# Self-similar spirals for the generalized surface quasi-geostrophic equations

Claudia García  Departament de Matemàtiques i Informàtica, Universitat de
Barcelona, Gran Via de les Corts Catalanes 585, 08007 Barcelona, Spain
<EMAIL_ADDRESS>and  Javier Gómez-Serrano  Department of Mathematics,
Brown University, Kassar House, 151 Thayer St., Providence, RI 02912, USA
Departament de Matemàtiques i Informàtica, Universitat de Barcelona, Gran Via
de les Corts Catalanes 585, 08007 Barcelona, Spain Centre de Recerca
Matemàtica, Edifici C, Campus Bellaterra, 08193 Bellaterra, Spain
<EMAIL_ADDRESS><EMAIL_ADDRESS>
(Date: August 27, 2024)

###### Abstract.

In this paper we construct a large class of non-trivial (non-radial) self-
similar solutions of the generalized surface quasi-geostrophic equation
(gSQG). To the best of our knowledge, this is the first rigorous construction
of any self-similar solution for these equations. The solutions are of spiral
type, locally integrable, and may have mixed sign. Moreover, they bear some
resemblance with the finite time singularity scenario numerically proposed by
Scott and Dritschel [74] in the SQG patch setting.

C.G. and J.G.-S. have been partially supported by the European Research
Council ERC-StG-852741 (CAPA) and the MICINN (Spain) research grant number
PID2021–125021NA–I00. C. G. has also been partially supported by the
MINECO–Feder (Spain) research grant number RTI2018–098850–B–I00, and the Junta
de Andalucía (Spain) Project FQM 954. This material is based upon work
supported by the National Science Foundation under Grant No. DMS–1929284 while
C.G. and J.G.-S. were in residence at the Institute for Computational and
Experimental Research in Mathematics in Providence, RI, during the program
“Hamiltonian Methods in Dispersive and Wave Evolution Equations”. This work is
supported by the Spanish State Research Agency, through the Severo Ochoa and
María de Maeztu Program for Centers and Units of Excellence in R&D
(CEX2020-001084-M)

###### Contents

  1. 1 Introduction
    1. 1.1 Main result
    2. 1.2 Spiral solutions for 2D Euler
    3. 1.3 Self-similar solutions, non-uniqueness and finite time singularities for SQG and gSQG
    4. 1.4 Main ideas of the proof
    5. 1.5 Organization of the paper
  2. 2 Self-similar solutions
    1. 2.1 Adapted coordinates
    2. 2.2 The equation in the new coordinates
    3. 2.3 Trivial solution
    4. 2.4 Formulation with the appropriate scaling
  3. 3 Functional regularity
    1. 3.1 Function spaces
    2. 3.2 Deformation of the Euclidean norm
    3. 3.3 Regularity of the nonlinear function
  4. 4 Linearized operator
    1. 4.1 Linearization at the equilibrium
    2. 4.2 Fredholmness at the trivial solution
    3. 4.3 Expression of the linearized operator in Fourier series
  5. 5 Invertibility of the linear operator
    1. 5.1 Kernel study
    2. 5.2 Mellin transform
  6. 6 $m$-fold symmetry and main theorem in the variables $(\beta,\phi)$
    1. 6.1 $m$-fold symmetry in the function spaces
    2. 6.2 Main theorem in the adapted coordinates
  7. 7 Recovering the solution in the original coordinates
    1. 7.1 Invertibility of the change of coordinates
    2. 7.2 Solution in the physical variables
    3. 7.3 Main result
  8. A Radial solutions
  9. B Special functions

##  1\. Introduction

The generalized surface quasi-geostrophic equations describing the evolution
of the potential temperature $\theta$ read as

$\left\\{\begin{array}[]{ll}\theta_{t}+(u\cdot\nabla)\theta=0,&\textnormal{in
}[0,+\infty)\times\operatorname{\mathbb{R}}^{2},\\\
u=-\nabla^{\perp}(-\Delta)^{-1+\frac{\gamma}{2}}\theta,&\textnormal{in
}[0,+\infty)\times\operatorname{\mathbb{R}}^{2},\\\
\theta(0,x)=\theta_{0}(x),&\textnormal{with
}x\in\operatorname{\mathbb{R}}^{2}.\end{array}\right.$ (1)

In this system $u$ refers to the velocity field,
$\nabla^{\perp}=(-\partial_{2},\partial_{1})$ and $\gamma\in[0,2)$. The
velocity field is linked to the potential temperature $\theta$ via the
operator $-(-\Delta)^{-1+\frac{\gamma}{2}}$ agreeing with
$u=\nabla^{\perp}\psi$ and

$\psi(t,x)=-(-\Delta)^{-1+\frac{\gamma}{2}}\theta(x)=\int_{\operatorname{\mathbb{R}}^{2}}K_{\gamma}(x-y)\theta(t,y)dy,$
(2)

where

$K_{\gamma}(x)=\left\\{\begin{array}[]{ll}\frac{1}{2\pi}\log|x|,&\gamma=0,\\\
-\frac{C_{\gamma}}{2\pi}\frac{1}{|x|^{\gamma}},&\gamma\in(0,2),\end{array}\right.$

and
$C_{\gamma}=\frac{\Gamma(\gamma/2)}{2^{1-\gamma}\Gamma\left(\frac{1-\gamma}{2}\right)}$.
The case $\gamma=0$ corresponds to the Euler equations and $\gamma=1$ to the
surface quasi-geostrophic (SQG) model. In this work we will be interested in
the case $\gamma\in(0,1)$.

The SQG equation $(\gamma=1)$ models the evolution of the temperature from a
general quasi-geostrophic system for atmospheric and oceanic flows (see [20,
50, 67, 60] for more details). Constantin–Majda–Tabak pointed out the analogy
between the SQG equation and the incompressible 3D Euler equations [20] and
carried out the first numerical and analytical study of the equation. The gSQG
or (SQG)γ model (1) was proposed by Córdoba–Fontelos–Mancho–Rodrigo in [24] as
an interpolation between Euler and surface quasi–geostrophic equations. Yet,
almost 30 years after [20], the global existence vs finite time singularity
question remains open, even in the case $\gamma>0$. As a first step towards
the finite time singularity direction, this paper aims to construct self-
similar solutions, which will have infinite energy. We leave for future work
the transformation of the aforementioned self-similar solutions into finite
energy ones that may develop a finite time singularity.

###  1.1. Main result

Our main task is to construct spiral solutions that are perturbations of a
stationary solution. Indeed, we will perturb appropriately the following
radial one

$\theta(t,x)=\theta_{0}(x)=|x|^{-\frac{1}{\mu}},$ (3)

for $\mu>0$. The singularity of the stationary solution (3) depends on the
parameter $\mu$. However, we need some condition on $\mu$ to ensure the
existence of the associated stream function, that is, the integrability of the
singular operator (2). In particular, assuming that $\theta$ agrees with (3),
one needs that

$\frac{1}{2}<\mu<\frac{1}{2-\gamma},$ (4)

to have that $\psi(x)=C_{0}|x|^{2-\gamma-\frac{1}{\mu}}$, where $C_{0}$ is
defined in (A.2).

In this work, we will be interested in self-similar solutions. That means

$\theta(t,x)=\lambda^{1+\alpha-\gamma}\theta(\lambda^{1+\alpha}t,\lambda
x),\quad\forall\lambda>0.$ (5)

By taking $\lambda=1/t^{\frac{1}{1+\alpha}}$, the problem reduces to find a
self-similar profile $\hat{\theta}$ such that

$\theta(t,x)=\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}}}\hat{\theta}\left(\frac{x}{t^{\frac{1}{1+\alpha}}}\right).$
(6)

Moreover, in order to have that the trivial singular radial solution (3) is
self-similar, one needs the following condition on $\mu$ and $\alpha$:

$\frac{1}{\mu}=1+\alpha-\gamma,$ (7)

which, together with (4), implies that $\alpha\in(1,1+\gamma)$.

Our main theorem reads as follows.

######  Theorem 1.1.

Let $\gamma\in(0,1)$ and $\alpha\in(1,1+\gamma)$. Then, there exists
$\varepsilon>0$ such that for any $\frac{2\pi}{m}$-periodic $\tilde{\Omega}\in
B_{L^{p}([0,2\pi])}(1,\varepsilon)$ and $p>\frac{1}{1-\gamma}$, the initial
condition

$\theta_{0}(re^{i\vartheta})=r^{-(1+\alpha-\gamma)}\tilde{\Omega}(\vartheta),$

describes a self-similar solution of the type (6) for
$\textnormal{(SQG)}_{\gamma}$, for any $m\geq 1$.

We refer to Figure 1 to illustrate the kind of self-similar profiles obtained
in Theorem 1.1. Let us remark that comparing to the Euler case studied in [4,
33, 34], condition (4) gives us that $\theta_{0}\in L^{1}_{\textnormal{loc}}$.
More precisely, we can prove that the initial vorticity $\theta_{0}$ belongs
to $L^{q}_{\textnormal{loc}}$, with $q<\min\\{p,2\mu\\}$, provided that
$\tilde{\Omega}\in L^{p}$.

Figure 1. Left: the trivial self-similar radial solution (3). Right: the
perturbed solution of Theorem 1.1.

We now summarize the previous literature related to this problem. We will
first discuss the 2D Euler equation, then discuss the SQG and gSQG equations.

###  1.2. Spiral solutions for 2D Euler

The study of spiraling solutions for the 2D Euler equation has a long history
due to its connections to turbulence and non-uniqueness, which starts from the
seminal work of Kaden [57] and Pullin [68] in the context of vortex sheets.
These vortex spiral solutions exhibit non-uniqueness phenomena. Elling
constructed in [33] spiral solutions close to constant in a high symmetry
class, and later upgraded them to sign-changing solutions [34]. Lately,
Cieslak–Kokocki–Ozanski [17, 18] have proved local existence for logarithmic
spiral vortex sheets. See also [36] where spirals are reported in the context
of water waves.

Bressan, in a series of papers together with Shen and Murray [4, 5] has laid
out a very ambitious program to prove non-uniqueness by constructing an
initial data that can be continued as either one spiral or two spirals. The
idea developed in these works amounts to proving some a priori estimates and
establishing the existence of some of the building blocks of the program, as
well as providing compelling numerical evidence of it.

We remark that in our case, in contrast to the previous literature, we are
able to achieve solutions in any symmetry class due to the study of the full
linear operator, as opposed to: 1. Either a Neumann series in the symmetry
class, where the contribution of one of the pieces of the linear operator
dominates over the other whenever the solution is sufficiently symmetric [33,
34], or 2. The case of an inner domain where the problem has less constraints
[4]). Moreover, let us emphasize one fundamental difference between the Euler
case and the gSQG equations: in [4, 33, 34] the authors really exploit the
fact that the relation between the stream function and the vorticity is local,
which does not hold for the gSQG equations, and hence new techniques must be
developed here.

In [30], Elgindi-Jeong obtained global well-posedness for a class of patches
with boundary singular at the origin which includes spirals. Later, in [31],
the authors constructed a patch solution whose boundary forms filaments which
spirals around the center of the patch, as time goes to infinity, which was
suggested numerically by [29].

###  1.3. Self-similar solutions, non-uniqueness and finite time singularities
for SQG and gSQG

Resnick, in his thesis [71], showed global existence of weak solutions in
$L^{2}$ using the oddness of the Riesz transform to obtain an extra
cancellation. Marchand [62] extended Resnick’s result to the class of initial
data belonging to $L^{p}$ with $p>4/3$.

The question of non-uniqueness for weak solutions of SQG is still a very
challenging problem which has received a lot of attention lately. Starting
from [2, 56] up to the breakthrough by Buckmaster–Shkoller–Vicol [6],
alternative newer proofs by Isett and Ma [55] and the study of the stationary
problem by Cheng–Kwon–Li [16]. It is likely that the scenario crafted by
Bressan and his collaborators for 2D Euler also holds in the gSQG case.

The problem of whether the SQG and gSQG system presents finite time
singularities or there is global existence is one of the most outstanding
problems in mathematical fluid mechanics. Kiselev and Nazarov [59] constructed
solutions for which there was norm inflation, and Friedlander and Shvydkoy
[37] showed the existence of unstable eigenvalues of the spectrum. We also
refer to the construction of Castro and Córdoba of singular solutions with
infinite energy in [9].

Aside from the analytical work, there have been many numerical results in this
framework. Constantin–Majda–Tabak [20] reported a possible singularity in the
form of a hyperbolic saddle closing in finite time. In contrast, Ohkitani and
Yamada [64] and Constantin–Nie–Schorghofer [21] suggested that the growth was
double exponential instead. Córdoba, in his thesis [22] bounded the growth by
a quadruple exponential (see also later work by Córdoba and Fefferman on a
double exponential bound [23] and numerical simulations by Deng–Hou–Li–Yu
[28]), thus settling the question. 20 years later, using improved algorithms
and more computational resources, Constantin et al. [19] recalculated the
hyperbolic saddle scenario, finding no evidence of blowup. Scott, in [75],
starting from elliptical configurations, proposed a candidate that develops
filamentation and after a few cascades, blowup of $\nabla\theta$. This is the
only scenario that still stands up to date in the smooth setting.

A simpler setting for the SQG and gSQG problem is the so-called patch setting,
where the scalar takes the form of a characteristic function of a time-
dependent domain with smooth boundary. Rodrigo (in a $C^{\infty}$ space) [72]
and Gancedo [39] and Constantin et al [15] (in a Sobolev space) proved local
existence for the case $0<\gamma\leq 1$ and $1<\gamma$ respectively. Some 20
years ago, Córdoba–Fontelos–Mancho–Rodrigo [24] found numerically strong
evidence of finite time singularity formation in the form of a self-
intersecting patch where at the point of intersection the curvature blows up
(the curve has to lose regularity at the self-intersection point due to
Gancedo–Strain [38]). They conducted experiments for different values of
$\gamma\in(0,1]$ and reported blowup for all of them, see also [61]. Moreover,
they numerically checked that the singularity was of asymptotically self-
similar type. After them, Scott–Dritschel performed a very careful study using
ellipses as initial condition: On the one hand, in [73], after taking an
ellipse with aspect ratio $a=0.16$, a self-similar cascade of filamentation is
reported, where the filament width collapses to zero in a finite time. On the
other hand, in [74], a more detailed analysis is done (for initial aspect
ratios $a=0.094,0.132,0.180$), and the boundary is split into three scales: an
inner (faster) scale that develops a self-similar corner, a middle scale that
develops a self-similar spiral, and the outermost region. We note that there
is some resemblance between the middle scale of this analysis and the
solutions constructed in this paper which are also self-similar spirals.

Finally, we also mention the formation of singularities in other geometries
(in the presence of a boundary) by Kiselev–Ryzhik–Yao–Zlatoš [58] (for the
case $0<\gamma<\frac{1}{12}$) and Gancedo–Patel [40] (for the case
$0<\gamma<\frac{1}{3}$). We refer also the reader to [1, 10, 13, 25, 27, 41,
42, 46, 47, 48, 49, 51, 54, 53, 63, 70] for other results concerning global
solutions for the gSQG equations.

Regarding negative results of self-similar solutions (i.e. whenever the only
possible solution is the trivial one) Cannone–Xue [8] and Chae [14] proved
that there are no locally self-similar solutions under suitable technical
assumptions, one of them being integrable. Although our solutions decay at
infinity, they do not decay sufficiently fast to be in $L^{q}$ with $q>2$ as
needed in [8].

###  1.4. Main ideas of the proof

Let us give the main ideas of the proof. By inserting the ansatz (6) in the
gSQG equations, we find the equivalent equation for $\hat{\theta}$:

$\left\\{\frac{1+\alpha-\gamma}{1+\alpha}-\frac{2}{1+\alpha}\right\\}\hat{\theta}(z)-\nabla\cdot\left\\{\left(\hat{u}(z)-\frac{z}{1+\alpha}\right)\hat{\theta}(z)\right\\}=0,$
(8)

where the vector field

$q(z):=\hat{u}(z)-\frac{z}{1+\alpha}=\nabla^{\perp}\hat{\psi}(z)-\frac{z}{1+\alpha},$
(9)

is called the pseudo-velocity, and its integral curves are the pseudo-
streamlines. Hence, in order to find nontrivial self-similar solutions one
needs to solve (5).

