Particle Models with Self Sustained Current

Abstract

We present some computer simulations run on a stochastic cellular automaton (CA). The CA simulates a gas of particles which are in a channel,the interval [1, L] in $\mathbb Z$, but also in “reservoirs” $\mathcal R_1$ and $\mathcal R_2$. The evolution in the channel simulates a lattice gas with Kawasaki dynamics with attractive Kac interactions; the temperature is chosen smaller than the mean field critical one. There are also exchanges of particles between the channel and the reservoirs and among reservoirs. When the rate of exchanges among reservoirs is in a suitable interval the CA reaches an apparently stationary state with a non zero current; for different choices of the initial condition the current changes sign. We have a quite satisfactory theory of the phenomenon but we miss a full mathematical proof.

Appendices

Appendix 1: Estimates on the Current Between Reservoirs

Recalling (2.9) we have

$$\begin{aligned} j_{\mathcal R_2\rightarrow \mathcal R_1}(t) =j_t := \zeta _t \sum _{i_+,i_-}\mathbf 1_{\xi _t=(i_+,i_-)}[\theta ''_t(i_+)- \theta ''_t(i_-)] \end{aligned}$$

(10.1)

where $\zeta _t$ and $\xi _t$ are random variables independent of the process till time t and of $\theta ''_t$, they are also independent of each other. $\zeta _t$ takes value 1 with probability ${ \gamma }p$ and value 0 with probability $1-{ \gamma }p$; the values of $\xi _t$ are pairs $(i_+,i_-)$, $i_+\in \mathcal R_2$, $i_-\in \mathcal R_1$ and $P(\xi _t=(i_+,i_-))=\frac{1}{R^2}$. The sum $\sum _{i_+,i_-}$ is over $i_+\in \mathcal R_2$ and $i_-\in \mathcal R_1$. We first estimate the expected value of $j_{\mathcal R_2\rightarrow \mathcal R_1}(t)$:

$$\begin{aligned} E_{ \gamma }\left[ j_{\mathcal R_2\rightarrow \mathcal R_1}(t)]= E_{ \gamma }[\frac{N''_{\mathcal R_2}(t)-N''_{\mathcal R_1}(t)}{R}\right] { \gamma }p \end{aligned}$$

(10.2)

where

$$\begin{aligned} N''_{\mathcal R_i}(t)=\sum _{i\in \mathcal R_i}\theta ''_t(i),\quad i=1,2 \end{aligned}$$

(10.3)

Since $|N''_{\mathcal R_i}(t)-N_{\mathcal R_i}(t)|\le 2$ for all t we have

$$\begin{aligned} \left| E_{ \gamma }[ j_{\mathcal R_2\rightarrow \mathcal R_1}(t)]-{ \gamma }p E_{ \gamma }\left[ \frac{N_{\mathcal R_2}(t)-N_{\mathcal R_1}(t)}{R}\right] \right| \le { \gamma }p\frac{4}{R} \end{aligned}$$

(10.4)

We will next prove (4.1). Since $\theta ''$ has values 0, 1 we have from (10.1)

$$\begin{aligned} E[ j_t ]\le { \gamma }p \end{aligned}$$

(10.5)

By (2.5) the left hand site of (4.1), can be written as

$$\begin{aligned} A_T:=E\left[ \left[ \left\{ \frac{1}{T} \sum _{t=0}^{T-1} [j_t- { \gamma }p R^{-1}(N_{+,t}-N_{-,t}) \right] \right\} ^2\right] \end{aligned}$$

(10.6)

where

$$\begin{aligned} N_{+,t}= \sum _{i_+}\eta _t(i_+)=N_{\mathcal R_2}(t), \quad N_{-,t}= \sum _{i_-}\eta _t(i_-)=N_{\mathcal R_1} (t) \end{aligned}$$

(10.7)

Define $N''_{\pm ,t}$ as in (10.7) but with $\theta ''_t$ instead of $\eta _t$ and $A''_T$ as in (10.6) but with $N''_{\pm ,t}$.

