Growth Competitions on Spherically Symmetric Riemannian Manifolds

We propose a model for a growth competition between two subsets of a Riemannian manifold. The sets grow at two different rates, avoiding each other. It is shown that if the competition takes place on a surface which is rotationally symmetric about the starting point of the slower set, then if the surface is conformally equivalent to the Euclidean plane, the slower set remains in a bounded region, while if the surface is nonpositively curved and conformally equivalent to the hyperbolic plane, both sets may keep growing indefinitely.


Introduction
Consider two subsets A, B of, say, the Euclidean plane, which evolve over time, A = A t , B = B t , according to the following simple rules: both sets begin as singletons, A 0 = {q}, B 0 = {p} for some points p = q, and expand at rates λ > 1 and 1 respectively, without intersecting each other. If a point belongs to one of the sets at a certain time, then it remains in the set forever. What will the sets A ∞ := t A t and B ∞ := t B t look like?
This growth competition can take place on essentially any metric space. Its precise formulation is given in the following section, and its existence and uniquness on complete Riemannian manifolds is established. We then study growth competitions on Riemannian manifolds which are spherically symmetric about the point p. The cases of the Euclidean plane and the hyperbolic plane exhibit contrasting behaviors: on the Euclidean plane, the faster set A will eventually trap the set B in a bounded region, while on the hyperbolic plane, if the two sets begin sufficiently far apart, then there is coexistence, i.e., both sets keep expanding forever. In fact, we show: Theorem 1. Let M be a two-dimensional, complete, non-compact Riemannian manifold, which is rotationally symmetric about p ∈ M . Let q ∈ M and λ > 1, and let {A t , B t } be the corresponding growth competition. Then (1) If M is parabolic, then B ∞ is bounded.
(2) If M is hyperbolic and nonpositively curved, then B ∞ is unbounded if d(p, q) is sufficiently large.
The problem of determining the shapes of the competing sets was suggested by Itai Benjamini [1], and is loosely inspired by probabilistic competitions on Z d and other graphs, see [2], [3]. A related problem, concerning a strategy to control a forest fire on the Euclidean plane, was introduced by Bressan, see [4], [5], [6]. In [7], Bressan's fire confinement problem is treated using the apparatus of viscosity solutions to the Hamilton-Jacobi equation. This could be the appropriate framework for dealing with problems such as the one introduced here. However, as we are interested mostly in the relationship between coexistence and the underlying geometry, we have managed to define the competition in a way which enables an elementary proof of existence and uniqueness of the solution, while still capturing the essence of the problem. Given the connection presented here between the conformal type of a simply-connected surface, and the outcome of growth competitions on it, it is now natural to ask: Problem. Prove or disprove each of the following statements.
(1) On a hyperbolic surface, for every λ > 1 there is some choice of p, q such that B ∞ is unbounded. (2) On a parabolic surface, for every λ > 1 and every L > 0, there is some choice of p, q such that d(p, q) > L and B ∞ is bounded.
Acknowledgements: I would like to express my gratitude to Itai Benjamini for offering me this problem and assisting in writing the paper, and to Bo'az Klartag for pointing me to the paper [8], as well as reviewing the proof thoroughly and suggesting improvements.
Partially supported by a grant from the Israel Science Foundation (ISF).

Growth competitions
Let X be a metric space, let p = q ∈ X, and let λ > 1. Let {A t , B t } t≥0 be two families of subsets of X. Say that a path γ : . The pair A t , B t will be called a growth competition between p and q, if and for all t > 0: where the union on the left is over paths avoiding B, and the union on the right is over paths avoiding A.
Denote also In the rest of this section, we prove existence and uniqueness of growth competitions on complete Riemannian manifolds. Fix a manifold (M, g), two points p = q ∈ M , and λ > 1.
The following lemma is quite evident from the definitions.
Lemma 1. Let {A t , B t } be a growth competition between p and q.
