# Mean-Field Limits for Stochastic Interacting Particles on Digraph Measures

Christian Kuehn and Carlos Pulido

###### Abstract

Many natural phenomena are effectively described by interacting particle
systems, which can be modeled using either deterministic or stochastic
differential equations (SDEs). In this study, we specifically investigate
particle systems modeled by SDEs, wherein the mean field limit converges to a
Vlasov-Fokker-Planck-type equation. Departing from conventional approaches in
stochastic analysis, we explore the network connectivity between particles
using diagraph measures (DGMs). DGMs are one possible tool to capture sparse,
intermediate and dense network/graph interactions in the mean-field thereby
going beyond more classical approaches such as graphons. Since the main goal
is to capture large classes of mean-field limits, we set up our approach using
measure-theoretic arguments and combine them with suitable moment estimates to
ensure approximation results for the mean-field.

##  1 Introduction

Consider a physical system composed of $N$ particles that interact with each
other through a network/graph of connections, where information about the
connectivity between particles can be represented by a graph $G$. The vertices
of this graph represent each particle, and the edges of the graph model
pairwise interactions between them.

The study of these systems, particularly when $N$ is a very large number, has
been approached in various ways, depending on different mathematical
treatments used to represent the graph. To set the context, let us begin by
introducing the equations that determine the system’s dynamics. Let
$X_{i}(t)\in\mathbb{R}^{d}$ denote the state of particle $i$, and consider the
following stochastic differential equations (SDEs) that define the evolution
of all particles in the system.

$\displaystyle\mathrm{d}X_{i}(t)=f(X_{i}(t))\mathrm{d}t+\frac{1}{N}\sum_{j=1}^{N}A^{N}_{ij}g(X_{i},X_{j})\mathrm{d}t+\frac{1}{N}\sum_{j=1}^{N}\hat{A}^{N}_{ij}h(X_{i},X_{j})\mathrm{d}B_{t}^{i},$
$\displaystyle X_{i}(0)=X_{i}^{0},\quad i=1,\ldots,N.$ (1)

Here, $A_{i,j}^{N}$ is the adjacency matrix of the system, representing the
basic deterministic network interaction between particles $i$ and $j$, and
$\hat{A}^{N}_{ij}$ serves as an adjacency matrix to represent, how
interactions between particles $i$ and $j$ influence the noise term.
${A}^{N}_{ij}$ and $\hat{A}^{N}_{ij}$ are not necessarily identical, which
already provides a useful generalization to existing mean-fields that aim to
include network coupling. The functions $f$, $g$ and $h$ are Lipschitz and
bounded. The noise term contains a sequence $\left\\{B^{i}_{t}\right\\}$, for
$1\leq i\leq N$, of independent and identically distributed Brownian motions
in $\mathbb{X}\subseteq\mathbb{R}^{d}$. The initial conditions
$\left\\{X_{i}^{0}\right\\}_{i=1,..,N}$ form a sample of independently and
identically distributed random variables with a probability distribution
$\mu^{0}\in\mathcal{P}(\mathbb{X})$. $X_{i}(t)$ represents, at the microscopic
level, the trajectory of particle $i$. As the number of particles in the
system becomes very large, the study of the system’s dynamics becomes
increasingly complex. Therefore, to analyze these issues, various techniques
have been introduced, focusing on the study of the density of a typical
particle and its evolution [5, 10, 20, 22, 14]. These techniques were
classically developed on an all-to-all coupled graphs.

The main objective of this work is to determine the mean-field limit for the
interacting particle system (1). The main challenge is that we have to
consider also a suitable notion of graph limits and incorporate the graph
limit object in our analysis. For dense graphs, it is understood that graphons
provide a suitable tool, and one can stay within a familiar setting of
functions and norms as graphons are just functions themselves. Yet, to capture
sparse and intermediate density graph limits, it is necessary to work with
graph limits that are only measures. In this work we show that mean-fields can
be obtained on a type of graph representation called diagraph measure (DGM).

The study of the mean-field limit for equations similar to (1) has been
approached in different ways, depending on how the adjacency matrices have
been handled in the analysis.

First, in cases where the stochastic terms absent, numerous studies explore
the mean-field limit by considering $A_{ij}=1$, which are considered mean-
field limits for systems of completely exchangeable particles, as studied in
e.g. [16, 5, 10, 20]. On the other hand, when considering systems of non-
exchangeable particles and accounting for graph properties, the mean-field
limit has been derived in various frameworks, e.g., in [9, 18, 13, 17, 15, 1,
21]. Each one employs different graph representation techniques such as
graphons [15, 18, 9], diagraph measures [17], or graphops [13]. In all these
cases, it is demonstrated that in the limit, the system satisfies a Vlasov
equation.

Second, when considering the stochastic terms, different perspectives have
been studied. For example, [6, 3, 4] have addressed various treatments of the
matrix, ranging from $A_{ij}=1$ to considering the graph as a graphon. Here we
focus on studying the graph as a diagraph measure (DGM) and derive the
resulting Vlasov-Fokker-Plank equation in the stochastic case. To approach
this, we will concentrate on the coupling method [7, 8], and utilize the
techniques of Sznitmann [22], along with those used in [3, 4], to study the
mean-field limit of the particle system (1).

As our goal is to obtain a mean-field limit, we aim to determine a measure of
the form $\mu(t,x)$ representing the mesoscopic/typical particle evolution in
time and space of the original microscopic model. Since the SDEs for the
particles lead to a stochastic process, the density $\mu$ must be a
probability measure. However, as we are working with DGMs, this measure will
also depend on an additional variable representing the heterogeneity of the
graph, which we denote as $u\in I$, where $I$ is the set of all possible
values that the graph variable can take. Our probability measure will take the
form $\mu(t,x,u)$, and we will define
$\bar{\mu}(t,x)=\int_{I}\mu(t,x,u)\leavevmode\nobreak\ \mathrm{d}u$. For
simplicity, we will also use the following notation: $\mu(t,x)=\mu_{t}(x)$ and
$\mu(t,x,u)=\mu_{u,t}(x)$.

Our objective is, therefore, to establish, from (1), a probability measure
whose limit converges in a suitable metric space to a (weak) solution of some
Vlasov-Fokker-Planck equation. To achieve this, we will represent the law of
the entire system through the empirical measure, which will play a central
role. The empirical measure is defined as follows:

$\mu^{N}_{t}=\frac{1}{N}\sum_{i=1}^{N}\delta_{X_{i}(t)},\text{ for
}t\in[0,T].$

The next step, once the empirical measure is defined, is to identify a problem
with sufficiently good properties, whose solution can be compared with ours.
This is the basis of coupling methods. We seek a problem where the dynamics of
each particle is suitably independent of the others, and we are going to prove
the existence of a solution to this problem. Once the existence results is
established, we are then going to show that as the number of particles tends
to infinity, both problems exhibit similar behavior. Mathematically, this is
expressed by proving that the distances between the empirical measure and a
suitable limiting measure tend to zero, as $N\rightarrow\infty$, in some
metric space on probability measures that we have to select.

To start, let us introduce the system of independent processes that we are
going to consider:

$\begin{split}X_{u}(t)=&X_{u}(0)+\int_{0}^{t}f(X_{u}(s))\leavevmode\nobreak\
\mathrm{d}s+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}g(X_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\mathrm{d}s\\\
&+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}h(X_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}B_{s}^{u},\end{split}$
(2)

where $\mu_{u,t}=\mathcal{L}(X(t)|U=u)$, and $U$ is a uniform random variable
on $I$. The initial conditions $X_{u}(0)=X_{u}^{0}$, for $u\in I$, form a
sample of independently and identically distributed random variables with a
probability distribution $\bar{\mu}^{0}\in\mathcal{P}(\mathbb{X})$. Note that
two new measures appear, which are finite positive Borel measures
$\eta,\hat{\eta}\in\mathcal{B}(I,\mathcal{M}_{+}(I))$. The measures
$\eta,\hat{\eta}$ are the fiber measures associated to the adjacency matrices
of the graphs $A^{N}_{i,j}$ and $\hat{A}^{N}_{i,j}$. The fiber measures are
defined on a vertex space $I$. In this way, these measures represents the
edges between different $u$ and the rest of the vertices of $I$. This
representation of graphs by a measure is known as a diagraph measure (DGM);
the concept is a step towards potentially covering even broader classes of
graph limits, e.g., those given by graphops (graph operators). For background
on graph limits and DGMs we refer to [2, 17]. The Brownian motion satisfies
the same properties as before.

Next, observe in (2) that the particle dynamics, for a given $u$, are
independent of the dynamics of the other particles. Therefore, the system
represents an independent particle process. Furthermore, the graph’s
heterogeneity variable is taken as a random variable, uniformly distributed
over the interval $I$. Similarly to the previous section, we can define the
probability measure associated with the solution $X$ as
$\bar{\mu}=\int_{I}\mu_{u,t}\,\textnormal{d}u$. Since the aim was to compare
the two systems as the number of particles becomes sufficiently large, we want
to determine whether there is a metric space with metric $d$ such that:

$d(\bar{\mu},\mu_{N})\rightarrow 0,\text{ as }N\rightarrow\infty.$

Since we are dealing with graphs, the graphs $A^{N}_{i,j}$ and
$\hat{A}^{N}_{i,j}$ will also change with the number of particles, and we must
also be able to represent their convergence as $N$ tends to infinity.
Moreover, as we are comparing it with the independent system, we must employ a
technique of representing matrices/operators in the form of measures and
determine a space in which we can establish the convergence.

In this paper, we will focus on proving that if our two graphs, each
represented by a directed graph measure (DGM), converge towards certain
measures, then under specific assumptions (Assumptions $H$ and $\tilde{H}$
below), the empirical measure describing the dynamics of the states of the
particles will converge to a probability measure $\mu_{u,t}$. This limit
satisfies the following Vlasov-Fokker-Plank equation:

$\begin{split}&\partial_{t}\bar{\mu}_{t}+\partial_{x}\left(\bar{\mu}_{t}f(x)+\int_{I}{\mu}_{u,t}\int_{I}\int_{\mathbb{X}}g(x,y)\leavevmode\nobreak\
\mu_{v,t}(\text{d}y)\eta^{u}(\text{d}v)\text{d}u\right)\\\
&+\frac{1}{2}\partial_{x}^{2}\left(\int_{I}{\mu}_{u,t}\left[\int_{I}\int_{\mathbb{X}}h(x,y)\leavevmode\nobreak\
\mu_{v,t}(\text{d}y)\hat{\eta}^{u}(\text{d}v)\right]^{2}\leavevmode\nobreak\
\text{d}u\right)=0.\end{split}$ (3)

This will be proved in Theorems 5 and 7, where in each theorem, we will handle
the limit measure $\mu_{u,t}$ differently. However, in both cases, we will
obtain the same Vlasov-Fokker-Plank equation.

The paper is organized as follows. In Section 2, we introduce various measure
and probability spaces, along with the associated metrics, that will serve as
the foundation for establishing the convergence of the empirical measure and
of DGMs. Section 3 states the key results and interpretations of the work,
along with the necessary assumptions for proving these results. The proof of
Theorem 5 is presented in Section 4, where we focus on studying the
independent process (2). The existence and uniqueness of the solution to this
problem will be explored in the same section. In Section 5, we employ coupling
methods to illustrate the convergence in law of the empirical measure towards
the solution of (2), thereby proving Theorem 5. Section 6 is dedicated to
proving Theorem 7 using techniques similar to those employed in the preceding
sections. Finally, Section 7 covers examples of Vlasov-Fokker-Plank equations
for some DGMs.

##  2 Metric Measure Spaces and Digraph Measures

We are going to introduce various spaces and notation that we will use. We are
concerned with the dynamics of particles in a finite time interval, denoted as
$[0,T]$, where we fix $T>0$ as a parameter. Our analysis takes place within
the framework of a filtered probability space
$(\Omega,\mathcal{F},\\{F_{t}\\}_{t\in[0,T]},\mathbb{P})$, where $\\{F_{t}\\}$
denotes a filtration that complies with standard conditions. The particle
trajectories $X(t)$ take values in $\mathbb{X}\subseteq\mathbb{R}^{d}$.

First of all, let us determine the measurement space over which we are going
to define our digraphs. Let $I$ be a complete metric space, and consider the
space $\mathcal{B}(I,\mathcal{M}_{+}(I))$, which is the space of bounded
measurable functions from $I$ to $\mathcal{M}_{+}(I)$, where
$\mathcal{M}_{+}(I)$ is the set of all finite Borel positive measures on $I$.
Also, consider $\mathcal{C}(I,\mathcal{M}_{+}(I))$ as the space of continuous
functions from $I$ to $\mathcal{M}_{+}(I$). Let us define
$\mathcal{B}(\mathbb{X})$ the space of bounded measurable functions from
$\mathbb{X}$ to $\mathbb{R}^{d}$ and

$L_{f}:=\sup_{x,y\in\mathbb{X},x\neq y}\frac{|f(x)-f(y)|}{|x-y|},$

$\text{the Lipschitz constant of }f\in\mathcal{C}\left(\mathbb{X}\right)$. Let
$\mathcal{BL}\left(\mathbb{X}\right)$ be the space of bounded Lipschitz
continuous functions and
$\mathcal{BL}_{1}\left(\mathbb{X}\right)=\left\\{f\in\mathcal{B}\mathcal{L}\left(\mathbb{X}\right):\mathcal{B}\mathcal{L}(f):=\|f\|_{\infty}+L_{f}\leq
1\right\\}$.

