Second order local minimal-time mean field games

Abstract

The paper considers a forward–backward system of parabolic PDEs arising in a Mean Field Game (MFG) model where every agent controls the drift of a trajectory subject to Brownian diffusion, trying to escape a given bounded domain $\Omega$ in minimal expected time. Agents are constrained by a bound on the drift depending on the density of other agents at their location. Existence for a finite time horizon T is proven via a fixed point argument, but the natural setting for this problem is in infinite time horizon. Estimates are needed to treat the limit $T\rightarrow \infty$, and the asymptotic behavior of the solution obtained in this way is also studied. This passes through classical parabolic arguments and specific computations for MFGs. Both the Fokker–Planck equation on the density of agents and the Hamilton–Jacobi–Bellman equation on the value function display Dirichlet boundary conditions as a consequence of the fact that agents stop as soon as they reach $\partial \Omega$. The initial datum for the density is given, and the long-time limit of the value function is characterized as the solution of a stationary problem.

Appendix A: Regularizing effects of parabolic equations

This appendix is concerned with the regularizing properties of a class of parabolic equations including both the Fokker–Planck and the Hamilton–Jacobi–Bellman equations we consider in this paper. More precisely, we consider the increase of the exponent p of the $L^p$ integrability in space of the solution of the system. As recalled in the introduction, the computations and results presented here are very similar to those from the appendix of [7], the main difference lying in the boundary condition. The main result of this appendix is the following.

Proposition A.1

Let $T\in (0,+\infty ]$. Let $V, F, f, g, u \in C^\infty ((0,T)\times \Omega )$ with $V, g \in L^\infty ((0, T) \times \Omega )$, $u \ge 0$, $u=0$ on $\partial \Omega$, such that

$$\begin{aligned} \partial _t u - \nu \Delta u + \nabla \cdot (u V) + \nabla \cdot F + f + g \cdot \nabla u \le 0,\quad \text {on} \quad (0,T) \times \Omega . \end{aligned}$$

(A.1)

Then, for every $p>1$, every number $a \in (0, 1)$ and $t_1,t_2$ such that $0<t_1<t_2<T$ and $a<{|}{t_1-t_2}{|}<a^{-1}$, there is $C>0$, depending only on $p,a, {\Vert }{V}{\Vert }_{L^\infty }, {\Vert }{g}{\Vert }_{L^{\infty }}$ such that

$$\begin{aligned} {\Vert }{u(t_2,\cdot )}{\Vert }_{L^{\infty }} \le C \left( {\Vert }{u(t_1,\cdot )}{\Vert }_{L^p} + {\Vert }{F}{\Vert }_{L^\infty ((t_1,t_2)\times \Omega )} + {\Vert }{f}{\Vert }_{L^\infty ((t_1,t_2)\times \Omega )}\right) . \end{aligned}$$

The same result is true omitting the assumption $u\ge 0$ if the PDE (A.1) is satisfied as an equality instead of an inequality.

The proof follows a standard method based on Moser’s iterations that will be detailed here. This appendix is included for completeness: the experienced reader will recognize well-known computations, which are simplified in this setting thanks in particular to the Dirichlet boundary conditions we use.

Proof

Let u be as in the proposition. For $k>1$, we define

$$\begin{aligned} m_k(t) := \int _{\Omega } u^k(t,x){{\,\mathrm{d\!}\,}}x. \end{aligned}$$

We also define $\alpha := \frac{2^\star }{2} = \frac{n}{n-2}$ if $n>2$ (here $2^\star$ is the Sobolev exponent in dimension n). When $n=1, 2$ we set $\alpha := 2$ (but any number larger than 1 and smaller than $+\infty$ could be used here). Moreover, we set

$$\begin{aligned} M := {\Vert }{F}{\Vert }_{L^\infty ((t_1,t_2)\times \Omega )}+ {\Vert }{f}{\Vert }_{L^\infty ((t_1,t_2)\times \Omega )} \end{aligned}$$

Step 1. $L^p$ estimates.

