Existence and nonexistence to exterior Dirichlet problem for Monge–Ampère equation

Abstract

We consider the exterior Dirichlet problem for Monge–Ampère equation with prescribed asymptotic behavior. Based on earlier work by Caffarelli and the first named author, we complete the characterization of the existence and nonexistence of solutions in terms of their asymptotic behaviors.

Appendix

In this “Appendix”, we prove the following lemma.

Lemma 3.5

Let u, v be two viscosity solutions of (3.1), then for any $0<\alpha <1$, $\alpha u+(1-\alpha )v$ is a viscosity subsolution of (3.1).

Proof

To begin with, let us recall the definition of $\epsilon$-upper envelope of u, see e.g. [15],

$$\begin{aligned} u^\epsilon (x):=\max _{y\in \bar{\Omega }}\{u(y)-\frac{1}{\epsilon }|y-x|^2\},\quad x\in \bar{\Omega }. \end{aligned}$$

Then

$$\begin{aligned} u^\epsilon \rightarrow u, \quad \epsilon \rightarrow 0^+, \end{aligned}$$

in $C^0_{loc}(\Omega )$.

And $u^\epsilon$ is a viscosity subsolution of (3.1).

Moreover $u^\epsilon$ is second order differentiable almost everythere and

$$\begin{aligned} D^2u^\epsilon \ge -\frac{2}{\epsilon }I,\quad a.e. \quad in\quad \Omega . \end{aligned}$$

Let $\bar{x}\in \Omega$ be a point where $u^\epsilon$ is second order differentiable, say $\bar{x}=0$. For $\delta >0$ small, define

$$\begin{aligned} \phi (x)=u^\epsilon (0)+\nabla u^\epsilon (0)x+\frac{1}{2}x^\prime (D^2 u^\epsilon (0)+\delta ) x. \end{aligned}$$

Then $\phi (x)\ge u^\epsilon (x)$ near 0 and $\phi (0)=u^\epsilon (0)$.

Since $u^\epsilon$ is a viscosity subsolution, we have $\lambda (D^2\varphi )(0)\in \bar{V}$. Sending $\delta$ to 0, we have $\lambda (D^2u^\epsilon )(0)\in \bar{V}$. Thus $\lambda (D^2u^\epsilon )\in \bar{V}$ a.e.

Similarly, $\lambda (D^2v^\epsilon )\in \bar{V}$ a.e., where $v^\epsilon$ is the $\epsilon$-upper envelope of v.

Define

$$\begin{aligned} w^\epsilon _\alpha =\alpha u^\epsilon +(1-\alpha )v^\epsilon . \end{aligned}$$

Since $\bar{V}$ is convex, it follows that $\lambda (D^2 w^\epsilon _\alpha )\in \bar{V}$ a.e.

We now prove that $w^\epsilon _\alpha$ is a viscosity subsolution in $\Omega$.

Let $\bar{x}\in \Omega$ be an arbitrary point, say $\bar{x}=0$. For any $\phi \in C^2$, such that $\phi (0)=w^\epsilon _\alpha (0)$ and $\phi (x)\ge w^\epsilon _\alpha (x)$ for x near 0.

For $\delta >0$ small, define $\phi _\delta (x)=\phi (x)+\delta |x|^2$, then

$$\begin{aligned} \phi _\delta (x)\ge w^\epsilon _\alpha (x)+\delta ^3,\quad for\quad |x|=\delta . \end{aligned}$$

Consider

$$\begin{aligned} \xi (x)=\phi _\delta (x)-w^\epsilon _\alpha (x)-\delta ^4. \end{aligned}$$

Then $\xi (0)=-\delta ^4$ and $\xi (x)> 0$ for $|x|=\delta$. Since $D^2u^\epsilon \ge -\frac{2}{\epsilon }I$, $D^2v^\epsilon \ge -\frac{2}{\epsilon }I$ almost everywhere, we have $D^2\xi \le C(\epsilon )I$ almost everywhere. It follows from the Alexandrov-Bakelman-Pucci inequality that

$$\begin{aligned} \delta ^4\le C\left( \int _{\{\xi =\Gamma _\xi \}}\det (D^2\Gamma _\xi )\right) ^{\frac{1}{n}}, \end{aligned}$$

where $\Gamma _\xi$ is the convex envelope of $\xi$.

As in the previous section, there exists some $x\in \{\xi =\Gamma _\xi \}\cap B_\delta (0)$, where $w^\epsilon _\alpha (x)$ is second order differentiable, $\lambda (D^2w^\epsilon _\alpha )(x)\in \bar{V}$, and $D^2\xi (x)\ge 0$, i.e.

$$\begin{aligned} D^2\phi _\delta (x)\ge D^2w^\epsilon _\alpha (x). \end{aligned}$$

It follows that $\lambda ( D^2\phi _\delta )(x)\in \bar{V}$. Sending $\delta$ to 0, we have $\lambda ( D^2\varphi )(0)\in \bar{V}$. Thus $w^\epsilon _\alpha$ is a viscosity subsolution. Since $w^\epsilon _\alpha \rightarrow \alpha u+(1-\alpha )v$ in $C^0_{loc}(\Omega )$, sending $\epsilon$ to 0, it follows that $\alpha u+(1-\alpha )v$ is a viscosity subsolution.