Kardar–Parisi–Zhang Equation and Large Deviations for Random Walks in Weak Random Environments

Abstract

We consider the transition probabilities for random walks in $1+1$ dimensional space-time random environments (RWRE). For critically tuned weak disorder we prove a sharp large deviation result: after appropriate rescaling, the transition probabilities for the RWRE evaluated in the large deviation regime, converge to the solution to the stochastic heat equation (SHE) with multiplicative noise (the logarithm of which is the KPZ equation). We apply this to the exactly solvable Beta RWRE and additionally present a formal derivation of the convergence of certain moment formulas for that model to those for the SHE.

Appendix 1: Sharp Large Deviation for Symmetric Simple Random Walk

Recall that

$$\begin{aligned} \mathscr {P}_\varepsilon (t_\varepsilon ,x_\varepsilon ,m_1,m_2)= \mathbf {P}^{0}\left( S_{\frac{t_\varepsilon }{\varepsilon ^2}+m_1}=v\frac{t_{\varepsilon }}{\varepsilon ^2}+\frac{x_\varepsilon }{\varepsilon }+m_2\right) , \end{aligned}$$

$p_{\sigma ^2}(t,x)$ is the density of $N(0,\sigma ^2t)$, and assume that $m_1,m_2\in \mathbb {Z}$, $|m_1|,|m_2|\leqslant 1$ and $m_1-m_2$ is even.

Lemma 4.1

For fixed $t>0,x\in \mathbb {R}$, we have

$$\begin{aligned} \frac{1}{\varepsilon }e^{\frac{t_\varepsilon }{\varepsilon ^2}\mathcal {I}(v)+\frac{x_\varepsilon }{\varepsilon }\mathcal {I}'(v)}\mathscr {P}_\varepsilon (t_\varepsilon ,x_\varepsilon ,m_1,m_2)\rightarrow \frac{2p_{1-v^2}(t,x)}{(1+v)^{\frac{m_1+m_2}{2}}(1-v)^{\frac{m_1-m_2}{2}}} \end{aligned}$$

(4.1)

as $\varepsilon \rightarrow 0$, and there exists a constant $C>0$ such that for all $t>0,x\in \mathbb {R}$,

$$\begin{aligned}&\frac{1}{\varepsilon }e^{\frac{t_\varepsilon }{\varepsilon ^2}\mathcal {I}(v){+}\frac{x_\varepsilon }{\varepsilon }\mathcal {I}'(v)}\mathscr {P}_\varepsilon (t_\varepsilon ,x_\varepsilon ,m_1,m_2) \leqslant C\left( \frac{1}{\varepsilon }\mathbf {1}\{t\leqslant 100\varepsilon ^2,|x|\leqslant C\varepsilon \}\right. \nonumber \\&\quad \left. +\mathbf {1}\{t>100\varepsilon ^2\}\frac{1}{\sqrt{t}}e^{-\frac{x^2}{Ct}}\right) . \end{aligned}$$

(4.2)

Proof

To simplify the notation, let

$$\begin{aligned} n=\frac{t_\varepsilon }{\varepsilon ^2}+m_1, \ \ m=v\frac{t_\varepsilon }{\varepsilon ^2}+\frac{x_\varepsilon }{\varepsilon }+m_2, \end{aligned}$$

(4.3)

so that $\mathscr {P}_\varepsilon (t_\varepsilon ,x_\varepsilon ,m_1,m_2)=\mathbf {P}^{0}(S_n=m)$. We will utilize the notation $\lesssim$ when the left-hand side is bounded by a constant (independent of $\varepsilon$ and t and x) times the right-hand side for all $\varepsilon$ sufficiently small. We first prove the estimate (4.2) that is uniform in $t>0,x\in \mathbb {R}$, then prove the convergence in (4.1) for fixed $t>0,x\in \mathbb {R}$.

Case 1 $t\leqslant 100\varepsilon ^2$. It is clear that $t_\varepsilon \leqslant 100\varepsilon ^2$, and by the fact that $|m|\leqslant n$ (or else $\mathbf {P}^{0}(S_n=m)=0$), there exists $C>0$ so that $|x_\varepsilon |\leqslant C\varepsilon$. Thus, we have

$$\begin{aligned} \frac{1}{\varepsilon }e^{\frac{t_\varepsilon }{\varepsilon ^2}\mathcal {I}(v)+\frac{x_\varepsilon }{\varepsilon }\mathcal {I}'(v)}\mathscr {P}_\varepsilon (t_\varepsilon ,x_\varepsilon ,m_1,m_2)\lesssim \frac{1}{\varepsilon } \mathbf {1}\{|x_\varepsilon |\leqslant C\varepsilon \}. \end{aligned}$$

This proves the first bound on the right-hand side of (4.2).

