Frank Merle
email: merle@math.u-cergy.fr

Pierre Raphaël
email: praphael@maths.cam.ac.uk

AND Igor Rodnianski

Jeremie Szeftel
email: jeremie.szeftel@upmc.fr

ON THE IMPLOSION OF A COMPRESSIBLE FLUID I: Smooth self-similar inviscid profiles

In this paper and its sequel, we construct a set of finite energy smooth initial data for which the corresponding solutions to the compressible three dimensional Navier-Stokes and Euler equations implode (with infinite density) at a later time at a point, and completely describe the associated formation of singularity. This paper is concerned with existence of smooth self-similar profiles for the barotropic Euler equations in dimension d ≥ 2 with decaying density at spatial infinity. The phase portrait of the nonlinear ode governing the equation for spherically symmetric self-similar solutions has been introduced in the pioneering work of Guderley. It allows to construct global profiles of the selfsimilar problem, which however turn out to be generically non-smooth across the associated acoustic cone. In a suitable range of barotropic laws and for a sequence of quantized speeds accumulating to a critical value, we prove the existence of non-generic C ∞ self-similar solutions with suitable decay at infinity. The C ∞ regularity is used in a fundamental way in our companion paper (part II) in the analysis of the associated linearized operator and leads, in turn, to the construction of finite energy blow up solutions of the compressible Euler and Navier-Stokes equations in dimensions d = 2, 3.

1. Introduction 1.1. Setting of the problem 1.2. The self-similar equation 1.3. Emden transform and Guderley's phase portrait 1.4. Regularity at P 2 and the reconnection problem 1.5. Statement of the main result 1.6. Further qualitative properties of the solution 1.7. Organization of the paper 1.8. Acknowledgements 2. The geometry of the phase portrait 2.1. Roots of ∆, ∆ 1 ∆ 2 2.2. Double roots 2.3. Relative positions of P 2 , P 3 , P 5 2.4. Slopes of w 2 (σ) and w - 2 (σ) at P 2 3. General properties of the dynamical system (1.9) 3.1. The spherically symmetric solution emerging from the origin 3.2. Solutions crossing red between P 2 and P o 3.3. Diagonalized system at P 2 3.4. Integral curves passing through P 2 4. Renormalization of the flow near P Setting of the problem. In this paper and its sequel [START_REF] Merle | On the implosion of a three dimensional compressible fluid II: Singularity formation[END_REF], we construct a set (of finite co-dimension) of finite energy smooth initial data for which the corresponding solutions to the compressible three dimensional Navier-Stokes and Euler equations implode (with infinite density) at a point, and completely describe the associated formation of singularity. The first paper will deal exclusively with the isentropic compressible Euler equations in dimension d ≥ 2, y ∈ R d , (Euler)

∂ t ρ + ∇ • (ρu) = 0 ρ∂ t u + ρu • ∇u + ∇p = 0 p = γ-1
γ ρ γ ρ(t, y) > 0.

(1.1) Our goal is the construction of a family of smooth, global in space, self-similar profiles corresponding to spherically symmetric solutions of (1.1) which arise from smooth initial data and blow up at a chosen point (T, 0). We are specifically interested in solutions which do not exhibit growth at spatial infinity. In fact, our solutions will obey lim |y|→+∞ (ρ(t, y), u(t, y)) = 0.

(

Existence of self-similar solutions with spherical symmetry for the Euler equation with a continuum of admissible blow up speeds has been known since the pioneering work of Guderley [START_REF] Guderley | Starke kugelige und zylindrische Verdichtungsstösse in der Nähe des Kugelmittelpunktes bzw. der Zylinderachse[END_REF] and Sedov [START_REF] Sedov | Similarity and dimensional methods in mechanics[END_REF]. However, the known solutions are either nonglobal or non-smooth (in self-similar variables). The former solutions are ubiquitous in the physics literature and describe physical phenomena of denotation, implosion, flame propagation etc. and by design contain a shock, discontinuously connecting a smooth solution to another state. We refer to them as non-global even if, more accurately, they should be called discontinuous, to differentiate them from the second class of solutions -non-smooth ones. The latter appear to be much less known but can be easily constructed from the phase portrait analysis introduced in [START_REF] Guderley | Starke kugelige und zylindrische Verdichtungsstösse in der Nähe des Kugelmittelpunktes bzw. der Zylinderachse[END_REF].

The solutions are global in a sense that they connect the behavior (1.2) at infinity with the regular behavior at the center of symmetry. In the process, they have to cross the so called sonic line -a point on it representing a backward acoustic cone from the singular point in the original variables. It turns out that generically this crossing is non-smooth. The regularity of the associated solution depends on the values of various parameters (dimension, equation of state, scaling) but the standard Lyapunov analysis suggests that it is always finite (although it can be made arbitrarily high for a particular choice of parameters.) In principle, such solutions, together with the finite speed of propagation, immediately lead to the existence of finite energy well localized blow up solutions of the Euler equations. Note however that for the reason explained above these solutions do not arise from smooth initial data. But, in principle, such solutions would correspond to a formation of a strong singularity, in which the solution concentrates at a point and the density blows up. In the previous rigorous work on compressible Euler and Navier-Stokes equations such phenomenon was not known and the singularity formation was tied to either formation of shocks (in the Euler case) or to qualitative arguments based on the virial type arguments (for both Euler and Navier-Stokes). For a more detailed discussion see our second paper in the series [START_REF] Merle | On the implosion of a three dimensional compressible fluid II: Singularity formation[END_REF]. The aim of this paper is to show the existence of non-generic C ∞ self-similar solutions for quantized values of the blow up speed in the vicinity of a certain critical value. This is contrary to the Lyapunov analysis which would suggest that it might never be possible. In the companion paper [START_REF] Merle | On the implosion of a three dimensional compressible fluid II: Singularity formation[END_REF], we will use these solutions as the leading order blow up profiles for the compressible Navier-Stokes equation (as well as its inviscid Euler limit) to produce (a finite co-dimension set of) blowing up solutions arising from smooth initial data. The C ∞ regularity of the profile is needed not only for the regularity of initial data but much more importantly, in fact, crucially, for the stability analysis. The stability analysis itself is needed not only to establish existence of a whole finite co-dimensional manifold (in the moduli space of initial data) of blowing up solutions but also for the existence result of even just one such solution. These profiles are merely approximate solutions for the Navier-Stokes problem and their stability and thus their C ∞ regularity are essential to ensuring that the approximation holds until the blow up time.

Remark 1.1. In [START_REF] Merle | On blow up for the energy super critical defocusing NLS[END_REF], we will also use these C ∞ self-similar solutions as the leading order profiles to produce blowing up solutions arising from smooth initial data of finite energy for the energy super-critical defocusing nonlinear Schrödinger equations.

Self-similar motion has long been recognized as an important concept in hydrodynamics (see e.g. [START_REF] Sedov | Similarity and dimensional methods in mechanics[END_REF] and the references therein). It could be said that it originated in the dimensional analysis of Reynolds and eventually crystallized as a model for both the simplest and universal behaviors in fluid and gas dynamics. The simplicity stems from the fact that the assumption of self-similar motion (together with spherical or cylindrical symmetry in higher dimensions) reduces the Euler equations to a system of ode's. The universality is supported by the ubiquity of self-similar solutions as well as the belief that self-similar motions act as an attractor for many different phenomena in hydro/gas dynamics. In that respect, two types of selfsimilar motions have been discussed in the physics literature, [START_REF] Zel'dovich | Physics of shock waves and high temperature hydrodynamic phenomena[END_REF]. In the first kind, all self-similar parameters are determined from the dimensional analysis, while in the second kind, an undetermined (free) parameter is fixed by the boundary conditions or some other physical requirements on the solution. Self-similar solutions have been extensively used and analyzed in the study of problems involving detonation and implosion waves, combustion, reflected shocks, etc. The current approach has been pioneered by Guderley [4] and has been given numerous treatments, see [START_REF] Courant | Supersonic flow and shock waves[END_REF][START_REF] Sedov | Similarity and dimensional methods in mechanics[END_REF]. In that approach, the study of self-similar (spherically symmetric) solutions of the Euler equations is reduced to the system

dw dx = -∆ 1 ∆ dσ dx = -∆ 2 ∆ (1.3)
where ∆, ∆ 1,2 are polynomials in w, σ and the similarity variable Z = e x is related to the original (t, y) via

Z = |y| (T -t) 1 r
The 1/r parameter is the similarity exponent (free parameter) and is the inverse of what in this paper we call speed r. The equations (1.3) are an autonomous system of ode's. Its phase portrait and, specifically, the set where ∆, ∆ 1 , ∆ 2 vanish determine the qualitative properties of all solutions. A self-similar profile is a stationary solution to (1.6):

(r -1)ρ + Λρ + ∇ • (ρû) = 0 (r -1)û + Λû + û • ∇û + ∇(ρ γ-1 ) = 0 (1.7)

which produces a blow up solution for (1.1) with the rate of concentration

λ(t) = λ 0 (T -t) 1 r , ν(t) = ν 0 r(T -t).
1.3. Emden transform and Guderley's phase portrait. In the pioneering work [START_REF] Guderley | Starke kugelige und zylindrische Verdichtungsstösse in der Nähe des Kugelmittelpunktes bzw. der Zylinderachse[END_REF], see also [START_REF] Sedov | Similarity and dimensional methods in mechanics[END_REF][START_REF] Meyer-Ten-Vehn | Selfsimilar compression waves in gas dynamics[END_REF], all solutions to (1.7) with spherical symmetry are mapped through the Emden transform (ρ(Z))

γ-1 2 = 2 Zσ(x) û(Z) = -Zw(x) Z = e x (1.8) with w the component along the radial outgoing normal vector, onto the autonomous system of nonlinear ode's:

(w -1) dw dx + σ dσ dx + (w 2 -rw + σ 2 ) = 0

σ dw dx + (w -1) dσ dx + σ w d + 1 -r = 0 ⇔ ∆ dw dx = -∆ 1 ∆ dσ dx = -∆ 2 (1.9)
with the explicit

∆ = (w -1) 2 -σ 2 ∆ 1 = w(w -1)(w -r) -d(w -w e )σ 2 ∆ 2 = σ ( + d -1)w 2 -w( + d + r -r) + r -σ 2
(1.10)

and w e = (r -1) d .

(1.11)

The above system can be fully analyzed through the phase portrait in the (σ, w) plane. The shape of the phase portrait is highly dependent on the values of the parameters (r, , d). Let us introduce the following critical speeds which will play a fundamental role in the forthcoming analysis

r * (d, ) = d+ + √ d , r + (d, ) = 1 + d-1 (1+ √ ) 2 , (1.12)
where an explicit computation 1 we will prove that the phase portrait contains five fundamental points (P i ) 1≤i≤5 , see figure 1, figure 2 and section 2, as well as two sonic lines.

Sonic lines w -1 = ±σ. This is exactly the set where ∆ = 0. In the original variables of the Euler equations, the sonic lines correspond to the equation

|y| r(T -t) = -(u ± c) , c = ∂p ∂ρ (1.16)
where c is the sound speed and the right hand side is a function of

|y| λ 0 (T -t) 1 r
. On the other hand, the equation

d|y| dt = -(u ± c)
describes solutions (t, |y|(t)) -acoustical cones (radial characteristics) of the metric associated with a solution of the compressible Euler equations. It then follows that for any point Z * on the sonic line, the set (T -t, Z * λ 0 (T -t)

1 r
) is the backward acoustical cone from the singular point (0, 0). On the phase diagram, the points to the right of the sonic line w -1 = σ will correspond to the (t, y) points in the interior of the acoustical (light) cone, while the points to the left of the sonic line lie in the exterior of the cone. Moreover, the absolute value of characteristic speed u + c -particle velocity plus the sound speed -is smaller in the exterior and larger in the interior of the cone. P 5 point. The point

P 5 = σ 5 = r √ d d + , w 5 = r d +
is an endpoint of the dynamical system (1.9), i.e. ∆ 1 (P 5 ) = ∆ 2 (P 5 ) = 0 ∆(P 5 ) = 0 and the relative position of P 5 with respect to the sonic line is given by w 5 + σ 5 < 1 ⇔ r < r * (d, ).

Points P 2 , P 3 . Trajectories can only cross the sonic line at the triple points, where

∆ = ∆ 1 = ∆ 2 ,
which are (0, 0), (1, 0), (r, 0) and two other points on the sonic line w + σ = 1 which we refer to as P 2 , P 3 and which exist thank to the constraint (1.15). P 6 point. The point P 6 = (w = w e , σ = +∞) is a saddle point at +∞ and corresponds to x = -∞, i.e., Z = 0. P 4 point. The point P 4 = (0, 0) attracts solutions which vanish near x → +∞, i.e., Z → +∞.

A classical analysis of the phase portrait, figure 1, figure 2, yields the following result (see Lemma 3.1 and 3.2 for a proof). Lemma 1.2 (General structure of spherically symmetric self-similar solutions). Assume (1.15). Then,

(1) Solutions near the origin: there exists a unique trajectory of (1.9) which connects P 6 to P 2 . This trajectory corresponds to the unique (up to scaling 2 ) (local) spherically symmetric solution to (1.7) which is C ∞ on the ball |Z| ∈ [0, Z 2 ) with 0 corresponding to P 6 and Z 2 to P 2 .

(2) Solutions near infinity: there exists a one parameter family (indexed by the scaling of (1.7)) of trajectories which connect P 2 to P 4 , and one can fix the scaling so that P 2 is reached at the same Z = Z 2 . The corresponding curve then corresponds to the spherically symmetric solution to (1.7) and is in C ∞ on |Z| ∈ (Z 2 , ∞) with ∞ corresponding to P 4 .

(3) Connection at P 2 : in both cases, P 2 is reached for a finite value of Z, i.e., 0 < Z 2 < ∞, and the solutions constructed in ( 1) and ( 2) can be glued continuously to each other at Z = Z 2 .

In other words, the unique 3 , spherically symmetric solution (ρ(Z), û(Z)) to (1.7), which is smooth at Z = 0, extends to the point Z 2 (which corresponds to P 2 on the phase diagram) where it can be glued to any of the one parameter family of solutions (ρ(Z), û(Z)) defined for [Z 2 , ∞) (which correspond to the trajectories connecting P 2 to P 4 ). This procedure yields spherically symmetric solutions to (1.7) which are C ∞ (R d \{Z = Z 2 }) and vanish as Z → +∞.

1.4. Regularity at P 2 and the reconnection problem. It remains to understand the regularity at P 2 . The above gluing procedure produces a solution with limited C k(r) regularity 4 at the degenerate point P 2 , see Remark 3.7. As it turns out, this limited regularity has a dramatic effect on the spectral structure of the linearized operator for (1.6) close to this self-similar profile, and we do not know how to use these non-C ∞ solutions to produce finite energy imploding solutions to the viscous perturbations of (1.1) studied in our second paper [START_REF] Merle | On the implosion of a three dimensional compressible fluid II: Singularity formation[END_REF].

On the other hand, P 2 is a regular singular point of (1.9). As a result, there always 5 exists one trajectory which is C ∞ at P 2 . Thus, the problem becomes: can we find parameters (d, , r) for which the P 6 -P 2 solution is C ∞ at P 2 and can be glued to a P 2 -P 4 solution which is also C ∞ at P 2 (see figure 1)? Such a solution would produce a C ∞ self-similar profile with vanishing density and velocity as Z → +∞.

The Lyapunov type (linear) analysis at P 2 predicts that the regularity of solutions passing through is determined by the eigenvalues λ ± of the Jacobian matrix. Both λ -< λ + < 0, so that P 2 is a stable node. As a result, all curves through P 2 but one will have limited regularity C λλ + . The exceptional solution is C ∞ but it does 2 note that (1.7) is invariant under the scaling (ρ(Z), û(Z)) → (λ -γ-1 2 ρ(λZ), λ -1 û(λZ)) for any λ > 0. This scaling invariance in Z for (1.7) corresponds to a translation invariance in x at the level of (1.9). 3 up to scale invariance 4 with k(r) ∼ c(d, ) r o (d, )-r as r ↑ r o (d, ). 5 at least away from critical integer values, see Lemma 3.6. not go to P 6 and is thus inadmissible as a global profile. The C λλ + regularity is sharply insufficient for our purposes (linear stability analysis of these solutions as solutions of the Euler equations misses exactly one derivative, see [START_REF] Merle | On blow up for the energy super critical defocusing NLS[END_REF]). Instead, we consider the regime where λ + ↑ 0 (and λ -stays uniformly bounded away from 0 and -∞) which corresponds to a limiting degeneracy of the phase portrait for the critical speed r ↑ r o (d, ) given by (1.14) where a vanishing "o" structure appears to the left of P 2 . We perform a careful nonlinear and global analysis near the eye configuration to show the existence of a discrete sequence of C ∞ solutions as r = r n ↑ r o .

1.5. Statement of the main result. Our main result in this paper is the existence of such C ∞ profiles for quantized values of the speed r near the critical speed whose value depends on given by (1.4). We introduce the set O d ⊂ (0, +∞) \ {d} defined as the union of the interval (d, +∞) and a subset O * d of (0, d) defined by the condition that ∈ O d for < d if the value of the function ν ∞ ( , d) in (F.10) is > 0. In dimensions d = 2, 3, O d = (0, +∞) \ {d} and for 5 ≤ d ≤ 9 the subset O * d is a finite collection of subintervals and it is non-empty, see (F.9). such that (1.7) with r = r n admits a global C ∞ spherically symmetric solution (ρ(Z), û(Z)) which terminates at P 4 at spatial infinity (i.e. as Z → +∞), see figure 1.

Corollary 1.4. Under the assumptions of Theorem 1.3, the Euler equations (1.1) admit a family of spherically symmetric self-similar solutions, smooth away from the concentration point (T, 0):

ρ(t, y) = 1 (T -t) 2(r-1) r(γ-1) ρ(Z) u(t, y) = 1 (T -t) (r-1) r û(Z) Z = y λ 0 (T -t) 1 r
for any λ 0 > 0 with the asymptotics , for some ρ * > 0. In particular, these solutions decay at infinity but do not have finite energy.

ρ(t, y) = ρ * (1 + o |Z|→+∞ ( 1
The function S ∞ (d, ) appears in the asymptotic analysis of the flow near P 2 . It can be explicitly expressed as a normally convergent series

S ∞ (d, ) = +∞ n=0 u n (d, ), |u n (d, )| ≤ c d, 1 + n 2
where the series u n (d, ) satisfies an explicit though complicated non linear induction relation. The proof of convergence of the series yields the analyticity of the mapping → S ∞ (d, ) in a suitable open set of the complex plane.

More precisely, the function S ∞ (d, ) is in fact composed of two different functions S * ∞ (d, ), defined for the critical value of the speed r * on the subset O * d ⊂ (0, d), and S + ∞ (d, ), defined for the critical value of the speed r + for > d. It turns out that the explicit definitions of these functions allow us to extend them holomorphically. Moreover, since u n (d, ) is given by an explicit induction relation on the coefficients, we can perform an elementary numerical computation of the series. We first give the results in dimension d = 2 and d = 3 which will be used for the study of the compressible Euler and Navier-Stokes equations in [START_REF] Merle | On the implosion of a three dimensional compressible fluid II: Singularity formation[END_REF], see Appendix I. The assertion in (4) allows us to check the non-vanishing condition for small values of only. (1.19)

Numerical claim

The case of higher dimensions will be relevant for the study of the energy super critical defocusing (NLS) equation in [START_REF] Merle | On blow up for the energy super critical defocusing NLS[END_REF], for which the power nonlinearity involves the real number p given by p = 1 + 4 .

(1.20)

We numerically check the non degeneracy S ∞ (d, ) = 0 for a range of dimensions and integer nonlinearities, see Appendix I.

Numerical claim [Numerical study of the zeroes of S ∞ (d, ), case d ≥ 5] Let p( ) be given by (1.20). Then the condition (1.17) holds for (d, p) ∈ {(5, 9), [START_REF] Lazarus | Self-similar solutions for converging shocks and collapsing cavities[END_REF][START_REF] Krieger | Large global solutions for energy supercritical nonlinear wave equations on R 3+1[END_REF], [START_REF] Lepin | Self-similar solutions of a semilinear heat equation[END_REF][START_REF] Guderley | Starke kugelige und zylindrische Verdichtungsstösse in der Nähe des Kugelmittelpunktes bzw. der Zylinderachse[END_REF], [START_REF] Malgrange | Sur le théorème de Maillet[END_REF][START_REF] Courant | Supersonic flow and shock waves[END_REF], (9, 3)}. Comments on the results.

1. The set O d . It is defined by the requirement ν ∞ (d, ) > 0 with ν ∞ explicitly given by the formulas (F.10) for < d and (F.15) for > d. It turns out that ν ∞ > 0 for all ∈ R * + \{d} in dimension d = 2, 3. In higher dimensions, O d also includes the interval > d and its intersection with < d consists of a finite union of open intervals. The latter intersection is non-empty for 5 ≤ d ≤ 9. The condition ν ∞ (d, ) > 0 is convenient for our analysis but is clearly not sharp and could be relaxed. Let us emphasize that (1.18), (1.21) merely provide explicit examples of admissible couples (d, ).

2. The asymptotic analysis r ↑ r o (d, ). The proof of Theorem 1.3 involves a careful renormalization of the flow (1.9) for |r -r o | 1, r < r o . The reason for this choice is the presence of an "eye" of the phase portrait at the left of P 2 : as r ↑ r o , the critical point at the left of P 2 which is P 5 for < d and P 3 for > d, converges to P 2 , and this convergence happens with a suitable ordering of the points P 2 , P 3 , P 5 which depends on whether > d or < d, see Lemma 2.6. The choice of the critical values r + or r * in (1.14) is dictated by this geometry of the phase portrait. The fact that two of the points P 2 , P 3 , P 5 collide at r o induces a degeneracy of the flow at the critical value which is the starting point of a renomalization process in terms of a small parameter b = o r↑r o [START_REF] Bizoń | Self-similar solutions of semilinear wave equations with a focusing nonlinearity[END_REF]. We would however like to stress the fact that, given the smallness parameter b, the understanding of the C ∞ regularity of the solution involves formally the expansion of the solution to the order 1 b which is too large for a WKB type analysis. Instead, we rely on a holomorphic expansion of the solution at P 2 to extract a formal limit involving the limiting series S ∞ (d, ). The condition S ∞ (d, ) = 0 turns out to be sufficient to prove the existence of a C ∞ profile, but we do not know if it is also necessary, as this would require pushing the asymptotic expansions to the next order. We refer to section 4.1 for a detailed strategy of the proof of Theorem 1.3 and the description of the role played by the limiting series S ∞ (d, ).

The sequence r n .

There is an abundant literature devoted to the existence of self-similar solutions in various semilinear problems. Let us for example consider the semilinear focusing wave equation ∂ 2 t u = ∆u + u p . Then, in dimension d ≥ 3, for a suitable range of p, the equation admits a discrete family of exact selfsimilar solutions of the form

u n (t, x) = 1 (T -t) 2 p-1 U n (y), y = x T -t , where U n (y) is C ∞ in y and decays like U n (y) ∼ |y| -2 p-1 as |y| → +∞, see [1] 6 .
Let us emphasize that the blow up speed in this case is dictated by the scaling of the equation, and the profiles U n are automatically C ∞ across the light cone. The quasilinear situation of Theorem 1.3 is however different: the Euler equations possess a 2-parameter family of scalings, and, as a result, the blow up speed r of the self-similar profile (1.5) is a free parameter. The choice of the sequence r n in Theorem 1.3 is related to the requirement that the profile is C ∞ through the acoustic light cone which is no longer automatic. Theorem 1.3 thus provides a family of 6 see also [START_REF] Krieger | Large global solutions for energy supercritical nonlinear wave equations on R 3+1[END_REF] for an extension to the defocusing case, and [START_REF] Collot | On the stability of type I blow up for the energy super critical heat equation[END_REF][START_REF] Lepin | Self-similar solutions of a semilinear heat equation[END_REF][START_REF] Troy | The existence of bounded solutions of a semilinear heat equation[END_REF] for the case of the semilinear heat equation. admissible 7 blow up profiles which accumulate at the critical speed r o (d, ). This is, to our knowledge, a completely new phenomenon.

4.

The case = d. The case d = = 3 is γ = 5 3 , i.e., the law for the monoatomic gas, and is always degenerate since for d = :

r = r * (d, d) = r + (d, d) ⇒ P 2 = P 3 = P 5
and the phase portrait has a triple point degeneracy which requires a separate treatment.

5. Decay of self-similar solutions. As stated in Corollary 1.4, the constructed selfsimilar solutions (ρ(t, y), u(t, y)) decay as y → ∞ but do not have finite energy. Nonetheless, this decay is sufficient for us to construct, in [START_REF] Merle | On the implosion of a three dimensional compressible fluid II: Singularity formation[END_REF], solutions to the Euler and Navier-Stokes equations which arise from smooth, well localized initial data (in particular, of finite energy) and form a singularity in finite time via a profile given in this paper. We also are able to utilize these profiles to produce blowing up solutions to the super-critical defocusing nonlinear Schrödinger equations (in d ≥ 5) which also arise from smooth, well localized (in particular, finite energy) initial data, [START_REF] Merle | On blow up for the energy super critical defocusing NLS[END_REF].

1.6. Further qualitative properties of the solution. We emphasize again that the main motivation of Theorem 1.3 is the dynamical study of (1.6), central for the construction of well localized smooth blow up solutions in the second paper [START_REF] Merle | On the implosion of a three dimensional compressible fluid II: Singularity formation[END_REF], as well as in [START_REF] Merle | On blow up for the energy super critical defocusing NLS[END_REF]. The study of the linearized operator close to the solution given by Theorem 1.3 and performed in [START_REF] Merle | On blow up for the energy super critical defocusing NLS[END_REF] requires, in an essential way, the C ∞ regularity at P 2 as well as suitable positivity properties which we now collect. These properties are responsible for the coercivity of the linearized operator, which is why we refer to them as repulsivity.

Positivity inside the light cone. We first claim the following positivity property inside the light cone σ > σ(P 2 ) (Z < Z 2 ) for the P 6 -P 2 trajectory. Lemma 1.6 (Repulsivity inside the light cone). Let d ≥ 2 then there exists 0 < ε 1 such that for all ( , r) in the range

d ≥ 2 0 < < d r * (d, ) -ε(d, ) < r < r * (d, ) or d = 2, 3 > d r + (d, ) -ε(d, ) < r < + (d, ) (1.22)
there exists c r > 0 such that the P 6 -P 2 trajectory (σ(x), w(x)) given by Lemma 1.2 satisfies the following bound. Let

F = σ + dσ dx , (1.23) then ∀σ ≥ σ(P 2 ), 1 -w -dw dx 2 -F 2 ≥ c r 1 -w -dw dx -(1-w)F σ ≥ c r 1 -w -dw dx ≥ c r (1.24)
Taking r = r n for n large enough ensures that (1.24) holds for the solution of Theorem 1.3. Property (1.24) is sharp: the constant c r → 0 as r ↑ r o . Note that the fact that the repulsivity property (1.24), which is at the heart of the stability 7 selected by the fact that their C ∞ regularity and the decay at +∞ guarantee that the linearized operator displays dynamical stability properties, [START_REF] Merle | On blow up for the energy super critical defocusing NLS[END_REF].

of the linearized flow, can be proved in the full range > 0 is surprising, and will require a substantial amount of algebra.

Positivity outside the light cone. We claim another positivity property in the exterior of the light cone (Z > Z 2 ) in a range of parameters which includes the solutions of Theorem 1.3. (1.25)

Let r n be given by Theorem 1.3 and (w, σ) be the corresponding C ∞ integral curve.

Then for all n large enough, there exists c rn > 0 such that

∀0 < σ ≤ σ(P 2 ), 1 -w -dw dx 2 -F 2 > c rn , 1 -w -dw dx > c rn .
(1.26) 1.7. Organization of the paper. In section 2, we establish the main geometric properties of the phase diagram 1, in particular related to the location of the roots of the polynomials ∆, ∆ 1 , ∆ 2 in the suitable range of parameters. In section 3, we recall the main dynamical properties of solutions to (1.9), and in particular discuss the existence of the P 6 -P 2 trajectory, and the regularity of integral curves passing through P 2 . Both sections 2 and 3 are classical and we recall most details for the convenience of the reader. In section 4, we start the semi classical analysis r ↑ r o (d, ), provide an overview of the strategy of the proof of Theorem 1.

The geometry of the phase portrait

The aim of this section is to start the analysis of the nonlinear ODE system (1.9)

(w -1) dw dx + σ dσ dx + (w 2 -rw + σ 2 ) = 0 σ dw dx + (w -1) dσ dx + σ w d + 1 -r = 0 ⇔ ∆ dw dx = -∆ 1 ∆ dσ dx = -∆ 2 (2.1)
by examining the roots of the polynomials ∆, ∆ 1 , ∆ 2 given by (1.10)

∆ = (w -1) 2 -σ 2 ∆ 1 = w(w -1)(w -r) -d(w -w e )σ 2 ∆ 2 = σ ( + d -1)w 2 -w( + d + r -r) + r -σ 2 (2.2
)

w e = (r -1) d . (2.3)
Their location is heavily dependent on the values of the parameters (d, , r). Let

w -= w e (r + ) = (r + -1) d = (d -1) d(1 + √ ) 2 .
We observe w -< 1 and from the beginning we restrict to the case

d ≥ 2 0 < w e < w -⇔ 1 < r < r + (d, ) (2.4)
where we recall (1.11), (1.12). Note also that r * ≤ r + , see (1.13), so that the above condition is not a restriction in view of r < r o .

2.1. Roots of ∆, ∆ 1 ∆ 2 . ∆ has been normalized to vanish on the sonic lines

{∆ = 0} = {w = 1 + σ} ∪ {w = 1 -σ}
which are independent of the parameters. We now study the roots of ∆ 2 .

Lemma 2.1 (Roots of ∆ 2 ). Assume (2.4). There exists σ (0) 2 (d, ) ∈ [0, +∞) such that the roots of ∆ 2 in the range σ ≥ 0 are given by

w ± 2 (σ) = 1 2( +d-1) 2 + d -1 -dwe(1-) ± I(σ) σ ≥ σ (0) 2 (2.5)
where

J(w e ) = d 2 1 -2 w 2 e - 2d(d -1)( + 1) w e + (d -1) 2 , I(σ) = J(w e ) -4dw e + 4 ( + d -1)σ 2 .
(2.6)

Moreover, ∀σ > σ (0) 2 , (w - 2 ) (σ) < 0, (w + 2 ) (σ) > 0. (2.7) Remark 2.2. The value σ (0)
2 is explicitly given by (2.8) if J(w e ) -4dw e < 0 and σ (0) 2 = 0 otherwise. In the range of parameters (2.4), both cases σ (0) 2 = 0 and σ (0) 2 > 0 are possible which means that the parabola defining the set of zeroes of ∆ 2 may or may not touch the line σ = 0 in figure 1. This will play no role in our qualitative study of the flow.

Proof of Lemma 2.1. We solve ∆ 2 = 0 for σ = 0 which is

0 = ( + d -1)w 2 -( + d + r -r)w + r -σ 2 = ( + d -1)w 2 - + d + 1 + dw e -1 - dw e w + 1 + dw e -σ 2 = ( + d -1)w 2 -2 + d -1 + dw e 1 - 1 w + ( + dw e ) -σ 2 .
The discriminant is given by

I(σ) = 2 + d -1 + dw e 1 - 1 2 -4 ( + dw e ) -σ 2 ( + d -1) = 4 ( + d -1)σ 2 + d 2 (1 -) 2 2 w 2 e - 2d ( d + + d -1)w e + (d -1) 2 = J(w e ) -4dw e + 4 ( + d -1)σ 2
which justifies the formula for w ± 2 (σ). We now study the sign of J(w e ) -4dw e . The equation J(w e ) -4dw e = 0 has real roots given by

w * * ,± = d(1 -) 2 d + + d -1 ± 2 d(d + -1) .
Hence J(w e ) -4dw e < 0 for w * * ,-< w e < w * * ,+ in which case I(σ) ≥ 0 for

σ ≥ σ (0) 2 =
4dw e -J(w e ) 4 ( + d -1) .

(2.8)

Next J(w e ) -4dw e ≥ 0 for w e ≥ w * * ,+ or w e ≤ w * * ,-in which case I(σ) ≥ 0 for all σ ≥ σ (0) 2 = 0. The monotonicity (2.7) follows directly from the fact that we have I (σ) > 0 for all σ > 0.

Lemma 2.3 (Roots of ∆ 1 ). Assume (2.4). For all σ ≥ 0, the equation ∆ 1 (w, σ) = 0 has exactly three distinct root branches w 1 (σ) < w 2 (σ) < w 3 (σ) which satisfy the following: relative positions: ∀σ ≥ 0,

w 1 (σ) ≤ 0 < w e < w 2 (σ) ≤ 1 < r ≤ w 3 (σ).
(2.9)

monotonicity: ∀σ > 0, w 1 (σ) < 0, w 2 (σ) < 0, w 3 (σ) > 0.
(2.10) asymptotics:

w 1 (σ) = - dw e r σ 2 + O(σ 3 ), w 2 (σ) = 1 - d(1 -w e )σ 2 r -1 + O(σ 3 ), w 3 (σ) = r + d(r -w e ) r(r -1) σ 2 + O(σ 3 ), as σ → 0 (2.11) and w 1 (σ) = - √ dσ + O(1), w 2 (σ) = w e + O(σ -2 ), w 3 (σ) = √ dσ + O(1)
as σ → +∞.

(2.12)

Proof of Lemma 2.3. At σ = 0, we have the three obvious roots

w 1 (0) = 0, w 2 (0) = 1, w 3 (0) = r
where we recall that (2.4), (1.11) imply r > 1.

(2.13) Also, for each fixed σ > 0, we have

lim w→+∞ ∆ 1 (w, σ) = +∞, lim w→-∞ ∆ 1 (w, σ) = -∞ ∆ 1 (0, σ) = dw e σ 2 > 0, ∆ 1 (w e , σ) = w e (w e -1)(w e -r) > 0 ∆ 1 (1, σ) = -d(1 -w e )σ 2 < 0, ∆ 1 (r, σ) = -d(r -w e )σ 2 < 0,
where we used from (1.11), (2.13) that w e < 1 < r. Thus, for all σ ≥ 0, ∆ 1 (w, σ) has exactly three distinct simple roots which then satisfy (2.9) and

∂ w ∆ 1 (w 1 (σ), σ) > 0, ∂ w ∆ 1 (w 2 (σ), σ) < 0, ∂ w ∆ 1 (w 3 (σ), σ) > 0. (2.14)
We may now apply the implicit function theorem and conclude that the roots w 1 (σ), w 2 (σ), w 3 (σ) are smooth function of σ for σ ≥ 0. Furthermore, we have the formula for j = 1, 2, 3

w j (σ) = - ∂ σ ∆ 1 (w j (σ), σ) ∂ w ∆ 1 (w j (σ), σ) = 2dσ(w j (σ) -w e ) ∂ w ∆ 1 (w j (σ), σ) , σ ≥ 0 (2.15)
which together with (2.14) and the location of the roots (2.9) ensures the monotonicity (2.10). We now compute the limiting asymptotics. Near σ = 0, we compute from (2.15):

w j (0) = 0, w j (0) = 2d(w j (0) -w e ) ∂ w ∆ 1 (w j (0), 0) , j = 1, 2, 3,
which together with the above explicit values w j (0), j = 1, 2, 3, and the fact that

∂ w ∆ 1 (w, 0) = w(w -1) + w(w -r) + (w -1)(w -r)
yields the expansion (2.11) as σ → 0. To compute the expansion of the three roots near +∞, we notice that

∆ 1 (± √ dσ + c 1 , σ) = d(2c 1 -r -1 + w e )σ 2 + O(σ), ∆ 1 (w e + c 2 σ -2 , σ) = w e (w e -1)(w e -r) -dc 2 + O(σ -2 ),
and an appropriate choice of the constants c 1 , c 2 yields (2.12) from the mean value theorem.

2.2. Double roots. We now discuss the double roots ∆ 1 = ∆ 2 = 0 which play a fundamental role in the study of (1.9).

Lemma 2.4 (Double roots). Assume (2.4). The solutions to ∆ 1 = ∆ 2 = 0 are:

P 1 = (0, 1), P 2 = (1 -w -, w -) P 3 = (1 -w + , w + ), P 4 = (0, 0), P 5 = (σ 5 , w 5 ), P 5 = (0, r), (2.16)
where the points are defined as follows: P 5 point.

P 5 = w 5 = r d + , σ 5 = r √ d d +
(2.17) P 2 , P 3 points. Let J(w e ) given (2.6), then

w ± = 1 2(d -1) dw e + d -1 - dw e ± J(w e ) (2.18)
with w e < w ± < 1.

(2.19) Location. P 2 , P 3 , P 5 are located on the curve of the middle root (σ, w 2 (σ)) of ∆ 1 . Moreover, P 2 , P 5 are on the curve of the lower root w - 2 of ∆ 2 .

Position of the middle root. Let w e < w -and w 2 (σ) be the middle root of ∆ 1 , then the relative position of the middle root with respect to the sonic line is:

σ + w 2 (σ) > 1 for 0 < σ < σ(P 3 ) < 1 for σ(P 3 ) < σ < σ(P 2 ) > 1 for σ > σ(P 2 ).
(2.20)

Proof of Lemma 2.4. It relies on the factorization

∆ = a 1 b 2 -b 1 a 2 ∆ 1 = -b 1 d 2 + b 2 d 1 ∆ 2 = d 2 a 1 -d 1 a 2 .
(2.21)

with a 1 = w -1, b 1 = σ, d 1 = w 2 -rw + σ 2 , a 2 = σ , b 2 = w -1, d 2 = σ 1 + d w -r . (2.22)
step 1 Computation of the triple points. From (2.21):

(∆ 1 = ∆ 2 = 0) ⇔ (b 1 d 2 = b 2 d 1 and d 2 a 1 = d 1 a 2 ) . ( 2
0 = d 2 = σ 1 + d -r = σd (1 -w e ) = 0, a contradiction.
Case

a 1 = b 2 = 0. If σ = 0, then d 2 = b 1 = 0 and hence 0 = d 1 = w(w -r)
, and hence the points P 4 = (0, 0) and P 5 = (0, r).

If σ = 0, then a 1 , b 2 , b 1 , a 2 = 0. If d 2 = 0, then d 1 = 0 and hence w = w 5 = r 1+ d , σ 2 5 = w 5 (r-w 5 ) = dr 2 (d+ ) 2 , σ 5 = r √ d d+ .
We observe

w 5 < 1 ⇔ w e < 1
and hence (2.4) and (2.9) implies that P 5 lies on the middle root

w 2 (σ) of ∆ 1 . If d 2 = 0 then a 1 a 2 = b 1 b 2 = d 1 d 2 , ∆ 1 = ∆ 2 = ∆ = 0.
Hence w = 1 + σ or w + σ = 1 which we consider separately.

Points on the lower sonic line, i.e., w = 1 -σ. First, note that it suffices to consider the solutions to ∆ 2 (σ, 1 -σ) = 0. Indeed, since a 1 = b 2 = 0 and d 2 = 0, then, since we have

∆ 2 = ∆ = 0, we infer b 1 b 2 = a 1 a 2 = d 1 d 2 .
It then follows that ∆ 1 = 0. Thus, we now consider the solutions to ∆ 2 (σ, 1-σ) = 0. Since σ = 0:

0 = P (w) = ( + d -1)w 2 -( + d + r -r)w + r -(1 -w) 2 = ( + d -1)w 2 -( + d + r -r)w + r -(w 2 -2w + 1) = (d -1)w 2 -( (r -1) + d -r)w + (r -1) = (d -1)w 2 -dw e + d -1 - dw e w + dw e .
(2.24)

The roots of P are real iff

J = dw e + d -1 - dw e 2 -4d(d -1)w e (2.25) = d 2 1 -2 w 2 e - 2d(d -1)( + 1) w e + (d -1) 2 ≥ 0.
The discriminant of J, as a second order polynomial in w e , is > 0, and the roots of J are given by

w ± = 2 2d 2 (1 -) 2 2d(d -1)( + 1) ± 4d(d -1) √ = (d -1) d(1 -) 2 + 1 ± 2 √ .
Hence J ≥ 0 ⇐⇒ (w e ≤ w -or w e ≥ w + ).

(2.26) If J ≥ 0, the roots of P are given by

w ± = 1 2(d -1) dw e + d -1 - dw e ± √ J .
We compute from (2.24):

P (1) = r -1 > 0 P (1) = d -1 -( -1)(r -1)
and

P (0) = (r -1) > 0 P (0) = -[d -1 + ( -1)(r -1)
] Since P is a second order polynomial with non negative second order term, we conclude

0 < w ± < 1 ⇔ P (0) < 0 P (1) > 0 ⇔ d -1 + ( -1)(r -1) > 0 d -1 -( -1)(r -1) > 0 (2.27)
This holds for = 1, and we now distinguish the following two cases: case < 1. We want

d -1 -(1 -)(r -1) > 0 ⇔ r < 1 + d -1
1and we observe from (2.4)

r < r + = 1 + d -1 (1 + √ ) 2 < 1 + d -1 1 - . case > 1. We want d -1 -( -1)(r -1) > 0 ⇔ r < 1 + d -1
-1 and we observe from (2.4)

r < r + = 1 + d -1 (1 + √ ) 2 < 1 + d -1 -1 .
This concludes the proof of the inequality

0 < w ± < 1.
This implies from (2.9) that P 2 , P 3 lie on the middle root w 2 (σ) of ∆ 1 which is a non increasing function of σ from (2.10), and hence necessarily w e < w ± .

Points on the upper sonic line, i.e., w = 1 + σ. We consider the solutions to ∆ 2 (σ, 1+ σ) = 0 or, equivalently,

Q(σ) = ( + d -1)(1 + σ) 2 -( + d + r -r)(1 + σ) + r -σ 2 = 0. We compute Q(0) = d + r -1, Q (0) = d + r -(r -1) > r
from (2.4), and hence the second order polynomial Q is > 0 for σ > 0 and there is no intersection point on the upper sonic line.

step 2 Location of P 2 , P 3 , P 5 . We have established that P 2 , P 3 , P 5 lie on the middle root of ∆ 1 . Next, note that w = 1 -σ is decreasing and w + 2 (σ) in increasing by (2.7), so they can intersect at most once, and hence at least one point among P 2 , P 3 must be on the root (σ, w - 2 (σ)) of ∆ 2 . Since w -< w + , we infer that P 2 is on the root (σ, w - 2 (σ)) of ∆ 2 .

step 3 Proof of (2.20). Assume now w e < w -, then J(w e ) > 0 ensures w -< w + and, since P 2 , P 3 lie on the sonic line w + σ = 1,

σ(P 3 ) < σ(P 2 ).
Hence 0 < σ(P 3 ) < σ(P 2 ) are three distinct roots of

R(σ) = ∆ 1 (1 -σ, σ) = (1 -σ)(-σ)(1 -σ -r) -d(1 -σ -w e )σ 2
which is an order three polynomial, and hence these are the only roots which are simple. Since w 2 (σ) lies above the sonic line σ +w = 1 near σ = 0 and near σ = +∞ from (2.11), the ordering (2.20) follows 2.3. Relative positions of P 2 , P 3 , P 5 . We now discuss a very important property for our forthcoming analysis regarding the relative position of P 2 and P 5 . We recall the definition (1.12), (1.14) of the critical speed exponents. Let us start with comparing these values. (2.28)

Proof of Lemma 2.5. We compute from (2.19):

w -= (d -1) d(1 -) 2 + 1 ± 2 √ = (d -1) d(1 + √ ) 2 < 1
and hence from (1.12):

w e < w -⇔ (r -1) d < (d -1) d(1 + √ ) 2 ⇔ r ≤ r + (d, ).
Then

r + (d, ) -r * (d, ) = 1 + d -1 (1 + √ ) 2 - d + + √ d = (d -1)( + √ d) -(d - √ d)(1 + 2 √ + ) ( + √ d)(1 + √ ) 2 = ( √ d -1) ( √ - √ d) 2 ( + √ d)(1 + √ ) 2 > 0.
(2.29)

We now design an admissible portrait as follows.

Lemma 2.6 (Admissible phase portrait, see figure 3). Assume

d ≥ 2, > 0, 1 < r ≤ r + (d, ).
(2.30)

then the conclusions of Lemma 2.4 hold with 0 < w e ≤ w -and P 2 , P 3 , P 5 are well defined. More precisely, denoting σ 2 = σ(P 2 ), σ 3 = σ(P 3 ) and σ 5 = σ(P 5 ):

1. below the r * speed: for 1 < r < r * (d, ), P 5 lies strictly below the sonic line 0 < w 5 + σ 5 < 1 and

0 < σ 3 < σ 5 < σ 2 .
(2.31)

2. r * speed: for r = r * (d, ),

< d ⇔ P 5 = P 2 > d ⇔ P 5 = P 3 = d ⇔ P 5 = P 2 = P 3 .
(2.32)

3. below the r + speed: for r * (d, ) < r < r + (d, ) and > d,

σ 5 < σ 3 < σ 2 (2.33)
and

P 3 → P 2 as r ↑ r + .
Remark 2.7. We note here the fundamental role payed by the case = d which corresponds to γ = 1 + 2 d and is a degenerate triple point configuration

r = r * (d, d) = r + (d, d) ⇔ P 2 = P 3 = P 5 .
Proof of Lemma 2.6. We observe

w -< 1 ⇔ (d -1)( √ -1) 2 d(1 -) 2 < 1 ⇔ (d -1) d(1 + √ ) 2 < 1 ⇔ d(2 √ + 1) + > 0
which holds, and this together with (2.28) ensures that the conclusions of Lemma 2.4 hold and P 2 , P 3 are well defined, distinct and on the sonic line. Subcritical speed. For 1 < r < r * (d, ), we compute from (2.17):

w 5 + σ 5 = r d + + r √ d d + = r r * (d, ) < 1 (2.34)
and hence P 5 lies strictly below the sonic line w + σ = 1. Since P 5 lies on the curve (σ, w 2 (σ)) where w 2 (σ) is the middle root of ∆ 1 , the ordering (2.20) implies (2.31).

Critical speed. Let now r = r * (d, ). Then P 5 is on the sonic line and since (0, 1), P 2 , P 3 are the only intersections of w 2 (σ) with the sonic line, P 5 coincides necessarily with P 2 or P 3 . For d = ,

r * (d, ) = r + (d, ) ⇒ w e = w -⇒ P 2 = P 3
and hence P 2 = P 3 = P 5 . For d = , let J = J(w e ) ≥ 0 be given by (2.6) and, from (2.18), let

σ 2 ≡ σ(P 2 ) = 1 -w -= 1 - 1 2(d -1) dw e + d -1 - dw e - √ J = 1 2(d -1) d -1 + d 1 -1 w e + √ J and
σ 3 ≡ σ(P 3 ) = 1 -w + = 1 - 1 2(d -1) dw e + d -1 - dw e + √ J = 1 2(d -1) d -1 + d 1 -1 w e - √ J
so that from (2.17):

σ 2 -σ 5 = √ J-A 2(d-1) σ 3 -σ 5 = - √ J-A 2(d-1) (2.35) with A = 2(d -1) √ d r d+ -d + 1 -(1 -)(r -1) = 2(d-1) √ d d+ -1 + (r -r 0 ( )) r 0 ( ) = d+ -2 2(d-1) √ d d+ -1+ > 0,
where we used that

∀ > 0, ∀d ≥ 2, 2(d -1) √ d d + -1 + > 0.
We compute

r * (d, ) -r 0 = d + √ d + - d + -2 2(d-1) √ d d+ + -1 = d + ( √ d + )(2(d -1) √ d + ( -1)(d + )) 2(d -1) √ d + ( -1)(d + ) -( √ d + )(d + -2) = ( √ d -1)(d + ) ( √ d + )(2(d -1) √ d + ( -1)(d + )) (d -)
Therefore,

A(r * ) > 0 for < d < 0 for > d.
(2.36)

Since P 5 coincides with P 2 or P 3 , (2.35) implies

J(r * ) = A 2 (r * ) (2.37)
and hence the sign of A(r * ) is given by (2.36) and (2.35) yield (2.32).

step 3 We now turn to the case r * < r < r + for > d. We claim

r + -r 0 < 0 (2.38) and J -A 2 = -c 1 (d, )(r + c 2 (d, ))(r -r * ), c 1 , c 2 > 0.
(2.39) Assume (2.38), (2.39), then for r * < r < r + , A < 0 and J -A 2 < 0 implies from (2.35) that σ 3 -σ 5 > 0, and since σ 2 > σ 3 by definition, (2.33) is proved. Proof of (2.38). We compute

r * (d, ) -r 0 = d + √ d + - d + -2 2(d-1) √ d d+ + -1 = d + ( √ d + )(2(d -1) √ d + ( -1)(d + )) 2(d -1) √ d + ( -1)(d + ) -( √ d + )(d + -2) = ( √ d -1)(d + ) ( √ d + )(2(d -1) √ d + ( -1)(d + )) (d -)
and hence recalling (2.29):

r + -r 0 = r + -r * + r * -r 0 = ( √ d -1)(d + )(d -) ( √ d + )(2(d -1) √ d + ( -1)(d + )) + ( √ d -1) ( √ - √ d) 2 ( + √ d)(1 + √ ) 2 = ( √ d -1))( √ d - √ ) + √ d (d + )( √ + √ d) 2(d -1) √ d + ( -1)(d + )) - √ - √ d (1 + √ ) 2
and hence the sign is dictated by

P (d, ) = (d + )( √ + √ d)(1 + √ ) 2 -( √ - √ d) 2(d -1) √ d + ( -1)(d + )) = (d + ) ( √ + √ d)(1 + 2 √ + ) -( √ - √ d)( -1) -2(d -1) √ d( √ - √ d) = (d + ) √ d + √ + 2 √ √ d + 2 + √ d + √ -( √ - √ - √ d + √ d) -2(d -1) √ d( √ - √ d) = (d + )[(2 √ d + 2) √ + ( √ d + 3) ] -2(d -1) √ d( √ - √ d) = (d + )( √ d + 3) + 2 √ d(d -1) + √ [(d + )(2 √ d + 2) -2(d -1) √ d] > 0 and (2.38) is proved.
Proof of (2.39). We compute

J = d 2 1 -2 w 2 e - 2d(d -1)( + 1) w e + (d -1) 2 = (1 -) 2 (r -1) 2 -2(d -1)( + 1)(r -1) + (d -1) 2 = (1 -) 2 r 2 -2r (1 -) 2 + ( + 1)(d -1) + (d -1) 2 + (1 -) 2 + 2(d -1)( + 1).
and hence injecting the value of A:

J -A 2 = -    r 2   2(d -1) √ d d + -1 + 2 -(1 -) 2   + 2r   (1 -) 2 + ( + 1)(d -1) - 2(d -1) √ d d + -1 + 2 r 0   r -(d -1) 2 -(1 -) 2 -2(d -1)( + 1) + 2(d -1) √ d d + -1 + 2 r 2 0    = -(ar 2 + 2br -c) with using > d > 1: a = 2(d -1) √ d d + + -1 2 -( -1) 2 = 4(d -1) √ d d + (d -1) √ d d + + -1 > 0 b = (1 -) 2 + ( + 1)(d -1) - 2(d -1) √ d d + -1 + 2 r 0 = (1 -) 2 + ( + 1)(d -1) -(d + -2) 2(d -1) √ d d + -1 + = 1 -2 + 2 + d -+ d -1 - 2(d -1)(d + -2) √ d d + -( d + 2 -2 -d -+ 2) = 2(d -1) 1 - (d + -2) √ d d + and c = (d -1) 2 + (1 -) 2 + 2(d -1)( + 1) - 2(d -1) √ d d + -1 + 2 r 2 0 = (d -1) 2 + (1 -) 2 + 2(d -1)( + 1) -(d + -2) 2 = 4(d -1).
Since a, c > 0, the roots are given by

r * ± = 1 a (-b ± b 2 + ac), r * -< 0.
We now observe that (2.37) implies

r * + = r * (2.40)
which can also be checked directly, and (2.39) is proved.

