Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 39A / b / Solution

Use a Fourier mode $u_m^n={\zeta }^n{e}^{im\theta }$. Substitution givesIts roots have product one. They lie on the unit circle for all spatial frequencies precisely when $|1-2\mu {\sin }^2(\theta /2)|\le 1$, givingFor $\mu >1$, the Nyquist frequency has a real reciprocal root pair with one root of modulus greater than one, giving exponential growth and instability.

The endpoint convention needs care in this two-step wave scheme: repeated unit roots occur at zero frequency, and also at the Nyquist frequency when $\mu =1$. They are not power-bounded for arbitrary two independent displacement sequences with a mesh-independent difference. With the usual initialization from displacement and velocity,each Fourier component satisfies ${\hat{u}}^n=\cos (n\eta ){\hat{u}}^0+k[\sin (n\eta )/\sin \eta ]{\hat{v}}^0$, where $\cos \eta =1-2\mu {\sin }^2(\theta /2)$. Taking the limiting values at repeated roots and using $|\sin (n\eta )|\le n|\sin \eta |$ gives $|{\hat{u}}^n|\le |{\hat{u}}^0|+nk|{\hat{v}}^0|$. This is a uniform finite-time stability bound in the physical displacement/velocity data and includes $\mu =1$. A strict root-condition test for arbitrary two-level data must record the double-root qualification rather than calling that endpoint unconditionally power-bounded.