Annealed Asymptotics for Brownian Motion of Renormalized Potential in Mobile Random Medium

Abstract

Motivated by the study of the directed polymer model with mobile Poissonian traps or catalysts and the stochastic parabolic Anderson model with time-dependent potential, we investigate the asymptotic behavior of

$$\begin{aligned} \mathbb {E}\otimes \mathbb {E}_0\exp \left\{ \pm \ \theta \int \limits ^t_0\bar{V}(s,B_s)\hbox {d}s\right\} \quad (t\rightarrow \infty ) \end{aligned}$$

where $\theta >0$ is a constant, $\overline{V}$ is the renormalized Poisson potential of the form

$$\begin{aligned} \overline{V}(s,x)=\int \limits _{\mathbb {R}^d}\frac{1}{|y-x|^p}\left( \omega _s(\hbox {d}y)-\hbox {d}y\right) , \end{aligned}$$

and $\omega _s$ is the measure-valued process consisting of independent Brownian particles whose initial positions form a Poisson random measure on $\mathbb {R}^d$ with Lebesgue measure as its intensity. Different scaling limits are obtained according to the parameter $p$ and dimension $d$. For the logarithm of the negative exponential moment, the range of $\frac{d}{2}<p<d$ is divided into five regions with various scaling rates of the orders $t^{d/p}$, $t^{3/2}$, $t^{(4-d-2p)/2}$, $t\log t$ and $t$, respectively. For the positive exponential moment, the limiting behavior is studied according to the parameters $p$ and $d$ in three regions. In the subcritical region ($p<2$), the double logarithm of the exponential moment has a rate of $t$. In the critical region ($p=2$), it has different behavior over two parts decided according to the comparison of $\theta$ with the best constant in the Hardy inequality. In the supercritical region $(p>2)$, the exponential moments become infinite for all $t>0$.

Appendix

In this section, we prove some technical results used in the main body of the paper.

1.1 Hardy Inequality

Hardy’s inequality plays an important role in this paper. Searching in literature, we have found large amount of versions of Hardy’s inequality (i.e., [29, 33]) but the form needed in this paper. For reader’s convenience, we state Hardy’s inequality for $d=3$ in the following lemma and provide a short proof.

Lemma 6.1

For any $f\in W^{1,2}(\mathbb {R}^3)$,

$$\begin{aligned} \int \limits _{\mathbb {R}^3}{f^2(x)\over \vert x\vert ^2}\hbox {d}x\le 4\int \limits _{\mathbb {R}^3}\vert \nabla f(x)\vert ^2 \hbox {d}x. \end{aligned}$$

(6.1)

Further, the number 4 is the best constant in the sense that for any $\epsilon >0$ one can find a function $f_\epsilon \in W^{1,2}(\mathbb {R}^3)$ with compact support such that

$$\begin{aligned} \int \limits _{\mathbb {R}^3}{f_\epsilon ^2(x)\over \vert x\vert ^2}\hbox {d}x> (4-\epsilon ) \int \limits _{\mathbb {R}^3}\vert \nabla f_\epsilon (x)\vert ^2 \hbox {d}x. \end{aligned}$$

(6.2)

Proof

Write $x=(x_1,x_2,x_3)$. Using integration by parts

$$\begin{aligned} \int \limits _{\mathbb {R}^3}{f^2(x)\over \vert x\vert ^2}\hbox {d}x =\int \limits _{\mathbb {R}^3}x_j\Big [{2x_i\over \vert x\vert ^4}f^2(x)-{2\over \vert x\vert ^2} f(x){\partial f\over \partial x_j}\Big ]\hbox {d}x\quad j=1,2,3. \end{aligned}$$

Summing over $j$ on the both sides

$$\begin{aligned} 3\int \limits _{\mathbb {R}^3}{f^2(x)\over \vert x\vert ^2}\hbox {d}x= 2\int \limits _{\mathbb {R}^3}\Big [{f^2(x)\over \vert x\vert ^2}- {\nabla f\cdot x\over \vert x\vert ^2}f(x)\Big ]\hbox {d}x. \end{aligned}$$

Thus,

$$\begin{aligned} \int \limits _{\mathbb {R}^3}{f^2(x)\over \vert x\vert ^2}\hbox {d}x =-2\int \limits _{\mathbb {R}^3}{\nabla f\cdot x\over \vert x\vert }{f(x)\over \vert x\vert }\hbox {d}x \le 2\bigg (\int \limits _{\mathbb {R}^3}{\vert \nabla f\cdot x\vert ^2\over \vert x\vert ^2} \hbox {d}x\bigg )^{1/2}\bigg (\int \limits _{\mathbb {R}^3}{f^2(x)\over \vert x\vert ^2}\hbox {d}x\bigg )^{1/2}. \end{aligned}$$

Therefore,

$$\begin{aligned} \int \limits _{\mathbb {R}^3}{f^2(x)\over \vert x\vert ^2}\hbox {d}x\le 4 \int \limits _{\mathbb {R}^3}{\vert \nabla f\cdot x\vert ^2\over \vert x\vert ^2}\hbox {d}x \le 4\int \limits _{\mathbb {R}^3}\vert \nabla f(x)\vert ^2\hbox {d}x. \end{aligned}$$

