Critical branching Brownian motion with absorption: survival probability

Abstract

We consider branching Brownian motion on the real line with absorption at zero, in which particles move according to independent Brownian motions with the critical drift of $-\sqrt{2}$. Kesten (Stoch Process 7:9–47, 1978) showed that almost surely this process eventually dies out. Here we obtain upper and lower bounds on the probability that the process survives until some large time $t$. These bounds improve upon results of Kesten (Stoch Process 7:9–47, 1978), and partially confirm nonrigorous predictions of Derrida and Simon (EPL 78:60006, 2007).

1 Introduction

1.1 Main results

We consider branching Brownian motion with absorption, which is constructed as follows. At time zero, there is a single particle at $x > 0$. Each particle moves independently according to one-dimensional Brownian motion with a drift of $-\mu$, and each particle independently splits into two at rate 1. Particles are killed when they reach the origin. This process was first studied in 1978 by Kesten [21], who showed that almost surely all particles are eventually killed if $\mu \ge \sqrt{2}$, whereas with positive probability there are particles alive at all times if $\mu < \sqrt{2}$. Thus, $\mu = \sqrt{2}$ is the critical value for the drift parameter.

Harris et al. [17] obtained an asymptotic result for the survival probability of this process when $\mu < \sqrt{2}$. Harris and Harris [16] focused on the subcritical case $\mu > \sqrt{2}$ and estimated the probability that the process survives until time $t$ for large values of $t$. Results about the survival probability in the nearly critical case when $\mu$ is just slightly larger than $\sqrt{2}$ were obtained in [4, 9, 24]. Questions about the survival probability have likewise been studied for branching random walks in which particles are killed when they get below a barrier. See [1, 3, 12, 13, 19] for recent progress in this area. These questions are indeed of interest in a wide range of problems going well beyond branching process theory. Let us mention for instance: particle systems and in particular the so-called Fleming–Viot systems proposed by Burdzy et al. [10, 11]; theoretical population genetics, where the Brownian trajectories of particles represent the fitness of individuals in this population, and the killing is a crude model for the effect of natural selection (this was suggested most notably by Brunet et al. in [7, 8]); and nonlinear partial differential equations, as the equation satisfied by survival probabilities of branching Brownian motions with killing give rise to reaction-diffusion equations of the Fisher-KPP type which are the subject of much current research (see e.g. [14, 15] and references therein).

In this paper, we consider the critical case in which $\mu = \sqrt{2}$. Let $\zeta$ be the time when the process becomes extinct, which we know is almost surely finite. Kesten showed (see Theorem 1.3 of [21]) that there exists $K > 0$ such that for all $x > 0$, we have

$$\begin{aligned} x e^{\sqrt{2}x - K (\log t)^2 - (3 \pi ^2 t)^{1/3}} \le {{\mathrm{\mathbb {P}}}}_x(\zeta > t) \le (1 + x) e^{\sqrt{2}x + K(\log t)^2 - (3 \pi ^2 t)^{1/3}} \end{aligned}$$

for sufficiently large $t$, where ${{\mathrm{\mathbb {P}}}}_x$ denotes the law of the process started from one particle at $x>0$. Our main result, which is Theorem 1 below, improves upon this result. For this result, and throughout the rest of the paper, we let

$$\begin{aligned} \tau = \frac{2 \sqrt{2}}{3 \pi ^2}, \quad c = \tau ^{-1/3} = \left( \frac{3 \pi ^2}{2 \sqrt{2}} \right) ^{1/3}. \end{aligned}$$

(1)

Theorem 1

There exist positive constants $C_1$ and $C_2$ such that

$$\begin{aligned} C_1 e^{\sqrt{2} x} \sin \left( {\pi x\over ct^{1/3}}\right) t^{1/3}e^{-(3 \pi ^2 t)^{1/3}} \le {{\mathrm{\mathbb {P}}}}_x(\zeta >t) \le C_2 e^{\sqrt{2} x} \sin \left( {\pi x\over ct^{1/3}}\right) t^{1/3}e^{-(3 \pi ^2 t)^{1/3}}\nonumber \\ \end{aligned}$$

(2)

for any $x > 0$ and $t > 0$ such that $x<ct^{1/3}-1.$ In particular, there exist positive constants $C_3$ and $C_4$ such that for any fixed $x>0$, we have

$$\begin{aligned} C_3 x e^{\sqrt{2} x} e^{-(3 \pi ^2 t)^{1/3}} \le {{\mathrm{\mathbb {P}}}}_x(\zeta > t) \le C_4 x e^{\sqrt{2} x} e^{-(3 \pi ^2 t)^{1/3}} \end{aligned}$$

(3)

for sufficiently large $t$.

The main novelty in Theorem 1 is that the terms $e^{\pm K (\log t)^2}$ in Kesten’s upper and lower bounds may be replaced by constants $C_1$ and $C_2$ respectively. Nonrigorous work of Derrida and Simon [9] indicates that it should be possible to obtain a result even sharper than Theorem 1. Indeed, equation (13) of [9] indicates that for each fixed $x$, we should have

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}_x(\zeta > t) \sim C e^{-(3 \pi ^2 t)^{1/3}} \end{aligned}$$

as $t \rightarrow \infty$, where $C$ is a constant depending on $x$.

Note that the result (2) is only valid when $0 < x < ct^{1/3} - 1$. However, when $x=ct^{1/3}-1$, Eq. (2) shows that the survival probability up to $t$ is already of order 1. It is an open question whether there exists a function $\phi : \mathbb {R}\mapsto [0,1]$ such that

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}_{ct^{1/3}+x} (\zeta >t) \rightarrow \phi (x) \end{aligned}$$

as $t\rightarrow \infty$, where ${{\mathrm{\mathbb {P}}}}_z$ denotes probabilities for branching Brownian motion started from a single particle at $z$.

A major step in the proof of Theorem 1 is to obtain sharp estimates on the extinction time $\zeta$ when the position $x$ of the initial particle tends to infinity. For this question we obtain the following result of independent interest.

Theorem 2

Let $\varepsilon > 0$. Then there exists a positive number $\beta > 0$, depending on $\varepsilon$, such that for sufficiently large $x$,

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}_x (\tau x^3 - \beta x^2 < \zeta < \tau x^3 + \beta x^2) \ge 1 - \varepsilon . \end{aligned}$$

Let $x>0$ and let $t = \tau x^3$. Thus Theorem 2 says that if there is initially one particle at $x$, the extinction time of the process will be close to $t$ (if $x$ is large). Conversely, fix $t>0$ and define a function

$$\begin{aligned} L(s) = c (t-s)^{1/3}. \end{aligned}$$

(4)

From Theorem 2, we see that if a particle reaches $L(s)$ at time $s \in (0, t)$, then there is a good chance that a descendant of this particle will survive until time $t$. Our strategy for proving Theorem 1 will be to estimate the probability that a particle reaches $L(s)$ for some $s \in (0,t)$, and then argue that, up to a constant, this is the same as the probability that the process survives until time $t$. We note that the importance of the curve $(L(s), 0\le s \le t)$ was already apparent in the work of Kesten [21], who already considered the strategy of counting the number of particles hitting that curve.

Theorem 1 gives an estimate of the probability that the process started with one particle at $x > 0$ survives until some large time $t$. An important open question is to determine, conditional on survival up to a large time $t$, what the configuration of particles will look like before time $t$. The complete description of the configuration of particles, conditionally upon survival up to a large time $t$, is known as the Yaglom conditional limit. This is in turn related to a main conjecture concerning the limiting behaviour of the Fleming–Viot process proposed by Burdzy et al. [10, 11]. See [2] for a recent discussion and verification in a particular case of that conjecture.

This is the first in a series of two papers concerning the properties of critical branching Brownian motion with absorption. In the companion paper [6], we use ideas developed in this paper to obtain a precise description of the particle configuration at times $0 \le s \le t$, when the position $x$ of the initial particle tends to infinity and $t = \tau x^3$. It seems likely that the results and methods of [6] will also shed some light on the behavior of the process conditioned to survive for a long time.

1.2 Organization of the paper

In Sects. 2 and 3, we collect some general results about branching Brownian motion killed at the boundaries of a strip. Theorems 1 and 2 are proved in Sect. 4. Throughout the paper, $C$ will denote a positive constant whose value may change from line to line, and $\asymp$ will mean that the ratio of the two sides is bounded above and below by positive constants.

2 Branching Brownian motion in a strip

We collect in this section some results pertaining to branching Brownian motion in a strip. Consider branching Brownian motion in which each particle drifts to the left at rate $-\sqrt{2}$, and each particle independently splits into two at rate 1. Particles are killed if either they reach 0 or if they reach $L(s)$ at time $s$, where $L(s) \ge 0$ for all $s$. We assume that the initial configuration of particles is deterministic, with all particles located between 0 and $L(0)$.

Let $N(s)$ be the number of particles at time $s$, and denote the positions of the particles at time $s$ by $X_1(s) \ge X_2(s) \ge \cdots \ge X_{N(s)}(s)$. Let

$$\begin{aligned} Z(s) = \sum _{i=1}^{N(s)} e^{\sqrt{2} X_i(s)} \sin \left( \frac{\pi X_i(s)}{L(s)} \right) \!. \end{aligned}$$

Let $(\mathcal{F}_s, s \ge 0)$ denote the natural filtration associated with the branching Brownian motion.

Let $q_s(x,y)$ denote the density of the branching Brownian motion, meaning that if initially there is a single particle at $x$ and $A$ is a Borel subset of $(0,L(s))$, then the expected number of particles in $A$ at time $s$ is

$$\begin{aligned} \int _A q_s(x,y) \, dy. \end{aligned}$$

2.1 A constant right boundary

We first consider briefly the case in which $L(s) = L$ for all $s$, which was studied in [5]. The following result is Lemma 5 of [5], which relies only on standard and elementary estimates about Brownian motion.

Lemma 3

For $s > 0$ and $x,y \in (0, L)$, let

$$\begin{aligned} p_s(x,y) = \frac{2}{L} e^{-\pi ^2 s/2L^2} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L} \right) e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L} \right) \!, \end{aligned}$$

and define $D_s(x,y)$ so that

$$\begin{aligned} q_s(x,y) = p_s(x,y)(1 + D_s(x,y)). \end{aligned}$$

Then for all $x,y \in (0,L)$, we have

$$\begin{aligned} |D_s(x,y)| \le \sum _{n=2}^{\infty } n^2 e^{-\pi ^2(n^2 - 1)s/2L^2}. \end{aligned}$$

Lemma 3 allows us to approximate $q_s(x,y)$ by $p_s(x,y)$ when $s$ is sufficiently large. We will also use the following result, which follows from (28) and (51) of [5] and is proved using Green’s function estimates for Brownian motion in a strip.

Lemma 4

For all $s \ge 0$ and all $x, y \in (0, L)$, we have

$$\begin{aligned} \int _0^{\infty } q_s(x,y) \, ds \le \frac{2 e^{\sqrt{2}(x-y)} x (L - y)}{L}. \end{aligned}$$

2.2 A piecewise linear right boundary

Fix $m > 0$, and fix $0 < K < L$. Also, let $t > 0$. We consider here the case in which

$$\begin{aligned} L(s) = \left\{ \begin{array}{l@{\quad }l} L &{} \hbox {if }\,\, 0 \le s \le t - m^{-1}(L-K) \\ K + m(t-s) &{} \hbox {if }\,\, t - m^{-1}(L-K) \le s \le t. \end{array} \right. \end{aligned}$$

We will assume that $m^{-1}(L-K) \le t/2$. Thus, the right boundary stays at $L$ from time 0 until at least time $t/2$, but eventually moves to the left at a linear rate, reaching $K$ at time $t$.

