Non uniform rotating vortices and periodic orbits for the two-dimensional Euler Equations
Claudia García Taoufik Hmidi Juan Soler



This paper concerns the study  of  some special ordered   structures   in turbulent  flows. 
In particular,  a systematic and relevant methodology    is proposed to construct non trivial and non radial rotating vortices  with non necessarily uniform densities and with different m–fold symmetries, m≥ 1.   In particular, a complete  study is provided for the truncated quadratic density (A|x|^2+B)1_(x), with  the unit disc. We exhibit different behaviors with respect to the coefficients A and B describing the rarefaction of bifurcating  curves. 




𝒫𝒯 symmetry and anti-symmetry by anti-Hermitian wave coupling and nonlinear optical interactions
    Stefano Longhi
    December 30, 2023
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§ INTRODUCTION


The main goal of this paper is to investigate  the emergence of  some special ordered   structures   in turbulent  flows. 
The search for Euler and Navier-Sokes solutions  is a classical problem of permanent relevance that seeks to understand the complexity and dynamics of certain singular structures in Fluid Mechanics. Only a few solutions are known without much information about their dynamics. 

We will focus on the two-dimensional Euler equations that  can be  written in  the velocity–vorticity formulation as follows

    {[   ω_t+(v·∇) ω=0,   in [0,+∞)×ℝ^2,;           v=K*ω,   in [0,+∞)×ℝ^2,; ω(t=0,x)=ω_0(x),      with x∈ℝ^2. ].

 The second equation  links the velocity to the vorticity through the so called Biot–Savart  law, where K(x)=1/2πx^⊥/|x|^2.
This is a Hamiltonian system that develops   various  interesting behaviors at different levels, which are in the center of intensive  research activities.  Lots of studies have been devoted to the existence and stability of relative equilibria (in general, translating and rotating steady-state solutions called V-states).
We point out  that  despite the complexity of the motion  and the deformation process that the vorticity undergoes, some special vortices  subsist  without any deformation and keep their shape during the motion. These fascinating   and intriguing  structures   illustrate somehow the emergence of the order from disordered and chaotic  motion.
 The first known example in the literature goes back to Kirchhoff who discovered that a vorticity uniformly distributed inside an elliptic shape performs a uniform rotation about its center  with constant angular velocity. Notice  that the solutions of vortex patch type (solutions with piecewise constant vorticity)  have motivated important mathematical achievements in recent years. For example, the  existence of global solutions in this setting  is rigorously obtained by Yudovich  <cit.>. The L^1 assumption can be replaced by a m–fold condition of symmetry thanks to the work of Elgindi and Jeong <cit.>.
The main feature of  the vortex patch problem is the persistance of this  structure due to the  transportation  of the vorticity by the flow.  However, the  regularity persistence of the boundary with 𝒞^1,α-regularity is delicate and  was first
shown by Chemin <cit.>, and then via different techniques by Bertozzi and Constantin <cit.> and Serfati <cit.>.  Coming back to the emergence of relative equilibria, uniformly rotating m-fold patches with lower symmetries generalizing Kirchhoff ellipses  were discovered numerically  by Deem and Zabusky <cit.>. 
Having this kind of  V-states solutions in mind, Burbea <cit.> designed a  rigorous approach  to generate them close to a Rankine vortex through  complex analytical tools and bifurcation theory. Later this idea was improved and extended to different directions: regularity of the boundary, various topologies, effects of the rigid boundary, and different nonlinear transport equations. For the first subject, the  regularity of the contour was analyzed in <cit.>. There, it was proved that close to the unit disc the boundary of the rotating patches are not only 𝒞^∞ but analytic. As to the second point,  similar results with   richer structures  have been obtained  for  doubly connected patches <cit.>. The existence of small loops in the bifurcation diagram has been  achieved very recently in <cit.>.  For disconnected patches, the existence of co-rotating and counter–rotating vortex pairs was discussed in 
<cit.>. We mention that the bifurcation theory is so robust that partial results have been extended to different models such as the generalized surface quasi-geostrophic equations <cit.> or Shallow-water quasi-geostrophic equations <cit.>, but the computations turn  out to be much more involved in those cases.

It should be noted that the particularity of the  rotating patches is that the dynamics is reduced to the motion of a finite number of  curves in the complex plane, and therefore the implementation of the bifurcation is straightforward. However, the construction of smooth rotating vortices is much more intricate due to the size of the kernel of the linearized operator, which is in general infinite dimensional because it contains at least every radial function. Some strategies  have been elaborated in order to capture some non trivial rotating smooth solutions. The first result amounts  to Castro, Córdoba  and Gómez-Serrano
 <cit.> who established for the SQG and Euler equations the existence of 3-fold smooth rotating vortices using a reformulation of the equation through the level sets of the vorticity. However the spectral study turns to be highly complex and they use  computer–assisted proofs  to check  the suitable  spectral properties. In a recent paper 
<cit.> the same   authors removed the computer assistance part and proved the existence of 𝒞^2 rotating vorticity with m-fold symmetry, for any m≥2. The  proof relies on the desingularization and bifurcation from the vortex patch problem. We point out that the profile of the  vorticity is constant outside a very thin region where the transition occurs, and the  thickness of this region serves as a bifurcation parameter. Remark  that different   variational arguments  were developed in <cit.>.

The  main objective of this paper is to construct a systematic scheme  which turns to be relevant  to detect non trivial  rotating vortices    with non uniform densities, far from  the patches but close to some known  radial profiles. Actually, we are looking for compactly supported rotating vortices  in the form

    ω(t,x)=ω_0(e^-itΩ x),   ω_0=(f∘Φ^-1) 1_D,   ∀ x∈^2,

where Ω is the angular velocity, 1_D is  the characteristic function of  a smooth simply connected domain D, the real function f:→ denotes  the density profile  and Φ:→ D is a conformal mapping from the unit disc  into  D. It is a known fact that an initial vorticity ω_0 with velocity v_0 generates a rotating solution, with constant angular velocity Ω, if and only if

    (v_0(x)-Ω x^⊥)·∇ω_0(x)=0,  ∀ x∈^2.

Thus the ansatz (<ref>) is a solution of Euler equations (<ref>)   if and only if the following  equations 

    (v(x)-Ω x^⊥)·∇ (f∘Φ^-1)(x)   =   0,  in   D,
    
    (v(x)-Ω x^⊥)·(f∘Φ^-1)(x)n⃗(x)   =   0,  on  ∂ D,

are  simultaneously satisfied, where n⃗ is the upward unit normal vector  to the boundary ∂ D. Regarding its relationship with the issue of finding vortex patches, the problem presented here exhibits a greater complexity. While a rotating vortex patch solution can be described by the boundary equation (<ref>),   here we also need to work with the corresponding coupled density equation (<ref>). One major problem that one should face in order to make the bifurcation argument useful is related to the size of the kernel of the linearized operator which is in general infinite-dimensional. In  the vortex patch framework  we overcome this difficulty using  the contour dynamics equation  and by imposing a suitable symmetry on the V-states: they should be invariant by the dihedral group D_m.  In this manner we guarantee that the linearized operator becomes a Fredholm operator with zero index. 
In the current context,  we note that all smooth radial functions belong to the kernel. One possible strategy that one could implement is to  filter those non desirable functions from the structure of the function spaces by removing the mode  zero. However, this attempt fails because the space will not be stable by the nonlinearity especially for the density equation  (<ref>): the frequency zero can be obtained from a resonant regime, for example the square of a non vanishing  function  on the disc generates always the zero mode. Even though, if we assume that we were able to  solve this technical problem  by some special  fine  tricks, a  second but more delicate one arises   with the formulation (<ref>).  The linearized operator around any radial solution is not of Fredholm type: it is smoothing in the radial component. In fact, if ω_0 is radial, then the linearized operator associated with the nonlinear map

    F(ω)(x)=(v(x)-Ω x^⊥)·∇ω(x),

is given in polar coordinates  by

    ℒ h=(v_θ^0/r-Ω)∂_θ h+K(h)·∇ω_0,   K(h)(x)=1/2π∫_^2(x-y)^⊥/|x-y|^2h(y)dy.
 
The loss of information in the radial direction can not be compensated by the operator K which is compact. This means that when using standard function spaces,  the  range of the linearized operator will be of infinite codimension. This discussion illustrates the limitation of working directly with the model  (<ref>). 
Thus, we should  first proceed  with   reformulating  differently the equation  (<ref>) in order to avoid the preceding technical problems  and capture non radial solutions by a bifurcation argument. We point out that the main obstacle comes  from the density equation (<ref>) and the elementary key observation is that a solution to this  equation means that the density is constant along the level sets of the relative stream function. This can be guaranteed if one looks for solutions to the restricted problem,

    G(Ω,f,Φ)(z)≜ℳ(Ω,f(z))+1/2π∫_Dlog|Φ(z)-Φ(y)|f(y)||Φ^'(y)|^2 dy-1/2Ω|Φ(z)|^2
    =0,
 
for every z∈, and for some suitable real function ℳ. The free function ℳ can be fixed  so that the radial profile is a solution. For instance, as it will be shown in Section <ref>, for the radial profile 

    f_0(r)=Ar^2+B

we get the explicit form

    ℳ(Ω,s)=4Ω-B/8A s-1/16 A s^2+3B^2+A^2+4AB-8Ω B/16 A.

Moreover, with this reformulation, we can ensure that no other radial solution can be captured around the radial profile except for a singular value, see Proposition <ref>.

Before stating our result we need to introduce the following set, which is nothing but the singular set introduced later, see (<ref>), in the case of the quadratic profile,

    𝒮_sing={A/4+B/2-A(n+1)/2n(n+2)-B/2n,     n∈^⋆∪{+∞}}.

The  main result of this paper concerning the quadratic profile is the following. 
 
Let A>0, B∈ and m a positive integer. Then the following results hold true.


  * If A+B<0, then there is    m_0∈ (depending only on A and B)  such that for any m≥ m_0, there exists a branch of non radial rotating solutions 
with m-fold symmetry  for the Euler equation, bifurcating from the radial solution  (<ref>)   at some given Ω_m>A+2B/4.

  * If B>A, then for any integer m∈[1, B/A+1/8] or m∈[1,2B/A-9/2] there exists a branch of non radial rotating solutions  with m-fold symmetry  for the Euler equation, bifurcating from the radial solution (<ref>) at some given 0≤Ω_m<B/2. However, there is no bifurcation for any symmetry .

  * If B>0 or B≤-A/1+ϵ for some 0,0581<ϵ<1, then there exists a branch of non radial  1-fold symmetry rotating solutions    for the Euler equation, bifurcating from the radial solution (<ref>)   at Ω_1=0. 

  * If 
- A/2< B< 0 and Ω∉𝒮_sing,   then there is no solutions to (<ref>) close to the quadratic profile.

  * In the frame of the rotating vortices constructed in (1),(2) and (3), the particle trajectories inside their supports are concentric periodic orbits around the origin.  
 

This  theorem will be fully detailed in Theorem <ref>, Theorem <ref> and Theorem <ref>. Before giving some details about the main ideas of the proofs, we wish to draw some useful comments:


  * The upcoming Theorem <ref> states that the orbits associated with (<ref>) are periodic with smooth period, and at any time the flow is invariant by a rotation of angle 2π/m. Moreover, it generates a group of diffeomorphisms of the closed unit disc.

  * The V-states  constructed in the above theorem have the form f∘Φ^-11_D. Also, it is proved  that the density f is 𝒞^1,α() and the boundary ∂ D is 𝒞^2,α with  α∈ (0,1). We believe that by implementing the techniques used in <cit.> it could be shown  that the density and the domain are analytic. An indication  supporting this intuition   is provided by the generator of the kernel associated with the density equation, see (<ref>), which is analytic up to the boundary.

  * The dynamics of the  1–fold symmetric V-states is rich and  very interesting. The branch can survive even in the region where no other symmetry is allowed. It should be noted  that in the context of vortex patches the bifurcation from the disc or the annulus  occurs only  with symmetry m≥2 and never with the symmetry 1.  The only examples that we know in the literature about the emergence of the symmetry one is the bifurcation from Kirchhoff ellipses <cit.> or the presence of the boundary effects <cit.>.   

  * From the homogeneity of Euler equations the transformation (A,B,Ω)↦ (-A,-B,-Ω) leads to the same class of solutions  in Theorem <ref>. This observation allows including in the main theorem the case A<0.

  * The assumptions on A and B seen in  Theorem <ref>-(1)-(2) about the bifurcation cases imply that the radial profile f_0 is not changing the sign in the unit disc.  However in the point  (3) the profile can change the sign.

  * The bifurcation with m-fold symmetry, m≥1,  when B∈(-A,-A/2)  is not well understood. We only know that  we can obtain a branch of 1-fold symmetric solutions bifurcating from Ω_1=0 for B∈(-A,-A/1+ϵ) for some ϵ∈(0,1), nothing is known for other symmetries. We expect that similarly to the result of Theorem <ref>-(2), they do exist but only for lower frequencies, and the bifurcation curves are rarefied when B approaches -A/2.

  * Let us remark the existence of solutions with lower m-fold symmetry  coming from the second point of the above Theorem. Fixing A, the number of allowed symmetries increases when |B| increases. We guess that there is a smooth curve when passing from one symmetry to another one, see Fig.1. 





Let us briefly outline the general strategy we follow to prove the main result and that could be implemented for more general profiles. Using the conformal mapping Φ we can translate the equations (<ref>)–(<ref>) into the disc  and its boundary . Equations (<ref>)–(<ref>) depend functionally on the parameters (Ω, f, Φ), so that we can write them  as

    {[ G(Ω,f,Φ)(z)           =  0,   ∀ z∈,; F(Ω,f,Φ)(w)           =  0,  ∀ w∈ , ].

with

    F(Ω,f,Φ)(w)≜Im[ (Ω Φ(w)-1/2π∫_f(y)/Φ(w)-Φ(y)|Φ^'(y)|^2 dA(y))Φ^'(w)w]
    =0,  ∀ w∈ ,

where the functional G is described in (<ref>).
The aim is to parametrize the solutions in (Ω,f,Φ) close to some initial radial solution (Ω, f_0,Id), with f_0 being a radial profile and Id  the identity map.   Then, we will deal with the unknowns g and ϕ defined by

    f=f_0+g,   Φ=Id+ϕ.

Thus,  the equations in (<ref>) are parametrized in the form G(Ω,g,ϕ)(z)=0 and F(Ω,g,ϕ)(w)=0, where   G(Ω,0,0)(z)=0 and F(Ω,0,0)(w)=0. The idea is to start by solving the boundary equation, which would reduce a variable  through a mapping (Ω,f)↦Φ=𝒩(Ω,f), i.e. to prove that under some restrictions  F(Ω,g,ϕ)=0 is equivalent to ϕ=𝒩(Ω,g). However, the argument stumbles when we realize that this can only be done outside  a set of singular values

    𝒮_sing≜{Ω : ∂_ϕ F(Ω,0,0)  },

for which the Implicit Function Theorem can be applied. Then, we prove that there exists  an open interval I for Ω such that  I⊂\𝒮_sing and 𝒩 is well–defined in appropriated spaces, which will be subspaces of Hölder–continuous functions. Under the hypothesis that Ω∈ I, the problem of finding solutions of (<ref>) is reduced to solve

    G(Ω,g)(z)≜ G(Ω,g,𝒩(Ω,g))(z)= 0,   ∀ z∈ .

In order to find time dependent non radial rotating solutions to (<ref>) we use the procedure developed in <cit.> that suggests the bifurcation theory as a tool to generate solutions from a stationary one via the Crandall–Rabinowitz Theorem. The values Ω that could lead to the bifurcation to non trivial solutions are located in the dispersion set

    𝒮_disp≜{Ω : Ker  D_g G(Ω, 0)≠{0}}.

The problem then consists in verifying  that the singular (<ref>) and dispersion (<ref>) sets are well-separated, for a correct definition of the interval I. Achieving this objective together with the analysis of the dimension properties of the kernel and the codimension of the range of D_gG(Ω,0), as well as  verifying the transversality property requires a complex and precise spectral and asymptotic analysis.
Although our discussion is quite general, we  focus  our attention  on the  special case of  quadratic profiles (<ref>).  In this case   we obtain  a compact representation of the dispersion set. Indeed, as we shall see in Section <ref>, the resolution of the kernel equation leads to a Volterra type integro-differential equation   that one may solve through transforming it into an ordinary differential equation of second order with polynomial coefficients. Surprisingly, the new  equation can be solved explicitly through  variation of the constant and is connected to  Gauss hypergeometric functions.
The structure of the dispersion set is very subtle and appears to be very sensitive to the parameters A and B. Our analysis allows us to highlight   some special regimes on A and B, see Proposition <ref> and Proposition <ref>.  

Let us emphasize that the  techniques developed in the quadratic profile   are robust and could be extended to  other profiles. In this direction, we first  provide in Section <ref> the explicit expression of the function ℳ  when the density  admits a polynomial or  Gaussian distribution. In general, the explicit resolution of the kernel equations may turn out  to be a very challenging problem. Second, we will notice  in Remark <ref>  that when f_0=Ar^2m+B with m∈^⋆, explicit formulas are expected through some elementary transformations and the  kernel elements  are linked also to hypergeometric equations.



In Section <ref> we shall be concerned with the proof of the point (5) of Theorem <ref>  concerning the planar trajectories of  the particles located inside the support of the rotating vortices. 
We analyze the properties of periodicity and  symmetries of the solutions  via the study of the associated dynamical Hamiltonian structure in Eulerian coordinates, which was highlighted by Arnold <cit.>. This Hamiltonian nature   of the Euler equations has been the idea behind the study of conservation laws in the hydrodynamics of an ideal fluid <cit.>, as well as  in a certain sense to justify Boltzmanns principle from classical mechanics <cit.>.

We shall give   in Theorem <ref> a precise statement and  prove that close to the quadratic  profile  all the trajectories are periodic orbits located inside  the support of the V-states, enclosing a simply connected domain containing the origin, and are symmetric with respect to the real axis. In addition,  every orbit is invariant by a rotation of angle 2π/m, as it has been proved for the branch of bifurcated solutions, where the parameter m is determined by the spectral properties. The periodicity of the orbits follows from the Hamiltonian structure of the autonomous dynamical system,

    ∂_t ψ(t,z)=W(Ω,f,Φ)(ψ(t,z)),  ψ(0,z)=z∈,

where

    W(Ω,f,Φ)(z)=(i/2π∫_f(y)/Φ(z)-Φ(y)|Φ^'(y)|^2 dy-iΩΦ(z))Φ^'(z),    z∈.

Notice  that W is nothing but the pull-back of  the vector filed v(x)-Ω x^⊥ by the conformal mapping Φ.  This vector field  remains Hamiltonian and is tangential to the boundary . Moreover,  we check  that close to the radial profile, it has only one critical point located at the origin which must be  a center. As a consequence, the trajectories near the origin are organized through periodic  orbits. Since  the trajectories are located in the level sets of the energy functional given by the relative stream function, then using simple arguments we show the limit cycles  are excluded and thus all the trajectories are periodic enclosing the origin which is the only fixed point, which, together with the trajectories defined above, is a way of solving the hyperbolic system (<ref>).  This allows to define  the period map  z∈↦ T_z, whose regularity will be at the same level as the profiles. As a by-product we find the following  equivalent reformulation of the density equation
    f(z)-1/T_z∫_0^T_z f(ψ(τ,z))dτ=0,   ∀ z∈.



Other approaches to study stationary solutions, and thus to explore the possibilities of bifurcating them, have been proposed through the study of characteristic trajectories associated with stationary velocities

    ∂ X(t)/∂ t = v(X(t)),    X(0) =x ∈^2,

in connection with the elliptic equation -Δψ = ω = F̃(ψ) (ψ being the current function ∇^⊥ψ =v), see <cit.>.
In this context, the idea of looking for smooth stationary solutions was first developed by Nadirashvili in <cit.>, where the geometry (curvature) of streamlines was studied. Luo  and  Shvydkoy <cit.> provide a classification of some kind of homogeneous stationary smooth solutions  with locally finite energy, where new solutions having hyperbolic, parabolic and elliptic structure of streamlines appear. Choffrut and Šverák <cit.> showed analogies  of  finite-dimensional  results  in  the  infinite-dimensional  setting  of Eulers equations, under some non-degeneracy assumptions, proving a local one-to-one correspondence between steady-states  and co-adjoint orbits. Then,  Choffrut and Székelyhidi <cit.> showed, using an  h-principle <cit.>, that there is  an abundant set of weak, bounded stationary solutions in the neighborhood of any smooth stationary solution. Kiselev and Šverák <cit.> construct an example of initial data in the disc such that the corresponding solutions for the 2D Euler equation exhibit double exponential growth in the gradient of vorticity, which is related to the lack of Lipschitz regularity, and to an example of the singular stationary solution provided by Bahouri and Chemin <cit.>, which produces a flow map whose Hölder regularity decreases in time.

We finally comment on three recent approaches to the analysis of rotating solutions. The first one concerns  rotating vortex patches. In <cit.>, Hassania, Masmoudi and Wheeler construct continuous curves of rotating vortex patch solutions, where the minimum along the interface of the angular fluid velocity in the rotating frame becomes
arbitrarily small, which agrees with the conjecture about singular limiting patches with 90^∘ corners <cit.>.
In the second contribution <cit.>, it was studied the existence of  smooth rotating vortices desingularized from a vortex patch, as it was mentioned before. The techniques are  based on the analysis  of the level sets of the vorticity of a global rotating solution. Since the level sets z(α,ρ,t) rotate with constant angular velocity, they satisfy
ω(z(α,ρ,t)),t) =f(ρ). Thus, in <cit.> is studied the problem of bifurcating it for some specific choice of f.
In a  broad sense, this result connects with that developed in this paper about the study of orbits and their periodicity. Finally, Bedrossian, Coti Zelati and Vicol in <cit.> analyze the incompressible 2D Euler equations linearized around a radially symmetric, strictly monotone decreasing vorticity distribution. For sufficiently regular data,  inviscid damping of the θ-dependent radial and angular velocity fields is proved. In this case, the vorticity weakly converges back to radial symmetry as t
→∞, a phenomenon known as vortex axisymetrization. Also they show that the
θ-dependent angular Fourier modes in the vorticity are ejected from the origin as t
→∞, resulting in faster inviscid damping rates than those possible with passive scalar evolution (vorticity depletion).


The results and techniques presented in this paper are powerful enough to be extended to other situations and equations such as SQG equations, co–rotating time–dependent solutions, solutions depending only on one variable,...



§ PRELIMINARIES AND STATEMENT OF THE PROBLEM

The aim of this section is to formulate  the equations governing general rotating solutions of the  Euler equations. 
We will also set down some of the tools that we use throughout the paper such as the functional setting or some properties about the extension of Cauchy integrals.



 §.§ Equation for  rotating vortices

Let us begin with the equations for compactly supported  rotating solutions (<ref>)–(<ref>) and assume  that  f is not vanishing on the boundary. In the opposite case,  the equation   (<ref>) degenerates and becomes trivial, which implies that we just have one equation to analyze. Thus, (<ref>) becomes 

    (v(x)-Ω x^⊥)·n⃗(x)=0,  on  ∂ D.

 We will rewrite these equations in the unit disc through the  use of the conformal map Φ:→ D. Note that from now on and  for the sake of simplicity we will identify the Euclidean and the complex planes. Then, we write the velocity field as
 
    v(x)=i/2π∫_D (f∘Φ^-1)(y)/x-ydA(y),   ∀ x∈,

where dA refers to the planar Lebesgue measure, getting
 
    v(Φ(z))=i/2π∫_f(y)/Φ(z)-Φ(y)|Φ'(y)|^2dA(y),   ∀ z∈.

Using the conformal parametrization  θ∈[0,2π]↦Φ(e^iθ) of ∂ D, we find that a normal vector to the boundary is given by n⃗(Φ(w))=wΦ^'(w), with w∈. In order to deal with (<ref>), we need to  transform carefully the term ∇ (f∘Φ^-1)(Φ(z)) coming from the density equation. Recall that for  any complex function φ:→ of class 𝒞^1 seen as a function of ^2, we can define 

    ∂_zφ(z)≜1/2(∂_1φ(z)+i∂_2φ(z))  and  ∂_zφ(z)≜1/2(∂_1φ(z)-i∂_2φ(z)),

which are known in the literature  as Wirtinger derivatives. Let us state  some of their basic properties:

    ∂_zφ=∂_zφ,  ∂_zφ=∂_zφ.

Given two complex functions φ_1, φ_2:→ of class 𝒞^1 in the Euclidean coordinates, the chain rule comes as follows

    ∂_z (φ_1∘φ_2)   =   (∂_z φ_1∘φ_2)∂_zφ_2+(∂_zφ_1∘φ_2)∂_zφ_2,
    ∂_z (φ_1∘φ_2)   =   (∂_z φ_1∘φ_2)∂_zφ_2+(∂_zφ_1∘φ_2)∂_zφ_2.

Moreover, since Φ is a conformal map, one has that ∂_zΦ=0. Identifying  the gradient with the operator 2∂_z  leads to

    ∇ (f∘Φ^-1)=2∂_z(f∘Φ^-1).

From  straightforward computations using   the holomorphic structure of Φ^-1, combined with  the previous properties of the Wirtinger derivatives, we get

    ∂_x (f∘Φ^-1)(x)   =   (∂_x f)(Φ^-1(x))∂_xΦ^-1(x)+(∂_x f)(Φ^-1(x))∂_xΦ^-1(x)
       =   (∂_x f)(Φ^-1(x))(Φ^-1)^'(x),

where the prime notation ^' for Φ^-1 denotes the complex  derivative  in the holomorphic case.
Using that (Φ^-1∘Φ)(z)=z and differentiating it we obtain

    (Φ^-1)^'(Φ(z))=Φ'(z)/|Φ^'(z)|^2,

which implies

    ∂_z (f∘Φ^-1)(Φ(z))=∂_z f(z) Φ^'(z)/|Φ^'(z)|^2.

Putting everything together in (<ref>)-(<ref>) we find the following equivalent expression
 
    W(Ω,f,Φ)·∇ f   =   0,  in  ,
     
    W(Ω,f,Φ)·n⃗   =   0,  on  ,

where W is given by

    W(Ω,f,Φ)(z)=(i/2π∫_f(y)/Φ(z)-Φ(y)|Φ^'(y)|^2 dy-iΩΦ(z))Φ^'(z).

Then, the vector field W(Ω,f,Φ) is incompressible. This fact is a consequence  of the  lemma below. Given a vector field X:→ of class 𝒞^1 in the Euclidean variables, let us associate the divergence operator with the Wirtinger derivatives as follows

    div X(z)=2Re[∂_z X(z)].



Given X:D_1→ an incompressible vector field, Φ:D_2→ D_1 a conformal map, where D_1,D_2⊂, then
(X∘Φ)Φ':D_2→
is incompressible.


Using (<ref>) we have that
Re[∂_x X(x)]=0, for any x∈ D_1.

The properties of the Wirtinger derivatives lead to

    ∂_z [X(Φ(z))Φ'(z)]   =   ∂_z[X(Φ(z))]Φ^'(z)+X(Φ(z))∂_z Φ^'(z)
       =   (∂_z X) (Φ(z))Φ^'(z)Φ^'(z)+(∂_z X)(Φ(z))∂_z Φ(z) Φ^'(z)
          +X(Φ(z))∂_z Φ^'(z)
       =   (∂_z X)(Φ(z))|Φ^'(z)|^2,  ∀ z∈ D_2.

Hence, we have that

    Re[∂_z (X(Φ(z))Φ'(z))]=|Φ'(z)|^2 Re[(∂_z X)(Φ(z))]=0,  ∀ z∈ D_2,

 and (X∘Φ)Φ' is incompressible.





Let us remark that the equation associated to the  V-states in <cit.> is nothing but the boundary equation (<ref>). In <cit.>,  V-states close to a trivial solution are obtained by means of a perturbation of the domain via a conformal mapping. Since we are perturbing also the initial density, we must analyze one more equation: the density equation (<ref>). In order to apply the Crandall–Rabinowitz Theorem we will not deal with (<ref>) because  it seems not to be suitable when studying the linearized operator. Hence, we will reformulate this equation in Section <ref>. Moreover, we will provide an alternative way of writing (<ref>) in Section <ref> to understand the behavior of the orbits of the dynamical system associated to it.




 §.§ Function spaces

The right choice of the function spaces will be crucial  in order to construct non radial rotating solutions different to the vortex patches.

Before going into further details we  introduce the classical Hölder spaces in the unit disc . Let us denote 𝒞^0,α() as the set of continuous functions such that

    f_𝒞^0,α()≜f_L^∞()+sup_z_1≠ z_2∈|f(z_1)-f(z_2)|/|z_1-z_2|^α<+∞,

for any α∈(0,1). By 𝒞^k,α(), with k∈, we denote the 𝒞^k functions whose k-order derivative lies   Recall the Lipschitz space with the semi-norm defined as

    f_Lip()≜sup_z_1≠ z_2∈|f(z_1)-f(z_2)|/|z_1-z_2|.

Similarly, we define the Hölder spaces 𝒞^k,α() in the unit circle . Let us  supplement these spaces with additional symmetry structures:

    𝒞^k,α_s()≜{ g: →∈𝒞^k,α(),   g(re^iθ)=∑_n≥ 0 g_n(r)cos(nθ),  g_n∈, ∀ z=re^iθ∈},


    𝒞^k,α_a()≜{ρ: →∈𝒞^k,α(),  ρ(e^iθ)=∑_n≥ 1ρ_nsin(nθ),    ∀ w=e^iθ∈}.

These spaces  are equipped with the usual  norm ·_𝒞^k,α. One can easily check that if the functions  g∈𝒞^k,α_s() and ρ∈𝒞^k,α_a(), then they satisfy the following properties

    g(z)=g(z),   ρ(w)=-ρ(w),   ∀ z∈, ∀   w∈.

The space 𝒞_s^k,α() will contain the perturbations of the initial radial density. The condition on g means that this perturbation  is invariant by reflexion on the real axis. Let us remark that we introduce also a radial perturbation coming from the frequency n=0, this fact will be a key point in the bifurcation argument.

The second kind of function spaces is ℋ𝒞^k,α(), which is the set  of holomorphic functions ϕ in   belonging to 𝒞^k,α() and satisfying  

    ϕ(0)=0,  ϕ'(0)=0  and  ϕ(z)=ϕ(z),   ∀ z∈.

With these properties, the function ϕ admits the following expansion

    ϕ(z)= z∑_n≥ 1 a_n z^n,    a_n∈.

Thus, we have

    ℋ𝒞^k,α()≜{ϕ∈𝒞^k,α(),  ϕ(w)=w∑_n≥ 1a_nw^n,  a_n∈, ∀ w∈}.

Notice that if Φ≜Id+ϕ is conformal then  Φ() is a simply connected domain,  symmetric with respect to the real axis and whose boundary is 𝒞^k,α. The space ℋ𝒞^k,α() is a closed subspace of 𝒞^k,α() equipped with  the same norm, so it is complete. In the bifurcation argument, we will perturb also the initial domain  via a conformal map that will lie in this space.

The last condition on ϕ, given by (<ref>), together with the symmetry condition for the density (<ref>), means that we are looking for rotating  initial data which admit at least one axis of symmetry. For the rotating patch problem this is the minimal requirement that we should impose and up to now we do not know whether such structures without any prescribed symmetry could exist.

Now, we introduce  the following trace problem concerning  the extension of Cauchy integrals, which is a classical result in complex analysis  and potential  theory. It is directly linked to <cit.> and <cit.>) and for the convenience of the reader we give a proof.

Let k∈ and 
α∈ (0,1). Denote by   𝒞:ℋ𝒞^k,α()→ℋ𝒞^k,α()  the linear map defined  by

    ϕ(w)=w∑_n∈a_nw^n,  ∀ w∈⟹𝒞(ϕ)=∑_n∈a_nz^n,   ∀ z∈.

Then,  𝒞(ϕ) is well-defined and continuous.


First, it is a simple matter to check that the map 𝒞 is well-defined. Thus, it remains to check the continuity. We recall from   <cit.>) the following estimates  on the modulus of continuity

    sup_  z_1,z_2∈ |z_1-z_2|≤δ |𝒞(ϕ)(z_1)-𝒞(ϕ)(z_2)|≤ 3sup_w_1,w_2∈ |w_1-w_2|≤δ |ϕ(w_1)-ϕ(w_2)|,

for any δ<π/2 and for any  continuous function 𝒞(ϕ) in , analytic in   and having trace function ϕ on the unit circle.
Therefore,  given z_1,z_2∈ with |z_1-z_2|≤1, we obtain

    |𝒞(ϕ)(z_1)-𝒞(ϕ)(z_2)|/|z_1-z_2|^α   ≤    3sup_w_1,w_2∈ |w_1-w_2|≤|z_1-z_2|  |ϕ(w_1)-ϕ(w_2)|/|z_1-z_2|^α
       ≤    3sup_w_1,w_2∈ |w_1-w_2|≤1  |ϕ(w_1)-ϕ(w_2)|/|w_1-w_2|^αsup_w_1,w_2∈ |w_1-w_2|≤|z_1-z_2|  |w_1-w_2|^α/|z_1-z_2|^α
       ≤   3sup_w_1,w_2∈ |w_1-w_2|≤1  |ϕ(w_1)-ϕ(w_2)|/|w_1-w_2|^α.

Now, let z∈ and w∈ such that |z-w|≤1, then we also get from (<ref>)

    |𝒞(ϕ)(z)|≤ |ϕ(w)|+3sup_w_1,w_2∈ |w_1-w_2|≤ 1  |ϕ(w_1)-ϕ(w_2)|
    ≤  |ϕ(w)|+3sup_w_1,w_2∈ |w_1-w_2|≤1  |ϕ(w_1)-ϕ(w_2)|/|w_1-w_2|^α,

which implies that

    𝒞(ϕ)_L^∞()≤ 3ϕ_𝒞^0,α().

Combining the preceding estimates,  we deduce that 

    𝒞(ϕ)_𝒞^0,α()≤ 6ϕ_𝒞^0,α().

Note that this estimate can be extended to higher derivatives and thus we obtain 

    𝒞(ϕ)_𝒞^k,α()≤ 6ϕ_𝒞^k,α(),

which completes the proof.




§ BOUNDARY EQUATION

This section focuses  on studying the second equation (<ref>) concerning the boundary equation and prove that we can parametrize the solutions in (Ω,f,Φ) close to the initial radial solution (f_0,Id), with f_0 being a radial profile,  through a mapping (Ω,f)↦Φ=𝒩(Ω,f).  We will deal with the unknowns g and ϕ defined by

    f=f_0+g,   Φ=Id+ϕ.

Equation (<ref>) can be written in the following way

    F(Ω,g,ϕ)(w)≜Im[ (Ω Φ(w)-1/2π∫_f(y)/Φ(w)-Φ(y)|Φ^'(y)|^2 dA(y))Φ^'(w)w]
     =0,

for any w∈. Notice that  from this formulation we can retrieve the fact that 

    F(Ω,0,0)=0,   ∀ Ω∈,

which is compatible with the fact that any radial initial data leads to a stationary solution of the Euler equations. Indeed, this identity follows from  Proposition <ref> which implies that

    1/2π∫_f_0(y)/w-y dA(y)=  w∫_0^1 s f_0(s)ds,   ∀  w∈.

The idea to solve the nonlinear equation (<ref>)  is to apply  the Implicit Function Theorem.
Define the open balls

    {[                  B_𝒞_s^k,α(g_0,ε)                                 = {g∈𝒞_s^k,α()  s.t.  g-g_0_k,α< },;                    B_ℋ𝒞^k,α(ϕ_0,)                                 =  {ϕ∈ℋ𝒞^k,α()  s.t.  ϕ-ϕ_0_k,α< }, ].

for >0, g_0∈𝒞_s^k,α() and  ϕ_0∈𝒞^k,α().
The first result  concerns the well–definition and regularity of the functional  F introduced in (<ref>). 

Let ∈ (0,1), then
F:× B_𝒞_s^1,α(0,)× B_ℋ𝒞^2,α(0,)⟶𝒞^1,α_a() 
 is well-defined and  of class 𝒞^1. 


If ϕ∈ B_ℋ𝒞^k,α(0,) with <1, then Φ=Id+ϕ is conformal and bi-Lipschitz. 


Let us show that F∈𝒞^1,α(). Since Φ,Φ'∈𝒞^1,α() it remains to study the integral term. This is a consequence of Lemma <ref> in  Appendix <ref>, which yields F∈𝒞^1,α().

Let us turn to the persistence of the symmetry. According to (<ref>) one has to check that

    F(Ω, g, ϕ)(w)=-F(Ω, g, ϕ)(w),   ∀  w∈.
 
Using the symmetry properties of the density and the conformal mapping we write

    F(Ω, g, ϕ)(w)   =   Im[ (ΩΦ(w)-1/2π∫_f(y)/Φ(w)-Φ(y)|Φ^'(y)|^2 dA(y))Φ^'(w)w]
       =   Im[ (ΩΦ(w)-1/2π∫_f(y)/Φ(w)-Φ(y)|Φ^'(y)|^2 dA(y))Φ^'(w)w]
       =   Im[ (ΩΦ(w)-1/2π∫_f(y)/Φ(w)-Φ(y)|Φ^'(y)|^2 dA(y))Φ^'(w)w].

Therefore, we get (<ref>).
This concludes that F(Ω,g,ϕ) is in 𝒞^1,α_a().  Notice that the dependance with respect to Ω is smooth and we will focus on the Gâteaux derivatives of  F with respect to g and ϕ.  Straightforward computations lead to 

    D_g F(Ω,g,ϕ)h(w)   =   -Im[ wΦ'(w)/2π∫_h(y)/Φ(w)-Φ(y)|Φ'(y)|^2dA(y) ],
     D_ϕ F(Ω,g,ϕ)k(w)   =   Im[Ωk(w)Φ'(w)w+ΩΦ(w)k'(w)w
          -wk'(w)/2π∫_f(y)/Φ(w)-Φ(y)|Φ'(y)|^2dA(y)
          +wΦ'(w)/2π∫_k(w)-k(y)/(Φ(w)-Φ(y))^2f(y)|Φ'(y)|^2dA(y)
          -wΦ'(w)/π∫_f(y)/Φ(w)-Φ(y)Re[Φ'(y)k'(y)]dA(y)].

Let us use the operator ℱ[Φ] defined in (<ref>). Although in Lemma <ref> ℱ[Φ] is defined in  we can extend it up to the boundary  getting the same result. Hence, all the above expressions can be written through this operator as

    D_g F(Ω,g,ϕ)h(w)   =   -Im[ wΦ'(w)/2πℱ[Φ](h)(w) ],
     D_ϕ F(Ω,g,ϕ)k(w)   =   Im[Ωk(w)Φ'(w)w+ΩΦ(w)k'(w)w-wk'(w)/2πℱ[Φ](f)(w)
          .+wΦ'(w)/2π∫_k(w)-k(y)/(Φ(w)-Φ(y))^2f(y)|Φ'(y)|^2dA(y).
          -.wΦ'(w)/πℱ[Φ](Re[Φ'(·)k'(·)]/|Φ'(·)|^2)(w)].

Since Re[Φ'(·)k'(·)]/|Φ'(·)|^2, Φ', Φ,k'∈𝒞^1,α() and are continuous with respect to Φ, Lemma <ref> entails that all the terms except the integral one
lie in 𝒞^1,α() and they are continuous with respect to Φ. The continuity with respect to f comes also from the same result. Note that although our unknowns are (g,ϕ), studying the continuity with respect to (g,ϕ) is equivalent to doing it with respect to (f,Φ).
We  shall now focus our attention on the integral term by splitting it as follows

    ∫_k(w)-k(y)/(Φ(w)-Φ(y))^2f(y)|Φ'(y)|^2dA(y)   =   ∫_(k(w)-k(y))(f(y)-f(w))/(Φ(w)-Φ(y))^2|Φ'(y)|^2dA(y)
          +f(w)∫_k(w)-k(y)/(Φ(w)-Φ(y))^2|Φ'(y)|^2dA(y)
       ≜   𝒥_1[Φ]f(z)+f(w)𝒥_2[Φ].

First, we deal with 𝒥_1[Φ]. Clearly

    |𝒥_1[Φ](w)|≤ Cf_𝒞^1,α(),

and we define

    K(w,y)   ≜   ∇_w (k(w)-k(y))(f(y)-f(w))/(Φ(w)-Φ(y))^2
       =   -(k(w)-k(y))/(Φ(w)-Φ(y))^2∇_w f(w) +(f(y)-f(w))∇_w (k(w)-k(y))/(Φ(w)-Φ(y))^2
       ≜   -∇_w f(w)K_1(w,y)+K_2(w,y).

Using the same argument as in  (<ref>), we can check that K_1 and K_2 verify both the hypotheses of Lemma <ref>. This implies that 𝒥_1[Φ] lies in 𝒞^1,α(). Taking two conformal maps Φ_1 and Φ_2 and estimating 𝒥_1[Φ_1]-𝒥_1[Φ_2], we find integrals similar to those treated in Lemma <ref>.



Concerning the second integral 𝒥_2[Φ], which seems to be more singular, we use the Cauchy–Pompeiu's formula (<ref>) to find

    ∫_k(w)-k(y)/Φ(w)-Φ(y)|Φ'(y)|^2dA(y)=1/2i∫_k(w)-k(ξ)/Φ(w)-Φ(ξ)Φ(ξ)Φ'(ξ)dξ.

Differentiating it, we deduce 

    ∫_k(w)-k(y)/(Φ(w)-Φ(y))^2|Φ'(y)|^2dA(y)   =   k'(w)Φ(w)π/2Φ'(w)+1/2i∫_k(ξ)-k(w)/(Φ(ξ)-Φ(w))^2Φ(ξ)Φ'(ξ)dξ
       ≜   k'(w)Φ(w)π/2Φ'(w)+1/2i𝒯[Φ](w).

The first term is in 𝒞^1,α() and is clearly continuous with respect to Φ. Integration by parts in the second term 𝒯[Φ] leads to

    𝒯[Φ](w)=-∫_k'(ξ)Φ(ξ)/Φ(ξ)-Φ(w)dξ+∫_k(ξ)-k(w)/Φ(ξ)-Φ(w)ξ^2Φ'(ξ)dξ.

