Hydrodynamics for Asymmetric Simple Exclusion on a Finite Segment with Glauber-Type Source

Abstract

We consider an open interacting particle system on a finite lattice. The particles perform asymmetric simple exclusion and are randomly created or destroyed at all sites, with rates that grow rapidly near the boundaries. We study the hydrodynamic limit for the particle density at the hyperbolic space-time scale and obtain the entropy solution to a boundary-driven quasilinear conservation law with a source term. Different from the usual boundary conditions introduced in Bardos et al (Commun Partial Differ Equ 4(9):1017–1034, https://doi.org/10.1080/03605307908820117, 1979) and Otto (C R Acad Sci Paris 322(1):729–734, 1996), discontinuity (boundary layer) does not formulate at the boundaries due to the strong relaxation scheme.

1 Introduction

In the past decades, hydrodynamic limit for interacting particle system with boundary effect has attracted a lot of attention [1, 2, 4,5,6, 8, 11, 15, 24]. The limit captures the evolution of the conserved field of the microscopic dynamics as hydrodynamic equation, at the macroscopic time scale that the dynamics is equilibrated locally. Most of these works focus on symmetric dynamics and the Fick’s law of diffusion or fractional diffusion with various types of boundary conditions, see, e.g., [2, 4,5,6, 8, 15].

For asymmetric mass-conserving systems, the dynamics reaches local equilibrium at the hyperbolic time scale, and the hydrodynamic equations are given by hyperbolic transport equations [22]. When nonlinear interaction exists, these equations are featured by discontinuous phenomenon both inside the domain (shock wave) and at the boundary (boundary layer). The non-regularity becomes the main obstacle in deducing the hydrodynamic limit. Asymmetric simple exclusion process (ASEP) with open boundaries is the simplest model. In its dynamics, each particle performs an asymmetric random walk on the finite lattice $\{1,\ldots ,N-1\}$ under the exclusion rule: two particles cannot occupy the same site simultaneously. Particles are created and annihilated randomly at sites 1 and $N-1$, modeling the exchange of mass between the system and two external reservoirs at given densities. In [1, 24, 25], the hydrodynamic limit for the mass density of open ASEP is proved to be Burgers equation with boundary conditions introduced in [3, 20]. Due to the discontinuous nature, these boundary conditions do not prescribe the density at boundary, even when the reservoir dynamics is overwhelmingly accelerated compared to the exclusion [25]. Instead, they impose a set of possible values for the boundary density. The hydrostatic limit for the same dynamics is studied in [1, 9]: the stationary density profile is the stationary solution to the hydrodynamic equation. It is determined by the boundary data through a variational property [21]: the stationary flux is maximized if the density gradient is opposite to the drift, and is minimized otherwise. The result is generalized in [7] to the quasi-static transform: if the reservoir densities are changing slowly at a time scale that is larger than the hyperbolic one, the profile evolves with the corresponding quasi-static solution [17].

The motivation of this article is to study the hydrodynamic limit for a hyperbolic system disturbed by a nonlocal external field. We are particularly interested in the macroscopic behavior when the perturbation is extremely strong at the boundary. Consider the ASEP on $\{1,\ldots ,N-1\}$ where particles are created (resp. annihilated) at each site i with rate $V_i\rho _i$ (resp. $V_i(1-\rho _i)$). Assume two profiles $(V,\rho ):(0,1)\rightarrow {\mathbb {R}}_+\times [0,1]$ such that

$$\begin{aligned} (V_i,\rho _i)=(V,\rho ) \left( \tfrac{i}{N} \right) , \quad \lim _{x\rightarrow 0+} V(x) = \lim _{x\rightarrow 1-} V(x) = \infty . \end{aligned}$$

(1.1)

In other words, a reservoir of density $\rho _i$ is placed at each site i, and the system exchanges particles with it with frequency $V_i$ that is growing rapidly near the boundaries. When the exclusion dynamics is accelerated by N, the density profile shall evolve with the $L^\infty$ entropy solution to the following quasilinear balance law in the [0, 1]-interval:

$$\begin{aligned} \partial _tu + \partial _x(u(1-u)) + G(x,u) = 0, \quad G=V(x)(u-\rho (x)), \end{aligned}$$

(1.2)

with proper boundary conditions. We prove in Theorem 2.8 that, when the integrals of V are infinity around both 0 and 1, the boundary conditions are $u|_{x=0}=\rho (0)$, $u|_{x=1}=\rho (1)$. In sharp contrast to the equations obtained in [1, 24, 25], the boundary values of u are fixed by $\rho$ in a weak sense, see Proposition 2.11. Hence, any shock wave is attenuated while approaching the boundaries, and no boundary layer is observable at any positive macroscopic time. A consequence of the hydrodynamic limit is the $L^1$-weak continuity in time of the entropy solution obtained in Corollary 2.10.

The term G in (1.2) acts as a source (resp. sink) where u is less (resp. greater) than $\rho$, so it can be viewed as a relaxation scheme to the profile $\rho$. When $\rho$ is a constant, it is a conservation system with relaxation introduced in [16], with the first component degenerated to a stationary solution. Such system is widely used to model non-equilibrium transport in kinetic theory and fluid dynamics. In our situation, the entropy solution to the initial-boundary problem of (1.2) is constructed in different ways depending on the integrability of V, see Definition 2.1 and 2.7. We focus on the non-integrable case and discuss the integrable case briefly in Sect. 2.4.

The proof in this article is proceeded in two main steps. First, we prove that in the space-time scaling limit, the empirical Young measure of the particle field is concentrated on the space of Dirac-type Young measures. Then, we show that the limit is a measure-valued entropy solution to (1.2) with proper boundary conditions. The hydrodynamic limit then follows from the uniqueness of the entropy solution. Both steps are proved through delicate analyses of the microscopic entropy production associated with Lax entropy–flux pairs.

The use of Young measure and microscopic entropy production is present in the seminal paper [22]. It is combined with the compensated compactness method to prove the concentration property of the Young measure in [12, 13]. To use this method, additional oscillating dynamics is added to ASEP to create microscopic viscosity. Finally, we point out that although the process studied in this article is attractive, we cannot apply the coupling argument used in [1] because the invariant measure is not product in general.

2 Model and Results

2.1 Model

For a scaling parameter $N \in {\mathbb {N}}_+$, consider the configuration space

$$\begin{aligned}\Omega _N:= \big \{\eta =(\eta _i)_{0 \le i \le N}, \eta _i \in \{0,1\}\big \}.\end{aligned}$$

The dynamics on $\Omega _N$ consists of three parts: the nearest-neighbor asymmetric exclusion, the external Glauber field and the boundary dynamics. The exclusion is generated by

$$\begin{aligned} L_{\textrm{exc}} f (\eta ) = \sum _{i=0}^{N-1} \left( c_{i,i+1}(\eta )+\frac{\sigma _N}{2} \right) \Big [f (\eta ^{i,i+1}) - f(\eta )\Big ], \end{aligned}$$

for any function f on $\Omega _N$, where, for constant $p\in (\tfrac{1}{2},1]$,

$$\begin{aligned} c_{i,i+1} (\eta ) = p\eta _i (1-\eta _{i+1}) + (1-p) \eta _{i+1} (1-\eta _i), \end{aligned}$$

(2.1)

$\sigma _N$ is a parameter that grows to infinity slower than N, and $\eta ^{i,i+1}$ is the configuration obtained from $\eta$ by swap** the values of $\eta _i$ and $\eta _{i+1}$. The factor $\sigma _N$ stands for a strong microscopic viscosity, which is necessary for the technique used in Sect. 5. The Glauber dynamics is generated by

$$\begin{aligned} L_{\textrm{G}} f (\eta )= \frac{1}{N}\sum _{i=1}^{N-1} c_{i,\textrm{G}} (\eta ) \Big [ f (\eta ^{i}) - f(\eta )\Big ], \end{aligned}$$

where, for parameters $V_i>0$ and $\rho _i\in (0,1)$,

$$\begin{aligned} c_{i,\textrm{G}}(\eta ):=V_i[\rho _i(1-\eta _i)+(1-\rho _i)\eta _i], \end{aligned}$$

(2.2)

and $\eta ^{i}$ is the configuration obtained from $\eta$ by flip** the value of $\eta _i$. Finally, the sites $i=0$ and N are attached to two extra birth-and-death dynamics, interpreted as boundary reservoirs. The corresponding generator reads

$$\begin{aligned} L_{\textrm{bd}}f(\eta ) = c_0(\eta ) \Big [ f (\eta ^{0}) - f(\eta )\Big ] + c_{N} (\eta ) \Big [ f (\eta ^{N}) - f(\eta )\Big ], \end{aligned}$$

where, for boundary rates $c_{\textrm{in}}^\pm$, $c_{\textrm{out}}^\pm \ge 0$,

$$\begin{aligned} c_0 (\eta ) = c_{\textrm{in}}^- (1-\eta _0) + c_{\textrm{out}}^- \eta _0, \quad c_N (\eta ) = c_{\textrm{in}}^+(1-\eta _N) + c_{\textrm{out}}^+ \eta _N. \end{aligned}$$

(2.3)

Assume two profiles $V:(0,1)\rightarrow {\mathbb {R}}_+$ and $\rho :[0,1]\rightarrow (0,1)$ such that $V_i=V(\tfrac{i}{N})$ and $\rho _i=\rho (\tfrac{i}{N})$ for $i=1$,..., $N-1$. Suppose that $V\in {\mathcal {C}}^1((0,1);{\mathbb {R}}_+)$, $V\rightarrow +\infty$ as $x\rightarrow 0$, 1, and $\rho \in {\mathcal {C}}^1([0,1];(0,1))$ with Lipschitz continuous $\rho '$. In particular,

$$\begin{aligned} \inf _{x\in (0,1)} V(x)>0, \quad \inf _{x\in [0,1]} \rho (x) > 0, \quad \sup _{x\in [0,1]} \rho (x) < 1. \end{aligned}$$

(2.4)

The generator of the process then reads

$$\begin{aligned} L_N = N\big (L_{\textrm{exc}} + L_{\textrm{G}} + L_{\textrm{bd}}\big ), \end{aligned}$$

(2.5)

where the factor N corresponds to the hyperbolic time scale.

2.2 Scalar Balance Law in a Bounded Domain

In this part, we introduce the partial differential equation that is obtained in the hydrodynamic limit for the model defined in the previous section. Let

$$\begin{aligned} J(u):= (2p-1) u (1-u), \quad G(x,u):= V(x) (u-\rho (x)) \end{aligned}$$

(2.6)

be the macroscopic flux and the source term corresponding to $L_\textrm{exc}$ and $L_{\textrm{G}}$, respectively. Given measurable function $u_0:(0,1)\rightarrow [0,1]$, consider the following balance law: for $(t,x)\in \Sigma :={\mathbb {R}}_+\times (0,1)$,

$$\begin{aligned} \partial _{t} u(t,x)+\partial _{x}[J(u(t,x))]+G(x,u(t,x)) =0, \quad u|_{t=0}=u_0, \end{aligned}$$

(2.7)

with proper boundary conditions that will be specified later.

The weak solution to (2.7) is in general not unique, so we are forced to consider the entropy solution. Recalling (1.1), our aim is to examine the case when the strength of the source is extremely strong at the boundaries. We see in Definitions 2.1 and 2.7 that the definition of entropy solution is different when V is integrable or not at the boundaries.

We begin with the case that V is non-integrable at 0 and 1, i.e., for any small y,

$$\begin{aligned} \int _0^y V(x)\,dx = +\infty , \qquad \int _{1-y}^1 V(x)\,dx = +\infty . \end{aligned}$$

(2.8)

Recall that a Lax entropy–flux pair of (2.7) is a pair of functions f, $q \in \mathcal C^2({\mathbb {R}})$ such that $f''\ge 0$ and $q'=J'f'=(2p-1)(1-2u)f'(u)$ for all $u\in {\mathbb {R}}$.

Definition 2.1

Suppose that V satisfies (2.8). We call $u=u(t,x)$ an entropy solution to (2.7) with the compatible boundary conditions

$$\begin{aligned} u(\cdot ,0)=\rho (0), \qquad u(\cdot ,1)=\rho (1), \end{aligned}$$

(2.9)

if $u:\Sigma \rightarrow [0,1]$ is measurable and satisfies the generalized entropy inequality

$$\begin{aligned}{} & {} \int _{0}^{1} f(u_{0}) \varphi (0,\cdot )\,dx+\iint _{\Sigma }\big [f (u) \partial _{t} \varphi +q (u) \partial _{x} \varphi \big ]\,d x d t \nonumber \\{} & {} \quad \ge \iint _{\Sigma } f'(u)V(x)(u-\rho ) \varphi \,d x d t. \end{aligned}$$

(2.10)

for any Lax entropy–flux pair (f, q) and any $\varphi \in {\mathcal {C}}_c^2 ({\mathbb {R}}\times (0,1))$, $\varphi \ge 0$.

Remark 2.2

When $\rho \in C^1$, u in Definition 2.1 satisfies the energy estimate

$$\begin{aligned} \int _{0}^{T} \int _{0}^{1} V(x)[u(t, x)-\rho (x)]^{2}\,d x d t<\infty , \quad \forall \,T>0. \end{aligned}$$

(2.11)

Indeed, suppose that $\rho$ is smooth. For any $\varepsilon >0$, choose $\psi _\varepsilon \in {\mathcal {C}}_c^\infty ((0,1))$ such that $\psi _\varepsilon (x)\in [0,1]$, $\psi _\varepsilon |_{[\varepsilon ,1-\varepsilon ]}\equiv 1$ and $|\psi '_\varepsilon (x)|\le 2\varepsilon ^{-1}$. Fixing any $\phi \in {\mathcal {C}}_c^\infty ({\mathbb {R}})$ such that $\phi \ge 0$ and applying (2.10) with $f_1=\tfrac{1}{2}u^2$, $\varphi _1=\phi (t)\psi _\varepsilon (x)$ and $f_2=-u$, $\varphi _2=\varphi _1\rho$ respectively, we obtain the upper bound

$$\begin{aligned}&\iint _\Sigma V(x)[u(t,x)-\rho (x)]^2\phi (t)\psi _\varepsilon (x)dxdt\\&\quad =\iint _\Sigma f'_1(u)V(x)(u-\rho )\varphi _1\,dxdt + \iint _\Sigma f'_2(u)V(x)(u-\rho )\varphi _2\,dxdt\\&\quad \le \phi (0)\int _0^1 \big [f_1(u_0)-u_0\rho \big ]\psi _\varepsilon \,dx\\&\qquad + \iint _\Sigma \big [f_1(u)\partial _t\varphi _1+q_1(u)\partial _x\varphi _1 - u\partial _t\varphi _2-J(u)\partial _x\varphi _2\big ]dxdt, \end{aligned}$$

where $q_1$ is the flux corresponding to $f_1$. Since $|\psi _\varepsilon |\le 1$, the first term on the right-hand side is bounded by $|\phi |_\infty \Vert f_1(u_0)-u_0\rho \Vert _{L^\infty }$. The second term reads

$$\begin{aligned} \iint _\Sigma&\big [(f_1(u)-u\rho )\partial _t\varphi _1 + (q_1(u)-J(u)\rho )\partial _x\varphi _1 - J(u)\rho '\varphi _1\big ]dxdt\\ \le \;&C \iint _\Sigma \big (|\phi '(t)\psi _\varepsilon (x)| + |\psi '_\varepsilon (x)\phi (t)| + |\rho '(x)\phi (t)\psi _\varepsilon (x)|\big )dxdt, \end{aligned}$$

where $C=\Vert f_1(u)-u\rho \Vert _{L^\infty } + \Vert q_1(u)-J(u)\rho \Vert _{L^\infty } + \Vert J(u)\Vert _{L^\infty }$. Since $|\psi _\varepsilon |\le 1$, $|\psi '_\varepsilon |\le 2\varepsilon ^{-1}$ and is non-zero if and only if $x\in (0,\epsilon )\cup (1-\epsilon ,1)$, it is bounded by $C_\phi (1+|\rho '|_\infty )$ with a constant $C_\phi$ that is independent of $\varepsilon$. Taking $\varepsilon \rightarrow 0$ and using monotone convergence theorem,

$$\begin{aligned} \iint _\Sigma V(x)[u(t,x)-\rho (x)]^2\phi (t)dxdt \le C_\phi (1+|\rho '|_\infty ). \end{aligned}$$

Since $\phi \in {\mathcal {C}}_c^\infty ({\mathbb {R}};{\mathbb {R}}_+)$ is arbitrary, (2.11) holds for any finite $T>0$. By standard argument of compactness, the estimate can be extended to any $\rho \in C^1([0,1])$.

