1 Introduction

The study of the singular perturbation of control systems has a long history going back to [5,6,7] and references therein. Such a problem concerns the analysis of systems where some state variables evolve at a much faster time scale than the others. Generally, the solution of a typical singular perturbation problem leads to the elimination of the fast state variable and, consequently, to a reduction in the dimension of the system. Clearly, the limit problem keeps some information on the fast part.

Besides classical control systems, other types of singular perturbation problems have been studied and we refer, for instance, to homogenization (e.g., [27, 33]) and the long time behavior (e.g., [22, 23, 10]). More recently, such analysis was extended to the case of differential games (e.g., [2, 3, 29, 35]) and of that mean field games (MFG) (e.g., [12, 13, 16, 17, 19, 34]). Based on this recent literature, in this paper we make a step further. Indeed, the goal of this work is twofold: first, we show a connection between the classical MFG system, where the underlying payoff is a calculus of variation problem, and the MFG with control of acceleration; secondly, we show how a MFG of control system can be recovered from a MFG system with control of acceleration. We will extend such analysis to the case of singular perturbation problems associated with sub-Riemannian structure in a future work.

We recall that MFG were introduced in [30,31,32] and [25, 26] in order to describe the behavior of Nash equilibria in problems with infinitely many rational agents (we refer to [15] and references therein for more details). Since these pioneering works, the MFG theory has grown very fast: we refer for instance to the survey papers and the monographs [11, 18, 24]. The classical MFG system introduced in [30,31,32] describes systems in which each of the typical payoff is represented by deterministic calculus of variation problem. MFG systems with control of acceleration, first introduced in [1, 14], describe models where agents control their acceleration and the cost functional to minimize depends on higher order derivatives of admissible trajectories. Such problems naturally appear in the study of agent-based models which describe the collective behavior of various animal populations (e.g., [28, 36]) or crowd dynamics (e.g., [20, 21]). The study of the singular perturbation problem provided in this paper finds a lot of applications as, for instance, to a MFG system with Cucker–Smale type dynamics ( [9]) to describe the behavior of a flock in which the control is increasingly weaker.

We now describe the problems we address in this paper.

  1. (1)

    Convergence to the classical MFG system. We study the limit of the solution to the system

    $$\begin{aligned} {\left\{ \begin{array}{ll} -\partial _{t} u^{{{\,\mathrm{\varepsilon }\,}}} +\frac{1}{2{{\,\mathrm{\varepsilon }\,}}}|D_{v}u^{{{\,\mathrm{\varepsilon }\,}}}|^{2} - \langle D_{x}u^{{{\,\mathrm{\varepsilon }\,}}}, v \rangle -L_{0}(x, v, m^{{{\,\mathrm{\varepsilon }\,}}}_{t})= 0, &{} (t,x,v) \in [0,T] \times \mathbb {R}^{2d}\\ \partial _{t}\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t} - \langle D_{x}\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t},v \rangle - \frac{1}{{{\,\mathrm{\varepsilon }\,}}}{{\,\textrm{div}\,}}_{v}\left( \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}D_{v}u^{{{\,\mathrm{\varepsilon }\,}}} \right) =0, &{} (t,x,v) \in [0,T] \times \mathbb {R}^{2d}\\ \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{0}=\mu _{0}, \quad u^{{{\,\mathrm{\varepsilon }\,}}}(T,x,v)=g(x,m^{{{\,\mathrm{\varepsilon }\,}}}_{T}), &{} (x,v) \in \mathbb {R}^{2d} \end{array}\right. } \end{aligned}$$
    (1.1)

    as the parameter \({{\,\mathrm{\varepsilon }\,}}\) goes to zero. Heuristically, the state equation associated with the above PDEs system is given by

    $$\begin{aligned} {\left\{ \begin{array}{ll} \dot{x}(t)=v(t)\\ \dot{v}(t)= \frac{1}{{{\,\mathrm{\varepsilon }\,}}} \alpha (t) \end{array}\right. } \end{aligned}$$
    (1.2)

    where \(\alpha : [0,T] \rightarrow \mathbb {R}^{d}\) is a measurable control function and, from [1, 14], we have that for any \({{\,\mathrm{\varepsilon }\,}}> 0\) a typical player aims to minimize a cost functional of the form

    $$\begin{aligned} \int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }(s)|^{2} + L_{0}(\gamma (s), {\dot{\gamma }}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s} + g(\gamma (T), m^{{{\,\mathrm{\varepsilon }\,}}}_{T}). \end{aligned}$$

    Moreover, still from [1, 14], under suitable assumptions (listed below), we have that for any \({{\,\mathrm{\varepsilon }\,}}> 0\) there exists a unique solution \((u^{{{\,\mathrm{\varepsilon }\,}}}, \mu ^{{{\,\mathrm{\varepsilon }\,}}})\) to (1.1).

    Following the previous considerations on a typical singular perturbation problem, in case of control of acceleration, we expect that the fast variable, in this case the velocity of each player, is eliminated in the limit and all the information are captured by the behavior of the space variable. Moreover, since the aim of such analysis is to establish a rigorous mathematical connection between the classical MFG system and the MFG system with control of acceleration, the data in (1.3) have a particular structure. Indeed, we observe that the function \(L_{0}\) and the terminal costs g only depend on the space marginal of the measures \(\{\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\}_{t \in [0,T]}\). Such a marginal flow posses all the information of the behavior of the fast variable in the limit, and it is also the object of investigation in classical MFG since it represents the distribution of players in space at each time \(t \in [0,T]\).

  2. (2)

    Convergence to MFG of control system. In the second part, we analyze the limit of the solution to the system

    $$\begin{aligned} {\left\{ \begin{array}{ll} -\partial _{t} u^{{{\,\mathrm{\varepsilon }\,}}} +\frac{1}{2{{\,\mathrm{\varepsilon }\,}}}|D_{v}u^{{{\,\mathrm{\varepsilon }\,}}}|^{2} - \langle D_{x}u^{{{\,\mathrm{\varepsilon }\,}}}, v \rangle -L_{0}(x, v, \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t})= 0, &{} (t,x,v) \in [0,T] \times \mathbb {R}^{2d}\\ \partial _{t}\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t} - \langle D_{x}\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t},v \rangle - \frac{1}{{{\,\mathrm{\varepsilon }\,}}} {{\,\textrm{div}\,}}_{v}\left( \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}D_{v}u^{{{\,\mathrm{\varepsilon }\,}}} \right) =0, &{} (t,x,v) \in [0,T] \times \mathbb {R}^{2d}\\ \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{0}=\mu _{0}, \quad u^{{{\,\mathrm{\varepsilon }\,}}}(T,x,v)=g(x,m^{{{\,\mathrm{\varepsilon }\,}}}_{T}), &{} (x,v) \in \mathbb {R}^{2d} \end{array}\right. } \end{aligned}$$
    (1.3)

    still as the parameter \({{\,\mathrm{\varepsilon }\,}}\) goes to zero. The main issue here is that both the data L and g depend on \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}\) and we have to deal with the convergence of the whole measure. Note that, even though the limit control problem does not depend on velocity as a state variable we have that the second marginal of the limit measure, and so the Lagrangian function still depends on it. For this reason, we expect the limit system to be of mean field game of control type.

Next, we briefly explain the main results and the methods of proof.

  1. (1)

    Toward the classical MFG system. We prove that \((u^{{{\,\mathrm{\varepsilon }\,}}}, m^{{{\,\mathrm{\varepsilon }\,}}})\), where \(m^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) is the space marginal of the solution \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) for any \(t \in [0,T]\), converges (up to subsequence) to a solution \((u^{0}, m^{0})\) of the classical MFG system

    $$\begin{aligned} {\left\{ \begin{array}{ll} (i)\,\, -\partial _{t} u^{0}(t,x) + H_{0}(x, D_{x}u^{0}(t,x), m^{0}_{t})=0, &{} \quad (t,x) \in [0,T] \times \mathbb {R}^{d} \\ (ii)\,\, \partial _{t} m^{0}_{t} - {{\,\textrm{div}\,}}\Big ( m^{0}_{t}D_{p}H_{0}(x, D_{x}u^{0}(t,x), m^{0}_{t}) \Big )=0, &{} \quad (t,x) \in [0,T] \times \mathbb {R}^{d} \\ m^{0}_{0}= \pi _{1} \sharp \mu _{0},\,\, u^{0}(T,x)=g(x,m^{0}_{T}), &{} \quad x \in \mathbb {R}^{d} \end{array}\right. } \end{aligned}$$
    (1.4)

    where \(H_{0}:\mathbb {R}^d \times \mathbb {R}^d \rightarrow \mathbb {R}\) is the Legendre Transform of the function \(L_0\). Observe that, the main difference between our result and the existing one concerning the homogenization problem in MFG [9, 19, 34] is that the limit system is still of MFG type. Indeed, in [19, 34] and [9] it has been proved that in the limit the MFG structure of the problem is lost and, in particular, an explicit example of MFG system with potential coupling function is constructed in [34].

    In order to prove our first main convergence result, we begin by showing that \(u^{{{\,\mathrm{\varepsilon }\,}}}\) is equibounded and \(m^{{{\,\mathrm{\varepsilon }\,}}}\) is tight (see Lemma 4.1 and Theorem 4.4). Thus, as a first consequence we get that, up to a subsequence, there exists \(m^{0} \in C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) such that \(m^{{{\,\mathrm{\varepsilon }\,}}} \rightarrow m^{0}\) in \(C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\). Then, we proceed with the analysis of the value function \(u^{{{\,\mathrm{\varepsilon }\,}}}\): we show that \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,\cdot ,v)\) is equi-Lipschitz continuous, \(u^{{{\,\mathrm{\varepsilon }\,}}}(\cdot ,x,v)\) is equicontinuous and \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,\cdot )\) has decreasing oscillation w.r.t. \({{\,\mathrm{\varepsilon }\,}}\) (see Lemma 4.6 and Proposition 4.7). We finally address the locally uniform convergence of \(u^{{{\,\mathrm{\varepsilon }\,}}}\), showing that there exists a subsequence \({{\,\mathrm{\varepsilon }\,}}_{k} \downarrow 0\) such that \((u^{{{\,\mathrm{\varepsilon }\,}}_{k}}, m^{{{\,\mathrm{\varepsilon }\,}}_{k}})\) converges to a solution \((u^{0}, m^{0})\) of (1.4) (see Theorem 4.9, Proposition 4.10 and Corollary 4.12). The main issues in proving the above results are due to the lack of strict convexity and the lack of coercivity of the Hamiltonian in system (1.1). The technic we use to study our singular perturbation problem is a combination of variational approach to Hamilton–Jacobi equation and optimal transport in order to overcome the issues mentioned above.

  2. (2)

    Toward MFG of control system. Just for simplicity of notation, we restrict the attention to a Lagrangian of the form

    $$\begin{aligned} L(x,v,w, m)= \frac{1}{2}|w|^2 +\frac{1}{2}|v|^2 + L_{0} (x, m). \end{aligned}$$

    In this setting, we prove that \((u^{{{\,\mathrm{\varepsilon }\,}}}, \mu ^{{{\,\mathrm{\varepsilon }\,}}})\) converges (up to subsequence) to a solution \((u^{0}, \mu ^{0})\) to the MFG of control system

    $$\begin{aligned} {\left\{ \begin{array}{ll} (i)\,\, -\partial _{t} u^{0}(t,x) + \frac{1}{2} |D_{x}u^{0}(t,x)|^{2} - L_{0}(x,\mu ^{0}_{t})=0, &{} \quad (t,x) \in [0,T] \times \mathbb {R}^{d}\\ (ii)\,\, \partial _{t} m^{0}_{t} - {{\,\textrm{div}\,}}\big ( m^{0}_{t} D_{x}u^{0}(t,x) \big )=0, &{} \quad (t,x) \in [0,T] \times \mathbb {R}^{d}\\ (iii)\,\, \mu ^{0}_{t} = (\text {Id}(\cdot ), Du^{0}(t, \cdot )) \sharp m^{0}_{t}\\ \mu ^{0}_{0}= \mu _{0},\,\, u^{0}(T,x)=g(x, \mu ^{0}_{T}), &{} \quad x \in \mathbb {R}^{d} \end{array}\right. } \end{aligned}$$
    (1.5)

    where \(H_{0}:\mathbb {R}^d \times \mathbb {R}^d \rightarrow \mathbb {R}\) is the Legendre Transform of the function \(L_0\), \(m^{0}_{t} = \pi _1 \sharp \mu ^{0}_{t}\) and \(\text {Id}(\cdot )\) denotes the identity function. As observed before, the main difference with the previous study is the convergence of the whole measure \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}\) which requires a finer study of the Euler–Lagrange flow associated with the problem of control of acceleration (Proposition 5.2, Proposition 5.5 and Proposition 5.6). Indeed, we show again that the limit system has a MFG structure and, in particular, equations (i), (ii) are in common with system (1.4) and they differ only in the measure argument of the function \(L_{0}\). However, system (1.5) has a third equation, (iii), which describes the evolution of the flow \(\{\mu ^{0}_{t}\}_{t \in [0,T]}\): the second marginal, that is the one w.r.t. the velocity variable, is given by the push-forward of the optimal feedback function \(Du^{0}\) by the first marginal \(\{m^{0}_{t}\}_{t \in [0,T]}\). Heuristically, such an equation describes the evolution of the density distribution of controls w.r.t. the state of a typical player. For this reason, system (1.5) is called MFG system of control. In conclusion, we stress that the result can be generalized to any Lagrangian following the same arguments but with heavy notation that leads to an hard presentation of the ideas.

The paper is organized as follows. In Sect. 2, we fix the notation that will be used throughout the paper and we recall the main definitions and results from measure theory. In Sect. 3, we introduce the MFG system associated with the singular perturbation problem, we give the standing assumptions on the data and finally, we state the main results (Theorems 3.2 and 3.5). Sections 4 and 5 are devoted to the proofs of preliminary results needed to demonstrate Theorems 3.2 and 3.5, respectively.

