The critical barrier for the survival of branching random walk with absorption

www.imstat.org/aihp DOI:10.1214/11-AIHP453

© Association des Publications de l’Institut Henri Poincaré, 2012

The critical barrier for the survival of branching random walk with absorption

Bruno Jaffuel

LPMA, Université Paris VI, 4 Place Jussieu, F-75252 Paris Cedex 05, France. E-mail:[email protected] Received 9 November 2009; revised 6 July 2011; accepted 18 August 2011

Abstract. We study a branching random walk onRwith an absorbing barrier. The position of the barrier depends on the generation.

In each generation, only the individuals born below the barrier survive and reproduce. Given a reproduction law, Biggins et al. [Ann.

Appl. Probab.1(1991) 573–581] determined whether a linear barrier allows the process to survive. In this paper, we refine their result: in the boundary case in which the speed of the barrier matches the speed of the minimal position of a particle in a given generation, we add a second order terman^1/3to the position of the barrier for thenth generation and find an explicit critical value acsuch that the process dies whena < acand survives whena > ac. We also obtain the rate of extinction whena < acand a lower bound for the population when it survives.

Résumé. Nous étudions une marche aléatoire branchante surRavec une barrière absorbante. La position de la barrière dépend de la génération. À chaque génération, seuls les individus nés sous la barrière survivent et se reproduisent. Étant donnée une loi de reproduction, Biggins et al. [Ann. Appl. Probab.1(1991) 573–581] ont déterminé, pour une barrière linéaire, si le processus survit ou s’éteint. Dans cet article, nous affinons ce résultat : dans le cas frontière où la vitesse de la barrière correspond à la vitesse de la particule la plus à gauche d’une génération donnée, nous allons à l’ordre suivant en ajoutant un termean^1/3à la position de la barrière pour lanième génération et obtenons une valeur critique explicitea_ctelle que le processus s’éteint quanda < a_cet survit quanda > a_c. Nous obtenons aussi le taux d’extinction lorsquea < a_cet une borne inférieure sur la taille de la population lorsqu’il survit.

MSC:60J80

Keywords:branching random walk; survival probability

1. Introduction

We study a discrete-time branching random walk onR. The population forms a well-known Galton–Watson treeT, and some extra information is added: to each individualu∈T we attach a displacementξu∈Rfrom the position of her parent. We set the initial ancestorat the origin, hence the individualuhas position

V (u)=

<v≤u

ξ_v=

|u|

i=1

ξ_ui,

where|u|is the generation ofuandui the ancestor ofuin generationi. We denote byTn:= {u∈T: |u| =n}the population at timen. We define an infinite pathuthroughT as a sequence of individualsu=(u_i)_i∈Nsuch that

∀i∈N, |u_i| =i and u_i< u_i+1. We denote their collection byT∞.

Now we explain how the displacements ξ_u, u∈T, are distributed. A simple choice, with very nice properties would be to take them i.i.d. but actually everything still works in a more general setting. All individuals still reproduce independently and the same way, but we allow correlations in the number and displacements of the children of every single individual. If we writeΓ (u)for the set of children ofu, our requirement is that the point processes{ξ_v, v∈ Γ (u)}(withurunning over all the potential individuals of the random treeT) are i.i.d.

We define a barrier as a functionϕ:N→R. In the branching random walk with absorption, the individualsusuch thatV (u) > ϕ(|u|), i.e. born above the barrier are removed: they are immediately killed and do not reproduce.

Kesten [12], Derrida and Simon [8,9], Harris and Harris [11] have studied the continuous analog of this process, the branching Brownian motion with absorption. The understanding of what happens in the continuous setting, more convenient to handle from technical point of view, greatly helps us in the discrete one. In particular, we borrow here some ideas from Kesten [12].

Biggins et al. [7] introduced the branching random walk with an absorbing barrier in order to answer questions about parallel simulations. Pemantle [14] and Gantert et al. [10] also studied this model.

A natural question that arises is whether the process survives. This obviously depends on the walk as well as on the barrier. The case of the linear barriers has been solved by Biggins et al. [7].

Before stating their result, we need to introduce some notation:

We denote the intensity measure of the point process byμ, and its Laplace–Stieljes transform byΦ:

Φ(t )=E

|u|=1

e^{−t ξu}

=

Re^{−t z}μ(dz).