Lemma 1

$$\begin{aligned} A_{T} \le A''_T + \frac{16}{R} ({ \gamma }p)^2 + \frac{16}{R^2} ({ \gamma }p)^2 \end{aligned}$$

(10.8)

Proof

Call

$$\begin{aligned} a_{t}= & {} j_t - { \gamma }p R^{-1}(N''_{+,t}-N''_{-,t})\end{aligned}$$

(10.9)

$$\begin{aligned} b_{t}= & {} { \gamma }p R^{-1}\{(N''_{+,t}-N''_{-,t}) - (N_{+,t}-N_{-,t})\} \end{aligned}$$

(10.10)

Then

$$\begin{aligned} A_{T} = E\left[ \frac{1}{T^2} \sum _{s,t} (a_{t}-b_{t})(a_{s}-b_{s})\right] \end{aligned}$$

(10.11)

Hence

$$\begin{aligned} A_{T} \le A''_{T} +2 E\left[ \frac{1}{T^2} \sum _{s,t}|a_{t}||b_{s}|\right] + E\left[ \frac{1}{T^2} \sum _{s,t}|b_{s}| |b_t|\right] \end{aligned}$$

(10.12)

$|b_t|\le { \gamma }p \frac{4}{R}$ because $|N''_{+,t} - N''_{-,t}| \le R$ and $|N''_{\pm ,t} - N_{\pm ,t}| \le 2$. By (10.5) and $|N''_{+,t} - N''_{-,t}| \le R$ we get $E_{ \gamma }[|a_t|] \le 2 { \gamma }p$, therefore

$$\begin{aligned} A_{T} \le A''_{T} +2 { \gamma }p \frac{8}{R} { \gamma }p + [{ \gamma }p \frac{4}{R}]^2 \end{aligned}$$

(10.13)

$\square$

Lemma 2

Let $s<t$ and $a_t$ as in (10.9) then

$$\begin{aligned} E[a_s a_t] =0 \end{aligned}$$

(10.14)

Proof

By the independence properties of $\zeta _t$ and $\xi _t$:

$$\begin{aligned} E [ a_s j_t ] = E \left[ a_s { \gamma }p \sum _{i_+,i_-} R^{-2}[\theta ''_t(i_+)- \theta ''_t(i_-)]\right] = E \Big [ a_s { \gamma }p R^{-1}[N''_{+,t}- N''_{-,t}]\Big ] \end{aligned}$$

(10.15)

$\square$

As a consequence

$$\begin{aligned} A''_T = \frac{1}{T^2} \sum _{t=0}^{T-1} E[a_t^2 ] \end{aligned}$$

(10.16)

We expand the square in $E[a_t^2 ]$, the first term is

$$\begin{aligned} E\left[ \zeta _t \sum _{i_+,i_-}\sum _{i'_+,i'_-}\mathbf 1_{\xi _t=(i_+,i_-)} \mathbf 1_{\xi _t=(i'_+,i'_-)}[\theta ''_t(i_+)- \theta ''_t(i_-)][\theta ''_t(i'_+)- \theta ''_t(i'_-)]\right] \end{aligned}$$

Due to the characteristic functions $i_{\pm }=i'_{\pm }$ so that the above is bounded by ${ \gamma }p$. The double product in the expansion of $E[a_t^2 ]$ is bounded by $2({ \gamma }p)^2$ and the third term by $({ \gamma }p)^2$, so that

$$\begin{aligned} A''_T \le \frac{1}{T} \{{ \gamma }p + 3({ \gamma }p)^2\} \end{aligned}$$

(10.17)

Going back to (10.13) we get

$$\begin{aligned} A_{T} \le \frac{1}{T} \{{ \gamma }p + 3({ \gamma }p)^2\} +16 \frac{({ \gamma }p)^2}{R} + 16[ \frac{{ \gamma }p}{R}]^2 \end{aligned}$$

(10.18)

which concludes the proof of (4.1).

Appendix 2: Proof of Theorems 1 and 2

1.1 Proof of (5.6)

Here we prove that m(r, t) satisfies (5.6) both in the CC-CA and in the OS-CA.

Let $u(r,t)=m(r,t)+1$ then m satisfies (5.6) if and only if u satisfies

$$\begin{aligned}&\frac{\partial }{\partial t}u(r,t) = \frac{1}{2} \frac{\partial ^2 u}{\partial r^2} - C \frac{\partial }{\partial r} \left\{ [u(2- u]\int _{r}^{r+1}[u(r+\xi ,t) - u(r-\xi ,t)] d\xi \right\} \qquad \quad \end{aligned}$$

(10.19)

with $u(r+\xi ,t)= u_+(t)=m_+(t)+1$ if $r+\xi \ge \ell$ and $u(r-\xi ,t)= u_-(t)=m_-(t)+1$ if $r-\xi \le 0$. In the OS-CA $m_\pm (t)\equiv m_\pm$.