Let $\eta^{x},\nu^{x}\in\mathcal{M}_{+}(I)$, then we can define the bounded
Lipschitz distance as:

$d_{BL}(\eta^{x},\nu^{x}):=\sup_{f\in\mathcal{BL}_{1}(\mathbb{X})}\int_{\mathbb{X}}f(y)\leavevmode\nobreak\
\left(\eta^{x}(\mathrm{d}y)-\nu^{x}(\mathrm{d}y)\right).$

Hence, given $\eta,\nu\in\mathcal{B}(I,\mathcal{M}_{+}(I))$, define the
uniform bounded Lipschitz metric:

$d_{\infty}(\eta,\nu):=\sup_{x\in\mathbb{X}}d_{BL}(\eta^{x},\nu^{x}).$

and the bounded Lipschitz norm (on the space of all finite signed Borel
measures):

$\|\eta\|:=\sup_{x\in\mathbb{X}}\sup_{f\in\mathcal{BL}_{1}(\mathbb{X})}\int_{\mathbb{X}}f(y)\leavevmode\nobreak\
\eta^{x}(\mathrm{d}y).$

We have the following Proposition [17, Proposition 2.6]:

######  Proposition 1.

Let $I$ be a complete separable metric space. Assume $I$ is compact. Then
$(\mathcal{B}(I,\mathcal{M}_{+}(I)),d_{\infty})$ and
$(\mathcal{C}(I,\mathcal{M}_{+}(I)),d_{\infty})$ are complete metric spaces.

In our problem, we will consider $I=[0,1]$. Since we are working with DGMs,
and in our problem we have a graph represented by a matrix, we need to
approximate this graph by a DGM. Consider a partition of the interval
$I=[0,1]$ as given by $I_{i}^{N}=]\frac{i-1}{N},\frac{i}{N}]$, for $1<i\leq
N$, and $I_{1}^{N}=[0,\frac{1}{N}]$. Let us build the diagraph measures
associated with the matrices $A^{N}_{i,j}$ and $\hat{A}^{N}_{i,j}$ as follows

$\begin{split}\eta^{u}_{A^{N}}(v):=\sum_{i=1}^{N}\textbf{1}_{I_{i}^{N}}(u)\sum_{j=1}^{N}\frac{A^{N}_{i,j}}{N}\delta_{\frac{j}{N}}(v),\\\
\eta^{u}_{\hat{A}^{N}}(v):=\sum_{i=1}^{N}\textbf{1}_{I_{i}^{N}}(u)\sum_{j=1}^{N}\frac{\hat{A}^{N}_{i,j}}{N}\delta_{\frac{j}{N}}(v),\end{split}$
(4)

where $\textbf{1}_{\frac{j}{N}}(x)$ is the indicator function and
${\frac{j}{N}}$ serves as the representative of the set $I_{j}^{N}$ and . We
recall that the sequence of graphs $\\{A^{N}\\}_{N\geq 1}$ converges to the
diagraph measure $\eta$ if and only if
$d_{\infty}(\eta_{A^{N}},\eta)\rightarrow 0$, as $N\rightarrow\infty$.
Similarly, we aim for $\hat{A}^{N}$ to converge to the diagraph measure
$\hat{\eta}$. Observe that
$\eta_{A^{N}}\in\mathcal{B}\left(I,\mathcal{M}_{+}(I)\right)\bigcap\mathcal{C}\left(I,\mathcal{M}_{+}(I)\right)$.

In order to analyze the collection of probability laws, let us consider the
following space of probability measures

$\mathcal{N}:=\left\\{\nu\in\mathcal{P}(I\times\mathcal{C}([0,T],\mathbb{X})):\pi_{1}\circ\nu=\lambda\right\\},$

where $\lambda$ denotes the Lebesgue measure on $I$ and $\pi_{1}$ is the
projection map associated to the first coordinate. To compare probability
measures we make use of the following Wasserstein-2 metric:

$W_{2}(\mu,\nu):=\left(\inf\left\\{\mathbb{E}|X-\tilde{X}|^{2}:\mathcal{L}(X)=\mu,\mathcal{L}(\tilde{X})=\nu\right\\}\right)^{\frac{1}{2}},\;\mu,\nu\in\mathcal{P}(\mathbb{R}^{d}).$

As the probability measures also depend on time, we can define the following
Wasserstein-2 metric:

$\displaystyle W_{2,t}(\mu,\nu):=$
$\displaystyle\left(\inf\left\\{\mathbb{E}\|X-\tilde{X}\|_{*,t}^{2}:\mathcal{L}(X)=\mu,\mathcal{L}(\tilde{X})=\nu\right\\}\right)^{\frac{1}{2}},$
$\displaystyle
t\in[0,T],\quad\mu,\nu\in\mathcal{P}(\mathcal{C}([0,T],\mathbb{X})),$

where $\|X\|_{*,t}:=\sup_{0\leq s\leq t}|X_{s}|$ for
$X\in\mathcal{C}([0,T],\mathbb{X})$, $t\in[0,T]$. Additionally, as we are also
working with diagraph measures, we must also measure distances taking into
account the graph’s heterogeneity variable. For this purpose, we have the
following two measures:

$W^{\mathcal{N},2}_{2,t}(\mu,\nu):=\left(\int_{I}\left[W_{2,t}(\mu_{u},\nu_{u})\right]^{2}\leavevmode\nobreak\
\mathrm{d}u\right)^{\frac{1}{2}},\;t\in[0,T],\;\mu,\nu\in\mathcal{N},$
$W^{\mathcal{N},\infty}_{2,t}(\mu,\nu):=\sup_{u\in
I}W_{2,t}(\mu_{u},\nu_{u}),\;t\in[0,T],\;\mu,\nu\in\mathcal{N}.$

Since $I=[0,1]$, the following inequality holds:

$W^{\mathcal{N},2}_{2,t}(\mu,\nu)\leq W^{\mathcal{N},\infty}_{2,t}(\mu,\nu).$

We will work in the space $\mathcal{N}$ which, equipped with the distance
$W^{\mathcal{N},2}_{2,t}(\mu,\nu)$, is a complete metric space.

######  Remark 2.

Notice that the Wasserstein metric can be compared with the bounded Lipschitz
metric, denoted as $d_{\mathcal{BL}}$, because the following inequality holds:

$W_{2,t}(\mu_{u},\nu_{u})\geq\sup_{f\in\mathcal{BL}_{1}}\left|\int_{\mathbb{R}^{d}}f(x)\leavevmode\nobreak\
\mu_{u,t}(\mathrm{d}x)-\int_{\mathbb{R}^{d}}f(x)\leavevmode\nobreak\
\nu_{u,t}(\mathrm{d}x)\right|,\;\mu,\nu\in\mathcal{N},$

i.e. $d_{\mathcal{BL}}(\mu_{u,t},\nu_{u,t})\leq W_{2,t}(\mu_{u},\nu_{u})$ (see
[12]).

##  3 Main Results

Before stating our results, we introduce our main assumptions.

### Assumptions ($H$)

Let us assume that $f$, $h$ and $g$ are bounded and Lipschitz continuous
functions. We denote $B_{f}=\sup_{x\in\mathbb{X}}|f(x)|$,
$B_{h}=\sup_{(x,y)\in\mathbb{X}^{2}}|h(x,y)|$ and
$B_{g}=\sup_{(x,y)\in\mathbb{X}^{2}}|g(x,y)|$, and $|f(x)-f(y)|\leq
L_{f}|x-y|$ holds for $x,y\in\mathbb{X}$, $|h(x_{1},y_{1})-h(x_{2},y_{2})|\leq
L_{h}(|x_{1}-x_{2}|+|y_{1}-y_{2}|)$ and $|g(x_{1},y_{1})-g(x_{2},y_{2})|\leq
L_{g}(|x_{1}-x_{2}|+|y_{1}-y_{2}|)$ for $(x_{i},y_{i})\in\mathbb{X}^{2}$,
where $L_{f}$, $L_{h}$ and $L_{g}$ are the respective Lipschitz constants.

We remark that the Lipschitz assumptions are in many systems not a severe
restriction, e.g., if the system is globally dissipative, one may use cut-off
arguments to make sufficiently smooth nonlinearities Lipschitz in all parts of
phase space relevant for the long-term dynamics. Of course, there are
situations, where Lipschitz assumptions fail, e.g., for singular interaction
terms but we are not going to pursue this direction here.

Let $[0,T]$ be the time domain for the system dynamics for some $T>0$.
Consider a filtered probability space
$(\Omega,\mathcal{F},\\{F_{t}\\}_{t\in[0,T]},\mathbb{P})$, where $\\{F_{t}\\}$
denotes a filtration that complies with standard conditions. The particle
trajectories $X(t)$ take values in $\mathbb{X}$. The random variable $U$ is
given on a compact Polish probability space $(I,\mathfrak{B}(I),\mu_{I})$,
which we select as $I=[0,1]$ with the usual Lebesgue measure here but point
out that more general choices might turn out to be useful as well. Indeed,
recall that $I$ will be space to track the node labels upon passing to the
mean-field limit, so choosing a more geometric space for
$(I,\mathfrak{B}(I),\mu_{I})$ could track distances on the graph but here we
work measure-theoretically so the Borel-Maraham theorem guarantees that no
important measure-theoretic information is lost by considering $I=[0,1]$.
Regarding the assumptions about the graph, we assume that $\sup_{1\leq i\leq
N}\sum_{j=1}^{N}(A^{N}_{i,j})^{2}=\mathcal{O}(N)$ and $\sup_{1\leq i\leq
N}\sum_{j=1}^{N}(\hat{A}^{N}_{i,j})^{2}=\mathcal{O}(N)$ as
$N\rightarrow\infty$.

Finally, we need to impose conditions on the initial data. We assume that the
initial data ${X_{i}(0)}$, for $1\leq i\leq N$, are independent and
identically distributed, and we have finite second moments initially
$\mathbb{E}|X_{i}(0)^{2}|<+\infty$, for all $i\in\\{1,2,\ldots,N\\}$.
Moreover, assume that

$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}|X_{i}^{0}-X_{\frac{i}{N}}^{0}|^{2}=0,$

for any random variable $X_{i}^{0}$ and $X^{0}_{\frac{i}{N}}$, for
$i=1,\ldots,N$, where $\mu^{0}=\mathcal{L}(X_{i}^{0})$ and
$\bar{\mu}^{0}=\mathcal{L}(X_{\frac{i}{N}}^{0})$, for $i=1,\ldots,N$.

After establishing the key assumptions of our problem (1), we proceed to
present our main theorem, where we establish the convergence in law of the
empirical measure towards a probability measure that satisfies the Vlasov-
Fokker-Planck equation. The proof of this theorem will be provided in Section
5.

######  Theorem 3.

Under the assumptions ($H$), let us consider sequences of graphs
$\\{A_{i,j}^{N}\\}_{N\geq 1}$ and $\\{\hat{A}_{i,j}^{N}\\}_{N\geq 1}$. Assume
that there exists $\eta$ and $\hat{\eta}$ (DGM) such that $\eta_{A^{N}}$
converges to $\eta$ and $\eta_{\hat{A}^{N}}$ converges to $\hat{\eta}$. In
this case, the empirical measure $\mu_{N}$ converges in probability to a
measure $\bar{\mu}\in\mathcal{P}(C([0,T],\mathbb{X}))$, where $\bar{\mu}$
represents the law of the solution $X$ of equation (2). And $\bar{\mu}$ solves
the Vlasov-Fokker-Plank equation in the weak sense

$\begin{split}\partial_{t}\bar{\mu}_{t}&+\partial_{x}\left(\bar{\mu}_{t}f(x)+\int_{I}{\mu}_{u,t}\int_{I}\int_{\mathbb{X}}g(x,y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\mathrm{d}u\right)=\\\
&+\frac{1}{2}\partial^{2}_{x}\left(\int_{I}{\mu}_{u,t}\left[\int_{I}\int_{\mathbb{X}}h(x,y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\right]^{2}\leavevmode\nobreak\
\mathrm{d}u\right).\end{split}$ (5)

Note that when studying this problem, we have assumed that the heterogeneity
variable of the graph is a random variable given by a uniform distribution. We
can modify this hypothesis and obtain better properties regarding the law
$\mu_{u,t}$.

As we did before, let us consider the following system of independent
processes:

$\begin{split}X_{u}(t)=&X_{u}(0)+\int_{0}^{t}f(X_{u}(s))\leavevmode\nobreak\
\mathrm{d}s+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}g(X_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta_{u}(\mathrm{d}v)\mathrm{d}s\\\
&+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}h(X_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}B_{s}^{u},\end{split}$
(6)

where $\mu_{u,t}=\mathcal{L}(X_{u}(t))$, for $u\in I$. Since we do not take
$u$ as a random variable, we need to modify the way we understand the law of
$X_{u}(t)$ for each $u$ and impose more properties on this law. To do this, we
will modify our space, and instead of working in $\mathcal{N}$, we will work
in the following space:

$\displaystyle\tilde{\mathcal{N}}:=\left\\{\nu=(\mu_{u},u\in
I)\in(\mathcal{P}(\mathcal{C}([0,T],\mathbb{X})))^{I}:u\rightarrow\mu_{u}\text{
is measurable},\right.$ $\displaystyle\left.\sup_{u\in
I}\int_{\mathcal{C}_{d}}\|x\|_{*,T}^{2}\leavevmode\nobreak\
\mu_{u}(\mathrm{d}x)<\infty\right\\}.$

This space equipped with the distance $W^{\mathcal{N},\infty}_{2,t}$ is a
complete metric space.

