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\def\rmfcintilla{{\it Rev.\ Mex.\ Fis.\/}  {\bf 65} (5)  529--535}
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\markboth{M.S.\ Hashemi, M.\ Inc, and M.\ Bayram}
{Symmetry properties and exact solutions of the time fractional\ldots}
\begin{document}
\title{Symmetry properties and exact solutions of the time fractional Kolmogorov-Petrovskii-Piskunov equation
\vspace{-6pt}}
\author{M.S.\ Hashemi$^{a}$, M.\ Inc$^{b}$, and M.\ Bayram$^{c}$}
\address{$^{a}$Department of Mathematics, Basic Science Faculty, 
\\
University of Bonab, Bonab 55517, Iran.  \\
$^{b}$Department of Mathematics, Science Faculty,
\\
 Firat University, 23119 Elazig, Turkey.  \\
$^{c}$Department of Computer Engineering, 
\\
Istanbul Gelisim University, Istanbul, Turkey. }
\maketitle
\recibido{17 January 2019}{19 February 2019
\vspace{-12pt}}
\begin{abstract}
In this paper, the time fractional Kolmogorov-Petrovskii-Piskunov
(TFKPP) equation is analyzed by means of Lie symmetry approach. The
TFKPP is reduced to ordinary differential equation of fractional
order via the attained point symmetries. Moreover, the simplest
equation method is used in construct the exact solutions of
underlying equation with recently introduced conformable
fractional derivative.
\end{abstract}
\keys{ Time fractional Kolmogorov-Petrovskii-Piskunov equation;\ Lie
symmetry analysis;\ Erd\'{e}lyi-Kober fractional derivative;\ 
Riemann-Liouville derivative;\ conformable fractional derivative;\ 
simplest equation method. \vspace{-4pt}}
\pacs{02.30.Uu;\ 04.20.Jb;\ 05.40.Fb;\ 05.60.-k \hspace{90pt} DOI: https://doi.org/10.31349/RevMexFis.65.529  \vspace{-4pt}}
\begin{multicols}{2}


\section{Introduction}


The fractional calculus (FC) began to wind up exceptionally famous
in a few parts of science and engineering.  Numerous  important
event, that is,  acoustics, anomalous diffusion, chemistry,
control processing, electro-magnetics,  and visco-elasticity have
been expressed by FC.  It is known that a systematic method for
extracting the analytical solution of both ordinary differential equations (ODEs) and partial differential equations (PDEs) was first
proposed by the Norwegian mathematician Sophus Lie in the early
19th century. The fundamental overview of this strategy is the
estimation of variable changes that can leave differential
condition unchanged. Therefore, a vital role in the field of FC is
to attain the Lie symmetries and the solutions of the equations
with the FC derivatives. There have been some properties of the
fractional sense that could not be found in classical sense, owing
to this we feel motivated to establish the symmetries of TFKPP
equation. This equation has
the following generalized form~[1-4]
\begin{align}\label{1}
\partial_t^\alpha u & =u_{xx}+\lambda u+\mu u^2+\gamma u^3,\
\lambda+\mu+\gamma=0,\ 
\nb
\psi^2 & =\mu^2-4\lambda\gamma\geq0,
\end{align}
where  $\partial_t^\alpha u:=\mathcal{D}_t^\alpha u$ stands for
Riemann-Liouville of order $\alpha$, expressed as [5]
\begin{align}\label{2}
\mathcal{D}_t^\alpha u(x,t)=\left\{
  \begin{array}{c}
    \frac{1}{\Gamma (n-\alpha)}\frac{\partial^n}{\partial t^n} \int\limits_0^t (t-\xi)^{n-\alpha-1} u(x,\xi) d\xi,  \\ 
 {} \qquad\qquad\qquad\qquad n-1<\alpha<n    \\
    \frac{\partial^n u}{\partial t^n},\qquad\qquad  \ \alpha=n\in\mathbb{N}\\
  \end{array}
\right.
\end{align}

The TFKPP Eq. (\ref{1}), has a large application and includes
as particular cases the time fractional Fitzhugh-Nagumo equation
($\lambda=-c,\ \mu=c+1,\ \gamma=-1,\ 0<c<1$), which is used in
population genetics, the time fractional Newell-Whitehead equation
($\lambda=1,\ \mu=0,\ \gamma=-1$). Recently, the homotopy
perturbation method and homotopy analysis method have utilized to
consider the TFKPP equation by Gepreel [1] and
Hariharan [2], respectly
with $\lambda=\mu=0$ and $\gamma=-2.$

In FC,  there are large amount of differential derivatives were
defined \textit{e.g}. [6-9]. In the calculus,
the chain rule is a useful and an applicable. It  is also hold for
conformable fractional derivatives.

As far as we know, every proposed fractional derivative has some
disadvantages. Therefore, Khalil \textit{et al.,}
[9], proposed a new definitions:


\textbf{Definition 1.1.}
Surmise that $f:[a,b]\times(0,\infty)\rightarrow\mathbb{R}$, then
the conformable fractional derivative of $f$ is given by
\begin{align}\label{a}
_t\mathbb{T}_\alpha(f)(x,t) & =\lim_{\epsilon\rightarrow
0}\frac{f\left(x,t+\epsilon t^{1-\alpha}\right)-f(x,t)}{\epsilon},
\nb
 \alpha & \in (0,1],
\end{align}
for all $t>0$.
 
