\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 243, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/243\hfil Higher order PDE with Caputo-Fabrizio derivative]
{Higher order multi-term time-fractional partial differential
equations involving Caputo-Fabrizio derivative}

\author[E. Karimov, S. Pirnafasov \hfil EJDE-2017/243\hfilneg]
{Erkinjon Karimov, Sardor Pirnafasov}

\address{Erkinjon Karimov \newline
Institute of Mathematics named after V. I. Romanovsky,
Uzbekistan Academy of Sciences, Tashkent 100125, Uzbekistan}
\email{erkinjon@gmail.com}

\address{Sardor Pirnafasov \newline
Institute of Mathematics named after V. I. Romanovsky,
Uzbekistan Academy of Sciences, Tashkent 100125, Uzbekistan}
\email{sardor.pirnafasov@gmail.com}

\dedicatory{Communicated by Mokhtar Kirane}

\thanks{Submitted August 10, 2017. Published October 5, 2017.}
\subjclass[2010]{33E12}
\keywords{Fractional differential equations; separation of variables;
\hfill\break\indent  Caputo-Fabrizio derivative}

\begin{abstract}
 In this work we discuss higher order multi-term partial differential
 equation (PDE) with the Caputo-Fabrizio fractional derivative in time.
 Using method of separation of variables, we reduce fractional order
 partial differential equation to the integer order. We represent
 explicit solution of formulated problem in particular case by Fourier series.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction}

Consideration of new fractional derivative with non-singular kernel was
initiated by Caputo and Fabrizio in their work \cite{CF1}.
Motivation came from application. Precisely,  new fractional derivatives
can better describe material heterogeneities and structures with different
scales. Special role of spatial fractional derivative in the study of
the macroscopic behaviors of some materials, related with nonlocal interactions,
which are prevalent in determining the properties of the material was also
highlighted. In their next work \cite{CF2}, authors represented some
applications of the introduced fractional derivative. Nieto and Losada \cite{NL}
studied some properties of this fractional derivative naming it as
Caputo-Fabrizio (CF) derivative. They introduced fractional integral associated
 with the CF derivative, applying it to the solution of linear and nonlinear
differential equations involving CF derivative.

Later, many authors showed interest in the CF derivative and as a result,
several applications  were discovered. For instance, in groundwater modeling,
in electrical circuits, in controlling the wave movement, in nonlinear
Fisher's reaction-diffusion equation, in modeling of a mass-spring-damper
system, etc. We note also some recent works related with CF derivative
 \cite{GA1,GA2,GY,ASh,GA3,H}.

Different methods were applied for solving differential equations involving
CF derivative. Namely, Laplace transform, reduction to integral equations,
and reduction to  integer order differential equations.
Last two methods were used in the works \cite{K1,K2}.

In this paper, we aim to show an algorithm to reduce initial value problem (IVP)
for multi-term fractional DE with CF derivative to the IVP for integer order
 DE and using this result to prove a unique solvability of a boundary-value
problem (BVP) for PDE involving CF derivative on time-variable. First,
we give preliminary information on CF derivative and then we formulate
our main problem. Representing formal solution of the formulated problem
by infinite series, in particular case, we prove uniform convergence
of that infinite series.

\section{Preliminaries}

The fractional derivative of order $\alpha\, (0<\alpha<1)$ in CF sense \cite{CF2}
is defined as
\begin{equation}
{}_{CF}D_{at}^{\alpha}g(t)=\frac{1}{1-\alpha}
\int_a^t g'(s)e^{-\frac{\alpha}{1-\alpha}(t-s)}ds\,.
\label{eqn2.1}
\end{equation}
This operator is well defined on the space
\[
W^{\alpha,1}=\big\{g(t)\in L^1(a,\infty):
(g(t)-g_a(s))e^{-\frac{\alpha}{1-\alpha}(t-s)}\in L^1(a,t)\times L^1(a,\infty)\big\},
\]
whose norm, for $\alpha\neq 1$ is given by
\[
\|g(t)\|_{W^{\alpha,1}}
=\int_a^\infty |g(t)dt+\frac{\alpha}{1-\alpha}\int_a^\infty\int_{-\infty}^t
|g_s(s)|e^{-\frac{\alpha}{1-\alpha}(t-s)}ds dt,
\]
where $g_a(t)=g(t)$, $t\geq a$, $g_a(t)=0$, $-\infty<t<a$ \cite{CF2}.
The following equality is shown in \cite{GA1},
\[
{}_{CF}D_{at}^{\alpha+n}g(t)={}_{CF}D_{at}^\alpha\left({}_{CF}D_{at}^n g(t)\right)\,.
\]

\section{Formulation of a problem and formal solution}

Consider the time-fractional PDE
\begin{equation}
\sum_{n=0}^{k} \lambda_n \,\cdot\,{}_{CF}D_{0t}^{\alpha+n}u(t,x)-u_{xx}(t,x)=f(t,x)
 \label{eqn3.1}
\end{equation}
in a domain $\Omega=\{(t,x):\, 0<t<q,\,0<x<1\}$.
Here $\lambda_n$ are given real numbers, $f(t,x)$ is a given function,
$k\in \mathbf{N}_0$, $q\in \mathbb{R}^+$.

