 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/13c/c/solution

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

Put

y = k^{2} \geq 0

. At the threshold of a Turing instability, the maximum of

σ (y) = - Dy - r + ρ \frac{y}{1 + y}

touches zero. Hence

σ (y_{c}) = 0, \frac{d σ}{d y} (y_{c}) = - D + \frac{ρ}{( 1 + y _{c} ) ^{2}} = 0.

ρ = D (1 + y_{c})^{2}

. Substitution in the first gives

0 = - D y_{c} - r + D y_{c} (1 + y_{c}) = D y_{c}^{2} - r .

Therefore

y_{c} = \frac{r}{D}, k_{c} = (\frac{r}{D})^{1/4},

and

ρ_{c} (r) = D (1 + \frac{r}{D})^{2} = (r + D)^{2} .

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