The Convergence Problem in Mean Field Games with Local Coupling

Abstract

The paper studies the convergence, as N tends to infinity, of a system of N coupled Hamilton–Jacobi equations, the Nash system, when the coupling between the players becomes increasingly singular. The limit equation turns out to be a mean field game system with a local coupling.

Appendix

In the appendix, we state an estimate for equations of the form:

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t w - \Delta w + V(t,x)\cdot Dw =f \qquad \mathrm{in}\; [0,T]\times \mathbb {T}^d,\\ \displaystyle w(0,x)=w_0(x)\qquad \mathrm{in}\; \mathbb {T}^d, \end{array}\right. \end{aligned}$$

(53)

where V is a fixed bounded vector field.

Proposition 5.1

If w is a solution to the above equation with $w_0\in C^{1+\alpha }$, then

$$\begin{aligned} \sup _{t\in [0,T]}\Vert w(t)\Vert _{1+\alpha }\le C\left[ \Vert w_0\Vert _{1+\alpha }+\Vert f\Vert _\infty \right] , \end{aligned}$$

where C depends on $\Vert V\Vert _\infty$, T, $\alpha$ and d only.

This kind of estimate is standard in the literature: for instance Theorem IV.9.1 of [17] (and its Corollary) states that Dw is bounded in $C^{\beta /2,\beta }$ for any $\beta \in (0,1)$. However the bound might depend on the vector field V and not only on its norm. We only check this is not the case.

Proof

Let us first check that the result holds for an homogenous initial datum. More precisely, we prove in a first step that, if w solve (53) with $w(0,\cdot )=0$, then $\Vert Dw\Vert _\infty \le C \Vert f\Vert _\infty$, where the constant C depends on $\Vert V\Vert _\infty$, T and d only. For this we argue by contradiction and assume for a while that there exists $V_n$ and $f_n$, bounded in $L^\infty$, and $w_n$ such that

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t w_n - \Delta w_n + V_n\cdot Dw_n =f_n \qquad \mathrm{in}\; [0,T]\times \mathbb {T}^d\\ \displaystyle w_n(0,x)=0\qquad \mathrm{in}\; \mathbb {T}^d. \end{array}\right. \end{aligned}$$

with $k_n:=\Vert Dw_n\Vert _\infty \rightarrow +\infty$. We set $\tilde{w}_n:= w_n/k_n$, $\tilde{f}_n:= f_n/k_n$. Then $\tilde{w}_n$ solves the heat equation with a right-hand side $\tilde{f}_n-V_n\cdot D\tilde{w}_n$ which is bounded in $L^\infty$. By standard estimates on the heat potential (see (3.2) of Chapter 3 in [17]), $\partial _t \tilde{w}_n$ and $D^2\tilde{w}_n$ are bounded in $L^p$ for any p independently of n. Then a Sobolev type inequality (Lemma II.3.3 in [17]) implies that $D\tilde{w}_n$ is bounded in $C^{\beta /2,\beta }$ independently of n for any $\beta \in (0,1)$. On the other hand, $(\tilde{f}_n)$ tends to 0 in $L^2$ and, by standard energy estimates, $(D\tilde{w}_n)$ tends to 0 in $L^2$. This is in contradiction with the fact that $\Vert D\tilde{w}_n\Vert _\infty =1$ and that $D\tilde{w}_n$ is bounded in $C^{\beta /2,\beta }$. So we have proved that there exists a constant C, depending on $\Vert V\Vert _\infty$, d and T only, such that the solution to (53) with $w(0,\cdot )=0$ satisfies $\Vert Dw\Vert _\infty \le C\Vert f\Vert _\infty$. Using the same argument on the the heat potential as above yields to

$$\begin{aligned} \Vert Dw\Vert _{C^{\beta /2,\beta }}\le C_\beta \Vert f-V\cdot Dw\Vert _\infty \le C_\beta \Vert f\Vert _\infty , \end{aligned}$$

where $C_\beta$ depends on $\Vert V\Vert$, d, T and $\beta$ only.

We now remove the assumption that $w_0=0$. We rewrite w as the sum $w=w_1+w_2$ where $w_1$ solves the heat equation with initial condition $w_0$ and $w_2$ solves equation (53) with right-hand side $f-V\cdot Dw_1$ and initial condition $w_2(0,\cdot )=0$. By maximum principle, we have

$$\begin{aligned} \sup _{t\in [0,T]} \Vert Dw_1(t)\Vert _{\alpha }\le C\Vert Dw_0\Vert _{\alpha }. \end{aligned}$$

By the first step of the proof, we also have, for any $\beta \in (0,1)$,

$$\begin{aligned} \Vert Dw_2\Vert _{C^{\beta /2,\beta }}\le C_\beta \Vert f-V\cdot Dw_1\Vert _\infty \le C_\beta (\Vert f\Vert _\infty + \Vert Dw_0\Vert _{\alpha }). \end{aligned}$$

Choosing $\beta =\alpha$ then gives the result. $\square$