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Maximizing the pp-th moment of exit time of planar Brownian motion from a given domain

Maher Boudabra    Greg Markowsky
Abstract

In this paper we address the question of finding the point which maximizes the pp-th moment of the exit time of planar Brownian motion from a given domain. We present a geometrical method of excluding parts of the domain from consideration which makes use of a coupling argument and the conformal invariance of Brownian motion. In many cases the maximizing point can be localized to a relatively small region. Several illustrative examples are presented.

2010 Mathematics subject classification: 60J65.

Keywords: Planar Brownian motion, exit time.

1 Introduction

Let Zt:=Xt+iYtZ_{t}:=X_{t}+iY_{t} be a planar Brownian motion starting at a point aa in a domain UU. We will let τU=τU(a)\tau_{U}=\tau_{U}(a) be the first time that ZtZ_{t} exits UU, and we will use the standard notation EaE_{a} to denote expectation conditioned on Z0=aZ_{0}=a a.s. The focus of this paper is the following optimization problem.

For a given domain in the plane and 0<p<0<p<\infty, find the point aa which maximizes the quantity Ea[(τU)p]E_{a}[(\tau_{U})^{p}].

We will refer to such a point as a pp-th center of UU; it is not in general unique, as the easy example of an infinite strip shows. For many domains, even simple ones such as an isosceles triangle, it is difficult to find any of the pp-th centers, however we will show how elementary coupling arguments and the conformal invariance of Brownian motion in many cases allows us to locate a small region in UU which must contain all pp-th centers. In certain cases in which the domain in question has a high degree of symmetry, it will allow us to locate all pp-th centers.

Before describing our methods, we present a brief overview of some earlier works related to this problem. The case p=1p=1 is commonly referred to as the "torsion problem" due to its connection with mechanics, and is naturally the most tractable. The function h(a)=Ea[τU]h(a)=E_{a}[\tau_{U}] satisfies Δh=2\Delta h=-2, and therefore p.d.e. techniques can be employed to great effect. [1, Ch. 6] contains a good account of this problem and methods of attacking it in special cases, such as when the domain in question is convex. Further results along the same lines, focussing in particular on convex domains, can be found in [2, 3, 4].

Other interesting related problems have been tackled by p.d.e. methods. For example, in the famous paper [5] (see also the related work [6]) eigenvalue techniques are used to demonstrate relationships between Ea[τU]E_{a}[\tau_{U}] and geometric qualities of the domain, such as the size of the hyperbolic density and the inradius (the radius of the largest disk contained in the domain). The methods developed there have been extended by other authors in a number of different directions. For example, in [7] a number of related stochastic domination results were proved concerning convex domains in n{\mathbb{R}}^{n} and various types of symmetrizations. These results allow conclusions to be reached concerning the comparison of pp-th moments of the exit times from these domains. One striking consequence of the eigenvalue methods is the fact that over all domains with a given area the disk maximizes the pp-th moment of the exit time of Brownian motion for all pp. The recent work [8] contains a discussion and refinements of this result.

Our results differ from those described above in the following ways. We have not employed p.d.e. methods at all, choosing instead to work with an elementary coupling method. Perhaps as a consequence of this, convexity plays little role in our discussion, although a weaker concept called Δ\Delta-convexity (defined below) will be important. The type of coupling we will use is not entirely new, and has found a number of uses in related topics, for instance in investigations into the "hot spots" conjecture such as [9, 10, 11]. However, we believe that it has not been applied directly in the manner that we do before. Furthermore, we restrict our attention to two dimensions, which allows conformal mappings to take prominence and to extend the standard notion of coupling. We present several methods for localizing the pp-centers of a domain, and then consider a number of specific domains, showing in each case how our methods can be used to localize the pp-th centers of the domain. In what follows we assume pp is a fixed positive number; however, in order to reduce the qualifications needed to state our results, for any planar domain UU for which we are interested in maximizing the pp-th moment we will assume that Ea[(τU)p]<E_{a}[(\tau_{U})^{p}]<\infty for all points aUa\in U; this would follow if Ea[(τU)p]<E_{a}[(\tau_{U})^{p}]<\infty for any aUa\in U, as is shown in [12].

2 Partial symmetry and convexity with respect to a line.

Definition 1.

Let UU be a domain of \mathbb{C}. We say that a line Δ:ax+by+c=0\Delta:ax+by+c=0 is a partial symmetry axis for UU if one of the two sets U+:=U{ax+by+c>0}U^{+}:=U\cap\left\{ax+by+c>0\right\} or U:=U{ax+by+c<0}U^{-}:=U\cap\left\{ax+by+c<0\right\} can be folded over Δ\Delta and fits into UU, more precisely if σΔ(U+)\sigma_{\Delta}(U^{+}) or σΔ(U)\sigma_{\Delta}(U^{-}) remains inside UU, where σΔ\sigma_{\Delta} denotes the symmetry over Δ\Delta. The subset among U±U^{\pm} that satisfies this property (that is, the smaller side with respect to the symmetry) is called symmetric side of UU over Δ\Delta. So, for instance, any line intersecting 𝔻={|z|<1}{\mathbb{D}}=\{|z|<1\} is a partial symmetry axis for 𝔻{\mathbb{D}}, but the line y=2xy=2x is not one for the square {|x|<1,|y|<1}\{|x|<1,|y|<1\} since the reflection over this line of the point (1,1)(1,1) is the point (15,75)(\frac{1}{5},\frac{7}{5}), which is not in the closure of the square. Notice that both of U±U^{\pm} are symmetric sides if and only if Δ\Delta is a symmetry axis for UU.

