# Exponential–Weierstrass type, exponential–Jacobi type and solitary type
solutions to some conformable fractional equations

Sirendaoreji
College of Mathematical Science,Inner Mongolia Normal University,
Huhhot 010022,Inner Mongolia, P.R. China

###### Abstract

A new algebraic method to find two special types of exact traveling wave
solutions and the solitary type solutions to some conformable fractional
partial differential equations is proposed. The two special types of solutions
given by the product of exponential and Weierstrass elliptic functions, and
the product of exponential and Jacobi elliptic functions are new and they
cannot be obtained by using the existing algebraic methods.

Key Words: fractional derivative, conformable fractional derivative,
fractional partial differential equation, traveling wave solution.

MSC 2010:35R11;34A08; 35C07; 35A20.

##  1 Introduction

In recent years,the problem of finding exact solutions to nonlinear fractional
partial differential equations (FPDEs) has become an active research area in
nonlinear science due to the wide applicability of nonlinear FPDEs to describe
many important phenomena and dynamic processes in fluid mechanics,biological
and chemical processes, signal processing, control systems and so on. Based on
the product rule and the chain rule for fractional derivatives, several direct
methods have been applied to find the exact solutions to some nonlinear
FPDEs[1, 2, 6, 7, 8]. However,the recent studies showed that the fractional
derivatives, such as the Riemann–Liouville derivative and the Caputo
derivative etc. do not satisfy the product rule and the chain rule[13, 14].
Therefore,the previous works of using these fractional derivatives to solve
those nonlinear FPDEs were not correct. But a so-called $\alpha$–order
conformable fractional derivative (CFD) of a function
$f:\left[0,\infty\right)\rightarrow{{R}}$ given by

$D_{t}{{}^{\alpha}}f(t)=\lim_{\varepsilon\rightarrow{0}}{\frac{f(t+\varepsilon{t^{1-\alpha}})-f(t)}{\varepsilon}},t>0,\alpha\in\left(0,1\right],$
(1.1)

which was defined by Khalil et al. [10] just satisfies the required properties
such as [10, 16]

$\displaystyle
D_{t}^{\alpha}(af+bg)=aD_{t}^{\alpha}{f}+bD_{t}^{\alpha}{g},\forall{x}\in{R},\mbox{(Linearity)}~{}$
(1.2) $\displaystyle D_{t}^{\alpha}{t^{\beta}}=\beta{t^{\beta-\alpha}},$ (1.3)
$\displaystyle
D_{t}^{\alpha}(fg)=gD_{t}^{\alpha}{f}+fD_{t}^{\alpha}{g},\mbox{(Product
rule)},$ (1.4) $\displaystyle
D_{t}^{\alpha}{f(g(t))}=f^{\prime}_{g}(g(t))D_{t}^{\alpha}{g},\mbox{(Chain
rule)},$ (1.5) $\displaystyle
D_{t}^{\alpha}{f}=t^{1-\alpha}f^{\prime}(t),f\,\mbox{is differentiable}.$
(1.6)

Therefore, it is an acceptable choice to use CFD to solve the conformable
fractional partial differential equations (CFPDEs). As a matter of fact,the
CFD (1.1) with its properties (1.4),(1.5) and (1.6) were successfully used to
solve some CFPDEs by means of the Riccati equation expansion [3, 15],the
sine–cosine method [4], the $\exp\left(\phi(\varepsilon)\right)$–expansion
[9], the modified Kudryashov method [9, 12] and the Jacobi elliptic function
expansion method[18], etc. However,we find that all these existing direct
methods cannot be used to find two types of exact traveling wave solutions of
CFPDEs when their original equations have the non–integrable background.The
first type solution,namely the exponential–Weierstrass type solution,is
expressed by the product of exponential function and Weierstrass elliptic
function. The second type solution,namely the exponential–Jacobi type
solution, is expressed by the product of exponential function and Jacobi
elliptic function.Motivated by this problem,in this paper, we shall propose a
direct algebraic method for constructing these two special types of solutions
for CFPDEs. It is seen that the method introduced in this paper is effective
to find the exponential–Weierstrass type,the exponential–Jacobi type and the
solitary type solutions of the CFPDEs.

