# Construction of a stable blow-up solution for a class of strongly perturbed
semilinear heat equations

V. T. Nguyen and H. Zaag 111This author is supported by the ERC Advanced Grant
no. 291214, BLOWDISOL and by the ANR project ANAÉ ref. ANR-13-BS01-0010-03.
Université Paris 13, Sorbonne Paris Cité,
LAGA, CNRS (UMR 7539), F-93430, Villetaneuse, France.

###### Abstract

We construct a solution for a class of strongly perturbed semilinear heat
equations which blows up in finite time with a prescribed blow-up profile. The
construction relies on the reduction of the problem to a finite dimensional
one and the use of index theory to conclude.

Keywords: Blow-up profile, finite-time blow-up, stability, semilinear heat
equations.

##  1 Introduction

We are interested in the following nonlinear parabolic equation:

$\left\\{\begin{array}[]{rcl}u_{t}&=&\Delta u+|u|^{p-1}u+h(u),\\\
u(0)&=&u_{0}\in L^{\infty}(\mathbb{R}^{n}),\end{array}\right.$ (1)

where $u$ is defined for $(x,t)\in\mathbb{R}^{n}\times[0,T)$, $1<p$ and
$p<\frac{n+2}{n-2}$ if $n\geq 3$, the function $h$ is in
$\mathcal{C}^{1}(\mathbb{R},\mathbb{R})$ satisfying

$j=0,1,\;\;|h^{(j)}(z)|\leq
M\left(\dfrac{|z|^{p-j}}{\log^{a}(2+z^{2})}+1\right)\;\;\;\text{with}\;\;a>1,\;M>0,$
(2)

or

$h(z)=\mu\dfrac{|z|^{p-1}z}{\log^{a}(2+z^{2})}\;\;\;\text{with}\;\;a>0,\;\mu\in\mathbb{R}.$
(3)

By standard results, the Cauchy problem for equation (1) can be solved in
$L^{\infty}(\mathbb{R}^{n})$. The solution $u(t)$ of (1) would exist either on
$[0,+\infty)$ (global existence) or only on $[0,T)$, with $0<T<+\infty$. In
this case, we say that $u(t)$ blows up in finite time $T$, namely

$\lim_{t\to T}\|u(t)\|_{L^{\infty}(\mathbb{R}^{n})}=+\infty.$

Here $T$ is called the blow-up time, and a point $x_{0}\in\mathbb{R}^{n}$ is
called a blow-up point if and only if there exist $(x_{n},t_{n})\to(x_{0},T)$
such that $|u(x_{n},t_{n})|\to+\infty$ as $n\to+\infty$. In this paper, we are
interested in the finite time blow-up for equation (1).

When $h\equiv 0$, the blow-up result for equation (1) is largely well-
understood. The existence of blow-up solutions has been proved by several
authors (see Fujita [7], Ball [1], Levine [13]). We have a lots of results
concerning the asymptotic blow-up behavior, locally near a given blow-up point
(see Giga and Kohn [10], Weissler [25], Filippas, Kohn and Liu [5], [6],
Herrero and Velázquez [11], [12], [23], [24], Merle and Zaag [17], [18],
[19]). The notion of asymptotic profile appears also in various papers (see
Bricmont and Kupiainen [3], Merle and Zaag [16], Berger and Kohn [2], Nguyen
[20] for numerical studies).

Given $b$ a blow-up point of $u$, we study the behavior of $u$ near the
singularity $(b,T)$ through the following _similarity variables_ introduced by
Giga and Kohn [8, 9, 10]:

$y=\frac{x-b}{\sqrt{T-t}},\quad s=-\log(T-t),\quad
w_{b}(y,s)=(T-t)^{\frac{1}{p-1}}u(x,t),$ (4)

and $w_{b}$ satisfies for all $(y,s)\in\mathbb{R}^{n}\times[-\log T,+\infty)$,

$\partial_{s}w_{b}=(\Delta-\frac{y}{2}\cdot\nabla+1)w_{b}-\frac{p}{p-1}w_{b}+|w_{b}|^{p-1}w_{b}+e^{-\frac{ps}{p-1}}h\left(e^{\frac{s}{p-1}}w_{b}\right).$
(5)

In [21], the author showed that if $w_{b}$ does not approach $\phi$
exponentially fast, where $\phi$ is the positive solution of the associated
ordinary differential equation of equation (5),

$\phi_{s}=-\frac{\phi}{p-1}+\phi^{p}+e^{-\frac{ps}{p-1}}h(e^{\frac{s}{p-1}}\phi)\quad\text{such
that}\quad\phi(s)\to\kappa\quad\text{as}\quad s\to+\infty,$ (6)

then the solution $u$ of (1) would approach an explicit universal profile as
follows:

$(T-t)^{\frac{1}{p-1}}u(b+z\sqrt{(T-t)|\log(T-t)|},t)\to
f(z)\quad\text{as}\quad t\to T,$ (7)

in $L_{loc}^{\infty}$ and in the case $a>1$, where

$f(z)=\kappa\left(1+c_{p}|z|^{2}\right)^{-\frac{1}{p-1}},\quad\text{with}\;\;c_{p}=\frac{p-1}{4p}.$
(8)

The goal of this work is to show that the behavior (7) does occur. More
precisely, we construct a blow-up solution of equation (1) satisfying the
behavior described in (7). This is our main result:

######  Theorem 1 (Existence of a blow-up solution for equation (1) with the
description of its profile).

There exists $T>0$ such that equation (1) has a solution $u(x,t)$ in
$\mathbb{R}^{n}\times[0,T)$ satisfying:
$i)$ the solution $u$ blows up in finite-time $T$ at the point $b=0$,
$ii)$

$\left\|(T-t)^{\frac{1}{p-1}}u(\cdot\sqrt{T-t},t)-f\left(\frac{\cdot}{\sqrt{|\log(T-t)|}}\right)\right\|_{W^{1,\infty}(\mathbb{R}^{n})}\leq\frac{C}{|\log(T-t)|^{\varrho}},$
(9)

for all $\varrho\in(0,\nu)$ with $\nu=\min\\{a-1,\frac{1}{2}\\}$ in the case
(2) and $\nu=\min\\{a,\frac{1}{2}\\}$ in the case (3), $C$ is some positive
constant and $f$ is defined in (8).
$iii)$ There exists
$u_{*}\in\mathcal{C}(\mathbb{R}^{n}\setminus\\{0\\},\mathbb{R})$ such that
$u(x,t)\to u_{*}(x)$ as $t\to T$ uniformly on compact subsets of
$\mathbb{R}^{n}\setminus\\{0\\}$, where

$u_{*}(x)\sim\left(\frac{8p|\log|x||}{(p-1)^{2}|x|^{2}}\right)^{\frac{1}{p-1}}\quad\text{as}\quad
x\to 0.$

######  Remark 1.

Note that $i)$ directly follows from $ii)$. Indeed, $ii)$ implies that
$u(0,t)\sim\kappa(T-t)^{-\frac{1}{p-1}}\to+\infty$ as $t\to T$, which means
that $u$ blows up in finite-time $T$ at the point $0$. From $iii)$, we see
that $u$ blows up only at the point $b=0$.

######  Remark 2.

Note that the profile $f$ is the same as in the nonlinear heat equation
without the perturbation ($h\equiv 0$), see Bricmont and Kupiainen [3], Merle
and Zaag [16].
The estimate (9) holds in $W^{1,\infty}$ and uniformly in
$z\in\mathbb{R}^{n}$. In the previous work, Ebde and Zaag [4] gives such a
uniform convergence in the case $h$ involving a nonlinear gradient term. In
fact, the convergence in $W^{1,\infty}$ comes from a parabolic regularity
estimate for equation (5) (see Proposition 5 below). Dealing with the case
$h\equiv 0$, Bricmont and Kupiainen [3], Merle and Zaag [16] also give such a
uniform convergence but only in $L^{\infty}(\mathbb{R}^{n})$. In most papers,
the same kind convergence is proved, but only uniformly on a smaller subsets,
$|z|\leq K\sqrt{|\log(T-t)|}$ (see Velázquez [23]).

The proof of Theorem 1 bases on techniques developed by Bricmont and Kupiainen
in [3] and Merle and Zaag in [16] for the semilinear heat equation

$u_{t}=\Delta u+|u|^{p-1}u.$ (10)

Note that the perturbation term $h$ certainly impacts on the construction of
solutions of (1) satisfying (9). This causes some crucial modifications in
[16] in order to totally control the term $h$. Although these modifications do
not affect the general framework developed in [16], they lay in 3 crucial
places:
\- We modify the profile around which we study equation (5), so that we go
beyond the order $\frac{1}{s^{a}}$ generated by the perturbation term. Indeed,
for small $a>0$ and with the same profile as in [16], the order
$\frac{1}{s^{a}}$ will become too strong and will not allow us to close our
estimates. See Section 2 below, particularly definition (12), which enables us
to reach the order $\frac{1}{s^{a+1}}$.
\- In order to handle the order $\frac{1}{s^{a+1}}$, we need to modify the
definition of the shrinking set near the profile. See Section 3 and
particularly Proposition 3 below.
\- A sharp understanding of the dynamics of the linearized operator of (5)
around the profile (12), and which allows to handle the new definition of the
shrinking set. See Lemma (7) below.

For that reason, we will stress only the main parts of the proof of Theorem 1
and put forward the novelties of our argument. In particular, the proof relies
on the understanding of the dynamics of the self-similar version of equation
(5) around the profile (8). Following the work by Merle and Zaag [16], the
proof will be divided into 2 steps:
\- In the first step, we reduce the problem to a finite-dimensional problem:
we will show that it is enough to control a finite-dimensional variable in
order to control the solution near the profile.
\- In the second step, we proceed by contradiction to solve the finite-
dimensional problem and conclude using index theory.

We would like to mention that Masmoudi and Zaag [14] adapted the method of
[16] for the following Ginzburg-Landau equation:

$u_{t}=(1+\imath\beta)\Delta u+(1+\imath\delta)|u|^{p-1}u,$ (11)

where $p-\delta^{2}-\beta\delta(p+1)>0$ and
$u:\mathbb{R}^{n}\times[0,T)\to\mathbb{C}$. Note that the case $\beta=0$ and
$\delta\in\mathbb{R}$ small has been studied earlier by Zaag [26].
In [22], Nouaili and Zaag successfully used the method of [16] for the
following complex valued semilinear heat equation:

$u_{t}=\Delta u+u^{2},$

where $u(t):x\in\mathbb{R}^{n}\to\mathbb{C}$.

As in [16], [26] and [14], it is possible to make the interpretation of the
finite-dimensional variable in terms of the blow-up time and the blow-up
point. This allows us to derive the stability of the profile $f$ in Theorem 1
with respect to perturbations in the initial data. More precisely, we have the
following:

######  Theorem 2 (Stability of the solution constructed in Theorem 1).

Let us denote by $\hat{u}(x,t)$ the solution constructed in Theorem 1 and by
$\hat{T}$ its blow-up time. Then, there exists a neighborhood
$\mathcal{V}_{0}$ of $\hat{u}(x,0)$ in $W^{1,\infty}$ such that for any
$u_{0}\in\mathcal{V}_{0}$, equation (1) has a unique solution $u(x,t)$ with
initial data $u_{0}$, and $u(x,t)$ blows up in finite time $T(u_{0})$ at one
single blow-up point $b(u_{0})$. Moreover, estimate (7) is satisfied by
$u(x-b,t)$ and

$T(u_{0})\to\hat{T},\quad b(u_{0})\to 0\quad\text{as $u_{0}\to\hat{u}_{0}$ in
$W^{1,\infty}(\mathbb{R}^{n})$}.$

######  Remark 3.

We will not give the proof of Theorem 2 because the stability result follows
from the reduction to a finite dimensional case as in [16] with the same
proof. Hence, we only prove the reduction and refer to [16] for the stability.
Note that from the parabolic regularity, our stability result holds in the
larger space $L^{\infty}(\mathbb{R}^{n})$.

##  2 Formulation of the problem

As in [16], [3], we give the proof in one dimension ($n=1$). The proof remains
the same for higher dimensions ($n\geq 2$). We would like to find $u_{0}$
initial data such that the solution $u$ of equation (1) blows up in finite
time $T$ and satisfies the estimate (9). Using similarity variables (4), this
is equivalent to finding $s_{0}>0$ and $w_{0}(y)\equiv w(y,s_{0})$ such that
the solution $w$ of equation (5) with initial data $w_{0}$ satisfies

$\lim_{s\to+\infty}\|w(y,s)-f(\frac{y}{\sqrt{s}})\|_{W^{1,\infty}}=0,$

where $f$ is given in (8).
In order to prove this, we will not linearize equation (5) around
$f+\frac{\kappa}{2ps}$ as in [16]. We will instead introduce

$q=w-\varphi,\quad\text{where}\quad\varphi=\frac{\phi(s)}{\kappa}\left(f(\frac{y}{\sqrt{s}})+\frac{\kappa}{2ps}\right),$
(12)

with $\phi$ and $f$ are introduced in (6) and (8). Then, the problem is
reduced to constructing a function $q$ such that

$\lim_{s\to+\infty}\|q(y,s)\|_{W^{1,\infty}}=0$

and $q$ is a solution of the following equation for all
$(y,s)\in\mathbb{R}\times[s_{0},+\infty)$,

$q_{s}=(\mathcal{L}+V)q+B(q)+R(y,s)+N(y,s),$ (13)

where $\mathcal{L}=\Delta-\frac{y}{2}\cdot\nabla+1$ and

$\displaystyle V(y,s)$
$\displaystyle=p\left(\varphi(y,s)^{p-1}-\frac{1}{p-1}\right)+\imath
e^{-s}h^{\prime}(e^{\frac{s}{p-1}}\varphi),$ (14) $\displaystyle B(q)$
$\displaystyle=|\varphi+q|^{p-1}(\varphi+q)-\varphi^{p}-p\varphi^{p-1}q,$ (15)
$\displaystyle R(y,s)$
$\displaystyle=-\varphi_{s}+\Delta\varphi-\frac{y}{2}\cdot\nabla\varphi-\frac{\varphi}{p-1}+\varphi^{p}+e^{\frac{-ps}{p-1}}h\left(e^{\frac{s}{p-1}}\varphi\right),$
(16) $\displaystyle N(q,s)$
$\displaystyle=e^{\frac{-ps}{p-1}}\left[h\left(e^{\frac{s}{p-1}}(\varphi+q)\right)-h\left(e^{\frac{s}{p-1}}\varphi\right)-\imath
e^{\frac{s}{p-1}}h^{\prime}\left(e^{\frac{s}{p-1}}\varphi\right)q\right],$
(17)

with $\imath=0$ in the case (2) and $\imath=1$ in the case (3).

One can remark that we don’t linearize (5) around
$\tilde{\varphi}=f(\frac{y}{\sqrt{s}})+\frac{\kappa}{2ps}$ as in the case of
equation (10) treated in [16]. In fact, if we do the same, we may obtain some
terms like $\frac{1}{s^{a}}$ coming from the strong perturbation $h$ in
equation (5), and we may not be able to control these terms in the case $a<3$.
To extend the range of $a$, we multiply the factor $\frac{\phi(s)}{\kappa}$ to
$\tilde{\varphi}$ in order to go beyond the order $\frac{1}{s^{a}}$ and reach
at the order $\frac{1}{s^{a+1}}$. Linearizing around $\varphi$ given in (12)
is a major novelty in our approach.