Theorem 2.

Let SS be the symmetric side of UU over a partial symmetry axis Δ\Delta. Then, for any aSa\in S we can find Brownian motions ZtZ_{t} starting at aa and Z~t\tilde{Z}_{t} starting at σΔ(a)\sigma_{\Delta}(a) defined on the same probability space such that τUτ~U\tau_{U}\leq\tilde{\tau}_{U} a.s. (where τ~U\tilde{\tau}_{U} is the exit time from UU of Z~\tilde{Z}). Furthermore, if σΔ(S)\sigma_{\Delta}(S) is strictly contained in U(SΔ)U\setminus(S\cup\Delta) then P(τU<τ~U)>0P(\tau_{U}<\tilde{\tau}_{U})>0. In particular, Ea[τUp]EσΔ(a)[τ~Up]E_{a}[\tau_{U}^{p}]\leq E_{\sigma_{\Delta}(a)}[\tilde{\tau}_{U}^{p}] (with strict inequality if σΔ(S)\sigma_{\Delta}(S) is strictly contained in U(SΔ)U\setminus(S\cup\Delta)).

Proof.

This follows from a coupling argument. Let ZtZ_{t} start at aSa\in S, and let HΔH_{\Delta} be its hitting time of the line Δ\Delta. Form the process Z~t\tilde{Z}_{t} by the rule

Z~t={σΔ(Zt)if t<HΔZtif tHΔ.\tilde{Z}_{t}=\left\{\begin{array}[]{ll}\sigma_{\Delta}(Z_{t})&\qquad\mbox{if }t<H_{\Delta}\\ Z_{t}&\qquad\mbox{if }t\geq H_{\Delta}\;.\end{array}\right.

It follows from the Strong Markov property and the reflection invariance of Brownian motion that Z~t\tilde{Z}_{t} is a Brownian motion. Clearly τU=τ~U\tau_{U}=\tilde{\tau}_{U} on the set {τUHΔ}\{\tau_{U}\geq H_{\Delta}\}, and our conditions on SS imply τUτ~U\tau_{U}\leq\tilde{\tau}_{U} on the set {τU<HΔ}\{\tau_{U}<H_{\Delta}\}. Furthermore, if σΔ(S)\sigma_{\Delta}(S) is strictly contained in U(SΔ)U\setminus(S\cup\Delta), then ZtZ_{t} has some positive probability of leaving UU before Z~t\tilde{Z}_{t} does; this is implied for instance by [13, Thm. I.6.6 ]. The result follows. ∎

This theorem allows us in essence to exclude the symmetric side of any partial symmetry axis for UU in our search for pp-th centers. The only exception to this rule is when σΔ(S)=U(SΔ)\sigma_{\Delta}(S)=U\setminus(S\cup\Delta), i.e. when Δ\Delta is a symmetry axis of UU; however in most cases a symmetry axis will contain all pp-th centers. To see why this is so, we need another definition.

Definition 3.

Let Δ\Delta be a symmetry axis of UU. We say that UU is Δ\Delta-convex if

(z,t)U×[0,1]tz+(1t)σΔ(z)U.\forall\left(z,t\right)\in U\times[0,1]\,\,\,tz+(1-t)\sigma_{\Delta}(z)\in U.

In other words, for every zUz\in U, the segment joining zz and σΔ(z)\sigma_{\Delta}(z) remains inside UU.

It is clear that any convex UU is Δ\Delta-convex for any symmetry axis Δ\Delta, and a less trivial example can be given by {f(x)<y<f(x)}\{-f(x)<y<f(x)\}, where ff is a positive continuous function on the real line, which is Δ\Delta-convex with Δ=\Delta={\mathbb{R}}. A domain which is not Δ\Delta-convex with respect to a symmetry axis can be given by Wϵ={ϵ<y<ϵ,|x|<1}{|z1|<12}{|z+1|<12}W_{\epsilon}=\{-\epsilon<y<\epsilon,|x|<1\}\cup\{|z-1|<\frac{1}{2}\}\cup\{|z+1|<\frac{1}{2}\} with ϵ<12\epsilon<\frac{1}{2}; this has the real and imaginary axes as symmetry axes but is not Δ\Delta-convex with respect to the imaginary axis (though it is with respect to the real line). As will be seen below, this domain also shows why Δ\Delta-convexity is required in the following proposition.

Proposition 4.

Suppose UU is Δ\Delta-convex with respect to a symmetry axis Δ\Delta. Then all pp-th centers of UU lie on Δ\Delta.

Proof.

