Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 341 / 2

The characteristic polynomials of a linear multistep method areandFor the order conditions for a linear multistep method, put $\alpha =(a,-1-a,1)$ and $\beta =(-(1+5a)/12,,8(1-a)/12,,(5+a)/12)$. The defectsvanish for $q=0,1,2,3$, whileThereforeIndeed, at $a=-1$ one has $C_4=0$ but $C_5=-4/3\ne 0$.

Every member of the family has order at least three, so it is consistent. By the Dahlquist equivalence theorem, convergence is therefore equivalent to zero-stability, which is characterized by the root condition for a multistep method.

The roots of $\rho$ are $1$ and $a$. They lie in the closed unit disk exactly when $|a|\le 1$. At $a=1$, however, the unit root $1$ is repeated, whereas at $a=-1$ the distinct unit roots $1$ and $-1$ are both simple. Consequently

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 $a=1$,Thus $\zeta =1$ is an amplification root for every $z=h\lambda$ with $\mathrm{Re}z<0$, whereas the multistep A-stability criterion requires all such roots to lie strictly inside the unit disk. Hence