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Aug 11, 2022 at 9:03 comment added node @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.
Aug 7, 2022 at 21:34 comment added Willie Wong @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.
Aug 6, 2022 at 19:31 comment added Math604 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)
Aug 5, 2022 at 10:54 history edited node
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Aug 5, 2022 at 10:40 history asked node CC BY-SA 4.0