Here is the problem: Let $U \subset \mathbb{R}^n$ be a connected open set with regular boundary, and $f:\mathbb{R}\to\mathbb{R}$ a function such that $tf(t)\geq0$ for all $t\in\mathbb{R}$. Show that every solution $u\in C^2(\overline{U})$ of the problem:
\begin{equation} \begin{cases} \Delta u=f(u) \hspace{0.1in} \text{in } U \\ \frac{\partial u}{\partial \hat{n}} = 0\hspace{0.1in} \text{in } \partial U \end{cases} \end{equation}
Is necessarily constant, where $\hat{n}$ denotes the unitary normal to $\partial U$. Furthermore, state an additional condition on $f$ that guarantees that $u$ vanishes identically on $U$.
My idea was to use the Divergence Theorem, which in this case implies that:
\begin{equation} \int_U \Delta u = \int_{\partial U} \frac{\partial u}{\partial \hat{n}} = 0 \end{equation}
Hence:
\begin{equation} \int_U f\circ u = 0 \end{equation}
Now, the condition on $f$ just means that $f(t)$ has the same sign as $t$, and at $0$ continuity might fail. And, since $U$ is connected, if $u$ is not constant, it assumes every value between each two distinct values it assumes.
I tried to use these facts to get a contradiction with the vanishing of the last integral, but I got stuck. Any help or hints are welcome.
I'll assume your $\Delta$ takes the sign convention $\operatorname{div}\circ\operatorname{grad}$ rather than $-\operatorname{div}\circ\operatorname{grad}$.
By Green's first identity/Divergence theorem/..., $$ \int_U (\underbrace{u\Delta u}_{=uf(u)\geq 0}+\underbrace{\lvert\nabla u\rvert^2}_{\geq 0})=\int_U\nabla\cdot(u\nabla u)=\int_{\partial U}u\frac{\partial u}{\partial n}=0, $$ so we must have $\nabla u=0$ on $U$. Can you see how to finish this and do the second part?