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Fisher-KPP equation with small data and the extremal process of branching Brownian motion

Leonid Mytnik111Faculty of Industrial Engineering and Management, Technion, Technion City, Haifa 3200003, Israel; email: [email protected]    Jean-Michel Roquejoffre222Institut de Mathématiques de Toulouse; UMR 5219, Université de Toulouse; CNRS, UPS IMT, F-31062 Toulouse Cedex 9, France; e-mail: [email protected]    Lenya Ryzhik333Department of Mathematics, Stanford University, Stanford CA 94350, USA; email: [email protected]
Abstract

We consider the limiting extremal process 𝒳\mathcal{X} of the particles of the binary branching Brownian motion. We show that after a shift by the logarithm of the derivative martingale ZZ, the rescaled ”density” of particles, which are at distance n+xn+x from a position close to the tip of 𝒳\mathcal{X}, converges in probability to a multiple of the exponential exe^{x} as n+n\to+\infty. We also show that the fluctuations of the density, after another scaling and an additional random but explicit shift, converge to a 11-stable random variable. Our approach uses analytic techniques and is motivated by the connection between the properties of the branching Brownian motion and the Bramson shift of the solutions to the Fisher-KPP equation with some specific initial conditions initiated in [9, 10] and further developed in the present paper. The proofs of the limit theorems for 𝒳\mathcal{X} rely crucially on the fine asymptotics of the behavior of the Bramson shift for the Fisher-KPP equation starting with initial conditions of ”size” 0<ε10<\varepsilon\ll 1, up to terms of the order [(logε1)]1γ[{(\log\varepsilon^{-1})]^{-1-\gamma}}, with some γ>0\gamma>0.

1 Introduction

The BBM connection to the Fisher-KPP equation

The standard binary branching Brownian motion (BBM) on {\mathbb{R}} is a process that starts at the initial time t=0t=0 with one particle at the position x=0x=0. The particle performs a Brownian motion until a random, exponentially distributed with parameter 11, time τ>0\tau>0 when it splits into two off-spring particles. Each of the new particles starts an independent Brownian motion at the branching point. They have independent, again exponentially distributed with parameter 11, clocks attached to them that ring at their respective branching times. At the branching time, the particle that is branching produces two independent off-spring, and the process continues, so that at a time t>0t>0 we have a random number NtN_{t} of Brownian particles. It will be convenient for us to assume that the variance of each individual Brownian motion is not tt but 2t2t.

A remarkable observation by McKean [19] is a connection between the location of the maximal particle for the BBM and the Fisher-KPP equation

ut=2ux2+uu2,\dfrac{\partial{u}}{\partial{t}}=\frac{\partial^{2}u}{\partial x^{2}}+u-u^{2}, (1.1)

with the initial condition u(0,x)=𝟙(x<0)u(0,x)=\mathbbm{1}(x<0). Let x1(t)x2(t)xNt(t)x_{1}(t)\geq x_{2}(t)\geq\ldots\geq x_{N_{t}}(t) be the positions of the BBM particles at time tt. McKean has discovered an exact formula

u(t,x)=[x1(t)>x].u(t,x)=\mathbb{P}[x_{1}(t)>x]. (1.2)

More generally, given a sufficiently regular function g(x)g(x), the solution to the Fisher-KPP equation (1.1) with the initial condition u(0,x)=g(x)u(0,x)=g(x) can be written as

u(t,x)\displaystyle u(t,x) =1𝔼x[j=1Nt(1g(xj(t))]=1𝔼0[j=1Nt(1g(x+xj(t))]\displaystyle=1-{\mathbb{E}}_{x}\Big{[}\prod_{j=1}^{N_{t}}(1-g(x_{j}(t))\Big{]}=1-{\mathbb{E}}_{0}\Big{[}\prod_{j=1}^{N_{t}}(1-g(x+x_{j}(t))\Big{]} (1.3)
=1𝔼0[j=1Nt(1g(xxj(t))]=1𝔼x[j=1Nt(1g(xj(t))],\displaystyle=1-{\mathbb{E}}_{0}\Big{[}\prod_{j=1}^{N_{t}}(1-g(x-x_{j}(t))\Big{]}=1-{\mathbb{E}}_{-x}\Big{[}\prod_{j=1}^{N_{t}}(1-g(-x_{j}(t))\Big{]},

so that (1.1) is a special case of (1.3) with g(x)=𝟙(x<0)g(x)=\mathbbm{1}(x<0). Here, 𝔼x{\mathbb{E}}_{x} refers to a BBM that starts at t=0t=0 with a single particle at the position xx\in{\mathbb{R}} and not at x=0x=0. The slightly unusual form of (1.3) in the right side, with the process starting at (x)(-x), will be convenient when we look at the limiting measure of the BBM, also with a flipped sign, as in (1.10) below.

McKean’s interpretation has been used by Bramson [7, 8] to establish the following result on the long term behavior of the solutions to the Fisher-KPP equation. It has been known since the pioneering work of Fisher [12] and Kolmogorov, Petrovskii and Piskunov [16] that the Fisher-KPP equation admits traveling wave solutions of the form u(t,x)=Uc(xct)u(t,x)=U_{c}(x-ct) that satisfy

cUc=Uc′′+UcUc2,Uc()=1,Uc(+)=0,-cU_{c}^{\prime}=U_{c}^{\prime\prime}+U_{c}-U_{c}^{2},~{}~{}U_{c}(-\infty)=1,~{}~{}U_{c}(+\infty)=0, (1.4)

for all cc=2c\geq c_{*}=2. We will denote by U(x)U(x) the traveling wave U2(x)U_{2}(x) that corresponds to the minimal speed c=2c_{*}=2:

2U=U′′+UU2,U()=1,U(+)=0.-2U^{\prime}=U^{\prime\prime}+U-U^{2},~{}~{}U(-\infty)=1,~{}~{}U(+\infty)=0. (1.5)

Solutions to (1.5) are defined up to a translation in space. We fix the particular translate by requiring that the traveling wave has the asymptotics

U(x)(x+k0)ex, as x+,U(x)\sim(x+k_{0})e^{-x},\hbox{ as $x\to+\infty$}, (1.6)

with the pre-factor in front of the right side equal to one, and some k0k_{0}\in{\mathbb{R}} that is not, to the best of our knowledge, explicit. Let now u(t,x)u(t,x) be the solution to (1.1) with the initial condition g(x)g(x) such that 0g(x)10\leq g(x)\leq 1 for all xx\in{\mathbb{R}}, and g(x)g(x) is compactly supported on the right – there exists L0L_{0} such that g(x)=0g(x)=0 for all xL0x\geq L_{0}. It was already shown in [16], for the particular example of g(x)=𝟙(x0)g(x)=\mathbbm{1}(x\leq 0), that there exists a reference frame mkpp(t)m_{kpp}(t) such that mkpp(t)/t2m_{kpp}(t)/t\to 2 as t+t\to+\infty and

u(t,x+mkpp(t))U(x) as t+,u(t,x+m_{kpp}(t))\to U(x)\hbox{ as $t\to+\infty$}, (1.7)

uniformly on semi-infinite intervals of the form xKx\geq K, for each KK\in{\mathbb{R}} fixed. Bramson has refined this result, showing that there exists a constant s^[g]\hat{s}[g] that is known as the Bramson shift corresponding to the initial condition gg, such that

u(t,x+m(t))U(x+s^[g]) as t+,u(t,x+m(t))\to U(x+\hat{s}[g])\hbox{ as $t\to+\infty$}, (1.8)

uniformly on semi-infinite intervals of the form xKx\geq K, for each KK\in{\mathbb{R}} fixed, with

m(t)=2t+32logt.m(t)=2t+\frac{3}{2}\log t. (1.9)

Note that we have chosen the sign of s^[g]\hat{s}[g] in (1.8) in the way that makes the shift positive for ”small” initial conditions that we will consider later. In that sense, at a time t1t\gg 1, the solution is located at the position m(t)m(t). A shorter probabilistic proof of this convergence was given recently in [24], and PDE proofs of various versions on Bramson’s results have been obtained in [14, 18, 22, 25], with further refinements in [13, 15, 23] and especially in the recent fascinating paper [5].

The limiting extremal process of BBM and its connection to the Bramson shift

Motivated by the above discussion, one may consider not just the maximal particle but the statistics of the BBM process re-centered at the location m(t)m(t), and ask if it can also be connected to the solutions to the Fisher-KPP equation. Let x1(t)x2(t)x_{1}(t)\geq x_{2}(t)\geq\ldots be the positions of the BBM particles at time tt, and consider the BBM measure seen from m(t)m(t):

𝒳t=kNtδm(t)xk(t).\displaystyle\mathcal{X}_{t}=\sum_{k\leq N_{t}}\delta_{m(t)-x_{k}(t)}. (1.10)

Recall that NtN_{t} is the number of particles alive at time tt. It was shown in [1, 2, 3, 10] that there exists a point process 𝒳\mathcal{X} so that we have

𝒳t𝒳=kδχk as t+,\displaystyle\mathcal{X}_{t}\Rightarrow\mathcal{X}=\sum_{k}\delta_{\chi_{k}}~{}~{}\hbox{ as $t\to+\infty$}, (1.11)

with χ1χ2\chi_{1}\leq\chi_{2}\leq\ldots, so that χ1\chi_{1} corresponds to the maximal particle in the BBM, χ2\chi_{2} to the second largest, and so on. In what follows we will call 𝒳\mathcal{X} the limiting extremal process or simply extremal process. Sometimes it is also called in the literature the decorated Poisson point process, see e.g. [1]. The properties of the limit measure 𝒳\mathcal{X} are closely related to the long time limit of the derivative martingale introduced in [17]:

Zt=kNt(2txk(t))e(2txk(t))Z as t+-a.s.Z_{t}=\sum_{k\leq N_{t}}(2t-x_{k}(t))e^{-(2t-x_{k}(t))}\rightarrow Z~{}\hbox{ as $t\to+\infty$,~{} ${\mathbb{P}}$-a.s.} (1.12)

It turns out that there exists a direct connection between 𝒳\mathcal{X}, ZZ and the Bramson shift via the Laplace transform of 𝒳\mathcal{X}. As explained in Appendix C of [10], see also [3], the results of [17] imply that for any test function ψ0\psi\geq 0 that is compactly supported on the right, we have the identity

𝔼[e𝒳(ψ)]=𝔼[eZes^[ψ^]],ψ^(x)=1eψ(x).{\mathbb{E}}\left[e^{-\mathcal{X}(\psi)}\right]={\mathbb{E}}\Big{[}e^{-Ze^{-\hat{s}[\hat{\psi}]}}\Big{]},~{}~{}\hat{\psi}(x)=1-e^{-\psi(x)}. (1.13)

Here, for a measure μ\mu and a function ff we use the notation

μ(f)=f(x)μ(dx).\mu(f)=\int_{\mathbb{R}}f(x)\mu(dx).

Note that 0ψ^(x)10\leq\hat{\psi}(x)\leq 1, and ψ^(x)\hat{\psi}(x) is also compactly supported on the right, so that its Bramson shift s^[ψ^]\hat{s}[\hat{\psi}] is well-defined and finite. In addition to a special case of (1.13) with ψ^(x)=𝟙(x0)\hat{\psi}(x)=\mathbbm{1}(x\leq 0), it was shown in [17] that for each yy\in{\mathbb{R}} we have

𝔼[eZey]=1U(y).\displaystyle{\mathbb{E}}\Big{[}e^{-Ze^{-y}}\Big{]}=1-U(y). (1.14)

This together with (1.13), characterizes the Laplace transform of 𝒳\mathcal{X}: for any test function ψ0\psi\geq 0 compactly supported on the right, we have the duality identity

𝔼[e𝒳(ψ)]=1U(s^[ψ^]),ψ^=1eψ.\displaystyle{\mathbb{E}}\left[e^{-\mathcal{X}(\psi)}\right]=1-U(\hat{s}[{\hat{\psi}}]),~{}~{}\hat{\psi}=1-e^{-\psi}. (1.15)

Let us note that the normalization of the traveling wave in (1.6) implies, in particular, that extra shifts appear neither in (1.14), nor in (1.15). A helpful discussion of the normalization constants in these identities can be found in Chapter 2 of [6]. We also explain this in Section 2.

It is important to note that, in fact, the results in [1], [3] and in Appendix C of [10] imply the conditional version of (1.13), see Lemma 2.1 below:

𝔼[e𝒳(ψ)|Z]=eZes^[ψ^].{\mathbb{E}}\left[e^{-\mathcal{X}(\psi)}|Z\right]=e^{-Ze^{-\hat{s}[\hat{\psi}]}}. (1.16)

In principle, (1.16) completely characterizes the conditional distribution of the measure 𝒳\mathcal{X} in terms of its conditional Laplace transform. However, the Bramson shift is a very implicit function of the initial condition, and making the direct use of (1.16) is by no means straightforward. One of the goals of the present paper is precisely to make use of this connection to obtain new results about the extremal process 𝒳\mathcal{X}.

Let us first illustrate what kind of results on the Bramson shift we may need on the example of the asymptotic growth of 𝒳\mathcal{X}, Theorem 1.1 in [11], originally conjectured in [10]. This result says that there exists A0>0A_{0}>0 so that

1A0Zxex𝒳((,x])1, as x+, in probability.\frac{1}{A_{0}Zxe^{x}}\mathcal{X}((-\infty,x])\to 1,~{}~{}\hbox{ as $x\to+\infty$, in probability.} (1.17)

The proof of (1.17) in [11] uses purely probabilistic tools. In order to relate this result to the Bramson shift and the realm of PDE, we can do the following. Consider the shifted and rescaled version of the measure 𝒳\mathcal{X}, as in (1.17):

Yn(dx)=n1en𝒳n(dx),\displaystyle Y_{n}(dx)=n^{-1}e^{-n}\mathcal{X}_{n}(dx), (1.18)

where

𝒳n=kδχkn,\displaystyle\mathcal{X}_{n}=\sum_{k}\delta_{\chi_{k}-n}, (1.19)

so that

1Znen𝒳((,n])=1Znen𝒳n((;0])=1ZYn((;0]).\frac{1}{Zne^{n}}\mathcal{X}((-\infty,n])=\frac{1}{Zne^{n}}\mathcal{X}_{n}((-\infty;0])=\frac{1}{Z}Y_{n}((-\infty;0]). (1.20)

We may analyze the conditional on ZZ Laplace transform of YnY_{n} using (1.16): given a non-negative function ϕ0(x)\phi_{0}(x) compactly supported on the right, we have

𝔼[eYn(ϕ0)|Z]=𝔼[en1en𝒳n(ϕ0)|Z]=𝔼[e𝒳(ϕn)|Z]=exp{Zes^[ψn]},{\mathbb{E}}\left[e^{-Y_{n}(\phi_{0})}|Z\right]={\mathbb{E}}\left[e^{-n^{-1}e^{-n}\mathcal{X}_{n}(\phi_{0})}|Z\right]={\mathbb{E}}\left[e^{-\mathcal{X}(\phi_{n})}|Z\right]=\exp\big{\{}-Ze^{-\hat{s}[\psi_{n}]}\big{\}}, (1.21)

with

ϕn(x)=n1enϕ0(xn),ψn(x)=1exp{ϕn(x)}.\phi_{n}(x)=n^{-1}e^{-n}\phi_{0}(x-n),~{}~{}~{}\psi_{n}(x)=1-\exp\{-\phi_{n}(x)\}. (1.22)

Note that ψn\psi_{n} is also compactly supported on the right, so that its Bramson shift is well-defined. Furthermore, for n1n\gg 1 the function ψn(x)\psi_{n}(x) is small: it is of the size O(n1en)O(n^{-1}e^{-n}), as is ϕn(x)\phi_{n}(x). Thus, (1.21) relates the understanding of the conditional on ZZ weak limit of YnY_{n} to the asymptotics of the Bramson shift for small initial conditions for the Fisher-KPP equation (1.1), and this is the strategy we will exploit in this paper to obtain limit theorems for the process 𝒳\mathcal{X}. Let us stress that a connection between the limiting statistics of BBM and the Bramson shift for small initial conditions was already made in [10], though with a slightly different objective in mind, and in a different way.

The Bramson shift for small initial conditions

We now state the results for the Bramson shift of the solutions to the Fisher-KPP equation

uεt=2uεx2+uεuε2,\dfrac{\partial{u_{\varepsilon}}}{\partial{t}}=\frac{\partial^{2}u_{\varepsilon}}{\partial x^{2}}+u_{\varepsilon}-u_{\varepsilon}^{2}, (1.23)

with a small initial condition

uε(0,x)=εϕ0(x),u_{\varepsilon}(0,x)=\varepsilon\phi_{0}(x), (1.24)

that we will need for studying the limiting behavior of 𝒳\mathcal{X}. Here, ε1\varepsilon\ll 1 is a small parameter, and the function ϕ0(x)\phi_{0}(x) is non-negative, bounded and compactly supported on the right: there exists L0L_{0}\in{\mathbb{R}} such that ϕ0(x)=0\phi_{0}(x)=0 for xL0x\geq L_{0}. We will use the notation xε=s^[εϕ0]x_{\varepsilon}=\hat{s}[\varepsilon\phi_{0}] for the Bramson shift of εϕ0\varepsilon\phi_{0}:

|uε(t,x+m(t)|U(x+xε)0 as t+. uniformly on compact intervals in x,|u_{\varepsilon}(t,x+m(t)|\to U(x+x_{\varepsilon})\to 0\hbox{ as $t\to+\infty$. uniformly on compact intervals in $x$,} (1.25)

We chose the sign of xεx_{\varepsilon} in (1.25) so that xε>0x_{\varepsilon}>0 for ε>0\varepsilon>0 sufficiently small. The following proposition gives the asymptotic behavior for xεx_{\varepsilon} for small ε>0\varepsilon>0 that is sufficiently precise to recover (1.17).

Proposition 1.1

Under the above assumptions on ϕ0\phi_{0}, we have

|xεlogε1+loglogε1+logc¯|0 as ε0,|x_{\varepsilon}-\log\varepsilon^{-1}+\log\log\varepsilon^{-1}+\log\bar{c}|\to 0\hbox{ as $\varepsilon\downarrow 0$,} (1.26)

with

c¯=14πexϕ0(x)𝑑x.\bar{c}=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}e^{x}\phi_{0}(x)dx. (1.27)

To formulate the convergence result, let 𝒞c\mathcal{C}_{c} (resp. 𝒞c+\mathcal{C}^{+}_{c}) be the space of continuous (resp. non-negative continuous) compactly supported functions on {\mathbb{R}}, and 𝒞bc\mathcal{C}_{bc} (resp. 𝒞bc+\mathcal{C}^{+}_{bc}) be the space of bounded continuous (resp. non-negative bounded continuous) functions on {\mathbb{R}} compactly supported on the right. Let v\mathcal{M}_{v} (resp. v+\mathcal{M}_{v}^{+}) be the space of signed (resp. non-negative) Radon measures on {\mathbb{R}} such that |μ|((,0])<|\mu|((-\infty,0])<\infty, equipped with the topology τv\tau_{v} generated by

μnnτvμμn(f)nμ(f)f𝒞bc.\mu_{n}\underset{n\rightarrow\infty}{\overset{\tau_{v}}{\longrightarrow}}\mu\iff\mu_{n}(f)\underset{n\rightarrow\infty}{\longrightarrow}\mu(f)\quad\forall f\in{\mathcal{C}}_{bc}.

An immediate corollary of Proposition 1.1 is the following version of (1.17). We set

μ(dx)=14πexdx,\mu(dx)=\frac{1}{\sqrt{4\pi}}e^{x}\,dx, (1.28)

so that

c¯=μ(ϕ0).\bar{c}=\mu(\phi_{0}). (1.29)
Theorem 1.2

Conditionally on ZZ, we have

Yn(dx)nZμ(dx)in v+ in probability.Y_{n}(dx)\underset{n\rightarrow\infty}{\longrightarrow}Z\mu(dx)~{}~{}\hbox{{in ${\mathcal{M}}^{+}_{v}$} in probability.} (1.30)

In other words, Yn(dx)Y_{n}(dx) looks like an exponential shifted by logZ\log Z to the left. This theorem also explicitly identifies the constant A0A_{0} in (1.17). Accordingly, we can reformulate Theorem 1.2 as follows: consider the measures 𝒳n\mathcal{X}^{*}_{n} shifted by logZ\log Z:

𝒳nkδχk+logZ,\mathcal{X}^{*}_{n}\equiv\sum_{k}\delta_{\chi_{k}+\log Z}, (1.31)

and

Yn(dx)=n1en𝒳n(dx).Y^{*}_{n}(dx)=n^{-1}e^{-n}\mathcal{X}^{*}_{n}(dx). (1.32)

This gives the following version of Theorem 1.2:

Corollary 1.3

We have

Yn(dx)nμ(dx),in v+ in probability.Y^{*}_{n}(dx)\underset{n\rightarrow\infty}{\longrightarrow}\mu(dx),~{}~{}\hbox{{in ${\mathcal{M}}^{+}_{v}$} in probability.} (1.33)

As we have mentioned, Theorem 1.2 and Corollary 1.3 are not really new, except for identifying the constant A0A_{0}, even though the approach via the asymtptotics of the Bramson shift produces an analytic rather than a probabilistic proof. In order to obtain genuinely new results on the fluctuations of Yn(dx)Y_{n}(dx) around Zμ(dx)Z\mu(dx), we will need a finer asymptotics for the shift xεx_{\varepsilon} than in Proposition 1.1. Let us define the constant

c¯1\displaystyle\bar{c}_{1} =14πxexϕ0(x)𝑑x,\displaystyle=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}xe^{x}\phi_{0}(x)dx, (1.34)

that depends on the initial condition ϕ0\phi_{0}, as does c¯\bar{c} in (1.27), and universal constants

g=01ez2/4zey2/4𝑑y𝑑z21ez2/4z1y2ey2/4𝑑y,g_{\infty}=\int_{0}^{1}e^{z^{2}/4}\int_{z}^{\infty}e^{-y^{2}/4}dydz-2\int_{1}^{\infty}e^{z^{2}/4}\int_{z}^{\infty}\frac{1}{y^{2}}e^{-y^{2}/4}dy, (1.35)

and

m1=32g+k0+12,m_{1}=\frac{3}{2}g_{\infty}+k_{0}+\frac{1}{2}, (1.36)

that do not depend on ϕ0\phi_{0}. Here, k0k_{0} is the constant that appears in the asymptotics (1.6) for U(x)U(x). The following theorem allows us to obtain convergence in law of the fluctuations of YnY_{n}.

Theorem 1.4

Under the above assumptions on ϕ0\phi_{0}, we have the asymptotics

xε=logε1loglogε1logc¯2loglogε1logε1(m1logc¯+c¯1c¯)1logε1+O(1(logε1)1+γ),x_{\varepsilon}=\log\varepsilon^{-1}-\log\log\varepsilon^{-1}-\log\bar{c}-\frac{2\log\log\varepsilon^{-1}}{\log\varepsilon^{-1}}-\Big{(}m_{1}-\log\bar{c}+\frac{\bar{c}_{1}}{\bar{c}}\Big{)}\frac{1}{\log\varepsilon^{-1}}+O\Big{(}\frac{1}{(\log\varepsilon^{-1})^{1+\gamma}}\Big{)}, (1.37)

as ε0\varepsilon\downarrow 0, with some γ>0\gamma>0.

The first two terms in (1.26) and (1.37) have been predicted in [10] in addressing a different BBM question, using an informal Tauberian type argument that we were not able to make rigorous. The rest of terms have not been predicted, to the best of our knowledge. The proof in the current paper does not seem to be directly related to the arguments of [10] but the general approach to the statistics of BBM via the Bramson shift asymptotics for small initial conditions comes from [10].

Weak convergence of the fluctuations of the extremal process

Theorem 1.2 indicates that to obtain the limiting behavior of the fluctuations of Yn(dx)Y_{n}(dx) one should consider a rescaling of the signed measure

Yn(dx)Zμ(dx)=Yn(dx)14πe(x+logZ)dx.Y_{n}(dx)-Z\mu(dx)=Y_{n}(dx)-\frac{1}{\sqrt{4\pi}}e^{(x+\log Z)}dx. (1.38)

It turns out, however, that there is an extra small deterministic shift of the exponential profile that needs to be performed before the rescaling: a better object to rescale is not as in (1.38) but

Yn(dx)14πe(x+logZ+en)dx,Y_{n}(dx)-\frac{1}{\sqrt{4\pi}}e^{(x+\log Z+e_{n})}dx, (1.39)

with an extra deterministic correction

en=log(1+2lognn)2lognn.e_{n}=\log\Big{(}1+\frac{2\log n}{n}\Big{)}\approx\frac{2\log n}{n}. (1.40)

The properly rescaled measures are

Vn(ϕ0)=n(Yn(ϕ0)(1+2lognn)Zμ(ϕ0)),\displaystyle V_{n}(\phi_{0})=n\Big{(}Y_{n}(\phi_{0})-(1+\frac{2\log n}{n})Z\mu(\phi_{0})\Big{)}, (1.41)
Vn(ϕ0)=n(Yn(ϕ0)(1+2lognn)μ(ϕ0)).\displaystyle V^{*}_{n}(\phi_{0})=n\Big{(}Y^{*}_{n}(\phi_{0})-(1+\frac{2\log n}{n})\mu(\phi_{0})\Big{)}.

Let us set

ν(dx)=(4π)1/2xexdx,\nu(dx)=(4\pi)^{-1/2}xe^{x}\,dx, (1.42)

and let {Rt,t0}\{R_{t},t\geq 0\} be a spectrally positive 11-stable stochastic process with the Laplace transform

𝔼[eλRt]=etλlogλ,λ>0,t0,{\mathbb{E}}\left[e^{-\lambda R_{t}}\right]=e^{t\lambda\log\lambda},\quad\forall\lambda>0,\;t\geq 0,

such that {Rt,t0}\{R_{t},t\geq 0\} is independent of ZZ. The main probabilistic result of this paper is the following theorem describing the limiting behavior of fluctuations of the extremal process. We use the notation \Rightarrow for the convergence in distribution.

Theorem 1.5

(i) Conditionally on ZZ, we have

Vn𝕃Z(dx)in v, as n.\displaystyle V_{n}\Rightarrow\mathbb{L}_{Z}(dx)\quad\hbox{in ${\mathcal{M}}_{v}$, as $n\rightarrow\infty$.} (1.43)

Here, 𝕃Z\mathbb{L}_{Z} is a random measure such that

𝕃Z(dx)=RZμ(dx)+Z(m1μ(dx)+ν(dx)),\mathbb{L}_{Z}(dx)=R_{Z}\mu(dx)+Z(m_{1}\mu(dx)+\nu(dx)), (1.44)

and m1m_{1} is the constant that appears in (1.37).
(ii) We also have

Vn(dx)𝕃1(dx)in v, as n.\displaystyle V^{*}_{n}(dx)\Rightarrow\mathbb{L}_{1}(dx)\quad\hbox{in ${\mathcal{M}}_{v}$, as $n\rightarrow\infty$.} (1.45)

Here, 𝕃1(dx)\mathbb{L}_{1}(dx) is a random measure such that

𝕃1(dx)=R1μ(dx)+(m1μ(dx)+ν(dx)).\mathbb{L}_{1}(dx)=R_{1}\mu(dx)+(m_{1}\mu(dx)+\nu(dx)). (1.46)

One can immediately deduce from the above theorem, that, conditionally on ZZ, for any test function ϕ0𝒞bc+\phi_{0}\in{\mathcal{C}}^{+}_{bc}𝕃Z(ϕ0)\mathbb{L}_{Z}(\phi_{0}) is a spectrally positive 11-stable random variable with the Laplace transform:

𝔼[eλ𝕃Z(ϕ0)|Z]=eZμ(ϕ0)λlogλ+λZμ(ϕ0)logμ(ϕ0)λZ(m1μ(ϕ0)+ν(ϕ0)),λ>0,\displaystyle{\mathbb{E}}\left[e^{-\lambda\mathbb{L}_{Z}(\phi_{0})}\big{|}Z\right]=e^{Z\mu(\phi_{0})\lambda\log\lambda+\lambda Z\mu(\phi_{0})\log\mu(\phi_{0})-\lambda{Z}(m_{1}\mu(\phi_{0})+\nu(\phi_{0}))},\;\forall\lambda>0, (1.47)

We note that, in the study of BBM, both the 11-stable process behavior and a small deterministic correction have appeared in a related but different context for the convergence (1.12) of ZtZ_{t} to ZZ as t+t\to+\infty in [21]. There, the correction is of the form logt/t\log t/\sqrt{t}, and is close in spirit to enlogn/ne_{n}\sim\log n/n due to the diffusive nature of the space-time scaling, though the exact translation of the corrections appearing here and in [21] is not quite clear. The logt/t\log t/t correction to the front location has also been observed in [5] and proved in [13]. The 11-stable like tails for ZZ itself have been observed already in [4] for the BBM, and in [20] for the branching random walk.