2.4. Slopes of w 2 (σ) and w - 2 (σ) at P 2 . The point P 2 will play a fundamental role in the proof of Theorem 1.3. In this section we collect the main geometric properties of the phase portrait near P 2 in the regime (2.30). We note

σ 2 = σ(P 2 ), w 2 = 1 -σ 2 .
Definition of the slopes We compute the slopes of root curves w 2 (σ) and w - 2 (σ) at P 2 by defining the following coefficients

c 1 = ∂ w ∆ 1 (P 2 ) = 3w 2 2 -2(r + 1)w 2 + r -dσ 2 2 c 2 = ∂ w ∆ 2 (P 2 ) = σ 2 [2w 2 ( + d -1) -( + d + r -r)] c 3 = ∂ σ ∆ 1 (P 2 ) = -2dσ 2 w 2 + 2 (r -1)σ 2 c 4 = ∂ σ ∆ 2 (P 2 ) = -2σ 2 2 ,
(2.41) so that the corresponding slopes arec 3 c 1 andc 4 c 2 . We now claim: Lemma 2.8 (Sign of the slopes). Assume d ≥ 2 and

1 < r < r * (d, ) for < d r * (d, ) < r < r + (d, ) for > d (2.42) then c i < 0, 1 ≤ i ≤ 4 c 2 c 3 -c 1 c 4 < 0. (2.43)
Proof of Lemma 2.8. The argument relies solely on the consideration of the relative positions of the red and green curves locally near P 2 which is the same in the range (2.42). Indeed, in view of the discussion of the roots of ∆ 1 , and since we have established that P 2 is on the middle root of ∆ 1 , we have

c 1 = ∂ ω ∆ 1 (P 2 ) < 0.
Since P 2 corresponds to the smallest root w - 2 of ∆ 2 and ∆ 2 is a second order polynomial in w with a strictly positive coefficient in front of the w 2 term, we have

c 2 = ∂ ω ∆ 2 (P 2 ) < 0.
Also, since σ > 0 at P 2 , and since w e < w -< 1, we have

c 3 = ∂ σ ∆ 1 (P 2 ) = -2d(w --w e )(1 -w -) < 0, c 4 = ∂ σ ∆ 2 (P 2 ) = -2(1 -w -) 2 < 0.
Finally, we compute

c 2 c 3 -c 1 c 4 = ∂ ω ∆ 2 (P 2 )∂ σ ∆ 1 (P 2 ) -∂ ω ∆ 1 (P 2 )∂ σ ∆ 2 (P 2 ) = ∂ ω ∆ 2 (P 2 )∂ ω ∆ 1 (P 2 ) ∂ σ ∆ 1 (P 2 ) ∂ ω ∆ 1 (P 2 ) - ∂ σ ∆ 2 (P 2 ) ∂ ω ∆ 2 (P 2 ) = c 1 c 2 (w - 2 ) (σ) -w 1 (σ) | σ=1-w -
where we used the fact that σ = 1 -w -and w = w -at P 2 , the fact that w 2 is both the middle root of ∆ 1 , i.e., w 1 , and the smallest root of ∆ 2 , i.e., w - 2 , with the formula for w - 2 (σ) and w 1 (σ) following from the implicit function theorem. Now, at P 2 , the slope of ∆ 2 is strictly more negative than the slope of ∆ 1 since P 2 is the last intersection in σ and ∆ 1 asymptotes to w e while ∆ 2 goes to -∞. Thus, we have (w

- 2 ) (1 -w -) -w 2 (1 -w -) < 0 and hence c 2 c 3 -c 1 c 4 < 0 as desired.
Slopes and eigenvalues. In additions to the slopes of the root curves w 2 (σ) and w - 2 (σ) we will also compute the slopes of any integral curve passing through P 2 . it turns out that there are only two possible values:

c ± = c 4 -c 1 ± (c 1 -c 4 ) 2 + 4c 2 c 3 2|c 2 | (2.44)
This follows since c ± are the solutions of the equation

c ± = c 1 c ± + c 3 c 2 c ± + c 4 .
(2.45)

The characteristic matrix

A(P 2 ) = c 1 c 3 c 2 c 4 ,
coming from the linearization of (1.9) at P 2 (see (3.16) (3.17)), possesses the following eigenvalues:

λ ± = c 1 + c 4 ± (c 1 -c 4 ) 2 + 4c 2 c 3 2 (2.46)
It may be diagonalized as follows:

P -1 [A(P 2 )] P = λ + 0 0 λ - with P = c -c + 1 1 , P -1 = 1 c + -c - -1 c + 1 -c - .
(2.47) Lemma 2.9 (Estimates on the slopes). Assume (2.42) and let

A = λ - λ + = c 1 c 4 -c 2 c 3 (c 4 + c 2 c -) 2 , (2.48) Then, c -< 0 < c + c 4 + c 2 c -< 0 A > 1 -1 < -c 4 c 2 < c -< -c 3 c 1 < 0 λ -< λ + < 0 (2.49)
Proof of Lemma 2.9. From c 2 c 3 > 0:

c 4 -c 1 -(c 1 -c 4 ) 2 + 4c 2 c 3 2|c 2 | < 0 < c 4 -c 1 + (c 1 -c 4 ) 2 + 4c 2 c 3 2|c 2 | and hence c -< 0 < c + .
Next, we compute

c 4 + c 2 c -= c 4 + c 1 + (c 1 -c 4 ) 2 + 4c 2 c 3 2 = 2(c 2 c 3 -c 1 c 4 ) -c 4 -c 1 + (c 1 -c 4 ) 2 + 4c 2 c 3 < 0.
We now observe

c 2 c -+ c 4 = c 4 -c 4 -c 1 - √ ∆ 2 = c 4 +c 1 + √ ∆ 2 = λ + c 2 c + + c 4 = c 4 -c 4 -c 1 + √ ∆ 2 = c 4 +c 1 - √ ∆ 2 = λ -.
(2.50) and hence λ -< λ + < 0. We now estimate A:

c 4 c 1 -c 3 c 2 -(c 4 + c 2 c -) 2 = c 4 c 1 -c 3 c 2 - 4(c 2 c 3 -c 1 c 4 ) 2 (-c 4 -c 1 + (c 1 -c 4 ) 2 + 4c 2 c 3 ) 2 = -(c 4 + c 2 c -) (-c 4 -c 1 + (c 1 -c 4 ) 2 + 4c 2 c 3 ) 2 -4(c 4 c 1 -c 2 c 3 ) 2(-c 4 -c 1 + (c 1 -c 4 ) 2 + 4c 2 c 3 ) = -(c 4 + c 2 c -) (c 4 + c 1 ) 2 + 2(-c 4 -c 1 ) (c 1 -c 4 ) 2 + 4c 2 c 3 + (c 1 -c 4 ) 2 + 4c 2 c 3 -4(c 4 c 1 -c 2 c 3 ) 2(-c 4 -c 1 + (c 1 -c 4 ) 2 + 4c 2 c 3 ) = -(c 4 + c 2 c -) (-c 4 -c 1 ) (c 1 -c 4 ) 2 + 4c 2 c 3 + (c 1 -c 4 ) 2 + 4c 2 c 3 (-c 4 -c 1 + (c 1 -c 4 ) 2 + 4c 2 c 3 ) 2 and hence A -1 = c 1 c 4 -c 2 c 3 (c 4 + c 2 c -) 2 -1 = - (-c 4 -c 1 ) (c 1 -c 4 ) 2 + 4c 2 c 3 + (c 1 -c 4 ) 2 + 4c 2 c 3 (c 4 + c 2 c -)(-c 4 -c 1 + (c 1 -c 4 ) 2 + 4c 2 c 3 ) .
Using in particular that c 1 , c 2 , c 3 , c 4 < 0, this yields

A > 1.
Next, note that we have

- c 3 c 1 < 0
since c 3 < 0 and c 1 < 0. Also, since {∆ 2 = 0} intersects w = 1 -σ at P 2 and P 3 , (σ, w - 2 (σ)) is above w = 1 -σ for σ > 1 -w -and below for σ < 1 -w -. Since -c 4 /c 2 is the slope of w = w - 2 (σ) at P 2 , we infer

- c 4 c 2 > -1.
It remains to compare c -to -c 3 /c 1 . We compute

c -+ c 3 c 1 = c -+ |c 3 | |c 1 | = |c 1 |(c 4 -c 1 -(c 1 -c 4 ) 2 + 4c 2 c 3 ) + 2c 2 c 3 2|c 1 ||c 2 | = c 1 (c 1 -c 4 ) + 2c 2 c 3 -|c 1 | (c 1 -c 4 ) 2 + 4c 2 c 3 2|c 1 ||c 2 | . Now, we have c 1 (c 1 -c 4 ) + 2c 2 c 3 2 -|c 1 | (c 1 -c 4 ) 2 + 4c 2 c 3 2 = c 2 1 (c 1 -c 4 ) 2 + 4c 2 c 3 c 1 (c 1 -c 4 ) + 4c 2 2 c 2 3 -c 2 1 (c 1 -c 4 ) 2 -4c 2 1 c 2 c 3 = 4c 2 c 3 (c 2 c 3 -c 1 c 4 ) < 0 and hence c -+ c 3 c 1 < 0.
This concludes the proof of (2.49).

3. General properties of the dynamical system (1.9)

In this section we establish the general properties of Lemma 1.2 for the dynamical system (1.9). We assume

d ≥ 2, 1 < r < r * ( , d) for < d r * (d, ) < r < r + (d, ) for > d (3.1)
so that the shape of the phase portrait is given by respectively figure 1 or figure 2. We recall that w 2 (σ) is the middle root of ∆ 1 and w - 2 (σ) is the smallest root of ∆ 2 given by (2.5). The arguments in this section are classical and are given for the reader's convenience. We note

P o = P 5 for < d P 3 for > d.
3.1. The spherically symmetric solution emerging from the origin. We first claim the existence and uniqueness (up to the scaling symmetry) of a spherically symmetric solution to (1.7) which exists on the interval [0, Z 2 ] and, in the variables of Emden transform, corresponds to the integral curve P 6 -P 2 .

Lemma 3.1 (The solution emerging from P 6 ). Assume (3.1).

(1) Existence: there is σ 0 > 0 large enough and a unique curve w(σ) solution to (1.9) on (σ 0 , +∞) with It admits the asymptotic expansion as σ → +∞:

w(σ) = w e + w e (w e -1)(w e -r) d + 2 1 σ 2 + O 1 σ 4 . (3.3)
(2) Original variables: The curve corresponds to the unique (up to scaling invariance) spherically symmetric solution of (1.7) defined on the interval |Z| ∈ [0, Z 0 ]. This solution belongs to C ∞ (|Z| ≤ Z 0 ).

(3) Reaching P 2 : w exists in fact on (σ 2 , +∞) and we have w

∈ C ∞ (σ 2 , +∞) with ∀σ 2 < σ < +∞, w - 2 (σ) < w(σ) < w 2 (σ) lim σ↓σ 2 w(σ) = w 2 , (3.4)
see figure 1. Then, w satisfies

d w dσ - d σ w = w(w -1)(w -r) -d wσ 2 σ ( + d -1)w 2 -w( + d + r -r) + r -σ 2 - d σ w = w -w(w-1)(w-r) dσ 2 1 + -( +d-1)w 2 +w( +d+ r-r)-r σ 2 -w d σ =   -w(w -1)(w -r) + d ( + d -1)w 2 -w( + d + r -r) + r w 1 + -( +d-1)w 2 +w( +d+ r-r)-r σ 2   1 σ 3
and hence

d dσ w σ d = 1 σ d d w dσ - d σ w (3.5) =   -w(w -1)(w -r) + d ( + d -1)w 2 -w( + d + r -r) + r w 1 + -( +d-1)w 2 +w( +d+ r-r)-r σ 2   1 σ d+3 .
Existence. We solve

w (3.6) = -σ d +∞ σ   -w(w -1)(w -r) + d ( + d -1)w 2 -w( + d + r -r) + r w 1 + -( +d-1)w 2 +w( +d+ r-r)-r σ 2   dσ σ d+3 .
using an elementary fixed point argument which yields the existence and uniqueness on σ 0 ≤ σ ≤ +∞ of w such that

| w| 1 σ 2 for σ ≥ σ 0 (3.7)
provided σ 0 has been chosen large enough.

Uniqueness. Let now w(σ) be a solution to (3.5) on [σ 0 , +∞) with lim σ→+∞ w(σ) = w e . We integrate (3.5). The integral of the RHS of (3.5) converges at +∞ from the a priori bound |w| 1, and hence w satisfies (3.6) and the a priori bound (3.7).

The uniqueness claim follows.

Asymptotics as σ → +∞. The integral equation (3.6) for w and the fact that = w e (w e -1)(w e -r) d + 2

-w(w -1)(w -r) + d ( + d -1)w 2 -w( + d + r -r) + r w 1 + -( +d-1)w 2 +w( +d+ r-r)-r
1 σ 2 + O 1 σ 4 .

yield (3.3).

Spherical symmetry and regularity at the origin. From (3.3), on the solution, as σ → +∞:

∆ = -σ 2 1 + O 1 σ 2 ∆ 2 = -σ 3 1 + O 1 σ 2
As a result,

dx dσ = - ∆ ∆ 2 = - 1 σ 1 + O 1 σ 2 .
Up to a constant

x = -logσ + O 1 σ 2 ⇒ σ = e -x 1 + O(e 2x ) as x → -∞.
Recalling the Emden transform formula (1.8), we obtain from (3.3) the asymptotics as Z = e x → 0:

(ρ(Z)) γ-1 2 = 2 Zσ(x) = 2 (1 + O(Z 2 )) û(Z) = -Zw(x) = -Z w e + O(Z 2 ) (3.8)
We also observe that the system (2.1) can be rewritten with respect to the variables

s := Z 2 , f := w -w e , g = Zσ in the form df ds = -d 2s f + F(s, f, g)
dg ds = G(s, f, g) for some functions F, G with the property that they are smooth functions of all the variables in the neighborhood of the point (0, 0, 2 ). We may rewrite the above in the form

f (s) = s -d 2 s 0 (s) d 2 F(s, f, g)ds g(s) = 2 + s 0 G(s, f, g)ds
This produces a unique fixed point solution which is C 1 in s and, obviously, coincides with the solution constructed from (3.6). Rewriting the above again as

f (s) = s 1 0 (κ) d 2 F(sκ, f (sκ), g(sκ))dκ g(s) = 2 + s 1 0 G(κ, f (sκ), g(sκ))dκ implies that f (s) = 1 0 (κ) d 2 F(sκ, f (sκ), g(sκ))dκ + s 1 0 (κ) d 2 d ds (F(sκ, f (sκ), g(sκ))) dκ = 1 0 (κ) d 2 F(sκ, f (sκ), g(sκ))dκ + 1 0 (κ) d 2 +1 d dκ (F(sκ, f (sκ), g(sκ))) dκ = 1 0 (κ) d 2 F(sκ, f (sκ), g(sκ))dκ + F(s, f (s), g(s)
)

- d + 2 2 1 0 (κ) d 2 F(sκ, f (sκ), g(sκ))dκ
Similarly, for g. This expresses f (s) (and g (s)) in terms of regular kernels involving f and g and immediately implies (by step by step differentiation) that both f and g are C ∞ with respect to s including at s = 0. Recalling that s = Z 2 and the Emden transform relation (3.8) to the original variables û, ρ, yields the desired local (defined in a neighborhood of the origin) C ∞ (R d , R 2 ) spherically symmetric solution.

step 2 Reaching P 2 . Let w r (σ) be the unique curve entering P 6 constructed in step 1. From (3.3),

∆ 1 (σ, w r (σ)) = w r (σ)(w r (σ) -1)(w r (σ) -r) -d(w r (σ) -w e )σ 2
= w e (w e -1)(w e -r) 1 -

d d + 2 + O 1 σ 2
= w e (w e -1)(w e -r)

2 d + 2 + O 1 σ 2 > 0
near +∞. By Lemma 2.3, ∆ 1 < 0 for w < w 1 (σ) and for w ∈ (w 2 (σ), w 3 (σ)), and ∆ 1 > 0 for w > w 3 (σ) and for w ∈ (w 1 (σ), w 2 (σ)). Similarly, by Lemma 2.1, ∆ 2 > 0 for w ∈ (w - 2 (σ), w + 2 (σ)) and ∆ 2 < 0 for w < w - 2 (σ) and for w > w + 2 (σ). From the asymptotic expansion of w r (σ) we can conclude that w r (σ) < 0 for σ > σ 0 . Given that w ± 2 (σ) → ±∞ as σ → ∞, it follows that w r (σ) ∈ (w - 2 (σ), w + 2 (σ)) for σ > σ 0 and thus ∆ 2 (σ, w r (σ)) > 0. Since,

0 > w r (σ) = ∆ 1 ∆ 2 ,
it follows that ∆ 1 (σ, w r (σ)) < 0. Combining this with the fact that w r (σ) → w e = lim σ→∞ w 2 (σ) and examining the regions of constant signs of ∆ 1 described above, imply that w r (σ) < w 2 (σ) for all σ > σ 0 . Thus, we have obtained w - 2 (σ) < w r (σ) < w 2 (σ) for σ ≥ σ 0 . In view of the phase portrait of figure 3 and the strict monotonicity w 2 < 0 given by (2.10), this implies that the curve w r reaches P 2 with ∀σ > σ 2 , w - 2 (σ) < w r (σ) < w 2 (σ).

(3.9)

Let us give a quick proof. From (3.9), we consider the region below the middle root of ∆ 1 , above the smallest root of ∆ 2 and to the right of P 2 .

R r = {(σ, w), w - 2 (σ) ≤ w ≤ w 2 (σ), σ ≥ 1 -w -}
The point P 2 is the utmost right joint root of ∆ 1 and ∆ 2 . The curves (σ, w 2 (σ)) and (σ, w - 2 (σ)) intersect there and lim σ→∞ w 2 (σ) = w e , lim σ→∞ w - 2 (σ) = -∞. Therefore, w - 2 (σ) < w 2 (σ) to the right of P 2 . Moreover, in that region w 1 (σ) < w - 2 (σ) and w 2 (σ) < w + 2 (σ). As a consequence of the above, in the region R r

∆ 1 ≤ 0, ∆ 2 ≥ 0.
Now, assume by contradiction that there exists

σ 1 > 1 -w -such that (σ, w r (σ)) is in R r for σ > σ 1 and (σ 1 , w r (σ 1 )
) is the last point in that region. Then either (σ 1 , w r (σ 1 )) is on the middle root of ∆ 1 to the right of P 2 , in which case we have

w r (σ 1 ) = 0
But in view of (2.10), w 2 (σ 1 ) < 0, and we have a contradiction since w r (σ) < w 2 (σ) for all σ > σ 1 . Or (σ 1 , w r (σ 1 )) is on the smallest root of ∆ 2 to the right of P 2 , in which case we have

w r (σ 1 ) = -∞
This is again a contradiction since w r (σ) > w - 2 (σ) all σ > σ 1 . Therefore, the curve can only exit the region R r at σ = 1 -w -, i.e., through P 2 . Since there are no other accumulation points in this zone, the curve must approach P 2 as σ ↓ σ 2 . Therefore, the curve w r (σ) is defined on (σ 2 , +∞), is C ∞ on this interval and satisfies (3.4).

3.2.

Solutions crossing red between P 2 and P o . We now analyze trajectories that cross the middle root w 2 (σ) of ∆ 1 between 8 P 2 and P o . Lemma 3.2 (Solutions crossing red between P 2 and P o ). Assume (3.1). Let σ 5 < σ * < σ 2 and w(σ) be the solution to (1.9) with the data w(σ * ) = w 2 (σ * ). Then:

1. backward flow: w ∈ C ∞ (0, σ * ] and lim σ↓0 w(σ) = 0. Moreover, σ → 0 corresponds to x → +∞ and there exist (w ∞ , σ ∞ ) ∈ R × R * + such that for x → +∞ σ(x) = σ ∞ e -rx (1 + O(e -rx )) , w(x) = w ∞ e -rx (1 + O(e -rx )) . (3.10)
8 Since the roots of ∆1 are depicted in red on the various figures of this paper, we will sometimes say that such solutions cross red between P2 and P o .

forward flow

: w ∈ C ∞ [σ * , σ 2 ) and ∀σ * < σ < σ 2 , w 2 (σ) < w(σ) < w - 2 (σ) lim σ↑σ 2 w(σ) = w 2 , (3.11)
see figure 1.

Remark 3.3. The above lemma shows that all such solutions provide admissible P 2 -P 4 connections.

Proof of Lemma 3.2. The fact that the solution generates a P 4 -P 2 connection with forward flow trapped in the region (3.11) follows again directly from the phase portrait of figure 3 and the monotonicity (2.10). We leave that to the reader, while we focus on the proof of the asymptotic expansion (3.10) near σ = 0.

As σ → 0, we have

dx dσ = - ∆(σ, w(σ)) ∆ 2 (σ, w(σ)) = - (w(σ) -1) 2 -σ 2 σ ( + d -1)w(σ) 2 -( + d + r -r)w(σ) + r -σ 2 = - 1 + o(1)
rσ so that x → +∞ as σ → 0. It thus suffices to consider the flow for x large enough.

Next, note that for (w, σ) → 0, we have

∆ 1 ∆ -rw + ∆ 2 ∆ -rσ w 2 + σ 2 .
Decomposing w and σ under the form

w(x) = w ∞ e -rx + w, σ(x) = σ ∞ e -rx + σ, we have d w dx + r w = - ∆ 1 ∆ -rw , dσ dx + rσ = - ∆ 2 ∆ -rσ . Assuming w(x), σ(x) = o(e -rx ), x → +∞, we infer w(x) = e -rx +∞ x e ry ∆ 1 ∆ -rw dy, σ(x) = e -rx +∞ x e ry ∆ 2 ∆ -rσ dy.
An elementary fixed point argument then allows to obtain for any

(w ∞ , σ ∞ ) with |w ∞ | + |σ ∞ | = 1 a unique corresponding solution (w(x), σ(x)) satisfying (3.10),
which is defined for e -rx ≥ δ( , r, d) with δ( , r, d) > 0 small enough, and hence for |w| + |σ| ≤ δ( , r, d). We have thus described the behavior of all solution curves (w, σ) converging to P 4 as σ → 0, and shown that they all satisfy (3.10).

The results of the previous two sections provide the proof of all the statements of Lemma 1.2. We summarize them as follows. We have constructed the unique spherically symmetric smooth solution of (1.7) on the interval [0, Z 2 ) and a one parameter of spherically symmetric smooth solutions of (1.7) on the interval (Z 2 , ∞). These solutions agree at Z 2 and thus can be glued to each other continuously. Our goal however is to construct a global C ∞ solution. At this point it is already clear that the crux of the matter is the point P 2 .

3.3. Diagonalized system at P 2 . The point P 2 will play an essential role in the proof of Theorem 1.3. The dynamical properties of this point in the regime (3.1) can only be seen after passing to the diagonalized variables (2.47). We recall the values of the slopes (2.41), (2.44), (2.46), the diagonalization matrices (2.47) and the non degeneracy properties of Lemma 2.8 and Lemma 2.9 in the range (3.1). We rewrite the system in coordinates which diagonalize its linear part. We will also introduce a time variable and recast the system as a dynamical flow approaching the point P 2 (from either side) as t → ∞. Lemma 3.4 (Equations in the diagonal form). Assume (3.1). Let

w = w 2 + W σ = σ 2 + Σ dt dx = -1 ∆ , X = W Σ , Y = P -1 X = W Σ (3.12)
then (1.9) becomes:

dY dt = 1 c + -c - G 1 G 2 (3.13)
with

G 1 = (c + -c -)λ + W + d20 W 2 + d11 W Σ + d02 Σ2 + d30 W 3 + d21 W 2 Σ + d12 W Σ2 + d03 Σ 3 = -∆ 1 + c + ∆ 2 , (3.14)
G 2 = (c + -c -)λ -Σ + ẽ20 W 2 + ẽ11 W Σ + ẽ02 Σ2 + ẽ30 W 3 + ẽ21 W 2 Σ + ẽ12 W Σ2 + ẽ03 Σ 3 = ∆ 1 -c -∆ 2 (3.15)
and where the values of the coefficients are collected in (F.1), (F.2).

Proof of Lemma 3.4. This is a direct computation.

step 1 Reexpressing ∆, ∆ 1 , ∆ 2 . Let w = w 2 + W , σ = σ 2 + Σ, we compute the nonlinear terms ∆ 1 = w 3 -(r + 1)w 2 + rw -dwσ 2 + (r -1)σ 2 = w 3 2 + 3w 2 2 W + 3w 2 W 2 + W 3 -(r + 1)(w 2 2 + 2w 2 W + W 2 ) + r(w 2 + W ) -d(w 2 + W )(σ 2 2 + 2σ 2 Σ + Σ 2 ) + (r -1)(σ 2 2 + 2σ 2 Σ + Σ 2 ) = W (3w 2 2 -2(r + 1)w 2 + r -dσ 2 2 ) + Σ(-2dσ 2 w 2 + 2 (r -1)σ 2 ) + W 2 (3w 2 -(r + 1)) + Σ 2 ( (r -1) -dw 2 ) + ΣW (-2dσ 2 ) + W 3 -dW Σ 2 = c 1 W + c 3 Σ + d 20 W 2 + d 11 W Σ + d 02 Σ 2 + W 3 -dW Σ 2 (3.16)
and

∆ 2 = σ ( + d -1)w 2 -w( + d + r -r) + r -σ 2 = σ 2 + Σ ( + d -1)(w 2 2 + 2w 2 W + W 2 ) -( + d + r -r)(w 2 + W ) + r -(σ 2 2 + 2σ 2 Σ + Σ 2 ) = σ 2 + Σ ( + d -1)(2w 2 W + W 2 ) -( + d + r -r)W -(2σ 2 Σ + Σ 2 ) = W σ 2 [2w 2 ( + d -1) -( + d + r -r)] -2σ 2 2 Σ + W 2 σ 2 ( + d -1) + Σ 2 [-3σ 2 ] + W Σ 2w 2 ( + d -1) -( + d + r -r) + W 2 Σ + d -1 + Σ 3 [-1] = c 2 W + c 4 Σ + e 20 W 2 + e 11 W Σ + e 02 Σ 2 + e 21 W 2 Σ + Σ 3 [-1] (3.17)
where the parameters are given by (F.1).

step 2 Y variable. We now pass to the Y variable:

∆ 1 (X) = c 1 W + c 3 Σ + d 20 W 2 + d 11 W Σ + d 02 Σ 2 + W 3 -dW Σ 2 = c 1 (c -W + c + Σ) + c 3 ( W + Σ) + d 20 (c -W + c + Σ) 2 + d 11 (c -W + c + Σ)( W + Σ) + d 02 ( W + Σ) 2 + (c -W + c + Σ) 3 -d(c -W + c + Σ)( W + Σ) 2 = (c 1 c -+ c 3 ) W + (c 1 c + + c 3 ) Σ + d 20 (c 2 - W 2 + 2c -c + W Σ + c 2 + Σ2 ) + d 11 (c - W 2 + (c -+ c + ) W Σ + c + Σ2 ) + d 02 ( W 2 + 2 W Σ + Σ2 ) + (c 3 - W 3 + 3c 2 - W 2 c + Σ + 3c -W c 2 + Σ2 + c 3 + Σ3 ) -d(c -W + c + Σ)( W 2 + 2 W Σ + Σ2 ) = (c 1 c -+ c 3 ) W + (c 1 c + + c 3 ) Σ + (d 20 c 2 -+ d 11 c -+ d 02 ) W 2 + (2c -c + d 20 + (c -+ c + )d 11 + 2d 02 ) W Σ + (d 20 c 2 + + d 11 c + + d 02 ) Σ2 + (c 3 --dc -) W 3 + (3c 2 -c + -2dc --dc + ) W 2 Σ + (3c -c 2 + -dc --2dc + ) W Σ2 + (c 3 + -dc + )Σ 3 and ∆ 2 (X) = c 2 W + c 4 Σ + e 20 W 2 + e 11 W Σ + e 02 Σ 2 + + d -1 W 2 Σ -Σ 3 = c 2 (c -W + c + Σ) + c 4 ( W + Σ) + e 20 (c -W + c + Σ) 2 + e 11 (c -W + c + Σ)( W + Σ) + e 02 ( W + Σ) 2 + + d -1 (c -W + c + Σ) 2 ( W + Σ) -( W + Σ) 3 = (c 2 c -+ c 4 ) W + (c 2 c + + c 4 ) Σ + e 20 (c 2 - W 2 + 2c -c + W Σ + c 2 + Σ2 ) + e 11 (c - W 2 + (c -+ c + ) W Σ + c + Σ2 ) + e 02 ( W 2 + 2 W Σ + Σ2 ) + + d -1 c 2 - W 3 + (c 2 -+ 2c + c -) W 2 Σ + (2c -c + + c 2 + ) W Σ2 + c 2 + Σ3 -( W 3 + 3 W 2 Σ + 3 W Σ2 + Σ3 ) = (c 2 c -+ c 4 ) W + (c 2 c + + c 4 ) Σ + (
+ + d -1 c 2 --1 W 3 + + d -1 (c 2 -+ 2c -c + ) -3 W 2 Σ + + d -1 (2c -c + + c 2 + ) -3 W Σ2 + + d -1 c 2 + -1 Σ3 .
Linear terms yield

1 c + -c - -((c 1 c -+ c 3 ) W + (c 1 c + + c 3 ) Σ) + c + [(c 2 c -+ c 4 ) W + (c 2 c + + c 4 ) Σ] (c 1 c -+ c 3 ) W + (c 1 c + + c 3 ) Σ -c -[(c 2 c -+ c 4 ) W + (c 2 c + + c 4 ) Σ] = 1 c + -c - [-(c 1 c -+ c 3 ) + c + (c 2 c -+ c 4 )] W + [-(c 1 c + + c 3 ) + c + (c 2 c + + c 4 )] Σ ((c 1 c -+ c 3 ) -c -(c 2 c -+ c 4 )) W + [(c 1 c + + c 3 ) -c -(c 2 c + + c 4 )] Σ
We then recall the c ± equation (2.44) which kills the off diagonal term

-(c 1 c + + c 3 ) + c + (c 2 c + + c 4 ) = 0 (c 1 c -+ c 3 ) -c -(c 2 c -+ c 4 ) = 0
and compute using (2.50):

-(c 1 c -+c 3 )+c + (c 2 c -+c 4 ) c + -c - = -c -(c 2 c -+c 4 )++c + (c 2 c -+c 4 ) c + -c - = c 2 c -+ c 4 = λ + (c 1 c + +c 3 )-c -(c 2 c + +c 4 ) c + -c - = c + (c 2 c + +c 4 )-c -(c 2 c + +c 4 ) c + -c - = c 2 c + + c 4 = λ -
For the quadratic terms, we compute for the first coordinate

-dc -) + c + + d -1 c 2 --1 and d21 = -(3c 2 -c + -2dc --dc + ) + c + + d -1 (c 2 -+ 2c -c + ) -3 and d12 = -(3c -c 2 + -dc --2dc + ) + c + + d -1 (2c -c + + c 2 + ) -3 .
Similarly for the second coordinate:

-c + -2dc --dc + ) -c - + d -1 (c 2 -+ 2c -c + ) -3 , this is (F.2).
Introducing the time variable t (we note that t → +∞ corresponds to both Z ↑↓ Z 2 ):

dt dx = - 1 ∆ yields dY dt = (c + -c -)λ + W + d20 W 2 + d11 W Σ+ d02 Σ2 + d30 W 3 + d21 W 2 Σ+ d12 W Σ2 + d03 Σ 3 c + -c - (c + -c -)λ -Σ+ẽ 20 W 2 +ẽ 11 W Σ+ẽ 02 Σ2 +ẽ 30 W 3 +ẽ 21 W 2 Σ+ẽ 12 W Σ2 +ẽ 03 Σ 3 c + -c - .
This is (3.14), (3.15).

3.4.

Integral curves passing through P 2 .

Lemma 3.5 (Slope of the curves converging to P 2 ). Assume (3.1). Let c + , c -, A be given by (2.44), (2.48). Then, (1) all integral curves of (1.9) converging to P 2 have slope given either by c + or c -, where the slope refers to the limit of dw dσ as σ → σ 2 , (2) there are only two curves with slope c + , with one converging to P 2 from the region ∆ 1 > 0 and the other converging to P 2 from the region ∆ 1 < 0, while all the other curves have slope c -, (3) the unique curve converging to P 2 from the region ∆ 1 > 0 with the slope c + exists on 0 < σ < σ 2 and converges to P 4 as σ → 0.

Proof. The Jacobian matrix of the autonomous system (3.13) at the equilibrium ( W , Σ) = (0, 0) is diagonal with two negative eigenvalues λ -< λ + < 0. Thus, standard results imply that (0, 0) is an asymptotically stable node, and the following holds for the trajectories converging to (0, 0) as t → +∞ (1) there exists exactly two trajectories tangent to the eigenvector of the Jacobian matrix corresponding to the smallest eigenvalue λ -, i.e. these trajectories satisfy

lim t→+∞ W Σ = 0,
(2) all the other trajectories are tangent to the eigenvector of the Jacobian matrix corresponding to the largest eigenvalue λ + , i.e. these trajectories satisfy

lim t→+∞ Σ W = 0.
Coming back to (W, Σ), we infer that the slope of any curve converging to P 2 is either c + or c -, and there are exactly two with slope c + , while all the others have slope c -. Also, since c + > 0 while the slope w 2 (σ 2 ) of the middle root of ∆ 1 is negative, see Lemma 2.9, among the two curves converging to P 2 with c + slope, one converges to P 2 from the region ∆ 1 > 0 and the other converges to P 2 from the region ∆ 1 < 0.

The fact that the unique curve converging to P 2 from the region ∆ 1 > 0 with slope c + is defined all the way to σ = 0 and attracted to P 4 is a straightforward consequence of the phase portrait in Figure 3. Lemma 3.6. Assume (3.1). Let c -, A be given by (2.44), (2.48) and assume that

A = K + α, K ∈ N\{0, 1}, 0 < α < 1.
Then, there exists a unique solution curve (w, σ) to (1.9) which is a C ∞ function of x at P 2 with slope c -. This solution also satisfies that w(σ) is a C ∞ function of σ at σ 2 .

Remark 3.7. One can show that all the other curves converging to P 2 with slope c -, see Lemma 3.5, are C K+α both on the right and on the left of P 2 .

Proof. Recall from Lemma 3.5 that the curves converging to P 2 with slope c -satisfy W = o( Σ) near P 2 , and hence, we have at P 2 , i.e., at W = 0,

Ψ := Σ W , Ψ(0) = 0. (3.20) Now W d Ψ d W + Ψ = d Σ d W = (c + -c -)λ -Σ + ẽ20 W 2 + ẽ11 W Σ + ẽ02 Σ2 + ẽ30 W 3 + ẽ21 W 2 Σ + ẽ12 W Σ2 + ẽ03 Σ 3 (c + -c -)λ + W + d20 W 2 + d11 W Σ + d02 Σ2 + d30 W 3 + d21 W 2 Σ + d12 W Σ2 + d03 Σ 3 = (c + -c -)λ -Ψ + ẽ20 W + ẽ11 W Ψ + ẽ02 Ψ2 W + ẽ30 W 2 + ẽ21 W 2 Ψ + ẽ12 W 2 Ψ2 + ẽ03 Ψ3 W 2 (c + -c -)λ + + d20 W + d11 W Ψ + d02 Ψ2 W + d30 W 2 + d21 W 2 Ψ + d12 W 2 Ψ 2 + d03 Ψ3 W 2
and hence

W d Ψ d W - λ - λ + -1 Ψ = W G 1 ( W , Ψ) G 2 ( W , Ψ) with G 2 (0, 0) = 0. Since A = λ -/λ + , we infer W d Ψ d W -[A -1] Ψ = W G 1 ( W , Ψ) G 2 ( W , Ψ) . Let k ≥ 1.
Assuming that W is C ∞ , we differentiate k times, and evaluate at W = 0.

We infer

[k + 1 -A] Ψ(k) (0) = k G 1 ( W , Ψ) G 2 ( W , Ψ) (k-1) | ( W , Ψ)=(0,0) , k ≥ 1, Ψ(0) = 0. (3.21)
Since A / ∈ N, (3.21) yields the uniqueness of the Taylor expansion of Ψ at any order. Let k ∈ N * with k > A -2. We denote by P k ( W ) the unique Taylor polynomial of degree k provided by solving the iteration (3.21

) for j = 1, • • • , k. Then, P k ( W ) satisfies W dP k d W -[A -1] P k = W G 1 ( W , P k ( W )) G 2 ( W , P k ( W )) + W k+1 Q k ,
where Q k is a polynomial uniquely prescribed by k, A, G 1 and G 2 . We decompose

Ψ = P k + εk , εk = O( W k+1 ).
Plugging this decomposition into the equation for Ψ, we infer

W dε k d W -[A -1] εk = W G 1 ( W , P k ( W ) + εk ) G 2 ( W , P k ( W ) + εk ) - W dP k d W -[A -1] P k = -W k+1 Q k + W εk H k ( W , εk )
where H k is a smooth function near (0, 0) uniquely determined by k, A, G 1 and G 2 . We deduce

d d W W 1-A εk = -W k+1-A Q k + εk W A-1 H k ( W , εk ).
Now, since εk = O( W k+1 ) and k > A -2, we deduce that if a solution curve is C ∞ at P 2 with slope c -, then it must satisfy

εk = W A-1 W 0 -wk+1-A Q k + εk wA-1 H k ( w, εk ) d w. (3.22)
Using a fixed point argument, there exists a unique solution to (3.22) defined in a neighborhood of W = 0. Uniqueness follows.

We now focus on the existence of a solution curve C ∞ at P 2 with slope c -. Let

Ψ := P A -2 + ε A -2 ,
where ε A -2 is the unique solution to the fixed point argument fixed (3.22) for

k = A -2 > A -2.
With this definition, Ψ yields a solution curve with slope c - at P 2 , and it remains to prove that it is C ∞ at P 2 . In fact, it suffices to prove that

P k + εk = P A -2 + ε A -2 for all k ≥ A -2, (3.23)
as it will then imply that Ψ has a Taylor expansion at P 2 at any order and is hence C ∞ . Next, we focus on proving (3.23). For k > A -2, we define δ Ψk,A by δ Ψk,A :

= P k -P A -2 + εk -ε A -2
which satisfies in view of the properties of

P A -2 , P k , ε A -2 and εk |δ Ψk,A | = O( W A -1 )
and

W dδ Ψk,A d W -[A -1] δ Ψk,A = W δ Ψk,A J k,A ( W , P k , P A -2 , εk , ε A -2 )
where J k,A is a smooth function near (0, 0) uniquely prescribed by k, A, G 1 and G 2 .

In view of the above control for δ Ψk,A , we deduce

|δ Ψk,A | W A-1 W 0 |δ Ψk,A | wA-1 d w.
Hence δ Ψk,A vanishes identically, so that (3.23) holds as desired.

We have thus obtained a unique solution ( Σ, W ) such that Ψ is C ∞ at W = 0. Now, proceeding as in the proof of Lemma 3.4, we easily obtain from (2.47) (3.12)

∆ = W c ∆ + W F 0 ( W , Ψ) + ΨG 0 ( W , Ψ) , ∆ 1 = W c ∆ 1 + W F 1 ( W , Ψ) + ΨG 1 ( W , Ψ) , ∆ 2 = W c ∆ 2 + W F 2 ( W , Ψ) + ΨG 2 ( W , Ψ) ,
where F j , G j , j = 0, 1, 2 are polynomials in ( W , Ψ) and c ∆ j , j = 0, 1, 2 are nonzero constants. Together with (3.14), this implies that

G 1 ∆ , ∆ 1 ∆ 2 ,
are C ∞ and nonzero at W = 0. In view of (3.13), we have

d W dx = 1 c + -c - G 1 dt dx = - 1 c + -c - G 1 ∆ so that W is a C ∞ function of x at P 2 .
Then Ψ and hence Σ are also C ∞ function of x at P 2 , and hence, (w, σ) is a solution to (1.9) which is a C ∞ function of x at

P 2 . Finally, since ∆ 1 ∆ 2 is C ∞ and nonzero at W = 0, w(σ) is a C ∞ function of σ at σ 2 .

Renormalization of the flow near P 2

Our aim in this section is to start the proof of Theorem 1.3 with the renormalization of the flow (1.9) for 0 < r o -r 1 near P 2 with r o given by (1.14). We first detail the strategy of the proof in section 1.3, and then proceed to the expected renormalization. The proof of Theorem 1.3 will then be completed in sections 5, 6, 7. For the rest of this paper, we assume

∈ R * \{d}, 0 < r o (d, ) -r 1.
We will also denote Here we use a fundamental degeneracy of the phase portrait, see Lemma 2.6, which gives the eye property lim

P o = P 5 for < d P 3 for < d.
r↑r o |P o (r) -P 2 (r)| = 0, (4.1)
and the slopes of w 2 (σ) and w - 2 (σ) converge to the same value, see figure 3,4. Passing to the diagonalized variables (2.47) and after explicit suitable reductions, this degeneracy leads to a quadratic cancellation, (4.13). Solving for w(σ), we are left with the study of a problem of the form 9 :

u(1 -u) dΘ du -[γ -2 + (ν b + 3)u] Θ = - G b . (4.2)
Here we introduced the parameters which appear in the renormalization process

γ = c∞(d, )(1+o b→0 (1)) b , c ∞ > 0 ν b = ν ∞ (d, ) + o b→0 (1), ν ∞ = 0 (4.3)
The trajectory is Θ(u) with u = 0 at σ 2 , u = 1 at σ o . G is an explicit nonlinear term. An analogous reduction can be performed to the right of P 2 .

step 2 Main part of the solution. The nonlinear ode (4.2) has a regular singular point at the origin u = 0. Therefore, it admits a C ∞ solution with an holomorphic expansion at the origin

Θ(u) = +∞ k=0 θ k (b, d, )u k
where θ k (b, d, ) is given by an explicit b dependent induction relation. We let

γ -1 = K + α γ , 0 < α γ < 1 (4.4)
and truncate the holomorphic expansion at the critical frequency 10 :

Θ(u) = K-2 k=0 θ k u k + θ K-1 u K-1 + (-1) K-1 S K-1 Θ main (u) -T (r G ) (4.5)
where

Θ main = O(u K
) is an explicit integral, and T (r G ) is a remainder which is of higher order. Our first fundamental observation is that there exists a strong limit

lim b→0 S K-1 = S ∞ (d, ). (4.6)
The proof relies on bounding the formal series solution to a limiting problem with b = 0 first. This is done in Proposition 5.1, which belongs to the realm of nonlinear Maillet theorems, [START_REF] Malgrange | Sur le théorème de Maillet[END_REF][START_REF] Sibuya | Linear Differential Equations in the Complex Domain: Problems of Analytic Continuation[END_REF]. The original problem (4.2) can be thought of as a b-deformation of the limiting problem. The challenge however is that we need uniform estimates for all frequencies up to the critical value K which itself is of size ∼ 1 b 9 observe that the notation u for the variable appearing in (4.2) should not be confused with the one for the solution to the Euler equation (1.1). 10 which corresponds to the limit of regularity of a generic solution, see Lemma 3.6.

step 3 Non vanishing of S ∞ (d, ). The proof of finiteness of S ∞ (d, ) implies the analycity of the mapping → S ∞ (d, ) on suitable domains of the complex plane (see Lemma 1.5). This number can be reexpressed as an explicit normally convergent series, but we do not know how to prove analytically that it is non zero. We therefore perform a numerical study of this convergent series which allows us to provide windows of parameters (d, ) for which

S ∞ (d, ) = 0. ( 4
Θ(u * ) = Θ * , 0 < u * < 1. (4.8)
The second fundamental observation is that the function Θ main can be analyzed explicitly near u = 0

Θ main = Γ(α γ )Γ(1 -α γ )K ν b +3-αγ u K-1 (4.9) × [1 + o b→0 (1)] 1 Γ(1 -α γ ) + K + ν b + 2 Γ(2 -α γ ) u + lot .
In a boundary layer close to the integer values

α γ ∈ (ε 1 , 2ε 1 ) ∪ (1 -2ε 2 , 1 -ε 2 ), ε i = o b→0 (1), (4.10)
we can ensure that (4.8) happens for a small 0 < u * (α γ ) 1. For the interval 2ε 1 < α γ < 1 -2ε 2 , we need to understand Θ main away from u = 0 where the truncated Taylor expansion no longer dominates, and here we use the explicit integral representation of Θ main to show that (4.8) happens for u * (α γ ) < 1 2 . The conclusion is that for every α γ ∈ (0, 1) except maybe a very small b dependent boundary layer around the integer values, the solution to (4.5) reaches (4.8) in time 0 < u * < 1 2 . This means that we are leaving a large neighborhood around the P 2 -P 5 separatrix 11 with a prescribed sign Θ * 1. A further use of monotonicity properties of the flow (1.9) allows us to conclude that the integral curve will intersect either the root branch w 2 (σ) or w - 2 (σ) for some σ 5 < σ * < σ 2 . The (-1) K prefactor in (4.5) dictates that the former happens when K is even while the latter holds when K is odd (if S ∞ > 0 and the other way around if S ∞ < 0.) Once the trajectory reaches w 2 (σ), by Lemma 3.2, it then continues on to P 4 , as desired.

step 5 Oscillations at the right of P 2 . The analysis of the flow to the right of P 2 produces the same decomposition (4.9) but with u < 0. We then observe since lim αγ ↑1 Γ(1-α γ ) = +∞ that by choosing α γ in a boundary layer close to respectively 0 or 1, the sign u < 0 allows us to reach

Θ(u * ) = Θ * for ε 1 < α γ < 2ε 1 -Θ * for ε 2 < 1 -α γ < 2ε 2 , u * < 0, |u * | 1.
Then, if K is even, the curve will exit through w 2 (σ) in the first case, in which case we say it exits red, and through the branch w - 2 (σ) in the second case, in which case we say it exits green. For odd K the situation is reversed (Again, this holds for S ∞ > 0. For S ∞ < 0 the picture is reversed.) step 6 Conclusion by continuity. Given K with the suitable parity (depending on the sign of S ∞ (d, ) = 0), we vary the parameter α γ ∈ (ε 1 , 1 -ε 2 ) continuously and conclude that the C ∞ curve through P 2 crosses the red at the left of P 2 for all α γ ∈ (e 1 , 1 -ε 2 ), while it also crosses the red at the right of P 2 at the beginning of the α γ interval, green at the end. Hence an elementary continuity argument implies the existence of at least one value α * γ ∈ (ε 1 , 1 -ε 2 ) such that the solution curve intersects the P 2 -P 6 trajectory given by Lemma 3.1. The two curves intersect each other away from singular points and thus, by uniqueness, must coincide. It follows easily that this constructed solution satisfies the conclusions of Theorem 1.3. In other words, as long as the non degeneracy condition (4.7) is satisfied, the integer interval γ ∈ [K + 1, K + 2], with K given by (4.4) large enough and of suitable parity, contains at least one C ∞ P 6 -P 4 solution. Since γ is related to b through (4.3) and b = o r↑r o (1), the above construction produces an infinite family of global C ∞ solutions parametrized by the speeds

r k ↑ r o (d, ) for each (d, ) such that S ∞ (d, ) = 0.
4.2. Degeneracy of the geometry at r o (d, ). We use Lemma 3.4 and the Y variable (3.12) to map (1.9) onto (3.13). The starting point of the renormalization procedure is the following fundamental degeneracy property as r ↑ r o (d, ).

Remark 4.1 (Notation for the parameters). From now on and for the rest of this paper we adopt the following notation: all slopes, characteristic eigenvalues and geometrical parameters involved in the renormalization of the flow, Lemma 3.4, depend of r, and will be noted with an ∞ subscript when evaluated at r = r o . The non degeneracy and signs of some of these limiting values will be crucial in the forthcoming analysis, and all relevant values are collected in Appendix F. and define

µ + = λ + b (4.12)
then, denoting by Wo and Σo the value at P o of W and Σ defined in (3.12), we have

Wo = -b (c + -c -)µ + d20 + O(b 2 ) Σo = -ẽ20 W 2 o (c + -c -)λ -+ O(b 3 ) µ + = µ ∞ + + O(b) (4.13)
where the non degenerate limiting values are computed in Appendix F.

Proof. The value of σ 2 (r) is computed from (2.18) and hence σ 2 ∈ C ∞ (1, r + ). Moreover, J(r) is from (2.25) a second order polynomial with roots r

+ = 1 + d-1 (1+ √ ) 2 < r -= 1 + d-1 (1- √
) 2 and hence the root r + is simple. Since r * < r + , we conclude that with the definition (4.11), σ 2 (r) and the slopes coefficients c i (r) given by (2.41)

are smooth functions of b on [0, b * ], 0 < b * (d, )
1 universal small enough. We now explicitly check that the determinant (c 1 -c 4 ) 2 + 4c 2 c 3 which appears in the definition of the slopes and eigenfunctions (2.44), (2.46) is non degenerate at r o , see limiting values in Appendix F and the non degeneracy of λ -, which ensures that c ± , λ ± are smooth functions of b all the way to b = 0. The eye property (4.1) thus implies

|W o | + |Σ o | = |w o -w 2 | + |σ eye -σ 2 | Cb.
Since the coefficients of the matrix P -1 given by (2.47) do not degenerate at r o from direct check, we conclude

| Wo | + | Σo | ≤ Cb.
The coefficients dij , ẽij of the polynomials of the RHS of (3.13) are computed from (F.2) and are O(1) at r o . Moreover G 1 , G 2 vanish at P o from (3.14), (3.15), and this forces (4.13).

4.3.

Renormalization. We now proceed to the renormalization of (3.13) for 0 < b 1.