Let us start with proving that, for $k_0>1$, there is $C>0$ depending on $k_0$ and on the $L^\infty$ norms of V, g, such that, for every $k>k_0>1$,

$$\begin{aligned} \frac{{{\,\mathrm{d\!}\,}}}{{{\,\mathrm{d\!}\,}}t}(m_k e^{-Ck^2 t}) + \frac{1}{C} m_{\alpha k}^{\frac{1}{\alpha }} e^{-Ck^2 t} \le Ce^{-Ck^2 t} k^2 M^k. \end{aligned}$$

(A.2)

In order to do so, we differentiate $m_k$ with respect to t, to get

$$\begin{aligned}&m_k^\prime (t) \le k\int _{\Omega }(\nu \Delta u - \nabla \cdot (u V)-g\cdot \nabla u - \nabla \cdot F - f ) u^{k-1} \\&\quad \le - k(k-1) \nu \int _{\Omega } \vert \nabla u \vert ^2 u^{k-2} + k(k-1) \int _{\Omega }(V\cdot \nabla u) u^{k-1} - k\int _\Omega (g \cdot \nabla u) u^{k-1} \\&\quad \quad + k(k-1)\int _{\Omega } (F\cdot \nabla u) u^{k-2} - k\int _{\Omega }f u^{k-1}. \end{aligned}$$

Now, owing to Young’s inequality, we can find $C_1,C_2,C_3>0$ depending only on ${\Vert }{V}{\Vert }_{L^\infty }$, ${\Vert }{g}{\Vert }_{L^{\infty }}$, $k_0$, $\nu$ such that

$$\begin{aligned} m_k^\prime (t) \le - C_1 k^2\int _{\Omega } \vert \nabla u \vert ^2 u^{k-2} + C_2k^2 \int _{\Omega }u^{k} + C_3 k^2 \int _{\Omega } \vert F\vert ^2 u^{k-2} + k \int _{\Omega }\vert f \vert u^{k-1} \end{aligned}$$

(note that we replaced the coefficient $k(k-1)$ with $k^2$, as these two numbers are equivalent up to multiplicative constants as far as $k> k_0>1$). Moreover, thanks again to a Young inequality, we have

$$\begin{aligned} \vert F\vert ^2 u^{k-2} \le \frac{2}{k} \vert F\vert ^k + \frac{k-2}{k} u^{k} \quad \text { and }\quad \vert f\vert u^{k-1} \le \frac{1}{k}\vert f \vert ^k + \frac{k-1}{k}u^{k}. \end{aligned}$$

Therefore, up to increasing $C_2,C_3$, we get

$$\begin{aligned}&m_k^\prime (t) \le - C_1 k^2\int _{\Omega } \vert \nabla u \vert ^2 u^{k-2} + C_2k^2 \int _{\Omega }u^{k} + C_3 k^2 \int _{\Omega } \vert F\vert ^k + k \int _{\Omega }\vert f \vert ^k \\&\quad \le - C_1 k^2\int _{\Omega } \vert \nabla u \vert ^2 u^{k-2} + C_2 k^2 \int _{\Omega }u^{k} + C_3 k^2 M^k. \end{aligned}$$

Now, owing to the Gagliardo–Nirenberg–Sobolev inequality, we have, for some $C_4>0$,

$$\begin{aligned} k^2\int _{\Omega } \vert \nabla u \vert ^2 u^{k-2} = 4\int _{\Omega }\vert \nabla ( u^\frac{k}{2} )\vert ^2 \ge C_4 \left( \int _{\Omega }u^{k\alpha }\right) ^{\frac{1}{\alpha }} = C_4 m_{k\alpha }^\frac{1}{\alpha }. \end{aligned}$$

Hence

$$\begin{aligned} m_k^\prime (t) + C_1C_4 m_{\alpha k}^{\frac{1}{\alpha }}\le C_2k^2 m_k +C_3 k^2 M^k. \end{aligned}$$

Let us denote $C:= \max \{\frac{1}{C_1C_4},C_2,C_3\}$. Then, the above equation gives us

$$\begin{aligned} m_k^\prime (t) -Ck^2 m_k + \frac{1}{C} m_{\alpha k}^{\frac{1}{\alpha }}\le C k^2 M^k, \end{aligned}$$

which we can rewrite in order to get (A.2).

Step 2. Estimates on $m_{\alpha k}$.

