Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 12 / 5 / i

Solution

Let

B_{r} (x_{0})

have closure contained in the domain of a harmonic function

u

. Put

ω_{n} = ∣ B_{1} (0) ∣

, so

∣ B_{r} ∣ = ω_{n} r^{n}

and

∣ \partial B_{r} ∣ = n ω_{n} r^{n - 1}

u (x_{0}) = \frac{1}{∣ \partial B _{r} ∣} \int_{\partial B_{r} (x_{0})} u d S = \frac{1}{∣ B _{r} ∣} \int_{B_{r} (x_{0})} u d x .

M (r) = \frac{1}{n ω _{n}} \int_{S^{n - 1}} u (x_{0} + r θ) d S_{θ} .

Differentiating under the integral and applying the divergence theorem gives

M^{'} (r) = \frac{1}{n ω _{n} r ^{n - 1}} \int_{\partial B_{r} (x_{0})} \partial_{ν} u d S = \frac{1}{n ω _{n} r ^{n - 1}} \int_{B_{r} (x_{0})} Δ u d x = 0.

Continuity gives

M (r) \to u (x_{0})

r ↓ 0

, proving the spherical average identity. Integrating that identity over radii with the polar-coordinate weight yields

\int_{B_{r} (x_{0})} u = \int_{0}^{r} n ω_{n} t^{n - 1} M (t) d t = ω_{n} r^{n} u (x_{0}),

which proves the ball identity. For

n = 1

, the sphere average is the average of the two endpoints, and the same formulas hold with the counting surface measure.

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