We assume that the expected number of children Φ(0) is finite and that negative displacements occur, i.e. that μ((−∞,0)) >0.

We also defineΨ =logΦ, this is a strictly convex function that takes values in(−∞,+∞]. We call critical the case where

Φ(1)=E

|u|=1

e^−ξu

=1 and Φ(1):=E

|u|=1

ξue^−ξu

=0.

This can also be writtenΨ (1)=0 andΨ(1)=0.

Theorem 1.1 (Biggins et al. [7]). In the critical case,we have:

P

∃u∈T_∞,∀i≥1, V (ui)≤iε =0 ifε≤0,

>0 ifε >0.

The aim of this article is to refine this result by replacing the linear barrieri→iε with a more general barrier i→ϕ(i).

Given a barrierϕ we do not know in general whetherP(∃u∈T∞,∀i≥1, V (ui)≤ϕ(i))=0 or not. We assume from now on that we are in the critical case. It is well known that many noncritical random walks can be transformed into critical ones by a linear modification of the displacements, so we do not loose much in generality. Theorem1.1 leads us to focus on barriers such that^ϕ(i)_i →0.

We introduce the parameter σ²:=Φ(1)=E

|u|=1

ξ_u²e^ξu

and assume through the following that it is finite.

Some specific technical difficulties arise in the computation of the second moment (that we use in order to give a lower bound for the survival probability or to prove survival) when dealing with Galton–Watson trees of unbounded degree. Actually individuals with many children may cause trouble. In order to have a sufficient control, we assume

from now on that the number of children of a single individual is uniformly bounded or that the following condition holds:

∃δ₁>0, Φ(1+δ₁) <+∞ and ∃δ₂>0, E #T₁^1+δ2

<+∞. (1.1)

Under these assumptions, we obtain the following result:

Theorem 1.2. Leta_c=³₂(3π²σ²)^1/3.We have:

P

∃u∈T_∞,∀i≥1, V (ui)≤ai^1/3 =0 ifa < ac,

>0 ifa > ac.

Unfortunately, we are not able to conclude in the casea=a_c, nor to give a necessary and sufficient condition on a general barrier for a line of descent to survive below it.

Theorem1.2has the following corollary:

Corollary 1.3. Under the hypothesis of Theorem1.2,we have,almost surely,on the set of ultimate survival of the underlying Galton–Watson process,

uinf∈T∞lim sup

n→∞

V (un) n^1/3 =a_c.

While proving Theorem1.2, we actually obtain stronger results. The two following propositions together imply the theorem.

Proposition 1.4. If a > ac,then the equation a=b+ ^3π_2b^2σ2² has two solutions inb,let ba be the one such that b_a>^2a₃^c.For anyε >0,for anyN∈Nlarge enough,we have with positive probability:

∀k≥1, #

u∈TN^k: ∀i≤N^k, (a−b_a)i^1/3≤V (u_i)≤ai^1/3

≥exp

N^k/3(b_a−ε) .

Proposition 1.5. Ifa < ac,then there exists some constantc >0such that 1

n^1/3logP

∃u∈Tn,∀i≤n, V (u_i)≤ai^1/3

→ −c.

The constantc, which depends ona, is determined in Section5.

Whena < a_c, extinction means that the total progenyZis almost surely finite. This random variable has infinite mean, since the expected number of surviving individuals in generationnis exp(an^1/3(1+o(1))). We can estimate the tail of the distribution ofZ:

Proposition 1.6. Ifa < ac,then letgbe the optimal function determined in Section5,c:=g(0)andd:=max_[0,1]g.

P(Z > k)=k^{−(c/d)(1+o(1))}.

Remark 1.7. In the casea≤0,gis decreasing,henced=cand the claim of Proposition1.6is weaker than a known result,conjectured by Aldous and proved by Addario-Berry and Broutin[1],and improved by Aidékon[2]that for a=0,E[Z]<+∞andE[ZlogZ] = +∞.Exponents less than−1are obtained(see Aidékon,Hu and Zindy[3])for linear barriersi→ −εi,which corresponds to what is often referred as the subcritical case.

Consider a general barrierϕ:N→R. We define a⁺:=lim sup

n→∞

ϕ(n)

n^1/3 and a⁻:=lim inf

n→∞

ϕ(n) n^1/3.

