaystyle
u(x,t)=\left[-{\frac{\omega}{3}}+{\frac{2m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\right]\mathrm{nd}^{2}\left(\rho(x-{\omega}t,m\right)$
$\displaystyle\quad+\left[{\frac{m^{2}\omega}{3}}+{\frac{m^{2}(m^{2}-2)}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\right]\mathrm{sd}^{2}(\rho(x-{\omega}t),m),$
$\displaystyle\qquad\rho={\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(103)

(7) The solution of Eq.(83) induced by the condition of $F_{33}$ reads

$\displaystyle
a_{0}=-{\frac{\omega}{3}}+{\frac{2m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}={\frac{2c_{3}^{2}}{m^{2}-1}}\left({\frac{m^{4}-m^{2}+1}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4(m^{2}-1)}}\left({\frac{m^{4}-m^{2}+1}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

from which we obtain the Jacobian elliptic function solution of Eq.(80) as

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}-{\frac{m^{2}-2}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}-\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{dn}^{2}\left(\rho(x-{\omega}t,m\right),$
$\displaystyle{\quad}\rho=-{\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(104)

(8) The condition of $F_{34}$ leads the solution of Eq.(83) as following

$\displaystyle
a_{0}=-{\frac{\omega}{3}}+{\frac{2m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}={\frac{2c_{3}^{2}}{m^{2}-1}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4(m^{2}-1)}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

which gives the Jacobian elliptic function solution of Eq.(80) in the form

$\displaystyle
u(x,t)=\left[-{\frac{\omega}{3}}+{\frac{2m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\right]nd^{2}\left(\rho(x-{\omega}t,m)\right)$
$\displaystyle\quad+\left[{\frac{m^{2}\omega}{3}}+{\frac{m^{2}(m^{2}-2)}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\right]\mathrm{sd}^{2}(\rho(x-{\omega}t,m),$
$\displaystyle\qquad\rho=-{\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(105)

(9) The solution of Eq.(83) obtained by the condition of $F_{35}$ reads

$\displaystyle
a_{0}=-{\frac{\omega}{3}}-{\frac{m^{2}-2}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}={\frac{2c_{3}^{2}}{m^{2}-1}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4(m^{2}-1)}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

from which we obtain the Jacobian elliptic function solution of Eq.(80) as

$\displaystyle
u(x,t)=\left[-{\frac{\omega}{3}}-{\frac{m^{2}-2}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\right]\mathrm{nc}^{2}\left(\rho(x-{\omega}t,m)\right)$
$\displaystyle\quad+\left[{\frac{\omega}{3}}-{\frac{2m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\right]\sc^{2}(\rho(x-{\omega}t,m),$
$\displaystyle\quad\rho=-{\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(106)

(10) By the condition of $F_{36}$ we get the solution of Eq.(83) as following

$\displaystyle
a_{0}=-{\frac{\omega}{3}}-{\frac{m^{2}-2}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}={\frac{2c_{3}^{2}}{1-m^{2}}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4(1-m^{2})}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

which leads the Jacobian elliptic function solution of Eq.(80) in the form

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}+\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{ns}^{2}(\rho(x-{\omega}t),m),$
$\displaystyle\quad\rho=-{\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(107)

(11) The condition of $F_{37}$ gives the solution of Eq.(83) as following

$\displaystyle
a_{0}=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}=2c_{3}^{2}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

which gives the Jacobian snoidal wave solution of Eq.(80) in the form

$\displaystyle
u(x,t)=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}+m^{2}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\mathrm{sn}^{2}(\rho(x-{\omega}t),m),$
$\displaystyle{\quad}\rho={\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(108)

(12) The solution of Eq.(83) obtained by using the condition of $F_{38}$ reads

$\displaystyle
a_{0}=-{\frac{\omega}{3}}-{\frac{m^{2}+1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}},a_{1}=4c_{3},$
$\displaystyle{\quad}a_{2}=2c_{3}^{2}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},c_{4}={\frac{c_{3}^{2}}{4}}\left({\frac{2(m^{4}-m^{2}+1)}{\omega(3-\omega)}}\right)^{\frac{1}{2}},$

from which we obtain the Jacobian elliptic function solution of Eq.(80) as

$\displaystyle
u(x,t)=\left[-{\frac{\omega}{3}}+{\frac{2m^{2}-1}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\right]\mathrm{nd}^{2}\left(\rho(x-{\omega}t,m)\right)$
$\displaystyle\quad+\left[{\frac{m^{2}\omega}{3}}+{\frac{m^{2}(m^{2}-2)}{3}}\left({\frac{2\omega(3-\omega)}{m^{4}-m^{2}+1}}\right)^{\frac{1}{2}}\right]\mathrm{sd}^{2}(\rho(x-{\omega}t,m),$
$\displaystyle\quad\rho={\frac{1}{2}}\left({\frac{\omega(3-\omega)}{2(m^{4}-m^{2}+1)}}\right)^{\frac{1}{4}},0<\omega<3.$
(109)

##  6 Concluding remarks

We constructed the BT and the SF of two kinds of sub-equations for Eq.(1) in
terms of the indirect mapping method for simplicity.As a matter of fact, we
can establish the BT and the SF for most of those sub-equations for Eq.(1) by
the elementary integration method which were not considered here and will be
reported elsewhere.It is also clear that the new solutions of Eq.(1) obtained
from the BT or SF can also be used to find new solutions of NLEEs which is
under consideration.But an open problem as how to construct multiple traveling
wave solutions of NLEEs from the solutions of NLEEs obtained by using the
auxiliary equations is to be solved.

