Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 4 / b / Solution

For a Fourier mode, the amplification polynomial isPut $a=1-2\mu$ and $\xi =e^{i\theta /2}\eta$. The roots becomeIf $|a|<1$, the square root is real, both roots have modulus one, and their separation is at least $2\sqrt{1-a^2}>0$. The companion matrix has eigenvectors $({\xi }_{\pm },1{)}^T$, so their uniform separation gives a mesh-independent bound on its powers. By power boundedness of a two-level Fourier scheme and the discrete Fourier transform, this proves stability for $0<\mu <1$.

If $|a|>1$, choose $\theta =\pi$. The roots have product of modulus one and unequal moduli, so one has modulus greater than one. The scheme is unstable.

The boundary cases require more than a root-modulus check. At $\theta =\pi$, $\mu =0$ gives $(\xi -1{)}^2=0$, while $\mu =1$ gives $(\xi +1{)}^2=0$. The companion matrices have nontrivial Jordan blocks and solutions of the form $(A+Bn)(\pm 1{)}^n$. Their linear growth rules out a uniform stability bound. This mode occurs on even periodic grids, which are enough to disprove mesh-uniform stability at the endpoints.

Therefore the stable range is exactlyWithin this range, second-order convergence additionally requires two appropriately accurate starting time levels. The unit-modulus parasitic root near $\theta =0$ does not decay, so poor initialization cannot be repaired by the scheme.