To obtain an estimate of $q_s(x,y)$, we will need the following result for the probability that a Brownian bridge crosses a line. This result is well-known and follows immediately, for example, from Proposition 3 of [23]. We let $B^{br}_{x,y,t} = (B^{br}_{x,y,t}(s), 0 \le s \le t)$ denote the Brownian bridge from $x$ to $y$ of length $t$.

Lemma 5

If $x < a$ and $y < a + bt$, then

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}(B^{br}_{x,y,t}(s) \ge a + bs \text{ for } \text{ some } s \in [0,t]) = \exp \left( -\frac{2(a-x)(a + bt - y)}{t}\right) \!.\quad \quad \end{aligned}$$

(5)

If $x > a$ and $y > a + bt$, then

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}(B^{br}_{x,y,t}(s) \le a + bs \text{ for } \text{ some } s \in [0,t]) = \exp \left( -\frac{2(x-a)(y - a - bt)}{t}\right) \!.\quad \quad \end{aligned}$$

(6)

Proof

Proposition 3 of [23] states that if $a > 0$ and $y < a + bt$, then

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}(B^{br}_{0,y,t}(s) \ge a + bs \text{ for } \text{ some } s \in [0,t]) = \exp \left( -\frac{2a(a + bt - y)}{t}\right) \!. \end{aligned}$$

The result (5) follows because $(B^{br}_{0,y,t}(s) + x(t-s)/t, 0 \le s \le t)$ is a Brownian bridge of length $t$ from $x$ to $y$. Then (6) follows because $(-B^{br}_{x,y,t}(s), 0 \le s \le t)$ is a Brownian bridge of length $t$ from $-x$ to $-y$. $\square$

Lemma 6

There exists a positive constant $C$ such that if $t > 0$ and $K + mt/2 \le 2L$, then for all $x \in [0,L]$ and all $y \in [0,K]$, we have

$$\begin{aligned} q_t(x,y) \le \frac{CL^4}{t^{5/2}} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L} \right) e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{K} \right) \!. \end{aligned}$$

Proof

First, we claim that

$$\begin{aligned} q_t(x,y)&= \frac{1}{\sqrt{2 \pi t}} e^{-(x - y)^2/2t} \cdot e^{\sqrt{2}(x-y) - t} \cdot e^t \cdot {{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{x,y,t}(s)\le L(s) \text{ for } \text{ all } s \in [0,t]). \end{aligned}$$

To see this, observe that the first factor is the density for standard Brownian motion, the second factor is a Girsanov term that relates Brownian motion with drift to standard Brownian motion, the third factor of $e^t$ accounts for the branching at rate 1 (this corresponds to the so-called Many-to-One Lemma), and the fourth factor is the probability that a Brownian particle that starts at $x$ and ends at $y$ avoids being killed at one of the boundaries. Therefore,

$$\begin{aligned} q_t(x,y) \le \frac{C e^{\sqrt{2}(x-y)}}{\sqrt{t}} {{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{x,y,t}(s) \le L(s) \text{ for } \text{ all } s \in [0,t]). \end{aligned}$$

(7)

Let $g$ denote the density of $B^{br}_{x,y,t}(t/2)$. Then

$$\begin{aligned}&{{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{x,y,t}(s) \le L(s) \text{ for } \text{ all } s \in [0,t]) \nonumber \\&\quad = \int _0^{L(t/2)} {{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{x,z,t/2}(s) \le L(s) \text{ for } \text{ all } s \in [0, t/2]) \nonumber \\&\qquad \times {{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{z,y,t/2} \le L(t/2 + s) \text{ for } \text{ all } s \in [0, t/2]) g(z) \, dz. \end{aligned}$$

(8)

Recall that $L(s) = L$ for all $s \in [0, t/2]$. Therefore, if $0 \le x \le L/2$ and $0 \le z \le L$, then by (6) with $a = b = 0$,

$$\begin{aligned}&{{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{x,z,t/2}(s) \le L(s) \text{ for } \text{ all } s \in [0, t/2]) \nonumber \\&\quad \le {{\mathrm{\mathbb {P}}}}(B^{br}_{x,z,t/2}(s) \ge 0 \text{ for } \text{ all } s \in [0, t/2]) \nonumber \\&\quad = 1 - {{\mathrm{\mathbb {P}}}}(B^{br}_{x,z,t/2}(s) \le 0 \text{ for } \text{ some } s \in [0, t/2]) \nonumber \\&\quad = 1 - \exp \left( - \frac{4xz}{t} \right) \nonumber \\&\quad \le \frac{4xL}{t}. \end{aligned}$$

(9)

If $L/2 \le x \le L$ and $0 \le z \le L$, then by (5) with $a = L$ and $b = 0$,

$$\begin{aligned}&{{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{x,z,t/2}(s) \le L(s) \text{ for } \text{ all } s \in [0, t/2])\nonumber \\&\quad \le {{\mathrm{\mathbb {P}}}}(B^{br}_{x,z,t/2}(s) \le L \text{ for } \text{ all } s \in [0, t/2]) \nonumber \\&\quad = 1 - {{\mathrm{\mathbb {P}}}}(B^{br}_{x,z,t/2}(s) \ge L \text{ for } \text{ some } s \in [0, t/2]) \nonumber \\&\quad = 1 - \exp \left( - \frac{4(L-x)(L-z)}{t} \right) \nonumber \\&\quad \le \frac{4(L-x)L}{t}. \end{aligned}$$

(10)

Combining (9) and (10), we get

$$\begin{aligned}&{{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{x,z,t/2}(s) \le L(s) \text{ for } \text{ all } s \in [0, t/2]) \nonumber \\&\quad \le \frac{4L}{t} \min \{x, L-x\} \le \frac{CL^2}{t} \sin \left( \frac{\pi x}{L} \right) \!. \end{aligned}$$

(11)

If $0 \le y \le K/2$ and $0 \le z \le L$, then using the same reasoning as in (9),

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{z,y,t/2}(s) \le L(t/2 + s) \text{ for } \text{ all } s \in [0, t/2])\nonumber \\ \le {{\mathrm{\mathbb {P}}}}(B^{br}_{z,y,t/2}(s) \ge 0 \text{ for } \text{ all } s \in [0, t/2]) \le \frac{4yL}{t}. \end{aligned}$$

(12)

If $K/2 \le y \le K$, then by (5) with $a = K + mt/2$ and $b = -m$,

$$\begin{aligned}&{{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{z,y,t/2}(s) \le L(t/2 + s) \text{ for } \text{ all } s \in [0, t/2]) \nonumber \\&\quad \le {{\mathrm{\mathbb {P}}}}(B^{br}_{z,y,t/2}(s) \le K + m(t/2 - s) \text{ for } \text{ all } s \in [0, t/2]) \nonumber \\&\quad = 1 - {{\mathrm{\mathbb {P}}}}(B^{br}_{z,y,t/2}(s) \ge K + m(t/2 - s) \text{ for } \text{ some } s \in [0, t/2]) \nonumber \\&\quad = 1 - \exp \left( \frac{4(K + mt/2 - z)(K - y)}{t} \right) \!. \nonumber \\&\quad \le \frac{4(K + mt/2)(K-y)}{t}. \end{aligned}$$

(13)

From (12) and (13) and the assumption that $K + mt/2 \le 2L$, we get

$$\begin{aligned}&{{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{z, y, t/2}(s) \le L(t/2 + s) \text{ for } \text{ all } s \in [0, t/2]) \nonumber \\&\quad \le \frac{8L}{t} \min \{y, K-y\} \le \frac{CL^2}{t} \sin \left( \frac{\pi y}{K} \right) \!. \end{aligned}$$

(14)

By (8), (11), and (14),

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}(0 \le B^{br}_{x,y,t}(s) \le L(s) \text{ for } \text{ all } s \in [0,t])&\le \frac{CL^4}{t^2} \sin \left( \frac{\pi x}{L} \right) \sin \left( \frac{\pi y}{K} \right) \int _0^{L(t/2)} g(z) \, dz \\&\le \frac{CL^4}{t^2} \sin \left( \frac{\pi x}{L} \right) \sin \left( \frac{\pi y}{K} \right) \!. \end{aligned}$$

The lemma follows by combining this result with (7). $\square$

2.3 A curved right boundary

We now consider the more general case in which the right boundary may change over time, which was studied in detail in [18]. Harris and Roberts [18] considered branching Brownian motion restricted to stay between $f(s) - L(s)$ and $f(s) + L(s)$, which is equivalent to our setting when both $f(s)$ and $L(s)$ are set equal to what we have denoted by $L(s)/2$. Assume that $s \mapsto L(s)$ is twice continuously differentiable.

Fix a point $x$ such that $0 < x < L(0)$. Following the analysis in [18], let $(\xi _t)_{t \ge 0}$ be a standard Brownian motion started at $x$, and define

$$\begin{aligned} G(s)&= \exp \left( \frac{1}{2} \int _0^s L'(u) \, d\xi _u - \frac{1}{8} \int _0^s L'(u)^2 \, du + \int _0^s \frac{\pi ^2}{2L(u)^2} \, du \right) \\&\times \exp \left( \frac{L'(s)}{2L(s)} (\xi _s -L(s)/2)^2 -\int _0^s \left( \frac{L''(u)}{2 L(u)} (\xi _u - L(u)/2)^2 + \frac{L'(u)}{2L(u)} \right) du \right) \!. \end{aligned}$$

Also, define

$$\begin{aligned} V(s) = G(s) \sin \left( \frac{\pi \xi _s}{L(s)}\right) \mathbf{1}_{\{ 0 < \xi _u < L(u) \, \forall u \le s\}}. \end{aligned}$$

(15)

It is shown in [18] using Itô’s formula (see Lemma 4.2 of [18] and the discussion immediately following that result) that the process $(V(s), s \ge 0)$ is a martingale.

We now write $G(s)$ as a product of three terms $G(s) = A(s) B(s) C(s)$ as follows:

$$\begin{aligned} A(s)&= \exp \left( \frac{1}{2} \int _0^s L'(u) \, d\xi _u - \frac{1}{8} \int _0^s L'(u)^2 \, du \right) \\ B(s)&= \exp \left( \, \int _0^s \frac{\pi ^2}{2L(u)^2} \, du - \int _0^s \frac{L'(u)}{2L(u)} \, du \right) \\ C(s)&= \exp \left( \frac{L'(s)}{2L(s)} (\xi _s -L(s)/2)^2 -\int _0^s \frac{L''(u)}{2L(u)} (\xi _u - L(u)/2)^2 \, du \right) \!. \end{aligned}$$

This leads to the following result about the expectation of $Z(s)$.

