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Past exam of the mathematics course of the University of Cambridge

2025

Paper 4

At the first loss of stability, the minimum of $\Delta(q)$ just touches zero. The threshold condition from part (b) is
$$
\sqrt\beta=1+\alpha\sqrt{D_c},
$$
so
$$
D_c=\frac{(\sqrt\beta-1)^2}{\alpha^2}.
$$
The minimizing squared wavenumber is
$$
q_*=-\frac{D_c\alpha^2-(\beta-1)}{2D_c}
=\frac{\beta-1-D_c\alpha^2}{2D_c}.
$$
At threshold, $\beta-1-D_c\alpha^2=2\alpha\sqrt{D_c}$, and therefore
$$
k_*^2=q_*=\frac{\alpha}{\sqrt{D_c}}
=\frac{\alpha^2}{\sqrt\beta-1}.
$$
Thus the <Turing threshold of the Brusselator> selects
$$
\boxed{k_*=\frac{\alpha}{\sqrt{\sqrt\beta-1}}}.
$$

Solution

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

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

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

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