 Solution

The relevant one of the characteristic polynomials of a linear multistep method isThe coefficients on the right are chosen to give third order, so consistency of a numerical method is already available. By the Dahlquist equivalence theorem, convergence with convergent starting values is equivalent to zero-stability: every root of $\rho$ must satisfy $|w|\le 1$, and every root on the unit circle must be simple.

If $a<0$, the quadratic is negative at $w=-1$ and tends to $+\infty$ as $w\to -\infty$, so it has a root less than $-1$. If $a>1$, its root product is $a$, so at least one root has modulus greater than one. For $0\le a\le 1/4$, its real roots $(-1\pm \sqrt{1-4a})/2$ lie in $[-1,0]$. For $1/4<a\le 1$, its conjugate roots have modulus $\sqrt{a}\le 1$.

At $a=0$, the three roots are $1,0,-1$, with the unit-modulus roots simple. At $a=1$, they are $1,e^{2\pi i/3},e^{-2\pi i/3}$, again simple. The repeated quadratic root at $a=1/4$ is $-1/2$, strictly inside the unit circle, which the root condition for a multistep method permits. ThereforeThis is the root stability of a cubic multistep polynomial; oscillatory parasitic roots at the endpoints do not invalidate convergence when their starting amplitudes tend to zero.