Differentiating and integrating it  by parts again one obtain the following expression

    𝒯[Φ]'(w)   =   Φ'(w)_∂_ξ(k'(ξ)Φ(ξ)/Φ'(ξ))/Φ(ξ)-Φ(w)dξ-Φ'(w)_∂_ξ((k(ξ)-k(w))ξ^2Φ'(ξ)/Φ'(ξ))/Φ(ξ)-Φ(w)dξ
       =   Φ'(w)_∂_ξ(k'(ξ)Φ(ξ)/Φ'(ξ))/Φ(ξ)-Φ(w)dξ-Φ'(w)_k'(ξ)ξ^2Φ'(ξ)/Φ'(ξ)^2/Φ(ξ)-Φ(w)Φ'(ξ)dξ
          -Φ'(w)_(k(ξ)-k(w))∂_ξ(ξ^2Φ'(ξ)/Φ'(ξ))/Φ(ξ)-Φ(w)dξ
       =   Φ'(w)ℐ[Φ](∂(k'(·)Φ(·)/Φ'(·))/Φ'(·))(w)-Φ'(w)ℐ[Φ](k'(·)(·)^2Φ'(·)/Φ'(·)^2)(w)
          -Φ'(w)ℐ[Φ](k(·)∂(·^2Φ'(·)/Φ'(·)))(w)+Φ'(w)k(w)ℐ[Φ](∂(·)^2Φ'(·)/Φ'(·))(w),

where ℐ[Φ] is the operator defined in (<ref>). Since the functions

    ∂(k'(·)Φ(·)/Φ'(·))/Φ'(·),  k'(·)(·)^2Φ'(·)/Φ'(·)^2,   k,  ∂((·)^2Φ'(·)/Φ'(·))∈𝒞^0,α(),

and are continuous with respect to Φ, we can use Lemma <ref>. Hence, these terms lie in 𝒞^0,α(). Moreover, the same argument gives us the continuity with respect to Φ. To conclude, we use the fact that the Gâteaux derivatives are continuous with respect to (g,ϕ) and so they are in fact Fréchet derivatives. 

The next task is to  implement the Implicit Function Theorem in order to solve the boundary equation (<ref>) through a two-parameters  curve solutions in infinite-dimensional spaces.  
Given a radial function f_0∈𝒞^1,α(), we associate to it  the singular set

    𝒮_sing≜{Ω_n≜∫_0^1sf_0(s)ds-n+1/n∫_0^1s^2n+1f_0(s)ds,  ∀ n∈^⋆∪{+∞}}.

This terminology will be later justified in the proof of the next proposition. Actually, this set corresponds to the location of  the points Ω where the partial  linearized operator ∂_ϕ F(Ω,0,0) is not invertible.
Let us establish the  following result.

Let  f_0:→ be a radial function in 𝒞^1,α(). Let I be an open interval such that  I⊂\𝒮_sing.
Then, there exists ε>0 and  a 𝒞^1 function  

    𝒩: I× B_𝒞_ s^1,α(0,ε)⟶ B_ℋ𝒞^2,α(0,ε),

 with the following property:

    F(Ω,g,ϕ)=0⟺ϕ=𝒩(Ω,g),

 for any (Ω,g,ϕ)∈ I× B_𝒞_s^1,α(0,ε)× B_ℋ𝒞^2,α(0,ε).
In addition,  we obtain the identity 

    D_g𝒩(Ω, 0)h(z)=z∑_n≥ 1 A_n z^n,

for any h∈𝒞_s^1,α(), with  h(re^iθ)=∑_n≥ 0h_n(r)cos(nθ), and 

    A_n=∫_0^1s^n+1h_n(s)ds/2n(Ω_n-Ω),

for any n≥ 1 where Ω_n is defined in (<ref>).  Moreover, we have

    𝒩(Ω, 0)h_𝒞^2,α()≤  Ch_𝒞^1,α().



From  the definition of the function space 𝒞^1,α_s() we are adding also a radial perturbation of the initial radial part given by the first mode n=0. However, from the expression of D_g𝒩(Ω, 0)h(z) the first frequency  disappears and the sum starts at n=1. This is an expected fact because   (Ω,g,0) is a  solution of F(Ω,g,ϕ)  for any radial smooth function g. This means that 𝒩(Ω, g)=0, and hence  D_g𝒩(Ω,0)h is vanishing when  h is radial.  


Applying the Implicit Function Theorem consists in checking that 

    D_ϕ F(Ω, 0, 0):ℋ𝒞^2,α()⟶𝒞^1,α_a(),
 
is an isomorphism. A combination of (<ref>) with Proposition <ref> allow us to compute explicitly the differential of F(Ω, g,ϕ) on the initial solution as follows

    D_ϕ F(Ω, 0, 0)k(w)   =   Im[Ωk(w)w+Ωwk'(w)w-wk'(w)/2π∫_f_0(y)/w-ydA(y)
          .+w/2π∫_k(w)-k(y)/(w-y)^2f_0(y)dA(y).-w/π∫_f_0(y)/w-yRe[k'(y)]dA(y)]
       =   ∑_n≥ 1a_n Im[Ωw^n+Ω(n+1)w^n-(n+1)w^n∫_0^1sf_0(s)ds
          +w^n∫_0^1sf_0(s)ds-(n+1)w^n∫_0^1s^2n+1f_0(s)ds]
       =   ∑_n≥ 1a_nn{Ω-∫_0^1sf_0(s)ds+n+1/n∫_0^1s^2n+1f_0(s)ds}sin(nθ).

Similarly, we get 

    D_g F(Ω,0,0)h(w)   =   -Im[w/2π∫_h(y)/w-ydA(y)]
    =-π/2π∑_n≥ 1Im[w w^n+1∫_0^1s^n+1h_n(s)ds]
       =   1/2∑_n≥ 1∫_0^1s^n+1h_n(s)dssin(nθ),

where 

    z↦ k(z)=∑_n≥1 a_n z^n+1∈ℋ𝒞^2,α()  and   z↦ h(z)=∑_n≥0 h_n(r) cos(nθ)∈𝒞^1,α_s(),
 
are given as in (<ref>) and (<ref>), respectively. Then,  we have that D_ϕ F(Ω,0,0) is one–to–one linear mapping and is continuous  according to  Proposition <ref>, for any  Ω∈I. Using the Banach Theorem it suffices to check that this mapping is onto. Notice that at the formal level the inverse operator can be easily computed  from the expression of D_ϕ F(Ω,0,0) and it is given by
 
    D_ϕ F(Ω,0,0)^-1ρ(z)=z∑_n≥ 1ρ_n/n(Ω-Ω_n)z^n,

for any ρ(e^iθ)=∑_n≥ 1ρ_nsin(nθ). Thus the problem reduces to check that
D_ϕ F(Ω,0,0)^-1ρ∈ℋ𝒞^2,α().
 First, we will prove that this function is holomorphic inside the unit disc . For this purpose we use that
 
    ρ_n=1/π∫_0^2πρ(e^iθ)sin(nθ) dθ.

 Since ρ∈ L^∞(),  we obtain that the coefficients sequences  (ρ_n)∈ℓ^∞. Using the facts that lim_n→∞Ω_n=Ω_∞ and that Ω is far away from the singular set, then we deduce that the Fourier coefficients  of D_ϕ F(Ω,0,0)^-1ρ are bounded. Consequentely, this function is holomorphic inside the unit disc. It remains to check that this function belongs to  𝒞^2,α(). By virtue of    Lemma <ref>  it is enough to check that the restriction on the boundary belongs to 𝒞^1,α().  First, we must notice that if ρ∈𝒞^1,α(), then 

    w↦ρ_+(w)≜∑_n≥ 1ρ_nw^n∈𝒞^1,α().

For this purpose, let us write ρ in the form

    ρ(w)=1/2i∑_n∈ℤρ_nw^n,  with  ρ_-n=-ρ_n.

Hence, ρ_+ is nothing but  the  Szegö projection of ρ, which is continuous on 𝒞^1,α().  Note that this latter property is based upon the fact that ρ_+ can be expressed from  ρ through the Cauchy integral operator

    ρ_+(w)=1/π∫_ρ(ξ)/ξ-wdξ,
 
and one may  use  T(1)-Theorem of Wittmann for Hölder spaces, see for instance <cit.> and <cit.> or Lemma <ref>. Secondly, we will prove that D_ϕ F(Ω,0,0)^-1ρ∈𝒞^2,α(). We define

    D_ϕ F(Ω,0,0)^-1ρ(w)≜ wq(w).

Let us show that   q is bounded. Using (<ref>) and  integration by parts, we have

    |ρ_n|≤ 2ρ^'_L^∞()/n,

which implies that q is bounded. To prove higher regularity, we write q as a convolution

    q=ρ_+⋆ K_1,

where 

    K_1(w)=∑_n≥ 1w^n/n(Ω-Ω_n).

Since ρ_+∈𝒞^1,α, we just need to check  that K_1∈ L^1. To do that, we use the Parseval's identity, which provides that K_1∈ L^2():

    K_1_2^2=∑_n≥ 11/n^2(Ω-Ω_n)^2≤ C∑_n≥ 11/n^2
    <+∞,

where C is a constant connected to the distance between Ω and  the singular set 𝒮_sing defined in (<ref>).  To study its derivative, let us write it as

    q'(w)=w∑_n≥ 1ρ_n/Ω-Ω_nw^n
    =w[∑_n≥ 1ρ_n/βw^n+∑_n≥ 1ρ_n(1/β+u_n-1/β)w^n]
    ≜w[1/βρ_+(w)+S],

where

    β=Ω-Ω_∞,    u_n=n+1/n∫_0^1s^2n+1f_0(s)ds.

From the foregoing discussion, we have seen that ρ_+∈𝒞^1,α(). As to the term  S, it  can be written in convolution form

    S=ρ_+⋆ K_2,

with

    K_2(w)=-∑_n≥ 1u_n/(β+u_n)βw^n.

Since ρ_+∈𝒞^1,α(),  then we just need to check that K_2∈ L^1() to conclude. Using Parseval's identity we have that K_2∈ L^2 because

    K_2_2^2=1/β^2∑_n≥ 1u_n^2/(β+u_n)^2   ≤    C∑_n≥ 1(n+1)^2/n^2(∫_0^1s^2n+1f_0(s)ds)^2
       ≤    C∑_n≥ 1(n+1)^2/n^2(2n+1)^2 
    <  +∞.

This achieves that D_ϕ F(Ω,0,0)^-1ρ∈ℋ𝒞^2,α() and consequently the linearized operator D_ϕ F(Ω,0,0)  is an isomorphism.
Hence, the Implicit Function Theorem can be used and it ensures  the existence of a  𝒞^1–function 𝒩 such that 

    F(Ω,g,ϕ)=0⟺ϕ=𝒩(Ω,g),

for any  (Ω,g,ϕ)∈ I× B_𝒞_s^1,α(0,ε)× B_ℋ𝒞^2,α(0,ε). Differentiating with respect to g, we obtain 

    D_g F(Ω, g, 𝒩(Ω,g))=∂_g F(Ω, g, ϕ)+∂_ϕ F(Ω,g,ϕ)∘∂_g𝒩(Ω,g)=0,

which yields

    ∂_g 𝒩(Ω,0)h(z)=-∂_ϕ F(Ω,0,0)^-1∘∂_g F(Ω,0,0)h(z).

Then, using (<ref>) and (<ref>), straightforward computations  show that

    D_g𝒩(Ω, 0)h(z)=-z∑_n≥ 1∫_0^1s^n+1h_n(s)ds/2n(Ω-Ω_n) z^n.

This concludes the proof of the announced result. 



§ DENSITY EQUATION

This section aims at studying the density equation (<ref>) in order to get non radial rotating solutions via the Crandall–Rabinowitz Theorem. We will reformulate it in a more convenient way since we are not able to use the original expression (<ref>) due to the  structural defect on  its linearized operator as it has been pointed out previously. We must have in mind that under suitable assumptions, the conformal map is recovered from the angular velocity Ω and the density function via Proposition <ref>.



 §.§ Reformulation of the density equation

Taking  an initial data in the form (<ref>) and noting that if  the density f is fixed close to f_0 and Ω does not lie in  the singular set 𝒮_sing, then the conformal mapping is uniquely determined as a consequence of Proposition <ref>. Now, we turn to the analysis of  the first equation of (<ref>)
that we intend to solve  for a restricted class of  initial densities. The strategy to implement it is to look for  solutions satisfying  the specific  equation 

    ∇ (f∘Φ^-1)(x)   =   μ(Ω, (f∘Φ^-1)(x)) (v(x)-Ω x^⊥)^⊥
       =   μ(Ω, (f∘Φ^-1)(x)) (v^⊥(x)+Ω x),  ∀ x∈ D,

for some scalar function μ. One can easily check that any solution of (<ref>) is a solution of the initial density equation (<ref>) but the reversed is not in general true. Remark that from this latter equation we are looking for  particular solutions due to the precise dependence of the scalar function μ with respect to f.
The scalar function μ must be fixed in such a way that the radial profile f_0, around which we look for non trivial solutions, is also a solution of  (<ref>). Therefore, for any initial radial profile candidate to be bifurcated, we will obtain a different density equation. Notice also that it is not necessary  in general to impose to μ to be well–defined  on  but just on some open interval containing the image of   by f_0.

 Now, let us show how to construct  concretely the function μ. By virtue of Proposition <ref>, the associated conformal map to any radial profile is the identity map. Therefore,  it is obvious that a smooth  radial profile f_0 is a solution of (<ref>) if and only if

    f_0'(r)z/r   =   μ(Ω,f_0(z))  [-1/2π∫_z-y/|z-y|^2f_0(y)dA(y)+Ω z]
       =   μ(Ω,f_0(z)) [-1/2π∫_f_0(y)/z-ydA(y)+Ω z]
       =   μ(Ω,f_0(z))[-1/r^2∫_0^rsf_0(s)ds+Ω]z,  ∀ z∈,  r=|z|,

 where we have used the explicit computations given in Proposition <ref>. Thus, we infer  that the function μ must satisfy  the compatibility condition

    μ(Ω,f_0(r))=1/rf_0'(r)/Ω-1/r^2∫_0^rsf_0(s)ds,  ∀   r∈ (0,1].

We emphasize  that not all the radial profiles verify the last equation. In fact, we can violate this equation by working with non monotonic profiles. Taking f_0 verifying (<ref>), let us go through the above procedure and see how to reformulate the density equation. 
Consider the function

    ℳ_f_0(Ω,)=∫_t_0^1/μ(Ω,s)ds,

for some t_0∈.  We use the subscript f_0 in order to stress that the above function depends on the choice of the initial profile f_0. This rigidity is very relevant in our study  and enables us to include the structure of the solution into the formulation. By this way, we expect  to remove the pathological behavior  of the old formulation and  to prepare the problem for the bifurcation arguments.   From the expression of the velocity field, it is obvious that

    v^⊥(x)+Ω x=-∇(1/2π∫_D log|x-y|(f∘Φ^-1)(y)dA(y)-1/2Ω|x|^2).

Since  D is a simply connected domain, then integrating   (<ref>)  yields to the equivalent form

    ℳ_f_0(Ω,(f∘Φ^-1)(x))+1/2π∫_D log|x-y| (f∘Φ^-1)(y)dA(y)-1/2Ω|x|^2=λ,   ∀ x∈ D,

for some constant λ. Using  a change of variable through the conformal map Φ:→ D, we obtain the equivalent formulation in the unit disc

    ℳ_f_0(Ω,f(z))+1/2π∫_log|Φ(z)-Φ(y)|f(y)|Φ'(y)|^2dA(y)-1/2Ω|Φ(z)|^2=λ,  ∀ z∈.

It remains to   fix the constant λ by using that the initial radial profile should be  a solution of (<ref>). Thus the last integral identity in Proposition <ref> entails that

    λ=ℳ_f_0(f_0(r))-∫_r^11/τ∫_0^τ sf_0(s)dsdτ-1/2Ω r^2.

Notice that λ does not depend on r since  f_0 verifies  (<ref>).
Then, we finally arrive at the following reformulation for the density equation

    G_f_0(Ω,g,ϕ)(z)≜ℳ_f_0(Ω,f(z))+1/2π∫_log|Φ(z)-Φ(y)|f(y)|Φ'(y)|^2dA(y)-Ω/2|Φ(z)|^2-λ
    =0,

for any z∈. The above expression yields 

    G_f_0(Ω,0,0)=0,  ∀Ω∈.
  Thanks to Proposition <ref>, the conformal mapping is parametrized  outside the singular set by Ω and g and thus the  equation for  the density becomes  
 
    G_f_0(Ω,g)≜ G_f_0(Ω,g, 𝒩(Ω,g))=0.

Next, let us  analyze the constraint   (<ref>), for  some particular examples. Since we are looking for smooth solutions, it is  convenient to deal with smooth radial profiles. Then, one stands

    f_0(r)=f_0(r^2),
 
and thus  (<ref>) becomes 

    μ(Ω,f_0(r))=4 rf_0^'(r)/2Ω r-∫_0^rf_0(s)ds,  ∀   r∈ (0,1].

At this stage, there are two ways to proceed. The first one is to start with f_0 and reconstruct μ, and the second one is to impose μ and solve the nonlinear differential equation on f_0. This last approach is implicit and more delicate to implement. Therefore, let us proceed with the first approach and apply it to   some special examples.   
 


  §.§.§ Quadratic profiles

The first example  is the quadratic   profile of the type

    f_0(r)=Ar^2+B,

where A,B∈. In this case, f_0(r)=Ar+B and thus  (<ref>) agrees with 

    μ(Ω,f_0(r))
    =4Ar/2Ω r-Ar/2-Br
    =8A/4Ω-B-f_0(r),  ∀ r∈(0,1].

Then, we find

    μ_(Ω,)=8A/4Ω-B-,

which implies from (<ref>) that  

    ℳ_f_0(Ω,)=4Ω-B/8A-1/16 A^2.

Thus, using (<ref>), we deduce that

    λ   =   4Ω-B/8A f_0(r)-1/16 A f_0(r)^2-∫_r^11/τ∫_0^τ sf_0(s)dsdτ-Ω r^2/2
       =   8Ω B-3B^2-A^2-4AB/16 A.

As we have  mentioned before, the conformal mapping is determined by Ω and g and so the last equation  takes the form (<ref>).
The subscript f_0 will be omitted when we refer to this equation  with the quadratic profile if there is no confusion.

Let us remark some comparison to the vortex patch problem. The case A=0 agrees with a vortex patch of the type f_0(r)=B. It was mentioned before that the boundary equation studied in Section <ref> is the one studied in <cit.> when analyzing the vortex patch problem.  Here we have one more equation in  given by the density equation. This amounts to look for solutions of the type

    ω_0(x)=(B+g)(Φ^-1(x))  1_Φ()(x),

which implies that the initial vorticity B of the vortex patch is perturbed by a function that could not to be constant. However, using (<ref>) and evaluating in  f_0(r)=B, for any r∈[0,1], one gets that for this case μ≡ 0. Then if you perturb with g the equation to be studied is

    ∇ ((f_0+g)∘Φ^-1)(x)=0,   ∀ x∈Φ().

Using the conformal map Φ and changing the variables we arrive at 

    ∇ ((f_0+g)∘Φ^-1)(Φ(z))=0,  ∀ z∈.

By virtue of (<ref>), the above equation leads to 

    ∇ (f_0+g)(z)=0,

which gives us that g must be a constant. Hence, using our approach we get that starting with a vortex patch we just can obtain another vortex patch solution.



  §.§.§ Polynomial profiles

The second example is to consider a general polynomial profile of the type

    f_0(r)= Ar^m+B,    m∈^⋆,    A≥0,    B∈.

From  (<ref>) we obtain that

    μ(Ω,)=4m(m+1) A^1/m(-B)^m-1/m/2Ω(m+1)-mB-,   ∀≥ B.

Consequently, we find that

    ℳ_f_0(Ω,)=1/4m(m+1) A^1/m[(2Ω(m+1)-mB)∫_B^(s-B)^1-m/mds-∫_B^ s(s-B)^1-m/mds].

Remark  that for m=1 we recover the previous quadratic profiles. The discussion developed later about  the quadratic profile can be also extended to this polynomial profile as we shall  comment in detail in  Remark <ref>.




  §.§.§ Gaussian profiles

Another example which is relevant is given by  the Gaussian distribution,

    f_0(r)=e^A r,    A∈^⋆.

Inserting  f_0 into (<ref>) one obtains that

    μ(Ω,f_0(r))=4A^2 r/2Ω A r-e^Ar+1,

and thus

    μ(Ω,)=4A ln/1-+2Ωln.

Then the formula (<ref>) allows us to get

    ℳ_f_0(Ω,)=1/4A[2Ω+∫_1^1-s/ln s ds].





 §.§ Functional regularity
 
In this section, we will be interested in  the regularity of the functional  G obtained in (<ref>) for a the quadratic profile. Notice that in the case of the quadratic profile (<ref>), the singular set (<ref>) becomes

    𝒮_sing={Ω_n≜A/4+B/2-A(n+1)/2n(n+2)-B/2n,  ∀ n∈^⋆∪{+∞}}.


Let  f_0 be the quadratic profile given by (<ref>) and I be an open interval with  I⊂\𝒮_sing. Then, there exists ε>0 such that

    G:I× B_𝒞_s^1,α()(0,)→𝒞_s^1,α(),
 
is well–defined and of class 𝒞^1, where G is defined in (<ref>) and   B_𝒞_s^1,α()(0,) in (<ref>).


Let us show that G(Ω,g)∈𝒞^1,α(). Clearly, ℳ(Ω,f) is polynomial in f and by the algebra  structure of Hölder spaces we deduce that ℳ(Ω,f)∈𝒞^1,α(). Since Φ∈𝒞^2,α(), then the  only term which deserves attention is the integral one. It is clear that

    ∫_log|Φ(·)-Φ(y)|f(y)|Φ'(y)|^2dA(y)∈𝒞^0().

To estimate its derivative, we note that

    ∇_z log|Φ(z)-Φ(y)|=(Φ(z)-Φ(y))Φ'(z)/|Φ(z)-Φ(y)|^2
    =Φ'(z)/Φ(z)-Φ(y),

which implies that  

    ∇_z ∫_log|Φ(z)-Φ(y)|f(y)|Φ'(y)|^2dA(y)   =   Φ'(z)∫_f(y)/Φ(z)-Φ(y)|Φ'(y)|dA(y)
       =   Φ'(z) ℱ[Φ](f)(z),

where the operator ℱ[Φ] is defined in (<ref>). Thus, we can use Lemma <ref> obtaining that  ℱ[Φ] belongs to  𝒞^1,α(). Since Φ'∈𝒞^1,α() we deduce  that the integral term of G(Ω,g)  lies in the space  𝒞^2,α() and is continuous with respect to (f,Φ). 

Let us check the symmetry property. Take g  and ϕ satisfying  g(z)=g(z)  and  ϕ(z)=ϕ(z).
It is a simple matter to verify that
 
    ℳ(Ω,f(z))=ℳ(Ω,f(z))  and   |Φ(z)|^2=|Φ(z)|^2,  ∀  z∈.
 
 For the Newtonian potential, the change of variables y↦y leads to

    1/2π∫_log|Φ(z)-Φ(y)| f(y)|Φ^'(y)|^2 dA(y)
       =   1/2π∫_log|Φ(z)-Φ(y)| f(y)|Φ^'(y)|^2 dA(y)
       =   1/2π∫_log|Φ(z)-Φ(y)| f(y)|Φ^'(y)|^2 dA(y).

Let us turn to the computations of the Gâteaux derivatives, that can be computed as

    D_gG(Ω,g)h(z)=∂_g G(Ω,g,ϕ)h(z)+∂_ϕ G(Ω,g,ϕ) ∘∂_g 𝒩(Ω,g)h(z).

By virtue of Proposition <ref>, it is known that 𝒩 is 𝒞^1, which implies that ∂_g𝒩(Ω,g) is continuous.  Gâteaux derivatives are given by

    D_g G(Ω,g,ϕ)h(z)   =   D_g ℳ(Ω,f(z))h(z)+1/2π∫_log|Φ(z)-Φ(y)|h(y)|Φ'(y)|^2dA(y)
       =   4Ω-B/8Ah(z)-1/8Af(z)h(z)
    +1/2π∫_log|Φ(z)-Φ(y)|h(y)|Φ'(y)|^2dA(y),

and

    D_ϕ G(Ω,g,ϕ)k(z)   =   Re/2π∫_k(z)-k(y)/Φ(z)-Φ(y)f(y)|Φ'(y)|^2dA(y)-ΩRe[Φ(z)k(z)]
          +1/π∫_log|Φ(z)-Φ(y)|f(y)Re[Φ^'(y)k'(y)]dA(y).

We focus our attention on the integral terms. The operator ℱ[Φ] in (<ref>) allows us to write 

    ∫_k(·)-k(y)/Φ(·)-Φ(y)f(y)|Φ'(y)|^2dA(y)=k(z)ℱ[Φ](f)-ℱ[Φ](kf)(z).

Thus,  Lemma <ref> concludes that this term lies in 𝒞^1,α() and is continuous with respect  For the other terms involving the logarithm, we can compute its gradient as before; for instance

    ∇_z ∫_log|Φ(z)-Φ(y)|f(y)Re[Φ^'(y)k'(y)]dA(y)
              = Φ'(z)∫_f(y)/Φ(z)-Φ(y)Re[Φ'(y)k'(y)]dA(y)
     =Φ'(z) ℱ[Φ](Re[Φ'(·)k'(·)]/|Φ'(·)|^2)(z).

Since Re[Φ'(·)k'(·)]/|Φ'(·)|^2∈𝒞^1,α() and is continuous with respect to Φ, Lemma <ref> concludes that this term lies in 𝒞^1,α() and is continuous with respect to Φ. For the other terms involving a logarithmic part, the same procedure can be done. Trivially, both D_gG(Ω,g,ϕ) and D_ϕ G(Ω,g,ϕ) are continuous with respect to g. We have obtained that the Gateaux derivatives are continuous with respect to (g,ϕ) and hence they are Fréchet derivatives.
 
 


Although we have done the previous discussion for the quadratic profile, the same argument may be applied for any radial profile f_0. Note that the only difference with the quadratic profile is that the function ℳ_f_0 and the constant λ_f_0 will depend on f_0. Hence, we just have to study the regularity of function ℳ_f_0 in order to give a similar result.


 

 §.§ Radial solutions

The main goal of this sections is the resolution of  in the class of radial functions but in a small neighborhood  of  the quadratic  We establish that except  for one singular value for Ω,  no radial solutions different from f_0  may be found around it. This discussion is essential in order to ensure that with the new reformulation  we avoid the main defect of  the old one (<ref>): the kernel is infinite-dimensional and  contains radial solutions.
As it was observed before, Proposition <ref> gives us that  the  associated conformal mapping of any radial function is the identity map, and therefore  (<ref>) becomes

    G(Ω,f-f_0)(z)=4Ω-B/8Af(|z|)-1/16Af^2(|z|)+1/2π∫_log|z-y|f(|y|)dA(y)-Ω|z|^2/2-λ=0,

for any z∈ where λ is given by (<ref>). Thus the last integral identity of Proposition <ref> gives

    4Ω-B/8Af(r)-1/16Af^2(r)-∫_r^11/τ∫_0^τ sf(s)dsdτ-1/2Ω r^2-λ
    =0,  ∀ r∈[0,1].

Introduce the function G_rad: ×𝒞([0,1];) →𝒞([0,1];), defined by 

    G_rad(Ω,f)(r)=4Ω-B/8Af(r)-1/16Af^2(r)-∫_r^11/τ∫_0^τ sf(s)dsdτ-Ω r^2/2-λ,   ∀ r∈ [0,1].

 It is obvious that G_rad
is well–defined and furthermore it satisfies 

    G_rad(Ω,f_0)=0,   ∀  Ω ∈.

Through this work,  it will be more convenient to work  with the variable x instead of Ω defined as

    1/x=4/A(Ω-B/2).

Before stating our result, some properties of the hypergeometric function
x∈(-1,1)↦ F(1-√(2),1+√(2); 1;x),
are needed.  A brief account on some useful properties of Gauss hypergeometric functions will be  discussed later in the Appendix <ref>. In view of (<ref>)  we obtain   the identity

    F(1-√(2),1+√(2); 1;x)=1/1-x F(-√(2),√(2); 1;x).

According to Appendix <ref>,  we have F(1-√(2),1+√(2);1;0)=1, and it diverges to -∞  This implies that there is at least one root in (0,1). Combined with the fact that its derivative is negative according to (<ref>),  we may show that this root is unique. Denote this zero by x_0∈ (0,1)  and   set

    Ω_0≜B/2+A/4 x_0 .

Setting  the ball

    B(f_0,)={f∈𝒞([0,1];),  f-f_0_L^∞≤},

for any >0, the first result can be stated as follows.

Let f_0 be the quadratic profile  (<ref>), with A∈^⋆, B∈.  Then,  there exists >0 such that

    G_rad(Ω,f)=0 ⟺ f=f_0,

for any (Ω,f)∈ I× B(f_0,) and  any  bounded  with I∩([B/2,B/2+A/4]∪{Ω_0})=∅.


We remark that G_rad is a 𝒞^1 function on (Ω,f). The idea is to apply the Implicit Function Theorem to deduce the result.  By differentiation with respect to f, one gets  that 

    D_fG_rad(Ω,f_0)h(r)
       =   4Ω-B/8Ah(r)-1/8Af_0(r)h(r)-∫_r^11/τ∫_0^τ sh(s)dsdτ
       =   4Ω-Ar^2-2B/8Ah(r)-∫_r^11/τ∫_0^τ sh(s)dsdτ,

for any h∈𝒞([0,1];). Now we shall look for the kernel of this operator, which consists of elements h solving a Volterra  integro-differential  equation of the type

    4Ω-Ar^2-2B/8Ah(r)-∫_r^11/τ∫_0^τ sh(s)dsdτ=0,  ∀ r∈[0,1].

The assumption Ω∉[B/2,B/2+A/4]  implies  that r∈[0,1]↦4Ω-Ar^2-2B/8A is not vanishing and smooth.  Thus from the regularization of the integral, one can check that  any element of the kernel is actually  𝒞^∞. Our purpose is to derive a differential equation by differentiating successively this integral equation. With the notation (<ref>) the kernel equation can be written in the form  

    ℒh(r)≜(1/x-r^2) h(r)-8∫_r^11/τ∫_0^τ sh(s)dsdτ= 0,  ∀ r∈[0,1].

Remark that the assumptions on Ω can be translated into x, as x∈(-∞,1) and x≠ 0. 
Differentiating the function ℒh  yields

    (ℒh)^'(r)=(1/x-r^2)h^'(r)-2r h(r)+8/r∫_0^rsh(s)ds.

Multiplying by r and differentiating again we deduce that 

    [r(ℒh)^'(r)]^'/r=(1/x-r^2)h^''(r)+(1/x_0-5r^2) h^'(r)/r+4h(r)=0,  ∀ r ∈(0,1).

In order to solve the above equation we look for solutions in the form

    h(r)=ρ( x r^2).

This ansatz can be justified a posteriori by evoking the uniqueness principle   for ODEs.
Doing the change of variables y=x r^2, we  transform the preceding equation to 

    y(1-y)ρ^''(y)+(1-3y)ρ^'(y)+ρ(y)=0.

Appendix <ref> leads to assure that the  only bounded solutions close to zero  to this hypergeometric equation are given by

    ρ(y)   =   γ  F(1+√(2),1-√(2);1;y),  ∀γ∈,

and thus

    h(r)=γ  F(1+√(2),1-√(2);1;x r^2),  ∀γ∈.

It is important to note that from  the integral representation (<ref>) of hypergeometric functions, we can extend the above solution to   x∈(-∞,1). Coming back to the equation  (<ref>) and integrating two times, we obtain two real numbers α, β∈ such that

    ℒh(r)=αln r+β,  ∀ r∈(0,1].

Since ℒh∈𝒞([0,1],), we obtain that α=0 and thus 
ℒh(r)=β.
By definition one has  ℒh(1)=(1/x-1) h(1). The fact that x≠ 1 implies that ℒh=0 if and only if
h(1)=0. 
According to (<ref>), this condition is equivalent to 
γ F(1-√(2),1+√(2);1;x)=0.
It follows that the kernel is trivial (γ=0) if and only if x≠ x_0, with x_0 being the only zero of F(1-√(2),1+√(2);1;·). However, for x=x_0 the kernel is one-dimensional and is generated by this hypergeometric function. Those claims will be made more rigorous in what follows.

∙ Case x≠ x_0. As we have mentioned before, the kernel is trivial and it remains to check that ℒ is an isomorphism.  With this aim, it suffices to prove that ℒ is a Fredholm operator of zero index.  First, we  can split ℒ as follows

    ℒ≜ℒ_0+𝒦,  ℒ_0≜(1/x-r^2)Id  and  𝒦 h(r)≜-8∫_r^11/τ∫_0^τ sh(s)dsdτ.

Second, it is obvious that ℒ_0: 𝒞([0,1];)→𝒞([0,1];) is an isomorphism, it is a Fredholm operator of zero index. Now, since the Fredholm operators with given index are stable by compact perturbation (for more details, see Appendix <ref>), then to  check that ℒ has zero index it is enough to establish  that 

    𝒦:𝒞([0,1];)→𝒞([0,1];),
 is compact. One can easily obtain that for h∈𝒞([0,1];) the function 𝒦h belongs to  𝒞^1([0,1];). Furthermore, by change of variables

    (𝒦h)^'(r)=r∫_0^1sh(rs) ds  and   (𝒦h)^'(0)=0,   ∀ r∈(0,1],

 which implies that 

    𝒦h_𝒞^1≤ Ch_L^∞.

Since the embedding 𝒞^1([0,1];)↪𝒞([0,1];) is compact, we find that 𝒦 is a compact operator. Finally, we get that ℒ is an isomorphism. This ensures  that D_fG_rad(Ω,f_0) is an isomorphism, and therefore the Implicit  Function Theorem together with (<ref>) allow us to deduce that the only solutions of G(Ω,f)=0  in  I× B(f_0,) are given by the trivial 

∙ Case x= x_0. In this special case the kernel of ℒ is one–dimensional and is generated by

    Kerℒ=⟨ F(1-√(2),1+√(2);1;x  (·)^2)⟩.

This case will be deeply discussed  below in Proposition <ref>.



Let us focus on the case Ω=Ω_0. From (<ref>),  the kernel of the linearized operator  is one–dimensional, and we will be able to implement the  Crandall–Rabinowitz Theorem. Our result reads as follows.

Let f_0 be the quadratic profile  (<ref>) with A∈^⋆, B∈ and fix Ω_0 as   Then, there exists an open  neighborhood U of (Ω_0,f_0) in ×𝒞([0,1];) and a continuous curve ξ∈ (-a,a)↦ (Ω_ξ, f_ξ)∈ U with a>0 such that

    G_rad(Ω_ξ,f_ξ)=0,   ∀ξ∈ (-a,a).



We must check that the hypotheses of the Crandall–Rabinowitz Theorem are achived. It is clear that

    G_rad(Ω,f_0)=0,   ∀ Ω∈.

It is not difficult to show that the mapping   (Ω,f)↦ G_rad(Ω,f) is  𝒞^1. In addition,  we have seen  in the foregoing discussion that  D_f G_rad(Ω_0,f_0) is a Fredholm operator with zero index and its  kernel  is one–dimensional.
Therefore to apply the bifurcation arguments it remains  just to check the transversality condition in the Crandall-Rabinowitz Theorem. Having this in mind, we should first find a practical  characterization for the range  of the linearized operator.   We note that an element  d∈𝒞([0,1];) belongs to the range of D_f G_rad(Ω_0,f_0) if   the equation

    1/x_0-r^2/8h(r)-∫_r^11/τ∫_0^τ sh(s)dsdτ=d(r),  ∀ r∈[0,1],

admits a solution h in 𝒞([0,1];),  
where x_0 is given  by (<ref>). Consider the auxiliary  

    H(r)≜∫_r^11/τ∫_0^τ sh(s)dsdτ,  ∀  r∈[0,1].

Then   H belongs to  𝒞^1([0,1];) and it satisfies the boundary condition  

    H(1)=0  and   H^'(0)=0.

In order to write down  an ordinary differential equation for H, let us define the linear operator

    ℒh(r)≜∫_r^11/τ∫_0^τ sh(s)dsdτ,  ∀ r∈[0,1],

for any h∈𝒞([0,1];). Then, we derive successively,

    (ℒh)'(r)=-1/r∫_0^rsh(s)ds,

and 

    r(ℒh)”(r)+(ℒh)'(r)=-rh(r).

Applying this identity to H one arrives at

    rH”(r)+H'(r)=-8rx_0/1-x_0r^2[H(r)+d(r)].

Thus,  H solves the second order differential equation

    r(1-x_0r^2)H”(r)+(1-x_0r^2)H'(r)+8rx_0H(r)=-8rx_0d(r),

supplemented with  boundary conditions (<ref>). The argument is to come back to the original equation (<ref>) and show that the candidate 

    h(r)≜8x_0/1-x_0r^2[H(r)+d(r)],

is actually a solution to this equation. Then, we need to  check that 
ℒh=H.
By setting H≜ℒh-H, we deduce that

    r(H)”(r)+(H)'(r)=0,

with  the boundary conditions H(1)=0 and H^'(0)=0, which come from (<ref>). The solution of this differential equation is

    H(r)=λ_0+λ_1ln r,  ∀λ_0,λ_1∈

Since ℒh and H belong to 𝒞([0,1];), then necessarily  λ_1=0,  and from the boundary  condition we find λ_0=0. This implies that ℒh=H, and it shows finally that solving  (<ref>)  is equivalent to solving  (<ref>). Now, let us focus on the resolution of (<ref>). For this purpose, we proceed by finding a particular solution for the homogeneous equation and use later the method of  variation of  constants. 
Looking for a solution to the homogeneous equation in the form
H_0(r)=ρ(x_0r^2),
and using the variable y=x_0r^2, one arrives at

    (1-y)yρ^''(y)+(1-y)ρ^'(y)+2ρ(y)=0.

This is a hypergeometric equation, and one  solution  is given by
y↦ F(-√(2),√(2);1;y).
Thus, a particular solution to the homogeneous equation is 
H_0:  r∈ [0,1]↦ F(-√(2),√(2);1;x_0r^2).
Then, the general solutions for (<ref>) are given through the formula

    H(r)=H_0(r)[K_2+∫_δ^r1/τH_0^2(τ)[K_1-8x_0∫_0^τsH_0(s)/1-x_0s^2d(s)ds]dτ],  ∀ r∈[0,1],

where K_1,K_2 are real constants and δ∈(0,1) is any given number. Since H_0 is smooth on the interval  [0,1], with H_0(0)=1, one can check that H admits a singular term close to zero taking the form K_1ln r. This forces K_1 to vanish because   H  is continuous up to the origin. Therefore, we infer that

    H(r)=H_0(r)[K_2-8x_0∫_δ^r1/τH_0^2(τ)∫_0^τsH_0(s)/1-x_0s^2d(s)dsdτ],   ∀ r∈[0,1].

The last integral term is convergent at the origin and one may take δ=0. This implies that
 
    H(r)=H_0(r)[K_2-8x_0∫_0^r1/τH_0^2(τ)∫_0^τsH_0(s)/1-x_0s^2d(s)dsdτ],   ∀ r∈[0,1].

From this  expression, we deduce the second condition of (<ref>). For the first condition, H(1)=0, we first note from (<ref>) that F(-√(2),√(2);1;x_0)=0. Then, we can compute the limit at r=1 via l'Hôpital's rule leading to 

    H(1)=-8x_0lim_r→ 1^-H_0(r)∫_0^r1/τH_0^2(τ)∫_0^τsH_0(s)/1-x_0s^2d(s)dsdτ
    =8x_0∫_0^1sH_0(s)/1-x_0s^2d(s)ds/H_0^'(1).

From the expression of H_0 and (<ref>) we recover that
H_0^'(1)=-4x_0F(1-√(2),1+√(2);2;x_0).
We point out that this quantity is not vanishing. This can be proved by differentiating the relation (<ref>), which implies that

    F(1-√(2),1+√(2);2;x_0)=(x_0-1)F(2-√(2),2+√(2);2;x_0),

and the latter term is not vanishing from the definition of hypergeometric functions.
It follows that the condition H(1)=0 is equivalent to 

    ∫_0^1sH_0(s)/1-x_0s^2d(s)ds=0.

This characterizes the elements of the range of the linearized operator. 
Now, we are in a position  to check the transversality condition. According to the expression (<ref>), one gets by differentiating with respect to Ω that

    D_Ω,f G_rad(Ω_0,f_0)h(r)=1/2Ah(r).

Recall from (<ref>) and the relation (<ref>) that the kernel is generated by 
r∈[0,1]↦H_0(r)/1-x_0 r^2.
Hence, from (<ref>) the transversality assumption in the Crandall–Rabinowitz Theorem becomes

    ∫_0^1sH_0^2(s)/(1-x_0s^2)^2ds≠ 0,
 
which is trivially satisfied, and concludes  the announced result.




§ LINEARIZED OPERATOR FOR THE DENSITY EQUATION
 
This section is devoted to the study of the linearized operator of the density equation (<ref>). First, we will compute it with a general f_0 and provide a suitable formula in  the case of  quadratic profiles.  Second, we shall prove  that the  linearized operator is a Fredholm operator of index zero because it takes the form of  a compact perturbation of an invertible operator. More details about Fredholm operators can be found in Appendix <ref>.
 Later we will focus our attention on the algebraic structure of the kernel and the range and give explicit expressions by using hypergeometric functions. We point out that the kernel  description  is done through the resolution of  a Volterra integro-differential equation.


 §.§ General formula and Fredholm index of the linearized operator

Let f_0 be  an arbitrary  smooth radial function satisfying  (<ref>) and let  us compute the linearized operator of the functional G_f_0 given by  (<ref>). First, using Proposition <ref> one gets

    D_g𝒩(Ω, 0)h(z)=z∑_n≥ 1 A_n z^n   ≜    k(z),

where A_n is given in (<ref>) and h∈𝒞^1,α_s(). Therefore, differentiating with respect to g yields

    D_gG_f_0(Ω,0)h   =   ∂_g G_f_0(Ω,0,0)h+∂_ϕG_f_0(Ω,0,0) ∂_g𝒩h(Ω,0)
    
       =   ∂_g G_f_0(Ω,0,0)h+∂_ϕG_f_0(Ω,0,0) k.

Using the Fréchet derivatives from Proposition <ref>, we have that

    D_gG_f_0(Ω,0)h(z)   =   D_g ℳ_f_0(Ω,f_0(z))h(z)+1/2π∫_log |z-y|h(y)dA(y)-ΩRe[zk(z)]
          +1/2π∫_Re[k(z)-k(y)/z-y]f_0(y)dA(y)
          +1/π∫_log|z-y|f_0(y)Re[k'(y)]dA(y).

From the definition of ℳ_f_0 in (<ref>)  we infer that

    D_g ℳ_f_0(Ω,f_0(z))h(z)=h(z)/μ(Ω,f_0(|z|)),  ∀  z∈.

