# Behavior rigidity near non-isolated blow-up points for the semilinear heat
equation

Frank Merle
CY Cergy Paris Université and IHES
Hatem Zaag
Université Sorbonne Paris Nord, LAGA, CNRS (UMR 7539), F-93430, Villetaneuse,
France

###### Abstract

We consider the semilinear heat equation with Sobolev subcritical power
nonlinearity in dimension $N=2$, and $u(x,t)$ a solution which blows up in
finite time $T$. Given a non isolated blow-up point $a$, we assume that the
Taylor expansion of the solution near $(a,T)$ obeys some degenerate situation
labeled by some even integer $m(a)\geq 4$. If we have a sequence $a_{n}\to a$
as $n\to\infty$, we show after a change of coordinates and the extraction of a
subsequence that either ${a_{n,1}}-a_{1}=o((a_{n,2}-a_{2})^{2})$ or
$|a_{n,1}-a_{1}||a_{n,2}-a_{2}|^{-\beta}|\log|a_{n,2}-a_{2}||^{-\alpha}\to
L>0$ for some $L>0$, where $\alpha$ and $\beta$ enjoy a finite number of
rational values with $\beta\in(0,2]$ and $L$ is a solution of a polynomial
equation depending on the coefficients of the Taylor expansion of the
solution. If $m(a)=4$, then $\alpha=0$ and either $\beta=3/2$ or $\beta=2$.

MSC 2010 Classification: 35L05, 35K10, 35K58, 35B44, 35B40

Keywords: Semilinear heat equation, blow-up behavior, blow-up set.

##  1 Introduction

We consider the semilinear heat equation in space dimension $N\geq 1$ with a
Sobolev subcritical power nonlinearity:

$\left\\{\begin{array}[]{l}\partial_{t}u=\Delta u+|u|^{p-1}u,\\\ u(0)=u_{0}\in
L^{\infty}(\mathbbm{R}^{N})\end{array}\right.$ (1.1)

where

$p>1,\;\;(N-2)p<N+2.$ (1.2)

Other phenomena arise in the Sobolev critical and supercritical cases (see
from example Merle, Raphaël and Szeftel [11], Schweyer [14] and the references
therein).

We consider $u(t):x\in{\mathbbm{R}}^{N}\rightarrow u(x,t)\in{\mathbbm{R}}$ a
solution which blows up at time $T>0$ and introduce the set of its blow-up
points

$\mathscr{S}=\\{a\in\mathbbm{R}^{N}\;\;|\;\;|u(a,t)|\to+\infty\mbox{ as }t\to
T\\}$

(note that $u(x,t)\to u^{*}(x)$ as $t\to T$ whenever $x\not\in\mathscr{S}$).
In the literature, we know examples of blow-up solutions where $\mathscr{S}$
is finite or the union of some concentric spheres.

In this paper, we are interested in non-isolated blow-up points. No results
are available in this context except for curve singularity (see Zaag [17, 18,
19, 20] and Ghoul, Nguyen and Zaag [4]). Our goal is to introduce new
techniques to track this kind of question.

When $N=1$, we know from Chen and Matano [1] and Herrero and Velázquez [8]
that all the blow-up points are isolated. As we are interested in the
asymptotic behavior near non-isolated blow-up points, we need to assume that
$N\geq 2$.

Let us note that all our statements do hold for unsigned solutions
(Proposition 1 together with Theorems 2 and 3). However, for simplicity, we
reduce in the presentation and the proofs to the case of nonnegative
solutions, without loss of generality. Indeed, we know from Corollary 2 page
108 in Merle and Zaag [13] that the solution has a constant sign in some
neighborhood of any given blow-up point, which means that in similarity
variables (1.3), the unsigned case can be seen as a perturbation of the
nonnegative case with arbitrarily small exponential terms. Although the proof
needs a crucial blow-up criterion given below in Proposition 3.2 and which is
valid only for nonnegative solutions, one should keep in mind that we have a
twin version valid for unsigned solutions and given in Proposition 1.2 page
111 in [13].

Given $a\in\mathscr{S}$, it is convenient to study the local behavior of
$u(x,t)$ near $(a,T)$ in the similarity variables version $w_{a}(y,s)$ first
introduced by Giga and Kohn in [5] by

$w_{a}(y,s)=(T-t)^{\frac{1}{p-1}}u(x,t)\mbox{ where
}y=\frac{x-a}{\sqrt{T-t}}\mbox{ and }s=-\log(T-t).$ (1.3)

Using (1.1), we see that $w_{a}$ (or $w$ for short) satisfies the following
PDE for all $s\geq-\log T$ and $y\in\mathbbm{R}^{N}$:

$\partial_{s}w=\Delta w-\frac{1}{2}y\cdot\nabla w-\frac{w}{p-1}+|w|^{p-1}w.$
(1.4)

From Giga and Kohn [6], we know that

$w_{a}(y,s)\to\kappa\equiv(p-1)^{-\frac{1}{p-1}}\mbox{ as }s\to\infty,$ (1.5)

uniformly on compact sets and also in $L^{2}_{\rho}(\mathbbm{R}^{N})$, the
$L^{2}$ space with respect to the measure density

$\rho(y)=\exp\left(-\frac{|y|^{2}}{4}\right)/(4\pi)^{N/2}.$ (1.6)

According to Velázquez [16] (see also Filippas and Kohn [2] together with
Filippas and Liu [3]) , we may refine that convergence and obtain the
following first order classification:
\- either

$w_{a}(y,s)-\kappa\sim-\frac{\kappa}{4ps}\sum_{i=1}^{l}h_{2}(y_{l})$ (1.7)

where $l=1,\dots,N$, after a rotation of coordinates;
\- or

$w_{a}(y,s)-\kappa\sim
e^{-(\frac{m}{2}-1)s}\sum_{j_{1}+\dots+j_{N}=m}C_{m,j_{2},,\dots,j_{N}}h_{j_{1}}(y_{1})\dots
h_{j_{N}}(y_{N})$ (1.8)

as $s\to\infty$, for some even integer $m=m(a)\geq 4$, where
$y=(y_{1},\dots,y_{N})$, $h_{j}(\xi)$ is the rescaled Hermite polynomial
defined by

$h_{j}(\xi)=\sum_{i=0}^{\big{[}j/2\big{]}}\frac{j!}{i!(j-2i)!}(-1)^{i}\xi^{j-2i},$
(1.9)

and the multilinear form

$\sum_{j_{1}+\dots+j_{N}=m}C_{m,j_{2},\dots,j_{N}}y_{1}^{j_{1}}\dots
y_{N}^{j_{N}}$ (1.10)

is non zero and nonpositive. Extending the definition of $m(a)$ by $2$ if
(1.7) holds, Khenissy, Rebai and Zaag called $m(a)$ the “profile order at $a$”
in [10].

From Velázquez [15], we know that this expansion may indicate whether $a$ is
an isolated blow-up point or not. Indeed, from Theorem 2 page 1570 in [15], we
have the following:

(i) If (1.7) holds with $l=N$ or (1.8) holds with the multilinear form in
(1.10) negative, then $a$ is an isolated blow-up point.

(ii) More generally, for any $\epsilon>0$, there exists $\delta>0$ such that
$\mathscr{S}\cap B(0,\delta)\subset\Omega_{\epsilon}$ where:
\-
$\Omega_{\epsilon}\equiv\\{y\in\mathbbm{R}^{N}\;|\;\sum_{i=1}^{l}y_{i}^{2}\leq\epsilon|y|^{2}\\}$
if (1.7) holds,
\-
$\Omega_{\epsilon}\equiv\\{y\in\mathbbm{R}^{N}\;|\;\left|\sum_{j_{1}+\dots+j_{N}=m}C_{m,j_{2},\dots,j_{N}}y_{1}^{j_{1}}\dots
y_{N}^{j_{N}}\right|\leq\epsilon|y|^{m}\\}$ if (1.8) holds.

Expressing item (ii) differently, we may say that the set
$\Omega_{0}\cap\\{|y|=1\\}$ (which is finite) gives indications on the
location of neighboring blow-up points. Indeed, if
$\Omega_{0}\cap\\{|y|=1\\}=\emptyset$, then we are in Case (i), and $a$ is an
isolated blow-up point ; if $\Omega_{0}\cap\\{|y|=1\\}\neq\emptyset$, then we
cannot assert whether $a$ is isolated or not (in fact, we believe the converse
of item (i) to be very hard); if we further assume that $a$ is non isolated,
then for sure the neighboring blow-up points are located “along” the
directions of the non-zero elements of $\Omega_{0}\cap\\{|y|=1\\}$.

In this paper, our first goal is to refine the expansion (1.7)-(1.8) up to the
second order, and more if possible. In fact, we don’t consider the case (1.7),
where more refined expansions were obtained in a series of papers (see Zaag
[17, 18, 19] and Ghoul, Nguyen and Zaag [4]). We will instead focus on the
case (1.8), where no refinement is available, up to our knowledge. In
addition, the exponential decay observed in the case (1.8) is more
advantageous than the polynomial interaction of case (1.7), which allows us to
better handle the interactions between the various components of the solution
in Theorem 3 below.

Our first result states that in fact such an expansion is possible up to any
order:

######  Proposition 1 (Asymptotic expansion in similarity variables for $m\geq
4$).

Consider $w(y,s)$ a solution to equation (1.4) defined for all
$y\in\mathbbm{R}^{N}$ and $s\geq s_{0}$ for some $s_{0}$ and assume that it
satisfies the expansion (1.8) for some even integer $m\geq 4$. Then, for any
integer $M\geq 2m$,

$w(y,s)=\kappa+\sum_{{\scriptsize\begin{array}[]{l}j=2m,\dots,M\\\
l=1,\dots,\alpha_{j}\\\ (i_{1},\dots,i_{N})\in
E_{j,l}\end{array}}}e^{-\frac{j}{2}s}s^{l}h_{i_{1}}(y_{1})\dots
h_{i_{N}}(y_{N})+o(e^{-\frac{M}{2}s}s^{\alpha_{M}})$

as $s\to\infty$, uniformly on compact sets and in $L^{q}_{\rho}$ for any
$q\geq 2$, where $\alpha_{j}\in\mathbbm{N}$ and
$E_{j,l}\subset\mathbbm{N}^{N}$ is finite.

If the multilinear form in (1.10) has some degenerate directions, assuming
furthermore that $w=w_{a}$ where $a$ is a non isolated blow-up point of (1.1),
we may uncover some rigidity in the expansion, in the sense that we show that
some coefficients of the Taylor expansion are zero. In order to simplify the
presentation, we assume in the following that

$N=2,\;\;p=2,\mbox{ and }u_{0}\geq 0.$ (1.11)

The general case where $N=2$, $p>1$ and $u_{0}$ has no sign follows with the
same proof and natural adaptations of the statements. More precisely, we claim
the following:

######  Theorem 2 (Second order refined asymptotic expansion near a non
isolated point, when $m\geq 4$).

Consider $u(x,t)$ a solution of equation (1.1) blowing up at time $T>0$.
Assume in addition that the origin is a non isolated blow-up point with
$m(0)=m\geq 4$. Then:
(i) Up to some rotation of coordinates, it holds that

$\displaystyle w_{0}(y,s)=1+$ $\displaystyle
e^{(1-\frac{m}{2})s}\sum_{j=0}^{m-2}C_{m,j}h_{m-j}(y_{1})h_{j}(y_{2})$ (1.12)
$\displaystyle+e^{\frac{1-m}{2}s}\sum_{j=0}^{m}C_{m+1,j}h_{m+1-j}(y_{1})h_{j}(y_{2})+O(se^{-\frac{m}{2}s})$

as $s\to\infty$ in $L^{q}_{\rho}$ for any $q\geq 2$, for some real
coefficients $C_{i,j}$ such that the multilinear form in (1.10) is
nonpositive, where $w_{0}(y,s)$ is defined in (1.3).
(ii) If $m=4$, then

$(C_{4,0},C_{4,1},C_{4,2})\neq(0,0,0),\;\;C_{4,0}\leq 0,\;C_{4,2}\leq 0\mbox{
and }C_{4,1}^{2}-4C_{4,0}C_{4,2}\leq 0.$ (1.13)

######  Remark.

If the origin is an isolated blow-up point, we expect no rigidity in the
coefficients of the Taylor expansion.

######  Remark.

Several higher order improvements of (1.12) (showing cancelations of
coefficients) are available in the proof.

Our second statement concerns the local geometry of the blow-up set:

######  Theorem 3 (Rigidity in the blow-up set near a non-isolated blow-up
point when $m\geq 4$).

Consider under the hypotheses of Theorem 2 some sequence
$a_{n}=({a_{n,1}},{a_{n,2}})$ of non-zero blow-up points converging to the
origin. Then,
(i) It holds that

$a_{n}\cdot e^{\bot}=o(a_{n}\cdot e)\mbox{ as }n\to\infty,$

up to extracting a subsequence, where $e\in\mathbbm{R}^{2}$ is a unitary
vector of a degenerate direction of the multilinear form (1.10), $e^{\bot}$ is
unitary and $e\cdot e^{\bot}=0$.
(ii) Up to a rotation and a symmetry of the axes, and up to a subsequence, it
holds that ${a_{n,1}}\geq 0$, ${a_{n,2}}\geq 0$ and (1.12) still holds with
possibly different constants, with

$\mbox{either }{a_{n,1}}=o({a_{n,2}}^{2})\mbox{ or }{a_{n,1}}\sim
L|{a_{n,2}}|^{\beta}|\log{a_{n,2}}|^{\alpha},$

for some $\beta$ and $\alpha$ enjoying a finite number of rational values with
$0<\beta\leq 2$ (see the proof for a finer description of the localization of
$\beta$).
Moreover, $L$ is a solution of a polynomial equation involving the
coefficients of the Taylor expansion (1.12) of $w_{0}(y,s)$ or one of its
higher order refinements.
(iii) When $m=4$, the only possibilities for $e$ in item (i) are $e=(0,1)$ and
$e=(-2C_{4,2},C_{4,1})$, the second possibility occurring only if
$C_{4,1}^{2}-4C_{4,0}C_{4,2}=0$ and $C_{4,2}\neq 0$. In addition, we have the
following simple statement for the behavior of $a_{n}$ in item (ii):

$\mbox{either }{a_{n,1}}=o({a_{n,2}}^{2}),\mbox{ or }{a_{n,1}}\sim
L{a_{n,2}}^{2}\mbox{ or }{a_{n,1}}\sim L{a_{n,2}}^{3/2}\mbox{ with }L>0.$

######  Remark.

The possible values taken by $\beta$ are such that $\frac{\beta-1}{2}\in
E_{1}\cup E_{2}$ where the 2 sets $E_{1}$ and $E_{2}$ are defined respectively
in (5.18) and (5.19) below in the proof.

######  Remark.

Our strategy to find the precise subquadratic regimes when $m=4$ can be
carried out for any fixed $m\geq 6$. For example, when $m=6$, we found the
following values for $\beta$ : $4/3$, $3/2$ and $5/3$, with $\alpha=0$ each
time.

######  Remark.

Note that in the case $m=2$, we think that a similar result also holds.
However, the proof should be very different, since we are in polynomial scales
of time $s$ (see (1.7)), unlike the case $m\geq 4$ where we are in exponential
scales (see (1.8)).

######  Remark.

We believe that our techniques should apply in the case $N\geq 3$ and yield a
similar result.

Let us briefly explain our strategy in this paper. Using Proposition 1, we
first make a Taylor expansion of $w_{0}$. Given $b$ a nearby blow-up point, we
derive from this a Taylor expansion for $w_{b}$ (use (3.5) given below). Since
$w_{b}$ is uniformly bounded in $L^{\infty}$, its components in the expansion
(2.4) cannot grow, which implies some cancelations in the Taylor coefficients
of $w_{0}$, justifying (1.12). In a second step, by the same argument, we
derive some constraints on the location of the neighboring blow-up points,
leading to Theorems 2 and 3.

Note that for simplicity, we give the proofs only under assumption (1.11). The
general case follows by the same proof. The only delicate point is to replace
the blow-up criterion given below in Proposition 3.2 by its twin version valid
for unsigned solutions and given in Proposition 2.1 page 111 of Merle and Zaag
[13].

##  2 Existence of an expansion of the solution in similarity variables up to
any order

We prove Proposition 1 in this section.

###### Proof of Proposition 1.

Note that the case $m=4$ is harder than the case $m\geq 6$. Indeed, the
quadratic term in the equation induces more interactions in the former than in
the latter case. For that reason, we only give the proof in the harder case,
namely when $m=4$.

Introducing

$v=w-1,$ (2.1)

we see from equation (1.4) that $v$ satisfies the following PDE for all
$s\geq-\log T$ and $y\in\mathbbm{R}^{2}$:

$\partial_{s}v=\mathscr{L}v+v^{2}$ (2.2)

where

$\mathscr{L}v=\Delta v-\frac{1}{2}y\cdot\nabla v+v=\frac{1}{\rho}\mathop{\rm
div}(\rho\nabla v)+v,$ (2.3)

and $\rho(y)$ is defined in (1.6). The operator $\mathscr{L}$ is self-adjoint
in $L^{2}_{\rho}(\mathbbm{R}^{2})$, and its spectrum is given by the set
$\\{1-\frac{j}{2}\;|\;j\in\mathbbm{N}\\}$, which consists only in eigenvalues,
having $h_{n}(y_{1})h_{l}(y_{2})$ (1.9) as eigenfunctions, in the sense that

$\mathscr{L}(h_{n}(y_{1})h_{l}(y_{2}))=\left(1-\frac{n+l}{2}\right)h_{n}(y_{1})h_{m}(y_{2}).$

Expanding $v$ as follows:

$v(y,s)=\sum_{i\in\mathbbm{N},\;0\leq j\leq
i}v_{i,j}(s)h_{i-j}(y_{1})h_{j}(y_{2})$ (2.4)

and introducing for any $l\in\mathbbm{N}$, $v_{-,l}$ and the
$L^{2}_{\rho}(\mathbbm{R}^{2})$ projector $P_{-,l}$ defined by

$v_{-,l}(y,s)=P_{-,l}(v)\equiv\sum_{i\geq l,\;0\leq j\leq
i}v_{i,j}(s)h_{i-j}(y_{1})h_{j}(y_{2}),$ (2.5)

we write the following equations satisfied by $v_{i,j}$ and $v_{-,l}$:

$\displaystyle v_{i,j}^{\prime}(s)$
$\displaystyle=\left(1-\frac{i}{2}\right)v_{i,j}(s)+\int
v(y,s)^{2}k_{i-j}(y_{1})k_{j}(y_{2})\rho(y)dy,$ (2.6)
$\displaystyle\partial_{s}v_{-,l}$
$\displaystyle=\mathscr{L}v_{-,l}+P_{-,l}(v^{2}),$ (2.7)

where

$k_{j}(\xi)=h_{j}(\xi)/\|h_{j}\|_{L^{2}_{\rho}}^{2}.$ (2.8)

This way, we are in a position to perform the first iteration, in order to
refine the asymptotic expansion in (1.8).

Part 1: The first iteration

Like the following iterations, we proceed in 3 steps in order to get the next
terms in the expansion, starting from (1.8):
\- we first use parabolic regularity to improve the convergence in (1.8), from
$L^{2}_{\rho}(\mathbbm{R}^{2})$ to $L^{q}_{\rho}(\mathbbm{R}^{2})$ for any
$q\geq 2$;
\- then, we use the improved convergence to expand the quadratic term in
equation (2.2), and write an ODE satisfied by the component $v_{i,j}$ defined
in (2.4); solving that ODE gives an estimate on $v_{i,j}$, better than what
(1.8) states;
\- finally, we use again the estimate on the quadratic term of equation (2.2)
and write from (2.7) a differential inequality satisfied by $v_{-,l}$ defined
in (2.5); integrating that inequality gives an improved estimated on
$v_{-,l}$.

Step 1: Parabolic regularity to improve the convergence in (1.8)

The following parabolic regularity estimate is crucial for the improvement:

######  Lemma 2.1.

(Regularizing effect of the operator $\mathscr{L}$)
(i) (Herrero and Velázquez [9]) For any $q>1$, $r>1$, $\psi_{0}\in
L^{q}_{\rho}(\mathbbm{R}^{N})$ and
$s>\max\left(0,-\log(\frac{q-1}{r-1})\right)$, it holds that

$\|e^{\mathscr{L}s}\psi_{0}\|_{L^{r}_{\rho}(\mathbbm{R}^{N})}\leq\frac{C(q,r)e^{s}e^{-\frac{Ns}{2r}}}{(1-e^{-s})^{\frac{N}{2q}}(q-1-e^{-s}(r-1))^{\frac{N}{2r}}}\|\psi_{0}\|_{L^{q}_{\rho}(\mathbbm{R}^{N})}.$

(ii) Consider $r\geq 2$ and $v_{0}\in L^{r}_{\rho}(\mathbbm{R}^{N})$ such that
$|v_{0}(y)|+|\nabla v_{0}(y)|\leq C(1+|y|^{k})$ for some $k\in\mathbbm{N}$.
Then, for all $s\geq 0$, we have
$\|e^{\mathscr{L}s}v_{0}\|_{L^{r}_{\rho}(\mathbbm{R}^{N})}\leq
e^{s}\|v_{0}\|_{L^{r}_{\rho}(\mathbbm{R}^{N})}$.

######  Remark.

Although we are working in two space dimensions, we felt it better to state
the result for any $N\geq 1$, for future purpose.

