Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 341 / Section A / 2

For the nodes $c_1=\alpha$ and $c_2=1$, the Lagrange basis isDirect integration givesand thereforeThe Collocation Runge-Kutta method also requiresThis exposes a sign error in the printed tableau: its lower-right entry is shown as $(-1+2\alpha )/[2(1-\alpha )]$. With $1-2\alpha$ in that position, the tableau is exactly the claimed collocation method. Taken literally, the printed weights satisfy $b^{T}1=\alpha /(1-\alpha )$, so the method is not even consistent unless $\alpha =1/2$ and cannot be a collocation method for general $\alpha$.

For the intended collocation weights, interpolation makes the associated quadrature rule exact for every polynomial of degree at most one, so the method has order at least two. It has order at least three precisely when the node polynomial is orthogonal to constants:ThusAt $\alpha =1/3$ this is the two-stage Radau IIA method. Since one node is fixed at the endpoint, no value of $\alpha$ makes the two-node quadrature exact through degree three, so order four cannot occur.

For completeness, the tableau exactly as printed has order zero when $\alpha \ne 1/2$, since $b^{T}1\ne 1$. At $\alpha =1/2$ the two signs coincide because the second weight vanishes, and the resulting method has order two.

For algebraic stability of a Runge-Kutta method, the weights must be nonnegative andmust be positive semidefinite. With the intended collocation weights,Positive semidefiniteness is therefore possible only at $\alpha =1/3$; substitution gives nonnegative weights and a positive-semidefinite $M$. Hence the intended family is algebraically stable exactly whenWith the sign printed in the paper, one instead obtainsEquality forces $\alpha =1/2$, where $M$ still has nonzero off-diagonal entries and is indefinite. The literal printed tableau is consequently algebraically stable for no value of $\alpha$.