#  Profile of a touch-down solution to a nonlocal MEMS Model

###### Abstract.

In this paper, we are interested in the mathematical model of MEMS devices
which is presented by the following equation on $(0,T)\times\Omega:$

$\displaystyle\partial_{t}u=\Delta
u+\displaystyle\frac{\lambda}{(1-u)^{2}\left(1+\displaystyle\gamma\int_{\Omega}\frac{1}{1-u}dx\right)^{2}},\quad
0\leq u<1,$

where $\Omega$ is a bounded domain in $\mathbb{R}^{n}$ and $\lambda,\gamma>0$.
In this work, we have succeeded to construct a solution which quenches in
finite time T only at one interior point $a\in\Omega$. In particular, we give
a description of the quenching behavior according to the following final
profile

$1-u(x,T)\sim\theta^{*}\left[\frac{|x-a|^{2}}{|\ln|x-a||}\right]^{\frac{1}{3}}\text{
as }x\to a,\theta^{*}>0.$

The construction relies on some connection between the quenching phemonenon
and the blowup phenomenon. More precisely, we change our problem to the
construction of a blowup solution for a related PDE and describe its behavior.
The method is inspired by the work of Merle and Zaag [MZ97a] with a suitable
modification. In addition to that, the proof relies on two main steps: A
reduction to a finite dimensional problem and  a topological argument based on
Index theory. The main difficulty and novelty of this work is that we handle
the nonlocal integral term in the above equation. The interpretation of the
finite dimensional parameters in terms of the blowup point and the blowup time
allows to derive the stability of the constructed solution with respect to
initial data.

###### Key words and phrases:

Blowup solution, Blowup profile, Stability, Semilinear heat equation, non
variation heat equation

###### 1991 Mathematics Subject Classification:

Primary: 35K50, 35B40; Secondary: 35K55, 35K57.

Giao Ky DUONG 111 G. K. Duong is supported by the project INSPIRE. This
project has received funding from the European Union’s Horizon 2020 research
and innovation programme under the Marie Skłodowska-Curie grant agreement No
665850. and Hatem ZAAG

Université Paris 13, Sorbonne Paris Cité, LAGA, CNRS UMR 7539, Villetaneuse,
France.

##  1\. Introduction.

We are interested in the motion of some elastic membranes which is usually
found in Micro-Electro Mechanical System (MEMS) devices, which are available
in a variety of electronic devices, such as microphones, transducers, sensors,
actuators and so on. Described briefly, MEMS devices contain an elastic
membrane which is hanged above a rigid ground plate connected in series with a
fixed voltage source and a fixed capacitor. For more details on the physical
background and possible applications, we refer the reader to [FMPS07],
[Guo14], [PB02] and [PT01].

For a MEMS device (in [GS15] and [Guo14]), the distance between the rigid
ground plate and the elastic membrane changes with time. It is refered to as
the deflection of the membrane. Here, we assume that this distance is very
small compared to the device. In fact, we can fully describe the behavior of
the deflection by the following hyperbolic equation

$\left\\{\begin{array}[]{rcl}\varepsilon^{2}\partial_{tt}u+\partial_{t}u&=&\Delta
u+\displaystyle\frac{\lambda
f(x,t)}{(1-u)^{2}\left(1+\gamma\displaystyle\int_{\Omega}\frac{1}{1-u}dx\right)^{2}},\quad
x\in\Omega,t>0,\\\\[14.22636pt] \\\ u(x,t)&=&0,x\in\partial\Omega,t>0,\\\ \\\
\\\ u(x,0)&=&u_{0}(x),x\in\bar{\Omega}.\end{array}\right.$ (1.1)

where $\Omega$ is considered as the domain of the rigid plate, $u$ is the
deflection of the membrane to the plate, $\lambda>0,\gamma>0$ and $f$ is
continuous. Here, the distance between the rest position of the membrane and
the rigid plate is normalized to 1. When the device is under voltage, $u$ will
vary in the interval $[0,1)$. In addition to that, the parameter $\lambda$
represents the ratio of the reference electrostatic force to the reference
elastic force and $\varepsilon$ is the ratio of the interaction of the
inertial and damping terms in our model. Moreover, the function $f$ represents
the varying dielectric properties of the membrane. See [GK12] for more
details.

In fact, we are interested in a simpler case of (1.1) considered in the
following parabolic equation:

$\left\\{\begin{array}[]{rcl}\partial_{t}u&=&\Delta
u+\displaystyle\frac{\lambda}{(1-u)^{2}\left(1+\displaystyle\gamma\int_{\Omega}\frac{1}{1-u}dx\right)^{2}},\quad
x\in\Omega,t>0,\\\\[19.91684pt] u(x,t)&=&0,x\in\partial\Omega,t>0,\\\
u(x,0)&=&u_{0}(x),x\in\Omega.\end{array}\right.$ (1.2)

Moreover, we are also interested in the following generalization of problem
(1.2):

$\left\\{\begin{array}[]{rcl}\partial_{t}u&=&\Delta
u+\displaystyle\frac{\lambda}{(1-u)^{p}\left(1+\displaystyle\gamma\int_{\Omega}\frac{1}{1-u}dx\right)^{q}},\quad
x\in\Omega,t>0,\\\\[19.91684pt] u(x,t)&=&0,x\in\partial\Omega,t>0,\\\
u(x,0)&=&u_{0}(x),x\in\Omega,\end{array}\right.$ (1.3)

where $p,q>0$. Introducing

$Q_{T}=(0,T)\times\Omega,\text{ where }T>0,$ (1.4)

we say that $u$ is a classical solution of (1.2) (in the sense of Proposition
1.2.2 page 13 in Kavallaris and Suzuki [KS18]) if $u$ is a function in
$C^{2,1}(Q_{T})\cap C(\bar{Q}_{T})$ that satisfies (1.2) at every point in
$Q_{T}$ as well as the boundary and initial conditions, with

$u(x,t)\in[0,1),\forall x\in\Omega,t\in(0,T).$

According to the above mentioned reference in [KS18], the local Cauchy problem
of (1.2) is solved. Then, either our solution is global in time or there
exists $T>0$ such that

$\liminf_{t\to T}\left[\min_{x\in\bar{\Omega}}\\{1-u(t,x)\\}\right]=0.$ (1.5)

We can see that if the above condition occurs, the right-hand side of (1.2)
may become singular. This phenomenon is refered to as  touch-down  in finite
time $T$ in reference to the physical phenomenon, where the membrane ”touches”
the rigid ground plate which is placed below. In fact, in our setting, we
follow the literature and place the regid plate at $u=1$, above the membrane
which is located at $u(x,t)$. Note that in case of touch-down, the MEMS device
breaks down.

Mathematically, we may refer to the behavior in (1.5) as finite-time
quenching. Moreover, $a\in\Omega$ is a quenching point if and only if there
exist sequences $(a_{n},t_{n})\in\Omega\times(0,T)$ such that

$u(a_{n},t_{n})\to 1,\text{ as }n\to+\infty.$

The  touch-down phenomenon has been strongly studied in recent decades. In one
space dimension, we would like to mention the paper by Guo, Hu and Wang in
[GHW09] who gave a sufficient condition for quenching, and also a lower bound
on the quenching final profile (see Remark 1.5 below). There is also the paper
by Guo and Hu in [GH18] who find a constant limit for the similarity variables
version valid only on compact sets, and yielding the quenching rate.

In higher dimensions, let us for example mention the following result by Guo
and Kavallaris in [GK12]:

Consider $\Omega$ such that $|\Omega|\leq\frac{1}{2}$. Then, for all
$\lambda>0$ fixed and $\gamma>0$, there exist initial data with a small energy
such that problem (1.2) has a solution which quenches in finite time.

In our paper, we are interested in proving a general quenching result with no
restriction on any $\lambda>0,\gamma>0$ and $C^{2}$ bounded domain $\Omega$.
In fact, we do much better that [GH18] and [GHW09], and give a sharp
decription of the asymptotics of the solution near the quenching region. The
following is the main result:

######  Theorem 1.1 (Existence of a touch-down solution).

Consider $\lambda>0,\gamma>0$ and $\Omega$ a $C^{2}$ bounded domain in
$\mathbb{R}^{n},$ containing the origin. Then, there exist initial data
$u_{0}\in C^{\infty}(\bar{\Omega})$ such that the solution of (1.2) quenches
in finite time $T=T(u_{0})>0$ only at the origin. In particular, the following
holds:

  * $(i)$

The intermidiate profile: For all $t\in[0,T)$

$\left\|\frac{(T-t)^{\frac{1}{3}}}{1-u(.,t)}-\theta^{*}\left(3+\frac{9}{8}\frac{|.|^{2}}{\sqrt{(T-t)|\ln(T-t)|}}\right)^{-\frac{1}{3}}\right\|_{L^{\infty}(\Omega)}\leq\frac{C}{\sqrt{|\ln(T-t)|}},$
(1.6)

for some $\theta^{*}=\theta^{*}(\lambda,\gamma,\Omega,T)>0$.

  * $(ii)$

The final profile: There exists $u^{*}\in C^{2}(\Omega)\cap C(\bar{\Omega})$
such that $u$ uniformly converges to $u^{*}$ as $t\to T,$ and

$1-u^{*}(x)\sim\theta^{*}\left[\frac{9}{16}\frac{|x|^{2}}{|\ln|x||}\right]^{\frac{1}{3}}\text{
as }x\to 0.$ (1.7)

######  Remark 1.2.

Note that when $\gamma=0$, our problem coincides with the work of Filippas and
Guo [FG93] and also Merle and Zaag [MZ97a]. Our paper is then meaningful when
$\gamma\neq 0$, and the whole issue is how to control the non local term. Note
that [FG93] derived the final quenching profile, however, only in one space
dimension, whereas [MZ97a] constructed a quenching solution in higher
dimensions, proved its stability with respect to initial data, and gave its
intermediate and final profiles.

######  Remark 1.3.

For simplicity, we choose to write our result when the solution quenches at
the origin. Of course, we can make it quenches at any arbitrary $a\in\Omega$,
simply replace $x$ by $x-a$ in the statement.

######  Remark 1.4.

In Theorem 1.1, we can describe the evolution of our solution at $x=0$ as
follows:

$1-u(0,t)\sim\frac{(T-t)^{\frac{1}{3}}}{\sqrt[3]{3}\theta^{*}},\text{ as }t\to
T.$

######  Remark 1.5.

From (1.7), we see that the final profile $u^{*}$ has a cusp at the origin
which is equivalent to

$\frac{C_{0}|x|^{\frac{2}{3}}}{|\ln|x||^{\frac{1}{3}}}.$

This description is in fact much better than the result of Guo, Hu and Wang in
[GHW09] who gave some sufficient conditions for quenching in one space
dimension, and proved the existence of a cusp at the quenching point bounded
from below by $C(\beta)|x|^{\beta}$ for any
$\beta\in\left(\frac{2}{3},1\right)$, which is less accurate than our estimate
(1.7).

######  Remark 1.6.

Note that we can explicitly write the formula of the initial data

$u(x,0)=\frac{\bar{u}(x,0)}{\bar{u}(x,0)+1},$ (1.8)

where

$\bar{u}(x,0)=\frac{\bar{\theta}(0)}{\lambda^{\frac{1}{3}}}U(x,0),$

with

$U(x,0)=T^{-\frac{1}{3}}\left[\varphi(\frac{x}{\sqrt{T}},-\ln
T)+\left(d_{0}+d_{1}\cdot
z\right)\chi_{0}\left(\frac{16|z|}{K_{0}^{2}}\right)\right]\chi_{1}(x,0)+(1-\chi_{1}(x,0))H^{*}(x),$
$z=\frac{x}{\sqrt{T|\ln T|}},$
$\chi_{1}(x,0)=\chi_{0}\left(\frac{|x|}{\sqrt{T}|\ln T|}\right),$

and $\bar{\theta}(0)$ is the unique positive solution of the following
equation

$\bar{\theta}(0)=\left(1+\gamma|\Omega|+\frac{\gamma}{\sqrt[3]{\lambda}}\bar{\theta}(0)\int_{\Omega}U(0)dx\right)^{\frac{2}{3}},$

and $\chi_{0},\varphi,H^{*}$ are defined in (2.18), (2.23), (3.12),
respectively. Here, $T$ is small enough and the parameters $d_{0},d_{1}$ are
fine-tuned in order to get the desired behavior.

######  Remark 1.7 (An open question).

How big can $\theta^{*}$ be? This question is related to the work of Merle and
Zaag in [MZ97a] (see the Theorem on page 1499), which corresponds to the case
where $\gamma=0$. For that case, the answer is
$\theta^{*}=\frac{1}{\sqrt[3]{\lambda}}$. It is very interesting to answer the
question in the general case. By a glance to (2.8), we know that $\theta^{*}$
is stricly greater than
$\frac{(1+\gamma|\Omega|)^{\frac{2}{3}}}{\sqrt[3]{\lambda}}$. Let us define

$\mathcal{T}_{max}=\left(\frac{(1+\gamma|\Omega|)^{\frac{2}{3}}}{\sqrt[3]{\lambda}},+\infty\right),$

and

$\mathcal{T}=\left\\{\theta^{*}\in\mathbb{R}\text{ such that }\eqref{profile-
intermidiate}\text{ holds with }u\text{ a positive solution to }\eqref{equa-
mems-u},\text{ for some }T>0\right\\}.$

Then, by a fine modification in the proof, we can construct a solution such
that $\theta^{*}$ arbitrarily takes large values in $\mathcal{T}_{max}$. In
particular, we can prove that $\mathcal{T}$ is a dense subset of
$\mathcal{T}_{max}$. We would like to make the following conjecture

$\mathcal{T}=\mathcal{T}_{max}.$

Now, we would like to mention that our proof of Theorem 1.1 holds in a more
general setting. More precisely, if we consider problem (1.3) in the following
regime

$n-\frac{2}{p+1}>0,\text{ and }q>0\text{ and }n\geq 1,$ (1.9)

then, Theorem 1.1 changes as follows:

######  Theorem 1.8 (Existence of a touch-down solution to (1.3)).

Consider $\lambda,\gamma>0,$ and $\Omega$ a $C^{2}$ bounded domain in
$\mathbb{R}^{n}$ and condition (1.9) holds. Then, there exist initial data
$\hat{u}_{0}$ in $C^{\infty}(\bar{\Omega})$ such that the solution of equation
(1.3) touches down in finite time only at the origin. In particular, the
following holds:

  * $(i)$

The intermidiate profile, for all $t\in[0,T)$

$\left\|\frac{(T-t)^{\frac{1}{p+1}}}{1-u(.,t)}-\hat{\theta}^{*}\left(p+1+\frac{(p+1)^{2}}{4p}\frac{|.|^{2}}{\sqrt{(T-t)|\ln(T-t)|}}\right)^{-\frac{1}{p+1}}\right\|_{L^{\infty}(\Omega)}\leq\frac{C}{\sqrt{|\ln(T-t)|}},$
(1.10)

for some $\hat{\theta}^{*}(\lambda,\gamma,\Omega,p,q)>0$.

  * $(ii)$

The exists $\hat{u}^{*}\in C^{2}(\Omega)\cap C(\bar{\Omega})$ such that $u$
uniformly converges to $\hat{u}^{*}$ as $t\to T,$ and

$1-\hat{u}^{*}(x)\sim\hat{\theta}^{*}\left[\frac{(p+1)^{2}}{8p}\frac{|x|^{2}}{|\ln|x||}\right]^{\frac{1}{p+1}}\text{
as }x\to 0.$ (1.11)

######  Remark 1.9.

When $\gamma=0$, we have no nonlocal term in the equation, and this result was
already proved in [MZ97a]. We also mention that in one space dimension, the
final profile in (1.11) was also derived in [FG93].

######  Remark 1.10.

We don’t give the proof of Theorem 1.8 here because the techniques are the
same as for Theorem 1.1. In fact, for simplicity, we will only give the proof
for the MEMS case

$p=q=2,$

considered in equation (1.2) and Theorem 1.1. Of course, all our estimates can
be carried on for the general case.

In addition to that, we can apply the techniques of Merle in [Mer92] to create
a soluiton which quenches at arbitrary given points.

######  Corollary 1.11.

For any $k$ points $a_{1},a_{2},....,a_{k}$ in $\Omega,$ there exist initial
data such that (1.3) has a solution which quenches exactly at
$a_{1},...,a_{k}$. Moreover, the local behavior at each $a_{i}$ is also given
by (1.10), (1.11) by replacing $x$ by $x-a_{i}$ and $L^{\infty}(\Omega)$ by
$L^{\infty}(|x-a_{i}|\leq\omega_{0}),$ for some $\omega_{0}>0,$ small enough.

As a consequence of our techniques, we can derive the stability of the
quenching solution which we constructed in Theorem 1.8 under the perturbations
of initial data.

######  Theorem 1.12 (Stability of the constructed solution).

Let us consider $\hat{u},$ the solution which we constructed in Theorem 1.8
and we also define $\hat{T}$ as the quenching time of the solution and
$\hat{\theta}^{*}$ as the coefficient in front of the profiles (1.10) and
(1.11). Then, there exists an open subset $\hat{\mathcal{U}}_{0}$ in
$C(\bar{\Omega}),$ containing $\hat{u}(0)$ such that for all initial data
$u_{0}\in\hat{\mathcal{U}}_{0},$ equation (1.3) has a unique solution $u$
quenching in finite time $T(u_{0})$ at only one quenching point $a(u_{0})$.
Moreover, the asymptotics (1.10) and (1.11) hold by replacing $\hat{u}(x,t)$
by $u(x-a(u_{0}),t),$ and $\hat{\theta}^{*}$ by some $\theta^{*}(u_{0})$ .
Note that, we have

$(a(u_{0}),T(u_{0}),\theta^{*}(u_{0}))\to(0,\hat{T},\hat{\theta}),\text{ as
}\|u_{0}-\hat{u}_{0}\|_{C(\bar{\Omega})}\to 0.$

Let us now comment on the organization of the paper. As we have stated
earlier, Theorem 1.1 is a special case of Theorem 1.8. For simplicity in the
notations, we only prove Theorem 1.1. The interested reader may derive the
general case by inspection. Moreover, we don’t prove Corollary 1.11 and
Theorem 1.12, since the former follows from Theorem 1.8 and the techniques of
Merle in [Mer92], and the latter follows also from Theorem 1.8 by the method
of Merle and Zaag in [MZ97b]. In conclusion, we only prove Theorem 1.1 in this
paper.

The paper is organized as follows:

\- In Section 2, we give a different formulation of the problem, and show how
the profile in (1.6) arises naturally.

\- In Section 3, we give the proof without technical details.

\- In Section 4, we prove the technical details.

Some appendices are added at the end.

######  Acknowledgment 1.13.

We would like to thank the referees for the careful reading of the paper and
for pointing out various earlier references.

##  2\.  Setting of the problem

###  2.1.  Our main idea

We aim in this subsection at explaining our key idea in this paper. The
rigorous proof will be given later. Introducing

$\alpha(t)=\displaystyle\frac{\lambda}{\left(1+\gamma\displaystyle\int_{\Omega}\frac{1}{1-u(t)}dx\right)^{2}},$
(2.1)

we rewrite (1.2) as the following

$\partial_{t}u=\Delta u+\frac{\alpha(t)}{(1-u)^{2}}.$ (2.2)

Under this general form, we see our equation (1.2) as a step by step
generalization, starting from a much simpler context:

\- Problem 1: Case where $\alpha(t)\equiv\alpha_{0}$. This case was considered
by Merle and Zaag in [MZ97a] where, the authors constructed a solution
$u_{\alpha_{0}}$ satisfying

$u_{\alpha_{0}}(x,t)\to 1\text{ as }(x,t)\to(x_{0},T),$

for some $T>0,$ and $x_{0}\in\Omega$. In particular, they gave a sharp
description for the quenching profile. Technically, the authors in that work
introduced

$\bar{u}=\frac{1}{1-u}-1=\frac{u}{1-u},$

and constructed a blowup solution for the following equation derived from
(2.2):

$\partial_{t}\bar{u}=\Delta\bar{u}-2\frac{|\nabla\bar{u}|^{2}}{\bar{u}}+\alpha_{0}\bar{u}^{4},\text{
with }\alpha(t)\equiv\alpha_{0},$ (2.3)

(see equation (III), page 1500 in [MZ97a] for more details).

\- Problem 2: Case where $0<\alpha_{1}\leq\alpha(t)\leq\alpha_{2}$ for all
$t>0$ for some $0<\alpha_{1}<\alpha_{2}$. This case is indeed a reasonable
generalization which follows with no difficulty from the stategy of [MZ97a]
for Problem 1.

\- Problem 3: Equation (1.2). Our idea here is to see (1.2) as a coupled
system between Problem 2 and (2.1):

$\displaystyle\left\\{\begin{array}[]{rcl}\partial_{t}u&=\Delta
u+\frac{\alpha(t)}{(1-u)^{2}},\\\\[11.38092pt]
\alpha(t)&=\frac{\lambda}{\left(1+\gamma\int_{\Omega}\frac{1}{1-u}dx\right)^{2}}.\end{array}\right.$

A simple idea would be to try a kind of fixed-point argument starting from
some solution to Problem 1, then defining $\alpha(t)$ according to (2.1)
defined with this solution, then solving Problem 2 with this $\alpha(t)$, then
defining a new $\alpha(t)$ with the new solution, and so forth.

In order to make this method to work, one has to check whether the iterated
$\alpha(t)$ stay away from $0$ and $+\infty$, as requested in the context of
Problem 2. We checked whether this holds when $u$ solves Problem 1.
Fortunately, this was the case, and this gave us a serious hint to treat our
equation (1.2) as a pertubation of Problem 1.

In fact, our proof uses no interation, and we diredly apply the stategy of
Merle and Zaag in [MZ97a] to control the various terms (including the nonlocal
term), in order to find a solution which stays near the desired behavior.