2 Notations and preliminaries

2.1 Notation

We write below a list of symbols used throughout this paper.

  • Denote by \(\mathbb {N}\) the set of positive integers, by \(\mathbb {R}^d\) the d-dimensional real Euclidean space, by \(\langle \cdot ,\cdot \rangle \) the Euclidean scalar product, by \(|\cdot |\) the usual norm in \(\mathbb {R}^d\), and by \(B_{R}\) the open ball with center 0 and radius R.

  • For a Lebesgue-measurable subset A of \(\mathbb {R}^d\), we let \(\mathcal {L}^{d}(A)\) be the d-dimensional Lebesgue measure of A and \(\textbf{1}_{A}:\mathbb {R}^n\rightarrow \{0,1\}\) be the characteristic function of A, i.e.,

    $$\begin{aligned} \textbf{1}_{A}(x)= {\left\{ \begin{array}{ll} 1 \ \ \ {} &{}x\in A,\\ 0 &{}x \not \in A. \end{array}\right. } \end{aligned}$$

    We denote by \(L^p(A)\) (for \(1\le p\le \infty \)) the space of Lebesgue-measurable functions f with \(\Vert f\Vert _{p,A}<\infty \), where

    $$\begin{aligned}&\Vert f\Vert _{\infty , A}:={{\,\mathrm{ess\ sup}\,}}_{x \in A} |f(x)|,\\&\Vert f\Vert _{p,A}:=\left( \int _{A}|f|^{p}\ dx\right) ^{\frac{1}{p}}, \quad 1\le p<\infty . \end{aligned}$$

    For brevity, \(\Vert f\Vert _{\infty }\) and \(\Vert f\Vert _{p}\) stand for \(\Vert f\Vert _{\infty ,\mathbb {R}^d}\) and \(\Vert f\Vert _{p,\mathbb {R}^d}\), respectively.

  • \(C_b(\mathbb {R}^d)\) stands for the function space of bounded uniformly continuous functions on \(\mathbb {R}^d\). \(C^{2}_{b}(\mathbb {R}^{d})\) stands for the space of bounded functions on \(\mathbb {R}^d\) with bounded uniformly continuous first and second derivatives. \(C^k(\mathbb {R}^{d})\) (\(k\in \mathbb {N}\)) stands for the function space of k-times continuously differentiable functions on \(\mathbb {R}^d\), and \(C^\infty (\mathbb {R}^{d}):=\cap _{k=0}^\infty C^k(\mathbb {R}^{d})\). \(C_c^\infty (\mathbb {R}^{d})\) stands for the space of functions in \(C^\infty (\mathbb {R}^{d})\) with compact support. Let \(a<b\in \mathbb {R}\). \(AC([a,b];\mathbb {R}^d)\) denotes the space of absolutely continuous maps \([a,b]\rightarrow \mathbb {R}^d\).

  • For \(f \in C^{1}(\mathbb {R}^{d})\), the gradient of f is denoted by \(Df=(D_{x_{1}}f,..., D_{x_{n}}f)\), where \(D_{x_{i}}f=\frac{\partial f}{\partial x_{i}}\), \(i=1,2,\ldots ,d\). Let k be a nonnegative integer, and let \(\alpha =(\alpha _1,\ldots ,\alpha _d)\) be a multi-index of order k, i.e., \(k=|\alpha |=\alpha _1+\ldots +\alpha _d\), where each component \(\alpha _i\) is a nonnegative integer. For \(f \in C^{k}(\mathbb {R}^{d})\), define \(D^{\alpha }f:= D_{x_{1}}^{\alpha _{1}} \cdot \cdot \cdot D^{\alpha _{d}}_{x_{d}}f\).

2.2 The Wasserstein spaces

We recall here the notations and definitions of Wasserstein spaces and Wasserstein distance, for more details we refer to [4, 37].

Let \((X,\textbf{d})\) be a metric space (in the paper, we use \(X= \mathbb {R}^d\) or \(X= \mathbb {R}^d\times \mathbb {R}^d\)). Denote by \(\mathcal {B}(X)\) the Borel \(\sigma \)-algebra on X and by \(\mathcal {P}(X)\) the space of Borel probability measures on X. The support of a measure \(\mu \in \mathcal {P}(X)\), denoted by \({{\,\textrm{spt}\,}}(\mu )\), is the closed set defined by

$$\begin{aligned} {{\,\textrm{spt}\,}}(\mu ):= \Big \{x \in X: \mu (V_x)>0\ \text {for each open neighborhood } V_x \ \text { of }\ x\Big \}. \end{aligned}$$

We say that a sequence \(\{\mu _k\}_{k\in \mathbb {N}}\subset \mathcal {P}(X)\) is weakly-\(*\) convergent to \(\mu \in \mathcal {P}(X)\), denoted by \(\mu _k {\mathop {\longrightarrow }\limits ^{w^*}}\mu \), if

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _{X} f(x)\,\textrm{d}\mu _n(x)=\int _{X} f(x) \,\textrm{d}\mu (x), \quad \forall f \in C_b(X). \end{aligned}$$

For \(p\in [1,+\infty )\), the Wasserstein space of order p is defined as

$$\begin{aligned} \mathcal {P}_p(X):=\left\{ m\in \mathcal {P}(X): \int _{X} \textrm{d}(x_0,x)^p\,\textrm{d}m(x) <+\infty \right\} , \end{aligned}$$

for some (and thus all) \(x_0 \in X\). Given any two measures m and \(m^{\prime }\) in \(\mathcal {P}_p(X)\), define

$$\begin{aligned} \Pi (m,m'):=\Big \{\lambda \in \mathcal {P}(X \times X): \lambda (A\times X)=m(A),\ \lambda (X \times A)=m'(A),\ \forall A\in \mathcal {B}(X)\Big \}.\nonumber \\ \end{aligned}$$
(2.1)

The Wasserstein distance of order p between m and \(m'\) is defined by

$$\begin{aligned} d_p(m,m')=\inf _{\lambda \in \Pi (m,m')}\left( \int _{X\times X}\textrm{d}(x,y)^p\,\textrm{d}\lambda (x,y) \right) ^{1/p}. \end{aligned}$$

The distance \(d_1\) is also commonly called the Kantorovich–Rubinstein distance and can be characterized by a useful duality formula (see, for instance, [37]) as follows

$$\begin{aligned} d_1(m,m')=\sup \left\{ \int _{X} f(x)\,\textrm{d}m(x)-\int _{X} f(x)\,\textrm{d}m'(x) \ |\ f:X\rightarrow \mathbb {R} \ \ \text {is}\ 1\text {-Lipschitz}\right\} ,\nonumber \\ \end{aligned}$$
(2.2)

for all m, \(m'\in \mathcal {P}_1(X)\).

Let \(X_{1}\), \(X_{2}\) be metric spaces, let \(\mu \in \mathcal {P}(X_{1})\) and let \(f: X_{1} \rightarrow X_{2}\) be a \(\mu \) measurable map. Then, we denote by \(f \sharp \mu \in \mathcal {P}(X_{2})\) the push-forward of \(\mu \) through f defined by

$$\begin{aligned} f \sharp \mu (B):= \mu (f^{-1}(B)), \quad \forall \ B \in \mathcal {B}(X_{2}). \end{aligned}$$

More generally, in integral form, it reads as

$$\begin{aligned} \int _{X_{1}}{\varphi (f(x))\ \mu (\textrm{d}x)} = \int _{X_{2}}{\varphi (y)\ f \sharp \mu (\textrm{d}y)}. \end{aligned}$$

3 Setting and main results

3.1 Convergence to classical MFG system

Let \(L_{0}: \mathbb {R}^{2d} \times \mathcal {P}_{1}(\mathbb {R}^{d}) \rightarrow \mathbb {R}\) satisfy the following.

(M1):

\(L_{0}\) is continuous w.r.t. all variables and for any \(m \in \mathcal {P}_{1}(\mathbb {R}^{d})\) the map \((x,v) \mapsto L_{0}(x,v,m)\) belongs to \(C^{1}(\mathbb {R}^{2d})\).

(M2):

The map \(v \mapsto L_0(x, v, m)\) belongs to \(C^2(\mathbb {R}^d)\), and there exists \(M_{0} > 0\) such that for any \((x,v,m) \in \mathbb {R}^{2d} \times \mathcal {P}_{1}(\mathbb {R}^{d})\)

$$\begin{aligned}&|L_0(x, 0, m)| \le \ M_0, \end{aligned}$$
(3.1)
$$\begin{aligned}&|D_{x}L_{0}(x,v,m)| \le \ M_{0}\big (1+|v|^{2} \big ),&\end{aligned}$$
(3.2)
$$\begin{aligned}&|D_{v}L_{0}(x,v,m)| \le \ M_{0}\big ( 1 + |v|\big ), \end{aligned}$$
(3.3)
$$\begin{aligned}&\frac{1}{M_{0}}\text {Id} \le \ D^{2}_{v}L_{0}(x,v,m) \le \ M_0 \text {Id}. \end{aligned}$$
(3.4)
(M3):

There exist two moduli \(\theta : \mathbb {R}_{+} \rightarrow \mathbb {R}_{+}\) and \(\omega _{0}: \mathbb {R}_{+} \rightarrow \mathbb {R}_{+}\) such that

$$\begin{aligned} |L_{0}(x,v,m_{1})-L_{0}(x,v,m_{2})| \le \theta (|x|) \omega _{0}(d_{1}(m_{1},m_{2})), \end{aligned}$$

for any \((x,v) \in \mathbb {R}^{2d}\) and \(m_{1}\), \(m_{2} \in \mathcal {P}_{1}(\mathbb {R}^{d})\).

Observe that from (M2) one easily obtains

$$\begin{aligned} \frac{1}{4M_{0}}|v|^{2}-M_{0} \le L_{0}(x,v,m) \le \ 4M_{0}(1+|v|^{2}), \end{aligned}$$
(3.5)

and, without loss of generality, \(L_{0}(x,v,m) \ge 0\) for any \((x,v,m) \in \mathbb {R}^{2d} \times \mathcal {P}_{1}(\mathbb {R}^{d})\). Let \(H_{0}\) be the Legendre Transform of the function \(L_{0}\), i.e.,

$$\begin{aligned} H_{0}(x,p,m)=\sup _{ v \in \mathbb {R}^{d}} \big \{-\langle p,v \rangle - L_{0}(x,v,m) \big \}. \end{aligned}$$

We consider the MFG system

$$\begin{aligned} {\left\{ \begin{array}{ll} -\partial _{t} u^{{{\,\mathrm{\varepsilon }\,}}} +\frac{1}{2{{\,\mathrm{\varepsilon }\,}}}|D_{v}u^{{{\,\mathrm{\varepsilon }\,}}}|^{2} - \langle D_{x}u^{{{\,\mathrm{\varepsilon }\,}}}, v \rangle -L_{0}(x, v, m^{{{\,\mathrm{\varepsilon }\,}}}_{t})= 0, &{} (t,x,v) \in [0,T] \times \mathbb {R}^{2d}\\ \partial _{t}\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t} - \langle D_{x}\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}, v \rangle - \frac{1}{{{\,\mathrm{\varepsilon }\,}}}{{\,\textrm{div}\,}}_{v}\left( \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}D_{v}u^{{{\,\mathrm{\varepsilon }\,}}} \right) =0, &{} (t,x,v) \in [0,T] \times \mathbb {R}^{2d}\\ \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{0}=\mu _{0}, \quad u^{{{\,\mathrm{\varepsilon }\,}}}(T,x,v)=g(x,m^{{{\,\mathrm{\varepsilon }\,}}}_{T}), &{} (x,v) \in \mathbb {R}^{2d} \end{array}\right. } \end{aligned}$$
(3.6)

where \(\pi _{1}: \mathbb {R}^{2d} \rightarrow \mathbb {R}^{d}\) denotes the projection onto the first factor, i.e., \(\pi _{1} (x,v)= x\) and, so, \(m^{{{\,\mathrm{\varepsilon }\,}}}_{t}=\pi _{1} \sharp \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) represents the space marginal. We assume the following on the boundary data of the system.

(BC1):

The measure \(\mu _{0} \in \mathcal {P}(\mathbb {R}^{2d})\) is absolutely continuous w.r.t. Lebesgue measure, we still denote by \(\mu _{0}\) its density, and it has compact support.

(BC2):

The terminal costs \(g(\cdot , m)\) belong to \(C^{1}_{b}(\mathbb {R}^{d})\) such that \(M_{0} \ge \max \{\frac{1}{2}, \frac{1}{2}\Vert Dg(\cdot , m)\Vert _{\infty , \mathbb {R}^{d}}\}\) and \(g(x,\cdot )\) uniformly continuous w.r.t. space.

We also recall that \(m_{0}:= \pi _{1} \sharp \mu _{0}\).

