The Escape model on a homogeneous tree

There are two types of particles interacting on a homogeneous tree of degree d + 1. The particles of the first type colonize the empty space with exponential rate 1, but cannot take over the vertices that are occupied by the second type. The particles of the second type spread with exponential rate \lambda. They colonize the neighboring vertices that are either vacant or occupied by the representatives of the opposite type, and annihilate the particles of the type 1 as they reach them. There exists a critical value \lambda_c =(2d - 1) + \sqrt{(2d -1)^2 -1} such that the first type survives with positive probability for \lambda<\lambda_c, and dies out with probability one for \lambda>\lambda_c. We also find the growth profile which characterizes the rate of growth of the type 1 in the space-time on the event of survival.


Introduction
We consider a model of a predator-prey type which we call the Escape model. There are two entities growing on the vertices of a homogeneous tree T d = T of degree d + 1. The entities may be thought of as biological species, political parties or manufacturers competing on a market. The second entity dominates the first in the sense that the representatives of the second entity can take over the vertices occupied by the representatives of the first entity but not vice versa. It is also assumed that the second entity grows faster. We are interested in the possibility of the long-term coexistence of the species which occurs when the first species survives.
At each time t each site of the tree is occupied by at most one representative of either the two entities. We refer to the representatives of the entities as particles of types 1 and 2 respectively. If a site is not occupied, it is said to be vacant. The dynamics of the process is specified by exponential transition rates. A vacant site gets colonized by a particle of type 1 with exponential rate equal to the number of nearest neighbors of type 1. The sites that are either vacant or occupied by type 1 flip to 2 with the rate λ > 1 times the number of neighbors of type 2. If a vertex is occupied by a particle of type 2, the particle stays at the vertex forever. We assume that at time zero there are finitely many particles of each type.
The sets of sites occupied by particles of types 1 and 2 at time t ≥ 0 are denoted by A(t) and B(t) respectively. The event ∩ t≥0 {A(t) = ∅} is referred to as the event of survival of type 1. If at time zero the cluster A(0) is surrounded by B(0), then all particles of type 1 eventually die out with probability one. Thus, we are only interested in the initial configurations for which there exist a vertex x and an infinite geodesic segment γ x,∞ such that x ∈ A(0) and γ x,∞ ∩ B(0) = ∅. All such configurations are referred to as non-trivial configurations. The question of interest is for what values of λ type 1 survives with positive probability in the long term. Theorem 1. There exists a critical value λ c = (2d − 1) + (2d − 1) 2 − 1 such that, for all λ ∈ (1, λ c ) and for all non-trivial finite configurations (A(0), B(0)), type 1 survives with positive probability. For all λ ∈ (λ c , ∞), type 1 dies out with probability one.
For c > 0 denote by M n (n/c) the number of vertices x ∈ A(n/c) at distance n from the root ρ. From the results of Sections 4 and 5, it follows that lim 1 n log (E M n (n/c)) = −g(c). where The function g(c) is referred to as the growth profile of type 1. The growth profile was introduced in Lalley [3] (in a slightly different form) to study the weakly supercritical contact process on a homogeneous tree. The function g(c) has a unique minimum at c 0 , is strictly decreasing on (0, c 0 ) and strictly increasing on (c 0 , ∞). For all λ ∈ (1, λ c ), we have g(c 0 ) < 0, and let r 1 and r 2 , with 0 < r 1 < r 2 , be the two solutions of g(c) = 0.
Theorem 2. Let λ ∈ (1, λ c ). For every ǫ > 0 and all large t, the particles of type 1 are concentrated in the annulus of radii (r 1 − ǫ)t and (r 2 + ǫ)t centered at the root. For every c ∈ (r 1 , r 2 ), almost surely on the event of survival of type 1, To investigate the Escape model, we consider the simplest growth model, the Richardson model, presented in Section 3.

Preliminaries.
