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 Incoming links: Product rule for the Laplacian

Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 76 / 5 / Solution 2026-10-03

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 37A / b / i / Solution 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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