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\markboth{\footnotesize \emph{\emph{International Journal of Applied Mathematical Research}}}{\footnotesize \emph{\emph{International Journal of Applied Mathematical Research}}}
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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}}\\
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\centerline {\huge{\bf A New Difference Scheme for }}
\centerline{}
\centerline{\huge{\bf Fractional Cable Equation and Stability
Analysis}}
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\centerline{\bf {Ibrahim Karatay and Nurdane Kale${*}$ }}
\centerline{} {\small \centerline{\emph{ Fatih University }}
\centerline{}
\centerline{\emph{*nurdaneguduk@fatih.edu.tr}}}
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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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\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.}}
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\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) {,(00,$
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