Putting together  the preceding formulas we obtain

    D_g G_f_0(Ω,0)h(z)   =   h(z)/μ(Ω,f_0(r))+1/2π∫_log |z-y|h(y)dA(y)-ΩRe[zk(z)]
          +1/2π∫_Re[k(z)-k(y)/z-y]f_0(y)dA(y)
          +1/π∫_log|z-y|f_0(y)Re[k'(y)]dA(y),

where  μ is given by the compatibility condition (<ref>). 
Next, we shall rewrite  the linearized operator for the quadratic profile, and we will omit the subscript f_0  for the sake of simplicity. Taking  μ as in (<ref>), we get

    D_g G(Ω,0)h(z)   =   1/8(1/x-r^2)h(z)+1/2π∫_log |z-y|h(y)dA(y)-ΩRe[zk(z)]
          +1/2π∫_Re[k(z)-k(y)/z-y]f_0(y)dA(y)
          +1/π∫_log|z-y|f_0(y)Re[k'(y)]dA(y),

where x is given by (<ref>).

In the following result we show that the linearized operator associated to a quadratic profile is a Fredholm operator with  index zero. Similar result may be obtained in the general case imposing suitable conditions on the profiles.
 
 Let f_0 be the profile (<ref>), with A∈^⋆ and B∈. Assume that   Ω∉[B/2,B/2+A/4]∪𝒮_sing. Then,  
D_g G(Ω,0):𝒞^1,α_s()→𝒞^1,α_s() is a Fredholm operator with zero index.  


Using (<ref>) we have

    D_g G(Ω,0)h(z)   =   1/8(1/x-r^2)h(z)+1/2π∫_log |z-y|h(y)dA(y)-ΩRe[zk(z)]
          +1/2π∫_Re[k(z)-k(y)/z-y]f_0(y)dA(y)
          +1/π∫_log|z-y|f_0(y)Re[k'(y)]dA(y)
       ≜   [1/8(1/x-r^2)Id+𝒦]h(z),

where k is related to h through (<ref>). The assumption on Ω entails that the smooth function z∈↦1/x-|z|^2 is not vanishing on the closed unit disc. Then the operator 

    1/8(1/x-|z|^2)Id:𝒞^1,α_s()→𝒞^1,α_s(),

 is an isomorphism. Hence, it is a Fredholm operator with zero index. To check that ℒ is also a Fredholm operator with zero index, it suffices to prove that the operator 𝒦: 𝒞^1,α_s()→𝒞^1,α_s() is compact. To do that, we will prove that 𝒦: 𝒞^1,α()→𝒞^1,γ(), for any  γ∈ (0,1). 
We split 𝒦 as follows

    𝒦h=∑_j=1^4𝒦_jh,

with

    𝒦_1h= -ΩRe[zk(z)],  𝒦_2h(z)=1/2π∫_log |z-y|h(y)dA(y),


    𝒦_3h(z)=1/π∫_log|z-y|f_0(y)Re[k'(y)]dA(y),  𝒦_4h=1/2π∫_Re[k(z)-k(y)/z-y]f_0(y)dA(y).

The estimate of the first term 𝒦_1h follows from (<ref>) and (<ref>), leading to 

    𝒦_1h_𝒞^2,α()≤ Ch_𝒞^1,α().

Concerning the term 𝒦_2 we note that 

    𝒦_2h_L^∞()≤ Ch_L^∞(),

and differentiating it, we obtain

    ∇_z𝒦_2h(z)= 1/2π∫_h(y)/z-ydA(y).

Lemma <ref> yields 

    ∇_z 𝒦_2h_𝒞^0,γ()≤ Ch_L^∞(),

for any γ∈(0,1), and thus

    𝒦_2h_𝒞^1,γ()≤ Ch_L^∞().

The estimate of 𝒦_3 is similar to that of 𝒦_2, and using (<ref>) and (<ref>)  we find that

    𝒦_3h_𝒞^1,γ()≤  Cf_0_L^∞() k'_L^∞()≤ Cf_0_L^∞()h_𝒞^1,α().

Setting

    K(z,y)=k(z)-k(y)/z-y,   ∀  z≠ y,

it is obvious that
|K(z,y)|≤k^'_L^∞().
Therefore, we have

    𝒦_4h_L^∞≤  Cf_0_L^∞k^'_L^∞().

Moreover, by differentiation we find

    ∇_z𝒦_4h(z)=1/2πRe∫_∇_zK(z,y)f_0(y)dA(y).

Straightforward computations show that

    |∇_zK(z,y)|   ≤    Ck^'_L^∞()|z-y|^-1,
    
    |∇_zK(z_1,y)-∇_zK(z_2,y)|   ≤    C|z_1-z_2|[k^''_L^∞()/|z_1-y|+k^'_L^∞()/|z_1-y||z_2-y|].

Thus, hypotheses (<ref>) are satisfied and we can use Lemma <ref>  and (<ref>) to find

    ∇_z𝒦_4h_𝒞^0,γ()≤ Cf_0_L^∞k_𝒞^2()≤ Cf_0_L^∞h_𝒞^1,α(),

obtaining

    𝒦_4h_C^1,γ()≤  Cf_0_L^∞()h_𝒞^1,α().

Combining the estimates (<ref>),(<ref>),(<ref>) and (<ref>), we deduce 

    𝒦h_𝒞^1,γ()≤  Ch_𝒞^1,α(),

which concludes the proof.

To end  this subsection, we give a more explicit form of the linearized operator. Coming back to the general expression in (<ref>) and using Proposition <ref>  we get that

    D_g G_f_0(Ω,0)h(z)   =   ∑_n≥ 1cos(nθ)[h_n(r)/μ(Ω,f_0(r))-r/n(A_nG_n(r)+1/2r^n+1H_n(r))]
          +h_0(r)/μ(Ω,f_0(r))-∫_r^11/τ∫_0^τ sh_0(s)dsdτ,

for 

    r e^iθ↦ h(re^iθ)=∑_n∈h_n(r)cos(nθ)∈𝒞^1,α_s(),

where

    G_n(r)   ≜   nΩ r^n+1+r^n-1∫_0^1sf_0(s)ds-(n+1)r^n-1∫_0^rsf_0(s)ds
    +n+1/r^n+1∫_0^rs^2n+1f_0(s)ds,
    
    H_n(r)   ≜   r^2n∫_r^11/s^n-1h_n(s)ds+∫_0^rs^n+1h_n(s)ds,

 for any n≥1. The value of A_n  is given by (<ref>) and recall that it was derived  from the expression ∂_g 𝒩(Ω,0) when studying the boundary equation.
Moreover, there is another useful expression for A_n coming from the value of G_n(1)

    G_n(1)=n[Ω-∫_0^1sf_0(s)ds+n+1/n∫_0^1s^2n+1f_0(s)ds]
    =n(Ω-Ω_n).

Those preceding identities agrees with

    A_n=-H_n(1)/2G_n(1),   ∀ n≥ 1.

In the special case of f_0 being a quadratic profiles of the type (<ref>),
straightforward computations imply that

    D_g G(Ω,0)h(z)   =   ∑_n≥ 1cos(nθ) [1/x-r^2/8h_n(r)-r/n(A_nG_n(r)+1/2r^n+1H_n(r))]
          +1/x-r^2/8h_0(r)-∫_r^11/τ∫_0^τ sh_0(s)dsdτ,

with 

    G_n(r)   =   -An(n+1)/4(n+2)r^n-1P_n(r^2),
    
    H_n(r)   =   r^2n∫_r^11/s^n-1h_n(s)ds+∫_0^rs^n+1h_n(s)ds,
    
    P_n(r)   =   r^2-n+2/n+11/xr-A+2B/An+2/n(n+1),
    
     G_n(1)   =   -An(n+1)/4(n+2)P_n(1)
    =n[A/4(1/x-1)+A(n+1)/2n(n+2)+B/2n],
    
    A_n   =   -H_n(1)/2G_n(1)
    = H_n(1)/2n{Ω_n-Ω}.

Remark  that G_n(1)≠ 0  since we are assuming that Ω∉𝒮_sing,  the singular set  defined 

From now on we will work only with the quadratic profiles.  Similar study could be implemented with general profiles but the analysis may turn out to be  very difficult because the spectral  study is intimately related to the distribution of the selected  profile.




 §.§ Kernel structure and negative results

The current objective is to conduct a precise study for the kernel structure of the linearized operator (<ref>). We must identify 
the master equation describing the dispersion relation. As a by-product we   connect the dimension of  the kernel to the number of roots of the master equation. We shall distinguish in this study between the regular case corresponding to x∈(-∞,1) and the singular case associated to x>1. For this latter case we prove that the equation (<ref>) has no solution close to the trivial one. 


  §.§.§ Regular case

Let us start with a preliminary result devoted to the explicit resolution of a second order differential equation with polynomial coefficients taking the form

    (1-xr^2)rF”(r)-(1-xr^2)(2n-1)F'(r)+8rxF(r)=g(r),  ∀ r∈[0,1],

 This will be applied later to  the study of the kernel and the range. Before stating our result we need to introduce some functions
 
    F_n(r)=F(a_n, b_n; c_n; r),   a_n=n-√(n^2+8)/2,   
    b_n=n+√(n^2+8)/2,   
    c_n=n+1.
 
where r∈[0,1)↦ F(a,b;c;r) denotes  the Gauss hypergeometric function defined in (<ref>). 
Let  n≥ 1 be an integer,  x∈(-∞,1) and g∈𝒞([0,1];). Then, the general continuous solutions of equation (<ref>)
supplemented with the initial condition F(0)=0, are given by a one–parameter curve 

    F(r)=r^2nF_n(xr^2)[F(1)/F_n(x)-x^n-1/4∫_xr^2^x1/τ^n+1F_n^2(τ)∫_0^τF_n(s)/1-s(x/s)^1/2g((s/x)^1/2)dsdτ].



Consider the auxiliary function
F(r)=F(xr^2)
and set y=xr^2. Note that y∈[0,x] when x>0 and y∈[x,0] if x<0, then r=(y/x)^1/2 in both cases. Hence, the equation governing  this new function is

    (1-y)yF^''(y)-(n-1)(1-y)F^'(y)+2F(y)=1/4x(x/y)^1/2g((y/x)^1/2),

with the boundary condition F(0)=0. The strategy  to be followed consists in solving the homogeneous equation and using later the  method of variation of constants. The homogeneous problem is given by 

    (1-y)yF_0^''(y)-(n-1)(1-y)F_0^'(y)+2F_0(y)=0.

Comparing it with the general differential equation (<ref>), we  obviously find that F_0 satisfies a hypergeometric equation with the  parameters

    a=-n-√(n^2+8)/2,  b=-n+√(n^2+8)/2,   c=1-n.

The general theory of hypergeometric functions gives us that this differential equation is degenerate because  c is a negative integer, see discussion in Appendix <ref>.  However, we still get  two independent solutions generating the class of solutions to this differential equation:  one is smooth and the second  is singular and contains a logarithmic singularity at the origin. The smooth one is  given by 

    y∈(-∞,1)↦ y^1-cF(1+a-c,1+b-c,2-c,y).

With the special parameters (<ref>), it becomes 

    y∈(-∞,1)↦ y^nF(a_n,b_n;c_n;y)= y^nF_n(y).

It is important to note that,  by Taylor  expansion, the hypergeometric function initially  defined in the unit disc   admits an analytic continuation in the complex plane cut along the  real axis from 1 to +∞ . This comes from  the integral representation (<ref>).

Next, we  use  the method of variation of  constants with the smooth homogeneous solution  and set

    F_0: (-∞,1)∈↦ y^nF_n(y).

We wish to mention that when using  the method of variation of  constants with the smooth solution we also find the trace of the singular solution. As we will notice in the next step, this singular part will not contribute for the full inhomogeneous problem due  to the required regularity and   the boundary   Now, we solve the equation (<ref>) by  looking  for solutions  in the form
F(y)=F_0(y)K(y).
By setting  K≜ K^', one has that

    K^'(y)+[2F_0^'(y)/F_0(y)-n-1/y]K(y)=1/4x(x/y)^1/2g((y/x)^1/2)/y(1-y)F_0(y),

which can be integrated in the following way

    K(y)=y^n-1/F_0^2(y){K_1+1/4x∫_0^yF_0(s)/s^n(1-s)(x/s)^1/2g((s/x)^1/2)ds},

where K_1 is a constant. Thus integrating successively  we  find  K and  F and from  the expression of F_0 we deduce

    F(y)=y^nF_n(y)[K_2-∫_y^sign y1/τ^n+1F_n^2(τ){K_1+1/4x∫_0^τF_n(s)/1-s(x/s)^1/2g((s/x)^1/2)ds}dτ],

for any y∈ (-∞,1), with K_2 a constant and where sign is the sign function. 
From straightforward computations using integration by parts we get 

    F(0)=
    -K_1lim_y→ 0 y^nF_n(y)∫_y^sign y1/τ^n+1F_n^2(τ)dτ=-K_1/n.

Combined with
the initial condition F(0)=0 we obtain 
 K_1=0.  Coming back to  the original function  
F(r)=F(xr^2),
we obtain

    F(r)=
    x^nr^2nF_n(xr^2)[K_2-1/4x∫_xr^2^sign x1/τ^n+1F_n^2(τ)∫_0^τF_n(s)/1-s(x/s)^1/2g((s/x)^1/2)dsdτ].

The constant K_2 can be computed by evaluating the preceding expression at r=1. We finally get 

    F(r)=r^2nF_n(xr^2)[F(1)/F_n(x)-x^n-1/4∫_xr^2^x1/τ^n+1F_n^2(τ)∫_0^τF_n(s)/1-s√(x/s) g(√(s/x))dsdτ],

for r∈[0,1]. Observe  from the integral representation (<ref>) that the function F_n does not vanish on (-∞,1) for n≥1. Hence, (<ref>) is well-defined and F is C^∞ in [0,1] when x<1.  
 

The next goal is to  give the kernel structure of the linearized operator D_g G(Ω,0).  We emphasize that according to Proposition <ref>, this is a Fredholm operator of zero index, which implies in particular  that its kernel is finite–dimensional.  Before that, we introduce the singular set for x connected to the singular set of Ω through the relations (<ref>) and (<ref>)

    𝒮_sing={x_n=A/4(Ω_n-B/2),   Ω_n∈𝒮_sing}.

For any n≥ 1, consider the following sequences of functions 

    ζ_n(x)≜  F_n(x)[1-x+A+2B/A(n+1) x]+∫_0^1 F_n(τ x) τ^n[-1+2xτ] dτ,  ∀ x∈(-∞,1],

 where F_n has been introduced in (<ref>). Then we prove the following. 

Let  A∈^⋆, B∈ and  
x∈ (-∞,1)\{𝒮_sing∪{0,x_0}}, with (<ref>) and  (<ref>). Define the set

    𝒜_x≜{n∈^⋆,  ζ_n(x)=0}.

Then, the kernel of D_gG(Ω,0) is finite–dimensional  and  generated by the 𝒞^∞ functions
{h_n,    n∈𝒜_x},
with
h_n: z∈↦Re[𝒢_n(|z|^2)z^n]
and 

    𝒢_n()=1/1-x [-P_n()/P_n(1)+F_n(x)/F_n(x)-2xF_n(x)/P_n(1)∫_^11/τ^n+1F_n^2(xτ)∫_0^τs^nF_n(xs)/1-xsP_n(s)dsdτ].

As a consequence,  dim Ker  D_gG(Ω,0)=Card 𝒜_x.  The functions P_n and F_n are defined in (<ref>) and (<ref>), respectively.



Notice that the set 𝒜_x can  be  empty;  in that case  the kernel of D_gG(Ω,0) is trivial. Otherwise, the set 𝒜_x is  finite.


To analyze the kernel structure, we return  to (<ref>) and solve the equations keeping in mind the relations (<ref>). Thus we should solve 

    h_n(r)+4r/n(r^2-1/x)[-H_n(1)/G_n(1)G_n(r)+H_n(r)/r^n+1]   =0,  ∀ r∈[0,1],  ∀ n∈^⋆,
    1/x-r^2/8h_0(r)-∫_τ^11/τ∫_0^τ sh_0(s)dsdτ   =0,  ∀ r∈[0,1],

where the functions involved in the last expressions are given in (<ref>)-(<ref>). The term (r^2-1/x) is not vanishing from the assumptions on x. Note that the last equation for n=0 has been already studied in Proposition <ref>, which  implies that  if x≠ x_0, then  the zero function is the only  solution. Hence, let us focus on the case n≥ 1 and solve the associated equation. To deal with this equation we  write down a  differential equation for H_n and use Lemma <ref>.  Firstly, we define the linear operator 

    ℒh(r)≜ r^2n∫_r^11/s^n-1h(s)ds+∫_0^rs^n+1h(s)ds,

for any h∈𝒞([0,1];). Then,  by differentiation we obtain

    (ℒh)^'(r)=2nr^2n-1∫_r^11/s^n-1h(s)ds
    =2n/r[ℒh(r)-∫_0^rs^n+1h(s)ds].

It is important to precise at this stage that ℒh satisfies the boundary conditions

    (ℒh)(0)=(ℒh)^'(1)=0.

Indeed, the second condition is obvious and to get the first one we use  that h is bounded:

    |ℒh(r)|≤h_L^∞(|r^2n-r^2+n|/|n-2|+r^n+2/n+2),

for any n≠ 2. In the case n=2 we have

    |ℒh(r)|≤h_L^∞(r^4|ln r|+r^4/4),

and for n=1 it is clearly verified.
Differentiating again  we obtain

    1/2n[r(ℒh)^'(r)]^'-(ℒh)^'(r)=-r^n+1h(r).

Since   H_n=ℒh_n, one has 

    1/2n[rH_n^'(r)]^'-H_n^'(r)=-r^n+1h_n(r).

 Using Equation (<ref>), we deduce that H_n satisfies the following differential equation

    (1-xr^2)rH_n”(r)-(1-xr^2)(2n-1)H_n'(r)+8rxH_n(r)=8H_n(1)x/G_n(1)r^n+2G_n(r),

complemented with  the boundary conditions (<ref>). 
Let us show how to recover the full solutions of (<ref>) from this equation.  Assume that we have constructed all  the  solutions H_n of (<ref>),   with the boundary conditions (<ref>). Then, to obtain  the solutions of  (<ref>), we should check the compatibility condition
ℒ h_n=H_n,
by setting

    h_n(r)≜4r/n(r^2-1/x)[H_n(1)G_n(r)/G_n(1)-H_n(r)/r^n+1].

Combining (<ref>) and (<ref>), we deduce that H≜ℒh_n-H_n satisfies 

    1/2n[r H^'(r)]^'-H^'(r)=0.

By solving this differential equation we obtain the existence of two real constants  λ_0 and λ_1  such that
H(r)=λ_0+λ_1 r^2n, for any r∈ [0,1].
Since both ℒh_n and H_n satisfy (<ref>), then H satisfies also these conditions. Hence, we find
H(r)=0, for any r∈[0,1],
and this concludes that h_n, given by (<ref>), is a solution of  (<ref>).
We emphasize that  h_n satisfies the compatibility condition

    H_n(1)=∫_0^1 r^n+1h_n(r)dr.
 
 Indeed, integrating  (<ref>) from 0 to 1, we obtain

    H_n^'(1)/2n-(H_n(1)-H_n(0))=-∫_0^1 r^n+1h_n(r)dr.

Thus, if H_n satisfies the boundary conditions (<ref>),  then the compatibility condition is automatically verified.
Now, let us come back  to the resolution of  (<ref>). Since  H_n(0)=0,  then one  can apply  Lemma <ref>  with

    g(r)=8H_n(1)x/G_n(1) r^n+2G_n(r).

Therefore, we obtain after a  change of variables  that 

    H_n(r)   =   r^2nF_n(xr^2)H_n(1)[1/F_n(x).
     
          .-2x^n/G_n(1)∫_xr^2^x1/τ^n+1F_n^2(τ)∫_0^τF_n(s)/1-s(s/x)^n+1/2G_n((s/x)^1/2)dsdτ]
     
       =     H_n(1)r^2nF_n(xr^2)[1/F_n(x)
          -8x/G_n(1)∫_r^11/τ^2n+1F_n^2(xτ^2)∫_0^τs^n+2F_n(xs^2)/1-xs^2G_n(s)dsdτ].

It remains to check the second initial condition: H_n^'(1)=0. From straightforward computations using (<ref>)  and (<ref>) we find that

    H_n^'(1)   =   H_n(1)/F_n(x)[2nF_n(x)+2xF_n^'(x)]+8xH_n(1)/G_n(1)F_n(x)∫_0^1s^n+2F_n(xs^2)/1-xs^2G_n(s)ds
       =   2H_n(1)/F_n(x)[φ_n(x)-An(n+1)x/(n+2)G_n(1)∫_0^1s^2n+1F_n(xs^2)/1-xs^2P_n(s^2)ds]
       =   2n H_n(1)/F_n(x)G_n(1)Ψ_n(x),

where 

    Ψ_n(x)≜φ_n(x)[A/4(1/x-1)+A/2n+1/n^2+2n+B/2n]-A(n+1)x/n+2∫_0^1s^2n+1 F_n(xs^2)/1-xs^2P_n(s^2)ds,

and

    φ_n(x)≜ nF_n(x)+xF_n^'(x).

Note that G_n(1)≠ 0 because Ω∉𝒮_sing. Let us link Ψ_n  to the function ζ_n. Recall from (<ref>) and (<ref>) that 

    x(1-x)F_n^''+(n+1)(1-x)F_n^'(x)+2F_n=0.

Differentiating the function φ_n(x)  and using the differential equation for F_n we realize that

    φ_n^'(x)   =   (n+1)F_n'(x)+xF_n”(x)
       =   1/1-x[(1-x)(n+1)F_n'(x)+(1-x)xF_n”(x)]
    =-2F_n(x)/1-x.

The change of variables xs^2↦τ in the integral term yields

    ∫_0^1s^2n+1 F_n(xs^2)/1-xs^2P_n(s^2)ds=1/2x^n+1∫_0^xτ^nF_n(τ)/1-τP_n(τ/x)dτ.

Therefore we get

    Ψ_n(x)=φ_n(x)[A/4(1/x-1)+A/2n+1/n^2+2n+B/2n]+A(n+1)/4(n+2) x^n∫_0^x φ^'_n(τ)τ^n P_n(τ/x)dτ.

From Definition (<ref>) we find the identity

    ∫_0^x φ^'_n(τ)τ^n P_n(τ/x)dτ=1/x^2∫_0^x φ^'_n(τ)τ^n [τ^2-τn+2/n+1-A+2B/An+2/n(n+1) x^2]dτ.

Integrating by parts, we deduce 

    A(n+1)/4(n+2) x^n∫_0^x φ^'_n(τ)τ^n P_n(τ/x)dτ   =   A/4φ_n(x) (n+1/n+2-1/x-A+2B/A n)
          -A(n+1)/4 x^n+2∫_0^xφ_n(τ) τ^n-1( τ^2-τ-A+2B/A(n+1)x^2 ) dτ.

By virtue of the following identity

    A/4(1/x-1)+A/2n+1/n^2+2n+B/2n+A/4(n+1/n+2-1/x-A+2B/A n)=0,

the boundary term in the integral is canceled  with the first part of Ψ_n. Thus

    Ψ_n(x)=-A(n+1)/4 x^n+2∫_0^xφ_n(τ) τ^n-1( τ^2-τ-A+2B/A(n+1)x^2 ) dτ,

where, after a change of variables in the integral term, we get  that

    Ψ_n(x)=A(n+1)/4 x∫_0^1φ_n(τ x) τ^n-1( -xτ^2+τ+A+2B/A(n+1)x ) dτ .

Setting 

    ζ_n(x)=∫_0^1φ_n(τ x) τ^n-1(-x τ^2+τ+A+2B/A(n+1) x) dτ,

we find the relation

    Ψ_n(x)=A(n+1)/4 xζ_n(x).

Observe  first that the zeroes of Ψ_n and ζ_n are the same.
Coming back to (<ref>) and 
integrating by parts we get the equivalent form

    ζ_n(x)= F_n(x)(1-x+A+2B/A(n+1) x)+∫_0^1 F_n(τ x) τ^n(-1+2xτ) dτ.

This gives  (<ref>).
According to (<ref>), the constraint  H_n'(1)=0  is equivalent to  H_n(1)=0 or ζ_n(x)=0. In the first case,  we get  from (<ref>) that H_n≡ 0 and inserting this into (<ref>) we find
h_n(r)=0, for any r∈[0,1].
Thus, for  n∉𝒜_x, where 𝒜_x is given by  (<ref>), we obtain that  there is only one  solution for the kernel equation, which is  the trivial one.
As to the second condition ζ_n(x)=0, which agrees with n∈𝒜_x, one gets from (<ref>)  and (<ref>)  that the kernel of D_gG(Ω,0) restricted to the level  frequency n is generated by 

    h_n(re^iθ)=h^⋆_n(r)cos(nθ),

with

    h^⋆_n(r)=1/1-xr^2[-rG_n(r)/G_n(1)+ r^nF_n(xr^2)/F_n(x)-8xr^nF_n(xr^2)_r^1∫_0^τs^n+2F_n(xs^2)/1-xs^2G_n(s)/G_n(1)ds/τ^2n+1F_n^2(xτ^2)dτ].

The fact that h_n=4x/nH_n(1)  h^⋆_n together with  H_n(1)=∫_0^1s^n+1h_n(s)ds imply

    ∫_0^1s^n+1h_n^⋆(s)ds=n/4x.

Using

    rG_n(r)/G_n(1)=r^nP_n(r^2)/P_n(1),

and a suitable change of variables allows getting the formula,

    _r^1∫_0^τs^n+2F_n(xs^2)/1-xs^2G_n(s)/G_n(1)ds/τ^2n+1F_n^2(xτ^2)dτ   =   1/P_n(1)_r^1∫_0^τs^2n+1F_n(xs^2)/1-xs^2P_n(s^2)ds/τ^2n+1F_n^2(xτ^2)dτ
       =   1/4P_n(1)_r^2^1∫_0^τs^nF_n(xs)/1-xsP_n(s)ds/τ^n+1F_n^2(xτ)dτ.

We have

    h^⋆_n(r)=r^n/1-xr^2[-P_n(r^2)/P_n(1)+ F_n(xr^2)/F_n(x)-2x F_n(xr^2)/P_n(1)∫_r^2^11/τ^n+1F_n^2(xτ)∫_0^τs^nF_n(xs)/1-xsP_n(s)ds dτ].

Setting

    𝒢_n()=1/1-x[-P_n()/P_n(1)+F_n(x)/F_n(x)-2xF_n(x)/P_n(1)∫_^11/τ^n+1F_n^2(xτ)∫_0^τs^nF_n(xs)/1-xsP_n(s)ds dτ],

we deduce that

    h_n(z)=𝒢_n(|z|^2)|z|^ncos(nθ)
    =[𝒢_n(|z|^2)z^n],   ∀   z∈.

We intend to check that h_n belongs to 𝒞^∞(;). To get this it is enough to   verify that 𝒢 belongs to 𝒞^∞([0,1];). Since x∈ (-∞,1), then τ∈[0,1]↦ F_n(xτ) is in 𝒞^∞([0,1];). The change of variables s=τθ implies 

    _^1∫_0^τs^nF_n(xs)/1-xsP_n(s)ds/τ^n+1F_n^2(xτ)dτ =_^1∫_0^1θ^nF_n(xτθ) P_n(τθ)/1-xτθdθ/F_n^2(xτ)dτ.

Since F_n does not vanish on (-∞,1), then the mapping τ∈[0,1]↦1/F_n(xτ) belongs to  𝒞^∞([0,1];).  It suffices to observe that the integral function is 𝒞^∞ on [0,1].  Then,  we have an independent element of the kernel given by h_n^⋆(·), for any n∈𝒜_x. This concludes the announced result.




The hypergeometric function F_n for n=1 can be computed as
F_1(r)=F(a_1,b_1;c_1;r)=1-r.
Hence, the function (<ref>) becomes

    ζ_1(x)   =    F_1(x)[1-x+A+2B/2Ax]+∫_0^1F_1(τ x)τ[-1+2xτ]dτ
       =   (1-x)[1-x+A+2B/2Ax]+∫_0^1(1-τ x)τ[-1+2xτ]dτ
    =(1-x)(B/Ax+1/2).

The root x=1 is not allowed since x∉𝒮_sing. Therefore, the unique root is x=-A/2B. Coming back to Ω using (<ref>), one has that Ω=0.





  §.§.§ Singular case

The singular case x∈(1,+∞) is studied in this section. Notice that from  (<ref>), we obtain 

    B/2<Ω<B/2+A/4.

It is worthy to point out that this  case is degenerate  because the leading terms of the  equations of the linearized operator (<ref>)  vanish inside the unit disc. 
To understand this operator one should deal with a second differential equation of hypergeometric type with a singularity. Thus, the first  difficulty amounts to solving  those equations across the singularity and invert the operator. This can be done in a straightforward way  getting   that the operator is injective with an explicit representation of its formal inverse. However, it is not an isomorphism and undergoes  a loss of regularity in the Hölder class. Despite this  bad behavior, one would  expect at least  the persistence of the injectivity for the nonlinear problem.  This  problem appears in different contexts, for instance in the inverse backscattering problem <cit.>. The idea to overcome this difficulty is to prove two key ingredients. The first one concerns the coercivity of the linearized operator with a quantified loss in the Hölder class. The second point is to  use the Taylor expansion and to establish  a soft  estimate  for the reminder combined with  an interpolation argument.  This argument leads to the following result.

Let 0<α<1, A>0 and B∈ such that B/A∉[-1,-1/2]. Assume that  Ω satisfies
 (<ref>) and Ω∉𝒮_sing, where this latter set is defined in (<ref>). Then, there exists a small  of the origin in 𝒞^2,α_s() such that the nonlinear equation (<ref>) has no solution in V, except the origin. Notice that in the case B<-A,  the condition Ω∉𝒮_sing follows automatically  from (<ref>).

The proof of this theorem will be given at the end of this section. Before we should develop some tools. Let us start with solving the kernel equations, for this reason we introduce some auxiliary functions. Set

    F_n(x)=F(-a_n,b_n;b_n-a_n+1;x),

and define the functions

    ℱ_K_1,K_2(y)=    {[                      y^nF_n(y)[K_1/F_n(1)-∫_y^1∫_0^τF_n(s)/1-sℛ(s/x)ds/τ^n+1F_n^2(τ)dτ],    y∈[0,1],; y^a_nF_n(1/y)[K_1/F_n(1)+∫_1/y^1K_2-∫_τ^1s^n-a_n-1F_n(s)/1-sℛ(1/sx)ds/τ^n+1-2a_nF_n^2(τ)dτ],    y≥1, ].

and

    ℱ_K_1,K_2(y)=    {[                                 y(1-y)[K_1+∫_0^y∫_0^τℛ(s/x)ds/τ^2(1-τ)^2dτ],    y∈[0,1],; 1/yF_1(1/y)[1/3∫_0^1ℛ(s/x)ds+∫_1/y^1K_2-∫_τ^1sF_1(s)/1-sℛ(1/sx)ds/τ^4F_1^2(τ)dτ],   y≥1, ].

with

    ℛ(y)=1/4xy^-1/2g(y^1/2),

where g is the source term in (<ref>).
Our first result reads as follows.

Let  n≥ 2 be an integer,  x∈(1,+∞) and g∈𝒞([0,1];). Then, the  continuous solutions in [0,+∞) to the equation (<ref>), 
such that F(0)=0, are given by the  two–parameters curve 

    r∈[0,1]↦ F(r)=ℱ_K_1,K_2(xr^2),   K_1,K_2∈,

with the constraint 

    ℛ(1/x)=0.

Moreover, if g∈𝒞^μ([0,1];), for some μ>0, then the  above solutions  are  𝒞^1 on [0,1] if and only if the following conditions hold true:

    K_1=0  and   K_2=F_n(1)/F_n(1)∫_0^1F_n(s)ℛ(s/x)/1-sds.

For n=1, if F(0)=0 and g∈𝒞([0,1];), with the condition (<ref>), then
the continuous solutions are given by the two-parameters curve

    r∈[0,1]↦ F(r)=ℱ_K_1,K_2(xr^2),   K_1,K_2∈.

If g∈𝒞^μ([0,1];), for some μ>0,  then this solution  is 𝒞^1  if and only if

    ℛ(0)=∫_0^1/xℛ(τ)dτ=0,

and K_1 and K_2 satisfy

    K_2   =-3(K_1+∫_0^1∫_0^τℛ(s/x)ds/τ^2(1-τ)^2dτ).

Moreover, if F'(1)=0, we have the additional constraint

    K_2   (1/x^2..[F_1(1/x)+xF_1'(1/x)]+x/F_1(1/x))
    =x/F_1(1/x)∫_1/x^1sF_1(s)/1-sℛ(1/sx)ds
       +1/x^2[F_1(1/x)+1/xF_1'(1/x)] ×∫_1/x^1∫_τ^1sF_1(s)/1-sℛ(1/sx)ds/τ^4F_1^2(τ)dτ.




We proceed as in the proof of Lemma <ref>. The resolution of the equation (<ref>),  in the interval [0,1], is exactly the same and we find

    F(y)=ℱ_K_1,K_2(y),   ∀ y∈[0,1].

In the interval [1,x], we first solve the homogeneous equation associated . By virtue of Appendix <ref>, one gets two independent solutions, one of them is described by

    F_0(y)= y^a_nF_n(1/y),   ∀ y∈[1,x].

Using the method of variation of constants, we obtain that the general solutions to the equation  (<ref>) in this interval  take the form

    F(y)=ℱ_K_1,K_2(y),   ∀ y∈[1,x].

Since K_1 coincides in both sides  of (<ref>) and the integrals converge at 1 by (<ref>), then we deduce that F is continuous in [0,x]. Therefore, we get the first part of the Lemma by using simply that F(r)=F(x r^2), for r∈ [0,1].

Let us now  select  in this class those solutions who are  𝒞^1. Notice that the solutions ℱ  are 𝒞^1 in [0,x]\{1}. So it remains to study the derivatives from the left and the right on y=1. Since F_n and F_n have no derivatives on the left at 1 and verify

    |F_n^'(y)|∼ Cln(1-y)  and   |F_n^'(y)|∼ Cln  (1-y),

for any y∈[0,1), see (<ref>), then the first members of (<ref>) have no derivatives at 1. This implies necessary that  K_1=0. Moreover, one gets

    ℱ_K_1,K_2^'(1^-)=∫_0^1F_n(s)/1-sℛ(s/x)ds/F_n(1)·

Since F_n(1)>0 and ℛ is Hölder continuous, then the convergence of this integral is equivalent to  the condition ℛ(1/x)=0. Now, using (<ref>), we deduce that the right derivative of ℱ_K_1,K_2 at 1 is given by

    ℱ_K_1,K_2^'(1^+)=K_2/F_n(1)·

Combining (<ref>) with  (<ref>), we deduce that F_n admits a derivative at 1 if and only if 

    K_2=F_n(1)/F_n(1)∫_0^1F_n(s)/1-sℛ(s/x)ds.

Thus,   the 𝒞^1-solution to (<ref>) is given by 

    F(y)= y^nF_n(y)_1^y_0^τF_n(s)/1-sℛ(s/x)ds/τ^n+1F_n^2(τ)dτ 1_[0,1](y)+
          y^a_nF_n(1/y)_1/y^1F_n(1)/F_n(1)_0^1F_n(s)/1-sℛ(s/x)ds-_τ^1s^n-a_n-1F_n(s)/1-sℛ(1/sx)ds/τ^n+1-2a_nF_n^2(τ)dτ 1_[1,∞)(y),

and, therefore, the solution to (<ref>) takes the form

    F(r)=F(xr^2).

This implies in particular that there is only one 𝒞^1 solution   to (<ref>) and satisfies 
F(x^-1/2)=0.
The case n=1 is  very special  since F_1(x)=1-x and hence F_1 vanishes at 1. As in the previous discussion, the solution of (<ref>) in [0,1] is given by

    F(y)=   
              	y(1-y)[K_1+∫_0^y∫_0^τℛ(s/x)ds/τ^2(1-τ)^2dτ],

whereas for y∈[1,∞) the solution reads as

    F(y)=1/yF_1(1/y)[K_3+∫_1/y^1K_2-∫_τ^1sF_1(s)/1-sℛ(1/sx)ds/τ^4F_1^2(τ)dτ],

where K_1, K_2 and K_3  are constants. We can check that the continuity of F at 1 is satisfied if and only if 

    K_3   =   1/F̂_1(1)∫_0^1ℛ(s/x)ds
       =   1/3∫_0^1ℛ(s/x)ds,

by using in the last line the explicit expression of F̂_1(1) coming from (<ref>):

    F̂_1(1)=F(1,2;4;1)=Γ(4)Γ(1)/Γ(3)Γ(2)=3.

Note that we need (<ref>) in order to ensure the convergence of the integral

    ∫_τ^1sF_1(s)/1-sℛ(1/sx)ds.

Let us deal with the derivative, given by

    F'(y)={[                                 (1-2y)[K_1+∫_0^y∫_0^τℛ(s/x)ds/τ^2(1-τ)^2dτ]+1/y(1-y)∫_0^yℛ(s/x)ds,                                                                                           y∈[0,1],;                                                     -1/y^2[F_1(1/y)+1/yF_1'(1/y)][1/3∫_0^1ℛ(s/x)ds; +∫_1/y^1K_2-∫_τ^1sF_1(s)/1-sℛ(1/sx)ds/τ^4F_1^2(τ)dτ]
    +y/F_1(1/y)[K_2-∫_1/y^1sF_1(s)/1-sℛ(1/sx)ds],                                                                                              y≥ 1. ].

The convergence in 0 and 1 of the first part comes from ℛ(0)=ℛ(1/x)=∫_0^1ℛ(s/x)dτ=0. In which case, one gets

    F'(1^-)=-K_1-∫_0^1∫_0^τℛ(s/x)ds/τ^2(1-τ)^2dτ.

For the second part of F, one needs ∫_0^1ℛ(s/x)dτ=0 since F' is singular at 1 as it was mentioned before. Then,

    F'(1^+)=1/3K_2.

Clearly, we have the constraint

    K_2=-3(K_1+∫_0^1∫_0^τℛ(s/x)ds/τ^2(1-τ)^2dτ),

in order to obtain 𝒞^1 solutions.
If, in addition, F'(1)=0, which agrees with F'(x)=0, we obtain the following additional equation for K_2:

    K_2(1/x^2[F_1(1/x)+xF_1'(1/x)]+x/F_1(1/x))
    =x/F_1(1/x)∫_1/x^1sF_1(s)/1-sℛ(1/sx)ds+
          1/x^2[F_1(1/x)+1/xF_1'(1/x)]∫_1/x^1∫_τ^1sF_1(s)/1-sℛ(1/sx)ds/τ^4F_1^2(τ)dτ.

Notice that

    1/x^2[F_1(1/x)+xF_1'(1/x)]+x/F_1(1/x)≠ 0,

using that F_1(z)=F(1,2;4;z), for z∈(0,1), which is a positive and increasing function. Then, we can obtain the exact value of K_2 and then K_1 via relation (<ref>). Using that F(r)=F(xr^2). The proof is now achieved.




Let A>0 and B∈ such that B/A∉[-1,-1/2]. Let Ω satisfy (<ref>) with Ω∉𝒮_sing, where the last set is defined in (<ref>). Then, the following holds true:


  * The kernel of  D_gG(Ω,0) is trivial.
 
  * If  h and d are smooth enough with
 
    D_gG(Ω,0)h=d,    h(re^iθ)=∑_n∈h_n(r)cos(nθ),    d(re^iθ)=∑_n∈d_n(r)cos(nθ),

 then, there exists an absolute constant C>0 such that 
 
    h_n_𝒞^0([0,1])≤ C d_n_𝒞^1([0,1]),   ∀  n∈.

 
  * Coercivity with loss of derivative: for any α∈(0,1), there exists C>0 such that
 
    h_𝒞^0()≤ CD_gG(Ω,0) h_𝒞_s^2,α().

 


(1)  First, note that x≠ x_0, where x_0 is defined in (<ref>), because x_0∈(0,1). Then, Proposition <ref> implies  that the last equation in (<ref>) admits only the zero function as a solution.
We will check how the condition (<ref>) gives us that  there are not nontrivial smooth solutions for n=1. This can be done easily with the explicit expression of g given in (<ref>) for n=1. Since

    G_1(r)=-2[r^4-3/2xr^2-3/2(1-1/x)],
 
one obtains that

    g(r)=-16H_1(1)xr^3[r^4-3/2xr^2-3/2(1-1/x)].

Then, condition (<ref>) is equivalent to

    3x^2-3x+1= 0   or   H_1(1)=0.

Since x>1,  one has  H_1(1)=0 and g≡ 0. Then, ℛ≡ 0. By Lemma <ref>, one has that the solution of (<ref>) with H_1(0)=0 and H_1'(1)=0 is the trivial one: H_1(r)≡ 0. Coming back to (<ref>), we deduce that h_1≡ 0.  Therefore, the only smooth solution is the zero function, which  implies finally  that the kernel is trivial.

 Let us now deal with n≥2.  Applying Lemma <ref> to  the equation (<ref>) with (<ref>) we get that this equation admits a 𝒞^1 solution if and only if

    H_n(1)G_n(x^-1/2)=0.

Using the expression (<ref>), we find that 

    G_n(x^-1/2)=-An(n+1)/4(n+2)x^1-n/2P_n(x^-1),

and from (<ref>) one has

    P_n(x^-1)=-1/n+1(1/x^2+A+2B/An+2/n).

With the assumptions A+2B/A∉[-1,0] and x>1 one gets

    P_n(x^-1)≠0,   ∀ n≥ 2,

obtaining from (<ref>) that

    H_n(1)=0,   ∀ n≥ 2.

Coming back to (<ref>) we find that the source term is vanishing everywhere. Now, from  and (<ref>) we infer that 

    H_n(r)=0,  ∀ n≥ 2, r∈[0,1].

Inserting this  into (<ref>), we obtain h_n≡ 0 for any n∈^⋆. Finally, this implies that the vanishing function is the only element of the kernel.

(2) To get this result we should derive   a priori estimates for the solutions to  the equation

    D_gG(Ω,0)h=d.

