The upper threshold in ballistic annihilation
aa r X i v : . [ m a t h . P R ] M a y THE UPPER THRESHOLD IN BALLISTIC ANNIHILATION
DEBBIE BURDINSKI, SHREY GUPTA, AND MATTHEW JUNGE
Abstract.
Three-speed ballistic annihilation starts with infinitely many particles on thereal line. Each is independently assigned either speed-0 with probability p , or speed- ± p ≤ p c = 1 / p c ≤ . p c ≤ . . p c > Introduction
Two decades ago, physicists devoted considerable attention to a simple but difficult toanalyze process called ballistic annihilation [EF85, CPY90, BF95, DRFP95, MP94, KRL95,Red97, Tri02]. Particles are placed on the real line according to a unit intensity Poisson pointprocess and each is independently assigned a speed according to a probability measure ν .After this assignment, the model is deterministic; particles move at their speed and mutuallyannihilate upon colliding. The canonical example is with speeds from {− , , } sampledaccording to the symmetric measure ν = 1 − p δ − + pδ + 1 − p δ . (1)Physicists refer to this as an A + A → ν [BNRL93]. The authors’motivation was to understand “intriguing features” in the decay kinetics of irreversibleaggregation , A i + A j → A i + j , which has been used to model coalescence of fluid vortices[Mcw84] and planet formation by accretion [WD89]. The authors give heuristics for thedecay rate of particles, and conjecture it responds continuously to perturbations in thespeed measure.The followup work [KRL95] predicts more interesting behavior in ballistic annihilationwith discrete speeds. It is thought that the process undergoes an abrupt phase transition.Consider ballistic annihilation with the measure from (1). We will call speed-0 particles inert and speed- ± active . Let θ t ( p ) be the probability an inert particle survivesup to time t , and θ ( p ) = θ ∞ ( p ) be the probability it is never annihilated. Krapivsky et. al.infer in [KRL95] that there is a critical value p c such that θ ( p ) = 0 for p ≤ p c , and above p c it holds that θ ( p ) >
0. They conjecture p c = 1 / θ t : θ t ( p ) ∼ C p t − , p < p c Ct − / , p = p c − p − / , p > p c , with C p = 2 p (1 − p ) π and C = 2 / / . A simple heuristic is given in [DRFP95] for why p c = 1 /
4. Active particles move towardsone another at relative speed 2, while the gap between a moving and inert particle is coveredat relative speed 1. Thus, collisions between active particles ought to occur twice as often asthose between inert and active particles. If we look at all of the collisions in a large interval,then each is one of three possibilities:(0 , − , (1 , , and (1 , − . (2)Doubling the (1 , − p = 1 / + ± p . For p < / − /
2. At criticality it is claimed to be −
1, and for p > p c they infer that thesurvival time decays at an exponential rate. Droz et. al. in [DRFP95] provide a nearly com-plete proof of these conjectures (including those for θ t ( p ) predicted by [KRL95]). However,some steps are not rigorous and the argument gives little intuition. These formulas comefrom a complicated differential equation involving the distance between neighbor particlesat time t . Krapivsky et. al. point out in [KRL95] that there is still need for methods “thatwould provide better intuitive insights into the intriguing qualitative features of ballisticannihilation.” For example, there is no probabilistic proof that p c = 0, let alone is equal to1 / p c ≤ . + p c ≤ /
3, andoutline an approach to prove p c ≤ . p , inert particles donot survive.In this article, we consider ballistic annihilation with an inert particle at the origin, andparticles placed on Z + with i.i.d. speeds sampled according to ν from (1). To distinguishour case from ballistic annihilation with exponentially distributed spacings, we define ψ ( p )as the probability the speed-0 particle at the origin is never annihilated, and p ′ c = inf { p : ψ ( p ) > } . HE UPPER THRESHOLD IN BALLISTIC ANNIHILATION 3
Understanding survival of inert particles appears to be equally interesting and challengingwhether the spacings are deterministic, or exponential(1) distributed. This is supportedby recent findings [BM17] for the closely related bullet process with finitely many particlesand a non-atomic probability measure on speeds. They find that the law for the number ofsurviving particles is independent of the initial spacings. Our main result is an improvedbound for p ′ c . Theorem 1. p ′ c ≤ . . For ballistic annihilation with unit spacings, triple collisions may slightly change thecritical threshold. Let q = (1 − p ) /
2, so that the probability three consecutive particles triplecollide is pq (i.e. a (1 , , −
1) configuration). This is ≈ .