Let \(\Gamma \) be the set of \(C^{1}\) curves \(\gamma :[0,T] \rightarrow \mathbb {R}^{d}\), endowed with the local uniform convergence of the curve and its derivative, and given \((t,x,v) \in [0,T] \times \mathbb {R}^{2d}\) let \(\Gamma _{t}(x,v)\) be the subset of \(\Gamma \) such that \(\gamma (t)=x\), \({\dot{\gamma }}(t)=v\). Similarly, let \(\Gamma _{t}(x)\) be the subset of \(\Gamma \) such that \(\gamma (t)=x\). Define the functional \(J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}: \Gamma \rightarrow \mathbb {R}\)

$$\begin{aligned} J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}(\gamma )=&\int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }(s)|^{2} + L_{0}(\gamma (s), {\dot{\gamma }}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s} + g(\gamma (T),m^{{{\,\mathrm{\varepsilon }\,}}}_{T}), \quad \text {if}\,\, \gamma \in H^{2}(0,T; \mathbb {R}^{d}) \end{aligned}$$

and set \(J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}(\gamma )=+\infty \) if \(\gamma \not \in H^{2}(0,T;\mathbb {R}^{d})\). Then, from [1, 14] we know that there exist a solution \((u^{{{\,\mathrm{\varepsilon }\,}}}, \mu ^{{{\,\mathrm{\varepsilon }\,}}}) \in W^{1,\infty }_{loc}([0,T] \times \mathbb {R}^{2d}) \times C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{2d}))\) to system (3.6) such that

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)= \inf _{\gamma \in \Gamma _{t}(x,v)} J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}(\gamma ) \end{aligned}$$
(3.7)

and for any \(t \in [0,T]\) the probability measure \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) is the image of \(\mu _{0}\) under the flow

$$\begin{aligned} {\left\{ \begin{array}{ll} {\dot{\gamma }}(t) = v(t)\\ \dot{v}(t) = - \frac{1}{{{\,\mathrm{\varepsilon }\,}}}D_{v}u^{{{\,\mathrm{\varepsilon }\,}}}(t,\gamma (t), v(t)). \end{array}\right. } \end{aligned}$$
(3.8)

That is, \(u^{{{\,\mathrm{\varepsilon }\,}}}\) solves the Hamilton–Jacobi equation in the viscosity sense and \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}\) solves the continuity equation in the sense of distributions.

Remark 3.1

Note that for a.e. \((x,v) \in \mathbb {R}^{2d}\) there exists a unique solution to system (3.8), which we will denote by \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}_{(x,v)}\), such that \(\gamma _{(x,v)}^{{{\,\mathrm{\varepsilon }\,}}}(0)=x\) and \({\dot{\gamma }}_{(x,v)}^{{{\,\mathrm{\varepsilon }\,}}}(0)=v\). Moreover, \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}_{(x,v)}(\cdot )\) is optimal for \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)\) and satisfies \(\gamma _{(x,v)}^{{{\,\mathrm{\varepsilon }\,}}}(t)=x\), \({\dot{\gamma }}_{(x,v)}^{{{\,\mathrm{\varepsilon }\,}}}(t)=v\).

Theorem 3.2

(Main result 1). Assume (M1)(M3) and (BC1), (BC2). Let \((u^{{{\,\mathrm{\varepsilon }\,}}}, \mu ^{{{\,\mathrm{\varepsilon }\,}}})\) be a solution to (3.10) and let \(m^{{{\,\mathrm{\varepsilon }\,}}}_{t}=\pi _{1} \sharp \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) for any \(t \in [0,T]\). Then, there exists a sequence \(\{{{\,\mathrm{\varepsilon }\,}}_{k}\}_{k \in \mathbb {N}}\) with \({{\,\mathrm{\varepsilon }\,}}_{k} \downarrow 0\), as \(k \rightarrow \infty \), a function \(u^{0} \in W^{1,\infty }_{loc}([0,T] \times \mathbb {R}^{d})\) and a flow of probability measures \(\{m^{0}_{t}\}_{t \in [0,T]} \in C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) such that for any \(R \ge 0\)

$$\begin{aligned} \lim _{k \rightarrow \infty } u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v) = u^{0}(t,x), \quad \text {uniformly on}\,\, [0,T] \times \overline{B}_{R} \times \overline{B}_{R} \end{aligned}$$

and

$$\begin{aligned} \lim _{k \rightarrow \infty } m^{{{\,\mathrm{\varepsilon }\,}}_{k}}= m^{0}, \quad \text {in}\,\, C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d})). \end{aligned}$$

Moreover, the following holds.

(i):

\((u^{0}, m^{0}) \in W^{1,\infty }_{loc}([0,T] \times \mathbb {R}^{d}) \times C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) is a solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} -\partial _{t} u^{0}(t,x) + H_{0}(x, D_{x}u^{0}(t,x), m^{0}_{t})=0, &{} \quad (t,x) \in [0,T] \times \mathbb {R}^{d} \\ \partial _{t} m^{0}_{t} - {{\,\textrm{div}\,}}\Big ( m^{0}_{t}D_{p}H_{0}(x, D_{x}u^{0}(t,x), m^{0}_{t}) \Big )=0, &{} \quad (t,x) \in [0,T] \times \mathbb {R}^{d} \\ m^{0}_{0}= m_{0},\,\, u^{0}(T,x)=g(x,m^{0}_{T}), &{} \quad x \in \mathbb {R}^{d}, \end{array}\right. } \end{aligned}$$
(3.9)

that is, \(u^{0}\) solves the Hamilton–Jacobi equation in the viscosity sense and \(m^{0}\) is a solution of the continuity equation in the sense of distributions.

(ii):

For any \(t \in [0,T]\), the probability measure \(m^{0}_{t}\) is the image of \(m_{0}\) under the Euler flow associated with \(L_{0}\).

Remark 3.3

Note that, assumption (M2) is needed to guarantee the well-posedness of the limit system (3.9). Indeed, following [14] and [1], (M2) can be weaken in the analysis of (3.6).

Remark 3.4

Let \((u^{{{\,\mathrm{\varepsilon }\,}}}, \mu ^{{{\,\mathrm{\varepsilon }\,}}})\) be a solution to (3.10). Assume that \(H_{0}\) is of separated form, i.e., there exists a coupling function \(F:\mathbb {R}^{d} \times \mathcal {P}_{1}(\mathbb {R}^{d}) \rightarrow \mathbb {R}\) such that

$$\begin{aligned} H_{0}(x,p,m)=H(x,p)-F(x,m), \quad \forall (x,p,m) \in \mathbb {R}^{2d} \times \mathcal {P}_{1}(\mathbb {R}^{d}). \end{aligned}$$

Moreover, assume that F is continuous w.r.t. all variables, that the map \(x \mapsto F(x,m)\) belongs to \(C^{1}_{b}(\mathbb {R}^{d})\) and that the functions F, g are monotone in the sense of Lasry–Lions, i.e.,

$$\begin{aligned} \int _{\mathbb {R}^{d}}{\big (F(x,m_{1})-F(x,m_{2}) \big )\ (m_{1}(dx)-m_{2}(dx))} \ge 0,&\quad \forall \ m_{1}, m_{2} \in \mathcal {P}_{1}(\mathbb {R}^{d})\\ \int _{\mathbb {R}^{d}}{\big (g(x,m_{1})-g(x,m_{2}) \big )\ (m_{1}(dx)-m_{2}(dx))} \ge 0, \quad&\forall \ m_{1}, m_{2} \in \mathcal {P}_{1}(\mathbb {R}^{d}). \end{aligned}$$

Then, from [1, 14] we know that that there exists a unique solution \((u^{0}, m^{0}) \in W^{1,\infty }_{loc}([0,T] \times \mathbb {R}^{d}) \times C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) of (3.13) and thus, as \((u^{{{\,\mathrm{\varepsilon }\,}}}, m^{{{\,\mathrm{\varepsilon }\,}}})\) is relatively compact then convergence of \((u^{{{\,\mathrm{\varepsilon }\,}}}, m^{{{\,\mathrm{\varepsilon }\,}}})\) holds for the whole sequence.

3.2 Convergence to MFG of control

We now consider the function \(L_{0}: \mathbb {R}^{d} \times \mathcal {P}_{1}(\mathbb {R}^{2d}) \rightarrow \mathbb {R}\), and we assume the following.

(C1):

\(L_{0}\) is continuous w.r.t. all variables and for any \(\mu \in \mathcal {P}_{1}(\mathbb {R}^{2d})\) the map \( x \mapsto L_{0}(x, \mu )\) belongs to \(C^{1}(\mathbb {R}^{d})\).

(C2):

There exist two moduli \(\theta : \mathbb {R}_{+} \rightarrow \mathbb {R}_{+}\) and \(\omega _{0}: \mathbb {R}_{+} \rightarrow \mathbb {R}_{+}\) such that

$$\begin{aligned} |L_{0}(x,\mu _{1})-L_{0}(x,\mu _{2})| \le \theta (|x|) \omega _{0}(d_{1}(\mu _{1},\mu _{2})), \end{aligned}$$

for any \(x \in \mathbb {R}^{d}\) and \(\mu _{1}\), \(\mu _{2} \in \mathcal {P}_{1}(\mathbb {R}^{2d})\).

We consider the MFG system

$$\begin{aligned} {\left\{ \begin{array}{ll} -\partial _{t} u^{{{\,\mathrm{\varepsilon }\,}}} +\frac{1}{2{{\,\mathrm{\varepsilon }\,}}}|D_{v}u^{{{\,\mathrm{\varepsilon }\,}}}|^{2} - \langle D_{x}u^{{{\,\mathrm{\varepsilon }\,}}}, v \rangle - \frac{1}{2}|v|^{2} - L_{0} (x,\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t})= 0, &{} (t,x,v) \in [0,T] \times \mathbb {R}^{2d}\\ \partial _{t}\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t} - \langle D_{x}\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t},v \rangle - \frac{1}{{{\,\mathrm{\varepsilon }\,}}}{{\,\textrm{div}\,}}_{v}\left( \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}D_{v}u^{{{\,\mathrm{\varepsilon }\,}}} \right) =0, &{} (t,x,v) \in [0,T] \times \mathbb {R}^{2d}\\ \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{0}=\mu _{0}, \quad u^{{{\,\mathrm{\varepsilon }\,}}}(T,x,v)=g(x,\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{T}), &{} (x,v) \in \mathbb {R}^{2d} \end{array}\right. } \end{aligned}$$
(3.10)

and we assume

(A1):

the measure \(\mu _{0} \in \mathcal {P}(\mathbb {R}^{2d})\) is absolutely continuous w.r.t. Lebesgue measure, we still denote by \(\mu _{0}\) its density, and it has compact support.

(A2):

The terminal costs \(g(\cdot , \mu )\) belong to \(C^{1}_{b}(\mathbb {R}^{d})\), \(g(x,\cdot )\) uniformly continuous w.r.t. space, and we have that \(M_{0} \ge \max \{\frac{1}{2}, \frac{1}{2}\Vert Dg(\cdot , \mu )\Vert _{\infty , \mathbb {R}^{d}}\}\).

Similarly to the previous part, we define the functional \(J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}: \Gamma \rightarrow \mathbb {R}\)

$$\begin{aligned} J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}(\gamma )=&\int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }(s)|^{2} + \frac{1}{2}|{\dot{\gamma }}(s)|^{2}+L_{0}(\gamma (s), \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s}\\ {}&+ g(\gamma (T),\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{T}), \quad \text {if}\,\, \gamma \in H^{2}(0,T; \mathbb {R}^{d}) \end{aligned}$$

and set \(J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}(\gamma )=+\infty \) if \(\gamma \not \in H^{2}(0,T;\mathbb {R}^{d})\). Then, from [1, 14] we know that there exist a solution \((u^{{{\,\mathrm{\varepsilon }\,}}}, \mu ^{{{\,\mathrm{\varepsilon }\,}}}) \in W^{1,\infty }_{loc}([0,T] \times \mathbb {R}^{2d}) \times C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{2d}))\) to system (3.6) such that

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)= \inf _{\gamma \in \Gamma _{t}(x,v)} J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}(\gamma ) \end{aligned}$$
(3.11)

and for any \(t \in [0,T]\) the probability measure \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) is the image of \(\mu _{0}\) under the flow

$$\begin{aligned} {\left\{ \begin{array}{ll} {\dot{\gamma }}(t) = v(t)\\ \dot{v}(t) = - \frac{1}{{{\,\mathrm{\varepsilon }\,}}}D_{v}u^{{{\,\mathrm{\varepsilon }\,}}}(t,\gamma (t), v(t)). \end{array}\right. } \end{aligned}$$
(3.12)

That is, \(u^{{{\,\mathrm{\varepsilon }\,}}}\) solves the Hamilton–Jacobi equation in the viscosity sense and \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}\) solves the continuity equation in the sense of distributions.

Theorem 3.5

(Main result 2). Assume (C1)(C3) and (A1), (A2). Let \((u^{{{\,\mathrm{\varepsilon }\,}}}, \mu ^{{{\,\mathrm{\varepsilon }\,}}})\) be a solution to (3.10). Then, there exists a sequence \(\{{{\,\mathrm{\varepsilon }\,}}_{k}\}_{k \in \mathbb {N}}\) with \({{\,\mathrm{\varepsilon }\,}}_{k} \downarrow 0\), as \(k \rightarrow \infty \), a function \(u^{0} \in W^{1,\infty }_{loc}([0,T] \times \mathbb {R}^{d})\) and a flow of probability measures \(\{\mu ^{0}_{t}\}_{t \in [0,T]} \in C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{2d}))\) such that for any \(R \ge 0\)

$$\begin{aligned} \lim _{k \rightarrow \infty } u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v) = u^{0}(t,x), \quad \text {uniformly on}\,\, [0,T] \times \overline{B}_{R} \times \overline{B}_{R} \end{aligned}$$

and

$$\begin{aligned} \lim _{k \rightarrow \infty } \mu ^{{{\,\mathrm{\varepsilon }\,}}_{k}}= \mu ^{0}, \quad \text {in}\,\, C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{2d})). \end{aligned}$$

Moreover, we have that the pair \((u^{0}, \mu ^{0}) \in W^{1,\infty }_{loc}([0,T] \times \mathbb {R}^{d}) \times C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) is a solution of the MFG of control system

$$\begin{aligned} {\left\{ \begin{array}{ll} (i)\,\, -\partial _{t} u^{0}(t,x) + \frac{1}{2}|D_{x}u^{0}(t,x)|^{2}- L_{0}(x, \mu ^{0}_{t})=0, &{} \quad (t,x) \in [0,T] \times \mathbb {R}^{d}\\ (ii)\,\, \partial _{t} m^{0}_{t} - {{\,\textrm{div}\,}}\big ( m^{0}_{t}D_{x}u^{0}(t,x) \big )=0, &{} \quad (t,x) \in [0,T] \times \mathbb {R}^{d}\\ (iii)\,\, \mu ^{0}_{t} = (\text {Id}(\cdot ), D_{x}u^{0}(t, \cdot )) \sharp m^{0}_{t}, &{} \quad t \in [0,T]\\ m^{0}_{0}= m_{0},\,\, u^{0}(T,x)=g(x,\mu ^{0}_{T}), &{} \quad x \in \mathbb {R}^{d}, \end{array}\right. } \end{aligned}$$
(3.13)

that is: \(u^{0}\) solves the Hamilton–Jacobi equation in the viscosity sense, \(m^{0}_{t}=\pi _{1} \sharp \mu ^{0}_{t}\), for all \(t \in [0,T]\), is a solution of the continuity equation in the sense of distributions and the measure \(\mu ^0\) satisfies equality (iii) for any \(t \in [0,T]\).

