 Solution

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

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

J_{i} = A_{i} P - \frac{1}{2} \frac{\partial}{\partial x _{j}} (B_{ij} P) .

Then

- \frac{\partial J _{i}}{\partial x _{i}} = - \frac{\partial}{\partial x _{i}} (A_{i} P) + \frac{1}{2} \frac{\partial ^{2}}{\partial x _{i} \partial x _{j}} (B_{ij} P),

\partial_{t} P + \nabla \cdot J = 0

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