Well-posedness of the Mean Field PDE and Particle System for Stein Variational Gradient Descent

We consider the following interacting particle system in $\mathbb{R}^d$:

$$
\begin{aligned}
\dot{x}_i(t) &= -\frac{1}{N}\sum _{j=1}^{N}\nabla K(x_i(t)-x_j(t)) -\frac{1}{N} \sum _{j=1} ^{N}K\big(x_i(t)-x_j(t)\bigr)\nabla V (x_j(t)), \\\
x_i(0) &= x_i ^0 \in \mathbb R^d, \qquad i=1,\cdots,N.
\end{aligned} \tag{1}
$$

We refer to each of the $N$ functions $x_i(\cdot) \in \mathbb{R}^d$ as a particle. The function $K : \mathbb{R}^d \mapsto \mathbb{R}$ is a smooth, symmetric, and positive definite kernel. The function $V : \mathbb{R}^d \to \mathbb{R}$ is a smooth potential such that $e^{-V(x)}$ is integrable. More specific assumptions about $K$ and $V$ are given below.

We are interested in the macroscopic behavior of the particle system $(1)$ as $N \to \infty$ in the framework of mean field limit. Formally this mean field limit is described by the following non-local, nonlinear partial differential equation (PDE):
$$
\begin{aligned}
&\partial_t \rho = \nabla \cdot \left(\rho\left(K \ast (\nabla \rho+\nabla V\rho)\right)\right), \\\
&\rho(0,\cdot) = \rho_0(\cdot).
\end{aligned} \tag{2}
$$

In this article, we will prove the well-posedness of (1) and (2).

Assumptions and Notations

Throughout the paper we assume that the kernel $K$ satisfies the following:

Assumption 1. $K : \mathbb{R}^d \mapsto \mathbb{R}$ is at least $C^4$ with bounded derivatives. In addition, $K(x-y)$ is symmetric and positive definite, meaning that
$$
\sum_{i=1}^{m}\sum_{j=1}^{m}K(x_i-x_j)\xi_i\xi_j \geq 0,
\qquad
\forall\ x_i \in \mathbb{R}^d,\ \xi_i \in \mathbb{R},\ m \in \mathbb{N}.
$$

A canonical choice of $K$ satisfying Assumption 1 is a Gaussian kernel, e.g.

$$
K(x)=\frac{1}{(4\pi)^{d/2}}\exp\left(-\frac{|x|^2}{4}\right).
$$

For the potential function $V : \mathbb{R}^d \mapsto \mathbb{R}$, we will assume the following:

Assumption 2. $(A1)$ $V \in C^\infty(\mathbb{R}^d), V \geq 0$ and $V(x) \to +\infty$ if $|x| \to +\infty$.

$(A2)$ There exists a constant $C_V > 0$ and some index $q > 1$ such that

$$
|\nabla V(x)|^q \leq C_V(1+V(x)) \quad \text{for every } x \in \mathbb{R}^d
$$

and that

$$
\sup_{\theta \in [0,1]}
\left|
\nabla^2 V(\theta x+(1-\theta)y)
\right|^q
\leq
C_V(1+V(x)+V(y)).
$$

$(A3)$ For any $\alpha,\beta > 0$, there exists a constant $C_{\alpha,\beta} > 0$ such that if $|y| \leq \alpha |x|+\beta$, then

$$
(1+|x|)\left(|\nabla V(y)|+|\nabla^2 V(y)|\right)
\leq
C_{\alpha,\beta}(1+V(x)).
$$

Remark. By setting $\alpha=1,\beta=0$ and $y=x$ in $(A3)$, we have that

$$
(1+|x|)\left(|\nabla V(x)|+|\nabla^2 V(x)|\right)
\leq
C_1(1+V(x))
$$

for some constant $C_1 > 0$. These assumptions are by no means sharp, but proves to be sufficient for the validity of our theorems. Assumption 2 $(A2)$ implies that there is $C_0$ such that

$$
V(x) \leq C_0(1+|x|^{q^\star})
\qquad
\forall\ x \in \mathbb{R}^d\tag{3}
$$

where $q^\star=\frac{q}{q-1}$. Indeed, this follows from

$$
\frac{d}{dt}(1+V(t\hat{n}))^{\frac{q-1}{q}}
=
\left(\frac{q-1}{q}\right)
\frac{\hat{n}\cdot \nabla V(t\hat{n})}
{(1+V(t\hat{n}))^{1/q}}
\leq
\left(1-\frac{1}{q}\right)C_V^{1/q}.
$$

where $\hat{n}=x/|x|$, and then integrating from $t=0$ to $t=|x|$. It is also easy to check that Assumption 2 is fulfilled by even polynomials up to order $q^\star$.