Case 2 $t>100\varepsilon ^2$. Recall that $|m_1|,|m_2|\leqslant 1$, so $n\geqslant 90$—we will implicitly use the largeness of n in some of the bounds below. By (4.3),

$$\begin{aligned} \frac{t_\varepsilon }{\varepsilon ^2}\mathcal {I}(v)+\frac{x_\varepsilon }{\varepsilon }\mathcal {I}'(v)=\frac{1}{2} n\log (1-v^2)+\frac{1}{2}m\log \frac{1+v}{1-v} -m_1\mathcal {I}(v)+(vm_1-m_2)\mathcal {I}'(v), \end{aligned}$$

which implies

$$\begin{aligned} e^{\frac{t_\varepsilon }{\varepsilon ^2}\mathcal {I}(v)+\frac{x_\varepsilon }{\varepsilon }\mathcal {I}'(v)}\lesssim e^{\frac{1}{2} n\log (1-v^2)+\frac{1}{2}m\log \frac{1+v}{1-v}}. \end{aligned}$$

Using Stirling’s approximation,

$$\begin{aligned} \sqrt{2\pi } N^{N+\frac{1}{2}}e^{-N}\leqslant N!\leqslant eN^{N+\frac{1}{2}}e^{-N}, \end{aligned}$$

we derive the bound

$$\begin{aligned} \mathbf {P}^{0}(S_n=m)\lesssim \sqrt{\frac{n}{n^2-m^2}} e^{-\frac{n+m}{2}\log (1+\frac{m}{n})-\frac{n-m}{2}\log (1-\frac{m}{n})} \end{aligned}$$

when $|m|<n$.

Now we consider three different subcases of case 2. Fix $0<\tau \ll 1$, and let $k=m/n$.

Case 2.i $|k|=1$. We have $P(S_n=m)=2^{-n}$, so

$$\begin{aligned} \frac{1}{\varepsilon }e^{\frac{t_\varepsilon }{\varepsilon ^2}\mathcal {I}(v)+\frac{x_\varepsilon }{\varepsilon }\mathcal {I}'(v)}\mathscr {P}_\varepsilon (t_\varepsilon ,x_\varepsilon ,m_1,m_2) \lesssim \frac{1}{\varepsilon }e^{\frac{1}{2} n\log (1-v^2)+\frac{1}{2}m\log \frac{1+v}{1-v}-n\log 2}. \end{aligned}$$

In both cases of $m=n$ and $m=-n$, using the fact that $v\in (0,1)$, we have

$$\begin{aligned} \frac{1}{\varepsilon }e^{\frac{1}{2} n\log (1-v^2)+\frac{1}{2}m\log \frac{1+v}{1-v}-n\log 2}\leqslant \frac{1}{\varepsilon } e^{-\delta n} \end{aligned}$$

for some $\delta >0$ depending on v. Furthermore,

$$\begin{aligned} \frac{1}{\varepsilon }e^{-\delta n}=\frac{1}{\sqrt{\varepsilon ^2 n}}\sqrt{n}e^{-\frac{\delta n}{2}}e^{-\frac{\delta n}{2}} \lesssim \frac{1}{\sqrt{t}}e^{-\frac{\delta n}{2}}, \end{aligned}$$

so it remains to show that $x^2/t\lesssim n$. By the fact that $|m|\leqslant n$, we have

$$\begin{aligned} -t_\varepsilon \lesssim \varepsilon x_\varepsilon \lesssim t_\varepsilon , \end{aligned}$$

which implies $-t\lesssim \varepsilon x\lesssim t$. This implies that in this case, the second bound on the right-hand side of (4.2) holds.

Case 2.ii $|k|<1$ and $|k-v|\geqslant \tau$. We have

$$\begin{aligned} \frac{1}{\varepsilon }e^{\frac{t_\varepsilon }{\varepsilon ^2}\mathcal {I}(v)+\frac{x_\varepsilon }{\varepsilon }\mathcal {I}'(v)}\mathscr {P}_\varepsilon (t_\varepsilon ,x_\varepsilon ,m_1,m_2) \lesssim \frac{1}{\varepsilon } \sqrt{\frac{n}{n^2-m^2}}e^{-\frac{n}{2}F(k)}, \end{aligned}$$

(4.4)

where

$$\begin{aligned} F(k)=(1+k)\log \frac{1+k}{1+v}+(1-k)\log \frac{1-k}{1-v}, \ \ k \in (-1,1). \end{aligned}$$

(4.5)