Remark 2.3

If u is continuous in space, (2.11) together with (2.8) implies that $u(t,0)=\rho (0)$ and $u(t,1)=\rho (1)$ for almost all $t>0$. Hence, (2.9) turns out to be the reasonable choice of the boundary conditions, see also Proposition 2.11 below.

The following uniqueness criteria is taken from [26, Theorem 2.12].

Proposition 2.4

Assume further that

$$\begin{aligned}&\limsup _{y \rightarrow 0+} \frac{1}{y^2} \int _{0}^y \left[ \frac{1}{V(x)} + \frac{1}{V(1-x)} \right] \,dx < + \infty ,\end{aligned}$$

(2.12)

$$\begin{aligned}&\lim _{y \rightarrow 0+} \left[ \int _0^y V(x) \big [\rho (x) - \rho (0)\big ]^2 dx + \int _{1-y}^1 V(x) \big [\rho (x) - \rho (1)\big ]^2 dx \right] = 0. \end{aligned}$$

(2.13)

Then, there is at most one function $u \in L^\infty (\Sigma )$ that fulfills Definition 2.1.

Remark 2.5

Suppose that $V>0$ satisfies (2.12). By Cauchy–Schwarz inequality,

$$\begin{aligned}\liminf _{y\rightarrow 0+} \int _0^y V(x)dx \ge \liminf _{y\rightarrow 0+} \bigg (\frac{1}{y^2}\int _0^y \frac{1}{V(x)}dx\bigg )^{-1} > 0,\end{aligned}$$

which means that V is not integrable at 0. The same argument holds for the integration on $(1-y,1)$. Therefore, (2.12) contains the non-integrable condition (2.8).

Now we turn to the integrable case: $V \in L^1((0,1))$. The next definition is first introduced by F. Otto, see [20, Eq. 9].

Definition 2.6

We call $(F, Q) \in {\mathcal {C}}^{2}([0,1]^2; {\mathbb {R}}^{2})$ a boundary entropy–flux pair if

• (i) for all $k \in [0,1]$, $(F,Q) (\cdot , k)$ is a Lax entropy-flux pair, i.e., $\partial _uQ(\cdot ,k)=J'\partial _uF(\cdot ,k)$;

• (ii) for all $k \in [0,1]$, $F(k, k)=\partial _{u} F(u, k)|_{u=k}=Q(k, k)=0$.

For V integrable, the definition of entropy solution is the same as [20, Proposition 2] for $V\equiv 0$ and [18, Definition 1] for V bounded and smooth.

Definition 2.7

Let $\alpha$, $\beta \in [0,1]$ be two constants and suppose that $V \in L^1 ((0,1))$. We call $u=u(t,x)$ an entropy solution to (2.7) with the boundary conditions given by

$$\begin{aligned} u(t,0)=\alpha , \qquad u(t,1)=\beta , \end{aligned}$$

(2.14)

if $u: \Sigma \rightarrow [0,1]$ is a measurable function such that for any boundary entropy–flux pair (F, Q), any $k \in [0,1]$, and any $\varphi \in {\mathcal {C}}_c^2 ({\mathbb {R}}^2)$ such that $\varphi \ge 0$,

$$\begin{aligned} \begin{aligned}&\int _{0}^{1} f_k(u_{0}) \varphi (0,\cdot )\,dx + \iint _{\Sigma } \big [f_k(u) \partial _{t} \varphi + q_k(u) \partial _{x} \varphi \big ]\,d x d t \\&\quad \ge \iint _{\Sigma } f'_k(u) V(x)(u-\rho ) \varphi \,d x d t - \int _{0}^{\infty }\big [f_k(\beta )\varphi (\cdot , 1)+ f_k(\alpha )\varphi (\cdot , 0)\big ]\,d t, \end{aligned} \end{aligned}$$

(2.15)

where $(f_k,q_k):=(F,Q)(\cdot ,k)$.

Since the integrable case is not the focus of this paper, we omit the uniqueness and other properties and refer to [26] and the references therein.

2.3 Hydrodynamic Limit

Let $\{\eta ^N(t)\in \Omega _N;t\ge 0\}$ be the Markov process generated by $L_N$ in (2.5) and initial distribution $\mu _N$. Through this article, the superscript N in $\eta ^N$ is omitted when there is no confusion. Denote by ${\mathbb {P}}_{\mu _N}$ the distribution of $\eta (\cdot )$ on ${\mathcal {D}}([0,\infty ),\Omega _N)$, the space of all càdlàg paths on $\Omega _N$, and by ${\mathbb {E}}_{\mu _N}$ the expectation of ${\mathbb {P}}_{\mu _N}$.

Suppose that the sequence of $\mu _N$ is associated with a measurable function $u_0:(0,1)\rightarrow [0,1]$ in the following sense: for any $\psi \in {\mathcal {C}}({\mathbb {R}})$,

$$\begin{aligned} \lim _{N \rightarrow \infty } \mu _N \left\{ \left| \frac{1}{N} \sum _{i=0}^N \eta _i(0)\psi \left( \tfrac{i}{N} \right) - \int _0^1 u_0 (x) \psi (x) dx \right|> \delta \right\} = 0, \quad \forall \,\delta >0. \end{aligned}$$

(2.16)

Our main result shows that in the non-integrable case, the empirical density of the particles converges, as $N\rightarrow \infty$, to the entropy solution to (2.7) and (2.9).

Theorem 2.8

Assume (2.12), (2.13) and (2.16). Also assume that

$$\begin{aligned} \lim _{N\rightarrow \infty } N^{-1}\sigma _N^2 = \infty , \qquad \lim _{N\rightarrow \infty } N^{-1}\sigma _N = 0. \end{aligned}$$

(2.17)

Then, for any $\psi \in {\mathcal {C}}({\mathbb {R}})$ and almost all $t > 0$,

$$\begin{aligned} \lim _{N\rightarrow \infty } {\mathbb {P}}_{\mu _N} \left\{ \left| \frac{1}{N} \sum _{i=0}^N \eta _i (t) \psi \left( \tfrac{i}{N} \right) - \int _0^1 u(t,x)\psi (x) dx \right|>\delta \right\} = 0, \quad \forall \,\delta >0, \end{aligned}$$

(2.18)

where u is the unique entropy solution in Definition 2.1.

Remark 2.9

Below we list two important remarks concerning Theorem 2.8.

• (i) We assume (2.12) and (2.13) only for the uniqueness in Proposition 2.4. If V satisfies only (2.8), our argument proves that the empirical distribution of $\eta ^N$ is tight and all limit points are concentrated on the possible entropy solutions.

• (ii) Observe that the rates in $L_{\textrm{bd}}$ do not appear in the limit. Indeed, let $F_\epsilon (\eta ) = \sum _{0 \le i \le \epsilon N} \eta _i$ be the cumulative mass on $\{\eta _0,\ldots ,\eta _{[\epsilon N]}\}$. Then,

  $$\begin{aligned} L_N F_\epsilon (\eta ) =\,&N\big [c_{\textrm{in}}^- - (c_{\textrm{in}}^-+c_{\textrm{out}}^-)\eta _0\big ] -NJ_{[\epsilon N]}\\&- N(1-p+\sigma _N)(\eta _{[\epsilon N]}-\eta _{[\epsilon N]+1}) + \sum _{1 \le i \le \epsilon N} V_i(\rho _i-\eta _i), \end{aligned}$$

  where $J_i=(2p-1)\eta _i(1-\eta _{i+1})$. From (2.8), $\sum _{1 \le i \le \epsilon N} V_i \gg N$. Hence, to make the contribution of the last term be of order ${\mathcal {O}}(N)$, the mass density of the boundary block $\{1,\ldots ,\epsilon N\}$ should be near to $\rho (0)$.

As a corollary of Theorem 2.8, the regularity of the entropy solution is improved.

Corollary 2.10

Assume (2.12) and (2.13). Let u be the unique entropy solution to (2.7) and (2.9) in Definition 2.1. Then,

$$\begin{aligned}u \in L^\infty ((0,\infty )\times (0,1)) \cap {\mathcal {C}}([0,\infty );L^1),\end{aligned}$$

where $L^1=L^1((0,1))$ is endowed with the weak topology. In particular, the convergence in Theorem 2.8 holds for all $t>0$.

Under (2.12), the macroscopic density near the boundary is prescribed by the reservoir in the following sense: for any $t>0$,

$$\begin{aligned} \lim _{y\rightarrow 0+} \lim _{N\rightarrow \infty } \int _0^t \frac{1}{yN} \sum _{i=0}^{\lfloor yN \rfloor } \eta _i (s)ds = t\rho (0) \quad \text {in }{\mathbb {P}}_{\mu _N}-probability, \end{aligned}$$

(2.19)

and similarly for the right boundary. Indeed, for any $t>0$, $y\in (0,1)$ and $\delta >0$,

$$\begin{aligned} \begin{aligned}&\frac{1}{y}\int _0^t \int _0^y |u(s,x)-\rho (x)|\,dxds\\ \le \,&\frac{1}{4\delta }\int _0^t \int _0^y V(x)(u-\rho )^2dxds + \frac{t\delta }{y^2}\int _0^y \frac{1}{V(x)}dx. \end{aligned} \end{aligned}$$

Taking $y\rightarrow 0+$, (2.11) together with (2.12) yields that

$$\begin{aligned} \lim _{y\rightarrow 0+} \frac{1}{y}\int _0^t \int _0^y |u(s,x)-\rho (x)|\,dxds \le Ct\delta . \end{aligned}$$

As $\delta$ is arbitrary, the limit is 0. Recall that $\rho$ is continuous, so we have

$$\begin{aligned} \lim _{y\rightarrow 0+} \frac{1}{y}\int _0^t \int _0^y |u(s,x)-\rho (0)|\,dxds = 0, \quad \forall \,t>0. \end{aligned}$$

Combining this with Theorem 2.8, we obtain (2.19) for all positive time t. These limits can be derived directly from the microscopic dynamics by imposing a slightly stronger growth condition on V, see the next proposition.

Proposition 2.11

Suppose that V satisfies the following condition:

$$\begin{aligned} \lim _{y\rightarrow 0+} \Big \{y\inf _{x\in (0,y)} V(x)\Big \} = \infty . \end{aligned}$$

(2.20)

Then, (2.19) and the similar limit for the right boundary hold for all $t > 0$.

Example

Fix some $\gamma >0$, $\rho _0$ and $\rho _1\in (0,1)$. By taking

$$\begin{aligned}V(x)=\frac{1}{x^\gamma }+\frac{1}{(1-x)^\gamma }, \quad \rho (x)=\frac{\rho _0(1-x)^\gamma +\rho _1x^\gamma }{(1-x)^\gamma +x^\gamma }, \end{aligned}$$

we obtain the source term given by

$$\begin{aligned}G(x,u)=\frac{u-\rho _0}{x^\gamma } + \frac{u-\rho _1}{(1-x)^\gamma }. \end{aligned}$$

In this case, the dynamics of $L_{\textrm{G}}$ can be interpreted as two infinitely extended reservoirs [4,5,6] placed respectively at the sites $\{-1,-2,\ldots ,\}$ and $\{N+1,N+2,\ldots \}$. When $\gamma \ge 1$, V satisfies (2.12), so the hydrodynamic limit can apply.

2.4 Discussion on the Integrable Case

When $V \in L^1 ((0,1))$, we expect that Theorem 2.8 holds with the entropy solution in Definition 2.7. Since the dynamics of $L_{\textrm{G}}$ is no more dominating at the boundaries, the boundary data $(\alpha ,\beta )$ may depend on $c_{\textrm{in}}^\pm$, $c_{\textrm{out}}^\pm$ as well as V, $\rho$. In particular when $V=0$, we expect that $\alpha$, $\beta$ are determined by

$$\begin{aligned}J(\alpha ) = c_{\textrm{in}}^-(1-\alpha ) - c_{\textrm{out}}^-\alpha , \qquad J(\beta ) = c_{\textrm{out}}^-\beta - c_{\textrm{in}}^+(1-\beta ).\end{aligned}$$

This is proved in [1] for microscopic dynamics without extra symmetric regularization and the special choice of reservoirs such that

$$\begin{aligned}c_{\textrm{in}}^-=p\alpha , \quad c_{\textrm{out}}^-=(1-p)(1-\alpha ), \quad c_{\textrm{in}}^+=(1-p)\beta , \quad c_{\textrm{out}}^+=p(1-\beta ).\end{aligned}$$

We underline that the problem remains open for general reservoirs even when $V=0$.

The situation is easier when further speed-up is imposed on the boundary reservoirs. Let $V \in L^1((0,1))$ satisfy (1.1) and assume the compatibility conditions

$$\begin{aligned}\rho (0) = \frac{c_{\textrm{in}}^-}{c_{\textrm{in}}^- + c_{\textrm{out}}^-}, \qquad \rho (1) = \frac{c_{\textrm{in}}^+}{c_{\textrm{in}}^+ + c_\textrm{out}^+}.\end{aligned}$$

Fix $a>0$ and consider the process generated by $L'_N:= N(L_{\textrm{exc}} + L_{\textrm{G}} + N^aL_{\textrm{bd}})$. In this case, the hydrodynamic equation is still given by (2.7) and (2.9), but the solution should be understood in the sense of Definition 2.7. This can be proved with the argument in [25].

3 Outline of the Proof

Hereafter, we fix an arbitrary $T>0$ and restrict the argument within the finite time horizon [0, T]. Let ${\mathcal {M}}_+([0,1])$ be the space of finite, positive Radon measures on [0, 1], endowed with the weak topology. Define the empirical distribution $\pi ^N=\pi ^N(t,dx)$ as

$$\begin{aligned} \pi ^N(t,dx):= \frac{1}{N}\sum _{i=0}^N \eta _i(t)\delta _{\frac{i}{N}}(dx), \quad \forall \,t\in [0,T], \end{aligned}$$

(3.1)

where $\delta _u(dx)$ stands for the Dirac measure at u. Denote by ${\mathcal {D}}={\mathcal {D}}([0,T];{\mathcal {M}}_+([0,1]))$ the space of càdlàg paths on ${\mathcal {M}}_+([0,1])$ endowed with the Skorokhod topology. To prove Theorem 2.8, it suffices to show that the distribution of $\pi ^N$ on ${\mathcal {D}}$ converges weakly as $N\rightarrow \infty$ and the limit is concentrated on the single path $\pi (t,dx)=u(t,x)dx$. However, to formulate the evolution equation (2.7) of u we need a type of convergence that also applies to nonlinear functions. The idea is to introduce the Young measure corresponding to the mesoscopic block average, cf. [12, 13, 22] and [14, Chapter 8].