We use $\mathscr P_V$ and $\mathscr P_p$ denote the set of Borel probability measures $\mu$ on $\mathbb{R}^d$ satisfying

$$
\Vert\mu\Vert_{\mathscr P_V} =
\int_{\mathbb{R}^d}(1+V(x))\ d\mu < \infty
\qquad
\text{or}
\qquad
\Vert\mu\Vert_{\mathscr P_p}
=
\int_{\mathbb{R}^d}|x|^p\ d\mu(x) < \infty,
$$

respectively. Thanks to $(3)$, we have $\mathscr P_p \subset \mathscr P_V$ for any $p \geq q^\star=\frac{q}{q-1}$.

Well-posedness of the PDE and the Particle System

Observe that (2) is a nonlinear transport equation of the form $\partial_t \rho + \nabla \cdot (\rho U[\rho]) = 0$, where $U[\rho]$ is the vector field

$$
U[\rho] (x)=-(\nabla K * \rho)(x)-(K * (\nabla V \rho))(x), \qquad x \in \mathbb R^d. \tag{4}
$$

Given a measure $\rho \in \mathscr P_V$, $U[\rho]$ is well-defined. In fact, due to Assumption 2 $(A2)$, $U[\rho] (x)$ is Lipschitz continuous and bounded over $\mathbb R^d$:

$$
\begin{aligned}
|U[\rho] (x)| &\leq \Vert\nabla K\Vert_{\infty}+\Vert K\Vert_{\infty} C \Vert\rho\Vert_{\mathcal P_V}, \\\
|\nabla U[\rho] (x)| &\leq \Vert D^2 K\Vert_{\infty}+\Vert DK\Vert_{\infty} C \Vert\rho\Vert_{\mathcal P_V}.
\end{aligned} \tag{5}
$$

We say that a measure-valued function $\rho \in C([0,\infty);\mathscr P)$ (where $\mathscr P$ is given the topology of weak convergence) is a weak solution to (2) with initial condition $\rho_0=\nu \in \mathscr P_V$ if

$$
\sup_{t \in [0,T]} \Vert\rho_t\Vert_{\mathscr P_V} < \infty, \qquad \forall\ T > 0 \tag{6}
$$

and

$$
\int_{0}^{\infty} \int_{\mathbb R^d} \left(\partial_t \phi(t,x)+\nabla \phi(t,x) \cdot U[\rho_t]\ (x)\right)\rho_t(dx)\ dt+\int_{\mathbb R^d}\phi(0,x)\nu(dx)=0 \tag{7}
$$

holds for all $\phi \in C_0^\infty([0,\infty)\times \mathbb R^d)$. Recall that $\rho_t=\rho(t,\cdot)$.

Well-posedness of the Mean Field PDE

Theorem 1. Let $V$ satisfy Assumption 2. For any $\nu \in \mathscr P_V$, there is a unique $\rho \in C([0,\infty);\mathscr P_V)$ which is a weak solution to (2) with initial condition $\rho_0=\nu$. Moreover there is $C_1>0$ (depending on $K$ and $V$) such that

$$
\Vert\rho_t\Vert_{\mathscr P_V} \leq e^{C_1 t}\Vert\nu\Vert_{\mathscr P_V}, \qquad t \geq 0. \tag{8}
$$

If $\nu \in \mathscr P_p \cap \mathscr P_V$, then $\rho \in C([0,\infty);\mathscr P_p)$, as well, with $\Vert\rho_t\Vert_{\mathscr P_p} \leq e^{C_2 t}\Vert\nu\Vert_{\mathscr P_p}$.

In order to prove Theorem 1, we introduce the mean field characteristic flow for the PDE (2).

