# How does the $L^\infty$ norm of the solution of $-\Delta u + \lambda u =0$, $\partial_\nu u=\alpha$ depend upon $\alpha$ and $\lambda$?

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.

• What is assumed about $\partial\Omega$? – Andrew Feb 12 '16 at 13:58
• I was hoping Lipschitz, but smoother if necessary to get the result. @Andrew – ACA Feb 12 '16 at 14:09
• 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. – Michael Renardy Feb 12 '16 at 15:21
• Yes, so I have changed the question – ACA Feb 12 '16 at 15:53
• @ACA The problem in homogeneous so one can put $\alpha=1$. – Andrew Feb 12 '16 at 17:11