###  2.2. Formulation of the problem

In this section, we aim at setting the mathematical framework of our problem.
The rigorous proof will be given later. Our aim is to construct a solution for
equation (1.2), defined for all $(x,t)\in\Omega\times[0,T),$ for some $T>0$
with $0\leq u(x,t)<1,$ and

$u(x,t)\to 1\text{ as }(x,t)\to(x_{0},T),$

for some $x_{0}\in\Omega$. Without loss of generality, we assume that

$x_{0}=0\in\Omega.$

Introducing,

$\bar{u}=\frac{1}{1-u}-1=\frac{u}{1-u}\in[0,+\infty),$ (2.5)

we derive from (1.2) the following equation on $\bar{u}$

$\left\\{\begin{array}[]{rcl}\displaystyle\partial_{t}\bar{u}&=&\Delta\bar{u}-2\frac{|\nabla\bar{u}|^{2}}{\bar{u}+1}+\displaystyle\frac{\lambda(\bar{u}+1)^{4}}{(1+\gamma|\Omega|+\displaystyle\gamma\int_{\Omega}\bar{u}dx)^{2}},x\in\Omega,t>0,\\\\[14.22636pt]
\bar{u}(x,t)&=&0,x\in\partial\Omega,t>0,\\\\[5.69046pt]
\bar{u}(x,0)&=&\bar{u}_{0}(x),\quad x\in\bar{\Omega}.\end{array}\right.$ (2.6)

Our aim becomes then to construct a blowup solution for equation (2.6) such
that

$\bar{u}(0,t)\to+\infty\text{ as }t\to T.$

In order to see our equation as a (not so small) perturbation of the standard
case in (2.3), we suggest to make one more scaling by introducing

$U(x,t)=\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}\bar{u}(x,t),\quad
U(x,t)\geq 0,\quad\forall(x,t)\in\Omega\times[0,T),$ (2.7)

with

$\bar{\theta}(t)=\left(1+\gamma|\Omega|+\gamma\int_{\Omega}\bar{u}(t)dx\right)^{\frac{2}{3}}.$
(2.8)

Then, thanks to equation (2.6), we deduce the following equation to be
satisfied by $U$:

$\left\\{\begin{array}[]{rcl}\displaystyle\partial_{t}U&=&\Delta
U-2\frac{|\nabla
U|^{2}}{U+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}}+\left(U+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}\right)^{4}-\frac{\bar{\theta}^{\prime}(t)}{\bar{\theta}(t)}U,x\in\Omega,t>0,\\\\[14.22636pt]
U(x,t)&=&0,x\in\partial\Omega,t>0,\\\\[5.69046pt]
U(x,0)&=&U_{0}(x),x\in\bar{\Omega}.\end{array}\right.$ (2.9)

Note that in the blowup regime, which is our focus, $U$ is large and equation
(2.9) appears indeed as a perturbation of equation (2.3).

Introducing the following notation

$\displaystyle\bar{\mu}(t)$ $\displaystyle=$
$\displaystyle\int_{\Omega}U(t)dx,$ (2.10)

we may rewrite (2.8) as the following equation

$\bar{\theta}(t)=\left(1+\gamma|\Omega|+\frac{\gamma}{\lambda^{\frac{1}{3}}}\bar{\theta}(t)\bar{\mu}(t)\right)^{\frac{2}{3}}.$
(2.11)

This implies that $\bar{\theta}(t)$ solves the following cubic equation

$\theta^{3}(t)=\left(1+\gamma|\Omega|+\frac{\gamma}{\lambda^{\frac{1}{3}}}\bar{\theta}(t)\bar{\mu}(t)\right)^{2}=(A+B(t)\bar{\theta}(t))^{2}=A^{2}+2AB(t)\bar{\theta}(t)+B^{2}(t)\bar{\theta}^{2}(t),$
(2.12)

where

$A=1+\gamma|\Omega|\text{ and
}B(t)=\frac{\gamma}{\lambda^{\frac{1}{3}}}\bar{\mu}(t).$

Since it happens that $\bar{\theta}(t)$ is the unique positive solution of
(2.12), we may solve (2.12) and express $\bar{\theta}(t)$ in terms of
$\bar{\mu}(t)$ as follows

$\displaystyle\bar{\theta}(t)$ $\displaystyle=$
$\displaystyle\frac{\sqrt[3]{27A^{2}+3\sqrt[3]{3}\sqrt{27A^{2}+4A^{3}B^{3}(t)}+18AB^{3}(t)+2B^{6}(t)}}{3\sqrt[3]{2}}+\frac{B^{3}(t)}{3}$
$\displaystyle+$
$\displaystyle\frac{\sqrt[3]{2}(6AB(t)+B^{4}(t))}{\sqrt[3]{27A^{2}+3\sqrt[3]{3}\sqrt{27A^{2}+4A^{3}B^{3}(t)}+18AB^{3}(t)+2B^{6}(t)}}.$

Particularly, we show here the equivalence between equation (2.6) and (2.9).

######  Lemma 2.1 (Equivalence between (2.6) and (2.9)).

Consider $\lambda>0,\gamma>0$ and $\Omega$ a bounded domain in
$\mathbb{R}^{n}$. Then, the following holds:

$(i)$ We consider $\bar{u}$ a solution of equation (2.6) on $[0,T),$ for some
$T>0$ and introduce

$U(t)=\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}\bar{u}(t),$

where
$\bar{\theta}(t)=\left(1+\gamma|\Omega|+\gamma\int_{\Omega}\bar{u}(t)dx\right)^{\frac{2}{3}}$.
Then, $U$ is a solution of equation (2.9) on $[0,T)$.

$(ii)$ Otherwise, we consider $U$ a solution of equation (2.9) on $[0,T),$ for
some $T>0$ and introduce

$\bar{u}(t)=\frac{\bar{\theta}(t)}{\lambda^{\frac{1}{3}}}U(t),\forall
t\in[0,T),$

where $\bar{\theta}(t)$ is defined as in relation (2.11), then $\bar{u}$ is
the solution of equation (2.6) on $[0,T)$. In particular, the uniqueness of
the solution is preserved.

###### Proof.

The proof is easily deduced from the definition in this lemma. We kindly ask
the reader to self-check. ∎

######  Remark 2.2.

From settings (2.5) and (2.7) and the local well-posedness of equation (1.2)
in the sense of classical solutions (see Proposition 1.2.2 at page 12 in
Kavallaris and Suzuki [KS18]), we can derive the local existence and
uniqueness of classical solutions of equations (2.6) and (2.9). Since
nonnegativity is preserved for these equation, we will assume that $\bar{u}$
and $U$ are nonegative.

Thanks to Lemma 2.1, our problem is reduced to constructing a nonnegative
solution to (2.9), which blows up in finite time only at the origin. We also
aim at describing its asymptotics at the singularity.

Since we defined $U$ in (2.7) on purpose so that (2.9) appears as a
perturbation of equation (2.3) for $U$ large, it is reasonable to make the
following hypotheses:

  * $(i)$

$1\leq\bar{\theta}(t)\leq C_{0}$ for some $C_{0}>0$. Note that from (2.11), we
have $\bar{\theta}(t)\geq 1.$

  * $(ii)$

$|\bar{\theta}^{\prime}(t)|\ll U^{3}(t)\text{ when }U\text{ large}.$

It is then reasonable to expect for equation (2.9) the same profile as the one
constructed in [MZ97a] for equation (2.3). So, it is natural to follow that
work by introducing the following Similarity-Variables:

$W(y,s)=(T-t)^{\frac{1}{3}}U(x,t),\text{ and }s=-\ln(T-t)\text{ and
}y=\frac{x}{\sqrt{T-t}}.$ (2.14)

Using equation (2.9), we write the equation of $W$ in $(y,s)$ as follows

$\left\\{\begin{array}[]{rcl}\partial_{s}W&=&\Delta W-\frac{1}{2}y\cdot\nabla
W-\frac{W}{3}-2\frac{\left|\nabla
W\right|^{2}}{W+\frac{\lambda^{\frac{1}{3}}e^{-\frac{s}{3}}}{\theta(s)}}+\left(W+\frac{\lambda^{\frac{1}{3}}e^{-\frac{s}{3}}}{\theta(s)}\right)^{4}-\frac{\theta^{\prime}(s)}{\theta(s)}W,\\\\[14.22636pt]
W(y,s)&=&0,y\in\partial\Omega_{s},s>-\ln T,\\\\[5.69046pt] W(y,-\ln
T)&=&W_{0}(y),y\in\bar{\Omega}_{s},\end{array}\right.$ (2.15)

where

$\theta(s)=\bar{\theta}(t(s))=\bar{\theta}(T-e^{-s}),$ (2.16)

and

$\Omega_{s}=e^{\frac{s}{2}}\Omega,$ (2.17)

with $\bar{\theta}$ satisfies (2.11) and (2.2).

We observe in equation (2.15) that $\Omega_{s}$ changes as $s\to+\infty$. This
is a major difficulty in comparison with the situation where
$\Omega=\mathbb{R}^{n}$. In order to overcome this difficulty, we intend to
introduce some cut-off of the solution, so that we reduce to the case
$\Omega=\mathbb{R}^{n}$. Of course, there is a price to pay, in the sense that
we will need to handle some cut-off terms. Our model for this will be the work
made by Mahmoudi, Nouaili and Zaag in [MNZ16] for the construction of a blowup
solution to the semilinear heat equation defined on the circle. Let us note
that the situation with $\Omega$ bounded was already mentioned in [MZ97a].
However, the authors in that work avoided the problem by giving the proof only
in the case where $\Omega=\mathbb{R}^{n}$. In this work, we are happy to
handle the case with a bounded $\Omega$, following the ideas of Mahmoudi,
Nouaili and Zaag in [MNZ16]. Let us mention that Velázquez was also faced in
[Vel93] by the question of reducing a problem defined on a bounded interval to
a problem considered on the whole real line. He made the reduction thanks to
the extension of the solution defined on the interval to another solution
defined on the whole line, thanks to some truely $1$-$d$ techniques. In our
case, given that we work in higher dimensions, we use a different method,
based on the localization of the equation, thanks to some cut-off functions.

More precisely, we introduce the following cut-off function

$\chi_{0}\in C_{0}^{\infty}([0,+\infty)),\quad
supp(\chi_{0})\subset[0,2],\quad 0\leq\chi_{0}(x)\leq 1,\forall x\text{ and
}\chi_{0}(x)=1,\forall x\in[0,1].$ (2.18)

Then, we define the following function

$\psi_{M_{0}}(y,s)=\chi_{0}\left(M_{0}ye^{-\frac{s}{2}}\right),\text{ for some
}M_{0}>0.$ (2.19)

Let us introduce

$w(y,s)=\left\\{\begin{array}[]{rcl}W(y,s)\psi_{M_{0}}(y,s)&\text{ if
}&y\in\Omega_{s},\\\\[8.5359pt] 0&&\text{otherwise }.\end{array}\right.$
(2.20)

We remark that $w$ is defined on $\mathbb{R}^{n}$ and $s\geq-\ln T$ and
$w\equiv 0$ whenever $|y|\geq\frac{2}{M_{0}}e^{\frac{s}{2}}$. Note that
$M_{0}$ will be fixed large enough together with others parameters at the end
of our proof.

Using equation (2.15), we derive from (2.11) the equation satisfied by $w$ as
follows

$\partial_{s}w=\Delta w-\frac{1}{2}y\cdot\nabla w-\frac{1}{3}w-2\frac{|\nabla
w|^{2}}{w+\frac{\lambda^{\frac{1}{2}}e^{-\frac{s}{3}}}{\theta(s)}}+\left(w+\frac{\lambda^{\frac{1}{3}}e^{-\frac{s}{3}}}{\theta(s)}\right)^{4}-\frac{\theta^{\prime}(s)}{\theta(s)}w+F(w,W),$
(2.21)

where $F(w,W)$ encapsulates the cut-off terms and is defined as follows

$F(w,W)=\left\\{\begin{array}[]{rcl}&&W\left[\partial_{s}\psi_{M_{0}}-\Delta\psi_{M_{0}}+\frac{1}{2}y\cdot\nabla\psi_{M_{0}}\right]-2\nabla\psi_{M_{0}}\cdot\nabla
W\\\\[8.5359pt] &&+2\frac{|\nabla
w|^{2}}{w+\frac{\lambda^{\frac{1}{2}}e^{-\frac{s}{3}}}{\theta(s)}}-2\frac{|\nabla
W|^{2}\psi_{M_{0}}}{W+\frac{\lambda^{\frac{1}{2}}e^{-\frac{s}{3}}}{\theta(s)}}+\psi_{M_{0}}\left(W+\frac{\lambda^{\frac{1}{3}}e^{-\frac{s}{3}}}{\theta(s)}\right)^{4}-\left(w+\frac{\lambda^{\frac{1}{3}}e^{-\frac{s}{3}}}{\theta(s)}\right)^{4}\\\\[14.22636pt]
&&\text{ if }y\in\Omega e^{\frac{s}{2}},\\\\[11.38092pt] &&0\text{
otherwise}.\end{array}\right.$ (2.22)

We remark that $F\equiv 0$ on the region
$\\{y\in\mathbb{R}^{n}\left|\right.|y|\leq\frac{1}{M_{0}}e^{\frac{s}{2}}\text{
or }|y|\geq\frac{2}{M_{0}}e^{\frac{s}{2}}\\}$ and that we have from the
conditions $(i)$ and $(ii)$ on $\bar{\theta}(t)$ on page $(i)$ that

$1\leq\theta(s)\leq C_{0},\text{ and }|\theta^{\prime}(s)|\ll W^{3}(y,s).$

Making the further assumption that

$\theta^{\prime}(s)\to 0,$

we see that equation (2.25) is almost the same as equation (15) at page 1502
in [MZ97a] at least when $|y|\leq\frac{e^{\frac{s}{2}}}{M_{0}}$ . Hence, it is
reasonable to expect for equation (2.21) the same profile as the authors found
in [MZ97a] for equation (15) in that work, namely

$\varphi(y,s)=\left(3+\frac{9}{8}\frac{|y|^{2}}{s}\right)^{-\frac{1}{3}}+\frac{(3)^{-\frac{1}{3}}n}{4s},$
(2.23)

(note that, this profile was also defined in [MZ97a] for a general $p>2$, and
that here we need to take $p=4$ hence $\kappa=(3)^{-\frac{1}{3}}$). In
particular, we would like to construct $w$ as a perturbation of $\varphi$. So,
we introduce the following function

$q=w-\varphi.$ (2.24)

Using equation (2.15), we easily write the equation of $q$

$\partial_{s}q=(\mathcal{L}+V)q+T(q)+B(q)+N(q)+R(y,s)+F(w,W),$ (2.25)

where

$\displaystyle\mathcal{L}$ $\displaystyle=$
$\displaystyle\Delta-\frac{1}{2}y\cdot\nabla+Id,$ (2.26) $\displaystyle
V(y,s)$ $\displaystyle=$ $\displaystyle
4\left(\varphi^{3}(y,s)-\frac{1}{3}\right),$ (2.27) $\displaystyle
J(q,\theta(s))$ $\displaystyle=$ $\displaystyle-2\frac{\left|\nabla
q+\nabla\varphi\right|^{2}}{q+\varphi+\frac{\lambda^{\frac{1}{3}}e^{-\frac{s}{3}}}{\theta(s)}}+2\frac{|\nabla\varphi|^{2}}{\varphi+\frac{\lambda^{\frac{1}{3}}e^{-\frac{s}{3}}}{\theta(s)}},$
(2.28) $\displaystyle B(q)$ $\displaystyle=$
$\displaystyle\left(q+\varphi+\frac{\lambda^{\frac{1}{3}}e^{-\frac{s}{3}}}{\theta(s)}\right)^{4}-\varphi^{4}-4\varphi^{3}q,$
(2.29) $\displaystyle R(y,s)$ $\displaystyle=$
$\displaystyle-\partial_{s}\varphi+\Delta\varphi-\frac{1}{2}y\cdot\nabla\varphi-\frac{\varphi}{3}+\varphi^{4}-2\frac{|\nabla\varphi|^{2}}{\varphi+\frac{\lambda^{\frac{1}{3}}e^{-\frac{s}{3}}}{\theta(s)}},$
(2.30) $\displaystyle N(q)$ $\displaystyle=$
$\displaystyle-\frac{\theta^{\prime}(s)}{\theta(s)}\left(q+\varphi\right),$
(2.31)

with $\theta(s)$ defined in (2.16) and $F(w,W)$ given in (2.22).

In particular, we assume that $U$ and $q$ have good conditions such that
Lemmas D.2, D.3, D.4, D.5 and D.6 hold. Then, it is easy to see that all terms
in the right-hand side of (2.25) become very small, except for
$(\mathcal{L}+V)q$. As a matter of fact, this term plays the most important
role in our analysis. Therefore, we show here some main properties on the
linear operator $\mathcal{L}$ and the potential $V$ (see more details in
[DNZ18], [Duoed])

\- Operator $\mathcal{L}$: This operator is self-adjoint in
$\mathcal{D}(\mathcal{L})\subset L^{2}_{\rho}(\mathbb{R}^{n}),$ where
$L^{2}_{\rho}$ is defined as follows

$L^{2}_{\rho}(\mathbb{R}^{n})=\left\\{f\in
L^{2}_{loc}(\mathbb{R}^{n})\left|\right.\int_{\mathbb{R}^{n}}|f(y)|^{2}\rho(y)dy<+\infty\right\\},$

and

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

This is the spectrum set of operator $\mathcal{L}$

$\text{Spec}(\mathcal{L})=\left\\{1-\frac{m}{2}\left|\right.m\in\mathbb{N}\right\\}.$

The eigenspace which corresponds to $\lambda_{m}=1-\frac{m}{2}$ is given

$\mathcal{E}_{m}=\left\\{h_{m_{1}}(y_{1}).h_{m_{2}}(y_{2})....h_{m_{n}}(y_{n})\left|\right.m_{1}+...+m_{n}=m\right\\},$

where $h_{m_{i}}$ is the (rescaled ) Hermite polynomial in one dimension.

\-  Potential $V$: It has two impotant properties:

  * $(i)$

The potential $V(.,s)\to 0$ in $L^{2}_{\rho}(\mathbb{R}^{n})$ as
$s\to+\infty$: In particular, in the region $|y|\leq K_{0}\sqrt{s}$ ( the
singular domain), $V$ has some weak perturbations on the effect of operator
$\mathcal{L}$.

  * $(ii)$

$V(y,s)$ is almost a constant on the region $|y|\geq K_{0}\sqrt{s}$: For all
$\epsilon>0$, there exists $\mathcal{C}_{\epsilon}>0$ and $s_{\epsilon}$ such
that

$\sup_{s\geq
s_{\epsilon},\frac{|y|}{\sqrt{s}}\geq\mathcal{C}_{\epsilon}}\left|V(y,s)-\left(-\frac{4}{3}\right)\right|\leq\epsilon.$

Note that $-\frac{4}{3}<-1$ and that the largest eigenvalue of $\mathcal{L}$
is $1$. Hence, roughly speaking, we may assume that $\mathcal{L}+V$ admits a
strictly negative spectrum. Thus, we can easily control our solution in the
region $\\{|y|\geq K_{0}\sqrt{s}\\}$ with $K_{0}$ large enough.

From these properties, it appears that the behavior of $\mathcal{L}+V$ is not
the same inside and outside of the singular domain $\\{|y|\leq
K_{0}\sqrt{s}\\}$. Therefore, it is natural to decompose every $r\in
L^{\infty}(\mathbb{R}^{n})$ as follows:

$r(y)=r_{b}(y)+r_{e}(y)\equiv\chi(y,s)r(y)+(1-\chi(y,s))r(y),$ (2.32)

where $\chi(y,s)$ is defined as follows

$\chi(y,s)=\chi_{0}\left(\frac{|y|}{K_{0}\sqrt{s}}\right),$ (2.33)

and $\chi_{0}$ is given in (2.18). From the above decomposition, we
immediately have the following:

$\displaystyle\text{Supp }(r_{b})$ $\displaystyle\subset$
$\displaystyle\\{|y|\leq 2K_{0}\sqrt{s}\\},$ $\displaystyle\text{Supp
}(r_{e})$ $\displaystyle\subset$ $\displaystyle\\{|y|\geq K_{0}\sqrt{s}\\}.$

In addition to that, we are interested in expanding $r_{b}$ in
$L^{2}_{\rho}\left(\mathbb{R}^{n}\right)$ according to the basis which is
created by the eigenfunctions of operator $\mathcal{L}$:

$\displaystyle r_{b}(y)$ $\displaystyle=$ $\displaystyle r_{0}+r_{1}\cdot
y+y^{T}\cdot r_{2}\cdot y-2\text{ Tr}(r_{2})+r_{-}(y),$ or $\displaystyle
r_{b}(y)$ $\displaystyle=$ $\displaystyle r_{0}+r_{1}\cdot y+r_{\perp}(y),$

where

$r_{i}=\left(P_{\beta}(r_{b})\right)_{\beta\in\mathbb{N}^{n},|\beta|=i},\forall
i\geq 0,$ (2.34)

with $P_{\beta}(r_{b})$ being the projection of $r_{b}$ on the eigenfunction
$h_{\beta}$ defined as follows:

$P_{\beta}(r)=\int_{\mathbb{R}^{n}}r_{b}\frac{h_{\beta}}{\|h_{\beta}\|_{L^{2}_{\rho}}}\rho
dy,\forall\beta\in\mathbb{N}^{n},$ (2.35)

and

$r_{\perp}=P_{\perp}(r)=\sum_{\beta\in\mathbb{R}^{n},|\beta|\geq
2}h_{\beta}P_{\beta}(R_{b}),$ (2.36)

and

$r_{-}=\sum_{\beta\in\mathbb{R}^{n},|\beta|\geq 3}h_{\beta}P_{\beta}(r_{b}).$
(2.37)

In other words, $r_{\perp}$ is the part of $r_{b}$ which is orthogonal to the
eigenfunctions corresponding to eigenvalues $0$ and $1$ and $r_{-}$ is
orthogonal to the eigenfunctions corresponding to eigenvalues $1,\frac{1}{2}$
and $0$. We should note that $r_{0}$ is a scalar, $r_{1}$ is a vector and
$r_{2}$ is a square matrix of order $n$ and that they are the components of
$r_{b}$ not $r$. Finally, we write $r$ as follows

$\displaystyle r(y)$ $\displaystyle=$ $\displaystyle r_{0}+r_{1}\cdot
y+y^{T}\cdot r_{2}\cdot y-2\text{ Tr}(r_{2})+r_{-}(y)+r_{e}(y),$ or
$\displaystyle r(y)$ $\displaystyle=$ $\displaystyle r_{0}+r_{1}\cdot
y+r_{\perp}(y)+r_{e}(y).$ (2.39)

A summary of our problem: Even though we created many extra functions from $U$
to $q$, we always concentrate on solution $U$ to equation (2.9). More
precisely, we aim at constructing $U$ blowing up in finite time. Then, we will
use equation (2.25) as a crucial formulation in our proof. Indeed, in order to
control $U$ blowing up in finite time, it is enough to control the transform
function $q$ of $U$ (see definitions (2.14), (2.20) and (2.24)) satisfying

$\left\|q(.,s)\right\|_{L^{\infty}(\mathbb{R}^{n})}\to 0,\text{ as
}s\to+\infty.$ (2.40)

##  3\. The proof of the existence result assuming technical details

In this section, we aim at giving a proof without technical details to Theorem
1.1. We would like to summarize the structure of this section as follows:

\- Construction of a shrinking set: We rely here on the ideas of the Merle and
Zaag’s work in [MZ97a] to introduce a shrinking set that will guarantee the
convergence to zero for $q$ defined in (2.24). This set will constrain our
solution as we want. Once our solution is trapped in, we may show the main
asymptotics of our solution. In particular, (2.40) holds and our result
follows.