This paper is organized as follows.In Sec. 2, by introducing two new second
order auxiliary nonlinear ordinary differential equations and their solutions
we shall propose a new algebraic method for solving CFPDEs. In Sec. 3,we shall
use our suggested method to find the exact traveling wave solutions of some
CFPDEs. Finally, Sec. 4 offers a discussion.

##  2 Description of the method

Now let us simplify describe our direct algebraic method for finding traveling
wave solutions for CFPDEs. Suppose that a CFPDE is given by

$P(u,D_{t}^{\alpha}{u},D_{x}^{\beta}{u},D_{t}^{2\alpha}u,{D_{t}^{\alpha}}D_{x}^{\beta}{u},D_{x}^{2\beta}u,\cdots)=0,$
(2.7)

where $P$ is a polynomial of its unknowns and $0<\alpha,\beta\leq{1}$. The key
steps of our method are outlined as follows.

Step 1: Making the wave transformation

$u(x,t)=u(\xi),\xi={\frac{x^{\beta}}{\beta}}+\omega{\frac{t^{\alpha}}{\alpha}},$
(2.8)

and using the properties of CFD we can convert Eq.(2.7) into the following ODE

$Q(u,{\frac{du}{d\xi}},{\frac{d^{2}u}{d\xi^{2}}},\cdots)=0.$ (2.9)

Step 2: (a) To seek the exponential–Weierstrass type and solitary type exact
traveling wave solutions,we assume that Eq. (2.9) has the solution of the form

$u(\xi)=a_{0}+a_{1}F(\xi),$ (2.10)

where $a_{0},a_{1}$ are constants, $F(\xi)$ satisfies the following second
order auxiliary ODE[19]

$F^{\prime\prime}(\xi)=bF^{2}(\xi)-6a^{2}F(\xi)+5aF^{\prime}(\xi).$ (2.11)

This equations admits the following solutions

$F(\xi)=\left\\{\begin{aligned}
&{\frac{6}{b}}e^{2a\xi}\wp\left({\frac{1}{a}}e^{a\xi}+c_{1},0,g_{3}\right),\\\
&{\frac{3a^{2}}{2b}}\left[1+\tanh\left({\frac{a}{2}}\xi\right)\right]^{2},\\\
&{\frac{3a^{2}}{2b}}\left[1+\coth\left({\frac{a}{2}}\xi\right)\right]^{2},\end{aligned}\right.$
(2.12)

(b) To find the exponential–Jacobi type and the solitary type exact traveling
wave solutions,we can take the solution of Eq. (2.9) of the form

$u(\xi)=a_{1}F(\xi),$ (2.13)

where $a_{1}$ is an undermined constant, $F(\xi)$ satisfies the following
second order auxiliary ODE[19]

$F^{\prime\prime}(\xi)=cF^{3}(\xi)-2a^{2}F(\xi)-3aF^{\prime}(\xi).$ (2.14)

This equation has the following solutions

$F(\xi)=\left\\{\begin{array}[]{l}{\varepsilon}ae^{-a\xi}\mathrm{ds}\left(e^{-a\xi}+c_{2},{\frac{\sqrt{2}}{2}}\right),c=2,\\\
{\varepsilon}ae^{-a\xi}\mathrm{nc}\left(\sqrt{2}e^{-a\xi}+c_{2},{\frac{\sqrt{2}}{2}}\right),c=2,\\\
{\frac{{\varepsilon}a}{2}}\left[1-\tanh\left({\frac{a}{2}}\xi\right)\right],c=2,\\\
{\frac{{\varepsilon}a}{2}}\left[1-\coth\left({\frac{a}{2}}\xi\right)\right],c=2,\\\
{\varepsilon}ae^{-a\xi}\mathrm{cn}\left(\sqrt{2}e^{-a\xi}+c_{3},{\frac{\sqrt{2}}{2}}\right),c=-2,\\\
{\frac{\sqrt{2}}{2}}{\varepsilon}ae^{-a\xi}\mathrm{sd}\left(\sqrt{2}e^{-a\xi}+c_{3},{\frac{\sqrt{2}}{2}}\right),c=-2,\end{array}\right.$
(2.15)

where $\wp$ expresses the Weierstras elliptic function,
$\mathrm{ds},\mathrm{nc},\mathrm{cn},\mathrm{sd}$ are the Jacobi eliptic
functions,$\varepsilon=\pm{1}$, and $a,b,c_{1},c_{2},c_{3},g_{3}$ are
constants.

