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\markboth{\footnotesize \emph{\emph{International Journal of Applied Mathematical Research}}}{\footnotesize \emph{\emph{International Journal of Applied Mathematical Research}}}
\date{}
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\multirow{8}{*}{\includegraphics[width=1.4cm]{logo}}&\\&{\scriptsize\emph{\textbf{International Journal of Applied Mathematical Research},  3 (X) (2014) XXX-XXX}}\\
 &{\scriptsize\emph{\copyright Science Publishing Corporation}}\\
 &{\scriptsize\emph{www.sciencepubco.com/index.php/IJAMR}}\\
 &{\scriptsize\emph{doi: }}\\

 &{\scriptsize\emph {Research paper, Short communication, Review, Technical paper}}\\
                              &{\scriptsize\emph {\textbf{Please keep only one of the types above and delete the others, including this phrase!}}}
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\centerline {\huge{\bf A New Difference Scheme for  }}

\centerline{}

\centerline{\huge{\bf Fractional Cable Equation and Stability
Analysis}}

\centerline{}
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\centerline{\bf {Ibrahim Karatay and Nurdane Kale${*}$ }}

\centerline{} {\small \centerline{\emph{ Fatih University }}



\centerline{}

\centerline{\emph{*nurdaneguduk@fatih.edu.tr}}}

\centerline{}
\centerline{}

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\newtheorem{Theorem}{Theorem}[section]

\newtheorem{Definition}[Theorem]{Definition}

\newtheorem{Corollary}[Theorem]{Corollary}

\newtheorem{Lemma}[Theorem]{Lemma}

\newtheorem{Example}[Theorem]{Example}
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\noindent \hspace{-3 pt}{\scriptsize \textbf{ Copyright \copyright 2014 Author. This is an open access article distributed under the \href{http://creativecommons.org/licenses/by/3.0/}{%
Creative Commons Attribution License} Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.}}
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\smallskip

\noindent
\hrulefill


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent \textbf{Abstract}\\
\centerline{}

We consider the fractional cable equation. For solution of fractional Cable equation involving Caputo fractional derivative, a new difference scheme is constructed based on Crank Nicholson difference scheme. We prove that the proposed method is unconditionally stable by using spectral stability technique. \\
\centerline{} \noindent {\footnotesize \emph{\textbf{Keywords}}:
\emph{Caputo fractional derivative, Difference scheme, Stability.}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent\hrulefill
%=============================
\section{Introduction}
%=============================
\qquad In this study, we consider the following time fractional
cable equation;
\begin{equation}
\left\{ {{%
\begin{array}{l}
{\frac{\partial ^{\alpha }u(x,t)}{\partial t^{\alpha
}}=}\frac{\partial ^{2}u\left( x,t\right) }{\partial x^{2}}-\mu
^{2}u\left( x,t\right) +f\left(
x,t\right) {,(0<x<1,0<t<1),}\hfill  \\
{u(x,0)=r(x),}\text{ }{0<x<1,} \\
{u(0,t)=0,}\text{ \ }{u(1,t)=0,}\text{ \ }{0\leq t\leq 1.}%
\end{array}%
}}\right.   \label{eq1zz}
\end{equation}%
Here, the term ${\frac{\partial ^{\alpha }u(t,x)}{\partial t^{\alpha
}}}$ denotes $\alpha $-order Caputo derivative with the formula:
\begin{equation}
{\frac{\partial ^{\alpha }u(x,t)}{\partial t^{\alpha
}}=}\frac{1}{\Gamma
(1-\alpha )}\int\limits_{0}^{t}\frac{u_{t}(x,\tau )}{(t-\tau )^{\alpha }}%
d\tau ,\text{ where }0<\alpha <1,  \label{eq2}
\end{equation}%
where $\Gamma (.)$ is the Gamma function.
%==============================
\section{Discretization of Problem}
%==============================

\qquad We introduce the basic ideas for the numerical solution of
the Time Fractional Cable equation by Crank-Nicholson difference
scheme.

