 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/3/a/solution

Δ_{r} G (r, r^{'}) = δ (r - r^{'}) in V, G (r, r^{'}) = 0 on S .

The normal derivative is taken with respect to the integration variable and the outward normal of

V

\int_{V} (u Δ G - G Δ u) d V = \oint_{S} (u \partial_{n} G - G \partial_{n} u) d S .

r^{'}

. Substituting

Δ u = ρ

u = f

S

and

G = 0

there yields

u (r^{'}) - \int_{V} G (r, r^{'}) ρ (r) d V = \oint_{S} f (r) \partial_{n} G (r, r^{'}) d S .

Hence

u (r^{'}) = \int_{V} ρ (r) G (r, r^{'}) d V + \oint_{S} f (r) \partial_{n} G (r, r^{'}) d S .

One can justify applying the identity to the singular Green function by removing a small sphere around the source and then taking its radius to zero.

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