The paper is organized is follows. Section 2 explains how the the probabilistic statements, Theorem 1.2 and Theorem 1.5, follow from the correspomnding asymptoptics for the Bramson shift xεx_{\varepsilon} in Proposition 1.1 and Theorem 1.4, as well as the the duality identities (1.14) and (1.15). The rest of the paper is devoted to the proof of Proposition 1.1 and Theorem 1.4. Section 3 introduces the self-similar variables that are used throughout the proofs, and reformulates the required results as Propositions 3.2 and 3.3. We also explain in that section how the values of the specific constants that appear in the asymptotics for xεx_{\varepsilon} come about. Section 4 contains the analysis of the linear Dirichlet problem that appears throughout the PDE approach to the Bramson shift [14, 22, 23]. More specifically, we describe an approximate solution to the adjoint linear problem that is one reason for the logarithmic correction in ene_{n} in (1.40). We should stress that this is not the only contribution to the logn/n\log n/n term – the second one comes from the nonlinear term in the Fisher-KPP equation. Section 5 contains the proof of Proposition 3.2. Of course, this Proposition is just a weaker version of Proposition 3.3, in the same vein as Proposition 1.1 is an immediate consequence of Theorem 1.4. However, the proof of Proposition 3.2 is much shorter than that of Proposition 3.3, so we present it separately for the convenience of the reader. Section 6 contains the proof of Proposition 3.3, with some intermediate steps proved in Section 7. Finally, Appendix A contains an auxiliary lemma on the continuity of the Bramson shift.

We use the notation CC, CC^{\prime}, C0C_{0}, C0C_{0}^{\prime}, etc. for various constants that do not depend on ε\varepsilon, and can change from line to line.

Acknowledgment. We are deeply indebted to Lisa Hartung for illuminating discussions during the early stage of this work, and are extremely grateful to Éric Brunet and Julien Berestycki for generously sharing their deep understanding of various aspects of the BBM and the Bramson shift, without which this work would not be possible. The work of JMR was supported by from the European Research Council under the European Union’s Seventh Framework Programme (FP/2007-2013) ERC Grant Agreement n. 321186 - ReaDi, and from the ANR NONLOCAL project (ANR-14-CE25-0013). The work or LM and LR was supported by a US-Israel BSF grant, LR was partially supported by NSF grants DMS-1613603 and DMS-1910023, and ONR grant N00014-17-1-2145.

2 The proof of the probabilistic statements

In this section, we explain how the probabilistic results, Theorem 1.2 and Theorem 1.5, follow from Proposition 1.1 and Theorem 1.4, respectively.

The duality identity

We first briefly explain the duality identities (1.14) and (1.15), and, in particular, the fact that no extra shift of the wave is needed in these relations, as soon as the wave normalization (1.6) is fixed. To see this, note that in the formal limit ψ(x)𝟙(xy)\psi(x)\to\infty\cdot\mathbbm{1}(x\leq y), we have ψ^(x)=𝟙(xy)\hat{\psi}(x)=\mathbbm{1}(x\leq y), and (1.15) reduces it to

[χ1y]=1U(s^[χ0]+y),χ0(x)=𝟙(x0).\mathbb{P}[\chi_{1}\leq y]=1-U(\hat{s}[\chi_{0}]+y),~{}~{}\chi_{0}(x)=\mathbbm{1}(x\leq 0). (2.1)

This is simply the definition of the Bramson shift, combined with the probabilistic interpretation (1.2) of the solution to (1.1) with the initial condition u(0,x)=𝟙(x0)u(0,x)=\mathbbm{1}(x\leq 0). Thus, no extra shift is needed neither in (2.1), nor in (1.15). The generalization of (2.1) to (1.15) is explained in Appendix C of [10]. Some normalization constants appear in the discussion there, but as we see, they are not needed once the wave is normalized by (1.6) rather than by

xU~(x)𝑑x=0,\int_{-\infty}^{\infty}x\tilde{U}^{\prime}(x)dx=0, (2.2)

as in [10].

As for the derivative martingale identity (1.14), it is simply an immediate consequence of expressions (10) and (11) in [17], combined with the normalization (1.6) that implies that C=1C=1 in both of these equations in [17].

Proof of Theorem 1.2

We now prove Theorem 1.2. For an arbitrary ϕ0𝒞bc+\phi_{0}\in\mathcal{C}^{+}_{bc}, we set, as in (1.22),

ϕn(x)=n1enϕ0(xn),ϕ~n(x)=n1enϕ0(x),\phi_{n}(x)=n^{-1}e^{-n}\phi_{0}(x-n),~{}~{}~{}\tilde{\phi}_{n}(x)=n^{-1}e^{-n}\phi_{0}(x), (2.3)

and

ψn(x)=1exp{ϕn(x)},ψ~n(x)=1exp{ϕ~n(x)},\psi_{n}(x)=1-\exp\{-\phi_{n}(x)\},~{}~{}\tilde{\psi}_{n}(x)=1-\exp\{-\tilde{\phi}_{n}(x)\}, (2.4)

with

ψ0(x)=1exp{ϕ0(x)}.\psi_{0}(x)=1-\exp\{-\phi_{0}(x)\}. (2.5)

Note that both ψn\psi_{n} and ψ~n\tilde{\psi}_{n} look like ”small step” initial conditions:

ψn(x)ϕn(x),ψ~n(x)ϕ~n(x) as n+.\psi_{n}(x)\approx\phi_{n}(x),~{}~{}\tilde{\psi}_{n}(x)\approx\tilde{\phi}_{n}(x)\hbox{ as $n\to+\infty$}.

We need the following result that, in fact, follows easily from [1], [3] and Appendix C in [10]. We provide the proof for the sake of completeness we provide the proof.

Lemma 2.1

For any ϕ0𝒞bc+\phi_{0}\in\mathcal{C}^{+}_{bc}, we have

𝔼[e𝒳(ϕ0)|Z]=exp{Zes^[ψ0]},{\mathbb{E}}\Big{[}e^{-\mathcal{X}(\phi_{0})}|Z\Big{]}=\exp\big{\{}-Ze^{-\hat{s}[\psi_{0}]}\big{\}}, (2.6)

and hence

𝔼[eYn(ϕ0)|Z]=exp{Zes^[ψn]}.\displaystyle{\mathbb{E}}\Big{[}e^{-Y_{n}(\phi_{0})}|Z\Big{]}=\exp\big{\{}-Ze^{-\hat{s}[\psi_{n}]}\big{\}}. (2.7)

Proof. Fix an arbitrary ϕ0𝒞c+\phi_{0}\in\mathcal{C}^{+}_{c} and let {t}t0\{\mathcal{F}_{t}\}_{t\geq 0} be a filtration generated by {𝒳t}t0\{\mathcal{X}_{t}\}_{t\geq 0}. Then, by the definition (1.10) of {𝒳t}t0\{\mathcal{X}_{t}\}_{t\geq 0}, the Markov property and (1.3), we get (with some ambiguity of notation we set 𝔼=𝔼0{\mathbb{E}}={\mathbb{E}}_{0}: the expectation for BBMs starting with one particle at 0)

𝔼[e𝒳t(ϕ0)\displaystyle{\mathbb{E}}\Big{[}e^{-\mathcal{X}_{t}(\phi_{0})} |s]=𝔼[j=1Nteϕ0(2t(3/2)logtxj(t)|s]\displaystyle|\mathcal{F}_{s}\Big{]}={\mathbb{E}}\Big{[}\prod_{j=1}^{N_{t}}e^{-\phi_{0}(2t-(3/2)\log t-x_{j}(t)}|\mathcal{F}_{s}\Big{]} (2.8)
=j=1Ns𝔼xj(s)[i=1Nts(1(1eϕ0(2(ts)(3/2)log(ts)xi(ts)+2s+(3/2)log((ts)/t)))]\displaystyle=\prod_{j=1}^{N_{s}}{\mathbb{E}}_{x_{j}(s)}\Big{[}\prod_{i=1}^{N_{t-s}}\Big{(}1-(1-e^{-\phi_{0}(2(t-s)-({3}/{2})\log(t-s)-x_{i}(t-s)+2s+({3}/{2})\log((t-{s})/{t}))}\Big{)}\Big{]}
=j=1Ns(1u(ts,m(ts)xj(s)+2s+32log(1st))),a.s.\displaystyle=\prod_{j=1}^{N_{s}}(1-u(t-s,m(t-s)-x_{j}(s)+2s+\frac{3}{2}\log(1-\frac{s}{t}))),\quad{\mathbb{P}}-\text{a.s.}

Here, u(t,x)u(t,x) is the solution to (1.1) with the initial condition u(0,x)=ψ0(x)=1exp(ϕ0(x))u(0,x)=\psi_{0}(x)=1-\exp(-\phi_{0}(x)). Next, we use (1.8) to get

limt\displaystyle\lim_{t\rightarrow\infty} j=1Ns(1u(ts,m(ts)xj(s)+2s+32log(1st)))=j=1Ns(1U(s^[ψ0]xj(s)+2s))\displaystyle\prod_{j=1}^{N_{s}}(1-u(t-s,m(t-s)-x_{j}(s)+2s+\frac{3}{2}\log(1-\frac{s}{t})))=\prod_{j=1}^{N_{s}}(1-U(\hat{s}[\psi_{0}]-x_{j}(s)+2s)) (2.9)
=exp{j=1Nslog(1U(s^[ψ0]xj(s)+2s)}.\displaystyle=\exp\Big{\{}-\sum_{j=1}^{N_{s}}-\log(1-U(\hat{s}[\psi_{0}]-x_{j}(s)+2s)\Big{\}}.

Following the derivations in (2.4.9)-(2.4.12) in [6] (see also (23)-(25) in [17]) and using (1.6) which gives C=1C=1 in the above references we obtain

limsexp{j=1Nslog(1U(s^[ψ0]xj(s)+2s)}=exp{Zes^[ψ0]}, a.s.\lim_{s\rightarrow\infty}\exp\Big{\{}-\sum_{j=1}^{N_{s}}-\log(1-U(\hat{s}[\psi_{0}]-x_{j}(s)+2s)\Big{\}}=\exp\Big{\{}-Ze^{-\hat{s}[\psi_{0}]}\Big{\}},\quad{\mathbb{P}}-\text{~{}}{\rm a.s.} (2.10)

Fix an arbitrary bounded continuous function hh. Putting together (LABEL:eq:28_05_1), (LABEL:eq:28_05_2), and (2.10), we get,

lims\displaystyle\lim_{s\rightarrow\infty} limt𝔼[h(Zs)e𝒳t(ϕ0)]=limslimt𝔼[h(Zs)𝔼[e𝒳t(ϕ0)|s]]\displaystyle\lim_{t\rightarrow\infty}{\mathbb{E}}\Big{[}h(Z_{s})e^{-\mathcal{X}_{t}(\phi_{0})}\Big{]}=\lim_{s\rightarrow\infty}\lim_{t\rightarrow\infty}{\mathbb{E}}\Big{[}h(Z_{s}){\mathbb{E}}\Big{[}e^{-\mathcal{X}_{t}(\phi_{0})}|\mathcal{F}_{s}\Big{]}\Big{]} (2.11)
=lims𝔼[h(Zs)ej=1Nslog(1U(s^[ψ0]xj(s)+2s)]=𝔼[h(Z)eZes^[ψ0]].\displaystyle=\lim_{s\rightarrow\infty}{\mathbb{E}}\Big{[}h(Z_{s})e^{-\sum_{j=1}^{N_{s}}-\log(1-U(\hat{s}[\psi_{0}]-x_{j}(s)+2s)}\Big{]}={\mathbb{E}}\Big{[}h(Z)e^{-Ze^{-\hat{s}[\psi_{0}]}}\Big{]}.

We used the bounded convergence theorem in the last equality.

Recall, that by Theorem 2.1 in [1], the pair (¯𝒳t,Zt)(\bar{}\mathcal{X}_{t},Z_{t}) converges in distribution to (¯𝒳,Z)(\bar{}\mathcal{X},Z) as tt\rightarrow\infty, where ¯𝒳t\bar{}\mathcal{X}_{t} (resp. ¯𝒳\bar{}\mathcal{X}) is just the measure 𝒳t\mathcal{X}_{t} (resp. 𝒳\mathcal{X}) shifted by logZ\log Z. This together with a.s. convergence of ZtZ_{t} to ZZ implies also convergence in distribution of (𝒳t,Zt)(\mathcal{X}_{t},Z_{t}) to (𝒳,Z)(\mathcal{X},Z) as tt\rightarrow\infty, and thus

𝔼[h(Z)e𝒳(ϕ0)]=limt𝔼[h(Zt)e𝒳t(ϕ0)].\displaystyle{\mathbb{E}}\Big{[}h(Z)e^{-\mathcal{X}(\phi_{0})}\Big{]}=\lim_{t\rightarrow\infty}{\mathbb{E}}\Big{[}h(Z_{t})e^{-\mathcal{X}_{t}(\phi_{0})}\Big{]}.

From this we get

|𝔼[h(Z)e𝒳(ϕ0)]𝔼[h(Z)eZes^[ψ0]]|=|limt𝔼[h(Zt)e𝒳t(ϕ0)]𝔼[h(Z)eZes^[ψ0]]|\displaystyle\left|{\mathbb{E}}\Big{[}h(Z)e^{-\mathcal{X}(\phi_{0})}\Big{]}-{\mathbb{E}}\Big{[}h(Z)e^{-Ze^{-\hat{s}[\psi_{0}]}}\Big{]}\right|=\left|\lim_{t\rightarrow\infty}{\mathbb{E}}\Big{[}h(Z_{t})e^{-\mathcal{X}_{t}(\phi_{0})}\Big{]}-{\mathbb{E}}\Big{[}h(Z)e^{-Ze^{-\hat{s}[\psi_{0}]}}\Big{]}\right|
|limt𝔼[(h(Zt)h(Zs))e𝒳t(ϕ0)]|+|limt𝔼[h(Zs)e𝒳t(ϕ0)]𝔼[h(Z)eZes^[ψ0]]|,s>0.\displaystyle\leq\left|\lim_{t\rightarrow\infty}{\mathbb{E}}\Big{[}(h(Z_{t})-h(Z_{s}))e^{-\mathcal{X}_{t}(\phi_{0})}\Big{]}\right|+\left|\lim_{t\rightarrow\infty}{\mathbb{E}}\Big{[}h(Z_{s})e^{-\mathcal{X}_{t}(\phi_{0})}\Big{]}-{\mathbb{E}}\Big{[}h(Z)e^{-Ze^{-\hat{s}[\psi_{0}]}}\Big{]}\right|,\;\forall s>0.

Therefore, we have

|𝔼[h(Z)e𝒳(ϕ0)]𝔼[h(Z)eZes^[ψ0]]|\displaystyle\left|{\mathbb{E}}\Big{[}h(Z)e^{-\mathcal{X}(\phi_{0})}\Big{]}-{\mathbb{E}}\Big{[}h(Z)e^{-Ze^{-\hat{s}[\psi_{0}]}}\Big{]}\right| lim supslimt𝔼[|h(Zt)h(Zs)|]\displaystyle\leq\limsup_{s\rightarrow\infty}\lim_{t\rightarrow\infty}{\mathbb{E}}\Big{[}\left|h(Z_{t})-h(Z_{s})\right|\Big{]}
+|lim supslimt𝔼[h(Zs)e𝒳t(ϕ0)]𝔼[h(Z)eZes^[ψ0]]|=0,\displaystyle+\left|\limsup_{s\rightarrow\infty}\lim_{t\rightarrow\infty}{\mathbb{E}}\Big{[}h(Z_{s})e^{-\mathcal{X}_{t}(\phi_{0})}\Big{]}-{\mathbb{E}}\Big{[}h(Z)e^{-Ze^{-\hat{s}[\psi_{0}]}}\Big{]}\right|=0,

where convergence to zero of the first term follows from a.s. convergence of ZtZ_{t} to ZZ and the bounded convergence theorem. The second term converges to zero by (2.11). Hence, we have

𝔼[h(Z)e𝒳(ϕ0)]=𝔼[h(Z)eZes^[ψ0]],{\mathbb{E}}\Big{[}h(Z)e^{-\mathcal{X}(\phi_{0})}\Big{]}={\mathbb{E}}\Big{[}h(Z)e^{-Ze^{-\hat{s}[\psi_{0}]}}\Big{]},

for any ϕ0𝒞c+\phi_{0}\in\mathcal{C}^{+}_{c} and any bounded continuous function hh, which implies (2.6) for any ϕ0𝒞c+\phi_{0}\in\mathcal{C}^{+}_{c}. The extension to ϕ0𝒞bc+\phi_{0}\in\mathcal{C}^{+}_{bc} follows via approximation and a kind of continuity of s^[ψ]\hat{s}[\psi] in functions in 𝒞bc+\mathcal{C}^{+}_{bc} and bounded by 11, made precise in Lemma A.1 in Appendix A. \Box

We continue the proof of Theorem 1.2. Recall that we fixed an arbitrary ϕ0𝒞bc+\phi_{0}\in\mathcal{C}^{+}_{bc}. It follows from Proposition 1.1, with εn=n1exp(n)\varepsilon_{n}=n^{-1}\exp(-n) that the Bramson shift appearing in the right side of (2.7) has the asymptotics

s^[ψn]=n+s^[ψ~n]=n+log(nen)log(n+logn)logc¯+o(1)logμ(ϕ0), as n+,\hat{s}[\psi_{n}]=-n+\hat{s}[\tilde{\psi}_{n}]=-n+\log(ne^{n})-\log(n+\log n)-\log\bar{c}+o(1)\to-\log\mu(\phi_{0}),\hbox{ as $n\to+\infty$,} (2.12)

with the measure μ\mu defined in (1.28), and ψ~n\tilde{\psi}_{n} as in (2.4).

To get the weak convergence of the measures Yn(dx)Y_{n}(dx), it is sufficient to get the convergence of Yn(ϕ0)Y_{n}(\phi_{0}) for any ϕ0𝒞bc+\phi_{0}\in\mathcal{C}^{+}_{bc}. Thus, we will check convergence of the Laplace transforms of Yn(ϕ0)Y_{n}(\phi_{0}) (conditionally on ZZ). As a consequence of Lemma 2.1 and (2.12), we obtain

limn𝔼[eYn(ϕ0)|Z]=limneZes^[ψn]=eZμ(ϕ0).\lim_{n\rightarrow\infty}{\mathbb{E}}\left[e^{-Y_{n}(\phi_{0})}|Z\right]=\lim_{n\rightarrow\infty}e^{-Ze^{-\hat{s}[\psi_{n}]}}=e^{-Z\mu(\phi_{0})}. (2.13)

Since ϕ0𝒞bc+\phi_{0}\in\mathcal{C}_{bc}^{+} was arbitrary, this implies that, conditionally on ZZ,

YnZμ,in v+,asn.Y_{n}\Rightarrow Z\mu,\quad\hbox{in ${\mathcal{M}}^{+}_{v}$},\;\mbox{as}\;n\rightarrow\infty. (2.14)

Therefore, conditionally on ZZ, we have

Yn(dx)Zμ(dx)in v+,asn,in probability,Y_{n}(dx)\rightarrow Z\mu(dx)~{}~{}\hbox{in ${\mathcal{M}}^{+}_{v}$},\;\mbox{as}\;n\rightarrow\infty,\hbox{in probability,} (2.15)

finishing the proof of Theorem 1.2\Box

Corollary 1.3 is an immediate consequence of Theorem 1.2.

Proof of Theorem 1.5

We will prove only part (i) since the proof of (ii) goes along the same lines. Once again, to get the weak convergence of measures Vn(dx)V_{n}(dx), it is sufficient to get the convergence of Vn(ϕ0)V_{n}(\phi_{0}) for any ϕ0𝒞bc+\phi_{0}\in\mathcal{C}^{+}_{bc}, and we will check the convergence of the Laplace transforms of Vn(ϕ0)V_{n}(\phi_{0}) (conditionally on ZZ). Fix an arbitrary ϕ0𝒞bc+\phi_{0}\in\mathcal{C}^{+}_{bc}. We have

𝔼[eVn(ϕ0)|Z]\displaystyle{\mathbb{E}}\Big{[}e^{-V_{n}(\phi_{0})}\big{|}Z\Big{]} =𝔼[exp(n(Yn(ϕ0)(1+2lognn)Zμ(ϕ0)))|Z]\displaystyle={\mathbb{E}}\Big{[}\exp\Big{(}-n\Big{(}Y_{n}(\phi_{0})-(1+\frac{2\log n}{n})Z\mu(\phi_{0})\Big{)}\Big{)}\big{|}Z\Big{]} (2.16)
=𝔼[exp(nYn(ϕ0))|Z]exp[(n+2logn)Zμ(ϕ0)]\displaystyle={\mathbb{E}}\Big{[}\exp\big{(}-nY_{n}(\phi_{0})\big{)}\big{|}Z\Big{]}\exp\Big{[}\big{(}n+2\log n)Z\mu(\phi_{0})\Big{]}
=exp[Zes^[ηn]]exp[(n+2logn)Zc¯].\displaystyle=\exp\Big{[}-Ze^{-\hat{s}[\eta_{n}]}\Big{]}\exp\Big{[}\big{(}n+2\log n)Z\bar{c}\Big{]}.

We used expression (1.29) for c¯\bar{c} and identity (2.7) above, with ϕ0\phi_{0} replaced by nϕ0n\phi_{0}, and

ηn(x)=1eenϕ0(xn).\eta_{n}(x)=1-e^{-e^{-n}\phi_{0}(x-n)}.

Let us also introduce

ζn(x)=nϕn(x)=enϕ0(xn),xn=s^[ζn],x~n=s^[ηn],\zeta_{n}(x)=n\phi_{n}(x)=e^{-n}\phi_{0}(x-n),~{}~{}x_{n}=\hat{s}[{\zeta_{n}}],~{}~{}\tilde{x}_{n}=\hat{s}[\eta_{n}],

so that (2.16) can be written as

𝔼[eVn(ϕ0)|Z]\displaystyle{\mathbb{E}}\Big{[}e^{-V_{n}(\phi_{0})}\big{|}Z\Big{]} =exp[Z(ex~n(n+2logn)c¯)].\displaystyle=\exp\Big{[}-Z\big{(}e^{-\tilde{x}_{n}}-\big{(}n+2\log n)\bar{c}\big{)}\Big{]}. (2.17)

Note that by Theorem 1.4 with εn=en\varepsilon_{n}=e^{-n} we get

x~n\displaystyle\tilde{x}_{n} =xn+O(enϕ0)=n+logεn1loglogεn1logc¯2loglogεn1logεn1\displaystyle=x_{n}+O(e^{-n}\left\|\phi_{0}\right\|_{\infty})=-n+\log\varepsilon_{n}^{-1}-\log\log\varepsilon_{n}^{-1}-\log\bar{c}-\frac{2\log\log\varepsilon_{n}^{-1}}{\log\varepsilon_{n}^{-1}} (2.18)
(m1logc¯+c¯1c¯)1logεn1+O(1(logεn1)1+δ)\displaystyle-\Big{(}m_{1}-\log\bar{c}+\frac{\bar{c}_{1}}{\bar{c}}\Big{)}\frac{1}{\log\varepsilon_{n}^{-1}}+O\Big{(}\frac{1}{(\log\varepsilon_{n}^{-1})^{1+\delta}}\Big{)}
=lognlogc¯2lognn1n(m1logc¯+c¯1c¯)+O(n1δ).\displaystyle=-\log n-\log\bar{c}-2\frac{\log n}{n}-\frac{1}{n}\Big{(}m_{1}-\log\bar{c}+\frac{\bar{c}_{1}}{\bar{c}}\Big{)}+O(n^{-1-\delta}).

Using this in (2.17) gives

𝔼[eVn(ϕ0)|Z]\displaystyle{\mathbb{E}}\Big{[}e^{-V_{n}(\phi_{0})}\big{|}Z\Big{]} =exp[Z(ex~nc¯n2c¯logn)]\displaystyle=\exp\Big{[}-Z\big{(}e^{-\tilde{x}_{n}}-\bar{c}n-2\bar{c}\log n\big{)}\Big{]} (2.19)
=exp[c¯Zn{exp[2lognn+1n(m1logc¯+c¯1c¯)+O(n1δ)]12lognn}]\displaystyle=\exp\Big{[}-\bar{c}Zn\Big{\{}\exp\Big{[}2\frac{\log n}{n}+\frac{1}{n}\Big{(}m_{1}-\log\bar{c}+\frac{\bar{c}_{1}}{\bar{c}}\Big{)}+O(n^{-1-\delta})\Big{]}-1-\frac{2\log n}{n}\Big{\}}\Big{]}
=exp[c¯Zn{1n(m1logc¯+c¯1c¯)+O(n1δ)}]\displaystyle=\exp\Big{[}-\bar{c}Zn\Big{\{}\frac{1}{n}\Big{(}m_{1}-\log\bar{c}+\frac{\bar{c}_{1}}{\bar{c}}\Big{)}+O(n^{-1-\delta})\Big{\}}\Big{]}
exp[c¯Z(m1logc¯+c¯1c¯)]=exp[Z(c¯logc¯c¯m1c¯1)], as n+.\displaystyle\to\exp\Big{[}-\bar{c}Z\Big{(}m_{1}-\log\bar{c}+\frac{\bar{c}_{1}}{\bar{c}}\Big{)}\Big{]}=\exp\Big{[}Z\Big{(}\bar{c}\log\bar{c}-\bar{c}m_{1}-\bar{c}_{1}\Big{)}\Big{]},~{}~{}\hbox{ as $n\to+\infty$.}

By the definition of 𝕃1\mathbb{L}_{1}, c¯\bar{c} and c¯1\bar{c}_{1} we get

𝔼[e𝕃1(ϕ0)|Z]=exp[Zc¯logc¯Z(c¯m1+c¯1)],\displaystyle{\mathbb{E}}\Big{[}e^{-\mathbb{L}_{1}(\phi_{0})}\big{|}Z\Big{]}=\exp\Big{[}Z\bar{c}\log\bar{c}-Z(\bar{c}m_{1}+\bar{c}_{1})\Big{]},

and since ϕ0𝒞bc+\phi_{0}\in\mathcal{C}^{+}_{bc} was arbitrary we are done. \Box

3 The solution asymptotics in self-similar variables

Proposition 1.1 is a consequence of the following two steps. The first result connects the Bramson shift of a solution to the Fisher-KPP equation with a small initial condition to the asymptotics of the solution to a problem in the self-similar variables with an initial condition shifted far to the right.

Proposition 3.1

Let rr_{\ell} be the solution to

rτη2rη2rη2r+32eτ/2rη+e3τ/2ηexp(τ/2)r2=0,τ>0,η,\displaystyle\dfrac{\partial{r_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{r_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{r_{\ell}}}{\partial{\eta}^{2}}-r_{\ell}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{r_{\ell}}}{\partial{\eta}}+e^{3\tau/2-\eta\exp(\tau/2)}r_{\ell}^{2}=0,~{}~{}\tau>0,~{}~{}\eta\in{\mathbb{R}}, (3.1)

with the initial condition r(0,η)=ψ0(η)r_{\ell}(0,\eta)=\psi_{0}(\eta-\ell), where ψ0(η)=eηϕ0(η)\psi_{0}(\eta)=e^{\eta}\phi_{0}(\eta). Then, for each >0\ell>0, the function r(τ,η)r_{\ell}(\tau,\eta) has the asymptotics

r(τ,η)r()ηeη2/4, as τ+, for η>0.r_{\ell}(\tau,\eta)\sim r_{\infty}(\ell)\eta e^{-\eta^{2}/4},\hbox{ as $\tau\to+\infty$, for $\eta>0$}. (3.2)

Furthermore, the Bramson shift that appears in Proposition 1.1 is given by

xε=logε1logr(ε), with ε=logε1.x_{\varepsilon}=\log\varepsilon^{-1}-\log r_{\infty}(\ell_{\varepsilon}),\hbox{ with $\ell_{\varepsilon}=\log\varepsilon^{-1}$.} (3.3)

The second result, at the core of the proof of Proposition 1.1, describes the asymptotics of r()r_{\infty}(\ell) for large \ell.