By (5.2)

$$\begin{aligned} \lim _{{ \gamma }\rightarrow 0} \lim _{{ \gamma }x\rightarrow r, { \gamma }^{2}t\rightarrow \tau } E_{ \gamma }[ \eta (x,v,t)]= & {} \frac{1}{2} u(x,t), \quad v\in \{-1,1\}\nonumber \\ \lim _{{ \gamma }\rightarrow 0} \lim _{{ \gamma }x\rightarrow r, { \gamma }^{2}t\rightarrow \tau } u_{ \gamma }(x,t)= & {} u(x,t),\quad u_{ \gamma }(x,t)=E_{ \gamma }[ \eta (x,t)] \end{aligned}$$

(10.20)

So that we need to prove that the limit of $u_{ \gamma }$ satisfies (10.19).

By assumption u(r, t) is smooth so that it is enough to prove weak convergence namely that for any smooth test function f(r, t) with compact support in $(0,\ell )\times (0,\infty )$,

$$\begin{aligned}&\int u(r,t)\frac{\partial f(r,t)}{\partial t} dr dt \nonumber \\&\quad \quad = - \frac{1}{2} \int u(r,t)\frac{\partial ^2 f(r,t)}{\partial r^2} dr dt \nonumber \\&\quad \quad - \int \frac{\partial f(r,t)}{\partial r} C \left\{ [u(2- u]\int _{r}^{r+1}[u(r+\xi ,t) - u(r-\xi ,t)] \right\} dr dt \end{aligned}$$

(10.21)

By an integration by parts

$$\begin{aligned}&\int u(r,t)\frac{\partial f(r,t)}{\partial t} dr dt=-\lim _{{ \gamma }\rightarrow 0} { \gamma }^3\sum _{x,t} f({ \gamma }x,{ \gamma }^2t) { \gamma }^{-2}\{u_{ \gamma }(x;t+1)- u_{ \gamma }(x;t)\} \end{aligned}$$

We will next consider $u_{ \gamma }(x;t+1)- u_{ \gamma }(x;t)$. Recalling that $j_{x,x+1}(t)$ is the number of particles which in the time step $t, t+1$ cross the bond $(x,x+1)$, $x\in \{1,\ldots ,L-1\}$ (counting as positive those which jump from x to $x+1$ and as negative those from $x+1$ to x), we have

$$\begin{aligned} u_{ \gamma }(x;t+1)- u_{ \gamma }(x;t)=E_{ \gamma }[j_{x-1,x}(t)]-E_{ \gamma }[j_{x,x+1}(t)] \end{aligned}$$

We then have denoting by $\nabla _{ \gamma }$ the discrete derivative ($\nabla _{ \gamma }\varphi (x)=\varphi (x+1)-\varphi (x)$),

$$\begin{aligned} \int u(r,t)\frac{\partial f(r,t)}{\partial t} dr dt=-\lim _{{ \gamma }\rightarrow 0} { \gamma }^3\sum _{x,t} { \gamma }^{-1} \nabla _{ \gamma }f({ \gamma }x,{ \gamma }^2t) { \gamma }^{-1}E_{ \gamma }[j_{x,x+1}(t)] \end{aligned}$$

(10.22)

Lemma 3

$$\begin{aligned} E_{ \gamma }[j_{x,x+1}(t)]= \frac{1}{2}[u_{ \gamma }(x;t)- u_{ \gamma }(x+1;t)+E_{ \gamma }\Big [\chi _{x,{ \gamma };t} { \epsilon }_{x,{ \gamma };t}+\chi _{x+1,{ \gamma };s} { \epsilon }_{x+1,{ \gamma };t}\Big ] \end{aligned}$$

(10.23)

where ${ \epsilon }_{x,{ \gamma };t}$ is ${ \epsilon }_{x,{ \gamma }}$ computed at time t and

$$\begin{aligned} \chi _{x,{ \gamma };t} = \eta (x,1;t)\Big (1- \eta (x,-1;t)\Big )+ \eta (x,-1;t)\Big (1- \eta (x,1;t)\Big ) \end{aligned}$$