Lemma 7

Suppose initially there is a single particle at $x$. Then

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_x[Z(s)] = e^{\sqrt{2} x}B(s)^{-1} {{\mathrm{\mathbb {E}}}}_x[V(s) A(s)^{-1} C(s)^{-1}]. \end{aligned}$$

Proof

Recall that $(\xi _t)_{t \ge 0}$ is standard Brownian motion with $\xi _0 = x$. By the well-known Many-to-One Lemma for branching Brownian motion (see, for example, equation (3) of [16]),

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_x[Z(s)] = e^s {{\mathrm{\mathbb {E}}}}_x\left[ e^{\sqrt{2} (\xi _s -\sqrt{2}s) } \sin \left( \frac{\pi (\xi _s -\sqrt{2} s)}{L(s)} \right) \mathbf{1}_{\{0 < \xi _u - \sqrt{2}u < L(u) \, \forall u \le s\}} \right] \!. \end{aligned}$$

Using Girsanov’s Theorem to relate Brownian motion with drift to standard Brownian motion,

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_x[Z(s)]&= e^s {{\mathrm{\mathbb {E}}}}_x\left[ e^{-s - \sqrt{2}(\xi _s - x)} \cdot e^{\sqrt{2} \xi _s} \sin \left( \frac{\pi \xi _s}{L(s)} \right) \mathbf{1}_{\{0 < \xi _u < L(u) \, \forall u \le s\}} \right] \\&= e^{\sqrt{2} x} {{\mathrm{\mathbb {E}}}}_x \left[ \sin \left( \frac{\pi \xi _s}{L(s)} \right) \mathbf{1}_{\{0 < \xi _u < L(u) \, \forall u \le s\}} \right] \\&= e^{\sqrt{2} x} {{\mathrm{\mathbb {E}}}}_x \left[ \frac{V(s)}{G(s)} \right] \\&= e^{\sqrt{2} x} B(s)^{-1} {{\mathrm{\mathbb {E}}}}_x[V(s) A(s)^{-1} C(s)^{-1}], \end{aligned}$$

as claimed. $\square$

3 The case $L(s) = c(t-s)^{1/3}$

Fix any time $t > 0$, and for $0 \le s \le t$, define

$$\begin{aligned} L(s) = c (t-s)^{1/3}, \end{aligned}$$

where $c$ was defined in (1). This right boundary was previously considered by Kesten [21]. Note that for $0 < s < t$,

$$\begin{aligned} L'(s) = - \frac{c}{3} (t - s)^{-2/3} \end{aligned}$$

and

$$\begin{aligned} L''(s) = - \frac{2c}{9} (t - s)^{-5/3}. \end{aligned}$$

Also, a straightforward calculation gives

$$\begin{aligned} B(s)^{-1} = \exp ( -(3 \pi ^2)^{1/3}(t^{1/3} - (t - s)^{1/3})) \left( \frac{t-s}{t} \right) ^{1/6}\!. \end{aligned}$$

We consider in this section branching Brownian motion with drift $-\sqrt{2}$ in which particles are killed if they reach $0$ or $L(s)$ at time $s$. All particles will be killed by time $t$ because $L(t) = 0$. We define $X_i(s), \,N(s)$, and $Z(s)$ as in Sect. 2.

3.1 Estimating ${{\mathrm{\mathbb {E}}}}_x[Z(s)]$

In this section, we will estimate ${{\mathrm{\mathbb {E}}}}_x[Z(s)]$ when $0 < s < t$. In view of Lemma 7, this will require bounds on $A(s)$ and $C(s)$, which we present in Lemmas 8 and 9 below. Note that the constants $c_1, \ldots , c_6$ in these lemmas and in Proposition 10 do not depend on the initial position $x$ of the Brownian motion $(\xi _t)_{t \ge 0}$.

Lemma 8

There exist positive constants $c_1$ and $c_2$ such that for all $s \in (0, t)$, almost surely on the event $\{0 < \xi _u < L(u) \, \forall u \le s\}$ we have

$$\begin{aligned} \exp (-c_1(t-s)^{-1/3}) \le C(s) \le \exp (c_2 (t-s)^{-1/3}). \end{aligned}$$

Proof

On the event $\{0 < \xi _u < L(u) \, \forall u \le s\}$, we have

$$\begin{aligned} C(s)&\le \exp \left( \, \int _0^s \left| \frac{L''(u)}{2 L(u)} (\xi _u - L(u)/2)^2 \right| du \right) \nonumber \\&\le \exp \left( \, \int _0^s \left| \frac{L''(u) L(u)}{8} \right| du \right) \nonumber \\&= \exp \left( \frac{c^2}{36} \int _0^s (t-u)^{-4/3} \, du \right) \nonumber \\&\le \exp \left( \frac{c^2}{12} (t-s)^{-1/3} \right) \!. \end{aligned}$$

(16)

On the other hand, on the event $\{0 < \xi _u < L(u) \, \forall u \le s\}$,

$$\begin{aligned} C(s)&\ge \exp \left( \frac{L'(s)}{2L(s) } (\xi _s -L(s)/2)^2 \right) \nonumber \\&\ge \exp \left( -\frac{c^2}{24} (t-s)^{-1/3} \right) \!. \end{aligned}$$

(17)

The result follows from (16) and (17). $\square$

Lemma 9

There exist positive constants $c_3$ and $c_4$ such that for all $s \in (0, t)$, almost surely on the event $\{0 < \xi _u < L(u) \, \forall u \le s\}$ we have

$$\begin{aligned} \exp (- c_3 (t-s)^{-1/3}) \le A(s) \le \exp (c_4 (t-s)^{-1/3}). \end{aligned}$$

Proof

Observe that

$$\begin{aligned} \int _0^s L'(u)^2 \, du = \frac{c^2}{3} \left( (t - s)^{-1/3} - t^{-1/3} \right) \le \frac{c^2}{3} (t - s)^{-1/3}. \end{aligned}$$

Therefore,

$$\begin{aligned} \exp \left( \frac{1}{2} \int _0^s L'(u) \, d\xi _u \right) \exp \left( - \frac{c^2}{24} (t-s)^{-1/3} \right) \le A(s) \le \exp \left( \frac{1}{2} \int _0^s L'(u) \, d\xi _u \right) \!, \end{aligned}$$

so it suffices to prove the result with $\exp (\frac{1}{2} \int _0^s L'(u) \, d\xi _u)$ in place of $A(s)$.

Using the (stochastic) Integration by Parts Formula and the fact that $L'$ has finite variation,

$$\begin{aligned} \int _0^s L'(u) \, d\xi _u = L'(s)\xi _s -L'(0)\xi _0 -\int _0^s L''(u) \xi _u \, du. \end{aligned}$$

On the event $\{0 < \xi _u < L(u) \, \forall u \le s\}$, we have $0 \le - L'(s)\xi _s \le \frac{c^2}{3}(t-s)^{-1/3}$, which is also valid for $s = 0$, and

$$\begin{aligned} 0 \le - \int _0^s L''(u) \xi _u \, du \le \frac{2c^2}{9} \int _0^s (t-u)^{-4/3} \, du \le \frac{2c^2}{3} (t-s)^{-1/3}. \end{aligned}$$

These inequalities yield the conclusion. $\square$

Proposition 10

There exist positive constants $c_5$ and $c_6$ such that for all $s \in (0, t)$,

$$\begin{aligned} Z(0) B(s)^{-1} \exp (- c_5 (t-s)^{-1/3}) \le {{\mathrm{\mathbb {E}}}}_x[Z(s)] \le Z(0) B(s)^{-1} \exp (c_6 (t-s)^{-1/3}). \end{aligned}$$

Proof

First, suppose that initially there is a single particle at $x$ with $0 < x < L(0)$. Recall the definition of $V(s)$ from (15). Because $V(s) = 0$ outside of the event $\{0 < \xi _u < L(u) \, \forall u \le s\}$, it follows from Lemmas 7, 8, and 9 that there are constants $c_7$ and $c_8$ such that

$$\begin{aligned}&e^{\sqrt{2} x} B(s)^{-1} {{\mathrm{\mathbb {E}}}}_x[V(s)] \exp \,(-c_7 (t-s)^{-1/3})\\&\quad \le {{\mathrm{\mathbb {E}}}}_x[Z(s)] \le e^{\sqrt{2} x} B(s)^{-1} {{\mathrm{\mathbb {E}}}}_x[V(s)] \exp (c_8 (t-s)^{-1/3}). \end{aligned}$$

Because $(V(s), s \ge 0)$ is a martingale,

$$\begin{aligned} e^{\sqrt{2}x}{{\mathrm{\mathbb {E}}}}_x[V(s)] = e^{\sqrt{2}x}V(0) = e^{\sqrt{2} x} G(0) \sin \left( \frac{\pi x}{L(0)} \right) = Z(0) G(0). \end{aligned}$$

The result follows because

$$\begin{aligned} 1 \ge G(0) = \exp \left( \frac{L'(0)}{2L(0)} (\xi _0 - L(0)/2)^2 \right) \ge \exp \left( \frac{L'(0) L(0)}{8} \right) = \exp \left( - \frac{c^2}{24} t^{-1/3} \right) \!. \end{aligned}$$

$\square$

Note that because $B(s)$ and the constants $c_5$ and $c_6$ do not depend on the position $x$ of the initial particle, Proposition 10 and the corollary below hold for general initial configurations by summing over the particles.

Corollary 11

Let $(\mathcal{F}_u, u \ge 0)$ be the natural filtration associated with the branching Brownian motion. Let $0 < r < s < t$. Let

$$\begin{aligned} B_r(s)&= \exp \left( \int _r^s \frac{\pi ^2}{2 L(u)^2} \, du - \int _r^s \frac{L'(u)}{2 L(u)} \, du \right) \\&= \exp \left( (3 \pi ^2)^{1/3} ((t-r)^{1/3} - (t-s)^{1/3}) \right) \left( \frac{t-r}{t-s} \right) ^{1/6}\!. \end{aligned}$$

Then

$$\begin{aligned} Z(r) B_r(s)^{-1} \exp (-c_5 (t - s)^{-1/3}) \le {{\mathrm{\mathbb {E}}}}[Z(s)|\mathcal{F}_r] \le Z(r) B_r(s)^{-1} \exp (c_6(t - s)^{-1/3}), \end{aligned}$$

where $c_5$ and $c_6$ are the constants from Proposition 10.