Step 3: Substituting (2.10) with (2.11), and (2.13) with (2.14) separately
into Eq. (2.9) and setting the coefficients of like powers of
$F^{i}(F^{\prime})^{j}$ to zero, we get a set of algebraic equations for
unknowns $a_{0},a_{1},a,b,\omega$ ,and for unknowns
$a_{1},a,\omega$,respectively. The system of algebraic equations is solved by
using a computer algebraic system,then the values of these unknowns can be
obtained.

Step 4: The exact traveling wave solutions of the Eq.(2.7) can be obtained by
putting the values of unknowns obtained in Step 3 with (2.12) into (2.10),and
(2.15) into (2.13),respectively.

##  3 Applications of the method

Now we consider some illustrative examples to show the effectiveness of our
proposed method.

###### Example 1

The space–time fractional KdV–Burgers equation[5]

$\displaystyle
D_{t}^{\alpha}{u}+{\lambda}uD_{x}^{\beta}{u}+{\mu}{D_{x}^{2\beta}u}+\nu{D_{x}^{3\beta}u}=0,$
(3.16)

in which $\lambda,\mu,\nu$ are constants.

Taking (2.8) into (3.16) we obtain the following ODE

${\omega}u^{\prime}+{\lambda}uu^{\prime}+{\mu}u^{\prime\prime}+{\nu}u^{\prime\prime\prime}=0.$
(3.17)

Substituting (2.10) with (2.11) into (3.17) and setting the coefficients of
$F,F^{2}$, $F^{\prime},FF^{\prime}$ to be zero,we obtain a set of algebraic
equations

$\displaystyle\left\\{\begin{array}[]{l}2b{\nu}a_{1}+{\lambda}a_{1}^{2}=0,\\\
-30a^{3}{\nu}a_{1}-6a^{2}{\mu}a_{1}=0,\\\ 5ab{\nu}a_{1}+b{\mu}a_{1}=0,\\\
19a^{2}{\nu}a_{1}+5a{\mu}a_{1}+{\lambda}a_{0}a_{1}+{\omega}a_{1}=0.\end{array}\right.$

It solves that

$a=-{\frac{\mu}{5\nu}},a_{0}={\frac{6\mu^{2}-25\omega\nu}{25\lambda\nu}},a_{1}=-{\frac{2b\nu}{\lambda}}.$
(3.19)

Inserting (2.12) with (3.19) into (2.10) we get the exponential–Weierstrass
type and solitary type exact traveling wave solutions of Eq.(3.16) as
following

$\begin{array}[]{l}u_{1}(x,t)=-{\frac{12}{\lambda}}e^{-\frac{2\mu}{5\nu}\left({\frac{x^{\beta}}{\beta}}+\omega{\frac{t^{\alpha}}{\alpha}}\right)}\wp\left({\frac{5\nu}{\mu}}e^{-\frac{\mu}{5\nu}\left({\frac{x^{\beta}}{\beta}}+\omega{\frac{t^{\alpha}}{\alpha}}\right)}+c_{1},0,g_{3}\right)+{\frac{6\mu^{2}-25\omega\nu}{25\lambda\nu}},\\\
u_{2}(x,t)=-{\frac{3\mu^{2}}{25\lambda\nu}}\left[1-\tanh{\frac{\mu}{10\nu}}\left({\frac{x^{\beta}}{\beta}}+\omega{\frac{t^{\alpha}}{\alpha}}\right)\right]^{2}+{\frac{6\mu^{2}-25\omega\nu}{25\lambda\nu}},\\\
u_{3}(x,t)=-{\frac{3\mu^{2}}{25\lambda\nu}}\left[1-\coth{\frac{\mu}{10\nu}}\left({\frac{x^{\beta}}{\beta}}+\omega{\frac{t^{\alpha}}{\alpha}}\right)\right]^{2}+{\frac{6\mu^{2}-25\omega\nu}{25\lambda\nu}},\end{array}$

where $c_{1},g_{3}$ are arbitrary constants.