\textbf{Problem.} Find a solution of \eqref{eqn3.1} satisfying the following
conditions:
\begin{gather}
u(t,x)\in C^2(\Omega),\quad
\frac{\partial^n u(t,x)}{\partial t^n}\in W^{\alpha,1}(0,q), \label{eqn3.2} \\
u(t,0)=u(t,1)=0,\quad \frac{\partial^i u(t,x)}{\partial t^i}\big|_{t=0}=\tilde{C}_i,
\quad i=0,1,2,\dots ,k,\label{eqn3.3}
\end{gather}
where $\tilde{C}_i$ are any real numbers.

\subsection{Solution of higher order multi-term fractional ordinary
differential equations}
We expand $u(t,x)$ as the Fourier series
\begin{equation}
u(t,x)=\sum_{m=1}^\infty T_m(t)\sin m\pi x, \label{eqn3.4}
\end{equation}
where $T_m(t)$ are the Fourier coefficients of $u(t,x)$.

Substituting this representation in \eqref{eqn3.1} and considering the
initial conditions \eqref{eqn3.3}, we get the following IVP with respect
to time-variable:
\begin{equation}
\begin{gathered}
\sum_{n=0}^{k}\lambda_n\,\cdot\, {}_{CF}D_{0t}^{\alpha+n}T_m(t)+(m\pi)^2T_m(t)
=f_m(t),\\
T_m^{(i)}(0)=C_i,\quad i=0,1,2,\dots ,k\\
\end{gathered} \label{eqn3.5}
\end{equation}
where $f_m(t)$ are Fourier coefficients of $f(t,x)$.

Based on definition \eqref{eqn2.1} and initial conditions \eqref{eqn3.3},
after  integrating by parts, we rewrite the fractional derivatives as follows:
\begin{gather*}
{}_{CF}D_{0t}^{\alpha}T_m(t)=\frac{1}{1-\alpha}T_m(t)
 -\frac{\alpha}{(1-\alpha)^2}\int_0^t T_m(s)e^{-\frac{\alpha}{1-\alpha}(t-s)}ds
-\frac{T_m(0)}{1-\alpha}e^{-\frac{\alpha}{1-\alpha}t},\\
\begin{aligned}
{}_{CF}D_{0t}^{\alpha+1}T_m(t)
&=\frac{1}{1-\alpha}T_m'(t)-\frac{\alpha}{(1-\alpha)^2}T_m(t)+
\frac{\alpha^2}{(1-\alpha)^3}\int_0^t T_m(s)e^{-\frac{\alpha}{1-\alpha}(t-s)}ds\\
&\quad +\frac{\alpha T_m(0)}{(1-\alpha)^2}e^{-\frac{\alpha}{1-\alpha}t}
 -\frac{T_m'(0)}{1-\alpha}e^{-\frac{\alpha}{1-\alpha}t},
\end{aligned} \\
\begin{aligned}
{}_{CF}D_{0t}^{\alpha+2}T_m(t)
&=\frac{1}{1-\alpha}T_m''(t)-\frac{\alpha}{(1-\alpha)^2}T_m'(t)+
\frac{\alpha^2}{(1-\alpha)^3}T_m(t)\\
&\quad -\frac{\alpha^3}{(1-\alpha)^4}\int_0^t T_m(s)e^{-\frac{\alpha}{1-\alpha}(t-s)}ds-
\frac{\alpha^2 T_m(0)}{(1-\alpha)^3}e^{-\frac{\alpha}{1-\alpha}t} \\
&\quad +\frac{\alpha T_m'(0)}{(1-\alpha)^2}e^{-\frac{\alpha}{1-\alpha}t}
 -\frac{T_m''(0)}{1-\alpha}e^{-\frac{\alpha}{1-\alpha}t}\,.
\end{aligned}
\end{gather*}
Continuing this procedure, we  find for $n\geq 1$ the  formula
\begin{equation}
\begin{aligned}
{}_{CF}D_{0t}^{\alpha+n}T_m(t)
&=\frac{1}{1-\alpha}\Big\{\sum_{i=0}^n \big(-\frac{\alpha}{1-\alpha}\big)^i
\big[ T_m^{(n-i)}(t)-T_m^{(n-i)}(0)e^{-\frac{\alpha}{1-\alpha}t}\big] \\
&\quad +\big(-\frac{\alpha}{1-\alpha}\big)^{n+1}
 \int_0^t T_m(s)e^{-\frac{\alpha}{1-\alpha}(t-s)}ds\Big\}.
\end{aligned}\label{eqn3.6}
\end{equation}
We substitute \eqref{eqn3.6} into \eqref{eqn3.5} and deduce
\begin{align*}
&\sum_{n=0}^k\frac{\lambda_n}{(1-\alpha)}
\Big\{\sum_{i=0}^{n}\big(-\frac{\alpha}{1-\alpha}\big)^i
\big[T_m^{(n-i)}(t)-T_m^{(n-i)}(0)e^{-\frac{\alpha}{1-\alpha}t}\big] \\
& +\big(-\frac{\alpha}{1-\alpha}\big)^{n+1}
 \int_0^t T_m(s)e^{-\frac{\alpha}{1-\alpha}(t-s)}ds\Big\}+(m\pi)^2T_m(t)\\
&=f_m(t).
\end{align*}
We multiply this equality by $(1-\alpha)e^{\frac{\alpha}{1-\alpha}t}$:
\begin{equation}
\begin{aligned}
&\sum_{n=0}^k \lambda_n\Big\{\sum_{i=0}^n\big(-\frac{\alpha}{1-\alpha}\big)^i
\big[T_m^{(n-i)}(t)e^{\frac{\alpha}{1-\alpha}t}-T_m^{(n-i)}(0)\big] \\
&+\big(-\frac{\alpha}{1-\alpha}\big)^{n+1}
 \int_0^t T_m(s)e^{\frac{\alpha}{1-\alpha}s}ds\Big\}
 +(m\pi)^2(1-\alpha)T_m(t)e^{\frac{\alpha}{1-\alpha}t}\\
&=(1-\alpha)e^{\frac{\alpha}{1-\alpha}t}f_m(t).
\end{aligned} \label{eqn3.7}
\end{equation}
Introducing new function $\tilde{T}_m(t)=T_m(t)e^{\frac{\alpha}{1-\alpha}t}$,
we rewrite some items of \eqref{eqn3.7}:
\begin{equation}
\begin{gathered}
T_m'(t)e^{\frac{\alpha}{1-\alpha}t}=\tilde{T}_m'(t)
-\frac{\alpha}{1-\alpha}\tilde{T}_m(t), \\
T_m''(t)e^{\frac{\alpha}{1-\alpha}t}=\tilde{T}_m''(t)
-\frac{2\alpha}{1-\alpha}\tilde{T}_m'(t)+\big(\frac{\alpha}{1-\alpha}\big)^2
\tilde{T}_m(t), \\
\dots \\
T_m^{(n)}(t)e^{\frac{\alpha}{1-\alpha}t}=\sum_{j=0}^n (-1)^{n-j}
\frac{n!}{j!(n-j)!}\big(\frac{\alpha}{1-\alpha}\big)^{n-j}\tilde{T}_m^{(j)}(t).
\end{gathered}\label{eqn3.8}
\end{equation}
We note that $\tilde{T}_m^{(0)}(t)=\tilde{T}_m(t)$.