In addition to the assumptions ($H$), we must establish one more set of
assumptions:

### Assumptions ($\tilde{H}$)

Let us assume that the law of probability $\bar{\mu}_{0}$ of the initial
condition is measurable with respect to the heterogeneity variable of the
graph. In other words, the mapping $u\in I\rightarrow\bar{\mu}^{0}_{u}\in
P(\mathbb{X})$ is measurable. Moreover, assume that
$W_{2}\left(\bar{\mu}_{u_{1}}^{0},\bar{\mu}_{u_{2}}^{0}\right)\leq\alpha\left|u_{1}-u_{2}\right|$,
for $u_{1},u_{2}\in I$ and $\alpha\in\mathbb{R}_{+}$.

With the additional assumptions about our system, we obtain the following
result:

######  Theorem 4.

Under the assumptions (H) and ($\tilde{H}$), let us consider a sequence of
graphs $\\{A_{i,j}^{N}\\}_{N\geq 1}$ and $\\{\hat{A}_{i,j}^{N}\\}_{N\geq 1}$.
Assume that there exists $\eta$ and $\hat{\eta}$ (DGM) such that
$\eta_{A^{N}}$ converges to $\eta$ and $\eta_{\hat{A}^{N}}$ converges to
$\hat{\eta}$. In this case, the empirical measure $\mu_{N}$ converges in
probability in $\mathcal{P}(C([0,T],\mathbb{X}))$ to the measure $\bar{\mu}$.
And $\bar{\mu}$ solves the Vlasov-Fokker-Plank equation in the weak sense

$\begin{split}\partial_{t}\bar{\mu}_{t}&+\partial_{x}\left(\bar{\mu}_{t}f(x)+\int_{I}{\mu}_{u,t}\int_{I}\int_{\mathbb{X}}g(x,y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\mathrm{d}u\right)=\\\
&+\frac{1}{2}\partial^{2}_{x}\left(\int_{I}{\mu}_{u,t}\left[\int_{I}\int_{\mathbb{X}}h(x,y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\right]^{2}\leavevmode\nobreak\
\mathrm{d}u\right).\end{split}$ (7)

Observing that in both theorems, we obtain similar results regarding the
probability $\bar{\mu}$ that is, in both, we demonstrate that our empirical
measure converges to it in probability and satisfies the Vlasov-Fokker-Planck
equation. However, in each case, the law represents a different concept.

In the first case, $\mu_{u,t}(x)$ belongs to the space $\mathcal{P}(I\times
C([0,t],\mathbb{X}))$, so we are considering that our probability measure
determines the probability of finding a particle at position $x$ with
heterogeneity variable $u$, for each time instant $t\in[0,T]$. In contrast, in
the second case, for each fixed $u\in I$, the probability measure
$\mu_{u,t}(x)$ determines the probability of finding the particle at position
$x$.

These different ways of understanding each probability measure lead us to
observe $\bar{\mu}$ in different ways: In the first case, it is understood as
a marginal as we integrate over all possible values that $u$ can take, giving
us $\bar{\mu}$ as the probability law of $X$. Meanwhile, in the second case,
since we have a probability law for each $u$, by integrating with respect to
$u$, we are measuring the average probability of finding particles at position
$X$.

##  4 Existence of Solutions

First of all, we are going to prove the existence and uniqueness of solution
of a solution to (2).

######  Proposition 5.

Under the assumptions ($H$), for every random variable U on I there exists a
unique solution to (2).

###### Proof.

To prove the existence of solution for the equation (2), we will consider an
operator defined on the space $\mathcal{N}$, and look for a fixed point of
this operator. That is, let us consider the mapping
$\mu\in\mathcal{N}\mapsto\mathcal{F}(\mu)\in\mathcal{N}$, where
$\mathcal{F}(\mu)$ is the law associated to the solution of the equation

$\begin{split}X^{\mu}_{u}(t)=&X^{\mu}_{u}(0)+\int_{0}^{t}f(X^{\mu}_{u}(s))\leavevmode\nobreak\
\mathrm{d}s+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}g(X^{\mu}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\mathrm{d}s\\\
&+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}h(X^{\mu}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}B_{s}^{u}.\end{split}$
(8)

Note that if we have a fixed point, then
$\mathcal{F}(\mu)=\mathcal{L}(X^{\mu})=\mu$, so we would prove the existence
of solution. First of all, we must verify that the operator is well defined
and that for each $\mu\in\mathcal{N}$ we have existence of solution of the
problem. Let us first prove that the operator is well defined. To do this, let
us take $\mu\in\mathcal{N}$, and $X^{0}_{u}(t)=X_{u}(0)$, $\forall t\in[0,T]$
and $u\in I$. Consider the following recurrence equation

$\displaystyle X^{n}_{u}(t)$ $\displaystyle=$ $\displaystyle
X^{n-1}_{u}(0)+\int_{0}^{t}f(X^{n-1}_{u}(s))\mathrm{d}s+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}g(X^{n-1}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\mathrm{d}s$ (9)
$\displaystyle+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}h(X^{n-1}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}B_{s}^{u},$

where $X^{k}_{u}(0)=X^{0}_{u}(0)$, for all $k\geq 1$. Let us prove that
$\left\\{X^{n}_{u}\right\\}_{n\geq 0}$ is Cauchy. Since
$X^{n+1}_{u}-X^{n}_{u}$ is a martingale, we can use Burkholder-Davis-Gundy’s
inequality ( [11]) and obtain

$\mathbb{E}\|X^{n+1}_{u}-X^{n}_{u}\|^{2}_{*,t}\leq
K_{BDG}\mathbb{E}[X^{n+1}_{u}-X^{n}_{u}]_{t}.$ (10)

In this case the $K_{BDG}=4$. Estimating the right-hand term we have

$\displaystyle\mathbb{E}|X^{n+1}_{u}(t)-X^{n}_{u}(t)|^{2}\leq
3\mathbb{E}\int_{0}^{t}|f(X^{n}_{u}(s))-f(X^{n-1}_{u}(s))|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+3\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}g(X^{n}_{u}(s),y)\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)-g(X^{n-1}_{u}(s),y)\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\right|^{2}\mathrm{d}s$
$\displaystyle+3\mathbb{E}\left|\int_{0}^{t}\int_{I}\int_{\mathbb{X}}\left(h(X^{n}_{u}(s),y)\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)-h(X^{n-1}_{u}(s),y)\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\right)\mathrm{d}B_{s}^{u}\right|^{2}.$

Using properties of stochastic integrals [19], i.e., by Itô isometry, we can
rewrite the last integral term to convert it to a deterministic integral and
obtain

$\displaystyle\mathbb{E}$ $\displaystyle|X^{n+1}_{u}(t)-X^{n}_{u}(t)|^{2}\leq
3\mathbb{E}\int_{0}^{t}|f(X^{n}_{u}(s))-f(X^{n-1}_{u}(s))|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+3\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}g(X^{n}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)-g(X^{n-1}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\right|^{2}\mathrm{d}s$
$\displaystyle+3\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}h(X^{n}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)-h(X^{n-1}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\right|^{2}\mathrm{d}s.$

Using that the functions $g$, $h$ and $f$ are Lipschitz, we get

$\displaystyle\mathbb{E}|X^{n+1}_{u}(t)-X^{n}_{u}(t)|^{2}$ $\displaystyle\leq
3\mathbb{E}\int_{0}^{t}L_{f}^{2}|X^{n}_{u}(s)-X^{n-1}_{u}(s)|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+3\mathbb{E}\int_{0}^{t}\left(\int_{I}\int_{\mathbb{X}}L_{g}|X^{n-1}_{u}(s)-X^{n}_{u}(s)|\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\right)^{2}\mathrm{d}s$
$\displaystyle+3\mathbb{E}\int_{0}^{t}\left(\int_{I}\int_{\mathbb{X}}L_{h}|X^{n-1}_{u}(s)-X^{n}_{u}(s)|\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\right)^{2}\mathrm{d}s.$

Therefore, we get

$\mathbb{E}|X^{n+1}_{u}(t)-X^{n}_{u}(t)|^{2}\leq
3\left(L_{f}^{2}+L_{g}^{2}\|\eta\|^{2}+L_{h}^{2}\|\hat{\eta}\|^{2}\right)\mathbb{E}\int_{0}^{t}|X^{n}_{u}(s)-X^{n-1}_{u}(s)|^{2}\leavevmode\nobreak\
\mathrm{d}s.$

By taking the supremum in $s$ of the difference between $X^{n}$ and $X^{n-1}$,
in the right-hand side integral, it follows

$\mathbb{E}|X^{n+1}_{u}(t)-X^{n}_{u}(t)|^{2}\leq
3\left(L_{f}^{2}+L_{g}^{2}\|\eta\|^{2}+L_{h}^{2}\|\hat{\eta}\|^{2}\right)\mathbb{E}\int_{0}^{t}\|X^{n}_{u}-X^{n-1}_{u}\|_{*,s}^{2}\leavevmode\nobreak\
\mathrm{d}s.$

By using (10), we arrive at the expression:

$\mathbb{E}\|X^{n+1}_{u}-X^{n}_{u}\|^{2}_{*,t}\leq
12\left(L_{f}^{2}+L_{g}^{2}\|\eta\|^{2}+L_{h}^{2}\|\hat{\eta}\|^{2}\right)\int_{0}^{t}\mathbb{E}\|X^{n}_{u}-X^{n-1}_{u}\|_{*,s}^{2}\leavevmode\nobreak\
\mathrm{d}s.$

Defining
$C=12\left(L_{f}^{2}+L_{g}^{2}\|\eta\|^{2}+L_{h}^{2}\|\hat{\eta}\|^{2}\right)$
and iterating this, we get that:

$\mathbb{E}\|X^{n+1}_{u}-X^{n}_{u}\|^{2}_{*,t}\leq
C^{n}\frac{T^{n}}{n!}\mathbb{E}\|X^{1}_{u}-X^{0}_{u}\|_{*,s}^{2},$

where $\mathbb{E}\|X^{1}_{u}-X^{0}_{u}\|_{*,s}^{2}$ is bounded, due to the
assumptions on the initial data and the fact that the functions $f$,$g$ and
$h$ are bounded. It follows that the sequence
$\left\\{X^{n}_{u}\right\\}_{n\geq 0}$ is Cauchy and converges uniformly in
probability at $u\in I$ to $X^{\mu}_{u}$, which satisfies the equation (8).
Therefore, we have that $\mathcal{F}$ is a well-defined map from $\mathcal{N}$
to $\mathcal{N}$. It remains to be seen that the application $\mathcal{F}$ has
a fixed point in this space. To do this, let us consider
$\mu,\nu\in\mathcal{N}$, with $X^{\mu}_{u}$ and $X^{\nu}_{u}$ their respective
solutions to (8), and let us estimate
$\mathbb{E}|X^{\mu}_{u}-X^{\nu}_{u}|^{2}$ as follows

$\displaystyle\mathbb{E}|X^{\mu}_{u}-X^{\nu}_{u}|^{2}$ $\displaystyle\leq
3\mathbb{E}\int_{0}^{t}|f(X^{\mu}_{u}(s))-f(X^{\nu}_{u}(s))|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+3\mathbb{E}\int_{0}^{t}\int_{I}\left|\int_{\mathbb{X}}g(X^{\mu}_{u}(s),y)\mu_{v,s}(\mathrm{d}y)-g(X^{\nu}_{u}(s),y)\nu_{v,s}(\mathrm{d}y)\right|^{2}\eta^{u}(\mathrm{d}v)\mathrm{d}s$
$\displaystyle+3\mathbb{E}\int_{0}^{t}\int_{I}\left|\int_{\mathbb{X}}h(X^{\mu}_{u}(s),y)\mu_{v,s}(\mathrm{d}y)-h(X^{\nu}_{u}(s),y)\nu_{v,s}(\mathrm{d}y)\right|^{2}\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}s,$

where we have used again Itô’s isometry. By adding and subtracting
$g(X^{\mu}_{u}(s),y)\nu_{v,s}(\mathrm{d}y)$ in the second term, and
$h(X^{\mu}_{u}(s),y)\nu_{v,s}(\mathrm{d}y)$ we can bound the above expression
by the sum of the following terms.