Lemma 4.3 (Renormalization and quasilinear formulation). Let

W = -b w Σ = b 2 σ , ψ = σ w = ψo φ w = wo u , φ(u) = u + (1 -u)Ψ(u) (4.14)
where ψo and wo denote respectively the value at P o of ψ and w. Then (3.13) is mapped to the quasilinear problem

1 + H 2 + G 2 Ψ + NL 2 u(1 -u) dΨ du (4.15) + (1 -2u)(1 + H 2 + G 2 Ψ + NL 2 ) -γ(1 + G 1 ) + 2uG 2 Ψ = u γbH 1 -2(1 + H 2 ) + γb NL 1 bu -2 NL 2
where

H 1 , H 2 , G 1 , G 2 are explicit polynomials in (b, u)
given by (G.2), (G.4), and the nonlinear terms ( NL i ) i=1,2 are given by (G.5). Moreover, We compute from (4.13), (F.4), (F.12) the expansion as b → 0:

wo = (c ∞ + -c ∞ -)µ ∞ + d∞ 20 + O(b) ψo = - ẽ∞ 20 µ ∞ + d∞ 20 λ ∞ - + O(b). ( 4
µ + (r) = λ + (r) b = µ ∞ + + O(b) < 0 wo (r) = -Wo (r) b = (c ∞ + -c ∞ -)µ ∞ + d∞ 20 + O(b) ψo = σo wo = -Σo b 2 Wo b = ẽ20 Wo b(c + -c -)λ -= - ẽ∞ 20 µ ∞ + d∞ 20 λ ∞ - + O(b)
and (4.16) is proved. We now compute the flow from (3.13). First equation. We compute

G 1 c + -c - = -b 2 d w dτ = -b 2 (c + -c -)µ + w + b 2 d20 w2 -b 3 d11 wσ + b 4 d02 σ2 -b 3 d30 w3 + b 4 d21 w2 σ -b 5 d12 wσ 2 + b 6 d03 σ3 c + -c - i.e. d w dτ = -(c + -c -)|µ + | w + | d20 | w2 + b( d11 wσ + d30 w3 ) -b 2 ( d02 σ2 + d21 w2 σ) + b 3 d12 wσ 2 -b 4 d03 σ3 c + -c - .
We insert (4.17) and compute:

F 2 = - G 1 b 2 (4.18) = -(c + -c -)|µ + | w + | d20 | w2 + b( d11 wσ + d30 w3 ) -b 2 ( d02 σ2 + d21 w2 σ) + b 3 d12 wσ 2 -b 4 d03 σ3 = w -(c + -c -)|µ + | + | d20 | w + b( d11 w ψ + d30 w2 ) -b 2 ( d02 ψ2 w + d21 w2 ψ) + b 3 d12 w2 ψ2 -b 4 d03 w2 ψ3 = w -(c + -c -)|µ + | + | d20 | + b d11 ψ -b 2 d02 ψ2 w + b d30 -b 2 d21 ψ + b 3 d12 ψ2 -b 4 d03 ψ3 w2
Second equation. We compute

G 2 c + -c - = b 3 dσ dτ = b 2 (c + -c -)λ -σ + b 2 ẽ20 w2 -b 3 ẽ11 wσ + b 4 ẽ02 σ2 -b 3 ẽ30 w3 + b 4 ẽ21 w2 σ -b 5 ẽ12 wσ 2 + b 6 ẽ03 σ 3 c + -c - .
and hence

b dσ dτ = 1 c + -c - × -(c + -c -)|λ -|σ + ẽ20 w2 -b(ẽ 11 wσ + ẽ30 w3 ) + b 2 (ẽ 02 σ2 + ẽ21 w2 σ) -b 3 ẽ12 wσ 2 + b 4 ẽ03 σ3
and

F 1 = G 2 b 2 (4.19) = -(c + -c -)|λ -|σ + ẽ20 w2 -b(ẽ 11 wσ + ẽ30 w3 ) + b 2 (ẽ 02 σ2 + ẽ21 w2 σ) -b 3 ẽ12 wσ 2 + b 4 ẽ03 σ3 = w -(c + -c -)|λ -| ψ + ẽ20 w -b(ẽ 11 w ψ + ẽ30 w2 ) + b 2 (ẽ 02 w ψ2 + ẽ21 w2 ψ) -b 3 ẽ12 w2 ψ2 + b 4 ẽ03 w2 ψ3 ) = w -(c + -c -)|λ -| ψ + ẽ20 -bẽ 11 ψ + b 2 ẽ02 ψ2 w + -bẽ 30 + b 2 ẽ21 ψ -b 3 ẽ12 ψ2 + b 4 ẽ03 ψ3 w2 .
Conclusion. We have obtained the system

d w dτ = F 2 c + -c - (4.20) = w -(c + -c -)|µ + | + | d20 | + b d11 ψ -b 2 d02 ψ2 w + b d30 -b 2 d21 ψ + b 3 d12 ψ2 -b 4 d03 ψ3 w2 c + -c - and b dσ dτ = F 1 c + -c - (4.21) = w -(c + -c -)|λ -| ψ + ẽ20 -bẽ 11 ψ + b 2 ẽ02 ψ2 w + -bẽ 30 + b 2 ẽ21 ψ -b 3 ẽ12 ψ2 + b 4 ẽ03 ψ3 w2 c + -c - .
Remark 4.6. Note that from (4.13)

ψo = - Σo b Wo = O b→0 (1)
(4.21) then forces the relations for b = 0:

ψ∞ o = ẽ∞ 20 (c ∞ + -c ∞ -)|λ ∞ -| w∞ o |µ ∞ + | = | d∞ 20 | c ∞ + -c ∞ - w∞ o . (4.22)
Note also that we have the signs, valid for all d ≥ 2 and = d:

d∞ 20 < 0, ẽ∞ 20 > 0, see (F.8).
step 2 Normalization of w. We further renormalize the flow to obtain the leading order size 1 constants leading the nonlinear dynamics as b → 0. Let

w = wo u, ψ = ψo φ (4.23)
and define

γ = |λ -| |λ + | = |λ -| b|µ + | (4.24)
then recalling (4.20)

du dτ = 1 wo d w dτ = F 2 wo (c + -c -) = |µ + |u (4.25) × -1 + | d20 | + b d11 ψo φ -b 2 d02 ψ2 o φ 2 |µ + |(c + -c -) wo u + b d30 -b 2 d21 ψo φ + b 3 d12 ψ2 o φ 2 -b 4 d03 ψ3 o φ 3 |µ + |(c + -c -) w2 o u 2 (4.26) = |µ + |u × -1 + | D20 | + b D11 φ -b 2 D02 φ 2 u + b D30 -b D21 φ + b 2 D12 φ 2 -b 3 D03 φ 3 u 2
with Dij given by (G.1). Observe that by definition of P o :

-1 + | D20 | + b D11 -b 2 D02 + b D30 -b 2 D21 + b 3 D12 -b 4 D03 = 0.
We then compute

d ψ dτ = 1 w dσ dτ - σ w2 d w dτ = 1 w dσ dτ - ψ 1 w d w dτ ⇔ dφ dτ + 1 u du dτ φ = 1 ψo
w dσ dτ and recalling (4.21):

b w dσ dτ = F 1 (c + -c -) w = |λ -| ψo (4.27) × -φ + ẽ20 -bẽ 11 ψo φ + b 2 ẽ02 ψ2 o φ 2 (c + -c -)|λ -| ψo wo u + -bẽ 30 + b 2 ẽ21 ψo φ -b 3 ẽ12 ψ2 o φ 2 + b 4 ẽ03 ψ3 o φ 3 (c + -c -)|λ -| ψo w2 o u 2 (4.28) ⇔ 1 |µ + | ψo γ 1 w dσ dτ = -φ + Ẽ20 -b Ẽ11 φ + b 2 Ẽ02 )φ 2 u + -b Ẽ30 + b 2 Ẽ21 φ -b 3 Ẽ12 φ 2 + b 4 Ẽ03 φ 3 u 2
with Ẽij given by (G.1), and again by definition of P o :

-1 + Ẽ20 -b Ẽ11 + b 2 Ẽ02 -b Ẽ30 + b 2 Ẽ21 -b 3 Ẽ12 + b 4 Ẽ03 = 0.
This yields the renormalized φ equation

dφ dτ + 1 u du dτ φ = |µ + |γ -φ + Ẽ20 -b Ẽ11 φ + b 2 Ẽ02 φ 2 u + b -Ẽ30 + b Ẽ21 φ -b 2 Ẽ12 φ 2 + b 3 Ẽ03 φ 3 u 2 . step 3 Quasilinear formulation. Let Λ = Z d dZ = - 1 |µ + | d dτ ,
then equivalently:

Λu = u 1 -| D20 | + b D11 φ -b 2 D02 φ 2 u -b D30 -b D21 φ + b 2 D12 φ 2 -b 3 D03 φ 3 u 2 Λφ + Λu u φ = γ φ -Ẽ20 -b Ẽ11 φ + b 2 Ẽ02 φ 2 u -b -Ẽ30 + b Ẽ21 φ -b 2 Ẽ12 φ 2 + b 3 Ẽ03 φ 3 u 2 .
with the relation on the parameters:

-1 + | D20 | + b D11 -b 2 D02 + b D30 -b 2 D21 + b 3 D12 -b 4 D03 = 0 -1 + Ẽ20 -b Ẽ11 + b 2 Ẽ02 -b Ẽ30 + b 2 Ẽ21 -b 3 Ẽ12 + b 4 Ẽ03 = 0. (4.29) Let F 1 (u, φ) = φ -Ẽ20 -b Ẽ11 φ + b 2 Ẽ02 φ 2 u -b -Ẽ30 + b Ẽ21 φ -b 2 Ẽ12 φ 2 + b 3 Ẽ03 φ 3 u 2 F 2 (u, φ) = 1 -| D20 | + b D11 φ -b 2 D02 φ 2 u -b D30 -b D21 φ + b 2 D12 φ 2 -b 3 D03 φ 3 u 2 then this is Λu = uF 2 (u, φ) Λφ + Λu u φ = γF 1 (u, φ).
(4.30)

We have from (4.27), (4.25):

F 1 = |λ -| ψo wo (c + -c -)u(-F 1 ) F 2 = wo (c + -c -)|µ + |u(-F 2 ) (4.31) Then Λφ = Z dφ dZ = Z dφ du du dZ = Λu dφ du
and hence the φ(u) renormalized quasi linear formulation

dφ du + φ u = γF 1 (φ,u) uF 2 (φ,u) lim u→0 φ = 0, lim u→1 φ = 1. (4.32)
Reexpression of the nonlinear terms. From (4.29):

Ẽ20 = 1 + b( Ẽ11 + Ẽ30 ) -b 2 ( Ẽ02 + Ẽ21 ) + b 3 Ẽ12 -b 4 Ẽ03
Then,

F 1 (u, φ) = φ -1 + b( Ẽ11 + Ẽ30 ) -b 2 ( Ẽ02 + Ẽ21 ) + b 3 Ẽ12 -b 4 Ẽ03 -b Ẽ11 φ + b 2 Ẽ02 φ 2 u -b -Ẽ30 + b Ẽ21 φ -b 2 Ẽ12 φ 2 + b 3 Ẽ03 φ 3 u 2 = φ -u + b -( Ẽ11 + Ẽ30 )u + Ẽ11 φu + Ẽ30 u 2 + b 2 ( Ẽ02 + Ẽ21 )u -Ẽ02 φ 2 u -Ẽ21 φu 2 + b 3 -Ẽ12 u + Ẽ12 φ 2 u 2 + b 4 Ẽ03 u -Ẽ03 φ 3 u 2 .
Similarly,

| D20 | = 1 -b( D11 + D30 ) + b 2 ( D02 + D21 ) -b 3 D12 + b 4 D03 = 0 and F 2 (u, φ) = 1 -1 -b( D11 + D30 ) + b 2 ( D02 + D21 ) -b 3 D12 + b 4 D03 + b D11 φ -b 2 D02 φ 2 u -b D30 -b D21 φ + b 2 D12 φ 2 -b 3 D03 φ 3 u 2 = 1 -u + b ( D11 + D30 )u -D11 φu -D30 u 2 + b 2 -( D02 + D21 )u + D02 φ 2 u + D21 φu 2 + b 3 D12 u -D12 φ 2 u 2 + b 4 -D03 u + D03 φ 3 u 2 .
step 5 Changing variables. We change variables to make the critical points of the ode appear explicitly. Let

Φ = φ -u. (4.33) Reexpressing F 1 . Recall F 1 (u, φ) = φ -u + b -( Ẽ11 + Ẽ30 )u + Ẽ11 φu + Ẽ30 u 2 + b 2 ( Ẽ02 + Ẽ21 )u -Ẽ02 φ 2 u -Ẽ21 φu 2 + b 3 -Ẽ12 u + Ẽ12 φ 2 u 2 + b 4 Ẽ03 u -Ẽ03 φ 3 u 2 then -( Ẽ11 + Ẽ30 )u + Ẽ11 φu + Ẽ30 u 2 = -( Ẽ11 + Ẽ30 )u(1 -u) + Ẽ11 u Φ and ( Ẽ02 + Ẽ21 )u -Ẽ02 φ 2 u -Ẽ21 φu 2 = ( Ẽ02 + Ẽ21 )u -Ẽ02 (u + Φ) 2 u -Ẽ21 u 2 (u + Φ) = ( Ẽ02 + Ẽ21 )u(1 -u)(1 + u) -u 2 (2 Ẽ02 + Ẽ21 ) Φ -Ẽ02 u Φ 2 and -Ẽ12 u + Ẽ12 φ 2 u 2 = -Ẽ12 u + Ẽ12 (u + Φ) 2 u 2 = -Ẽ12 u(1 -u)(1 + u + u 2 ) + 2 Ẽ12 u 3 Φ + Ẽ12 u 2 Φ 2 and Ẽ03 u -Ẽ03 φ 3 u 2 = Ẽ03 u -Ẽ03 (u + Φ) 3 u 2 = Ẽ03 u(1 -u)(1 + u + u 2 + u 3 ) -Ẽ03 (3u 4 Φ + 3u 3 Φ 2 + u 2 Φ 3 ) Thus, F 1 (u, φ) = Φ + b -( Ẽ11 + Ẽ30 )u(1 -u) + Ẽ11 u Φ + b 2 ( Ẽ02 + Ẽ21 )u(1 -u)(1 + u) -u 2 (2 Ẽ02 + Ẽ21 ) Φ -Ẽ02 u Φ 2 + b 3 -Ẽ12 u(1 -u)(1 + u + u 2 ) + 2 Ẽ12 u 3 Φ + Ẽ12 u 2 Φ 2 + b 4 Ẽ03 u(1 -u)(1 + u + u 2 + u 3 ) -Ẽ03 (3u 4 Φ + 3u 3 Φ 2 + u 2 Φ 3 ) = u(1 -u) -b( Ẽ11 + Ẽ30 ) + b 2 ( Ẽ02 + Ẽ21 )(1 + u) -b 3 Ẽ12 (1 + u + u 2 ) + b 4 Ẽ03 (1 + u + u 2 + u 3 ) + Φ 1 + Ẽ11 bu -(2 Ẽ02 + Ẽ21 )b 2 u 2 + 2 Ẽ12 b 3 u 3 -3 Ẽ03 b 4 u 4 + b Φ 2 -Ẽ02 bu + Ẽ12 b 2 u 2 -3 Ẽ03 b 3 u 3 -b 2 Φ 3 Ẽ03 b 2 u 2 = u(1 -u)bH 1 (b, u) + (1 + G 1 (bu)) Φ + NL 1 (u, Φ) (4.34) with H 1 (b, u) = -( Ẽ11 + Ẽ30 ) + b( Ẽ02 + Ẽ21 )(1 + u) -b 2 Ẽ12 (1 + u + u 2 ) + b 3 Ẽ03 (1 + u + u 2 + u 3 ) G 1 (x) = Ẽ11 x -(2 Ẽ02 + Ẽ21 )x 2 + 2 Ẽ12 x 3 -3 Ẽ03 x 4 NL 1 (u, Φ) = b Φ 2 -Ẽ02 bu + Ẽ12 b 2 u 2 -3 Ẽ03 b 3 u 3 -b 2 Φ 3 Ẽ03 b 2 u 2 (4.35) Reexpressing F 2 . Recall F 2 (u, φ) = 1 -u + b ( D11 + D30 )u -D11 φu -D30 u 2 + b 2 -( D02 + D21 )u + D02 φ 2 u + D21 φu 2 + b 3 D12 u -D12 φ 2 u 2 + b 4 -D03 u + D03 φ 3 u 2 then ( D11 + D30 )u -D11 φu -D30 u 2 = ( D11 + D30 )u(1 -u) -D11 u Φ and -( D02 + D21 )u + D02 φ 2 u + D21 φu 2 = -( D02 + D21 )u + D02 (u + Φ) 2 u + D21 (u + Φ)u 2 = -( D02 + D21 )u(1 -u)(1 + u) + u 2 (2 D02 + D21 ) Φ + D02 u Φ 2 and D12 u -D12 φ 2 u 2 = D12 u -D12 (u + Φ) 2 u 2 = D12 u(1 -u)(1 + u + u 2 ) -2 D12 u 3 Φ -D12 u 2 Φ 2 and -D03 u + D03 φ 3 u 2 = -D03 u + D03 (u + Φ) 3 u 2 = -D03 u(1 -u)(1 + u + u 2 + u 3 ) + D03 (3u 4 Φ + 3u 3 Φ 2 + u 2 Φ 3 ).
Then,

F 2 (u, φ) = 1 -u + b ( D11 + D30 )u(1 -u) -D11 u Φ + b 2 -( D02 + D21 )u(1 -u)(1 + u) + u 2 (2 D02 + D21 ) Φ + D02 u Φ 2 + b 3 D12 u(1 -u)(1 + u + u 2 ) -2 D12 u 3 Φ -D12 u 2 Φ 2 + b 4 -D03 u(1 -u)(1 + u + u 2 + u 3 ) + D03 (3u 4 Φ + 3u 3 Φ 2 + u 2 Φ 3 ) = (1 -u) 1 + b( D11 + D30 )u -b 2 ( D02 + D21 )u(1 + u) + b 3 D12 u(1 + u + u 2 ) -b 4 D03 u(1 + u + u 2 + u 3 ) + Φ -D11 bu + (2 D02 + D21 )b 2 u 2 -2 D12 b 3 u 3 + 3 D03 b 4 u 4 + b Φ 2 D02 bu -D12 b 2 u 2 + 3 D03 b 3 u 3 + b 2 Φ 3 D03 b 2 u 2 = (1 -u) [1 + H 2 (b, u)] + G 2 (bu) Φ + NL 2 (u, Φ) (4.36) with H 2 (b, u) = b( D11 + D30 )u -b 2 ( D02 + D21 )u(1 + u) + b 3 D12 u(1 + u + u 2 ) -b 4 D03 u(1 + u + u 2 + u 3 ) (4.37)
and

G 2 (x) = -D11 x + (2 D02 + D21 )x 2 -2 D12 x 3 + 3 D03 x 4 NL 2 (u, Φ) = b Φ 2 D02 bu -D12 b 2 u 2 + 3 D03 b 3 u 3 + b 2 Φ 3 D03 b 2 u 2 (4.38)
step 6 Final change of variables. We now reexpress (4.32) as

F 2 (φ, u) u dφ du + φ = γF 1 ⇔ (1 -u)(1 + H 2 ) + G 2 Φ + NL 2 u d Φ du + Φ + 2u = γ u(1 -u)bH 1 + (1 + G 1 ) Φ + NL 1
We change variables

Φ = (1 -u)Ψ (4.39)
and define x = bu.

Nonlinear terms. We rewrite from (4.35)

NL 1 = bM 11 (x) Φ 2 + b 2 M 12 (x) Φ 3 M 11 = -Ẽ02 x + Ẽ12 x 2 -3 Ẽ03 x 3 M 12 = Ẽ03 x 2
and from (4.38):

NL 2 = bM 21 (x) Φ 2 + b 2 M 22 (x) Φ 3 M 21 = D02 x -D12 x 2 + 3 D03 x 3 M 22 = D03 x 2 (4.40) Then, NL 1 = bM 21 (1 -u) 2 Ψ 2 + b 2 M 12 (1 -u) 3 Ψ 3 = (1 -u) (b -x)M 11 (x)Ψ 2 + (b 2 -2bx + x 2 )M 12 Ψ 3 ≡ (1 -u) NL 1 (4.41)
and

NL 2 = bM 21 (1 -u) 2 Ψ 2 + b 2 M 22 (1 -u) 3 Ψ 3 = (1 -u) (b -x)M 21 (x)Ψ 2 + (b 2 -2bx + x 2 )M 22 Ψ 3 ≡ (1 -u) NL 2 (4.42)
Ψ equation. We compute

u d Φ du + Φ + 2u = u (1 -u) dΨ du -Ψ + (1 -u)Ψ + 2u = u(1 -u) dΨ du + (1 -2u)Ψ + 2u.
Then,

(1 -u)(1 + H 2 ) + G 2 Φ + NL 2 = (1 -u) 1 + H 2 + G 2 Ψ + NL 2 u(1 -u)bH 1 + (1 + G 1 ) Φ + NL 1 = (1 -u) buH 1 + (1 + G 1 )Ψ + NL 1 .
This gives the Ψ equation:

1 + H 2 + G 2 Ψ + NL 2 u(1 -u) dΨ du + (1 -2u)Ψ + 2u = γ xH 1 + (1 + G 1 )Ψ + NL 1 = u γbH 1 + γ u + γbG 1 x Ψ + γb NL 1 x .
Equivalently:

1 + H 2 + G 2 Ψ + NL 2 u(1 -u) dΨ du + (1 -2u)Ψ = u γbH 1 + γ u + γbG 1 x Ψ + γb NL 1 x -2 1 + H 2 + G 2 Ψ + NL 2 i.e., 1 + H 2 + G 2 Ψ + NL 2 u(1 -u) dΨ du + (1 -2u)(1 + H 2 + G 2 Ψ + NL 2 ) -γ(1 + G 1 ) + 2uG 2 Ψ = u γbH 1 -2(1 + H 2 ) + γb NL 1 x -2 NL 2 .
Reordering terms. We split H 2 :

H 2 (b, u) = b( D11 + D30 )u -b 2 ( D02 + D21 )u(1 + u) + b 3 D12 u(1 + u + u 2 ) -b 4 D03 u(1 + u + u 2 + u 3 ) = 3 j=0 b j H 2,j (x),
and similarly

H 1 (b, u) = -( Ẽ11 + Ẽ30 ) + b( Ẽ02 + Ẽ21 )(1 + u) -b 2 Ẽ12 (1 + u + u 2 ) + b 3 Ẽ03 (1 + u + u 2 + u 3 ) = 3 j=0 b j H 1,j (bu)
with H i,j given by (G.3). We reorder the nonlinear terms using the same rule:

NL 1 = (b -x)M 11 (x)Ψ 2 + (b 2 -2bx + x 2 )M 12 Ψ 3 = 2 j=0 b j NL 1j NL 2 = (b -x)M 21 (x)Ψ 2 + (b 2 -2bx + x 2 )M 22 Ψ 3 = 2 j=0 b j NL 2j (4.43)
with (G.5).

Bounding the Taylor series of the formal limit problem

We now start the analysis of the non linear ode (4.15) for 0 < u < 1. It has a regular singular point at the origin and our first task is to estimate the growth of the Taylor coefficients of solutions' expansions at u = 0. This will be done in two steps. First, in this section we estimate the growth of the coefficients for a formal b = 0 limiting system, Proposition 5.1. This will make appear the function S ∞ (d, ). Then, in section 6 we will obtain uniform bounds in b for the Taylor coefficients associated to (4.15) for frequencies k 1 b . 5.1. Formal limit b = 0. Recall (4.15) and let 12

Ψ(u) = Ψ(x), x = bu then 1 + H 2 + G 2 Ψ + NL 2 x b 1 - x b b d Ψ dx + 1 - 2x b (1 + H 2 + G 2 Ψ + NL 2 ) -γ(1 + G 1 ) + 2 x b G 2 Ψ = x b γbH 1 -2(1 + H 2 ) + γb NL 1 x -2 NL 2 ⇔ 1 + H 2 + G 2 Ψ + NL 2 x(b -x) d Ψ dx (5.1) + (b -2x)(1 + H 2 + G 2 Ψ + NL 2 ) -bγ(1 + G 1 ) + 2xG 2 Ψ = x γbH 1 -2(1 + H 2 ) + γb NL 1 x -2 NL 2 .
We introduce the parameters

a = γb = |λ -| |µ + | > 0 ν = -γb( D11 + D30 -Ẽ11 ), (5.2)
which have a well defined limit as b → 0 noted a ∞ , ν ∞ , and assume

ν ∞ (d, ) > 0 (5.3)
which is checked in sections F.3 and F.5. In view of the explicit formulas of Appendix G and remark G.1, the formal limit b → 0 is:

1 + H ∞ 20 + G ∞ 2 Ψ + NL ∞ 20 (-x 2 ) d Ψ dx (5.4) + -2x(1 + H ∞ 20 + G ∞ 2 Ψ + NL ∞ 20 ) -a ∞ (1 + G ∞ 1 ) + 2xG ∞ 2 Ψ = x a ∞ H ∞ 10 -2(1 + H ∞ 20 ) + a ∞ NL ∞ 10 x -2 NL ∞ 20
where we recall 13 that the subscript ∞ means that we compute all parameters ( Dij , Ẽij ) given by (G.1) in their well defined limit b = 0. Let us stress the fact that this is not a dynamical limit, since the change variables x = bu maps the original flow on the set x = 0. Our claim is that for fixed order k, the Taylor coefficients associated to (5.4) are the b = 0 limit of the Taylor coefficients associated to (4.15) up to a suitable renormalization, see (6.16).

5.2.

Boundedness of the limiting series. The condition (5.3) holds, by a direct examination, at r + and > d from (F.15), and at r * on a collection of subintervals of (0, d) from (F.10). In fact, (F.9) implies that in the latter set is non-empty and for d = 2, 3 coincides with (0, d). We also note the fact that the condition (5.3) is not necessary for the following arguments. We also make the following note that in the case of r + , in principle, all the arguments below can be extended to the values of ∈ (0, d). In particular, from (F.15) the function ν ∞ is still positive there. This extension will be important for the analyticity argument in Appendix E.

Our aim in this section is to prove the following bound.

Proposition 5.1 (Boundedness for (5.4)). Assume (5.3). Then there exists c ν∞,a∞ > 0 such that the following holds. Let Ψ be the unique C ∞ local solution to (5.4) on [0, x 0 ], then the sequence

ψk = Ψ(k) (0) k! satisfies | ψk | ≤ c ν,a Γ(k + ν ∞ + 2) a k ∞ .
(5.5)

Remark 5.2. The bound (5.5) falls within the range of nonlinear Maillet type theorems, see [START_REF] Malgrange | Sur le théorème de Maillet[END_REF][START_REF] Ramis | Dévissage Gevrey[END_REF][START_REF] Sibuya | Linear Differential Equations in the Complex Domain: Problems of Analytic Continuation[END_REF]. We shall give a self contained proof which will allow us to obtain quantitative bounds. The latter is crucial for future uniform b independent bounds for all frequencies k 1 b for the full problem, see (6.34). Note also that the toy model problem

x 2 Θ +Θ = 1+x with Ψ = xΘ leads to the bound |Θ (k) (0)| k! = Γ(k)
suggesting that (5.5) is essentially optimal. 5.3. Conjugation formula. We start by conjugating (5.4) to an explicitly solvable (at the linear level) problem.

Lemma 5.3 (Conjugation).

There exist functions ξ(x), (µ j (x), ν j (x)) 1≤j≤4 , holomorphic in a neighborhood of x = 0, dependent on (d, ), such that the change of variables

Ψ(x) = M (x)Φ(x), M (x) = e -x 0 ξ(y)dy Θ(x) = Φ(x)
x (5.6) 13 Remark 4.1 maps (5.4) to

x 2 dΘ dx + [a ∞ + (ν ∞ + 3)x]Θ = µ 0 + x   x 4 j=1 µ j Θ j + x 2 x dΘ dx 4 j=1 ν j Θ j   . (5.7)
Proof of Lemma 5.3. This is an explicit computation.

step 1 Linear conjugation. We solve the linear problem

[x + xH ∞ 20 ] (-x) d Ψ dx + [-2x(1 + H ∞ 20 ) -a ∞ (1 + G ∞ 1 ) + 2xG ∞ 2 ] Ψ = xF. (5.8)
by conjugating it back to an explicitly solvable problem:

(5.8) ⇔ d Ψ dx + 2x(1 + H ∞ 20 ) + a ∞ (1 + G ∞ 1 ) -2xG ∞ 2 x 2 (1 + H 20 ) Ψ = - xF x 2 (1 + H 20 ) ⇔ d Ψ dx + 2 x + a ∞ (1 + Ẽ∞ 11 x) x 2 (1 + H ∞ 20 ) + a ∞ (G ∞ 1 -Ẽ∞ 11 x) -2xG ∞ 2 x 2 (1 + H ∞ 20 ) Ψ = - F x(1 + H ∞ 20 )
.

We have

a ∞ (1 + Ẽ∞ 11 x) x 2 (1 + H ∞ 20 ) = a ∞ x 2 (1 + Ẽ∞ 11 x) 1 -( D∞ 11 + D∞ 30 )x + 1 1 + H ∞ 20 -1 + ( D∞ 11 + D∞ 30 )x = a ∞ x 2 (1 + Ẽ∞ 11 x) 1 1 + H ∞ 20 -1 + ( D∞ 11 + D∞ 30 )x -a Ẽ∞ 11 ( D∞ 11 + D∞ 30 ) + a ∞ x 2 + a ∞ [ Ẽ∞ 11 -( D∞ 11 + D∞ 30 )] x Let ξ(x) = a ∞ x 2 (1 + Ẽ∞ 11 x) 1 1 + H ∞ 20 -1 + ( D∞ 11 + D∞ 30 )x -a ∞ Ẽ∞ 11 ( D∞ 11 + D∞ 30 ) + a ∞ (G ∞ 1 -Ẽ∞ 11 x) -2xG ∞ 2 x 2 (1 + H ∞ 20 ) Observe that H ∞ 20 -( D∞ 11 + D∞ 30 )x and G ∞ 1 -Ẽ∞ 11 x are polynomials in x starting with x 2 , while G ∞
2 is a polynomial beginning with x. The function ξ(x) is then holomorphic in a neighborhood of x = 0. We have

(5.8) ⇔ d Ψ dx + a ∞ x 2 + ν ∞ + 2 x + ξ(x) Ψ = - F x(1 + H ∞ 20 ) . Define M (x) = e -x 0 ξ(y)dy
(5.9) a holomorphic function in a neighborhood of x = 0, and introduce the change of variables

Ψ(x) = M (x)Φ(x), G(x) = - F (x) M (x)(1 + H ∞ 20 (x))
.

We have obtained the conjugation formula:

(5.8) ⇔ dΦ dx + a ∞ x 2 + ν ∞ + 2 x Φ = G x .
(5.10) step 2 Conjugation for the nonlinear problem. The equation (5.4) is in the form (5.8)

[x + xH ∞ 20 ] (-x) d Ψ dx + [-2x(1 + H ∞ 20 ) -a ∞ (1 + G ∞ 1 ) + 2xG ∞ 2 ] Ψ = x(F 0 + F ) for the source term F 0 = a ∞ H ∞ 10 -2(1 + H ∞ 20
) and the nonlinear term

F = a ∞ NL ∞ 10 x -2 NL ∞ 20 + G ∞ 2 Ψ + NL ∞ 20 2 Ψ + x d Ψ dx
We conjugate this using (5.10):

Ψ(x) = M (x)Φ(x), G(x) = - F (x)+F 0 M (x)(1+H ∞ 20 (x)) x 2 dΦ dx + [a ∞ + (ν ∞ + 2)x] Φ = xG
We express the nonlinearity in terms of Φ and obtain a representation:

G = ξ 0 + 4 j=2 ξ j Φ j + (xΦ ) 3 j=1 ( ξj Φ j )
where (ξ j , ξj ) are explicit holomorphic functions in a neighborhood of the origin. We have obtained the equivalent nonlinear problem:

x 2 dΦ dx + [a ∞ + (ν ∞ + 2)x] Φ = x   ξ 0 + 4 j=2 ξ j Φ j + x dΦ dx 3 j=1 ( ξj Φ j )   . (5.11)
A direct computation shows Φ(0) = 0 and we let

Θ = Φ x , (5.12
) so that:

(5.11) ⇔ x 2 x dΘ dx + Θ + [a ∞ + (ν ∞ + 2)x] xΘ = x   ξ 0 + 4 j=2 ξ j x j Θ j + x(xΘ + Θ) 3 j=1 ( ξj x j Θ j )   ⇔ x 2 dΘ dx + [a ∞ + (ν ∞ + 3)x]Θ = ξ 0 + 4 j=2 ξ j x j Θ j + x x dΘ dx 3 j=1 ( ξj x j Θ j ) + 3 j=1 ξj (xΘ) j+1 ⇔ x 2 dΘ dx + [a ∞ + (ν ∞ + 3)x]Θ = µ 0 + x   x 4 j=1 µ j Θ j + x 2 x dΘ dx 4 j=1 ν j Θ j  
where µ j , ν j are holomorphic functions of x in a neighborhood of 0, and (5.7) is proved.

5.4.

The nonlinear induction relation. The uniqueness of a local C ∞ solution near x = 0 to (5.7), and thus (5.4), follows from an elementary fixed point argument which is left to the reader. We let

µ jk = µ (k) j (0) k! , ν jk = ν (k) j (0) k! , θ k = Θ (k) (0) k! , φ k = Φ (k) (0)
k! and claim the following fundamental nonlinear bound. Lemma 5.4 (Bound on the limiting sequence). For some large enough universal constant c(ν ∞ , a ∞ ) > 0:

|φ k | ≤ c(ν ∞ , a ∞ ) Γ(k + ν ∞ + 2) a k ∞ , ∀k ≥ 1.
(5.13) Also, we have

+∞ k=0 a k ∞ g k Γ(ν ∞ + k + 3) ≤ c(ν ∞ , a ∞ ) < +∞ (5.14)
where g k is defined below, see (5.15).

Proof of Lemma 5.4. We compute the nonlinear induction relation and estimate the sequence using convolution estimates for the Γ function.

step 1 The induction relation. We formally expand 14

Θ = +∞ k=0 θ k x k G = µ 0 + x x 4 j=1 µ j Θ j + x 2 (xΘ ) 4 j=1 ν j Θ j = +∞ k=0 g k x k (5.15)
and obtain, from (5.7), the induction relation:

x 2 dΘ dx + (a ∞ + (ν ∞ + 3)x)Θ = G ⇔ +∞ k=1 kθ k x k+1 + +∞ k=0 a ∞ θ k x k + +∞ k=0 (ν ∞ + 3)θ k x k+1 = +∞ k=0 g k x k ⇔ +∞ k=2 (k -1)θ k-1 x k + +∞ k=0 a ∞ θ k x k + +∞ k=1 (ν ∞ + 3)x k θ k-1 = +∞ k=0 g k x k ⇔ a ∞ θ 0 = g 0 (ν ∞ + 3)θ 0 + a ∞ θ 1 = g 1 (k + ν ∞ + 2)θ k-1 + a ∞ θ k = g k , k ≥ 2 ⇔ a ∞ θ 0 = g 0 (k + ν ∞ + 3)θ k + a ∞ θ k+1 = g k+1 , k ≥ 0 (5.16)
We now compute g k from (5.15). By Leibniz

(f g) k = 1 k! d k dx k (f g)(0) = 1 k! k 1 +k 2 =k k! k 1 !k 2 ! f (k 1 ) (0)g (k 2 ) (0) = k 1 +k 2 =k f k 1 g k 2 Therefore, (x 2 µ j Θ j ) k+1 = (µ j Θ j ) k-1 = k 1 +...k j+1 =k-1 µ jk 1 θ k 2 . . . θ k j+1 .
Then

x dΘ dx k = kθ k yields x 3 x dΘ dx ν j Θ j k+1 = x dΘ dx ν j Θ j k-2 = k 1 +...k j+2 =k-2 ν jk 1 θ k 2 . . . θ k j+1 (k j+2 θ k j+2 )
14 Θ being smooth but not analytic, the expansions are only formal and the equalities below correspond to equalities between Taylor expansions at any order. and (5.16) yields the induction relation for k ≥ 5:

(k + ν ∞ + 3)θ k + a ∞ θ k+1 = (µ 0 ) k+1 + 4 j=1 k 1 +...k j+1 =k-1 µ jk 1 θ k 2 . . . θ k j+1 + 4 j=1 k 1 +...k j+2 =k-2 ν jk 1 θ k 2 . . . θ k j+1 (k j+2 θ k j+2 ) (5.17)
step 2 Renormalization of the sequence. We claim the bound:

|θ k | ≤ c(ν ∞ , a ∞ ) Γ(k + ν ∞ + 3) a k ∞ , ∀k ≥ 1 (5.18)
and

|(k + ν ∞ + 2)θ k-1 + a ∞ θ k | ≤ c(ν, a) Γ(k + ν ∞ + 3) (1 + k 2 )a k ∞ , ∀k ≥ 1. (5.19)
These imply, from (5.12) and (5.16)

|φ k | = |(xΘ) k | = |θ k-1 | ≤ c(ν ∞ , a ∞ ) Γ(k-1+ν∞+3) a k ∞ , a k ∞ g k Γ(ν∞+k+3) ≤ c(ν∞,a∞) 1+k 2 ,
and (5.13) and (5.14) follow. We therefore focus on the proof of (5.18), (5.19). We start by renormalizing the sequence. Let

w k = a k ∞ Γ(ν∞+k+3) θ k (h 0 ) k = a k ∞ Γ(ν∞+k+3) (µ 0 ) k+1 h k = a k ∞ Γ(ν∞+k+3) µ k hk = a k ∞ Γ(ν∞+k+3) ν k .
(5.20) then (5.17) becomes

Γ(ν ∞ + k + 4)a ∞ a k+1 ∞ w k+1 + (k + ν ∞ + 3) Γ(ν ∞ + k + 3) a k ∞ w k = Γ(ν ∞ + k + 3) a k ∞ (h 0 ) k + 1 a k-1 ∞ 4 j=1 k 1 +...k j+1 =k-1 h jk 1 w k 2 . . . w k j+1 Π j+1 i=1 Γ(ν ∞ + k i + 3) + 1 a k-2 ∞ 4 j=1 k 1 +...k j+2 =k-2 hjk 1 w k 2 . . . w k j+1 (k j+2 w k j+2 )Π j+2 i=1 Γ(ν ∞ + k i + 3) ⇔ w k+1 + w k = 1 ν ∞ + k + 3 (h 0 ) k + a ∞ (k + ν ∞ + 3)(k + ν ∞ + 2) 4 j=1 k 1 +...k j+1 =k-1 h jk 1 w k 2 . . . w k j+1 Π j+1 i=1 Γ(ν ∞ + k i + 3) Γ(k -1 + ν ∞ + 3) + a 2 ∞ Π 3 j=1 (k + ν ∞ + j) 4 j=1 k 1 +...k j+2 =k-2 hjk 1 w k 2 . . . w k j+1 (k j+2 w k j+2 ) Π j+2 i=1 Γ(ν ∞ + k i + 3) Γ(k -2 + ν ∞ + 3)
We obtain the upper bound

|w k+1 | < |w k | + c ν∞,a∞ k + 1 (h 0 ) k (5.21) + c ν∞,a∞ 1 + k 2 4 j=1 k 1 +...k j+1 =k-1 |h jk 1 w k 2 . . . w k j+1 | Π j+1 i=1 Γ(ν ∞ + k i + 3) Γ(k -1 + ν ∞ + 3) + c ν∞,a∞ 1 + k 2 4 j=1 k 1 +...k j+2 =k-2 | hjk 1 w k 2 . . . w k j+1 w k j+2 | Π j+2 i=1 Γ(ν ∞ + k i + 3) Γ(k -2 + ν ∞ + 3) .
We now proceed via a bootstrap argument and iteratively improve the control of the sequence w k from a large exponential bound to boundedness, the key being the use of suitable uniform convolution estimates. The latter are proved in Appendix B.

step 3 Large exponential bound. Since the functions h 0 , h, h are holomorphic and (5.20), there exists

A = A(ν ∞ , a ∞ ) > 0 such that ∀k ≥ 1, |(h 0 ) k | + |h k | + | hk | ≤ A k Γ(k + 1) . (5.22) Pick C 0 ≥ C 0 (ν ∞ , a ∞ ) 1 large enough, we claim |w k | ≤ C k 0 . (5.23)
The bound holds for k ≤ 5 by choosing C 0 large enough. We assume the bound for j ≤ k and prove it for k + 1. We have from (5.22) the rough bound

|(h 0 ) k | + |h k | + | hk | < C k 0
provided C 0 has been chosen large enough. Then, from (5.21), (B.5):

|w k+1 | < C k 0 + c ν∞,a∞ C k 0 1 + k + c ν∞,a∞ 1 + k 2 4 j=1 C j+2 ν C k-1 0 and |w k+1 | C k+1 0 < 1 C 0 + c a∞,ν∞ C 0 (1 + k) < 1 for k ≥ 1 and C 0 > C 0 (ν ∞ , a ∞ ) large enough. (5.23) is proved.
step 4 Improvement of the exponential bound. Let C 0 (ν, a) be a large enough constant such that (5.23) holds. Pick a small enough universal constant

δ(ν ∞ , a ∞ ) > 0 such that δ(ν ∞ , a ∞ ) (C 0 (ν ∞ , a ∞ )) -1 .
We claim the following: assume that there exist

K n > 1 e δ ≤C n ≤ C 0 (5.24) such that ∀k ≥ 1, |w k | < K n C k n , (5.25) then there exists K n+1 = K n+1 (K n , C n ) > 1 such that ∀k ≥ 1, |w k | < K n+1 C n e -δ k . (5.26) Pick k * (K n , ν ∞ , a ∞ ) large enough, then the bound (5.25) implies (5.26) for k ≤ k * provided we choose K n+1 (K n , ν ∞ , a ∞ ) ≥ K n e k * δ
. We now bootstrap the bound (5.26) by induction for k ≥ k * , assuming it for j ≤ k and proving it for k + 1.

Assume

k 1 ≤ k 2 • • • ≤ k j+1 ,
and let m = k 1 + • • • + k j and use the rough upper bound from (5.22) to estimate

|h jk 1 ||w k 2 | . . . |w k j+1 || Π j+1 i=1 Γ(ν ∞ + k i + 3) Γ(k -1 + ν ∞ + 3) (K n C k 1 n )|w k 2 | . . . |w k j+1 | Π j+1 i=1 Γ(ν ∞ + k i + 3) Γ(k -1 + ν ∞ + 3) (K n C k 1 n )|w k 2 | . . . |w k j+1 | Π j i=1 Γ(ν ∞ + k i + 3) Γ(ν ∞ + m + 3) Γ(ν ∞ + m + 3)Γ(ν ∞ + k j+1 + 3) Γ(ν ∞ + k -1 + 3) . case m ≤ k j+1 In this case, we let m = x(k -1), k j+1 = (1 -x)(k -1), x ≤ 1 2 .
For δ < x ≤ 1 2 , from (B.2) and the monotonicity of Φ

k-1 :

Φ (2) k-1 (x) ≥ Φ (2) k-1 (δ) ≥ 1 2 δ|logδ|
Then, using (5.25), (B.1), (B.5):

c ν∞,a∞ 1 + k 2 k 1 +•••+k j =m δ<x≤ 1 2 (K n C k 1 n )|w k 2 | . . . |w k j+1 | × Π j i=1 Γ(ν ∞ + k i + 3) Γ(ν ∞ + m + 3) Γ(ν ∞ + m + 3)(Γ(ν ∞ + k j+1 + 3) Γ(ν ∞ + k -1 + 3) c ν∞,a∞ 1 + k 2 k 1 +•••+k j =m K j n C m n Π j i=1 Γ(ν ∞ + k i + 3) Γ(ν ∞ + m + 3) δ<x≤ 1 2 K n C k j+1 n e -δ|logδ| 4 k ≤ c ν∞,a∞ 1 + k 2 K 5 n C k-1 n e -δ|logδ| 8 k k 1 +•••+k j =m Π j i=1 Γ(ν ∞ + k i + 3) Γ(ν ∞ + m + 3) ≤ c ν∞,a∞ 1 + k 2 K 5 n C n e -δ|logδ| 8 k For Rν k < x < δ, with R ν = R ν∞ introduced in Lemma B.1, we have in view of (B.2), Φ (2) k (x) ≥ x|logx| 2 ≥ x|logδ| 2
and thus, since

0 < δ C -1 0 , √ 1 + mC m n e -kΦ (2) k (x) ≤ √ 1 + me mlogCn-m|logδ| 4 ≤ √ 1 + me mlogC 0 -m|logδ| 4 ≤ e -m|logδ| 8
Using the bootstrap bounds (5.25) and (5.26) for k j+1 ≤ k -1:

c ν∞,a∞ 1 + k 2 k 1 +•••+k j =m Rν k <x≤δ (K n C k 1 n )|w k 2 | . . . |w k j+1 | × Π j i=1 Γ(ν ∞ + k i + 3) Γ(ν ∞ + m + 3) Γ(ν ∞ + m + 3)(Γ(ν ∞ + k j+1 + 3) Γ(ν ∞ + k -1 + 3) ≤ c ν∞,a∞ 1 + k 2 k 1 +•••+k j =m K j n C m n Π j i=1 Γ(ν ∞ + k i + 3) Γ(ν ∞ + m + 3) Rν k <x≤δ K n+1 C n e -δ k j+1 e -m|logδ| 8 C m n ≤ c ν∞,a∞ K 5 n K n+1 1 + k 2 C n e -δ k k 1 +•••+k j =m Π j i=1 Γ(ν ∞ + k i + 3) Γ(ν ∞ + m + 3) ≤ c ν∞,a∞ K 5 n K n+1 1 + k 2 C n e -δ k
where we used (B.5) in the last step. For x ≤ Rν k , there are m≤xk ≤ R ν terms in the sum and k j ≤ m≤kx ≤ R ν fixed. We therefore use Φ

c ν,a 1 + k 2 k 1 +•••+k j =m x≤ Rν k (K n C k 1 n )|w k 2 | . . . |w k j+1 | × Π j i=1 Γ(ν ∞ + k i + 2) Γ(ν ∞ + m + 3) Γ(ν ∞ + m + 3)(Γ(ν ∞ + k j+1 + 3) Γ(ν ∞ + k -1 + 3) ≤ c ν∞,a∞ K n+1 1 + k 2 C n e -δ k j+1 k 1 +•••+k j =m Π j i=1 Γ(ν ∞ + k i + 3) Γ(ν + m + 3) ≤ c ν,a K n+1 1 + k 2 C n e -δ k .
case k j+1 ≤ m In this case, we let δ > 0 as in Lemma B.1, and let

k j+1 = x(k -1), m = (1 -x)(k -1), x ≤ 1 2 . Since k 1 ≤ k 2 ≤ . . . k j+1 , we have k -1 = m + k j+1 ≤ (j + 1)k j+1
and we are always in the range δ < 1 j+2 ≤ x ≤ 1 2 . We then use verbatim the same chain of estimates as above: from (B.2) and the monotonicity of

Φ (2) k-1 : Φ (2) k-1 (x) ≥ Φ (2) k-1 (δ) ≥ 1 2 δ|logδ|
Therefore, using (5.25), (B.5):

c ν∞,a∞ 1 + k 2 k 1 +•••+k j =m δ<x≤ 1 2 (K n C k 1 n )|w k 2 | . . . |w k j+1 ||w k j+1 | × Π j i=1 Γ(ν ∞ + k i + 3) Γ(ν ∞ + m + 3) Γ(ν ∞ + m + 3)(Γ(ν ∞ + k j+2 + 3) Γ(ν ∞ + k -1 + 3) ≤ c ν∞,a∞ 1 + k 2 k 1 +•••+k j =m K j+1 n C k-1 n Π j i=1 Γ(ν ∞ + k i + 3) Γ(ν ∞ + m + 3) δ<x≤ 1 2 e -δ|logδ| 4 k ≤ c ν∞,a∞ 1 + k 2 K 5 n C k-1 n e -δ|logδ| 8 k k 1 +•••+k j =m Π j i=1 Γ(ν ∞ + k i + 2) Γ(ν ∞ + m + 2) ≤ c ν∞,a∞ 1 + k 2 K 5 n C n e -δ|logδ| 8 k
Conclusion The collection of above bounds inserted into (5.21) ensures:

|w k+1 | < |w k | + c ν∞,a∞ A k Γ(k + 1) + c ν∞,a∞ K 5 n K n+1 1 + k 2 C n e -δ k
Then, provided that k > k * (K n , ν, a) has been chosen large enough:

w k+1 K n+1 (C n e -δ ) k+1 < K n K n+1 C n e -δ + c ν,a A k Γ(k + 1)K n+1 (C n e -δ ) k+1 + c ν,a K 5 n (1 + k 2 )(C n e -δ ) < 1 2 + c ν∞,a∞ K 5 n 1 + k 2 < 1
and (5.26) is proved.

step 5 Boundedness. The initial bound (5.23) allows us to apply (5.25), (5.26) iteratively a finite number of times 15 to obtain the bound for some large enough constant C > 1 depending on ν ∞ and a ∞ :

∀k ≥ 1, |w k | < C
and (5.18) is proved. As a byproduct, we obtain

a k ∞ g k+1 Γ(ν ∞ + k + 1 + 3) = |w k + w k+1 | < c ν∞,a∞ k + 1 (h 0 ) k + c ν∞,a∞ 1 + k 2 4 j=1 k 1 +...k j+1 =k-1 |h jk 1 w k 2 . . . w k j+1 | Π j+1 i=1 Γ(ν ∞ + k i + 3) Γ(k -1 + ν ∞ + 3) + c ν∞,a∞ 1 + k 2 4 j=1 k 1 +...k j+2 =k-2 | hjk 1 w k 2 . . . w k j+1 w k j+2 | Π j+2 i=1 Γ(ν ∞ + k i + 3) Γ(k -2 + ν ∞ + 3) < c ν∞,a∞ A k Γ(k + 1) + c ν∞,a∞ 1 + k 2
and (5.16) follows as well. 15 Recall that Cn+1 = e -δ Cn so that Cn = e -nδ C0. Thus, we need Cn = C0e -nδ = 1, i.e. it suffices to choose the number n of iterations as

n(ν∞, a∞) = 1 δ(ν∞, a∞) log C0(ν∞, a∞)
where have adjusted the choice of δ to ensure that the formula provides an integer value for n.

5.5. Proof of Proposition 5.1. In view of (5.13) and the conjugation formula (5.6), the bound (5.5) directly follows from the following continuity lemma.

Lemma 5.5 (Continuity). Let M (x) be holomorphic in a neighborhood of 0, then there exists c M > 0 such that for all functions Φ which are C ∞ at the origin,

sup k≥0 a k ∞ |(M Φ) k | Γ(k + ν ∞ + 2) ≤ c M sup k≥0 a k ∞ |φ k | Γ(k + ν ∞ + 2)
.

(5.27)

Proof of Lemma 5.5. Since M is holomorphic in a neighborhood of the origin

|m k | = 1 k! d k M dx k (0) ≤ C k and then sup k≥0 a k ∞ |m k | Γ(k + ν ∞ + 2) ≤ c M .
We estimate using (B.5):

a k ∞ |(M Φ) k | Γ(k + ν + 2) = a k ∞ k 1 +k 2 =k |m k 1 φ k 2 | Γ(k + ν ∞ + 2) ≤ sup k≥0 a k ∞ |φ k | Γ(k + ν ∞ + 2) sup k≥0 a k ∞ |m k | Γ(k + ν ∞ + 2) k 1 +k 2 =k Γ(k 1 + ν ∞ + 2)Γ(k 2 + ν ∞ + 2) Γ(k + ν ∞ + 2) ≤ c M sup k≥0 a k ∞ |φ k | Γ(k + ν ∞ + 2)
and (5.27) is proved.

Bounding the Taylor series for (4.15)

We now start the study of the full problem (4.15). We first aim at obtaining uniform in b bounds on the coefficients of the Taylor series as well as the convergence to the limiting problem as b → 0. We recall the notation

γ -1 = K + α γ , K ∈ N * , 0 < α γ < 1.
6.1. b dependent conjugation. We conjugate the b dependent problem (4.15) to an explicitly solvable (at the linear level) b-dependent equation.

Lemma 6.1 (Linear conjugation).

There exist a function ξb (x) which is holomorphic in a b-independent neighborhood of x = 0 such that the conjugation

Ψ(u) = M b (x)Φ(u), G = F M b (1 + H 20 ) (6.1) with M b (x) = e ξb (x) (6.2) and ν b = ν + ξ(b) ν = -γb( D11 + D30 -Ẽ11 ) (6.3) maps 16 [1 + H 20 ] u(1 -u) dΨ du + [(1 -2u)(1 + H 20 ) -γ(1 + G 1 ) + 2uG 2 ] Ψ = -uF (6.4
) 16 The link between (4.15) and (6.4) will be given later, see (6.20) (6.21). Also, see (G.3) for the definition of H20.

to dΦ du - γ -1 u + γ + ν b + 1 1 -u Φ = - G 1 -u . (6.5)
Remark 6.2. We see immediately the fundamental difference between (6.5) and (5.10): for the b-dependent problem, the point u = 0 is a regular singular point, while it is a singular singular point for b = 0. As a result, we will see a change in the behavior of the Taylor series for large frequencies, which will be reflected in the nature of the weight w γ,ν , see Lemma 6.3.

Proof of Lemma 6.1. This is a direct computation.

step 1 Linear conjugation. We rewrite (6.4)

dΨ du - dζ b du ζ b Ψ = - F (1 -u)(1 + H 2,0 ) (6.6)
with

dζ b du ζ b = γ(1 + G 1 ) u(1 -u)(1 + H 2,0 ) - 2G 2 (1 -u)(1 + H 2,0 ) - 1 -2u u(1 -u) = γ(1 + G 1 ) u(1 -u)(1 + H 2,0 ) - 2G 2 (1 -u)(1 + H 2,0 ) - 1 u + 1 1 -u From (4.35): G 1 (x) = Ẽ11 x + x G1 G1 = -(2 Ẽ02 + Ẽ21 )x + 2 Ẽ12 x 2 -3 Ẽ03 x 3 and γ(1 + G 1 ) u(1 -u) = γ(1 + Ẽ11 bu + bu G1 ) u(1 -u) = γ u(1 -u) + Ẽ11 bγ 1 -u + γb G1 1 -u yields dζ b du ζ b = γ u(1 -u) + Ẽ11 bγ 1 -u 1 1 + H 2,0 + γb G1 -2G 2 (1 -u)(1 + H 2,0 ) - 1 u + 1 1 -u .
We recall that x = bu and rewrite

γ u(1 -u)(1 + H 2,0 ) = γ u(1 -u) 1 -( D11 + D30 )bu + 1 1 + H 2,0 -1 + ( D11 + D30 )x = γ 1 u + 1 1 -u - γb( D11 + D30 ) 1 -u + γb 1 -u 1 x 1 1 + H 2,0 -1 + ( D11 + D30 )x and Ẽ11 bγ (1 -u)(1 + H 2,0 ) = Ẽ11 bγ 1 -u + Ẽ11 bγ 1 -u 1 1 + H 2,0 -1 .
We have therefore obtained the formula:

dζ b du ζ b = γ 1 u + 1 1 -u - 1 u + 1 1 -u - γb( D11 + D30 ) 1 -u + Ẽ11 bγ 1 -u + ξ(x) 1 -u with ξ(x) = γb G1 -2G 2 1 + H 2,0 + γb x 1 1 + H 2,0 -1 + ( D11 + D30 )x + γb Ẽ11 1 1 + H 2,0 -1 .
We recall from (5.2)

ν = -γb( D11 + D30 -Ẽ11 ),
and rewrite:

dζ b du ζ b = γ -1 u + γ + ν + 1 1 -u + ξ(x) 1 -u . (6.7)
step 2 Computation of the kernel. We now use the analyticity of ξ in |x| < 1 C and ξ(0) = 0 to compute:

ξ(x) 1 -u = 1 1 -u +∞ k=1 ξ k x k = 1 1 -u +∞ k=1 ξ k b k u k = +∞ k=1 ξ k b k 1 -u - +∞ k=1 ξ k b k 1 -u k 1 -u = ξ(b) 1 -u - +∞ k=1 ξ k b k k-1 j=0 u j .
Let ν b be given by ( 6.3), we have obtained:

dζ b du ζ b = γ -1 u + γ + ν b + 1 1 -u - +∞ k=1 ξ k b k k-1 j=0 u j .
We compute a primitive of the remaining term:

+∞ k=1 ξ k b k k-1 j=0 u j+1 j + 1 = +∞ j=0 u j+1 j + 1 +∞ k=j+1 ξ k b k = +∞ j=1 u j j +∞ k=j ξ k b k = +∞ j=1 ξb,j b j u j = +∞ j=1 ξb,j x j with ξb, j = 1 j +∞ k=j ξ k b k-j = 1 j +∞ k=0 ξ k+j b k . The holomorphic bound |ξ j | ≤ C j ensures | ξb,j | ≤ 1 j +∞ k=0 b k C k+j ≤ C j j 1 1 -bC ≤ C j j (6.8)
for 0 < b < b * universal small enough. We have therefore obtained the formula

dζ b du ζ b = γ -1 u + γ + ν b + 1 1 -u + d du ξb (x) (6.9)
where

ξb (x) = - +∞ j=1 ξb,j x j , | ξb,j | ≤ C j (6.10)
is holomorphic in a neighborhood of x = 0 independent of b.

step 3 Conclusion. From (6.6), (6.9):

dΨ du - γ -1 u + γ + ν b + 1 1 -u + d du ξb (x) Ψ = - F (1 -u)(1 + H 20 )
and (6.5) follows.