Proof

Apply the Markov Property at time $r$, and then apply Proposition 10 with $t^* = t - r$ and $L^*(u) = c(t^* - u)^{1/3} = c(t - r - u)^{1/3} = L(u + r)$. $\square$

3.2 Bounding the density

We now use the estimate of ${{\mathrm{\mathbb {E}}}}[Z(s)]$ from Proposition 10 to obtain bounds on the density. For $0 \le r < s < t$, let $q_{r,s}(x,y)$ represent the density of particles at time $s$ that are descended from a particle at the location $x$ at time $r$. That is, if $A$ is a Borel subset of $(0, L(s))$, then the expected number of particles in $A$ at time $s$ descended from the particle which is at $x$ at time $r$ is

$$\begin{aligned} \int _A q_{r,s}(x,y) \, dy. \end{aligned}$$

Note that $q_s(x,y) = q_{0,s}(x,y)$. For $x,y>0$ and $0 \le r\le s \le t$, let

$$\begin{aligned} \psi _{r,s}(x,y)&= \frac{1}{L(s)} e^{-(3 \pi ^2)^{1/3}((t-r)^{1/3} - (t-s)^{1/3})} \left( \frac{t-s}{t-r} \right) ^{1/6}\\&\times ~e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(r)} \right) e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)} \right) \!. \end{aligned}$$

This expression becomes simpler if we view the process from time $t$, as we get

$$\begin{aligned} \psi _{t-u,t-v}(x,y) = \frac{1}{c} e^{-(3 \pi ^2)^{1/3} (u^{1/3} - v^{1/3})}\left( \frac{1}{uv}\right) ^{1/6} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{cu^{1/3}} \right) e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{cv^{1/3}} \right) \!. \end{aligned}$$

Proposition 12

Fix a positive constant $b$. There exists a constant $A > 0$ and positive constants $C'$ and $C''$, with $C''$ depending on $b$, such that if $r + L(r)^2 \le s \le t - A$, then

$$\begin{aligned} q_{r,s}(x,y) \ge C' \psi _{r,s}(x,y), \end{aligned}$$

(18)

and if $r + bL(r)^2 \le s \le t - A$, then

$$\begin{aligned} q_{r,s}(x,y) \le C'' \psi _{r,s}(x,y). \end{aligned}$$

(19)

Proof

Let ${{\mathrm{\mathbb {E}}}}_{r,x}$ denote expectation for the process starting from a single particle at $x$ at time $r$. Note that if $r < u < s$, then

$$\begin{aligned} q_{r,s}(x,y) = \int _0^{L(u)} q_{r,u}(x,z) q_{u,s}(z,y) \, dz. \end{aligned}$$

(20)

We first prove the upper bound. We may assume $b \le 1$. Assume $r + bL(r)^2 \le s \le t - A$. Let $u = s - { b}L(s)^2$. Note that $u > r$ because $L(s) < L(r)$. Let $m = -2L'(s) = (2c/3)(t-s)^{-2/3}$. For $u \le v \le s$, let

$$\begin{aligned} {\hat{L}}(v) = \left\{ \begin{array}{l@{\quad }l} L(u) &{} \text{ if } u \le v \le s - m^{-1}(L(u) - L(s)) \\ L(s) + m(s-v) &{} \text{ if } s - m^{-1}(L(u)-L(s)) \le v \le s. \end{array} \right. \end{aligned}$$

Note that ${\hat{L}}(v) \ge L(v)$ for all $v \in [u, s]$ (see Fig. 1). Therefore, if we define ${\hat{q}}_{u,s}(z,y)$ in the same way as $q_{u,s}(z,y)$, except that for $v \in [u,s]$, particles are killed when they reach ${\hat{L}}(v)$ instead of when they reach $L(v)$, then

$$\begin{aligned} q_{u,s}(z,y) \le {\hat{q}}_{u,s}(z,y). \end{aligned}$$

(21)

We now wish to apply Lemma 6 with $K = L(s), \,L = L(u)$ and $t = s - u$. We need to check first that $L(s)+m(s-u)/2\le 2L(u)$ and second that $m^{-1}(L(u)-L(s))\le (s-u)/2$. For the first condition, as long as $A$ is chosen to be large enough that $L(t - A) \ge c^3/3$, we have

$$\begin{aligned} L(s) + \frac{m(s-u)}{2} = L(s) + \frac{mbL(s)^2}{2} = L(s) + \frac{b c^3}{3} \le 2L(s) \le 2L(u). \end{aligned}$$

The second condition also holds because

$$\begin{aligned} m^{-1}(L(u) - L(s)) \le m^{-1}|L'(s)|(s - u) = \frac{s-u}{2}. \end{aligned}$$

Therefore, by Lemma 6,

$$\begin{aligned} {\hat{q}}_{u,s}(z,y) \le \frac{C L(u)^4}{(b L(s)^2)^{5/2}} e^{\sqrt{2} z} \sin \left( \frac{\pi z}{L(u)} \right) e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)} \right) \!. \end{aligned}$$

(22)

Note that

$$\begin{aligned} L(u) - L(s) \le -L'(s)(s - u) = \frac{b c^3}{3}. \end{aligned}$$

(23)

Therefore, if $A$ is large enough that $L(t - A) \ge c^3/3$, then $L(u) \le 2 L(s)$, so combining (20), (21), (22), we get

$$\begin{aligned} q_{r,s}(x,y)&\le \frac{C}{L(s)} e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)} \right) \int _0^{L(u)} e^{\sqrt{2} z} \sin \left( \frac{\pi z}{L(u)} \right) q_{r,u}(x,z) \, dz \\&= \frac{C}{L(s)} e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)} \right) {{\mathrm{\mathbb {E}}}}_{r,x}[Z(u)], \end{aligned}$$

where above, we recall that the subscript indicates that we start with one particle at time $r$ and at position $x$. Therefore, using Corollary 11 to bound ${{\mathrm{\mathbb {E}}}}_{r,x}[Z(u)]$,

$$\begin{aligned} q_{r,s}(x,y)&\le \frac{C}{L(s)} e^{-(3 \pi ^2)^{1/3}((t - r)^{1/3} - (t - u)^{1/3})} \left( \frac{t-u}{t-r} \right) ^{1/6}\\&\times ~ e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(r)} \right) e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)} \right) \!. \end{aligned}$$

The upper bound (19) now follows because $(t - u)^{1/3} \le (t - s)^{1/3} + bc^2/3$ by (23) and $t - u = (t - s) + (s - u) \le C(t - s)$.

Fig. 1

The function $\hat{L}$

We next prove the lower bound. Assume that $r + L(r)^2 \le s \le t - A$. Let $u = s - L(s)^2/2$. Note that $u > r$ because $L(s) < L(r)$. For $0 \le z \le L(s)$, define ${\tilde{q}}_{u,s}(z,y)$ in the same way as $q_{u,s}(z,y)$ except that for $v \in [u,s]$, particles are killed when they reach $L(s)$ instead of when they reach $L(v)$. Then

$$\begin{aligned} q_{u,s}(z,y) \ge {\tilde{q}}_{u,s}(z,y). \end{aligned}$$

(24)

By Lemma 3, if $0 \le z \le L(s)$, then because

$$\begin{aligned} \sum _{n=2}^{\infty } n^2 e^{-\pi ^2 (n^2 - 1)(s-u)/2L(s)^2} = \sum _{n=2}^{\infty } n^2 e^{-\pi ^2 (n^2 - 1)/4} < 1, \end{aligned}$$

we have

$$\begin{aligned} {\tilde{q}}_{u,s}(z,y)\ge \frac{C}{L(s)} e^{-\pi ^2 (s-u)/2L(s)^2} e^{\sqrt{2} z} \sin \left( \frac{\pi z}{L(s)} \right) e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)} \right) \!.\quad \quad \end{aligned}$$

(25)

By (20), (24), and (25),

$$\begin{aligned} q_{r,s}(x,y) \ge \frac{C}{L(s)} e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)} \right) \int _0^{L(s)} e^{\sqrt{2} z} \sin \left( \frac{\pi z}{L(s)} \right) q_{r,u}(x,z) \, dz. \end{aligned}$$

Using (23) with $b=1/2$, we get $L(u) - L(s) \le c^3/6$. Therefore, there is a positive constant $C$ such that $\sin (\pi z/L(s)) \ge C \sin (\pi z/L(u))$ for all $z \le L(u) - c^3$. It follows that

$$\begin{aligned}&q_{r,s}(x,y)\nonumber \\&\quad \ge \frac{C}{L(s)} e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)} \right) \left( {{\mathrm{\mathbb {E}}}}_{r,x}[Z(u)] - \int _{L(u) - c^3}^{L(u)} e^{\sqrt{2} z} \sin \left( \frac{\pi z}{L(u)} \right) q_{r,u}(x,z) \, dz \right) \!.\nonumber \\ \end{aligned}$$

(26)

By Corollary 11,

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{r,x}[Z(u)] \ge C e^{-(3 \pi ^2)^{1/3}((t-r)^{1/3} - (t-u)^{1/3})} \left( \frac{t-u}{t-r} \right) ^{1/6} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(r)} \right) \!.\quad \quad \end{aligned}$$

(27)

Also, because $u - r = s - L(s)^2/2 - r \ge L(r)^2 - L(s)^2/2 \ge L(r)^2/2$, we can apply the upper bound (19) to get

$$\begin{aligned}&\int _{L(u) - c^3}^{L(u)} e^{\sqrt{2} z} \sin \left( \frac{\pi z}{L(u)} \right) q_{r,u}(x,z) \, dz \nonumber \\&\quad \le \frac{C}{L(u)} e^{-(3 \pi ^2)^{1/3}((t - r)^{1/3} - (t-u)^{1/3})} \left( \frac{t-u}{t-r} \right) ^{1/6}\nonumber \\&\quad \quad \times ~e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(r)} \right) \int _{L(u) - c^3}^{L(u)} \sin \left( \frac{\pi z}{L(u)} \right) ^2 \, dz \nonumber \\&\quad \le \frac{C}{L(u)^3} e^{-(3 \pi ^2)^{1/3}((t - r)^{1/3} - (t-u)^{1/3})} \left( \frac{t-u}{t-r} \right) ^{1/6} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(r)} \right) \!.\quad \quad \end{aligned}$$

(28)

Choosing $A$ sufficiently large, the lower bound (18) now follows from (26), (27), (28), and the fact that $t - u \ge t - s$. $\square$

3.3 Particles hitting the right boundary

For $0 \le s < u \le t$, let $R_{s,u}$ denote the number of particles that are killed at $L(r)$ for some $r \in [s, u]$. Let ${{\mathrm{\mathbb {E}}}}_{s,x}$ denote expectation for the process started from a single particle at $x$ at time $s$.

Lemma 13

If $0 \le s < u < t$, then

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{s,x}[R_{s, u}] \le \frac{x e^{\sqrt{2} x} e^{-\sqrt{2} L(u)}}{L(u)}. \end{aligned}$$

Proof

For branching Brownian motion with absorption only at the origin, if we define

$$\begin{aligned} M(s) = \sum _{i=1}^{N(s)} X_i(s) e^{\sqrt{2} X_i(s)}, \end{aligned}$$

then it is well-known (see, for example, Lemma 2 of [16]) that the process $(M(s), s \ge 0)$ is a martingale. Now, for $u \in [s, t]$, let

$$\begin{aligned} M_s(u) = \sum _{i=1}^{N(u)} X_i(u) e^{\sqrt{2} X_i(u)} + L(u) e^{\sqrt{2} L(u)} R_{s,u}. \end{aligned}$$

(29)

We claim that the process $(M_s(u), s \le u \le t)$ is a supermartingale for branching Brownian motion with killing both at the origin and at the right boundary $L(\cdot ).$ To see this, observe that because the process $(M(s), s \ge 0)$ is a martingale when there is no killing at the right boundary, this process would still be a martingale if particles were stopped, but not killed, upon reaching the right boundary. (Essentially, this follows from the simple fact that a martingale, stopped at a stopping time, is still a martingale. See for instance Theorem 5.1 and Corollary 5.4 in [20] for a related statement and proof.) Because the function $u \mapsto L(u)$ is decreasing and because $x \mapsto x e^{\sqrt{2}x}$ is increasing, the process becomes a supermartingale if particles, after hitting the right boundary, follow the right boundary until time $t$. This is the process defined in (29) because there will be $R_{s,u}$ particles at $L(u)$ at time $u$.