Let $\Sigma _T=(0,T)\times (0,1)$. Recall that a Young measure on $\Sigma _T$ is a measurable map $\nu :\Sigma _T\rightarrow {\mathcal {P}}({\mathbb {R}})$, where ${\mathcal {P}}({\mathbb {R}})$ is the space of probability measures on ${\mathbb {R}}$ endowed with the topology defined by the weak convergence. Denote by ${\mathcal {Y}}={\mathcal {Y}}(\Sigma _T)$ the set of all Young measures on $\Sigma _T$, and by $\nu =\{\nu _{t,x};(t,x)\in \Sigma _T\}$ the element in ${\mathcal {Y}}$. A sequence $\{\nu ^n;n\ge 1\}$ of Young measures is said to converge to $\nu \in {\mathcal {Y}}$ if for any bounded and continuous function f on $\Sigma _T\times {\mathbb {R}}$,

$$\begin{aligned} \lim _{n\rightarrow \infty } \iint _{\Sigma _T} dxdt \int _{\mathbb {R}}f(t,x,\lambda )\nu _{t,x}^n(d\lambda ) = \iint _{\Sigma _T} dxdt \int _{\mathbb {R}}f(t,x,\lambda )\nu _{t,x}(d\lambda ). \end{aligned}$$

(3.2)

Any measurable function u on $\Sigma _T$ is naturally viewed as a Young measure:

$$\begin{aligned} \nu _{t,x}(d\lambda ):=\delta _{u(t,x)}(d\lambda ), \quad \forall \,(t,x)\in \Sigma _T. \end{aligned}$$

(3.3)

Denote by ${\mathcal {Y}}_d$ the set of all $\nu \in {\mathcal {Y}}$ of this kind.

Hereafter, we fix some mesoscopic scale $K = K (N)$ such that

$$\begin{aligned} K \ll \sigma _N, \quad N \sigma _N \ll K^3, \quad \sigma _N^2 \ll NK. \end{aligned}$$

(3.4)

The existence of such K is guaranteed by (2.17). For $\eta \in \Omega _N$ and $i=K$,..., $N-K$, define the smoothly weighted block average as

$$\begin{aligned} {\hat{\eta }}_{i,K}:= \sum _{j=-K+1}^{K-1} w_{j} \eta _{i-j}, \quad w_j:= \frac{K-|j|}{K^2}. \end{aligned}$$

(3.5)

Consider the space-time empirical density

$$\begin{aligned} u^N(t,x):= \sum _{i=K}^{N-K} {\hat{\eta }}_{i,K}(t)\chi _{N,i}(x), \quad \forall \,(t,x)\in \Sigma _T, \end{aligned}$$

(3.6)

where $\chi _{N,i} (\cdot )$ is the indicator function of the interval $[\tfrac{i}{N} - \tfrac{1}{2N}, \tfrac{i}{N} + \tfrac{1}{2N})$.

Lemma 3.1

(Tightness) Let ${\mathbb {Q}}_N$ be the distribution of $(\pi ^N,\nu ^N)$, where $\pi ^N$ is defined in (3.1) and $\nu ^N$ is the Young measure corresponding to $u^N$ in (3.6) in the sense of (3.3). Then, the sequence of ${\mathbb {Q}}_N$ is tight with respect to the product topology on $\mathcal D\times {\mathcal {Y}}$.

Let ${\mathbb {Q}}$ be a limit point of ${\mathbb {Q}}_N$. With some abuse of notations, we denote the subsequence converging to ${\mathbb {Q}}$ still by ${\mathbb {Q}}_N$. Below we characterize ${\mathbb {Q}}$ by three propositions.

Proposition 3.2

The following holds for ${\mathbb {Q}}$-almost every $(\pi ,\nu )$.

(i):

$\pi (t,dx)=\varpi (t,x)dx$ for every $t\in [0,T]$ with some $\varpi (t,\cdot ) \in L^1((0,1))$, and $t\mapsto \varpi (t,\cdot )$ is a continuous map with respect to the weak topology of $L^1$.

(ii):

$\nu _{t,x}([0,1])=1$ for almost all $(t,x)\in \Sigma _T$.

(iii):

$\varpi (t,x)=\int \lambda \nu _{t,x}(d\lambda )$ for almost all $(t,x)\in \Sigma _T$.

Proposition 3.3

${\mathbb {Q}}({\mathcal {D}} \times {\mathcal {Y}}_d)=1$, where ${\mathcal {Y}}_d$ is the set of delta-Young measures in (3.3).

To state the last proposition, define the entropy production

$$\begin{aligned} X^{(f,q)} (\nu ,\varphi ):= - \iint _{\Sigma _T} dxdt \left[ \partial _t\varphi \int _{\mathbb {R}}f\,d\nu _{t,x} + \partial _x\varphi \int _{\mathbb {R}}q\,d\nu _{t,x} \right] , \end{aligned}$$

(3.7)

for $\nu \in {\mathcal {Y}}$, $\varphi \in {\mathcal {C}}^1 ({\mathbb {R}}^2)$ and Lax entropy–flux pair (f, q).

Proposition 3.4

It holds ${\mathbb {Q}}$-almost surely that

$$\begin{aligned} X^{(f,q)}(\nu ,\varphi ) + \iint _{\Sigma _T} dxdt \left[ \varphi \int _{\mathbb {R}}f'(\lambda )G(x,\lambda )\nu _{t,x}(d\lambda ) \right] \le \int _0^1 f(u_0)\varphi (0,x)dx, \end{aligned}$$

(3.8)

for any Lax entropy–flux pair (f, q) and any $\varphi \in \mathcal C_c^2([0,T)\times (0,1))$ such that $\varphi \ge 0$, where $G(x,\lambda )=V(x)[\lambda -\rho (x)]$.

Remark 3.5

Similarly to Remark 2.2, one can obtain a measure-valued energy bound: it holds ${\mathbb {Q}}$-almost surely that

$$\begin{aligned} \iint _{\Sigma _T} dxdt \left[ V(x)\int _{\mathbb {R}}\big [\lambda -\rho (x)\big ]^2d\nu _{t,x} \right] < \infty . \end{aligned}$$

(3.9)

This can be derived directly from the microscopic dynamics, see Sect. 6.

Remark 3.6

The arguments we used to prove Lemma 3.1 and Proposition 3.2, 3.3 and 3.4 do apply to all $V \in {\mathcal {C}}^1((0,1))$, bounded or unbounded. However, only in the non-integrable case are they sufficient to identify the limit equation.

We organize the remaining contents as follows. Lemma 3.1 and Proposition 3.2 are proved in Sect. 4. Proposition 3.3 is proved in Sect. 5 and Proposition 3.4 is proved in Sect. 6.1. With these results, the proofs of Theorem 2.8 and Corollary 2.10 are straightforward and are stated right below. The direct proofs of Proposition 2.11 and (3.9) using the relative entropy method [27] are stated in Sect. 6.2 and 6.3.

Proof of Theorem 2.8 and Corollary 2.10

Recall that ${\mathbb {Q}}$ is a probability measure on $\mathcal D\times {\mathcal {Y}}$. In view of Proposition 3.2 (i) and 3.3, $\pi (t)=\varpi (t,x)dx$ and $\nu _{t,x}=\delta _{u(t,x)}$, ${\mathbb {Q}}$-almost surely. Proposition 3.2 (iii) then yields that ${\mathbb {Q}}$ is concentrated on the trajectories such that $\varpi =u$.

To prove Theorem 2.8, we need to show that u, and hence $\varpi$, is the entropy solution to (2.7) and (2.9). By Proposition 3.2 (ii), $u(t,x)\in [0,1]$ so that $u \in L^\infty (\Sigma _T)$. Furthermore, by substituting $\nu _{t,x}=\delta _{u(t,x)}(d\lambda )$ in Proposition 3.4, we obtain that u satisfies the generalized entropy inequality in Definition 2.1. The proof is then concluded by the uniqueness of the entropy solution, see Proposition 2.4.

Finally, Corollary 2.10 follows directly from the argument above and the sample path regularity of $\varpi$ obtained in Proposition 3.2 (i). $\square$

We close this section with some useful notations. For a function $\varphi =\varphi (t,x)$, let

$$\begin{aligned} \varphi _i(t):= \varphi \left( t,\tfrac{i}{N} - \tfrac{1}{2N} \right) , \quad {\bar{\varphi }}_i(t):= N \int _{\tfrac{i}{N} - \tfrac{1}{2N}}^{\tfrac{i}{N} + \tfrac{1}{2N}} \varphi (t,x) dx. \end{aligned}$$

(3.10)

Recall the mesoscopic scale $K=K(N)$ in (3.4). For a sequence $\{a_i;i=0,\ldots ,N\}$, ${\hat{a}}_{i,K}$ stands for the smoothly weighted average in (3.5). Since K is fixed through the paper, we write ${\hat{a}}_i$ when there is no confusion. We shall frequently use the notions of discrete gradient and Laplacian operators, which are defined as usual:

$$\begin{aligned} \nabla a_i =a_{i+1} - a_i, \quad \nabla ^* a_i = a_{i-1} - a_i, \quad \Delta a_i = a_{i+1} - 2 a_i + a_{i-1}. \end{aligned}$$

(3.11)

Notice that $\Delta =-\nabla \nabla ^*=-\nabla ^*\nabla$.

4 Tightness

Recall that ${\mathbb {Q}}_N$ is the distribution of $(\pi ^N,\nu ^N)$ on ${\mathcal {D}}\times {\mathcal {Y}}$. Here ${\mathcal {D}}=\mathcal D([0,T],{\mathcal {M}}_+([0,1]))$ is the space of càdlàg paths endowed with the Skorokhod topology and ${\mathcal {Y}}=\mathcal Y(\Sigma _T)$ is the space of Young measures endowed with the topology defined by (3.2).

Proof of Lemma 3.1

It suffices to show that both $\{\pi ^N\}$ and $\{\nu ^N\}$ are tight. The coordinate $\nu ^N$ is easy. Since $u^N\in [0,1]$, for all N we have ${\mathbb {Q}}_N\{\nu ^N\in {\mathcal {Y}}_*\}=1$, where

$$\begin{aligned}{\mathcal {Y}}_*:= \big \{\nu \in {\mathcal {Y}} \,\big |\, \nu _{t,x}([0,1])=1 \text { for all } (t,x)\in \Sigma _T\big \}.\end{aligned}$$

Because ${\mathcal {Y}}_*$ is compact in ${\mathcal {Y}}$, $\{\nu ^N\}$ is tight.

For the coordinate $\pi ^N$, by [14, Chapter 4, Theorem 1.3 & Proposition 1.7], we only need to show that for any $\psi \in {\mathcal {C}}([0,1])$, some constant $C_\psi$ and any $\delta >0$,

$$\begin{aligned}&\sup _N {\mathbb {Q}}_N \left\{ \sup _{t\in [0,T]} \big |\langle \pi ^N(t),\psi \rangle \big | < C_\psi \right\} = 1,\end{aligned}$$

(4.1)

$$\begin{aligned}&\lim _{\varepsilon \downarrow 0} \lim _{N\rightarrow \infty } {\mathbb {Q}}_N \left\{ \sup _{|t-s|<\varepsilon } \big |\langle \pi ^N(t),\psi \rangle - \langle \pi ^N(s),\psi \rangle \big | > \delta \right\} = 0, \end{aligned}$$

(4.2)

where $\langle \,\cdot ,\cdot \,\rangle$ is the scalar product between ${\mathcal {M}}_+$ and ${\mathcal {C}}([0,1])$. The first one is obvious:

$$\begin{aligned} \big |\langle \pi ^N(t),\psi \rangle \big | = \frac{1}{N} \left| \sum _{i=0}^N \eta _i(t)\psi \left( \tfrac{i}{N} \right) \right| \le \frac{1}{N}\sum _{i=0}^N \left| \psi \left( \tfrac{i}{N} \right) \right| , \quad \forall \,t\in [0,T]. \end{aligned}$$

(4.3)

For the second one, choose $\psi _*\in {\mathcal {C}}_c^2((0,1))$ such that $\Vert \psi -\psi _*\Vert _{L^1}<4^{-1}\delta$. Note that

$$\begin{aligned}\big |\langle \pi ^N(t),\psi -\psi _* \rangle - \langle \pi ^N(s),\psi -\psi _* \rangle \big | \le \frac{1}{N}\sum _{i=0}^N \left| \psi \left( \tfrac{i}{N} \right) - \psi _* \left( \tfrac{i}{N} \right) \right| < \frac{\delta }{2},\end{aligned}$$

uniformly in s, t and all sample paths. Hence, it suffices to show (4.2) with $\psi$ replaced by $\psi _*$. Without loss of generality, let $s<t$. Then,

$$\begin{aligned} \langle \pi ^N(t),\psi _* \rangle - \langle \pi ^N(s),\psi _* \rangle = \int _s^t L_N\big [\langle \pi ^N(r),\psi _* \rangle \big ]dr + M_{N,\psi _*}(t-s), \end{aligned}$$

(4.4)

where $M_{N,\psi _*}$ is the Dynkin’s martingale. As $\psi _*$ is compactly supported, $\eta _0$ or $\eta _N$ does not appear in $\langle \pi ^N(r),\psi _* \rangle$, so that

$$\begin{aligned} L_N\big [\langle \pi ^N,\psi _* \rangle \big ] =\,&(1-p+\sigma _N) \sum _{i=1}^{N-1} \eta _i \left[ \psi _* \left( \tfrac{i+1}{N} \right) + \psi _* \left( \tfrac{i-1}{N} \right) - 2\psi _* \left( \tfrac{i}{N} \right) \right] \\&+\,(2p-1)\sum _{i=0}^{N-1} \eta _i(1-\eta _{i+1}) \left[ \psi _* \left( \tfrac{i+1}{N} \right) - \psi _* \left( \tfrac{i}{N} \right) \right] \\&+\,\frac{1}{N}\sum _{i=0}^N V \left( \tfrac{i}{N} \right) \left[ \eta _i-\rho \left( \tfrac{i}{N} \right) \right] \psi _* \left( \tfrac{i}{N} \right) . \end{aligned}$$

Using the fact that $\sigma _N \ll N$ and $\psi _*\in \mathcal C_c^2((0,1))$, $L_N[\langle \pi ^N,\psi _* \rangle ]$ is uniformly bounded. Therefore, the first term in (4.4) vanishes uniformly when $|t-s|\rightarrow 0$. We are left with the martingale in (4.4). Dy Dynkin’s formula, the quadratic variation reads

$$\begin{aligned}\langle M_{N,\psi _*} \rangle (t-s) = \int _s^t \Big (L_N\big [\langle \pi ^N,\psi _* \rangle ^2\big ] - 2 \langle \pi ^N,\psi _* \rangle L_N\big [\langle \pi ^N,\psi _* \rangle \big ]\Big )\,dr.\end{aligned}$$

Recall that $\psi _*\in {\mathcal {C}}_c^2((0,1))$, direct calculation shows that

$$\begin{aligned} L_N\big [\langle \pi ^N,\psi _* \rangle ^2\big ] - 2 \langle \pi ^N,\psi _* \rangle L_N\big [\langle \pi ^N,\psi _* \rangle \big ] = \frac{1}{N^2}\sum _{i=0}^{N-1} c_{i,\textrm{G}} (\eta ) (1-2\eta _i)^2 \psi _*^2 \left( \tfrac{i}{N} \right) \\ +\,\frac{1}{N}\sum _{i=0}^{N-1} \left( c_{i,i+1} (\eta ) + \frac{\sigma _N}{2} \right) (\eta _i - \eta _{i+1})^2 \left[ \psi _* \left( \tfrac{i+1}{N} \right) - \psi _* \left( \tfrac{i}{N} \right) \right] ^2. \end{aligned}$$

By (2.17), it is bounded from above by $CN^{-1}$. Therefore,

$$\begin{aligned} E^{{\mathbb {Q}}_N} \big [|\langle M_{N,\psi _*} \rangle (t-s)|\big ] \le C(t-s)N^{-1}. \end{aligned}$$

(4.5)

We only need to apply Doob’s inequality.

