= Solution
Consistency was established in part (a). By the <Dahlquist equivalence theorem>, convergence additionally requires the <root condition for a multistep method>. Since $\rho(w)=(w-1)(w-\alpha)$, its roots lie in the closed <unit disk> with each unit-modulus root simple exactly for
$$
\boxed{-1\leq\alpha<1.}
$$
The endpoint $\alpha=-1$ is <zero-stability> although it has a simple parasitic root at minus one; $\alpha=1$ has a repeated root at one and is excluded.
For every parameter in this convergent range, the leading coefficient of $\sigma$ is $(5+\alpha)/12>0$, while
$$
\sigma(-1)=-(1-\alpha)/3<0.
$$
Since $\sigma(w)\to+\infty$ as $w\to-\infty$, it has a real root $w_*<-1$. Its other real root lies between minus one and one because $\sigma(1)=1-\alpha>0$, so the exterior root is simple.
On the <Dahlquist test equation>, the <amplification polynomial of a multistep method> is $P_z(w)=\rho(w)-z\sigma(w)$, with $z=h\lambda$. For $|z|\to\infty$, the equivalent polynomial $-P_z/z=\sigma-\rho/z$ tends coefficientwise to $\sigma$, and its leading coefficient stays nonzero. Choose a small closed disk about $w_*$ lying wholly outside the <unit disk> and containing no other root of $\sigma$. On its boundary $|\sigma|$ has a positive minimum, while $|\rho/z|$ is smaller for every sufficiently large $|z|$. <Rouché's theorem> then puts one root of $P_z$ inside this exterior disk for all such complex $z$.
That root violates the stability <root condition for a multistep method>. Hence there is a finite radius $R_\alpha$ beyond which no point belongs to the <absolute stability> domain:
$$
\boxed{\mathcal S_\alpha\subseteq\{z\in\mathbb C:|z|\leq R_\alpha\},\qquad-1\leq\alpha<1.}
$$
This proves boundedness in every complex direction, rather than only excluding <A-stability>. It is an instance of <exterior roots of the derivative polynomial bound multistep stability>.
Back to article page
