Question about boundness of derivatives.

My doubt is in the paper: Further qualitative properties for elliptic equations in unbounded domains, by Berestycki, Caffarelli and Niremberg (page: 93)

My question is simples. For any direction $\xi\in S^1$ we have $$\Delta\frac{\partial u}{\partial\xi}+f'(u(x))\frac{\partial u}{\partial\xi}=0, \ \ \ \mbox{in} \ \ \mathbb{R}^2.$$ We consider a positive bounded function $u\in C^2(\mathbb{R}^2)$. Why all the derivatives $\frac{\partial u}{\partial\xi}$ are bounded in the plane? In the paper, he just claim (by elliptic theory).

Thank you.

• is there any info about $f'(u(x))$, is it non-positive? – Yimin Feb 24 '13 at 8:31
• I just know that $f$ is a $C^1$ positive function. – José Carlos Feb 25 '13 at 17:17

The paper assumes that $$\Delta u + f(u) = 0$$ and you're given that $u$ is uniformly bounded, and $f$ is $C^1$, gives you a good place to start.
Consider any unit ball, and on that ball, you first have that, since $u$ is bounded, $\Delta u = -f(u)$ is uniformly bounded.
Elliptic estimates (your choice of potential theory or weak solutions or any other you like) then tell you that $u$ is uniformly bounded in $C^1$ on a smaller ball, say $B_{\frac{1}{2}}$. Since the ball was arbitrary, this estimate on the $C^1$ norm of $u$ applies at any point.
• Could you show me this elliptic estimate? I was thinking about the elliptic estimate of the Poission equation (Chapter 4 - Gilbarg Trudinger), but it needs the distance of $B$ to $\partial\Omega$, and my set $\Omega$ is unbounded. Can you show the estimate that I can use? Thank you very much one more time @Ray Yang. – José Carlos Feb 26 '13 at 3:55
• Jose, the argument, if you want to use Newtonian potentials, is not explicitly in Gilbarg-Trudinger, but what you do is you work to see that, if , say $w$ is the Newtonian potential for $f(u)$ restricted to a ball, that is to say $w(x) = \frac{-1}{2\pi} \int_{\mathbb{R}^2} \log |x-y| (f(u(y)\chi_B(y)) dy$, then the derivatives of $w$ are bounded in terms of the bound on $f(u)$, and so is $w$ itself, that is to say, $w$ is bounded in $C^1$. The difference on $B$ between $w$ and $u$ is thus a harmonic function, so $|Dw - Du|$ can be bounded on $B_{\frac{1}{2}}$, and hence so can $u$. – Ray Yang Feb 26 '13 at 12:38