 \textbf{Theorem 1.1}  [9] \label{t1}
\textit{Suppose that} $a,b\in\mathbb{R}$ \textit{and} $\alpha\in(0,1]$, \textit{then}

(i) $_t\mathbb{T}_\alpha(au+bv)=a\  _t\mathbb{T}_\alpha(u)+b\
_t\mathbb{T}_\alpha(v),$

\vspace{5pt}
(ii) $_t\mathbb{T}_\alpha(t^\lambda)=\lambda t^{\lambda-\alpha},\ \ \lambda\in\mathbb{R}$,

\vspace{5pt}
(iii) $_t\mathbb{T}_\alpha(uv)=u\ _t\mathbb{T}_\alpha(v)+v\ _t\mathbb{T}_\alpha(u)$,

\vspace{5pt}
(iv) $_t\mathbb{T}_\alpha \left( \dfrac{u}{v} \right)=\frac{u\ _t\mathbb{T}_\alpha(v)-v\ _t\mathbb{T}_\alpha(u)}{v^2}$,

\vspace{5pt}
(v) $_t\mathbb{T}_\alpha(u)(t)=t^{1-\alpha}u'(t),\ \ u\in C^1.$


\vspace{5pt}
More than that, the chain rule is valid for conformable fractional
derivatives, shown by Abdeljawad [10].

 \textbf{Theorem 1.2.} \label{t2}
\textit{Surmise that} $f:(0,\infty)\rightarrow\mathbb{R}$ \textit{is a real
differentiable,} $\alpha-$\textit{differentiable function. Assume that} $g$
\textit{is a function defined in the range of} $f$ \textit{and also differentiable;
then, one has the following rule:}
\begin{equation}
_t\mathbb{T}_\alpha(fog)(t)=t^{1-\alpha}g'(t)f'(g(t)).
\end{equation}


There are many investigation about conformable fractional derivatives
[11-14] and also some physical interpretations of this newly introduced 
fractional derivative are described in [15]. 

The organization of the manuscript is given below:
In Sec. 2, we provide some preliminaries. Section 3, is devoted
to the description of Lie symmetry analysis of TFKPP 
Eq.~(\ref{1}). General similarity forms and symmetry reductions are
established. In Sec. 4, exact solutions to the TFKPP equation
with conformable fractional derivative are investigated. Finally,
the last section is devoted to conclusions.




