Long Time Average of First Order Mean Field Games and Weak KAM Theory

Abstract

We show that the long time average of solutions of first order mean field game systems in finite horizon is governed by an ergodic system of mean field game type. The well-posedness of the latter system and the uniqueness of the ergodic constant rely on weak KAM theory.

Appendix: Proof of the existence and uniqueness result

The following result is stated in [10]. For convenience of the reader we recall the main ingredients of the proof.

Theorem 5.1

[10]

Let H and F satisfy conditions (4) and (7). Then (1) has a solution. If moreover the following inequality holds:

$$\int_{\mathbb {T}^d} \bigl(F(x,m_1)-F(x,m_2) \bigr)d(m_1-m_2) \geq0\quad\forall m_1,m_2 \in P\bigl(\mathbb {T}^d\bigr), $$

then the solution of (1) is unique.

Proof

The proof is based on a vanishing viscosity argument. Let ϵ>0 and (u ^ϵ,m ^ϵ) be the solution to

$$ \left \{ \begin{array}{r@{\quad}l@{\quad}l} (\mathrm{i})& -\partial_t u^\epsilon -\epsilon \Delta u^\epsilon +H\bigl(x,Du^\epsilon \bigr) =F\bigl(x,m^\epsilon (t)\bigr)&\mbox{in}\ (0,T)\times \mathbb{R}^d,\\ (\mathrm{ii}) & \partial_t m^\epsilon -\epsilon \Delta m^\epsilon -\operatorname{div} \bigl(m^\epsilon D_pH\bigl(x,Du^\epsilon \bigr)\bigr)=0& \mbox{in}\ (0,T)\times \mathbb{R}^d,\\ (\mathrm{iii})& m^T(0)=m_0, \qquad u^\epsilon (x,T)=u^f(x)& \operatorname{in}\ \mathbb{R}^d. \end{array} \right . $$

(19)

By (7) and using standard regularity results for parabolic equations and fixed point arguments, it is not difficult to check that system (19) has at least one classical solution. Following Lemma 5.2 below, the (u ^ϵ) are uniformly Lipschitz continuous and semi-concave in space with a semi-concavity modulus independent of ϵ. We need two other estimates: the first one is for the (m ^ϵ) and the second one for the time dependence of the (u ^ϵ).

Let us first show that the (m ^ϵ) are uniformly bounded. For this we note that

$$\begin{aligned} &\operatorname{div} \bigl(m^\epsilon D_pH\bigl(x,Du^\epsilon \bigr)\bigr)\\ &\quad= \big \langle Dm^\epsilon , D_pH\bigl(x,Du^\epsilon \bigr)\big \rangle + m^\epsilon \operatorname{Tr} \bigl(D^2_{xp}H\bigl(x,Du^\epsilon \bigr)+ D^2_{pp}H\bigl(x,Du^\epsilon \bigr)D^2u^\epsilon \bigr)\\ &\quad\leq\big \langle Dm^\epsilon , D_pH\bigl(x,Du^\epsilon \bigr)\big \rangle + Cm^\epsilon \end{aligned}$$

because $D^{2}_{xp}H(x,Du^{\epsilon })$ is bounded thanks to the regularity of H and the uniform Lipschitz continuity of u ^ϵ, and $\operatorname{Tr} (D^{2}_{pp}H(x,Du^{\epsilon })D^{2}u^{\epsilon })$ is bounded above because $D^{2}_{pp}H$ is positive and u ^ϵ is uniformly semi-concave. So m ^ϵ is a subsolution of the transport equation

$$\partial_t m^\epsilon -\epsilon \Delta m^\epsilon -\big \langle Dm^\epsilon , D_pH\bigl(x,Du^\epsilon \bigr) \big \rangle -Cm^\epsilon =0\quad \mbox{in}\ (0,T)\times \mathbb{R}^d. $$