\- Preparation of initial data: We introduce a familly of initial data to
equation (2.9) depending on some finite set parameters. As a matter of fact,
we will choose these parameters such that our solution belongs to the
shrinking set for all $t\in[0,T)$.

\- The existence of a trapped solution: Using a reduction to a finite
dimensional problem (corresponding to the finite parameters introduced in our
initial data) and a topological argument, we can derive the existence of a
blowup solution in finite time, trapped in the shrinking set. More precisely,
we show in this part that there exist initial data in that family of initial
data such that our solution is completely confined in the shrinking set.

\- The conclusion of Theorem 1.1: Finally, we rely on the existence of a
blowup solution, trapped in the shrinking set to get the conclusion of Theorem
1.1.

###  3.1.  Shrinking set

In order to control the solution $U$ blowing up in finite time and satisfying
(2.40), we adopt the general ideas given by Merle and Zaag in [MZ97a]. For
each $K_{0}>0,\epsilon_{0}>0,\alpha_{0}>0$ and $t\in[0,T)$ with $T>0$, we
define

$\displaystyle P_{1}(t)$ $\displaystyle=$
$\displaystyle\left\\{x\in\mathbb{R}^{n}\left|\right.|x|\leq
K_{0}\sqrt{(T-t)|\ln(T-t)|}\right\\},$ (3.1) $\displaystyle P_{2}(t)$
$\displaystyle=$
$\displaystyle\left\\{x\in\mathbb{R}^{n}\left|\right.\frac{K_{0}}{4}\sqrt{(T-t)|\ln(T-t)|}\leq|x|\leq\epsilon_{0}\right\\},$
(3.2) $\displaystyle P_{3}(t)$ $\displaystyle=$
$\displaystyle\left\\{x\in\mathbb{R}^{n}\left|\right.|x|\geq\frac{\epsilon_{0}}{4}\right\\}.$
(3.3)

As a matter of fact, we have

$\Omega\subset\mathbb{R}^{n}=P_{1}(t)\cup P_{2}(t)\cup P_{3}(t),\text{ for all
}t\in[0,T).$

We aim at controlling our problem on $P_{1}(t),P_{2}(t)$ and $P_{3}(t)$ as
follows:

\- On region $P_{1}(t)(\text{{blowup region})}$: We control $w$ (see (2.14))
instead of $U$. More precisely, we show that $w$ is a perturbation of the
profile $\varphi$ (the blowup profile, introduced in (2.23)). Then, (2.40)
will follow from the control of $w$.

\- On region $P_{2}(t)(\text{{intermidiate region})}$: We control a rescaled
function $\mathcal{U}$ instead of $U$. More precisely, $\mathcal{U}$ is
defined as follows: For all $x\in
P_{2}(t),\xi\in(T-t(x))^{-\frac{1}{2}}(\bar{\Omega}-x)$ and
$\tau\in\left[-\frac{t(x)}{T-t(x)},1\right),$ we define

$\mathcal{U}(x,\xi,\tau)=\left(T-t(x)\right)^{\frac{1}{3}}U\left(x+\xi\sqrt{T-t(x)},(T-t(x))\tau+t(x)\right),$
(3.4)

where $t(x)$ is defined as the solution of the following equation

$\displaystyle|x|$ $\displaystyle=$
$\displaystyle\frac{K_{0}}{4}\sqrt{(T-t(x))|\ln(T-t(x))|}\text{ and }t(x)<T.$
(3.5)

We remark that if $\epsilon_{0}$ is small enough, then $t(x)$ is well defined
for all $x$ in $P_{2}(t)$. In addition to that, using (3.5), we have the
following asymptotic

$t(x)\to T,\text{ as }x\to 0.$

For convenience, we introduce

$\varrho(x)=T-t(x).$ (3.6)

Then, the following holds

$\varrho(x)\to 0\text{ as }x\to 0.$

As a matter of fact, using (2.9), we write the equation satisfied by
$\mathcal{U}$ in
$(\xi,\tau)\in\varrho^{-\frac{1}{2}}(x)(\bar{\Omega}-x)\times\left[-\frac{t(x)}{\varrho(x)},1\right)$
as follows:

$\partial_{\tau}\mathcal{U}=\Delta_{\xi}\mathcal{U}-2\frac{\left|\nabla\mathcal{U}\right|^{2}}{\mathcal{U}+\frac{\lambda^{\frac{1}{3}}\varrho^{\frac{1}{3}}(x)}{\tilde{\theta}(\tau)}}+\left(\mathcal{U}+\frac{\lambda^{\frac{1}{3}}\varrho^{\frac{1}{3}}(x)}{\tilde{\theta}(\tau)}\right)^{4}-\frac{\tilde{\theta}_{\tau}^{\prime}(\tau)}{\tilde{\theta}(\tau)}\mathcal{U},$
(3.7)

where

$\tilde{\theta}(\tau)=\bar{\theta}(\tau\varrho(x)+t(x)),$ (3.8)

with $\bar{\theta}(t)$ defined in (3.6). We now consider the following domain

$|\xi|\leq\alpha_{0}\sqrt{|\ln(\varrho(x))|}\text{ and
}\tau\in\left[-\frac{t(x)}{\varrho(x)},1\right).$

When $\tau=0$, we are in region $P_{1}(t(x))$, in fact (note that
$P_{1}(t(x))$ and $P_{2}(t(x))$ have some overlapping by definition). From our
constraints in $P_{1}(t(x))$, we derive that $\mathcal{U}(x,\xi,0)$ is flat in
the sense that

$\mathcal{U}(x,\xi,0)\sim\left(3+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}.$

Our idea is to show that this flatness will be conserved for all
$\tau\in[0,1)$ (that is for all $t\in[t(x),T)$), in the sense that the
solution will not depend that much on space. In one word, $\mathcal{U}$ is
regarded as a perturbation of $\hat{\mathcal{U}}(\tau),$ where
$\hat{\mathcal{U}}(\tau)$ is defined as follows

$\left\\{\begin{array}[]{rcl}\partial_{\tau}\hat{\mathcal{U}}(\tau)&=&\hat{\mathcal{U}}^{4}(\tau),\\\
\hat{\mathcal{U}}(0)&=&\left(3+\displaystyle\frac{9}{8}\displaystyle\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}.\end{array}\right.$
(3.9)

Note that, we can give an explicit formula to the solution of equation (3.9)

$\hat{\mathcal{U}}(\tau)=\left(3(1-\tau)+\displaystyle\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}.$
(3.10)

\- On region $P_{3}(t)(\text{{regular region})}$: Thanks to the well-posedness
property of the Cauchy problem for equation (2.25), we control the solution
$U$ as a perturbation of initial data $U(0).$ Indeed, the blowup time $T$ will
be chosen small in our analysis.

Relying on those ideas, we give in the following the definition of our
shrinking set:

######  Definition 3.1 (Definition of $S(t)$).

Let us consider
$T>0,K_{0}>0,\epsilon_{0}>0,\alpha_{0}>0,A>0,\delta_{0}>0,C_{0}>0,\eta_{0}>0$
and $t\in[0,T).$ Then, we introduce the following set

$S(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},\eta_{0},C_{0},t)\quad(S(t)\text{
for short}),$

as a subset of $C^{2,1}\left(\Omega\times(0,t)\right)\cap
C(\bar{\Omega}\times[0,t]),$ containing all functions $U$ satisfying the
following conditions:

  * $(i)$

Estimates in $P_{1}(t)$: We have $q(s)\in V_{K_{0},A}(s),$ where $q(s)$ is
introduced in (2.24), $s=-\ln(T-t)$ and $V_{K_{0},A}(s)$ is a subset of all
function $r$ in $L^{\infty}(\mathbb{R}^{n}),$ satisfying the following
estimates:

$\displaystyle|r_{i}|$ $\displaystyle\leq$
$\displaystyle\frac{A}{s^{2}},(i=0,1),\text{ and }|r_{2}|\leq\frac{A^{2}\ln
s}{s^{2}},$ $\displaystyle|r_{-}(y)|$ $\displaystyle\leq$
$\displaystyle\frac{A^{2}}{s^{2}}(1+|y|^{3}),\text{ and
}\|r_{e}\|_{L^{\infty}(\mathbb{R}^{n})}\leq\frac{A^{2}}{\sqrt{s}},$
$\displaystyle\left|\left(\nabla r\right)_{\perp}\right|$ $\displaystyle\leq$
$\displaystyle\frac{A}{s^{2}}(1+|y|^{3}),\forall y\in\mathbb{R}^{n},$

where the definitions of $r_{i},r_{-},\left(\nabla r\right)_{\perp}$ are given
in (2.34), (2.36) and (2.37), respectively.

  * $(ii)$

Estimates in $P_{2}(t)$: For all
$|x|\in\left[\frac{K_{0}}{4}\sqrt{(T-t)|\ln(T-t)|},\epsilon_{0}\right],\tau(x,t)=\frac{t-t(x)}{\varrho(x)}$
and $|\xi|\leq\alpha_{0}\sqrt{|\ln\varrho(x)|},$ we have the following

$\displaystyle\left|\mathcal{U}(x,\xi,\tau(x,t))-\hat{\mathcal{U}}(\tau(x,t))\right|$
$\displaystyle\leq$ $\displaystyle\delta_{0},$
$\displaystyle\left|\nabla_{\xi}\mathcal{U}(x,\xi,\tau(x,t))\right|$
$\displaystyle\leq$ $\displaystyle\frac{C_{0}}{\sqrt{|\ln\varrho(x)|}},$

where $\mathcal{U},$ $\hat{\mathcal{U}}$ and $\varrho(x)$ are given (3.4),
(3.6) and (3.10), respectively.

  * $(iii)$

Estimates in $P_{3}(t)$: For all
$x\in\\{|x|\geq\frac{\epsilon_{0}}{4}\\}\cap\Omega,$ we have

$\displaystyle\left|U(x,t)-U(x,0)\right|$ $\displaystyle\leq$
$\displaystyle\eta_{0},$ $\displaystyle\left|\nabla U(x,t)-\nabla
e^{t\Delta}U(x,0)\right|$ $\displaystyle\leq$ $\displaystyle\eta_{0}.$

In addition to that, we would like to introduce the set
$S^{*}(K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},T)$ as
follows:

######  Definition 3.2.

For all $T>0,K_{0}>0,\epsilon_{0}>0,\alpha_{0}>0,A>0,\delta_{0}>0,C_{0}>0,$
and $\eta_{0}>0$, we introduce
$S^{*}(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0})$
($S^{*}(T)$ for short) as the subset of all functions $U$ in
$C^{2,1}(\Omega\times(0,T))\cap C(\bar{\Omega}\times[0,T)),$ satisfying the
following: for all $t\in[0,T),$ we have

$U\in S(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},t).$

######  Remark 3.3.

The shrinking set $S(t)$ is inspired by the work of Merle and Zaag in [MZ97a].
However, we’ve made two major changes:

\- A simplification, by removing an unnecessary condition on the second
derivative in space in region $P_{2}(t)$.

\- A smart change in region $P_{3}(t)$, by replacing $\nabla U$ by $\nabla
e^{t\Delta}U(0)$. This change is crucial since we are working on a bounded
domain $\Omega$.

######  Remark 3.4.

The conditions in $P_{2}$ and $P_{3}$ in Definition 3.1 are designed to make
our solution more regular and these conditions help us to control $U$ in
$P_{1}$ and $q(s)\in V_{K_{0},A}(s)$. Finally, the main purpose is to satisfy
(2.40). In other words, the control $U$ in $P_{1}$ is the main issue.

######  Remark 3.5.

In our paper, we use a lot of parameters to control our solution. However,
they will be fixed at the end of the proof. In addition to that, we would like
to give some conventions on the universal constant in our paper: We use $C$
for universal constants which depend only $n,\Omega,\gamma,\lambda$ and we
write $C(K_{0},\epsilon_{0},...)$ for constants which depend
$K_{0},\epsilon_{0},...$, respectively.

As we mentioned in Remark 3.4, we would like to show here some estimates of
the sizes of $q$ and $\nabla q$, where $q$ is the transformed function of $U$
when $U\in S(t)$.

######  Lemma 3.6 (Sizes of $q$ and $\nabla q$).

Let us consider $K_{0}\geq 1$ and $\epsilon_{0}>0$. Then, there exist
$T_{1}(K_{0},\epsilon_{0})$ and $\eta_{1}(\epsilon_{0})$ such that for all
$\alpha_{0}>0,A>0,\delta_{0}\leq\frac{1}{2}\hat{\mathcal{U}}(0)$ (see (3.10)),
$C_{0}>0,\eta_{0}\leq\eta_{1},T\leq T_{1}$ and $t\in[0,T)$: if $U\in
S(K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},t),$ then, the
following holds:

  * $(i)$

The estimates on $q$: For all $y\in\mathbb{R}^{n}$ and $s=-\ln(T-t),$ we have

$|q(y,s)|\leq\frac{C(K_{0})A^{2}}{\sqrt{s}}\text{ and
}\left|q(y,s)\right|\leq\frac{C(K_{0})A^{2}\ln s}{s^{2}}(1+|y|^{3}).$

  * $(ii)$

The estimates on $\nabla q$: For all $y\in\mathbb{R}^{n}$, we have

$|\nabla q(y,s)|\leq\frac{C(K_{0},C_{0})A^{2}}{\sqrt{s}},\quad\left|\nabla
q(y,s)\right|\leq\frac{C(K_{0},C_{0})A^{2}\ln s}{s^{2}}(1+|y|^{3}),$

and

$\left|(1-\chi(y,s))\nabla q(y,s)\right|\leq\frac{C(K_{0})}{\sqrt{s}}.$

###### Proof.

The conclusion directly follows from the definition of the shrinking set
$S(t)$ and $V_{K_{0},A}(s)$. In addition to that, these definitions are almost
the same as in [MZ97a]. Therefore, we kindly refer the reader to see Lemma B.1
at page 1537 in [MZ97a]. ∎

###  3.2. Initial data

In this subsection, we will concentrate on introducing our initial data to
equation (2.9) so that it is trapped in $S(0)$. In order to do that, we first
introduce the following cut-off function:

$\displaystyle\chi_{1}(x)$ $\displaystyle=$
$\displaystyle\chi_{0}\left(\frac{|x|}{\sqrt{T}|\ln T|}\right),$ (3.11)

where $\chi_{0}$ is given in (2.18). In addition to that, we introduce $H^{*}$
as a function in $C^{\infty}_{0}(\mathbb{R}^{n}\setminus\\{0\\})$ satisfying

$H^{*}(x)=\left\\{\begin{array}[]{rcl}&&\left[\displaystyle\frac{9}{16}\frac{|x|^{2}}{|\ln|x||}\right]^{-\frac{1}{3}},\quad\forall|x|\leq\min\left(\frac{1}{4}d(0,\partial\Omega),\frac{1}{2}\right),x\neq
0,\\\\[19.91684pt]
&&0,\quad\forall|x|\geq\frac{1}{2}d(0,\partial\Omega),\\\\[14.22636pt]
\end{array}\right.$ (3.12)

and for all $x\in\mathbb{R}^{n},x\neq 0,$ the following condition holds

$0\leq
H^{*}(x)\leq\left[\frac{9}{16}\frac{|x|^{2}}{|\ln|x||}\right]^{-\frac{1}{3}}.$

We now give the definition of our initial data corresponding to equation
(2.9): For all $(d_{0},d_{1})\in\mathbb{R}^{1+n}$, we define

$\displaystyle U_{d_{0},d_{1}}(x,0)$ $\displaystyle=$ $\displaystyle
T^{-\frac{1}{3}}\left[\varphi\left(\frac{x}{\sqrt{T}},-\ln
s_{0}\right)+(d_{0}+d_{1}\cdot
z)\chi_{0}\left(\frac{|z|}{\frac{K_{0}}{32}}\right)\right]\chi_{1}(x)$
$\displaystyle+$ $\displaystyle H^{*}(x)\left(1-\chi_{1}(x)\right),$

where $z=\displaystyle\frac{x}{\sqrt{T|\ln T|}}$ and
$\varphi,\chi_{0},\chi_{1},H^{*}$ are defined as in (2.23), (2.18), (3.11) and
(3.12), respectively.

From (3.2), we would like to give the definition of initial data corresponding
to equation (2.25), $q_{d_{0},d_{1}}(s_{0})$ with $s_{0}=-\ln T$:

$q_{d_{0},d_{1}}(y,s_{0})=e^{-\frac{s_{0}}{3}}U_{d_{0},d_{1}}\left(ye^{-\frac{s_{0}}{3}},0\right)\psi_{M_{0}}(y,s_{0})-\varphi(y,s_{0}),$
(3.14)

where and $\psi_{M_{0}},\varphi$ and $U_{d_{0},d_{1}}$ are introduced in
(2.19), (2.23) and (3.2), respectively.

######  Remark 3.7.

We would like to explain in brief how our initial data $U_{d_{0},d_{1}}$ has
naturally the form shown in (3.2). As we mentiond at the beginning of this
section, our purpose is to control initial data in $S(0)$. More precisely, our
inital data have to satisfy items $(i)$ and $(ii)$ in Definition 3.1. As a
matter of fact, when $T$ is small enough, the second term in the right hand
side of (3.2) is zero on $P_{1}(0)$. Then, our initial data has only the first
term and we adopt the idea given in [MZ97b] (see also [MZ97a], [GNZ17]), we
use $d_{0},d_{1}$ in order to control $q(s_{0})$ in $V_{K_{0},A}(s_{0})$. In
addition to that, we would like to mention that Propostion 3.13 below states
that when $q$ is trapped in $V_{K_{0},A}(s)$, it has only two components
$(q_{0},q_{1})(s)$ which may attain their upper bound, the others being
strictly less than their upper bound specified in the definition of
$V_{K_{0},A}(s)$. This is indeed the reason to use $(d_{0},d_{1})$ in our
initial data. More precisely, these $1+n$ parameters allows us to a reduction
to a finite dimensional problem. We now mention the control in $P_{2}$. In
that region, $|x|$ is small enough and we may consider that $U$ is near the
final profile

$\left[\displaystyle\frac{9}{16}\frac{|x|^{2}}{|\ln|x||}\right]^{-\frac{1}{3}}.$

As a matter of fact, it is reasonable to introduce $H^{*}$ as the main
asymptotic of our initial data in $P_{2}(0)$. Using some priori estimates, we
can derive good estimates in $P_{2}(0)$. More precisely, the following
proposition is our statement:

######  Proposition 3.8 (Preparation of initial data).

There exists $K_{2}>0$ such that for all $K_{0}\geq K_{2}$ and $\delta_{2}>0,$
there exist $\alpha_{2}(K_{0},\delta_{2})>0$ and $C_{2}(K_{0})>0$ such that
for every $\alpha_{0}\in(0,\alpha_{2}],$ there exists
$\epsilon_{2}(K_{0},\delta_{2},\alpha_{0})>0$ such that for every
$\epsilon_{0}\in(0,\epsilon_{2}]$ and $A\geq 1,$ there exists
$T_{2}(K_{0},\delta_{2},\epsilon_{0},A,C_{2})>0$ such that for all $T\leq
T_{2}$ and $s_{0}=-\ln T$. The following holds:

(I) We can find a set $\mathcal{D}_{A}\subset[-2,2]\times[-2,2]^{n}$ such that
if we define the following mapping

$\displaystyle\Gamma:\mathbb{R}\times\mathbb{R}^{n}$ $\displaystyle\to$
$\displaystyle\mathbb{R}\times\mathbb{R}^{n}$ $\displaystyle(d_{0},d_{1})$
$\displaystyle\mapsto$ $\displaystyle\left(q_{0},q_{1}\right)(s_{0}),$

then, $\Gamma$ is affine, one to one from $\mathcal{D}_{A}$ to
$\hat{\mathcal{V}}_{A}(s_{0}),$ where $\hat{\mathcal{V}}_{A}(s)$ is defined as
follows

$\hat{\mathcal{V}}_{A}(s)=\left[-\frac{A}{s^{2}},\frac{A}{s^{2}}\right]^{1+n}.$
(3.15)

Moreover, we have

$\Gamma\left|{}_{\partial\mathcal{D}_{A}}\right.\subset\partial\hat{\mathcal{V}}_{A}(s_{0}),$

and

$\text{deg}\left(\Gamma\left.\right|_{\partial\mathcal{D}_{A}}\right)\neq 0,$
(3.16)

where $q_{0},q_{1}$ are defined as in (2.2), considered as the components of
$q_{d_{1},d_{1}}(s_{0}),$ which is a transform function of
$U_{d_{0},d_{1}}(0),$ given in (2.24).

(II) We now consider $(d_{0},d_{1})\in\mathcal{D}_{A}$. Then, initial data
$U_{d_{0},d_{1}}(0)$ belongs to

$S(K_{0},\epsilon_{0},\alpha_{0},A,\delta_{2},C_{2},0,0)=S(0),$

where $S(0)$ is defined in Definition 3.1. Moreover, the following astimates
hold

  * $(i)$

Estimates in $P_{1}(0):$ We have
$q_{d_{0},d_{1}}(s_{0})\in\mathcal{V}_{K_{0},A}(s_{0})$ and

$|q_{0}(s_{0})|\leq\frac{A}{s_{0}^{2}},\quad|q_{1,j}(s_{0})|\leq\frac{A}{s_{0}^{2}},\quad|q_{2,i,j}(s_{0})|\leq\frac{\ln
s_{0}}{s_{0}^{2}},\forall i,j\in\\{1,...,n\\},$
$|q_{-}(y,s_{0})|\leq\frac{1}{s_{0}^{2}}(|y|^{3}+1),\quad|\nabla
q_{\perp}(y,s_{0})|\leq\frac{1}{s_{0}^{2}}(|y|^{3}+1),\forall
y\in\mathbb{R}^{n},$

and

$q_{e}\equiv 0,$

where the components of $q_{d_{0},d_{1}}(s_{0})$ are defined in (2.36).