Let aUΔa\in U\setminus\Delta, and let a^=12a+12σΔ(a)\hat{a}=\frac{1}{2}a+\frac{1}{2}\sigma_{\Delta}(a) be the orthogonal projection of aa onto Δ\Delta. Let LL be the line parallel to Δ\Delta which passes through the point 12a+12a^\frac{1}{2}a+\frac{1}{2}\hat{a}; speaking informally, this is the line halfway between aa and Δ\Delta. Δ\Delta-convexity implies that LL is a partial symmetry axis of UU, and if SS is the component of ULU\setminus L containing aa then σL(S)\sigma_{L}(S) is strictly contained in U(SL)U\setminus(S\cup L). It therefore follows from Theorem 2 that Ea[τUp]<Ea^[τUp]E_{a}[\tau_{U}^{p}]<E_{\hat{a}}[\tau_{U}^{p}]. The result follows. ∎

Note that this proposition completely solves our problem in the case that our domain is Δ\Delta-convex with respect to two or more non-parallel symmetry axes, since all pp-th centers must lie at their point of intersection, and we have also incidentally proved the purely geometrical fact that all such symmetry axes must coincide at a unique point; more on this in the final section. Thus, for instance, all pp-centers of any regular polygon, a circle, an ellipse, a rhombus, and any number of other easily constructed examples must lie at their natural centers. To see an example of a domain with intersecting symmetry axes but where the point of intersection is not a pp-th center, let us return to the domains WϵW_{\epsilon} described immediately before this proposition. Proposition 4 implies that all pp-th centers lie on the real line, but it is easy to see that if we make ϵ\epsilon sufficiently small then 0, the intersection point of the two symmetry axes, will not be a pp-th center (clearly τWϵ(1)τ{|z1|<1/2}(1)\tau_{W_{\epsilon}}(1)\geq\tau_{\{|z-1|<1/2\}}(1), so that E1[τWϵp]E_{1}[\tau_{W_{\epsilon}}^{p}] remains always greater than a positive constant, but τWϵ(0)\tau_{W_{\epsilon}}(0) decreases monotonically to 0 a.s. as ϵ0\epsilon\searrow 0, so that E1[τWϵp]0E_{1}[\tau_{W_{\epsilon}}^{p}]\searrow 0).

Let us now look at an example that shows the use of the results proved to this point, but also their limitations. Suppose UU is the isosceles right triangle with vertices at 1,1,-1,1, and ii. The imaginary axis is an axis of symmetry, and UU is Δ\Delta-convex with respect to this axis, so all pp-th centers must lie on the imaginary axis. The line {y=12}\{y=\frac{1}{2}\} is a partial symmetry axis for UU, with U{y>12}U\cap\{y>\frac{1}{2}\} the symmetric side, so all pp-centers must lie on {x=0,y12}\{x=0,y\leq\frac{1}{2}\}. Now, common sense tells us that the pp-th centers can’t be too close to the real axis as well, because this is a boundary component, but there is no good partial symmetry axis to apply to conclude that rigorously. The way out of this difficulty is to extend our method of reflection to curves more general than straight lines. For this, we will need to utilize the conformal invariance of Brownian motion, via the following famous theorem of Lévy (see [13] or [14]).

Theorem 5.

If ff is a holomorphic function then f(Zt)f(Z_{t}) is a time changed Brownian motion. More precisely, f(Zκ1(t))f(Z_{\kappa^{-1}(t)}) is a Brownian motion where

κ(t):=0t|f(Zs)|2𝑑s\kappa(t):=\int_{0}^{t}|f^{\prime}(Z_{s})|^{2}ds

for t0t\geq 0.

This allows us to extend Theorem 2, as follows.

Proposition 6.

Suppose UU is a domain with an axis of symmetry Δ\Delta, and suppose ff is a conformal map defined on UU with the property that |f(z)||f(σΔ(z))||f^{\prime}(z)|\geq|f^{\prime}(\sigma_{\Delta}(z))| for all zAz\in A, where AA is one component of U\ΔU\backslash\Delta and σΔ\sigma_{\Delta} is the symmetry over Δ\Delta. Then, for any aAa\in A, we can find Brownian motions ZtZ_{t} starting at f(a)f(a) and Z~t\tilde{Z}_{t} starting at f(σΔ(a))f(\sigma_{\Delta}(a)) defined on the same probability space such that τf(U)(f(a))τ~f(U)(f(σΔ(a)))\tau_{f(U)}(f(a))\geq\tilde{\tau}_{f(U)}(f(\sigma_{\Delta}(a))) a.s. (where τ~U\tilde{\tau}_{U} is the exit time from UU of Z~\tilde{Z}). In particular, Ef(a)[τf(U)p]Ef(σΔ(a))[τf(U)p]E_{f(a)}[\tau_{f(U)}^{p}]\geq E_{f(\sigma_{\Delta}(a))}[\tau_{f(U)}^{p}]. If there is any point in AA at which |f(z)|>|f(σΔ(z))||f^{\prime}(z)|>|f^{\prime}(\sigma_{\Delta}(z))| then Ef(a)[τf(U)p]>Ef(σΔ(a))[τf(U)p]E_{f(a)}[\tau_{f(U)}^{p}]>E_{f(\sigma_{\Delta}(a))}[\tau_{f(U)}^{p}] (for this statement we recall the assumption that Ew[τf(U)p]<E_{w}[\tau_{f(U)}^{p}]<\infty for any wf(U)w\in f(U)).

Proof.

