Local Equilibrium in Inhomogeneous Stochastic Models of Heat Transport

Abstract

We extend the duality of Kipnis et al. (J Stat Phys 27:65–74, 1982) to inhomogeneous lattice gas systems where either the components have different degrees of freedom or the rate of interaction depends on the spatial location. Then the dual process is applied to prove local equilibrium in the hydrodynamic limit for some inhomogeneous high dimensional systems and in the nonequilibrium steady state for one dimensional systems with arbitrary inhomogeneity.

Appendix

Here, we prove Lemma 6. We will use the notations of Sect. 6. The proof uses similar coupling to the one in the proof of Lemma 5. This will suffice in case of (a), as $\Vert Z\Vert$ is small and in case of (d), as we only need a weak estimate. However, we need to perform the coupling on a mesoscopic timescale to complete the proof of (b) and (c).

Proof of (d) we prove the following slightly stronger statement.

For every $\eta >0$ there is $\varepsilon = \varepsilon (\eta ) >0$ such that if $\log L - \log \log L< m < \log L + \log \varepsilon$ and L is large enough, then $\mathcal E_{L,M} < \eta .$

With some fixed $C_1,$ we couple $Z(k),\, k \in [s_0,\, (s_0 + C_1M^2)\wedge s_1]$ to a SSRW $W(k),\, k \in [s_0,\, (s_0 + C_1M^2) \wedge s_1],\, W(s_0) = Z(s_0).$ We fix $C_1,$ as we can by Lemma 5, such that $\mathbb P(s_1 > s_0 + C_1M^2) < \eta /10.$ Second, Lemma 4 implies that by choosing $C_2 = C_2(C_1)$ large enough,

$$\begin{aligned} \mathbb P\left( \Vert W(k ) - Z( k) \Vert > C_2 M^3 / L^2\,\text {for some}\,k< \left( s_0 + C_1M^2\right) \wedge s_1\right) < \eta /10. \end{aligned}$$

Now let us define

$$\begin{aligned}&\underline{s}_1 = \min \{ k{\text {:}}\, \Vert W(k) \Vert< (1/2 + \zeta ) M\,\text {or}\, \Vert W(k) \Vert> (2 - \zeta ) M \}, \\&\overline{s}_1 = \min \{ k{\text {:}}\, \Vert W(k) \Vert < (1/2 - \zeta ) M\,\text {or}\,\Vert W(k) \Vert > (2 + \zeta ) M \}, \end{aligned}$$

where $\zeta$ is fixed in such a way that the following events concerning a planar Brownian motion $\mathcal W$ have probability at least $1 - \eta /10{\text {:}}$

• (A) if $\Vert \mathcal W (0) \Vert = 1/2 + \zeta$ then $\mathcal W$ reaches $B(0,\,1/2- \zeta )$ before reaching $\mathbb R^2 \setminus B(0,\,2 - \zeta )$ and

• (B) if $\Vert \mathcal W (0) \Vert = 2 - \zeta$ then $\mathcal W$ reaches $\mathbb R^2 \setminus B(0,\, 2 + \zeta )$ before reaching $B(0,\, 1/2 + \zeta ).$

By the fact that $\log \Vert \mathcal W\Vert$ is martingale and by Donsker’s theorem, we can also assume

$$\begin{aligned} \left| \mathbb P \left( \left\| W \left( \underline{s}_1\right) \right\|< (1/2 + \zeta ) M \right) - 1/2\right| < \eta /10, \end{aligned}$$

possibly by further reducing $\zeta$ and by choosing L large. Finally, we choose $\varepsilon = \sqrt{ \zeta / C_2}$ so as $\zeta M > C_2 M^3 /L^2.$ Lemma 6(d) follows.

Proof of (a) if $2 \le \Vert Z(k + s_0)\Vert$ for some $k < L^{13/10} \wedge \underline{\tau }_{3/5},$ then using Lemma 4 we can couple $Z(k) - Z(k-1)$ to a step of a SSRW $W(k) - W(k-1)$ with probability $O(L^{-7/5}).$ Thus the probability

$$\begin{aligned} \mathbb P\left( Z(k) - Z(k-1) = W(k) - W(k-1)\,\text {for all}\, k < L^{13/10},\, 2 \le \left\| Z\left( k + s_0\right) \right\| \right) , \end{aligned}$$

is $O(L^{-1/10}).$ Now Lemma 6(a) follows from the proof of Lemmas 10 and 11 in [17].