4 Proof of Theorem 3.2

In order to prove Theorem 3.2, we proceed by steps analyzing the behavior of the value function \(u^{{{\,\mathrm{\varepsilon }\,}}}\) and that of the flow of probability measures \(\{m^{{{\,\mathrm{\varepsilon }\,}}}_{t}\}_{t \in [0,T]}\) separately. First, we show that \(u^{{{\,\mathrm{\varepsilon }\,}}}\) is equibounded and that, up to a subsequence, \(m^{{{\,\mathrm{\varepsilon }\,}}}\) converges to a flow of probability measure in \(C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\). Then, we address the problem of convergence of the value function, up to a subsequence, to a solution of a suitable Hamilton–Jacobi equation and we study the limit of its minimizing trajectories. Consequently, we are able to characterize the limit flow of measures as solution of a continuity equation coupled with the Hamilton–Jacobi equation, previously constructed, and together they define the limit MFG system (3.9).

Lemma 4.1

Assume (M1)(M3) and (BC1), (BC2). Then, we have that

$$\begin{aligned} -TM_{0}-\Vert g(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}} \le u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v) \le 5M_{0}T(1+|v|^{2}) + \Vert g(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}}, \end{aligned}$$

for any \((t,x,v) \in [0,T] \times \mathbb {R}^{2d}\) and for any \({{\,\mathrm{\varepsilon }\,}}> 0\).

Proof

First, since \(u^{{{\,\mathrm{\varepsilon }\,}}}\) satisfy (3.7), from (3.5) and (BC) follows that for any \((t,x,v) \in [0,T] \times \mathbb {R}^{2d}\) there holds

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v) \ge - C_{0}T-\Vert g(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}}. \end{aligned}$$

On the other hand, let us recall that \(u^{{{\,\mathrm{\varepsilon }\,}}}\) solves the Hamilton–Jacobi equation

$$\begin{aligned} -\partial _{t} u^{{{\,\mathrm{\varepsilon }\,}}} +\frac{1}{2{{\,\mathrm{\varepsilon }\,}}}|D_{v}u^{{{\,\mathrm{\varepsilon }\,}}}|^{2} - \langle D_{x}u^{{{\,\mathrm{\varepsilon }\,}}}, v \rangle -L_{0}(x, v, m^{{{\,\mathrm{\varepsilon }\,}}}_{t})= 0. \end{aligned}$$
(4.1)

Then, the function

$$\begin{aligned} \zeta (t,x,v)= g(x,m^{{{\,\mathrm{\varepsilon }\,}}}_{T}) +C(1+|v|^{2})(T-t), \quad (t,x,v) \in [0,T] \times \mathbb {R}^{2d} \end{aligned}$$

is a supersolution to (4.1) for a suitable choice of the real constant \(C \ge 0\). Indeed, we have that

$$\begin{aligned} - \ {}&\partial _{t}\zeta (t,x,v) + \frac{1}{2{{\,\mathrm{\varepsilon }\,}}}|D_{v}\zeta (t,x,v)|^{2} - \langle D_{x}\zeta (t,x,v), v \rangle -L_{0}(x,v)\\ \ge \ {}&C(1+|v|^{2}) + 2\frac{(T-t)^{2}C^{2}}{{{\,\mathrm{\varepsilon }\,}}}|v|^{2} -\langle D_{x}g(x,m^{{{\,\mathrm{\varepsilon }\,}}}_{T}), v\rangle - 4M_{0}(1+|v|^{2})\\ \ge \ {}&C(1+|v|^{2}) -\frac{1}{2}\Vert Dg(\cdot ,m^{{{\,\mathrm{\varepsilon }\,}}}_{T}) \Vert _{\infty ,\mathbb {R}^{d}} - \frac{1}{2}|v|^{2} - 4M_{0}(1+|v|^{2}) \end{aligned}$$

where the last inequality holds by Young’s inequality. Thus, taking \(C=5M_{0}\) by (BC) we obtain

$$\begin{aligned} M_{0}(1+|v|^{2}) -\frac{1}{2}\Vert Dg(\cdot ,m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty ,\mathbb {R}^{d}} - \frac{1}{2}|v|^{2} \ge 0. \end{aligned}$$

So, we get the result by Comparison Theorem [8, Theorem 2.12]. \(\square \)

Corollary 4.2

Assume (M1)(M3) and (BC1), (BC2). Let \((t,x,v) \in [0,T] \times \mathbb {R}^{2d}\), and let \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) be a minimizer for \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)\). Then, there exists a constant \(Q_{1} \ge 0\) such that

$$\begin{aligned} \int _{t}^{T}{|{\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s)|^{2}\ \textrm{d}s} \le Q_{1}(1+|v|^{2}), \quad \forall \ {{\,\mathrm{\varepsilon }\,}}> 0. \end{aligned}$$

where \(Q_{1}\) is independent of \({{\,\mathrm{\varepsilon }\,}}\), t, x and v.

Proof

On the one hand, from Lemma 4.1 we know that

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v) \le {5M}_{0}T\big (1+|v|^{2}\big ) + \Vert g(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}}, \quad \forall \ (t,x,v) \in [0,T] \times \mathbb {R}^{2d}. \end{aligned}$$

On the other hand, let \((t,x,v) \in [0,T] \times \mathbb {R}^{2d}\) and let \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) be a minimizer for \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)\). Then, by (3.5) we have that

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)&=\ \int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}}(s)|^{2} + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s} + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T),m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\\ \ge&\int _{t}^{T}{L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s})\ \textrm{d}s} -\Vert g(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}}\\ \ge \ {}&\int _{t}^{T}{\left( \frac{1}{4M_{0}}|{\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s)|^{2} - M_{0} \right) \ \textrm{d}s} -\Vert g(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}}. \end{aligned}$$

Therefore, combining the above inequalities we get

$$\begin{aligned} \int _{t}^{T}{|{\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s)|^{2}\ \textrm{d}s} \le 4M_{0}\big (\Vert g(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}} + 5M_{0}T(1+|v|^{2})\big )=:Q_{1}(1+|v|^{2}) \end{aligned}$$

where \(Q_{1}\) depends only on \(M_{0}\), T and \(\Vert g(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}}\) which is bounded uniformly in \(m^{{{\,\mathrm{\varepsilon }\,}}}_{T}\). \(\square \)

Corollary 4.3

Assume (M1)(M3) and (BC1), (BC2). Then, there exists a constant \(Q_{2} \ge 0\) such that for any \(s_{1}\), \(s_{2} \in [0,T]\) with \(s_{1} \le s_{2}\) there holds

$$\begin{aligned} d_{1}(m^{{{\,\mathrm{\varepsilon }\,}}}_{s_{2}}, m^{{{\,\mathrm{\varepsilon }\,}}}_{s_{1}}) \le Q_{2} |s_{1}-s_{2}|^{\frac{1}{2}}, \quad \forall \ {{\,\mathrm{\varepsilon }\,}}> 0 \end{aligned}$$

where \(Q_{2}\) is independent of \({{\,\mathrm{\varepsilon }\,}}\).

Proof

We first recall that for any \(t \in [0,T]\) we know that \(m^{{{\,\mathrm{\varepsilon }\,}}}_{t}=\pi _{1} \sharp \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) where \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) is the image of \(\mu _{0}\) under the flow (3.8) whose space marginal we denote by \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}_{(x,v)}\) for \((x,v) \in \mathbb {R}^{2d}\).

Let \(s_{1}\), \(s_{2} \in [0,T]\) be such that \(s_{1} \le s_{2}\). Then, by (2.2) we have that

$$\begin{aligned} d_{1}(m^{{{\,\mathrm{\varepsilon }\,}}}_{s_{1}}, m^{{{\,\mathrm{\varepsilon }\,}}}_{s_{2}}) \le \int _{\mathbb {R}^{d}}{|\gamma ^{{{\,\mathrm{\varepsilon }\,}}}_{(x,v)}(s_{1})-\gamma ^{{{\,\mathrm{\varepsilon }\,}}}_{(x,v)}(s_{2})|\ \mu _{0}(\textrm{d}x,\textrm{d}v)} \end{aligned}$$

and thus, appealing to Corollary 4.2 and the Hölder inequality we obtain

$$\begin{aligned} d_{1}(m^{{{\,\mathrm{\varepsilon }\,}}}_{s_{1}}, m^{{{\,\mathrm{\varepsilon }\,}}}_{s_{2}}) \le |s_{1}-s_{2}|^{\frac{1}{2}}\left( \int _{\mathbb {R}^{2d}}{Q_{1}(1+|v|^{2})\ \mu _{0}(\textrm{d}x,\textrm{d}v)}\right) ^{\frac{1}{2}}. \end{aligned}$$

So, since \(\mu _{0}\) has compact support we get the result setting

$$\begin{aligned} Q_{2}=\left( \int _{\mathbb {R}^{2d}}{Q_{1}(1+|v|^{2})\ \mu _{0}(\textrm{d}x,\textrm{d}v)}\right) ^{\frac{1}{2}}. \end{aligned}$$

\(\square \)

We are now ready to prove that the flow of probability measures \(m^{{{\,\mathrm{\varepsilon }\,}}}\) converges, up to a subsequence. First, we recall that for any \(t \in [0,T]\) the measure \(m_{t}^{{{\,\mathrm{\varepsilon }\,}}}\) is the space marginal of \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) which is given by the push-forward of the initial distribution \(\mu _{0}\) by the optimal flow (3.8), that is

$$\begin{aligned} {\left\{ \begin{array}{ll} {\dot{\gamma }}_{(x,v)}(t) = v(t), &{} \gamma _{(x,v)}(0)=x\\ \dot{v}(t) = - \frac{1}{{{\,\mathrm{\varepsilon }\,}}}D_{v}u^{{{\,\mathrm{\varepsilon }\,}}}(t,\gamma _{(x,v)}(t), v(t)), &{} v(0)=v. \end{array}\right. } \end{aligned}$$

Theorem 4.4

Assume (M1)(M3) and (BC1), (BC2). Then, the flow of measures \(\{m^{{{\,\mathrm{\varepsilon }\,}}}_{t}\}_{t \in [0,T]}\) is tight and there exists a sequence \(\{{{\,\mathrm{\varepsilon }\,}}_{k}\}_{k \in \mathbb {N}}\) such that \(m^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) converges to some probability measure \(m^{0}\) in \(C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\).

Proof

Since \(m_{t}^{{{\,\mathrm{\varepsilon }\,}}}=\pi _{1} \sharp \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\), for any \(t \in [0,T]\), where \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t} \) is given by push-forward of \(\mu _{0}\) under the flow (3.8), we know that

$$\begin{aligned} \int _{\mathbb {R}^{d}}{|x|^{2}\ m_{t}^{{{\,\mathrm{\varepsilon }\,}}}(\textrm{d}x)}=\int _{\mathbb {R}^{2d}}{|\gamma _{(x,v)}^{{{\,\mathrm{\varepsilon }\,}}}(t)|^{2}\ \mu _{0}(\textrm{d}x,\textrm{d}v)}. \end{aligned}$$

So, we are interested in estimating the curve \(\gamma _{(x,v)}^{{{\,\mathrm{\varepsilon }\,}}}\) for any (xv), uniformly in \({{\,\mathrm{\varepsilon }\,}}>0\). In order to get it, from Corollary 4.2 we immediately deduce that

$$\begin{aligned} |\gamma ^{{{\,\mathrm{\varepsilon }\,}}}_{(x,v)}(s)| \le |x| + \sqrt{T}\sqrt{Q_{1}}(1+|v|^{2})^{\frac{1}{2}}, \quad \forall \ s \in [0,T]. \end{aligned}$$

Hence, for any \(t \ge 0\) we have that

$$\begin{aligned} \int _{\mathbb {R}^{d}}{|x|^{2}\ m^{{{\,\mathrm{\varepsilon }\,}}}_{t}(\textrm{d}x)}&= \int _{\mathbb {R}^{d}}{|\gamma ^{{{\,\mathrm{\varepsilon }\,}}}_{(x,v)}(t)|^{2}\ \mu _{0}(\textrm{d}x,\textrm{d}v)}\\&\le \int _{\mathbb {R}^{2d}}{C_{0}\big (|x|^{2} + TQ_{1}(1+|v|^{2})\big )\ \mu _{0}(\textrm{d}x,\textrm{d}v)} \end{aligned}$$

for some constant \(C_{0} \ge 0\). Thus, since \(\mu _{0}\) has compact support, we deduce that \(\{m^{{{\,\mathrm{\varepsilon }\,}}}_{t}\}_{t \in [0,T]}\) has bounded second-order momentum, uniformly in \({{\,\mathrm{\varepsilon }\,}}> 0\) and, consequently, \(\{m^{{{\,\mathrm{\varepsilon }\,}}}_{t}\}_{t \in [0,T]}\) is tight. Therefore, by Prokhorov Theorem and Ascoli–Arzela Theorem, \(\mu _{0}\) has uniformly bounded support and by Corollary 4.3\(m^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) is equicontinuous in time, there exists a sequence \(\{{{\,\mathrm{\varepsilon }\,}}_{k}\}_{k \in \mathbb {N}}\) and measure \(m^{0} \in C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) such that \(m^{{{\,\mathrm{\varepsilon }\,}}_{k}} \rightarrow m^{0}\) in \(C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\).\(\square \)

Next, we turn to the convergence of the value function \(u^{{{\,\mathrm{\varepsilon }\,}}}\). Before proving it, we need preliminary estimates on the oscillation of the value function w.r.t. velocity variable and then, w.r.t. time and space variable. In particular, we will show that the function \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,\cdot )\) has decreasing oscillation w.r.t. \({{\,\mathrm{\varepsilon }\,}}\), which will allow us to conclude that the limit function does not depend on v.

