Fluctuations of the solutions to the KPZ equation in dimensions three and higher

Abstract

We prove, using probabilistic techniques and analysis on the Wiener space, that the large scale fluctuations of the KPZ equation in $d\ge 3$ with a small coupling constant, driven by a white in time and colored in space noise, are given by the Edwards-Wilkinson model. This gives an alternative proof, that avoids perturbation expansions, to the results of Magnen and Unterberger (J Stat Phys 171:543–598, 2018).

Appendices

Auxiliary lemmas

Lemma A.1

For any $t>0,x\in \mathbb {R}^d$, we have

$$\begin{aligned} D\log Z(t,x)=\frac{DZ(t,x)}{Z(t,x)}\in L^2(\Omega ;H). \end{aligned}$$

Proof

Recall that

so, for each t and x fixed, we have

(A.1)

for any $n\in \mathbb {Z}_+$, where we recall that H is the $L^2(\mathbb {R}^{d+1})$-space with respect to the s, y-variables. To deal with the logarithm function, which is singular at the origin and grows at infinity, we use an approximation $f_n\in \mathcal {C}_c^\infty (\mathbb {R})$ such that $f_n(x)=\log x$ for $x\in [1/n,n]$ and $|f_n'(x)|\le |x|^{-1}$. It is clear that

$$\begin{aligned} D f_n(Z(t,x))=f_n'(Z(t,x)) DZ(t,x)\in L^2(\Omega ; H), \end{aligned}$$

and the error

as $n\rightarrow \infty$, where we used Proposition 2.3 in the last step, together with (A.1). By [35, Proposition 1.2.1], the proof is complete. $\square$

Lemma A.2

There exists $\beta _0>0$ such that if $\beta <\beta _0$, we have in $d\ge 3$ that

$$\begin{aligned} \sup _{x\in \mathbb {R}^d} \mathbb {E}_B\left[ \exp \left\{ \beta \int _0^\infty R(x+B_s)ds\right\} \right] <\infty , \end{aligned}$$

(A.2)

and

$$\begin{aligned} \sup _{t>0, x,y\in \mathbb {R}^d} \mathbb {E}_B\left[ \exp \left\{ \beta \int _0^t R(x+B_s)ds\right\} \bigg | B_t=y\right] <\infty . \end{aligned}$$

(A.3)

Proof

The statement in (A.2) follows from Portenko’s lemma, see [33, (3.1), (3.2)].

We turn to proving (A.3). Conditioned on $B_t=y$, the process $\{B_s\}_{s\le t}$ is a Brownian bridge. In particular, it has a Markovian representation. Thus, again by Portenko’s lemma, it is enough to show that

$$\begin{aligned} \beta \sup _{t>0, x,y\in \mathbb {R}^d} \mathbb {E}_B\left[ \int _0^{t} R(x+B_s)ds \bigg | B_t=y\right] < 1, \end{aligned}$$

(A.4)

for all $\beta$ small enough. By symmetry, as $X_s$ is a Brownian bridge, it suffices to show that

$$\begin{aligned} \sup _{t>0, x,y\in \mathbb {R}^d} \mathbb {E}_B\left[ \int _0^{t/2} R(x+B_s)ds \bigg | B_t=y\right] < \infty . \end{aligned}$$

(A.5)

Note that $X_s$ has mean sy / t and variance $s(t-s)/t$, which in our range is larger than Cs. In particular, for $s\le t/2$, we have

$$\begin{aligned} \sup _{x,y\in \mathbb {R}^d}\mathbb {P}_B\left[ |B_s-x|\le 1\bigg | B_t=y\right] \le C{(1+}s)^{-d/2}. \end{aligned}$$

Since $s^{-d/2}$ is integrable as $s\rightarrow +\infty$, this yields (A.5) and completes the proof of the lemma. $\square$

Lemma A.3

Let B be a standard Brownian motion in $d\ge 3$. For any $\alpha \in (0,2)$, $t>0$ and compact set $A\subset \mathbb {R}^d$ with $0\notin A$, we have

$$\begin{aligned} \sup _{w\in A}\mathbb {E}_{B}\left[ \left( \int _{t/\varepsilon ^\alpha }^{t/\varepsilon ^2}R(B_s)ds\right) ^2 \,\bigg |\, B_{t/\varepsilon ^2}=\frac{w}{\varepsilon }\right] \lesssim \varepsilon ^{\alpha \left( \frac{d}{2}-1\right) }. \end{aligned}$$

