Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 41C (Numerical Analysis)

Multiplication by $e^{-im\theta }$ and summation over $m\in \mathbb{Z}$ giveThus the recurrence becomeswhereAssuming $A(\theta )\ne 0$, the amplification factor of a two-sided one-step stencil is

Iteration givesIf $|H(\theta )|\le 1$ everywhere, the Parseval identity givesConversely, if $|H({\theta }_0)|>1$, continuity supplies a neighbourhood on which $|H|\ge 1+ϵ$. Choose nonzero square-integrable Fourier data supported there. Its norm then grows at least as $(1+ϵ{)}^n$, contradicting boundedness. HenceThis is the one-step Von Neumann stability analysis criterion.

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.