 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-3/14a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

u

G_{f s}

V

with a small ball about

r_{0}

removed. Both functions are harmonic there, so only boundary terms remain. The contribution from the small sphere tends to

u (r_{0})

because

G_{f s} (r; r_{0}) = - \frac{1}{4 π ∣ r - r _{0} ∣}

has unit delta source, while the remaining small-sphere term vanishes. Taking the radius to zero gives Green's third identity

u (r_{0}) = \int_{S} (u \frac{\partial G _{f s}}{\partial n} - G_{f s} \frac{\partial u}{\partial n}) d S .

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