Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 41C / b

The first scheme hasAt $\theta =\pi$,Since the Courant number obeys $\mu \ge 0$, stability therefore holds exactly for

For the second scheme,A direct calculation givesThis is at most one for every $\theta$ exactly when

The centred second difference has Fourier symbolWriting $q={\sin }^2(\theta /2)$ givesFor the denominator to stay positive and $|H|\le 1$, one needsThe right inequality follows from $\mu \ge 0$, while the left is equivalent to $2-4\alpha q\ge 0$. Hence stability for every Fourier mode isapart from the singular endpoint $(\mu ,\alpha )=(0,1/2)$, where both sides annihilate the $\theta =\pi$ mode and the recurrence does not determine its next value.