Averaging on Macroscopic Scales with Application to Smoluchowski–Kramers Approximation

Abstract

This paper develops an averaging approach on macroscopic scales to derive Smoluchowski–Kramers approximation for a Langevin equation with state dependent friction in d-dimensional space. In this approach we couple the microscopic dynamics to the macroscopic scales. The weak convergence rate is also presented.

1 Introduction

The Smoluchowski–Kramers (SK) approximation is useful to describe the motion of a particle with small mass which has been studied in lots of works beginning with Smoluchowski [20] and Kramers [17]. The motion of a particle with mass $0< \epsilon \ll 1$ in ${{\mathbb {R}}}^{d}$ ($d\ge 1$) is described by the following Langevin equation

$$\begin{aligned} \epsilon \ddot{x}^{\epsilon }_{t}+\alpha {\dot{x}}_{t}^{\epsilon }=F(x_{t}^{\epsilon })+\sigma (x_{t}^{\epsilon }) {\dot{B}}_{t}\,, \quad x^{\epsilon }(0)=x_{0}\,,\quad {\dot{x}}^{\epsilon }(0)=v_{0}, \end{aligned}$$

where constant friction $\alpha >0$ , $F(x): {\mathbb {R}}^d\rightarrow {\mathbb {R}}^{d}$, $\sigma (x): {\mathbb {R}}^d\rightarrow {\mathbb {R}}^{d\times k}$ and $\{B_t\}$ is k-dimensional standard Wiener process. The classical SK approximation states that for every $T>0$

$$\begin{aligned} \lim _{\epsilon \rightarrow 0} {\mathbb {E}}\sup _{0\le t\le T}\Vert x^{\epsilon }_{t}-x_{t}\Vert _{{{\mathbb {R}}}^{d}} =0, \end{aligned}$$

with

$$\begin{aligned} \alpha {\dot{x}}_{t}=F(x_{t})+\sigma (x_{t}){\dot{B}}_{t}\,,\quad x(0)=x_{0}\,. \end{aligned}$$

For more detail one can refer to [8] . The above limit equation, with just letting $\epsilon =0$ , is not surprising for such constant friction $\alpha$. However a noise-induced drift was observed in experiment [21] for the case of state dependent friction, which implies the limit equation can not be obtained by letting $\epsilon =0$ . Recent work by Hottovy et al. [14] presented a mathematical explanation, but lack of some intuition, by a theory of the convergence of stochastic integral with respect to semimartingale, for such experimental observation.

In this paper we present a new approach which makes the limit equation more intuitively, although in a weak sense. We consider the following Langevin equation with state dependent friction,

$$\begin{aligned}{} & {} \epsilon \ddot{x}^{\epsilon }_t+\alpha ({x}^{\epsilon }_t){\dot{x}}^{\epsilon }_t=F({x}^{\epsilon }_t)+\sigma ({x}^{\epsilon }_t){\dot{B}}_t, \end{aligned}$$

(1.1)

$$\begin{aligned}{} & {} x_0^{\epsilon }={x}_{0}\;\;,{\dot{x}}_0^{\epsilon }={v}_{0},\;\;{x}_{0},{v}_{0} \in {\mathbb {R}}^d, \end{aligned}$$

(1.2)

where $\alpha (x): {\mathbb {R}}^d\rightarrow {\mathbb {R}}^{d\times d}$ is a $d\times d$ invertible matrix-valued function. Our idea is to consider the limit of $\rho _t^\epsilon$, the law of $x^\epsilon _t$ , as $\epsilon \rightarrow 0$ . For this we first write out the equations solved by $\rho _t^\epsilon$ (see (1.6)–(1.7)). However, these equations are not closed, we couple the equations (1.1)–(1.2) to (1.6)–(1.7) . Then we pass the limit $\epsilon \rightarrow 0$ in equations (1.6)–(1.7) via an averaging approach.

Typically, write the equation (1.1) into the following equivalent form

$$\begin{aligned}{} & {} {\dot{x}}_t^\epsilon =v_t^\epsilon , \end{aligned}$$

(1.3)

$$\begin{aligned}{} & {} \epsilon {\dot{v}}_t^\epsilon =-\alpha (x_t^\epsilon )v_t^\epsilon +F(x_t^\epsilon )+\sigma (x_t^\epsilon ){\dot{B}}_t. \end{aligned}$$

(1.4)

First it is known that the law $f_t^\epsilon \in {\mathcal {P}}({\mathbb {R}}^d\times {\mathbb {R}}^d)$, the set consisting of all probability measures on ${{\mathbb {R}}}^{d}\times {{\mathbb {R}}}^{d}$ , of $({x}^{\epsilon }_t,{\dot{x}}^{\epsilon }_t)$ satisfies the Fokker–Planck equation

$$\begin{aligned}{} & {} \partial _t f_t^\epsilon +v\cdot \nabla _x f_t^\epsilon -\frac{1}{\epsilon }\nabla _v\cdot (\alpha (x)vf_t^\epsilon -F(x)f_t^\epsilon )=\frac{1}{\epsilon ^2}\sum _{i=1}^d\sum _{j=1}^d\partial _{v_i}\partial _{v_j}\left( a_{ij}(x)f_t^\epsilon \right) \end{aligned}$$

(1.5)

in the weak sense, where $a(x)=\sigma (x)\sigma ^\top (x)$ and $\sigma ^\top (x)$ is the transpose of $\sigma (x)$, that is for $\varphi \in C_0^\infty ({\mathbb {R}}^d\times {\mathbb {R}}^d)$,

$$\begin{aligned}{} & {} \langle f_t^\epsilon , \varphi \rangle -\langle f_0^\epsilon , \varphi \rangle \\= & {} \int _0^t\int _{{\mathbb {R}}^d\times {\mathbb {R}}^d}\Big [v\cdot \nabla _x\varphi \\{} & {} -\frac{1}{\epsilon }\left( \alpha (x)v-F(x)\right) \cdot \nabla _v\varphi +\frac{1}{\epsilon ^2}\sum _{i=1}^d\sum _{j=1}^da_{ij}(x)\partial _{v_i}\partial _{v_j}\varphi \Big ]f_s^\epsilon (dx,dv)ds. \end{aligned}$$

The law of $x^\epsilon _t$ is

$$\begin{aligned} \rho _t^\epsilon (x)=\int _{{{\mathbb {R}}}^{d}} f_t^\epsilon (x,v)dv, \end{aligned}$$

and define

$$\begin{aligned} Y_t^\epsilon (x)=\int _{{{\mathbb {R}}}^{d}} vf_t^\epsilon (x,v)dv. \end{aligned}$$

Integrating both sides of the equation (1.5) with respect to v, and multiplying both sides of (1.5) by v , then integrating with respect to v, we get

$$\begin{aligned}{} & {} \partial _t\rho _t^\epsilon (x)= -\nabla _x\cdot Y_t^\epsilon (x), \end{aligned}$$

(1.6)

$$\begin{aligned}{} & {} \partial _t Y_t^\epsilon (x)=-\left[ \nabla _x\cdot \left( \int v\otimes v f_t^\epsilon dv\right) \right] ^\top -\frac{1}{\epsilon }\alpha (x)Y_t^\epsilon (x)+\frac{1}{\epsilon }F(x)\rho _t^\epsilon (x). \end{aligned}$$

(1.7)

Notice that

$$\begin{aligned} \int v\otimes v f_t^\epsilon dv=\rho _t^\epsilon (x)\int v\otimes v \frac{f_t^\epsilon }{\rho _t^\epsilon (x)} dv=\rho _t^\epsilon (x){\mathbb {E}}^x(v_t^\epsilon \otimes v_t^\epsilon ), \end{aligned}$$

providing $\rho _{t}^{\epsilon }\ne 0$ . Here ${\mathbb {E}}^x(v_t^\epsilon \otimes v_t^\epsilon )$ is the expectation with fixing $x^\epsilon =x$ in equation (1.4) . Thus, we obtain the following closed system

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t\rho _t^\epsilon (x)= -\nabla _x\cdot Y_t^\epsilon (x),\\ \partial _t Y_t^\epsilon (x)=-\frac{1}{\epsilon }\alpha (x)Y_t^\epsilon (x)+\frac{1}{\epsilon }F(x)\rho _t^\epsilon (x)-\left[ \nabla _x\cdot (\rho _t^\epsilon (x){\mathbb {E}}^x(v_t^\epsilon \otimes v_t^\epsilon ))\right] ^\top ,\\ {\dot{x}}_t^\epsilon =v_t^\epsilon ,\\ \epsilon v_t^\epsilon =-\alpha (x_t^\epsilon )v_t^\epsilon +F(x_t^\epsilon )+\sigma (x_t^\epsilon ){\dot{B}}_t. \end{array} \right. \end{aligned}$$

(1.8)

The above equations (1.8) is in the form of a slow-fast system with slow component $(\rho _t^\epsilon , x^{\epsilon }_{t})$ and fast component $(Y_t^\epsilon , v^{\epsilon }_{t})$. So an averaging method is applicable to pass the limit $\epsilon \rightarrow 0$ [7, 13, 18, 19, e.g.]. In fact, by the idea of averaging approach [7, Chapter 5], fixing $\rho ^{\epsilon }_{t}$ to a probability measure $\rho$ (see equation (4.1) and Remark 4.1), we have $Y^\epsilon _t$, as $\epsilon \rightarrow 0$ , converges weakly to (see Lemma 4.1)

$$\begin{aligned} Y^{*,\rho }(x)=\alpha ^{-1}(x)F(x)\rho (x)-\alpha ^{-1}(x)\left[ \nabla _x\cdot (\rho (x)J(x))\right] ^\top . \end{aligned}$$

Then we formally derive the limit equation (2.4) by replacing $Y^\epsilon _t$ by $Y^{*,\rho }$ in the first equation of (1.8). We call the above an averaging principle on macroscopic scale.