The pre-image equation  is equivalent to solve the infinite-dimensional system

    1/x-r^2/8h_n(r)-r/n[A_nG_n(r)+1/2r^n+1H_n(r)]   =d_n(r),  ∀ n≥ 1,
    1/x-r^2/8h_0(r)-∫_r^11/τ∫_0^τ sh_0(s)dsdτ   =d_0(r),

where we use the notations of (<ref>), (<ref>) and (<ref>). 
We first analyze the case of large  n, for which we can apply the contraction principle and  get the desired   estimates. Later, we will deal with low frequencies, which is more delicate and requires the integral representation  (<ref>).  Let us first work with large values of n. Observe that  the first equation of (<ref>) can be transformed into

    h_n(r)=8/1/x-r^2[A_n/nrG_n(r)+H_n(r)/2nr^n+d_n(r)].

The estimate of   A_n, defined by (<ref>) and (<ref>), can be done as follows
 
    |A_n|≤∫_0^1s^n+1|h_n(s)|ds/2n|Ω_n-Ω|.
  
Keeping in mind the relation (<ref>) and the assumption x∉𝒮_sing, we obtain

    inf_n≥11/|Ω_n-Ω|<+ ∞,

and, on the other hand, it is obvious that

    ∫_0^1s^n+1|h_n(s)|ds≤h_n_𝒞^0([0,1])/n+2.
  
Combining  the preceding estimates, we find

    |A_n|≤C/n^2h_n_𝒞^0([0,1]),

for some  constant C independent of n.  Let us remark that according to   the equation (<ref>), one should get the compatibility condition

    A_n/nx^-1/2G_n(x^-1/2)+H_n(x^-1/2)/2nx^-n/2+d_n(x^-1/2)=0.
 
Hence, applying  the Mean Value Theorem, combined with (<ref>), we obtain

    h_n_𝒞^0([0,1])   ≤    Ch_n_𝒞^0([0,1])/n^3(rG_n(r))^'_𝒞^0([0,1])
          +1/2n(H_n(r)/r^n)^'_𝒞^0([0,1])+d_n^'_𝒞^0([0,1]),

where in this inequality C may depend  on x but not on n. Now, it is straightforward to check  that

    |(rG_n(r))^'|≤ C n^2,  ∀  r∈ [0,1],

and second

    (H_n(r)/r^n)^'=n r^n-1∫_r^1h_n(s)/s^n-1ds-n/r^n+1∫_0^rs^n+1 h_n(s)ds.

From this latter identity, we infer that 

    |(H_n(r)/r^n)^'|   ≤   (n/n-2+n/n+2)h_n_𝒞^0([0,1])≤ Ch_n_𝒞^0([0,1]),  ∀  r∈ [0,1],

for any n≥3. Consequently, we get

    (H_n(r)/r^n)^'_𝒞^0([0,1])≤ Ch_n_𝒞^0([0,1]).

Plugging  (<ref>) and (<ref>) into (<ref>), we find

    h_n_𝒞^0([0,1])≤C/nh_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1]).

Hence, choosing n_0 large enough we deduce  that 
    h_n_𝒞^0([0,1])≤d_n^'_𝒞^0([0,1]),  ∀  n≥ n_0.

Next, we deal with the cases 1≤ n≤ n_0. The preceding argument fails and to invert the operator we recover explicitly the solution h_n from d_n according to the integral representation (<ref>). For this purpose we will proceed as  in the range  study  in Subsection <ref>. By virtue of  (<ref>), we find that H_n satisfies an equation of the type (<ref>). Thus, using  (<ref>), we deduce

    H_n(r)   =     x^nr^2nF_n(xr^2)_1^xr^2_0^τF_n(s)/1-sℛ(s/x)ds/τ^n+1F_n^2(τ)dτ 1_[0,x^-1/2](r) +	x^a_nr^2a_nF_n(1/xr^2)
          ×_1/xr^2^1F_n(1)/F_n(1)_0^1F_n(s)/1-sℛ(s/x)ds-_τ^1s^n-a_n-1F_n(s)/1-sℛ(1/sx)ds/τ^n+1-2a_nF_n^2(τ)dτ 1_(x^-1/2,1](r),

for n≥ 2, and

    H_1(r)=    	xr^2(1-xr^2)[K_1-∫_xr^2^0∫_0^τℛ(s/x)ds/τ^2(1-τ)^2dτ] 1_[0,x^-1/2](r)
       +  	1/xr^2F_1(1/xr^2)[1/3∫_0^1ℛ(s/x)ds+∫_1/xr^2^1K_2-∫_τ^1sF_1(s)/1-sℛ(1/sx)ds/τ^4F_1^2(τ)dτ]    1_(x^-1/2,1](r),

 where K_1 and K_2 are given in (<ref>)-(<ref>). From (<ref>), we get

    ℛ(y)=1/4x[A A_nn(n+1)/n+2 y^nP_n(y)-4 x ny^n/2d_n(y^1/2)],

where P_n is defined in (<ref>). 

Let us relate A_n with d_n.  This can be obtained from the , which implies 
 
    A_n=(n+2) x^n/2+1d_n(x^-1/2)/A(n+1) P_n(x^-1).

Let us remark that this relation is different from (<ref>), which is not useful for low frequencies.  Consequently, using (<ref>) we infer that 
 
    | A_n|≤ Cd_n_𝒞^0([0,1]),   ∀ n∈[1,n_0].

Concerning ℛ, we can get successively

    |ℛ(y)|≤ C[|A_n|y^n+y^n/2|d_n(y^1/2)|],  ∀  y∈[0,1],

which implies that

    ||ℛ||_𝒞^0([0,1])≤ C[|A_n|+||d_n||_𝒞^0([0,1])]≤ C||d_n||_𝒞^0([0,1]),

for any n≥ 1. In the case n≥ 2, we also obtain that

    |ℛ^'(y)|≤ C[|A_n|y^n-1+y^n/2-1(|d_n(y^1/2)|+|d_n^'(y^1/2)|)],   ∀  y∈[0,1],

which amounts to

    ℛ^'_𝒞^0([0,1])≤ C ( d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])),  ∀ n≥2.

Note that this last estimate can not be used for n=1 since ℛ is only Hölder continuous. Then, we can find in this case that

    ||ℛ||_𝒞^0,γ([0,1])≤ C||d_n||_𝒞^0,γ([0,1]),

for μ=min(1/2,α).

Let us begin with n≥ 2. Using  the  boundedness property of F_n, which we shall see later in Lemma <ref>, combined  with an integration by parts imply 

    |_1^xr^2_0^τF_n(s)/1-sℛ(s/x)ds/τ^n+1F_n^2(τ)dτ|   ≤   C| _1^xr^2_0^τ|ℛ(s/x)|/1-sds/τ^n+1dτ|
       ≤   C∫_0^1|ℛ(s/x)|/1-sds+Cr^-2n∫_0^xr^2|ℛ(s/x)|/1-sds
          +|∫_1^xr^2τ^-n|ℛ(τ/x)|/1-τdτ|.

We discuss first the case  r∈[0,1/2x^-1/2]. From the compatibility assumption (<ref>), we infer that ℛ(1/x)=0, and therefore we deduce from (<ref>) and (<ref>) that

    ∫_0^1|ℛ(s/x)|/1-sds   +r^-2n∫_0^xr^2|ℛ(s/x)|/1-sds≤∫_0^1|ℛ(s/x)|/1-sds+Cr^-2n∫_0^xr^2|ℛ(s/x)|ds
       ≤ C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])+ r^-n+2d_n_𝒞^0([0,1])),

 for 2≤ n≤ n_0. In a similar way to  the last integral term of (<ref>), we  split it  as follows using the estimates (<ref>) and (<ref>)

    _xr^2^1τ^-n|ℛ(τ/x)|/1-τdτ   =   _1/2^1τ^-n|ℛ(τ/x)|/1-τdτ+_xr^2^1/2τ^-n|ℛ(τ/x)|/1-τdτ
       ≤   C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1]))+C_xr^2^1/2τ^-n|ℛ(τ/x)|dτ
       ≤   C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])+Cr^-n+2d_n_𝒞^0([0,1])).

Putting together the preceding estimates we get that 

    |_1^xr^2_0^τF_n(s)/1-sℛ(s/x)ds/τ^n+1F_n^2(τ)dτ|≤ C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])+Cr^-n+2d_n_𝒞^0([0,1])),

for r∈[0,1/2x^-1/2]. Plugging this into  (<ref>) yields

    |H_n(r)|≤ C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])) r^2n+Cr^n+2d_n_𝒞^0([0,1]),

 for r∈[0,1/2x^-1/2] and 2≤ n≤ n_0. Now, we wish  to estimate the derivative of H_n(r)/r^n. Coming back to (<ref>), we deduce from elementary computations that 
 
    H_n^'(r)=H_n(r)(2n/r+2xr F_n^'(xr^2)/F_n(xr^2))+2/rF_n(xr^2)_0^xr^2F_n(s)/1-sℛ(s/x)ds,

. By (<ref>), we get
 
    |H_n(r) (2n/r+2xr F_n^'(xr^2)/F_n(xr^2))|    ≤     C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])) r^2n-1
          +Cr^n+1d_n_𝒞^0([0,1]).

 Concerning the integral term, it suffices to  apply (<ref>) in order to get
  
    2/rF_n(xr^2)_0^xr^2F_n(s)/1-s|ℛ(s/x)|ds≤C/r_0^xr^2|ℛ(s/x)|ds≤  C (|A_n| r^2n+1+ r^n+1d_n_𝒞^0([0,1])).

 Hence, combining the preceding estimates leads to
 
    |H_n^'(r)|≤  C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])) r^2n-1+Cr^n+1d_n_𝒞^0([0,1]).

This estimate together with  (<ref>)  allows getting
 
    |(H_n(r)/r^n)^'|≤  C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])),   ∀  r∈[0,1/2x^-1/2],

 for  2≤ n≤ n_0. The case n=1 can be done using similar ideas since we only have the singularity at 0 in this interval. Note that K_1 and K_2 can be estimated in terms of ℛ having

    |K_1|, |K_2|≤ ||ℛ||_𝒞^0,γ([0,1]).
 

Let us now move to the intermediate  case x∈ [1/2x^-1/2,x^-1/2]. Then, there is no singularity in this range  except for r=x^-1/2 due to the logarithmic behavior  of F_n close to this point. This logarithmic divergence can be controlled   from the smallness of the integral term in H_n. Let us show the idea. When we differentiate H_n, we obtain one term of the type

    2x^n+1r^2n+1 F_n'(xr^2)_1^xr^2_0^τF_n(s)/1-sℛ(s/x)ds/τ^n+1F_n^2(τ)dτ,

where we notice the logarithmic singularity coming from F_n' at 1. However, one has

    |2x^n+1r^2n+1 F_n'(xr^2)_1^xr^2_0^τF_n(s)/1-sℛ(s/x)ds/τ^n+1F_n^2(τ)dτ|   ≤ C|F_n'(xr^2)(1-xr^2)_0^xr^2F_n(s)/1-sℛ(s/x)ds/(xr^2)^n+1F_n^2(xr^2)|
       ≤ C(||ℛ||_𝒞^0([0,1])+||ℛ'||_𝒞^0([0,1]))
       ≤ C(||d_n||_𝒞^0([0,1])+||d_n'||_𝒞^0([0,1])).

Therefore, after straightforward efforts on (<ref>) using (<ref>), it implies that 
 
    |H_n^'(r)|≤  C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])),

for 2≤ n≤ n_0. Hence, we obtain 
 
    |(H_n(r)/r^n)^'|≤  C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])),   ∀  r∈[1/2x^-1/2,x^-1/2],

for 2≤ n≤ n_0. In this interval, the case n=1 is different. This is because we have not singularity coming from the hypergeometric function but we do have it coming from the integral. Hence, some more manipulations are needed. In this case, H_1 reads as

    H_1(r)=xr^2(1-xr^2)[K_1+∫_0^1/2∫_0^τℛ(s/x)ds/τ^2(1-τ)^2dτ+∫_1/2^xr^2∫_0^τℛ(s/x)ds/τ^2(1-τ)^2dτ].

The first integral term can be treated as in the previous computations in the interval [0,1/2 x^-1/2]. Let us focus on the singular integral term. By a change of variables, one has

    |∫_1/2^xr^2∫_0^τℛ(s/x)ds/τ^2(1-τ)^2dτ|
       = | ∫_0^xr^2ℛ(s/x)ds/x^2r^4(1-xr^2)-8∫_0^1/2ℛ(s/x)ds-∫_1/2^xr^2τℛ(τ/x)-2∫_0^τℛ(s/x)ds/τ^3(1-τ)dτ|
       ≤  C( ||ℛ||_𝒞^0([0,1])+||ℛ||_𝒞^0,γ([0,1])),

where we have used (<ref>)-(<ref>). Then, we obtain

    |H_1(r)|≤ C(1-xr^2) ||ℛ||_𝒞^0,γ([0,1]).

Similar arguments can be done to find that

    |(H_1(r)/r)^'|≤  C ||d_n||_𝒞^0,γ([0,1]),

for any r∈[1/2 x^-1/2, x^-1/2].

It remains to establish similar results for the case r∈[x^-1/2,1]. With this aim we use the second integral in (<ref>). Notice that the only singular point is r=x^-1/2 due to the logarithmic singularity of the hypergeometric function F_n defined in (<ref>). One can check that this function is strictly increasing, positive and satisfies

    1≤F_n(r)≤sup_n∈F_n(1)<+∞,   ∀ r∈[0,1].

As in the previous interval, the smallness of the integral term controls this singularity. The same happens for the case n=1.
Hence, we get 
 
    |(H_n(r)/r^n)^'|≤  C (|A_n|+ d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])),   ∀  r∈ [x^-1/2,1],

for 1≤ n≤ n_0. Therefore,   in  all the cases we have 

    |(H_n(r)/r^n)^'|≤  C ( d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])),  ∀  r∈ [0,1],

for 1≤ n≤ n_0. Applying the Mean Value Theorem to (<ref>), and using (<ref>) and (<ref>), allow us to obtain

    h_n_𝒞^0([0,1])≤  C ( d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])),   ∀ n∈[1,n_0].

Similar arguments can be done in order to deal with the equation for n=0. Note that the resolution of this equation is similar to the work done in Proposition <ref>.

Then, combining all the estimates, it yields

    h_n_𝒞^0([0,1])≤  C ( d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])),  ∀ n∈.

This achieves the proof of the announced result.

(3) Let us recall the formula for the Fourier coefficients

    d_n(r)=1/π∫_0^2πd(rcosθ,rsinθ) cos(nθ)dθ.

We can prove that

    d_n_𝒞^0([0,1])≤C/n^1+αd_𝒞^1,α(),

for n≥1. This can be done integrating by parts as

    d_n(r)=-r/nπ∫_0^2π∇ d(rcosθ,rsinθ) · (-sinθ, cosθ) sin(nθ)dθ,

and writing it as

    d_n(r)=   r/nπ∫_0^2π∇ d(rcos(θ+π/n),rsin(θ+π/n)) ·(-sin(θ+π/n), cos(θ+π/n)) sin(nθ)dθ
    
    =   r/2nπ∫_0^2π[∇ d(rcos(θ+π/n),rsin(θ+π/n)) ·(-sin(θ+π/n), cos(θ+π/n)) 
       -∇ d(rcosθ,rsinθ) · (-sinθ, cosθ)]sin(nθ)dθ.

Consequently,

    |d_n(r)|≤C/n^1+α||d||_𝒞^1,α().

With similar arguments, one achieves that 

    d_n_𝒞^0([0,1])+d_n^'_𝒞^0([0,1])≤C/n^1+αd_𝒞^2,α(),

and then

    h_n_𝒞^0([0,1])≤C/n^1+αd_𝒞^2,α().

Therefore, we obtain

    h_𝒞^0()≤∑_n∈h_n_𝒞^0([0,1])≤ Cd_𝒞^2,α(),

which completes the proof.


The next target is  to provide  the proof of Theorem <ref>.

In a small neighborhood of the origin we have the  following  decomposition  through Taylor expansion at the second order 

    G(Ω,h)=D_gG(Ω,0)(h)+1/2 D^2_g,gG(Ω,0)(h,h)+ℛ_2(Ω, h),

where ℛ_2(Ω, h) is the remainder term, which verifies

    ℛ_2(Ω, h)_𝒞^2,α()≤1/6D^3_g,g,gG(Ω,g)(h,h,h)_𝒞^2,α(),

where g=th for some t∈(0,1). We intend to show the following,

    1/2 D^2_g,gG(Ω,0)(h,h)+ℛ_2(Ω, h)_𝒞^2,α()≤ Ch_𝒞^2,α()^δ_1h_𝒞^0()^δ_2,

for some δ_1>0 and δ_2≥ 1, getting the last bound will  be crucial in our argument. First, let us deal with the second derivative of G. Straightforward computations, similar to what was done in Proposition <ref>, lead to

    D^2_g,gG(Ω,0)   (h,h)(z)=-h(z)^2/8A+Re/π∫_∂_g ϕ(Ω,0)(h)(z)-∂_g ϕ(Ω,0)(h)(y)/z-yh(y)dA(y)
       +2/π∫_log|z-y|h(y)Re[∂_g ϕ(Ω,0)(h)'(y)]dA(y)
       -Re/2π∫_(∂_g ϕ(Ω,0)(h)(z)-∂_g ϕ(Ω,0)(h)(y))^2/(z-y)^2f_0(y)dA(y)
       +2Re/π∫_∂_g ϕ(Ω,0)(h)(z)-∂_g ϕ(Ω,0)(h)(y)/z-yf_0(y)Re[∂_g ϕ(Ω,0)(h)'(y)]dA(y)
       -Ω|∂_g ϕ(Ω,0)(h)(z)|^2+1/π∫_log|z-y|f_0(y)|∂_g ϕ(Ω,0)(h)'(y)|^2dA(y)
       -ΩRe[∂^2_g,gϕ(Ω,0)(h,h)(z)]
       +Re/2π∫_∂^2_g,gϕ(Ω,0)(h,h)(z)-∂^2_g,gϕ(Ω,0)(h,h)(y)/z-yf_0(y)dA(y)
       +1/π∫_log|z-y|f_0(y)Re[∂^2_g,gϕ(Ω,0)(h,h)'(y)]dA(y).

Recall the relation between ∂_gϕ(Ω,0)(h) and h 

    ∂_gϕ(Ω,0)(h)(z)=z∑_n≥ 1 A_n z^n,   A_n=∫_0^1s^n+1h_n(s)ds/2n(Ω_n-Ω)·

By Proposition <ref>, one has that

    ∂_gϕ(Ω, 0)h_𝒞^2,α()≤  Ch_𝒞^1,α().

We claim that one can reach the limit case, 

    ∂_gϕ(Ω, 0)h_𝒞^1()≤ ||h||_𝒞^0().

The last estimate can be done using Proposition <ref> in order to work in  as follows

    |∂_gϕ(Ω, 0)h'(e^iθ)|   =|∑_n≥ 1∫_0^1 s^n+1h_n(s)ds e^inθ|
       = 1/2| ∑_n≥ 1∫_0^2π∫_0^1 s^n+1h(se^iθ')cos(nθ')e^inθdsdθ'|
       = 1/4|∑_n≥ 1∫_0^2π∫_0^1 s^n+1h(se^iθ')(e^inθ'+e^-inθ')e^inθdsdθ'|
       = 1/4|∑_n≥ 1∫_0^2π∫_0^1 s^n+1h(se^iθ')(e^in(θ+θ')+e^in(θ-θ'))dsdθ'|.

Using Fubini we deduce that

    |∂_gϕ(Ω, 0)h'(e^iθ)|   =   1/4|∫_0^2π∫_0^1 s^2h(se^iθ')(e^i(θ+θ')/1-se^i(θ+θ')+e^i(θ-θ')/1-se^i(θ-θ'))dsdθ'|
       ≤    C||h||_𝒞^0().

The latter estimate follows from the convergence of the double integral

    ∫_0^2π∫_0^1ds dθ/|1-se^iθ|<∞.

The next step is to deal with ∂^2_g,gϕ(Ω,0)(h,h). By differentiating it, similarly to the proof of  in Proposition <ref>, we obtain

    ∂^2_g,gϕ(Ω,0)(h,h)=   -1/2∂_ϕ F(Ω,0,0)^-1[∂^2_g,gF(Ω,0,0)(h,h)+2∂^2_g,ϕF(Ω,0,0)(h, ∂_gϕ(Ω,0)h).
       . +∂^2_ϕ,ϕF(Ω,0,0)(∂_gϕ(Ω,0)h,∂_gϕ(Ω,0)h)]
    
    =    -1/2∑_n≥ 1ρ_n/n(Ω-Ω_n)z^n+1,

where ρ(w)=∑_n≥ 1ρ_nsin(nθ) and

    ρ(w)=   ∂^2_g,gF(Ω,0,0)(h,h)(w)+2∂^2_g,ϕF(Ω,0,0)(h, ∂_gϕ(Ω,0)h)(w)
       +∂^2_ϕ,ϕF(Ω,0,0)(∂_gϕ(Ω,0)h,∂_gϕ(Ω,0)h)(w)
    
    =   Im[ -w∂_gϕ(Ω,0)(h)'(w)/π∫_h(y)/w-ydA(y)
    
       +w/π∫_∂_gϕ(Ω,0)(h)(w)-∂_gϕ(Ω,0)(h)(y)/(w-y)^2h(y)dA(y)
       .-2w/π∫_h(y)/w-yRe[∂_gϕ(Ω,0)(h)'(y)]dA(y).
    . +2Ω ∂_gϕ(Ω,0)(h)(w)∂_gϕ(Ω,0)(h)'(w)w.
       . +w∂_gϕ(Ω,0)(h)'(w)/π∫_∂_gϕ(Ω,0)(h)(w)-∂_gϕ(Ω,0)(h)(y)/(w-y)^2f_0(y)dA(y).
       .-2w∂_gϕ(Ω,0)(h)'(w)/π∫_f_0(y)/w-yRe[∂_gϕ(Ω,0)(h)'(y)]dA(y).
       . -w/π∫_[∂_gϕ(Ω,0)(h)(w)-∂_gϕ(Ω,0)(h)(y)]^2/(w-y)^3f_0(y)dA(y).
       .+2w/π∫_∂_gϕ(Ω,0)(h)(w)-∂_gϕ(Ω,0)(h)(y)/(w-y)^2f_0(y) Re[∂_gϕ(Ω,0)(h)'(y)]dA(y).
       -w/π∫_f_0(y)/w-y|∂_gϕ(Ω,0)(h)'(y)|^2dA(y) ].

By Proposition <ref> and due to the fact that  ∂^2_g,gϕ(Ω,0) can be seen as a convolution operator, one has that

    ||∂^2_g,gϕ(Ω,0)||_𝒞^k,α()≤ C ||ρ||_𝒞^k,α().

Moreover, we claim that

    ||ρ||_𝒞^k,α()≤ C||h||_𝒞^2,α()^σ_1||h||_𝒞^0()^σ_2,    k=0,1,2,

with  σ_2≥ 1. First, we use the interpolation inequalities for Hölder spaces

    ||h||_𝒞^k,α()≤ C ||h||_𝒞^k_1,α_1()^β||h||_𝒞^k_2,α_2()^1-β,

for k, k_1 and k_2 non negative integers, 0≤α_1,α_2≤ 1 and

    k+α=β(k_1+α_1)+(1-β)(k_2+α_2),

where β∈(0,1). The proof of the interpolation inequality can be found in <cit.>.  In order to get the announced results, we would need to use some classical results in Potential Theory dealing with the Newtonian potential and the Beurling transform, see Appendix <ref> or for instance <cit.>.
Now, let us show the idea behind (<ref>). To estimate  the first term of ρ, we combine  (<ref>) with the law products in Hölder spaces, as follows,

    ∂_gϕ(Ω,0)(h)'(·)   ∫_h(y)/(·)-ydA(y)_𝒞^k,α()≤ C ∂_gϕ(Ω,0)(h)'(·)_𝒞^0()∫_h(y)/(·)-ydA(y)_𝒞^k,α()
       +C∂_gϕ(Ω,0)(h)'(·)_𝒞^k,α()∫_h(y)/(·)-ydA(y)_𝒞^0()
    ≤    C( h_𝒞^0()h_𝒞^k,α()+h_𝒞^k,α()h_𝒞^0()).

Then, (<ref>) is satisfied for the first term. Let us deal with the second term of ρ. For k=0, one has

    ∫_   ∂_gϕ(Ω,0)(h)(·)-∂_gϕ(Ω,0)(h)(y)/((·)-y)^2h(y)dA(y)_𝒞^0,α()
       ≤  C∂_gϕ(Ω,0)(h)(·) p.v.∫_h(y)/((·)-y)^2dA(y)_𝒞^0,α()
                       +C p.v.∫_∂_gϕ(Ω,0)(h)(y)  h(y)/((·)-y)^2dA(y)_𝒞^0,α()
    ≤    C∂_gϕ(Ω,0)(h)(·)_𝒞^0,α() p.v.∫_h(y)/((·)-y)^2dA(y)_𝒞^0()+C∂_gϕ(Ω,0)(h)  h_𝒞^0,α()
                       +C∂_gϕ(Ω,0)(h)(·)_𝒞^0() p.v.∫_h(y)/((·)-y)^2dA(y)_𝒞^0,α()
    ≤   Ch_𝒞^0()h_𝒞^0,α().

For k=1, 2, we would need the use of the interpolation inequalities:

    ∫_   ∂_gϕ(Ω,0)(h)(·)-∂_gϕ(Ω,0)(h)(y)/((·)-y)^2h(y)dA(y)_𝒞^k,α()
    ≤    C∂_gϕ(Ω,0)(h)(·) p.v.∫_h(y)/((·)-y)^2dA(y)_𝒞^k,α()
       +C p.v.∫_∂_gϕ(Ω,0)(h)(y)  h(y)/((·)-y)^2dA(y)_𝒞^k,α()
    ≤    C∂_gϕ(Ω,0)(h)(·)_𝒞^k,α() p.v.∫_h(y)/((·)-y)^2dA(y)_𝒞^0()
       +C∂_gϕ(Ω,0)(h)(·)_𝒞^0() p.v.∫_h(y)/((·)-y)^2dA(y)_𝒞^k,α()+C∂_gϕ(Ω,0)(h)  h_𝒞^k,α()
    ≤   h_𝒞^k-1,α()h_𝒞^0,α()+Ch_𝒞^0()h_𝒞^k,α().

It remains to use the interpolation inequalities in order to conclude. For the case k=1, we need to use (<ref>) in order to get

    h_𝒞^0,α()h_𝒞^0,α()≤ C h_𝒞^0()^2-2α/2+αh_𝒞^2,α()^2α/2+α.

We use again (<ref>) for k=2: 

    h_𝒞^1,α()h_𝒞^0,α()≤ C h_𝒞^0()^3/2+αh_𝒞^2,α()^1+2α/2+α.

Note that in every case, the exponent of the 𝒞^0-norm is bigger than 1. As to  the remaining  terms of ρ,  we develop similar estimates with the same order of difficulties leading to the announced inequality (<ref>).

Once we have these preliminaries estimates, we can check that (<ref>) holds true. For example, let us illustrate the basic  idea to implement through the second term.

    ∫_log|(·)-y|h(y)Re[∂_g ϕ(Ω,0)(h)'(y)]dA(y)_𝒞^2,α()≤   C h∂_g ϕ(Ω,0)(h)'_𝒞^0,α().

Using the classical law products and (<ref>) we find 

    ∫_log|(·)-y|h(y)Re[∂_g ϕ(Ω,0)(h)'(y)]dA(y)_𝒞^2,α()≤   C h_𝒞^0()∂_g ϕ(Ω,0)(h)'_𝒞^0,α()
       + Ch_𝒞^0,α()∂_g ϕ(Ω,0)(h)'_𝒞^0()
    ≤    Ch_𝒞^0,α()h_𝒞^0().

The other terms can be estimated in a similar way, achieving (<ref>). The same arguments  applied to the remainder term lead to

    ℛ_2(Ω, h)_𝒞^2,α()≤ Ch_𝒞^2,α()^δ_1h_𝒞^0()^δ_2.

The computations are  long here but the analysis is straightforward.


Let us see how to achieve the argument. Assume that h is a zero to G in a small neighborhood of the origin, then 

    D_gG(Ω,0)h=-1/2 D^2_g,gG(Ω,0)(h,h)-ℛ_2(Ω, h).

Applying Lemma <ref>–(3) we deduce that 

    h_𝒞^0()   ≤ CD_gG(Ω,0)h_𝒞^2,α()
       ≤ C1/2 D^2_g,gG(Ω,0)(h,h)+ℛ_2(Ω, h)_𝒞^2,α()
       ≤ Ch_𝒞^2,α^δ_1h_𝒞^0()^δ_2,

Consequently, if h_𝒞^2,α()^δ_1<C^-1, then necessary h_𝒞^0()=0 since δ_2≥ 1. Therefore, we deduce that  there is only the trivial solution in this ball.


The   quadratic profiles are  particular cases of the polynomial profiles studied in Section <ref>; f_0(r)=Ar^2m+B. Here, we briefly show how to develop this case. Studying the kernel for this case is equivalent to study the equations

    h_n(r)+r/nm(2m+2)r^2m-2/(r^2m-1/x_m)[-H_n(1)/G_n(1)G_n(r)+H_n(r)/r^n+1]   =   0,  ∀ r∈[0,1],  ∀ n∈^⋆,
    1/x_m-r^2/8h_0(r)-∫_τ^11/τ∫_0^τ sh_0(s)dsdτ   =   0,  ∀ r∈[0,1],
 
where the functions H_n and G_n are defined in (<ref>)-(<ref>) and 

    1/x_m=2m+2/A(Ω-B/2).

Thus H_n verifies the following equation

    r(1-x_mr^2m)H_n”(r)   -   (2n-1)(1-x_mr^2m)H_n'(r)
          +2m(2m+2)r^2m-1x_mH_n(r)=2m(2m+2)x_mr^n+2mH_n(1)/G_n(1)G_n(r).

Using the change of variables y=x_mr^2m and setting H_n(r)=ℱ(x_mr^2m), one has that

    y(1-y)ℱ”(y)   +   m-n/m(1-y)ℱ'(y)+m+1/mℱ(y)
          =m+1/m(y/x_m)^n+1/2mH_n(1)/G_n(1)G_n((y/x_m)^1/2m).

The homogeneous equation of the last differential equation can be solved in terms of hypergeometric functions as it was done in the quadratic profile. Then, similar arguments can be applied to this case.




 §.§ Range structure

Here, we provide an algebraic description of the range. This will be useful when studying the transversality assumption of the Crandall-Rabinowitz Theorem. Our result reads as follows.

Let  A∈^⋆, B∈ and x_0 be given by (<ref>). Let   x∈ (-∞,1)\{𝒮_sing∪{0,x_0}}, where  the set 𝒮_sing is defined in (<ref>). Then

    Im  D_gG(Ω,0)= {d∈𝒞_s^1,α():  ∫_ d(z)K_n(z)dz=0,    n∈𝒜_x},

where

    K_n(z)=Re[F_n(x|z|^2)/1-x|z|^2z^n],  Ω=B/2+A/4x,

and the set 𝒜_x is defined by (<ref>).


In order to describe the range of the D_gG(Ω,0):𝒞_s^1,α()→𝒞_s^1,α(), we should solve the equation 

    D_gG(Ω,0)h=d,   h(r e^iθ)=
    ∑_n≥ 0h_n(r)cos(nθ),     d(r e^iθ)=
    ∑_n≥ 0d_n(r)cos(nθ).

From the structure  of the linearized operator seen in (<ref>), this problem is equivalent to

    1/x-r^2/8h_n(r)-r/n[A_nG_n(r)+1/2r^n+1H_n(r)]   =d_n(r),  ∀ n≥ 1
    1/x-r^2/8h_0(r)-∫_r^11/τ∫_0^τ sh_0(s)dsdτ   =d_0(r),

where the functions involved in the last expressions are defined in (<ref>)-(<ref>).  By Proposition <ref>, the case n=0 can be analyzed through the Inverse Function Theorem getting  a unique  solution.  Let us focus on the case n≥ 1 and proceed as in the  preceding study for  the kernel. We use  the linear operator defined in (<ref>),


    ℒh= r^2n∫_r^11/s^n-1h(s)ds+∫_0^rs^n+1h(s)ds,

for any h∈𝒞([0,1];), which satisfies the boundary conditions  in (<ref>) and 

    1/2n(r(ℒh)^'(r))^'-(ℒh)^'(r)=-r^n+1h(r).

Taking H_n≜ℒh_n and using  (<ref>) we find  that H_n solves

    (1-xr^2)rH_n”(r)-(1-xr^2)(2n-1)H_n'(r)   +8rxH_n(r)
       =-16A_nxr^n+2G_n(r)-16xnr^n+1d_n(r),

complemented with the  boundary conditions H_n(0)=H_n'(1)=0. This  differential equation is equivalent to (<ref>).
Once we have a solution of the differential equation (<ref>) we have to verify that ℒh_n=H_n, where

    h_n(r)≜8x/1-xr^2[d_n(r)+A_nr/nG_n(r)+1/2nr^nH_n(r)].

Denote H≜ℒh_n-H_n, then it satisfies 

    1/2n[r H^'(r)]^'-H^'(r)=0.

From the boundary conditions one obtains that H=0 and thus  ℒh_n=H_n. 
Now, since H_n(0)=0, Lemma <ref> can be applied with

    g(r)=-16A_nxr^n+2G_n(r)-16xnr^n+1d_n(r).

Thus, the solutions are given by

    H_n(r)   =   r^2nF_n(xr^2)[H_n(1)/F_n(x)+4A_nx^n∫_xr^2^x1/τ^n+1F_n^2(τ)∫_0^τF_n(s)/1-s(s/x)^n+1/2G_n((s/x)^1/2)dsdτ.
          +.4nx^n∫_xr^2^x1/τ^n+1F_n^2(τ)∫_0^τF_n(s)/1-s(s/x)^n/2d_n((s/x)^1/2)dsdτ].

A change of variables combined with (<ref>) yield

    H_n(r)   =   H_n(1)r^2nF_n(xr^2)[1/F_n(x)-8x/G_n(1)∫_r^11/τ^2n+1F_n^2(xτ^2)∫_0^τs^n+2F_n(xs^2)/1-xs^2G_n(s)dsdτ]
          +16nxr^2nF_n(xr^2)∫_r^11/τ^2n+1F_n^2(xτ^2)∫_0^τs^n+1F_n(xs^2)/1-xs^2d_n(s)dsdτ.

Note that when d_n≡ 0, the function H_n agrees with the one obtained for  the kernel. It remains to check the boundary conditions. Clearly H_n(0)=0, then we focus on  proving H_n^'(1)=0. Following  the  computations leading to (<ref>), we obtain

    H_n^'(1)=2n H_n(1)/F_n(x)G_n(1)Ψ_n(x)-16nx/F_n(x)∫_0^1s^n+1F_n(xs^2)/1-xs^2d_n(s)ds.

We will distinguish two cases. In the first one  n∉𝒜_x. By virtue of Proposition <ref> and Proposition <ref>, we obtain that  D_gG(Ω,0)  is an isomorphism. Otherwise,  we have  Ψ_n(x)=0 and the boundary condition is equivalent to

    ∫_0^1r^n+1F_n(xr^2)/1-xr^2d_n(r)dr=0.

Define
z∈↦K_n(z)=Re[F_n(x|z|^2)/1-x|z|^2z^n],
and consider  the linear form 
T_K_n: 𝒞_s^1,α()↦ given by

    T_K_n d≜∫_ d(y)K_n(y)dA(y).

Condition (<ref>) leads to T_K_n d=0, which follows from 

    ∫_0^2π d(re^iθ)cos( nθ)dθ=π d_n(r).

Since K_n belongs to  𝒞^∞(;), we deduce that T_K_n is continuous. Thus, Ker T_K_n is closed and of co-dimension one. In addition, from the preceding analysis one has that

    Im  D_gG(Ω,0)⊆{d∈𝒞_s^1,α()  :  ∫_ d(z)K_n(z)dz=0,    n∈𝒜_x}⊆⋂_ n∈𝒜_xKer T_K_n .

The elements of the family {K_n:   n∈𝒜_x} are independent, and thus  ⋂_ n∈𝒜_xKer T_K_n is  closed  and of co–dimension  card𝒜_x. As a consequence of Proposition <ref>,  Ker D_gG(Ω,0) is of dimension card𝒜_x. Using   Proposition <ref>,  D_gG(Ω,0) is a Fredholm  operator of index zero. Consequently, Im D_gG(Ω,0) is of co–dimension card𝒜_x, and thus

    Im  D_gG(Ω,0)=⋂_ n∈𝒜_xKer T_K_n.

This achieves the proof of the announced result.





§ SPECTRAL STUDY

The aim of this section is to study some qualitative properties of the roots  of the spectral function (<ref>) that will be needed when we apply bifurcation arguments.   For instance, to identify the eigenvalues and  explore the kernel structure of the linearized operator, we should  carefully  analyze   the existence and uniqueness  of  roots  x_n of   (<ref>) at each frequency level n and  study  their  monotonicity.  This part is highly  technical and requires  cautious  manipulations on hypergeometric functions and their asymptotics with respect to n. Notice that for some special  regime in A and B, the monotonicity turns to be very  intricate  and   it is only established for higher frequencies through   refined expansions of the eigenvalues x_n with respect to n. Another problem that one should face is connected to the separation between the eigenvalues set and the singular set associated to (<ref>). It seems that the two sequences admit the same leading term and the separation is obtained  at  the second asymptotics  level, which requires much more efforts because the sequence {x_n} converges to 1, which is a singular point for the hypergeometric function involved in  (<ref>). Recall that n and m are non negative integers.



 §.§ Reformulations of the dispersion  equation


In what follows, we intend  to write down various formulations for the dispersion equation (<ref>) describing the set
(<ref>).
This set is given by the zeroes of (<ref>) and the elements of this set are called “eigenvalues”.  As we shall notice, the study of some qualitative behavior of the zeroes  will be much more tractable through the use of different representations  connected to some specific algebraic  structure of the hypergeometric equations. Recall the use of the notation F_n(x)=F(a_n,b_n;c_n;x), where the coefficients are given by (<ref>). The Kummer quadratic transformations introduced in Appendix <ref> leads to the following result:


 The following identities hold true:

    ζ_n(x)=   [1+x(A+2B/A(n+1)-1)]F(a_n,b_n; n+1;x)
       +2x-1/n+1F(a_n,b_n; n+2;x)-2x/(n+1)(n+2)F(a_n,b_n; n+3;x),


    ζ_n(x)=   A+2B/A(n+1) xF(a_n,b_n;n+1;x)
       +n-(n+1)x/n+1F(a_n,b_n;n+2;x)
    +2nx/(n+1)(n+2)F(a_n,b_n;n+3;x),

for any n∈ and x∈(-∞,1), where we have used the notations  (<ref>).


Let us begin with (<ref>). 
The integral term in (<ref>) can be written  as follows

    ∫_0^1 F_n(τ x) τ^n[-1+2xτ] dτ =(2x-1)∫_0^1 F_n(τ x) τ^n dτ-2x∫_0^1 F_n(τ x) τ^n(1-τ) dτ.

This leads to

    ζ_n(x)   =    F_n(x)[1-x+A+2B/A(n+1) x]+(2x-1)∫_0^1 F_n(τ x) τ^n dτ
          -2x∫_0^1 F_n(τ x) τ^n(1-τ) dτ.

We use (<ref>) in order to get successively

    ∫_0^1 F(a_n,b_n;n+1;τ x) τ^n dτ=F(a_n,b_n; n+2;x)/n+1

and

    ∫_0^1 F(a_n,b_n;n+1;τ x) τ^n[1-τ] dτ=F(a_n,b_n; n+3;x)/(n+1)(n+2),

 for any x∈(-∞,1). Taking into account these identities, we can rewrite (<ref>)  as (<ref>).
 
 
In oder to obtain (<ref>), we use (<ref>) with a=a_n,  b=b_n and c=n+1, which yields

    F(a_n,b_n;n+1)(x-1)   =   (n+3)x-(n+1)/n+1F(a_n,b_n;n+2;x)
          +(a_n-(n+2))((n+2)-b_n)x/(n+1)(n+2)F(a_n,b_n;n+3;x)
       =   (n+3)x-(n+1)/n+1F(a_n,b_n;n+2;x)
          -(2n+2)x/(n+1)(n+2)F(a_n,b_n;n+3;x),

where we have taken into account  the identities a_n+b_n=n and a_nb_n=-2. By virtue of  the first assertion of this lemma we obtain

    ζ_n(x)   =   A+2B/A(n+1) xF(a_n,b_n;n+1;x)
          +(n+1)-(n+3)x/n+1F(a_n,b_n;n+2;x)
    +(2n+2)x/(n+1)(n+2)F(a_n,b_n;n+3;x)
          +2x-1/n+1F(a_n,b_n;n+2;x)-2x/(n+1)(n+2)F(a_n,b_n;n+3;x)
       =   A+2B/A(n+1) xF(a_n,b_n;n+1;x)
          +n-(n+1)x/n+1F(a_n,b_n;n+2;x)
    +2nx/(n+1)(n+2)F(a_n,b_n;n+3;x).

This achieves the proof of the second identity (<ref>).


Let us also remark that using  (<ref>) we can deduce another useful equivalent expression for ζ_n

    ζ_n(x)=I_n^1(x)F(a_n,b_n;n+1;x)+I_n^2(x)F(a_n+1,b_n;n+2;x)+I_n^3(x)F(a_n,b_n;n+3;x),

where

    I_n^1(x)   ≜   n-a_n/n+1-a_n+x(A+2B/A(n+1)-n-1+a_n/n+1-a_n),
    
    I_n^2(x)   ≜   -a_n(2x-1)/(n+1)(n+1-a_n),
    
    I_n^3(x)   ≜   -2x/(n+1)(n+2).


 





 §.§ Qualitative properties of  hypergeometric functions
 
The main task of this section is to provide  suitable properties  about the analytic continuation of  the mapping  (n,x)↦ F(a_n,b_n;c_n;x) and some partial monotonicity behavior.
First, applying  the integral representation  (<ref>) with  the special coefficients (<ref>), we find

    F(a_n,b_n;c_n;x)=Γ(n+1)/Γ(n-a_n)Γ(1+a_n)∫_0^1 τ^n-a_n-1 (1-τ)^a_n (1-τ x)^-a_ndτ,

for x∈(-∞,1].
Notice that, due  to (<ref>) we can evaluate it at 1, obtaining

    F(a_n,b_n;c_n;1)=Γ(n+1)/Γ(n-a_n+1)Γ(1+a_n),

for any n≥ 2, where we have used the identity Γ(x+1)=xΓ(x).  We observe that the representation  (<ref>) fails for the case n=1  because a_1=-1.  This does not matter since as we have already mentioned in Remark <ref>, the case n=1 is explicit and the study of ζ_1 can be done by hand. 
It is a well-known fact  that the Gamma function  can be extended analytically to \{0,-1,-2,…}. Therefore, the  map n∈^⋆\{1}↦ F(a_n,b_n;c_n;1) admits a 𝒞^∞- extension given by

    ℱ:∈]1,+∞[↦Γ(+1)/Γ(-a_+1)Γ(1+a_),

with a_=-4/+√(^2+8).


