Hydrodynamics of the Generalized N-urn Ehrenfest Model

Abstract

In this paper we are concerned with a generalized N-urn Ehrenfest model, where balls perform independent random walks between N boxes uniformly laid on [0,1]. After a proper scaling of the transition rates function of the aforesaid random walk, we derive the hydrodynamic limit of the model, i.e., the law of large numbers which the empirical measure of the model follows, under an assumption where the initial number of balls in each box independently follows a Poisson distribution. We show that the empirical measure of the model converges weakly to a deterministic measure with density driven by an integral equation. Furthermore, we derive non-equilibrium fluctuation of the model, i.e, the central limit theorem from the above hydrodynamic limit. We show that the non-equilibrium fluctuation of the model is driven by a time-inhomogeneous generalized O-U process on the dual of C[0,1]. At last, we prove a large deviation principle from the hydrodynamic limit under an assumption where the transition rates function from [0,1] × [0,1] to $[0, +\infty )$ of the aforesaid random walk is a product of two marginal functions from [0,1] to $[0, +\infty )$.

Appendices

A Appendix

A.1 Proof of Lemma 2.2

Proof of Lemma 2.2

We first construct a weak solution μ of Eq. 2.1 to show the existence. We use $\|f\|_{\infty }$ to denote the $l_{\infty }$-norm of f for any f ∈ C[0, 1] and $\|\lambda \|_{\infty }$ to denote $\sup _{0\leq x, y\leq 1}\lambda (x, y)$. For each f ∈ C[0, 1], we define

$$ \left( P_{3}f\right)(x)={{\int}_{0}^{1}}\lambda(y, x)f(y)dy $$

for any x ∈ [0, 1]. Since $\|P_{2}f\|_{\infty }, \|P_{3}f\|_{\infty }\leq \|\lambda \|_{\infty }\|f\|_{\infty }$ for any f ∈ C[0, 1], the linear C[0, 1]-valued ODE

$$ \begin{cases} \frac{d}{dt}\rho_{t}=(P_{3}-P_{2})\rho_{t}, &\text{~}0\leq t\leq T,\\ \rho_{0}(x)=\phi(x), &\text{~}0\leq x\leq 1, \end{cases} $$

satisfies Lipschitz’ condition and hence has the unique solution

$$ \rho_{t}=e^{t(P_{3}-P_{2})}\phi=\sum\limits_{k=0}^{+\infty}\frac{t^{k}\left( P_{3}-P_{2}\right)^{k}}{k!}\phi. $$

Let μ_t(dx) = ρ_t(x)dx for any 0 ≤ t ≤ T, then μ = {μ_t}_0≤t≤T is a weak solution of Eq. 2.1 since

$$ {{\int}_{0}^{1}}f(x)\left( P_{3}\rho_{t}\right)(x)dx={{\int}_{0}^{1}} \left( P_{1}f\right)(x)\rho_{t}(x)dx. $$

Now we only need to show the uniqueness. Let μ¹,μ² be two weak solutions of Eq. 2.1 and ν = μ¹ − μ², then

$$ \nu_{t}(f)={{\int}_{0}^{t}} \nu_{s}\left( (P_{1}-P_{2})f\right)ds $$

for any 0 ≤ t ≤ T. For any ${\mathscr{A}}\in \mathcal {S}$, we define $\|{\mathscr{A}}\|_{\infty }=\sup \left \{{\mathscr{A}}(f):~\|f\|_{\infty }=1\right \}$. Then, for any f satisfying $\|f\|_{\infty }=1$,

$$ |\nu_{s}\left( (P_{1}-P_{2})f\right)|\leq \|\nu_{s}\|_{\infty}\|(P_{1}-P_{2})f\|_{1}\leq 2\|\nu_{s}\|_{\infty}\|\lambda\|_{\infty}. $$

Therefore, for any 0 ≤ t ≤ T,

$$ \|\nu_{t}\|_{\infty}\leq 2\|\lambda\|_{\infty}{{\int}_{0}^{t}} \|\nu_{s}\|_{\infty}ds. $$

Then, by Gronwall’s inequality, $\|\nu _{t}\|_{\infty }\leq 0e^{2t\|\lambda \|_{\infty }}=0$ for any 0 ≤ t ≤ T and hence μ¹ = μ². □

A.2 The uniqueness of solution to Eq. 2.4

In this subsection we give an outline of the proof of the uniqueness of solution to Eq. 2.4. Our proof follows an analysis similar with that given in the proof of Theorem 1.4 of [15].