We show in this step that, for $k> k_0>1$ and for $0<t_1<t_2<T$, we have

$$\begin{aligned} m_{\alpha k}^{\frac{1}{\alpha }}(t_2) \le e^{C\alpha k^2(t_2-t_1)}\frac{1}{t_2-t_1}e^{C k^2t_ 2}\int _{t_1}^{t_2}m_{\alpha k}^{\frac{1}{\alpha }}(s)e^{-Ck^2s}{{\,\mathrm{d\!}\,}}s + e^{C\alpha k^2(t_2-t_1)}M^k,\nonumber \\ \end{aligned}$$

(A.3)

for some C depending on $k_0$ and on the $L^\infty$ norms of V, g.

The relation (A.2) provides

$$\begin{aligned} \frac{{{\,\mathrm{d\!}\,}}}{{{\,\mathrm{d\!}\,}}t}(m_k e^{-Ck^2 t}) \le Ce^{-Ck^2 t} k^2 M^k. \end{aligned}$$

Let us take $s \in (t_1,t_2)$. We integrate the above inequality for $t\in (s,t_2)$ to get:

$$\begin{aligned}&m_k(t_2) e^{-Ck^2 t_2} \le m_k(s) e^{-Ck^2 s} + CM^k\int _s^{t_2} e^{-Ck^2 t}{{\,\mathrm{d\!}\,}}t\\&\quad \le m_k(s) e^{-Ck^2 s} + M^k e^{-C k^2 s}. \end{aligned}$$

Taking the power $\frac{1}{\alpha }<1$ and using its subadditivity yields

$$\begin{aligned} m_k^{\frac{1}{\alpha }}(t_2) e^{-\frac{C}{\alpha }k^2 t_2} \le m_k^{\frac{1}{\alpha }}(s)e^{-\frac{C}{\alpha }k^2 s} + M^{\frac{k}{\alpha }}e^{-\frac{C}{\alpha }k^2 s} . \end{aligned}$$

We replace k by $\alpha k$ so as to re-write the above inequality as

$$\begin{aligned} m_{\alpha k}^{\frac{1}{\alpha }}(t_2) e^{-C\alpha k^2 t_2} \le m_{\alpha k}^{\frac{1}{\alpha }}(s)e^{-C\alpha k^2 s} + M^ke^{-C\alpha k^2 s} . \end{aligned}$$

We multiply by $e^{Ck^2s(\alpha -1)}$ and integrate this for $s \in (t_1,t_2)$ in order to obtain

$$\begin{aligned}&m_{\alpha k}^{\frac{1}{\alpha }}(t_2) \int _{t_1}^{t_2}e^{C\alpha k^2 (s-t_2)}e^{-Ck^2s}{{\,\mathrm{d\!}\,}}s \\&\quad \le \int _{t_1}^{t_2}m_{\alpha k}^{\frac{1}{\alpha }}(s)e^{-Ck^2 s}{{\,\mathrm{d\!}\,}}s + M^k\int _{t_1}^{t_2}e^{-Ck^2 s} {{\,\mathrm{d\!}\,}}s . \end{aligned}$$

We then use $e^{C\alpha k^2 (s-t_2)}\ge e^{C\alpha k^2 (t_1-t_2)}$ in order to obtain

$$\begin{aligned} m_{\alpha k}^{\frac{1}{\alpha }}(t_2) \le e^{C\alpha k^2 (t_2-t_1)} \left( \int _{t_1}^{t_2}e^{-Ck^2s}{{\,\mathrm{d\!}\,}}s\right) ^{-1}\int _{t_1}^{t_2}m_{\alpha k}^{\frac{1}{\alpha }}(s)e^{-Ck^2 s}{{\,\mathrm{d\!}\,}}s + e^{C\alpha k^2 (t_2-t_1)}M^k \end{aligned}$$

and finally we use $\int _{t_1}^{t_2}e^{-Ck^2s}{{\,\mathrm{d\!}\,}}s\ge (t_2-t_1)e^{-Ck^2t_2}$, which provides the desired inequality.

Step 3. Higher integrability estimates.

