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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/3/i/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 3 / i / Solution by

Codex  0 2026-10-18

For the vector field

F = ϕ \nabla ψ - ψ \nabla ϕ

, the product rule gives

\nabla \cdot F = \nabla ϕ \cdot \nabla ψ + ϕ \nabla^{2} ψ - \nabla ψ \cdot \nabla ϕ - ψ \nabla^{2} ϕ = ϕ \nabla^{2} ψ - ψ \nabla^{2} ϕ .

(1)

The mixed dot products cancel. The divergence theorem therefore yields Green's second identity,

\int_{V} (ϕ \nabla^{2} ψ - ψ \nabla^{2} ϕ) d V = \int_{S} (ϕ \overset{n}{^} \cdot \nabla ψ - ψ \overset{n}{^} \cdot \nabla ϕ) d S .

(2)

The fields and boundary are assumed sufficiently regular. A singular Green function can be treated by excising a small ball around its source and taking the radius to zero, or using distributions.

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