# Boundary regularity for the Monge-Ampère equation $\det D^2u=1$

Let $$\Delta$$ be an open triangle in $$\mathbb{R}^2$$ and $$u\in C^0(\overline{\Delta})\cap C^\infty(\Delta)$$ be the convex function satisfying $$\det D^2u=1,\quad u|_{\partial\Delta}=0.$$ Classical results on Monge-Ampère equations imply that there exists a unique such $$u$$. I need informations on the boundary regularity of $$u$$, but could not find any literature on regularity of such Monge-Ampère equations near a boundary line segment.

Question 1. What can be said about the boundary regularity of $$u$$? In particular, is $$\|\nabla u(x)\|$$ bounded as $$x\in\Delta$$ approaches a point on $$\partial \Delta$$ which is not a vertex?

Remarks.

1. The second question is equivalent to asking whether $$u$$ has infinite slope at a non-vertex point $$p\in\partial\Delta$$, or more precisely, whether the limit $$\lim_{t\rightarrow 0^+}\frac{u((1-t)p+tq)}{t}\in [-\infty,0)$$ equals $$-\infty$$ for some $$q\in\Delta$$ (which implies the same limit for every $$q\in \Delta$$).

2. It can be shown $$u$$ has finite slope at a vertex (namely, the above limit is finite if $$p\in\partial\Delta$$ is a vertex): assuming $$p=0$$ w.l.o.g., by Comparison Principle, $$u$$ is minorized by a convex function $$v$$ of the form $$v(x)=c|x|^\alpha-L(x),$$ where the constant $$\alpha\in (1,2)$$ is arbitrary, $$c>0$$ is a constant to make $$\det D^2v=c^2\alpha^2(\alpha-1)|x|^{2\alpha-4}$$ bigger than $$1$$ on $$\Delta$$, and $$L$$ is a linear function on $$\mathbb{R}^2$$ to make $$v|_{\partial\Delta}\leq 0$$; whereas $$v$$ has value $$0$$ and finite slope at $$x=0$$.

3. Attempting to apply Comparison Principle near non-vertex boundary points, one is lead to the following problem:

Question 2. Let $$D=\{x\in\mathbb{R}^2\mid |x|<1,\, x_1>0\}$$ be the half-disk. Is there a nonnegative convex function $$u\in C^0(\overline{D})$$ which vanishes on the boundary line segment $$\{x_1=0,\, |x_2|\leq 1\}$$, such that $$\det D^2u\geq \lambda>0$$?

I tried to construct such a $$u$$ of the form $$u(x_1,x_2)=f(x_1)g(x_2)$$ (with $$f$$ and $$g$$ nonnegative, convex and $$C^2$$) but was lead to the conclusion that such $$u$$ can never satisfy the hypotheses.

## 1 Answer

The answer to both questions is no.

For Question $$2$$: After subtracting a function of the form $$cx_1$$ we may assume that $$u \geq 0$$ and $$u(t,0) = o(t)$$. (Take $$c$$ to be the slope of the tangent line to $$u(t,0)$$ at $$t = 0$$; then $$u - cx_1 \geq 0$$ on the $$x_1$$ axis and vanishes on the $$x_2$$ axis, so by convexity is nonnegative in $$D$$). Then $$\{u < \epsilon\}$$ contains the triangle $$T_{\epsilon}$$ with vertices $$\pm e_2$$ and $$4l_{\epsilon}e_1$$ where $$\epsilon^{-1}l_{\epsilon} \rightarrow \infty$$ as $$\epsilon \rightarrow 0$$. It is straightforward to check that the paraboloid $$P_{\epsilon} = 2\epsilon(l_{\epsilon}^{-2}(x_1 - l_{\epsilon})^2 + 4x_2^2)$$ satisfies $$P_{\epsilon} > \epsilon \geq u$$ on $$\partial T_{\epsilon}$$, vanishes at $$(l_{\epsilon},0)$$, and $$\det D^2P_{\epsilon} \rightarrow 0$$ as $$\epsilon \rightarrow 0$$. For $$\epsilon$$ small this violates the comparison principle.

Question $$1$$ can be reduced to Question $$2$$ as follows: if there exists a sequence of points approaching the flat boundary where the graph of $$u$$ has supporting planes of uniformly bounded slope, then in the limit we get a supporting plane to the graph of $$u$$ with bounded slope that vanishes on one of the edges. Then subtract the plane, translate, rotate and rescale.

However, near the flat boundary I suspect there is some expansion of $$u$$ in powers of "singular building blocks," maybe of the form $$x_1(-\log x_1)^{1/2}$$ and $$x_2$$. (This is motivated by the observation that $$w := x_1(-\log x_1)^{1/2}(x_2^2-1)$$ satisfies $$0 < \lambda < \det D^2w < \Lambda$$ in the domain $$\{|x_2| < c(-\log (x_1))^{-1/2}\}$$ near the origin, which has boundary that separates from the $$x_2$$ axis slower than any polynomial).

• That's really a great argument that I should have known earlier! By the way, I need to further consider the equation with right-hand side $|x|^\alpha$ instead of $1$ (still with zero boundary value, on a triangle $\Delta$ with $0\in\mathbb{R}^2$ as a vertex) and determine whether the derivative of the solution $u$ at $0$ (along a vector pointing inside $\Delta$) is finite. I can show that it is finite if $\alpha>-2$ and infinite if $\alpha<-2$, but have trouble with the $\alpha=-2$ case. Could you maybe shed some light on this? – Xin Nie Nov 9 '18 at 23:04
• The gradient will blow up at the vertex in the case $\alpha = -2$. To see this, consider the barrier $w = -y(\log(1/y))^{1/2} + x^2(\log(1/y))^{1/2}/y$ in the quarter space $\{y > |x|\}$. A computation gives $\det D^2w = y^{-2}(1-x^2/y^2 + O(1/|\log y|))$, so $w$ is a super solution with gradient that blows up logarithmically at $0$. – Connor Mooney Nov 15 '18 at 17:57
• My computations show that $\Omega:=\{\det D^2w>0\}$ is the raindrop-shaped region $\left\{0<y<e^{1/2},\ x^2<\frac{y^2\left(\frac{1}{2}-\log y\right)}{\frac{3}{2}-\log y}\right\}\subset\{y>|x|\}$. Do you mean using $\tilde{w}=w+cy$ (with $c>0$ large enough to ensure $\tilde{w}\geq 0$ on $\partial\Omega$) as a barrier on $\Omega$? – Xin Nie Nov 16 '18 at 11:50
• That is right, thanks for doing the computation carefully- in any corner of angle $< \pi$ we can take an affine rescaling of $w$, multiply by a small constant, and add a large multiple of $y$ to obtain a super-solution that is nonnegative on the boundary of a region where $u \leq 0$ and thus show blowup of the gradient at the vertex. – Connor Mooney Nov 16 '18 at 15:08