Let us now show that, for $k>k_0$, there is $C>0$ (depending on the same quantities as in the previous steps), such that

$$\begin{aligned} m_{\alpha k}^{\frac{1}{\alpha k}}(t_2) \le \frac{e^{Ck(t_2-t_1)}}{(C^{-1}|t_2-t_1|)^{1/k}}\left( m_k(t_1) + M^k\right) ^{\frac{1}{k}}. \end{aligned}$$

(A.4)

First of all, integrating (A.2) for $t\in (t_1,t_2)$, and discharging the final value $m_k(t_2)e^{-Ct_2}$, we obtain

$$\begin{aligned}&\frac{1}{C}\int _{t_1}^{t_2} m_{\alpha k}^{\frac{1}{\alpha }}(t) e^{-C k^2 t}{{\,\mathrm{d\!}\,}}t \le m_k(t_1) e^{-Ck^2 t_1} + M^k\int _{t_1}^{t_2}Ck^2e^{-Ck^2 t}{{\,\mathrm{d\!}\,}}t \\&\quad \le e^{-Ck^2t_1}\left( M^k+m_k(t_1)\right) . \end{aligned}$$

Combining this with (A.3), we get

$$\begin{aligned} m_{\alpha k}^{\frac{1}{\alpha }}(t_2) \le e^{C\alpha k^2(t_ 2-t_1)}\left( C\frac{e^{Ck^2(t_2-t_1)}}{t_2-t_1}(m_k(t_1) +M^k) + M^k \right) . \end{aligned}$$

Up to enlarging the constant C and using $0< t_2-t_1<a^{-1}$, we can write the above inequality in a simpler form, i.e.

$$\begin{aligned} m_{\alpha k}^{\frac{1}{\alpha }}(t_2) \le \frac{e^{C(\alpha +1)k^2(t_2-t_1)}}{C^{-1}(t_2-t_1)}\left( m_k(t_1) + M^k\right) , \end{aligned}$$

hence, (A.4) holds true, after raising to the power 1/k and including $\alpha +1$ in the constant C.

Step 4. Iterations.

We conclude the proof in this step by proving that, for $p,t_1,t_2$ as in the statement of the proposition, there is $C>0$ such that

$$\begin{aligned} {\Vert }{u(t_2,\cdot )}{\Vert }_{L^{\infty }} \le C ( {\Vert }{u(t_1,\cdot )}{\Vert }_{L^p} + {\Vert }{F}{\Vert }_{L^\infty } + {\Vert }{f}{\Vert }_{L^{\infty }}). \end{aligned}$$

(A.5)

We denote

$$\begin{aligned} s_n := t_2 - \frac{t_2-t_1}{(2\alpha )^n}, \ k_n := \alpha ^n p, \ \beta _n := \frac{e^{Ck_n(s_{n+1}-s_n)}}{(C^{-1}(s_{n+1} - s_{n}))^{\frac{1}{k_n}}}, \end{aligned}$$

and

$$\begin{aligned} a_n := m_{k_n}^{\frac{1}{k_n}}(s_n),\quad {\tilde{a}}_n:=\max \{a_n, M\}. \end{aligned}$$

Then, (A.4) gives us that

$$\begin{aligned} a_{n+1} \le \beta _n (a_n^{k_n} +M^{k_n})^{\frac{1}{k_n}}\le \beta _n 2^{\frac{1}{k_n}}{\tilde{a}}_n. \end{aligned}$$

Hence, up to replacing the constant C with a larger one so as to suppose $\beta _n 2^{\frac{1}{k_n}}\ge 1$, we find

$$\begin{aligned} {\tilde{a}}_{n+1} \le \beta _n 2^{\frac{1}{k_n}}{\tilde{a}}_n. \end{aligned}$$

We observe that we have $\prod _{n=0}^{+\infty } \beta _n 2^{\frac{1}{k_n}} < +\infty$ as a consequence of the logarithmic estimate

$$\begin{aligned}&\sum _{n=0}^{+\infty } \log (\beta _n 2^{\frac{1}{k_n}}) \le \sum _{n=0}^{+\infty } Ck_n\frac{t_2-t_1}{(2\alpha )^n}\\&\quad +\frac{1}{k_n}(\log 2+n\log (2\alpha )-\log (t_2-t_1)+\log C) < +\infty . \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \max \{ \lim _{n\rightarrow +\infty } a_n , M\} \le \left( \prod _{n=0}^{+\infty } \beta _n 2^{\frac{1}{k_n}}\right) \max \{a_0 , M \}\le C(a_0+M). \end{aligned}$$