Considering \eqref{eqn3.8}, from \eqref{eqn3.7} we deduce
\begin{equation}
\begin{aligned}
&\sum_{n=0}^k \lambda_n\Big\{\sum_{i=0}^n\big(-\frac{\alpha}{1-\alpha}\big)^i
\sum_{j=0}^{n-i}\frac{(n-i)!}{j!(n-i-j)!}\big(-\frac{\alpha}{1-\alpha}\big)^{n-i-j}\\
&\times\big[\tilde{T}_m^{(j)}(t)-\tilde{T}_m^{(j)}(0)\big]
 +\big(-\frac{\alpha}{1-\alpha}\big)^{n+1}\int_0^t \tilde{T}_m(s)ds\Big\}
 +(m\pi)^2(1-\alpha)\tilde{T}_m(t)\\
&=(1-\alpha)e^{\frac{\alpha}{1-\alpha}t}f_m(t).
\end{aligned}\label{eqn3.9}
\end{equation}
Differentiating \eqref{eqn3.9} once by $t$, we will get $(k+1)$th order DE.
 Using its general solution and applying initial conditions, one can get
 explicit form of functions $T_m(t)$, consequently formal solution of the
formulated problem is represented by infinite series \eqref{eqn3.4}.
 Imposing certain conditions to the given functions, we prove uniform
convergence of infinite series, which will complete the proof of the unique
solvability of the formulated problem.

In the next section we show the complete steps in a particular case.
We note that even this particular case was not considered before.

\section{Particular case}

In this subsection we consider the  case $k=2$,  to show the complete steps.
 In this case, after differentiating  \eqref{eqn3.9} once with respect to $t$,
 we  get the following third order ordinary DE,
\begin{equation}
\tilde{T}_m'''(t)+A_1\tilde{T}_m''(t)+A_2\tilde{T}_m'(t)+A_3\tilde{T}_m=g_m(t)
\label{eqn4.1}
\end{equation}
where $\tilde{T}_m(t)=T_m(t)e^{\frac{\alpha}{1-\alpha}t}$,
\begin{equation}
\begin{gathered}
A_1=-\frac{3\alpha}{1-\alpha}+\frac{\lambda_1}{\lambda_2} ,\\
A_2=3\big(-\frac{\alpha}{1-\alpha}\big)^2+\frac{\lambda_0
-\frac{2\alpha\lambda_1}{1-\alpha}+(m\pi)^2(1-\alpha)}{\lambda_2}, \\
A_3=\big(-\frac{\alpha}{1-\alpha}\big)^3
+\frac{\big(-\frac{\alpha}{1-\alpha}\big)^2\lambda_1
-\frac{\alpha}{1-\alpha}\lambda_0}{\lambda_2}, \\
g_m(t)=[\alpha{f}_m(t)+(1-\alpha){f_m}'(t)]e^{\frac{\alpha}{1-\alpha}t}.
\end{gathered}\label{eqn4.2}
\end{equation}

\subsection{General solution}

The characteristic equation of \eqref{eqn4.1} is
$$
\mu^3+A_1\mu^2+A_2\mu+A_3=0,
$$
whose discriminant is
$$
\Delta_m=-4A_1^3A_3+A_1^2A_2^2-4A_2^3+18A_1A_2A_3-27A_3^2.
$$
According to the general theory, form of solutions depends on the sign
of the discriminant. Below we will give explicit forms of solutions case by case.