Proof

Observe that the expected number of particles that goes from x to $x+1$ is

$$\begin{aligned} E_{ \gamma }\left[ \eta (x,1;t)\eta (x,-1,t)+\chi _{x,{ \gamma };t}\left( \frac{1}{2}+{ \epsilon }_{x,{ \gamma };t}\right) \right] =\frac{1}{2} u_{ \gamma }(x,t)+ E_{ \gamma }\Big [\chi _{x,{ \gamma };t}{ \epsilon }_{x,{ \gamma };t}\Big ] \end{aligned}$$

The expected number of particles that goes from $x+1$ to x is

$$\begin{aligned} E_{ \gamma }\left[ \eta (x+1,1;t)\eta (x+1,-1,t)+\chi _{x+1,{ \gamma };t}\left( \frac{1}{2}-{ \epsilon }_{x+1,{ \gamma };t}\right) \right]= & {} \frac{1}{2} u_{ \gamma }(x+1,t)\\&- E_{ \gamma }\big [\chi _{x+1,{ \gamma };t}{ \epsilon }_{x+1,{ \gamma };t}\big ] \end{aligned}$$

so that we get (10.23). $\square$

We insert (10.23) in (10.22) and, denoting by $\Delta _{ \gamma }$ the discrete laplacian, we get

$$\begin{aligned}&{ \gamma }^3\sum _{x,t} { \gamma }^{-1} \nabla _{ \gamma }f({ \gamma }x,{ \gamma }^2t) { \gamma }^{-1}j_{ \gamma }(x,x+1,t)\nonumber \\&\quad \quad = \frac{1}{2} { \gamma }^3\sum _{x,t}{ \gamma }^{-2} \Delta _{ \gamma }f({ \gamma }x,{ \gamma }^2t) u_{ \gamma }(x,t)\nonumber \\&\quad \quad +\, { \gamma }^3\sum _{x,t} { \gamma }^{-1} 2 f'({ \gamma }x,{ \gamma }^2t)E_{ \gamma }[\chi _{x,{ \gamma };t}] E_{ \gamma }[{ \gamma }^{-1}{ \epsilon }_{x,{ \gamma };t}] +R_{ \gamma }\end{aligned}$$

(10.24)

where $2 f'({ \gamma }x,{ \gamma }^2t)= [\nabla _{ \gamma }f({ \gamma }x,{ \gamma }^2t)+ \nabla _{ \gamma }f({ \gamma }(x-1),{ \gamma }^2t)]$ and

$$\begin{aligned} R_{ \gamma }:=2 { \gamma }^3\sum _{x,t} { \gamma }^{-1} 2 f'({ \gamma }x,{ \gamma }^2t) E_{ \gamma }\left[ \chi _{x,{ \gamma };t}\Big ({ \gamma }^{-1}{ \epsilon }_{x,{ \gamma };t}- E_{ \gamma }[{ \gamma }^{-1}{ \epsilon }_{x,{ \gamma };t}]\Big )\right] \end{aligned}$$

By (10.20) and (5.4)

$$\begin{aligned}&\lim _{{ \gamma }\rightarrow 0} \lim _{{ \gamma }x\rightarrow r, { \gamma }^{2}t\rightarrow \tau } E_{ \gamma }\left[ \chi _{x,{ \gamma };t}]=\frac{1}{2} u(r,t)[2-u(r,t)\right] \nonumber \\&\lim _{{ \gamma }\rightarrow 0} \lim _{{ \gamma }x\rightarrow r, { \gamma }^{2}t\rightarrow \tau } E_{ \gamma }[{ \gamma }^{-1}{ \epsilon }_{x,{ \gamma };t}] = \int _{r}^{r+1} C[u(r+\xi ,t) - u(r-\xi ,t)]d\xi \end{aligned}$$

(10.25)

We postpone the proof of

$$\begin{aligned} \lim _{{ \gamma }\rightarrow 0} { \gamma }^3\sum _{x=2}^{{ \gamma }^{-1}\ell -1} \sum _{t=1}^{{ \gamma }^{-2}T} E_{ \gamma }\Big [\big |{ \gamma }^{-1}{ \epsilon }_{x,{ \gamma };t}- E_{ \gamma }[{ \gamma }^{-1}{ \epsilon }_{x,{ \gamma };t}]\big | \Big ] = 0 \end{aligned}$$

(10.26)

where $(0,\ell )\times (0,T)$ contains the support of f(r, t).

