Suppose $u \in C^2(\Omega)\cap C^1(\bar\Omega)$ satisfies
$$ -\Delta u+u^3=0 \quad\text{ in } \Omega $$ $$ \frac{\partial}{\partial\eta}u+\alpha u=\varphi \quad\text{ on }\partial \Omega $$ for continous functions $\alpha$ and $\varphi$ on $\Omega$. With $\alpha(x) \geq \alpha_0>0$. Prove that $$\sup_{\Omega}u \leq \frac{1}{\alpha_0}\sup_{\partial \Omega}|\varphi|$$
where $\eta$ is the unite outward normal vector field along $\partial \Omega$.
On
The proof is simply based on the maximum principle. We note that operator $$L:=-\Delta+u^2\cdot$$ is strictly elliptic in $\Omega$, and $Lu\geq(=)0$ in $\Omega$. Then, the maximum principle yields that $u$ achieves its maximum at some boundary point $x_0\in\partial\Omega.$ At $x_0,$ we have $\partial_\nu u\geq0$, where $\nu$ is the unit outward normal vector of $\partial\Omega$ at $x_0.$ Thus, the boundary value condition at $x_0$ yields the desired estimate.
Let $\,\Omega\,$ be a bounded domain with $\,\partial\Omega\in C^2$, and let $\,u\in C^2(\Omega)\cap C^1(\overline{\Omega})\,$ be any solution of the nonlinear boundary value problem $$ \begin{cases} -\Delta u+u^3=0, \quad x\in\Omega\subset\mathbb{R}^n,\;n\geqslant 2,\\ \bigl(\frac{\partial u}{\partial\nu}+\alpha u\bigr)\bigr|_{\partial\Omega}= \varphi,\quad \alpha,\varphi\in C(\partial\Omega), \end{cases}\tag{1} $$ with $\,\nu\,$ being an outward unit normal to $\,\partial\Omega$, where $\,\alpha_0\overset{\rm def}{=}\inf\limits_{\partial\Omega}\alpha>0$. Then $$ \sup_{\Omega} u\leqslant \sup_{\partial\Omega}\frac{|\varphi|}{\alpha}. \tag{2} $$ To prove $(2)$, notice that solution $\,u\in C^2(\overline{\Omega})\,$ since $\,\Delta u=u^3\in C^1(\overline{\Omega})$. Multiplying equation $(1)$ by $\,v\in H^1(\Omega)$, and integrating by parts yields $$ \int\limits_{\Omega}\bigl(\nabla u,\nabla v\bigr)\,dx+ \int\limits_{\Omega}u^3 v\,dx+\int\limits_{\partial\Omega} (\alpha u-\varphi)v\,ds=0\quad \forall\, v\in H^1(\Omega)\tag{3} $$ Proof of $(2)$ is based on employing a one parameter function $\,\eta_{m}\in C^1(\mathbb{R})\,$ of the form $$ \eta_{m}(\xi)= \begin{cases} 0,\quad \xi\leqslant m,\\ \xi-m,\quad \xi>m. \end{cases} $$ It is important that $\,\eta_m\,$ is Lipschitz on $\mathbb{R}$. Now choose $$ m\overset{\rm def}{=}\sup_{\partial\Omega} \frac{|\varphi|}{\alpha}\geqslant 0 $$ and notice that for any $\,u\in C^1(\overline{\Omega})\,$, a function $\,\eta_m(u)\in H^1(\Omega)\,$ since it is subject to the chain rule $\,\nabla \eta_m(u)=\eta'_m(u)\nabla u\,$ while $\,\Omega\,$ is bounded. Choosing the test function $\,v=\eta_m(u)\,$ in $(3)$ yields $$ \int\limits_{\Omega}|\nabla u|^2\eta'_m(u)\,dx+ \int\limits_{\Omega}u^3\eta_m(u)\,dx+ \int\limits_{\partial\Omega}(\alpha u-\varphi)\eta_m(u)\,ds=0. $$ It is clear that $$ |\nabla u|^2\eta'_m(u)\geqslant 0,\quad u^3\eta_m(u)\geqslant 0, \quad (\alpha u-\varphi)\eta_m(u)\geqslant 0 $$ on $\,\Omega$, hence $$ |\nabla u|^2\eta'_m(u)=u^3\eta_m(u)=(\alpha u-\varphi)\eta_m(u)=0 $$ on $\,\Omega$. Multiplying equality $\,|\nabla u|^2\eta'_m(u)=0\,$ by $\,\eta'_m(u)\,$ results in equality $\,|\nabla\eta_m(u)|^2=0$, whence follows $\,\nabla\eta_m(u)=0\,$ on $\,\Omega$, i.e., $\,\eta_m(u)= C=const\,$ on $\,\Omega$. Hence $\,u^3 C=0\,$ on $\,\Omega$. Should $\,C\neq 0$, equality $\,u=0\,$ on $\,\Omega\,$ would imply $\,\varphi=0\,$ on $\,\partial\Omega$, and there would be nothing to prove. Otherwise, $C=0\,$ which immediately implies $\,\eta_m(u)=0\,$ on $\,\Omega$, i.e., $\,u\leqslant m\,$ on $\,\Omega$. Q.E.D.
Remark. In case of a bounded domain $\,\Omega\subset\mathbb{R}^n\,$ is Lipschitz, basically the same argument results in the inequality $$ \underset{\Omega}{\rm ess\,sup}\, u\leqslant \underset{\partial\Omega}{\rm ess\,sup}\,\frac{|\varphi|}{\alpha} $$ for any weak solution $\,u\in W^{1,p}(\Omega)\,$ of the nonlinear problem $(1)$, where $p\geqslant \max\{2, \frac{4n}{n+4}\}$, $n\geqslant 2$, while essential suprema are taken w.r.t. the $n$-dimensional or $(n-1)$-dimensional Lebesgue measure accordingly.