Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 63 / 4

Let $g(y)=f_y(y)f(y)$. The chain rule gives $y^{\prime \prime }=g(y)$ along a smooth solution. Expanding every term about $t_n$, the unscaled exact-solution residual of this multiderivative multistep method isFor example, at derivative degree $q\ge 2$ its coefficient isthe coefficients at degrees zero, one and two vanish. Thus the formal order of a numerical method is two for $a\ne 11/5$, and three for $a=11/5$, where the fourth-degree coefficient is $-4/15$. This order statement describes the exact-solution defect; convergence also needs zero-stability.

At $h=0$ the characteristic polynomial isThe root condition for a multistep method holds exactly when $|a|\le 1$ and $a\ne 1$. In particular $a=-1$ has two distinct unit roots and is admissible, whereas $a=1$ has a repeated unit root. Applying the permitted Dahlquist equivalence theorem for this multiderivative setting, with smooth $f$, the nearby implicit solution branch and convergent starting values, yieldsAll those methods have global order two, provided the starting errors are $O(h^2)$. The formally third-order choice $a=11/5$ is not zero-stable and therefore is not convergent as a method for general initial-value problems.

One can see the failure without any nonlinear difficulty. For $y^{\prime }=0$, an error mode $e_n=a^n{e}_0$ grows exponentially if $|a|>1$. At $a=1$, $e_n=C+Dn$; choosing $e_0=0$, $e_1=h$ gives vanishing starting errors but $e_{T/h}=T$. This also explains why a small residual alone does not rescue that degenerate choice: at $a=1$ the first-derivative term disappears and the formula has no zero-stable first-order evolution interpretation.

Apply the $a=0$ method to the Dahlquist test equation and write $z=h\lambda$. The recurrence becomesso every amplification root must satisfyChoose $z=-1$, which lies strictly in the left half-plane. The polynomial then gives ${\xi }^2+2\xi -2=0$, henceOne root has modulus $1+\sqrt{3}>1$. Generic initial perturbations excite that growing recurrence mode even though the exact scalar solution decays. The $a=0$ method is not A-stable. There is also a singular implicit coefficient at $z=-\sqrt{2/3}$, another obstruction inside the left half-plane.