  * $(ii)$

Estimates in $P_{2}(0)$: For every $|x|\in\left[\frac{K_{0}}{4}\sqrt{T|\ln
T|},\epsilon_{0}\right],\tau_{0}(x)=-\frac{t(x)}{\varrho(x)}$ and
$|\xi|\leq\alpha_{0}\sqrt{|\ln\varrho(x)|},$ we have

$\left|\mathcal{U}(x,\xi,\tau_{0}(x))-\hat{\mathcal{U}}(\tau_{0}(x))\right|\leq\delta_{2}\text{
and
}|\nabla_{\xi}\mathcal{U}(x,\xi,\tau_{0}(x))|\leq\frac{C_{2}}{\sqrt{|\ln\varrho(x)|}},$

where $\mathcal{U},\hat{\mathcal{U}},$ and $\varrho(x)$ are defined as in
(3.4), (3.10) and (3.6) respectively.

###### Proof.

The proof of Proposition 3.8 will be given in Appendix A. We now assume that
this proposition holds and continue to get to the conclusion of Theorem 1.1. ∎

###  3.3.  Existence of a solution trapped in $S^{*}(T)$

In this subsection, we would like to derive the existence of a blowup solution
$U$ to equation (2.9), trapped in $S^{*}(T)$. As we said earlier, our proof
will be a (non trivial) adaptation of the proof designed by Merle and Zaag in
[MZ97a] for the more standard case (2.3). However, in comparision with (2.3),
we observe in equation (2.9) a new feature, the nonlocal term involving
$\bar{\theta}(t)$. As a matter of fact, it is important to study this term and
it derivative. In particular, in the works which we used to make the main idea
for our work (such as [MZ97a], [MZ97b], [GNZ17]), the authors only studied for
constant coefficients parabolic equations. Hence, it makes a main highlight in
our work. For that reason, we show here some estimates on $\bar{\theta}(t)$
(also on $\bar{\mu}(t)$). The following is our statement:

######  Proposition 3.9 (Some estimates of $\bar{\theta}(t)$ and
$\bar{\mu}(t)$).

Let us consider $\lambda>0,\gamma>0$ and $\Omega$ a $C^{2}$ bounded domain.
Then, there exists $K_{3}>0$ such that for all $K_{0}\geq K_{3},\delta_{0}>0,$
there exist $\alpha_{3}(K_{0},\delta_{0})>0$ such that for all
$\alpha_{0}\leq\alpha_{3},$ there exists
$\epsilon_{3}(K_{0},\delta_{0},\alpha_{0})>0$ such that for all
$\epsilon_{0}\leq\epsilon_{3}$ and $A\geq 1,C_{0}>0,\eta_{0}>0$, there exists
$T_{3}>0$ such that for all $T\leq T_{3}$ the following holds: Assuming $U$ is
a non negative solution of equation (2.9) on $[0,t_{1}],$ for some $t_{1}<T$,
$U\in S(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},t)=S(t)$
for all $t\in[0,t_{1}]$ and $U(0)=U_{d_{0},d_{1}}(0),$ given in (3.2) with
some $(d_{0},d_{1})\in\mathbb{R}^{1+n},|d_{0}|,|d_{1}|\leq 2$, the following
statements holds:

  * $(i)$

For all $t\in[0,t_{1}],$ $\bar{\mu}(t)$ and $\bar{\theta}(t)$ are positive and
these estimates hold

$\displaystyle 0\leq\bar{\mu}(t)$ $\displaystyle\leq$ $\displaystyle C,$
(3.17) $\displaystyle 1\leq\bar{\theta}(t)$ $\displaystyle\leq$ $\displaystyle
C.$ (3.18)

Moreover, for all $t\in(0,t_{1})$, we have

$\displaystyle|\bar{\mu}^{\prime}(t)|$ $\displaystyle\leq$ $\displaystyle
C(T-t)^{\frac{3n-8}{6}}|\ln(T-t)|^{n},$ (3.19)
$\displaystyle\left|\bar{\theta}^{\prime}(t)\right|$ $\displaystyle\leq$
$\displaystyle C(T-t)^{\frac{3n-8}{6}}|\ln(T-t)|^{n}.$ (3.20)

  * $(ii)$

In particular, if $U\in S(t)$ for all $t\in[0,T)$, then $\bar{\mu}(t)$ and
$\bar{\theta}(t)$ converge respectively to $\bar{\mu}_{T}$ and
$\bar{\theta}_{T}\in\mathbb{R}_{+}^{*}$ as $t\to T$.

######  Remark 3.10.

Although we know from item (ii) that $\bar{\theta}(t)$ converges to
$\bar{\theta}_{T},$ we don’t know how big is $\bar{\theta}_{T}$. In
particular, the dependence of these constants on $\gamma,\lambda,\Omega$ and
$T,\epsilon_{0},...,$ is not clear yet.

###### Proof.

We can see that item $(ii)$ is a direct consequence of $(i)$. So, we give only
the proof of item $(i)$. Using (2.10) and the fact that $U(t)\geq 0$ for all
$t$, we derive the following

$\bar{\mu}(t)=\int_{\Omega}U(t)dx\geq 0.$

In addition to that, we write

$\bar{\mu}(t)\leq\int_{\Omega}U(t)dx\leq\int_{P_{1}(t)}U(t)dx+\int_{P_{2}(t)}U(t)dx+\int_{P_{3}(t)}U(t)dx,$
(3.21)

where $P_{1}(t),P_{2}(t),P_{3}(t)$ are given in (3.1), (3.2) and (3.3),
respectively. Remembering $\varrho(x),$ defined in (3.6), we see that the
following holds

$\varrho(x)\sim\frac{8}{K_{0}^{2}}\frac{|x|^{2}}{|\ln|x||}\text{ as }x\to 0.$

In particular, using Definition 3.1, we get the following estimates: for all
$t\in[0,t_{1}]$

$\displaystyle\text{ On }P_{1}(t),|U(x,t)|$ $\displaystyle\leq$
$\displaystyle(T-t)^{-\frac{1}{3}}\left[\frac{CA^{2}}{\sqrt{|\ln(T-t)|}}+|\varphi(0,-\ln(T-t))|\right]\leq
C(T-t)^{-\frac{1}{3}},$ $\displaystyle\text{ On }P_{2}(t),|U(x,t)|$
$\displaystyle\leq$
$\displaystyle\varrho^{-\frac{1}{3}}(x)\left[\hat{\mathcal{U}}(\tau(x,t))+\delta_{0}\right]\leq
C\left[\frac{|x|^{2}}{|\ln|x||}\right]^{-\frac{1}{3}},$ $\displaystyle\text{
On }P_{3}(t),|U(x,t)|$ $\displaystyle\leq$
$\displaystyle|U(x,0)|+\eta_{0}\leq|U(x,0)|+1,$

provided that $K_{0}\geq K_{3,1}$ $\delta_{0}\leq 1$ and $\eta_{0}\leq 1$.
Integrating $U$ on each $P_{i}(t),i=1,2,3$, we obtain the following

$\displaystyle\int_{P_{1}(t)}U(t)dx$ $\displaystyle\leq$ $\displaystyle
C(K_{0})(T-t)^{\frac{n}{2}-\frac{1}{3}}|\ln(T-t)|^{\frac{n}{2}},$
$\displaystyle\int_{P_{2}(t)}U(t)dx$ $\displaystyle\leq$ $\displaystyle
C\int_{|x|\leq\epsilon_{0}}\left[\frac{|x|^{2}}{|\ln|x||}\right]^{-\frac{1}{3}}dx\leq
C\epsilon_{0}^{n-\frac{2}{3}}|\ln\epsilon_{0}|^{\frac{1}{3}},$
$\displaystyle\int_{P_{3}(t)}U(t)$ $\displaystyle\leq$
$\displaystyle\left[\int_{\frac{\epsilon_{0}}{4}\leq|x|,x\in\Omega}\left[H^{*}+1\right]dx\right],$

where $H^{*}$ is defined in (3.12). Using (3.21) and the above estimates, it
is easy to obtain the following estimate

$\mu(t)\leq C,\text{ for all }t\in[0,t_{1}],$

provided that $K_{0}\geq
K_{3,2}(\lambda,\gamma),\epsilon_{0}\leq\epsilon_{3,1}(\lambda,\gamma),\eta_{0}\leq\eta_{3,1}(\lambda,\gamma)$
and $T\leq T_{3,1}(K_{0},\lambda,\gamma)$. This yields (3.17) and (3.18) also
follows by using (2.11) and (3.17). We now give a proof to (3.19). Integrating
two sides of equation (2.9), we get the following ODE

$\bar{\mu}^{\prime}(t)+\frac{\bar{\theta}^{\prime}(t)}{\bar{\theta}(t)}\bar{\mu}(t)=\int_{\Omega}\Delta
U(t)dx+\int_{\Omega}\left(\left(U(t)+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}\right)^{4}-2\frac{|\nabla
U(t)|^{2}}{U(t)+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}}\right)dx.$
(3.22)

We aim at proving the following estimate

$\left|\int_{\Omega}\left(\left(U(t)+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}\right)^{4}-2\frac{|\nabla
U(t)|^{2}}{U(t)+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}}\right)dx\right|\leq
C(T-t)^{\frac{3n-8}{6}}|\ln(T-t)|^{n}.$ (3.23)

In order to do so, we first prove that

$\displaystyle\int_{\Omega}U^{4}(t)dx$ $\displaystyle\leq$ $\displaystyle
C(T-t)^{\frac{3n-8}{6}}|\ln(T-t)|^{n},$ (3.24)
$\displaystyle\int_{\Omega}\frac{\left|\nabla
U(t)\right|^{2}}{U(t)+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}}dx$
$\displaystyle\leq$ $\displaystyle C(T-t)^{\frac{3n-8}{6}}|\ln(T-t)|^{n}.$
(3.25)

The techniques of proofs (3.24) and (3.25) are the same. Therefore, we only
give here the proof of (3.25). Let us consider

$I(x,t)=\frac{\left|\nabla
U(x,t)\right|^{2}}{U(x,t)+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}}.$

Then,

$\int_{\Omega}I(x,t)dx\leq\int_{P_{1}(t)}I(x,t)dx+\int_{P_{2}(t)}I(x,t)dx+\int_{P_{3}(t)}I(x,t)dx.$

Now we claim the following lemma:

######  Lemma 3.11.

Under the hypothesis in Proposition 3.9, for all $t\in(0,t_{1}]$, the
following estimates hold:

$\displaystyle\text{ On }P_{1}(t):I(x,t)$ $\displaystyle\leq$ $\displaystyle
C(K_{0})(T-t)^{-\frac{4}{3}},$ (3.26) $\displaystyle\text{ On
}P_{2}(t):I(x,t)$ $\displaystyle\leq$ $\displaystyle
C(K_{0})\varrho^{-\frac{4}{3}}(x)\leq
C(K_{0})\left[\frac{|x|^{2}}{|\ln|x||}\right]^{-\frac{4}{3}},$ (3.27)
$\displaystyle\text{ On }P_{3}(t):I(x,t)$ $\displaystyle\leq$ $\displaystyle
C\left(\left|\nabla U(x,0)\right|^{2}+\eta_{0}^{2}\right)=C(|\nabla
H^{*}(x)|+\eta_{0}^{2}).$ (3.28)

###### Proof.

From the definition of $S(t),$ we easily derive (3.28). So, we only give here
the proofs of (3.26) and (3.27). We now start with (3.26). Let us consider
$x\in P_{1}(t)$ and we use the condition of $U$ in $P_{1}(t)$ to get the
following

$\frac{1}{C(K_{0})}(T-t)^{-\frac{1}{3}}\leq U(x,t)\leq
C(K_{0})(T-t)^{-\frac{1}{3}}.$ (3.29)

In addition to that, thanks to item $(ii)$ in Lemma 3.6, we get

$|\nabla_{y}W\left(\frac{x}{\sqrt{T-t}},-\ln(T-t)\right)|\leq\frac{C(K_{0})A^{2}}{\sqrt{|\ln(T-t)|}},$

which yields

$\left|\nabla U(x,t)\right|\leq C(K_{0})(T-t)^{-\frac{5}{6}}.$ (3.30)

Then, (3.26) follows by (3.29) and (3.30).

We now consider $x\in P_{2}(t)$. It is easy to derive from item $(ii)$ in
Definition 3.1 that

$\displaystyle\frac{1}{C(K_{0})}\varrho^{\frac{1}{3}}(x)$ $\displaystyle\leq$
$\displaystyle U(x,t)\leq C(K_{0})\varrho^{\frac{1}{3}}(x),$
$\displaystyle\left|\nabla U(x,t)\right|$ $\displaystyle\leq$ $\displaystyle
C\varrho^{-\frac{5}{6}}(x),$

provided that $\delta_{0}\leq\delta_{3,1}$ and
$\epsilon_{0}\leq\epsilon_{3,2}$. This gives (3.27) and concludes the proof of
Lemma 3.11. ∎

We now continue the proof of Proposition 3.9. Considering $t\in(0,t_{1})$ and
taking the integral on two sides of (3.26), we write

$\displaystyle\int_{P_{1}(t)}|I(x,t)|dx$ $\displaystyle\leq$ $\displaystyle
C(K_{0})\int_{|x|\leq K_{0}\sqrt{(T-t)|\ln(T-t)|}}(T-t)^{-\frac{4}{3}}dx$
$\displaystyle\leq$ $\displaystyle
C(K_{0})(T-t)^{\frac{n}{2}-\frac{4}{3}}\left|\ln(T-t)\right|^{\frac{n}{2}}.$

Integrating the two sides of (3.27) and using the following fact

$\varrho(x)\sim\frac{8}{K_{0}^{2}}\frac{|x|^{2}}{|\ln|x||}\text{ as }x\to 0,$

we obtain the following

$\displaystyle\int_{P_{2}(t)}\left|I(x,t)\right|\leq
C(K_{0})\left[\epsilon_{0}^{n-\frac{8}{3}}|\ln\epsilon_{0}|^{\frac{4}{3}}-((T-t)|\ln(T-t)|)^{\frac{3n-8}{6}}|\ln((T-t)|\ln(T-t)|)|^{\frac{4}{3}}\right].$

In addition to that, from (3.28), we have

$\int_{P_{3}(t)}|I(x,t)|dx\leq C.$

Hence, (3.25) holds.

In addition to that, using (F.3), we can derive that

$\int_{\Omega}\Delta U(t)dx<\infty,\forall t\in(0,t_{1}).$

Therefore, we have

$\lim_{\upsilon\to 0}\int_{\\{x,d(x,\partial\Omega)>\upsilon\\}}\Delta
Udx=\int_{\Omega}\Delta U(t)dx.$

Moreover, for all $\upsilon>0$ small enough and from item $(iii)$ of
Definition (3.1), we have

$\left|\int_{\\{x,d(x,\partial\Omega)>\upsilon\\}}\Delta
Udx\right|=\left|\int_{\partial\\{x,d(x,\partial\Omega)>\upsilon\\}}\nu(x)\cdot\nabla
U(x,t)dS\right|\leq C.$ (3.31)

This implies that

$\int_{\Omega}\Delta U(t)dx\leq C.$ (3.32)

Hence, from (3.22), (3.23) and (3.32), we derive the following

$\left|\bar{\mu}^{\prime}(t)+\frac{\bar{\theta}^{\prime}(t)}{\bar{\theta}(t)}\bar{\mu}(t)\right|\leq
C(T-t)^{\frac{3n-8}{6}}|\ln(T-t)|^{n}.$ (3.33)

In addition to that, from the relation between $\bar{\mu}$ and $\bar{\theta}$
in (2.11), we write

$\displaystyle\frac{\bar{\theta}^{\prime}(t)}{\bar{\theta}(t)}=\frac{2\gamma}{3\lambda^{\frac{1}{3}}}\left(1-\frac{\frac{2\gamma}{3\lambda^{\frac{1}{3}}}\bar{\mu}(t)}{\left(1+\gamma|\Omega|+\frac{\gamma}{\lambda^{\frac{1}{3}}}\bar{\theta}(t)\bar{\mu}(t)\right)^{\frac{1}{3}}}\right)^{-1}\left(1+\gamma|\Omega|+\frac{\gamma}{\lambda^{\frac{1}{3}}}\bar{\theta}(t)\bar{\mu}(t)\right)^{-\frac{1}{3}}\mu^{\prime}(t).$

We also have the fact that

$\sqrt{\bar{\theta}(t)}\geq\frac{\gamma}{\lambda^{\frac{1}{3}}}\bar{\mu}(t),$

which yields that

$1\leq\left(1-\frac{\frac{2\gamma}{3\lambda^{\frac{1}{3}}}\bar{\mu}(t)}{\left(1+\gamma|\Omega|+\frac{\gamma}{\lambda^{\frac{1}{3}}}\bar{\theta}(t)\bar{\mu}(t)\right)^{\frac{1}{3}}}\right)^{-1}\leq
3.$

Hence, $\bar{\theta}^{\prime}(t)$ and $\bar{\mu}^{\prime}(t)$ have the same
sign and we can use (3.33) to conclude that

$\left|\bar{\mu}^{\prime}(t)\right|\leq
C(T-t)^{\frac{3n-8}{6}}|\ln(T-t)|^{n}.$ (3.34)

This yields (3.19) and (3.20). Thus, we get the conclusion of the proof of
Proposition 3.9. ∎

######  Proposition 3.12 (Existence of a solution to equation (2.9), confined
in $S^{*}$).

We can find parameters
$T>0,K_{0}>0,\epsilon_{0}>0,\alpha_{0}>0,A>0,\delta_{0}>0,C_{0}>0,\eta_{0}>0$
such that there exist $(d_{0},d_{1})\in\mathbb{R}\times\mathbb{R}^{n}$ such
that with initial data $U_{d_{0},d_{1}}(0)$(given in (3.2)), the solution $U$
of equation (2.9) exists on $\Omega\times[0,T)$ and

$U\in S^{*}(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},T),$

where $S^{*}(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},T)$
given in (3.2).

###### Proof.

As a matter of fact, this Proposition plays a central role in our problem. In
other words, it will imply Theorem 1.1 (see subsection 3.4 below). The proof
of this Proposition will be presented in two steps:

\- First step: We use a reduction of our problem to a finite dimensional one.
More precisely, we prove that the controling $U$ in $S(t)$ for all $t\in[0,T)$
is reduced to the control of $(q_{0},q_{1})(s)$ in $\hat{\mathcal{V}}_{A}(s)$
(see Proposition 3.13 below).

\- Second step: In this step, we aim at proving that there exist
$(d_{0},d_{1})\in\mathbb{R}^{1+n}$ such that $U\in
S^{*}(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},T)$ with
suitable parameters. Then, the conclusion follows from a topological argument
based on Index theory.

We now give two main steps with more technical details:

$a)$ Reduction to a finite dimensional problem: In this step, we derive that
the control of $U\in S(t)$ with $t\in[0,T)$ is reduced to the control of the
transform function $q(s)$ such that two first components $(q_{0},q_{1})(s)$
are trapped in $\hat{\mathcal{V}}_{A}(s)$ (see (3.15)), where $s=-\ln(T-t)$.
More precisely, the following proposition is our statement:

######  Proposition 3.13 (Reduction to a finite dimensional problem).

There exist
$T>0,K_{0}>0,\epsilon_{0}>0,\alpha_{0}>0,A>0,\delta_{0}>0,C_{0}>0,\eta_{0}>0$
such that the following holds: We consider $U$ a solution of equation (2.9)
that exists on $[0,t_{1}],$ for some $t_{1}<T,$ with initial data
$U_{d_{0},d_{1}}(0)$ given in (3.2), for some
$(d_{0},d_{1})\in\mathcal{D}_{A}$. We also assume that we have $U\in S(t)$ for
all $\forall t\in[0,t_{1}]\text{ and }U\in\partial S(t_{1})$ (see the
definiton of
$S(t)=S(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},t)$ in
Definition 3.1 and the set $\mathcal{D}_{A}$ given in Proposition 3.8). Then,
the following statments hold:

  * $(i)$

We have $(q_{0},q_{1})(s_{1})\in\partial\hat{\mathcal{V}}_{A}(s_{1}),$ where
$(q_{0},q_{1})(s)$ are components of $q(s)$ given in (2.2) and $q(s)$ is the
transform function of $U$ defined in (2.24) and $s_{1}=\ln(T-t_{1})$.

  * $(ii)$

There exists $\nu_{0}>0$ such that for all $\nu\in(0,\nu_{0}),$ we have

$(q_{0},q_{1})(s_{1}+\nu)\notin\hat{\mathcal{V}}_{A}(s_{1}+\nu).$

Consequently, there exists $\nu_{1}>0$ such that

$U\notin S(t_{1}+\nu),\forall\nu\in(0,\nu_{1}).$

The idea of the proof is inspired (in a non trivial way) by the ideas given by
Merle and Zaag in [MZ97a]. Since the proof is long and technical, we leave it
to Section 4. Therefore, we assume here that Proposition 3.13 holds and go
forward to the conclusion of Proposition 3.12.