Proposition 3.2

The function r()r_{\infty}(\ell) satisfies the following asymptotics:

r()=c¯+O(log), as +,r_{\infty}(\ell)=\bar{c}\ell+O(\log\ell),~{}~{}\hbox{ as $\ell\to+\infty$}, (3.4)

with the constant c¯\bar{c} as in (1.27).

To prove Theorem 1.4, we refine Proposition 3.2 to the following.

Proposition 3.3

The function r()r_{\infty}(\ell) satisfies the following asymptotics:

r()=c¯+2c¯log+m1c¯+c¯1c¯logc¯+O(δ),r_{\infty}(\ell)=\bar{c}\ell+2\bar{c}\log\ell+m_{1}\bar{c}+\bar{c}_{1}-\bar{c}\log\bar{c}+O(\ell^{-\delta}), (3.5)

with the constants c¯\bar{c}, c¯1\bar{c}_{1} and m1m_{1} as in (1.27), (1.34) and (1.36).

Using (3.3), we obtain from Proposition 3.3 that

xε\displaystyle x_{\varepsilon} =εlogr(ε)=εlog(c¯ε+2c¯logε+m1c¯c¯logc¯+c¯1+O(εδ))\displaystyle=\ell_{\varepsilon}-\log r_{\infty}(\ell_{\varepsilon})=\ell_{\varepsilon}-\log\Big{(}\bar{c}\ell_{\varepsilon}+2\bar{c}\log\ell_{\varepsilon}+m_{1}\bar{c}-\bar{c}\log\bar{c}+\bar{c}_{1}+O(\ell_{\varepsilon}^{-\delta})\Big{)} (3.6)
=εlogεlogc¯2logεεm1ε+logc¯εc¯1c¯ε+O(ε1δ),\displaystyle=\ell_{\varepsilon}-\log\ell_{\varepsilon}-\log\bar{c}-2\frac{\log\ell_{\varepsilon}}{\ell_{\varepsilon}}-\frac{m_{1}}{\ell_{\varepsilon}}+\frac{\log\bar{c}}{\ell_{\varepsilon}}-\frac{\bar{c}_{1}}{\bar{c}\ell_{\varepsilon}}+O(\ell_{\varepsilon}^{-1-\delta}),

which proves Theorem 1.4. Thus, our goal is to prove Proposition 3.3.

Of course, Proposition 3.2 in an immediate consequence of Proposition 3.3. However, as its proof is both much shorter and helpful in the proof of the latter, we present its proof separately in Section 5.

Common sense scaling arguments

In order to verify that the constants in Proposition 3.3 are plausible, let us see assume that we have the asymptotics

r()=c¯+m0c¯log+m1c¯+m2c¯logc¯+m3c¯1+O(δ), as +,r_{\infty}(\ell)=\bar{c}\ell+m_{0}\bar{c}\log\ell+m_{1}\bar{c}+m_{2}\bar{c}\log\bar{c}+m_{3}\bar{c}_{1}+O(\ell^{-\delta}),~{}~{}\hbox{ as $\ell\to+\infty$}, (3.7)

and see what simple arguments say about the possible values of the coefficients m0m_{0}, m1m_{1}, m2m_{2} and m3m_{3}. First, consider a shifted initial condition ϕ0L(x)=ϕ0(xL)\phi_{0}^{L}(x)=\phi_{0}(x-L). Then, the shift xεx_{\varepsilon} given by (3.6) should also change by LL, so that

xεL=xεL.x_{\varepsilon}^{L}=x_{\varepsilon}-L. (3.8)

Note that

c¯L=14πexϕ0(xL)𝑑x=eLc¯,\displaystyle\bar{c}_{L}=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}e^{x}\phi_{0}(x-L)dx=e^{L}\bar{c},
c¯1L=14πxexϕ0(xL)𝑑x=eLc¯1+LeLc¯.\displaystyle\bar{c}_{1}^{L}=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}xe^{x}\phi_{0}(x-L)dx=e^{L}\bar{c}_{1}+Le^{L}\bar{c}.

Using (3.7) gives

xεL\displaystyle x_{\varepsilon}^{L} =logε1logrL(ε)\displaystyle=\log\varepsilon^{-1}-\log r_{\infty}^{L}(\ell_{\varepsilon})
=logε1loglogε1logc¯Lm0loglogε1logε1m1logε1m2logc¯Llogε1m3c¯1Lc¯Llogε1+O(ε1δ)\displaystyle=\log\varepsilon^{-1}-\log\log\varepsilon^{-1}-\log\bar{c}_{L}-m_{0}\frac{\log\log\varepsilon^{-1}}{\log\varepsilon^{-1}}-\frac{m_{1}}{\log\varepsilon^{-1}}-\frac{m_{2}\log\bar{c}_{L}}{\log\varepsilon^{-1}}-\frac{m_{3}\bar{c}_{1}^{L}}{\bar{c}_{L}\log\varepsilon^{-1}}+O(\ell_{\varepsilon}^{-1-\delta})
=logε1loglogε1logc¯Lm0loglogε1logε1m1logε1m2logc¯logε1m2Llogε1\displaystyle=\log\varepsilon^{-1}-\log\log\varepsilon^{-1}-\log\bar{c}-L-m_{0}\frac{\log\log\varepsilon^{-1}}{\log\varepsilon^{-1}}-\frac{m_{1}}{\log\varepsilon^{-1}}-\frac{m_{2}\log\bar{c}}{\log\varepsilon^{-1}}-\frac{m_{2}L}{\log\varepsilon^{-1}}
m3eLc¯1+LeLc¯c¯eLlogε1+O(ε1δ)=xεLm2Llogε1m3Llogε1+O(ε1δ).\displaystyle-m_{3}\frac{e^{L}\bar{c}_{1}+Le^{L}\bar{c}}{\bar{c}e^{L}\log\varepsilon^{-1}}+O(\ell_{\varepsilon}^{-1-\delta})=x_{\varepsilon}-L-\frac{m_{2}L}{\log\varepsilon^{-1}}-\frac{m_{3}L}{\log\varepsilon^{-1}}+O(\ell_{\varepsilon}^{-1-\delta}).

This means that for (3.8) to hold we must have

m3=m2.m_{3}=-m_{2}. (3.9)

Note that in (3.5) we have m2=1m_{2}=-1 and m3=1m_{3}=1, so that (3.9) holds.

The second invariance is to consider an initial condition ϕ0λ(x)=λϕ0\phi_{0}^{\lambda}(x)=\lambda\phi_{0}. This is equivalent to keeping ϕ0\phi_{0} intact and replacing ε\varepsilon by ελ=ελ\varepsilon_{\lambda}=\varepsilon\lambda. If we replace ϕ0\phi_{0} by λϕ0\lambda\phi_{0} in (3.6) and keep ε\varepsilon unchanged, this corresponds to replacing c¯\bar{c} by λc¯\lambda\bar{c} and c¯1\bar{c}_{1} by λc¯1\lambda\bar{c}_{1}, which gives

xελ\displaystyle x_{\varepsilon}^{\lambda} =logε1loglogε1logc¯logλm0loglogε1logε1m1logε1m2logc¯logε1m2logλlogε1\displaystyle=\log\varepsilon^{-1}-\log\log\varepsilon^{-1}-\log\bar{c}-\log\lambda-m_{0}\frac{\log\log\varepsilon^{-1}}{\log\varepsilon^{-1}}-\frac{m_{1}}{\log\varepsilon^{-1}}-\frac{m_{2}\log\bar{c}}{\log\varepsilon^{-1}}-\frac{m_{2}\log\lambda}{\log\varepsilon^{-1}} (3.10)
m3c¯1c¯logε1+O(ε1δ)=xεlogλm2logλlogε1+O(ε1δ).\displaystyle-m_{3}\frac{\bar{c}_{1}}{\bar{c}\log\varepsilon^{-1}}+O(\ell_{\varepsilon}^{-1-\delta})=x_{\varepsilon}-\log\lambda-\frac{m_{2}\log\lambda}{\log\varepsilon^{-1}}+O(\ell_{\varepsilon}^{-1-\delta}).

If, instead, we replace ε\varepsilon by ελ=ελ\varepsilon_{\lambda}=\varepsilon\lambda and keep ϕ0\phi_{0} intact, we get from (3.6)

xελ\displaystyle x_{\varepsilon}^{\lambda} =logε1+logλ1log(logε1+logλ1)logc¯m0log(logε1+logλ1)logε1+logλ1\displaystyle=\log\varepsilon^{-1}+\log\lambda^{-1}-\log(\log\varepsilon^{-1}+\log\lambda^{-1})-\log\bar{c}-m_{0}\frac{\log(\log\varepsilon^{-1}+\log\lambda^{-1})}{\log\varepsilon^{-1}+\log\lambda^{-1}} (3.11)
m1logε1+logλ1m2logc¯logε1+logλ1m3c¯1c¯(logε1+logλ1)+O(ε1δ)\displaystyle-\frac{m_{1}}{\log\varepsilon^{-1}+\log\lambda^{-1}}-\frac{m_{2}\log\bar{c}}{\log\varepsilon^{-1}+\log\lambda^{-1}}-\frac{m_{3}\bar{c}_{1}}{\bar{c}(\log\varepsilon^{-1}+\log\lambda^{-1})}+O(\ell_{\varepsilon}^{-1-\delta})
=xεlogλlogλ1logε1+O(ε1δ).\displaystyle=x_{\varepsilon}-\log\lambda-\frac{\log\lambda^{-1}}{\log\varepsilon^{-1}}+O(\ell_{\varepsilon}^{-1-\delta}).

Comparing (3.10) and (3.11) we see that we should have

m2=1,m_{2}=-1, (3.12)

that, in view of (3.9), implies that m3=1m_{3}=1.

Some preliminary transformations and the self-similar variables

The conclusion of Proposition 3.1 follows from a series of changes of variables that we now describe. We first go into the moving frame, writing solution to (1.23)-(1.24) as

uε(t,x)=u~ε(t,x2t+32log(t+1)).u_{\varepsilon}(t,x)=\tilde{u}_{\varepsilon}(t,x-2t+\frac{3}{2}\log(t+1)). (3.13)

The function u~ε(t,x)\tilde{u}_{\varepsilon}(t,x) satisfies

u~εt(232(t+1))u~εx=2u~εx2+u~εu~ε2.\displaystyle\dfrac{\partial{\tilde{u}_{\varepsilon}}}{\partial{t}}-\Big{(}2-\frac{3}{2(t+1)}\Big{)}\dfrac{\partial{\tilde{u}_{\varepsilon}}}{\partial{x}}=\dfrac{\partial^{2}{\tilde{u}_{\varepsilon}}}{\partial{x}^{2}}+\tilde{u}_{\varepsilon}-\tilde{u}_{\varepsilon}^{2}. (3.14)

Next, we take out the exponential decay factor, writing

u~ε(t,x)=exzε(t,x),\tilde{u}_{\varepsilon}(t,x)=e^{-x}z_{\varepsilon}(t,x), (3.15)

which gives

zεt32(t+1)(zεzεx)=2zεx2exzε2.\displaystyle\dfrac{\partial{z_{\varepsilon}}}{\partial{t}}-\frac{3}{2(t+1)}\Big{(}z_{\varepsilon}-\dfrac{\partial{z_{\varepsilon}}}{\partial{x}}\Big{)}=\dfrac{\partial^{2}{z_{\varepsilon}}}{\partial{x}^{2}}-e^{-x}z_{\varepsilon}^{2}. (3.16)

As (3.16) is a perturbation of the standard heat equation, it is helpful to pass to the self-similar variables:

zε(t,x)=vε(log(t+1),xt+1).z_{\varepsilon}(t,x)=v_{\varepsilon}(\log(t+1),\frac{x}{\sqrt{t+1}}). (3.17)

The function vε(τ,η)v_{\varepsilon}(\tau,\eta) is the solution of

vετη2vεη2vεη232vε+32eτ/2vεη+eτηexp(τ/2)vε2=0,\displaystyle\dfrac{\partial{v_{\varepsilon}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{v_{\varepsilon}}}{\partial{\eta}}-\dfrac{\partial^{2}{v_{\varepsilon}}}{\partial{\eta}^{2}}-\frac{3}{2}v_{\varepsilon}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{v_{\varepsilon}}}{\partial{\eta}}+e^{\tau-\eta\exp(\tau/2)}v_{\varepsilon}^{2}=0, (3.18)

with the initial condition

vε(0,η)=εeηϕ0(η).v_{\varepsilon}(0,\eta)=\varepsilon e^{\eta}\phi_{0}(\eta). (3.19)

In order to get rid of the pre-factor ε\varepsilon in the initial condition (3.19), and also to adjust the zero-order term in (3.18), it is convenient to represent vε(τ,η)v_{\varepsilon}(\tau,\eta) as

vε(τ,η)=εv1(τ,η)eτ/2.v_{\varepsilon}(\tau,\eta)=\varepsilon v_{1}(\tau,\eta)e^{\tau/2}. (3.20)

Here, v1(τ,η)v_{1}(\tau,\eta) is the solution of

v1τη2v1η2v1η2v1+32eτ/2v1η+εe3τ/2ηexp(τ/2)v12=0,\displaystyle\dfrac{\partial{v_{1}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{v_{1}}}{\partial{\eta}}-\dfrac{\partial^{2}{v_{1}}}{\partial{\eta}^{2}}-v_{1}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{v_{1}}}{\partial{\eta}}+\varepsilon e^{3\tau/2-\eta\exp(\tau/2)}v_{1}^{2}=0, (3.21)

with the initial condition

v1(0,η)=eηϕ0(η).v_{1}(0,\eta)=e^{\eta}\phi_{0}(\eta). (3.22)

The next, and last, in this chain of preliminary transformations is to eliminate the pre-factor ε\varepsilon in the last term in (3.21). We choose

β(τ)=eτ/2logε,\beta(\tau)=e^{-\tau/2}\log\varepsilon, (3.23)

so that

εe3τ/2ηexp(τ/2)=e3τ/2(ηβ(τ))exp(τ/2),\varepsilon e^{3\tau/2-\eta\exp(\tau/2)}=e^{3\tau/2-(\eta-\beta(\tau))\exp(\tau/2)}, (3.24)

and make a change of the spatial variable:

v1(τ,η)=rε(τ,ηβ(τ)).v_{1}(\tau,\eta)=r_{\varepsilon}(\tau,\eta-\beta(\tau)). (3.25)

The function rεr_{\varepsilon} satisfies:

rετη2rεη2rεη2rε+32eτ/2rεη+e3τ/2ηexp(τ/2)rε2=0,\displaystyle\dfrac{\partial{r_{\varepsilon}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{r_{\varepsilon}}}{\partial{\eta}}-\dfrac{\partial^{2}{r_{\varepsilon}}}{\partial{\eta}^{2}}-r_{\varepsilon}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{r_{\varepsilon}}}{\partial{\eta}}+e^{3\tau/2-\eta\exp(\tau/2)}r_{\varepsilon}^{2}=0, (3.26)

with the initial condition

rε(0,η)=ψ0(ηε),r_{\varepsilon}(0,\eta)=\psi_{0}(\eta-\ell_{\varepsilon}), (3.27)

with ε\ell_{\varepsilon} as in (3.3), and

ψ0(η)=eηϕ0(η).\psi_{0}(\eta)=e^{\eta}\phi_{0}(\eta). (3.28)

This, with a slight abuse of notation, is exactly (3.1). Note that rεr_{\varepsilon} depends on ε\varepsilon only through ε\ell_{\varepsilon} as it appears in the initial condition. We will interchangeably, with some abuse of notation use rε(t,x)r_{\varepsilon}(t,x) and rlε(t,x)r_{l_{\varepsilon}}(t,x) for the same object.

As far as the asymptotics of rε(τ,η)r_{\varepsilon}(\tau,\eta) and its connection to the Bramson shift are concerned, it was shown in [22] that there exists a constant v(ε)>0v_{\infty}(\varepsilon)>0 so that the solution vε(τ,η)v_{\varepsilon}(\tau,\eta) of (3.18) has the asymptotics

vε(τ,η)v(ε)ηeη2/4eτ/2, as τ+, for η>0.v_{\varepsilon}(\tau,\eta)\sim v_{\infty}(\varepsilon)\eta e^{-\eta^{2}/4}e^{\tau/2},\hbox{ as $\tau\to+\infty$, for $\eta>0$}. (3.29)

and the Bramson shift is given by

xε=logv(ε).x_{\varepsilon}=-\log v_{\infty}(\varepsilon). (3.30)

The corrseponding long-time asymptotics for the function v1(τ,η)v_{1}(\tau,\eta), the solution to (3.21) is

v1(τ,η)v~(ε)ηeη2/4, as τ+, for η>0,v_{1}(\tau,\eta)\sim\tilde{v}_{\infty}(\varepsilon)\eta e^{-\eta^{2}/4},\hbox{ as $\tau\to+\infty$, for $\eta>0$}, (3.31)

with

v~(ε)=εv(ε),\tilde{v}_{\infty}(\varepsilon)=\varepsilon v_{\infty}(\varepsilon), (3.32)

and the asymptotics for rεr_{\varepsilon} is

rε(τ,η)=v1(τ,η+β(τ))v~(ε)(η+β(τ))e(η+β(τ)2/4v~(ε)ηeη2/4, as τ+, for η>0,r_{\varepsilon}(\tau,\eta)=v_{1}(\tau,\eta+\beta(\tau))\sim\tilde{v}_{\infty}(\varepsilon)(\eta+\beta(\tau))e^{-(\eta+\beta(\tau)^{2}/4}\sim\tilde{v}_{\infty}(\varepsilon)\eta e^{-\eta^{2}/4},\hbox{ as $\tau\to+\infty$, for $\eta>0$}, (3.33)

so that

r(ε)=v~(ε)=εv(ε),r_{\infty}(\ell_{\varepsilon})=\tilde{v}_{\infty}(\varepsilon)=\varepsilon v_{\infty}(\varepsilon), (3.34)

and the Bramson shift is

xε=logv(ε)=logε1logr(ε).x_{\varepsilon}=-\log v_{\infty}(\varepsilon)=\log\varepsilon^{-1}-\log r_{\infty}(\ell_{\varepsilon}). (3.35)

This finishes the proof of Proposition 3.1\Box

4 Connection to the linear Dirichlet problem

Before giving the proof of Proposition 3.2, let us recall the intuition that leads to the long-time asymptotics (3.2) for the solution of (3.1), and also explain how the asymptotics (3.4) comes about. The key point is that we may think of (3.1) as a linear equation with the factor

e3τ/2ηexp(τ/2)r(τ,η)e^{3\tau/2-\eta\exp(\tau/2)}r_{\ell}(\tau,\eta) (4.1)

in the last term in its right side playing the role of an absorption coefficient. Disregarding our lack of information about r(τ,η)r_{\ell}(\tau,\eta) that enters (4.1), we expect that when τ1\tau\gg 1 this term is ”extremely large” for η<0\eta<0 and ”extremely small” for η>0\eta>0. Thinking again of (3.1) as a linear equation for r(τ,η)r_{\ell}(\tau,\eta), the former means that r(τ,η)r_{\ell}(\tau,\eta) is very small for η<0\eta<0, while the latter indicates that r(τ,η)r_{\ell}(\tau,\eta) essentially solves a linear problem for η>0\eta>0. The drift term in (3.1) with the pre-factor eτ/2e^{-\tau/2} is also very small at large times. Thus, if we take some T1T\gg 1, then for τT\tau\geq T, a good approximation to (3.1) is the linear Dirichlet problem

ζτη2ζη2ζη2ζ=0,τ>T,η>0\displaystyle\dfrac{\partial{\zeta_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{\zeta_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{\zeta_{\ell}}}{\partial{\eta}^{2}}-\zeta_{\ell}=0,~{}~{}\tau>T,~{}\eta>0 (4.2)
ζ(τ,0)=0,\displaystyle\zeta_{\ell}(\tau,0)=0,
ζ(T,η)=r(T,η).\displaystyle\zeta_{\ell}(T,\eta)=r_{\ell}(T,\eta).

In other words, one would solve the full nonlinear problem on the whole line only until a large time T1T\gg 1, and for τ>T\tau>T simply solve the linear Dirichlet problem (4.2). It is easy to see that

ζ¯(η)=ηeη2/4,\bar{\zeta}(\eta)=\eta e^{-\eta^{2}/4}, (4.3)

is a steady solution to (4.2). In addition, the operator

u=2uη2+η2uη+u,η>0,{\cal L}u=\frac{\partial^{2}u}{\partial\eta^{2}}+\frac{\eta}{2}\dfrac{\partial{u}}{\partial{\eta}}+u,~{}~{}\eta>0, (4.4)

with the Dirichlet boundary condition at η=0\eta=0 has a discrete spectrum. It follows that ζ(τ,η)\zeta_{\ell}(\tau,\eta) has the long time asymptotics

ζ(τ,η)ζ()ηeη2/4,τ+.\zeta_{\ell}(\tau,\eta)\sim\zeta_{\infty}(\ell)\eta e^{-\eta^{2}/4},~{}~{}\hbox{$\tau\to+\infty$}. (4.5)

As the integral

0ηζ(τ,η)𝑑η=0ηζ(T,η)𝑑η\int_{0}^{\infty}\eta\zeta_{\ell}(\tau,\eta)d\eta=\int_{0}^{\infty}\eta\zeta_{\ell}(T,\eta)d\eta (4.6)

is conserved, the coefficient ζ()\zeta_{\infty}(\ell) is determined by the relation

ζ()0η2eη2/4𝑑η=0ηζ(T,η)𝑑η,\zeta_{\infty}(\ell)\int_{0}^{\infty}\eta^{2}e^{-\eta^{2}/4}d\eta=\int_{0}^{\infty}\eta\zeta_{\ell}(T,\eta)d\eta, (4.7)

so that

ζ()=14π0ηζ(T,η)𝑑η=14π0ηr(T,η)𝑑η.\zeta_{\infty}(\ell)=\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta\zeta_{\ell}(T,\eta)d\eta=\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta r_{\ell}(T,\eta)d\eta. (4.8)

As we expect ζ(τ,η)\zeta_{\ell}(\tau,\eta) and r(τ,η)r_{\ell}(\tau,\eta) to be close if TT is sufficiently large, we should have an approximation

ζ()r(),\zeta_{\infty}(\ell)\approx r_{\infty}(\ell), (4.9)

if T1T\gg 1. This, in turn, implies that

r()=limτ+14π0ηr(τ,η)𝑑η.r_{\infty}(\ell)=\lim_{\tau\to+\infty}\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta r_{\ell}(\tau,\eta)d\eta. (4.10)

This informal argument is made rigorous in [22].

The limit in the right side of (4.10) is an implicit functional of the initial conditions for the nonlinear problem (3.1), and the evolution of the solution in the initial time layer, before the linear approximation kicks in, is difficult to control, so that there is no explicit expression for r()r_{\infty}(\ell). In the present setting, however, the initial condition r(0,η)r_{\ell}(0,\eta) in (3.1) is shifted to the right by 1\ell\gg 1. Therefore, at small times the solution is concentrated at η1\eta\gg 1, a region where the factor in front of the nonlinear term in (3.1)

exp(3τ2ηeτ/2)1\exp\Big{(}\frac{3\tau}{2}-\eta e^{\tau/2}\Big{)}\ll 1 (4.11)

is very small even for τ=O(1)\tau=O(1). Hence, solutions to the nonlinear equation (3.1) with the initial conditions (3.27) should be well approximated, to the leading order, by the linear problem

r~τη2r~η2r~η2r~+32eτ/2r~η=0,r~(0,η)=r(0,η),\displaystyle\dfrac{\partial{\tilde{r}_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{\tilde{r}_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{\tilde{r}_{\ell}}}{\partial{\eta}^{2}}-\tilde{r}_{\ell}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{\tilde{r}_{\ell}}}{\partial{\eta}}=0,~{}~{}\tilde{r}_{\ell}(0,\eta)=r_{\ell}(0,\eta), (4.12)

even for small times. However, the solution ”does not yet know” for ”small” τ\tau that there is a large dissipative term in the nonlinear equation, or the Dirichlet boundary condition in the linear version, and evolves ”as if (4.12) is posed for η\eta\in{\mathbb{R}}”. This leads to exponential growth in τ\tau until the solution spreads sufficiently far to the left, close to η=0\eta=0 and ”discovers” the Dirichlet boundary condition (or the nonlinearity in the full nonlinear version). During this ”short time” evolution we have

ddτηr~(τ,η)𝑑η=32eτ/2r~(τ,η)𝑑η.\frac{d}{d\tau}\int\eta\tilde{r}_{\ell}(\tau,\eta)d\eta=\frac{3}{2}e^{-\tau/2}\int\tilde{r}_{\ell}(\tau,\eta)d\eta. (4.13)

Unlike the first moment, the total mass in the right side does not grow as \ell gets larger – the shift of the initial condition to the right increases the first moment but not the mass. Thus, the first moment of r(τ,η)r_{\ell}(\tau,\eta) will only change by a factor that is o(1)o(1) during the ”short time” evolution, so that it is conserved to the leading order in \ell. The ”long time” evolution following this initial time layer is well approximated by the linear Dirichlet problem (4.2) that preserves the first moment. Thus, altogether, the first moment will not change to the leading order if 1\ell\gg 1 is large, so that

limτ+0ηr(τ,η)𝑑η=(1+o(1))0ηr(0,η)𝑑η, as ε0,\lim_{\tau\to+\infty}\int_{0}^{\infty}\eta r_{\ell}(\tau,\eta)d\eta=(1+o(1))\int_{0}^{\infty}\eta r_{\ell}(0,\eta)d\eta,~{}~{}\hbox{ as $\varepsilon\to 0$}, (4.14)

which leads to the explicit expression for r()r_{\infty}(\ell) in terms of the initial first moment:

r()\displaystyle r_{\infty}(\ell) =(1+o(1))14π0ηr(0,η)𝑑η=(1+o(1))14π0ηeηϕ0(η)𝑑η\displaystyle=(1+o(1))\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta r_{\ell}(0,\eta)d\eta=(1+o(1))\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta e^{\eta-\ell}\phi_{0}(\eta-\ell)d\eta (4.15)
=(1+o(1))14π(η+)eηϕ0(η)𝑑η=(1+o(1))4πeηϕ0(η)𝑑η\displaystyle=(1+o(1))\frac{1}{\sqrt{4\pi}}\int_{-\ell}^{\infty}(\eta+\ell)e^{\eta}\phi_{0}(\eta)d\eta=(1+o(1))\frac{\ell}{\sqrt{4\pi}}\int_{-\infty}^{\infty}e^{\eta}\phi_{0}(\eta)d\eta
=c¯(1+o(1)),\displaystyle=\bar{c}(1+o(1))\ell,

which is (3.4). This very informal argument is behind the reason why we can describe the Bramson shift so explicitly for ε1\varepsilon\ll 1, which corresponds to 1\ell\gg 1. The rest of the proof of Proposition 3.2 formalizes this argument by providing matching upper and lower bounds on the limit in the right side of (4.10).

