Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 38C (Numerical Analysis)

Substitute the exact smooth solution into the leapfrog advection scheme and expand the centered differences. With $\tau =\Delta t$ and $h=\Delta x$, the normalized local truncation error isBecause $u_t=u_x$, repeated differentiation gives $u_{ttt}=u_{xxx}$. HenceEquivalently the unnormalized one-update residual is $\tau ({\tau }^2-h^2)u_{xxx}/3+O({\tau }^5+\tau h^4)$. Thus the scheme is second order in space and time. When $\tau =h$, the exact translating solution $u(x,t)=F(x+t)$ satisfies the grid recurrence exactly; initialization still matters for the independent computational branch.

Insert the Fourier mode $u_m^n=g^n{e}^{im\theta }$ into the leapfrog advection scheme. The amplification roots satisfyFor $0<\mu <1$, both roots have modulus one and remain uniformly separated by at least $2\sqrt{1-{\mu }^2}$. The two-level Fourier amplification matrices are therefore uniformly diagonalizable, and their powers stay bounded. The Parseval identity gives stability in the discrete two-norm. For $\mu >1$, some frequencies have $|\mu \sin \theta |>1$ and one root has modulus greater than one, giving instability.

For two-level stability at the leapfrog Courant boundary, at $\mu =1$ and $\theta =\pi /2$, the repeated root is $i$ and the recurrence has solutions $a_n=(A+Bn)i^n$. For example $a_0=0,a_1=1$ gives $a_n=n{i}^{n-1}$. Thus arbitrary two-level perturbations grow without a uniform power bound. The precise answer for unrestricted starting data isThe often-used root-modulus CFL condition is $\mu \le 1$, but by itself it misses this boundary Jordan growth. A startup that selects the exact translating branch at $\mu =1$ removes that growth for the selected data; it does not make the unrestricted two-level update power-bounded.