There are a lot of literature about SK approximation in case of variable friction. Freidlin and Hu [9] considered the SK approximation for (1.1) by regularizing the noise. Freidlin, Hu and Wentzell [10], also by regularization method, considered the SK approximation with some degenerating friction. There are also some works on SK approximation of infinite dimensional system with constant damping [3] and state–dependent damping [5, 6, e.g.] and some related problem, large deviation e.g. [4].

The rest of this paper is organized as follows. In Sect. 2, we give some preliminaries, assumptions and the main result. The tightness of $\{\rho ^{\epsilon }_{t}\}$ is shown in Sect. 3, then the averaging procedure is implemented in the last section. It should be clarified that the positive constant C and $C_T$ may be different from line to line in the proofs.

2 Preliminaries and Main Result

Let $(\Omega ,{\mathcal {F}},{\mathbb {P}})$ be a complete probability space, and ${\mathbb {E}}$ denote the expectation with respect to ${\mathbb {P}}$ . Denote by $|\cdot |$ the norm on ${\mathbb {R}}^d$ and $\langle \cdot ,\cdot \rangle$ the inner product in space $L^2({{\mathbb {R}}}^{d})$ .

We make the following assumptions.

$(\mathbf {H_1})$ $\alpha (x): {\mathbb {R}}^d\rightarrow {\mathbb {R}}^{d\times d}$ is continuous differentiable function. The smallest eigenvalue $\lambda _1(x)$ of $\frac{1}{2}(\alpha +\alpha ^{\top })$ is positive uniformly with respect to x, i.e. for some constant $C_{\lambda _\alpha }>0$ ,

$$\begin{aligned} \lambda _1(x)\ge C_{\lambda _\alpha }>0\,. \end{aligned}$$

(2.1)

$(\mathbf {H_2})$ F(x) and $\sigma (x)$ are continuous differentiable and Lipschitz functions with Lipschitz constant $C_F$ and $C_{\sigma }$ respectively, i.e., for $x,y\in {\mathbb {R}}^d$,

$$\begin{aligned}{} & {} |F(x)-F(y)|\le C_F |x-y|, \end{aligned}$$

(2.2)

$$\begin{aligned}{} & {} |\sigma (x)-\sigma (y)|\le C_{\sigma } |x-y|. \end{aligned}$$

(2.3)

$(\mathbf {H_3})$ There is a constant $C>0$ such that $|\partial _{x_{k}}\alpha _{ij}(x)|\le C$, for all $1\le i,j, k \le d$, $x\in {\mathbb {R}}^d$ .

Remark 2.1

Since the global Lipschitz condition implies linear growth, from (2.2), we have $|F(x)|\le C_F(1+|x|)$. Here we keep the same notation $C_F$ for simplicity. In the following, we also use $|F(x)|\le C_F\sqrt{1+|x|^2}$. A similar bound holds for $\sigma (x)$.

Hottovy et al. [14] assumed that the solutions are tight, that they just needed some continuity property of F and $\sigma$, to pass the limit $\epsilon \rightarrow 0$. Here we pose the Lipschitz assumption on F and $\sigma$ to show the tightness of the solutions.

Next we present our main result.

Theorem 2.1

Under the assumptions of $(\mathbf {H_1})$–$(\mathbf {H_3})$, for every $t>0$ , $\rho _t^\epsilon$, the solution to equation (1.8), converges weakly to $\rho _t$ solving the following equation in weak sense

$$\begin{aligned}{} & {} \partial _t \rho _t(x)=-\nabla _x\cdot (\alpha ^{-1}(x)F(x)\rho _t(x)+\alpha ^{-1}(x)(\nabla _x\cdot (\rho _t(x) J(x)))^\top ), \end{aligned}$$

(2.4)

which corresponds to the following stochastic differential equation (SDE)

$$\begin{aligned}{} & {} {\dot{x}}_t=\alpha ^{-1}(x_t)F(x_t)+S(x_t)+\alpha ^{-1}(x_t)\sigma (x_t){\dot{B}}_t\,. \end{aligned}$$

(2.5)

Here

$$\begin{aligned} S_i(x)=\frac{\partial }{\partial x_k}(\alpha ^{-1}(x))_{ij}J_{jk}(x). \end{aligned}$$

and J(x) is the solution of the Lyapunov equation

$$\begin{aligned}{} & {} J(x)\alpha ^\top (x)+\alpha (x)J(x)=\sigma (x)\sigma ^\top (x). \end{aligned}$$

(2.6)

Furthermore, there is a constant $C_T>0$ such that for every $t\in (0,T)$ and $\psi \in C_0^\infty ({\mathbb {R}}^d)$

$$\begin{aligned} |\langle \rho _t^\epsilon (x)-\rho _t(x),\psi \rangle |\le \epsilon C_T\Vert \nabla \psi \Vert _{Lip}, \end{aligned}$$

(2.7)

where $\Vert \cdot \Vert _{Lip}$ denotes the Lipschitz norm defined by

$$\begin{aligned} \Vert f\Vert _{Lip}=\Vert f\Vert _{\infty }+\sup _{x\ne y}\frac{|f(x)-f(y)|}{|x-y|}. \end{aligned}$$

Remark 2.2

The above convergence rate in (2.7), from the estimate (4.11) in the proof, is sharp. So an interesting problem is the higher order correction to $\rho _{t}^{\epsilon }$, that is what is the limit of

$$\begin{aligned} \frac{ 1}{\epsilon } (\rho ^{\epsilon }_{t}-\rho _{t}) \end{aligned}$$

as $\epsilon \rightarrow 0$ . To determine the limit we have to give a more detail estimation than that in Lemma 4.1. This will be considered in our future work.

Remark 2.3

The unique solution to equation (2.6) has the following explicit expression [2, Page 179]

$$\begin{aligned} J(x)=\int _0^\infty e^{-\alpha (x)t}\sigma (x)\sigma ^\top (x)e^{-\alpha ^\top (x)t}dt. \end{aligned}$$

In fact the matrix J(x) is the limit, as $\epsilon \rightarrow 0$ , of the covariance of $\sqrt{\epsilon }v^\epsilon _t$ with freezing $x^\epsilon =x$ (Lemma 3.3) .

Remark 2.4

To see the relationship between (2.4) and (2.5), by Einstein summation notation, write (2.4) as

$$\begin{aligned} \partial _t \rho _t(x)= & {} -\nabla _x\cdot (\alpha ^{-1}(x)F(x)\rho _t(x))+\partial _{x_i}(\alpha _{ik}^{-1}\partial _{x_j}(\rho _t(x)J_{jk}))\\= & {} -\nabla _x\cdot (\alpha ^{-1}(x)F(x)\rho _t(x))+\partial _{x_i}(\alpha _{ik}^{-1}(\partial _{x_j}\rho _t(x)J_{jk}+\rho _t(x)\partial _{x_j}J_{jk}))\\= & {} -\nabla _x\cdot (\alpha ^{-1}(x)F(x)\rho _t(x))+\partial _{x_i}(\partial _{x_j}\rho _t(x)\alpha _{ik}^{-1}J_{jk}+\rho _t(x)\alpha _{ik}^{-1}\partial _{x_j}J_{jk})\\= & {} -\nabla _x\cdot (\alpha ^{-1}(x)F(x)\rho _t(x))+\partial _{x_i}(\partial _{x_j}(\rho _t(x)\alpha _{ik}^{-1}J_{jk})-\rho _t(x)J_{jk}\partial _{x_j}\alpha _{ik}^{-1}). \end{aligned}$$

From (2.6), we have

$$\begin{aligned} \alpha ^{-1}(x)J(x)+J(x)[\alpha ^{-1}(x)]^\top =\alpha ^{-1}(x)\sigma (x)\sigma ^{\top }(x)[\alpha ^{-1}(x)]^\top . \end{aligned}$$

(2.8)

Denote the right hand side of (2.8) by A, and extract the (i, j) element of both sides,

$$\begin{aligned} \alpha _{ik}^{-1}J_{kj}+J_{ik}\alpha _{jk}^{-1}= A_{ij}. \end{aligned}$$

By the symmetry of J(x), we get

$$\begin{aligned} \partial _t \rho _t(x)= & {} -\nabla _x\cdot (\alpha ^{-1}(x)F(x)\rho _t(x))+\partial _{x_i}(\frac{1}{2}\partial _{x_j}(A_{ij}\rho _t(x))-\rho _t(x)J_{jk}\partial _{x_j}\alpha _{ik}^{-1})\\= & {} -\nabla _x\cdot (\alpha ^{-1}(x)F(x)\rho _t(x)+S(x_t)\rho _t(x))+\frac{1}{2}\partial _{x_i}\partial _{x_j}(A_{ij}\rho _t(x)), \end{aligned}$$

which corresponds to SDE (2.5).

