Phase transition in 3-speed ballistic annihilation

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Phase transition in 3-speed ballistic annihilation

Laurent TOURNIER(LAGA, Université Paris 13) Joint work with John HASLEGRAVEand Vladas SIDORAVICIUS

Probability seminar NYU Shanghai — May 14, 2019

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Plan of the talk – Part 1

1 Introduction

2 Results

3 Sketch of proof

4 About “universality”

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Introduction – Ballistic annihilation

Ballistic annihilation model, from physics literature (1990’s) LetVn,n∈Zbe i.i.d. random variables, with distributionµonR.

From each locationxn,n∈Z, of a point process onR, a particle starts moving at constant speedVn. When two particles collide, they annihilate.

Time

Space

0 x1 x2x3· · ·

x−1

x−2

x−3

· · ·

→Speed of decay of density of particles ?

→Ifµhas atoms, are there surviving particles ?

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Introduction – Discrete cases

Two-speed model

From each integer, a particle is released with random speed±1. They annihilate upon collision.

Time

Space

0 1 2 3 · · ·

−2 −1

· · ·

Simple combinatorics. Densityc(t) =P(return time of SRW>2t)∼ct⁻^1/2 Description of “flocks of particles” : Belitzky–Ferrari ’95

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Introduction – Discrete cases

Two-speed model

From each integer, a particle is released with random speed±1. They annihilate upon collision.

Simple combinatorics. Densityc(t) =P(return time of SRW>2t)∼ct⁻^1/2 Description of “flocks of particles” : Belitzky–Ferrari ’95

Three-speed model (Ben-Naim–Redner–Leyvraz ’93, Piasecki ’95)

From each location of a Poisson point process, a particle starts with random speed among−1,0,+1, with symmetric distribution. Annihilation upon collision.

Time

Space

0 x1 x2x3· · ·

x−1

x−2

x−3

· · ·

Combinatorics becomevery intricate: no simple rule to check survival, long range dependences in both directions, dependence in interdistances, no monotonicity...

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Introduction – Three-speed ballistic annihilation

Velocities are sampled according toµ= ¹⁻₂^pδ₋1+pδ₀+¹⁻₂^pδ+1. Simulations forp=0.24,p=0.25,p=0.26 :

Transition atpc=¹₄“computed” by Piaseckiet al.’95, and asymptotics for densities.

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Introduction – Three-speed ballistic annihilation

Velocities are sampled according toµ= ¹⁻₂^pδ₋1+pδ₀+¹⁻₂^pδ+1. Simulations forp=0.24,p=0.25,p=0.26 :

Transition atpc=¹₄“computed” by Piaseckiet al.’95, and asymptotics for densities.

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Plan of the talk – Part 2

1 Introduction

2 Results

3 Sketch of proof

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Results

Consider the ballistic annihilation model, where interdistances have an atomless distribution, velocities are







−1 with probability(1−p)/2 0 with probabilityp +1 with probability(1−p)/2.

Defineθ(p) =P(the particle at 0 survives indefinitely).

Theorem (Haslegrave-Sidoravicius-T. ’18+)

The model undergoes a phase transition at pc= ¹₄:θ(p)>0⇔p>1/4.

Moreover,

for all p>1

4, θ(p) = (2√ p−1)².

0 0.5 1

0 0.25 0.5 0.75 1p

%-

%↑

↑- ↑

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Results – continued

Denote byc0(t)the density of stationary particles present at timet.

Denote byc₊(t)the density of (+1)-particles present at timet.

Assume further that interdistances are exponentially integrable, with unit expectation.

We have the following asymptotics, as t→∞: for some c=c(p)>0,

c₀(t) =











2p

π(1−4p)+o(1)

t⁻¹ if p<1/4,

2^2/3

4Γ(2/3)²+o(1)

t⁻^2/3 if p=1/4, (2√p−1)²+o(e⁻^ct) if p>1/4, and

c+(t) =











√1 π

√1−4p+o(1)

t⁻^1/2 if p<1/4,

2^2/3

8Γ(2/3)²+_8Γ(1/3)³ +o(1)

t⁻^2/3 if p=1/4,

o(e⁻^ct) if p>1/4.