Proof of Proposition 3.2

For (i), notice that (4.3) and the weak convergence yield that for any fixed $\psi \in {\mathcal {C}}([0,1])$, ${\mathbb {Q}}\{\sup _{t\in [0,T]} |\langle \pi (t),\psi \rangle | \le C\Vert \psi \Vert _{L^1}\}=1$. Since ${\mathcal {C}}([0,1])$ is separable, by standard density argument it holds ${\mathbb {Q}}$-almost surely that

$$\begin{aligned}\sup _{t\in [0,T]} \big |\langle \pi (t),\psi \rangle \big | \le C\Vert \psi \Vert _{L^1}, \quad \forall \,\psi \in {\mathcal {C}}([0,1]).\end{aligned}$$

So $\pi (t)$ is absolutely continuous with respect to the Lebesgue measure on [0, 1] and can then be written as $\varpi (t,x)dx$. Moreover, (4.2) assures that ${\mathbb {Q}}$ is concentrated on continuous paths, see [14, Chapter 4, Remark 1.5]. The continuity of $t \mapsto \varpi (t,\cdot )$ is then proved.

Since (ii) is a direct result from the proof of the tightness of $\{\nu ^N\}$, we are left with (iii). Pick $\varphi \in {\mathcal {C}}^1(\Sigma _T)$. From the definition of $\nu _{t,x}^N$,

$$\begin{aligned}\int _0^1 \varphi (t,x)dx \left[ \int _0^1 \lambda \nu _{t,x}^N(d\lambda ) \right] = \frac{1}{N}\sum _{i=K}^{N-K} {\hat{\eta }}_{i,K}(t){{\bar{\varphi }}}_i(t),\end{aligned}$$

where ${{\bar{\varphi }}}_i(t)$ is given by (3.10). By the regularity of $\varphi$,

$$\begin{aligned}\left| \iint _{\Sigma _T} dxdt \left[ \varphi \int _0^1 \lambda \nu _{t,x}^N(d\lambda ) \right] - \int _0^T \langle \pi ^N,\varphi (t,\cdot ) \rangle dt \right| \le \frac{C_\varphi TK}{N}.\end{aligned}$$

Since $K = K(N) \ll N$, the right-hand side above vanishes as $N\rightarrow \infty$. From the weak convergence, this yields that for any $\delta >0$ and fixed $\varphi \in {\mathcal {C}}^1(\Sigma _T)$,

$$\begin{aligned} {\mathbb {Q}}\left\{ \left| \iint _{\Sigma _T} dxdt \left[ \varphi \int _0^1 \lambda \nu _{t,x}(d\lambda ) \right] - \int _0^T \langle \pi ,\varphi (t,\cdot ) \rangle dt \right| > \delta \right\} = 0. \end{aligned}$$

By choosing a countable and dense subset of ${\mathcal {C}}^1(\Sigma _T)$ and applying (i),

$$\begin{aligned} {\mathbb {Q}}\left\{ \iint _{\Sigma _T} \varphi \left[ \int _0^1 \lambda \nu _{t,x}(d\lambda ) - \varpi \right] dxdt = 0, \ \forall \,\varphi \in {\mathcal {C}}^1(\Sigma _T) \right\} = 1. \end{aligned}$$

The conclusion in (iii) then follows. $\square$

5 Compensated Compactness

Given a Lax entropy–flux pair (f, q), recall the entropy production defined in (3.7). To simplify the notations, we will write $X(\nu ,\varphi )$ when the choice of (f, q) is clear. Without loss of generality, we also fix $p=1$ to shorten the formulas.

This section is devoted to the proof of Proposition 3.3. Note that ${\mathcal {Y}}_d$, the subset of delta-type Young measures, is not closed in ${\mathcal {Y}}$, so ${\mathbb {Q}}_N({\mathcal {Y}}_d)=1$ for all N does not guarantee that ${\mathbb {Q}}({\mathcal {Y}}_d)=1$. From [12, Proposition 2.1 & Lemma 5.1], Proposition 3.3 follows from the next result, see also [13, Section 5.6].

Proposition 5.1

Fix an arbitrary Lax entropy–flux pair (f, q). Let $\varphi =\phi \psi$ with $\phi \in {\mathcal {C}}_c^\infty (\Sigma _T)$ and $\psi \in {\mathcal {C}}^\infty ({\mathbb {R}}^2)$. Then, we have the following decomposition:

$$\begin{aligned}X (\nu ^N,\varphi ) = Y_N (\varphi ) + Z_N (\varphi ),\end{aligned}$$

and there exist random variables $A_{N,\phi }$, $B_{N,\phi }$ independent of $\psi$, such that

$$\begin{aligned}&|Y_N(\varphi )| \le A_{N,\phi }\Vert \psi \Vert _{H^1}, \quad \limsup _{N\rightarrow \infty } {\mathbb {E}}_{\mu _N} [A_{N,\phi }] = 0;\end{aligned}$$

(5.1)

$$\begin{aligned}&|Z_N(\varphi )| \le B_{N,\phi }\Vert \psi \Vert _{L^\infty }, \quad \sup _{N\ge 1} {\mathbb {E}}_{\mu _N} [B_{N,\phi }] < \infty . \end{aligned}$$

(5.2)

Here, $\Vert \cdot \Vert _{H^1}$ and $\Vert \cdot \Vert _{L^\infty }$ are the $H^1$- and $L^\infty$-norm computed on $\Sigma _T=(0,T)\times (0,1)$.

Remark 5.2

Proposition 5.1 is a microscopic (stochastic) synthesis of the Murat–Tartar theory established in [19, 23]. The role of the function $\phi \in \mathcal C_c^\infty (\Sigma _T)$ is to localize the estimates in (5.1) and (5.2) away from the boundaries where the potential V is unbounded.

5.1 Basic Decomposition

We first prove a basic decomposition for the microscopic entropy production. Given a bounded function $\varphi =\varphi (t,x)$, recall the notations $\varphi _i=\varphi _i(t)$ and ${{\bar{\varphi }}}_i={{\bar{\varphi }}}_i(t)$ defined in (3.10). Denote by ${\hat{\eta _i}}(t)={\hat{\eta }}_{i,K}(t)$, ${\hat{J}}_i(t)={\hat{J}}_{i,K}(t)$ and ${\hat{G}}_i(t)={\hat{G}}_{i,K}(t)$ the smoothly weighted averaged averages introduced in (3.5) of $\eta _i(t)$, $J_i(t)=\eta _i(t)(1-\eta _{i+1}(t))$ and $G_i(t)=V_i(\eta _i(t)-\rho _i)$. We shall abbreviate them to ${\hat{\eta _i}}$, ${\hat{J}}_i$ and ${\hat{G}}_i$ when there is no confusion.

Lemma 5.3

Fix a Lax entropy–flux pair (f, q). For $\varphi \in \mathcal C^1({\mathbb {R}}^2)$ such that $\varphi (T,\cdot )=0$,

$$\begin{aligned} X(\nu ^N,\varphi ) = \frac{1}{N} \sum _{i=K}^{N-K} f ({\hat{\eta }}_i (0)) {\bar{\varphi }}_i (0) + {\mathcal {A}}_N + {\mathcal {S}}_N + {\mathcal {G}}_N + {\mathcal {M}}_N + {\mathcal {E}}_N. \end{aligned}$$

(5.3)

The terms in (5.3) are defined below. ${\mathcal {A}}_N$, ${\mathcal {S}}_N$ and ${\mathcal {G}}_N$ are given by

$$\begin{aligned}&{\mathcal {A}}_N={\mathcal {A}}_N (\varphi ) :=\int _0^T \sum _{i=K}^{N-K} {\bar{\varphi }}_i(t) f^\prime ({\hat{\eta }}_{i}) \nabla ^* \left[ \hat{J}_{i} - J ({\hat{\eta }}_{i}) \right] dt,\\&{\mathcal {S}}_N={\mathcal {S}}_N (\varphi ) := \sigma _N \int _0^T \sum _{i=K}^{N-K} {\bar{\varphi }}_i(t) f^\prime ({\hat{\eta }}_{i}) \Delta {\hat{\eta }}_{i} dt,\\&{\mathcal {G}}_N={\mathcal {G}}_N (\varphi ) := -\int _0^T \frac{1}{N} \sum _{i=K}^{N-K} {{\bar{\varphi }}}_i(t) f' ({\hat{\eta }}_{i}) {\hat{G}}_i\,dt. \end{aligned}$$

${\mathcal {M}}_N$ is a martingale given by

$$\begin{aligned}{\mathcal {M}}_N={\mathcal {M}}_N (\varphi ):= -\int _0^T \frac{1}{N} \sum _{i=K}^{N-K} {\bar{\varphi }}'_i(t)M_i(t)dt,\end{aligned}$$

where $M_i=M_i(t)$ is the Dynkin’s martingale associated with $f({\hat{\eta _i}}(t))$, see (5.5) below. Finally, ${\mathcal {E}}_N={\mathcal {E}}_{N,1}+{\mathcal {E}}_{N,2}+{\mathcal {E}}_{N,3}$ is defined through

$$\begin{aligned} {\mathcal {E}}_{N,1}={\mathcal {E}}_{N,1} (\varphi )&:= -f(0)\iint _{B_{T,N}} \partial _t\varphi \,dxdt - q(0)\iint _{B_{T,N}} \partial _x\varphi \,dxdt,\\ {\mathcal {E}}_{N,2}={\mathcal {E}}_{N,2} (\varphi )&:= \int _0^T \sum _{i=K}^{N-K} \big [ {\bar{\varphi }}_i \nabla ^* q ({\hat{\eta }}_i) - q ({\hat{\eta }}_i) \nabla \varphi _i \big ] dt,\\ {\mathcal {E}}_{N,3}={\mathcal {E}}_{N,3} (\varphi )&:= \int _0^T \frac{1}{N}\sum _{i=K}^{N-K} {\bar{\varphi }}_i \left[ \epsilon _{i,K}^{(1)} + \epsilon _{i,K}^{(2)} \right] dt, \end{aligned}$$

where $B_{T,N}$ is the region $t\in [0,T]$ and $x\in [0,\tfrac{2K-1}{2N})\cup [1-\tfrac{2K-1}{2N},1]$,

$$\begin{aligned}\epsilon _{i,K}^{(1)} = L_N f ({\hat{\eta }}_i) - f^\prime ({\hat{\eta }}_i) L_N {\hat{\eta }}_i, \quad \epsilon _{i,K}^{(2)} = N\big [f^\prime ({\hat{\eta }}_i)\nabla ^* J({\hat{\eta }}_i) - \nabla ^* q ({\hat{\eta }}_i)\big ].\end{aligned}$$

Proof

By the definition of $\nu ^N$,

$$\begin{aligned} X(\nu ^N,\varphi ) = - \int _0^T \frac{1}{N} \sum _{i=K}^{N-K} f ({\hat{\eta }}_i) {\bar{\varphi }}_i^\prime \,dt - \int _0^T \sum _{i=K}^{N-K} q ({\hat{\eta }}_i)\nabla \varphi _i\,dt + {\mathcal {E}}_{N,1} (\varphi ). \end{aligned}$$

(5.4)

By Dynkin’s formula, for $K \le i \le N-K$,

$$\begin{aligned} M_i (t):= f ({\hat{\eta }}_i (t)) - f ({\hat{\eta }}_i (0)) - \int _0^t L_N \big [f ({\hat{\eta }}_i (s))\big ] ds, \quad t\in [0,T], \end{aligned}$$

(5.5)

defines a martingale. Since $\varphi$ vanishes at $t=T$, ${\mathcal {M}}_N(\varphi )$ satisfies that

$$\begin{aligned} \begin{aligned} - \int _0^T \frac{1}{N}\sum _{i=K}^{N-K} f({\hat{\eta _i}}){{\bar{\varphi }}}'_i\,dt =\,&\frac{1}{N}\sum _{i=K}^{N-K} f({\hat{\eta _i}}(0)){{\bar{\varphi }}}_i(0)\\&+ \int _0^T \frac{1}{N}\sum _{i=K}^{N-K} {{\bar{\varphi }}}_i L_N \big [f({\hat{\eta _i}})\big ]dt + {\mathcal {M}}_N(\varphi ). \end{aligned} \end{aligned}$$

(5.6)

Recall the definition of $L_N$ in (2.5). Note that, for $K \le i \le N-K$, ${\hat{\eta }}_i$ does not depend on $\eta _{0}$ or $\eta _{N}$, and thus $L_{\textrm{bd}} {\hat{\eta }}_i = L_\textrm{bd} {\hat{\eta }}_i = 0$. Notice that

$$\begin{aligned}L_N [{\hat{\eta }}_i] = N \nabla ^* \hat{J}_i + N\sigma _N \Delta {\hat{\eta }}_i - {\hat{G}}_i, \quad \forall \,i=K,\ldots ,N-K.\end{aligned}$$

Therefore, for $i=K$,..., $N-K$, $L_N[f({\hat{\eta _i}})]$ is equal to

$$\begin{aligned} \epsilon _{i,K}^{(1)} + f^\prime ({\hat{\eta }}_i)&\left( N \nabla ^* \hat{J}_i + N\sigma _N \Delta {\hat{\eta }}_i - {\hat{G}}_i \right) = \epsilon _{i,K}^{(1)} + \epsilon _{i,K}^{(2)}\\&+\,Nf^\prime ({\hat{\eta }}_i) \nabla ^* \left[ \hat{J}_i-J({\hat{\eta _i}}) \right] + N \nabla ^* q({\hat{\eta _i}}) + f^\prime ({\hat{\eta }}_i) \left( N\sigma _N \Delta {\hat{\eta }}_i - {\hat{G}}_i \right) . \end{aligned}$$

From the above formula of $L_N[f({\hat{\eta _i}})]$,

$$\begin{aligned} \begin{aligned} \int _0^T \frac{1}{N}\sum _{i=K}^{N-K} {{\bar{\varphi }}}_i L_N \big [f({\hat{\eta _i}})\big ]dt =&\int _0^T \sum _{i=K}^{N-K} {{\bar{\varphi }}}_i\nabla ^*q({\hat{\eta _i}})\,dt \\&+\,{\mathcal {A}}_N(\varphi ) + {\mathcal {S}}_N(\varphi ) + {\mathcal {G}}_N(\varphi ) + {\mathcal {E}}_{N,3}(\varphi ). \end{aligned} \end{aligned}$$

(5.7)

We then conclude the proof by inserting (5.6) and (5.7) into the first term on the right-hand side of (5.4). $\square$

5.2 Dirichlet Forms

Given a function $\alpha :[0,1]\rightarrow (0,1)$, denote by $\nu ^N_{\alpha (\cdot )}$ the product measure on $\Omega _N$ with marginals

$$\begin{aligned}\nu ^N_{\alpha (\cdot )} (\eta (i)=1) = \alpha (i/N), \quad \forall \,i = 0,1, \ldots ,N.\end{aligned}$$

