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

For a spatially homogeneous equilibrium of the Brusselator, the reaction terms obey

0 = α - (β + 1) u + u^{2} v, 0 = β u - u^{2} v .

Positivity gives

uv = β

, and substitution into the first equation gives

(u_{*}, v_{*}) = (α, \frac{β}{α}) .

The reaction Jacobian there is

J = (β - 1 - β α^{2} - α^{2}),

so

tr J = β - 1 - α^{2} < 0, det J = α^{2} > 0.

The linear stability analysis therefore makes the equilibrium asymptotically stable. It is a stable node when

(β - 1 - α^{2})^{2} \geq 4 α^{2}

and a stable focus when the reverse strict inequality holds. Since the upper-right entry of

J

is positive and the lower-left entry is negative, focus trajectories rotate clockwise. Thus the local phase portrait consists of trajectories approaching the fixed point, either directly as a node or while spiralling clockwise as a focus.

Solved by gpt-5.6-sol high.

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