###### Proof.

(i) See Section 2 page 139 of [9]. Although the proof in that paper was given
for $N=1$, it extends to higher dimensions with no difficulty.
(ii) From Lemma 4 page 555 in Bricmont and Kupiainen, we know that
$|e^{\mathscr{L}s}(1+|y|^{k})|\leq Ce^{s}(1+|y|^{k})$. Therefore, if
$\psi(y,s)=e^{\mathscr{L}s}v_{0}(y)$, we see from the hypotheses that
$|\psi(y)|+|\nabla\psi(y)|\leq Ce^{s}(1+|y|^{k})$. Using the linear equation
satisfied by $\psi$, we justify that

$\displaystyle\frac{d}{ds}\int|\psi(y,s)|^{r}\rho(y)dy$
$\displaystyle=-r(r-1)\int|\psi(y,s)|^{r-2}|\nabla\psi(y,s)|^{2}\rho(y)dy+\int|\psi(y,s)|^{r}\rho(y)dy$
$\displaystyle\leq\int|\psi(y,s)|^{r}\rho(y)dy,$

and the result follows. ∎

Now, if

$\bar{v}(y,s)=e^{-s}\sum_{j=0}^{4}C_{4,j}h_{4-j}(y_{1})h_{j}(y_{2})\mbox{ and
}g=v-\bar{v}$ (2.9)

we see from (1.8) and the definition (2.1) of $v(y,s)$ that

$\|g(s)\|_{L^{2}_{\rho}}=o(e^{-s})\mbox{ as }s\to\infty.$ (2.10)

Moreover, $\bar{v}$ is an approximate solution of equation (2.2), up to some
error term, in the sense that

$\partial_{s}\bar{v}=\mathscr{L}\bar{v}+\bar{v}^{2}+O\left(e^{-2s}(1+|y|^{8})\right)$

(in fact, in this first iteration, unlike the following iterations, $\bar{v}$
solves the linear equation, but we felt it better to write that equation as a
perturbation of the nonlinear equation (2.2), since this latter fact will be
always true in the following iterations). Using (2.2), we write the following
equation satisfied by $g(y,s)$:

$\partial_{s}g=\mathscr{L}g+(v+\bar{v})g+O\left(e^{-2s}(1+|y|^{8})\right).$
(2.11)

Recalling the following estimate from Giga and Kohn [6] and Giga, Matsui and
Sasayama [7]:

$\|u(t)\|_{L^{\infty}}\leq\frac{C}{T-t},$

we see by definitions (1.3) and (2.1) of $w$ and $v$ that

$\|v(s)\|_{L^{\infty}}\leq C.$ (2.12)

Furthermore, by defintion (2.9) of $\bar{v}$, we see that

$|\bar{v}(y,s)|\leq C,\mbox{ if }|y|<e^{\frac{s}{4}}.$ (2.13)

Consider then $q\geq 2$. Using the Duhamel formulation of equation (2.11)
together with (2.12) and (2.13), we write for some $M>0$ and for some $s^{*}$
to be taken fixed large enough,

$\displaystyle\|g(s)\|_{L^{q}_{\rho}}$
$\displaystyle\leq\|e^{s^{*}(\mathscr{L}+M\operatorname{\mathrm{Id}})}g(s-s*)\|_{L^{q}_{\rho}}+\int_{s-s^{*}}^{s}\|e^{(s-\sigma)(\mathscr{L}+M\operatorname{\mathrm{Id}})}1_{\\{|x|>e^{\frac{\sigma}{4}}\\}}\bar{v}g(\sigma)\|_{L^{q}_{\rho}}d\sigma$
$\displaystyle+C\int_{s-s^{*}}^{s}\|e^{(s-\sigma)(\mathscr{L}+M\operatorname{\mathrm{Id}})}e^{-2\sigma}(1+|x|^{8})\|_{L^{q}_{\rho}}d\sigma.$

Fixing $s^{*}>\log(q-1)$, we write from Lemma 2.1

$\|g(s)\|_{L^{q}_{\rho}}\leq
C\|g(s-s*)\|_{L^{2}_{\rho}}+\int_{s-s^{*}}^{s}\|1_{\\{|x|>e^{\frac{\sigma}{4}}\\}}\bar{v}g(\sigma)\|_{L^{q}_{\rho}}d\sigma+Ce^{-2s}\int_{s-s^{*}}^{s}\|(1+|x|^{8})\|_{L^{q}_{\rho}}d\sigma.$
(2.14)

Note that $\|g(s-s*)\|_{L^{2}_{\rho}}=o(e^{-s})$ as $s\to\infty$ from (2.10)
and $\|(1+|x|^{8})\|_{L^{q}_{\rho}}\leq C$. It remains then to estimate the
middle term in the right-hand side of (2.14) in order to conclude.
Since for $\sigma\in[s-s^{*},s]$ and $|y|>e^{\frac{\sigma}{4}}$, we have by
definition (2.9) of $g$ and $\bar{v}$ and estimate (2.12),

$|\bar{v}(y,\sigma)g(y,\sigma)|=|\bar{v}(y,\sigma)(v(y,\sigma)-\bar{v}(y,\sigma)|\leq
Ce^{-s}(1+|y|^{4})+Ce^{-2s}(1+|y|^{8}),$

and

$0\leq\rho(y)\leq Ce^{-|y|^{2}/8}\times e^{-|y|^{2}/8}\leq
Ce^{-e^{(s-s^{*})/2}/8}e^{-|y|^{2}/8},$

it follows that

$\|1_{\\{|x|>e^{\frac{\sigma}{4}}\\}}\bar{v}g(\sigma)d\sigma\|_{L^{q}_{\rho}}\leq
Ce^{-s-e^{(s-s^{*})/2}/8}.$

Gathering the above-mentioned bounds, it follows that

$\|g(s)\|_{L^{q}_{\rho}}=o(e^{-s})\mbox{ as }s\to\infty,$

which is precisely the goal of Step 1.

Step 2: Refinement of the behavior of $v_{i,j}$

Since estimate (1.8) holds in $L^{4}_{\rho}$ by Step 1, we may use it to
refine the ODE (2.6) satisfied by $v_{i,j}$ and write:

$\left|v_{i,j}^{\prime}(s)-\left(1-\frac{i}{2}\right)v_{i,j}(s)\right|\leq
C\|v(s)\|_{L^{4}_{\rho}}^{2}\leq Ce^{-2s}.\\\ $ (2.15)

Since $|v_{i,j}(s)|\leq Ce^{-s}$ by (2.1) and (1.8), using elementary ODE
techniques, we derive for $s$ large enough:

$\sup_{i\leq 5}|v_{i,j}(s)-C_{i,j}e^{(1-\frac{i}{2})s}|\leq Cse^{-2s},$ (2.16)

with

$C_{i,j}=0\mbox{ if }i\leq 3.$

Note that the constant $C$ in this step may depend on $(i,j)$.

Step 3: Refinement of the behavior of $v_{-,l}$

Since
$\|P_{-,l}(v^{2})\|_{L^{2}_{\rho}}\leq\|v^{2}\|_{L^{2}_{\rho}}=\|v\|_{L^{4}_{\rho}}^{2}\leq
Ce^{-2s}$ by the regularity in Step 1, we see from equation (2.7) that

$\|v_{-,l}\|_{L^{2}_{\rho}}^{\prime}\leq\left(1-\frac{l}{2}\right)\|v_{-,l}\|_{L^{2}_{\rho}}+Ce^{-2s},$
(2.17)

where $C$ may depend on $l$. Taking $l=6$ and integrating this inequality
yields

$\|v_{-,6}\|_{L^{2}_{\rho}}\leq Cse^{-2s}.$ (2.18)

Gathering (2.16) and (2.18), we see that

$v(y,s)=e^{-s}\sum_{j=0}^{4}C_{4,j}h_{4-j}(y_{1})h_{j}(y_{2})+e^{-\frac{3}{2}s}\sum_{j=0}^{5}C_{5,j}h_{5-j}(y_{1})h_{j}(y_{2})+O(se^{-2s})$
(2.19)

as $s\to\infty$, in $L^{2}_{\rho}(\mathbbm{R}^{2})$. This gives the first
terms of the expansion announced in Proposition 1.

Part 2: the following iterations

The strategy developed in the first iteration works here, iteratively! Let us
see how the second iteration works: using the parabolic regularity developed
in Step 1 of the first iteration, one can show that estimate (2.19) holds also
in $L^{q}_{\rho}$, for any $q\geq 2$, and also uniformly on compact sets.
Using this, one can refine the quadratic term in equations (2.6) and (2.7) up
to $O(se^{-3s})$. Integrating those equations (with $i\leq 7$ and $l=8$), we
refine the expansion of $v_{,j}$ and $v_{-,l}$ up to that order, resulting in
an expansion of $v$ in $L^{2}_{\rho}$, up to $O(s^{2}e^{-3s})$, and showing
terms like $e^{-s}$, $e^{-\frac{3}{2}s}$, $se^{-2s}$ and $e^{-2s}$. Iterating
the process, we may get an expansion for $v$ valid up to any order, in any
$L^{q}_{\rho}$ space with $q\geq 2$, and also uniformly on compact sets. This
concludes the proof of Proposition 1, when $p=2$, the solution is nonnegative
and $m=4$. As we explained in the beginning of the proof, this is the harder
case, and the adaptation to the general case is straightforward. ∎

##  3 Rigidity in the Taylor expansion for general $m\geq 4$

This section is devoted to the proof of Theorem 2.

###### Proof of Theorem 2.

Consider $u(x,t)$ a solution of equation (1.1) blowing up at time $T>0$.
Assume that the origin is a non isolated blow-up point with $m(0)=m\geq 4$ is
even. We proceed in 4 steps:
\- First, we recall from Velázquez [16] the first order Taylor expansion of
$w_{0}(y,s)$ and show that two coefficients are zero.
\- Second, we introduce some geometrical transformation as a crucial tool in
the proof.
\- Third, we give the next order in the Taylor expansion and show that one
coefficient is zero.
\- Forth, we take $m=4$ and justify (1.13).

Step 1: First order Taylor expansion

As stated in (1.8), we know from Velázquez [16] that

$w_{0}(y,s)=1+e^{(1-\frac{m}{2})s}\sum_{j=0}^{m}C_{m,j}h_{m-j}(y_{1})h_{j}(y_{2})+o(e^{(1-\frac{m}{2})s})$
(3.1)

for some real coefficients $C_{m,j}$ for $j=0,\dots,m$ such that the
multilinear form

$\sum_{j=0}^{m}C_{m,j}y_{1}^{m-j}y_{2}^{j}$ (3.2)

is non zero and nonpositive. From the alternative due to Velázquez [16] and
given on page 1, we know that this multiform has (at least) one direction of
degeneracy. Up to making a rotation of coordinates, we may assume that the
degeneracy direction is the axis $\\{y_{1}=0\\}$, which means that

$C_{m,m}=0.$ (3.3)

Moreover, in order to guarantee the nonpositivity of the multilinear form, we
directly see that

$C_{m,m-1}=0,$ (3.4)

otherwise, for $|y_{2}|$ large enough, the multilinear form may achieve
positive values.

Step 2: A geometrical transformation

As already written in the introduction, a crucial idea lays at the heart of
our strategy. It consists in remarking that any Taylor expansion of
$w_{0}(y,s)$ can be translated into a Taylor expansion of $w_{b}(z,s)$, for
any other point $b\in\mathbbm{R}^{2}$ (not necessarily a blow-up point),
thanks to the following relation which follows from the similarity variables
definition (1.4):

$w_{b}(y_{b},s)=w_{0}(y,s)\mbox{ where }y=be^{\frac{s}{2}}+y_{b}.$ (3.5)

With a suitable choice of $b$, the uniform boundedness of $w_{b}$ in
$L^{\infty}$ induces some cancelations of the coefficients appearing in the
Taylor expansion of $w_{0}$.

Step 3: Second order Taylor expansion

Following the first order expansion (3.1) together with (3.3) and (3.4), we
may use Proposition 1 to find the next order in the Taylor expansion:

$\displaystyle w_{0}(y,s)=$ $\displaystyle
1+e^{(1-\frac{m}{2})s}\sum_{j=0}^{m-2}C_{m,j}h_{m-j}(y_{1})h_{j}(y_{2})$ (3.6)
$\displaystyle+e^{\frac{1-m}{2}s}\sum_{j=0}^{m+1}C_{m+1,j}h_{m+1-j}(y_{1})h_{j}(y_{2})+\bar{v}_{0}(y,s)$

with

$\|\bar{v}_{0}(s)\|_{L^{q}_{\rho}}=O(se^{-\frac{m}{2}s})\mbox{ as
}s\to\infty,$ (3.7)

for any $q\geq 2$, for some real constants $C_{m+1,j}$ for $j=0,\dots,m+1$. We
claim that $C_{m+1,m+1}=0$. Indeed, let us assume by contradiction that
$C_{m+1,m+1}\neq 0$ and find a contradiction. Following Step 2 and choosing

$b=(0,Ae^{\frac{s_{0}}{2}}),$ (3.8)

where $A$ and $s_{0}$ will be taken large enough (in modulus for $A$), we show
the following (note that $b$ may or may not be a blow-up point):

######  Lemma 3.1.

For any $A\in\mathbbm{R}$ such that $|A|\geq 1$, it holds that

$w_{b,0,0}(s_{0})-1\sim C_{m+1,m+1}A^{m+1}e^{\frac{(1-m)}{2}s_{0}}\mbox{ as
}s_{0}\to\infty,$

where $b$ is given by (3.8), and $w_{b,0,0}(s_{0})$ is the coordinate of
$w_{b}(y,s_{0})$ along the polynomial $h_{0}(y_{1})h_{0}(y_{2})\equiv 1$ as in
(2.4).

Let us first use this lemma to find a contradiction, then prove the lemma.
Choosing $A$ of the sign of $C_{m+1,m+1}$, then taking $s_{0}$ large enough,
we see that

$w_{b,0,0}(s_{0})>1.$ (3.9)

In other words, $w_{b}$ satisfies the following blow-up criterion we proved in
[12]:

######  Proposition 3.2 (Blow-up criterion for nonnegative solutions of
(1.4)).

Let $w$ be a nonnegative solution of (1.4) which satisfies

$w_{0,0}(s_{1})>1$

for some $s_{1}\in\mathbbm{R}$. Then, $w$ cannot be defined for all
$(y,s)\in\mathbbm{R}^{2}\times[s_{1},\infty)$.

######  Remark.

This proposition is valid only for nonnegative solutions. For unsigned
solutions, it should be replaced by a twin statement given in Proposition 1.2
page 111 in [13].

###### Proof.

See Proposition 3.8 page 164 in [12]. ∎

Since $w_{b}$ exists by definition (1.3) for all
$(y,s)\in\mathbbm{R}^{2}\times[-\log T,+\infty)$, a contradiction follows from
(3.9) and this proposition. Thus, $C_{m+1,m+1}=0$, provided that we prove
Lemma 3.1.

###### Proof of Lemma 3.1.

Using the transformation (3.5) together with the definition (3.8) of $b$, we
see that

$v_{b}(y_{b},s_{0})=v_{0}(y,s_{0})\mbox{ with
}y_{1}=y_{b,1},\;\;y_{2}=y_{b,2}+A,$ (3.10)

hence, we write from (3.6)

$\displaystyle v_{b}(y_{b},s_{0})$
$\displaystyle=e^{(1-\frac{m}{2})s_{0}}\sum_{j=0}^{m-2}C_{m,j}h_{m-j}(y_{b,1})h_{j}(y_{b,2}+A)$
(3.11)
$\displaystyle+e^{\frac{1-m}{2}s_{0}}\sum_{j=0}^{m+1}C_{m+1,j}h_{m+1-j}(y_{b,1})h_{j}(y_{b,2}+A)+\bar{v}_{b}(y_{b},s_{0})$
(3.12)

with

$\bar{v}_{b}(y_{b},s_{0})=\bar{v}_{0}(y,s_{0}).$ (3.13)

Recalling the classical relation

$h_{j}^{\prime}=jh_{j-1}$

related to the Hermite polynomials (1.9), we may use a Taylor expansion to
derive the following binomial relation

$h_{j}(\xi+A)=\sum_{m=0}^{j}\binom{j}{m}A^{j}h_{j-m}(\xi).$ (3.14)

Using this relation and projecting the expression (3.12), we derive the
following expression

$v_{b,0,0}(s_{0})=A^{m+1}C_{m+1,m+1}e^{\frac{1-m}{2}s_{0}}+\bar{v}_{b,0,0}(s_{0})$

where we have used the classical orthogonality relation

$\int_{\mathbbm{R}}h_{l}(\xi)h_{j}(\xi)\rho(\xi)d\xi=2^{j}j!\delta_{l,j}.$
(3.15)

We need to estimate $\int\bar{v}_{b}(y_{b},s_{0})\rho(y_{b})dy_{b}$ in order
to conclude. By definition, we write

$\displaystyle\bar{v}_{b,0,0}(s_{0})$
$\displaystyle=\int\bar{v}_{0}(y_{1},y_{2},s_{0})\rho(y-(0,A))dy.$

Now, let us compute

$\rho(y-(0,A))=\rho(y)\exp\left(\frac{A}{2}y_{2}\right)\exp\left(-\frac{A^{2}}{4}\right).$

Since $\int\exp(Ay_{2})\rho(y)dy\leq Ce^{A^{2}}$, using the Cauchy-Schwartz
inequality together with (3.7), we write

$|\bar{v}_{b,0,0}(s_{0})|\leq
Ce^{\frac{A^{2}}{4}}\left(\int\bar{v}_{0}(y,s_{0})^{2}\rho(y)dy\right)^{1/2}\leq
Ce^{\frac{A^{2}}{4}}s_{0}e^{-\frac{m}{2}s_{0}}.$ (3.16)

This concludes the proof of Lemma 3.1. ∎

This also concludes the proof of the fact that $C_{m+1,m+1}=0$. From (3.6), se
see that the Taylor expansion (1.12) holds.

Step 4: Proof of (1.13) when $m=4$

Using (1.12), we see that the 4-form introduced in (3.2) reads as follows:

$C_{4,0}x_{1}^{4}+C_{4,1}x_{1}^{3}x_{2}+C_{4,2}x_{1}^{2}x_{2}^{2}.$

Since this form is non-zero and nonpositive, dividing by $x_{1}^{2}$, we see
that (1.13) follows.

This concludes the proof of Theorem 2. ∎

##  4 Rigidity in the geometry of the blow-up set when $m=4$

This section is devoted to the proof of Theorem 3 when $m=4$. The proof when
$m\geq 6$ is different and will be given in Section 5.

Consider $u(x,t)$ a solution of equation (1.1) blowing up at time $T>0$.
Assume that the origin is a non isolated blow-up point with $m(0)=4$. Consider
$a_{n}$ a sequence of non-zero blow-up points $a_{n}$ converging to the
origin.

We proceed in 4 subsections to give the proof:
\- In Subsection 4.1, we prove item (i) and the first part of item (iii). In
particular, after a change of variables, we identify 3 possible regimes for
the convergence of $a_{n}$: superquadratic, quadratic and subquadratic.
\- In Subsection 4.2, we show that the only subquadratic regime follows a
$3/2$ law, and determine its constant.
\- In Subsection 4.3, we focus on the quadratic regime, and show that its
“constant” enjoys at most 3 possible values. This will give the second part of
item (iii), which clearly implies item (ii).
\- Finally, in Subsection 4.4, we give the proof of some propositions which
were used in the previous subsections, and which are devoted to some
refinements of the behavior of the solution.

###  4.1 Superlinear behavior for neighboring points

Recalling the alternative of Velázquez [16] given on page 1, we see that
locally near the origin, the blow-up set is located along some degeneracy
directions of the multilinear form (1.10). From the choice of the axes we made
in Theorem 2 and (1.12), we know that the axis $\\{y_{1}=0\\}$ is a degeneracy
direction. Moreover, using (1.13), we remark that if

$C_{4,1}^{2}-4C_{4,0}C_{4,2}=0\mbox{ and }C_{4,2}\neq 0,$

then, we may have another (different) degeneracy direction for the multilinear
form (1.10), namely the line of equation $C_{4,1}x_{1}+2C_{4,2}x_{2}=0$, no
more. Thus, item (i) holds for the sequence $a_{n}$. Since these two straight
lines are orthogonal (respectively) to $(0,1)$ and $(C_{4,1},2C_{4,2})$, the
first part of item (iii) follows too.

Up to a rotation and symmetry of the axes, together with the extraction of a
subsequence still denoted the same, we may assume that $a_{n}$ converges along
the $x_{2}$-axis, (which means that

${a_{n,1}}=o({a_{n,2}})$ (4.1)

as $n\to\infty$), and that

$\forall n\in\mathbbm{N},\;{a_{n,1}}\geq 0\mbox{ and }{a_{n,2}}\geq 0.$ (4.2)

Note that after such a change of variables, the Taylor expansion (1.12)
remains valid, with possibly different coefficients, denoted $C_{4,j}$ for
$j=0,1,2$ and $C_{5,j}$ for $j=1,\dots,4$.

Up to further extracting a subsequence still denoted the same, we are in one
of the following cases as $n\to\infty$:
\- (subquadratic regime) with ${a_{n,1}}\gg{a_{n,2}}^{2}$,
\- (quadratic regime) with ${a_{n,1}}\sim L{a_{n,2}}^{2}$ for some $L>0$,
\- (superquadratic regime) with ${a_{n,1}}\ll{a_{n,2}}^{2}$.

In the two following subsections, we investigate the subquadratic then the
quadratic regimes for the sequence $a_{n}$, in order to conclude the proof of
Theorem 3.

###  4.2  Subquadratic regimes for neighboring blow-up points

In this subsection, we assume that

$a_{n,1}\gg{a_{n,2}}^{2}\mbox{ as }n\to\infty.$ (4.3)

We will show that the sequence $a_{n}$ necessarily follows a $3/2$ power law.
We will also determine the “constant” in front of such a law.

Since $a_{n}\neq 0$, we see from (4.1) and (4.3) that

${a_{n,2}}>0\mbox{ and }{a_{n,1}}>0.$ (4.4)

We proceed in 4 parts to prove the $3/2$ law for $a_{n}$:
\- In Part 1, we prove that $C_{4,2}=C_{4,1}=0$ and $C_{4,0}<0$.
\- In Part 2. we prove that $C_{5,4}=0$ and
$a_{n,1}=O\left({a_{n,2}}^{\frac{3}{2}}\right)$ as $n\to\infty$.
\- In Part 3, assuming that $\frac{{a_{n,1}}}{{a_{n,2}}^{\frac{3}{2}}}$ has a
non-zero limit, we determine that limit in terms of the Taylor coefficients of
the solution.
\- In Part 4, we rule out the case where
${a_{n,1}}=o({a_{n,2}}^{\frac{3}{2}})$.

Part 1: Proof of the fact that $C_{4,2}=C_{4,1}=0$ and $C_{4,0}<0$

We claim that it is enough to prove that

$C_{4,2}=0.$ (4.5)

Indeed, if this holds, recalling that $C_{4,4}=C_{4,3}=0$ in the multilinear
form (1.10), thanks to (1.12), we get the following simpler expression for the
form:

$C_{4,0}y_{1}^{4}+C_{4,1}y_{1}^{3}y_{2}.$

Since the is nonpositive and non-zero as stated right after (1.10), we
necessarily have

$C_{4,1}=0$

and

$C_{4,0}<0.$ (4.6)

Let us then focus on the proof of (4.5). Proceeding by contradiction, we
assume that

$C_{4,2}\neq 0.$ (4.7)

Let us reach a contradiction in this case.

