# The Intersection Probability of Brownian Motion and SLEκ

Authors: Shizhong Zhou; Shiyi Lan
Journal: Advances in Mathematical Physics
 (2015)

---

## Abstract

By using excursion measure Poisson kernel method, we obtain a second-order differential equation of the intersection probability of Brownian motion andSLEκ. Moreover, we find a transformation such that the second-order differential equation transforms into a hypergeometric differential equation. Then, by solving the hypergeometric differential equation, we obtain the explicit formula of the intersection probability for the trace of the chordal SLEκ and planar Brownian motion started from distinct points in an upper half-plane H-.


---

## Body

## 1. Introduction

Shortly after Schramm [1] in 2000 introduced the percolation exploration path and SLE curve, Lawler et al. [2–5] proved in succession the intersection exponents of Brownian motion in the physics literature: half-plane exponents, plane exponents, and two-side exponents. That is, they proved that the intersection exponents of the two independent planar Brownian motions started from distinct points. Naturally, one can ask what the intersection exponents (or probability) of Brownian motion and SLEκ are in the half-plane. Not only is this question significant itself, but it also can probe more into the property of SLEκ. Kozdron [6, 7] derived the intersection probability of Brownian motion and SLE2. The aim of this paper is to derive the intersection probability of Brownian motion and SLEκ in a half-plane.Chordal
S
L
E
κ. The chordal Schramm-Loewner evolution SLEκ [8, 9] are growth process defined via conformal maps [10, 11] which are solutions of the ordinary differential equation as follows:(1)∂tgtz=2gtz-Wt,where g0(z)=z and Wt is a one-dimensional Brownian motion. For any fixed T>0, z0∈H-, and all z in a neighborhood of z0, either there exists a solution of (1) for t∈[0,T] or, for some t0∈(0,T], there exists a solution for t∈[0,T) such that limt↗t0gt(z)=Wt0. Let ΩT denote the (open) set of z∈H- such that the first holds and KT the set of z∈H- such that the second is true. We call the process (gt,t≥0) the chordal Schramm-Loewner evolution process with parameter κ (abbreviated as SLEκ). It is easy to see that the hulls of chordal SLEκ are the hulls of a continuous path:(2)γt≔limz→0⁡gt-1z+κWt,which is called the trace of the SLEκ process.Brownian Excursions Poisson Kernel. For a simply connected domain D⊂C and when ∂D is locally analytic at points x and y, the Brownian excursion Poisson kernel is defined as the normal derivative of the usual Poisson kernel. Since Brownian excursion measure is the scaling limit of simple random walk excursion measure, the excursion Poisson kernel is also the mass of the Brownian excursion measure. More precisely speaking, the excursion Poisson kernel is defined as (3)H∂Dx,y≔limε→0⁡HDx+εnx,yε,where nx is the unit normal at x pointing into the domain D and HD(z,y) is the usual Poisson kernel. Brownian excursion Poisson kernel satisfied the conformal covariance property as follows.Lemma 1 ([7, Proposition 2.11]).
Supposef:D→D~ is a conformal transformation where D~⊂C is a simply connected Jordan domain and ∂D~ is locally analytic at points f(x) and f(y); then(4)H∂Dx,y=f′xf′yH∂D~fx,fy.Particularly, when D=U (the unit disc), (5)H∂Ux,y=1πy-x2=12π1-cos⁡arg⁡y-arg⁡x.And when D=H (the upper half-plane),(6)H∂Ux,y=1πy-x2.


## 2. Intersection Probability

In this section, our main goal is now to compute the intersection probability ofSLEκ and Brownian motion started from distinct points in a half-plane H. In order to determinethe probability, our strategy for establishing this result will be as follows. By using the methods of the excursion measure Poisson kernel, we will first determine an explicit differential equation for P{γ[0,∞)∩β[0,tβ]=⌀}, which is a hypergeometric differential equation. Then, we obtain the general solution of the differential equation.