6.2. The b-dependent discrete weight. We study the discrete weight associated to (6.5). Lemma 6.3 (Properties of the weight). Let .11) then:

w γ,ν (k) = Γ(γ -1 -k)Γ(ν + k + 2) Γ(γ -1) , k ∈ N. ( 6
(1) value for k ≥ K: ∀j ≥ 0,

w γ,ν b (K + j) = (-1) j Γ(α γ )Γ(1 -α γ ) Γ(K + j + ν b + 2) Γ(K + α γ )Γ(j + 1 -α γ )
. (6.12)

(2) ν dependence:

∀k ≥ 0, γw γ,ν b (k + 1) w γ-1,ν b +1 (k) = γ γ -2 (6.13)
(3) induction property:

∀k ≥ 1, (γ -k -2)w γ-1,ν b +1 (k) -(k + ν b + 2)w γ-1,ν b +1 (k -1) = 0. (6.14)
Proof. (6.12) directly follows from (6.11), (A.10).

We then compute for k ≥ 0:

γw γ,ν b (k + 1) w γ-1,ν b +1 (k) = γ Γ(γ -1 -(k + 1))Γ(k + 1 + ν b + 2) Γ(γ -1) Γ(γ-2-k)Γ(k+ν b +3) Γ(γ-2) = γΓ(γ -2) Γ(γ -1) = γ γ -2
and (6.13) is proved. We now turn to the induction formula:

for k ≥ 1, (γ -k -2)w γ-1,ν b +1 (k) -(k + ν b + 2)w γ-1,ν b +1 (k -1) = (γ -k -2) Γ(γ -2 -k)Γ(k + ν b + 1 + 2) Γ(γ -2) -(k + ν b + 2) Γ(γ -2 -(k -1))Γ(k -1 + ν b + 1 + 2) Γ(γ -2) = Γ(γ -k -1)Γ(k + ν b + 3) Γ(γ -2) - Γ(γ -k -1)Γ(k + ν b + 3) Γ(γ -2) = 0.
6.3. Boundedness of the sequence. We claim the following b-dependent nonlinear bound which, in a certain sense, is a deformation of (5.5).

Proposition 6.4 (b-dependent boundedness).

There exist universal constants c ν,a > 0 and 0 < b * 1 such that the following holds for all 0 < b < b * . Let Ψ be a solutions of (4.15) and define

ψ k = 1 k! d k Ψ du k (0), then ∀ 0 ≤ k ≤ K, |ψ k | ≤ c ν,a w γ,ν b (k). (6.15)
Moreover, let Ψ∞ be the unique local C ∞ solution to the limiting problem (5.4) and

ψ∞ k = 1 k! d k Ψ∞ dx k (0), then ∀k ≥ 0, lim b→0 ψ k b k = ψ∞ k . (6.16)
The rest of this section is devoted to the proof of Proposition 6.4.

6.4. Proof of (6.15) for frequencies k γ. We claim the following small frequency bound. Lemma 6.5 (Uniform small frequency bound). The limit (6.16) holds. Moreover, there exists c ν,a > 0 such that for all k * ≥ 0, there exists 0 < b * (k * ) 1 such that

∀0 < b < b * (k * ), ∀0 ≤ k ≤ k * , |ψ k | ≤ c ν,a w γ,ν b (k).
(6.17)

Proof of Lemma 6.5. Recall

Ψ(u) = Ψ(x), x = bu so that ψ k b k = 1 k!b k d k Ψ du k (0) = 1 k! d k Ψ
dx k (0) = ψk and (6.16) is equivalent to

lim b→0 ψk = ψ∞ k .
This follows immediately by passing to the b → 0 limit for fixed k in the induction relation for the ψk from (5.1). The details are straightforward and left to the reader.

Similarly, note that we also have

lim b→0 g k b k = g ∞ k . (6.18)
We conclude from (5.5) that there exists c ν,a such that

∀k ≥ 0, | ψ∞ k | ≤ c ν,a Γ(ν + k + 2) a k . ( 6

.19)

Pick now an arbitrary k * ≥ 0, then from (6.16), (6.19):

∀0 < b < b * (k * ), ∀0 ≤ k ≤ k * , | ψk | ≤ 2c ν,a Γ(ν + k + 2) a k .
We now estimate using (C.1):

|ψ k | w γ,ν b (k) ≤ c 1 |ψ k |γ k Γ(ν b + k + 2) = c 1 | ψk |b k γ k Γ(ν b + k + 2) = c 1 | ψk |a k Γ(ν b + k + 2) ≤ 2c 1 c ν,a Γ(ν + k + 2) Γ(ν b + k + 2) .
Since ν b → ν as b → 0, we may choose b * (k * ) small enough so that

∀0 < b < b * (k * ), ∀0 ≤ k ≤ k * , Γ(ν + k + 2) Γ(ν b + k + 2) ≤ 2.
Then,

|ψ k | w γ,ν b (k) ≤ 4c 1 c ν,a
and (6.17) is proved.

1 + H 2 + G 2 Ψ + NL 2 u(1 -u) dΨ du + (1 -2u)(1 + H 2 + G 2 Ψ + NL 2 ) -γ(1 + G 1 ) + 2uG 2 Ψ = u γbH 1 -2(1 + H 2 ) + γb NL 1 x -2 NL 2 ⇔ [1 + H 20 ] u(1 -u) dΨ du + [(1 -2u)(1 + H 20 ) -γ(1 + G 1 ) + 2uG 2 ] Ψ = uF (6.20)
with recalling 1 u = b x :

F = γbH 1 -2(1 + H 2 ) (6.21) + γb NL 1 x -2 NL 2 -(1 -u)   3 j=1 b j H 2j (x) + G 2 Ψ + NL 2   dΨ du +    - b x (G 2 Ψ + NL 2 ) + 2(G 2 Ψ + NL 2 ) - b(1 -2u) x 3 j=1 b j H 2j (x)    Ψ
From (6.1), (6.5) we now obtain the nonlinear conjugated problem:

dΦ du - γ -1 u + γ + ν b + 1 1 -u Φ = - G 1 -u (6.22)
step 2 Computation of G. We plug (6.1) into (6.21) and decompose

G = G 0 + L(Φ) + NL(Φ) (6.23)
as follows.

Source term. We have

G 0 (b, x)= γbH 1 -2(1 + H 2 ) M b (1 + H 20 )
which, from (6.10), admits a holomorphic expansion in a neighborhood (independent of b) of x = 0 i.e.,

G 0 (x) = +∞ k=0 g 0k x k (6.24)
for some b-dependent coefficients g 0k satisfying

|g 0k | ≤ C k (6.25)
for some C > 0 independent of b.

Small linear term. It is given explicitly by

L(Φ) = 1 M b (1 + H 20 )    -(1 -u)   3 j=1 b j H 2j (x)   (M b (x)Φ) - b(1 -2u) x 3 j=1 b j H 2j (x)M b Φ    = 1 M b (1 + H 20 )      -bM b (x)(1 -u) 3 j=1 b j H 2j (x) - 3 j=1 b j+1 H 2j (x) x -2b j H 2j   Φ -(1 -u)M b (x)   3 j=1 b j H 2j (x)   Φ    (6.26)
We rewrite

(1 -u)Φ = 1 -u u uΦ = b x -1 uΦ
and obtain

L(Φ) = 1 M b (1 + H 20 )      -M b (x)(b -x) 3 j=1 b j H 2j (x) - 3 j=1 b j+1 H 2j (x) x -2b j H 2j   Φ -M b (x) 3 j=1 b j+1 H 2j (x) x -b j H 2j (x) uΦ    Therefore, L(Φ) = b (bh 1 (x) + xh 2 (x))Φ + (bh 3 (x) + xh 4 (x))uΦ
where

h j (x) = +∞ k=0 h jk x k , |h jk | ≤ C k
for some C > 0 independent of b. Nonlinear term. We have NL(Φ) (6.27)

= 1 M b (1 + H 20 ) γb NL 1 x -2 NL 2 -(1 -u) G 2 Ψ + NL 2 Ψ + (G 2 Ψ + NL 2 ) 2 - b x Ψ .
We rewrite

(1 -u) G 2 Ψ + NL 2 Ψ = 1 -u u G 2 Ψ + NL 2 uΨ = b x -1 G 2 Ψ + NL 2 uΨ
and thus NL(Φ) is given, structurally, by

NL(Φ) = x 4 j=2 m (1) j Φ j + b 4 j=2 m (2) j Φ j +   x 3 j=1 m (3) j Φ j + b 3 j=1 m (4) j Φ j   uΦ with m ( ) j (x) = +∞ k=0 m ( ) jk x k , |m ( ) jk | ≤ C k .
Conclusion. We have obtained the conjugated nonlinear problem (6.22) with

G = G 0 + b (bh 1 (x) + xh 2 (x))Φ + (bh 3 (x) + xh 4 (x)
)uΦ (6.28)

+ x 4 j=2 m (1)
j Φ j + b 4 j=2 m (2)
j Φ j +   x 3 j=1 m (3) j Φ j + b 3 j=1 m (4) j Φ j   uΦ .
step 3 Final change of variables. We now let

Φ = buΘ = xΘ (6.29)
so that (6.22) becomes:

Φ - γ -1 u + γ + ν b + 1 1 -u Φ = - G 1 -u ⇔ buΘ + bΘ - γ -1 u + γ + ν b + 1 1 -u buΘ = - G 1 -u ⇔ Θ - γ -2 u + γ + ν b + 1 1 -u Θ = - G bu(1 -u) ⇔ u(1 -u)Θ -[γ -2 + (ν b + 3)u] Θ = - G b . (6.30)
We now express G in terms of Θ and track the orders of vanishing in x. We compute

uΦ = u[buΘ + bΘ] = x(uΘ + Θ).
Then,

b (bh 1 (x) + xh 2 (x))Φ + (bh 3 (x) + xh 4 (x))uΦ = b 2 h 1 + bxh 2 xΘ + (b 2 h 3 + bxh 4 )[uxΘ + xΘ] = b 2 x(h 1 + h 3 ) + bx 2 (h 2 + h 4 ) Θ + b 2 xh 3 + x 2 h 4 uΘ = b 2 x h1 + bx 2 h2 Θ + b 2 x h3 + bx 2 h4 uΘ .
For the nonlinear term:

x 4 j=2 m (1)
j Φ j + b 4 j=2 m (2)
j Φ j +   x 3 j=1 m (3) j Φ j + b 3 j=1 m (4) j Φ j   uΦ = x 4 j=2
x j m

(1)

j Θ j + b 4 j=2 m (2)
j x j Θ j +   x 3 j=1 m (3)
j x j Θ j + b 3 j=1 m (4) j x j Θ j   (xuΘ + xΘ) =   4 j=2
x j+1 m

(1)

j Θ j + 3 j=1 m (3) j x j+2 Θ j+1 + b 4 j=2 m (2) j x j Θ j + b 3 j=1 m (4) j x j+1 Θ j+1   +   3 j=1 m (3)
j x j+2 Θ j + b 3 j=1 m (4) j x j+1 Θ j   (uΘ ) = 4 j=2 x j+1 m(1) j Θ j + b 4 j=2 m (2)
j x j Θ j +   3 j=1 m(3) j x j+2 Θ j + b 3 j=1 m(4) j x j+1 Θ j   (uΘ ).
We now rewrite

G = G 0 + b 2 x h1 + bx 2 h2 Θ + b 2 x h3 + bx 2 h4 uΘ (6.31) + 4 j=2 x j+1 m(1) j Θ j + b 4 j=2 m (2)
j x j Θ j +   3 j=1 m(3) j x j+2 Θ j + b 3 j=1 m(4) j x j+1 Θ j   (uΘ )
where, for some large enough universal constant

C = C ν,a > 0 independent of b < b * and all k ≥ 0, |(G 0 ) k | + |( hl ) k | + |( m(l) j ) k | ≤ C k b k (6.32)
6.6. Bounding the sequence θ k and proof of Proposition 6.4. We let 17

φ k = 1 k! d k Φ du k (0), θ k = 1 k! d k Θ du k (0), g k = 1 k! d k G du k (
0) so that from (6.29):

φ 0 = 0 φ k = bθ k-1 , k ≥ 1. ( 6
|θ k | ≤ |w γ-1,ν b +1 (k)| (6.34)
(This implies (6.15).) Moreover,

∀0 ≤ k ≤ K, |g k | ≤ c ν,a |w γ,ν b (k)| 1 + k . (6.35)
Proof of Lemma 6.6. This is a direct consequence of the form of G given in (6.31) (6.32).

step 1 Small frequency universal bound.

≤ k * ≤ K, max 0≤k≤k * |(hφ) k | w γ,ν b (k) ≤ C h max 0≤k≤k * |φ k | w γ,ν b (k) . ( 6

.36)

Proof of Lemma 6.7. First observe that (C.2) implies the lower bound

∀0 ≤ k ≤ K, w γ,ν b (k) = Γ(γ -1 -k) Γ(γ -1) Γ(ν b + k + 2) ≥ Γ(ν b + k + 2) (γ -1) k (6.37)
which then gives the upper bound 18 :

b k C k w γ,ν b (k) ≤ (bC(γ -1)) k Γ(ν b + k + 2) ≤ (2Ca) k Γ(k + ν b + 2) ≤ c(C) 1 + k 2 . (6.38) for all 0 ≤ k ≤ K. Therefore, b k |h k | |w γ,ν b (k)| ≤ b k C k |w γ,ν b (k)| ≤ C h 1 + k 2 .
(6.39)

17 Observe that the coefficients φ k and θ k appearing here should not be confused with the ones of section 5.4 for the formal limit problem. 18 We use in particular ν b > 0 and the fact that Γ is increasing on [2, +∞).

for all 0 ≤ k ≤ K. We then estimate using (C.5), for k≤k * ≤ K:

(hφ) k w γ,ν b (k) = 1 w γ,ν b (k) k 1 +k 2 =k b k 1 h k 1 φ k 2 ≤ k 1 +k 2 =k b k 1 |h k 1 | w γ,ν b (k 1 ) |φ k 2 | w γ,ν b (k 2 ) w γ,ν b (k 1 )w γ,ν b (k 2 ) w γ,ν b (k) ≤ C h max 0≤k<k * |φ k | w γ,ν b (k) k 1 +k 2 =k w γ,ν b (k 1 )w γ,ν b (k 2 ) w γ,ν b (k) ≤ C h max 0≤k<k * |φ k | w γ,ν b (k)
and (6.36) is proved.

We conclude from (6.1), (6.17) that there exists c ν,a such that for all k * > 1,

there exists 0 < b * (k * ) 1 such that for 0 < b < b * (k * ) and 0 ≤ k ≤ k * , |φ k | ≤ c ν,a w γ,ν b (k). (6.40)
Thus, from (6.33):

|θ k | = 1 b |φ k+1 | ≤ γ a c ν,a w γ,ν b (k + 1).
Together with (6.13), we deduce the existence of M ν,a such that for all k * ≥ 1, for all 0 <b < b * (k * ),

∀0 ≤ k ≤ k * , |θ k | w γ-1,ν b +1 (k) ≤ M ν,a . (6.41)
step 2 Induction relation. We now compute from (6.30) the induction relation satisfied by derivatives at the origin. We formally expand

Θ = +∞ k=0 θ k u k , G = +∞ k=0 g k u k
and obtain from (6.30):

u(1 -u)Θ -[γ -2 + (ν b + 3)u] Θ = - G b ⇔ +∞ k=1 (u -u 2 )kθ k u k-1 -(γ -2) +∞ k=0 θ k u k -(ν b + 3) +∞ k=0 θ k u k+1 = - 1 b +∞ k=0 g k u k ⇔ +∞ k=1 kθ k u k - +∞ k=2 (k -1)θ k-1 u k -(γ -2) +∞ k=0 θ k u k -(ν b + 3) +∞ k=1 θ k-1 u k = - 1 b +∞ k=0 g k u k ⇔ -(γ -2)θ 0 = -g 0 b (k -γ + 2)θ k -(k + ν b + 2)θ k-1 = -g k b , k ≥ 1 ⇔ θ 0 = γ a(γ-2) g 0 (γ -k -2)θ k + (k + ν b + 2)θ k-1 = g k b , k ≥ 1 (6.42) Let ζ k = θ k w γ-1,ν b +1 (k)
, then (6.14), (6.13) yield:

(γ -k -2)w γ-1,ν b +1 (k)ζ k + (k + ν b + 2)w γ-1,ν b +1 (k -1)ζ k-1 = g k b ⇔ (k + ν b + 2)w γ-1,ν b +1 (k -1)(ζ k + ζ k-1 ) = γg k a ⇔ ζ k + ζ k-1 = γ a(γ -2) g k (k + ν b + 2)w γ,ν b (k)
. (6.43) step 3 Bootstrap bound. Let M ν,a be the universal constant in (6.41), we now bootstrap the bound

∀k * ≤ k ≤ K -1, |ζ k | ≤ 2M ν,a (6.44)
which, by (6.41), holds for 0 ≤ k < k * arbitrarily large and 0 < b < b * (k * ). We argue by induction, assuming the claim for 0 ≤ j ≤ k ind -1 and proving it for k ind with k ind ≤ K -1. We claim the following crucial nonlinear bound for g k

∀0 ≤ k ≤ k ind , |g k | |w γ,ν b (k)| ≤ c ν,a M 4 ν,a 1 + k . (6.45)
Source term. The estimates (6.38) and (6.32) yield the following uniform in b, k ≤ K:

|(G 0 ) k | w γ,ν b (k) ≤ (bC) k w γ,ν b (k) ≤ c ν,a 1 + k .
Small linear term. Let 1 ≤ j ≤ 4 and j ≤ k ≤ k ind , then from (6.41) (6.44):

|θ k-j | ≤ M ν,a w γ-1,ν b +1 (k -j) (6.46)
We now observe the relation for 0 ≤ j ≤ 4 from (6.13):

b j |w γ-1,ν b +1 (k -j)| |w γ,ν b (k)| = b j (γ -2)|w γ,ν b (k -j + 1)| |w γ,ν b (k)| ≤ c ν,a b j-1 |w γν b (k -(j -1))| |w γ,ν b (k)|
and from (6.14) for 0 ≤ j ≤ 4:

w γ,ν b (k) = k + ν b + 1 γ -k -1 w γ,ν b (k -1) = k -1 + ν b + 2 γ -2 -(k -1) w γ,ν b (k -1) = w γ,ν b (k -j)Π j m=1 k -m + ν b + 2 γ -2 -(k -m)
which leads to the estimate:

|w γ,ν b (k -m)| |w γ,ν b (k)| ≤ Π m j=1 γ -2 -k + j |k -j + ν b + 2| ≤ c ν γ m (1 + k) m (6.47)
and then

b j |w γ-1,ν b +1 (k -j)| |w γ,ν b (k)| ≤ c ν,a b j-1 |w γ,ν b (k -(j -1))| |w γ,ν b (k)| ≤ c ν,a a j-1 (1 + k) j-1 .
(6.48)

Moreover, from (6.38):

b k C k |w γ-1,ν b +1 (k)| ≤ c ν,a . (6.49)
We estimate the worst term using (6.32), (6.46), (6.48), (6.49), (C.5):

for k ≤ k ind , bx 2 h4 uΘ k = bb 2 u 2 h4 uΘ k = bb 2 h4 uΘ k-2 = bb 2 k 1 +k 2 =k-2 ( h4 ) k 1 (k 2 θ k 2 ) ≤ bb 2 kc ν,a M ν,a k 1 +k 2 =k-2 |w γ-1,ν b +1 (k 1 )w γ-1,ν b +1 (k 2 )| ≤ bkc ν,a M ν,a b 2 w γ-1,ν b +1 (k -2) ≤ c ν,a M ν,a |w γ,ν b | 1 + k since bk ≤ bγ = a. We estimate similarly for k ≤ k ind : b 2 x h3 uΘ k = b 3 h3 uΘ k-1 ≤ b 3 kc ν,a M ν,a |w γ-1,ν b +1 (k -1)| ≤ b 2 kc ν,a M ν,a w γ,ν b (k) ≤ c ν,a M ν,a |w γ,ν b (k)| 1 + k .
Remaining linear terms do not have the k loss of (uΦ ) k and are easier to estimate.

Nonlinear term. The worst nonlinear term is for 1 ≤ j ≤ 3:

m(3) j x j+2 Θ j (uΘ ) k = b j+2 u j+2 m(3) j Θ j (uΘ ) k = b j+2 m(3) j Θ j (uΘ ) k- (j+2)
.

Then for all ≤ k ind -1 from (6.32), (6.46), (6.49), (6.44) and (C.6):

m(3) j Θ j (uΘ ) = k 1 +•••+k j+2 = m(3) jk 1 θ k 2 . . . θ k j+1 (k j+2 θ k j+2 ) ≤ c ν,a M j+1 ν,a k 1 +•••+k j+2 = Π j+2 i=1 |w γ-1,ν b +1 (k i )| ≤ c ν,a M j+1 ν,a |w γ-1,ν b +1 ( )|
Therefore, recalling (6.48), for 0 ≤ k ≤ k ind :

m(3) j x j+2 Θ j (uΘ ) k = b j+2 m(3) j Θ j (uΘ ) k-(j+2) ≤ kc ν,a M j+1 ν,a b j+2 |w γ-1,ν b +1 (k -(j + 2))| ≤ c ν,a M j+1 ν,a k (1 + k) j+2-1 |w γ,ν b (k)| ≤ c ν,a M j+1 ν,a |w γ,ν b (k)| (1 + k) j ≤ c ν,a M 4 ν,a 1 + k |w γ,ν b (k)| since j ≥ 1. Similarly, for 1 ≤ j ≤ 3: b m(4) j x j+1 Θ j (uΘ ) k = bb j+1 m(4) j Θ j (uΘ ) k-j-1 ≤ bkc ν,a M j+1 ν,a b j+1 |w γ-1,ν b +1 (k -j-1)| ≤ bkc ν,a M j+1 ν,a |w γ,ν b (k)| (1 + k) j ≤ c ν,a M 4 ν,a 1 + k |w γ,ν b (k)|
since bk ≤ bγ = a and j ≥ 1. The two remaining nonlinear terms in (6.31) do not have the k loss of (uΘ ) k and are thus better by a factor of 1 1+k . The collection of above bounds concludes the proof of (6.45). step 4 Closing (6.44). From (6.45), (6.43), we have for k ≤ k ind :

|ζ k | < |ζ k-1 | + c ν,a M 4 ν,a
1 + k 2 . We sum over k ∈ {k * , k ind } and conclude from (6.41):

|ζ k ind | ≤ |ζ k * | + c ν,a M 4 ν,a k ind k=k * 1 k 2 ≤ M ν,a + c ν,a M 4 ν,a k * < 2M ν,a provided k * > k * (M ν,a
) has been chosen large enough. (6.44) is proved. This also concludes the proof of (6.35) and of Lemma 6.6.

The proof of Proposition 6.4 follows immediately from (6.33), (6.34), (6.13), (6.1), and Lemma 6.7.

Quantitative study of the C ∞ solution

We now turn to the qualitative of the C ∞ solution of (4.15). Understanding of the Taylor expansion at u = 0 is not sufficient to analyze the solution away from u = 0. Our main goal is to show that truncating the Taylor series at k = K yields the dominant terms in the solution which, together with a remainder, can be computed and estimated thanks to an explicit integral representation.

We study the C ∞ solution. We define the operator

T (G) = u γ-2 (1 -u) γ+ν b +1 u 0 (1 -v) γ+ν b v γ-1 G b dv. (7.1)
Remark 7.1. Recall from Lemma 3.6 that there exists a unique solution curve which is C ∞ at P 2 , i.e at u = 0. Recall also from Remark 3.7 that all other solutions have only finite regularity at u = 0. The operator T in (7.1) will be used in the fixed point formulation (7.20) to identify the unique C ∞ solution at u = 0. Note that the non smooth solutions exhibit a u γ-2 singularity at u = 0.

7.1. Remainder function. We introduce several special functions defined via the integral operator (7.1). These will be fundamental in understanding the leading order terms which appear when the Taylor expansion no longer dominates.

Lemma 7.2 (Definition and properties of the first remainder function). Let

M 0 (u) = (K + ν b + 2)w γ-1,ν b +1 (K -1)T (bu K ). (7.2) Then M 0 (u) = (1 + o K→+∞ (1))Γ(α γ )K ν b +4-αγ T (bu K ). (7.3)
Moreover, there exist universal constants 0 < c ν,1 < c ν,2 such that for all 0 < b < b * (ν), the following holds:

behavior for small u: for 0 ≤ u ≤ b,

c ν,1 ≤ M 0 (u) Γ(α γ )Γ(1 -α γ )K ν b +4-αγ u K ≤ c ν,2 (7.4)
behavior for large u: for b ≤ u < 1 2 :

c ν,1 ≤ M 0 (u) Γ(α γ )Γ(1 -α γ )K ν b +3 u 1-u K-1 u αγ ≤ c ν,2 . (7.5)
control of the iterate: let 1 ≤ j ≤ 5, then

T (u j M 0 ) M 0 L ∞ (u≤ 1 2 ) ≤ c ν b . (7.6)
control of the derivative:

∀0 ≤ u ≤ 1 2 , |uM 0 | M 0 ≤ c ν b . (7.7)
Proof of Lemma 7.2. This follows from the explicit integral representation (7.1).

step 1 Proof of (7.3). By definition:

M 0 (u) = (K + ν b + 2)w γ-1,ν b +1 (K -1)T (bu K ) = (K + ν b + 2) Γ(α γ )Γ(K + ν b + 2) Γ(γ -2) T (bu K ) = Γ(α γ )Γ(K + ν b + 3) Γ(K -1 + α γ ) T (bu K ) = (1 + o K→+∞ (1))Γ(α γ )K ν b +4-αγ T (bu K )
where we used (A.8) in the last step. 1) .

step 2 Estimate for 0 ≤ u ≤ b. For u ≤ b, (1 -u) K = e Klog(1-u) = e -Ku+O(Ku 2 ) = e O(
Then, from (7.3):

M 0 (u) = (1 + o K→+∞ (1))Γ(α γ )K ν b +4-αγ u γ-2 (1 -u) γ+ν b +1 u 0 (1 -v) γ+ν b v γ-1 v K dv = e O(1) Γ(α γ )K ν b +4-αγ u K-1+αγ u 0 dv v αγ = e Oν (1) Γ(α γ ) 1 -α γ K ν b +4-αγ u K-1+αγ u 1-αγ = e O(1) Γ(α γ )Γ(1 -α γ )K ν b +4-αγ u K
where we used the fact that 19

1 2 ≤ xΓ(x) ≤ 1 for 0 < x ≤ 1,
and (7.4) is proved.

19 Indeed, we have xΓ(x) = Γ(x + 1) and the well known bound 0.88 ≤ Γ(x) ≤ 1 on 1 ≤ x ≤ 2.

step 3 Estimate for b ≤ u ≤ 1 2 . First, from (7.3), (A.1), (A.8), we have the global bound

M 0 (u) ≤ c ν Γ(α γ )K ν b +4-αγ u γ-2 (1 -u) γ+ν b +1 1 0 (1 -v) γ+ν b v γ-1 v K dv ≤ c ν Γ(α γ )K ν b +4-αγ u γ-2 (1 -u) γ+ν b +1 1 0 (1 -v) K+1+αγ +ν b v -αγ dv ≤ c ν Γ(α γ )K ν b +4-αγ u K-1+αγ (1 -u) K+1+αγ +ν b +1 B(1 -α γ , K + α γ + ν b + 2) ≤ c ν Γ(α γ )K ν b +4-αγ u K-1+αγ (1 -u) K+1+αγ +ν b +1 Γ(1 -α γ )Γ(K + ν b + α γ + 2) Γ(K + ν b + 3) ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +4-αγ u 1 -u K-1 u αγ K 1-αγ ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +3 u 1 -u K-1 u αγ .
This gives the upper bound in (7.5). For the lower bound, we use:

u 0 (1 -v) K+1+αγ +ν b v -αγ dv ≥ b 0 (1 -v) K+1+αγ +ν b v -αγ dv = b 0 e [-(K+1+αγ+νb)v+O(Kv 2 )] dv v αγ ≥ c ν b 0 dv v αγ ≥ c ν b 1-αγ 1 -α γ ≥ c ν Γ(1 -α γ ) K 1-αγ
Then, (7.3) gives

M 0 (u) ≥ c ν Γ(α γ )K ν b +4-αγ u K-1+αγ (1 -u) K+1+αγ +ν b +1 Γ(1 -α γ ) K 1-αγ ≥ c ν Γ(α γ )Γ(1 -α γ )K ν b +3 u 1 -u K-1
u αγ and (7.5) is proved. step 4 Control of the iterate. For u ≤ b, and since j ≥ 1, we estimate from (7.4):

T (u j M 0 ) M 0 ≤ c ν u K u γ-2 (1 -u) γ+ν b +1 u 0 (1 -v) γ+ν b v γ-1 v K+j b dv ≤ c ν u 1-αγ u 0 v j-αγ dv b ≤ c ν u j b(j + 1 -α γ ) ≤ c ν b j b(2 -α γ ) ≤ c ν For b ≤ u ≤ 1 2 , we estimate from (7.4): b 0 (1 -v) γ+ν b v γ-1 v j M 0 (v)dv ≤ b 0 Γ(α γ )Γ(1 -α γ )K ν b +4-αγ v K+j v γ-1 dv ≤ Γ(α γ )Γ(1 -α γ )K ν b +4-αγ b j+1-αγ j + 1 -α γ ≤ b j Γ(α γ )Γ(1 -α γ )K ν b +3
and from (7.5)

u b (1 -v) γ+ν b v γ-1 v j M 0 (v)dv ≤ c ν u b (1 -v) γ+ν b v γ-1 Γ(α γ )Γ(1 -α γ )K ν b +3 v 1 -v K-1 v αγ v j dv ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +3 u b v j-1 dv ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +3 . Thus, T (u j M 0 ) M 0 ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +3 u 1-u K-1 u αγ × u γ-2 (1 -u) γ+ν b +1 1 b Γ(α γ )Γ(1 -α γ )K ν b +3
≤ c ν b and (7.6) is proved.

step 5 Control of the derivative. From (7.1):

u [T (G)] = 1 1 -u G b + [(γ -2) + (ν b + 3)u]T (G) (7.8)
which yields, recalling (7.3):

uM 0 = (1 + o K→+∞ (1))Γ(α γ )K ν b +4-αγ u T (bu K ) = (1 + o K→+∞ (1))Γ(α γ )K ν b +4-αγ 1 -u bu K b + [(γ -2) + (ν b + 3)u]T (bu k ) = (1 + o K→+∞ (1))Γ(α γ )K ν b +4-αγ 1 -u u K + [(γ -2) + (ν b + 3)u]M 0 1 -u . (7.9)
From (7.4), (7.9), we estimate for u ≤ b:

|uM 0 | M 0 ≤ c ν Γ(α γ )K ν b +4-αγ u K Γ(α γ )Γ(1 -α γ )K ν b +4-αγ u K + c ν b ≤ c ν b
and for b ≤ u ≤ 1 2 from (7.5), (7.9):

|uM 0 | M 0 ≤ Γ(α γ )K ν b +4-αγ u K Γ(α γ )Γ(1 -α γ )K ν b +3 u 1-u K-1 u αγ + c ν b ≤ c ν b
and (7.7) is proved.

We now establish additional estimates for the remainder functions.

Lemma 7.3 (Holomorphic representation and bounds). Let j ≥ 0 and

M j (u) = (K + j + ν b + 2)(-1) j w γ-1,ν b +1 (K -1 + j)T (bu K+j ), (7.10)
then we have the following convergent series representation for |u| < 1:

M j (u) = +∞ m=j+1 (-1) m w γ-1,ν b +1 (K -1 + m)u K-1+m . (7.11)
Moreover, let 1 ≤ j ≤ 5, then there exist universal constants 0 < c ν,1 < c ν,2 such that: behavior for small u: for 0 ≤ u ≤ b,

c ν,1 ≤ M j (u) Γ(α γ )Γ(1 -α γ )K ν b +j+4-αγ u K+j ≤ c ν,2 (7.12)
behavior for large u: for b ≤ u < 1 2 :

c ν,1 ≤ M j (u) Γ(α γ )Γ(1 -α γ )K ν b +3 u 1-u K-1 u αγ ≤ c ν,2 . (7.13) Proof. step 1 Holomorphic representation. Given a C ∞ function G with G = O u→0 (u K ), Θ = T (G) satisfies the linear equation u(1 -u)Θ -[γ -2 + (ν b + 3)u] Θ = G b (7.14)
which can be formally solved via a series representation:

G = +∞ k=0 g k u k Θ = +∞ k=0 θ k u k i.e., u(1 -u) ∞ k=0 θ k u k -(γ -2 + (ν b + 3)u) +∞ k=0 θ k u k = +∞ k=1 (kθ k u k -kθ k u k+1 ) - +∞ k=0 (γ -2)θ k u k -(ν b + 3) +∞ k=0 θ k u k+1 = -(γ -2)θ 0 + +∞ k=1 (k -γ + 2)θ k u k - +∞ k=1 (k + ν b + 2)θ k-1 u k = -(γ -2)θ 0 + +∞ k=1 [(k -γ + 2)θ k -(k + ν b + 2)θ k-1 ] u k
with the induction relation

θ 0 = -g 0 b(γ-2) (γ -k -2)θ k + (k + ν b + 2)θ k-1 = -g k b , k ≥ 1. Let ζ k = θ k w γ-1,ν b +1 (k) ,
then as in (6.43):

ζ k + ζ k-1 = - γ a(γ -2) g k (k + ν b + 2)w γ,ν b (k) , k ≥ 1 which yields ζ k = - γ a(γ -2) (-1) k k j=0 (-1) j g j (j + ν b + 2)w γ,ν b (j) , k ≥ 1 and thus θ k = (-1) k w γ-1,ν b +1 (k)S k , k ≥ 0 S k = -γ a(γ-2) k j=0
(-1) j g j (j+ν b +2)wγ,ν b (j) .

(7.15)

Given j ≥ 0 and G = u K+j this yields:

θ k = 0 for k ≤ K + j -1 (-1) k w γ-1,ν b +1 (k)S K+j for k ≥ K + j
Therefore, the representation is a normally convergent series 20 for |u| < 1. Using (6.13):

T (u K+j ) = S K+j +∞ k=K+j (-1) k w γ-1,ν b +1 (k)u k = - γ a(γ -2) (-1) K+j (K + j + ν b + 2)w γ,ν b (K + j) +∞ k=K+j (-1) k w γ-1,ν b +1 (k)u k = 1 b (-1) K-1+j (K + j + ν b + 2)w γ-1,ν b +1 (K + j -1) +∞ k=K+j (-1) k w γ-1,ν b +1 (k)u k gives M j (u) = (-1) K-1 +∞ k=K+j (-1) k w γ-1,ν b +1 (k)u k = +∞ m=j+1 (-1) m w γ-1,ν b +1 (K-1+m)u K-1+m
and (7.11) is proved. We now assume j ≥ 1.

step 2 Estimate for 0 ≤ u ≤ b. From (6.12) (6.13) and (A.8), we have

(-1) j w γ-1,ν b +1 (K -1 + j) = Γ(α γ )Γ(1 -α γ ) Γ(K + j + ν b + 2) Γ(K -1 + α γ )Γ(j + 1 -α γ ) = Γ(α γ )Γ(1 -α γ ) Γ(j + 1 -α γ ) [1 + o b→0 (1)] K ν b +j+3-αγ = e O(1) Γ(α γ )Γ(1 -α γ )K ν b +j+3-αγ (7.16)
where we used j ≤ 5 γ in the second step and j ≥ 1 in the last step. For u ≤ b, (1) and therefore:

(1 -u) K = e Klog(1-u) = e -Ku+O(Ku 2 ) = e O
M j (u) = e O(1) (K + j + ν b + 2)Γ(α γ )Γ(1 -α γ )K ν b +j+3-αγ × u γ-2 (1 -u) γ+ν b +1 u 0 (1 -v) γ+ν b v γ-1 v K+j dv = e O(1) Γ(α γ )Γ(1 -α γ )K ν b +j+4-αγ u K-1+αγ u 0 v j dv v αγ = e O(1) Γ(α γ )Γ(1 -α γ )K ν b +j+4-αγ u K+j .
20 From (6.12) (6.13) and (A.8), we have

wγ-1,ν b +1(k) = O(k K+ν b +3+αγ ) as k → +∞
which implies that the series converges for |u| < 1.

step 3 Estimate for b ≤ u ≤ 1 2 . We have the global bound from (7.16), (A.1), (A.8):

M j (u) ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +j+4-αγ u γ-2 (1 -u) γ+ν b +1 1 0 (1 -v) γ+ν b v γ-1 v K+j dv ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +j+4-αγ u γ-2 (1 -u) γ+ν b +1 1 0 (1 -v) K+1+αγ +ν b v j-αγ dv ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +j+4-αγ u K-1+αγ (1 -u) K+1+αγ +ν b +1 B(j + 1 -α γ , K + α γ + ν b + 2) ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +j+4-αγ u K-1+αγ (1 -u) K+1+αγ +ν b +1 Γ(j + 1 -α γ )Γ(K + ν b + α γ + 2) Γ(K + j + α γ + ν b + 3 -α γ ) ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +j+4-αγ u 1 -u K-1 u αγ K j+1-αγ ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +3 u 1 -u K-1 u αγ
which yields the upper bound in (7.12). For the lower bound, we use:

u 0 (1 -v) K+1+αγ +ν b v j-αγ dv ≥ b 0 (1 -v) K+1+αγ +ν b v j-αγ dv = b 0 e [-(K+1+αγ+νb)v+O(Kv 2 )] v j dv v αγ ≥ c ν b 0 v j dv v αγ ≥ c ν b j+1-αγ (7.16
) then gives the lower bound:

M j (u) ≥ c ν Γ(α γ )Γ(1 -α γ )K ν b +j+4-αγ u K-1+αγ (1 -u) K+1+αγ +ν b +1 b j+1-αγ ≥ c ν Γ(α γ )Γ(1 -α γ )K ν b +3 u 1 -u K-1
u αγ and (7.12) is proved. 7.2. Fixed point formulation of the C ∞ solution. For a given function F with sufficient regularity at the origin, we denote

r F (u) = F (u) - K-1 k=0 f k u k , f k = f (k) (0) k! . ( 7
.17) Lemma 7.4 (Fixed point formulation for the C ∞ solution). Let G be given by (6.31) and consider the decomposition

G = K-1 k=0 g k u k + r G , Θ = K-1 k=0 θ k u k + r Θ , (7.18)
where

θ k = (-1) k w γ-1,ν b +1 (k)S k , 0 ≤ k ≤ K -1, S k = γ a(γ-2) k j=0 (-1) j g j (j+ν b +2)wγ,ν b (j) , (7.19)
then the unique solution to the fixed point problem 21

r Θ = (-1) K-1 S K-1 M 0 (u) -T (r G ).
(7.20) 21 The fixed point procedure based on (7.20) will be performed in Lemma 7.8, see (7.49) for the space on rΘ which will be used.

generates the unique solution Θ to (6.30) which satisfies:

∀ 0 ≤ k ≤ K -1, lim u↓0 Θ (k) (u) k! = θ k (7.21)
where (θ k ) k≥0 is computed by induction from (6.42).

Proof of Lemma 7.4. Recall (6.30):

u(1 -u)Θ -[γ -2 + (ν b + 3)u] Θ = - G b .
We let

G = K-1 k=0 g k u k + r G Θ = K-1 k=1 θ k u k + r Θ step 1 Polynomial cancellations. We compute: u(1 -u) K-1 k=0 θ k u k -(γ -2 + (ν b + 3)u) K-1 k=0 θ k u k = K-1 k=1 (kθ k u k -kθ k u k+1 ) - K-1 k=0 (γ -2)θ k u k -(ν b + 3) K-1 k=0 θ k u k+1 = -(γ -2)θ 0 + K-1 k=1 (k -γ + 2)θ k u k - K k=1 (k + ν b + 2)θ k-1 u k = -(γ -2)θ 0 + K-1 k=1 [(k -γ + 2)θ k -(k + ν b + 2)θ k-1 ] u k -(K + ν b + 2)θ K-1 u K Therefore, u(1 -u) K-1 k=0 θ k u k -(γ -2 + (ν b + 3)u) K-1 k=0 θ k u k + K-1 k=0 g k b u k = K-1 k=1 (k -γ + 2)θ k -(k + ν b + 2)θ k-1 + g k b u k -(γ -2)θ 0 + g 0 b -(K + ν b + 2)θ K-1 u K
which yields the induction relation

θ 0 = g 0 b(γ-2) (γ -k -2)θ k + (k + ν b + 2)θ k-1 = g k b , 1 ≤ k ≤ K -1. Let ζ k = θ k w γ-1,ν b +1 (k) ,
then equivalently from (6.43):

ζ k + ζ k-1 = γ a(γ -2) g k (k + ν b + 2)w γ,ν b (k) , 1 ≤ k ≤ K -1 which yields ζ k = γ a(γ -2) (-1) k k j=0 (-1) j g j (j + ν b + 2)w γ,ν b (j) , 0 ≤ k ≤ K -1
and (7.19) follows.

step 2 Equation for the remainder. We conclude:

u(1 -u)Θ -[γ -2 + (ν b + 3)u] Θ = - G b ⇔ -(K + ν b + 2)θ K-1 u K + u(1 -u)r Θ -[γ -2 + (ν b + 3)u] r Θ = - r G b ⇔ r Θ - γ -2 + (ν b + 3)u u(1 -u) r Θ = 1 u(1 -u) (K + ν b + 2)θ K-1 u K - r G b ⇔ r Θ - γ -2 u + γ + ν b + 1 1 -u r Θ = 1 u(1 -u) (K + ν b + 2)θ K-1 u K - r G b ⇔ d du (1 -u) γ+ν b +1 u γ-2 r Θ = (1 -u) γ+ν b +1 u γ-2 1 u(1 -u) (K + ν b + 2)θ K-1 u K - r G b
and thus, any C ∞ solution must be the unique solution to the fixed point equation:

r Θ = u γ-2 (1 -u) γ+ν b +1 u 0 (1 -v) γ+ν b v γ-1 (K + ν b + 2)θ K-1 v K - r G b dv
with (7.21) forced by the Taylor expansion 22 . We now recall (7.2):

M 0 (u) = (K + ν b + 2)w γ-1,ν b +1 (K -1)T (bu K ) = (K + ν b + 2)w γ-1,ν b +1 (K -1) u γ-2 (1 -u) γ+ν b +1 u 0 (1 -v) γ+ν b v γ-1 v K dv and thus r Θ = θ K-1 w γ-1,ν b +1 (K -1) M 0 (u) -T (r G ) = (-1) K-1 S K-1 M 0 (u) -T (r G ),
this is (7.20).

7.3.

Convergence of the leading order Taylor coefficient. The truncation of the Taylor series produces the leading order term, provided the last Taylor coefficient is non zero. This is the S ∞ (d, ) = 0 condition.

Lemma 7.5. We have

S K-1 = (1 + o b→0 (1))S ∞ (7.22)
where S ∞ is given by

S ∞ (d, ) := 1 a +∞ j=0 (-1) j a j g ∞ j Γ(ν + j + 3) (7.23)
and where g ∞ j corresponds to the limiting problem (5.4) and is given by (5.15). Proof of Lemma 7.5. Since we have from (5.16)

+∞ j=0 (-1) j a j g ∞ j Γ(ν + j + 3) ≤ c ν   +∞ j=0 1 1 + j 2   ≤ c ν < +∞,
and from (6.35), 22 The statement on existence and uniqueness of the fixed point, the fact that the corresponding solution is smooth, and the fact that (7.21) holds has in fact already been proved in a more general case in Lemma 3.6. its suffices to prove the convergence term by term as b → 0. Now, recall (6.18),

+∞ j=0 (-1) j g j (j + ν b + 2)w γ,ν b (j) ≤ c ν   +∞ j=0 1 1 + j 2   ≤ c ν < +∞,
lim b→0 g j b j = g ∞ j .
Since from the definition of w γ,ν b (j),

(j + ν b + 2)w γ,ν b (j) = Γ(ν b + j + 3) Γ(γ -1 -j) Γ(γ -1) = Γ(ν + j + 3) γ j (1 + o b→0 (1)) = Γ(ν + j + 3) b j a j (1 + o b→0 (1)) we obtain lim b→0 (j + ν b + 2) w γ,ν b (j) b j = Γ(ν + j + 3)
a j from which we deduce the convergence term by term when b → 0, as desired.

7.4. The Θ main leading order term. We may now extract the leading order terms in Θ. From (7.18), (7.19), (7.20) and (7.22),

Θ(u) = K-2 k=0 θ k u k + θ K-1 u K-1 + (-1) K-1 S K-1 M 0 (u) -T (r G ) = K-2 k=0 θ k u k + (-1) K-1 S K-1 w γ-1,ν b +1 (K -1)u K-1 + M 0 (u) -T (r G ) = K-2 k=0 θ k u k + (-1) K-1 S ∞ [1 + o b→0 (1)] Θ main (u) -T (r G ) (7.24)
and from (7.11)

Θ main (u) = w γ-1,ν b +1 (K -1)u K-1 + M 0 (u) = w γ-1,ν b +1 (K -1)u K-1 + +∞ m=1 (-1) m w γ-1,ν b +1 (K -1 + m)u K-1+m = +∞ j=0 (-1) j w γ-1,ν b +1 (K -1 + j)u K-1+j = Γ(α γ )Γ(1 -α γ ) +∞ j=0 Γ(K + j + ν b + 2) Γ(K -1 + α γ )Γ(j + 1 -α γ ) u K-1+j . (7.25)
Equivalently, from (7.16) and (7.11):

Θ main (u) = Γ(α γ )Γ(1 -α γ ) Γ(K + ν b + 2) Γ(K -1 + α γ )Γ(1 -α γ ) u K-1 + Γ(α γ )Γ(1 -α γ ) Γ(K + ν b + 3) Γ(K -1 + α γ )Γ(2 -α γ ) u K + M 1 (u) = Γ(α γ ) Γ(K + ν b + 2) Γ(K -1 + α γ ) u K-1 1 + K + ν b + 2 Γ(2 -α γ ) Γ(1 -α γ )u + M 1 (u) = Γ(α γ )Γ(1 -α γ )K ν b +3-αγ u K-1 (7.26) × [1 + o b→0 (1)] 1 Γ(1 -α γ ) + K + ν b + 2 Γ(2 -α γ ) u + (uK) αγ M 1 (u) u K-1 Γ(α γ )Γ(1 -α γ )K ν b +3 u αγ .
We now turn to the study of Θ main . Lemma 7.6 (Properties of Θ main ). The function Θ main is positive and strictly increasing on [0, 1 2 ]. Moreover, pick universal constants 1 δ , Θ * 1 then for all 0 < b < b * (Θ * , δ), there exists a unique solution u * (α γ ) to

Θ main (u * (α γ )) = Θ * , u * (α γ ) ∈ 0, 1 2 (7.27)
which satisfies the following bounds:

first boundary close to integers. If α γ is such that

α γ = K ν b +3 (σb) K-1 Θ * , with δ < σ < 1 δ then Γ(α γ )Γ(1 -α γ )K ν b +3-αγ (u * (α γ )) K-1 = Θ * e Oν (σ
α γ = 1 - K ν b +3 (σb) K-1 Θ * , with δ < σ < 1 δ , then (7
.29) holds and

Γ(α γ )Γ(1 -α γ )K ν b +3-αγ K + ν b + 2 Γ(2 -α γ ) (u * (α γ )) K = Θ * e Oν (σ) . (7.30)
away from the integers. If α γ is such that (1) (7.31)

K ν b +3 b δ K-1 Θ * < α γ < 1 - K ν b +3 b δ K-1 Θ * then Γ(α γ )Γ(1 -α γ )K ν b +3 u * (α γ ) 1 -u * (α γ ) K-1 (u * (α γ )) αγ = Θ * e Oν
and b 2δ < u * (α γ ) < 1 2 . (7.32)
Proof of Lemma 7.6. The strict monotonicity and positivity follows from (7.25), since Θ main is given by a series with positive coefficients. From (7.26) (7.13),

Θ main 1 2 ≥M 1 1 2 ≥ c ν,1 Γ(α γ )Γ(1 -α γ )K ν b +3 ≥ c ν,1 uK 3 Θ *
which together with Θ main (0) = 0 ensures that that there is a unique solution of the equation

Θ main (u * (α γ )) = Θ * , 0 < u * (α γ ) < 1 2 .
We now aim at estimating the size of the unique solution u * (α γ ) in various regimes of the parameter α γ .

step 1 First boundary layer. We introduce the following function

M * (u) := Γ(α γ )Γ(1 -α γ )K ν b +3-αγ u K-1 .
Assume that

α γ = K ν b +3 (σb) K-1 Θ * for some δ < σ < 1 δ .
Then α γ |logb| 1 for 0 < b < b * (δ) and we compute:

u = M * (u) K ν b +3-αγ Γ(α γ )Γ(1 -α γ ) 1 K-1 = α γ M * (u) K ν b +3 (1 + O(α γ |logb|)) 1 K-1 (7.33) = K ν b +3 (σb) K-1 Θ * M * (u) K ν b +3 (1 + o b→0 (1)) 1 K-1 = (1 + o b→0 (1)) M * (u) Θ * 1 K-1 σb.
Next, we bound from below and above the function Θ main (u)/M * (u) for u in the interval bδ ≤ u ≤ b/δ. First, in view of (7.26), we have

Θ main (u) M * (u) = [1 + o b→0 (1)] 1 Γ(1 -α γ ) + K + ν b + 2 Γ(2 -α γ ) u +(uK) αγ M 1 (u) u K-1 Γ(α γ )Γ(1 -α γ )K ν b +3 u αγ . (7.34)
Since all three terms in (7.34) are positive and in view of the range of α γ , we deduce

Θ main (u) M * (u) ≥ [1 + o b→0 (1)] 1 Γ(1 -α γ ) ≥ 1 2 .
From (7.13) with j = 1, we estimate for b ≤ u ≤ b δ , using also the fact that |log(1 -u)| ≤ u:

(uK) αγ M 1 (u) M * (u) ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +3 u 1-u K-1 u αγ u K-1 Γ(α γ )Γ(1 -α γ )K ν b +3 u αγ ≤ c ν e (K-1) b δ ≤ c ν e cν δ . (7.35)
Using (7.12), the estimate also holds for δb ≤ u ≤ b. We conclude that for δb ≤ u ≤ b δ :

1 2 M * (u) Θ * ≤ Θ main (u) Θ * ≤ e cν δ M * (u) Θ * .
Since M * ((0, +∞)) = (0, +∞), there exists 0 < u 1 < u 2 < +∞ such that

1 2 M * (u 2 ) Θ * ≥ 1, e cν δ
M * (u 1 ) Θ * ≤ 1. Note also that for any u such that M * (u) = O ν (δ -1 )Θ * , we have from (7.33)

u = (1 + o b→0 (1)) M * (u) Θ * 1 K-1 σb = σb e Oν (δ -1 ) 1 K-1 = σb(1 + o b→0 (1))
so that u belongs to the range b ≤ u ≤ b δ . This is in particular true for u 1 and u 2 , and therefore

Θ main (u 2 ) Θ * ≥ 1, Θ main (u 1 ) Θ * ≤ 1, bδ ≤ u 1 < u 2 ≤ b δ
M * * (u) := Γ(α γ )Γ(1 -α γ )K ν b +3-αγ u K-1 K + ν b + 2 Γ(2 -α γ ) u.
Assume that we have

α γ = 1 - K ν b +3 (σb) K-1
Θ * for some δ < σ < 1 δ and compute:

u = M * * (u) Γ(α γ )Γ(1 -α γ ) K+ν b +2 Γ(2-αγ ) K ν b +3-αγ 1 K = M * * (u)(1 -α γ )(1 + o b→0 (1)) K ν b +3 1 K = M * * (u) Θ * 1 K (σb) K-1 K (1 + o b→0 (1)) = M * * (u) Θ * 1 K σb(1 + o b→0 (1)) (7.36)
Next, we bound from below and above the function Θ main (u)/M * * (u) for u in the interval bδ ≤ u ≤ b/δ. First, in view of (7.26), we have

Θ main (u) M * * (u) = [1 + o b→0 (1)] 1 + Γ(2 -α γ ) Γ(1 -α γ )(K + ν b + 2)u + Γ(2 -α γ ) (K + ν b + 2)u (uK) αγ M 1 (u) u K-1 Γ(α γ )Γ(1 -α γ )K ν b +3 u αγ .(7.37)
Since all three terms in (7.37) are positive, we deduce

Θ main (u) M * * (u) ≥ [1 + o b→0 (1)] ≥ 1 2 .
For δb ≤ u ≤ b/δ, we have, recalling (7.35):

M 1 (u) ≤ c ν Γ(α γ )Γ(1 -α γ )K ν b +3-αγ u K-1 e cν δ
and, since

Γ(2 -α γ ) (K + ν b + 2)u ≤ c ν bu ≤ c ν δ we have 1 2 M * * (u) Θ * ≤ Θ main (u) Θ * ≤ e cν δ
M * * (u) Θ * We may conclude, as in step 1, that u * (α γ ) belongs to the range δb ≤ u ≤ b/δ, and (7.29) (7.30) hold. step 3 Away from the integer boundary layer. We now rewrite (7.26) using (A.8):

Θ main (u) = Γ(α γ ) Γ(K + ν b + 2) Γ(K -1 + α γ ) u K-1 1 + K + ν b + 2 Γ(2 -α γ ) Γ(1 -α γ )u + M 1 (u) = Γ(α γ )Γ(1 -α γ )K ν b +3 u 1 -u K-1 u αγ    M 1 (u) Γ(α γ )Γ(1 -α γ )K ν b +3 u 1-u K-1 u αγ + + (1 + o b→0 (1))(1 -u) K-1 (uK) αγ Γ(1 -α γ ) + (1 + o b→0 (1))uK(1 -u) K-1 (uK) αγ Γ(2-α γ ) .
For u = σb, σ ≥ 1 δ , we have using the bound log(1 -u) ≤ -u:

(uK) -αγ (1 -u) K-1 Γ(1 -α γ ) + (uK) 1-αγ (1 -u) K-1 Γ(2 -α γ ) ≤ c ν σ 1-αγ e -(K-1)σb ≤ c ν σ 1-αγ e -cν σ = o δ→0 (1)
We then obtain the representation formula in this zone

Θ main (u) = Γ(α γ )Γ(1 -α γ )K ν b +3 u 1 -u K-1 u αγ    M 1 (u) Γ(α γ )Γ(1 -α γ )K ν b +3 u 1-u K-1 u αγ + o δ→0 (1)    .
We conclude from (7.13): (1) and (7.31) follows. Also, we have

Θ main (u) = Θ * ⇔ Γ(α γ )Γ(1 -α γ )K ν b +3 u 1 -u K-1 u αγ = Θ * e Oν
Θ * Γ(α γ )Γ(1 -α γ )K ν b +3 = Θ * α γ (Γ(1 -α γ )) 2 K ν b +3 = Θ * (1 -α γ ) Γ(α γ )Γ(2 -α γ )K ν b +3
Thus, in view of the range of α γ , we infer

Θ * Γ(α γ )Γ(1 -α γ )K ν b +3 ≥ 1 Γ 1 2 max Γ 1 2 , Γ 3 2 b δ K-1 so that u 1 -u K-1 u αγ ≥ b δ K-1
e Oν (1) and (7.32) easily follows.

