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1、中国科学技术大学 硕士学位论文 图上Fokker-Planck方程解趋于全局平衡态的收敛性 姓名：车睿 申请学位级别：硕士 专业：数学系;基础数学 指导教师：黄文;邵松 2011-04 S.N. Chow,W. Huang, Y. LiH.M. Zhou34?k?gdU MarkovLFokker-planck?X“?N 2?y?A ?Fokker-Planck3dV?Riemannian6/?N?5 |“L?Su?L?Fokker-Planck“z Fokker-Planckk?Gibbs“3?k Fokker-Planck?)u?,(?y?)U ?“ ii Abstract In 4, C
2、how, Huang, Li and Zhou consider Fokker-planck equations for a free energy function or Markov process defi ned on a graph with a fi nite number of vertices and edges. If N 2 is the number of vertices of the graph, they shown that the corresponding Fokker-Planck equation is a system of N nonlinear or
3、dinary diff erential equations defi ned on a Riemannian manifold of probability distributions. There have diff erent choices for inner products on the space of probability distributions result in diff erent Fokker-Planck equations for the same process. Each of these Fokker-Planck equations has uniqu
4、e global equilibrium: Gibbs distribution. In this paper we study the speed of convergence towards global equilibrium for the solution of these Fokker-Planck equations on a graph, and prove the convergence is exponential. iii 3L!n!z!)?+y?SK ?k?“?LVm?zdFokker-Planck ? d 1 2 I E a b?G“ ? = (i)N i=1D3:8
5、V?i:ai? 0 xD“ M = = (i)N i=1 R N| N X i=1 i= 1i 0i = 1,2, ,N, V?V?mpz = iN i=1 mV?V=i 3ai?V N X i=1 i= 1,i 0. KgdUkXe?/ F() = N X i=1 ii+ N X i=1 ilogi,(0.0.1) pi:ai?“gdUk? Gibbs?,(L i = 1 K ei/,K = N X i=1 ei/.(0.0.2) lgdU?Chow?43mM?idT d?9?(?“?XLi6/(M,d)gdU (0.0.1)?F6 ?M?Fokker-Planck: di dt = X j
6、N(i),ji (j+ logj) (i+ logi)j + X jN(i),ji (j+ logj) (i+ logi)j + X jN(i),j 0 (0.0.3)(0.0.4)Pk?5(4?n4.1n5.2): 1.VmMgdU?F6?)? 3mM?“ 2. (0.0.2)?Gibbs= ( i)Ni=1 3M?-?.?,gd UF3X?Gibbs?. 3.?G?0 M,Pk?)“D0 M,(t)? ? (t) : 0,) M v: (a)gdUF(t)mt4, (b)3RNFokker-Planck?)3Euclideane(t) ?t +. p(0.0.3)(0.0.4)3?Fokk
7、er-PlanckI (0.0.3)dim(M,d) gdUF?F6?,?MarkovLvk(?X“ ?, Fokker-PlanckII (0.0.4)duMarkovL/xD06?“ w,m(M,d )(4?n5.2)?gdU? /F 60 “ ?(M,d )%1w?“ Chow?y?3eMU?yk ?d 3z1w?“ L?Gibbs= ( i)Ni=1 M(0.0.3)(0.0.4) ?“(0.0.3)(0.0.4)?)(t)3RNi?em u?“g,?Ku(0.0.3)(0.0.4)?)u? ?OAO/U?3?O? ?Fokker-Planck?)u? y?“ 1uFokker-Pla
8、nck?;(J 3!?Fokker-Planck?;(J,AOChow?34 ?“3YmRNugdUFokker-Planck9L? n3?X. M = ( = iN i=1 R N| N X i=1 i= 1,i 0 N Y i=1 i= 0 ) 3 M?mTM TM = ( = (i)N i=1 R N| N X i=1 i= 0 ) Fokker-Planck)u?57 ?dRNEuclidean“Kd M?Riemannian ? : (M,d) (RN,d)(1.0.7) () = (i()N i=1, M ?1wN?“ u? = (i)N i=1 M ?TM#?S“RN?dX: p
9、 q?=?p1 q1= p2 q2= = pN qN. WmRN/ “Ku?dW?TM?N?Xe up = (pi)N i=1 W (p) = (i)Ni=1 TM: i= X jN(i),ji (pi pj)j+ X jN(i),ji (j+ logj) (i+ logi)j + X jN(i),j 0Gibbs= ( i)Ni=1 L i = 1 K ei/,K = N X i=1 ei/(1.0.14) T(0.0.3)3M-?gdUF3Gibbs? ?“ 3.? 00 M(0.0.3)3) (t) : 0,) M (t)v (a)gdUF(t)t4 (b)3e(t) ?t +. y.
10、1).u 0M?gdUF: F() = N X i=1 ii+ N X i=1 ilogi(1.0.15) = iN i=1 M“?(M,g)gdUF?F6 d dt = gradF(),(1.0.16) 10 I E a pgradF() TM“ g (d dt ,) = diff F(). TM.(1.0.17) ?(1.0.17)? g (d dt ,) = N X i=1 di dt pi(1.0.18) (1.0.17)?m diff F(). = N X i=1 (i+ (1 + logi)i.(1.0.19) d i= X jN(i),ji (pi pj)j+ X jN(i),j
11、i (j i)j+ (logj logi)j) + X jN(i),ji (j i)j+ (logj logi)j) + X jN(i),ji (i j)j+ (logi logj)j) + X jN(i),ji (i j)j+ (logi logj)j) + X jN(i),j F(1) ?F(1) = maxF() : (0)g“1= “= maxF() : (0) = F(). qduF?: (0) = ?(3)?y“? lL?Chow?4?uV?MarkovL?/xD06“ Vm?u?M,Fokker-Planck=Fokker- PlanckII(0.0.4)“u?/?Fokker-Planck Ym?/?,“?Fokker-PlanckII,MgdUF?F 6Chow, Huang, LiZhou()?1UEk ?Fokker-Planck Iaq?5 =?() ()“ d?Sg )?m(M,d )“ ? du()6udN? M 7 g Yl?(M,g )Riemannian6/“ ?(M,g )3F6“+XduMm?AX ?2?F6/Xe g (d dt (t),) = diff F(t). TM,(1.0.22) diff