\section{Lie symmetry analysis of fractional partial differential equations}


Here, some description for solving fractional partial differential
equations (FPDEs) via Lie symmetry analysis will be provided.
Surmise that FPDE having as in [16-26]
\begin{equation}\label{b}
 \partial_t^{\alpha}u=F(x,t,u,u_{x},u_{xx}),\qquad  0<\alpha<1.
 \end{equation}
If (\ref{b}) is invariant under a one parameter Lie group of point
transformations
\begin{equation}\label{d}
\bar{t}=\bar{t}(x,t,u;\epsilon),\ \ \
\bar{x}=\bar{x}(x,t,u;\epsilon),\ \ \
\bar{u}=\bar{u}(x,t,u;\epsilon),
\end{equation}
the vector field of an evolution type of equation is as follows:
\begin{equation}\label{c}
V=\xi^{t}(x,t,u)\frac{\partial}{\partial
t}+\xi^{x}(x,t,u)\frac{\partial}{\partial
x}+\phi(x,t,u)\frac{\partial}{\partial u},
\end{equation}
where the coefficients $\xi^{t},\ \xi^{x}$ and $\phi$ of the
vector field are to be determined. When $V$  satisfy the Lie
symmetry condition,  the vector field (\ref{c}) generates a
symmetry of (\ref{b}),
\begin{equation}
pr^{(\alpha,2)}V(\Delta)|_{\Delta=0}=0,\ \
\Delta=\partial_t^\alpha u-F.\nonumber
\end{equation}
Thus the extension operator take the form
\begin{equation}
pr^{(\alpha,2)}V=V+\phi_\alpha^0\partial_{\partial_t^\alpha
u}+\phi^x\partial_{u_x}+\phi^{xx}\partial_{u_{xx}},\nonumber
\end{equation}
where
\begin{align*}
\phi^{x}&=\mathcal{D}_{x}(\phi)-u_{x}\mathcal{D}_{x}(\xi^{x})-u_{t}\mathcal{D}_{x}(\xi^{t}),\ \ \ \nonumber\\
\phi^{xx}&=\mathcal{D}_{x}(\phi^{x})-u_{xt}\mathcal{D}_{x}(\xi^{t})-u_{xx}\mathcal{D}_{x}(\xi^{x}),\ \nonumber\\
\phi_\alpha^0&=\mathcal{D}_t^\alpha(\phi)+\xi^{x}
\mathcal{D}_t^\alpha(u_x)-\mathcal{D}_t^\alpha(\xi^{x} u_x)
\na
+\mathcal{D}_t^\alpha(\mathcal{D}_t(\xi^{t})u)-\mathcal{D}_t^{\alpha+1}(\xi^{t}
u)+\xi^{t} \mathcal{D}_t^{\alpha+1}(u).
\end{align*}
The condition of invariance
\begin{equation}\label{f}
 \xi^{t}(x,t,u)|_{t=0}=0,\nonumber
\end{equation}
%%%%%%%%%%%%%%%%%%%%%
is inevitable for the (\ref{d}), due to the (\ref{2}).\\
The $\alpha^{th}$ extended infinitesimal is presented as:
\begin{align}\label{e}
\phi_\alpha^0 & =\mathcal{D}_t^\alpha(\phi)
+\xi^{x}\mathcal{D}_t^\alpha(u_x) 
-\mathcal{D}_t^\alpha(\xi^{x} u_x)
+\mathcal{D}_t^\alpha(\mathcal{D}_t(\xi^{t})u)
\na
-\mathcal{D}_t^{\alpha+1}(\xi^{t}u)
+\xi^{t} \mathcal{D}_t^{\alpha+1}(u),
 \end{align}
where $\mathcal{D}_{t}^{\alpha}$ exhibits the total fractional derivative
operator.  The fractional generalized Leibnitz rule is expressed
as
\begin{align}\label{j}
\mathcal{D}_t^\alpha\big[u(t)v(t)\big] {=} \sum_{n=0}^\infty\left(
 \begin{array}{c}
\hspace{-3pt} \alpha \hspace{-3pt}\\
\hspace{-3pt} n \hspace{-3pt}\\
\end{array}
 \right)\mathcal{D}_t^{\alpha-n}u(t)\mathcal{D}_t^nv(t),\
 \ \alpha>0,
 \end{align}
here
\begin{equation}\label{h}
\left(
  \begin{array}{c}
\hspace{-3pt}    \alpha \hspace{-3pt}\\
\hspace{-3pt}    n \hspace{-3pt}\\
  \end{array}
\right)=\frac{(-1)^{n-1}\alpha\Gamma(n-\alpha)}{\Gamma(1-\alpha)\Gamma(n+1)}.\nonumber
 \end{equation}
Therefore using (\ref{j}) one can represent (\ref{e}) as
\begin {align}\label{k}
\phi_\alpha^0 & =\mathcal{D}_t^\alpha(\phi)-\alpha
\mathcal{D}_t(\xi^{t})\frac{\partial ^\alpha u}{\partial
t^\alpha}-\sum_{n=1}^\infty \left(
\begin{array}{c}
\hspace{-3pt} \alpha \hspace{-3pt}\\
\hspace{-3pt}  n \hspace{-3pt}\\
 \end{array}
 \right)\mathcal{D}_t^n(\xi^{x})\mathcal{D}_t^{\alpha-n}(u_x)
 \nonumber\\
 &-\sum_{n=1}^\infty \left(
 \begin{array}{c}
\hspace{-3pt} \alpha \hspace{-3pt}\\
\hspace{-3pt} n+1 \hspace{-3pt}\\
 \end{array}
 \right)\mathcal{D}_t^{n+1}(\xi^{t})\mathcal{D}_t^{\alpha-n}(u).
 \end{align}
Using chain rule
\begin{align*} \label{g}
\frac{d^m f(g(t))}{dt^m} & =\sum_{k=0}^m\sum_{r=0}^k\left(
  \begin{array}{c}
 k \\
 r \\
\end{array}
 \right)
 \nt
 \frac{1}{k!}[-g(t)]^r\frac{d^m}{dt^m}[g(t)^{k-r}]\frac{d^k f(g)}{dg^k},\nonumber
 \end{align*}
and setting $f(t)=1,$ one can get
\begin{align*}
\mathcal{D}_t^\alpha(\phi) & =\frac{\partial^\alpha \phi}{\partial
t^\alpha}+\phi_u\frac{\partial^\alpha u}{\partial
t^\alpha}-u\frac{\partial^\alpha \phi_u}{\partial
t^\alpha}
\na
+\sum_{n=1}^\infty \left(
 \begin{array}{c}
\hspace{-3pt} \alpha \hspace{-3pt}\\
\hspace{-3pt} n \hspace{-3pt}\\
 \end{array}
\right)\frac{\partial ^n\phi_u}{\partial
 t^n}\mathcal{D}_t^{\alpha-n}(u)+\vartheta,\nonumber
 \end{align*}
%%%%%%%%%%%%%%%%%
where
\begin{align*}
\vartheta & =\sum_{n=2}^\infty\sum_{m=2}^n\sum_{k=2}^m\sum_{r=0}^{k-1}\left(
 \begin{array}{c}
\hspace{-3pt} \alpha \hspace{-3pt}\\
\hspace{-3pt} n \hspace{-3pt}\\
\end{array}
 \right)\left(
 \begin{array}{c}
\hspace{-3pt}  n \hspace{-3pt}\\
\hspace{-3pt} m \hspace{-3pt}\\
 \end{array}
 \right)\left(
 \begin{array}{c}
\hspace{-3pt} k \hspace{-3pt}\\
\hspace{-3pt}  r \hspace{-3pt}\\
 \end{array}
 \right)\frac{1}{k!}
\nt
\frac{t^{n-\alpha}}{\Gamma(n+1-\alpha)}[-u]^r\frac{\partial^m}{\partial t^m}
 [u^{k-r}]\frac{\partial^{n-m+k}\phi}{\partial t^{n-m}\partial u^k}.\nonumber
 \end{align*}
Therefore
\begin {align*}
 \phi_\alpha^0 & =\frac{\partial^\alpha\phi}{\partial
 t^\alpha}+\left(\phi_u-\alpha \mathcal{D}_t(\xi^{t})\right)\frac{\partial^\alpha u}{\partial
 t^\alpha}-u \frac{\partial^\alpha\phi_u}{\partial t^\alpha}+\vartheta
 \\ &
+\sum_{n=1}^\infty \big[\left(
 \begin{array}{c}
 \alpha \\
 n \\
\end{array}
\right)\frac{\partial^\alpha\phi_u}{\partial
  t^\alpha}-\left(
  \begin{array}{c}
 \alpha \\
 n+1 \\
 \end{array}
 \right)\mathcal{D}_t^{n+1}(\xi^{t})
 \big]\mathcal{D}_t^{\alpha-n}(u)
 \\  & 
 -\sum_{n=1}^\infty\left(
  \begin{array}{c}
\alpha \\
 n \\
 \end{array}
\right)\mathcal{D}_t^n(\xi^{x})\mathcal{D}_t^{\alpha-n}(u_x).
 \end{align*}




