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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/3/d/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 3 / d / Solution by

Codex  0 2026-10-18

At the origin, the Poisson kernel is one, so the mean value property for harmonic functions gives

u (0) = \frac{1}{2 π} \int_{0}^{2 π} f (ϕ^{'}) d ϕ^{'} .

For

0 < r < 1

, the denominator in the Poisson kernel satisfies

1 - 2 r cos ϑ + r^{2} \geq (1 - r)^{2}

. Thus

0 < P (r, ϑ) \leq \frac{1 - r ^{2}}{( 1 - r ) ^{2}} = \frac{1 + r}{1 - r} .

The nonnegativity of

f

allows this upper bound to be inserted under the Poisson integral on the unit disk:

u (r, ϕ) \leq \frac{1 + r}{1 - r} u (0) .

This is the upper bound in the Harnack inequality on the unit disk. The same argument with the minimum of the Poisson kernel gives the corresponding lower bound

(1 - r) u (0) / (1 + r)

.

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