Because the process defined in (29) is a supermartingale, we have

$$\begin{aligned} xe^{\sqrt{2} x}&= {{\mathrm{\mathbb {E}}}}_{s,x}[M_s(s)] \ge {{\mathrm{\mathbb {E}}}}_{s,x}[M_s(u)] \ge {{\mathrm{\mathbb {E}}}}_{s,x}[L(u) e^{\sqrt{2} L(u)} R_{s,u}]\\&= L(u) e^{\sqrt{2} L(u)} {{\mathrm{\mathbb {E}}}}_{s,x}[R_{s,u}]. \end{aligned}$$

The result follows. $\square$

Lemma 14

There is a constant $A > 0$ such that for all $s, \,u$, and $x$ such that $s \ge 0, \,0 < x < L(s)$, and $s + L(s)^2 \le u \le t - A$, we have

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{s,x}[R_{u, u+1}] \asymp \frac{1}{L(u)^2} e^{-(3 \pi ^2)^{1/3} (t-s)^{1/3}} \left( \frac{t-u}{t-s} \right) ^{1/6} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(s)} \right) \!. \end{aligned}$$

Proof

We adapt ideas from the proofs of Lemma 15 and Proposition 16 in [5]. By applying the Markov property at time $u$, we get

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{s,x}[R_{u, u+1}] = \int _0^{L(u)} q_{s,u}(x,y) {{\mathrm{\mathbb {E}}}}_{u,y}[R_{u, u+1}] \, dy. \end{aligned}$$

(30)

Let $(\xi _r)_{r \ge 0}$ be standard Brownian motion with $\xi _0 = 0$. Because a particle at time $u$ will have on average $e$ descendants at time $u+1$ if no particles are killed, the expectation ${{\mathrm{\mathbb {E}}}}_{u,y}[R_{u,u+1}]$ is bounded above by $e$ times the probability that a particle started from $y$ at time $u$ is to the right of $L(u+1)$ at some time before time $u+1$. Therefore, it follows from the Reflection Principle and the inequality

$$\begin{aligned} \int _z^{\infty } e^{-x^2/2} \, dx \le z^{-1} e^{-z^2/2}, \end{aligned}$$

valid for $z>0$, that if $y < L(u+1)$, then

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{u,y}[R_{u, u+1}]&\le e\,{{\mathrm{\mathbb {P}}}}\left( \max _{0 \le r \le 1} (\xi _r - \sqrt{2} r) \ge L(u+1) - y\right) \\&\le e\,{{\mathrm{\mathbb {P}}}}\left( \max _{0 \le r \le 1} \xi _r \ge L(u+1) - y\right) \\&\le 2e\,{{\mathrm{\mathbb {P}}}}(\xi _1 \ge L(u+1) - y) \\&\le \frac{C}{L(u+1) - y} e^{-(L(u+1) - y)^2/2}. \end{aligned}$$

Therefore, letting $\alpha = L(u) - L(u+1)$ and requiring $A$ to be large enough that $L(t-A+1) > 1$, we have that $0 \le \alpha \le C$ and (using the change of variable $z=L(u)-y)$

$$\begin{aligned}&\int _0^{L(u) - \alpha - 1} e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(u)} \right) {{\mathrm{\mathbb {E}}}}_{u,y}[R_{u, u+1}] \, dy \nonumber \\&\quad \le C \int _0^{L(u) - \alpha - 1} e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(u)} \right) \frac{1}{L(u+1) - y} e^{-(L(u+1) - y)^2/2} \, dy \nonumber \\&\quad \le C e^{-\sqrt{2} L(u)} \int _{\alpha + 1}^{L(u)} e^{\sqrt{2} z} \cdot \frac{\pi z}{L(u)} \cdot \frac{1}{z - \alpha } e^{-(z - \alpha )^2/2} \, dz \nonumber \\&\quad \le \frac{C e^{-\sqrt{2} L(u)}}{L(u)}. \end{aligned}$$

(31)

Using the bound ${{\mathrm{\mathbb {E}}}}_{u,y}[R_{u, u+1}] \le e$, we get

$$\begin{aligned} \int _{L(u) - \alpha - 1}^{L(u)} e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(u)} \right) {{\mathrm{\mathbb {E}}}}_{u,y}[R_{u, u+1}] \, dy \le \frac{C e^{-\sqrt{2} L(u)}}{L(u)}. \end{aligned}$$

(32)

Combining (31) and (32) with (30) and Proposition 12, and using the fact that $e^{-\sqrt{2} L(u)} = e^{-(3 \pi ^2)^{1/3} (t - u)^{1/3}}$, we get, for $A$ large enough,

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{s,x}[R_{u, u+1}] \le \frac{C}{L(u)^2} e^{-(3 \pi ^2)^{1/3}(t - s)^{1/3}} \left( \frac{t-u}{t-s} \right) ^{1/6} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(s)} \right) \!, \end{aligned}$$

which is the upper bound in the statement of the lemma.

Next, observe that for $y \in [L(u) - 1, L(u)]$, we have

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{u,y}[R_{u, u+1}] \ge {{\mathrm{\mathbb {P}}}}(\xi _1 - \sqrt{2} \ge L(u+1) - y) \ge {{\mathrm{\mathbb {P}}}}(\xi _1 \ge 1 + \sqrt{2}) \ge C. \end{aligned}$$

Thus, by (30) and Proposition 12,

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{s,x}[R_{u,u+1}]&\ge \frac{C}{L(u)} e^{-(3 \pi ^2)^{1/3}((t-s)^{1/3} - (t-u)^{1/3})} \left( \frac{t-u}{t-s} \right) ^{1/6} \\&\times ~e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(s)} \right) \int _{L(u) - 1}^{L(u)} e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(u)} \right) \, dy \\&\ge \frac{C}{L(u)^2} e^{-(3 \pi ^2)^{1/3}(t - s)^{1/3}} \left( \frac{t-u}{t-s} \right) ^{1/6} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(s)} \right) \!, \end{aligned}$$

which gives the required lower bound. $\square$

Lemma 15

There is a constant $A_0 > 0$ and positive constants $C'$ and $C''$ such that if $0 \le s \le t - A_0$ and $0 < x < L(s)$, then

$$\begin{aligned} C' h(s,x) \le {{\mathrm{\mathbb {E}}}}_{s,x}[R_{s,t}] \le C''(h(s,x) + j(s,x)), \end{aligned}$$

(33)

where

$$\begin{aligned} h(s,x) = e^{\sqrt{2}x} \sin \left( \frac{\pi x}{L(s)}\right) (t-s)^{1/3} \exp (- (3\pi ^2(t-s))^{1/3}) \end{aligned}$$

(34)

and

$$\begin{aligned} j(s,x) = x e^{\sqrt{2}x} (t-s)^{-1/3} \exp (- (3\pi ^2(t-s))^{1/3}). \end{aligned}$$

Also, if $0 < \alpha < \beta < 1$, then

$$\begin{aligned} C' h(s,x) \le {{\mathrm{\mathbb {E}}}}_{s,x}[R_{s + \alpha (t-s), s + \beta (t-s)}] \le C'' h(s,x), \end{aligned}$$

(35)

where the constants $C'$ and $C''$ depend on $\alpha$ and $\beta$.

Proof

If $u = s + L(s)^2$, then for sufficiently large $A_0$,

$$\begin{aligned} L(s) - L(u) \le -L'(u)(u - s) = \frac{c^3}{3} \left( \frac{t-s}{t-u} \right) ^{2/3} \le C. \end{aligned}$$

Therefore, by Lemma 13, using that $\sqrt{2} L(s) = (3 \pi ^2(t-s))^{1/3}$,

$$\begin{aligned} 0 \le {{\mathrm{\mathbb {E}}}}_{s,x}[R_{s, s + L(s)^2}] \le \frac{C x e^{\sqrt{2} x} e^{-\sqrt{2} L(s)}}{L(s)} \le C j(s,x). \end{aligned}$$

(36)

We may choose $A_0$ to be large enough that $s + L(s)^2 \le t - A - 1$ whenever $0 \le s \le t - A_0$, where $A$ is the constant from Lemma 14. By Lemma 14,

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{s,x}[R_{s+L(s)^2, t-A}]&\asymp&\frac{e^{-(3 \pi ^2)^{1/3}(t-s)^{1/3}}}{(t-s)^{1/6}} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(s)} \right) \int _{s + L(s)^2}^{t-A} \frac{(t-u)^{1/6}}{L(u)^2} \, du \nonumber \\&\asymp&\frac{e^{-(3 \pi ^2)^{1/3}(t-s)^{1/3}}}{(t-s)^{1/6}} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(s)} \right) \int _{s + L(s)^2}^{t - A} \frac{1}{(t-u)^{1/2}} \, du \nonumber \\&\asymp&h(s,x). \end{aligned}$$

(37)

Because particles branch at rate one, ${{\mathrm{\mathbb {E}}}}_{s,x}[R_{t-A, t}]$ is at most $e^A$ times the expected number of particles between 0 and $L(t-A)$ at time $t-A$. Therefore, by Proposition 12,

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{s,x}[R_{t-A, t}]&\le e^A \int _0^{L(t-A)} q_{s,t-A}(x,y) \, dy \nonumber \\&\le \frac{C e^{-(3 \pi ^2)^{1/3}(t-s)^{1/3}}}{(t-s)^{1/6}} e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(s)} \right) \int _0^{L(t-A)} e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)} \right) \, dy \nonumber \\&\le \frac{C h(s,x)}{(t-s)^{5/6}}. \end{aligned}$$

(38)

The result (33) follows from (36), (37), and (38). The result (35) follows from the reasoning in (37), using $s + \alpha (t-s)$ and $s + \beta (t-s)$ as the limits of integration.

$\square$

Lemma 16

Let $0 < \alpha < \beta < 1$. Let $A_0$ be the constant defined in Lemma 15. Then there exist positive constants $C'$ and $C''$ depending on $\alpha$ and $\beta$ such that if $t \ge A_0$ and $0 < x < L(0) - 1$, then

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{0,x}[R_{\alpha t, \beta t}^2] \le C h(0,x). \end{aligned}$$

Proof

The proof is similar to the proof of Proposition 18 in [5]. Throughout this proof, we write $R = R_{\alpha t, \beta t}$. Note that $R^2 = R + 2Y$, where $Y$ is the number of distinct pairs of particles that reach $L(s)$ for some $s \in [\alpha t, \beta t]$. A branching event at the location $y$ at time $s$ produces, on average, $({{\mathrm{\mathbb {E}}}}_{s,y}[R])^2$ pairs of particles that reach the right boundary and have their most recent common ancestor at time $s$. Therefore, by Lemma 15, we may write

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}_{0,x}[R^2]&= {{\mathrm{\mathbb {E}}}}_{0,x}[R] + 2 \int _0^{\beta t} \int _0^{L(s)} q_{0,s}(x,y) ({{\mathrm{\mathbb {E}}}}_{s,y}[R])^2 \, dy \, ds \nonumber \\&\le {{\mathrm{\mathbb {E}}}}_{0,x}[R] + C \int _0^{\beta t} \int _0^{L(s)} q_{0,s}(x,y)(h(s,y)^2 + j(s,y)^2) \, dy \, ds.\quad \quad \end{aligned}$$

(39)

We bound separately the term involving $h(s,y)^2$ and the term involving $j(s,y)^2$. We also treat separately the cases $s \le L(0)^2$ and $s \ge L(0)^2$.