7.5. Bilinear estimate for T . We now develop the set of nonlinear estimates to control the fixed point equation (7.20).

Lemma 7.7 (Pointwise bilinear estimate). Let

F = K-1 k=0 f k u k + r F , G = K-1 k=0 g k u k + r G ,
and

A F = sup 0≤k≤K-1 |f k | |w γ-1,ν b +1 (k)| , A G = sup 0≤k≤K-1 |g k | |w γ-1,ν b +1 (k)| ,
then we have the following pointwise bounds for 0 ≤ u ≤ 1 2 :

(1) Bound for T r. Let 1 ≤ j ≤ 5, then

T (r u j F G ) M 0 (u) ≤ c ν,a A F A G 1 + u K+j-2 Γ(α γ )K ν b +3-αγ (7.38) + c ν,a b 1 + u K-1 Γ(α γ )K ν b +3-αγ A F r G M 0 L ∞ (v≤u) + A G r F M 0 L ∞ (v≤u) + c ν,a b r F M 0 L ∞ (v≤u) r G M 0 L ∞ (v≤u) M 0 L ∞ (v≤u) .
(2) Bound for r. Let 0 ≤ j ≤ 5,

r u j F G M 0 (u) ≤ c ν,a A F A G 1+u K-2 Γ(α γ )K ν b +2-αγ + u K-1 Γ(α γ )K ν b +3-αγ + c ν,a 1 + u K-1 Γ(α γ )K ν b +3-αγ A F r G M 0 L ∞ (v≤u) + A G r F M 0 L ∞ (v≤u) + c ν,a r F M 0 L ∞ (v≤u) r G M 0 L ∞ (v≤u) M 0 L ∞ (v≤u) . (7.39)
Proof of Lemma 7.7. We compute

u j F G = 2(K-1) k=0   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 f k 1 g k 2   u k+j + u j r G K-1 k=0 f k u k +u j r F K-1 k=0 g k u k + u j r F r G .
Then, recalling (7.17):

r u j F G = 2(K-1) k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 f k 1 g k 2   u k+j +u j r G K-1 k=0 f k u k + r F K-1 k=0 g k u k + r F r G .
We now apply T and estimate all the terms in the corresponding identity.

step 1 Polynomial terms. We compute from (7.2):

1 M 0 T     2(K-1) k=K-j k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 f k 1 g k 2   u k+j   = 2(K-1) k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 f k 1 g k 2   T (u k+j ) (K + ν b + 2)w γ-1,ν b +1 (K -1)T (bu K )
.

We now observe the bound for m ≥ 0 and 0 ≤ u < 1:

0 ≤ T (u K+m ) T (u K ) = u 0 (1-v) γ+ν b v γ-1 v K+m dv u 0 (1-v) γ+ν b v K v γ-1 dv ≤ u m (7.40)
which gives the estimate

1 M 0 T   2(K-1) k=K-j k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 f k 1 g k 2   u k+j ≤ sup 0≤k≤K-1 |f k | |w γ-1,ν b +1 (k)| sup 0≤k≤K-1 |g k | |w γ-1,ν b +1 (k)| 1 b(K + ν b + 2)w γ-1,ν b +1 (K -1) × 2(K-1) k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K ≤ A F A G w γ-1,ν b +1 (K -1) 2(K-1) k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K . case K -j ≤ k ≤ K -1:
we use the convolution bound (C.5) which gives the estimate

K-1 k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K ≤ c ν,a K-1 k=K-j w γ-1,ν b +1 (k)u k+j-K = c ν,a j-1 m=0 w γ-1,ν b +1 (K -1 -(j -1 -m))u m .
We now recall (6.13) (6.47) which ensures that for m ≤ j -1 ≤ 5:

w γ-1,ν b +1 (K -1 -(j -1 -m)) w γ-1,ν b +1 (K -1) = w γ,ν b (K -(j -1 -m)) w γ,ν b (K) ≤ c ν γ 1 + K j-1-m ≤ c ν
and leads to the bound:

1 w γ-1,ν b +1 (K -1) K-1 k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K ≤ c ν,a j-1 m=0 u j ≤ c ν,a . (7.41)
case k ≥ K: we use (6.13) and the truncated convolution bound (C.17):

1 w γ-1,ν b +1 (K -1) 2(K-1) k=K   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K ≤ c ν,a w γ-1,ν b +1 (K -1) 2(K-1) k=K w γ-1,ν b +1 (k -(K -1))w γ-1,ν b +1 (K -1)u k+j-K ≤ c ν,a u j K-2 m=0 w γ-1,ν b +1 (m + 1)u m ≤ c ν,a u j-1 K-1 k=1 w γ-1,ν b +1 (k)u k . (7.42)
We now claim the uniform bound for u ≤ 1 2 :

K-2 k=0 w γ-1,ν b +1 (k)u k ≤ c ν,a (7.43)
which is proved below. Since from (A.8)

w γ-1,ν b +1 (K -1) = Γ(α γ )Γ(K -1 + ν b + 1 + 2) Γ(γ -2) = Γ(α γ )Γ(K + ν b + 2) Γ(K -1 + α γ ) = [1 + o K→∞ (1)] Γ(α γ )K ν b +3-αγ , (7.44)
we conclude:

u j-1 K-1 k=1 w γ-1,ν b +1 (k)u k ≤ c ν,a u j-1 1 + u K-1 Γ(α γ )K ν b +3-αγ , (7.45)
and the collection of the above bounds yields, using also j ≥ 1,

1 M 0 T   2(K-1) k=K-j k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 f k 1 g k 2   u k+j ≤ A F A G w γ-1,ν b +1 (K -1) 2(K-1) k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K ≤ c ν,a A F A G 1 + u K+j-2 Γ(α γ )K ν b +3-αγ . (7.46)
Proof of (7.43). From (6.13) and (C.3), for some large enough K ν :

K-Kν k=0 w γ-1,ν b +1 (k)u k ≤ K-Kν k=0 w γ-1,ν b +1 (k)= (γ -2) K+1-Kν k=1 w γ,ν b (k) ≤ c ν
and from (6.13) and (C.4):

K-2 k=K-Kν +1 w γ-1,ν b +1 (k)u k = K-2 k=K-Kν +1 (γ -2)w γ,ν b (k + 1)u k ≤ c ν K-2 k=K-Kν +1 (γ -2)Γ(γ -2 -k)γ ν b +2-(γ-2-k) 1 2 k ≤ K cν 2 K ≤ c ν and (7.43) is proved.
step 2 Cross terms. We estimate pointwise from (7.6), (7.43), (7.44):

T u j r G ( K-1 k=0 f k u k ) M 0 ≤ A F sup v≤u K-1 k=0 w γ-1,ν b +1 (k)v k r G M 0 L ∞ (v≤u) T (u j M 0 ) M 0 ≤ c ν,a b A F 1 + u K-1 Γ(α γ )K ν b +3-αγ r G M 0 L ∞ (v≤u)
and similarly for the other cross term.

step 3 Nonlinear term. We estimate directly from (7.6):

T (u j r F r G ) M 0 (u) ≤ r F M 0 L ∞ (v≤u) r G M 0 L ∞ (v≤u) M 0 L ∞ (v≤u) T (u j M 0 ) M 0 ≤ c ν,a b r F M 0 L ∞ (v≤u) r G M 0 L ∞ (v≤u) M 0 L ∞ (v≤u) .
step 4 Estimate for r. We now revisit the above estimates allowing for j = 0. We start with the polynomial term. We have:

2(K-1) k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 f k 1 g k 2   u k+j ≤ c ν,a A F A G u K 2(K-1) k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K .
For j ≥ 0 and K -j ≤ k ≤ K -1, we use (7.41) and (7.44) which imply:

u K K-1 k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K ≤ c ν,a w γ-1,ν b +1 (K -1)u K ≤ c ν,a Γ(α γ )K ν b +3-αγ u K .
For K ≤ k ≤ 2(K -1) and j ≥ 0, we argue differently, depending on u.

case u ≤ b. We recall (7.42) which yields:

u K 2(K-1) k=K   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K ≤ c ν,a w γ-1,ν b +1 (K -1)u K u j K-2 m=0 w γ-1,ν b +1 (m + 1)u m .
We now have the rough bound from (6.14):

w γ-1,ν b +1 (m+1) = m + ν b + 3 γ -m -3 w γ-1,ν b +1 (m) = m + ν b + 3 K + α γ -m -2 w γ-1,ν b +1 (m) ≤ c ν b w γ-1,ν b +1 (m).
for m ≤ K -3. Then, from (7.43) and (7.44):

u K 2(K-1) k=K   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K ≤ c ν,a Γ(α γ )K ν b +3-αγ b u K+j K-3 m=0 w γ-1,ν b +1 (m)u m +c ν,a Γ(α γ )K ν b +3-αγ u K u j w γ-1,ν b +1 (K -1)u K-2 ≤ c ν,a Γ(α γ )K ν b +3-αγ b u K+j +c ν,a Γ(α γ )K ν b +3-αγ u K u j Γ(α γ )K ν b +3-αγ u K-2
which yields the bound:

2(K-1) k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 f k 1 g k 2   u k+j ≤ c ν,a b A F A G Γ(α γ )K ν b +3-αγ u K +A F A G c ν,a Γ(α γ )K ν b +3-αγ u K Γ(α γ )K ν b +3-αγ u K-2 .
Therefore, from (7.4) for u ≤ b:

1 M 0 2(K-1) k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 f k 1 g k 2   u k+j ≤ cν,a b A F A G Γ(α γ )K ν b +3-αγ u K +A F A G c ν,a Γ(α γ )K ν b +3-αγ u K Γ(α γ )K ν b +3-αγ u K-2 Γ(α γ )Γ(1 -α γ )K ν b +4-αγ u K ≤ c ν A F A G +c ν A F A G Γ(α γ )K ν b +2-αγ u K-2 ≤ c ν A F A G 1 + u K-2 Γ(α γ )K ν b +2-αγ .
case u ≥ b. We recall (7.42) (7.44) (7.45) which yield

u K 2(K-1) k=K   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 |w γ-1,ν b +1 (k 1 )||w γ-1,ν b +1 (k 2 )|   u k+j-K ≤ c ν,a u j-1 Γ(α γ )K ν b +3-αγ u K u j-1 1 + u K-1 Γ(α γ )K ν b +3-αγ
and thus, for b ≤ u ≤ 1 2 , from (7.5):

1 M 0 2(K-1) k=K-j   k 1 +k 2 =k,0≤k 1 ,k 2 ≤K-1 f k 1 g k 2   u k+j ≤ A F A G Γ(α γ )K ν b +3-αγ u K u j-1 1 + u K-1 Γ(α γ )K ν b +3-αγ Γ(α γ )Γ(1 -α γ )K ν b +3 u 1-u K-1 u αγ ≤ c ν A F A G 1 + u K-1 Γ(α γ )K ν b +3-αγ .
We estimate the cross term from (7.43), (7.44)

u j r G ( K-1 k=0 f k u k ) M 0 ≤ A F sup v≤u K-1 k=0 w γ-1,ν b +1 (k)v k r G M 0 L ∞ (v≤u) ≤ c ν,a A F 1 + u K-1 Γ(α γ )K ν b +3-αγ r G M 0 L ∞ (v≤u)
and the nonlinear term

u j r F r G M 0 (u) ≤ r F M 0 L ∞ (v≤u) r G M 0 L ∞ (v≤u) M 0 L ∞ (v≤u) .
7.6. Controlling the final remainder. We are now in position to prove the exit condition for the C ∞ solution for a large enough range of parameters.

Lemma 7.8 (Uniform control of the final remainder). Pick universal constants 1 δ , Θ * 1, then for all 0 < b < b * (δ, Θ * ) 1 small enough, the following holds. Let

K ν b +3 (bδ) K-1 Θ * < α γ < 1 - K ν b +3 (bδ) K-1 Θ * (7.47)
and let u * (α γ ) be the solution to (7.27) described by Lemma 7.6. Then the C ∞ solution to (7.20) satisfies for all 0 ≤ u ≤ u * (α γ ) the bound:

|r G | M 0 + |T r G | M 0 < √ b. (7.48)
Proof of Lemma 7.8. We recall the fixed point formulation (7.24) of the C ∞ solution and the expression (6.31) for the nonlinear term:

G = G 0 + b 2 x h1 + bx 2 h2 Θ + b 2 x h3 + bx 2 h4 uΘ + 4 j=2 x j+1 m(1) j Θ j + b 4 j=2 m (2) j x j Θ j +   3 j=1 m(3) j x j+2 Θ j + b 3 j=1 m(4) j x j+1 Θ j   (uΘ ).
We now bootstrap the bound

|r Θ (u)| M 0 < Θ * . (7.49)
Let us first check that (7.49) holds for u small enough. From (7.24), we have

|r Θ (u)| M 0 ≤ c ν,a + |T (r G )(u)| M 0 ≤ c ν,a + sup v≤u |G(v)| v K |T (u K )| M 0 ≤ c ν,a + sup v≤u |G(v)| v K 1 b(K + ν b + 2)w γ-1,ν b +1 (K -1)
where we have used (7.2) in the last inequality. Also, we have, using (6.35) and (6.13),

lim u→0 sup v≤u |G(v)| v K = |g K | ≤ c ν,a |w γ,ν b (K)| 1 + K ≤ c ν,a |w γ-1,ν b +1 (K -1)| (1 + K)(γ -2) .
We infer for u small enough that

|r Θ (u)| M 0 ≤ c ν,a < Θ *
so that (7.49) indeed holds for u small enough. We therefore work on the interval u ∈ [0, u boot ] with 0 <u boot ≤ u * (αγ) where (7.49) holds, and aim at improving (7.49).

step 1 Uniform bounds for 0 ≤ u ≤ u * (α γ ). By definition (7.25):

Θ main (u) = w γ-1,ν b +1 (K -1)u K-1 + M 0 (u)
and hence, since Θ main is non decreasing:

∀u ∈ [0, u * (α γ )], 0 ≤ M 0 (u) ≤ Θ main (u) ≤ Θ main (u * (α γ )) = Θ * . (7.50)
Observe that in the regime (7.28):

Γ(α γ )Γ(1 -α γ )K ν b +3-αγ (u * (α γ )) K-1 = C Θ * ,δ ,
in the regime (7.30), recalling (7.29):

Γ(α γ )Γ(1 -α γ )K ν b +3-αγ (u * (α γ )) K-1 ≤ Γ(2 -α γ )Θ * e Oν (σ) u * (α γ )(K + ν b + 2) ≤ Θ * e Oν (σ) δ ≤ C Θ * ,δ ,
and in the regime (7.31), recalling (7.32):

Γ(α γ )Γ(1 -α γ )K ν b +3-αγ (u * (α γ )) K-1 ≤ (1 -u * (α γ )) K-1 (Ku * (α γ )) αγ Θ * e Oν (1) ≤ C Θ * ,δ
In all three cases

∀u ∈ [0, u * (α γ )], Γ(α γ )Γ(1 -α γ )K ν b +3-αγ u K-1 ≤ C Θ * ,δ . (7.51)
Also, we have, using (7.51),

Γ(α γ )Γ(1 -α γ )K ν b +2-αγ u K-2 ≤ Γ(α γ )Γ(1 -α γ )K ν b +2-αγ (u * (α γ )) K-2 ≤ C Θ * ,δ Ku * (α γ )
and thus, using (7.29) (7.32), we deduce

∀u ∈ [0, u * (α γ )], Γ(α γ )Γ(1 -α γ )K ν b +2-αγ u K-2 ≤ C Θ * ,δ . (7.52)
We now decompose r G according to the decomposition (6.31) of G and estimate all the terms.

step 2 Source term. For any holomorphic function H(x), we have from (7.17):

|r H | = +∞ k=K h k u k ≤ +∞ k=K (bC h u) k ≤ (bC h ) K u K .
We therefore estimate for u ≤ b from (7.4):

|r H | M 0 ≤ (bC H ) K u K Γ(α γ )Γ(1 -α γ )K ν b +4-αγ u K ≤ (bC H ) K Γ(α γ )Γ(1 -α γ )K ν b +3
and for b ≤ u ≤ 1 2 from (7.5):

|r H | M 0 ≤ (bC H ) K u K Γ(α γ )Γ(1 -α γ )K ν b +3 u 1-u K-1 u αγ ≤ (bC H ) K Γ(α γ )Γ(1 -α γ )K ν b +3 It implies the rough bound r H M 0 L ∞ (0≤u≤ 1 2 ) ≤ b 4 . (7.53)
In view of (6.32) this bound can be applied to the source term G 0 .

step 3 Derivative term. Let

Θ = K-1 k=0 θ k u k + r Θ , then uΘ = K-1 k=1 kθ k u k + ur Θ .
Therefore,

(uΘ ) k = kθ k , 0 ≤ k ≤ K -1 r uΘ = ur Θ . (7.54)
Moreover, from (7.20):

ur Θ = (-1) K-1 S K-1 uM 0 (u) -u [T (r G )]
We recall from (7.8):

u [T (G)] = 1 1 -u G b + [(γ -2) + (ν b + 3)u]T (G)
which yields

r uΘ = (-1) K-1 S K-1 uM 0 (u) - 1 1 -u r G b + [(γ -2) + (ν b + 3)u]T (r G )
We obtain the estimate, using (7.7) and (6.34):

|r uΘ | M 0 ≤ cν b 1 + |r G | M 0 + |T (r G )| M 0 sup 0≤k≤K-1 |(uΘ ) k | w γ-1,ν b +1 (k) ≤ cν b .
(7.55) step 4 Linear term. Recall from (6.36) that for a holomorphic function:

∀0 ≤ k ≤ K -1, |(HΘ) k | ≤ c H w γ-1,ν b +1 (k).
no derivative term. We apply (7.38) with the bounds (7.53), (7.50), (7.51) (7.52) and (6.34) to derive for 1 ≤ j ≤ 5:

T (r u j HΘ ) M 0 (u) ≤ c ν,a A Θ A H 1 + u K+j-2 Γ(α γ )K ν b +3-αγ + c ν,a b 1 + u K-1 Γ(α γ )K ν b +3-αγ A Θ r H M 0 L ∞ (v≤u) + A H r Θ M 0 L ∞ (v≤u) + c ν,a b r Θ M 0 L ∞ (v≤u) r H M 0 L ∞ (v≤u) M 0 L ∞ (v≤u) ≤ c ν,a + C Θ * ,δ b
where we used the bootstrap bound (7.49) in the last step. Now, writing

b 2 x h1 Θ = b 3 u( h1 Θ) bx 2 h2 Θ = b 3 u 2 ( h2 Θ) gives T r [b 2 x h1 +bx 2h 2] Θ M 0 ≤ b 3 c ν,a + C Θ * ,δ b ≤ b 2 C Θ * ,δ .
Similarly, from (7.39), (7.49), (7.53), (7.50), (7.51) (7.52) and (6.34):

r u j HΘ M 0 (u) ≤ C Θ * ,δ Therefore, r [b 2 x h1 +bx 2h 2] Θ M 0 ≤ b 3 C Θ * ,δ .
derivative term. We first use (7.55), (7.39) to estimate for a holomorphic function H:

r u j H(uΘ ) M 0 (u) ≤ c ν,a b 1+u K-1 Γ(α γ )K ν b +3-αγ + u K-2 Γ(α γ )K ν b +2-αγ + c ν,a 1 + u K-1 Γ(α γ )K ν b +3-αγ 1 + r uΘ M 0 L ∞ (v≤u) + C Θ * ,δ r uΘ M 0 L ∞ (v≤u) ≤ C Θ * ,δ 1 b + r uΘ M 0 L ∞ (v≤u)
and from (7.55), (7.38):

T (r u j H(uΘ ) ) M 0 (u) ≤ C Θ * ,δ 1 b 2 + 1 b r uΘ M 0 L ∞ (v≤u) .
We conclude, using (7.55):

T r [b 2 x h1 +bx 2h 2] uΘ M 0 ≤ b 3 C Θ * ,δ 1 b 2 + 1 b r uΘ M 0 L ∞ (v≤u) ≤ C Θ * ,δ b + b 2 r uΘ M 0 L ∞ (v≤u) ≤ C Θ * ,δ b + b r G M 0 L ∞ (v≤u) + b T (r G ) M 0 L ∞ (v≤u)
and

r [b 2 x h1 +bx 2h 2] uΘ M 0 ≤ b 3 C Θ * ,δ 1 b + r uΘ M 0 L ∞ (v≤u) ≤ C Θ * ,δ b 2 + b 2 r G M 0 L ∞ (v≤u) + b 2 T (r G ) M 0 L ∞ (v≤u)
step 5 Nonlinear term.

no derivative term. First, we have in view of (6.34) (C.6) that for 1 ≤ m ≤ 5 and 0 ≤ k ≤ K -1:

|(Θ m ) k | ≤ k 1 +•••km=k |Θ k 1 | • • • |Θ km | ≤ c ν,a k 1 +•••km=k w γ-1,ν b +1 (k 1 ) • • • w γ-1,ν b +1 (k m ) ≤ c ν,a w γ-1,ν b +1 (k)
We then estimate, using (7.39) iteratively in m for 2 ≤ m ≤ 5, 0 ≤ j ≤ 5, and using also (7.49), (7.53), (7.50), (7.51) (7.52):

r u j Θ m M 0 (u) ≤ c ν,a A u j A Θ + r Θ M 0 L ∞ (v≤u)   m =1 A Θ + A Θ + r Θ M 0 L ∞ (v≤u) m-2   × 1 + u K-1 Γ(α γ )K ν b +3-αγ +u K-2 Γ(α γ )K ν b +2-αγ m-1 ≤ C Θ * ,δ
Similarly, for a holomorphic function H:

r u j HΘ m M 0 (u) ≤ C Θ * ,δ (7.56)
which implies that

r 4 j=2 x j+1 m(1) j Θ j +b 4 j=2 m (2) j x j Θ j M 0 ≤ C Θ * ,δ b 3 .
Similarly, for 1 ≤ j ≤ 5 from (7.49) (7.53), (7.50), (7.51):

T (r u j HΘ j ) M 0 (u) ≤ C Θ * ,δ b gives T r 4 j=2 x j+1 m(1) j Θ j +b 4 j=2 m (2) j x j Θ j M 0 ≤ b 3 C Θ * ,δ b ≤ C Θ * ,δ b 2 .
derivative term. We estimate from (7.56), (7.39), (7.55), (7.50), (7.51) (7.52):

r u j HΘ m (uΘ ) M 0 (u) ≤ C Θ * ,δ 1 b + r uΘ M 0 L ∞ (v≤u) ≤ C Θ * ,δ 1 b + 1 b r G M 0 L ∞ (v≤u) + T (r G ) M 0 L ∞ (v≤u) .
Similarly,

T r u j HΘ m (uΘ ) M 0 (u) ≤ C Θ * ,δ b 1 b + r uΘ M 0 L ∞ (v≤u) ≤ C Θ * ,δ b 2 1 + r G M 0 L ∞ (v≤u) + T (r G ) M 0 L ∞ (v≤u) .
Therefore,

|r 3 j=1 m(3) j x j+2 Θ j +b 3 j=1 m(4) j x j+1 Θ j (uΘ ) | M 0 ≤ b 3 C Θ * ,δ 1 b + 1 b r G M 0 L ∞ (v≤u) + T (r G ) M 0 L ∞ (v≤u) ≤ C Θ * ,δ b 2 1 + r G M 0 L ∞ (v≤u) + T (r G ) M 0 L ∞ (v≤u) and |T r 3 j=1 m(3) j x j+2 Θ j +b 3 j=1 m(4) j x j+1 Θ j (uΘ ) | M 0 ≤ b 3 C Θ * ,δ b 2 1+ r G M 0 L ∞ (v≤u) + T (r G ) M 0 L ∞ (v≤u) ≤ bC Θ * ,δ 1 + r G M 0 L ∞ (v≤u) + T (r G ) M 0 L ∞ (v≤u) .
step 6 Conclusion. The collection of the above bounds yields:

|r G | M 0 ≤ C Θ * ,δ b 2 1 + r G M 0 L ∞ (v≤u) + T (r G ) M 0 L ∞ (v≤u)
and

|T r G | M 0 ≤ bC Θ * ,δ 1 + r G M 0 L ∞ (v≤u) + T (r G ) M 0 L ∞ (v≤u)
which imply for 0 < b < b * (Θ * , δ) small enough

|r G | M 0 + |T r G | M 0 < √ b
which, reinserted into (7.20), yields:

|r Θ | M 0 ≤ c ν,a + √ b < Θ * 2 ,
and (7.49), (7.48) are proved.

7.7. Exit on the left of P 2 . We are now in position to establish the fundamental exit property of the C ∞ solution to the left of P 2 . Recall from Lemma 3.6 that the unique C ∞ solution has a slope c -at P 2 . Recall also from (2.49) that c -is strictly between the slope of the middle root w 2 (σ) of ∆ 1 at P 2 and the slope of the smallest root w - 2 (σ) of ∆ 2 at P 2 . Thus, at the left of P 2 , for u small enough, the C ∞ solution lies strictly above the middle root of ∆ 1 , and strictly below the smallest root of ∆ 2 . The goal of Lemma 7.9 below is to show that the C ∞ solution exists this region, i.e. that there exist a value 0 < U * < 1 such that at u = U * , the solution either touches the middle root of ∆ 1 , or touches the smallest root of ∆ 2 . Indeed, since at such a point we have respectively w (σ) = 0 or w (σ) = -∞, and since (w - 2 ) (σ) < 0 and w 2 (σ) < 0 respectively by (2.7) and (2.10), touching either root implies that the C ∞ solution exists the region for u > U * .

We assume without loss of generality that

S ∞ (d, ) > 0 (7.57)
and need only to reverse the parity of K in Lemma 7.9 below if S ∞ (d, ) < 0.

Lemma 7.9 (Exit on the left). Pick universal constants 1 δ , Θ * 1 large enough as in Lemma 7.8, then for all 0 < b < b * (Θ * , δ) small enough and α γ in the range (7.47), the C ∞ solution exits on the left of P 2 at u = U * , where 0 < U * < 3 4 , by crossing the smallest root of ∆ 2 for K odd, and by crossing the middle root of ∆ 1 for K even.

Proof of Lemma 7.9. step 1 Reaching Θ * . Recall (7.24)

Θ(u) = K-2 k=0 θ k u k + (-1) K-1 S ∞ [1 + o b→0 (1)] Θ main (u) -T (r G )
then, provided Θ * has been chosen large enough, we conclude from (7.43), (7.48), (7.50) that for 0 < b < b * (Θ * , δ) small enough, for all u ∈ [0, u * (α γ )]:

|Θ(u) -(-1) K-1 S ∞ Θ main (u)| ≤ S ∞ Θ * 10 . (7.58) Therefore, Θ * 2 ≤ (-1) K-1 Θ(u * (α γ )) S ∞ ≤ 2Θ * . (7.59) step 2 Computation of ∆ 1 , ∆ 2 .
We now unfold our changes of variables and show that, depending on the sign of (-1) K-1 S ∞ , we must have passed through either the green or the red curves. To this end, we examine ∆ 1 and ∆ 2 under the assumption

|u| ≤ 3 4 , |Ψ(u)| ≤ 2Θ * . (7.60)
From (6.29), (6.1), (4.39)

Φ = (1 -u)Ψ Ψ(u) = M b (x)Φ(u) Φ = buΘ
where M b is bounded and given by (6.2). Moreover, from (3.15), (4.19), (4.31), (4.34), (4.22):

∆ 1 -c -∆ 2 = G 2 = b 2 F 1 = -b 2 |λ -| ψo wo (c + -c -)uF 1 = -b 2 |λ -| ψo wo (c + -c -)u u(1 -u)bH 1 (b, u) + (1 + G 1 (bu)) Φ + NL 1 (u, Φ) = -b 2 w2 o ẽ20 (1 + O(b))u u(1 -u)bH 1 (b, u) + (1 + G 1 (bu))bu(1 -u)M b Θ + NL 1 (u, Φ) = -b 3 w2 o ẽ20 (1 + O(b))u 2 (1 -u) H 1 (b, u) + (1 + G 1 (bu))M b Θ + NL 1 (u, Φ) bu(1 -u)
and from (3.14), (4.18), (4.31), (4.36), (4.22):

-∆ 1 + c + ∆ 2 = G 1 = -b 2 F 2 = b 2 wo (c + -c -)|µ + |uF 2 = b 2 wo (c + -c -)|µ + |u (1 -u) [1 + H 2 (b, u)] + G 2 (bu) Φ + NL 2 (u, Φ) = b 2 | d20 | w2 o (1 + O(b))u(1 -u) 1 + H 2 (b, u) + G 2 (bu)M b Φ + NL 2 (u, Φ) 1 -u
From (6.2), for x = bu and u ≤ 1,

M b (x) = 1 + O(b), (7.61)
and from (4.35), (4.37), (4.38) and ( 7.60)

H 1 (b, u) = -( Ẽ11 + Ẽ30 ) + O(b) G 1 = b Ẽ11 u + O(b 2 ) NL 1 = O(b 2 u) H 2 (b, u) = O(b) G 2 (x) = O(b) NL 2 = O(b 2 u)
It implies that, as long as Θ(u) = 0, we have:

∆ 1 -c -∆ 2 = -Θ(u)bu w2 o ẽ20 b 2 u(1 -u) 1+O 1 Θ(u) + O(b) -∆ 1 + c + ∆ 2 = | d20 | w2 o b 2 u(1 -u)(1 + O(b)) i.e., ∆ 1 = 1 c + -c - w2 o b 2 u(1 -u) 1+O 1 Θ(u) + b -c + ẽ20 Θ(u)bu+c -| d20 | ∆ 2 = 1 c + -c - w2 o b 2 u(1 -u) 1+O 1 Θ(u) + b -ẽ 20 Θ(u)bu + | d20 |
or, equivalently,

∆ 1 = -1 c + -c - w2 o b 2 u(1 -u) 1+O 1 Θ(u) + b c + ẽ20 Φ(u) + |c -|| d20 | ∆ 2 = 1 c + -c - w2 o b 2 u(1 -u) 1+O 1 Θ(u) + b -ẽ 20 Φ(u) + | d20 | . (7.62)
Note that we have used the signs, valid for all d ≥ 2 and 0 ≤ ≤ d:

d20 < 0, ẽ20 > 0, see (F.8). step 3 Touching ∆ 1 = 0 or ∆ 2 = 0. At u = u * (α γ )
, by (7.59) we have

Φ(u * ) = bu * Θ(u * ), (-1) K-1 Θ(u * ) S ∞ ≥ Θ * 2 .
We now claim that there exists

u * ≤ U * ≤ 3 4 , Φ(U * ) = (-1) K-1 Θ *
Then, from (7.62), since d20 = 0 and ẽ20 > 0 by (F.8), and since Φ(0) = 0, we must have crossed ∆ 1 = 0 or ∆ 2 = 0, depending on wether K is even or odd. Assume K odd, so that from (7.59) Θ(u * ) > 0. The case K even can be treated similarly. Since u ≤ 3 4 and M b (u) = 1 + O(b):

Ψ(u * ) = bu * Θ(u * )(1 + O(b)), Ψ(U * ) = Θ * .
In view of (4.15)

1 + H 2 + G 2 Ψ + NL 2 u(1 -u)Ψ + (1 -2u)(1 + H 2 + G 2 Ψ + NL 2 ) -γ(1 + G 1 ) + 2uG 2 Ψ = u γbH 1 -2(1 + H 2 ) + γb NL 1 x -2 NL 2 .
Then, on the interval u ∈ [u

H 1 (b, u) + ( Ẽ11 + Ẽ30 ) + |H 2 | + |G 2 | + |G 1 | ≤ C Θ * b | NL 2 | + | NL 1 x | ≤ C Θ * b
We insert this into (4.15) and conclude from (7.63), provided Θ * > 0 has been chosen large enough,

uΨ ≥ γ 4 Ψ. (7.64) Therefore, Ψ(u) ≥ Ψ(u * ) u u * γ 4 ≥ |S ∞ |bu * Θ * 4 u u * γ 4 ≥ Θ * for u u * γ 4 ≥ 4 |S ∞ |bu * Θ * , u ≥ u * 4 |S ∞ |bu * Θ * 4 γ = u * (1 + O(b|logb|))
where we have used the fact that u * ≥ bδ in view of (7.29) (7.32). Since u * ≤ 1 2 , we established that the contact happens before u = 3 4 . 7.8. Exit on the right. Below, we obtain an analog of Lemma 7.9 on the right, albeit in a significantly more restricted range of α γ in (0, 1). We assume (7.57). Lemma 7.10 (Exit on the right). Pick universal constants 1 δ , Θ * 1 large enough as in Lemma 7.8, then for all 0 < b < b * (Θ * , δ) small enough and α γ given by

α γ = K ν b +3 b √ δ K-1 Θ * (7.65) or α γ = 1 - K ν b +3 b √ δ K-1 Θ * , (7.66)
the C ∞ solution exits on the right of P 2 at u = U * , where 0 < U * < 3 4 , by crossing ∆ 1 = 0 in the case (7.65) and by crossing ∆ 2 = 0 in the case (7.66).

Proof of Lemma 7.10. For α γ given by (7.65) or (7.66) we only need to consider u in the the range -b ≤ u ≤ 0. In that range, the proof of Lemma 7.10 follows very closely the one of Lemma 7.9 for the case 0 ≤ u ≤ b. The C ∞ regularity at the right of P 2 all the way to u = 0 together with the property (7.21) follow again from the explicit integral representation of the remainder function. We focus on the exit behavior.

step 1 Bounds on M j . For -b ≤ u ≤ b, we have (1) so that the cases 0 ≤ u ≤ b and -b ≤ u ≤ 0 can be treated similarly in the definition of T and, as a consequence, in M 0 and M j . In particular, the proof of (7.4), (7.6), (7.7) and (7.12) obtained for 0 ≤ u ≤ b immediately extends to the case -b ≤ u ≤ 0, i.e., we have for

(1 -u) K = e Klog(1-u) = e O
-b ≤ u ≤ 0 c ν,1 ≤ M 0 (u) Γ(α γ )Γ(1 -α γ )K ν b +4-αγ u K ≤ c ν,2 , (7.67) for 1 ≤ j ≤ 5, T (u j M 0 ) M 0 L ∞ (-b≤u≤0) ≤ c ν b , (7.68) ∀ -b ≤ u ≤ 0, |uM 0 | M 0 ≤ c ν b . (7.69)
and for 1 ≤ j ≤ 5 and -b ≤ u ≤ 0

c ν,1 ≤ M j (u) Γ(α γ )Γ(1 -α γ )K ν b +j+4-αγ u K+j ≤ c ν,2 .
(7.70) step 2 Estimate on Θ main . Recall from (7.26)

Θ main (u) = Γ(α γ )Γ(1 -α γ )K ν b +3-αγ u K-1 × [1 + o b→0 (1)] 1 Γ(1 -α γ ) + K + ν b + 2 Γ(2 -α γ ) u + (uK) αγ M 1 (u) u K-1 Γ(α γ )Γ(1 -α γ )K ν b +3 u αγ .
In view of (7.70), we deduce for -bδ

1 3 ≤ u ≤ -bδ, Θ main (u) = Γ(α γ )Γ(1 -α γ )K ν b +3-αγ u K-1 (7.71) × [1 + o b→0 (1)] 1 Γ(1 -α γ ) + K + ν b + 2 Γ(2 -α γ ) u + O c ν δ 2 3
. step 3 Boundary layer. In view of (7.71), we easily obtain the following analog of the first two cases of Lemma 7.6 for α γ given by (7.65) or (7.66). Pick universal constants 1 δ , Θ * 1 then for all 0 < b < b * (Θ * , δ), we have: first layer: if α γ is given by (7.65), then there exist a solution to

Θ main (u * (α γ )) = (-1) K-1 Θ * , (7.72)
satisfying the following bounds

Γ(α γ )Γ(1 -α γ )K ν b +3-αγ (u * (α γ )) K-1 = Θ * e Oν (δ -1 ) (7.73) and u * (α γ ) = - √ δb(1 + o b→0 (1)). (7.74)
second layer: if α γ is given by (7.66), then there exist a solution to Θ main (u * (α γ )) = (-1) K Θ * , (7.75) satisfying (7.74) and

Γ(α γ )Γ(1 -α γ )K ν b +3-αγ K + ν b + 2 Γ(2 -α γ ) (u * (α γ )) K = Θ * e Oν (δ -1
) . (7.76) step 4 Estimate on r G . In view of (7.67) and (7.74), we have for u * (α γ ) defined in step 3

∀u ∈ [u * (α γ ), 0], 0 ≤ |M 0 (u)| ≤ c ν |M 0 (u * (α γ ))| ≤ c ν Θ * √ δ = C Θ * ,δ . (7.77)
Also, proceeding as in the proof of (7.51) and ( 7.52), we obtain the following analogs

∀u ∈ [u * (α γ ), 0], Γ(α γ )Γ(1 -α γ )K ν b +3-αγ |u| K-1 ≤ C Θ * ,δ . (7.78) and ∀u ∈ [u * (α γ ), 0], Γ(α γ )Γ(1 -α γ )K ν b +2-αγ |u| K-2 ≤ C Θ * ,δ . (7.79)
The estimate (7.48) holds for -b ≤ u ≤ 0, i.e. for -b ≤ u ≤ 0, and we now claim:

|r G | M 0 + |T r G | M 0 < √ b. (7.80)
Indeed, the proof of (7.48) for the range 0 ≤ u ≤ b does not use the sign of u, and all estimates hold by replacing everywhere u with |u|. Using also (7.77) (7.78) (7.79), the proof immediately extends to the range -b ≤ u ≤ 0. Thus, (7.80) holds for -b ≤ u ≤ 0.

step 6 Conclusion. Recall (7.24)

Θ(u) = K-2 k=0 θ k u k + (-1) K-1 S ∞ [1 + o b→0 (1)] Θ main (u) -T (r G )
then, provided Θ * has been chosen large enough, we conclude from (7.43) (with u replaced by |u|), (7.80), (7.77) that for 0 < b < b * (Θ * , δ) small enough, for all u ∈ [u * (α γ ), 0]:

|Θ(u) -(-1) K-1 S ∞ Θ main (u)| ≤ S ∞ Θ * 10 .
Therefore, if α γ is given by (7.65)

Θ * 2 ≤ Θ(u * (α γ )) S ∞ ≤ 2Θ * , (7.81)
and, if α γ is given by (7.66),

Θ * 2 ≤ -Θ(u * (α γ )) S ∞ ≤ 2Θ * . (7.82)
We now note that (7.62) holds independently of the sign of u ∈ (-1, 1). If α γ is given by (7.65), then, by (7.81),

Φ(u * ) = bu * Θ(u * ), Θ(u * ) S ∞ ≥ Θ * 2 ,
and, if α γ is given by (7.66), then, by (7.82),

Φ(u * ) = bu * Θ(u * ), -Θ(u * ) S ∞ ≥ Θ * 2 .
The rest of the argument of step 3 in the proof of Lemma 7.9 extends to the case u < 0. Therefore, first layer: if α γ is given by (7.65), there exists U * (α γ ) such that

- 3 4 ≤ U * ≤ u * , Φ(U * ) = -Θ * , F    K ν b +3 b √ δ K-1 Θ *    > 0. (7.84)
Also, for α γ given by (7.66), since Φ (rad) [K, α γ ](u) lives between ∆ 1 = 0 and ∆ 2 = 0 curves for any u < 0, and, since Φ[K, α γ ](u) has crossed ∆ 2 = 0 before u = -3/4 and cannot cross ∆ 2 = 0 twice, we deduce

F   1 - K ν b +3 b √ δ K-1 Θ *    < 0. (7.85)
Continuous dependence of the ode on the parameter α γ ∈ (0, 1) implies the continuity of F . We then infer by the mean value theorem the existence of α K γ such that

F α K γ = 0, α K γ ∈    K ν b +3 b √ δ K-1 Θ * , 1 - K ν b +3 b √ δ K-1 Θ *    . (7.86)
Then, by the uniqueness of solutions to the ode at u

= 3/4, Φ[K, α K γ ] must coincide with Φ (rad) [K, α K γ ].
To the left of P 2 , with α K γ in the range (7.47), we have that Φ[K, α K γ ] crosses ∆ 1 = 0 before reaching P o . Thus, we have obtained for any even K large enough the existence of α K γ in (0, 1) such that the smooth profile Φ[K, α K γ ](u) coincides with the P 2 -P 6 solution to the right of P 2 and exits on the left of P 2 by crossing ∆ 1 = 0 before reaching P 5. The constructed solution is C ∞ to the right and the left of P 2 , with derivatives satisfying (7.21) on both sides. step 2 Conclusion. Since the curve crosses ∆ 1 = 0 for σ 2 < σ < σ 5 , it is attracted to P 4 by Lemma 3.2. It remains to show that in the (σ(x), w(x)) parametrization, the point P 2 is reached in finite time. Indeed,

lim Σ→0 W Σ = c -
and thus, from (3.17), (2.50), (F.3):

dΣ dx = dσ dx = - ∆ 2 ∆ = - (c 2 c -+ c 4 )Σ(1 + o(1)) -2σ 2 (1 + c -)Σ(1 + o(1)) = λ + 2σ 2 (1 + c -) (1 + o(1)) > 0
which proves the claim. The resulting C ∞ solution corresponds to the global P 6 -P 2 -P 4 trajectory.

7.10. Uniform control of the P 2 -P o separatrix. We conclude this section with a simple uniform estimate on the separatrix P 2 -P o which will be required in the proof of the exterior positivity property, see Lemma 9.9. Recall that all integral curves of (1.9) that have a slope c -at P 2 are parametrized by Θ with respect to the u variable. In view of Lemma 3.5, the separatrix P 2 -P o is one of these curves, and we denote by Θ S its Θ parametrization. The following lemma provides a bound on Θ S . Lemma 7.11 (Computing Θ S ). There exist universal constants C, b * such that for all 0 < b < b * , the unique separatrix curve in the eye P 2 -P o satisfies:

∀0 ≤ u ≤ 1, Θ S + Ẽ11 + Ẽ30 + 2 a < Cb. (7.87)
Proof of Lemma 7.11. This follows from the fixed point representation of the separatrix.

step 1 Bound on the separatrix. We first claim for the separatrix

|Θ S | 1 (7.88)
uniformly as b → 0. Indeed, there holds from (6.30):

Θ - γ -2 u + γ + ν b + 1 1 -u Θ = - G bu(1 -u) (7.89) ⇔ (1 -u) γ+ν b +1 u γ-2 Θ = - G(1 -u) ν b +γ+1 bu(1 -u)u γ-2
and the separatrix is the unique solution which reaches u = 1:

Θ S = u γ-2 (1 -u) γ+ν b +1 1 u G bσ(1 -σ) (1 -σ) γ+ν b +1 σ γ-2 dσ. (7.90)
Let the operator

T (G)(u) = u γ-2 (1 -u) γ+ν b +1 1 u G bσ(1 -σ) (1 -σ) γ+ν b +1 σ γ-2 dσ, we claim, recalling a = γb, T (G) L ∞ [0,1] a G L ∞ ([0,1]) (7.91)
Assume (7.91), then the source term is given by (6.23):

G = G 0 + L(Φ) + NL(Φ) Φ = buΘ G 0 = γbH 1 -2(1+H 2 ) M b (1+H 20 )
which is just bounded and (7.88) follows by an elementary fixed point argument.

Proof of (7.91). We estimate for 1 2 ≤ u ≤ 1:

u γ-2 (1 -u) γ+ν b +1 1 u 1 bσ(1 -σ) (1 -σ) γ+ν b +1 σ γ-2 dσ ≤ u γ-2 (1 -u) γ+ν b +1 1 bu γ-1 1 u (1 -σ) γ+ν b dσ ≤ 2 b(γ + ν b ) ≤ 2 a and for 0 < u < 1 2 : u γ-2 (1 -u) γ+ν b +1 1 u 1 bσ(1 -σ) (1 -σ) γ+ν b +1 σ γ-2 dσ ≤ u γ-2 (1 -u) γ+ν b +1 (1 -u) γ+ν b b 1 u dσ σ γ-1 ≤ 2 b(γ -2) ≤ 3 a
and (7.91) follows.

step 2 Next term. We now extract the main term. Since H 2 = O(b) we have

G 0 = γbH 1 -2(1 + H 2 ) M b (1 + H 20 ) = g 0 + O(b), g 0 = -a( Ẽ11 + Ẽ30 ) -2 a = γb = |λ -| |µ + | . Let Θ = g 0 a + Θ
then from (7.89):

Θ - γ -2 u + γ + ν b + 1 1 -u Θ = - G bu(1 -u) with G = G -bu(1 -u) γ -2 u + γ + ν b + 1 1 -u g 0 a = G -bγ g 0 a + O(b) = O(b)
and hence (7.91) ensures Θ L ∞ ([0,1]) b which is (7.87).

Interior positivity

This section is devoted to the proof of Lemma 1.6. This is the first positivity property at the heart of the control of the linearized operator in [START_REF] Merle | On blow up for the energy super critical defocusing NLS[END_REF]. In this section we fix (d, , r) and assume that we are in the range (1.5) for some sufficiently small 0 < ε(d, )

1. We let (σ(x), w(x)) be the P 6 -P 2 trajectory given in Lemma 3.1, and aim at proving the positivity property (1.24) in the region σ ≥ σ 2 , where we recall the definition (1.23).

The proof relies on classical maximum principle arguments for ODEs along the P 6 -P 2 curve. We will need to compute various numerical constants depending on (d, , r). Their signs will be essential and checked at r = r o (d, ) only, using (1.22) and ε(d, ) small enough. We will work in the variables

W = w -w 2 , Σ = σ -σ 2 , Φ =
W Σ and use the algebra which follows from (3.16):

∆ 1 = Σ c 1 Φ + c 3 + [d 20 Φ 2 + d 11 Φ + d 02 ]Σ + [Φ 3 -dΦ]Σ 2 ∆ 2 = Σ c 2 Φ + c 4 + [e 20 Φ 2 + e 11 Φ + e 02 ]Σ + [e 21 Φ 2 -1]Σ 2 ∆ = -Σ 2σ 2 (1 + Φ) + Σ(1 -Φ 2 ) (8.1)
8.1. Sharp bound on the slope function Φ. We recall the notations

w e = (r -1) d , w 2 = 1 -σ 2 .
Lemma 8.1 (Sharp bound on the slope function). Assume (1.22). Let

α(r) = |c -| w 2 -we α ∞ = α(r o ) = 2d( + √ d) d( √ d-1)+ ( √ d+1) for r = r * d(1+ √ ) 2 d √ + for r = r + (8.2) then for Σ ≥ 0: c - 1 + α(r)Σ ≤ Φ < 0. (8.3)
Remark 8.2. The bound (8.3) is saturated at P 2 .

Proof of Lemma 8.1. We compute

w 2 (r * ) -w e (r * ) = 1 - √ d + √ d - d+ + √ d -1 d = + √ d - ( √ d -1) √ d( + √ d) = √ d( + √ d) (8.4)
and

w 2 (r + ) -w e (r + ) = √ 1 + √ - (d -1) d(1 + √ ) 2 = √ d(1 + √ ) -(d -1) d(1 + √ ) 2 = d √ + d(1 + √ ) 2 . ( 8.5)
It implies that w 2 > w e for r close enough to r o ( ). For 0 < α < α ∞ let us consider the function Φ of Σ given by

Φ = (1 + αΣ)Φ = Φ + αW then Φ(0) = c - Φ(+∞) = -α(w 2 -w e ) (8.6)
From (8.2) and (F.3), (F.11) which imply that c -< 0,

c -< -α(w 2 -w e ) ⇔ α < |c -| w 2 -w e = α(r).
We compute:

dW dΣ = Σ dΦ dΣ + Φ = ∆ 1 ∆ 2 ⇔ dΦ dΣ = ∆ 1 -Φ∆ 2 Σ∆ 2 Thus, d Φ dΣ = dΦ dΣ + α dW dΣ = ∆ 1 -Φ∆ 2 Σ∆ 2 + α ∆ 1 ∆ 2 = ∆ 1 -Φ∆ 2 + αΣ∆ 1 Σ∆ 2 = (1 + αΣ) 2 ∆ 1 -Φ∆ 2 Σ(1 + αΣ)∆ 2 .
step 1 Repulsivity. We assume that there exists a point Σ where Φ = c -with Σ > 0. First, we claim that

(1 + αΣ) 2 ∆ 1 -Φ∆ 2 > 0. (8.7)
Assuming that, since ∆ 2 < 0 in this zone, we obtain

d Φ dΣ < 0.
Since the Φ(∞) > c -, this leads to a contradiction and implies that c - 1+αΣ < Φ < 0. (8.3) follows by passing to the limit α → α(r). Proof of (8.7).