An approximate solution to the adjoint linear problem

In order to improve on the approximate conservation law (4.13) let us make the following observation. Let us set

Qk(τ,η)=η+kψ¯(η)eτ/2,Q_{k}(\tau,\eta)=\eta+k\bar{\psi}(\eta)e^{-\tau/2}, (4.16)

with

ψ¯(η)=0ηez2/4zey2/4𝑑y𝑑z,\bar{\psi}(\eta)=\int_{0}^{\eta}e^{z^{2}/4}\int_{z}^{\infty}e^{-y^{2}/4}dydz, (4.17)

and consider a solution to the linear Dirichlet problem

pτη2pη2pη2p+keτ/2pη=f(τ,η),p(τ,0)=0.\displaystyle\dfrac{\partial{p}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{p}}{\partial{\eta}}-\dfrac{\partial^{2}{p}}{\partial{\eta}^{2}}-p+ke^{-\tau/2}\dfrac{\partial{p}}{\partial{\eta}}=f(\tau,\eta),~{}~{}p(\tau,0)=0. (4.18)
Lemma 4.1

We have

ddτ0Qk(τ,η)p(τ,η)𝑑η=k2eτ0ψ¯(η)ηp(τ,η)𝑑η+0f(τ,η)Qk(τ,η)𝑑η.\frac{d}{d\tau}\int_{0}^{\infty}Q_{k}(\tau,\eta)p(\tau,\eta)d\eta=k^{2}e^{-\tau}\int_{0}^{\infty}\dfrac{\partial{\bar{\psi}(\eta)}}{\partial{\eta}}p(\tau,\eta)d\eta+\int_{0}^{\infty}f(\tau,\eta)Q_{k}(\tau,\eta)d\eta. (4.19)

Proof. Note that

ddτ\displaystyle\frac{d}{d\tau} 0Qk(τ,η)p(τ,η)𝑑η=0(Qkτp+Qk[η2pη+2pη2+pkeτ/2pη+f])𝑑η\displaystyle\int_{0}^{\infty}\!Q_{k}(\tau,\eta)p(\tau,\eta)d\eta=\int_{0}^{\infty}\Big{(}\dfrac{\partial{Q_{k}}}{\partial{\tau}}p+Q_{k}\Big{[}\frac{\eta}{2}\dfrac{\partial{p}}{\partial{\eta}}+\dfrac{\partial^{2}{p}}{\partial{\eta}^{2}}+p-ke^{-\tau/2}\dfrac{\partial{p}}{\partial{\eta}}+f\Big{]}\Big{)}d\eta (4.20)
=0(p[Qkτ+2Qkη2η(η2Qk)+Qk+keτ/2Qkη]+fQk)𝑑η.\displaystyle=\int_{0}^{\infty}\Big{(}p\Big{[}\dfrac{\partial{Q_{k}}}{\partial{\tau}}+\dfrac{\partial^{2}{Q_{k}}}{\partial{\eta}^{2}}-\dfrac{\partial{}}{\partial{\eta}}\Big{(}\frac{\eta}{2}Q_{k}\Big{)}+Q_{k}+ke^{-\tau/2}\dfrac{\partial{Q_{k}}}{\partial{\eta}}\Big{]}+fQ_{k}\Big{)}d\eta.

It is easy to check that the function ψ¯(η)\bar{\psi}(\eta) is a solution to

η2ψ¯η2ψ¯η21=0,ψ¯(0)=0.\displaystyle\frac{\eta}{2}\dfrac{\partial{\bar{\psi}}}{\partial{\eta}}-\dfrac{\partial^{2}{\bar{\psi}}}{\partial{\eta}^{2}}-1=0,~{}~{}\bar{\psi}(0)=0. (4.21)

With this, we can compute that the function Qk(τ,η)Q_{k}(\tau,\eta) satisfies

Qkτη(η2Qk)+2Qkη2+Qk+keτ/2Qkη=keτ/2k2eτ/2ψ¯(η)k2ψ¯(η)eτ/2\displaystyle\dfrac{\partial{Q_{k}}}{\partial{\tau}}-\dfrac{\partial{}}{\partial{\eta}}\Big{(}\frac{\eta}{2}Q_{k}\Big{)}+\dfrac{\partial^{2}{Q_{k}}}{\partial{\eta}^{2}}+Q_{k}+ke^{-\tau/2}\dfrac{\partial{Q_{k}}}{\partial{\eta}}=ke^{-\tau/2}-\frac{k}{2}e^{-\tau/2}\bar{\psi}(\eta)-\frac{k}{2}\bar{\psi}(\eta)e^{-\tau/2} (4.22)
kη2ψ¯(η)ηeτ/2+k2ψ¯(η)η2eτ/2+kψ¯(η)eτ/2+k2ψ¯(η)ηeτ\displaystyle-\frac{k\eta}{2}\dfrac{\partial{\bar{\psi}(\eta)}}{\partial{\eta}}e^{-\tau/2}+k\dfrac{\partial^{2}{\bar{\psi}(\eta)}}{\partial{\eta}^{2}}e^{-\tau/2}+k\bar{\psi}(\eta)e^{-\tau/2}+k^{2}\dfrac{\partial{\bar{\psi}(\eta)}}{\partial{\eta}}e^{-\tau}
=k(1η2ψ¯(η)η+2ψ¯(η)η2)eτ/2+k2ψ¯(η)ηeτ=k2ψ¯(η)ηeτ.\displaystyle=k\Big{(}1-\frac{\eta}{2}\dfrac{\partial{\bar{\psi}(\eta)}}{\partial{\eta}}+\dfrac{\partial^{2}{\bar{\psi}(\eta)}}{\partial{\eta}^{2}}\Big{)}e^{-\tau/2}+k^{2}\dfrac{\partial{\bar{\psi}(\eta)}}{\partial{\eta}}e^{-\tau}=k^{2}\dfrac{\partial{\bar{\psi}(\eta)}}{\partial{\eta}}e^{-\tau}.

Using this in (4.20) gives (4.19). \Box

Note that for 0η10\leq\eta\leq 1 we have

ψ¯(η)Cη.\bar{\psi}(\eta)\leq C\eta. (4.23)

For the asymptotics of ψ¯(η)\bar{\psi}(\eta) for large η1\eta\gg 1, note that

zey2/4𝑑y=z2y(ey2/4)𝑑y=2zez2/4z2y2ey2/4𝑑y.\int_{z}^{\infty}e^{-y^{2}/4}dy=-\int_{z}^{\infty}\frac{2}{y}\Big{(}e^{-y^{2}/4}\Big{)}^{\prime}dy=\frac{2}{z}e^{-z^{2}/4}-\int_{z}^{\infty}\frac{2}{y^{2}}e^{-y^{2}/4}dy. (4.24)

It follows that for η1\eta\geq 1 we have

ψ¯(η)\displaystyle\bar{\psi}(\eta) =ψ¯(1)+1ηez2/4zey2/4𝑑y𝑑z=ψ¯(1)+2logη21ηez2/4z1y2ey2/4𝑑y\displaystyle=\bar{\psi}(1)+\int_{1}^{\eta}e^{z^{2}/4}\int_{z}^{\infty}e^{-y^{2}/4}dydz=\bar{\psi}(1)+2\log\eta-2\int_{1}^{\eta}e^{z^{2}/4}\int_{z}^{\infty}\frac{1}{y^{2}}e^{-y^{2}/4}dy (4.25)
=2logη+g(η),η1,\displaystyle=2\log\eta+g(\eta),~{}~{}\eta\geq 1,

where g(η)g(\eta) is a bounded function such that

g=limη+g(η)=01ez2/4zey2/4𝑑y𝑑z21ez2/4z1y2ey2/4𝑑y.g_{\infty}=\lim_{\eta+\infty}g(\eta)=\int_{0}^{1}e^{z^{2}/4}\int_{z}^{\infty}e^{-y^{2}/4}dydz-2\int_{1}^{\infty}e^{z^{2}/4}\int_{z}^{\infty}\frac{1}{y^{2}}e^{-y^{2}/4}dy. (4.26)

This is how the constant gg_{\infty} appears in (1.35) and in Theorem 1.4.

5 The proof of Proposition 3.2

5.1 An upper bound for the first moment

In this section, we prove an upper bound for r()r_{\infty}(\ell).

Lemma 5.1

There exists K>0K>0 so that we have

r()=lim supτ+14π0ηr(τ,η)𝑑ηc¯+Klogfor 2,r_{\infty}(\ell)=\limsup_{\tau\to+\infty}\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta r_{\ell}(\tau,\eta)d\eta\leq\bar{c}\ell+K\log\ell~{}~{}\hbox{for $\ell\geq 2$}, (5.1)

with c¯\bar{c} as in (1.27).

Reduction to a Dirichlet problem

We first bound the solution to (3.1) by a solution to the linear Dirichlet problem, up to a relatively small error. We start with two observations. First, the solution of the original KPP problem (1.23) satisfies u(t,x)1u(t,x)\leq 1, hence the function vε(τ,η)v_{\varepsilon}(\tau,\eta) defined in (3.17) satisfies

vε(τ,η)eηeτ/2.v_{\varepsilon}(\tau,\eta)\leq e^{\eta e^{\tau/2}}.

Retracing our changes of variables, we deduce that v1(τ,η)v_{1}(\tau,\eta) defined in (3.20) satisfies

v1(τ,η)=ε1vε(τ,η)eτ/2ε1eτ/2+ηeτ/2,v_{1}(\tau,\eta)=\varepsilon^{-1}v_{\varepsilon}(\tau,\eta)e^{-\tau/2}\leq\varepsilon^{-1}e^{-\tau/2+\eta e^{\tau/2}}, (5.2)

and rε(t,x)r_{\varepsilon}(t,x) introduced in (3.25) obeys

rε(τ,η)=v1(τ,η+β(τ))ε1e(ηeτ/2logε1)eτ/2eτ/2=eηeτ/2τ/2.r_{\ell_{\varepsilon}}(\tau,\eta)=v_{1}(\tau,\eta+\beta(\tau))\leq\varepsilon^{-1}e^{(\eta-e^{-\tau/2}\log\varepsilon^{-1})e^{\tau/2}}e^{-\tau/2}=e^{\eta e^{\tau/2}-\tau/2}. (5.3)

It follows that at the boundary η=0\eta=0 we have

0<r(τ,0)eτ/2, for all τ>0,0<r_{\ell}(\tau,0)\leq e^{-\tau/2},~{}~{}\hbox{ for all $\tau>0$,} (5.4)

so that for τ1\tau\gg 1 the function rr_{\ell} does satisfy an approximate Dirichlet boundary condition at η=0\eta=0. However, the bound (5.4) is very poor for τ=O(1)\tau=O(1) – recall that the initial condition is located at distance 1\ell\gg 1 away from the origin, so the solution remains small near η=0\eta=0 for some time τ1\tau\gg 1. In particular, as a first step, we can bound r(τ,η)r_{\ell}(\tau,\eta) from above by the solution to the linear problem on the whole line:

R¯τη2R¯η2R¯η2R¯+32eτ/2R¯η=0,η,\displaystyle\dfrac{\partial{\bar{R}_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{\bar{R}_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{\bar{R}_{\ell}}}{\partial{\eta}^{2}}-\bar{R}_{\ell}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{\bar{R}_{\ell}}}{\partial{\eta}}=0,~{}~{}\eta\in{\mathbb{R}},
R¯(0,η)=r(0,η).\displaystyle\bar{R}_{\ell}(0,\eta)=r_{\ell}(0,\eta). (5.5)

A change of variables

R¯(τ,η)=Q(τ,η32τeτ/2)\bar{R}_{\ell}(\tau,\eta)=Q_{\ell}(\tau,\eta-\frac{3}{2}\tau e^{-\tau/2})

leads to the standard heat equation in the self-similar variables

Qτη2Qη2Qη2Q=0,η,\displaystyle\dfrac{\partial{Q_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{Q_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{Q_{\ell}}}{\partial{\eta}^{2}}-Q_{\ell}=0,~{}~{}\eta\in{\mathbb{R}},
Q(0,η)=r(0,η).\displaystyle Q_{\ell}(0,\eta)=r_{\ell}(0,\eta). (5.6)

Thus, the function R¯(τ,η)\bar{R}_{\ell}(\tau,\eta) can be written explicitly as

R¯(τ,η)\displaystyle\bar{R}_{\ell}(\tau,\eta) =eτG(eτ1,(η32τeτ/2)eτ/2y)r(0,y)𝑑y\displaystyle=e^{\tau}\int G(e^{\tau}-1,(\eta-\frac{3}{2}\tau e^{-\tau/2})e^{\tau/2}-y)r_{\ell}(0,y)dy (5.7)
=eτG(eτ1,ηeτ/232τy)r(0,y)𝑑y.\displaystyle=e^{\tau}\int G(e^{\tau}-1,\eta e^{\tau/2}-\frac{3}{2}\tau-y)r_{\ell}(0,y)dy.

Here, G(t,x)G(t,x) is the standard heat kernel:

G(t,x)=14πte|x|2/(4t).G(t,x)=\frac{1}{\sqrt{4\pi t}}e^{-|x|^{2}/(4t)}. (5.8)

As r(0,η)=ψ0(η)r_{\ell}(0,\eta)=\psi_{0}(\eta-\ell), and ψ0(η)\psi_{0}(\eta) satisfies ψ0(η)=0\psi_{0}(\eta)=0 for η>L0\eta>L_{0} and ψ0(η)Ceη\psi_{0}(\eta)\leq Ce^{\eta} for η<0\eta<0, we have

R¯(τ,0)\displaystyle\bar{R}_{\ell}(\tau,0) =eτG(eτ1,y+32τ)r(0,y)𝑑y=eτG(eτ1,y++32τ)ψ0(y)𝑑y\displaystyle=e^{\tau}\int G(e^{\tau}-1,y+\frac{3}{2}\tau)r_{\ell}(0,y)dy=e^{\tau}\int G(e^{\tau}-1,y+\ell+\frac{3}{2}\tau)\psi_{0}(y)dy (5.9)
CeτL0G(eτ1,y++32τ)ey𝑑y=Ceτe(3/2)τL0++(3τ/2)G(eτ1,y)ey𝑑y.\displaystyle\leq Ce^{\tau}\int_{-\infty}^{L_{0}}G(e^{\tau}-1,y+\ell+\frac{3}{2}\tau)e^{y}dy=Ce^{\tau}e^{-\ell-(3/2)\tau}\int_{-\infty}^{L_{0}+\ell+(3\tau/2)}G(e^{\tau}-1,y)e^{y}dy.

Note that for any LL\in{\mathbb{R}} we have

LG(t,y)ey𝑑y\displaystyle\int_{-\infty}^{L}G(t,y)e^{y}dy =14πtLexp(y24t+y)𝑑y=14πL/texp(y24+ytt+t)𝑑y\displaystyle=\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{L}\exp\Big{(}-\frac{y^{2}}{4t}+y\Big{)}dy=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{L/\sqrt{t}}\exp\Big{(}-\frac{y^{2}}{4}+y\sqrt{t}-t+t\Big{)}dy (5.10)
=et4πL/texp((y2t)2)𝑑y=et4πL/t2texp(y24)𝑑y.\displaystyle=\frac{e^{t}}{\sqrt{4\pi}}\int_{-\infty}^{L/\sqrt{t}}\exp\Big{(}-\Big{(}\frac{y}{2}-\sqrt{t}\Big{)}^{2}\Big{)}dy=\frac{e^{t}}{\sqrt{4\pi}}\int_{-\infty}^{L/\sqrt{t}-2\sqrt{t}}\exp\Big{(}-\frac{y^{2}}{4}\Big{)}dy.

We take δ>0\delta>0 sufficiently small, and consider two cases. First, if

τ<τ1=loglog(2δ),\tau<\tau_{1}=\log\ell-\log(2-\delta),

then we have, from (5.9) and (5.10), for for 0<τ<τ10<\tau<\tau_{1}:

R¯(τ,0)\displaystyle\bar{R}_{\ell}(\tau,0) Ceeτ/2eeτ1Ceeτ1/2eeτ11=Ceeτ1/2e=C1/2C1/4eτ/4.\displaystyle\leq Ce^{-\ell}e^{-\tau/2}e^{e^{\tau}-1}\leq Ce^{-\ell}e^{-\tau_{1}/2}e^{e^{\tau_{1}}-1}=Ce^{-\ell}e^{-\tau_{1}/2}e^{{\ell}}=C\ell^{-1/2}\leq C\ell^{-1/4}e^{-\tau/4}. (5.11)

On the other hand, if τ>τ1\tau>\tau_{1}, then, taking

t=eτ1,L=L0++3τ2,t=e^{\tau}-1,~{}~{}L=L_{0}+\ell+\frac{3\tau}{2}, (5.12)

in (5.10), we see that for \ell sufficiently large, the upper limit of integration

Lt2t\displaystyle\frac{L}{\sqrt{t}}-2\sqrt{t} =1eτ1(L0++3τ22eτ+2)=1eτ1(L0+(2δ)eτ1+3τ22eτ+2)\displaystyle=\frac{1}{\sqrt{e^{\tau}-1}}\Big{(}L_{0}+\ell+\frac{3\tau}{2}-2e^{\tau}+2\Big{)}=\frac{1}{\sqrt{e^{\tau}-1}}\Big{(}L_{0}+(2-\delta)e^{\tau_{1}}+\frac{3\tau}{2}-2e^{\tau}+2\Big{)} (5.13)
δ2eτ/2\displaystyle\leq-\frac{\delta}{2}e^{\tau/2}

is very negative. In particular, the integral in the right side of (5.10) can be estimated as

L/t2texp(y24)𝑑yC|Lt2t|1exp(14(Lt2t)2)\displaystyle\int_{-\infty}^{L/\sqrt{t}-2\sqrt{t}}\exp\Big{(}-\frac{y^{2}}{4}\Big{)}dy\leq C\Big{|}\frac{L}{\sqrt{t}}-2\sqrt{t}\Big{|}^{-1}\exp\Big{(}-\frac{1}{4}\Big{(}\frac{L}{\sqrt{t}}-2\sqrt{t}\Big{)}^{2}\Big{)} (5.14)
=C|Lt2t|1eL2/(4t)eLet.\displaystyle=C\Big{|}\frac{L}{\sqrt{t}}-2\sqrt{t}\Big{|}^{-1}e^{-L^{2}/(4t)}e^{L}e^{-t}.

Then, we have from (5.9), (5.10), (5.13) and (5.14)

R¯(τ,0)Ceτ/2et|Lt2t|1eL2/(4t)eLet\displaystyle\bar{R}_{\ell}(\tau,0)\leq Ce^{-\ell-\tau/2}e^{t}\Big{|}\frac{L}{\sqrt{t}}-2\sqrt{t}\Big{|}^{-1}e^{-L^{2}/(4t)}e^{L}e^{-t} (5.15)
Cδeτ/2eτ/2eL0++3τ/2e(L0++3τ/2)2/(4(eτ1))Cδeτ/2e(L0++3τ/2)2/(4(eτ1))\displaystyle\leq C_{\delta}e^{-\ell-\tau/2}e^{-\tau/2}e^{L_{0}+\ell+3\tau/2}e^{-(L_{0}+\ell+3\tau/2)^{2}/(4(e^{\tau}-1))}\leq C_{\delta}e^{\tau/2}e^{-(L_{0}+\ell+3\tau/2)^{2}/(4(e^{\tau}-1))}
Ceτ/2Ceτ/4eτ1/4C1/4eτ/4,\displaystyle\leq Ce^{-\tau/2}\leq Ce^{-\tau/4}e^{-\tau_{1}/4}\leq C\ell^{-1/4}e^{-\tau/4},

provided that ττ1\tau\geq\tau_{1} and

τ(L0++3τ/2)2/(4(eτ1)).\tau\leq(L_{0}+\ell+3\tau/2)^{2}/(4(e^{\tau}-1)). (5.16)

In particular, we can take

ττ2\displaystyle\tau\leq\tau_{2} =2logloglog3,\displaystyle=2\log\ell-\log\log\ell-3, (5.17)

as long as \ell is sufficiently large, because then we have

4τ(eτ1)4τeτ2log42e3log2,4\tau(e^{\tau}-1)\leq 4\tau e^{\tau}\leq 2\log\ell\frac{4\ell^{2}}{e^{3}\log\ell}\leq\ell^{2}, (5.18)

so that (5.16) holds. It follows that

r(τ,0)C1/4eτ/4,0ττ2.r_{\ell}(\tau,0)\leq C\ell^{-1/4}e^{-\tau/4},~{}~{}~{}0\leq\tau\leq\tau_{2}. (5.19)

Taking into account (5.4), we deduce that we also have

r(τ,0)eτ/2eτ2/4eτ/4=C(log)1/41/2eτ/4C1/4eτ/4,for τ>τ2.r_{\ell}(\tau,0)\leq e^{-\tau/2}\leq e^{-\tau_{2}/4}e^{-\tau/4}=\frac{C(\log\ell)^{1/4}}{\ell^{1/2}}e^{-\tau/4}\leq\frac{C}{\ell^{1/4}}e^{-\tau/4},~{}~{}\hbox{for $\tau>\tau_{2}$}. (5.20)

It follows that the function r(τ,η)r_{\ell}(\tau,\eta), the solution to (3.1), is bounded from above by the solution to the linear half-line problem

r¯τη2r¯η2r¯η2r¯+32eτ/2r¯η=0,η>0,\displaystyle\dfrac{\partial{\bar{r}_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{\bar{r}_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{\bar{r}_{\ell}}}{\partial{\eta}^{2}}-\bar{r}_{\ell}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{\bar{r}_{\ell}}}{\partial{\eta}}=0,~{}~{}\eta>0,
r¯(τ,0)=C1/4eτ/4,\displaystyle\bar{r}_{\ell}(\tau,0)=C\ell^{-1/4}e^{-\tau/4}, (5.21)

with the initial condition

r¯(0,η)=ψ0(η).\bar{r}_{\ell}(0,\eta)=\psi_{0}(\eta-\ell). (5.22)

In order to deal with the small but non-zero boundary condition in (5.1), we make yet another change of variables:

r¯(τ,η)=q¯(τ,η)+C1/4eτ/4g(η),\bar{r}_{\ell}(\tau,\eta)=\bar{q}_{\ell}(\tau,\eta)+C\ell^{-1/4}e^{-\tau/4}g(\eta), (5.23)

with a smooth function g(η)g(\eta) such that g(0)=1g(0)=1, g(η)0g(\eta)\geq 0 and g(η)=0g(\eta)=0 for η1\eta\geq 1. This leads to

q¯τ+32eτ/2q¯η=q¯+C1/4G(τ,η)eτ/4,η>0,\displaystyle\dfrac{\partial{\bar{q}_{\ell}}}{\partial{\tau}}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{\bar{q}_{\ell}}}{\partial{\eta}}={\cal L}\bar{q}_{\ell}+C\ell^{-1/4}G(\tau,\eta)e^{-\tau/4},~{}~{}\eta>0,
q¯(τ,0)=0,\displaystyle\bar{q}_{\ell}(\tau,0)=0, (5.24)

with a uniformly bounded function G(τ,η)G(\tau,\eta) that is supported in η[0,1]\eta\in[0,1] and is independent of \ell. We recall that the operator {\cal L} is defined in (4.4). This change of variable does not affect the asymptotics of the first moment:

lim supτ+0ηr(τ,η)𝑑ηlim supτ+0ηr¯(τ,η)𝑑η=lim supτ+0ηq¯(τ,η)𝑑η.\limsup_{\tau\to+\infty}\int_{0}^{\infty}\eta r_{\ell}(\tau,\eta)d\eta\leq\limsup_{\tau\to+\infty}\int_{0}^{\infty}\eta\bar{r}_{\ell}(\tau,\eta)d\eta=\limsup_{\tau\to+\infty}\int_{0}^{\infty}\eta\bar{q}_{\ell}(\tau,\eta)d\eta. (5.25)

The initial condition for q¯\bar{q}_{\ell} is

q¯(0,η)=r¯(0,η)C1/4g(η).\bar{q}_{\ell}(0,\eta)=\bar{r}_{\ell}(0,\eta)-C\ell^{-1/4}g(\eta). (5.26)

It is easy to see that

0ηq¯(τ,η)𝑑η0ηp¯(τ,η)𝑑η+C01/4.\int_{0}^{\infty}\eta\bar{q}_{\ell}(\tau,\eta)d\eta\leq\int_{0}^{\infty}\eta\bar{p}_{\ell}(\tau,\eta)d\eta+C_{0}\ell^{-1/4}. (5.27)

Here, C0C_{0} is a constant that depends neither on \ell nor on ϕ0\phi_{0}, and p¯(τ,η)\bar{p}_{\ell}(\tau,\eta) is the solution to the homogeneous problem

p¯τ+32eτ/2p¯η=p¯,η>0,\displaystyle\dfrac{\partial{\bar{p}_{\ell}}}{\partial{\tau}}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{\bar{p}_{\ell}}}{\partial{\eta}}={\cal L}\bar{p}_{\ell},~{}~{}\eta>0,
p¯(τ,0)=0,\displaystyle\bar{p}_{\ell}(\tau,0)=0, (5.28)
p¯(0,η)=ψ0(η).\displaystyle\bar{p}_{\ell}(0,\eta)=\psi_{0}(\eta-\ell).

Let us now use Lemma 4.1, with k=3/2k=3/2: multiply (5.1) by Q3/2(τ,η)Q_{3/2}(\tau,\eta) and integrate:

ddτ\displaystyle\frac{d}{d\tau} 0Q3/2(τ,η)p¯(τ,η)𝑑η=94eτ0p¯(τ,η)ψ¯(η)η𝑑ηCeτ0p¯(τ,η)1+η𝑑η\displaystyle\int_{0}^{\infty}Q_{3/2}(\tau,\eta)\bar{p}_{\ell}(\tau,\eta)d\eta=\frac{9}{4}e^{-\tau}\int_{0}^{\infty}\bar{p}_{\ell}(\tau,\eta)\dfrac{\partial{\bar{\psi}(\eta)}}{\partial{\eta}}d\eta\leq Ce^{-\tau}\int_{0}^{\infty}\frac{\bar{p}_{\ell}(\tau,\eta)}{1+\eta}d\eta (5.29)
Ceτ0p¯(τ,η)𝑑η.\displaystyle\leq Ce^{-\tau}\int_{0}^{\infty}\bar{p}_{\ell}(\tau,\eta)d\eta.

Note, however, that integrating (5.1) gives a simple upper bound

ddτ0p¯(τ,η)𝑑η=p¯(τ,0)η+120p¯(τ,η)𝑑η120p¯(τ,η)𝑑η,\displaystyle\frac{d}{d\tau}\int_{0}^{\infty}\bar{p}_{\ell}(\tau,\eta)d\eta=-\dfrac{\partial{\bar{p}_{\ell}(\tau,0)}}{\partial{\eta}}+\frac{1}{2}\int_{0}^{\infty}\bar{p}_{\ell}(\tau,\eta)d\eta\leq\frac{1}{2}\int_{0}^{\infty}\bar{p}_{\ell}(\tau,\eta)d\eta, (5.30)

implying a trivial and very poor upper bound

0p¯(τ,η)𝑑ηC0eτ/2.\displaystyle\int_{0}^{\infty}\bar{p}_{\ell}(\tau,\eta)d\eta\leq C_{0}e^{\tau/2}. (5.31)

We remind that we use the notation C0C_{0} for various constants that do not depend on ε\varepsilon. Using this estimate in the right side of (5.29) gives

0Q3/2(τ,η)p¯(τ,η)𝑑η0Q3/2(0,η)p¯(0,η)𝑑η+C0.\displaystyle\int_{0}^{\infty}Q_{3/2}(\tau,\eta)\bar{p}_{\ell}(\tau,\eta)d\eta\leq\int_{0}^{\infty}Q_{3/2}(0,\eta)\bar{p}_{\ell}(0,\eta)d\eta+C_{0}. (5.32)

Recalling the definition (4.16) of Qk(τ,η)Q_{k}(\tau,\eta) and passing to the limit τ+\tau\to+\infty gives

p¯()\displaystyle\bar{p}_{\infty}(\ell) =limτ+14π0ηp¯(τ,η)𝑑η14π0[η+32ψ¯(η)]ψ0(η)𝑑η+C0\displaystyle=\lim_{\tau\to+\infty}\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta\bar{p}_{\ell}(\tau,\eta)d\eta\leq\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}[\eta+\frac{3}{2}\bar{\psi}(\eta)]\psi_{0}(\eta-\ell)d\eta+C_{0} (5.33)
=14πηeηϕ0(η)𝑑η+4πeηϕ0(η)𝑑η+32ψ¯(η+)eηϕ0(η)dη4π+C0\displaystyle=\frac{1}{\sqrt{4\pi}}\int_{-\ell}^{\infty}\eta e^{\eta}\phi_{0}(\eta)d\eta+\frac{\ell}{\sqrt{4\pi}}\int_{-\ell}^{\infty}e^{\eta}\phi_{0}(\eta)d\eta+\frac{3}{2}\int_{-\ell}^{\infty}\bar{\psi}(\eta+\ell)e^{\eta}\phi_{0}(\eta)\frac{d\eta}{\sqrt{4\pi}}+C_{0}
c¯+3c¯log+C1, for 2,\displaystyle\leq\bar{c}\ell+{3\bar{c}}\log\ell+C_{1},\hbox{ for $\ell\geq 2$},

with c¯\bar{c} as in (1.27), and a constant C1C_{1} that may depend on ϕ0\phi_{0}. We used (4.25) in the last inequality above. The conclusion of Lemma 5.1 now follows.