###### Example 2

The time fractional Fisher equation[17]

$D_{t}^{\alpha}{u}=u_{xx}+6u\left(1-u\right).$ (3.20)

Substituting (2.8) with $\beta=1$ into (3.20), we can convert the Eq. (3.20)
into the following ODE

${\omega}u^{\prime}=u^{\prime\prime}+6u\left(1-u\right),$ (3.21)

Substituting (2.10) and (2.11) into (3.21) and setting the coefficients of
$F^{\prime},F^{j}~{}(j=0,1,2)$ to zero,we have

$\displaystyle\left\\{\begin{array}[]{l}6a_{0}^{2}-6a_{0}=0,\\\
6a_{1}^{2}-a_{1}b=0,\\\ -5aa_{1}+a_{1}\omega=0,\\\
6a^{2}a_{1}+12a_{0}a_{1}-6a_{1}=0.\end{array}\right.$

This algebraic equations is solved that

$\displaystyle a_{0}=0,a_{1}={\frac{b}{6}},a=1,\omega=5,$ (3.23)
$\displaystyle a_{0}=0,a_{1}={\frac{b}{6}},a=-1,\omega=-5.$ (3.24)

Now the exponential–Weierstrass type and the solitary type exact traveling
wave solutions for Eq. (3.20) can be obtained by taking (2.12) with (3.23) and
(3.24) into (2.10),respectively,they are

$\begin{array}[]{l}u_{1}(x,t)=e^{2x+{\frac{10t^{\alpha}}{\alpha}}}\wp\left(e^{x+{\frac{5t^{\alpha}}{\alpha}}}+c_{1},0,g_{3}\right),\\\
u_{2}(x,t)={\frac{1}{4}}\left[1+\tanh{\frac{1}{2}}\left(x+{\frac{5t^{\alpha}}{\alpha}}\right)\right]^{2},\\\
u_{3}(x,t)={\frac{1}{4}}\left[1+\coth{\frac{1}{2}}\left(x+{\frac{5t^{\alpha}}{\alpha}}\right)\right]^{2},\\\
u_{4}(x,t)=e^{-2x+{\frac{10t^{\alpha}}{\alpha}}}\wp\left(e^{-x+{\frac{5t^{\alpha}}{\alpha}}}+c_{1},0,g_{3}\right),\\\
u_{5}(x,t)={\frac{1}{4}}\left[1-\tanh{\frac{1}{2}}\left(x-{\frac{5t^{\alpha}}{\alpha}}\right)\right]^{2},\\\
u_{6}(x,t)={\frac{1}{4}}\left[1-\coth{\frac{1}{2}}\left(x-{\frac{5t^{\alpha}}{\alpha}}\right)\right]^{2},\end{array}$

where $c_{1},g_{3}$ are arbitrary constants.

###### Example 3

The time fractional RLW–Burgers equation[12]

$D_{t}^{\alpha}{u}+pu_{x}+quu_{x}+ru_{xx}+su_{xxt}=0.$ (3.25)

Inserting (2.8) with $\beta=1$ into (3.25) leads the following ODE

$\left(\omega+p\right)u^{\prime}+quu^{\prime}+ru^{\prime\prime}+{\omega}su^{\prime\prime\prime}=0.$
(3.26)

Taking (2.10) with (2.11) into (3.26) and setting the coefficients of
$F,F^{2},F^{\prime},FF^{\prime}$ to be zero,then we obtain

$\left\\{\begin{array}[]{l}-30a^{3}{\omega}sa_{1}-6a^{2}ra_{1}=0,\\\
2b{\omega}sa_{1}+qa_{1}^{2}=0,\\\ 5ab{\omega}sa_{1}+bra_{1}=0,\\\
19a^{2}{\omega}sa_{1}+5ara_{1}+qa_{0}a_{1}+\left(\omega+p\right)a_{1}=0.\end{array}\right.$