Proof of (b) and (c) the idea of the proof of these cases is borrowed from [9]. First, we fix some positive $\xi <3/20,$ write $K= \lfloor L^{\xi } \rfloor$ and consider the process $(Z(s_0+ jK))_{j \ge 1},$ stopped upon reaching $B(0,\,M/2 - K)$ or $\mathbb R^{2} \setminus B(0,\, 2M + K).$ Let us denote by $\bar{s}$ the corresponding stopping time, i.e., $\bar{s} = sK$ with the smallest s such that $Z(s_0+ sK)$ is stopped. We will also write $\mathfrak z = Z(s_0+ jK) - Z(s_0+ (j-1)K)$ and $\mathfrak x_j = \log \Vert Z(s_0+ jK) \Vert ^2.$ Now we have

$$\begin{aligned} \left\| Z\left( s_0+ (j+1)K\right) \right\| ^2 = \left\| Z\left( s_0+ jK\right) \right\| ^2 + 2\left( Z\left( s_0+ jK\right) ,\, \mathfrak z\right) + \Vert \mathfrak z,\,\mathfrak z \Vert ^2, \end{aligned}$$

and

$$\begin{aligned} \mathfrak x_{j+1} - \mathfrak x_j =\log \left( 1+ \frac{2(Z(s_0+ jK),\,\mathfrak z) +\Vert \mathfrak z,\,\mathfrak z \Vert ^2 }{\Vert Z(s_0+ jK) \Vert ^2} \right) . \end{aligned}$$

By Taylor expansion,

$$\begin{aligned}&\mathfrak x_{j+1} - \mathfrak x_j \\= & {} \frac{2(Z(s_0+ jK),\, \mathfrak z)}{\Vert Z(s_0+ jK) \Vert ^2} + \frac{\Vert \mathfrak z\Vert ^2}{\Vert Z(s_0+ jK) \Vert ^2} -\frac{2(Z(s_0+ jK),\, \mathfrak z)^2}{\Vert Z(s_0+ jK) \Vert ^4} + O\left( L^{3 \xi } M^{-3}\right) \\= & {} I + II + III + O\left( L^{3 \xi } M^{-3}\right) . \end{aligned}$$

As before, we couple $\mathfrak z$ to a SSRW W(K) by Lemma 4 such that if $\mathfrak y = \mathfrak z - W(K),$ then for all positive integer n, 

$$\begin{aligned} \mathbb P ( \Vert \mathfrak y \Vert =k ) = O\left( L^{n\xi -2n} M^n\right) . \end{aligned}$$

If $\mathcal F_k$ is the filtration generated by $\tilde{\varvec{Z}}_i (s_0 + l),\,l \le k,\, i=1,\,2,$ then

$$\begin{aligned} \mathbb E ( I) = \mathbb E \left( \frac{ \mathbb E ( {2(Z(s_0+ jK),\, W(K))} | \mathcal F_j )}{\Vert Z(s_0+ jK) \Vert ^2} \right) + \mathbb E \left( \frac{ \mathbb E ( {2(Z(s_0+ jK),\,\mathfrak y)} | \mathcal F_j )}{\Vert Z(s_0+ jK) \Vert ^2} \right) . \end{aligned}$$

Here, the first term is zero by symmetry of the SSRW W,  and the second is estimated by the Cauchy–Schwarz inequality in cases $1 \le \Vert \mathfrak y \Vert \le 2$ and $\Vert \mathfrak y \Vert \ge 3.$ We conclude $\mathbb E (I) = L^{\xi -2}.$ A similar argument shows that

$$\begin{aligned} \mathbb E \left( \Vert \mathfrak z\Vert ^2\right) = K + O\left( M L^{3\xi -2}\right) . \end{aligned}$$

(22)