By Proposition 12 and (34),

$$\begin{aligned}&\int _{L(0)^2}^{\beta t} \int _0^{L(s)} q_{0,s}(x,y) h(s,y)^2 \, dy \, ds \nonumber \\&\quad \le C e^{-(3 \pi ^2)^{1/3}t^{1/3}} e^{\sqrt{2}x} \sin \left( \frac{\pi x}{L(0)} \right) \int _{L(0)^2}^{\beta t} \int _0^{L(s)} \frac{1}{L(s)} \left( \frac{t-s}{t} \right) ^{1/6} e^{(3 \pi ^2)^{1/3}(t-s)^{1/3}} \nonumber \\&\qquad \times e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)}\right) \left\{ (t-s)^{1/3} e^{-(3 \pi ^2)^{1/3}(t-s)^{1/3}} e^{\sqrt{2}y} \sin \left( \frac{\pi y}{L(s)}\right) \right\} ^2 \, dy \, ds \nonumber \\&\quad \le \frac{C e^{-(3 \pi ^2)^{1/3}t^{1/3}}}{t^{1/6}} e^{\sqrt{2}x} \sin \left( \frac{\pi x}{L(0)}\right) \int _{L(0)^2}^{\beta t} e^{-(3 \pi ^2)^{1/3}(t-s)^{1/3}} (t - s)^{1/2} \nonumber \\&\qquad \times \int _0^{L(s)} e^{\sqrt{2}y} \sin ^3\left( \frac{\pi y}{L(s)}\right) \, dy \, ds \nonumber \\&\quad \le \frac{C e^{-(3 \pi ^2)^{1/3}t^{1/3}}}{t^{1/6}} e^{\sqrt{2}x} \sin \left( \frac{\pi x}{L(0)}\right) \int _{L(0)^2}^{\beta t} e^{-(3 \pi ^2)^{1/3}(t - s)^{1/3}} (t - s)^{1/2} \frac{e^{\sqrt{2} L(s)}}{L(s)^3} \, ds \nonumber \\&\quad \le \frac{C e^{-(3 \pi ^2)^{1/3}t^{1/3}}}{t^{1/6}} e^{\sqrt{2}x} \sin \left( \frac{\pi x}{L(0)}\right) \int _{L(0)^2}^{\beta t} \frac{1}{(t - s)^{1/2}} \, ds \nonumber \\&\quad \le C h(0,x). \end{aligned}$$

(40)

A similar computation gives

$$\begin{aligned}&\int _{L(0)^2}^{\beta t} \int _0^{L(s)} q_{0,s}(x,y) j(s,y)^2 \, dy \, ds \nonumber \\&\quad \le C e^{-(3 \pi ^2)^{1/3}t^{1/3}} e^{\sqrt{2}x} \sin \left( \frac{\pi x}{L(0)} \right) \int _{L(0)^2}^{\beta t} \int _0^{L(s)} \frac{1}{L(s)} \left( \frac{t-s}{t} \right) ^{1/6} e^{(3 \pi ^2)^{1/3}(t - s)^{1/3}} \nonumber \\&\qquad \times e^{-\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)}\right) \left\{ (t-s)^{-1/3} e^{-(3 \pi ^2)^{1/3}(t-s)^{1/3}} y e^{\sqrt{2}y} \right\} ^2 \, dy \, ds \nonumber \\&\quad \le \frac{C e^{-(3 \pi ^2)^{1/3}t^{1/3}}}{t^{1/6}} e^{\sqrt{2}x} \sin \left( \frac{\pi x}{L(0)}\right) \int _{L(0)^2}^{\beta t} e^{-(3 \pi ^2)^{1/3}(t-s)^{1/3}} \frac{1}{(t-s)^{5/6}} \nonumber \\&\qquad \times \int _0^{L(s)} e^{\sqrt{2} y} y^2 \sin \left( \frac{\pi y}{L(s)} \right) \, ds \nonumber \\&\quad \le \frac{C e^{-(3 \pi ^2)^{1/3}t^{1/3}}}{t^{1/6}} e^{\sqrt{2}x} \sin \left( \frac{\pi x}{L(0)}\right) \int _{L(0)^2}^{\beta t} \frac{1}{(t - s)^{1/2}} \, ds \nonumber \\&\quad \le C h(0,x). \end{aligned}$$

(41)

It remains to bound from above the two integrals between $0$ and $L(0)^2$. If $0 \le s \le L(0)^2$, then $t^{1/3} - (t-s)^{1/3} \le C$, and $\sin (\pi y/L(s)) \le C \sin (\pi y/L(0))$ for all $y \in [0, L(s)]$. Also, because $q_{0,s}(x,y)$ is bounded above by the density that would be obtained if particles were killed at $L(0)$, rather than $L(r)$, for $r \in [0,s]$, Lemma 4 implies that

$$\begin{aligned} \int _0^{L(0)^2} q_{0,s}(x,y) \, ds \le \frac{2 e^{\sqrt{2}(x-y)} x (L(0) - y)}{L(0)}. \end{aligned}$$

Thus

$$\begin{aligned}&\int _0^{L(0)^2} \int _0^{L(s)} q_{0,s}(x,y) h(s,y)^2 \, dy \, ds \\&\quad \le C \int _0^{L(0)^2} \int _0^{L(s)} q_{0,s}(x,y) \left\{ e^{\sqrt{2} y} \sin \left( \frac{\pi y}{L(s)} \right) (t - s)^{1/3} e^{-(3 \pi ^2)^{1/3}(t-s)^{1/3}} \right\} ^2 \, dy \, ds \\&\quad \le C e^{-2(3 \pi ^2)^{1/3} t^{1/3}} t^{2/3} \int _0^{L(0)} e^{2 \sqrt{2} y} \sin ^2 \left( \frac{\pi y}{L(0)} \right) \left( \int _0^{L(0)^2} q_{0,s}(x,y) \, ds \right) \, dy \\&\quad \le C x e^{\sqrt{2} x} e^{-2(3 \pi ^2)^{1/3} t^{1/3}} t^{2/3} \int _0^{L(0)} e^{\sqrt{2} y} \sin ^2 \left( \frac{\pi y}{L(0)} \right) \frac{L(0) - y}{L(0)} \, dy \\&\quad \le C x e^{\sqrt{2} x} e^{-2(3 \pi ^2)^{1/3} t^{1/3}} t^{2/3} \cdot \frac{e^{\sqrt{2} L(0)}}{L(0)^3} \\&\quad \le C x e^{\sqrt{2} x} e^{-(3 \pi ^2)^{1/3} t^{1/3}} t^{-1/3}. \end{aligned}$$

Because

$$\begin{aligned} x t^{-1/3} \le C t^{1/3} \sin (\pi x/L(0)) \end{aligned}$$

(42)

when $0 < x < L(0) - 1$, it follows that

$$\begin{aligned} \int _0^{L(0)^2} \int _0^{L(s)} q_{0,s}(x,y) h(s,y)^2 \, dy \, ds \le C h(0,x). \end{aligned}$$

(43)

Likewise, using that $y (t-s)^{-1/3} \le C$ for $y \le L(s)$, we get

$$\begin{aligned}&\int _0^{L(0)^2} \int _0^{L(s)} q_{0,s}(x,y) j(s,y)^2 \, dy \, ds\nonumber \\&\quad \le C \int _0^{L(0)^2} \int _0^{L(s)} q_{0,s}(x,y) \left\{ e^{\sqrt{2} y} e^{-(3 \pi ^2)^{1/3}(t-s)^{1/3}} \right\} ^2 \, dy \, ds \nonumber \\&\quad \le C e^{-2(3 \pi ^2)^{1/3} t^{1/3}} \int _0^{L(0)} e^{2 \sqrt{2} y} \left( \int _0^{L(0)^2} q_{0,s}(x,y) \, ds \right) \, dy \nonumber \\&\quad \le C x e^{\sqrt{2} x} e^{-2(3 \pi ^2)^{1/3} t^{1/3}} \int _0^{L(0)} e^{\sqrt{2} y} \cdot \frac{L(0) - y}{L(0)} \, dy \nonumber \\&\quad \le C x e^{\sqrt{2} x} e^{-(3 \pi ^2)^{1/3} t^{1/3}} t^{-1/3}. \nonumber \\&\quad \le C h(0,x). \end{aligned}$$

(44)

The result follows from (39), (40), (41), (43), (44), and Lemma 15. $\square$

Corollary 17

Let $A_0$ be the constant defined in Lemma 15. If there is a single particle at $x$ at time zero, where $0 < x < L(0) - 1$, then for $t \ge A_0$,

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}_x(R_{0,t} > 0) \asymp e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(0)} \right) t^{1/3} \exp (-(3\pi ^2t)^{1/3}). \end{aligned}$$

Likewise, if $0 < \alpha < \beta < 1$, then there are positive constants $C'_{\alpha , \beta }$ and $C''_{\alpha , \beta }$, depending on $\alpha$ and $\beta$ such that for all $t \ge A_0$,

$$\begin{aligned}&C'_{\alpha , \beta } e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(0)} \right) t^{1/3} \exp (-(3\pi ^2t)^{1/3})\\&\quad \le {{\mathrm{\mathbb {P}}}}_x(R_{\alpha t, \beta t} > 0) \le C''_{\alpha , \beta } e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(0)} \right) t^{1/3} \exp (-(3\pi ^2t)^{1/3}). \end{aligned}$$

Proof

Note that $j(0,x) \le C h(0,x)$ when $x < L(0) - 1$ by (42). Therefore, by Lemma 15 with $s = 0$ and Markov’s Inequality,

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}_x(R_{\alpha t, \beta t} > 0) \le {{\mathrm{\mathbb {P}}}}_x(R_{0,t} > 0) \le {{\mathrm{\mathbb {E}}}}_x[R_{0,t}] \le C(h(0,x) + j(0,x)) \le C h(0,x). \end{aligned}$$

For the lower bound, we use a standard second moment argument and apply Lemmas 15 and 16 to get

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}_x(R_{0,t} > 0) \ge {{\mathrm{\mathbb {P}}}}_x(R_{\alpha t, \beta t} > 0) \ge \frac{({{\mathrm{\mathbb {E}}}}_{0,x}[R_{\alpha t, \beta t}])^2}{{{\mathrm{\mathbb {E}}}}_{0,x}[R_{\alpha t, \beta t}^2]} \ge \frac{C h(0,x)^2}{h(0,x)} = C h(0,x). \end{aligned}$$

The result follows. $\square$

4 Proofs of main results

In this section, we prove Theorems 1 and 2. The key to these proofs is Proposition 20 below. We first recall the following result due to Neveu [22].

