### 4.1 Fractional sub-equation method

Now, we outline the main steps of the fractional sub-equation method for solving fractional differential equations.

For a given NFDE, consider two variables *x* and *t*,

P(u,{u}_{t},{u}_{x},{D}_{t}^{\alpha}u,{D}_{x}^{\alpha}u,\dots )=0,\phantom{\rule{1em}{0ex}}0<\alpha \le 1,

(41)

where {D}_{x}^{\alpha}u and {D}_{t}^{\alpha}u are the modified Riemann-Liouville derivatives of *u* with respect to *t* and *x*, respectively.

Step 1: By making use of the traveling wave transformation

u(x,t)=u(\xi ),\phantom{\rule{1em}{0ex}}\xi =x+ct,

(42)

where *c* is a nonzero constant to be determined later, (41) can be reduced to a nonlinear fractional ordinary differential equation (NFODE)

P(u,c{u}^{\prime},{u}^{\prime},{c}^{\alpha}{D}_{\xi}^{\alpha}u,{D}_{\xi}^{\alpha}u,\dots )=0,\phantom{\rule{1em}{0ex}}0<\alpha \le 1.

(43)

Step 2: Suppose that Eq. (43) has the following solution:

u(\xi )={a}_{0}+\sum _{i=1}^{n}{a}_{i}{(\psi (\xi ))}^{i},

(44)

where {a}_{i} (i=1,\dots ,n) are constants to be determined later, positive integer *n* can be determined by balancing the highest order derivatives and nonlinear terms in Eq. (41) or Eq. (43). The function \psi (\xi ) satisfies the following Bäklund transformation of the fractional Riccati equation [38]:

\psi (\xi )=\frac{-\sigma B+D\varphi (\xi )}{D+B\varphi (\xi )},

(45)

where *B*, *D* are arbitrary parameters, and B\ne 0. Meanwhile, \varphi (\xi ) are decided by

{D}_{\xi}^{\alpha}\varphi (\xi )=\sigma +{\varphi}^{2}(\xi ),

(46)

where *σ* is a constant. Eq. (46) has the following solutions:

\varphi (\xi )=\{\begin{array}{cc}-\sqrt{-\sigma}tanh(\sqrt{-\sigma}\xi ,\alpha ),\hfill & \sigma <0,\hfill \\ -\sqrt{-\sigma}coth(\sqrt{-\sigma}\xi ,\alpha ),\hfill & \sigma <0,\hfill \\ \sqrt{\sigma}tan(\sqrt{\sigma}\xi ,\alpha ),\hfill & \sigma >0,\hfill \\ -\sqrt{\sigma}cot(\sqrt{\sigma}\xi ,\alpha ),\hfill & \sigma >0,\hfill \\ -\frac{\mathrm{\Gamma}(1+\alpha )}{{\xi}^{\alpha}+\omega},\hfill & \omega \text{is constant},\sigma =0,\hfill \end{array}

(47)

with the generalized hyperbolic and trigonometric functions

\begin{array}{r}{sin}_{\alpha}(\xi )=\frac{{E}_{\alpha}(i{\xi}^{\alpha})-{E}_{\alpha}(-i{\xi}^{\alpha})}{2i},\phantom{\rule{2em}{0ex}}{cos}_{\alpha}(\xi )=\frac{{E}_{\alpha}(i{\xi}^{\alpha})+{E}_{\alpha}(-i{\xi}^{\alpha})}{2i},\\ {sinh}_{\alpha}(\xi )=\frac{{E}_{\alpha}({\xi}^{\alpha})-{E}_{\alpha}(-{\xi}^{\alpha})}{2},\phantom{\rule{2em}{0ex}}{cosh}_{\alpha}(\xi )=\frac{{E}_{\alpha}({\xi}^{\alpha})+{E}_{\alpha}(-{\xi}^{\alpha})}{2},\\ {tan}_{\alpha}(\xi )=\frac{{sin}_{\alpha}(\xi )}{{cos}_{\alpha}(\xi )},\phantom{\rule{2em}{0ex}}{cot}_{\alpha}(\xi )=\frac{{cos}_{\alpha}(\xi )}{{sin}_{\alpha}(\xi )},\\ {tanh}_{\alpha}(\xi )=\frac{{sinh}_{\alpha}(\xi )}{{cosh}_{\alpha}(\xi )},\phantom{\rule{2em}{0ex}}{coth}_{\alpha}(\xi )=\frac{{cosh}_{\alpha}(\xi )}{{sinh}_{\alpha}(\xi )},\end{array}

(48)

here {E}_{\alpha}(\xi )={\sum}_{k=0}^{\mathrm{\infty}}\frac{{\xi}^{k}}{\mathrm{\Gamma}(1+k\alpha )} (\alpha >0) is the Mittag-Leffler function in one parameter.