Proof. Let 0 < t < ∞ and let x ∈ A t . There exists a path γ : [0, t] → M avoiding B such that x = γ(t 0 ) for some 0 ≤ t 0 ≤ t, and since γ| [0,t0] is also a curve avoiding B, x ∈ A t0 . Let η : [0, t ) → M be a curve avoiding A. We claim that η(s) = x for all s ∈ [0, t ). Indeed, Thus η(s) = x for all s ∈ [0, t ), and since η is an arbitrary curve avoiding A, x / ∈ B t . This proves that A t , B t are disjoint for all t, t ≥ 0, and it follows that A ∞ and B ∞ are disjoint.
Denote by d the distance function of (M, g). For x ∈ M and R > 0, denote by B(x, R) = {y ∈ M | d(x, y) < R} the open ball of radius R centered at x. Another fact which follows trivially from the definitions is the following.
We shall now construct a certain subset of M and prove that it must coincide with B ∞ . This will leave us with a unique candidate for a growth competition. If the set S is open, then by the Arzelà-Ascoli theorem, the metric space (M \ S, d S ) is a geodesic metric space, i.e., for each x, y ∈ M \ S there exists a path γ realizing their intrinsic distance. If M \ S is not path-connected, then d S might attain the value ∞.
A subset S ⊆ M is said to be star-shaped about a point x 0 ∈ S if for every x ∈ S, any minimizing geodesic joining x 0 and x lies inside S.
Proof. For n = 0 there is nothing to prove. Let n ≥ 1 and let x ∈ Ω n . We must show that a minimizing geodesic joining p and x lies in Ω n . By induction we may assume that x / ∈ Ω n−1 , so that d Ωn−1 (x, q) > λd(x, p). Let γ be a unit-speed minimizing geodesic from p to x. By induction, there exists some t 0 > 0 such that γ(t) ∈ Ω n−1 exactly when t < t 0 , and The following lemma states the key property of Ω.

Lemma 6.
For all x ∈ M \ Ω, with equality if and only if x ∈ ∂Ω.
Proof. Let x ∈ ∂Ω. For each n ≥ 1, let x n ∈ Ω n be a point of minimal distance to x. Clearly x n → x as n → ∞. The point x n must lie in ∂Ω n , for otherwise we would have x ∈ Ω n ⊆ Ω. Now, on one hand, by (6), and taking n → ∞ we get λd(x, p) ≤ d Ω (x, q). On the other hand, again by (6), there are curves γ n joining q and x n of length λd(x n , p) and lying outside Ω n−1 , and by the Arzelá-Ascoli theorem, a subsequence of them converges to a curve γ lying outside Ω n−1 for all n, and thus outside Ω, and of length λd(x, p). This implies that d Ω (x, q) ≤ λd(x, p), and together with (8) we have equality. Now let x ∈ M \ Ω, and let γ by a unit-speed minimizing geodesic from p to x. Since Ω is star-shaped, we there exist 0 < t 0 < t 1 such that γ(t 1 ) = x and γ(t) ∈ Ω exactly when t < t 0 . Let x = γ(t 0 ). Then x ∈ ∂Ω, and since γ| [t0,t1] lies outside Ω, we get We now have what we need in order to argue that Ω coincides with B ∞ , and write down the solution to the competition.
Proof. For the inclusion Ω ⊆ B ∞ , we must show that Ω n ⊆ B ∞ for all n ∈ N. For n = 0 there is nothing to prove. Let n ≥ 1. Let γ be a unit speed geodesic emanating from p, and suppose that γ does not avoid A.
Since Ω n is star-shaped about p, this proves that Ω n ⊆ B ∞ for all n ∈ N whence Ω ⊆ B ∞ .
We have shown that B ∞ = Ω. Lemma 6 implies that M \ Ω is path-connected, so for every x ∈ M \ Ω, there is a curve η lying outside Ω and joining q to x. Since B ∞ ⊆ Ω, the curve η avoids B and therefore x ∈ A ∞ . On the other hand, by Lemma 1, A ∞ ⊆ M \ Ω.
