Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 71 / 2

For a real skew-symmetric matrix, $P^T=-P$. A scalar quadratic form satisfies $y^{T}Py=(y^{T}Py{)}^T=y^T{P}^{T}y=-y^{T}Py$, so it is zero. ThereforeThe finite-dimensional linear system has the global solution $e^{tP}{y}_0$. Equivalently, $(e^{tP}{)}^T{e}^{tP}=e^{-tP}{e}^{tP}=I$, so its exact flow is an orthogonal matrix and preserves the Euclidean norm.

The Runge-Kutta method has update $y_{n+1}=y_n+h{\sum }_i{b}_i{k}_i$. Expanding its squared Euclidean norm gives the exact identityThe stated per-step relation makes this difference zero. Induction from the initial vector then provesFor a positive step size the relation is also necessary for exact per-step norm conservation. This algebraic identity applies whenever the stage equations define a step; it does not silently assume solvability of arbitrary implicit stages.

Define the stage states $Y_i=y_n+h{\sum }_j{a}_{ij}{k}_j$, so $k_i=P{Y}_i$. Skew symmetry gives $Y_i^T{k}_i=Y_i^{T}P{Y}_i=0$, and henceInsert this into the preceding norm identity and symmetrize the double sum, using $k_i^T{k}_j=k_j^T{k}_i$. The result is the Runge-Kutta conservation of quadratic invariants identityThuswhich proves the sufficient condition for every well-defined step and then for all $n$ by induction. For example the implicit midpoint rule, with $a_{11}=1/2$ and $b_1=1$, has $M_{11}=0$ and preserves this norm exactly. The condition is sufficient, and need not be necessary for a particular matrix or a particular solution.