\section{Symmetry representation of TFKPP equation}


In view of the Lie theory, we have:
\begin{equation}\label{cond}
\phi_\alpha^0=\phi^{xx}+\lambda\phi+2\mu\phi u+3\gamma\phi u^2.
\end{equation}
Substituting  (\ref{k}) into (\ref{cond}), the determining
equations for Eq.~(\ref{1}) is attained,consequently, we have
\begin{align*}
\xi^{t} & =4tc_3,\qquad  \xi^{x}=c_1+2\alpha xc_3,
\nb
\phi & =c_2u+(3\alpha-2)uc_3+\mathcal{C}(x,t),
\end{align*}
where $c_1,c_2$ and $c_3$ are constants and $\mathcal{C}(x,t)$ is
a solution of Eq.~(\ref{1}). Therefore, the algebra $\mathrm{g}$
of Eq.~(\ref{1}) can be written as
\begin{align*}
V_1& =\frac{\partial}{\partial x},\ V_2=u\frac{\partial}{\partial u},\ \
\nb
 V_3 & =4t\frac{\partial}{\partial t}+2\alpha
x\frac{\partial}{\partial x}+(3\alpha-2) u\frac{\partial}{\partial u},\ 
\nb
V_4 & =\mathcal{C}(x,t)\frac{\partial}{\partial u}.\nonumber
\end{align*}
For $V_3$, one can write
\begin{equation}
\frac{dt}{4t}=\frac{dx}{2\alpha x}=\frac{du}{(3\alpha-2) u},\nonumber
\end{equation}
and this give
\begin{equation}\label{similarities}
\zeta=xt^{\frac{-\alpha}{2}},\ \
u(x,t)=t^{\frac{3\alpha-2}{4}}\mathcal{F}(\zeta).
\end{equation}




\textbf{Theorem 3.1.}  \textit{The transformation} (\ref{similarities}) \textit{reduces} (\ref{1}) \textit{to the
following:}
\begin{equation}\label{the1}
\left(\mathcal{P}_{\frac{2}{\alpha}}^{-\frac{\alpha}{4}+\frac{1}{2},\alpha}\mathcal{F}\right)(\zeta)=\mathcal{F}''+\lambda\mathcal{F}+\mu\mathcal{F}^2+\gamma\mathcal{F}^3,
\end{equation}
\textit{with the Erd\'{e}lyi-Kober (EK) fractional differential operator}
$\mathcal{P}_\beta^{\tau,\alpha}$ \textit{defined by}
\begin {align}
\left(\mathcal{P}_\beta^{\tau,\alpha}\mathcal{F}\right):
 & =\prod_{j=0}^{n-1}
\left(\tau+j-\frac{1}{\beta}\zeta\frac{d}{d\zeta}\right)
\left(\mathcal{K}_\beta^{\tau+\alpha,n-\alpha}\mathcal{F}\right)(\zeta),
\nb
\ n & =\left\{
      \begin{array}{c}
        [\alpha]+1,\ \ \alpha\notin \mathbb{N} \\
        \alpha,\ \ \ \ \alpha\in \mathbb{N} \\
      \end{array}
    \right.\nonumber
\end{align}
\textit{where}
\begin{equation}
\left(\mathcal{K}_\beta^{\tau,\alpha}\mathcal{F}\right)(\zeta):=\left\{
 \begin{array}{c}
\hspace{-5pt} \frac{1}{\Gamma(\alpha)}\int\limits_1^\infty (u-1)^{\alpha-1}u^{-(\tau+\alpha)}
\mathcal{F}(\zeta u^{\frac{1}{\beta}})du, \\
 \hspace{-3pt} \mathcal{F}(\zeta),\ \ \alpha=0, \\
 \end{array}
\right.\nonumber
\end{equation}
\textit{is the EK fractional integral operator.}