Motivated by numerous works in the literature about the existence of
nontrivial steady vortex patches in different active scalar equations [7, 26,
27, 41, 42, 43, 44, 45, 46, 48, 51, 52, 70], we will solve (8) via
perturbative arguments by means of the infinite dimensional Implicit Function
theorem. That means that we will look for nontrivial self-similar solutions
which are close to the trivial stationary solution (3). However, perturbative
arguments can not be directly applied to (8) since the linearized operator at
a trivial radial solution (3) is not Fredholm. That comes from the fact that
the solution is singular at $0$. However the kind of singularity at the linear
level of the terms with the radial and angular derivatives is different, which
complicates the choice of the appropriate weighted space. That problem also
appeared when perturbing radial functions in the Euler and gSQG equations [12,
11, 45]. The authors overcame such issue by either performing an appropriate
change of variables by means of the level sets or by working in a suitable
class of solutions.

In this work, we will tackle the lack of Fredholmness of the linarized
operator when working directly with (8) in the physical variables, by using
the adapted coordiantes introduced by Elling [33]. Hence, we will transform
the polar coordiantes $(r,\vartheta)$ into some new ones $(\beta,\phi)$. Such
transformation is a nonlinear change of coordinates which is implicit in the
sense that it depends on the solution itself. Here $\beta\in(0,\infty)$ is the
angle along a particular pseudo-streamline. Moreover $\beta=0$ will correspond
to the spatial infinity while $\beta=+\infty$ approaches the spiral center. On
the other hand, $\phi$ is the polar angle at $r\rightarrow+\infty$. In order
to define $(\beta,\phi)$ we assume that $\partial_{\beta}$ is tangential to
its pseudo-streamlines. Define the angle $\phi$ as follows

$\vartheta=\beta+\phi,\quad\phi\in\operatorname{\mathbb{T}}.$

We will fix $\beta$ such that the pseudo-velocity (9) has vanishing
$\phi$-component. Defining

$\Theta(\beta,\phi):=\hat{\theta}(z(\beta,\phi)),\quad\Psi(\beta,\phi):=\hat{\psi}(z(\beta,\phi)),$

we achieve that

$\displaystyle\left(\nabla^{\perp}\hat{\psi}-\frac{z}{1+\alpha}\right)\cdot\nabla_{z}\hat{\theta}=$
$\displaystyle\left((\beta_{z_{2}}\phi_{z_{1}}-\beta_{z_{1}}\phi_{z_{2}})\Psi_{\phi}-\frac{1}{1+\alpha}(z_{1}\beta_{z_{1}}+z_{2}\beta_{z_{2}}),\right.$
$\displaystyle\left.(\beta_{z_{1}}\phi_{z_{2}}-\beta_{z_{2}}\phi_{z_{1}})\Psi_{\beta}-\frac{1}{1+\alpha}(z_{1}\phi_{z_{1}}+z_{2}\phi_{z_{2}})\right)\cdot\nabla_{\beta,\phi}\Theta.$

Hence, we fix $\beta$ such that $\phi$-component of the pseudo-velocity is 0,
that is:

$(\beta_{z_{1}}\phi_{z_{2}}-\beta_{z_{2}}\phi_{z_{1}})\Psi_{\beta}-\frac{1}{1+\alpha}(z_{1}\phi_{z_{1}}+z_{2}\phi_{z_{2}})=0,$
(10)

which, after some computations agrees with

$r=(-(1+\alpha)\Psi_{\beta})^{\frac{1}{2}}.$

$r$
---
$(z_{1},z_{2})$
---
$(z_{1},z_{2})$
---
$\vartheta$
---
Figure 2. Comparison between the polar coordinates in the $(z_{1},z_{2})$
plane with the $(\beta,\phi)$ adapted coordinates.

We refer to Figure 2 to illustrate the adapted coordinates. Miraculously, when
assuming (10) we have that

$\displaystyle\left(\nabla^{\perp}\hat{\psi}-\frac{z}{1+\alpha}\right)$
$\displaystyle\cdot\nabla_{z}\hat{\theta}=\left\\{(\beta_{z_{2}}\phi_{z_{1}}-\beta_{z_{1}}\phi_{z_{2}})\Psi_{\phi}-\frac{1}{1+\alpha}(z_{1}\beta_{z_{1}}+z_{2}\beta_{z_{2}})\right\\}\Theta_{\beta},$
(11)

where only $\partial_{\beta}$ survives there. That makes that the vorticity
(8) written in the adapted coordinates depend only on the $\beta$-derivative
of $\Theta$. Thanks to that, a trivial solution to (8) is found as

$\Theta(\beta,\phi)=(\Psi_{\varphi})^{-\frac{1+\alpha-\gamma}{2}}\Omega(\phi)=(\Psi_{\varphi})^{-\frac{1/\mu}{2}}\Omega(\phi),$
(12)

for any $\Omega$. However, $\Theta$ depends on the stream function $\Psi$ and
it remains to check the elliptic relation (2) between these two functions.
Written in the adapted coordinates and using

$\Psi(\beta,\phi)=-C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\beta^{\frac{\alpha-1}{1+\alpha}}(1+f(\beta,\phi)),\quad\Omega(\phi)=(1+\alpha)^{-\frac{1}{2\mu}}\tilde{\Omega}(\phi),$

the relation amounts to

$\tilde{F}(f,\tilde{\Omega})(\beta,\phi)=0,\quad(\beta,\phi)\in[0,\infty)\times\operatorname{\mathbb{T}},$

where

$\displaystyle\tilde{F}(f,\tilde{\Omega})(\beta,\phi)=1+f(\beta,\phi)$
$\displaystyle-(-1)^{\frac{-1}{2\mu}}\frac{C_{\gamma}(\alpha-1)^{\frac{\alpha-1}{2}}}{4\pi
C_{0}(1+\alpha)^{\frac{\alpha-1}{2}}}\beta^{-\frac{\alpha-1}{1+\alpha}}$
$\displaystyle\times\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f(b,\Phi)\right\\}^{-\frac{1}{2\mu}}\left\\{\frac{2(\alpha-1)}{(1+\alpha)^{2}}+\partial_{\overline{\beta\varphi}}f(b,\Phi)\right\\}\tilde{\Omega}(\Phi)dbd\Phi}{D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}},$
(13)

and

$\displaystyle D(f)(\beta,\phi,b,\Phi):=$
$\displaystyle\tilde{\Psi}_{\beta}+\tilde{\Psi}_{b}-2\tilde{\Psi}_{\beta}^{\frac{1}{2}}\tilde{\Psi}_{b}^{\frac{1}{2}}\cos(\phi+\beta-\Phi-b)$
(14) $\displaystyle=$
$\displaystyle\beta^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}+b^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}$
$\displaystyle-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\cos(\phi+\beta-\Phi-b)$
$\displaystyle=$
$\displaystyle\left(\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}$
$\displaystyle+2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}$
$\displaystyle\times(1-\cos(\phi+\beta-\Phi-b)).$

There, $\partial_{\varphi}:=\partial_{\phi}-\partial_{\beta}$ and notice that
the previous expression does not depend on $\partial_{\phi}$ which is crucial
for our analysis. Indeed, the choice of the adapted coordinates is the key
point of this work due to two facts: first we can solve the vorticity equation
with (12), and second the nonlinear equation only depends on one derivative as
opposite to (8).

Then, the proof of the main theorem relies on the application of the infinite
dimensional Implicit Function theorem to $\tilde{F}$ in some appropriate
spaces. The choice of the function spaces is a key point in this work since we
should add some appropriate decay for $f$ as $\beta$ approaches $0$ and
$\infty$. Let us point out that the funcional $\tilde{F}$ differs to the
formulation used in [4, 33, 34] when working with the Euler equations since
there the relation between the vorticity and the streamfunction is local, as
opposite to the gSQG equations.

In order to apply the Implicit Function theorem to $\tilde{F}$, we should
check that the linear operator around the trivial solution is an isomorphism.
In particular, we shall check that its kernel is trivial. To that aim, we
apply the Mellin transform to the kernel equation as it was motivated by some
works on coagulation and fragmentation models (see [3, 35]) to obtain a
recurrence equation with singular coefficients. Later, we relate such
recurrence equation with the moment problem concluding that the function in
the kernel is singular and indeed does not belong to our function spaces.
Finally, after showing that the Implicit Function theorem can be applied to
the formulation with the adapted coordiantes, we shall prove the invertibility
of the adapted coordinates when we are close to the trivial solution to come
back to the physical variables.

###  1.5. Organization of the paper

The paper is organized in the following way: Section 2 sets the problem up and
writes the equations in the adapted coordinates. Section 3 defines the
functional spaces in which we will work on and performs estimates at the
nonlinear level. On the contrary, in Section 4 we perform regularity estimates
of the linear operator. Section 5 is devoted to the spectral study and in
particular the invertibility of the linear operator. Finally Sections 6 and 7
comprise the reduction to $m$-fold symmetry classes and the translation of
Theorem 6.2 into standard coordinates. We also include two Appendices:
Appendix A concerns basic results on radial solutions of the fractional
Laplacian, whereas Appendix B deals with basic properties of some special
functions needed throughout the text.

##  2\. Self-similar solutions

In this section, we aim to provide the definition of a self-similar solution
to the gSQG equations together with the formulation that characterizes them.
Note that solutions to the gSQG equations (1) are invariant under the
following scaling: for $\alpha>-1$,

$\theta(t,x)\mapsto\theta_{\lambda}(t,x)=\lambda^{1+\alpha-\gamma}\theta(\lambda^{1+\alpha}t,\lambda
x);\quad u(t,x)\mapsto
u_{\lambda}(t,x)=\lambda^{\alpha}u(\lambda^{1+\alpha}t,\lambda
x),\quad\forall\lambda>0.$

We will say that $\theta$ solution of (1) is self-similar if
$\theta=\theta_{\lambda}$, for any $\lambda>0$. That is,

$\theta(t,x)=\lambda^{1+\alpha-\gamma}\theta(\lambda^{1+\alpha}t,\lambda
x),\quad\forall\lambda>0.$ (15)

Hence, in order to have that the trivial solution (3) is self-similar in the
above way, we have the following condition for the singularity parameter $\mu$
depending on $\alpha$ and $\gamma$:

$\frac{1}{\mu}=1+\alpha-\gamma.$ (16)

By (4) and the relation (16) we find that

$1<\alpha<1+\gamma.$ (17)

Now, by taking $\lambda=1/t^{\frac{1}{1+\alpha}}$ in (15), our aim is to look
for self-similar solutions of the form:

$\theta(t,x)=\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}}}\hat{\theta}\left(\frac{x}{t^{\frac{1}{1+\alpha}}}\right),$
(18)

for some self-similar profile $\hat{\theta}$.

In the following proposition, we show the expression of $u$ with the self-
similar ansatz.

######  Proposition 2.1.

If $\theta$ takes the form (18), then

$u(t,x)=\frac{1}{t^{\frac{\alpha}{1+\alpha}}}\hat{u}\left(\frac{x}{t^{\frac{1}{1+\alpha}}}\right),$

where $\hat{u}=-\nabla^{\perp}(-\Delta)^{-1+\frac{\gamma}{2}}\hat{\theta}$.
Moreover, (1) is equivalent to

$\frac{1+\alpha-\gamma}{1+\alpha}\hat{\theta}(z)+\frac{1}{1+\alpha}z\cdot\nabla\hat{\theta}(z)-\hat{u}\cdot\nabla\hat{\theta}(z)=0,$
(19)

where $z=\frac{x}{t^{\frac{1}{1+\alpha}}}$. In a similar way, let us define
$\hat{\psi}$ as

$\hat{\psi}(z)=-(-\Delta)^{-1+\frac{\gamma}{2}}\hat{\theta}(z).$

###### Proof.

First, let us compute $u$ assuming (18):

$\displaystyle u(t,x)=$
$\displaystyle-\nabla^{\perp}(-\Delta)^{-1+\frac{\gamma}{2}}\theta(t,x)$
$\displaystyle=$
$\displaystyle-\frac{C_{\gamma}\gamma}{4\pi}\int_{\operatorname{\mathbb{R}}^{2}}\frac{(x-y)^{\perp}}{|x-y|^{\gamma+2}}\theta(t,y)dA(y)$
$\displaystyle=$
$\displaystyle\frac{C_{\gamma}\gamma}{4\pi}\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}}}\int_{\operatorname{\mathbb{R}}^{2}}\frac{(x-y)^{\perp}}{|x-y|^{\gamma+2}}\hat{\theta}\left(\frac{y}{t^{\frac{1}{1+\alpha}}}\right)dA(y)$
$\displaystyle=$
$\displaystyle\frac{C_{\gamma}\gamma}{4\pi}\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}}}\frac{t^{\frac{1}{1+\alpha}}}{t^{\frac{\gamma+2}{1+\alpha}}}\int_{\operatorname{\mathbb{R}}^{2}}\frac{\left(\frac{x}{t^{\frac{1}{1+\alpha}}}-\frac{y}{t^{\frac{1}{1+\alpha}}}\right)^{\perp}}{\left|\frac{x}{t^{\frac{1}{1+\alpha}}}-\frac{y}{t^{\frac{1}{1+\alpha}}}\right|^{\gamma+2}}\hat{\theta}\left(\frac{y}{t^{\frac{1}{1+\alpha}}}\right)dA(y)$
$\displaystyle=$
$\displaystyle\frac{C_{\gamma}\gamma}{4\pi}\frac{1}{t^{\frac{\alpha}{1+\alpha}}}\int_{\operatorname{\mathbb{R}}^{2}}\frac{\left(\frac{x}{t^{\frac{1}{1+\alpha}}}-y\right)^{\perp}}{\left|\frac{x}{t^{\frac{1}{1+\alpha}}}-y\right|^{\gamma+2}}\hat{\theta}(y)dA(y)$
$\displaystyle=$
$\displaystyle\frac{1}{t^{\frac{\alpha}{1+\alpha}}}\hat{u}\left(\frac{x}{t^{\frac{1}{1+\alpha}}}\right),$

where $\hat{u}=-\nabla^{\perp}(-\Delta)^{-1+\frac{\gamma}{2}}\hat{\theta}$.

Let us now check (19). Note that

$\displaystyle\partial_{t}\theta(t,x)=$
$\displaystyle-{\frac{1+\alpha-\gamma}{1+\alpha}}\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}+1}}\hat{\theta}\left(\frac{x}{t^{\frac{1}{1+\alpha}}}\right)-\frac{1}{1+\alpha}\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}+1}}\frac{x}{t^{\frac{1}{1+\alpha}}}\cdot(\nabla\hat{\theta})\left(\frac{x}{t^{\frac{1}{1+\alpha}}}\right)$
$\displaystyle=$
$\displaystyle-\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}+1}}\left\\{\frac{1+\alpha-\gamma}{1+\alpha}\hat{\theta}(z)+\frac{1}{1+\alpha}z\cdot\nabla\hat{\theta}(z)\right\\},$
$\displaystyle\nabla\theta(t,x)=$
$\displaystyle\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}}}\frac{1}{t^{\frac{1}{1+\alpha}}}(\nabla\hat{\theta})\left(\frac{x}{t^{\frac{1}{1+\alpha}}}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}}}\frac{1}{t^{\frac{1}{1+\alpha}}}\nabla\hat{\theta}(z).$

Then,

$\displaystyle\partial_{t}\theta(t,x)+(u(t,x)\cdot\nabla)\theta(t,x)=-\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}+1}}\left\\{\frac{1+\alpha-\gamma}{1+\alpha}\hat{\theta}(z)+\frac{1}{1+\alpha}z\cdot\nabla\hat{\theta}(z)-\hat{u}(z)\cdot\nabla\hat{\theta}(z)\right\\},$

yielding to the equivalent equation (19):

$\frac{1+\alpha-\gamma}{1+\alpha}\hat{\theta}(z)+\frac{1}{1+\alpha}z\cdot\nabla\hat{\theta}(z)-\hat{u}(z)\cdot\nabla\hat{\theta}(z)=0.$

Note that since $\nabla\cdot\hat{u}=0$, the previous equation can be written
as

$\left\\{\frac{1+\alpha-\gamma}{1+\alpha}-\frac{2}{1+\alpha}\right\\}\hat{\theta}(z)-\nabla\cdot\left\\{\left(\hat{u}(z)-\frac{z}{1+\alpha}\right)\hat{\theta}(z)\right\\}=0.$

∎

The vector field

$q(z):=\hat{u}(z)-\frac{z}{1+\alpha}=\nabla^{\perp}\hat{\psi}(z)-\frac{z}{1+\alpha},$
(20)

is called the pseudo-velocity, and its integral curves are the pseudo-
streamlines.

