Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 67 / 1

Introduce one stage $k=f(t_n+h/2,Y)$ with $Y=y_n+hk/2$, and update $y_{n+1}=y_n+hk$. This is the implicit midpoint rule as a one-stage Runge-Kutta method, with Butcher tableauFor a smooth vector field, write $\Delta =y_{n+1}-y_n$. Expanding its stage equation giveswhich matches the exact Taylor expansion through degree two. The method therefore has order at least two. On the Dahlquist test equation $y^{\prime }=\lambda y$, its stability function isThe third coefficient differs from the exponential's $1/6$, so the order is exactly two.

There is no pole in $\mathrm{Re}z\le 0$, andThus $|R(z)|\le 1$ throughout the closed left half-plane:This conclusion is A-stability, not strong damping of arbitrarily stiff modes: $R(z)\to -1$ along the negative real axis.

Let $Q(y)=y^{T}Sy$. In the usual pointwise meaning of an ODE invariant, differentiation along an exact solution through an arbitrary state givesFor a nonautonomous equation this identity must hold at the stage time as well as at the initial time. It follows directly from invariance for solutions started at any time and state, or from the stated initial-time invariance when the exact flow onto the stage-time domain is invertible. This is the first-integral condition used below.

Put $Y=(y_n+y_{n+1})/2$ and $t_{\ast }=t_n+h/2$. Since $S$ is symmetric,Induction over all well-defined implicit steps givesNo positive-definiteness assumption on $S$ was used. Nor is uniqueness of the stage essential to this algebraic identity: any stage solution satisfying the method and the pointwise invariant condition preserves $Q$. This is the one-stage case of Runge-Kutta conservation of quadratic invariants.