Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 2

For the harmonic extension in a circular sector, write a separated harmonic function as $F(r)H(\theta )$. The Laplace equation in polar coordinates yieldsThe angular Dirichlet boundary conditions select the Fourier sine basisIndeed, the angular equation has no nonzero solution with nonpositive ${\nu }^2$: multiplying $-H^{\prime \prime }={\nu }^{2}H$ by $H$ and integrating gives $\int (H^{\prime }{)}^2={\nu }^2\int H^2$. The radial Cauchy-Euler equation has solutions $r^{{\nu }_n}$ and $r^{-{\nu }_n}$. Boundedness at the vertex excludes the latter, so superposition gives

The Fourier orthogonality relation on this angular interval isProjecting the outer boundary condition onto each angular eigenfunction givesThe solution is unique in the class of bounded harmonic functions with the prescribed boundary trace. For continuous boundary data, the difference of two solutions is zero on the boundary and bounded at the vertex, so the maximum principle for harmonic functions makes it zero throughout the sector. Equivalently all its Fourier sine series coefficients vanish. The usual convergence interpretation applies if the data are only piecewise smooth.

Reflecting the solution across $\theta =\pi \alpha$ preserves its Dirichlet boundary conditions and outer boundary data, so uniqueness makes the solution reflection-symmetric. Sinceonly odd angular Fourier modes survive. Put $\rho =(r/R{)}^{1/(2\alpha )}$ and $\psi =\theta /(2\alpha )$. On the symmetry axis $\psi =\pi /2$, comparison of the convergent power series givesThusThis is the imaginary part of $\sin (\rho e^{i\psi })$, so a useful closed form isAt either straight edge $\sin \psi =0$, while on the symmetry axis the formula is $\sinh \rho$. Every term is harmonic in the sector, and the factorial denominator gives convergence up to $r=R$, verifying the full solution.