Non-equilibrium Stationary Properties of the Boundary Driven Zero-Range Process with Long Jumps

Abstract

We consider the zero-range process with long jumps and in contact with infinitely extended reservoirs in its non-equilibrium stationary state. We derive the hydrostatic limit and the Fick’s law, which are a consequence of a static relationship between the exclusion process and the zero-range process. We also obtain the large deviation principle for the empirical density, i.e. we compute the non-equilibrium free energy.

Appendix A: Macroscopic Properties of NESS in Boundary Driven Long Range Exclusion

In this section, we present some results for the stationary behaviour of the open boundary exclusion process with long jumps based on previous results from [5, 10, 13]. Recall (2.12) and note that we consider the process speeded up in the time scale $\Theta (N)$ as in (2.11). Recall that $\tilde{\alpha }, \tilde{\beta }\in (0,1)$.

1.1 A.1. Hydrostatic Limit

Let ${{\mathcal M}}^+$, be the space of positive measures on [0, 1] with total mass bounded by 1 and equipped with the weak $\star$-topology. For any $\eta \in \Omega ^N$ the empirical measure ${ \pi }^N_{ex}:={ \pi }^N_{ex} (\eta ) \in {{\mathcal M}}^+$ is defined by

$$\begin{aligned} {\pi }^N_{ex} (\eta ) ={\dfrac{1}{\# \Lambda _N}} \sum _{x\in \Lambda _N} \eta _x \delta _{\tfrac{x}{N}}(du). \end{aligned}$$

(A.1)

Let $P^N$ be the probability measure on ${{\mathcal M}}^+$ obtained as the pushforward of $\mu _{ss}^N$ by $\pi ^N_{ex}$. We denote the action of $\pi _{ex}^N \in {{\mathcal M}}^+$ on a continuous function $G:[0,1]\rightarrow {{\mathbb {R}}}$ by

$$\begin{aligned} \langle \pi ^N_{ex} , G \rangle = \int _{[0,1]} G(u) \ \pi ^N_{ex} (du). \end{aligned}$$

Theorem A.1

(Hydrostatic limit in mean) For any $G \in C^0 ([0,1])$ we have

$$\begin{aligned} \lim _{N\rightarrow \infty } E_{\mu _{ss}^N}\left[ \langle \pi ^N_{ex},G\rangle -\int _0^1\bar{\rho }(u)G(u)du\right] =0, \end{aligned}$$

where

• (a) for $\theta <0$ and $\gamma \in (0,2)$, ${\bar{\rho }} (u) = \dfrac{V_{0}(u)}{V_{1}(u)}.$

• (b) for $\theta =0$ and $\gamma \in (0,2)-\{1\}$, $\bar{\rho }(\cdot )$ is the unique weak solution of (2.23), with $\hat{\kappa }=\kappa$.

• (c) for $\theta \in (0,\gamma -1)$ and $\gamma \in (1,2)$, $\bar{\rho }(\cdot )$ is the unique weak solution of (2.24).

• (d) for $\theta = \gamma -1$ and $\gamma \in (1,2)$, $\bar{\rho }(\cdot )$ is the weak solution of (2.25) with $\hat{\kappa }=\kappa d$.

• (e) for $\theta > 0$ and $\gamma \in (0, 1]$ or for $\theta > \gamma - 1$ and $\gamma \in (1, 2)$, $\bar{\rho }(\cdot )$ is the weak solution of (2.27) with ${\hat{M}} =\tfrac{\tilde{\alpha }+\tilde{\beta }}{2}$, i.e. ${\bar{\rho }}=\tfrac{\tilde{\alpha }+\tilde{\beta }}{2}$.

Proof

We start by noting that a simple computation based on the fact that the mass of the system is finite, shows that the sequence $\{P^N\}_{N\ge 2}$ is tight and that all limit points are concentrated on measures $\pi (du)$ which are absolutely continuous with respect to Lebesgue measure on [0, 1], i.e. $\pi (du)=\rho (u)du$. Let us also introduce $\{\bar{\pi }^N\}_{N\ge 2}= \left\{ E_{\mu _{ss}^N} \left[ \pi _{ex}^N\right] \right\} _{N \ge 2}$ which forms a sequentially compact sequence of ${{\mathcal {M}}}^+$ whose limit points $\bar{\pi }(du)$ are absolutely continuous with respect to the Lebesgue measure on [0, 1], i.e. $\bar{\pi }(du)=\bar{\rho }(u)du$. Let $\bar{\pi }(du)=\bar{\rho }(u)du$ be a limit point of $\{\bar{\pi }^N\}_{N\ge 2}$. Without loss of generality we can consider a subsequence for which $\{P^N\}_{N\ge 2}$ is also converging to a limit point denoted by $P^*$ (the corresponding expectation is denoted by ${E} ^*$). To lighten notation, in the sequel, we assume that we are taking the limit according to this subsequence, even if it is not specified. Observe that $\bar{\rho }= E^* [\rho ]$. Our goal is to show that $\bar{\rho }$ is unique and given as in Theorem A.1.

