A Capture Problem in Brownian Motion and Eigenvalues of Spherical Domains

We resolve a question of Bramson and Griffeath by showing that the expected capture time of four independent Brownian predators pursuing one Brownian prey on a line is finite. Our main tool is an eigenvalue estimate for a particular spherical domain, which we obtain by a coning construction and domain perturbation.

In this paper, we examine the expected capture time of a single Brownian prey pursued by n independent Brownian predators. All motion is restricted to a line. Bramson and Griffeath [BG] first considered this problem, and estimated the capture time in various circumstances. In particular, they showed that if at time t = 0 there are predators on both sides of the prey then the expected capture time is finite. For this reason, we will assume that the initial position of the prey is x 0 (0) = 1 and the initial positions of the predators are x 1 (0) = · · · = x n (0) = 0. In this case, [BG] showed that the expected capture time is infinite for n = 1, 2, 3, and conjectured that it is finite for n ≥ 4 (as indicated by simulations). Li and Shao [LS] showed that the expected capture time is finite for n ≥ 5. Using a similar method, we resolve the remaining case by showing that the expected capture time is finite for n = 4.
One key difference between Li and Shao's method in [LS] and our method here is that they consider a difference process, while we do a geometric splitting in spherical polar coordinates to reduce the dimension of the problem. See Section 2 for more details on our dimension reduction.
One can reformulate the capture problem described above in terms of exit of a Brownian particle in R n+1 from a specific cone. We denote the position of the prey at time t as x 0 (t) and the position of the jth predator at time t as x j (t). By our choice of initial conditions, the initial position of the Brownian particle x(t) = (x 0 (t), x 1 (t), . . . , x n (t)) is x(0) = (1, 0, . . . , 0). The event of capture is then equivalent to the Brownian particle x(t) first leaving the cone C n+1 := {(x 0 , x 1 , . . . , x n ) | x 0 ≥ x j , j = 1, . . . , n}, with x(0) = (1, 0, . . . , 0), and so we must estimate the expected first exit time of a Brownian particle from the cone C n+1 , with the starting position (1, 0, . . . , 0).
DeBlassie [DB] developed the theory of estimating exit times for Brownian motion from cones in Euclidean space. Let C = {(r, θ) | r ≥ 0, θ ∈ D ⊂ S n } be the cone over a domain D ⊂ S n . Also let τ x be the exit time from C of a Brownian particle with starting position x, and let P(τ x > t) be the probability that τ x > t. DeBlassie showed P(τ Here λ 1 (D) is the first Dirichlet eigenvalue of D. In the particular case we are interested in, the expected exit time of a Brownian particle from C n+1 is finite if and only if a(n) > 1, which reduces to where D n = C n+1 ∩ S n . Our method for proving Theorem 1 is to estimate the first eigenvalue of D n using the monotonicity property of eigenvalues, a coning construction, and domain perturbation. The rest of the paper proceeds as follows. In Section 1 we discuss the geometry of C n+1 , D n , and related regions. Section 2 contains the separation of variables background one needs to estimate the expected capture time. For the reader's convenience, we also include the proof that the expected capture time is infinite for n = 1, 2, 3 predators in this section. We prove the n = 4 eigenvalue estimate in Section 3. Finally, we describe a numerical computation of λ 1 (D 3 ) and a lower bound for λ 1 (D 4 ) in Section 4.

Geometry of the cone C
The cone C n+1 and its spherical angle D n have much symmetry. First observe that C n+1 contains the line L spanned by (1, 1, . . . , 1); this is the line where all the inequalities x 0 ≥ x j , j = 1, . . . , n are equalities. Thus we can split C n+1 as a sum of a line and lower dimensional cone where S n−1 is the unit sphere in L ⊥ = span{V j }. The domain D n is a double cone over T n−1 . More precisely, let N be one of the intersection points in S n ∩ L (there are two such points), and let (r, θ) be polar coordinates in S n , centered at N . Then where we have identified S n−1 with the unit sphere in the tangent space T N S n . In Figure 1 we show T 1 and D 2 .
In later sections, we will use a generalization of this type of spherical cone. In general, let Ω be a domain in the equatorial S n−1 of S n , and let r 0 ∈ (0, π]. Then we define the truncated cone We abbreviate T C(Ω, π) = T C(Ω). In this notation, D n = T C(T n−1 ) = T C(T n−1 , π).
The domain T n−1 has symmetry. If we let then we see C n+1 = C 0 n+1 and C j n+1 are pairwise congruent. Thus T n−1 is a face of the regular (n + 1)-hedral tesselation of the standard S n−1 one obtains by connecting the vertices of a regular (n + 1)-simplex with great circle arcs. In particular, one can compute the diameter of T n−1 as the distance from a vertex to the center of the opposite face, which is δ(n − 1) = arccos(− n − 1 2n ). Moreover, the spherical angle of T n−1 at a vertex is T n−2 . So we can construct a succession of comparison domains for T 1 , T 2 , . . . starting with T 1 and using the coning construction described above. To this end, we letT 2π 3 ] = T 1 ,T n := T C(T n−1 , δ(n)).