It solves that

$a=-{\frac{r}{5s\omega}},a_{0}=-{\frac{25s\omega^{2}+25ps\omega-6r^{2}}{25qs\omega}},a_{1}=-{\frac{2bs\omega}{q}}.$
(3.27)

The exponential–Weierstrass type and the solitary type exact solutions of the
Eq.(3.25) are obtained by substituting (2.12) with (3.27) into (2.10),they are
the following

$\begin{array}[]{l}u_{1}(x,t)=-{\frac{12}{q}}e^{-\frac{2r}{5s\omega}\left(x+{\frac{{\omega}t^{\alpha}}{\alpha}}\right)}\wp\left({\frac{5s\omega}{r}}e^{-\frac{r}{5s\omega}\left(x+{\frac{{\omega}t^{\alpha}}{\alpha}}\right)}+c_{1},0,g_{3}\right)\\\
\qquad-{\frac{25s\omega^{2}+25ps\omega-6r^{2}}{25qs\omega}},\\\
u_{2}(x,t)=-{\frac{3r^{2}}{25qs\omega}}\left[1-\tanh{\frac{r}{10s\omega}}\left(x+{\frac{\omega{t^{\alpha}}}{\alpha}}\right)\right]^{2}-{\frac{25s\omega^{2}+25ps\omega-6r^{2}}{25qs\omega}},\\\
u_{3}(x,t)=-{\frac{3r^{2}}{25qs\omega}}\left[1-\coth{\frac{r}{10s\omega}}\left(x+{\frac{\omega{t^{\alpha}}}{\alpha}}\right)\right]^{2}-{\frac{25s\omega^{2}+25ps\omega-6r^{2}}{25qs\omega}},\end{array}$

where $c_{1},g_{3}$ are free parameters.

###### Example 4

The time fractional Cahn–Allen equation[11]

$D_{t}^{\alpha}{u}-u_{xx}-u+u^{3}=0.$ (3.28)

Substituting (2.8) with $\beta=1$ into (3.28) we get the following ODE

${\omega}u^{\prime}-u^{\prime\prime}-u+u^{3}=0.$ (3.29)

Substituting (2.13) with (2.14) into (3.29) and setting the coefficients of
$F,F^{3},F^{\prime}$ to zero gives the following ODE

$\left\\{\begin{array}[]{l}2a^{2}a_{1}-a_{1}=0,\\\ a_{1}^{3}-ca_{1}=0,\\\
3aa_{1}+{\omega}a_{1}=0.\end{array}\right.$

Solving this algebraic equation we obtain that

$a=\pm{\frac{\sqrt{2}}{2}},a_{1}=\pm\sqrt{c},\omega=\mp{\frac{3\sqrt{2}}{2}}.$
(3.30)

The exponential–Jacobi type and the solitary type exact solutions of Eq.
(3.28) are obtained by taking (3.30) and (2.15) with $c=2$ into (2.13),they
are now given by

$\displaystyle\begin{array}[]{l}u_{1}(x,t)={\varepsilon}e^{\mp{\frac{\sqrt{2}}{2}}\left(x\mp{\frac{3\sqrt{2}}{2\alpha}t^{\alpha}}\right)}\mathrm{ds}\left(e^{\mp{\frac{\sqrt{2}}{2}}\left(x\mp{\frac{3\sqrt{2}}{2\alpha}t^{\alpha}}\right)}+c_{2},{\frac{\sqrt{2}}{2}}\right),\\\
u_{2}(x,t)={\varepsilon}e^{\mp{\frac{\sqrt{2}}{2}}\left(x\mp{\frac{3\sqrt{2}}{2\alpha}t^{\alpha}}\right)}\mathrm{nc}\left(\sqrt{2}e^{\mp{\frac{\sqrt{2}}{2}}\left(x\mp{\frac{3\sqrt{2}}{2\alpha}t^{\alpha}}\right)}+c_{2},{\frac{\sqrt{2}}{2}}\right),\\\
u_{3}(x,t)={\frac{\varepsilon}{2}}\left[1\mp\tanh{\frac{\sqrt{2}}{4}}\left(x\mp{\frac{3\sqrt{2}t^{\alpha}}{2\alpha}}\right)\right],\\\
u_{4}(x,t)={\frac{\varepsilon}{2}}\left[1\mp\coth{\frac{\sqrt{2}}{4}}\left(x\mp{\frac{3\sqrt{2}t^{\alpha}}{2\alpha}}\right)\right],\end{array}$

where $c_{2}$ is a free parameter.