Case $\Delta_m>0$. The characteristic equation has 3 different real roots
($\mu_1,\mu_2,\mu_3$), hence based on general solution we can find explicit
form of $T_m(t)$ as
\begin{equation}
\begin{aligned}
T_m(t)
&=C_1 e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t} +C_2 e^{(\mu _2
-\frac{\alpha }{1-\alpha } )t} +C_3 e^{(\mu _3 -\frac{\alpha }{1-\alpha } )t}  \\
&\quad +\frac{1}{(\mu _2 \mu _3^2 -\mu _2^2 \mu _3
 -\mu _1 \mu _3^2 +\mu _1^2 \mu _3 +\mu _1 \mu _2^2
 -\mu _1^2 \mu _2 )} \\
&\quad\times \Big[e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t} (\mu _3 -\mu _2 )
\int \big(\alpha g_m(z)+(1-\alpha )g_m'(z)\big)e^{(\frac{\alpha }{1-\alpha }
 -\mu _1 )z} \,dz \\
&\quad + e^{(\mu _2 -\frac{\alpha }{1-\alpha } )t} (\mu _1 -\mu _3 )
 \int (\alpha g_m(z)+(1-\alpha )g_m'(z))e^{(\frac{\alpha }{1-\alpha }
  -\mu _2 )z} \,dz\\
&\quad +e^{\left(\mu _3 -\frac{\alpha }{1-\alpha } \right)t}
  (\mu _2 -\mu _1 )\int (\alpha g_m(z)+(1-\alpha )g_m'(z))
 e^{(\frac{\alpha }{1-\alpha } -\mu _3 )z}\,dz   \Big].
\end{aligned}\label{eqn4.3}
\end{equation}

Case $\Delta_m<0$. The characteristic equation has one real ($\mu_1$)
and two complex-conjugate roots ($\mu_2=\mu_{21}\pm i\mu_{22}$).
Therefore, $T_m(t)$ has the form
\begin{align*}
T_m(t)
&=C_1 e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t} +(C_2 \cos \mu _{22} t
 +C_3 \sin \mu _{22} t)e^{(\mu _{21} -\frac{\alpha }{1-\alpha } )t}  \\
&\quad +\frac{1}{\mu _{22}^2 -3\mu _{21}^2 -2\mu _1 \mu _{21} +\mu _1^2 }
  \Big[e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t}
 \int (\alpha g_m(z) \\
&\quad +(1-\alpha )g_m'(z))e^{(\frac{\alpha }{1-\alpha } -\mu _1 )z} \,dz
  +\frac{1}{\mu _{22} } e^{(\mu _{21} -\frac{\alpha }{1-\alpha } )t}
 \cos \mu _{22} t\int (\mu _1 \sin \mu _{22} z \\
&\quad -\mu _{21} \sin \mu _{22} z  -\mu _{22} \cos \mu _{22} z)
  \left(\alpha g_m(z)+(1-\alpha )g_m'(z)\right)
 e^{(\frac{\alpha }{1-\alpha } -\mu _{21} )z} \,dz  \\
&\quad + \frac{1}{\mu _{22} } e^{(\mu _{21} -\frac{\alpha }{1-\alpha } )t}
 \sin \mu _{22} t  \int (\mu _{21} \cos \mu _{22} z-\mu _{22} \sin \mu _{22} z \\
&\quad -\mu _1 \cos \mu _{22} z)
  \left(\alpha g_m(z)+(1-\alpha )g_m'(z)\right)e^{(\frac{\alpha }{1-\alpha }
 -\mu _{21} )z} \,dz \Big].
  \end{align*} %\label{eqn4.4}


Case $\Delta_m=0$. We have have 2 sub-cases:

(a) Three real roots, two of which are equal, third one is different
 ($\mu_1=\mu_2,\,\mu_3$):
\begin{equation}
\begin{aligned}
T_m(t)
&=C_1 e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t} +C_2 te^{(\mu _1
 -\frac{\alpha }{1-\alpha } )t} +C_3 e^{(\mu _3 -\frac{\alpha }{1-\alpha } )t}  \\
&\quad +\frac{1}{(\mu _1^2 -2\mu _1 \mu _3 +\mu _3 )} \Big[e^{(\mu _1
  -\frac{\alpha }{1-\alpha } )t} \int (\mu _3 z-1-\mu _1 z)(\alpha g_m(z)\\
&\quad +(1-\alpha )g_m'(z))    e^{(\frac{\alpha }{1-\alpha } -\mu _1 )z} \,dz \\
&\quad +te^{(\mu _1 -\frac{\alpha }{1-\alpha } )t} (\mu _1 -\mu _3 )
 \int (\alpha g_m(z)+(1-\alpha )g_m'(z))e^{(\frac{\alpha }{1-\alpha } -\mu _1 )z} 
\,dz
 \\
&\quad + e^{(\mu _3 -\frac{\alpha }{1-\alpha } )t}
 \int (\alpha g_m(z)+(1-\alpha )g_m'(z))e^{(\frac{\alpha }{1-\alpha } -\mu _3 )z}
 \,dz \Big];
 \end{aligned} \label{eqn4.5}
\end{equation}