In following analysis, we will use the following integral form of equation
(13): for each $s\geq\sigma\geq s_{0}$:

$q(s)=\mathcal{K}(s,\sigma)q(\sigma)+\int_{\sigma}^{s}\mathcal{K}(s,\tau)\left[B(q(\tau))+R(\tau)+N(q(\tau),\tau)\right]d\tau,$
(18)

where $\mathcal{K}$ is the fundamental solution of the linear operator
$\mathcal{L}+V$ defined for each $\sigma>0$ and for each $s\geq\sigma$,

$\partial_{s}\mathcal{K}(s,\sigma)=(\mathcal{L}+V)\mathcal{K}(s,\sigma),\quad\mathcal{K}(\sigma,\sigma)=Identity.$
(19)

Since the dynamics of equation (13) are influenced by the linear part, we
first need to recall some properties of the operator $\mathcal{L}$ from
Bricmont and Kupiainen [3]. The operator $\mathcal{L}$ is self-adjoint in
$L^{2}_{\rho}(\mathbb{R}^{n})$, where $L^{2}_{\rho}$ is the weighted $L^{2}$
space associated with the weight $\rho$ defined by

$\rho(y)=\left(\frac{1}{4\pi}\right)^{n/2}e^{-\frac{|y|^{2}}{4}}.$

Its spectrum is given by

$spec(\mathcal{L})=\\{1-\frac{m}{2},\;m\in\mathbb{N}\\},$

and its eigenfunctions are derived from Hermite polynomials.
If $n=1$, the eigenfunction corresponding to $1-\frac{m}{2}$ is

$h_{m}(y)=\sum_{k=0}^{\left[\frac{m}{2}\right]}\frac{m!}{k!(m-2k)!}(-1)^{k}y^{m-2k}.$
(20)

We also denote $k_{m}(y)=\frac{h_{m}(y)}{\|h_{m}(y)\|_{L^{2}_{\rho}}^{2}}$.
If $n\geq 2$, we write the spectrum of $\mathcal{L}$ as
$spec(\mathcal{L})=\\{1-\frac{|m|}{2},\;|m|=m_{1}+\dots+m_{n},\;(m_{1},\dots,m_{n})\in\mathbb{N}^{n}\\}$.
Given $m=(m_{1},\dots,m_{n})\in\mathbb{N}^{n}$, the eigenfunction
corresponding to $1-\frac{|m|}{2}$ is

$H_{m}(y)=h_{m_{1}}(y_{1})\dots h_{m_{n}}(y_{n}),\quad\text{where $h_{m}$ is
defined in \eqref{equ:eigenfu1}.}$ (21)

The potential $V(y,s)$ has two fundamental properties:
$i)$ $V(\cdot,s)\to 0$ in $L^{2}_{\rho}$ as $s\to+\infty$. In particular, the
effect of $V$ on the bounded sets or in the ”blow-up” region ($|y|\leq
K\sqrt{s}$) is regarded as a perturbation of the effect of $\mathcal{L}$.
$ii)$ outside of the ”blow-up” region, we have the following property: for all
$\epsilon>0$, there exist $C_{\epsilon}>0$ and $s_{\epsilon}$ such that

$\sup_{s\geq s_{\epsilon},|y|\geq
C_{\epsilon}\sqrt{s}}|V(y,s)-(-\frac{p}{p-1})|\leq\epsilon.$ (22)

This means that $\mathcal{L}+V$ behaves like $\mathcal{L}-\frac{p}{p-1}$ in
the region $|y|\geq K\sqrt{s}$. Because $1$ is the biggest eigenvalue of
$\mathcal{L}$, the operator $\mathcal{L}-\frac{p}{p-1}$ has purely negative
spectrum. Therefore, the control of $q(y,s)$ in $L^{\infty}$ outside of the
”blow-up” region will be done without difficulties.
Since the behavior of $V$ inside and outside of the ”blow-up” region are
different, let us decompose $q$ as following: Let
$\chi_{0}\in\mathcal{C}_{0}^{\infty}([0,+\infty))$ with
$supp(\chi_{0})\subset[0,2]$ and $\chi_{0}\equiv 1$ on $[0,1]$. We define

$\chi(y,s)=\chi_{0}(\frac{|y|}{K\sqrt{s}}),$ (23)

where $K>0$ to be fixed large enough, and write

$q(y,s)=q_{b}(y,s)+q_{e}(y,s),$ (24)

where $q_{b}(y,s)=\chi(y,s)q(y,s)$ and $q_{e}(y,s)=(1-\chi(y,s))q(y,s)$. Note
that $supp(q_{b}(s))\subset\mathbf{B}(0,2K\sqrt{s})$ and
$supp(q_{e}(s))\subset\mathbb{R}\setminus\mathbf{B}(0,K\sqrt{s})$.
In order to control $q_{b}$, we expand it with respect to the spectrum of
$\mathcal{L}$ in $L^{2}_{\rho}$. More precisely, we write $q$ into 5
components as follows:

$q(y,s)=\sum_{m=0}^{2}q_{m}(s)h_{m}(y)+q_{-}(y,s)+q_{e}(y,s),$ (25)

where $q_{m},q_{-}$ are coordinates of $q_{b}$ (not of $q$), namely that
$q_{m}$ is the projection of $q_{b}$ in $h_{m}$ and $q_{-}=P_{-}(q_{b})$ with
$P_{-}$ being the projector on the negative subspace of $\mathcal{L}$.

##  3 Proof of the existence of a blow-up solution with the given blow-up
profile

In this section, we use the framework developed in [16] in order to prove
Theorem 1. We proceed in 4 steps:
\- In the first step, we define a shrinking set $V_{A}(s)$ and translate our
goal of making $q(s)$ go to $0$ in $L^{\infty}(\mathbb{R})$ in terms of
belonging to $V_{A}(s)$. We also exhibit a two parameter initial data family
for equation (13) whose coordinates are very small (with respect to the
requirements of $V_{A}(s)$), except the two first $q_{0}$ and $q_{1}$. Note
that the set $V_{A}(s)$ is different from the corresponding one in [16], and
this makes the second major novelty of our work, in addition to the
modification of the profile in (12).
\- In the second step, using the spectral properties of equation (13), we
reduce our goal from the control of $q(s)$ (an infinite dimensional variable)
in $V_{A}(s)$ to the control of its two first components $(q_{0}(s),q_{1}(s))$
(a two-dimensional variable) in
$\left[-\frac{A}{s^{1+\nu}},\frac{A}{s^{1+\nu}}\right]^{2}$ with $\nu>0$.
\- In the third step, we solve the local in time Cauchy problem for equation
(13).
\- In the last step, we solve the finite dimensional problem using index
theory and conclude the proof of Theorem 1.

In what follows, the constant $C$ denotes a universal one independent of
variables, only depending upon constants of the problems such as $a$, $p$,
$M$, $\mu$ and $K$ in (23).

###  3.1 Definition of a shrinking set $V_{A}(s)$ and preparation of initial
data

Let first introduce the following proposition:

######  Proposition 3 (A shrinking set to zero).

Let $\nu=\min\\{a-1,\frac{1}{2}\\}$ in the case (2) and
$\nu=\min\\{a,\frac{1}{2}\\}$ in the case (3), we fix $\varrho\in(0,\nu)$. For
each $A>0$, for each $s>0$, we define
$\hat{V}_{A}(s)=\left[-\frac{A}{s^{1+\nu}},\frac{A}{s^{1+\nu}}\right]^{2}\subset\mathbb{R}^{2}$,
and $V_{A}(s)$ as being the set of all functions $g$ in $L^{\infty}$ such
that:

$m=0,1,\;\;|g_{m}(s)|\leq\frac{A}{s^{1+\nu}},\quad|g_{2}(s)|\leq\frac{A^{2}}{s^{1+\nu}},$
$\forall
y\in\mathbb{R},\quad|g_{-}(y,s)|\leq\frac{A}{s^{3/2+\varrho}}(1+|y|^{3}),\quad\|g_{e}(s)\|_{L^{\infty}}\leq\frac{A^{2}}{s^{\varrho}},$

where $g_{m},g_{-}$ and $g_{e}$ are defined in (25). Then we have for all
$s\geq e$ and $g\in V_{A}(s)$,

$\forall
y\in\mathbb{R},\;\;|g(y,s)|\leq\frac{CA^{2}}{s^{3/2+\varrho}}(1+|y|^{3})+\frac{CA^{2}}{s^{1+\nu}}(1+|y|^{2})\quad\text{and}\quad\|g(s)\|_{L^{\infty}}\leq\frac{CA^{2}}{s^{\varrho}}.$
(26)

###### Proof.

From the definition of $V_{A}(s)$ and the fact that
$\left|\frac{1-\chi(y,s)}{1+|y|^{3}}\right|\leq\frac{C}{s^{3/2}}$, the
conclusion of Proposition 3 simply follows. ∎

Initial data (at time $s_{0}=-\log T$) for the equation (13) will depend on
two real parameters $d_{0}$ and $d_{1}$ as given in the following proposition:

######  Lemma 4 (Decomposition of initial data on the different components)).

For each $A>1$, there exists $\delta_{1}(A)>0$ such that for all
$s_{0}\geq\delta_{1}(A)$: If we consider the following function as initial
data for equation (13):

$q_{d_{0},d_{1}}(y,s_{0})=\frac{\phi(s_{0})}{\kappa}\left(f^{p}(z)(d_{0}+d_{1}z)-\frac{\kappa}{2ps_{0}}\right),$
(27)

where $z=\frac{y}{\sqrt{s_{0}}}$, $f$ and $\phi$ are defined in (8) and (6),
then
$i)$ There exists a constant $C=C(p)>0$ such that the components of
$q_{d_{0},d_{1}}(s_{0})$ (or $q(s_{0})$ for short) satisfy:

$\displaystyle q_{0}(s_{0})$
$\displaystyle=d_{0}a_{0}(s_{0})+b_{0}(s_{0}),\quad\text{with}\quad
a_{0}(s_{0})\sim C,\quad|b_{0}(s_{0})|\leq\frac{C}{s_{0}},$ (28)
$\displaystyle q_{1}(s_{0})$
$\displaystyle=d_{1}a_{1}(s_{0})+b_{1}(s_{0}),\quad\text{with}\quad
a_{1}(s_{0})\sim\frac{C}{\sqrt{s_{0}}},\quad|b_{1}(s_{0})|\leq\frac{C}{s_{0}^{2}},$
(29)

and

$\displaystyle|q_{2}(s_{0})|\leq\frac{C|d_{0}|}{s_{0}}+Ce^{-s_{0}},\quad|q_{-}(y,s_{0})|\leq\left(\frac{C|d_{0}|}{s_{0}}+\frac{C|d_{1}|}{s_{0}\sqrt{s_{0}}}\right)(1+|y|^{3}),$
$\displaystyle\|q_{e}(s_{0})\|_{L^{\infty}}\leq
C|d_{0}|+\frac{C|d_{1}|}{\sqrt{s_{0}}},\quad\|\nabla
q(s_{0})\|_{L^{\infty}}\leq\frac{C(|d_{0}|+|d_{1}|)}{\sqrt{s_{0}}}.$

$ii)$ For each $A>0$, if $(d_{0},d_{1})$ is chosen so that
$(q_{0},q_{1})(s_{0})\in\hat{V}_{A}(s_{0})$, then

$\displaystyle|d_{0}|+|d_{1}|\leq\frac{C}{s_{0}},$
$\displaystyle|q_{2}(s_{0})|\leq\frac{C}{s_{0}^{2}},\;\;\left\|\frac{q_{-}(y,s_{0})}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{C}{s_{0}^{2}},\;\;\|q_{e}(s_{0})\|_{L^{\infty}}\leq\frac{C}{s_{0}},$
$\displaystyle q(s_{0})\in V_{A}(s_{0}),\;\;\|\nabla
q(s_{0})\|_{L^{\infty}}\leq\frac{C}{s_{0}\sqrt{s_{0}}},$

where the statement $q(s_{0})\in V_{A}(s_{0})$ holds with ”strict
inequalities”, except for $(q_{0},q_{1})(s_{0})$, in the sense that

$m=0,1,\;\;|q_{m}(s)|\leq\frac{A}{s^{1+\nu}},\quad|q_{2}(s)|<\frac{A^{2}}{s^{1+\nu}},$
$\forall
y\in\mathbb{R},\quad|q_{-}(y,s)|<\frac{A}{s^{3/2+\varrho}}(1+|y|^{3}),\quad\|q_{e}(s)\|_{L^{\infty}}<\frac{A^{2}}{s^{\varrho}}.$

$iii)$ There exists a rectangle
$\mathcal{D}_{s_{0}}\subset\left[-\frac{C}{s_{0}},\frac{C}{s_{0}}\right]^{2}$
such that the mapping $(d_{0},d_{1})\mapsto(q_{0},q_{1})(s_{0})$ is linear and
one to one from $\mathcal{D}_{s_{0}}$ onto
$\left[-\frac{A}{s_{0}^{1+\nu}},\frac{A}{s_{0}^{1+\nu}}\right]^{2}$ and maps
$\partial\mathcal{D}_{s_{0}}$ into
$\partial\left[-\frac{A}{s_{0}^{1+\nu}},\frac{A}{s_{0}^{1+\nu}}\right]^{2}$.
Moreover, it is of degree one on the boundary and the following equivalence
holds:

$q(s_{0})\in V_{A}(s_{0})\;\;\text{if and only
if}\;\;(d_{0},d_{1})\in\mathcal{D}_{s_{0}}.$

###### Proof.

$i)$ Since we have the similar expression of initial data (27) as in [16], we
refer the reader to Lemma 3.5 of [16], except for the bound on $\|\nabla
q(s_{0})\|_{L^{\infty}}$. Note that although $i)$ is not stated explicitly in
Lemma 3.5 of [16], they are clearly written in its proof. For $\|\nabla
q(s_{0})\|_{L^{\infty}}$, we use (27) and the fact that
$f^{\prime}(z)=-\frac{p-1}{2p}zf^{p}(z)$, $f^{p}(z),zf^{p-1}(z)$ and
$z^{2}f^{p-1}(z)$ are in $L^{\infty}(\mathbb{R})$ to derive

$\displaystyle|\nabla q(y,s_{0})|$
$\displaystyle\leq\left|\frac{\phi(s_{0})}{\kappa}\right|\left|\frac{f^{p}(z)}{\sqrt{s_{0}}}\left(pd_{0}zf^{p-1}(z)+d_{1}+pd_{1}z^{2}f^{p-1}(z)\right)\right|$
$\displaystyle\leq\frac{C}{\sqrt{s_{0}}}(|d_{0}|+|d_{1}|).$

$ii)$ We see from (28) and (29) that if $(d_{0},d_{1})$ is chosen so that
$(q_{0},q_{1})(s_{0})\in\left[-\frac{A}{s_{0}^{1+\nu}},\frac{A}{s_{0}^{1+\nu}}\right]^{2}$,
then $|d_{0}|$ and $|d_{1}|$ are bounded by $\frac{C}{s_{0}}$. Substituting
these bounds into the estimates stated in $i)$, we immediately derive $ii)$.
$iii)$ It follows from (28) and (29), part $ii)$ and the definition of $V_{A}$
given in Proposition 3. This ends the proof of Lemma 4. ∎

As stated in Theorem 1, the convergence holds in $W^{1,\infty}(\mathbb{R})$,
we need the following parabolic regularity estimate for equation (13), with
$q(s_{0})$ given by (18) and $q(s)\in V_{A}(s)$. More precisely, we have the
following:

######  Proposition 5.

For each $A\geq 1$, there exists $\delta_{2}(A)>0$ such that for all
$s_{0}\geq\delta_{2}(A)$: if $q(s)$ is a solution of equation (13) on
$[s_{0},s_{1}]$ with initial data at $s=s_{0}$, $q_{d_{0},d_{1}}(s_{0})$ given
in (18) where $(d_{0},d_{1})\in\mathcal{D}_{s_{0}}$, assume in addition that
$q(s)\in V_{A}(s)$ for $s\in[s_{0},s_{1}]$, then

$\|\nabla q(s)\|_{L^{\infty}}\leq\frac{CA^{2}}{s^{\varrho}},\quad\forall
s\in[s_{0},s_{1}],$

for some positive constant $C$.