If $\gamma \in (0,2)$, the energy estimates of Sect. 3.3 of [10] show that for $\theta \ge 0$

$$\begin{aligned} E^*\left[ \Vert \rho \Vert _{\gamma /2}^2\right] <\infty \end{aligned}$$

and for $\theta \le 0$

$$\begin{aligned} E^*\left[ \int _0^1 \left\{ \frac{(\tilde{\alpha }-\rho (u))^2}{u^\gamma } + \frac{(\tilde{\beta }-\rho (u))^2}{(1-u)^\gamma } \right\} du \right] < \infty . \end{aligned}$$

From Jensen’s inequality, we have also

$$\begin{aligned} \Vert \bar{\rho }\Vert _{\gamma /2}<\infty , \quad \int _0^1 \left\{ \frac{(\tilde{\alpha }-\bar{\rho }(u))^2}{u^\gamma } + \frac{(\tilde{\beta }-\bar{\rho }(u))^2}{(1-u)^\gamma } \right\} du < \infty . \end{aligned}$$

It remains now to check that $\bar{\rho }$ satisfies the other conditions in the notions of stationary weak solutions.

Recall (2.11). Note that

$$\begin{aligned} \begin{aligned} \Theta (N) \mathcal {L}_N\left\langle {\pi }^N_{ex}, G\right\rangle&=\frac{\Theta (N)}{\# \Lambda _N} \sum _{x \in \Lambda _{N}}\left( \mathbb {L}_{N} G\right) \left( \tfrac{x}{N}\right) \eta (x) \\&\quad +\frac{\kappa \Theta (N)}{ N^{\theta } \# \Lambda _N} \sum _{x \in \Lambda _{N}} G\left( \tfrac{x}{N}\right) \left\{ r_{N}^{-}\left( \tfrac{x}{N}\right) \left( \tilde{\alpha }-\eta (x)\right) +r_{N}^{+}\left( \tfrac{x}{N}\right) \left( \tilde{\beta }-\eta (x)\right) \right\} \end{aligned} \end{aligned}$$

(A.2)

where the action of $\mathbb {L}_{N}$ on functions G is defined by

$$\begin{aligned} \left( \mathbb {L}_{N} G\right) \left( \tfrac{x}{N}\right) =\sum _{y \in \Lambda _{N}} p(y-x)\left[ G\left( \tfrac{y}{N}\right) -G\left( \tfrac{x}{N}\right) \right] . \end{aligned}$$

(A.3)

Taking the expectation with respect to $\mu _{ss}^N$ on (A.2), we get, from stationarity, that

$$\begin{aligned} \begin{aligned} 0&=\frac{\Theta (N)}{\# \Lambda _N} \sum _{x \in \Lambda _{N}}\left( \mathbb {L}_{N} G\right) \left( \tfrac{x}{N}\right) E_{\mu _{ss}^N}[\eta (x)] \\&\quad +\frac{\kappa \Theta (N)}{ N^{\theta } \# \Lambda _N} \sum _{x \in \Lambda _{N}} G\left( \tfrac{x}{N}\right) \left\{ r_{N}^{-}\left( \tfrac{x}{N}\right) \left( \tilde{\alpha }-E_{\mu _{ss}^N}[\eta (x)]\right) +r_{N}^{+}\left( \tfrac{x}{N}\right) \left( \tilde{\beta }-E_{\mu _{ss}^N}[\eta (x)]\right) \right\} . \end{aligned} \end{aligned}$$

(A.4)

Recall (2.9). We define the functions $r_{N}^{\pm }: [0,1]\rightarrow {{\mathbb {R}}}$ as the linear interpolation of $r_N^- \left( \tfrac{x}{N}\right)$ and $r_N^+ \left( \tfrac{x}{N}\right)$ for all $x \in \Lambda _{N}$ with $r_{N}^{\pm }(0) = r_{N}^{\pm }\left( \tfrac{1}{N}\right)$ and $r_{N}^{\pm }(1) = r_{N}^{\pm }\left( \tfrac{N-1}{N}\right)$. By Lemma 3.3 in [13], for $0<\gamma <2$ we have that

$$\begin{aligned} \begin{aligned}&\lim _{N\rightarrow \infty } N^{\gamma } \left( r_N^-\right) (u) =r^-(u),\,\,\,\lim _{N\rightarrow \infty } N^{\gamma } \left( r_N^+\right) (u) =r^+(u) \end{aligned} \end{aligned}$$

(A.5)

uniformly in $[a,1-a]$ for $a\in (0,1)$ and from that lemma it also follows that

$$\begin{aligned} \lim _{N\rightarrow \infty }N^{\gamma }({{\mathbb L}}_N G)(u) = ({{\mathbb {L}}}G)(u) \end{aligned}$$

(A.6)

uniformly in $[a,1-a]$, for all functions G with compact support included in $[a,1-a]$.

Now, we split the analysis by taking into account the value of $\theta$.