To establish (6.2), for each large $M>0$, we define $g_M\in W^{1,2}(\mathbb {R}^3)$ as following:

$$\begin{aligned} g_M(x)=\left\{ \begin{array}{ll} M^{1/2} &{} 0\le \vert x\vert \le M^{-1}\\ \\ \vert x\vert ^{-1/2} &{} M^{-1}<\vert x\vert \le M\\ \\ \displaystyle {2M-\vert x\vert \over M^{3/2}} &{} M<\vert x\vert \le 2M\\ \\ 0 &{} \vert x\vert > 2M.\end{array}\right. \end{aligned}$$

It is straightforward to exam that $g_M$ is locally supported and

$$\begin{aligned} \int \limits _{\mathbb {R}^3}{g_M^2(x)\over \vert x\vert ^2}\hbox {d}x= \bigg \{4-28\Big ({7\over 3}+{1\over 2}\log M\Big )^{-1}\bigg \} \int \limits _{\mathbb {R}^3}\vert \nabla g_M(x)\vert ^2\hbox {d}x. \end{aligned}$$

For each $\epsilon >0$, take $M>0$ sufficiently large so

$$\begin{aligned} 28\Big ({7\over 3}+{1\over 2}\log M\Big )^{-1}<\epsilon \end{aligned}$$

and let $f_\epsilon (x)=g_M(x)$. $\square$

What has been used in this paper is the following version of Hardy’s inequality.

Lemma 6.2

For any $\theta >0$,

$$\begin{aligned} \sup _{g\in \mathcal{F}_3}\bigg \{\theta \int \limits _{\mathbb {R}^3}{g^2(x)\over \vert x\vert ^2}\hbox {d}x -{1\over 2}\int \limits _{\mathbb {R}^3}\vert \nabla g(x)\vert ^2\hbox {d}x\bigg \} = \left\{ \begin{array}{ll} 0 &{} \text {if } \theta \le 1/8,\\ \infty &{} \text {if } \theta > 1/8. \end{array}\right. \end{aligned}$$

(6.3)

Proof

By Hardy’s inequality, the left-hand side of (6.3) is non-positive when $\theta <1/8$. On the other hand, it is no less than

$$\begin{aligned} -{1\over 2}\inf _{g\in \mathcal{F}_3}\int \limits _{\mathbb {R}^3}\vert \nabla g(x)\vert ^2\hbox {d}x \end{aligned}$$

which is equal to zero. Thus, for $\theta \le 1/8$,

$$\begin{aligned} \sup _{g\in \mathcal{F}_3}\bigg \{\theta \int \limits _{\mathbb {R}^3}{g^2(x)\over \vert x\vert ^2}\hbox {d}x -{1\over 2}\int \limits _{\mathbb {R}^3}\vert \nabla g(x)\vert ^2\hbox {d}x\bigg \}=0. \end{aligned}$$

Assume $\theta > 1/8$. By the optimality of Hardy’s inequality described in (6.2),

$$\begin{aligned} H(\theta )\equiv \sup _{g\in \mathcal{F}_3} \bigg \{\theta \int \limits _{\mathbb {R}^3}{g^2(x)\over \vert x\vert ^2}\hbox {d}x -{1\over 2}\int \limits _{\mathbb {R}^3}\vert \nabla g(x)\vert ^2\hbox {d}x\bigg \}>0. \end{aligned}$$

Given $a>0$, the substitution $g(x)=a^{3/2}f(ax)$ leads to $H(\theta )=a^2H(\theta )$. So $M(\theta )=\infty$. $\square$

1.2 An Auxiliary Limit Result

Let

$$\begin{aligned} Q(b)=\int \limits _{\{\vert x\vert \ge b\}} {1\over \vert x\vert ^p}\mathbb {E}{1\{\vert x+U\vert \ge b\}\over \vert x+U\vert ^p}\hbox {d}x \quad (b\ge 0) \end{aligned}$$

(6.4)

with $U\sim N(0, I_d)$. In this subsection, we give the limiting behaviors of $Q(b)$ as $b\rightarrow 0+$ and as $b\rightarrow \infty$, respectively.

Lemma 6.3

$$\begin{aligned} \lim _{b\rightarrow 0^+}Q(b)&=\mathbb {E}\int \limits _{\mathbb {R}^d}{1\over \vert x\vert ^p}{1\over \vert x+U\vert ^p}\hbox {d}x=C(d,p)\mathbb {E}\vert U\vert ^{-(2p-d)}\nonumber \\&=C(d,p)2^{d-p-1}\hbox {d}\omega _d(2\pi )^{-d/2}\Gamma (d-p), \end{aligned}$$

(6.5)

where

$$\begin{aligned} C(d,p)=\pi ^{d/2}{\displaystyle \Gamma ^2\Big ({d-p\over 2}\Big ) \Gamma \Big ({2p-d\over 2}\Big )\over \displaystyle \Gamma ^2\Big ({p\over 2}\Big )\Gamma (d-p)}. \end{aligned}$$

(6.6)