$b)$ Topological argument and the conclusion of Proposition 3.12: In this
step, by using Proposition 3.13 and a topological argument based on Index
theory, we conclude Proposition 3.12. More precisely, we prove that there
exist $T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0}$ and
$(d_{0},d_{1})\in\mathcal{D}_{A}$ such that with initial data
$U_{d_{0},d_{1}}(0)$ (defined in (3.2)), the solution of equation (2.9) exists
on $[0,T)$ and belongs to $S^{*}(T)$ where $S^{*}(T)$ is defined in Definiton
3.2. Indeed, let us consider parameters
$T>0,K_{0}>0,\epsilon_{0}>0,\alpha_{0}>0,A>0,\delta_{0}>0,C_{0},\eta_{0}>0$
such that Propositions 3.8 and 3.13 hold. Using Proposition 3.8, we have the
following

$\forall(d_{0},d_{1})\in\mathcal{D}_{A},\quad U_{d_{0},d_{1}}(0)\in S(0).$

In particular, it follows from Proposition 1.2.2 page 12 in Kavallaris and
Suzuki [KS18] together with Lemma 2.1 that equation (2.9) is locally in time
well-posed in $C^{2,1}(\Omega\times(0,T_{0}))\subset
C(\bar{\Omega}\times[0,T_{0}])$, for some $T_{0}>0$ . Therefore, for every
$(d_{0},d_{1})\in\mathcal{D}_{A}$, we define $t^{*}(d_{0},d_{1})\in[0,T)$ as
the maximum time, satisfying

$U_{d_{0},d_{1}}\in S(t),\forall t\in[0,t^{*}(d_{0},d_{1})),$

where $U_{d_{0},d_{1}}$ is the solution of (2.9) corresponding to initial data
$U_{d_{0},d_{1}}(0),$ introduced in (3.2). Then, we have two possible cases:

$a)$ Either $t^{*}(d_{0},d_{1})=T$ for some $(d_{0},d_{1})\in\mathcal{D}_{A},$
then, we get the conclusion of the proof.

$b)$ Or $t^{*}(d_{0},d_{1})<T,$ for all $(d_{0},d_{1})\in\mathcal{D}_{A}$.
This case in fact never occurs, as we will show in the following.

Indeed, assuming by contradiction that case $b)$ hold and using the continuity
of the solution in time and the definition of the maximun time
$t^{*}(d_{0},d_{1}),$ we have

$U_{d_{0},d_{1}}(t^{*}(d_{0},d_{1}))\in\partial S(t^{*}(d_{0},d_{1})).$

Thanks to the finite dimensional reduction property given in item $(i)$ of
Proposition 3.13, we derive the following

$\left(q_{0},q_{1}\right)(s_{*}(d_{1},d_{2}))\in\partial\hat{\mathcal{V}}_{A}(s_{*}(d_{0},d_{1})),$

where $q_{0},q_{1}$ are defined in (2.2) as the components of
$q_{d_{0},d_{1}},$ which is a transformed function of $U_{d_{0},d_{1}}$ (see
(2.24)) and $s_{*}(d_{0},d_{1})=-\ln(T-t^{*}(d_{0},d_{1})).$ Then, we may
define the following mapping

$\displaystyle\Lambda:\mathcal{D}_{A}$ $\displaystyle\to$
$\displaystyle\left([-1,1]\times[-1,1]^{n}\right)$
$\displaystyle\left(d_{0},d_{1}\right)$ $\displaystyle\mapsto$
$\displaystyle\frac{s_{*}^{2}(d_{0},d_{1})}{A}\left(q_{0},q_{1}\right)\left(s_{*}(d_{0},d_{1})\right).$

From the definition of $t^{*}(d_{1},d_{2}),$ the components $(q_{0},q_{1})$
and the transversal crossing property given in item $(ii)$ in Proposition
3.13, we see that $\Lambda$ is continuous on $\mathcal{D}_{A}$. In addition to
that, from item $(i)$ of Proposition 3.8, we can derive that for all
$(d_{0},d_{1})\in\partial\mathcal{D}_{A}$

$\left(q_{0},q_{1}\right)(s_{0})\in\partial\hat{\mathcal{V}}_{A}(s_{0}),\quad
s_{0}=-\ln T.$

However, using item $(ii)$ of Proposition 3.13 again and the definition of
$t^{*}(d_{0},d_{1})$ we deduce that

$t^{*}(d_{0},d_{1})=0,$

which yields

$s_{*}(d_{0},d_{1})=s_{0}\text{ and
}\Lambda(d_{0},d_{1})=\frac{s_{0}^{2}}{A}\Gamma(d_{0},d_{1}),$

where $\Gamma$ is defined in item $(I)$ of Proposition 3.8. Hence, thanks to
(3.16), we conclude

$\text{ deg }\left(\Lambda\left.\right|_{\mathcal{D}_{A}}\right)\neq 0.$

In fact, such a mapping $\Lambda$ can not exist by using Index theory. Hence,
case $b)$ doesn’t occur only case $a)$ occurs. Thus, the conclusion of
Proposition 3.12 follows. ∎

###  3.4. The conclusion of Theorem 1.1

In this subsection, we would like to give a complete proof of Theorem 1.1. We
now consider the solution $U$ which has been constructed in Proposition 3.12.
Then, $U$ exists on $[0,T)$ and

$U(t)\in S(t),\forall t\in[0,T).$

Using item $(i)$ in Definition 3.1, we have the following

$q\text{ exists on }[-\ln T,+\infty)\text{ and
}\|q(.,s)\|_{L^{\infty}(\mathbb{R}^{n})}\leq\frac{C}{\sqrt{s}},\forall
s\in[-\ln T,+\infty),$ (3.35)

for some constant $C>0$. Thanks to (2.5), (2.7), (2.14) and (2.20), we have

$\left\|\frac{(T-t)^{\frac{1}{3}}\lambda^{\frac{1}{3}}}{\bar{\theta}(t)(1-u(.,t))}-\left(3+\frac{9}{8}\frac{|.|^{2}}{(T-t)|\ln(T-t)|}\right)^{-\frac{1}{3}}\right\|_{L^{\infty}(\Omega)}\leq\frac{C}{\sqrt{|\ln(T-t)|}}.$
(3.36)

Using (3.18) and (3.20), we can derive that $\bar{\theta}(t)$ converges to
$\theta_{T}>0$ with

$\left|\bar{\theta}(t)-\theta_{T}\right|\leq C(T-t)^{\frac{1}{12}},\forall
t\in[0,T).$

This implies (1.6) with $\theta^{*}=\frac{\theta_{T}}{\lambda^{\frac{1}{3}}}.$
Then, item $(i)$ of Theorem 1.1 follows. We now prove that $u$ quenches only
at $0$. Indeed, from the above estimate, we can derive that $0$ is a quenching
point of $u$. Now, we aim at proving that $x\in\Omega\setminus\\{0\\}$ are not
quenching points of $u$. In fact, relying on relations (2.5) and (2.7), it is
enough to prove the following Lemma:

######  Proposition 3.14.

The solution $U$ satisfies the following statements:

  * $(i)$

For all $x\in\Omega\setminus\\{0\\},$ there exits $\nu(x)>0$ such that

$\limsup_{t\to T}\sup_{|x^{\prime}-x|\leq\nu(x)}U(x^{\prime},t)<+\infty.$
(3.37)

  * $(ii)$

For all $x\in\Omega\setminus\\{0\\}$, $\lim_{t\to T}U(x,t)$ exists. In
particular, if we define $U^{*}(x)=\lim_{t\to T}U(x,t),$ for all
$x\in\Omega\diagdown\\{0\\}$, then $u^{*}\in C(\bar{\Omega}\setminus\\{0\\}),$
and $U(t)$ uniformly converges to $u^{*}$ on every compact subset of
$\bar{\Omega}\setminus\\{0\\}$. In particular, we have the following
asymptotic behavior

$U^{*}(x)\sim\left[\frac{9}{32}\frac{|x|^{2}}{|\ln|x||}\right]^{-\frac{1}{3}},\text{
as }x\to 0.$ (3.38)

###### Proof.

We consider $U$ the solution constructed in Proposition 3.12. The proof will
be given in two parts:

\- The proof of item $(i)$: The proof follows from the definition of shrinking
set $S(t)$. Let us consider two cases: $|x|>\frac{\epsilon_{0}}{4},x\in\Omega$
and $|x|\leq\frac{\epsilon_{0}}{4},x\in\Omega$.

\+ The case where $|x|>\frac{\epsilon_{0}}{4},x\in\Omega$: Using item $(iii)$
of Definition 3.1, we conclude that for all $t\in[0,T)$,

$U(x,t)\leq U(x,0)+\eta_{0}<+\infty.$

Then, (3.37) follows.

\+ The case where $|x|\leq\frac{\epsilon_{0}}{4},x\in\Omega$: For every $x$ in
that region, we can find $t_{x}$ close to $T$ such that
$|x|\in\left[\frac{K_{0}}{4}\sqrt{(T-t_{x})|\ln(T-t_{x})|},\epsilon_{0}\right]$.
Moreover, we derive that
$|x|\in\left[\frac{K_{0}}{4}\sqrt{(T-t)|\ln(T-t)|},\epsilon_{0}\right]$ for
all $t\in[t_{x},T)$. Considering $t\in[t_{x},T)$ and using item $(ii)$ in
Definition 3.1, we derive the following

$U(x+\xi\sqrt{\varrho(x)},t)\leq\varrho^{-\frac{1}{3}}(x)\left[\hat{\mathcal{U}}(\tau(x,t))+\delta_{0}\right],\forall|\xi|\leq\alpha_{0}\sqrt{|\ln\varrho(x)|}.$

This estimate directly implies (3.37).

\- The proof of item $(ii)$: By using parabolic regularity and the technique
given by Merle in [Mer92], item $(i)$ and Lemma F.1, we may derive that there
exists a function $U^{*}\in C(\bar{\Omega}\setminus\\{0\\})$ such that
$U(x,t)\to U^{*}(x),$ as $t\to T$, for all $x\in\bar{\Omega},x\neq 0$.
Moreover, one can prove that the convergence is uniform on every compact
subset of $\bar{\Omega}\backslash\\{0\\}$. It remains to give the asymptotic
behavior (3.38). We consider $x_{0}\in\Omega$ such that $|x_{0}|$ is small
enough. We first introduce the following functions:
$\mathcal{U}(x_{0},\xi,\tau)$ is defined in (3.4) and

$\displaystyle\mathcal{V}(x_{0},\xi,\tau)$ $\displaystyle=$
$\displaystyle\nabla_{\xi}\mathcal{U}(x_{0},\xi,\tau),$ (3.39)

where $\xi\in\varrho^{-\frac{1}{3}}(x_{0})(\Omega-x_{0})\subset\mathbb{R}^{n}$
and $\tau\in\left[-\frac{t(x_{0})}{\varrho(x_{0})},1\right)$, where $t(x_{0})$
and $\varrho(x_{0})$ are defined as in (3.5) and (3.6), respectively. We aim
at proving the following estimates:

$\sup_{\tau\in[0,1),|\xi|\leq|\ln(\varrho(x_{0}))|^{\frac{1}{4}}}\left|\mathcal{U}(x_{0},\xi,\tau)-\hat{\mathcal{U}}(\tau)\right|\leq\frac{C}{|\ln(\varrho(x_{0}))|^{\frac{1}{4}}},$
(3.40) $\sup_{\tau\in[0,1),|\xi|\leq
2|\ln(\varrho(x_{0}))|^{\frac{1}{4}}}|\mathcal{V}(x_{0},\xi,\tau)|\leq\frac{C}{|\ln(\varrho(x_{0}))|^{\frac{1}{4}}},$
(3.41)

and

$\displaystyle\sup_{\tau\in[\tau_{0},1),|\xi|\leq\frac{1}{2}|\ln(\varrho(x_{0}))|^{\frac{1}{4}}}\left|\partial_{\tau}\mathcal{U}(x_{0},\xi,\tau)\right|\leq
C(x_{0}),$ (3.42)

for some $\tau_{0}\in(0,1)$, fixed, and we also recall that
$\hat{\mathcal{U}}(\tau)$ is introduced in (3.10).

We see that (3.41) follows from the fact that $U\in S(t),\forall t\in[0,T)$
and item $(ii)$ of Definition 3.1. Thus, we only need to give the proofs of
(3.40) and (3.42).

\- The proof of (3.40): We write here the equation of $\mathcal{U}$ from (3.9)

$\partial_{\tau}\mathcal{U}=\Delta_{\xi}\mathcal{U}-2\frac{\left|\nabla_{\xi}\mathcal{U}\right|^{2}}{\mathcal{U}+\frac{\lambda^{\frac{1}{3}}\varrho^{\frac{1}{3}}(x_{0})}{\tilde{\theta}(\tau)}}+\left(\mathcal{U}+\frac{\lambda^{\frac{1}{3}}\varrho^{\frac{1}{3}}(x_{0})}{\tilde{\theta}(\tau)}\right)^{4}-\frac{\tilde{\theta}^{\prime}(\tau)}{\tilde{\theta}(\tau)}\mathcal{U},$
(3.43)

where $\tilde{\theta}(\tau)=\bar{\theta}(\tau\varrho(x_{0})+t(x_{0}))$ is
given in (3.8). From (3.36) with $t=t(x_{0})$, we derive that

$\displaystyle\sup_{|\xi|\leq
6|\ln(\varrho(x_{0}))|^{\frac{1}{4}}}\left|\mathcal{U}(x_{0},\xi,0)-\hat{\mathcal{U}}(0)\right|$
$\displaystyle\leq$
$\displaystyle\frac{C}{|\ln(\varrho(x_{0}))|^{\frac{1}{4}}}.$ (3.44)

In addition to that, from item $(ii)$ of Definition 3.1, we have for all
$|\xi|\leq 6|\ln\varrho(x_{0})|^{\frac{1}{4}}$ and $\tau\in[0,1)$:

$\displaystyle\mathcal{U}(x_{0},\xi,\tau)\geq\frac{1}{2}\hat{\mathcal{U}}(0),$
(3.45)
$\displaystyle\mathcal{U}(x_{0},\xi,\tau)\leq\frac{3}{2}\hat{\mathcal{U}}(1),$
(3.46)

provided that $\delta_{0}\leq\frac{1}{2}\hat{\mathcal{U}}(0)$. We now consider
$\mathbb{U}(\xi,\tau)$ as follows

$\mathbb{U}(\xi,\tau)=\mathcal{U}(x_{0},\xi,\tau)-\hat{\mathcal{U}}(\tau),\text{
where }\xi\in\varrho^{-\frac{1}{3}}(x_{0})(\Omega-x_{0})\text{ and
}\tau\in[0,1).$

We then derive an equation satisfied by $\mathbb{U}$

$\partial_{\tau}\mathbb{U}=\Delta_{\xi}\mathbb{U}+G_{1}+G_{2},$ (3.47)

where $G_{1},G_{2}$ are defined as follows

$\displaystyle G_{1}(\xi,\tau)$ $\displaystyle=$
$\displaystyle-2\frac{|\nabla\mathcal{U}|^{2}}{\mathcal{U}+\frac{\lambda^{\frac{1}{3}}\varrho^{\frac{1}{3}}(x_{0})}{\tilde{\theta}(\tau)}}-\frac{\tilde{\theta}^{\prime}(\tau)}{\tilde{\theta}(\tau)}\mathcal{U},$
$\displaystyle G_{2}(\xi,\tau)$ $\displaystyle=$
$\displaystyle\left(\mathcal{U}+\frac{\lambda^{\frac{1}{3}}\varrho^{\frac{1}{3}}(x_{0})}{\tilde{\theta}(\tau)}\right)^{4}-\hat{\mathcal{U}}^{4}(\tau).$

Next, we derive from the definition of $\tilde{\theta}(\tau)$, Proposition 3.9
and the fact that for all $\tau\in\left(0,1\right)$,

$\left|\tilde{\theta}^{\prime}(\tau)\right|\leq
C\varrho^{\frac{1}{12}}(x_{0})(1-\tau)^{-\frac{11}{12}},$

and

$1\leq\tilde{\theta}(\tau)\leq C.$

Hence, from (3.41), (3.45) and (3.46), we deduce that for all
$\tau\in[0,1),|\xi|\leq 2|\ln\varrho(x_{0})|^{\frac{1}{4}}$

$\displaystyle\left|G_{1}(\xi,\tau)\right|$ $\displaystyle=$
$\displaystyle\left|-2\frac{|\nabla\mathcal{U}(x_{0},\xi,\tau)|^{2}}{\mathcal{U}(x_{0},\xi,\tau)+\frac{\lambda^{\frac{1}{3}}\varrho^{\frac{1}{3}}(x_{0})}{\tilde{\theta}(\tau)}}-\frac{\tilde{\theta}^{\prime}(\tau)}{\tilde{\theta}(\tau)}\left(\mathcal{U}(x_{0},\xi,\tau)\right)\right|$
$\displaystyle\leq$
$\displaystyle\frac{C}{|\ln\varrho(x_{0})|^{\frac{1}{4}}}\left((1-\tau)^{-\frac{11}{12}}+1\right).$

In addition to that, we derive from (3.46) that

$|G_{2}(\xi,\tau)|\leq
C|\mathbb{U}(x_{0},\xi,\tau)|+\frac{C}{|\ln\varrho(x_{0})|^{\frac{1}{4}}}.$

We now recall the cut-off function $\chi_{0},$ defined as in (2.18) , then, we
introduce

$\phi_{1}(\xi)=\chi_{0}\left(\frac{|\xi|}{\left|\ln(\varrho(x_{0}))\right|^{\frac{1}{4}}}\right).$

As a matter of fact, we have some rough estimates on $\phi_{1}$

$\|\nabla_{\xi}\phi_{1}\|_{L^{\infty}(\mathbb{R}^{n})}\leq\frac{C}{|\ln(\varrho(x_{0}))|^{\frac{1}{4}}}\text{
and
}\|\Delta_{\xi}\phi_{1}\|_{L^{\infty}(\mathbb{R}^{n})}\leq\frac{C}{\left|\ln(\varrho(x_{0}))\right|^{\frac{1}{2}}}.$
(3.48)

Let us define $\mathbb{U}_{1}(\xi,\tau)=\phi_{1}(\xi)\mathbb{U}(\xi,\tau),$
for all $\xi\in\mathbb{R}^{n}$ and $\tau\in[0,1)$. Then, $\mathbb{U}_{1}$
satisfies the following equation

$\partial_{t}\mathbb{U}_{1}=\Delta\mathbb{U}_{1}-2\nabla\phi_{1}\cdot\nabla\mathbb{U}-\Delta\phi_{1}\mathbb{U}+\phi_{1}G_{1}(\xi,\tau)+\phi_{1}G_{2}(\xi,\tau).$

Using Duhamel’s principal, we write an integral equation satisfied by
$\mathbb{U}_{1}$

$\mathbb{U}_{1}(\tau)=e^{\tau\Delta}\mathbb{U}_{1}(0)+\int_{0}^{\tau}e^{(\tau-\sigma)\Delta}\left[-2\nabla\phi_{1}\cdot\nabla\mathbb{U}-\Delta\phi_{1}\mathbb{U}+\phi_{1}G_{1}+\phi_{1}G_{2}\right](\sigma)d\sigma.$

This implies that for all $\tau\in[0,1),$ we have

$\displaystyle\|\mathbb{U}_{1}(.,\tau)\|_{L^{\infty}(\mathbb{R}^{n})}\leq\frac{C}{|\ln(\varrho(x_{0}))|^{\frac{1}{4}}}+C\int_{0}^{\tau}\|\mathbb{U}_{1}(.,\sigma)\|_{L^{\infty}(\mathbb{R}^{n})}d\sigma.$

Thanks to Granwall’s inequality, we get the following

$\|\mathbb{U}_{1}(.,\tau)\|_{L^{\infty}(\mathbb{R}^{n})}\leq\frac{C}{|\ln(\varrho(x_{0}))|^{\frac{1}{4}}},\forall\tau\in\left[0,1\right),$

which yields (3.40).

Using (3.42), we can derive that the limit $\lim_{\tau\to
1}\mathcal{U}(x_{0},0,\tau)$ exists. In addition to that, we derive from
(3.40) that

$U^{*}(x_{0})=\lim_{\tau\to
1}\frac{\mathcal{U}(x_{0},0,\tau)}{\varrho^{\frac{1}{3}}(x_{0})}\sim\left(\frac{9}{8}\frac{K_{0}^{2}}{16}\varrho(x_{0})\right)^{-\frac{1}{3}}\sim\left[\frac{9}{16}\frac{|x_{0}|^{2}}{|\ln|x_{0}||}\right]^{-\frac{1}{3}}\text{
as }x_{0}\to 0.$

This is the conclusion of (3.38). So, we get the proof in Proposition 3.14 and
we also get the complete conclusion of Theorem 1.1. ∎

##  4\.  Reduction to a finite dimensional problem

This section plays a central role in our analysis. In fact, it is devoted to
the proof of Proposition 3.13. More precisely, this section has two parts:

\- In the first subsection, we prove  priori estimates  on $U$ in
$P_{1}(t),P_{2}(t)$ and $P_{3}(t)$ when $U$ is trapped in $S(t)$.

\- The second subsection is devoted to the conclusion of Proposition 3.13. In
fact, we use the first subsection to derive that $U$ satisfies almost all the
conditions in $S(t)$ with strict bounds, except for the bounds on $q_{0}(s)$
and $q_{1}(s)$, with $s=-\ln(T-t)$. This means that in order to control $U$ in
$S(t)$, we need to control only $(q_{0},q_{1})(s)$ in
$\hat{\mathcal{V}}_{A}(s),$ defined in (3.15). In addition to that, we also
prove the outgoing transversal crossing property. It means that if the
solution $U$ touches the boundary of $S(t_{1})$ for some $t_{1}\in(0,T)$,
then, $U$ will be outside $S(t)$ for all $t\in(t_{1},t_{1}+\nu)$ with $\nu$
small enough. In one word, this is the reduction to a finite dimensional
problem: the control of two components $(q_{0},q_{1})(s)$ in
$\hat{\mathcal{V}}_{A}(s)$.

###  4.1. A priori estimates

We proceed in 3 steps: $a,b$ and $c)$, respectively devoted to parts
$P_{1}(t),P_{2}(t)$ and $P_{3}(t)$.

a) We aim in the following Proposition at proving a priori estimates for $U$
in $P_{1}(t)$:

######  Lemma 4.1.

There exists $K_{4}>0,A_{4}>0$ such that for all $K_{0}\geq K_{4},A\geq A_{4}$
and $l^{*}>0$ there exists $T_{4}(K_{0},A,l^{*})$ such that for all
$\epsilon_{0}>0,\alpha_{0}>0,\delta_{0}>0,\eta_{0}>0,C_{0}>0,T\leq T_{4}$ and
for all $l\in[0,l^{*}]$, the following holds: Assume that we have the
following conditions:

  * -

We consider initial data $U(0)=U_{d_{0},d_{1}}(0)$, given in (3.2) and
$(d_{0},d_{1})\in\mathcal{D}_{A},$ given in Proposition 3.8 such that
$(q_{0},q_{1})(s_{0})$ belongs to $\hat{\mathcal{V}}_{A}(s_{0}),$ where
$s_{0}=-\ln T$, $\hat{\mathcal{V}}_{A}(s)$ is defined in (3.15) and
$q_{0},q_{1}$ are components of $q_{d_{0},d_{1}}(s_{0}),$ a transform function
of $U,$ defined in (2.24).