Hence, thanks to $\lim _{n\rightarrow +\infty }a_n = {\Vert }{u(t_2,\cdot )}{\Vert }_{L^{\infty }}$, we obtain (A.5). This concludes the proof. $\square$

Corollary A.2

Let $T\in (0,+\infty ]$. Let $V \in L^{\infty }((0,T)\times \Omega )$. Let $u\in L^1((0,T)\times \Omega )$ be a positive distributional solution of

$$\begin{aligned} \partial _t u - \nu \Delta u -\nabla \cdot (u V) \le 0,\quad \text { on } \ (0,T)\times \Omega , \end{aligned}$$

satisfying the following mild regularity assumption: u is obtained as a measurable curve $(u_t)_t$ of functions of the x variable, which is such that $t\mapsto \int _\Omega \eta (x) u_t(x) {{\,\mathrm{d\!}\,}}x$ is continuous in time for every $\eta \in C^\infty (\Omega )$ (note that we do not restrict to $\eta \in C^\infty _c(\Omega )$). Then, for every $p>1$ and $a \in (0, 1)$, there is $C>0$, depending only on p, a, ${\Vert }{V}{\Vert }_{L^\infty }$, such that we have

$$\begin{aligned} {\Vert }{u(t_2,\cdot )}{\Vert }_{L^{\infty }} \le C {\Vert }{u(t_1,\cdot )}{\Vert }_{L^p} \end{aligned}$$

for every $0<t_1<t_2<T$ with $a<{|}{t_2-t_1}{|}<a^{-1}$.

Proof

To prove this estimate the only important point is to regularize the equation so as to apply Proposition A.1. In order for the proof to be self-contained, we detail a two-step approximation argument.

We convolve the equation by an approximation of the identity and to apply Proposition A.1. However, convolving will not preserve the Dirichlet boundary conditions, so we first have to extend u by zero on a bigger set.

We define $\Omega ^+$ to be a open bounded regular set such that $\Omega + B_1 \subset \Omega ^+$, where $B_1$ is unit ball in $\mathbb {R}^N$.

We define $u^+(t,x) := u(t,x)$ if $x\in \Omega$, and $u^+(t,x) = 0$ elsewhere. Let $\eta _\varepsilon (x)$ be an approximation of the identity whose support is included in $B_1$. We define $u_\varepsilon := u^+\star \eta _\varepsilon$ (here, $\star$ is the convolution in space only). It is a function which is smooth in x and continuous in t. We then convolve in time as well, taking $\chi _\delta (t)$ an approximation of the identity whose support is included in $\mathbb {R}_+$. Defining $u_{\varepsilon ,\delta }:=\chi _\delta \star u_\varepsilon$ we have now a function which is smooth in time and space. It satisfies, in the classical sense,

$$\begin{aligned} \partial _t u_{\varepsilon ,\delta } -\nu \Delta u_{\varepsilon ,\delta } -\nabla \cdot ( u_{\varepsilon ,\delta } V_{\varepsilon ,\delta } )\le 0, \quad \text { for } \ t\in (0,T), \ x\in \Omega ^+, \end{aligned}$$

with $V_{\varepsilon ,\delta } := \frac{\chi _\delta \star \eta _\varepsilon \star (u V)}{ u_{\varepsilon ,\delta }} \in C^\infty$. Moreover, the $L^\infty$ norm of $V_{\varepsilon ,\delta }$ is bounded independently of $\varepsilon$ and $\delta$. Then, $u_{\varepsilon ,\delta }$ is positive, regular and is a (classical) subsolution of a Fokker–Planck equation with regular coefficients, hence we can apply Proposition A.1. We then take the limit $\delta \rightarrow 0$, and we observe that we have

$$\begin{aligned} {\Vert }{u_\varepsilon (t,\cdot )}{\Vert }_{L^p}=\lim _{\delta \rightarrow 0}{\Vert }{u_{\varepsilon ,\delta }(t,\cdot )}{\Vert }_{L^p} \end{aligned}$$

for every t, since $u_\varepsilon$ is continuous. Then, we have

$$\begin{aligned} {\Vert }{u(t,\cdot )}{\Vert }_{L^p}=\lim _{\varepsilon \rightarrow 0}{\Vert }{u_{\varepsilon }(t,\cdot )}{\Vert }_{L^p} \end{aligned}$$

from standard properties of convolutions (with the possibility, of course, that this limit and this norm take the value $+\infty$). $\square$

Corollary A.3

Let $T \in (0,+\infty ]$. Let $f, g\in L^\infty$ and let $u\in L^\infty ((0,T);L^2(\Omega ))\cap L^2((0,T); H^1_0(\Omega ))$ be solution (in the weak sense) of

$$\begin{aligned} \partial _t u - \nu \Delta u + f + g \cdot \nabla u = 0,\quad \text {on} \quad (0,T) \times \Omega , \end{aligned}$$

with Dirichlet boundary conditions and initial datum $u(0,\cdot ) = u_0 \in L^2$.