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Plan of the talk – Part 3

1 Introduction

2 Results

3 Sketch of proof

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Proof – First remarks

Let us prove the first theorem :pc= ¹₄andθ(p) = (2√p−1)²₊. A few remarks :

by symmetry and independence, it suffices to consider the system on(0,+∞) and to evaluateq=P(0,∞)(0←•)~ ; then we haveθ(p) =p(1−q)².

a (−1)-particle is never caught by a particle on its right. Therefore, for allk∈N, the eventP(0,∞)(0←•k~ )only depends on the finite system of the firstkparticles.

the distribution of the system is invariant under mirroring the piece of configuration between particleskandl(for anyk<l)

a(+1)-particle almost surely collides with another particle : if not, then (by ergodicity) almost surely infinitely many would survive forever in the process onR; but by symmetry the same holds for(−1)-particles...

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Proof – Identities in pictures

1 − q

or or or

= p(1 − q) pq(1 − q) α(1 − q)

x1 x1 x1 x1

+ +

+

=

Ifq6=1 , 1^stequation givesα. Inject into 2^nd, useq6=1. Get q=√¹p−1 .

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Proof – Identities in pictures

1 − q

or or or

= p(1 − q) pq(1 − q) α(1 − q) pq(1 − q)

x1 x1 x1 x1

+ +

+

=

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Proof – Identities in pictures

1 − q

or or or

= p(1 − q) pq(1 − q) α(1 − q) pq(1 − q)

x1 x1 x1 x1

+ +

+

=

q

or or or

=

^1−p

2

pq

²

αq

x1 x1 x1 x1

+ +

+

=

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Proof – Identities in pictures

1 − q

or or or

= p(1 − q) pq(1 − q) α(1 − q) pq(1 − q)

x1 x1 x1 x1

+ +

+

=

q

or or or

=

^1−p

2

pq

²

αq

¹₂

pq

²

x1 x1 x1 x1

+ +

+

=

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Proof – Identities in pictures

1 − q

or or or

= p(1 − q) pq(1 − q) α(1 − q) pq(1 − q)

x1 x1 x1 x1

+ +

+

=

q

or or or

=

^1−p

2

pq

²

αq

¹₂

pq

²

x1 x1 x1 x1

+ +

+

=

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Consequences

0 1 2

p

√1p−1

1

4 1

Eitherq=1 orq=√¹p−1.

Ifp≤¹4, then necessarilyq=1. Also, clearlyq=0 atp=1.

What else ?

Needsa prioriregularity onq.

qis lower semi-continuous :P(∃i<k,0←•ⁱ~ )%q In particular,

{p>1

4:q=1}={p>1 4:q> 1

√p−1}is open

we can also (less directly, and not monotonically) approximate thesuper-critical phase by finite conditions

“ϕk>0” and get {p>1

4:q<1}=^[

k

{p>1

4:ϕk(p)>0}is open

These together imply by connectedness that the supercritical phase covers the whole interval(¹₄,1].

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Consequences

0 1 2

p

√1p−1

1

4 1

What else ? Needsa prioriregularity onq.

{p>1

4:q=1}={p>1 4:q> 1

√p−1}is open

4:q<1}=^[

k

{p>1

4:ϕk(p)>0}is open

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Consequences

0 1 2

p

√1p−1

1

4 1

{p>1

4:q=1}={p>1 4:q> 1

√p−1}is open

4:q<1}=^[

k

{p>1

4:ϕk(p)>0}is open

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Consequences

0 1 2

p

√1p−1

1

4 1

{p>1

4:q=1}={p>1 4:q> 1

√p−1}is open

“ϕ_k>0” and get {p>1

4:q<1}=^[

k

{p> 1

4:ϕk(p)>0}is open

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Consequences

0 1 2

p

√1p−1

1

4 1

{p>1

4:q=1}={p>1 4:q> 1

√p−1}is open

“ϕ_k>0” and get {p>1

4:q<1}=^[

k

{p> 1

4:ϕk(p)>0}is open

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Effective characterization of the survival phase

Explore the configuration from left to right, one particle at a time.