We deduce from Theorem1.2that there is extinction whena⁺< a_c and survival whena⁻> a_c. Making some modifications to the computations of Section3, we can prove the following result:

Theorem 1.8. Assumea⁺≥a_c.The equationa⁺=b+^3π_2b^2σ2² admits a unique solutionb_ac=^2a₃^c ifa⁺=a_c,and two solutions ifa > a_c.Letb_a+≥^2a₃^c be the larger solution.

Ifa⁻<^3π2σ2

2b²_a+ ,then there is extinction.We have a partial converse:ifa⁺≥ac anda⁻are reals such thata⁻<

3π²σ²

2b²_a+ ,then we can find a corresponding barrierϕ such that the absorbed branching random walk survives with positive probability.

The rest of the paper is organized as follows:

Section2introduces the tools we will use in the proof of our main results.

Section3is devoted to the proof of the upper bound in Proposition1.5, which contains the first part of Theorem1.2.

In Section4, we prove Proposition1.4which implies the second part of Theorem1.2.

In Section 5, we complete the proof of Proposition 1.5. We skip many details of technical arguments already exposed in Section4to obtain the lower bound and go back over some results of Section3in order to prove that the two bounds agree.

In Section6, we prove Theorem1.8, Proposition1.6and Corollary1.3.

2. Some preliminaries

2.1. Many-to-one lemma SinceE[

|u|=1e^−ξu] =1, we can define the law of a random variableXsuch that for any measurable nonnegative functionf,

E f (X)

=E

|u|=1

e^−ξuf (ξ_u)

.

ThenE[X] =E[

|u|=1ξ_ue^−ξu]so thatXis centered by hypothesis.

We writeN^∗for the set of positive integers. Let(X_i)_i∈N∗be a i.i.d. sequence of copies ofX. Write for anyn∈N, S_n:=

0<i≤nX_i.Sis then a mean-zero random walk starting from the origin.

We can now state the many-to-one lemma (this is exactly Lemma 4.1(iii) of Biggins and Kyprianou [6]):

Lemma 2.1 (Biggins and Kyprianou [6]). For anyn≥1and any measurable functionF:Rⁿ→ [0,+∞), E

|u|=n

e^{−V (u)}F

V (u_i),1≤i≤n

=EF (S_i,1≤i≤n) .

The proof of the lower bound for the survival probability also requires the following bivariate version of the many- to-one lemma.

Lemma 2.2 (Gantert, Hu and Shi [10]). Let(X, ν)be a random variable taking values inR×N^∗such that for any measurable nonnegative functionf,

E f (X, ν)

=E

|u|=1

e^−ξuf (ξu,#T1)

.

Let n≥1 and (X_i, ν_i)_1≤i≤n be i.i.d. copies of(X, ν). Write for any 0≤k≤n, S_k:=

0<i≤kX_i. Then for any measurable functionF:(R×N^∗)ⁿ→ [0,+∞),

E

|u|=n

e^{−V (u)}F

V (u_i),#Γ (ui−1),1≤i≤n

=E F (S_i, ν_i,1≤i≤n) .

The proof, very similar to the one of Lemma2.1, is omitted.

2.2. Mogul’skii’s estimate

LetF[0,1](respectivelyC[0,1]) be the set of functions (respectively continuous functions)[0,1] →R.

For anyL,L∈F[0,1], we writeL <Lwhen∀t∈ [0,1], L(t ) <L(t )andL≤Lwhen∀t∈ [0,1], L(t )≤L(t ).

Ifn≥1, we writeL <_nLwhen∀1≤k≤n,L(^k_n) <L(_n^k)andL≤nLwhen∀1≤k≤n,L(_n^k)≤L(^k_n).

Theorem 2.3 (Mogul’skii). Letξ₁,ξ₂,. . .be i.i.d.random variables such thatE[ξ₁] =0andσ²:=E[ξ₁²]<∞.Let (x_n, n≥0)be a sequence of positive numbers such that

nlim→∞x_n= +∞,

nlim→∞

x_n

√n=0.

Define for anyn≥0

S_n:=S₀+ξ₁+ξ₂+ · · · +ξ_n,

whereS₀=zalmost surely under the probabilityP^z(z∈R).