Proof

By a direct calculation, we have

$$\begin{aligned} \begin{aligned}&\mathbb {E}_{B}\left[ \left( \int _{t/\varepsilon ^\alpha }^{t/\varepsilon ^2}R(B_s)ds\right) ^2 \,\bigg |\, B_{t/\varepsilon ^2}=\frac{w}{\varepsilon }\right] \\&\quad =2\int _{[t/\varepsilon ^\alpha \le s_1 \le s_2 \le t/\varepsilon ^2]}\int _{\mathbb {R}^{2d}}R(x)R(y)G_{s_1}(x)G_{s_2-s_1}(y-x)G_{t/\varepsilon ^2-s_2}\\&\qquad \left( \frac{w}{\varepsilon }-y\right) G_{t/\varepsilon ^2}\left( \frac{w}{\varepsilon }\right) ^{-1}dxdyds_1ds_2. \end{aligned} \end{aligned}$$

By the fact that $w\in A$ and $0\notin A$, we have

$$\begin{aligned} \frac{G_{t/\varepsilon ^2-s_2}\left( \frac{w}{\varepsilon }-y\right) }{G_{t/\varepsilon ^2}\left( \frac{w}{\varepsilon }\right) }=\frac{G_{t-\varepsilon ^2s_2}(w-\varepsilon y)}{G_t(w)} \lesssim 1, \end{aligned}$$

uniformly in $s_2\le t/\varepsilon ^2, w\in A, \, y\in \mathrm {supp}(R)$ and $\varepsilon \ll 1$. This shows that we can remove the conditional expectation and derive

$$\begin{aligned} \sup _{w\in A} \mathbb {E}_{B}\left[ \left( \int _{t/\varepsilon ^\alpha }^{t/\varepsilon ^2}R(B_s)ds\right) ^2 \,\bigg |\, B_{t/\varepsilon ^2}=\frac{w}{\varepsilon }\right] \lesssim \mathbb {E}_B\left[ \left( \int _{t/\varepsilon ^\alpha }^{t/\varepsilon ^2}R(B_s)ds\right) ^2\right] \lesssim \varepsilon ^{\alpha \left( \frac{d}{2}-1\right) }, \end{aligned}$$

which completes the proof. $\square$

Negative moments of Z(t, x)

We now prove Proposition 2.3. The goal is to show there exists $\beta _0>0$ such that if $\beta <\beta _0$ and $n\in \mathbb {Z}_+$, we have

$$\begin{aligned} \sup _{t>0} \mathbb {E}[ Z(t,x)^{-n}] \le C_{\beta ,n}, \end{aligned}$$

(B.1)

with some constant $C_{\beta ,n}>0$. We adapt to our setting the proof of [24, Corollary 4.8], which deals with the case when the noise is also singular in space. The same proof applies to our situation, and we only present the details for the convenience of the readers.

Since Z(t, x) has the same distribution as u(t, x), it suffices to estimate the small ball probability $\mathbb {P}[u(t,x)\le r]$ for $r\ll 1$. We define an approximation of the spacetime white noise

$$\begin{aligned} W_\varepsilon (t,x)=e^{-\varepsilon (t^2+|x|^2)}\int _{\mathbb {R}^{d+1}} \phi _\varepsilon (t-s,x-y)dW(s,y), \end{aligned}$$

where $\phi _\varepsilon (t,x)=\varepsilon ^{-d-2}\phi (t/\varepsilon ^2,x/\varepsilon )$ with $\phi \in \mathcal {C}_c^\infty (\mathbb {R}^{d+1})$ such that $\phi \ge 0$ is even and $\Vert \phi \Vert _{L^1}=1$. It is clear that for fixed $\varepsilon >0$, $W_\varepsilon \in L^2(\mathbb {R}^{d+1})\cap \mathcal {C}^\infty (\mathbb {R}^{d+1})$ almost surely. We will use $\Vert \cdot \Vert _2$ to denote the $L^2(\mathbb {R}^{d+1})$ norm. Define

$$\begin{aligned} V_\varepsilon (t,x)=\int _{\mathbb {R}^d} \varphi (x-y)W_\varepsilon (t,y)dy, \ \ \mathscr {R}_\varepsilon (t,s,x,y)=\mathbb {E}[V_\varepsilon (t,x)V_\varepsilon (s,y)], \end{aligned}$$