### 2.1. The Explicit Differential Equation of Intersection Probability

We note that ifWt is a standard one-dimensional Brownian motion, -Wt is also a standard one-dimensional Brownian motion. Hence, gt(z) satisfies the ordinary differential equation as follows:(7)∂tgtz=2/κgtz+Wt,g0z=0.Letting Xt=gt(x)+Wt and Yt=gt(y)+Wt, we have the following.Theorem 2.
The intersection probability for the trace of the chordalSLEκ and planar Brownian motion started from distinct points in a half-plane H satisfies(8)Pγ0,∞∩β0,tβ=⌀=Eexp⁡-∫0∞1Xs-1Ys2ds.Proof.
Suppose thatHt is the slit-plane Ht=H∖γ(0,t] for any 0<t<∞. This implies that(9)Pγ0,t∩β0,tβ=⌀=EH∂Htx,yH∂Hx,y.Letting t→∞, this yields (10)Pγ0,∞∩β0,tβ=⌀=limt→∞⁡EH∂Htx,yH∂Hx,y=Elimt→∞⁡H∂Htx,yH∂Hx,y.By using expression (7) and conformal covariance, we can obtain(11)H∂Htx,y=gt′xgt′yH∂Hgtx,gty,H∂Htx,yH∂Hx,y=gt′xgt′yH∂Hgtx,gtyH∂Hx,y=gt′xgt′yx-ygtx-gty2.We define as(12)Jt=gt′xgt′yx-ygtx-gty2,J∞=limt→∞⁡Jt.Hence, it yields(13)Pγ0,∞∩β0,tβ=⌀=limt→∞⁡EJt=Elimt→∞⁡Jt=EJ∞.Recalling Xt=gt(x)+Wt and Yt=gt(y)+Wt, we have(14)dXt=2κXtdt+dWt,dYt=2κYtdt+dWt.We also note that(15)∂∂tlog⁡gt′x=∂t∂xgtxgt′x=∂x∂tgtxgt′x=∂x2/κXtgt′x=-2κXt2,∂∂tlog⁡gt′y=-2κYt2.Then, we can obtain(16)∂∂tlog⁡gtx-log⁡gty=∂/∂tgtx-gtygtx-gty=∂/∂tgtx-∂/∂tgtygtx+Wt-gty+Wt=2/κXt-2/κYtXt-Yt=-2κXtYt.Because of(17)J0=g0′xg0′yx-yg0x-g0y2=1,we finally obtain(18)Jt=J0exp⁡∫0t∂∂slog⁡Jsds=exp⁡-∫0∞1Xs-1Ys2ds.Hence, we have(19)Pγ0,∞∩β0,tβ=⌀=Eexp⁡-∫0∞1Xs-1Ys2ds.This completes the proof of the theorem.