Let ZtZ_{t} be a Brownian motion starting at aa, and let Z~t\tilde{Z}_{t} be defined as in Theorem 2. According to Theorem 5, the processes f(Zt)f(Z_{t}) and f(Z~t)f(\tilde{Z}_{t}) are time-changed Brownian motions, and the time changes are given by κ1(s)\kappa^{-1}(s) and κ~1(s)\tilde{\kappa}^{-1}(s), respectively, where

κ(t)=0tf(Zt)2𝑑tκ~(t)=0tf(Z~t)2𝑑t}t<τU.\left.\begin{aligned} \kappa(t)&=\int_{0}^{t}\mid f^{\prime}(Z_{t})\mid^{2}dt\\ \tilde{\kappa}(t)&=\int_{0}^{t}\mid f^{\prime}(\tilde{Z}_{t})\mid^{2}dt\end{aligned}\;\right\}\;\;t<\tau_{U}.

Now, our assumptions imply |f(Zt)||f(Z~t)||f^{\prime}(Z_{t})|\geq|f^{\prime}(\tilde{Z}_{t})| a.s. for all t<τUt<\tau_{U}, and thus κ(t)κ~(t)\kappa(t)\geq\tilde{\kappa}(t) a.s. for all t<τUt<\tau_{U}. It follows from this that ττ~\tau\geq\tilde{\tau} a.s., where τ\tau and τ~\tilde{\tau} are the exit times from f(U)f(U) of the Brownian motions f(Zκ1(s))f(Z_{\kappa^{-1}(s)}) and f(Z~κ~1(s))f(\tilde{Z}_{\tilde{\kappa}^{-1}(s)}) which begin at f(a)f(a) and f(σΔ(a))f(\sigma_{\Delta}(a)), respectively. The result follows. ∎

We can obtain a corollary which will be useful for the isosceles triangle and in other cases by taking f(z)=z0+R(z+izi)f(z)=z_{0}+R\Big{(}\frac{z+i}{z-i}\Big{)} for z0z_{0}\in{\mathbb{C}} and R>0R>0, which takes the real axis to the circle C={|zz0|=R}C=\{|z-z_{0}|=R\}, the upper half-plane to the outside of CC, and the lower half-plane to the inside. We have |f(z)|=2R|zi|2|f^{\prime}(z)|=\frac{2R}{|z-i|^{2}}, and it is easy to check that |f(z)|>|f(z¯)||f^{\prime}(z)|>|f^{\prime}(\bar{z})| for all zz in the upper half-plane. Applying Proposition 6 and working through the implications yields the following.

Corollary 7.

Let C={|zz0|=R}C=\{|z-z_{0}|=R\} be a circle in {\mathbb{C}}, with inside {\cal I} and outside 𝒪{\cal O}. If UU is a domain such that σC(U)(U𝒪)\sigma_{C}(U\cap{\cal I})\subseteq(U\cap{\cal O}), then no pp-th center of UU lies in {\cal I}.

Note that here σC\sigma_{C} denotes reflection over the circular arc CC, that is, σC(z)=z0+R2zz0\sigma_{C}(z)=z_{0}+\frac{R^{2}}{\stackrel{{\scriptstyle\line(1,0){18.0}}}{{z-z_{0}}}}. We remark further that the singularity that ff has at ii does not cause a problem in this result, because P0(Zt=i for some t0)=0P_{0}(Z_{t}=i\mbox{ for some }t\geq 0)=0, so a.s. a Brownian motion starting at 0 will not hit the singularity, anyway. The compact set {Bt:0tτU}\{B_{t}:0\leq t\leq\tau_{U}\} is therefore bounded away from ii a.s., and the result goes through.

Let us now apply this corollary to the isosceles right triangle UU above. If CC is the circle passing through 1-1 and 11 which intersects the real axis at angles of π8\frac{\pi}{8}, then the reflection of the set 𝒜=U{\cal A}={\cal I}\cap U will be the region {\cal B} in the upper half-plane bounded by CC and the circle which passes through 1-1 and 11 and intersects the real axis at angles of π4\frac{\pi}{4}; this can be seen by noting that the transformation σC\sigma_{C} preserves angles and and also preserves the class of circles on the Riemann sphere (which includes lines, interpreted as circles through \infty).

Refer to caption
Figure 2.1: Reflection over a circle

As this region lies within UU, we conclude that no pp-th centers lie within 𝒜{\cal A}. A bit of Euclidean geometry shows that CC intersects the imaginary axis at csc(π8)cot(π8).20\csc(\frac{\pi}{8})-\cot(\frac{\pi}{8})\approx.20, and coupled with our observations above we see that all pp-th centers must lie on the imaginary axis between the points .2i.2i and .5i.5i. In fact, the upper bound of .5i.5i can be improved by using the angle bisector of the angles at 11 or 1-1, see Example 2 in Section 3. The reader may also have observed that the reflected circular domain does not do a good job of filling the triangle, and therefore it stands to reason that the lower bound may also be improved; more on this in the final section.

Finally, the following result can be useful when UU is mapped to itself by an antiholomorphic function f¯\bar{f} (this means that the conjugate of f¯\bar{f}, f(z)f(z), is holomorphic). We will denote the derivative of this function (with respect to z¯\bar{z} by f(z)¯\overline{f^{\prime}(z)}). An example of this is when UU is an annulus, as will be explored in Section 3.

