Is there a solution to the following problem? $-\Delta u = 1$ in $\Omega$ and $\frac{\partial u}{\partial \nu} = 0$ on $\partial \Omega$. where $\Omega$ is bounded.
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1$\begingroup$ Try it on an interval first? $\endgroup$– Otis ChodoshCommented May 30, 2012 at 14:37
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$\begingroup$ Look up "strong maximum principle". For example around page 33 or 34 of Gilbarg and Trudinger, Elliptic Partial Differential Equations of Second Order $\endgroup$– Willie WongCommented May 30, 2012 at 14:50
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1 Answer
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I $\Omega$ is a bounded domain, the answer is NO. Because of $$\int_{\Omega}\Delta v dx=\int_{\partial\Omega}\frac{\partial v}{\partial\nu}ds.$$
answered May 30, 2012 at 14:50