Consider the half space $\Omega=\{x=(x_1,...,x_N)\in\mathbb{R}^N:x_N>0\}$. Let $u\in C^2(\Omega)\cap C(\overline{\Omega})$ a positive bounded solution of $$ \begin{eqnarray*} \Delta u+f(x_N,u)=0, & \mbox{in} & \Omega\\ u=0, & \mbox{on} & \partial\Omega \end{eqnarray*} $$ Here $f:[0,\infty]\times[0,M]\rightarrow\mathbb{R}$, $M=\sup_{\Omega}u$, for any tfinite interval $f$ is Lipschitz in $u$ on $[0,M]$ and $$ f(x_N,u)\leq f(\infty,u)<0, \ \ \mbox{in} \ \ \Omega. $$ Then $$ \nabla u\leq C, $$ where $C$ depends only on $\max f(x_N,u)$. How can I obtain this estimate? Thank you.
1 Answer
A more general form of the problem is: $$\Delta u = g$$ with $g > 0$, $u \geq 0$ and $u(x',0) = 0$ for all $x' \in \mathbb{R}^{n1}$. Here are some things we can say:
1) If $0 < g < K$ then by $W^{2,p}$ estimates and Sobolev embedding we have an interior $C^{1,\alpha}$ estimate (for any $\alpha$), giving in particular the (scalinginvariant) estimate $$\nabla u_{B_{r/2}(x)} \leq \frac{C}{r}\u\_{L^{\infty}(B_r(x))}$$ for all $B_r(x) \subset \Omega$, where $C$ depends only on $K$.
2) If $0 \leq u \leq M$ for some $M$ apriori, then using barriers of the form $C(x(x_0',1)^{2n}1)$ for any $x_0'$ and $C$ large depending on $M$ we easily obtain $$0 \leq u \leq C(M)x_n.$$ Of course, we must have this apriori bound since quadratic growth like $x_n^2$ is allowed otherwise.
Thus, 2) gives a boundary gradient bound, and when coupled with the scaling invariant interior estimate of 1) we get a full gradient bound.
3) As a remark, if $g$ is bounded we in fact get that $u_n$ is $C^{\alpha}$ up to the boundary by a boundary Harnack inequality of Krylov, which involves using the Harnack inequality to "improve" the trapping planes from 2) at smaller scales. Philosophically, ellipticity gives that the nice boundary and boundary data have some influence when we step in.
4) As another remark, if $g \geq \delta > 0$ then $u$ must have quadratic growth. Indeed, if not, the rescalings $u_R(x) = \frac{1}{R^2}u(Rx)$ satisfy the same conditions but go to $0$ on $B_2^+$, hence the function $\frac{\delta}{2n}(xe_n^2  \frac{1}{2})$ would lie above $u_R$ in $B_2^+$ for $R$ large, contradicting that $u \geq 0$.

$\begingroup$ Connor, thank you for your patience and attention. I didn't understand some steps of your proof. First, observe that I have no information about the signal of $f$, and you assumed $g>0$. Second, could you explain me how to obtain the estimate $0\leq u\leq C(M)x_{n}$ using the barrier function? Finally, can you reexplain the steps $3$ and $4$? Thank you very much. $\endgroup$ May 21, 2014 at 22:33

$\begingroup$ Since you wrote $f(x_N,u) < 0$ it seems to me you do have an equation of the form $\Delta u = g$ with $g > 0$. $\endgroup$ May 31, 2014 at 10:48

$\begingroup$ The barriers in 2) lie above $u$ and are $0$ at $(x_0',0)$, and along $(x_0',t)$ for $t > 0$ it is easy to check that the barrier centered at $(x_0',1)$ lies below $Ct$. For the other questions feel free to shoot me an email. $\endgroup$ May 31, 2014 at 10:55