###  2.1. Adapted coordinates

Our aim is to perturb the stationary radial solution (3) to find non trivial
self similar solutions with a locally integrable singularity at the origin.
However, perturbation techniques, for instance the Implicit Function theorem
or Crandall-Rabinowitz theorem, of radial solutions to gSQG or Euler equations
seem to be difficult due to the very special structure of the equation.
Roughly speaking, the nonlinear operator and the linear one behave very
differently in terms of the singularity at the origin. That makes difficult
the choice of the appropriate function spaces and leads to main difficulties
in the Fredholm structure of the linearized operator. A similar issue appears
when dealing with uniformly rotating solutions, which could be solved in [11,
12, 45] by restricting the class of solutions or doing an appropriate change
of variables by using the level sets of the vorticity.

Although here we are not dealing with rotating solutions, our equation (19)
has a very similar structure close to the radial solution (3). Moreover, we
will have to deal with extra difficulties due to the singularity of (3) at $0$
and its behavior at $+\infty$. More precisely, we are now working in an
unbounded domain and we need to use some appropriate weights at $0$ and
$+\infty$. In order to overcome the problematic with the Fredholm structure of
the linearized operator, working with some adapted coordinates instead of the
usual polar coordinates seems to be more convenient in this case.

Let $(r,\vartheta)$ be the polar coordinates with $z=re^{i\vartheta}$. Here,
we would like to find some adapted coordinates $(\beta,\phi)$ such that the
pseudo-velocity (20) has vanishing $\phi$-component.

First, define $\phi$ as

$\vartheta=\beta+\phi,\quad\phi\in\operatorname{\mathbb{T}}.$ (21)

In the following proposition we find the nonlinear expression of $\beta$
depending on $(r,\vartheta)$ in order to have vanishing $\phi$-component of
the pseudo-velocity. We shall see that such nonlinear expression is indeed an
implicit function that depends on the solution itself. Denote

$\Theta(\beta,\phi):=\hat{\theta}(z(\beta,\phi)),\quad\Psi(\beta,\phi):=\hat{\psi}(z(\beta,\phi)).$

######  Proposition 2.2.

If $r=(-(1+\alpha)\Psi_{\beta})^{\frac{1}{2}}$ and $\vartheta=\beta+\phi$,
then the $\phi$-component of the pseudo-velocity, that is,
$\nabla^{\perp}\hat{\psi}-\frac{z}{1+\alpha}$, vanishes.

###### Proof.

Let us compute the term
$\left(\nabla^{\perp}\hat{\psi}-\frac{z}{1+\alpha}\right)\cdot\nabla_{z}\hat{\theta}$
in (19) in terms of $(\beta,\phi)$:

$\displaystyle\left(\nabla^{\perp}\hat{\psi}-\frac{z}{1+\alpha}\right)$
$\displaystyle\cdot\nabla_{z}\hat{\theta}=\left(-\hat{\psi}_{z_{2}}-\frac{z_{1}}{1+\alpha}\right)\hat{\theta}_{z_{1}}+\left(\hat{\psi}_{z_{1}}-\frac{z_{2}}{1+\alpha}\right)\hat{\theta}_{z_{2}}$
$\displaystyle=$
$\displaystyle\left\\{-\Psi_{\beta}\beta_{z_{2}}-\Psi_{\phi}\phi_{z_{2}-\frac{z_{1}}{1+\alpha}}\right\\}\left\\{\Theta_{\beta}\beta_{z_{1}}+\Theta_{\phi}\phi_{z_{1}}\right\\}$
$\displaystyle+\left\\{\Psi_{\beta}\beta_{z_{1}}+\Psi_{\phi}\phi_{z_{1}-\frac{z_{2}}{1+\alpha}}\right\\}\left\\{\Theta_{\beta}\beta_{z_{2}}+\Theta_{\phi}\phi_{z_{2}}\right\\}$
$\displaystyle=$
$\displaystyle\left\\{(\beta_{z_{1}}\phi_{z_{2}}-\beta_{z_{2}}\phi_{z_{1}})\Psi_{\beta}-\frac{1}{1+\alpha}(z_{1}\phi_{z_{1}}+z_{2}\phi_{z_{2}})\right\\}\Theta_{\phi}$
$\displaystyle+\left\\{(\beta_{z_{2}}\phi_{z_{1}}-\beta_{z_{1}}\phi_{z_{2}})\Psi_{\phi}-\frac{1}{1+\alpha}(z_{1}\beta_{z_{1}}+z_{2}\beta_{z_{2}})\right\\}\Theta_{\beta}$
$\displaystyle=$
$\displaystyle\left((\beta_{z_{2}}\phi_{z_{1}}-\beta_{z_{1}}\phi_{z_{2}})\Psi_{\phi}-\frac{1}{1+\alpha}(z_{1}\beta_{z_{1}}+z_{2}\beta_{z_{2}}),\right.$
$\displaystyle\left.(\beta_{z_{1}}\phi_{z_{2}}-\beta_{z_{2}}\phi_{z_{1}})\Psi_{\beta}-\frac{1}{1+\alpha}(z_{1}\phi_{z_{1}}+z_{2}\phi_{z_{2}})\right)\cdot\nabla_{\beta,\phi}\Theta.$

We want to fix $\beta$ such that the $\phi$-component of the pseudo-velocity
is 0, that is, $\beta$ will be defined through the following nonlinear
equation

$(\beta_{z_{1}}\phi_{z_{2}}-\beta_{z_{2}}\phi_{z_{1}})\Psi_{\beta}-\frac{1}{1+\alpha}(z_{1}\phi_{z_{1}}+z_{2}\phi_{z_{2}})=0.$
(22)

Note that assuming (22), we have that

$\displaystyle\left(\nabla^{\perp}\hat{\psi}-\frac{z}{1+\alpha}\right)$
$\displaystyle\cdot\nabla_{z}\hat{\theta}=\left\\{(\beta_{z_{2}}\phi_{z_{1}}-\beta_{z_{1}}\phi_{z_{2}})\Psi_{\phi}-\frac{1}{1+\alpha}(z_{1}\beta_{z_{1}}+z_{2}\beta_{z_{2}})\right\\}\Theta_{\beta},$

where only $\partial_{\beta}$ survives there, which will be the crucial in
order to find some trivial solution to the vorticity equation.

We can also compute the Jacobian of the transformation as follows

$\left[\begin{array}[]{cc}z_{1,\beta}&z_{1,\phi}\\\
z_{2,\beta}&z_{2,\phi}\end{array}\right]=\left[\begin{array}[]{cc}\beta_{z_{1}}&\beta_{z_{2}}\\\
\phi_{z_{1}}&\phi_{z_{2}}\end{array}\right]^{-1}=\frac{1}{\beta_{z_{1}}\phi_{z_{2}}-\beta_{z_{2}}\phi_{z_{1}}}\left[\begin{array}[]{cc}\phi_{z_{2}}&-\beta_{z_{2}}\\\
-\phi_{z_{1}}&\beta_{z_{1}}\end{array}\right].$ (23)

and using (22)–(23), we find

$\Psi_{\beta}+\frac{1}{1+\alpha}z\times
z_{\beta}=\Psi_{\beta}+\frac{1}{1+\alpha}(z_{1}z_{2,\beta}-z_{2}z_{1,\beta})=0.$

By virtue of the variables $\vartheta(\beta,\phi)=\beta+\phi$ and
$z=z(r,\vartheta)$, we find that

$z\times z_{\beta}=|z|^{2},$

which amounts to

$\Psi_{\beta}=-\frac{1}{(1+\alpha)}|z|^{2}.$

Hence, the map $(\beta,\phi)\mapsto(r,\vartheta)$ takes the form

$r=(-(1+\alpha)\Psi_{\beta})^{\frac{1}{2}},\quad\vartheta=\beta+\phi.$

That concludes the proof of this proposition. ∎

Let us remark again that the change of variables of the previous proposition
is a nonlinear change of variables that depends on the solution, indeed, it
depends on the stream function. Later, in Section 7.1 we will check that we
can come back to the physical variables and thus the change of variables is
invertible. Moreover, from now on define

$\partial_{\varphi}=\partial_{\phi}-\partial_{\beta}.$ (24)

In the following, let us compute the jacobian of the transformation.

######  Proposition 2.3.

The jacobian of the transformation $(\beta,\phi)\mapsto(z_{1},z_{2})$ is given
by

$|J|=\frac{1+\alpha}{2}\Psi_{\beta\varphi},$

where $\partial_{\varphi}=\partial_{\phi}-\partial_{\beta}$.

###### Proof.

Note that the matrix of the transformation is given by

$J=\left[\begin{array}[]{cc}z_{1,\beta}&z_{1,\phi}\\\
z_{2,\beta}&z_{2,\phi}\end{array}\right].$

From Proposition 2.2 we have that

$r=(-(1+\alpha)\Psi_{\beta})^{\frac{1}{2}},\quad\vartheta=\beta+\phi,$

and using such expression, we can compute the following

$\displaystyle\partial_{\phi}z_{1}=$
$\displaystyle\partial_{r}z_{1}\partial_{\phi}r+\partial_{\vartheta}z_{1}\partial_{\phi}\theta$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\cos(\vartheta)(-(1+\alpha)\Psi_{\beta})^{-\frac{1}{2}}(1+\alpha)\partial^{2}_{\phi\beta}\Psi-r\sin(\vartheta),$
$\displaystyle\partial_{\beta}z_{1}=$
$\displaystyle-\frac{1}{2}\cos(\vartheta)(-(1+\alpha)\Psi_{\beta})^{-\frac{1}{2}}(1+\alpha)\partial^{2}_{\beta\beta}\Psi-r\sin(\vartheta),$
$\displaystyle\partial_{\phi}z_{2}=$
$\displaystyle-\frac{1}{2}\sin(\vartheta)(-(1+\alpha)\Psi_{\beta})^{-\frac{1}{2}}(1+\alpha)\partial^{2}_{\phi\beta}\Psi+r\cos(\vartheta),$
$\displaystyle\partial_{\beta}z_{2}=$
$\displaystyle-\frac{1}{2}\sin(\vartheta)(-(1+\alpha)\Psi_{\beta})^{-\frac{1}{2}}(1+\alpha)\partial^{2}_{\beta\beta}\Psi+r\cos(\vartheta).$

Then, after some straightforward computations, we finally find that

$\displaystyle|J|=$ $\displaystyle
z_{1,\beta}z_{2,\phi}-z_{1,\phi}z_{2,\beta}=\frac{1+\alpha}{2}\Psi_{\beta\varphi}.$

∎

###  2.2. The equation in the new coordinates

In Section 2.1 we have defined the adapted coordinates that will be used
through this work. Recall that we have chosen such coordinates in order to
have that the $\phi$ component of the pseudo-velocity is vanishing. In this
way, we find

$\left(\nabla^{\perp}\hat{\psi}-\frac{z}{1+\alpha}\right)\cdot\nabla\hat{\theta}=\left\\{(\beta_{z_{2}}\phi_{z_{1}}-\beta_{z_{1}}\phi_{z_{2}})\Psi_{\phi}-\frac{1}{1+\alpha}(z_{1}\beta_{z_{1}}+z_{2}\beta_{z_{2}})\right\\}\Theta_{\beta},$

and then the vorticity equation (19) agrees with

$\frac{1+\alpha-\gamma}{1+\alpha}\Theta(\beta,\phi)=\left\\{(\beta_{z_{2}}\phi_{z_{1}}-\beta_{z_{1}}\phi_{z_{2}})\Psi_{\phi}-\frac{1}{1+\alpha}(z_{1}\beta_{z_{1}}+z_{2}\beta_{z_{2}})\right\\}\Theta_{\beta}.$
(25)

The very special structure of (25) involving only a $\beta$ derivative of
$\Theta$ gives us a trivial solution. That will be discussed in the following
result.

######  Proposition 2.4.

Equation (25) agrees with

$(1+\alpha-\gamma)\Theta(\beta,\phi)\Psi_{\beta\varphi}(\beta,\phi)+2\Psi_{\varphi}(\beta,\phi)\Theta_{\beta}(\beta,\phi)=0,$
(26)

which has the following trivial solution:

$\Theta(\beta,\phi)=(\Psi_{\varphi})^{-\frac{1+\alpha-\gamma}{2}}\Omega(\phi)=(\Psi_{\varphi})^{-\frac{1/\mu}{2}}\Omega(\phi),$
(27)

for any $\Omega$.

###### Proof.

Let us start from (25) taking into account that

$\beta_{z_{2}}\phi_{z_{1}}-\beta_{z_{1}}\phi_{z_{2}}=-|J^{-1}|=-\frac{1}{|J|}=-\frac{2}{(1+\alpha)\Psi_{\beta\varphi}},$

that is, (25) agrees with

$\frac{1+\alpha-\gamma}{1+\alpha}\Theta(\beta,\phi)\Psi_{\beta\varphi}+\left\\{\frac{2}{1+\alpha}\Psi_{\phi}+\frac{\Psi_{\beta\varphi}}{1+\alpha}(z_{1}\beta_{z_{1}}+z_{2}\beta_{z_{2}})\right\\}\Theta_{\beta}=0.$
(28)

Let us now compute
$z_{1}\beta_{z_{1}}+z_{2}\beta_{z_{2}}=(z_{1},z_{2})\cdot\nabla_{z}\beta$.
Note that

$J^{-1}=\left[\begin{array}[]{cc}\beta_{z_{1}}&\beta_{z_{2}}\\\
\phi_{z_{1}}&\phi_{z_{2}}\end{array}\right]=\frac{1}{|J|}\left[\begin{array}[]{cc}z_{2,\phi}&-z_{1,\phi}\\\
-z_{2,\beta}&z_{1,\beta}\end{array}\right].$ (29)

Then:

$\displaystyle\beta_{z_{1}}=$
$\displaystyle\frac{2}{(1+\alpha)\Psi_{\beta\varphi}}\left\\{-\frac{\sin(\vartheta)(1+\alpha)}{2r}\partial^{2}_{\phi\beta}\Psi+r\cos(\vartheta)\right\\},$
$\displaystyle\beta_{z_{2}}=$
$\displaystyle-\frac{2}{(1+\alpha)\Psi_{\beta\varphi}}\left\\{-\frac{\cos(\vartheta)(1+\alpha)}{2r}\partial^{2}_{\phi\beta}\Psi-r\sin(\vartheta)\right\\}.$

After some straightforward computations we find

$z_{1}\beta_{z_{1}}+z_{2}\beta_{z_{2}}=\frac{2r^{2}}{(1+\alpha)\Psi_{\beta\varphi}}=-2\frac{\Psi_{\beta}}{\Psi_{\beta\varphi}}.$

Putting everything together, one achieve that (28) is equivalent to (26), that
is:

$\displaystyle(1+\alpha-\gamma)\Theta(\beta,\phi)\Psi_{\beta\varphi}(\beta,\phi)+2\Psi_{\varphi}(\beta,\phi)\Theta_{\beta}(\beta,\phi)=0.$

Note that it has the following trivial solution:

$\Theta(\beta,\phi)=(\Psi_{\varphi})^{-\frac{1+\alpha-\gamma}{2}}\Omega(\phi),$

for any function $\Omega$. ∎

Via the adapted coordinates, we have simplified the nonlinear equation (19)
into (26) where a trivial solution is found by (27). However, it does not take
into account the relation between $\hat{\theta}$ and $\hat{\psi}$ via the
singular integral operator (2), and this will be our aim from now on. Note
that

$\hat{\psi}(z)=-\frac{C_{\gamma}}{2\pi}\int_{\operatorname{\mathbb{R}}^{2}}\frac{\hat{\theta}(y)}{|x-y|^{\gamma}}dA(y).$
(30)

By writing (30) with the adapted coordinates with the help of the previous
computation on the jacobian of the transformation in Proposition 2.3 together
with the trivial solution from in Proposition 2.4, we find that it agrees with

$\Psi(\beta,\phi)=\frac{C_{\gamma}(1+\alpha)^{1-\gamma/2}}{4\pi}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{(\Psi_{\varphi}(b,\Phi))^{-\frac{1+\alpha-\gamma}{2}}\Omega(\Phi)\Psi_{b\varphi}(b,\Phi)dbd\Phi}{\\{-\Psi_{\beta}-\Psi_{b}+2\Psi_{\beta}^{\frac{1}{2}}\Psi_{b}^{\frac{1}{2}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}.$
(31)

In the previous expression we need to use that $r=0$ corresponds to
$\beta=+\infty$ and $r=+\infty$ agrees with $\beta=0$ as we shall see later,
which is the reason we have a change of sign in the integral (31). Hence, our
aim is to find functions $\Psi$ and $\Omega$ such that

$\displaystyle F(\Psi,\Omega)$ $\displaystyle(\beta,\phi):=\Psi(\beta,\phi)$
$\displaystyle-\frac{C_{\gamma}(1+\alpha)^{1-\gamma/2}}{4\pi}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{(\Psi_{\varphi}(b,\Phi))^{-\frac{1}{2\mu}}\Omega(\Phi)\Psi_{b\varphi}(b,\Phi)dbd\Phi}{\\{-\Psi_{\beta}-\Psi_{b}+2\Psi_{\beta}^{\frac{1}{2}}\Psi_{b}^{\frac{1}{2}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}=0,$
(32)

for any $\beta\in[0,+\infty)$ and $\phi\in\operatorname{\mathbb{T}}$. In the
previous expression we have used that $\frac{1}{\mu}=1+\alpha-\gamma$ coming
from (16).