5.2 A lower bound for the first moment

We now prove a lower bound for the first moment matching the upper bound in Lemma 5.1, to the leading order in \ell, which will finish the proof of the asymptotics (3.4) in Proposition 3.2.

Lemma 5.2

There exists a constant K1K_{1} so that

lim infτ+14π0ηr(τ,η)𝑑ηc¯+K1log,for 2.\liminf_{\tau\to+\infty}\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta r_{\ell}(\tau,\eta)d\eta\geq\bar{c}\ell+K_{1}\log\ell,~{}~{}\hbox{for $\ell\geq 2$.} (5.34)

Let us go back to (3.1):

rτη2rη2rη2r+32eτ/2rη+e3τ/2ηexp(τ/2)r2=0,\displaystyle\dfrac{\partial{r_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{r_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{r_{\ell}}}{\partial{\eta}^{2}}-r_{\ell}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{r_{\ell}}}{\partial{\eta}}+e^{3\tau/2-\eta\exp(\tau/2)}r_{\ell}^{2}=0, (5.35)

with the initial condition

r(0,η)=eηϕ0(η).r_{\ell}(0,\eta)=e^{\eta-\ell}\phi_{0}(\eta-\ell). (5.36)

We would like first to get rid of the nonlinearity in the last term in the left side of (5.35). This is done as follows: recall that we have

r(τ,η)R¯(τ,η),r_{\ell}(\tau,\eta)\leq\bar{R}(\tau,\eta), (5.37)

where R¯(τ,η)\bar{R}_{\ell}(\tau,\eta) is the solution to the linear whole line problem (5.1). Note that (5.7) implies that

R¯(τ,η)Ceτ/2ψ0(y)𝑑y=Ceτ/2.\bar{R}_{\ell}(\tau,\eta)\leq Ce^{\tau/2}\int\psi_{0}(y)dy=Ce^{\tau/2}. (5.38)

Thus, a lower bound for rr_{\ell} is the solution to

r~τη2r~η2r~ε2r~+32eτ/2r~η+Ce2τηexp(τ/2)r~=0,η>mτeτ/2,\displaystyle\dfrac{\partial{\tilde{r}_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{\tilde{r}_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{\tilde{r}_{\varepsilon}}}{\partial{\ell}^{2}}-\tilde{r}_{\ell}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{\tilde{r}_{\ell}}}{\partial{\eta}}+Ce^{2\tau-\eta\exp(\tau/2)}\tilde{r}_{\ell}=0,~{}~{}~{}\eta>m\tau e^{-\tau/2}, (5.39)

with the boundary condition

r~(τ,mτeτ/2)=0,\tilde{r}_{\ell}(\tau,m\tau e^{\tau/2})=0, (5.40)

and the initial condition

r~(0,η)=r(0,η),\tilde{r}_{\ell}(0,\eta)=r_{\ell}(0,\eta), (5.41)

and with m>2m>2. Next, we shift, setting

p~(τ,η)=h(τ,ηmτeτ/2),\tilde{p}_{\ell}(\tau,\eta)=h_{\ell}(\tau,\eta-m\tau e^{-\tau/2}),

so that h(τ,η)h_{\ell}(\tau,\eta) satisfies

hτη2hη2hη2h+(32m)eτ/2hη+Ce(2m)τηeτ/2h=0,\dfrac{\partial{h_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{h_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{h_{\ell}}}{\partial{\eta}^{2}}-h_{\ell}+\Big{(}\frac{3}{2}-m\Big{)}e^{-\tau/2}\dfrac{\partial{h_{\ell}}}{\partial{\eta}}+Ce^{(2-m)\tau-\eta e^{\tau/2}}h_{\ell}=0, (5.42)

with the boundary condition

h(τ,0)=0,h_{\ell}(\tau,0)=0, (5.43)

and the initial condition

h(0,η)=r(0,η),η>0.h_{\ell}(0,\eta)=r_{\ell}(0,\eta),~{}~{}\eta>0. (5.44)

We multiply (5.42) by

Q~m(τ,η)=η+(32m)ψ¯(η)eτ/2,\tilde{Q}_{m}(\tau,\eta)=\eta+\Big{(}\frac{3}{2}-m\Big{)}\bar{\psi}(\eta)e^{-\tau/2}, (5.45)

and integrate, using Lemma 4.1:

ddτ\displaystyle\frac{d}{d\tau} 0Q~m(τ,η)h(τ,η)𝑑η=C0e(2m)τηeτ/2Qm(τ,η)h(τ,η)𝑑η\displaystyle\int_{0}^{\infty}\tilde{Q}_{m}(\tau,\eta)h_{\ell}(\tau,\eta)d\eta=-C\int_{0}^{\infty}e^{(2-m)\tau-\eta e^{\tau/2}}Q_{m}(\tau,\eta)h_{\ell}(\tau,\eta)d\eta (5.46)
+(32m)2eτ0h(τ,η)ψ¯(η)η𝑑ηCe(3/2m)τ0ηeτ/2eηeτ/2h(τ,η)𝑑η\displaystyle+\Big{(}\frac{3}{2}-m\Big{)}^{2}e^{-\tau}\int_{0}^{\infty}h_{\ell}(\tau,\eta)\dfrac{\partial{\bar{\psi}(\eta)}}{\partial{\eta}}d\eta\geq-Ce^{(3/2-m)\tau}\int_{0}^{\infty}\eta e^{\tau/2}e^{-\eta e^{\tau/2}}h_{\ell}(\tau,\eta)d\eta
Ce(3/2m)τ0h(τ,η)𝑑η.\displaystyle\geq-Ce^{(3/2-m)\tau}\int_{0}^{\infty}h_{\ell}(\tau,\eta)d\eta.

We used the fact that m>2m>2 in the second inequality above, and that ψ¯\bar{\psi} is non-negative and increasing.

Integrating (5.42) using positivity of h(τ,η)h(\tau,\eta) for η>0\eta>0 gives

ddτ0h(τ,η)𝑑η120h(τ,η)𝑑η+h(τ,0)η0.\frac{d}{d\tau}\int_{0}^{\infty}h_{\ell}(\tau,\eta)d\eta-\frac{1}{2}\int_{0}^{\infty}h_{\ell}(\tau,\eta)d\eta+\dfrac{\partial{h(\tau,0)}}{\partial{\eta}}\leq 0.

As h(τ,0)/η>0\partial h(\tau,0)/\partial\eta>0, this implies a trivial bound

0h(τ,η)𝑑ηeτ/20h(0,η)𝑑η=c¯eτ/2.\int_{0}^{\infty}h_{\ell}(\tau,\eta)d\eta\leq e^{\tau/2}\int_{0}^{\infty}h_{\ell}(0,\eta)d\eta=\bar{c}e^{\tau/2}. (5.47)

As m>2m>2 in (5.46), we then obtain

0Q~m(τ,η)h(τ,η)𝑑η0Q~m(0,η)h(0,η)𝑑ηC,\int_{0}^{\infty}\tilde{Q}_{m}(\tau,\eta)h_{\ell}(\tau,\eta)d\eta\geq\int_{0}^{\infty}\tilde{Q}_{m}(0,\eta)h_{\ell}(0,\eta)d\eta-C, (5.48)

so that

p¯()\displaystyle\bar{p}_{\infty}(\ell) =limτ+14π0ηp¯(τ,η)𝑑ηlim infτ+14π0ηh(τ,η)𝑑η\displaystyle=\lim_{\tau\to+\infty}\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta\bar{p}_{\ell}(\tau,\eta)d\eta\geq\liminf_{\tau\to+\infty}\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta h_{\ell}(\tau,\eta)d\eta (5.49)
=lim infτ+14π0Q~m(τ,η)h(τ,η)𝑑η14π0[η+(32m)ψ¯(η)]ψ0(η)𝑑ηC0\displaystyle=\liminf_{\tau\to+\infty}\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\tilde{Q}_{m}(\tau,\eta)h_{\ell}(\tau,\eta)d\eta\geq\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}[\eta+(\frac{3}{2}-m)\bar{\psi}(\eta)]\psi_{0}(\eta-\ell)d\eta-C_{0}
c¯KlogC1, for 2,\displaystyle\geq\bar{c}\ell-K\log\ell-C_{1},\hbox{ for $\ell\geq 2$},

with c¯\bar{c} as in (1.27), and an appropriate KK\in{\mathbb{R}}. This finishes the proof of the lower bound in Lemma 5.2, and also completes the proof of estimate (3.4) in Proposition 3.2\Box

6 The proof of Proposition 3.3

The proof of Proposition 3.3 is much more involved than that of Proposition 3.2. In this section, we outline the main steps and state the auxiliary results needed in the proof.

We start with (3.26):

rτη2rη2rη2r+32eτ/2rη+e3τ/2ηexp(τ/2)r2=0,\displaystyle\dfrac{\partial{r_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{r_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{r_{\ell}}}{\partial{\eta}^{2}}-r_{\ell}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{r_{\ell}}}{\partial{\eta}}+e^{3\tau/2-\eta\exp(\tau/2)}r_{\ell}^{2}=0, (6.1)

with the initial condition r(0,η)=ψ0(η)r_{\ell}(0,\eta)=\psi_{0}(\eta-\ell). Recall that we are interested in

r()=limτ+14π0ηr(τ,η)𝑑η=limτ+14π0Q(τ,η)r(τ,η)𝑑η.r_{\infty}(\ell)=\lim_{\tau\to+\infty}\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\eta r_{\ell}(\tau,\eta)d\eta=\lim_{\tau\to+\infty}\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}Q(\tau,\eta)r_{\ell}(\tau,\eta)d\eta. (6.2)

Here, Q(τ,η)=Q3/2(τ,η)Q(\tau,\eta)=Q_{3/2}(\tau,\eta) is the approximate solution to the adjoint linear problem, defined in (4.16) with k=3/2k=3/2. Multiplying (6.1) by Q(τ,η)Q(\tau,\eta) and integrating in η\eta gives, according to Lemma 4.1:

ddτ0Q(τ,η)r(τ,η)𝑑η=0e3τ/2ηexp(τ/2)r2(τ,η)Q(τ,η)𝑑η+E(τ),\frac{d}{d\tau}\int_{0}^{\infty}Q(\tau,\eta)r_{\ell}(\tau,\eta)d\eta=-\int_{0}^{\infty}e^{3\tau/2-\eta\exp(\tau/2)}r_{\ell}^{2}(\tau,\eta)Q(\tau,\eta)d\eta+E_{\ell}(\tau), (6.3)

so that

r()\displaystyle r_{\infty}(\ell) =Q+Y+,\displaystyle=Q_{\ell}+Y_{\ell}+{\cal E}_{\ell}, (6.4)

with

Q\displaystyle Q_{\ell} =14π0Q(0,η)r(0,η)𝑑η,\displaystyle=\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}Q(0,\eta)r_{\ell}(0,\eta)d\eta, (6.5)
Y\displaystyle Y_{\ell} =14π00e3τ/2ηexp(τ/2)r2(τ,η)Q(τ,η)𝑑τ𝑑η\displaystyle=-\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\int_{0}^{\infty}e^{3\tau/2-\eta\exp(\tau/2)}r_{\ell}^{2}(\tau,\eta)Q(\tau,\eta)d\tau d\eta
\displaystyle{\cal E}_{\ell} =14π0E(τ)𝑑τ.\displaystyle=\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}E_{\ell}(\tau)d\tau.

The error term

E(τ)=E(1)(τ)+E(2)(τ)E_{\ell}(\tau)=E_{\ell}^{(1)}(\tau)+E_{\ell}^{(2)}(\tau) (6.6)

has two contributions. The first

E(1)(τ)=r(τ,0)Q(τ,0)ηE_{\ell}^{(1)}(\tau)=r_{\ell}(\tau,0)\dfrac{\partial{Q(\tau,0)}}{\partial{\eta}} (6.7)

comes from the boundary at η=0\eta=0 since we do not have r(τ,0)=0r_{\ell}(\tau,0)=0 but only that r(τ,0)r_{\ell}(\tau,0) is small. The second comes from the error term in Lemma 4.1, because Q(τ,η)Q(\tau,\eta) is only an approximate solution to the adjoint problem and not an exact one:

E(2)(τ)=94eτ0ψ¯(η)ηr(τ,η)𝑑ηCeτ0r(τ,η)1+η𝑑η.E_{\ell}^{(2)}(\tau)=\frac{9}{4}e^{-\tau}\int_{0}^{\infty}\dfrac{\partial{\bar{\psi}(\eta)}}{\partial{\eta}}r_{\ell}(\tau,\eta)d\eta\leq Ce^{-\tau}\int_{0}^{\infty}\frac{r_{\ell}(\tau,\eta)}{1+\eta}d\eta. (6.8)

The linear term and the error term in (6.4) are quite straightforward to evaluate and estimate, respectively. The main difficulty will be in finding the precise asymptotics of the nonlinear term in the right side of (6.4).

The error term bound

The error term in (6.4) is bounded by the following lemma.

Lemma 6.1

There exists C>0C>0 so that

()C1/8.{\cal E}_{\infty}(\ell)\leq\frac{C}{\ell^{1/8}}. (6.9)

Proof. As we have seen in the proof of Lemma 5.1 – see (5.19) and (5.20), we have an upper bound

r(τ,0)r¯(τ,0)C1/4eτ/4r_{\ell}(\tau,0)\leq\bar{r}_{\ell}(\tau,0)\leq\frac{C}{\ell^{1/4}}e^{-\tau/4} (6.10)

hence E(1)E_{\ell}^{(1)} can be bounded as

0E(1)(τ)𝑑τC1/4.\int_{0}^{\infty}E_{\ell}^{(1)}(\tau)d\tau\leq\frac{C}{\ell^{1/4}}. (6.11)

We now estimate E(2)E_{\ell}^{(2)}. As in the proof of the upper bound in Lemma 5.1, see (5.23)-(5.1), we deduce from (6.10) that

E(2)(τ)C1/4eτ+Ceτ0p¯(τ,η)1+η𝑑η,E_{\ell}^{(2)}(\tau)\leq\frac{C}{\ell^{1/4}}e^{-\tau}+Ce^{-\tau}\int_{0}^{\infty}\frac{\bar{p}_{\ell}(\tau,\eta)}{1+\eta}d\eta, (6.12)

where p¯(τ,η)\bar{p}_{\ell}(\tau,\eta) is the solution to (5.1):

p¯τ+32eτ/2p¯η=2p¯η2+η2p¯η+p¯,η>0,\displaystyle\dfrac{\partial{\bar{p}_{\ell}}}{\partial{\tau}}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{\bar{p}_{\ell}}}{\partial{\eta}}=\dfrac{\partial^{2}{\bar{p}_{\ell}}}{\partial{\eta}^{2}}+\frac{\eta}{2}\dfrac{\partial{\bar{p}_{\ell}}}{\partial{\eta}}+\bar{p}_{\ell},~{}~{}\eta>0, (6.13)
p¯(τ,0)=0,\displaystyle\bar{p}_{\ell}(\tau,0)=0,
p¯(0,η)=ψ0(η)=eηϕ0(η).\displaystyle\bar{p}_{\ell}(0,\eta)=\psi_{0}(\eta-\ell)=e^{\eta-\ell}\phi_{0}(\eta-\ell).

The simple-minded bound (5.31) implies that if we fix any T>0T>0, then

eτ0p¯(τ,η)1+η𝑑ηCeτ/2Ceτ/4eT/4 for all τ>T.e^{-\tau}\int_{0}^{\infty}\frac{\bar{p}_{\ell}(\tau,\eta)}{1+\eta}d\eta\leq Ce^{-\tau/2}\leq Ce^{-\tau/4}e^{-T/4}\hbox{ for all $\tau>T$.} (6.14)

For short times τ<T\tau<T we can take L>0L>0 and use (5.31) to write

eτ0p¯(τ,η)1+η𝑑η\displaystyle e^{-\tau}\int_{0}^{\infty}\frac{\bar{p}_{\ell}(\tau,\eta)}{1+\eta}d\eta Ceτ(supη[0,L]p¯(τ,η))logL+CLeτ/2\displaystyle\leq Ce^{-\tau}\Big{(}\sup_{\eta\in[0,L]}\bar{p}_{\ell}(\tau,\eta)\Big{)}\log L+\frac{C}{L}e^{-\tau/2} (6.15)
Ceτ(supη[0,L]R¯(τ,η))logL+CLeτ/2.\displaystyle\leq Ce^{-\tau}\Big{(}\sup_{\eta\in[0,L]}\bar{R}_{\ell}(\tau,\eta)\Big{)}\log L+\frac{C}{L}e^{-\tau/2}.

Here, R¯(τ,η)\bar{R}_{\ell}(\tau,\eta) is the solution to the whole line problem (5.1):

R¯τη2R¯η2R¯η2R¯+32eτ/2R¯η=0,η,\displaystyle\dfrac{\partial{\bar{R}_{\ell}}}{\partial{\tau}}-\frac{\eta}{2}\dfrac{\partial{\bar{R}_{\ell}}}{\partial{\eta}}-\dfrac{\partial^{2}{\bar{R}_{\ell}}}{\partial{\eta}^{2}}-\bar{R}_{\ell}+\frac{3}{2}e^{-\tau/2}\dfrac{\partial{\bar{R}_{\ell}}}{\partial{\eta}}=0,~{}~{}\eta\in{\mathbb{R}},
R¯(0,η)=r(0,η),\displaystyle\bar{R}_{\ell}(0,\eta)=r_{\ell}(0,\eta), (6.16)

given explicitly by (5.7):

R¯(τ,η)\displaystyle\bar{R}_{\ell}(\tau,\eta) =eτG(eτ1,ηeτ/232τy)r(0,y)𝑑y.\displaystyle=e^{\tau}\int G(e^{\tau}-1,\eta e^{\tau/2}-\frac{3}{2}\tau-y)r_{\ell}(0,y)dy. (6.17)

In the proof of Lemma 5.1 we only looked for the bound on R¯(τ,η)\bar{R}_{\ell}(\tau,\eta) at η=0\eta=0 but now we need to consider η[0,L]\eta\in[0,L]. Note that for TT and LL not too large, the function R¯(τ,η)\bar{R}(\tau,\eta) is increasing in η\eta for η[0,L]\eta\in[0,L] and 0<τ<T0<\tau<T. Hence, we have

R¯(τ,η)\displaystyle\bar{R}_{\ell}(\tau,\eta) R¯(τ,L)=eτG(eτ1,Leτ/232τy)r(0,y)𝑑y, for η[0,L] and 0<τ<T.\displaystyle\leq\bar{R}_{\ell}(\tau,L)=e^{\tau}\int G(e^{\tau}-1,Le^{\tau/2}-\frac{3}{2}\tau-y)r_{\ell}(0,y)dy,~{}~{}\hbox{ for~{}$\eta\in[0,L]$ and $0<\tau<T$.} (6.18)

As in (5.9) and (5.11), we get, as ψ0(y)=0\psi_{0}(y)=0 for yL0y\geq L_{0}:

R¯(τ,L)\displaystyle\bar{R}_{\ell}(\tau,L) =eτG(eτ1,y+32τLeτ/2)r(0,y)𝑑y=eτG(eτ1,y++32τLeτ/2)ψ0(y)𝑑y\displaystyle=e^{\tau}\int G(e^{\tau}-1,y+\frac{3}{2}\tau-Le^{\tau/2})r_{\ell}(0,y)dy=e^{\tau}\int G(e^{\tau}-1,y+\ell+\frac{3}{2}\tau-Le^{\tau/2})\psi_{0}(y)dy (6.19)
CeτL0G(eτ1,y++32τLeτ/2)ey𝑑y\displaystyle\leq Ce^{\tau}\int_{-\infty}^{L_{0}}G(e^{\tau}-1,y+\ell+\frac{3}{2}\tau-Le^{\tau/2})e^{y}dy
=Ceτe(3/2)τ+Leτ/2L0++(3τ/2)Leτ/2G(eτ1,y)ey𝑑y\displaystyle=Ce^{\tau}e^{-\ell-(3/2)\tau+Le^{\tau/2}}\int_{-\infty}^{L_{0}+\ell+(3\tau/2)-Le^{\tau/2}}G(e^{\tau}-1,y)e^{y}dy
Ceτ/2+Leτ/2eeτ1CeT/2+LeT/2eeT1.\displaystyle\leq Ce^{-\ell-\tau/2+Le^{\tau/2}}e^{e^{\tau}-1}\leq Ce^{-\ell-T/2+Le^{T/2}}e^{e^{T}-1}.

Let us take T=(1/2)logT=(1/2)\log\ell and L=1/2L=\ell^{1/2}, which gives

R¯(τ,η)\displaystyle\bar{R}_{\ell}(\tau,\eta) Celog/4+3/4+1/21Ce/2.\displaystyle\leq Ce^{-\ell-\log\ell/4+\ell^{3/4}+\ell^{1/2}-1}\leq Ce^{-\ell/2}. (6.20)

Using this in (6.15), we obtain for τ<T\tau<T:

eτ0p¯(τ,η)1+η𝑑ηCeτe/2log+C1/2eτ/2 for all τ<T,\displaystyle e^{-\tau}\int_{0}^{\infty}\frac{\bar{p}_{\ell}(\tau,\eta)}{1+\eta}d\eta\leq Ce^{-\tau}e^{-\ell/2}\log\ell+\frac{C}{\ell^{1/2}}e^{-\tau/2}\hbox{ for all $\tau<T$}, (6.21)

while (6.14) becomes

eτ0p¯(τ,η)1+η𝑑ηCeτ/41/8 for all τ>T.e^{-\tau}\int_{0}^{\infty}\frac{\bar{p}_{\ell}(\tau,\eta)}{1+\eta}d\eta\leq Ce^{-\tau/4}\ell^{-1/8}\hbox{ for all $\tau>T$.} (6.22)

It follows that

0E(2)(τ)𝑑τC1/8,\int_{0}^{\infty}E_{\ell}^{(2)}(\tau)d\tau\leq\frac{C}{\ell^{1/8}}, (6.23)

and the conclusion of Lemma 6.1 follows. \Box

The linear contribution

The term QQ_{\ell} in (6.5) can be computed explicitly from the asymptotics for Q(0,η)Q(0,\eta) as η\eta\to\infty coming from (4.25):

Q(0,η)=η+3logη+32g+O(1η) as η+,Q(0,\eta)=\eta+3\log\eta+\frac{3}{2}g_{\infty}+O\Big{(}\frac{1}{\eta}\Big{)}\hbox{ as $\eta\to+\infty$}, (6.24)

with gg_{\infty} defined in (4.26). This gives

Q=14π0(η+3logη+32g)ψ0(η)𝑑η+O(1)=c¯+3c¯log+32gc¯+c¯1+O(1),\displaystyle Q_{\ell}=\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}(\eta+3\log\eta+\frac{3}{2}g_{\infty})\psi_{0}(\eta-\ell)d\eta+O\Big{(}\frac{1}{\ell}\Big{)}=\bar{c}\ell+3\bar{c}\log\ell+\frac{3}{2}g_{\infty}\bar{c}+\bar{c}_{1}+O(\ell^{-1}), (6.25)

with c¯\bar{c} and c¯1\bar{c}_{1} defined in (1.27) and (1.34), respectively. In particular, this is how the term c¯1\bar{c}_{1} appears in Theorem 1.4.

The nonlinear contribution

It remains to find the asymptotics of the term YY_{\ell} in (6.5), and that computation is quite long. The first step is a series of simplifications. We start by approximating Q(τ,η)Q(\tau,\eta) in (6.5) by η\eta: set

Y¯=14π00e3τ/2ηexp(τ/2)r2(τ,η)η𝑑τ𝑑η.\displaystyle\bar{Y}_{\ell}=-\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\int_{0}^{\infty}e^{3\tau/2-\eta\exp(\tau/2)}r_{\ell}^{2}(\tau,\eta)\eta d\tau d\eta. (6.26)
Lemma 6.2

There exists γ>0\gamma>0 so that

YY¯=O(γ), as +.Y_{\ell}-\bar{Y}_{\ell}=O(\ell^{-\gamma}),\hbox{ as $\ell\to+\infty$}. (6.27)

This lemma is proved in Section 7.5. The next approximation involves going back to the original space-time variables and restricting the spatial integration to ”relatively short” distances x(t+1)δx\leq(t+1)^{\delta} with δ>0\delta>0 small.

Lemma 6.3

There exists δ>0\delta>0 and γ>0\gamma>0 so that

|Y¯Y¯(1)|=O(γ),as +,|\bar{Y}_{\ell}-\bar{Y}_{\ell}^{(1)}|=O(\ell^{-\gamma}),~{}~{}\hbox{as $\ell\to+\infty$,} (6.28)

where

Y¯(1)\displaystyle\bar{Y}_{\ell}^{(1)} =14π00(t+1)δxexz2(t,x)dxdt(1+t)3/2,\displaystyle=-\frac{1}{\sqrt{4\pi}}\int_{0}^{\infty}\int_{0}^{(t+1)^{\delta}}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{{(1+t)^{3/2}}}, (6.29)

and z(t,x)z_{\ell}(t,x) is the solution to

zt2zx232(t+1)(zzx)+exz2=0,\dfrac{\partial{z_{\ell}}}{\partial{t}}-\dfrac{\partial^{2}{z_{\ell}}}{\partial{x}^{2}}-\frac{3}{2(t+1)}\Big{(}z_{\ell}-\dfrac{\partial{z_{\ell}}}{\partial{x}}\Big{)}+e^{-x}z_{\ell}^{2}=0, (6.30)

with z(0,x)=ψ0(x)z_{\ell}(0,x)=\psi_{0}(x-\ell).

Note that (6.30) is simply (3.16) with a slightly different notation. The functions z(t,x)z_{\ell}(t,x) and r(τ,η)r_{\ell}(\tau,\eta) are related by

z(t,x)=t+1r(log(t+1),x/t+1).z_{\ell}(t,x)=\sqrt{t+1}r_{\ell}(\log(t+1),x/\sqrt{t+1}). (6.31)

This lemma is proved in Section 7.1 below. It is easy to see why (6.28) should be true: we expect that the function z(t,x)z_{\ell}(t,x) behaves roughly as

z(t,x)r()xc¯x, for x1 and t1.z_{\ell}(t,x)\sim r_{\infty}(\ell)x\sim\bar{c}\ell x,\hbox{ for $x\gg 1$ and $t\gg 1$.} (6.32)

With this input, a back of the envelope computation shows that the integral in (6.29) over x(t+1)δx\geq(t+1)^{\delta} is, indeed, small, due to the exponential decay factor.

The next approximation is to discard the ”short times” from the time integration, looking only at times t2αt\geq\ell^{2-\alpha} with some α>0\alpha>0 sufficiently small. This is, again, quite expected: as the integration is now over the region x(t+1)δx\leq(t+1)^{\delta}, and z(t,x)z_{\ell}(t,x) is initially supported at x=x=\ell, it takes the time of roughly t2t\sim\ell^{2} to populate the domain of integration, so shorter times can be discarded.

Lemma 6.4

Let α>0\alpha>0 sufficiently small. Then, there exists γ>0\gamma>0 so that

|Y¯(2)Y¯(1)|=O(γ),as +,|\bar{Y}_{\ell}^{(2)}-\bar{Y}_{\ell}^{(1)}|=O(\ell^{-\gamma}),~{}~{}\hbox{as $\ell\to+\infty$,} (6.33)

where

Y¯(2)\displaystyle\bar{Y}_{\ell}^{(2)} =14π2α0(t+1)δxexz2(t,x)dxdt(1+t)3/2,\displaystyle=-\frac{1}{\sqrt{4\pi}}\int_{\ell^{2-\alpha}}^{\infty}\int_{0}^{(t+1)^{\delta}}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{{(1+t)^{3/2}}}, (6.34)

This lemma is proved in Section 7.2 below.

