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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 14C / a / Solution

Let

M = (f_{u} g_{u} f_{v} g_{v})

be the reaction Jacobian at the homogeneous equilibrium. Stability without diffusion gives

f_{u} + g_{v} < 0, J = det M = f_{u} g_{v} - f_{v} g_{u} > 0.

For a spatial Fourier mode with

- \nabla^{2}

eigenvalue

k^{2}

, the linearized matrix is

M - k^{2} (D_{u} 0 0 D_{v}) .

Its trace is even more negative than

tr M

, so instability occurs exactly when its determinant becomes negative. Put

z = D_{v} k^{2}

and

d = D_{u} / D_{v}

. Then

Δ (z) = J - (f_{u} + d g_{v}) z + d z^{2} .

This upward-opening quadratic is negative for some

z > 0

exactly when its minimum occurs at positive

z

and lies below zero:

f_{u} + d g_{v} > 0, J - \frac{( f _{u} + d g _{v} ) ^{2}}{4 d} < 0.

The second strict inequality implies the first when written with a positive square root, so the two-species diffusion-driven instability criterion becomes

f_{u} + d g_{v} > 2 dJ .

Solved by gpt-5.6-sol high.

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