When $\alpha (\cdot ) \equiv \alpha$ is a constant, we shorten the notation as $\nu ^N_{\alpha (\cdot )} = \nu ^N_{\alpha }$. Given two probability measures $\nu$ and $\mu$ on $\Omega _N$, let $f:=\mu /\nu$ be the density function. Define

$$\begin{aligned} D_{\textrm{exc}}^N (\mu ;\nu )&:= \frac{1}{2} \sum _{\eta \in \Omega _N} \sum _{i=0}^{N-1} \Big ( \sqrt{f(\eta ^{i,i+1})} - \sqrt{f(\eta )} \Big )^2 \nu (\eta ), \end{aligned}$$

(5.8)

$$\begin{aligned} D_{\textrm{G}}^N (\mu ;\nu )&:= \frac{1}{2N} \sum _{\eta \in \Omega _N} \sum _{i=1}^{N-1} c_{i,\textrm{G}} (\eta ) \Big ( \sqrt{f(\eta ^{i})} - \sqrt{f(\eta )} \Big )^2 \nu (\eta ),\end{aligned}$$

(5.9)

$$\begin{aligned} D_-^N (\mu ;\nu )&:= \frac{1}{2} \sum _{\eta \in \Omega _N} c_0 (\eta ) \Big ( \sqrt{f(\eta ^{0})} - \sqrt{f(\eta )} \Big )^2 \nu (\eta ),\end{aligned}$$

(5.10)

$$\begin{aligned} D_+^N (\mu ;\nu )&:= \frac{1}{2} \sum _{\eta \in \Omega _N} c_N (\eta ) \Big ( \sqrt{f(\eta ^{N})} - \sqrt{f(\eta )} \Big )^2 \nu (\eta ), \end{aligned}$$

(5.11)

with $c_{i,\textrm{G}}$ in (2.2) and $c_0$, $c_N$ in (2.3). Note that $D_-^N\equiv 0$ if $c_{\textrm{in}}^-=c_\textrm{out}^-=0$, and similarly for $D_+^N$. Let $\mu _t^N$ be the distribution of the process at time t. Define

$$\begin{aligned}D_{\textrm{exc}}^N(t) = D_{\textrm{exc}}^N (\mu _t^N;\nu _{\frac{1}{2}}^N), \quad D_{\textrm{G}}^N(t) = D_{\textrm{G}}^N(\mu _t^N;\nu _{\rho (\cdot )}^N), \quad \forall \,t\ge 0.\end{aligned}$$

Let $c_-=c_{\textrm{in}}^-(c_{\textrm{in}}^-+c_\textrm{out}^-)^{-1}$ when at least one of $c_{\textrm{in}}^-$ and $c_\textrm{out}^-$ is positive. Define

$$\begin{aligned}D_-^N(t) = D_-^N (\mu _t^N;\nu _{c_-}^N), \quad \forall \,t\ge 0.\end{aligned}$$

If $c_{\textrm{in}}^-=c_{\textrm{out}}^-=0$ we fix $D_-^N(t)\equiv 0$. Let $c_+$ and $D_+^N(t)$ be defined similarly.

Lemma 5.4

For any $t>0$, there is a constant C independent of N, such that

$$\begin{aligned}\int _0^t \Big [\sigma _ND_{\textrm{exc}}^N(s) + D_{\textrm{G}}^N(s) + D_-^N(s) + D_+^N(s)\Big ]\,ds \le C.\end{aligned}$$

Proof

Let $\nu =\nu _\alpha ^N$ with $\alpha \equiv \tfrac{1}{2}$. For a probability measure $\mu =f\nu$ on $\Omega _N$, from the calculation in Appendix A,

$$\begin{aligned} \big \langle f, L_{\textrm{exc}}[\log f] \big \rangle _\nu&\le -(\sigma _N+1)D_{\textrm{exc}}^N(\mu ;\nu ) + C,\end{aligned}$$

(5.12)

$$\begin{aligned} \big \langle f,L_{\textrm{G}}[\log f] \big \rangle _\nu&\le -2D_{\textrm{G}}^N(\mu ;\nu _{\rho (\cdot )}^N) + \frac{1}{N}\sum _{i=1}^{N-1} V_i \log \left( \tfrac{\rho _i}{1-\rho _i} \right) (\rho _i-E^\mu [\eta _i]),\end{aligned}$$

(5.13)

$$\begin{aligned} \big \langle f, L_{\textrm{bd}}[\log f] \big \rangle _\nu&\le -2\big [D_-^N(\mu ;\nu _{c_-}^N) + D_+^N(\mu ;\nu _{c_+}^N)\big ] + C. \end{aligned}$$

(5.14)

For $s\in [0,t]$, let $f_s^N:=\mu _s^N/\nu$, then

$$\begin{aligned} \big \langle f_s^N,L_N \log f_s^N \big \rangle _\nu \le \,&- 2N\sigma _ND^N_{\textrm{exc}}(s) - 2ND_{\textrm{G}}^N(s) - 2N\big [D_-^N(s)+D_+^N(s)\big ]\\&+ \sum _{i=1}^{N-1} V_i \log \left( \tfrac{\rho _i}{1-\rho _i} \right) \big (\rho _i-{\mathbb {E}}_{\mu _N} [\eta _i(s)]\big ) + CN. \end{aligned}$$

Applying Lemma 5.5 below, we obtain the estimate

$$\begin{aligned}\int _0^t \Big [\sigma _ND_{\textrm{exc}}^N(s) + D_{\textrm{G}}^N(s) + D_-^N(s) + D_+^N(s)\Big ]\,ds \le \int _0^t \frac{\langle f_s^N,-L_N [\log f_s^N] \rangle _\nu }{2N}\,ds + C.\end{aligned}$$

Standard manipulation gives that

$$\begin{aligned}\int _0^t \big \langle f_s^N,-L_N [\log f_s^N] \big \rangle _\nu \,ds = \sum _{\eta \in \Omega _N} \log [f_0^N(\eta )]\mu _0^N(\eta ) - \sum _{\eta \in \Omega _N} \log [f_t^N(\eta )]\mu _t^N(\eta )\end{aligned}$$

is bounded by $C'N$, so we conclude the proof. $\square$

The following a priori bound is used in the previous proof.

Lemma 5.5

Suppose that $a\in {\mathcal {C}}^1([0,1])$ has Lipschitz continuous derivative. Let $a_i=a_i^N=a(\tfrac{i}{N})$. Then, there is a constant C independent of N, such that

$$\begin{aligned}{\mathbb {E}}_{\mu _N} \left[ \int _0^t \frac{1}{N}\sum _{i=1}^{N-1} V_ia_i\big (\rho _i-\eta _i(s)\big )\,ds \right] \le C.\end{aligned}$$

In particular, the profile $a(x)=\log [\rho (x)]-\log [1-\rho (x)]$ satisfies the condition in the lemma.

Proof

By Dynkin’s formula,

$$\begin{aligned}{\mathbb {E}}_{\mu _N} \left[ \frac{1}{N}\sum _{i=0}^N a_i \big (\eta _{i}(t) - \eta _{i}(0)\big ) - \int _0^t \frac{1}{N}\sum _{i=0}^N a_i L_N \eta _{i}(s)\,ds \right] = 0.\end{aligned}$$

From the definition of $L_N$,

$$\begin{aligned} \frac{1}{N}\sum _{i=0}^N a_iL_N \eta _{i} =\,&\frac{1}{N}\sum _{i=1}^{N-1} V_ia_i(\rho _i - \eta _i) + \sum _{i=0}^{N-1} (\nabla a_i)j_{i,i+1}\\&+ a_0\big [c_{\textrm{in}}^- - (c_{\textrm{in}}^-+c_{\textrm{out}}^-)\eta _0\big ] + a_N\big [c_{\textrm{in}}^+ - (c_{\textrm{in}}^++c_{\textrm{out}}^+)\eta _N\big ], \end{aligned}$$

where $j_{i,i+1}=\eta _i(1-\eta _{i+1})+\sigma _N(\eta _i-\eta _{i+1})$. As $|a_i| \le |a|_\infty$ and $|\eta _i| \le 1$,

$$\begin{aligned}{\mathbb {E}}_{\mu _N} \left[ \int _0^t \frac{1}{N}\sum _{i=1}^{N-1} V_ia_i(\rho _i - \eta _i)\,ds + \int _0^t \sum _{i=0}^{N-1} (\nabla a_i)j_{i,i+1}\,ds \right] \le C.\end{aligned}$$

Furthermore, using sum-by-parts formula,

$$\begin{aligned} \sum _{i=0}^{N-1} (\nabla a_i)j_{i,i+1} =\,&\sum _{i=0}^{N-1} (\nabla a_i)\eta _i(1-\eta _{i+1})\\&+ \sigma _N \left[ \sum _{i=1}^{N-1} (\Delta a_i)\eta _i + (\nabla a_0)\eta _0 - (\nabla a_{N-1})\eta _N \right] . \end{aligned}$$

Since $\sigma _N = o(N)$, one can conclude from the regularity of $a(\cdot )$. $\square$

5.3 Proof of Proposition 5.1

We fix $\phi \in {\mathcal {C}}_c^\infty (\Sigma _T)$ and prove Proposition 5.1 by estimating each term in the decomposition (5.3) uniformly in $\psi$. Compared to the proofs of [25, Proposition 6.1] and [24, Lemma 4.3], extra effort is needed to take care of the term with respect to V. In the following contents, C is constant that may depend on (f, q) and T but is independent of $(\phi ,\psi )$, while $C_\phi$ is constant that also depends on $\phi$ but is independent of $\psi$.

We first show that the error term ${\mathcal {E}}_N$ vanishes in the limit.

Lemma 5.6

${\mathcal {E}}_{N,1}\equiv 0$ for sufficiently large N. Moreover, as $N \rightarrow \infty$,

(i):

${\mathcal {E}}_{N,2}$ vanishes uniformly in $H^{-1} (\Sigma _{T})$, thus satisfies (5.1);

(ii):

${\mathcal {E}}_{N,3}$ vanishes uniformly in ${\mathcal {M}}(\Sigma _T)$, thus satisfies (5.2).

Proof

For $\phi \in {\mathcal {C}}_c^1(\Sigma _T)$, there is $\delta _\phi >0$ such that $\phi (t,x)=0$ for $x\notin (\delta _\phi ,1-\delta _\phi )$. Note that $K \ll N$, so we can find $N_\phi$ depending only on $\phi$ such that $\varphi =\phi \psi =0$ on $B_{T,N}$ for all $\psi$ and $N>N_\phi$. The vanishment of ${\mathcal {E}}_{N,1}$ then follows.

We first prove (i). For $N>N_\phi$, we can perform summation by parts without generating boundary term:

$$\begin{aligned} {\mathcal {E}}_{N,2} (\psi ) = \int _0^T \sum _{i=K}^{N-K} \big ( {\bar{\varphi }}_i - \varphi _i \big ) \nabla ^* q ({\hat{\eta }}_i)dt. \end{aligned}$$

Note that for some $\theta _x \in ((2i-1)/2N,x)$,

$$\begin{aligned} |{\bar{\varphi }}_i - \varphi _i| \le \int _{\tfrac{i}{N} - \frac{1}{2N}}^{{\tfrac{i}{N} + \frac{1}{2N}}} |\partial _x\varphi (\theta _x)| dx, \quad |\nabla ^* q ({\hat{\eta }}_i)| \le C\big |{\hat{\eta }}_{i-1}-{\hat{\eta }}_i\big | \le CK^{-1}. \end{aligned}$$

Thus, by Cauchy-Schwarz inequality and that $\varphi =\psi \phi$,

$$\begin{aligned} |{\mathcal {E}}_{N,2} (\varphi ) | \le CK^{-1}\Vert \varphi \Vert _{H^1} \le C_\phi K^{-1}\Vert \psi \Vert _{H^1}. \end{aligned}$$

To prove (ii), Taylor’s expansion gives that

$$\begin{aligned} \left| \epsilon _{i,K}^{(1)} \right| \le C\sum _{j=i-K}^{i+K} \left[ N \sigma _N ({\hat{\eta }}^{j,j+1}_i - {\hat{\eta }}_i)^2 + V \left( \tfrac{j}{N} \right) ({\hat{\eta }}^j_i - {\hat{\eta }}_i)^2 \right] . \end{aligned}$$

By the definition of ${\hat{\eta _i}}$ in (3.5), $|{\hat{\eta }}_i^{j,j+1}-{\hat{\eta _i}}| = K^{-2}$ and $|{\hat{\eta }}_i^j-{\hat{\eta _i}}| = w_{j-i}$, so

$$\begin{aligned} \left| \varepsilon _{i,K}^{(1)} \right| \le CN\sigma _NK^{-3} + C\sum _{|j|<K} V \left( \tfrac{i+j}{N} \right) w_j^2. \end{aligned}$$

As $\varphi =\phi \psi$, the definition (3.10) of ${{\bar{\varphi }}}_i$ yields that

$$\begin{aligned} \frac{1}{N}\sum _{i=K}^{N-K} {{\bar{\varphi }}}_i \sum _{|j|<K} V \left( \tfrac{i+j}{N} \right) w_j^2 \le \Vert \psi \Vert _{L^\infty }\sum _{|j|<K} w_j^2 \sum _{i=K}^{N-K} V \left( \tfrac{i+j}{N} \right) \int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} |\phi |\,dx. \end{aligned}$$

Recall that $\phi =0$ for $x\notin (\delta _\phi ,1-\delta _\phi )$. Let $N'_\phi$ be such that $KN^{-1}<2^{-1}\delta _\phi$ for $N>N'_\phi$. Then, for every $N>N'_\phi$, the expression above is bounded by

$$\begin{aligned}\Vert \psi \Vert _{L^\infty }\sum _{|j|<K} w_j^2 \left( \Vert \phi V\Vert _{L^1} + \Vert \phi \Vert _{L^1}\sup \left\{ |V'|; x\in \left[ \frac{\delta _\phi }{2},1-\frac{\delta _\phi }{2} \right] \right\} \frac{2K+1}{2N} \right) .\end{aligned}$$

From the choice of $w_j$ in (3.5), it is bounded by $C_\phi K^{-1}\Vert \psi \Vert _{L^\infty }$. Similarly, using Taylor’s expansion and the relation $f'J'=q'$,

$$\begin{aligned}\left| \epsilon _{i,K}^{(2)} \right| \le CN({\hat{\eta }}_{i-1} - {\hat{\eta }}_i)^2 \le CNK^{-2}.\end{aligned}$$

Putting the estimates above together,

$$\begin{aligned}|{\mathcal {E}}_{N,3} (\psi ) | \le C_\phi \left( \frac{N\sigma _N}{K^3} + \frac{N}{K^2} + \frac{1}{K} \right) \Vert \psi \Vert _{L^\infty }.\end{aligned}$$

The proof is then concluded by the choice of K in (3.4). $\square$

Similarly, with the compactness of $\phi$ we can carry out the estimate for ${\mathcal {G}}_N$.

Lemma 5.7

The functional ${\mathcal {G}}_N$ satisfies (5.2).

Proof

Recall that $G_i=V_i(\eta _i-\rho _i)$, so $|G_i| \le C|V(\tfrac{i}{N})|$. Then,

$$\begin{aligned}|{\mathcal {G}}_N (\varphi )| \le C\int _0^T \frac{1}{N}\sum _{i=K}^{N-K} |{{\bar{\varphi }}}_i|\sum _{|j|<K} \left| V \left( \tfrac{i+j}{N} \right) \right| w_j\,dt.\end{aligned}$$

By the argument in Lemma 5.6 (ii), it is bounded by $C_\phi \Vert \psi \Vert _{L^\infty }$ uniformly in N. $\square$

Now, we deal with the martingale term.