$\displaystyle\mathbb{E}|X^{\mu}_{u}-X^{\nu}_{u}|^{2}$ $\displaystyle\leq
3\mathbb{E}\int_{0}^{t}|f(X^{\mu}_{u}(s))-f(X^{\nu}_{u}(s))|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+6\mathbb{E}\int_{0}^{t}\int_{I}\left|\int_{\mathbb{X}}g(X^{\mu}_{u}(s),y)\leavevmode\nobreak\
(\mu_{v,s}(\mathrm{d}y)-\nu_{v,s}(\mathrm{d}y))\right|^{2}\eta^{u}(\mathrm{d}v)\mathrm{d}s$
$\displaystyle+6\mathbb{E}\int_{0}^{t}\int_{I}\left|\int_{\mathbb{X}}(g(X^{\mu}_{u}(s),y)-g(X^{\nu}_{u}(s),y))\leavevmode\nobreak\
\nu_{v,s}(\mathrm{d}y)\right|^{2}\eta^{u}(\mathrm{d}v)\mathrm{d}s$
$\displaystyle+6\mathbb{E}\int_{0}^{t}\int_{I}\left|\int_{\mathbb{X}}h(X^{\mu}_{u}(s),y)(\mu_{v,s}(\mathrm{d}y)-\nu_{v,s}(\mathrm{d}y))\right|^{2}\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}s$
$\displaystyle+6\mathbb{E}\int_{0}^{t}\int_{I}\left|\int_{\mathbb{X}}(h(X^{\mu}_{u}(s),y)-h(X^{\nu}_{u}(s),y))\nu_{v,s}(\mathrm{d}y)\right|^{2}\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}s.$

Using the Lipschitz condition of the functions $f$, $g$ and $h$, and the
Remark 2, we have

$\displaystyle\mathbb{E}|X^{\mu}_{u}-X^{\nu}_{u}|^{2}$
$\displaystyle\leq(3L_{f}^{2}+6(L_{g}^{2}\|\eta\|+L_{h}^{2}\|\hat{\eta}\|))\mathbb{E}\int_{0}^{t}|X^{\mu}_{u}(s)-X^{\nu}_{u}(s)|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+6B_{g}^{2}\int_{0}^{t}\int_{I}|W_{2,s}(\mu_{v,s},\nu_{v,s})|^{2}\leavevmode\nobreak\
\eta^{u}(\mathrm{d}v)\mathrm{d}s$
$\displaystyle+6B_{h}^{2}\int_{0}^{t}\int_{I}|W_{2,s}(\mu_{v,s},\nu_{v,s})|^{2}\leavevmode\nobreak\
\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}s.$

Taking the supremum for $v\in I$ in the 2-Wasserstein metric, we arrive at the
following expression:

$\displaystyle\mathbb{E}|X^{\mu}_{u}-X^{\nu}_{u}|^{2}$
$\displaystyle\leq(3L_{f}^{2}+6(L_{g}^{2}\|\eta\|+L_{h}^{2}\|\hat{\eta}\|))\mathbb{E}\int_{0}^{t}|X^{\mu}_{u}(s)-X^{\nu}_{u}(s)|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+6(B_{g}^{2}\|\eta\|+B_{h}^{2}\|\hat{\eta}\|)\int_{0}^{t}|W^{\mathcal{N},\infty}_{2,s}(\mu_{s},\nu_{s})|^{2}\leavevmode\nobreak\
\mathrm{d}s.$

It then, follows from Grönwall’s inequality that

$\mathbb{E}|X^{\mu}_{u}-X^{\nu}_{u}|^{2}\leq
6(B_{g}^{2}\|\eta\|+B_{h}^{2}\|\hat{\eta}\|)e^{(3L_{f}^{2}+6(L_{g}^{2}\|\eta\|+L_{h}^{2}\|\hat{\eta}\|))T}\int_{0}^{t}|W^{\mathcal{N,\infty}}_{2,s}(\mu,\nu)|^{2}\leavevmode\nobreak\
\mathrm{d}s.$

By this inequality and using the Burkholder-Davis-Gundy (BDG) inequality
again, we get that

$\mathbb{E}\|X^{\mu}_{u}-X^{\nu}_{u}\|_{*,t}^{2}\leq
24(B_{g}^{2}\|\eta\|+B_{h}^{2}\|\hat{\eta}\|)e^{(3L_{f}^{2}+6(L_{g}^{2}\|\eta\|+L_{h}^{2}\|\hat{\eta}\|))T}\int_{0}^{t}|W^{\mathcal{N,\infty}}_{2,s}(\mu,\nu)|^{2}\leavevmode\nobreak\
\mathrm{d}s.$

This means that

$|W^{\mathcal{N},\infty}_{2,t}(\mathcal{F}(\mu),\mathcal{F}(\nu))|^{2}\leq
C\int_{0}^{t}|W^{\mathcal{N},\infty}_{2,s}(\mu,\nu)|^{2}\leavevmode\nobreak\
\mathrm{d}s,$ (11)

where
$C=24(B_{g}^{2}\|\eta\|+B_{h}^{2}\|\hat{\eta}\|)e^{(3L_{f}^{2}+6(L_{g}^{2}\|\eta\|+L_{h}^{2}\|\hat{\eta}\|))T}$.
This expression gives the path-wise uniqueness of the solution, and also allow
us to prove the existence of solution. For this purpose we will build an
iterative process, as follows. Consider $\nu=(\mathcal{L}(Z_{u}),u\in I)$,
where $Z_{u}(t)=X_{u}(0)$ for all $u\in I$ and $t\in[0,T]$. Iterating this,
and using (11), we get

$W^{\mathcal{N},2}_{2,T}(\mathcal{F}^{n+1}(\nu),\mathcal{F}^{n}(\nu))\leq\frac{C^{n}T^{n}}{n!}|W^{\mathcal{N},\infty}_{2,T}(\mathcal{F}(\nu),\nu)|.$

It follows that the sequence $\mathcal{F}^{n}(\nu)$ is Cauchy for $n$ large
enough, where we have used that
$W^{\mathcal{N},\infty}_{2,T}(\mathcal{F}(\nu),\nu)<\infty$, due to the
assumptions on the initial data and the fact that the functions $f$, $g$, and
$h$ are bounded. This sequence will have a limit since $\mathcal{N}$ is a
complete metric space, and hence there exists $\mu=(\mathcal{L}(X_{u}),u\in
I)\in\mathcal{N}$ solution of (2), for the initial data $X_{u}(0)$. ∎

######  Lema 6.

Let $\bar{\mu}$ be the law associated to the solution $X$ of (2), with
$\mu_{u,t}=\mathcal{L}(X(s)|U=u)$ being the conditional probability with
respect to the random variable $U$. Then $\mu_{u,t}$ is a weak solution to the
following Vlasov-Fokker-Planck equation:

$\begin{split}\partial_{t}{\mu}_{u,t}&+\partial_{x}\left({\mu}_{u,t}f(x)+{\mu}_{u,t}\int_{I}\int_{\mathbb{X}}g(x,y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\right)=\\\
&+\frac{1}{2}\partial^{2}_{x}\left({\mu}_{u,t}\left[\int_{I}\int_{\mathbb{X}}h(x,y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\right]^{2}\right).\end{split}$
(12)

###### Proof.

Consider a test function $\phi\in C^{\infty}_{\textnormal{c}}(\mathbb{X})$,
and let us compute the time derivative of $\phi(X_{u}(t))$, where $u$ is a
sample of the random variable $U$. To do that, let us make use of Itô’s
formula. We have

$\begin{split}\mathrm{d}\phi=&\left(\partial_{x}\phi
f(X_{u}(t))+\partial_{x}\phi\int_{I}\int_{\mathbb{X}}g(X_{u}(t),y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\right.\\\
&\left.+\frac{1}{2}\left[\int_{I}\int_{\mathbb{X}}h(x,y)\mu_{v,t}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\right]^{2}\partial^{2}_{x}\phi\right)\mathrm{d}t\\\
&+\partial_{x}\phi\int_{I}\int_{\mathbb{X}}h(x,y)\mu_{v,t}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}B^{u}_{t}.\end{split}$
(13)

If we calculate the expectation of this expression and integrate by parts, we
obtain the weak formulation (12). ∎

With the objective of demonstrating the convergence of the solution, it is
crucial to investigate, how the solution depends on the digraph measure
$\eta$. Specifically, we want to establish suitable continuity properties.
Such continuity proofs will lay the foundation for later proving the
convergence to a mean-field limit.

######  Proposition 7.

Given $\hat{\eta}_{1},\hat{\eta}_{2}\in\mathcal{BC}(I,\mathcal{M}_{+}(I))$,
let $\mu_{1}$ and $\mu_{2}$ be the laws of the solutions of (2) for the DGMs
$\hat{\eta}_{1}$ and $\hat{\eta}_{2}$. Then,

$\left[W_{2,s}^{\mathcal{N},2}(\mu^{1},\mu^{2})\right]^{2}\leq\hat{C}_{1}\hat{C}_{2}\left[d_{\infty}(\hat{\eta}_{1},\hat{\eta}_{2})\right]^{2}e^{\hat{C}_{1}T},$
(14)

where
$\hat{C}_{1}=3\left[2B_{g}^{2}\|\eta\|^{2}+3B_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]e^{\left(3L_{f}^{2}+3\left[2L_{g}^{2}\|\eta\|^{2}+3L_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]\right)T}$
and
$\hat{C}_{2}=\frac{3B^{2}_{h}T}{\left[2B_{g}^{2}\|\eta\|^{2}+3B_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]}$.

Given $\eta_{1},\eta_{2}\in\mathcal{BC}(I,\mathcal{M}_{+}(I))$, let $\mu_{1}$
and $\mu_{2}$ be the laws of the solutions of (2) for the DGMs $\eta_{1}$ and
$\eta_{2}$. Then,

$\left[W_{2,s}^{\mathcal{N},2}(\mu^{1},\mu^{2})\right]^{2}\leq
C_{1}C_{2}\left[d_{\infty}(\eta_{1},\eta_{2})\right]^{2}e^{C_{1}T},$ (15)

where
$C_{1}=3\left[3B_{g}^{2}\|\eta_{1}\|^{2}+2B_{h}^{2}\|\hat{\eta}\|^{2}\right]e^{\left(3L_{f}^{2}+3\left[3L_{g}^{2}\|\eta_{1}\|^{2}+2L_{h}^{2}\|\hat{\eta}\|^{2}\right]\right)T}$
and
$C_{2}=\frac{3B^{2}_{g}T}{\left[3B_{g}^{2}\|\eta_{1}\|^{2}+2B_{h}^{2}\|\hat{\eta}\|^{2}\right]}$.

###### Proof.

We will proceed to prove only the first statement of the proposition, as the
proof idea is the same for both cases. Consider two different DGMs
$\hat{\eta}_{1}$ and
$\hat{\eta}_{2}\in\mathcal{B}(I,\mathcal{M}_{+}(I))\bigcap\mathcal{C}(I,\mathcal{M}_{+}(I))$
with the respective solutions they induce denoted by $X^{1}$ and $X^{2}$. We
have

$\displaystyle\mathbb{E}|X_{u}^{1}(t)-X_{u}^{2}(t)|^{2}$ $\displaystyle\leq
3\mathbb{E}\int_{0}^{t}|f(X_{u}^{1}(s))-f(X_{u}^{2}(s))|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+3\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}g(X^{1}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}^{1}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)-g(X^{2}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}^{2}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\right|^{2}\,\mathrm{d}s$
$\displaystyle+3\mathbb{E}\left|\int_{0}^{t}\int_{I}\int_{\mathbb{X}}\left(h(X^{1}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}^{1}(\mathrm{d}y)\hat{\eta}^{u}_{1}(\mathrm{d}v)-h(X^{2}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}^{2}(\mathrm{d}y)\hat{\eta}^{u}_{2}(\mathrm{d}v)\right)\,\mathrm{d}B^{u}_{s}\right|^{2}.$

Let us start with the term related to Brownian motion. We can rewrite this
term again using standard properties of stochastic integrals. Furthermore, by
adding and subtracting terms, we get

$\displaystyle 3\mathbb{E}$
$\displaystyle\left|\int_{0}^{t}\int_{I}\int_{\mathbb{X}}\left(h(X^{1}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}^{1}(\mathrm{d}y)\hat{\eta}^{u}_{1}(\mathrm{d}v)-h(X^{2}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}^{2}(\mathrm{d}y)\hat{\eta}^{u}_{2}(\mathrm{d}v)\right)\,\mathrm{d}B^{u}_{s}\right|^{2}$
$\displaystyle\leq$
$\displaystyle+9\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}\left[h(X^{1}_{u}(s),y)-h(X^{2}_{u}(s),y)\right]\mu_{v,s}^{1}(\mathrm{d}y)\hat{\eta}^{u}_{1}(\mathrm{d}v)\right|^{2}\,\mathrm{d}s$
$\displaystyle+9\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}h(X^{2}_{u}(s),y)\left[\mu_{v,s}^{1}(\mathrm{d}y)-\mu_{v,s}^{2}(\mathrm{d}y)\right]\hat{\eta}^{u}_{1}(\mathrm{d}v)\right|^{2}\,\mathrm{d}s$
$\displaystyle+9\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}h(X^{2}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}^{2}(\mathrm{d}y)\left[\hat{\eta}^{u}_{1}(\mathrm{d}v)-\hat{\eta}^{u}_{2}(\mathrm{d}v)\right]\right|^{2}\,\mathrm{d}s.$

Using the properties of $h$, and just like we did in the previous proof, we
can bound these terms as follow

$\displaystyle 3\mathbb{E}$
$\displaystyle\left|\int_{0}^{t}\int_{I}\int_{\mathbb{X}}\left(h(X^{1}_{u}(s),y)\mu_{v,s}^{1}(\mathrm{d}y)\hat{\eta}^{u}_{1}(\mathrm{d}v)-h(X^{2}_{u}(s),y)\mu_{v,s}^{2}(\mathrm{d}y)\hat{\eta}^{u}_{2}(\mathrm{d}v)\right)\,\mathrm{d}B^{u}_{s}\right|^{2}$
$\displaystyle\leq$
$\displaystyle+9L_{h}^{2}\|\hat{\eta}_{1}\|^{2}\mathbb{E}\int_{0}^{t}\left|X^{1}_{u}(s)-X^{2}_{u}(s)\right|^{2}\,\mathrm{d}s$
$\displaystyle+9B_{h}^{2}\|\hat{\eta}_{1}\|^{2}\int_{0}^{t}\left|W_{2,s}^{\mathcal{N},\infty}(\mu^{1},\mu^{2})\right|^{2}\,\mathrm{d}s$
$\displaystyle+9B_{h}^{2}T\left[d_{\infty}(\hat{\eta}_{1},\hat{\eta}_{2})\right]^{2}.$

If we go back to the original inequality and bound the remainder terms using
the same methodology as we did for the Brownian term, we obtain the following
expression

$\displaystyle\mathbb{E}|X_{u}^{1}(t)-X_{u}^{2}(t)|^{2}$
$\displaystyle\leq\left(3L_{f}^{2}+3\left[2L_{g}^{2}\|\eta\|^{2}+3L_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]\right)\mathbb{E}\int_{0}^{t}|X_{u}^{1}-X_{u}^{2}|_{*,s}^{2}\,\mathrm{d}s$
$\displaystyle+3\left[2B_{g}^{2}\|\eta\|^{2}+3B_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]\int_{0}^{t}\left[W_{2,s}^{\mathcal{N},\infty}(\mu^{1},\mu^{2})\right]^{2}\,\mathrm{d}s$
$\displaystyle+9B_{h}^{2}T\left[d_{\infty}(\hat{\eta}_{1},\hat{\eta}_{2})\right]^{2}.$