Whenz=0,writeP:=P⁰and define,for anyt∈ [0,1], s_n(t):=S_{t n}

xn =ξ₁+ξ₂+ · · · +ξ_k xn

fork/n≤t < (k+1)/n.

Then,for anyL1,L2∈C[0,1],with

L₁< L₂ and L₁(0) <0< L₂(0), (2.1)

we have,asn→ ∞, log

P(L₁< s_n< L₂)

∼ −C_L1,L2nx_n⁻², where

C_L1,L2:=π²σ² 2

₁

0

dt

(L₂(t)−L₁(t))².

We keep the notations and assumptions of Theorem2.3throughout this section. For the proof, we refer to [13].

We actually need more sophisticated versions of this estimate. For the proofs of the following results, we refer to [4]. In all this section, changing strict inequalities into weak ones in the definition of the events we are interested (but not in (2.1)) does not change the estimate of the probability.

Lemma 2.4 (Lemma 4.4 of [4]). SetL₁andL₂like in Theorem2.3.For any sequences(Lⁿ₁)_nand(Lⁿ₂)_nofF[0,1]

such thatLⁿ₁−L_1∞→0andLⁿ₂−L_2∞→0asn→ ∞,we have log

P

Lⁿ₁<_ns_n<_nLⁿ₂

∼log P

Lⁿ₁< s_n< Lⁿ₂

∼ −nx_n⁻²C_L1,L2.

From now on, we set:

∀n≥1, x_n:=n^1/3. (2.2)

Proposition 2.5 (Proposition 4.7 of [4]). SetL₁,L₂∈C[0,1],with

L₁(0)≤0≤L₂(0) and ∀t∈ [0,1], L₁(t)≤L₂(t). (2.3) Let(Lⁿ₁)_nand(Lⁿ₂)_nbe sequences ofF[0,1]such thatLⁿ₁−L_1∞→0andLⁿ₂−L_2∞→0asn→ ∞.We assume BandC are mappings[0,1] ×N^∗→N^∗,nondecreasing in the first component and such that,for anyα∈ [0,1],the sequences(B(α, n)−αn)nand(C(α, n)−αn)nare bounded.

Uniformly in0≤β < γ ≤1,we have lim sup

n→∞

1 n^1/3log

sup

z P^z

Lⁿ₁ k

n

<S_k−B(β,n) xn

< Lⁿ₂ k

n

,∀B(β, n) < k≤C(γ , n)

≤ −C_L^β,γ

1,L₂:= −π²σ² 2

γ β

dt

(L2(t)−L1(t))²,

where thesup_zis over thez∈Rsuch thatxnLⁿ₁(^B(β,n)_n )≤z≤xnLⁿ₂(^B(β,n)_n ).

Remark 2.6. The upper bound in Theorem2.3and Lemma2.4is still valid with condition(2.1)replaced by(2.3).

In order to deal with Galton–Watson trees of infinite degree, we borrow Lemma 2.1 from [10]. Combined with the arguments leading to Proposition2.5, it gives us the following estimate.

Proposition 2.7. For eachn≥1,letX_i⁽ⁿ⁾, 1≤i≤nbe i.i.d.real-valued random variables.We defineS⁽ⁿ⁾_i =S₀⁽ⁿ⁾+ X₁⁽ⁿ⁾+ · · · +X⁽ⁿ⁾_i for1≤i≤n.

Assume that there exist constantsδ >0andσ²>0such that sup

n≥1

E X⁽ⁿ⁾

1 ^2+δ

<+∞, EX⁽ⁿ⁾₁

=o n^−2/3

and Var X⁽ⁿ⁾₁

→σ². (2.4)

LetL₁,L₂,(Lⁿ₁)_nand(Lⁿ₂)_nbe like in Lemma2.4.Letβandγbe real numbers such that0≤β < γ≤1.Let(B(n))_n and(C(n))_nbe sequences of reals such that the sequences(B(n)−βn)_nand(C(n)−γ n)_nare bounded and∀n≥1, 1≤B(n) < C(n)≤n.

Letu^∗andv^∗be real numbers such thatL₁(β) < u^∗< v^∗< L₂(β).Letu_nandv_nbe sequences of real numbers such that

u_n

x_n →u^∗, v_n

x_n →v^∗, Lⁿ₁ B(n)

n

x_n≤u_n≤v_n≤Lⁿ₂ B(n)

n

x_n ∀n≥1.