(b) all 3 real roots are the same ($\mu_1=\mu_2=\mu_3$):
\begin{equation}
\begin{aligned}
T_m(t)
&=C_1 e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t}
 +C_2 te^{(\mu _1 -\frac{\alpha }{1-\alpha } )t}
 +C_3 t^2 e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t} \\
&\quad +\frac{1}{2} e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t}
 \int z^2 (\alpha g_m(z)+(1-\alpha )g_m'(z))e^{(\frac{\alpha }{1-\alpha }
  -\mu _1 )z} \,dz  \\
&\quad -te^{(\mu _1 -\frac{\alpha }{1-\alpha } )t}
 \int z(\alpha g_m(z)+(1-\alpha )g_m'(z))e^{(\frac{\alpha }{1-\alpha }
 -\mu _1 )z} \,dz  \\
&\quad +\frac{1}{2} t^2 e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t}
 \int (\alpha g_m(z)+(1-\alpha )g_m'(z))e^{(\frac{\alpha }{1-\alpha }
 -\mu _1 )z} \,dz .
\end{aligned}\label{eqn4.6}
\end{equation}
Here $C_j$ ($j=\overline{1,3}$) are any constants, which will be defined
using initial conditions.

\subsection{Convergence part}

We consider the case $\Delta_m>0$ in details. Found solution we satisfy to the
initial conditions \eqref{eqn3.5}. Without losing generality, we assume that
$\tilde{C}_i=0$ ($i=\overline{0,2}$). Regarding the $C_j$ we will get
 the following algebraic system of equations
\begin{gather*}
C_1 +C_2 +C_3 =-d_1,  \\
C_1 (\mu _1 -\frac{\alpha }{1-\alpha } )
 +C_2 (\mu _2 -\frac{\alpha }{1-\alpha } )
 +C_3 (\mu _3 -\frac{\alpha }{1-\alpha } )=-d_2,  \\
C_1 (\mu _1 -\frac{\alpha }{1-\alpha } )^2
 +C_2 (\mu _2 -\frac{\alpha }{1-\alpha } )^2
 +C_3 (\mu _3 -\frac{\alpha }{1-\alpha } )^2 =-d_3,
\end{gather*}
where
\begin{equation}
\begin{aligned}
&d_1\\
&=\frac{1}{(\mu _2 \mu _3^2 -\mu _2^2 \mu _3 -\mu _1 \mu _3^2 +\mu _1^2 \mu _3
 +\mu _1 \mu _2^2 -\mu _1^2 \mu _2 )}
 \Big[(\mu _3 -\mu _2 )\int^{t}\big(\alpha g_m(z) \\
&\quad +(1-\alpha )g_m'(z)\big)\,dz\big|_{t=0}
 +(\mu _1 -\mu _3 )\int^{t} \big(\alpha g_m(z)+(1-\alpha )g_m'(z)\big)\,dz
 \big|_{t=0} \\
 &\quad +(\mu _2 -\mu _1 )\int^{t} \big(\alpha g_m(z)+(1-\alpha )g_m'(z)\big)
 \,dz\big|_{t=0}  \Big],
 \end{aligned}  \label{eqn4.7}
 \end{equation}

  \begin{equation}
 \begin{aligned}
 d_2
&=\frac{1}{(\mu _2 \mu _3^2 -\mu _2^2 \mu _3 -\mu _1 \mu _3^2 +\mu _1^2 \mu _3
 +\mu _1 \mu _2^2 -\mu _1^2 \mu _2 )}  \\
&\quad\times [(\mu _1 -\frac{\alpha }{1-\alpha } )(\mu _3 -\mu _2 )
  \int^{t} \big(\alpha g_m(z)+(1-\alpha )g_m'(z)\big)\,dz\big|_{t=0} \\
&\quad +(\mu _3 -\mu _2 )(\alpha g_m(0)+(1-\alpha )g_m'(0))\\
&\quad +(\mu _2 -\frac{\alpha }{1-\alpha } )(\mu _1 -\mu _3 )
\int^{t} \big(\alpha g_m(z)+(1-\alpha )g_m'(z)\big)\,dz\big|_{t=0}\\
&\quad +(\mu _1 -\mu _3 )(\alpha g_m(0)+(1-\alpha )g_m'(0))
 +(\mu _3 -\frac{\alpha }{1-\alpha } )(\mu _2 -\mu _1 )\\
&\quad \times\int^{t}  \big(\alpha g_m(z)+(1-\alpha )g_m'(z)\big)\,dz\big|_{t=0}\\
&\quad +(\mu _2 -\mu _1 )(\alpha g_m(0)+(1-\alpha )g_m'(0))\Big],
  \end{aligned}  \label{eqn4.8}
  \end{equation}

