Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 40E / c

Insert the Fourier modeinto the leapfrog advection scheme. This givesand hence the amplification polynomialIts roots areIf $0<\mu <1$, the square root is real for every $\theta$, andIf $\mu >1$, choosing $|\sin \theta |>1/\mu$ makes the roots non-real multiples of $i$ with reciprocal moduli, one of which exceeds one, so the method is unstable.

The endpoint $\mu =1$ requires care. At $\theta =\pi /2$ the polynomial isso a generic mode has the form $(c_1+c_{2}n)i^n$ and grows linearly. It satisfies the weaker test $|{\zeta }_{\pm }|\le 1$ often quoted in elementary Fourier analysis, but it violates the root condition for a multistep method because the unit-modulus root is repeated. Therefore the rigorous uniform stability range for arbitrary two-level initial data isIf “stable” is being used only for the non-amplification condition on root moduli, the conventionally quoted range is $0<\mu \le 1$, with $\mu =1$ a defective marginal endpoint rather than a uniformly stable case.