Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 2

The tableau is the three-stage Lobatto IIIA method. For its stage matrix $A$, weights $b$, nodes $c$ and $C=\mathrm{diag}(c)$, the Butcher order conditions through order four arewhere $c^{\circ j}$ means coordinatewise powers. Direct substitution verifies all eight identities. Equivalently, this is a collocation Runge-Kutta method at the three Lobatto quadrature nodes, whose weights integrate cubic polynomials exactly.

Order five would require $b^T{c}^{\circ 4}=1/5$, but here $b^T{c}^{\circ 4}=5/24$. Therefore the method has exactly order four.

The stability function of a Runge-Kutta method is $R(z)=1+z{b}^T(I-zA{)}^{-1}1$. Substitution givesWriting the numerator and denominator as $N(z)$ and $D(z)$,The denominator has zeros $3\pm i\sqrt{3}$, both in the right half-plane. Therefore $|R(z)|\le 1$ throughout the closed left half-plane, with strict inequality in its interior. The method is A-stable. Since $R(z)\to 1$ at infinity, it is not L-stable.

For algebraic stability, the weights must be nonnegative and the real symmetric matrix $M=BA+A^{T}B-b{b}^T$, where $B=\mathrm{diag}(b)$, must be a positive semidefinite matrix.

The weights are positive, but $a_{11}=0$ and $b_1=1/6$, soA positive semidefinite matrix cannot have a negative diagonal entry, as its quadratic form at the first standard basis vector would be negative. Thus this Lobatto IIIA method is not algebraically stable, despite its A-stability.