 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/13c/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Assume again that all boundary terms vanish. Multiplying the Fokker-Planck equation by

x_{k}

and integrating by parts gives

\frac{d}{d t} ⟨ x_{k} ⟩ = - \int x_{k} \partial_{i} (A_{i} P) d V + \frac{1}{2} \int x_{k} \partial_{i} \partial_{j} (B_{ij} P) d V = \int δ_{ik} A_{i} P d V = ⟨ A_{k} ⟩ .

Similarly,

\partial_{i} (x_{k} x_{l}) = δ_{ik} x_{l} + δ_{i l} x_{k}

and

\partial_{i} \partial_{j} (x_{k} x_{l}) = δ_{ik} δ_{j l} + δ_{i l} δ_{jk} .

Because

B

is symmetric,

\frac{d}{d t} ⟨ x_{k} x_{l} ⟩ = ⟨ x_{l} A_{k} + x_{k} A_{l} + B_{k l} ⟩ .

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