Count+1 (resp.−1) for each “fresh”↑(resp.-)

As long as↑outnumber-, 0 is not hit. Thus,p>¹₃impliesq<1. More generally,

Explore the configuration from left to right,kparticles at a time.

First “resolve” the inner interactions of thesekparticles, then explore until the first

“fresh” site in an analogous sense, and repeat. As long as

ϕk=E[#(surviving↑inkparticles)−#(surviving-inkparticles)]>0, 0 has positive chance not to be hit :q<1.

→In fact, this is equivalent :q<1⇔ ∃k,ϕk>0.

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Effective characterization of the survival phase

+1

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Effective characterization of the survival phase

+1

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Effective characterization of the survival phase

+1 firstannihilator (left-to-right)

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Effective characterization of the survival phase

+1 −1

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Effective characterization of the survival phase

+1 −1

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Effective characterization of the survival phase

+1 −1 −1

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Effective characterization of the survival phase

+1 −1 −1+1+1

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Effective characterization of the survival phase

+1 −1 −1+1+1 +1−1−1

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Effective characterization of the survival phase

+1 −1 −1+1+1 +1−1−1 +1+1

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Effective characterization of the survival phase

+1 −1 −1+1+1 +1−1−1 +1+1

As long as↑outnumber-, 0 is not hit. Thus,p>¹₃impliesq<1.

More generally,

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Effective characterization of the survival phase

+1 −1 −1+1+1 +1−1−1 +1+1

More generally,

“fresh” site in an analogous sense, and repeat.

As long as

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Effective characterization of the survival phase

+1 −1 −1+1+1 +1−1−1 +1+1

More generally,

“fresh” site in an analogous sense, and repeat.

As long as

→In fact, this is equivalent :q<1⇔ ∃k,ϕ >0.

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Plan of the talk – Part 4

1 Introduction

2 Results

3 Sketch of proof

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On the distribution of interdistances

Let us denote bymthe distribution of interdistances in the initial configuration.

The proof, hence the result, doesn’t depend onm, besides being atomless Yet,

The model genuinely depends onm: not only probabilities of configurations can vary, but even some configurations are possible or not, depending onm.

(consider exponential distribution, vs. uniform distribution on[1,2]) no possible coupling between models for different choices ofm

→Even though the law of the pairingdoesdepend onm, sub/supercriticality doesn’t !

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A universal distribution (given p)

A stronger universality property holds true.

Denote byAtheindex(inN) of the first particle hitting 0 on(0,∞), if there is any, and letA=∞otherwise.

The distribution of A does not depend on m (provided m is atomless).

It can even be “computed” : for0≤p≤1, for x∈[−1,1], the generating series fp(x) =E[x^A1_{_A<∞}] =

∞ n=1

∑

P(A=n)xⁿ satisfies

pxfp(x)⁴−(1+2p)xfp(x)²+2fp(x)−(1−p)x=0. (1) NB. Sinceq=P(0,∞)(A<∞) =fp(1)we recover a polynomial equation forq.

In particular, we can extendfanalytically to the whole planeCexcept for slits ; and we can then find asymptotics forP(A=n)asn→∞bysingularity analysis.

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A universal distribution (given p)

A stronger universality property holds true.

Denote byAtheindex(inN) of the first particle hitting 0 on(0,∞), if there is any, and letA=∞otherwise.

The distribution of A does not depend on m (provided m is atomless).

It can even be “computed” : for0≤p≤1, for x∈[−1,1], the generating series fp(x) =E[x^A1_{_A<∞}] =

∞ n=1

∑

P(A=n)xⁿ satisfies

pxfp(x)⁴−(1+2p)xfp(x)²+2fp(x)−(1−p)x=0. (1) NB. Sinceq=P(0,∞)(A<∞) =fp(1)we recover a polynomial equation forq.

In particular, we can extendfanalytically to the whole planeCexcept for slits ; and we can then find asymptotics forP(A=n)asn→∞bysingularity analysis.