The first result that we should discuss concerns some  useful asymptotic behaviors for  ↦ℱ().

The following properties are satisfied:


  * Let ≥1, then the function x∈(-∞,1]↦ F(a_,b_;c_;x) is positive and strictly decreasing.

  * For large ≫ 1, we have

    ℱ()
    =1-2ln/-2γ/+O(ln^2 /^2)

and

    ℱ^'()
    =2ln/^2+2(γ-1)/^2+O( ln^2 /^3),

where  γ  is  the Euler constant.
In particular, we have the asymptotics 

    F(a_n,b_n;c_n;1)=1-2ln n/n-2γ/n+O(ln^2 n/n^2)

and

    d/dnF(a_n,b_n;c_n;1)
    =2ln n/n^2+2(γ-1)/n^2+O( ln^2 n/n^3),

for large n. 



(1) 
The case =1 follows obviously  from the explicit expression given in Remark <ref>. Now let us  consider >1. According to (<ref>), we can differentiate   F  with respect to x:

    F^'(a_,b_;c_;x)=a_ b_/c_F(a_+1,b_+1;c_+1;x),  ∀ x∈ (-∞,1).

Using the integral representation  (<ref>) and the positivity of Gamma function, one has that  for any c>b>0

    F(a,b;c,x)>0,  ∀ x∈ (-∞,1).

Since a_∈(-1,0), b_, c_>0, then we deduce that F(a_+1,b_+1;c_+1;x)>0 and thus

    F^'(a_,b_;c_;x)<0,  ∀ x∈ (-∞,1).

This implies  that  x↦ F(a_n,b_n;c_n;x) is strictly decreasing and together with (<ref>) we obtain  

    F(a,b;c,x)≥ F(a_,b_;c_,1)>0,  ∀ x∈ (-∞,1).




(2) The  following asymptotic behavior

    Γ(+α)/Γ(+β)=∑_n∈ C_n(α-β,β) ^α-β-n,

holds as → +∞ by using <cit.>, where the coefficients C_n(α-β,β) can be obtained recursively and are polynomials on the variables α,β. In addition, the first coefficients can be calculated explicitly

    C_0(α-β,β)=1  and   C_1(α-β,β)=1/2(α-β)(α+β-1).

Taking  α=1 and β=1-a_ we deduce

    Γ(+1)/Γ(+1-a_)= ^a_+1/2 a_(1-a_)^a_-1+O(1/^2)
    =  ^a_+O(1/^2),

where we have used that a_∼ -2/. From the  following expansion

    ^a_=e^a_ln
    = 1-2ln/+O(ln^2 /^2),

we get

    Γ(+1)/Γ(+1-a_)   =   1-2ln/+O(ln^2 /^2).

Using again  Taylor expansion, we find
Γ(1+a_)=1+a_Γ^'(1)+O(a_^2).
Therefore, combining this  with Γ^'(1)=-γ and a_∼ -2/ yields

    Γ(1+a_)=1+2γ/+O(1/^2).

Consequently, it follows that  ℱ admits the following asymptotic behavior at infinity

    ℱ()=1-2ln/+O(^-2ln^2 )/1+2γ/+O(^-2)
    
    = 1-2ln/-2γ/+O(ln^2 /^2).

Since Γ is real  analytic, we have that  ℱ is also real analytic and one may deduce the asymptotics at +∞ of the derivative ℱ^'  through the  differentiation  term by term the asymptotics of ℱ. Thus, we obtain the second expansion in assertion (2).

Our next purpose is  to provide some useful estimates  for   F(a_n,b_n;c_n;x) and its partial  derivatives.  More precisely, we state the following result.


With the notations (<ref>), the following assertions hold  true.


  * The sequence n∈[1,+∞)↦ F(a_n,b_n;c_n;x) is strictly increasing, for any x∈ (0,1], and strictly decreasing, for any x∈(-∞,0).

  *  Given n≥1 we have

    |∂_n F_n(x)|≤-2x F_n(x)/(n+1)^2,

for any x∈(-∞,0].

  * There exists C>0 such that

    |∂_xF(a_n,b_n;c_n;x)|≤ C+C|ln (1-x)|,

for any x∈ [0,1], and n≥ 2.

  * There exists C>0 such that

    |∂_n F(a_n,b_n;c_n;x)|≤ Cln n/n^2,

for any x∈ [0,1], and n≥2.

  * There exists C>0 such that

    |∂_xxF(a_n,b_n;c_n;x)|≤C/1-x,

for any x∈ [0,1] and n≥2.



(1)
Recall that F_n solves the equation

    x(1-x)F_n”(x)+(n+1)(1-x)F_n'(x)+2F_n(x)=0,

with F_n(0)=1 and F_n'(0)=a_nb_n/c_n=-2/n+1. As we have mentioned in the beginning of this  section the dependence with respect to n is smooth, here we use n as a continuous parameter instead  Then, differentiating  with respect to n we get

    x(1-x)(∂_nF_n)”(x)+(n+1)(1-x)(∂_nF_n)'(x)+2(∂_nF_n)=-(1-x)F_n'(x),

with (∂_nF_n)(0)=0 and (∂_nF_n)'(0)=2/(n+1)^2. We can explicitly solve the last differential equation by using the variation of the constant and keeping in mind that  x↦ F_n(x) is a homogeneous solution. Thus, we obtain

    ∂_n F_n(x)=F_n(x)[K_2-∫_x^11/F_n^2(τ)τ^n+1(K_1-∫_0^τ F_n'(s)F_n(s)s^nds)dτ],

where the constant K_1 must be zero to remove the singularity at 0, in a similar way to the proof of Lemma  <ref>. Since (∂_nF_n)(0)=0, we deduce that

    K_2=-∫_0^11/F_n^2(τ)τ^n+1∫_0^τ F_n'(s)F_n(s)s^n ds   dτ ,

and then

    ∂_n F_n(x)=-F_n(x)∫_0^x1/F_n^2(τ)τ^n+1∫_0^τ F_n'(s)F_n(s)s^ndsdτ.

The change of variables s=τθ leads to 

    ∂_n F_n(x)=-F_n(x)∫_0^x1/F_n^2(τ)∫_0^1 F_n'(τθ)F_n(τθ)θ^ndθ dτ.

Hence,  it is clear that ∂_n F_n(x)>0, for x∈[0,1), using Lemma <ref>-(1)  . In the case x∈(-∞,0] we similarly get ∂_n F_n(x)<0.  Let us observe that  the compatibility condition  (∂_nF_n)'(0)=2/(n+1)^2 can be directly checked from the preceding representation. Indeed, one has

    (∂_nF_n)'(0)   =   lim_x→0∂_nF_n(x)/x
    =-lim_x→ 0^+F_n'(0)∫_0^1θ ^ndθ/F_n(0)
    =2/(n+1)^2.



(2)
First, notice that

    F^'_n(x)=-2/n+1F(a_n+1,b_n+1,c_n+1,x),

and from (<ref>) we deduce that

    |F^'_n(x)|≤2/n+1F(a_n+1,b_n+1,c_n+1,x).



Now studying the variation of   x∈(-∞,1)↦ F(a_n+1,b_n+1,c_n+1,x) by means of the integral representation (<ref>), we can show that it is  strictly increasing and positive, which implies in turn that

    0<F(a_n+1,b_n+1,c_n+1,x)≤ F(a_n+1,b_n+1,c_n+1,0)=1,  ∀ x∈ (-∞,0].

This allows us to get,

    0≤ -F^'_n(x)≤2/n+1,  ∀ x∈ (-∞,0].

Lemma <ref>-(1) implies in particular that

    F_n(x)≥1,  ∀ x∈ (-∞,1),

and coming back  to (<ref>) we find 

    |∂_n F_n(x)|   ≤   2 F_n(x)/n+1∫_x^01/F_n^2(τ)∫_0^1 F_n(τθ)θ^ndθ dτ≤2 F_n(x)/(n+1)^2∫_x^0dτ/F_n(τ)≤-2x F_n(x)/(n+1)^2,


for x∈(-∞,0]. This achieves the proof of the announced inequality.


(3) From previous computations we have

    ∂_xF(a_n,b_n;c_n;x)=-2/n+1 F(1+a_n,n+1-a_n; n+2;x),

which admits the integral representation

    ∂_xF(a_n,b_n;c_n;x)
       =   -2Γ(n+2)/(n+1)Γ(n+1-a_n)Γ(1+a_n)∫_0^1 τ^n-a_n (1-τ)^a_n (1-τ x)^-1-a_ndτ
       =   -2(n+2)/n+1ℱ(n) ∫_0^1 τ^n-a_n (1-τ)^a_n (1-τ x)^-1-a_ndτ
       =   -2(n+2)/n+1ℱ(n) J_n (x),

where

    J_n(x)≜∫_0^1 τ^n-a_n (1-τ)^a_n (1-τ x)^-1-a_ndτ.

Using

    sup_n≥12(n+2)/n+1≤ 3,

and the first assertion of Lemma <ref>, we have ℱ(n)∈[0,1], and

    sup_n≥12(n+2)ℱ(n)/n+1≤ 3.

Consequently,

    |∂_xF(a_n,b_n;c_n;x)|≤ 3J_n.

To estimate J_n we simply write

    J_n ≤∫_0^1 (1-τ)^a_n(1-τ/2)^-1-a_ndτ≤ C∫_0^1 (1-τ)^a_n dτ≤C/1+a_n≤ C,

for some constant C independent of n≥2, and for x∈[0,1/2]. In the case x∈[1/2,1), making the change of variable

    τ=1-1-x/xτ',

and denoting  the new variable again by  τ, we obtain

    J_n   ≤   x^-a_n-1∫_0^x/1-xτ^a_n(1+τ)^-1-a_ndτ
       ≤   x^-a_n-1∫_0^1τ^a_n(1+τ)^-1-a_ndτ+x^-a_n-1∫_1^x/1-xτ^a_n(1+τ)^-1-a_ndτ
       ≤   C+C∫_1^x/1-x(1+τ/τ)^-a_n(1+τ)^-1dτ≤ C+C|ln(1-x)|,

which  achieves the proof.

(4) According to (<ref>) and Lemma <ref>-(1) we may write 

    |∂_n F_n(x)|   ≤    F_n(x)∫_0^x1/F_n^2(τ)τ^n+1∫_0^τ |F_n'(s)|F_n(s)s^ndsdτ
       ≤   1/F_n^2(1)∫_0^1τ^-n-1∫_0^τ |F_n'(s)s^ndsdτ,

for x∈[0,1] and n≥2. Using (<ref>), the definition of J_n and the fact that 0<-a_n<1,  we obtain 

    |F_n'(x)|≤ 3∫_0^1τ^n-a_n(1-τ)^a_n(1-τ x)^-1-a_n dτ≤3/1-x∫_0^1τ^n-a_n(1-τ)^a_n dτ.

Now, recall the classical result on the Beta function B defined as follows

    ∫_0^1τ^n-a_n(1-τ)^a_n dτ = B(n+1-a_n,1+a_n)
    =Γ(n+1-a_n)Γ(1+a_n)/Γ(n+2),

which implies in view of (<ref>) that
 
    ∫_0^1τ^n-a_n(1-τ)^a_n dτ=1/(n+1)F_n(1).

 Consequently
 
    |F_n'(x)|≤3/(1-x)(n+1)F_n(1).

 Inserting this inequality into (<ref>) we deduce that
 
    |∂_n F_n(x)|
       ≤   1/(n+1)F_n^3(1)∫_0^1τ^-n-1∫_0^τs^n/1-sdsdτ,

and integrating by parts we find
  
    ∫_0^1τ^-n-1∫_0^τs^n/1-sdsdτ   =   [1-τ^-n/n∫_0^τs^n/1-s ds]_0^1+1/n∫_0^1s^n-1/s-1 ds
       =   1/n∫_0^1∑_k=0^n-1s^k ds
    =1/n∑_k=1^n1/k.

Thus, it follows from  the classical inequality
 ∑_k=1^n1/k≤ 1+ln n
that

    |∂_n F_n(x)|
    ≤1+ln n/n^2F_n^3(1),

 for n≥2. Since F_n(1)>0 and converges to 1, as n goes to ∞, then one can find an absolute constant C>0 such that
 
    |∂_n F_n(x)|
    ≤ Cln n/n^2,

for any n≥2, which achieves the proof of the estimate.


(5) Differentiating the integral representation (<ref>) again with respect to x we obtain

    ∂_xxF(a_n,b_n;c_n;x)
    = (a_n+1)a_n(n-a_n)(n+2)/n+1ℱ(n) J_n(x),

with

    J_n(x)≜∫_0^1 τ^n-a_n+1 (1-τ)^a_n (1-τ x)^-2-a_ndτ.

According to (<ref>) one finds that

    |∂_xxF(a_n,b_n;c_n;x)|
    ≤ 4J_n(x).

The procedure for estimating J_n matches the one given for J_n in the previous assertion. Indeed,  we have the uniform bound
J_n(x)≤ C,
for x∈[0,1/2]. Now,  the change of variables (<ref>) leads to

    J_n(x) ≤∫_0^1 (1-τ)^a_n (1-τ x)^-2-a_ndτ≤x^-a_n-1/1-x∫_0^x/1-xτ^a_n(1+τ)^-2-a_ndτ≤C/1-x,

where  we have used the bounds 0<-a_n<2/1+√(3)<1, which are verified for any n≥2 and  x∈[1/2,1). Inserting this estimate into (<ref>) we obtain the announced inequality.



Next we shall  prove the following.

There exists C>0 such that  

    |F(a_n,b_n;n+1;x)-1|    ≤    Cln n/n, 
    
    1≤ F(a_n+1,b_n;n+2;x)    ≤    C n, 
    
    |F(a_n,b_n;n+3;x)-1|    ≤    Cln n/n,

for any  n≥ 2 and any x∈[0,1].


The estimate  (<ref>) follows easily from  the second assertion of Lemma <ref>, combined with the monotonicity of F_n and Lemma <ref>. Indeed,

    |F(a_n,b_n;c_n;0)-F(a_n,b_n;c_n;x)|≤ F(a_n,b_n;c_n;0)-F(a_n,b_n;c_n;1)
    ≤ Cln n/n.


In the case (<ref>), applying 
 similar arguments as in the first assertion of Lemma <ref>, we conclude that the function x∈[0,1]→ F(a_n+1,b_n;n+2;x) is positive and strictly increasing. Hence,

    1≤ F(a_n+1,b_n;n+2;x) ≤  F(a_n+1,b_n;n+2;1).

Combining  (<ref>) and (<ref>) we obtain the estimate:

    1≤ F(a_n+1,b_n;n+2;x) ≤  F(a_n+1,b_n;n+2;1)
    ≤Γ(n+2)/Γ(n+1-a_n)Γ(2+a_n)≤  C n.


To check (<ref>), we use 
the first assertion of Lemma <ref>, 

    0≤ 1-F(a_n,b_n;n+3;x)≤ 1-F(a_n,b_n;n+3;1).

Moreover, by virtue of  (<ref>) one has

    F(a_n,b_n;n+3;1)=Γ(n+3)/Γ(n+3-a_n)Γ(3)/Γ(3+a_n).

As a consequence of (<ref>) and  a_n∼ -2/n, we obtain

    Γ(n+3)/Γ(n+3-a_n)=1-2ln n/n+O(1/n)     Γ(3)/Γ(3+a_n)=1+O(1/n).

Therefore, the following asymptotic expansion

    F(a_n,b_n;n+3;1)=1-2ln n/n+O(1/n)

holds and the estimate follows easily.



Another useful property deals with the behavior of the hypergeometric function with respect to the third variable c.


Let n≥1, then 
the mapping c∈ (b_n,∞)↦ F(a_n,b_n;c;x) is strictly increasing for x∈(0,1) and strictly decreasing  for x∈(-∞,0).




First, we check the case n=1 that comes by

    F(a_1,b_1;c;x)=1-2/c x.

Let n>1 and recall that the hypergeometric function F(a_n,b_n;c;x) solves  the differential equation

    x(1-x)∂^2_xxF(a_n,b_n;c;x)+[c-(n+1)x]∂_xF(a_n,b_n;c;x)+2F(a_n,b_n;c;x)=0,

with F(a_n,b_n;c;0)=1 and ∂_xF(a_n,b_n;c;0)=-2/c. Hence by differentiation  it is easy to check that ∂_c F(a_n,b_n;c;x) solves

    x(1-x)∂^2_xx(∂_c F(a_n,b_n;c;x))+[c-(n+1)x]∂_x(∂_c F(a_n,b_n;c;x))+2(∂_ cF(a_n,b_n;c;x))
          =-∂_x F(a_n,b_n;c;x),

with initial conditions 

    ∂_ cF(a_n,b_n;c;0)=0  and  ∂_x(∂_ cF(a_n,b_n;c;0))=2/c^2.
 
Note that a homogeneous solution of the last differential equation is F(a_n,b_n;c;x). By the variation of constant method one can look for the full solution to the differential equation in the form  

    ∂_ cF(a_n,b_n;c;x)=K(x)F(a_n,b_n;c;x),
 and from straightforward computations we find that  T=K' solves the first order differential equation, 

    T'(x)+[2∂_x F(a_n,b_n;c;x)/F(a_n,b_n;c;x)+c-(n+1)x/x(1-x)]  T(x)=-∂_x F(a_n,b_n;c;x)/x(1-x)F(a_n,b_n;c;x).

The general solution to this latter equation is given by 

    T(x)=(1-x)^c-(n+1)/F(a_n,b_n;c;x)^2|x|^c[K_1-∫_0^x |s|^c s^-1F(a_n,b_n;c;s)∂_sF(a_n,b_n;c;s)/(1-s)^c-nds],

for x∈ (-∞,1) 
where K_1∈ is a real constant.
 Thus

    ∂_c F(a_n,b_n;c;x)=F(a_n,b_n;c;x)[K_2+
                ∫_x_0^x(1-τ)^c-(n+1)/F(a_n,b_n;c;τ)^2|τ|^c{K_1-∫_0^τ|s|^cs^-1F(a_n,b_n;c;s)∂_sF(a_n,b_n;c;s)/(1-s)^c-nds}dτ],

where K_1, K_2∈ and x_0∈(-1,1). Since ∂_c F(a_n,b_n;c;x) is not singular at x=0, we get that K_1=0. Then, changing the constant K_2 one can take x_0=0 getting

    ∂_c F(a_n,b_n;c;x)=F(a_n,b_n;c;x)[K_2-
                .∫_0^x(1-τ)^c-(n+1)/F(a_n,b_n;c;τ)^2|τ|^c∫_0^τ|s|^cs^-1F(a_n,b_n;c;s)∂_sF(a_n,b_n;c;s)/(1-s)^c-ndsdτ].

The initial condition ∂_c F(a_n,b_n;c;0)=0 implies that K_2=0 and hence

    ∂_c F(a_n,b_n;c;x)   =   -F(a_n,b_n;c;x)×
          ∫_0^x(1-τ)^c-(n+1)∫_0^τ|s|^cs^-1F(a_n,b_n;c;s)∂_sF(a_n,b_n;c;s)/|τ|^cF(a_n,b_n;c;τ)^2(1-s)^c-ndsdτ.

Similarly to the proof of Lemma <ref>-(1) one may obtain that  

    F(a_n,b_n;c;x)>0  and  ∂_x F(a_n,b_n;c;x)<0,
 
for any c>b_n and x∈(-∞,1) . This  entails that ∂_c F(a_n,b_n;c;x) is positive for x∈(0,1), and negative when x∈(-∞,0), which  concludes the proof.




 §.§ Eigenvalues

The existence of eigenvalues, that are the elements of the dispersion set  defined in (<ref>), is connected to the problem of studying the roots of the equation 
introduced in (<ref>). 
Here, we will develop different cases illustrating strong discrepancy on the structure of the dispersion set. Assuming A>0 and B<-A, we find that the dispersion set is infinite. However, for the case A>0 and B≥ -A/2, the dispersion set is finite.  Notice that the transient regime corresponding to -A≤ B≤ -A/2 is not covered by the current study and turns to be more complicate due to the complex structure  of  the spectral function (<ref>).


Let us begin with studying   the cases

    A>0,      A+B<0,
        A>0,      A+2B≤ 0,
        A>0,      A+2B≥ 0.



Our first main result reads as follows. 

The following assertions hold true:


  * Given A, B satisfying (<ref>), there exist n_0∈^⋆, depending only on A and B, and a unique root x_n∈ (0,1) of  (<ref>), i.e.
ζ_n(x_n)=0,
for any n≥ n_0.
In addition,

    x_n∈(0,1+A+B/An),

and the sequence n∈[n_0,+∞)↦x_n is strictly increasing.

  * Given A, B satisfying the weak condition (<ref>) and   n∈^⋆, then ζ_n has no solution in (-∞,0].

  * Given  A, B satisfying (<ref>), then ζ_n has no solution in [0,1], for n≥2.



(1) The expression of the spectral equation (<ref>) agrees with

    ζ_n(x)=I_n^1(x)F(a_n,b_n;n+1;x)+I_n^2(x)F(a_n+1,b_n;n+2;x)+I_n^3(x)F(a_n,b_n;n+3;x),

where  I_n^1, I_n^2 and I_n^3 are defined in Lemma <ref>.
From this  expression we get  ζ_n(0)=n/n+1.  To find a solution in (0,1) we shall apply the Intermediate Value Theorem, and for this purpose we need to check that  ζ_n(1)<0. 
Applying  (<ref>) we get

    ζ_n(1)   =   [1/n+1-a_n+A+2B/A(n+1)]Γ(n+1)/Γ(n+1-a_n)Γ(1+a_n)
          -a_nΓ(n+1)/Γ(n+2-a_n)Γ(2+a_n)-4Γ(n)/Γ(n+3-a_n) Γ(3+a_n).

Using the following  expansion for large n≫1

    1/n+1-a_n=1/n+1+O(1/n^3),

and (<ref>), we find

    ζ_n(1)=2A+B/n+1+O(ln n/n^2).

Thus, under the hypothesis (<ref>),  there  exists n_0∈^⋆, depending on A,B,  such that

    ζ_n(1)<0,  ∀ n≥ n_0.

This proves the existence of  at least one  solution x_n∈ (0,1) to the equation
ζ_n(x_n)=0,
for any n≥ n_0.
The next objective is  to localize this root and show that  x_n∈(0,1+A+B/An). For this goal it suffices to verify that

    ζ_n(1-)<0,    ∀∈(0,-A+B/An).

Let us begin with the first term I_n^1(x) in the expression (<ref>) which implies that

    I_n^1(1-)   =   [1-2a_n/n+1-a_n+A+2B/A(n+1)]-[A+2B/A(n+1)-n-1+a_n/n+1-a_n].

Now, it is straightforward to check the  following asymptotic expansions,

    1-2a_n/n+1-a_n+A+2B/A(n+1)=2A+B/A(n+1)-(2n+1)a_n/(n+1)(n+1-a_n)
    =2A+B/A(n+1)+O(1/n^2),

and

    -[A+2B/A(n+1)-n-1+a_n/n+1-a_n]   =    -[-1+A+2B/A(n+1)+2-2a_n/n+1-a_n]
       ≤   +O(1/n^2)
    ≤   -A+B/An+O(1/n^2),

Therefore, we obtain

    I_n^1(1-)≤A+B/An+O(1/n^2).

Thanks to (<ref>), we deduce

    | 1-F(a_n,b_n;n+1;1-)| 
    ≤  Cln n/n,

which yields in turn

    I_n^1(1-)F(a_n,b_n;n+1;1-)≤A+B/A(n+1)+O(ln n/n^2).

Next, we will deal with the second term I_n^2 of (<ref>). Directly from (<ref>) we get

    |I_n^2(x)|F(a_n+1,b_n;n+2;x)=|a_n(2x-1)|/(n+1)(n+1-a_n)F(a_n+1,b_n;n+2;x)
    ≤C/n^2 .

Similarly, the estimate (<ref>) implies that

    |I_n^3(x)|F(a_n,b_n; n+3;x)=2xF(a_n,b_n;n+3;x)/(n+1)(n+2)≤C/n^2.

Inserting (<ref>), (<ref>) and (<ref>) into the expression of ζ_n we find

    ζ_n(1-)≤A+B/A(n+1)+O(ln n/n^2),

for any ∈(0,-A+B/An).
From this we deduce the existence of n_0 depending on A and B such that

    ζ_n(1-)≤A+B/2A(n+1)<0,

for any n≥ n_0. Then ζ_n has no zero in (1+A+B/An,1) and this achieves the proof of the first result.

Next we shall prove that x_n is the only zero of ζ_n in (0,1). For this purpose it appears to be more convenient to  use the expression for ζ_n given by (<ref>). Let us differentiate  ζ_n with respect to x as follows

    ∂_xζ_n(x)   =   F_n^'(x)[1-x+A+2B/A(n+1) x]+F_n(x)[-1+A+2B/A(n+1)]
           + ∫_0^1 F_n^'(τ x) τ^n+1[-1+2xτ] dτ+2∫_0^1 F_n(τ x) τ^n+1 dτ.

From Lemma <ref>-(1), we recall that  F_n>0 and F_n^'<0. Hence for A+2B<0 and x∈(0,1) we get

    ∂_xζ_n(x)≤ F_n^'(x)A+2B/A(n+1) x-F_n(x)
    + ∫_0^1 F_n^'(τ x) τ^n+1[-1+2xτ] dτ
    +2∫_0^1 F_n(τ x) τ^n+1 dτ.

Applying the  third assertion of Lemma <ref> we find 

    |F_n^'(x)|≤ Cln n,  ∀ n≥2,

for any x∈ (0,1+A+B/An) and with C a constant depending only on  A and B. It follows that

    ∂_xζ_n(x)≤ C|A+2B|/Aln n/n-F_n(x)+Cln n/n+C/n≤ -1+C2A+2|B|/Aln n/n+(1-F_n(x)),

for x∈[0,1+A+B/An), which implies according to (<ref>) that

    ∂_xζ_n(x)
    ≤-1+CA+|B|/Aln n/n.

 Hence, there exists n_0 such that

    ∂_xζ_n(x)≤-1/2,   ∀ x∈[0,1+A+B/An),  ∀ n≥ n_0 .

Thus, the function x∈[0,1+A+B/An)↦ζ_n(x) is strictly decreasing  and admits only one zero that we have   denoted by x_n.

It remains to show that n∈[n_0,+∞)↦x_n is strictly increasing, which implies in particular that

    ζ_m(x_n)≠0,   ∀ n≠ m≥ n_0.

For this aim, it suffices to show that  the mapping  n∈[n_0,+∞)↦ζ_n(x) is strictly increasing, for any x∈ (0,1). Setting

    F_n(x)=1+ρ_n(x),

we can write

    ζ_n(x)   =   n/n+1-n/n+2 x+A+2B/A(n+1)x+ρ_n(x)[1-x+A+2B/A(n+1)x]
           +∫_0^1ρ_n(τ x) τ^n[-1+2xτ]dτ≜n/n+1-n/n+2 x+A+2B/A(n+1)x+R_n(x) .

Since F_n is analytic with respect to its parameters and we can think in n as a continuous parameter,  n↦ζ_n(x) is also analytic. Therefore,
differentiating with respect to n, we deduce that

    ∂_nζ_n(x)=1/(n+1)^2-2x/(n+2)^2 -A+2B/A(n+1)^2x+∂_nR_n(x).

Consequently,

    ∂_nζ_n(x)≥1/(n+1)^2[1-2x -A+2B/Ax]+∂_nR_n(x)
    ≥1/(n+1)^2[1-3A+2B/Ax]+∂_nR_n(x).

We use the following trivial bound

    1 -3A+2B/Ax≥min(1,κ),   ∀ x∈ [0,1],

where κ=-2A+B/A is strictly positive due to the assumptions (<ref>).
Therefore, we can rewrite the bound for ∂_nζ_n(x) as follows

    ∂_nζ_n(x)≥min(1,κ)/(n+1)^2+∂_nR_n(x).

To estimate ∂_nR_n(x) we shall differentiate (<ref>)  with respect to n, 

    ∂_nR_n(x)   =   ∂_nF_n(x)(1-x)+∂_nF_n(x)A+2B/A(n+1)x
    -ρ_n(x)A+2B/A(n+1)^2x
          + ∫_0^1(∂_nF_n(τ x)) τ^n[-1+2xτ]dτ
    +∫_0^1ρ_n(τ x) τ^nlnτ[-1+2xτ]dτ.

From (<ref>) and Lemma <ref>, we deduce

    ∂_nR_n(x)   ≥   ∂_nF_n(x)A+2B/A(n+1)x-ρ_n(x)A+2B/A(n+1)^2x
           + ∫_0^1∂_nF_n(τ x)  τ^n[-1+2xτ]dτ+∫_0^1ρ_n(τ x) τ^nlnτ[-1+2xτ]dτ,

and

    |∂_nF_n(x)A+2B/A(n+1)x-ρ_n(x)A+2B/A(n+1)^2x|   ≤    C|A+2B|/Aln n/n^3.

Observe that the  first integral can be bounded as follows

    |∫_0^1∂_nF_n(τ x)  τ^n[-1+2xτ]dτ| ≤ Cln n/n^2∫_0^1 τ^n|-1+2xτ|dτ≤ Cln n/n^3,

while for the second one we have 

    |∫_0^1ρ_n(τ x) τ^nlnτ[-1+2xτ]dτ| ≤ Cln n/n∫_0^1τ^n|lnτ| dτ
    ≤ Cln n/n^3,

where we have used (<ref>) and Lemma <ref>. Plugging  these estimates  into (<ref>), we find

    ∂_nζ_n(x)≥min(1,κ)/(n+1)^2-C|A+2B|/Aln (n+1)/(n+1)^3,   ∀ x∈ [0,1].

Then, there exists n_0 depending only on A,B such that

    ∂_nζ_n(x)≥min(1,κ)/2(n+1)^2,

for any n≥ n_0 and any x∈ [0,1]. This implies that n∈ [n_0,+∞[↦ζ_n(x) is strictly increasing and thus (<ref>) holds.


(2)  From   (<ref>) one has 

    ζ_n(x)   =   A+2B/A(n+1)xF(a_n,b_n;n+1;x)-x/n+1F(a_n,b_n;n+2;x)
          + n(1-x)/n+1F(a_n,b_n;n+2;x)+2nx/(n+1)(n+2)F(a_n,b_n;n+3;x).

Remark that  the involved  hypergeometric functions are strictly positive, which implies that

    ζ_n(x)>
     n/n+1((1-x)F(a_n,b_n;n+2;x)+2x/n+2F(a_n,b_n;n+3;x)).

To get the announced result, it is enough to check that

    (1-x)F(a_n,b_n;n+2;x)+2x/(n+2)F(a_n,b_n;n+3;x)≥1,  ∀ x∈(-∞,0),

which follows from  Lemma <ref>:

    (1-x)F(a_n,b_n;n+2;x)   +   2x/(n+2)F(a_n,b_n;n+3;x)
       ≥   F(a_n,b_n;n+2;x)(1-n x/n+2)≥F(a_n,b_n;n+2;x)≥1,

 for  any x≤0. Thus, ζ_n(x)>0 for any x∈(-∞,0) and this concludes the proof.


(3) Let us use the expression of ζ_n given in (<ref>) obtaining

    ζ_n(x)   =   A+2B/A(n+1) xF(a_n,b_n;n+1;x)+n(1-x)/n+1F(a_n,b_n;n+2;x)
          -x/n+1F(a_n,b_n;n+2;x)+2nx/(n+1)(n+2)F(a_n,b_n;n+3;x).

Then

    ζ_n(x)>
    x/n+1(-F(a_n,b_n;n+2;x)+2n/n+2F(a_n,b_n;n+3;x)),

 for any x∈ [0,1). From Lemma <ref> we deduce

    -F(a_n,b_n;n+2;x)+2n/(n+2)F(a_n,b_n;n+3;x) ≥n-2/n+2F(a_n,b_n;n+2;x)
    ≥ 0,

 for any n≥2 and x∈(0,1). This implies that ζ_n(x)>0, for x∈(0,1) and n≥2. The case n=1 can be checked directly by the explicit expression stated in Remark <ref>. 

In the following result, we investigate more the case  (<ref>). We mention that according to  there are no eigenvalues in (0,1). Thus, it remains to explore the region (-∞,0) and study whether one can find eigenvalues there. Our result reads as follows.

Let n≥2 and  A, B∈ satisfying (<ref>). Then, the following assertions hold true:


  * If 
n≤B/A+1/8,
there exists a unique x_n∈(-1,0) such that ζ_n(x_n)=0.

  * If 
n≤2B/A,
there exists a unique x_n∈(-∞,0) such that ζ_n(x_n)=0, with

    1/1-A+2B/A(n+1)<x_n<0.

In addition,  the map x∈(-∞,0]↦ζ_n(x) is strictly increasing.

  * If 
n≥B/A+1,
then ζ_n has no solution in [-1,0].


  * If 
n≥2B/A+2,
then ζ_n has no solution in (-∞,0].




(1)
Thanks to  (<ref>) we have that
ζ_n(0)=n/n+1>0. So to apply the Intermediate Value Theorem and prove that ζ_n admits a solution in [-1,0] it suffices to guarantee that
ζ_n(-1)<0. Now coming back to (<ref>) and using that x∈ (-1,1)↦ F_n(x) is strictly decreasing we get

    ζ_n(-1)   =   F_n(-1)(2-A+2B/A(n+1))-∫_0^1F_n(-τ)τ^n(1+2τ) dτ
       <   F_n(-1)(2-A+2B/A(n+1))-∫_0^1τ^n(1+2τ) dτ
       <   F_n(-1)(2-A+2B/A(n+1))-3n+4/(n+1)(n+2),   ∀ x∈ [-1,0).

Consequently, to get ζ_n(-1)<0 we impose the condition

    2-A+2B/A(n+1)≤3n+4/(n+1)(n+2)F_n(-1).

Coming back to the integral representation, one gets 

    F_n(-1)≤ F_n(0) 2^-a_n≤ 2,

due to the fact a_n∈ (-1,0). In addition, it is easy to check that 

    3n+4/n+2≥5/2,  ∀ n≥2,

and the assumption (<ref>) is satisfied if 

    2-A+2B/A(n+1)≤5/4(n+1)

holds, or equivalently, if

    2≤ n≤B/A+1/8.

In conclusion, under the assumption (<ref>), the function ζ_n admits a solution x_n∈ (-1,0). 
Now, we localize this zero.
Since F_n is strictly positive  in [-1,1], then the  second term in (<ref>) is always strictly negative. Let us analyze the sign of the first term

    F_n(x)[1-x+A+2B/A(n+1)x],

which has a unique root 

    x_c=1/1-A+2B/A(n+1).

This root belongs to (-∞,0) if and only if n<2B/A, which follows  automatically from (<ref>). Moreover the mapping x↦ 1-x+A+2B/A(n+1)x will be strictly increasing.  Hence, , then x_n>x_c. So let us assume that x_c∈(-1,0), then 

    F_n(x)[1-x+A+2B/A(n+1)x]<0,  ∀ x∈ [-1,x_c],

which  implies that

    ζ_n(x)<0,  ∀ x∈ [-1,x_c].

Therefore, the solution x_n must belong to (x_c,0), and equivalently

    1/1-A+2B/A(n+1)<x_n<0.

The uniqueness of this solutions comes directly from the second assertion.

(2)
As in the previous argument we have that
ζ_n(0)=n/n+1>0 and the idea is to apply also the Intermediate Value Theorem. We intend to find the asymptotic behavior of ζ_n for x going to -∞. We first  find an asymptotic behavior of F_n. For 
this purpose we use the identity (<ref>), which implies that

    F_n(x)=(1-x)^-a_nF(a_n,a_n+1,n+1,x/x-1),  ∀ x≤0.

Setting

    φ_n(y)≜ F(a_n,a_n+1,n+1,y),  ∀ y∈ [0,1],

we obtain

    φ_n^'(y)=a_n(1+a_n)/n+1F(a_n+1,a_n+2,n+2,y),  ∀ y∈ [0,1)

By a monotonicity argument we deduce that

    |φ_n^'(y)|≤ C|F(a_n+1,a_n+2,n+2,y)|
    ≤ C|F(a_n+1,a_n+2,n+2,1)|
    ≤ C_n,  ∀  y∈[0,1],

 with C_n a constant depending on n. However, the dependence with respect to n does not matter because we are interested in the asymptotics for large negative x  but for  a fixed n. Then, let us drop n from the subscript of the constant C_n. Applying the Mean Value Theorem we get
 
    |φ_n(y)-φ_n(1)|≤ C(1-y),  ∀  y∈[0,1].

 Combining this estimate with (<ref>) and a_n∈(-1,0), we obtain  
 
 
    |F_n(x)-(1-x)^-a_nφ_n(1)|≤ C (1-x)^-a_n-1≤ C,  ∀ x≤ 0,

 which implies in turn that
  
    |F_n(x)-(1-x)^-a_nφ_n(1)|
    ≤ C,  ∀ x≤ -1.

 Consequently, we deduce that
 
    ∫_0^1F_n(τ x)τ^n(-1+2xτ) dτ   ∼    2x(-x)^-a_nφ_n(1)∫_0^1τ^-a_n+n+1dτ
       ∼   2/n+2-a_nx(-x)^-a_nφ_n(1),   ∀  -x≫ 1.

Coming back to (<ref>) and using once again (<ref>)  we get the asymptotic behavior
 
    ζ_n(x)   ∼   φ_n(1)  (-x)^-a_n(1-x+A+2B/A(n+1) x+ 2/n+2-a_nx)
       ∼   φ_n(1) (A+2B/A(n+1)-1+ 2/n+2-a_n)(-x)^-a_n x,  ∀  -x≫ 1.

The condition

    n≤2B/A,

 implies that 

    A+2B/A(n+1)-1+ 2/n+2-a_n>0.

Since φ_n(1)>0 , we deduce that

    lim_x→-∞ζ_n(x)=-∞.

Therefore we deduce from  the Intermediate Value Theorem that  under the assumption (<ref>), the function ζ_n admits a solution x_n∈ (-∞,0). Moreover, by the previous proof we get (<ref>).

It remains to prove the uniqueness of this solution. For this goal we check that the mapping x∈(-∞,0)↦ζ_n(x) is strictly increasing when n∈[1,2B/A].
Differentiating  ζ_n with respect to x yields

    ζ_n'(x)   =   F_n'(x)[1-x+A+2B/A(n+1)x]+F_n(x)[A+2B/A(n+1)-1]
          +∫_0^1F_n'(τ x)τ^n+1(-1+2xτ)dτ+2∫_0^1F_n(τ x)τ^n+1dτ.

From Lemma <ref>-(1) we infer   for x∈(-∞,0), and therefore we get 

    ζ_n^'(x)>0,  ∀  x∈ (-∞,x_c).
 
Let  x∈(x_c,0), then by a monotonicity argument we get

    0≤ 1-x+A+2B/A(n+1)x≤ 1,

and  thus

    F_n^'(x)[1-x+A+2B/A(n+1)x]+2∫_0^1F_n(τ x)τ^n+1dτ   ≥   F_n'(0)+2/n+2≥-2/(n+1)(n+2),

by using that F_n^'(0)=-2/n+1 and  that F_n^'(x) is decreasing and negative in (-∞,1). From the assumption (<ref>) and the positivity of F_n we get

    F_n(x)(A+2B/A(n+1)-1)≥0,  ∀ x∈ (-∞,0].

Therefore, putting together the preceding estimates we deduce that

    ζ_n^'(x)   >   -2/(n+1)(n+2)+∫_0^1F_n'(τ x)τ^n+1(-1+2xτ)dτ
       >   -2/(n+1)(n+2)-∫_0^1F_n^'(τ x)τ^n+1dτ,  ∀ x∈ (-∞,0).

At this stage it suffices to make appeal to (<ref>) in order to obtain

    -∫_0^1F_n^'(τ x)τ^n+1dτ≥2/(n+1)(n+2),  ∀ x∈ (-∞,0],

from which  it follows

    ζ_n^'(x)>0,  ∀ x∈ (-∞,0],

which implies that ζ_n is strictly increasing in (-∞,0], and thus x_n is the only solution in this interval.

(3) 
Using the definition of ζ_n in (<ref>) and the monotonicity of F_n, one has

    ζ_n(x)>F_n(x)[1-x+A+2B/A(n+1)x-1/n+1+2x/n+1].

Now it is easy to check that 

    1-x+A+2B/A(n+1)x-1/n+1+2x/n+1≥min{n/n+1, 2n-1/n+1-A+2B/A(n+1)},

for any x∈[-1,0]. This claim can be derived from the fact that the left-hand-side term is polynomial in x with degree one.
Consequently, if we assume 

    n≥ 1+B/A,

we get 2n-1/n+1-A+2B/A(n+1)≥0, and therefore (<ref>) implies

    ζ_n(x)>0,  ∀ x∈ [-1,0].
 
Then, ζ_n has no solution in [-1,0].

(4)
Using the expression of ζ_n in (<ref>) and the monotonicity of F_n, one has

    ζ_n(x)   >   F_n(x)[1-x+A+2B/A(n+1)x-1/n+1+2x/n+1]
       ≥   F_n(x)[n/n+1-x(n-1/n+1-A+2B/A(n+1))],  ∀  x∈(-∞,0).

The assumption

    n≥ 2+2B/A,

yields n-1/n+1-A+2B/A(n+1)≥0, and therefore (<ref>) implies

    ζ_n(x)>0,  ∀ x∈ (-∞,0].
 
Thus, ζ_n has no solution in (-∞,0].


In the next task we discuss the localization of the  zeroes of ζ_n and, in particular, we improve the lower bound  (<ref>). Notice that B>0 in order to get solutions of ζ_n in (-∞,0] in the case A>0 and  n≥2, by using Proposition <ref>-(4). Our result reads as follows.

Let  A, B>0 and n≥2. If x_n∈(-∞,0) is any solution of ζ_n, then the following properties are satisfied:


  * We have 
𝒫_n(x_n)<0,   with 𝒫_n(x)= n/n+1+x[-n/n+2+A+2B/A(n+1)].


  * If  x_n∈(-1,0), then x_⋆≜-2n+1/2(n+1)1/A+2B/A(n+1)-n+1/n+2<x_n.


  * We always have  x_n<-A/2B.





