Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 69 / 5

Along the exact ordinary differential equation solution, the chain rule gives $g(y)=f^{\prime }(y)f(y)=y^{\prime \prime }$. Insert the exact solution into the symmetric two-step two-derivative formula and expand about the middle time $t=t_{n+1}$. Its unnormalized local truncation error isThe coefficients of $h{y}^{\prime }$, $h^3{y}^{\prime \prime \prime }$ and $h^5{y}^{(5)}$ are respectivelyThe centered residual is an odd function of $h$, so every even power cancels. The next coefficient isThus the method has order six. Its zero-stability roots at $h=0$ are the simple roots $1,-1$. With sufficiently accurate starts and smooth derivative evaluation maps, convergence of a zero-stable multiderivative method gives sixth-order error on fixed time intervals. The first Dahlquist barrier for ordinary linear multistep methods does not apply, because this is a multiderivative multistep method.

The stability convention matters. The Cambridge numerical analysis notes define a strict linear stability domain through decay $y_n\to 0$. For a multistep recurrence, robust decay requires every amplification root to lie strictly inside the unit disk. With that convention we can prove the stronger result that this method's strict linear stability domain is empty, and therefore bounded. Mere boundedness of oscillatory sequences gives a different answer, described below.

On the Dahlquist test equation, $g(y)={\lambda }^{2}y$. With $z=h\lambda$, the amplification polynomial of a multistep method isExclude $D=0$, where the intended two-step update cannot be solved. If both polynomial roots have modulus below one, their product requires $|A|<|D|$. Butso necessarily $\mathrm{Re}z<0$. The complex quadratic Schur criterion also requiresIts left-hand side equals $32|\mathrm{Re}z|(|z{|}^2+15)$, while for $\mathrm{Re}z<0$ its right-hand side is only $28|\mathrm{Re}z|(|z{|}^2+15)$. The inequality is impossible. ThusFor comparison, the bounded-root stability set of the symmetric two-derivative formula is unbounded. On $z=iy$, rotate the amplification polynomial into a real reciprocal quadratic. Its two roots are distinct and on the unit circle precisely whenequivalently $y^2<(45-15\sqrt{5})/2$ or $y^2>(45+15\sqrt{5})/2$. In particular, arbitrarily large imaginary parameters give bounded, undamped numerical oscillations. At the two equality thresholds, repeated unit-circle roots produce unbounded solutions. Hence the question's bounded-domain claim holds for the strict decay convention; it does not hold if the phrase is interpreted using the non-growing root condition for a multistep method.