To prove Theorem 2.1, we first show the tightness of $\{x_t^\epsilon \}$ in Sect. 3, then for all sequences of $\{\rho ^\epsilon _{\cdot }\}$ there exits a subsequence $\{\rho ^{\epsilon _k}_{\cdot }\}$ converges weakly to $\{\rho _{\cdot }\}$ as $\epsilon _k\rightarrow 0$. Then, for the convergent subsequence $\rho ^{\epsilon _k}_{\cdot }$ , we apply the averaging approach (Sect. 4) to the slow-fast system (1.8) to derive the limit equation.

The following lemma is used to give an explicit representation for the covariance of $\sqrt{\epsilon } v^\epsilon _t$ with freezing $x^\epsilon =x$ in Lemma 3.3 .

Lemma 2.1

[1, Theorem 2] Let $I\subseteq {\mathbb {R}}$ be an open interval with $t_0\in I$, $A\in {\mathbb {C}}^{n\times n}$, $B\in {\mathbb {C}}^{m\times m}$, $C\in {\mathcal {C}}(I, {\mathbb {C}}^{n\times m})$ and $D\in {\mathbb {C}}^{m\times n}$. The Lyapunov differential equation

$$\begin{aligned} {\dot{X}}(t)=AX(t)+X(t)B+C(t),~X(t_0)=D, \end{aligned}$$

has the unique solution

$$\begin{aligned} X(t)=e^{A(t-t_0)}D e^{B(t-t_0)}+\int _{t_0}^t e^{A(t-s)}C(s)e^{B(t-s)}\textrm{d}s. \end{aligned}$$

The following lemma is important in the averaging approach in Sect. 4 .

Lemma 2.2

[15, p. 120] Let ${\textbf{A}}=(a_{ij})_{1\le i,j\le d}$ and ${\textbf{u}}=(u_i)_{1\le i\le d}$ be $d\times d$ matrix and $d\times 1$ vector respectively. Each element of ${\textbf{A}}$ and ${\textbf{u}}$ is a function of ${\textbf{x}}=(x_1,x_2,\ldots ,x_d)$, then

$$\begin{aligned} \nabla \cdot ({\textbf{A}}{\textbf{u}})=(\nabla \cdot {\textbf{A}}){\textbf{u}}+\textrm{tr} ({\textbf{A}}~\textrm{grad}~{\textbf{u}}), \end{aligned}$$

where $\textrm{grad}~{\textbf{u}}=\left( \frac{\partial u_i}{\partial x_j}\right) _{1\le i,j\le d}$ and $(\nabla \cdot {\textbf{A}})_j=\nabla \cdot {\textbf{A}}_j=\sum _{i=1}^d\frac{\partial a_{ij}}{\partial x_i}$ where ${\textbf{A}}_j$ is the j-th column of ${\textbf{A}}$.

3 Tightness of $\{x^\epsilon \}_\epsilon$

To prove the tightness of $\{x^\epsilon \}_\epsilon$ in space $C(0, T; {{\mathbb {R}}}^d)$, we need to show the boundedness in $C(0, T; {{\mathbb {R}}}^d)$ and the Hölder continuity of $\{x^\epsilon \}_\epsilon$ .

Lemma 3.1

Under assumptions $(\mathbf {H_1})$ and $(\mathbf {H_2})$, for all $T > 0$,

$$\begin{aligned} {\mathbb {E}}\sup _{0\le t\le T}|x_t^\epsilon |^2\le C_T, \end{aligned}$$

(3.1)

and for $0\le t_1$ , $t_2\le T$,

$$\begin{aligned} {\mathbb {E}}|x^\epsilon _{t_2}-x^\epsilon _{t_1}|^2\le C|t_2-t_1|. \end{aligned}$$

(3.2)

Proof

We intend to write an expression of $x^\epsilon _t$ in a mild formulation. Due to the state dependent friction, this is of some difficulty. For this we first consider the linear part of the $v^\epsilon$-equation (1.4), that is the following equation

$$\begin{aligned}{} & {} \dot{y_t}=-\frac{1}{\epsilon }\alpha (x_t^\epsilon )y_t,\\{} & {} y_0=I_d. \end{aligned}$$

Then

$$\begin{aligned} \frac{d}{dt}(y_t^{-1}{\dot{x}}_t^\epsilon )= & {} \frac{d}{dt}y_t^{-1}{\dot{x}}_t^\epsilon +y_t^{-1}\ddot{x}_t^\epsilon \\= & {} -y_t^{-1}\frac{dy_t}{dt}y_t^{-1}{\dot{x}}_t^\epsilon +y_t^{-1}\ddot{x}_t^\epsilon \\= & {} \frac{1}{\epsilon }y_t^{-1}\alpha (x_t^\epsilon ){\dot{x}}_t^\epsilon +y_t^{-1}\left( -\frac{\alpha (x_t^\epsilon )}{\epsilon }{\dot{x}}_t^\epsilon +\frac{1}{\epsilon }F(x_t^\epsilon )+\frac{1}{\epsilon }\sigma (x_t^\epsilon )\dot{B_t}\right) \\= & {} \frac{1}{\epsilon }y_t^{-1}F(x_t^\epsilon )+\frac{1}{\epsilon }y_t^{-1}\sigma (x_t^\epsilon )\dot{B_t}. \end{aligned}$$

Integrating from 0 to $\tau$ yields

$$\begin{aligned} y_\tau ^{-1}{\dot{x}}_\tau ^\epsilon =v_0+\frac{1}{\epsilon }\int _0^\tau y_s^{-1}F(x_s^\epsilon )ds+\frac{1}{\epsilon }\int _0^\tau y_s^{-1}\sigma (x_s^\epsilon )d{B_s}, \end{aligned}$$

and

$$\begin{aligned} {\dot{x}}_\tau ^\epsilon =y_\tau v_0+\frac{1}{\epsilon }\int _0^\tau y_\tau y_s^{-1}F(x_s^\epsilon )ds+\frac{1}{\epsilon }\int _0^\tau y_\tau y_s^{-1}\sigma (x_s^\epsilon )d{B_s}. \end{aligned}$$

(3.3)

Integrating from 0 to t for equation (3.3) yields

$$\begin{aligned}{} & {} x_t^\epsilon =x_0+\int _0^t y_\tau v_0 d\tau \\{} & {} \quad \quad \quad +\frac{1}{\epsilon }\int _0^t\int _0^\tau y_\tau y_s^{-1}F(x_s^\epsilon )ds d\tau \\{} & {} \quad \quad \quad +\frac{1}{\epsilon }\int _0^t\int _0^\tau y_\tau y_s^{-1}\sigma (x_s^\epsilon )d{B_s}d\tau \triangleq x_0+\sum _{i=1}^3I_i(t). \end{aligned}$$

Now define

$$\begin{aligned} z_\tau =y_\tau y_s^{-1}F(x_s^\epsilon ), ~\tau \ge s, \end{aligned}$$

(3.4)

then

$$\begin{aligned} z_s =F(x_s^\epsilon ), \end{aligned}$$

and

$$\begin{aligned} \frac{dz_\tau }{d\tau }= & {} \frac{dy_\tau }{d\tau }y_s^{-1}F(x_s^\epsilon )\\= & {} -\frac{1}{\epsilon }\alpha (x_\tau ^\epsilon )y_\tau y_s^{-1}F(x_s^\epsilon )\\= & {} -\frac{1}{\epsilon }\alpha (x_\tau ^\epsilon )z_\tau . \end{aligned}$$

Thus we have [12, Lemma 4.2 of Chpter IV],

$$\begin{aligned} |z_\tau |\le |F(x_s^\epsilon )|e^{-\int _s^\tau \frac{1}{\epsilon }\lambda _1(x)dt}\le |F(x_s^\epsilon )|e^{-\frac{1}{\epsilon }C_{\lambda _{\alpha }}(\tau -s)}. \end{aligned}$$

Similarly,

$$\begin{aligned} |y_\tau y_s^{-1}\sigma (x_s^\epsilon )|\le |\sigma (x_s^\epsilon )|e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(\tau -s)}, \end{aligned}$$

and

$$\begin{aligned} |y_\tau v_0|\le |v_0|e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }\tau }. \end{aligned}$$

Then we derive

$$\begin{aligned} {\mathbb {E}}\sup _{0\le t\le T}|I_1(t)|^2= & {} {\mathbb {E}}\sup _{0\le t\le T}\left| \int _0^t y_\tau v_0 d\tau \right| ^2\nonumber \\\le & {} |v_0|^2{\mathbb {E}}\sup _{0\le t\le T}\left| \int _0^t e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }\tau } d\tau \right| ^2\nonumber \\\le & {} |v_0|^2\frac{\epsilon ^2}{C_{\lambda _\alpha }^2}. \end{aligned}$$

(3.5)

