Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 72 / 4 / c

For the first Butcher tableau, $b=(1/2,1/2{)}^T$ and the off-diagonal entries sum to $1/2$, while each diagonal entry is $1/4$. ThusThe positive weights and zero matrix prove that method 1 is algebraically stable. It is the two-stage Gauss collocation method.

The second method is explicit and has $b=(1/4,3/8,3/8{)}^T$. Its first diagonal algebraic-stability entry isA positive semidefinite matrix cannot have a negative diagonal quadratic form, so method 2 is not algebraically stable. The negative value supplies a direct test; its order is irrelevant to this conclusion.

For the third method, $A=\left(\begin{array}{cc}1/4& -1/4\\ 1/4& 5/12\end{array}\right)$ and $b=(1/4,3/4{)}^T$. Direct computation givesBoth weights are positive, so method 3 is algebraically stable.