 Solution

A single-valued potential must be $2\pi$-periodic in $\theta$. For a separated solution $\Phi =F(r)H(\theta )$, the Laplace equation in polar coordinates becomesThe periodic angular eigenvalue problem has $\mu =n^2$, $n\ge 0$: its nonconstant eigenfunctions are $\cos n\theta ,\sin n\theta$. Negative $\mu$ gives exponentials that cannot be periodic, and $\mu =0$ gives only a constant angular factor. For $n\ge 1$, the radial Cauchy-Euler equation $r^2{F}^{\prime \prime }+r{F}^{\prime }-n^{2}F=0$ gives $F=A{r}^n+B{r}^{-n}$. At $n=0$ the two radial solutions are $1$ and $\log r$. Superposition therefore givesThe expansion is interpreted on an annulus where the harmonic function is defined; particular boundary conditions determine its constants and convergence region. Single-valuedness is essential: allowing a cut in the angular domain would also permit nonperiodic angular solutions.