Step 3: Substituting (44), (45) and (46) into (43) and setting the coefficients of the powers of {(\varphi (\xi ))}^{i} to be zero, one can obtain an over-determined nonlinear algebraic system in {a}_{i} (i=1,\dots ,n) and *c*.

Step 4: With the aid of Maple, solving the nonlinear algebraic system yields the explicit expressions of the parameters {a}_{i} (i=1,\dots ,n) and *c*. Then substituting these constants and the solutions of Eq. (47) into Eq. (44), we can get the exact and explicit solutions of the nonlinear fractional partial differential equation (NFPDE) (41).

### 4.2 Applications to the time fractional KdV equation

According to the above steps, firstly, we introduce the following transformations:

u(x,t)=u(\xi ),\phantom{\rule{1em}{0ex}}\xi =x+ct,

(49)

where *c* is a constant. Substituting (49) into (1), then (1) can be reduced to the following nonlinear fractional ordinary differential equation (NFODE):

{c}^{\alpha}{D}_{\xi}^{\alpha}u+au{u}_{\xi}+b{u}_{\xi \xi \xi}=0.

(50)

We suppose that Eq. (50) has the following solution:

u(\xi )={a}_{0}+\sum _{i=1}^{n}{a}_{i}{(\psi (\xi ))}^{i},

(51)

where {a}_{i} (i=1,\dots ,n) are constants to be determined later. Balancing the highest order derivative terms with nonlinear terms in Eq. (50), we get

u(\xi )={a}_{0}+{a}_{1}(\psi )+{a}_{2}{(\psi )}^{2}.

(52)

Substituting (52) along with (45) into (50) and then letting the coefficients of {(\varphi )}^{i} to zero, one can get some algebraic equations about *c*, {a}_{0}, {a}_{1} and {a}_{2}. Solving the algebraic equations by Maple, one can get the following.

Case 1:

\begin{array}{r}B=B,\phantom{\rule{2em}{0ex}}D=D,\phantom{\rule{2em}{0ex}}a=a,\phantom{\rule{2em}{0ex}}b=b,\phantom{\rule{2em}{0ex}}c={(-a{a}_{0})}^{\frac{1}{\alpha}},\\ \alpha =\alpha ,\phantom{\rule{2em}{0ex}}\sigma =0,\phantom{\rule{2em}{0ex}}{a}_{0}={a}_{0},\phantom{\rule{2em}{0ex}}{a}_{1}=0,\phantom{\rule{2em}{0ex}}{a}_{2}=-12\frac{b}{a}.\end{array}

(53)

Case 2:

\begin{array}{r}B=B,\phantom{\rule{2em}{0ex}}D=D,\phantom{\rule{2em}{0ex}}a=a,\phantom{\rule{2em}{0ex}}b=b,\phantom{\rule{2em}{0ex}}c=c,\phantom{\rule{2em}{0ex}}\alpha =\alpha ,\\ \sigma =-\frac{{D}^{2}}{{B}^{2}},\phantom{\rule{2em}{0ex}}{a}_{0}={a}_{0},\phantom{\rule{2em}{0ex}}{a}_{1}={a}_{1},\phantom{\rule{2em}{0ex}}{a}_{2}={a}_{2}.\end{array}

(54)

Case 3:

\begin{array}{r}B=B,\phantom{\rule{2em}{0ex}}D=D,\phantom{\rule{2em}{0ex}}a=-12\frac{b}{{a}_{2}},\phantom{\rule{2em}{0ex}}b=b,\phantom{\rule{2em}{0ex}}c={(-4\frac{b(2{a}_{2}\sigma -3{a}_{0})}{{a}_{2}})}^{\frac{1}{\alpha}},\\ \alpha =\alpha ,\phantom{\rule{2em}{0ex}}\sigma =\sigma ,\phantom{\rule{2em}{0ex}}{a}_{0}={a}_{0},\phantom{\rule{2em}{0ex}}{a}_{1}=0,\phantom{\rule{2em}{0ex}}{a}_{2}={a}_{2}.\end{array}

(55)

Case 4:

\begin{array}{r}B=B,\phantom{\rule{2em}{0ex}}D=0,\phantom{\rule{2em}{0ex}}a=a,\phantom{\rule{2em}{0ex}}b=-\frac{1}{12}a{a}_{2},\phantom{\rule{2em}{0ex}}c={(\frac{2}{3}a\sigma {a}_{2}-a{a}_{0})}^{\frac{1}{\alpha}},\\ \alpha =\alpha ,\phantom{\rule{2em}{0ex}}\sigma =\sigma ,\phantom{\rule{2em}{0ex}}{a}_{0}={a}_{0},\phantom{\rule{2em}{0ex}}{a}_{1}=0,\phantom{\rule{2em}{0ex}}{a}_{2}={a}_{2}.\end{array}

(56)