{\bf Proof:} \textit{Let} $n-1<\alpha<n,\ \ n=1,2,3,...$. \textit{By means of
Reimann-Liouville, one reaches}
\begin{align}\label{p1}
\frac{\partial^\alpha u}{\partial
t^\alpha} & =\frac{\partial^n}{\partial
t^n}
\bigg[\frac{1}{\Gamma(n-\alpha)}
\nt
\int\limits_0^t(t-s)^{n-\alpha-1}s^{\frac{3\alpha-2}{4}}\mathcal{F}\left(xs^\frac{-\alpha}{2}\right)ds\bigg].
\end{align}
Letting $\varrho= {t}/ {s}$, one can get
$ds=- ({t}/{\varrho^2})d\varrho$, therefore (\ref{p1}) can be
written as
\begin{equation}\label{p2}
\frac{\partial^\alpha u}{\partial
t^\alpha}=\frac{\partial^n}{\partial
t^n}\bigg[t^{n-\frac{\alpha}{4}-\frac{1}{2}}\left(\mathcal{K}_{\frac{2}{\alpha}}^{\frac{3\alpha+2}{4},n-\alpha}
\mathcal{F}\right)(\zeta)\bigg].\nonumber
\end{equation}
Taking into account the relation $(\zeta=xt^{ {-\alpha}/{2}})$,
we can obtain
\begin{equation}\label{p3}
t\frac{\partial}{\partial
t}\phi(\zeta)=t\frac{\partial\zeta}{\partial t}\frac{d
\phi(\zeta)}{d\zeta}=-\frac{\alpha}{2}\zeta\frac{d
\phi(\zeta)}{d\zeta}.\nonumber
\end{equation}
Therefore one can get
\begin {align}\label{p4}
\frac{\partial^n}{\partial t^n} & 
\bigg[t^{n-\frac{\alpha}{4}-\frac{1}{2}}
\left(\mathcal{K}_{\frac{2}{\alpha}}^{\frac{3\alpha+2}{4},n-\alpha} \mathcal{F}\right)(\zeta)\bigg]
\nonumber\\&
=\frac{\partial^{n-1}}{\partial
t^{n-1}}
\bigg[\frac{\partial}{\partial t}\left(t^{n-\frac{\alpha}{4}-\frac{1}{2}}
\left(\mathcal{K}_{\frac{2}{\alpha}}^{\frac{3\alpha+2}{4},n-\alpha}
\mathcal{F}\right)(\zeta)\right)\bigg]
\nonumber\\&
=\frac{\partial^{n-1}}{\partial
t^{n-1}}\bigg[t^{n-\frac{\alpha}{4}-\frac{3}{2}}\left(n-\frac{\alpha}{4}-\frac{1}{2}
-\frac{\alpha}{2}\zeta\frac{d}{d\zeta}\right)
\nt
\left(\mathcal{K}_{\frac{2}{\alpha}}^{\frac{3\alpha+2}{4},n-\alpha}
\mathcal{F}\right)(\zeta)\bigg]
\nonumber\\&
= \ldots =t^{-\frac{\alpha}{4}-\frac{1}{2}}\prod_{j=0}^{n-1}\left(-\frac{\alpha}{4}+\frac{1}{2}+j
-\frac{\alpha}{2}\zeta\frac{d}{d\zeta}\right)
\nt
\left(\mathcal{K}_{\frac{2}{\alpha}}^{\frac{3\alpha+2}{4},n-\alpha}
\mathcal{F}\right)(\zeta)
\nonumber\\& 
=t^{-\frac{\alpha}{4}-\frac{1}{2}}\left(\mathcal{P}_{\frac{2}{\alpha}}^{-\frac{\alpha}{4}+\frac{1}{2},\alpha}
\mathcal{F}\right)(\zeta).\nonumber
\end{align}
This completes the proof.



Also, for the symmetry of $V_1+V_2+V_3$, one can write
\begin{equation}
\frac{dt}{4t}=\frac{dx}{2\alpha x+1}=\frac{du}{(3\alpha-1) u},\nonumber
\end{equation}
which yields
\begin{equation}\label{similarities2}
\zeta=\frac{2\alpha x+1}{2\alpha}t^{\frac{-\alpha}{2}},\ \
u(x,t)=t^{\frac{3\alpha-1}{4}}\mathcal{F}(\zeta).
\end{equation}





\textbf{Theorem 3.2.}
\textit{The transformation} (\ref{similarities2}) \textit{reduces} (\ref{1}) \textit{to the
following nonlinear ordinary differential equation of fractional
order:}
\begin{equation}\label{the2}
\left(\mathcal{P}_{\frac{2}{\alpha}}^{\frac{3-\alpha}{4},\alpha}\mathcal{F}\right)(\zeta)=\mathcal{F}''+\lambda\mathcal{F}+\mu\mathcal{F}^2+\gamma\mathcal{F}^3.
\end{equation}

{\bf Proof:} Similar to the proof of previous theorem.




\section{Exact Solutions of TFKPP equation}

Symmetry analysis of differential equations gives many information about geometric properties of various differential equations. For example, it is possible to extract vector fields, infinitesimals, conservation laws and reductions of differential equations.   Reduction procedure of differential equations allows us  to reduce dimension of these equations by one less. In two dimensional partial differential equations (PDEs), reduction procedure gives an ordinary differential equation (ODE). So, solving this ODE concludes exact solution of original PDE. However, in FPDEs with Riemann-Liouville fractional derivatives we get ODEs with the EK derivatives which there is not a systematic method to find their exact solution. Therefore, after reduction of TFKPP equation with the Riemann-Liouville fractional derivative we obtain Eqs.~(\ref{the1}) and (\ref{the2}) which it is not possible to find analytical solutions. However, we can obtain exact solution of Eq.~(\ref{1}) with   $\partial_t^\alpha u:=\ _t\mathbb{T}_\alpha(u)$. In this section, we investigate the exact solutions of TFKPP equation with conformable fractional derivative.  