07 when p = 1 /
4. With exponentialspacings, the process loses one inert and one active particle whenever this configurationoccurs. However, with the triple collision, an extra active particle is destroyed. The heuristicat (2) predicts that, after one time unit, the density of 0-particles is w = p − pq + pq . The 2 pq term accounts for the configurations (1 ,
0) and (0 , − pq to preventdouble counting triple collisions. Similarly, the density of active particles is z = 2( q − q − pq ) . The heuristic at (2) suggest we solve w = z , which yields p ≈ . . Accordingly, weconjecture that p ′ c < . < p c . We do not think that p ′ c is much smaller than . , · , · , , · , · , − p and q , and so thecontribution will be very small.We prove this by relating survival in ballistic annihilation to survival of a Galton-Watsonprocess. The idea is that there is a random index η for which (i) when restricted to justparticles in [0 , η ] only inert particles survive, and (ii) the speeds of particles beyond η areindependent. If Z is the number of surviving inert particles in the process restricted to [0 , η ],then (ii) guarantees that each of these particles will independently spawn Z -distributedmore surviving inert particles. Because each new generation is determined independently,we obtain a Galton-Watson process that counts surviving particles.The construction of η is described in the proof of Proposition 5, and the equivalence of ψ ( p ) > ψ ( p ) > ⇐⇒ E Z > p -dependence in the expectation and write E Z in place of E p Z . Thesame reasoning we use to obtain (3) can be applied to the usual ballistic annihilation withexponential spacings. Thus, a similar equivalence holds for θ ( p ). In Remark 6, we brieflysketch how to adapt Proposition 5 to this setting.Once we have the equivalence at (3), we turn our attention to lower bounding E Z . Thereare certain random times before η at which we can stochastically lower bound Z . We thenuse a computer to estimate these probabilities. This is completely rigorous, but requires toomany calculations to be done by hand. DEBBIE BURDINSKI, SHREY GUPTA, AND MATTHEW JUNGE
Notation.
Depending on how specific we need to be, we will refer to particles with speeds ± active or as ± -particles . Because it will correspond to the original parentin a Galton-Watson process, we refer to the inert particle at the origin as the seed . Therandomness in ballistic annihilation is an initial vector X = ( X i ) ∞ i =0 with X = 0 and the X i for i ≥ ν from (1). These are the particle speeds.Let a i denote the particle initially at i . We will let a i ↔ a j mean that particles at i and j mutually annihilate. With discrete speeds and spacings, it is possible that three particlescollide simultaneously. Denote this with a i ↔ a j ↔ a k . We emphasize that writing a i ↔ a j does not preclude a third particle also being annihilated. That is { a i ↔ a j } ∩ { a i ↔ a j ↔ a k } 6 = ∅ . We will sometimes use the more specific notation a i a j for when the activeparticle from i annihilates the inert particle at j .2. An embedded Galton-Watson process
To show that inert particles survive with positive probability, it suffices to establish thatthe seed is never annihilated with positive probability. This is because an inert particle, say a n , has n particles to its left, and thus has some positive probability of not being annihilatedfrom that side. For example, all left particles are inert with probability p n − . Conditionalon surviving from the left, the probability that n is never annihilated by a particle from theright is the same as the probability the seed survives. This follows via the coupling thataligns the speeds to the right of the seed with those to the right of a n in two independentprocesses.That said, in this section, we develop a framework that lets us relate survival of the seedto non-extinction of a Galton-Watson process that counts surviving inert particles. This ismade explicit in Proposition 7. The construction hinges upon a renewal structure that hasnice monotonicity properties. It rests upon the following two lemmas. Lemma 2.
Let i < j and X j = − . The random variables ( X k ) k>j are ν -distributed andindependent of the event { a i ↔ a j } .Proof. The particles to the right of a j cannot influence the event { a i ↔ a j } , thus the speedsare independent of the event. (cid:3) If an active particle destroys an inert particle, then this induces a short range dependence.We know that all of the active particles that could reach the inert particle before its destroyerarrives must be annihilated. However, if we look sufficiently far away the particle speedsare once again independent.
Lemma 3.
