The Effect of Gaussian White Noise on Dynamical Systems Part I: - PowerPoint PPT Presentation

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling The Effect of Gaussian White Noise on Dynamical Systems Part I: Diffusion Exit from a Domain Barbara Gentz University of Bielefeld, Germany Department of Mathematical Sciences, Seoul National University 17 March 2014 Barbara Gentz gentz@math.uni-bielefeld.de http://www.math.uni-bielefeld.de/ ˜ gentz

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling Introduction: A Brownian particle in a potential Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 1 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling Small random perturbations Gradient dynamics (ODE) x det = −∇ V ( x det ˙ ) t t Random perturbation by Gaussian white noise (SDE) √ z d x ε t ( ω ) = −∇ V ( x ε t ( ω )) d t + 2 ε d B t ( ω ) x Equivalent notation y √ x ε t ( ω ) = −∇ V ( x ε ˙ t ( ω )) + 2 ε ξ t ( ω ) with ⊲ V : R d → R : confining potential, growth condition at infinity ⊲ { B t ( ω ) } t ≥ 0 : d -dimensional Brownian motion ⊲ { ξ t ( ω ) } t ≥ 0 : Gaussian white noise, � ξ t � = 0, � ξ t ξ s � = δ ( t − s ) Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 2 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling Fokker–Planck equation Stochastic differential equation (SDE) of gradient type √ d x ε t ( ω ) = −∇ V ( x ε t ( ω )) d t + 2 ε d B t ( ω ) Kolmogorov’s forward or Fokker–Planck equation ⊲ Solution { x ε t ( ω ) } t is a (time-homogenous) Markov process ⊲ Transition probability densities ( x , t ) �→ p ( x , t | y , s ) satisfy � � ∂ ∂ t p = L ε p = ∇ · ∇ V ( x ) p + ε ∆ p ⊲ If { x ε t ( ω ) } t admits an invariant density p 0 , then L ε p 0 = 0 ⊲ Easy to verify (for gradient systems) � p 0 ( x ) = 1 R d e − V ( x ) /ε d x e − V ( x ) /ε with Z ε = Z ε Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 3 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling Equilibrium distribution ⊲ Invariant measure or equilibrium distribution µ ε ( dx ) = 1 e − V ( x ) /ε dx Z ε ⊲ System is reversible w.r.t. µ ε (detailed balance) p ( y , t | x , 0) e − V ( x ) /ε = p ( x , t | y , 0) e − V ( y ) /ε ⊲ For small ε , the invariant measure µ ε concentrates in the minima of V ε = 1 / 4 2.0 ε = 1 / 10 2.0 ε = 1 / 100 2.0 1.5 1.5 1.5 1.0 1.0 1.0 0.5 0.5 0.5 � 3 � 2 � 1 0 1 2 3 � 3 � 2 � 1 0 1 2 3 � 3 � 2 � 1 0 1 2 3 Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 4 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling Timescales Let V be a double-well potential as before, start in x ε 0 = x ⋆ − = left-hand well How long does it take until x ε t is well described by its invariant distribution? ⊲ For ε small, paths will stay in the left-hand well for a long time ⊲ x ε t first approaches a Gaussian distribution, centered in x ⋆ − , 1 1 T relax = − ) = ( d =1) V ′′ ( x ⋆ curvature at the bottom of the well ⊲ With overwhelming probability, paths will remain inside left-hand well, for all times significantly shorter than Kramers’ time T Kramers = e H /ε , where H = V ( z ⋆ ) − V ( x ⋆ − ) = barrier height ⊲ Only for t ≫ T Kramers , the distribution of x ε t approaches p 0 The dynamics is thus very different on the different timescales Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 5 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling Diffusion exit from a domain Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 6 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling The more general picture: Diffusion exit from a domain √ d x ε t = b ( x ε 2 ε g ( x ε x 0 ∈ R d t ) d t + t ) d W t , General case: b not necessarily derived from a potential Consider bounded domain D ∋ x 0 (with smooth boundary) ⊲ First-exit time: τ = τ ε D = inf { t > 0: x ε t �∈ D} ⊲ First-exit location: x ε τ ∈ ∂ D Questions ⊲ Does x ε t leave D ? ⊲ If so: When and where? ⊲ Expected time of first exit? ⊲ Concentration of first-exit time and location? ⊲ Distribution of τ and x ε τ ? Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 7 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling First case: Deterministic dynamics leaves D If x t leaves D in finite time, so will x ε t . Show that deviation x ε t − x t is small: Assume b Lipschitz continuous and g = Id (isotropic noise) � t √ � x ε � x ε t − x t � ≤ L s − x s � d s + 2 ε � W t � 0 By Gronwall’s lemma, for fixed realization of noise ω √ � x ε � W s � e Lt s − x s � ≤ sup 2 ε sup 0 ≤ s ≤ t 0 ≤ s ≤ t ⊲ d = 1: Use Andr´ e’s reflection principle � � � � � � ≤ 2 e − r 2 / 2 t sup | W s | ≥ r ≤ 2 P sup W s ≥ r ≤ 4 P W t ≥ r P 0 ≤ s ≤ t 0 ≤ s ≤ t ⊲ d > 1: Reduce to d = 1 using independence ⊲ General case: Use large-deviation principle Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 8 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling Second case: Deterministic dynamics does not leave D Assume D positively invariant under deterministic flow: Study noise-induced exit √ d x ε t = b ( x ε 2 ε g ( x ε x 0 ∈ R d t ) d t + t ) d W t , ⊲ b , g locally Lipschitz continuous, bounded-growth condition ⊲ a ( x ) = g ( x ) g ( x ) T ≥ 1 M Id (uniform ellipticity) � � 1 Infinitesimal generator A ε of diffusion x ε t : A ε v ( x ) = lim E x v ( x t ) − v ( x ) t t ց 0 � d ∂ 2 A ε v ( x ) = ε a ij ( x ) v ( x ) + � b ( x ) , ∇ v ( x ) � ∂ x i ∂ x j i , j =1 Compare to Fokker–Planck operator: L ε is the adjoint operator of A ε Approaches to the exit problem ⊲ Mean first-exit times and locations via PDEs ⊲ Exponential asymptotics via Wentzell–Freidlin theory Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 9 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling Diffusion exit from a domain: Relation to PDEs Theorem � ⊲ Poisson problem: A ε u = − 1 in D E x { τ ε D } is the unique solution of u = 0 on ∂ D � ⊲ Dirichlet problem: A ε w = 0 in D E x { f ( x ε D ) } is the unique solution of τ ε on ∂ D w = f (for f : ∂ D → R continuous) Remarks ⊲ Expected first-exit times and distribution of first-exit locations obtained exactly from PDEs Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 10 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling Diffusion exit from a domain: Relation to PDEs Theorem � ⊲ Poisson problem: A ε u = − 1 in D E x { τ ε D } is the unique solution of u = 0 on ∂ D � ⊲ Dirichlet problem: A ε w = 0 in D E x { f ( x ε D ) } is the unique solution of τ ε on ∂ D w = f (for f : ∂ D → R continuous) Remarks ⊲ Expected first-exit times and distribution of first-exit locations obtained exactly from PDEs ⊲ In principle . . . Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 10 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling Diffusion exit from a domain: Relation to PDEs Theorem � ⊲ Poisson problem: A ε u = − 1 in D E x { τ ε D } is the unique solution of u = 0 on ∂ D � ⊲ Dirichlet problem: A ε w = 0 in D E x { f ( x ε D ) } is the unique solution of τ ε on ∂ D w = f (for f : ∂ D → R continuous) Remarks ⊲ Expected first-exit times and distribution of first-exit locations obtained exactly from PDEs ⊲ In principle . . . ⊲ Smoothness assumption for ∂ D can be relaxed to “exterior-ball condition” Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 10 / 32

Brownian particle Diffusion exit Wentzell–Freidlin theory Kramers law and beyond Cycling An example in d = 1 Motion of a Brownian particle in a quadratic single-well potential √ d x ε t = b ( x ε t ) d t + 2 ε d W t where b ( x ) = −∇ V ( x ), V ( x ) = ax 2 / 2 with a > 0 ⊲ Drift pushes particle towards bottom at x = 0 ⊲ Probability of x ε t leaving D = ( α 1 , α 2 ) ∋ 0 through α 1 ? Solve the (one-dimensional) Dirichlet problem � � A ε w = 0 in D 1 for x = α 1 with f ( x ) = = f on ∂ D 0 for x = α 2 w � α 2 � � α 2 � � e V ( y ) /ε d y e V ( y ) /ε d y x ε = E x f ( x ε P x D = α 1 D ) = w ( x ) = τ ε τ ε x α 1 Diffusion Exit from a Domain Barbara Gentz SNU, 17 March 2014 11 / 32