### 2.2. Intersection Probability

Theorem 3.
The intersection probability for the trace of the chordalSLEκ and planar Brownian motion started from distinct points in a half-plane H satisfies(20)Pγ0,∞∩β0,tβ=⌀=Γ4/κΓ4/κ+b^κΓ1+b^κΓ8/κ-1xy1/2-2/κ+b^κ/2F121-4κ,1-4κ+b^κ,1+b^κ;xy,0<κ<8,1,κ≥8,where b^κ=9κ2-8κ+16/κ.Proof.
In [8], the authors obtained the following deep theorem.
In the rangeκ∈(0,4], there exists a random simple curve γ:[0,∞)→H- with γ(0)=0 and γ(0,∞)⊂H. In the range κ∈(4,8), there exists a random simple curve γ:[0,∞)→H-. These curves have double points and they hit R, but they never cross themselves. In the range κ∈[8,∞), there exists a random simple curve γ:[0,∞)→H-. These curves are the space-filling curves, but they never cross themselves.
Whenκ∈[8,∞), these random curves are the space-filling curves. It is easy to obtain that (21)Pγ0,∞∩β0,tβ=⌀=1.In order to prove Theorem 2, therefore, it is sufficient that we only need to prove the theorem when κ∈(0,8).
Recalling(22)H~∗x,y=Pγ0,∞∩β0,tβ=⌀=Eexp⁡-∫0∞1Xs-1Ys2ds=Jt,Jt=gt′xgt′ygty-gtx2,by using Markov property, it is easy to see that JtH~∗(Xt,Yt) is a martingale. That is, Mt is a martingale if we define Mt≔E[J∞∣Ft]. Hence, we have(23)Mt=EJ∞∣Ft=Eexp⁡-∫0∞1Xs-1Ys2ds∣Ft=exp⁡-∫0∞1Xs-1Ys2ds·Eexp⁡-∫t∞1Xs-1Ys2ds∣Ft=JtH~∗Xt,Yt.
By using Itô formula [12] at t=0, this implies that(24)∂2H~∗∂x∂y+2κx∂2H~∗∂x2+2κy∂2H~∗∂y2+1x∂H~∗∂x+1y∂H~∗∂y-1x-1y2H~∗=0.Because the intersection probability in our question only depends on the ratio x/y, we define the transformation H~∗(x,y)=fx/y, where f is some second-order derivative function. Thus, we have(25)12y2+x2y4-xy3f′′xy+2κxy-2κxy3+xy3-1y2f′xy-1x-1y2fxy=0.Multiplying by 2y2 and letting u=x/y, this yields(26)1+u2-2uf′′u+22κu-2uκ+u-1f′u-2u-12fu=0.Multiplying by u2 and combining terms, we obtain(27)u21-u2f′′u+2u1-u2κ+2κ-1uf′u-21-u2fu=0.Or equivalently (because 0<u<1)(28)u21-uf′′u+2u2κ+2κ-1uf′u-21-ufu=0.It is easy to know that the second-order differential equation (28) has regular singular points at 0, 1, and ∞. Hence, it is possible to transform it into a hypergeometric differential equation [13] by writing (28) as(29)f′′u-21-4/κ1-u-2/κuf′u-2u2fu=0.
In order to solve (29), our main goal is to find a transformation such that it is a hypergeometric differential equation. We find that the transformation as follows just is our expectation:(30)fu=uαϕu,where α=1/2-2/κ+9κ2-8κ+16/2κ.
By (30), we have(31)f′u=αuα-1ϕu+uαϕ′u,f′′u=αα-1uα-2ϕu+2αuα-1ϕ′u+uαϕ′′u.By substituting (30) and (31) into (29), we finally obtain(32)
u
1
-
u
ϕ
′
′
u
+
1
+
9
κ
2
-
8
κ
+
16
κ
-
3
-
8
κ
+
9
κ
2
-
8
κ
+
16
κ
u
ϕ
′
u
+
4
κ
-
1
1
-
4
κ
+
9
κ
2
-
8
κ
+
16
κ
ϕ
u
=
0
.It is easy to see that (32) is now a well-known hypergeometric differential equation whose general solution is given by(33)
ϕ
u
=
C
1
F
1
2
1
-
4
κ
,
1
-
4
κ
+
9
κ
2
-
8
κ
+
16
κ
,
1
+
9
κ
2
-
8
κ
+
16
κ
;
u
+
C
2
u
-
9
κ
2
-
8
κ
+
16
/
κ
F
1
2
1
-
4
κ
,
1
-
4
κ
-
9
κ
2
-
8
κ
+
16
κ
,
1
-
9
κ
2
-
8
κ
+
16
κ
;
u
.Hence, we finally obtain that the general solution of (28) is(34)
f
u
=
C
1
u
α
F
1
2
1
-
4
κ
,
1
-
4
κ
+
9
κ
2
-
8
κ
+
16
κ
,
1
+
9
κ
2
-
8
κ
+
16
κ
;
u
+
C
2
u
α
-
9
κ
2
-
8
κ
+
16
/
κ
F
1
2
1
-
4
κ
,
1
-
4
κ
-
9
κ
2
-
8
κ
+
16
κ
,
1
-
9
κ
2
-
8
κ
+
16
κ
;
u
.
Furthermore, physical considerations imply thatf(u)→0 as u→0+ (or y→∞) and f(u)→1 as u→1- (or y→x). This yields (35)C2=0,C1=Γ4/κΓ4+9κ2-8κ+16/κΓ1+9κ2-8κ+16/κΓ8/κ-1.Thus, (36)
f
u
=
Γ
4
/
κ
Γ
4
+
9
κ
2
-
8
κ
+
16
/
κ
Γ
1
+
9
κ
2
-
8
κ
+
16
/
κ
Γ
8
/
κ
-
1
·
u
α
F
1
2
1
-
4
κ
,
1
-
4
κ
+
9
κ
2
-
8
κ
+
16
κ
,
1
+
9
κ
2
-
8
κ
+
16
κ
;
u
.This completes the proof of Theorem 3.By using the knowledge of the conformal transformation, For any bounded, simply connected planar domain, we have a similar theorem.Theorem 4.
LetD⊂C be a bounded, simply connected domain and let ω1, ω2, ω3, and ω4 be four points ordered counterclockwise around ∂D. The intersection probability for the trance of the chordal SLEκ from ω1 to ω4 in D and planar Brownian motion started from ω2 to ω3 in D satisfies(37)Pγ0,∞∩β0,tβ=⌀=Γ4/κΓ4/κ+b^κΓ1+b^κΓ8/κ-1ϕω21/2-2/κ+b^κ/2F121-4κ,1-4κ+b^κ,1+b^κ;ϕω2,0<κ<8,1,κ≥8,where b^κ=9κ2-8κ+16/κ and ϕ:D→H is the conformal transformation with ϕ(ω1)=0, ϕ(ω3)=1, and ϕ(ω4)=∞.


