arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 2307.02481math.PRcond-mat.stat-mechmath-phmath.MP
收藏

对称排除过程与储层耦合的非平衡稳态

Non-equilibrium steady state of the symmetric exclusion process with reservoirs

Institute of Mathematics, Statistics and Computer Science, University of São Paulo(圣保罗大学数学、统计和计算机科学研究所);Mathematical Institute, University of Oxford(牛津大学数学研究所);School of Mathematical and Statistical Sciences and the Center for Mechanisms of Evolution (Biodesign Institute), Arizona State University(亚利桑那州立大学数学与统计科学学院及进化机制中心(生物设计研究所))

Jhon Astoquillca, Simone Floreani, Adrián González Casanova

AI总结:

研究连通图上带左右储层的开放对称排除过程,证明其非平衡稳态为伯努利测度的加权混合,权重由吸收对偶过程给出,并在齐次线段上显式计算。

摘要:

考虑一个连通图上的开放对称排除过程,其顶点位于 $[N-1]:=\{1,\ldots, N-1\}$ 中,其中点 $1$ 和 $N-1$ 分别连接到一个左储层和一个右储层,密度分别为 $\rho_L,\rho_R\in(0,1)$。我们证明该系统的非平衡稳态为 $$\mu_{\text{stat}} = \sum_{I\subset \mathcal P([N-1]) }F(I)\bigg(\otimes_{x\in I}\rm{Bernoulli}(\rho_R)\otimes_{y\in [N-1]\setminus I}\rm{Bernoulli}(\rho_L) \bigg).$$ 在上述公式中,$\mathcal P([N-1])$ 表示 $[N-1]$ 的幂集,而数 $F(I)>0$ 满足 $\sum_{I\subset \mathcal P([N-1]) }F(I)=1$,并且由吸收随机对偶过程的吸收概率给出。通过概率论论证,当图为齐次线段时,我们显式地计算了因子 $F(I)$。

英文摘要:

Consider the open symmetric exclusion process on a connected graph with vertexes in $[N-1]:=\{1,\ldots, N-1\}$ where points $1$ and $N-1$ are connected, respectively, to a left reservoir and a right reservoir with densities $ρ_L,ρ_R\in(0,1)$. We prove that the non-equilibrium steady state of such system is $$μ_{\text{stat}} = \sum_{I\subset \mathcal P([N-1]) }F(I)\bigg(\otimes_{x\in I}\rm{Bernoulli}(ρ_R)\otimes_{y\in [N-1]\setminus I}\rm{Bernoulli}(ρ_L) \bigg).$$ In the formula above $ \mathcal P([N-1])$ denotes the power set of $[N-1]$ while the numbers $F(I)> 0$ are such that $\sum_{I\subset \mathcal P([N-1]) }F(I)=1$ and given in terms of absorption probabilities of the absorbing stochastic dual process. Via probabilistic arguments we compute explicitly the factors $F(I)$ when the graph is a homogeneous segment.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

↑