 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-353/2/b/i/solution

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

The independent additive noises give diagonal diffusion

κ = diag (b_{1}^{2}, b_{2}^{2})

. Thus

\partial_{t} P = - \partial_{x_{1}} [(r_{1} x_{1} - r_{1} c_{1} x_{1} x_{2}^{2} - d_{1} x_{1}^{2}) P] + \frac{b _{1}^{2}}{2} \partial_{x_{1}}^{2} P - \partial_{x_{2}} [(r_{2} x_{2} - r_{2} c_{2} x_{2} x_{1}^{2} - d_{2} x_{2}^{2}) P] + \frac{b _{2}^{2}}{2} \partial_{x_{2}}^{2} P .

The mixed derivatives of

κ^{- 1} a

are

\partial_{x_{2}} \frac{a _{1}}{b _{1}^{2}} = - \frac{2 r _{1} c _{1} x _{1} x _{2}}{b _{1}^{2}}, \partial_{x_{1}} \frac{a _{2}}{b _{2}^{2}} = - \frac{2 r _{2} c _{2} x _{1} x _{2}}{b _{2}^{2}} .

They agree precisely when

r_{1} c_{1} b_{2}^{2} = r_{2} c_{2} b_{1}^{2} .

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