  \begin{equation}
  \begin{aligned}
  d_3 
&=\frac{1}{(\mu _2 \mu _3^2 -\mu _2^2 \mu _3 -\mu _1 \mu _3^2 +\mu _1^2 \mu _3 
 +\mu _1 \mu _2^2 -\mu _1^2 \mu _2 )} \\
&\quad \times \Big[(\mu _1 -\frac{\alpha }{1-\alpha } )^2 (\mu _3 -\mu _2 )
 \int^{t} \big(\alpha g_m(z)+(1-\alpha )g_m'(z)\big)\,dz\big|_{t=0} \\
&\quad  +(\mu _1 -\frac{\alpha }{1-\alpha } )(\mu _3 -\mu _2 )
 (\alpha g_m(0)+(1-\alpha )g_m'(0))\\
&\quad +(\mu _3 -\mu _2 )(\alpha g_m'(0)+(1-\alpha )g_m''(0))
  +(\mu _2 -\frac{\alpha }{1-\alpha } )^2 (\mu _1 -\mu _3 )\\
&\quad  \times \int^{t} \left(\alpha g_m(z)+(1-\alpha )g_m'(z)\right)
\,dz\big|_{t=0}  +(\mu _2 -\frac{\alpha }{1-\alpha } )(\mu _1 -\mu _3 )\\
&\quad\times   (\alpha g_m(0)+(1-\alpha )g_m'(0))
  +(\mu _1 -\mu _3 )(\alpha g_m'(0)+(1-\alpha )g_m''(0))\\
&\quad  +(\mu _3 -\frac{\alpha }{1-\alpha } )^2 (\mu _2 -\mu _1 )\int^{t}
 \left(\alpha g_m(z)+(1-\alpha )g_m'(z)\right)\,dz\big|_{t=0}  \\
&\quad +(\mu _3 -\frac{\alpha }{1-\alpha } )(\mu _2 -\mu _1 )
 (\alpha g_m(0)+(1-\alpha )g_m'(0))\\
&\quad +(\mu _2 -\mu _1 )(\alpha g_m'(0)+(1-\alpha )g_m''(0))\Big].
  \end{aligned}   \label{eqn4.9}
  \end{equation}
Solving this system, we get
\begin{align*}
C_1
& =-d_1-\frac{-d_2 (\mu _1 -\frac{\alpha }{1-\alpha } )+d_3}
{(\mu _2 -\frac{\alpha }{1-\alpha } )(\mu _1 -\frac{\alpha }{1-\alpha } )
-(\mu _2 -\frac{\alpha }{1-\alpha } )^2 } \\
&\quad -\Big(-d_1(\mu _1 -\frac{\alpha }{1-\alpha } )
 (\mu _2 -\frac{\alpha }{1-\alpha } )-d_2 (\mu _2 -\frac{\alpha }{1-\alpha } )
 +d_2 (\mu _1 -\frac{\alpha }{1-\alpha } )-d_3 \Big)\\
&\quad\div \Big(\big[\mu _1 \mu _2 -\mu _2 \mu _3 
-\mu _3 \mu _1 -\mu _3^2
 -2(\frac{\alpha }{1-\alpha }) ^2 \big]
 \big[(\mu _2 -\frac{\alpha }{1-\alpha } )(\mu _1 -\frac{\alpha }{1-\alpha } )\\
 &\quad -(\mu _2 -\frac{\alpha }{1-\alpha } )^2 \big]\Big) 
\\
&\quad -\frac{-d_1 (\mu _1 -\frac{\alpha }{1-\alpha } )(\mu _2 -\frac{\alpha }{1-\alpha } )
 +d_2 (\mu _2 -\frac{\alpha }{1-\alpha } )+d_2 (\mu _1 -\frac{\alpha }{1-\alpha } )
 -d_3 }{\mu _1 \mu _2 -\mu _2 \mu _3 -\mu _3 \mu _1 -\mu _3^2 
 -2(\frac{\alpha }{1-\alpha } )^2 } 
\end{align*}
\begin{align*}
C_2 &=\frac{-d_2 (\mu _1 -\frac{\alpha }{1-\alpha } )+d_3 }
 {((\mu _2 -\frac{\alpha }{1-\alpha } )
 (\mu _1 -\frac{\alpha }{1-\alpha } )-(\mu _2 -\frac{\alpha }{1-\alpha } )^2 )} \\
&\quad -\Big((-d_1 )(\mu _1 -\frac{\alpha }{1-\alpha } )
 (\mu _2 -\frac{\alpha }{1-\alpha } )
 +d_2 (\mu _2 -\frac{\alpha }{1-\alpha } )
 +d_2 (\mu _1 -\frac{\alpha }{1-\alpha } )-d_3 \Big) \\
&\quad\div\Big(\big[\mu _1 \mu _2 -\mu _2 \mu _3 -\mu _3 \mu _1 -\mu _3^2 
 -2(\frac{\alpha }{1-\alpha } )^2 \big]
 \big[(\mu _2 -\frac{\alpha }{1-\alpha } )(\mu _1 -\frac{\alpha }{1-\alpha } )\\
&\quad -(\mu _2 -\frac{\alpha }{1-\alpha } )^2 \big] \Big) 
\end{align*}
\[
C_3 
=\frac{(-d_1 )(\mu _1 -\frac{\alpha }{1-\alpha } )
 (\mu _2 -\frac{\alpha }{1-\alpha } )
 +d_2 (\mu _2 -\frac{\alpha }{1-\alpha } )
 +d_2 (\mu _1 -\frac{\alpha }{1-\alpha } )-d_3 }
 {\mu _1 \mu _2 -\mu _2 \mu _3 -\mu _3 \mu _1 -\mu _3^2 
 -2(\frac{\alpha }{1-\alpha } )^2 }  
\]
In general, we can write
$$
|C_j|\leq M_1|d_1|+M_2|d_2|+M_3|d_3|.
$$
Hence, we need the following estimations in order to provide convergence 
of used series:
\begin{align*}
|d_1 |
&\le \frac{1}{(m\pi )^{4} } \Big|M_{4}  \int^{t}
 (\alpha  f_{4,0} (z)+(1-\alpha )f_{4,1} (z))e^{(\frac{\alpha }{1-\alpha } 
 -\mu _1 )z}  \,dz\big|_{t=0}  \\
&\quad +M_{5} \int^{t} (\alpha  f_{4,0} (z)+(1-\alpha )f_{4,1} (z))
 e^{(\frac{\alpha }{1-\alpha } -\mu _2 )z}  \,dz\big|_{t=0}\\
&\quad +M_{6} \int^{t} (\alpha  f_{4,0} (z)+(1-\alpha )f_{4,1} (z))
 e^{(\frac{\alpha }{1-\alpha } -\mu _3 )z} \,dz\big| _{t=0} \Big|\\
&\le \frac{1}{(m\pi )^{4} } M_{7} 
 \end{align*} %\label{eqn4.10}