For some positive integers $M$ and $N$, the grid sizes in space and
time for the finite difference algorithm are defined by $\ h=$ $1/M$
and $\tau =1/N,$
respectively. The grid points in the space interval $[0,1]$ are the numbers $%
x_{j}=jh$, $j=0,1,2,...,M$, and the grid points in the time interval
$[0,1]$ are labeled $t_{k}=k\tau ,k=0,1,2,...,N$. The values of the
functions $u$
and $f$ at the grid points are denoted $u_{j}^{k}=u(x_{j},t_{k})$ and $%
f_{j}^{k}=f(x_{j},t_{k}),$ respectively. Let $u(x,t),u_{t}(x,t)$ and $%
u_{tt}(x,t)$ are continuous on $[0,1]$.

As in the classical Crank-Nicholson difference scheme, a discrete
approximation to the fractional derivative $\frac{\partial ^{\alpha }u(x,t)}{%
\partial t^{\alpha }}$ at $(x_{j},t_{k+\frac{1}{2}})$ can be obtained by the
following approximation\cite{12}:

\begin{align}
\frac{\partial^{\alpha}u(x_{j},t_{k+\frac{1}{2}})}{\partial t^{\alpha}} & =%
\left[ w_{1}u^{k}+\overset{k-1}{\underset{m=1}{\sum}}\left(
w_{k-m+1}-w_{k-m}\right) u^{m}-w_{k}u^{0}+\sigma\frac{(u_{j}^{k+1}-u_{j}^{k})%
}{2^{1-\alpha}}\right]  \label{eq3} \\
& +O(\tau^{2-\alpha}).  \notag
\end{align}

Where $\sigma=\frac{1}{\Gamma(2-\alpha)}\frac{1}{\tau^{\alpha}}$ and
$w_{j}=$ $\sigma\left( (j+1/2)^{1-\alpha}-(j-1/2)^{1-\alpha}\right)
$ In addition for k = 0 there is no these terms $w_{1}u_{k}$ and
$w_{k}u_{0}$. On the other hand, we have
\begin{equation}
\frac{\partial^{2}u(x_{j},t_{k+\frac{1}{2}})}{\partial x^{2}}=\frac{1}{2}%
\left[ \frac{u_{j+1}^{k+1}-2u_{j}^{k+1}+u_{j-1}^{k+1}}{h^{2}}+\frac{%
u_{j+1}^{k}-2u_{j}^{k}+u_{j-1}^{k}}{h^{2}}\right] +O(h^{2}).
\label{eq3.5}
\end{equation}
%=============================================
\section{The Proposed Difference Scheme}
%=============================================
\qquad Using these approximations (3) and (4) into (1), we obtain
the following difference scheme for (1) which is accurate of order
$O(\tau ^{2-\alpha }+h^{2})$; \newline

$\bigskip $

$w_{1}u^{k}+\overset{k-1}{\underset{m=1}{\sum }}\left(
w_{k-m+1}-w_{k-m}\right) u^{m}-w_{k}u^{0}+\sigma \frac{%
(u_{j}^{k+1}-u_{j}^{k})}{2^{1-\alpha }}=\frac{1}{2}\left[ \frac{%
u_{j+1}^{k+1}-2u_{j}^{k+1}+u_{j-1}^{k+1}}{h^{2}}+\frac{%
u_{j+1}^{k}-2u_{j}^{k}+u_{j-1}^{k}}{h^{2}}\right] $