Case $\theta < 0$: In this regime we take $G\in C_c^\infty ((0,1))$ and $\Theta (N) = N^{\theta +\gamma }$. Observe that (A.4) can be written as

$$\begin{aligned} \begin{aligned} 0&=\langle \bar{\pi }^N,N^{\gamma +\theta }\mathbb {L}_{N} G\rangle -\kappa \langle \bar{\pi }^N, G N^\gamma (r_N^-+r_N^+)\rangle \\&\quad +\frac{\kappa }{\# \Lambda _N } \sum _{x \in \Lambda _{N}} G\left( \tfrac{x}{N}\right) N^\gamma \left\{ r_{N}^{-}\left( \tfrac{x}{N}\right) \tilde{\alpha }+r_{N}^{+}\left( \tfrac{x}{N}\right) \tilde{\beta }\right\} \end{aligned} \end{aligned}$$

(A.7)

From the weak convergence together with (A.5) and (A.6), last display converges, as $N\rightarrow +\infty ,$ to

$$\begin{aligned} -\kappa \int _0^1 G(u) \left\{ \bar{\rho }(u)V_1(u) du - V_0(u)\right\} du =0 \end{aligned}$$

which implies that $\bar{\rho }(u)=\frac{V_0(u)}{V_1(u)}$ for almost every u (see also Remark (2.14) in [10]).

Case $\theta = 0$: In this regime we take $G\in C_c^\infty ((0,1))$ and $\Theta (N) = N^{\gamma }$. From weak convergence together with (A.5) and (A.6) we get that

$$\begin{aligned} \begin{aligned} 0=\int _0^1 (\mathbb {L}G)(u)\bar{\rho }(u)du -\kappa \int _0^1 G(u)\bar{\rho }(u)V_1(u) du +\kappa \int _0^1 G(u) V_0(u) du. \end{aligned} \end{aligned}$$

Hence we have that $F_{RD} (\bar{\rho }, G)=0$ for any $G\in C_c^\infty ((0,1))$ (with ${\hat{\kappa }}=\kappa )$.

Case $\theta \in (0,\gamma -1)$ and $\gamma \in (1,2)$: In this regime we take $G\in C_c^\infty ((0,1))$ and $\Theta (N) = N^{\gamma }$. From the weak convergence, the rightmost term in the first line of (A.4) converges, as $N\rightarrow +\infty$, to

$$\begin{aligned} \int _0^1 (\mathbb {L}G)(u)\bar{\rho }(u)du. \end{aligned}$$

Moreover, the term on the second line of (A.2) can be bounded from above by a constant times

$$\begin{aligned} N^{\gamma -\theta -1}\sum _{x\in \Lambda _N} x^{-\gamma }G\left( \tfrac{x}{N}\right) \lesssim N^{-\theta } \end{aligned}$$

plus lower order terms in N which vanish as $N\rightarrow +\infty$. Hence we have that $F_{Dir} (\bar{\rho }, G)=0$ for any $G\in C_c^\infty ((0,1))$.

Case $\theta = \gamma -1$ and $\gamma \in (1,2)$: In this regime we take $G\in C^\infty ([0,1])$ and $\Theta (N) = N^{\gamma }$. We start by noting that from Lemma 5.1 of [5], we have that

$$\begin{aligned} \lim _{N \rightarrow \infty } N^{-1} \sum _{x \in \Lambda _{N}}\left| N^{\gamma }\left( \mathbb {L}_{N} G\right) \left( \tfrac{x}{N}\right) -(\mathbb {L} G)\left( \tfrac{x}{N}\right) \right| =0 \end{aligned}$$

for functions $G\in C^\infty ([0,1])$ and $\gamma \in (0,2).$ Moreover, the first term on the second line of (A.4) can be rewritten as

$$\begin{aligned} \begin{aligned}&\frac{\kappa N^{\gamma }}{ N^{\gamma -1} \# \Lambda _N}\left( \tilde{\alpha }-E_{\mu _{ss}^N}\Big [{A}[\eta ,I_{\varepsilon N}(0)]\Big ]\right) \sum _{x \in \Lambda _{N}} G\left( \tfrac{x}{N}\right) r_{N}^{-}\left( \tfrac{x}{N}\right) \\&\quad +\frac{\kappa N^{\gamma }}{ N^{\gamma -1} \# \Lambda _N} \sum _{x \in \Lambda _{N}} G\left( \tfrac{x}{N}\right) r_{N}^{-}\left( \tfrac{x}{N}\right) \left( E_{\mu _{ss}^N}\Big [ {A}[\eta ,I_{\varepsilon N}(0)]\Big ]-\eta (x)\right) \end{aligned} \end{aligned}$$

