 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/2/b/solution

On the boundary, integrate the expansion from part (a) over one angular period. Every nonconstant sine and cosine has zero average, so

\int_{0}^{2 π} f (ϕ) d ϕ = 2 π A_{0} .

All positive-power radial terms vanish at

r = 0

, leaving

Ψ (0) = A_{0}

. Hence

Ψ (0) = \frac{1}{2 π} \int_{0}^{2 π} f (ϕ) d ϕ = \frac{1}{2 π R} \int_{r = R} f d s .

This is the mean value property for harmonic functions. It is an average over arclength as well as over angle, since

d s = R d ϕ

. The usual continuity or integrable-boundary-data hypotheses justify taking this angular average.

 New to topics? Read the docs here!