$-\mu ^{2}\left( \frac{u_{j}^{k}+u_{j}^{k+1}}{2}\right) +f(x_{j},t_{k}+\frac{%
\tau }{2})$

\begin{equation*}
\left\{
\begin{array}{l}
\left[ w_{1}u_{j}^{k}+\overset{k-1}{\underset{m=1}{\sum }}\left(
w_{k-m+1}-w_{k-m}\right) u_{j}^{m}-w_{k}u_{j}^{0}+\sigma \frac{%
(u_{j}^{k+1}-u_{j}^{k})}{2^{1-\alpha }}\right]  \\
-\left[ \frac{u_{j+1}^{k+1}-2u_{j}^{k+1}+u_{j-1}^{k+1}}{2h^{2}}+\frac{%
u_{j+1}^{k}-2u_{j}^{k}+u_{j-1}^{k}}{2h^{2}}\right] +\mu ^{2}\left( \frac{%
u_{j}^{k}+u_{j}^{k+1}}{2}\right) =f(x_{j},t_{k}+\frac{\tau }{2}),\text{ } \\
0\leq k\leq N-1,\text{ }1\leq j\leq M-1,\text{ \ \ \ } \\
\\
u_{j}^{0}=r(x_{j}),\text{ }1\leq j\leq M,\text{\ \ \ \ \ \ \ \ } \\
u_{0}^{k}=0,\text{ \ }u_{M}^{k}=0,\text{ }0\leq k\leq N.%
\end{array}%
\right.
\end{equation*}

\begin{equation*}
\left\{
\begin{array}{l}
\left[ \left( -\frac{1}{2h^{2}}\right) u_{j+1}^{k+1}+\left( \frac{\sigma }{%
2^{1-\alpha }}+\frac{1}{h^{2}}+\frac{\mu ^{2}}{2}\right) u_{j}^{k+1}+\left( -%
\frac{1}{2h^{2}}\right) u_{j-1}^{k+1}\right]  \\
+\left[ \left( -\frac{1}{2h^{2}}\right) u_{j+1}^{k}+\left( -\frac{\sigma }{%
2^{1-\alpha }}+\frac{1}{h^{2}}+\frac{\mu ^{2}}{2}\right) u_{j}^{k}+\left( -%
\frac{1}{2h^{2}}\right) u_{j-1}^{k}\right]  \\
+\left[ w_{1}u_{j}^{k}+\overset{k-1}{\underset{m=1}{\sum }}\left(
w_{k-m+1}-w_{k-m}\right) u_{j}^{m}-w_{k}u_{j}^{0}\right]  \\
=f(x_{j},t_{k}+\frac{\tau }{2}),\text{ \ }0\leq k\leq N-1,\text{
}1\leq
j\leq M-1,\text{ } \\
u_{j}^{0}=r(x_{j}),\text{ }1\leq j\leq M,\text{\ } \\
u_{0}^{k}=0,\text{ \ }u_{M}^{k}=0,\text{ }0\leq k\leq N.%
\end{array}%
\right.
\end{equation*}

\bigskip We can arrange the system \ above to obtain

\begin{equation*}
\left\{
\begin{array}{l}
\left[ \left( -\frac{1}{2h^{2}}\right) u_{j+1}^{k+1}+\left( \frac{\sigma }{%
2^{1-\alpha }}+\frac{1}{h^{2}}+\frac{\mu ^{2}}{2}\right) u_{j}^{k+1}+\left( -%
\frac{1}{2h^{2}}\right) u_{j-1}^{k+1}\right]  \\
+\left[ \left( -\frac{1}{2h^{2}}\right) u_{j+1}^{k}+\left( -\frac{\sigma }{%
2^{1-\alpha }}+\frac{1}{h^{2}}+\frac{\mu ^{2}}{2}\right) u_{j}^{k}+\left( -%
\frac{1}{2h^{2}}\right) u_{j-1}^{k}\right]  \\
+\left[ \overset{k}{\underset{m=1}{\sum }}w_{m}\left(
u_{j}^{k-m+1}-u_{j}^{k-m}\right) \right]  \\
=f(x_{j},t_{k}+\frac{\tau }{2}),\text{ \ }0\leq k\leq N-1,\text{
}1\leq
j\leq M-1,\text{ } \\
u_{j}^{0}=r(x_{j}),\text{ }1\leq j\leq M,\text{\ } \\
u_{0}^{k}=0,\text{ \ }u_{M}^{k}=0,\text{ }0\leq k\leq N.%
\end{array}%
\right.
\end{equation*}