(A.8)

plus analogous terms with respect to the right boundary. The expectation of the second term above converges to zero as $N\rightarrow +\infty$ from Lemma 5.9 of [5] (see Remark 5.10 there). We note that, to apply Lemma 5.9 of [5], which is written with the time integral and in the $L^1$ sense, in last argument, we need to use the fact that $\mu _{ss}^N$ is a stationary measure, to introduce the time integral in the rightmost term of (A.8) and then use the aforementioned lemma. Now, for the first term, we can perform a Taylor expansion on G obtaining the following expression,

$$\begin{aligned} \begin{aligned}&\frac{\kappa N^{\gamma }G(0)\left( \tilde{\alpha }- E_{\mu _{ss}^N}\Big [ {A}[\eta ,I_{\varepsilon N}(0)]\Big ]\right) }{N^{\gamma -1} \ \# \Lambda _N} \sum _{x \in \Lambda _{N}} r_{N}^{-}\left( \tfrac{x}{N}\right) \\&+\frac{\kappa G^{\prime }(0)\left( \tilde{\alpha }-E_{\mu _{ss}^N}\Big [ {A}[\eta ,I_{\varepsilon N}(0)]\Big ]\right) }{\# \Lambda _N} \sum _{x \in \Lambda _{N}} x r_{N}^{-}\left( \tfrac{x}{N}\right) \end{aligned} \end{aligned}$$

plus lower order terms in N. Observe that

$$\begin{aligned} \begin{aligned} \dfrac{\kappa \left| G^{\prime }(0)\left( \tilde{\alpha }-E_{\mu _{ss}^N}\Big [ {A}[\eta ,I_{\varepsilon N}(0)]\Big ]\right) \right| }{\# \Lambda _N} \sum _{x \in \Lambda _{N}} x r_{N}^{-}\left( \tfrac{x}{N}\right) \lesssim&\frac{1}{\# \Lambda _N} \sum _{x\in \Lambda _N} x^{1-\gamma } = \mathcal {O}(N^{1-\gamma }), \end{aligned} \end{aligned}$$

which goes to zero as N goes to infinity. Finally, the remaining term can be treated by using the weak convergence, the fact that the limiting measure $\bar{\pi }(du)$ is absolutely continuous with respect to the Lebesgue measure with density $\bar{\rho }$ and also that

$$\begin{aligned} \lim _{N\rightarrow +\infty }\sum _{x\in \Lambda _N}r_N^\pm \left( \tfrac{x}{N}\right) =d. \end{aligned}$$

Hence we get that $F_{Rob} (\bar{\rho }, G)=0$ (with $\hat{\kappa }=\kappa d$) for any $G\in C^\infty ([0,1])$.

Case $\theta >\gamma -1$ and $\gamma \in (1,2)$: The proof in this regime is completely analogous to the previous case. Nevertheless, we could also obtain the result as a consequence of the hydrostatic limit, with a convergence in probability, by following the same strategy described in [51]. Since we do not need this stronger convergence to attain our results, we did not pursue this issue here.

Case $\theta > 0$ and $\gamma \in (0,1)$: In this case the analysis of the first term of (A.4) is given by an approximation argument of the operator $\mathbb {L}$ given in Lemma 5.1 of [5]. Since G(x/N) and $\eta (x)$ are bounded we know that the second term of (A.4) is bounded by a constant times

$$\begin{aligned} N^{\gamma -\theta -1} \sum _{x \in \Lambda _{N}} \left\{ r_{N}^{-}\left( \tfrac{x}{N}\right) +r_{N}^{+}\left( \tfrac{x}{N}\right) \right\} . \end{aligned}$$

Since $\gamma \in (0,1)$, we lose the convergence of the partial sum above. However, it is not difficult to see that

$$\begin{aligned} \begin{aligned}&N^{\gamma -\theta -1} \sum _{x \in \Lambda _{N}} \left\{ r_{N}^{-}\left( \tfrac{x}{N}\right) +r_{N}^{+}\left( \tfrac{x}{N}\right) \right\} \lesssim N^{-\theta }, \end{aligned} \end{aligned}$$

(A.9)

which vanishes as N goes to $\infty$. Hence $F_{Neu} (\bar{\rho }, G)=0$ for any $G\in C^\infty ([0,1])$.

In the two last cases, we have also to show that $\int _0^1 {\bar{\rho }} (u) du =\tfrac{\tilde{\alpha }+ \tilde{\beta }}{2}$. In fact, since in these cases, $F_{Neu} (\bar{\rho }, G)=0$ for any $G\in C^\infty ([0,1])$, we can conclude that $\bar{\rho }$ is equal to a constant M by using the same argument as in Lemma 2.14 when showing that $\Vert {\bar{\rho }}\Vert _{\gamma /2}=0$.

Recall the definition of ${\bar{\pi }}^N$ given at the beginning of the proof. We consider a sequence $(\chi ^\varepsilon )_{0<\varepsilon < 1/4}$ of smooth functions with values in [0, 1], symmetric with respect to 1/2, equal to 0 on $[0,\varepsilon /2]$ and to 1 on $[\varepsilon ,1/2]$. Observe that $(\chi ^\varepsilon )_\varepsilon$ converges in $L^1$ as $\varepsilon$ goes to 0 to the constant function equal to 1. Taking $G=\chi ^\varepsilon$ in (A.4) we get

$$\begin{aligned} \begin{aligned}&\frac{1}{N}\sum _{x \in \Lambda _{N}} \chi ^\varepsilon \left( \tfrac{x}{N}\right) \left\{ \tilde{\alpha }N^\gamma r_{N}^{-}\left( \tfrac{x}{N}\right) + \tilde{\beta }N^\gamma r_{N}^{+}\left( \tfrac{x}{N}\right) \right\} \\&\quad =\int _0^1 \chi ^\varepsilon (u) (N^\gamma r_N^- (u) +N^\gamma r_N^+ (u)) {\bar{\pi }}^N (du) + N^{-\theta } \frac{1}{N} \sum _{x \in \Lambda _N} (N^\gamma {\mathbb {L}}_N \chi ^\varepsilon ) \left( \tfrac{x}{N} \right) E_{\mu _{ss}^N} \left[ \eta (x)\right] . \end{aligned} \end{aligned}$$

