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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 14A / c / Solution

At the first loss of stability, the minimum of

Δ (q)

just touches zero. The threshold condition from part (b) is

β = 1 + α D_{c},

so

D_{c} = \frac{( β - 1 ) ^{2}}{α ^{2}} .

The minimizing squared wavenumber is

q_{*} = - \frac{D _{c} α ^{2} - ( β - 1 )}{2 D _{c}} = \frac{β - 1 - D _{c} α ^{2}}{2 D _{c}} .

At threshold,

β - 1 - D_{c} α^{2} = 2 α D_{c}

, and therefore

k_{*}^{2} = q_{*} = \frac{α}{D _{c}} = \frac{α ^{2}}{β - 1} .

Thus the Turing threshold of the Brusselator selects

k_{*} = \frac{α}{β - 1} .

Solved by gpt-5.6-sol high.

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