Proposition 8.

Let f¯:UU\overline{f}:U\longrightarrow U be antiholomorphic and consider the two following sets

Ω\displaystyle\Omega :={zU||f¯(z)|<1}\displaystyle:=\{z\in U|\,\,|\bar{f}^{\prime}(z)|<1\}
Λ\displaystyle\Lambda :={zU||f¯(z)|=1}.\displaystyle:=\{z\in U|\,\,|\bar{f}^{\prime}(z)|=1\}.

If f¯(UΩ)Ω\bar{f}(U\setminus\Omega)\subset\Omega and f¯|Λ=idΛ\bar{f}_{|\Lambda}=id_{\Lambda}, then all pp-th centers are contained in Ω\Omega.

Proof.

Let ZtZ_{t} be a Brownian motion starting at zUΩz\in U\setminus\Omega and HΛH_{\Lambda} its hitting time of Λ\Lambda, and consider WtW_{t} the Brownian motion derived from f(Zt)f(Z_{t}), i.e

Wt:=f(Zκ1(t))W_{t}:=f(Z_{\kappa^{-1}(t)})

where

κ(t)=0t|f¯(Zs)|2𝑑s.\kappa(t)=\int_{0}^{t}|\bar{f}^{\prime}(Z_{s})|^{2}ds.

Now, we are going to construct two Brownian motions Z~t\widetilde{Z}_{t} and W~t\widetilde{W}_{t} starting respectively at zz and w:=f(z)w:=f(z) such that W~t\widetilde{W}_{t} leaves UU before Z~t\widetilde{Z}_{t}. as follows :

  1. 1.

    If HΛ<τUZH_{\Lambda}<\tau_{U}^{{\scriptscriptstyle Z}} then run an independent Brownian motion, say BtB_{t}, starting at ZHΛZ_{H_{\Lambda}}and set

    Z~t\displaystyle\widetilde{Z}_{t} :=Zt𝟏{tHΛ}+BtHΛ𝟏{HΛ<t}\displaystyle:=Z_{t}\mathbf{1}_{\{t\leq H_{\Lambda}\}}+B_{t-H_{\Lambda}}\mathbf{1}_{\{H_{\Lambda}<t\}}
    W~t\displaystyle\widetilde{W}_{t} :=Wt𝟏{tκ(HΛ)}+Btκ(HΛ)𝟏{κ(HΛ)<t}.\displaystyle:=W_{t}\mathbf{1}_{\{t\leq\kappa(H_{\Lambda})\}}+B_{t-\kappa(H_{\Lambda})}\mathbf{1}_{\{\kappa(H_{\Lambda})<t\}}.

    W~t\widetilde{W}_{t} and Z~t\widetilde{Z}_{t} are well defined Brownian motions as

    W~κ(HΛ)=Wκ(HΛ)=f(ZHΛ)=ZHΛ=Z~HΛ.\widetilde{W}_{\kappa(H_{\Lambda})}=W_{\kappa(H_{\Lambda})}=f(Z_{H_{\Lambda}})=Z_{H_{\Lambda}}=\widetilde{Z}_{H_{\Lambda}}.

    Now notice that

    τUZ~\displaystyle\tau_{U}^{{\scriptscriptstyle\widetilde{Z}}} =HΛ+inf{t,BtU|B0=ZHΛ}\displaystyle=H_{\Lambda}+\inf\{t,\,\,B_{t}\notin U|B_{0}=Z_{H_{\Lambda}}\}
    κ(HΛ)+inf{t,BtU|B0=ZHΛ}\displaystyle\leq\kappa(H_{\Lambda})+\inf\{t,\,\,B_{t}\notin U|B_{0}=Z_{H_{\Lambda}}\}
    =τUW~.\displaystyle=\tau_{U}^{{\scriptscriptstyle\widetilde{W}}}.
  2. 2.

    If HΛτUZH_{\Lambda}\geq\tau_{U}^{{\scriptscriptstyle Z}} then just set Z~t=Zt\widetilde{Z}_{t}=Z_{t} and W~t=Wt\widetilde{W}_{t}=W_{t}. Therefore

    τUZ~κ(τUZ~)=τUW~.\tau_{U}^{{\scriptscriptstyle\widetilde{Z}}}\leq\kappa(\tau_{U}^{{\scriptscriptstyle\widetilde{Z}}})=\tau_{U}^{{\scriptscriptstyle\widetilde{W}}}.

In both cases, we have

τUZ~a.sτUW~.\tau_{U}^{{\scriptscriptstyle\widetilde{Z}}}\overset{{\scriptstyle a.s}}{\leq}\tau_{U}^{{\scriptscriptstyle\widetilde{W}}}.

Hence

Ez((τUZ~)p)Ew((τUW~)p),E_{z}((\tau_{U}^{{\scriptscriptstyle\widetilde{Z}}})^{p})\leq E_{w}((\tau_{U}^{{\scriptscriptstyle\widetilde{W}}})^{p}),

which ends the proof. ∎

Remark: Reflection over the circle, obtained above as a corollary of Proposition 6, can just as easily be deduced as a corollary of Proposition 8.

3 Applications

In this section we work through a series of examples that show how our results may be applied.