Lemma 5.8

The martingale ${\mathcal {M}}_N (\psi )$ satisfies (5.1).

Proof

Since $\partial _t\varphi =\phi \partial _t\psi +\psi \partial _t\phi$, using Cauchy–Schwarz inequality,

$$\begin{aligned} |{\mathcal {M}}_N(\varphi )|^2 \le \Vert \psi \Vert _{H^1}^2 \int _0^T \sum _{i=K}^{N-K} \int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} \big [\phi ^2+(\partial _t\phi )^2\big ]M_i^2(t)dxdt. \end{aligned}$$

Through Dynkin’s formula, $M_i$ satisfies that

$$\begin{aligned} {\mathbb {E}}_{\mu _N} \big [M_i^2(t)\big ]&= {\mathbb {E}}_{\mu _N} \left[ \int _0^t \sum _{j=i-K}^{i+K} N\big (\eta _j(1-\eta _{j+1})+\sigma _N\big ) \big (f({\hat{\eta }}_i^{j,j+1})-f({\hat{\eta }}_i)\big )^2ds \right. \\&\quad + \left. \int _0^t \sum _{j=i-K+1}^{i+K-1} V_j\big [\eta _j(1-\rho _j)+\rho _j(1-\eta _j)\big ] \big (f({\hat{\eta }}_i^j)-f({\hat{\eta }}_i)\big )^2ds \right] \\&\le CN\sigma _N\sum _{j=i-K}^{i+K} {\mathbb {E}}_{\mu _N} \big [({\hat{\eta }}_i^{j,j+1}-{\hat{\eta }}_i)^2 \big ] + C\sum _{|j|<K} V \left( \tfrac{i+j}{N} \right) w_j^2. \end{aligned}$$

Similarly to the estimate in Lemma 5.6(ii), $|{\mathcal {M}}_N(\varphi )| \le a_{N,\phi }\Vert \psi \Vert _{H^1}$ with

$$\begin{aligned} {\mathbb {E}}_{\mu _N} [a_{N,\phi }] \le C_\phi \sqrt{\frac{N\sigma _N}{K^3} + \frac{1}{K}}. \end{aligned}$$

The proof is then concluded by the choice of K in (3.4). $\square$

To deal with ${\mathcal {A}}_N$ and ${\mathcal {S}}_N$, we need the block estimates stated below. They follow from the upper bound of $D_\textrm{exc}^N$ in Lemma 5.4 and the logarithmic Sobolev inequality for exclusion process [28]. The proofs are the same as [25, Proposition 6.4 & 6.5] and [24, Proposition 4.6 & 4.7]. For this reason, we omit the details here.

Lemma 5.9

There exists some finite constant C independent of N, such that

$$\begin{aligned}&{\mathbb {E}}_{\mu _N} \Big [\int _{0}^{T} \sum _{i=K}^{N-K} \Big (\hat{J}_i (t) - J ({\hat{\eta }}_i (t))\Big )^2 dt\Big ] \le C \Big (\frac{K^2}{\sigma _N} + \frac{N}{K}\Big ),\end{aligned}$$

(5.15)

$$\begin{aligned}&{\mathbb {E}}_{\mu _N} \Big [\int _{0}^{T} \sum _{i=K}^{N-K} \Big (\nabla {\hat{\eta }}_i (t)\Big )^2 dt\Big ] \le C \Big (\frac{1}{\sigma _N} + \frac{N}{K^3}\Big ). \end{aligned}$$

(5.16)

Using Lemma 5.9, we can conclude the decompositions of ${\mathcal {A}}_N$ and ${\mathcal {S}}_N$.

Lemma 5.10

Define functionals ${\mathcal {A}}_{N,1}$ and ${\mathcal {S}}_{N,1}$ respectively by

$$\begin{aligned}&{\mathcal {A}}_{N,1}(\varphi ) := \int _0^T \sum _{i=K}^{N-K} {{\bar{\varphi }}}_i \left[ \hat{J}_{i} - J ({\hat{\eta }}_{i}) \right] \nabla f'({\hat{\eta _i}})dt,\end{aligned}$$

(5.17)

$$\begin{aligned}&{\mathcal {S}}_{N,1}(\varphi ) := -\,\sigma _N\int _0^T \sum _{i=K}^{N-K} {{\bar{\varphi }}}_i\nabla {\hat{\eta _i\nabla }} f'({\hat{\eta _i}})dt. \end{aligned}$$

(5.18)

Then, ${\mathcal {A}}_N-{\mathcal {A}}_{N,1}$ and ${\mathcal {S}}_N-{\mathcal {S}}_{N,1}$ satisfy (5.1), while ${\mathcal {A}}_{N,1}$ and $\mathcal S_{N,1}$ satisfy (5.2).

The proof of Lemma 5.10 follows [25, Lemma 6.6 & 6.7] almost line by line, so we only sketch the difference. It is worth noting that, ${\mathcal {S}}_{N,1}$ turns out to be the only term that survives in the limit, eventually generates the non-zero macroscopic entropy in (2.10).

Proof

We first treat ${\mathcal {A}}_N$. Since $\varphi =\phi \psi$ with $\phi \in {\mathcal {C}}_c^\infty (\Sigma _T)$ being fixed, similarly to the proof of Lemma 5.6 (i), we have for $N>N_\phi$ that

$$\begin{aligned}({\mathcal {A}}_N-{\mathcal {A}}_{N,1})(\varphi ) = \int _0^T \sum _{i=K}^{N-K} f'({\hat{\eta }}_{i+1}) \left[ \hat{J}_{i} - J ({\hat{\eta }}_{i}) \right] \nabla {{\bar{\varphi }}}_i\,dt.\end{aligned}$$

Applying Cauchy–Schwarz inequality and Lemma 5.9, we obtain that $|({\mathcal {A}}_N-{\mathcal {A}}_{N,1})(\varphi )| \le a_N\Vert \varphi \Vert _{H^1}$ and $|{\mathcal {A}}_{N,1}(\varphi )| \le b_N\Vert \varphi \Vert _{L^\infty }$, where $(a_N,b_N)$ are random variables such that

$$\begin{aligned}{\mathbb {E}}_{\mu _N} [a_N] \le C \sqrt{\frac{K^2}{N\sigma _N} + \frac{1}{K}}, \quad {\mathbb {E}}_{\mu _N} [b_N] \le C \left( \frac{K}{\sigma _N} + \frac{N}{K^2} \right) .\end{aligned}$$

Noting that $\Vert \varphi \Vert _{H^1} \le C_\phi \Vert \psi \Vert _{H^1}$, $\Vert \varphi \Vert _{L^\infty } \le C_\phi \Vert \psi \Vert _{L^\infty }$, the conclusion follows from (3.4).

The proof for ${\mathcal {S}}_N$ is similar. For $N>N_\phi$,

$$\begin{aligned}({\mathcal {S}}_N-{\mathcal {S}}_{N,1})(\varphi ) = -\sigma _N\int _0^T \sum _{i=K}^{N-K} f'({\hat{\eta }}_{i+1})\nabla {{\bar{\varphi }}}_i\nabla {\hat{\eta }}_i\,dt. \end{aligned}$$

By Cauchy–Schwarz inequality and Lemma 5.9, $|(\mathcal S_N-{\mathcal {S}}_{N,1})(\varphi )| \le a'_N\Vert \varphi \Vert _{H^1}$ and $|{\mathcal {S}}_{N,1}(\varphi )| \le b'_N\Vert \varphi \Vert _{L^\infty }$ with random variables $(a'_N,b'_N)$ satisfying

$$\begin{aligned} {\mathbb {E}}_{\mu _N} [a'_N] \le C\sqrt{\frac{\sigma _N}{N}+\frac{\sigma _N^2}{K^3}}, \quad {\mathbb {E}}_{\mu _N} [b'_N] \le C \left( 1+\frac{N\sigma _N}{K^3} \right) . \end{aligned}$$

The conclusion follows similarly. $\square$

6 Measure-Valued Entropy Solution

We prove that under ${\mathbb {Q}}$, $\nu$ satisfies (3.8) with probability 1. We call such a Young measure a measure-valued entropy solution to the initial-boundary value problem (2.7) and (2.9). Thanks to Proposition 3.3, $\nu$ is essentially the entropy solution.

6.1 Proof of (3.8)

Fix an arbitrary Lax entropy–flux pair (f, q) and $\varphi \in {\mathcal {C}}_c^2([0,T)\times (0,1))$ such that $\varphi \ge 0$. As in Sect. 5, we shall let $N\rightarrow \infty$ and examine the limit of each term in the decomposition (5.3). The main difference is that the proof of Proposition 5.1 requires uniform estimate in the test function, which is no more necessary here.

First, by Lemma 5.6, 5.8 and 5.10, for any $\delta >0$,

$$\begin{aligned}\lim _{N\rightarrow \infty } {\mathbb {Q}}_N \Big \{ |{\mathcal {A}}_N(\varphi )| + |\mathcal S_N(\varphi )-{\mathcal {S}}_{N,1}(\varphi )| + |{\mathcal {M}}_N(\varphi )| + |{\mathcal {E}}_N(\varphi )| > \delta \Big \} = 0.\end{aligned}$$

Meanwhile, the convexity of f ensures that $\nabla {\hat{\eta _i\nabla }} f'({\hat{\eta _i}}) \ge 0$, so ${\mathcal {S}}_{N,1}(\varphi )$ given by (5.18) is non-positive everywhere. Therefore, (3.8) holds ${\mathbb {Q}}$-almost surely if we can show that

$$\begin{aligned} \limsup _{N\rightarrow \infty } {\mathbb {Q}}_N \left\{ \left| {\mathcal {G}}_N(\varphi ) + \iint _{\Sigma _T} \varphi (t,x)\int _{\mathbb {R}}f'(\lambda )G(x,\lambda )\,\nu _{t,x}^N(d\lambda )\,dxdt \right| > \delta \right\} = 0,\nonumber \\ \end{aligned}$$

(6.1)

for any $\delta >0$, where $G(x,\lambda )=V(x)(\lambda -\rho (x))$. Recall the definition of ${\mathcal {G}}_N(\varphi )$ in Lemma 5.3. As $\varphi$ is compactly supported, for sufficiently large N we have

$$\begin{aligned} {\mathcal {G}}_N(\varphi ) =\,&- \iint _{\Sigma _T} \varphi (t,x)\int _{\mathbb {R}}f'(\lambda )G(x,\lambda )\,\nu _{t,x}^N(d\lambda )\,dxdt\\&- \int _0^T \sum _{i=K}^{N-K} f'({\hat{\eta _i}})\int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} \varphi \big [{\hat{G}}_i-G(\cdot ,{\hat{\eta _i}})\big ]\,dxdt, \end{aligned}$$

where ${\hat{G}}_i$ is the smoothly weighted average of $G_i=V(\tfrac{i}{N})(\eta _i-\rho (\tfrac{i}{N}))$. Straightforward computation shows that

$$\begin{aligned}\left| \int _0^T \sum _{i=K}^{N-K} f'({\hat{\eta _i}})\int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} \varphi \big [{\hat{G}}_i-G(\cdot ,{\hat{\eta _i}})\big ]\,dxdt \right| \le \frac{CK}{N},\end{aligned}$$

which vanishes uniformly as $N\rightarrow \infty$. We can then conclude (6.1).

6.2 Direct Proof of (3.9)

Notice that (3.9) holds if for some constant $C_0$,

$$\begin{aligned} {\mathbb {Q}}\left\{ \sup _{g\in {\mathcal {C}}_c^\infty ({\mathbb {R}}\times (0,1))} \left\{ \iint _{\Sigma _T} ({\bar{u}}-\rho )gV\,dxdt - C_0\iint _{\Sigma _T} g^2V\,dxdt \right\} < \infty \right\} = 1, \end{aligned}$$

(6.2)

where ${\bar{u}}={\bar{u}}(t,x):=\int _{\mathbb {R}}\lambda \nu _{t,x}(d\lambda )$ is a measurable function on $\Sigma _T$.

Let $\{g^j;j \ge 1\}$ be a countable subset of ${\mathcal {C}}^\infty _c ({\mathbb {R}}\times (0,1))$ which is dense in both the $L^1$ and the $L^2$ norm induced by V, i.e., for any $g\in {\mathcal {C}}^\infty _c ({\mathbb {R}}\times (0,1))$, there are $g^{j_n}$, $n\ge 1$, such that

$$\begin{aligned}\lim _{n\rightarrow \infty } \iint _{\Sigma _T} \big (g-g^{j_n}\big )^2V\,dxdt + \lim _{n\rightarrow \infty } \iint _{\Sigma _T} \big |g-g^{j_n}\big |V\,dxdt = 0.\end{aligned}$$

For $\ell \ge 1$, consider the functional $\Phi _\ell :{\mathcal {Y}}\rightarrow {\mathbb {R}}$ defined as

$$\begin{aligned}\Phi _\ell (\nu ):= \max _{1 \le j \le \ell } \left\{ \iint _{\Sigma _T} g^jV \left[ \int _{\mathbb {R}}(\lambda -\rho )\nu _{t,x} (d\lambda ) \right] dxdt - C_0\iint _{\Sigma _T} (g^j)^2V\,dxdt \right\} .\end{aligned}$$

Note that $\Phi _\ell$ is continuous for each fixed $\ell$. Lemma 6.1 below together with the weak convergence of ${\mathbb {Q}}_N$ then shows that there is a constant $C_1$ independent of $\ell$, such that

$$\begin{aligned}E^{{\mathbb {Q}}} \big [\Phi _\ell \big ] = \lim _{N\rightarrow \infty } E^{{\mathbb {Q}}_N} \big [\Phi _\ell \big ] \le C_1, \quad \forall \,\ell \ge 1.\end{aligned}$$

Taking $\ell \rightarrow \infty$ and applying the monotone convergence theorem,

$$\begin{aligned}E^{{\mathbb {Q}}} \left[ \sup _{j\ge 1} \left\{ \iint _{\Sigma _T} (\bar{u}-\rho )g^jV\,dxdt - C_0\iint _{\Sigma _T} (g^j)^2V\,dxdt \right\} \right] \le C_1. \end{aligned}$$

The condition (6.2) then follows from the dense property of $\{g^j;j\ge 1\}$.