By (A.6) and (A.5), we can then replace in the previous expression $N^\gamma {\mathbb {L}}_N$ by ${\mathbb {L}}$ and $N^\gamma r_N^\pm$ by $r^\pm$. Recall that ${\bar{\pi }}^N (du)$ converges weakly to $\bar{\rho }(u) du=M du$. Since $\theta >0$ we get

$$\begin{aligned} \int _0^1 \chi ^\varepsilon (u) V_1 (u) du = M \int _0^1 \chi ^\varepsilon (u) V_0 (u) du. \end{aligned}$$

(A.10)

In the case $\gamma \in (0,1)$ the functions $V_0$ and $V_1$ are integrable and we get by sending $\varepsilon$ to 0 that

$$\begin{aligned} M=\frac{\int _0^1 V_1 (u) du}{\int _0^1 V_0 (u) du}=\frac{\tilde{\alpha }+ \tilde{\beta }}{2}. \end{aligned}$$

In the case $\gamma \in (1,2)$ the integrals are diverging but

$$\begin{aligned} \int _0^1 \chi ^\varepsilon (u) V_1 (u) du \sim \frac{c_\gamma }{\gamma (\gamma -1)}\varepsilon ^{1-\gamma }\left[ \tilde{\alpha }+\tilde{\beta }\right] , \quad \int _0^1 \chi ^\varepsilon (u) V_0 (u) du \sim 2 \frac{c_\gamma }{\gamma (\gamma -1)}\varepsilon ^{1-\gamma }\ \end{aligned}$$

therefore we get again that

$$\begin{aligned} M=\frac{\tilde{\alpha }+ \tilde{\beta }}{2}. \end{aligned}$$

Conclusion: All limit points of the sequence $\{\bar{\pi }^N (du) \}_{N\ge 2}$ are in the form $\bar{\rho }(u) du$ where $\bar{\rho }$ is a weak solution of the hydrostatic equation. By uniqueness of weak solutions for these equations, $\bar{\rho }$ is unique and therefore the sequence is converging, without extracting a subsequence, to this unique weak solution. $\square$

Lemma A.2

The profiles $\bar{\rho }$ in Theorem A.1 are continuous in (0, 1).

Proof

In the case $\theta <0$ the claim follows easily since the profile is explicit. We consider now $\theta \ge 0$. If $\gamma \in (1,2)$, by definition of a weak solution, we know that $\bar{\rho }$ is bounded and belongs to ${\mathcal {H}}^{\gamma /2}$ and from Theorem 8.2 of [24], we conclude that $\bar{\rho }$ is $\tfrac{\gamma -1}{2}$-Hölder in [0, 1], therefore continuous in (0, 1). If $\gamma \in (0,1)$ and $\theta >0$, the profile is constant and therefore continuous. The only missing case is $\gamma \in (0,1)$ and $\theta =0$. We have hence to prove that the stationary solution $\rho$ of the regional fractional reaction–diffusion equation (2.23) is continuous in (0, 1) when $\gamma \in (0,1)$. It is known that if $\gamma \in (0,1]$, the condition $f \in {{\mathcal {H}}}^{\gamma /2}$ does not guarantee, contrarily to the case $\gamma \in (1,2)$, that f is continuous. Therefore the continuity property of $\rho$ can only result from the fact that $\rho$ satisfies the weak formulation of (2.23). This property is a consequence of potential theory for (fractional) Shrödinger theory developed in [17], more exactly of Proposition 6.1 of that article that we restate in our particular context. Before doing so, we introduce a few notations.

The fractional Laplacian $\vert \Delta \vert ^{\gamma /2}$ on ${\mathbb {R}}$ is the operator acting on functions $f:{\mathbb {R}} \rightarrow {{\mathbb {R}}}$ such that

$$\begin{aligned} \int _{{\mathbb {R}}} \frac{|f(u)|}{(1 +|u|)^{1+\gamma }} du < \infty \end{aligned}$$

(A.11)

as

$$\begin{aligned} -\left( \vert \Delta \vert ^{\gamma /2} f\right) (u) = c_\gamma \lim _{\epsilon \rightarrow 0} \int _{{\mathbb {R}}} {{\mathbb 1}}_{|u-v| \ge \epsilon } \frac{f(v) -f(u)}{|u-v|^{1+\gamma }} dv, \end{aligned}$$

(A.12)

for any $u \in {\mathbb {R}}$ if the limit exists (which is for example the case for smooth compactly supported functions). It can be extended into the weak fractional Laplacian (that we denote abusively by the same notation) by duality: For any f satisfying (A.11), $\vert \Delta \vert ^{\gamma /2} f$ is the distribution (or generalized function) on ${\mathbb {R}}$ satisfying the identity $\langle \vert \Delta \vert ^{\gamma /2} f, h \rangle = \langle f, \vert \Delta \vert ^{\gamma /2} h \rangle$, for any compactly supported function h on ${\mathbb {R}}$.

