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 \([0, +\infty )\).
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Acknowledgements
The author is grateful to Dr. Zhengyao Sun, Dr. Linjie Zhao, Prof. Minzhi Zhao and Prof. Qiang Yao for useful suggestions and comments. The author is grateful to the reviewers. Their comments are great help for the improvement of this paper. The author is grateful to the financial support from the National Natural Science Foundation of China with grant number 11501542.
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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
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
satisfies Lipschitz’ condition and hence has the unique solution
Let μt(dx) = ρt(x)dx for any 0 ≤ t ≤ T, then μ = {μt}0≤t≤T is a weak solution of Eq. 2.1 since
Now we only need to show the uniqueness. Let μ1,μ2 be two weak solutions of Eq. 2.1 and ν = μ1 − μ2, then
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 \(\|f\|_{\infty }=1\),
Therefore, for any 0 ≤ t ≤ T,
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
and
According to the second statement in the definition of {Vt}t≥ 0, it is easy to check that {At(H)}t≥ 0 and {Ξt(H)}t≥ 0 are both local martingales. Since
by Itô’s formula,
and hence
is a local martingale. Then, since {Ξt}t≥ 0 is also a local martingale,
Let \(i=\sqrt {-1}\) and
for any t ≥ 0, then {Jt(H)}0≤t≤T is a local martingale according to Itô’s formula and hence a martingale since {Jt(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
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., {Ht}t≥ 0 is the solution to
It is easy to check that {Ht}0≤t≤T is continuous in C([0, 1]) and hence
almost surely. Furthermore, since
we have
almost surely, where
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,
in L1. Repeatedly utilizing (A.1) by n times, we have
and hence
for any 0 ≤ t1 < t2 ≤ T and H ∈ C([0, 1]). The uniqueness of {Vt}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 μt≥ 0 satisfies that {μt}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, μt(dx) = ρ(t,x)dx, where
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
Let \(\{\mathcal {R}_{t}\}_{t\geq 0}\) be the Markov process with state space [0, 1] and generator \(\mathcal {A}\) given by
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,
It is easy to check that π(x)dx is a reversible distribution of \(\{\mathcal {R}_{t}\}_{t\geq 0}\), where
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 \({{\int \limits }_{0}^{1}}f(x)dx=1\),
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
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,
where \(q(x)=x\log x\) and fν is the probability density of ν. For any t ≥ 0, let νt be the probability measure of \(\mathcal {R}_{t}\) conditioned on \(\mathcal {R}_{0}\) with distribution ν. By Chapman-Kolmogorov equation, νt is also absolutely continuous with respect to Lebesgue measure and its probability density {fν(t,x) : x ∈ [0, 1]} is the solution to the equation
For any y ∈ [0, 1], let pt(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
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 fm(t,x) is the solution to Eq. A.6 with initial condition fm(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
for any Borel-measurable \(A, B\subseteq [0, 1]\). Since
for any h ∈ C[0, 1], we have
and especially
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,
As a result, \(\mathcal {E}(\nu _{t})\) is decreasing with t and hence
exists. Furthermore, by Chapman-Kolmogorov equation,
where δy(du) is the Dirac measure concentrated on y and l1(t,y),l2(t,y,⋅) is the solution to the equation
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),
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 {νt}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 {νt}t≥ 0. By Eq. A.6,
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 {fν(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 {fν(t,⋅)}t≥ 0. Since q is continuous on [0, 1],
For any given s > 0, \(\tilde {\nu }_{s}\) is the weak-limit of the subsequence \(\{\nu _{t_{k}+s}\}_{k\geq 1}\) of {νt}t≥ 0. As a result,
and hence \(\tilde {\nu }(dx)=\pi (x)dx\). As a result, νt converges weakly to π(x)dx as \(t\rightarrow +\infty \) and the proof is complete. □
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Xue, X. Hydrodynamics of the Generalized N-urn Ehrenfest Model. Potential Anal 59, 613–649 (2023). https://doi.org/10.1007/s11118-021-09980-7
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DOI: https://doi.org/10.1007/s11118-021-09980-7