Lemma 4.5

Assume (M1)(M3) and (BC1), (BC2). Let \(R \ge 0\) and let \((x,v_{0})\), \((x,v) \in \mathbb {R}^{d} \times \overline{B}_{R}\). Then, there exists \(C_{R} \ge 0\) and a parametric curve \(\sigma : [0,\sqrt{{{\,\mathrm{\varepsilon }\,}}}] \rightarrow \mathbb {R}^{d}\) such that

$$\begin{aligned} \sigma (0)=\sigma (\sqrt{{{\,\mathrm{\varepsilon }\,}}}) = x, \quad {\dot{\sigma }}(0)=v_{0}, \quad {\dot{\sigma }}(\sqrt{{{\,\mathrm{\varepsilon }\,}}})=v \end{aligned}$$

and

$$\begin{aligned} \frac{1}{\sqrt{{{\,\mathrm{\varepsilon }\,}}}}\int _{0}^{\sqrt{{{\,\mathrm{\varepsilon }\,}}}}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\sigma }(s)|^{2} + L_{0}(\sigma (s), {\dot{\sigma }}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s} \le C_{R} \end{aligned}$$

where \(C_{R}\) is independent of \({{\,\mathrm{\varepsilon }\,}}\), x, v and \(v_{0}\).

Proof

Let \(R \ge 0\) and let \((x,v_{0})\), \((x,v) \in \mathbb {R}^{d} \times \overline{B}_{R}\). Define the curve \(\sigma :[0,\sqrt{{{\,\mathrm{\varepsilon }\,}}}] \rightarrow \mathbb {R}^{d}\) by

$$\begin{aligned} \sigma (t)=x+v_{0}t+Bt^{2} + At^{3} \end{aligned}$$

with A, \(B \in \mathbb {R}\) satisfying the following conditions

$$\begin{aligned} \sigma (0)=\sigma (\sqrt{{{\,\mathrm{\varepsilon }\,}}}) = x, \quad {\dot{\sigma }}(0)=v_{0}, \quad {\dot{\sigma }}(\sqrt{{{\,\mathrm{\varepsilon }\,}}})=v. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} B= -(2v_{0}+v){{\,\mathrm{\varepsilon }\,}}^{-\frac{1}{2}},\\ A= (v+v_{0}){{\,\mathrm{\varepsilon }\,}}^{-1}. \end{array}\right. } \end{aligned}$$

Hence, we get

$$\begin{aligned}&\int _{0}^{\sqrt{{{\,\mathrm{\varepsilon }\,}}}}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\sigma }(s)|^{2} + L_{0}(\sigma (s), {\dot{\sigma }}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s})\right) \ \textrm{d}s} \\&\quad \le \int _{0}^{\sqrt{{{\,\mathrm{\varepsilon }\,}}}}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|2B+6At|^{2} + 4M_{0}(1+|v+2tB+3t^{2}A|^{2})\right) \ \textrm{d}s} \le \widehat{C}\sqrt{{{\,\mathrm{\varepsilon }\,}}}R^{2} \end{aligned}$$

for some positive constant \(\widehat{C}\) and the proof is thus complete. \(\square \)

Lemma 4.6

Assume (M1), (M2) and (BC). Let \(R \ge 0\), let \(T > 1\) and \({{\,\mathrm{\varepsilon }\,}}> 0\). Then, there exists \(\widehat{C}_{R}({{\,\mathrm{\varepsilon }\,}}) \ge 0\) such that for any \(t \in [0,T]\), any \(x \in \mathbb {R}^{d}\), and any v, w in \(\overline{B}_{R}\) there holds

$$\begin{aligned} |u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)-u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,w)| \le \widehat{C}_{R}({{\,\mathrm{\varepsilon }\,}}) \end{aligned}$$

and \(\widehat{C}_{R}({{\,\mathrm{\varepsilon }\,}}) \rightarrow 0\) as \({{\,\mathrm{\varepsilon }\,}}\downarrow 0\).

Proof

Fix \(R \ge 0\) and take (xv), \((x,w) \in \mathbb {R}^{d} \times \overline{B}_{R}\). Let \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) be a minimizer for \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)\) and define the curve

$$\begin{aligned} {\widehat{\gamma }}(s)= {\left\{ \begin{array}{ll} \sigma (s-t), &{} \quad s \in [t,t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}]\\ \gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), &{} \quad s \in [t+\sqrt{{{\,\mathrm{\varepsilon }\,}}},T] \end{array}\right. } \end{aligned}$$

where \(\sigma : [0,\sqrt{{{\,\mathrm{\varepsilon }\,}}}] \rightarrow \mathbb {R}^{2d}\) connects, in the sense of Lemma 4.5, (xw) with (xv). Then, we obtain

$$\begin{aligned}&u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,w) - u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v) \le \ \int _{t}^{t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\sigma }(s-t)|^{2} + L_{0}(\sigma (s-t), {\dot{\sigma }}(s-t), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s}\\&\quad +\ \int _{t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}})|^{2} + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s}\\&\quad +\ g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), m^{{{\,\mathrm{\varepsilon }\,}}}_{T}) - u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v) \\&=\ \int _{t}^{t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\sigma }(s-t)|^{2} + L_{0}(\sigma (s-t), {\dot{\sigma }}(s-t), m^{{{\,\mathrm{\varepsilon }\,}}}_{s-t}) \right) \ \textrm{d}s}\\&\quad +\ \int _{t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}})|^{2} + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), m^{{{\,\mathrm{\varepsilon }\,}}}_{s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}}) \right) \ \textrm{d}s}\\&\quad +\ g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T), m^{{{\,\mathrm{\varepsilon }\,}}}_{T}) + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), m^{{{\,\mathrm{\varepsilon }\,}}}_{T})-g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T), m^{{{\,\mathrm{\varepsilon }\,}}}_{T}) - u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)\\&\quad +\ \int _{t}^{t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}}{\big (L_{0}(\sigma (s-t), {\dot{\sigma }}(s-t), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) -L_{0}(\sigma (s-t), {\dot{\sigma }}(s-t), m^{{{\,\mathrm{\varepsilon }\,}}}_{s-t})\big )\ \textrm{d}s}\\&\quad +\ \int _{t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}}^{T}{\big (L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) - L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), m^{{{\,\mathrm{\varepsilon }\,}}}_{s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}} \big )\ \textrm{d}s}. \end{aligned}$$

Now, from Lemma 4.5 we know that

$$\begin{aligned} \int _{t}^{t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\sigma }(s-t)|^{2} + L_{0}(\sigma (s-t), {\dot{\sigma }}(s-t), m^{{{\,\mathrm{\varepsilon }\,}}}_{s-t}) \right) \ \textrm{d}s} \le C_{R}\sqrt{{{\,\mathrm{\varepsilon }\,}}}, \end{aligned}$$
(4.2)

and, moreover, from the optimality of \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) we get

$$\begin{aligned} \begin{aligned}&\int _{t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}})|^{2} + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), m^{{{\,\mathrm{\varepsilon }\,}}}_{s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}}) \right) \ \textrm{d}s} -u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)\\&\quad \le - \int _{T-\sqrt{{{\,\mathrm{\varepsilon }\,}}}}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}}(s)|^{2}+L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s} \le 0. \end{aligned} \end{aligned}$$
(4.3)

Then, as observed before from Corollary 4.2 we have that

$$\begin{aligned} |\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s)| \le |x| + \sqrt{T}\sqrt{Q_{1}}(1+|v|^{2})^{\frac{1}{2}}, \quad \forall \ s \in [0,T] \end{aligned}$$

and also that the curve \(\sigma \) is bounded. Hence, by (M3) and Corollary 4.3 we deduce that there exists \(P({{\,\mathrm{\varepsilon }\,}}) \ge 0\), with \(P({{\,\mathrm{\varepsilon }\,}}) \rightarrow 0\) as \({{\,\mathrm{\varepsilon }\,}}\downarrow 0\), such that

$$\begin{aligned} \begin{aligned}&\int _{t}^{t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}}{\big (L_{0}(\sigma (s-t), {\dot{\sigma }}(s-t), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) -L_{0}(\sigma (s-t), {\dot{\sigma }}(s-t), m^{{{\,\mathrm{\varepsilon }\,}}}_{s-t})\big )\ \textrm{d}s}\\&\quad + \int _{t+\sqrt{{{\,\mathrm{\varepsilon }\,}}}}^{T}{\big (L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) - L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), m^{{{\,\mathrm{\varepsilon }\,}}}_{s-\sqrt{{{\,\mathrm{\varepsilon }\,}}}} \big )\ \textrm{d}s}\\&\quad + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T-\sqrt{{{\,\mathrm{\varepsilon }\,}}}), m^{{{\,\mathrm{\varepsilon }\,}}}_{T})-g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T), m^{{{\,\mathrm{\varepsilon }\,}}}_{T}) \le P({{\,\mathrm{\varepsilon }\,}}) \end{aligned} \end{aligned}$$
(4.4)

where we have used that the modulus \(\theta \) in (M3) is bounded from the boundedness of \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) and \(\sigma \). Therefore, combining (4.2), (4.3) and (4.4) we get the result. \(\square \)

Proposition 4.7

Assume (M1)(M3) and (BC1), (BC2). Then, for any \(R \ge 0\) there exists a modulus \(\omega _{R}:\mathbb {R}_{+} \rightarrow \mathbb {R}_{+}\) and a constant \(C_{1} \ge 0\), independent of R, such that for any \({{\,\mathrm{\varepsilon }\,}}> 0\) the following holds:

$$\begin{aligned} |u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v) - u^{{{\,\mathrm{\varepsilon }\,}}}(s,x,v)| \le \ {}&\omega _{R}(|t-s|), \quad \forall \ (t,s,x,v) \in [0,T] \times [0,T] \times \overline{B}_{R} \times \overline{B}_{R} \end{aligned}$$
(4.5)
$$\begin{aligned} |D_{x}u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)| \le \ {}&C_{1} T(1+|v|^{2}), \quad \text {a.e.}\,\, (t,x,v) \in [0,T] \times \mathbb {R}^{d} \times \mathbb {R}^{d}. \end{aligned}$$
(4.6)

Proof

We begin by proving (4.6). Let \((t,x,v) \in [0,T] \times \mathbb {R}^{d} \times \mathbb {R}^{d}\) and let \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) be a minimizer for \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)\). Then, from (3.2) we get

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}}(t,x+h,v)&\le \int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}}(s)|^{2} + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s)+h, {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s} + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T)+h, m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\\&= u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v) + \int _{t}^{T}{\big (L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s)+h, {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s})-L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s})\big )\ \textrm{d}s}\\&\quad + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T)+h, m^{{{\,\mathrm{\varepsilon }\,}}}_{T})- g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T), m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\\&\le u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v) + \int _{t}^{T}{4M_{0}|h|(1+|{\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s) |^{2})\ \textrm{d}s} + \Vert Dg(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}}|h|. \end{aligned}$$

Hence, Corollary 4.2 yields to the conclusion.

Next, we proceed to show (4.5). Let \(R \ge 0\) and take \((t,x,v) \in [0,T] \times \overline{B}_{R} \times \overline{B}_{R}\). Let \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) be a minimizer for \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)\) and let \(h \in [0,T-t]\). Then, we have that

$$\begin{aligned}&u^{{{\,\mathrm{\varepsilon }\,}}}(t+h,x,v) \\&\quad \le \int _{t+h}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}}(s-h)|^{2} + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-h), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-h), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s} + g(\gamma (T-h), m^{{{\,\mathrm{\varepsilon }\,}}}_{T}) \\&\quad = \int _{t+h}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}}(s-h)|^{2} + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-h), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-h), m^{{{\,\mathrm{\varepsilon }\,}}}_{s-h}) \right) \ \textrm{d}s} + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T), m^{{{\,\mathrm{\varepsilon }\,}}}_{T}) \\&\qquad +\int _{t+h}^{T}{\big (L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-h), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-h), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) - L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-h), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s-h), m^{{{\,\mathrm{\varepsilon }\,}}}_{s-h}) \big )\ \textrm{d}s} \\&\qquad + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T-h), m^{{{\,\mathrm{\varepsilon }\,}}}_{T})-g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(T), m^{{{\,\mathrm{\varepsilon }\,}}}_{T}) \\&\quad \le u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v) + \int _{t+h}^{T}{\theta (|\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s-h)|)\omega _{0}(d_{1}(m^{{{\,\mathrm{\varepsilon }\,}}}_{s}, m^{{{\,\mathrm{\varepsilon }\,}}}_{s-h}))\ \textrm{d}s} + \Vert Dg(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}}|h| \end{aligned}$$

where the last inequality holds by (M3). Hence, from Corollary 4.2 we know that

$$\begin{aligned} |\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s)| \le |x| + \sqrt{T}\sqrt{Q_{1}}(1+|v|^{2})^{\frac{1}{2}}, \quad \forall \ s \in [0,T] \end{aligned}$$

and thus, \(\theta (\cdot )\) turns out to be bounded. Therefore, appealing to Corollary 4.3 we obtain

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}}(t+h,x,v) - u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v) \le T\theta (R)\omega _{0}(|h|^{\frac{1}{2}}) + \Vert Dg(\cdot , m^{{{\,\mathrm{\varepsilon }\,}}}_{T})\Vert _{\infty , \mathbb {R}^{d}}|h|. \end{aligned}$$
(4.7)