It should be noticed that the four lemmas presented in Sec.3 can be widely
used to prove the equivalence relations of numerous solutions for other
auxiliary equations.At the same time,the four lemmas can also be applied to
prove the equivalence relations of some direct methods. Therefore,the proof of
the equivalence relations of solutions for the auxiliary equations is more
important than that of discussing the equivalence relations of those obtained
solutions for a given nonlinear equation.This is the reason why we studied the
equivalence relations of solutions for Eq.(1) here in detail.

In Sec.5 we obtained some exact traveling wave solutions of wave speed defined
on the region $\omega\in(0,3)$ and $\omega\in(-\infty,0)\cup(3,+\infty)$ for
mCH equation that was not considered and obtained before so they are the new
solutions for mCH equation.Although some of them are same in form but they are
obtained by using different conditions and therefore they can be regarded as
new solutions as well.In the limit of $m\rightarrow{1}$,the solitary wave
solutions for mCH equation can be obtained from the Jacobian elliptic function
solutions constructed in Sec.5, and which were omitted.

## References

  * [1] S.K.Liu,Z.T. Fu,S.D.Liu,and Q. Zhao,“Jacobi elliptic function expansion method and periodic wave solutions of nonlinear wave equations,” Phys. Lett. A 289,69–74 (2001).
  * [2] M.L.Wang and X.Z.Li,“Applications of F–expansion to periodic wave solutions for a new Hamiltonian amplitude equation,” Chaos, Solitons and Fractals 24,1257–1268 (2005).
  * [3] Sirendaoreji and J.Sun,“Auxiliary equation method for solving nonlinear partial differential equations,” Phys. Lett. A 309,387–396 (2003).
  * [4] E.G.Fan,“Uniformly constructing a series of explicit exact solutions to nonlinear equations in mathematical physics,” Chaos,Solitons and Fractals 16,819–839 (2003).
  * [5] E.Yomba,“The extended Fan’s sub-equation method and its application to KdV-MKdV, BKK and variant Boussinesq equations,” Phys. Lett. A 336,463–476 (2005).
  * [6] A.A.Soliman and M.A.Abdou,“Exact travelling wave solutions of nonlinear partial differential equations,” Chaos, Solitons and Fractals 32,808–815 (2007).
  * [7] S.K.Liu and S.D.Liu,“Nonlinear Equations in Physics,” 32–53. Peking University Press, Peking (2000).
  * [8] E.G.Fan,“Integrable Systems and Computer Algebra,” 29–55. Science Press, Peking (2004).
  * [9] Z.B.Li,“Traveling Wave Solutions of Nonlinear Equations in Mathematical Physics,” 147–153.Science Press, Peking (2007).
  * [10] Y.C.Guo,“Nonlinear Partial Differential Equations,” 223–226. Tsinghua University Press, Peking (2008).
  * [11] N.Taghizadeh,Q.Zhou,M.Ekici,and M.Mirzazadeh,“Soliton solutions for Davydov solitons in $\alpha$–helix proteins,” Superlattices and Microstructures 102,323–341 (2017).
  * [12] M.M.El–Borai,H.M.El–Owaidy,H.M.Ahmed,and A.H.Arnous,“Exact and soliton solutions to nonlinear transmission line model,” Nonlinear Dyn. 87, 767–773 (2017).
  * [13] Sirendaoreji,“New exact traveling wave solutions for the Kawahara and the modified Kawahara equations,” Chaos, Solitons and Fractals 19, 147–150 (2004).
  * [14] Sirendaoreji,“Auxiliary equation method and new solutions of Klein-Gordon equations,” Chaos, Solitons and Fractals 31: 943–950 (2007).
  * [15] E.Yomba,“The sub-ODE method for finding exact travelling wave solutions of generalized nonlinear Camassa-Holm,and generalized nonlinear Schrödinger equations,” Phys. Lett. A 372,215–222 (2008).
  * [16] M.A.Abdou,“A generalized auxiliary equation method and its applications,” Nonlinear Dyn. 52: 95–102 (2008).
  * [17] N.Cheemaa and M.Younis,“New and more exact traveling wave solutions to integrable (2+1)-dimensional Maccari system,” Nonlinear Dyn. 83:1395–1401 (2016).
  * [18] Sirendaoreji,“Exact travelling wave solutions for four forms of nonlinear Klein–Gordon equations,” Phys. Lett. A 363: 440–447 (2007).
  * [19] H.Triki,H.Leblond,and D.Mihalache,“Soliton solutions of nonlinear diffusion–reaction–type equations with time–dependent coefficients accounting for long–range diffusion,” Nonlinear Dyn. 86:2115–2126 (2016).
