Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 39A (Numerical Analysis)

Put $\tau =\Delta t$ and $h=\Delta x$, with $h^2=\tau /\mu$ for fixed $\mu >0$. Substitute a smooth exact solution of the heat equation into the Adams-Bashforth method. Define its one-step defect $R$ as the left side minus the right side of the unscaled update. The centered finite difference satisfiesThe Taylor expansion in time givesSince $u_t=u_{xx}$ and $u_{tt}=u_{xxt}$, the first two terms cancel, leavingThis is the requested local error of the unscaled update. If “local truncation error” denotes the defect divided by $\Delta t$, it is instead $O(\Delta t)$ under fixed $\mu$: the centered spatial error is then $O(h^2)=O(\Delta t)$. The scheme is second order in time at fixed spatial discretization, which is a separate order statement.

For a Fourier stability analysis, substitute $u_j^n=g_n{e}^{ij\theta }$. The spatial second difference multiplies by $-4{\sin }^2(\theta /2)$. Set $a=4\mu {\sin }^2(\theta /2)\ge 0$. The mode satisfiesFor $a>0$ the two roots are real of opposite signs. Because $P(0)<0$ and $P(1)=a>0$, the positive root lies strictly between zero and one. The negative root is at least $-1$ exactly when $P(-1)=2-2a\ge 0$, namely $a\le 1$. At $a=0$ the roots are $1,0$; at $a=1$ they are $1/2,-1$, all simple. Thus the root condition for a multistep method for a stable two-step scheme holds exactly for $0\le a\le 1$. The discriminant $1-a+(9/4)a^2$ is bounded away from zero on that interval, so the mode powers are uniformly bounded, including the endpoints.

The maximum value of $a$ over all frequencies is $4\mu$, attained at $\theta =\pi$. ThereforeFor $\mu >1/4$, the highest-frequency mode has a negative amplification root of modulus greater than one and grows exponentially in the step number.

The leapfrog finite-difference scheme for the diffusion equation uses centered time and space differences:A Fourier mode with amplification factor $z$ givesWriting $b=4\mu {\sin }^2(\theta /2)$, the roots are $z_{\pm }=-b\pm \sqrt{1+b^2}$. For any $\mu >0$ and any nonzero spatial frequency, $b>0$ andThusAlthough the other root decays, the growing parasitic root is excited by generic starting data or perturbations, so it cannot be discarded in a stability test.