## 2.1. The Explicit Differential Equation of Intersection Probability

We note that ifWt is a standard one-dimensional Brownian motion, -Wt is also a standard one-dimensional Brownian motion. Hence, gt(z) satisfies the ordinary differential equation as follows:(7)∂tgtz=2/κgtz+Wt,g0z=0.Letting Xt=gt(x)+Wt and Yt=gt(y)+Wt, we have the following.Theorem 2.
The intersection probability for the trace of the chordalSLEκ and planar Brownian motion started from distinct points in a half-plane H satisfies(8)Pγ0,∞∩β0,tβ=⌀=Eexp⁡-∫0∞1Xs-1Ys2ds.Proof.
Suppose thatHt is the slit-plane Ht=H∖γ(0,t] for any 0<t<∞. This implies that(9)Pγ0,t∩β0,tβ=⌀=EH∂Htx,yH∂Hx,y.Letting t→∞, this yields (10)Pγ0,∞∩β0,tβ=⌀=limt→∞⁡EH∂Htx,yH∂Hx,y=Elimt→∞⁡H∂Htx,yH∂Hx,y.By using expression (7) and conformal covariance, we can obtain(11)H∂Htx,y=gt′xgt′yH∂Hgtx,gty,H∂Htx,yH∂Hx,y=gt′xgt′yH∂Hgtx,gtyH∂Hx,y=gt′xgt′yx-ygtx-gty2.We define as(12)Jt=gt′xgt′yx-ygtx-gty2,J∞=limt→∞⁡Jt.Hence, it yields(13)Pγ0,∞∩β0,tβ=⌀=limt→∞⁡EJt=Elimt→∞⁡Jt=EJ∞.Recalling Xt=gt(x)+Wt and Yt=gt(y)+Wt, we have(14)dXt=2κXtdt+dWt,dYt=2κYtdt+dWt.We also note that(15)∂∂tlog⁡gt′x=∂t∂xgtxgt′x=∂x∂tgtxgt′x=∂x2/κXtgt′x=-2κXt2,∂∂tlog⁡gt′y=-2κYt2.Then, we can obtain(16)∂∂tlog⁡gtx-log⁡gty=∂/∂tgtx-gtygtx-gty=∂/∂tgtx-∂/∂tgtygtx+Wt-gty+Wt=2/κXt-2/κYtXt-Yt=-2κXtYt.Because of(17)J0=g0′xg0′yx-yg0x-g0y2=1,we finally obtain(18)Jt=J0exp⁡∫0t∂∂slog⁡Jsds=exp⁡-∫0∞1Xs-1Ys2ds.Hence, we have(19)Pγ0,∞∩β0,tβ=⌀=Eexp⁡-∫0∞1Xs-1Ys2ds.This completes the proof of the theorem.