Separating variables
We discuss two types of separation of variables in this section. The first type is the separation of variables used in [DB] to estimate expected exit times of Brownian motion from Euclidean cones, and the second is the separation of variables one performs to estimate eigenvalues of a spherical domain with a conical structure. One needs the domains in question to have a certain amount of regularity in order to separate variables. In particular, these domains are piecewise C 1 , satisfy an exterior cone condition, and are well approximated (in terms of Hausdorff distance) by finite volume, C 1 domains. See the introduction to [DB] for a precise statement. It is straight-forward to verify that all of the domains we consider satisfy DeBlassie's hypotheses. We will refer to these domains as nice.

DeBlassie's separations of variables
We first review DeBlassie's [DB] argument. Consider the cone C ∈ R n+1 over a domain D ⊂ S n : Let τ x be the time it takes for a Brownian particle to exit C, with starting position x, and let u(x, t) = P(τ x > t) be the probability that τ x > t. Then u satisfies the heat equation In polar coordinates (r, θ, t), the PDE becomes Moreover, the solution scales as u(r, θ, t) = u(βr, θ, β 2 t), so we can separate variables and look for a solution of the form u = R(ξ)U (θ), where ξ = r 2 /2t. Then we get a positive separation constant λ j (D), and which has solutions of the form ρ(ξ) = f (a/2, a + (n + 1)/2, ξ), where f is the confluent hypergeometric function (see e.g., [L]). Using f and restricting to K ⊂ D compact and T > 0, one obtains (see [DB]) an expansion for u of the form where U j is the jth Dirichlet eigenfunction of ∆ S n on D and

This sum converges uniformly in
, which yields the conclusion of inequality (2).

Separating variables on the sphere
Next we separate variables to relate the eigenvalues λ 1 (D n ) and λ 1 (T n−1 ). First recall that we can write the Laplacian for S n in polar coordinates as where ∆ θ is the Laplacian on the equatorial S n−1 . This lemma gives us a dimension reduction.

Lemma 2
Let Ω be a nice domain in an equatorial S n−1 with first eigenvalue λ = λ 1 (Ω), and let D = T C(Ω) be the double cone over Ω. Then the first Dirichlet eigenvalues of D and Ω are related by In particular, λ 1 (D) > 2n + 2 whenever λ 1 (Ω) > 2n.
Then u is an eigenfunction on D with eigenvalue µ precisely when TR + (n − 1) cot r TṘ + csc 2 r R∆ θ T = −µT R.
Remark 1 A second solution to equation (5) has the form sin m r cos r, where m is again given by equation (6) but This eigenfunction vanishes on {π/2} × Ω, so it corresponds to a higher eigenvalue.
At this point, we can prove that the expected capture time for n = 1, 2, 3 predators is infinite. To prove the expected capture time is infinite, by inequality (2) we need to show λ 1 (D n ) ≤ 2n+ 2, or, equivalently, that λ 1 (T n−1 ) ≤ 2n. In the case of n = 1, we have D 1 = [−3π/4, π/4], and so λ 1 (D 1 ) = 1 < 4. In the case n = 2, we have T 1 = [0, 2π/3], and so λ 1 (T 1 ) = 9/4 < 4. We cannot compute λ 1 (T 2 ) so easily, but we can find a test function to show that λ 1 (T 2 ) < 6. We shall find λ 1 (T 2 ) numerically in Section 4. Recall the Rayleigh characterization of the first eigenvalue of a domain Ω: To show that λ 1 (T 2 ) < 6, it suffices to find f 0 ∈ H 1 0 (T 2 ) so that T2 |df 0 | 2 / T2 f 2 0 < 6. Let f 0 (x) = sin(dist(x, ∂T 2 )), and observe that |df 0 | 2 = 1 − f 2 0 off the set of focal points of ∂T 2 , which is a set of measure zero. A computation shows One can generalize the eigenvalue relationship (4) to spherical cones of the form T C(Ω, r 0 ), for 0 < r 0 < π, using the hypergeometric function Later we will use the following lemma to relate the eigenvalues ofT n−1 andT n , which we will then estimate to bound the asymptotic decay rates of probabalistic exit times.