We proceed as we did in Step 3 in the Proof of Theorem 2 to show that
$C_{m+1,m+1}=0$: we will derive information related to $w_{b}$ defined by
(3.5) in Step 2 of the Proof of Theorem 2. However, we will not use the
expression (3.8) for $b$. We will instead take

$b=a_{n}.$ (4.8)

There is also another difference with the proof of the fact that
$C_{m+1,m+1}=0$, in the sense that the contradiction will follow from the
behavior of $v_{b,1,0}$ instead of $v_{b,0,0}$. From (4.4) and (4.1), we
introduce $B_{n}$ such that

$a_{n,1}=B_{n}a_{n,2}\mbox{ with }B_{n}=o(1)\mbox{ as }n\to\infty.$ (4.9)

Following (4.4) and the fact that $a_{n}\to 0$, given any $A>0$ (to be take
large enough later), we also introduce

${s_{A,n}}\to+\infty\mbox{ as }n\to\infty$ (4.10)

such that

$a_{n,2}=Ae^{-\frac{{s_{A,n}}}{2}},\mbox{ hence
}a_{n,1}=B_{n}a_{n,2}=AB_{n}e^{-\frac{{s_{A,n}}}{2}}.$ (4.11)

In particular, the transformation (3.5) reads here as

$w_{b}(y_{b},s)=w_{0}(y,s)\mbox{ with
}y_{1}=y_{b,1}+AB_{n}e^{\frac{\tau}{2}},\;\;y_{2}=y_{b,2}+Ae^{\frac{\tau}{2}}\mbox{
and }\tau=s-{s_{A,n}}.$ (4.12)

Using (1.12), we write a new version of (3.12) adapted to our new choice of
$b$ in (4.8):

$\displaystyle v_{b}(y_{b},{s_{A,n}})$
$\displaystyle=e^{-{s_{A,n}}}\sum_{j=0}^{2}C_{4,j}h_{4-j}(y_{b,1}+AB_{n})h_{j}(y_{b,2}+A)$
$\displaystyle+e^{-\frac{3}{2}{s_{A,n}}}\sum_{j=0}^{4}C_{5,j}h_{5-j}(y_{b,1}+AB_{n})h_{j}(y_{b,2}+A)+\bar{v}_{b}(y_{b},{s_{A,n}})$
(4.13)

where $\bar{v}_{b}(y_{b},{s_{A,n}})$ is defined in (3.13) (with $s_{0}$
replaced by ${s_{A,n}}$). Using the binomial relation (3.14) together with the
orthogonality relation given in (3.15), we may derive from (4.13) a long
formula for $v_{b}(y_{b},{s_{A,n}})$. In fact, we won’t do that, since our
argument uses only the projections of $v_{b}(y_{b},{s_{A,n}})$ on
$h_{0}h_{0}$, $h_{1}h_{0}$ and $h_{0}h_{1}$, with the notation

$h_{l}h_{j}\equiv h_{l}(y_{b,1})h_{j}(y_{b,2}),$

i.e. the projections on the expanding modes $\lambda=1$ and
$\lambda=\frac{1}{2}$ of the linear operator $\mathscr{L}$ (2.3). For that
reason, we will introduce a visual transcription of those 3 projections, in
the form of a table whose columns are $v_{b,0,0}({s_{A,n}})$,
$v_{b,1,0}({s_{A,n}})$ and $v_{b,0,1}({s_{A,n}})$, and whose lines bear the
name of the coeffients $C_{i,j}$ present in (4.13), together with the rest
term $\bar{v}_{b}(y_{b},{s_{A,n}})$. More precisely, this is the table:

| $v_{b,0,0}({s_{A,n}})$ | $v_{b,1,0}({s_{A,n}})$ | $v_{b,0,1}({s_{A,n}})$
---|---|---|---
$C_{4,i}$ | $C_{4,i}A^{4}B_{n}^{4-i}e^{-{s_{A,n}}}$ | $(4-i)C_{4,i}A^{3}B_{n}^{3-i}e^{-{s_{A,n}}}$ | $iC_{4,i}A^{3}B_{n}^{4-i}e^{-{s_{A,n}}}$
$C_{5,i}$ | $C_{5,i}A^{5}B_{n}^{5-i}e^{-\frac{3}{2}{s_{A,n}}}$ | $(5-i)C_{5,i}A^{4}B_{n}^{4-i}e^{-\frac{3}{2}{s_{A,n}}}$ | $iC_{5,i}A^{4}B_{n}^{5-i}e^{-\frac{3}{2}{s_{A,n}}}$
$\bar{v}_{b}(y_{b},{s_{A,n}})$ | $\bar{v}_{b,0,0}({s_{A,n}})$ | $\bar{v}_{b,0,1}({s_{A,n}})$ | $\bar{v}_{b,1,0}({s_{A,n}})$

Arguing as with (3.16), we see that

$|\bar{v}_{b,i,j}({s_{A,n}})|\leq C(A){s_{A,n}}e^{-2{s_{A,n}}}\mbox{ with
}(i,j)=(0,0),\;\;(1,0)\mbox{ or }(0,1).$ (4.14)

Using this, together with the table and the smallness of $B_{n}$ written in
(4.9), we see that

$\displaystyle v_{b,1,0}({s_{A,n}})=$ $\displaystyle
2A^{3}B_{n}e^{-{s_{A,n}}}(C_{4,2}+O(B_{n}))+A^{4}e^{-\frac{3}{2}{s_{A,n}}}(C_{5,4}+O(B_{n}))$
(4.15) $\displaystyle+O(C(A){s_{A,n}}e^{-2{s_{A,n}}})$

as $n\to\infty$. In order to know which term is dominant between the two
appearing in the expansion of $v_{b,1,0}({s_{A,n}})$, let us introduce the
ratio

$R_{n}=\frac{A^{3}B_{n}e^{-{s_{A,n}}}}{A^{4}e^{-\frac{3}{2}{s_{A,n}}}}=\frac{B_{n}}{Ae^{-\frac{{s_{A,n}}}{2}}}=\frac{a_{n,1}}{a_{n,2}^{2}}\to\infty$
(4.16)

from (4.9), (4.11) and (4.3). In particular, it holds that

$B_{n}\gg Ae^{-\frac{{s_{A,n}}}{2}}\mbox{ as }n\to\infty.$ (4.17)

Since $C_{4,2}\neq 0$ by the contradiction hypothesis (4.7), using (4.15), we
write

$v_{b,1,0}({s_{A,n}})\sim 2A^{3}B_{n}C_{4,2}e^{-{s_{A,n}}}\mbox{ as
}n\to\infty.$ (4.18)

Let us explain now how we reach a contradiction, first formally, then in a
rigorous way.

Proceeding formally first, we restart the argument from (4.12) with
$\tau=s-{s_{A,n}}\geq 0$ this time. All the estimates run smoothly and we end-
up with the following modification of the table (note that we evaluate
functions at $s={s_{A,n}}+\tau$, hence exponential factors appear in the
estimates, complying with the expanding nature of the considered projections):

| $v_{b,0,0}({s_{A,n}}+\tau)$ | $v_{b,1,0}({s_{A,n}}+\tau)$ | $v_{b,0,1}({s_{A,n}}+\tau)$
---|---|---|---
$C_{4,i}$ | $C_{4,i}e^{\tau}A^{4}B_{n}^{4-i}e^{-{s_{A,n}}}$ | $(4-i)C_{4,i}e^{\frac{\tau}{2}}A^{3}B_{n}^{3-i}e^{-{s_{A,n}}}$ | $iC_{4,i}e^{\frac{\tau}{2}}A^{3}B_{n}^{4-i}e^{-{s_{A,n}}}$
$C_{5,i}$ | $C_{5,i}e^{\tau}A^{5}B_{n}^{5-i}e^{-\frac{3}{2}{s_{A,n}}}$ | $(5-i)C_{5,i}e^{\frac{\tau}{2}}A^{4}B_{n}^{4-i}e^{-\frac{3}{2}{s_{A,n}}}$ | $iC_{5,i}e^{\frac{\tau}{2}}A^{4}B_{n}^{5-i}e^{-\frac{3}{2}{s_{A,n}}}$
$\bar{v}_{b}(y_{b},{s_{A,n}}+\tau)$ | $\bar{v}_{b,0,0}({s_{A,n}}+\tau)$ | $\bar{v}_{b,0,1}({s_{A,n}}+\tau)$ | $\bar{v}_{b,1,0}({s_{A,n}}+\tau)$

Selecting the column on $v_{b,1,0}({s_{A,n}}+\tau)$, we write

$v_{b,1,0}({s_{A,n}}+\tau)\sim
2e^{\frac{\tau}{2}}A^{3}B_{n}C_{4,2}e^{-{s_{A,n}}}\mbox{ as }n\to\infty,$
(4.19)

which shows an increase in the size of $v_{b,1,0}({s_{A,n}}+\tau)$ with
$\tau$. Taking an arbitrarily small $\delta_{0}>0$ then choosing
$\tau=\tau_{n}$ such that

$2e^{\frac{\tau_{n}}{2}}|A^{3}B_{n}C_{4,2}|e^{-{s_{A,n}}}=\delta_{0},$

which means that

$\tau_{n}=2{s_{A,n}}+2\log\frac{\delta_{0}}{2|A^{3}B_{n}C_{4,2}|}\to\infty$
(4.20)

(use (4.10) and (4.9)), we see that

$|v_{b,1,0}({s_{A,n}}+\tau_{n})|\sim\delta_{0},$ (4.21)

and this violates the following convergence, uniform with the respect to the
blow-up point:

######  Proposition 4.1 (Uniform convergence of $v_{a}(y,s)$ to $0$ when $a$
is a blow-up point).

It holds that

$\sup_{a\in\mathscr{S}}\|v_{a}(s)\|_{L^{2}_{\rho}}\to 0\mbox{ as }s\to\infty.$
(4.22)

In particular, for any $i\in\mathbbm{N}$ and $j=0,\dots,i$ with
$(i,j)\neq(0,0)$, we have

$\sup_{a\in\mathscr{S}}|v_{a,i,j}(s)|\to 0\mbox{ as }s\to\infty.$ (4.23)

###### Proof.

Clearly, (4.23) directly follows from (4.22), thanks to the $L^{2}_{\rho}$
projection defined in (2.4). As for (4.22), it is a direct consequence of the
Liouville theorem of [12] and [13]. For a proof, one may adapt with no
difficulty the proof of Proposition 2.2 page 11 in Khenissy, Rebai and Zaag
[10]. ∎

Clearly, a contradiction follows from (4.21) and (4.23) (note that
${s_{A,n}}+\tau_{n}\to\infty$ by (4.10) and (4.20)), justifying that
$C_{4,2}=0$. Unfortunately, the change of variables starting from (4.12) can
be justified only when $\tau$ stays bound, and this is not the case in (4.20).

Fortunately, there is a rigorous way to justify (4.19). One has simply to take
(4.18) as initial data at $s={s_{A,n}}$, then integrate for $s\geq{s_{A,n}}$
the following ODE satisfied by $v_{b,1,0}(s)$ which we recall from (2.6):

$v_{b,1,0}^{\prime}(s)=\frac{1}{2}v_{b,1,0}(s)+\int
v_{b}(y,s)^{2}k_{1}(y_{1})\rho(y)dy,$ (4.24)

where $k_{1}(y_{1})=\frac{y_{1}}{2}$ was already introduced in (2.8).

In order to do so, we need to evaluate the size of $v_{b}(y,s)$, and this is
possible if we evaluate it at $s={s_{A,n}}$, then integrate the PDE satisfied
by $v_{b}(y,s)$ for $s\geq s_{n}$. Let us start then by evaluating the size of
$v_{b}(y,s)$ at $s={s_{A,n}}$.
Before that, recalling that (1.12) holds in $L^{4}_{\rho}$, we may use the
trick based on the Cauchy-Schwarz inequality we used for (3.16) to prove that
$\bar{v}_{b}({s_{A,n}})$ defined in (3.13) and appearing in (4.13) satisfies

$\|\bar{v}_{b}({s_{A,n}})\|_{L^{2}_{\rho}}\leq C(A){s_{A,n}}e^{-2{s_{A,n}}}.$
(4.25)

Hence, we may use (4.13) together with the binomial relation (3.14) in order
to show that

$\|v_{b}(y_{b},{s_{A,n}})\|_{L^{2}_{\rho}}\leq CA^{2}e^{-{s_{A,n}}},$ (4.26)

for $n$ large enough. Then, integrating the PDE (2.2), we show that for
$s\geq{s_{A,n}}$, the solution will stay as small as its value at
$s={s_{A,n}}$. More precisely, we claim the following:

######  Proposition 4.2 (Uniform bound for $v_{a_{n}}$).

There exists a universal constant ${C}_{0}>0$ such that for all $A\geq 1$ and
$D\geq 1$, for $n$ large enough, we have

$\forall
s\in[{s_{A,n}},D{s_{A,n}}],\;\;\|v_{a_{n}}(s)\|_{L^{2}_{\rho}}\leq{C}_{0}A^{2}e^{-{s_{A,n}}}.$

###### Proof.

See subsection 4.4 below. ∎

######  Remark.

The result can not hold if one replaces $v_{a_{n}}$ by $v_{x_{0}}$ where
$x_{0}$ is not a blow-up point. Indeed, from Giga and Kohn [6], we know in
that case that $w_{x_{0}}(s)\to 0$, hence $v_{x_{0}}(s)\to-1$ as $s\to\infty$.

######  Remark.

In fact, our next argument needs the same estimate with better norms, namely
the $L^{4}_{\rho}$. In order to justify that, we give two facts:
(i) First, given an arbitrary $\tau_{0}>0$, note that estimate (4.26) actually
holds for all $s\in[{s_{A,n}}-\tau_{0},{s_{A,n}}]$, by the same argument
(based on the table given right before (4.19), which holds uniformly for
$\tau\in[-\tau_{0},0]$ with the same proof). This will provide the
$L^{2}_{\rho}$ bound of Proposition 4.2 for any $s\in[{s_{A,n}},D{s_{A,n}}]$.
(ii) Second, using Lemma 2.1, note that the $L^{4}_{\rho}$ estimate follows
from the $L^{2}_{\rho}$ estimate, after some time shift
$\tau^{*}=\tau^{*}(4,2)$ (by the way, using the nonlinear equation (2.2)
satisfied by $v_{a_{n}}$ and the $L^{\infty}$ bound (2.12), we see that
$\partial_{s}|v|\leq(\mathscr{L}+C)|v|$ is the distribution sense, hence Lemma
2.1 applies). Choosing the constant $\tau_{0}$ of item (i) equal to $\tau^{*}$
gives the $L^{4}_{\rho}$ estimate for any $s\geq{s_{A,n}}$.

Note first from the definition (4.20) of $\tau_{n}$ and (4.17) that there
exists $D\geq 1$ such that ${s_{A,n}}+\tau_{n}\in[{s_{A,n}},D{s_{A,n}}]$.
Plugging the estimate of Proposition 4.2 in the quadratic term of (4.24) (we
need in fact the $L^{4}_{\rho}$ estimate, which holds thanks to the remark
following Proposition 4.2), then integrating for $s^{\prime}\in[{s_{A,n}},s]$
(remember that $|v_{a_{n},1,0}(s^{\prime})|$ is uniformly bounded from
(2.12)), we obtain (remember that $b=a_{n}$ from (4.8)):

$\forall
s\in[{s_{A,n}},D{s_{A,n}}],\;\;|v_{b,1,0}(s)-e^{\frac{\tau}{2}}v_{b,1,0}({s_{A,n}})|\leq
Ce^{\frac{\tau}{2}}A^{4}e^{-2{s_{A,n}}},\mbox{ where }\tau=s-{s_{A,n}}.$
(4.27)

Using (4.18) and (4.17), we see that (4.19) is justified. Taking
$\tau=\tau_{n}$ defined in (4.20), we see that (4.21) holds, which contradicts
Proposition 4.2. Therefore, (4.7) doesn’t hold. Thus, $C_{4,2}=0$ and also
$C_{4,1}=0$ and $C_{4,0}<0$ by the argument presented at the beginning of this
part.

Part 2: Proof of the fact that $a_{n,1}=O\left(a_{n,2}^{\frac{3}{2}}\right)$
as $n\to\infty$

This estimate will be achieved through 4 lemmas below, numbered 4.3 to 4.6,
where we successively prove that
$a_{n,1}=O\left(a_{n,2}^{\frac{4}{3}}\right)$, then
$a_{n,1}=o\left(a_{n,2}^{\frac{4}{3}}\right)$, then $C_{5,4}=0$ and finally
that $a_{n,1}=O\left(a_{n,2}^{\frac{3}{2}}\right)$ as $n\to\infty$. Let us
state the first lemma:

######  Lemma 4.3.

It holds that $a_{n,1}=O\left(a_{n,2}^{\frac{4}{3}}\right)$ as $n\to\infty$.

###### Proof.

Since $C_{4,2}=C_{4,1}=0$ by Part 1, arguing as in that part, we can get an
improved version of (4.15), thanks to the table given right before that
estimate, the smallness of $B_{n}$ written in (4.9) together with (4.14):

$v_{b,1,0}({s_{A,n}})=4A^{3}C_{4,0}B_{n}^{3}e^{-{s_{A,n}}}+A^{4}e^{-\frac{3}{2}{s_{A,n}}}(C_{5,4}+O(B_{n}))+O({s_{A,n}}e^{-2{s_{A,n}}})$

as $n\to\infty$. Here, a discussion arises, according to the ratio of the two
terms appearing in this expression:

$\frac{A^{3}B_{n}^{3}e^{-{s_{A,n}}}}{A^{4}e^{-\frac{3}{2}{s_{A,n}}}}=\frac{B_{n}^{3}}{Ae^{-\frac{{s_{A,n}}}{2}}}=\frac{a_{n,1}^{3}}{a_{n,2}^{4}},$

by definitions (4.9) and (4.11) of $B_{n}$ and ${s_{A,n}}$. Arguing by
contradiction to prove Lemma 4.3 and recalling that $C_{4,0}\neq 0$ by (4.6),
we see that we have

$B_{n}^{3}\gg Ae^{-\frac{{s_{A,n}}}{2}}\mbox{ and }v_{b,1,0}({s_{A,n}})\sim
4A^{3}C_{4,0}B_{n}^{3}e^{-{s_{A,n}}}\mbox{ as }n\to\infty,$ (4.28)

at least for a subsequence denoted the same. Arguing as we did for (4.19) and
using this time the table right before that estimate, we may show that

$v_{b,1,0}({s_{A,n}}+\tau)\sim
4e^{\frac{\tau}{2}}A^{3}B_{n}^{3}C_{4,0}e^{-{s_{A,n}}}\mbox{ as }n\to\infty,$
(4.29)

then, derive a contradiction as in (4.21), with

$\tau_{n}=-2\log\frac{4|C_{4,0}|A^{3}B_{n}^{3}e^{-{s_{A,n}}}}{\delta_{0}}$

which satisfies $\tau_{n}\to\infty$ from (4.9), hence
${s_{A,n}}+\tau_{n}\to\infty$ from (4.10) and also ${s_{A,n}}+\tau_{n}\leq
D{s_{A,n}}$ for some $D\geq 1$, from (4.28). The question then reduces to
justify (4.29), for this choice of $\tau_{n}$, and this can be done exactly as
in Part 1, relying on Proposition 4.2. Thus, Lemma 4.3 is proved. ∎

Now, we state the second lemma of this part:

######  Lemma 4.4.

It holds that $a_{n,1}=o\left(a_{n,2}^{\frac{4}{3}}\right)$ as $n\to\infty$.

###### Proof.

Arguing by contradiction, we assume that up to some subsequence (still denoted
the same), we have

$\frac{a_{n,1}}{a_{n,2}^{\frac{4}{3}}}\to L\mbox{ as }n\to\infty,$ (4.30)

for some $L>0$. By definitions (4.9) and (4.11) of $B_{n}$ and ${s_{A,n}}$, we
see that

$B_{n}\sim LA^{1/3}e^{-\frac{{s_{A,n}}}{6}}\mbox{ as }n\to\infty.$ (4.31)

Using (4.13) as in Part 1 and Lemma 4.3, we may use (4.14) and the table right
before it together with (4.31) to derive the following expression for another
coordinate, namely $v_{b,0,1}({s_{A,n}})$ (there is a great advantage in
considering this coordinate instead of its twin $v_{b,1,0}({s_{A,n}})$, in the
sense that with $v_{b,0,1}({s_{A,n}})$, there is no contribution of $C_{4,0}$)
:

$\displaystyle v_{b,0,1}({s_{A,n}})$
$\displaystyle=A^{4}B_{n}e^{-\frac{3}{2}{s_{A,n}}}(4C_{5,4}+O(B_{n}))+O\left(C(A){s_{A,n}}e^{-2{s_{A,n}}}\right)$
$\displaystyle=4A^{4+\frac{1}{3}}LC_{5,4}e^{-\frac{5}{3}{s_{A,n}}}+O\left(e^{-\frac{11}{6}{s_{A,n}}}\right).$

By the same argument as in Part 1 and Claim 4.3, we see that $C_{5,4}=0$ (the
justification follows also in the same way, thanks to Proposition 4.2).

Repeating the same argument presented here in this step, we derive the
following expression for the other coordinate (use the table right before
(4.14)):

$\displaystyle v_{b,1,0}({s_{A,n}})$
$\displaystyle=e^{\frac{\tau}{2}}\left\\{4A^{3}C_{4,0}B_{n}^{3}e^{-{s_{A,n}}}+O(A^{4}B_{n}e^{-\frac{3}{2}{s_{A,n}}})+O\left(C(A){s_{A,n}}e^{-2{s_{A,n}}}\right)\right\\}$
$\displaystyle\sim
4e^{\frac{\tau}{2}}A^{4}L^{3}C_{4,0}e^{-\frac{3}{2}{s_{A,n}}}.$

Again, the growing factor implies that $C_{4,0}=0$, which is a contradiction
by (4.6). Thus, (4.30) doesn’t hold and Lemma 4.4 holds. As for the rigorous
proof, it follows exactly as in Part 1, thanks to Proposition 4.2. This
concludes the proof of Lemma 4.4. ∎

Now, this is the third lemma of the part:

######  Lemma 4.5.

It holds that $C_{5,4}=0$.

