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The early seminal result of Bernstein in 1914 for $n=2$ is the well-known Bernstein theorem:

The only entire solutions to the minimal surface equation in $\mathbb R^3$ are the affine functions $$u(x,y)=ax+by+c,$$ where $a, b, c\in\mathbb{R}$.

Actually, Bernstein obtained his result as an application of the so called Bernstein’s geometric theorem:

If the Gauss curvature of the graph of $u\in C^\infty(\mathbb R^2)$ in $\mathbb R^3$ satisfies $K\leq 0$ everywhere and $K<0$ at some point, then $u$ cannot be bounded.

As a corollary, Bernstein proved a very general Liouville theorem:

Suppose $u$ is a solution to the elliptic equation $$\sum_{i,j=1}^2 a_{ij}u_{ij}=0\quad\text{in }\mathbb R^2$$ such that $$|u(x)|=o(|x|) \text{ as }|x|\to+\infty.$$ Then $u$ is a constant.

Note that in the above Liouville theorem, the equation doesn't need to be uniformly elliptic, hence it is a very powerful result. What I want to know if this result has a half space version, which is like harmonic functions. More precisely, I want to obtain the following propersition:

Suppose $u\geq 0$ is a solution to the elliptic equation $$\left\{\begin{aligned}\sum_{i,j=1}^2 a_{ij}u_{ij}&=0\quad\text{in }\mathbb R^2_+,\\ u(x,0)&=0\quad \text{on }\mathbb R.\end{aligned}\right.$$ Then $u$ is a linear function of form $$u(x,y)=Ay,\quad A\geq 0.$$

Note that in the question, $a_{ij}$ could be degenerate or sigular at $\infty$.

This question is motivated by seeing Mooney's notes: The Monge-Ampère equations. He used partial Legendre transform to investigate the Liouville theorem for Monge-Ampère equation in half space, and one of steps in his proof used the similar propersition for harmonic functions, and which can be proved by boundary Harnack inequality and odd extension of $u$. But it is failed for the case without uniform ellipticity.

I have searched on the internet, and I didn't find any references about this propersition. Is this propersiton true? And if there are some references that I missed? Thanks in advance.

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1 Answer 1

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Here is a counterexample: let $$u(x,y) = e^{-x^2}\sinh(y).$$ Then $$\det D^2u = -2e^{-2x^2}(\sinh^2(y) + 2x^2) < 0 \text{ ơn } \mathbb{R}^2 \backslash \{0\},$$ and the equation $$u_{xx} + (2-4x^2)u_{yy} = 0$$ is uniformly elliptic in a neighborhood of the origin.

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  • $\begingroup$ Thanks for your answer, it’s so excited for me to have your answer. I am using your notes to study Monge-ampere equation. As for the question, I want to know if it is right for $a_{ij}$ just strictly elliptic but not uniformly elliptic. In your counterexample, the equation is degenerate at $x=\pm\sqrt{2}/2$. Is it could be true if we allow the degeneration only at $\infty$? I think it maybe right, but I have no idea how to prove it. Thank you again for your answer! $\endgroup$
    – ling
    May 15 at 2:47
  • $\begingroup$ Since $\det D^2u < 0$ outside of $B_{1/4}$, $u$ solves a locally uniformly elliptic equation on the whole plane. $\endgroup$ May 15 at 2:55
  • $\begingroup$ To clarify, the positive and negative Hessian eigenvalues of $u$ are locally bounded away from $0$ and $\pm \infty$ outside $B_{1/4}$ (where the equation is already uniformly elliptic), so one can cook up locally uniformly elliptic coefficients on the plane such that $u$ solves the corresponding equation. $\endgroup$ May 15 at 3:26
  • $\begingroup$ Oh, I understand what you mean. In your example, $a_{ij}$ could be discontinuous and we could guling coefficients to construct a locally uniformly elliptic equation. Then there comes a question is that if we can assume some conditions on $a_{ij}$ to make this problem still true, is it must be uniformly elliptic? Thank you very much for your warm answer. $\endgroup$
    – ling
    May 15 at 4:30

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