###### Proof.

The proof is the same as Proposition 3.3 of [4]. We would like to mention that
the proof bases on a Gronwall’s argument and the following properties of the
kernel $e^{\theta\mathcal{L}}$ defined in (48):

$\forall g\in
L^{\infty},\;\;\|\nabla(e^{\theta\mathcal{L}}g)\|_{L^{\infty}}\leq\frac{Ce^{\theta/2}\|g\|_{L^{\infty}}}{\sqrt{1-e^{-\theta}}},$

and

$\forall f\in
W^{1,\infty},\;\;\|\nabla(e^{\theta\mathcal{L}}f)\|_{L^{\infty}}\leq
Ce^{\theta/2}\|\nabla f\|_{L^{\infty}}.$

Although the definition of $V_{A}$ is slightly different from the one defined
in [4], the reader will have absolutely no difficulty to adapt their proof to
the new situation. For that reason, we refer the reader to [4] for details of
the proof. ∎

###  3.2 Reduction to a finite dimensional problem

We are going to the crucial step of the proof of Theorem 1. In this step, we
will show that through a priori estimates, the control of $q(s)$ in $V_{A}$
reduces to the control of $(q_{0},q_{1})(s)$ in $\hat{V}_{A}(s)$. As presented
in [16] (see also[26],[14]), we would like to emphasize that this step make
the heart of the contribution. Even more, here lays another major contribution
of ours, in the sense that we understand better the dynamics of the
fundamental solution $\mathcal{K}(s,\sigma)$ defined in (19). Our sharper
estimates are given in Lemma 7 below. In fact all that we do is to rewrite the
corresponding estimates of Bricmont and Kupiainen [3] without taking into
account the particular form of the shrinking set they used. Furthermore,
because of the difference in the definition (12) of $\varphi$ and the
difference in the definition of $V_{A}$, the proof is far from being an
adaptation of the proof written in [16]. We therefore need some involved
arguments to control the components of $q$ and conclude the reduction to a
finite dimensional problem.

We mainly claim the following:

######  Proposition 6 (Control of $q(s)$ by $(q_{0},q_{1})(s)$ in $V_{A}(s)$).

There exist $A_{3}>0$ such that for each $A\geq A_{3}$, there exists
$\delta_{3}(A)>0$ such that for each $s_{0}\geq\delta_{3}(A)$, we have the
following properties:
\- if $(d_{0},d_{1})$ is chosen so that
$(q_{0},q_{1})(s_{0})\in\hat{V}_{A}(s_{0})$, and
\- if for all $s\in[s_{0},s_{1}]$, $q(s)\in V_{A}(s)$ and $q(s_{1})\in\partial
V_{A}(s_{1})$ for some $s_{1}\geq s_{0}$, then:
$i)\;$ (Reduction to a finite dimensional problem)
$\;\;(q_{0},q_{1})(s_{1})\in\partial\hat{V}_{A}(s_{1})$,
$ii)$ (Transversality) there exists $\eta_{0}>0$ such that for all
$\eta\in(0,\eta_{0})$,
$\;(q_{0},q_{1})(s_{1}+\eta)\not\in\partial\hat{V}_{A}(s_{1}+\eta)$ (hence,
$q(s_{1}+\eta)\not\in V_{A}(s_{1}+\eta)$).

The proof follows the general ideas of [16] and we proceed in three steps:
\- Step 1: we give a priori estimates on $q(s)$ in $V_{A}(s)$: assume that for
given $A>0$ lager, $\lambda>0$ and an initial time
$s_{0}\geq\sigma_{2}(A,\lambda)\geq 1$, we have $q(s)\in V_{A}(s)$ for each
$s\in[\tau,\tau+\lambda]$ where $\tau\geq s_{0}$, then using the integral form
(18) of $q(s)$, we derive new bounds on $q_{2}(s),q_{-}(s)$ and $q_{e}(s)$ for
$s\in[\tau,\tau+\lambda]$.
\- Step 2: we show that these new bounds are better than those defining
$V_{A}(s)$. It then remains to control $q_{0}(s)$ and $q_{1}(s)$. This means
that the problem is reduced to the control of a two dimensional variable
$(q_{0},q_{1})(s)$ and we then conclude $i)$ of Proposition 6.
\- Step 3: we use dynamics of $(q_{0},q_{1})(s)$ to show its transversality on
$\partial V_{A}(s)$, which corresponds to part $ii)$ of Proposition 6.

#### Step 1: A priori estimates on $q(s)$ in $V_{A}(s)$

As indicated above, the derivation of the new bounds on the components of
$q(s)$ bases on the integral formula (18). It is clear to see the strong
influence of the kernel $\mathcal{K}$ in this formula. Therefore, it is
convenient to give the following result from Bricmont and Kupiainen in [3]
which gives the dynamics of the linear operator $\mathcal{L}+V$:

######  Lemma 7 (Bricmont and Kupiainen [3]).

For all $\lambda>0$, there exists $\sigma_{0}=\sigma_{0}(\lambda)$ such that
if $\sigma\geq\sigma_{0}\geq 1$ and $\psi(\sigma)$ satisfies

$\sum_{m=0}^{2}|\psi_{m}(\sigma)|+\left\|\frac{\psi_{-}(y,\sigma)}{1+|y|^{3}}\right\|_{L^{\infty}}+\|\psi_{e}(\sigma)\|_{L^{\infty}}<+\infty,$
(30)

then, $\theta(s)=\mathcal{K}(s,\sigma)\psi(\sigma)$ satisfies for all
$s\in[\sigma,\sigma+\lambda]$,

$\displaystyle|\theta_{2}(s)|$
$\displaystyle\leq\left(\frac{\sigma}{s}\right)^{2}|\psi_{2}(\sigma)|+\frac{C(s-\sigma)}{s}\left(\sum_{l=0}^{2}|\psi_{l}(\sigma)|+\left\|\frac{\psi_{-}(y,\sigma)}{1+|y|^{3}}\right\|_{L^{\infty}}\right)$
$\displaystyle\qquad+C(s-\sigma)e^{-s/2}\|\psi_{e}(\sigma)\|_{L^{\infty}},$
(31)
$\displaystyle\left\|\frac{\theta_{-}(y,s)}{1+|y|^{3}}\right\|_{L^{\infty}}$
$\displaystyle\leq\frac{Ce^{s-\sigma}\left((s-\sigma)^{2}+1\right)}{s}\left(|\psi_{0}(\sigma)|+|\psi_{1}(\sigma)|+\sqrt{s}|\psi_{2}(\sigma)|\right)$
$\displaystyle\qquad+Ce^{-\frac{(s-\sigma)}{2}}\left\|\frac{\psi_{-}(y,\sigma)}{1+|y|^{3}}\right\|_{L^{\infty}}+\frac{Ce^{-(s-\sigma)^{2}}}{s^{3/2}}\|\psi_{e}(\sigma)\|_{L^{\infty}},$
(32) $\displaystyle\|\theta_{e}(s)\|_{L^{\infty}}$ $\displaystyle\leq
Ce^{s-\sigma}\left(\sum_{l=0}^{2}s^{l/2}|\psi_{l}(\sigma)|+s^{3/2}\left\|\frac{\psi_{-}(y,\sigma)}{1+|y|^{3}}\right\|_{L^{\infty}}\right)$
$\displaystyle\qquad+Ce^{-\frac{(s-\sigma)}{p}}\|\psi_{e}(\sigma)\|_{L^{\infty}}.$
(33)

where $C=C(\lambda,K)>0$ ($K$ is given in (23)), $\psi_{m},\psi_{-},\psi_{e}$
and $\theta_{m},\theta_{-},\theta_{e}$ are defined by (24) and (25).

######  Remark 4.

In view of the formula (18), we see that Lemma 7 will play an important role
in deriving the new bounds on the components of $q(s)$ and making our proof
simpler. This means that, given bounds on the components of
$q(\sigma),B(q(\tau)),R(\tau),N(q(\tau),\tau)$, we directly apply Lemma 7 with
$\mathcal{K}(s,\sigma)$ replaced by $\mathcal{K}(s,\tau)$ and then integrate
over $\tau$ to obtain estimates on the components of $q$.

###### Proof.

Let us mention that Lemma 7 relays mainly on the understanding of the behavior
of the kernel $\mathcal{K}(s,\sigma)$. The proof is essentially the same as in
[3], but those estimates did not present explicitly the dependence on all the
components of $\psi(\sigma)$ which is less convenient for our analysis below.
Because the proof is long and technical, we leave it to Appendix B. ∎

We now assume that for each $\lambda>0$, for each
$s\in[\sigma,\sigma+\lambda]$, we have $q(s)\in V_{A}(s)$ with $\sigma\geq
s_{0}$. Applying Lemma 7, we get new bounds on all terms in the right hand
side of (18), and then on $q$. More precisely, we claim the following:

######  Lemma 8.

There exists $A_{2}>0$ such that for each $A\geq A_{2}$, $\lambda^{*}>0$,
there exists $\sigma_{2}(A,\lambda^{*})>0$ with the following property: for
all $s_{0}\geq\sigma_{2}(A,\lambda^{*})$, for all $\lambda\leq\lambda^{*}$,
assume that for all $s\in[\sigma,\sigma+\lambda]$, $q(s)\in V_{A}(s)$ with
$\sigma\geq s_{0}$, then we have for all $s\in[\sigma,\sigma+\lambda]$,
$i)\;$ (linear term)

$\displaystyle|\alpha_{2}(s)|$
$\displaystyle\leq\left(\frac{\sigma}{s}\right)^{1-\nu}\frac{A^{2}}{s^{1+\nu}}+\frac{CA^{2}(s-\sigma)}{s^{2+\nu}},$
$\displaystyle\left\|\frac{\alpha_{-}(y,s)}{1+|y|^{3}}\right\|_{L^{\infty}}$
$\displaystyle\leq\frac{C}{s^{3/2+\varrho}}+\frac{C}{s^{3/2+\varrho}}\left(Ae^{-\frac{s-\sigma}{2}}+A^{2}e^{-(s-\sigma)^{2}}\right),$
$\displaystyle\|\alpha_{e}(s)\|_{L^{\infty}}$
$\displaystyle\leq\frac{C}{s^{\varrho}}+\frac{C}{s^{\varrho}}\left(Ae^{s-\sigma}+A^{2}e^{-\frac{s-\sigma}{p}}\right),$

where

$\mathcal{K}(s,\sigma)q(\sigma)=\alpha(y,s)=\sum_{m=0}^{2}\alpha_{m}(s)h_{m}(y)+\alpha_{-}(y,s)+\alpha_{e}(y,s).$

If $\sigma=s_{0}$, we assume in addition that $(d_{0},d_{1})$ is chosen so
that $(q_{0},q_{1})(s_{0})\in\hat{V}_{A}(s_{0})$. Then for all
$s\in[s_{0},s_{0}+\lambda]$, we have

$\displaystyle|\alpha_{2}(s)|\leq\frac{C}{s^{2}},\quad\left\|\frac{\alpha_{-}(y,s)}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{C}{s^{2}},\quad\|\alpha_{e}(s)\|_{L^{\infty}}\leq\frac{Ce^{s-s_{0}}}{\sqrt{s}}.$

$ii)\,$ (remaining terms)

$\displaystyle|\beta_{2}(s)|\leq\frac{C(s-\sigma)}{s^{2+\nu}},\quad\left\|\frac{\beta_{-}(y,s)}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{C}{s^{3/2+\varrho}},\quad\|\beta_{e}(s)\|_{L^{\infty}}\leq\frac{C}{s^{\varrho}},$

where

$\displaystyle\int_{\sigma}^{s}\mathcal{K}(s,\tau)\left[B(q(\tau))+R(\tau)+N(q(\tau),\tau)\right]d\tau$
$\displaystyle\qquad\qquad=\beta(y,s)=\sum_{m=0}^{2}\beta_{m}(s)h_{m}(y)+\beta_{-}(y,s)+\beta_{e}(y,s).$

###### Proof.

$i)$ It immediately follows from the definition of $V_{A}(\sigma)$ and Lemma
7. Note that in the case $\sigma=s_{0}$, we use in addition part $ii)$ of
Lemma 4 to have the conclusion. For part $ii)$, all what we need to do is to
find the estimates on the components of different terms appearing in equation
(13), then we use Lemma 7 and the linearity to have the conclusion. We claim
the following:

######  Lemma 9.

We have the following properties:
$i)$ (Estimates on $B(q)$) For all $A>0$, there exists $\sigma_{3}(A)$ such
that for all $\tau\geq\sigma_{3}(A)$, $q(\tau)\in V_{A}(\tau)$ implies

$m=0,1,2,\;|B_{m}(\tau)|\leq\frac{CA^{4}}{\tau^{2+2\nu}},\;\left|\frac{B_{-}(y,\tau)}{1+|y|^{3}}\right|\leq\frac{CA^{4}}{\tau^{3/2+2\varrho}},\;\|B_{e}(\tau)\|_{L^{\infty}}\leq\frac{CA^{2p^{\prime}}}{\tau^{\varrho
p^{\prime}}},$ (34)

where $p^{\prime}=\min\\{p,2\\}$.
$ii)$ (Estimates on $R$) There exists $\sigma_{4}>0$ such that for all
$\tau\geq\sigma_{4}$,

$\displaystyle m=0,1,\;$
$\displaystyle|R_{m}(\tau)|\leq\frac{C}{\tau^{2}},\quad|R_{2}(\tau)|\leq\frac{C}{\tau^{2+\nu}},$
and
$\displaystyle\left\|\frac{R_{-}(y,\tau)}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{C}{\tau^{2}},\quad\|R_{e}(\tau)\|_{L^{\infty}}\leq\frac{C}{\tau^{\nu}}.$
(35)