Definition 1. Given a probability measure $\nu$, we say that the map

$$
X(t,x,\nu):[0,\infty)\times \mathbb R^d \to \mathbb R^d
$$

is a mean field characteristic flow associated to the particle system (1) or to the mean field PDE $(2)$ if $X$ is $C^1$ in time and solves the following problem

$$
\begin{aligned}
&\partial_t X(t,x,\nu) = -(\nabla K * \mu_t)(X(t,x,\nu))-(K*(\nabla V\mu_t))(X(t,x,\nu)), \\\
&\mu_t = X(t,\cdot,\nu)_\sharp\nu, \\\
&X(0,x,\nu) = x.
\end{aligned} \tag{9}
$$

We first prove in the theorem below that the mean field characteristic flow (9) is well-defined. To this end, define the set of functions

$$
Y:=\left\{u \in C(\mathbb R^d;\mathbb R^d)\ \middle|\ \sup_{x\in\mathbb R^d}|u(x)-x|<\infty\right\}, \tag{10}
$$

which is a complete metric space with the uniform metric $d_Y(u,v)=\sup_x |u(x)-v(x)|$.

Theorem 2. Assume the conditions of Theorem 1 hold, and $\nu \in \mathscr P_V$. For any $T>0$, there exists a unique solution $X(\cdot,\cdot,\nu)\in C^1([0,T];Y)$ to the problem (9). Moreover, the measure $\mu_t=X(t,\cdot,\nu)_\sharp\nu$ satisfies

$$
\Vert\mu_t\Vert_{\mathscr P_V}\leq e^{C\Vert\nabla K\Vert_\infty t}\Vert\nu\Vert_{\mathscr P_V}, \tag{11}
$$

some constant $C$ that is independent of $\nu$.

Proof. The proof of the theorem consists of two steps.

Step 1 (local well-posedness): Fix $r>0$, and define

$$
Y_r:=\left\{u\in Y\ \middle|\ \sup_{x\in\mathbb R^d}|u(x)-x|\leq r\right\}.
$$

We prove that there exists $T_0>0$ such that the problem (9) has a unique solution $X(t,x)$ in the set

$$
S_r=C([0,T_0];Y_r)
$$

which is a complete metric space, with metric

$$
d_S(u,v)=\sup_{t\in[0,T_0]}d_Y(u(t,\cdot),v(t,\cdot)).
$$

By the definition of mean field characteristic flow,

$$
\begin{aligned}
X(t,x,\nu)-x &= \int_0^t \partial_s X(s,x,\nu)\ ds \\\
&= -\int_0^t \left((\nabla K * \mu_s)(X(s,x,\nu))+(K * \nabla V\mu_s)(X(s,x,\nu))\right)\ ds \\\
&= -\int_0^t \int_{\mathbb R^d} \nabla K(X(s,x,\nu)-y)\ \mu_s(dy)\ ds-\int_0^t \int_{\mathbb R^d} K(X(s,x,\nu)-y)\nabla V(y)\ \mu_s(dy)\ ds \\\
&= -\int_0^t \int_{\mathbb R^d} \nabla K(X(s,x,\nu)-X(s,x’,\nu))\ \nu(dx’)\ ds \\\
&\quad -\int_0^t \int_{\mathbb R^d} K(X(s,x,\nu)-X(s,x’,\nu))\nabla V(X(s,x’,\nu))\ \nu(dx’)\ ds.
\end{aligned} \tag{12}
$$

Let us define the operator $\mathcal F:u(t,\cdot)\mapsto \mathcal F(u)(t,\cdot)$ by

$$
\begin{aligned}
\mathcal F(u)(t,x) &:= x-\int_0^t\int_{\mathbb R^d}\nabla K(u(s,x)-u(s,x’))\nu(dx’)ds \\\
&\quad -\int_0^t\int_{\mathbb R^d}K(u(s,x)-u(s,x’))\nabla V(u(s,x’))\nu(dx’)ds.
\end{aligned} \tag{13}
$$

Our goal is to show that $\mathcal F$ is a contraction in $S_r$, and thus has a unique fixed point.