  * [20] J.P.Yu,D.S.Wang,,Y.L.Sun,and S.P.Wu,“Modified method of simplest equation for obtaining exact solutions of the Zakharov–Kuznetsov equation, the modified Zakharov-Kuznetsov equation, and their generalized forms,” Nonlinear Dyn. 85,2449–2465 (2016).
  * [21] X.Z.Li and M.L.Wang,“A sub–ODE method for finding exact solutions of a generalized KdV–mKdV equation with high-order nonlinear terms,” Phys.Lett. A 361, 115–118 (2007).
  * [22] C.S.Liu,“Applications of complete discrimination system for polynomial for classifications of traveling wave solutions to nonlinear differential equations,” Comput.Phys. Commun. 181,317–324 (2010).
  * [23] J.Y.Hu,“Classification of single traveling wave solutions to the nonlinear dispersion Drinfel’d–Sokolov system,” Appl. Math. Comput. 219, 2017–2025 (2012).
  * [24] Taogetusang,“Several auxiliary equations and infinite sequence exact solutions to nonlinear evolution equations,” Acta Phys. Sin. 60,050201:1–13 (2011).
  * [25] X.Zeng and X.L.Yong,“A new mapping method and its applications to nonlinear partial differential equations,” Phys. Lett. A 372,6602–6607 (2008).
  * [26] S.Zhang,“A generalized auxiliary equation method and its application to (2 + 1)–dimensional Korteweg–de Vries equations,” Comput. Math. Appl. 54,1028–1038 (2007).
  * [27] S.Zhang and T.C.Xia,“A generalized auxiliary equation method and its application to (2+1)-dimensional asymmetric Nizhnik–Novikov–Vesselov equations,” J. Phys. A: Math. Theor. 40: 227–248 (2007).
  * [28] D.J.Hunag,D.S.Li and H.Q.Zhang,“Explicit and exact travelling wave solutions for the generalized derivative Shrödinger equation,” Chaos, Solitons and Fractals 31,586–593 (2007).
  * [29] G.Q.Xu,“Extended auxiliary equation method and its applications to three generalized NLS equations,” Abstract and Appl. Anal. 2014, 1–7 (2014).
  * [30] E.M.E.Zayed and K.A.E.Alurrfi,“Extended auxiliary equation method and its applications for finding the exact solutions for a class of nonlinear Schröinger–type equations,” Appl. Math. Comput. 289,111-131 (2016).
  * [31] Z.Pinar and T.Q̈zis,“An observation on the periodic solutions to nonlinear physical models by means of the auxiliary equation with a sixth-degree nonlinear term. Commun. Nonl. Sci. Numer. Simulat. 18, 2177–2187 (2013).
  * [32] Sirendaoreji,“Unified Riccati equation expansion method and its application to two new classes of Benjamin–Bona–Mahony equations. Nonlinear Dyn. 89,333–344 (2017)
  * [33] C.P.Liu and X.P.Liu,“A note on the auxiliary equation method for solving nonlinear partial differential equations,” Phys. Lett. A 348,222–227 (2006).
  * [34] X.P.Liu and C.P.Liu,“The relationship among the solutions of two auxiliary ordinary differential equations,” Chaos Solitons and Fractals 39,1915–1919 (2009).
  * [35] C.D.Tian,“The rational solutions for a kind of mCH equation,” J. Qilu Norm. Univ. 26,119–123 (2011).
  * [36] X.L.Yang and J.S.Tan,“New travelling wave solutions for combined KdV–mKdV equation and (2+1)–dimensional Broer–Kaup–Kupershmidt system,” Chin.Phys. 16,310–317 (2007).
  * [37] A.H.Lu and X.K.Shao,“Using the elliptical auxiliary equation method to obtain exact solutions of the variable coefficients Boussinesq equation,” J. Chongqing Univ. Art. Sci.33,11–13 (2014).
  * [38] Taogetusang and Sirendaoreji,“ New type of exact solitary wave solutions for dispersive long–wave equation and Benjamin equation,” Acta Physica Sinica 55,3246–3254 (2006).
  * [39] A.M.Wazwaz,“Solitary wave solutions for modified forms of Degasperis–Procesi and Camassa–Holm equations,” Phys. Lett. A, 352,500-504 (2006).
  * [40] A.M.Wazwaz,“New solitary wave solutions to the modified forms of Degasperis-Procesi and Camassa-Holm equations,” Appl. Math. Comput. 186,130–141 (2007).
  * [41] Q.D.Wang and M.Y.Tang,“New exact solutions for two nonlinear equations,” Phys. Lett. A,372,2995–3000 (2008).
  * [42] S.X.Liang and D.J.Jeffrey,“New traveling wave solutions to modified CH and DP equations,” Comp.Phys.Commun.180,1429–1433 (2009).