###### Example 5

The space–time fractional mKdV–Burgers equation [15]

$D_{t}^{\alpha}{u}+{\lambda}u^{2}D_{x}^{\alpha}{u}+rD_{x}^{2\alpha}u+sD_{x}^{3\alpha}u=0.$
(3.32)

Inserting (2.8) with $\beta=\alpha$ into (3.32) leads the following ODE

${\omega}u^{\prime}+{\lambda}u^{2}u^{\prime}+ru^{\prime\prime}+su^{\prime\prime\prime}=0.$
(3.33)

Taking (2.13) with (2.14) into (3.33) and setting the coefficients of
$F,F^{2},F^{\prime},F^{2}F^{\prime}$ to zero,we obtain a set of algebraic
equations

$\left\\{\begin{array}[]{l}6a^{3}sa_{1}-2a^{2}ra_{1}=0,\\\
{\lambda}a_{1}^{3}+3csa_{1}=0,\\\ -3acsa_{1}+cra_{1}=0,\\\
7a^{2}sa_{1}-3ara_{1}+{\omega}a_{1}=0.\end{array}\right.$

Its solution is found to be

$a={\frac{r}{3s}},a_{1}=\pm\sqrt{-\frac{3cs}{\lambda}},\omega={\frac{2r^{2}}{9s}}.$
(3.34)

The exponential–Jacobi type and the solitary type exact traveling wave
solutions of Eq.(3.32) obtained by taking (3.34) and (2.15) with $c=2$ into
(2.13) are

$\displaystyle\begin{array}[]{l}u_{1}(x,t)={\frac{r\varepsilon}{3s}}\sqrt{-\frac{6s}{\lambda}}e^{-{\frac{r}{3s}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{2r^{2}t^{\alpha}}{9s\alpha}}\right)}\mathrm{ds}\left(e^{-{\frac{r}{3s}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{2r^{2}t^{\alpha}}{9s\alpha}}\right)}+c_{2},{\frac{\sqrt{2}}{2}}\right),s\lambda<0,\\\
u_{2}(x,t)={\frac{r\varepsilon}{3s}}\sqrt{-\frac{6s}{\lambda}}e^{-{\frac{r}{3s}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{2r^{2}t^{\alpha}}{9s\alpha}}\right)}\mathrm{nc}\left(\sqrt{2}e^{-{\frac{r}{3s}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{2r^{2}t^{\alpha}}{9s\alpha}}\right)}+c_{2},{\frac{\sqrt{2}}{2}}\right),s\lambda<0,\\\
u_{3}(x,t)=-{\frac{r\varepsilon}{6s}}\sqrt{-\frac{6s}{\lambda}}\left[1-\tanh{\frac{r}{6s}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{2r^{2}t^{\alpha}}{9s\alpha}}\right)\right],s\lambda<0,\\\
u_{4}(x,t)=-{\frac{r\varepsilon}{6s}}\sqrt{-\frac{6s}{\lambda}}\left[1-\coth{\frac{r}{6s}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{2r^{2}t^{\alpha}}{9s\alpha}}\right)\right],s\lambda<0,\end{array}$

where $c_{2}$ is an arbitrary constant.

By inserting (3.34) and (2.15) with $c=-2$ into (2.13) we get the following
exponential–Jacobi type and solitary type traveling wave solutions of
Eq.(3.28)

$\displaystyle
u_{5}(x,t)={\frac{r\varepsilon}{3s}}\sqrt{\frac{6s}{\lambda}}e^{-{\frac{r}{3s}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{2r^{2}t^{\alpha}}{9s\alpha}}\right)}\mathrm{cn}\left(\sqrt{2}e^{-{\frac{r}{3s}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{2r^{2}t^{\alpha}}{9s\alpha}}\right)}+c_{2},{\frac{\sqrt{2}}{2}}\right),s\lambda>0,$
$\displaystyle
u_{6}(x,t)={\frac{r\varepsilon}{3s}}\sqrt{\frac{3s}{\lambda}}e^{-{\frac{r}{3s}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{2r^{2}t^{\alpha}}{9s\alpha}}\right)}\mathrm{sd}\left(\sqrt{2}e^{-{\frac{r}{3s}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{2r^{2}t^{\alpha}}{9s\alpha}}\right)}+c_{2},{\frac{\sqrt{2}}{2}}\right),s\lambda<0,$

in which $c_{2}$ is a free parameter.