\subsection{Simplest equation method and its applications to time fractional differential equations}


This approach was proposed in [27,28]. The
steps for the approach is stated as follows:

Let the TFDE is given by
\begin{equation}\label{TFDE}
P\left(u,\ _t\mathbb{T}_\alpha(u),u_x,u_{xx},...\right)=0.\ \
0<\alpha\leq 1,
\end{equation}
Then the modified version of simplest equation method procedure
have
the following steps:

{\bf Step 1:} We utilize the following
\begin{equation}\label{travel}
u(x,t)=\Theta(\xi),\ \
\xi=A\left(x-\nu\frac{t^\alpha}{\alpha}\right),
\end{equation}
where $A$ and $\nu$ are nonzero constants to be determined later.

Consequently we attain with parameters $A$ and $\nu$ the following
\begin{equation}\label{step1ode}
P\left(\Theta,-A\nu\Theta',A\Theta',A^2\Theta'',...\right)=0.
\end{equation}

{\bf Step 2:} Suppose that Eq.~(\ref{step1ode}) possesses
\begin{equation}\label{theta1}
\Theta(\xi)=\sum_{i=0}^Na_i[z(\xi)]^i,
\end{equation}
where $a_i,\ i=0,1,...,N$, are constants to be determined later.
The positive value of $N$ in (\ref{theta1}), which the pole order
for the general solution of Eq.~(\ref{step1ode}), can be
determined by substituting $\Theta(\xi)=\xi^{-m},\ (m>0)$.

In the present paper,  we use the Bernoulli and Riccati equations
which their solutions can be expressed by elementary functions.
For the Bernoulli equation:$^i$
\begin{equation}\label{Bernoulli}
\frac{dz}{d\xi}=az(\xi)+b[z(\xi)]^k,\
k\in\mathbb{N}\backslash\{1\},\nonumber
\end{equation}
we use the solutions
\begin{equation}\label{B1}
z(\xi)=\sqrt[k-1]{\frac{a\exp\big[a(k-1)(\xi+\xi_0)\big]}{1-b\exp\big[a(k-1)(\xi+\xi_0)\big]}},\nonumber
\end{equation}
for the case $a>0,\ b<0$ and
\begin{equation}\label{B2}
z(\xi)=\sqrt[k-1]{-\frac{a\exp\big[a(k-1)(\xi+\xi_0)\big]}{1+b\exp\big[a(k-1)(\xi+\xi_0)\big]}},\nonumber
\end{equation}
for the case $a<0,\ b>0$ and $\xi_0$ is a constant of integration.
For the Riccati equation
\begin{equation}\label{Riccati}
\frac{dz}{d\xi}=a+b[z(\xi)]^2,\nonumber
\end{equation}
which admits the following exact solutions:
\begin{equation}\label{R1}
z(\xi)=-\frac{\sqrt{-ab}}{b}\tanh\bigg[\sqrt{-ab}\xi-\frac{\epsilon\ln(\xi_0)}{2}\bigg],\
\xi_0>0,\ \epsilon=\pm 1,\nonumber
\end{equation}
when $ab<0$ and
\begin{equation}\label{R2}
z(\xi)=\frac{\sqrt{ab}}{b}\tan\big[\sqrt{ab}\xi+\xi_0\big],\
\xi_0=Const.,\nonumber
\end{equation}
when $ab>0$.

{\bf Step 3:} Plugging (\ref{theta1}) into (\ref{step1ode}) and
equating the coefficients of $z^i$ to zero, one can obtain an
algebraic system in $A,\ \nu$
and $a_i,\ i=0,...,N$.

\subsection{Application to the TFKPP equation}

The transformation
\begin{equation}\label{similarity}
u(x,t)=\Theta(\xi),\ \ \xi=A\left(x-\nu\frac{
t^\alpha}{\alpha}\right),
\end{equation}
changes Eq.~(\ref{1}) with   $\partial_t^\alpha u=\ _t\mathbb{T}_\alpha(u)$ to:
\begin{equation}\label{ode1}
A^2\Theta''+\nu
A\Theta'+\lambda\Theta+\mu\Theta^2+\gamma\Theta^3=0.
\end{equation}

We suppose that Eq.~(\ref{ode1}) has solution of the form
(\ref{theta1}). Balancing the highest order derivative terms with
nonlinear terms in Eq.~(\ref{ode1}), we get $N=1$, and hence
\begin{equation}\label{n1sol}
\Theta(\xi)=a_0+a_1z(\xi),\ a_1\neq 0.
\end{equation}
Substituting (\ref{n1sol}) along with (\ref{Bernoulli}) into
Eq.~(\ref{ode1}) and then vanishing the coefficients of $z^i$, one
can get some algebraic equations about $a_0,a_1,A$ and $\nu$,
which solving them by \textit{Maple}, concludes:\\


$\bullet$ Case 1:\\
\begin{align*}
a_0&=0,\ a_1=\frac{b\left(\mu^2\mp|\mu|\psi\right)}{2a\mu\gamma},
\\
A&=\frac{\sqrt{-\mu^2+2\lambda\gamma\pm|\mu|\psi}}{2a\sqrt{\gamma}},
\\
\nu & =\frac{\mu^2-6\lambda\gamma\mp|\mu|\psi}{2\sqrt{\gamma\left(-\mu^2+2\lambda\gamma\pm|\mu|\psi\right)}},
\end{align*}
where $\psi=\sqrt{\mu^2-4\lambda\gamma}$. In this case, the exact
solutions of Eq.~(\ref{ode1}) are:



