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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.
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    $\begingroup$ Might want to not use $D$ for two different things... $\endgroup$
    – paul garrett
    Commented Oct 6, 2022 at 16:36

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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.

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  • $\begingroup$ I don't completely follow; let's say $L = 2\pi$ to simplify notation. The gradient of $u$ is $Du(x,y,z) = (y \operatorname{sin}(z),x\operatorname{sin}(z),xy\operatorname{cos}(z)).$ In particular this vanishes when $x = y = 0$, so that the whole line $x = y = 0$ is singular. $\endgroup$
    – Leo Moos
    Commented Oct 7, 2022 at 12:44
  • $\begingroup$ Of course you're right, sorry! $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 7, 2022 at 13:07
  • $\begingroup$ Next time I will think before I type. :-) Anyway, it looks like there is a simple explicit example — see the edited answer. I hope this time it is okay. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 7, 2022 at 14:46
  • $\begingroup$ Thanks, this is nice - I didn't expect a rotationally symmetric solution! To find it, did you start with the ansatz $1 - \operatorname{cos}(z)v(r,\theta)$ and then you recognised the Bessel ODE for $v$? Was it clear from the start that there has to be a solution of this form? PS: minor detail - I think you might be off by a factor of two for your $L$, because I get $\partial_z u(x,y,z) = I_0(r) \operatorname{sin}(z)$. $\endgroup$
    – Leo Moos
    Commented Oct 7, 2022 at 15:48
  • $\begingroup$ Oh, right, corrected. To find it, I just looked at harmonic functions with separated variables $z$ and $r$, and took the which has $x^2 + y^2 - 2 z^2$ as the second-order term in its Taylor expansion. (This harmonic polynomial is of course very similar to your $p(x,y,z)$.) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 7, 2022 at 20:03

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