###  2.3. Trivial solution

The problem reduces to study the nonlinear equation (2.2) which is nothing but
the relation between $\Omega$ and $\Psi$. In this section, we look for trivial
solutions of such equation. That trivial solutions comes from the fact that
any radial function $\theta$ provides a stationary solution of (1), where the
stream function is also radial. Here, we will use the computations done in
Appendix A for the fractional Laplacian.

######  Proposition 2.5.

If

$\Omega^{0}=\frac{1}{(1+\alpha)^{\frac{1}{2\mu}}},$

and

$\Psi^{0}(\beta)=-C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\beta^{-\frac{1-\alpha}{1+\alpha}}.$
(33)

then $F(\Psi^{0},\Omega^{0})(\beta,\phi)=0$, for any $\beta\in[0,\infty)$ and
$\phi\in\operatorname{\mathbb{T}}$, where $C_{0}$ is defined in (A.2).
Moreover, the adapted coordinates $(r,\vartheta)\mapsto(\beta,\phi)$ at the
trivial solution reads as follows

$\beta=C_{0}(\alpha-1)r^{-1-\alpha},\quad\phi=\vartheta-
C_{0}(\alpha-1)r^{-1-\alpha}.$

###### Proof.

The trivial solution that we will construct comes from the stationary one

$\theta(x)=|x|^{-\frac{1}{\mu}},$

for $\frac{1}{2}<\mu<\frac{1}{2-\gamma}$. From Corollary A.2 we have that
$\psi$ is also radial and is given by

$\psi(x)=-C_{0}|x|^{2-\gamma-\frac{1}{\mu}}=-C_{0}|x|^{1-\alpha},$

by using (16) and where $C_{0}$ is defined in (A.2). Using (18), we have that
$\hat{\theta}(z)=|z|^{-\frac{1}{\mu}}$ and
$\hat{\psi}(z)=-C_{0}|z|^{1-\alpha}$. Let us now obtain such solution in terms
of $(\beta,\phi)$.

First, let us find the explicit relation between $\beta$ and $r$ via the
nonlinear change of variables in Proposition 2.2. Indeed we have that
$\Psi_{\beta}=-\frac{r^{2}}{1+\alpha}$ and then

$\displaystyle\partial_{r}\hat{\psi}(z)=C_{0}(\alpha-1)r^{-\alpha}=\partial_{r}\beta\partial_{\beta}\Psi=-\partial_{r}\beta\frac{r^{2}}{1+\alpha}.$

Hence

$\partial_{r}\beta=-C_{0}(1+\alpha)(\alpha-1)r^{-2-\alpha},$

implying

$\beta=C_{0}(\alpha-1)r^{-1-\alpha},$ (34)

which amounts to

$r=\left(C_{0}(\alpha-1)\right)^{\frac{1}{1+\alpha}}\beta^{-\frac{1}{1+\alpha}}.$

Since $\alpha>1$ and $r\in[0,\infty)$ we get that $\beta\in[0,\infty)$. In
this way, we can compute $\Psi$ depending on $\beta$ as

$\Psi^{0}(\beta)=-C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\beta^{-\frac{1-\alpha}{1+\alpha}}.$

Let us now compute $\Omega$. Note that

$\Theta(\beta)=\left(\frac{1}{C_{0}(\alpha-1)}\right)^{\frac{1/\mu}{1+\alpha}}\beta^{\frac{1/\mu}{1+\alpha}},$

and using the relation (27), we find

$\Omega^{0}=\frac{1}{(1+\alpha)^{\frac{1}{2\mu}}}.$

That concludes the proof. ∎

######  Remark 2.6.

The solution that we aim to construct are perturbations of the singular radial
stationary solution $\theta(x)=|x|^{-\frac{1}{\mu}}$. The associated initial
stream function is also radial and can be computed via Corollary A.2 as
$\psi(x)=-C_{0}|x|^{2-\gamma-\frac{1}{\mu}}=-C_{0}|x|^{1-\alpha}$. Note that
both the potential temperature $\theta$ and the associated stream function
have a singularity when $|x|=0$ since $\alpha>1$ via (17). However, the change
of variables $(r,\vartheta)\mapsto(\beta,\phi)$ at the trivial solution gives
us $\beta=C_{0}(1-\alpha)r^{-1-\alpha}$. Hence, $\theta$ and $\psi$ written in
terms of $\beta$ are not anymore singular at $\beta=0$, but as
$\beta\rightarrow+\infty$.

###  2.4. Formulation with the appropriate scaling

Our aim is to construct solutions which are perturbations of the trivial one
(33). Since $\alpha>1$ one has that $\Psi^{0}$ in (33) is vanishing at $0$ and
diverges at $+\infty$. We shall use similar scaling for the perturbations.
First, we shall simplify the constant in $\Psi$ and $\Omega$ by working with
$\tilde{\Psi}$ and $\tilde{\Omega}$ instead:

$\Psi(\beta,\phi)=-C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\tilde{\Psi}(\beta,\phi),$

and

$\Omega(\phi)=(1+\alpha)^{-\frac{1}{2\mu}}\tilde{\Omega}(\phi).$

In that way, (2.2) reads as

$\displaystyle\frac{F(\Psi,\Omega)(\beta,\phi)}{-C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}}=\tilde{\Psi}(\beta,\phi)$
$\displaystyle-(-1)^{-\frac{1}{2\mu}+\frac{\gamma}{2}}\frac{C_{\gamma}(1+\alpha)^{-\frac{\alpha-1}{2}}(\alpha-1)^{\frac{\alpha-1}{2}}}{4\pi
C_{0}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{(\tilde{\Psi}_{\varphi}(b,\Phi))^{-\frac{1}{2\mu}}\tilde{\Omega}(\Phi)\tilde{\Psi}_{b\varphi}(b,\Phi)dbd\Phi}{\\{-\tilde{\Psi}_{\beta}-\tilde{\Psi}_{b}+2\tilde{\Psi}_{\beta}^{\frac{1}{2}}\tilde{\Psi}_{b}^{\frac{1}{2}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}=0.$

Hence, now take $f$ as

$\tilde{\Psi}(\beta,\phi)=\beta^{\frac{\alpha-1}{1+\alpha}}\left(1+f(\beta,\phi)\right),$

and we divide all the equation by $\beta^{\frac{\alpha-1}{1+\alpha}}$ getting
the equivalence formulation

$\displaystyle\tilde{F}(f,\tilde{\Omega})(\beta,\phi):=\beta^{-\frac{\alpha-1}{1+\alpha}}\frac{F(\Psi,\Omega)(\beta,\phi)}{-C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}}=1+f(\beta,\phi)$
$\displaystyle-(-1)^{-\frac{1}{2\mu}+\frac{\gamma}{2}}\frac{C_{\gamma}(1+\alpha)^{-\frac{\alpha-1}{2}}(\alpha-1)^{\frac{\alpha-1}{2}}}{4\pi
C_{0}}\beta^{-\frac{\alpha-1}{1+\alpha}}$
$\displaystyle\times\int_{0}^{2\pi}\int_{0}^{\infty}\frac{(\tilde{\Psi}_{\varphi}(b,\Phi))^{-\frac{1}{2\mu}}\tilde{\Omega}(\Phi)\tilde{\Psi}_{b\varphi}(b,\Phi)dbd\Phi}{\\{-\tilde{\Psi}_{\beta}-\tilde{\Psi}_{b}+2\tilde{\Psi}_{\beta}^{\frac{1}{2}}\tilde{\Psi}_{b}^{\frac{1}{2}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}=0.$

We can compute $\partial_{\varphi}\tilde{\Psi}$,
$\partial_{\beta}\tilde{\Psi}$ and $\partial_{\beta\varphi}\tilde{\Psi}$ in
terms of $f$ as follows

$\displaystyle\partial_{\beta}\tilde{\Psi}(\beta,\phi)=$
$\displaystyle\beta^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\},$
$\displaystyle\partial_{\varphi}\tilde{\Psi}(\beta,\phi)=$
$\displaystyle\beta^{\frac{-2}{1+\alpha}}\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f(\beta,\phi)\right\\},$
$\displaystyle\partial_{\beta\varphi}\tilde{\Psi}(\beta,\phi)=$
$\displaystyle\beta^{-1}\beta^{\frac{-2}{1+\alpha}}\left\\{\frac{2(\alpha-1)}{(1+\alpha)^{2}}+\partial_{\overline{\beta\varphi}}f\right\\},$
(35)

where the linear operators
$\partial_{\overline{\beta}},\partial_{\overline{\varphi}}$ and
$\partial_{\overline{\beta\varphi}}$ reads as

$\displaystyle\partial_{\overline{\beta}}f(\beta,\phi):=$
$\displaystyle\frac{\alpha-1}{1+\alpha}f(\beta,\phi)+\beta
f_{\beta}(\beta,\phi),$ (36)
$\displaystyle\partial_{\overline{\varphi}}f(\beta,\phi):=$
$\displaystyle-\frac{\alpha-1}{1+\alpha}f(\beta,\phi)+\beta
f_{\varphi}(\beta,\phi),$ (37)
$\displaystyle\partial_{\overline{\beta\varphi}}f:=$
$\displaystyle\frac{\alpha-1}{1+\alpha}f+\partial_{\overline{\beta}}\partial_{\overline{\varphi}}f$
$\displaystyle=$
$\displaystyle\frac{2(\alpha-1)}{(1+\alpha)^{2}}f+\frac{\alpha-1}{1+\alpha}\beta(f_{\varphi}-f_{\beta})+\beta^{2}f_{\beta\varphi}.$
(38)

Hence, $\tilde{F}$ reads as follows

$\displaystyle\tilde{F}(f,\tilde{\Omega})(\beta,\phi)=1+f(\beta,\phi)$
$\displaystyle-(-1)^{\frac{-1}{2\mu}}\frac{C_{\gamma}(\alpha-1)^{\frac{\alpha-1}{2}}}{4\pi
C_{0}(1+\alpha)^{\frac{\alpha-1}{2}}}\beta^{-\frac{\alpha-1}{1+\alpha}}$
$\displaystyle\times\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f(b,\Phi)\right\\}^{-\frac{1}{2\mu}}\left\\{\frac{2(\alpha-1)}{(1+\alpha)^{2}}+\partial_{\overline{\beta\varphi}}f(b,\Phi)\right\\}\tilde{\Omega}(\Phi)dbd\Phi}{D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}},$
(39)

where

$\displaystyle D(f)(\beta,\phi,b,\Phi):=$
$\displaystyle\tilde{\Psi}_{\beta}+\tilde{\Psi}_{b}-2\tilde{\Psi}_{\beta}^{\frac{1}{2}}\tilde{\Psi}_{b}^{\frac{1}{2}}\cos(\phi+\beta-\Phi-b)$
(40) $\displaystyle=$
$\displaystyle\beta^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}+b^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}$
$\displaystyle-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\cos(\phi+\beta-\Phi-b)$
$\displaystyle=$
$\displaystyle\left(\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}$
$\displaystyle+2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}$
$\displaystyle\times(1-\cos(\phi+\beta-\Phi-b)).$

Let us remark that in the expression of $\tilde{F}$ no derivatives in $\phi$
appear. Indeed, it only depends on $\partial_{\beta}$ and
$\partial_{\varphi}$. In that way, we will not assume any boundedness of
$\partial_{\phi}$ in the function spaces. Moreover, since we are working the
an unbounded space $[0,\infty)$ and the trivial solution is singular at
$\beta=0$, we shall insert some weights in the function spaces as we will see
in the following section.

##  3\. Functional regularity

###  3.1. Function spaces

Let us describe the function spaces for the scaled perturbation $f$. Indeed,
we will take that $f$ is bounded and has some appropriate decay at $0$ and
$+\infty$. We need such a decay for $f$ in order to tackle the singularity at
$\beta=0$ and $\beta\rightarrow+\infty$ in the nonlinear equation.

For $\sigma>0$ (which will be fixed later), let us define the weighted norm

$||f||_{L^{\infty}_{\sigma}}:=\sup_{\beta\in[0,+\infty),\phi\in[0,2\pi]}\frac{(1+\beta)^{2\sigma}}{\beta^{\sigma}}|f(\beta,\phi)|.$
(41)

We will assume that $f$ is Hölder continuous in the variable $\beta$ and then
define

$||f||_{C^{\delta}_{\beta}}:=\sup_{\phi\in[0,2\pi]}\sup_{|\beta-b|<1}|\beta+b|^{\delta}\frac{|f(\beta,\phi)-f(b,\phi)|}{|\beta-b|^{\delta}},$
(42)

and Lipschitz continuous in the periodic variable $\phi$:

$||f||_{\textnormal{Lip}_{\phi}}:=\sup_{\beta\in[0,+\infty)}\sup_{|\phi-\Phi|<1}\frac{|f(\beta,\phi)-f(\beta,\Phi)|}{\sin((\Phi-\phi)/2)}.$
(43)

Hence, let us define

$\mathcal{X}^{\sigma,\delta}:=\left\\{f:(\beta,\phi)\in[0,+\infty)\times[0,2\pi]\rightarrow\operatorname{\mathbb{C}},\quad||f||_{\mathcal{X}^{\sigma,\delta}}:=||f||_{L^{\infty}_{\sigma}}+||f||_{C^{\delta}_{\beta}}<+\infty\right\\}.$
(44)

Note that the functions $f$ lying in $\mathcal{X}^{\sigma,\delta}$ satisfy

$f\sim_{0}\beta^{\sigma},\quad f\sim_{\infty}\beta^{-\sigma},$

which will be crucial to work with the singularity at $0$ and $+\infty$.

