Put $c_1=1/2$ and $c_2=c\ne1/2$. The weights and stage matrix become
$$
b=(1,0)^T,
\qquad
A=\begin{pmatrix}
\dfrac{4c-1}{4(2c-1)}&-\dfrac1{4(2c-1)}\\[5pt]
\dfrac{c^2}{2c-1}&\dfrac{c(c-1)}{2c-1}
\end{pmatrix}.
$$
For <algebraic stability of a Runge-Kutta method>,
$$
M_{ij}=b_ia_{ij}+b_ja_{ji}-b_ib_j,
$$
so
$$
M=\begin{pmatrix}
\dfrac1{2(2c-1)}&-\dfrac1{4(2c-1)}\\[5pt]
-\dfrac1{4(2c-1)}&0
\end{pmatrix}.
$$
A positive-semidefinite matrix with a zero diagonal entry must have every entry in that row and column equal to zero. Here the off-diagonal entry never vanishes for finite $c\ne1/2$. Therefore there is no admissible value of $c_2$ for which the method is algebraically stable.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 341 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-341/2/c/solution