$iii)$ (Estimates on $N(q,\tau)$) For all $A>0$, there exists $\sigma_{5}(A)$
such that for all $\tau\geq\sigma_{5}(A)$, $q(\tau)\in V_{A}(\tau)$ implies

$m=0,1,2,\;|N_{m}(\tau)|\leq\frac{CA^{4}}{\tau^{2+2\nu}},\;\left\|\frac{N_{-}(y,\tau)}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{CA^{4}}{\tau^{2+2\varrho}},\;\|N_{e}(\tau)\|_{L^{\infty}}\leq\frac{CA^{4}}{\tau^{2\varrho}}.$
(36)

###### Proof.

Since the proof is technical, we leave it to Appendix C. ∎

Substituting the estimates stated in Lemma 9 into Lemma 7, then integrating
over $[\sigma,s]$ with respect to $\tau$, and taking
$\sigma_{2}(A,\lambda^{*})\geq\max\\{\sigma_{3},\sigma_{4},\sigma_{5}\\}$ such
that

$\forall
s\geq\sigma_{2}(A,\lambda^{*}),\;\;(A^{4}+1)e^{\lambda^{*}}((\lambda^{*})^{3}+1)\left(s^{-\varrho(p^{\prime}-1)}+s^{-(\nu-\varrho)}\right)\leq
1,$

with $p^{\prime}=\min\\{p,2\\}$, we have the conclusion. This ends the proof
of Lemma 8. ∎

Thanks to Lemma 9, we obtain the following equations satisfied by the
expanding modes:

######  Lemma 10 (ODE satisfied by the expanding modes).

For all $A>0$, there exists $\sigma_{6}(A)$ such that for all
$s\geq\sigma_{6}(A)$, $q(s)\in V_{A}(s)$ implies that for all
$s\geq\sigma_{6}(A)$,

$m=0,1,\quad\left|q_{m}^{\prime}(s)-(1-\frac{m}{2})q_{m}(s)\right|\leq\frac{C}{s^{3/2+\nu}},$
(37)

and

$\left|q_{2}^{\prime}(s)+\frac{2}{s}\;q_{2}(s)\right|\leq\frac{C}{s^{2+\nu}}.$
(38)

###### Proof.

The proof is very close to that in [16]. We therefore give the sketch of the
proof. By the definition (25), we write

$m=0,1,2,\;\;\frac{dq_{m}(s)}{ds}=\int\frac{\partial\chi(y,s)}{\partial
s}q(s)k_{m}\rho dy+\int\chi(y,s)\frac{\partial q(s)}{\partial s}k_{m}\rho
dy:=I+II.$

Since the support of $\frac{\partial\chi(y,s)}{\partial s}$ is the set
$K\sqrt{s}\leq|y|\leq 2K\sqrt{s}$ (see (23)), using the fact that
$\|q(s)\|_{L^{\infty}}\leq\frac{CA^{2}}{s^{\varrho}}$ (see (26)), we obtain

$|I|\leq\int\left|\frac{\partial\chi(y,s)}{\partial s}\right||q(s)||k_{m}|\rho
dy\leq CA^{2}e^{-s}s^{-\varrho},$

for $K$ large enough.
For $II$, we have by equation (13),

$\displaystyle II$ $\displaystyle=\int\chi(y,s)\mathcal{L}q(s)k_{m}\rho
dy+\int\chi(y,s)V(s)q(s)k_{m}\rho dy$
$\displaystyle+\int\chi(y,s)\left[B(q(s))+R(s)+N(q(s),s)\right]k_{m}\rho
dy:=IIa+IIb+IIc.$

Since $\mathcal{L}$ is self-adjoint on $L^{2}_{\rho}$ and
$\mathcal{L}(\chi(y,s)k_{m})=(1-\frac{m}{2})\chi(y,s)k_{m}+\frac{\partial^{2}\chi(y,s)}{\partial
s^{2}}k_{m}+\frac{\partial\chi(y,s)}{\partial s}(2\frac{\partial
k_{m}}{dy}-\frac{y}{2}k_{m})$, we obtain

$IIa=\int\mathcal{L}(\chi(y,s)k_{m})q(s)\rho
dy=(1-\frac{m}{2})q_{m}(s)+\mathcal{O}(CA^{2}e^{-s}),$

where $\mathcal{O}(r)$ stands for quantity whose absolute value is bounded
precisely by $r$ and not $Cr$.
Recalling from part $c)$ of Lemma B.1 that
$|V(y,s)|\leq\frac{C}{s}(1+|y|^{2})$ and from (26) that
$|q(y,s)|\leq\frac{CA^{2}}{s^{1+\nu}}(1+|y|^{3})$, we derive

$m=0,1,\;\;|IIb|\leq\frac{CA^{2}}{s^{2+\nu}}\int(1+|y|^{5})|k_{m}|\rho
dy\leq\frac{CA^{2}}{s^{2+\nu}}.$

For $m=2$, using the second estimate in part $c)$ of Lemma B.1, namely that
$V(y,s)=-\frac{h_{2}(y)}{4s}+\mathcal{O}\left(\frac{C(1+|y|^{4})}{s^{1+\bar{a}}}\right)$
with $\bar{a}=\min\\{a-1,a\\}$ in the case (2) and $\bar{a}=\min\\{a,1\\}$ in
the case (3), simultaneously noting that $\int h_{2}^{2}\rho dy=8$, $\int
h_{2}^{3}\rho dy=64$ and $2+\bar{a}+\nu\geq 2+2\nu$, we obtain

$m=2,\;\;IIb=-\frac{2}{s}\;q_{2}(s)+\mathcal{O}\left(\frac{CA^{2}}{s^{2+2\nu}}\right).$

The bound for $IIc$ already obtained from (34), (35) and (36). Adding all
these bounds and taking $\sigma_{6}(A)$ large enough such that for all
$s\geq\sigma_{6}(A)$, $A^{4}s^{-\nu}+A^{2}s^{2+\nu}e^{-s}\leq 1$, we then have
the conclusion. This ends the proof of Lemma 10. ∎

#### Step 2: Deriving conclusion $i)$ of Proposition 6

Here we use Lemma 8 in order to derive conclusion $i)$ of Proposition 6.
Indeed, from equation (18) and Lemma 8, we derive new bounds on $|q_{2}(s)|$,
$\left\|\frac{q_{-}(y,s)}{1+|y|^{3}}\right\|_{L^{\infty}}$ and
$\|q_{e}(s)\|_{L^{\infty}}$, assuming that for all
$s\in[\sigma,\sigma+\lambda]$, $q(s)\in V_{A}(s)$, for
$\lambda\leq\lambda^{*}$ and $\sigma\geq s_{0}\geq\sigma_{1}(A,\lambda^{*})$
($\sigma_{1}$ is given in Lemma 8). The key estimate is to show that for
$s=\sigma+\lambda$ (or $s\in[\sigma,\sigma+\lambda]$ if $\sigma=s_{0}$), these
bounds are better than those defining $V_{A}(s)$, provided that
$\lambda\leq\lambda^{*}(A)$. More precisely, we claim that following
proposition which directly follows $i)$ of Proposition 6:

######  Proposition 11 (Control of $q(s)$ by $(q_{0},q_{1})(s)$ in
$V_{A}(s)$).

There exist $A_{4}>1$ such that for each $A\geq A_{4}$, there exists
$\delta_{4}(A)>0$ such that for each $s_{0}\geq\delta_{4}(A)$, we have the
following properties:
\- if $(d_{0},d_{1})$ is chosen so that
$(q_{0},q_{1})(s_{0})\in\hat{V}_{A}(s_{0})$, and
\- if for all $s\in[s_{0},s_{1}]$, $q(s)\in V_{A}(s)$ for some $s_{1}\geq
s_{0}$, then: for all $s\in[s_{0},s_{1}]$,

$|q_{2}(s)|<\frac{A^{2}}{s^{1+\nu}},\quad\left\|\frac{q_{-}(y,s)}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{A}{2s^{3/2+\varrho}},\quad\|q_{e}(s)\|_{L^{\infty}}\leq\frac{A^{2}}{2s^{\varrho}}.$
(39)

Let us now derive the conclusion $i)$ of Proposition 6 from Proposition 11,
and we then prove it later.

###### Proof of $i)$ of Proposition 6.

Indeed, if $q(s_{1})\in\partial V_{A}(s_{1})$, we see from (39) and the
definition of $V_{A}(s)$ given in Proposition 3 that the first two components
of $q(s_{1})$ must be in $\partial\hat{V}_{A}(s_{1})$, which is the conclusion
of part $i)$ of Proposition 6, assuming Proposition 11 holds. ∎

We now give the proof of Proposition 11 in order to conclude the proof of part
$i)$ of Proposition 6.

###### Proof of Proposition 11.

Note that the conclusion of this proposition is very similar to Proposition
3.7, pages 157 in [16]. However, its proof is far from being an adaptation of
the proof given in the case of the semilinear heat equation treated in [16]
because of the difference of the definition of $V_{A}(s)$ and the presence of
the strong perturbation term. In fact, the argument given in [16] does not
work here to control $|q_{2}(s)|$ in this new situation, we use instead
equation (38) to handle this term.

Let $\lambda_{1}\geq\lambda_{2}$ be two positive numbers which will be fixed
in term of $A$ later. It is enough to show that (39) holds in two cases:
$s-s_{0}\leq\lambda_{1}$ and $s-s_{0}\geq\lambda_{2}$. In both cases, we use
Lemma 8 and equation and suppose $A\geq A_{2}>0$,
$s_{0}\geq\max\\{\sigma_{2}(A,\lambda_{1}),\sigma_{2}(A,\lambda_{2}),\sigma_{6}(A),1\\}$.

Case $s-s_{0}\leq\lambda_{1}$: Since we have for all $\tau\in[s_{0},s]$,
$q(\tau)\in V_{A}(\tau)$, we apply Lemma 8 with $A$ and
$\lambda^{*}=\lambda_{1}$, and $\lambda=s-s_{0}$. From (18) and Lemma 8, we
have

$|q_{2}(s)|\leq\frac{C}{s^{2}}+\frac{C\lambda_{1}}{s^{2+\nu}},\quad\quad\left\|\frac{q_{-}(y,s)}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{C}{s^{3/2+\varrho}},\quad\|q_{e}(s)\|_{L^{\infty}}\leq\frac{Ce^{\lambda_{1}}}{\sqrt{s}}+\frac{C}{s^{\varrho}}.$

If we fix $\lambda_{1}=\frac{3}{2}\log A$ and $A$ large enough, then (39)
satisfies.

Case $s-s_{0}\geq\lambda_{2}$: Since we have for all $\tau\in[\sigma,s]$,
$q(\tau)\in V_{A}(\tau)$, we apply Lemma 8 with $A$,
$\lambda=\lambda^{*}=\lambda_{2}$, $\sigma=s-\lambda_{2}$. From (18) and Lemma
8, we have

$\displaystyle\left\|\frac{q_{-}(y,s)}{1+|y|^{3}}\right\|_{L^{\infty}}$
$\displaystyle\leq\frac{C}{s^{3/2+\varrho}}\left(1+Ae^{-\frac{\lambda_{2}}{2}}+A^{2}e^{-\lambda_{2}^{2}}\right),$
$\displaystyle\|q_{e}(s)\|_{L^{\infty}}$
$\displaystyle\leq\frac{C}{s^{\varrho}}\left(1+Ae^{\lambda_{2}}+A^{2}e^{-\frac{\lambda_{2}}{p}}\right).$

To obtain (39), except for $|q_{2}(s)|$ which will be treated later, it is
enough to have $A\geq 4C$ and

$\displaystyle
C\left(Ae^{-\frac{\lambda_{2}}{2}}+A^{2}e^{-\lambda_{2}^{2}}\right)\leq\frac{A}{4},$
$\displaystyle
C\left(Ae^{\lambda_{2}}+A^{2}e^{-\frac{\lambda_{2}}{p}}\right)\leq\frac{A^{2}}{4}.$

If we fix $\lambda_{2}=\log(A/8C)$ and take $A$ large enough, we see that
these requests are satisfied. This follows the last two estimates in (39).

It remains to show that if $q(s)\in V_{A}(s)$ for all $s\in[s_{0},s_{1}]$ then
$|q_{2}(s)|<\frac{A^{2}}{s^{1+\nu}}$ for all $s\in[s_{0},s_{1}]$. We proceed
by contradiction, assume that for all $s\in[s_{0},s_{*})$,
$|q_{2}(s)|<\frac{A^{2}}{s^{1+\nu}}$ and
$|q_{2}(s_{*})|=\frac{A^{2}}{s_{*}^{1+\nu}}$. Considering the case
$q_{2}(s_{*})=-\frac{A^{2}}{s_{*}^{1+\nu}}$, we have

$q_{2}^{\prime}(s_{*})\leq\frac{d}{ds}\left(\frac{-A^{2}}{s_{*}^{1+\nu}}\right)\leq\frac{(1+\nu)A^{2}}{s_{*}^{2+\nu}}.$
(40)

On the other hand, we have from (38),

$q_{2}^{\prime}(s_{*})\geq-\frac{2}{s}\;q_{2}(s_{*})-\frac{C}{s_{*}^{2+\nu}}=\frac{2A^{2}-C}{s_{*}^{2+\nu}},$

which contradicts (40) if we take $A$ large enough.
Using the same argument in the case where
$q_{2}(s_{*})=\frac{A^{2}}{s_{*}^{1+\nu}}$, we also have a contradiction. This
completes the proof of Proposition 11 and part $i)$ of Proposition 6 too. ∎

#### Step 3: Deriving conclusion $ii)$ of Proposition 6.

We prove part $ii)$ of Proposition 6 here. In order to prove this, we follow
the ideas of [16] to show that for each $m\in\\{0,1\\}$ and each
$\iota\in\\{-1,1\\}$, if $q_{m}(s_{1})=\frac{\iota A}{s_{1}^{3/2+\nu}}$, then
$\frac{dq_{m}}{ds}(s_{1})$ has the opposite sign of
$\frac{d}{ds}\left(\frac{\iota A}{s^{3/2+\nu}}\right)(s_{1})$ so that
$(q_{0},q_{1})(s_{1})$ actually leaves $\hat{V}_{A}$ at $s_{1}$ for $s_{1}\geq
s_{0}$ where $s_{0}$ will be large enough. Indeed, from equation (37), we take
$A=2C+1$. If $\iota=1$, then $\frac{dq_{m}}{ds}(s_{1})>0$ and if $\iota=-1$,
then $\frac{dq_{m}}{ds}(s_{1})<0$. This implies that
$(q_{0},q_{1})(s_{1}+\eta)\not\in\partial\hat{V}_{A}(s_{1}+\eta)$ which yields
conclusion $ii)$ of Proposition 6.

###  3.3 Local in time solution of equation (13)

In the following, we find a local in time solution for equation (13).

######  Proposition 12 (Local in time solution of equation (13)).

For all $A>1$, there exists $\delta_{5}(A)$ such that for all
$s_{0}\geq\delta_{5}(A)$, the following holds: For all
$(d_{0},d_{1})\in\mathcal{D}_{s_{0}}$, there exists
$s_{max}(d_{0},d_{1})>s_{0}$ such that equation (13) with initial data
$q(s_{0})$ given in (27) has a unique solution satisfying $q(s)\in V_{A+1}(s)$
for all $s\in[s_{0},s_{max})$.

