Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 71 / 1

Along a smooth exact solution, the chain rule gives $g(y)=f^{\prime }(y)f(y)=y^{\prime \prime }$. Expand the unscaled defect of the multiderivative multistep method about the central time $t_n$. The antisymmetric difference of the solution and second derivative, and the symmetric sum of first derivatives, giveAll derivatives are evaluated at $t_n$. Reflection symmetry removes every even power. The fifth-order conditions require the coefficients of $h,h^3,h^5$ to vanish. The last two equations give $\beta =1/15$ and $\alpha =7/15$, and the first then gives $\gamma =16/15$. ThusThese determine the symmetric two-step two-derivative formula. Substitution into the next coefficient givesThe requested fifth-order conditions actually produce formal order six. In the usual maximal-order terminology there is no member of this symmetric family with exact order five: cancelling the fifth-power defect also cancels the automatically zero sixth-power defect. Interpreting the printed request as order at least five gives precisely the boxed parameters.

At zero step size the recurrence polynomial is $w^2-1$, with simple roots $1$ and $-1$, so it satisfies the root condition for a multistep method and is zero-stable. With sufficiently smooth locally Lipschitz derivative maps, the nearby implicit solution branch and starting values of matching accuracy, the convergence of a zero-stable multiderivative method gives global order six. Formal order by itself does not imply decay of its additional numerical mode.

Apply the method to the Dahlquist test equation $y^{\prime }=\lambda y$, so $g(y)={\lambda }^{2}y$, and put $z=h\lambda$. A trial numerical mode $y_n=w^n$ satisfiesFor every real $z<0$, the leading coefficient is positive, whileThe intermediate value theorem gives a real amplification root $w<-1$. Its modulus exceeds one although the exact scalar solution is decaying. ThereforeFor a concrete choice, $z=-1$ gives $23w^2+16w-9=0$, whose negative root is $(-8-\sqrt{271})/23<-1$. This proves negative-real instability of the symmetric two-derivative formula directly, without using either an order barrier or the supplied quadratic criterion. The unstable branch is the parasitic amplification root issuing from $w=-1$ at $z=0$; its small-step expansion is $w=-1+z/15+O(z^2)$, so a negative $z$ moves it outside the unit circle immediately.