Property 6.1 of [17] claims that if $f:{\mathbb {R}} \rightarrow {\mathbb {R}}$ is a solution, in the distributional sense⁵, on (0, 1) of the equation

$$\begin{aligned} \vert \Delta \vert ^{\gamma /2} f + q f =0 \end{aligned}$$

(A.13)

then f is continuous on (0, 1) as soon as the function $q:(0,1) \rightarrow {\mathbb {R}}$ belongs to the (local) Kato class of exponent $\gamma$, i.e.

$$\begin{aligned} \lim _{r\rightarrow 0}\, \sup _{u\in {\mathbb {R}}} \int _{u-r}^{u+r} \frac{ ({{\mathbb 1}}_{[\varepsilon , 1-\varepsilon ]} q) (u) }{|u-v|^{1-\gamma }} \ dv =0 \end{aligned}$$

for any $\varepsilon \in (0,1/2)$. It is not difficult to see that if $\rho$ is the weak solution of (2.23) and is extended by 0 outside of (0, 1), then $\rho$ satisfies, in the distributional sense, (A.13) on (0, 1) for q given as a linear combination of $r^-$ and $r^+$ (defined by (2.22)). Hence to conclude the proof of the lemma, it is sufficient to prove that $r^\pm$ belong to the (local) Kato class of exponent $\gamma$. This exercise is left to the interested reader. $\square$

1.2 A.2. Fractional Fick’s Law of the Boundary Driven Exclusion

By adapting the strategy of [13] we can obtain the “fractional Fick’s Law” which is given in the next theorem. The expression of the current is given by

$$\begin{aligned} \begin{aligned} W_x^{ex}(\eta )=&\sum _{\begin{array}{c} 1 \le y\le x-1 \\ x-1<z\le N-1 \end{array}} p(z-y) (\eta (y) -\eta (z) ) \\&\quad +{\frac{\kappa }{ N^{\theta }}}\Bigg [\sum _{x \le z \le N-1} r_N^-\left( \tfrac{z}{N}\right) ( \tilde{\alpha }-\eta (z)) -\sum _{1 \le y\le x-1}r_N^+\left( \tfrac{y}{N}\right) (\tilde{\beta }-\eta (y))\Bigg ]. \end{aligned} \end{aligned}$$

(A.14)

Theorem A.3

(Fractional Fick’s law) Let $\bar{\rho }(\cdot )$ be the hydrostatic profile of the boundary driven exclusion given in Theorem A.1. For $u\in (0,1)$ the following fractional Fick’s law holds, apart from the case $\theta =0$ and $\gamma =1$ :

$$\begin{aligned} \lim _{N\rightarrow \infty }\frac{1}{B_N(\theta )} {E}_{ \mu _{ss}^N}\left[ W_{[uN]}^{ex}\right] = \int _0^1 h_\theta (u)\bar{\rho }(u)du + C(\tilde{\alpha },\tilde{\beta },\theta ),\end{aligned}$$

(A.15)

where

$$\begin{aligned} B_N(\theta ): = N^{1-\gamma }{\mathbb 1}_{\theta \ge 0}+ {N^{1-\theta -\gamma }}{\mathbb 1}_{\theta < 0}, \end{aligned}$$

(A.16)

the function $h_\theta :(0,1)\rightarrow \mathbb {R}$ is given by

$$\begin{aligned} h_\theta (u)={\left\{ \begin{array}{ll} c_\gamma \left( \dfrac{\kappa }{\gamma }{} {\textbf {1}}_{\theta \le 0} {+}\dfrac{1}{1-\gamma }{} {\textbf {1}}_{\theta \ge 0}\right) [(1-u)^{1-\gamma }-u^{1-\gamma }], \quad \text {if} \quad \gamma \ne 1 ,\\ c_\gamma {\textbf {1}}_{\theta \ge 0}[\log (1-u)-\log (u)], \quad \text {if} \quad \gamma = 1 \end{array}\right. }\end{aligned}$$

(A.17)

and

$$\begin{aligned} C(\tilde{\alpha },\tilde{\beta }, \theta ) = \dfrac{c_\gamma \kappa (\tilde{\alpha }-\tilde{\beta })}{\gamma (2-\gamma )}{} {\textbf {1}}_{ \theta \le 0, } \end{aligned}$$

(A.18)

This implies that

• (a) for $\theta < 0$,

  $$\begin{aligned} \begin{aligned} \lim _{N \rightarrow \infty }\frac{1}{N^{1-\theta -\gamma }}{{\mathbb {E}}}_{ \mu _{ss}^N}\left[ {W_{[uN]}^{ex}}\right]&= \kappa \int _{u}^{1}(\tilde{\alpha }-\bar{\rho }(v))r^{-}(v)dv- \kappa \int _{0}^{u}(\tilde{\beta }-\bar{\rho }(v))r^{+}(v)dv\\&=\kappa c_\gamma \gamma ^{-1} \int _0^1\dfrac{\tilde{\alpha }-\tilde{\beta }}{v^\gamma +(1-v)^\gamma }dv; \end{aligned} \end{aligned}$$