Now, we can define the function spaces for $f$:

${X}_{1}^{\sigma,\delta}:=\\{f,\quad f,\beta f_{\beta},\beta
f_{\varphi},\beta^{2}f_{\beta\varphi}\in\mathcal{X}^{\sigma,\delta},||f||_{\textnormal{Lip}_{\phi}}+||\beta
f_{\beta}||_{\textnormal{Lip}_{\phi}}<+\infty,f(\beta,\phi)=\sum_{k\in\operatorname{\mathbb{Z}}}f_{k}(\beta)e^{ik\phi}\\},$
(45)

with the norm

$||f||_{X_{1}^{\sigma,\delta}}:=||f||_{\mathcal{X}^{\sigma,\delta}}+||\beta
f_{\beta}||_{\mathcal{X}^{\sigma,\delta}}+||\beta
f_{\varphi}||_{\mathcal{X}^{\sigma,\delta}}+||\beta^{2}f_{\varphi\beta}||_{\mathcal{X}^{\sigma,\delta}}+||f||_{\textnormal{Lip}_{\phi}}+||\beta
f_{\beta}||_{\textnormal{Lip}_{\phi}}.$ (46)

Notice that the expression of the nonlinear operator (2.4) only involves
derivatives of the type $\partial_{\varphi}$ and $\partial_{\beta}$, and not
$\partial_{\phi}$. For that reason, we only consider such derivatives in the
function space for the perturbations (45) and we will take into account that
$\partial_{\phi}=\partial_{\varphi}+\partial_{\beta}$. However, we shall ask
for Lipschitz continuity in the $\phi$ variable in order to tackle the
singularity of the denominator in (2.4).

Moreover, let us denote by $B_{X_{1}^{\sigma,\delta}}(\epsilon)$ the ball in
$X_{1}^{\sigma,\delta}$ centered at 0 and radius $\epsilon$. Let us remark
that $1\notin X_{1}^{\sigma,\delta}$ (this is the associated trivial solution
after the scaling in (2.4)) because it does not have the same decay at $0$ and
$+\infty$. Hence, we will have to subtract the contribution from the trivial
solution from the nonlinear operator to obtain its well posedness.

######  Remark 3.1.

For $f\in X_{1}^{\sigma,\delta}$, one has that
$\partial_{\overline{\beta}}f,\partial_{\overline{\varphi}}f,\partial_{\overline{\beta\varphi}}f\in\mathcal{X}^{\sigma,\delta}$,
where those linear operators are defined in (36)–(38).

We will assume that $\Omega$ lies in the Lebesgue space $L^{p}$ for some $p$
that will be fixed later. Then, define

$X_{2}^{p}:=\\{f\in L^{p}(0,2\pi),\quad f(\phi)=1+\sum_{0\neq
k\in\operatorname{\mathbb{Z}}}f_{k}e^{ik\phi}\\}.$ (47)

######  Remark 3.2.

Our goal is to prove self-similar solutions for a large family of
perturbations of the initial vorticity. Hence, we need to assume that $\Omega$
lies in a large space. Due to the regularity of the nonlinear function
$\tilde{F}$, we realized that the minimum condition to have that all the
analysis works is that $\Omega\in L^{p}$, for some appropriate $p$ that will
depend on $\gamma$.

Finally, we set the range space:

$Y^{\sigma,\delta}:=\\{f,\quad f,\beta
f_{\varphi},\in\mathcal{X}^{\sigma,\delta},||f||_{\textnormal{Lip}_{\phi}}<+\infty\\},$
(48)

with the norm

$||f||_{Y^{\sigma,\delta}}:=||f||_{\mathcal{X}^{\sigma,\delta}}+||\beta
f_{\varphi}|_{\mathcal{X}^{\sigma,\delta}}+||f||_{\textnormal{Lip}_{\phi}}.$

From now on, let us denote by $B_{X}(c,r)$ the ball in $X$ centered at $c$ of
radius $r$.

######  Remark 3.3.

Note that any function $f\in X_{1}^{\sigma,\delta}$ can be decompose in
Fourier series in $\phi$, that is, there exists $f_{k}(\beta)$ such that

$f(\beta,\phi)=\sum_{k\in\operatorname{\mathbb{Z}}}f_{k}(\beta)e^{ik\phi}.$

Although we are looking for a real function $f$, which will be the
perturbation of the initial stream function, we will consider it initially as
a complex function. Later, we can just include in the function spaces that $f$
is real.

Moreover, we can translate the decay properties of $f$ into the coefficients
$f_{k}$. In that way, we find that $f_{k}\in\mathcal{X}_{0}$ where

$\displaystyle\mathcal{X}_{0}:=$ $\displaystyle\left\\{f,\quad f,\beta
f^{\prime}\in\mathcal{X}^{\sigma,\delta}\right\\}.$ (49)

However, let us remark that functions with Fourier coefficients in
$\mathcal{X}_{0}$ are not necessary in $X_{1}^{\sigma,\delta}$. We can do the
same with $f\in Y^{\sigma,\delta}$ having that the associated Fourier
coefficients lie in the following space:

$\displaystyle\mathcal{Y}_{0}:=$ $\displaystyle\left\\{f,\quad
f\in\mathcal{X}^{\sigma,\delta}\right\\}.$ (50)

These spaces will be used in order to check that the kernel of the linearized
operator is trivial.

######  Remark 3.4.

The function spaces used in [4, 33, 34] for the Euler equations can not be
used here. This is due to the different formulation of the nonlinear function
$F$ used in these papers. Note that the relation between the stream function
$\psi$ and the vorticity $\omega$ in the Euler equations is given via a local
relation. There, there is no need to use the nonlocal expression via the
singular integral, as we have to do for gSQG equations.

######  Remark 3.5.

We consider that $\Omega$ is of the form

$\Omega(\phi)=1+\sum_{0\neq k\in\operatorname{\mathbb{Z}}}g_{k}e^{ik\phi},$

where $1$ is the trivial solution and then no perturbation of the mode $k=0$
is done there. That ensures us that we do not obtain more trivial solutions
and that the solution is non radial.

######  Remark 3.6.

Function spaces with weights are commonly used when working in an unbounded
space. Moreover, such weight must be consider when taking any derivative. That
happens also in [32], where the authors needs to include some weight even for
the Hölder norm. Moreover, note that in the case that $f$ only depends on the
angle $\phi$ (and not on $\beta$), then these are the usual spaces
$C^{1},L^{\infty}$,…

###  3.2. Deformation of the Euclidean norm

In order to study the regularity of the nonlinear function (2.4), we shall
work with the denominator given in (40). Such denominator is nothing but the
euclidean norm deformed by the adapted coordinates and the scaling. That
happened also in [43, 44], where the authors performed some cylindrical
coordinates amounting to a deformation of the euclidean norm. Hence,
estimating by below the denominator, i.e.,

$\displaystyle D(f)(\beta,\phi,b,\Phi):=$
$\displaystyle\tilde{\Psi}_{\beta}+\tilde{\Psi}_{b}-2\tilde{\Psi}_{\beta}^{\frac{1}{2}}\tilde{\Psi}_{b}^{\frac{1}{2}}\cos(\phi+\beta-\Phi-b)$
(51) $\displaystyle=$
$\displaystyle\left(\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}$
$\displaystyle+2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}$
$\displaystyle\times(1-\cos(\phi+\beta-\Phi-b)),$

is crucial to study the regularity of (2.4). That is the goal of the following
proposition.

######  Proposition 3.7.

There exists $\epsilon>0$ such that if $f\in
B_{X_{1}^{\sigma,\delta}}(0,\epsilon)$ then

$\displaystyle D(f)(\beta,\phi,b,\Phi)\geq$ $\displaystyle
C(\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}})\left\\{\frac{|\beta-b|^{2}}{|\beta+b|^{2}}+\sin^{2}\left((\beta-b+\phi-\Phi)/2\right)\right\\}.$
(52)

######  Remark 3.8.

In particular, we find that $D(f)(\beta,\phi,b,\Phi)\geq
CD(0)(\beta,\phi,b,\Phi)$. Moreover, we also get that
$D(f)(\beta,\phi,b,\Phi)\leq CD(0)(\beta,\phi,b,\Phi)$.

###### Proof.

Take $f$ such that $f\in B_{X_{1}^{\sigma,\delta}}(\epsilon)$. In particular
$\partial_{\overline{\beta}}f\in\mathcal{X}^{\sigma,\delta}$ and
$||\partial_{\overline{\beta}}f||_{\textnormal{Lip}_{\phi}}<+\infty$.

Hence, we have that

$C^{-1}<\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}<C,$

and

$C^{-1}<\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}<C.$

That, by symmetry, implies

$\displaystyle D(f)(\beta,\phi,b,\Phi)\geq$
$\displaystyle\left(\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}$
$\displaystyle+C^{-1}(\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}})(1-\cos(\phi+\beta-\Phi-b)),$

for $C$ uniformly in $b,\beta,\phi,\Phi$. Let us then work with the first
part. Note that

$\displaystyle\left|\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right|$
$\displaystyle=\left|\frac{\beta^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}-b^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}}{\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}+b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}}\right|$
$\displaystyle\gtrsim\frac{\left|\beta^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}-b^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}\right|}{\beta^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}}}.$

On the other hand, we have

$\displaystyle\left|\beta^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}-b^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}\right|\gtrsim|\beta^{\frac{-2}{1+\alpha}}-b^{\frac{-2}{1+\alpha}}|$
$\displaystyle-(\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}})|\partial_{\overline{\beta}}f(\beta,\phi)-\partial_{\overline{b}}f(b,\Phi)|$
$\displaystyle\gtrsim$
$\displaystyle|\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}|\left\\{\frac{|\beta-b|}{\beta+b}-|\partial_{\overline{\beta}}f(\beta,\phi)-\partial_{\overline{b}}f(b,\Phi)|\right\\}.$

Then, we find

$\displaystyle\left|\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right|$
$\displaystyle\gtrsim|\beta^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}}|\left\\{\frac{|\beta-b|}{\beta+b}-|\partial_{\overline{\beta}}f(\beta,\phi)-\partial_{\overline{b}}f(b,\Phi)|\right\\},$

which implies

$\displaystyle\left|\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right|^{2}$
$\displaystyle\gtrsim|\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}|\left\\{\frac{|\beta-b|^{2}}{|\beta+b|^{2}}-|\partial_{\overline{\beta}}f(\beta,\phi)-\partial_{\overline{b}}f(b,\Phi)|^{2}\right\\}.$

That is,

$\displaystyle D(f)(\beta,\phi,b,\Phi)\gtrsim$
$\displaystyle(\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}})\left\\{\frac{|\beta-b|^{2}}{|\beta+b|^{2}}+(1-\cos(\phi+\beta-\Phi-b))-|\partial_{\overline{\beta}}f(\beta,\phi)-\partial_{\overline{b}}f(b,\Phi)|^{2}\right\\}.$
(53)

It remains to work with

$|\partial_{\overline{\beta}}f(\beta,\phi)-\partial_{\overline{b}}f(b,\Phi)|.$

Since we should bound this quantity by $\frac{|\beta-b|}{|\beta+b|}$ or
$1-\cos(\phi+\beta-\Phi-b)$, we need to define the auxiliary function

$g(\beta,x)=\partial_{\overline{\beta}}f(\beta,x-\beta).$

In that way, taking $x=\beta+\phi$ and $y=b+\Phi$ one has

$|\partial_{\overline{\beta}}f(\beta,\phi)-\partial_{\overline{b}}f(b,\Phi)|=|g(\beta,x)-g(b,y)|\leq|g(\beta,x)-g(b,x)|+|g(b,x)-g(b,y)|.$
(54)

For the first term we use the mean value theorem. Note that
$\partial_{\beta}g$ behaves as $\partial_{\varphi}f$ and
$\beta\partial_{\beta\varphi}f$, by the definition of
$\partial_{\overline{\beta}}$. Then, from the definition of
$X_{1}^{\sigma,\delta}$ in (45) we have that $\beta\partial_{\varphi}g$ is
bounded. In that way, one finds

$|g(\beta,x)-g(b,x)|\leq
C||f||_{X_{1}^{\sigma,\delta}}\frac{|\beta-b|}{\beta+b}.$

For the second term, we use the Lipschitz continuity of $f$ and $\beta
f_{\beta}$ (and then $\partial_{\overline{\beta}}f$) with respect to $\phi$
obtaining

$|g(b,x)-g(b,y)|\leq C||f||_{X_{1}^{\sigma,\delta}}|\sin((x-y)/2)|.$

By using that $x=\beta+\phi$ and $y=b+\Phi$ we find

$|\partial_{\overline{\beta}}f(\beta,\phi)-\partial_{\overline{b}}f(b,\Phi)|\leq
C||f||_{X_{1}^{\sigma,\delta}}\left\\{\frac{|\beta-b|}{\beta+b}+\sin^{2}\left((\beta-b+\phi-\Phi)/2\right)\right\\}.$

Coming back to (53) and using that $||f||_{X_{1}^{\sigma,\delta}}$ is small
enough we achieve (52):

$\displaystyle D(f)(\beta,\phi,b,\Phi)\geq$
$\displaystyle(\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}})\left\\{\frac{|\beta-b|^{2}}{|\beta+b|^{2}}+\sin^{2}\left((\beta-b+\phi-\Phi)/2\right)\right\\}.$

That concludes the proof. ∎

###  3.3. Regularity of the nonlinear function

In the following proposition, we deal with singular operators defined as

$\mathcal{H}_{f}(h_{1},h_{2})(\beta,\phi):=\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)dbd\Phi}{D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}}.$
(55)

This will be crucial in order to prove later the persistence regularity of the
nonlinear functional (2.2).

######  Proposition 3.9.

Let $\gamma,\sigma\in(0,1)$ and $\alpha\in(1,2)$. Then, there exists
$\epsilon<1$ such that if $f\in B_{X_{1}^{\sigma,\delta}}(0,\epsilon)$ and

$0<\sigma<\min\left\\{\frac{\alpha-1}{1+\alpha},\frac{\gamma-\alpha+1}{1+\alpha}\right\\},\quad
1<\alpha<1+\gamma,$ (56)

then

$\displaystyle\left\|\mathcal{H}(h_{1},h_{2})\right\|_{Y^{\sigma,\delta}}\leq
C||h_{1}||_{\mathcal{X}^{\sigma,\delta}}||h_{2}||_{X_{2}^{p}},$ (57)

for any $p>\frac{1}{1-\gamma}$ and $\delta<\min\\{1-\gamma,\sigma\\}$.

###### Proof.

The proof is very technical and will be divided in different steps. We will
provide the key points and the general ideas.

$\bullet$ $\mathcal{H}_{f}\in L^{\infty}_{\sigma}$.