Example 1.

Let UU be the upper half-disc {|z|<1,Im(z)>0}\{|z|<1,Im(z)>0\}. The imaginary axis is an axis of symmetry, and UU is Δ\Delta-convex with respect to this axis, so all pp-th centers lie on the imaginary axis. The line Δ=:{Im(z)=12}\Delta=:\{Im(z)=\frac{1}{2}\} is clearly a partial symmetry axis with symmetric part U{Im(z)>12}U\cap\{Im(z)>\frac{1}{2}\}, and UU is Δ\Delta-convex as well. Thus, all pp-centers belong to the set {Re(z)=0,0Im(z)12}\{Re(z)=0,0\leq Im(z)\leq\frac{1}{2}\}. Now, if we let CC be the circle {|z+i|=2}\{|z+i|=\sqrt{2}\}, then CC passes through 11 and 1-1, making an angle of π4\frac{\pi}{4} at each point with the real axis. If we let 𝒜=U{\cal A}=U\cap{\cal I} as before, with {\cal I} the inside of CC, then σC(𝒜)=\sigma_{C}({\cal A})={\cal B}, where =U𝒪{\cal B}=U\cap{\cal O} and 𝒪{\cal O} is the outside of CC. By Corollary 7, no pp-th center lies in 𝒜{\cal A}. Thus, all pp-th centers lie on the line segment {Re(z)=0,21Im(z)12}\{Re(z)=0,\sqrt{2}-1\leq Im(z)\leq\frac{1}{2}\}, which is in bold in the figure below.

Refer to caption
Figure 3.1: Reflection over a circle
Example 2.

Now let UU be an isosceles triangle with vertices at 1,1-1,1, and NiNi with N>0N>0. It will be convenient for us to index UU by the angles at 11 and 1-1, so if we let θ\theta be this angle then N=tanθN=\tan\theta. Proposition 4 tells us that all pp-th centers lie on the imaginary axis. We have seen already from the example discussed in connection with Proposition 6 that all pp-th centers must lie below M2i\frac{M}{2}i, however we will show now how this can be improved. Let BB be the angle bisector of one of the base angles of UU. BB is a partial symmetry axis of UU, with symmetric side given by the component of U\BU\backslash B corresponding to the shorter side of the triangle. Thus, if θ>π3\theta>\frac{\pi}{3}, then all pp-th centers must lie above BB, while if θ<π3\theta<\frac{\pi}{3}, then all pp-th centers must lie below BB. Now let MM be the perpendicular bisector of the edge connecting 11 to MiMi (this is often referred to as the mediator). This is also a partial symmetry axis of UU, and the symmetric side is the component of U\MU\backslash M which does not contain 1-1. Thus, if θ>π3\theta>\frac{\pi}{3}, then all pp-th centers must lie below MM, while if θ<π3\theta<\frac{\pi}{3}, then all pp-th centers must lie above MM. Thus, the intersections of MM and BB with the imaginary axis provide upper and lower bounds for all pp-th centers, although which is the upper bound and which is the lower bound depends on θ\theta. The following figure demonstrates this phenomenon (Δ\Delta denotes the imaginary axis).

Refer to caption
Figure 3.2: The angle bisector and mediator

Naturally they coincide at θ=π3\theta=\frac{\pi}{3}. It can be checked that, regardless of θ\theta, this gives a better upper bound than N2\frac{N}{2}, which was given by reflection over {Im(z)=N2}\{Im(z)=\frac{N}{2}\}. A bit of Euclidean geometry shows that the intersection of BB with the imaginary axis is at the point tan(θ2)i\tan(\frac{\theta}{2})i, and the intersection of MM with the imaginary axis is at the point (1tanθ1sin2θ)i(\frac{1}{\tan\theta}-\frac{1}{\sin 2\theta})i. Furthermore, we always have as a lower bound the intersection of the imaginary axis and the circle passing through 11 and 1-1, making an angle of θ2\frac{\theta}{2} with the real axis; this follows from Corollary 7 as above. This point is 1cos(θ2)sin(θ2)\frac{1-\cos(\frac{\theta}{2})}{\sin(\frac{\theta}{2})}. The following graph shows these upper and lower bounds; all pp-th centers must lie in the regions labelled Ω\Omega.

Refer to caption
Figure 3.3: Upper and lower bounds for pp-th centers

Remark: We believe that a better lower bound can be achieved through numerical conformal mapping; more on this in the final section.

Example 3.

Let 𝒜r,R\mathcal{A}_{r,R} be the annulus {r<|z|<R}\{r<|z|<R\}. Then all pp-centers lie in {rR<|z|<R+r2}\{\sqrt{{\scriptstyle rR}}<|z|<\frac{R+r}{2}\}

Proof.

Consider f(z)=rRz¯f(z)=\frac{rR}{\overline{z}} which maps 𝒜r,R\mathcal{A}_{r,R} to itself. Under the same notation as in Proposition 8 we have

Ω\displaystyle\Omega :={rR<|z|<R}\displaystyle:=\{\sqrt{rR}<|z|<R\}
Λ\displaystyle\Lambda :={|z|=rR}\displaystyle:=\{|z|=\sqrt{rR}\}

and we can check easily that ff satisfies the requirements of Proposition 8). Therefore we can eliminate UΩU\setminus\Omega from consideration, and we obtain the lower bound rR\sqrt{{\scriptstyle rR}}. In order to get the upper bound R+r2\frac{R+r}{2} we can see, as illustrated by Fig. (3.4), that the line Δ\Delta is a partial symmetry axis. The result follows.

