Mean field games with state constraints: from mild to pointwise solutions of the PDE system

Abstract

Mean Field Games with state constraints are differential games with infinitely many agents, each agent facing a constraint on his state. The aim of this paper is to provide a meaning of the PDE system associated with these games, the so-called Mean Field Game system with state constraints. For this, we show a global semiconvavity property of the value function associated with optimal control problems with state constraints.

Appendix: proof of Lemma 2.1

The idea of the proof is based on [11, Theorem 4.5]. Let $x\in \partial \Omega$ and let $\nu (x)$ be the outward unit normal to $\partial \Omega$ in x. Let $\theta \in {\mathbb {R}}^n$ be such that $\langle \theta ,\nu (x)\rangle \le 0$. Let us set

$$\begin{aligned} M(\theta ,x)=\min _{ p\in D^+u(x)}\langle p,\theta \rangle . \end{aligned}$$

It suffices to prove that

$$\begin{aligned} \limsup _{\begin{array}{c} h\rightarrow 0^+\\ \theta '\rightarrow \theta \\ x+h\theta '\in {\overline{\Omega }} \end{array}} \frac{u(x+h\theta ')-u(x)}{h}\le M(\theta ,x)\le \liminf _{\begin{array}{c} h\rightarrow 0^+\\ \theta '\rightarrow \theta \\ x+h\theta '\in {\overline{\Omega }} \end{array}} \frac{u(x+h\theta ')-u(x)}{h}. \end{aligned}$$

(5.1)

The first inequality in (5.1) is straightforward. Indeed, for any $p\in D^+u(x)$,

$$\begin{aligned} \limsup _{\begin{array}{c} h\rightarrow 0^+\\ \theta '\rightarrow \theta \\ x+h\theta '\in {\overline{\Omega }} \end{array}} \frac{u(x+h\theta ')-u(x) - \langle p,h\theta '\rangle }{h}\le 0. \end{aligned}$$

So,

$$\begin{aligned} \limsup _{\begin{array}{c} h\rightarrow 0^+\\ \theta '\rightarrow \theta \\ x+h\theta '\in {\overline{\Omega }} \end{array}}\frac{u(x+h\theta )-u(x)}{h}\le \langle p,\theta \rangle , \ \ \ \forall \ p\in D^+u(x). \end{aligned}$$

In order to prove the last inequality in (5.1), pick sequences $h_k\rightarrow 0$ and $\theta _k\rightarrow \theta$ such that $x+h_k\theta _k\in {\overline{\Omega }}$ and

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{u(x+h_k\theta _k)-u(x)}{h_k}=\liminf _{\begin{array}{c} h\rightarrow 0^+\\ \theta '\rightarrow \theta \\ x+h\theta '\in {\overline{\Omega }} \end{array}}\frac{u(x+h\theta ')-u(x)}{h}. \end{aligned}$$

(5.2)

Let us define

$$\begin{aligned} {\mathcal {Q}}(x,\theta _k)=\Big \{x'\in \Omega : \langle x'-x,\theta _k\rangle >0, |\langle x'-x,\theta _k\rangle \theta _k-(x'-x)|\le |x'-x|^2\Big \}. \end{aligned}$$

We observe that the interior of ${\mathcal {Q}}(x,\theta _k)$ is nonempty. Since u is Lipschitz there exists a sequence $x_k$ such that

• (i) $x_k\in {\mathcal {Q}}(x,\theta _k)$, $x_k\rightarrow x$ as $k\rightarrow \infty$;

• (ii) u is differentiable at $x_k$ and there exists ${\overline{p}}\in D^+u(x)$ such that $D u(x_k)\rightarrow {\overline{p}}$ as $k\rightarrow \infty$;

• (iii) $|s_k-h_k|\le h_k^2$, where $s_k=\langle x_k-x,\theta _k\rangle$.

By the Lipschitz continuity of u, we note that (iii) yields

$$\begin{aligned}&\Big | \frac{u(x+h_k\theta _k)-u(x)}{h_k}-\frac{u(x+s_k\theta _k)-u(x)}{s_k}\Big |\le \frac{|u(x+h_k\theta _k)-u(x+s_k\theta _k)|}{h_k}\\&\quad +\left| \frac{1}{h_k}-\frac{1}{s_k}\right| \big [|u(x+s_k\theta _k)-u(x)|\big ]\le 2\mathrm{Lip}(u) h_k. \end{aligned}$$

So, by (5.2) we have that

$$\begin{aligned} \lim _{k\rightarrow \infty } \frac{u(x+s_k\theta _k)-u(x)}{s_k}=\liminf _{\begin{array}{c} h\rightarrow 0^+\\ \theta '\rightarrow \theta \\ x+h\theta '\in {\overline{\Omega }} \end{array}}\frac{u(x+h\theta ')-u(x)}{h}. \end{aligned}$$

(5.3)

Moreover,

$$\begin{aligned}&u(x+s_k\theta _k)-u(x)=[u(x+s_k\theta _k)-u(x_k)]+ [u(x_k)-u(x)-\langle D u(x_k),x_k-x\rangle ]\\&\quad +\langle D u(x_k),x_k-x-s_k\theta _k\rangle + \langle s_k D u(x_k),\theta _k\rangle . \end{aligned}$$

Since u is locally Lipschitz and $x_k\in {\mathcal {Q}}(x,\theta _k)$, one has that

$$\begin{aligned}&\big |u(x+s_k\theta _k)-u(x_k)\big |+ \big |\langle D u(x_k),x_k-x-s_k\theta _k\rangle \big |\le 2\mathrm{Lip}(u)\big |x_k-x-s_k\theta _k\big |\\&\quad \le 2\mathrm{Lip}(u) |x_k-x|^2. \end{aligned}$$

Since u is semiconcave we deduce that

$$\begin{aligned} u(x_k)-u(x)-\langle D u(x_k),x_k-x\rangle \ge -C|x_k-x|\omega (|x_k-x|), \end{aligned}$$

for some constant $C>0$. Therefore

$$\begin{aligned} \frac{u(x+s_k\theta _k)-u(x)}{s_k}\ge \langle D u(x_k),\theta _k \rangle - \frac{2\mathrm{Lip}(u)|x_k-x|^2+C|x_k-x|\omega (|x_k-x|)}{s_k}. \end{aligned}$$

By the definition of ${\mathcal {Q}}(x,\theta _k)$ one has that $s_k|\theta _k|\ge |x_k-x|-|x_k-x|^2$, so that, as $x_k\rightarrow x$, $|x_k-x|\le 2s_k$ for k large enough. Recalling (ii), (5.3), and the fact that $\theta _k\rightarrow \theta$, we conclude that

$$\begin{aligned} \liminf _{\begin{array}{c} h\rightarrow 0^+\\ \theta '\rightarrow \theta \\ x+h\theta '\in {\overline{\Omega }} \end{array}}\frac{u(x+h\theta ')-u(x)}{h}\ge \langle {\overline{p}}, \theta \rangle \ge M(\theta ,x). \end{aligned}$$

(5.4)

This completes the proof.