Past exam of the numerical solution of differential equations course of the University of Oxford / 2011 / A / 2

In the normalization used by the supplied coefficient expansion, the local truncation error of the linear multistep method isThe nonzero sum in the denominator is an explicit hypothesis. Consistency of a numerical method requires $T_n\to 0$ uniformly on a fixed finite time interval for every sufficiently smooth exact solution as $h\to 0$. Equivalently, the two polynomial identities areThese follow by inserting constant and affine functions, or by setting the first two Taylor coefficients $C_0,C_1$ to zero.

Zero-stability means that perturbations of the starting values of the zero-step recurrence remain bounded by a fixed multiple of their initial size, uniformly as the number of steps tends to infinity on a fixed time interval. In particular solutions of ${\sum }_{j=0}^k{\alpha }_j{e}_{n+j}=0$ obey ${\max }_{n\le N}|e_n|\le C{\max }_{j<k}|e_j|$ with $C$ independent of $N$.

The equivalent root condition for a multistep method isA root outside the unit circle gives exponential growth of a recurrence relation mode; a repeated root on it gives polynomial growth. Repeated roots strictly inside it are harmless because their polynomial factor is dominated by decay.

The first characteristic polynomial factors asThe two quadratic roots have product one. Both can lie in the closed unit disk only if both have modulus one, requiring $|\alpha +1|\le 2$. At $\alpha =-3$ the root one is triple and $\beta =0$; at $\alpha =1$ the root minus one is double. Both fail the root condition for a multistep method. In the intervening open interval the two quadratic roots are distinct conjugate unit-modulus roots, different from one. ThusUsing $2\beta =\alpha +3$, the supplied expansion givesThroughout the zero-stable range, $C_3\ne 0$, so every such method has order two. There is a unique higher-order choice:Here $C_3=C_4=0$ and $C_5=1/10\ne 0$, giving exactly order four, with normalized leading truncation error $h^4{y}^{(5)}/120$. But the extra roots are $-5\pm 2\sqrt{6}$, one of modulus greater than one. By the Dahlquist equivalence theorem this consistent but non-zero-stable method is not convergent for arbitrary consistent starting values.