We first show that $\mathcal F$ maps $S_r$ into $S_r$. Checking that $(t,x) \mapsto \mathcal F(u)(t,x)$ is continuous is straightforward; we need to establish a bound on $|\mathcal F(u)(t,x)-x|$. If $u \in S_r$, then for any $s \in [0,T_0]$ and $x’ \in \mathbb R^d$,

$$
|u(s,x’)| \leq |x’| + |u(s,x’)-x’| \leq |x’| + r. \tag{14}
$$

Then according to Assumption 2 (A3), there exists a positive constant $C_r$ such that

$$
|\nabla V(u(s,x’))| \leq C_r(1+V(x’)), \qquad \forall x’ \in \mathbb R^d. \tag{15}
$$

As a consequence, we have

$$
|\mathcal F(u)(t,x)-x| \leq t \Vert\nabla K\Vert_\infty + t C \Vert K\Vert_\infty \int_{\mathbb R^d} (1+V(x’)) \nu(dx’) \leq \tilde{C} t. \tag{16}
$$

Therefore,

$$
\sup_{t\in[0,T_0]} \sup_{x\in\mathbb R^d} |\mathcal F(u)(t,x)-x| \leq \tilde{C} T_0 \leq r \tag{17}
$$

if $T_0 \leq r/\tilde{C}$. This shows that $\mathcal F$ maps from $S_r$ to $S_r$, if $T_0$ is sufficiently small.

Next, we show that $\mathcal F$ is indeed a contraction on $S_r$. If $u,v \in S_r$, then for any $t \in [0,T_0]$ and $x \in \mathbb R^d$,

$$
\begin{aligned}
|\mathcal F(u)(t,x)-\mathcal F(v)(t,x)| &\leq \int_0^{T_0} \Big| \int_{\mathbb R^d} \nabla K(u(s,x)-u(s,x’)) - \nabla K(v(s,x)-v(s,x’)) \ \nu(dx’) \Big| ds \\\
&\quad + \int_0^{T_0} \Big| \int_{\mathbb R^d} \big(K(u(s,x)-u(s,x’)) - K(v(s,x)-v(s,x’))\big) \nabla V(u(s,x’)) \ \nu(dx’) \Big| ds \\\
&\quad + \int_0^{T_0} \Big| \int_{\mathbb R^d} K(v(s,x)-v(s,x’)) \big(\nabla V(u(s,x’)) - \nabla V(v(s,x’))\big) \ \nu(dx’) \Big| ds.
\end{aligned} \tag{18}
$$

The first term on the right side above can be bounded from above by

$$
T_0 \Vert D^2 K\Vert_\infty 2 d_S(u,v).
$$

Thanks to (14) and Assumption 2, the second term can be bounded from above by

$$
T_0 \Vert\nabla K\Vert_\infty 2 d_S(u,v) \int_{\mathbb R^d} C_r (1+V(x’)) \nu(dx’). \tag{19}
$$

To bound the last term on the right side of (18), using Assumption 2 (A2) one obtains that

$$
|\nabla V(u(s,x’)) - \nabla V(v(s,x’))| \leq \max_{\theta \in [0,1]} |\nabla^2 V(\theta u(s,x’) + (1-\theta)v(s,x’))| d_S(u,v) \leq C_r (1+V(x’)) d_S(u,v), \tag{20}
$$

where in the last inequality we have used the fact that $\theta u + (1-\theta)v \in S_r$ so that $\theta u + (1-\theta)v$ also satisfies the inequality (14), which enables us to apply (A3) of Assumption 2. Plugging (20) into the integral of the last term on the right side of (18), we can bound the last term by

$$
T_0 C_r \Vert K\Vert_\infty \int_{\mathbb R^d} (1+V(x’)) \nu(dx’) d_S(u,v). \tag{21}
$$

Combining the estimates above leads to

$$
d_S(\mathcal F(u),\mathcal F(v)) \leq T_0 \left( 2 \Vert D^2 K\Vert_\infty + 2 C_r (\Vert K\Vert_\infty + \Vert\nabla K\Vert_\infty) \right) \int_{\mathbb R^d} (1+V(x’)) \nu(dx’)\ d_S(u,v). \tag{22}
$$

which implies that $\mathcal F$ is a contraction on $S_r$ when $T_0$ is small enough. By the contraction mapping theorem, $\mathcal F$ has a unique fixed point $X(\cdot,\cdot,\nu) \in S_r$. After defining $\mu_t = X(t,\cdot,\nu)_\sharp\nu$, one sees that $X(t,x,\nu)$ solves (9) in the small time interval $[0,T_0]$.