\begin{align*}
|d_2|
&\le \frac{1}{(m\pi )^{4} } \Big|M_{8}  e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t}
 \int^{t} (\alpha  f_{4,0} (z)+(1-\alpha )f_{4,1} (z))
 e^{(\frac{\alpha }{1-\alpha } -\mu _1 )z}\,dz\big|_{t=0} \\
&\quad +M_{9} e^{(\mu _2 -\frac{\alpha }{1-\alpha } )t} \int^{t}
 (\alpha  f_{4,0} (z)+(1-\alpha )f_{4,1} (z))e^{(\frac{\alpha }{1-\alpha } 
 -\mu _2 )z}\,dz\big|_{t=0} \\
&\quad +M_{10} e^{(\mu _3 -\frac{\alpha }{1-\alpha } )t} \int^{t}
 (\alpha  f_{4,0} (z)+(1-\alpha )f_{4,1} (z))e^{(\frac{\alpha }{1-\alpha } 
 -\mu _3 )z}\,dz\big|_{t=0} \\
&\quad +M_{11} (\alpha f_{4,0} (z)+(1-\alpha )f_{4,1} (z))\Big| \\
&\le \frac{1}{(m\pi )^{4} } M_{12}  
\end{align*} %\label{eqn4.11}

\begin{align*}
|d_3|
&\le \frac{1}{(m\pi )^{4} } \Big|M_{13} e^{(\mu _1 -\frac{\alpha }{1-\alpha } )t} 
\int^{t} (\alpha  f_{4,0} (z)+(1-\alpha )f_{4,1} (z))
 e^{(\frac{\alpha }{1-\alpha } -\mu _1 )z}\,dz|_{t=0}  \\
&\quad +M_{14} e^{(\mu _2 -\frac{\alpha }{1-\alpha } )t} \int^{t}
 (\alpha  f_{4,0} (z)+(1-\alpha )f_{4,1} (z))e^{(\frac{\alpha }{1-\alpha }
  -\mu _2 )z} \,dz\big|_{t=0}  \\
&\quad +M_{15} e^{(\mu _3 -\frac{\alpha }{1-\alpha } )t} \int^{t}
  (\alpha  f_{4,0} (z)+(1-\alpha )f_{4,1} (z))e^{(\frac{\alpha }{1-\alpha } 
 -\mu _3 )z}\,dz\big|_{t=0}  \\
&\quad  +M_{16} (\alpha f_{4,0} (z)+(1-\alpha )f_{4,1} (z))+M_{17} 
 (\alpha f_{4,1} (z)+(1-\alpha )f_{4,2} (z))\Big|\\
&\le \frac{1}{(m\pi )^{4} } M_{18} 
 \end{align*} %\label{eqn4.12}

Here $M_i$ ($i=\overline{1,18}$) are  positive constants,
\begin{gather*}
f_m(t)=\int _{0}^{1}f(t,x)\sin m\pi x\,dx
 =\frac{1}{(m\pi )^{4} }  f_{4,0} (t),\\
f_m'(t)=\frac{1}{(m\pi )^{4} } \int _{0}^{1}\frac{\partial }{\partial {t} } 
\big(\frac{\partial ^{4} }{\partial {x}^{4} } f(t,x)\big)\sin m\pi x\,dx
=\frac{1}{(m\pi )^{4} }  f_{4,1} (t),\\
f_m''(t)=\frac{1}{(m\pi )^{4} } \int _{0}^{1}\frac{\partial }{\partial^2 {t} } 
\big(\frac{\partial ^{4} }{\partial {x}^{4} } f(t,x)\big)\sin m\pi x\,dx
=\frac{1}{(m\pi )^{4} }  f_{4,2} (t), \\
%\label{eqn4.13}
f_{4,0} (t)=\int _{0}^{1}\frac{\partial ^{4} f(t,x)}{\partial {x}^{4} } 
 \sin m\pi x\,dx, \\
f_{4,1} (t)=\, \int _{0}^{1}\frac{\partial ^{5} f(t,x)}{\partial {t} 
\partial {x}^{4} }  \sin m\pi x\,dx,
\\
f_{4,2} (t)= \int _{0}^{1}\frac{\partial ^{6} f(t,x)}
 {\partial {t}^2 \partial {x}^{4} }  \sin m\pi x\,dx.
%\label{eqn4.14}
\end{gather*}
We note that for above-given estimations, we need to impose  the 
following conditions to the given function $f(t,x)$:
\begin{equation}
\begin{gathered}
\frac{\partial{f}}{\partial{t}}|_{t=0}=0, \quad 
\frac{\partial^2{f}}{\partial{t}^2}|_{t=0}=0, \quad 
\frac{\partial^3{f}}{\partial{t}^3}|_{t=0}=0, \\
f(t,1)=f(t,0)=0,\quad 
\frac{\partial ^2 f(t,1)}{\partial{x}^2 } 
=\frac{\partial ^2 f(t,0)}{\partial {x}^2 } =0.
\end{gathered}\label{eqn4.15}
\end{equation}

