Local Thermal Equilibrium for Certain Stochastic Models of Heat Transport

Abstract

This paper is about nonequilibrium steady states (NESS) of a class of stochastic models in which particles exchange energy with their “local environments” rather than directly with one another. The physical domain of the system can be a bounded region of $\mathbb R^d$ for any $d \ge 1$. We assume that the temperature at the boundary of the domain is prescribed and is nonconstant, so that the system is forced out of equilibrium. Our main result is local thermal equilibrium in the infinite volume limit. In the Hamiltonian context, this would mean that at any location x in the domain, local marginal distributions of NESS tend to a probability with density $\frac{1}{Z} e^{-\beta (x) H}$, permitting one to define the local temperature at x to be $\beta (x)^{-1}$. We prove also that in the infinite volume limit, the mean energy profile of NESS satisfies Laplace’s equation for the prescribed boundary condition. Our method of proof is duality: by reversing the sample paths of particle movements, we convert the problem of studying local marginal energy distributions at x to that of joint hitting distributions of certain random walks starting from x, and prove that the walks in question become increasingly independent as system size tends to infinity.

Appendix: Proof of Proposition 3

Our proof uses the continuous mapping theorem, which can be stated as follows (see e.g. Theorem 5.1 in [1]). As before, weak convergence is denoted by “$\Rightarrow$”, and for a mapping f and measure $\mu$, $f_*\mu$ is the measure given by $f_*\mu (A)=\mu (f^{-1}(A))$.

Theorem 6

(Continuous mapping theorem) Let X and $X'$ be separable metric spaces, and let P(X) denote the set of Borel probability measures on X. We consider a Borel measurable mapping $f:X \rightarrow X'$ with discontinuity set $D_f$, and let $\mu _n, \mu \in P(X)$ be such that (i) $\mu _n \Rightarrow \mu$ as $n \rightarrow \infty$ and (ii) $\mu (D_f) = 0$. Then $f_*{\mu _n} \Rightarrow f_*\mu$ as $n \rightarrow \infty$.

To fix some notation, for $\mathfrak T \in \mathbb R^+$ we let

$$\begin{aligned} X = C([0, \mathfrak T], \mathbb R^d) \end{aligned}$$

be the set of continuous maps from $[0, \mathfrak T]$ to $\mathbb R^d$ endowed with the sup norm, making it a separable metric space. For $a \in \mathbb R^d$, we let $B^a$ be the standard Brownian motion starting from $a \in \mathbb R^d$ up to time $\mathfrak T$, and with a slight abuse of notation, we use $B^a$ to denote also the corresponding measure on X. Let $W^a_n$ be the rescaled SSRW up to time $\lfloor n\mathfrak T \rfloor$ starting from $a \in \mathbb R^d$, i.e. if $\hat{S}_k$ is a d-dimensional SSRW with $\hat{S}_0=\langle a \sqrt{n} \rangle$, then $W^a_n(k/n) = \frac{\hat{S}_k}{\sqrt{n}}$ for $k = 0,1,\ldots , \lfloor n\mathfrak T \rfloor$, and $W^a_n(t)$ is obtained by interpolating linearly between $t = k/n$ and $t = (k+1)/n$. As with $B^a$, we use $W^a_n$ to denote also the corresponding measure on X. By the invariance principle,

$$\begin{aligned} W^a_n \Rightarrow B^a \quad \text{ as } n \rightarrow \infty \end{aligned}$$

(34)

on any interval $[0, \mathfrak T]$. The convergence in (34) differs from that asserted in Proposition 3 in that the latter is for paths that terminate not at a fixed time but upon reaching $\partial \mathcal D$.

Proof of Proposition 3

Let $x \in \mathcal D$ and $\varepsilon >0$ be given. We write

$$\begin{aligned} \mathcal D^1 = \{ y \in \mathbb R^d | \exists z \in \mathcal D, |y-z|<1\}, \end{aligned}$$

and let $\mathfrak T = \mathfrak T (\varepsilon )$ be such that a Brownian motion starting from x reaches $\partial \mathcal D^1$ before time $\mathfrak T$ with probability at least $1-\varepsilon /(2\Vert T\Vert _{\infty })$. For $\omega \in X=C([0, \mathfrak T], \mathbb R^d)$, we define $\tau (\omega ) = \min \{ t \in [0, \mathfrak T]: \omega (t) \in \partial \mathcal D \}$ if such a t exists, $= \infty$ if it does not. Then we define $f: X \rightarrow \mathbb R$ by

$$\begin{aligned} f( \omega ) \ = \ \left\{ \begin{array}{ll} T(\omega (\tau )) &{}\quad \text{ if } \tau \le \mathfrak T \\ \max _{y \in \partial \mathcal D} T(y) &{}\quad \text{ if } \tau = \infty . \end{array} \right. \end{aligned}$$

Lemma 18

$B^x(D_f) = 0$

We first finish the proof assuming the result of this lemma. Via a rigid translation, we may assume $x =0$ (so that $xL \in \mathbb Z^d$ for all L). Then with $S_0=xL$, $S_n/L$ in the proposition is $W^0_{L^2} (n/L^2)$ in the notation above. Observe that we are now in the setting of the continuous mapping theorem: $W^0_{L^2} \Rightarrow B^0$ as $L \rightarrow \infty$ is condition (i) in Theorem 6, and the assertion in Lemma 18 is condition (ii). Thus the theorem applies, and its conclusion together with our choice of $\mathfrak T$ gives exactly (5) in the case $S_0=xL$.

