= Solution
Use <separation of variables> with $\Phi=M(\mu)N(\nu)$. A nonzero product gives
$$
\frac{M''}{M}=-\frac{N''}{N}=\lambda,\qquad M''-\lambda M=0,\qquad N''+\lambda N=0.
$$
The 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> yields
$$
\boxed{\begin{aligned}
\Phi(\mu,\nu)={}&A_0+B_0\mu\\
&+\sum_{n=1}^\infty\Big\{[A_n\cosh(n\mu)+B_n\sinh(n\mu)]\cos(n\nu)\\
&\hspace{45mm}+[C_n\cosh(n\mu)+D_n\sinh(n\mu)]\sin(n\nu)\Big\}.
\end{aligned}}
$$
To 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>.
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