Proof

The first equality follows from the monotone convergence theorem. The second follows from the identity (see p. 118, [15] or p. 118, (8), [37])

$$\begin{aligned} \int \limits _{\mathbb {R}^d}{1\over \vert x-y\vert ^p}{1\over \vert x-z\vert ^p}\hbox {d}x =C(d,p){1\over \vert y-z\vert ^{2p-d}}\quad y,z\in \mathbb {R}^d. \end{aligned}$$

(6.7)

$\square$

Lemma 6.4

Under $d/2<p<d$, we have

$$\begin{aligned} \lim _{b\rightarrow \infty }b^{2p-d}Q(b)={\hbox {d}\omega _d\over 2p-d}. \end{aligned}$$

(6.8)

Proof

By variable substitution, we have,

$$\begin{aligned}&\int \limits _{\{\vert x\vert \ge b\}} {1\over \vert x\vert ^p}{1\{\vert x+ U\vert \ge b\}\over \vert x+U\vert ^p}\hbox {d}x\\&\quad =\vert U\vert ^{-(2p-d)}\int \limits _{\{\vert x\vert \ge b\vert U\vert ^{-1}\}} {1\over \vert x\vert ^p}{1\{\vert x+\vert U\vert ^{-1}U\vert \ge b\vert U\vert ^{-1}\}\over \vert x+\vert U\vert ^{-1}U\vert ^p}\hbox {d}x\\&\quad =\vert U\vert ^{-(2p-d)}H\big (b\vert U\vert ^{-1}\big ), \end{aligned}$$

where

$$\begin{aligned} H(b)=\int \limits _{\{\vert x\vert \ge b\}}{1\over \vert x\vert ^p} {1\{\vert x+x_0\vert \ge b\}\over \vert x+x_0\vert ^p}\hbox {d}x, \end{aligned}$$

$x_0$ is a fixed point with $\vert x_0\vert =1$, and the last step follows from the fact that $H(b)$ does not depend on the location of $x_0$ on the unit sphere. Thus,

$$\begin{aligned}&Q(b)=\mathbb {E}\vert U\vert ^{-(2p-d)}H\big (b\vert U\vert ^{-1}\big )\\&\quad =\int \limits _{\mathbb {R}^d}{\hbox {d}x\over \vert x\vert ^p\vert x+x_0\vert ^p} \bigg [\int \limits _{\mathbb {R}^d}{1\{\vert y\vert \ge b\vert x\vert ^{-1}\} 1\{\vert y\vert \ge b\vert x+x_0\vert ^{-1}\}\over \vert y\vert ^{2p-d}} p_1(y) \hbox {d}y\bigg ]\hbox {d}x\\&\quad \sim \int \limits _{\{\vert x\vert \ge C\}}{\hbox {d}x\over \vert x\vert ^p\vert x+x_0\vert ^p} \bigg [\int \limits _{\mathbb {R}^d}{1\{\vert y\vert \ge b\vert x\vert ^{-1}\} 1\{\vert y\vert \ge b\vert x+x_0\vert ^{-1}\}\over \vert y\vert ^{2p-d}} p_1(y)\hbox {d}y\bigg ]\hbox {d}x, \end{aligned}$$

where $p_1(x)$ is the density of the $d$-dimensional standard normal distribution, $C>0$ is a large but fixed constant. By the fact that $C\gg 1=\vert x_0\vert$, for $b\rightarrow \infty$, we have

$$\begin{aligned}&Q(b)\sim q(C)\int \limits _{\{\vert x\vert \ge C\}} {\hbox {d}x\over \vert x\vert ^{2p}} \bigg [\int \limits _{\{\vert y\vert \ge b\vert x\vert ^{-1}\}} {1\over \vert y\vert ^{2p-d}}p_1(y)\hbox {d}y\bigg ]\\&\quad =q(C)\int \limits _{\mathbb {R}^d}{1\over \vert y\vert ^{2p-d}}p_1(y)\bigg [ \int \limits _{\{\vert x\vert \ge \max \{C, b\vert y\vert ^{-1}\}\}}{\hbox {d}x\over \vert x\vert ^{2p}}\bigg ] \hbox {d}y\\&\quad =q(C){\hbox {d}\omega _d\over 2p-d}\int \limits _{\mathbb {R}^d}{1\over \vert y\vert ^{2p-d}}p_1(y) \min \bigg \{{1\over C^{2p-d}}, \Big ({\vert y\vert \over b}\Big )^{2p-d}\bigg \}\hbox {d}y\\&\quad =q(C){\hbox {d}\omega _d\over 2p-d}{1\over C^{2p-d}}\int \limits _{\{\vert y\vert \ge C^{-1}b\}} {1\over \vert y\vert ^{2p-d}}p_1(y)\hbox {d}y\\&\quad +\,q(C){\hbox {d}\omega _d\over 2p-d}{1\over b^{2p-d}} \int \limits _{\{\vert y\vert \le C^{-1}b\}}p_1(y)\hbox {d}y, \end{aligned}$$

where $q(C)\rightarrow 1$ as $C\rightarrow \infty$. The first term on the right-hand side is obviously negligible. $\square$