3.3 Stability analysis of the downwind method

\[ C_{i}^{n+1}=C_{i}^{n}-\frac {u\tau }{h}\left ( C_{i+1}^{n}-C_{i}^{n}\right ) \]

Substitute the trial solution \(A^{n}e^{jkih}\) into the above

\begin{align*} A^{n+1}e^{jkih} & =A^{n}e^{jkih}-\frac {u\tau }{h}\left ( A^{n}e^{jk\left ( i+1\right ) h}-A^{n}e^{jkih}\right ) \\ \xi & =1-\frac {u\tau }{h}\left ( e^{jkh}-1\right ) \\ & =1+\frac {u\tau }{h}-\frac {u\tau }{h}e^{jkh}\\ & =1+\frac {u\tau }{h}-\frac {u\tau }{h}\left ( \cos \left ( kh\right ) +j\sin \left ( kh\right ) \right ) \\ & =1+\frac {u\tau }{h}\left ( 1-\cos kh\right ) -j\frac {u\tau }{h}\sin kh \end{align*}

Let \(\frac {u\tau }{h}=\lambda \)

Hence

\[ \xi =1+\lambda \left ( 1-\cos kh\right ) -j\lambda \sin kh \]

\begin{align*} \left \vert \xi \right \vert ^{2} & =\left ( 1+\lambda \left ( 1-\cos kh\right ) \right ) ^{2}+\left ( \lambda \sin kh\right ) ^{2}\\ & =1+2\lambda \left ( 1-\cos kh\right ) +\lambda ^{2}\left ( 1-\cos kh\right ) ^{2}+\lambda ^{2}\sin ^{2}kh\\ & =1+2\lambda \left ( 1-\cos kh\right ) +\lambda ^{2}\left ( 1-2\cos kh+\cos ^{2}kh\right ) +\lambda ^{2}\sin ^{2}kh\\ & =1+2\lambda -2\lambda \cos kh+\lambda ^{2}-2\lambda ^{2}\cos kh+\lambda ^{2}\cos ^{2}kh+\lambda ^{2}\sin ^{2}kh\\ & =1+2\lambda -2\lambda \cos kh+2\lambda ^{2}-2\lambda ^{2}\cos kh\\ & =1+2\lambda \left ( 1+\lambda \right ) \left ( 1-\cos kh\right ) \end{align*}

Hence for stability it is required that

\[ \left \vert 1+2\lambda \left ( 1+\lambda \right ) \left ( 1-\cos kh\right ) \right \vert \leq 1 \]

or

\[ 2\lambda \left ( 1+\lambda \right ) \left ( 1-\cos kh\right ) \leq 0 \]

since \(\lambda =\frac {u\tau }{h}\), a positive quantity, then the above condition can not be satisfied. Hence the downwind method is unconditionally unstable.