Applying the BDG inequality and Grönwall’s inequality we get

$\mathbb{E}|X_{u}^{1}-X_{u}^{2}|_{*,t}^{2}\leq
C_{1}\left(\int_{0}^{t}\left[W_{2,s}^{\mathcal{N},\infty}(\mu^{1},\mu^{2})\right]^{2}\,\mathrm{d}s+C_{2}\left[d_{\infty}(\hat{\eta}_{1},\hat{\eta}_{2})\right]^{2}\right),$

where
$C_{1}=3\left[2B_{g}^{2}\|\eta\|^{2}+3B_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]e^{\left(3L_{f}^{2}+3\left[2L_{g}^{2}\|\eta\|^{2}+3L_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]\right)T}$
and
$C_{2}=\frac{3B^{2}_{h}T}{\left[2B_{g}^{2}\|\eta\|^{2}+3B_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]}$.
By definition of $W_{2,t}^{\mathcal{N},\infty}(\mu^{1},\mu^{2})$, we have the
following:

$\left[W_{2,s}^{\mathcal{N},\infty}(\mu^{1},\mu^{2})\right]^{2}\leq
C_{1}\left(\int_{0}^{t}\left[W_{2,s}^{\mathcal{N},\infty}(\mu^{1},\mu^{2})\right]^{2}\,\mathrm{d}s+C_{2}\left[d_{\infty}(\hat{\eta}_{1},\hat{\eta}_{2})\right]^{2}\right).$

Applying Grönwall’s inequality, we find

$\left[W_{2,s}^{\mathcal{N},\infty}(\mu^{1},\mu^{2})\right]^{2}\leq
C_{1}C_{2}\left[d_{\infty}(\hat{\eta}_{1},\hat{\eta}_{2})\right]^{2}e^{C_{1}T}.$

And thus, we obtain the desired inequality

$\left[W_{2,s}^{\mathcal{N},2}(\mu^{1},\mu^{2})\right]^{2}\leq
C_{1}C_{2}\left[d_{\infty}(\hat{\eta}_{1},\hat{\eta}_{2})\right]^{2}e^{C_{1}T}.$

This finishes the proof. ∎

##  5 Proof of Theorem 5

Once the existence of the solution of the independent problem has been proven
and the continuity of the solution with respect to the DGM has been shown, the
next step is to determine how close the solutions of the problem (1) are to
the solutions of the independent problem (2). To compare trajectories of
solutions between equation (1) and equation (2), we will associate the
particles $X_{i}$ with the particles $X_{u(i)}$. To achieve this, we will
consider a partition of the interval $I$ defined by $I^{N}_{i}$, as introduced
previously. For each interval, we will select a representative
$u(i)=\frac{i}{N}$. Consequently, we will compare the trajectories of
$X_{i}^{N}$ and $X_{\frac{i}{N}}$. To proceed with this matter, it is
important to consider that each one is described by a distinct stochastic
process. However, to identify both particles as similar, we make the
assumption that for our choice of the random variable $u=\frac{i}{N}$, both
particles exhibit the same Brownian motions, i.e.,
$B^{i}_{t}=B^{\frac{i}{N}}_{t}$. This strategy of aligning the trajectories
based on $u$ enables us to evaluate the similarity between the two particles,
despite their different stochastic processes.

######  Proposition 8.

Under the hypotheses of Theorem 5, the following inequality

$\displaystyle\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}\|X_{i}^{N}(t)-$
$\displaystyle X_{\frac{i}{N}}\|^{2}_{*,t}\leq
C\frac{16}{N}\sum_{i=1}^{N}\mathbb{E}|X_{i}^{0}-X_{\frac{i}{N}}^{0}|^{2}$
$\displaystyle+C\left(\frac{2}{N}(B_{g}^{2}+B_{h}^{2})\mathcal{O}(1)+B_{g}^{2}\left[d_{\infty}\left(\eta_{A^{N}},\eta\right)\right]^{2}+B_{h}^{2}\left[d_{\infty}\left(\eta_{\hat{A}^{N}},\hat{\eta}\right)\right]^{2}\right),$

holds, where $C=48Te^{(16L_{f}^{2}+96(L_{g}^{2}+L_{h}^{2})\mathcal{O}(1))T}$.

###### Proof.

Fix $t\in[0,T]$. Let us compare $X_{i}^{N}$ and $X_{\frac{i}{N}}$, solutions
of (1) and (2), respectively.

$\displaystyle\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}\|X_{i}^{N}(t)-X_{\frac{i}{N}}\|^{2}_{*,t}\leq\frac{16}{N}\sum_{i=1}^{N}\mathbb{E}|X_{i}^{0}-X_{\frac{i}{N}}^{0}|^{2}$
$\displaystyle+\frac{16}{N}\sum_{i=1}^{N}\mathbb{E}\int_{0}^{t}|f(X_{i}^{N}(s))-f(X_{\frac{i}{N}}(s))|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+\frac{16}{N}\sum_{i=1}^{N}\mathbb{E}\int_{0}^{t}\left|\frac{1}{N}\sum_{i=1}^{N}A_{i,j}^{N}g(X_{i}^{N}(s),X_{j}^{N}(s))-\int_{I}\int_{\mathbb{X}}g(X_{\frac{i}{N}}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{\frac{i}{N}}(\mathrm{d}v)\right|^{2}\mathrm{d}s$
$\displaystyle+\frac{16}{N}\sum_{i=1}^{N}\mathbb{E}\int_{0}^{t}\left|\frac{1}{N}\sum_{j=1}^{N}\hat{A}^{N}_{ij}h(X_{i}^{N}(s),X_{j}^{N}(s))\mathrm{d}s-\int_{I}\int_{\mathbb{X}}h(X_{\frac{i}{N}}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{\frac{i}{N}}(\mathrm{d}v)\right|^{2}\mathrm{d}s.$

Hence, we get that
$\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}\|X_{i}^{N}(t)-X_{\frac{i}{N}}\|^{2}_{*,t}\leq\mathrm{E}_{0}+\mathrm{E}_{f}+\mathrm{E}_{g}+\mathrm{E}_{h}$,
so let us estimate each term. Notice that, by the construction of the
solution, we have assumed that $B^{i}_{t}=B^{\frac{i}{N}}$. Let us start with
$E_{f}$. We can bound the term depending on the function $f$ in a similar way
to what we have been doing throughout this work. Using the fact that the
function $f$ is Lipschitz,

$\displaystyle\mathrm{E}_{f}$ $\displaystyle\leq
L_{f}^{2}\frac{16}{N}\sum_{i=1}^{N}\int_{0}^{t}\mathbb{E}|X_{i}^{N}(s)-X_{\frac{i}{N}}(s)|^{2}\leavevmode\nobreak\
\mathrm{d}s.$

To estimate $E_{g}$, we will add and subtract different similar terms to find

$\begin{split}\mathrm{E}_{g}\leq\frac{48}{N}&\Bigg{(}\int_{0}^{t}\sum_{i=1}^{N}\mathbb{E}\left|\frac{1}{N}\sum_{j=1}^{N}A^{N}_{i,j}(g(X_{i}^{N}(s),X_{j}^{N}(s))-g(X_{\frac{i}{N}}(s),X_{\frac{j}{N}}(s)))\right|^{2}\mathrm{d}s\\\
&+\int_{0}^{t}\sum_{i=1}^{N}\mathbb{E}\left|\frac{1}{N}\sum_{j=1}^{N}A^{N}_{i,j}\left(g(X_{\frac{i}{N}}(s),X_{\frac{j}{N}}(s))-\int_{\mathbb{X}}g(X_{\frac{i}{N}}(s),y)\leavevmode\nobreak\
\mu_{\frac{j}{N},s}(\mathrm{d}y)\right)\right|^{2}\mathrm{d}s\\\
&+\int_{0}^{t}\sum_{i=1}^{N}\mathbb{E}\left|\int_{I}\int_{\mathbb{X}}g(X_{\frac{i}{N}}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\left(\eta_{A^{N}}^{\frac{i}{N}}(\mathrm{d}v)-\eta^{\frac{i}{N}}(\mathrm{d}v)\right)\right|^{2}\leavevmode\nobreak\
\mathrm{d}s\Bigg{)},\end{split}$

where $\eta^{\frac{i}{N}}_{A^{N}}$ is given by (4). We can denote the previous
estimate as
$\mathrm{E}_{g}\leq\mathrm{E}^{1}_{g}+\mathrm{E}^{2}_{g}+\mathrm{E}^{3}_{g}$.
Let us bound each term separately. Using the Lipschitz property of $g$, the
term $\mathrm{E}^{1}_{g}$ can be bounded as follows:

$\displaystyle\mathrm{E}^{1}_{g}$ $\displaystyle\leq
L_{g}^{2}\frac{48}{N^{3}}\int_{0}^{t}\sum_{i=1}^{N}\left[\sum_{j=1}^{N}(A^{N}_{i,j})^{2}\cdot\mathbb{E}\left[\sum_{j=1}^{N}\left(|X_{i}^{N}(s)-X_{\frac{i}{N}}(s)|+|X_{j}^{N}(s)-X_{\frac{j}{N}}(s)|\right)^{2}\right]\right]\mathrm{d}s$
$\displaystyle\leq
L_{g}^{2}\frac{96}{N^{3}}\int_{0}^{t}\sum_{i=1}^{N}\left[\sum_{j=1}^{N}(A^{N}_{i,j})^{2}\cdot\sum_{j=1}^{N}\left(\mathbb{E}|X_{i}^{N}(s)-X_{\frac{i}{N}}(s)|^{2}+\mathbb{E}|X_{j}^{N}(s)-X_{\frac{j}{N}}(s)|^{2}\right)\right]\mathrm{d}s$
$\displaystyle\leq\frac{96}{N}L_{g}^{2}\mathcal{O}(1)\int_{0}^{t}\sum_{i=1}^{N}\mathbb{E}|X_{i}^{N}(s)-X_{\frac{i}{N}}(s)|^{2}\mathrm{d}s,$

where we have used that $\sum_{j=1}^{N}(A^{N}_{i,j})^{2}=\mathcal{O}(N)$, for
all $1\leq i\leq N$. Similarly, the term $\mathrm{E}_{g}^{3}$ can be bounded
by using the fact that $g$ is a Lipschitz and bounded function:

$\mathrm{E}_{g}^{3}\leq
48TB_{g}^{2}\left[d_{\infty}(\eta_{A^{N}},\eta)\right]^{2}.$ (16)

Finally, we only need to estimate $\mathrm{E}_{g}^{2}$. Note that this term
can be rewritten by expanding the square as follows:

$\displaystyle\mathrm{E}_{g}^{2}$
$\displaystyle=\frac{48}{N^{3}}\int_{0}^{t}\sum_{i=1}^{N}\sum_{j=1}^{N}\sum_{k=1}^{N}\mathbb{E}\left[\left(A^{N}_{i,j}g(X_{\frac{i}{N}}(s),X_{\frac{j}{N}}(s))-\int_{\mathbb{X}}g(X_{\frac{i}{N}}(s),y)A^{N}_{i,j}\leavevmode\nobreak\
\mu_{\frac{j}{N},s}(\mathrm{d}y)\right)\right.$
$\displaystyle\left.\left(A^{N}_{i,k}g(X_{\frac{i}{N}}(s),X_{\frac{k}{N}}(s))-\int_{\mathbb{X}}g(X_{\frac{i}{N}}(s),y)A^{N}_{i,k}\leavevmode\nobreak\
\mu_{\frac{k}{N},s}(\mathrm{d}y)\right)\right]\mathrm{d}s.$ (17)

We observe that, due to the independence of the $X_{\frac{i}{N}}$ as
constructed in (2), all terms will be 0 except when $k=j$ or $k=i$. Therefore,
by making use of the boundedness of $g$ we obtain

$\mathrm{E}_{g}^{2}\leq\frac{96}{N}B_{g}^{2}\int_{0}^{t}\sum_{i=1}^{N}\frac{1}{N^{2}}\sum_{j=1}^{N}(A^{N}_{i,j})^{2}.$

Using that $\sum_{j=1}^{N}(A^{N}_{i,j})^{2}=\mathcal{O}(N)$, for all $1\leq
i\leq N$, we have

$\mathrm{E}_{g}^{2}\leq\frac{96}{N}B_{g}^{2}T\mathcal{O}(1).$

The bounds for the term $E_{h}$ are calculated in the same way as the previous
term. Putting all the estimates together, we obtain:

$\displaystyle\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}\|X_{i}^{N}(t)-X_{\frac{i}{N}}\|^{2}_{*,t}$
$\displaystyle\leq\left(16L_{f}^{2}+96(L_{g}^{2}+L_{h}^{2})\mathcal{O}(1)\right)\int_{0}^{t}\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}|X_{i}^{N}(s)-X_{\frac{i}{N}}(s)|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+\frac{96}{N}(B_{g}^{2}+B_{h}^{2})T\mathcal{O}(1)+\frac{16}{N}\sum_{i=1}^{N}\mathbb{E}|X_{i}^{0}-X_{\frac{i}{N}}^{0}|^{2}$
$\displaystyle+48T\left(B_{g}^{2}\left[d_{\infty}\left(\eta_{A^{N}},\eta\right)\right]^{2}+B_{h}^{2}\left[d_{\infty}\left(\eta_{\hat{A}^{N}},\hat{\eta}\right)\right]^{2}\right).$

Grönwall’s inequality leads to

$\displaystyle\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}\|X_{i}^{N}(t)-$
$\displaystyle X_{\frac{i}{N}}\|^{2}_{*,t}\leq
C\frac{16}{N}\sum_{i=1}^{N}\mathbb{E}|X_{i}^{0}-X_{\frac{i}{N}}^{0}|^{2}$
$\displaystyle+C\left(\frac{2}{N}(B_{g}^{2}+B_{h}^{2})\mathcal{O}(1)+B_{g}^{2}\left[d_{\infty}\left(\eta_{A^{N}},\eta\right)\right]^{2}+B_{h}^{2}\left[d_{\infty}\left(\eta_{\hat{A}^{N}},\hat{\eta}\right)\right]^{2}\right),$

where $C=48Te^{(16L_{f}^{2}+96(L_{g}^{2}+L_{h}^{2})\mathcal{O}(1))T}$. ∎

To prove Theorem 5, we have to demonstrate that the empirical measure
$\mu_{N}=\frac{1}{N}\sum_{i=1}^{N}\delta_{X_{i}^{N}}$ converges in probability
to $\bar{\mu}=\int_{I}\mu_{u}du$.