and

$$\begin{aligned} \mathscr {U}_\varepsilon (t,x)=\mathbb {E}_B\left[ e^{\mathcal {V}_t^\varepsilon (B) }\right] , \end{aligned}$$

with

$$\begin{aligned} \mathcal {V}_t^\varepsilon (B)=\beta \int _0^t V_\varepsilon (t-s,x+B_s)ds-\frac{1}{2}\beta ^2\mathcal {Q}_\varepsilon (t,x,x,B,B), \end{aligned}$$

where

$$\begin{aligned} \mathcal {Q}_\varepsilon (t,x,y,B^1,B^2)=\int _0^t\int _0^t \mathscr {R}_\varepsilon (t-s,t-\ell ,x+B_s^1,y+B_{\ell }^2)dsd\ell . \end{aligned}$$

By [24, Proposition 4.2], $\mathscr {U}_\varepsilon (t,x)\rightarrow u(t,x)$ in probability so we only need to estimate $\mathbb {P}[\mathscr {U}_\varepsilon (t,x)\le r]$ for $r\ll 1$.

With any given $W_\varepsilon$, define the expectation

$$\begin{aligned} \mathbb {E}^{W_\varepsilon }_B[ F(B^1,B^2)]=\frac{\mathbb {E}_B[ F(B^1,B^2)e^{\mathcal {V}_t^\varepsilon (B^1)+\mathcal {V}_t^\varepsilon (B^2)}]}{\mathbb {E}_B[ e^{\mathcal {V}_t^\varepsilon (B^1)+\mathcal {V}_t^\varepsilon (B^2)}]}. \end{aligned}$$

To emphasize the dependence of $\mathscr {U}_\varepsilon$ on $W_\varepsilon$, we write $\mathscr {U}_\varepsilon (t,x)=\mathscr {U}_\varepsilon (t,x,W_\varepsilon )$. For any $\lambda >0$, define the set

$$\begin{aligned} A_\lambda (t,x)=\left\{ W_\varepsilon : \mathscr {U}_\varepsilon (t,x,W_\varepsilon )>\frac{1}{2}, \int _0^t\mathbb {E}_B^{W_\varepsilon }[ R(B^1_s-B^2_s)]ds\le \lambda \right\} . \end{aligned}$$

Lemma B.1

For any $\tilde{W}_\varepsilon \in A_\lambda (t,x)$, we have

$$\begin{aligned} \mathscr {U}_\varepsilon (t,x,W_\varepsilon )\ge \frac{1}{2}e^{-\sqrt{\lambda } \Vert W_\varepsilon -\tilde{W}_\varepsilon \Vert _2}, \end{aligned}$$

with $\Vert \cdot \Vert _2$ denoting the $L^2(\mathbb {R}^{d+1})$ norm.

Proof

We write

$$\begin{aligned} \begin{aligned} \mathscr {U}_\varepsilon (t,x,W_\varepsilon )=\mathbb {E}_B[ e^{\mathcal {V}_t^\varepsilon (B)}]&=\mathbb {E}_B[e^{{\tilde{\mathcal {V}}}_t^\varepsilon (B)}] \frac{ \mathbb {E}_B[e^{\mathcal {V}_t^\varepsilon (B)-{\tilde{\mathcal {V}}}_t^\varepsilon (B)} e^{{\tilde{\mathcal {V}}}_t^\varepsilon (B)}]}{\mathbb {E}_B[ e^{{\tilde{\mathcal {V}}}_t^\varepsilon (B)}]}\\&=\mathscr {U}_\varepsilon (t,x,\tilde{W}_\varepsilon ) \mathbb {E}_B^{\tilde{W}_\varepsilon }[ e^{\mathcal {V}_t^\varepsilon (B)-{\tilde{\mathcal {V}}}_t^\varepsilon (B)}], \end{aligned} \end{aligned}$$

where ${\tilde{\mathcal {V}}}_t^\varepsilon$ is obtained by replacing $W_\varepsilon \mapsto \tilde{W}_\varepsilon$ in the expression of $\mathcal {V}_t^\varepsilon$. Since $\tilde{W}_\varepsilon \in A_\lambda$, by Jensen’s inequality we have

$$\begin{aligned} \mathscr {U}_\varepsilon (t,x,W_\varepsilon )\ge \frac{1}{2} \exp \left( \mathbb {E}_B^{\tilde{W}_\varepsilon }[\mathcal {V}_t^\varepsilon (B)-{\tilde{\mathcal {V}}}_t^\varepsilon (B)]\right) . \end{aligned}$$

