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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-107/1/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 107 / 1 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

For the spherical average

M (r)

, rescaling the sphere and applying the divergence theorem gives

M (r) = \frac{1}{n ω _{n}} \int_{S^{n - 1}} u (y + r θ) d S_{θ}, M^{'} (r) = \frac{1}{n ω _{n} r ^{n - 1}} \int_{\partial B_{r} (y)} \partial_{ν} u = \frac{1}{n ω _{n} r ^{n - 1}} \int_{B_{r} (y)} Δ u = 0.

M (r) \to u (y)

r ↓ 0

. Integrating these spherical averages in radius proves the mean value property for harmonic functions:

u (y) = \frac{1}{n ω _{n} r ^{n - 1}} \int_{\partial B_{r} (y)} u = \frac{1}{ω _{n} r ^{n}} \int_{B_{r} (y)} u .

For the strong maximum principle for harmonic functions, suppose an interior point attains the maximum

m

on a connected domain. The continuous nonnegative function

m - u

has zero average on each sufficiently small ball around that point, so it vanishes there. The maximum set is therefore open as well as closed, and connectedness makes it the entire domain. Apply the same argument to

- u

for a minimum.

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