###### Proof.

(of Theorem 5) The convergence in probability, follows from the convergence
for both random variables: $U$ and the one generated by the Brownian motion.
To do this, it is sufficient to establish that:

$\lim_{N\rightarrow+\infty}E_{u}\times\mathbb{E}\left|\int_{\mathbb{X}}f(y)\leavevmode\nobreak\
\bar{\mu}_{N}(\mathrm{d}y)-\int_{\mathbb{X}}f(y)\leavevmode\nobreak\
\bar{\mu}(\mathrm{d}y)\right|^{2}=0,$ (18)

for every bounded and Lipschitz function $f$, where $E_{u}$ denotes the
expectation related to the random variable $U$. We know that
$\mu_{i/N,t}=\mathcal{L}(X(t)|U=i/N)$, is the probability measure
$\bar{\mu}_{t}$ conditioned on $U=i/N$. Therefore we obtain

$\displaystyle\lim_{N\rightarrow+\infty}E_{u}\times\mathbb{E}\left|\int_{\mathbb{X}}f(y)\leavevmode\nobreak\
\bar{\mu}_{N}(\mathrm{d}y)-\int_{\mathbb{X}}f(y)\leavevmode\nobreak\
\bar{\mu}(\mathrm{d}y)\right|^{2}$
$\displaystyle\quad\leq\lim_{N\rightarrow+\infty}\frac{1}{N}\sum_{i=1}^{N}E_{u}\times\mathbb{E}\left|f(X_{i})-f\left(X_{\frac{i}{N}}\right)\right|^{2}.$

Taking into account the Lipschitz property of f and using Proposition 8, we
can conclude

$\displaystyle\lim_{N\rightarrow+\infty}E_{u}\times\mathbb{E}\left|\int_{\mathbb{X}}f(y)\leavevmode\nobreak\
\bar{\mu}_{N}(\mathrm{d}y)-\int_{\mathbb{X}}f(y)\leavevmode\nobreak\
\bar{\mu}(\mathrm{d}y)\right|^{2}$
$\displaystyle\quad\leq\lim_{N\rightarrow+\infty}\frac{1}{N}\sum_{i=1}^{N}E_{u}\times\mathbb{E}\left|X_{i}-X_{\frac{i}{N}}\right|_{*,t}^{2}=0.$

This finishes the proof of Theorem 5. ∎

##  6 Proof of Theorem 7

Finally, let us prove Theorem 7. Unlike before, since our probability measure
will now take actual input values $u\in I$, instead of considering $u$ as a
random variable, we will work in the space $\tilde{\mathcal{N}}$, in which we
must prove the measurability with respect to $u$. The overall proof idea of
Theorem 7 will follow very similar ideas as before, yet certain details that
need to be modified in some proofs due to working with measurability in the
variable $u$. Therefore, in this section, we will not go into detail in all
the proofs but will only introduce those steps that are different from the
previous ones. Let us begin by demonstrating the existence and uniqueness of
the solution to (6).

######  Proposition 9.

Under de assumptions $H$ and $\tilde{H}$, there exist a unique solution
$X_{u}$, for $u\in I$, to (6), where $\mu_{t}=(\mathcal{L}(X_{u}):u\in
I)\in\tilde{\mathcal{N}}$. Moreover, $\mu_{u,t}$ is a weak solution of the
Vlasov-Fokker-Plank equation:

$\begin{split}\partial_{t}{\mu}_{u,t}&+\partial_{x}\left({\mu}_{u,t}f(x)+{\mu}_{u,t}\int_{I}\int_{\mathbb{X}}g(x,y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\right)=\\\
&+\frac{1}{2}\partial^{2}_{x}\left({\mu}_{u,t}\left[\int_{I}\int_{\mathbb{X}}h(x,y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\right]^{2}\right),\forall
u\in I.\end{split}$ (19)

###### Proof.

The proof will be carried out in the same way as we did in Proposition 5. We
will consider an operator $\tilde{\mathcal{F}}$ defined in the space
$\mathcal{L}$ and seek a fixed point. To do this, let us consider the mapping
$\mu\in\tilde{\mathcal{N}}\mapsto\tilde{\mathcal{F}}(\mu)\in\tilde{\mathcal{N}}$,
where $\tilde{\mathcal{F}}(\mu)$ is the law associated with the solution of
the equation:

$\begin{split}X^{\mu}_{u}(t)=&X^{\mu}_{u}(0)+\int_{0}^{t}f(X^{\mu}_{u}(s))\leavevmode\nobreak\
\mathrm{d}s+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}g(X^{\mu}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\mathrm{d}s\\\
&+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}h(X^{\mu}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}v_{s}^{u}.\end{split}$
(20)

Note that if we have a fixed point then
$\tilde{\mathcal{F}}(\mu)=\mathcal{L}(X^{\mu})=\mu$, so this would prove the
existence of solution. First, we must prove that the operator is well-defined,
meaning that there exists a solution for (20), for every
$\mu\in\tilde{\mathcal{N}}$, and furthermore, the solutions are in
$\tilde{\mathcal{N}}$. To do this, let us take $\mu\in\mathcal{N}$, and let
$X^{0}_{u}(t)=X_{u}(0)$, $\forall t\in[0,T]$ and $u\in I$. Consider the
following recurrence equation:

$\displaystyle X^{n}_{u}(t)$ $\displaystyle=$ $\displaystyle
X^{n-1}_{u}(0)+\int_{0}^{t}f(X^{n-1}_{u}(s))\leavevmode\nobreak\
\mathrm{d}s+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}g(X^{n-1}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\mathrm{d}s$ (21)
$\displaystyle+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}h(X^{n-1}_{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}B_{s}^{u},$

where $X^{k}_{u}(0)=X^{0}_{u}(0)$, for all $k\geq 1$. Notice that if we prove
that the sequence $\left\\{X^{n}_{u}\right\\}_{n\geq 1}$ is Cauchy, and
moreover, for each $n\in\mathbb{N}$, $\mathcal{L}(X^{n}_{u}:u\in
I)\in\tilde{\mathcal{N}}$, then we will have that the sequence has a limit,
and this limit belongs to $\tilde{\mathcal{N}}$ since this space is complete
with the Wasserstein distance $W_{2,t}^{\tilde{\mathcal{N}},\infty}$.

The proof that $\left\\{X^{n}_{u}\right\\}_{n\geq 0}$ is Cauchy, is identical
to the one carried out in Proposition 5, therefore, we will not develop it.
Let us focus on proving that for each $n\in\mathbb{N}$, the mapping $u\in
I\mapsto\mathcal{L}(X^{n}_{u},B^{u})$ is measurable. We will prove it by
induction. By construction and assumptions $\tilde{H}$, measurability holds
for $n=0$. Suppose it holds for $n=0,\ldots,N-1$. Let us now establish that
measurability holds for $n=N$. To show this, it is sufficient to show that it
is measurable for any collection of times, that is,

$I\ni
u\mapsto\mathcal{L}\left({X}_{u}^{N}\left(t_{1}\right),B^{u}\left(t_{1}\right),\ldots,{X}_{u}^{N}\left(t_{m}\right),B^{u}\left(t_{m}\right)\right)\in\mathcal{P}\left(\left(\mathbb{R}^{d}\times\mathbb{R}^{d}\right)^{m}\right).$

is measurable for all $0\leq t_{1}\leq\cdots\leq t_{m}\leq T$ and
$m\in\mathbb{N}$. It further suffices to demonstrate that

$I\ni
u\mapsto\mathbb{E}\left[\prod_{i=1}^{m}\left(\alpha_{i}\left({X}_{u}^{N}\left(t_{i}\right)\right)\beta_{i}\left(B^{u}\left(t_{i}\right)\right)\right)\right]\in\mathbb{R},$

is measurable, for all $0\leq t_{1}\leq\cdots\leq t_{m}\leq T,m\in\mathbb{N}$
and bounded and continuous functions
$\left\\{\alpha_{i},\alpha_{i}:i=1,\ldots,m\right\\}$ on $\mathbb{R}^{d}$. Let
us now establish that $X_{u}^{N}(t)$ is measurable. To achieve this, consider
$X_{u}^{N,\delta}(t)$ as a solution to the following auxiliary process

$\displaystyle{X}_{u}^{N,\delta}(t)=$
$\displaystyle{X}_{u}^{N-1}(0)+\int_{0}^{t}f({X}_{u}^{N-1}\left(\left\lfloor\frac{s}{\delta}\right\rfloor\delta\right))\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}g\left({X}_{u}^{N-1}\left(\left\lfloor\frac{s}{\delta}\right\rfloor\delta\right),y\right)\leavevmode\nobreak\
\mu_{v,\left\lfloor\frac{s}{\delta}\right\rfloor\delta}(dy)\eta^{u}(\mathrm{d}v)\mathrm{d}s$
$\displaystyle+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}h\left({X}_{u}^{N-1}\left(\left\lfloor\frac{s}{\delta}\right\rfloor\delta\right),x\right)\leavevmode\nobreak\
\mu_{v,\left\lfloor\frac{s}{\delta}\right\rfloor\delta}(dx)\hat{\eta}^{u}(\mathrm{d}v)dB^{u}_{s},$

where $\delta\in(0,1)$. Note that, due to the construction of the auxiliary
process, for each $\delta>0$, we can express $X_{u}^{N,\delta}(t)$ as a finite
sum of terms that depend on

$\left\\{{X}_{u}^{N-1}(0),{X}_{u}^{N-1}(\delta),\ldots,{X}_{u}^{N-1}\left(\left\lfloor\frac{t}{\delta}\right\rfloor\delta\right)\right\\}\quad\text{and}\quad\left\\{B^{u}_{0},B^{u}_{\delta},\ldots,B^{u}_{\left(\left\lfloor\frac{t}{\delta}\right\rfloor\delta\right)}\right\\}.$

Furthermore, ${X}_{u}^{N,\delta}(t)$ converges to ${X}_{u}^{k}(t)$ in
probability as $\delta\rightarrow 0$, for each $u\in I$. So it suffices to
prove that

$I\ni
u\mapsto\mathbb{E}\left[\prod_{i=1}^{m}\left(\alpha_{i}\left({X}_{u}^{k,\delta}\left(t_{i}\right)\right)\beta_{i}\left(B^{u}_{t_{i}}\right)\right)\right]\in\mathbb{R},$

is measurable, for all $0\leq t_{1}\leq\cdots\leq t_{m}\leq T,m\in\mathbb{N}$
and bounded and continuous functions
$\left\\{\alpha_{i},\beta_{i}:i=1,\ldots,m\right\\}$ on $\mathbb{R}^{d}$. Fix
$t\in[0,\bar{T}]$. Since the measurability holds for $N-1$, it further
suffices to show that

${X}_{u}^{N,\delta}(t)=\gamma\left(u,{X}_{u}^{N-1},B^{u}\right),$

for some measurable function
$\gamma:I\times\mathcal{C}([0,T])\times\mathcal{C}([0,T])\rightarrow\mathbb{R}$.
Using that ${X}_{u}^{N,\delta}(t)$ is a finite sum of terms depending on
$\left\\{{X}_{u}^{N-1}(0),{X}_{u}^{N-1}(\delta),\ldots,{X}_{u}^{N-1}\left(\left\lfloor\frac{t}{\delta}\right\rfloor\delta\right)\right\\}$
and
$\left\\{B^{u}_{0},B^{u}_{\delta},\ldots,B^{u}_{\left(\left\lfloor\frac{t}{\delta}\right\rfloor\delta\right)}\right\\}$
continuously, we have that $\gamma(u,\cdot,\cdot)$ is continuous on
$\mathcal{C}([0,T])\times\mathcal{C}([0,T])$, for each $u\in I$, and that
$h(\cdot,x,w)$ is measurable on $I$ for each
$(x,w)\in\mathcal{C}([0,T])\times\mathcal{C}([0,T])$. Therefore, $\gamma$ is
measurable and this proves the mesurability for $n=N$. In summary, we have
thus proven that the operator $\tilde{\mathcal{F}}$ is well-defined. Now, we
need to analyze the existence of a fixed point for the operator. However, this
is nothing more than proving that given two laws $\mu$ and $\nu$ in
$\tilde{\mathcal{N}}$, it holds that:

$|W^{\tilde{\mathcal{N}},\infty}_{2,t}(\mathcal{F}(\mu),\mathcal{F}(\nu))|^{2}\leq
C\int_{0}^{t}|W^{\tilde{\mathcal{N}},\infty}_{2,s}(\mu,\nu)|^{2}\leavevmode\nobreak\
\mathrm{d}s,$ (22)

for a certain constant $C>0$. The proof of this inequality is similar to the
one carried out in Proposition 5, so we omit the details. The last inequality
gives the path-wise uniqueness of the solution, and also allow us to prove the
existence of solution. For this purpose we will build an iterative process, as
follows. Consider $\nu=(\mathcal{L}(Z_{u}):u\in I)$, where $Z_{u}(t)=X_{u}(0)$
for all $u\in I$ and $t\in[0,T]$. Iterating this, and using (11), we get:

$W^{\tilde{\mathcal{N}},\infty}_{2,T}(\tilde{\mathcal{F}}^{n+1}(\nu),\tilde{\mathcal{F}}^{n}(\nu))\leq\frac{C^{n}T^{n}}{n!}|W^{\tilde{\mathcal{N}},\infty}_{2,T}(\tilde{\mathcal{F}}(\nu),\nu)|.$

It follows that the sequence
$\left\\{\tilde{\mathcal{F}}^{n}(\nu)\right\\}_{n}$ is Cauchy for $n$ large
enough, where we have used that
$W^{\tilde{\mathcal{N}},\infty}_{2,T}(\tilde{\mathcal{F}}(\nu),\nu)<\infty$,
due to the assumptions on the initial data and the fact that the functions
$f$, $g$, and $h$ are bounded. This sequence will have a limit since
$\tilde{\mathcal{N}}$ is a complete metric space, and hence there exists
$\mu=(\mathcal{L}(X_{u}),u\in I)\in\tilde{\mathcal{N}}$ solution of (2), for
the initial data $X_{u}(0)$. ∎

Once the existence and uniqueness of the solution to (6) have been
established, the next step is to study how these solutions depend on the
graph. In other words, what properties the solutions have with respect to the
DGMs: $\eta$ and $\hat{\eta}$, and with respect to the graph’s heterogeneity
variable $u$. For this purpose, we have the following result.