By Hölder inequality and Fubini theorem

$$\begin{aligned} {\mathbb {E}}\sup _{0\le t\le T}|I_2(t)|^2= & {} {\mathbb {E}}\sup _{0\le t\le T}\left| \frac{1}{\epsilon }\int _0^t\int _0^\tau z_\tau ds d\tau \right| ^2 \nonumber \\\le & {} \frac{1}{\epsilon ^2}{\mathbb {E}}\sup _{0\le t\le T}\left| \int _0^t\int _0^\tau \Vert F(x_t^\epsilon )\Vert e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(\tau -s)}ds d\tau \right| ^2\nonumber \\= & {} \frac{1}{\epsilon ^2}{\mathbb {E}}\sup _{0\le t\le T}\left| \int _0^t\int _s^t \Vert F(x_t^\epsilon )\Vert e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(\tau -s)} d\tau ds\right| ^2\nonumber \\\le & {} {\mathbb {E}}\sup _{0\le t\le T}\left| \int _0^t\Vert F(x_t^\epsilon )\Vert \frac{1}{C_{\lambda _\alpha }}\right| ^2ds\nonumber \\\le & {} \frac{C_T}{C_{\lambda _\alpha }^2}\left( 1+\int _0^T{\mathbb {E}}\sup _{0\le r\le s}|x_r^\epsilon |^2ds\right) . \end{aligned}$$

(3.6)

By Doob’s maximal inequality, Fubini theorem and Hölder inequality, we obtain

$$\begin{aligned} {\mathbb {E}}\sup _{0\le t\le T}|I_3(t)|^2= & {} {\mathbb {E}}\sup _{0\le t\le T}\left| \frac{1}{\epsilon }\int _0^t\int _0^\tau y_\tau y_s^{-1}\sigma (x_s^\epsilon )d{B_s}d\tau \right| ^2\nonumber \\= & {} {\mathbb {E}}\sup _{0\le t\le T}\left| \frac{1}{\epsilon }\int _0^t\int _s^ty_\tau y_s^{-1}\sigma (x_s^\epsilon )d\tau d{B_s}\right| ^2\nonumber \\\le & {} \frac{4}{\epsilon ^2}{\mathbb {E}}\left| \int _0^T\int _s^ty_\tau y_s^{-1}\sigma (x_s^\epsilon )d\tau d{B_s}\right| ^2\nonumber \\\le & {} \frac{4}{\epsilon ^2}{\mathbb {E}}\int _0^T\left( \int _s^t|\sigma (x_s^\epsilon )|e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(\tau -s)d\tau }\right) ^2 ds\nonumber \\\le & {} \frac{C_T}{C_{\lambda _\alpha }^2}\left( 1+\int _0^T{\mathbb {E}}\sup _{0\le r\le s}|x_r^\epsilon |^2ds\right) . \end{aligned}$$

(3.7)

Combining (3.5), (3.6) with (3.7) yields

$$\begin{aligned} {\mathbb {E}}\sup _{0\le t\le T}|x_t^\epsilon |^2\le C\left( |x_0|^2+|v_0|^2\frac{\epsilon ^2}{C_{\lambda _\alpha }^2}+\frac{C_T}{C_{\lambda _\alpha }^2}\left( 1+\int _0^T{\mathbb {E}}\sup _{0\le r\le s}|x_r^\epsilon |^2ds\right) \right) . \end{aligned}$$

Then Gronwall inequality yields

$$\begin{aligned} {\mathbb {E}}\sup _{0\le t\le T}|x_t^\epsilon |^2\le C\left( |x_0|^2+|v_0|^2\frac{\epsilon ^2}{C_{\lambda _\alpha }^2}+\frac{C_T}{C_{\lambda _\alpha }^2}\right) e^{\frac{C_T}{C_{\lambda _\alpha }^2}}\le C_T. \end{aligned}$$

Next, let $0\le t_1 < t_2\le T$,

$$\begin{aligned} x_{t_2}^\epsilon -x_{t_1}^\epsilon= & {} \int _{t_1}^{t_2} y_\tau v_0 d\tau +\frac{1}{\epsilon }\int _{t_1}^{t_2}\int _0^\tau y_\tau y_s^{-1}F(x_s^\epsilon )ds d\tau \\{} & {} {}+\frac{1}{\epsilon }\int _{t_1}^{t_2}\int _0^\tau y_\tau y_s^{-1}\sigma (x_s^\epsilon )d{B_s}d\tau \\\triangleq & {} \sum _{i=1}^3J_i. \end{aligned}$$

First, Hölder inequality yields

$$\begin{aligned} {\mathbb {E}}|J_1|^2={\mathbb {E}}\left| \int _{t_1}^{t_2} y_\tau v_0 d\tau \right| ^2\le |v_0|^2{\mathbb {E}}\left| \int _{t_1}^{t_2} e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }\tau } d\tau \right| ^2 \le C|t_2-t_1|^2. \end{aligned}$$

(3.8)

Further by Hölder inequality, Fubini theorem and integral median theorem we have

$$\begin{aligned} {\mathbb {E}}|J_2|^2= & {} \frac{1}{\epsilon ^2}{\mathbb {E}}\left| \int _{t_1}^{t_2}\int _0^\tau y_\tau y_s^{-1}F(x_s^\epsilon )dsd\tau \right| ^2\nonumber \\= & {} \frac{1}{\epsilon ^2}{\mathbb {E}}\left| \int _{0}^{t_1}\int _{t_1}^{t_2} y_\tau y_s^{-1}F(x_s^\epsilon )d\tau ds+ \int _{t_1}^{t_2}\int _{s}^{t_2} y_\tau y_s^{-1}F(x_s^\epsilon )d\tau ds \right| ^2\nonumber \\\le & {} \frac{2}{\epsilon ^2}{\mathbb {E}}\left| \int _{0}^{t_1}\int _{t_1}^{t_2} y_\tau y_s^{-1}F(x_s^\epsilon )d\tau ds\right| ^2+\frac{2}{\epsilon ^2}\left| \int _{t_1}^{t_2}\int _{s}^{t_2} y_\tau y_s^{-1}F(x_s^\epsilon )d\tau ds \right| ^2\nonumber \\\le & {} \frac{2}{\epsilon ^2}{\mathbb {E}}\left( \int _{0}^{t_1}\int _{t_1}^{t_2} |F(x_s^\epsilon )| e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(\tau -s)}d\tau ds\right) ^2\nonumber \\{} & {} +\frac{2}{\epsilon ^2}{\mathbb {E}}\left( \int _{t_1}^{t_2}\int _{s}^{t_2} |F(x_s^\epsilon )| e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(\tau -s)}d\tau ds \right) ^2\nonumber \\= & {} \frac{2}{\epsilon ^2}{\mathbb {E}}\left( \int _{0}^{t_1} |F(x_s^\epsilon )| e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(\xi -s)}|t_2-t_1|ds\right) ^2\nonumber \\{} & {} +\frac{2}{C_{\lambda _\alpha }^2}{\mathbb {E}}\left( \int _{t_1}^{t_2} |F(x_s^\epsilon )|(1- e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(t_2-s)}) ds \right) ^2\nonumber \\\le & {} \frac{2}{\epsilon ^2}|t_2-t_1|^2{\mathbb {E}}\int _0^{t_1}|F(x_s^\epsilon )|^2ds{\mathbb {E}} \int _0^{t_1}e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(\xi -s)}ds\nonumber \\{} & {} +\frac{2}{C_{\lambda _\alpha }^2}{\mathbb {E}}\int _{t_1}^{t_2}|F(x_s^\epsilon )|^2ds|t_2-t_1|\nonumber \\\le & {} \frac{C_F^2}{\epsilon C_{\lambda _\alpha }}|t_2-t_1|^2\int _0^{t_1}(1+{\mathbb {E}}\sup _{0< u\le s}|x_u^\epsilon |^2)ds(e^{-\frac{2}{\epsilon }C_{\lambda _\alpha }(\xi -t_1)}-e^{-\frac{2}{\epsilon }C_{\lambda _\alpha }\xi })\nonumber \\{} & {} \quad \quad +\frac{2C}{C_{\lambda _\alpha }^2}\int _{t_1}^{t_2}(1+{\mathbb {E}}\sup _{0< u\le s}|x_u^\epsilon |^2)ds|t_2-t_1|\nonumber \\\le & {} \frac{C}{\epsilon }|t_2-t_1|^2(e^{-\frac{2}{\epsilon }C_{\lambda _\alpha }(\xi -t_1)}+e^{-\frac{2}{\epsilon }C_{\lambda _\alpha }\xi }) +C |t_2-t_1|^2\nonumber \\\le & {} C |t_2-t_1|^2. \end{aligned}$$

(3.9)