\begin {align}
\Theta(\xi)&=\frac{b(-\mu^2\pm|\mu|\psi)\exp\big[a(\xi+\xi_0)\big]}{2\gamma\mu\left(-1+
b\exp\big[a(\xi+\xi_0)\big]\right)},\
\ a>0,\ b<0,
\nonumber\\
\Theta(\xi)&=\frac{b(-\mu^2\pm|\mu|\psi)\exp\big[a(\xi+\xi_0)\big]}{2\gamma\mu\left(1+
b\exp\big[a(\xi+\xi_0)\big]\right)},\ \ a<0,\ b>0,\nonumber
\end{align}
and using the substitution in (\ref{travel}) we get the final
solutions:
\begin{align*}
 u(x,t) & =\frac{b(-\mu^2\pm|\mu|\psi\exp\big[\eta(x,t)\big])}{2\gamma\mu\left(-1+b\exp\big[\eta(x,t)\big]\right)},\
\ a>0,\ b<0,\nonumber\\
 u(x,t) & =\frac{b(-\mu^2\pm|\mu|\psi)\exp\big[\eta(x,t)\big]}{2\gamma\mu\left(1+
b\exp\big[\eta(x,t)\big]\right)},\ \ a<0,\ b>0, 
\end{align*}
\ru \columnbreak \noindent
where
\begin {align*}
\eta(x,t) &=\frac{2x\sqrt{\gamma(-\mu^2+2\gamma\lambda\pm|\mu|\psi)}}{4\gamma}
\\
 &+\frac{(-\mu^2+6\gamma\lambda\pm|\mu|\psi)t^\alpha+4\xi_0\gamma
a\alpha}{4\gamma\alpha}.
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$\bullet$ Case 2:\\
\begin{align*}
a_0 &=\frac{\psi-\mu}{2\gamma},\
a_1=\frac{b(3\mu^2-12\lambda\gamma-\mu\psi)}{\gamma
a(3\psi-\mu)},
\\
 A & =\frac{\sqrt{2}\psi}{2a\sqrt{-\gamma}},\
 \nu=\frac{\sqrt{2}\mu}{2\sqrt{-\gamma}}.
\end{align*}
\end{multicols}

\noindent
 In this case, the exact
solutions of Eq.~(\ref{ode1}) are:
\begin {align*}
\Theta(\xi)&=\frac{-2\mu^2+2\mu\psi+6\lambda\gamma+b\left(6\lambda\gamma-\mu^2-\mu\psi\right)\exp\big[a(\xi+\xi_0)\big]}{\gamma(3\psi-\mu)\left(-1+
b\exp\big[a(\xi+\xi_0)\big]\right)}, \qquad   a>0, b<0,
\nonumber\\
\Theta(\xi)& =\frac{2\mu^2+2\mu\psi-6\lambda\gamma+b\left(6\lambda\gamma-\mu^2-\mu\psi\right)\exp\big[a(\xi+\xi_0)\big]}{\gamma(3\psi-\mu)\left(1+ b\exp\big[a(\xi+\xi_0)\big]\right)}, \qquad   a<0, b>0, 
\end{align*}
or equivalently
\begin {align*}
 u(x,t) & =\frac{-2\mu^2+2\mu\psi+6\lambda\gamma+b\left(6\lambda\gamma-\mu^2-\mu\psi\right)\exp\big[\eta(x,t)\big]}{\gamma(3\psi-\mu)\left(-1+
b\exp\big[\eta(x,t)\big]\right)},\qquad
\ a>0,\ b<0,
\nonumber\\
 u(x,t) & =\frac{2\mu^2+2\mu\psi-6\lambda\gamma+b\left(6\lambda\gamma-\mu^2-\mu\psi\right)\exp\big[\eta(x,t)\big]}{\gamma(3\psi-\mu)\left(1+ b\exp\big[\eta(x,t)\big]\right)},\qquad a<0,\ b>0, 
\end{align*}
where
\begin{equation}
\eta(x,t)=\xi_0a+\frac{x\psi\sqrt{-2\gamma}}{2\gamma}-\frac{\mu\psi
t^\alpha}{2\gamma\alpha}.\nonumber
\end{equation}


\begin{multicols}{2}

$\bullet$ Case 3:\\
\begin {align}
a_0& =\frac{\psi-\mu}{2\gamma},\quad
a_1=\frac{2b(\mu^2-3\lambda\gamma-\mu\psi)}{\gamma a(3\psi-\mu)},
\nonumber\\
 A& =\frac{\sqrt{2\lambda\gamma-\mu^2+\mu\psi}}{2a\sqrt{\gamma}},\quad
 \nu=\frac{6\lambda\gamma-\mu^2+\mu\psi}{2\sqrt{\gamma\left(2\lambda\gamma-\mu^2+\mu\psi\right)}}.\nonumber
\end{align}
In this case, we can obtain
\begin {align}
\Theta(\xi)&=\frac{2\left(-\mu^2+3\lambda\gamma
+\mu\psi\right)}{\gamma(3\psi-\mu)\left(-1+
b\exp\big[a(\xi+\xi_0)\big]\right)},\
\na
\ a>0,\ b<0,
\nonumber\\
\Theta(\xi)&=\frac{2\left(\mu^2-3\lambda\gamma-\mu\psi\right)}{\gamma(3\psi-\mu)\left(1+
b\exp\big[a(\xi+\xi_0)\big]\right)},\ 
\na 
a<0,\ b>0,  \nonumber
\end{align}
\columnbreak \rd \noindent
and using the substitution in (\ref{travel}) we have
\begin {align}
u(x,t)&=\frac{2\left(-\mu^2+3\lambda\gamma+\mu\psi\right)}{\gamma(3\psi-\mu)\left(-1+
b\exp\big[\eta(x,t)\big]\right)},\
\na 
\ a>0,\ b<0,
\nonumber\\
u(x,t)&=\frac{2\left(\mu^2-3\lambda\gamma
-\mu\psi\right)}{\gamma(3\psi-\mu)\left(1+
b\exp\big[\eta(x,t)\big]\right)},
\na 
a<0,\ b>0,
\end{align}
where
\begin{align}
\eta(x,t) & =\xi_0
a+\frac{x\sqrt{-\mu^2+\mu\psi+2\gamma\lambda}}{2\sqrt{\gamma}}
\na
-\frac{-\mu^2+\mu\psi+6\gamma\lambda}{4\gamma\alpha}t^\alpha.\nonumber
\end{align}