  (A.19)

• (b) for $\theta = 0$,

  $$\begin{aligned} \begin{aligned} \lim _{N \rightarrow \infty } \frac{1}{N^{1-\gamma }} {{\mathbb {E}}}_{\mu _{ss}^N}\left[ {W_{[uN]}^{ex}}\right] =c_{\gamma } \int _{0}^u \; \int _{u}^{1} \, \frac{{ \bar{\rho }} (v) -{ \bar{\rho }} (w)}{(w-v)^{1+\gamma }}\, dw dv&+ \kappa \int _{u}^{1}(\tilde{\alpha }-\bar{\rho }(v))r^{-}(v)dv\\&- \kappa \int _{0}^{u}(\tilde{\beta }-\bar{\rho }(v))r^{+}(v)dv; \end{aligned} \end{aligned}$$

  (A.20)

• (c) for $\theta >0$,

  $$\begin{aligned} \begin{aligned} \lim _{N \rightarrow \infty } \frac{1}{N^{1-\gamma }}{{\mathbb {E}}}_{\mu _{ss}^N}\left[ {W_{[uN]}^{ex}}\right]&=c_{\gamma } \int _{0}^u \; \int _{u}^{1} \, \frac{{ \bar{\rho }} (v) -{ \bar{\rho }} (w)}{(w-v)^{1+\gamma }}\, dw dv. \end{aligned} \end{aligned}$$

  (A.21)

Proof

Since the measure $\mu _{ss}^N$ is stationary and ${\mathcal {L}}_N \eta _ x = W_x^{ex}(\eta )-W_{x+1}^{ex}(\eta )$ for all $x\in \Lambda _N$ and for all $\eta$, it follows that ${E}_{\mu _{ss}^N}[ W_x^{ex}] = {E}_{\mu _{ss}^N}[ W_1^{ex}],$ for all $x\in \Lambda _N$. Then we can write

$$\begin{aligned} \begin{aligned} {E}_{\mu _{ss}^N}[ W_1^{ex}]&= \frac{\kappa N^{-\theta }}{\# \Lambda _N}\sum _{z\in \Lambda _N} z \left[ \tilde{\alpha }- {E}_{\mu _{ss}^N}[\eta (z)] \right] \sum _{y\le 0}p(z-y) \\&\quad + \frac{\kappa N^{-\theta }}{\# \Lambda _N}\sum _{y\in \Lambda _N} (N-1-y) \left[ {E}_{\mu _{ss}^N}[\eta (y)]-\tilde{\beta }\right] \sum _{z\ge N}p(z-y) \\&\quad +\frac{1}{\# \Lambda _N}\sum _{z\in \Lambda _N}\sum _{y=1}^{z-1}p(z-y) (z-y)\left[ {E}_{\mu _{ss}^N}[\eta (y)]-{E}_{\mu _{ss}^N}[\eta (z)]\right] . \end{aligned} \end{aligned}$$

(A.22)

We define the linear interpolation functions ${\tilde{r}}_{N}^{\pm } :[0,1]\rightarrow {{\mathbb {R}}}$, such that for all $z\in \Lambda _N$ we have that

$$\begin{aligned} {\tilde{r}}_{N}^{-}\left( \tfrac{z}{N}\right) =\sum _{y \ge z} y p(y) \quad \text {and} \quad {\tilde{r}}_{N}^{+}\left( \tfrac{z}{N}\right) =-\sum _{y \le z-N} y p(y).\end{aligned}$$

(A.23)

Using (A.22) it is not difficult to see that

$$\begin{aligned} \frac{1}{B_N(\theta )} {E}_{\mu _{ss}^N}[ W_1^{ex}] =\frac{1}{\# \Lambda _N} \sum _{x\in \Lambda _N}h_\theta ^N\left( \tfrac{x}{N}\right) E_{\mu _{ss}^N}[\eta (x)] + \dfrac{ C_\theta ^N}{B_N(\theta )}, \end{aligned}$$

(A.24)

where

$$\begin{aligned} h_\theta ^N \left( \tfrac{x}{N}\right) = \frac{\kappa N^{1-\theta }}{B_N(\theta )}\left[ -\tfrac{x}{N}r^-_N\left( \tfrac{x}{N}\right) +\left( \tfrac{N-1-x}{N}\right) r^+_N\left( \tfrac{x}{N}\right) \right] +\frac{{{\tilde{r}}}^-_N\left( \tfrac{x}{N}\right) - {{\tilde{r}}}^+_N\left( \tfrac{x}{N}\right) }{B_N(\theta )}, \end{aligned}$$

and

$$\begin{aligned} C_{\theta }^N:={\frac{\kappa }{ N^{\theta }\ \# \Lambda _N}}\left[ \tilde{\alpha }\sum _{z\in \Lambda _N}\sum _{y\le 0}z p(z-y) -\tilde{\beta }\sum _{y\in \Lambda _N}\sum _{z\ge N}(N-1-y)p(z-y)\right] . \end{aligned}$$