In the last step of (3.9), we have used the fact that $f(x)=xe^{-ax},a>0,x\in (0,+\infty )$ is bounded. Similarly, we have

$$\begin{aligned}{} & {} {\mathbb {E}}|J_3|^2=\frac{1}{\epsilon ^2}{\mathbb {E}}\left| \int _{t_1}^{t_2}\int _0^\tau y_\tau y_s^{-1}\sigma (x_s^\epsilon )d{B_s}d\tau \right| ^2\nonumber \\{} & {} =\frac{1}{\epsilon ^2}{\mathbb {E}}\left| \int _{0}^{t_1}\int _{t_1}^{t_2} y_\tau y_s^{-1}\sigma (x_s^\epsilon )d\tau d{B_s}+\int _{t_1}^{t_2}\int _{s}^{t_2} y_\tau y_s^{-1}\sigma (x_s^\epsilon )d\tau d{B_s}\right| ^2\nonumber \\{} & {} \le \frac{2}{\epsilon ^2}{\mathbb {E}}\left| \int _{0}^{t_1}\int _{t_1}^{t_2} y_\tau y_s^{-1}\sigma (x_s^\epsilon )d\tau d{B_s}\right| ^2+\frac{2}{\epsilon ^2}{\mathbb {E}}\left| \int _{t_1}^{t_2}\int _{s}^{t_2} y_\tau y_s^{-1}\sigma (x_s^\epsilon )d\tau d{B_s}\right| ^2\nonumber \\{} & {} \le \frac{2}{\epsilon ^2}\int _0^{t_1}{\mathbb {E}}\left( \int _{t_1}^{t_2}|\sigma (x_s^\epsilon )|e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(\tau -s)}d\tau \right) ^2ds +\frac{2}{\epsilon ^2}\int _{t_1}^{t_2}{\mathbb {E}}\left( \int _{s}^{t_2}|\sigma (x_s^\epsilon )|e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(\tau -s)}d\tau \right) ^2ds\nonumber \\{} & {} \le \frac{2}{\epsilon ^2}\int _0^{t_1}{\mathbb {E}}|\sigma (x_s^\epsilon )|^2e^{-\frac{2}{\epsilon }C_{\lambda _\alpha }(\xi -s)}|t_2-t_1|^2ds +\frac{2}{C_{\lambda _\alpha ^2}}\int _{t_1}^{t_2}{\mathbb {E}}|\sigma (x_s^\epsilon )|^2(1-e^{-\frac{1}{\epsilon }C_{\lambda _\alpha }(t_2-s)})^2ds\nonumber \\{} & {} \le \frac{2C_\sigma ^2}{\epsilon ^2}|t_2-t_1|^2\int _0^{t_1}(1+{\mathbb {E}}\sup _{0< u\le s}|x_u^\epsilon |^2)(e^{-\frac{2}{\epsilon }C_{\lambda _\alpha }(\xi -s)})ds +\frac{2C_\sigma ^2}{C_{\lambda _\alpha }^2}\int _{t_1}^{t_2}(1+{\mathbb {E}}\sup _{0< u\le s}|x_u^\epsilon |^2)ds\nonumber \\{} & {} \le \frac{C}{\epsilon }|t_2-t_1|^2(e^{-\frac{2}{\epsilon }C_{\lambda _\alpha }(\xi -t_1)}-e^{-\frac{2}{\epsilon }C_{\lambda _\alpha }\xi })+C|t_2-t_1|\nonumber \\{} & {} \le C|t_2-t_1|. \end{aligned}$$

(3.10)

Now (3.8)–(3.10) yields (3.2). The proof is complete. $\square$

Now by Lemma  3.1 and the Garcia–Rademich–Rumsey theorem [11], we have the tightness of solutions.

Lemma 3.2

The process $\{x^\epsilon \}_\epsilon$ is tight in space $C(0,T,{\mathbb {R}}^d)$ for all $T>0$ .

Remark 3.1

The above result is assumed by Hottovy et al. [14, Assumption 3].

Next we show the limit of the covariance of $\sqrt{\epsilon }v_t^\epsilon$ as $\epsilon \rightarrow 0$ with frozen $x^\epsilon =x$ . For this we consider the following linear equation for $x\in {{\mathbb {R}}}^d$,

$$\begin{aligned} \epsilon {\dot{v}}_t^{\epsilon , x}=-\alpha (x)v_t^{\epsilon ,x}+F(x)+\sigma (x){\dot{B}}_t. \end{aligned}$$

Then we have

Lemma 3.3

Assume $(\mathbf {H_1})$ and $(\mathbf {H_2})$ hold, for $x\in {{\mathbb {R}}}^d$

$$\begin{aligned} \epsilon {\mathbb {E}} v^{\epsilon ,x}_t\otimes v^{\epsilon ,x}_t=J(x)+\epsilon C(x,t), \end{aligned}$$

where $|C(x,t)|\le C(1+|x|^2)$ and J(x) solves (2.6).

Proof

First by the Itô’s formula,

$$\begin{aligned} \frac{d}{dt}{\mathbb {E}}(\epsilon v_t^{\epsilon ,x}\otimes v_t^{\epsilon ,x})= & {} -\frac{\alpha (x)}{\epsilon }{\mathbb {E}}(\epsilon v_t^{\epsilon ,x}\otimes v_t^{\epsilon ,x})\nonumber \\{} & {} {}+F(x)\otimes {\mathbb {E}}v_t^{\epsilon ,x}-\frac{1}{\epsilon }{\mathbb {E}}(\epsilon v_t^{\epsilon ,x}\otimes v_t^{\epsilon ,x})\alpha ^\top (x)\nonumber \\{} & {} {}+{\mathbb {E}}v_t^{\epsilon ,x}\otimes F(x)+\frac{1}{\epsilon }\sigma (x)\sigma ^\top (x), \end{aligned}$$

(3.11)

and

$$\begin{aligned} \frac{d}{dt}{\mathbb {E}} v_t^{\epsilon ,x}=-\frac{1}{\epsilon }\alpha (x){\mathbb {E}}v_t^{\epsilon ,x}+\frac{1}{\epsilon }F(x). \end{aligned}$$

(3.12)

Applying Lemme 2.1 to equation (3.11) and the Duhamel’s principle to equation (3.12) respectively, yields

$$\begin{aligned}{} & {} {\mathbb {E}}(\epsilon v_t^{\epsilon ,x}\otimes v_t^{\epsilon ,x})\\ {}= & {} e^{-\frac{\alpha (x)}{\epsilon }t}{\mathbb {E}}(\epsilon v_0\otimes v_0)e^{-\frac{\alpha ^\top (x)}{\epsilon }t}\\{} & {} {}+\int _0^te^{-\frac{\alpha (x)}{\epsilon }(t-s)}\left( F(x)\otimes {\mathbb {E}}v_t^{\epsilon ,x}+{\mathbb {E}}v_t^{\epsilon ,x}\otimes F(x)+\frac{1}{\epsilon }\sigma (x)\sigma ^\top (x)\right) e^{-\frac{\alpha ^\top (x)}{\epsilon }(t-s)}ds, \end{aligned}$$

and

$$\begin{aligned} {\mathbb {E}}v_t^{\epsilon ,x}= & {} e^{-\frac{\alpha (x)}{\epsilon }t}v_0+\frac{1}{\epsilon }\int _0^te^{-\frac{\alpha (x)}{\epsilon }(t-s)}F(x)ds\\= & {} e^{-\frac{\alpha (x)}{\epsilon }t}v_0+\alpha ^{-1}(x)(I-e^{-\frac{\alpha (x)}{\epsilon }t})F(x). \end{aligned}$$

Thus

$$\begin{aligned} |{\mathbb {E}}v_t^{\epsilon ,x}|\le & {} |e^{-\frac{\alpha (x)}{\epsilon }t}v_0|+|\alpha ^{-1}(x)(I-e^{-\frac{\alpha (x)}{\epsilon }t})F(x)|\\\le & {} C+\frac{C_F}{C_{\lambda _{\alpha }}}(1+|x|)\\\le & {} C(1+|x|), \end{aligned}$$

and then $|F(x)\otimes {\mathbb {E}}v_t^{\epsilon ,x}| \le C(1+|x|^2)$. Now let $\tau =\frac{t-s}{\epsilon }$, we have

$$\begin{aligned}{} & {} \int _0^te^{-\frac{\alpha (x)}{\epsilon }(t-s)}\left( F(x)\otimes {\mathbb {E}}v_t^{\epsilon ,x}+{\mathbb {E}}v_t^{\epsilon ,x}\otimes F(x)+\frac{1}{\epsilon }\sigma (x)\sigma ^\top (x)\right) e^{-\frac{\alpha ^\top (x)}{\epsilon }(t-s)}ds \\ {}= & {} \int _0^{\frac{t}{\epsilon }}e^{-\alpha (x)\tau }(\epsilon F(x)\otimes {\mathbb {E}}v_t^{\epsilon ,x}+\epsilon {\mathbb {E}}v_t^{\epsilon ,x}\otimes F(x)+\sigma (x)\sigma ^\top (x))e^{-\alpha ^\top (x)\tau }d\tau \\ {}= & {} J(x)+\epsilon \int _0^{\frac{t}{\epsilon }}e^{-\alpha (x)\tau }(F(x)\otimes {\mathbb {E}}v_t^{\epsilon ,x}+ {\mathbb {E}}v_t^{\epsilon ,x}\otimes F(x))e^{-\alpha ^\top (x)\tau }d\tau \\ {}{} & {} {}-\int _{\frac{t}{\epsilon }}^\infty e^{-\alpha (x)\tau }\sigma (x)\sigma ^\top (x)e^{-\alpha ^\top (x)\tau }d\tau \\= & {} J(x)+\epsilon C(x,t)\,. \end{aligned}$$

The proof is complete. $\square$

Remark 3.2

Here we point out that an important step is to estimate $\epsilon {\mathbb {E}}|v_t^\epsilon |^2$ in the work of Hottovy et al. [14] . However, in our approach we need the estimate of $\epsilon {\mathbb {E}}v^{\epsilon ,x}_t\otimes v^{\epsilon , x}_t$ instead with fixed x .