Proof of uniqueness to Eq. 2.4

For each t ≥ 0 and H ∈ C([0, 1]), we define

$$ A_{t}(H)=V_{t}(H)-V_{0}(H)-{{\int}_{0}^{t}} V_{s}\left( (P_{1}-P_{2})H\right)ds $$

and

$$ {\Xi}_{t}(H)={V_{t}^{2}}(H)-{V_{0}^{2}}(H)-{{\int}_{0}^{t}}2V_{s}(H)V_{s}\left( (P_{1}-P_{2})H\right)ds-{{\int}_{0}^{t}}\|b_{s}H\|_{2}^{2}ds. $$

According to the second statement in the definition of {V_t}_t≥ 0, it is easy to check that {A_t(H)}_t≥ 0 and {Ξ_t(H)}_t≥ 0 are both local martingales. Since

$$ dV_{t}(H)=V_{t}(\left( P_{1}-P_{2}\right)H)dt+dA_{t}(H), $$

by Itô’s formula,

$$ \begin{array}{@{}rcl@{}} d{V_{t}^{2}}(H)&=&2V_{t}(H)dV_{t}(H)+d\langle A(H)\rangle_{t}\\ &=&2V_{t}(H)V_{t}(\left( P_{1}-P_{2}\right)H)dt+d\langle A(H)\rangle_{t}+2V_{t}(H)dA_{t}(H) \end{array} $$

and hence

$$ \left\{{V_{t}^{2}}(H)-{V_{0}^{2}}(H)-{{\int}_{0}^{t}}2V_{s}(H)V_{s}\left( (P_{1}-P_{2})H\right)ds-\langle A(H)\rangle_{t}\right\}_{t\geq 0} $$

is a local martingale. Then, since {Ξ_t}_t≥ 0 is also a local martingale,

$$ \langle A(H)\rangle_{t}={{\int}_{0}^{t}} \|b_{s}H\|_{2}^{2}ds. $$

Let $i=\sqrt {-1}$ and

$$ J_{t}(H)=\exp\left\{iA_{t}(H)+\frac{1}{2}\langle A(H)\rangle_{t}\right\}=\exp\left\{iA_{t}(H)+\frac{1}{2}{{\int}_{0}^{t}}\|b_{s}H\|_{2}^{2}ds\right\} $$

for any t ≥ 0, then {J_t(H)}_0≤t≤T is a local martingale according to Itô’s formula and hence a martingale since {J_t(H)}_0≤t≤T are uniformly bounded. As a result, let ${{\Upsilon }_{s}^{t}}(H)=\frac {J_{t}(H)}{J_{s}(H)}$ for any 0 ≤ s < t, then

$$ \mathbb{E}\left( {{\Upsilon}_{s}^{t}}(H)\big|V_{u}, u\leq s\right)=1 $$

(A.1)

for any 0 ≤ s < t and H ∈ C([0, 1]). For given 0 ≤ t₁ < t₂ ≤ T and integers 0 ≤ k ≤ n, let $\sigma _{n,k}=t_{1}+\frac {k(t_{2}-t_{1})}{n}$. For any t ≥ 0, let $H_{t}=e^{t(P_{1}-P_{2})}H$, i.e., {H_t}_t≥ 0 is the solution to

$$ \begin{cases} &\frac{d}{dt}H_{t}=(P_{1}-P_{2})H_{t}, \\ &H_{0}=H. \end{cases} $$