(1)
Since  F_n(τ x)< F_n(x), for any τ∈[0,1) and x∈(-∞,0), then we deduce from the expression  (<ref>) that

    ζ_n(x)> F_n(x)[1-x+A+2B/A(n+1)x-1/n+1+2x/n+2]=F_n(x)𝒫_n(x).
  
As F_n is strictly positive in (-∞,1), then 
𝒫_n(x_n)<0, 
for any root of ζ_n.

(2)
Recall from Proposition <ref>–(3) that if ζ_n admits a solution in (-1,0) with n≥2 then necessary 2≤ n≤ 1+B/A. This implies that n≤2B/A and hence the mapping x↦ 1-x+A+2B/A(n+1)x is increasing. Combined with  the definition of  (<ref>)  and (<ref>) we deduce that

    1-x+A+2B/A(n+1)x≥ 0,	  ∀ x∈ (x_c,0)

and x_n∈(x_c,0). Using the monotonicity of F_n combined with the bound (<ref>) we find from   (<ref>) 

    ζ_n(x)<2[1-x+A+2B/A(n+1)x]-1/n+1+2x/n+2,  ∀ x∈ (x_c,0).

Evaluating at any root x_n we obtain

    -2n+1/n+1<2x_n[A+2B/A(n+1)-n+1/n+2].

 Keeping  in mind that n≤2B/A, we get 
A+2B/A(n+1)-n+1/n+2>0, and therefore 
we find the announced lower bound for x_n.
 

(3)
In a similar way to the upper bound for x_n, we turn to (<ref>), and evaluate this inequality at -A/2B. Then we find

    ζ_n(-A/2B)>F_n(-A/2B)𝒫_n(-A/2B)
     >𝒫_n(-A/2B).

Explicit computations yield

    𝒫_n(-A/2B)=n-1/n+1+A/2B[n/n+2-1/n+1]

Since A/2B>0 and n≥2, then we infer that 𝒫_n(-A/2B)>0 and ζ_n(-A/2B)>0. Now we recall from Proposition <ref>-(2) that x∈(-∞,0)↦ζ_n(x) is strictly increasing. Thus combined this property with the preceding one we deduce that

    x_n<-A/2B,

which achieves the proof.


Notice that from Proposition <ref>-(4) when B≤ 0, the function ζ_n has no solution in (-∞,0] for any n≥2. Moreover, in the case that 0<B≤A/4, Proposition <ref>-(4) and Proposition <ref>-(1) give us again that  ζ_n has no solution in (-∞,0] for any n≥2. Combining these facts 
with  Proposition <ref>–(3),  we immediately get the following result.

Let A>0 and B satisfying

    -A/2≤ B≤A/4.

Then, the function  ζ_n has no solution   in (-∞,1] for any n≥2. However,  the function ζ_1 admits the solution x_1=-A/2B. Notice that this latter solution belongs to (-∞,1) if and only if B∉[ -A/2,0].



In the next result, we study the case when x_1∈(0,1+ϵ/2] for some 0<ϵ<1, showing that there is no intersection with other eigenvalues.


Let A>0. There exists ϵ∈(0,1) such that if B≤-A/1+ϵ, then ζ_n(x_1)≠ 0 for any n≥ 2 and x_1=-A/2B, with
ϵ≈ 0,0581.


From (<ref>), one has

    ζ_n(x_1)=F_n(x_1)n/n+1(1-x_1)+∫_0^1F_n(x_1τ)τ^n[-1+2x_1τ]dτ.

By the integral representation of F_n given in (<ref>), we obtain that

    F_n(x)>Γ(n+1)/Γ(n-a_n)Γ(a_n+1)∫_0^1 ^n-a_n-1(1-)^a_nd(1-x)^-a_n=(1-x)^-a_n,

for any x∈(0,1), using the Beta function (<ref>).
By the monotonicity of F_n(x) with respect to x and the above estimate, we find that

    (1-x_1)^-a_n< F_n(x_1)<1,

for any n≥ 2 and x_1∈(0,1), which agrees with the hypothesis on A and B. Hence, we have that

    ζ_n(x_1)>n/n+1(1-x_1)^1-a_n+2x_1(1-x_1)^-a_n/n+2-1/n+1.

The above expression is increasing with respect to n, which implies that

    ζ_n(x_1)>2/3(1-x_1)^1-a_2+x_1/2(1-x_1)^-a_2-1/3.

Since a_2=1-√(3), we get

    ζ_n(x_1)>2/3(1-x_1)^√(3)+x_1/2(1-x_1)^√(3)-1-1/3≜𝒫(x_1).

The function 𝒫 decreases in (0,1) and admits a unique  root x whose approximate value is given by
x=0,52907.
Hence, 𝒫(x_1)≥0 for x_1∈(0,x], and consequently we get

    ζ_n(x_1)>0,

for B≤-A/2x , achieving the announced result.



We finish this section by the following result concerning  the monotonicity of the eigenvalues.

Let  A>0 and 2B>A. Then, the following assertions hold true: 



  * Let x∈ (-∞,0), then n∈[1,2B/A]↦ζ_n(x) is strictly increasing. In addition, we have 

    {x∈ (-∞,0],  ζ_n(x)= 0}∩{x∈ (-∞,0],  ζ_m(x)=0}=∅,

for any n≠ m∈[1,2B/A], and each set contains at most one element. 

  * The sequence n∈[1, B/A+1/8]↦ x_n is strictly decreasing, where the {x_n} are constructed in Proposition <ref>.

  * If   m∈[1, 2B/A-2], then 

    ζ_n(x_m)≠ 0,   ∀ n∈^⋆\{ m}.





(1)
We shall  prove that  the mapping n∈[1,2B/A]↦ζ_n(x) is strictly increasing for fixed x∈ (-∞,0] and 2B≥ A. 
Differentiating (<ref>) with respect to n we get

    ∂_nζ_n(x)   =   ∂_nF_n(x)[1-x+A+2B/A(n+1) x]+∫_0^1 ∂_nF_n(τ x) τ^n[-1+2xτ] dτ
          -F_n(x)A+2B/A(n+1)^2 x+∫_0^1 F_n(τ x) τ^n lnτ[-1+2xτ] dτ.

Using Lemma <ref>-(1) and the positivity of F_n, we deduce that 

    ∫_0^1 ∂_nF_n(τ x) τ^n[-1+2xτ] dτ>0,  ∫_0^1 F_n(τ x) τ^n lnτ[-1+2xτ] dτ>0,  ∀  x∈ (-∞,0).

Due to the assumption   n∈[1,2B/A], we have that x_c<0, where  x_c is defined in (<ref>).
If x∈(-∞,x_c], we find that

    1-x+A+2B/A(n+1) x≤ 0,

which implies

    ∂_n ζ_n (x)>0,  ∀ x∈(-1,x_c),∀ n≥ 1.

We obtain 

    0<1-x+A+2B/A(n+1) x<1,

 for x∈(x_c,0), which yields in view of  Lemma <ref>-(2)

    ∂_nζ_n(x)   >   ∂_nF_n(x)[1-x+A+2B/A(n+1) x]-F_n(x)A+2B/A(n+1)^2 x
       >   ∂_nF_n(x)-F_n(x)A+2B/A(n+1)^2 x
    >-2xF_n(x)/(n+1)^2[-1+A+2B/2A].

Taking into account 2B≥ A, one gets
∂_nζ_n(x)>0, ∀  x∈(-∞,0].
It remains to discuss the  case x_c≤-1. Remark that the estimate (<ref>) is satisfied for any x∈(-∞,0), and then the foregoing  inequality holds, and one gets finally

    ∂_nζ_n(x)>0,  ∀  x∈[-1,0],∀ n∈[1,2B/A].

Consequently, we deduce that the mapping n∈[1,2B/A]↦ζ_n(x) is strictly increasing for any x∈ (-∞,0). This implies in particular that the functions ζ_n and ζ_m have no common zero in (-∞,0) for n≠ m∈[1,2B/A].

(2) This follows  by combining that x∈ (-∞,0)↦ζ_n(x) and n∈[1,2B/A]↦ζ_n(x) are strictly increasing, proved in Proposition <ref>-(1) and Proposition <ref>-(1).

(3) By the last assertions, this is clear for n≤2B/A and it is also true for n≥2B/A+2, since ζ_n has not roots in (-∞,1), by Proposition <ref> and Proposition <ref>. Then, let us study the case n∈(2B/A,2B/A+2). First, using (<ref>), we get that x_m, which is a solution of ζ_m=0 with m≤2B/A-2, verifies

    x_m≥ -2B-A/2A.

Now, the strategy is to show that 𝒫_n(x_m)>0 for n∈(2B/A,2B/A+2), and then  Proposition <ref> will imply that ζ_n(x_m)≠ 0. By definition, we have that

    𝒫_n(x_m)=n/n+1+x_m[-n/n+2+A+2B/A(n+1)].

If -n/n+2+A+2B/A(n+1)≤ 0, then 𝒫_n(x_m)>0. Otherwise, we use (<ref>) getting

    𝒫_n(x_m)   >   n/n+1-2B-A/2A[-n/n+2+A+2B/A(n+1)]
       =   1/2(n+1)(n+2)[n^2(1+2B/A)+n(4+2B/A-4B^2/A^2)+2(1-4B^2/A^2)].

Straightforward computations yield that the above parabola is increasing in n∈(2B/A,2B/A+2). Evaluating at n=2B/A in the parabola, we find

    𝒫_n(x_m)>8B/A+2/2(n+1)(n+2)>0.





 §.§ Asymptotic expansion of the eigenvalues

When solving  the boundary equation in  Proposition <ref>, one requires that the angular velocity is located outside the singular set (<ref>).  Consequently, in order to  apply the bifurcation  argument for the density equation we should check that  the eigenvalues  {x_n} constructed in Proposition <ref>  do not intersect the singular set. This problem sounds to be very technical and in the case A+B<0, where we know that the dispersion set is infinite, we reduce  the problem to  studying the asymptotic behavior of each sequence.  Let us start with a preliminary result.

Let (x_n)_n∈ be a sequence of real  numbers in (-1,1) such that  x_n=1-κ/n+o(1/n), for some strictly positive number κ. Then the following asymptotics

    F(a_n+1,b_n;n+2;x_n)=n(η + o(1)),

holds, with η=κ∫_0^+∞τ e^-κτ/1+τdτ.


The integral representation of hypergeometric functions (<ref>) allows us to write

    F(a_n+1,b_n;n+2;x_n)=Γ(n+2)/Γ(n-a_n)Γ(2+a_n)ℓ_n
    =n(n+1)Γ(n)/Γ(n-a_n)Γ(2+a_n)ℓ_n,

where

    ℓ_n≜∫_0^1τ^n-a_n-1(1-τ)^1+a_n(1-τ x_n)^-1-a_ndτ.

Set
ε_n=1-x_n/x_n,  making the change of variables
τ=1-ε_n τ^', and  keeping the same notation τ to the new variable, we obtain

    ℓ_n=x_n^-a_n-1ε_n∫_0^1/ε_n(1-ε_nτ)^n-a_n-1τ^1+a_n(1+τ)^-a_n-1dτ.

From the first order expansion of x_n,f one has the pointwise convergence

    lim_n→+∞(1-ε_nτ)^n-a_n-1=e^-κτ,

for any τ>0. Since  the sequence n↦(1-κ/n)^n is increasing,  the Lebesgue Theorem leads to

    lim_n→+∞∫_0^1/ε_n(1-ε_nτ)^n-a_n-1τ^1+a_n(1+τ)^-a_n-1dτ=∫_0^+∞e^-κττ/1+τdτ.

Therefore,  we obtain the equivalence
ℓ_n∼η/n.
Combining the previous estimates with (<ref>) we find the announced estimate.

The next objective is to give the asymptotic expansion of the eigenvalues.


Let A and B be such that (<ref>) holds. Then, the sequence  {x_n, n≥ n_0},   constructed in Proposition <ref>, admits the following asymptotic behavior

    x_n=1-κ/n+c_κ/n^2+o(1/n^2),

where

    κ=-2A+B/A  and  c_κ=κ^2-2+2∫_0^+∞e^-κτ/(1+τ)^2dτ.



Let us consider the ansatz

    x_n=1-κ/n+1-u_n,   u_n=o(1/n).

From (<ref>), we know that  x_n satisfies the equation

    ζ_n(x_n)   =   I_n^1(x_n)F(a_n,b_n;n+1;x_n)+I_n^2(x_n)F(a_n+1,b_n;n+2;x_n)
          +I_n^3(x_n)F(a_n,b_n;n+3;x_n)
     =0,

with

    I_n^1(x_n)=n-a_n/n+1-a_n-(1-κ/n+1-u_n)(1+κ/n+1+n-1-a_n/n+1-a_n),


    I_n^2=-a_n(2x_n-1)/(n+1)(n+1-a_n)  and   I_n^3=-2x_n/(n+1)(n+2).

It is a simple matter to have

    I_n^1=-κ(1-κ)/(n+1)^2+u_n+o(u_n)+O(1/n^3),


    I_n^2=2/(n+1)^3+o(1/n^3)
        
    I_n^3=-2/(n+1)^2+O(1/n^3),

where we have used a_n∼ -2/n. By virtue of the above estimates, Lemma <ref> and Lemma <ref>, the expansion of u_n reads as

    u_n= κ(1-κ)+2-2η/(n+1)^2+o(1/n^2)
    = κ-κ^2+2-2η/n^2+o(1/n^2).

Using
1/n+1=1/n-1/n^2+o(1/n^2),
we find

    x_n= 1-κ/n+κ^2-2+2η/n^2+o(1/n^2).

The final expression holds as a consequence of the following integration by parts 

    η = 1-κ∫_0^+∞e^-κτ/1+τdτ
    = ∫_0^+∞e^-κτ/(1+τ)^2dτ.




 §.§ Separation of  the singular and dispersion sets

In Section <ref> we have established some conditions in order to solve the boundary equation. If we want to apply Proposition <ref>, we must verify that Ω does not lie in the singular set 𝒮_sing given in (<ref>). Moreover, from the last analysis, we have checked that the dispersion set 𝒮, defined in (<ref>), contains different sets depending on the assumptions on A and B.  We will prove the following results.


Let A and B such that (<ref>) holds. Denote by

    Ω_n≜B/2+A/4 x_n,
 
where the sequence (x_n)_n≥ n_0 has been defined in Proposition <ref>. If  n_0 is large enough depending on A and B, then

    Ω_n≠Ω_np,     ∀  n≥ n_0,  ∀ p∈^⋆,

where Ω_np belongs to the set 𝒮_sing  introduced in (<ref>). 
Moreover, there exists κ_c>0 such that for any κ>κ_c we find n_0∈ such that

    Ω_n≠Ω_m,     ∀  n,m≥ n_0.

The number κ_c∈(0,2) is the unique  solution of the equation

    κ_c-2∫_0^+∞e^-κ_cτ/(1+τ)^2 dτ=0.



It is a simple matter to have

    4/A(Ω_n-B/2)=1+κ/n+2/n^2+O(1/n^3).

Setting
x_n=1/4/A(Ω_n-B/2),
we obtain

    x_n=1-κ/n+κ^2-2/n^2+O(1/n^3).

Thus, condition  Ω_n≠Ω_m  is equivalent to x_n≠x_m. According to Proposition <ref>, we have

    x_n=1-κ/n+c_κ/n^2+o(1/n^2),

for n≥ n_0, where

    c_κ=κ^2-2+2∫_0^+∞e^-κτ/(1+τ)^2dτ.

This implies that x_n≠x_n for large n.
Moreover,

    x_np=1-κ/n p+κ^2-2/n^2 p^2+O(1/n^3)
    =1-κ/n p+O(1/n^2),

for any p∈^⋆ and  O(1/n^2) being  uniform on p. Therefore,  we get that
x_np>x_n, for any k≥ 2,
with n≥ n_0t and n_0 large enough. Consequently, we deduce that
x_n≠x_np, for any n≥ n_0, and p∈^⋆.
To establish the second assertion, we will use an asymptotic expansion for x_n+1. Thanks to (<ref>) we can write

    x_n+1=1-κ/n+κ^2+κ-2/n^2+O(1/n^3).

Then, since (x_n)_n≥ n_0 is strictly increasing, to prove x_n≠x_n it suffices just to check that
x_n+1> x_n, which leads to

    g(κ)≜κ- 2∫_0^+∞e^-κτ/(1+τ)^2dτ>0.

Since g is strictly increasing on [0,+∞) and satisfies
g(0)=-2 and g(2)>0,
there exists only one solution κ_c∈(0,2) for the equation g(κ)=0. This concludes the  proof.

The next task is to discuss the separation problem when the dispersion set is finite.


Let A>0, B∈ and  𝒮_sing being the set defined in (<ref>). Then the following assertions hold true:


  * If B∉[-A/2,-A/4], then
x_1=-A/2B∉𝒮_sing.

  * If B>A, then  the sequence m∈[2, 2B/A]↦ x_m  defined  in Proposition <ref> satisfies

    x_m≠x_nm,  ∀ n≥ 1.





Recall from the definition of the set 𝒮_sing given in  (<ref>) that

    1/x_n=1-2(n+1)/n(n+2)-2B/An,

where we have used (<ref>) and (<ref>). Notice  that when A and B are positive then  n↦1/x_n is strictly  increasing.

(1)
Let us prove that

    -2B/A≠ 1-2(n+1)/n(n+2)-2B/An,  ∀ n≥1.

 Note that for n=1 this constraint is always satisfied since  we get

    -1/3-2B/A≠ -2B/A.

Thus (<ref>)  is equivalent to 

    n^2-2/n^2+n-2≠ -2B/A,   ∀  n≥2.

One can easily check that left part is strictly increasing on [2,+∞) and so

    1/2≤n^2-2/n^2+n-2≤1,  ∀  n≥2.

Consequently, if -2B/A∉ [1/2,1] then the condition (<ref>) is satisfied and this ensures the first point.

(2) From the expression of 𝒫_n given in Proposition <ref>, we may write  

    𝒫_n(x_m)=n/n+1(1-x_m/x_n),

and  𝒫_m(x_m)<0, which implies
1/x_m<1/x_m.
By the monotonicity of 1/x_n, it is enough to prove

    1/x_m<1/x_2m,

in order to conclude. According to (<ref>), we have
1/x_m<1-A+2B/A(m+1).
To obtain (<ref>), it is enough to establish

    1-A+2B/A(m+1)≤1/x_2m=1-2m+1/2m(m+1)-B/Am,

which is equivalent to

    A+2B/A(m+1)≥2m+1/2m(m+1)+B/Am.

This latter one agrees with

    B/A≥1/2(m-1),

which holds true   since B/A≥ 1.




 §.§ Transversal property

This section is devoted to the transversality assumption concerning the fourth hypothesis of the Crandall–Rabinowitz Theorem <ref>. We shall reformulate an equivalent tractable statement, where the problem reduces to check the non-vanishing of a suitable integral. However, it is slightly hard to check this property for all the eigenvalues.  We give positive results for higher frequencies using the asymptotics, which have been developed in the preceding sections for some  special  regimes on A and B. The first result in this direction is summarized as follows.


Let  A>0 and  B∈, x∈ (-∞,1)\{𝒮_sing∪{0, x_0}} and n∈𝒜_x where all the elements involved can be found in  (<ref>) (<ref>)-(<ref>) and (<ref>). Let 

    r e^iθ∈↦h_n(re^iθ)=h_n^⋆(r)cos(nθ) ∈Ker D_gG(Ω,0) ,

Then

    D_Ω,gG(Ω,0)h_n∉Im  D_gG(Ω,0)

if and only if h_n^⋆ satisfies 

    ∫_0^1s^n+1F_n(xs^2)/1-xs^2[h^⋆_n(s)/2A-A_ns^n+2+A_nsG_n(s)/G_n(1)]ds≠0,

where

    A_n=-∫_0^1s^n+1h_n^⋆(s)ds/2G_n(1),

and G_n is defined by (<ref>).

 
Differentiating the expression of the linearized operator (<ref>)  with respect to Ω we obtain

    D_Ω,gG(Ω,0)h(re^iθ)=∑_n≥ 1cos(nθ)[1/2Ah_n(r)-r/n((∂_Ω A_n)G_n(r)+A_n ∂_Ω G_n(r))]
    
    +1/2Ah_0(r).

Differentiating  the identity (<ref>) with respect to Ω yields

    ∂_Ω A_n = H_n(1)/2n(Ω-Ω_n)^2
    = -nA_n/G_n(1).

Similarly, we get from (<ref>), (<ref>) and  the relation (<ref>) between x and Ω that

    ∂_Ω G_n(r)=A n/4r^n+1∂_Ω(1/x)
    =n r^n+1.

Putting together the preceding identities we find

    D_Ω,gG(Ω,0)h(re^iθ)
       =   ∑_n≥ 1cos(nθ)[1/2Ah_n(r)+A_nrG_n(r)/G_n(1)-A_nr^n+2]+1/2Ah_0(r).

Evaluating this formula at h_n yields

    D_Ω,gG(Ω,0)h_n(re^iθ)   =   [h_n^⋆(r)/2A-A_nr^n+2 +A_nrG_n(r)/G_n(1)]cos(nθ),

where A_n is related to h_n^⋆ via (<ref>). Now applying Proposition <ref> we obtain that this element  does not belong to Im D_gG(Ω,0)  if and only if the function 

    r∈[0,1]↦ d_n^⋆(r)≜h_n^⋆(r)/2A-A_nr^n+2 +A_nrG_n(r)/G_n(1)

verifies

    ∫_0^1s^n+1F_n(xs^2)/1-xs^2d^⋆_n(s)ds≠0,

which gives the announced result.

The next goal is to check the condition  (<ref>) for large n in the regime (<ref>). We need first to rearrange the function d_n^⋆ defined above and use the explicit expression of h_n given in (<ref>). From  (<ref>) we get  

    A_n=-n/8xG_n(1).
 
Then, multiplying d_n^⋆  by 1/A_n, we obtain 

    1/A_n[h^⋆_n(r)/2A-A_nr^n+2+A_nrG_n(r)/G_n(1)]=[h_n^⋆(r)/2AA_n-r^n+2 +rG_n(r)/G_n(1)]
    ≜ r^nℋ(r^2),

where the function ℋ takes the form 


    ℋ()   =   4xG_n(1)/An(1-x)[P_n()/P_n(1)-F_n(x)/F_n(x)+2xF_n(x)/P_n(1)∫_^11/τ^n+1F_n^2(xτ)∫_0^τs^nF_n(xs)/1-xsP_n(s)dsdτ]
                                    -+P_n()/P_n(1).

With the change of variables s s^2 in the integral, condition (<ref>) is equivalent to

    ∫_0^1s^nF_n(xs)/1-xsℋ(s)ds≠ 0.




  §.§.§ Regime A>0 and A+B<0

Here, we study the transversality assumption in the regime (<ref>). Notice that the existence of infinite countable set of eigenvalues has been already established in Proposition <ref>. However, due to the complex structure  of the integrand in (<ref>) it appears   quite difficult to check  the non-vanishing of the integral for a given frequency. Thus, we have to overcome this difficulty  using an asymptotic behavior of the integral and checking by this way the transversality only for high frequencies. More precisely, we prove the following result.

Let A and B satisfying  (<ref>) and {x_n} the sequence  constructed in Proposition <ref>-(1). Then there exists n_0∈ such that 

    ∫_0^1s^nF_n(x_ns)/1-x_nsℋ(s)ds=-n/2(1+o(1)),  ∀ n≥ n_0.





We proceed with studying  the asymptotic behavior of the above integral for large frequencies n. We write  ℋ=ℋ_1+ℋ_2+ℋ_3,
with

    ℋ_1()   =   4x_nG_n(1)/An(1-x_n)[P_n()/P_n(1)-F_n(x_n)/F_n(x_n)],
    ℋ_2()   =   4x_nG_n(1)/An(1-x_n)2x_nF_n(x_n)/P_n(1)∫_^11/τ^n+1F_n^2(x_nτ)∫_0^τs^nF_n(x_ns)/1-x_nsP_n(s)dsdτ,
    ℋ_3()   =   -+P_n()/P_n(1).

Let us start with the function ℋ_1. Proposition <ref> leads to

    1/x_n   =   1+κ/n-c_κ-κ^2/n^2+O(1/n^3),

which, together with the expression of G_n(1) given in (<ref>), imply

    G_n(1)∼-A/4n(c_κ-κ^2+2).

Recall that the inequality 
F_n(1)≤ F_n(x_n)≤1
holds for any ∈[0,1], and thus  Lemma <ref> gives  that

    1/2≤ F_n(1),   ∀ n≥ n_0.

Hence,  ℋ_1 can be bounded as follows

    |ℋ_1()|≤C/n^2(1-x_n)[|P_n()|/|P_n(1)|+1],  ∀∈[0,1],

with C a constant depending in A and B. 
From the definition of P_n in (<ref>) we may obtain

    |P_n()|≤ C [1-x_n+1/n],  ∀∈[0,1].

Moreover, plugging (<ref>) into (<ref>) we deduce that

    P_n(1)∼ -4/AnG_n(1)
    ∼c_κ-κ^2+2/n^2 .

Putting everything together one gets 

    |ℋ_1()|≤ C[1/n(1-x_n)+1],  ∀ ∈[0,1],

from which we infer that

    |∫_0^1s^nF_n(x_ns)/1-x_nsℋ_1(s)ds| ≤C/n∫_0^1s^n/(1-x_ns)^2ds+C∫_0^1s^n/1-x_nsds
     ≜ C(I_1/n+I_2).
 
To estimate the  second integral we use the change of variables s=1-ε_nτ, with ε_n=1-x_n/x_n, leading to the asymptotic behavior

    I_2=1/x_n∫_0^1/ε_n(1-ε_nτ)^n/1+τdτ∼∫_0^n/κ(1-κτ/n)^n/1+τ∼∫_0^+∞e^-κτ/1+τdτ,

where we have used the expansion of x_n given by  Proposition <ref>. 
As  to the first integral  we just have to integrate by parts  and use the previous computations,

    I_1=1/x_n1/1-x_n-n/x_n∫_0^1s^n-1/1-x_nsds=(1/κ-∫_0^∞e^-κτ/1+τdτ)n +o(n),

and consequently,

    sup_n≥ n_0|∫_0^1s^nF_n(x_ns)/1-x_nsℋ_1(s)ds|<+∞.

The estimate ℋ_2 it  is straightforward. Indeed, from (<ref>) we may write

    |ℋ_2()|≤C/1-x_n∫_^11/τ^n+1∫_0^τs^n/1-x_ns|P_n(s)|dsdτ.

Using once again (<ref>) and Proposition <ref>  we get

    ∫_0^τs^n/1-x_ns|P_n(s)|ds   ≤   C/n∫_0^τs^n/1-x_nsds+C∫_0^τs^nds 
       ≤   C/n(1-x_n)τ^n+1/n+1+Cτ^n+1/n+1≤C/nτ^n+1,   ∀τ∈[0,1].

Hence, we deduce that

    ∫_^11/τ^n+1∫_0^τs^n/1-x_ns|P_n(s)|ds≤ C1-/n.

Since  the function ∈[0,1]↦1-/1-x_n is strictly decreasing , then we have

    0≤1-/1-x_n ≤1,  ∀∈[0,1].

Consequently, inserting the preceding two estimates into  (<ref>), we obtain

    |ℋ_2(t)|≤C/n,  ∀∈[0,1].

Thus we infer

    |∫_0^1s^nF_n(x_ns)/1-x_nsℋ_2(s)ds|   ≤   C/n∫_0^1 s^n/1-x_nsds,

which implies 

    sup_n≥ n_0|∫_0^1s^nF_n(x_ns)/1-x_nsℋ_2(s)ds|≤C/n,

where we have used  (<ref>). It remains to estimate the integral term associated with  ℋ_3 which takes the form

    ∫_0^1s^nF_n(x_ns)/1-x_nsℋ_3(s)ds=-∫_0^1s^n+1F_n(x_ns)/1-x_nsds+1/P_n(1)∫_0^1s^nF_n(x_ns)/1-x_nsP_n(s)ds.

Similarly to (<ref>), one has

    sup_n≥ n_0∫_0^1s^n+1F_n(x_ns)/1-x_nsds<+∞.

To finish we just have to deal with the second integral term. Observe from (<ref>)  that P_n is a monic  polynomial of degree two, and thus from Taylor formula one gets

    P_n()=P_n(1/x_n)+(-1/x_n)P_n^'(1/x_n)+(-1/x_n)^2.
 
It is easy to  check the following behaviors 

    P_n(1/x_n)∼κ/n,   P_n'(1/x_n)∼ 1.

Hence, we obtain 

    ∫_0^1s^nF_n(x_ns)/1-x_nsP_n(s)ds   =   P(1/x_n)∫_0^1s^nF_n(x_ns)/1-x_nsds-1/x_nP_n'(1/x_n)∫_0^1s^nF_n(x_ns)ds
          +1/x_n^2∫_0^1s^nF_n(x_ns)(1-x_ns)ds.

Concerning the last term we use the asymtotics of x_n, leading to

    1/x_n^2∫_0^1s^nF_n(x_ns)(1-x_ns)ds ≤ C ∫_0^1s^n(1-x_ns)ds
    ≤ C (1/n+1-x_n/n+2)
    ≤C/n^2.

For the first and second terms we use the estimate  |F_n(x_nt)-1|<Cln n/n coming from Lemma <ref>. Hence, we find 

    -1/x_nP_n^'(1/x_n)∫_0^1s^nF_n(x_ns)ds=-1/n(1+o(1)).

Again from (<ref>) we find

    ∫_0^1s^nF_n(x_ns)/1-x_nsds = ∫_0^1s^n/1-x_nsds+o(1)
     = ∫_0^+∞e^-κτ/1+τdτ+o(1).

Putting together the preceding estimates, we obtain

    ∫_0^1s^nF_n(x_ns)/1-x_nsP_n(s)ds=κ∫_0^+∞e^-κτ/1+τdτ-1/n+o(1/n),

and combining this estimate with (<ref>), we infer

    ∫_0^1s^nF_n(x_ns)/1-x_nsℋ_3(s)ds= κ∫_0^+∞e^-κτ/1+τdτ-1/c_κ-κ^2+2n+o(n).

Using the explicit expression of  above constants  defined in Proposition <ref>, we get that

    κ∫_0^+∞e^-κτ/1+τdτ-1/c_κ-κ^2+2=-1/2,

and therefore 

    ∫_0^1s^nF_n(x_ns)/1-x_nsℋ_3(s)ds= -n/2+o(n).

Combining this estimate with (<ref>) and (<ref>), we deduce that

    ∫_0^1s^nF_n(x_ns)/1-sx_nℋ(s)ds= -n/2 +o(n),

which achieves the proof of the announced result.



  §.§.§ Regime B>A>0

In this special regime there is only a finite number of eigenvalues that can be indexed by a decreasing sequence, see Proposition <ref>-(1).   In what follows we shall prove that the transversality assumption is always satisfied without any additional constraint on the parameters. More precisely, we prove the following result.

Let B>A>0. Then, the transversal property (<ref>) holds, for every subsequence {x_n; n∈[2, 𝒩_A,B]} defined in  Proposition <ref>, where

    𝒩_A,B≜max(B/A+1/8,2B/A-9/2).




Let us start with the case n∈[2, B/A+1/8].  
Using the expression of P_n introduced in  (<ref>) one has

    x_nP_n(1)=x_n[1-A+2B/An+2/n(n+1)]-n+2/n+1.

Moreover, from the  definition of 𝒫 seen in Proposition <ref>, we get

    𝒫_n(x_n)   =   n/n+1+x_n[-n/n+2+A+2B/A(n+1)]
       =   n/n+2(n+2/n+1+x_n[-1+(A+2B)(n+2)/An(n+1)])
    =-n/n+2x_n P_n(1)<0.

This implies that  P_n(1)<0. Since ∈[0,1]↦ P_n() is strictly increasing with x_n<0, we deduce that  

    P_n()<0,   ∀∈[0,1].

 Therefore, we get from  (<ref>) that  G_n(1)>0.
Let us study every term involved in (<ref>) by using the decomposition ℋ=ℋ_1+ℋ_2+ℋ_3  of   Proposition <ref>. From the preceding properties, it is clear that
ℋ_2()>0, for ∈(0,1).
In addition, we also have that ∈[0,1]↦ -+P_n()/P_n(1) is strictly decreasing and thus

    ℋ_3()=-+P_n()/P_n(1)≥ℋ_3(1)=0,   ∀∈[0,1].

Concerning ℋ_1 we first note that the mapping ∈[0,1]↦P_n()/P_n(1)-F_n(x_n)/F_n(x_n) is strictly decreasing which follows from (<ref>) and the  fact that F_n is decreasing and P_n is increasing. Thus,

    P_n()/P_n(1)-F_n(x_n)/F_n(x_n)   ≥   P_n(1)/P_n(1)-F_n(x_n)/F_n(x_n)=0,   ∀∈[0,1].

Combining (<ref>) with (<ref>), we deduce 

    ℋ_1()=4x_nG_n(1)/An(1-x_n)[P_n()/P_n(1)-F_n(x_n)/F_n(x_n)]<0,  ∈(0,1).

We continue our analysis assuming that

    |4x_nG_n(1)/An(1-x_n)|≤ 1,  ∀ ∈[0,1),

holds, we see how to conclude with. Since ℋ_2 is always positive  then ℋ will be strictly positive if one can show  that 
ℋ_1()+ℋ_3()>0, for any ∈(0,1).
With (<ref>) in mind, one gets

    ℋ_1()+ℋ_3()≥-[P_n()/P_n(1)-F_n(x_n)/F_n(x_n)]-+P_n()/P_n(1)=F_n(x_n)/F_n(x_n)-,  ∀∈[0,1].

Computing the derivatives of the  function in the right-hand side term, we find

    ∂_(F_n(x_n)/F_n(x_n)-)=x_nF_n'(x_n)/F_n(x_n)-1,  ∂^2_(F_n(x_n)/F_n(x_n)-)=x_n^2F_n”(x_n)/F_n(x_n)<0.

The latter fact implies that the  first derivative is decreasing, and  thus

    ∂_(F_n(x_n)/F_n(x_n)-)≤x_nF_n'(0)/F_n(x_n)-1
    ≤-2x_n/(n+1)F_n(x_n)-1
    ≤2/n+1-1<0,  ∀∈[0,1],

where we have used  Lemma <ref>-(1).
Therefore, we conclude that the mapping 
∈[0,1]↦F_n(x_n)/F_n(x_n)-t decreases and, since it  vanishes at =1, we get

    ℋ_1()+ℋ_3()≥F_n(x_n)/F_n(x_n)->0,  ∀∈[0,1).

This implies that ℋ()>0 for any ∈[0,1) and hence the transversality assumption (<ref>) is satisfied.
Let us now  turn to the proof of (<ref>) and observe that from (<ref>)

    |4x_nG_n(1)/An(1-x_nt)|   =   1/1-x_ntn+1/n+2x_nP_n(1)
       ≤   n+1/n+2{x_n[1-A+2B/An+2/n(n+1)]-n+2/n+1}.

Using x_⋆, defined in Proposition <ref>, and the fact that A+2B/An+2/n(n+1)-1>0, we obtain

    |4x_nG_n(1)/An(1-x_nt)|   ≤   n+1/n+2{x_⋆[1-A+2B/An+2/n(n+1)]-n+2/n+1}
       ≤   n+1/n+2{2n+1/2(n+1)A+2B/An+2/n(n+1)-1/A+2B/A(n+1)-n+1/n+2-n+2/n+1}.

Consequently (<ref>) is satisfied provided that

    A+2B/An+2/n(n+1)-1/A+2B/A(n+1)-n+1/n+2≤ 4n+2/2n+1.

Since A+2B/A(n+1)-n+1/n+2>0, then the preceding inequality is true if and only if

    A+2B/A(n+1)≥2n^2+3n/2n^2+3n-2.

It is easy to check that the sequence n≥2↦2n^2+3n/2n^2+3n-2 is decreasing, and then the foregoing inequality is satisfied for any n≥2  if

    A+2B/A(n+1)≥7/6.

Fom the assumption n+1≤B/A+9/8, the above inequality holds if

    B/A≥5/16,

which follows from the condition B>A.


Let us now move on the case  n∈[2, 2B/A-9/2]. Note that we cannot use x_⋆ coming from Proposition <ref> since we do not know if x_n∈(-1,0). In fact Proposition <ref> gives us that x_n<-1, for n≥B/A+1. Hence, we should slightly modify the arguments used to (<ref>). From (<ref>), we deduce that  (<ref>) is satisfied if

    x_n≥2n+2/n+1/1-A+2B/An+2/n(n+1).

By (<ref>), it is enough to prove that 

    1/1-A+2B/A(n+1)≥2n+2/n+1/1-A+2B/An+2/n(n+1).

Straightforward computations give that the last inequality is equivalent to

    A+2B/A(n+1)≥(n+3)n/(n+2)(n-1).

Therefore, if  n satisfies 2≤ n≤2B/A-9/2, then we have

    n+1+9/2/n+1≤A+2B/A(n+1).

Hence, one can check that

    n+1+9/2/n+1≥(n+3)n/(n+2)(n-1),

for any  n≥2, and thus  (<ref>) is verified.



  §.§.§ One-fold case

The main objective is to make a complete study in the case n=1. It is very particular because F_1 is explicit according to Remark <ref> and, therefore, we can get a compact formula for the integral of (<ref>).   Our main result reads as follows.

Let n=1 and  x=-A/2B, then we have the formula

    ∫_0^1sF_1(xs)/1-xsℋ(s)ds=x-1/2x.

In particular the transversal assumption  (<ref>) is satisfied  if and only if
x∉{0,1}.



Note that from (<ref>)-(<ref>) one has

    P_1()=^2-3/2x-3/2[1-1/x],    P_n(1)=-1/2,    G_1(1)=A/12.

Moreover,  we get F_1()=1- using   Remark <ref>, and thus

    ℋ()   =   1/3x/(1-x)[P_1()/P_1(1)-F_1(x)/F_1(x)+2xF_1(x)/P_1(1)∫_^11/τ^2F_1^2(xτ)∫_0^τsF_1(xs)/1-xsP_1(s)dsdτ]
          -+P_1()/P_1(1)
       ≜   1/3x/(1-x)ℋ()-+P_1()/P_1(1).

From straightforward computations we deduce 

    ∫_^11/τ^2F_1^2(xτ)∫_0^τsF_1(xs)/1-xsP_1(s)dsdτ   =   ∫_^1τ^2/4-1/2xτ-3/4(1-1/x)/(1-τ x)^2dτ

Denoting by φ(τ)=τ^2/4-1/2xτ-3/4(1-1/x) and integrating by parts, we get

    ∫_^1φ(τ)/(1-τ x)^2dτ=1/x[φ(1)/1-x-φ()/1- x]+1-/2x^2
    =1/x[1-2x/4x(1-x)+1-/2x-φ()/1- x].

Therefore, after standard computations,  we get the simplified  formula

    ℋ()=3(1- x)(-1)/x.

Inserting this into (<ref>), we find

    ℋ()=-1+P_1()/P_1(1)
    =-2^2+3/x+2-3/x.

Plugging it into the integral of (<ref>), it yields

    ∫_0^1sℋ(s)ds=x-1/2x.





§ EXISTENCE OF NON–RADIAL TIME–DEPENDENT ROTATING SOLUTIONS


At this stage we are able to give the full statement of our main result, by using the analysis of the previous sections. In order to apply the Crandall–Rabinowitz Theorem, let us introduce the m–fold symmetries in our spaces. 

    𝒞^k,α_s,m()   ≜   {g∈𝒞^k,α_s() :   g(e^i2π/mz)=g(z),  ∀ z∈},
    𝒞^k,α_a,m()   ≜   {ρ∈𝒞^k,α_a() :   ρ(e^i2π/mw)=ρ(w),  ∀ w∈},
    ℋ𝒞^k,α_m()   ≜   {ϕ∈ℋ𝒞^k,α() :   ϕ(e^i2π/mz)=e^i2π/mϕ(z),  ∀ z∈}.

Note that the functions g∈𝒞^k,α_s,m(), ρ∈𝒞^k,α_a,m() and ϕ∈ℋ𝒞^k,α_m() admit the following representation:

    g(re^iθ)=∑_n≥ 0 g_nm(r)cos(nmθ),  ρ(e^iθ)=∑_n≥ 0ρ_nsin(nmθ)  and  ϕ(z)=z∑_n≥ 1 z^nm,

where z∈,  r∈[0,1] and θ∈[0,2π]. With these spaces, the functional F defined in (<ref>), concerning the boundary equation, is also well–defined:


Let ∈ (0,1), then 
 
    F: × B_𝒞_s,m^1,α(0,)× B_ℋ𝒞^2,α_m(0,)↦𝒞^1,α_a,m()
 is well–defined and of class 𝒞^1,  where the balls were defined in (<ref>).


Thanks to Proposition <ref>, it remains to prove that F(Ω,g,ϕ) satisfies F(Ω,g,ϕ)(e^i2π/mw)=F(w), with w∈:

    F(Ω,g,ϕ)(e^i2π/mw)   =   Im[ (Ω Φ(e^i2π/mw)-1/2π∫_f(y)|Φ^'(y)|^2/Φ(e^i2π/mw)-Φ(y) dA(y))Φ^'(e^i2π/mw)e^i2π/mw]
       =   Im[ (Ω e^-i2π/mΦ(w)-1/2π∫_f(e^i2π/my)|Φ^'(e^i2π/my)|^2/Φ(e^i2π/mw)-Φ(e^i2π/my) dA(y))Φ^'(w)e^i2π/mw]
       =   Im[ (Ω e^-i2π/mΦ(w)-e^-i2π/m/2π∫_f(y)|Φ^'(y)|^2 /Φ(w)-Φ(y)dA(y))Φ^'(w)e^i2π/mw]
       =   F(Ω,g,ϕ)(w),

where we have used that ϕ(e^i2π/mz)=e^i2π/mϕ(z), ϕ'(e^i2π/mz)=ϕ'(z) and  g(e^i2π/mz)=g(z). 



We must define the singular set (<ref>) once we have introduced the symmetry in the spaces. Fixing f_0 as a quadratic profile (<ref>), the singular set (<ref>) becomes

    𝒮^m_sing≜{Ω_mn≜A/4+B/2-A(nm+1)/2nm(nm+2)-B/2nm  ,    n∈^⋆∪{+∞}}.


For the density equation defined in (<ref>) and the new spaces we obtain the following result.


Let I be an open interval with  I⊂\𝒮^m_sing. Then, there exists ε>0 such that

    G:I× B_𝒞_s,m^1,α()(0,)→𝒞_s,m^1,α()
 
is well–defined and of class 𝒞^1, where  G is defined in (<ref>) and   B_𝒞_s^1,α()(0,)  in (<ref>).




