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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/14c/d/solution

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

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

M = (- 1 1 + δ - 1 1 - δ),

we have

tr M = - δ < 0, J = det M = 2 δ > 0,

so the homogeneous equilibrium is stable for

δ > 0

. The Turing condition is

- 1 + d (1 - δ) > 2 2 δ d .

y = d

, equality at threshold is

(1 - δ) y^{2} - 2 2 δ y - 1 = 0.

Its positive root is

y_{c} = \frac{2 δ + 1 + δ}{1 - δ}, d_{c} = (\frac{2 δ + 1 + δ}{1 - δ})^{2} .

For

d > d_{c}

sufficiently close to the threshold the strict instability condition holds, while

d_{c} = 1 + 2 2 δ + O (δ) ⟶ 1 (δ \to 0^{+}) .

Thus unequal diffusivities are necessary, but their ratio can be arbitrarily close to one when the stable reaction matrix is itself arbitrarily close to marginal stability. This is the near-unity diffusivity ratio for a Turing instability.

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