The above leading term is $\mathbb E ( \Vert W (K) \Vert ^2) = K,$ which can be easily proved by leveraging the fact that $(W_{(1)}(k) - W_{(1)}(k-1))_{1 \le k \le K}$ and $(W_{(2)}(k) - W_{(2)}(k-1))_{1 \le k \le K}$ are two independent one dimensional SSRW’s, where $v_{(1)} = ((1/\sqrt{2},\,1/\sqrt{2}),\, v)$ and $v_{(2)} = ((1/\sqrt{2},\, - 1/\sqrt{2}),\, v).$ Finally,

$$\begin{aligned}&\mathbb E\left( 2\left( Z\left( s_0+ jK\right) ,\,\mathfrak z\right) ^2 | \mathcal F_j\right) \\= & {} \sum _{i,\,j=1}^2 2 Z_{(i)}\left( s_0+ jK\right) Z_{(j)}\left( s_0+ jK\right) \mathbb E \left( \mathfrak z_{(i)} \mathfrak z_{(j)} \right) \\= & {} \sum _{i=1}^2 2 \left( Z_{(i)}\left( s_0+ jK\right) \right) ^2 \mathbb E \left[ \left( W_{(i)}(K)\right) ^2\right] + O\left( M^2 L^{3 \xi -2}\right) \\= & {} \left\| Z\left( s_0+ jK\right) \right\| ^2K + O\left( M^2 L^{3 \xi -2}\right) . \end{aligned}$$

We conclude that

$$\begin{aligned} \mathbb E \left( \left( \mathfrak x_{j+1} - \mathfrak x_j \right) 1_{\bar{s} > jK}\right) = O\left( L^{3 \xi } M^{-3} + L^{\xi -2}\right) . \end{aligned}$$

(23)

Thus

$$\begin{aligned} \mathbb E \left( \mathfrak x_{\min \{ \bar{s}/ K, A/ K \}} \right) = 2m + O\left( A L^{2 \xi } M^{-3} +A_M L^{-2}\right) . \end{aligned}$$

(24)

Recall that we are given some $\gamma <1.$

• In case of Lemma 6(b), let us choose $A = M^{2+ \frac{1-\gamma }{\gamma }}.$ Assuming, as we can that $\xi < 1-\gamma$ gives $A L^{-2}< L^{-\xi }.$ If $\xi$ is small (say smaller than 1 / 20), then $A L^{2 \xi } M^{-3}< L^{-\xi }$ holds as well.

• In case of Lemma 6(c) let us choose $A =L^2 \log ^{-3/2} L.$ Since both $1-\gamma$ and $\xi$ are small, we have $A L^{2 \xi } M^{-3} +A_M L^{-2} = O(\log ^{-3/2} L).$

Thus the error term in (24) can be replaced by O(E),  where $E =L^{-\xi }$ in case of Lemma 6(b) and $E = \log ^{-3/2} L$ in case of Lemma 6(c). Next, by Lemma 5,

$$\begin{aligned} \mathfrak x_{\min \{ \bar{s}/ K, A/ K \}} = 2 m \pm 2 + O\left( L^{\xi } M^{-1}\right) = 2 m \pm 2 + O\left( L^{-\xi }\right) , \end{aligned}$$

(25)

holds except on a set of probability $O(\theta ^{A/M^2}).$ On this exceptional set, we bound the left hand side of (25) by $O(\log M).$ In case of Lemma 6(b), $\theta ^{A/M^2} \log M$ is superpolynomially small in L. In case of Lemma 6(c), $\theta ^{A/M^2} \log M< \theta ^{\sqrt{\log L}} \log L= O(\log ^{-3/2} L).$ It follows that

$$\begin{aligned} \mathbb P \left( \left\| Z_{s_0 + \bar{s} K} \right\| \le M/2 - K \right) = \frac{1}{2} - O(E). \end{aligned}$$

Since $\Vert Z_{s_0 + \bar{s} K} \Vert \le M/2 - K$ implies $Z(s_1) \in \mathcal C_{m-1},$ we conclude

$$\begin{aligned} \mathbb P \left( Z\left( s_1\right) \in \mathcal C_{m-1}\right) \ge \frac{1}{2} - O(E). \end{aligned}$$

An analogous argument shows

$$\begin{aligned} \mathbb P \left( Z\left( s_1\right) \in \mathcal C_{m+1} \right) \ge \frac{1}{2} - O(E). \end{aligned}$$

Lemma 6(b) and (c) follow.