Also, in the use of Riccati equation, substituting (\ref{n1sol})
along with (\ref{Riccati}) into Eq.~(\ref{ode1}) and then
vanishing the coefficients of $z^i$, we  can obtain  some
algebraic equations
about $a_0,a_1,A$ and $\nu$, that  solving them by Computer algebra technique , concludes:\\



$\bullet$ Case 1:\\
\begin{align*}
a_0 & =-\frac{\mu}{2\gamma},\quad
a_1=\pm\frac{ib\psi}{2\gamma\sqrt{ab}},
\\
A & =\pm\frac{\sqrt{2}\psi}{4\sqrt{\gamma ab}},\quad
\nu=\frac{i\sqrt{2}\mu}{2\sqrt{\gamma}}.
\end{align*}
In this case, the exact solutions of Eq.~(\ref{ode1}) are:
\begin{align*}
\Theta(\xi)& =\frac{-\mu\pm
i\psi\tan\big[\sqrt{ab}\xi+\xi_0\big]}{2\gamma},\
\na   ab>0,\nonumber\\
\Theta(\xi) & =\frac{-\mu\pm
\psi\tanh\big[\sqrt{-ab}\xi-\frac{\epsilon\ln(\xi_0)}{2}\big]}{2\gamma},
\na
 ab<0,\nonumber
\end{align*}
and using the substitution in (\ref{travel}) we get the following
final solutions:
\begin{align*}
u(x,t) & =-\frac{1}{2\gamma}
\bigg(\mu
\na
+ i\psi\tan\bigg[\frac{-\sqrt{2\gamma}\psi
x\alpha\mp4\gamma\xi_0\alpha+i\mu\psi
t^\alpha}{4\gamma\alpha}\bigg]\bigg),\nonumber
\end{align*}
when $ab>0$ and
\begin{align*}
u(x,t) & =-\frac{1}{2\gamma}\bigg(\mu
\na
+
\psi\tanh\bigg[\frac{-i\sqrt{2\gamma}\psi
x\alpha\pm2\gamma\epsilon\ln(\xi_0)\alpha-\mu\psi
t^\alpha}{4\gamma\alpha}\bigg]\bigg),\nonumber
\end{align*}
when $ab<0$.



$\bullet$ Case 2:\\
\begin{align*}
a_0 & =-\frac{\mu}{2\gamma},\
a_1=\mp\frac{ib\psi}{2\gamma\sqrt{ab}},
\nb
A & =\pm\frac{\sqrt{2}\psi}{4\sqrt{\gamma ab}},\
\nu=-\frac{i\sqrt{2}\mu}{2\sqrt{\gamma}},\nonumber
\end{align*}
Exact solutions of Eq.~(\ref{ode1}) extracted from this case are:
\begin{align*}
\Theta(\xi) & =\frac{-\mu\mp
i\psi\tan\big[\sqrt{ab}\xi+\xi_0\big]}{2\gamma},\
\ ab>0,
 \\
\Theta(\xi) & =\frac{-\mu\mp
\psi\tanh\big[\sqrt{-ab}\xi-\frac{\epsilon\ln(\xi_0)}{2}\big]}{2\gamma},\
\ ab<0, 
\end{align*}
or equivalently
\begin{align*}
u(x,t) & =-\frac{1}{2\gamma}\bigg(\mu
\na
+i\psi\tan\bigg[\frac{\sqrt{2\gamma}\psi
x\alpha\pm4\gamma\xi_0\alpha+i\mu\psi
t^\alpha}{4\gamma\alpha}\bigg]\bigg), 
\end{align*}
when $ab>0$ and
\begin{align*}
u(x,t) & =-\frac{1}{2\gamma}\bigg(\mu
\na
+\psi\tanh\bigg[\frac{i\sqrt{2\gamma}\psi
x\alpha\mp2\gamma\epsilon\ln(\xi_0)\alpha-\mu\psi
t^\alpha}{4\gamma\alpha}\bigg]\bigg), 
\end{align*}
when $ab<0$.


\section{Conclusion}

In this study, the Lie group analysis method was successfully
applied to investigate the reduction and symmetry properties of
the TFKPP equation. Moreover,  we have arrived to some exact
solutions of the conformable  TFKPP equation, thanks to the
application of simplest equation method. The results of this study
undoubtedly offer helpful information about the TFKPP equation.

\section*{Acknowledgments}

The authors would like to thank the referees for the helpful
suggestions.


\end{multicols}
\medline
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\end{thebibliography}
\end{multicols}
\end{document}
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