Step 2 (Extension of local solution): Considering the bounds in the previous step, it is clear that the local solution may be extended beyond time $T_0$ as long as the quantity

$$
\Vert\mu_t\Vert_{\mathscr P_V} = \int_{\mathbb R^d} (1+V(X(t,x,\nu))) \nu(dx) \tag{23}
$$

remains finite. We now establish an a priori bound on this quantity, showing that the local solution may be extended for all $t>0$.

$$
\begin{aligned}
\partial_t \int_{\mathbb R^d} (1+V(X(t,x,\nu))) \nu(dx) &= - \int_{\mathbb R^d} \int_{\mathbb R^d} \nabla V(X(t,x,\nu)) \cdot \nabla K(X(t,x,\nu)-X(t,x’,\nu)) \nu(dx’) \nu(dx) \\\
&\quad - \int_{\mathbb R^d} \int_{\mathbb R^d} K(X(t,x,\nu)-X(t,x’,\nu)) \nabla V(X(t,x,\nu)) \cdot \nabla V(X(t,x’,\nu)) \nu(dx’) \nu(dx) \\\
&\leq \Vert\nabla K\Vert_\infty \int_{\mathbb R^d} |\nabla V(X(t,x,\nu))| \nu(dx) \\\
&\leq C_r \Vert\nabla K\Vert_{\infty} \int_{\mathbb R^d} (1+V(X(t,x,\nu))) \nu(dx).
\end{aligned} \tag{24}
$$

The last inequality follows from Assumption 2 (A3) and the fact that $K$ is positive definite so that the second line above is non-positive. As a consequence,

$$
\int_{\mathbb R^d} (1+V(X(t,x,\nu))) \nu(dx) \leq e^{C_r \Vert\nabla K\Vert_\infty t} \int_{\mathbb R^d} (1+V(x)) \nu(dx) \tag{25}
$$

holds for all $t \in [0,T_0]$. With this bound, one can iterate the argument to extend the local solution defined on $[0,T_0]\times \mathbb R^d$ to all of $[0,\infty)\times \mathbb R^d$, so that

$$
\Vert\mu_t\Vert_{\mathscr P_V} \leq e^{C_r \Vert\nabla K\Vert_\infty t} \Vert\nu\Vert_{\mathscr P_V} \tag{26}
$$

holds for all $t>0$. Similarly, we can claim that there is $C>0$ (depending on $r$) such that $d_Y(X(t,x,\nu),x) \leq Ce^{Ct}$ holds for all $t>0$, which implies that $X(t,\cdot,\nu)\in Y$. In fact, by definition of mean field characteristic flow and (5) and (26), we have

$$
\begin{aligned}
|X(t,x,\nu)-x| &\leq \int_0^t |U[\mu_s] (X(s,x,\nu))|\ ds \\\
&\leq \int_0^t \left(\Vert\nabla K\Vert_\infty+\Vert K\Vert_\infty C\Vert\mu_s\Vert_{\mathscr P_V}\right)\ ds \\\
&\leq \int_0^t \left(\Vert\nabla K\Vert_\infty+\Vert K\Vert_\infty C e^{C_r\Vert\nabla K\Vert_\infty s}\Vert\nu\Vert_{\mathscr P_V}\right)\ ds \\\
&\leq C e^{Ct}.
\end{aligned} \tag{27}
$$

Finally, thanks to the integral formulation $\partial_t X$ is continuous on $[0,\infty)\times \mathbb R^d$. The proof is complete. $\square$

Proof of Theorem 1.

Existence of the weak solution: By the existence of mean field characteristic flow, there exists $X(t,x,\nu)$ such that

$$
\frac{d}{dt}X(t,x,\nu)=U[\mu_t] (X(t,x,\nu)).
$$

For all $\phi\in C_c^\infty([0,\infty)\times\mathbb R^d)$, we have

$$
\frac{d}{dt}\phi(t,X(t,x,\nu)) = \partial_t\phi(t,X(t,x,\nu)) + \nabla\phi(t,X(t,x,\nu)) \cdot \partial_tX(t,x,\nu),
$$

then

$$
\frac{d}{dt}\phi(t,X(t,x,\nu)) = \partial_t\phi(t,X(t,x,\nu)) + \nabla\phi(t,X(t,x,\nu)) \cdot U[\mu_t] (X(t,x,\nu)).
$$