It is easy to check that {H_t}_0≤t≤T is continuous in C([0, 1]) and hence

$$ \begin{array}{@{}rcl@{}} \lim\limits_{n\rightarrow+\infty} \Bigg(\sum\limits_{k=0}^{n-1}{\int}_{\sigma_{n,k}}^{\sigma_{n, k+1}}V_{u}\left( (P_{1}-P_{2})H_{t_{2}-\sigma_{n, k+1}}\right)du\\ -\sum\limits_{k=0}^{n-1}{\int}_{\sigma_{n,k}}^{\sigma_{n, k+1}}V_{\sigma_{n,k}}\left( (P_{1}-P_{2})H_{t_{2}-u}\right)du\Bigg)=0 \end{array} $$

almost surely. Furthermore, since

$$ \begin{array}{@{}rcl@{}} \sum\limits_{k=0}^{n-1}{\int}_{\sigma_{n,k}}^{\sigma_{n, k+1}}V_{\sigma_{n,k}}\left( (P_{1}-P_{2})H_{t_{2}-u}\right)du &=&\sum\limits_{k=0}^{n-1}V_{\sigma_{n,k}}\left( {\int}_{\sigma_{n,k}}^{\sigma_{n, k+1}}(P_{1}-P_{2})H_{t_{2}-u}du\right)\\ &=&-\sum\limits_{k=0}^{n-1}V_{\sigma_{n,k}}\left( H_{t_{2}-\sigma_{n, k+1}}-H_{t_{2}-\sigma_{n,k}}\right), \end{array} $$

we have

$$ \lim\limits_{n\rightarrow+\infty}\prod\limits_{k=0}^{n-1}{\Upsilon}_{\sigma_{n,k}}^{\sigma_{n, k+1}}\left( H_{t_{2}-\sigma_{n, k+1}}\right) =\frac{Y_{t_{2}}}{Y_{t_{1}}} $$

almost surely, where

$$ Y_{t}=\exp\left\{iV_{t}(H_{t_{2}-t})+\frac{1}{2}{{\int}_{0}^{t}}\|b_{u}H_{t_{2}-u}\|_{2}^{2}du\right\} $$

for 0 ≤ t ≤ t₂. Since $\{\prod_{k=0}^{n-1}{\Upsilon }_{\sigma _{n,k}}^{\sigma _{n, k+1}}\left (H_{t_{2}-\sigma _{n, k+1}}\right )\}_{n\geq 1}$ are uniformly bounded, the convergence above is also in L¹. As a result,

$$ \mathbb{E}\big(Y_{t_{2}}\big|V_{u}, u\leq t_{1}\big)=\lim_{n\rightarrow+\infty}\mathbb{E}\big(Y_{t_{1}}\prod\limits_{k=0}^{n-1}{\Upsilon}_{\sigma_{n,k}}^{\sigma_{n, k+1}}\left( H_{t_{2}-\sigma_{n, k+1}}\right)\big|V_{u}, u\leq t_{1}\big) $$

in L¹. Repeatedly utilizing (A.1) by n times, we have

$$ \mathbb{E}\big(Y_{t_{1}}\prod\limits_{k=0}^{n-1}{\Upsilon}_{\sigma_{n,k}}^{\sigma_{n, k+1}}\left( H_{t_{2}-\sigma_{n, k+1}}\right)\big|V_{u}, u\leq t_{1}\big) =Y_{t_{1}} $$

and hence

$$ \mathbb{E}\left( \exp\{iV_{t_{2}}(H)\}\big|V_{u}, u\leq t_{1}\right)=e^{iV_{t_{1}}(H_{t_{2}-t_{1}})-\frac{1}{2}{\int}_{t_{1}}^{t_{2}}\|b_{u}H_{t_{2}-u}\|_{2}^{2}du} $$

(A.2)

for any 0 ≤ t₁ < t₂ ≤ T and H ∈ C([0, 1]). The uniqueness of {V_t}_t≥ 0 follows from Eq. A.2 and the given distribution of V₀. □

A.3 A generalized N-urn Ehrenfest model with a metastable state

Throughout this subsection we assume that our model is under Assumptions (A) and (B) given in Section 2. We will show that in this case the model has a metastable state, i.e, $\lim _{t\rightarrow +\infty }\mu _{t}=\mu _{\infty }$ for some $\mu _{\infty }\in \mathcal {S}$, where μ_t≥ 0 satisfies that {μ_t}_0≤t≤T is the solution to Eq. 2.1 for any T > 0.