## 2.2. Intersection Probability

Theorem 3.
The intersection probability for the trace of the chordalSLEκ and planar Brownian motion started from distinct points in a half-plane H satisfies(20)Pγ0,∞∩β0,tβ=⌀=Γ4/κΓ4/κ+b^κΓ1+b^κΓ8/κ-1xy1/2-2/κ+b^κ/2F121-4κ,1-4κ+b^κ,1+b^κ;xy,0<κ<8,1,κ≥8,where b^κ=9κ2-8κ+16/κ.Proof.
In [8], the authors obtained the following deep theorem.
In the rangeκ∈(0,4], there exists a random simple curve γ:[0,∞)→H- with γ(0)=0 and γ(0,∞)⊂H. In the range κ∈(4,8), there exists a random simple curve γ:[0,∞)→H-. These curves have double points and they hit R, but they never cross themselves. In the range κ∈[8,∞), there exists a random simple curve γ:[0,∞)→H-. These curves are the space-filling curves, but they never cross themselves.
Whenκ∈[8,∞), these random curves are the space-filling curves. It is easy to obtain that (21)Pγ0,∞∩β0,tβ=⌀=1.In order to prove Theorem 2, therefore, it is sufficient that we only need to prove the theorem when κ∈(0,8).
Recalling(22)H~∗x,y=Pγ0,∞∩β0,tβ=⌀=Eexp⁡-∫0∞1Xs-1Ys2ds=Jt,Jt=gt′xgt′ygty-gtx2,by using Markov property, it is easy to see that JtH~∗(Xt,Yt) is a martingale. That is, Mt is a martingale if we define Mt≔E[J∞∣Ft]. Hence, we have(23)Mt=EJ∞∣Ft=Eexp⁡-∫0∞1Xs-1Ys2ds∣Ft=exp⁡-∫0∞1Xs-1Ys2ds·Eexp⁡-∫t∞1Xs-1Ys2ds∣Ft=JtH~∗Xt,Yt.
By using Itô formula [12] at t=0, this implies that(24)∂2H~∗∂x∂y+2κx∂2H~∗∂x2+2κy∂2H~∗∂y2+1x∂H~∗∂x+1y∂H~∗∂y-1x-1y2H~∗=0.Because the intersection probability in our question only depends on the ratio x/y, we define the transformation H~∗(x,y)=fx/y, where f is some second-order derivative function. Thus, we have(25)12y2+x2y4-xy3f′′xy+2κxy-2κxy3+xy3-1y2f′xy-1x-1y2fxy=0.Multiplying by 2y2 and letting u=x/y, this yields(26)1+u2-2uf′′u+22κu-2uκ+u-1f′u-2u-12fu=0.Multiplying by u2 and combining terms, we obtain(27)u21-u2f′′u+2u1-u2κ+2κ-1uf′u-21-u2fu=0.Or equivalently (because 0<u<1)(28)u21-uf′′u+2u2κ+2κ-1uf′u-21-ufu=0.It is easy to know that the second-order differential equation (28) has regular singular points at 0, 1, and ∞. Hence, it is possible to transform it into a hypergeometric differential equation [13] by writing (28) as(29)f′′u-21-4/κ1-u-2/κuf′u-2u2fu=0.
