Maximizing the -th moment of exit time of planar Brownian motion from a given domain
Abstract
In this paper we address the question of finding the point which maximizes the -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 be a planar Brownian motion starting at a point in a domain . We will let be the first time that exits , and we will use the standard notation to denote expectation conditioned on a.s. The focus of this paper is the following optimization problem.
For a given domain in the plane and , find the point which maximizes the quantity .
We will refer to such a point as a -th center of ; 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 -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 which must contain all -th centers. In certain cases in which the domain in question has a high degree of symmetry, it will allow us to locate all -th centers.
Before describing our methods, we present a brief overview of some earlier works related to this problem. The case is commonly referred to as the "torsion problem" due to its connection with mechanics, and is naturally the most tractable. The function satisfies , 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 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 and various types of symmetrizations. These results allow conclusions to be reached concerning the comparison of -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 -th moment of the exit time of Brownian motion for all . 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 -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 -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 -th centers of the domain. In what follows we assume is a fixed positive number; however, in order to reduce the qualifications needed to state our results, for any planar domain for which we are interested in maximizing the -th moment we will assume that for all points ; this would follow if for any , as is shown in [12].
2 Partial symmetry and convexity with respect to a line.
Definition 1.
Let be a domain of . We say that a line is a partial symmetry axis for if one of the two sets or can be folded over and fits into , more precisely if or remains inside , where denotes the symmetry over . The subset among that satisfies this property (that is, the smaller side with respect to the symmetry) is called symmetric side of over . So, for instance, any line intersecting is a partial symmetry axis for , but the line is not one for the square since the reflection over this line of the point is the point , which is not in the closure of the square. Notice that both of are symmetric sides if and only if is a symmetry axis for .
Theorem 2.
Let be the symmetric side of over a partial symmetry axis . Then, for any we can find Brownian motions starting at and starting at defined on the same probability space such that a.s. (where is the exit time from of ). Furthermore, if is strictly contained in then . In particular, (with strict inequality if is strictly contained in ).
Proof.
This follows from a coupling argument. Let start at , and let be its hitting time of the line . Form the process by the rule
It follows from the Strong Markov property and the reflection invariance of Brownian motion that is a Brownian motion. Clearly on the set , and our conditions on imply on the set . Furthermore, if is strictly contained in , then has some positive probability of leaving before 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 in our search for -th centers. The only exception to this rule is when , i.e. when is a symmetry axis of ; however in most cases a symmetry axis will contain all -th centers. To see why this is so, we need another definition.
Definition 3.
Let be a symmetry axis of . We say that is -convex if
In other words, for every , the segment joining and remains inside .
It is clear that any convex is -convex for any symmetry axis , and a less trivial example can be given by , where is a positive continuous function on the real line, which is -convex with . A domain which is not -convex with respect to a symmetry axis can be given by with ; this has the real and imaginary axes as symmetry axes but is not -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 -convexity is required in the following proposition.
Proposition 4.
Suppose is -convex with respect to a symmetry axis . Then all -th centers of lie on .
Proof.
Let , and let be the orthogonal projection of onto . Let be the line parallel to which passes through the point ; speaking informally, this is the line halfway between and . -convexity implies that is a partial symmetry axis of , and if is the component of containing then is strictly contained in . It therefore follows from Theorem 2 that . The result follows. ∎
Note that this proposition completely solves our problem in the case that our domain is -convex with respect to two or more non-parallel symmetry axes, since all -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 -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 -th center, let us return to the domains described immediately before this proposition. Proposition 4 implies that all -th centers lie on the real line, but it is easy to see that if we make sufficiently small then 0, the intersection point of the two symmetry axes, will not be a -th center (clearly , so that remains always greater than a positive constant, but decreases monotonically to 0 a.s. as , so that ).
Let us now look at an example that shows the use of the results proved to this point, but also their limitations. Suppose is the isosceles right triangle with vertices at and . The imaginary axis is an axis of symmetry, and is -convex with respect to this axis, so all -th centers must lie on the imaginary axis. The line is a partial symmetry axis for , with the symmetric side, so all -centers must lie on . Now, common sense tells us that the -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 is a holomorphic function then is a time changed Brownian motion. More precisely, is a Brownian motion where
for .
This allows us to extend Theorem 2, as follows.
Proposition 6.
Suppose is a domain with an axis of symmetry , and suppose is a conformal map defined on with the property that for all , where is one component of and is the symmetry over . Then, for any , we can find Brownian motions starting at and starting at defined on the same probability space such that a.s. (where is the exit time from of ). In particular, . If there is any point in at which then (for this statement we recall the assumption that for any ).
Proof.
Let be a Brownian motion starting at , and let be defined as in Theorem 2. According to Theorem 5, the processes and are time-changed Brownian motions, and the time changes are given by and , respectively, where
Now, our assumptions imply a.s. for all , and thus a.s. for all . It follows from this that a.s., where and are the exit times from of the Brownian motions and which begin at and , respectively. The result follows. ∎
We can obtain a corollary which will be useful for the isosceles triangle and in other cases by taking for and , which takes the real axis to the circle , the upper half-plane to the outside of , and the lower half-plane to the inside. We have , and it is easy to check that for all in the upper half-plane. Applying Proposition 6 and working through the implications yields the following.
Corollary 7.
Let be a circle in , with inside and outside . If is a domain such that , then no -th center of lies in .
Note that here denotes reflection over the circular arc , that is, . We remark further that the singularity that has at does not cause a problem in this result, because , so a.s. a Brownian motion starting at will not hit the singularity, anyway. The compact set is therefore bounded away from a.s., and the result goes through.
Let us now apply this corollary to the isosceles right triangle above. If is the circle passing through and which intersects the real axis at angles of , then the reflection of the set will be the region in the upper half-plane bounded by and the circle which passes through and and intersects the real axis at angles of ; this can be seen by noting that the transformation preserves angles and and also preserves the class of circles on the Riemann sphere (which includes lines, interpreted as circles through ).