######  Proposition 10.

Given $\hat{\eta}_{1},\hat{\eta}_{2}\in\mathcal{BC}(I,\mathcal{M}_{+}(I))$,
let $\mu_{1}$ and $\mu_{2}$ be the laws of the solutions of (6) for the DGMs
$\hat{\eta}_{1}$ and $\hat{\eta}_{2}$. Then,

$\left[W_{2,s}^{\tilde{\mathcal{N}},\infty}(\mu^{1},\mu^{2})\right]^{2}\leq\hat{C}_{1}\hat{C}_{2}\left[d_{\infty}(\hat{\eta}_{1},\hat{\eta}_{2})\right]^{2}e^{\hat{C}_{1}T},$
(23)

where
$\hat{C}_{1}=3\left[2B_{g}^{2}\|\eta\|^{2}+3B_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]e^{\left(3L_{f}^{2}+3\left[2L_{g}^{2}\|\eta\|^{2}+3L_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]\right)T}$
and
$\hat{C}_{2}=\frac{3B^{2}_{h}T}{\left[2B_{g}^{2}\|\eta\|^{2}+3B_{h}^{2}\|\hat{\eta}_{1}\|^{2}\right]}$.

Given $\eta_{1},\eta_{2}\in\mathcal{BC}(I,\mathcal{M}_{+}(I))$, let $\mu_{1}$
and $\mu_{2}$ be the laws of the solutions of (6) for the DGMs $\eta_{1}$ and
$\eta_{2}$. Then,

$\left[W_{2,s}^{\tilde{\mathcal{N}},\infty}(\mu^{1},\mu^{2})\right]^{2}\leq
C_{1}C_{2}\left[d_{\infty}(\eta_{1},\eta_{2})\right]^{2}e^{C_{1}T},$ (24)

where
$C_{1}=3\left[3B_{g}^{2}\|\eta_{1}\|^{2}+2B_{h}^{2}\|\hat{\eta}\|^{2}\right]e^{\left(3L_{f}^{2}+3\left[3L_{g}^{2}\|\eta_{1}\|^{2}+2L_{h}^{2}\|\hat{\eta}\|^{2}\right]\right)T}$
and
$C_{2}=\frac{3B^{2}_{g}T}{\left[3B_{g}^{2}\|\eta_{1}\|^{2}+2B_{h}^{2}\|\hat{\eta}\|^{2}\right]}$.

Given $u_{1},u_{2}\in I$, let $\mu_{u}$ be the law of the solution of (6).
Then,

$\displaystyle\left[W_{2,T}(\mu_{u_{1}},\mu_{u_{2}})\right]^{2}$
$\displaystyle\leq
C\left(\left[W_{2}(\bar{\mu}^{0}_{u_{1}},\bar{\mu}^{0}_{u_{2}})\right]^{2}\right.$
$\displaystyle\left.+2T\left[B_{g}^{2}d^{2}_{BL}(\eta^{u_{1}},\eta^{u_{2}})+B_{h}^{2}d^{2}_{BL}(\hat{\eta}^{u_{1}},\hat{\eta}^{u_{2}})\right]\right),$

where
$C=16e^{16\left(L^{2}_{f}+2L^{2}_{g}|\eta|^{2}+2L^{2}_{h}|\hat{\eta}|^{2}\right)T}$.

###### Proof.

The proof of the first two results is analogous to the one carried out in
Proposition 7, so we will not go into detail. Just note that since the
structure of the two equations, (2) and (6), is analogous, the bounds will be
the same thanks to the relationship between the measures $W^{.,2}$ and
$W^{.,\infty}$ and the fact that the set over which $u$ takes variables is the
interval $I=[0,1]$.

Let $u_{1},u_{2}\in I$, and let $\mu$ be the law of the solution to (6). We
know that for each $u\in I$, $X_{u}$ satisfies the equation:

$\displaystyle X^{u}(t)=$ $\displaystyle
X^{u}(0)+\int_{0}^{t}f(X^{u}(s))\leavevmode\nobreak\
\mathrm{d}s+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}g(X^{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\mathrm{d}s$
$\displaystyle+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}h(X^{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}B_{s}^{u}.$ (25)

To remove the dependence of $B^{u}_{s}$ with respect to $u$, we will employ
the coupling method (see [7]). This method allows us to relate our stochastic
process with the following system, as in the limit both systems satisfy the
same dynamics:

$\displaystyle X^{u}(t)=$ $\displaystyle
X^{u}(0)+\int_{0}^{t}f(X^{u}(s))\leavevmode\nobreak\
\mathrm{d}s+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}g(X^{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\mathrm{d}s$
$\displaystyle+\int_{0}^{t}\int_{I}\int_{\mathbb{X}}h(X^{u}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\mathrm{d}B_{s}.$ (26)

Therefore, since we are interested in analyzing what happens in the limit, it
will be enough to prove the property for the second system. Let us proceed to
bound $\mathbb{E}|X_{u_{1}}(t)-X_{u_{2}}(t)|^{2}$:

$\displaystyle\mathbb{E}|X_{u_{1}}(t)-X_{u_{2}}(t)|^{2}\leq
4\mathbb{E}|X_{u_{1}}(0)-X_{u_{2}}(0)|^{2}+4\mathbb{E}\int_{0}^{t}|f(X_{u_{1}}(s))-f(X_{u_{2}}(s))|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+4\mathbb{E}\int_{0}^{t}|\int_{I}\int_{\mathbb{X}}g(X^{u_{1}}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u_{1}}(\mathrm{d}v)-g(X^{u_{2}}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u_{2}}(\mathrm{d}v)|^{2}\mathrm{d}s$
$\displaystyle+4\mathbb{E}\int_{0}^{t}|\int_{I}\int_{\mathbb{X}}h(X^{u_{1}}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u_{1}}(\mathrm{d}v)-h(X^{u_{2}}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u_{2}}(\mathrm{d}v)|^{2}\mathrm{d}s.$

We have used the Hölder inequality and the properties of stochastic integrals.
By adding and subtracting terms in the last two integrals, we obtain the
following:

$\displaystyle\mathbb{E}|X_{u_{1}}(t)-X_{u_{2}}(t$ $\displaystyle)|^{2}\leq
4\mathbb{E}|X_{u_{1}}(0)-X_{u_{2}}(0)|^{2}+4\mathbb{E}\int_{0}^{t}|f(X_{u_{1}}(s))-f(X_{u_{2}}(s))|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+8\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}\left(g(X^{u_{1}}(s),y)-g(X^{u_{2}}(s),y)\right)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\eta^{u_{1}}(\mathrm{d}v)\right|^{2}\mathrm{d}s$
$\displaystyle+8\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}g(X^{u_{1}}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\left(\eta^{u_{1}}(\mathrm{d}v)-\eta^{u_{2}}(\mathrm{d}v)\right)\right|^{2}\mathrm{d}s$
$\displaystyle+8\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}\left(h(X^{u_{1}}(s),y)-h(X^{u_{2}}(s),y)\right)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\hat{\eta}^{u_{1}}(\mathrm{d}v)\right|^{2}\mathrm{d}s$
$\displaystyle+8\mathbb{E}\int_{0}^{t}\left|\int_{I}\int_{\mathbb{X}}h(X^{u_{1}}(s),y)\leavevmode\nobreak\
\mu_{v,s}(\mathrm{d}y)\left(\eta^{u_{1}}(\mathrm{d}v)-\hat{\eta}^{u_{2}}(\mathrm{d}v)\right)\right|^{2}\mathrm{d}s.$

Using the definition of bounded Lipschitz distance for the DGM and leveraging
the Lipschitz and boundedness properties of the functions $f$, $h$, and $g$,
we obtain the following estimate:

$\displaystyle\mathbb{E}|X_{u_{1}}(t)-X_{u_{2}}(t)|^{2}$ $\displaystyle\leq
4\mathbb{E}|X_{u_{1}}(0)-X_{u_{2}}(0)|^{2}$
$\displaystyle+4\left(L^{2}_{f}+2L^{2}_{g}\|\eta\|^{2}+2L^{2}_{h}\|\hat{\eta}\|^{2}\right)\mathbb{E}\int_{0}^{t}|X_{u_{1}}(s)-X_{u_{2}}(s)|^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+8T\left[B_{g}^{2}d^{2}_{BL}(\eta^{u_{1}},\eta^{u_{2}})+B_{h}^{2}d^{2}_{BL}(\hat{\eta}^{u_{1}},\hat{\eta}^{u_{2}})\right].$

Taking the supremum in $s$ of the difference $|X_{u_{1}}(s)-X_{u_{2}}(s)|$,
and using the BDG inequality, we get:

$\displaystyle\mathbb{E}\|X_{u_{1}}-X_{u_{2}}\|_{*,t}^{2}$ $\displaystyle\leq
16\mathbb{E}|X_{u_{1}}(0)-X_{u_{2}}(0)|^{2}$
$\displaystyle+16\left(L^{2}_{f}+2L^{2}_{g}\|\eta\|^{2}+2L^{2}_{h}\|\hat{\eta}\|^{2}\right)\mathbb{E}\int_{0}^{t}\|X_{u_{1}}-X_{u_{2}}\|_{*,s}^{2}\leavevmode\nobreak\
\mathrm{d}s$
$\displaystyle+32T\left[B_{g}^{2}d^{2}_{BL}(\eta^{u_{1}},\eta^{u_{2}})+B_{h}^{2}d^{2}_{BL}(\hat{\eta}^{u_{1}},\hat{\eta}^{u_{2}})\right].$

Using the Grönwall’s inequality yields

$\displaystyle\mathbb{E}\|X_{u_{1}}-X_{u_{2}}\|_{*,t}^{2}$ $\displaystyle\leq
C\left(\mathbb{E}|X_{u_{1}}(0)-X_{u_{2}}(0)|^{2}\right.$
$\displaystyle\left.+2T\left[B_{g}^{2}d^{2}_{BL}(\eta^{u_{1}},\eta^{u_{2}})+B_{h}^{2}d^{2}_{BL}(\hat{\eta}^{u_{1}},\hat{\eta}^{u_{2}})\right]\right),$

where
$C=16e^{16\left(L^{2}_{f}+2L^{2}_{g}|\eta|^{2}+2L^{2}_{h}|\hat{\eta}|^{2}\right)T}$.
By the definition of the Wasserstein distance, we have:

$\displaystyle\left[W_{2,T}(\mu_{u_{1}},\mu_{u_{2}})\right]^{2}$
$\displaystyle\leq C\left(\mathbb{E}|X_{u_{1}}(0)-X_{u_{2}}(0)|^{2}\right.$
$\displaystyle\left.+2T\left[B_{g}^{2}d^{2}_{BL}(\eta^{u_{1}},\eta^{u_{2}})+B_{h}^{2}d^{2}_{BL}(\hat{\eta}^{u_{1}},\hat{\eta}^{u_{2}})\right]\right).$

As this inequality holds for any random variable $X_{u}(0)$, if we take the
infimum, we get:

$\displaystyle\left[W_{2,T}(\mu_{u_{1}},\mu_{u_{2}})\right]^{2}$
$\displaystyle\leq
C\left(\left[W_{2}(\bar{\mu}^{0}_{u_{1}},\bar{\mu}^{0}_{u_{2}})\right]^{2}\right.$
$\displaystyle\left.+2T\left[B_{g}^{2}d^{2}_{BL}(\eta^{u_{1}},\eta^{u_{2}})+B_{h}^{2}d^{2}_{BL}(\hat{\eta}^{u_{1}},\hat{\eta}^{u_{2}})\right]\right).$

Thus, we obtain the desired inequality and conclude the proof. ∎

After establishing the existence of solution for (6) and studying the
properties of it, let us proceed to determine how close the solutions of (1)
are to those of (6). To accomplish this, as we did in the previous section, we
will compare the trajectories of solutions between equation (1) and equation
(2), associating the particles $X_{i}$ with the particles $X_{u}(i)$. To
facilitate this comparison, we will consider a partition of the interval $I$
defined by $I^{N}_{i}$, as introduced earlier. Within each interval, we will
select a representative $u(i)=\frac{i}{N}$. Consequently, we will compare the
trajectories of $X_{i}^{N}$ and $X_{\frac{i}{N}}$. It is important to point
out that each one is described by a distinct stochastic process. However, to
identify both particles as similar, we make the assumption that for our choice
of the random variable $u=\frac{i}{N}$, both particles exhibit the same
Brownian motions, i.e., $B^{i}_{t}=B^{\frac{i}{N}}_{t}$.