###### Proof.

Proceeding by contradiction, we assume that

$C_{5,4}\neq 0.$ (4.32)

This time, the argument follows from the behavior of $v_{b,1,0}(s)$. Using
(4.17) and Lemma 4.4, we see by definitions (4.9) and (4.11) of $B_{n}$ and
${s_{A,n}}$ that

$Ae^{-\frac{{s_{A,n}}}{2}}\ll B_{n}\ll
A^{\frac{1}{3}}e^{-\frac{{s_{A,n}}}{6}}\mbox{ as }n\to\infty.$ (4.33)

Proceeding as before, we may use the table right before (4.19) to derive that

$v_{b,1,0}({s_{A,n}}+\tau)\sim
A^{4}C_{5,4}e^{\frac{\tau}{2}}e^{-\frac{3}{2}{s_{A,n}}}\mbox{ as }n\to\infty$

(recall that in (4.13), $C_{4,1}=C_{4,2}=0$ from Part 1). The growth factor
$e^{\frac{\tau}{2}}$ allows to get a contradiction as in previous steps. Thus,
(4.32) doesn’t hold and $C_{5,4}=0$. The rigorous justification again follows
exactly as in Part 1, thanks to Proposition 4.2. This concludes the proof of
Lemma 4.5. ∎

Now, this is the final lemma of the part:

######  Lemma 4.6.

It holds that ${a_{n,1}}=O({a_{n,2}}^{3/2})$ as $n\to\infty$.

###### Proof.

Since $C_{5,4}=0$ from Lemma 4.5, we go back again to the table right before
(4.19) to derive the following expansion

$v_{b,1,0}({s_{A,n}}+\tau)=e^{\frac{\tau}{2}}\left\\{4A^{3}C_{4,0}B_{n}^{3}e^{-{s_{A,n}}}+2A^{4}B_{n}e^{-\frac{3}{2}{s_{A,n}}}(C_{5,3}+O(B_{n}))+O\left({s_{A,n}}e^{-2{s_{A,n}}}\right)\right\\}$
(4.34)

as $n\to\infty$. Making the ratio between the coefficients of the first two
terms, we get

$\frac{A^{3}B_{n}^{3}e^{-{s_{A,n}}}}{A^{4}B_{n}e^{-\frac{3}{2}{s_{A,n}}}}=\frac{B_{n}^{2}}{Ae^{-\frac{{s_{A,n}}}{2}}}=\frac{{a_{n,1}}^{2}}{{a_{n,2}}^{3}},$
(4.35)

by definitions (4.9) and (4.11) of $B_{n}$ and ${s_{A,n}}$. Assuming by
contradiction that Lemma 4.6 doesn’t hold for some subsequence still denoted
the same, we see that $B_{n}\gg\sqrt{A}e^{-\frac{{s_{A,n}}}{4}}$, hence,

$v_{b,1,0}({s_{A,n}}+\tau)\sim
4e^{\frac{\tau}{2}}A^{3}C_{4,0}B_{n}^{3}e^{-{s_{A,n}}}\mbox{ as }n\to\infty,$

and the growth factor shows that $C_{4,0}=0$, which is a contradiction from
(4.6). The justification of this step follows as in Part 1. This concludes the
proof of Lemma 4.6. ∎

Part 3: Possible limits for $\frac{{a_{n,1}}}{{a_{n,2}}^{\frac{3}{2}}}$

From Lemma 4.6, we may extract a subsequence (still denoted the same) such
that

$\frac{{a_{n,1}}}{{a_{n,2}}^{\frac{3}{2}}}\to L_{0}\mbox{ as }n\to\infty,$
(4.36)

for some $L_{0}\geq 0$. Let us assume that $L_{0}>0$ (note that the case
$L_{0}=0$ will be ruled out in Part 4). Using (4.35), we see that

$B_{n}\sim L_{0}\sqrt{A}e^{-\frac{{s_{A,n}}}{4}}\mbox{ as }n\to\infty.$

From (4.34), we see that

$v_{b,1,0}({s_{A,n}}+\tau)\sim
e^{\frac{\tau}{2}}A^{4}B_{n}e^{-\frac{3}{2}{s_{A,n}}}\left(4C_{4,0}L_{0}^{2}+2C_{5,3}+O({s_{A,n}}e^{-\frac{{s_{A,n}}}{2}})\right)\mbox{
as }n\to\infty.$

As before, we see from the growth factor that

$4C_{4,0}L_{0}^{2}+2C_{5,3}=0.$

Since $C_{4,0}<0$ from (4.6) and $L_{0}>0$, we necessarily see that
$C_{5,3}>0$ and

$L_{0}=\sqrt{-\frac{C_{5,3}}{2C_{4,0}}}.$ (4.37)

The justification of this part can be done exactly as Part 1.

Part 4: Ruling out the case where ${a_{n,1}}=o({a_{n,2}}^{\frac{3}{2}})$ as
$n\to\infty$

Let us assume by contradiction that the limit defined in (4.36) is zero. In
other words,

$L_{0}=0.$ (4.38)

From (4.17) and (4.35), we see that

$Ae^{-\frac{{s_{A,n}}}{2}}\ll B_{n}\ll
A^{\frac{1}{2}}e^{-\frac{{s_{A,n}}}{4}}\mbox{ as }n\to\infty.$ (4.39)

At the end of this part, we will reach a contradiction, ruling this case out.
In fact, our argument will follow from 3 steps:
\- In Step 1, we show that $a_{n,1}=O\left(a_{n,2}^{2}|\log a_{n,2}|\right)$
as $n\to\infty$.
\- In Step 2, we prove that $C_{6,6}=C_{5,3}=C_{6,5}=0$.
\- In Step 3, we reach a contradiction and finish the argument of Part 4.

Step 1: Proof of the fact that $a_{n,1}=O\left(a_{n,2}^{2}|\log
a_{n,2}|\right)$ as $n\to\infty$

Let us first show how the ratio between $a_{n,1}$ and $a_{n,2}^{2}|\log
a_{n,2}|$ naturally arises in our argument. Using (4.34) together with (4.39),
we write:

$v_{b,1,0}({s_{A,n}}+\tau)=e^{\frac{\tau}{2}}\left\\{2A^{4}B_{n}e^{-\frac{3}{2}{s_{A,n}}}(C_{5,3}+O(B_{n}))+O\left({s_{A,n}}e^{-2{s_{A,n}}}\right)\right\\}\mbox{
as }n\to\infty.$ (4.40)

Clearly, the balance between the $C_{5,3}$ term and the error term is
important. Considering the ratio between the two, we write

$\frac{A^{4}B_{n}e^{-\frac{3}{2}{s_{A,n}}}}{{s_{A,n}}e^{-2{s_{A,n}}}}=A^{5}\frac{B_{n}}{{s_{A,n}}Ae^{-\frac{{s_{A,n}}}{2}}}\sim\frac{A^{5}}{2}\frac{a_{n,1}}{a_{n,2}^{2}|\log
a_{n,2}|}\mbox{ as }n\to\infty$ (4.41)

thanks to the definitions (4.9) and (4.11) of $B_{n}$ and ${s_{A,n}}$. Thus,
the ratio in the title of Step 1 appears.

Proceeding by contradiction, we assume that (for a subsequence denoted the
same), we have

$a_{n,1}\gg a_{n,2}^{2}|\log a_{n,2}|\mbox{ as }n\to\infty.$ (4.42)

Using (4.41) and (4.39), we see that

${s_{A,n}}e^{-\frac{{s_{A,n}}}{2}}\ll B_{n}\ll e^{-\frac{{s_{A,n}}}{4}}\mbox{
as }n\to\infty.$ (4.43)

Using (4.40), we see that

$v_{b,1,0}({s_{A,n}}+\tau)\sim
2e^{\frac{\tau}{2}}A^{4}B_{n}e^{-\frac{3}{2}{s_{A,n}}}(C_{5,3}+o(1))\mbox{ as
}n\to\infty,$

and the growth factor implies that

$C_{5,3}=0.$ (4.44)

The rigorous justification is exactly the same as in earlier steps, relying on
Proposition 4.2.

At this stage, we need to further refine the Taylor expansion of $v_{0}(y,s)$
given in (1.12), taking into account that with respect to the expansion
(4.13), we already have $C_{4,1}=C_{4,2}=C_{5,3}=C_{5,4}=0$, from Part 1,
Lemma 4.5 and (4.44). Using the strategy described in Proposition 1, we write:

$\displaystyle v_{0}(y,s)=$
$\displaystyle-\frac{e^{-2s}}{3}C_{4,0}^{2}\gamma_{4,4,0}h_{0}h_{0}-\frac{e^{-2s}}{2}C_{4,0}^{2}\gamma_{4,4,2}h_{2}h_{0}+\left[C_{4,0}e^{-s}-e^{-2s}C_{4,0}^{2}\gamma_{4,4,4}\right]h_{4}h_{0}$
$\displaystyle+e^{-\frac{3}{2}s}\sum_{j=0}^{2}C_{5,j}h_{5-j}h_{j}+se^{-2s}C_{4,0}^{2}\gamma_{4,4,6}h_{6}h_{0}+e^{-2s}\sum_{j=0}^{6}C_{6,j}h_{6-j}h_{j}$
$\displaystyle+e^{-2s}C_{4,0}^{2}h_{8}h_{0}+O\left(se^{-\frac{5}{2}s}\right),$
(4.45)

for some real coefficients $C_{6,j}$ with $j=0,\dots,6$, where we have used
the following notation

$\gamma_{l,m,n}=\int_{\mathbbm{R}}h_{l}(\xi)h_{m}(\xi)k_{n}(\xi)\rho(\xi)d\xi,$

with the polynomials defined in (1.9) and (2.8), the weight $\rho$ in (1.6)
and the convention

$\gamma_{l,m,n}=0\mbox{ if }l,\;m\mbox{ or }n\mbox{ is negative },$

together with the notation

$h_{l}h_{m}\mbox{ which stands for }h_{l}(y_{1})h_{m}(y_{2}).$

Following this new refined expansion, we give first the following cancelation:

######  Claim 4.7.

It holds that $C_{6,6}=C_{6,5}=0$.

###### Proof.

Using the expansion (4.45), we give the following table, which is an update of
the former table given right before (4.14):

| $v_{b,0,0}({s_{A,n}}+\tau)$ | $v_{b,1,0}({s_{A,n}}+\tau)$ | $v_{b,0,1}({s_{A,n}}+\tau)$
---|---|---|---
$C_{4,0}^{2}\gamma_{4,4,0}$ | $-\frac{C_{4,0}^{2}}{3}\gamma_{4,4,0}e^{-2\tau}e^{-2{s_{A,n}}}$ | 0 | 0
$C_{4,0}^{2}\gamma_{4,4,2}$ | $-\frac{C_{4,0}^{2}}{2}\gamma_{4,4,2}e^{-\tau}A^{2}B_{n}^{2}e^{-2{s_{A,n}}}$ | $-C_{4,0}^{2}\gamma_{4,4,2}e^{-\frac{3}{2}\tau}AB_{n}e^{-2{s_{A,n}}}$ | 0
$C_{4,0}$ | $C_{4,0}e^{\tau}A^{4}B_{n}^{4}e^{-{s_{A,n}}}$ | $4C_{4,0}e^{\frac{\tau}{2}}A^{3}B_{n}^{3}e^{-{s_{A,n}}}$ | 0
$C_{4,0}^{2}\gamma_{4,4,4}$ | $-C_{4,0}^{2}\gamma_{4,4,4}A^{4}B_{n}^{4}e^{-2{s_{A,n}}}$ | $-4C_{4,0}^{2}\gamma_{4,4,4}e^{-\frac{\tau}{2}}A^{3}B_{n}^{3}e^{-2{s_{A,n}}}$ | 0
$C_{5,i}$ | $C_{5,i}e^{\tau}A^{5}B_{n}^{5-i}e^{-\frac{3}{2}{s_{A,n}}}$ | $(5-i)C_{5,i}e^{\frac{\tau}{2}}A^{4}B_{n}^{4-i}e^{-\frac{3}{2}{s_{A,n}}}$ | $iC_{5,i}e^{\frac{\tau}{2}}A^{4}B_{n}^{5-i}e^{-\frac{3}{2}{s_{A,n}}}$
$C_{4,0}^{2}\gamma_{4,4,6}$ | $C_{4,0}^{2}\gamma_{4,4,6}e^{\tau}A^{6}B_{n}^{6}({s_{A,n}}+\tau)e^{-2{s_{A,n}}}$ | $6C_{4,0}^{2}\gamma_{4,4,6}e^{\frac{\tau}{2}}A^{5}B_{n}^{5}({s_{A,n}}+\tau)e^{-2{s_{A,n}}}$ | 0
$C_{6,i}$ | $C_{6,i}e^{\tau}A^{6}B_{n}^{6-i}e^{-2{s_{A,n}}}$ | $(6-i)C_{6,i}e^{\frac{\tau}{2}}A^{5}B_{n}^{5-i}e^{-2{s_{A,n}}}$ | $iC_{6,i}e^{\frac{\tau}{2}}A^{5}B_{n}^{6-i}e^{-2{s_{A,n}}}$
$C_{4,0}^{2}$ | $C_{4,0}^{2}e^{2\tau}A^{8}B_{n}^{8}e^{-2{s_{A,n}}}$ | $8C_{4,0}^{2}e^{\frac{3}{2}\tau}A^{7}B_{n}^{7}e^{-2{s_{A,n}}}$ | 0
$O(se^{-\frac{5}{2}s})$ | $e^{\tau}O(({s_{A,n}}+\tau)e^{-\frac{5}{2}{s_{A,n}}})$ | $e^{\frac{\tau}{2}}O(({s_{A,n}}+\tau)e^{-\frac{5}{2}{s_{A,n}}})$ | $e^{\frac{\tau}{2}}O(({s_{A,n}}+\tau)e^{-\frac{5}{2}{s_{A,n}}})$

Please keep in mind that this table is fully justified if $\tau$ is zero or
bounded. It happens that for our argument, we will need $\tau=\tau_{n}$ to go
to infinity as $n\to\infty$. In that case, note that the terms bearing the
coefficients $e^{2\tau}$ and $e^{3\tau/2}$ are “artificial” and should not be
taken into account in our argument, including the formal one. This way, we
will ignore those terms in our formal argument. Fortunately, when it comes to
the rigorous argument, this table will be useful for $\tau=0$ or bounded, and
the conclusion will follow from the integration of the PDE for $\tau$ ranging
between $0$ and suitably large values.

\- Proof of the fact that $C_{6,6}=0$.
Using (4.43) and the previous table, we may derive the following estimate for
large $\tau$:

$v_{b,0,0}({s_{A,n}}+\tau)=e^{\tau}(A^{6}C_{6,6}+o(1))e^{-2{s_{A,n}}}\mbox{ as
}n\to\infty,$ (4.46)

and the growth factor implies that

$C_{6,6}=0.$ (4.47)

As we have just written after the previous table, we don’t take into account
the term with $e^{2\tau}$ is the formal derivation of (4.46). Now, as for the
rigorous justification, it goes as in previous steps, relying on the previous
table only for bounded $\tau$, together with Proposition 4.2. However, the
rigorous proof is a little more delicate than earlier, since we need to take
$A$ large enough in order to conclude, as we explain in the following:
\- first, relying on the previous table and (4.43), we write the following
expansion for $\tau=0$ (note that it is different from (4.46), since we have
lower order exponential terms in $\tau$ in the table which cannot be neglected
for small $\tau$):

$v_{b,0,0}({s_{A,n}})=\left[-\frac{C_{4,0}^{2}}{3}\gamma_{4,4,0}+C_{6,6}A^{6}+o(1)\right]e^{-2{s_{A,n}}}\mbox{
as }n\to\infty.$ (4.48)

\- second, using Proposition 4.2 (with $D=4$) and proceeding as for (4.27), we
integrate the ODE (2.6) satisfied by $v_{b,0,0}$ and write:

$\forall
s\in[{s_{A,n}},D{s_{A,n}}],\;\;|v_{b,0,0}(s)-e^{\tau}v_{b,0,0}({s_{A,n}})|\leq
Ce^{\tau}A^{4}e^{-2{s_{A,n}}},\mbox{ where }\tau=s-{s_{A,n}}.$ (4.49)

Using (4.48), and taking $A$ large enough so that $A^{6}$ term in (4.48)
dominates its neighbor with $C_{4,0}^{2}$, and also the $A^{4}$ term in the
right-hand side of (4.49), we see that (4.46) is justified.

\- Proof of the fact that $C_{6,5}=0$.
Proceeding as before, we may write the following expansion:

$\displaystyle v_{b,0,1}({s_{A,n}}+\tau)=e^{\frac{\tau}{2}}$
$\displaystyle\left\\{2A^{4}B_{n}^{3}e^{-\frac{3}{2}{s_{A,n}}}\left(C_{5,2}+O(B_{n})\right)\right.$
(4.50)
$\displaystyle\left.+5A^{5}B_{n}e^{-2{s_{A,n}}}(C_{6,5}+O(B_{n}))+O\left({s_{A,n}}e^{-\frac{5}{2}{s_{A,n}}}\right)\right\\}.$

Assuming by contradiction that $C_{6,5}\neq 0$, we see from (4.43) that

$v_{b,0,1}({s_{A,n}}+\tau)\sim
5e^{\frac{\tau}{2}}A^{5}B_{n}e^{-2{s_{A,n}}}C_{6,5},$ (4.51)

and the growth factor leads to a contradiction as usual. Thus, $C_{6,5}=0$. As
for the rigorous justification, it is a little more complicated than usual.
Indeed, when $\tau=0$, we have $|v_{b,0,1}({s_{A,n}})|\ll e^{-2{s_{A,n}}}$,
less than $e^{-2{s_{A,n}}}$, which is the bound Proposition 4.2 gives on the
quadratic term in the equation (2.6) satisfied by $v_{b,0,1}$. In other words,
Proposition 4.2 is not enough to show that the linear term dominates the
quadratic term when we integrate equation (2.6). Thus, we need to refine
Proposition 4.2 in the following:

######  Proposition 4.8 (Uniform expansion for $v_{a_{n}}$).

Following Proposition 4.2 and assuming that

$v_{0}(y,s)=C_{4,0}e^{-s}h_{4}h_{0}+O(e^{-\frac{3}{2}s})\mbox{ as }s\to\infty$
(4.52)

(in $L^{q}_{\rho}$ for any $q\geq 2$), we claim that for all $A\geq 1$ and
$D\geq 1$, for $n$ large enough, we have for all $s\in[{s_{A,n}},D{s_{A,n}}]$,

$\|v_{a_{n}}(y,s)-C_{4,0}e^{-s}h_{4}(y_{1})\|_{L^{2}_{\rho}}\leq
C(A)e^{-\frac{3}{2}{s_{A,n}}}.$

######  Remark.

As we wrote right after Proposition 4.2, here also, we have the same estimate
valid in $L^{4}_{\rho}$, by the same parabolic regularity argument.

###### Proof.

See Subsection 4.4. ∎

Let us use this proposition to justify (4.51) (note from (4.45) that (4.52)
holds here).
Note first from (4.43) that the time $\tau_{n}$ which allows (4.51) to be
equal to some $\delta_{0}>$ satisfies $\tau_{n}\in[0,\bar{D}{s_{A,n}}]$, for
some $\bar{D}>0$. Let us then apply Proposition 4.8 with $D=\bar{D}+1$.
First, when $s={s_{A,n}}$, we may use (4.50) (which is fully justified when
$\tau=0$) and (4.43) to write (remember that $b=a_{n}$ from (4.8)):

$v_{b,0,1}({s_{A,n}})=5A^{5}B_{n}e^{-2{s_{A,n}}}C_{6,5}+O(B_{n}^{3}e^{-\frac{3}{2}{s_{A,n}}})+O({s_{A,n}}e^{-\frac{5}{2}{s_{A,n}}}).$
(4.53)

Now, we recall the ODE (2.6) satisfied by $v_{b,0,1}$:

$\forall s\geq{s_{A,n}},\;\;v_{b,0,1}^{\prime}(s)=\frac{1}{2}v_{b,0,1}(s)+\int
v_{b}(y,s)^{2}k_{1}(y_{2})\rho(y)dy.$

Using Proposition 4.8 with the $L^{4}_{\rho}$ norm (use the remark following
the proposition), we may estimate the quadratic term as follows (note that the
contribution of the term $[C_{4,0}e^{-s}h_{4}(y_{1})]^{2}$ vanishes thanks to
the orthogonality between $h_{4}(y_{1})^{2}$ and $h_{1}(y_{2})$ in
$L^{2}_{\rho}$), and write the following differential inequality on
$v_{b,0,1}(s)$:

$\displaystyle\forall
s\geq{s_{A,n}},\;\;|v_{b,0,1}^{\prime}(s)-\frac{1}{2}v_{b,0,1}(s)|\leq$
$\displaystyle
C(s-{s_{A,n}})e^{-s}A^{4}e^{-2{s_{A,n}}}+C(A)e^{-s}e^{-\frac{3}{2}{s_{A,n}}}$
$\displaystyle+$ $\displaystyle
C(s-{s_{A,n}})^{2}A^{8}e^{-4{s_{A,n}}}+C(A)^{2}e^{-3{s_{A,n}}}.$

Integrating this equation, we see that

$\forall\tau\geq
0,\;\;|v_{b,0,1}(\tau+{s_{A,n}})-e^{\frac{\tau}{2}}v_{b,0,1}({s_{A,n}})|\leq
C(A)e^{\frac{\tau}{2}}e^{-\frac{5}{2}{s_{A,n}}},$

for $n$ large enough. Using (4.53) and (4.43), we see that (4.51) is
justified. Thus, $C_{6,5}=0$. This concludes the proof of Claim 4.7. ∎

Now, we are ready to give the final argument of Step 1.