###### Proof.

Using the definition (12) of $q$ and the equivalent formulation (4), we see
that the Cauchy problem of (13) is equivalent to the Cauchy problem of
equation (1). Note that the initial data for (1) is derived the initial data
for (13) at $s=s_{0}$ given in (27), namely

$u_{d_{0},d_{1}}(x)=\frac{T^{-\frac{1}{p-1}}\phi(-\log
T)}{\kappa}\left\\{f(z)\left(1+\frac{d_{0}+d_{1}z}{p-1+\frac{(p-1)^{2}}{4p}z^{2}}\right)\right\\},$

where $f$ is defined in (8), $T=e^{-s_{0}}$ and $z=\frac{x}{\sqrt{T|\log
T|}}$.
This initial data is belong to $L^{\infty}(\mathbb{R})$ which insures the
local existence of $u$ in $L^{\infty}(\mathbb{R})$ (see the introduction).
From part $iii)$ of Lemma 4, we have $q_{d_{0},d_{1}}(s_{0})\in
V_{A}(s_{0})\subseteq V_{A+1}(s_{0})$. Then there exists $s_{max}$ such that
for all $s\in[s_{0},s_{max})$, we have $q(s)\in V_{A+1}(s)$. This concludes
the proof of Proposition 12. ∎

###  3.4 Deriving conclusion $ii)$ of Theorem 1

In this subsection, we derive conclusion $ii)$ of Theorem 1 using the previous
subsections. Although the derivation of the conclusion is the same as in [16],
we would like to give details of its proof for the reader’s convenience and
for explaining the two-point strategy: reduction to a finite dimensional
problem and the conclusion $ii)$ of Theorem 1 using index theory.

###### Proof of $ii)$ of Theorem 1.

We first solve the finite-dimensional problem and show the existence of $A>1$,
$s_{0}>0$ and $(d_{0},d_{1})\in\\\ mathcal{D}_{s_{0}}$ such that problem (13)
with initial data at $s=s_{0}$, $q_{d_{0},d_{1}}(s_{0})$ is given in (27) has
a solution $q(s)$ defined for all $s\in[s_{0},+\infty)$ such that

$q(s)\in V_{A}(s),\quad\forall s\in[s_{0},+\infty).$ (41)

For this purpose, let us take $A\geq A_{1}$ and $s_{0}\geq\delta_{3}$, where
$A_{1}$ and $\delta_{3}$ are given in Proposition 6, we will find the
parameter $(d_{0},d_{1})$ in the set $\mathcal{D}_{s_{0}}$ defined in Lemma 4
such that (41) holds. We proceed by contradiction and assume from $iii)$ of
Lemma 4 that for all $(d_{0},d_{1})\in\mathcal{D}_{s_{0}}$, there exists
$s_{*}(d_{0},d_{1})\geq s_{0}$ such that $q_{d_{0},d_{1}}(s)\in V_{A}(s)$ for
all $s\in[s_{0},s_{*}]$ and $q_{d_{0},d_{1}}(s_{*})\in\partial V_{A}(s_{*})$.
Applying Proposition 6, we see that $q_{d_{0},d_{1}}(s_{*})$ can leave
$V_{A}(s_{*})$ only by its first two components, hence,

$(q_{0},q_{1})(s_{*})\in\partial\hat{V}_{A}(s_{*}),$

(see Proposition 3 for the definition of $\hat{V}_{A}$). Therefore, we can
define the following function:

$\displaystyle\Phi\;:\mathcal{D}_{s_{0}}$
$\displaystyle\mapsto\partial([-1,1]^{2})$ $\displaystyle(d_{0},d_{1})$
$\displaystyle\to\frac{s_{*}^{1+\nu}}{A}(q_{0},q_{1})(s_{*}).$

Since $q(y,s_{0})$ is continuous in $(d_{0},d_{1})$ (see Lemma 4), we have
that $(q_{0},q_{1})(s)$ is continuous with respect to $(d_{0},d_{1},s)$. Then
using the transversality property of $(q_{0},q_{1})$ on $\partial\hat{V}_{A}$
(part $ii)$ of Proposition 6), we claim that $s_{*}(d_{0},d_{1})$ is
continuous. Therefore, $\Phi$ is continuous.
If we manage to prove that $\Phi$ is of degree one on the boundary, then we
have a contradiction from the degree theory. Let us prove that. From Lemma 4,
we see that if $(d_{0},d_{1})$ is on the boundary of $\mathcal{D}_{s_{0}}$,
then

$(q_{0},q_{1})(s_{0})\in\partial\hat{V}_{A}(s_{0}),\quad\text{and}\quad
q(s_{0})\in V_{A}(s_{0}),$

where the statement $q(s)\in V_{A}(s)$ holds with strict inequalities for
$q_{2},q_{-}$ and $q_{e}$. Using again $ii)$ of Proposition 6, we see that
$q(s)$ can leave $V_{A}(s)$ at $s=s_{0}$, hence $s_{*}=s_{0}$. Using $iii)$ of
Lemma 4, we have that the restriction of $\Phi$ to the boundary is of degree
$1$. This gives us a contradiction (by the index theory). Thus, there exists
$(d_{0},d_{1})\in\mathcal{D}_{s_{0}}$ such that for all $s\geq s_{0}$,
$q_{d_{0},d_{1}}(s)\in V_{A}(s)$, which is the conclusion (41).
Since $q_{d_{0},d_{1}}(s)$ satisfies (41), we use the parabolic estimate in
Proposition 5, the transformations (12) and (4) and the fact that
$\frac{\phi(s)}{\kappa}=1+\mathcal{O}(s^{-a})$ with $a>0$ to derive estimate
(9). This concludes the proof of $ii)$ of Theorem 1. ∎

###  3.5 Deriving conclusion $iii)$ of Theorem 1

We give the proof of part $iii)$ of Theorem 1 in this subsection. We consider
$u(t)$ solution of equation (1) which blows-up in finite-time $T>0$ at only
one blow-up point $x=0$ and satisfies (9). Adapting the techniques used by
Merle in [15] to equation (1) without the perturbation ($h\equiv 0$), we show
the existence of a profile
$u_{*}\in\mathcal{C}(\mathbb{R}\setminus\\{0\\},\mathbb{R})$ such that
$u(x,t)\to u_{*}(x)$ as $t\to T$ uniformly on compact subsets of
$\mathbb{R}\setminus\\{0\\}$, where $u_{*}$ is given in $iii)$ of Theorem 1.
Note that Zaag [26], Masmoudi and Zaag [14] successfully used these techniques
to equation (11). Since the proof is very similar to that written in [26] and
[14], and no new idea is needed, we just give the key argument and kindly
refer the reader to see Section 4 in [26] for details.

For each $x_{0}\in\mathbb{R}\setminus\\{0\\}$ small enough, we define for all
$(\xi,\tau)\in\mathbb{R}\times\left[-\frac{t(x_{0})}{T-t(x_{0})},1\right)$ the
following function:

$v(x_{0},\xi,\tau)=(T-t(x_{0}))^{\frac{1}{p-1}}u(x_{0}+\xi\sqrt{T-t(x_{0})},t(x_{0})+(T-t(x_{0}))\tau),$
(42)

where $t(x_{0})$ is uniquely defined by

$|x_{0}|=K_{0}\sqrt{(T-t(x_{0}))|\log(T-t(x_{0}))|},$ (43)

with $K_{0}>0$ to be fixed large enough later.
Note that $v$ blows up at time $\tau=1$ at only one blow-up point $x_{0}=0$.
From (1) and (42), we see that $v(x_{0},\xi,\tau)$ satisfies the following
equation: for all $\tau\in\left[-\frac{t(x_{0})}{T-t(x_{0})},1\right)$,

$\frac{\partial
v}{\partial\tau}=\Delta_{\xi}v+|v|^{p-1}v+(T-t(x_{0}))^{\frac{p}{p-1}}h\left((T-t(x_{0}))^{-\frac{1}{p-1}}v\right).$
(44)

From estimate (9), the definition (42) of $v$ and (43), we have the following:

$\sup_{|\xi|<|\log(T-t(x_{0}))|^{\frac{\varrho}{2}}}|v(x_{0},\xi,0)-f(K_{0})|\leq\frac{C}{|\log(T-t(x_{0}))|^{\frac{\varrho}{2}}}\to
0\quad\text{as}\;\;x_{0}\to 0,$

where $f$ is given in (8).
Using the continuity with respect to initial data for equation (1) (also for
equation (44)) associated to a space-localization in the ball
$B(0,|\xi|<|\log(T-t(x_{0}))|^{\frac{\varrho}{2}})$, it is showed in Section 4
of [26] that

$\sup_{|\xi|<|\log(T-t(x_{0}))|^{\frac{\varrho}{2}},0\leq\tau<1}|v(x_{0},\xi,\tau)-\hat{f}_{K_{0}}(\tau)|\leq\epsilon(x_{0})\to
0\quad\text{as}\;\;x_{0}\to 0,$

where
$\hat{f}_{K_{0}}(\tau)=\kappa(1-\tau+\frac{p-1}{4p}K_{0}^{2})^{-\frac{1}{p-1}}$.
Then letting $\tau\to 1$ and using the definition (42) of $v$, we have

$\displaystyle u_{*}(x_{0})=\lim_{t^{\prime}\to T}u(x,t^{\prime})$
$\displaystyle=(T-t(x_{0}))^{-\frac{1}{p-1}}\lim_{\tau\to 1}v(x_{0},0,\tau)$
$\displaystyle\sim(T-t(x_{0}))^{-\frac{1}{p-1}}\hat{f}_{K_{0}}(1),\quad\text{as}\;\;x_{0}\to
0.$

From (43), we have

$(T-t(x_{0}))^{-\frac{1}{p-1}}\sim\left(\frac{|x_{0}|^{2}}{2K_{0}^{2}|\log
x_{0}|}\right)^{-\frac{1}{p-1}},\quad\text{as}\;\;x_{0}\to 0.$

Hence,

$u_{*}(x_{0})\sim\left(\frac{8p|\log
x_{0}|}{(p-1)^{2}|x_{0}|^{2}}\right)^{\frac{1}{p-1}},\quad\text{as}\;\;x_{0}\to
0,$

which concludes the proof of part $iii)$ of Theorem 1.

##  Appendix A Appendix A

We claim the following:

######  Lemma A.1.

Let $\varepsilon\in(0,p]$, there exist $C=C(a,p,\mu,M)>0$ and
$s_{0}=s_{0}(a,\varepsilon)>0$ such that for all $s\geq s_{0}$,
$i)$ if $h$ is given by (2),

$j=0,1,\;\;e^{-\frac{(p-j)s}{p-1}}\left|h^{(j)}\left(e^{\frac{s}{p-1}}w\right)\right|\leq
Cs^{-a}\left(|w|^{p-j}+1\right),$

$ii)$ if $h$ is given by (3),

$\sum_{j=0}^{3}e^{-\frac{(p-j)s}{p-1}}|w|^{j}\left|h^{(j)}\left(e^{\frac{s}{p-1}}w\right)\right|\leq
Cs^{-a}(|w|^{p}+|w|^{p-\varepsilon}).$

###### Proof.

We see that the proof directly follows from the following key estimate:

$\frac{|w|^{\varepsilon}}{\log^{a}\left(2+e^{\frac{2s}{p-1}}w^{2}\right)}\leq\frac{C}{s^{a}}(|w|^{\varepsilon}+1),\quad\forall
s\geq s_{0}(a,\varepsilon).$ (45)

Indeed, considering the case $w^{2}e^{\frac{s}{p-1}}\geq 4$, we have

$\frac{|w|^{\varepsilon}}{\log^{a}\left(2+e^{\frac{2s}{p-1}}w^{2}\right)}\leq\frac{|w|^{\varepsilon}}{\log^{a}\left(4e^{\frac{s}{p-1}}\right)}\leq\frac{C|w|^{\varepsilon}}{s^{a}},$

then the case $w^{2}e^{\frac{s}{p-1}}\leq 4$ which follows that

$\frac{|w|^{\varepsilon}}{\log^{a}\left(2+e^{\frac{2s}{p-1}}w^{2}\right)}\leq\frac{|w|^{\varepsilon}}{\log^{a}(2)}\leq
Ce^{-\frac{\varepsilon s}{p-1}}\leq Cs^{-a}.$

This concludes the proof of (45) and the proof of Lemma A.1 also. ∎

The following lemma shows the existence of solutions of the associated ODE of
equation (5):

######  Lemma A.2.

Let $\phi$ be a positive solution of the following ordinary differential
equation:

$\phi_{s}=-\frac{\phi}{p-1}+\phi^{p}+e^{-\frac{ps}{p-1}}h\left(e^{\frac{s}{p-1}}\phi\right).$
(46)

Then $\phi(s)\to\kappa$ as $s\to+\infty$ and $\phi(s)$ is given by

$\phi(s)=\kappa(1+\eta_{a}(s))^{-\frac{1}{p-1}},\quad\text{where}\quad\eta_{a}(s)=\mathcal{O}\left(\frac{1}{s^{a}}\right).$
(47)

If $h(x)=\mu\frac{|x|^{p-1}x}{\log^{a}(2+x^{2})}$, then

$\eta_{a}(s)\sim
C_{0}\int_{s}^{+\infty}\frac{e^{s-\tau}}{\tau^{a}}d\tau=\frac{C_{0}}{s^{a}}\left(1+\sum_{j\geq
1}\frac{b_{j}}{s^{j}}\right),$

where $C_{0}=\mu\left(\frac{p-1}{2}\right)^{a}$ and
$b_{j}=(-1)^{j}\prod_{i=0}^{j-1}(a+i)$.

