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Consider Poisson equation with Neumann boundary condition but the right hand side of boundary condition is in term of the unknown function $u$. How we can solve it?

$\Delta u(x) = f(x)\quad in~ \Omega$

$\frac{\partial u(x)}{\partial n }=g(u(x))\quad on~\partial \Omega$

where n is outward normal vector.

For special case let $g=\sqrt u$.

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Neumann not Newman – mathphysicist Oct 8 '10 at 23:52
Where does this come up? Do you have a general form for $g$? (Also, I would hesitate calling this a Neumann BC problem. If $g$ were a linear function of its argument, then this would be a classic Robin type boundary condition.) – Willie Wong Oct 9 '10 at 0:01
The answer depends very much on the form of $g$. For example if $g=0$ then solutions will exist only if $\int_{\Omega} f dx = 0$. If $g(u)=u$ then as Willy pointed out this is a Robin condition and by Fredholm there will exist a unique solution for each $f$. For general $g$ you will need again some sort of compatibility condition such as $\int_{\Omega} f(x) dx = \int_{\partial \Omega} g(u(x))$. I think this should only make sense if $g$ is continuous. Anyway the short answer to this quesiton is that it depends very much on the form of $g$ and so I think one needs to reduce it to cases. – Dorian Oct 9 '10 at 2:17
up vote 1 down vote accepted

Set $G$ a primitive of $g$. Then the solution is a critical point of the functional $E:H^1(\Omega)\rightarrow{\mathbb R}$ defined by $$E[u]:=\int_\Omega \left(\frac12|\nabla u|^2+fu\right)dx-\int_{\partial\Omega}G(u)ds.$$ If $G(\pm\infty)=-\infty$, you may look for a minimum of $E$ over $H^1(\Omega)$. When $g$ is a decreasing function, $G$ is concave and $E$ is strictly convex: your solution is unique.

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Hi, Could you introduce me the references of your answer? Best – Reza Oct 9 '10 at 11:46
An extremely well-known reference is in French. J.-L. Lions, Quelques méthodes de résolution des problèmes aux limites non linéaires. Dunod; Gauthier-Villars, Paris 1969 – Denis Serre Oct 9 '10 at 12:19
No Ref. in English? – Reza Oct 9 '10 at 13:23
There is still the issue of coercitivity is there not? I think one would need in addition some sort of lower bound on $G$ of the form $|G| \geq C|$ (maybe that can be improved a bit). – Dorian Oct 9 '10 at 13:43
@Dorian. Coercivity is ensured by $G(\pm\infty)=-\infty$. Of course, if $g$ is decreasing but either of $G(\pm\infty)$ is bounded, there might be a problem for the existence. – Denis Serre Oct 9 '10 at 17:12

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