Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 60 / 5

Write $h=\Delta x$, $k=\Delta t=\mu h$. Every smooth exact solution of the advection equation here is $u(x,t)=F(x+t)$. At $s=x+t$, substitution into the scheme gives the unscaled residualThe constant, linear and quadratic Taylor coefficients vanish. Its cubic coefficient isFor a general smooth function the first-order residual is $2k(u_t-u_x)$, so the proper normalized local truncation error on an exact solution isHence the scheme has formal order two for fixed general positive $\mu$. At $\mu =1/2$ and $\mu =1$, the displayed residual vanishes identically for every $F$, not just to one higher Taylor order: the recurrence transports exactly sampled characteristics when both starting levels are compatible with the exact solution. This formal exactness does not imply stability. At $\mu =0$ there is no positive time step. The two-level recurrence also requires a starting procedure for $U^1$; one initial solution profile alone is not two independent starting levels.

Insert a Fourier mode $U_m^n=G^n{e}^{im\theta }$. With $r=1-2\mu$ and $\eta =e^{i\theta }$, the amplification polynomial isIts two roots have product $-\eta$, of modulus one. Thus if both roots are bounded by one, both must have modulus exactly one. Set $G=e^{i\theta /2}\lambda$. The equation becomesWhen $|r|<1$, both roots have modulus one and their separation obeys $|G_{+}-G_{-}|\ge 2\sqrt{1-r^2}>0$ for every frequency. The companion matrix has bounded entries and can be diagonalized with the two eigenvectors $(G_{+},1{)}^T,(G_{-},1{)}^T$. The uniform separation bounds the inverse eigenvector matrix. Hence every matrix power is bounded uniformly in $n$ and $\theta$, and the Parseval identity proves L2 norm stability for the two starting levels on the whole grid.

If $|r|>1$, frequencies near $\theta =\pi$ give one root outside the unit circle, so the scheme is unstable. If $|r|=1$, the Nyquist polynomial is $(G-r{)}^2$, a repeated unit root. The corresponding companion matrix has a nontrivial Jordan block and solutions containing $n{r}^n$. Thus the endpoints are unstable as well. On the whole line the individual Nyquist plane wave is not square-summable, but Fourier packets localized arbitrarily close to that frequency realize the unbounded power norm by continuity. Therefore the uniform stability of a shifted two-level advection scheme criterion isIn particular $\mu =1/2$ is stably exact with compatible starts, whereas $\mu =1$ is formally exact but unstable under arbitrary perturbations. Checking only the moduli of the roots and retaining both endpoints would miss this essential distinction.