###### Example 6

The space–time fractional telegraph equation[15]

$D_{t}^{2\alpha}u-D_{x}^{2\alpha}u+D_{t}^{\alpha}{u}+{\mu}u+{\nu}u^{3}=0.$
(3.36)

Taking (2.8) with $\beta=\alpha$ into (3.36) we obtain the following ODE

$\left(\omega^{2}-1\right)u^{\prime\prime}+{\omega}u^{\prime}+{\mu}u+{\nu}u^{3}=0.$
(3.37)

Substituting (2.13) with (2.14) into (3.37) and setting the coefficients of
$F,F^{3},F^{\prime}$ to zero we obtain

$\left\\{\begin{array}[]{l}-3a\omega^{2}a_{1}+3aa_{1}+{\omega}a_{1}=0,\\\
-2a^{2}\omega^{2}a_{1}+2a^{2}a_{1}+{\mu}a_{1}=0,\\\
{\nu}a_{1}^{3}+c\omega^{2}a_{1}-ca_{1}=0.\end{array}\right.$

Solutions of this algebraic equations are found to be

$\displaystyle
a={\frac{9\mu-2}{2}}\sqrt{\frac{\mu}{9\mu-2}},a_{1}=\pm\sqrt{-\frac{2c}{\nu\left(9\mu-2\right)}},\omega=3\sqrt{\frac{\mu}{9\mu-2}},$
(3.38) $\displaystyle
a=-{\frac{9\mu-2}{2}}\sqrt{\frac{\mu}{9\mu-2}},a_{1}=\pm\sqrt{-\frac{2c}{\nu\left(9\mu-2\right)}},\omega=-3\sqrt{\frac{\mu}{9\mu-2}},$
(3.39)

When $c=2$ and $c=-2$,by substituting (3.38) and (2.15) into
(2.13),respectively,we get the exponential–Jacobi type and the solitary type
traveling wave solutions of Eq.(3.36) as following

$\begin{array}[]{l}u_{1}(x,t)=\varepsilon\sqrt{-\frac{\mu}{\nu}}e^{-\eta^{+}}\mathrm{ds}\left(e^{-\eta^{+}}+c_{2},{\frac{\sqrt{2}}{2}}\right),\mu<0,\nu>0,\mbox{or}\;\mu>{\frac{2}{9}},\nu<0,\\\
u_{2}(x,t)=\varepsilon\sqrt{-\frac{\mu}{\nu}}e^{-\eta^{+}}\mathrm{nc}\left(\sqrt{2}e^{-\eta^{+}}+c_{2},{\frac{\sqrt{2}}{2}}\right),\mu<0,\nu>0,\mbox{or}\;\mu>{\frac{2}{9}},\nu<0,\\\
u_{3}(x,t)={\frac{\varepsilon}{2}}\sqrt{-\frac{\mu}{\nu}}\left[1-\tanh\eta^{+}\right],\mu<0,\nu>0,\mbox{or}\;\mu>{\frac{2}{9}},\nu<0,\\\
u_{4}(x,t)={\frac{\varepsilon}{2}}\sqrt{-\frac{\mu}{\nu}}\left[1-\coth\eta^{+}\right],\mu<0,\nu>0,\mbox{or}\;\mu>{\frac{2}{9}},\nu<0,\\\
u_{5}(x,t)=\varepsilon\sqrt{\frac{\mu}{\nu}}e^{-\eta^{+}}\mathrm{cn}\left(\sqrt{2}e^{-\eta^{+}}+c_{2},{\frac{\sqrt{2}}{2}}\right),\mu<0,\nu<0,\mbox{or}\;\mu>{\frac{2}{9}},\nu>0,\\\
u_{6}(x,t)={\frac{\varepsilon}{2}}\sqrt{\frac{2\mu}{\nu}}e^{-\eta^{+}}\mathrm{sd}\left(\sqrt{2}e^{-\eta^{+}}+c_{2},{\frac{\sqrt{2}}{2}}\right),\mu<0,\nu<0,\mbox{or}\;\mu>{\frac{2}{9}},\nu>0,\end{array}$