  * -

We have $U\in
S(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},t)$ for all
$t\in[T-e^{-\sigma},T-e^{-(\sigma+l)}]$, for some $\sigma\geq s_{0}$ and
$l\in[0,l^{*}]$.

Then, the following estimates hold:

  * $(i)$

For all $s\in[\sigma,\sigma+l]$, we have

$\displaystyle\left|q_{0}^{\prime}(s)-q_{0}(s)\right|+\left|q^{\prime}_{1,i}(s)-\frac{1}{2}q_{1,i}(s)\right|\leq\frac{C}{s^{2}},\forall
i\in\\{1,...,n\\},$ (4.1)

and

$\left|q^{\prime}_{2,i,j}(s)+\frac{2}{s}q_{2,i,j}(s)\right|\leq\frac{CA}{s^{3}},\forall
i,j\in\\{1,...,n\\},$ (4.2)

where $q_{1}=(q_{1,j})_{1\leq i\leq n},q_{2}=(q_{2,i,j})_{1\leq i,j\leq n}$
and $q_{1},q_{2}$ are defined in (2.34).

  * $(ii)$

Control of $q_{-}(s)$: For all $s\in[\sigma,\sigma+l],y\in\mathbb{R}^{n}$ we
have the two following cases:

    * -

The case where $\sigma\geq s_{0}:$

$\displaystyle\left|q_{-}(y,s)\right|\leq
C\left(Ae^{-\frac{s-\sigma}{2}}+A^{2}e^{-(s-\sigma)^{2}}+(s-\sigma)\right)\frac{(1+|y|^{3})}{s^{2}},$
(4.3)

    * -

The case where $\sigma=s_{0}$

$\left|q_{-}(y,s)\right|\leq C(1+(s-\sigma))\frac{(1+|y|^{3})}{s^{2}}.$ (4.4)

  * $(iii)$

Control of the gradient term of $q$: For all
$s\in[\sigma,\sigma+l],y\in\mathbb{R}^{n}$, we have the two following cases:

    * -

The case where $\sigma\geq s_{0}:$

$\displaystyle\left|(\nabla q)_{\perp}(y,s)\right|\leq
C\left(Ae^{-\frac{s-\sigma}{2}}+e^{-(s-\sigma)^{2}}+(s-\sigma)+\sqrt{s-\sigma}\right)\frac{(1+|y|^{3})}{s^{2}},$
(4.5)

    * -

The case where $\sigma=s_{0}$

$\left|(\nabla q)_{\perp}(y,s)\right|\leq
C\left(1+(s-\sigma)+\sqrt{s-\sigma}\right)\frac{(1+|y|^{3})}{s^{2}}.$ (4.6)

  * $(iii)$

Control of the outside part $q_{e}$: For all $s\in[\sigma,\sigma+\lambda],$ we
have the two following cases:

    * -

The case where $\sigma\geq s_{0}:$

$\displaystyle\left\|q_{e}(.,s)\right\|_{L^{\infty}(\mathbb{R}^{n})}\leq
C\left(A^{2}e^{-\frac{s-\sigma}{2}}+Ae^{(s-\sigma)}+1+(s-\sigma)\right)\frac{1}{\sqrt{s}},$
(4.7)

    * -

The case where $\sigma=s_{0}$

$\left\|q_{e}(.,s)\right\|_{L^{\infty}(\mathbb{R}^{n})}\leq
C\left(1+(s-\sigma)\right)\frac{1}{\sqrt{s}}.$ (4.8)

###### Proof.

The proof of this proposition relies completely on techniques given by Merle
and Zaag in [MZ97a]. As a matter of fact, the equation (2.25) is quite the
same as in that paper if we ignore some perturbations which will be very small
in our analysis. More precisely, thanks to Lemmas D.1, D.2, D.3, D.4, D.5 and
D.6, we assert that the techniques in [MZ97a] hold in our case. Hence, we
kindly refer the reader to Lemma 3.2 at page 1523 in [MZ97a] for more details.
∎

This implies a priori estimates in $P_{1}(t)$ as follows:

######  Proposition 4.2 (A priori estimates in $P_{1}(t)$).

There exist $K_{5}\geq 1$ and $A_{5}\geq 1$ such that for all $K_{0}\geq
K_{5},A\geq
A_{5},\epsilon_{0}>0,\alpha_{0}>0,\delta_{0}\leq\frac{1}{2}\hat{\mathcal{U}}(0),C_{0}>0,\eta_{0}>0$,
there exists
$T_{5}(K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0})$ such that
for all $T\leq T_{5}$, the following holds: If $U$ a nonnegative solution of
equation (2.9) satisfying $U\in
S(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},t)$ for all
$t\in[0,t_{5}]$ for some $t_{5}\in[0,T)$, and initial data
$U(0)=U_{d_{0},d_{1}}$ given in (3.2) for some $d_{0},d_{1}\in\mathcal{D}_{A}$
given in Proposition 3.8, then, for all $s\in[-\ln T,-\ln(T-t_{5})]$, we have
the following:

$\displaystyle\forall i,j\in\\{1,\cdots,n\\},\quad|q_{2,i,j}(s)|$
$\displaystyle\leq$ $\displaystyle\frac{A^{2}\ln s}{2s^{2}},$
$\displaystyle\left\|\frac{q_{,-}(.,s)}{1+|y|^{3}}\right\|_{L^{\infty}}$
$\displaystyle\leq$ $\displaystyle\frac{A}{2s^{2}},\left\|\frac{(\nabla
q(.,s))_{\perp}}{1+|y|^{3}}\right\|_{L^{\infty}}\leq\frac{A}{2s^{2}}\text{ and
}\|q_{e}(s)\|_{L^{\infty}}\leq\frac{A^{2}}{2\sqrt{s}},$

where $q$ is a transformed function of $U$ given in (2.24).

###### Proof.

The proof is a consequence of Lemma 4.1. In particular, the proof is the same
as in the work of Merle and Zaag in [MZ97b]. Hence, we refer the reader to
Proposition 3.7, page 157 in that work. ∎

b) We now show a priori estimate on $U$ in $P_{2}(t)$. We start with the
following lemma:

######  Lemma 4.3 (A priori estimates in the intermediate region).

There exists $K_{6}$ and $A_{6}>0,$ such that for all $K_{0}\geq K_{6},A\geq
A_{6},\delta_{6}>0$, there exists
$\alpha_{6}(K_{0},\delta_{6})>0,C_{6}(K_{0},A)>0$ such that for all
$\alpha_{0}\leq\alpha_{6},C_{0}>0$, there exists
$\epsilon_{6}(\alpha_{0},A,\delta_{6},C_{0})$ such that for all
$\epsilon_{0}\leq\epsilon_{6}$, there exists
$T_{6}(\epsilon_{0},A,\delta_{6},C_{0})$ and
$\eta_{6}(\epsilon_{0},A,\delta_{0},C_{0})>0$ such that for all $T\leq
T_{6},\eta_{0}\leq\eta_{6},\delta_{0}\leq\frac{1}{2}\left(3+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}$,
the following holds: if $U\in
S(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},t)$ for all
$t\in[0,t_{*}],$ for some $t_{*}\in[0,T)$, then, for all
$|x|\in\left[\frac{K_{0}}{4}\sqrt{(T-t_{*})|\ln(T-t_{*})|},\epsilon_{0}\right]$,
we have:

  * $(i)$

For all $|\xi|\leq\frac{7}{4}\alpha_{0}\sqrt{|\ln\varrho(x)|}$ and
$\tau\in\left[\max\left(0,-\frac{t(x)}{\varrho(x)}\right),\frac{t_{*}-t(x)}{\varrho(x)}\right]$,
the transformed function $\mathcal{U}(x,\xi,\tau)$ defined in (3.4) satisfies
the following:

$\displaystyle\left|\nabla_{\xi}\mathcal{U}(x,\xi,\tau)\right|$
$\displaystyle\leq$ $\displaystyle\frac{2C_{0}}{\sqrt{|\ln\varrho(x)|}},$
(4.9) $\displaystyle\mathcal{U}(x,\xi,\tau)$ $\displaystyle\geq$
$\displaystyle\frac{1}{4}\left(3+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}},$
(4.10) $\displaystyle\left|\mathcal{U}(x,\xi,\tau)\right|$ $\displaystyle\leq$
$\displaystyle 4.$ (4.11)

  * $(ii)$

For all $|\xi|\leq 2\alpha_{0}\sqrt{|\ln\varrho(x)|}$ and
$\tau_{0}=\max\left(0,-\frac{t(x)}{\varrho(x)}\right)$: we have

$\left|\mathcal{U}(x,\xi,\tau_{0})-\hat{\mathcal{U}}(\tau_{0})\right|\leq\delta_{6}\text{
and
}\left|\nabla_{\xi}\mathcal{U}(x,\xi,\tau_{0})\right|\leq\frac{C_{6}}{\sqrt{|\ln\varrho(x)|}}.$

###### Proof.

We leave the proof to Appendix B. ∎

Using the above lemma, we now give a priori estimates in $P_{2}(t).$ The
following is our statement:

######  Proposition 4.4 (A priori estimates in $P_{2}(t)$).

There exists $K_{7}>0$ and $A_{7}>0$ such that for all $K_{0}\geq K_{7},A\geq
A_{7}$, there exists $\delta_{7}\leq\frac{1}{2}\hat{\mathcal{U}}(0)$ and
$C_{7}(K_{0},A)$ such that for all $\delta_{0}\leq\delta_{7},C_{0}\geq C_{7}$
there exists $\alpha_{7}(K_{0},\delta_{0})$ such that for all
$\alpha_{0}\leq\alpha_{7}$, there exist
$\epsilon_{7}(K_{0},\delta_{0},C_{0})>0$ such that for all
$\epsilon_{0}\leq\epsilon_{7}$, there exists
$T_{7}(\epsilon_{0},A,\delta_{0},C_{0})>0$ such that for all $T\leq T_{7}$ the
following holds: We assume that we have $U\in
S(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},t)$ for all
$t\in[0,t_{7}]$ for some $t_{7}\in[0,T)$, then, for all
$|x|\in\left[\frac{K_{0}}{4}\sqrt{(T-t_{*})|\ln(T-t_{*})|},\epsilon_{0}\right],$
$|\xi|\leq\alpha_{0}\sqrt{|\ln\varrho(x)|}$ and
$\tau\in\left[\max\left(-\frac{t(x)}{\varrho(x)},0\right),\frac{t_{7}-t(x)}{\varrho(x)}\right]$,
we have

$\left|\mathcal{U}(x,\xi,\tau_{*})-\hat{\mathcal{U}}(x,\xi,\tau_{*})\right|\leq\frac{\delta_{0}}{2}\text{
and
}\left|\nabla\mathcal{U}(x,\xi,\tau)\right|\leq\frac{C_{0}}{2\sqrt{|\ln\varrho(x)|}},$

where $\varrho(x)=T-t(x)$.

###### Proof.

We leave the proof to Appendix C. ∎

######  Remark 4.5.

Unlike what Merle and Zaag did in [MZ97a], we don’t require any condition in
$\nabla^{2}\mathcal{U}$ in $P_{2}(t)$ (see Definition 3.1), as we have
aldready stated in Remark 3.3. Accordingly, our a priori estimates in
$P_{2}(t)$ will be simpler than those of [MZ97a], as one may see from the
proof given in Appendix C.

c) We now give a priori estimates on $U$ in $P_{3}(t)$:

######  Proposition 4.6 (A priori estimates in $P_{3}$).

Let us consider
$K_{0}>0,\epsilon_{0}>0,\alpha_{0}>0,A>0,\delta_{0}\in[0,\frac{1}{2}\hat{\mathcal{U}}(0)],C_{0}>0,\eta_{0}>0$.
Then, there exists $T_{8}(\eta_{0})>0$ such that for all $T\leq T_{8}$, the
following holds: We assume that $U$ is a nonnegative solution of (2.9) on
$[0,t_{8}]$ for some $t_{8}<T$, and $U\in
S(K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},t)$ for all
$t\in[0,t_{8}]$ and initial data $U(0)=U_{d_{0},d_{1}}$ given in (3.2) with
$|d_{0}|,|d_{1}|\leq 2$. Then, for all $|x|\geq\frac{\epsilon_{0}}{4}$ and
$t\in(0,t_{8}],$

$\displaystyle\left|U(x,t)-U(x,0)\right|\leq\frac{\eta_{0}}{2},$ (4.12)
$\displaystyle\left|\nabla U(x,t)-\nabla
e^{t\Delta}U(x,0)\right|\leq\frac{\eta_{0}}{2}.$ (4.13)

######  Remark 4.7.

As we have mentioned in Remark 3.3, we draw the attention of the reader to the
change we have made with respect to the work of Merle and Zaag in [MZ97a]: We
compare $\nabla U(t)$ to $\nabla e^{t\Delta}U(0)$ and not to $\nabla U(0)$ in
[MZ97a] and this is crucial, since we are working on a bounded domain.

Following the remark, we have just stated, we give in the following a crucial
parabolic estimate for the free Dirichlet heat semi-group in $\Omega$:

######  Lemma 4.8 (A parabolic regularity on the linear problem).

Let us consider initial data $U_{d_{0},d_{1}},$ given in (3.2), for some
$|d_{0}|,|d_{1}|\leq 2$. If we define

$L(t)=e^{t\Delta}U_{d_{0},d_{1}},t\in(0,T].$

Then, $L(t)\in C(\bar{\Omega}\times[0,T])\cap C^{\infty}(\Omega\times(0,T])$.
Moreover, the following holds

$\|\nabla_{x}L(t)\|_{L^{\infty}\left(|x|\geq\frac{\epsilon_{0}}{8},x\in\Omega\right)}\leq
C(\epsilon_{0}),\forall[0,T],$ (4.14)

where $\epsilon_{0}$ introduced in the definition of $U_{d_{0},d_{1}}$.

###### Proof.

See Appendix E ∎

###### The proof of Proposition 4.6.

We rewrite the equation satisfied by $U$ as follows

$\partial_{t}U=\Delta U+G(U),$

where

$G(U)=-2\frac{|\nabla
U|^{2}}{U+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}}+\left(U+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}\right)^{4}-\frac{\bar{\theta}^{\prime}(t)}{\bar{\theta}(t)}U.$

We remark that in order to get the conclusion, it is enough to prove that for
all $x\in\Omega,|x|\geq\frac{\epsilon_{0}}{4}$ and $t\in(0,t_{8}],$ we have
the following estimates

$\displaystyle\left|U_{1}(x,t)-U_{1}(x,0)\right|\leq\frac{\eta_{0}}{2},$
(4.15) $\displaystyle\left|\nabla U_{1}(x,t)-\nabla
e^{t\Delta}U_{1}(x,0)\right|\leq\frac{\eta_{0}}{2},$ (4.16)

where
$U_{1}(x,t)=\exp\left(\displaystyle\int_{0}^{t}\frac{\bar{\theta}^{\prime}(s)}{\bar{\theta}(s)}ds\right)U(x,t).$
Using the equation satisfied by $U$, we may derive an equation satisfied by
$U_{1}$ as follows:

$\partial_{t}U_{1}=\Delta U_{1}+G_{1},$ (4.17)

where
$G_{1}(t)=\exp\left(\displaystyle\int_{0}^{t}\frac{\bar{\theta}^{\prime}(s)}{\bar{\theta}(s)}ds\right)\left[-2\frac{|\nabla
U|^{2}}{U+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}}+\left(U+\frac{\lambda^{\frac{1}{3}}}{\bar{\theta}(t)}\right)^{4}\right]$.
In particular, from the fact that $U\in S(t)$ and Proposition 3.9, we can
derive the following

$\left|\exp\left(\pm\int_{0}^{t}\frac{\bar{\theta}^{\prime}(s)}{\bar{\theta}(s)}\right)ds\right|\leq
2.$

Moreover, from item $(iii)$ of Definition of 3.1 and Lemma 4.8, we derive the
following:

$|G_{1}(x,t)|\leq
C(K_{0},\epsilon_{0},\eta_{0}),\forall|x|\geq\frac{\epsilon_{0}}{8}\text{ and
}\forall t\in(0,t_{8}].$

In the following, we first prove (4.15) then (4.16).

\+ The proof of (4.15): We consider a cut-off function $\chi_{2}\in
C^{\infty}_{0}(\bar{\Omega})$ such that $\chi_{2}=1$ for all
$|x|\geq\frac{\epsilon_{0}}{6},x\in\bar{\Omega}$ and $\chi_{2}=0$ for all
$|x|\leq\frac{\epsilon_{0}}{8}$ and $|\nabla\chi_{2}|+|\Delta\chi_{2}|\leq
C(\epsilon_{0}).$ If we define $U_{2}=U_{1}\chi_{2},$ then $U_{2}$ satisfies
the following

$\partial_{t}U_{2}=\Delta U_{2}+G_{2},$

where

$G_{2}(U)=-2\nabla
U_{1}\cdot\nabla\chi_{2}-\Delta\chi_{2}U_{1}-\chi_{2}G_{1}.$

Using the estimate of $G_{1}$ and the following fact

$|\nabla U_{1}(x,t)|+|U_{1}(x,t)|\leq
C(K_{0},\epsilon_{0},\eta_{0}),\forall|x|\geq\frac{\epsilon_{0}}{8}\text{ and
}t\in[0,t_{8}],$

which is a consequence of the fact that $U\in S(t)$ (particularly items $(i)$
and $(iii)$ in Definition 3.1), we conclude the following

$\left\|G_{2}(x,t)\right\|_{L^{\infty}(\Omega)}\leq
C(K_{0},\epsilon_{0},C_{0},\eta_{0}),\forall|x|\geq\frac{\epsilon_{0}}{8}\text{
and }\forall t\in[0,t_{8}].$

We now use a Duhamel formula to write $U_{2}$ as follows

$U_{2}(t)=e^{t\Delta}U_{2}(0)+\int_{0}^{t}e^{(t-\tau)\Delta}\left(G_{2}(U(\tau))\right)d\tau,$
(4.18)

where $e^{t\Delta}$ stands for the Dirichlet heat semi-group on $\Omega$ (see
more in Appendix E). In particular, we have for all $U_{0}\in
L^{\infty}(\Omega),$

$\displaystyle\left\|e^{t\Delta}U_{0}\right\|_{L^{\infty}(\Omega)}\leq\|U_{0}\|_{L^{\infty}(\Omega)}.$

Therefore,

$\displaystyle\left|U_{2}(t)-U_{2}(0)\right|$ $\displaystyle\leq$
$\displaystyle\left|U_{2}(t)-e^{t\Delta}U_{2}(0)\right|+\left|e^{t\Delta}U_{2}(0)-U_{2}(0)\right|$
$\displaystyle\leq$
$\displaystyle\left|\int_{0}^{t}e^{(t-s)\Delta}G_{2}(s)ds\right|+\left|e^{t\Delta}U_{2}(0)-U_{2}(0)\right|$
$\displaystyle\leq$ $\displaystyle
C(K_{0},\epsilon_{0},C_{0},\eta_{0})T+\left\|e^{t\Delta}(U_{2}(0))-U_{2}(0)\right\|_{L^{\infty}(\Omega)}.$

In addition to that, because $U_{2}(0)$ is smooth and has a compact support in
$\Omega$, we can prove that

$\left\|e^{t\Delta}(U_{2}(0))-U_{2}(0)\right\|_{L^{\infty}(\Omega)}\to 0\text{
as }t\to 0,$

which yields the fact that

$\left\|U_{2}(t)-U_{2}(0)\right\|_{L^{\infty}(\Omega)}\leq\frac{\eta_{0}}{2},$

provided that $T\leq T_{8,1}(K_{0},\epsilon_{0},C_{0},\eta_{0}).$ This
concludes the proof of (4.15).

\+ The proof of (4.16): We derive from (4.18) the following fact:

$\nabla U_{2}(t)=\nabla e^{t\Delta}U_{2}(0)+\int_{0}^{t}\nabla
e^{(t-\tau)\Delta}G_{2}(\tau)d\tau.$

This implies that

$\displaystyle\left|\nabla U_{2}(t)-\nabla
e^{t\Delta}U_{1}(0)\right|\leq\left|\nabla e^{t\Delta}U_{2}(0)-\nabla
e^{t\Delta}U_{1}(0)\right|+\left|\int_{0}^{t}\nabla
e^{(t-\tau)\Delta}G_{2}(\tau)d\tau\right|.$

Using (E.3) and Lemma (E.1) below, we derive that

$\left|\int_{0}^{t}\nabla e^{(t-\tau)\Delta}G_{2}(\tau)d\tau\right|\leq
C(K_{0},\epsilon_{0},C_{0},\eta_{0})\int_{0}^{t}\frac{1}{\sqrt{t-\tau}}d\tau\leq
C(K_{0},\epsilon_{0},C_{0},\eta_{0})\sqrt{T}.$

In order to finish the proof, it is enough to prove that for all
$|x|\geq\frac{\epsilon_{0}}{4}$, we have

$\left|\nabla e^{t\Delta}U_{2}(0)-\nabla
e^{t\Delta}U_{1}(0)\right|\leq\frac{\eta_{0}}{4},$ (4.19)

provided that $T\leq T_{8,2}$. Indeed, using the definition of the Dirichlet
heat semi-group and Lemma E.1 below, we may write the following:

$\displaystyle\left|\nabla e^{t\Delta}U_{2}(0)-\nabla
e^{t\Delta}U_{1}(0)\right|$ $\displaystyle=$
$\displaystyle\left|\int_{\Omega}\nabla_{x}G(x,y,t,0)(1-\chi_{2}(y))U_{d_{0},d_{1}}(y)dy\right|$
$\displaystyle\leq$ $\displaystyle
C\int_{|y|\leq\frac{\epsilon_{0}}{6}}\frac{\exp\left(-\frac{|x-y|^{2}}{t}\right)}{t^{\frac{n+1}{2}}}|U_{d_{0},d_{1}}(y)|dy$
$\displaystyle\leq$ $\displaystyle
C\int_{|y|\leq\frac{\epsilon_{0}}{6}}\exp\left(-\frac{|x-y|^{2}}{t}\right)\left(\frac{|x-y|}{\sqrt{t}}\right)^{n+2}\frac{\sqrt{t}}{|x-y|^{n+2}}|U_{d_{0},d_{1}}(y)|dy$
$\displaystyle\leq$ $\displaystyle
C(\epsilon)\sqrt{t}\int_{|y|\leq\frac{\epsilon_{0}}{6}}|U_{d_{0},d_{1}}(y)|dy$
$\displaystyle\leq$ $\displaystyle
C(\epsilon_{0})\sqrt{t}\|U_{d_{0},d_{1}}\|_{L^{1}(\Omega)}\leq
C(\epsilon_{0})\sqrt{T}.$

This yields (4.19), provided that $T\leq T_{8,3}(\epsilon_{0})$. In
particular, from the defintions of $U_{2}$ and $U_{1}$, we can derive (4.16).
Finally, we get the conclusion of Proposition 4.6. ∎

###  4.2.  The conclusion of the proof of Proposition 3.13

It this part, we aim at giving a complete proof to Proposition 3.13:

###### The proof of Proposition 3.13 .

We first choose the parameters
$K_{0},\epsilon_{0}>0,\alpha_{0}>0,A>0,\delta_{0}>0,\delta_{1}>0,C_{0}>0,\eta_{0}>0$
and $T>0$ such that Propositions 3.8, 4.2, 4.4 and 4.6 hold. In particular,
the constant $T$ will be fixed small later. Then, the conclusion of the proof
follows as we will show in the following. We now consider $U,$ a solution of
equation (2.9), with initial data $U_{d_{0},d_{1}}(0)$, defined in Definition
3.2 and satisfying the following:

$U\in S(T,K_{0},\alpha_{0},\epsilon_{0},A,\delta_{0},C_{0},\eta_{0},t)=S(t),$

for all $t\in[0,t_{*}]$ for some $t_{*}\in(0,T)$ and

$u\in\partial S(t_{*}).$

$(i)$ Using Propositions 4.2, 4.4 and 4.6, we can derive that

$(q_{1},q_{2})(s_{*})\in\partial\hat{V}_{A}(s_{*}),$ (4.20)

where $s_{*}=\ln(T-t_{*})$.