We have,for anyε >0,

lim inf

n→∞

1 n^1/3log

infz P^z

∀B(n) < k≤C(n), Lⁿ₁ k

n

<

S_k⁽ⁿ⁾_−B(n) n^1/3 < Lⁿ₂

k n

;

Lⁿ₂ C(n)

n

−ε <S_C(n)⁽ⁿ⁾ _−B(n) n^1/3 < Lⁿ₂

C(n) n

≥ −C_L^β,γ

1,L₂, where theinfzis over thez∈Rsuch thatu_n≤z≤v_n.

2.3. A rough estimate

In Section4, we need a lower bound in a particular case withL1(0) <0=L2(0)andL2≥0 on[0,1]. In order to apply the results above, the following lemma will be useful:

Lemma 2.8. There are constants M≥1, and ε1>0 such that, with k := ε2n^1/3, such that the probability P_n(M, ε₁, ε₂)defined as

P

∃u∈Tk,∀i < k,#Γ (ui)≤M, L₁ i

n

≤V (u_i) n^1/3 ≤L₂

i n

; −Mε₂≤V (u_k)

n^1/3 ≤ −ε₁ε₂

satisfies

εlim2→0lim inf

n→∞

1

n^1/3logP_n(M, ε₁, ε₂)=0.

Proof. Letε₁>0 and such that for someM≥1, p:=P

#T1≤M; ∃u∈T1,−M≤V (u)≤ −ε₁

>0.

By independence, P

∃u∈Tk,∀i < k,#Γ (ui)≤M,∀i≤k,−iM≤V (ui)≤ −iε1

≥p^k.

Letε2>0 such thatMε2<−L1(0). For any integer nlarge enough, we takek:= ε2n^1/3. Hence, forε2>0 small enough, we have

P

∃u∈Tk,∀i < k,#Γ (ui)≤M;L1

i n

≤ V (u_i) n^1/3 ≤L2

i n

; −Mε2≤V (u_k)

n^1/3 ≤ −ε1ε2

≥p^k.

3. Upper bound for the survival probability

3.1. Splitting the survival probability Fixa >0. Obviously,

P

∃u∈T_∞,∀i, V (u_i)≤ai^1/3

= lim

n→∞P

∃u∈Tn,∀i≤n, V (u_i)≤ai^1/3 .

From now on,n≥1 is fixed.

We set a second barrieri→ai^1/3−b_i,n(withb_i,n>0 for 1≤i≤nyet to be determined) below the first one i→ai^1/3: if a particle crosses it, then its descendants will be likely to stay below the first one until generationn.

LetH (u):=inf{k≤n: V (uk) < ak^1/3−bk,n}be the first time the line of descent of a particleu∈Tncrosses this second barrier (H (u)= ∞if the particle stays between the barriers until timen). We split the sum accordingly:

P

∃u∈Tn,∀i≤n, V (ui)≤ai^1/3

≤R_∞+ n j=1

Rj, (3.1)

where Rj=P

∃u∈Tn, H (u)=j,∀i≤n, V (ui)≤ai^1/3

forj=1, . . . , n,∞.

By Chebyshev’s inequality and then Lemma2.1, we get R_∞≤E

u∈Tn

1_{∀i≤n,ai1/3−bi,n≤V (ui)≤ai^1/3}

≤E e^Sn1_{∀i≤n,ai1/3−b_i,n≤S_i≤ai^1/3}

≤e^an1/3P

∀i≤n, ai^1/3−b_i,n≤S_i≤ai^1/3

. (3.2)

For 1≤j≤n, R_j ≤E

v∈Tj

1_{∀i<j,ai^1/3_{−bi,n≤V (vi)≤ai}^1/3,V (v)<aj^1/3−b_j,n}

≤E e^Sj1_{∀i<j,ai1/3−bi,n≤Si≤ai^1/3,V (Sj)<aj^1/3−bj,n}

≤e^aj1/3−bj,nP

∀i < j, ai^1/3−bi,n≤Si≤ai^1/3

. (3.3)

