# $\Delta u = f(u) \mbox{ in$\Omega$}\\\frac{\partial u}{\partial \vec n} = 0\mbox{ in$\partial\Omega$}$, show $f$ is constant

Let $\Omega\subset \mathbb{R}^N$ be an open with boundary class $C^1$ and connected. Let, also, $f:\mathbb{R}\to\mathbb{R}$ satisfying $tf(t)\ge 0$ for all $t\in\mathbb{R}$

Show that every $u\in C^2(\Omega)$ satisfying the problem below is constant. Find also $f$ that makes the solution $0$ everywhere

$$\Delta u = f(u) \mbox{ in \Omega}\\\frac{\partial u}{\partial \vec n} = 0\mbox{ in \partial\Omega}$$

I tried to use the green identity

$$\int_{\Omega}v\Delta u + \nabla v\cdot\nabla u\ d\sigma(y) = \int_{\partial\Omega} u(y)\frac{\partial u}{\partial \vec{n}}d\sigma(y)$$

If we take $v=1$ then $\nabla v = 0$ and since $\frac{\partial u}{\partial \vec n}$ on the boundary:

$$\int_{\Omega} \Delta u\ d\sigma(y) = \int_{\Omega} f(u)\ d\sigma(y) = 0$$

But I don't think I can take anything useful from this integral. I know that $tf(t)\ge 0$ but I don't know how it helps.

I've been thinking about this for a while. It looks like cauchy or neumann problems but I couldn't find anything useful that could help me solving this.

• Well, $t f(t)$ might lead naturally to considering $u f(u) = u \Delta u$, so perhaps a different choice of $v$ is in order. – user296602 Aug 22 '18 at 3:57

As indicated in my comment, it's natural to consider $v = u$ instead, because it leads to a natural usage of the fact about $f$. Following Green's identity, we have
$$\int_{\Omega} (u \Delta u + |\nabla u|^2) \, dV = \int_{\partial \Omega} u \frac{\partial u}{\partial \vec n} d\sigma = 0.$$
Now since the first term in the integral above is nonnegative by the assumption on $f$, we can conclude that $\nabla u = 0$ almost everywhere. One can continue from here.