Then we rewrite the equation following type%
\begin{equation}
\left\{
\begin{array}{l}
\left[ \left( -\frac{1}{2h^{2}}\right) u_{j+1}^{k}+\left( -\frac{1}{2h^{2}}%
\right) u_{j+1}^{k+1}\right]  \\
+\left[ \left( -\frac{\sigma }{2^{1-\alpha }}+\frac{1}{h^{2}}+\frac{\mu ^{2}%
}{2}\right) u_{j}^{k}+\left( \frac{\sigma }{2^{1-\alpha }}+\frac{1}{h^{2}}+%
\frac{\mu ^{2}}{2}\right) u_{j}^{k+1}+\overset{k}{\underset{m=1}{\sum }}%
w_{m}\left( u_{j}^{k-m+1}-u_{j}^{k-m}\right) \right]  \\
+\left[ \left( -\frac{1}{2h^{2}}\right) u_{j-1}^{k}+\left( -\frac{1}{2h^{2}}%
\right) u_{j-1}^{k+1}\right]  \\
=f(x_{j},t_{k}+\frac{\tau }{2}),\text{ \ }0\leq k\leq N-1,\text{
}1\leq
j\leq M-1,\text{ } \\
u_{j}^{0}=r(x_{j}),\text{ }1\leq j\leq M,\text{\ } \\
u_{0}^{k}=0,\text{ \ }u_{M}^{k}=0,\text{ }0\leq k\leq N.%
\end{array}%
\right.   \label{scheme}
\end{equation}
%==================================
\subsection{Spectral Stability of the Difference Method}
%==================================

The difference scheme above (\ref{scheme}) can be written in matrix
form:

$DU_{j+1}+EU_{j}+DU_{j-1}=\varphi_{j}$ where $\varphi_{j}=\left[
\varphi
_{j}^{0},\varphi_{j}^{1},\varphi_{j}^{2},...,\varphi_{j}^{N}\right]
^{T},\varphi_{j}^{0}=r(x_{j}),\varphi_{j}^{k}=f(x_{j},t_{k+\frac{1}{2}%
}),1\leq j\leq M,1\leq k\leq N,$ and $U_{j}=\left[
U_{J}^{0},U_{J}^{1},U_{J}^{2},...,U_{J}^{N}\right] ^{T}.$

Here $D_{_{\left( N+1\right) \text{x}\left( N+1\right) }}$and
$E_{\left(
N+1\right) \text{x}\left( N+1\right) }$are the matrices of the form%
\begin{equation*}
D=\left( -\frac{1}{2h^{2}}\right) \left[
\begin{array}{cccccc}
0 &  &  &  &  &  \\
1 & 1 &  &  &  &  \\
& 1 & 1 &  &  &  \\
&  & \ddots & \ddots &  &  \\
&  &  & \ddots & \ddots &  \\
&  &  &  & 1 & 1%
\end{array}
\right]
\end{equation*}

\begin{equation*}
E=\left[
\begin{array}{cccccc}
1 &  &  &  &  &  \\
b & a &  &  &  &  \\
-w_{1} & b+w_{1} & a &  &  &  \\
-w_{2} & w_{2}-w_{1} & b+w_{1} & a &  &  \\
\vdots &  & \ddots & \ddots & \ddots &  \\
-w_{N-1} & w_{N-1}-w_{N-2} & \cdots & w_{2}-w_{1} & b+w_{1} & a%
\end{array}%
\right]
\end{equation*}%
\newline
where $a=\frac{\sigma }{2^{1-\alpha }}+\frac{1}{h^{2}}+\frac{\mu
^{2}}{2}$ , $b=-\frac{\sigma }{2^{1-\alpha
}}+\frac{1}{h^{2}}+\frac{\mu ^{2}}{2}$

Using the idea on the modified Gauss-Elimination method, we can
convert into the following form:

$U_{j}=\psi_{j+1}U_{j+1}+\mu_{j+1},j=M,...,2,1,0.$ \newline

Then, we write

$D+E\psi_{j+1}+D\psi_{j}\psi_{j+1}=0,$

$E\mu_{j+1}+D\psi_{j}\mu_{j+1}+D\mu_{j}=\varphi_{j},$ where $1\leq
j\leq M.$

So, we obtain the following pair of formulas:

$\psi_{j+1}=-\left( E+D\psi_{j}\right) ^{-1}D,$ $\mu_{j+1}=\left(
E+D\psi_{j}\right) ^{-1}\left( \varphi_{j}-D\mu_{j}\right) ,$ where
$1\leq j\leq M.$

We will prove that $\rho $ $\left( \psi _{j}\right) $ $<1$, $1\leq
j\leq M$ , by induction. Since $\psi _{1}$ is a zero matrix $\rho $
$\left( \psi _{1}\right) $ $=0<1.$Moreover, $\psi _{2}=-E^{-1}D,\rho
\left( \psi
_{2}\right) $ $=\rho \left( -E^{-1}D\right) =\left\vert \frac{-1}{\frac{%
\sigma }{2^{1-\alpha }}+\frac{1}{h^{2}}+\frac{\mu
^{2}}{2}}\right\vert
.\left\vert \left( -\frac{1}{2h^{2}}\right) \right\vert =\frac{1/h^{2}}{%
2\left( \frac{\sigma }{2^{1-\alpha }}+\frac{1}{h^{2}}+\frac{\mu ^{2}}{2}%
\right) },$since $\psi _{2}$ is of the form%
\begin{equation*}
\psi _{2}=\left[
\begin{array}{ccccc}
0 &  &  &  &  \\
\ast & \frac{1/h^{2}}{2\left( \frac{\sigma }{2^{1-\alpha }}+\frac{1}{h^{2}}+%
\frac{\mu ^{2}}{2}\right) } &  &  &  \\
\ast & \ast & \frac{1/h^{2}}{2\left( \frac{\sigma }{2^{1-\alpha }}+\frac{1}{%
h^{2}}+\frac{\mu ^{2}}{2}\right) } &  &  \\
&  &  & \ddots &  \\
\ast & \ast & \ast & \ast & \frac{1/h^{2}}{2\left( \frac{\sigma }{%
2^{1-\alpha }}+\frac{1}{h^{2}}+\frac{\mu ^{2}}{2}\right) }%
\end{array}%
\right] _{\left( M+1\right) x\left( M+1\right) }
\end{equation*}%
$\sigma =\frac{1}{\Gamma (2-\alpha )}\frac{1}{\tau ^{\alpha }}>0,$
therefore, $\rho \left( \psi _{2}\right) $ $<1.$