Using Proposition 3.7, we can bound $\mathcal{H}$ as

$\displaystyle|\mathcal{H}_{f}(h_{1},h_{2})(\beta,\phi)|\leq$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}\beta^{\frac{\gamma}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}b^{\frac{-2}{1+\alpha}}\frac{|h_{1}(b,\phi+\beta-\eta-b)||h_{2}(\phi+\beta-\eta-b)|dbd\eta}{(\beta^{\frac{\gamma}{1+\alpha}}+b^{\frac{\gamma}{1+\alpha}})\left\\{\frac{|\beta-b|^{2}}{|\beta+b|^{2}}+\sin^{2}\left(\frac{\eta}{2}\right)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle\leq$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}\beta^{\frac{\gamma}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+b)^{2\sigma}|h_{2}(\phi+\beta-\eta-b)|}{(\beta^{\frac{\gamma}{1+\alpha}}+b^{\frac{\gamma}{1+\alpha}})\left\\{\frac{|\beta-b|^{2}}{|\beta+b|^{2}}+\sin^{2}\left(\frac{\eta}{2}\right)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle\times\frac{(1+b)^{2\sigma}}{b^{\sigma}}|h_{1}(b,\phi+\beta-\eta-b)|dbd\eta$
$\displaystyle=$ $\displaystyle
C\beta^{\sigma}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{z^{\frac{-2}{1+\alpha}}z^{\sigma}/(1+\beta
z)^{2\sigma}|h_{2}(\phi+\beta-\eta-\beta
z)|}{(1+z^{\frac{\gamma}{1+\alpha}})\left\\{\frac{|1-z|^{2}}{|1+z|^{2}}+\sin^{2}\left(\frac{\eta}{2}\right)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle\times\frac{(1+\beta z)^{2\sigma}}{(\beta
z)^{\sigma}}|h_{1}(\beta z,\phi+\beta-\eta-\beta z)|dzd\eta,$

where we have done the change of variables $b=\beta z$ in the last equality.
Then, we find

$\displaystyle||\mathcal{H}(h_{1},h_{2})||_{L^{\infty}_{\sigma}}\leq$
$\displaystyle
C||h_{1}||_{L^{\infty}_{\sigma}}||h_{2}||_{L^{1}}\sup_{\beta\in[0,+\infty)}\int_{0}^{\infty}\frac{(1+\beta)^{2\sigma}}{(1+\beta
z)^{2\sigma}}\frac{z^{\frac{-2}{1+\alpha}}z^{\sigma}|1+z|^{\gamma}}{|1-z|^{\gamma}|1+z|^{\frac{\gamma}{1+\alpha}}}dz.$

Recall that $\alpha>1$ by (7). In order to estimate the above integral, we
split in cases. Take $\varepsilon>0$, then let us work with
$\beta\in[0,\varepsilon)$ and $\beta\in[\varepsilon,+\infty)$ separately.
First,

$\sup_{\beta\in[0,\varepsilon)}\int_{0}^{\infty}\frac{(1+\beta)^{2\sigma}}{(1+\beta
z)^{2\sigma}}\frac{z^{\frac{-2}{1+\alpha}}z^{\sigma}|1+z|^{\gamma}}{|1-z|^{\gamma}|1+z|^{\frac{\gamma}{1+\alpha}}}dz\leq
C\sup_{\beta\in[0,\varepsilon)}\int_{0}^{\infty}\frac{z^{\frac{-2}{1+\alpha}}z^{\sigma}|1+z|^{\gamma}}{|1-z|^{\gamma}|1+z|^{\frac{\gamma}{1+\alpha}}}dz\leq
C,$

if

$0<\sigma<\frac{\gamma-\alpha+1}{1+\alpha},\quad 1<\alpha<1+\gamma.$ (58)

In the other case:

$\sup_{\beta\in[\varepsilon,+\infty)}\int_{0}^{\infty}\frac{(1+\beta)^{2\sigma}}{(1+\beta
z)^{2\sigma}}\frac{z^{\frac{-2}{1+\alpha}}z^{\sigma}|1+z|^{\gamma}}{|1-z|^{\gamma}|1+z|^{\frac{\gamma}{1+\alpha}}}dz\leq
C\sup_{\beta\in[\varepsilon,+\infty)}\int_{0}^{\infty}\frac{1}{z^{\sigma}}\frac{z^{\frac{-2}{1+\alpha}}|1+z|^{\gamma}}{|1-z|^{\gamma}|1+z|^{\frac{\gamma}{1+\alpha}}}dz\leq
C,$

if

$0<\sigma<\frac{\alpha-1}{1+\alpha}.$ (59)

Hence, we find that

$0<\sigma<\min\left\\{\frac{\alpha-1}{1+\alpha},\frac{\gamma-\alpha+1}{1+\alpha}\right\\},\quad
1<\alpha<1+\gamma.$ (60)

$\bullet$ $\mathcal{H}_{f}\in C^{\delta}_{\beta}$.

Let us now check the Hölder semi-norm. Take $\beta_{1}<\beta_{2}$ and then

$\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}<\beta_{1}^{-\frac{\alpha-1}{1+\alpha}}.$
(61)

Doing the difference of $\mathcal{H}_{f}$ at two points we arrive at

$\displaystyle|\mathcal{H}_{f}(h_{1},h_{2})(\beta_{1},\phi)-\mathcal{H}_{f}(h_{1},h_{2})(\beta_{2},\phi)|=\Bigg{|}\beta_{1}^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)dbd\Phi}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}$
(62)
$\displaystyle-\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)dbd\Phi}{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\Bigg{|}$
$\displaystyle\leq$ $\displaystyle
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}\left(\beta_{1}^{-\frac{\alpha-1}{1+\alpha}}+\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\right){\Bigg{|}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)dbd\Phi}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}\Bigg{|}}$
$\displaystyle+\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\Bigg{|}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)dbd\Phi}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}-\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)dbd\Phi}{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\Bigg{|}.$

Due to (61) and using the previous analysis to bound $\mathcal{H}_{f}$ in
$L^{\infty}_{\sigma}$ we get that

$\left(\beta_{1}^{-\frac{\alpha-1}{1+\alpha}}+\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\right)\Bigg{|}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)dbd\Phi}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}\Bigg{|}<C.$

Hence, we need to estimate the second term in (62). Due to its expression, we
will need to work with the difference of the denominator at two points. Thus

$\displaystyle
D(f)(\beta_{1},\phi,b,\Phi)-D(f)(\beta_{2},\phi,b,\Phi)=\left(\beta_{1}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}$
$\displaystyle-\left(\beta_{2}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{2},\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}$
$\displaystyle+2\beta_{1}^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}(1-\cos(\phi+\beta_{1}-\Phi-b))$
$\displaystyle-2\beta_{2}^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{2},\phi)\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}(1-\cos(\phi+\beta_{2}-\Phi-b)).$
(63)

Let us estimate it in different steps. First

$\displaystyle\Big{|}\left(\beta_{1}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}$
$\displaystyle-\left(\beta_{2}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{2},\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}\Big{|}$
$\displaystyle\leq$
$\displaystyle\left|\beta_{1}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}^{\frac{1}{2}}-\beta_{2}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{2},\Phi)\right\\}^{\frac{1}{2}}\right|$
$\displaystyle\times\Big{|}\left(\beta_{1}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)$
$\displaystyle+\left(\beta_{2}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{2},\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)\Big{|},$

where

$\displaystyle\left|\beta_{1}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}^{\frac{1}{2}}-\beta_{2}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{2},\Phi)\right\\}^{\frac{1}{2}}\right|$
$\displaystyle\leq$ $\displaystyle
C\left|\frac{\beta_{1}^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}-\beta_{2}^{\frac{-2}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{2},\Phi)\right\\}}{\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}}}\right|$
$\displaystyle\leq$ $\displaystyle
C\left|\frac{(\beta_{1}^{\frac{-2}{1+\alpha}}-\beta_{2}^{\frac{-2}{1+\alpha}})\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}+\beta_{2}^{\frac{-2}{1+\alpha}}\left\\{\partial_{\overline{\beta}}f(\beta_{1},\Phi)-\partial_{\overline{\beta}}f(\beta_{2},\Phi)\right\\}}{\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}}}\right|$
$\displaystyle\leq$ $\displaystyle
C\frac{\beta_{1}^{\frac{-2}{1+\alpha}}+\beta_{2}^{\frac{-2}{1+\alpha}}}{\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}}}\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}$
$\displaystyle\leq$ $\displaystyle
C\left\\{\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}}\right\\}\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}.$

Hence

$\displaystyle\Big{|}\left(\beta_{1}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}$
$\displaystyle-\left(\beta_{2}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{2},\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}\Big{|}$
$\displaystyle\leq$ $\displaystyle
C\Big{|}\left(\beta_{1}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)$
$\displaystyle+\left(\beta_{2}^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{2},\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)\Big{|}$
$\displaystyle\times\left\\{\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}}\right\\}\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}$
$\displaystyle\leq$ $\displaystyle
C(D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}+D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}})\left\\{\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}}\right\\}\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}.$

Now let us estimate the second part of (3.3) using similar ideas

$\displaystyle\Big{|}\beta_{1}^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{1},\phi)\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}(1-\cos(\phi+\beta_{1}-\Phi-b))$
$\displaystyle-\beta_{2}^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta_{2},\phi)\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}(1-\cos(\phi+\beta_{2}-\Phi-b))\Big{|}$
$\displaystyle\leq$ $\displaystyle
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}})b^{\frac{-1}{1+\alpha}}\left\\{(\sin^{2}((\phi+\beta_{1}-\Phi-b)/2))+(\sin^{2}((\phi+\beta_{2}-\Phi-b)/2))\right\\}$
$\displaystyle+C|\sin^{\delta}((\beta_{1}-\beta_{2})/2)|(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}})b^{\frac{-1}{1+\alpha}}$
$\displaystyle\times\left\\{(\sin^{2-\delta}((\phi+\beta_{1}-\Phi-b)/2))+(\sin^{2-\delta}((\phi+\beta_{2}-\Phi-b)/2))\right\\}$
$\displaystyle\leq$ $\displaystyle
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}})b^{\frac{-1}{1+\alpha}}\left\\{(\sin^{2}((\phi+\beta_{1}-\Phi-b)/2))+(\sin^{2}((\phi+\beta_{2}-\Phi-b)/2))\right\\}$
$\displaystyle+C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}(\beta_{1}^{\frac{-1}{1+\alpha}+\delta}+\beta_{2}^{\frac{-1}{1+\alpha}+\delta})b^{\frac{-1}{1+\alpha}}\left\\{(\sin^{2-\delta}((\phi+\beta_{1}-\Phi-b)/2))+(\sin^{2-\delta}((\phi+\beta_{2}-\Phi-b)/2))\right\\}$
$\displaystyle\leq$ $\displaystyle
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}})(D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}+D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}})$
$\displaystyle+C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}(\beta_{1}^{\frac{-1}{1+\alpha}+\delta}+\beta_{2}^{\frac{-1}{1+\alpha}+\delta})\left\\{(\sin^{1-\delta}((\phi+\beta_{1}-\Phi-b)/2))+(\sin^{1-\delta}((\phi+\beta_{2}-\Phi-b)/2))\right\\}$
$\displaystyle\times(D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}+D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}}).$

Then, at the end we get

$\displaystyle|D(f)(\beta_{2},\phi,b,\Phi)-D(f)(\beta_{1},\phi,b,\Phi)|\leq$
$\displaystyle
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}}+\beta_{1}^{\frac{-1}{1+\alpha}+\delta}+\beta_{2}^{\frac{-1}{1+\alpha}+\delta})$
$\displaystyle\times(D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}+D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}}).$
(64)

Hence

$\displaystyle\left|\frac{1}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}-\frac{1}{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\right|\leq\frac{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}-D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}$
$\displaystyle\lesssim$
$\displaystyle\frac{|D(f)(\beta_{2},\phi,b,\Phi)-D(f)(\beta_{1},\phi,b,\Phi)|}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\frac{1}{(D(f)(\beta_{2},\phi,b,\Phi)^{\frac{2-\gamma}{2}}+D(f)(\beta_{1},\phi,b,\Phi)^{\frac{2-\gamma}{2}})},$

and using (3.3) we find

$\displaystyle\left|\frac{1}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}-\frac{1}{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\right|$
(65) $\displaystyle\leq$ $\displaystyle
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}\frac{\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}}+\beta_{1}^{\frac{-1}{1+\alpha}+\delta}+\beta_{2}^{\frac{-1}{1+\alpha}+\delta}}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}+D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}}D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}.$

Let us come back to (62) to achieve

$\displaystyle\left|\mathcal{H}_{f}(h_{1},h_{2})(\beta_{1},\phi)-\mathcal{H}_{f}(h_{1},h_{2})(\beta_{2},\phi)\right|\leq
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}$
$\displaystyle+C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}|h_{1}(b,\Phi)||h_{2}(\Phi)|dbd\Phi$
$\displaystyle\times\frac{\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}}+\beta_{1}^{\frac{-1}{1+\alpha}+\delta}+\beta_{2}^{\frac{-1}{1+\alpha}+\delta}}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}+D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}}D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}dbd\Phi.$

By virtue of Proposition 3.7 we can bound the denominator as

$\displaystyle\left|\mathcal{H}_{f}(h_{1},h_{2})(\beta_{1},\phi)-\mathcal{H}_{f}(h_{1},h_{2})(\beta_{2},\phi)\right|\leq
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}$
$\displaystyle+C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}\int_{0}^{2\pi}\int_{0}^{\infty}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}|h_{1}(b,\Phi)||h_{2}(\Phi)|$
$\displaystyle\times\Bigg{(}\frac{\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\beta_{1}^{\frac{-1}{1+\alpha}}}{(\beta_{1}^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})(\beta_{2}^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}})\left\\{\frac{|\beta_{1}-b|^{2}}{|\beta_{1}+b|^{2}}+\sin^{2}((\beta_{1}+\phi-b-\Phi)/2)\right\\}^{\frac{1}{2}}\left\\{\frac{|\beta_{2}-b|^{2}}{|\beta_{2}+b|^{2}}+\sin^{2}((\beta_{2}+\phi-b-\Phi)/2)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle+\frac{\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\beta_{2}^{\frac{-1}{1+\alpha}}}{(\beta_{2}^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})(\beta_{1}^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}})\left\\{\frac{|\beta_{2}-b|^{2}}{|\beta_{2}+b|^{2}}+\sin^{2}((\beta_{2}+\phi-b-\Phi)/2)\right\\}^{\frac{1}{2}}\left\\{\frac{|\beta_{1}-b|^{2}}{|\beta_{1}+b|^{2}}+\sin^{2}((\beta_{1}+\phi-b-\Phi)/2)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle+\frac{\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\beta_{1}^{\frac{-1}{1+\alpha}}\beta_{1}^{\delta}}{(\beta_{1}^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})(\beta_{2}^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}})\left\\{\frac{|\beta_{1}-b|^{2}}{|\beta_{1}+b|^{2}}+\sin^{2}((\beta_{1}+\phi-b-\Phi)/2)\right\\}^{\frac{1}{2}}\left\\{\frac{|\beta_{2}-b|^{2}}{|\beta_{2}+b|^{2}}+\sin^{2}((\beta_{2}+\phi-b-\Phi)/2)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle+\frac{\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\beta_{2}^{\frac{-1}{1+\alpha}}\beta_{2}^{\delta}}{(\beta_{2}^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})(\beta_{1}^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}})\left\\{\frac{|\beta_{2}-b|^{2}}{|\beta_{2}+b|^{2}}+\sin^{2}((\beta_{2}+\phi-b-\Phi)/2)\right\\}^{\frac{1}{2}}\left\\{\frac{|\beta_{1}-b|^{2}}{|\beta_{1}+b|^{2}}+\sin^{2}((\beta_{1}+\phi-b-\Phi)/2)\right\\}^{\frac{\gamma}{2}}}\Bigg{)}dbd\Phi.$

Let us show the bound for the third integral, and the others will follow
similarly. Using now that $h_{1}\in\mathcal{X}^{\sigma,\delta}$ and then

$\displaystyle\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\beta_{1}^{\frac{-1}{1+\alpha}+\delta}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}|h_{1}(b,\Phi)||h_{2}(\Phi)|dbd\Phi}{(\beta_{1}^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})(\beta_{2}^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}})\left\\{\frac{|\beta_{1}-b|^{2}}{|\beta_{1}+b|^{2}}+\sin^{2}((\beta_{1}+\phi-b-\Phi)/2)\right\\}^{\frac{1}{2}}\left\\{\frac{|\beta_{2}-b|^{2}}{|\beta_{2}+b|^{2}}+\sin^{2}((\beta_{2}+\phi-b-\Phi)/2)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle\leq
C||h_{1}||_{L^{\infty}_{\sigma}}\beta_{1}^{\delta}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+b)^{2\sigma}|h_{2}(\Phi)|dbd\Phi}{(\beta_{2}^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}})\left\\{\frac{|\beta_{1}-b|^{2}}{|\beta_{1}+b|^{2}}+\sin^{2}((\beta_{1}+\phi-b-\Phi)/2)\right\\}^{\frac{1}{2}}\left\\{\frac{|\beta_{2}-b|^{2}}{|\beta_{2}+b|^{2}}+\sin^{2}((\beta_{2}+\phi-b-\Phi)/2)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle\leq
C||h_{1}||_{L^{\infty}_{\sigma}}\beta_{1}^{\delta}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta_{2}^{\frac{\gamma}{1+\alpha}}\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+b)^{2\sigma}|h_{2}(\Phi)|dbd\Phi}{(\beta_{2}^{\frac{\gamma}{1+\alpha}}+b^{\frac{\gamma}{1+\alpha}})\frac{|\beta_{1}-b|^{1-\nu}}{|\beta_{1}+b|^{1-\nu}}|\sin((\beta_{1}+\phi-b-\Phi)/2)|^{\nu}\frac{|\beta_{2}-b|^{\gamma}}{|\beta_{2}+b|^{\gamma}}}$

for $\nu\in(0,1)$ with

${1-\nu+\gamma\in(0,1).}$ (66)