Observe that (10.22), (10.24), (10.25) and (10.26) yield (10.21) concluding the proof of $\square$

1.2 Proof of (10.26)

By Cauchy-Schwartz it is enough to prove that

$$\begin{aligned} \lim _{{ \gamma }\rightarrow 0} { \gamma }^3\sum _{x=2}^{{ \gamma }^{-1}\ell -1} \sum _{t=1}^{{ \gamma }^{-2}T} E_{ \gamma }\Big [\big |{ \gamma }^{-1}{ \epsilon }_{x,{ \gamma };t}- E_{ \gamma }[{ \gamma }^{-1}{ \epsilon }_{x,{ \gamma };t}]\big |^2 \Big ] = 0 \end{aligned}$$

(10.27)

We thus need to compute the limit of

$$\begin{aligned} { \gamma }^5\sum _{r,r',r''\in { \gamma }\mathbb Z} \sum _{\tau \in { \gamma }^2 \mathbb Z} g_{ \gamma }(r,r',r'',\tau ) \end{aligned}$$

(10.28)

where ${ \gamma }^{-1}r \in [2,{ \gamma }^{-1}\ell -1]$, $|r'-r| \le 1$, $|r''-r| \le 1$, ${ \gamma }^{-2}\tau \in [1,{ \gamma }^{-2}T]$ and

$$\begin{aligned} g_{ \gamma }(r,r',r'',\tau ) = C^2E_{ \gamma }[\tilde{\eta }_{{ \gamma }^{-2}\tau } ({ \gamma }^{-1} (r'-r)) \tilde{\eta }_{{ \gamma }^{-2}\tau } ({ \gamma }^{-1} (r''-r)) ] \end{aligned}$$

(10.29)

where $\tilde{\eta }_t(x) = \eta (x,t)-E_{ \gamma }[\eta (x,t)]$ if $x\in [1,L]$, otherwise it is $=\frac{2 N_{\mathcal R_i}}{R}- E_{ \gamma }[\frac{2 N_{\mathcal R_i}}{R}]$ where $i=2$ if $x>L$ and $i=1$ if $x<1$ otherwise in the OS-CA is equal to $m_\pm$ respectively. By (5.4) and (5.5), (10.28) vanishes as ${ \gamma }\rightarrow 0$. $\square$

1.3 Proof of (5.11)

We call

$$\begin{aligned} I_{x,{ \gamma }}^T={ \gamma }\sum _{t=0}^{T-1} E_{ \gamma }[j_{x,x+1}(t)] \end{aligned}$$

(10.30)

Lemma 4

There are c and $c'$ so that for all $r'<r''$ in $(0,\ell )$

$$\begin{aligned} \left| \frac{1}{x''-x'}\sum _{y=x'}^{x''}I_{y,{ \gamma }}^T\right| \le c,\quad x'=[{ \gamma }^{-1}r'],\quad x''=[{ \gamma }^{-1}r''] \end{aligned}$$

(10.31)

$$\begin{aligned} \Big |I_{x'',{ \gamma }}^T-I_{x',{ \gamma }}^T\Big |\le c' |r''-r'| \end{aligned}$$

(10.32)

Proof

By (10.23), using that $|\chi _{x,{ \gamma };t}|\le 2$ and $|{ \epsilon }_{x,{ \gamma };t}| \le 2C{ \gamma }$ for all x and t and after telescopic cancellations we get

$$\begin{aligned} \left| \frac{1}{x''-x'}\sum _{y=x'}^{x''}I_{y,{ \gamma }}^T\right| \le \left| { \gamma }\sum _{s=0}^{T-1}\frac{1}{x''-x'} E_{ \gamma }\left[ \frac{1}{2}(\eta (x',s)-\eta (x''+1,s))\right] \right| + 8 C { \gamma }^2T \end{aligned}$$

The right hand side converges to $\frac{1}{r''-r'}\int _0^\tau \frac{1}{2} [m(r'.s)-m(r'',s)] ds+8C^2\tau$ which, by the smoothness of m, proves (10.31).