By the maximum principle we get ∥m ^ϵ∥_∞≤∥m ₀∥_∞ e ^CT. Next we claim that the map t→m ^ϵ(t) (as a map from [0,T] to $\mathbb {P}(\mathbb {T}^{d})$) is uniformly Hölder continuous: indeed, if we multiply (19)-(ii) by m ^ϵ and integrate in time-space, we get

$$\epsilon \int_0^T\int_{\mathbb {T}^d} \big|Dm^\epsilon \big|^2 \leq\frac{1}{2} \int_{\mathbb {T}^d} \bigl(\big|m^2(T)\big|+\big|m^2(0)\big|\bigr) + \int_0^T \int_{\mathbb {T}^d} m^\epsilon \big|Dm^\epsilon \big|\big|D_pH \bigl(x,Du^\epsilon \bigr)\big|. $$

As m ^ϵ and D _p H(x,Du ^ϵ) are uniformly bounded, this implies that

$$ \epsilon \biggl(\int_0^T\int _{\mathbb {T}^d} \big|Dm^\epsilon \big|^2 \biggr)^{\frac{1}{2}} \leq C. $$

(20)

Then, for any smooth test function $\varphi:\mathbb {T}^{d}\to \mathbb{R}$ and for any 0≤t ₁≤t ₂≤T, we have by (19)-(ii):

$$\begin{aligned} \int_{\mathbb {T}^d} \varphi\bigl(m^\epsilon (t_2)-m^\epsilon (t_1) \bigr) =& - \int_{t_1}^{t_2}\int_{\mathbb {T}^d} \epsilon \big \langle Dm^\epsilon , D\phi\big \rangle + m^\epsilon \big \langle D_pH(x,Du), D\phi\big \rangle \\ \leq& C(t_2-t_1)^{\frac{1}{2}}\|D\varphi \|_\infty, \end{aligned}$$

because Du ^ϵ and m ^ϵ are uniformly bounded and (20) holds. Taking the supremum over all 1-Lipschitz continuous maps φ gives $\mathbf{d}_{1}(m(t_{1}),m(t_{2}))\leq C(t_{2}-t_{1})^{\frac{1}{2}}$.

Our last estimate is a uniform continuity in time of the (u ^ϵ). For this we note that, as u ^f(⋅) is bounded in ${\mathcal{C}}^{2}$, the maps w ^±(x,t)=u ^f(x)±C ₁(T−t) are sub (for ±=−) and super (for ±=+) solution of (19)-(i) for C ₁ sufficiently large (but not depending of ϵ). Hence ∥u ^ϵ(⋅,t)−u ^f(⋅)∥_∞≤C ₁(T−t). For h>0, we consider $u^{\epsilon }_{h}(x,t)= u(x, t-h)$. Because of the uniform Hölder regularity of the map t→m ^ϵ(t) in $P(\mathbb {T}^{d})$ and the uniform continuity of F, there is η(h)→0 as h→0 uniformly in ϵ, such that

$$\sup_{t\in[h,T]}\bigl \Vert F\bigl(\cdot,m^\epsilon (t-h) \bigr)-F\bigl(\cdot,m^\epsilon (t)\bigr)\bigr \Vert _\infty\leq \eta(h). $$

So u _h satisfies

$$-\partial_t u^\epsilon _h -\epsilon \Delta u^\epsilon _h +H\bigl(x,Du^\epsilon _h\bigr) =F\bigl(x,m^\epsilon (t-h)\bigr)\leq F\bigl(x,m^\epsilon (t)\bigr)+ \eta(h) $$

with terminal condition $u^{\epsilon }_{h}(x,T)= u^{\epsilon }(T-h,x)\leq u^{f}(x)+C_{1}h$. By comparison, we get $u^{\epsilon }_{h}(x,t)-\eta(h)(T-t)-C_{1}h\leq u^{\epsilon }(t,x)$, i.e.,

$$u^\epsilon (x,t-h)\leq u^\epsilon (t,x)+\eta(h) (T-t)+C_1h. $$

In a symmetric way,

$$u^\epsilon (x,t-h)\geq u^\epsilon (t,x)-\eta(h) (T-t)-C_1h. $$

Therefore ∥u ^ϵ(⋅,t−h)−u ^ϵ(⋅,t)∥_∞≤η(h)T+C ₁ h, which proves the uniform continuity of u ^ϵ.

