 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-v/2/b/i/solution

False. The constant function

u = 1

is a harmonic function and has zero normal derivative on every smooth boundary, but it is nonzero in

V

∭_{V} ∣\nabla u ∣^{2} d V = \iint_{\partial V} u \partial_{n} u d S - ∭_{V} u Δ u d V = 0.

Hence

\nabla u = 0

u

constant throughout

V

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