3.2. Asymptotics forR_∞

In order to apply Lemma2.4(combined with Remark2.6), we setb_i,n:=n^1/3g(_nⁱ)for some continuous function g:[0,1] → [0,+∞). We take for anyt∈ [0,1],g2(t):=at^1/3andg1(t)=g2(t)−g(t). Then we have

lim sup

n→∞

logP(∀i≤n, ai^1/3−b_i,n≤S_i≤ai^1/3)

n^1/3 ≤ −C_g1,g2. (3.4)

Putting together equations (3.2) and (3.4), we get lim sup

n→∞

logR_∞

n^1/3 ≤ −s1, (3.5)

where

s₁:= −a+C_g1,g2= −a+π²σ² 2

₁

0

dt

g(t)². (3.6)

3.3. Asymptotics forR_j

We defineB andC byB(α, n):=0 andC(α, n):= αn +1 and write, for anyα∈(0,1),j :=C(α, n). Proposi- tion2.5yields that, uniformly inα∈(0,1),

lim sup

n→∞

1 n^1/3logP

∀i < j, ai^1/3−n^1/3g i

n

≤Si≤ai^1/3

≤ −C_g^0,α

1,g₂. (3.7)

Putting together equations (3.3) and (3.7), we get that, uniformly inα∈(0,1), lim sup

n→∞

1

n^1/3logRj≤aα^1/3−g(α)−C_g^0,α

1,g₂. (3.8)

Obviously, for anyn≥1, n

j=1

Rj(n)≤n sup

1≤j≤n

Rj(n)=n sup

0<α<1

RC(α,n)(n). (3.9)

As a consequence, lim sup

n→∞

1 n^1/3log

n j=1

R_j(n)≤ −s₂, (3.10)

where

s2:= min

0≤α≤1

−aα^1/3+g(α)+C_g^0,α

1,g₂

= min

0≤α≤1

−aα^1/3+g(α)+π²σ² 2

_α

0

dt g(t)²

. (3.11)

Combining (3.10) with (3.5) and (3.1), we obtain lim sup

n→∞

1 n^1/3logP

∃u∈Tn,∀i≤n, V (ui)≤ai^1/3

≤ −s, wheres:=min(s1, s₂).

3.4. Choice ofgfor the upper bound

Seta∈(0, ac). We are looking for a functiongsuch thats >0. The existence of such a function implies extinction and ends the proof the first part of Theorem1.2.

We add the constraintg(1)=0 (but assume1 0

du

g(u)² <∞). Takingα=1, we see from (3.11) and (3.6) that this impliess₂≤s₁and, as a result,s=s₂.

We choose g in such a way that the quantity−aα^1/3+g(α)+^π2₂^σ2_α

0 dt

g(t)² which appears in (3.11) does not depend onα. Hencegis defined as the solution of the equation:

∀t∈ [0,1], −at^1/3+f (t )+π²σ² 2

_t

0

du

f (u)² =s, (3.12)

wheresis some positive constant, the value of which is to be set later in such a way thatf (1−)=0. According to the computations above, this value ofswill give a bound for the rate of decay of the survival probability.

Equivalently, equation (3.12) may be writtenf (0)=sand∀t∈(0,1), f(t)=a

3t^−2/3− π²σ²

2f (t )². (3.13)

By the Picard–Lindelöf theorem (see for example [5]), such an ordinary differential equation admits a unique maximal solutionf defined on an interval[0, tmax)witht_max∈(0,+∞]. And ift_max<+∞, thenf has limit 0 or+∞when t goes tot_max.

Remark 3.1. The fact thatf(0)does not exist here is not troublesome at all since the proof of the theorem,using Picard iterates,actually relies on equation(3.12).

In order to prove that there exists an initial valuessuch thatt_max=1 and limt→1f (t )=0, we get a closer look at the differential equation.

First we state three simple results specific to this differential equation.

Proposition 3.2. Letλ >0andf a continuous function[0, t0)→(0,+∞).Definefλ:(0, λ⁻¹t0)→(0,+∞)by fλ(t)=λ^−1/3f (λt ).

Thenf satisfies equation(3.13)on(0, t0)if and only iffλdoes on(0, λ⁻¹t0).

Proof. Assume thatf satisfies equation (3.13) for any 0< t < t₀. Then for any 0< t < λ⁻¹t₀, f_λ(t)=λ^2/3f(λt )

=λ^2/3 a

3(λt )^−2/3− π²σ² 2f (λt )²

=a

3t^−2/3− π²σ² 2fλ(t)².

