For <algebraic stability of a Runge-Kutta method>, the weights must be nonnegative and
$$
M=BA+A^TB-bb^T,
\qquad B=\operatorname{diag}(b_1,b_2),
$$
must be <positive semidefinite matrix>[positive semidefinite]. With the intended collocation weights,
$$
\det M=-\frac{(3\alpha-1)^2}{16(\alpha-1)^2}.
$$
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 when
$$
\boxed{\alpha=\frac13.}
$$
With the sign printed in the paper, one instead obtains
$$
M_{22}=-\frac{3(2\alpha-1)^2}{4(\alpha-1)^2}\leq0.
$$
Equality 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 $\alpha$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-341/section-a/2/c/solution