As this region lies within , we conclude that no -th centers lie within . A bit of Euclidean geometry shows that intersects the imaginary axis at , and coupled with our observations above we see that all -th centers must lie on the imaginary axis between the points and . In fact, the upper bound of can be improved by using the angle bisector of the angles at or , 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 is mapped to itself by an antiholomorphic function (this means that the conjugate of , , is holomorphic). We will denote the derivative of this function (with respect to by ). An example of this is when is an annulus, as will be explored in Section 3.
Proposition 8.
Let be antiholomorphic and consider the two following sets
If and , then all -th centers are contained in .
Proof.
Let be a Brownian motion starting at and its hitting time of , and consider the Brownian motion derived from , i.e
where
Now, we are going to construct two Brownian motions and starting respectively at and such that leaves before . as follows :
-
1.
If then run an independent Brownian motion, say , starting at and set
and are well defined Brownian motions as
Now notice that
-
2.
If then just set and . Therefore
In both cases, we have
Hence
which ends the proof. ∎
3 Applications
In this section we work through a series of examples that show how our results may be applied.
Example 1.
Let be the upper half-disc . The imaginary axis is an axis of symmetry, and is -convex with respect to this axis, so all -th centers lie on the imaginary axis. The line is clearly a partial symmetry axis with symmetric part , and is -convex as well. Thus, all -centers belong to the set . Now, if we let be the circle , then passes through and , making an angle of at each point with the real axis. If we let as before, with the inside of , then , where and is the outside of . By Corollary 7, no -th center lies in . Thus, all -th centers lie on the line segment , which is in bold in the figure below.

