Moderate Deviations for the SSEP with a Slow Bond

Abstract

We consider the one dimensional symmetric simple exclusion process with a slow bond. In this model, particles cross each bond at rate $N^2$, except one particular bond, the slow bond, where the rate is N. Above, N is the scaling parameter. This model has been considered in the context of hydrodynamic limits, fluctuations and large deviations. We investigate moderate deviations from hydrodynamics and obtain a moderate deviation principle.

Appendix A

1.1 A.1: Lemma 1.1

Proof of Lemma 1.1

We use $e_{-n}$ to denote $-(2\pi n)^2$ for $n\ge 0$ and use $e_n$ to denote $-k_n^2$ for $n\ge 1$. Let

$$\begin{aligned} \widehat{\mathscr {G}}_0=\left\{ G\in C^2[0,1]:~G^\prime (0)=G^\prime (1)=G(0)-G(1)\right\} , \end{aligned}$$

then, as we will show at the end of this proof, $\{e_n\}_{-\infty<n<+\infty }$ are all the eigenvalues of $\tilde{\Delta }$ limited on $\widehat{\mathscr {G}}_0$. Moreover, $\theta _{-n}(x)=\cos \big (2n\pi x\big )$ is the eigenvector with respect to $e_{-n}$ and $\theta _n(x)=\sin \big (k_n(x-\frac{1}{2})\big )$ is the eigenvector with respect to $e_n$.

According to the definition of the operator $\frac{d}{dx}\frac{d}{dW}$ introduced in [10], when W(dx) equals to Lebesgue measure plus the Dirac measure at 1, it is easy to check that the domain $\mathscr {D}_W$ of $\frac{d}{dx}\frac{d}{dW}$ includes $\widehat{\mathscr {G}}_0$ while $\tilde{\Delta }\Big |_{\hat{\mathscr {G}}_0}=\frac{d}{dx}\frac{d}{dW}\Big |_{\hat{\mathscr {G}}_0}$. By [10, Theorem 1], all the eigenvalues of $\frac{d}{dx}\frac{d}{dW}$ form an orthogonal basis of $L^2[0,1]$. As a result, to complete this proof, we only need to check the following two claims,

• 1. $\{e_n\}_{-\infty<n<+\infty }$ are all the eigenvalues of $\tilde{\Delta }$ limited on $\widehat{\mathscr {G}}_0$;

• 2. if f is an eigenvector of $\frac{d}{dx}\frac{d}{dW}$, then $f\in \hat{\mathscr {G}}_0$.

For the first claim, it is obviously that $G\equiv 1$ is the eigenvector of $\tilde{\Delta }\Big |_{\hat{\mathscr {G}}_0}$ with respect to $e_0=0$. So, from now on, we assume that $\lambda \ne 0$ is an eigenvalue of $\tilde{\Delta }\Big |_{\hat{\mathscr {G}}_0}$ while $G\ne 0$ is an eigenvector of $\tilde{\Delta }\Big |_{\hat{\mathscr {G}}_0}$ with respect to $\lambda$. We further let $H(x)=G(x+\frac{1}{2})$ for $-\frac{1}{2}\le x\le \frac{1}{2}$. If $\lambda >0$, let $c=\sqrt{\lambda }$, then, since $\tilde{\Delta }G=\lambda G$ and $G\in \hat{\mathscr {G}_0}$,

$$\begin{aligned} H(x)=a_1e^{cx}+a_2e^{-cx} \end{aligned}$$

for some $a_1, a_2\in \mathbb {R}$ while $H^{\prime }\left( \frac{1}{2}\right) =H^{\prime }\left( -\frac{1}{2}\right) =H\left( -\frac{1}{2}\right) -H\left( \frac{1}{2}\right)$. Therefore, $-a_2=a_1=a$ for some $a\in \mathbb {R}$ while

$$\begin{aligned} ac(e^{\frac{c}{2}}+e^{-\frac{c}{2}})=2a(e^{-\frac{c}{2}}-e^{\frac{c}{2}}). \end{aligned}$$

If $a\ne 0$, then

$$\begin{aligned} 0<c(e^{\frac{c}{2}}+e^{-\frac{c}{2}})=2(e^{-\frac{c}{2}}-e^{\frac{c}{2}})<0 \end{aligned}$$

since $c>0$, which is contradictory. Hence, we have $a=0$ and $G=0$, which is also contradictory. Therefore, $\lambda <0$. Let $c=\sqrt{-\lambda }$, then

$$\begin{aligned} H(x)=a_1\sin (cx)+a_2\cos (cx) \end{aligned}$$

for some $a_1, a_2\in \mathbb {R}$ while $H^{\prime }\left( \frac{1}{2}\right) =H^{\prime }\left( -\frac{1}{2}\right) =H\left( -\frac{1}{2}\right) -H\left( \frac{1}{2}\right)$. Therefore,

$$\begin{aligned} ca_2\sin \left( \frac{c}{2}\right) =0 \text {~and~} ca_1\cos \left( \frac{c}{2}\right) =-2a_1\sin \left( \frac{c}{2}\right) . \end{aligned}$$

As a result, if $a_2=0$, then $a_1\ne 0$ while c is a root of the equation $-\frac{x}{2}=\tan \frac{x}{2}$. Else if $a_2\ne 0$, then $\sin \left( \frac{c}{2}\right) =0$ and hence $c=2n\pi$ for some integer $n\ge 1$. Consequently, $\lambda =-(2n\pi )^2$ or $-k_n^2$ for some integer $n\ge 1$.

