 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-5/3/a/v/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

v = 1

is an admissible

H^{1} (U)

test for the Neumann Poisson problem. Its weak gradient is zero, so the weak identity immediately gives

\int_{U} f d x = 0.

This is necessary, and the preceding Hilbert-space construction proves sufficiency for

f \in L^{2} (U)

. It is the balance condition corresponding to zero total boundary flux.

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