The eigenvalue estimate
In this section, we complete the proof of Theorem 1. First observe that it suffices to show λ 1 (T 3 ) > 8. In fact, it suffices to find an effective lower bound for λ 1 (T 2 ) using the following scheme. We define λ cr by If we construct a domain G 2 ⊂ S 2 such that T 2 ⊂ G 2 and λ 1 (G 2 ) > λ cr , then by domain monotonicity and equation (8), λ 1 (T 3 ) ≥ λ 1 (T C(T 2 , δ(3)))) ≥ λ 1 (T C(G 2 , δ(3))) > 8, completing the proof of Theorem 1. As the last step in our proof, we construct a domain G 2 ⊂ S 2 as a perturbation of T 2 , such that T 2 ⊂ G 2 and λ 1 (G 2 ) = 5.102 > λ cr . Figure 2 shows sketches of the domains T 2 , G 2 andT 2 , after stereographic projection to the plane. Previous work of Rayleigh [R] and Pólya-Szegö [PS] motivates us to consider this type of domain perturbation. They studied the eigenvalue of a planar domain which has the form {(r, θ) | 0 ≤ r ≤ c + ǫf (θ)} in polar coordinates, for some small ǫ > 0, and gave an expression for the eigenvalue λ 1 in terms of ǫf . In our case, we fix λ = λ 1 and find a domain G 2 with λ as its first eigenvalue.
The domains T 2 , G 2 andT 2 all have three boundary curves, and two of these boundary curves for each domain lie along the great circle arcs θ = 0, 2π/3. For convenience, we convert to a planar domain using stereographic projection, with the south pole corresponding to r = 0 in polar coordinates on S 2 . The relationship between polar coordinates (r, θ) on S 2 and polar coordinates (ρ, θ) on the plane is ρ = tan(r/2). Under stereographic projection, the great circle arcs θ = 0, 2π/3 correspond to rays emanating from the origin at angles 0, 2π/3. The third boundary curve of T 2 is given by the circular arc which we can rewrite as Thus showing T 2 ⊂ G 2 is equivalent to showing H(r, θ) > 0 along the arc given by (ρ = β(θ), θ) for 0 ≤ θ ≤ 2π/3. One can see from Figure 2 that H > 0 along this third boundary component of T 2 . One strategy for a rigorous proof that H > 0 is the following. We first evaluate H at a point θ 0 on the curve, checking H > 0 at (ρ = β(θ 0 ), θ 0 ), and bound the derivative H θ on an interval containing θ 0 . Our bound H θ ≥ M gives us a lower bound H > H(θ 0 )−M (θ−θ 0 ). Thus H > 0 on a possibly smaller neighborhood of θ 0 . We then repeat this procedure with each endpoint of this (smaller) interval. One can simplify the computations by observing T 2 and G 2 are symmetric about the ray θ = π/3. Then it suffices to evaluate H and its derivative at θ = 0, 1/2, 2/3, 2π/9, π/3. . Given our lower bound λ 1 (T 2 ) ≥ λ 1 (G 2 ) = 5.102, one can use equation (8) to show λ 1 (T 3 ) ≥ 8.00087815. This in turn gives λ 1 (D 4 ) ≥ 10.001024501 and a(4) ≥ 1.00007318. In the same way, the lower bound λ 1 (T 2 ) ≥ 5.102 gives a(3) ≥ .90671950.
The most important property of sinc functions is that they are well-suited to approximating integrals in a strip. In particular, one can increase accuracy in a numerical computation by increasing the number of evaluation points, without recomputing grids (as in finite elements). In order to take advantage of this simplicity, we will conformally map a subdomain of T 2 to a half-infinite strip.
