3.2 Stability analysis for FTCS

Using Von Neumann method, the following trial solution to the PDE is assumed

\[ c\left ( x,t\right ) =A\left ( t\right ) e^{jkx}\]

where \(j=\sqrt {-1}\) and \(k\) is the wave number  and \(A\) is the amplitude of the wave, as a function of time.

Hence the solution at time step \(n\) and at \(x=x_{i}=ih\) is written as

\begin{equation} A^{n}e^{jkih} \tag {2}\end{equation}

Substitute this trial solution (2) into the (1) results in

\begin{equation} A^{n+1}e^{jkih}=A^{n}e^{jkih}-\frac {u\tau }{2h}\left ( A^{n}e^{jk\left ( i+1\right ) h}-A^{n}e^{jk\left ( i-1\right ) h}\right ) \tag {3}\end{equation}

Let \(\xi \) be the ratio of the amplitude of the wave at time step \(n+1\) relative to that at time step \(n\). hence

\[ \xi =\frac {A^{n+1}}{A^{n}}\]

Divide (3) by \(A^{n}\) results in

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

Divide the above by \(e^{jkih}\)

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

Hence

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

This implies that \(\left \vert \xi \right \vert \geq 1\) regardless of the time step \(\tau \) selected or the space step \(h\), hence

\[ \fbox {FTCS is unconditionally unstable.}\]

For a fixed speed \(u\), the instability can be delayed by making \(\frac {\tau }{h}\) smaller, but it could not be prevented. Eventually this numerical solution will blow up. This will be illustrated below in an animation. See case 3 and 4 as examples.

The instability can be delayed by making \(\tau \) smaller for a fixed \(h,\) or by making \(h\) larger for a fixed \(\tau \).