The proof of the existence of $\mu _{\infty }$

By Lemma 2.2, μ_t(dx) = ρ(t,x)dx, where

$$ \frac{d}{dt}\rho(t,x)=-m\lambda_{1}(x)\rho(t,x)+\lambda_{2}(x)r(t) $$

under Assumption (B), where $m={{\int \limits }_{0}^{1}}\lambda _{2}(y)dy$ and $r(t)={{\int \limits }_{0}^{1}}\lambda _{1}(y)\rho (t,y)dy$. Let $K(t,x)=\frac {\lambda _{1}(x)}{\lambda _{2}(x)}\rho (t,x)$, $\hat {\lambda }_{1}(x)=m\lambda _{1}(x)$, $\hat {\lambda }_{2}(x)=\frac {\lambda _{2}(x)}{m}$, then

$$ \frac{d}{dt}K(t,x)=\hat{\lambda}_{1}(x){{\int}_{0}^{1}}\left( K(t,y)-K(t,x)\right)\hat{\lambda}_{2}(y)dy. $$

(A.3)

Let $\{\mathcal {R}_{t}\}_{t\geq 0}$ be the Markov process with state space [0, 1] and generator $\mathcal {A}$ given by

$$ \mathcal{A}f(x)=\hat{\lambda}_{1}(x){{\int}_{0}^{1}} \left( f(y)-f(x)\right)\hat{\lambda}_{2}(y)dy $$

for any f ∈ C[0, 1] and x ∈ [0, 1], i.e, conditioned on $\mathcal {R}_{t}=x$, the process waits for an exponential time with rate $\hat {\lambda }_{1}(x)$ to escape from x and choose the next state y according to the probability measure $\hat {\lambda }_{2}(y)dy$. Then, by Eq. A.3,

$$ K(t,x)=\mathbb{E}_{x}K(0, \mathcal{R}_{t}). $$

(A.4)

It is easy to check that π(x)dx is a reversible distribution of $\{\mathcal {R}_{t}\}_{t\geq 0}$, where

$$ \pi(x)=\frac{\frac{\hat{\lambda}_{2}(x)}{\hat{\lambda}_{1}(x)}}{{{\int}_{0}^{1}}\frac{\hat{\lambda}_{2}(y)}{\hat{\lambda}_{1}(y)}dy} $$

for any x ∈ [0, 1]. For any probability measure ν on [0, 1] absolutely continuous with respect to Lebesgue measure, we claim that $\mathcal {R}_{t}$ converges weakly to π(x)dx as $t\rightarrow +\infty$ conditioned on $\mathcal {R}_{0}$ is with distribution ν. We prove this claim later. As a result, for any positive f ∈ C[0, 1] with ${{\int \limits }_{0}^{1}}f(x)dx=1$,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{t\rightarrow+\infty}{{\int}_{0}^{1}}f(x)K(t,x)dx&=\lim_{t\rightarrow+\infty}{{\int}_{0}^{1}}f(x)\mathbb{E}_{x}K(0, \mathcal{R}_{t})dx\\ &={{\int}_{0}^{1}}\pi(x)K(0, x)dx \end{array} $$

(A.5)

by Eq. A.4. Since $K(0, x)=\frac {\lambda _{1}(x)}{\lambda _{2}(x)}\rho (0, x)=\frac {\lambda _{1}(x)}{\lambda _{2}(x)}\phi (x)$ and $\frac {h(x)}{{{\int \limits }_{0}^{1}}h(x)dx}$ is a probability density for any positive h ∈ C[0, 1], Eq. A.5 implies that $\lim _{t\rightarrow +\infty }\mu _{t}=\mu _{\infty }$, where

$$ \mu_{\infty}(dx)=\frac{{{\int}_{0}^{1}}\phi(u)du}{{{\int}_{0}^{1}}\frac{\lambda_{2}(u)}{\lambda_{1}(u)}du}\frac{\lambda_{2}(x)}{\lambda_{1}(x)}dx. $$

