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Let $\Omega \subset \mathbb{R}^{3}$ be a $C^{1}$-domain, not necessarily bounded. Consider solutions $\phi : \overline{\Omega} \to \mathbb{R}$, $\phi \in C^{\infty}(\Omega) \cap C^{1}(\overline{\Omega})$, $\nabla\phi \in L^{2}(\Omega)$, to the Neumann problem \begin{align} \Delta \phi (x) &= 0 &&\text{ in } \Omega \\ \nabla_{\! n} \phi(x) &= g(x) &&\text{ in } \partial \Omega \end{align} I want to prove that $\nabla \phi$ is unique in this case.

The standard approach to proving this in bounded domains is to take two solutions to the above problem, say $\phi$ and $\psi$, and investigate the convergent integral \begin{align} I = \int_{\Omega} \lvert \nabla \phi - \nabla \psi \rvert^{2} \, d^{3}x. \end{align} Using the divergence theorem, we find that \begin{align} I = \int_{\partial \Omega} (\phi - \psi) (\nabla_{\! n} \phi - \nabla_{\! n} \psi) \, d^{2}x = 0, \end{align} such that we must have $\nabla \phi = \nabla \psi$ in $\Omega$, and by continuity, in $\overline{\Omega}$.

Is this proof valid when $\Omega$ is unbounded?

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    $\begingroup$ consider the case of $\Omega= R \times (0,\pi)$ with $ \partial_\nu \phi=0$ on $ \partial \Omega$. You can find nonzero solutions; if you impose some bounds on the solution then you can prove $ \phi=0$ is the only solution. (I mean a constant is the only solution) $\endgroup$
    – Math604
    Commented Aug 6, 2022 at 19:31
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    $\begingroup$ @Math604: i interpret the OP's question to be precisely whether $\nabla \phi \in L^2(\Omega)$ is a strong enough bound to guarantee uniqueness. $\endgroup$
    – Willie Wong
    Commented Aug 7, 2022 at 21:34
  • $\begingroup$ @WillieWong: Yes, that is what I'm asking. When $\Omega$ is bounded, the above proof holds. When $\Omega$ is unbounded with compact boundary, then one can also apply the above reasoning by taking a sufficiently large ball, integrating, and letting the radius of the ball approach infinity. This utilizes some further results on the decay of harmonic functions. For general unbounded domains, I am however unsure how this would work. $\endgroup$
    – node
    Commented Aug 11, 2022 at 9:03

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