Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 39C (Numerical Analysis)

Write $h=\Delta x$, $k=\Delta t$ and apply the recurrence to an exact smooth solution. The centred second difference is $h^2{u}_{xx}+h^4{u}_{xxxx}/12+O(h^6)$. Expanding the previous-time second difference and using $u_t=u_{xx}$ gives the one-step residualwhere $h^2=k/\mu$ with fixed positive $\mu$. Thus the local one-step error is $O((\Delta t{)}^2)$ as stated. If local truncation error is defined after division by the step, it is instead $O(\Delta t)$ in this coupled refinement; the spatial error is only second order in $h$. This distinguishes the printed local-error convention from the second-order time accuracy of the underlying Adams-Bashforth method at fixed spatial discretization.

Use a Fourier mode $u_m^n=z^n{e}^{im\theta }$. Put $a=4\mu {\sin }^2(\theta /2)\ge 0$. Its amplification polynomial of a multistep method isThe two roots are real, one nonnegative and one nonpositive. For $a>0$, evaluation at $1$ gives $a>0$ and at zero gives $-a/2<0$, so the positive root lies below one. The negative root is at least $-1$ exactly when the value at $-1$, namely $2-2a$, is nonnegative. Thus both roots have modulus at most one exactly for $0\le a\le 1$. At $a=0$ they are $1,0$, and at $a=1$ they are $1/2,-1$; all unit-modulus roots are simple, satisfying the multistep root condition for a multistep method. Since the largest available $a$ is $4\mu$, the Adams-Bashforth stability for centered diffusion condition isFor $\mu >1/4$, the highest-frequency mode has a root less than $-1$ and grows, proving necessity as well as sufficiency.