To prove the full statement of Proposition 3, we observe that if the result was false, there would be a sequence $x_k \in \mathbb R^d$ with $x_k \rightarrow x$ and a sequence $L_k \rightarrow \infty$ such that if $S_n^{(k)}$ is the SSRW on $\mathcal D_{L_k}$ with $S^{(k)}_0 = x_k L_k$, then

$$\begin{aligned} \left| \mathbb E \left( T \left( \frac{S^{(k)}_{\tau }}{L_k} \right) \right) -u(x)\right| > \varepsilon \end{aligned}$$

where $\tau$ is the smallest n such that $S^{(k)}_n \in \mathcal B_{L_k}$. Such a scenario cannot occur: Since $W^{x_k}_{L_k^2} = (x_k-x) + W^x_{L_k^2}$, it follows from (34) that $W^{x_k}_{L_k^2} \Rightarrow B^x$ on $[0, \mathfrak T]$, and the argument in the last paragraph with $W^{x_k}_{L_k^2}$ in the place of $W^0_{L^2}$ gives the opposite inequality. $\square$

To complete the proof, it remains to show that the discontinuity set of f has zero Wiener measure.

Proof of Lemma 18

First, we identify the discontinuity set $D_f$. If $\tau (\omega ) = \infty$, then the trajectory of $\omega$ up to time $\mathfrak T$ is bounded away from $\partial \mathcal D$, hence f is continuous at $\omega$. If $\tau (\omega ) < \mathfrak T$, then $\liminf _{\omega ' \rightarrow \omega } \tau (\omega ') \ge \tau (\omega )$ for the same reason, but the corresponding $\limsup$ can be strictly greater than $\tau (\omega )$ if the trajectory of $\omega$ does not cross to the other side of $\partial \mathcal D$ immediately following $\tau (\omega )$. More precisely, we have deduced that $D_f = \{\tau = \mathfrak T\} \cup E$ where

$$\begin{aligned} E = \{\omega : \tau (\omega ) < \mathfrak T \text{ and } \exists \eta =\eta (\omega ) >0 \text{ s.t. } \omega ((\tau , \tau +\eta )) \subset \bar{\mathcal D}\} \end{aligned}$$

where $\bar{\mathcal D}$ is the closure of D.

Clearly, $\{ \tau = \mathfrak T\}$ has measure 0, so it suffices to show $B^x(E)=0$.

Since harmonic measure is absolutely continuous, the set of $\omega$ for which $\omega (\tau )$ lies at a point at which $\partial \mathcal D$ is not $C^2$ differentiable has measure zero. Let $\omega$ be outside of this measure zero set, and fix an orthonormal basis $\{e_1, \ldots , e_d\}$ of $\mathbb R^d$ such that $e_1$ is the outward normal to $\partial \mathcal D$ at $\omega (\tau )$. Then there exist $K>0$ and a neighborhood U of $\omega (\tau )$ in $\mathbb R^d$ such that

$$\begin{aligned} y \in \bar{\mathcal D} \quad \text { implies } \quad y_1 < K \Vert (y_2, \ldots , y_d)\Vert ^2. \end{aligned}$$

(35)

Recall that by the strong Markov property, $B^x$ starting from the stopping time $\tau$ is a Brownian motion. In particular, by projecting this Brownian motion, which we call $\hat{B}(t)$, to the line parallel to $e_1$ and to the hyperplane spanned by $e_2, \ldots , e_d$, we obtain two independent Brownian motions, $\hat{B}_1(t)$ and $\hat{B}_{d-1}(t)$. Since $\hat{B}_1(t)/\sqrt{t}$ and $\hat{B}_{d-1}(t)/\sqrt{t}$ have standard normal distributions,

$$\begin{aligned} \mathbb P(|\hat{B}_1(t)| < t^{2/3}) \rightarrow 0 \quad \text { and } \quad \mathbb P(\Vert \hat{B}_{d-1}(t)\Vert > t^{1/3}/\sqrt{K}) \rightarrow 0 \end{aligned}$$

as $t \rightarrow 0$. Choosing $t_n \downarrow 0$ so that

$$\begin{aligned} \sum _n \mathbb P(|\hat{B}_1(t_n)| < t_n^{2/3}), \ \ \sum _n \mathbb P(\Vert \hat{B}_{d-1}(t_n)\Vert > t_n^{1/3}/\sqrt{K}) \ < \ \infty , \end{aligned}$$

it follows from the Borel–Cantelli lemma that $|\hat{B}_1(t_n)| > t_n^{2/3}$ and $\Vert \hat{B}_{d-1}(t_n)\Vert < t_n^{1/3}/\sqrt{K}$ hold for all but finitely many n. For $t_n$ for which these inequalities hold, we are guaranteed that $\hat{B}(t_n) \not \in \bar{\mathcal D}$ if $\hat{B}_1(t_n)>0$ and $\hat{B}(t_n) \in U$.

Now it is a well known fact that on the time interval $[0, \eta ]$ for every $\eta >0$, a 1D Brownian motion starting from 0 makes infinitely many excursions from 0, and each excursion is positive with probability 1 / 2 independently of other excursions. Applying this fact to $\hat{B}_1(t)$, and assuming (as we may) that each $t_n$ lies in a different excursion, it follows that with probability 1, $\hat{B}_1(t_n)>0$ for infinitely many n. Since $\mathbb P(\hat{B}(t) \in U, t \in [0, \eta ]) \rightarrow 1$ as $\eta \rightarrow 0$, we have proved that $B^x (E) = 0$. $\square$