Refer to caption
Figure 3.4: Bounds for the annulus

Remarks:

  • Another way to get the same lower bound as above is to note that Proposition 6 extends to non-injective maps in suitable situations. We may use the map

    f:{lnr<Re(z)<lnR}𝒜r,Rzezf:\begin{aligned} \left\{\ln r<Re(z)<\ln R\right\}&\longrightarrow\mathcal{A}_{r,R}\\ z&\longmapsto e^{z}\end{aligned}

    and apply this extension of Proposition 6 with reflection axis {Re(z)=lnR+lnr2}\{Re(z)=\frac{\ln R+\ln r}{2}\} in order to obtain the result.

  • It should be mentioned that an explicit formula for the first moment can be obtained by Dynkin’s formula, and it is

    Ez(τ𝒜r,R)=R2ln(|z|r)r2ln(|z|R)lnRr|z|2.E_{z}(\tau_{\mathcal{A}_{r,R}})=\frac{R^{2}\ln\left(\frac{|z|}{r}\right)-r^{2}\ln\left(\frac{|z|}{R}\right)}{\ln\frac{R}{r}}-|z|^{2}.

    This can be shown to be maximal at |z|=R2r22lnRr|z|=\sqrt{\frac{R^{2}-r^{2}}{2\ln\frac{R}{r}}}. Our estimates are therefore not necessary for the first moments, but as an aside we obtain the inequality

    Rr<R2r22lnRr<R+r2.\sqrt{Rr}<\sqrt{\frac{R^{2}-r^{2}}{2\ln\frac{R}{r}}}<\frac{R+r}{2}.

    Setting a=R2a=R^{2}, b=r2b=r^{2}, and squaring the inequalities gives the following:

    ab<ablnab<a+b+2ab4a+b2.\sqrt{ab}<\frac{a-b}{\ln\frac{a}{b}}<\frac{a+b+2\sqrt{ab}}{4}\leq\frac{a+b}{2}.

    The quantity ablnab\frac{a-b}{\ln\frac{a}{b}} is of great importance in the study of heat flow, and in that context it is known as the logarithmic mean temperature difference, or LMTD (see [15]). We have therefore given a new proof of the fundamental fact that the LMTD lies between the arithmetic and geometric means, and in fact have proved that the upper bound can be lowered to the arithmetic mean of the arithmetic and geometric means.

Example 4.

Let \mathscr{H} be the region {|x|>|y|,x2y2<1}\{|x|>|y|,x^{2}-y^{2}<1\}; this is the region bounded by the lines y=±xy=\pm x and the hyperbola x2y2=1x^{2}-y^{2}=1, see the figure below.

It is perhaps not obvious for which pp we have Ew[τp]<E_{w}[\tau_{\mathscr{H}}^{p}]<\infty, however we can show that Ew[τp]<E_{w}[\tau_{\mathscr{H}}^{p}]<\infty for any p>0p>0 and ww\in\mathscr{H}, as follows. \mathscr{H} is contained in the union of two infinite strips which are othogonal. Any strip has all moments of its exit time finite: Brownian motion is rotation invariant, so the moments are the same as for a horizontal strip, and these moments in turn are the same as for a one-dimensional Brownian motion from a bounded interval, since that is what we obtain when we project the Brownian motion onto the imaginary axis; these moments are well known to be finite for all pp, and in fact they can be calculated explicitly for integer pp using the Hermite polynomials. We would like to conclude that the union of these two strips must then have finite pp-th moment, however easy examples show that it is not necessarily the case that the union of two domains with finite pp-th moment must itself have finite pp-th moment. A method does exist for reaching the desired conclusion, though, and it is contained in Theorem 3 and Lemmas 1 and 2 of [16]. It is strightforward to verify that our infinite strips satisfy the required conditions: their interesection is bounded, and boundary arcs intersect at non-zero angles. Therefore the exit time for their union has finite pp-th moments for all pp, and thus so does \mathscr{H}. See [16] for details.

Now let us see how our methods can be used to localize the pp-th centers. The real axis is an axis of symmetry, however the domain is not Δ\Delta-convex, so we may not apply Proposition 4. However, all pp-centers do lie on the real axis, and we may prove this as follows. The map f(z)=zf(z)=\sqrt{z} maps the strip {0<Re(z)<1}\{0<Re(z)<1\} conformally onto \mathscr{H}. Any horizontal line can be used in Proposition 6, and we note that |f(z)|=12|z||f^{\prime}(z)|=\frac{1}{2\sqrt{|z|}} is monotone decreasing in |z||z| and therefore in |Im(z)||Im(z)|. Thus, all pp-th centers must lie on the image of the real axis under ff, which is again the real axis. So we need only consider point on {\mathbb{R}}. The line {Re(z)=12}\{Re(z)=\frac{1}{2}\} is a partial symmetry axis, which gives a lower bound of 12\frac{1}{2} for all pp-th centers. For an upper bound, note that {Re(z)=12}\{Re(z)=\frac{1}{2}\} is another axis of symmetry of {0<Re(z)<1}\{0<Re(z)<1\}, and the monotonicity of the derivative shows again via Proposition 6 that we only need to look in the region f({0<Re(z)<12})f(\{0<Re(z)<\frac{1}{2}\}), which is the region {x2y2<12}\{x^{2}-y^{2}<\frac{1}{2}\} inside UU. This gives an upper bound of 12\frac{1}{\sqrt{2}} on the real axis. Thus, all pp-centers lie on {Im(z)=0,12<Re(z)<12}\{Im(z)=0,\frac{1}{2}<Re(z)<\frac{1}{\sqrt{2}}\}. This set is in bold below.