Hence

$\displaystyle\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\beta_{1}^{\frac{-1}{1+\alpha}+\delta}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}|h_{1}(b,\Phi)||h_{2}(\Phi)|dbd\Phi}{(\beta_{1}^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})(\beta_{2}^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}})\left\\{\frac{|\beta_{1}-b|^{2}}{|\beta_{1}+b|^{2}}+\sin^{2}((\beta_{1}+\phi-b-\Phi)/2)\right\\}^{\frac{1}{2}}\left\\{\frac{|\beta_{2}-b|^{2}}{|\beta_{2}+b|^{2}}+\sin^{2}((\beta_{2}+\phi-b-\Phi)/2)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle\leq
C||h_{1}||_{L^{\infty}_{\sigma}}\beta_{1}^{\delta}\int_{0}^{\infty}\frac{\beta_{2}^{\frac{\gamma}{1+\alpha}}\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+b)^{2\sigma}}{(\beta_{2}^{\frac{\gamma}{1+\alpha}}+b^{\frac{\gamma}{1+\alpha}})\frac{|\beta_{1}-b|^{1-\nu}}{|\beta_{1}+b|^{1-\nu}}\frac{|\beta_{2}-b|^{\gamma}}{|\beta_{2}+b|^{\gamma}}}\int_{0}^{2\pi}|\sin^{-\nu}((\beta_{2}+\phi-b-\Phi)/2)||h_{2}(\Phi)|d\Phi
db$

Note that by interpolation we get

$\displaystyle\int_{0}^{2\pi}|\sin^{-\nu}((\beta_{2}+\phi-b-\Phi)/2)||h_{2}(\Phi)|d\Phi\leq$
$\displaystyle
C\left(\int_{0}^{2\pi}|h_{2}(\Phi)|^{p}d\Phi\right)^{\frac{1}{p}}\left(\int_{0}^{2\pi}|\sin^{-\nu}((\Phi)/2)|^{q}d\Phi\right)^{\frac{1}{q}}$
$\displaystyle\leq$ $\displaystyle C||h_{2}||_{X_{2}^{p}},$ (67)

provided that $\nu q<1$ and that $\frac{1}{p}+\frac{1}{q}=1$. By using (66),
we can choose any $p$ such that

$\frac{1}{1-\gamma}<p.$ (68)

Finally, using the change of variables $b=\beta_{2}z$ we get:

$\displaystyle\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\beta_{1}^{\frac{-1}{1+\alpha}+\delta}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}|h_{1}(b,\Phi)||h_{2}(\Phi)|dbd\Phi}{(\beta_{1}^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})(\beta_{2}^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}})\left\\{\frac{|\beta_{1}-b|^{2}}{|\beta_{1}+b|^{2}}+\sin^{2}((\beta_{1}+\phi-b-\Phi)/2)\right\\}^{\frac{1}{2}}\left\\{\frac{|\beta_{2}-b|^{2}}{|\beta_{2}+b|^{2}}+\sin^{2}((\beta_{2}+\phi-b-\Phi)/2)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle\leq
C||h_{1}||_{L^{\infty}_{\sigma}}||h_{2}||_{X_{2}^{p}}\beta_{1}^{\delta}\int_{0}^{\infty}\frac{\beta_{2}^{\frac{\gamma}{1+\alpha}}\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+b)^{2\sigma}}{(\beta_{2}^{\frac{\gamma}{1+\alpha}}+b^{\frac{\gamma}{1+\alpha}})\frac{|\beta_{1}-b|^{1-\nu}}{|\beta_{1}+b|^{1-\nu}}\frac{|\beta_{2}-b|^{\gamma}}{|\beta_{2}+b|^{\gamma}}}db$
(69) $\displaystyle\leq
C||h_{1}||_{L^{\infty}_{\sigma}}||h_{2}||_{X_{2}^{p}}\beta_{2}^{\delta}\int_{0}^{\infty}\frac{z^{\frac{-2}{1+\alpha}}(z\beta_{2})^{\sigma}/(1+z\beta_{2})^{2\sigma}}{(1+z^{\frac{\gamma}{1+\alpha}})\frac{|\beta_{1}-z\beta_{2}|^{1-\nu}}{|\beta_{1}+z\beta_{2}|^{1-\nu}}\frac{|1-z|^{\gamma}}{|1+z|^{\gamma}}}dz$
$\displaystyle\leq C||h_{1}||_{L^{\infty}_{\sigma}}||h_{2}||_{X_{2}^{p}},$

where we have used that $\beta_{1}<\beta_{2}$, and which is bounded provided
that $\sigma$ verifies (58) and (59) and

$\delta<\sigma,$

in order to get the boundedness as $\beta_{2}$ goes to $\infty$. Hence,
putting everything together we find that

$|\mathcal{H}_{f}(\beta_{1},\phi,b,\Phi)-\mathcal{H}_{f}(\beta_{2},\phi,b,\Phi)|\leq
C||h_{1}||_{L^{\infty}_{\sigma}}||h_{2}||_{X_{2}^{p}}\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}},$

implying that $H_{f}\in C^{\delta}_{\beta}$.

$\bullet$ $\mathcal{H}_{f}\in\textnormal{Lip}_{\phi}$.

We shall prove that

$\displaystyle|\mathcal{H}_{f}(\beta,\phi_{1},b,\Phi)-\mathcal{H}_{f}(\beta,\phi_{2},b,\Phi)|\leq
C|\sin((\phi_{1}-\phi_{2})/2)|.$ (70)

For that we need to estimate the difference of the denominator at two points:

$\displaystyle
D(f)(\beta,\phi_{1},b,\Phi)-D(f)(\beta,\phi_{2},b,\Phi)=\left(\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi_{1})\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}$
$\displaystyle-\left(\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi_{2})\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)^{2}$
$\displaystyle+2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi_{1})\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}(1-\cos(\phi_{1}+\beta-\Phi-b))$
$\displaystyle-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi_{2})\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}(1-\cos(\phi_{2}+\beta-\Phi-b)).$

Hence we can achieve that

$\displaystyle|D(f)(\beta,\phi_{1},b,\Phi)-D(f)(\beta,\phi_{2},b,\Phi)|\leq
C|\sin((\phi_{1}-\phi_{2})/2)|{(\beta^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})}$
$\displaystyle\times(D(f)(\beta,\phi_{1},b,\Phi)^{\frac{1}{2}}+D(f)(\beta,\phi_{2},b,\Phi)^{\frac{1}{2}}).$
(71)

Hence

$\displaystyle\left|\frac{1}{D(f)(\beta,\phi_{1},b,\Phi)^{\frac{\gamma}{2}}}-\frac{1}{D(f)(\beta,\phi_{2},b,\Phi)^{\frac{\gamma}{2}}}\right|\leq\frac{D(f)(\beta,\phi_{1},b,\Phi)^{\frac{\gamma}{2}}-D(f)(\beta,\phi_{2},b,\Phi)^{\frac{\gamma}{2}}}{D(f)(\beta,\phi_{1},b,\Phi)^{\frac{\gamma}{2}}D(f)(\beta,\phi_{2},b,\Phi)^{\frac{\gamma}{2}}}$
$\displaystyle\leq$
$\displaystyle\frac{|D(f)(\beta,\phi_{1},b,\Phi)-D(f)(\beta,\phi_{2},b,\Phi)|}{D(f)(\beta,\phi_{1},b,\Phi)^{\frac{\gamma}{2}}D(f)(\beta,\phi_{2},b,\Phi)^{\frac{\gamma}{2}}}\frac{1}{D(f)(\beta,\phi_{2},b,\Phi)^{\frac{2-\gamma}{2}}+D(f)(\beta,\phi_{1},b,\Phi)^{\frac{2-\gamma}{2}}},$

and using (3.3) we find

$\displaystyle\left|\frac{1}{D(f)(\beta,\phi_{1},b,\Phi)^{\frac{\gamma}{2}}}-\frac{1}{D(f)(\beta,\phi_{2},b,\Phi)^{\frac{\gamma}{2}}}\right|$
$\displaystyle\leq$ $\displaystyle
C|\sin((\phi_{1}-\phi_{2})/2)|\frac{{(\beta^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})}}{D(f)(\beta,\phi_{1},b,\Phi)^{\frac{1}{2}}D(f)(\beta,\phi_{2},b,\Phi)^{\frac{\gamma}{2}}+D(f)(\beta,\phi_{2},b,\Phi)^{\frac{1}{2}}D(f)(\beta,\phi_{1},b,\Phi)^{\frac{\gamma}{2}}}.$

In that way, we obtain that

$\displaystyle|\mathcal{H}_{f}(\beta,\phi_{1},b,\Phi)-\mathcal{H}_{f}(\beta,\phi_{2},b,\Phi)|$
$\displaystyle\leq$ $\displaystyle
C|\sin((\phi_{1}-\phi_{2})/2)|\beta^{-\frac{\alpha-1}{1+\alpha}}$
$\displaystyle\times\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}{(\beta^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})}|h_{1}(b,\Phi)|h_{2}(\Phi)|dbd\Phi}{D(f)(\beta,\phi_{1},b,\Phi)^{\frac{1}{2}}D(f)(\beta,\phi_{2},b,\Phi)^{\frac{\gamma}{2}}+D(f)(\beta,\phi_{2},b,\Phi)^{\frac{1}{2}}D(f)(\beta,\phi_{1},b,\Phi)^{\frac{\gamma}{2}}}$
$\displaystyle\leq$ $\displaystyle
C|\sin((\phi_{1}-\phi_{2})/2)|\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}{(\beta^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})}|h_{1}(b,\Phi)|h_{2}(\Phi)|dbd\Phi}{D(f)(\beta,\phi_{1},b,\Phi)^{\frac{1}{2}}D(f)(\beta,\phi_{2},b,\Phi)^{\frac{\gamma}{2}}}$
$\displaystyle\leq$ $\displaystyle
C|\sin((\phi_{1}-\phi_{2})/2)|\beta^{-\frac{\alpha-1}{1+\alpha}}$
$\displaystyle\times\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}{(\beta^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})}|h_{1}(b,\Phi)|h_{2}(\Phi)|dbd\Phi}{\left\\{\beta^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}}\right\\}\left\\{\beta^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}}\right\\}\frac{|\beta-b|^{\gamma}}{|\beta+b|^{\gamma}}\left\\{\frac{|\beta-b|^{2}}{|\beta+b|^{2}}+\sin^{2}((\beta-b+\phi_{1}-\Phi)/2)\right\\}^{\frac{1}{2}}}$
$\displaystyle\leq$ $\displaystyle
C|\sin((\phi_{1}-\phi_{2})/2)|\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}{b^{\frac{-2}{1+\alpha}}}|h_{1}(b,\Phi)||h_{2}(\Phi)|dbd\Phi}{\left\\{\beta^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}}\right\\}\frac{|\beta-b|^{\gamma+1-\nu}}{|\beta+b|^{\gamma+1-\nu}}|\sin((\beta-b+\phi_{1}-\Phi)/2)|^{\nu}}.$

Using now (3.3) we get

$\displaystyle|\mathcal{H}_{f}(\beta,\phi_{1},b,\Phi)-\mathcal{H}_{f}(\beta,\phi_{2},b,\Phi)|$
$\displaystyle\leq$ $\displaystyle
C||h_{1}||_{X_{1}^{\sigma,\delta}}||h_{2}||_{X_{2}^{p}}|\sin((\phi_{1}-\phi_{2})/2)|\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+b)^{2\sigma}db}{\left\\{\beta^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}}\right\\}\frac{|\beta-b|^{\gamma+1-\nu}}{|\beta+b|^{\gamma+1-\nu}}},$

for $p$ with $\frac{1}{p}+\frac{1}{q}=1$ and $q\nu<1$ as done in (3.3). Doing
now the change of variable $b=z\beta$ amounts to

$\displaystyle|\mathcal{H}_{f}(\beta,\phi_{1},b,\Phi)-\mathcal{H}_{f}(\beta,\phi_{2},b,\Phi)|$
$\displaystyle\leq$ $\displaystyle
C||h_{1}||_{\mathcal{X}^{\sigma,\delta}}||h_{2}||_{X_{2}^{p}}|\sin((\phi_{1}-\phi_{2})/2)|\beta^{\sigma}\int_{0}^{\infty}\frac{z^{\frac{-2}{1+\alpha}}z^{\sigma}/(1+z\beta)^{2\sigma}dz}{\left\\{1+z^{\frac{\gamma}{1+\alpha}}\right\\}\frac{|1-z|^{\gamma+1-\nu}}{|1+z|^{\gamma+1-\nu}}}$
$\displaystyle\ \leq$ $\displaystyle
C||h_{1}||_{\mathcal{X}^{\sigma,\delta}}||h_{2}||_{X_{2}^{p}},$

provided that $\gamma+1-\nu<1$, $p$ satisfies (68) and $\sigma$ satisfies (58)
and (59).

$\bullet$ $\beta\partial_{\varphi}\mathcal{H}_{f}\in L^{\infty}_{\sigma}$.

By using the expression of $\mathcal{H}_{f}$ we find

$\displaystyle\partial_{\varphi}\mathcal{H}_{f}(h_{1},h_{2})(\beta,\phi)=$
$\displaystyle\frac{\alpha-1}{1+\alpha}\frac{1}{\beta}\mathcal{H}_{f}(h_{1},h_{2})(\beta,\phi)$
$\displaystyle-\frac{\gamma}{2}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)\partial_{\varphi}D(f)(\beta,\phi,b,\Phi)}{D(f)(\beta,\phi,b,\Phi)^{\frac{2+\gamma}{2}}}dbd\Phi.$

Note that

$\displaystyle\beta\partial_{\varphi}\mathcal{H}_{f}(h_{1},h_{2})(\beta,\phi)=$
$\displaystyle\frac{\alpha-1}{1+\alpha}\mathcal{H}_{f}(h_{1},h_{2})(\beta,\phi)$
$\displaystyle-\frac{\gamma}{2}\beta\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)\partial_{\varphi}D(f)(\beta,\phi,b,\Phi)}{D(f)(\beta,\phi,b,\Phi)^{\frac{2+\gamma}{2}}}dbd\Phi$
$\displaystyle=:$ $\displaystyle\mathcal{H}_{1}+\mathcal{H}_{2},$ (72)

where the first term was already studied in the previous steps. For the second
term, let us compute the derivative of the denominator as follows:

$\displaystyle\partial_{\varphi}D(f)(\beta,\phi,b,\Phi)=2\left(\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)$
$\displaystyle\times\beta^{\frac{-1}{1+\alpha}}\frac{\partial_{\varphi}\partial_{\overline{\beta}}f(\beta,\phi)}{2\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}}$
$\displaystyle+\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{-1}{2}}\partial_{\varphi}\partial_{\overline{\beta}}f(\beta,\phi)\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}(1-\cos(\phi+\beta-\Phi-b))$
$\displaystyle-\frac{2}{1+\alpha}\frac{1}{\beta}\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}(1-\cos(\phi+\beta-\Phi-b)),$
(73)

implying that

$\displaystyle|\partial_{\varphi}D(f)(\beta,\phi,b,\Phi)|\leq
CD(f)(\beta,\phi,b,\Phi)^{\frac{1}{2}}\beta^{\frac{-1}{1+\alpha}}\frac{1}{\beta}.$
(74)

In this way, we find that

$\displaystyle\left|\beta\partial_{\varphi}\mathcal{H}(h_{1},h_{2})(\beta,\phi)\right|\leq
C|\mathcal{H}_{f}(h_{1},h_{2})(\beta,\phi)|$
$\displaystyle+C\beta^{\frac{-1}{1+\alpha}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}|h_{1}(b,\phi+\beta-\eta-b)||h_{2}(\phi+\beta-\eta-b)|}{D(f)(\beta,\phi,b,\phi+\beta-\eta-b)^{\frac{1+\gamma}{2}}}dbd\eta$
$\displaystyle\leq$ $\displaystyle
C|\mathcal{H}_{f}(h_{1},h_{2})(\beta,\phi)|$
$\displaystyle+C||h_{1}||_{X_{1}^{\sigma,\delta}}\beta^{\sigma}\beta^{\frac{-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{z^{-\frac{\gamma}{1+\alpha}}z^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+\beta
z)^{2\sigma}|h_{2}(\phi+\beta-\eta-b)|}{(\beta^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}})(\beta^{\frac{-1}{1+\alpha}}+b^{\frac{-1}{1+\alpha}})\left\\{\frac{|\beta-b|^{2}}{|\beta+b|^{2}}+\sin^{2}(\eta/2)\right\\}^{\frac{1+\gamma}{2}}}dbd\eta.$

Using (3.3) we get

$\displaystyle\left|\beta\partial_{\varphi}\mathcal{H}(h_{1},h_{2})(\beta,\phi)\right|\leq$
$\displaystyle C|\mathcal{H}_{f}(h_{1},h_{2})(\beta,\phi)|$
$\displaystyle+C||h_{1}||_{\mathcal{X}^{\sigma,\delta}}||h_{2}||_{X_{2}^{p}}\beta^{\sigma}\int_{0}^{\infty}\frac{z^{-\frac{\gamma}{1+\alpha}}z^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+\beta
z)^{2\sigma}}{(\beta^{\frac{-\gamma}{1+\alpha}}+b^{\frac{-\gamma}{1+\alpha}})\frac{|\beta-b|^{1+\gamma-\nu}}{|\beta+b|^{1+\gamma-\nu}}}db.$

Then:

$\displaystyle||\beta\partial_{\varphi}\mathcal{H}(h_{1},h_{2})||_{\sigma}\leq
C||h_{1}||_{\mathcal{X}^{\sigma,\delta}}||h_{2}||_{X_{2}^{p}},$ (75)

provided that (58) and (59) hold, as well as (68) and $1+\gamma-\nu<1$.

