Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 69 / 2

The coefficients are those of the three-stage Lobatto IIIA method. Set $e=(1,1,1{)}^T$, $c=Ae=(0,1/2,1{)}^T$ and $C=\mathrm{diag}(c)$. Direct calculation gives all eight fourth-order conditions for a Runge-Kutta method:Powers of $c$ are componentwise. These Butcher order conditions prove order at least four for a smooth general nonlinear vector field; matching a scalar test equation alone would not suffice. A necessary fifth-order condition fails:Hence the method has order exactly four, with one-step local truncation error $O(h^5)$ and, under stability and smoothness hypotheses, global error $O(h^4)$.

For the Dahlquist test equation with $z=h\lambda$, the stage equations give the stability function $R(z)=1+z{b}^T(I-zA{)}^{-1}e$. Substitution simplifies it toThe denominator zeros are $3\pm i\sqrt{3}$, so no pole lies in the closed left half-plane. Moreover,Thus $|R(z)|\le 1$ throughout that half-plane: the method is A-stable. On the imaginary axis its amplification modulus is one. Since $R(z)\to 1$ as $z\to -\infty$, it is not L-stable; linear stability does not guarantee strong damping of very stiff modes.

Algebraic stability of a Runge-Kutta method requires $b_i\ge 0$ and a positive-semidefinite symmetric matrixAlthough all weights are positive, its first diagonal entry isThe quadratic form on the first coordinate vector is negative, so $M$ cannot be positive semidefinite. The method is not algebraically stable. This distinguishes nonlinear contractivity from the A-stability proved in part 2.