The Second Boundary Value Problem for a Discrete Monge–Ampère Equation

Abstract

In this work we propose a discretization of the second boundary condition for the Monge–Ampère equation arising in geometric optics and optimal transport. The discretization we propose is the natural generalization of the popular Oliker–Prussner method proposed in 1988. For the discretization of the differential operator, we use a discrete analogue of the subdifferential. Existence, unicity and stability of the solutions to the discrete problem are established. Convergence results to the continuous problem are given.

Appendix

We gave a geometric proof for Theorem 8 based on Lemma 8. Here we give an analytical proof based on infimal convolution. The epigraph of the infimal convolution illustrates with an analytical argument Lemma 8.

Let v be a continuous convex function on a closed convex set ${\widetilde{S}}$ with non empty interior. Let S denote the epigraph of v. Here S is unbounded unlike in Lemma 3. Let us consider another extension of v to $\mathbb {R}^d$ as an extended value function

$$\begin{aligned} v_{\infty }(x) = \left\{ \begin{array}{ll} v(x) &{} \text { if } x \in {\widetilde{S}}\\ +\infty &{} \text { otherwise }. \end{array} \right. \end{aligned}$$

Recall the function $k_{\varOmega ^*}$ from (22). The infimal convolution of $v_{\infty }$ and $k_{\varOmega ^*}$ is a function $v_{\infty } \square k_{\varOmega ^*}: \mathbb {R}^d \rightarrow \mathbb {R}^d \cup \{ \, +\infty \, \}$ defined as

$$\begin{aligned} v_{\infty } \square k_{\varOmega ^*}(x) = \inf _{y \in \mathbb {R}^d} v_{\infty }(y) + k_{\varOmega ^*} (x-y). \end{aligned}$$

Since $v_{\infty }(y) = +\infty$ for $y \notin {\widetilde{S}}$, we have

$$\begin{aligned} v_{\infty } \square k_{\varOmega ^*}(x) = \inf _{y \in {\widetilde{S}} } v(y) + k_{\varOmega ^*} (x-y). \end{aligned}$$

Let ${\text {epi}}u$ denotes the epigraph of a function u. Note that ${\text {epi}}v = {\text {epi}}v_{\infty }$ as $+\infty \notin \mathbb {R}$. For given functions $\phi _1$ and $\phi _2$ from $\mathbb {R}^d$ to $\mathbb {R}^d \cup \{ \, +\infty \, \}$ we have ${\text {epi}}\phi _1 + {\text {epi}}\phi _2 \subset {\text {epi}}\phi _1 \square \phi _2$. The infimal convolution is said to be exact at $x \in \mathbb {R}^d$ if there exists $y \in \mathbb {R}^d$ such that $\phi _1 \square \phi _2(x) = \phi _1(y) + \phi _2(x-y)$. If $\phi _1 \square \phi _2$ is exact at all $x \in \mathbb {R}^d$, ${\text {epi}}\phi _1 + {\text {epi}}\phi _2 = {\text {epi}}\phi _1 \square \phi _2$, [20, Lemma 2.8].

Given $x \in \mathbb {R}^d$, the function $y \mapsto v(y) + k_{\varOmega ^*} (x-y)$ is continuous on ${\widetilde{S}}$ and hence has a minimum on ${\widetilde{S}}$. Thus $v_{\infty } \square k_{\varOmega ^*}$ is exact at all points $x \in \mathbb {R}^d$ and we conclude that

$$\begin{aligned} {\text {epi}}v_{\infty } \square k_{\varOmega ^*} = {\text {epi}}v + {\text {epi}}k_{\varOmega ^*}, \end{aligned}$$

i.e. $M=S+K_{\varOmega ^*}$ where $M={\text {epi}}v_{\infty } \square k_{\varOmega ^*}$. This is essentially the content of Lemma 8.

Theorem 28

A necessary and sufficient condition for $v_{\infty } \square k_{\varOmega ^*}$ to be a convex extension of v is that $\partial v(({\widetilde{S}})^\circ ) \subset \overline{\varOmega ^*}$.

Proof

Recall that a function $\phi$ defined on $\mathbb {R}^d$ is proper if there exists $x_0 \in \mathbb {R}^d$ such that $\phi (x_0)< +\infty$ and $\phi (x) > - \infty$ for all $x \in \mathbb {R}^d$. As $v_{\infty }$ and $k_{\varOmega ^*}$ are proper convex functions, $v_{\infty } \square k_{\varOmega ^*}$ is a convex function by [19, Proposition 2.56].

Recall that $\partial k_{\varOmega ^*}(\mathbb {R}^d) = \overline{\varOmega ^*}$. Let us first assume that $v_{\infty } \square k_{\varOmega ^*}=v$ on ${\widetilde{S}}$. Then for all $x \in ({\widetilde{S}})^\circ$, $\partial v(x) = \partial v_{\infty } \square k_{\varOmega ^*} (x)$. This follows from the locality of the subdifferential c.f. [23, Exercise 1].

By [10, Proposition 16.48 (i) ], we have for $x \in ({\widetilde{S}})^\circ$, $\partial v(x) = \partial v_{\infty } \square k_{\varOmega ^*} (x) = \partial v_{\infty }(y) \cap \partial k_{\varOmega ^*} (x-y)$, where $y \in {\widetilde{S}}$ with $v_{\infty } \square k_{\varOmega ^*} (x) = v_{\infty }(y) + k_{\varOmega ^*} (x-y)$. Here $y=x$ and $\partial k_{\varOmega ^*} (0) = \overline{\varOmega ^*}$. We conclude that $\partial v(({\widetilde{S}})^\circ ) \subset \overline{\varOmega ^*}$.

Let us now assume that $\partial v(({\widetilde{S}})^\circ ) \subset \overline{\varOmega ^*}$. We show that $v_{\infty } \square k_{\varOmega ^*}$ is a convex extension of v. Let $x \in ({\widetilde{S}})^\circ$. We have $v_{\infty } \square k_{\varOmega ^*}(x)\le v(x)$. Assume by contradiction that $v_{\infty } \square k_{\varOmega ^*}(x) < v(x)$. This means that we can find $y \in {\widetilde{S}}$ such that

$$\begin{aligned} v(y)+k_{\varOmega ^*}(x-y) < v(x). \end{aligned}$$

(48)

Let now $p \in \partial v(x)$. We have $p \in \overline{\varOmega ^*}$. By definition, $v(y) \ge v(x) + p (y-x)$. Thus, by (48)

$$\begin{aligned} v(y) > v(y)+k_{\varOmega ^*}(x-y) + p \cdot (y-x). \end{aligned}$$

It follows that $p \cdot (x-y) > k_{\varOmega ^*}(x-y) = \sup _{p \in \overline{\varOmega ^*}} p \cdot (x-y)$ This contradicts $p \in \overline{\varOmega ^*}$. We conclude that $v= v_{\infty } \square k_{\varOmega ^*}$ on $({\widetilde{S}})^\circ$. Recall that v is continuous on ${\widetilde{S}}$. Also, $v_{\infty } \square k_{\varOmega ^*}$ is a proper convex function which is bounded above on ${\widetilde{S}}$, and hence continuous on ${\widetilde{S}}$, c.f. [6, Lemma 2]. It follows that $v_{\infty } \square k_{\varOmega ^*} = v$ on ${\widetilde{S}}$. $\square$