4 Averaging Approach

In this section we just consider a convergent subsequence $\rho ^{\epsilon _k}_{\cdot }$ and for simplicity we still write it as $\rho ^\epsilon _{\cdot }$ . Let $\rho _{\cdot }$ be the limit of $\rho ^\epsilon _{\cdot }$ . Next we determine the equation for $\rho _{\cdot }$ by an averaging approach.

Averaging is effective to study the approximation for a slow-fast system [7, 13, 16]. Here we apply the Khasminskii’s scheme [16] to (1.8). For small $\epsilon$, $\rho ^\epsilon _t$ evolves slow, so we can consider the fast part $Y^\epsilon$ by freezing the slow part $\rho ^\epsilon _t$ to be some $\rho \in {\mathcal {P}}({{\mathbb {R}}}^d)$ and fix $t=\tau$ in ${\mathbb {E}}^x(v_t^\epsilon \otimes v_t^\epsilon )$ . For this we introduce ${\tilde{Y}}_t^{\epsilon ,\rho ,\tau }(x)$ the solution of the following equation

$$\begin{aligned}{} & {} \partial _t {\tilde{Y}}_t^{\epsilon ,\rho ,\tau }(x)=-\frac{\alpha (x)}{\epsilon }{\tilde{Y}}_t^{\epsilon ,\rho ,\tau }(x)+\frac{1}{\epsilon }F(x)\rho (x)-[\nabla _x\cdot (\rho (x){\mathbb {E}}^x(v_\tau ^\epsilon \otimes v_\tau ^\epsilon ))]^\top , \end{aligned}$$

(4.1)

with ${\tilde{Y}}_0^{\epsilon ,\rho ,\tau }(x)=Y_0$. The following lemma shows that the fast part converges uniformly in $\tau$ to some vector with frozen slow part as $\epsilon \rightarrow 0$ .

Lemma 4.1

For every fixed $t_{*}>0$ , under the assumptions $(\mathbf {H_1})$–$(\mathbf {H_3})$, fix $\rho _t^{\epsilon }=\rho \in {\mathcal {P}}({{\mathbb {R}}}^{d})$ with $\int |x|^2\rho (x) dx$ and $\Vert |Y_0|\Vert _{L^1}$ bounded, there is a constant $C_T>0$ such that for $\varphi \in C_0^\infty ({\mathbb {R}}^d, {{\mathbb {R}}}^{d})$, and $t\ge t_{*}$,

$$\begin{aligned} |\langle {\tilde{Y}}_t^{\epsilon ,\rho ,\tau }(x)-Y^{*,\rho }(x),\varphi \rangle |\le C_T \epsilon \Vert \varphi \Vert _{Lip}, \end{aligned}$$

where

$$\begin{aligned} Y^{*,\rho }(x)=\alpha ^{-1}(x)F(x)\rho (x)-\alpha ^{-1}(x)\left[ \nabla _x\cdot (\rho (x)J(x))\right] ^\top . \end{aligned}$$

Proof

Applying Duhamel’s principle to equation (4.1) yields

$$\begin{aligned} {\tilde{Y}}_t^{\epsilon ,\rho ,\tau }(x)= & {} e^{-\frac{\alpha (x)}{\epsilon }t}Y_0+\frac{1}{\epsilon }\int _0^te^{-\frac{\alpha (x)}{\epsilon }(t-s)}F(x)\rho (x)ds\nonumber \\ {}{} & {} {}-\frac{1}{\epsilon }\int _0^te^{-\frac{\alpha (x)}{\epsilon }(t-s)}[\nabla _x\cdot (\rho (x){\mathbb {E}}^x(\epsilon v_\tau ^\epsilon \otimes v_\tau ^\epsilon ))]^\top ds. \end{aligned}$$

(4.2)

Multiplying both sides of the equation (4.2) by the test function $\varphi$ yields

$$\begin{aligned} \langle {\tilde{Y}}_t^{\epsilon ,\rho ,\tau }(x),\varphi \rangle= & {} \left\langle e^{-\frac{\alpha (x)}{\epsilon }t}Y_0,\varphi \right\rangle +\frac{1}{\epsilon }\int _0^t\left\langle e^{-\frac{\alpha (x)}{\epsilon }(t-s)}F(x)\rho (x), \varphi \right\rangle ds\\ {}{} & {} {} -\frac{1}{\epsilon }\int _0^t\left\langle e^{-\frac{\alpha (x)}{\epsilon }(t-s)}[\nabla _x\cdot (\rho (x){\mathbb {E}}^x(\epsilon v_\tau ^\epsilon \otimes v_\tau ^\epsilon ))]^\top ,\varphi \right\rangle ds\\\triangleq & {} J_1+J_2+J_3. \end{aligned}$$

By Hölder inequality,

$$\begin{aligned} |J_1|=|\langle e^{-\frac{\alpha (x)}{\epsilon }t}Y_0,\varphi \rangle |\le e^{-\frac{C_{\lambda _{\alpha }}}{\epsilon }t_*}\Vert |Y_0|\Vert _{L^1}\Vert \varphi \Vert _{Lip} \le C\epsilon \Vert \varphi \Vert _{Lip}. \end{aligned}$$

(4.3)

Next,

$$\begin{aligned}{} & {} J_2=\left\langle \alpha ^{-1}(x)\left( I-e^{-\frac{\alpha (x)}{\epsilon }t}\right) F(x)\rho (x),\varphi \right\rangle \\{} & {} \quad \;=\langle \alpha ^{-1}(x)F(x)\rho (x),\varphi \rangle -\left\langle \alpha ^{-1}(x)e^{-\frac{\alpha (x)}{\epsilon }t}F(x)\rho (x),\varphi \right\rangle , \end{aligned}$$

by $\mathbf {(H_1)}$ and $\mathbf {(H_2)}$,

$$\begin{aligned} |J_2-\langle \alpha ^{-1}(x)F(x)\rho (x),\varphi \rangle |\le C\epsilon \Vert F(x)\sqrt{\rho (x)}\Vert _{L^2}\Vert \varphi \sqrt{\rho (x)}\Vert _{L^2} \le C_T\epsilon \Vert \varphi \Vert _{Lip}.~~~~~~~~~ \end{aligned}$$

(4.4)

At last, by Lemma 3.3,

$$\begin{aligned} J_3= & {} -\left\langle \alpha ^{-1}(x)\left( I-e^{-\frac{\alpha (x)}{\epsilon }t}\right) [\nabla _x\cdot (\rho (x){\mathbb {E}}^x(v_\tau ^\epsilon \otimes v_\tau ^\epsilon ))]^\top ,\varphi \right\rangle \quad \; \\ {}= & {} -\langle \alpha ^{-1}(x)\left( I-e^{-\frac{\alpha (x)}{\epsilon }t}\right) [\nabla _x\cdot (\rho (x)(J(x)+\epsilon C(x,\tau )))]^\top ,\varphi \rangle \quad \; \\ {}= & {} -\langle \alpha ^{-1}(x)(\nabla _x\cdot (\rho (x)J(x)))^\top ,\varphi \rangle -\epsilon \langle \alpha ^{-1}(x)[\nabla _x\cdot (\rho (x) C(x,\tau ))]^\top ,\varphi \rangle \quad \\ {}{} & {} {}+\langle \alpha ^{-1}(x)e^{-\frac{\alpha (x)}{\epsilon }t}[\nabla _x\cdot (\rho (x)(J(x)+\epsilon C(x,\tau )))]^\top ,\varphi \rangle . \end{aligned}$$

By Gaussian property and the definition of J(x), $|\nabla _x\cdot (\rho (x)J(x))|\le C(1+|x|)\rho (x)$, then

$$\begin{aligned} |\langle \alpha ^{-1}(x)e^{-\frac{\alpha (x)}{\epsilon }t}[\nabla _x\cdot (\rho (x)J(x))]^\top ,\varphi \rangle |\le C\epsilon \Vert (1+|x|)\rho (x)\Vert _{L^1}\Vert \varphi \Vert _{Lip} \le C\epsilon \Vert \varphi \Vert _{Lip}, \end{aligned}$$

and

$$\begin{aligned} |\langle \alpha ^{-1}(x)e^{-\frac{\alpha (x)}{\epsilon }t}[\nabla _x\cdot (\epsilon \rho (x) C(x,\tau ))]^\top ,\varphi \rangle |= & {} \epsilon |\langle \textrm{tr}(\rho (x) C(x,\tau )\textrm{grad}(\alpha ^{-1}(x)e^{-\frac{\alpha (x)}{\epsilon }t}\varphi ))\rangle |\\\le & {} C\epsilon \Vert (1+|x|^2)\rho (x)\Vert _{L^1}\Vert \varphi \Vert _{Lip}\\\le & {} C\epsilon \Vert \varphi \Vert _{Lip}. \end{aligned}$$

Thus

$$\begin{aligned}{} & {} |J_3+\langle \alpha ^{-1}(x)(\nabla _x\cdot (\rho (x)J(x)))^\top ,\varphi \rangle | \nonumber \\\le & {} \epsilon |\langle \textrm{tr}(\rho (x) C(x,\tau )\textrm{grad}(\alpha ^{-1}(x)\varphi ))\rangle |+ C\epsilon \Vert \varphi \Vert _{Lip}\nonumber \\\le & {} C\epsilon \Vert (1+|x|^2)\rho (x)\Vert _{L^1}\Vert \varphi \Vert _{Lip} \nonumber \\\le & {} C_T \epsilon \Vert \varphi \Vert _{Lip}. \end{aligned}$$

(4.5)

By (4.3), (4.4) and (4.5), the proof is complete. $\square$

Remark 4.1

As we have mentioned in the Introduction, one can derive the limit equation formally for $\rho _{t}$ by replacing $Y^{\epsilon }_{t}$ by $Y^{*, \rho }$ in the first equation of (1.8) .