Lemma 18

Consider branching Brownian motion with drift $-\sqrt{2}$ and no absorption, started with a single particle at the origin. For each $y \ge 0$, let $K(y)$ be the number of particles that reach $-y$ in a modified process in which particles are killed upon reaching $-y$. Then there exists a random variable $W$, with ${{\mathrm{\mathbb {P}}}}(0 < W < \infty ) = 1$ and ${{\mathrm{\mathbb {E}}}}[W] = \infty$, such that

$$\begin{aligned} \lim _{y \rightarrow \infty } y e^{-\sqrt{2} y} K(y) = W\,a.s. \end{aligned}$$

To prove Proposition 20, we will use the following result about the survival probability of a Galton–Watson process, which is Lemma 13 of [4].

Lemma 19

Let $(p_k)_{k=0}^{\infty }$ be a sequence of nonnegative numbers that sum to 1, and let $X$ be a random variable such that ${{\mathrm{\mathbb {P}}}}(X = k) = p_k$ for all nonnegative integers $k$. Let $q$ be the extinction probability of a Galton–Watson process with offspring distribution $(p_k)_{k=1}^{\infty }$ started with a single individual. Then

$$\begin{aligned} 1 - q \ge \frac{2({{\mathrm{\mathbb {E}}}}[X] - 1)}{{{\mathrm{\mathbb {E}}}}[X(X-1)]}. \end{aligned}$$

Proposition 20

Fix $t > 0$, and suppose that initially there is a single particle at $x = ct^{1/3}$. Then there are constants $A > 0$ and $C > 0$ such that if $t \ge A$, the probability that there is at least one particle remaining at time $t$ is at least $C$.

Proof

We prove this result by constructing a branching process that resembles a discrete-time Galton–Watson process but allows individuals to have different offspring distributions. We will show that the probability that this branching process survives is bounded below by a positive constant, and that survival of this branching process implies that the branching Brownian motion survives until at least time $t - A$. This will in turn give the branching Brownian motion a positive probability of surviving until time $t$, which will imply the result.

Let $C' = C'_{1/3, 2/3}$, where $C'_{1/3, 2/3}$ is the constant from Corollary 17 with $\alpha = 1/3$ and $\beta = 2/3$. Consider the setting of Lemma 18, in which we have branching Brownian motion with drift $-\sqrt{2}$ and no absorption. For $y > 0$, let $K(y)$ denote the number of particles that reach $-y$, if particles are killed upon reaching $-y$. For $\xi > 0$, let $K_{\xi }(y)$ be the number of these particles that reach $y$ before time $\xi$. Because the random variable $W$ in Lemma 18 has infinite expected value, it follows from Lemma 18 and Fatou’s Lemma that we can choose $y > 0$ sufficiently large that

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}[y e^{-\sqrt{2} y} K(y)] \ge \frac{3 \cdot 2^{1/3} c}{C'}. \end{aligned}$$

We can then choose a real number $\xi > 0$ and a positive integer $M$ sufficiently large that

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}[y e^{-\sqrt{2} y} (K_{\xi }(y) \wedge M)] \ge \frac{2 \cdot 2^{1/3} c}{C'}. \end{aligned}$$

(45)

Let $A_0$ be defined as in Corollary 17. Choose $A$ to be large enough that the following hold:

$$\begin{aligned}&A \ge \max \{A_0 + \xi , 2 \xi \} \end{aligned}$$

(46)

$$\begin{aligned}&cA^{1/3} \ge 2y \end{aligned}$$

(47)

$$\begin{aligned}&cA^{1/3} - c(A - \xi )^{1/3} \le \frac{y}{2}. \end{aligned}$$

(48)

Let $t \ge A$, and let $L(s) = c(t-s)^{1/3}$ for $0 \le s \le t$.

We now construct the branching process inductively (see Fig. 2). Let $T_0 = \{0\}$. Suppose that $T_n = \{t_{n,1}, t_{n,2}, \ldots , t_{n,m_n}\}$, which will imply that at the $n$th stage of the process, there are particles at positions $L(t_{n,1}), \ldots , L(t_{n, m_n})$ at times $t_{n,1}, \ldots , t_{n,m_n}$. For $j = 1, 2, \ldots , m_n$, if $t_{n,j} \ge t - A$, then we put $t_{n,j}$ in the set $T_{n+1}$. If $t_{n,j} < t - A$, then we follow the trajectories after time $t_{n,j}$ of the descendants of the particle that reached $L(t_{n,j})$ at time $t_{n,j}$ until either time $t_{n,j} + \xi$, or until the descendant particles reach $L(t_{n,j}) - y$, which is positive by (47). Denote the times, before time $t_{n,j} + \xi$, at which descendant particles reach $L(t_{n,j}) - y$ by $u_{n,j,1} < \cdots < u_{n,j,\ell _{n,j}}$. For $\ell = 1, \ldots , \ell _{n,j} \wedge M$, if at least one descendant of the particle that reaches $L(t_{n,j}) - y$ at time $u_{n,j,\ell }$ later reaches $L(s)$ at some time $s \in [u_{n,j,\ell } + (t - u_{n,j,\ell })/3, u_{n,j,\ell } + 2(t - u_{n,j,\ell })/3]$, then we put the smallest time $s$ at which this occurs in the set $T_{n+1}$. For $n \ge 0$, let $Z_n$ be the cardinality of $T_n$.

Fig. 2

Construction of the branching process $T_n, n\ge 1$. Here we look at particle $j$ in generation $n$ alive at time $t_{n,j}$. It has three descendants that hit level $L(t_{n,j})-y$ figured by three squares. The second particle has a descendant that hits $L$ between time $u_{n,j,2}+(t-u_{n,j,2})/3$ and $u_{n,j,2}+2(t-u_{n,j,2})/3)$. The first of these descendants, indicated by a black dot, belongs to $T_{n+1}$

The next step is to obtain bounds on the moments of $Z_1$ which are valid for all $t \ge A$. Write $u_i = u_{0,1,i}$. Then particles reach $L(0) - y$ at times $u_1, \ldots , u_{\ell _{0,1}}$. Observe that

$$\begin{aligned} Z_1 = \xi _1 + \cdots + \xi _{\ell _{0,1} \wedge M}, \end{aligned}$$

(49)

where $\xi _i = 1$ if the particle that reaches $L(0) - y$ at time $u_i$ has a descendant that reaches $L(s)$ at some time $s \in [u_i + (t - u_i)/3, u_i + 2(t - u_i)/3]$ and $\xi _i = 0$ otherwise. Let $\mathcal{G}$ be the $\sigma$-field generated by $u_1, \ldots , u_{\ell _{0,1}}$. By Corollary 17, if $t \ge A$, then

$$\begin{aligned}&C'e^{\sqrt{2}(x - y)} \sin \left( \frac{\pi (x-y)}{L(u_i)} \right) (t - u_i)^{1/3} \exp (-(3 \pi ^2(t-u_i))^{1/3}) \nonumber \\&\quad \! \le {{\mathrm{\mathbb {P}}}}(\xi _i = 1|\mathcal{G}) \le C e^{\sqrt{2}(x - y)} \sin \left( \frac{\pi (x-y)}{L(u_i)} \right) (t - u_i)^{1/3} \exp (-(3 \pi ^2(t-u_i))^{1/3}).\nonumber \\ \end{aligned}$$

(50)

Because $A \ge A_0 + \xi$ by (46), there is a constant $C$ such that if $t \ge A$ then

$$\begin{aligned} 1&= e^{\sqrt{2} x} \exp (-(3 \pi ^2 t)^{1/3}) \le e^{\sqrt{2} x} \exp (-(3 \pi ^2(t - u_i))^{1/3})\nonumber \\&\le e^{\sqrt{2} x} \exp (-(3 \pi ^2(t - \xi ))^{1/3}) \le C. \end{aligned}$$

(51)

Because $A \ge 2 \xi$ by (46), if $t \ge A$ then

$$\begin{aligned} (t/2)^{1/3} \le (t - \xi )^{1/3} \le (t - u_i)^{1/3} \le t^{1/3}. \end{aligned}$$

(52)

Therefore, using again that $A \ge 2 \xi$, we get, when $t \ge A$,

$$\begin{aligned} \sin \left( \frac{\pi (x-y)}{L(u_i)} \right)&= \sin \left( \frac{\pi (L(u_i) - x + y)}{L(u_i)} \right) \le \frac{\pi (L(u_i) - x + y)}{L(u_i)}\nonumber \\&\le \frac{\pi y}{L(u_i)} \le \frac{\pi y}{c(t - \xi )^{1/3}} \le \frac{2^{1/3} \pi y}{c t^{1/3}}. \end{aligned}$$

(53)

By (48),

$$\begin{aligned} x - L(u_i) \le L(0) - L(\xi ) = ct^{1/3} - c(t - \xi )^{1/3} \le y/2 \end{aligned}$$

for $t \ge A$. Using this result and the fact that $\sin (x) \ge 2x/\pi$ for $0 \le x \le \pi /2$, we get

$$\begin{aligned} \sin \left( \frac{\pi (x-y)}{L(u_i)} \right) = \sin \left( \frac{\pi (L(u_i) - x + y)}{L(u_i)} \right) \ge \frac{2(L(u_i) - x + y)}{L(u_i)} \ge \frac{y}{L(u_i)} \ge \frac{y}{ct^{1/3}}.\nonumber \\ \end{aligned}$$

(54)

Combining (50), (51), (52), (53), and (54), we get

$$\begin{aligned} \frac{C'}{2^{1/3} c} ye^{-\sqrt{2} y} \le {{\mathrm{\mathbb {P}}}}(\xi _i = 1|\mathcal{G}) \le C y e^{-\sqrt{2} y}. \end{aligned}$$

(55)

Because $\ell _{0,1}$ has the same distribution as $K_{\xi }(y)$, it follows from (45), (49), and (55) that

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}[Z_1] \ge \frac{C'}{2^{1/3} c} y e^{-\sqrt{2} y} {{\mathrm{\mathbb {E}}}}[K_{\xi }(y) \wedge M] \ge 2. \end{aligned}$$

(56)

From (49), we see that $Z_1 \le M$ so

$$\begin{aligned} {{\mathrm{\mathbb {E}}}}[Z_1^2] \le M^2 \le C. \end{aligned}$$

(57)

For $n \ge 0$, let $q_{t,n} = {{\mathrm{\mathbb {P}}}}(T_n = \emptyset )$. Let $q_t = \lim _{n \rightarrow \infty } q_{t,n} = {{\mathrm{\mathbb {P}}}}(T_n = \emptyset \text{ for } \text{ some } n)$. Let $p_t(k) = {{\mathrm{\mathbb {P}}}}(Z_1 = k)$. For $z \in [0,1]$, let

$$\begin{aligned} \varphi _t(z) = \sum _{k=0}^{\infty } p_t(k)z^k. \end{aligned}$$

Let $q_{t,*} = \min \{q \in [0,1]: \varphi _t(q) = q\}$, which is the probability that a Galton–Watson branching process goes extinct if each individual independently has $k$ offspring with probability $p_t(k)$.