At last we prove our claim. The proof follows the entropy method given in Section 2.4 of [22] and hence we only give an outline. For any ν absolutely continuous with respect to Lebesgue measure, let $\mathcal {E}(\nu )$ be the entropy of ν relative to π, i.e,

$$ \mathcal{E}(\nu)={{\int}_{0}^{1}}q\left( \frac{f_{\nu}(x)}{\pi(x)}\right)\pi(x)dx, $$

where $q(x)=x\log x$ and f_ν is the probability density of ν. For any t ≥ 0, let ν_t be the probability measure of $\mathcal {R}_{t}$ conditioned on $\mathcal {R}_{0}$ with distribution ν. By Chapman-Kolmogorov equation, ν_t is also absolutely continuous with respect to Lebesgue measure and its probability density {f_ν(t,x) : x ∈ [0, 1]} is the solution to the equation

$$ \begin{cases} \frac{d}{dt}f_{\nu}(t,x)=-\hat{\lambda}_{1}(x)f_{\nu}(t,x)+\hat{\lambda}_{2}(x){{\int}_{0}^{1}}f_{\nu}(t,y)\hat{\lambda}_{1}(y)dy,\\ f_{\nu}(0, x)=f_{\nu}(x). \end{cases} $$

(A.6)

For any y ∈ [0, 1], let p_t(y,du) be the probability measure of $\mathcal {R}_{t}$ conditioned on $\mathcal {R}_{0}=y$, then for any Borel-measurable set B ⊂ [0, 1], we define

$$ m(t, B)={{\int}_{0}^{1}}p_{t}(y, B)dy, $$

i.e, m(t,du) is the probability measure of $\mathcal {R}_{t}$ conditioned on $\mathcal {R}_{0}$ is uniformly distributed on [0, 1]. Then, m(t,du) is absolutely continuous with respect to Lebesgue measure and its probability density f_m(t,x) is the solution to Eq. A.6 with initial condition f_m(0,x) = 1 for all x ∈ [0, 1]. By classic theory of measures on product spaces, for each u ∈ [0, 1], there exists probability measure m₂(t,u,dy) such that

$$ {\int}_{A}p_{t}(y, B)dy={\int}_{B}m_{2}(t, u, A)m(t, du)={\int}_{B}m_{2}(t, u, A)f_{m}(t,u)du $$

for any Borel-measurable $A, B\subseteq [0, 1]$. Since

$$ \mathbb{E}_{\nu} h(\mathcal{R}_{t})={{\int}_{0}^{1}} f_{\nu}(y)\left( {\int}_{[0, 1]}h(u)p_{t}(y, du)\right)dy $$

for any h ∈ C[0, 1], we have

$$ f_{\nu}(t,x)=f_{m}(t,x){\int}_{[0, 1]}f_{\nu}(y)m_{2}(t, x, dy) $$

(A.7)

and especially

$$ \pi(x)=f_{m}(t,x){\int}_{[0, 1]}\pi(y)m_{2}(t, x, dy) $$

(A.8)

since π(x)dx is an invariant measure of $\{\mathcal {R}_{t}\}_{t\geq 0}$. By Eqs. A.7, A.8 and Jensen’s inequality,

$$ \begin{array}{@{}rcl@{}} \mathcal{E}(\nu_{t})&=&{{\int}_{0}^{1}}q\left( \frac{f_{\nu}(t, x)}{\pi(x)}\right)\pi(x)dx\\ &=&{{\int}_{0}^{1}}q\left( {\int}_{[0, 1]}\frac{f_{m}(t,x)f_{\nu}(y)}{\pi(x)}m_{2}(t, x, dy)\right)\pi(x)dx\\ &=&{{\int}_{0}^{1}}q\left( {\int}_{[0, 1]}f_{m}(t,x)\frac{\pi(y)}{\pi(x)}\frac{f_{\nu}(y)}{\pi(y)}m_{2}(t,x,dy)\right)\pi(x)dx\\ &\leq& {{\int}_{0}^{1}}\left( {\int}_{[0, 1]}f_{m}(t, x)\frac{\pi(y)}{\pi(x)}q\left( \frac{f_{\nu}(y)}{\pi(y)}\right)m_{2}(t, x, dy)\right)\pi(x)dx\\ &=&{\int}_{[0, 1]}\left( {\int}_{[0, 1]}\pi(y)q\left( \frac{f_{\nu}(y)}{\pi(y)}\right)m_{2}(t, x, dy)\right)m(t, dx)\\ &=&{{\int}_{0}^{1}}\pi(y)q\left( \frac{f_{\nu}(y)}{\pi(y)}\right)\left( {\int}_{[0, 1]}1 p_{t}(y, dx)\right)dy\\ &=&{{\int}_{0}^{1}}\pi(y)q\left( \frac{f_{\nu}(y)}{\pi(y)}\right)dy=\mathcal{E}(\nu). \end{array} $$