2.1. A homogeneous tree. A homogeneous tree T d = T of degree d + 1 is an infinite tree such that every vertex has exactly d + 1 nearest neighbors. A distinguished vertex is called root and denoted by ρ. For every two vertices x and y of the tree, denote by |x, y| the number of edges in the shortest path from x to y (the path without loops). If x is the root, then we simply write |y|. Note that |·, ·| is a metric on T . For every vertex x ∈ T , a geodesic segment γ x,∞ is an infinite path in T beginning at x and having no loops. Define by T + (x), the set of all vertices y such that the shortest path connecting y with the root ρ goes through x. Consider a geodesic segment γ x,∞ such that γ x,∞ ∈ T + (x). Given an integer m, consider also a sequence of vertices (y k ) k≥0 on the geodesic segment γ x,∞ such that |y k | = mk + |x| for all integers k ≥ 0. For every k ≥ 1, y k−1 is called the m-predecessor of y k in T . Let D(x, r) = {y ∈ T : |x, y| ≤ r} be the closed disk of radius r ∈ (0, ∞) centered at x, and let C(x, r) = {y ∈ T : |x, y| = r} be the circumference of that disk. If x is the root, then we write D r and C r . Note that for all integers n ≥ 1, the number of vertices in C n is (d + 1)d n .

2.2.
Construction of the Escape process. The Escape process can be built using a percolation structure as follows. For each ordered pair of neighboring vertices x and y in T , define two independent Poisson processes with respective rates 1 and λ − 1, and respective occurrence times {T x,y n : n ≥ 1} and {U x,y n : n ≥ 1}. Set T x,y 0 = 0 and U x,y 0 = 0, and make these Poisson processes independent from pair to pair. Consider T × R + . Arrows are drawn from x to y at the occurrence times T x,y n and U x,y n . We say that there is a directed path in T × R + from (x 0 , s 0 ) to (x n , s n+1 ) if there is a sequence of times s 0 < s 1 < .. < s n+1 and sequence of vertices x 0 , x 1 , .., x n so that for each j, 1 < j ≤ n, there is an arrow from x j−1 to x j at time s j . A type 1 path is a directed path that uses only arrows generated by the Poisson processes T n . Fix an initial configuration A(0), B(0) and erase all arrows that lie only on paths that begin at points (x, 0) such that x ∈ (A(0) ∪ B(0)). For every vertex y we say that y ∈ B(t) if and only if there is a vertex z ∈ B(0) and directed path from (z, 0) to (y, t) in the modified percolation structure. Define A(t) to be the set of vertices y such that y ∈ B(t) and there is a type 1 path (in the modified percolation structure) that ends at (y, t) and starts at (z, 0) for some z ∈ A(0).

The Richardson model on a homogeneous tree
The Richardson process on T with parameter λ > 0 is a continuous time Markov process R(t) on the set of finite subsets of T . We say that a vertex x is infected (or occupied) at time t if x ∈ R(t), and is vacant otherwise. The process develops according to the following rules: if a vertex gets infected, it stays infected forever and starts infecting unoccupied neighboring vertices with rate λ, i.e. that the infection times have exponential distributions with parameter λ. The infection times are all independent. Consequently, a vacant site becomes infected with the rate λ · (number of infected neighbors).
The model with parameter λ = c can be obtained from the model with parameter λ = 1 by running the later process at speed c. Therefore we treat just the case λ = 1.
Without loss of generality consider the initial configuration where at time zero the only occupied site is the root R(0) = {ρ}. The main questions were how fast the infected set grows and what limiting shape the infected set has. For the tree T the number of vertices in the disk D n grows exponentially with n (whereas for Z d it has polynomial growth). Consequently, the behavior of the Richardson process on T is different from the behavior of the model on the integer lattice (see [4] and [1]). The infected region R(t) still grows linearly with time, but there are constants a and b, with a < b, such that as t goes to infinity we can classify two subregions: a completely infected subregion, having approximately the shape of a ball D ta , and a partially infected subregion, having approximately the shape of a ring D tb \ D ta . Consequently, there are two speeds: the speed of invasion b, indicating how fast the infection spreads, and the speed of occupation a, governing the rate of growth of the region that is completely covered by the infection.
for c ∈ (0, ∞), and let a and b, with 0 < a < 1 < b, be the two roots of the equation f (c) = 0. Then a = sup{a ′ } and b = inf{b ′ }, where the sup and inf are taken over all a ′ and b ′ satisfying Let N n (t) be the number of vertices at distance n from the root that are infected at time t, and let F n (t) be the number of vertices in C n that are not infected at time t. We compute asymptotic values of N n (t) and F n (t) as n goes to infinity and t = n/c for different values of c.