Recall that T 2 is an equilateral triangle on S 2 with all its interior angles equal to 2π/3. Let T 1 , T 2 , T 3 be the vertices of T 2 , let S j be the midpoint of the side opposite T j , and let F be the center of mass of T 2 . Observe that T 2 is invariant under reflection through the lines F T j , which divide T 2 into six congruent subtriangles (see Figure 3). Thus the first eigenfunction is invariant under these reflections, and we can recover it by restricting to the smaller triangle Ω, which has vertices F, T 1 , S 3 . The triangle Ω has a right angle at the vertex S 3 , and an angle of π/3 at the vertices F and T 1 . The first Dirichlet eigenfunction of T 2 , restricted to Ω, will have Dirichlet data on the edge T 1 S 3 and Neumann data on the edges F T 1 and F S 3 .
We first transform Ω to a planar domain (which we again denote as Ω) using stereographic project, sending S 3 to 0. This transformation sends F to 1 2 ( √ 6 − √ 2) and T 1 to i 2 ( √ 6 − √ 2). Now Ω is bounded by the two straight line segments joining 0 to 1 2 ( √ 6 − √ 2) and 0 to i 2 ( √ 6 − √ 2) and the circular arc joining 1 2 ( √ 6 − √ 2) to i 2 ( √ 6 − √ 2) which makes an angle of π/3 with the axes. Next we find a conformal transformation f which takes Ω to the half-infinite strip Under this transformation, the first eigenfunction u satisfies if ℜe z = 0 and ℑm z > 0, ∂u ∂n = 0, if 0 < ℜe z < π 2 and ℑm z = 0, or if ℜe z = π 2 and ℑm z > 0; where ∆ * = f * ∆ = ∆ z is the pulled back Laplacian. There are formulae for conformally mapping domains bounded by finitely many circular arcs [N] generalizing the Schwarz-Christoffel formula. Following Forsythe [F], the Schwarz triangle mapping z ∈ D or from sin 2 z ∈ {ℑz > 0} of the upper halfplane to w ∈ Ω is given by The group generated by reflections of Ω along its edges in fact tiles the sphere. The inverse function w → z 2 extends to a single sheeted cover and is invariant under the symmetry group, thus is a rational function of the plane. Thus we may compute f . Writing g(z) = cos 2/3 z, Pulling back under w = f (z), the conformal weight takes the form The branch cuts for the square and cube roots may be taken above the negative real axis. Thus g(D) lies in the fourth quadrant so the denominator in f is nonvanishing. Let G(z, z ′ ) denote the Green's function for the problem on D if ℜe z = 0 and ℑm z > 0, ∂u ∂n = 0, if 0 < ℜe z < π/2 and ℑm z = 0 or if ℜe z = π/2 and ℑm z > 0.
Pulling back by f , the eigenvalue problem for the triangle may be restated as finding an eigenvalue for the integral operator The key point is to notice that the operator has logarithmic and algebraic singularities at the points z ′ = 0, π/2 and z = z ′ . Thus we need to handle these singularities. The solution is zero along the imaginary axis, but free along the other two sides, which may be extended to functions to the plane which have odd reflection symmetry along the imaginary axis and even reflection symmetry along the other sides. We shall approximate u(z) in an m-dimensional space X m with the same symmetries. Also, noticing that the eigenfunction on T 2 behaves like 1 − dist 2 (z, T 2 ) at the vertex F , we actually have u ∈ Lip(D) and u decays algebraically at ∞. Choosing a basis {φ 1 , . . . , φ m } of X m , we shall compute the matrix of the transformation A ℓk = P ℓ Aφ k , whose largest eigenvalue µ m → 1/λ as m → ∞ and which is an upper bound λ ≤ 1/µ m [St]. The integral operator shall be computed numerically via sinc quadrature, which can handle such mild singularities.
The computation shows that λ 1 (T 2 ) ≈ 5.159 . . . which is above the critical value. This is computational evidence that the exit time has finite expectation. We present a table of a few the computed eigenvalues for approximations in spaces of given dimension, coming from the Sinc-Galerkin collocation scheme described.