However the slow part $\rho ^\epsilon _t$ does evolve, in order to approximate $Y^\epsilon _t$ we follow the Khasminskii’s scheme. For this we restrict the system in a small time interval, for example $[t_k, t_{k+1}]$ and freeze the slow part to be $\rho ^\epsilon _{t_k}$ . We show that (Lemma 4.2) $Y^\epsilon _t$ is approximated well by ${\hat{Y}}^\epsilon _t$ with frozen $\rho ^\epsilon _t=\rho ^\epsilon _{t_k}$ if the length of time interval $[t_k, t_{k+1}]$ is small. For this we divide the time interval [0, T] into small intervals of size $\delta >0$, i.e. $0=t_0<t_1<\ldots <t_{\lfloor T/\delta \rfloor }+1=T$, $t_k=k\delta$, $k=0,1,\ldots , \lfloor T/\delta \rfloor$. For $t\in [t_k,t_{k+1}]$, we define the auxiliary process $\{ {\hat{\rho }}_t^\epsilon (x),{\hat{Y}}_t^\epsilon (x)\}_{0\le t \le T}$ satisfying

$$\begin{aligned}{} & {} \partial _t {\hat{\rho }}_t^\epsilon (x)=-\nabla _x\cdot {\hat{Y}}_t^\epsilon (x),\\{} & {} \partial _t {\hat{Y}}_t^\epsilon (x)=-\frac{1}{\epsilon }\alpha (x){\hat{Y}}_t^\epsilon (x)+\frac{1}{\epsilon }F(x)\rho _{t_k}^\epsilon (x)-\left[ \nabla _x\cdot (\rho _{t_k}^\epsilon (x){\mathbb {E}}^x(v_{t_k}^\epsilon \otimes v_{t_k}^\epsilon )\right] ^\top ,\\{} & {} {\hat{\rho }}_{t_k}^\epsilon (x)=\rho _{t_k}^\epsilon (x),\;\;{\hat{Y}}_{0}^\epsilon (x)=Y_{0}. \end{aligned}$$

Remark 4.2

One can see that ${\hat{Y}}^\epsilon _t={\tilde{Y}}_t^{\epsilon , \rho ^\epsilon _{t_k}, t_k}$ .

Lemma 4.2

Assume $(\mathbf {H_1})$–$(\mathbf {H_3})$ hold, for $T>0$ and $\varphi \in C_0^\infty ({\mathbb {R}}^d, {{\mathbb {R}}}^{d})$,

$$\begin{aligned} \sup _{0\le t\le T}|\langle Y_t^\epsilon (x)-{\hat{Y}}_t^\epsilon (x),\varphi \rangle |\le C_T\left( \frac{\delta }{\epsilon ^2}+\frac{\delta }{\epsilon }\right) \Vert \varphi \Vert _{Lip}\,. \end{aligned}$$

Proof

Let $Z_t^\epsilon (x)=Y_t^\epsilon (x)-{\hat{Y}}_t^\epsilon (x)$. For all $t\in [t_k, t_{k+1}]$, we have

$$\begin{aligned} \partial _t{Z}_t^\epsilon (x)= & {} -\frac{1}{\epsilon }\alpha (x)Z_t^\epsilon (x)+\frac{1}{\epsilon }F(x)(\rho _{t}^\epsilon (x)-\rho _{t_k}^\epsilon (x))\\ {}{} & {} {}-\left[ \nabla _x\cdot (\rho _{t}^\epsilon (x){\mathbb {E}}^x(v_{t}^\epsilon \otimes v_{t}^\epsilon )-\nabla _x\cdot (\rho _{t_k}^\epsilon (x){\mathbb {E}}^x(v_{t_k}^\epsilon \otimes v_{t_k}^\epsilon )\right] ^\top . \end{aligned}$$

By Duhamel’s principle,

$$\begin{aligned}{} & {} Z_t^\epsilon (x)=\frac{1}{\epsilon }\int _{t_k}^te^{-\frac{1}{\epsilon }\alpha (x)(t-s)}F(x)(\rho _{s}^\epsilon (x)-\rho _{t_k}^\epsilon (x))ds\\ {}{} & {} \quad \quad \quad \quad -\int _{t_k}^te^{-\frac{1}{\epsilon }\alpha (x)(t-s)}[\nabla _x\cdot (\rho _{s}^\epsilon (x){\mathbb {E}}^x(v_{s}^\epsilon \otimes v_{s}^\epsilon )-\nabla _x\cdot (\rho _{t_k}^\epsilon (x){\mathbb {E}}^x(v_{t_k}^\epsilon \otimes v_{t_k}^\epsilon )]^\top ds. \end{aligned}$$

For $\varphi \in C_0^\infty ({\mathbb {R}}^d, {{\mathbb {R}}}^{d})$, we obtain

$$\begin{aligned}{} & {} \langle Z_t^\epsilon (x),\varphi \rangle \\= & {} \frac{1}{\epsilon }\int _{t_k}^t\langle e^{-\frac{1}{\epsilon }\alpha (x)(t-s)}F(x)(\rho _{s}^\epsilon (x)-\rho _{t_k}^\epsilon (x)),\varphi \rangle ds\\{} & {} {}-\int _{t_k}^t\langle e^{-\frac{1}{\epsilon }\alpha (x)(t-s)}[\nabla _x\cdot (\rho _{s}^\epsilon (x){\mathbb {E}}^x(v_{s}^\epsilon \otimes v_{s}^\epsilon )-\nabla _x\cdot (\rho _{t_k}^\epsilon (x){\mathbb {E}}^x(v_{t_k}^\epsilon \otimes v_{t_k}^\epsilon )]^\top ,\varphi \rangle ds\\\triangleq & {} I_1+I_2. \end{aligned}$$

First

$$\begin{aligned} |I_1|= & {} \frac{1}{\epsilon }\left| \int _{t_k}^t[{\mathbb {E}}(e^{-\frac{1}{\epsilon }\alpha (x_s^\epsilon )(t-s)}F(x_s^\epsilon )\varphi (x_s^\epsilon )) -{\mathbb {E}}(e^{-\frac{1}{\epsilon }\alpha (x_{t_k}^\epsilon )(t-s)}F(x_{t_k}^\epsilon )\varphi (x_{t_k}^\epsilon ))]ds\right| \\\le & {} \frac{1}{\epsilon }\int _{t_k}^t[|{\mathbb {E}}(e^{-\frac{1}{\epsilon }\alpha (x_s^\epsilon )(t-s)}F(x_s^\epsilon )\varphi (x_s^\epsilon ))| +|{\mathbb {E}}(e^{-\frac{1}{\epsilon }\alpha (x_{t_k}^\epsilon )(t-s)}F(x_{t_k}^\epsilon )\varphi (x_{t_k}^\epsilon ))|]ds\\\le & {} \frac{1}{\epsilon }\int _{t_k}^t C_F \Vert \varphi \Vert _{Lip}(1+{\mathbb {E}}|x_s|)ds+\frac{1}{\epsilon }\int _{t_k}^t C_F \Vert \varphi \Vert _{Lip}(1+{\mathbb {E}}|x_{t_k}|)ds\,. \end{aligned}$$

Then, by Lemma 3.1, we have

$$\begin{aligned} |I_1|\le C_T \frac{\delta }{\epsilon }\Vert \varphi \Vert _{Lip}. \end{aligned}$$

(4.6)

Further by Lemma 2.2,

$$\begin{aligned} I_2= & {} -\int _{t_k}^t\langle [\nabla _x\cdot (\rho _{s}^\epsilon (x){\mathbb {E}}^x(v_{s}^\epsilon \otimes v_{s}^\epsilon )-\nabla _x\cdot (\rho _{t_k}^\epsilon (x){\mathbb {E}}^x(v_{t_k}^\epsilon \otimes v_{t_k}^\epsilon )],e^{-\frac{1}{\epsilon }\alpha ^\top (x)(t-s)}\varphi \rangle ds\\ {}= & {} \int _{t_k}^t\int _{{\mathbb {R}}^d} \textrm{tr} \left[ (\rho _{s}^\epsilon (x){\mathbb {E}}^x(v_{s}^\epsilon \otimes v_{s}^\epsilon )-\rho _{t_k}^\epsilon (x){\mathbb {E}}^x(v_{t_k}^\epsilon \otimes v_{t_k}^\epsilon ))\textrm{grad}\left( e^{-\frac{1}{\epsilon }\alpha ^\top (x)(t-s)}\varphi \right) \right] dx ds\,. \end{aligned}$$