###### Proof.

See Lemma A.3 in [21]. ∎

##  Appendix B Proof of Lemma 7

In this appendix, we give the proof of Lemma 7. The proof follows from the
techniques of Bricmont and Kupiainen [3] with some additional care, since we
have a different profile function $\varphi$ defined in (12), and since we give
the explicit dependence of the bounds in terms of all the components of
initial data. As mentioned early, the proof relies mainly on the understanding
of the behavior of the kernel $\mathcal{K}(s,\sigma,y,x)$ (see (19)). This
behavior follows from a perturbation method around
$e^{(s-\sigma)\mathcal{L}}(y,s)$, where the kernel of $e^{t\mathcal{L}}$ is
given by Mehler’s formula:

$e^{t\mathcal{L}}(y,x)=\frac{e^{t}}{\sqrt{4\pi(1-e^{-t})}}exp\left[-\frac{(ye^{-t/2}-x)^{2}}{4(1-e^{-t})}\right].$
(48)

By definition (19) of $\mathcal{K}$, we use a Feynman-Kac representation for
$\mathcal{K}$:

$\mathcal{K}(s,\sigma,y,x)=e^{(s-\sigma)\mathcal{L}}(y,x)\int
d\mu_{yx}^{s-\sigma}(\omega)e^{\int_{0}^{s-\sigma}V(\omega(\tau),\sigma+\tau)d\tau},$
(49)

where $d\mu_{yx}^{s-\sigma}$ is the oscillator measure on the continuous paths
$\omega:[0,s-\sigma]\to\mathbb{R}$ with $\omega(0)=x$, $\omega(s-\sigma)=y$,
i.e. the Gaussian probability measure with covariance kernel

$\displaystyle\Gamma(\tau,\tau^{\prime})$
$\displaystyle=\omega_{0}(\tau)\omega_{0}(\tau^{\prime})$
$\displaystyle+2\left(e^{-\frac{1}{2}|\tau-\tau^{\prime}|}-e^{-\frac{1}{2}|\tau+\tau^{\prime}|}+e^{-\frac{1}{2}|2(s-\sigma)+\tau-\tau^{\prime}|}-e^{-\frac{1}{2}|2(s-\sigma)-\tau-\tau^{\prime}|}\right),$

which yields $\int d\mu_{yx}^{s-\sigma}\omega(\tau)=\omega_{0}(\tau)$, with

$\omega_{0}(\tau)=\left(\sinh((s-\sigma)/2)\right)^{-1}\left(y\sinh(\frac{\tau}{2})+x\sinh(\frac{s-\sigma-\tau}{2})\right).$

In view of (49), we can consider the expression for $\mathcal{K}$ as a
perturbation of $e^{(s-\sigma)\mathcal{L}}$. Since our profile $\varphi$
defined in (12) is different from the one defined in [3], we have here a
potential $V$ defined in (14) which is different as well. Thus, we first
estimate the potential $V$, then we restate some basic properties of the
kernel $\mathcal{K}$.

######  Lemma B.1 (Estimates on the potential $V$).

For $s$ large enough, we have
$a)\;\;$ $V(y,s)\leq\frac{C}{s^{a^{\prime}}}$ with $a^{\prime}=\min\\{a,1\\}$.
$b)$ $\;\;\left|\frac{d^{m}V(y,s)}{dy^{m}}\right|\leq\frac{C}{s^{m/2}}$ for
$m=0,1,2$.
$c)\;\;$ $|V(y,s)|\leq\frac{C}{s}(1+|y|^{2})$,
$\;\;V(y,s)=-\frac{h_{2}(y)}{4s}+\tilde{V}(y,s)$,
where
$|\tilde{V}(y,s)|=\mathcal{O}\left(\frac{1+|y|^{4}}{s^{2}}\right)+\mathcal{O}\left(\frac{1}{s^{a}}\right)$
in the case (2) and
$|\tilde{V}(y,s)|=\mathcal{O}\left(\frac{1+|y|^{4}}{s^{2}}\right)+\mathcal{O}\left(\frac{1+|y|^{2}}{s^{a+1}}\right)$
in the case (3). In particular, in both cases
$|\tilde{V}(y,s)|\leq\frac{C(1+|y|^{4})}{s^{1+\bar{a}}}$, where
$\bar{a}=\min\\{a-1,1\\}$ in the case (2) and $\bar{a}=\min\\{a,1\\}$ in the
case (3).

###### Proof.

$a)$ From the definition (14) of $V$, we see that

$V(y,s)\leq
p(\varphi(0,s)^{p-1}-\kappa^{p-1})+\imath\left|e^{-s}h^{\prime}\left(e^{\frac{s}{p-1}}\varphi\right)\right|,$

where $\imath$ is defined in (14). From Lemma A.2, we have

$\varphi(0,s)^{p-1}-\kappa^{p-1}=\kappa^{p-1}\left[(1+\eta_{a}(s))^{-1}\left(1+\frac{1}{2ps}\right)^{p-1}-1\right]\leq\frac{C}{s^{a^{\prime}}}.$

Since $|\varphi|$ is bounded, Lemma A.1 yields
$\imath\left|e^{-s}h^{\prime}\left(e^{\frac{s}{p-1}}\varphi\right)\right|\leq\frac{\imath
C}{s^{a}}$. This concludes part $a)$.
$b)$ We introduce $W(z,s)=V(y,s)$ with $z=\frac{y}{\sqrt{s}}$. In order to
derive part $b)$, it is enough to show that $|\frac{d^{m}W}{dz^{m}}|\leq C$
for $m=0,1,2$, which follows easily from Lemma A.1 and the following key
estimates

$\frac{\partial f(z)}{\partial z}=-\frac{zf(z)}{2p(1+c_{p}z^{2})},$

where $f$ and $c_{p}$ are defined in (8).
$c)$ Since $|V(y,s)|\leq C$ for all $y\in\mathbb{R}$ and $s\geq 1$,
considering the cases $|y|\leq\sqrt{s}$, then $|y|\geq\sqrt{s}$, we directly
see that the first estimate follows from the second. Hence, we only prove the
second. To do so, we introduce $\tilde{W}(Z,s)=V(y,s)$ with
$Z=\frac{|y|^{2}}{s}$. By the definition (12) and by a direct calculation, we
find that

$\displaystyle\frac{d^{2}\tilde{W}(Z,s)}{dZ^{2}}$
$\displaystyle=p(p-1)(p-2)\varphi^{p-3}(Z,s)\left(\frac{d\varphi(Z,s)}{dZ}\right)^{2}$
$\displaystyle+\imath
e^{-\frac{(p-3)s}{p-1}}h^{\prime\prime\prime}\left(e^{\frac{s}{p-1}}\varphi(Z,s)\right)\left(\frac{d\varphi(Z,s)}{dZ}\right)^{2}$
$\displaystyle+\left[p(p-1)\varphi^{p-2}(Z,s)+\imath
e^{-\frac{(p-2)s}{p-1}}h^{\prime\prime}\left(e^{\frac{s}{p-1}}\varphi(Z,s)\right)\right]\frac{d^{2}\varphi(Z,s)}{dZ^{2}}.$

Applying Lemma A.1 with $\varepsilon=\frac{p-1}{2}$, we see that

$\displaystyle\left|\frac{d^{2}\tilde{W}(Z,s)}{dZ^{2}}\right|$
$\displaystyle\leq
C\left(\varphi^{p-3}(Z,s)+\imath\varphi^{p-3-\frac{p-1}{2}}(Z,s)\right)\left(\frac{d\varphi(Z,s)}{dZ}\right)^{2}$
$\displaystyle+C\left(\varphi^{p-2}(Z,s)+\imath\varphi^{p-2-\frac{p-1}{2}}(Z,s)\right)\left|\frac{d^{2}\varphi(Z,s)}{dZ^{2}}\right|,\quad\forall
s\geq s_{0}.$

From the definition (12) of $\varphi$, we note that
$\frac{d\varphi}{dZ}=-\frac{c_{p}\phi}{\kappa}F^{p}(Z)$, where
$c_{p}=\frac{p-1}{4p}$ and $F(Z)=\kappa(1+c_{p}Z)^{-\frac{1}{p-1}}$, we derive

$\varphi(Z,s)^{p-3-\frac{p-1}{2}}\left(\frac{d\varphi(Z,s)}{dZ}\right)^{2}\leq
C\left(F+\frac{\kappa}{2ps}\right)^{p-3-\frac{p-1}{2}}F^{2p}\leq 2C,$

and

$\varphi(Z,s)^{p-2-\frac{p-1}{2}}\left|\frac{d^{2}\varphi(Z,s)}{dZ^{2}}\right|\leq
C\left(F+\frac{\kappa}{2ps}\right)^{p-2-\frac{p-1}{2}}F^{2p}\leq 2C.$

Hence, $\left|\frac{d^{2}\tilde{W}(Z,s)}{dZ^{2}}\right|$ is bounded for all
$Z\in[0,+\infty)$ and for all $s\geq s_{0}$. Then by a Taylor expansion, we
have

$\left|\tilde{W}(Z,s)-\tilde{W}(0,s)-Z\frac{\partial\tilde{W}(0,s)}{\partial
Z}\right|\leq CZ^{2},\quad\forall Z\in[0,+\infty),\;\;\forall s\geq s_{0}.$

From the definition (12) of $\varphi$ and from Lemma A.2, we have

$\displaystyle W(0,s)$
$\displaystyle=\frac{p}{p-1}\left[\frac{1}{(1+\eta_{a})}\left(1+\frac{1}{2ps}\right)^{p-1}-1\right]+\imath
e^{-s}h^{\prime}\left(e^{\frac{s}{p-1}}(\phi+\frac{\phi}{2ps})\right)$
$\displaystyle=\frac{1}{2s}-\frac{p}{p-1}\left(\frac{\eta_{a}(s)}{1+\eta_{a}(s)}\right)+\imath
e^{-s}h^{\prime}\left(e^{\frac{s}{p-1}}\phi\right)+\mathcal{O}\left(\frac{1}{s^{a+1}}\right)+\mathcal{O}\left(\frac{1}{s^{2}}\right).$

Recalling from Lemma A.2 that
$\eta_{a}(s)=\mathcal{O}\left(\frac{1}{s^{a}}\right)$, this immediately yields
$W(0,s)=\frac{1}{2s}+\mathcal{O}\left(\frac{1}{s^{a}}\right)+\mathcal{O}\left(\frac{1}{s^{2}}\right)$
in the case (2). In the case (3), we obtain by a direct calculation,

$\displaystyle\left|-\frac{p}{p-1}\left(\frac{\eta_{a}(s)}{1+\eta_{a}(s)}\right)+\imath
e^{-s}h^{\prime}\left(e^{\frac{s}{p-1}}\phi\right)\right|$
$\displaystyle\qquad\qquad=\left|-\frac{p}{(p-1)(1+\eta_{a}(s))}\left(\eta_{a}(s)-\frac{C_{0}}{s^{a}}\right)\right|+\mathcal{O}\left(\frac{1}{s^{a+1}}\right)=\mathcal{O}\left(\frac{1}{s^{a+1}}\right).$

In the last estimate, we used that fact that
$\eta_{a}(s)=\frac{C_{0}}{s^{a}}+\mathcal{O}\left(\frac{1}{s^{a+1}}\right)$ in
the case (3) (see Lemma A.2). Hence,
$W(0,s)=\frac{1}{2s}+\mathcal{O}\left(\frac{1}{s^{a+1}}\right)+\mathcal{O}\left(\frac{1}{s^{2}}\right)$
in the case (3).
For $\frac{\partial W(0,s)}{\partial Z}$, we use Lemmas A.1 and A.2 to derive

$\displaystyle\frac{\partial W(0,s)}{\partial Z}$
$\displaystyle=-\frac{1}{4(1+\eta_{a}(s))}\left(1+\frac{1}{2ps}\right)^{p-2}-\imath
e^{-\frac{(p-2)s}{p-1}}\frac{\phi}{4p}h^{\prime\prime}\left(e^{\frac{s}{p-1}}(\phi+\frac{\phi}{2ps})\right)$
$\displaystyle=-\frac{1}{4}+\mathcal{O}\left(\frac{1}{s^{a}}\right)+\mathcal{O}\left(\frac{1}{s}\right).$

Returning to $V$, we conclude part $c)$. This ends the proof of Lemma B.1. ∎

In what follows, we denote $\int f(y)g(y)\rho(y)dy$ by $\left\langle
f,g\right\rangle$ and write $\chi(y,s)=\chi(s)$ ($\chi$ is defined in (23)).
Let us now recall some basic properties of the kernel $\mathcal{K}$ stated in
[3]:

######  Lemma B.2 (Bricmont and Kupiainen [3]).

For all $s\geq\sigma\geq\max\\{s_{0},1\\}$ with $s\leq 2\sigma$ and $s_{0}$
given in Lemma A.1, for all $(y,x)\in\mathbb{R}^{2}$,
$a)$ $|\mathcal{K}(s,\sigma,y,x)|\leq Ce^{(s-\sigma)\mathcal{L}}(y,x)$.
$b)$
$\mathcal{K}(s,\sigma,y,x)=e^{(s-\sigma)\mathcal{L}}(y,x)\left(1+P_{2}(y,x)+P_{4}(y,x)\right)$,
where

$|P_{2}(y,x)|\leq\frac{C(s-\sigma)}{s}(1+|y|+|x|)^{2},$
$\text{and}\;\;|P_{4}(y,x)|\leq\frac{C(s-\sigma)(1+s-\sigma)}{s^{2}}(1+|y|+|x|)^{4}.$

Moreover, $\left|\left\langle
k_{2},\left(\mathcal{K}(s,\sigma)-\left(\frac{\sigma}{s}\right)^{2}\right)h_{2}\right\rangle\right|\leq\frac{C(s-\sigma)(1+s-\sigma)}{s^{1+\bar{a}}}$,
with $\bar{a}=\min\\{a-1,1\\}$ in the case (2) and $\bar{a}=\min\\{a,1\\}$ in
the case (3).
$c)$ $\|\mathcal{K}(s,\sigma)(1-\chi)\|_{L^{\infty}}\leq
Ce^{-\frac{(s-\sigma)}{p}}$.

###### Proof.

$a)$ From part $a)$ of Lemma B.1 and the definition (49) of $\mathcal{K}$, we
have

$\displaystyle|\mathcal{K}(s,\sigma,y,x)|$ $\displaystyle\leq
e^{(s-\sigma)\mathcal{L}}(y,x)\int
d\mu_{yx}^{s-\sigma}(\omega)e^{\int_{0}^{s-\sigma}C(\sigma+\tau)^{-a^{\prime}}d\tau}$
$\displaystyle\leq Ce^{(s-\sigma)\mathcal{L}}(y,x)\int
d\mu_{yx}^{s-\sigma}(\omega)\leq Ce^{(s-\sigma)\mathcal{L}}(y,x),$

since $s\leq 2\sigma$ and $d\mu_{yx}^{s-\sigma}$ is a probability.
$b)$ The proof is exactly the same as the corresponding one written in [3].
Although there is the difference of $\tilde{V}(y,s)$ given in part $c)$ of
Lemma B.1, this change does not affect the argument given in [3]. For that
reason, we refer the reader to Lemma 5, page 555 in [3] for details of the
proof.
$c)$ Our potential $V$ given in (14) has the same behavior as the potential in
[3] for $\frac{|y|^{2}}{s}$ and $s$ large (see (22)). For that reason, we
refer to Lemma, page 559 in [3] for its proof. ∎

Before going to the proof of Lemma 7, we would like to state some basic
estimates which will be used frequently in the proof.

######  Lemma B.3.

For $K$ large enough, we have the following estimates:
$a)$ For any polynomial $P$,

$\int P(y)\mathbf{1}_{\\{|y|\geq K\sqrt{s}\\}}\rho(y)dy\leq C(P)e^{-s}.$ (50)

$b)$ Let $p\geq 0$ and $|f(x)|\leq(1+|x|)^{p}$, then

$|(e^{t\mathcal{L}}f)(y)|\leq Ce^{t}(1+e^{-t/2}|y|)^{p},$ (51)

###### Proof.

$i)$ follows from a direct calculation. $ii)$ follows from the explicit
expression (48) by a simple change of variable. ∎

Let us now give the proof of Lemma 7.

