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

Table of contents
- Solution c

Solution

Put

c_{1} = 1/2

and

c_{2} = c \neq = 1/2

. The weights and stage matrix become

b = (1, 0)^{T}, A = \frac{4 c - 1}{4 ( 2 c - 1 )} \frac{c ^{2}}{2 c - 1} - \frac{1}{4 ( 2 c - 1 )} \frac{c ( c - 1 )}{2 c - 1} .

For algebraic stability of a Runge-Kutta method,

M_{ij} = b_{i} a_{ij} + b_{j} a_{ji} - b_{i} b_{j},

so

M = \frac{1}{2 ( 2 c - 1 )} - \frac{1}{4 ( 2 c - 1 )} - \frac{1}{4 ( 2 c - 1 )} 0 .

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 \neq = 1/2

. Therefore there is no admissible value of

c_{2}

for which the method is algebraically stable.

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