We have that

$$\begin{aligned} \left| { \gamma }\sum _{t=0}^{T-1} j_{x',x'+1}(t)]-{ \gamma }\sum _{t=0}^{T-1} j_{x'',x''+1}(t)\right| \le c' { \gamma }|x''-x'| \end{aligned}$$

because the particles which contribute to the left hand site are: (1) those which reach for the first time $x'+1$ jumping from $x'$ and at the final time are in $[x'+1,x'']$; (2) those which reach for the first time $x''$ jumping from $x''+1$ and at the final time are in $[x'+1,x'']$; (3) those initially in $[x'+1,x'']$ and which leave this interval for the last time jumping to $x''+1$; (4) those initially in $[x'+1,x'']$ and which leave this interval for the last time jumping to $x'$. $\square$

The family $\{I_{x,{ \gamma }}^T\}$ thought as functions of $r={ \gamma }x$ are equibounded and equicontinuous in any compact of $(0,\ell )$, thus they converge pointwise by subsequences. We will then prove (5.11) by identifying the limit. By continuity it will be enough to prove

$$\begin{aligned} \lim _{{ \gamma }\rightarrow 0}\frac{1}{x''-x'}\sum _{y=x'}^{x''} I_{x,{ \gamma }}^T=\frac{1}{r''-r'} \int _{r'}^{r''}dr\int _0^\tau I(r,s)ds \end{aligned}$$

(10.33)

By (10.23)

$$\begin{aligned} \frac{1}{x''-x'}\sum _{y=x'}^{x''} I_{x,{ \gamma }}^T= & {} { \gamma }\sum _{s=0}^{T-1}\Bigg \{\frac{1}{x''-x'} E_{ \gamma }\left[ \frac{1}{2}(\eta (x',s)-\eta (x''+1,s))\right] \\&+ \frac{1}{x''-x'}\sum _{y=x'}^{x''}E_{ \gamma }[\chi _{x,{ \gamma };s}{ \epsilon }_{x,{ \gamma };s}+\chi _{x+1,{ \gamma };s}{ \epsilon }_{x+1,{ \gamma };s}]\Bigg \} \end{aligned}$$

The first term converges to

$$\begin{aligned} \frac{1}{r''-r'} \frac{1}{2}\int _0^\tau [m(r',s)-m(r'',s)]ds= \frac{1}{r''-r'} \frac{1}{2}\int _0^\tau ds\int _{r'}^{r''}dr\frac{\partial m(r,s)}{\partial s} \end{aligned}$$

(10.34)

By (10.25) and (10.26) the second one converges to

$$\begin{aligned} -\frac{1}{r''-r'} \int _0^\tau \int _{r'}^{r''} C [1- m^2]\int _{r}^{r+1}[m(r+\xi ,s) - m(r-\xi ,s)] d\xi dr ds \end{aligned}$$

1.4 Proof of (5.12)

As the two are similar, we just prove the second equality in (5.12). The same proof as the one for (10.32) shows that

$$\begin{aligned} \Big |I_{x,{ \gamma }}^T-I_{\mathrm{ch}\rightarrow \mathcal R_2,{ \gamma }}^T\Big |\le c' |\ell -r|,\quad x=[{ \gamma }^{-1}r] \end{aligned}$$

(10.35)

where

$$\begin{aligned} I_{\mathrm{ch}\rightarrow \mathcal R_2,{ \gamma }}^T= { \gamma }\sum _{t=0}^{T-1} E_{ \gamma }[j_{\mathrm{ch}\rightarrow \mathcal R_2}(t)] \end{aligned}$$

Let $\tilde{I}$ be a limit point of $I_{\mathrm{ch}\rightarrow \mathcal R_2,{ \gamma }}^T$ as ${ \gamma }\rightarrow 0$ then

$$\begin{aligned} \left| \int _0^\tau I(r,s)ds -\tilde{I}\right| \le c' |\ell -r| \end{aligned}$$

Using the expression (5.6) for I(r, t) and the continuity of m, we get in the limit $r\rightarrow \ell$ that $\tilde{I}=\int _0^\tau I(\ell ,s)ds$.