######  Proposition 11.

Under the hypotheses of Theorem 7, the following inequality

$\displaystyle\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}\|X_{i}^{N}(t)-$
$\displaystyle X_{\frac{i}{N}}\|^{2}_{*,t}\leq
C\frac{16}{N}\sum_{i=1}^{N}\mathbb{E}|X_{i}^{0}-X_{\frac{i}{N}}^{0}|^{2}$
$\displaystyle+C\left(\frac{2}{N}(B_{g}^{2}+B_{h}^{2})\mathcal{O}(1)+B_{g}^{2}\left[d_{\infty}\left(\eta_{A^{N}},\eta\right)\right]^{2}+B_{h}^{2}\left[d_{\infty}\left(\eta_{\hat{A}^{N}},\hat{\eta}\right)\right]^{2}\right),$

holds, where $C=48Te^{(16L_{f}^{2}+96(L_{g}^{2}+L_{h}^{2})\mathcal{O}(1))T}$.

###### Proof.

Since the proof of this result is identical to the one carried out in the
previous section, we will not go into detail about the proof. ∎

We can now proceed to prove Theorem 7.

###### Proof.

We want to prove that the empirical measure $\mu^{N}$ converges in probability
to the measure $\bar{\mu}$. To do this, it is sufficient to establish that:

$\lim_{N\rightarrow+\infty}\mathbb{E}\left|\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
{\mu}_{N}(\mathrm{d}y)-\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
\bar{\mu}(\mathrm{d}y)\right|^{2}=0,$

for every bounded and Lipschitz function $k$. To show that this tends to zero,
let us add and subtract some term:

$\displaystyle\mathbb{E}\left|\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
{\mu}^{N}(\mathrm{d}y)-\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
\bar{\mu}(\mathrm{d}y)\right|^{2}$
$\displaystyle\leq\mathbb{E}\left|\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
{\mu}^{N}(\mathrm{d}y)-\frac{1}{N}\sum_{i=1}^{N}\int_{\mathbb{X}}k(y)\delta_{X_{\frac{i}{N}}}(y)\right|^{2}$
$\displaystyle+\mathbb{E}\left|\frac{1}{N}\sum_{i=1}^{N}\left(\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
\delta_{X_{\frac{i}{N}}}(\mathrm{d}y)-\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
\mu_{\frac{i}{N}}(\mathrm{d}y)\right)\right|^{2}$
$\displaystyle+\mathbb{E}\left|\frac{1}{N}\sum_{i=1}^{N}\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
\mu_{\frac{i}{N}}(\mathrm{d}y)-\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
\bar{\mu}(\mathrm{d}y)\right|^{2}.$ (27)

Let us see that each term converges to zero. If we expand the first term, we
get:

$\displaystyle\mathbb{E}$
$\displaystyle\left|\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
{\mu}^{N}(\mathrm{d}y)-\frac{1}{N}\sum_{i=1}^{N}\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
\delta_{X_{\frac{i}{N}}}(y)\right|^{2}$
$\displaystyle\leq\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}|k(X_{i})-k(X_{\frac{i}{N}})|^{2}.$

Using $k$ is Lipschitz and by Proposition 11, we deduce that it tends to zero.
Let us proceed with the second term of (27).

$\mathbb{E}\left|\frac{1}{N}\sum_{i=1}^{N}\left(\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
\delta_{X_{\frac{i}{N}}}(\mathrm{d}y)-\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
\mu_{\frac{i}{N}}(\mathrm{d}y)\right)\right|^{2}\leq\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}\left|k(X_{\frac{i}{N}})-\mathbb{E}|k(X_{\frac{i}{N}})|\right|^{2}.$

Using that $k$ is bounded, we prove that this term converges to zero. Finally,
for the last term of (27), we can write it as:

$\mathbb{E}\left|\frac{1}{N}\sum_{i=1}^{N}\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
\mu_{\frac{i}{N}}(\mathrm{d}y)-\int_{I}\int_{\mathbb{X}}k(y)\leavevmode\nobreak\
{\mu}_{u}(\mathrm{d}y)du\right|^{2}.$

Using Proposition 10, which establishes the continuity of $\hat{\mu}_{u}$ with
respect to $u$, allows us to demonstrate that this term tends to zero.
Consequently, $\mu^{N}\rightarrow\bar{\mu}$ in probability. ∎

##  7 Examples

As demonstrated earlier, under certain conditions, the empirical measure of
the stochastic process (1) converges to the measure $\bar{\mu}(t,x)$ that
satisfies the differential equation:

$\begin{split}\partial_{t}\bar{\mu}_{t}&+\partial_{x}\left(\bar{\mu}_{t}f(x)+\int_{I}{\mu}_{u,t}\int_{I}\int_{\mathbb{X}}g(x,y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\eta^{u}(\mathrm{d}v)\mathrm{d}u\right)=\\\
&+\frac{1}{2}\partial^{2}_{x}\left(\int_{I}{\mu}_{u,t}\left[\int_{I}\int_{\mathbb{X}}h(x,y)\leavevmode\nobreak\
\mu_{v,t}(\mathrm{d}y)\hat{\eta}^{u}(\mathrm{d}v)\right]^{2}\mathrm{d}u\right).\end{split}$
(28)

Now, let us very briefly explore different examples of stochastic processes
and how this theorem would apply to these processes, along with the
corresponding limiting probability equation.

Consider first the case where the Brownian term is only multiplied by a
constant, and the graph connects all nodes with each other, i.e., the
adjacency matrix $A_{ij}=1$ for $i=1,..,N$. We obtain the following stochastic
process:

$\displaystyle\mathrm{d}X_{i}(t)$
$\displaystyle=f(X_{i}(t))\mathrm{d}t+\frac{1}{N}\sum_{j=1}^{N}g(X_{i},X_{j})\mathrm{d}t+\sigma\mathrm{d}B_{t}^{i}.$
(29)

In this case, the limit differential equation is:

$\begin{split}\partial_{t}\bar{\mu}_{t}&+\partial_{x}\left(\bar{\mu}_{t}f(x)+\bar{\mu}_{t}\int_{\mathbb{X}}g(x,y)\leavevmode\nobreak\
\mu_{t}(\mathrm{d}y)\right)=\frac{\sigma^{2}}{2}\partial^{2}_{x}\bar{\mu}_{t}.\end{split}$
(30)

Now, assume that the limiting DGMs $\eta$ and $\hat{\eta}$ have sufficiently
good properties, and $\eta$,
$\hat{\eta}\in\mathcal{B}(I,\mathcal{M}_{+,abs}(I))\bigcap\mathcal{C}(I,\mathcal{M}_{+,abs}(I))$,
where $\mathcal{M}_{+,abs}(I)$ is the set of all finite Borel positive
measures on $I$ absolutely continuous with respect to the Lebesgue measure
$m$. In this case, denoting $W$ and $\hat{W}$ as the respective Radon-Nikodym
derivatives of the previous DGMs, the system can be seen as a representation
through graphons. In this space, the adjacency matrices are represented
through graphons, and the limiting equation has the form

$\begin{split}\partial_{t}\bar{\mu}_{t}&+\partial_{x}\left(\bar{\mu}_{t}f(x)+\int_{I}{\mu}_{u,t}\int_{I}\int_{\mathbb{X}}W(u,v)g(x,y)\mu_{v,t}(\mathrm{d}y)\mathrm{d}v\mathrm{d}u\right)=\\\
&+\frac{1}{2}\partial^{2}_{x}\left(\int_{I}{\mu}_{u,t}\left[\int_{I}\int_{\mathbb{X}}\hat{W}(u,v)h(x,y)\mu_{v,t}(\mathrm{d}y)\mathrm{d}v\right]^{2}du\right).\end{split}$
(31)

This differential equation is consistent with the results obtained in other
works [3, 4] for graphons. Yet, our result is a clear generalization as there
are many DGMs that have no graphon representation as we do not assume any
absolute continuity of the graph limit. For example, we can apply our result
to certain sparse graphs, such as certain ring networks, where the limiting
DGM would be $2\delta_{u}(v)$. The resulting differential equation is then:

$\begin{split}\partial_{t}\bar{\mu}_{t}&+2\partial_{x}\left(\bar{\mu}_{t}f(x)+\int_{I}{\mu}_{u,t}\int_{\mathbb{X}}g(x,y)\leavevmode\nobreak\
\mu_{u,t}(\mathrm{d}y)\mathrm{d}u\right)=\\\
&+\partial^{2}_{x}\left(\int_{I}{\mu}_{u,t}\left[\int_{\mathbb{X}}h(x,y)\leavevmode\nobreak\
\mu_{u,t}(\mathrm{d}y)\right]^{2}\mathrm{d}u\right).\end{split}$ (32)

In particular, all the DGM examples discussed for the deterministic particle
system case in [17] can be carried over now to the stochastic setting, which
provides also a clear indication that our approach to mean-field limits via
DGMs is quite robust.

## Acknowledgments

This paper (CP) has been partially supported by Grant PID2022-137228OB-I00
funded by the Spanish Ministerio de Ciencia, Innovacion y Universidades,
MICIU/AEI/10.13039/501100011033 & “ERDF/EU A way of making Europe”, by Grant
C-EXP-265-UGR23 funded by Consejeria de Universidad, Investigacion e
Innovacion & ERDF/EU Andalusia Program, and by Modeling Nature Research Unit,
project QUAL21-011.

## References

  * [1] N. Ayi and N. Pouradier Duteil.  Mean-field and graph limits for collective dynamics models with time-varying weights.  Journal of Differential Equations, 299:65–110, 10 2021.
  * [2] A. Backhausz and B. Szegedy.  Action convergence of operators and graphs.  Canad. J. Math., 74(1):72–121, 2022.
  * [3] E. Bayraktar, S. Chakraborty, and R. Wu.  Graphon mean field systems.  The Annals of Applied Probability, 33(5):3587–3619, 2023.
  * [4] G. Bet, F. Coppini, and F.R. Nardi.  Weakly interacting oscillators on dense random graphs.  Journal of Applied Probability, page 1–24, 2023.
  * [5] W. Braun and K. Hepp.  The Vlasov dynamics and its fluctuation in the $1/n$ limit of interacting particles.  Communications in Mathematical Physics, 56:101–113, 06 1977.
  * [6] D. Bresch, P.-E. Jabin, and J. Soler.  A new approach to the mean-field limit of Vlasov-Fokker-Planck equations.  arXiv 2203.15747, 2023.
  * [7] L.-P. Chaintron and A. Diez.  Propagation of chaos: a review of models, methods and applications. I. models and methods.  Kinetic and Related Models, 15(6):895–1015, 2022.
  * [8] L.-P. Chaintron and A. Diez.  Propagation of chaos: A review of models, methods and applications.ii. applications.  Kinetic and Related Models, 15(6):1017–1173, 2022.
  * [9] H. Chiba and G. Medvedev.  The mean field analysis of the kuramoto model on graphs i. the mean field equation and transition point formulas.  Discrete and Continuous Dynamical Systems - A, 39, 12 2016.
  * [10] R.L. Dobrushin.  Vlasov equations.  Functional Analysis and Its Applications, 13(2):115–123, Apr 1979\.
  * [11] Lawrence Evans.  An introduction to stochastic differential equations.  AMS, 2012.
  * [12] A.L. Gibbs and F.E. Su.  On choosing and bounding probability metrics.  International Statistical Review, 70(3):419–435, 2002.
  * [13] M. Gkogkas and C. Kuehn.  Graphop mean-field limits for kuramoto-type models.  SIAM Journal on Applied Dynamical Systems, 21:248–283, 02 2022\.
  * [14] F. Golse.  On the Dynamics of Large Particle Systems in the Mean Field Limit, pages 1–144.  01 2016.
  * [15] P.-E. Jabin, D. Poyato, and J. Soler.  Mean-field limit of non-exchangeable systems.  arXiv preprint, arXiv:2112.15406., 2021.
  * [16] P.-E. Jabin and Z. Wang.  Mean field limit and propagation of chaos for Vlasov systems with bounded forces.  Journal of Functional Analysis, 271(12), 2016.
  * [17] C. Kuehn and C. Xu.  Vlasov equations on digraph measures.  Journal of Differential Equations, 339:261–349, 2022.
  * [18] G. Medvedev.  The nonlinear heat equation on dense graphs and graph limits.  SIAM Journal on Mathematical Analysis, 46(4):2743–2766, 2014.
  * [19] P. Mörters and Y. Peres.  Brownian Motion.  Cambridge Series in Statistical and Probabilistic Mathematics. Cambridge University Press, 2010.
  * [20] H. Neunzert.  An Introduction to the nonlinear Boltzmann-Vlasov Equation, volume 1048, pages 60–110.  11 2006.
  * [21] T. Paul and E. Trélat.  From microscopic to macroscopic scale equations: mean field, hydrodynamic and graph limits.  arXiv preprint, arXiv:2209.08832, 2022.
  * [22] A.S. Sznitman.  Topics in propagation of chaos.  In P.-L. Hennequin, editor, Ecole d’Eté de Probabilités de Saint-Flour XIX - 1989, volume 1464 of Lecture Notes in Mathematics, pages 165–251. Springer, 1991.