= Solution
The <Second Dahlquist barrier> states that an <A-stable> linear <multistep method> has order at most two. Part a shows that every method in this family has order at least three, so no member can be A-stable.
The exceptional reducible case does not evade the conclusion. At $a=1$,
$$
\rho(\zeta)-z\sigma(\zeta)
=(\zeta-1)\left[(\zeta-1)-\frac z2(\zeta+1)\right].
$$
Thus $\zeta=1$ is an amplification root for every $z=h\lambda$ with $\operatorname{Re}z<0$, whereas the multistep A-stability criterion requires all such roots to lie strictly inside the unit disk. Hence
$$
\boxed{\text{there is no real value of }a\text{ for which the method is A-stable}.}
$$
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