For the second claim, if $f\in \mathscr {D}_W$ is an eigenvector of $\frac{d}{dx}\frac{d}{dW}$, then there exist $a, b, \lambda \in \mathbb {R}$ while $\mathfrak {f}\in L^2[0, 1)$ such that

$$\begin{aligned} \int _0^1\mathfrak {f}(z)dz=0, \text {~}\int _{(0, 1]}W(dy)\left( b+\int _0^y\mathfrak {f}(z)dz\right) =0 \end{aligned}$$

while

$$\begin{aligned} f(x)=a+bW(x)+\int _{(0,x]}W(dy)\int _0^y\mathfrak {f}(z)dz \end{aligned}$$

for $0\le x<1$ and $\frac{d}{dx}\frac{d}{dW}f=\mathfrak {f}=\lambda f$, where

$$\begin{aligned} W(x)= {\left\{ \begin{array}{ll} x &{} \text {~if~}0\le x<1,\\ 2 &{} \text {~if~}x=1. \end{array}\right. } \end{aligned}$$

Hence, $f(x)=a+bx+\int _0^x\left( \int _0^y \mathfrak {f}(x)dz\right) dy$ for $0\le x<1$. Supplementarily define

$$\begin{aligned} f(1)=f(1-)=\lim _{x\uparrow 1}f(x)=a+b+\int _0^1\left( \int _0^y \mathfrak {f}(x)dz\right) dy, \end{aligned}$$

then $f\in C[0, 1]$ with $f(0)=a$ and $f(1)=a-b$ since

$$\begin{aligned} \int _0^1\left( \int _0^y \mathfrak {f}(x)dz\right) dy&=\int _{(0, 1]}W(dy)\left( b+\int _0^y\mathfrak {f}(z)dz\right) -2b-\int _0^1 \mathfrak {f}(y)dy=-2b. \end{aligned}$$

Since $f(x)=a+bx+\int _0^x\left( \int _0^y \mathfrak {f}(x)dz\right) dy$ for $0\le x\le 1$,

$$\begin{aligned} f^\prime (x)=b+\int _0^x\mathfrak {f}(y) dy \end{aligned}$$

for $0\le x\le 1$. Therefore, $f\in C^1[0, 1]$ with $f^\prime (0)=b$ and $f^\prime (1)=b$ since

$$\begin{aligned} \int _0^1\mathfrak {f}(z)dz=0. \end{aligned}$$

As a result,

$$\begin{aligned} f^\prime (0)=f^\prime (1)=f(0)-f(1). \end{aligned}$$

(4.12)

Since $\mathfrak {f}=\lambda f$ in $L^2[0, 1)$ while $f\in C[0, 1]$ as we have shown above, we can choose a continuous version of $\mathfrak {f}$ and supplementarily define $\mathfrak {f}(1)=\lambda f(1)$ such that $\mathfrak {f}\in C[0, 1]$. Since $\mathfrak {f}\in C[0, 1]$ and $f^\prime (x)=b+\int _0^x\mathfrak {f}(y) dy$ for $0\le x\le 1$,

$$\begin{aligned} f^{\prime \prime }(x)=\mathfrak {f}(x)=\lambda f(x) \end{aligned}$$

for $0\le x\le 1$ and hence $f^{\prime \prime }\in C[0, 1]$, implying $f\in C^2[0, 1]$. Consequently, by Eq. (4.12), $f\in \hat{\mathscr {G}}_0$. This completes the proof. $\square$

1.2 A.2: Existence and Uniqueness of Solution to Eq. (4.3)

Proof of the existence

We directly construct a solution to Eq. (4.3). For $-\infty<n<+\infty$, let $\{x_t^n\}_{0\le t\le T}$ be the unique solution to the ODE

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\frac{d}{dt}x_t^n=e_nx_t^n+\langle \theta _n|G_t\rangle ,\\ &{}x_0^n=\int _0^1\phi (x)\theta _n(x)dx, \end{array}\right. } \end{aligned}$$

where $e_n$ is defined as in the proof of Lemma 1.1. That is to say,

$$\begin{aligned} x_t^n=e^{e_nt}\int _0^1 \phi (x)\theta _n(x)dx+\int _0^te^{e_n(t-s)}\langle \theta _n|G_s\rangle ds. \end{aligned}$$

For any $f=\sum _{-\infty<n<+\infty }C_n(f)\theta _n\in \mathscr {G}_0$ and $t\ge 0$, we define

$$\begin{aligned} \mu _t^G(f)=\sum _{-\infty<n<+\infty }C_n(f)x_t^n. \end{aligned}$$

Note that the coefficients $\{C_n(f)\}_{-\infty<n<+\infty }$ are unique according to Lemma 1.1 and hence the definition of $\mu ^G$ is reasonable. Since

$$\begin{aligned} \mu _t^G(\theta _n)=x_t^n \quad \text {and} \quad \mu _t^G(\tilde{\Delta }\theta _n)=\mu _t^G(e_n\theta _n)=e_nx_t^n, \end{aligned}$$

it is easy to check that $\mu ^G$ is the solution to Eq. (4.3). $\square$

Proof of the Uniqueness

Assuming that $\mu$ and $\nu$ are both solutions to Eq. (4.3), then

$$\begin{aligned} |\mu _t(\theta _n)-\nu _t(\theta _n)|\le |e_n|\int _0^t |\mu _s(\theta _n)-\nu _s(\theta _n)|ds. \end{aligned}$$

By Grownwall’s inequality,

$$\begin{aligned} |\mu _t(\theta _n)-\nu _t(\theta _n)|\le 0e^{|e_n|t}=0 \end{aligned}$$

for any $0\le t\le T$ and $n\ge 1$. Hence, $\mu =\nu$ and the proof is complete. $\square$