Past exam of the numerical analysis course of the University of Oxford / 2021 / 3 / c

No. A necessary condition for consistency of a numerical method is reproduction of the constant solution $y(x)=1$ of the ordinary differential equation $y^{\prime }=0$. In a linear multistep method this givesBut the reverse triangle inequality gives, under the strict coefficient inequality,Thus the necessary order conditions for a linear multistep method already fail at order zero, regardless of the values of the ${\beta }_j$.

As appropriate for a genuine $k$-step linear multistep method, take ${\alpha }_k\ne 0$. Suppose the characteristic polynomial has a polynomial root $\xi$ with modulus $|\xi |>1$. From $\rho (\xi )=0$ and the triangle inequality,Dividing by the positive number $|{\alpha }_k||\xi {|}^{k-1}$ gives $|\xi |\le 1$, a contradiction. HenceThe nonzero leading coefficient matters: if ${\alpha }_k=0$, the assumed inequality forces all the ${\alpha }_j$ to vanish, so $\rho$ is the zero polynomial rather than a characteristic polynomial with $k$ roots. This degenerate case does not define the method in the question.

Yes: for $k=2$ the stated inequality guarantees zero-stability. Part (ii) puts every polynomial root in the closed unit disc. The remaining root condition for a multistep method is that every root on the unit circle has multiplicity of a root one.

If this failed for a quadratic polynomial, it would have a repeated root $\xi$ with $|\xi |=1$, andConsequently $|{\alpha }_1|=2|{\alpha }_2|$ and $|{\alpha }_0|=|{\alpha }_2|$, contradictingbecause ${\alpha }_2\ne 0$. Thus all roots on the unit circle are simple andThis conclusion does not assert consistency of a numerical method: part (i) rules out consistency of a numerical method whenever the coefficient inequality is strict. Zero-stability and consistency of a numerical method are separate requirements.