In view of (54), one can get new types of explicit solutions of Eq. (1) as follows:

\begin{array}{rl}u=& {a}_{0}+{a}_{1}\left(\frac{-\sigma B-D\sqrt{-\sigma}tanh(\sqrt{-\sigma}\xi ,\alpha )}{D-B\sqrt{-\sigma}tanh(\sqrt{-\sigma}\xi ,\alpha )}\right)\\ +{a}_{2}{\left(\frac{-\sigma B-D\sqrt{-\sigma}tanh(\sqrt{-\sigma}\xi ,\alpha )}{D-B\sqrt{-\sigma}tanh(\sqrt{-\sigma}\xi ,\alpha )}\right)}^{2},\end{array}

(57)

where \sigma <0, \xi =x+ct,

\begin{array}{rl}u=& {a}_{0}+{a}_{1}\left(\frac{-\sigma B-D\sqrt{-\sigma}coth(\sqrt{-\sigma}\xi ,\alpha )}{D-B\sqrt{-\sigma}coth(\sqrt{-\sigma}\xi ,\alpha )}\right)\\ +{a}_{2}{\left(\frac{-\sigma B-D\sqrt{-\sigma}coth(\sqrt{-\sigma}\xi ,\alpha )}{D-B\sqrt{-\sigma}coth(\sqrt{-\sigma}\xi ,\alpha )}\right)}^{2},\end{array}

(58)

where \sigma <0, \xi =x+ct,

\begin{array}{rl}u=& {a}_{0}+{a}_{1}\left(\frac{-\sigma B+D\sqrt{\sigma}tan(\sqrt{\sigma}\xi ,\alpha )}{D+B\sqrt{\sigma}tan(\sqrt{\sigma}\xi ,\alpha )}\right)\\ +{a}_{2}{\left(\frac{-\sigma B+D\sqrt{\sigma}tan(\sqrt{\sigma}\xi ,\alpha )}{D+B\sqrt{\sigma}tan(\sqrt{\sigma}\xi ,\alpha )}\right)}^{2},\end{array}

(59)

where \sigma >0, \xi =x+ct,

\begin{array}{rl}u=& {a}_{0}+{a}_{1}\left(\frac{-\sigma B-D\sqrt{\sigma}cot(\sqrt{\sigma}\xi ,\alpha )}{D-B\sqrt{\sigma}cot(\sqrt{\sigma}\xi ,\alpha )}\right)\\ +{a}_{2}{\left(\frac{-\sigma B-D\sqrt{\sigma}cot(\sqrt{\sigma}\xi ,\alpha )}{D-B\sqrt{\sigma}cot(\sqrt{\sigma}\xi ,\alpha )}\right)}^{2},\end{array}

(60)

where \sigma >0, \xi =x+ct.

If \sigma =0, one can get u=0.

Using (55), one can get new types of explicit solutions of Eq. (1) as follows:

u={a}_{0}+{a}_{2}{\left(\frac{-\sigma B-D\sqrt{-\sigma}tanh(\sqrt{-\sigma}\xi ,\alpha )}{D-B\sqrt{-\sigma}tanh(\sqrt{-\sigma}\xi ,\alpha )}\right)}^{2},

(61)

where \sigma <0, \xi =x+ct,

u={a}_{0}+{a}_{2}{\left(\frac{-\sigma B-D\sqrt{-\sigma}coth(\sqrt{-\sigma}\xi ,\alpha )}{D-B\sqrt{-\sigma}coth(\sqrt{-\sigma}\xi ,\alpha )}\right)}^{2},

(62)

where \sigma <0, \xi =x+ct,

u={a}_{0}+{a}_{2}{\left(\frac{-\sigma B+D\sqrt{\sigma}tan(\sqrt{\sigma}\xi ,\alpha )}{D+B\sqrt{\sigma}tan(\sqrt{\sigma}\xi ,\alpha )}\right)}^{2},

(63)

where \sigma >0, \xi =x+ct,

u={a}_{0}+{a}_{2}{\left(\frac{-\sigma B-D\sqrt{\sigma}cot(\sqrt{\sigma}\xi ,\alpha )}{D-B\sqrt{\sigma}cot(\sqrt{\sigma}\xi ,\alpha )}\right)}^{2},

(64)

where \sigma >0, \xi =x+ct,

u={a}_{0}+{a}_{2}{\left(\frac{D\mathrm{\Gamma}(1+\alpha )}{-D({\xi}^{\alpha}+\omega )+B\mathrm{\Gamma}(1+\alpha )}\right)}^{2},

(65)

where \sigma =0, \xi =x+ct.

**Remark 2** Using (53) and (56), we can also get other exact solutions of (1). Here we do not list all of them.

**Remark 3** To the best of our knowledge, the solutions obtained in this paper have not been reported in previous literature. Therefore, these solutions are new.

**Remark 4** It is interesting to note that if \alpha =1, FDEs (1) can be reduced to the conventional integer order KdV equation, and the obtained exact solutions can be reduced to the conventional hyperbolic and trigonometric functions.