Now, assume $\rho\left( \psi_{j}\right) $ $<1$. We find that

$\psi _{j+1}=-\left( E+D\psi _{j}\right) ^{-1}D$%
\begin{equation*}
=\left( \frac{1}{2h^{2}}\right) \left[
\begin{array}{ccccc}
0 &  &  &  &  \\
\ast  & \frac{1}{E_{2,2}-\left( 1/2h^{2}\right) \psi _{j_{2,2}}} &  &  &  \\
\ast  & \ast  & \frac{1}{E_{3,3}-\left( 1/2h^{2}\right) \psi
_{j_{3,3}}} &
&  \\
&  &  & \ddots  &  \\
\ast  & \ast  & \ast  & \ast  & \frac{1}{E_{M+1,M+1}-\left(
1/2h^{2}\right)
\psi _{j_{M+1,M+1}}}%
\end{array}%
\right]
\end{equation*}%
and we already know that $E_{j,j}=\frac{\sigma }{2^{1-\alpha }}+\frac{1}{%
h^{2}}+\frac{\mu ^{2}}{2}$ and $\psi _{j_{r,r}}=\rho \left( \psi
_{j}\right)
$ for $2\leq r\leq M+1:$%
\begin{equation*}
\rho \left( \psi _{j+1}\right) =\left\vert \frac{1/2h^{2}}{\frac{\sigma }{%
2^{1-\alpha }}+\frac{1}{h^{2}}+\frac{\mu
^{2}}{2}-\frac{1}{2h^{2}}\rho
\left( \psi _{j}\right) }\right\vert =\frac{M^{2}}{2\left[ M^{2}\left( 1-%
\frac{\rho \left( \psi _{j}\right) }{2}\right) +\frac{\sigma }{2^{1-\alpha }}%
+\frac{\mu ^{2}}{2}\right] }
\end{equation*}%
Since $0\leq \rho \left( \psi _{j}\right) <1$, it follows that $\rho
\left( \psi _{j+1}\right) <1$. So, $\rho \left( \psi _{j}\right) <1$
for any j, where $1\leq j\leq M.$
%==================================
\subsection{Numerical Example}
%==================================
Consider this problem,
\begin{equation*}
\left\{
\begin{array}{l}
{\frac{\partial ^{\alpha }u(t,x)}{\partial t^{\alpha
}}=\frac{\partial
^{2}u(t,x)}{\partial x^{2}}}-u(t,x)+\frac{2t^{2-\alpha }}{\Gamma (3-\alpha )}%
(1-x)\sin (x)+2t^{2}\left[ \cos (x)+(1-x)\sin (x))\right] {,} \\
{(0<x<1,0<t<1),} \\
{u(0,x)=0,}\text{ }{0\leq x\leq 1,} \\
{u(t,0)=0,}\text{ \ }{u(t,1)=0,}\text{ \ }{0\leq t\leq 1.}\hfill
\end{array}%
\right.
\end{equation*}%
\bigskip Exact solution of this problem is $u(t,x)=t^{2}(1-x)\sin (x)$. The
errors for some M and N are given in figure 1. The errors when
solving this problem are listed in the table1 for various values of
time and space nodes.

\begin{figure}[h!]
\begin{center}
\includegraphics[bb=0 0 560 420,scale=0.4]{cable.jpg}
\end{center}
\caption{The errors when t=1 for some M and N} \label{fig1}
\end{figure}





\begin{table}[h!]
\caption{The errors for some values of M, N and $\protect\alpha $ }
\label{1}
\bigskip {%
\begin{tabular}{l|l|l|l}
\hline\hline & $\alpha=0.3$ & $\alpha=0.5$ & $\alpha=0.8$ \\ \hline
\begin{tabular}{l}
$N$ \\ \hline
$8$ \\
$16$ \\
$32$%
\end{tabular}
\begin{tabular}{l}
$M$ \\ \hline
$32$ \\
$32$ \\
$32$%
\end{tabular}
&
\begin{tabular}{l}
$Error(\alpha,\tau)$ \\ \hline
0.001811212 \\
0.000449950 \\
0.000111687%
\end{tabular}%
\begin{tabular}{l}
Err. rate \\ \hline
- \\
4.02 \\
4.02%
\end{tabular}
&
\begin{tabular}{l}
$Error(\alpha,\tau)$ \\ \hline
0.001688126 \\
0.000409875 \\
0.000099150%
\end{tabular}%
\begin{tabular}{l}
Err. rate \\ \hline
- \\
4.1 \\
4.1%
\end{tabular}
&
\begin{tabular}{l}
$Error(\alpha,\tau)$ \\ \hline
0.001265365 \\
0.000301407 \\
0.000086960%
\end{tabular}%
\begin{tabular}{l}
Err. rate \\ \hline
- \\
4.1 \\
3.46%
\end{tabular}
\\ \hline\hline
\end{tabular}
}
\end{table}




%==========================
\section{Conclusion}
%==========================
\qquad In this work, $O(\tau ^{2-\alpha }+h^{2})$ order
approximation for the Caputo fractional derivative based on the
Crank-Nicholson difference scheme was successfully applied to solve
the time-fractional cable equation. It is proven that the
time-fractional Crank-Nicholson difference scheme is unconditionally
stable by spectral stability analysis.






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\end{document}
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