\- Final argument of Step 1. Focusing on $v_{b,1,0}(s)$, we write the
following from the table given right before (4.47):

$\displaystyle v_{b,1,0}({s_{A,n}}+\tau)=e^{\frac{\tau}{2}}$
$\displaystyle\left\\{4A^{3}B_{n}^{3}C_{4,0}e^{-{s_{A,n}}}+3A^{4}B_{n}^{2}e^{-\frac{3}{2}{s_{A,n}}}(C_{5,2}+O(B_{n}))\right.$
$\displaystyle\left.+2A^{5}B_{n}e^{-2{s_{A,n}}}(C_{6,4}+O(B_{n}))+O\left({s_{A,n}}e^{-\frac{5}{2}{s_{A,n}}}\right)\right\\}.$

Here again, the terms coming with $e^{3\tau/2}$ in the table should not be
taken into account at the formal level, as we have already explained following
that table. Since $C_{4,0}\neq 0$ from (4.6), using (4.43), we see that

$v_{b,1,0}({s_{A,n}}+\tau)\sim
4e^{\frac{\tau}{2}}A^{3}B_{n}^{3}C_{4,0}e^{-{s_{A,n}}}\mbox{ as }n\to\infty,$

hence, a contradiction follows from the growth factor $e^{\frac{\tau}{2}}$
(the rigorous justification comes exactly as with the proof of the fact that
$C_{6,5}=0$ in Claim 4.7). This concludes the argument of Step 1, proving that
${a_{n,1}}=O({a_{n,2}}^{2}|\log{a_{n,2}}|)$ as $n\to\infty$.

Step 2: Proof of the fact that $C_{6,6}=C_{5,3}=C_{6,5}=0$.

We will prove these 3 cancelations in the order they appear. Before doing
that, let us recall that the Taylor expansion (4.13) still holds. Even better,
we already have some cancelations in that expansion, namely:
$C_{4,2}=C_{4,1}=0$ and $C_{5,4}=0$ (use Part 1 and Lemma 4.5). One more thing
we should keep in mind: using (4.3) and Step 1 we know that

${a_{n,2}}^{2}\ll{a_{n,1}}\leq C{a_{n,2}}^{2}|\log{a_{n,2}}|\mbox{ as
}n\to\infty.$

Using the definitions (4.9) and (4.11) of $B_{n}$ and ${s_{A,n}}$, we see that

$Ae^{-\frac{{s_{A,n}}}{2}}\ll B_{n}\leq
CA{s_{A,n}}e^{-\frac{{s_{A,n}}}{2}}\mbox{ as }n\to\infty.$ (4.54)

With these facts in mind, we may use the strategy of Proposition 1 and refine
the Taylor expansion (1.12) written for $v_{0}(y,s)$ and prove that:

$\displaystyle v_{0}(y,s)=$
$\displaystyle-\frac{e^{-2s}}{3}C_{4,0}^{2}\gamma_{4,4,0}h_{0}h_{0}-\frac{e^{-2s}}{2}C_{4,0}^{2}\gamma_{4,4,2}h_{2}h_{0}+\left[C_{4,0}e^{-s}-e^{-2s}C_{4,0}^{2}\gamma_{4,4,4}\right]h_{4}h_{0}$
$\displaystyle+e^{-\frac{3}{2}s}\sum_{j=0}^{3}C_{5,j}h_{5-j}h_{j}+se^{-2s}C_{4,0}^{2}\gamma_{4,4,6}h_{6}h_{0}+e^{-2s}\sum_{j=0}^{6}C_{6,j}h_{6-j}h_{j}$
$\displaystyle+e^{-2s}C_{4,0}^{2}h_{8}h_{0}+O\left(se^{-\frac{5}{2}s}\right),$
(4.55)

for some real coefficients $C_{6,j}$ with $j=0,\dots,6$, with the same
notations following (4.45). Let us note that for the formal arguments based on
this expansion, the table given right before (4.47) is valid here.

In the following, we successively prove the 3 cancelations mentioned in the
title of this step.

\-  Proof of the fact that $C_{6,6}=0$.
Using the same strategy as usual, we justify that

$\displaystyle v_{b,0,0}({s_{A,n}}+\tau)=e^{\tau}$
$\displaystyle\left\\{A^{4}B_{n}^{4}e^{-{s_{A,n}}}C_{4,0}+A^{5}B_{n}^{2}e^{-\frac{3}{2}{s_{A,n}}}(C_{5,3}+O(B_{n}))\right.$
$\displaystyle+\left.A^{6}e^{-2{s_{A,n}}}(C_{6,6}+O(B_{n}))+O\left({s_{A,n}}e^{-\frac{5}{2}{s_{A,n}}}\right)\right\\}.$

As we have written following the table given right before (4.47), we ignore
the terms coming with $e^{2\tau}$ in the derivation of this expansion.
Assuming by contradiction that $C_{6,6}\neq 0$ and using (4.54), we see that

$v_{b,0,0}({s_{A,n}}+\tau)\sim e^{\tau}A^{6}e^{-2{s_{A,n}}}C_{6,6}\mbox{ as
}n\to\infty.$

Then, the growth factor $e^{\tau}$ leads to a contradiction (for the rigorous
justification, Proposition 4.2 is enough here, provided that one takes $A$
large enough; for a similar argument see the proof of the fact that
$C_{6,6}=0$ in Claim 4.7 above). Thus, $C_{6,6}=0$.

\- Proof of the fact that $C_{5,3}=0$.
Using the same strategy as before, and taking into account the size of $B_{n}$
given in (4.54) and assuming by contradiction that $C_{5,3}\neq 0$, we see
that

$v_{b,1,0}({s_{A,n}}+\tau)\sim
2e^{\frac{\tau}{2}}A^{4}B_{n}e^{-\frac{3}{2}{s_{A,n}}}C_{5,3}\mbox{ as
}n\to\infty,$

and the growth factor $e^{\frac{\tau}{2}}$ leads to a contradiction, as usual.
Once again, the terms with $e^{3\tau/2}$ in the table given right before
(4.47) are ignored in this expansion. As for the rigorous argument,
Proposition 4.2 is sufficient for the justification. Thus, $C_{5,3}=0$.

\- Proof of the fact that $C_{6,5}=0$.
By the same strategy, taking into account the size of $B_{n}$ given in (4.54)
and assuming by contradiction that $C_{6,5}\neq 0$, we see that

$v_{b,1,0}({s_{A,n}}+\tau)\sim
e^{\frac{\tau}{2}}A^{5}e^{-2{s_{A,n}}}C_{6,5}\mbox{ as }n\to\infty,$

and the growth factor $e^{\frac{\tau}{2}}$ leads to a contradiction, as
before. Note that we ignored the terms involving $e^{3\tau/2}$ in the table
given right before (4.47). As for the rigorous justification, it goes as in
the proof of the fact $C_{6,6}=0$ in Claim 4.7; in particular, the power of
$A$ is crucial. Thus, $C_{6,5}=0$. This concludes the proof of the
cancelations in the title of Step 2.

Step 3: Final argument of Part 4

Recalling the Taylor expansion (4.55) and the cancelations of Step 2, we may
use the strategy of Proposition 1 and get to the next order:

$\displaystyle
v_{0}(y,s)=-\frac{e^{-2s}}{3}C_{4,0}^{2}\gamma_{4,4,0}h_{0}h_{0}-\frac{2e^{-\frac{5}{2}s}}{3}C_{4,0}\sum_{j=0}^{1}C_{5,j}\gamma_{4,5-j,1-j}h_{1-j}h_{j}$
(4.56)
$\displaystyle-\frac{e^{-2s}}{2}C_{4,0}^{2}\gamma_{4,4,2}h_{2}h_{0}-e^{-\frac{5}{2}s}C_{4,0}\sum_{j=0}^{2}C_{5,j}\gamma_{4,5-j,3-j}h_{3-j}h_{j}$
$\displaystyle+\left[C_{4,0}e^{-s}-e^{-2s}C_{4,0}^{2}\gamma_{4,4,4}\right]h_{4}h_{0}+\sum_{j=0}^{2}[C_{5,j}e^{-\frac{3}{2}s}-2e^{-\frac{5}{2}s}C_{4,0}C_{5,j}\gamma_{4,5-j,5-j}]h_{5-j}h_{j}$
$\displaystyle+se^{-2s}C_{4,0}^{2}\gamma_{4,4,6}h_{6}h_{0}+e^{-2s}\sum_{j=0}^{4}C_{6,j}h_{6-j}h_{j}+\sum_{j=0}^{2}2se^{-\frac{5}{2}s}C_{4,0}C_{5,j}\gamma_{4,5-j,7-j}h_{7-j}h_{j}$
$\displaystyle+\sum_{j=0}^{7}C_{7,j}e^{-\frac{5}{2}s}h_{7-j}h_{j}+e^{-2s}C_{4,0}^{2}h_{8}h_{0}+2C_{4,0}e^{-\frac{5}{2}s}\sum_{j=0}^{2}C_{5,j}h_{9-j}h_{j}+O\left(s^{2}e^{-3s}\right),$

for some real coefficients $C_{7,j}$ with $j=0,\dots,7$, with the notations
given after (4.45).

With this expansion, we are ready to find the contradiction which will show
that (4.38) doesn’t hold, concluding thus the argument of Part 4.

Starting from the Taylor expansion (4.56) and using the usual strategy, we may
update the tool-table before (4.47) and write the following (note that we
include only the terms generating $e^{\alpha\tau}$ with $\alpha>0$ and that we
change the notation by avoiding to reproduce the coefficients of the first
column in the following; in other words, the second to forth column have to be
multiplied by the first in order to get the desired expression):

| $v_{b,0,0}({s_{A,n}}+\tau)$ | $v_{b,1,0}({s_{A,n}}+\tau)$ | $v_{b,0,1}({s_{A,n}}+\tau)$
---|---|---|---
$C_{4,0}$ | $e^{\tau}A^{4}B_{n}^{4}e^{-{s_{A,n}}}$ | $4e^{\frac{\tau}{2}}A^{3}B_{n}^{3}e^{-{s_{A,n}}}$ | 0
$C_{5,i}$ | $e^{\tau}A^{5}B_{n}^{5-i}e^{-\frac{3}{2}{s_{A,n}}}$ | $(5-i)e^{\frac{\tau}{2}}A^{4}B_{n}^{4-i}e^{-{s_{A,n}}}$ | $ie^{\frac{\tau}{2}}A^{4}B_{n}^{5-i}e^{-{s_{A,n}}}$
$C_{4,0}^{2}\gamma_{4,4,6}$ | $e^{\tau}A^{6}B_{n}^{6}({s_{A,n}}+\tau)e^{-2{s_{A,n}}}$ | $6e^{\frac{\tau}{2}}A^{5}B_{n}^{5}({s_{A,n}}+\tau)e^{-2{s_{A,n}}}$ | 0
$C_{6,i}$ | $e^{\tau}A^{6}B_{n}^{6-i}e^{-2{s_{A,n}}}$ | $(6-i)e^{\frac{\tau}{2}}A^{5}B_{n}^{5-i}e^{-2{s_{A,n}}}$ | $ie^{\frac{\tau}{2}}A^{5}B_{n}^{6-i}e^{-2{s_{A,n}}}$
$C_{4,0}C_{5,i}\gamma_{4,5-i,7-i}$ | $e^{\tau}A^{7}B_{n}^{7-i}({s_{A,n}}+\tau)e^{-\frac{5}{2}{s_{A,n}}}$ | $(7-i)e^{\frac{\tau}{2}}A^{6}B_{n}^{6-i}({s_{A,n}}+\tau)e^{-\frac{5}{2}{s_{A,n}}}$ | $ie^{\frac{\tau}{2}}A^{6}B_{n}^{7-i}({s_{A,n}}+\tau)e^{-\frac{5}{2}{s_{A,n}}}$
$C_{7,i}$ | $e^{\tau}A^{7}B_{n}^{7-i}e^{-\frac{5}{2}{s_{A,n}}}$ | $(7-i)e^{\frac{\tau}{2}}A^{6}B_{n}^{6-i}e^{-\frac{5}{2}{s_{A,n}}}$ | $ie^{\frac{\tau}{2}}A^{6}B_{n}^{7-i}e^{-\frac{5}{2}{s_{A,n}}}$
$C_{4,0}^{2}$ | $e^{2\tau}A^{8}B_{n}^{8}e^{-2{s_{A,n}}}$ | $e^{\frac{3}{2}\tau}A^{7}B_{n}^{7}e^{-2{s_{A,n}}}$ | 0
$C_{4,0}C_{5,i}$ | $e^{2\tau}A^{9}B_{n}^{9-i}e^{-\frac{5}{2}{s_{A,n}}}$ | $e^{\frac{3}{2}\tau}A^{8}B_{n}^{8-i}e^{-\frac{5}{2}{s_{A,n}}}$ | $e^{\frac{3}{2}\tau}A^{8}B_{n}^{9-i}e^{-\frac{5}{2}{s_{A,n}}}$
$O(s^{2}e^{-3s})$ | $e^{\tau}O({s_{A,n}}+\tau)^{2}e^{-3{s_{A,n}}}$ | $e^{\frac{\tau}{2}}O({s_{A,n}}+\tau)^{2}e^{-3{s_{A,n}}}$ | $e^{\frac{\tau}{2}}O({s_{A,n}}+\tau)^{2}e^{-3{s_{A,n}}}$

As we have written right after the previous table (given right before (4.47)),
the terms coming with $e^{2\tau}$ and $e^{3\tau/2}$ are not relevant for our
formal argument.

From this table, and focusing only on the terms bearing $e^{\frac{\tau}{2}}$
(remember, this is formal), we may write the following expansion:

$\displaystyle
v_{b,1,0}({s_{A,n}}+\tau)=e^{\frac{\tau}{2}}\left\\{4A^{3}B_{n}^{3}C_{4,0}e^{-{s_{A,n}}}+3A^{4}B_{n}^{2}e^{-\frac{3}{2}{s_{A,n}}}(C_{5,2}+O(B_{n}))\right.$
$\displaystyle\left.+2A^{5}B_{n}e^{-2{s_{A,n}}}(C_{6,4}+O(B_{n}))+A^{6}e^{-\frac{5}{2}{s_{A,n}}}(C_{7,6}+O(B_{n}))+O({s_{A,n}}^{2}e^{-3{s_{A,n}}})\right\\}.$

Since $B_{n}\gg e^{-\frac{{s_{A,n}}}{2}}$ by (4.3) (translated in particular
in (4.54)), we clearly see that

$v_{b,1,0}({s_{A,n}}+\tau)\sim
4e^{\frac{\tau}{2}}A^{3}B_{n}^{3}C_{4,0}e^{-{s_{A,n}}}\mbox{ as }n\to\infty,$

since $C_{4,0}\neq 0$ by (4.6). Choosing a suitable $\tau$, we reach a
contradiction with (4.23) as usual, at least formally. Now, on a rigorous
level, Proposition 4.2 is not enough, and one needs to use Proposition 4.8 and
the following argument to conclude (note that Proposition 4.8 can indeed be
applied here, since the hypothesis (4.52) is fulfilled thanks to (4.56)).
Therefore, the limit $L_{0}$ given in (4.36) is zero.

As a conclusion to Subsection 4.2, whenever we are in the superquadratic case
(4.3), only the $3/2$ power regime is allowed, with a constant $L_{0}$ given
by (4.37). In the next subsection, we will investigate the quadratic regime.

###  4.3  The quadratic regime

This part is devoted to the study of the quadratic regime mentioned on page
4.1. Precisely, we assume that

${a_{n,1}}\sim L{a_{n,2}}^{2}\mbox{ as }n\to\infty,$ (4.57)

for some $L>0$. We will show that $L$ can enjoy only a finite number of
values, which depend on the Taylor expansion of the solution.

Our strategy in the same as before: use the Taylor expansion of $v_{0}(y,s)$
to derive an expansion for $v_{b}(y,s)$ with $b=a_{n}$. The non-growth
condition imposed on the three components $v_{b,0,0}$, $v_{b,1,0}$ and
$v_{b,0,1}$ then implies some conditions on the coefficients as usual. It
happens that those conditions take the form of polynomial equations on $L$,
whose coefficients are derived from the Taylor expansion.
We proceed in 3 steps, where we successively improve the Taylor expansion of
the solution, in order to find the values of $L$.

Step 1: Expansion of order $e^{-\frac{3}{2}{s_{A,n}}}$ in $v_{b,1,0}$

To begin with, let us recall that the expansion (1.12) is valid here. In
particular, the table given right before (4.19) will be useful in our
argument. Noting that

$B_{n}\sim LAe^{-\frac{{s_{A,n}}}{2}}\mbox{ as }n\to\infty$ (4.58)

from (4.57) and (4.11), then, using our usual strategy, we prove formally that

$v_{b,1,0}({s_{A,n}}+\tau)=e^{\frac{\tau}{2}}A^{4}e^{-\frac{3}{2}{s_{A,n}}}\left(2C_{4,2}L+C_{5,4}+o(1)\right)\mbox{
as }n\to\infty.$

Given the exponential growth factor, this implies that

$2C_{4,2}L+C_{5,4}=0$ (4.59)

(in order to justify rigorously this estimate, simply note that Proposition
4.2 holds here, since it only uses the fact that (1.12) holds, which is the
case here; using the ODE argument given after Proposition 4.2 allows us to
conclude).
If $C_{4,2}\neq 0$, then $C_{4,2}<0$ from (1.13), and (4.59) implies that

$L=-\frac{C_{5,4}}{2C_{4,2}}.$

Since $L>0$, this implies that $C_{5,4}>0$. Thus, we assume in the following
that

$C_{4,2}=0.$ (4.60)

Since the multilinear form in (1.10) is nonnegative, this implies that

$C_{4,1}=0.$ (4.61)

Using (4.59), we also derive that

$C_{5,4}=0.$ (4.62)

Step 2: Expansion of order $e^{-2{s_{A,n}}}$ in the 3 expanding components

Starting from the second order expansion (1.12) and using (4.60), (4.61) and
(4.62), we may use the strategy of Proposition 1 to derive the following third
order Taylor expansion, analogous to (4.45) above (with the same notations):

$\displaystyle v_{0}(y,s)=$
$\displaystyle-\frac{e^{-2s}}{3}C_{4,0}^{2}\gamma_{4,4,0}h_{0}h_{0}-\frac{e^{-2s}}{2}C_{4,0}^{2}\gamma_{4,4,2}h_{2}h_{0}+\left[C_{4,0}e^{-s}-e^{-2s}C_{4,0}^{2}\gamma_{4,4,4}\right]h_{4}h_{0}$
$\displaystyle+e^{-\frac{3}{2}s}\sum_{j=0}^{3}C_{5,j}h_{5-j}h_{j}+se^{-2s}C_{4,0}^{2}\gamma_{4,4,6}h_{6}h_{0}+e^{-2s}\sum_{j=0}^{6}C_{6,j}h_{6-j}h_{j}$
$\displaystyle+e^{-2s}C_{4,0}^{2}h_{8}h_{0}+O\left(se^{-\frac{5}{2}s}\right),$
(4.63)

for some real coefficients $C_{6,j}$ with $j=0,\dots,6$.
Using our usual strategy, we derive the following expansion:

$v_{b,1,0}({s_{A,n}}+\tau)=e^{\frac{\tau}{2}}A^{5}e^{-2{s_{A,n}}}(2LC_{5,3}+C_{6,5}+o(1)),$

at least on a formal level (this time, the table right before (4.46) is useful
to derive this, together with the estimate (4.58) on $B_{n}$). The exponential
growth factor then implies that

$2LC_{5,3}+C_{6,5}=0$ (4.64)

(as for the rigorous justification, again, Proposition 4.2 is sufficient,
since the power of $A$ in the linear term in the ODE solution is $5$, larger
that $4$, which is the power of the quadratic contribution).
If $C_{5,3}\neq 0$, then we see that

$L=-\frac{C_{6,5}}{2C_{5,3}}.$

Thus, we assume in the following that

$C_{5,3}=0.$ (4.65)

Using (4.64), we see that

$C_{6,5}=0.$ (4.66)

Writing the expansion of $v_{b,0,0}$ this time, we formally see from (4.58)
and the table given before (4.46) that

$v_{b,0,0}({s_{A,n}}+\tau)=e^{\tau}A^{6}e^{-2{s_{A,n}}}(C_{6,6}+o(1)),$

and the growth factor implies that

$C_{6,6}=0$ (4.67)

(again, the rigorous justification uses Proposition 4.2 and the difference of
the powers of $A$ between the linear term and the quadratic term in the
solution of the ODE ($6$ against $4$)).

Step 3: Expansion of order $e^{-\frac{5}{2}{s_{A,n}}}$ in two expanding
components

At this stage, we need to further improve expansion (4.63) up to the forth
order. Starting from (4.63) and using the strategy of Proposition 1, together
with (4.65), (4.66) and (4.67), we write a similar expansion to (4.56):

$\displaystyle
v_{0}(y,s)=-\frac{e^{-2s}}{3}C_{4,0}^{2}\gamma_{4,4,0}h_{0}h_{0}-\frac{2e^{-\frac{5}{2}s}}{3}C_{4,0}\sum_{j=0}^{1}C_{5,j}\gamma_{4,5-j,1-j}h_{1-j}h_{j}$
$\displaystyle-\frac{e^{-2s}}{2}C_{4,0}^{2}\gamma_{4,4,2}h_{2}h_{0}-e^{-\frac{5}{2}s}C_{4,0}\sum_{j=0}^{2}C_{5,j}\gamma_{4,5-j,3-j}h_{3-j}h_{j}$
$\displaystyle+\left[C_{4,0}e^{-s}-e^{-2s}C_{4,0}^{2}\gamma_{4,4,4}\right]h_{4}h_{0}$
$\displaystyle+\sum_{j=0}^{2}[C_{5,j}e^{-\frac{3}{2}s}-2e^{-\frac{5}{2}s}C_{4,0}C_{5,j}\gamma_{4,5-j,5-j}]h_{5-j}h_{j}$
(4.68)
$\displaystyle+se^{-2s}C_{4,0}^{2}\gamma_{4,4,6}h_{6}h_{0}+e^{-2s}\sum_{j=0}^{4}C_{6,j}h_{6-j}h_{j}+\sum_{j=0}^{2}2se^{-\frac{5}{2}s}C_{4,0}C_{5,j}\gamma_{4,5-j,7-j}h_{7-j}h_{j}$
$\displaystyle+\sum_{j=0}^{7}C_{7,j}e^{-\frac{5}{2}s}h_{7-j}h_{j}+e^{-2s}C_{4,0}^{2}h_{8}h_{0}+2C_{4,0}e^{-\frac{5}{2}s}\sum_{j=0}^{2}C_{5,j}h_{9-j}h_{j}+O\left(s^{2}e^{-3s}\right),$

for some real coefficients $C_{7,j}$ with $j=0,\dots,7$, with the notations
given after (4.45).