$(ii)$ Using item $(i)$, we derive that either

$\left|q_{0}(s_{*})\right|=\frac{A}{s^{2}_{*}},$

or there exists $j_{0}\in\\{1,...,n\\}$ such that

$\left|q_{1,j_{0}}(s_{*})\right|=\frac{A}{s^{2}_{*}}.$

Then, without loss of generality, we can suppose that the first case occurs,
because the argument is the same in other cases. Hence, using (4.1) in Lemma
4.1, we see that

$\left|q^{\prime}_{0}(s)-q_{0}(s)\right|\leq\frac{C}{s^{2}}.$

Therefore, we obtain that the sign of $q_{0}^{\prime}(s_{*})$ is opposite to
the sign of

$\frac{d}{ds}\left(\epsilon_{0}\frac{A}{s^{2}}\right)(s_{*}),$

provided that $A\geq 2C$, where $\varepsilon_{0}=\pm 1$ and
$q_{0}(s_{*})=\epsilon_{0}\frac{A}{s_{*}^{2}}$. This means that the flow of
$q_{0}$ is transverse outgoing on the bounds of the shrinking set

$-\frac{A}{s^{2}}\leq q_{0}(s)\leq\frac{A}{s^{2}}.$

It follows then that $(q_{0},q_{1})(s)$ leaves $\hat{\mathcal{V}}(s)$ at
$s_{*}$. Thus, we conclude item $(ii)$. Finally, we get the conclusion of
Proposition 3.13 ∎

##  Appendix A  Preparation of initial data

In this section, we give the proof of Proposition 3.8. More precisely, we aim
at proving the following lemma which directly implies Proposition 3.8:

######  Lemma A.1.

There exists $K_{2}>0$ such that for all $K_{0}\geq K_{2},\delta_{2}>0,$ there
exist $\alpha_{2}(K_{0},\delta_{2})>0,C_{2}>0$ such that for all
$\alpha_{0}\in(0,\alpha_{2}]$ there exists
$\epsilon_{2}(K_{0},\delta_{2},\alpha_{0})>0$ such that for all
$\epsilon_{0}\in(0,\epsilon_{2}]$ and $A\geq 1$, there exists
$T_{2}(K_{0},\delta_{2},\epsilon_{0},A,C_{2})>0$ such that for all
$T\in(0,T_{2}],$ there exists a subset $\mathcal{D}_{A}\subset[-2,2]^{1+n}$
such that the following properties hold: Asumme that initial data
$U_{d_{0},d_{1}}(0)$ is given as in (3.2), then:

$A)$ For all $(d_{0},d_{1})\in\mathcal{D}_{A}$, we have initial data
$U(0)=U_{d_{0},d_{1}}(0)\in
S(T,K_{0},\epsilon_{0},\alpha_{0},A,\delta_{2},0,C_{2},0,0)$. In particular,
we have the following:

  * $(i)$

Estimates in $P_{1}(0)$: we have the transformed function $q(s_{0})$ of
$U_{d_{0},d_{1}}(0),$ trapped in $V_{K_{0},A}(s_{0}),$ where $s_{0}=-\ln T$
and we have also the following estimates:

$\left|q_{0}(s_{0})-\frac{Ad_{0}}{s_{0}^{2}}\right|+\left|q_{1,j}(s_{0})-\frac{Ad_{1,j}}{s_{0}^{2}}\right|\leq
Ce^{-s_{0}},\text{ for all }j\in\\{1,...,n\\},$
$\left|q_{2,i,j}(s_{0})\right|\leq\frac{\ln s_{0}}{s_{0}^{2}},\text{ for all
}i,j\in\\{1,...,n\\},$
$\left|q_{-}(y,s_{0})\right|\leq\frac{1}{s_{0}^{2}}(1+|y|^{3}),\quad\left|(\nabla_{y}q)_{\perp}(y,s_{0})\right|\leq\frac{1}{s_{0}^{2}}(1+|y|^{3}),\text{
for all }y\in\mathbb{R}^{n},$

and

$q_{e}(s_{0})\equiv 0,$

where the components of $q$ are defined in (2.39).

  * $(ii)$

Estimates in $P_{2}(0)$: For all $|x|\in\left[\frac{K_{0}}{4}\sqrt{T|\ln
T|},\epsilon_{0}\right]$ and and $|\xi|\leq\alpha_{0}\sqrt{|\ln\varrho(x)|}$,
we have

$\left|\mathcal{U}(x,\xi,\tau_{0}(x))-\hat{\mathcal{U}}(\tau_{0}(x))\right|\leq\delta_{2},\text{
and
}\left|\nabla_{\xi}\mathcal{U}(x,\xi,\tau_{0}(x))\right|\leq\frac{C_{2}}{\sqrt{|\ln\varrho(x)|}},$

where $\tau_{0}(x)=-\frac{t(x)}{\varrho(x)}$ and
$\mathcal{U},\hat{\mathcal{U}},t(x),\varrho(x)$ are given in (3.4), (3.5),
(3.6) and (3.10).

$B)$ We have the following

$(d_{0},d_{1})\in\mathcal{D}_{A}\text{ if and only if
}(q_{0},q_{1})(s_{0})\in\hat{\mathcal{V}}_{A}(s_{0})$

and

$(d_{0},d_{1})\in\partial\mathcal{D}_{A}\text{ if and only if
}(q_{0},q_{1})(s_{0})\in\partial\hat{\mathcal{V}}_{A}(s_{0}),$

where $\hat{\mathcal{V}}_{A}(s)$ given in (3.15).

###### Proof.

We see that part $B)$ directly follows from item $(i)$ of part $A)$. In
addition to that, our definition is almost the same as in [TZ19] (see also the
work of Ghoul, Nguyen and Zaag [GNZ17], the works of Merle and Zaag [MZ97a],
[MZ97b]). So, we kindly refer the reader to see the proofs of the existence of
the set $\mathcal{D}_{A}$, item $i$ in $A)$ and part $B)$ in Proposition 4.5
in [TZ19]. Here we only give the proof of item $(ii)$ in part $A)$. We now
consider
$T>0,K_{0}>0,\epsilon_{0}>0,\alpha_{0}>0,\delta_{2}>0,C_{2}>0,\eta_{0}>0$. We
aim at proving that if these constants are suitably chosen, then for all
$x\in\left[\frac{K_{0}}{4}\sqrt{T|\ln T|},\epsilon_{0}\right]$ and $|\xi|\leq
2\alpha_{0}\sqrt{|\ln\varrho(x)|},$ where $\varrho(x)$ given in (3.5), we have
the following

$\left|\mathcal{U}(x,\xi,\tau_{0}(x))-\hat{\mathcal{U}}(\tau_{0}(x))\right|\leq\delta_{2},\quad\left|\nabla_{\xi}\mathcal{U}(x,\xi,\tau_{0}(x))\right|\leq\frac{C_{2}}{\sqrt{|\ln\varrho(x)|}}.$

We observe from the definition of $t(x)$ given in (3.5) that if
$\alpha_{0}\leq\alpha_{2,1}$ and $\epsilon_{0}\leq\epsilon_{2,1}$, then, for
all $x\in\left[\frac{K_{0}}{4}\sqrt{T\ln T},\epsilon_{0}\right]$ and
$|\xi|\leq 2\alpha_{0}\sqrt{|\ln\varrho(x)|}$, we have

$\left|\xi\sqrt{\varrho(x)}\right|\leq\frac{|x|}{2},$

which yields

$\displaystyle\frac{r_{0}}{2}\leq\frac{|x|}{2}\leq\left|x+\xi\sqrt{T(x)}\right|\leq\frac{3}{2}|x|,\text{
with }r_{0}=\frac{K_{0}}{4}\sqrt{T|\ln T|}.$ (A.1)

Hence, for all $x\in\left[\frac{K_{0}}{4}\sqrt{T|\ln T|},\epsilon_{0}\right],$
we have

$\chi\left(16(x+\xi\sqrt{\varrho(x)})\sqrt{T},-\ln
T\right)\chi_{1}(x+\xi\sqrt{\varrho(x)})=0,$

where $\chi$ and $\chi_{1}$ are defined in (2.33) and (3.11), respectively.
Therefore, from (3.2) and the definition of $\mathcal{U}$ in (3.4), we may
derive that for all $x\in\left[\frac{K_{0}}{4}\sqrt{T|\ln
T|},\epsilon_{0}\right]$ and $|\xi|\leq 2\alpha_{0}\sqrt{|\ln\varrho(x)|},$

$\displaystyle\mathcal{U}(x,\xi,\tau_{0})=(I)\chi_{1}\left(x+\xi\sqrt{\varrho(x)}\right)+(II)\left(1-\chi_{1}(x+\xi\sqrt{\varrho(x)})\right),$

where

$\displaystyle(I)$ $\displaystyle=$
$\displaystyle\left(\frac{\varrho(x)}{T}\right)^{\frac{1}{3}}\left(3+\frac{9}{8}\frac{|x+\xi\sqrt{\varrho(x)}|^{2}}{T|\ln
T|}\right)^{-\frac{1}{3}},$ and $\displaystyle(II)$ $\displaystyle=$
$\displaystyle\varrho^{\frac{1}{3}}(x)H^{*}(x+\xi\sqrt{\varrho(x)}),$

with $H^{*}(x)$ given in (3.12). In addition to that, from the definition of
$\varrho(x)$, given in (3.6), we obtain the following asymptotics

$\displaystyle\ln\varrho(x)\sim 2\ln|x|\text{ and
}\varrho(x)\sim\frac{8}{K_{0}^{2}}\frac{|x|^{2}}{|\ln|x||}\text{ as }|x|\to
0.$ (A.2)

Besides that, we introduce $r_{0}=\frac{K_{0}}{4}\sqrt{T|\ln T|}$ and
$R_{0}=\sqrt{T}|\ln T|.$ Then, the following holds

$\displaystyle\varrho(r_{0})\sim T,\text{ and
}\varrho(R_{0})\sim\frac{16}{K_{0}^{2}}T|\ln T|\text{ and
}\varrho(2R_{0})\sim\frac{64}{K_{0}^{2}}T|\ln T|\text{ as }T\to 0.$ (A.3)

We aim in the following at giving some estimates on
$\mathcal{U}(x,\xi,\tau_{0}(x))$ and
$\nabla_{\xi}\mathcal{U}(x,\xi,\tau_{0}(x))$.

\- Estimate on $\mathcal{U}$: From the definition of the cut-off function
$\chi_{1}$ given in (3.11), it is enough to prove that for all
$|x|\in\left[r_{0},(2+\frac{1}{100})R_{0}\right]$ and $|\xi|\leq
2\alpha_{0}\sqrt{|\ln\varrho(x)|}$, we have

$\displaystyle\left|I_{1}-\hat{\mathcal{U}}(\tau_{0})\right|\leq\frac{\delta_{2}}{2},$
(A.4)

on one hand and also that for all
$|x|\in\left[\frac{99}{100}R_{0},\epsilon_{0}\right]$ and $|\xi|\leq
2\alpha_{0}\sqrt{|\ln\varrho(x)|},$ we have

$\displaystyle\left|I_{2}-\hat{\mathcal{U}}(\tau_{0})\right|\leq\frac{\delta_{2}}{2},$
(A.5)

on the other hand. Indeed, let us start with the proof of (A.4): We consider
$|x|\in\left[r_{0},(2+\frac{1}{100})R_{0}\right]$ and $|\xi|\leq
2\alpha_{0}\sqrt{|\ln\varrho(x)|}$. Then, we write the following:

$\displaystyle\left|I_{1}-\hat{\mathcal{U}}(\tau_{0}(x))\right|$
$\displaystyle=$
$\displaystyle\left|\left(3\frac{T}{\varrho(x)}+\frac{9}{8}\frac{|x+\xi\sqrt{\varrho(x)}|^{2}}{\varrho(x)|\ln
T|}\right)^{-\frac{1}{3}}-\left(3\frac{T}{\varrho(x)}+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}\right|$
$\displaystyle=$
$\displaystyle\left|\left(3\frac{T}{\varrho(x)}+\frac{9}{8}\frac{K_{0}^{2}}{16}+\frac{9}{8}\left[\frac{|x+\xi\sqrt{\varrho(x)}|^{2}}{\varrho(x)|\ln
T|}-\frac{K_{0}^{2}}{16}\right]\right)^{-\frac{1}{3}}-\left(3\frac{T}{\varrho(x)}+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}\right|.$

In addition to that, we have

$\displaystyle\frac{|x+\xi\sqrt{\varrho(x)}|^{2}}{\varrho(x)|\ln
T|}-\frac{K_{0}^{2}}{16}$ $\displaystyle=$
$\displaystyle\frac{|x|^{2}}{\varrho(x)|\ln
T|}\left(1+2\frac{x\cdot\xi}{|x|^{2}}\sqrt{\varrho(x)}+\frac{|\xi|^{2}\varrho(x)}{|x|^{2}}\right)-\frac{K_{0}^{2}}{16}$
$\displaystyle=$ $\displaystyle\frac{K_{0}^{2}}{16}\frac{|\ln\varrho(x)|}{|\ln
T|}\left(1+2\frac{x\cdot\xi}{|x|^{2}}\sqrt{\varrho(x)}+\frac{|\xi|^{2}\varrho(x)}{|x|^{2}}\right)-\frac{K_{0}^{2}}{16}.$

Besides that, we also have the following:

$\displaystyle\left|\frac{x\cdot\xi}{|x|^{2}}|\sqrt{\varrho(x)}|\right|\leq
4\alpha_{0},$
$\displaystyle\left|\frac{|\xi|^{2}}{|x|^{2}}\varrho(x)\right|\leq
4\alpha_{0}^{2}.$

Moreover, for all
$|x|\in\left[r_{0},\left(2+\frac{1}{100}\right)R_{0}\right]$, we derive from
(A.3) that

$\frac{|\ln\varrho(x)|}{|\ln T|}\sim 1,\text{ as }T\to 0.$

So, the following holds

$\left|\frac{|x+\xi\sqrt{\varrho(x)}|^{2}}{\varrho(x)|\ln
T|}-\frac{K_{0}^{2}}{16}\right|\to 0,$

as $(\alpha_{0},T)\to(0,0)$. From this fact, we can derive that if $T\leq
T_{2,1}(K_{0},\delta_{2}),\alpha_{0}\leq\alpha_{2,2}(K_{0},\delta_{2})$, we
have

$\displaystyle\left|I_{1}-\hat{\mathcal{U}}(\tau_{0}(x))\right|$
$\displaystyle=$
$\displaystyle\left|\left(3\frac{T}{\varrho(x)}+\frac{9}{8}\frac{|x+\xi\sqrt{\varrho(x)}|^{2}}{\varrho(x)|\ln
T|}\right)^{-\frac{1}{3}}-\left(3\frac{T}{\varrho(x)}+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}\right|$
$\displaystyle\leq$ $\displaystyle
C(K_{0})\left|\frac{|x+\xi\sqrt{T(x)}|^{2}}{T(x)|\ln
T|}-\frac{K_{0}^{2}}{16}\right|\leq\frac{\delta_{1}}{2}.$

This concludes the proof of (A.4).

We now aim at proving (A.5). We consider
$|x|\in\left[\frac{99}{100}R_{0},\epsilon_{0}\right]$ and $|\xi|\leq
2\alpha_{0}\sqrt{|\ln\varrho(x)|}.$ Using the definition of $(II),$ we write
as follows

$\displaystyle\left|(II)-\hat{\mathcal{U}}(\tau_{0}(x))\right|$
$\displaystyle=$
$\displaystyle\left|\left(\frac{9}{16}\frac{\left|x+\xi\sqrt{\varrho(x)}\right|^{2}}{\varrho(x)|\ln|x+\xi\sqrt{\varrho(x)}||}\right)^{-\frac{1}{3}}-\left(3\frac{T}{\varrho(x)}+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}\right|$
$\displaystyle=$
$\displaystyle\left|\left(\frac{9}{8}\frac{K_{0}^{2}}{16}+\frac{9}{16}\left(\frac{\left|x+\xi\sqrt{\varrho(x)}\right|^{2}}{\varrho(x)|\ln|x+\xi\sqrt{\varrho(x)}||}-\frac{K_{0}^{2}}{8}\right)\right)^{-\frac{1}{3}}-\left(\frac{9}{8}\frac{K_{0}^{2}}{16}+3\frac{T}{\varrho(x)}\right)^{-\frac{1}{3}}\right|.$

Besides that, the function $\varrho(x)$ is radial in $x$, and increasing in
$|x|$ when $|x|$ is small enough. Then, for all
$\epsilon_{0}\leq\epsilon_{2,1}$ and
$|x|\in\left[\frac{99}{100}R_{0},\epsilon_{0}\right]$, we have

$\displaystyle\left|\frac{T}{\varrho(x)}\right|\leq\left|\frac{T}{\varrho\left(\frac{99}{100}R_{0}\right)}\right|\leq
C(K_{0})|\ln T|^{-1}\to 0\text{ as }T\to 0.$ (A.6)

In addition to that, we have

$\displaystyle\frac{\left|x+\xi\sqrt{\varrho(x)}\right|^{2}}{\varrho(x)|\ln|x+\xi\sqrt{\varrho(x)}||}-\frac{K_{0}^{2}}{8}$
$\displaystyle=$
$\displaystyle\frac{1}{\varrho(x)|\ln|x+\xi\sqrt{\varrho(x)}||}\left[|x|^{2}+2x\cdot\xi\sqrt{\varrho(x)}+|\xi|^{2}\varrho(x)\right]-\frac{K_{0}^{2}}{8}$
$\displaystyle=$
$\displaystyle\frac{K_{0}^{2}}{16}\left[\frac{|\ln\varrho(x)|}{|\ln|x+\xi\sqrt{\varrho(x)}||}-2+4\alpha_{0}\frac{|\ln\varrho(x)|}{|\ln|x+\xi\sqrt{\varrho(x)}||}\right.$
$\displaystyle+$
$\displaystyle\left.4\alpha_{0}^{2}\frac{|\ln\varrho(x)|}{|\ln|x+\xi\sqrt{\varrho(x)}||}\right].$

In particular, we have the following fact

$\displaystyle\ln\varrho(x)$ $\displaystyle\sim$ $\displaystyle 2\ln|x|,\text{
as }|x|\sim 0,$ $\displaystyle\frac{1}{|\ln|x+\xi\sqrt{\varrho(x)}||}$
$\displaystyle\sim$ $\displaystyle\frac{1}{|\ln|x||},\text{ as }\alpha_{0}\to
0.$

This yields

$\displaystyle\left|\frac{\left|x+\xi\sqrt{\varrho(x)}\right|^{2}}{\varrho(x)|\ln|x+\xi\sqrt{\varrho(x)}||}-\frac{K_{0}^{2}}{8}\right|\to
0\text{ as }(\epsilon_{0},\alpha_{0})\to(0,0).$ (A.7)

From (A.6) and (A.7), we derive that

$\displaystyle\left|(II)-\hat{\mathcal{U}}(\tau_{0}(x))\right|\leq
C(K_{0})\left[\left|\frac{\left|x+\xi\sqrt{\varrho(x)}\right|^{2}}{\varrho(x)|\ln|x+\xi\sqrt{\varrho(x)}||}-\frac{K_{0}^{2}}{8}\right|+\frac{T}{\varrho(x)}\right]\leq\frac{\delta_{2}}{2},$

provided that
$\alpha\leq\alpha_{2,3}(K_{0},\delta_{2}),\epsilon_{0}\leq\alpha_{2,2}(K_{0},\delta_{2},\alpha_{0})$
and $T\leq T_{2,3}$. Thus, $\eqref{prove-2-delta-1-2}$ holds. Finally, we get
the conclusion that for all $|x|\in\left[\frac{K_{0}}{4}\sqrt{T|\ln
T|},\epsilon_{0}\right]$ and $|\xi|\leq 2\alpha_{0}\sqrt{|\ln\varrho(x)|}$, we
have

$\left|\mathcal{U}(x,\xi,\tau_{0}(x))-\hat{\mathcal{U}}(\tau_{0}(x))\right|\leq\delta_{2}.$