T = (1 + αΣ) 2 ∆ 1 -c -∆ 2 Σ = (1 + αΣ) 2 c 1 c -+ c 3 + [d 20 c 2 -+ d 11 c -+ d 02 ]Σ + [c 3 --dc -]Σ 2 -c -c 2 c -+ c 4 + [e 20 c 2 -+ e 11 c -+ e 02 ]Σ + [e 21 c 2 --1]Σ 2 From (2.45), c 1 c -+ c 3 -c -(c 2 c -+ c 4 ) = 0. At r o (d, )
we also have the additional cancellation (c 2 c -+ c 4 ) = λ + = 0. Therefore,

T Σ = (1 + αΣ) 2 d 20 c 2 -+ d 11 c -+ d 02 + (c 3 --dc -)Σ -c -e 20 c 2 -+ e 11 c -+ e 02 + (e 21 c 2 --1)Σ = (1 + 2αΣ + α 2 Σ 2 ) d 20 c 2 -+ d 11 c -+ d 02 + (c 3 --dc -)Σ -c -e 20 c 2 -+ e 11 c -+ e 02 + (e 21 c 2 --1)Σ = 3 i=0 Ã∞ i (1 + o b→0 (1))Σ i
where ̰ i corresponds to the limiting values at r = r o (d, ). We claim

̰ i > 0 (8.8)
which concludes the proof of (8.7) for r close enough to r * (d, ).

step 2 ̰ 0 . We compute:

Ã∞ 0 = d ∞ 20 (c ∞ -) 2 + d ∞ 11 c ∞ -+ d ∞ 02 -c ∞ -(e ∞ 20 (c ∞ -) 2 + e ∞ 11 c ∞ -+ e ∞ 02 ) = -e ∞ 20 (c ∞ -) 3 + (c ∞ -) 2 (d ∞ 20 -e ∞ 11 ) + (d ∞ 11 -e ∞ 02 )c ∞ -+ d ∞ 02 = ẽ∞ 20 > 0 from (F.8
), (F.13).

step 3 ̰ 3 . We compute:

Ã∞ 3 = (α ∞ ) 2 (c ∞ -) 3 -dc ∞ -= -c ∞ -(α ∞ ) 2 (d -(c ∞ -) 2 ). case r = r * : d -(c ∞ -) 2 = d - 4 2 d d( √ d -1) + ( √ d + 1) 2 = d d( √ d -1) + ( √ d + 1) 2 d( √ d -1) + ( √ d + 1) 2 -4 2 = d d( √ d -1) + ( √ d + 1) -2 d( √ d -1) + ( √ d + 1) + 2 d( √ d -1) + ( √ d + 1) 2 = d d( √ d -1) + ( √ d -1) d( √ d -1) + ( √ d + 3) d( √ d -1) + ( √ d + 1) 2 > 0 case r = r + : Ã3 = -c ∞ -(α ∞ ) 2 (d -(c ∞ -) 2 ) = (α ∞ ) 2 (d -1).
Hence ̰ 3 > 0.

step 4 ̰ 1 . We compute:

Ã∞ 1 = (c ∞ -) 3 -dc ∞ --c ∞ -(e ∞ 21 (c ∞ -) 2 -1) + 2α ∞ (d ∞ 20 (c ∞ -) 2 + d ∞ 11 c ∞ -+ d ∞ 02 ) = -(d -1)c ∞ --(c ∞ -) 3 d -1 + 2α ∞ (d ∞ 20 (c ∞ -) 2 + d ∞ 11 c ∞ -+ d ∞ 02 ). case r = r * . d ∞ 20 c ∞ -+ d ∞ 11 = √ d + d - + √ d 2 √ d d( √ d -1) + ( √ d + 1) - 2d √ d + √ d = 2 √ d ( + √ d)(d( √ d -1) + ( √ d + 1)) -d d( √ d -1) + ( √ d + 1) + ( √ d + d -) = 2 √ d ( + √ d)(d( √ d -1) + ( √ d + 1)) -d 2 ( √ d -1) - √ d(d -1) -2 < 0 It implies d ∞ 20 (c ∞ -) 2 + d ∞ 11 c ∞ -+ d ∞ 02 = c ∞ -(d ∞ 20 c ∞ -+ d ∞ 11 ) - √ d + √ d = - 2 √ d d( √ d -1) + ( √ d + 1) 2 √ d ( + √ d)[d( √ d -1) + ( √ d + 1)] -d 2 ( √ d -1) - √ d(d -1) -2 - √ d + √ d = √ d d( √ d -1) + ( √ d + 1) and 4 √ d d 2 ( √ d -1) + √ d(d -1) + 2 -d( √ d -1) + ( √ d + 1) 2 = d 2 ( √ d -1)[4 √ d -( √ d -1)] + [4d(d -1) -2d(d -1)] + 2 4 √ d -(d + 2 √ d + 1) = d 2 ( √ d -1)(3 √ d + 1) + 2d(d -1) -2 (d + 1 -2 √ d)
Therefore,

d ∞ 20 (c ∞ -) 2 + d ∞ 11 c ∞ -+ d ∞ 02 (8.9) = √ d[d 2 ( √ d -1)(3 √ d + 1) + 2d(d -1) -2 (d + 1 -2 √ d)] d( √ d -1) + ( √ d + 1) 2 ( + √ d)
.

In the numerator, the polynomial in is increasing on 0 < < d; it is > 0 at = 0, so that it is strictly positive on 0 < < d. Therefore,

d ∞ 20 (c ∞ -) 2 + d ∞ 11 c ∞ -+ d ∞ 02 > 0.
Hence Ã1 > 0 near r * . case r = r + . We compute

d ∞ 11 -d ∞ 20 -d ∞ 02 = -2d 1 + √ - - √ -d -1 (1 + √ ) 2 + + d √ (1 + √ ) 2 (8.10)
= -2d(1 + √ ) + + d √ -+ √ + d + 1 (1 + √ ) 2 = √ (-d + 1) -d + 1 (1 + √ ) 2 = -(d -1) 1 + √ and hence Ã∞ 1 = -(d -1)c ∞ --(c ∞ -) 3 d -1 + 2α ∞ (d ∞ 20 (c ∞ -) 2 + d ∞ 11 c ∞ -+ d ∞ 02 ) = d -1 + d -1 + 2α(d ∞ 20 -d ∞ 11 + d ∞ 02 ) > 0. step 5 Ã2 . We compute Ã∞ 2 = 2α ∞ (c ∞ -) 3 -dc ∞ -+ (α ∞ ) 2 (d ∞ 20 (c ∞ -) 2 + d ∞ 11 c ∞ -+ d ∞ 02 ) = -2c ∞ -α ∞ (d -(c ∞ -) 2 ) + (α ∞ ) 2 (d ∞ 20 (c ∞ -) 2 + d ∞ 11 c ∞ -+ d ∞ 02 ) > 0 Since -1 ≤ c ∞ -< 0, α ∞ > 0 and d ∞ 20 (c ∞ -) 2 + d ∞ 11 c ∞ -+ d ∞ 02 > 0, we infer Ã2 > 0.
This concludes the proof of (8.7). 8.2. Formula for F . The function F given by (1.23) has a special structure. Lemma 8.3 (Formula for F). Assume (1.22). Then,

F = - (d -1)σ ∆ (w -w -)(w -w + ) (8.11)
where (w -, w + ) are the w-coordinates of P 2 , P 3 .

Proof of Lemma 8.3. We compute

F = σ + dσ dx = 1 ∆ [σ∆ -∆ 2 ] = σ ∆ (w 2 -2w + 1 -σ 2 ) -( + d -1)w 2 + ( + d + r -r)w -r + σ 2 = - σ ∆ (d -1)w 2 -(d -r + (r -1))w + (r -1) = -(d -1) σ ∆ (w -w -)(w -w + ) (8.12)
where (w -, w + ) are the ordinates of P 2 , P 3 from (2.24).

8.3. Positivity to the right of P 2 . We claim the following fundamental lower bound at the right of P 2 .

Lemma 8.4 (Positivity to the right of P 2 ). Assume (1.22). Then, for Σ ≥ 0:

1 -w - dw dx -F ≥ c > 0. (8.13)
Proof of Lemma 8.4. The key here is the sharp bound (8.3).

step 1 Value at P 2 . We compute at P 2 :

F = σ + dσ dx = σ - ∆ 2 ∆ = σ 2 + Σ - Σ c 2 Φ + c 4 + [e 20 Φ 2 + e 11 Φ + e 02 ]Σ + [e 21 Φ 2 -1]Σ 2 -Σ [2σ 2 (1 + Φ) + Σ(1 -Φ 2 )]
and therefore since Φ(P 2 ) = c -and -1 < c -< 0, c 2 < c 4 < 0 from (2.49):

F (P 2 ) = σ 2 + c 2 c -+ c 4 2σ 2 (1 + c -) = c -(c 2 -c 4 ) 2σ 2 (1 + c -) > 0 and 1 -w 2 -F (P 2 ) = - λ + 2σ 2 (1 + c -) > 0.
Then,

dw dx = - ∆ 1 ∆ = c 1 c -+ c 3 2σ 2 (1 + c -) + O(Σ) = c -λ + 2σ 2 (1 + c -) + O(Σ) and
1 -w - dw dx -F (P 2 ) = - λ + 2σ 2 (1 + c -) - c -λ + 2σ 2 (1 + c -) = - λ + 2σ 2 > 0. (8.14)
step 2 Bad set and no contact condition. We now study the null equation:

∆ 1 -w - dw dx -F = 0 ⇔ ∆ 1 -w + ∆ 1 ∆ -σ + ∆ 2 ∆ = 0 ⇔ (1 -w) 2 -σ 2 (1 -w -σ) + ∆ 1 + ∆ 2 = 0 ⇔ (1 -w) 3 -(1 -w) 2 σ -(1 -w)σ 2 + σ 3 + w(1 -w)(r -w) -d(w -w e )σ 2 + σ ( + d -1)w 2 -w( + d + r -r) + r -σ 2 = 0 ⇔ (1 -w) (1 -w) 2 + w(r -w) + (d + -1)w 2 -(1 -w) 2 + r - w ( + d + r -r) σ -[d(w -w e ) + 1 -w] σ 2 = 0 ⇔ [d(w -w e ) + 1 -w] σ 2 - (d -1)w 2 + ( + r -d -r)w+ (r -1) σ -(1 -w) (1 -w) 2 + w(r -w) = 0.
To the right of P 2 , between the red and green curves, we have w e < w < w 2 < 1.

As a result, the coefficients α(w), γ(w) of the above quadratic equation

α(w)σ 2 + β(w)σ -γ(w) = 0
are positive and therefore there is exactly one positive root σ 0 (w). The dependence of σ 0 (w) on w is continuous and σ 0 (w 2 ) = σ 2 . In the (Σ, W ) plane the null set is represented by the continuous null curve

(Σ 0 (W ), W ) with W ∈ [w e -w 2 , 0].
We now compute the derivative at P 2 recalling (2.41):

(1 -w -σ 0 (w))∆ + ∆ 1 (w, σ 0 (w)) + ∆ 2 (w, σ 0 (w)) = 0 ⇒ c 1 + σ 0 c 3 + c 2 + σ 0 c 4 = 0 ⇒ σ 0 (w 2 ) = - c 1 + c 2 c 3 + c 4 ⇒ w 0 (σ 2 ) = - c 3 + c 4 c 1 + c 2
and hence the local representation in the (Σ, W ) plane of the null set:

W 0 (Σ) = - c 3 + c 4 c 1 + c 2 Σ + O(Σ 2 ).
On the other hand, from (2.45), (F.3):

- c 3 + c 4 c 1 + c 2 < c -⇔ c 3 + c 4 < -(c 1 + c 2 )c -⇔ c 1 c -+ c 3 + c 2 c -+ c 4 < 0 ⇔ (1 + c -)λ + < 0.
Therefore, the curve Φ = c - 1+α(r)Σ , which is equivalent to W = c -Σ 1+α(r)Σ , lies for Σ > 0 small, strictly above the null curve (Σ 0 (W ), W ). We now claim that, if we can show that

∀Σ > 0, 1 -w - dw dx -F Σ, c - 1 + α(r)Σ > 0, (8.15)
it will imply that (1 -w -w -F ) > 0 ∀Σ ≥ 0 along the solution curve. We argue by contradiction. First, using (8.14), we could find a positive value Σ 0 such that (1 -w -w -F ) (Σ 0 , W (Σ 0 )) = 0. Therefore, the point (Σ 0 , W (Σ 0 )) belongs to the null set. Since along the solution curve w e < w(σ) < w 2 for any σ > σ 2 , the point (Σ 0 , W (Σ 0 )) must lie on the null curve (Σ 0 (W ), W ). That is Σ 0 (W (Σ 0 )) = Σ 0 . We now follow this curve from the value W (Σ 0 ) to W = 0. For W sufficiently small the null curve lies below the curve W = c -Σ 1+α(r)Σ , while at the point (Σ 0 , W (Σ 0 )), which belongs to the solution curve, it must be above W = c -Σ 1+α(r)Σ . Therefore, the curve W = c -Σ 1+α(r)Σ and the null curve must intersect, which is impossible in view of the claim (8.15). From now on, we focus on the proof of (8.15). step 3 Computation of 1 -w -w . We compute in (W, Σ) coordinates:

∆(1 -w -w ) = ∆(1 -w) + ∆ 1 = -(σ 2 -W )Σ 2σ 2 (1 + Φ) + Σ(1 -Φ 2 ) + Σ c 1 Φ + c 3 + [d 20 Φ 2 + d 11 Φ + d 02 ]Σ + [Φ 3 -dΦ]Σ 2 = Σ -σ 2 2σ 2 (1 + Φ) + Σ(1 -Φ 2 ) + ΦΣ 2σ 2 (1 + Φ) + Σ(1 -Φ 2 ) + c 1 Φ + c 3 + [d 20 Φ 2 + d 11 Φ + d 02 ]Σ + [Φ 3 -dΦ]Σ 2 = Σ B 0 + B 1 Σ + B 2 Σ 2 with B 0 = c 1 Φ + c 3 -2σ 2 2 (1 + Φ) B 1 = d 20 Φ 2 + d 11 Φ + d 02 -σ 2 (1 -Φ 2 ) + 2σ 2 Φ(1 + Φ) B 2 = Φ 3 -dΦ + Φ(1 -Φ 2 ) = -(d -1)Φ Let K(Σ, Φ) = B 0 (Φ) + B 1 (Φ)Σ + B 2 (Φ)Σ 2 ,
then we have obtained:

1 -w -w = ∆(1 -w) + ∆ 1 ∆ = ΣK(Σ, Φ) ∆ .
To ease the notations, we let α = α(r) and compute

K Σ, c - 1 + αΣ = c 1 c - 1 + αΣ + c 3 -2σ 2 2 1 + c - 1 + αΣ + (d 20 + 3σ 2 ) c - 1 + αΣ 2 + (d 11 + 2σ 2 ) c - 1 + αΣ + d 02 -σ 2 Σ -(d -1) c - 1 + αΣ Σ 2 = 1 (1 + αΣ) 2 (c 1 c --2σ 2 2 c -)(1 + αΣ) + (c 3 -2σ 2 2 )(1 + αΣ) 2 + (d 20 + 3σ 2 )c 2 -Σ + c -(d 11 + 2σ 2 )Σ(1 + αΣ) + (d 02 -σ 2 )Σ(1 + αΣ) 2 -(d -1)c -Σ 2 (1 + αΣ) = 3 i=0 Bi Σ i (1 + αΣ) 2
which yields the formula:

(1 -w -w ) Σ, c - 1 + αΣ = Σ( 3 i=0 Bi Σ i ) ∆(1 + αΣ) 2 .
(8.16)

We now compute the coefficients Bi . For B0 , we need to keep the exact structure at r, but for B1,2,3 is suffices to compute the approximate values at r = r o (d, ) which is done below.

step 4 B0 . We compute for all r:

B0 = c 1 c --2σ 2 2 c -+ c 3 -2σ 2 2 = c -λ + -2σ 2 2 (1 + c -). step 5 B∞ 1 . We compute B∞ 1 = (c ∞ 1 c ∞ --2(σ ∞ 2 ) 2 c ∞ -)α ∞ + 2α ∞ (c ∞ 3 -2(σ ∞ 2 ) 2 ) + (c ∞ -) 2 (d ∞ 20 + 3σ ∞ 2 ) + c ∞ -(d ∞ 11 + 2σ ∞ 2 ) + d ∞ 02 -σ ∞ 2 . case r = r * . We compute using c ∞ 1 c ∞ -+ c ∞ 3 = 0: c ∞ 1 c ∞ --2(σ ∞ 2 ) 2 c ∞ -+ 2(c ∞ 3 -2(σ ∞ 2 ) 2 ) = c ∞ 3 -2(σ ∞ 2 ) 2 (2 + c ∞ -) = c ∞ 3 -2(σ ∞ 2 ) 2 -2(σ ∞ 2 ) 2 (1 + c ∞ -) = - 2d( + 1) ( + √ d) 2 -2σ 2 2 (1 + c -) = - 2d( + 1) ( + √ d) 2 - 2d ( + √ d) 2 ( √ d -1)(d -) d( √ d -1) + ( √ d + 1) = - 2d ( + √ d) 2 d( √ d -1) + ( √ d + 1) ( + 1)[d( √ d -1) + ( √ d + 1)] + (d -)( √ d -1) = - 2d ( + √ d) 2 d( √ d -1) + ( √ d + 1) 2d( √ d -1) + (d √ d -d + 2) + ( √ d + 1) 2
and hence

(c ∞ 1 c ∞ --2(σ ∞ 2 ) 2 c ∞ -)α ∞ + 2α ∞ (c ∞ 3 -2(σ ∞ 2 ) 2 ) = - 2d ( + √ d) 2 d( √ d -1) + ( √ d + 1) 2d( √ d -1) + (d
√ d -d + 2) + ( √ d + 1) 2 × 2d( + √ d) d( √ d -1) + ( √ d + 1) = - 4d 2 2d( √ d -1) + (d √ d -d + 2) + ( √ d + 1) 2 ( + √ d) d( √ d -1) + ( √ d + 1) 2 .
Then,

3(c ∞ -) 2 σ ∞ 2 + 2σ ∞ 2 c ∞ --σ ∞ 2 = σ ∞ 2 (3(c ∞ -) 2 + 2c ∞ --1) = σ ∞ 2 (1 + c ∞ -)(3c ∞ --1) = - √ d + √ d ( √ d -1)(d -) d( √ d -1) + ( √ d + 1) 1 + 6 √ d d( √ d -1) + ( √ d + 1) = - √ d( √ d -1)(d -) d( √ d -1) + (7 √ d + 1) ( + √ d) d( √ d -1) + ( √ d + 1) 2 .
Now, recalling (8.9):

B∞ 1 = - P B1 ( ) ( + √ d) d( √ d -1) + ( √ d + 1) 2 with P B1 ( ) = 4d 2 2d( √ d -1) + (d √ d -d + 2) + ( √ d + 1) 2 + √ d( √ d -1)(d -) d( √ d -1) + (7 √ d + 1) - √ d[d 2 ( √ d -1)(3 √ d + 1) + 2d(d -1) -2 (d + 1 -2 √ d)] = √ d( √ d -1)(d -) d( √ d -1) + (7 √ d + 1) + 8d 3 ( √ d -1) + d 3 √ d-2d 3 + d 2 √ d + 8d 2 + 2 2d 2 √ d + 4d 2 + 2d √ d + 3 √ d(d + 1 -2 √ d) > 0
and thus B∞ 1 < 0. We can fully expand:

(d -) d( √ d -1) + (7 √ d + 1) = d 2 ( √ d -1) + (6d √ d + 2d) -(7 √ d + 1) 2 and P B1 ( ) = 8d 3 ( √ d -1) + √ d( √ d -1)d 2 ( √ d -1) + d 3 √ d-2d 3 + d 2 √ d + 8d 2 + √ d( √ d -1)(6d √ d + 2d) + 2d 2 √ d + 4d 2 + 2d √ d - √ d( √ d -1)(7 √ d + 1) 2 + √ d(d + 1 -2 √ d) 3 = 9d 3 √ d -10d 3 + d 2 √ d + d 3 √ d-2d 3 + 7d 2 √ d + 4d 2 -2d √ d + 2d 2 √ d + 4d 2 -5d √ d + 6d + √ d 2 + d √ d -2d + √ d 3 . case r = r + . We compute using c ∞ 1 c ∞ -+ c ∞ 3 = 0: c ∞ 1 c ∞ --2(σ ∞ 2 ) 2 c ∞ -+ 2(c ∞ 3 -2(σ ∞ 2 ) 2 ) = c ∞ 3 -2(σ ∞ 2 ) 2 (2 + c ∞ -) = c ∞ 3 -2(σ ∞ 2 ) 2 -2(σ ∞ 2 ) 2 (1 + c ∞ -) = c ∞ 3 -2(σ ∞ 2 ) 2 = - 2 √ (d + √ ) (1 + √ ) 3 - 2 (1 + √ ) 2 = - 2[ √ (d + √ ) + 1 + √ ] (1 + √ ) 3 = -2 -2(d + 1) √ -2 (1 + √ ) 3
and hence

(c ∞ 1 c ∞ --2(σ ∞ 2 ) 2 c ∞ -)α ∞ + 2α ∞ (c ∞ 3 -2(σ ∞ 2 ) 2 ) = -2 -2(d + 1) √ -2 (1 + √ ) 3 d(1 + √ ) 2 d √ + = -2d -2d(d + 1) √ -2d (1 + √ )(d √ + ) . Then 3(c ∞ -) 2 σ ∞ 2 + 2σ ∞ 2 c ∞ --σ ∞ 2 = σ ∞ 2 (3(c ∞ -) 2 + 2c ∞ --1) = 0.
Finally recalling (8.10):

(c ∞ -) 2 d ∞ 20 + c ∞ -d ∞ 11 + d ∞ 02 = d ∞ 20 -d ∞ 11 + d ∞ 02 = d -1 1 + √ and hence B∞ 1 = -2d -2d(d + 1) √ -2d (1 + √ )(d √ + ) + d -1 1 + √ = -2d -2d(d + 1) √ -2d + (d -1)(d √ + ) (1 + √ )(d √ + ) = - (d + 1) + (d 2 + 3d) √ + 2d (1 + √ )(d √ + ) < 0.
step 6 B2 . case r = r * . We compute at r = r * (d, ):

B∞ 2 = (c ∞ 3 -2(σ ∞ 2 ) 2 )(α ∞ ) 2 + c ∞ -α ∞ (d ∞ 11 + 2σ ∞ 2 ) + 2α ∞ (d ∞ 02 -σ ∞ 2 ) -(d -1)c ∞ - = -2 d ( + √ d) 2 - 2d ( + √ d) 2 α 2 + c -α - 2d √ d + √ d + 2 √ d + √ d + 2α - ( + 1) √ d + √ d -(d -1)c - = - (2d + 2 d)4d 2 d( √ d -1) + ( √ d + 1) 2 - 2α √ d + √ d [ + 1 + c -(d -1)] -(d -1)c - = - (2d + 2 d)4d 2 d( √ d -1) + ( √ d + 1) 2 - 4d √ d d( √ d -1) + ( √ d + 1) ( + 1 + c -(d -1)) -(d -1)c - = - (2d + 2 d)4d 2 d( √ d -1) + ( √ d + 1) 2 - 4d √ d d( √ d -1) + ( √ d + 1) + 1 - 2 (d -1) √ d d( √ d -1) + ( √ d + 1) + 2 √ d(d -1) d( √ d -1) + ( √ d + 1) = - P B2 ( ) d( √ d -1) + ( √ d + 1) 2 with P B2 ( ) = 8d 3 (d + ) + 4d √ d ( + 1) d( √ d -1) + ( √ d + 1) -2 (d -1) √ d -2 √ d(d -1) d( √ d -1) + ( √ d + 1) = 8d 4 + 4d 2 (d - √ d) + 2d 3 -2d 2 √ d + 14d 2 + 2d √ d + 2d 2 + 2d √ d + 2d + 2 √ d 2
> 0 which implies P B2 ( ) > 0 and B∞ 2 < 0.

case r = r + . We compute at r = r + ( ):

B∞ 2 = (c ∞ 3 -2(σ ∞ 2 ) 2 )(α ∞ ) 2 + c ∞ -α ∞ (d ∞ 11 + 2σ ∞ 2 ) + 2α ∞ (d ∞ 02 -σ ∞ 2 ) -(d -1)c ∞ - = (c ∞ 3 -2(σ ∞ 2 ) 2 )(α ∞ ) 2 + (2d ∞ 02 -d ∞ 11 -4σ ∞ 2 )α ∞ + d -1 = - 2 √ (d + √ ) (1 + √ ) 3 - 2 (1 + √ ) 2 d(1 + √ ) 2 d √ + 2 + 2 --d √ (1 + √ ) 2 + 2d -4 1 + √ d(1 + √ ) 2 d √ + + d -1 = - d 2 (1 + √ ) (d √ + ) 2 2 √ (d + √ ) + 2(1 + √ ) + d d √ + -2 -2d √ + (2d -4)(1 + √ ) + d -1 = - d 2 (1 + √ )[2 + (2d + 2) √ + 2] (d √ + ) 2 + d(-2 -4 √ + 2d -4) d √ + + d -1 = - P B2 ( ) (d √ + ) 2 with P B2 ( ) = d 2 (1 + √ )[2 + (2d + 2) √ + 2] + d(2 + 4 √ -2d + 4)(d √ + ) -(d -1)(d √ + ) 2 = (d + 1) 2 + (2d 2 + 6d) √ + (d 3 + 7d 2 + 4d) + (8d 2 ) √ + 2d 2
and hence P B2 ( ) > 0 and B∞ 2 < 0.

step 7 B3 . case r = r * . We compute at r = r * (d, ): B3 = (α ∞ ) 2 (d ∞ 02 -σ ∞ 2 ) -(d -1)c ∞ -α ∞ = α ∞ 2d( + √ d) d( √ d -1) + ( √ d + 1) - √ d + √ d - √ d + √ d + (d -1) 2 √ d d( √ d -1) + ( √ d + 1) = - 2α √ d d( √ d -1) + ( √ d + 1) [d( + 1) -(d -1) ] = - 2α(d + ) √ d d( √ d -1) + ( √ d + 1) < 0. case r = r + . We compute at r = r + ( ): B3 = (α ∞ ) 2 (d ∞ 02 -σ ∞ 2 ) -(d -1)c ∞ -α ∞ = α d(1 + √ ) 2 d √ + --d √ (1 + √ ) 2 - 1 1 + √ + d -1 = α d(1 + √ ) 2 d √ + --d √ -1 - √ (1 + √ ) 2 + d -1 = α d √ + d --(d + 1) √ -1 + (d -1)(d √ + ) = α d √ + [--2d √ -d] = - d(1 + √ ) 2 ( + 2d √ + d) (d √ + ) 2 < 0 step 7 Computation of F . F = σ + σ = σ - ∆ 2 ∆ = σ∆ -∆ 2 ∆ = - Σ ∆ (σ 2 + Σ)[2σ 2 (1 + Φ) + Σ(1 -Φ 2 )] + c 2 Φ + c 4 + [e 20 Φ 2 + e 11 Φ + e 02 ]Σ + [e 21 Φ 2 -1]Σ 2 = - Σ ∆ 2σ 2 2 (1 + Φ) + c 2 Φ + c 4 + 2σ 2 (1 + Φ) + σ 2 (1 -Φ 2 ) + e 20 Φ 2 + e 11 Φ + e 02 Σ + 1 -Φ 2 + e 21 Φ 2 -1 Σ 2 = Σ (2σ 2 2 + c 2 )Φ + (2σ 2 + e 11 )Φ + (e 20 -σ 2 )Φ 2 Σ + (e 21 -1)Φ 2 Σ 2 (-∆)
where we used that for all r,

c 4 = -2σ 2 2 , e 02 = -3σ 2 .
We therefore obtain

F Σ, c - 1 + αΣ = Σ -∆ (2σ 2 2 + c 2 ) c - 1 + αΣ + (2σ 2 + e 11 ) c - 1 + αΣ + (e 20 -σ 2 ) c - 1 + αΣ 2 Σ + (e 21 -1) c - 1 + αΣ 2 Σ 2 = Σ -∆(1 + αΣ) 2 (2σ 2 2 + c 2 )c -(1 + αΣ) + (2σ 2 + e 11 )c -Σ(1 + αΣ) + (e 20 -σ 2 )(c -) 2 Σ + (e 21 -1)(c -) 2 Σ 2 = Σ 2 i=0 A i Σ i (-∆)(1 + αΣ) 2 .
(8.17)

As for the Bi , we compute A 0 for all r but A 1,2 at r = r * (d, ) only.

step 8 A 0 . We compute:

A 0 = (2σ 2 2 + c 2 )c -. step 9 A ∞ 1 . We compute A ∞ 1 = α ∞ (2(σ ∞ 2 ) 2 + c ∞ 2 )c ∞ -+ (2σ ∞ 2 + e ∞ 11 )c ∞ -+ (e ∞ 20 -σ ∞ 2 )(c ∞ -) 2 case r = r * (d, ). First, 2σ ∞ 2 + e ∞ 11 = 2 √ d + √ d - d( √ d -1) + (1 + √ d) ( + √ d) = 2 √ d -d( √ d -1) -(1 + √ d) ( + √ d) = -d( √ d -1) + ( √ d -1) ( + √ d) = - ( √ d -1)(d -) ( + √ d)
and

α ∞ (2(σ ∞ 2 ) 2 + c ∞ 2 ) = 2d( + √ d) d( √ d -1) + ( √ d + 1) 2d ( + √ d) 2 - √ d ( + √ d) 2 d( √ d -1) + ( √ d + 1) = 2d √ d As a result, α ∞ (2(σ ∞ 2 ) 2 + c ∞ 2 )c ∞ -+ (2σ ∞ 2 + e ∞ 11 )c ∞ - =   - 2d √ d( √ d -1)(d -) ( + √ d) d( √ d -1) + ( √ d + 1) - ( √ d -1)(d -) ( + √ d)   c - = 2 √ d d( √ d -1) + ( √ d + 1) ( √ d -1)(d -)[2d √ d + d( √ d -1) + ( √ d + 1)] ( + √ d) d( √ d -1) + ( √ d + 1) = 2 √ d( √ d -1)(d -)[3d √ d -d + ( √ d + 1)] ( + √ d) d( √ d -1) + ( √ d + 1) 2 .
We expand

(d -)[3d √ d -d + ( √ d + 1)] = d 2 (3 √ d -1) + (d √ d + d -3d √ d+d) -( √ d + 1) 2 = d 2 (3 √ d -1) + (-2d √ d + 2d) -( √ d + 1) 2
and obtain

α ∞ (2(σ ∞ 2 ) 2 + c ∞ 2 )c ∞ -+ (2σ ∞ 2 + e ∞ 11 )c ∞ - = 2 √ d( √ d -1) d 2 (3 √ d -1) + (-2d √ d + 2d) -( √ d + 1) 2 ( + √ d) d( √ d -1) + ( √ d + 1) 2 = 6d 3 √ d -8d 3 + 2d 2 √ d + -4d 2 √ d + 8d 2 -4d √ d -2 √ d(d -1) 2 ( + √ d) d( √ d -1) + ( √ d + 1) 2 .
We now add

(e ∞ 20 -σ ∞ 2 )(c ∞ -) 2 = √ d(d -1 + ) ( + √ d) - √ d + √ d c 2 -= √ d(d -1) ( + √ d) 2 √ d d( √ d -1) + ( √ d + 1) 2 = 4 d √ d(d -1) ( + √ d) d( √ d -1) + ( √ d + 1)
2 and get the formula:

A ∞ 1 = P A 1 ( + √ d) d( √ d -1) + ( √ d + 1) 2 with P A 1 = 6d 3 √ d -8d 3 + 2d 2 √ d + -4d 2 √ d + 8d 2 -4d √ d + 4d 2 √ d -4d √ d -2 √ d(d -1) 2 = 6d 3 √ d -8d 3 + 2d 2 √ d + 8d 2 -8d √ d -2 √ d(d -1) 2 case r = r + (d, ). Recall A ∞ 1 = α ∞ (2(σ ∞ 2 ) 2 + c ∞ 2 )c ∞ -+ (2σ ∞ 2 + e ∞ 11 )c ∞ -+ (e ∞ 20 -σ ∞ 2 )(c ∞ -) 2 . Observe 2σ ∞ 2 + e ∞ 11 = 2 1 + √ - 2 1 + √ = 0 and 2(σ ∞ 2 ) 2 + c ∞ 2 = 2 (1 + √ ) 2 - 2 (1 + √ ) 2 = 0
and hence

A ∞ 1 = e ∞ 20 -σ ∞ 2 = + d -1 (1 + √ ) - 1 1 + √ = d -1 (1 + √ ) . step 10 A ∞ 2 . We have A ∞ 2 = α ∞ (2σ ∞ 2 + e ∞ 11 )c ∞ -+ (e ∞ 21 -1)(c ∞ -) 2 . case r = r * (d, ). First, (e ∞ 21 -1)(c ∞ -) 2 = + d -1 -1 2 √ d d( √ d -1) + ( √ d + 1) 2 = 4 d(d -1) [d( √ d -1) + ( √ d + 1)] 2 and α ∞ (2σ ∞ 2 + e ∞ 11 )c ∞ - = 2d( + √ d) d( √ d -1) + ( √ d + 1) - ( √ d -1)(d -) ( + √ d) - 2 √ d d( √ d -1) + ( √ d + 1) = 4d √ d( √ d -1)(d -) [d( √ d -1) + ( √ d + 1)] 2
which implies

A ∞ 2 = P A 2 d( √ d -1) + ( √ d + 1)
A ∞ 2 = α ∞ 2 1 + √ - 2 1 + √ + + d -1 -1 = d -1 .
step 11 Conclusion. We are now in position to prove (8.15). From (8.17), (8.16):

(1 -w -w -F ) Σ, c - 1 + αΣ = Σ( 3 i=0 Bi Σ i ) ∆(1 + αΣ) 2 - Σ 2 i=0 A i Σ i (-∆)(1 + αΣ) 2 = Σ ∆(1 + αΣ) 2 i=0 3 ( Bi + A i )Σ i with A 3 = 0. We claim for 0 < r o (d, ) -r 1: Bi + Ãi < 0, 0 ≤ i ≤ 3 (8.18)
Since ∆ < 0 at the right of P 2 where Σ > 0, (8.18) gives

(1 -w -w -F ) Σ, c - 1 + αΣ = Σ ∆(1 + αΣ) 2 3 i=0 ( Bi + A i )Σ i > 0
and (8.15) is proved. step 12 Proof of (8.18) for r o = r * . B0 + A 0 . We compute using c 4 = -2σ 2 2 for all r

A 0 = (2σ 2 2 + c 2 )c -= 2σ 2 2 (1 + c -) + c 2 c -+ c 4 = 2σ 2 2 (1 + c -) + λ + and hence A 0 + B0 = 2σ 2 2 (1 + c -) + λ + + c -λ + -2σ 2 2 (1 + c -) = (1 + c -)λ + < 0. B∞ 1 + A ∞ 1 . We compute at r * B∞ 1 + A ∞ 1 = - P B1 ( ) ( + √ d) d( √ d -1) + ( √ d + 1) 2 + P A 1 ( + √ d) d( √ d -1) + ( √ d + 1) 2 = - P B1 -P A 1 ( + √ d) d( √ d -1) + ( √ d + 1) 2 with P B1 -P A 1 = 9d 3 √ d -10d 3 + d 2 √ d + d 3 √ d-2d 3 + 7d 2 √ d + 4d 2 -2d √ d + 2d 2 √ d + 4d 2 -5d √ d + 6d + √ d 2 + d √ d -2d + √ d 3 - 6d 3 √ d -8d 3 + 2d 2 √ d + 8d 2 -8d √ d -2 √ d(d -1) 2 = 3d 3 √ d -2d 3 -d 2 √ d + d 3 √ d -2d 3 + 7d 2 √ d -4d 2 + 6d √ d + 2d 2 √ d + 4d 2 -3d √ d + 6d - √ d 2 + d √ d -2d + √ d 3 > 0 for d ≥ 2 and therefore B∞ 1 + A ∞ 1 < 0. B∞ 2 + A ∞ 2 .
We compute:

B2 + A 2 = - P B2 -P A 2 d( √ d -1) + ( √ d + 1) 2 with P B2 -P A 2 = 8d 4 + 4d 2 (d - √ d) + 2d 3 -2d 2 √ d + 14d 2 + 2d √ d + 2d 2 + 2d √ d + 2d + 2 √ d 2 -4d 2 (d - √ d) + (4d √ d -4d) = 8d 4 + 2d 3 -2d 2 √ d + 14d 2 -2d √ d + 4d + 2d 2 +2d √ d + 2d + 2 √ d 2 > 0 for d ≥ 2, which implies B2 + A 2 < 0. B∞ 3 + A ∞ 3 .
We have B3 + A 3 = B3 < 0. This concludes the proof of (8.18) for r o = r * . step 13 Proof of (8.18) for r o = r + . B0 + A 0 . We have verbatim as above

A 0 + B0 = λ + < 0. B∞ 1 + A ∞ 1 .
We compute:

B∞ 1 + A ∞ 1 = - (d + 1) + (d 2 + 3d) √ + 2d (1 + √ )(d √ + ) + d -1 (1 + √ ) = - (d + 1) + (d 2 + 3d) √ + 2d -(d -1)(d √ + ) (1 + √ )(d √ + ) = - Q 1 ( ) (1 + √ )(d √ + )
with

Q 1 ( ) = (d + 1) 2 + (d 2 + 3d) √ + (d + 1) -d(d -1) √ > 0 for > d, d = 2, 3. B∞ 2 + A ∞ 2 .
We compute:

B∞ 2 + A ∞ 2 = - (d + 1) 2 + (2d 2 + 6d) √ + (d 3 + 7d 2 + 4d) + (8d 2 ) √ + 2d 2 (d √ + ) 2 + d -1 = - Q 2 ( ) (d √ + ) 2 with Q 2 = (d + 1) 3 + (2d 2 + 6d) 2 √ + (d 3 + 7d 2 + 4d) 2 + (8d 2 ) √ + 2d 2 -(d -1)(d 2 + 2d √ + 2 ) = (d + 1) 3 + (2d 2 + 6d) 2 √ + (d 3 + 7d 2 + 3d + 1) 2 + (6d 2 + 2d) √ + (3d 2 -d 3 ) > 0. B∞ 3 + A ∞ 3 . We have B∞ 3 + A ∞ 3 = B∞ 3 < 0.
This concludes the proof of (8.18) for r o = r + . 8.4. Proof of Lemma 1.6. We are now in position to finish the proof of Lemma 1.6. We need to show (1.24). To the right of P 2 , w > 0 and F > 0 from (8.11). Therefore, the first statement in (1.24) follows from (8.13). For the second, to the right of P 2 we have σ ≥ σ 2 and ∆ < 0, i.e., σ > 1 -w. These, together with (8.13), imply

1 -w -w - 1 -w σ F ≥ 1 -w -w -F ≥ c > 0
The third statement follows from F > 0 at the right of P 2 . (1.24) is proved.

Exterior positivity

We now turn to the proof of Lemma 1.7. We will establish the following result which gives a precise range of validity of (1.26) and enlarges the set of admissible parameters. Let Then there exists ε(d, ) such that for all

r o (d, ) -ε(d, ) < r < r o (d, ) (9.3)
and for any P 2 -P 4 trajectory with c -slope at P 2 as in Lemma 3.2, there exists c r > 0 such that

∀0 < σ ≤ σ 2 , (1 -w -w ) 2 -F 2 > c r 1 -w -w > c r . (9.4)
Remark 9.2. We note that

< 2 (d) ⇔ r * (d, ) > 2 (9.5)
The latter is a fundamental property for the study of the defocusing (NLS) problem, [START_REF] Merle | On blow up for the energy super critical defocusing NLS[END_REF]. The lower bounds in (9.2) are actually given by the condition

> 1 (d) (9.6)
where

2( √ d + 1
) , see (9.13). We will see below that (9.6) is a necessary condition for (9.4) to hold near r * (d, ).

Remark 9.3. From (8.11), we have that along the solution curve for σ < σ 2 F < 0 for w < w 2 , F > 0 for w > w 2 .

We will first treat the case w < w 2 , in which case it suffices to prove 1-w -w +F > 0, and then the case w > w 2 , in which case it suffices to prove 1 -w -w -F > 0.

9.1. Bound on the slope for the P 2 -P 4 separatrix. The P 2 -P 4 separatrix23 , i.e., the unique solution connecting P 2 and P 4 with the slope c + at P 2 described in (3) of Lemma 3.5, furnishes a natural lower bound for the solution curves of Theorem 1.3. We start with a rough estimate on its slope24 .

Lemma 9.4 (Upper bound on the slope for the P 2 -P 4 separatrix). Under the assumptions of Lemma 9.1, the P 2 -P 4 separatrix given in (3) of Lemma 3.5 satisfies

0 < Φ S = W Σ ≤ c + (9.8)
where we recall that W = w -w 2 and Σ = σ -σ 2 .

Proof of Lemma 9.4. We have σ < σ 2 , i.e., Σ < 0 and W = w -w 2 < 0. Therefore,

Φ = W Σ > 0.
step 1 Setting up. We have

Φ S (P 2 ) = c + , Φ S (P 4 ) = -w 2 -σ 2 = 1 -σ 2 σ 2
and hence at r = r o :

Φ S (P ∞ 2 ) -Φ S (P ∞ 4 ) = -1-σ 2 σ 2 = + 1 -+ √ d √ d = ( √ d-1) √ d > 0 for r = r * √ (d+ √ ) 1+ √ - √ = (d-1) √ 1+ √
> 0 for r = r + which ensures Φ S (P 2 ) -Φ S (P 4 ) > 0. (9.9)

We now recall

dΦ S dΣ = ∆ 1 -Φ S ∆ 2 Σ∆ 2
and study the sign at a possible point of contact on the curve with the value

Φ S (Σ) = c + , Σ < 0.
From (8.1), at the point of contact:

∆ 1 -c + ∆ 2 Σ = c 1 c + + c
2 = A 0 + A 1 Σ + A 2 Σ 2 .
We have A 0 = 0 from (2.45). We now compute all coefficients at r o (d, ), and the associated non degeneracy claim will follow for r close enough to r o (d, ).

step 2 Sign of A ∞ 2 . case r = r * . We compute:

A ∞ 2 = 3 -d - + d -1 2 -1 = 3 -d -2 ( + d -1) + = [-d -(d -1) + 1] = -(d -1)(1 + ) < 0.
case r = r + . We compute:

A ∞ 2 = (c ∞ + ) 3 -dc ∞ + -c ∞ + (e ∞ 21 (c ∞ + ) 2 -1) = c ∞ + (1 -e ∞ 21 )(c ∞ + ) 2 -(d -1) = c ∞ + 1 - + d -1 (d + √ ) 2 (1 + √ ) 2 -(d -1) = -(d -1)c ∞ + (d + √ ) 2 (1 + √ ) 2 + 1 < 0. step 3 Sign of A ∞ 1 . case r = r * . We have A ∞ 1 = d ∞ 20 (c ∞ + ) 2 + d ∞ 11 c ∞ + + d ∞ 02 -c ∞ + (e ∞ 20 (c ∞ + ) 2 + e ∞ 11 c ∞ + + e ∞ 02 )
We compute:

e ∞ 20 (c ∞ + ) 2 + e ∞ 11 c ∞ + + e ∞ 02 = √ d(d -1 + ) ( + √ d) 2 - d( √ d -1) + (1 + √ d) ( + √ d) - 3 √ d + √ d = √ d(d -1 + ) -d √ d + d -(1 + √ d) -3 √ d + √ d = -d √ d + d -3 √ d + (d √ d - √ d -1 - √ d) + 2 √ d + √ d = -d √ d + d -3 √ d + (d √ d -2 √ d -1) + 2 √ d + √ d and d ∞ 20 (c ∞ + ) 2 + d ∞ 11 c ∞ + + d ∞ 02 = - √ d -(d -) + √ d 2 - 2d √ d + √ d - √ d + √ d = - + √ d ( √ d + d -) + 2d √ d + √ d = - + √ d -2 + ( √ d + d) + 2d √ d + √ d .
Therefore,

( + √ d)A ∞ 1 = 2 -( √ d + d) -2d √ d - √ d - -d √ d + d -3 √ d + (d √ d -2 √ d -1) + 2 √ d = case r = r + . We have A ∞ 1 = d ∞ 20 (c ∞ + ) 2 + d ∞ 11 c ∞ + + d ∞ 02 -c ∞ + (e ∞ 20 (c ∞ + ) 2 + e ∞ 11 c ∞ + + e ∞ 02 ) We compute: e ∞ 20 (c ∞ + ) 2 + e ∞ 11 c ∞ + + e ∞ 02 = + d -1 (1 + √ ) √ (d + √ ) 1 + √ 2 - 2 1 + √ √ (d + √ ) 1 + √ - 3 1 + √ = ( + d -1)(d + √ ) 2 -2 √ (d + √ )(1 + √ ) -3(1 + √ ) 2 (1 + √ ) 3 = ( + d -1)(d 2 + 2d √ + ) -2 √ (d + (d + 1) √ + ) -3(1 + 2 √ + ) (1 + √ ) 3 = 2 + (2d -2) √ + (d 2 -d -6) + (2d 2 -4d -6) √ + d 3 -d 2 -3 (1 + √ ) 3 = Q 1 ( ) (1 + √ ) 3 and d ∞ 20 (c ∞ + ) 2 + d ∞ 11 c ∞ + + d ∞ 02 = - √ -d -1 (1 + √ ) 2 √ (d + √ ) 1 + √ 2 - 2d 1 + √ √ (d + √ ) 1 + √ - + d √ (1 + √ ) 2 = ( - √ -d -1)(d + √ ) 2 -2d √ (d + √ )(1 + √ ) 2 -( + d √ )(1 + √ ) 2 (1 + √ ) 4
and

Q 2 ( ) = ( - √ -d -1)(d + √ ) 2 -2d √ (d + √ )(1 + √ ) 2 -( + d √ )(1 + √ ) 2 = [ 2 - √ -(d + 1) ](d 2 + 2d √ + ) -(2d 2 √ + 2d )(1 + 2 √ + ) -( + d √ )(1 + 2 √ + ) = 3 + (2d -1) 2 √ + (d 2 -5d -2) 2 + (-5d 2 -7d -2) √ + (-d 3 -5d 2 -4d -1) + (-2d 2 -d) √ .
Hence

A ∞ 1 = Q 2 ( ) (1 + √ ) 4 - √ (d + √ ) 1 + √ Q 1 ( ) (1 + √ ) 3 = - Q 3 (1 + √ ) 4 with Q 3 = √ (d + √ )Q 1 -Q 2 = (d √ + ) 2 + (2d -2) √ + (d 2 -d -6) + (2d 2 -4d -6) √ + d 3 -d 2 -3 -3 -(2d -1) 2 √ -(d 2 -5d -2) 2 + (5d 2 + 7d + 2) √ + (d 3 + 5d 2 + 4d + 1) + (2d 2 + d) √ = (d -1) 2 √ + (2d 2 + 2d -4) 2 + (d 3 + 6d 2 -3d -4) √ + (4d 3 -2d -2) + (d 4 -d 3 + 2d 2 -2d) √ > 0 and hence A ∞ 1 < 0.
Derivative at P 2 . Near P 2 we have the Taylor expansion:

dΦ dΣ = A 1 Σ + A 2 Σ 2 Σ [c 2 Φ + c 4 + O(Σ)] → A 1 c 2 c + + c 4 = A 1 λ - > 0. (9.10)
Conclusion. At the point of contact, we therefore obtain:

dΦ dΣ = ∆ 1 -c + ∆ 2 Σ∆ 2 = Σ(A 1 + A 2 Σ) ∆ 2 = - |Σ|(-|A 1 | + |A 2 ||Σ|) ∆ 2 = |Σ|(|A 1 | -|A 2 ||Σ|) ∆ 2 . Let Σ = -Σ, Φ( Σ) = Φ(Σ), then d Φ d Σ = - Σ(|A 1 | -|A 2 | Σ) ∆ 2 .
Assume now that there exists 0 < Σ * < σ 2 with Φ( Σ * ) > c + , then from (9.9), (9.10), the curve is strictly below c + for 0 < Σ 1 and for | Σ -σ 2 | 1. Therefore, there must exist Σ1 < Σ * < Σ2 with

Φ( Σ1 ) = Φ( Σ2 ) = c + Φ ( Σ1 ) ≥ 0 Φ ( Σ2 ) ≤ 0 Since ∆ 2 > 0 in this zone: |A 1 | -|A 2 | Σ1 ≤0, |A 1 | -|A 2 | Σ2 ≥ 0 which forces Σ2 ≤ |A 1 | |A 2 | ≤ Σ1 , a contradiction.
F S (P 2 ) < 0 (9.11)
Moreover, in the case < d, we have:

∃0 < ε(d, ) 1 such that ∀0 < ε < ε(d, ), (1 -w -w + F )(P 2 ) > 0 ⇔ > 1 (d) (9.12)
with 1 given by (9.7).

Remark 9.6. The necessary admissible range near r * is therefore

1 (d) < < 2 (d) = d -2 √ d.
We compute numerically:

1 (5) = 0.1023, 2 (5) = 0.5279, 1 (6)
F S = σ 2 - Σ c 2 Φ + c 4 + [e 20 Φ 2 + e 11 Φ + e 02 ]Σ + [e 21 Φ 2 -1]Σ 2 -Σ [2σ 2 (1 + Φ) + Σ(1 -Φ 2 )] + O(Σ) = σ 2 + c 2 c ± + c 4 2σ 2 (1 + c ± ) + O(Σ) = σ 2 + λ - 2σ 2 (1 + c + ) + O(Σ) and w = - ∆ 1 ∆ = - c 1 Φ + c 3 + [d 20 Φ 2 + d 11 Φ + d 02 ]Σ + [Φ 3 -dΦ]Σ 2 -[2σ 2 (1 + Φ) + Σ(1 -Φ 2 )] = c 1 c + + c 3 2σ 2 (1 + c + ) + O(Σ) = c + λ - 2σ 2 (1 + c + ) + O(Σ)
and we compute these quantities at P 2 and r = r o .

case r o = r * .