Starting from the table in Step 3 of Part 4 above, together with estimate
(4.58) on $B_{n}$, we formally obtain the following, thanks to our usual
strategy:

$\displaystyle v_{b,0,0}({s_{A,n}}+\tau)$
$\displaystyle=e^{\tau}A^{7}e^{-\frac{5}{2}{s_{A,n}}}(C_{7,7}+o(1)),$ (4.69)
$\displaystyle v_{b,1,0}({s_{A,n}}+\tau)$
$\displaystyle=e^{\frac{\tau}{2}}A^{6}e^{-\frac{5}{2}{s_{A,n}}}(4L^{3}C_{4,0}+3L^{2}C_{5,2}+2LC_{6,4}+C_{7,6}+o(1)).$

The growth factors both imply that (at least formally)

$C_{7,7}=0\mbox{ and }4L^{3}C_{4,0}+3L^{2}C_{5,2}+2LC_{6,4}+C_{7,6}=0.$ (4.70)

In particular, we see that $L$ solves an equation of degree 3 (remember that
$C_{4,0}\neq 0$ from (1.13)). Thus, it can at most bear 3 values. We need of
course to justify (4.70) rigorously. For that, we need to refine Proposition
4.8 up to the order $e^{-\frac{3}{2}{s_{A,n}}}$. More precisely, this is our
statement:

######  Proposition 4.9 (Uniform estimate of $v_{a_{n}}(y,{s_{A,n}})$ up to
the order $e^{-\frac{3}{2}{s_{A,n}}}$).

With all the information we have in this Step 3 of Subsection 4.3, we claim
that that for all $A\geq 1$ and $D\geq 1$, for $n$ large enough, we have for
all $s\in[{s_{A,n}},D{s_{A,n}}]$,

$\|v_{a_{n}}(y,s)-C_{4,0}e^{-s}h_{4}(y_{1})\|_{L^{2}_{\rho}}\leq
CA^{2}e^{-\frac{3}{2}{s_{A,n}}}$

###### Proof.

See Subsection 4.4. ∎

As one may see by comparing the 2 propositions, the new version replaces the
$C(A)$ constant appearing in Proposition 4.8 by $CA^{2}$. With this
improvement, the argument given right after that proposition works (thanks to
the difference between the powers of $A$), confirming the validity of (4.70).
Note however that expansion (4.69) is valid for $\tau$ large, but not for
$\tau=0$; for that reason, one has to carefully use (4.68) to derive that

$v_{b,0,0}({s_{A,n}})=-\frac{e^{-2{s_{A,n}}}}{3}C_{4,0}^{2}\gamma_{4,4,0}+e^{-\frac{5}{2}{s_{A,n}}}(-\frac{2}{3}AC_{4,0}C_{5,1}\gamma_{4,4,0}+A^{7}C_{7,7}+o(1)).$

Apart from that, the adaptation of the argument given right after Proposition
4.8 is straightforward.

This concludes the proof of Theorem 3 when $m=4$, if one assumes the result of
Propositions 4.2, 4.8 and 4.9. In order to finish the argument, those
propositions will be proved in the next subsection.

###  4.4 Uniform estimates for $w_{a_{n}}$

This subsection is dedicated to the proofs of Propositions 4.2, 4.8 and 4.9.
This is the only missing element to terminate the proof of Theorem 3 in the
case $m=4$, given in the previous subsections.

###### Proof of Proposition 4.2.

We will in fact sketch the idea, and avoid giving details.
Let us first start by recalling from Khenissy, Rebai and Zaag [10] that

$v_{a_{n}}(y,s)\sim P_{\\{2\leq i\leq 4\\}}(v_{a_{n}})\mbox{ as }s\to\infty,$
(4.71)

uniformly for $n$ large enough (see Theorem 1 page 4 in that paper). We also
know from (4.22) that

$\|v_{a_{n}}(s)\|_{L^{2}_{\rho}}\to 0\mbox{ as }s\to\infty,$

uniformly in $n$. From (4.71), our desired conclusion will follow if we fix
${C}_{0}>0$ such that for any $A\geq 1$ and $D\geq 1$,

$\forall i=2,3,4,\;\;\forall j=0,\dots,i,\;\;\forall
s\in[{s_{A,n}},D{s_{A,n}}],\;\;|v_{a_{n},i,j}(s)|\leq{C}_{0}A^{2}e^{-{s_{A,n}}},$
(4.72)

whenever $n$ is large enough.
We proceed in 2 steps to fix such a ${C}_{0}$: in Step 1, we initialize (4.72)
at $s={s_{A,n}}$, then, in Step 2, we proceed by contradiction to prove it for
$s\geq{s_{A,n}}$ (the integration of the projections of equation (2.2) on the
various coordinates will be crucial for our argument).
Step 1: Initialization of (4.72). Since there is a universal constant
$C_{2}\geq 1$ such that for any $v\in L^{2}_{\rho}$ and $i=2,3,4$,

$\frac{1}{C_{2}}\max_{j=0,\dots,i}|v_{i,j}|\leq\|P_{i}(v)\|_{L^{2}_{\rho}}\leq
C_{2}\max_{j=0,\dots,i}|v_{i,j}|,$ (4.73)

using (4.26) together with (4.71), we see that

$\forall i=2,3,4,\;\;\forall
j=0,\dots,i,\;\;|v_{a_{n},i,j}({s_{A,n}})|\leq\bar{C}A^{2}e^{-{s_{A,n}}},$
(4.74)

whenever $n$ is large enough, for some universal constant $\bar{C}>0$.
Fixing

${C}_{0}=2\bar{C},$ (4.75)

we guarantee from (4.74) that (4.72) holds at $s={s_{A,n}}$.
In particular, (4.72) holds at $s={s_{A,n}}$.
Step 2: The contradiction argument. Let us assume by contradiction that (4.72)
is true for all $s\in[{s_{A,n}},s^{*}_{n}]$, for some minimal
$s^{*}_{n}<D{s_{A,n}}$ and stops from being true at $s=s^{*}_{n}$. From Step 1
and continuity, it follows that $s_{n}^{*}>{s_{A,n}}$. This also implies that
we have an equality case at $s=s^{*}_{n}$, in the sense that

$|v_{a_{n},i,j}(s_{n}^{*})|={C}_{0}A^{2}e^{-{s_{A,n}}},\mbox{ for some
}i=2,3,4\mbox{ and }j=0,\dots,i.$ (4.76)

Starting from (4.75), we integrate the ODE (2.6) satisfied by $v_{a_{n},i,j}$,
with $2\leq i\leq 4$:

$\forall s\geq
s_{n},\;\;v_{a_{n},i,j}^{\prime}(s)=(1-\frac{i}{2})v_{a_{n},i,j}(s)+\int
k_{i-j}(y_{1})k_{j}(y_{2})v_{a_{n}}(y,s)^{2}\rho(y)dy.$

Since (4.72) is valid for all $s\in[{s_{A,n}},s^{*}_{n}]$, using (4.71), we
have

$|\int k_{i-j}(y_{1})k_{j}(y_{2})v_{a_{n}}(y,s)^{2}\rho(y)dy|\leq
C{C}_{0}^{2}A^{4}e^{-2{s_{A,n}}}$

(in fact, we need here an estimate in the $L^{4}_{\rho}$ norm and not just in
the $L^{2}_{\rho}$, and this is possible thanks to parabolic regularity, as we
have already explained right after the statement of Proposition 4.2).
Therefore, for all $s\in[{s_{A,n}},s^{*}_{n}]$, we have

$|v_{a_{n},i,j}^{\prime}(s)-(1-\frac{i}{2})v_{a_{n},i,j}(s)|\leq
C{C}_{0}^{2}A^{4}e^{-2{s_{A,n}}}.$

Recalling that $s_{n}^{*}\leq D{s_{A,n}}$, we integrate this differential
inequality, ending with

$\displaystyle|v_{a_{n},i,j}(s)|$ $\displaystyle\leq
e^{(1-\frac{i}{2})(s-{s_{A,n}})}|v_{a_{n},i,j}({s_{A,n}})|+C{C}_{0}^{2}A^{4}e^{-2{s_{A,n}}}(s-{s_{A,n}}),$
$\displaystyle\leq\frac{{C}_{0}}{2}A^{2}e^{-{s_{A,n}}}+C{C}_{0}^{2}A^{4}e^{-2{s_{A,n}}}(D-1){s_{A,n}}\leq\frac{3{C}_{0}}{4}A^{2}e^{-{s_{A,n}}},$
(4.77)

if $n$ is large enough. This means in particular that (4.76) is not true and
concludes the proof of Proposition 4.2. ∎

###### Proof of Proposition 4.8.

Following Proposition 4.2, we assume that (4.52) holds. We also recall from
(4.9) that $B_{n}\to 0$ as $n\to\infty$.
We proceed in 2 steps: we first initialize the estimate at $s={s_{A,n}}$,
then, we use the bound in Proposition 4.2 to derive then integrate
differential inequalities satisfied by the various components.
Step 1: Initialization at $s={s_{A,n}}$. Using the expansion (4.52) and using
our geometric transformation as usual, we may write the following estimates
for the various components of the solution at $s={s_{A,n}}$ (for the last
estimate, remember that (4.52) holds also in $L^{4}_{\rho}$ and use the
Cauchy-Schwarz identity as we did for (4.25)):

$\displaystyle v_{a_{n},4,0}({s_{A,n}})$
$\displaystyle=C_{4,0}e^{-{s_{A,n}}}+O(e^{-\frac{3}{2}{s_{A,n}}}),$ (4.78)
$\displaystyle|v_{a_{n},i,j}({s_{A,n}})|$ $\displaystyle\leq
C_{i}e^{-\frac{3}{2}{s_{A,n}}}\mbox{ if }(i,j)\neq(4,0),$ (4.79)
$\displaystyle\|P_{\\{\lambda\leq-\frac{3}{2}\\}}(v_{a_{n}}({s_{A,n}}))\|_{L^{2}_{\rho}}$
$\displaystyle\leq C(A)e^{-\frac{3}{2}{s_{A,n}}}.$ (4.80)

Step 2: Differential inequalities satisfied by the various components. Using
Proposition 4.2 whose conclusion also holds in $L^{4}_{\rho}$, as explained in
the remark following its statement, we may write the following differential
inequalities, for $n$ large enough and for all $s\in[{s_{A,n}},D{s_{A,n}}]$,

$\displaystyle\forall i\in\mathbbm{N},\;\;\forall
j=0,\dots,i,\;\;|v_{a_{n},i,j}^{\prime}(s)-(1-\frac{i}{2})v_{a_{n},i,j}(s)|$
$\displaystyle\leq C_{i}A^{4}e^{-2{s_{A,n}}},$
$\displaystyle\frac{d}{ds}\|P_{\\{\lambda\leq
1-\frac{i}{2}\\}}(v_{a_{n}}(s))\|_{L^{2}_{\rho}}^{2}\leq-2(1-\frac{i}{2})\|P_{\\{\lambda\leq
1-\frac{i}{2}\\}}(v_{a_{n}}(s))\|_{L^{2}_{\rho}}^{2}+$ $\displaystyle
C_{i}^{2}A^{8}e^{-4{s_{A,n}}}.$

Integrating these various differential inequalities, we get the following
estimates, for $n$ large enough and for all $s\in[{s_{A,n}},D{s_{A,n}}]$,

$\displaystyle\forall i=0,1,\;\;\forall j=0,\dots,i,\;\;|v_{a_{n},i,j}(s)|$
$\displaystyle\leq CA^{4}e^{-2{s_{A,n}}},$ (4.81) $\displaystyle\forall
j=0,1,2,\;\;|v_{a_{n},2,j}(s)-v_{a_{n},2,j}({s_{A,n}})|$ $\displaystyle\leq
C(s-{s_{A,n}})A^{4}e^{-2{s_{A,n}}},$ (4.82) $\displaystyle\forall
i=3,4,\;\;\forall
j=0,\dots,i,\;\;|v_{a_{n},i,j}(s)-e^{(1-\frac{i}{2})(s-{s_{A,n}})}v_{a_{n},i,j}({s_{A,n}})|$
$\displaystyle\leq CA^{4}e^{-2{s_{A,n}}},$ (4.83)
$\displaystyle\|P_{\\{\lambda\leq-\frac{3}{2}\\}}(v_{a_{n}}(s))\|_{L^{2}_{\rho}}\leq
e^{-\frac{3}{2}(s-{s_{A,n}})}\|P_{\\{\lambda\leq-\frac{3}{2}\\}}(v_{a_{n}}({s_{A,n}}))\|_{L^{2}_{\rho}}$
$\displaystyle+CA^{4}e^{-2{s_{A,n}}}.$ (4.84)

Using (4.78)-(4.80), and recalling that $s\leq D{s_{A,n}}$, we conclude the
proof of Proposition 4.8. ∎

###### Proof of Proposition 4.9.

Note first that we can use all the information available in Step 3 of
Subsection 4.3 For the proof, we follow the proof of Proposition 4.8 with only
one modification: we need to refine identities (4.78)-(4.80) so that we reach
the orders $e^{-\frac{3}{2}{s_{A,n}}}$. For that reason, we will keep only the
$e^{-s}$ and $e^{-\frac{3}{2}s}$ terms in the expansion (4.68) and write

$v_{0}(y,s)=C_{4,0}e^{-s}h_{4}h_{0}+e^{-\frac{3}{2}s}\sum_{j=0}^{2}C_{5,j}h_{5-j}h_{j}+O(se^{-2s})$

as $s\to\infty$, in $L^{q}_{\rho}(\mathbbm{R}^{N})$, for any $q\geq 2$. Then,
using the estimate (4.58) on $B_{n}$ together with the change of variables
(4.12) (with $\tau=0$), we may use the binomial relation (3.14) to write

$\displaystyle v_{b}(y_{b},{s_{A,n}})=$ $\displaystyle
C_{4,0}e^{-{s_{A,n}}}\left[h_{4}(y_{b,1})+4h_{3}(y_{b,1})AB_{n}+O(B_{n}^{2})\right]$
$\displaystyle+e^{-\frac{3}{2}{s_{A,n}}}\sum_{j=0}^{2}C_{5,j}h_{5-j}(y_{b,1})h_{j}(y_{b},2)+O(C(A){s_{A,n}}e^{-2{s_{A,n}}})$

in $L^{2}_{\rho}(\mathbbm{R}^{N})$ as $n\to\infty$, thanks to the same
justification as for (4.25). This can be translated as follows:

$\displaystyle v_{b,3,0}({s_{A,n}})$
$\displaystyle=4C_{4,0}LA^{2}e^{-\frac{3}{2}{s_{A,n}}}+o(e^{-\frac{3}{2}{s_{A,n}}}),$
$\displaystyle v_{b,4,0}({s_{A,n}})$
$\displaystyle=C_{4,0}e^{-{s_{A,n}}}+O\left(C(A){s_{A,n}}e^{-2{s_{A,n}}}\right),$
$\displaystyle\mbox{for }i=0,1,2\;\;v_{b,5,i}({s_{A,n}})$
$\displaystyle=C_{5,i}e^{-\frac{3}{2}{s_{A,n}}}+O\left(C(A){s_{A,n}}e^{-2{s_{A,n}}}\right),$
$\displaystyle\sup_{(i,j)\not\in\\{(3,0),(4,0),(5,0),(5,1),(5,2)\\}}|v_{b,i,j}({s_{A,n}})|$
$\displaystyle+\|P_{\\{\lambda\leq-\frac{3}{2}\\}}(v_{a_{n}}({s_{A,n}}))\|_{L^{2}_{\rho}}\leq
C(A){s_{A,n}}e^{-2{s_{A,n}}}$

Using these estimates as initial data, together with the estimates
(4.81)-(4.84), which hold here, since the hypotheses of Proposition 4.8 hold
also (because (4.52) follows from (4.68), which holds like all the information
of Step 3 in Subsection 4.3, we conclude the proof of Proposition 4.9 (we also
need Proposition 4.2). ∎

We have just proved the Propositions 4.2, 4.8 and 4.9 in this subsection,
finishing this way the proof of Theorem 3.

##  5 Rigidity in the geometry of the blow-up set when $m\geq 6$

This section is devoted to the proof of Theorem 3 when $m\geq 6$. As for
$m=4$, the proof uses the same strategy based on the geometric transformation
(3.5) given in Step 2 of the proof of Theorem 3. However, the number of steps
depends linearly on $m$. For that reason, we have to proceed differently, at
some point in the proof. In addition, we have more complicated formulas, and
of course in the outcome, the result for $m\geq 6$ is less explicit, hence,
less spectacular than for $m=4$. Accordingly, we only give the main steps of
the proof and don’t insist on details.

###### Proof of Theorem 3 when $m\geq 6$.

Consider $u(x,t)$ a solution of equation (1.1) blowing up at time $T>0$.
Assume that the origin is a non isolated blow-up point where $m(0)=m\geq 6$ is
even and consider an arbitrary sequence of non-zero blow-up points
$a_{n}=(a_{n,1},a_{n,2})$ converging to the origin as $n\to\infty$.
(i) This item follows exactly as in the case $m=4$ (see Subsection 4.1 above).
In particular, after extracting a subsequence and making a suitable change of
variables, we may assume that

${a_{n,1}}\geq 0\mbox{ and }a_{n,2}\geq 0\mbox{ for all }n\in\mathbbm{N}$
(5.1)

and

${a_{n,1}}=o({a_{n,2}}).$ (5.2)

Note that Theorem 2 holds here, and so does the expansion (1.12), and that the
multilinear form in (1.10) is non zero and nonnegative.
(ii) Following (5.2), after the extraction of a subsequence still denoted the
same, we identify 3 possible regimes for $a_{n}$: superquadratic, quadratic
and subquadratic, as with $m=4$. We then proceed in 3 parts:
\- Part 1 is dedicated to the superquadratic case, where
${a_{n,1}}\gg{a_{n,2}}^{2}$ as $n\to\infty$.
\- In Part 2, we deal with the quadratic case, where ${a_{n,1}}\sim
L{a_{n,2}}^{2}$, for some $L>0$.
\- Finally, in Part 3, we gather all the information and conclude the proof of
Theorem 3.

Part 1: Quantified superquadratic regimes for $a_{n}$

In this part, we assume that

$a_{n,1}\gg a_{n,2}^{2}\mbox{ as }n\to\infty.$ (5.3)

We then proceed in 4 steps:
\- In Step 1, we show that $C_{m,m-2}=C_{m,m-3}=0$, where those coefficients
appear in the expansion (1.12).
\- In Step 2, we introduce a new parameter $\theta$ to measure the convergence
of $B_{n}=a_{n,1}/a_{n,2}$ to $0$ in exponential scales of ${s_{A,n}}$ defined
in (4.11), and show that $\theta$ enjoys only a finite number of values in
$\mathbbm{Q}\backslash\\{0\\}$.
\- In Step 3, we make a refinement of Step 2, by introducing one further
parameter $\alpha$ to measure the convergence of $B_{n}$ to $0$ in polynomial
corrections of exponential scales, and show that $\alpha$ enjoys only a finite
number of values in $\mathbbm{Q}$.
\- In Step 4, following Steps 2 and 3, we make one further refinement, by
introducing a new variable $\psi_{n}$ to quantify the convergence of $B_{n}$
to $0$, and show that it converges (up to a subsequence) to some value related
to some root of a polynomial involving the coefficients of the Taylor
expansion of $v_{0}(y,s)$.

Step 1: Proof of the fact that $C_{m,m-2}=C_{m,m-3}=0$

Considering the expansion (1.12), we will now prove that
$C_{m,m-2}=C_{m,m-3}=0$. As in the case $m=4$, the nonnegativity of the
multilinear form (1.10) implies that it is enough to show that $C_{m,m-2}=0$.
Proceeding by contradiction, we assume that $C_{m,m-2}\neq 0$. Let us reach a
contradiction from the behavior of $w_{b}$ defined by (4.12), where $b=a_{n}$.
Using the notations and assumptions in (4.9)-(4.11), we see from (5.3), (4.16)
and (5.2) that

$1\gg B_{n}\gg Ae^{-\frac{{s_{A,n}}}{2}}\mbox{ as }n\to\infty.$ (5.4)

Using the expansion (1.12) and proceeding as in Part 1 of Subsection 4.2 when
$m=4$, both formally and rigorously, we readily see that

$v_{b,1,0}({s_{A,n}}+\tau)\sim
2e^{\frac{\tau}{2}}A^{m-1}B_{n}C_{m,m-2}e^{(1-\frac{m}{2}){s_{A,n}}}\mbox{ as
}n\to\infty.$

The growth factor leads to contradiction. Thus, $C_{m,m-2}=0$ and
$C_{m,m-3}=0$ too.

Step 2: Measuring $B_{n}$ in exponential scales of ${s_{A,n}}$.

From (5.4), we may assume that

$-\frac{\log B_{n}}{{s_{A,n}}}\to\theta\in[0,\frac{1}{2}]\mbox{ or
equivalently that }B_{n}=e^{-(\theta+o(1)){s_{A,n}}},\mbox{ as }n\to\infty,$
(5.5)

up to a extracting a subsequence, still denoted the same. Now, we are going to
prove the following three-fold statement:

######  Proposition 5.1.

(i) The limit $\theta\in[0,\frac{1}{2}]$ defined in (5.5) is rational.
(ii) It holds that $\theta\neq 0$.
(iii) In fact, $\theta$ enjoys only a finite number of rational values in
$E_{1}\cup E_{2}$ where the two sets $E_{1}$ and $E_{2}$ are defined below
respectively in (5.18) and (5.19).

######  Remark.

In fact, as we will see from the proof, items (ii) and (iii) are by-products
of the proof of item (i). For clarity, we dedicate the next step to the proof
of item (i), then, we explain how to derive items (ii) and (iii) in the
following steps.

######  Remark.

Note that

$E_{1}\cup E_{2}\subset\\{\frac{L}{2G}\;\;1\leq G\leq 2m-3\mbox{ and }1\leq
L\leq\min(G,m-2)\\}.$

###### Proof.

We will give the proof of the three items successively.

\- Proof of item (i) of Proposition 5.1.