Let i < j and X j = 0 . The random variables ( X k ) k>j +( j − i ) are independent ofthe event { a i a j } .Proof. It is elementary to work out that the particles beyond j + ( j − i ) cannot reach a j before a i does, thus their speeds are independent of the event. (cid:3) Remark . The speed X j +( j − i ) is independent of { a i a j } but possibly a j +( j − i ) triplecollides with a j .Conditioned on the event from Lemma 3 we will refer to [ i, j − i ] as a window of depen-dence . Given x = ( x , x , . . . , x n ), we define BA( x ) to be ballistic annihilation on R withparticles at 0 , , . . . , n where the particle at i has speed x i . Run BA( x ) until every collisionthat could occur has occurred (this takes at most n time units). Let ξ i ( x ) = x i { a i survives } + 2 · { a i is annihilated } (4) HE UPPER THRESHOLD IN BALLISTIC ANNIHILATION 5 and ξ ( x ) = ( ξ , . . . , ξ n ) so that the i th entry is the speed of a i if it survives, or 2 if a i isannihilated in BA( x ).We return to ballistic annihilation with the seed at the origin and a particle with a ν -distributed speed at each nonnegative integer. The last piece of notation we need is that X [ i, j ] = ( X i , . . . , X j ) is the restriction to the coordinates between i and j . We now explainthe renewal structure. The following proposition asserts that there exists a random index η such that only inert particles survive in BA( X [0 , η ]), and the particle speeds beyond η areindependent. Proposition 5.
There exists a random variable η ≥ such that either(i) η = ∞ and the seed survives, or(ii) η < ∞ and ξ ( X [0 , η ]) ⊆ { , } η and ( X i ) i>η are ν -distributed and independent of η and each other. Additionally, if η > , then the seed survives in BA( X [0 , η ]) .Proof. We remark that the last condition of (ii) for η > − η = 1, theparticle at η will not destroy the seed.We define η in terms of X and a random variable η for the distance we must look outfor the process to be completely independent of how a is annihilated when X = 1: η := { X = 1 } + { X = 1 } η . (5)Now we describe η . Consider ballistic annihilation with X = 0 and X = 1, and let γ be the index of the particle a ↔ a γ . First off, if γ = ∞ then set η = ∞ and the seedsurvives. Supposing γ < ∞ , if there is a triple collision, we take the larger of the twoindices (so necessarily X γ = − X γ = −
1, then we set η = γ . In thiscase, we have ξ ( X [0 , η ]) = { } × { } η , and the particle speeds beyond η are independentbecause of the renewal in Lemma 2.It gets more complicated when X γ = 0. Lemma 3 with i = 1 and j = γ tells us thatthere is an I := [1 , γ −
1] window of dependence on this event. Let X = X [0 , γ − τ = max { i ∈ I : ξ i ( X ) = 0 } be the starting location of the rightmost surviving inert particle. Note that the seed mustsurvive, so we know that τ is well defined and nonnegative. Also, let κ = min { τ ≤ i ≤ γ − ξ i ( X ) = 1 } be the first surviving 1-particle to the right of τ . If there is no such particle, set κ = 0 . We will consider this case separately in a moment.If κ >
0, then we let γ be such that a κ ↔ a γ . In words, γ is the index of the particlethat destroys the active particle a κ . As before, if there is a triple collision, then we takethe particle with larger index. If X γ = −
1, then the process renews and we set η = γ .If not, then a new window of dependence is induced by the event { a κ a γ } . We onceagain look at the furthest left surviving 1-particle and either set η equal to the index ofthe − η ; never halt, in which case η = ∞ ; or give some κ i = 0, which puts us in the casedescribed now for κ .If κ = 0, then there are no surviving active particles in BA( X ). However, it is possiblethat some 1-particles that annihilate inert particles in BA( X ) will survive longer when par-ticles from beyond X are introduced. For example, the survival of a 1-particle annihilated DEBBIE BURDINSKI, SHREY GUPTA, AND MATTHEW JUNGE by an inert particle at x is prolonged if a − X reaches x first.To account for this, we look at the collection of 1-particles from BA( X ) that are destroyedby inert particles J = { j : X j ∈ X , X j = 1 , a j a j ′ with X j ′ = 0 } . If J is empty, then we set η = 2 γ −
1. Otherwise, for each j ∈ J we augment X to alarger interval X ,j = X [ w j , z j ] defined to be the interval such that a − z j couldtriple collide (if it had a clear path) with a j and a j ′ . Set z = max j ∈ J z j to be the furthestright index that we must consider in order to know whether each collision a j a j ′ occursin the BA( X ).If all of the collisions counted by J occur, then we possibly gain some additional particlesin BA( X [0 , z ]). However, since all of the collisions in J still occurred we once again cannothave any surviving − X [0 , z ]). So, we repeat the procedure from some κ equal to the leftmost surviving 1-particle in BA( X [0 , z ]). This will either eventuallyterminate and we will obtain η , or never halt and set η = ∞ with the seed never beingreached by a − J does not occur because aparticle from [2 γ − , z ] reaches an inert particle first. Letting κ be the leftmost such1-particle, we then know that a κ will now survive longer when we introduce more particlesto the right. We set γ to be the index such that a κ ↔ a γ . This reinitiates the procedurewe have defined, which will either terminate at a renewal time η , or never terminate so that η = ∞ and the seed survives. (cid:3) Remark . A similar statement also holds when particles are placed according to a unitintensity Poisson process. Lemma 2 and Lemma 3 still hold in this setting, and thus we canfollow the same steps to obtain a renewal as in Proposition 5.Define the random variable Z = |{ i : ξ i ( X [0 , η ]) = 0 }| to be the number of surviving inertparticles. Because the process renews after η we can link the expected number of survivinginert particles in BA( X [0 , η ]) to ψ ( p ) via a Galton-Watson process. Proposition 7.