On the other hand, let \(R \ge 0\) and let \((t,x,v) \in [0,T] \times \overline{B}_{R}\times \overline{B}_{R}\). For \(h \in [0,T-t]\), define the curve \(\gamma :[t,t+h] \rightarrow \mathbb {R}^{d}\) by \(\gamma (s)=x+(s-t)v\). Then, by Dynamic Programming Principle we deduce that

$$\begin{aligned} \begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)&\le \int _{t}^{t+h}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }(s)|^{2} + L_{0}(\gamma (s), {\dot{\gamma }}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s} \\&\quad + u^{{{\,\mathrm{\varepsilon }\,}}}(t+h, \gamma (t+h), {\dot{\gamma }}(t+h)) \\&= \int _{t}^{t+h}{L_{0}(x+(s-t)v, v, m^{{{\,\mathrm{\varepsilon }\,}}}_{s})\ \textrm{d}s} + u^{{{\,\mathrm{\varepsilon }\,}}}(t+h, x+hv, v) \\&\le 4M_{0}(1+R^{2})|h| + u^{{{\,\mathrm{\varepsilon }\,}}}(t+h,x,v) + C_{1}T(1+R^{2})|h| \end{aligned} \end{aligned}$$
(4.8)

where we applied (3.5) and (4.6) to get the last inequality. Therefore, combining (4.7) and (4.8) the proof is complete. \(\square \)

Remark 4.8

Next, we study the behavior of the value function \(u^{{{\,\mathrm{\varepsilon }\,}}}\) as \({{\,\mathrm{\varepsilon }\,}}\rightarrow 0\) and before doing that we recall the following argument needed to get uniform convergence to a function which does not depend on v. Assume that there exists a nonnegative function \(\Theta (\delta _{0}, {{\,\mathrm{\varepsilon }\,}}_{0}, R_{0})\) such that

$$\begin{aligned} \Theta (\delta _{0}, \varepsilon _{0}, R_{0}) \rightarrow 0, \quad \text {as}\quad \varepsilon _{0}, \delta _{0} \downarrow 0, \end{aligned}$$

and assume that for any \(|t_{1}-t_{2}| + |x_{1} - x_{2}| \le \delta _{0}\), any \({{\,\mathrm{\varepsilon }\,}}\le {{\,\mathrm{\varepsilon }\,}}_{0}\) and any \(|x_{i}|\), \(|v_{i}| \le R_{0}\) (\(i=1,2\)) there holds

$$\begin{aligned} |u^{{{\,\mathrm{\varepsilon }\,}}}(t_{1}, x_{1}, v_{1}) - u^{{{\,\mathrm{\varepsilon }\,}}}(t_{2}, x_{2}, v_{2})| \le \Theta (\delta _{0}, \varepsilon _{0}, R_{0}). \end{aligned}$$

Then: if \(u^{{{\,\mathrm{\varepsilon }\,}}}\) converge point-wise, then \(u^{{{\,\mathrm{\varepsilon }\,}}}\) converges locally uniformly and the limit function does not depend on v. \(\square \)

Let \(m^{0} \in C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) be the flow of measures obtained in Theorem 4.4 as limit of the flow \(m^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) in \(C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) for some subsequence \({{\,\mathrm{\varepsilon }\,}}_{k} \downarrow 0\). Define the function \(u^{0}: [0,T] \times \mathbb {R}^{2d} \rightarrow \mathbb {R}\) by

$$\begin{aligned} u^{0}(t,x)= \inf _{\gamma \in \Gamma _{t}(x)} \left\{ \int _{t}^{T}{L_{0}(\gamma (s), {\dot{\gamma }}(s), m^{0}_{s})\ \textrm{d}s} + g(\gamma (T), m^{0}_{T})\right\} . \end{aligned}$$
(4.9)

We will prove now that for the subsequence \({{\,\mathrm{\varepsilon }\,}}_{k}\) the sequence of value functions \(u^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) locally uniformly converges to \(u^{0}\).

Theorem 4.9

Assume (M1)(M3) and (BC1), (BC2). Then, there exists a subsequence \({{\,\mathrm{\varepsilon }\,}}_{k} \downarrow 0\) such that \(u^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) locally uniformly converges to \(u^{0}\).

Proof

We proceed to show first the point-wise convergence of \(u^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) to \(u^{0}\), for some subsequence \({{\,\mathrm{\varepsilon }\,}}_{k} \downarrow 0\), and then, using Remark 4.8, i.e., constructing such a modulus \(\Theta \), we deduce that the convergence is locally uniform.

From Theorem 4.4, let \({{\,\mathrm{\varepsilon }\,}}_{k}\) be the subsequence such that \(m^{{{\,\mathrm{\varepsilon }\,}}_{k}} \rightarrow m^{0}\) in \(C([0,T];\mathcal {P}_{1}(\mathbb {R}^{d}))\) as \(k \rightarrow \infty \). Let \(R \ge 0\), let \((t,x,v) \in [0,T] \times \mathbb {R}^{d} \times \overline{B}_{R}\) and let \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) be a minimizer for \(u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v)\). Then, we have that

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v)&= \int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}_{k}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s)|^{2} + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{s})\right) \ \textrm{d}s} + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{T})\\&\ge \ \ \int _{t}^{T}{L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{s})\ \textrm{d}s}+ g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{T})\\&\ge \ \inf _{\gamma \in \Gamma _{t}(x)}\left\{ \int _{t}^{T}{L_{0}(\gamma (s), {\dot{\gamma }}(s), m^{0}_{s})\ \textrm{d}s} + g(\gamma (T), m^{0}_{T}) \right\} +\ g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T), m^{{{\,\mathrm{\varepsilon }\,}}}_{T}) - g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T), m^{0}_{T})\\&\quad +\ \int _{t}^{T}{\Big (L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{s}) - L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), m^{0}_{s}) \Big )\ \textrm{d}s} \ge \ u^{0}(t,x) - o(1) \end{aligned}$$

where the last inequality holds by (M1) and the convergence of \(m^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) in \(C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\).

On the other hand, let \(R \ge 0\) and take \((t,x,v) \in [0,T] \times \mathbb {R}^{d} \times \overline{B}_{R}\). Let \(\gamma ^{0} \in \Gamma _{t}(x)\) be a solution of

$$\begin{aligned} \inf _{\gamma \in \Gamma _{t}(x)} \left\{ \int _{t}^{T}{L_{0}(\gamma (s), {\dot{\gamma }}(s), m^{0}_{s})\ \textrm{d}s}+g(\gamma (T), m^{0}_{T})\right\} . \end{aligned}$$

Next, we distinguish two cases: first, when \({\dot{\gamma }}^{0}(t)=v\) and then when \({\dot{\gamma }}^{0}(t)\not =v\). Indeed, if \({\dot{\gamma }}^{0}(t)=v\), by the Euler equation and the \(C^{2}\)-regularity of \(L_{0}\), we have that \(\gamma \in C^{2}([0,T])\). Hence, we can use \(\gamma ^{0}\) as a competitor for \(u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v)\) and we get

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v)&\le \int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }^{0}(s)|^{2} + L_{0}(\gamma ^{0}(s), {\dot{\gamma }}^{0}(s), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{s})\right) \ \textrm{d}s} + g(\gamma ^{0}(T), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T))\nonumber \\&\le \ \int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }^{0}(s)|^{2} + L_{0}(\gamma ^{0}(s), {\dot{\gamma }}^{0}(s), m^{0}_{s})\right) \ \textrm{d}s} + g(\gamma ^{0}(T), m^{0}(T)) \nonumber \\&\quad +\ \int _{t}^{T}{\left( L_{0}(\gamma ^{0}(s), {\dot{\gamma }}^{0}(s), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{s}) - L_{0}(\gamma ^{0}(s), {\dot{\gamma }}^{0}(s), m^{0}_{s}) \right) \ \textrm{d}s}\nonumber \\&\quad +\ g(\gamma ^{0}(T), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T)) - g(\gamma ^{0}(T), m^{0}(T)) \le \ u^{0}(t,x) + o(1) \end{aligned}$$
(4.10)

where the last inequality follows again from the convergence of \(m^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) in \(C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\). If this is not the case, i.e., \({\dot{\gamma }}^{0}(t)\not =v\), from Lemma 4.6 we deduce that

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v)&=\ u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v) - u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x, {\dot{\gamma }}^{0}(t))\\&\quad + u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,{\dot{\gamma }}^{0}(t)) \le o(1) + u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,{\dot{\gamma }}^{0}(t)). \end{aligned}$$

Thus, in order to conclude it is enough to estimate \(u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,{\dot{\gamma }}^{0}(t))\) as in (4.10). Therefore, we obtain

$$\begin{aligned} u^{0}(t,x) - o(1) \le u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v) \le u^{0}(t,x) + o(1) \end{aligned}$$

which implies that \(u^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) point-wise converges to \(u^{0}\).

Finally, in order to conclude we need to show that the convergence is locally uniform. From (4.5), (4.6) and Lemma 4.6 we have that for any \(R \ge 0\) and any \((t_{1}, x_{1}, v_{1})\), \((t_{2}, x_{2}, v_{2}) \in [0,T] \times \overline{B}_{R} \times \overline{B}_{R}\) there holds

$$\begin{aligned} |u^{{{\,\mathrm{\varepsilon }\,}}}(t_{1}, x_{1}, v_{1}) - u^{{{\,\mathrm{\varepsilon }\,}}}(t_{2}, x_{2}, v_{2})| \le \omega _{R}(|t_{1}-t_{2}|) + C_{1}|x_{1}-x_{2}| + C_{R}\sqrt{{{\,\mathrm{\varepsilon }\,}}}. \end{aligned}$$

Therefore, setting

$$\begin{aligned} \Theta (\delta _{0}, \varepsilon _{0}, R_{0})= \omega _{R_{0}}(\delta _{0}) + C_{1}\delta _{0} + C_{R_{0}}\sqrt{\varepsilon _{0}} \end{aligned}$$

by Remark 4.8 we deduce that the convergence is locally uniform and the proof is thus complete. \(\square \)

After proving the convergence of \(u^{{{\,\mathrm{\varepsilon }\,}}}\), we go back to the analysis of the flow of measures and in particular we will characterize it in terms of the limit function \(u^{0}\). In order to do so, we study the convergence of minimizers for \(u^{{{\,\mathrm{\varepsilon }\,}}}\) and appealing to such a result we will show that \(m^{0} \in C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) solves a continuity equation with vector field \(D_{p}H_{0}(x,D_{x}u^{0})\), in the sense of distribution.

Proposition 4.10

Assume (M1)(M3) and (BC1), (BC2). Let \((t, x, v) \in [0,T] \times \mathbb {R}^{2d}\) be such that \(u^{0}\) is differentiable at (tx) and let \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) be a minimizer for \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)\). Then, \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) uniformly converges to a curve \(\gamma ^{0} \in \text {AC}([0,T]; \mathbb {R}^{d})\) and \(\gamma ^{0}\) is the unique minimizer for \(u^{0}(t,x)\) in (4.9).

Proof

Let us start by proving that \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) uniformly converges, up to a subsequence. By Corollary 4.2, we know that

$$\begin{aligned} \int _{t}^{T}{|{\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}}(s)|^{2}\ \textrm{d}s} \le Q_{1}(1+|v|^{2}). \end{aligned}$$

Thus, for any \(s \in [t,T]\), by Hölder inequality we have that

$$\begin{aligned} |\gamma ^{{{\,\mathrm{\varepsilon }\,}}}(s)| \le |x| + \sqrt{T}\sqrt{Q_{1}}(1+|v|^{2})^{\frac{1}{2}}. \end{aligned}$$

Therefore, \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) is bounded in \(H^{1}(0,T; \mathbb {R}^{d})\) which implies that by Ascoli–Arzela Theorem there exists a sequence \(\{{{\,\mathrm{\varepsilon }\,}}_{k}\}_{k \in \mathbb {N}}\) and a curve \(\gamma ^{0} \in \text {AC}([0,T]; \mathbb {R}^{d})\) such that \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) converges uniformly to \(\gamma ^{0}\).

We show now that such a limit \(\gamma ^{0}\) is a minimizer for \(u^{0}(t,x)\). First, we observe that

$$\begin{aligned}&\liminf _{k \rightarrow \infty } \left[ \int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}_{k}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s)|^2 + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), m^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s} + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{T})\right] \\&\quad \ge \liminf _{k \rightarrow \infty } \left[ \int _{t}^{T}{L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{s})\ \textrm{d}s} + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{T})\right] . \end{aligned}$$

Then, as observed at the beginning of this proof \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) is uniformly bounded in \(H^{1}(0,T)\). So by lower-semicontinuity of L and Theorem 4.4, we deduce that

$$\begin{aligned} \begin{aligned}&\liminf _{k \rightarrow \infty } \left[ \int _{t}^{T}{L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s))\ \textrm{d}s} + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{T}) \right] \\&\quad \ge \int _{t}^{T}{L_{0}(\gamma ^{0}(s), {\dot{\gamma }}^{0}(s), m^{0}_{s})\ \textrm{d}s} + g(\gamma ^{0}(T), m^{0}_{T}). \end{aligned} \end{aligned}$$
(4.11)

Moreover, for any \(R \ge 0\) taking \((t,x,v) \in [0,T] \times \mathbb {R}^{d} \times \overline{B}_{R}\), from Theorem 4.9 we obtain

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v) \le u^{0}(t,x) + o(1) \end{aligned}$$

and we recall that

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}_{k}}(t,x,v) = \int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}_{k}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s)|^{2} + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{s})\right) \ \textrm{d}s}+g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{T}). \end{aligned}$$

Hence, we get

$$\begin{aligned} o(1) + u^{0}(t,x) \ge \int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}_{k}}{2}|\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s)|^{2} + L_{0}(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}_{k}}(s), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{s})\right) \ \textrm{d}s} + g(\gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}(T), m^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{T}). \end{aligned}$$

Therefore, passing to the limit as \({{\,\mathrm{\varepsilon }\,}}\downarrow 0\) from (4.11) we obtain

$$\begin{aligned} u^{0}(t,x) \ge \int _{t}^{T}{L_{0}(\gamma ^{0}(s), {\dot{\gamma }}^{0}(s), m^{0}_{s})\ \textrm{d}s} + g(\gamma ^{0}(T), m^{0}_{T}) \end{aligned}$$

which proves that \(\gamma ^{0}\) is a minimizer for \(u^{0}(t,x)\). Since \(u^{0}\) is differentiable at \((t,x) \in \mathbb {R}^{d}\), there exists a unique minimizing trajectory and thus, we have that the uniform convergence of \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) holds for the whole sequence. \(\square \)

Remark 4.11

Since \(u^{0}\) is locally Lipschitz continuous w.r.t. time and Lipschitz continuous w.r.t. space, we have that Proposition 4.10 holds for a.e. \((t,x) \in [0,T] \times \mathbb {R}^{d}\).