This means thatf_λalso satisfies equation (3.13) for any 0< t < λ⁻¹t₀.

Conversely, assume thatf_λsatisfies equation (3.13) on(0, λ⁻¹t₀). We notice that ifλ>0, then(f_λ)_λ=f_λλ. We takeλ=λ⁻¹. Hence(f_λ)_λ=f also satisfies equation (3.13) for any 0< t < (λλ)⁻¹t₀=t₀. Proposition 3.3. Set0< a₁< a₂ands >0.Letf₁andf₂be functions[0, tmax)→(0,+∞)such that

∀0≤t < t_max,∀i∈ {1,2}, −a_it^1/3+f_i(t)+π²σ² 2

_t

0

du fi(u)²=s.

Then,for all0≤t < t_max,f₁(t)≤f₂(t).

Proof. It suffices to prove that, if 0≤t_start, 0< a₁< a₂and 0< x₁≤x₂, then there existt_next> t_startsuch that there are functionsf₁andf₂:[t_start, t_next)→(0,+∞)such that

∀t_start≤t < t_next,∀i∈ {1,2}, −a_i

t^1/3−t_start^1/3

+f_i(t)+π²σ² 2

_t

tstart

du fi(u)² =x_i; then, for anyt_start≤t < t_next,f₁(t)≤f₂(t).

We chooset_nextsuch that the Picard interatesf_iⁿdefined, fori∈ {1,2}, by:

∀t_start≤t < t_next, f_i⁰(t)=x_i;

∀n∈N,∀tstart≤t < tnext, f_iⁿ⁺¹(t)=f_iⁿ(tstart)+ai

t^1/3−t_start^1/3

−π²σ² 2

t tstart

du f_iⁿ(u)², exist and converge on[t_start, t_next). The limitsf_i are solutions of the integral equations fori∈ {1,2}.

It is easy to prove by induction onnthat

∀n∈N,∀tstart≤t < tnext, f₁ⁿ(t)≤f₂ⁿ(t).

Lettingntend to infinity gives us the desired conclusion.

Proposition 3.4. Letf be as above.Then we are in one of the following cases:

(A) t_max= +∞andf (t )→ +∞ast→ +∞;

(B) t_max<+∞andf (t )→0ast→t_max.

Proof. First notice that for any 0< t < t_max,f (t )≤s+at^1/3. A consequence of this inequality is that ift_max<+∞, then the limit off whent goes totmaxcan only be 0.

Now, suppose thattmax= +∞but thatf does not go to infinity. Then there areM >0 and a sequence(tn)n≥1 with limntn= +∞such that for anyn≥1,f (tn)≤M. We can choosensuch that ^a₃t_n^−2/3−^π_2M^2σ2² <0.

Then it is easy to see thatf decreases aftertn. Indeed, consider t_∗:=inf

t≥t_n, f(t) >a

3t_n^−2/3−π²σ² 2M²

.

We have, fort_n≤t≤t_∗, f(t)=a

3t^−2/3− π²σ² 2f (t )²≤a

3t_n^−2/3−π²σ² 2M² <0.

If we assumet_∗<+∞, thenf(t_∗) <0, thenf decreases in a neighborhood oft_∗and the inequalityf(t)≤^a₃t_n^−2/3−

π²σ²

2M² still holds on this neighborhood, which contradicts the definition oft_∗.

We have proved thatf(t)is less than a negative constant fort≥tn, which implies thatf reaches zero in finite

time.

Assume we are in the second case of Proposition3.4. We setλ:=t_max⁻¹ and define the functionf_λlike in Proposi- tion3.2(witht₀=t_max). We chooseg=f_λand setg(1)=0 so thatgis continuous over[0,1]and satisfies (3.13) for allt∈(0,1).

Remark 3.5. A consequence of Proposition3.2is that the choice of the valuesoff (0)does not matter at all.If we replaces >0with anothers >0,we then replaceλwithλ=λ(^s_s)³and finally get the sameg.

So we only have to prove that, whena < a_c, we are in case (B) of Proposition3.4, and we will deduce the upper bound in Theorem1.2. This is contained in the following:

Proposition 3.6. Letf be the solution of equation(3.13)with initial conditionf (0)=1.

