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Let $\lambda > 0$ be a constant and let $u$ be the weak solution on a bounded domain $\Omega$ of $$-\Delta u + \lambda u = 0 \quad\text{in $\Omega$}$$ $$\partial_\nu u = \alpha \quad \text{on $\partial\Omega$}$$ where $\partial_\nu u$ is the normal derivative and $\alpha > 0$ is a constant. Actually $u$ satisfies $$\int_\Omega \nabla u \nabla \varphi + \lambda u\varphi = \int_{\partial\Omega} \alpha \varphi$$ for all $\varphi \in H^1(\Omega)$.

In fact $u \in H^2(\Omega) \cap C^0(\bar \Omega)$ (see Salsa's book on PDEs in Action).

My question is, can $\lVert u \rVert_{L^\infty(\Omega)}$ bounded above by $\alpha$ and $\lambda$ in a simple explicit way (eg. maybe it's less than $\alpha$?)


The Salsa book gives $$\lVert u \rVert_{\infty} \leq C(\lVert \alpha\rVert_{L^q}, \lambda, \Omega)$$ but is imprecise about this constant -- and the result may not be sharp since the RHS depends on the $L^q$ norm of $\alpha$, but our $\alpha$ is much nicer than the general case.

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  • $\begingroup$ What is assumed about $\partial\Omega$? $\endgroup$ – Andrew Feb 12 '16 at 13:58
  • $\begingroup$ I was hoping Lipschitz, but smoother if necessary to get the result. @Andrew $\endgroup$ – ACA Feb 12 '16 at 14:09
  • $\begingroup$ Well, if $u\in H^2$, and the dimension is 3 or less, then $u\in L^\infty$ by Sobolev imbedding. So it seems like you answered your own question. $\endgroup$ – Michael Renardy Feb 12 '16 at 15:21
  • $\begingroup$ Yes, so I have changed the question $\endgroup$ – ACA Feb 12 '16 at 15:53
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    $\begingroup$ @ACA The problem in homogeneous so one can put $\alpha=1$. $\endgroup$ – Andrew Feb 12 '16 at 17:11

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