Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 30E / Solution

For the disc, put $u(r,\phi )=R_0(r)\Theta (\phi )$. The polar Laplace equation separates aswhere periodicity requires an integer $n\ge 0$. The radial ansatz $R_0=r^{\alpha }$ gives $\alpha =\pm n$; for $n=0$ the solutions are a constant and $\log r$. Regularity at the origin excludes negative powers and the logarithm.

Writing the boundary data as a Fourier serieswith $a_0=(2\pi {)}^{-1}{\int }_0^{2\pi }{u}_D$, $a_n={\pi }^{-1}{\int }_0^{2\pi }{u}_D\cos n\phi$, and the analogous formula for $b_n$, givesEquivalently it is the Poisson integral on the unit disk with kernelFor continuous boundary data this integral converges to the prescribed value at every boundary point, is smooth and harmonic inside, and uniqueness follows from the maximum principle for harmonic functions.

For the square, superpose four solutions, each carrying one side's data and zero data on the other three sides. Set $k_n=n\pi /a$ andSeparation of variables gives the explicit sumEach summand has zero Laplacian and the required zero values on three open sides. The Fourier sine series supplies the fourth side. For regular continuous data with matching values at the corners, the harmonic extension also takes the common corner values; termwise boundary evaluation of the sine series at the corners is not valid in general. For incompatible corner data there is no continuous solution on the closed square, but the displayed harmonic solution still attains the prescribed data on each open side.