Lemma 6.1

There exist constants $C_0$ and $C_1$ such that, for each $\ell \ge 1$,

$$\begin{aligned} \limsup _{N \rightarrow \infty } {\mathbb {E}}_{\mu _N} \left[ \max _{1 \le j \le \ell } \left\{ \iint _{\Sigma _T} \! (u^N-\rho )g^jV\,dxdt - C_0\!\iint _{\Sigma _T} (g^j)^2V\,dxdt \right\} \right] \le C_1.\nonumber \\ \end{aligned}$$

(6.3)

Proof

Only in this proof, to shorten the formulas we denote

$$\begin{aligned}\Vert g\Vert _V^2:= \iint _{\Sigma _T} g^2(t,x)V(x)\,dxdt, \quad \overline{(gV)}_i(t):= N\int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} g(t,x)V(x)\,dx.\end{aligned}$$

Since $g^j$ is compactly supported, for N sufficiently large,

$$\begin{aligned} \iint _{\Sigma _T} \big (u^N-\rho \big )g^jV\,dxdt = \int _0^T \frac{1}{N}\sum _{i=K}^{N-K} {\hat{\eta }}_i\overline{(g^jV)}_i\,dt - \iint _{\Sigma _T} \rho g^jV\,dxdt. \end{aligned}$$

To conclude the proof, it suffices to prove that

$$\begin{aligned}&\limsup _{N\rightarrow \infty } \left| \int _0^T \frac{1}{N}\sum _{i=K}^{N-K} {\hat{\rho }}_i\overline{(g^jV)}_i\,dt - \iint _{\Sigma _T} \rho g^jV\,dxdt \right| = 0, \quad \forall \,1 \le j \le \ell , \end{aligned}$$

(6.4)

$$\begin{aligned}&\limsup _{N\rightarrow \infty } {\mathbb {E}}_{\mu _N} \left[ \max _{j \le \ell } \left\{ \int _0^T \frac{1}{N}\sum _{i=K}^{N-K} ({\hat{\eta }}_i-{\hat{\rho }}_i)\overline{(g^jV)}_i\,dt - C_0\Vert g^j\Vert _V^2 \right\} \right] \le C_1. \end{aligned}$$

(6.5)

We begin with (6.4), which is completely deterministic. For some fixed $j\le \ell$, using the fact that $g^j$ is compactly supported, we only need to show that

$$\begin{aligned}\limsup _{N\rightarrow \infty } \left| \int _0^T dt\sum _{i=K}^{N-K} \int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} \big [{\hat{\rho _i}}-\rho (x)\big ]g^j(t,x)V(x)\,dx \right| = 0.\end{aligned}$$

This follows as $\rho$ is continuous and ${\hat{\rho _i}}$ is the smoothly weighted average of $\rho (\tfrac{i}{N})$.

Now we prove (6.5). First note that it suffices to prove it with K replaced by any other mesoscopic scale $n=n(N)$ such that $K \le n = o(N)$, since

$$\begin{aligned}\limsup _{N\rightarrow \infty } \max _{j\le \ell } \left\{ \int _0^T \frac{1}{N} \Big (\sum _{i=K}^{n-1} + \sum _{i=N-n+1}^{N-K}\Big ) \big |\overline{(g^jV)}_i\big | \right\} = 0.\end{aligned}$$

The choice of n is specified below in Lemma 6.2. Recall that $\nu _{\rho (\cdot )}^N$ is the product measure on $\Omega _N$ associated with the profile $\rho$. From the entropy inequality, the expectation in (6.5) is bounded by

$$\begin{aligned}\frac{H(\mu _N \,|\, \nu _{\rho (\cdot )}^N)}{N} + \frac{1}{N} \log {\mathbb {E}}_{\nu _{\rho (\cdot )}^N} \left[ \exp \left\{ \max _{j \le \ell } \Big \{ \int _0^T \sum _{i=n}^{N-n} ({\hat{\eta _i}}-{\hat{\rho _i}})\overline{(g^jV)}_i\,dt - C_0N\Vert g^j\Vert _V^2 \Big \} \right\} \right] .\end{aligned}$$

Due to (2.4), the relative entropy is bounded by CN, so the first term is uniformly bounded. Also notice that for any random variables $X_1$,... $X_\ell$,

$$\begin{aligned}\log E \big [e^{\max \{X_j;1 \le j \le \ell \}}\big ] \le \log E \Big [\sum _{1 \le j \le \ell } e^{X_j}\Big ] \le \log \ell + \max _{1 \le j \le \ell } \log E\big [e^{X_j}\big ],\end{aligned}$$

so we only need to find universal constants $C_0$ and $C_1$, such that

$$\begin{aligned} \log {\mathbb {E}}_{\nu _{\rho (\cdot )}^N} \left[ \exp \left\{ \int _0^T \sum _{i=n}^{N-n} ({\hat{\eta _i}}-{\hat{\rho _i}})\overline{(gV)}_i\,dt \right\} \right] \le \big (C_0\Vert g\Vert _V^2+C_1\big )N, \end{aligned}$$

(6.6)

for all $g\in {\mathcal {C}}_c^\infty ({\mathbb {R}}\times (0,1))$. By the Feynman–Kac formula (see, e.g., [2, Lemma 7.3]), the left-hand side of (6.6) is bounded from above by

$$\begin{aligned} \int _0^T \sup _{f} \left\{ \sum _{\eta \in \Omega _N} \sum _{i=n}^{N-n} \big ({\hat{\eta }}_i - {\hat{\rho }}_i\big ) \overline{(gV)}_i f(\eta ) \nu ^N_{\rho (\cdot )} (\eta ) + \big \langle \sqrt{f}, L_N\sqrt{f} \big \rangle _{\nu _{\rho (\cdot )}^N} \right\} dt, \end{aligned}$$

(6.7)

where the supremum is taken over all $\nu _{\rho (\cdot )}^N$-densities. For each j, performing the change of variables $\eta \mapsto \eta ^j$,

$$\begin{aligned}&\sum _{\eta \in \Omega _N} (\eta _j - \rho _j) f(\eta ) \nu ^N_{\rho (\cdot )} (\eta )\\ =\,&\frac{1}{2}\sum _{\eta \in \Omega _N} (\eta _j - \rho _j) f(\eta ) \nu ^N_{\rho (\cdot )} (\eta ) + \frac{1}{2}\sum _{\eta \in \Omega _N} (1- \eta _j - \rho _j) f(\eta ^j) \nu ^N_{\rho (\cdot )} (\eta ^j) \\ =\,&\frac{1}{2}\sum _{\eta \in \Omega _N} (\eta _j - \rho _j) \big [f(\eta )-f(\eta ^j)\big ] \nu ^N_{\rho (\cdot )} (\eta ), \end{aligned}$$

where the last line follows from the equality $(1- \eta _j - \rho _j) \nu ^N_{\rho (\cdot )} (\eta ^j) = - (\eta _j - \rho _j) \nu ^N_{\rho (\cdot )} (\eta )$. Thus, the first term inside the supremum in (6.7) reads

$$\begin{aligned}&\sum _{i=n}^{N-n} \overline{(gV)}_i \sum _{|j|<K} w_j \sum _{\eta \in \Omega _N} (\eta _{i+j}-\rho _{i+j}) f(\eta ) \nu _{\rho (\cdot )}^N(\eta )\\ =\,&\frac{1}{2}\sum _{i=n}^{N-n} \overline{(gV)}_i \sum _{|j|<K} w_j \sum _{\eta \in \Omega _N} (\eta _{i+j}-\rho _{i+j}) \big [f(\eta )-f(\eta ^{i+j})\big ] \nu ^N_{\rho (\cdot )}(\eta ). \end{aligned}$$

Using Cauchy–Schwarz inequality, we can bound it from above by $\mathcal I_1+{\mathcal {I}}_2$, where

$$\begin{aligned}&{\mathcal {I}}_1 := \frac{1}{2} \sum _{i=n}^{N-n} \sum _{|j|<K} \sum _{\eta \in \Omega _N} w_j c_{i+j,\textrm{G}}(\eta ) \Big (\sqrt{f(\eta )} - \sqrt{f(\eta ^{i+j})}\Big )^2 \nu ^N_{\rho (\cdot )}(\eta ),\\&{\mathcal {I}}_2 := \frac{1}{2} \sum _{i=n}^{N-n} \sum _{|j|<K} \sum _{\eta \in \Omega _N} w_j \frac{(\eta _{i+j} - \rho _{i+j})^2}{c_{i+j,\textrm{G}}(\eta )} \overline{(gV)}_i^2 \Big (\sqrt{f(\eta )} + \sqrt{f(\eta ^{i+j})}\Big )^2 \nu ^N_{\rho (\cdot )}(\eta ), \end{aligned}$$

where $c_{i,\textrm{G}}(\eta )=V_i[\rho _i(1-\eta _i)+\eta _i(1-\rho _i)]>0$ due to (2.4). Recall the Dirichlet form $D^N_\textrm{G}$ defined in (5.9). Since $\sum _{|j|<K} w_j=1$,

$$\begin{aligned}{\mathcal {I}}_1 \le \frac{1}{2}\sum _{i=1}^{N-1} \sum _{|j|<K} w_j \sum _{\eta \in \Omega _N} c_{i,\textrm{G}}(\eta ) \Big (\sqrt{f(\eta )} - \sqrt{f(\eta ^i)}\Big )^2 \nu ^N_{\rho (\cdot )}(\eta ) = N D^N_{\textrm{G}} (\mu ;\nu _{\rho (\cdot )}^N),\end{aligned}$$

where $\mu :=f\nu _{\rho (\cdot )}^N$. To estimate ${\mathcal {I}}_2$, notice that

$$\begin{aligned}\frac{(\eta _{i+j} - \rho _{i+j})^2}{c_{i+j,\textrm{G}}(\eta )}\overline{(gV)}_i^2 \le \frac{C}{V_{i+j}}\int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} V\,dx \cdot \bigg [\int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} V\,dx\bigg ]^{-1}\overline{(gV)}_i^2.\end{aligned}$$

From (2.4), Lemma 6.2 below and Cauchy–Schwarz inequality,

$$\begin{aligned}\frac{(\eta _{i+j} - \rho _{i+j})^2}{c_{i+j,\textrm{G}}(\eta )}\overline{(gV)}_i^2 \le CN\int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N} + \frac{1}{2N}} V(x) g(t,x)^2 dx,\end{aligned}$$

with some constant C independent of (i, j, g). Therefore, ${\mathcal {I}}_2 \le C_0N\Vert g\Vert _V^2$. Putting the estimates for ${\mathcal {I}}_1$, ${\mathcal {I}}_2$ into (6.7), we see that (6.6) holds if we can show

$$\begin{aligned} ND^N_{\textrm{G}} (\mu ;\nu _{\rho (\cdot )}^N) + \langle \sqrt{f}, L_N \sqrt{f}\rangle _{\nu _{\rho (\cdot )}^N} \le CN, \end{aligned}$$

(6.8)

for any N, any $\nu _{\rho (\cdot )}^N$-density f and $\mu =f\nu _{\rho (\cdot )}^N$

The proof of (6.8) is standard. Since $\nu _{\rho (\cdot )}^N$ is reversible for $L_{\textrm{G}}$,

$$\begin{aligned}D^N_{\textrm{G}} (\mu ;\nu _{\rho (\cdot )}^N) + \big \langle \sqrt{f}, L_{\textrm{G}} \sqrt{f}\big \rangle _{\nu _{\rho (\cdot )}^N} = 0.\end{aligned}$$

In Appendix A, we prove that (cf. (5.12) and (5.14))

$$\begin{aligned}&\big \langle \sqrt{f},L_{\textrm{exc}}\sqrt{f} \big \rangle _{\nu _{\rho (\cdot )}^N} \le -\tfrac{1}{4}(\sigma _N-1)D_\textrm{exc}^N(\mu ;\nu _{\rho (\cdot )}^N) + C,\end{aligned}$$

(6.9)

$$\begin{aligned}&\big \langle \sqrt{f},L_{\textrm{bd}}\sqrt{f} \big \rangle _{\nu _{\rho (\cdot )}^N} \le - D_-^N(\mu ;\nu _{\rho (\cdot )}^N) - D_+^N(\mu ;\nu _{\rho (\cdot )}^N) + C. \end{aligned}$$

(6.10)

Since the Dirichlet forms are non-negative, (6.8) follows. $\square$

Lemma 6.2

Suppose that $V\in {\mathcal {C}}^1((0,1))$ and $\inf _{(0,1)} V>0$. Define $n=n(N)$ as

$$\begin{aligned}n:= \inf \left\{ n \ge K;\,\sup \Big \{|V'(x)|; x\in \left[ \tfrac{n-K}{N},1-\tfrac{n-K}{N} \right] \Big \} \le \frac{N}{K} \right\} .\end{aligned}$$

Then, $n=o(N)$ as $N\rightarrow \infty$, and there is a constant $C=C(V)$ such that

$$\begin{aligned}\max _{n \le i \le N-n,|j|<K} \left\{ V \left( \tfrac{i+j}{N} \right) ^{-1} \int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} V\,dx \right\} \le \frac{C}{N}.\end{aligned}$$

Proof

We first prove that for any $\varepsilon >0$, $n\le \varepsilon N$ for sufficiently large N. Indeed, let $N_\varepsilon$ be such that $KN^{-1}<2^{-1}\varepsilon$ for all $N>N_\varepsilon$. Then, if $N>N_\varepsilon$,

$$\begin{aligned}\sup \Big \{|V'(x)|;x\in \left[ \tfrac{\varepsilon N-K}{N},1-\tfrac{\varepsilon -K}{N} \right] \Big \} \le \sup \Big \{|V'(x)|;x\in \left[ \tfrac{\varepsilon }{2},1-\tfrac{\varepsilon }{2} \right] \Big \} = C_\varepsilon .\end{aligned}$$

We can further choose $N_\varepsilon$ such that $KN^{-1}<C_\varepsilon ^{-1}$ for all $N>N_\varepsilon$, then $n\le \varepsilon N$.

For the second criterion, suppose that $|V(\tfrac{i+j}{N})|$ takes the minimum value for $|j|<K$ at some $j_{N,i}$. Then, for each $i=n$,..., $N-n$,

$$\begin{aligned} \left| N\int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} V\,dx - V \left( \tfrac{i+j_{N,i}}{N} \right) \right|&\le \sup \Big \{|V'|\textbf{1}_{[\frac{i-K}{N},\frac{i+K}{N}]} \Big \}\frac{|j_{N,i}|+1}{N}\\&\le \sup \Big \{|V'|\textbf{1}_{[\frac{n-K}{N},1-\frac{n-K}{N}]} \Big \}\frac{K}{N} \le 1. \end{aligned}$$

Therefore, for each i and j,

$$\begin{aligned}\big [V \left( \tfrac{i+j}{N} \right) \big ]^{-1}\int _{\frac{i}{N}-\frac{1}{2N}}^{\frac{i}{N}+\frac{1}{2N}} V\,dx \le \left[ V \left( \tfrac{i+j_{N,i}}{N} \right) \right] ^{-1} \left[ 1+V \left( \tfrac{i+j_{N,i}}{N} \right) \right] \le 1+\frac{1}{\inf _{(0,1)} V}.\end{aligned}$$

The second criterion then follows from (2.4). $\square$

6.3 Direct Proof of Proposition 2.11

By the continuity of $\rho (\cdot )$,

$$\begin{aligned}\lim _{\varepsilon \rightarrow 0+} \lim _{N \rightarrow \infty } \frac{1}{\varepsilon N} \sum _{i=1}^{\varepsilon N} \rho _i = \rho (0).\end{aligned}$$

Thus, we only need to show for fixed $\varepsilon >0$ that

$$\begin{aligned} \lim _{N \rightarrow \infty } \int _0^t \frac{1}{\varepsilon N} \sum _{i=1}^{\varepsilon N} \big (\eta _{i} (s) - \rho _i\big ) ds = 0 \quad \text {in} {\mathbb {P}}_{\mu _N}-probability. \end{aligned}$$

(6.11)

Take

$$\begin{aligned}A:= A(\varepsilon ):= \sqrt{\varepsilon / \inf \big \{ V(x);x\in (0,\varepsilon ) \big \} }.\end{aligned}$$

Let $\mu ^{(\varepsilon N)} (s)$ denote the distribution of $\{\eta _1 (s), \eta _2 (s), \ldots , \eta _{\varepsilon N} (s)\}$. By entropy inequality,

$$\begin{aligned}&{\mathbb {E}}_{\mu _N} \Big [ \Big | \int _0^t \frac{1}{\varepsilon N} \sum _{i=1}^{\varepsilon N} \big (\eta _{i} (s) - \rho _i\big ) ds \Big |\Big ] \le \int _0^t E^{\mu ^{(\varepsilon N)} (s)} \Big [ \Big | \frac{1}{\varepsilon N} \sum _{i=1}^{\varepsilon N} \big (\eta _{i} - \rho _i\big ) \Big |\Big ] ds \\&\le \int _0^t \frac{H(\mu ^{(\varepsilon N)} (s) | \nu ^N_{\rho (\cdot )})}{AN} ds + \frac{t}{AN} \log E^{\nu ^N_{\rho (\cdot )}} \Big [ \exp \Big \{ \Big | \frac{A}{\varepsilon } \sum _{i=1}^{\varepsilon N} \big (\eta _{i} - \rho _i\big ) \Big | \Big \}\Big ] =: \textrm{I} + \textrm{II}. \end{aligned}$$

We first bound $\textrm{II}$, which is simpler. Note that we could first remove the absolute value inside the exponential. Since $\nu ^N_{\rho (\cdot )}$ is a product measure, and by Taylor’s expansion, there exists some constant C such that

$$\begin{aligned} \textrm{II} \le \frac{t}{AN} \sum _{i=1}^{\varepsilon N} \log E^{\nu ^N_{\rho (\cdot )}} \Big [ \exp \Big \{ \frac{A}{\varepsilon } \big (\eta _{i} - \rho _i\big ) \Big \}\Big ] \le \frac{CtA}{\varepsilon }, \end{aligned}$$

which converges to zero as $\varepsilon \rightarrow 0$ by (2.20).