Because of the bounds on (u ^ϵ,m ^ϵ), we can assume that (up subsequences) u ^ϵ converges uniformly to some continuous map u. Moreover, by uniform semi-concavity, (Du ^ϵ) converges a.e. to Du. On another hand m ^ϵ converges in L ^∞-weak* and in ${\mathcal{C}}^{0}([0,T],P(\mathbb {T}^{d}))$ to some $m\in L^{\infty}\cap{\mathcal{C}}^{0}([0,T],P(\mathbb {T}^{d}))$. In particular, m(0)=m ₀. Using the continuity assumption (4), we also find that F(⋅,m ^ϵ(⋅)) converges uniformly to F(⋅,m(⋅)). By standard viscosity solutions arguments, we can conclude to the convergence of u ^ϵ to the unique solution of

$$\left \{ \begin{array}{l@{\quad}l} -\partial_t u+H(x,Du) =F\bigl(x,m(t)\bigr)& \mbox{in}\ (0,T)\times \mathbb{R}^d,\\ u(x,T)=u^f(x)& \mbox{in}\ \mathbb{R}^d. \end{array} \right . $$

Note that u is Lipschitz continuous. Next we turn to the limit of m ^ϵ: for a fixed test function $\varphi\in{\mathcal{C}}^{\infty}_{c}((0,T)\times \mathbb {T}^{d})$, we have

$$\int_0^T\int_{\mathbb {T}^d} \bigl(- \partial_t \varphi-\epsilon \Delta\varphi+\big \langle D_pH(x,Du_n),D \varphi(t,x)\big \rangle \bigr) m_n(t,x)=0, $$

where D _p H(x,Du _n ) is bounded and converges a.e. to D _p H(x,Du) while m ^ϵ converges weakly* to m. So we get as ϵ→+∞,

$$\int_0^T\int_{\mathbb {T}^d} \bigl(- \partial_t \varphi(t,x)+\big \langle D_pH(x,Du),D\varphi(t,x)\big \rangle \bigr) m(t,x)=0, $$

which shows that m is a solution of the continuity equation (1)-(ii). In conclusion, the pair (u,m) solves (1). The uniqueness for this system is established in full details in [10], so we omit the proof. □

Lemma 5.2

There exists a constant C>0 such that

$$\|Du^\epsilon \|_\infty\leq C\quad \mbox{\textit{and}} \quad D^2u^\epsilon \leq C\quad \forall \epsilon >0. $$

Proof

The proof uses Bernstein method. We first show the uniform Lipschitz continuity in space. Let $z=\frac{1}{2}|Du^{\epsilon }|^{2}$. From classical computations, we have

$$\partial_t z +\frac{1}{2}\Delta z + \frac{1}{2}\big \langle D_x H \bigl(x,Du^\epsilon \bigr), Du^\epsilon \big \rangle +\frac{1}{2}\big \langle D_p H\bigl(x,Du^\epsilon \bigr), Dz\big \rangle \leq\frac{1}{2} \big \langle D_xF \bigl(x,m^\epsilon (t)\bigr), Du^\epsilon \big \rangle . $$

Using the last part of assumption (7) and the smoothing assumption (5) on F, this implies that

$$\partial_t z +\frac{1}{2}\Delta z -\frac{1}{2}C_0(1+2z) + \frac{1}{2}\big \langle D_p H\bigl(x,Du^\epsilon \bigr), Dz\big \rangle \leq\frac{1}{2} \sup_m \big\|D_xF(\cdot,m)\big\|_\infty z^{\frac{1}{2}}. $$

As $z(T,x)= \frac{1}{2}|Du^{f}(x)|^{2}$, we obtain by comparison a uniform bound on z, i.e., on Du ^ϵ.

Next we prove the semi-concavity. Let us fix a direction $v\in \mathbb{R}^{d}$ with |v|=1 and compute the derivative of equation (19)-(i) twice with respect to v:

$$-\partial_t u^\epsilon _{vv} -\epsilon \Delta u^\epsilon _{vv} + \bigl[H\bigl(\cdot,Du^\epsilon \bigr) \bigr]_{vv} =F_{vv}\bigl(x,m^\epsilon (t)\bigr), $$

where $[H(\cdot,Du^{\epsilon }) ]_{v} = H_{v}+\sum_{j=1}^{d} H_{p_{j}} u^{\epsilon }_{vx_{j}}$, so that

