 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/37a/b/i/solution

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

The componentwise product rule for the Laplacian and the harmonicity of

Φ

give

\nabla^{2} (x \cdot Φ) = 2\nabla \cdot Φ + x \cdot \nabla^{2} Φ = 2\nabla \cdot Φ .

Hence

\nabla \cdot u = \nabla^{2} (x \cdot Φ + χ) - 2\nabla \cdot Φ = 2\nabla \cdot Φ + 0 - 2\nabla \cdot Φ = 0.

Thus

\nabla \cdot u = 0

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