= Solution
Only $-1<\alpha<1$ needs consideration. On the <Dahlquist test equation>, the <amplification polynomial of a multistep method> is $P_z(w)=\rho(w)-z\sigma(w)$, where $z=h\lambda$. <A-stability> requires all amplification roots to have modulus at most one for every $\operatorname{Re}z\leq0$, with simple unit-modulus roots.
Choose the negative real value $z=-2/(1-\alpha)$. The leading coefficient of $P_z$ is
$$
1-z(\alpha-1)=-1,
$$
whereas $P_z(1)=-z\sigma(1)=4>0$. Hence $P_z(w)\to-\infty$ as real $w\to+\infty$. The <intermediate value theorem> supplies a real amplification root $w>1$, so this negative test value is unstable. Therefore
$$
\boxed{\text{no parameter makes the method both convergent and A-stable}.}
$$
This is a direct <absolute stability> obstruction, independent of the <Second Dahlquist barrier>, which by itself would not rule out the convergent second-order members.
Back to article page