###### Proof of Lemma 7.

We consider $\lambda>0$, let $\sigma_{0}\geq\lambda$, $\sigma\geq\sigma_{0}$
and $\psi(\sigma)$ satisfying (30). We want to estimate some components of
$\theta(y,s)=\mathcal{K}(s,\sigma)\psi(\sigma)$ for each
$s\in[\sigma,\sigma+\lambda]$. Since $\sigma\geq\sigma_{0}\geq\lambda$, we
have

$\sigma\leq s\leq 2\sigma.$ (52)

Therefore, up to a multiplying constant, any power of any $\tau\in[\sigma,s]$
will be bounded systematically by the same power of $s$.

$a)$ Estimate of $\theta_{2}$: We first write

$\displaystyle\theta_{2}(s)$ $\displaystyle=\left\langle
k_{2},\chi(s)\mathcal{K}(s,\sigma)\psi(\sigma)\right\rangle$
$\displaystyle=\sigma^{2}s^{-2}\psi_{2}(\sigma)+\left\langle
k_{2},(\chi(s)-\chi(\sigma))\sigma^{2}s^{-2}\psi(\sigma)\right\rangle$
$\displaystyle\qquad+\left\langle
k_{2},\chi(s)(\mathcal{K}(s,\sigma)-\sigma^{2}s^{-2})\psi(\sigma)\right\rangle:=\sigma^{2}s^{-2}\psi_{2}(\sigma)+Ib+IIb.$

To bound $Ib$, we write
$\psi(x,\sigma)=\sum_{l=0}^{2}\psi_{l}(\sigma)h_{l}(x)+\frac{\psi_{-}(x,\sigma)}{1+|x|^{3}}(1+|x|^{3})+\psi_{e}(x,\sigma)$
and use (50) to derive

$|Ib|\leq
C(s-\sigma)e^{-s}\sigma^{2}s^{-2}\left(\sum_{l=0}^{2}|\psi_{l}(\sigma)|+\left\|\frac{\psi_{-}(x,\sigma)}{1+|x|^{3}}\right\|_{L^{\infty}}+\|\psi_{e}(\sigma)\|_{L^{\infty}}\right).$

For $IIb$, we write

$\displaystyle IIb$ $\displaystyle=\sum_{l=0}^{2}\left\langle
k_{2},\chi(s)(\mathcal{K}(s,\sigma)-\sigma^{2}s^{-2})h_{l}\right\rangle\psi_{l}(\sigma)$
$\displaystyle+\left\langle
k_{2},\chi(s)(\mathcal{K}(s,\sigma)-\sigma^{2}s^{-2})\psi_{-}(\sigma)\right\rangle$
$\displaystyle+\left\langle
k_{2},\chi(s)(\mathcal{K}(s,\sigma)-\sigma^{2}s^{-2})\psi_{e}(\sigma)\right\rangle:=IIb.1+IIb.2+IIb.3.$

Let us bound $IIb.1$. For $l=2$, we already get from part $b)$ of Lemma B.2
and (50) that

$\left|\left\langle
k_{2},\chi(s)(\mathcal{K}(s,\sigma)-\sigma^{2}s^{-2})h_{2}\right\rangle\psi_{2}(\sigma)\right|\leq\frac{C(s-\sigma)(1+s-\sigma)}{s^{1+\bar{a}}}|\psi_{2}(\sigma)|,$

with $\bar{a}>0$.
For $l=0$ or $1$, we use $b)$ of Lemma B.2, (51), (50) and the fact that
$\left\langle k_{2},h_{l}\right\rangle=0$ and
$e^{(s-\sigma)\mathcal{L}}h_{l}=e^{(1-l/2)(s-\sigma)}h_{l}$ to find that

$\displaystyle\left|\left\langle
k_{2},\chi(s)(\mathcal{K}(s,\sigma)-\sigma^{2}s^{-2})h_{l}\right\rangle\psi_{l}(\sigma)\right|$
$\displaystyle\leq\left|\left\langle
k_{2},\chi(s)\left(\mathcal{K}(s,\sigma)-e^{(s-\sigma)\mathcal{L}}\right)h_{l}\right\rangle\right||\psi_{l}(\sigma)|$
$\displaystyle+\left|\left\langle
k_{2},\chi(s)\left(e^{(s-\sigma)\mathcal{L}}-\sigma^{2}s^{-2}\right)h_{l}\right\rangle\right||\psi_{l}(\sigma)|$
$\displaystyle\leq C(s-\sigma)\left(s^{-1}+e^{-s}\right)|\psi_{l}(\sigma)|$
$\displaystyle\leq\frac{C(s-\sigma)}{s}|\psi_{l}(\sigma)|.$

This yields

$|IIb.1|\leq\frac{C(s-\sigma)}{s}\sum_{l=0}^{2}|\psi_{l}(\sigma)|.$

If we write
$\psi_{-}(x,\sigma)=\frac{\psi_{-}(x,\sigma)}{1+|x|^{3}}(1+|x|^{3})$ and use
the same arguments as for $l=0$, we obtain

$|IIb.2|\leq\frac{C(s-\sigma)}{s}\left\|\frac{\psi_{-}(x,\sigma)}{1+|x|^{3}}\right\|_{L^{\infty}}.$

For $IIb.3$, we write

$\displaystyle IIb.3$ $\displaystyle=\left\langle
k_{2},\chi(s)\left(\mathcal{K}(s,\sigma)-e^{(s-\sigma)\mathcal{L}}\right)\psi_{e}(\sigma)\right\rangle$
$\displaystyle+\left\langle
k_{2},\chi(s)\left(e^{(s-\sigma)\mathcal{L}}-1\right)\psi_{e}(\sigma)\right\rangle+\left\langle
k_{2},\chi(s)(1-\sigma^{2}s^{-2})\psi_{e}(\sigma)\right\rangle.$

Using (50), we bound the last term by
$C(s-\sigma)e^{-\sigma}\|\psi_{e}(\sigma)\|_{L^{\infty}}\leq
C(s-\sigma)e^{-s/2}\|\psi_{e}(\sigma)\|_{L^{\infty}}$ from (52). For the
second term, we write
$e^{(s-\sigma)\mathcal{L}}-1=\int_{0}^{s-\sigma}d\tau\mathcal{L}e^{\tau\mathcal{L}}$
and use the fact that

$\sup_{|y|\leq 2K\sqrt{s},|x|\geq
K\sqrt{\sigma}}e^{-\frac{|y|^{2}}{4}-\frac{(ye^{-\tau/2}-x)^{2}}{4(1-e^{-\tau})}}\leq
e^{-2s},$ (53)

for $K$ large enough, then it is also bounded by
$C(s-\sigma)e^{-s}\|\psi_{e}(\sigma)\|_{L^{\infty}}$. For the first term, we
use $b)$ of Lemma B.2, (51) and again (53) to bound it by
$C(s-\sigma)s^{-1}e^{-s}\|\psi_{e}(\sigma)\|_{L^{\infty}}$. This yields

$|IIb.3|\leq C(s-\sigma)e^{-s/2}\|\psi_{e}(\sigma)\|_{L^{\infty}}.$

Collecting all these bounds yields the bound for $\theta_{2}(s)$ as stated in
(31).

$b)$ Estimate of $\theta_{-}$: By definition,

$\displaystyle\theta_{-}(y,s)$
$\displaystyle=P_{-}\left[\chi(s)\mathcal{K}(s,\sigma)\psi(\sigma)\right]=\sum_{l=0}^{2}\psi_{l}(\sigma)P_{-}\left[\chi(s)\mathcal{K}(s,\sigma)h_{l}\right]$
$\displaystyle+P_{-}\left[\chi(s)\mathcal{K}(s,\sigma)\psi_{-}(\sigma)\right]+P_{-}\left[\chi(s)\mathcal{K}(s,\sigma)\psi_{e}(\sigma)\right]:=Ic+IIc+IIIc.$

In order to bound $Ic$, we write
$\mathcal{K}(s,\sigma)=\mathcal{K}(s,\sigma)-e^{(s-\sigma)\mathcal{L}}+e^{(s-\sigma)\mathcal{L}}$,
then we use the fact that
$e^{(s-\sigma)\mathcal{L}}h_{l}=e^{(1-l/2)(s-\sigma)}h_{l}$, part $b)$ of
Lemma B.2 and (51) to derive for $l=0,1,2$,

$\displaystyle\left|\left(\mathcal{K}(s,\sigma)-e^{(s-\sigma)(1-l/2)}\right)h_{l}\right|$
$\displaystyle=\left|e^{(s-\sigma)\mathcal{L}}\left(P_{2}+P_{4}\right)h_{l}\right|$
$\displaystyle\leq\frac{Ce^{s-\sigma}(s-\sigma)}{s}\left(1+e^{-(s-\sigma)/2}|y|\right)^{2+l}$
$\displaystyle+\frac{Ce^{s-\sigma}(s-\sigma)(1+s-\sigma)}{s^{2}}\left(1+e^{-(s-\sigma)/2}|y|\right)^{4+l}.$

On the support of $\chi(s)$, namely when $|y|\leq 2K\sqrt{s}$, we can bound
$s^{-k/2}|y|^{k}$ by $C$ for $k\in\mathbb{N}$. Then, from the easy-to-check
fact that

$\text{if}\;\;|f(y)|\leq
m(1+|y|^{3}),\;\;\text{then}\;\;P_{-}\left[f(y)\right]\leq Cm(1+|y|^{3}),$
(54)

we obtain

$\displaystyle l=0,1,\;\;\;$ $\displaystyle
P_{-}\left[\psi_{l}(\sigma)\chi(s)\mathcal{K}(s,\sigma)h_{l}-\psi_{l}(\sigma)e^{(s-\sigma)(1-l/2)}(\chi(s)h_{l})\right]$
$\displaystyle\qquad\qquad\qquad\qquad\leq\frac{Ce^{s-\sigma}(s-\sigma)(1+s-\sigma)}{s}(1+|y|^{3})|\psi_{l}(\sigma)|,$

and

$\displaystyle
P_{-}\left[\psi_{2}(\sigma)\chi(s)\mathcal{K}(s,\sigma)h_{2}-\psi_{2}(\sigma)e^{(s-\sigma)(1-l/2)}(\chi(s)h_{2})\right]$
$\displaystyle\qquad\qquad\qquad\qquad\leq\frac{Ce^{s-\sigma}(s-\sigma)(1+s-\sigma)}{\sqrt{s}}(1+|y|^{3})|\psi_{2}(\sigma)|.$

Since $P_{-}(h_{l})=0$ and $|(1-\chi(y,s))h_{l}(y)|\leq
Cs^{-3/2+l/2}(1+|y|^{3})$, we have

$l=0,1,2,\;\;\left|\psi_{l}(\sigma)e^{(s-\sigma)(1-l/2)}P_{-}\left[\chi(s)h_{l}(y)\right]\right|\leq\frac{Ce^{s-\sigma}}{s^{3/2-l/2}}|\psi_{l}(\sigma)|(1+|y|^{3}).$

Hence,

$|Ic|\leq\frac{Ce^{s-\sigma}\left((s-\sigma)^{2}+1\right)}{s}\left(|\psi_{0}(\sigma)|+|\psi_{1}(\sigma)|+\sqrt{s}|\psi_{2}(\sigma)|\right)(1+|y|^{3}).$

To bound $IIIc$, we use $a)$ of Lemma B.2 and the definition (48) of
$e^{(s-\sigma)\mathcal{L}}$ to write

$\displaystyle\left\|\frac{\chi(y,s)\mathcal{K}(s,\sigma)\psi_{e}(x,\sigma)}{1+|y|^{3}}\right\|_{L^{\infty}}$
$\displaystyle\leq Ce^{s-\sigma}\|\psi_{e}(\sigma)\|_{L^{\infty}}$
$\displaystyle\sup_{|y|\leq 2K\sqrt{s},|x|\geq
K\sqrt{\sigma}}e^{-\frac{1}{2}\frac{(ye^{-(s-\sigma)/2}-x)^{2}}{4(1-e^{-(s-\sigma)})}}(1+|y|^{3})^{-1}$
$\displaystyle\leq\left\\{\begin{array}[]{ll}Cs^{-3/2}\|\psi_{e}(\sigma)\|_{L^{\infty}}&\text{if}\;s-\sigma\leq
s_{*}\\\ Ce^{-s}\|\psi_{e}(\sigma)\|_{L^{\infty}}&\text{if}\;s-\sigma\geq
s_{*}\end{array}\right.$

for a suitable constant $s_{*}$.
Exploiting again (54), we obtain the bound on this term which can be written
as

$|IIIc|\leq
Cs^{-3/2}e^{-(s-\sigma)^{2}}\|\psi_{e}(\sigma)\|_{L^{\infty}}(1+|y|^{3})\;\;\text{for
$\sigma$ large enough}.$

We still have to consider $IIc$. In order to bound this term, we proceed as in
[3]. We write

$\mathcal{K}(s,\sigma)\psi_{-}(\sigma)=\int
dxe^{x^{2}/4}\mathcal{K}(s,\sigma)(\cdot,x)f(x)=\int
dxN(\cdot,x)E(\cdot,x)f(x),$ (55)

where $f(x)=e^{-x^{2}/4}\psi_{-}(x,\sigma)$ and

$N(y,x)=\frac{e^{s-\sigma}e^{x^{2}/4}}{\sqrt{4\pi(1-e^{-(s-\sigma)})}}e^{-\frac{(ye^{-(s-\sigma)/2}-x)^{2}}{4(1-e^{-(s-\sigma)})}},$
$E(y,x)=\int
d\mu^{s-\sigma}_{yx}(\omega)e^{\int_{0}^{s-\sigma}V(\omega(\tau),\sigma+\tau)d\tau}.$

Let $f^{0}=f$ and for $m\geq 1$,
$f^{(-m-1)}(y)=\int_{-\infty}^{y}dxf^{(-m)}(x)$, then we have the following:

######  Lemma B.4.

$|f^{(-m)}(y)|\leq
C\left\|\frac{\psi_{-}(x,\sigma)}{1+|x|^{3}}\right\|_{L^{\infty}}(1+|y|)^{(3-m)}e^{-y^{2}/4}$
for $m\leq 3$.