where

$\eta^{+}={\frac{9\mu-2}{2}}\sqrt{\frac{\mu}{9\mu-2}}\left({\frac{x^{\alpha}}{\alpha}}+{\frac{3}{\alpha}}\sqrt{{\frac{\mu}{9\mu-2}}}t^{\alpha}\right).$

When $c=2$ and $c=-2$,by substituting (3.39) and (2.15) into
(2.13),respectively,the exponential–Jacobi type and solitary type traveling
wave solutions of Eq.(3.36) are obtained as

$\begin{array}[]{l}u_{7}(x,t)=\varepsilon\sqrt{-\frac{\mu}{\nu}}e^{\eta^{-}}\mathrm{ds}\left(e^{\eta^{-}}+c_{2},{\frac{\sqrt{2}}{2}}\right),\mu<0,\nu>0\;\mbox{or}\;\mu>{\frac{2}{9}},\nu<0,\\\
u_{8}(x,t)=\varepsilon\sqrt{-\frac{\mu}{\nu}}e^{\eta^{-}}\mathrm{nc}\left(\sqrt{2}e^{\eta^{-}}+c_{2},{\frac{\sqrt{2}}{2}}\right),\mu<0,\nu>0\;\mbox{or}\;\mu>{\frac{2}{9}},\nu<0,\\\
u_{9}(x,t)={\frac{\varepsilon}{2}}\sqrt{-\frac{\mu}{\nu}}\left[1+\tanh\eta^{-}\right],\mu<0,\nu>0,\mbox{or}\;\mu>{\frac{2}{9}},\nu<0,\\\
u_{10}(x,t)={\frac{\varepsilon}{2}}\sqrt{-\frac{\mu}{\nu}}\left[1+\coth\eta^{-}\right],\mu<0,\nu>0,\mbox{or}\;\mu>{\frac{2}{9}},\nu<0,\\\
u_{11}(x,t)=\varepsilon\sqrt{\frac{\mu}{\nu}}e^{\eta^{-}}\mathrm{cn}\left(\sqrt{2}e^{\eta^{-}}+c_{2},{\frac{\sqrt{2}}{2}}\right),\mu<0,\nu<0\;\mbox{or}\;\mu>{\frac{2}{9}},\nu>0,\\\
u_{12}(x,t)={\frac{\varepsilon}{2}}\sqrt{\frac{2\mu}{\nu}}e^{\eta^{-}}\mathrm{sd}\left(\sqrt{2}e^{\eta^{-}}+c_{2},{\frac{\sqrt{2}}{2}}\right),\mu<0,\nu<0\;\mbox{or}\;\mu>{\frac{2}{9}},\nu>0,\end{array}$

where

$\eta^{-}={\frac{9\mu-2}{2}}\sqrt{\frac{\mu}{9\mu-2}}\left({\frac{x^{\alpha}}{\alpha}}-{\frac{3}{\alpha}}\sqrt{{\frac{\mu}{9\mu-2}}}t^{\alpha}\right).$

##  4 Discussion

The examples in Sec. 3 showed that our method is very effective to find the
exponential–Weierstrass type,the exponential–Jacobi type and solitary type
exact traveling wave solutions for CFPDEs. In our method, the auxiliary
equations (2.11) and (2.14) are new and they have not been previously used in
any direct methods. Therefore,all our exponential–Weierstrass type and
exponential–Jacobi type traveling wave solutions are new and cannot be found
by using the existing direct algebraic methods. It is also pointed out that
the original integer order nonlinear partial differential equations of the
CFPDEs considered in our examples are non–integrable. Therefore,our method is
restricted to solve those CFPDEs whose original equations are non–integrable.

## Acknowledgements

This work has been supported by the National Natural Science Foundation of
China under Grant No.11861050.

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