Death and Resurrection of a Current by Disorder, Interaction or Periodic Driving

Abstract

We present several models of biased transport on (random) comb-like structures and on percolating backbones, that allow full mathematical control. We show how power-law corrections in the distribution of trap sizes may lead to a discontinuity in the current-field characteristic: the current jumps to zero when the driving exceeds a threshold. The current may resurrect when the field is modulated in time, also discontinuously: a little shaking enables the current to jump up. Finally, exclusion between particles postpones or even prevents the current from dying, while attraction such as modeled in zero range processes may expedite it.

Appendices

Appendix A: Derivation of the First-Hitting Time Formula

In this section we derive the expression (7) for the expected time for a diffusion X(t) with $X(0)=x_j \in \mathbb {R}$ and evolving according to

$$\begin{aligned} \dot{X}_t=-V^{\prime }(X_t)+\sqrt{2}\,\xi _t \end{aligned}$$

(A1)

to first hit a given point $x_{j+1}>x_j$. Here, we replace the notation $x_j$ by $x\in \mathbb {R}$ and $x_{j+1}$ by $y\in \mathbb {R}$.

One proceeds by using Dynkin’s formula; for a smooth function f,

$$\begin{aligned} \mathbb {E}^x[f(X_{\tau })]=f(x)+\mathbb {E}^x\left[ \int _0^\tau (Lf)(X_s)\text {d}s\right] \end{aligned}$$

(A2)

wherein

• $\mathbb {E}^x$ is the expectation value conditioned on the initial condition $X(0)=x$;

• $\tau$ is the (random) first exit time from some open bounded interval surrounding x. We denote it $I_a=(a,y)\subset \mathbb {R}$ and keep in mind to let $a \rightarrow -\infty$ when we are ready;

• L is the backward generator associated to the Markov process (A1):

  $$\begin{aligned} (Lg)(.)=-V^{\prime }(.)g^{\prime }(.)+g^{\prime \prime }(.)=e^{V(.)}\left( e^{-V(.)}g^{\prime }(.)\right) ^{\prime } \end{aligned}$$

The first way we want to use Dynkin’s formula is by defining the function f as the solution to the following Dirichlet boundary value problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} (Lf)(x)\equiv -1 \\ f(a)=f(y)=0 \end{array}\right. } \end{aligned}$$

Note that for such function f the Dynkin formula (A2) reduces to

$$\begin{aligned} 0=f(x)+\mathbb {E}^x\left[ \int _0^\tau (-1)\text {d}s\right] =f(x)-\mathbb {E}^x[\tau ] \end{aligned}$$

So $f(x)=\mathbb {E}^x[\tau ]$ is the expected escape time required to leave the interval [a, y].

The (unique) solution to this problem is given by

$$\begin{aligned}&f_{(a)}(x)=\left( -C_a\int _x^b \text {d}y \,e^{V(y)-V(b)}+\int _x^b \text {d}y\int _{a}^{y} \text {d}z \,e^{V(y)-V(z)}\right) ,\nonumber \\&\quad C_a = \frac{\int _a^b \text {d}y\int _{a}^{y} \text {d}z \,e^{V(y)-V(z)}}{\int _a^b \text {d}y \,e^{V(y)-V(b)}} \end{aligned}$$

(A3)

The escape time increases if we lower a as can be verified from the expression (A3). In any case, once the probability that the diffusion diverges to $-\infty$ before hitting y is proven to be zero, we then know that the true average escape time $\langle t_{x \rightarrow y}\rangle$ from the interval $(-\infty ,y)$ is given by

$$\begin{aligned} \langle t_{x \rightarrow y}\rangle =\lim _{a \rightarrow -\infty }f_{(a)}(x)=\inf _{a<x}f_{(a)}(x) \end{aligned}$$

Letting $a \rightarrow -\infty$ in the expression (A3) yields $C_a \rightarrow 0$ provided

$$\begin{aligned} \exists C_1 >0,\,C_2 \in \mathbb {R}:\,\forall z \le y \le b:\, V(y)-V(z)\le - C_1 (y-z)+C_2. \end{aligned}$$

(A4)

which is the case with probability 1 in our setup, by the strong law of large numbers. Hence (A3) reduces to (7).

To finally show that the scenario of a divergence of the diffusion to $-\infty$ before hitting y has zero probability, we want to apply the Dynkin formula to the function $\tilde{f}$ defined through the boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} (L\tilde{f})(x)\equiv 0 \\ \tilde{f}(a)=1,\,\tilde{f}(y)=0 \end{array}\right. } \end{aligned}$$

(A5)

Inserting this into the Dynkin formula (A2) yields

$$\begin{aligned} \mathbb {E}^x(\tilde{f}(X_\tau ))=\tilde{f}(x) \end{aligned}$$

wherein the left-hand side expresses the probability that the diffusion will hit $x=a$ before hitting $x=y$. The (unique) solution to (A5) is given by

$$\begin{aligned} \tilde{f}_{(a)}(x)=\frac{\int _x^b \text {d}y \,e^{V(y)-V(b)}}{\int _a^b \text {d}z\,e^{V(z)-V(b)}}\underset{(A4)}{\le } \frac{\int _x^b \text {d}y \,e^{V(y)-V(b)}}{C_1^{-1}(1+e^{C_1(b-a)})e^{C_2}} \rightarrow 0 \end{aligned}$$

as $a \rightarrow -\infty$. This implies the requested result.