Let η be as in Proposition and Z ≥ be the number of surviving inertparticles from BA( X [0 , η ]) . It holds that ψ ( p ) > ⇐⇒ E Z > . Proof.
The random variable Z can be used as the offspring distribution for a Galton-Watsonprocess that counts surviving inert particles. Starting with the seed, Proposition 5 ensuresthat we have Z inert particles in [0 , η ] that survive from BA( X [0 , η ]). Moreover, the speeds( X i ) i>η are i.i.d. ν -distributed.We claim that each of these Z particles will (eventually) serve as a seed that spawns Z -distributed more surviving inert particles. We do this in a “depth first” manner. Considerthe rightmost surviving inert particle in [0 , η ]. Say it is at i . By construction there are nosurviving active particles in [ i, η ]. If i = η , then this is exactly the same initial configurationas with the seed. Even if i < η , we still obtain a Z -distributed number of surviving inertparticles before the process renews. This is because a − i . This happens irregardless of whether or not there is a gap ( i < η ) or not( i = η ). We continue generating Z offspring at each surviving inert particle and stop if theprocess goes extinct.The previous discussion implies that if the Z -distributed Galton-Watson process survivesthen the seed is never annihilated. Conversely, if ψ ( p ) >
0, then the seed survives with
HE UPPER THRESHOLD IN BALLISTIC ANNIHILATION 7 positive probability. This can only happen if the Galton-Watson process started from theseed does not going extinct. The result follows from the elementary fact that non-extinctionof a Galton-Watson process with positive probability is equivalent to E Z > (cid:3) Approximating E Z The complicated definition of η from Proposition 5 suggests it would be difficult tocalculate E Z explicitly. Even calculating the distribution of η seems beyond our reach.However, there are certain times before η at which we can obtain lower bounds on E Z . Ithelps to explain our approach in stages. First, we consider the effect of a .3.1. The effect of
BA( X [0 , . The simplest lower bound on Z is to look at what happensin BA( X [0 , X = −
1, then Z = 0, and if X = 0, then Z = 2. This gives Z (cid:23) { X = 0 } . (6)Thus, if p > /
2, we have E Z >
1. Equivalently, p ′ c ≤ /
2. This is a start, but not so inter-esting. The same statement could be proven by accounting for the number of inert particlesversus active particles with a p -biased random walk. Survival of the seed is equivalent tothe walk never returning to 0, which happens with positive probability when p > / η >
1. If X = 1, then the seed willnot be annihilated by any particles in [0 , η ] by construction of η . Thus, Z (cid:23) Z (cid:23) { X = 0 } { X = 1 } . (7)Taking expectation in (7) yields E Z ≥ p + − p . When p > /
3, this is larger than 1. Thus,we arrive easily at the bound p ′ c ≤ /
3, which is proven in [DJK +
16] and proven for p c in[ST17].3.2. The effect of a γ . We can do better by extracting some benefit from the 1-particleat 1. Let γ be the index of the particle that destroys a as in the proof of Proposition 5.Let Z be the number of surviving inert particles in the window of dependence induced bythe event { a ↔ a γ } . Depending on whether X γ = − , γ ]or [0 , γ − X [0 , η ]),and thus { X = 1 } Z (cid:22) { X = 1 } Z . This lets us improve (7) to the following dominancerelation Z (cid:23) { X = 0 } { X = 1 } Z . (8)Figure 1 depicts a realization in which Z = 2. To better understand Z , we decompose itrelative to the behavior of γ : Z = ∞ X n =2 { γ = n } Z = ∞ X n =2 { X γ = − , γ = n } + { X γ = 0 , γ = n } Z = 1 + ∞ X n =2 { X γ = 0 , γ = n } ( Z − . (9) DEBBIE BURDINSKI, SHREY GUPTA, AND MATTHEW JUNGE · · ·
Figure 1.