Refer to caption
Figure 3.5: Bounds for the hyperbolic region
Example 5.

Let 𝒞R\mathscr{C}_{R} be the crescent-like shape limited by the two circles {|z12|=12}\{|z-\frac{1}{2}|=\frac{1}{2}\} and {|zR2|=R2}\{|z-\frac{R}{2}|=\frac{R}{2}\} (see the figure below). 𝒞R\mathscr{C}_{R} is the image of the region {1R<Re(z)<1}\{\frac{1}{R}<Re(z)<1\} under the conformal map f(z)=1zf(z)=\frac{1}{z}. Note that |f(z)|=1|z|2|f^{\prime}(z)|=\frac{1}{|z|^{2}} is monotone decreasing in |z||z|, so by the same argument as in Example 4 all pp-th centers lie on the real axis. Furthermore {Re(z)=R+12}\{Re(z)=\frac{R+1}{2}\} is a partial symmetry axis for 𝒞R\mathscr{C}_{R}, and this allows us to eliminate the region {Re(z)>R+12\{Re(z)>\frac{R+1}{2} from consideration. We may also use axis of symmetry {Re(z)=1+1R2}\{Re(z)=\frac{1+\frac{1}{R}}{2}\} in order to conclude via Proposition 6 that we can exclude the region f({1R<Re(z)<1+1R2})f(\{\frac{1}{R}<Re(z)<\frac{1+\frac{1}{R}}{2}\}), which is the region 𝒞R{|zRR+1|<RR+1}\mathscr{C}_{R}\cap\{|z-\frac{R}{R+1}|<\frac{R}{R+1}\}, in the search for pp-th centers. We see that all pp-centers lie on the interval {Im(z)=0,2RR+1<Re(z)<R+12}\{Im(z)=0,\frac{2R}{R+1}<Re(z)<\frac{R+1}{2}\}, which is in bold below for R=2R=2.

Refer to caption
Figure 3.6: Bounds for the crescent region

4 Concluding remarks

As remarked earlier, in Figure 2.1 the regions 𝒜{\cal A} and {\cal B} do not fill all of UU, and it is natural to search for a a better bound by finding a conformal map that fills the entire domain. Let us consider the Schwarz-Christoffel transformation sending the unit disk to U(θ)U(\theta) given by (see [17, Ch. 2])

f(z)=A+C0z(1w)θπ1(1+w)θπ1(1+iw)2θπ𝑑wf(z)=A+C\int_{0}^{z}(1-w)^{\frac{\theta}{\pi}-1}(1+w)^{\frac{\theta}{\pi}-1}(1+iw)^{\frac{-2\theta}{\pi}}dw

for appropriate choices of constants AA and CC; note that this is chosen so that 11 and 1-1 are mapped to the base angles, and ii is mapped to the top angle. We have

|f(z)|=|C||1z|θπ1|1+z|θπ1|1+iz|2θπ.|f^{\prime}(z)|=|C||1-z|^{\frac{\theta}{\pi}-1}|1+z|^{\frac{\theta}{\pi}-1}|1+iz|^{\frac{-2\theta}{\pi}}.

It can be checked then that |f(z)|>|f(z¯)||f^{\prime}(z)|>|f^{\prime}(\bar{z})| whenever Re(z)>0Re(z)>0, so Proposition 6 implies that no pp-th centers can be found in the image of 𝔻{Re(z)<0}{\mathbb{D}}\cap\{Re(z)<0\}. From this point, a numerical method can be employed, and the resulting bound should improve the one we found, if desired.

As was mentioned in connection with Δ\Delta-convexity, there are some purely geometrical consequences of our results. In that context, the following may be proved.

Proposition 9.

Suppose a domain UU is Δ\Delta-convex with respect to two parallel symmetry axes. Then we can find a[,)a\in[-\infty,\infty) and b(,]b\in(-\infty,\infty] so that UU is a rotation of the domain {a<Re(z)<b}\{a<Re(z)<b\}; in other words, UU is all of {\mathbb{C}}, is a half-plane, or is an infinite strip.

As a corollary of this, and of our probabilistic results above, we obtain the following.

Corollary 10.

Suppose UU is a domain which is not all of {\mathbb{C}}, a half-plane, or an infinite strip. Then if there are multiple axes of symmetry to which UU is Δ\Delta-convex then they all meet at a unique point.

5 Acknowledgements

The authors would like to thank Paul Jung, Fuchang Gao, and Lance Smith for helpful conversations. We are also grateful to several anonymous referees for helpful comments.

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