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Singularity analysis

Main theorem (from Flajolet & Sedgewick’sAnalytic Combinatorics) Letfbe a holomorphic function on the unit disk. Assume that 1 is the unique singularity offon the unit circle. Assume furthermore thatfcan be extended analytically to a∆-domain :

0 1

Forα∈R\ {0,1,2,3, . . .},C∈C^∗, f(z)−f(1) ∼

z→1 z∈∆

C(1−z)^α ⇒ [zⁿ]f(z)_n∼

→∞

C

Γ(−α)n⁻^(α+1)

In our case,

We need to find the singularities offp;

Existence of analytic extension follows standarly (monodromy theorem). The implicit equationF(z,f(z)) =0 actually defines a multivalued analytic function f(z), or an analytic function on the algebraic Riemann surface{F(·,·) =0}. One easily finds singularities off; but are they singularities off?

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Singularity analysis

0 1

Forα∈R\ {0,1,2,3, . . .},C∈C^∗, f(z)−f(1) ∼

z→1 z∈∆

C(1−z)^α ⇒ [zⁿ]f(z)_n∼

→∞

C

Γ(−α)n⁻^(α+1) In our case,

Existence of analytic extension follows standarly (monodromy theorem).

The implicit equationF(z,f(z)) =0 actually defines a multivalued analytic function f(z), or an analytic function on the algebraic Riemann surface{F(·,·) =0}. One easily finds singularities off; but are they singularities off?

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Singularity analysis

0 1

Forα∈R\ {0,1,2,3, . . .},C∈C^∗, f(z)−f(1) ∼

z→1 z∈∆

C(1−z)^α ⇒ [zⁿ]f(z)_n∼

→∞

C

Γ(−α)n⁻^(α+1) In our case,

Existence of analytic extension follows standarly (monodromy theorem).

F(z,f(z)) =

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Looking for the singularities of f

fhas singularities at(z,w) = (0,∞),(±1,±1)and(±R,±W)whereR=q

3p 1−p. ifp<1/4, thenR<1 :fis smooth atR. Thus±1 are singularities off; ifp>1/4, thenR>1, andf(1) =q<1 (first theorem) so 1 is not a singularity off. Thus±Rare singularities off.

Standard computations then give the asymptotics forP(A=n).

NB. Asymptotics for densities need an extra approximation (indices→distances).

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Looking for the singularities of f

fhas singularities at(z,w) = (0,∞),(±1,±1)and(±R,±W)whereR=q

3p 1−p. ifp<1/4, thenR<1 :fis smooth atR. Thus±1 are singularities off; ifp>1/4, thenR>1, andf(1) =q<1 (first theorem) so 1 is not a singularity

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Concluding remarks

One can deduce that certain other quantities are universal.

For instance, in the supercritical regime, theskyline processis (shapes and number of vertices).

For distributions with atoms,triple collisionsmay happen.

Assume they resolve bytotal annihilation. The arguments still go through, but identities involveσ(p) =P(triple collision at 0), apparently not explicit. Ifm=δ1, extinction holds forp<0.2347 and survival forp>0.2405.

Assume triple collisions resolveuniformly at random among±1. Then the model still observes universality, and in particular changes phase at 1/4.

If distribution of speed is not symmetric (but still takes 3 values), then some of the analysis carries over (with adaptations) but involves too many unknowns to get uniqueness of phase transition ; still gives extinction below 1/4. Results due to Junge–Lyu ’18.

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Concluding remarks

One can deduce that certain other quantities are universal. For instance, in the supercritical regime, theskyline processis (shapes and number of vertices).

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Concluding remarks

{

{ {

{

# of particles

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Concluding remarks

{

{ {

{

# of particles

Assume they resolve bytotal annihilation. The arguments still go through, but identities involveσ(p) =P(triple collision at 0), apparently not explicit.

Ifm=δ1, extinction holds forp<0.2347 and survival forp>0.2405.

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Concluding remarks

{

{ {

{

# of particles

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Concluding remarks

{

{ {

{

# of particles

If distribution of speed is not symmetric (but still takes 3 values), then some of the analysis carries over (with adaptations) but involves too many unknowns to get uniqueness of phase transition ; still gives extinction below 1/4. Results due