Similarly to the previous result, from Proposition <ref> we just have to check that G(Ω,g)(e^i2π/mz)=G(Ω,g)(z), for z∈. From Proposition <ref>, there exists ε>0 such that the conformal map ϕ is given by (Ω,g), and lies in B_ℋ𝒞^2,α_m(0,ε). Then, using  G(Ω,g)=G(Ω,g,ϕ(Ω,g)), we have

    G(Ω,g, ϕ)(e^i2π/mz)   =   4Ω-B/8Af(e^i2π/mz)-f(e^i2π/mz)^2/16A
          +1/2π∫_log|Φ(e^i2π/mz)-Φ(y)| f(y)|Φ^'(y)|^2 dA(y)-Ω|Φ(e^i2π/mz)|^2/2-λ
       =   4Ω-B/8Af(z)-f(z)^2/16A
          +1/2π∫_log|Φ(e^i2π/mz)-Φ(e^i2π/my)| f(e^i2π/my)|Φ^'(e^i2π/my)|^2 dA(y)
          -Ω|e^i2π/mΦ(z)|^2/2-λ
       =    G(Ω,g,ϕ)(z),

where we have used the properties of functions g and ϕ.



We have now all the tools we need to prove the first three points of Theorem <ref>, which can be detailed as follows:


Let A>0, B∈ and f_0 a quadratic profile (<ref>). Then the following results hold true.


  * If  A+B<0, then there is m_0∈ (depending only on A and B)  such that for any m≥ m_0 there exists


  * Ω_m=A+2B/4+Aκ/4m+A/4κ^2-c_κ/m^2+o(1/m^2), 

  * V a neighborhood of (Ω_m,0,0) in ×𝒞^1,α_s,m()×ℋ𝒞_m^2,α(),

  * a continuous curve ξ∈(-a,a)↦(Ω_ξ,f_ξ, ϕ_ξ)∈ V, 

such that (<ref>),

    ω_0=(f∘Φ^-1)1_Φ(),   f=f_0+f_ξ,  Φ=Id+ϕ_ξ,

defines a curve of non radial solutions of Euler equations that rotates at constant angular velocity Ω_ξ. 
The constants κ and c_κ are defined in (<ref>).



  * If B>A>0, then for any integer m∈[1, 2B/A-9/2] or m∈[1, B/A+1/8]  there exists 


  * 0≤Ω_m<B/2,  

  * V a neighborhood of (Ω_m,0,0) in ×𝒞^1,α_s,m()×ℋ𝒞_m^2,α(),

  * a continuous curve ξ∈(-a,a)↦(Ω_ξ,f_ξ, ϕ_ξ)∈ V, 

such that (<ref>) defines a curve of non radial solutions of Euler equations. However, there is no bifurcation with any symmetry  .


  * If B>0 or B≤ -A/1+ϵ, for some 0,0581<ϵ<1, then there exists


  * V a neighborhood of (0,0,0) in ×𝒞^1,α_s()×ℋ𝒞^2,α(),

  * a continuous curve ξ∈(-a,a)↦(Ω_ξ,f_ξ, ϕ_ξ)∈ V, 

such that (<ref>) defines a curve of one-fold non radial solutions of Euler equations. 

  * If - A/2≤ B≤ 0, then there is no bifurcation with any  symmetry m≥1. However, in the case that 0<B< A/4, there is no bifurcation with any  symmetry m≥2.






 (1) Let us prove the first assertion in the case  A+B<0.
We will implement the Crandall–Rabinowitz Theorem (<ref>) to G, defined in (<ref>). First, we must concrete the domain of Ω. From Proposition <ref>, there exist n_0,1 and a unique solution x_m∈(0,1) of ζ_m(x)=0 for any m≥ n_0,1. Then, we fix Ω_m=A/4x_m+B/2>A/4+B/2. Note that one can take n_0,1 large enough to avoid Ω_0 defined in (<ref>). Moreover, Proposition <ref> gives us that Ω_m≠Ω̂_mn, for n∈, with Ω_mn∈𝒮^m_sing. Therefore, let I be an interval with Ω_m∈ I and

    I∩𝒮^m_sing=∅,   Ω_0∉I.
 
By virtue of Proposition <ref>, we know that there exists ε>0 and  a 𝒞^1 function  
𝒩: I× B_𝒞_ s,m^1,α(0,ε)⟶ B_ℋ𝒞_m^2,α(0,ε),
such that  

    F(Ω,g,ϕ)=0⟺ϕ=𝒩(Ω,g)

holds, for any (Ω,g,ϕ)∈ I× B_𝒞_s,m^1,α(0,ε)× B_ℋ𝒞_m^2,α(0,ε). Hence, the conformal map is defined through the density  for that ε. We define the density equation,

    G: I× B_𝒞_s,m^1,α(0,ε) →𝒞_s,m^1,α(),

with the expression given in (<ref>). Thanks to Proposition <ref>, the function G is well-defined in these spaces and is 𝒞^1 with respect to (Ω,g). It remains to check the spectral properties of the Crandall–Rabinowitz Theorem. Using Proposition <ref>, we know that the dimension of the kernel of the linearized operator D_gG(Ω,0) is given by the number of elements of the set 𝒜_x defined in (<ref>).
Note that we have introduced the symmetry m in our spaces, and consequently in all the functions involved in the linearized operator. Take n=1. We know that x_m is the unique root x_m of ζ_m(x), and the sequence n∈[n_0,+∞[↦ x_n is strictly increasing. Since we  fixed n=1, we get that
𝒜_x_m={m}.
As Ω_m∉𝒮^m_sing, we have that the kernel is one dimensional. Since D_gG(Ω,0) is a Fredholm operator of zero index, we have that the codimension of the range is also one. About the transversality condition, note that using Proposition <ref> and Proposition <ref> we have that there exists n_0,2 such that if m>n_0,2 the transversality condition is satisfied. Taking n_0=max{n_0,1,n_0,2}, the Crandall-Rabinowitz Theorem can be applied to G obtaining V a neighborhood of (Ω_m,f_0,Id) in ×𝒞^1,α_s,m()×ℋ𝒞_m^2,α(),
and a continuous curve ξ∈(-a,a)↦(Ω_ξ,f_ξ, ϕ_ξ)∈ V, 
solutions of G(Ω,g)=0. The conformal map ϕ_ξ is the one associated to f_ξ via Proposition <ref>. 
Moreover, thanks to Ω_ξ≠Ω_0, Proposition <ref> gives us that f_ξ can not be radial because of f_ξ≠ f_0.
Furthermore, by Proposition <ref> we know the asymtotics of x_m obtaining

    1/x_m=1+κ/m+κ^2-c_κ/m^2+o(1/m^2),

where κ and c_κ are defined in (<ref>).  Using the relation between x_m and Ω_m in (<ref>), we get  

    A+2B/4 < Ω_m=A+2B/4+Aκ/4m+A/4κ^2-c_κ/m^2+o(1/m^2).

 Therefore, we obtain that 
 
    ω_0=(f∘Φ^-1)1_Φ(),   f=f_0+f_ξ,  Φ=Id+ϕ_ξ,

  defines a non radial solution of Euler equation, which rotates at constant angular velocity Ω_ξ.  


(2)
Now, we are concerned with the existence of m–fold non radial solutions of the type (<ref>) in the case  B>A>0,  for any integer m∈[1, 2B/A-9/2] or m∈[1, B/A+1/8]. In this part of the theorem we also prove that  there is no bifurcation with the symmetry m, for any m≥2B/A+2.
As in Assertion (1), we check that the Crandall–Rabinowitz Theorem can be applied.  From Proposition <ref> and Proposition <ref>, there is a unique solution x_m∈(-∞,1) of ζ_m(x)=0. In fact, x_m<0. Then, we fix Ω_m=A/4x_m+B/2. Note that by (<ref>) and Proposition <ref> we get the bounds for Ω_m. Moreover, Proposition <ref> gives that  Ω_m∉𝒮^m_sing. Then, let I be an interval such that Ω_m∈ I and
I∩𝒮^m_sing=∅.
 
 Using again Proposition <ref> and Proposition <ref>, we get that G: I× B_𝒞_s,m^1,α(0,ε) →𝒞_s,m^1,α(),
is well-defined and 𝒞^1 with respect to (Ω,g). 

About the spectral properties, we have stated in the previous proof that the dimension of the kernel of the linearized operator is given by the roots of ζ_nm.  Taking n=1, we know that x_m is the unique root of ζ_m(x). By Corollary <ref> we get that ζ_nm(x_m)≠ 0, for any n≥ 2. Hence
𝒜_x_m={m}.
Due to Ω_m∉𝒮^m_sing, we have that the kernel is one dimensional. 
 D_gG(Ω,0) is a Fredholm operator of zero index, then we have that the codimension of the range is also one. About the transversal condition, note that using Proposition <ref>, we have that the transversal condition is satisfied. Similarly to the previous proof, the Crandall-Rabinowitz Theorem can be applied to G obtaining a curve ξ∈(-a,a)↦ (Ω_ξ, f_ξ) solutions of G(Ω,g)=0. 
 Moreover, thanks to Ω_ξ≠Ω_0∈(0,1), Proposition <ref> gives that f_ξ can not be radial since f_ξ≠ f_0. 

First,  note that  Ω∉[B/2, B/2+A/4] is equivalent to x<1. By Proposition <ref> and Proposition <ref> we get that ζ_n has not solutions in (-∞,1], for m≥2B/A+2, and then there is no bifurcation with that symmetry.  In the opposite case, x>1, there is no bifurcation according to 

(3) 
Here, we are concerning with the case A,B>0 or B≤ -A/1+ϵ for some ϵ∈(0,1), with -A/2B≠ x_0, where x_0 is defined through (<ref>). We work as in (1)-(2) checking the hypothesis of Crandall–Rabinowitz Theorem. Fixing Ω_1=0 agrees with x_1=-A/2B, where we use (<ref>).  Proposition <ref> allows us to have that x_1∉𝒮_sing. Then, we can take an interval I such that 0∈ I, and 

    I∩𝒮_sing=∅,   Ω_0∉I.

Again, Proposition <ref> and Proposition <ref> imply that

    G: I× B_𝒞_s^1,α(0,ε) →𝒞_s^1,α(),

is well-defined and is 𝒞^1 in (Ω,g).

We must check the spectral properties. Due to the assumptions on A and B, we get that x_1≤ 1. By Proposition <ref>, Proposition <ref> and Proposition <ref> we have that x_1≠ x_n if there exists x_n∈(-∞,1) solution of ζ_n. Note that such ϵ comes from the Proposition <ref>. Hence, by Corollary <ref>, we obtain that the kernel of D_g G(0,0) is one dimensional, and is generated by (<ref>), for n=1. Moreover, Proposition <ref> implies that D_g G(0,0) is a Fredholm operator of zero index, and then the codimension of the range is one. The transversal property is verified by virtue of Proposition <ref>. Then, Crandall–Rabinowitz Theorem can be applied obtaining the announced result. Note that the bifurcated solutions are not radial due to Proposition <ref>. 


(4) 
The first assertion concerning the non bifurcation result comes from Proposition <ref> and Proposition <ref> due to the fact that ζ_m has not solutions in (-∞,1], for m≥ 2. Moreover, by  Corollary <ref> and Theorem <ref> we get that there is no bifurcation for m=1 since the only possibility agrees with Ω=0, which satisfies (<ref>).   Finally, the bifurcation with x>1 is forbidden due again to Theorem <ref>. 

The last assertion follows from Corollary <ref> and Theorem <ref>.
  




§ DYNAMICAL SYSTEM AND ORBITAL ANALYSIS

In this section we wish to investigate the particle trajectories inside the support of the rotating vortices that we have constructed in Theorem <ref>. We will show that in the frame of these V-states the trajectories are organized through concentric periodic orbits  around the origin.  This allows to provide an equivalent  reformulation of the density equation (<ref>) via the study of the associated dynamical system. It is worth pointing out that some of the material developed in this section about periodic trajectories and the regularity of the period is partially known in the literature and for the convenience of the reader we will provide the complete proofs.

Assuming that (<ref>) is a solution of the Euler equations,  the level sets of Ψ(x)-Ω|x|^2/2, where Ψ is the stream function associated to (<ref>), are given by the collection of the  particle trajectories,

    ∂_tφ(t,x)   =   (v(φ(t,x))-Ωφ(t,x)^⊥)
    =∇^⊥(Ψ-Ω|·|^2/2)(φ(t,x)),
    φ(0,x)   =   x∈Φ().

In the same way we have translated the problem to the unit disc  using the conformal map Φ via the vector field W(Ω,f,Φ) in (<ref>),  we analyze the analogue in the level set context. 
We define the flow associated to W as

    ∂_t ψ(t,z)=W(Ω,f,Φ)(ψ(t,z)),  ψ(0,z)=z∈.

Since v(x)-Ω x^⊥ is divergence free, via Lemma <ref>, we obtain that W(Ω,f,Φ) is incompressible, and then the last system is  also Hamiltonian. In the following result, we highlight the relation between φ and ψ.


The following identity 

    φ(η_z(t),Φ(z))=Φ(ψ(t,z)),   ∀ z∈,

holds, where

    η_z'(t)=|Φ'(Φ^-1(φ(η_z(t),Φ(z)))|^2,  η_z(0)=0.





Let us check that Y(t,z)=Φ^-1(φ(t,Φ(z))) verifies a similar equation as ψ(t,z) sets,

    ∂_t Y(t,z)   =   (Φ^-1)'(Φ(Y(t,z))∂_t φ(t,Φ(z))
    =Φ'(Y(t,z))/|Φ'(Y(t,z))|^2(v(Φ(Y(t,z)))-ΩΦ(Y(t,z))^⊥)
       =   W(Ω,f,Φ)(Y(t,z))/|Φ'(Y(t,z))|^2.

Now, we rescale the time through the function η_z, and  Y(η_z(t),z) satisfies,

    ∂_t Y(η_z(t),z)=η_z'(t) (∂_t Y)(η_z(t),z))
    =η_z'(t)W(Ω,f,Φ)(Y(η(t),z))/|Φ'(Y(η(t),z))|^2
    =W(Ω,f,Φ)(Y(η(t),z)).

Since Y(η_z(0),z)=Y(0,z)=Φ^-1(φ(0,Φ(z))=z, we obtain the announced result.


The next  task  is to connect the solutions constructed in Theorem <ref> with the orbits of the associated dynamical system through the following result:



Let m≥ 1 and ξ∈(-a,a)↦(Ω_ξ,f_ξ, ϕ_ξ) be one of the  solutions constructed in Theorem <ref>. The flow ψ associated to W(Ω_ξ,f_ξ,Φ_ξ), defined in (<ref>), 
verifies the following properties:


  * ψ∈𝒞^1(, 𝒞^1,α()).

  * The trajectory t↦ψ(t,z) is T_z periodic, located inside the unit disc and invariant by the dihedral group D_m. Moreover, if m≥4 then  the period map z∈↦ T_z belongs to 𝒞^1,α(). 



  * The family (ψ(t))_t∈ generates a group of diffeomorphisms of the closed unit disc.


The  proof will be given in  Subsection <ref>. 




 §.§ Periodic orbits

Here we explore sufficient conditions for Hamiltonian vector fields defined on the unit disc whose orbits are all periodic.  More precisely, we shall establish the following result. 


Let W:𝔻→ be a vector field in 𝒞^1(𝔻) satisfying the following conditions:


  i) It has divergence free.

  ii) It is tangential to the boundary 𝕋, i.e.
Re(W(z)z)=0,   ∀  z∈.

  iii) It vanishes only at the origin.

Then, we have


  * All the trajectories are periodic orbits located inside the unit disc, enclosing a simply connected domain containing the origin.

  * The family (ψ(t))_t∈  generates a group of diffeomorphisms of  the  closed  unit  disc.

  * If W  is antisymmetric with respect to the real axis, that is,

    W(z)=-W(z),   ∀ z∈𝔻.

then the orbits are symmetric with respect to the real axis.


 
  * If W is invariant by a rotation centered at zero  with angle θ_0, i.e.
W(e^iθ_0 z)=e^iθ_0 W(z),  ∀ z∈𝔻,
then all the orbits are invariant by this rotation. 




(1)
Let ψ be the solution associated to the flux W

    {[ ∂_tψ(t,z)=W(ψ(t,z)),;          ψ(0,z)=z∈𝔻. ].

From the Cauchy–Lipschitz Theorem we know that  the trajectory t↦ψ(t,z) is defined in a maximal time interval (-T_⋆,T^⋆), with T_⋆, T^⋆>0, for  each  z∈𝔻. Note that when  z belongs to the boundary, then the second condition listed above implies necessarily that its trajectory  does not leave  the boundary. Since the vector field does not vanish anywhere  on the boundary according to the third condition,  the trajectory will cover all the unit disc. As the equation is autonomous, this ensures that  the unit disc is  a periodic orbit. 

By condition i) we get that (<ref>) is a Hamiltonian system. Let H be the Hamiltonian function such that W=∇^⊥ H. Since H is C^1 in  and constant on the boundary  according to the assumption ii), then from iii) the origin  corresponds to an extremum point.

Now, taking |z|<1, the solution is globally well–posed in time, that is, T_⋆=T^⋆=+∞. This follows easily from the fact that different orbits never  intersect and consequently we should get 

    |ψ(t,z)|<1,   ∀ t∈  (-T_⋆,T^⋆),

meaning that the solution is bounded and does not touch the boundary so it is globally defined according to a classical blow–up criterion.

We will check that all the orbits are periodic inside the unit disc. This follows from some straightforward considerations on the level sets of the Hamiltonian H. Indeed, the ω-limit of a point z≠0 cannot contain the origin because it is the only critical point and  the level sets of H around this point are periodic orbits. Thus we deduce from Poincar–Bendixon Theorem that  the ω-limit of z  will be a periodic orbit.  As the level sets cannot be limit cycles then we find that the trajectory of z coincides with the periodic orbit.  

(2) This follows from classical results on autonomous differential equation. In fact, we know that  the flow  
ψ: ×⟶
 is well–defined and 𝒞^1. For any t∈, it realizes   a bijection with
ψ^-1(t,·)=ψ(-t,·), 
and (ψ(t))_t∈ generates a group of diffeomorphisms on .


(3)
The symmetry of the orbits with respect to the real axis is a consequence of the following elementary fact.  Given z∈ and t↦ψ(t,z) its trajectory, then it follows that t↦ψ(-t,z) is also a  solution of the same Cauchy problem and by uniqueness we find the identity

    ψ(t,z)=ψ(-t,z),  ∀ t∈.



(4) Assume that W is invariant by the rotation R_θ_0  centered at zero and with angle θ_0. Let z∈𝔻, then we shall first check the identity

    e^iθ_0ψ(t,z)=ψ(t, e^iθ_0z),   ∀ t∈.

To do that, it suffices to verify that both functions satisfy the same differential equation with the same initial data, and thus the identity follows from the uniqueness of the Cauchy problem. Note that  (<ref>) means that the rotation of a trajectory is also a trajectory. Denote by D_z_0 and e^iθ_0D_z_0 the domains delimited by the curves t↦ψ(t,z_0) and  t↦ e^iθ_0ψ(t,z_0), respectively. Then, it is a classical result  that those domains are necessary simply connected and they contain the origin according to 1). Since different  trajectories never intersect, then we have only two possibilities: D_z_0⊂ e^iθ_0D_z_0 or the converse.  Since the rotation is a Lebesgue preserving  measure,  then D_z_0= e^iθ_0D_z_0, which implies that the periodic orbit t↦ψ(t,z_0) is invariant by the rotation R_θ_0.
 


 §.§ Reformulation with the trajectory map

In this section we discuss a new representation of solutions to the equations of the type

    W(z)·∇ f(z)=0,   ∀ z∈𝔻,

with  W a vector field as in Proposition <ref> and 
 f:𝔻→ a 𝒞^1 function.

Let W:𝔻→ be a vector field satisfying the assumptions i), ii) and iii) of Proposition <ref>. Then,  (<ref>) is equivalent to the formulation

    f(z)-1/T_z∫_0^T_zf(ψ(τ,z))dτ=0,   ∀  z∈𝔻,

with T_z being the period of the trajectory t↦ψ(t,z).


We  first check that (<ref>) is equivalent to

    f(ψ(t,z))=f(z),   ∀ t∈, ∀ |z|≤1.

Although for simplicity we can assume that the equivalence is done pointwise, where we need f ∈𝒞^1, the equivalence is perfectly valid in a weak sense without nothing more than assuming Hölder regularity  on f. Indeed, if f is a  𝒞^1 function satisfying  (<ref>), then by differentiating in time we get

    (W·∇ f)(ψ(t,z))=0,   ∀ t∈, ∀ |z|≤1.

According to Proposition <ref>, we have that (<ref>) is satisfied everywhere in the closed unit disc, for any t, ψ(t,𝔻)=𝔻.
Conversely, if f is a 𝒞^1 solution to  (<ref>), then differentiating with respect to t the function t↦ f(ψ(t,z)) we get

    d/dtf(ψ(t,z))=(W·∇ f)(ψ(t,z))=0.

Therefore,  we have (<ref>).
Now, we will verify that (<ref>) is in fact equivalent to (<ref>). The implication (<ref>)⟹(<ref>) is elementary. So it remains to check the converse. From (<ref>) one has

    f(ψ(t,z))-1/T_ψ(t,z)∫_0^T_ψ(t,z)f(ψ(τ,ψ(t,z))dτ=0.

Since the vector field is autonomous, then all the points located at the same orbit generate periodic trajectories with the same  period, and of course with the same orbit. Therefore, we have
T_ψ(t,z)=T_z.
Using 
ψ(τ,ψ(t,z))=ψ(t+τ,z),
a change of variables, and the T_z-periodicity of τ↦ f(ψ(τ,z)), then we deduce

    1/T_ψ(t,z)∫_0^T_ψ(t,z)f(ψ(τ,ψ(t,z))dτ=1/T_z∫_t^t+T_zf(ψ(τ,z))dτ
    =1/T_z∫_0^T_zf(ψ(τ,z))dτ
    =f(z).

Combining this with (<ref>),  we get  (<ref>). This completes the proof.




 §.§ Persistence of the symmetry

We shall consider a vector field W satisfying  the assumptions of  Proposition <ref> and (<ref>) and let ψ be its  associated  flow. We define the operator f↦ Sf by 

    Sf(z)=f(z)-1/T_z∫_0^T_zf(ψ(τ,z))dτ,   ∀ z∈𝔻.

We shall prove that f and Sf share the same planar group of invariance in the following sense.


Let f: 𝔻↦ be a smooth function. The following assertions hold true:


  * If f is invariant by reflection with respect to the real axis, then  Sf is invariant too. This means that

    f(z)=f(z),   ∀ z∈𝔻⟹  Sf(z)=Sf(z),   ∀ z∈𝔻.


  * If W and f are invariant  by  the rotation R_θ_0  centered at zero  with  angle  θ_0∈, then Sf commutes with the same rotation. This means that

    f(e^iθ_0z)=  f(z),   ∀ z∈𝔻⟹  (Sf)(e^iθ_0z)=Sf(z),   ∀ z∈𝔻.





(1) Let z∈𝔻, it is a simple matter to check that
ψ(t,z)=ψ(-t,z),
which implies
T_z=T_z,
and then 

    Sf(z)=f(z)-1/T_z∫_0^T_zf(ψ(-τ,z))dτ
    =f(z)-1/T_z∫_0^T_zf(ψ(-τ,z))dτ
    =Sf(z).


(2) According to Proposition (<ref>) and the fact that the vector-field W is invariant by the rotation R_θ_0,  then  we have that the orbits are symmetric with respect to this rotation and 

    T_e^iθ_0z=T_z  and  ψ(t,e^iθ_0z)=e^iθ_0ψ(t,z),

where we have used (<ref>), which implies that

    Sf(e^iθ_0z)=f(z)-1/T_z∫_0^T_zf(e^iθ_0ψ(τ,z))dτ
    =f(z)-1/T_z∫_0^T_zf(ψ(τ,z))dτ
    =Sf(z).

This concludes the proof.
 


 

 §.§ Analysis of the regularity


Next, we are interested in studying the  regularity of the the flow map  (<ref>) and the period map. The following result  is classical, see for instance  <cit.>.

Let α∈(0,1), W:𝔻↦^2  be a vector-field  in 𝒞^1,α(𝔻) satisfying the condition ii) of Proposition <ref> and ψ:×𝔻↦𝔻 its flow map. Then ψ∈𝒞^1(, 𝒞^1,α(𝔻)) and there exists C>0 such that

    ψ^± 1(t)_𝒞^1,α(𝔻)≤ e^CW_𝒞^1,α(𝔻|t|(1+W_𝒞^1,α(𝔻)|t|),  ∀ t∈,

holds.



Now we intend  to study the regularity of the function z=(x,y)∈𝔻↦ T_z. This is a classical subject in dynamical systems and several results are obtained in this direction for smooth Hamiltonians. Notice that in the most studies in the literature  the regularity is measured with respect to the energy and not with respect to the positions as we propose to do here.  

Let z∈𝔻 be a  given non equilibrium point, the orbit t↦ψ(t,z) is periodic and T_z is  the first strictly  positive time such that 

    ψ(T_z,z)-z=0.

This is an implicit function equation, from which we expect to deduce some regularity properties. Our result reads as follows.


Let W be a vector field in 𝒞^1( 𝔻),  satisfying the assumptions  of  and (<ref>) and such that

    W(z)=i zU(z),   ∀  z∈𝔻,

with 

    Re{U(z)}≠0,   ∀  z∈𝔻.
 Then the following assertions hold true:


  * The map z∈𝔻↦ T_z is continuous and verifies the upper bound

    0<T_z≤2π/inf_z∈𝔻|ReU(z)|,  ∀ z∈𝔻.



  * If  in addition U∈𝒞^1,α(𝔻), then z ↦ T_z  belongs to 𝒞^1,α(𝔻). 



Since the origin is an equilibrium point for the dynamical system, then its trajectory is periodic with any period. However, as we will see in the proof, there is a minimal strictly positive period denoted by T_z, for any curves passing through a non vanishing  The mapping z∈\{0}↦ T_z is not only well–defined, but it  can be extended continuously to zero. Thus we shall make the following convention

    T_0≡lim_z→ 0 T_z.



The upper bound in (<ref>) is “almost optimal” for radial profiles, where 
U(z)=U_0(|z|) ∈,
and explicit computations yield 

    T_z=2π/|U_0(|z|)|.




(1) We shall describe the trajectory parametrization using polar coordinates. Firstly, we may write for z=r e^iθ

    W(z)=[W^r(r,θ)+iW^θ(r,θ)] e^iθ,

with

    W^θ(r,θ)= r Re(U(r e^iθ))  and   W^r(r,θ)=- r Im(U(r e^iθ)).

Given 0<|z|≤1, we look for  a polar parametrization of the trajectory passing through z,

    ψ(t,z)=r(t) e^iθ(t),  r(0)=|z|,  θ(0)=Arg(z).

Inserting into (<ref>) we obtain the system

    ṙ(t)   =   -r(t)  Im{U(r(t) e^iθ(t))}≜ P(r(t),θ(t))
    θ̇(t)   =   Re{U(r(t) e^iθ(t))}≜ Q(r(t),θ(t)).

From   the assumption  (<ref>) we find

    U(z)=U(z),   ∀  z∈𝔻,

which implies in turn that

    P(r,-θ)=-P(r,θ)  and   Q(r,-θ)=Q(r,θ),   ∀ r∈[0,1],∀θ∈.

Thus, we have the Fourier expansions

    P(r,θ)=∑_n∈^⋆P_n(r)sin(nθ)  and   Q(r,θ)=∑_n∈Q_n(r)cos(nθ).

Denoting by T_n and U_n the classical Chebyshev polynomials that satisfy the identities

    cos(nθ))   =   T_n(cosθ)
    sin(nθ)   =   sin(θ) U_n-1(cosθ),

we obtain

    P(r,θ)   =   sinθ∑_n∈^⋆P_n(r)U_n-1(cosθ)
     ≡sinθ  F_1(r,cosθ).
    
    Q(r,θ)   =   ∑_n∈Q_n(r)T_n(cosθ)
    ≡ F_2(r,cosθ).

Coming back to  (<ref>), we get

    ṙ(t)   =sinθ(t)   F_1(r(t),cosθ(t))
    θ̇(t)   = F_2(r(t),cosθ(t)).

We look for  solutions in the form

    r(t)=f_z(cos(θ(t)),  with    f_z:[-1,1]→,

and then f satisfies the differential equation

    f_z^'(s)=-(F_1/F_2)(f_z(s),s), f(cosθ)=|z|.

Note that  the  preceding fraction is well–defined since the assumption (<ref>) is equivalent to

    F_2(r,cosθ)≠0,   ∀ r∈[0,1], θ∈.

Theorem <ref>-ii) agrees with

    F_1(0,s)=F_1(1,s)=0,   ∀ |s| ≤ 1,

which implies that the system (<ref>) admits a unique solution f_z:[-1,1]→_+ such that

    0≤ f_z(s)≤1,   ∀  s∈ [-1,1].

Hence, integrating the second equation of (<ref>) we find after a change of variable

    ∫_θ_0^θ(t)1/F_2(f_z(cos s), cos s)ds=t,

and, therefore, the following formula for the period is obtained

    T_z=|∫_θ_0^θ_0+2π1/F_2(f_z(cos s), cos s)ds|
    =∫_0^2π1/|F_2(f_z(cos s), cos s)|ds.

This gives the bound of the period stated in (<ref>). The continuity z↦ T_z follows from the same property of z↦ f_z, which can be derived from the continuous dependence with respect to  the initial conditions. 

(2)
Now, we will study the regularity of the period in 𝒞^1,α. Note that  (<ref>) involves the function f_z which is not smooth enough because  the initial condition z↦ f_z(cosθ) is only Lipschitz. So it seems  quite complicate to follow the regularity in 𝒞^1,α from that formula. The alternative way  is to study the regularity of the period using the implicit equation (<ref>). Thus, differentiating this equation
with respect to x and y  we obtain

    (∂_xT_z)∂_tψ(T_z,x,y)+∂_xψ(T_z,x,y)-1   =   0,
    
    (∂_yT_z)∂_tψ(T_z,x,y)+∂_yψ(T_z,x,y)-i   =   0.

From the flow equation and the periodicity condition we get

    ∂_tψ(T_z,z)=W(z),

which implies

    {[ (∂_xT_z)W(z)+∂_xψ(T_z,x,y)-1                            =                           0,; (∂_yT_z)W(z)+∂_yψ(T_z,x,y)-i                            =                           0. ].

Due to the assumption on W, the flow equation can be written as

    ∂_tψ(t,z)=iψ(t,z) U(ψ(t,z)),   ψ(0,z)=z,

which can be integrated, obtaining

    ψ(t,z)=z   e^i∫_0^tU(ψ(τ,z))dτ.

By differentiating this identity with respect to x, it yields

    ∂_xψ(t,z)= e^i∫_0^tU(ψ(τ,z))dτ[1+iz∫_0^t∂_x{U(ψ(τ,z))}dτ].

Since ψ(T_z,z)=z, we have

    e^i∫_0^T_zU(ψ(τ,z))dτ=1,

and thus

    ∂_xψ(T_z,z)= 1+iz∫_0^T_z∂_x{U(ψ(τ,z))}dτ.

Similarly, we find

    ∂_yψ(T_z,z)= i+iz∫_0^T_z∂_y{U(ψ(τ,z))}dτ.

Combining these identities with (<ref>), we obtain

    {[                (∂_xT_z)W(z)                           = -iz∫_0^T_z∂_x{U(ψ(τ,z))}dτ,;                (∂_yT_z)W(z)                           = -iz∫_0^T_z∂_y{U(ψ(τ,z))}dτ, ].

which, using the structure of W, reads as

    {[               (∂_xT_z)U(z)                          = -i∫_0^T_z∂_x{U(ψ(τ,z))}dτ,;               (∂_yT_z)U(z)                          = -i∫_0^T_z∂_y{U(ψ(τ,z))}dτ. ].

Now, notice that from  Theorem <ref>-iv), the vector field  W  vanishes only at zero and since U(0)≠ 0, we find 

    U(z)≠0,   ∀ z∈𝔻.

This implies that z∈𝔻↦1/U(z) is well–defined and belongs to 𝒞^1,α(𝔻). Therefore, we can write

    ∇_z T_z=-i/U(z)∫_0^T_z∇_z{U(ψ(τ,z))}dτ,

where we have used the notation ∇_z=(∂_x,∂_y). According to Proposition <ref> and classical composition laws, we obtain

    τ↦∇_z{U(ψ(τ,·))}∈𝒞(; 𝒞^0,α(𝔻)).

Since z↦ T_z is continuous, then  we find by composition that

    φ:z∈𝔻↦∫_0^T_z∇_z{U(ψ(τ,z))}dτ∈𝒞(𝔻).

Combining this information with (<ref>), we deduce that
z↦ T_z∈𝒞^1(𝔻).
Hence, we find in turn that
φ∈𝒞^0,α(𝔻)
by composition. Using (<ref>) again, it follows that
z↦∇_z T_z∈𝒞^0,α(𝔻).
Thus, z↦ T_z∈𝒞^1,α(𝔻). This achieves the proof.





 §.§ Application to the nonlinear problem

We intend in this section to prove   Let us point out that  from Proposition <ref>, the nonlinear vector field  W(Ω,f,Φ) is chosen in order to  be   tangent to the boundary everywhere. We will see
that not only this assumption  but all the  assumptions of  Proposition <ref> are satisfied  if f is chosen close  to a suitable radial profile. 




Let g∈𝒞_s,m^1,α() and ϕ∈ℋ𝒞_m^2,α(), then W(Ω,f,Φ)∈𝒞^1,α() and satisfies the symmetry properties (<ref>) and

    W(Ω,f,Φ)(e^i2π/mz)=e^i2π/mW(Ω,f,Φ)(z).
 
Moreover, if m≥4 then 

    W(Ω,f,Φ)(z)= iz U(z),

with U∈𝒞^1,α()




Using Proposition <ref> and Proposition <ref>, then the regularity and the symmetry properties of W(Ω,f,Φ) are verified.
Let us now check  (<ref>). Firstly, since Φ(0)=0 and ϕ∈ℋ𝒞^2,α(), we have

    Φ(z)=z Φ_1(z),   Φ_1∈𝒞^1,α().

In addition, Φ^'∈𝒞^1,α(), and thus we find

    iΩ Φ(z) Φ^'(z)=iz U_1(z)  with   U_1∈𝒞^1,α().

Now, to get (<ref>) it is enough to check that

    I(f,Φ)(z)=z U_2(z),  with   U_2∈𝒞^1,α(),

where

    I(f,Φ)(z)≜1/2π∫_f(y)/Φ(z)-Φ(y)|Φ^'(y)|^2 dA(y).

We  look for the first order Taylor expansion around the origin of I(f,Φ)(z). Using the m-fold symmetry of f and Φ, it is clear that

    I(f,Φ)(0)=- ∫_f(y)/Φ(y)|Φ^'(y)|^2 dA(y)
    =- ∫_f(e^i2π/my)/Φ(e^i2π/my)|Φ^'(e^i2π/my)|^2 dA(y)
    =e^i2π/mI(f,Φ)(0),

which leads to  

    I(f,Φ)(0)=0,

for m≥2. This implies that one can always write I(f,Φ)(z)=z U_2(z), but U_2 is only bounded:

    U_2_L^∞()≤∇_zI_2_L^∞()≤I_𝒞^1,α().

We shall see how the extra symmetry helps to get more regularity for U_2. According to 
Taylor expansion one gets

    I(f,Φ)(z)=a z+b z+c z^2+d z^2+ e |z|^2+l.o.t.  with   a,b,c,d,e∈.

Using the reflection invariance with respect to the real axis of f and Φ, we obtain

    I(f,Φ)(z)=I(z),

which implies that a,b,c,d,e∈. 
Now, the rotation invariance leads to

    I(f,Φ)(e^i2π/mz)=e^i2π/mI(z).

Then, we obtain

    e=0,   b(e^i4π/m-1)=0,    c(e^i2π/m-1)=0,  and   d(e^i6π/m-1)=0.

This implies that b=c=d=0, whenever m≥4. Thus, we have

    I(f,Φ)(z)=a z+h(z)

with h∈𝒞^2,α(), and 

    h(0)=0, ∇_zh(0)=0,  and  ∇_z^2h(0)=0.

From this we claim that

    h(z)=zk(z),  with   k∈𝒞^1,α(),

which concludes the proof.

Now we are in  a position to prove Theorem <ref>.

The existence of the conformal map Φ comes directly from Proposition <ref>, which gives the boundary equation (<ref>). Moreover, Lemma <ref> gives the decomposition (<ref>) and provides the necessary properties  to apply Proposition <ref>. Furthermore, we can use Proposition <ref> in order to obtain the equivalence between a rotating solution (<ref>) of the Euler equations   and the solution of  (<ref>). Proposition <ref> gives the regularity of the flow map, and it remains to check  (<ref>), using Proposition <ref>, in order to get the regularity of the period function. 
In the case of radial profile, we know that

    W(Ω,f_0,Id)(z)=izU_0(z),    U_0(z)=-Ω+1/r^2∫_0^r s f_0(s)ds,

which implies

    W(Ω,f,Φ)(z)-W(Ω,f_0,Id)(z)   =   -Ωϕ(z)(1+g^'(z))-Ω zg^'(z)+g^'(z) I(f,Φ)(z)
           + I(f,Φ)(z)-I(f_0,Id)(z).

It is easy to see that

    |ϕ(z)/z(1+g^'(z))|≤ϕ^'_L^∞(1+g^'_L^∞()).

From (<ref>) and Lemma <ref>, we have

    |g^'(z)I(f,Φ)(z)/z|≤g^'_L^∞()|∇_z I(f,Φ)_L^∞()≤ C(Φ)f_𝒞^1,α()g^'_L^∞().

Using again (<ref>), we deduce

    |I(f,Φ)(z)-I(f_0,Φ_0)(z)/z|   ≤   ∇_z [I(f,Φ)-I(f_0,Φ_0)]_L^∞().

Straightforward computations imply 

    I(f,Φ)(z)-I(f_0,Φ_0)(z)   =   1/2π∫_ g(y)/z-ydA(y)+1/2π∫_f(y)[ϕ(y)-ϕ(z)]/(y-z)(Φ(y)-Φ(z))dA(y)
           +1/2π∫_2Re{ϕ^'(y)}+|ϕ^'(y)|^2/Φ(z)-Φ(y) f(y)dA(y).

From this and using Lemma <ref>, we claim that

    ∇_z [I(f,Φ)-I(f_0,Φ_0)]_L^∞()   ≤    C(Φ)g_𝒞^1,α().

Combining the preceding estimates we find

    W(Ω,f,Φ)(z)=z[iU_0(z)+W(Ω,f,Φ)(z)-W(Ω,f_0,Id)(z)/z]
    ≡ z[iU_0(z)+W(Ω,f,Φ)(z)],

with

    W(Ω,f,Φ)_L^∞()≤ C(f,Φ)(h_𝒞^1,,α()+ϕ_𝒞^2,α()).

Now, we take h,ϕ small enough such that

    C(f,Φ)(h_𝒞^1,,α()+ϕ_𝒞^2,α())≤ε,

and ε verifies

    0<2ε≤inf_0< r≤1|Ω-1/r^2∫_0^r s f_0(s)ds|
    =  inf_z∈|U_0(z)|,

in order to have

    W(Ω,f,Φ)(z)=iz U(z),  with   2 |Re{U(z)}|≥inf_0< r≤1|Ω-1/r^2∫_0^r s f_0(s)ds|.

To end the proof let us check that this infimum is strictly positive  for the quadratic profile
f_0(r)=Ar^2+B, where A>0. Take  ξ∈(-a,a)↦(Ω_ξ,f_ξ, ϕ_ξ) the bifurcating curve from Theorem <ref>. Then  for a small enough

    Ω_ξ=B/2+A/4x_ξ, x_ξ<1  and  Ω_ξ∉𝒮_sing
  
Thus

    Ω_ξ-1/r^2∫_0^rsf_0(s)ds=A/4(1/x_ξ-r^2).

Consequently

    inf_r∈[0,1]|Ω_ξ-1/r^2∫_0^rsf_0(s)ds|=inf_r∈[0,1]A/4|1/x_ξ-r^2|
    > 0.

This achieves the proof.
 
 
 



The remaining part of this paper   is devoted to  some basic materials that we have used in the preceding proofs.


§ BIFURCATION THEORY AND FREDHOLM OPERATORS

 We  shall recall   Crandall-Rabinowitz Theorem which is a fundamental tool in bifurcation theory. Notice that  the main aim of this theory is to explore the topological transitions of the phase portrait through the variation of some parameters. A particular case is to understand this transition in  the equilibria set for   the stationary problem    F(λ,x)=0, where F:× X→ Y is a a smooth  function and the spaces X and Y are Banach spaces. Assuming that one has a particular solution,  F(λ,0)=0 for any λ∈, we would like to explore the bifurcation diagram close to this trivial solution,  and see whether we can find multiple branches of solutions bifurcating  from  a given point  (λ_0,0). When this occurs we say that the pair (λ_0,0) is a bifurcation point. When the linearized operator around this point generates a Fredholm operator, then one  may  use Lyapunov-Schmidt reduction in order to reduce the infinite-dimensional problem to a finite-dimensional one. For the latter problem we just formulate suitable assumptions so that the    Implicit Function Theorem can be applied. For more discussion in this subject, we refer to see <cit.>. 


In what follows we shall recall some basic results on Fredholm operators.

Let X and Y be  two Banach spaces. A continuous linear mapping T:X→ Y,  is a  Fredholm operator if it fulfills the following properties,


  * dim Ker   T<∞,

  * Im  T is closed in Y,

  * codim Im   T<∞.

The integer dim Ker  T-codim Im  T is called the Fredholm index of T.

Next, we shall discuss  the index persistence through compact perturbations, see  <cit.>.

The index of a Fredholm operator remains unchanged under compact perturbations.

Now, we recall the classical Crandall-Rabinowitz Theorem whose proof can be found  in <cit.>.


    Let X, Y be two Banach spaces, V be a neighborhood of 0 in X and F:ℝ× V→ Y be a function with the properties,
    
        
  * F(λ,0)=0 for all λ∈ℝ.
        
  * The partial derivatives  ∂_λ F_λ, ∂_fF and  ∂_λ∂_fF exist and are continuous.
        
  * The operator F_f(0,0) is Fedholm of zero index and Ker(F_f(0,0))=⟨ f_0⟩ is one-dimensional. 
                
  * Transversality assumption: ∂_λ∂_fF(0,0)f_0 ∉Im(∂_fF(0,0)).
    