\- Estimate on $\partial_{\xi}\mathcal{U}$:  From the definition of
$\mathcal{U}(x,\xi,\tau_{0}(x))=\mathcal{U}\left(x,\xi,-\frac{t(x)}{\varrho(x)}\right)$
given in (3.4) and the expression (3.2) of initial data, we decompose
$\nabla_{\xi}\mathcal{U}$ as follows

$\partial_{\xi}\mathcal{U}(x,\xi,\tau_{0}(x))=B_{1}+B_{2}+B_{3},$

where

$\displaystyle B_{1}$ $\displaystyle=$
$\displaystyle\left[-\frac{3}{4}\frac{\varrho^{\frac{5}{6}}(x)}{T^{\frac{4}{3}}|\ln
T|}(x+\xi\sqrt{\varrho(x)})\left(3+\frac{9}{8}\frac{\left|x+\xi\sqrt{\varrho(x)}\right|^{2}}{T|\ln
T|}\right)^{-\frac{4}{3}}\right]\chi_{1}(x+\xi\sqrt{\varrho(x)}),$ (A.8)
$\displaystyle B_{2}$ $\displaystyle=$
$\displaystyle\varrho^{\frac{5}{6}}(x)\nabla
H^{*}(x+\xi\sqrt{\varrho(x)})\left(1-\chi_{1}(x+\xi\sqrt{\varrho(x)})\right),$
(A.9) $\displaystyle B_{3}$ $\displaystyle=$
$\displaystyle\left[\left(\frac{\varrho(x)}{T}\right)^{\frac{1}{3}}\left(3+\frac{9}{8}\frac{|x+\xi\sqrt{\varrho(x)}|^{2}}{T|\ln
T|}\right)^{-\frac{1}{3}}+\frac{3^{-\frac{1}{3}}n}{|\ln
T|}\frac{\varrho^{\frac{1}{3}}(x)}{T^{\frac{1}{3}}}-\varrho^{\frac{1}{3}}(x)H^{*}(x+\xi\sqrt{\varrho(x)})\right]$
$\displaystyle\times$
$\displaystyle\sqrt{\varrho(x)}\nabla\chi_{1}(x+\xi\sqrt{\varrho(x)}).$

It is enough to prove the following estimates:

\- Estimate of $B_{1}:$ For all
$|x|\in\left[r_{0};(2+\frac{1}{100})R_{0}\right]$ and $|\xi|\leq
2\alpha_{0}\sqrt{|\ln\varrho(x)|}$ we have

$\displaystyle\left|B_{1}\right|\leq\frac{C(K_{0})}{\sqrt{|\ln\varrho(x)|}}.$
(A.11)

\- Estimate of $B_{2}:$ For all
$|x|\in\left[\frac{99}{100}R_{0},\epsilon_{0}\right]$ and $|\xi|\leq
2\alpha_{0}\sqrt{|\ln\varrho(x)|},$ we have

$\left|B_{2}\right|\leq\frac{C(K_{0})}{\sqrt{|\ln\varrho(x)|}}.$ (A.12)

\- Estimate of $B_{3}:$ For all
$|x|\in\left[\frac{99}{100}R_{0},(2+\frac{1}{100})R_{0}\right]$ and $|\xi|\leq
2\alpha_{0}\sqrt{|\ln\varrho(x)|},$ we have

$|B_{3}|\leq\frac{C(K_{0})}{\sqrt{|\ln\varrho(x)|}}.$ (A.13)

We now start the proof:

\- Estimate of $B_{1}$: We have the fact that for all $|z|\geq 1$

$\left(3+\frac{9}{8}|z|^{2}\right)^{-\frac{4}{3}}\leq C|z|^{-\frac{8}{3}}.$

Then,

$\displaystyle\left|B_{1}\right|$ $\displaystyle\leq$ $\displaystyle
C\frac{\varrho^{\frac{5}{6}}(x)}{T^{\frac{4}{3}}|\ln
T|}\frac{T^{\frac{4}{3}}|\ln
T|^{\frac{4}{3}}}{|x+\xi\sqrt{\varrho(x)}|^{\frac{5}{3}}}$ $\displaystyle\leq$
$\displaystyle C\frac{\varrho^{\frac{5}{6}}(x)|\ln
T|^{\frac{1}{3}}}{|x+\xi\sqrt{\varrho(x)}|^{\frac{5}{3}}}.$

Using (A.1), we obtain the following:

$\left|B_{1}\right|\leq C\frac{\varrho^{\frac{5}{6}}(x)|\ln
T|^{\frac{1}{3}}}{|x|^{\frac{5}{3}}}.$

In addition to that, for all
$|x|\in\left[r_{0},(2+\frac{1}{100})R_{0}\right]$, we have

$|\ln\varrho(x)|\sim|\ln T|,\text{ as }T\to 0.$

Then, we have

$|B_{1}|\leq\frac{C}{K_{0}^{2}}\frac{\varrho^{\frac{5}{6}}(x)|\ln
T|^{\frac{1}{3}}}{\varrho^{\frac{5}{6}}(x)|\ln\varrho(x)|^{\frac{5}{6}}}\leq\frac{C}{\sqrt{|\ln\varrho(x)|}},$

provided that $K_{0}\geq K_{2,3},T\leq T_{2,4}$. This yields (A.11).

\- Estimate of $B_{2}$: From the definition of $H^{*}(x),$ when
$|x|\leq\epsilon_{0}$, $\epsilon_{0}$ small enough, we have

$H^{*}(x)=\left[\frac{9}{16}\frac{|x|^{2}}{|\ln|x||}\right]^{-\frac{1}{3}}.$

This implies

$\left|\nabla H^{*}(x)\right|\leq
C\frac{|\ln|x||^{\frac{1}{3}}}{|x|^{\frac{5}{3}}}.$

Hence,

$\displaystyle|B_{2}|\leq
C\frac{\varrho^{\frac{5}{6}}(x)|\ln|x||^{\frac{1}{3}}}{|x|^{\frac{5}{3}}}\leq
C\frac{|\ln|x||^{\frac{1}{3}}}{|\ln\varrho(x)|^{\frac{1}{3}}}\frac{1}{\sqrt{|\ln\varrho(x)|}},$

on one hand. On the other hand, we have the following

$|\ln\varrho(x)|\sim 2|\ln|x||,\text{ as }x\to 0.$

Thus, (A.12) holds provided that $\epsilon_{0}\leq\epsilon_{2,4}(K_{0})$.

\- Estimate of $B_{3}$: We first use the definition of $\chi_{1}$ in (3.11) to
write

$\left|\nabla_{x}\chi_{1}(x)\right|\leq\frac{C}{\sqrt{T}|\ln T|}.$

We now consider
$|x|\in\left[\frac{99}{100}R_{0},(2+\frac{1}{100})R_{0}\right]$ and $|\xi|\leq
2\alpha_{0}\sqrt{|\ln\varrho(x)|}$. We define

$B_{4}=\left(3T+\frac{9}{8}\frac{|x+\xi\sqrt{\varrho(x)}|^{2}}{|\ln
T|}\right)^{-\frac{1}{3}}+\frac{3^{-\frac{1}{3}}n}{T^{\frac{1}{3}}|\ln
T|}-H^{*}(x+\xi\sqrt{\varrho(x)}).$

Then,

$B_{3}=B_{4}\varrho^{\frac{5}{6}}(x)\nabla\chi_{1}\left(x+\xi\sqrt{\varrho(x)}\right).$

We now aim at giving some estimates on $B_{4}$ as follows: Using the fact that
$|x|\in\left[\frac{99}{100}R_{0},(2+\frac{1}{100})R_{0}\right]$, $|\xi|\leq
2\alpha_{0}\sqrt{|\ln\varrho(x)|}$ and (A.1), we can derive that

$\displaystyle\frac{1}{C}T|\ln T|\leq\frac{|x+\xi\sqrt{\varrho(x)}|^{2}}{|\ln
T|}\leq CT|\ln T|,$ $\displaystyle\frac{1}{C}T|\ln
T|\leq\frac{|x+\xi\sqrt{\varrho(x)}|^{2}}{|\ln|x+\xi\sqrt{\varrho(x)}||}\leq
CT|\ln T|.$

This implies that

$|B_{4}|\leq C(T|\ln T|)^{-\frac{1}{3}}.$

Hence, we estimate $B_{3}$:

$\displaystyle\left|B_{3}\right|$ $\displaystyle\leq$ $\displaystyle C(T|\ln
T|)^{-\frac{1}{3}}\varrho^{\frac{5}{6}}(x)\nabla_{x}\chi_{1}\left(x+\xi\sqrt{\varrho(x)}\right)$
$\displaystyle\leq$ $\displaystyle
C\frac{\varrho^{\frac{5}{6}}(x)}{T^{\frac{5}{6}}}\frac{1}{|\ln
T|^{\frac{4}{3}}}$

In addition to that, for all
$|x|\in\left[\frac{99}{100}R_{0},(2+\frac{1}{100})R_{0}\right]$, we use (A.3)
to deduce that

$|\varrho(x)|\leq CT|\ln T|,$

and we also have the following fact

$\displaystyle|\ln\varrho(x)|$ $\displaystyle\sim$ $\displaystyle|\ln
T|,\text{ as }T\to 0.$

So, we conclude that

$|B_{3}|\leq\frac{C}{\sqrt{|\ln\varrho(x)|}}$

provided that $K_{0}\geq
K_{2,4},\epsilon_{0}\leq\epsilon_{2,5}(K_{0},\alpha_{0})$ and $T\leq
T_{2,5}(K_{0})$. Thus, we get the conclusion of (A.13). Finally, the
conclusion of Lemma A.1 follows. ∎

##  Appendix B A priori estimates in the intermediate region

In this section, we aim at giving the proof of Lemma 4.3. Because our
definitions are the same as in [MZ97a], estimates in this Proposition follow
in the same way as in that work. Hence, we kindly refer the reader to Lemma
2.6 in page 1515 in that work for the proof of (4.9) and item $(ii)$. It
happens that, although the authors in [MZ97a] gave a statement which is
similar to (4.10), they did not gave the proof. For that reason, we give here
the proof of (4.10) and (4.11).

\- The proof of (4.10): We consider
$|x|\in\left[\frac{K_{0}}{4}\sqrt{(T-t_{*})|\ln(T-t_{*})|},\epsilon_{0}\right],|\xi|\leq\frac{7}{4}\alpha_{0}\sqrt{|\ln\varrho(x)|}$
and
$\tau\in\left[\max\left(0,-\frac{t(x)}{\varrho(x)}\right),\frac{t_{*}-t(x)}{\varrho(x)}\right]$.
As a matter of fact, there exists $t\in[0,t_{*}]$ such that

$\tau=\frac{t-t(x)}{\varrho(x)}.$

Let us define

$X=x+\xi\sqrt{\varrho(x)}.$

We aim at considering the three following cases:

\+ The case where $|X|\leq\frac{K_{0}}{4}\sqrt{(T-t)|\ln(T-t)|}$. We write

$\mathcal{U}(x,\xi,\tau)=\varrho^{\frac{1}{3}}(x)U(X,t).$

We have the fact that $X\in P_{1}(t)$. Then, using item $(i)$ in Definition
3.1 together with item $(i)$ in Lemma 3.6, we get

$\displaystyle(T-t)^{-\frac{1}{3}}U(X,t)$ $\displaystyle=$ $\displaystyle
W(Y,s),\text{ where }Y=\frac{X}{\sqrt{T-t}},s=-\ln(T-t)$ $\displaystyle\geq$
$\displaystyle\left(3+\frac{|X|^{2}}{(T-t)|\ln(T-t)|}\right)^{-\frac{1}{3}}-\frac{CA^{2}}{\sqrt{s}}$
$\displaystyle\geq$
$\displaystyle\frac{1}{2}\left(3+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}},$

provided that $T\leq T_{6,1}(K_{0},A)$. This yields

$\displaystyle\mathcal{U}(x,\xi,\tau)\geq\left(\frac{\varrho(x)}{T-t}\right)^{-\frac{1}{3}}\frac{1}{2}\left(3+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}.$

In addition to that, the function $|x|\mapsto\varrho(x)$ is increasing when
$|x|$ is small enough. This implies that

$\varrho(x)\leq\varrho\left(\frac{K_{0}}{4(1-\frac{7}{4}\alpha_{0})}\sqrt{(T-t)|\ln(T-t)|}\right).$

From (3.5), (3.6) and (A.2), we derive that

$\varrho\left(\frac{K_{0}}{4(1-\frac{7}{4}\alpha_{0})}\sqrt{(T-t)|\ln(T-t)|}\right)\sim\frac{8}{K_{0}^{2}}\frac{K_{0}^{2}}{16(1-\frac{7}{4}\alpha_{0})^{2}}\frac{2(T-t)|\ln(T-t)|}{|\ln(T-t)|}=\frac{(T-t)}{(1-\frac{7}{4}\alpha_{0})^{2}},$

as $T\to 0$. Hence, we have

$\frac{\varrho(x)}{T-t}\leq 4,$

provided that $\alpha_{0}\leq\frac{2}{7},T\leq T_{6,2}$. Finally, we get

$\mathcal{U}(x,\xi,\tau)\geq\frac{1}{4}\left(3+\frac{8}{9}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}.$

\+ The case where
$|X|\in\left[\frac{K_{0}}{4}\sqrt{(T-t)|\ln(T-t)|},\epsilon_{0}\right]$. In
other words, we have $X\in P_{2}(t)$. We write as follows

$\mathcal{U}(x,\xi,\tau)=\varrho(x)U(X,t).$

In addition to that, using item $(ii)$ in the definition of $S(t)$ (see
Definition 3.1), we get the following:

$U(X,t)=\varrho^{-\frac{1}{3}}(X)\mathcal{U}(X,0,\frac{t-t(X)}{\varrho(X)})\geq\varrho^{-\frac{1}{3}}(X)\frac{1}{2}\left(3+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}},$

provided that
$\delta_{0}\leq\frac{1}{2}\left(3+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}.$
In particular, using the fact that

$(1-\frac{7}{4}\alpha_{0})|x|\leq|X|\leq(1+\frac{7}{4}\alpha_{0})|x|.$ (B.1)

Then, we get

$\left(\frac{\varrho(x)}{\varrho(X)}\right)^{\frac{1}{3}}\geq\frac{1}{2},$

provided that $\alpha_{0}\leq\alpha_{7,2}(K_{0})$ and
$|x|\leq\epsilon_{7,2}(K_{0},\alpha_{0})$. This yields that

$\mathcal{U}(x,\xi,\tau)\geq\frac{1}{4}\left(3+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}.$

\+ The case where $|X|\geq\epsilon_{0}$. This means $X\in P_{3}(t).$ We first
have the following fact

$\mathcal{U}(x,\xi,\tau)=\varrho^{\frac{1}{3}}(x)U(X,t)\geq\frac{1}{2}\varrho^{\frac{1}{3}}(x)U(X,0),$

provided that $\eta_{0}\leq\frac{1}{2}$ and $\epsilon_{0}\leq\epsilon_{6,3}$.
We remark also that
$|X|\leq(1+\frac{7}{4}\alpha_{0})|x|\leq(1+\frac{7}{4}\alpha_{0})\epsilon_{0}\leq\frac{3}{2}\epsilon_{0}$.
Then,

$U(X,0)=\left[\frac{9}{16}\frac{|X|^{2}}{|\ln|X||}\right]^{-\frac{1}{3}}.$

Moreover, using (A.2) and (B.1), we get

$\varrho^{\frac{1}{3}}(x)U(X,0)\geq\frac{1}{\sqrt[3]{2}}\left[\frac{9}{8}\frac{K_{0}^{2}}{16}\right]^{-\frac{1}{3}}\geq\frac{1}{2}\left(3+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}},$

provided that $\alpha_{0}\leq\alpha_{6,4},\epsilon_{0}\leq\epsilon_{6,3}$.

As a matter of fact, we obtain the following

$\mathcal{U}(x,\xi,\tau)\geq\frac{1}{4}\left(3+\frac{9}{8}\frac{K_{0}^{2}}{16}\right)^{-\frac{1}{3}}.$

This completely concludes the proof of (4.10).

\- The proof of (4.11): The idea of the proof is similar to the first one. We
also consider three cases

\+ The case where $|X|\leq\frac{K_{0}}{4}\sqrt{(T-t)|\ln(T-t)|}$. This implies
that $X\in P_{1}(t).$ We write here

$\mathcal{U}(x,\xi,\tau)=\varrho^{\frac{1}{3}}(x)U(X,t).$

Using item $(i)$ in the definition of $S(t)$(see Definition 3.1), together
with item $(i)$ in Lemma 3.6, we derive that

$|U(X,t)|\leq(T-t)^{-\frac{1}{3}}\left[\left(3+\frac{9}{8}\frac{|X|^{2}}{(T-t)|\ln(T-t)|}\right)^{-\frac{1}{3}}+\frac{CA^{2}}{\sqrt{|\ln(T-t)|}}\right]\leq
2(T-t)^{-\frac{1}{3}},$

provided that $T\leq T_{6,5}$. In addition to that, from the following fact

$\frac{K_{0}}{4}\sqrt{\varrho(X)|\ln\varrho(X)|}=|X|\leq\frac{K_{0}}{4}\sqrt{(T-t)|\ln(T-t)|},$

this yields that

$\varrho(X)\leq T-t.$

Then,

$\mathcal{U}(x,\xi,\tau)\leq
2\left(\frac{\varrho(x)}{\varrho(X)}\right)^{\frac{1}{3}}.$

On the other hand, using (B.1), we can derive

$\displaystyle\frac{\varrho(x)}{\varrho(X)}\leq 2,$ (B.2)

provided that $\alpha_{0}\leq\alpha_{6,4}$. This also yields that

$\mathcal{U}(x,\xi,\tau)\leq 4.$

\+ The case where
$|X|\in\left[\frac{K_{0}}{4}\sqrt{(T-t)|\ln(T-t)|},\epsilon_{0}\right]$. This
means $X\in P_{2}(t)$. We write

$\displaystyle\mathcal{U}(x,\xi,\tau)=\varrho^{\frac{1}{3}}(x)\varrho^{-\frac{1}{3}}(X)\mathcal{U}(X,0,\frac{t-t(X)}{\varrho(X)}).$

Hence, we derive from item $(ii)$ of Definition 3.1, the fact that $U\in S(t)$
and (B.1) that

$\mathcal{U}(x,\xi,\tau)\leq\left(\frac{\varrho(x)}{\varrho(X)}\right)^{\frac{1}{3}}\mathcal{U}(X,0,\frac{t-t(X)}{\varrho(X)})\leq
4,$

provided that $K_{0}\geq
K_{6,2},\alpha_{0}\leq\alpha_{6,4}(K_{0}),\delta_{0}\leq\delta_{6,1}$.

\+ The case where $|X|\geq\epsilon_{0}$. The result follows from item $(iii)$
of Definition 3.1.

Hence, (4.11) follows. Finally, we get the conclusion of Lemma 4.3.

##  Appendix C A priori estimate on $P_{2}(t)$

In this section, we aim at giving the proof of Proposition 4.4

###### The proof of Proposition 4.4 .

We first choose parameters
$K_{0},\epsilon_{0},\alpha_{0},A,\delta_{0},C_{0},\eta_{0},\delta_{6}$ such
that Lemma 4.3 holds. Then, items $(i)$ and $(ii)$ in that Lemma hold. We
would like to prove that: for all

$|x|\in\left[\frac{K_{0}}{4}\sqrt{(T-t_{7})|\ln(T-t_{7})|},\epsilon_{0}\right],|\xi|\leq\alpha_{0}\sqrt{|\ln\varrho(x)|},\tau\in\left[\max\left(0,-\frac{t(x)}{\varrho(x)}\right),\frac{t_{7}-t(x)}{\varrho(x)}\right]=[\tau_{0},\tau_{7}],$

the following holds

$\displaystyle\left|\mathcal{U}(x,\xi,\tau)-\hat{\mathcal{U}}(\tau)\right|$
$\displaystyle\leq$ $\displaystyle\frac{\delta_{0}}{2},$ (C.1)
$\displaystyle\left|\nabla_{\xi}\mathcal{U}(x,\xi,\tau)\right|$
$\displaystyle\leq$ $\displaystyle\frac{C_{0}}{2\sqrt{|\ln\varrho(x)|}}.$
(C.2)

We first recall equation (3.7)

$\partial_{\tau}\mathcal{U}=\Delta_{\xi}\mathcal{U}-2\frac{\left|\nabla\mathcal{U}\right|^{2}}{\mathcal{U}+\frac{\lambda^{\frac{1}{3}}\varrho^{\frac{1}{3}}(x)}{\tilde{\theta}(\tau)}}+\left(\mathcal{U}+\frac{\lambda^{\frac{1}{3}}\varrho^{\frac{1}{3}}(x)}{\tilde{\theta}(\tau)}\right)^{4}-\frac{\tilde{\theta}_{\tau}^{\prime}(\tau)}{\tilde{\theta}(\tau)}\mathcal{U}.$

\- The proof of (C.1): We first introduce the following function

$\mathcal{Z}(\xi,\tau)=\mathcal{U}(x,\xi,\tau)-\hat{\mathcal{U}}(\tau).$

Using (3.7), we write the following equation

$\partial_{\tau}\mathcal{Z}=\Delta\mathcal{Z}+\left(\mathcal{U}+\frac{\tilde{\theta}(\tau)\varrho^{\frac{1}{3}}(x)}{\lambda^{\frac{1}{3}}}\right)^{4}-\hat{\mathcal{U}}^{4}(\tau)+G(\xi,\tau),$

where

$G(\xi,\tau)=-2\frac{\left|\nabla\mathcal{U}\right|^{2}}{\mathcal{U}+\frac{\lambda^{\frac{1}{3}}\varrho^{\frac{1}{3}}(x)}{\tilde{\theta}(\tau)}}-\frac{\tilde{\theta}_{\tau}^{\prime}(\tau)}{\tilde{\theta}(\tau)}\mathcal{U}.$

Using Proposition 3.9 and the definition of $\tilde{\theta}(\tau)$ in (3.8),
we derive that

$\left|\tilde{\theta}^{\prime}(\tau)\right|\leq
C\varrho^{\frac{1}{12}}(x)(1-\tau)^{-\frac{11}{12}}.$ (C.3)

Hence, from Lemma 4.4, we derive the following: for all
$|\xi|\leq\frac{7}{4}\alpha_{0}\sqrt{|\ln\varrho(x)|}$ and
$\tau\in[\tau_{0},\tau_{7}]$,