Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 66 / 3

Let $p$ be the collocation polynomial of degree at most $s$ over one step, with $p(t_n)=y_n$, and let $Y_i=p(t_n+c_{i}h)$. Its derivative, of degree at most $s-1$, is fixed by its values at the distinct nodes. Lagrange interpolation therefore givesIntegrating from zero to $\tau$ yieldsAt $\tau =c_i$ this gives the stage equations of an implicit Runge-Kutta method; at $\tau =1$ it gives the update. ConsequentlyConversely, stages satisfying these equations define the integrated polynomial displayed above. It takes the stage values at the nodes and has the required derivative there, so it solves the collocation equations. This proves equivalence for every common solution branch, not merely equality on the scalar test equation. Also ${\sum }_j{a}_{ij}=c_i$, since the Lagrange polynomials sum to one. Stage existence or uniqueness requires the usual implicit-solvability assumptions; for a Lipschitz vector field a sufficiently small step gives a contraction. This is the collocation Runge-Kutta method construction.

The Lagrange interpolation polynomials for these nodes areTheir integrals give the Lobatto IIIA method tableauTo verify its nonlinear order, not just its scalar linear order, set $e=(1,1,1{)}^T$, $c=Ae$ and $C=\mathrm{diag}(c)$. Direct multiplication verifies the fourth-order conditions for a Runge-Kutta method:Powers of $c$ are componentwise. These eight Butcher order conditions establish order at least four for general smooth ODEs. On the Dahlquist test equation, elimination of the stages givesA nonzero fifth-order step defect rules out order five. Thus the method has exactly order four.

Write $P(z)=1+z/2+z^2/12$ and $Q(z)=1-z/2+z^2/12$. The zeros of $Q$ are $3\pm i\sqrt{3}$, so the stability function has no pole in the closed left half-plane. A direct modulus calculation givesFor $\mathrm{Re}z\le 0$ this is nonnegative, and hence $|R(z)|=|P/Q|\le 1$. Therefore the Lobatto IIIA method is A-stable. It is not L-stable, because $R(z)\to 1$ for large negative real $z$; unconditional scalar stability need not strongly damp the stiffest modes.