Excursion Reflected Brownian Motion and a Loewner Equation

Transcription

1 and a Loewner Equation Department of Mathematics University of Chicago Cornell Probability Summer School, 2011 and a Loewner Equation

2 The Chordal Loewner Equation Let γ : [0, ) C be a simple curve with γ(0) R and γ((0, )) H. For each t > 0 there is a unique map g t : H\γ((0, t)) H such that lim z g t (z) z = 0. It is a classical result of Loewner that g t (z) is a solution of the initial value problem ġ t (z) = ḃ(t) g t (z) U t, g 0 (z) = z, (1) where U t = g t (γ(t)) and b(t) = hcap(γ((0, t))). and a Loewner Equation

3 g t U t and a Loewner Equation

4 The Complex Poisson Kernel We define a function H H (z, x) = 1 z x on H R. It is easy to check that Im(H H (z, x)) = πh H (z, x) where H H is the Poisson kernel for Brownian motion. For this reason we call H H (z, x) the complex Poisson kernel for Brownian motion. Using this notation, (1) becomes ġ t (z) = ḃ(t)h H(g t (z), U t ). (2) and a Loewner Equation

5 The Problem A natural question to ask is whether an analog of (2) holds when we replace H with a finitely-connected domain. That is, if D H is a finitely-connected domain, can we find a strong Markov process in D and associated Poisson kernel so an analog of (2) holds? The answer is yes and leads us to consider a process called excursion reflected Brownian motion (ERBM). and a Loewner Equation

6 ERBM in C\D I We define ERBM in D = C\D to be a strong Markov process BD ER with state space D {D} satisfying the following properties: If we start the process at z D and let T = inf{t : BD ER (t) D}, then for 0 t T, BD ER (t) is a Brownian motion in D killed at D. Let C r be the circle of radius r > 1 centered at the origin and let τ r = inf{t : B ER D (t) C r} be the first time ERBM started at D hits C r. Then B ER D (τ r) has the uniform distribution. and a Loewner Equation

7 ERBM in C\D II The radial part of BC\D ER has the same distribution as the radial part of reflected Brownian motion in D. and a Loewner Equation

8 and a Loewner Equation

9 Existence of ERBM We can construct ERBM in C\D by explicitly defining a Feller-Dynkin semigroup and checking that the resulting process satisfies the necessary conditions. Once we have ERBM in C\D we can construct it in any domain with one hole (i.e. of the form C\A for a compact, connected set A) via conformal invariance. By piecing together countably many of these processes (this is deliberately vague!), we can construct ERBM in a domain with n holes. and a Loewner Equation

10 Poisson kernel for ERBM Let D = C\[A 0 A 1... A n ] where A 0, A 1,...A n are closed disjoint subsets of C with A i larger than a single point for i 1, the connected components of A 0 are simply connected, and A 1,..., A n are simply-connected and bounded. Let BD ER be ERBM in D excursion-reflected at A 1,..., A n and killed at A 0. If A 0 is piecewise analytic, then we can define a Poisson kernel (z, x) for ERBM in D. H ER D and a Loewner Equation

11 and a Loewner Equation

12 Conformal Mapping Using H ER D (, x) We call a domain a standard chordal domain if it is the upper-half plane with finitely many horizontal line segments removed. It is a classical result of complex analysis that any n-connected domain in C is conformally equivalent to a standard chordal domain and that this equivalence is unique up to a scaling and real translation. In fact, if f is such a map, then the imaginary part of f must be real multiple of HD ER (, x) for some x A 0. Using this insight, we can construct a new proof of this classical theorem. and a Loewner Equation

13 and a Loewner Equation

14 Complex Poisson Kernel for ERBM If D be a domain as above with A 0 = C\H, then there is a unique map HD ER : D R C satisfying the following: For each x R, z HD ER (z, x) is a conformal map onto a standard chordal domain. The imaginary part of HD ER(z, x) is πher D (z, x). (z, x) has certain asymtotics as z or z x. H ER D and a Loewner Equation

15 Setup Let D be a standard chordal domain and γ : [0, ) C be a simple curve with γ(0) = 0 and γ((0, )) D. Let D t = D\γ((0, t)). There is a unique conformal map h t : D t h(d t ) onto a standard chordal domain satisfying lim h t(z) z = 0. z We define Ũt to be h t (γ t ) (it is necessary to check that this definition makes sense). and a Loewner Equation

16 h t ~ U t and a Loewner Equation

17 Excursion Reflected Half Plane Capacity The map h t (z) has an expansion at infinity of the form h t (z) = z + b(t) z + O ( z 2). The quantity b(t) is called the ER-half plane capacity for γ((0, t)). and a Loewner Equation

18 The Main Result Theorem Let γ be as above and suppose that b(t) is C 1. Then for z D t, h t (z) satisfies ḣ t (z) = ḃ(t)her h t (D t ) (h t(z), Ũt) and a Loewner Equation

19 Outline of the Proof There are two main steps to proving the theorem. The first step is to show ht (z) z + b(t)hd ER(z, 0) c b(t) rad(γ((0,t))) z. The second step is to show the map t Ũt is a well-defined continuous function. and a Loewner Equation