Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 69 / 4

The characteristic polynomials of a linear multistep method areExpanding the exponential-symbol order criterion for a multistep method givesThus the formal local consistency order is two for $\alpha \ne 2$, and three at $\alpha =2$, since the fourth-degree coefficient there is $-1/12$. This exceptional formal order is degenerate: ${\rho }^{\prime }(1)=\sigma (1)=0$, so it is not a convergent third-order time integrator.

For zero-stability, factor $\rho (\zeta )=(\zeta -1)(\zeta -(\alpha -1))$. The root condition for a multistep method requires every root to have modulus at most one and every unit-modulus root to be simple. For real $\alpha$ it holds exactly forAt zero the second root is $-1$, simple and admissible; at two the root one is double and inadmissible. Outside the closed interval the second root has modulus greater than one. The Dahlquist equivalence theorem says that a consistent linear multistep method is convergent exactly when it is zero-stable, for a locally Lipschitz initial-value problem with consistent starting values and solvable implicit steps. Hence the displayed interval is also exactly the convergence range, and every convergent member has order two.

The common-factor cancellation defect in a multistep recurrence explains why the $\alpha =2$ exception cannot be repaired just by canceling $\zeta -1$. For $y^{\prime }=0$, its recurrence allows linear growth from a vanishing but insufficiently small starting perturbation: $y_0=0$, $y_1=\sqrt{h}$ gives $y_n=n\sqrt{h}$, diverging at a fixed positive time. Canceling the factor silently imposes an additional starting relation and changes the method.

Every convergent member is also A-stable. For $0<\alpha <2$, put $\zeta =e^{i\theta }$ and calculate its boundary-locus test for multistep A-stability:The denominator never vanishes on the unit circle in this parameter interval. It follows thatTherefore no amplification root can cross the unit circle as $z=h\lambda$ moves in the open left half-plane. For small negative real $z$, the root near one moves to $1+z+O(z^2)$, inside the circle; the other begins at $\alpha -1$, also strictly inside. The leading coefficient $1-(1-\alpha /4)z$ never vanishes in the left half-plane. Continuity of polynomial roots and connectedness of that half-plane now keep both roots inside everywhere. On its imaginary boundary the same locus has zero real part only at $\zeta =1$, corresponding to $z=0$; that root is simple because $2-\alpha >0$.

At $\alpha =0$, the two interlaced subsequences follow the trapezoidal rule with step $2h$:The right side has modulus at most one for $\mathrm{Re}z\le 0$. Its unit-circle roots are simple; the double zero at $z=-1$ lies strictly inside and is harmless. Thus the endpoint is also A-stable. Combining this with part 1, the requested range is $0\le \alpha <2$, including the BDF2 method at $\alpha =4/3$.