Then, for every $p>1$, and $a \in (0, 1)$ there is $C>0$, depending only on $p,a, {\Vert }{g}{\Vert }_{L^{\infty }}$ such that

$$\begin{aligned} {\Vert }{u(t_2,\cdot )}{\Vert }_{L^{\infty }} \le C \left( {\Vert }{u(t_1,\cdot )}{\Vert }_{L^p} + {\Vert }{f}{\Vert }_{L^{\infty }}\right) \end{aligned}$$

for every $t_1<t_2$ with $a<{|}{t_2-t_1}{|}<a^{-1}$.

Proof

Let $f_n,g_n$ be $C^\infty$ and such that $f_n \rightarrow f$ and $g_n \rightarrow g$ in the $L^2$ norm. Assume moreover that we have ${\Vert }{f_n}{\Vert }_{L^\infty }\rightarrow {\Vert }{f}{\Vert }_{L^\infty }$ and ${\Vert }{g_n}{\Vert }_{L^\infty }\rightarrow {\Vert }{g}{\Vert }_{L^\infty }$. Let $u_n$ be the solution of

$$\begin{aligned} \partial _t u_n - \nu \Delta u_n + f_n + g_n \cdot \nabla u_n = 0,\quad \text {on} \quad (0,+\infty ) \times \Omega , \end{aligned}$$

with Dirichlet boundary condition and with initial datum $u_0^n$, where $u_0^n$ is a smooth $L^2$ approximation of $u_0$.

Then, $u_n$ is smooth enough to apply Proposition A.1 to $u_n$, to get, for $p,t_1,t_2$ as in the statement of the corollary,

$$\begin{aligned} {\Vert }{u_n(t_2,\cdot )}{\Vert }_{L^{\infty }} \le C \left( {\Vert }{u_n(t_1,\cdot )}{\Vert }_{L^p} + {\Vert }{f_n}{\Vert }_{L^\infty } \right) . \end{aligned}$$

(A.6)

Then, as n goes to $+\infty$, $u_n$ converges (the arguments to prove this are standard and based on the weak $L^2$ convergence of $\nabla u_n$) to a solution (in the weak sense) of

$$\begin{aligned} \partial _t u - \nu \Delta u + f + g \cdot \nabla u = 0,\quad \text {on} \quad (0,+\infty ) \times \Omega , \end{aligned}$$

with Dirichlet boundary conditions and with initial datum $u_0$. The convergence is also strong in the $L^2$ sense. Because this solution is unique, it necessarily coincides with the original solution u of the statement. In order to obtain the result, we need to pass to the limit the inequality (A.6). The left-hand side can easily be dealt with by semicontinuity, while for the right-hand side, we suppose $p\le 2$ and we use strong $L^2$ convergence. Since this convergence is $L^2$ in space-time, we have convergence of the right-hand side only for a.e. $t_1$. Yet, using the fact that the solution u is continuous in time as a function valued into $L^2(\Omega )$, the result extends to all $t_1$. The inequality for $p=2$ implies that with $p>2$, up to modifying the constant in a way depending on ${|}{\Omega }{|}$. $\square$

The reader may observe that we used different regularization strategies to prove the two above corollaries. Indeed, the linear behavior of the Fokker–Planck equation allowed to directly regularize the solution (up to modifying the drift vector field: we convolve the solution and define a new drift vector field which preserves the same $L^\infty$ bound, a trick which is completely standard for curves in the Wasserstein space, see for instance [31, Chapter 5]). This is not possible for the Hamilton–Jacobi–Bellman equation. However, when uniqueness of the solution is known, regularizing the coefficients and the data of the equation is another option, and it is what we did in our last corollary.