Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 39A / ii / Solution

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.