3.7 Stability analysis of the Implicit FTCS

Replace the trial function from (2) in (5B) results in

\[ A^{n+1}e^{jkih}+\frac {u\tau }{2h}A^{n+1}e^{jk\left ( i+1\right ) h}-\frac {u\tau }{2h}A^{n+1}e^{jk\left ( i-1\right ) h}=A^{n}e^{jkih}\]

Divide by \(A^{n}e^{jkih}\)

\begin{align*} \xi +\frac {u\tau }{2h}\xi e^{jkh}-\frac {u\tau }{2h}\xi e^{-jkh} & =1\\ \xi \left ( 1+\frac {u\tau }{2h}e^{jkh}-\frac {u\tau }{2h}e^{-jkh}\right ) & =1\\ \xi \left ( 1+j\frac {u\tau }{h}\sin \left ( kh\right ) \right ) & =1\\ \xi & =\frac {1}{1+j\frac {u\tau }{h}\sin \left ( kh\right ) }=\frac {1-j\frac {u\tau }{h}\sin \left ( kh\right ) }{1+\frac {u\tau }{h}\sin \left ( kh\right ) }\end{align*}

Hence

\[ \left \vert \xi \right \vert =\frac {\sqrt {1+\left ( \frac {u\tau }{h}\right ) ^{2}\sin ^{2}\left ( kh\right ) }}{1+\frac {u\tau }{h}\sin \left ( kh\right ) }<1 \]

Hence this shows that the

\[ \fbox {Implicit FTCS\ method is unconditionally stable.}\]

This property is common to all implicit methods.

Even though the implicit FTCS is stable, it is not very accurate. See case 8 below for an example.