Let \(u^{0}\) be as in (4.9) and let \((\gamma ^{0}_{t}(\cdot ), {\dot{\gamma }}^{0}_{t}(\cdot ))\) be the flow of Euler–Lagrange equations associated with the minimization problem in (4.9). Note that, since \(u^{0}\) is Lipschitz continuous and \(\mu _{0}\) is absolutely continuous w.r.t. the Lebesgue measure, we have that on \({{\,\textrm{spt}\,}}(\mu _{0})\) the curve \((\gamma ^{0}_{t}(\cdot ), {\dot{\gamma }}^{0}_{t}(\cdot ))\) is a minimizer for \(u^{0}\). We also recall that the measure \(\mu ^{{{\,\mathrm{\varepsilon }\,}}}\) is the image of \(\mu _{0}\) under the flow (3.8), which is optimal as observed in Remark 3.1 for \(u^{{{\,\mathrm{\varepsilon }\,}}}(0,x,v)\) for a.e. \((x,v) \in \mathbb {R}^{2d}\), and thus, for any function \(\varphi \in C^{\infty }_{c}(\mathbb {R}^{d})\) the measure \(m^{{{\,\mathrm{\varepsilon }\,}}}_{t}\) is given by

$$\begin{aligned} \int _{\mathbb {R}^{d}}{\varphi (x)\ m^{{{\,\mathrm{\varepsilon }\,}}}_{t}(\textrm{d}x)} = \int _{\mathbb {R}^{2d}}{\varphi (\gamma ^{{{\,\mathrm{\varepsilon }\,}}}_{(x,v)}(t))\ \mu _{0}(\textrm{d}x,\textrm{d}v)}. \end{aligned}$$
(4.12)

We finally recall that by assumption \(\mu _{0}\) is absolutely continuous w.r.t. Lebesgue measure.

Corollary 4.12

Assume (M1)(M3) and (BC1), (BC2). Then, we have that

$$\begin{aligned} m^{0}_{t}=\gamma _{t}^{0}(\cdot ) \sharp m_{0}, \quad \forall \ t \in [0,T]. \end{aligned}$$
(4.13)

Moreover, \(m^{0} \in C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) solves

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _{t} m^{0}_{t} - {{\,\textrm{div}\,}}\Big (m^{0}_{t} D_{p}H_{0}(x, D_{x}u^{0}(t,x), m^{0}_{t})\Big )=0, &{} (t,x) \in [0,T] \times \mathbb {R}^{d}\\ m^{0}_{0}=m_{0}, &{} x \in \mathbb {R}^{d}, \end{array}\right. } \end{aligned}$$

in the sense of distributions.

Proof

From Theorem 4.4, let \({{\,\mathrm{\varepsilon }\,}}_{k} \downarrow 0\) be such that \(m^{{{\,\mathrm{\varepsilon }\,}}_{k}} \rightarrow m^{0}\) in \(C([0,T];\mathcal {P}_{1}(\mathbb {R}^{d}))\). Then, since \(\mu _{0}\) is absolutely continuous w.r.t. Lebesgue measure by Proposition 4.10, we have that

$$\begin{aligned} \gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{(x,v)}(t) \rightarrow \gamma ^{0}_{t}(x), \quad \mu _{0}\text {-a.e.}\, (x,v),\,\, \forall t \in [0,T]. \end{aligned}$$

Therefore, from (4.12), for \({{\,\mathrm{\varepsilon }\,}}={{\,\mathrm{\varepsilon }\,}}_{k}\), as \(k \rightarrow \infty \), we get

$$\begin{aligned} \int _{\mathbb {R}^{d}}{\varphi (x)\ m^{0}_{t}(\textrm{d}x)}= \int _{\mathbb {R}^{d}}{\varphi (\gamma _{t}^{0}(x))\ m_{0}(\textrm{d}x)}, \quad \forall \ t \in [0,T] \end{aligned}$$

which proves (4.13). Moreover, again by Proposition 4.10 we have that \(\gamma ^{0}_{t}\) is a minimizer for \(u^{0}(0,x)\) since it is the limit of \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}_{(x,v)}\) which is optimal \(u^{{{\,\mathrm{\varepsilon }\,}}}(0,x,v)\) and we are taking (xv) in a subset of full measure w.r.t. \(\mu _{0}\). Therefore, from the optimality of \(\gamma ^{0}\), we get

$$\begin{aligned} {\left\{ \begin{array}{ll} {\dot{\gamma }}^{0}_{t}(x)=D_{p}H_{0} (\gamma _{t}^{0}(x), Du^{0}(t,\gamma _{t}^{0}(x)), m^{0}_{t}), \quad t \in (0,T]\\ \gamma _{0}^{0}(x)=x. \end{array}\right. } \end{aligned}$$

Hence, for any \(\psi \in C^{\infty }_{c}([0,T) \times \mathbb {R}^{d})\) we obtain

$$\begin{aligned}&\frac{\textrm{d}}{\textrm{d}t}\int _{\mathbb {R}^{d}} {\psi (t,x)\ m^{0}_{t}(\textrm{d}x)}=\frac{\textrm{d}}{\textrm{d}t}\int _{\mathbb {R}^{d}}{\psi (t,\gamma _{t}^{0}(x))\ m_{0}(\textrm{d}x)} \\ =&\int _{\mathbb {R}^{d}}{\big (\partial _{t}\psi (t,\gamma _{t}^{0}(x)) + \langle D_{x}\psi (t, \gamma _{t}^{0}(x)), D_{p}H_{0}(\gamma _{t}^{0}(x), D_{x}\psi (t,\gamma _{t}^{0}(x)), m^{0}_{t})\big )\ m_{0}(\textrm{d}x)} \\ =&\int _{\mathbb {R}^{d}}{\big (\partial _{t}\psi (t,x) + \langle D_{x}\psi (t, x), D_{p}H_{0}(x, D_{x}\psi (t,x), m^{0}_{t})\big )\ m_{t}^{0}(dx)} \end{aligned}$$

and integrating, in time, over [0, T] we get the result. \(\square \)

We are now ready to prove the main result.

Proof of Theorem 3.2

Let \(\{{{\,\mathrm{\varepsilon }\,}}_{k}\}_{k \in \mathbb {N}}\) be such that \(m^{{{\,\mathrm{\varepsilon }\,}}_{k}} \rightarrow m^{0}\) in \(C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d})\) and \(u^{{{\,\mathrm{\varepsilon }\,}}_{k}} \rightarrow u^{0}\) locally uniformly on \([0,T] \times \mathbb {R}^{2d}\). Then, appealing to Theorem 4.9 and Corollary 4.12 we deduce that \((u^{0}, m^{0})\) is a solution to the MFG system

$$\begin{aligned} {\left\{ \begin{array}{ll} -\partial _{t} u^{0}(t,x) + H_{0}(x, D_{x}u^{0}(t,x), m^{0}_{t})=0, &{} (t,x) \in [0,T] \times \mathbb {R}^{d}\\ \partial _{t} m^{0}_{t}- {{\,\textrm{div}\,}}\big (m^{0}_{t} D_{p}H_{0}(x, D_{x}u^{0}(t,x), m^{0}_{t})\big )=0, &{} (t,x) \in [0,T] \times \mathbb {R}^{d}\\ m^{0}_{0}=m_{0}, \,\, u^{0}(T,x)=g(x,m^{0}_{T}) &{} x \in \mathbb {R}^{d} \end{array}\right. } \end{aligned}$$

which completes the proof. \(\square \)

5 Proof of Theorem 3.5

We recall that, in this section, we consider the MFG system

$$\begin{aligned} {\left\{ \begin{array}{ll} -\partial _{t} u^{{{\,\mathrm{\varepsilon }\,}}} +\frac{1}{2{{\,\mathrm{\varepsilon }\,}}}|D_{v}u^{{{\,\mathrm{\varepsilon }\,}}}|^{2} - \langle D_{x}u^{{{\,\mathrm{\varepsilon }\,}}}, v \rangle -\frac{1}{2}|v|^{2} - L_{0}(x, \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t})= 0, &{} (t,x,v) \in [0,T] \times \mathbb {R}^{2d}\\ \partial _{t}\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t} - \langle D_{x}\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t},v \rangle - \frac{1}{{{\,\mathrm{\varepsilon }\,}}} {{\,\textrm{div}\,}}_{v}\left( \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}D_{v}u^{{{\,\mathrm{\varepsilon }\,}}} \right) =0, &{} (t,x,v) \in [0,T] \times \mathbb {R}^{2d}\\ \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{0}=\mu _{0}, \quad u^{{{\,\mathrm{\varepsilon }\,}}}(T,x,v)=g(x,\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{T}), &{} (x,v) \in \mathbb {R}^{2d}. \end{array}\right. } \end{aligned}$$

So, the variational problem associated with such a system is given by

$$\begin{aligned} u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)= \inf _{\gamma \in \Gamma _{t}(x,v)} J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}(\gamma ) \end{aligned}$$

where

$$\begin{aligned} J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}(\gamma )=&\int _{t}^{T}{\left( \frac{{{\,\mathrm{\varepsilon }\,}}}{2}|\ddot{\gamma }(s)|^{2} + \frac{1}{2}|{\dot{\gamma }}(s)|^{2} +L_{0}(\gamma (s), \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{s}) \right) \ \textrm{d}s}\\ {}&+ g(\gamma (T),\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{T}), \quad \text {if}\,\, \gamma \in H^{2}(0,T; \mathbb {R}^{d}) \end{aligned}$$

with \(J^{{{\,\mathrm{\varepsilon }\,}}}_{t,T}(\gamma )=+\infty \) if \(\gamma \not \in H^{2}(0,T;\mathbb {R}^{d})\).

From the results on the previous section and the assumptions (C1)(C3) on \(L_{0}: \mathbb {R}^{d} \times \mathcal {P}(\mathbb {R}^{2d}) \rightarrow \mathbb {R}\) given above, we deduce that we only need to study the tightness of the flow of measures \(\{\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\}_{t \in [0,T]}\) w.r.t. the second marginal. This can be done by a finer analysis of the Euler–Lagrange flow.

Lemma 5.1

Let \((x,v) \in \mathbb {R}^{2d}\) and let \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) be a solution to the variational problem associated with \(u^{{{\,\mathrm{\varepsilon }\,}}}(0,x,v)\). Then, \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) is a solution of the Euler–Lagrange equation

$$\begin{aligned} -{{\,\mathrm{\varepsilon }\,}}\xi ^{(iv)}(t) +\ddot{\xi }^{{{\,\mathrm{\varepsilon }\,}}}(t) - D_{x}L_{0}(\xi (t), \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}) = 0 \end{aligned}$$

with boundary condition

$$\begin{aligned} \ddot{\xi }(T)= 0, \quad \text {and} \quad -{{\,\mathrm{\varepsilon }\,}}\xi ^{(iii)} (T) + {\dot{\xi }}(T) + D_{x}g(x, \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{T}). \end{aligned}$$

Proposition 5.2

Assume (C1)(C3) and (A1), (A2). Let \((x,v) \in \mathbb {R}^{2d}\) and let \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) be a solution to the variational problem associated with \(u^{{{\,\mathrm{\varepsilon }\,}}}(0,x,v)\). Then, there exists a constant \(C \ge 0\) such that for any \(\delta \in (0, 1)\) the following holds

$$\begin{aligned} \int _{\delta }^{T} |\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}}(s)|^{2}\ \textrm{d}s \le C. \end{aligned}$$

Proof

Fix \((x, v) \in \mathbb {R}^{2d}\) and a solution \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) to the problem associated with \(u^{{{\,\mathrm{\varepsilon }\,}}}(0,x,v)\). In the following, for simplicity of notation we drop \({{\,\mathrm{\varepsilon }\,}}\) setting \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}=\gamma \) and we drop the notation of the scalar product \(\langle \cdot , \cdot \rangle \).