    If Z is any complement of  Ker(F_f(0,0)) in X, then there is a neighborhood  U of (0,0) in ℝ× X, an interval  (-a,a), and two continuous functions Φ:(-a,a)→ℝ, β:(-a,a)→ Z such that Φ(0)=β(0)=0 and
    
    F^-1(0)∩ U={(Φ(ξ), ξ f_0+ξβ(ξ)) : |ξ|<a}∪{(λ,0): (λ,0)∈ U}.





§ POTENTIAL ESTIMATES

This appendix is devoted to  some classical estimates on potential theory that we have used before in Section <ref>. We shall deal in particular with truncated operators whose kernels are singular along  the diagonal. They have the form

    ℒ f(z)≜∫_K(z,y) f(y)dA(y), z∈

The action  of such operators over various function spaces and its connection to the singularity of the kernel  is  widely studied in the literature, see <cit.>. In the case of Caldron-Zygmund operator we refer to the recent papers <cit.> and the references therein. In what follows we shall establish some useful estimates whose proofs are classical  and  for the convenience  of the reader we decide to provide most  the  details.
The first result reads as follows.

Let α∈(0,1) and K:×→ is smooth off the diagonal and satisfies 

    |K(z_1,y)|≤C_0/|z_1-y|  and   |K(z_1,y)-K(z_2,y)|≤ C_0|z_1-z_2|/|z_1-y||z_2-y|,

for any z_1,z_2≠ y∈, with   C_0  a real positive constant. The operator defined in (<ref>) 
 
    ℒ: L^∞()→𝒞^0,α()
 
 is continuous, with the estimate

    ℒ f_𝒞^0,α()≤ C C_0 f_L^∞(),  ∀  f∈ L^∞(),

where C is a constant depending on α.


It is easy to see that

    ℒ f_L^∞()≤ CC_0f_L^∞().

Using  (<ref>) combined with an interpolation argument we may write

    |K(z_1,y)-K(z_2,y)|   ≤   |K(z_1,y)-K(z_2,y)|^α(|K(z_1,y)|^1-α+|K(z_2,y)|^1-α)
       ≤    CC_0|z_1-z_2|^α[1/|z_1-y|^1+α+1/|z_2-y|^1+α],

Thus from the inequality
 
    sup_z∈∫_|f(y)|/|z-y|^1+αdA(y)≤ CC_0f_L^∞()

we  deduce  the desired result.


Before giving the second result, we need to recall  Cauchy–Pompeiu's formula that we shall use later. Let  D  be a simply connected domain and φ:D→  be a 𝒞^1 complex function, then

    -1/π∫_D ∂_yφ(y)/w-ydA(y)=1/2π i∫_∂ Dφ(w)-φ(ξ)/w-ξdξ.



The following result deals with a specific type of integrals that we have already  encountered in Proposition <ref>. More precisely we shall be concerned with the integral

    ℱ[Φ](f)(z)≜∫_ f(y)/Φ(z)-Φ(y) |Φ^'(y)|^2 dA(y), z∈.


Let α∈(0,1) and Φ:→Φ()⊂ be a conformal bi-Lipschitz function  of class 𝒞^2,α(). Then

    ℱ[Φ]: 𝒞^1,α()→𝒞^1,α(),

is continuous.
 Moreover, the functional ℱ:Φ∈𝒰↦ℱ[Φ] is continuous, with
 
    𝒰≜{Φ∈𝒞^2,α() : Φ is bi-Lipschitz and conformal}.



We start with splitting ℱ[Φ] f as follows

    ℱ[Φ] f(z)   =   ∫_ f(y)-f(z)/Φ(z)-Φ(y) |Φ^'(y)|^2 dA(y)+ f(z)∫_ |Φ^'(y)|^2/Φ(z)-Φ(y)  dA(y)
       ≜   ℱ_1[Φ] f(z)+f(z) ℱ_2[Φ]f(z).

Let us estimate the first term ℱ_1[Φ] f. The L^∞() bound is straightforward  and comes from

    sup_z,y∈| f(y)-f(z)|/|Φ(z)-Φ(y)|≤f_Lip/Φ^-1_Lip·

Setting

    K[Φ](z,y)=∇_z ( f(y)-f(z)/Φ(z)-Φ(y)),

then one can easily check that K satisfies the assumptions 

    |K[Φ]( z,y)|   ≤    Cf_𝒞^1,α()1/|z-y|,
    
    |K[Φ](z_1,y)-K[Φ](z_2,y)|   ≤    Cf_𝒞^1,α()max_i,j∈{1,2}[|z_i-z_j|^α/|z_i-y|+|z_i-z_j|/|z_i-y||z_j-y|],

where the constant C depends on Φ.
Thus, Lemma (<ref>) yields

    ∇ℱ_1[Φ] f_𝒞^0,α()≤  Cf_𝒞^1,α()

and hence we find 

    ℱ_1[Φ] f_𝒞^1,α()≤  Cf_𝒞^1,α()

Let us now check the continuity of the operator ℱ_1:Φ∈𝒰↦ℱ_1[Φ].
Taking Φ_1,Φ_2∈𝒰 we may write,

    |ℱ_1[Φ_1] f(z)-ℱ_1[Φ_2] f(z)|   ≤   ∫_|f(y)-f(z)|/|Φ_1(y)-Φ_1(z)|||Φ_1'(y)|^2-|Φ_2'(y)|^2|dA(y)
       +   ∫_ |f(y)-f(z)||Φ_2'(y)||1/Φ_1(z)+Φ_1(y)-1/Φ_2(z)-Φ_2(y)|dA(y)
       ≤    Cf_Lip(Φ_1'-Φ_2'_L^∞()+Φ_1-Φ_2_L^∞()).

Similarly we get

    ∇ (ℱ_1[Φ_1] f(z)-ℱ_1[Φ_2] f(z))   =   ∫_D K[Φ_1]( z,y)(|Φ_1'(y)|^2-|Φ_2'(y)|^2)dA(y)
       +   ∫_(K[Φ_1]( z,y)-K[Φ_2]( z,y))|Φ_2'(y)|^2dA(y).

Now, performing the same arguments as for (<ref>) allows to get,

    ∫_D K[Φ_1]( ·,y)(|Φ_1'(y)|^2-|Φ_2'(y)|^2)dA(y)_𝒞^0,α()≤ CΦ_1'-Φ_2'_𝒞^0,α().

For the second integral term, we proceed first with splitting K[Φ] in the following way

    K[Φ](z,y)   =   -∇_z f(z)/Φ(z)-Φ(y)+(f(y)-f(z))∇_z(1/Φ(z)-Φ(y))
       ≜   -∇_z f(z) K_1[Φ](z,y)+K_2[Φ](z,y).

Let us check that K_1[Φ_1]-K_1[Φ_2] obeys to the assumptions  of Lemma <ref>. For the first one, it is clear from elementary computations that

    |K_1[Φ_1](z,y)-K_1[Φ_2](z,y)|≤ CΦ_1-Φ_2_Lip()|z-y|^-1.

Adding and subtracting adequately we obtain

    |(K_1[Φ_1]-K_1[Φ_2])      (z_1,y)-(K_1[Φ_1]-K_1[Φ_2])(z_2,y)|
       ≤   |((Φ_1-Φ_2)(z_1)-(Φ_1-Φ_2)(y))(Φ_1(z_2)-Φ_1(y))(Φ_2(z_2)-Φ_2(z_1))/(Φ_1(z_1)-Φ_1(y))(Φ_1(z_2)-Φ_1(y))(Φ_2(z_1)-Φ_2(y))(Φ_2(z_2)-Φ_2(y))|
       +   |(Φ_2(z_1)-Φ_2(y))(Φ_1(z_2)-Φ_1(y))((Φ_1-Φ_2)(z_1)-(Φ_1-Φ_2)(z_2))/(Φ_1(z_1)-Φ_1(y))(Φ_1(z_2)-Φ_1(y))(Φ_2(z_1)-Φ_2(y))(Φ_2(z_2)-Φ_2(y))|
       +   |((Φ_1-Φ_2)(z_2)-(Φ_1-Φ_2)(y))(Φ_2(z_1)-Φ_2(y))(Φ_1(z_1)-Φ_2(z_2))/(Φ_1(z_1)-Φ_1(y))(Φ_1(z_2)-Φ_1(y))(Φ_2(z_1)-Φ_2(y))(Φ_2(z_2)-Φ_2(y))|
       ≤    C Φ_1-Φ_2_Lip|z_1-z_2|/|z_1-y||z_2-y|·

Therefore, by applying Lemma <ref>  we deduce that

    ∫_(K_1[Φ_1]( ·,y)-K_1[Φ_2]( ·,y))|Φ_2'(y)|^2dA(y)_𝒞^0,α()≤ CΦ_1-Φ_2_𝒞^2,α().

Let us deal with  K_2. Note that 1/Φ(z)-Φ(y) is holomorphic, and then we can work with its complex derivative. We write it as

    K_2[Φ](z,y)   =   f(z)-f(y)/(Φ(z)-Φ(y))^2Φ^'(z).

We wish to apply once again Lemma <ref> and for this purpose we should check the suitable estimate for the kernel.
From straightforward computations we find that 

    |K_2[Φ_1](z,y)-K_2[Φ_2](z,y)|   ≤   |f(y)-f(z)/(Φ_1(z)-Φ_1(y))^2(Φ_1'(z)-Φ_2'(z))|
          +|(f(y)-f(z))Φ_2'(z)(1/(Φ_1(z)-Φ_1(y))^2-1/(Φ_2(z)-Φ_2(y))^2)|
       ≤    Cf_LipΦ_1-Φ_2_Lip|z-y|^-1

while for the second hypothesis we write

    |(K_2[Φ_1] -K_2[Φ_2])(z_1,y)-(K_2[Φ_1]-K_2[Φ_2])(z_2,y)|
       ≤   |f(y)-f(z_1)/(Φ_1(z_1)-Φ_1(y))^2(Φ_1'(z_1)-Φ_2'(z_1))-f(y)-f(z_2)/(Φ_1(z_2)-Φ_1(y))^2(Φ_1'(z_2)-Φ_2'(z_2))|
          +|(f(y)-f(z_1))Φ_2'(z_1)((Φ_1-Φ_2)(z_1)-(Φ_1-Φ_2)(y))((Φ_1+Φ_2)(z_1)-(Φ_1+Φ_2)(y))/(Φ_1(z_1)-Φ_1(y))^2(Φ_2(z_1)-Φ_2(y))^2.
          -.(f(y)-f(z_2))Φ_2'(z_2)((Φ_1-Φ_2)(z_2)-(Φ_1-Φ_2)(y))((Φ_1+Φ_2)(z_2)-(Φ_1+Φ_2)(y))/(Φ_1(z_2)-Φ_1(y))^2(Φ_2(z_2)-Φ_2(y))^2|
       ≤   Cf_Lip(Φ_1-Φ_2_Lip+Φ_1'-Φ_2'_Lip)|z_1-z_2|/|z_1-y||z_2-y|·

It follows from  Lemma <ref> that
 
    ∫_(K_2[Φ_1]( ·,y)-K_2[Φ_2]( ·,y))|Φ_2'(y)|^2dA(y)_𝒞^0,α()≤ CΦ_1-Φ_2_𝒞^2,α(),

which concludes the proof of  the continuity of ℱ_1 with respect to Φ.


Let us now move  to the second term ℱ_2. By using a change of variables one may write

    ℱ_2[Φ] f(z)=∫_Φ() 1/Φ(z)-y dA(y).

First, note that

    ℱ_2[Φ] f_L^∞()≤ C,

and

    ℱ_2[Φ_1] f-ℱ_2[Φ_2] f_L^∞()≤ CΦ_1-Φ_2_𝒞^2,α().

By Cauchy–Pompeiu's formula (<ref>) we get

    ℱ_2[Φ] f(z)   =   πΦ(z)-1/2i∫_Φ()ξ/ξ-Φ(z)  dξ
        =   πΦ(z)-1/2i∫_Φ(ξ)/Φ(ξ)-Φ(z)Φ^'(ξ)dξ
    =πΦ(z)-1/2i𝒞[Φ](z).

We observe that  the mapping Φ∈𝒞^2,α()↦Φ∈𝒞^2,α() is well-defined and  continuous. So it remains to check that 𝒞[Φ]∈𝒞^1,α() and prove  its continuity with respect to Φ. Note first that 𝒞[Φ] is holomorphic inside the unit disc and  its complex derivative is given by

    [𝒞[Φ]]^'(z)=Φ^'(z)∫_Φ(ξ)/(Φ(ξ)-Φ(z))^2Φ^'(ξ)dξ,  ∀   z∈.

Using a change of variables, we deduce that

    [𝒞[Φ]]^'(z)=-Φ^'(z)∫_Φ^'(ξ)  ξ^2/Φ(ξ)-Φ(z) dξ ,

where we have used the formula

    d/dξΦ(ξ)=-ξ^2Φ^'(ξ).

For this last integral we can use the upcoming Lemma <ref> to obtain that [𝒞[Φ]]'∈𝒞^0,α(). Although the last is clear, we show here an alternative procedure useful to check  the continuity with respect to Φ.
According to <cit.>, to show that [𝒞[Φ]]^'∈𝒞^0,α() it suffices to prove that

    |[𝒞[Φ]]^''(z)|≤ C(1-|z|)^α-1,   ∀ z∈.

Then, by differentiating we get

    [𝒞[Φ]]^''(z)   =   -Φ^''(z)_Φ^'(ξ)  ξ^2 /Φ(ξ)-Φ(z) dξ-(Φ^'(z))^2_Φ^'(ξ)  ξ^2 /(Φ(ξ)-Φ(z))^2  dξ
       ≜   -Φ^''(z)𝒞_1[Φ](z)-(Φ^'(z))^2𝒞_2[Φ](z).

For 𝒞_1[Φ] we simply write

    𝒞_1[Φ](z)   =   _ξ^2Φ^'(ξ)/Φ^'(ξ)-z^2Φ^'(z)/Φ^'(z)/Φ(ξ)-Φ(z)Φ^'(ξ) dξ+2iπΦ^'(z)/Φ^'(z)z^2.

Since  ξ∈↦ξ^2Φ^'(ξ) /Φ^'(ξ)∈𝒞^0,α(), then we have

    |𝒞_1[Φ](z)|≤ C(∫_ |z-ξ|^α/|z-ξ| |dξ|+1)
    ≤ C(1-|z|)^α-1.

It remains to estimate 𝒞_2[Φ]. Integration by parts implies

    𝒞_2[Φ](z)= ∫_ψ(ξ)/Φ(ξ)-Φ(z)Φ^'(ξ) dξ,

with

    ψ(ξ)=(Φ^'(ξ)ξ^2 /Φ^'(ξ))^'/Φ^'(ξ) ·

Since Φ∈𝒞^2,α() and is bi-Lipschitz, then  ψ∈𝒞^0,α(). Writing

    𝒞_2[Φ](z)=∫_ψ(ξ)-ψ(z)/Φ(ξ)-Φ(z)Φ^'(ξ) dξ+2iπψ(z),

implies that 

    |𝒞_2[Φ]|≤ C∫_|ψ(ξ)-ψ(z)|/|ξ-z| | dξ|+2πψ_L^∞≤ C∫_|ξ-z)|^α/|ξ-z| | dξ|+2πψ_L^∞.

Thus, we have

    |𝒞_2[Φ](z)|≤ C(1-|z|)^α-1.

Putting together the previous estimates, we deduce (<ref>) which implies ℱ_2[Φ]∈𝒞^1,α().


It remains to check the continuity of ℱ_2 with respect to Φ.
 Splitting ℱ_2[Φ_1]-ℱ_2[Φ_2] as 

    ℱ_2[Φ_1]f(z)-ℱ_2[Φ_2]f(z)   =   ∫_|Φ_1'(y)|^2-|Φ_2'(y)|^2/Φ_1(z)-Φ_1(y)dA(y)
          +∫_(Φ_2-Φ_1)(z)-(Φ_2-Φ_1)(y)/(Φ_1(z)-Φ_1(y))(Φ_2(z)-Φ_2(y))|Φ_2'(y)|^2dA(y),

combined with Lemma <ref> yield

    ℱ_2[Φ_1]-ℱ_2[Φ_2]_𝒞^0,α()≤ C Φ_1-Φ_2_𝒞^2,α().

Now, we need to prove a similar inequality for its derivative,

    ℱ_2'[Φ_1]-ℱ_2'[Φ_2]_𝒞^0,α()≤ C Φ_1-Φ_2_𝒞^2,α().

Since Φ∈𝒞^2,α()↦Φ^'∈𝒞^1,α() is clearly continuous, we just have to prove that

    |[𝒞[Φ_1]]”(z)-[𝒞[Φ_2]]”(z)|≤ C Φ_1-Φ_2_𝒞^2,α()(1-|z|)^α-1,   ∀ z∈.

It is enough to check the above estimate for 𝒞_1 and 𝒞_2. Let us show how  dealing with the first one, and the same arguments can be applied for 𝒞_2. We have

    |𝒞_1[Φ_1](z).   -   .𝒞_1[Φ_2](z)|≤|_ξ^2(Φ_1'(ξ)/Φ_1'(ξ)-Φ_2'(ξ)/Φ_2'(ξ))-z^2(Φ_1'(z)/Φ_1'(z)-Φ_2'(z)/Φ_2'(z))/Φ_1(ξ)-Φ_1(z)Φ_1'(ξ)dξ|
          +|_z^2Φ_2'(z)/Φ_2'(z)-ξ^2Φ_2'(ξ)/Φ_2'(ξ)/Φ_1(ξ)-Φ_1(z)(Φ_1'(ξ)-Φ_2'(ξ))dξ|
          +|_(Φ_1-Φ_2)(z)-(Φ_1-Φ_2)(ξ)/(Φ_1(ξ)-Φ_1(z))(Φ_2(ξ)-Φ_2(z))(z^2Φ_2'(z)/Φ_2'(z)-ξ^2Φ_2'(ξ)/Φ_2'(ξ))Φ_2'(ξ)dξ|
          +2π z^2|Φ_1'(z)-Φ_2'(z)/Φ_1'(z)+Φ_2'(z)Φ_2'(z)-Φ_1'(z)/Φ_1'(z)Φ_2'(z)|
       ≤    C Φ_1-Φ_2_𝒞^2,α()|(1-|z|)^1-α,

where  we have used that

    |ξ^2(Φ_1'(ξ)/Φ_1'(ξ)-Φ_2'(ξ)/Φ_2'(ξ))-z^2(Φ_1'(z)/Φ_1'(z)-Φ_2'(z)/Φ_2'(z))|≤ CΦ_1-Φ_2_𝒞^2,α(|z-ξ|.

Then, we obtain the  inequality for 𝒞_1[Φ], which concludes the continuity with respect to Φ. 

The following result deals with a Calderon-Zygmund type estimate, which will be necessary in the later development. The techniques used are related to the well-known T(1)-Theorem of Wittmann, see <cit.>. Let us define

    𝒦 f(z)≜∫_K(z,ξ) f(ξ)dξ,   ∀ z∈.



Let α∈(0,1) and K:×→ smooth outside the diagonal that satisfies 

    |K(z_1,y)|≤ C_0|z_1-y|^-1,
       |K(z_1,y)-K(z_2,y)|≤ C_0|z_1-z_2|/|z_1-y|^2,  if   2|z_1-z_2|≤ |z_1-y|,
       𝒦 (Id)∈𝒞^0,α(),
       |∫_∂ (∩ B_z_1(ρ)) K(z_1,ξ)dξ|<C_0,

for any z_1,z_2≠ y∈ and ρ>0, where C_0 is a positive constant that does not depend on z_1, z_2, y and ρ. Then,
 
    𝒦: 𝒞^0,α()→𝒞^0,α()
 
is continuous, with the estimate

    𝒦 f_𝒞^0,α()≤ C C_0f_𝒞^0,α(),

where C is a constant depending only on α.



From (<ref>) and (<ref>), we get easily that

    |𝒦f(z)|   ≤|∫_ K(z,ξ)(f(ξ)-f(z))dξ|+|f(z)∫_ K(z,ξ)dξ|
       ≤ C_0 f_𝒞^0,α()∫_|dξ|/|z-ξ|^1-α+Cf_𝒞^0,α()
       ≤  CC_0f_𝒞^0,α().

Taking z_1,z_2∈, we define d=|z_1-z_2|. We write

    ∫_ K(z_1,ξ)f(ξ)dξ   -∫_ K(z_2,ξ)f(ξ)dξ
    
    =   ∫_ K(z_1,ξ)(f(ξ)-f(z_1))dξ-∫_ K(z_2,ξ)(f(ξ)-f(z_1))dξ
       +f(z_1)∫_ K(z_1,ξ)dξ-f(z_1)∫_ K(z_2,ξ)dξ
    
    =   ∫_∩ B_z_1(3d) K(z_1,ξ)(f(ξ)-f(z_1))dξ-∫_∩ B_z_1(3d) K(z_2,ξ)(f(ξ)-f(z_1))dξ
       +∫_∩ B_z_1^c(3d)(K(z_1,ξ)-K(z_2,ξ))(f(ξ)-f(z_1))dξ
       +f(z_1)∫_ K(z_1,ξ)dξ-f(z_1)∫_ K(z_2,ξ)dξ
    ≜    I_1-I_2+I_3+f(z_1)(I_4-I_5).

Using (<ref>), we achieve

    |I_4-I_5|≤ C|z_1-z_2|^α.

Let us work with I_1 using the Layer Cake Lemma, see  <cit.>. We use that |∩ B_x(ρ)|≤ Cρ, for any ρ>0 and x∈^2, which means that it is 1-Ahlfors regular curve. In fact,  taking any z∈ and ρ>0 ones has that

    ∫_∩ B_z(ρ)|dξ|/|z-ξ|^1-α   =   ∫_0^∞|{ξ∈∩ B_z(ρ) : 1/|z-ξ|^1-α≥λ}|dλ,
       =   ∫_0^∞|{ξ∈∩ B_z(ρ) : |z-ξ|≤λ^-1/1-α}|dλ
       =   ∫_0^ρ^α-1|{ξ∈ : |z-ξ|≤ρ}|dλ
    +∫_ρ^α-1^+∞|{ξ∈ : |z-ξ|≤λ^-1/1-α}|dλ
       ≤    C(ρρ^α-1+∫_ρ^α-1^+∞λ^-1/1-αdλ)
    ≤ C ρ^α,

where |· | inside the integral denotes the arch length measure. Applying the last estimate to I_1, we find

    |I_1|≤ C_0f_𝒞^0,α()∫_∩ B_z_1(3d)|dξ|/|z_1-ξ|^1-α≤ C C_0f_𝒞^0,α()|z_1-z_2|^α.

 For the term I_3, we get
 
    |I_3|≤ C_0f_𝒞^0,α()|z_1-z_2|∫_∩ B_z_1(3d)^c|dξ|/|z_1-ξ|^2-α,

by (<ref>). Now, we use again the Layer Cake Lemma, obtaining

    ∫_∩ B_z(ρ)^c|dξ|/|z-ξ|^2-α   =∫_0^∞|{ξ∈ : 1/|z-ξ|^2-α≥λ, |z-ξ|≥ρ}|dλ,
       =∫_0^∞|{ξ∈ : |z-ξ|≤λ^-1/2-α, |z-ξ|≥ρ}|dλ
       =∫_0^ρ^α-2|{ξ∈ : ρ≤|z-ξ|≤λ^-1/2-α}|dλ
       ≤ C∫_0^ρ^α-2λ^-1/2-αdλ≤ C ρ^α-1.

Therefore,

    |I_3|≤ C_0f_𝒞^0,α()|z_1-z_2|^α.

It remains to estimate I_2. First, let us write it as

    I_2   =∫_∩ B_z_1(3d) K(z_2,ξ)(f(ξ)-f(z_2))dξ+(f(z_2)-f(z_1))∫_∩ B_z_1(3d) K(z_2,ξ)dξ
       ≜ H_1+(f(z_2)-f(z_1))H_2.

H_1 can be estimated as I_1 noting that B_z_1(3d)⊂ B_z_2(4d). To finish, we just need to check that H_2 is bounded. Decompose it as

    H_2=∫_∩ B_z_2(2d) K(z_2,ξ)dξ+∫_∩ B_z_1(3d)∩ B_z_2(2d)^c K(z_2,ξ)dξ≜ J_1+J_2,

since B_z_2(2d)⊂ B_z_1(3d). Note that 

    |J_2|≤ C_0 ∫_∩ B_z_1(3d)∩ B_z_2(2d)^c|dξ|/|z_2-ξ|≤ CC_0 d^-1|∩ B_z_1(3d)|≤ C C_0.

For the last term, we write

    J_1=∫_∂(∩ B_z_2(2d))  K(z_2,ξ)dξ-∫_∩∂ B_z_2(2d) K(z_2,ξ)dξ

By using condition (<ref>), we get that the first integral is bounded. For the second one, we obtain

    |∫_∩∂ B_z_2(2d) K(z_2,ξ)dξ|≤ C_0 ∫_∩∂ B_z_2(2d)|dξ|/|z_2-ξ|=1/2d|∩∂ B_z_2(2d)|≤ C.

Combining all the estimates, we achieved the announced result.



In the following result, we deal with the Cauchy integral defined as

    ℐ[Φ](f)(z)≜∫_f(ξ)Φ'(ξ)/Φ(z)-Φ(ξ)dξ.

Note that this classical operator is fully studied in <cit.> in the case that Φ=Id, then there we will adapt that proof.


Let α∈(0,1) and Φ:→Φ()⊂ be a conformal bi-Lipschitz function of class 𝒞^2,α(). Therefore, we have that

    ℐ[Φ]:𝒞^0,α()→𝒞^0,α(),

is continuous.
Moreover, ℐ:Φ∈𝒰↦ℐ[Φ] is continuous, where 𝒰 is defined in 


Note that 

    ℐ[Φ](f)(z)=∫_Φ()(f∘Φ^-1)(ξ)/Φ(z)-ξdξ=𝒞[f∘Φ^-1](Φ(z)),

where 𝒞 is the Cauchy Integral. Then, it is classical, see <cit.>, that

    ||ℐ[Φ](f)||_𝒞^0,α()=||𝒞[f∘Φ^-1]∘Φ||_𝒞^0,α()   ≤ C ||f∘Φ^-1||_𝒞^0,α()||Φ||_Lip()
       ≤ C ||f||_𝒞^0,α()||Φ^-1||_Lip()||Φ||_Lip().



To deal with the continuity with respect to the conformal map, we write

    ℐ[Φ_1]f(z)-ℐ[Φ_2]f(z)   =   ∫_ f(ξ)(Φ_1'(ξ)/Φ_1(z)-Φ_1(ξ)-Φ_2'(ξ)/Φ_2(z)-Φ_2(ξ))dξ≜∫_ f(ξ)K(z,ξ)dξ.

We will check that K verifies (<ref>)-(<ref>) in order to use Lemma <ref>. Straightforward computations yield

    |K(z_1,y)|≤ C ||Φ_1-Φ_2||_𝒞^1,α()|z_1-y|^-1,
       |K(z_1,y)-K(z_2,y)|≤ C ||Φ_1-Φ_2||_𝒞^1,α()|z_1-z_2|/|z_1-y|^2,  if   2|z_1-z_2|≤ |z_1-y|,

using that |z_2-y|≥ |z_1-y|-|z_1-z_2|≥1/2 |z_1-y| in the second property, which concerns  (<ref>)-(<ref>). Moreover,

    ∫_ K(z,ξ)dξ=∫_Φ_1()dξ/Φ_1(z)-ξ-∫_Φ_2()dξ/Φ_2(z)-ξ=0,

which implies (<ref>). In fact, 

    ∫_∂(∩ B_z(ρ)) K(z,ξ)dξ=∫_Φ_1(∂(∩ B_z(ρ)))dξ/Φ_1(z)-ξ-∫_Φ_2(∂(∩ B_z(ρ)))dξ/Φ_2(z)-ξ=C_0,

by applying the Residue Theorem, and where  C_0 that does not depend on ρ neither z, which agrees with (<ref>). Then, we achieve the proof using Lemma <ref>.


We give the explicit expressions of some integrals which appear in the analysis of  the linearized operator.


 Let α∈(0,1). Given h∈𝒞^1,α_s(),   k∈ℋ𝒞^2,α() and a  radial function f_0∈𝒞(), the following identities 

    ∫_k(z)-k(y)/(z-y)^2f_0(y)dA(y)=2π∑_n≥ 1A_n z^n-1[∫_0^|z|rf_0(r)dr-n∫_|z|^1rf_0(r)dr],
       ∫_f_0(y)/z-y[k'(y)]dA(y)=π∑_n≥ 1A_n(n+1) [-z^n-1∫_|z|^1rf_0(r)dr+z^n+1/|z|^2(n+1)∫_0^|z|r^2n+1f_0(r)dr ],
       ∫_f_0(y)/z-ydA(y)=2πz/|z|^2∫_0^|z|rf_0(r)dr,
       ∫_h(y)/z-ydA(y)=π∑_n≥ 1[-z^n-1∫_|z|^11/r^n-1h_n(r)dr+z^n+1/|z|^2n+2∫_0^|z|r^n+1h_n(r)dr],
       ∫_log |z-y|h(y)dA(y)=-π∑_n≥ 1cos(nθ)1/n[|z|^n∫_|z|^11/r^n-1h_n(r)dr+1/|z|^n∫_0^|z||z|^n+1h_n(r)dr],
       ∫_k(z)-k(y)/z-yf_0(y)dA(y)=2 π∑_n≥ 1A_nz^n∫_0^1sf_0(s)ds,
       ∫_log|z-y|f_0(y)Re[k'(y)]dA(y)=-π∑_n≥ 1A_nn+1/ncos(n θ)
       ×[|z|^n∫_|z|^1rf_0(r)dr+1/|z|^n∫_0^|z|r^2n+1f_0(r)dr],
       ∫_log |z-y|f_0(y)dA(y)=2π[∫_0^|z|1/τ∫_0^τ r f_0(r)dr-∫_0^11/τ∫_0^τ r f_0(r)dr]

holds for z∈.


Note that h and k can be given by

    h(z)=∑_n≥ 1h_n(r)cos(nθ),    k(z)=z∑_n≥ 1A_nz^n,

where z=re^iθ∈.


(1)  Using the expression for the function k,
the integral to be computed takes the form

    ∫_z^n+1-y^n+1/z-yf_0(y)/z-ydA(y).

An expansion of the function inside the integral provides

    ∫_z^n+1-y^n+1/z-yf_0(y)/z-ydA(y)=∑_k=0^nz^n-k∫_ y^kf_0(y)/z-ydA(y).

The use of polar coordinates yields

    ∫_ y^kf_0(y)/z-ydA(y)=i∫_0^1r^kf_0(r)∫_ξ^k-1/ξ-z/rdξ.

We split our study in the cases k=0 and k≥ 1 by making use of the Residue Theorem. For k=0, we obtain

    ∫_1/ξ1/ξ-z/rdξ={[         0,    |z|≤ r,; -2π i r/z,    |z|≥ r, ].

whereas, we find 

    ∫_ξ^k-1/ξ-z/rdξ={[ 2π iz^k-1/r^k-1,          |z|≤ r,;               0,          |z|≥ r, ].

for any k≥ 1. This allows us to have the following expression 

    ∫_z^n+1-y^n+1/z-yf_0(y)/z-ydA(y)   =   2π z^n-1∫_0^|z| rf_0(r)dr-2π∑_k=1^nz^n-k z^k-1∫_|z|^1rf_0(r)dr
       =   2π z^n-1[∫_0^|z|rf_0(r)dr-n∫_|z|^1rf_0(r)dr].


(2) Note that
k'(z)=∑_n≥ 1 A_n (n+1) z^n.
Then, the integral to be analyzed is

    ∫_f_0(y)/z-y[k'(y)]dA(y)=∑_n≥ 1A_n(n+1)/2∫_f_0(y)/z-y(y^n+y^n)dA(y).

We study the two terms in the integral by using polar coordinates and the Residue Theorem. For the first one we have

    ∫_f_0(y)/z-yy^ndA(y)=i∫_0^1r^nf_0(r)∫_ξ^n-1/ξ-z/rdξ dr
    =-2π z^n-1∫_|z|^1rf_0(r)dr.

In the same way, the second one can be written as

    ∫_f_0(y)/z-yy^ndA(y)   =   i∫_0^1f_0(r)r^n∫_1/ξ^n+11/ξ-z/rdξ dr
       =   2π1/z^n+1∫_0^|z|f_0(r)r^2n+1dr
    =2πz^n+1/|z|^2n+2∫_0^|z|f_0(r)r^2n+1dr,

that concludes the proof.



(3) This integral reads as

    ∫_f_0(y)/z-ydA(y)=i∫_0^1f_0(r)∫_1/ξ1/ξ-z/rdξ dr
    = 2π/z∫_0^|z|rf_0(r)dr
    =2πz/|z|^2∫_0^|z|rf_0(r)dr.


(4) We use the expression of  h(z) to deduce that

    ∫_h(y)/z-ydA(y)=1/2∑_n≥ 1∫_h_n(r)(e^inθ+e^-inθ)/z-ydA(y).

The two terms involved in the integral can be computed as follows

    ∫_h_n(r)e^inθ/z-ydA(y)   =   i∫_0^1h_n(r)∫_ξ^n-1/ξ-z/rdξ dr
    =-2π z^n-1∫_|z|^11/r^n-1h_n(r)dr,
    ∫_h_n(r)e^-inθ/z-ydA(y)   =   i∫_0^1h_n(r)∫_1/ξ^n+11/ξ-z/rdξ dr
       =   2π/z^n+1∫_0^|z|r^n+1h_n(r)dr
    =2πz^n+1/|z|^2(n+1)∫_0^|z|r^n+1h_n(r)dr.


(5) Let us differentiate with respect to r having that

    ∂_r ∫_log |re^iθ-y|h(y)dA(y)   =   Re[z/r∫_h(y)/z-ydA(y)]
       =   Re[z/r∑_n≥ 1π[-z^n-1∫_r^1 h_n(s)/s^n-1ds+1/z^n+1∫_0^rs^n+1h_n(s)ds]]
       =   π∑_n≥ 1cos(nθ)[-r^n-1∫_r^1 h_n(s)/s^n-1ds+1/r^n+1∫_0^rs^n+1h_n(s)ds].

This last integral was computed before by the Residue Theorem. Now, we realize that

    -∂_r1/n[r^n∫_r^1 1/s^n-1h_n(s)ds.   +   .1/r^n∫_0^rs^n+1h_n(s)ds]
          =-r^n-1∫_r^1 1/s^n-1h_n(s)ds+1/r^n+1∫_0^rs^n+1h_n(s)ds.

Then, we obtain 

    ∫_log |z-y|h(y)dA(y)=-π∑_n≥ 11/n[r^n∫_r^1h_n(s)/s^n-1ds+1/r^n∫_0^rs^n+1h_n(s)ds]cos(nθ)+H(θ),

where H is a function that only depends on θ. Taking r=0 we have that

    H(θ)=∫_log (|y|) h(y)dA(y)=0.

The last is equal to zero due to the form of the function h:
h(re^iθ)=∑_n≥ 1 h_n(r)cos(nθ).

(6)  This integral can be done by spliting it as follows

    ∫_k(z)-k(y)/z-yf_0(y)dA(y)   =   k(z)∫_f_0(y)/z-ydA(y)-∫_k(y)f_0(y)/z-ydA(y)
       =   ∑_n≥ 1 A_n[z^n+1∫_f_0(y)/z-ydA(y)-∫_y^n+1f_0(y)/z-ydA(y)].

Note that these integrals have be done before. Hence, we conclude using Integral (3) for the first one and (<ref>) for the second one.

(7) Similarly to Integral (5), we differentiate with respect to r

    ∂_r ∫_log |re^iθ   -    y|  f_0(y)Re[k'(y)]dA(y)=Re[z/r∫_f_0(y)Re[k'(y)]/z-ydA(y)]
       =   π∑_n≥ 1A_n(n+1)Re[z/r[-z^n-1∫_r^1sf_0(s)ds+z^n+1/r^2(n+1)∫_0^rs^2n+1f_0(s)ds ]],
       =   π∑_n≥ 1A_n(n+1) cos(nθ)[-r^n-1∫_r^1sf_0(s)ds+1/r^n+1∫_0^rs^2n+1f_0(s)ds],

where we use Integral (2). With the same argument than in Integral (5) we realize that

    -∂_r1/n[r^n∫_r^1 sf_0(s)ds.   +   .1/r^n∫_0^rs^2n+1f_0(s)ds]
          =-r^n-1∫_r^1sf_0(s)ds+1/r^n+1∫_0^rs^2n+1f_0(s)ds,

and hence

    ∫_log |re^iθ-y|      f_0(y)Re[k'(y)]dA(y)
          =-π∑_n≥ 1A_nn+1/ncos(nθ)[r^n∫_r^1sf_0(s)ds+1/r^n∫_0^rs^2n+1f_0(s)ds]+H(θ),

where H is a function that only depends on θ.  Evaluating in r=0 as in Integral (5), we get that H≡ 0,  obtaining the announced identity.


(8) 
As in Integral (5) and (7) we differentiate with respect to r having

    ∂_r ∫_log |re^iθ-y|f_0(y)dA(y)   =   Re[z/r∫_f_0(y)/z-ydA(y)]
    = 2π1/r∫_0^r sf_0(s)ds,

where the last integral is done in Integral (3).
Hence,

    ∫_log |re^iθ-y|f_0(y)dA(y)=2π∫_0^r 1/τ∫_0^τ sf_0(s)dsdτ +H(θ),

where H is a function that only depends on θ. Evaluating in r=0 we get that

    H(θ)=∫_0^1∫_0^2π slog s f_0(s)dsdθ=-2π∫_0^1 1/τ∫_0^τ sf_0(s)dsdτ,

concluding the proof.




§ GAUSS HYPERGEOMETRIC FUNCTION


We give a short  introduction to the Gauss hypergeometric functions and discuss some  of their basic properties. The formulae listed below were crucial in the computations of the linearized operator associated to the V-states equation and the analysis of its spectral study.  Recall that for any real numbers a,b∈ℝ,  c∈ℝ\(-ℕ) the hypergeometric function z↦ F(a,b;c;z) is defined on the open unit disc 𝔻 by the power series

    F(a,b;c;z)=∑_n=0^∞(a)_n(b)_n/(c)_nz^n/n!,   ∀ z∈𝔻.

The  Pochhammer symbol (x)_n is defined by

    (x)_n =    1,       n = 0, 
    
     x(x+1) ⋯ (x+n-1),     n ≥1,

and verifies

    (x)_n=x (1+x)_n-1,   (x)_n+1=(x+n) (x)_n.

The series converges absolutely for all values of |z|<1. For |z|=1 we have that it converges absolutely if Re (a+b-c)<0 and it diverges if 1≤Re(a+b-c). See <cit.> for more details.

We recall the integral representation of the hypergeometric function, see for instance  <cit.>. Assume that  Re(c) > Re(b) > 0, then we have

    F(a,b;c;z)=Γ(c)/Γ(b)Γ(c-b)∫_0^1 x^b-1 (1-x)^c-b-1(1-zx)^-a  dx,  ∀z∈\[1,+∞).

Notice that this representation shows that the hypergeometric function initially defined in the unit disc  admits an analytic continuation to the complex plane cut along  [1,+∞). Another useful identity is the following:  


    F(a,b;c;z)=(1-z)^-aF(a,c-b;c;z/z-1),  ∀  |arg(1-z)|<π,

for c> b>0. 

The function Γ: \{-}→ refers to the gamma function, which is the analytic continuation to the negative half plane of the usual gamma function defined on the positive . It is defined by the integral representation

    Γ(z)=∫_0^+∞τ^z-1e^-τ dτ,

and satisfies the relation
Γ(z+1)=z Γ(z),  ∀ z∈\(-).
From this we deduce the identities

    (x)_n=Γ(x+n)/Γ(x),   (x)_n=(-1)^nΓ(1-x)/Γ(1-x-n),

provided that all the quantities in the right terms are well-defined. 

We can differentiate the hypergeometric function  obtaining

    d^kF(a,b;c;z)/dz^k=(a)_k(b)_k/(c)_kF(a+k,b+k;c+k;z),

for k∈. Depending on the parameters, the hypergeometric function behaves differently at 1. When c> b>0 and (c-a-b)>0, it can be shown that it is absolutely convergent on the closed unit disc  and one finds the expression

    F (a,b;c;1)= Γ(c)Γ(c-a-b)/Γ(c-a)Γ(c-b) ,

whose  proof can be found in <cit.>. However, in the case a+b=c, the hypergeometric function exhibits a logarithmic singularity as follows
 
    lim_z→ 1-F(a,b;c;z)/-ln(1-z)=Γ(a+b)/Γ(a)Γ(b),

see for instance <cit.> for more details. Next we recall some Kummer's quadratic transformations of the hypergeometric series, see <cit.>,

    c F (a,b;c;z)-(c-a) F (a,b;c+1;z)-a F (a+1,b;c+1;z)   =   0, 
    
     (b-c)F(a,b-1;c;z)+(c-a-b)F(a,b;c;z)-a(z-1)F(a+1,b;c;z)   =   0,
    (2c-a-b+1)z-c/cF(a,b;c+1;z)+(a-c-1)(c-b+1)z/c(c+1) F(a,b;c+2;z)      
           = F(a,b;c;z)(z-1).

Other formulas  which have been used in the preceding sections  are

    {[          ∫_0^1 F(a,b;c;τ z)τ^c-1 dτ=1/cF(a,b;c+1;z),; ∫_0^1F(a,b;c;τ z)τ^c-1(1-τ) dτ=1/c(c+1)F(a,b;c+2;z). ].

The last point  that we wish to recall concerns the differential equation governing   the hypergeometric equation, which   is given by,

    z(1-z)F(z)”+(c-(a+b+1)z)F(z)'-abF(z)=0,

with  a, b,∈ and  c∈\(-) given. One of the two independent solutions of the last differential equation around z=0 is the hypergeometric function: F(a,b;c;z). It remains to identify the second independent solution. If none of c,c-a-b,a-b is an integer, then the other independent solution around the singularity z=0 is

    z^1-cF(a-c+1,b-c+1;2-c;z).

We will be interested in the critical case when c is a negative integer. In this case, the hypergeometric differential equation has as a smooth solution given by (<ref>). However, the second independent solution is singular and contains a logarithmic singularity, see <cit.> for more details. Real solutions around +∞ may be also obtained as it is done in <cit.>. In fact, the two independent solutions are given by

    z^-aF(a,a+1-c;a+1-b;1/z)   and   z^-bF(b,b+1-c;b+1-a;1/z).


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