While I was working on some theorems in PDEs, I encountered the following boundary value problem

$$\matrix{ {{\nabla ^2}H = 0} \hfill & {{\rm{in}}} \hfill & \Omega \hfill \cr {{\nabla ^2}H = 0} \hfill & {{\rm{on}}} \hfill & {r = a} \hfill \cr {{\nabla ^2}H = 0} \hfill & {{\rm{on}}} \hfill & {z = - \ell } \hfill \cr {{\nabla ^2}H = 0} \hfill & {{\rm{on}}} \hfill & {z = \ell } \hfill \cr } $$

where $H:\mathbb{R}^3\to\mathbb{R}$ is an infinitely differentiable scalar field $C^{\infty}(\mathbb{R}^3)$. The domain $\Omega$ is a cylinder defined as

$$\Omega = \left\{ {(r,\varphi ,z)|0 \le r \lt a,0 \lt \varphi < 2\pi , - \ell \lt z \lt \ell } \right\}$$

What can we say about the uniqueness of $H$? Specifically, as the Laplacian is prescribed over the boundary so the answer is not unique. But what is the degree of non-uniqueness?

Can we simply say that $H$ can be any harmonic function?

  • $\begingroup$ Did you mean to write $H=0$ on the various surfaces? $\endgroup$ – DisintegratingByParts Jan 19 '16 at 20:14
  • $\begingroup$ @TrialAndError: No it is the Laplacian of $H$, i.e., $\nabla^2H$. I think that I made myself clear in the title. Didn't I? :) $\endgroup$ – H. R. Jan 19 '16 at 20:36
  • $\begingroup$ How are you defining $\nabla^2 H$ on the boundary? You need differentiability at points on the boundary, and that typically requires an open region. Are you suggesting that $H$ is actually defined on an open neighborhood of the closed set $\Omega$? $\endgroup$ – DisintegratingByParts Jan 19 '16 at 20:45
  • $\begingroup$ @TrialAndError: OK, take it on the open neighborhood! :) $\endgroup$ – H. R. Jan 19 '16 at 20:46
  • 2
    $\begingroup$ Now, if you assume $H$ is defined and twice continuously differentiable on an open region containing $\Omega$, then your assumption that $\nabla^2H=0$ on the boundary of the region is equivalent to the assumption that $\nabla^2H=0$ on its interior. Right? $\endgroup$ – DisintegratingByParts Jan 19 '16 at 20:51

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