$$\begin{aligned} \bigl[H\bigl(\cdot,Du^\epsilon \bigr) \bigr]_{vv} = & H_{vv}+ \sum_{j=1}^d \bigl(2 H_{vp_j}u^\epsilon _{vx_j}+ H_{vp_j}u^\epsilon _{vvx_j} \bigr)+ \sum_{j,k=1}^d H_{p_kp_j}u^\epsilon _{vx_k}u_{vx_j} \\ \geq& -C\bigl(1+\big|D^2u^\epsilon \big|\bigr) +\big \langle D_pH_{v}, D u^\epsilon _{vv}\big \rangle +\frac {1}{\bar{C}} \big|D^2u^\epsilon \big|^2. \end{aligned}$$

In the last inequality, we have used the fact that H is of class ${\mathcal{C}}^{2}$ and that Du ^ϵ is uniformly bounded, and the strict convexity of H given in assumption (7). As F _vv (⋅,m ^ϵ) is bounded by assumption on F, and $-C(1+|D^{2}u^{\epsilon }|) +\frac{1}{\bar{C}} |D^{2}u^{\epsilon }|^{2}$ is bounded below by a constant independent of ϵ, we obtain

$$-\partial_t u^\epsilon _{vv} -\epsilon \Delta u^\epsilon _{vv} +\big \langle D_pH_{v}, D u^\epsilon _{vv}\big \rangle \leq C. $$

Since $u^{\epsilon }_{vv}(x,T)=u^{f}_{vv}(x)$ is bounded, we obtain by comparison a bound from above on $u^{\epsilon }_{vv}$ independent of ϵ and for any direction v. □

We complete the paper by a standard estimate on the continuity equation:

$$ \partial_t m +\operatorname{div}(m b)=0\quad \mbox{in}\ (0,T) \times \mathbb {T}^d. $$

(21)

Lemma 5.3

Assume that $b:(0,T)\times \mathbb {T}^{d}\to \mathbb{R}^{d}$ is a Borel vector field with ∥b∥_∞<+∞. If m satisfies (21), then m is Lipschitz continuous as a map from [0,T] to $P(\mathbb {T}^{d})$, with a Lipschitz constant bounded above by ∥b∥_∞.

Proof

Fix 0<t ₁<t ₂<T and let $h\in{\mathcal{C}}^{\infty}_{c}(\mathbb {T}^{d})$ be 1-Lipschitz continuous. Let ϵ>0 small and

$$\varphi_\epsilon (t,x)=\left \{ \begin{array}{l@{\quad}l} (t-t_1)h(x)/\epsilon & \mbox{if}\ t\in[t_1,t_1+\epsilon ],\\ h(x) & \mbox{if}\ t\in[t_1+\epsilon ,t_2-\epsilon ],\\ (t_2-t)h(x)/\epsilon & \mbox{if}\ t\in[t_2-\epsilon ,t_2],\\ 0 & \mbox{otherwise}. \end{array} \right . $$

As

$$\int_0^T \int_{\mathbb{R}^d} \bigl(- \partial_t\varphi_\epsilon +\langle m, D\varphi_\epsilon \rangle \bigr)m = 0, $$

we have

$$\int_{t_1}^{t_1+\epsilon } \int_{\mathbb{R}^d} \frac{hm }{\epsilon } +\int_{t_1+\epsilon }^{t_2-\epsilon } \int _{\mathbb{R}^d} \langle b, Dh\rangle m + \int_{t_2-\epsilon }^{t_2} \int_{\mathbb{R}^d} -\frac{hm}{\epsilon } = o(1). $$

Letting ϵ→0 gives for a.e. 0<t ₁<t ₂<T:

$$\int_{\mathbb{R}^d} h\, d\bigl(m(t_1)-m(t_2) \bigr) +\int_{t_1}^{t_2} \int_{\mathbb{R}^d} \langle b, Dh\rangle m=0 . $$

So

$$\int_{\mathbb{R}^d} h\, d\bigl(m(t_1)-m(t_2) \bigr) \leq\|b\|_\infty\| Dh\|_\infty |t_2-t_1| . $$

Taking the sup over h gives then

$$\mathbf{d}_1\bigl(m(t_1),m(t_2)\bigr) \leq \|b \|_\infty|t_2-t_1| . $$

 □