Using the strategy of Proposition 1, we may refine the Taylor expansion of
$v_{0}(y,s)$ (equal to $w_{0}(y,s)-1$ by definition (2.1)) given in (1.12)
(recalling that $C_{m,m-2}=C_{m,m-3}=0$ from Step 1), up to the order
$O(se^{(2-m)s})$:

$\displaystyle v_{0}(y,s)=$ $\displaystyle
e^{(1-\frac{m}{2})s}\sum_{j=0}^{m-4}C_{m,j}h_{m-j}(y_{1})h_{j}(y_{2})+e^{\frac{1-m}{2}s}\sum_{j=0}^{m}C_{m+1,j}h_{m+1-j}(y_{1})h_{j}(y_{2})$
$\displaystyle+e^{(1-\frac{k}{2})s}\sum_{k=m+2}^{2m-3}\sum_{j=0}^{k}C_{k,j}h_{k-j}(y_{1})h_{j}(y_{2})+\bar{v}_{0}(y,s)$
(5.6)

with $\bar{v}_{0}(y,s)=O(se^{(2-m)s})$ as $s\to\infty$. Note that we stopped
in this expansion when the quadratic term in the equation (2.2) satisfied by
$v_{0}(y,s)$ becomes relevant; this way, we only have “linear terms” in our
expansion.
Using the transformation (3.5), we may use the expansion in (5.6) to derive
the following expansion for $v_{b}(y_{b},{s_{A,n}}+\tau)$, which is analogous
to (4.13), with $\tau\geq 0$:

$\displaystyle v_{b}(y_{b},{s_{A,n}}+\tau)$
$\displaystyle=e^{(1-\frac{m}{2})({s_{A,n}}+\tau)}\sum_{j=0}^{m-2}C_{m,j}h_{m-j}(y_{b,1}+AB_{n}e^{\frac{\tau}{2}})h_{j}(y_{b,2}+Ae^{\frac{\tau}{2}})$
$\displaystyle+e^{\frac{1-m}{2}({s_{A,n}}+\tau)}\sum_{j=0}^{m}C_{m+1,j}h_{m+1-j}(y_{b,1}+AB_{n}e^{\frac{\tau}{2}})h_{j}(y_{b,2}+Ae^{\frac{\tau}{2}})$
$\displaystyle+e^{(1-\frac{k}{2})({s_{A,n}}+\tau)}\sum_{k=m+2}^{2m-3}\sum_{j=0}^{k}C_{k,j}h_{k-j}(y_{b,1}+AB_{n}e^{\frac{\tau}{2}})h_{j}(y_{b,2}+Ae^{\frac{\tau}{2}})$
$\displaystyle+\bar{v}_{b}(y_{b},{s_{A,n}}+\tau).$ (5.7)

As before, for more visibility, we may write the following table giving the
expansion for the 3 expanding components $v_{b,0,0}({s_{A,n}}+\tau)$,
$v_{b,0,1}({s_{A,n}}+\tau)$ and $v_{b,1,0}({s_{A,n}}+\tau)$, for $\tau\geq 0$:

| $v_{b,0,0}({s_{A,n}}+\tau)$ | $v_{b,1,0}({s_{A,n}}+\tau)$ | $v_{b,0,1}({s_{A,n}}+\tau)$
---|---|---|---
$C_{m,j}$ | $C_{m,j}e^{\tau}A^{m}B_{n}^{m-j}e^{(1-\frac{m}{2}){s_{A,n}}}$ | $(m-j)C_{m,j}e^{\frac{\tau}{2}}A^{m-1}B_{n}^{m-1-j}e^{(1-\frac{m}{2}){s_{A,n}}}$ | $jC_{m,j}e^{\frac{\tau}{2}}A^{m-1}B_{n}^{m-j}e^{(1-\frac{m}{2}){s_{A,n}}}$
$C_{m+1,j}$ | $C_{m+1,j}e^{\tau}A^{m+1}B_{n}^{m+1-j}e^{\frac{1-m}{2}{s_{A,n}}}$ | $(m+1-j)C_{m+1,j}e^{\frac{\tau}{2}}A^{m}B_{n}^{m-j}e^{\frac{1-m}{2}{s_{A,n}}}$ | $jC_{m+1,j}e^{\frac{\tau}{2}}A^{m}B_{n}^{m+1-j}e^{\frac{1-m}{2}{s_{A,n}}}$
$C_{k,j}$ | $C_{k,j}e^{\tau}A^{k}B_{n}^{k-j}e^{(1-\frac{k}{2}){s_{A,n}}}$ | $(k-j)C_{k,j}e^{\frac{\tau}{2}}A^{k-1}B_{n}^{k-1-j}e^{(1-\frac{k}{2}){s_{A,n}}}$ | $jC_{k,j}e^{\frac{\tau}{2}}A^{k-1}B_{n}^{k-j}e^{(1-\frac{k}{2}){s_{A,n}}}$
$\bar{v}_{b}(y_{b},{s_{A,n}}+\tau)$ | $\bar{v}_{b,0,0}({s_{A,n}}+\tau)$ | $\bar{v}_{b,0,1}({s_{A,n}}+\tau)$ | $\bar{v}_{b,1,0}({s_{A,n}}+\tau)$

with the rest terms satisfying

$|\bar{v}_{b,i,j}(y_{b},{s_{A,n}})|\leq C(A,i,j){s_{A,n}}e^{(2-m){s_{A,n}}},$
(5.8)

with the same argument as for (3.16).

Now, proceeding by contradiction, we assume that $\theta$ is not rational. Let
us first explain our argument. As in the case $m=4$, the contradiction will
follow from the behavior of one of the 3 components shown in the table above.
In fact, since the multilinear form in (1.10) is non zero, there is
$l=0,\dots,m$ such that $C_{m,l}\neq 0$. Therefore, it is convenient to choose
a component involving this $C_{m,l}$, hoping to reach a contradiction. Since
$l\in[0,m-4]$ from (1.12) and Step 1, we focus on the first two components,
since the third misses $C_{m,0}$. More precisely, we will choose the second
component, namely $v_{b,1,0}({s_{A,n}}+\tau)$, since it involves lower powers
of the small parameter $B_{n}$. Furthermore, we need the coefficient of the
term involving $C_{m,l}$, namely
$A^{m-1}e^{\frac{\tau}{2}}B_{n}^{m-1-l}e^{(1-\frac{m}{2}){s_{A,n}}}$, to be
dominant with respect to the error term whose size is shown in (5.8). Since we
have from (5.5) that

$A^{m-1}e^{\frac{\tau}{2}}B_{n}^{m-1-l}e^{(1-\frac{m}{2}){s_{A,n}}}\geq
A^{m-1}e^{\frac{\tau}{2}}B_{n}^{m-1}e^{(1-\frac{m}{2}){s_{A,n}}}=A^{m-1}e^{\frac{\tau}{2}}e^{(1-\frac{m}{2}-(m-1)\theta+o(1)){s_{A,n}}}$
(5.9)

as $n\to\infty$, a sufficient condition for this is to have

$1-\frac{m}{2}-(m-1)\theta>2-m,\mbox{ i.e. }\theta<\frac{m-2}{2(m-1)}.$ (5.10)

Accordingly, since $\theta\in[0,\frac{1}{2}]$ by (5.5), we consider two cases
in the following:

Case 1: $\theta\in[0,\frac{m-2}{2(m-1)})$. From the table above together with
(5.8), we consider the following expansion:

$v_{b,1,0}({s_{A,n}}+\tau)=e^{\frac{\tau}{2}}\left(\sum_{(k,j)\in
H_{1}}e^{(1-\frac{k}{2}){s_{A,n}}}A^{k-1}(k-j)C_{k,j}B_{n}^{k-1-j}+O({s_{A,n}}e^{(2-m){s_{A,n}}})\right),$
(5.11)

where

$\displaystyle H_{1}=\\{(k,j)\;\;|\;\;$ $\displaystyle m\leq k\leq
2m-3,\;\;0\leq j\leq k-1,\;\;$ (5.12) $\displaystyle C_{k,j}\neq 0\mbox{ and
}e^{(1-\frac{k}{2}){s_{A,n}}}B_{n}^{k-1-j}\gg{s_{A,n}}e^{(2-m){s_{A,n}}}\\}.$

We first note that $H_{1}\neq\emptyset$, since it contains $(m,l)$, by (5.9)
and (5.10). Second, the following non-codominance property between all terms
in the expansion (5.11) allows us to conclude:

######  Lemma 5.2 (No codominance of terms in the expansion of
$v_{b,1,0}({s_{A,n}}+\tau)$).

Consider two terms in the expansion (5.11), say
$(k-j)C_{k,j}A^{k-1}B_{n}^{k-1-j}e^{(1-\frac{k}{2}){s_{A,n}}}$ and
$(k^{\prime}-j^{\prime})C_{k^{\prime},j^{\prime}}A^{k^{\prime}-1}B_{n}^{k^{\prime}-1-j^{\prime}}e^{(1-\frac{k^{\prime}}{2}){s_{A,n}}}$
with $C_{k,j}C_{k^{\prime},j^{\prime}}\neq 0$, $m\leq k,k^{\prime}\leq 2m-3$,
$0\leq j\leq k-1$, $0\leq j^{\prime}\leq k^{\prime}-1$, and
$(k,j)\neq(k^{\prime},j^{\prime})$. Then, one of the terms dominates the other
as $n\to\infty$.

Indeed, if this lemma holds, recalling that $H_{1}$ is a non empty finite set,
we may consider $(\bar{k},\bar{j})\in H_{1}$ such that

$v_{b,1,0}({s_{A,n}}+\tau)\sim
e^{\frac{\tau}{2}}(\bar{k}-\bar{j})C_{\bar{k},\bar{j}}A^{\bar{k}-1}B_{n}^{\bar{k}-1-j}e^{(1-\frac{\bar{k}}{2}){s_{A,n}}}$
(5.13)

as $n\to\infty$, with $C_{\bar{k},\bar{j}}\neq 0$. From the growth factor
$e^{\frac{\tau}{2}}$, the coordinate $v_{b,1,0}({s_{A,n}}+\tau)$ will grow,
and a contradiction follows as usual, both for the formal and the rigorous
argument. It remains then to prove Lemma 5.2.

###### Proof of Lemma 5.2.

Using the expansion of $B_{n}$ in (5.5), we see that

$(k-j)C_{k,j}A^{k-1}B_{n}^{k-1-j}e^{(1-\frac{k}{2}){s_{A,n}}}=(k-j)C_{k,j}A^{k-1}e^{(1-\frac{k}{2}-(k-1-j)\theta+o(1)){s_{A,n}}}$
(5.14)

as $n\to\infty$, with a similar expansion with $(k^{\prime},j^{\prime})$.
Since $(k,j)\neq(k^{\prime},j^{\prime})$, it follows that
$(k,k-j)\neq(k^{\prime},k^{\prime}-j^{\prime})$. Therefore, recalling that
$\theta$ is not rational (this is in fact the contradiction hypothesis), it
follows that

$1-\frac{k}{2}-(k-1-j)\theta\neq
1-\frac{k^{\prime}}{2}-(k^{\prime}-1-j^{\prime})\theta,$

and the conclusion follows. ∎

######  Remark.

From this proof, we see that the hypothesis that $\theta$ is not rational is
too strong. In fact, the argument works whenever $\theta$ avoids the rationals
which are of the form $\frac{k^{\prime}-k}{2[(k-j)-(k^{\prime}-j^{\prime})]}$,
which make a finite collection of numbers, due to the boundedness of the
ranges where the parameters lay. This remark will show to be crucial below,
while adapting the present step in order to derive the proof of item (iii) of
Proposition 5.1.

Case 2: $\theta\in[\frac{m-2}{2(m-1)},\frac{1}{2}]$. It happens that the
argument of Case 1 works here, with small natural adaptations.
The first problem is that the terms in the expansion (5.11) may all be
dominated by the error term, therefore, we need to go further in the Taylor
expansion (5.6), and this is possible thanks to Proposition 1. The question
then is to know the order up to which we carry on the Taylor expansion.

To find that, note that focusing on terms with $C_{k,j}\neq 0$ is a convenient
way to have a relevant term in the expansion of $v_{b,1,0}({s_{A,n}}+\tau)$.
Such a term exists with $(k,j)=(m,l)$, since we know that the multilinear form
in (1.10) is non-zero. It remains then to guarantee that the order in front of
$C_{m,l}$, namely $e^{(1-\frac{m}{2}){s_{A,n}}}B_{n}^{m-1-l}$, is dominant
with respect to the error term. Since $l\geq 0$, hence
$e^{(1-\frac{m}{2}){s_{A,n}}}B_{n}^{m-1-l}\geq
e^{(1-\frac{m}{2}){s_{A,n}}}B_{n}^{m-1}=e^{(1-\frac{m}{2}-(m-1)\theta+o(1)){s_{A,n}}}$
as $n\to\infty$ by (5.5), we simply need to refine the the Taylor expansion
(5.6) of $v_{0}(y,s)$ up to the order $O(s^{\gamma}e^{(1-\frac{M+1}{2})s})$
for some $\gamma>0$, with $M\in\mathbbm{N}$ given by

$M=\lceil(2\theta+1)(m-1)\rceil,$ (5.15)

where the notation $\lceil\rceil$ stands for the ceiling (or upper integer
part) of a given number. This is possible, thanks to Proposition 1. However,
in comparison with the Taylor expansion given in (5.6), now we will see
“resonant” terms, of order $s^{i}e^{(1-\frac{k}{2})s}$, the first among them
occurs at $k=2m-2$, and corresponds to the effect of the quadratic term in
(2.2). More precisely, that term is of order $se^{2-m}$.

Now, using the geometric transformation given in Step 2 of Section 3, we may
write an expansion of $v_{b}(y_{b},{s_{A,n}}+\tau)$ analogous to (5.7). We may
also write a table similar to the one right before (5.8), giving an expansion
for $v_{b,1,0}({s_{A,n}}+\tau)$ analogous to (5.11), and which shows resonant
orders, as follows:

$\displaystyle
v_{b,1,0}({s_{A,n}}+\tau)=e^{\frac{\tau}{2}}\left(\sum_{(k,j,i)\in
H_{2}}\right.$
$\displaystyle({s_{A,n}}+\tau)^{i}e^{(1-\frac{k}{2}){s_{A,n}}}A^{k-1}(k-j)\tilde{C}_{k,j,i}B_{n}^{k-1-j}$
$\displaystyle\left.+O({s_{A,n}}^{\gamma}e^{(\frac{1-M}{2}){s_{A,n}}})\right),$
(5.16)

where $H_{2}\subset\mathbbm{N}^{3}$ is a natural adaptation of the set $H_{1}$
(5.12) as follows:

$\displaystyle H_{2}=\\{(k,j,i)\;\;|\;\;$ $\displaystyle m\leq k\leq
M,\;\;0\leq j\leq k-1,\;\;0\leq i\leq i_{k},$ (5.17)
$\displaystyle\tilde{C}_{k,j,i}\neq 0\mbox{ and
}{s_{A,n}}^{i}e^{(1-\frac{k}{2}){s_{A,n}}}B_{n}^{k-1-j}\gg{s_{A,n}}^{\gamma}e^{(\frac{1-M}{2}){s_{A,n}}}\\},$

where $i_{k}\in\mathbbm{N}$ is positive only at resonant orders.

As in Case 1, we first note that $H_{2}$ is non empty, thanks to the choice of
the order $M$ (5.15) up to which we made the Taylor expansion. Furthermore,
the non-codominance property holds here too!

Before proving that, we would like to comment on the time interval where we
investigate the codominance property. In Case 1, all the terms in the
expansion (5.11) are multiples of the sole function $e^{\frac{\tau}{2}}$.
Comparing them at $\tau=0$ or on any subinterval of $[0,\infty)$ gives the
same order. Here in Case 2, we have resonant terms, namely multiples of
$({s_{A,n}}+\tau)^{i}e^{\frac{\tau}{2}}$ with $i\in\mathbbm{N}$, and the
comparison at $\tau=0$ or on a larger interval may be different, depending on
the coefficient in front of the function and also on the size of the interval.
For that reason, we need to clearly fix some interval where we make the
comparison. The most natural choice is simply the time interval of validity of
the expansion (5.16), namely, the interval where all the functions stay less
than some fixed small $\delta_{0}>0$. More precisely, given some $(k,j,i)\in
H_{2}$, we take $\tau\in[0,\tau_{k,j,i.n}]$ such that

$({s_{A,n}}+\tau_{k,j,i,n})^{i}e^{\frac{\tau_{k,j,i,n}}{2}}e^{(1-\frac{k}{2}){s_{A,n}}}A^{k-1}(k-j)\tilde{C}_{k,j,i}B_{n}^{k-1-j}=\delta_{0}.$

Since the function $\tau\mapsto({s_{A,n}}+\tau)^{i}e^{\frac{\tau}{2}}$ is
increasing, we clearly have from (5.5) that $\tau_{k,j,i,n}\to\infty$ as
$n\to\infty$, and

$\tau_{k,j,i,n}\sim[k-2+2\theta(k-1-j)]{s_{A,n}}\mbox{ as }n\to\infty$

(note that the coefficient of ${s_{A,n}}$ is positive, since $\theta\geq 0$
and $k\geq m\geq 6$).
Since $6\leq m\leq k\leq M$ and $\theta\in[0,\frac{1}{2}]$, it follows that
$k-2+2\theta(k-1-j)\in[k-2,2k-3]\subset[m-2,2M-3]$, hence

$\tau_{k,j,in}\leq(2M-2){s_{A,n}}$

for $n$ large enough. Thus, we will consider $\tau\in[0,(2M-2){s_{A,n}}]$. On
that interval, we see that the resonant function
$({s_{A,n}}+\tau)^{i}e^{\frac{\tau}{2}}$ is comparable to a pure exponential
function, in the sense that

$\forall\tau\in[0,(2M-2){s_{A,n}}],\;\;{s_{A,n}}^{i}e^{\frac{\tau}{2}}\leq({s_{A,n}}+\tau)^{i}e^{\frac{\tau}{2}}\leq(2M-2)^{i}{s_{A,n}}^{i}e^{\frac{\tau}{2}}.$

Thus, the codominance property can be checked at $\tau=0$, as we did in Case
1.

Consider then two different $(k,j,i)$ and $(k^{\prime},j^{\prime},i^{\prime})$
in $H_{2}$, and let us show that either
$({s_{A,n}}+\tau)^{i}e^{\frac{\tau}{2}}e^{(1-\frac{k}{2}){s_{A,n}}}B_{n}^{k-1-j}$
or
$({s_{A,n}}+\tau)^{i^{\prime}}e^{\frac{\tau}{2}}e^{(1-\frac{k^{\prime}}{2}){s_{A,n}}}B_{n}^{k^{\prime}-1-j^{\prime}}$
dominates the other, for $n$ large and $\tau=0$, which is legitimate, from the
reduction we have just proved above. Note first from (5.5) that

${s_{A,n}}^{i}e^{(1-\frac{k}{2}){s_{A,n}}}B_{n}^{k-1-j}={s_{A,n}}^{i}e^{(1-\frac{k}{2}-(k-1-j)\theta+o(1))){s_{A,n}}},$

as $n\to\infty$, with a similar estimate for
$(k^{\prime},j^{\prime},i^{\prime})$.
If $(k,j)\neq(k^{\prime},j^{\prime})$, then
$(k,k-j)\neq(k^{\prime},k^{\prime}-j^{\prime})$, hence
$1-\frac{k}{2}-(k-1-j)\theta\neq
1-\frac{k^{\prime}}{2}-(k^{\prime}-1-j^{\prime})\theta$, since $\theta$ is not
rational. Taking $n$ large enough, we see that one term dominates the other.
Now, if $(k,j)=(k^{\prime},j^{\prime})$, then $i\neq i^{\prime}$, and the two
terms are different by the power of ${s_{A,n}}$ (their ratio is exactly
${s_{A,n}}^{i-i^{\prime}}$), and this implies that one term dominates the
other. Thus, we see that the non-codominance property holds in Case 2 too.
Since the set $H_{2}$ is finite and non empty, we may consider
$(\bar{k},\bar{j},\bar{i})\in H_{2}$ such that

$v_{b,1,0}({s_{A,n}}+\tau)\sim({s_{A,n}}+\tau)^{\bar{i}}e^{\frac{\tau}{2}}(\bar{k}-\bar{j})\tilde{C}_{\bar{k},\bar{j},\bar{i}}A^{\bar{k}-1}B_{n}^{\bar{k}-1-j}e^{(1-\frac{\bar{k}}{2}){s_{A,n}}}$

as $n\to\infty$, with $\tilde{C}_{\bar{k},\bar{j},\bar{i}}\neq 0$. From the
growth factor $e^{\frac{\tau}{2}}$, the coordinate $v_{b,1,0}({s_{A,n}}+\tau)$
will grow, and a contradiction follows as usual.

Of course, our argument in Cases 1 and 2 is formal, however, it can be made
rigorous as usual, like we did at the end of Part 1 of Subsection 4.2 in the
case where $m=4$. This finishes the proof of item (i) in Proposition 5.1.

\- Proof of item (ii) of Proposition 5.1.

The result comes from a small modification of the argument of the proof of
item (i). Assume by contradiction that $\theta$ defined in (5.5) is zero. All
the argument of the proof of item (i) holds here, and we naturally fall in
Case 1. In particular (5.11) holds and the finite set $H_{1}$ defined in
(5.12) is non empty. It remains just to check the non dominance property
stated in Lemma 5.2. Let us consider $(k,j)$ and $(k^{\prime},j^{\prime})$
with $C_{k,j}C_{k^{\prime},j^{\prime}}\neq 0$, $m\leq k,k^{\prime}\leq 2m-3$,
$0\leq j\leq k-1$, $0\leq j^{\prime}\leq k^{\prime}-1$, and
$(k,j)\neq(k^{\prime},j^{\prime})$, and show that either
$(k-j)C_{k,j}A^{k-1}B_{n}^{k-1-j}e^{(1-\frac{k}{2}){s_{A,n}}}$ or
$(k^{\prime}-j^{\prime})C_{k^{\prime},j^{\prime}}A^{k^{\prime}-1}B_{n}^{k^{\prime}-1-j^{\prime}}e^{(1-\frac{k^{\prime}}{2}){s_{A,n}}}$
dominates the other .
If $k\neq k^{\prime}$, then this is clear from (5.14).
If $k=k^{\prime}$, then, recalling that $(k,j)\neq(k^{\prime},j^{\prime})$, we
necessarily see that $j\neq j^{\prime}$, hence the power of $B_{n}$ is not the
same in the two terms. Since $B_{n}\to 0$ from (5.4), one term dominates the
other. Thus, Lemma 5.2 holds here too, and one can carry on the argument of
the proof of item (i) to derive that (5.13) holds, which yields a
contradiction from the exponential factor. Thus, $\theta\neq 0$.