Example 2.
Now let be an isosceles triangle with vertices at , and with . It will be convenient for us to index by the angles at and , so if we let be this angle then . Proposition 4 tells us that all -th centers lie on the imaginary axis. We have seen already from the example discussed in connection with Proposition 6 that all -th centers must lie below , however we will show now how this can be improved. Let be the angle bisector of one of the base angles of . is a partial symmetry axis of , with symmetric side given by the component of corresponding to the shorter side of the triangle. Thus, if , then all -th centers must lie above , while if , then all -th centers must lie below . Now let be the perpendicular bisector of the edge connecting to (this is often referred to as the mediator). This is also a partial symmetry axis of , and the symmetric side is the component of which does not contain . Thus, if , then all -th centers must lie below , while if , then all -th centers must lie above . Thus, the intersections of and with the imaginary axis provide upper and lower bounds for all -th centers, although which is the upper bound and which is the lower bound depends on . The following figure demonstrates this phenomenon ( denotes the imaginary axis).

Naturally they coincide at . It can be checked that, regardless of , this gives a better upper bound than , which was given by reflection over . A bit of Euclidean geometry shows that the intersection of with the imaginary axis is at the point , and the intersection of with the imaginary axis is at the point . Furthermore, we always have as a lower bound the intersection of the imaginary axis and the circle passing through and , making an angle of with the real axis; this follows from Corollary 7 as above. This point is . The following graph shows these upper and lower bounds; all -th centers must lie in the regions labelled .

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 be the annulus . Then all -centers lie in
Proof.
Consider which maps to itself. Under the same notation as in Proposition 8 we have
and we can check easily that satisfies the requirements of Proposition 8). Therefore we can eliminate from consideration, and we obtain the lower bound . In order to get the upper bound we can see, as illustrated by Fig. (3.4), that the line is a partial symmetry axis. The result follows.
∎

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
and apply this extension of Proposition 6 with reflection axis 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
This can be shown to be maximal at . Our estimates are therefore not necessary for the first moments, but as an aside we obtain the inequality
Setting , , and squaring the inequalities gives the following:
The quantity 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 be the region ; this is the region bounded by the lines and the hyperbola , see the figure below.
It is perhaps not obvious for which we have , however we can show that for any and , as follows. 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 , and in fact they can be calculated explicitly for integer using the Hermite polynomials. We would like to conclude that the union of these two strips must then have finite -th moment, however easy examples show that it is not necessarily the case that the union of two domains with finite -th moment must itself have finite -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 -th moments for all , and thus so does . See [16] for details.
Now let us see how our methods can be used to localize the -th centers. The real axis is an axis of symmetry, however the domain is not -convex, so we may not apply Proposition 4. However, all -centers do lie on the real axis, and we may prove this as follows. The map maps the strip conformally onto . Any horizontal line can be used in Proposition 6, and we note that is monotone decreasing in and therefore in . Thus, all -th centers must lie on the image of the real axis under , which is again the real axis. So we need only consider point on . The line is a partial symmetry axis, which gives a lower bound of for all -th centers. For an upper bound, note that is another axis of symmetry of , and the monotonicity of the derivative shows again via Proposition 6 that we only need to look in the region , which is the region inside . This gives an upper bound of on the real axis. Thus, all -centers lie on . This set is in bold below.

Example 5.
Let be the crescent-like shape limited by the two circles and (see the figure below). is the image of the region under the conformal map . Note that is monotone decreasing in , so by the same argument as in Example 4 all -th centers lie on the real axis. Furthermore is a partial symmetry axis for , and this allows us to eliminate the region from consideration. We may also use axis of symmetry in order to conclude via Proposition 6 that we can exclude the region , which is the region , in the search for -th centers. We see that all -centers lie on the interval , which is in bold below for .

4 Concluding remarks
As remarked earlier, in Figure 2.1 the regions and do not fill all of , 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 given by (see [17, Ch. 2])
for appropriate choices of constants and ; note that this is chosen so that and are mapped to the base angles, and is mapped to the top angle. We have
It can be checked then that whenever , so Proposition 6 implies that no -th centers can be found in the image of . 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 -convexity, there are some purely geometrical consequences of our results. In that context, the following may be proved.
Proposition 9.
Suppose a domain is -convex with respect to two parallel symmetry axes. Then we can find and so that is a rotation of the domain ; in other words, is all of , 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 is a domain which is not all of , a half-plane, or an infinite strip. Then if there are multiple axes of symmetry to which is -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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