Appendix B: Waiting Times on a Random Comb

Here we give details for better understanding of the equalities (12):

$$\begin{aligned} v= & {} \frac{1}{\mathbb {E}[\Delta t_0]}=\frac{\tanh ( E_1/2)}{\mathbb {E}[t_0^r]}=\frac{\tanh ( E_1/2)}{\sum _{j=0}^{\infty } \text {Prob}[L_0=j]\frac{e^{ (j+1) E_2}-1}{e^{ E_2}-1}}2\cosh ( E_1/2)\\= & {} 2\sinh ( E_1/2)\frac{e^{ E_2}-1}{ e^{ E_2}\langle e^{ L_0 E_2}\rangle -1}\nonumber \end{aligned}$$

(B1)

where $\mathbb {E}[\Delta t_0]$ is the average time required for the walker to reach the site (1, 0) for the first time when it starts at the site (0, 0). Likewise, $\mathbb {E}[t_0^r]$ is the average time required for the walker to reach either $(-1,0)$ or (1, 0) given that it started at (0, 0). A derivation of the first two equalities was already given in the body of the text. For the third equality, consider that

$$\begin{aligned} \mathbb {E}[t_0^r\left. \right| L_0=0]=\frac{1}{e^{ E_1/2}+e^{- E_1/2}} \end{aligned}$$

and for $n\in \mathbb {N}_0$,

$$\begin{aligned}&\mathbb {E}[t_0^r\left. \right| L_0=n ]\\&\quad =\sum _{j=0}^{\infty }\left( (j+1)t_0+j\tau _n\right) \text {Prob[ The walker makes }j\text { successive sorties from site }(0,0) \\&\qquad \text {into the dead-end }\{(0,j)\}_{1\le j \le n}\text { before finally jumping to }(\pm 1,0) ] \\&\quad =\sum _{j=0}^\infty \left( (j+1)t_0+j\tau _n\right) (1-p)^jp=\frac{1}{p}t_0+\frac{1-p}{p}\tau _n \end{aligned}$$

wherein

•  $p=\text {Prob}[\text {A random walk initialized at }(0,0)\text { hits } (\pm 1,0)\text { before hitting }(0,1)]$;

•  $t_0=\mathbb {E}[\text {Amount of time a walker initialized at } (0,0)\text { remains on site }(0,0)]$;

•  $\tau _n=\mathbb {E}[\text {time required for a random walk initialized at site }(0,1) \text { to hit }(0,0)]$.

One easily sees that $p=\frac{e^{ E_1/2}+e^{- E_1/2}}{e^{ E_1/2}+e^{- E_1/2}+e^{ E_2/2}}$ and $t_0=\frac{1}{e^{ E_1/2}+e^{- E_1/2}+e^{ E_2/2}}$.

$$\begin{aligned}= & {} \frac{e^{ E_1/2}+e^{- E_1/2}+e^{ E_2/2}}{e^{ E_1/2}+e^{- E_1/2}}\frac{1}{e^{ E_1/2}+e^{- E_1/2}+e^{ E_2/2}}+\frac{e^{ E_2/2}}{e^{ E_1/2}+e^{- E_1/2}}\tau _n \nonumber \\= & {} \frac{1}{2\cosh ( E_1/2)}\left( 1+e^{ E_2/2}\tau _n\right) \end{aligned}$$

(B2)

The $\tau _n$ can be computed by induction: clearly, $\tau _1=e^{ E_2/2}$ (that is simply the Poisson waiting time to jump back from (0, 1) to (0, 0)). We can compute $\tau _{n+1}$ from $\tau _n$ via the following consideration. In a dead-end with length $n+1>2$ a walker initialized at (0, 1) (i.e., the neck of the dead-end) will remain stuck on that site with an average waiting time $\frac{1}{e^{ E_2/2}+e^{- E_2/2}}$. After waiting till the first jump, the probability to jump to (0, 0) is $\frac{e^{- E_2/2}}{e^{ E_2/2}+e^{- E_2/2}}$ while the probability to jump deeper into the dead-end (i.e., into $\{(0,j)\}_{2\le j \le n+1}$) is $\frac{e^{E_2/2}}{e^{ E_2/2}+e^{- E_2/2}}$. In the latter case, the time required to return from $\{(0,j)\}_{2\le j \le n+1}$ to (0, 1) is $\tau _n$. Hence,

$$\begin{aligned}&\tau _{n+1}\\&\quad =\sum _{j=0}^{\infty }\left( (j+1)t_0^{\prime }+j\tau _n\right) \text {Prob[ The walker makes }j\text { successive sorties from site }(0,1) \\&\qquad \text {into the dead-end }\{(0,j)\}_{2\le j \le n+1}\text { before finally jumping to }(0,0) ] \\&\quad =\sum _{j=0}^\infty \left( (j+1)t_0^{\prime }+j\tau _n\right) (1-q)^jq=\frac{1}{q}t_0^{\prime }+\frac{1-q}{q}\tau _n \end{aligned}$$

wherein

• $q=\text {Prob}[\text {A random walk initialized at }(0,1)\text { hits } (0,0)\text { before hitting }(0,2)]$;

• $t_0^{\prime }=\mathbb {E}[\text {Amount of time a walker initialized at } (0,1)\text { remains on site }(0,1)]$.

Since $q=\frac{e^{- E_2/2}}{e^{ E_2/2}+e^{- E_2/2}}$ and $t_0^{\prime }=\frac{1}{e^{ E_2/2}+e^{- E_2/2}}$, we can solve the recursion for $\tau _n$. The result is $\tau _n=e^{ E_2/2}\frac{e^{ n E_2}-1}{e^{ E_2}-1}$. Plugging this into (B2) yields that

$$\begin{aligned} \mathbb {E}[t_0^r | L_0=n]=\frac{1}{2\cosh ( E_1/2)}\frac{e^{ (n+1) E_2}-1}{e^{ E_2}-1} \end{aligned}$$

Also for $n=0$, that formula turns out to be accurate. That in turn explains the third equality of (B1). The fourth equality of (B1) is a simple rewriting involving no calculations.