In this example, γ = 9. We are left with Z = 2 survivinginert particles in BA( X [0 , γ − γ ≥ γ when X = 1, and so the particle that destroys a must start at 2 or beyond. All of the cases where X γ = − X γ = 0. Let A n = { X γ = 0 , γ = n } ⊆ { } × { } × {− , , } n − (10)be the set of sub-configurations for which γ = n and X γ = 0. Notice we only need to knowthe entries up to 2 n − a a n .In light of (9), we would like to understand E [ { A n } ( Z − Z − ≥ , γ − n . We do so by countingsurviving inert particles in the final state of BA( X [0 , γ − n − possiblerealizations of particle speeds from { } × { } × {− , , } n − .Given x ∈ A n , let I ( x ) be the number of 0-entries in x . Simply by the definition of ν at(1), the probability x occurs is q n ( x ) := p I ( x ) (cid:18) − p (cid:19) n − − I ( x ) . Letting Z ( x ) be the number of surviving inert particles in BA( x ), we then have E [ { A n } ( Z − X x ∈ A n q n ( x )[ Z ( x ) − . (11)This is possible for a computer to calculate for small values of n . For instance, if wecompute for n ≤
18 (approximately 400 million cases) and take expectation in (9), weobtain the bound E Z ≥ X n =2 X x ∈ A n q n ( x )[ Z ( x ) −
1] = m ( p ) , (12)and thus by (8) we have E Z ≥ p + 1 − p m ( p ) . (13) HE UPPER THRESHOLD IN BALLISTIC ANNIHILATION 9
By numerically checking the boundary values of p , we find that E Z > p > . m ( . ≈ . ≈ . Using a surviving -particles from BA( X [0 , γ − . A simple way to optimizefurther is by re-using the previous calculation for configurations from A n in which there isa single surviving 1-particle at 2 n −
1, and otherwise only inert particles. Formally, let A ′ n = { x ∈ A n : ξ n − ( x ) = 1 , ξ i ( x ) ∈ { , } for 0 ≤ i ≤ n − } be the set of all such configurations. We claim that each x ′ ∈ A ′ n provides an independent Z − a γ − are independent. Thus, we can couple the number of surviving inert particlesinduced by a γ − to Z −
1. We subtract 1 so we do not double count the seed. Let b ( p ) = P n =1 P x ′ ∈ A ′ n q n ( x ′ ) be the probability of a configuration from A ′ n . Each time thisoccurs we obtain an expected m ( p ) − m ( p ) from (12). This willhappen geometric( b ( p ))-distributed many times, which has expectation b ( p ) / (1 − b ( p )). Itfollows that we can improve our bound from (13) to the following E Z ≥ p + 1 − p (cid:18) m ( p ) + b ( p )1 − b ( p ) ( m ( p ) − (cid:19) . It takes a computer about three hours to obtain an algebraic expression for b ( p ). Afterdoing so, we find that E Z > p > . . In this case we have m ( p ) ≈ . b ( p ) ≈ . . Further benefit from surviving active particles.
The bound we obtain on p ′ c is asfar as seemed reasonable to push our technique. More complicated calculations can be donewhere one considers the impact of other configurations of surviving 1-particles to the rightof γ . For example, one might consider the case that there is a single surviving 1-particleat 2 γ − k with k ≥ k = 1). However, we found that the improvementsto our bound were very small (about a .
004 to our bound on p c ) did not justify the addedcomplexity to the argument. It is our belief that the main benefit to estimating E Z wouldcome out of extending our approach and looking out to distances n >
18. When p = . P ( γ ≤
18) = . /
10 of the right tail,which ought to contain a significant number of surviving inert particles. However, withouta clever insight, extending much further appears computationally intractable.
Acknowledgements
We thank Rick Durrett and Laurent Tournier for invaluable comments and discussion.
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Debbie Burdinski
E-mail address : [email protected] Shrey Gupta
E-mail address : [email protected] Matthew Junge
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