Based on estimations \eqref{eqn4.7}-\eqref{eqn4.9}, we obtain
\begin{equation*}
|C_j|\leq \frac{M_{19}}{(m\pi)^4} %\label{eqn4.16}
\end{equation*}
and considering \eqref{eqn4.3}, finally we get
\begin{equation*}
|T_m|\leq \frac{M_{20}}{(m\pi)^4}. %\label{eqn4.17} 
\end{equation*}

Taking \eqref{eqn3.4} into account, on can easily deduce that
\begin{equation*}
|u(t,x)|\leq \frac{M_{21}}{(m\pi)^4},\,\,\,|u_{xx}(t,x)|
\leq \frac{M_{22}}{(m\pi)^2}. %\label{eqn4.18}
\end{equation*}

The required estimate
\begin{equation*}%\label{eqn4.19}
|{}_{CF}D_{0t}^\alpha u(t,x)|\leq \frac{M_{23}}{(m\pi)^4}
\end{equation*}
can be deduced easily, as well.

\begin{theorem} \label{thm4.1}
 If $f(t,x)\in C^2(\overline{\Omega})$,
$\frac{\partial^3 f(t,x)}{\partial t^3}\in C(\Omega)$ and 
$\frac{\partial^3 f(t,x)}{\partial t^3}$ is continuous up to $t=0$, 
$\frac{\partial ^{6} f(t,x)}{\partial {t}^2 \partial {x}^{4} }\in L^1(0,1)$ 
together with \eqref{eqn4.15}, then problem \eqref{eqn3.1}-\eqref{eqn3.3}, 
when $k=2$ has a unique solution represented by \eqref{eqn3.4}, where $T_m(t)$
 are defined by \eqref{eqn4.3}-\eqref{eqn4.6} depending on the sign of $\Delta_m$.
\end{theorem}

\begin{remark} \rm
Similar result can be obtained for general case, as well. For this, one needs 
to differentiate \eqref{eqn3.9} once by $t$ and write explicit form 
of $(k+1)$th order ordinary DE.
\end{remark}

\begin{remark} \rm
We note that used algorithm allows us to investigate fractional spectral problems
 such as
 \begin{gather*}
 {}_{CF}D_{0t}^{\alpha+1}T(t)+\mu T(t)=0,\\
 T(0)=0,\quad T(1)=0,\\
\end{gather*}
reducing it to the second order usual spectral problem.

 Using above-given algorithm, we obtain the following second order spectral problem
\begin{gather*}
 \bar{T}''(t)+(\mu-\frac{2\alpha}{1-\alpha})\bar{T}'(t)
+(\frac{\alpha}{1-\alpha})^2\bar{T}(t)=0\\
 \bar{T}(0)=0,\quad \bar{T}(1)=0,
 \end{gather*}
eigenvalues and corresponding eigenfunctions of which have a form
\begin{gather*}
 \mu_n=\frac{2\alpha}{1-\alpha}\pm 2\sqrt{(\frac{\alpha}{1-\alpha})^2-(n\pi)^2},
\quad n\in \mathbb{N}, \\
 \bar{T}_n(t)=e^{\frac{\mu_n-\frac{2\alpha}{1-\alpha}}{2}}\sin n\pi x,\quad
n\in \mathbb{N}.
\end{gather*}
\end{remark}

\begin{remark} \rm
 We can apply this approach for studying the  more general equation
 \begin{equation*}
 \sum_{n=0}^{k} \lambda_n(t,x) \cdot\,{}_{CF}D_{0t}^{\alpha+n}u(t,x)
-{}_{CF}D_{0t}^{\alpha+1}u(t,x)=f(t,x),
 \end{equation*}
 where $\lambda_n(t,x)$ might have singularity as well.
\end{remark}

\begin{remark} \rm
 We are able to consider another kind of fractional derivative without 
singularity with $\alpha$th order $(0<\alpha<1)$ such
 $$
 D_{0t}^\alpha T(t)=G(\alpha)\int_0^t T'(s)K(t,s,\alpha)ds,
 $$
 involving it in the  fractional DE
 $$
 \lambda_1 D_{0t}^\alpha T(t)+\lambda_2 T(t)=f(t).
 $$
 In this case, our kernel should satisfy to the  condition
 $$
 \frac{\partial K(t,s,\alpha)}{\partial s}\frac{\partial K(s,t,\alpha)}{\partial t}
=\big[\lambda_1G(\alpha)K(s,s,\alpha)+\lambda_2\big]
\big[\lambda_1G(\alpha)K(t,t,\alpha)+\lambda_2\big].
 $$
\end{remark}

\subsection*{Acknowledgments}
The authors are very grateful to the anonymous referees for their helpful
comments.

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\end{document}