We begin by multiplying the Euler–Lagrange equation

$$\begin{aligned} -{{\,\mathrm{\varepsilon }\,}}\gamma ^{(iv)}(t) +\ddot{\gamma }^{{{\,\mathrm{\varepsilon }\,}}}(t) - D_{x}L_{0}(\gamma (t), \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}) = 0 \end{aligned}$$

by \(s^{2}\ddot{\gamma }(t)\) and integrate by parts. We obtain

$$\begin{aligned}&-{{\,\mathrm{\varepsilon }\,}}\gamma ^{(iii)} (T) T \ddot{\gamma }(T)\\ +\ {}&\int _{0}^{T} \Big ({{\,\mathrm{\varepsilon }\,}}\gamma ^{(iii)}(s) \big (s \ddot{\gamma }(s) + s^{2} \gamma ^{(iii)}(s) \big ) + s^{2} \ddot{\gamma }(s)^{2} - D_{x}L_{0}(\gamma (s), \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{s}) s^{2}\ddot{\gamma }(s) \Big )\ \textrm{d}s = 0 \end{aligned}$$

which reduces to

$$\begin{aligned} \int _{0}^{T} \Big ({{\,\mathrm{\varepsilon }\,}}\gamma ^{(iii)}(s) \big (s \ddot{\gamma }(s) + s^{2} \gamma ^{(iii)}(s) \big ) + s^{2} \ddot{\gamma }(s)^{2} - D_{x}L_{0}(\gamma (s), \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{s}) s^{2}\ddot{\gamma }(s) \Big )\ \textrm{d}s = 0 \end{aligned}$$

by using the boundary condition on the Euler–Lagrange equation. From the Young’s inequality, we get

$$\begin{aligned}&\int _{0}^{T} \Big ({{\,\mathrm{\varepsilon }\,}}\gamma ^{(iii)}(s) \big (s \ddot{\gamma }(s) + s^{2} \gamma ^{(iii)}(s) \big ) + s^{2} \ddot{\gamma }(s)^{2}\Big )\ \textrm{d}s \\&\quad \le \ \int _{0}^{T} \frac{s^{2}}{2}D_{x}L_{0}(\gamma (s), \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{s})^{2}\ \textrm{d}s + \int _{0}^{T} \frac{s^{2}}{2}\ddot{\gamma }(s)^{2}\ \textrm{d}s \end{aligned}$$

which yields to

$$\begin{aligned} \int _{0}^{T} \Big ({{\,\mathrm{\varepsilon }\,}}\gamma ^{(iii)}(s) \big (s \ddot{\gamma }(s) + s^{2} \gamma ^{(iii)}(s) \big ) + \frac{s^{2}}{2} \ddot{\gamma }(s)^{2}\Big )\ \textrm{d}s \le \ \int _{0}^{T} \frac{s^{2}}{2}D_{x}L_{0}(\gamma (s), \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{s})^{2}\ \textrm{d}s. \end{aligned}$$

So, appealing to (C1) we have that there exists a constant \(C(T) \ge 0\) such that

$$\begin{aligned} \int _{0}^{T} \frac{s^{2}}{2}D_{x}L_{0}(\gamma (s), \mu ^{{{\,\mathrm{\varepsilon }\,}}}_{s})^{2}\ \textrm{d}s \le C(T) \end{aligned}$$

and, moreover, by the non-negativity of the term \(s^{2}\gamma ^{(iii)}(s)^{2}\) we finally get

$$\begin{aligned} \int _{0}^{T} \Big ({{\,\mathrm{\varepsilon }\,}}\gamma ^{(iii)}(s) s \ddot{\gamma }(s) + \frac{s^{2}}{2} \ddot{\gamma }(s)^{2}\Big )\ \textrm{d}s \le C(T). \end{aligned}$$
(5.1)

Integrating, again, by parts the term \( \int _{0}^{T} {{\,\mathrm{\varepsilon }\,}}\gamma ^{(iii)}(s) s \ddot{\gamma }(s)\ \textrm{d}s \) we obtain

$$\begin{aligned} \int _{0}^{T} {{\,\mathrm{\varepsilon }\,}}\gamma ^{(iii)}(s) s \ddot{\gamma }(s)\ \textrm{d}s = {{\,\mathrm{\varepsilon }\,}}T \ddot{\gamma }(T) - \int _{0}^{T} {{\,\mathrm{\varepsilon }\,}}\ddot{\gamma }(s) \big (\ddot{\gamma }(s) + s\gamma ^{(iii)}(s) \big )\ \textrm{d}s \end{aligned}$$

which implies that

$$\begin{aligned} - \int _{0}^{T} {{\,\mathrm{\varepsilon }\,}}\ddot{\gamma }(s)^{2}\ \textrm{d}s = 2{{\,\mathrm{\varepsilon }\,}}\int _{0}^{T} \gamma ^{(iii)}(s) s \ddot{\gamma }(s)\ \textrm{d}s \end{aligned}$$

and so

$$\begin{aligned} - \int _{0}^{T} \ddot{\gamma }(s)^{2}\ \textrm{d}s \le 2{{\,\mathrm{\varepsilon }\,}}\int _{0}^{T} \gamma ^{(iii)}(s) s \ddot{\gamma }(s)\ \textrm{d}s. \end{aligned}$$
(5.2)

Hence, combining (5.1) and (5.2), we get

$$\begin{aligned} \int _{0}^{T} s^{2} \ddot{\gamma }(s)^{2}\ \textrm{d}s \le C(T) \end{aligned}$$

which implies the result. \(\square \)

Theorem 5.3

Assume (C1)(C3) and (A1), (A2). The sequence \(\{\mu ^{{{\,\mathrm{\varepsilon }\,}}, \delta }_{t}\}_{\delta > 0}\) is tight, and the sequence \(\{\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\}_{{{\,\mathrm{\varepsilon }\,}}> 0}\) is relatively compact in \(C^{0}([0, T]; \mathcal {P}_{1}(\mathbb {R}^{2d}))\).

Proof

The tightness of the sequence \(\{\mu ^{{{\,\mathrm{\varepsilon }\,}}, \delta }_{t}\}_{\delta > 0}\) follows from Corollary 4.2 and Proposition 5.2.

Next, we show that \(\{\mu ^{{{\,\mathrm{\varepsilon }\,}}}_{t}\}_{{{\,\mathrm{\varepsilon }\,}}> 0}\) is relatively compact. Indeed, still from Corollary 4.3 and Proposition 5.2 we have that for any \(\delta > 0\) and any \(\delta \le s \le t \le T\)

$$\begin{aligned} d_{1}(\mu _{s}^{{{\,\mathrm{\varepsilon }\,}}}, \mu _{t}^{{{\,\mathrm{\varepsilon }\,}}})&= \int _{\mathbb {R}^{2d}} \left( |\gamma _{(x,v)}(t) - \gamma _{(x,v)}(s) | + |{\dot{\gamma }}_{(x,v)}(t) - \gamma _{(x,v)}(s)| \right) \ \mu _{0}(\textrm{d}x,\textrm{d}v)\\&\le (t-s)^{\frac{1}{2}} \left[ \left( \int _{\mathbb {R}^{2d}} |{\dot{\gamma }}_{(x,v)}(\tau )|^{2}\ \mu _{0}(\textrm{d}x,\textrm{d}v) \right) ^{\frac{1}{2}} + \left( \int _{\mathbb {R}^{2d}} |\ddot{\gamma }_{(x,v)}(\tau )|^{2}\ \mu _{0}(\textrm{d}x,\textrm{d}v) \right) ^{\frac{1}{2}} \right] \\&\le CQ_{2}(t-s)^{\frac{1}{2}} \end{aligned}$$

which completes the proof appealing to Prokhorov Theorem and Ascoli–Arzela Theorem. \(\square \)

Remark 5.4

Note that, following the above reasoning one easily deduce that the main result of this section is not uniform w.r.t. T.

Now, by using similar techniques of Theorem 4.9 and Proposition 4.10 one can prove the following.

Proposition 5.5

Assume (C1)(C3) and (A1), (A2). Then, we have that

(i):

there exists a subsequence \({{\,\mathrm{\varepsilon }\,}}_{k} \downarrow 0\) such that \(u^{{{\,\mathrm{\varepsilon }\,}}_{k}}\) locally uniformly converges to \(u^{0}\).

(ii):

Let \((t,x,v) \in [0,T] \times \mathbb {R}^{2d}\) be such that \(u^{0}\) is differentiable at (tx) and let \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) be a minimizer of \(u^{{{\,\mathrm{\varepsilon }\,}}}(t,x,v)\). Then, \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}\) uniformly converges to a curve \(\gamma ^{0}\) such that \({\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}} \rightarrow {\dot{\gamma }}^{0}\) as \({{\,\mathrm{\varepsilon }\,}}\downarrow 0\) and \(\gamma ^{0}\) is the unique minimizer for \(u^{0}\) in (4.9).

Proposition 5.6

Assume (C1)(C3) and (A1), (A2). Then, we have that

$$\begin{aligned} \mu ^{0}_{t}=(Id_{\cdot }, D_{x}u^{0}(t, \gamma _{t}^{0}(\cdot ))) \sharp m^{0}_{t}, \quad \forall \ t \in [0,T]. \end{aligned}$$

Moreover, \(m^{0} \in C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{d}))\) solves

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _{t} m^{0}_{t} - {{\,\textrm{div}\,}}\big (m^{0}_{t}D_{x}u^{0}(t,x)\big )=0, &{} (t,x) \in [0,T] \times \mathbb {R}^{d}\\ m^{0}_{0}=m_{0}, &{} x \in \mathbb {R}^{d}, \end{array}\right. } \end{aligned}$$

in the sense of distributions.

Proof

From Proposition 5.5, let \({{\,\mathrm{\varepsilon }\,}}_{k} \downarrow 0\) be such that \(\mu ^{{{\,\mathrm{\varepsilon }\,}}_{k}} \rightarrow \mu ^{0}\) in \(C([0,T];\mathcal {P}_{1}(\mathbb {R}^{2d}))\). Then, since \(\mu _{0}\) is absolutely continuous w.r.t. Lebesgue measure by Proposition 5.5 we have that

$$\begin{aligned} \gamma ^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{(x,v)}(t) \rightarrow \gamma ^{0}_{t}(x),\,\, \text {and}\,\, {\dot{\gamma }}^{{{\,\mathrm{\varepsilon }\,}}_{k}}_{(x,v)}(t) \rightarrow {\dot{\gamma }}^{0}_{t}(x) \quad \mu _{0}\text {-a.e.}\, (x,v),\,\, \forall t \in [0,T]. \end{aligned}$$

Therefore, from (4.12), for \({{\,\mathrm{\varepsilon }\,}}={{\,\mathrm{\varepsilon }\,}}_{k}\), as \(k \rightarrow \infty \) we get

$$\begin{aligned} \int _{\mathbb {R}^{d}}{\varphi (x)\ m^{0}_{t}(dx)}= \int _{\mathbb {R}^{d}}{\varphi (\gamma _{t}^{0}(x))\ m_{0}(dx)}, \quad \forall \ t \in [0,T]. \end{aligned}$$

Moreover, again by Proposition 5.5 we have that \(\gamma ^{0}_{t}\) is a minimizer for \(u^{0}(0,x)\) since it is the limit of \(\gamma ^{{{\,\mathrm{\varepsilon }\,}}}_{(x,v)}\) which is optimal \(u^{{{\,\mathrm{\varepsilon }\,}}}(0,x,v)\) and we are taking (xv) in a subset of full measure w.r.t. \(\mu _{0}\). Therefore, from the optimality of \(\gamma ^{0}\) we get

$$\begin{aligned} {\left\{ \begin{array}{ll} {\dot{\gamma }}^{0}_{t}(x)= Du^{0}(t,\gamma _{t}^{0}(x)), \quad t \in (0,T]\\ \gamma _{0}^{0}(x)=x. \end{array}\right. } \end{aligned}$$

Hence,

$$\begin{aligned} \mu ^{0}_{t}=(Id_{\cdot }, D_{x}u^{0}(t, \gamma _{t}^{0}(\cdot ))) \sharp m^{0}_{t}, \quad \forall \ t \in [0,T]. \end{aligned}$$

Then, for any \(\psi \in C^{\infty }_{c}([0,T) \times \mathbb {R}^{d})\) we obtain

$$\begin{aligned}&\frac{\textrm{d}}{\textrm{d}t}\int _{\mathbb {R}^{d}} {\psi (t,x)\ m^{0}_{t}(\textrm{d}x)}=\frac{d}{\textrm{d}t}\int _{\mathbb {R}^{d}}{\psi (t,\gamma _{t}^{0}(x))\ m_{0}(\textrm{d}x)}\\ =&\int _{\mathbb {R}^{d}}{\big (\partial _{t}\psi (t,\gamma _{t}^{0}(x)) + \langle D_{x}\psi (t, \gamma _{t}^{0}(x)), D_{x}\psi (t,\gamma _{t}^{0}(x))\big )\ m_{0}(\textrm{d}x)}\\ =&\int _{\mathbb {R}^{d}}{\big (\partial _{t}\psi (t,x) + \langle D_{x}\psi (t, x), D_{x}\psi (t,x)\big )\ m_{t}^{0}(\textrm{d}x)} \end{aligned}$$

and integrating, in time, over [0, T] we get the result. \(\square \)

Proof of Theorem 3.5

Let \(\{{{\,\mathrm{\varepsilon }\,}}_{k}\}_{k \in \mathbb {N}}\) be such that \(\mu ^{{{\,\mathrm{\varepsilon }\,}}_{k}} \rightarrow \mu ^{0}\) in \(C([0,T]; \mathcal {P}_{1}(\mathbb {R}^{2d})\) and \(u^{{{\,\mathrm{\varepsilon }\,}}_{k}} \rightarrow u^{0}\) locally uniformly on \([0,T] \times \mathbb {R}^{2d}\). Then, appealing to Proposition 5.5 and Proposition 5.6 we deduce that \((u^{0}, \mu ^{0})\) is a solution to the MFG system

$$\begin{aligned} {\left\{ \begin{array}{ll} -\partial _{t} u^{0}(t,x) + \frac{1}{2}|D_{x}u^{0}(t,x) |^{2}- L_{0}(x, \mu ^{0}_{t})=0, &{} \quad (t,x) \in [0,T] \times \mathbb {R}^{d}\\ \partial _{t} m^{0}_{t} - {{\,\textrm{div}\,}}\big ( m^{0}_{t}D_{x}u^{0}(t,x) \big )=0, &{} \quad (t,x) \in [0,T] \times \mathbb {R}^{d}\\ \mu ^{0}_{t} = (\text {Id}_{\cdot }, D_{x}u^{0}(t, \cdot )) \sharp m^{0}_{t}, &{} \quad t \in [0,T]\\ m^{0}_{0}= m_{0},\,\, u^{0}(T,x)=g(x,\mu ^{0}_{T}), &{} \quad x \in \mathbb {R}^{d}, \end{array}\right. } \end{aligned}$$

which completes the proof. \(\square \)