Corollary 1. For each p, q ∈ M and λ > 1, there is a unique growth competition, given by where B Ω (x, R) denotes the closed ball of radius R in the metric space (M \ Ω, d Ω ).
Proof. Proposition 1, Lemma 1 and the fact that Ω is star-shaped about p, imply that any growth competition must take the form (9). So it remains to show that this is indeed a growth competition. Clearly any path starting at p and staying inside Ω avoids A, and any path starting at q and not intersecting Ω avoids B. So we have the inclusions ⊆ in (1).
For the inclusions ⊇, first let γ be a curve avoiding A. The same argument as in the proof of Proposition 1 implies that γ remains inside Ω, and since it is 1-Lipschitz, it is contained in B(p, t). Now let γ : [0, t] → M be a curve avoiding B; we argue by induction that it does not enter Ω n . Again n = 0 is trivial. Let n ≥ 0 and assume that γ(t 0 ) ∈ Ω n+1 . By the definition of Ω n+1 , and since γ does not intersect Ω n by induction, Length(γ| [0,t0] ) > λd(x, p). Since γ is λ-Lipschitz, it follows that t 0 > d(x, p), whence γ(t 0 ) ∈ B t0 , which is a contradiction to the assumption that γ avoids B. Thus γ(s) / ∈ Ω n for all s ∈ [0, t] and all n ∈ N, whence γ(s) / ∈ Ω for all s ∈ [0, t] . Since γ is λ-Lispchitz, γ(s) ∈ B Ω (q, λt) for all s ∈ [0, t]. This finishes the proof of the inclusions ⊇ in (1).

Coexistence on spherically symmetric manifolds
We now restrict our attention to growth competitions taking place on complete, noncompact manifolds which are spherically symmetric about the point p. By spherically symmetric we mean that the metric takes the form (10) g = dr 2 + G(r) 2 dθ 2 where (r, θ) ∈ (0, R) × S n−1 are polar normal coordinates centered at p, dθ 2 is the standard metric on S p M ∼ = S n−1 , where S x M denotes the unit sphere of the tangent space to M at x, and G is a smooth positive function satisfying G → 0 and G/r → 1 as r → 0. Since M is noncompact, the coordinates (r, θ) are global, i.e. R = ∞, and M is diffeomorphic to R n . Set := d(p, q).
Let u 0 ∈ S p M denote the initial velocity vector of the unit-speed geodesic from p to q, so that in polar coordinates, q = ( , u 0 ). Coexistence in this setting is related to the convergence of the integral (11) dr.

Proposition 2.
Suppose that M is spherically symmetric about p, and that I = ∞. Then B ∞ is bounded.
Remark. In particular, if M = R n then B ∞ is bounded.
Proof. We prove that on every geodesic ray emanating from p, there is a point which can be reached by a path avoiding B within some time T independent of θ. Since B ∞ is star-shaped, the proposition will follow because then B ∞ ⊆ B(p, T ). Let θ ∈ S p M and let η be a unit-speed great circle in S p M with η(0) = u 0 and η(τ ) = θ for some τ ≤ π. Define a path γ : [0, T ] → M in polar coordinates by Since α tends to ∞ with t, we can choose T such that α(T ) = τ , and therefore γ(T ) = ( + T, η(τ )) = ( + t, θ) lies on the ray from p with direction θ. The path γ avoids B; indeed, γ(0) = ( , u 0 ) = q, γ is λ-Lipschitz because The condition I = ∞ is not necessary. First, observe that if is small, then the competition resembles a Euclidean competition and we cannot expect coexistence, no matter what G is. Second, having I < ∞ does not prevent M from containing spheres centered around p with arbitrarily large radius yet arbitrarily small surface area, and if q lies on such a sphere then the set A will conquer the entire sphere quickly, trapping B within the ball it bounds. Thus, at least as long as p remains at the origin, some extra assumption is needed in order to enable coexistence. Motivated by Itai Benjamini's observation [1] that coexistence is possible on Gromov-hyperbolic spaces, we add the assumption that M is nonpositively curved, i.e. that all its sectional curvatures are nonpositive. Remark. Note that if the sectional curvature of M is bounded from above by a negative constant then automatically I < ∞. In this case one can use the argument from [1] to prove that coexistence is possible.