###### Proof.

See Lemma 6, page 557 in [3]. ∎

We now rewrite (55) by integrating by parts as follows:

$\displaystyle\mathcal{K}(s,\sigma)\psi_{-}(\sigma)$
$\displaystyle=\sum_{l=0}^{2}(-1)^{l+1}\int
dx\partial_{x}^{l}N(y,x)\partial_{x}E(y,x)f^{(-l-1)}(x)$ $\displaystyle+\int
dx\partial_{x}^{3}N(y,x)E(y,x)f^{-3}(x).$ (56)

From the definition of $N(y,x)$, we have for $l=0,1,2,3$,

$|\partial_{x}^{l}N(y,x)|\leq
Ce^{-l(s-\sigma)/2}(1+|y|+|x|)^{l}e^{x^{2}/4}e^{(s-\sigma)\mathcal{L}}(y,x).$

Now using the integration by parts formula for Gaussian measures to write

$\displaystyle\partial_{x}E(y,x)=\frac{1}{2}\int_{0}^{s-\sigma}\int_{0}^{s-\sigma}d\tau
d\tau^{\prime}\partial_{x}\Gamma(\tau,\tau^{\prime})\int
d\mu_{yx}^{s-\sigma}(\omega)V^{\prime}(\omega(\tau),\sigma+\tau)$
$\displaystyle
V^{\prime}(\omega(\tau^{\prime}),\sigma+\tau^{\prime})e^{\int_{0}^{s-\sigma}d\tau^{\prime\prime}V(\omega(\tau^{\prime\prime}),\sigma+\tau^{\prime\prime})}$
$\displaystyle+\frac{1}{2}\int_{0}^{s-\sigma}d\tau\partial_{x}\Gamma(\tau,\tau)\int
d\mu_{yx}^{s-\sigma}(\omega)V^{\prime\prime}(\omega(\tau),\sigma+\tau)e^{\int_{0}^{s-\sigma}d\tau^{\prime\prime}V(\omega(\tau^{\prime\prime}),\sigma+\tau^{\prime\prime})}.$

Recalling from Lemma B.1 that $V(y,s)\leq\frac{C}{s^{a^{\prime}}}$ with
$a^{\prime}>0$ and
$\left|\frac{d^{m}V(y,s)}{dy^{m}}\right|\leq\frac{C}{s^{m/2}}$ for $m=0,1,2$.
Since $s\leq 2\sigma$, this yields
$\int_{0}^{s-\sigma}V(\omega(\tau),\sigma+\tau)d\tau\leq C$. Because
$d\mu^{s-\sigma}_{yx}$ is a probability, we then obtain

$|E(y,x)|\leq
C\quad\text{and}\quad|\partial_{x}E(y,x)|\leq\frac{C}{s}(s-\sigma)(1+s-\sigma)(|y|+|x|).$

Substituting all these bounds into (56), then using (51), Lemma B.4, the fact
that $s^{-1}(s-\sigma)(1+s-\sigma)\leq e^{-3/2(s-\sigma)}$ for $s$ large and
then (54), we derive

$|IIc|\leq
Ce^{-(s-\sigma)/2}\left\|\frac{\psi_{-}(x,\sigma)}{1+|x|^{3}}\right\|_{L^{\infty}}(1+|y|^{3}).$

Collecting all the bounds for $Ic,IIc$ and $IIIc$, we obtain the bound (32).

$c)$ Estimate for $\theta_{e}$: By definition, we write

$\theta_{e}(y,s)=(1-\chi(y,s))\mathcal{K}(s,\sigma)\psi(\sigma)=(1-\chi(y,s))\mathcal{K}(s,\sigma)\left(\psi_{b}(\sigma)+\psi_{e}(\sigma)\right).$

Using $c)$ of Lemma B.2, we have

$\|(1-\chi(y,s))\mathcal{K}(s,\sigma)\psi_{e}(\sigma)\|_{L^{\infty}}\leq
Ce^{-(s-\sigma)/p}\|\psi_{e}(\sigma)\|_{L^{\infty}}.$

It remains to bound $(1-\chi(y,s))\mathcal{K}(s,\sigma)\psi_{b}(\sigma)$. To
this end, we write

$\psi_{b}(x,\sigma)=\sum_{l=0}^{2}\psi_{l}(\sigma)h_{l}(x)+\frac{\psi_{-}(x,\sigma)}{1+|x|^{3}}(1+|x|^{3}),$

then we use $\chi(x,\sigma)|x|^{k}\leq C\sigma^{k/2}\leq Cs^{k/2}$ for
$k\in\mathbb{N}$, and $a)$ of Lemma B.2 to derive

$\displaystyle\|(1-\chi(y,s))\mathcal{K}(s,\sigma)\psi_{b}(x,\sigma)\|_{L^{\infty}}$
$\displaystyle\leq Ce^{s-\sigma}\sum_{l=0}^{2}s^{l/2}|\psi_{l}(\sigma)|$
$\displaystyle+Ce^{s-\sigma}s^{3/2}\left\|\frac{\psi_{-}(x,\sigma)}{1+|x|^{3}}\right\|_{L^{\infty}}.$

This yields the bound (33) and concludes the proof of Lemma 7. ∎

##  Appendix C Proof of Lemma 9

We give the proof of Lemma 9 here.

###### Proof of Lemma 9.

$i)$ From the definition (15) of $B$, we use a Taylor expansion and the
boundedness of $|\varphi|$ and $|q|$ to find that

$|\chi(\tau)B(q(\tau))|\leq C|q(\tau)|^{2}\quad\text{and}\quad|B(q(\tau))|\leq
C|q(\tau)|^{p^{\prime}},$ (57)

where $p^{\prime}=\min\\{2,p\\}$.
(Since we have the same definition of $B$ as in [16], we do not give the proof
of (57) and kindly refer the reader to Lemma 3.15, page 168 of [16] for its
proof.)
Using (57) and (26), we have

$\displaystyle|\chi(\tau)B(q(\tau))|\leq\frac{CA^{4}}{\tau^{3+2\varrho}}(1+|y|^{6})+\frac{CA^{4}}{\tau^{2+2\nu}}(1+|y|^{4}).$
(58)

From (58), we then derive for $m=0,1,2$,

$|B_{m}(\tau)|=\left|\int\chi(\tau)B(q(\tau))k_{m}\rho
dy\right|\leq\frac{CA^{4}}{\tau^{2+2\nu}}.$ (59)

Since $B_{-}(y,\tau)=\chi(\tau)B(q(\tau))-\sum_{m=0}^{2}B_{m}(\tau)h_{m}(y)$,
we have from (58) and (59),

$\displaystyle\left|\frac{B_{-}(y,\tau)}{1+|y|^{3}}\right|$
$\displaystyle\leq\left|\frac{\chi(\tau)B(q(\tau))}{1+|y|^{3}}\right|+\left|\frac{\sum_{m=0}^{2}B_{m}(\tau)h_{m}(y)}{1+|y|^{3}}\right|$
$\displaystyle\leq\chi(\tau)\left[\frac{CA^{4}}{\tau^{3+2\varrho}}(1+|y|^{3})+\frac{CA^{4}}{\tau^{2+2\nu}}(1+|y|)\right]+\frac{CA^{4}}{\tau^{2+2\nu}}\left(\frac{|\sum_{m=0}^{2}h_{m}(y)|}{1+|y|^{3}}\right)$

If we use $|y|^{l}\chi(y,\tau)\leq C\tau^{l/2}$ for $l\in\mathbb{N}$, and
$|\sum_{m=0}^{2}h_{m}(y)|\leq C(1+|y|^{2})$, then we obtain

$\left\|\frac{B_{-}(y,\tau)}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{CA^{4}}{\tau^{3/2+2\varrho}}.$

Using the second estimates in (57) and (26), we obviously obtain
$\|B(\tau)\|_{L^{\infty}}\leq\frac{CA^{2p^{\prime}}}{\tau^{\varrho
p^{\prime}}}$ which yields
$\|B_{e}(\tau)\|_{L^{\infty}}\leq\frac{CA^{2p^{\prime}}}{\tau^{\varrho
p^{\prime}}}$. This ends the proof of part $i)$.

$ii)$ From the definition (16) of $R$, we write
$\varphi(y,\tau)=\frac{\phi(\tau)}{\kappa}\vartheta(y,\tau)$ and
$R(y,\tau)=\frac{\phi(\tau)}{\kappa}Q+G$, where
$\vartheta(y,\tau)=f(\frac{y}{\sqrt{\tau}})+\frac{\kappa}{2p\tau}$ and

$\displaystyle Q(y,\tau)$
$\displaystyle=-\vartheta_{\tau}+\Delta\vartheta-\frac{y}{2}\nabla\vartheta-\frac{\vartheta}{p-1}+\vartheta^{p},$
(60) $\displaystyle G(y,\tau)$
$\displaystyle=-\frac{\phi^{\prime}}{\kappa}\vartheta-\frac{\phi}{\kappa}\vartheta^{p}+\phi^{p}\left(\frac{\vartheta}{\kappa}\right)^{p}+e^{\frac{-p\tau}{p-1}}h\left(e^{\frac{\tau}{p-1}}\frac{\phi}{\kappa}\vartheta\right).$
(61)

The conclusion of part $ii)$ is a direct consequence of the following:

######  Lemma C.1.

There exists $\sigma_{7}>0$ such that for all $\tau\geq\sigma_{7}$, we have
$i)$ (Estimates on $Q$)

$\displaystyle m=0,1,\;$
$\displaystyle|Q_{m}(\tau)|\leq\frac{C}{\tau^{2}},\quad|Q_{2}(\tau)|\leq\frac{C}{\tau^{3}},$
$\displaystyle\left\|\frac{Q_{-}(y,\tau)}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{C}{\tau^{2}},\quad\|Q_{e}(\tau)\|_{L^{\infty}}\leq\frac{C}{\sqrt{\tau}}.$
(62)

$ii)$ (Estimates on $G$)

$m=0,1,2,\;|G_{m}(\tau)|\leq\frac{C}{\tau^{1+a^{\prime}}},\;\left\|\frac{G_{-}(y,\tau)}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{C}{\tau^{1+a^{\prime}}},\;\|G_{e}(\tau)\|_{L^{\infty}}\leq\frac{C}{\tau^{a}},$
(63)

where $a^{\prime}=a>1$ in the case (2) and $a^{\prime}=a+1>1$ in the case (3).

###### Proof.

$i)$ See page 563 in [3]. For part $ii)$, one can see that it is a direct
consequence of the following:

$|G(y,\tau)|\leq\frac{C}{\tau^{a}}\quad\text{and}\quad|\chi(\tau)G(y,\tau)|\leq\frac{C}{\tau^{1+a^{\prime}}}(1+|y|^{2}).$
(64)

By the definition of $G_{m},G_{-}$ and $G_{e}$, part $ii)$ simply follows from
(64). By the linearity, this also concludes the proof of part $ii)$ of Lemma
9.

Let us now give the proof of (64). For the first estimate, we use the
definition (61) of $G$, Lemmas A.1 and A.2,

$\displaystyle|G(y,\tau)|\leq\left|\frac{\phi^{\prime}\vartheta}{\kappa}\right|+\left|\frac{\phi\vartheta}{\kappa}\right|\left|1-\frac{\phi^{p-1}}{\kappa^{p-1}}\right|+\left|e^{-\frac{ps}{p-1}}h\left(e^{\frac{s}{p-1}}\frac{\phi\vartheta}{\kappa}\right)\right|\leq\frac{C}{s^{a}}.$

For the second estimate in (64), we use the fact that $\phi$ satisfies (6) and
write

$\displaystyle G(y,\tau)$
$\displaystyle=\frac{\vartheta\phi}{\kappa^{p}}(\kappa^{p-1}-\phi^{p-1})(\kappa^{p-1}-\vartheta^{p-1})$
$\displaystyle+e^{-\frac{p\tau}{p-1}}\left[h\left(e^{\frac{\tau}{p-1}}\frac{\phi\vartheta}{\kappa}\right)-h\left(e^{\frac{\tau}{p-1}}\phi\right)\right]$
$\displaystyle+\left(1-\frac{\vartheta}{\kappa}\right)e^{-\frac{p\tau}{p-1}}h\left(e^{\frac{\tau}{p-1}}\phi\right):=\bar{G}+\tilde{G}+\hat{G}.$

Noting that
$\vartheta(y,\tau)=\kappa\left(1-\frac{h_{2}(y)}{4p\tau}+\mathcal{O}\left(\frac{|y|^{4}}{\tau^{2}}\right)\right)$
uniformly for $y\in\mathbb{R}$ and $\tau\geq 1$, and recalling from Lemma A.2
that $\phi(\tau)=\kappa(1+\eta_{a}(\tau))^{-\frac{1}{p-1}}$ where
$\eta_{a}(\tau)=\mathcal{O}(\tau^{-a})$, then using a Taylor expansion, we
derive

$\displaystyle\bar{G}(y,\tau)$
$\displaystyle=\frac{\phi\eta_{a}(\tau)}{1+\eta_{a}(\tau)}\left(\frac{h_{2}(y)}{4p\tau}+\mathcal{O}\left(\frac{|y|^{4}}{\tau^{2}}\right)\right),$
$\displaystyle\tilde{G}(y,\tau)$ $\displaystyle=-\phi
e^{-\tau}h^{\prime}\left(e^{\frac{\tau}{p-1}}\phi\right)\left(\frac{h_{2}(y)}{4p\tau}+\mathcal{O}\left(\frac{|y|^{4}}{\tau^{2}}\right)\right),$
$\displaystyle\hat{G}(y,\tau)$
$\displaystyle=e^{-\frac{p\tau}{p-1}}h\left(e^{\frac{\tau}{p-1}}\phi\right)\left(\frac{h_{2}(y)}{4p\tau}+\mathcal{O}\left(\frac{|y|^{4}}{\tau^{2}}\right)\right).$

This yields the second estimate in (64) in the case (2). If $h$ is given by
(3), we have furthermore

$\displaystyle\left|\frac{\phi\eta_{a}(\tau)}{1+\eta_{a}(\tau)}-e^{-s}h^{\prime}(e^{\frac{s}{p-1}}\phi)\phi+e^{-\frac{ps}{p-1}}h(e^{\frac{s}{p-1}}\phi)\right|$
$\displaystyle\leq\left|\frac{\phi}{1+\eta_{a}(\tau)}\right|\left|\eta_{a}(\tau)-\frac{\mu}{\log^{a}\left(2+e^{\frac{2\tau}{p-1}}\phi^{2}(\tau)\right)}\right|+\frac{C}{\tau^{a+1}}\leq\frac{2C}{\tau^{1+a}},$

which yields the second estimate in (64) in the case (3). This concludes the
proof of (64) and the proof of part $ii)$ of Lemma 9 also. ∎

$iii)$ From the definition (17) of $N$, we use a Taylor expansion for $N$ to
find that in the case (2),

$N(q(\tau),\tau)=e^{-\tau}h^{\prime}\left(e^{\frac{\tau}{p-1}}(\phi(\tau)+\theta_{1}q(\tau))\right)q(\tau)\quad\text{with}\quad\theta_{1}\in[0,1],$

and in the case (3),

$N(q(\tau),\tau)=e^{-\frac{(p-2)\tau}{p-1}}h^{\prime\prime}\left(e^{\frac{\tau}{p-1}}(\phi(\tau)+\theta_{2}q(\tau))\right)q^{2}(\tau)\quad\text{with}\quad\theta_{2}\in[0,1].$

Since $\varphi(\tau)\to\kappa$ and $\|q(\tau)\|_{L^{\infty}(\mathbb{R})}\to 0$
as $\tau\to+\infty$, this implies that there exists $\tau_{0}$ large enough
such that
$\frac{\kappa}{2}\leq|\phi(\tau)+\theta_{i}q(\tau)|\leq\frac{3\kappa}{2}$ for
all $\tau\geq\tau_{0}$ and $y\in\mathbb{R}$. Then by Lemma A.1, we have
$|N(q(\tau),\tau)|\leq\frac{C|q|^{\beta}}{\tau^{a}}$ where $\beta=1$ in the
case (2) and $\beta=2$ in the case (3), which implies part $iii)$ of Lemma 9.
This concludes the proof of Lemma 9. ∎

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