Lemmas 6.3 and 6.4 allow us to focus on the integration in the region 0<x<(t+1)δ0<x<(t+1)^{\delta} and times t>2αt>\ell^{2-\alpha}, and consider z(t,x)z_{\ell}(t,x) as the solution to (6.30) for 0<x<(t+1)δ0<x<(t+1)^{\delta}, with the initial condition z(0,x)=0z_{\ell}(0,x)=0 and the boundary condition at x=(t+1)δx=(t+1)^{\delta} coming from the ”outer” solution in the self-similar variables:

z(t,(1+t)δ)=(t+1)1/2r(log(t+1),(t+1)1/2+δ).z_{\ell}(t,(1+t)^{\delta})=(t+1)^{1/2}r_{\ell}(\log(t+1),(t+1)^{-1/2+\delta}). (6.35)

The analysis of the behavior of r(τ,η)r_{\ell}(\tau,\eta) for ηO(1)\eta\sim O(1) and τ+\tau\to+\infty that we have done so far, would only give information for xtx\sim\sqrt{t}, not xtδx\sim t^{\delta} with δ<1/2\delta<1/2, as this point corresponds to

ηt(1/2δ)=exp((1/2δ)τ)1.\eta\sim t^{-(1/2-\delta)}=\exp(-(1/2-\delta)\tau)\ll 1.

To bridge this gap, we need the following crucial lemma that is proved in Section 7.6. It refines the informal asymptotics in (6.32) by an extremely important Gaussian factor that interpolates the short time and the long time behavior.

Lemma 6.5

There exists α>0\alpha>0 sufficiently small and a constant K>0K>0 so that for all t>2αt>\ell^{2-\alpha} we have

z(t,(t+1)δ)(t+1)δ(c¯+K1δ)e2/4(t+1),\frac{z_{\ell}(t,(t+1)^{\delta})}{(t+1)^{\delta}}\leq(\bar{c}\ell+K\ell^{1-\delta})e^{-\ell^{2}/4(t+1)}, (6.36)

and for all 2α<t<2+α\ell^{2-\alpha}<t<\ell^{2+\alpha} we have

z(t,(t+1)δ)(t+1)δ(c¯K1δ)e2/4(t+1).\frac{z_{\ell}(t,(t+1)^{\delta})}{(t+1)^{\delta}}\geq(\bar{c}\ell-K\ell^{1-\delta})e^{-\ell^{2}/4(t+1)}. (6.37)

Lemma 6.5, proved in Section 7.6 allows us to look at upper and lower solutions z±z_{\pm} for zz_{\ell} in the region x(1+t)δx\leq(1+t)^{\delta} as the solutions to

z±t2z±x232(t+1)(z±z±x)+exz±2=0,x<(t+1)δ,\dfrac{\partial{z_{\pm}}}{\partial{t}}-\dfrac{\partial^{2}{z_{\pm}}}{\partial{x}^{2}}-\frac{3}{2(t+1)}\Big{(}z_{\pm}-\dfrac{\partial{z_{\pm}}}{\partial{x}}\Big{)}+e^{-x}z_{\pm}^{2}=0,~{}~{}x<(t+1)^{\delta}, (6.38)

with the boundary condition

z±(t,(1+t)δ)=(c¯±K1δ)e2/(4(t+1))(1+t)δ.z_{\pm}(t,(1+t)^{\delta})=\big{(}\bar{c}\ell\pm K\ell^{1-\delta}\big{)}e^{-\ell^{2}/(4(t+1))}(1+t)^{\delta}. (6.39)

The initial condition for z(t,x)z_{-}(t,x) at t=2αt=\ell^{2-\alpha} is z(2α,x)=0z_{-}(\ell^{2-\alpha},x)=0 for all x<(1+)δx<(1+\ell)^{\delta}, and for z+(t,x)z_{+}(t,x) it is z+(2α,x)=Cecαz_{+}(\ell^{2-\alpha},x)=Ce^{-c\ell^{\alpha^{\prime}}} for all x<(1+)δx<(1+\ell)^{\delta}, with some α>0\alpha^{\prime}>0, which comes from (7.14) below. We have then the inequality

Y¯+Y¯(2)Y¯,\bar{Y}_{\ell}^{+}\leq\bar{Y}_{\ell}^{(2)}\leq\bar{Y}_{\ell}^{-}, (6.40)

with

Y¯±=14π2α0(t+1)δxexz±2(t,x)dxdt(1+t)3/2.\bar{Y}_{\ell}^{\pm}=-\frac{1}{\sqrt{4\pi}}\int_{\ell^{2-\alpha}}^{\infty}\int_{0}^{(t+1)^{\delta}}xe^{-x}z_{\pm}^{2}(t,x)\frac{dxdt}{(1+t)^{3/2}}. (6.41)

The last step in the proof of Proposition 3.3. is to prove the following.

Lemma 6.6

There exists δ>0\delta>0 so that

Y¯±=c¯logc¯logc¯+k0c¯+c¯2+O(δ) as +,\bar{Y}_{\ell}^{\pm}=-\bar{c}\log\ell-\bar{c}\log\bar{c}+k_{0}\bar{c}+\frac{\bar{c}}{2}+O(\ell^{-\delta})\hbox{ as $\ell\to+\infty$,} (6.42)

with the constant k0k_{0} as in (1.6).

The proof of Lemma 6.6 is presented in Sections 7.3 and 7.4.

Now, putting together (6.25) and (6.42) gives

r()=c¯+2c¯log+32gc¯+c¯1c¯logc¯+k0c¯+c¯2+O(δ),r_{\infty}(\ell)=\bar{c}\ell+2\bar{c}\log\ell+\frac{3}{2}g_{\infty}\bar{c}+\bar{c}_{1}-\bar{c}\log\bar{c}+k_{0}\bar{c}+\frac{\bar{c}}{2}+O(\ell^{-\delta}), (6.43)

with some δ>0\delta>0, finishing the proof of Proposition 3.3.

7 Proofs of auxiliary lemmas

This section contains the proofs of the auxiliary lemmas used in Section 6 to prove Proposition 3.3, except for Lemma 6.5 that is proved in Section 7.6.

7.1 The proof of Lemma 6.3

First, we go back to the original, non-self-similar variables: write

r(τ,η)=w(eτ1,ηeτ/2),r_{\ell}(\tau,\eta)=w_{\ell}(e^{\tau}-1,\eta e^{\tau/2}),

so that the function w(t,x)w_{\ell}(t,x) satisfies

wt2wx2wt+1+32(t+1)wx+t+1exw2=0,\displaystyle\dfrac{\partial{w_{\ell}}}{\partial{t}}-\dfrac{\partial^{2}{w_{\ell}}}{\partial{x}^{2}}-\frac{w_{\ell}}{t+1}+\frac{3}{2(t+1)}\dfrac{\partial{w_{\ell}}}{\partial{x}}+\sqrt{t+1}e^{-x}w_{\ell}^{2}=0, (7.1)

with the initial condition w(0,x)=ψ0(x)w_{\ell}(0,x)=\psi_{0}(x-\ell). To revert back even more, we write w=z/t+1w_{\ell}=z_{\ell}/\sqrt{t+1}, with the function zz_{\ell} that satisfies (6.30) with z(0,x)=ψ0(x)z_{\ell}(0,x)=\psi_{0}(x-\ell). Then, we can re-write Y¯\bar{Y}_{\ell} as

Y¯\displaystyle\bar{Y}_{\ell} =00(t+1)3/2exw2(t,x)xt+1dtt+1dxt+1=00xexw2(t,x)dxdtt+1\displaystyle=-\int_{0}^{\infty}\int_{0}^{\infty}(t+1)^{3/2}e^{-x}w_{\ell}^{2}(t,x)\frac{x}{\sqrt{t+1}}\frac{dt}{t+1}\frac{dx}{\sqrt{t+1}}=-\int_{0}^{\infty}\int_{0}^{\infty}xe^{-x}w_{\ell}^{2}(t,x)\frac{dxdt}{\sqrt{t+1}} (7.2)
=00xexz2(t,x)dxdt(t+1)3/2.\displaystyle=-\int_{0}^{\infty}\int_{0}^{\infty}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{(t+1)^{3/2}}.

Next, we split Y¯\bar{Y}_{\ell} into two terms, corresponding to small and large xx:

Y¯\displaystyle\bar{Y}_{\ell} =00(t+1)δxexz2(t,x)dxdt(t+1)3/20(t+1)δxexz2(t,x)dxdt(t+1)3/2=Y¯(1)+Y¯(c),\displaystyle=-\int_{0}^{\infty}\int_{0}^{(t+1)^{\delta}}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{(t+1)^{3/2}}-\int_{0}^{\infty}\int_{(t+1)^{\delta}}^{\infty}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{(t+1)^{3/2}}=\bar{Y}_{\ell}^{(1)}+\bar{Y}_{\ell}^{(c)}, (7.3)

with δ(0,1/2)\delta\in(0,1/2) to be chosen, and split the second term again:

Y¯(c)\displaystyle\bar{Y}_{\ell}^{(c)} =0T(t+1)δxexz2(t,x)dxdt(t+1)3/2T(t+1)δxexz2(t,x)dxdt(t+1)3/2=Y¯(c1)+Y¯(c2),\displaystyle=-\int_{0}^{T}\int_{(t+1)^{\delta}}^{\infty}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{(t+1)^{3/2}}-\int_{T}^{\infty}\int_{(t+1)^{\delta}}^{\infty}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{(t+1)^{3/2}}=\bar{Y}_{\ell}^{(c1)}+\bar{Y}_{\ell}^{(c2)}, (7.4)

with a large TT to chosen. We estimate the term Y¯(c2)\bar{Y}_{\ell}^{(c2)} using the simple upper bound

z(t,x)C(t+1)3/2,z_{\ell}(t,x)\leq C(t+1)^{3/2}, (7.5)

so that

|Y¯(c2)|\displaystyle|\bar{Y}_{\ell}^{(c2)}| =T(t+1)δxexz2(t,x)dxdt(t+1)3/2CT(t+1)5/2e(t+1)δ𝑑tC(1+T)3exp(Tδ).\displaystyle=\int_{T}^{\infty}\int_{(t+1)^{\delta}}^{\infty}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{(t+1)^{3/2}}\leq C\int_{T}^{\infty}(t+1)^{5/2}e^{-(t+1)^{\delta}}dt\leq C(1+T)^{3}\exp(-T^{\delta}). (7.6)

To estimate Y¯(c1)\bar{Y}_{\ell}^{(c1)} we slightly refine (7.5) to

z(t,x)C(t+1)3/2z¯(t,x32log(t+1)).z_{\ell}(t,x)\leq C(t+1)^{3/2}\bar{z}(t,x-\frac{3}{2}\log(t+1)). (7.7)

Here, z¯(t,x)\bar{z}(t,x) is the solution to the heat equation on the whole line

z¯t=2z¯x2,z¯(0,x)=ψ0(x).\dfrac{\partial{\bar{z}}}{\partial{t}}=\dfrac{\partial^{2}{\bar{z}}}{\partial{x}^{2}},~{}~{}\bar{z}(0,x)=\psi_{0}(x-\ell). (7.8)

It follows that

z(t,x)C(t+1)ec, for 0t and 0<x</2,z_{\ell}(t,x)\leq C(t+1)e^{-c\ell},\hbox{ for $0\leq t\leq\ell$ and $0<x<\ell/2$}, (7.9)

so that for 0t0\leq t\leq\ell we have

(t+1)δxexz2(t,x)𝑑x0/2xexz2(t,x)𝑑x+/2xexz2(t,x)𝑑xC(t+1)3ec.\int_{(t+1)^{\delta}}^{\infty}xe^{-x}z_{\ell}^{2}(t,x)dx\leq\int_{0}^{\ell/2}xe^{-x}z_{\ell}^{2}(t,x)dx+\int_{\ell/2}^{\infty}xe^{-x}z_{\ell}^{2}(t,x)dx\leq C(t+1)^{3}e^{-c\ell}. (7.10)

Thus, if we take T=T=\ell, then

|Y¯(c1)|\displaystyle|\bar{Y}_{\ell}^{(c1)}| =0T(t+1)δxexz2(t,x)dxdt(t+1)3/2C0T(t+1)3eC𝑑tC(+1)4exp(c).\displaystyle=\int_{0}^{T}\int_{(t+1)^{\delta}}^{\infty}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{(t+1)^{3/2}}\leq C\int_{0}^{T}(t+1)^{3}e^{-C\ell}dt\leq C(\ell+1)^{4}\exp(-c\ell). (7.11)

Using T=T=\ell also in (7.6), we obtain

|Y¯(c)|\displaystyle|\bar{Y}_{\ell}^{(c)}| C(+1)4exp(cδ),\displaystyle\leq C(\ell+1)^{4}\exp(-c\ell^{\delta}), (7.12)

finishing the proof of Lemma 6.3.

7.2 The proof of Lemma 6.4

The difference

Y¯(1)Y¯(2)=14π02α0(t+1)δxexz2(t,x)dxdt(1+t)3/2\bar{Y}_{\ell}^{(1)}-\bar{Y}_{\ell}^{(2)}=-\frac{1}{\sqrt{4\pi}}\int_{0}^{\ell^{2-\alpha}}\int_{0}^{(t+1)^{\delta}}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{{(1+t)^{3/2}}} (7.13)

can be estimated using (7.7) and a variant of (7.9):

z(t,x)C(t+1)3/2ecα, for t2α and x(t+1)δ(2α)δz_{\ell}(t,x)\leq C(t+1)^{3/2}e^{-c\ell^{\alpha}},~{}\hbox{ for $t\leq\ell^{2-\alpha}$ and $x\leq(t+1)^{\delta}\leq\ell^{(2-\alpha)\delta}$. } (7.14)

This gives

|Y¯(1)Y¯(2)|\displaystyle|\bar{Y}_{\ell}^{(1)}-\bar{Y}_{\ell}^{(2)}| C02α0(t+1)δxex(1+t)3/2ecα𝑑x𝑑tC5ecα,\displaystyle\leq C\int_{0}^{\ell^{2-\alpha}}\int_{0}^{(t+1)^{\delta}}xe^{-x}{{(1+t)^{3/2}}}e^{-c\ell^{\alpha}}{dxdt}\leq C\ell^{5}e^{-c\ell^{\alpha}}, (7.15)

and (6.33) follows.

7.3 The proof of Lemma 6.6: an informal argument

Let us explain informally why Lemma 6.6 is true, before giving the proof in Section 7.4 below. We expect that the functions z±(t,x)z_{\pm}(t,x), solutions to (6.38)-(6.39), converge as t+t\to+\infty to z¯±(x)\bar{z}_{\pm}(x), the solutions to

z¯±′′+exz¯±2=0,-\bar{z}_{\pm}^{\prime\prime}+e^{-x}\bar{z}_{\pm}^{2}=0, (7.16)

with the boundary condition

z¯x(+)=r±:=c¯±K1δ,\bar{z}_{x}(+\infty)=r_{\pm}:=\bar{c}\ell\pm K\ell^{1-\delta}, (7.17)

with the constant KK as in Lemma 6.5. Note that z¯±(x)\bar{z}_{\pm}(x) has an explicit form

z¯±(x)=r±z¯0(xlogr±),\bar{z}_{\pm}(x)=r_{\pm}\bar{z}_{0}(x-\log r_{\pm}), (7.18)

with z¯0\bar{z}_{0} that solves

z¯0′′+exz¯02=0,-\bar{z}_{0}^{\prime\prime}+e^{-x}\bar{z}_{0}^{2}=0, (7.19)

with the slope at infinity

z¯0,x(+)=1.\bar{z}_{0,x}(+\infty)=1. (7.20)

More precisely, we expect that z(t,x)z(t,x) is well approximated by taking

r±(t)=r±e2/(4t),r_{\pm}(t)=r_{\pm}e^{-\ell^{2}/(4t)}, (7.21)

leading to the asymptotics

z±(t,x)r±(t)z¯0(xlogr±(t))=z¯±(x+24t)e2/(4t),z_{\pm}(t,x)\approx r_{\pm}(t)\bar{z}_{0}(x-\log r_{\pm}(t))=\bar{z}_{\pm}\Big{(}x+\frac{\ell^{2}}{4t}\Big{)}e^{-\ell^{2}/(4t)}, (7.22)

that holds for all t>0t>0, in the sense that

Y¯±\displaystyle\bar{Y}_{\ell}^{\pm} =14π2α0(t+1)δxexz¯±2(x+24t)e2/(2t)dxdt(t+1)3/2+O(γ)\displaystyle=-\frac{1}{\sqrt{4\pi}}\int_{\ell^{2-\alpha}}^{\infty}\int_{0}^{(t+1)^{\delta}}xe^{-x}\bar{z}_{\pm}^{2}\Big{(}x+\frac{\ell^{2}}{4t}\Big{)}e^{-\ell^{2}/(2t)}\frac{dxdt}{(t+1)^{3/2}}+O(\ell^{-\gamma}) (7.23)
=14πα0xexz¯±2(x+14t)e1/(2t)dxdtt3/2+O(γ)\displaystyle=-\frac{1}{\sqrt{4\pi}\ell}\int_{\ell^{-\alpha}}^{\infty}\int_{0}^{\infty}xe^{-x}\bar{z}_{\pm}^{2}\Big{(}x+\frac{1}{4t}\Big{)}e^{-1/(2t)}\frac{dxdt}{t^{3/2}}+O(\ell^{-\gamma})
=14π00xexz¯±2(x+s4)es/2dxdss+O(γ)\displaystyle=-\frac{1}{\sqrt{4\pi}\ell}\int_{0}^{\infty}\int_{0}^{\infty}xe^{-x}\bar{z}_{\pm}^{2}\Big{(}x+\frac{s}{4}\Big{)}e^{-s/2}\frac{dxds}{\sqrt{s}}+O(\ell^{-\gamma})
=14π0s/4(xs4)exz¯±2(x)es/4dxdss+O(γ)\displaystyle=-\frac{1}{\sqrt{4\pi}\ell}\int_{0}^{\infty}\int_{s/4}^{\infty}\Big{(}x-\frac{s}{4}\Big{)}e^{-x}\bar{z}_{\pm}^{2}(x)e^{-s/4}\frac{dxds}{\sqrt{s}}+O(\ell^{-\gamma})
=r±2π0s(xs)exz¯02(xlogr±)esdxdss+O(γ)\displaystyle=-\frac{r_{\pm}^{2}}{\sqrt{\pi}\ell}\int_{0}^{\infty}\int_{s}^{\infty}(x-s)e^{-x}\bar{z}_{0}^{2}(x-\log r_{\pm})e^{-s}\frac{dxds}{\sqrt{s}}+O(\ell^{-\gamma})
=r±π0slogr±(x+logr±s)exz¯02(x)esdxdss+O(γ)\displaystyle=-\frac{r_{\pm}}{\sqrt{\pi}\ell}\int_{0}^{\infty}\int_{s-\log r_{\pm}}^{\infty}(x+\log r_{\pm}-s)e^{-x}\bar{z}_{0}^{2}(x)e^{-s}\frac{dxds}{\sqrt{s}}+O(\ell^{-\gamma})
=r±π0G(logr±s)esdss+O(γ) as +,\displaystyle=-\frac{r_{\pm}}{\sqrt{\pi}\ell}\int_{0}^{\infty}G(\log r_{\pm}-s)e^{-s}\frac{ds}{\sqrt{s}}+O(\ell^{-\gamma})~{}~{}~{}\hbox{ as $\ell\to+\infty$},

with some γ>0\gamma>0, and

G(q)=q(x+q)exz¯02(x)𝑑x.G(q)=\int_{-q}^{\infty}(x+q)e^{-x}\bar{z}_{0}^{2}(x)dx. (7.24)

Note that

|G(q)|C(1+|q|).|G(q)|\leq C(1+|q|). (7.25)

It follows that

|logr±/2G(logr±s)esdss|Clogr±/2|logr±s|esdssCr±γCγ,\Big{|}\int_{\log r_{\pm}/2}^{\infty}G(\log r_{\pm}-s)e^{-s}\frac{ds}{\sqrt{s}}\Big{|}\leq C\int_{\log r_{\pm}/2}^{\infty}|\log r_{\pm}-s|e^{-s}\frac{ds}{\sqrt{s}}\leq\frac{C}{r_{\pm}^{\gamma}}\leq\frac{C}{\ell^{\gamma}}, (7.26)

with γ>0\gamma>0 sufficiently small. Therefore, we have

Y¯±\displaystyle\bar{Y}_{\ell}^{\pm} =r±π0logr±/2G(logr±s)esdss+O(γ) as +.\displaystyle=-\frac{r_{\pm}}{\sqrt{\pi}\ell}\int_{0}^{\log r_{\pm}/2}G(\log r_{\pm}-s)e^{-s}\frac{ds}{\sqrt{s}}+O(\ell^{-\gamma})~{}~{}~{}\hbox{ as $\ell\to+\infty$}. (7.27)

To analyze the asymptotic behavior of G(r)G(r) for r1r\gg 1, note that the solution to (7.19) with the normalization (7.20) is given by

z¯(x)=exU(x),\bar{z}(x)=e^{x}U(x), (7.28)

where U(x)U(x) is the Fisher-KPP minimal speed wave, solution to (1.5), with the normalization (1.6). Hence, the function z¯0(x)\bar{z}_{0}(x) has the asymptotics

z¯0(x)=x+k0+O(eωx), as x+,\bar{z}_{0}(x)=x+k_{0}+O(e^{-\omega x}),~{}~{}\hbox{ as $x\to+\infty$}, (7.29)

with k0k_{0} as in (1.6), and some ω>0\omega>0, and

z¯0(x)ex as x.\bar{z}_{0}(x)\sim e^{x}\hbox{ as $x\to-\infty$}. (7.30)

In addition, we have from (7.19)-(7.20) that

exz¯02(x)𝑑x=1,\int_{-\infty}^{\infty}e^{-x}\bar{z}_{0}^{2}(x)dx=1, (7.31)

and

xexz¯02(x)𝑑x\displaystyle\int_{-\infty}^{\infty}xe^{-x}\bar{z}_{0}^{2}(x)dx =xz¯0′′(x)𝑑x=limm+mmxz¯0′′(x)𝑑x\displaystyle=\int_{-\infty}^{\infty}x\bar{z}_{0}^{\prime\prime}(x)dx=\lim_{m\to+\infty}\int_{-m}^{m}x\bar{z}_{0}^{\prime\prime}(x)dx (7.32)
=limm+(mz¯0(m)+mz¯0(m)z¯0(m)+z¯0(m))=k0.\displaystyle=\lim_{m\to+\infty}\big{(}m\bar{z}_{0}^{\prime}(m)+m\bar{z}_{0}^{\prime}(-m)-\bar{z}_{0}(m)+\bar{z}_{0}(-m))=-k_{0}.

Therefore, we can write G(q)G(q) as

G(q)=(x+q)exz¯02(x)𝑑xq(x+q)exz¯02(x)𝑑x=qk0q(x+q)exz¯02(x)𝑑x.G(q)=\int_{-\infty}^{\infty}(x+q)e^{-x}\bar{z}_{0}^{2}(x)dx-\int_{-\infty}^{-q}(x+q)e^{-x}\bar{z}_{0}^{2}(x)dx=q-k_{0}-\int_{-\infty}^{-q}(x+q)e^{-x}\bar{z}_{0}^{2}(x)dx. (7.33)

For the last integral in the right side, we deduce from (7.30) that

q(x+q)exz¯02(x)𝑑x=q(x+q)ex𝑑x+O(e(1+δ)q)=eq+O(e(1+δ)q), as q+.\displaystyle\int_{-\infty}^{-q}(x+q)e^{-x}\bar{z}_{0}^{2}(x)dx=\int_{-\infty}^{-q}(x+q)e^{x}dx+O(e^{-(1+\delta)q})=-e^{-q}+O(e^{-(1+\delta)q}),~{}~{}\hbox{ as $q\to+\infty$}. (7.34)

It follows that for s<logr±/2s<\log r_{\pm}/2 we have

G(logr±s)=logr±sk01r±es+O(e(1+δ)(logr±s))=logr±sk0+O(γ).G(\log r_{\pm}-s)=\log r_{\pm}-s-k_{0}-\frac{1}{r_{\pm}}e^{s}+O(e^{-(1+\delta)(\log r_{\pm}-s)})=\log r_{\pm}-s-k_{0}+O(\ell^{-\gamma}). (7.35)

Hence, we have

Y¯±\displaystyle\bar{Y}_{\ell}^{\pm} =r±π0logr±/2[logr±sk0]esdss+O(γ)\displaystyle=-\frac{r_{\pm}}{\sqrt{\pi}\ell}\int_{0}^{\log r_{\pm}/2}[\log r_{\pm}-s-k_{0}]e^{-s}\frac{ds}{\sqrt{s}}+O(\ell^{-\gamma}) (7.36)
=r±π0[logr±sk0]esdss+O(γ)+O(γ)\displaystyle=-\frac{r_{\pm}}{\sqrt{\pi}\ell}\int_{0}^{\infty}[\log r_{\pm}-s-k_{0}]e^{-s}\frac{ds}{\sqrt{s}}+O(\ell^{-\gamma})+O(\ell^{-\gamma})
=r±π([logr±k0]Γ(1/2)Γ(3/2))+O(γ)=r±(logr±k012)+O(γ)\displaystyle=-\frac{r_{\pm}}{\sqrt{\pi}\ell}\big{(}[\log r_{\pm}-k_{0}]\Gamma(1/2)-\Gamma(3/2)\big{)}+O(\ell^{-\gamma})=-\frac{r_{\pm}}{\ell}\big{(}\log r_{\pm}-k_{0}-\frac{1}{2}\big{)}+O(\ell^{-\gamma})
=c¯logc¯logc¯+c¯k0+c¯2+O(γ).\displaystyle=-\bar{c}\log\ell-\bar{c}\log\bar{c}+\bar{c}k_{0}+\frac{\bar{c}}{2}+O(\ell^{-\gamma}).

We used above the fact that Γ(1/2)=π\Gamma(1/2)=\sqrt{\pi} and Γ(3/2)=π/2\Gamma(3/2)=\sqrt{\pi}/2.

7.4 The proof of Lemma 6.6

Let us now proceed with the actual proof of Lemma 6.6. The computation in the previous section relied crucially on approximation (7.22), and the main step is to justify it. The function z+(t,x)z_{+}(t,x) satisfies (6.38)-(6.39) (we drop the sub-script ++ in this section):

zt2zx232(t+1)(zzx)+exz2=0,x<(t+1)δ,\dfrac{\partial{z}}{\partial{t}}-\dfrac{\partial^{2}{z}}{\partial{x}^{2}}-\frac{3}{2(t+1)}\Big{(}z-\dfrac{\partial{z}}{\partial{x}}\Big{)}+e^{-x}z^{2}=0,~{}~{}x<(t+1)^{\delta}, (7.37)

for t2αt\geq\ell^{2-\alpha}, with the boundary condition

z(t,(t+1)δ)=r+e2/(4(t+1))(t+1)δ,z(t,(t+1)^{\delta})=r_{+}e^{-\ell^{2}/(4(t+1))}(t+1)^{\delta}, (7.38)

with r+=c¯+K1δr_{+}=\bar{c}\ell+K\ell^{1-\delta}, and the initial condition

z(t=2α,x)=ecα,x<(t+1)δ,z(t=\ell^{2-\alpha},x)=e^{-c\ell^{\alpha^{\prime}}},~{}~{}~{}~{}x<(t+1)^{\delta}, (7.39)

as in the upper bound (7.14), with some α>0\alpha^{\prime}>0.

Our goal is to find a more explicit super-solution than the solution to (7.37)-(7.39). Motivated by (7.18), let us define

ψ(t,x)=eζ(t)z¯0(xζ(t)),\psi(t,x)=e^{\zeta(t)}\bar{z}_{0}(x-\zeta(t)), (7.40)

with

ζ(t)=logr++log(1+logδ)24(t+1),\zeta(t)=\log r_{+}+\log\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}-\frac{\ell^{2}}{4(t+1)}, (7.41)

so that

ψ(t,x)\displaystyle\psi(t,x) =r+(1+logδ)e2/(4(t+1))z¯0(xlogr±log(1+logδ)+24(t+1))\displaystyle=r_{+}\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}e^{-\ell^{2}/(4(t+1))}\bar{z}_{0}\Big{(}x-\log r_{\pm}-\log\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}+\frac{\ell^{2}}{4(t+1)}\Big{)} (7.42)
=(1+logδ)e2/(4(t+1))z¯+(xlog(1+logδ)+24(t+1)),\displaystyle=\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}e^{-\ell^{2}/(4(t+1))}\bar{z}_{+}\Big{(}x-\log\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}+\frac{\ell^{2}}{4(t+1)}\Big{)},

with z¯+(x)\bar{z}_{+}(x), as in (7.18).