Let $g(x)=e^{-\frac{1}{\epsilon }\alpha ^\top (x)(t-s)}\varphi (x)$, by the chain rule,

$$\begin{aligned} \frac{\partial }{\partial x_i}g(x)=\frac{\partial }{\partial x_i}(e^{-\frac{1}{\epsilon }\alpha ^\top (x)(t-s)})\varphi (x)+e^{-\frac{1}{\epsilon }\alpha ^\top (x)(t-s)}\frac{\partial }{\partial x_i}\varphi (x)\,. \end{aligned}$$

Then, by assumptions $(\mathbf {H_1})$ and $(\mathbf {H_3})$,

$$\begin{aligned} |\frac{\partial }{\partial x_i}g(x)|\le C(\frac{1}{\epsilon }+1)\Vert \varphi \Vert _{Lip}, \end{aligned}$$

(4.7)

thus

$$\begin{aligned} |\textrm{grad}(g(x))|\le C(\frac{1}{\epsilon }+1)\Vert \varphi \Vert _{Lip}. \end{aligned}$$

(4.8)

By Lemma 3.3 and (4.8),

$$\begin{aligned}{} & {} \left| \int _{{\mathbb {R}}^d}\textrm{tr}(\rho _{s}^\epsilon (x){\mathbb {E}}^x(v_{s}^\epsilon \otimes v_{s}^\epsilon )\textrm{grad}(g(x)))dx\right| \\= & {} \frac{1}{\epsilon }|\textrm{tr}[{\mathbb {E}}({\mathbb {E}}^{x^\epsilon _s}(\epsilon v_{s}^\epsilon \otimes v_{s}^\epsilon )\textrm{grad}(g(x^\epsilon _s)))]|\\\le & {} \frac{C}{\epsilon } {\mathbb {E}}[(J(x^\epsilon _s)+\epsilon C(x^\epsilon _s,s))(\frac{1}{\epsilon }+1)\Vert \varphi \Vert _{Lip}\\\le & {} C_T (\frac{1}{\epsilon ^2}+\frac{1}{\epsilon })\Vert \varphi \Vert _{Lip}. \end{aligned}$$

Similarly,

$$\begin{aligned} \left| \int _{{\mathbb {R}}^d}\textrm{tr}(\rho _{t_k}^\epsilon (x){\mathbb {E}}^x(v_{t_k}^\epsilon \otimes v_{t_k}^\epsilon )\textrm{grad}(g(x)))dx\right| \le C_T (\frac{1}{\epsilon ^2}+\frac{1}{\epsilon })\Vert \varphi \Vert _{Lip}. \end{aligned}$$

Then we have

$$\begin{aligned} |I_2|\le C_T (\frac{\delta }{\epsilon ^2}+\frac{\delta }{\epsilon })\Vert \varphi \Vert _{Lip}. \end{aligned}$$

The proof is complete. $\square$

Proof of Theorem 2.1

From the first equation of (1.8), for $\psi \in C_0^\infty ({\mathbb {R}}^d)$ and $\varphi =\nabla \psi$ we derive

$$\begin{aligned}{} & {} \langle \rho _t^\epsilon (x),\psi \rangle =\langle \rho _{t_*},\psi \rangle +\int _{t_*}^t\langle Y_s^\epsilon (x),\varphi \rangle ds\nonumber \\ {}{} & {} \quad \quad \quad \quad \; =\langle \rho _{t_*},\psi \rangle +\int _{t_*}^t\langle Y^{*,\rho _s^\epsilon }(x),\varphi \rangle ds\nonumber \\ {}{} & {} \quad \quad \quad \quad \quad +\int _{t_*}^t\langle Y_s^\epsilon (x)-{\hat{Y}}_s^\epsilon (x),\varphi \rangle ds+\int _{t_*}^t\langle {\hat{Y}}_s^\epsilon (x)- Y^{*,\rho _s^\epsilon }(x),\varphi \rangle ds\\{} & {} \quad \quad \quad \quad \triangleq \langle \rho _{t_*},\psi \rangle +K_1+K_2+K_3. \end{aligned}$$

First, by the expression of $Y^{*,\rho }$,

$$\begin{aligned} K_1=\int _{t_0}^t\langle Y^{*,\rho _s^\epsilon }(x),\varphi \rangle ds\rightarrow \int _{t_0}^t\langle Y^{*,\rho _s}(x),\varphi \rangle ds. \end{aligned}$$

(4.9)

Next, by Lemma 4.2,

$$\begin{aligned} |K_2|\le C_T\left( \frac{\delta }{\epsilon ^2}+\frac{\delta }{\epsilon }\right) \Vert \varphi \Vert _{Lip}\rightarrow 0, \end{aligned}$$

(4.10)

by choosing $\delta =O(\epsilon ^3)$.

Note that on the time interval $[t_k,t_{k+1}]$, $\{{\hat{Y}}_t^\epsilon \}=\{{\tilde{Y}}_t^{\epsilon ,\rho _{t_k}^\epsilon ,t_k}\}$, let $t_*\in [t_{i_0},t_{i_0+1}]$,

$$\begin{aligned} K_3= & {} \int _{t_*}^{t_{i_0+1}}\langle {\tilde{Y}}_s^{\epsilon ,\rho _{t_{i_0}}^\epsilon ,t_{i_0}}-Y^{*,\rho _{t_{i_0}}^\epsilon },\varphi \rangle ds+ \sum _{k=i_0+1}^{\lfloor t/\delta \rfloor -1}\int _{t_k}^{t_{k+1}}\langle {\tilde{Y}}_s^{\epsilon ,\rho _{t_k}^\epsilon ,t_k}-Y^{*,\rho _{t_k}^\epsilon },\varphi \rangle ds\\ {}{} & {} {} +\int _{t_{\lfloor t/\delta \rfloor }}^{t}\langle {\tilde{Y}}_s^{\epsilon ,\rho _{t_{\lfloor t/\delta \rfloor }}^\epsilon ,t_{\lfloor t/\delta \rfloor }}-Y^{*,\rho _{t_{\lfloor t/\delta \rfloor }}^\epsilon },\varphi \rangle ds +\int _{t_*}^{t_{i_0+1}}\langle Y^{*,\rho _{t_{i_0+1}}^\epsilon }-Y^{*,\rho _{s}^\epsilon },\varphi \rangle ds\\ {}{} & {} {}+\sum _{k=i_0+1}^{\lfloor t/\delta \rfloor -1}\int _{t_k}^{t_{k+1}}\langle Y^{*,\rho _{t_k}^\epsilon }-Y^{*,\rho _{s}^\epsilon },\varphi \rangle ds +\int _{t_{\lfloor t/\delta \rfloor }}^{t}\langle Y^{*,\rho _{t_{\lfloor t/\delta \rfloor }}^\epsilon }-Y^{*,\rho _{s}^\epsilon },\varphi \rangle ds\\\triangleq & {} K_{31}+K_{32}+K_{33}+K_{34}+K_{35}+K_{36}. \end{aligned}$$

By Lemma 4.1,

$$\begin{aligned} |K_{31}+K_{32}+K_{33}|\le C_T \epsilon \Vert \varphi \Vert _{Lip}. \end{aligned}$$

(4.11)

By the defination of $Y^{*,\rho }$ and Lemma 3.1,

$$\begin{aligned} |K_{34}+K_{35}+K_{36}|\le & {} C_T\left( \int _{t_*}^{t_{i_0+1}} {\mathbb {E}}|x^{\epsilon }_{t_{i_0+1}}-x^{\epsilon }_{t_*}|^2ds+ \sum _{k=i_0+1}^{\lfloor t/\delta \rfloor -1}\int _{t_k}^{t_{k+1}}{\mathbb {E}}|x^{\epsilon }_{t_k}-x^{\epsilon }_s|^2ds \right. \nonumber \\{} & {} {}\left. +\int _{t_{\lfloor t/\delta \rfloor }}^{t}{\mathbb {E}}|x^{\epsilon }_{t_{\lfloor t/\delta \rfloor }}-x^{\epsilon }_s|^2ds\right) \Vert \varphi \Vert _{Lip}\nonumber \\\le & {} C_T \Vert \varphi \Vert _{Lip}\delta . \end{aligned}$$

(4.12)

By (4.9)–(4.12), passing the limit $\epsilon \rightarrow 0$ yields

$$\begin{aligned} \langle \rho _t^\epsilon (x),\psi \rangle \rightarrow \langle \rho _{t_*},\psi \rangle +\int _{t_*}^t\langle Y^{*,\rho _s}(x),\varphi \rangle ds. \end{aligned}$$

Since $\varphi =\nabla _x \psi$, it yields

$$\begin{aligned} \langle \partial _t \rho _t(x),\psi \rangle =\langle Y^{*,\rho _t}(x),\nabla _x\psi \rangle =-\langle \nabla _x\cdot Y^{*,\rho _t}(x),\psi \rangle , \end{aligned}$$

which is the weak form of

$$\begin{aligned} \partial _t \rho _t(x)=-\nabla _x\cdot (\alpha ^{-1}(x)F(x)\rho _t(x)+\alpha ^{-1}(x)(\nabla _x\cdot (\rho _t(x)J(x)))^\top ). \end{aligned}$$

The proof is complete. $\square$