Integrate over $t\in[0,\infty)$:

$$
0-\int_{\mathbb R^d}\phi(0,x)\ \nu(dx)= \int_0^\infty\int_{\mathbb R^d} \partial_t\phi(t,X(t,x,\nu)) + \nabla\phi(t,X(t,x,\nu)) \cdot U[\mu_t] (X(t,x,\nu)) \ dt.
$$

Integrate over $\nu$ and use the fact that $\mu_t=X(t,\cdot,\nu)_\sharp\nu$:

$$
-\int_{\mathbb R^d}\phi(0,x)\ \nu(dx)= \int_0^\infty\int_{\mathbb R^d} \partial_t\phi(t,y) + \nabla\phi(t,y)\cdot U[\mu_t] (y) \ \mu_t(dy)\ dt.
$$

which shows $\mu_t$ is just the weak solution.

Uniqueness of the weak solution: Suppose $q\in C([0,T];\mathscr P)$ is any other weak solution with $q_0=\nu \in \mathscr P_V$, which means

$$
\sup_{t \in [0,T]} \Vert q_t\Vert_{\mathscr P_V} < \infty, \qquad \forall\ T > 0 \tag{28}
$$

and

$$
\int_{0}^{\infty} \int_{\mathbb R^d} \left(\partial_t \phi(t,x)+\nabla \phi(t,x) \cdot U[q_t] (x)\right)q_t(dx)\ dt+\int_{\mathbb R^d}\phi(0,x)\nu(dx)=0 \tag{29}
$$

holds for all $\phi \in C_0^\infty([0,\infty)\times \mathbb R^d)$.

It follows that the vector field $(t,x)\mapsto U[q_t] (x)$ is bounded over $[0,T]\times \mathbb R^d$, continuous in $t$ and Lipschitz continuous in $x$. Then we can define a continuous family of maps $\tilde X(t,\cdot,\nu)$ by

$$
\partial_t \tilde X = U[q_t] (\tilde X), \tag{30}
$$

$$
\tilde X(0,x,\nu) = x. \tag{31}
$$

Then the measure $\tilde q_t = \tilde X(t,\cdot,\nu)_\sharp \nu$ is a weak solution to the linear transport equation

$$
\partial_t \tilde q + \nabla \cdot \big(\tilde q U[q_t] (x)\big) = 0 \tag{32}
$$

with initial condition $\tilde q_0 = \nu = q_0$. Uniqueness of the solution to this linear equation implies that $\tilde q_t = q_t$. That is, $\tilde X(t,\cdot,\nu)_\sharp \nu = q_t$, which means that $\tilde X(t,x,\nu)$ is the mean field characteristic flow for $\nu$. Uniqueness of the mean field characteristic flow implies that $\tilde X = X$, hence $q_t = \rho_t$. This proves that the weak solution is unique.

Norm control: By Theorem 2, we have known that

$$
\Vert\mu_t\Vert_{\mathscr P_V}\leq e^{C\Vert\nabla K\Vert_\infty t}\Vert\nu\Vert_{\mathscr P_V},
$$

Suppose that $\nu \in \mathscr P_p \cap \mathscr P_V$. As shown in the proof of Theorem 2, the map $X(t,x,\nu)$ is an element of the space $Y$ with $d_Y(X(t,x,\nu),x) \le Ce^{Ct}$. Therefore, since

$$
|X(t,x,\nu)|^p \le 2^p|x|^p + 2^p d_Y(X(t,x,\nu),x)^p,
$$

we have

$$
\int_{\mathbb R^d} |y|^p \ \rho_t(dy) = \int_{\mathbb R^d} |X(t,x,\nu)|^p \ \nu(dx) \le C \int_{\mathbb R^d} |x|^p \ \nu(dx) + C e^{Ct} \tag{33}
$$

for all $t>0$. Hence $\rho_t \in \mathscr P_p \cap \mathscr P_V$ for all $t>0$. $\square$

Reference

[LLN19] Jianfeng Lu, Yulong Lu, and James Nolen. Scaling limit of the stein variational gradient
descent: The mean field regime. SIAM Journal on Mathematical Analysis, 51(2):648–671, 2019.

The cover image of this article was taken in Tbilisi, Georgia.