For $\textrm{I}$, we consider the following Markov chain $X(t):= \{X_i(t)\}_{1\le i \le \varepsilon N}$, where $\{X_i (t)\}$, $1 \le i \le \varepsilon N$, are independent $\{0,1\}$-valued Markov chains, and the transition rates for $X_i (t)$ are given by

$$\begin{aligned}1 \rightarrow 0 \quad \text {rate} \quad V_i (1-\rho _i), \quad 0 \rightarrow 1 \quad \text {rate} \quad V_i \rho _i.\end{aligned}$$

Since $\rho (\cdot )$ is bounded away from zero and one, the logarithmic Sobolev constant for the chain $X_i (t)$ is of order $V_i$. By [10, Lemma 3.2], the logarithmic Sobolev constant for the chain X(t) has order

$$\begin{aligned}\min _{1 \le i \le \varepsilon N} V_i \ge \inf \{V(x); x \in (0,\varepsilon )\}.\end{aligned}$$

Therefore,

$$\begin{aligned}H(\mu ^{(\varepsilon N)} (s) | \nu ^N_{\rho (\cdot )}) \le \frac{CN}{\inf \{V(x); x \in (0,\varepsilon )\}} D^N_{\textrm{G}} (s).\end{aligned}$$

By Lemma 5.4, $\int _0^t D^N_{\textrm{G}} (s) ds \le C$. Thus,

$$\begin{aligned}\textrm{I} \le \frac{C}{A \inf \{V(x); x \in (0,\varepsilon )\}},\end{aligned}$$

which also converges to zero as $\varepsilon \rightarrow 0$.

Appendix A: Computations Concerning the Dirichlet Forms

Here we collect some fundamental estimates of the Dirichlet forms in (5.8)–(5.11).

1.1 A.1: Proof of (5.12)

From the definition of $L_{\textrm{exc}}$,

$$\begin{aligned} L_{\textrm{exc}} = \frac{\sigma _N+1}{2} S_{\textrm{exc}} + \frac{2p-1}{2} A_\textrm{exc}, \end{aligned}$$

(A.1)

where the operators $S_{\textrm{exc}}$ and $A_{\textrm{exc}}$ are respectively given by

$$\begin{aligned}S_{\textrm{exc}}g = \sum _{i=0}^{N-1} \big [g(\eta ^{i,i+1}) - g(\eta )\big ], \quad A_{\textrm{exc}}g = \sum _{i=0}^{N-1} (\eta _i-\eta _{i+1}) \big [g(\eta ^{i,i+1}) - g(\eta )\big ].\end{aligned}$$

Recall that $\nu =\nu _{\frac{1}{2}}^N$ and f is a $\nu$-density. By the basic inequality $x\log (y/x) \le 2\sqrt{x}(\sqrt{y}-\sqrt{x})$ for any x, $y\ge 0$,

$$\begin{aligned}\big \langle f,L_{\textrm{exc}}[\log f] \big \rangle _\nu \le 2 \big \langle \sqrt{f},L_{\textrm{exc}} [\sqrt{f}] \big \rangle _\nu .\end{aligned}$$

We compute $S_{\textrm{exc}}$ and $A_{\textrm{exc}}$ respectively. Since $\nu$ is reversible for $S_{\textrm{exc}}$,

$$\begin{aligned}\big \langle \sqrt{f},S_{\textrm{exc}} \sqrt{f} \big \rangle _\nu = -D_{\textrm{exc}}^N(\mu ;\nu ),\end{aligned}$$

where $\mu =f\nu$. For $A_{\textrm{exc}}$, we only need to observe that

$$\begin{aligned}\big \langle \sqrt{f},A_{\textrm{exc}} \sqrt{f} \big \rangle _\nu = \sum _{\eta \in \Omega _N} (\eta _N-\eta _0)f(\eta )\nu (\eta ) = E^\mu [\eta _N-\eta _0].\end{aligned}$$

The estimate is then concluded.

1.2 A.2: Proof of (5.13)

Let $f_*$ be the density of $\mu$ with respect to $\nu _{\rho (\cdot )}^N$, then

$$\begin{aligned} \log f(\eta ^i) - \log f(\eta )&= \log f_*(\eta ^i) - \log f_*(\eta ) + \log \big [\nu _{\rho (\cdot )}^N(\eta ^i)\big ] - \log \big [\nu _{\rho (\cdot )}^N(\eta )\big ]\\&= \log f_*(\eta ^i) - \log f_*(\eta ) + (1-2\eta _i) \log \left( \tfrac{\rho _i}{1-\rho _i} \right) . \end{aligned}$$

Therefore, $\langle f,L_{\textrm{G}}[\log f] \rangle _\nu$ equals to

$$\begin{aligned}\big \langle f_*,L_{\textrm{G}}[\log f_*] \big \rangle _{\nu _{\rho (\cdot )}^N} + \sum _{\eta \in \Omega _N} \frac{1}{N}\sum _{i=1}^{N-1} f(\eta ) c_{i,\textrm{G}}(\eta ) (1-2\eta _i) \log \left( \tfrac{\rho _i}{1-\rho _i} \right) \nu (\eta ).\end{aligned}$$

Since $\nu _{\rho (\cdot )}^N$ is reversible with respect to $L_\textrm{G}$, similarly as in Appendix A.1,

$$\begin{aligned}\big \langle f_*,L_{\textrm{G}}[\log f_*] \big \rangle _{\nu _{\rho (\cdot )}^N} \le 2 \big \langle \sqrt{f_*},L_{\textrm{G}}\sqrt{f_*}\big \rangle _{\nu _{\rho (\cdot )}^N} = -2D_{\textrm{G}}^N(\mu ;\nu _{\rho (\cdot )}^N).\end{aligned}$$

The estimate (5.13) then holds since $c_{i,\textrm{G}}(\eta )(1-2\eta _i) = V_i(\rho _i-\eta _i)$.

1.3 A.3: Proof of (5.14)

As in Appendix A.1, we begin with

$$\begin{aligned}\big \langle f,L_{\textrm{bd}}[\log f] \big \rangle _\nu \le 2 \big \langle \sqrt{f},L_{\textrm{bd}}\sqrt{f} \big \rangle _\nu .\end{aligned}$$

We compute the terms associated with $\eta _0$ as an example. Those associated with $\eta _N$ follow the same argument. Notice that

$$\begin{aligned}&2\sum _{\eta \in \Omega _N} c_0(\eta )\sqrt{f(\eta )} \left[ \sqrt{f(\eta ^0)} - \sqrt{f(\eta )} \right] \nu (\eta )\\ =\,&-\sum _{\eta \in \Omega _N} c_0(\eta ) \left[ \sqrt{f(\eta ^0)}-\sqrt{f(\eta )} \right] ^2 \nu (\eta ) + \sum _{\eta \in \Omega _N} c_0(\eta ) \big [f(\eta ^0)-f(\eta )\big ] \nu (\eta )\\ =\,&-2D_-^N(\mu ;\nu ) + \sum _{\eta \in \Omega _N} f(\eta ) \left[ c_0(\eta ^0)\frac{\nu (\eta ^0)}{\nu (\eta )}-c_0(\eta ) \right] \nu (\eta ). \end{aligned}$$

The conclusion then follows since $\nu (\eta ^0)=\nu (\eta )$ and $c_0(\eta ^0)-c_0(\eta ) = (c_{\textrm{out}}^--c_{\textrm{in}}^-)(1-2\eta _0)$.

1.4 A.4: Proof of (6.9)

Let f be a $\nu _{\rho (\cdot )}^N$-density function and denote $\mu =f\nu _{\rho (\cdot )}^N$. Observe that the main difference from (5.12) is that the reference measure $\nu _{\rho (\cdot )}^N$ is not reversible. Recall the decomposition (A.1) and we estimate $S_{\textrm{exc}}$ and $A_{\textrm{exc}}$ respectively.

We begin with the symmetric part $S_{\textrm{exc}}$. Let $g:=\sqrt{f}$. By dividing $g(\eta )$ into $2^{-1}(g(\eta )-g(\eta ^{i,i+1}))$ and $2^{-1}(g(\eta )+g(\eta ^{i,i+1}))$, we obtain

$$\begin{aligned}\langle g,S_{\textrm{exc}}g \rangle _{\nu _{\rho (\cdot )}^N} = -D_\textrm{exc}^N(\mu ;\nu _{\rho (\cdot )}^N) + \sum _\eta \sum _{i=0}^{N-1} \frac{g(\eta )+g(\eta ^{i,i+1})}{2} \big [g(\eta ^{i,i+1})-g(\eta )\big ] \nu _{\rho (\cdot )}^N(\eta ).\end{aligned}$$

Applying the change of variable $\eta ^{i,i+1}\rightarrow \eta$, the second term becomes

$$\begin{aligned}&\frac{1}{2}\sum _\eta \sum _{i=0}^{N-1} \frac{g(\eta )[g(\eta ^{i,i+1})-g(\eta )]}{2} \left[ \nu _{\rho (\cdot )}^N(\eta )-\nu _{\rho (\cdot )}^N(\eta ^{i,i+1}) \right] \\ \le \,&\frac{1}{2}D_{\textrm{exc}}^N(\mu ;\nu _{\rho (\cdot )}^N) + \frac{1}{4}\sum _{\eta \in \Omega _N} \sum _{i=0}^{N-1} g^2(\eta ) \bigg [1-\frac{\nu _{\rho (\cdot )}^N(\eta ^{i,i+1})}{\nu _{\rho (\cdot )}^N(\eta )}\bigg ]^2 \nu _{\rho (\cdot )}^N(\eta ). \end{aligned}$$

Since $\rho \in {\mathcal {C}}^1([0,1];(0,1))$, with a constant C independent of (i, N) we have

$$\begin{aligned} \left| 1-\frac{\nu _{\rho (\cdot )}^N(\eta ^{i,i+1})}{\nu _{\rho (\cdot )}^N(\eta )} \right| = \frac{|(\rho _{i+1}-\rho _i)(\eta _{i+1}-\eta _i)|}{\rho _i^{\eta _i}(1-\rho _i)^{1-\eta _i}\rho _{i+1}^{\eta _{i+1}}(1-\rho _{i+1})^{1-\eta _{i+1}}} \le \frac{C}{N}. \end{aligned}$$

(A.2)

Since $g^2\nu _{\rho (\cdot )}^N=\mu$ is a probability measure, we have

$$\begin{aligned}\langle g,S_{\textrm{exc}}g \rangle _{\nu _{\rho (\cdot )}^N} \le -\frac{1}{2}D_{\textrm{exc}}^N(\mu ;\nu _{\rho (\cdot )}^N) + \frac{C^2}{4N}.\end{aligned}$$

Now we treat the antisymmetric part $A_{\textrm{exc}}$. First observe that

$$\begin{aligned} \langle g,A_{\textrm{exc}}g \rangle _{\nu _{\rho (\cdot )}^N} =\,&-\frac{1}{2}\sum _\eta \sum _{i=0}^{N-1} (\eta _i-\eta _{i+1}) \big [g(\eta ^{i,i+1})-g(\eta )\big ]^2\nu _{\rho (\cdot )}^N(\eta )\\&+ \frac{1}{2}\sum _\eta \sum _{i=0}^{N-1} (\eta _i-\eta _{i+1})\big [g^2(\eta ^{i,i+1})-g^2(\eta )\big ]\nu _{\rho (\cdot )}^N(\eta ). \end{aligned}$$

As $|\eta _{i+1}-\eta _i|\le 1$, the first term is bounded by $D^N_\textrm{exc}(\mu ;\nu _{\rho (\cdot )}^N)$. The second term reads

$$\begin{aligned}\frac{1}{2}\sum _\eta \sum _{i=0}^{N-1} (\eta _{i+1}-\eta _i)f(\eta ) \nu _{\rho (\cdot )}^N(\eta ^{i,i+1}) + \frac{1}{2}\sum _\eta \sum _{i=0}^{N-1} (\eta _{i+1}-\eta _i) f(\eta )\nu _{\rho (\cdot )}^N(\eta ).\end{aligned}$$

Since $f(\eta )\nu _{\rho (\cdot )}^N(\eta )=\mu (\eta )$, its modulus is bounded by

$$\begin{aligned} \frac{1}{2}\sum _\eta \sum _{i=0}^{N-1} |\eta _{i+1}-\eta _i|\left| \frac{\nu _{\rho (\cdot )}^N(\eta ^{i,i+1})}{\nu _{\rho (\cdot )}^N(\eta )} - 1 \right| \mu (\eta ) + E^\mu [\eta _N-\eta _0], \end{aligned}$$

which is uniformly bounded due to (A.2). Therefore,

$$\begin{aligned}\langle g,A_{\textrm{exc}}g \rangle _{\nu _{\rho (\cdot )}^N} \le D_\textrm{exc}^N(\mu ;\nu _{\rho (\cdot )}^N) + C.\end{aligned}$$

Putting the two estimates into (A.1), we can conclude since $\sigma _N \ll N$.

1.5 A.5: Proof of (6.10)

Repeating the argument in the proof of (6.9) with $\nu$ replaced by $\nu _{\rho (\cdot )}^N$, we only need to bound

$$\begin{aligned}\frac{1}{2}\sum _{\eta \in \Omega _N} f(\eta ) \left[ c_0(\eta ^0)\frac{\nu _{\rho (\cdot )}^N(\eta ^0)}{\nu _{\rho (\cdot )}^N(\eta )}-c_0(\eta ) \right] \nu _{\rho (\cdot )}^N(\eta ).\end{aligned}$$

It is uniformly bounded since

$$\begin{aligned} \left| c_0(\eta ^0)\frac{\nu _{\rho (\cdot )}^N(\eta ^0)}{\nu _{\rho (\cdot )}^N(\eta )}-c_0(\eta ) \right| = \left| c_0(\eta ^0)\Big (\frac{\rho _0}{1-\rho _0}\Big )^{1-2\eta _0} - c_0(\eta ) \right| \le C.\end{aligned}$$