Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 66 / 2

The characteristic polynomials of a linear multistep method areTo determine formal order, substitute a smooth exact solution and expand about the first time level. The exponential-symbol order criterion for a multistep method collects precisely the same coefficients:The constant, linear and quadratic coefficients vanish for every $a$. The cubic coefficient vanishes only at $a=-1/5$, where the quartic coefficient is $-1/30\ne 0$. HenceHere order means the exact-solution step residual is $O(h^{p+1})$. It is a formal consistency result, not a convergence assertion. In particular at $a=1$ both ${\rho }^{\prime }(1)$ and $\sigma (1)$ vanish, and the double root at one destroys zero-stability; cancelling its common factor gives a different, first-order recurrence with an additional integration constant left unspecified by the original formula.

The Dahlquist equivalence theorem states that a consistent linear multistep method is convergent for suitably consistent starting values exactly when it is zero-stable. The root condition for a multistep method requires every root of $\rho$ to lie in the closed unit disk, with every unit-modulus root simple.

The roots are $1$ and $a$. Thus $|a|\le 1$ is necessary; $a=1$ is excluded because it gives a double root at one. At $a=-1$, the two unit roots are distinct, so the endpoint is allowed. Combined with part (a), this provesAssume a locally Lipschitz vector field, a smooth solution on the fixed time interval, a nearby solvable implicit branch and starting errors of the required order. The global order is three at $a=-1/5$ and two at the other convergent parameter values. Outside this interval, zero-step perturbations already grow through either an exterior root or a unit-root polynomial factor.

Use the standard A-stability convention including the root condition at $z=h\lambda =0$. The amplification polynomial of a multistep method isZero-stability first restricts the possible parameters to $-1\le a<1$. For $-1<a<0$, as $z$ tends to negative infinity, one amplification root tends to $(3a-1)/(1+a)$, whose modulus exceeds one. For $a=-1$, the characteristic equation is ${\zeta }^2-2z\zeta -1=0$; large negative real $z$ likewise gives an exterior root. Thus $a\ge 0$ is necessary.

For sufficiency take $0\le a<1$. The leading coefficient cannot vanish in the closed left half-plane. On the unit circle $\zeta =e^{i\theta }$ the boundary-locus test for multistep A-stability uses $z=\rho (\zeta )/\sigma (\zeta )$ and giveswhere the denominator is nonzero. When $a=0$, the exceptional zero of $\sigma$ at $\zeta =-1$ is not a zero of $\rho$ and therefore cannot be an amplification root for a finite $z$. Near $z=0$ on the negative real axis, the root issuing from one is $1+z+O(z^2)$ and is strictly inside the disk; the other root is close to $a$ and is also inside. Roots vary continuously, cannot escape through infinity because the leading coefficient is nonzero, and cannot cross the unit circle anywhere in the open left half-plane by the displayed boundary formula. Thus every root remains inside there. Continuity gives the boundary case; for $a>0$ a unit root on the imaginary axis can occur only at $z=0$, where it is simple. For $a=0$, cancellation of the harmless zero root leaves the trapezoidal rule, whose amplification factor has modulus at most one.

ThereforeThe third-order member is convergent but not A-stable, consistent with the Second Dahlquist barrier. At $a=1$, $P_z=(\zeta -1)[(1-z)\zeta -1]$ has a permanent unit root and a double root at $z=0$: the unreduced method is not zero-stable and is not A-stable under the stated convention. Testing only the open half-plane while overlooking its zero-step behavior would give a weaker conclusion.