 Solution

Use separation of variables with $\Phi =M(\mu )N(\nu )$. A nonzero product givesThe angular factor must be a periodic function of period $2\pi$. If $\lambda <0$, its exponential solutions cannot be periodic unless they vanish. If $\lambda =0$, $N=c+d\nu$ forces $d=0$, while $M=A_0+B_0\mu$. If $\lambda >0$, periodicity requires $\sqrt{\lambda }=n\in \mathbb{N}$, giving angular factors $\cos (n\nu ),\sin (n\nu )$ and radial factors $\cosh (n\mu ),\sinh (n\mu )$.

Superposing these separated harmonic functions yieldsTo see completeness, expand an arbitrary smooth harmonic function in a Fourier series in $\nu$ on each ellipse. Substitution into the equation forces every Fourier coefficient to obey the corresponding radial equation above. The series is interpreted with the convergence needed for a harmonic function, for example locally on compact subannuli. No boundedness condition at infinity or extension across $\mu =0$ has been imposed, so there is no reason to discard either radial branch in these elliptic coordinates.