It remains to show that

$$\begin{aligned} |\mathbb {E}_B^{\tilde{W}_\varepsilon }[\mathcal {V}_t^\varepsilon (B)-{\tilde{\mathcal {V}}}_t^\varepsilon (B)]| \le \sqrt{\lambda }\Vert W_\varepsilon -\tilde{W}_\varepsilon \Vert _2. \end{aligned}$$

(B.2)

We write

$$\begin{aligned} \begin{aligned} \mathcal {V}_t^\varepsilon (B)-{\tilde{\mathcal {V}}}_t^\varepsilon (B)&=\beta \int _0^t [V_\varepsilon (t-s,x+B_s)-\tilde{V}_\varepsilon (t-s,x+B_s)]ds\\&=\beta \int _0^t\int _{\mathbb {R}^d} \varphi (x+B_s-y)[W_\varepsilon (t-s,y)-\tilde{W}_\varepsilon (t-s,y)]dyds, \end{aligned} \end{aligned}$$

and apply the Cauchy-Schwarz inequality to get

$$\begin{aligned} \begin{aligned} |\mathbb {E}_B^{\tilde{W}_\varepsilon }[\mathcal {V}_t^\varepsilon (B)-{\tilde{\mathcal {V}}}_t^\varepsilon (B)]|&\le \beta \Vert W_\varepsilon -\tilde{W}_\varepsilon \Vert _2\sqrt{\int _0^t \int _{\mathbb {R}^d} |\mathbb {E}_B^{\tilde{W}_\varepsilon }[\varphi (x+B_s-y)] |^2 dyds} \\&\le \beta \Vert W_\varepsilon -\tilde{W}_\varepsilon \Vert _2\sqrt{\int _0^t \mathbb {E}_B^{\tilde{W}_\varepsilon }[R(B^1_s-B^2_s)]ds}\le \sqrt{\lambda } \Vert W_\varepsilon -\tilde{W}_\varepsilon \Vert _2, \end{aligned} \end{aligned}$$

which completes the proof. $\square$

Lemma B.2

There exists universal constants $\lambda ,c>0$ such that $\mathbb {P}[A_\lambda (t,x)]\ge c$.

Proof

We have

$$\begin{aligned} \mathbb {P}[A_\lambda (t,x)] \ge \mathbb {P}\left[ \mathscr {U}_\varepsilon (t,x,W_\varepsilon )> \frac{1}{2}\right] -\mathbb {P}[ B_\lambda (t,x)], \end{aligned}$$

with

$$\begin{aligned} B_\lambda (t,x)=\left\{ W_\varepsilon : \mathscr {U}_\varepsilon (t,x,W_\varepsilon )>\frac{1}{2}, \int _0^t\mathbb {E}_B^{W_\varepsilon }[ R(B^1_s-B^2_s)]ds> \lambda \right\} . \end{aligned}$$

Using the fact that $\mathbb {E}[\mathscr {U}_\varepsilon (t,x,W_\varepsilon )]=1$ and the Paley–Zygmund’s inequality, we have

$$\begin{aligned} \mathbb {P}\left[ \mathscr {U}_\varepsilon (t,x,W_\varepsilon )>\frac{1}{2}\right] \ge \frac{1}{4\mathbb {E}[\mathscr {U}_\varepsilon (t,x,W_\varepsilon )^2]}=\frac{1}{4\mathbb {E}_B[ e^{\beta ^2 \mathcal {Q}_\varepsilon (t,x,x,B^1,B^2)}]}. \end{aligned}$$

For $B_\lambda (t,x)$, we have, as $\mathscr {U}_\varepsilon (t,x,W_\varepsilon )>1/2$,

$$\begin{aligned} \begin{aligned} \mathbb {P}[B_\lambda (t,x)]&\le \mathbb {P}\left[ \int _0^t \mathbb {E}_B[ R(B_s^1-B_s^2) e^{\mathcal {V}_t^\varepsilon (B^1)+\mathcal {V}_t^\varepsilon (B^2)}] ds >\frac{\lambda }{4}\right] \\&\le \frac{4}{\lambda }\mathbb {E}_B\left[ e^{\beta ^2 \mathcal {Q}_\varepsilon (t,x,x,B^1,B^2)}\int _0^t R(B^1_s-B^2_s)ds\right] \\&\le \frac{4C}{\lambda }\mathbb {E}_B\left[ e^{2\beta ^2 \mathcal {Q}_\varepsilon (t,x,x,B^1,B^2)}\right] ^{1/2}, \end{aligned} \end{aligned}$$