Lemma 7.1

There exists C>0C>0 sufficiently large, and λ>0\lambda>0 sufficiently small, so that

z(t,x)ψ(t,x)+C1/4(t+1)λ, for t>tin=2α and |x|<(t+1)δ.z(t,x)\leq\psi(t,x)+\frac{C}{\ell^{1/4}(t+1)^{\lambda}},~{}~{}\hbox{ for $t>t_{in}=\ell^{2-\alpha}$ and $|x|<(t+1)^{\delta}$.} (7.43)

The proof of Lemma 7.1

Our goal will be to show that we can choose λ>0\lambda>0, γ>0\gamma>0 and C>0C>0 so that

s(t,x)=z(t,x)ψ(t,x)s¯(t,x)=C1/4(t+1)λcos(x(t+1)δ+γ),|x|<(t+1)δ.s(t,x)=z(t,x)-\psi(t,x)\leq\bar{s}(t,x)=\frac{C}{\ell^{1/4}(t+1)^{\lambda}}\cos\Big{(}\frac{x}{(t+1)^{\delta+\gamma}}\Big{)},~{}~{}|x|<(t+1)^{\delta}. (7.44)

Let us first show that with the choice of the shift ζ(t)\zeta(t) as in (7.41), at the boundary x=(t+1)δx=(t+1)^{\delta} we have

ψ(t,(t+1)δ)z(t,(t+1)δ),\psi(t,(t+1)^{\delta})\geq z(t,(t+1)^{\delta}), (7.45)

so that

s(t,(t+1)δ)0s¯(t,(t+1)δ) for tT0s(t,(t+1)^{\delta})\leq 0\leq\bar{s}(t,(t+1)^{\delta})\hbox{ for $t\geq T_{0}$. } (7.46)

At this point we have

ψ(t,(1+t)δ)=r+(1+logδ)e2/4(t+1)z¯0((t+1)δlogr+log(1+logδ)+24(t+1)).\psi(t,(1+t)^{\delta})=r_{+}\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}e^{-{\ell^{2}}/{4(t+1)}}\bar{z}_{0}\Big{(}(t+1)^{\delta}-\log r_{+}-\log\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}+\frac{\ell^{2}}{4(t+1)}\Big{)}. (7.47)

It follows from (7.29) that there exists k1k_{1} so that

z¯0(x)x+k1 for all x0.\bar{z}_{0}(x)\geq x+k_{1}\hbox{ for all $x\geq 0$}. (7.48)

Therefore, if

24(t+1)logr++log(1+logδ)k1,\frac{\ell^{2}}{4(t+1)}\geq\log r_{+}+\log\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}-k_{1},

then, as z¯0(x)\bar{z}_{0}(x) is increasing, we automatically have from (7.47) that

ψ(t,(1+t)δ)\displaystyle\psi(t,(1+t)^{\delta}) r+e2/4(t+1)((t+1)δlogr+log(1+logδ)+24(t+1)+k1)\displaystyle\geq r_{+}e^{-{\ell^{2}}/{4(t+1)}}\Big{(}(t+1)^{\delta}-\log r_{+}-\log\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}+\frac{\ell^{2}}{4(t+1)}+k_{1}\Big{)} (7.49)
r+e2/4(t+1)(t+1)δ.\displaystyle\geq r_{+}e^{-{\ell^{2}}/{4(t+1)}}(t+1)^{\delta}.

On the other hand, if

24(t+1)<logr++log(1+logδ)k1,\frac{\ell^{2}}{4(t+1)}<\log r_{+}+\log\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}-k_{1}, (7.50)

and \ell is sufficiently large, then we have

t>d2log,t>d\frac{\ell^{2}}{\log\ell}, (7.51)

with a sufficiently small d>0d>0, and (t+1)δ(t+1)^{\delta} dominates the other terms in the argument inside z¯0\bar{z}_{0} in the right side of (7.47). In particular, if (7.50) holds, since z¯0\bar{z}_{0} is increasing, we can use (7.48) to obtain

(1+logδ)\displaystyle\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)} z¯0((t+1)δlogr+log(1+logδ)+24(t+1))(1+logδ)z¯0((t+1)δ1logr+)\displaystyle\bar{z}_{0}\Big{(}\!(t+1)^{\delta}-\log r_{+}-\log\big{(}1+\frac{\log\ell}{\ell^{\delta}}\big{)}+\frac{\ell^{2}}{4(t+1)}\Big{)}\geq\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}\bar{z}_{0}\big{(}(t+1)^{\delta}-1-\log r_{+}\big{)} (7.52)
(1+logδ)((t+1)δ1logr+k1)(t+1)δ,\displaystyle\geq\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}\big{(}(t+1)^{\delta}-1-\log r_{+}-k_{1}\big{)}\geq(t+1)^{\delta},

because

logδ((t+1)δ1logr+k1)1logr+k1cδ/2logClog>0.\displaystyle\frac{\log\ell}{\ell^{\delta}}\Big{(}(t+1)^{\delta}-1-\log r_{+}-k_{1}\Big{)}-1-\log r_{+}-k_{1}\geq c\ell^{\delta/2}\log\ell-C\log\ell>0. (7.53)

We have used (7.51) above. Therefore, if we choose ζ(t)\zeta(t) as in (7.41), then the comparison at the boundary (7.45), indeed, holds.

To look at the other boundary point, x=(t+1)δx=-(t+1)^{\delta}, note that the function z¯(x)=Aex\bar{z}(x)=Ae^{x} is a super-solution to (7.37) for all A>1A>1 sufficiently large, and at x=(t+1)δx=(t+1)^{\delta} we have, due to Lemma 6.5:

z(t,(t+1)δ)C(t+1)δe2/(4(t+1)C(t+1)1/2+δAe(t+1)δ=z¯((t+1)δ),z(t,(t+1)^{\delta})\leq C\ell(t+1)^{\delta}e^{-\ell^{2}/(4(t+1)}\leq C(t+1)^{1/2+\delta}\leq Ae^{(t+1)^{\delta}}=\bar{z}((t+1)^{\delta}), (7.54)

if AA is sufficiently large. It follows that

z(t,x)Aex,x<(t+1)δ,z(t,x)\leq Ae^{x},~{}~{}x<(t+1)^{\delta},

and, in particular, we deduce that

z(t,(t+1)δ)Ae(t+1)δC1/4(t+1)λs¯(t,(t+1)δ),for t2α,z(t,-(t+1)^{\delta})\leq Ae^{-(t+1)^{\delta}}\leq\frac{C}{\ell^{1/4}(t+1)^{\lambda}}\leq\bar{s}(t,-(t+1)^{\delta}),~{}~{}\hbox{for $t\geq\ell^{2-\alpha}$,} (7.55)

and thus

s(t,(t+1)δ)s¯(t,(t+1)δ),t2α.s(t,-(t+1)^{\delta})\leq\bar{s}(t,-(t+1)^{\delta}),~{}~{}t\geq\ell^{2-\alpha}. (7.56)

At the initial time tin=2αt_{in}=\ell^{2-\alpha} we have the comparison

s(tin,x)z(tin,x)CecαC1/4(1+tin)λs¯(tin,x),for x<(tin+1)δ.s(t_{in},x)\leq z(t_{in},x)\leq Ce^{-c\ell^{\alpha^{\prime}}}\leq\frac{C}{\ell^{1/4}(1+t_{in})^{\lambda}}\leq\bar{s}(t_{in},x),~{}~{}\hbox{for $x<(t_{in}+1)^{\delta}$.} (7.57)

Having established the comparison at the boundary and the initial time, we now show that s¯(t,x)\bar{s}(t,x) is a super-solution for the equation for s(t,x)s(t,x). The function ψ(t,x)\psi(t,x) satisfies

ψt\displaystyle\dfrac{\partial{\psi}}{\partial{t}} 2ψx232(t+1)(ψψx)+exψ2=ζ˙(t)eζ(t)z¯0(xζ(t))ζ˙(t)eζ(t)z¯0(xζ(t))x\displaystyle-\dfrac{\partial^{2}{\psi}}{\partial{x}^{2}}-\frac{3}{2(t+1)}\Big{(}\psi-\dfrac{\partial{\psi}}{\partial{x}}\Big{)}+e^{-x}\psi^{2}=\dot{\zeta}(t)e^{\zeta(t)}\bar{z}_{0}(x-\zeta(t))-\dot{\zeta}(t)e^{\zeta(t)}\dfrac{\partial{\bar{z}_{0}(x-\zeta(t))}}{\partial{x}} (7.58)
eζ(t)2z¯0(xζ(t))x232(t+1)eζ(t)(z¯0(xζ(t))z¯0(xζ(t))x)+exe2ζ(t)z¯02(xζ(t))\displaystyle-e^{\zeta(t)}\dfrac{\partial^{2}{\bar{z}_{0}(x-\zeta(t))}}{\partial{x}^{2}}-\frac{3}{2(t+1)}e^{\zeta(t)}\Big{(}\bar{z}_{0}(x-\zeta(t))-\dfrac{\partial{\bar{z}_{0}(x-\zeta(t))}}{\partial{x}}\Big{)}+e^{-x}e^{2\zeta(t)}\bar{z}_{0}^{2}(x-\zeta(t))
=ζ˙(t)eζ(t)z¯0(xζ(t))ζ˙(t)eζ(t)z¯0(xζ(t))x3eζ(t)2(t+1)(z¯0(xζ(t))z¯0(xζ(t))x)\displaystyle=\dot{\zeta}(t)e^{\zeta(t)}\bar{z}_{0}(x-\zeta(t))-\dot{\zeta}(t)e^{\zeta(t)}\dfrac{\partial{\bar{z}_{0}(x-\zeta(t))}}{\partial{x}}-\frac{3e^{\zeta(t)}}{2(t+1)}\Big{(}\bar{z}_{0}(x-\zeta(t))-\dfrac{\partial{\bar{z}_{0}(x-\zeta(t))}}{\partial{x}}\Big{)}
=eζ(t)(ζ˙(t)32(t+1))(z¯0(xζ(t))z¯0(xζ(t))x)\displaystyle=e^{\zeta(t)}\Big{(}\dot{\zeta}(t)-\frac{3}{2(t+1)}\Big{)}\Big{(}\bar{z}_{0}(x-\zeta(t))-\dfrac{\partial{\bar{z}_{0}(x-\zeta(t))}}{\partial{x}}\Big{)}
eζ(t)(ζ˙(t)z¯0(xζ(t))x+32(t+1)z¯0(xζ(t))).\displaystyle\geq-e^{\zeta(t)}\Big{(}\dot{\zeta}(t)\dfrac{\partial{\bar{z}_{0}(x-\zeta(t))}}{\partial{x}}+\frac{3}{2(t+1)}\bar{z}_{0}(x-\zeta(t))\Big{)}.

We used the fact that the function ζ(t)\zeta(t) is increasing in tt and z¯0(x)\bar{z}_{0}(x) is increasing in xx in the last step above. Note that

ζ˙(t)eζ(t)\displaystyle\dot{\zeta}(t)e^{\zeta(t)} =r+(1+logδ)24(t+1)2e2/(4(t+1))C3(t+1)2e2/(4(t+1)),\displaystyle=r_{+}\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}\frac{\ell^{2}}{4(t+1)^{2}}e^{-\ell^{2}/(4(t+1))}\leq\frac{C\ell^{3}}{(t+1)^{2}}e^{-\ell^{2}/(4(t+1))}, (7.59)

hence

ζ˙(t)eζ(t)z¯0(xζ(t))x\displaystyle\dot{\zeta}(t)e^{\zeta(t)}\dfrac{\partial{\bar{z}_{0}(x-\zeta(t))}}{\partial{x}} C3(t+1)2e2/(4(t+1))C1/2(t+1)1/47/2(t+1)7/4e2/(4(t+1))\displaystyle\leq\frac{C\ell^{3}}{(t+1)^{2}}e^{-\ell^{2}/(4(t+1))}\leq\frac{C}{\ell^{1/2}(t+1)^{1/4}}\frac{\ell^{7/2}}{(t+1)^{7/4}}e^{-\ell^{2}/(4(t+1))} (7.60)
C1/2(t+1)1/4.\displaystyle\leq\frac{C}{\ell^{1/2}(t+1)^{1/4}}.

The second term in the right side of (7.58) for x<(t+1)δx<(t+1)^{\delta} is bounded by

32(t+1)eζ(t)z¯0(xζ(t))Ct+1e2/(4(t+1))((t+1)δ+2t+1)\displaystyle\frac{3}{2(t+1)}e^{\zeta(t)}\bar{z}_{0}(x-\zeta(t))\leq\frac{C\ell}{t+1}e^{-\ell^{2}/(4(t+1))}\Big{(}(t+1)^{\delta}+\frac{\ell^{2}}{t+1}\Big{)} (7.61)
=C((t+1)1δ+3(t+1)2)e2/(4(t+1))\displaystyle=C\Big{(}\frac{\ell}{(t+1)^{1-\delta}}+\frac{\ell^{3}}{(t+1)^{2}}\Big{)}e^{-\ell^{2}/(4(t+1))}
C(11/4(t+1)3/8δ5/4(t+1)5/8+11/2(t+1)1/47/2(t+1)7/4)e2/(4(t+1))C1/4(t+1)1/4.\displaystyle\leq C\Big{(}\frac{1}{\ell^{1/4}(t+1)^{3/8-\delta}}\cdot\frac{\ell^{5/4}}{(t+1)^{5/8}}+\frac{1}{\ell^{1/2}(t+1)^{1/4}}\cdot\frac{\ell^{7/2}}{(t+1)^{7/4}}\Big{)}e^{-\ell^{2}/(4(t+1))}\leq\frac{C}{\ell^{1/4}(t+1)^{1/4}}.

Therefore, the function s(t,x)=z(t,x)ψ(t,x)s(t,x)=z(t,x)-\psi(t,x) satisfies

st\displaystyle\dfrac{\partial{s}}{\partial{t}} 2sx232(t+1)(ssx)+ex(ψ(t,x)+z(t,x))sC1/4(t+1)1/4,x<(t+1)δ.\displaystyle-\dfrac{\partial^{2}{s}}{\partial{x}^{2}}-\frac{3}{2(t+1)}\Big{(}s-\dfrac{\partial{s}}{\partial{x}}\Big{)}+e^{-x}(\psi(t,x)+z(t,x))s\leq\frac{C}{\ell^{1/4}(t+1)^{1/4}},~{}~{}x<(t+1)^{\delta}. (7.62)

On the other hand, recall from [22] that for γ>0\gamma>0, the function s¯(t,x)\bar{s}(t,x) satisfies

s¯(t,x)t=λt+1s¯(t,x)+g(t,x),\dfrac{\partial{\bar{s}(t,x)}}{\partial{t}}=-\frac{\lambda}{t+1}\bar{s}(t,x)+g(t,x), (7.63)

with g(t,x)g(t,x) such that

|g(t,x)|C(δ+γ)|x|1/4(t+1)λ+δ+γ+1C(δ+γ)s¯(t,x)(t+1)γ+1, for |x|(t+1)δ.|g(t,x)|\leq\frac{C(\delta+\gamma)|x|}{\ell^{1/4}(t+1)^{\lambda+\delta+\gamma+1}}\leq\frac{C(\delta+\gamma)\bar{s}(t,x)}{(t+1)^{\gamma+1}},~{}~{}\hbox{ for $|x|\leq(t+1)^{\delta}$.} (7.64)

In addition, we have

2s¯(t,x)x2=1(t+1)2δ+2γs¯(t,x),\dfrac{\partial^{2}{\bar{s}(t,x)}}{\partial{x}^{2}}=-\frac{1}{(t+1)^{2\delta+2\gamma}}\bar{s}(t,x), (7.65)

and

32(t+1)(s¯(t,x)xs¯(t,x))Ct+1s¯(t,x), for |x|(t+1)δ.\frac{3}{2(t+1)}\Big{(}\dfrac{\partial{\bar{s}(t,x)}}{\partial{x}}-\bar{s}(t,x)\Big{)}\geq-\frac{C}{t+1}\bar{s}(t,x),~{}~{}\hbox{ for $|x|\leq(t+1)^{\delta}$.} (7.66)

We conclude that

s¯t\displaystyle\dfrac{\partial{\bar{s}}}{\partial{t}} 2s¯x232(t+1)(s¯s¯x)+ex(ψ(t,x)+z(t,x))s¯\displaystyle-\dfrac{\partial^{2}{\bar{s}}}{\partial{x}^{2}}-\frac{3}{2(t+1)}\Big{(}\bar{s}-\dfrac{\partial{\bar{s}}}{\partial{x}}\Big{)}+e^{-x}(\psi(t,x)+z(t,x))\bar{s} (7.67)
λ1+ts¯(t,x)+g(t,x)+1(1+t)2δ+2γs¯(t,x)Cs¯(t,x)(t+1)C(1+t)2δ+2γs¯(t,x)\displaystyle\geq-\frac{\lambda}{1+t}\bar{s}(t,x)+g(t,x)+\frac{1}{(1+t)^{2\delta+2\gamma}}\bar{s}(t,x)-\frac{C\bar{s}(t,x)}{(t+1)}\geq\frac{C}{(1+t)^{2\delta+2\gamma}}\bar{s}(t,x)
C1/4(1+t)λ+2δ+2γ, for t2α and |x|<(t+1)δ,\displaystyle\geq\frac{C}{\ell^{1/4}(1+t)^{\lambda+2\delta+2\gamma}},~{}~{}\hbox{ for $t\geq\ell^{2-\alpha}$ and $|x|<(t+1)^{\delta}$,}

provided that δ>0\delta>0 and γ>0\gamma>0 are sufficiently small. Now, putting together the boundary comparisons (7.46) and (7.56), the initial time comparison (7.57), as well as (7.62) and (7.67), we deduce from the comparison principle that if λ>0\lambda>0, δ>0\delta>0 and γ>0\gamma>0 are all sufficiently small, then

s(t,x)s¯(t,x) for t>tin=2α and |x|<(t+1)δ.s(t,x)\leq\bar{s}(t,x)\hbox{ for $t>t_{in}=\ell^{2-\alpha}$ and $|x|<(t+1)^{\delta}$.} (7.68)

This implies (7.44), and (7.43) follows, finishing the proof of Lemma 7.1\Box

The end of the proof of Lemma 6.6

Using (7.43) in (6.41) gives

Y¯(+)\displaystyle-\bar{Y}_{\ell}^{(+)} =14π2α0(t+1)δxexz2(t,x)dxdt(t+1)3/2\displaystyle=\frac{1}{\sqrt{4\pi}}\int_{\ell^{2-\alpha}}^{\infty}\int_{0}^{(t+1)^{\delta}}xe^{-x}z_{\ell}^{2}(t,x)\frac{dxdt}{{(t+1)^{3/2}}} (7.69)
14π2α0(t+1)δxex(ψ(t,x)+C1/4(t+1)λ)2dxdt(t+1)3/2=I+II+III,\displaystyle\leq\frac{1}{\sqrt{4\pi}}\int_{\ell^{2-\alpha}}^{\infty}\int_{0}^{(t+1)^{\delta}}xe^{-x}\Big{(}\psi(t,x)+\frac{C}{\ell^{1/4}(t+1)^{\lambda}}\Big{)}^{2}\frac{dxdt}{{(t+1)^{3/2}}}=I+II+III,

with the three terms coming from the expansion

(ψ(t,x)+C1/4(t+1)λ)2=ψ2(t,x)+2C1/4(t+1)λψ(t,x)+C21/2(t+1)2λ.\Big{(}\psi(t,x)+\frac{C}{\ell^{1/4}(t+1)^{\lambda}}\Big{)}^{2}=\psi^{2}(t,x)+\frac{2C}{\ell^{1/4}(t+1)^{\lambda}}\psi(t,x)+\frac{C^{2}}{\ell^{1/2}(t+1)^{2\lambda}}.

We have, clearly,

IIIC1/2.III\leq\frac{C}{\ell^{1/2}}. (7.70)

For the second term, we recall that for x>0x>0 we have

ψ(t,x)\displaystyle\psi(t,x) =eζ(t)z¯0(xζ(t))Ce2/(4(t+1))z¯0(x+24(t+1))\displaystyle=e^{\zeta(t)}\bar{z}_{0}(x-\zeta(t))\leq C\ell e^{-\ell^{2}/(4(t+1))}\bar{z}_{0}\Big{(}x+\frac{\ell^{2}}{4(t+1)}\Big{)} (7.71)
Ce2/(4(t+1))(x+1+24(t+1)),\displaystyle\leq C\ell e^{-\ell^{2}/(4(t+1))}\Big{(}x+1+\frac{\ell^{2}}{4(t+1)}\Big{)},

and use this to write

II\displaystyle II =C1/42α0(t+1)δxexψ(t,x)dxdt(t+1)3/2+λ\displaystyle=\frac{C}{\ell^{1/4}}\int_{\ell^{2-\alpha}}^{\infty}\int_{0}^{(t+1)^{\delta}}xe^{-x}\psi(t,x)\frac{dxdt}{{(t+1)^{3/2+\lambda}}} (7.72)
C1/400(t+1)δxexe2/(4(t+1))(x+1+24(t+1))dxdt(t+1)3/2+λ\displaystyle\leq\frac{C}{\ell^{1/4}}\int_{0}^{\infty}\int_{0}^{(t+1)^{\delta}}xe^{-x}\ell e^{-\ell^{2}/(4(t+1))}\Big{(}x+1+\frac{\ell^{2}}{4(t+1)}\Big{)}\frac{dxdt}{{(t+1)^{3/2+\lambda}}}
C1/41e2/(4t)(1+24t)dxdtt3/2+λC1/40e1/(4t)(1+14t)2dxdt(2t)3/2+λC1/4.\displaystyle\leq\frac{C}{\ell^{1/4}}\int_{1}^{\infty}\ell e^{-\ell^{2}/(4t)}\Big{(}1+\frac{\ell^{2}}{4t}\Big{)}\frac{dxdt}{{t^{3/2+\lambda}}}\leq\frac{C}{\ell^{1/4}}\int_{0}^{\infty}\ell e^{-1/(4t)}\Big{(}1+\frac{1}{4t}\Big{)}\frac{\ell^{2}dxdt}{{(\ell^{2}t)^{3/2+\lambda}}}\leq\frac{C}{\ell^{1/4}}.

For the main term, we need to compute more precisely, and we use expression (7.42) for ψ(t,x)\psi(t,x):

I=14π2α0(t+1)δxexψ2(t,x)dxdt(1+t)3/2\displaystyle I=\frac{1}{\sqrt{4\pi}}\int_{\ell^{2-\alpha}}^{\infty}\int_{0}^{(t+1)^{\delta}}xe^{-x}\psi^{2}(t,x)\frac{dxdt}{{(1+t)^{3/2}}} (7.73)
=(1+logδ)22α0(t+1)δxexe2/(2(t+1))z¯+2(xlog(1+logδ)+24(t+1))dxdt(t+1)3/2\displaystyle=\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}^{2}\!\int_{\ell^{2-\alpha}}^{\infty}\int_{0}^{(t+1)^{\delta}}\!\!\!\!\!\!\!\!xe^{-x}e^{-\ell^{2}/(2(t+1))}\bar{z}_{+}^{2}\Big{(}x-\log\Big{(}1+\frac{\log\ell}{\ell^{\delta}}\Big{)}+\frac{\ell^{2}}{4(t+1)}\Big{)}\frac{dxdt}{{(t+1)^{3/2}}}
=2α0(t+1)δxexe2/(2(t+1))z¯+2(x+24(t+1))dxdt(t+1)3/2+O(δ/2).\displaystyle=\!\int_{\ell^{2-\alpha}}^{\infty}\int_{0}^{(t+1)^{\delta}}\!\!\!\!\!\!\!\!xe^{-x}e^{-\ell^{2}/(2(t+1))}\bar{z}_{+}^{2}\Big{(}x+\frac{\ell^{2}}{4(t+1)}\Big{)}\frac{dxdt}{{(t+1)^{3/2}}}+O(\ell^{-\delta/2}).

This is simply expression (7.23) that we have computed in Section 7.3, leading to (7.36):

Y¯+c¯log+c¯logc¯c¯k0c¯2+O(γ)-\bar{Y}_{\ell}^{+}\leq\bar{c}\log\ell+\bar{c}\log\bar{c}-\bar{c}k_{0}-\frac{\bar{c}}{2}+O(\ell^{-\gamma}) (7.74)

This finishes the proof of the upper bound in Lemma 6.6.

Proceeding as in the proof of the upper bound, with some minor modifications, we can obtain a matching lower bound

Y¯c¯log+c¯logc¯c¯k0c¯2+O(γ)-\bar{Y}_{\ell}^{-}\geq\bar{c}\log\ell+\bar{c}\log\bar{c}-\bar{c}k_{0}-\frac{\bar{c}}{2}+O(\ell^{-\gamma}) (7.75)

which finishes the proof of Lemma 6.6.

7.5 The proof of Lemma 6.2

We need to show that

Y~\displaystyle\tilde{Y}_{\ell} =00e3τ/2ηexp(τ/2)r2(τ,η)ψ¯(η)eτ/2𝑑τ𝑑η=O(γ), as +.\displaystyle=-\int_{0}^{\infty}\int_{0}^{\infty}e^{3\tau/2-\eta\exp(\tau/2)}r_{\ell}^{2}(\tau,\eta)\bar{\psi}(\eta)e^{-\tau/2}d\tau d\eta=O(\ell^{-\gamma}),\hbox{ as $\ell\to+\infty$}. (7.76)

As in (7.2), we re-write this in terms of z(t,x)z_{\ell}(t,x) as

Y~\displaystyle\tilde{Y}_{\ell} =00(t+1)exw2(t,x)ψ¯(xt+1)dtt+1dxt+1\displaystyle=-\int_{0}^{\infty}\int_{0}^{\infty}(t+1)e^{-x}w_{\ell}^{2}(t,x)\bar{\psi}\Big{(}\frac{x}{\sqrt{t+1}}\Big{)}\frac{dt}{t+1}\frac{dx}{\sqrt{t+1}} (7.77)
=00exz2(t,x)ψ¯(xt+1)dxdt(t+1)3/2.\displaystyle=-\int_{0}^{\infty}\int_{0}^{\infty}e^{-x}z_{\ell}^{2}(t,x)\bar{\psi}\Big{(}\frac{x}{\sqrt{t+1}}\Big{)}\frac{dxdt}{(t+1)^{3/2}}.

We sketch the argument that can be made precise using Lemma 7.1 in a straightforward way. Let us use approximation (7.22) again:

z(t,x)z¯±(x)e2/(4(t+1)),z_{\ell}(t,x)\approx\bar{z}_{\pm}(x)e^{-\ell^{2}/(4(t+1))}, (7.78)

and insert this into (7.77). This would give

Y~\displaystyle\tilde{Y}_{\ell} 00exz¯±2(x)e2/(2(t+1))ψ¯(xt+1)dxdt(t+1)3/2\displaystyle\approx-\int_{0}^{\infty}\int_{0}^{\infty}e^{-x}\bar{z}_{\pm}^{2}(x)e^{-\ell^{2}/(2(t+1))}\bar{\psi}\Big{(}\frac{x}{\sqrt{t+1}}\Big{)}\frac{dxdt}{(t+1)^{3/2}} (7.79)
r±200exz¯02(xlogr±)e2/(2(t+1))ψ¯(xt+1)dxdt(t+1)3/2.\displaystyle\approx-r_{\pm}^{2}\int_{0}^{\infty}\int_{0}^{\infty}e^{-x}\bar{z}_{0}^{2}(x-\log r_{\pm})e^{-\ell^{2}/(2(t+1))}\bar{\psi}\Big{(}\frac{x}{\sqrt{t+1}}\Big{)}\frac{dxdt}{(t+1)^{3/2}}.