For all c ∈ (a, 1), Proof of Proposition 1. It is enough to verify that for every ǫ ∈ (0, a) Proof of equation (4). For every vertex x on the tree, let T (x) be the (random) time at which the vertex gets infected. Consider an arbitrary vertex at distance n from the root, and denote it by x n . Fix c > 1. Then we have the following estimate: where o(n)/n → 0 as n → ∞.
The estimate follows from Cramér's theorem for i.i.d. random variables (see [2]) and from the fact that T (x n ) is distributed as a sum of n i.i.d. exponentials with parameter 1. Observe that E N n (n/c) = (the total number of vertices in C n ) · P(x n ∈ R(n/c)) where f was defined in (1). Similarly, for c < 1 we have: Observe that f is strictly decreasing on (0, 1) and strictly increasing on (1, ∞), with unique minimum at c = 1. Moreover f (1) < 0 and f (0+) = f (∞−) = ∞. Thus, there are just two roots a and b of the equation f (c) = 0, such that a < 1 < b. Hence, for every ǫ > 0, we have that P ∃x ∈ C n : T (x) ≤ n b + ǫ ≤ E N n (n/(b + ǫ)).
Consequently the left side is summable, and by Borel-Cantelli lemma we have P ∃N < ∞ : ∀n > N and ∀x ∈ C n , T (x) > n b + ǫ = 1.
Finally, observe that this is equivalent to (the events are identical).
The limits lim d→∞ a(d) log d = 1, immediately follow from the fact that a and b are the roots of The phenomenon is easily anticipated. Since for larger d's there are more directions for the infection to spread around, it is natural that the invasion speed is strictly increasing with d. For the same reason, the occupation speed decreases to zero (the number of vertices in D n grows unboundedly with d).
Proof of Proposition 2. We only prove (2). The proof of (3) is identical. First we claim that for any ǫ > 0 lim sup 1 n log (N n (n/c)) ≤ −f (c) + ǫ almost surely. By Markov's inequality and (7), and the claim follows by Borel-Cantelli lemma. Let α = 1 c . To finish the proof, it is enough to show that, for an arbitrarily small ǫ > 0, with probability 1 Recall that 1 < c < b, so f (1/α) < 0. Assume that ǫ is small enough to satisfy f (1/α) + ǫ < 0. To make the notation less complicated, let By the continuity of f , there exists an ǫ 1 > 0 such that Then, for every fixed w > 1, there exists an integer m > 0 large enough such that Choose m to satisfy (12). Fix an arbitrary vertex x of the tree and consider a geodesic segment γ x,∞ ∈ T + (x). Consider a sequence of vertices (y i ) i≥0 on the geodesic segment γ x,∞ such that |y i | = mi + |x| for all integers i ≥ 0. Note that infection times T (y i ) are increasing in i. For every pair of non-negative integers n 1 and n 2 such that n 1 < n 2 , say that y n 2 is an m-descendant of y n 1 if, for all integers i ∈ [n 1 , n 2 ), Note that (Z k (x)) k≥0 is a Galton-Watson process with mean offspring number Claim 1. For every vertex x, almost surely on the event of (Z k (x)) k≥0 survival, there exists a (random) K < +∞ such that, for all k > K, Proof. The first inequality follows from the fact that for sufficiently large k's (α − ǫ 1 )(mk + |x|) ≥ (α − 2ǫ 1 )mk + T (x). To obtain the second inequality, observe that, for all large enough k, w k ≥ d m µ |x| , and hence, E Z k (x) > (µ m · w) k ≥ µ mk+|x| · d m . A standard theorem from the elementary theory of Galton-Watson processes states that if E Z 1 (x) > 1 and the variance of Z 1 (x) is finite, then on the event of survival Z 1 ) k converges almost surely to a positive random variable. Thus, the second inequality is obtained by direct application of (13).
