 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/3/b/i/solution

Put

s = ∣ r - r^{'} ∣

. Away from the source, the radial Laplacian in three dimensions gives

Δ_{r} G_{0} = \frac{1}{s ^{2}} \frac{d}{d s} (s^{2} \frac{d}{d s} [- \frac{1}{4 π s}]) = \frac{1}{s ^{2}} \frac{d}{d s} \frac{1}{4 π} = 0.

Thus

Δ G_{0} = 0 for r \neq = r^{'}

. At the source it is a distribution, not a smooth function; its outward gradient flux through a small sphere is

4 π s^{2} / (4 π s^{2}) = 1

Δ G_{0} = δ

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