Recall that ψ¯(η)Cη\bar{\psi}(\eta)\leq C\eta, so we can roughly estimate

|Y~|\displaystyle|\tilde{Y}_{\ell}| Cr±20exz¯02(xlogr±)e2/(2(t+1))|x|t+1dxdt(t+1)3/2\displaystyle\leq Cr_{\pm}^{2}\int_{0}^{\infty}\int_{-\infty}^{\infty}e^{-x}\bar{z}_{0}^{2}(x-\log r_{\pm})e^{-\ell^{2}/(2(t+1))}\frac{|x|}{\sqrt{t+1}}\frac{dxdt}{(t+1)^{3/2}} (7.80)
Cr±0exz¯02(x)e2/(2(t+1))|x|+logr±t+1dxdt(t+1)3/2\displaystyle\leq Cr_{\pm}\int_{0}^{\infty}\int_{-\infty}^{\infty}e^{-x}\bar{z}_{0}^{2}(x)e^{-\ell^{2}/(2(t+1))}\frac{|x|+\log r_{\pm}}{\sqrt{t+1}}\frac{dxdt}{(t+1)^{3/2}}
Cr±(1+logr±)0e2/(2(t+1))dt(t+1)2=Cr±(1+logr±)24Clog.\displaystyle\leq Cr_{\pm}(1+\log r_{\pm})\int_{0}^{\infty}e^{-\ell^{2}/(2(t+1))}\frac{dt}{(t+1)^{2}}=Cr_{\pm}(1+\log r_{\pm})\frac{\ell^{2}}{\ell^{4}}\leq\frac{C\log\ell}{\ell}.

This finishes the proof of Lemma 6.2\Box

7.6 The proof of Lemma 6.5

We now turn to the proof of Lemma 6.5. Let us go back to (6.30):

zlt2zlx232(t+1)(zlzlx)+exzl2=0,\dfrac{\partial{z_{l}}}{\partial{t}}-\dfrac{\partial^{2}{z_{l}}}{\partial{x}^{2}}-\frac{3}{2(t+1)}\Big{(}z_{l}-\dfrac{\partial{z_{l}}}{\partial{x}}\Big{)}+e^{-x}z_{l}^{2}=0, (7.81)

and undo yet another change of variables:

v(t,x)=\displaystyle v(t,x)= (t+1)3/2zl(t,x+32log(t+1)).\displaystyle(t+1)^{-3/2}z_{l}(t,x+\frac{3}{2}\log(t+1)). (7.82)

The function v(t,x)v(t,x) satisfies simply

vt2vx2+exv2=0\displaystyle\dfrac{\partial{v}}{\partial{t}}-\dfrac{\partial^{2}{v}}{\partial{x}^{2}}+e^{-x}v^{2}=0 (7.83)
v(0,x)=ψ0(x).\displaystyle v(0,x)=\psi_{0}(x-\ell).

Note that

z(t,(t+1)δ)(t+1)δ=(t+1)3/2v(t,(t+1)δ(3/2)log(t+1))(t+1)δ,\frac{z_{\ell}(t,(t+1)^{\delta})}{(t+1)^{\delta}}=\frac{(t+1)^{3/2}v(t,(t+1)^{\delta}-(3/2)\log(t+1))}{(t+1)^{\delta}}, (7.84)

so that our point of interest for v(t,x)v(t,x) is

x(t)=(t+1)δ32log(t+1).x_{\ell}(t)=(t+1)^{\delta}-\frac{3}{2}\log(t+1). (7.85)

We perform the standard parabolic scaling

p(s,ξ)=v(2s,ξ),p(s,\xi)=v(\ell^{2}s,\ell\xi), (7.86)

so that the function p(s,ξ)p(s,\xi) solves

ps2pξ2+2eξp2=0\displaystyle\dfrac{\partial{p}}{\partial{s}}-\dfrac{\partial^{2}{p}}{\partial{\xi}^{2}}+\ell^{2}e^{-\ell\xi}p^{2}=0 (7.87)
p(0,ξ)=ψ0((ξ1)).\displaystyle p(0,\xi)=\psi_{0}(\ell(\xi-1)).

In the new variables, the point of interest is

ξ(s)=x(2s)=(2s+1)δ(3/2)log(2s+1)sδ12δ,\xi_{\ell}(s)=\frac{x_{\ell}(\ell^{2}s)}{\ell}=\frac{(\ell^{2}s+1)^{\delta}-(3/2)\log(\ell^{2}s+1)}{\ell}\approx\frac{s^{\delta}}{\ell^{1-2\delta}}, (7.88)

and the important time scales are αsα\ell^{-\alpha}\ll s\ll\ell^{\alpha}, corresponding to 2αt2+α\ell^{2-\alpha}\ll t\ll\ell^{2+\alpha}. In particular, we have

ξ(s)=(1+O(γ))sδ12δ, for αsα.\xi_{\ell}(s)=(1+O(\ell^{-\gamma}))\frac{s^{\delta}}{\ell^{1-2\delta}},~{}~{}\hbox{ for $\ell^{-\alpha}\ll s\ll\ell^{\alpha}$}. (7.89)

Let us start with the proof of the lower bound (6.37). The maximum principle implies that

p(s,ξ)C0=ψ0L.p(s,\xi)\leq C_{0}=\|\psi_{0}\|_{L^{\infty}}.

Therefore, if we take 0<γδ0<\gamma\ll\delta and impose the Dirichlet boundary condition at ξ=(1γ)\xi=\ell^{-(1-\gamma)}, then we have

p(s,ξ)q(s,ξ),p(s,\xi)\geq q(s,\xi), (7.90)

where q(s,ξ)q(s,\xi) is the solution to

qs2qξ2+C02eγq=0,s>0,ξ>(1γ),\displaystyle\dfrac{\partial{q}}{\partial{s}}-\dfrac{\partial^{2}{q}}{\partial{\xi}^{2}}+C_{0}\ell^{2}e^{-\ell^{\gamma}}q=0,~{}~{}s>0,~{}~{}\xi>\ell^{-(1-\gamma)}, (7.91)
q(0,ξ)=ψ0(l(ξ1)),\displaystyle q(0,\xi)=\psi_{0}(l(\xi-1)),
q(0,(1γ))=0.\displaystyle q(0,\ell^{-(1-\gamma)})=0.

We see from (7.88) that, as γ<δ\gamma<\delta, we have

ξ(s)sδ212δαδ212δ>(1γ), for sα,\xi_{\ell}(s)\geq\frac{s^{\delta}}{2\ell^{1-2\delta}}\geq\frac{\ell^{-\alpha\delta}}{2\ell^{1-2\delta}}>\ell^{-(1-\gamma)},~{}~{}\hbox{ for $s\geq\ell^{-\alpha}$,} (7.92)

provided that γ\gamma is sufficiently small, so that we have not lost our point of interest in this approximation. An explicit formula for the solution to (7.91) gives

q(s,ξ(s))=eC02eγs4πs0+(e(ξ~(s)ζ)2/(4s)e(ξ~(s)+ζ)2/(4s))ψ0((ζ1))𝑑ζ, for sα,q(s,\xi_{\ell}(s))=\frac{e^{-C_{0}\ell^{2}e^{-\ell^{\gamma}}s}}{\sqrt{4\pi s}}\int_{0}^{+\infty}\Big{(}e^{-(\tilde{\xi}_{\ell}(s)-\zeta)^{2}/(4s)}-e^{-(\tilde{\xi}_{\ell}(s)+\zeta)^{2}/(4s)}\Big{)}\psi_{0}(\ell(\zeta-1))d\zeta,~{}~{}\hbox{ for $s\geq\ell^{-\alpha}$,} (7.93)

with

ξ~(s)=ξ(s)(1γ)sδ12δ11γsδ12δ, for sα.\tilde{\xi}_{\ell}(s)=\xi_{\ell}(s)-\ell^{-(1-\gamma)}\approx\frac{s^{\delta}}{\ell^{1-2\delta}}-\frac{1}{\ell^{1-\gamma}}\approx\frac{s^{\delta}}{\ell^{1-2\delta}},~{}~{}\hbox{ for $s\geq\ell^{-\alpha}$.} (7.94)

The restriction on ss in (7.93) comes from the requirement that (7.92) holds, so that ξ(s)\xi_{\ell}(s) is in the region where q(s,ξ)q(s,\xi) is defined.

The integrand in (7.93) is very small for ζ11\|\zeta-1\|\gg\ell^{-1} because of the exponential decay of the initial condition ψ0(x)\psi_{0}(x). With this in mind, we write the difference of the exponentials in (7.93) as

e(ξ~(s)ζ)2/(4s)e(ξ~(s)+ζ)2/(4s)=e(ξ~(s)ζ)2/(4s)(1eξ~(s)ζ/s).\displaystyle e^{-(\tilde{\xi}_{\ell}(s)-\zeta)^{2}/(4s)}-e^{-(\tilde{\xi}_{\ell}(s)+\zeta)^{2}/(4s)}=e^{-(\tilde{\xi}_{\ell}(s)-\zeta)^{2}/(4s)}(1-e^{-\tilde{\xi}_{\ell}(s)\zeta/s}). (7.95)

As ζ1\zeta\approx 1, we have

ξ~(s)ζs112δs1δ1 for s(12δ)/(1δ).\frac{\tilde{\xi}_{\ell}(s)\zeta}{s}\approx\frac{1}{\ell^{1-2\delta}s^{1-\delta}}\ll 1~{}~{}~{}\hbox{~{}~{}for $s\gg\ell^{(1-2\delta)/(1-\delta)}$.} (7.96)

Using the inequality

1euuu21-e^{-u}\geq u-u^{2} (7.97)

for sufficiently small uu we have then

q(s,ξ(s))esC0l2elγξ~(s)4πs3/2(1+o(1))0+e(ξ~(s)ζ)2/(4s)ζ(1ζξ~(s)s)ψ0((ζ1))𝑑ζ.q(s,\xi_{\ell}(s))\geq\frac{e^{-sC_{0}l^{2}e^{-l^{\gamma}}}\tilde{\xi}_{\ell}(s)}{\sqrt{4\pi}s^{3/2}}(1+o(\ell^{-1}))\int_{0}^{+\infty}e^{-(\tilde{\xi}_{\ell}(s)-\zeta)^{2}/(4s)}\zeta\Big{(}1-\frac{\zeta\tilde{\xi}_{\ell}(s)}{s}\Big{)}\psi_{0}(\ell(\zeta-1))d\zeta. (7.98)

The o(1)o(\ell^{-1}) correction in the pre-factor comes from ζ\zeta that violate (7.96), so that we can not use (7.97). Their contribution is extremely small since ψ0(x)\psi_{0}(x) is decaying exponentially as xx\to-\infty and has compact support for x>0x>0. Using the straightforward approximations for the Gaussian and the factor inside the parenthesis inside the integral in (7.98), as well as for the exponential factor in front of the integral, we obtain

q(s,ξ(s))\displaystyle q(s,\xi_{\ell}(s)) esC0l2elγξ~(s)4πs3/2(1+O(γ))0+e1/(4s)ζψ0((ζ1))𝑑ζ\displaystyle\geq\frac{e^{-sC_{0}l^{2}e^{-l^{\gamma}}}\tilde{\xi}_{\ell}(s)}{\sqrt{4\pi}s^{3/2}}(1+O(\ell^{-\gamma}))\int_{0}^{+\infty}e^{-1/(4s)}\zeta\psi_{0}(\ell(\zeta-1))d\zeta (7.99)
ξ~(s)e1/(4s)4πs3/2(1+O(γ))0+ζψ0((ζ1))𝑑ζ\displaystyle\geq\frac{\tilde{\xi}_{\ell}(s)e^{-1/(4s)}}{\sqrt{4\pi}s^{3/2}}(1+O(\ell^{-\gamma}))\int_{0}^{+\infty}\zeta\psi_{0}(\ell(\zeta-1))d\zeta
=ξ~(s)e1/(4s)4πs3/2(1+O(γ))+ψ0(ζ)𝑑ζ=c¯ξ~(s)e1/(4s)s3/2(1+O(γ)).\displaystyle=\frac{\tilde{\xi}_{\ell}(s)e^{-1/(4s)}}{\sqrt{4\pi}s^{3/2}\ell}(1+O(\ell^{-\gamma}))\int_{-\ell}^{+\infty}\psi_{0}(\zeta)d\zeta=\frac{\bar{c}\tilde{\xi}_{\ell}(s)e^{-1/(4s)}}{s^{3/2}\ell}(1+O(\ell^{-\gamma})).

Going back to (7.89) and (7.94), we note that

ξ~(s)=ξ(s)(1γ)=(1+O(γ))sδ12δ11γsδ12δ=(1+O(γ))ξ(s), for sα.\tilde{\xi}_{\ell}(s)=\xi_{\ell}(s)-\ell^{-(1-\gamma)}=(1+O(\ell^{-\gamma}))\frac{s^{\delta}}{\ell^{1-2\delta}}-\frac{1}{\ell^{1-\gamma}}\approx\frac{s^{\delta}}{\ell^{1-2\delta}}=(1+O(\ell^{-\gamma}))\xi_{\ell}(s),~{}~{}\hbox{ for $s\geq\ell^{-\alpha}$.} (7.100)

Using this in (7.99) gives

q(s,ξ(s))\displaystyle q(s,\xi_{\ell}(s)) c¯ξ(s)e1/(4s)s3/2(1+O(γ)), for sα.\displaystyle\geq\frac{\bar{c}\xi_{\ell}(s)e^{-1/(4s)}}{s^{3/2}\ell}(1+O(\ell^{-\gamma})),~{}~{}~{}~{}\hbox{ for $s\geq\ell^{-\alpha}$.} (7.101)

Unrolling the changes of variables (7.84), (7.86), and (7.88), and using the bound (7.90), as well as the approximation (7.89), we can re-write (7.101) as

z(t,(t+1)δ)(t+1)δ\displaystyle\frac{z_{\ell}(t,(t+1)^{\delta})}{(t+1)^{\delta}} =(t+1)3/2v(t,(t+1)δ(3/2)log(t+1))(t+1)δ\displaystyle=\frac{(t+1)^{3/2}v(t,(t+1)^{\delta}-(3/2)\log(t+1))}{(t+1)^{\delta}} (7.102)
=(t+1)3/2p(2t,1(t+1)δ(3/2)1log(t+1))(t+1)δ=(t+1)3/2p(2t,ξ(2t)(t+1)δ\displaystyle=\frac{(t+1)^{3/2}p(\ell^{-2}t,\ell^{-1}(t+1)^{\delta}-(3/2)\ell^{-1}\log(t+1))}{(t+1)^{\delta}}=\frac{(t+1)^{3/2}p(\ell^{-2}t,\xi_{\ell}(\ell^{-2}t)}{(t+1)^{\delta}}
(t+1)3/2q(2t,ξ(2t)(t+1)δ(t+1)3/2c¯ξ(2t)e2/(4t)3(t+1)δ(t+1)3/2(1+O(γ))\displaystyle\geq\frac{(t+1)^{3/2}q(\ell^{-2}t,\xi_{\ell}(\ell^{-2}t)}{(t+1)^{\delta}}\geq\frac{(t+1)^{3/2}\bar{c}\xi_{\ell}(\ell^{-2}t)e^{-\ell^{2}/(4t)}\ell^{3}}{(t+1)^{\delta}(t+1)^{3/2}\ell}(1+O(\ell^{-\gamma}))
=c¯2δtδe2/(4t)2(t+1)δ12δ(1+O(γ))=c¯(1+O(γ))e2/(4t), for 2αt2+α,\displaystyle=\frac{\bar{c}\ell^{-2\delta}t^{\delta}e^{-\ell^{2}/(4t)}\ell^{2}}{(t+1)^{\delta}\ell^{1-2\delta}}(1+O(\ell^{-\gamma}))=\bar{c}\ell(1+O(\ell^{-\gamma}))e^{-\ell^{2}/(4t)},\hbox{ for $\ell^{2-\alpha}\leq t\leq\ell^{2+\alpha}$},

which is the lower bound (6.37).

For the upper bound (6.36), we again look at the solution p(s,ξ)p(s,\xi) of (7.87). A simple upper bound for p(s,ξ)p(s,\xi) is by the solution to the heat equation on the whole real line:

p¯s2p¯ξ2=0,x,\displaystyle\dfrac{\partial{\bar{p}}}{\partial{s}}-\dfrac{\partial^{2}{\bar{p}}}{\partial{\xi}^{2}}=0,~{}~{}x\in{\mathbb{R}}, (7.103)
p¯(0,ξ)=ψ0((ξ1)).\displaystyle\bar{p}(0,\xi)=\psi_{0}(\ell(\xi-1)).

Accordingly, we set

ψ¯0(ξ)=p¯(sα,ξ)=14πsαe(ξζ)2/4sαψ0((ζ1))𝑑ζ,sα=α,\overline{\psi}_{0}(\xi)=\bar{p}(s_{\alpha},\xi)=\frac{1}{\sqrt{4\pi s_{\alpha}}}\int_{{\mathbb{R}}}e^{-(\xi-\zeta)^{2}/4s_{\alpha}}\psi_{0}(\ell(\zeta-1))d\zeta,~{}~{}s_{\alpha}=\ell^{-\alpha}, (7.104)

with a sufficiently small α\alpha to be chosen, depending on δ\delta. We also have the upper bound

p(s,ξ)eξ,p(s,\xi)\leq e^{\ell\xi}, (7.105)

that follows immediately from the maximum principle. In particular, we have

p(s,(1γ))eγ,p(s,-\ell^{-(1-\gamma)})\leq e^{-\ell^{\gamma}}, (7.106)

whence

p(s,ξ)eγ+p¯1(s,ξ).p(s,\xi)\leq e^{-\ell^{\gamma}}+\bar{p}_{1}(s,\xi). (7.107)

Here, p¯1(s,ξ)\bar{p}_{1}(s,\xi) is the solution to and v¯(τ,ξ)\overline{v}(\tau,\xi) solves

p¯1s2p¯1ξ2=0,s>sα,ξ>(1γ),\displaystyle\dfrac{\partial{\bar{p}_{1}}}{\partial{s}}-\dfrac{\partial^{2}{\bar{p}_{1}}}{\partial{\xi}^{2}}=0,~{}~{}s>s_{\alpha},~{}\xi>-\ell^{-(1-\gamma)}, (7.108)
p¯1(sα,ξ)=ψ¯0(ξ),\displaystyle\bar{p}_{1}(s_{\alpha},\xi)=\overline{\psi}_{0}(\xi),
p¯1(s,(1γ))=0.\displaystyle\bar{p}_{1}(s,-\ell^{-(1-\gamma)})=0.

Let us note the following properties of the initial condition ψ¯0\overline{\psi}_{0}. First, (7.104) implies that it is localized near ξ=1\xi=1, in the sense that

0<ψ¯0(ξ)Cα/21exp{α2β}, for |ξ1|β.0<\overline{\psi}_{0}(\xi)\leq C\ell^{-\alpha/2}\ell^{-1}\exp\big{\{}-\ell^{\alpha-2\beta}\big{\}},~{}~{}\hbox{ for $|\xi-1|\geq\ell^{-\beta}$.} (7.109)

Hence, as soon as ξ\xi departs from a very small neighborhood of 1, ψ¯0(ξ)\overline{\psi}_{0}(\xi) is exponentially small in \ell. Furthermore, the mass of ψ¯0(ξ)\overline{\psi}_{0}(\xi) is

ψ¯0(ξ)𝑑ξ=c¯,\displaystyle\int_{\mathbb{R}}\overline{\psi}_{0}(\xi)d\xi=\displaystyle\frac{\bar{c}}{\ell}, (7.110)

and its first moment is

ξψ¯0(ξ)𝑑ξ=c¯(1+O(1)),\displaystyle\int_{\mathbb{R}}\xi\overline{\psi}_{0}(\xi)d\xi=\displaystyle\frac{\bar{c}}{\ell}(1+O(\frac{1}{\ell})), (7.111)

because of (7.109).

It follows that the function p¯1(s,ξ)\bar{p}_{1}(s,\xi) can then be estimated along the same lines as q(s,ξ)q(s,\xi) in the proof of the lower bound. This eventually leads to (6.36). \Box

Appendix A Proof of an auxiliary lemma

Here we prove an elementary result used in the proof of Lemma 2.1.

Lemma A.1

Let ψ𝒞bc+\psi\in\mathcal{C}^{+}_{bc}, ψ()1\psi(\cdot)\leq 1, and gg be a continuous function such that g(x)=0g(x)=0 for x0x\leq 0, g(x)=1g(x)=1 for x1x\geq 1 and g(x)=xg(x)=x for x[0,1]x\in[0,1]. Define ψn(x)=g(x+n)ψ(x)𝒞c+\psi_{n}(x)=g(x+n)\psi(x)\in\mathcal{C}^{+}_{c}, then s^[ψn]s^[ψ]\hat{s}[\psi_{n}]\rightarrow\hat{s}[\psi] as nn\rightarrow\infty.

Proof. As ψn(x)ψ(x)\psi_{n}(x)\leq\psi(x), the comparison principle implies immediately that

s^[ψ]s^[ψn],\hat{s}[\psi]\leq\hat{s}[\psi_{n}], (A.1)

and we only need to verify an opposite bound. Let u(t,x)u(t,x), un(t,x)u_{n}(t,x) and u~n(t,x)\tilde{u}_{n}(t,x) be the solutions to (1.1) with the initial conditions

u(0,x)=ψ(x),un(t,x)=ψn(x),u~n(0,x)=ψ~n(x):=ψ(x)ψn(x).u(0,x)=\psi(x),~{}~{}~{}u_{n}(t,x)=\psi_{n}(x),~{}~{}~{}\tilde{u}_{n}(0,x)=\tilde{\psi}_{n}(x):=\psi(x)-\psi_{n}(x). (A.2)

The function vn=un+u~nv_{n}=u_{n}+\tilde{u}_{n} satisfies

vnt=2unx2+unun2+2u~nx2+u~nu~n22vnx2+vnvn2,\dfrac{\partial{v_{n}}}{\partial{t}}=\dfrac{\partial^{2}{u_{n}}}{\partial{x}^{2}}+u_{n}-u_{n}^{2}+\dfrac{\partial^{2}{\tilde{u}_{n}}}{\partial{x}^{2}}+\tilde{u}_{n}-\tilde{u}_{n}^{2}\geq\dfrac{\partial^{2}{v_{n}}}{\partial{x}^{2}}+v_{n}-v_{n}^{2}, (A.3)

with the initial condition vn(0,x)=u(0,x)v_{n}(0,x)=u(0,x). It follows from the maximum principle that

vn(t,x)u(t,x) for all t0 and x,\hbox{$v_{n}(t,x)\geq u(t,x)$ for all $t\geq 0$ and $x\in{\mathbb{R}}$},

and, in particular,

u(t,x+m(t))un(t,x+m(t))+u~n(t,x+m(t)).u(t,x+m(t))\leq u_{n}(t,x+m(t))+\tilde{u}_{n}(t,x+m(t)). (A.4)

Passing to the limit t+t\to+\infty, we obtain

U(x+s^[ψ])U(x+s^[ψn])+U(x+s^[ψ~n]),U(x+\hat{s}[\psi])\leq U(x+\hat{s}[\psi_{n}])+U(x+\hat{s}[\tilde{\psi}_{n}]), (A.5)

for each nn\in{\mathbb{N}} fixed and all xx\in{\mathbb{R}}. Dividing by xexp(x)x\exp(-x) and passing to the limit x+x\to+\infty, keeping nn\in{\mathbb{N}} fixed, gives

es^[ψ]es^[ψn]+es^[ψ~n].e^{-\hat{s}[\psi]}\leq e^{-\hat{s}[\psi_{n}]}+e^{-\hat{s}[\tilde{\psi}_{n}]}. (A.6)

As ψ~n(x)=0\tilde{\psi}_{n}(x)=0 for all xn+1x\geq-n+1, we know that s^[ψ~n]+\hat{s}[\tilde{\psi}_{n}]\to+\infty as n+n\to+\infty. Using this and passing to the limit n+n\to+\infty in (A.6) leads to

s^[ψ]lim supn+s^[ψn].\hat{s}[\psi]\geq\limsup_{n\to+\infty}\hat{s}[\psi_{n}]. (A.7)

This, together with (A.1) finishes the proof. \Box

References

  • [1] E. Aidékon, J. Berestycki, É. Brunet, Z. Shi, Branching Brownian motion seen from its tip, Probab. Theory Relat. Fields 157, 2013, 405–451.
  • [2] L.-P. Arguin, A. Bovier, and N. Kistler, Poissonian statistics in the extremal process of branching Brownian motion. Ann. Appl. Probab. 22, 2012, 1693–1711.
  • [3] L.-P. Arguin, A. Bovier, and N. Kistler, The extremal process of branching Brownian motion. Probab. Theory Relat. Fields 157, 2013, 535–574.
  • [4] J. Berestycki, N. Berestycki and J. Schweinsberg, The genealogy of branching Brownian motion with absorption, Ann. Probab. 41, 2013, 527–618.
  • [5] J. Berestycki, E. Brunet, and B. Derrida, Exact solution and precise asymptotics of a Fisher-KPP type front, J. Phys. A 51, 2018, 035204, 21 pp.
  • [6] A. Bovier, From spin glasses to branching Brownian motion – and back?, in ”Random Walks, Random Fields, and Disordered Systems” (Proceedings of the 2013 Prague Summer School on Mathematical Statistical Physics), M. Biskup, J. Cerny, R. Kotecky, Eds., Lecture Notes in Mathematics 2144, Springer, 2015.
  • [7] M.D. Bramson, Maximal displacement of branching Brownian motion, Comm. Pure Appl. Math. 31, 1978, 531–581.
  • [8] M.D. Bramson, Convergence of solutions of the Kolmogorov equation to travelling waves, Mem. Amer. Math. Soc. 44, 1983.
  • [9] E. Brunet and B. Derrida. Statistics at the tip of a branching random walk and the delay of traveling waves. Eur. Phys. Lett. 87, 2009, 60010.
  • [10] E. Brunet and B. Derrida. A branching random walk seen from the tip, Jour. Stat. Phys. 143, 2011, 420–446.
  • [11] A. Cortines, L. Hartung and O. Louidor, The Structure of Extreme Level Sets in Branching Brownian Motion, Ann. Probab. 47, 2019, 2257–2302.
  • [12] R.A. Fisher, The wave of advance of advantageous genes, Ann. Eugen., 7, 1937, 355–369.
  • [13] C. Graham, Precise asymptotics for Fisher KPP fronts, Nonlinearity, 32, 2019, 1967–1998.
  • [14] F. Hamel, J. Nolen, J.-M. Roquejoffre and L. Ryzhik, A short proof of the logarithmic Bramson correction in Fisher-KPP equation, Netw. Heterog. Media, 8, 2013, 275–289.
  • [15] C. Henderson, Population stabilization in branching Brownian motion with absorption and drift, Comm. Math. Sci. 14, 2016, 973–985.
  • [16] A.N. Kolmogorov, I.G. Petrovskii and N.S. Piskunov, Étude de l’équation de la diffusion avec croissance de la quantité de matière et son application à un problème biologique, Bull. Univ. État Moscou, Sér. Inter. A 1, 1937, 1–26.
  • [17] S. P. Lalley and T. Sellke A Conditional Limit Theorem for the Frontier of a Branching Brownian Motion Ann. Probab. Volume 15, Number 3 (1987), 1052-1061.
  • [18] K.-S. Lau, On the nonlinear diffusion equation of Kolmogorov, Petrovskii and Piskunov, J. Diff. Eqs. 59, 1985, 44-70.
  • [19] H.P. McKean, Application of Brownian motion to the equation of Kolmogorov-Petrovskii-Piskunov, Comm. Pure Appl. Math. 28 1975, 323–331.
  • [20] T. Madaule, The tail distribution of the Derivative martingale and the global minimum of the branching random walk, arXiv:1606.03211, 2016.
  • [21] P. Maillard and M. Pain, 1-stable fluctuations in branching Brownian motion at critical temperature I: the derivative martingale, Ann. Prob., 47, 2019, 2953–3002.
  • [22] J. Nolen, J.-M. Roquejoffre and L. Ryzhik, Convergence to a single wave in the Fisher-KPP equation, Chin. Ann. Math. Ser. B, 38, 2017, 629–646.
  • [23] J. Nolen, J.-M. Roquejoffre and L. Ryzhik, Refined long-time asymptotics for Fisher-KPP fronts, Comm. Contemp. Math., 2018, 1850072.
  • [24] M. Roberts, A simple path to asymptotics for the frontier of a branching Brownian motion, Ann. Prob. 41, 2013, 3518–3541.
  • [25] K. Uchiyama, The behavior of solutions of some nonlinear diffusion equations for large time, J. Math. Kyoto Univ. 18, 1978, 453–508.