(i) Ifa > a_c,thent_max= +∞andf (t )∼bt^1/3ast→ +∞withbdefined byb >^2a₃^c anda=b+^3π_2b^2σ2². (ii) Ifa=a_c,thent_max= +∞andf (t )∼^2a₃^ct^1/3ast→ +∞.

(iii) Ifa < a_c,thent_max<+∞andf (t )→0ast→t_max.

In the proof of the proposition, we will need the following lemma.

Lemma 3.7. Assume thatf is a solution on[0,+∞)of the differential equation and that:

lim sup

t→+∞

f (t ) t^1/3 ≤b.

Then we have

lim sup

t→+∞

f (t )

t^1/3 ≤b:=a−3π²σ² 2b² .

Proof. Letε >0. By hypothesis, for anytgreater than somet0, we havef (t )≤(b+ε)t^1/3. For some real constants c0andc₀ and anyt≥t0, we have, by equation (3.12):

f (t )≤c0+at^1/3− π²σ² 2(b+ε)²

t t0

du

u^2/3=c₀+

a− 3π²σ² 2(b+ε)²

t^1/3.

Hence lim sup

t→+∞

f (t ) t^1/3 ≤

a− 3π²σ² 2(b+ε)²

.

Lettingεtend to 0 ends the proof of the lemma.

Iterating Lemma3.7, we obtain:

Lemma 3.8. Assume that f is a solution on[0,+∞)of the differential equation and let b₀ be a real such that b0≥lim sup_t→+∞^{f (t)}_t1/3.We define the sequence(bn)_n∈Nrecursively bybn+1:=a−^3π_2b^2σ2²

n .Then

∀n≥1, lim sup

t→+∞

f (t ) t^1/3 ≤b_n. Proof of Proposition3.6.

(i) Assumea≥a_c and letbsuch thata=b+^3π_2b^2σ2². Define, for 0≤t≤t_max,f₀(t):=bt^1/3. Thenf₀satisfies equation (3.13) asf does, with initial conditionf₀(0)=0< f (0)=s. Hence

∀0≤t≤t_max, f (t )≥f₀(t).

This impliest_max= +∞. Now leth=f −f₀. Then, by equation (3.12), we have, fort≥0, h(t)=s+(a−b)t^1/3−

_t

0

π²σ²du 2f (u)²

=s+

a−b−3π²σ² 2b²

t^1/3+

_t

0

π²σ²du 2

1

f0(u)²− 1 f (u)²

.

Sincea=b+^3π_2b^2σ2², h(t )=s+

t 0

π²σ²du 2

1

f₀(u)²− 1 f (u)²

≤s+ t

0

π²σ² 2

2h(u)du f₀(u)³ . We apply Gronwall’s lemma and obtain, for any 0< t0< t,

h(t)≤h(t0)exp t

t0

π²σ²du b³u

=h(t0) t

t₀

_π²σ²/b³

. (3.14)

Notice that ^π2_b^σ3² =¹₃(^2a_3b^c)³. Then ifa > a_c andb >^2a₃^c, the exponent in the right-hand side of (3.14) will be less than¹₃. Hence inequality (3.14) implies (i).

(ii) Assumea=a_candb=^2a₃^c. This is the same as whena > a_c, except that the exponent in the right-hand side of (3.14) is exactly ¹₃, which means that for some constantb₀>^2a₃^c,

∀t≥t₀, f₀(t)≤f (t )≤b₀t^1/3.

Apply Lemma3.8. The result follows from that limnbn=^2a₃^c.

(iii) Assumea < a_c andt_max= +∞. Then, by (ii) and Proposition3.3, we have that any b₀> ^2a₃^c, for t large enough,

lim sup

t→+∞

f (t ) t^1/3 ≤b₀.

We apply Lemma3.8. Ifb0is close enough to ^2a₃^c, we will haveb1<^2a₃^c andb_n→ −∞asngoes to infinity, which is absurd. We conclude that the hypothesist_max= +∞is false, which proves the proposition.

4. Lower bound for the survival probability 4.1. Strategy of the estimate

The basic idea is to consider only the population between two barriers (belowi→ai^1/3but abovei→(a−b)i^1/3), estimate the first two moments of the number of individuals in generationn and then to use the Paley–Zygmund inequality to get the lower bound.