F S = 2(σ ∞ 2 ) 2 (1 + ) + λ - 2σ 2 (1 + ) = 1 2σ ∞ 2 (1 + )( + √ d) 2 2d(1 + ) -d(d - √ d) + 2d + (d + √ d) = - (d -)(d - √ d) 2 √ d(1 + )( + √ d) < 0 and 1 -w -w + F S = σ ∞ 2 - λ - 2σ ∞ 2 (1 + ) + σ ∞ 2 + λ - 2σ ∞ 2 (1 + ) = 2 σ ∞ 2 + λ - 2σ ∞ 2 (1 + ) - (1 + )λ - 2σ ∞ 2 (1 + ) = - 2(d -)(d - √ d) 2 √ d(1 + )( + √ d) + 1 + 2 √ d(1+ ) + √ d d(d - √ d) + 2d + (d + √ d) ( + √ d) 2 = 1 2 √ d(1 + )( + √ d) -2(d -)(d - √ d) + (1 + ) d(d - √ d) + 2d + (d + √ d) = P d ( ) 2 √ d(1 + )( + √ d) (9.14)
with

P d ( ) = -2(d -)(d - √ d) + (1 + ) d(d - √ d) + 2d + (d + √ d) = -2d(d - √ d) + d(d - √ d) + 2d + 2(d - √ d) + d(d - √ d) + 2d + d + √ d + 2 (d + √ d) = -d 2 + d √ d + 2d + d 2 -d √ d + 5d - √ d + (d + √ d) 2 = √ d -d √ d + d + 2 √ d + d √ d -d + 5 √ d -1 + ( √ d + 1) 2 (9.15)
Observe that

P d (1) = -2(d -1)(d - √ d) + 2 d(d - √ d) + 2d + (d + √ d) = 2(d - √ d) + 2 2d + (d + √ d) > 0 and P d (0) = -2d(d - √ d) + d(d - √ d) + 2d < 0 which implies the condition > 1 (d), 1 (d) < 1
with 1 given by (9.7). case r o = r + . We compute

F S = σ 2 + c 2 c + + c 4 2σ 2 (1 + c + ) = -c 4 (1 + c + ) + c 2 c + + c 4 2σ 2 (1 + c + ) = c + (c 2 -c 4 ) 2σ 2 (1 + c + ) < 0
from (2.49), and F (P ∞ 2 ) = 0 from (F.11). Then at r o and P ∞ 2 :

1 -w -w + F S = σ 2 - c + λ - 2σ 2 (1 + c + ) = 2σ 2 2 (1 + c + ) -c + λ - 2σ 2 (1 + c + ) = - (1 + c + )λ - 2σ 2 (1 + c + ) > 0 9.3.
Positivity in the region where w ≤ w 2 . We now consider any solution curve P 2 -P 4 with the c -slope at P 2 given in Lemma 3.2. Recall that such curves cross the middle root of ∆ 1 at 0 < σ * < σ 2 . In view of the signs of ∆, ∆ 1 and ∆ 2 , see figure 3, these curves are:

• decreasing on (σ * , σ 2 ),

• increasing on (σ 0 , σ * ), where 0 ≤ σ 0 < σ * is such that w(σ 0 ) = 0,

• w(σ) ≤ 0 on [0, σ 0 ].

In particular, on [0, σ 2 ], w(σ) can reach the value w 2 only on (σ 0 , σ * ) where it is increasing, and hence there exists a unique 0 < σ 1 < σ 2 such that we have w > w 2 for σ 1 < σ < σ 2 w < w 2 for 0 < σ < σ 1 (9.16) Lemma 9.7 (Positivity in the region where σ < σ 1 and w < w 2 ). Under the assumptions of Lemma 9.1, any P 2 -P 4 curve with c -slope at P 2 satisfies (9.4) in the region w < w 2 and 0 < σ < σ 1 , Proof of Lemma 9.7. Using (8.11), we have along the solution curve in the region w < w 2 : F < 0, w < w 2 < 1 and ∆ > 0. Note also that the solution curve has the slope c -at P 2 and can not intersect the separatrix curve (strictly) between P 2 and P 4 . Therefore, it must lie above the P 2 -P 4 separatrix. As a consequence its function Φ also satisfies (9.8):

0 < Φ ≤ c + ,
where the lower bound follows from the fact that we consider the region w ≤ w 2 . Thus we focus below on the region

0 < σ < σ 1 , w < w 2 , 0 < Φ ≤ c + . step 1 Study of 1 -w -w + F .
Remark 9.8. We will consider the expression for 1 -w -w + F

1 -w -w + F = (1 -w + σ) + ∆ 1 -∆ 2
∆ not just as a function on the solution curve but more generally as a function of σ and w.

We compute We study the roots of P σ (w) for 0 < w < w 2 = w -< w + < 1 < r which ensures:

(1 -w -w ) + F = - (d -1)σ ∆ (w -w -)(w -w + ) + (1 -w + ∆ 1 ∆ ) = 1 ∆ -(d -1)σ(w -w -)(w -w + ) + (1 -w)[(1 -w) 2 -σ 2 ] + w(1 -w)(r -w) -d(w -w e )σ 2 = - 1 ∆ [1 -w + d(w -w e )] σ 2 + (d -1)(w --w)(w + -w)σ -(1 -w)[(1 -w) 2 + w(r -w)] = - 1 ∆ (d -1)(w -w * )σ 2 + (d -1)(w --w)(w + -w)σ -(1 -w)[(1 -w) 2 + w(r -w)] = - 1 ∆ (d -1)(w -w * )σ 2 + (d -1)(w --w)(w + -w)σ -(1 -w)[1 + (r -2)w] = - P σ (w) ∆ (9.
(1 -w)[(1 -w) 2 + w(r -w)] > 0 (w + -w)(w --w) > 0 (9.20)
Then, at r * (d, ):

(d -1)(w ∞ 2 -(w * ) ∞ ) = (d -1)w ∞ 2 + 1 -(r * -1) = (d -1)(1 -σ ∞ 2 ) + 1 - d + + √ d -1 = d -(d -1) √ d + √ d - (d - √ d) + √ d = d( + √ d) -d √ d + √ d -d + √ d + √ d = √ d( + 1) + √ d > 0
and at r + (d, ):

(d -1)(w ∞ 2 -(w * ) ∞ ) = (d -1) √ 1 + √ + 1 - (d -1) (1 + √ ) 2 = d √ + 1 1 + √ - (d -1) (1 + √ ) 2 = + (d + 1) √ + 1 (1 + √ ) 2 > 0
and hence for r close enough to r o : The points P 2 = (σ 2 , w 2 ), P 3 = (σ + , w + ) and P 1 = (0, 1) are among the roots of P σ (w), and hence the curve σ 1 (w) must pass through these points and connects continuously (σ 2 , w 2 ) to (0, 1).

w 2 > w * .
step 3 Positivity on Φ = c + . Let the line

(D) = {W = c + Σ, 0 ≤ σ ≤ σ 2 } (9.22)
we claim that under the assumptions of Lemma 9.1:

1 -w -w + F > 0 on (D) (9.23)
Indeed, we compute:

∆(1 -w -w + F ) = (1 -w + σ)∆ + ∆ 1 -∆ 2 = (1 -w 2 -W + σ 2 + Σ) -Σ 2σ 2 (1 + Φ) + Σ(1 -Φ 2 ) + Σ c 1 Φ + c 3 + [d 20 Φ 2 + d 11 Φ + d 02 ]Σ + [Φ 3 -dΦ]Σ 2 -Σ c 2 Φ + c 4 + [e 20 Φ 2 + e 11 Φ + e 02 ]Σ + [e 21 Φ 2 -1]Σ 2 = -ΣG(Σ, Φ) with G(Σ, Φ) = [2σ 2 + Σ(1 -Φ)][2σ 2 (1 + Φ) + Σ(1 -Φ 2 )] -c 1 Φ -c 3 -[d 20 Φ 2 + d 11 Φ + d 02 ]Σ -[Φ 3 -dΦ]Σ 2 + c 2 Φ + c 4 + [e 20 Φ 2 + e 11 Φ + e 02 ]Σ + [e 21 Φ 2 -1]Σ 2 = 4σ 2 2 (1 + Φ) + Σ[2σ 2 (1 -Φ 2 ) + 2σ 2 (1 -Φ 2 )] + Σ 2 (1 -Φ) 2 (1 + Φ) -c 1 Φ -c 3 -[d 20 Φ 2 + d 11 Φ + d 02 ]Σ -[Φ 3 -dΦ]Σ 2 + c 2 Φ + c 4 + [e 20 Φ 2 + e 11 Φ + e 02 ]Σ + [e 21 Φ 2 -1]Σ 2 = 4σ 2 2 (1 + Φ) + c 2 Φ + c 4 -c 1 Φ -c 3 + [4σ 2 (1 -Φ 2 ) + e 20 Φ 2 + e 11 Φ + e 02 -d 20 Φ 2 -d 11 Φ -d 02 ]Σ + [(1 -Φ) 2 (1 + Φ) + e 21 Φ 2 -1 -Φ 3 + dΦ]Σ 2 = A 0 (Φ) + A 1 (Φ)Σ + A 2 (Φ)Σ 2 .
and we now distinguish r o = r * and r o = r + . step 4 Proof of (9.22) for r o = r * . We compute the A i on (D) at r = r * (d, ) i.e., Φ = c + = and evaluate the obtained sign of G. Computation of A ∞ 0 . We have from (9.14), (9.12):

A ∞ 0 = 4(σ ∞ 2 ) 2 (1+ )+λ ∞ --λ ∞ -= 2σ ∞ 2 (1+ ) P d ( ) 2 √ d(1 + )( + √ d) = P d ( ) ( + √ d) 2 > 0.
Computation of A ∞ 1 . We have

A ∞ 1 = A 11 + A 12 with A 11 = 4σ ∞ 2 (1 -2 ) + e ∞ 20 2 + e ∞ 11 + e ∞ 02 = 4 √ d + √ d (1 -2 ) + √ d(d -1 + ) ( + √ d) 2 - d( √ d -1) + (1 + √ d) ( + √ d) - 3 √ d + √ d = (4 √ d -4 2 √ d + √ d(d -1 + ) -d( √ d -1) -(1 + √ d) -3 √ d + √ d = √ d -d √ d + d + 2 (-4 √ d + √ d) + ( √ d(d -1) -1 - √ d) + √ d = - 3 √ d 2 -[(d -2) √ d -1] + √ d(d - √ d -1) + √ d
and

A 12 = -d ∞ 20 2 -d ∞ 11 -d ∞ 02 = √ d + (d -) + √ d 2 + 2d √ d + √ d + √ d + √ d = (2d √ d + √ d) + (d -+ √ d) 2 + √ d which implies A ∞ 1 = -3 √ d 2 + [(d -2) √ d -1] - √ d(d - √ d -1) + (2d √ d + √ d) + (d -+ √ d) 2 + √ d = (d -2 √ d -) 2 + (3d √ d - √ d -1) - √ d(d - √ d -1) + √ d = - Q d ( ) + √ d with Q d ( ) = √ d(d - √ d -1) -(3d √ d - √ d -1) -(d -2 √ d -) 2 .
Computation of A 2 . We have

A 2 (Φ) = (1 -Φ) 2 (1 + Φ) + e 21 Φ 2 -1 -Φ 3 + dΦ = (1 -Φ)(1 -Φ 2 ) + e 21 Φ 2 -1 -Φ 3 + dΦ = 1 -Φ 2 -Φ + Φ 3 + e 21 Φ 2 -1 -Φ 3 + dΦ = Φ [(d -1) + (e 21 -1)Φ] = Φ d -1 + + d -1 -1 Φ = (d -1)Φ( + Φ) (9.24)
which implies

A ∞ 2 = 2 (d -1). Discriminant Recall G(Σ, ) = A 0 + A 1 Σ + A 2 Σ 2 ,
we compute the discriminant at the critical value:

Discr d ( ) = (A ∞ 1 ) 2 -4A ∞ 0 A ∞ 2 = Q 2 d ( + √ d) 2 - 8 (d -1)P d ( + √ d) 2 = Q 2 d -8 (d -1)P d ( + √ d) 2 .
We collect the values

P d ( ) = -d 2 + d √ d + 2d + d 2 -d √ d + 5d - √ d + (d + √ d) 2 Q d ( ) = √ d(d - √ d -1) -(3d √ d - √ d -1) -(d -2 √ d -) 2 test(d, ) = Q 2 d -8 (d -

Conclusion. Since

A ∞ 2 > 0, we conclude G(Σ, Φ) = A 0 (Φ) + A 1 (Φ)Σ + A 2 (Φ)Σ 2 > 0 on (D)
given by (9.22). Therefore,

1 -w -w + F = - ΣG(Σ, Φ) ∆ > 0
on (D), and (9.23) is proved.

step 5 Proof of (9.22) for r o = r + . Computation of A 0 . We have since Φ = c + and recalling (2.45), (2.50):

A 0 = 4σ 2 2 (1 + c + ) + c 2 c + + c 4 -c 1 c + -c 3 = -2c 4 (1 + c + ) + c 2 c + + c 4 -c + λ - = -c 4 -c 2 c + -2(c 4 -c 2 )c + -c + λ -= -(1 + c + )λ --2(c 4 -c 2 )c + and hence A ∞ 0 = -(1 + c ∞ + )λ ∞ -> 0. We compute explicitly A 0 = 1 + √ (d + √ ) 1 + √ 2( + (d + 1) √ + 1 (1 + √ ) 3 = 2( + (d + 1) √ + 1) 2 (1 + √ ) 4
Computation of A ∞ 1 . We have

A ∞ 1 = A 11 + A 12 with A 11 = 4σ ∞ 2 (1 -(c ∞ + ) 2 ) + e ∞ 20 (c ∞ + ) 2 + e ∞ 11 c ∞ + + e ∞ 02 = 4 1 + √ 1 - (d + √ ) 2 (1 + √ ) 2 + + d -1 (1 + √ ) (d + √ ) 2 (1 + √ ) 2 - 2 1 + √ √ (d + √ ) 1 + √ - 3 1 + √ = Q 1 ( ) (1 + √ ) 3
with

Q 1 = 4 1 + 2 √ + -(d 2 + 2d √ + ) + ( + d -1)(d 2 + 2d √ + ) -2( + d √ )(1 + √ ) -3(1 + 2 √ + ) = 2 (-3) + √ (-6d -2) + (-3d 2 -d -2) + √ (2d 2 -4d + 2) + d 3 -d 2 + 1.
Then

A 12 = -d ∞ 20 (c ∞ + ) 2 -d ∞ 11 c ∞ + -d ∞ 02 = - - √ -d -1 (1 + √ ) 2 (d + √ ) 2 (1 + √ ) 2 + 2d 1 + √ √ (d + √ ) 1 + √ + + d √ (1 + √ ) 2 = Q 2 ( ) (1 + √ ) 4 with Q 2 = (d 2 + 2d √ + )(-2 + √ + (d + 1) ) + (2d + 2d 2 √ )(1 + 2 √ + ) + ( + d √ )(1 + 2 √ + ) = -3 + (-2d + 1) 2 √ + 2 (-d 2 + 5d + 2) + √ (5d 2 + 7d + 2) + (d 3 + 5d 2 + 4d + 1) + √ (d + 2d 2 ). Hence A ∞ 1 = Q 3 (1 + √ ) 4 with Q 3 = 2 (-3) + √ (-6d -2) + (-3d 2 -d -2) + √ (2d 2 -4d + 2) + d 3 -d 2 + 1 (1 + √ ) -3 + (-2d + 1) 2 √ + 2 (-d 2 + 5d + 2) + √ (5d 2 + 7d + 2) + (d 3 + 5d 2 + 4d + 1) + √ (d + 2d 2 ) = -3 -(2d + 2) 2 √ -(d 2 + d + 3) 2 + (2d 2 -2) √ + (d 3 + 4d 2 -d + 1) + (d 3 + 3d 2 -3d + 3) √ + d 3 -d 2 + 1
Computation of A 2 . We have A 2 (Φ) > 0 from (9.24) and explicitely

A ∞ 2 = (d -1)c + ( + c + ) = d -1 √ (d + √ ) 1 + √ + √ (d + √ ) 1 + √ = (d -1)(d + √ )( + 2 √ + d) (1 + √ ) 2
Conclusion. We are in the case d = 3. We numerically evaluate Q 3 ( ) and obtain

Q 3 ( ) < 0 for ≥ 4 in which case A ∞ 1 < 0 and since Σ < 0, G(Σ, Φ) = A 0 (Φ) + A 1 (Φ)Σ + A 2 (Φ)Σ 2 > 0. For 3 < < 4, we form the discriminant Discr = (A ∞ ) 2 1 -4A ∞ 0 A ∞ 2 = Q 2 3 (1 + √ ) 8 - 8( + (d + 1) √ + 1) 2 (1 + √ ) 4 (d -1)(d + √ )( + 2 √ + d) (1 + √ ) 2 = Q 4 (1 + √ ) 8 with Q 4 = Q 2 3 -8(d -1)(1 + √ ) 2 ( + (d + 1) √ + 1) 2 (d + √ )( + 2 √ + d)
and numerically evaluate Q 4 ( ) < 0 for 3 < < 4, and hence G(Σ, Φ) > 0. Hence

1 -w -w + F = - ΣG(Σ, Φ) ∆ > 0
on (D), and (9.23) is proved.

step 4 Proof of (9.4). Observe that

P σ (w 2 ) = (d -1)(w 2 -w * )σ 2 -(1 -w 2 )[1 + (r -2)w 2 ]
is a second order polynomial in σ with P σ 2 (w 2 ) = 0, positive highest order coeffieicent and such that at r * (d, ):

P 0 (w 2 ) = -(1 -w 2 )[1 + (r -2)w 2 ] with 1 + (r * (d, ) -2)w 2 = 1 + d + + √ d -2 1 - √ d + √ d = 1 + (d --2 √ d) ( + √ d) 2 = 2 + 2 √ d + d + (d --2 √ d) ( + √ d) 2 = (d + 1) ( + √ d) 2 and 1 + (r + (d, ) -2)w 2 = 1 + 1 + d -1 (1 + √ ) 2 -2 √ 1 + √ = 1 - √ 1 + √ 1 - d -1 (1 + √ ) 2 = 1 - √ [(1 + √ ) 2 -(d -1)] (1 + √ ) 3 = 1 + (d + 1) √ + (1 + √ ) 3 > 0.
step 1 Reduction to the control of the P 2 -P o separatrix. We compute

∆(1 -w -w -F ) = (1 -w-σ)∆ + ∆ 1 +∆ 2 = (1 -w 2 -W -σ 2 -Σ) -Σ 2σ 2 (1 + Φ) + Σ(1 -Φ 2 ) + Σ c 1 Φ + c 3 + [d 20 Φ 2 + d 11 Φ + d 02 ]Σ + [Φ 3 -dΦ]Σ 2 + Σ c 2 Φ + c 4 + [e 20 Φ 2 + e 11 Φ + e 02 ]Σ + [e 21 Φ 2 -1]Σ 2 = -ΣH(Σ, Φ) with H(Σ, Φ) = -Σ(1 + Φ)[2σ 2 (1 + Φ) + Σ(1 -Φ 2 )] -c 1 Φ -c 3 -[d 20 Φ 2 + d 11 Φ + d 02 ]Σ -[Φ 3 -dΦ]Σ 2 -c 2 Φ-c 4 -[e 20 Φ 2 + e 11 Φ + e 02 ]Σ-[e 21 Φ 2 -1]Σ 2 = -c 2 Φ-c 4 -c 1 Φ -c 3 + [-2σ 2 (1 + Φ) 2 -e 20 Φ 2 -e 11 Φ-e 02 -d 20 Φ 2 -d 11 Φ -d 02 ]Σ + [-(1 -Φ)(1 + Φ) 2 -e 21 Φ 2 +1 -Φ 3 + dΦ]Σ 2 .
We introduce the notation

Φ := Φ -c -.
We infer, using (2.45), (2.50):

H(Σ, Φ) = -(c 2 + c 1 ) Φ -c 2 c --c 4 -c 1 c --c 3 + [- 2σ
) Φ + (-2σ 2 -e 20 -d 20 ) Φ 2 ]Σ + [(d -1)c -+ (1 -e 21 )c 2 -+ (d -1) Φ + 2c -(1 -e 21 ) Φ + (1 -e 21 ) Φ 2 ]Σ 2 = F 0 (Σ) + F 1 (Σ) Φ + F 2 (Σ) Φ 2 where F 0 (Σ) := -λ + (1 + c -) + [-2σ 2 -e 02 -d 02 + (-4σ 2 -e 11 -d 11 )c -+ (-2σ 2 -e 20 -d 20 )c 2 -]Σ +[(d -1)c -+ (1 -e 21 )c 2 -]Σ 2 , F 1 (Σ) := -(c 2 + c 1 ) + [(-4σ 2 -e 11 -d 11 ) + 2c -(-2σ 2 -e 20 -d 20 )]Σ +[d -1 + 2c -(1 -e 21 )]Σ 2 , F 2 (Σ) := (-2σ 2 -e 20 -d 20 )Σ + (1 -e 21 )Σ 2 .
Sign of F 1 . Since -(c 2 + c 1 ) > 0 and does not degenerate as r → r o , we infer

F 1 (Σ) = |c 1 + c 2 | + O(Σ) > 0.
Sign of F 2 . We have

F 2 (Σ) = (2σ 2 + e 20 + d 20 )|Σ|(1 + O(Σ)) (9.26)
In the case r o = r * , we have

2σ 2 + e 20 + d 20 = 2 √ d + √ d + √ d(d -1 + ) ( + √ d) + - √ d -(d -) + √ d (9.27) = 2 √ d + √ d(d -1 + ) - √ d -(d -) ( + √ d) = 2 + (2 √ d -d) + √ d(d -1) ( + √ d)
The discriminant of the second order polynomial2 + (2

√ d -d) + √ d(d -1) is given by (2 √ d -d) 2 -4 √ d(d -1) = √ d √ d( √ d -2) 2 -4(d -1) = √ d d √ d -8d + 4 √ d + 4
which is negative for 3 ≤ d ≤ 9, so that the second order polynomial in has the sign of the main term. Therefore,

√ d(d -1) > 0 which implies F 2 (Σ) > 0 in the case r o = r * .
In the case r o = r + , we have

2σ 2 + e 20 + d 20 = 2 + 2 3 2 + 2 - √ + d(1 + √ -) -1 (1 + √ ) 2 > 0 in the case d = 3, > d, which implies F 2 (Σ) > 0 in that case as well.
Computation of the roots. We have

F 0 (Σ) = |λ + ||1 + c -| + O(Σ) = O(|r -r * ( )|).
Thus, the discriminant

F 1 (Σ) 2 -4F 2 (Σ)F 0 (Σ) = |c 2 + c 1 | + O(r -r * ( )) > 0.
Using also F 2 (Σ) > 0 and F 1 (Σ) > 0, the roots are given by

Φ ± = -|F 1 (Σ)| ± F 1 (Σ) 2 -4|F 2 (Σ)|F 0 (Σ) 2|F 2 (Σ)| .
We rewrite

Φ + = - 2F 0 (Σ) |F 1 (Σ)| + F 1 (Σ) 2 -4|F 2 (Σ)|F 0 (Σ) . (9.28) Conclusion. Since F 2 (Σ) > 0 and Φ - -1, we have 1 -w -w -F > 0 if and only if Φ > Φ + .
Since the curve connects P 2 to P 4 with c -slope, the solution must lie strictly below the P 2 -P o separatrix and hence Φ > Φ S and the conclusion follows from the lower bound along the separatrix curve:

Φ S > c -+ Φ + , ∀u ∈ 0, 3 4 . ( 9
0 dz (1 + z) K+2 z αγ dz = +∞ 0 z -αγ (1 + z) K+2 dz = B(1 -α γ , K + 1 + α γ ) = Γ(1 -α γ )Γ(K + 1 + α γ ) Γ(K + 2) . (A.2)
Recurrence formulas. We compute for k 1 ≤ k 2 :

Π k 2 j=k 1 (γ -j) = Π k 2 j=k 1 (α γ + K + 1 -j) = Π K+1-k 1 =K+1-k 2 (α j + ) = Π K+1-k 1 =1 (α j + ) Π K-k 2 =1 (α j + )
.

We now recall

Γ(x + 1) = xΓ(x), x ∈ C\Z - Γ(k + 1) = k! (A.3)
where Z -denotes the set of negative integers, from which

Π m =1 (x + ) = Γ(x + m + 1) Γ(x + 1) , x ∈ C\Z - (A.4)
which yields

Π k 2 j=k 1 (γ -j) = Γ(α γ + K + 2 -k 1 ) Γ(α γ + K -k 2 + 1) = Γ(γ + 1 -k 1 ) Γ(γ -k 2 ) (A.5) Therefore, Π K-1 j=0 (γ -j -2) = Π K+1 j=2 (γ -j) = Γ(α γ + K) Γ(α γ ) . (A.6)
Asymptotics. We recall Stirling's formula

Γ(x + 1) = (1 + o x→+∞ (1)) x e x √ 2πx. (A.7) Let |x| 1 γ,
this yields:

Γ(γ + x + 1) Γ(γ + 1) = (1 + o γ→+∞ (1)) γ+x e γ+x 2π(γ + x) γ e γ √ 2πγ = (1 + o γ→+∞ (1)) 1 e x e (γ+x) logγ+ x γ +O 1 γ 2 -γlogγ = (1 + o γ→+∞ (1)) 1 e x e xlogγ+x+Ox 1 γ = (1 + o γ→+∞ (1))γ x . (A.8)
Moreover,

Γ(γ + 1 + x) = (1 + o γ→+∞ (1)) γ + x e γ+x 2π(γ + x) = (1 + o γ→+∞ (1)) √ 2πγ e γ+x e (γ+x) logγ+ x γ +Ox 1 γ 2 = (1 + o γ→+∞ (1)) √ 2πγ e γ+x e γlogγ+x+xlogγ+Ox 1 γ = √ 2π(1 + o γ→+∞ (1)) γ γ+x+ 1 2 e γ . (A.9)
Value on R\N -.

there holds the bound

Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ≤ c ν 1 + k 1 e -k Φ (2) k (x) (B.1)
where the phase function satisfies

Φ (2) k (x) ≥ 0, (Φ (2)
k ) (x) ≥ 0. Moreover, for R k < x ≤ 1 2 , Φ (2)
k (x) ≥ 1 2 x|logx|. (B.2)
Proof of Lemma B.1. We assume k ≥ k * (R, ν) large enough and apply Stirling's formula

Γ(x + 1) = (1 + o x→+∞ (1)) x e x √ 2πx
to upper bound

Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ν k 1 +ν+1 e k 1 +ν+1 √ k 1 + ν + 1 k 2 +ν+1 e k 2 +ν+1 √ k 2 + ν + 1 k+ν+1 e k+ν+1 √ k + ν + 1 ν k 1 + ν + 1e -[(k+ν+1)log(k+ν+1)-(k 1 +ν+1)log(k 1 +ν+1)-(k 2 +ν+1)log(k 2 +ν+1)]
step 1 Study of the Stirling phase. We compute:

(k + ν + 1)log(k + ν + 1) -(k 1 + ν + 1)log(k 1 + ν + 1) -(k 2 + ν + 1)log(k 2 + ν + 1) = k 1 + ν + 1 k logk + log 1 + ν + 1 k -k x + ν + 1 k logk + log x + ν + 1 k -k (1 -x) + ν + 1 k logk + log (1 -x) + ν + 1 k = -(ν + 1)logk + k 1 + ν + 1 k log 1 + ν + 1 k -k x + ν + 1 k log x + ν + 1 k + (1 -x) + ν + 1 k log (1 -x) + ν + 1 k = -(ν + 1)log(ν + 1) + (ν + 1)log ν + 1 k + k 1 + ν + 1 k log 1 + ν + 1 k -k x + ν + 1 k log x + ν + 1 k + (1 -x) + ν + 1 k log (1 -x) + ν + 1 k = -(ν + 1)log(ν + 1) + kΦ (2) k (x)
which yields

Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ≤ c ν 1 + k 1 e -kΦ (2) k (x) . (B.3)
We have

Φ (2) k (0) = Φ (2)
k (1) = 0 and

∂ x Φ (2) k (x) = -log x + ν + 1 k + log (1 -x) + ν + 1 k = log (1 -x) + ν+1 k x + ν+1 k = log 1 + (1 -2x) x + ν+1 k ≥ 0 (B.4)
for x ∈ [0, 1 2 ]. We conclude that Φ

k (x) is a non negative non decreasing function of x ∈ [0, 1 2 ].

step 4 Small x lower bound. Assume R k < x ≤ 1 2 where R > R ν 1, then Φ (2) k (x) = ν + 1 k log ν + 1 k + 1 + ν + 1 k log 1 + ν + 1 k - x + ν + 1 k log x + ν + 1 k -(1 -x) + ν + 1 k log (1 -x) + ν + 1 k = - ν + 1 k log 1 + kx ν + 1 + 1 + ν + 1 k log 1 + ν + 1 k + x log x + ν + 1 k + (1 -x) + ν + 1 k log (1 -x) + ν + 1 k ≥ - ν + 1 k log 1 + R ν + 1 + x logx + log 1 + ν + 1 kx ≥ x|logx| 1 + O ν ν + 1 R log ν + 1 R ≥ 1 2 x|logx|.
Lemma B.2 (Uniform convolution bound). Let ν > 0, then there exists C ν > 0 such that for all k ≥ 1, for all j ≥ 2,

k 1 +•••+k j =k Π j i=1 Γ(k i + ν + 2) Γ(k + ν + 2) ≤ C j ν . (B.5)
Proof. We argue by induction on j.

step 1 Case j = 2. We apply Lemma B.1 with some large enough R ν > 1 and may without loss of generality assume k ≥ k * ν , with k * ν large enough universal constant. This leads to

Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ≤ c ν 1 + k 1 e -k Φ (2)
k (x) , where x = k 1 /k. We may assume k 1 ≤ k 2 so that x ≤ 1/2. For x ≤ Rν k , we use the lower bound Φ

(2) k (x) ≥ 0. For x ≥ Rν k the monotonicity of Φ (2) k on [0, 1/2] implies Φ (2) k (x) ≥ Φ (2) k R ν k ≥ + 1 2 R ν k log R ν k ≥ R ν 4k logk, (B.6)
where we required that R ν ≤ k * ν , from which

k 1 +k 2 =k Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ν 1 + k 1 +k 2 =k, Rν k ≤x≤ 1 2 1 + k 1 e -Rν logk 4 + k 1 +k 2 =k,0≤x≤ Rν k 1 + k 1 ν 1 + R 3 2 ν + k 3 2 k Rν 4 ν 1.
since the second sum has k 1 = xk ≤ R ν terms.

From Stirling's formula:

w γ,ν b (k) = Γ(γ -1 -k) Γ(γ -1) Γ(k + ν b + 2) ≤ c ν γ-2-k e γ-2-k k+ν b +1 e k+ν b +1 γ-2 e γ-2 (γ -2 -k)(k + ν b + 1) γ -2 1 2 ≤ c ν e -Φγ (k) (γ -2 -k)(k + ν b + 1) γ -2 1 2 with Φ γ (x) = (γ -2)log(γ -2) -(γ -2 -x)log(γ -2 -x) -(x + ν b + 1)log(x + ν b + 1).
We compute

Φ γ (x) = log(γ -2 -x) + 1 -log(x + ν b + 1) -1 = log γ -2 -x x + ν b + 1 = log 1 + γ -ν b -3 -2x x + ν b + 1 ≥ 0 for 1 ≤ x ≤ γ-ν b -3 2 ≤ 0 for γ-ν b -3 2 ≤ x ≤ K -2. Pick a large enough universal constant R ν 1, then for 1 ≤ x < R ν , Φ γ (x) = (γ -2)log(γ -2) -(γ -2) 1 - x γ -2 log(γ -2) + log 1 - x γ -2 -(x + ν b + 1)log(x + ν b + 1) = x γ -2 + O x 2 γ -2 + xlog(γ -2) -(x + ν b + 1)log(x + ν b + 1) = xlog(γ -2) 1 + O Rν 1 log(γ -2) + O ν (1).
This yields for 1 ≤ k ≤ R ν : Φ γ (x) ≥ log(γ -2) -C ν and, as a result,

w γ,ν b (k) ≤ c ν e -log(γ-2) (γ -2 -k)(k + ν + 1) γ -2 1 2 ≤ c ν γ . For R ν < x ≤ γ-ν b -3 2 Φ γ (x) ≥ Φ γ (R ν ) ≥ R ν 2 log(γ -2)
and thus,

w γ (k) ≤ c ν γ Rν 4
.

We obtain the bound

1≤k≤ γ-ν b -3 2 w γ,ν b (k) ≤ c ν,a b.
On the other hand,

Φ γ (γ -R ν ) = (γ -2)log(γ -2) -(R ν -2)log(R ν -2) -(γ -R ν + ν b + 1)log(γ -R ν + ν b + 1) = (γ -2)log(γ -2) -(γ -2) 1 - R ν -ν b -3 γ -2 log(γ -2) -log 1 - R ν -ν b -3 γ -2 + O(R ν ) ≥ R ν 4 log(γ -2).
The monotonicity of Φ γ then implies

Φ γ (k) ≥ Φ γ (γ -R ν ) ≥ R ν 4 log(γ -2) for γ -ν b -3 2 ≤ k ≤ γ -R ν which yields γ-ν b -3 2 ≤k≤γ-Kν w γ,ν b (k) ≤ c ν,a b
and concludes the proof of (C.3).

step 2 From Stirling's formula, for

K -K ν +1 ≤ k ≤ K -1: Γ(k + ν b + 2) Γ(γ -1) = (1 + o ν (1)) k+ν b +1 e k+ν b +1 γ-2 e γ-2 k + ν b + 1 γ -2 1 2 = e Oν (1) e -Φγ (k) with Φ γ (k) = (γ -2)log(γ -2) -(k + ν b + 1)log(k + ν b + 1). Let k = γ -x = K + 1 + α γ -x, 2 + α γ ≤ x ≤ K ν +α γ , then Φ γ (k) = (γ -2)log(γ -2) -(γ -x + ν b + 1)log(γ -x + ν b + 1) = (γ -2)log(γ -2) -(γ -2) 1 + ν b + 3 -x γ -2 log(γ -2) + log 1 + ν b + 3 -x γ -2 = (x -ν b -3)log(γ -2) + O ν (1)
and we obtain the formula

w γ,ν b (k) = Γ(γ -1 -k) e Oν (1) (γ -2) x-ν b -3 = e Oν (1) Γ(γ -1 -k)γ ν b +3-(γ-k) = e Oν (1) Γ(γ -1 -k)γ ν b +2-(γ-1-k) .
We now turn to the convolution type estimate on the weight.

Lemma C.3 (Convolution estimate).

There exist universal constants c ν,a > 0, 0 < b * (ν, a)

1 such that the following holds: for all 0 < b < b * , for all 0 ≤ k ≤ K,

k 1 +k 2 =k w γ,ν (k 1 )w γ,ν (k 2 ) ≤ c ν,a w γ,ν (k). (C.5)
More generally,

k 1 +•••+k j =k w γ,ν (k 1 ) . . . w γ,ν (k j ) ≤ c j ν,a w γ,ν (k). (C.6)
Proof of Lemma C.3. We prove (C.5). The proof of (C.6) follows by induction and is left to the reader. We parametrize

k 1 = kx, k 2 = (1 -x)k ⇔ γ -1 -k = (γ -1)(1 -α) γ -1 -k 1 = (γ -1)(1 -αx) γ -1 -k 2 = (γ -1)(1 -α(1 -x))
For k 1 = 0 or k 2 = 0, the claim is trivial and we therefore assume without loss of generality

1 ≤ k 1 ≤ k 2 < γ -2. We recall γ -1 -k = α γ + (K -k), 0 ≤ k ≤ K. step 1 Generic case. Assume first k ≤ K -2 (C.7) which ensures: γ -2 -k = α γ + (K -k -1) ≥ 1. (C.8) Representation formula. We have w γ,ν (k 1 )w γ,ν (k 2 ) w γ,ν (k) = Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(γ -1 -k)Γ(γ -1) Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) .
The function Γ(γ -1 -z) is holomorphic and bounded for 0 ≤ R(z) ≤ γ -1. Using the three line theorem, we obtain

Γ(γ -1 -k 1 ) ≤ Γ(γ -1) 1-x Γ(γ -1 -k) x , Γ(γ -1 -k 2 ) ≤ Γ(γ -1) x Γ(γ -1 -k) 1-x which implies that Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(γ -1 -k)Γ(γ -1) ≤ 1
Since the Γ function is strictly positive and bounded on [1, R] for all R > 0, we may use Stirling's formula to upper bound:

Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(γ -1 -k)Γ(γ -1) ≤ c ν γ-2-k 1 e γ-2-k 1 2π(γ -2 -k 1 ) γ-2-k 2 e γ-2-k 2 2π(γ -2 -k 2 ) γ-2-k e γ-2-k 2π(γ -2 -k) γ-2 e γ-2 2π(γ -2)
.

We compute the Stirling phase:

(γ -2)log(γ -2) + (γ -2 -k)log(γ -2 -k) -(γ -2 -k 1 )log(γ -2 -k 1 ) -(γ -2 -k 2 )log(γ -2 -k 2 ) = (γ -2)log(γ -2) + (γ -2 -k)log(γ -2 -k) -(γ -2 -kx)log(γ -2 -kx) -(γ -2 -k(1 -x))log(γ -2 -k(1 -x)) = γΦ (1) k,γ (x) (C.9)
Monotonicity for Φ

k,γ (x). We compute Φ

k,γ (0) = 0 and

Φ (1) k,γ (x) = klog(γ -2 -kx) + k -klog(γ -2 -k(1 -x)) -k = klog γ -2 -xk γ -2 -(1 -x)k = klog 1 + (1 -2x)k γ -2 -(1 -x)k > 0 (C.10) for 0 ≤ x < 1 2 .
In particular, Φ

(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ≤ c ν 1 + k 1 e -kΦ (2) k (x) ≤ c ν k Rν 4
and therefore, from (C.11):

w γ,ν (k 1 )w γ,ν (k 2 ) w γ,ν (k) = Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(γ -1 -k)Γ(γ -1) Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ≤ c ν k Rν 4 (γ -2 -k 1 )(γ -2 -k 2 ) (γ -2 -k)(γ -2) 1 2 If k ≤ γ-2 2 then k 1 , k 2 ≤ γ-2 2 and (γ -2 -k 1 )(γ -2 -k 2 ) (γ -2 -k)(γ -2) 1. Otherwise, γ -1 > k = k 1 + k 2 ≥ 2k 1 and, using (C.8): (γ -2 -k 1 )(γ -2 -k 2 ) (γ -2 -k)(γ -2) ≤ γ -2 -k 1 ≤ γ ≤ 2k which implies w γ,ν (k 1 )w γ,ν (k 2 ) w γ,ν (k) ≤ c ν k Rν 8
and gives

k 1 +k 2 =k,Rν ≤k 1 ≤ k 2 w γ,ν (k 1 )w γ,ν (k 2 ) w γ (k) ≤ c ν . (C.12)
case k 1 small. for x < Rν k i.e. k 1 < R ν , using Φ

k,γ ≥ 0, Φ

k ≥ 0, we have the bound

w γ,ν (k 1 )w γ,ν (k 2 ) w γ,ν (k) ≤ c ν 1 + k 1 (γ -2 -k 1 )(γ -2 -k 2 ) (γ -2 -k)(γ -2) 1 2
, and using (C.8):

(γ -2 -k 1 )(γ -2 -k 2 ) (γ -2 -k)(γ -2) ≤ γ -2 -k 2 γ -2 -k = γ -2 -k + k 1 γ -2 -k ≤ 1+ R ν γ -2 -k ≤ 1+R ν ≤ c ν
Since the sum has at most k 1 ≤ R ν terms, we obtain

k 1 +k 2 =k,0≤k 1 ≤Rν w γ,ν (k 1 )w γ,ν (k 2 ) w γ (k) ≤ c ν . (C.13)
This concludes the proof of (C.5) in the regime (C.7).

step 2 Boundary terms. We treat separately the cases

k ∈ {K -1, K}. (C.14)
For k 1 = 1, we have

Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(γ -1 -k)Γ(γ -1) = Γ(γ -2)Γ(γ -1 -(k -1)) Γ(K -k + α γ )Γ(γ -1) ≤ c γ Γ(K -k + α γ + 1) Γ(K -k + α γ ) ≤ c
Then, using Φ

k ≥ 0:

w γ,ν (k 1 )w γ,ν (k 2 ) w γ,ν (k) = Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(γ -1 -k)Γ(γ -1) Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ≤ c ν 1 + k 1 e -kΦ (2) k (x) ≤ c ν . We may therefore assume k 1 ≥ 2 (C.15) and therefore, γ -2 -k 2 ≥ γ -2 -(k -2) = α γ + (K -k) -1 + 2 ≥ 1 k 1 ≤ k 2 implies k 1 ≤ k 2 ≤ γ-1 2 and γ -1 -k 1 1. Moreover, Γ(γ -1 -k) = Γ((K -k) + α γ ) 1 independent of γ.
As a result, (B.1) yields the upper bound:

w γ,ν (k 1 )w γ,ν (k 2 ) w γ,ν (k) = Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(γ -1 -k)Γ(γ -1) Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ≤ c ν Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(γ -1) 1 + k 1 e -kΦ (2) k (x)
We compute the modified Stirling phase:

Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(γ -1) (C.16) ≤ c γ-2-k 1 e γ-2-k 1 2π(γ -2 -k 1 ) γ-2-k 2 e γ-2-k 2 2π(γ -2 -k 2 ) γ-2 e γ-2 2π(γ -2) ≤ c (γ -2 -k 1 )(γ -2 -k 2 ) γ -2 1 2 e -γΦ (3) γ (x) with γΦ (3) k,γ (x) = (γ -2)log(γ -2) -(γ -2 -k 1 )log(γ -2 -k 1 ) -(γ -2 -k 2 )log(γ -2 -k 2 ) = (γ -2)log(γ -2) -(γ -2 -kx)log(γ -2 -kx) -(γ -2 -k(1 -x))log(γ -2 -k(1 -x)).
We have verbatim as above

Φ (3) γ (x) ≥ 0 for 0 ≤ x ≤ 1 2 and, since x = k 1 k ≥ 2 k : Φ (3) γ (x) ≥ Φ (3) γ 2 k = (γ -2)log(γ -2) -(γ -2 -2)log(γ -2 -2) -(γ -2 -k + 2))log(γ -2 -k + 2)) = (γ -2)log(γ -2) -(γ -4)log(γ -4) -(K -k + α γ + 1)log(K -k + α γ + 1) ≥ -c
for some universal constant C independent of γ from (C.14). We obtain the bound

w γ,ν (k 1 )w γ,ν (k 2 ) w γ,ν (k) ≤ c (γ -2 -k 1 )(γ -2 -k 2 ) γ -2 1 2 1 + k 1 e -kΦ (2) k (x) .
case k 1 large. Arguing verbatim as before, we pick a large enough universal constant R ν 1 as in Lemma B.1 and estimate for x ≥ Rν k ,

w γ,ν (k 1 )w γ,ν (k 2 ) w γ,ν (k) ≤ c (γ -2 -k 1 )(γ -2 -k 2 ) γ -2 1 2 1 k Rν 4 and (γ -2 -k 1 )(γ -2 -k 2 ) γ -2 ≤ γ -2 ≤ 2k
(C.12) follows again.

case k 1 small. Since k 1 ≤ R ν , and in view of (C.14):

(γ -2 -k 1 )(γ -2 -k 2 ) γ -2 ≤ γ -2 -k 2 = γ -2 -(k -k 1 ) ≤ R ν + 2
and (C.13) follows again. This concludes the proof of (C.5).

Lemma C.4 (Truncated convolution estimate). For K + 1 ≤ k ≤ 2K:

k 1 +k 2 =k,0≤k 1 ,k 2 ≤K w γ,ν b (k 1 )w γ,ν b (k 2 ) ≤ c ν,a w γ,ν b (k -K)w γ,ν b (K). (C.17) Proof of Lemma C.4. Let K + 1 ≤ k ≤ 2K k 1 + k 2 = k 0 ≤ k 1 ≤ k 2 ≤ K then k -K ≤ k 1 ≤ k 2 .
We compute

Γ(ν b + 2 + k 1 )Γ(ν b + 2 + k 2 ) Γ(k -K)Γ(K) = Π k 1 j=0 (ν b + 2 + j)Π k 2 j=0 (ν b + 2 + j) Π K j=0 (ν b + 2 + j)Π k-K j=0 (ν b + 2 + j) = Π k 1 j=k-K+1 ν b + 2 + j ν b + 2 + j + (K -k 1 ) ≤ 1 since k 1 ≤ K. It follows that w γ,ν b (k 1 )w γ,ν b (k 2 ) w γ,ν b (k -K)w γ,ν b (K) ≤ Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(K + α γ -(k -K))Γ(α γ ) (C.18)
For k = 2K, we have k 1 = k 2 = K and from (C.18):

w γ,ν b (k 1 )w γ,ν b (k 2 ) w γ,ν b (k -K)w γ,ν b (K) ≤ Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(α γ )Γ(α γ ) = Γ(α γ )Γ(α γ ) (Γ(α γ )) 2 =1.
We therefore assume

k ≤ 2K -1 and
k 2 ≤ K -1, k 1 ≥ k -K + 1 Using (C.18): Γ(γ -1 -k 1 )Γ(γ -1 -k 2 ) Γ(K + α γ -(k -K))Γ(α γ ) = Γ(α γ + K -k 1 )Γ(α γ + K -k 2 ) Γ(2K + α γ -k)Γ(α γ ) ≤ 1 Γ(α γ ) × αγ +K-k 1 -1 e αγ +K-k 1 -1 αγ +K-k 2 -1 e αγ +K-k 2 -1 αγ +2K-k-1 e αγ +2K-k-1 (α γ + K -k 1 -1)(α γ + K -k 2 -1) α γ + 2K -k -1 1 2 ≤ c Γ(α γ ) (α γ + K -k 1 -1)(α γ + K -k 2 -1) α γ + 2K -k -1 1 2 e -Φ (5) γ,k (k 1 )
where

Φ (5) γ,k (x) = (α γ + 2K -k -1)log(α γ + 2K -k -1) -(α γ + K -x -1)log(α γ + K -x -1) -(α γ + K -(k -x) -1)log(α γ + K -(k -x) -1). satisfies Φ (5) γ,k (x) = log α γ + K -x -1 α γ + K -(k -x) -1 = log 1 + k -2x α γ + K -(k -x) -1 ≥ 0.
Moreover, for A > 0 and |y| A:

AlogA -(A + y)log(A + y) = AlogA -A 1 + y A logA + y A + O y A 2 = -y logA + 1 + O y A (C.19)
Pick a large enough universal constant K ν 1. For

x = k -K + y, 1 ≤ y≤ K ν (C.20)
we use (C.19) to estimate:

Φ (5) γ,k (x) = (α γ + 2K -k -1)log(α γ + 2K -k -1) -(α γ + 2K -k -1 -y)log(α γ + 2K -k -1 -y) -(α γ -1 + y)log(α γ -1 + y) = y [log (α γ + 2K -k -1) + O(1)] -(α γ -1 + y)log(α γ -1 + y) ≥ y 2 logγ (C.21) Similarly, Γ(ν b + k 1 + 2)Γ(ν b + k 2 + 2) Γ(k -K + ν b + 2)Γ(K + ν b + 2) ≤ (ν b + k 1 )(ν b + k 2 ) (k -K + ν b + 1)(K + ν b + 2) 1 2 e -Φ (6) γ,k (k 1 )
where

Φ (6) γ,k (x) = (k -K + ν b + 1)log(k -K + ν b + 1) + (K + ν b + 1)log(K + ν b + 1) -(ν b + x + 1)log(ν b + x + 1) -(ν b + k -x + 1)log(ν b + k -x + 1) satisfies Φ (6) γ,k (x) = log ν b + k -x + 1 ν b + x + 1 = log 1 + k -2x ν b + x + 1 ≥0. Moreover Φ (6) γ,k (x) = (k -K + ν b + 1)log(k -K + ν b + 1) + (K + ν b + 1)log(K + ν b + 1) -(ν b + x + 1)log(ν b + x + 1) -(ν b + k -x + 1)log(ν b + k -x + 1) = (k -K + ν b + 1)log(k -K + ν b + 1) + (K + ν b + 1)log(K + ν b + 1) -(ν b + k -K + 1 + y)log(ν b + k -K + 1 + y) -(ν b + K + 1 -y)log(ν b + K + 1 -y) Let φ(z) = (z + ν b + 1)log(z + ν b + 1) -(ν b + z + 1 + y)log(ν b + z + 1 + y) then φ (z) = log z + ν b + 1 ν b + z + 1 + y = log 1 - y ν b + z + 1 + y < 0 so that φ is deceasing. Then, φ(k -K) ≥ φ(K), which implies Φ (6) γ,k (x) ≥ (K + ν b + 1)log(K + ν b + 1) + (K + ν b + 1)log(K + ν b + 1) -(ν b + K + 1 + y)log(ν b + K + 1 + y) -(ν b + K + 1 -y)log(ν b + K + 1 -y)
Thus, in the regime (C.20), recalling (C.19):

Φ (6) γ,k (x) ≥ y log K + ν b + 1 K + ν b + 1 + O y K = O y 2 K . (C.22)
Also, we have the bound ν e -[(k+ν+1)log(k+ν+1)-(k 2 +ν+1)log(k 2 +ν+1)] We compute:

w γ,ν b (k 1 )w γ,ν b (k 2 ) w γ,ν b (k -K)w γ,ν b (K) ≤ c Γ(α γ ) (α γ + K -k 1 -1)(α γ + K -k 2 -1) α γ + 2K -k -1 (ν b + k 1 )(ν b + k 2 ) (k -K + ν b + 1)(K + ν b + 2)
(k + ν + 1)log(k + ν + 1) -(k 2 + ν + 1)log(k 2 + ν + 1) = k 1 + ν + 1 k logk + log 1 + ν + 1 k -k (1 -x) + ν + 1 k logk + log (1 -x) + ν + 1 k = kx logk + k 1 + ν + 1 k log 1 + ν + 1 k -k (1 -x) + ν + 1 k log (1 -x) + ν + 1 k .
Therefore,

[(k + ν + 1)log(k + ν + 1) -(k 2 + ν + 1)log(k 2 + ν + 1)] = k Φ (2)
k (x) which yields, using also that k 1 ≥ 0, Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ≤ c ν 1 + k 1 e -k Φ (2) k (x) . (D.3)

We have Φ

k (x) is a non-negative non-decreasing function of x ∈ [0, -(ν)/k]. Also, note that for k 1 + (ν) ≤ 0 and provided we choose R > R ν large enough, we have kx = k 1 ≤ |ν| < R so that we do not have to prove (D.2) in that case. This concludes the proof of the case k 1 + (ν) ≤ 0.

step 2 From now on, we focus on the case k 1 + (ν) > 0. In this case, we have

• either (ν) > 0 and x ∈ [0, 1 2 ],

• or (ν) ≤ 0 and x ∈ (- (ν) k , 1 2 ].

Since we also have k 2 + (ν) > 0 and k + (ν) > 0, we may thus apply Stirling's formula to upper bound We compute:

Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2)
(k + ν + 1)log(k + ν + 1) -(k 1 + ν + 1)log(k 1 + ν + 1) -(k 2 + ν + 1)log(k 2 + ν + 1)

= k 1 + ν + 1 k logk + log 1 + ν + 1 k -k x + ν + 1 k logk + log x + ν + 1 k -k (1 -x) + ν + 1 k logk + log (1 -x) + ν + 1 k = -(ν + 1)logk + k 1 + ν + 1 k log 1 + ν + 1 k -k x + ν + 1 k log x + ν + 1 k + (1 -x) + ν + 1 k log (1 -x) + ν + 1 k .
Therefore,

• if (ν) > 0, then

k (x) = ν + 1 k log ν + 1 k + 1 + ν + 1 k log 1 + ν + 1 k - x + ν + 1 k log x + ν + 1 k + (1 -x) + ν + 1 k log (1 -x) + ν + 1 k ,
• if (ν) ≤ 0, then

+ 1 + ν + 1 k log 1 + ν + 1 k - x + ν + 1 k log x + ν + 1 k + (1 -x) + ν + 1 k log (1 -x) + ν + 1 k .
Remark D.2. Note that the above choice for (ν) ≤ 0 is such that Φ

(2) k (x), which is defined in that case on x ∈ (- (ν) k , 1 2 ], and in Step 1 on x ∈ [0, -(ν) k ], is smooth on x ∈ [0, 1 2 ] \ {-(ν) k } and continuous at x = -(ν) k .