As a result, $\mathcal {E}(\nu _{t})$ is decreasing with t and hence

$$ \lim\limits_{t\rightarrow+\infty}\mathcal{E}(\nu_{t}) $$

exists. Furthermore, by Chapman-Kolmogorov equation,

$$ p_{t}(y, du)=l_{1}(t, y)\delta_{y}(du)+l_{2}(t, y, u)du, $$

where δ_y(du) is the Dirac measure concentrated on y and l₁(t,y),l₂(t,y,⋅) is the solution to the equation

$$ \begin{cases} &\frac{d}{dt}l_{1}(t, y)=-\hat{\lambda}_{1}(y)l_{1}(t,y),\\ &\frac{d}{dt}l_{2}(t, y, u)=\hat{\lambda}_{2}(u)l_{1}(t,y)\hat{\lambda}_{1}(y)+\hat{\lambda}_{2}(u){{\int}_{0}^{1}}l_{2}(t, y, x)\hat{\lambda}_{1}(x)dx-l_{2}(t, y, u)\hat{\lambda}_{1}(u),\\ &l_{1}(0, y)=1,\\ &l_{2}(0, y, u)=0. \end{cases} $$

Hence, $l_{1}(t, y)=e^{-\hat {\lambda }_{1}(y)t}$ and l₂(t,y,u) > 0 for any t > 0,u ∈ [0, 1]. According to the definition of m₂(t,x,dy),

$$ m_{2}(t, x, dy)f_{m}(t,x)=l_{1}(x)\delta_{x}(dy)+l_{2}(t, y, x)dy. $$

As a result, m₂(t,x,B) = 0 implies that the Lebesgue measure of B is 0 for any t > 0 and x ∈ [0, 1]. Consequently, $\mathcal {E}(\nu _{t})=\mathcal {E}(\nu )$ when and only when ν(dx) = π(x)dx. Since {ν_t}_t≥ 0 are probability measures on [0, 1], they are tight. Let $\tilde {\nu }$ be the weak limit of a subsequence $\{\nu _{t_{k}}\}_{k\geq 1}$ of {ν_t}_t≥ 0. By Eq. A.6,

$$ f_{\nu}(t, x)=f_{\nu}(x)e^{-\hat{\lambda}_{1}(x)t}+{{\int}_{0}^{t}}e^{\hat{\lambda}_{1}(x)(s-t)}\hat{\lambda}_{2}(x)\mathbb{F}_{\nu}(s)ds $$

(A.9)

where $\mathbb {F}_{\nu }(s)={{\int \limits }_{0}^{1}}\hat {\lambda }_{1}(y)f_{\nu }(s,y)dy$. By Eq. A.9, it is easy to check that {f_ν(t,⋅)}_t≥ 0 are uniformly bounded and equicontinuous. As a result, $\tilde {\nu }$ is absolutely continuous with respect to Lebesgue measure and its probability density $f_{\tilde {\nu }}$ is a C[0, 1]-limit of a subsequence of {f_ν(t,⋅)}_t≥ 0. Since q is continuous on [0, 1],

$$ \mathcal{E}(\tilde{\nu})=\lim\limits_{t\rightarrow+\infty}\mathcal{E}(\nu_{t}). $$

For any given s > 0, $\tilde {\nu }_{s}$ is the weak-limit of the subsequence $\{\nu _{t_{k}+s}\}_{k\geq 1}$ of {ν_t}_t≥ 0. As a result,

$$ \mathcal{E}(\tilde{\nu}_{s})=\lim_{t\rightarrow+\infty}\mathcal{E}(\nu_{t})=\mathcal{E}(\tilde{\nu}) $$

and hence $\tilde {\nu }(dx)=\pi (x)dx$. As a result, ν_t converges weakly to π(x)dx as $t\rightarrow +\infty$ and the proof is complete. □