It is straightforward to check that F(k) attains its minimum at v and $F''(k)$ is bounded from below by some positive constant. Since $|k-v|\geqslant \tau$, we have $F(k)\geqslant \delta \tau ^2$ with $\delta =\frac{1}{2}\min _{k\in (-1,1)} F''(k)$. In addition, since $|m|<n$, we have $-n+2\leqslant m\leqslant n-2$, so

$$\begin{aligned} \frac{1}{\varepsilon } \sqrt{\frac{n}{n^2-m^2}} \leqslant \frac{1}{\sqrt{\varepsilon ^2 n}}\frac{1}{\sqrt{1-\left( \frac{m}{n}\right) ^2}} \lesssim \frac{1}{\sqrt{t}}\sqrt{n}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \frac{1}{\varepsilon }e^{\frac{t_\varepsilon }{\varepsilon ^2}\mathcal {I}(v)+\frac{x_\varepsilon }{\varepsilon }\mathcal {I}'(v)}\mathscr {P}_\varepsilon (t_\varepsilon ,x_\varepsilon ,m_1,m_2) \lesssim \frac{1}{\sqrt{t}} \sqrt{n}e^{-\frac{ n}{2}\delta \tau ^2}\lesssim \frac{1}{\sqrt{t}} e^{-\frac{n}{4}\delta \tau ^2}, \end{aligned}$$

and the same discussion as in case 2.i shows $e^{-\frac{n}{4}\delta \tau ^2}\lesssim e^{-\frac{x^2}{Ct}}$ for some $C>0$, matching the second bound on the right-hand side of (4.2).

Case 2.iii $|k-v|<\tau$. We have

$$\begin{aligned} \frac{1}{\varepsilon } \sqrt{\frac{n}{n^2-m^2}}\lesssim \frac{1}{\sqrt{t}}. \end{aligned}$$

For the exponent, it is clear that $F(k)\geqslant \delta (k-v)^2$ with the same $\delta >0$ from case 2.ii, so

$$\begin{aligned} e^{-\frac{n}{2}F(k)}\leqslant e^{-\frac{n}{2}\delta (k-v)^2}. \end{aligned}$$

Since $k=m/n$ with m, n given in (4.3), we have

$$\begin{aligned} n(k-v)^2=\frac{(x_\varepsilon +(m_2-vm_1)\varepsilon )^2}{t_\varepsilon +m_1\varepsilon ^2}, \end{aligned}$$

(4.6)

so

$$\begin{aligned} e^{-\frac{n}{2}\delta (k-v)^2}\lesssim e^{-\frac{(|x|-\tilde{C}\varepsilon )^2}{Ct}}\leqslant e^{-\frac{x^2}{Ct}}e^{\frac{2\tilde{C}\varepsilon |x|}{Ct}} \end{aligned}$$

for some $C,\tilde{C}>0$. For $M>0$, if $|x|\leqslant M\varepsilon$, we have the desired estimate; if $|x|>M\varepsilon$, we have

$$\begin{aligned} e^{-\frac{x^2}{Ct}}e^{\frac{2\tilde{C}\varepsilon |x|}{Ct}}< e^{-\frac{x^2}{Ct}}e^{\frac{2\tilde{C}x^2}{MCt}}, \end{aligned}$$

so we only need to choose M sufficiently large to complete the proof of (4.2).

To prove (4.1), we note that for fixed $t>0,x\in \mathbb {R}$ and sufficiently small $\varepsilon$, $|k-v|\ll 1$ (we are in the region of case 2.iii). We use Stirling’s approximation and the fact that $n,\frac{n+m}{2},\frac{n-m}{2}\rightarrow \infty$ to obtain that

$$\begin{aligned} \frac{\mathbf {P}^{0}(S_n=m)}{\sqrt{\frac{2n}{\pi (n^2-m^2)}}e^{-\frac{n}{2}[(1+k)\log (1+k)+(1-k)\log (1-k)]}}\rightarrow 1 \end{aligned}$$

as $\varepsilon \rightarrow 0$. Thus we only need to analyze

$$\begin{aligned}&\frac{1}{\varepsilon }e^{\frac{t_\varepsilon }{\varepsilon ^2}\mathcal {I}(v)+\frac{x_\varepsilon }{\varepsilon }\mathcal {I}'(v)}\sqrt{\frac{2n}{\pi (n^2-m^2)}}e^{-\frac{n}{2}[(1+k)\log (1+k)+(1-k)\log (1-k)]}\\&=\sqrt{\frac{2n}{\pi (n^2-m^2)\varepsilon ^2}} \frac{e^{-\frac{n}{2}F(k)}}{(1+v)^{\frac{m_1+m_2}{2}}(1-v)^{\frac{m_1-m_2}{2}}}, \end{aligned}$$

with F(k) defined in (4.5). First, we have

$$\begin{aligned} \sqrt{\frac{2n}{\pi (n^2-m^2)\varepsilon ^2}}\rightarrow \sqrt{\frac{2}{\pi (1-v^2)t}}. \end{aligned}$$

Secondly, by (4.6), we have $|k-v|\lesssim \varepsilon$ for fixed $t>0,x\in \mathbb {R}$ and $n(k-v)^2\rightarrow \frac{x^2}{t}$ as $\varepsilon \rightarrow 0$. We expand $F(k)=\frac{1}{2}F''(v)(k-v)^2+O(|k-v|^3)$ and conclude that

$$\begin{aligned} \frac{n}{2}F(k)\rightarrow F''(v)\frac{x^2}{4t}=\frac{x^2}{2(1-v^2)t} \end{aligned}$$

as $\varepsilon \rightarrow 0$. This completes the proof of (4.1). $\square$