\- Proof of item (iii) of Proposition 5.1.

In this step, we explain how to derive item (iii) of Proposition 5.1 from the
proof of item (i).

As announced earlier, our argument is a small adaptation of the argument
already used for the proof of item (i). The key idea for the adaptation was
already mentioned in the remark following the proof of Lemma 5.2: having a non
rational $\theta$ is a too strong condition to guarantee non-codominance.
according to that remark and to the two cases mentioned in the proof of item
(i), we immediately see that $\theta$ should avoid the following two sets, in
order for the contradiction argument to work:

$\displaystyle E_{1}=[0,\frac{m-2}{2(m-1)}]\cap$
$\displaystyle\\{\frac{k^{\prime}-k}{2[(k-j)-(k^{\prime}-j^{\prime})]}\;\;|\;\;m\leq
k,k^{\prime}\leq 2m-3,$ (5.18) $\displaystyle 0\leq j\leq k-1,\;\;0\leq
j^{\prime}\leq k^{\prime}-1,(k,j)\neq(k^{\prime},j^{\prime})\mbox{ and
}k-j\neq k^{\prime}-j^{\prime}\\}.$ $\displaystyle
E_{2}=[\frac{m-2}{2(m-1)},\frac{1}{2}]\cap$
$\displaystyle\\{\frac{k^{\prime}-k}{2[(k-j)-(k^{\prime}-j^{\prime})]}\;\;|\;\;m\leq
k,k^{\prime}\leq 2m-2,$ (5.19) $\displaystyle 0\leq j\leq k-1,\;\;0\leq
j^{\prime}\leq k^{\prime}-1,(k,j)\neq(k^{\prime},j^{\prime})\mbox{ and
}k-j\neq k^{\prime}-j^{\prime}\\}.$

This concludes the proof of Proposition 5.1. ∎

Step 3: Polynomial corrections to the exponential decay of $B_{n}$.

In Step 2, we showed that

$B_{n}=e^{-\theta{s_{A,n}}}\varphi_{n}\mbox{ where
}\varphi_{n}=e^{o({s_{A,n}})}\mbox{ as }n\to\infty,$ (5.20)

for some $\theta\in(0,\frac{1}{2}]\cap\mathbbm{Q}$ enjoying only a finite
number of values. As one recalls from the proof, our starting point in the
proof lays in the Taylor expansion of the solution $v_{0}(y,s)$ provided by
Proposition 1. As that proposition allows the existence of “resonant” terms,
of the type $s^{i}e^{(1-\frac{k}{2})s}$, it is natural to expect that the
$o({s_{A,n}})$ in (5.20) is of order $\alpha\log{s_{A,n}}$ for some
$\alpha\in\mathbbm{R}$, resulting in $\varphi_{n}\sim{s_{A,n}}^{\alpha}$. This
is precisely the aim of this step. Let us then assume that for up some
subsequence denoted the same, we have

$\frac{\log\varphi_{n}}{\log{s_{A,n}}}\to\alpha\in[-\infty,+\infty].$ (5.21)

We claim the following:

######  Proposition 5.3.

(i) If $\theta\in[0,\frac{m-2}{2(m-1)})$, then $\alpha=0$.
(ii) If $\theta\in[\frac{m-2}{2(m-1)},\frac{1}{2}]$, then $\alpha\in E_{3}$
defined by

$\displaystyle
E_{3}\equiv\left\\{\frac{i^{\prime}-i}{k-j-(k^{\prime}-j^{\prime})}\;|\;m\leq
k,k^{\prime}\leq M,\;0\leq j\leq k-1,\;0\leq j^{\prime}\leq
k^{\prime}-1,\;\right.$ (5.22) $\displaystyle\left.0\leq i\leq i_{k},\;0\leq
i^{\prime}\leq
i_{k^{\prime}},(k,j,i)\neq(k^{\prime},j^{\prime},i^{\prime})\mbox{ and
}k-j\neq k^{\prime}-j^{\prime}\right\\},$

where $M$ is defined in (5.15) and $i_{k}$ right after (5.17).

###### Proof.

We proceed in 2 steps:
\- We first show that $\alpha$ is finite.
\- Then, we show that $\alpha$ enjoys a finite number of rational values.

\- Proof of the fact that $\alpha$ is finite.

Let us assume by contradiction that

$\mbox{either }\frac{\log\varphi_{n}}{\log{s_{A,n}}}\to-\infty\mbox{ or
}\frac{\log\varphi_{n}}{\log{s_{A,n}}}\to\infty\mbox{ as }n\to\infty.$ (5.23)

for a subsequence still denoted the same. In particular, this implies that

$\mbox{either }\varphi_{n}\to 0\mbox{ or }\varphi_{n}\to\infty,\mbox{ as
}n\to\infty.$ (5.24)

Our idea is to follow the pattern of the proof of item (i) of Proposition 5.1,
where we proved that $\theta$ is rational. Naturally, we consider the two
cases mentioned in that step.

Case 1: $\theta\in[0,\frac{m-2}{2(m-1)})$. Even though $\theta$ is rational
this time, the expansion (5.11) remains valid with the same finite set $H_{1}$
defined in (5.12), which is non empty, for the same reason. If we can show the
non-codominance property as in Lemma 5.2, then, we are done. Let us then prove
that lemma, in this new setting. Of course, we need a different argument,
since $\theta$ is rational this time, and the issue concerns the following
order term, involving the limit $\alpha$ defined in (5.21). Consider then two
terms in the expansion (5.11) of $v_{b,1,0}({s_{A,n}}+\tau)$, say
$(k-j)C_{k,j}A^{k-1}B_{n}^{k-1-j}e^{(1-\frac{k}{2}){s_{A,n}}}$ and
$(k^{\prime}-j^{\prime})C_{k^{\prime},j^{\prime}}A^{k^{\prime}-1}B_{n}^{k^{\prime}-1-j^{\prime}}e^{(1-\frac{k^{\prime}}{2}){s_{A,n}}}$
with $C_{k,j}C_{k^{\prime},j^{\prime}}\neq 0$, $m\leq k,k^{\prime}\leq 2m-3$,
$0\leq j\leq k-1$, $0\leq j^{\prime}\leq k^{\prime}-1$, and
$(k,j)\neq(k^{\prime},j^{\prime})$. By definition (5.20) of $\varphi_{n}$, we
see that

$(k-j)C_{k,j}A^{k-1}B_{n}^{k-1-j}e^{(1-\frac{k}{2}){s_{A,n}}}=(k-j)C_{k,j}A^{k-1}e^{(1-\frac{k}{2}-(k-1-j)\theta)){s_{A,n}}}\varphi_{n}^{k-1-j},$
(5.25)

with a similar expansion for $(k^{\prime},j^{\prime})$.
If $1-\frac{k}{2}-(k-1-j)\theta\neq
1-\frac{k^{\prime}}{2}-(k^{\prime}-1-j^{\prime})\theta$, recalling that
$\varphi_{n}=e^{o({s_{A,n}})}$ as stated in (5.20), we see that one of the two
terms dominates the other.
Now, if
$1-\frac{k}{2}-(k-1-j)\theta=1-\frac{k^{\prime}}{2}-(k^{\prime}-1-j^{\prime})\theta$
(and this may occur since $\theta$ is rational), recalling that
$(k,j)\neq(k^{\prime},j^{\prime})$ we necessarily have $k-1-j\neq
k^{\prime}-1-j^{\prime}$. In other words, the power of $\varphi_{n}$ is not
the same in the two terms. Since $\varphi_{n}\to 0$ or $\varphi_{n}\to\infty$,
as stated in (5.24), we see from the description (5.25) that one of the two
terms has to dominate the other. Thus, the non-codominance property holds, and
a contradiction follows as in Case 1 of the proof of item (i) of Proposition
5.1.

Case 2: $\theta\in[\frac{m-2}{2(m-1)},\frac{1}{2}]$. Again, the argument of
Case 2 in the proof of item (i) of Proposition 5.1 holds, and we may derive an
expansion of $v_{b,1,0}({s_{A,n}}+\tau)$ like in (5.16), with “resonant”
terms. The finite set $H_{2}$ is non empty, for the same reason. If we can
show the non-codominance property as before, then we are done.
Consider then two different $(k,j,i)$ and $(k^{\prime},j^{\prime},i^{\prime})$
in $H_{2}$, and let us show that either
$({s_{A,n}}+\tau)^{i}e^{\frac{\tau}{2}}e^{(1-\frac{k}{2}){s_{A,n}}}B_{n}^{k-1-j}$
or
$({s_{A,n}}+\tau)^{i^{\prime}}e^{\frac{\tau}{2}}e^{(1-\frac{k^{\prime}}{2}){s_{A,n}}}B_{n}^{k^{\prime}-1-j^{\prime}}$
dominates the other, for $n$ large. From the reduction we showed in Case 2 of
the proof of item (i) of Proposition 5.1 above, it is enough to check the
dominance at $\tau=0$. Note first from (5.20) that

${s_{A,n}}^{i}e^{(1-\frac{k}{2}){s_{A,n}}}B_{n}^{k-1-j}={s_{A,n}}^{i}e^{(1-\frac{k}{2}-(k-1-j)\theta)){s_{A,n}}}\varphi_{n}^{k-1-j},$
(5.26)

as $n\to\infty$, with a similar estimate for
$(k^{\prime},j^{\prime},i^{\prime})$.
If $1-\frac{k}{2}-(k-1-j)\theta\neq
1-\frac{k^{\prime}}{2}-(k^{\prime}-1-j^{\prime})\theta$, recalling that
$\varphi_{n}=e^{o({s_{A,n}})}$ from (5.20), we see that one term dominates the
other.
Assume then that
$1-\frac{k}{2}-(k-1-j)\theta=1-\frac{k^{\prime}}{2}-(k^{\prime}-1-j^{\prime})\theta$.
If $k-j=k^{\prime}-j^{\prime}$, then $k=k^{\prime}$, and since
$(k,j,i)\neq(k^{\prime},j^{\prime},i^{\prime})$, it follows that $i\neq
i^{\prime}$. From the expression (5.26), we see that the powers of ${s_{A,n}}$
are different, hence, one term dominates the other.
Now, if $k-j\neq k^{\prime}-j^{\prime}$, making the ratio between the two
terms, we find
${s_{A,n}}^{i-i^{\prime}}\varphi_{n}^{k-j-(k^{\prime}-j^{\prime})}$. Using
(5.20) and (5.24), we see that one term dominates the other.

Of course, our argument in Cases 1 and 2 is formal, however, it can be made
rigorous as usual, like we did at the end of Part 1 of Subsection 4.2 in the
case $m=4$. Thus, we have just proved that the parameter $\alpha$ defined in
(5.21) is finite.

\- Conclusion of the proof of Proposition 5.3.

In the previous step, we assumed that $\alpha=\pm\infty$ and reached a
contradiction. In fact, a careful check reveals that the contradiction can be
reached in Case 1, for any $\alpha\neq 0$, whereas in Case 2, we simply need
$\alpha$ to avoid the set $E_{3}$ defined in (5.22). This concludes the proof
of Proposition 5.3. ∎

Step 4: One further refinement in the behavior of $B_{n}$

So far, thanks to Propositions (5.1) and (5.3), we have proved that

$B_{n}=\exp(-\theta{s_{A,n}}){s_{A,n}}^{\alpha}\psi_{n}\mbox{ for some
}\psi_{n}={s_{A,n}}^{o(1)}$ (5.27)

as $n\to\infty$. This is the aim of this step:

######  Proposition 5.4.

Up to a subsequence, $\psi_{n}$ converges to some $L>0$, where $L$ enjoys a
finite number of values, all solutions of polynomials whose coefficients
depend on the coefficients that arise in the Taylor expansion of $v_{0}(y,s)$.

###### Proof.

Here again, we crucially use the geometric transformation introduced in Step 2
of Section 3. We proceed in 2 steps:
\- In Step (i), we show that $\psi_{n}$ is bounded away from $0$ and from
infinity.
\- In Step (ii), we show that up to a subsequence, $\psi_{n}$ converges to
some $L>0$, which is a solution of a polynomial whose coefficients depend on
$C_{k,j}$.

\- Step (i): $\psi_{n}$ is bounded away from $0$ and from infinity.

We proceed by contradiction, and assume that for a subsequence (still denoted
the same), we have

$\psi_{n}\to 0\mbox{ or }\psi_{n}\to\infty\mbox{ as }n\to\infty.$ (5.28)

As before, we follow the strategy of the proof of item (i) of Proposition 5.1,
with its two cases.
Starting by Case 1, where $\theta\in[0,\frac{m-2}{2(m-1)})$, we still see that
(5.11) holds, with $H_{1}$ which is still a non empty finite set. It remains
only to prove the non codominance property. Consider then two terms in the
expansion (5.11) of $v_{b,1,0}({s_{A,n}}+\tau)$, say
$(k-j)C_{k,j}A^{k-1}B_{n}^{k-1-j}e^{(1-\frac{k}{2}){s_{A,n}}}$ and
$(k^{\prime}-j^{\prime})C_{k^{\prime},j^{\prime}}A^{k^{\prime}-1}B_{n}^{k^{\prime}-1-j^{\prime}}e^{(1-\frac{k^{\prime}}{2}){s_{A,n}}}$
with $C_{k,j}C_{k^{\prime},j^{\prime}}\neq 0$, $m\leq k,k^{\prime}\leq 2m-3$,
$0\leq j\leq k-1$, $0\leq j^{\prime}\leq k^{\prime}-1$, and
$(k,j)\neq(k^{\prime},j^{\prime})$. Using (5.27), and recalling that
$\alpha=0$ from Proposition 5.3, we see that

$(k-j)C_{k,j}A^{k-1}B_{n}^{k-1-j}e^{(1-\frac{k}{2}){s_{A,n}}}=(k-j)C_{k,j}A^{k-1}e^{(1-\frac{k}{2}-(k-1-j)\theta){s_{A,n}}}\psi_{n}^{k-1-j},$
(5.29)

with a similar expansion for $(k^{\prime},j^{\prime})$.
If $1-\frac{k}{2}-(k-1-j)\theta\neq
1-\frac{k^{\prime}}{2}-(k^{\prime}-1-j^{\prime})\theta$, recalling that
$\psi_{n}={s_{A,n}}^{o(1)}=e^{o(\log{s_{A,n}})}$ as stated in (5.27), we see
that one of the two terms dominates the other.
Now, if
$1-\frac{k}{2}-(k-1-j)\theta=1-\frac{k^{\prime}}{2}-(k^{\prime}-1-j^{\prime})\theta$,
recalling that $(k,j)\neq(k^{\prime},j^{\prime})$ we necessarily have
$k-1-j\neq k^{\prime}-1-j^{\prime}$. In other words, the power of $\psi_{n}$
is not the same in the two terms. Since $\psi_{n}\to 0$ or
$\psi_{n}\to\infty$, as stated in (5.28), we see from the description (5.29)
that one of the two terms has to dominate the other. Thus, co-dominance holds
in this context, and a contradiction follows as in Case 1 of the proof of item
(i) of Proposition 5.1.
Now, moving to Case 2, where $\theta\in[\frac{m-2}{2(m-1)},\frac{1}{2}]$, we
may consider two terms in the expansion (5.16) of $v_{b,1,0}({s_{A,n}}+\tau)$,
say
$({s_{A,n}}+\tau)^{i}e^{\frac{\tau}{2}}e^{(1-\frac{k}{2}){s_{A,n}}}B_{n}^{k-1-j}$
and
$({s_{A,n}}+\tau)^{i^{\prime}}e^{\frac{\tau}{2}}e^{(1-\frac{k^{\prime}}{2}){s_{A,n}}}B_{n}^{k^{\prime}-1-j^{\prime}}$,
where $(k,j,i)$ and $(k^{\prime},j^{\prime},i^{\prime})$ belong to $H_{2}$
defined in (5.17), and prove that one dominates the other. From the reduction
we did in Case 2 of the proof of item (i) of Proposition 5.1, we may check the
dominance only at $\tau=0$. Using (5.27), we see that

${s_{A,n}}^{i}e^{(1-\frac{k}{2}){s_{A,n}}}B_{n}^{k-1-j}=e^{(1-\frac{k}{2}-(k-1-j)\theta){s_{A,n}}}{s_{A,n}}^{i+\alpha(k-1-j)}\psi_{n}^{k-1-j}$

as $n\to\infty$, with a similar estimate for
$(k^{\prime},j^{\prime},i^{\prime})$.
If $1-\frac{k}{2}-(k-1-j)\theta\neq
1-\frac{k^{\prime}}{2}-(k^{\prime}-1-j^{\prime})\theta$, recalling that
$\psi_{n}={s_{A,n}}^{o(1)}=e^{o(\log{s_{A,n}})}$ as stated in (5.27), we see
that one of the two terms dominates the other.
Assume then that
$1-\frac{k}{2}-(k-1-j)\theta=1-\frac{k^{\prime}}{2}-(k^{\prime}-1-j^{\prime})\theta$.
If $i+\alpha(k-1-j)\neq i^{\prime}+(k^{\prime}-1-j^{\prime})\alpha$, then,
using again the fact that $\psi_{n}={s_{A,n}}^{o(1)}=e^{o(\log{s_{A,n}})}$, we
see that the power of ${s_{A,n}}$ is different between the two terms, hence
one term dominates the other.
Now, if $i+\alpha(k-1-j)=i^{\prime}+(k^{\prime}-1-j^{\prime})\alpha$, then,
necessarily $k-1-j\neq k^{\prime}-1-j^{\prime}$, otherwise,
$(k,j,i)=(k^{\prime},j^{\prime},i^{\prime})$. Therefore, the power of
${s_{A,n}}$ is the same in the two terms, unlike the power of $\psi_{n}$.
Recalling that $\psi_{n}\to 0$ or $\psi_{n}\to\infty$, we see that one term
dominates the other.

Step (ii): $\psi_{n}$ converges (up to a subsequence)

From Step (i), we may assume that for some subsequence (still denoted the
same), we have

$\psi_{n}\to A^{2\theta}L\mbox{ as }n\to\infty$ (5.30)

for some $L>0$ (we write the limit in (5.30) as $A^{2\theta}L$ and not $L$,
since in this form, $L$ will be shown to be a solution of a polynomial whose
coefficients are independant of $A$).
We will distinguish two cases as in the proof of item (i) of Proposition 5.1.
Case 1: $\theta\in[0,\frac{m-2}{2(m-1)})$. Note that $\alpha=0$ by Proposition
5.3. In this case, we see that (5.11) still holds, with $H_{1}$ which is still
a non empty finite set. However, this time, we will have codominance, as we
show in the following. Since the number of terms is finite in the expansion
(5.11), we may consider $(\bar{k},\bar{j})\in H_{1}$ defined in (5.12) such
that the term corresponding to this parameter dominates all the others. Using
(5.11) and (5.27), this term reads

$e^{(1-\frac{\bar{k}}{2}){s_{A,n}}}A^{\bar{k}-1}(\bar{k}-\bar{j})C_{\bar{k},\bar{j}}B_{n}^{\bar{k}-1-\bar{j}}=A^{\bar{k}-1}C_{\bar{k},\bar{j}}(\bar{k}-\bar{j})e^{(1-\frac{\bar{k}}{2}-(\bar{k}-1-\bar{j})\theta){s_{A,n}}}\psi_{n}^{\bar{k}-1-\bar{j}}.$
(5.31)

As we wrote earlier, we expect here to have codominance. Let us then
characterize the set $\bar{E}$ of all $(k,j)\in H_{1}$ such that the
corresponding term is of the same size as the term corresponding to
$(\bar{k},\bar{j})$. From (5.31) and (5.30), this means that

$1-\frac{\bar{k}}{2}-(\bar{k}-1-\bar{j})\theta=1-\frac{k}{2}-(k-1-j)\theta.$
(5.32)

This way, we may keep only the dominant terms in (5.11), namely those coming
from $\bar{E}$, and write

$v_{b,1,0}({s_{A,n}}+\tau)=e^{\frac{\tau}{2}}\left(\sum_{(k,j)\in\bar{E}}e^{(1-\frac{k}{2}){s_{A,n}}}A^{k-1}(k-j)C_{k,j}B_{n}^{k-1-j}+O({s_{A,n}}e^{(2-m){s_{A,n}}})\right).$

Using (5.31) and (5.32), together with the convergence (5.30), we derive that

$\displaystyle v_{b,1,0}({s_{A,n}}+\tau)$
$\displaystyle=e^{\frac{\tau}{2}}e^{(1-\frac{\bar{k}}{2}-(\bar{k}-1-\bar{j})\theta){s_{A,n}}}\left(\sum_{(k,j)\in\bar{E}}A^{k-1}(k-j)C_{k,j}(A^{2\theta}L)^{k-1-j}+o(1)\right)$
$\displaystyle=e^{\frac{\tau}{2}}e^{(1-\frac{\bar{k}}{2}-(\bar{k}-1-\bar{j})\theta){s_{A,n}}}A^{\bar{k}-1+2\theta(\bar{k}-1-\bar{j})}\left(\sum_{(k,j)\in\bar{E}}(k-j)C_{k,j}L^{k-1-j}+o(1)\right)$

as $n\to\infty$. From the growth factor $e^{\frac{\tau}{2}}$, this implies
that

$\sum_{(k,j)\in\bar{E}}(k-j)C_{k,j}L^{k-1-j}=0.$

Since $L\neq 0$ and $C_{k,j}\neq 0$, for all $(k,j)\in\bar{E}$ (remember that
$\bar{E}\subset H_{1}$ defined in (5.12)), this sum contains at least two
terms, and this is precisely the desired polynomial relation. Remember that
$\bar{E}\neq\emptyset$, since it contains $(\bar{k},\bar{j})$, and that
$C_{k,j}\neq 0$, for any $(k,j)\in\bar{E}$, since $\bar{E}\subset H_{1}$
defined in (5.12). Note that the degree of this polynomial is bounded by
$k-1\leq 2m-4$ by definition (5.12) of $H_{1}$, hence, we have at most $2m-4$
possible values for $L$.
Case 2: $\theta\in[\frac{m-2}{2(m-1)},\frac{1}{2}]$. We are then in the
framework of Case 2 of the proof of item (i) of Proposition 5.1 above. In
particular, the finite set $H_{2}$ is still non empty, for the same reason.
However, we may have codominance in this context. Since the number of terms is