Recall that B ∞ = Ω by Proposition 1. Since M is nonpositively curved, any two points are joined by a unique minimizing geodesic. Say that a point x ∈ M is visible if the unique minimizing geodesic joining q and x does not intersect Ω, and that x is visible n if this geodesic does not intersect Ω n .
Since Ω is open and star-shaped, there is a function f : S p M → (0, ∞] such that Ω is given in polar normal coordinates by the relation (note that f may attain the value ∞). Similarly, let f 1 denote the function corresponding to the star-shaped set Ω 1 .
Proof. Let γ be a unit-speed geodesic emanating from q. The function is strictly convex since M is nonpositively curved, unless γ passes through p, in which case h(t) = λ| − t| − 1. In both cases, h vanishes at most twice, and is negative between its zeros; hence γ −1 (Ω 1 ) is an interval (possibly empty) and γ −1 (∂Ω 1 ) consists of at most two points; if it consists of exactly two points t 1 < t 2 , then only γ(t 1 ) is visible 1 , since γ| [0,t2] intersects Ω 1 and γ| [0,t1] does not. Let be the expression of γ in polar coordinates. Since M is spherically symmetric, the curve η lies on a great circle in S p M , and (η(t), η(0)) = (η(t), u 0 ) is strictly increasing. Therefore, if x 1 , x 2 are as in the statement of the lemma, and t i satisfy γ(t i ) = x i , then t 1 < t 2 , hence x 1 is visible 1 while x 2 is not.
Proof. Identifying M with the tangent space to M at p, the set Ω 1 is invariant under orthogonal transformations fixing the line through p and q. By this symmetry, it suffices to consider a unit-speed great circle η : [0, T ] → S p M joining −u 0 to u 0 and to show that there exists some t 0 ∈ [0, T ] such that (f 1 (η(t)), η(t)) is visible 1 exactly when t ≥ t 0 , and that (η(t 0 ), u 0 ) ≤ c(λ) < π.
We will show that provided that is larger that some L > 0 independent of θ. This proves that f (−u 0 ) = ∞, whence Ω(= B ∞ ) contains the ray {θ = −u 0 } and is therefore unbounded.
In the case dimM = 2, the surface M is either hyperbolic or parabolic (since it is rotationally-symmetric and non-compact). In the former case I < ∞, and in the latter I = ∞, see Milnor [8]. Theorem 1 now follows.
Remark. The proof of Proposition 3 enables us to identify the shape of Ω in the spherically symmetric, nonpositively curved case (both when I = ∞ and when I < ∞). Suppose for simplicity that dimM = 2, in which case, by passing to normal coordinates, we may assume that M = (R 2 , g) for some rotationally symmetric metric g, and p = (0, 0). There exists 0 ≤ a < π such that (f (θ) cos θ, f (θ) sin θ) ∈ ∂Ω is visible exactly when −a ≤ θ ≤ a, and ∂Ω ∩ {|θ| ≤ a} coincides with ∂Ω 1 . The remainder ∂Ω ∩ {a ≤ |θ| ≤ π} is the union of two speed-λ curves whose distance from the origin p increases at rate 1. In particular, if M is the Euclidean plane, then a = π/2, ∂Ω ∩ {x ≥ 0} = ∂Ω 1 ∩ {x ≥ 0} is a circular arc, and ∂Ω ∩ {x ≤ 0} is the union of two logarithmic spirals (see Figure 1). If M is the unit disc with the Poincaré metric, then the shape Ω 1 is not a circle, but the two curves are still logarithmic spirals.