 Solution

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/ii/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 / ii / Solution by

Codex  0 2026-10-18

Fix the evaluation point

r^{'}

inside

V

and put

ψ (r) = G (r, r^{'})

in Green's second identity. The two

k^{2}

terms cancel because

\nabla^{2} ϕ = - k^{2} ϕ

and

\nabla^{2} G = δ^{(3)} - k^{2} G

. The remaining volume integral is

\int ϕ (r) δ^{(3)} (r - r^{'}) d V = ϕ (r^{'})

. Thus

ϕ (r^{'}) = \int_{S} [ϕ (r) \partial_{n_{r}} G (r, r^{'}) - G (r, r^{'}) \partial_{n} ϕ (r)] d S .

The Dirac delta has a positive sign in this problem's Green-function equation, fixing the order of the two terms. All boundary derivatives act on the integration variable

r

.

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