Therefore we proved that, almost surely on the event of the survival of (Z k (x)) k≥0 , (11) is true. To show that (11) holds with probability one, observe that, for each integer j > 0, there are (d + 1)d mj−1 Galton-Watson processes (Z k (x)) k≥0 with |x| = mj. Let S j be the event of non-extinction for at least one of the processes. Since the Galton-Watson processes are independent, the probability of S j tends to 1 as j tends to +∞. Moreover, for each j > 0, S j ⊂ S j+1 which guarantees the almost sure result.

Two independent Richardson models.
To motivate the proof of Theorem 1 consider two independent Richardson processes R 1 (t) and R λ (t) with respective rates 1 and λ > 1. The processes have initial configurations R 1 (0) = R λ (0) = {ρ}, and are built on a homogeneous tree T . For c ∈ (0, ∞), estimate the expected number of vertices in (R 1 (n/c) \ R λ (n/c)) ∩ C n , that is the number of vertices at distance n from the root that are occupied by R 1 but not by R λ at time n/c. Let x n be a vertex with |x n | = n, u n (n/c) = P [x n ∈ {R 1 (n/c) \ R λ (n/c)}], and V n (n/c) = # {x n : x n ∈ R 1 (n/c) \ R λ (n/c)}.
Proof of Theorem 2 (sketch). Since A(t) can not grow faster than the infected set in the Richardson model with rate 1, from the results of Section 4 (use (15)) it follows that for any initial configuration (A(0), B(0)) and all sufficiently large n E M n (n/c) ≤ exp {−n · g(c) + nǫ} .
Thus, using the same lines of argument as in Proposition 3, it may be shown that for all large t Furthermore, for every c ∈ (r 1 , r 2 ) and any ǫ > 0 lim sup 1 n log (M n (n/c)) ≤ −g(c) + ǫ. Note that for every non-trivial configuration (A(0), B(0)), for all large t and all x ∈ A(t) we have T + (x) ∩ B(t) = ∅. Furthermore, almost surely on the event of type 1 survival, for every integer m there exist 0 < t < ∞ and a vertex x with m-predecessor y such that x ∈ A(t) and B(t) ∩ T + (y) = ∅. Define Z 1 (x) to be a subset of C |x|+m ∩ T + (x) such that z 1 ∈ Z 1 (x) if and only if there is a type 1 path from (x, t) to (z 1 , t + m/c) and there are no directed paths from (y, t) to (x, t + m/c). In general, for k ≥ 2, Z k (x) is defined as a subset of vertices in C |x|+mk ∩ T + (x) such that z k ∈ Z k (x) if and only if (1) z k−1 ∈ Z k−1 (x) where z k−1 is the m-predecessor of z k ; (2) there is a type 1 path from (z k−1 , t + (k − 1)m/c) to (z k , t + km/c); (3) there are no directed paths from (z k−2 , t + (k − 1)m/c) to (z k−1 , t + km/c) where z k−2 is the m-predecessor of z k−1 .
Let Z k (x) be the cardinality of the set Z k (x). It is clear from the definition that (Z k (x)) k≥0 is a Galton-Watson process with the mean offspring number Thus, for all sufficiently large m, E[Z 1 (x)] > 1. Note that M |x|+mk (t+km/c) dominates Z k (x), and hence, on the event of nonextinction of (Z k (x)) k≥0 , (16) holds. Observe that for every m, almost surely on the event of survival of the type 1, there are infinitely many vertices x at which the Galton-Watson processes (Z k (x)) k≥0 can be originated. Hence, (16) holds almost surely on the event of nonextinction of the first type. This finishes the proof of Theorem 2.
Proof of Theorem 1 (sketch). Fix λ ∈ (1, λ c ), and consider any non-trivial initial configuration. With positive probability, there exists a vertex x and a Galton-Watson process (Z k (x)) k≥0 (constructed in the proof of Theorem 2) with E Z 1 (x) > 1. Hence, the Galton-Watson process survives with positive probability and so does type 1. Consider the case λ ∈ (λ c , ∞). Since A(t) is dominated by the Richardson model with rate 1, by Proposition 3 in Section 4 type 1 dies out almost surely.