1.5 Proof of (5.8)

As the proofs are similar, we just prove the second equality in (5.8) for the CC-CA. Suppose by contradiction that there is $t>0$ such that $m(\ell ,t)\ne m_{+}(t)$ and for the sake of definiteness $m(\ell ,t)< m_{+}(t)$. Then there is $\delta >0$ and an interval $[t',t'']$ so that for $s\in [t',t'']$, $m_{+}(t)> m(\ell ,t) +\delta$. Recalling the Proof of Lemma 3

$$\begin{aligned} E_{ \gamma }[j_{\mathrm{ch}\rightarrow \mathcal R_2}(s)]= & {} E_{ \gamma }\left[ \frac{N_{\mathcal R_2}(s)}{R}\right] - \frac{1}{2} u_{ \gamma }(L,s)- E_{ \gamma }\Big [\chi _{L,{ \gamma };s}{ \epsilon }_{L,{ \gamma };s}\Big ] \\\ge & {} E_{ \gamma }\left[ \frac{N_{\mathcal R_2}(s)}{R}\right] -\frac{1}{2} u_{ \gamma }(L,s) -c{ \gamma }\end{aligned}$$

c a suitable constant, $c{ \gamma }$ bounding the term with ${ \epsilon }_{x,{ \gamma }}$. Then, recalling (2.3), (2.5), (2.6) and using the assumptions in Theorem 1 we get

$$\begin{aligned} \liminf _{{ \gamma }\rightarrow 0}{ \gamma }^2\sum _{s\in \mathbb Z \cap { \gamma }^{-2}[t',t'']} E_{ \gamma }[j_{\mathrm{ch}\rightarrow \mathcal R_2}(s)]\ge & {} \frac{1}{2} \int _{t'}^{t''}[m_+(s)- m(\ell ,s)]ds\ge \frac{\delta }{2} [t''-t'] \end{aligned}$$

which contradicts (5.12).

1.6 The Dynamics of the Reservoirs

We just prove (5.9). Let $\tau _0\ge 0$, $\tau >0$, $t_0=[{ \gamma }^{-2}\tau _0]$, $T=[{ \gamma }^{-2}\tau ]$, then

$$\begin{aligned} N_{\mathcal R_2}(t_0+T)-N_{\mathcal R_2}(t_0)=\sum _{t=t_0}^{t_0+T-1}\Big [ j_{\mathrm{ch}\rightarrow \mathcal R_2}(t)-j_{\mathcal R_2\rightarrow \mathcal R_1}(t)\big ] \end{aligned}$$

We take the expectation and we use (10.4) to get

$$\begin{aligned}&\left| E_{ \gamma }[N_{\mathcal R_2}(t_0+T)-N_{\mathcal R_2}(t_0)] -\sum _{t=t_0}^{t_0+T-1}\left[ E_{ \gamma }[j_{\mathrm{ch}\rightarrow \mathcal R_2}(t)]- E_{ \gamma }\left[ \frac{N_{\mathcal R_2}(t)-N_{\mathcal R_1}(t)}{R}\right] { \gamma }p \right] \right| \nonumber \\&\quad \le \frac{4{ \gamma }p}{R}T \end{aligned}$$

(10.36)

We then get

$$\begin{aligned} \frac{a}{2}[m_+(\tau _0+\tau )-m_+(\tau _0)]=\int _{\tau _0}^{\tau _0+\tau } I(\ell ,s)ds-p\int _{\tau _0}^{\tau _0+\tau } \frac{1}{2}[m_+(s)-m_-(s)]ds\nonumber \\ \end{aligned}$$

(10.37)

which is obtained from (10.36) by multiplying by ${ \gamma }$ and taking the limit ${ \gamma }\rightarrow 0$ after using that (1) $R=a{ \gamma }^{-1}$, (2) by (5.3)

$$\begin{aligned} \lim _{{ \gamma }\rightarrow 0}E_{ \gamma }\left[ \frac{N_{\mathcal R_2}(t)-N_{\mathcal R_1}(t)}{R}\right] =\frac{m_+(\tau )-m_-(\tau )}{2}, \quad t =[{ \gamma }^{-2}\tau ] \end{aligned}$$

(3) by (5.12)

$$\begin{aligned} \lim _{{ \gamma }\rightarrow 0}{ \gamma }\sum _{t=t_0}^{t_0+T-1} E_{ \gamma }[j_{\mathrm{ch}\rightarrow \mathcal R_2}(t)]=\int _{\tau _0}^{\tau _0+\tau } I(\ell ,s)ds \end{aligned}$$

Then (5.9) is obtained from (10.37) by dividing by $\tau$ and taking the limit $\tau \rightarrow 0$.