 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/3/a/solution

F = ϕ \nabla ψ - ψ \nabla ϕ

\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 .

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

The fields and boundary are assumed sufficiently regular; a Green-function singularity is handled by excising a small ball or by using distributions.

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