Let

$$\begin{aligned} q_* = \sup _{t > 0} q_{t,*}. \end{aligned}$$

We claim that for all $t > 0$ and all $n \ge 0$, we have $q_{t,n} \le q_*$. We prove this claim by induction on $n$. Because $q_{t,0} = 0$ for all $t > 0$, the claim is clear when $n = 0$. Suppose the claim holds for some $n$. Then by the induction hypothesis,

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}(T_{n+1} = \emptyset |T_1 = \{s_1, \ldots , s_k\}) = \prod _{j=1}^k q_{t-s_j, n} \le q_*^k. \end{aligned}$$

Taking expectations of both sides gives

$$\begin{aligned} q_{t,n+1} \le \sum _{k=0}^{\infty } p_t(k) q_*^k = \varphi _t(q_*). \end{aligned}$$

Because $\varphi _t(q_{t,*}) = q_{t,*}$ and $\varphi _t(1) = 1$, the fact that $z \mapsto \varphi _t(z)$ is nondecreasing and convex implies that if $z \ge q_{t,*}$, then $\varphi _t(z) \le z$. Therefore, since $q_* \ge q_{t,*}$, we have $\varphi _t(q_*) \le q_*$. Thus, $q_{t,n+1} \le q_*$, and the claim follows by induction.

The claim implies that $q_t \le q_*$ for all $t > 0$. If $0 < t \le A$, then $p_t(1) = 1$ and thus $q_{t,*} = 0$. If $t \ge A$, then by Lemma 19 and Eqs. (56) and (57),

$$\begin{aligned} 1 - q_{t,*} \ge \frac{2({{\mathrm{\mathbb {E}}}}[Z_1] - 1)}{{{\mathrm{\mathbb {E}}}}[Z_1(Z_1 - 1)]} \ge \frac{2({{\mathrm{\mathbb {E}}}}[Z_1] - 1)}{{{\mathrm{\mathbb {E}}}}[Z_1^2]} \ge C. \end{aligned}$$

It follows that $1 - q_* \ge C$, and therefore $1 - q_t \ge C$ for all $t \ge A$.

Thus, there is a constant $C$ such that, for all $t \ge A$, the probability that $T_n \ne \emptyset$ for all $n$ is at least $C$. However, if $T_n \ne \emptyset$ for all $n$, then eventually some particle must reach $L(s)$ for some $s \in [t-A, t-A/3]$. The probability that a particle reaching $L(s)$ for some $s \in [t-A, t-A/3]$ survives until time $t$ is bounded below by a constant. The result follows. $\square$

Proof of Theorem 2

We first obtain an upper bound for the extinction time. Let $\beta > 0$, and let $t_+ = t + \beta x^2$ where $t =\tau x^3$. For $0 \le s \le t_+$, let $L_+(s) = c(t_+ - s)^{1/3}$. Consider the process in which particles are killed at time $s$ if they reach $L_+(s)$. The probability that the original process survives until time $t_+$ is bounded above by the probability that a particle is killed at $L_+(s)$ for some $s \in [0,t_+]$. Note that $L_+(0) - x = c(t_+^{1/3} - t^{1/3}) \asymp \beta x^2 t^{-2/3} \asymp \beta$. Therefore, as soon as $x$ is large enough so that $t\ge A_0$ we can apply Corollary 17 to bound the probability that the original process survives until time $t_+$ by

$$\begin{aligned} C e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L_+(0)} \right) t_+^{1/3} e^{-(3 \pi ^2 t_+)^{1/3}}. \end{aligned}$$

(58)

Observe that furthermore

$$\begin{aligned} \sin \left( \frac{\pi x}{L_+(0)} \right) \le \frac{ \pi (L_+(0) - x)}{L_+(0)} \le C \beta t_+^{-1/3}. \end{aligned}$$

and

$$\begin{aligned} \exp ( \sqrt{2} x - (3 \pi ^2 t_+)^{1/3}) = \exp ( -(3 \pi ^2)^{1/3}(t_+^{1/3} - t^{1/3})) \le e^{-C' \beta }, \end{aligned}$$

for some positive constant $C'$. Therefore, the probability in (58) is at most $C \beta e^{-C' \beta }$, which is less than $\varepsilon /2$ for sufficiently large $\beta$. For such $\beta$, we have

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}_x(\zeta < t_+) \ge 1 - \frac{\varepsilon }{2} \end{aligned}$$

(59)

for sufficiently large $x$.

To obtain the lower bound on the extinction time, let $t_- = t - \beta x^2$. For $0 \le s \le t_-$, let $L_-(s) = c(t_- - s)^{1/3}$. For $y > 0$ and $\xi > 0$, let $K_{\xi }(y)$ denote the number of particles that would be killed, if particles were killed upon reaching $x-y$ before time $\xi$. By Lemma 18, we can choose $y$ and $\xi$ sufficiently large and $\gamma > 0$ sufficiently small that $y \ge 2c^3 \beta + 1$ and

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}(K_{\xi }(y) > \gamma y^{-1} e^{\sqrt{2} y}) > 1 - \frac{\varepsilon }{4}. \end{aligned}$$

(60)

Observe that for sufficiently large $x$,

$$\begin{aligned} t_- - \xi = t - \beta x^2 - \xi \ge \frac{t}{2}, \end{aligned}$$

(61)

which means for all $u \in (0, \xi )$,

$$\begin{aligned} x - L_-(u) = c[t^{1/3} - (t - \beta x^2 - u)^{1/3}] \le \frac{c}{3} \left( \frac{t}{2} \right) ^{-2/3} (\beta x^2 + \xi ) \le c^3 \beta \end{aligned}$$

for sufficiently large $x$. Because $y \ge c^3 \beta + 1$, it follows that

$$\begin{aligned} x - y \le L_-(u) - 1 \end{aligned}$$

(62)

for all $u \in (0, \xi )$, if $x$ is sufficiently large.

Now suppose a particle reaches $x - y$ at time $u \in (0, \xi )$. In view of (62), we can apply Corollary 17 to see that the probability that a descendant of this particle reaches $L(s)$ for some $s \in [u, u + (t_- - u)/2]$ is at least

$$\begin{aligned} C e^{\sqrt{2}(x - y)} \sin \left( \frac{\pi (x-y)}{L_-(u)} \right) (t_- - u)^{1/3} \exp (-(3 \pi ^2(t_- - u))^{1/3}). \end{aligned}$$

(63)

Using that $y \ge 2c^3\beta$ and that $\sin (x) \ge 2x/\pi$ for $0 \le x \le \pi /2$,

$$\begin{aligned} \sin \left( \frac{\pi (x-y)}{L_-(u)} \right)&= \sin \left( \frac{\pi (L_-(u) - x + y)}{L_-(u)} \right) \ge \frac{2(L_-(u) - x + y)}{L_-(u)}\nonumber \\&\ge \frac{2(y - c^3 \beta )}{c t^{1/3}} \ge \frac{y}{ct^{1/3}}. \end{aligned}$$

(64)

Also, for sufficiently large $x$, we have $t^{1/3} - (t - \beta x^2 - u)^{1/3} \ge (1/3)t^{-2/3} \cdot \beta x^2 = (c^2/3) \beta$, and so

$$\begin{aligned} \exp (-(3 \pi ^2(t_- - u))^{1/3})&= \exp (-(3 \pi ^2 t)^{1/3}) \exp ((3 \pi ^2)^{1/3}[t^{1/3} - (t - \beta x^2 - u)^{1/3}]) \nonumber \\&\ge \exp (-(3 \pi ^2 t)^{1/3}) \exp ((3 \pi ^2)^{1/3} c^2 \beta /3). \end{aligned}$$

(65)

Recall also that

$$\begin{aligned} e^{\sqrt{2} x} e^{-(3 \pi ^2 t)^{1/3}} = 1. \end{aligned}$$

(66)

By (61), (64), (65), and (66), for sufficiently large $x$, the probability in (63) is at least

$$\begin{aligned} C y e^{-\sqrt{2} y} e^{(3 \pi ^2)^{1/3} c^2 \beta /3}, \end{aligned}$$

(67)

where the constant $C$ does not depend on $\beta$. By Proposition 20, the probability that a descendant of this particle survives until time $t_-$ is also bounded below by (67), with a different positive constant $C$. Therefore, conditional on the event that $K_{\xi }(y) > \gamma y^{-1} e^{\sqrt{2} y}$, the probability that some particle survives until $t_-$ is at least

$$\begin{aligned} 1 - (1 - Cye^{-\sqrt{2} y} e^{(3 \pi ^2)^{1/3}c^2 \beta /3})^{\gamma y^{-1} e^{\sqrt{2} y}}. \end{aligned}$$

Using the inequality $1-a \le e^{-a}$ for $a \in \mathbb {R}$, we see that this expression is bounded below by

$$\begin{aligned} 1 - \exp ( -C \gamma e^{(3 \pi ^2)^{1/3}c^2 \beta /3}) \end{aligned}$$

and therefore is at least $1 - \varepsilon /4$ if $\beta$ is chosen to be large enough. Combining this result with (60) gives that for such $\beta$,

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}_x(\zeta > t_-) \ge 1 - \frac{\varepsilon }{2} \end{aligned}$$

(68)

for sufficiently large $x$. The result follows from (59) and (68). $\square$

Proof of Theorem 1

First, suppose that $t \ge \max \{A_0, 2A\}$, where $A_0$ is the constant from Corollary 17 and $A$ is the constant from Proposition 20. Suppose also that $0 < x < ct^{1/3} - 1$. For $0 \le s \le t$, let $L(s) = c(t-s)^{1/3}$. Consider a modification of the branching Brownian motion in which particles, in addition to getting killed at the origin, are killed if they reach $L(s)$ for some $s \in [0,t]$. Let $R_1$ be the number of particles that are killed at $L(s)$ for some $s \in (0,t)$, and let $R_2$ be the number of particles that are killed at $L(s)$ for some $s \in (0, t/2)$. By Corollary 17,

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}_x(R_1 > 0) \le C e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(0)} \right) t^{1/3} e^{-(3 \pi ^2 t)^{1/3}}. \end{aligned}$$

(69)

In this modified process, all particles disappear before time $t$. Therefore, the only way to have $\zeta > t$ is to have, in the modified process, a particle killed at $L(s)$ for some $s \in (0, t)$. The upper bound in (2) thus follows from the upper bound in (69).

Likewise, Corollary 17 implies that

$$\begin{aligned} {{\mathrm{\mathbb {P}}}}_x(R_2 > 0) \ge C e^{\sqrt{2} x} \sin \left( \frac{\pi x}{L(0)} \right) t^{1/3} e^{-(3 \pi ^2 t)^{1/3}}. \end{aligned}$$

By Proposition 20, a particle that reaches $L(s)$ at time $s \in (0, t/2)$ has a descendant alive at time $t$ with probability greater than $C$. This implies the lower bound in (2).

Next, suppose $0 < t < \max \{A_0, 2A\}$ and $0 < x < ct^{1/3} - 1$. Let $(B(s), s \ge 0)$ be standard Brownian motion with $B(0) = x$. The probability that the branching Brownian motion survives until time $t$ is bounded below by $P(B(s) > 0 \text{ for } \text{ all } s \in [0,t])$ and is bounded above by $e^t P(B(s) > 0 \text{ for } \text{ all } s \in [0,t])$. Because both $x$ and $t$ are bounded above by a positive constant, both of these expressions are of the order $x$, as are the expressions on the left-hand side and the right-hand side of (2). Consequently, (2) holds when $0 < t < \max \{A_0, 2A\}$.

Finally, (3) follows from (2) by fixing $x > 0$ and letting $t \rightarrow \infty$. $\square$