(B.3)

with some constant $C>0$. By Lemma B.3 below and choosing $\lambda$ large, there exists some constants $c,\lambda >0$ independent of $\varepsilon ,t,x$ such that $\mathbb {P}[A_\lambda (t,x)] \ge c$. $\square$

Lemma B.3

There exists $\beta _0>0$ such that if $\beta <\beta _0$, we have

$$\begin{aligned} 1\le \mathbb {E}_B\left[ e^{\beta \mathcal {Q}_\varepsilon (t,x,x,B^1,B^2)}\right] \le C_\beta . \end{aligned}$$

Proof

Recall that

$$\begin{aligned} \mathcal {Q}_\varepsilon (t,x,x,B^1,B^2)=\int _0^t\int _0^t \mathscr {R}_\varepsilon (t-s,t-\ell ,x+B_s^1,x+B_{\ell }^2)dsd\ell . \end{aligned}$$

We write $\mathscr {R}_\varepsilon$ explicitly:

$$\begin{aligned} \begin{aligned} \mathscr {R}_\varepsilon (t_1,t_2,x_1,x_2)&=\int _{\mathbb {R}^{2d}} \varphi (x_1-y_1)\varphi (x_2-y_2)\mathbb {E}[W_\varepsilon (t_1,y_1)W_\varepsilon (t_2,y_2)]dy_1dy_2\\&=\int _{\mathbb {R}^{2d}} \varphi (x_1-y_1)\varphi (x_2-y_2) e^{-\varepsilon (t_1^2+t_2^2+|y_1|^2+|y_2|^2)} \phi _\varepsilon \\&\qquad \star \phi _\varepsilon (t_1-t_2,y_1-y_2) dy_1dy_2\\&\le \int _{\mathbb {R}^{2d}} \varphi (x_1-y_1)\varphi (x_2-y_2) \phi _\varepsilon \star \phi _\varepsilon (t_1-t_2,y_1-y_2) dy_1dy_2, \end{aligned} \end{aligned}$$

with $\star$ denoting the convolution. By the fact that $\varphi ,\phi$ have compact supports, we have

for some $C>0$. Thus, $\mathcal {Q}_\varepsilon$ is essentially measuring the mutual “intersection” time of $B^1,B^2$. By [18, Corollary 4.4] and the fact that $d\ge 3$, the proof is complete. $\square$

Now we can write

$$\begin{aligned} \begin{aligned}&\mathbb {P}[\mathscr {U}_\varepsilon (t,x,W_\varepsilon )\le r] \le \mathbb {P}\left[ \frac{1}{2} e^{-\sqrt{\lambda } \mathrm {dist}(W_\varepsilon ,A_\lambda (t,x))}\le r\right] \\&\quad \le \mathbb {P}\left[ \mathrm {dist}(W_\varepsilon ,A_\lambda (t,x)) \ge \frac{\log (2r)^{-1}}{\sqrt{\lambda }} \right] , \end{aligned} \end{aligned}$$

(B.4)

where $\mathrm {dist}(W_\varepsilon ,A_\lambda (t,x))=\inf \{ \Vert W_\varepsilon -\tilde{W}_\varepsilon \Vert _2: \tilde{W}_\varepsilon \in A_\lambda (t,x)\}$. Now we can apply [24, Lemma 4.5] to obtain

$$\begin{aligned} \mathbb {P}\left[ \mathrm {dist}(W_\varepsilon ,A_\lambda (t,x)) \ge \tau +2 \sqrt{\log \frac{2}{c}} \right] \le 2e^{-\tau ^2/4} \end{aligned}$$

(B.5)

for all $\tau >0$, where $\lambda ,c>0$ are chosen as in Lemma B.2. Combining (B.4) and (B.5), we have

$$\begin{aligned} \mathbb {P}[\mathscr {U}_\varepsilon (t,x,W_\varepsilon )\le r] \le 2\exp \left( -\frac{1}{4}\left( \frac{\log (2r)}{\sqrt{\lambda }}+2\sqrt{\log \frac{2}{c}}\right) ^2\right) , \end{aligned}$$

which implies $\mathbb {E}[\mathscr {U}_\varepsilon (t,x,W_\varepsilon )^{-n}]\lesssim 1$ and completes the proof of (B.1).