The above choice for Φ for x ∈ [0, 1 2 ] if (ν)>0 and for x ∈ (-(ν) k , 1 2 ] if (ν) ≤ 0. Also, we have Φ

k at - (ν) k and its positivity on x ∈ [0, -(ν) k ] established in step 1. Together with step 1, we conclude that Φ

(2) k (x) is a non-negative non-decreasing function of x ∈ [0, 1 2 ].

step 3 Lower bound for small x. Assume

R k < x ≤ 1 2
where R > R ν 1. We start with the case (ν) > 0 for which we have in view of the definition of Φ [START_REF] Collot | On the stability of type I blow up for the energy super critical heat equation[END_REF] k in that case Lemma D.3 (Uniform convolution bound). Let ν ∈ C \ Z -. Then there exists C ν > 0 such that for all k ≥ 1, for all j ≥ 2,

Φ (2) k (x) = ν + 1 k log ν + 1 k + 1 + ν + 1 k log 1 + ν + 1 k - x + ν + 1 k log x + ν + 1 k -(1 -x) + ν + 1 k log (1 -x) + ν + 1 k = (ν) + 1 k log ν + 1 k + 1 + (ν) + 1 k log 1 + ν + 1 k - x + (ν) + 1 k log x + ν + 1 k -(1 -x) + (ν) + 1 k log (1 -x) + ν + 1 k - (ν) k log ν + 1 k + log 1 + ν + 1 k -log x + ν + 1 k -log (1 -x) + ν + 1 k = - (ν) + 1 k log x + ν+1 k ν+1 k + 1 + (ν) + 1 k log 1 + ν + 1 k -xlog x + ν + 1 k -(1 -x) + (ν) + 1 k log (1 -x) + ν + 1 k + O(1) (ν)
+ 1 + ν + 1 k log 1 + ν + 1 k - x + ν + 1 k log x + ν + 1 k + (1 -x) + ν + 1 k log (1 -x) + ν + 1 k = - (ν) logk k + 1 k log -(ν) + ν + 1 k + 1 + (ν) + 1 k log 1 + ν + 1 k - x + (ν) + 1 k log x + ν + 1 k -(1 -x) + (ν) + 1 k log (1 -x) + ν + 1 k - (ν) k log -(ν) + ν + 1 k + log 1 + ν + 1 k -log x + ν + 1 k -log (1 -x) + ν + 1 k = - (ν) k log x + ν+1 k 1 k - 1 k log   x + ν+1 k -(ν)+ν+1 k   + 1 + (ν) + 1 k log 1 + ν + 1 k -xlog x + ν + 1 k -(1 -x) + (ν) + 1 k log (1 -x) + ν + 1 k + O(1) (
k 1 +•••+k j =k Π j i=1 Γ(k i + ν + 2) Γ(k + ν + 2) ≤ C j ν . (D.6)
Proof. We argue by induction on j.

step 1 Case j = 2. We apply Lemma D.1 with some large enough R ν > 1 and assume, without loss of generality, k ≥ k * ν , with k * ν a universal, large enough constant. This leads to

Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ≤ c ν 1 + k 1 e -k Φ (2)
k (x) , where x = k 1 /k. We may assume k 1 ≤ k 2 so that x ≤ 1/2. For x ≤ Rν k , we use the lower bound Φ

k on [0, 1/2] implies

Φ (2) k (x) ≥ Φ (2) k R ν k ≥ + 1 2 R ν k log R ν k ≥ R ν 4k logk
Finally, it remains to prove that, for d = 2, 3, in the case r = r + , {0 < < d} and {d < < +∞} belong to the same connected component of Ω + d . First, recall from Lemma F.4 that, for d = 2, 3, in the case r = r + , ν > 0 on {0 < < d} and {d < < +∞} so that {0 < < d} and {d < < +∞} belong to Ω We then choose the suitable curve in the complex plane as

γ r 0 =    d + ν 0 (d) r 0 e i( π 2 +θ) , - π 2 ≤ θ ≤ π 2    , r 0 1.
For r 0 large enough, γ r 0 clearly has one end on {0 < < d} and the other on {d < < +∞} and is also a curve in C \ {d} ∪ R -. So, in view of the definition of Ω + d , it suffices to prove that ν / ∈ Z -for any ∈ γ r 0 for a suitable choice of r 0 . Now, in view of the above asymptotic for ν, we have for r 0 sufficiently large Proof of Lemma F.1. The proof of (F.3) follows from (2.17), (2.32) which ensure

σ ∞ 3 < σ ∞ 2 = σ ∞ 5 = r * (d, ) √ d d + = √ d + √ d .
Plugging this into (2.41) yields (F.3) via a direct computation. The computation of (F.4) can also be done analytically but is more involved and has been computed with Mathematica.

Observe that r = r * (d, ) for 0 < < d corresponds to parallel slopes at P 2 : (F.9)

c ∞ -= - c ∞ 3 c ∞ 1 = - c ∞
Proof. step 1 Quadratic terms. We compute the values at r * ( ) from (F.1): We then check the sign of N 0 numerically and confirm that ν > 0 for the above claimed range. with (g 1 , g 2 ) given by (H.1). Proof of Lemma H.2. We recall the formulas using the notation (H.6)

d ∞ 20 = - √ d-(d-) + √ d d ∞ 11 = -2d √ d + √ d d ∞ 02 = -d √ d+(d-) √ d + √ d
F 0 = f 00 + f 01 Σ + f 02 Σ 2 F 1 = f 10 + f 11 Σ + O(Σ 2 ) F 2 = f 21 Σ + O(Σ 2 )
and

Φ ± = -|F 1 (Σ)| ± F 1 (Σ) 2 -4|F 2 (Σ)|F 0 (Σ) 2|F 2 (Σ)| .
The difference between the case r o = r * and r o = r + is that f 00 = O(b) in the first case while f 00 = O(b 2 ) in the second case. The case r o = r * being similar and simpler is left to the reader. From now on, we focus the case r o = r + for which we need to prove (H.8). We compute explicitely

f 00 = O(b 2 ) f 10 = |c ∞ 1 + c ∞ 2 | + O(b) > 0 f 21 = -2σ ∞ 2 -e ∞ 20 -d ∞ 20 < 0
Then, remembering Σ < 0 and for 0 < b < b * small enough, yields F 2 > 0, F 1 < 0 for u ∈ [0, 1].

step 1 Taylor expansion. We compute using f 00 = O(b 2 ): We conclude and (H.8) is proved.

Φ + = - 2F 0 (Σ) F 1 (Σ) + F 1 (Σ) 2 -4F 2 (Σ)F 0 (Σ) = -2(f 00 + f 01 Σ + f 02 Σ 2 ) 2f 10 1 + f 11 f 10 Σ + O(b 2 ) = - 1 f 10 f 00 + f 01 Σ + f 02 Σ 2 1 - f 11 f 10 Σ + O(b 2 ) = - 1
H.3. The positivity condition and proof of (9.31) and (9.32). Again, the case r o = r * being similar and simpler is left to the reader, we focus on the case r o = r + for which we need to prove (9.32). We study the positivity condition Appendix I. Numerical computation of specific values of S ∞ (d, )

Φ S > Φ +
Let µ 0 , µ j and ν j , for j = 1, 2, 3, 4 the holomorphic functions introduced in (5.7) and hjk 1 w k 2 . . . w k j+1 (k j+2 w k j+2 ) Π j+2 i=1 Γ(ν + k i + 3) Γ(k -2 + ν + 3)

(h 0 ) k = a k Γ(ν+k+3) (µ 0 ) k h k = a k Γ(ν+k+3) µ k hk = a k Γ(ν+k+3) ν k , introduced in (5.

2 4 . 1 .

 41 Strategy of the proof of Theorem 1.3 4.2. Degeneracy of the geometry at r o (d, ) 4.3. Renormalization 1. Introduction 1.1.

1. 2 .

 2 The self-similar equation. on [0, T ) onto the global in time τ renormalized flow∂ τ ρ + (r -1)ρ + Λρ + ∇ • (ρû) = 0 ∂ τ û + (r -1)û + Λû + û • ∇û + ∇(ρ γ-1 ) = 0 Λ = Z • ∇ (1.6)

shows 1 <

 1 r * (d, ) ≤ r + (d, ) with equality iff = d. (1.13) Let us denote r o (d, ) = r * (d, ) for 0 < < d r + (d, ) for > d. (1.14) Then, in the range of parameters d ≥ 2, > 0, 1 < r < r * (d, ) for < d r * (d, ) < r < r + (d, ) for > d, (1.15)

Theorem 1 . 3 (

 13 Existence of C ∞ asymptotically vanishing self-similar profiles). Let d ≥ 2. Let the critical speed r o (d, ) be given by(1.14). Then there exists a functionS ∞ (d, ) : N * \{1} × O d → R such that for any ∈ O d obeying the condition S ∞ (d, ) = 0,(1.17)there exists a discrete sequencer n < r o (d, ), |r n -r o (d, )| 1 lim n→+∞ r n = r o (d, )

Lemma 1 . 5 ( 1 ) 2 )

 1512 Holomorphic extension). The following hold (For d ≥ 4 the function → S * ∞ (d, ) extends holomorphically to a complex neighborhood of O * d . (For d ≥ 4 the function → S + ∞ (d, ) extends holomorphically to a complex neighborhood of (d, +∞). (3) For d = 2, 3 the function → S * ∞ (d, ) extends holomorphically to a complex neighborhood of (0, d). (4) For d = 2, 3, the function → S + ∞ (d, ) extends holomorphically to an open set Ω + d of the complex plane, with R * + \{d} ⊂ Ω + d , and (0, d) and (d, +∞) belong to the same connected component of Ω + d . We do not know how to compute analytically the zeroes of S ∞ (d, ). However, for d = 2, 3, Lemma 1.5 ensures that, unless the function vanishes identically, the possible zeroes are isolated with possible accumulation points (0, d, +∞) only. For d ≥ 4 the same conclusions can be reached about each of the intervals in O * d and (d + ∞).

  [Numerical study of the zeroes of S ∞ (d, ), case d = 2, 3] In the case d = 2 and d = 3, we have S * ∞ (2, ) > 0 for = 0.1, S * ∞ (3, ) > 0 for = 0.1, (1.18) and S + ∞ (2, ) > 0 for = 0.1, S + ∞ (3, ) > 0 for = 0.1.

Figure 1 .

 1 Figure 1. Phase portrait in the range 1 < r < r * (d, ). Dashed curve is the trajectory of the solution constructed in Theorem 1.3.

Figure 2 .

 2 Figure 2. Phase portrait in the range r * (d, ) < r < r + (d, ), > d. Dashed curve is the trajectory of the solution constructed in Theorem 1.3.

Lemma 1 . 7 (

 17 Repulsivity outside the light cone). Let d = 3, > √ 3 satisfying (1.17), or (d, ) satisfies (1.21).

  .23) Also, by definition, a 1 = b 2 , and we argue below according to the cases a 1 = b 2 = 0 and a 1 = b 2 = 0. Case a 1 = b 2 = 0, i.e. w = 1. If σ = 0, we have the point P 1 . If σ = 0, b 1 = 0 and hence from (2.22) (2.23)

Lemma 2 . 5 (

 25 Comparison of r + and r * ). Let d ≥ 2, then w -= w e (r + ) < 1 r * (d, ) < r + (d, ). and w e < w -⇔ r ≤ r + (d, ).

Figure 3 .

 3 Figure 3. Shape of the phase portrait for 1 < r < r * ( , d)

Figure 4 . 1 , ∆ 2 .step 1

 4121 Figure 4. Shape of the phase portrait for r * ( , d) < r < r + (d, ), > d

σ 2 = 2 and-

 22 -w e (w e -1)(w e -r) + O 1 σ w e (w e -1)(w e -r)

d20 = -(d 20 c 2 -

 2 + d 11 c -+ d 02 ) + c + (e 20 c 2 -+ e 11 c -+ e 02 ) = (c + e 20 -d 20 )c 2 -+ (c + e 11 -d 11 )c -+ c + e 02 -d 02 , (3.18) and d11 = -(2c -c + d 20 + (c -+ c + )d 11 + 2d 02 ) + c + (2e 20 c -c + + e 11 (c -+ c + ) + 2e 02 ) = 2c -c + (c + e 20 -d 20 ) + (c -+ c + )(c + e 11 -d 11 ) + 2(c + e 02 -d 02 ) and d02 = -(d 20 c 2 + + d 11 c + + d 02 ) + c + (e 20 c 2 + + e 11 c + + e 02 ) = (c + e 20 -d 20 )c 2 + + (c + e 11 -d 11 )c + + c + e 02 -d 02 and d30 = -(c 3 -

ẽ20 = (d 20 c 2 -

 2 + d 11 c -+ d 02 ) -c -(e 20 c 2 -+ e 11 c -+ e 02 ) = (d 20 -c -e 20 )c 2 -+ (d 11 -c -e 11 )c -+ d 02 -c -e 02 (3.19) and ẽ11 = 2c -c + d 20 + (c -+ c + )d 11 + 2d 02 -c -(2e 20 c -c + + e 11 (c -+ c + ) + 2e 02 ) = 2c -c + (d 20 -c -e 20 ) + (c -+ c + )(d 11 -c -e 11 ) + 2(d 02 -c -e 02 ) and ẽ02 = d 20 c 2 + + d 11 c + + d 02 -c -(e 20 c 2 + + e 11 c + + e 02 ) = (d 20 -c -e 20 )c 2 + + (d 11 -c -e 11 )c + + d 02 -c -e 02 and ẽ21 = (3c 2

4. 1 .step 1

 11 Strategy of the proof of Theorem 1.3. We describe the main steps of the proof of Theorem 1.3. Renormalization. We introduce a suitable semi classical parameter, see(4.11), 0 < b = o r↑r o (1) and use the geometry of the "eye" P o , P 2 to produce a suitable renormalization.

Lemma 4 . 2 (

 42 Degeneracy in the diagonalized system). Let b = r * -r for < d √ r + -r for > d (4.11)

. 16 ) 4 . 4 .Remark 4 . 5 .

 164445 Remark Unfortunately, we need to keep track of all terms in (4.15) since they will create the limiting problem which, in turn, will give rise to the S ∞ (d, ) function, evaluated numerically. In the quasilinear formulation (4.14), (4.15), u = 0 is P 2 and u = 1 is P o .Proof of Lemma 4.3. This is a brute force computation. step 1 b renormalization. We renormalize (3.13): W = -b w Σ = b 2 σ λ + = bµ +

( 2 )

 2 k (x) ≥ 0 and the bootstrap bounds (5.23) and (5.26) for k j+1 ≤ k -1 to estimate:

6. 5 .step 1

 51 Nonlinear conjugation. The proof of (6.15) now requires a careful track of the b-dependencies in the full problem(4.15). The first step is to use Lemma 6.1 and analyze the nonlinear conjugated problem. Nonlinear conjugation. Recall (4.15):

  .33) Lemma 6.6 (Boundedness for the θ k sequence). There exists c ν,a > 0 and b * (ν, a) such that for all 0 < b < b * , for all 0 ≤ k ≤ K -1,

Lemma 6 . 7 (

 67 Stability by multiplication). Let h(u) = +∞ k=0 b k h k u k with the holomorphic bound |h k | ≤ C k . Then there exists C h and 0 < b * (C h ) 1 such that for all 0 < b < b * (C h ) and any 0

  so by the mean value theorem, u * (α γ ) belongs to the range δb ≤ u ≤ b/δ, and satisfies u * (α γ ) = σb(1 + o b→0 (1)) and (7.28), (7.29) are proved. step 2 Second boundary layer. We introduce the following function

2 withP A 2

 22 = 4 d(d -1) + 4d √ d( √ d -1)(d -) = 4d 2 (d -√ d) + (4d √ d -4d). case r = r + (d, ). Recall A 2 = α(2σ 2 + e 11 )c -+ (e 21 -1)c 2 and hence

2 (d) = d - 2 √ d for d ≥ 5 . ( 9 . 1 )√ 3 <

 25913 Also, recall the definition (1.20) of p. The following implies Lemma 1.7. Lemma 9.1 (Necessary/sufficient conditions for (1.26)). Assume d ≥ 3 and (1.22). For > d, r o = r * (d, ), let 2 (d) be given by (9.1), and assume moreover that ( , d) satisfy: for d = 3 0.11 < < 2 (5) = 0.53 for d = 5 i.e., 9 ≤ p ≤ 37, 0.2 < < 2 (6) = 1.10 for d = 6 i.e., 5 ≤ p ≤ 21, 0.3 < < 2 (7) = 1.71 for d = 7 i.e., 4 ≤ p ≤ 14, 0.45 < < 2 (8) = 2.34 for d = 8 i.e., 3 ≤ p ≤ 9, 0.65 < < 2 (9) = 3 for d = 9 i.e., 3 ≤ p ≤ 7; (9.2)

  3 -c + (c 2 c + + c 4 ) + d 20 c 2 + + d 11 c + + d 02 -c + (e 20 c 2 + + e 11 c + + e 02 ) Σ + c 3 + -dc + -c + (e 21 c 2 + -1) Σ

9. 2 .Lemma 9 . 5 (

 295 Study of F on the P 2 -P 4 separatrix. Value at P 2 ). Under the assumptions of Lemma 9.1, for the P 4 -P 2 separatrix and for r o ( ) -ε(d, ) < r < r o (d, ):

( 9 . 13 )

 913 = 0.1845, 2 (6) = 1.101, 1 (7) = 0.2525, 2 (7) = 1.7085, 1 (8) = 0.3098, 2 (8) = 2.3431, 1 (9) = 0.3589, 2 (9) = 3.Proof of Lemma 9.5. We compute all coefficients at r o (d, ) and argue by continuity for r close enough to r o (d, ). Near P 2 from (8.1), (2.45), (2.50):

  17

  ) with P σ (w) = [(d -1)σ + (r -2)]w 2 + a 1 (σ)w + a 2 (σ) (9.18) andw * = dw e -1 d -1 = (r -1) -1 d -1 . (9.19)

step 2

 2 Root for w 2 ≤ w ≤ 1. In this regime, we have from (9.20), since w ≥ w 2 > w * , only one positive root σ = σ 1 (w) = 1 2 (d -1)(w -w * ) -(d -1)(w --w)(w + -w) (9.21) + [(d -1)(w --w)(w + -w)] 2 + 4 (1 -w)[(1 -w) 2 + w(r -w)] (d -1)(w -w * ) .

5 ≤ d ≤ 12 :

 512 we find Discr d ( ) < 0 in the range (9.2). It fails numerically for d ≥ 13. for d = 3, 1 (3) < 0 and for √ 3 = 1.73 < < 3, Discr 3 ( ) < 0.

2 - 2 = 2 -

 222 e 02 -d 02 + (-4σ 2 -e 11 -d 11 )Φ + (-2σ 2 -e 20 -d 20 )Φ 2 ]Σ + [(d -1)Φ + (1 -e 21 )Φ 2 ]Σ -λ + (1 + c -) -(c 2 + c 1 ) Φ + [-2σ 2 -e 02 -d 02 + (-4σ 2 -e 11 -d 11 )c -+ (-2σ 2 -e 20 -d 20 )c +(-4σ 2 -e 11 -d 11 ) Φ + 2c -(-2σ 2 -e 20 -d 20

  .29) Assuming (9.29), we have 1 -w -w -F > 0, and (9.4) is proved.

+∞

1 2 ×

 12 e -(Φ (5) γ,k (x)+Φ

( 6 )c ν γ γ Rν 8 ≤ c ν . Case 1

 681 γ,k (x)) .Pick a large enough universal constant K ν 1 and recall x = k -K + y.Case y ≥ K ν . Using the monotonicity of the Stirling phases and (C.21), (C.22) we have the lower bound: for K ν universal large enough:k 1 +k 2 =k,y≥Kν ,0≤k 1 ≤k 2 ≤K w γ,ν b (k 1 )w γ,ν b (k 2 ) w γ,ν b (k -K)w γ,ν b (K) ≤ ≤ y ≤ K ν .We estimate:(α γ + K -k 1 -1)(α γ + K -k 2 -1) α γ + 2K -k -1 (ν b + k 1 )(ν b + k 2 ) (k -K + ν b + 1)(K + ν b + 2) = (α γ + 2K -k-y)(α γ + y -1) α γ + 2K -k -1 (ν b + k -K + y)(ν b + K -y) (k -K + ν b + 1)(K + ν b + 2) ≤ c ν α γ to upper bound, using also 0 ≤ k 1 ≤ |ν| and k 1 + ν + 2 ∈ C \ Z -, Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2)

k

  (x) = log(k) + 1 + log (1 -x) + ν + 1 k = log(k) + 1 + log (1 -x) + ν x ∈ [0, -(ν)/k].We conclude, in the case k 1 + (ν) ≤ 0, that Φ

ν k 1 +ν+1 e k 1 +ν+1 √ k 1 + ν + 1 k 2 +ν+1 e k 2 +ν+1 √ k 2 + ν + 1 k+ν+1 e k+ν+1 √ k + ν + 1 ν |k 1 +

 1211 ν + 1|e -[(k+ν+1)log(k+ν+1)-(k 1 +ν+1)log(k 1 +ν+1)-(k 2 +ν+1)log(k 2 +ν+1)]

[

  (k + ν + 1)log(k + ν + 1) -(k 2 + ν + 1)log(k 2 + ν + 1)] = -[(ν + 1)log(ν + 1)] + k Φ

[(k + ν + 1 )

 1 log(k + ν + 1) -(k 2 + ν + 1)log(k 2 + ν + 1)] = -logk -k -(ν) + ν + 1 k log -(ν) + ν + 1 k + k Φ (2) k (x) = -(-(ν) + ν + 1) log (-(ν) + ν + 1) + k Φ

( 2 )

 2 k (x) yields Γ(k 1 + ν + 2)Γ(k 2 + ν + 2) Γ(k + ν + 2) ≤ c ν 1 + k 1 e -k Φ (2) k (x) . (D.5)Recall that we are in the case k 1 + (ν) > 0, i.e. x > -(ν)/k. We have

  if (ν) > 0, and Φ

( 2 )

 2 k (-(ν) k ) ≥ 0 if (ν) ≤ 0 in view of the continuity of Φ

  kwhere we used in the last inequality that | (log(z))| ≤ π for (z) > 0 or (z) = 0. treat the case (ν) ≤ 0. Recall that since x > R k , we are in the case k 1 + (ν) > 0. Then, in view of the definition of Φ

2 +

 2 ν) kwhere we used in the last inequality that| (log(z))| ≤ π for (z) > 0 or (z) = 0. (ν) + 1) 2 + ( (ν)) 2 -1 kx log (R + (ν) + 1) 2 + ( (ν)) 2 1 + ( (ν))

( 2 )

 2 k (x) ≥ 0. For x ≥ Rν k the monotonicity of Φ

  d . It thus remains to prove that they belong to the same connected component of Ω d . To this end, Ω + d being open, it suffices to exhibit a continuous curve in Ω + d with one end on {0 < < d} and the other on {d < < +∞}. Now, from the explicit formula (F.15), we have the following asymptotic as → d ν= ν 0 (d) ( -d) 2 1 + a 1 (d)( -d) + a 2 (d)( -d) 2 + O ( -d)3 , ν 0 (d) > 0.

= 1 2 0 ..= 1 2 0= 2 + O k - 1 2

 101221 -e -2iθ r 0 + ia 1 (d) ν 0 (d)e iθ √ r 0 -a 2 (d)ν 0 (d)e 2iθ + O r -Taking the imaginary part, we see that the above expression crosses R -for θ = θ 0 with θ 0 satisfyingr 0 sin(2θ 0 ) -a 1 (d) ν 0 (d) cos(θ 0 )Plugging back in the above expression, we infer atθ = θ 0 -r 0 cos(2θ 0 ) -a 1 (d) ν 0 (d) sin(θ 0 ) √ r 0 + a 2 (d)ν 0 (d) + O r --r 0 + a 2 (d)ν 0 (d) + O r r 0 (k) := k + a 2 (d)ν 0 (d)ν 0 (d) r 0 (k) e i( π 2 +θ 0 (k)) = -k + 1and hence ν / ∈ Z -for any ∈ γ r 0 (k) provided k ∈ N is chosen large enough.Moreover, µ ∞ + (r * ) = -∂ r λ + (r * ) = -

F. 3 .. 8 )

 38 relation at r * (d, ):2(σ ∞ 2 ) 2 = c ∞ -(2(σ ∞ 2 ) 2 + λ -).(F.7) Signs of d20 , ẽ30 and ν at r * . We compute explicitly the sign of the coefficients d20 , ẽ30 and ν at r * ( ). Lemma F.2. For all d ≥ 2 and 0 < < d, we have d∞ 20 < 0, ẽ∞ 20 > 0. (FAlso recalling (1.20) p = 1 + 4 , (5.2): ν ∞ (d, ) > 0 for d = 2, 0 < < 2 d = 3, 0 < < 3 d = 5, p ≤ 10 d = 6, p ≤ 6 d = 7, p ≤ 4 d = 8, p ≤ 3 d = 9, p ≤ 3.

step 2 3 . 3 2 + d 2 )+ d 2 ( 4 - 12 √ d + 3d + 2d 3 2step 3 3 2 ( 2 - 5 √√ d + 2d -d 3 2 + d 2 )+ d 2 ( 4 - 12 √ d + 3d + 2d 3 2

 23322412333253224123 Computation of d∞ 20 , ẽ∞ 20 and ν ∞ . Recall (F.2) which together with (F.3), (F.6) yields at r * ( We now compute from (5.2), (G.1), (4.22):ν ∞ = -γb( D11 + D30 -Ẽ11 ) = -1 ( d20 ) 2 ẽ20 d11 + (c + -c -)|λ -| d30 -ẽ11 | d20 | = -(1 + √ d) 2( √ d -1)( + d)d(d -) 2 -1 × 4 ( √ d -1) 2 -3 d(1 + √ d) 2 -4 2 d(1 -√ d + 2d -d -d 2 ) + ( √ d -1) 2 d 3 (d -4) . (F.10) Sign of d∞ 20 , ẽ∞20 and ν ∞ . The sign d∞ 20 < 0 follows from its formula. Concerning ẽ∞ 20 , we see from its formula that -ẽ ∞ 20 has the same sign asE 20 := 2 (we infer for ≤ d that E 20 is decreasing and hence for 0 ≤ ≤ d, E 20 ( ) ≤ E 20 (0) = -d that ẽ20 > 0 for all d ≥ 2 and 0 ≤ ≤ d. Concerning ν, we see from its formula that ν has the opposite sign toN 0 := 4 ( √ d -1) 2 -3 d(1 + √ d) 2 -4 2 d(1 --d 2 ) + ( √ d -1) 2 d 3 (d -4).

F. 4 . 2 , 2 H 1 (G. 3 )Polynomials G 1 , G 2 G 1 3 , M 11 = 3 M 12 = Ẽ03 x 2 NLstep 2 + O(b 2 )Σo = -ẽ20 W 2 o 3 = 3 = ∆ 1 -⇒ 2 o 2 + 2 b 2 + O(b 3 )= a 1 b 1 +

 422131213113122222331222231 Slopes and eigenvalues at P 2 for r = r + (d, ). Lemma F.3 (Critical values of the slopes at P 2 ). Let > d, r = r + (d, ) = 1 + d -1 (1 + √ ) D11 = d11 ψo wo|µ + |(c + -c -) D02 = d02 ψ2 o wo |µ + |(c + -c -) D30 = d30 w2 o |µ + |(c + -c -) D21 = d21 ψo w2 o |µ + |(c + -c -) D12 = d12 ψ2 o w2 o |µ + |(c + -c -) D03 = d03 ψ3 o w2 o |µ + |(c + -c -) , Ẽ20 = ẽ20 wo (c + -c -)|λ -| ψo Ẽ11 = ẽ11 wo (c + -c -)|λ -| Ẽ02 = ẽ02 ψo wo (c + -c -)|λ -| Ẽ30 = ẽ30 w2 o (c + -c -)|λ -| ψo Ẽ21 = ẽ21 w2 o (c + -c -)|λ -| Ẽ12 = ẽ12 ψo w2 o (c + -c -)|λ -| Ẽ03 = ẽ03 ψ2 o w2 o (c + -c -)|λ -| (G.1) Remark G.1. Note that the coefficients Dij , Ẽij have a well defined limit D∞ ij , Ẽ∞ ij as b → 0 from (4.16). Polynomials H 1 , H (b, u) = 3 j=0 b j H 1,j (x) H 2 (b, u) = 3 j=0 b j H 2,j (x) (G.2)withH 1,0 (x) = -( Ẽ11 + Ẽ30 ) + ( Ẽ02 + Ẽ21 )x -Ẽ12 x 2 + Ẽ03 x 3 H 1,1 (x) = ( Ẽ02 + Ẽ21 ) -Ẽ12 x + Ẽ03 x 2 H 1,2 (x) = -Ẽ12 + Ẽ03 x H 1,3 (x) = Ẽ03 , H 2,0 (x) = ( D11 + D30 )x -( D02 + D21 )x 2 + D12 x 3 -D03 x 4 H 2,1 (x) = -( D02 + D21 )x + D12 x 2 -D03 x 3 H 2,2 (x) = D12 x -D03 x 2 H 2,3 = -D03 x. (b, u) = Ẽ11 x -(2 Ẽ02 + Ẽ21 )x 2 + 2 Ẽ12 x 3 -3 Ẽ03 x 4 G 2 (b, u) = -D11 x + (2 D02 + D21 )x 2 -2 D12 x 3 + 3 D03 x 4 (G.4) Nonlinear terms NL 1 = 2 j=0 b j NL 1j NL 2 = 2 j=0 b j NL 2j (G.5) with NL 10 = -xM 11 Ψ 2 + x 2 M 12 Ψ 3 NL 11 = M 11 Ψ 2 -2xM 12 Ψ 3 NL 12 = M 12 Ψ -Ẽ02 x + Ẽ12 x 2 -3 Ẽ03 x 20 = -xM 21 Ψ 2 + x 2 M 22 Ψ 3 NL 21 = M 21 Ψ 2 -2xM 22 Ψ 3 NL 22 = M 22 Ψ 3 , M 21 = D02 x -D12 x 2 + 3 D03 x 3 M 22 = D03 x 2 .Computation of ψo . Recall (4.13):Wo = -b (c + -c -)µ + d20 (c + -c -)λ -+ O(b 3) andG 1 = (c + -c -)λ + W + d20 W 2 + d11 W Σ + d02 Σ2 + d30 W 3 + d21 W 2 Σ + d12 W Σ2 + d03 Σ -∆ 1 + c + ∆ 2 , G 2 = (c + -c -)λ -Σ + ẽ20 W 2 + ẽ11 W Σ + ẽ02 Σ2 + ẽ30 W 3 + ẽ21 W 2 Σ + ẽ12 W Σ2 + ẽ03 Σ c -∆ 2 .This yieldsΣo [(c + -c -)λ -+ ẽ11 Wo ] = -ẽ 20 W 2 o -ẽ30 W 3 o + O(b 4 ) ⇒ Σo = -ẽ 20 W 2 o -ẽ30 W 3 o + O(b 4 ) (c + -c -)λ -+ ẽ11 Wo = -1 (c + -c -)λ - ẽ20 W 2 o + ẽ30 W 3 o + O(b 4 ) 1 -ẽ11 Wo (c + -c -)λ - + O(b 2 ) -c -)λ - Wo + O(b 2 )and(c + -c -)µ + b + d20 Wo + d11 Σo + d30 W 2 o + O(b 3 ) = 0 Wo = -1 d20 (c + -c -)µ + b -d11 ẽ20 W (c + -c -)λ - -c -)µ + b + -d11 ẽ20 (c + -c -)λ - + d30 b (c + -c -)µ + d20 O(b 3 ) = -(c + -c -)µ + d20 b --c -)λ - + d30 (c + -c -)µ + d20 a 2 b + O(b 2 )with (a 1 , a 2 ) given by (H.1). This yields -c -)λ - Wo + O(b 2 ) = a 1 ẽ20 (c + -c -)λ - 1 + a 2 b + O(b 2 ) 1 + ẽ30 ẽ20 -ẽ11 (c + -c -)λ - a 1 b + O(b 2 ) = g 1 + bg 2 + O(b 2 )

step 3 2 -f 10 =Φ + = - f 00 f 10 - f 01 f 10 a 1 a 1 bu + - f 01 a 1 a 2 f 10 u + f 01 f 11 -f 02 f 10 f 2 a 1 g 1 u 2 b 2 +

 32101011122 Computation of Φ S . We have for 0 ≤ u ≤ 1 recalling (H.1), (H.4):φ S = u + bu(1 -u)M b Θ S (u) = u + bu(1 -u)Θ 0 + O(ub 2 )and hence from (H.5):Φ S = c --b ψo (c + -c -)φ S -b 2 φ 2 S ψ2 o (c + -c -) + O(b 3 ) = c --b(c + -c -) g 1 + bg 2 + O(b 2 ) u + bu(1 -u)Θ 0 + O(ub 2 ) -b 2 (c + -c -) g 2 1 u 2 + O(b 3 ) = c --b(c + -c -)u g 1 + b (g 2 + g 1 Θ 0 (1 -u)) + O(b 2 ) -b 2 (c + -c -) g 2 1 u 2 + O(b 3 ) = c --b [(c + -c -)g 1 u] + b 2 -(c + -c -)[(g 2 + g 1 Θ 0 )u -g 1 Θ 0 u 2 ] -(c + -c -) g 2 1 u 2 + O(b 3 ) = c --b [(c + -c -)g 1 u] + b 2 -(c + -c -)(g 2 + g 1 Θ 0 )u + (c + -c -)g 1 Θ 0 -(c + -c -) g 2 1 u 2 + O(b 3 )and (H.2) is proved. H.2. Computation of Φ + . We now use the separatrix curve Φ S (u) to parametrize Σ in the eye and compute the root Φ + .Lemma H.2 (Computation of Φ + ). Let Φ + be given by (9.28). Letf 00 = -λ + (1 + c -) f 01 = -2σ 2 -e 02 -d 02 + (-4σ 2 -e 11 -d 11 )c -+ (-2σ 2 -e 20 -d 20 )c 2 f 02 = (d -1)c -+ (1 -e 21 )c -(c 2 + c 1 ) f 11 = (-4σ 2 -e 11 -d 11 ) + 2c -(-2σ 2 -e 20 -d 20 ) f 21 = -2σ 2 -e 20 -d 20 (H.6) then, uniformyl for 0 < b < b * 1 small enough and 0 ≤ u ≤ 1 for r o = r * : bu + O(b 2 ) (H.7) and for r o = r + : O(b 3 ).

F 1 (Σ) 2 -F 1 (

 121 4F 2 (Σ)F 0 (Σ) = (f 10 + f 11 Σ + O(Σ 2 )) 2 -4(f 21 Σ + O(Σ 2 )) f 00 + f 01 Σ + O(Σ ) = f 2 10 + 2f 10 f 11 Σ + O(b 2 ) = f 2 Σ) + F 1 (Σ) 2 -4F 2 (Σ)F 0 (Σ) = f 10 + f 11 Σ + O(b 2 ) + f 10 1 + 2f 11 f 10 Σ + O(b 2 ) = f 10 + f 11 Σ + O(b 2 ) + f 10 1 + f 11 f 10 Σ + O(b 2 ) = 2f 10 + 2f 11 Σ + O(b 2 ) = 2f 10 1 + f 11 f 10 Σ + O(b 2 ) .

f 10 ff 01 f 11 -f 02 f 10 f 2 10 Σ 2 +step 2 Φ + = - f 00 f 10 - f 01 f 10 [a 1 u]b + a 1 a 2 u -a 1 g 1 u 2 b 2 + f 01 f 11 -f 02 f 10 f 2 10 a 2 1 b 2 u 2 + O(b 3 a 1 bu + f 01 f 11 -f 02 f 10 f 2 10 a 2 1 u 2 - f 01 f 10 (a 1 a 2 u -a 1 g 1 u 2 ) b 2 + O(b 3 a 1 bu + - f 01 a 1 a 2 f 10 u + f 01 f 11 -f 02 f 10 f 2 a 1 g 1 u 2 b 2 +

 10111022102112311102231122 00 + f 01 Σ + -f 01 f 11 f 10 + f 02 Σ 2 + O(b 3 ) O(b 3). Expression in terms of u. We recall from (H.3):Σ = -b w∞ u + b 2 w∞ ψ∞ uφ = W∞ u -b W∞ ψ∞ uφ = a 1 b 1 + a 2 b + O(b 2 ) u -b 2 a 1 g 1 u 2 + O(b 3 ) = [a 1 u]b + a 1 a 2 u -a 1 g 1 u 2 b 2 + O(b 3 )and hence O(b 3 )

in the zone 0 ≤ u ≤ 1 u 2 > - f 00 f 10 - f 01 f 10 a 1 bu + - f 01 a 1 a 2 f 10 u + f 01 f 11 -f 02 f 10 f 2 a 1 g 1 u 2 b 2 +withA 0 = f 00 f 10 A 1 = 2 10 a 2 1 -- - f 01 f 10 c 1 +c 2 B 1 =ẽ20 λ -+ f 01 f 10 B 3

 12101122101211021103 which becomes-b [(c + -c -)g 1 u] + b 2 -(c + -c -)(g 2 + g 1 Θ 0 )u + (c + -c -)g 1 Θ 0 -(c + -c -) g 2 1 O(b 3 ) ⇔ A 0 + bA 1 u + b 2 (A 3 u + A 4 u 2 ) + O(b 3 ) > 0 -(c + -c -)g 1 + f 01 f 10 a 1 A 3 = f 01 f 10 a 1 a 2 -(c + -c -)(g 2 + g 1 Θ 0 ) A 4 = -f 01 f 10 a 1 g 1 + f 02 f 10 -f 01 f 11 f (c + -c -) g 2 1 + (c + -c -)g 1 Θ 0 We compute A 0 = f 00 f 10 = -λ + (1 + c -) -(c 1 + c 2 ) = bµ + 1 + c - c 1 + c 2 and A 1 = -(c + -c -)g 1 + f 01 f 10 a 1 = (c + -c -) ẽ20 µ + λ -d20 + f 01 f 10 -(c + -c -)µ + d20 = (c + -c -)µ + d20 ẽ20 λWe therefore divide by |µ + | = -µ + and obtain the conditionB 0 + bB 1 u + b 2 (B 3 u + B 4 u 2 ) + O(b 3 ) > 0 (H.9) with B 0 = -b(1+c -) c + -c - d20 -= A 3 |µ + | B 4 = A 4 |µ + | , (H.10)this is (9.32).

  20). Let (w k ) k≥0 be the sequence defined byw k+1 + w k = 1 a (h 0 ) k+1 + a (k + ν + 3)(k + ν + 2) 4 j=1 k 1 +...k j+1 =k-1 h jk 1 w k 2 . . . w k j+1 Π j+1 i=1 Γ(ν + k i + 3) Γ(k -1 + ν + 3) + a 2 Π 3 j=1 (k + ν + j) 4 j=1 k 1 +...k j+2 =k-2

  e 20 c 2 -+ e 11 c -+ e 02 ) W 2 + (2e 20 c -c + + e 11 (c -+ c + ) + 2e 02 ) W Σ + (e 20 c 2 + + e 11 c + + e 02 ) Σ2

  No oscillation at the left of P 2 . In the variable Θ, it is easily seen that the P 2 -P o separatrix satisfies |Θ| ,d 1, Lemma 7.11. Hence we pick a large enough (in absolute value) constant Θ *1 and aim at reaching the value

.7) step 4

see(2.29)

i.e. the only solution curve connecting P2 to P5.

x = logZ of the Emden transform (1.8).

( + √ d) 4 √ d d 2 ( √ d -1) + √ d(d -1) + 2 -d( √ d -1) + ( √ d + 1)2

( + √ d) d( √ d -1) + ( √ d + 1) 2 √ d -d( √ d -1) + ( √ d + 1) = -2d √ d( √ d -1)(d -) ( + √ d) d( √ d -1) + ( √ d + 1)

(d) (9.7) = -(d √ d -d + 5 √ d -1) + (d √ d -d + 5 √ d -1)

+ 4(d √ d -d -2 √ d)( √ d + 1)

We will sometime denote by a subscript S, e.g. ΦS

we mean here the slope of the line segments from P2 to points on the separatrix.

(1 -√ d) -( √ d + d + d √ d -2 √ d -1) -2d √ d -√ d + d √ d -d

+ 3 √ d = -( √ d -1) 2 + (d √ d + d -√ d -1) + d √ d + d -2 √ d < 0.

+ (2 √ d -d) +

Appendix C. Study of the weight w γ,ν b for k ≤ K Appendix D. Uniform convolution bounds for b = 0 with ν complex Appendix E. Proof of Lemma 1.5 Appendix F. Slopes and eigenvalues at P 2 near the critical value F.1. Values of the parameters F.2. Slopes and eigenvalues at P 2 for r = r * (d, ) F.3. Signs of d20 , ẽ30 and ν at r * F.4. Slopes and eigenvalues at P 2 for r = r + (d, ) F.5. Signs of d20 , ẽ30 and ν at r + Appendix G. Expansion of the functionals in (4.15) Appendix H. Proof of (9.31) and (9.32) H.1. Computation of Φ S H.2. Computation of Φ + H.3. The positivity condition and proof of (9.31) and (9.32) and the smooth solution crosses ∆ 1 = 0 for u ≥ -3/4. second layer: if α γ is given by (7.66), there exists U * (α γ ) such that

and the smooth solution crosses ∆ 2 = 0 for u ≥ -3/4. This concludes the proof of Lemma 7.10.

7.9. Proof of Theorem 1.3. We are now in position to conclude the proof of Theorem 1.3.

step 1 Continuous deformation of α γ . Let K be even and large enough, we claim that there exists α K γ in (0, 1) such that the C ∞ solution Φ[K, α K γ ](u) coincides with the unique P 6 -P 2 solution (to the right of P 2 ) and exits to the left of P 2 by crossing ∆ 1 = 0 before reaching P 5 (and, as a result, extends to P 4 .).

Indeed, let 1 δ , Θ * 1 large enough and 0 < b < b * (Θ * , δ) small enough as in Lemma 7.9, and α γ in the range (7.47). Assume also that K is even. Then, in view of Lemma 7.9, the C ∞ solution exits on the left of P 2 by crossing ∆ 1 = 0 before u = 3 4 . Consider then the C ∞ solution Φ[K, α γ ](u) on the right of P 2 . Let also Φ (rad) [K, α γ ](u) denote the unique solution, constructed earlier, which corresponds to the P 2 -P 6 trajectory. Then, let

Then, for α γ given by (7.65), since Φ (rad) [K, α γ ](u) is located between the ∆ 1 = 0 and ∆ 2 = 0 curves for any u < 0, and, since Φ[K, α γ ](u) has crossed ∆ 1 = 0 before u = -3/4 and cannot cross ∆ 1 = 0 twice, we deduce Hence P 0 (w 2 ) < 0 implies that

Let w 3 (σ) parametrize the line (w 3 (σ), σ) ∈ (D), then we also have P σ (w 3 (σ)) < 0 from (9.23). We now distinguish three cases: case (d -1)σ + (r -2) > 0. Since P σ (w) is second order in w with a positive highest order coefficient,

which, together with the fact that (9.8) holds on the solution curve, implies that 1 -w -w + F > 0 along the trajectory. The function 1-w -w +F is strictly positive at P 2 , converges to 1 as σ → 0 and can not vanish. It implies

Since, by the Remark 9.3, F ≤ 0 at the left of P 2 on the solution curve, (9.4) follows.

case (d -1)σ + (r -2) < 0. In this case, for each 0 ≤ σ < σ 2 , P σ (w 2 ) < 0. On the other hand, we also have the curve σ 1 (w) which, as we vary w ∈ [w 2 , 1] connects σ = σ 2 and σ = 0, and on which P σ (w) = 0. Since the point (0, 1) belongs to this curve and lies above the line w = w 2 , the whole curve σ 1 (w) must lie above the line w = w 2 . In particular, for each 0 ≤ σ < σ 2 we can find a (possibly non-unique) value w 1 (σ) such that P σ (w 1 ) = 0 and w 1 (σ) > w 2 . This implies P σ (w) < 0 for w < w 2 , thus, in particular, on the solution curve, and the conclusion follows as above.

case (d -1)σ + (r -2) = 0. Since P σ (w) is first order in w, this implies P σ (w) < 0 for w 3 (σ) ≤ w ≤ w 2 and we conclude as above. 9.4. Positivity in the region where w ≥ w 2 . We now are in position to conclude the proof of (9.4) and of Lemma 9.1. To this end, it suffices to prove the following lemma concerning the region w ≥ w 2 . Lemma 9.9 (Positivity in the region where σ 1 ≤ σ ≤ σ 2 and w ≥ w 2 ). Under the assumptions of Lemma 9.1, any P 2 -P 4 curve with c -slope at P 2 satisfies (9.4) in the region σ 1 ≤ σ ≤ σ 2 and w ≥ w 2 .

Proof. For w ≥ w 2 , in the region ∆ 1 ≥ 0, we have

since ∆ 2 > 0, ∆ > 0 and 1 -w-σ > 0 on the solution curve for Σ < 0. Thus, it suffices to consider the region Σ < 0 before the solution curve crosses the middle root of ∆ 1 = 0. Note in particular that this region is included in

step 2 Proof of (9.29). We reexpress (9.29) using the renormalization (4.14)

and for r o = r + :

Then, (9.29) follows from the statement: ∃c(d,

and for r o = r + :

The proof of (9.32) is detailed in Appendix H together with the explicit computation of the constants B 0 , B 1 , B 3 , B 4 given by (H.10) which allows us to conclude the proof of (9.34) below. case r o = r * (d, ). We have

The inequality above follows by a direct check. Then, uniformly in b small enough and u ∈ [0, 1]:

We compute:

and

We have that Q(d, , u) is strictly positive on 0 ≤ u < u( , d) where

Also, since → u( , d) is increasing for ≥ 0, we have

Now, d → u(0, d) attains its minimum for d = 16 and we find

. This case is degenerate since an explicit computation shows that

and, as a result, we need to consider the order b 2 . We compute, using Appendix H:

Then (9.34) follows from

which is immediate from the above formula for F ( , d, u).

Appendix A. Facts related to the Γ function

We collect various classical facts about the Γ function.

Euler β function. B(x, y) is defined for all (x, y) with (x) > 0 and (y) > 0 by

Proof. By definition

Π J j=0 (x + j) and thus, with J = K x -1 and using (A.4),

Appendix B. Uniform convolution bounds for b = 0

step 2 Induction. We assume j and prove j + 1:

Appendix C. Study of the weight w γ,ν b for k ≤ K

We derive estimates and convolution bounds for the weight w γ,ν b (k) given by (6.11). We start with estimating the weight.

Lemma C.1 (Estimates on the weight).

There exist a universal constant c > 0 and γ * 1 such that for al γ > γ * , the following holds. Let 0 ≤ k < γ -1, then:

Proof. From (6.11), we have

and thus

Lemma C.2 (Summation bound). For some c ν,a > 0, K ν 1 and all 0 < b < b * (ν) we have the following uniform bounds:

Proof. step 1 Away from the boundary. Pick a large enough universal integer K ν 1 and assume first k ≤ K -K ν . Then,

where we used 2K -k -1 ≥ 0. We then conclude, using the positivity of the total Stirling phase, since there are finitely many terms:

The collection of the above bounds yields (C.17).

Appendix D. Uniform convolution bounds for b = 0 with ν complex

This appendix is the extension of Appendix B to the case where ν is complex. We use the standard formulas log(re iθ ) = log(r) + iθ, arg(x + iy) = 2 arctan y

where we have chosen the principal branch of the logarithm, i.e., on C \ R -.

Lemma D.1 (Lower bound on the Stirling phase).

1 such that for all k > k * , for all

we have

where the phase function satisfies 25

Proof of Lemma D.1. We assume k ≥ k * (R, ν) large enough, and consider two cases:

step 1 We consider first the case k 1 + (ν) ≤ 0. In this case, we have

Note that, since k ≥ k * (R, ν) is large enough, we have k 2 + (ν) > 0 and k + (ν) > 0. We may thus apply Stirling's formula 26

25 In the case where (ν) > 0, Φ

k is smooth. In the case (ν) ≤ 0, Φ

is smooth except at the point x = -(ν) k where it is only continuous, see Remark D.2. In particular, at that point, ( Φ

k ) (x) ≥ 0 means that the left and right derivative, although not equal, are both positive. 26 In what follows, we apply Stirling's formula for z satisfying (z) > 0. In particular, √ z and log(z) are both defined.

for R ν ≤ k * ν , from which

since the second sum has k 1 = xk ≤ R ν terms.

step 2 Induction. We assume j and prove j + 1:

Appendix E. Proof of Lemma 1.5

This Appendix is devoted to the proof of Lemma 1.5 which is a direct consequence of analyticity. We study the function S ∞ (d, ) from (7.23) which arises from the limiting problem (5.4). We recall that there are in fact two different limiting problems which correspond to the parameters r = r * , r + associated to the respective ranges 0 < < d and > d. Each of the limiting problem gives rise to a collections of coefficients: ν ∞ , µ + , c -, , c + , ... each of which is a function of . In fact, each of them is a different function of depending on the case r * or r + , see Appendix F. Let us associate superscript * to the former and + to the later. Since r + corresponds to the range > d, the coefficients ν + ∞ , ... are originally defined for the same range of but, by the direct examination of Appendix F, can be extended through the same formula to the interval ∈ (0, d). In fact, they can be similarly extended as holomorphic functions to a small complex neighborhood of R + \ {d}. In the * case, the function ν * ∞ is originally positive on the subset O * d ⊂ (0, d). Again, by examining the formulas in Appendix F, we can conclude that ν * ∞ , ... are holomorphic functions of in a small complex neighborhood of O * d (we do not need to extend them to > d).

We We emphasize that the extensions above are not abstract but follow from extending the values of to the complex plane in explicit formulas.

Proof. Note from their definition that all the constants appearing in the definition of the holomorphic functions µ 0 , µ j and ν j , i.e.,

are rational functions of , √ and 1 4 , and hence are holomorphic in for ∈ C \ R - wherever the denominators do not vanish. These denominators are given by the following list

For real values of , these denominators vanish at = 0, = d, as well as at certain negative values of which explicitly depend on d. In particular, we deduce that all the above constants are holomorphic in in a small neighborhood of R + \ {d}. This applies both to the * and the + case and immediately implies that all the Taylor coefficients (µ j ) k , (ν j ) k and (µ 0 ) k of µ j , ν j and µ 0 are holomorphic in on the same set.

We now consider the set Ω + d obtained the intersection of the above small neighborhood of R + \ {d} with the set (ν + ∞ ) -1 C \ Z -. Note that since by (F.15) the function ν + ∞ > 0 for all > 0, the set Ω + d contains R + \ {d}.

In the * case, we define the set Ω * d to be simply a small complex neighborhood of O * d . Note that since ν * ∞ > 0 on O * d , the condition that ν ∈ C \ Z -is automatically satisfied on Ω * d , provided the neighborhood is chosen to be sufficiently small.

Our goal is now is to show that S * ∞ and S + ∞ are holomorphic respectively on Ω * d and Ω + d and that R + \ {d} belongs to the same connected component of Ω + d . The next argument applies to both, so will simply use the notations S ∞ and Ω d .

In view of the definitions of Ω d , in particular, for any k, ν

We deduce that all the Taylor coefficients (h j ) k , ( hj ) k and (h 0 ) k of h j , hj and h 0 are holomorphic in on Ω d . In view of the definition of w k , we conclude that for any k, w k is holomorphic in on Ω d . In particular, this is also true for (-1) k w k . As a result, for any k, S k is holomorphic in on Ω d .

To conclude that S ∞ (d, ) is holomorphic in on Ω d , it suffices to prove that S k (d, ) converges uniformly to S ∞ (d, ) on any compact of Ω d . Note that Lemma 5.4 implies that S k (d, ) converges uniformly to S ∞ (d, ) on any compact of 0 < < d in R, so it suffices to prove that the conclusion of Lemma 5.4 still holds on Ω d . Now, a quick inspection of the proof of Lemma 5.4 reveals that the only obstruction is that the conclusions of Appendix B do not hold if ν ∈ C \ Z -. The fact that these conclusions hold even if ν ∈ C \ Z -has been checked in Appendix D. Thus S ∞ (d, ) is holomorphic on Ω d .

Appendix F. Slopes and eigenvalues at P 2 near the critical value

In this appendix we collect the values of the parameters which appear in the computations near P 2 for r < r o close to r o . F.1. Values of the parameters. For 1 < r < r + (d, ): (F.2) F.2. Slopes and eigenvalues at P 2 for r = r * (d, ). We compute the slopes and eigenvalues at P 2 for the critical speed r = r * (d, ), 0 < < d.

Lemma F.1 (Critical values of the slopes at P 2 ). Let

then at P 2 :

(F.3)

Proof. This follows from a direct computation. (F.12) has been computed with Mathematica.

F.5. Signs of d20 , ẽ30 and ν at r + . We compute explicitly the sign of the coefficients d20 , ẽ30 and ν at r * ( ).

Lemma F.4. For all d ≥ 2 and > d, we have d∞ 20 < 0, ẽ∞ 20 > 0, ν ∞ > 0.

(F.13)

Proof. We collect the values

We compute directly

) . and

and the claim is proved. We note here that this function ν ∞ is positive for every value of > 0.

Appendix G. Expansion of the functionals in (4.15)

In this Appendix we collect all the formulas for all the terms appearing in (4.15). We recall x = bu.

Appendix H. Proof of (9.31) and (9.32) This section is devoted to the derivation of the estimates (9.31) and (9.32) by computing both the P 2 -P o separatrix Φ S and the root Φ + in the variables of the renormalization (4.14).

g 2 = a 1 ẽ20 (c + -c -)λa 2 + ẽ30 ẽ20 - We check numerically that S ∞ (d, ) = 0 in the following cases: cases < d.

(