In order to solve (29), our main goal is to find a transformation such that it is a hypergeometric differential equation. We find that the transformation as follows just is our expectation:(30)fu=uαϕu,where α=1/2-2/κ+9κ2-8κ+16/2κ.
By (30), we have(31)f′u=αuα-1ϕu+uαϕ′u,f′′u=αα-1uα-2ϕu+2αuα-1ϕ′u+uαϕ′′u.By substituting (30) and (31) into (29), we finally obtain(32)
u
1
-
u
ϕ
′
′
u
+
1
+
9
κ
2
-
8
κ
+
16
κ
-
3
-
8
κ
+
9
κ
2
-
8
κ
+
16
κ
u
ϕ
′
u
+
4
κ
-
1
1
-
4
κ
+
9
κ
2
-
8
κ
+
16
κ
ϕ
u
=
0
.It is easy to see that (32) is now a well-known hypergeometric differential equation whose general solution is given by(33)
ϕ
u
=
C
1
F
1
2
1
-
4
κ
,
1
-
4
κ
+
9
κ
2
-
8
κ
+
16
κ
,
1
+
9
κ
2
-
8
κ
+
16
κ
;
u
+
C
2
u
-
9
κ
2
-
8
κ
+
16
/
κ
F
1
2
1
-
4
κ
,
1
-
4
κ
-
9
κ
2
-
8
κ
+
16
κ
,
1
-
9
κ
2
-
8
κ
+
16
κ
;
u
.Hence, we finally obtain that the general solution of (28) is(34)
f
u
=
C
1
u
α
F
1
2
1
-
4
κ
,
1
-
4
κ
+
9
κ
2
-
8
κ
+
16
κ
,
1
+
9
κ
2
-
8
κ
+
16
κ
;
u
+
C
2
u
α
-
9
κ
2
-
8
κ
+
16
/
κ
F
1
2
1
-
4
κ
,
1
-
4
κ
-
9
κ
2
-
8
κ
+
16
κ
,
1
-
9
κ
2
-
8
κ
+
16
κ
;
u
.
Furthermore, physical considerations imply thatf(u)→0 as u→0+ (or y→∞) and f(u)→1 as u→1- (or y→x). This yields (35)C2=0,C1=Γ4/κΓ4+9κ2-8κ+16/κΓ1+9κ2-8κ+16/κΓ8/κ-1.Thus, (36)
f
u
=
Γ
4
/
κ
Γ
4
+
9
κ
2
-
8
κ
+
16
/
κ
Γ
1
+
9
κ
2
-
8
κ
+
16
/
κ
Γ
8
/
κ
-
1
·
u
α
F
1
2
1
-
4
κ
,
1
-
4
κ
+
9
κ
2
-
8
κ
+
16
κ
,
1
+
9
κ
2
-
8
κ
+
16
κ
;
u
.This completes the proof of Theorem 3.By using the knowledge of the conformal transformation, For any bounded, simply connected planar domain, we have a similar theorem.Theorem 4.
LetD⊂C be a bounded, simply connected domain and let ω1, ω2, ω3, and ω4 be four points ordered counterclockwise around ∂D. The intersection probability for the trance of the chordal SLEκ from ω1 to ω4 in D and planar Brownian motion started from ω2 to ω3 in D satisfies(37)Pγ0,∞∩β0,tβ=⌀=Γ4/κΓ4/κ+b^κΓ1+b^κΓ8/κ-1ϕω21/2-2/κ+b^κ/2F121-4κ,1-4κ+b^κ,1+b^κ;ϕω2,0<κ<8,1,κ≥8,where b^κ=9κ2-8κ+16/κ and ϕ:D→H is the conformal transformation with ϕ(ω1)=0, ϕ(ω3)=1, and ϕ(ω4)=∞.



---

*Source: 627423-2015-08-26.xml*