Let $D \subset \mathbf{R}^2$ be the unit disc, and $L > 0$. Let $u: D \times (-L,L) \to \mathbf{R}$ satisfy
\begin{equation}
\begin{cases}
\Delta u = 0 \quad \text{ on $D \times (-L,L)$ } \\
\frac{\partial u}{\partial z} = 0 \quad \text{ on $D \times \{ -L , L \}$}.
\end{cases}
\end{equation}


**Question.** Is it possible that the singular set $S(u) = \{ u = 0 \} \cap \{ \lvert Du \rvert = 0 \}$ is exactly $\{(0,0,0)\}$? Does the answer depend on $L$?


* Trying separation of variables is a natural reflex, but this produces $u$ with either no singularities or a much larger singular set.
* Without the Neumann-type boundary conditions on the ends of the cylinder there are examples: the harmonic polynomial $p(x,y,z) = 2x^3 - 3xy^2 - 3xz^2$ is one.



Is there a harmonic function with just one singular point?
**Yes**, this is possible. An explicit example is
$$u(x, y, z) = 1 - I\_0\left(\sqrt{x^2 + y^2}\right) \, \cos z$$
when $L = \pi$, and $u\big(\frac{\pi x}{L}, \frac{\pi x}{L}, \frac{\pi x}{L}\big)$ for a general $L$. Here $I\_0$ is the Bessel $I$ function.