From (A.23), we get the following convergence

$$\begin{aligned} \lim _{N\rightarrow \infty } \left| h_\theta ^N\left( \tfrac{[uN]}{N}\right) - h_\theta (u) \right| = 0 \end{aligned}$$

(A.25)

which holds uniformly for $u \in (a,1-a)$ (with $0<a<1$ fixed), and the function $h_\theta :(0,1)\rightarrow \mathbb {R}$ is given in (A.17). We note that $h_\theta$ is singular at $u=0$ and $u=1$ if $\gamma \in (1,2)$, but it is integrable in [0, 1] for $\gamma \in (0,2)$. Moreover, it is easy to see that, for $u\in (0,1)$ and for $\gamma \ne 1$ we have that

$$\begin{aligned} \vert h_\theta (u)\vert \lesssim u^{1-\gamma }+(1-u)^{1-\gamma }. \end{aligned}$$

(A.26)

Regarding the last term of (A.24), a simple computation shows that

$$\begin{aligned} \displaystyle \lim _{N\rightarrow \infty }\frac{1}{B_N(\theta )}C_\theta ^N=C(\tilde{\alpha },\tilde{\beta },\theta ), \end{aligned}$$

where $C(\tilde{\alpha },\tilde{\beta },\theta )$ was defined in (A.18). Now, note that from (A.25) and the fact that $| \eta (x)|\le 1$, we get that

$$\begin{aligned} \lim _{N\rightarrow \infty }\left| \frac{1}{\# \Lambda _N} \sum _{x\in N I(a)}\Big ( h_\theta ^N\left( \tfrac{x}{N}\right) -h_\theta \left( \tfrac{x}{N}\right) \Big )E_{\mu _{ss}^N}[\eta (x)] \right| = 0, \end{aligned}$$

(A.27)

where $I(a) = [a,1-a]$ and $NI(a)= [Na,N(1-a)]\cap \mathbb {N}.$ From (A.26) and the fact that $| \eta (x)|\le 1$ we get that

$$\begin{aligned} \left| \frac{1}{\# \Lambda _N} \sum _{x\in (N I(a))^{c}} h_\theta ^{N}\left( \tfrac{x}{N}\right) {E}_{\mu _{ss}^N} [ \eta (x) ] \right| \lesssim \left[ a^{2-\gamma }+(1-a)^{2-\gamma }\right] . \end{aligned}$$

Moreover, from Theorem A.1 and, for $h_\theta ^a$ a continuous extension of the function $h_\theta$ restricted to I(a), we get that

$$\begin{aligned} \lim _{N\rightarrow \infty }\left| \frac{1}{\# \Lambda _N} \sum _{x\in \Lambda _N}h_\theta ^a\left( \tfrac{x}{N}\right) E_{\mu _{ss}^N}[\eta (x)] -\int _0^1 h_\theta ^a(u)\bar{\rho }(u)du\right| = 0. \end{aligned}$$

(A.28)

Now, from (A.24) and (A.28) sending first $N\rightarrow \infty$ and then $a\rightarrow 0$ we obtain (A.15).

The other expressions of the limiting current given at the end of the theorem are achieved by using properties of the integrals and the fact that the limit does not depend on the variable u. To check them properly, let us consider for instance $\theta >0$ and $\gamma \ne 1$. Since the limit does not depend on u, we have that

$$\begin{aligned} c_{\gamma } \int _{0}^u \; \int _{u}^{1} \, \frac{{ \bar{\rho }} (v) -{ \bar{\rho }} (w)}{(w-v)^{1+\gamma }}\, dw dv=c_\gamma \int _0^1\int _0^u\int _u^1 \frac{\bar{\rho }(v)-\bar{\rho }(w)}{(w-v)^{\gamma +1}}dwdvdu. \end{aligned}$$

Using Fubini’s theorem twice, last display equals to

$$\begin{aligned} c_\gamma \int _0^1\int _v^1\int _v^w \frac{\bar{\rho }(v)-\bar{\rho }(w)}{(w-v)^{\gamma +1}}dudwdv=c_\gamma \int _0^1\int _v^1 \frac{\bar{\rho }(v)-\bar{\rho }(w)}{(w-v)^{\gamma }}dwdv. \end{aligned}$$

Finally, a simple computation, based again on Fubini’s theorem, shows that last display is equal to

$$\begin{aligned} \int _0^1 h_\theta (v)\bar{\rho }(v)dv . \end{aligned}$$

This ends the case $\theta >0$. The cases $\theta \le 0$ can be obtained by performing similar computations to the ones above, plus the fact that

$$\begin{aligned} \kappa \int _{0}^1 \int _{u}^{1}\tilde{\alpha }r^{-}(v)dv\, du- \kappa \int _{0}^{1}\int _0^u\tilde{\beta }r^{+}(v)dv\, du=\frac{c_\gamma \kappa (\tilde{\alpha }-\tilde{\beta })}{\gamma (2-\gamma )}. \end{aligned}$$

Finally, we note that the second equality in item a) is obtained by algebraic manipulations using the fact that $\bar{\rho }(u) = \dfrac{ V_0(u)}{V_1(u)}$. $\square$