$\bullet$ $\beta\partial_{\varphi}\mathcal{H}_{f}(h_{1},h_{2})\in
C^{\delta}_{\beta}$.

Since $\mathcal{H}_{f}(h_{1},h_{2})\in C^{\delta}_{\beta}$, the first term in
(3.3) does. Hence, it remains to check the second term in (3.3):
$\mathcal{H}_{2}$. Take $\beta_{1}<\beta_{2}$, and then they satisfy (61). Let
us write the difference of $\mathcal{H}_{2}$ in these two terms as follows

$\displaystyle\mathcal{H}_{2}(\beta_{1})-\mathcal{H}_{2}(\beta_{2})=C\beta_{1}^{\frac{2}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)\partial_{\varphi}D(f)(\beta_{1},\phi,b,\Phi)}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}}$
$\displaystyle\times\left\\{\frac{1}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1+\gamma}{2}}}-\frac{1}{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1+\gamma}{2}}}\right\\}dbd\Phi$
$\displaystyle+C\beta_{1}^{\frac{2}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)(\partial_{\varphi}D(f)(\beta_{1},\phi,b,\Phi)-\partial_{\varphi}D(f)(\beta_{2},\phi,b,\Phi))}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1+\gamma}{2}}}dbd\Phi$
$\displaystyle+C\beta_{1}^{\frac{2}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)\partial_{\varphi}D(f)(\beta_{2},\phi,b,\Phi)}{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1+\gamma}{2}}}$
$\displaystyle\times\left\\{\frac{1}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}}-\frac{1}{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}}}\right\\}dbd\Phi$
$\displaystyle+C(\beta_{1}^{\frac{2}{1+\alpha}}-\beta_{2}^{\frac{2}{1+\alpha}})\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}h_{1}(b,\Phi)h_{2}(\Phi)\partial_{\varphi}D(f)(\beta_{2},\phi,b,\Phi)}{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{2+\gamma}{2}}}dbd\Phi$
$\displaystyle=:$
$\displaystyle\mathcal{H}_{2,1}+\mathcal{H}_{2,2}+\mathcal{H}_{2,3}+\mathcal{H}_{2,4}.$

By using (65) and (74) we can bound $\mathcal{H}_{2,1}$ and
$\mathcal{H}_{2,3}$. Let us show the estimate for $\mathcal{H}_{2,1}$. Indeed,
from (3.3) we find for $\delta=1$ that

$\displaystyle|D(f)(\beta_{2},\phi,b,\Phi)-D(f)(\beta_{1},\phi,b,\Phi)|\leq$
$\displaystyle
C\frac{|\beta_{1}-\beta_{2}|}{|\beta_{1}+\beta_{2}|}(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}}+\beta_{1}^{\frac{-1}{1+\alpha}+1}+\beta_{2}^{\frac{-1}{1+\alpha}+1})$
$\displaystyle\times(D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}+D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}}).$

By interpolating with

$|D(f)(\beta_{2},\phi,b,\Phi)-D(f)(\beta_{1},\phi,b,\Phi)|\leq|D(f)(\beta_{2},\phi,b,\Phi)+D(f)(\beta_{1},\phi,b,\Phi)|,$

we find

$\displaystyle|D(f)(\beta_{2},\phi,b,\Phi)-D(f)(\beta_{1},\phi,b,\Phi)|\leq$
$\displaystyle
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}(\beta_{1}^{\frac{-\delta}{1+\alpha}}+\beta_{2}^{\frac{-\delta}{1+\alpha}}+\beta_{1}^{\frac{-\delta}{1+\alpha}+\delta}+\beta_{2}^{\frac{-\delta}{1+\alpha}+\delta})$
$\displaystyle\times(D(f)(\beta_{1},\phi,b,\Phi)^{\frac{2-\delta}{2}}+D(f)(\beta_{2},\phi,b,\Phi)^{\frac{2-\delta}{2}}).$
(76)

Hence using the ideas in (65) together with (3.3) we find

$\displaystyle\left|\frac{1}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1+\gamma}{2}}}-\frac{1}{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1+\gamma}{2}}}\right|\leq
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}$ (77)
$\displaystyle\times\frac{\beta_{1}^{\frac{-\delta}{1+\alpha}}+\beta_{2}^{\frac{-\delta}{1+\alpha}}+\beta_{1}^{\frac{-\delta}{1+\alpha}+\delta}+\beta_{2}^{\frac{-\delta}{1+\alpha}+\delta}}{D(f)(\beta_{1},\phi,b,\Phi)^{\frac{\delta}{2}}D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1+\gamma}{2}}+D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\delta}{2}}D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1+\gamma}{2}}}.$
(78)

Hence, (77) together with (74) amounts to

$\displaystyle|\mathcal{H}_{2,1}|\leq
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}\beta_{1}^{\frac{2}{1+\alpha}}\beta_{1}^{\frac{-1}{1+\alpha}}\beta_{1}^{-1}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}|h_{1}(b,\Phi)||h_{2}(\Phi)|(\beta_{1}^{\frac{-\delta}{1+\alpha}}+\beta_{1}^{\frac{-\delta}{1+\alpha}+\delta})dbd\Phi}{D(f)(\beta_{2},\phi,b,\Phi)^{\frac{\delta}{2}}D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1+\gamma}{2}}}$
$\displaystyle\leq$ $\displaystyle
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}\beta_{1}^{\frac{2}{1+\alpha}}\beta_{1}^{\frac{-1}{1+\alpha}}\beta_{1}^{-1}$
$\displaystyle\times\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}|h_{1}(b,\Phi)||h_{2}(\Phi)|(\beta_{1}^{\frac{-\delta}{1+\alpha}}+\beta_{1}^{\frac{-\delta}{1+\alpha}+\delta})dbd\Phi}{|\beta_{2}^{\frac{-\delta}{1+\alpha}}+b^{\frac{-\delta}{1+\alpha}}|\left\\{\frac{|\beta_{2}-b|^{2}}{|\beta_{2}+b|^{2}}+\sin^{2}((\beta_{2}-b+\phi-\Phi)/2)\right\\}^{\frac{\delta}{2}}|\beta_{1}^{\frac{-1-\gamma}{1+\alpha}}+b^{\frac{-1-\gamma}{1+\alpha}}|\left\\{\frac{|\beta_{1}-b|^{2}}{|\beta_{1}+b|^{2}}+\sin^{2}((\beta_{1}-b+\phi-\Phi)/2)\right\\}^{\frac{1+\gamma}{2}}}$

Using the ideas for (3.3) we finally get

$|\mathcal{H}_{2,1}|\leq
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}||h_{1}||_{L^{\infty}_{\sigma}}||h_{2}||_{L^{p}},$

for some constant $C$, provided that $\delta<1-\gamma$, $\delta<\sigma$ and
$p>\frac{1}{1-\gamma}$.

Let us move to $\mathcal{H}_{2,3}$ and show the main idea. In order to work
with that term, we should estimate the difference of $\partial_{\varphi}D(f)$
in two points. Note that

$\displaystyle\partial_{\varphi}D(f)(\beta,\phi,b,\Phi)=2\left(\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}\right)$
$\displaystyle\times\partial_{\varphi}\left[\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}\right]$
$\displaystyle+2\partial_{\varphi}\left[\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}\right]b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}(1-\cos(\phi+\beta-\Phi-b)).$

By working with that term as for (3.3) we obtain

$\displaystyle|\partial_{\varphi}D(f)(\beta_{1},\phi,b,\Phi)-\partial_{\varphi}D(f)(\beta_{2},\phi,b,\Phi)|\leq
C$ $\displaystyle\frac{|\beta_{1}-\beta_{2}|}{|\beta_{1}+\beta_{2}|}$ (79)
$\displaystyle\times(\beta_{1}^{-1}\beta_{1}^{\frac{-2}{1+\alpha}}+\beta_{2}^{-1}\beta_{2}^{\frac{-2}{1+\alpha}}+\beta_{1}^{-1}\beta_{1}^{\frac{-2}{1+\alpha}+1}+\beta_{2}^{-1}\beta_{2}^{\frac{-2}{1+\alpha}+1}),$

which interpolating with (74) agrees with

$\displaystyle|\partial_{\varphi}D(f)(\beta_{1},\phi,b,\Phi)-\partial_{\varphi}D(f)(\beta_{2},\phi,b,\Phi)|\leq
C\frac{|\beta_{1}-\beta_{2}|^{\delta}}{|\beta_{1}+\beta_{2}|^{\delta}}$ (80)
$\displaystyle\times(\beta_{1}^{-\delta}\beta_{1}^{\frac{-2\delta}{1+\alpha}}+\beta_{2}^{-\delta}\beta_{2}^{\frac{-2\delta}{1+\alpha}}+\beta_{1}^{-\delta}\beta_{1}^{\frac{-\delta}{1+\alpha}+\delta}+\beta_{2}^{-\delta}\beta_{2}^{\frac{-\delta}{1+\alpha}+\delta})$
$\displaystyle\times(\beta_{1}^{-\frac{1-\delta}{1+\alpha}}\beta_{1}^{-(1-\delta)}D(f)(\beta_{1},\phi,b,\Phi)^{\frac{1-\delta}{2}}+\beta_{2}^{-\frac{1-\delta}{1+\alpha}}\beta_{2}^{-(1-\delta)}D(f)(\beta_{2},\phi,b,\Phi)^{\frac{1-\delta}{2}}).$

Then, the key point to bound $\mathcal{H}_{2,2}$ is the use of (80) instead of
just (79).

Finally, the estimate for the last term $\mathcal{H}_{2,4}$ follows
identically as for $\beta\partial_{\varphi}\mathcal{H}_{f}\in
L^{\infty}_{\sigma}$ taking into account that

$|\beta_{1}^{\frac{2}{1+\alpha}}-\beta_{2}^{\frac{2}{1+\alpha}}|\leq
C\frac{|\beta_{1}-\beta_{2}|}{|\beta_{1}+\beta_{2}|}(\beta_{1}^{\frac{2}{1+\alpha}}+\beta_{2}^{\frac{2}{1+\alpha}})\leq
C\frac{|\beta_{1}-\beta_{2}|}{|\beta_{1}+\beta_{2}|}\beta_{2}^{\frac{2}{1+\alpha}}.$

The $C^{1}$ regularity of $F$ follows similarly by estimating the associated
Fréchet derivatives. ∎

In the next proposition, we give the well-definition of the nonlinear function
together with its $C^{1}$ regularity.

######  Proposition 3.10.

Let $\gamma,\sigma\in(0,1)$ and $\alpha\in(1,2)$ satisfying (56). There exists
$\epsilon<1$ such that $\tilde{F}:B_{X_{1}^{\sigma,\delta}}(\epsilon)\times
X_{2}^{p}\rightarrow Y^{\sigma,\delta}$ is well-defined and $C^{1}$, for any
$p>\frac{1}{1-\gamma}$ and $\delta<\min\\{1-\gamma,\sigma\\}$.

###### Proof.

Let us prove that $\tilde{F}$ is well-defined and $C^{1}$ in $f$, where we
shall use the expression of $\tilde{F}(f,\Omega)$ in (2.4). Note that the
first term in the expression of $\tilde{F}(f,\Omega)$ is just the constant
$1$, which is not in $Y^{\sigma,\delta}$ since it does not decay appropriately
at $0$ and $\infty$. For that reason, we need to add and subtract the
contribution of the trivial solution in $\tilde{F}$ obtaining that

$\displaystyle\tilde{F}(f,\Omega)(\beta,\phi)=f(\beta,\phi)-(-1)^{\frac{-1}{2\mu}}\frac{C_{\gamma}(\alpha-1)^{\frac{\alpha-1}{2}}}{4\pi
C_{0}(1+\alpha)^{\frac{\alpha-1}{2}}}(I_{1}(f,\Omega)+I_{2}(f,\Omega)+I_{3}(f,\Omega)+I_{4}(f,\Omega))(\beta,\phi),$

where

$\displaystyle
I_{1}(f,\Omega):=\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f(b,\Phi)\right\\}^{-\frac{1}{2\mu}}\partial_{\overline{\beta\varphi}}f(b,\Phi)\Omega(\Phi)dbd\Phi}{D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}},$
$\displaystyle
I_{2}(f,\Omega):=\frac{2(\alpha-1)}{(1+\alpha)^{2}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\left[\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f(b,\Phi)\right\\}^{-\frac{1}{2\mu}}-\left\\{-\frac{\alpha-1}{1+\alpha}\right\\}^{-\frac{1}{2\mu}}\right]\Omega(\Phi)dbd\Phi}{D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}},$
$\displaystyle
I_{3}(f,\Omega):=\frac{2(\alpha-1)}{(1+\alpha)^{2}}\left\\{-\frac{\alpha-1}{1+\alpha}\right\\}^{-\frac{1}{2\mu}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\Omega(\Phi)}{D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}}$
$\displaystyle\times\left\\{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}-D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}\right\\}dbd\Phi,$
$\displaystyle
I_{4}(f,\Omega):=\frac{2(\alpha-1)}{(1+\alpha)^{2}}\left\\{-\frac{\alpha-1}{1+\alpha}\right\\}^{-\frac{1}{2\mu}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\Omega(\Phi)-1)dbd\Phi}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}}.$

Note that now $f\in Y^{\sigma,\delta}$ and it remains to prove that $I_{i}\in
Y^{\sigma,\delta}$, for $i=1,2,3,4$.

By using the operator $\mathcal{H}$ defined in Proposition 3.9 we can write
$I_{1}$, $I_{2}$ and $I_{4}$ as:

$\displaystyle I_{1}(f,\Omega)=$
$\displaystyle\alpha_{1}\mathcal{H}\left(\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f\right\\}^{-\frac{1}{2\mu}}\partial_{\overline{\beta\varphi}}f,\Omega\right),$
$\displaystyle I_{2}(f,\Omega)=$
$\displaystyle\alpha_{2}\mathcal{H}\left(\left[\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f\right\\}^{-\frac{1}{2\mu}}-\left\\{-\frac{\alpha-1}{1+\alpha}\right\\}^{-\frac{1}{2\mu}}\right],\Omega\right),$

for some constants $\alpha_{1},\alpha_{2},\alpha_{3}$. Then, by Proposition
3.9 we find

$\displaystyle||I_{1}||_{Y^{\sigma,\delta}}+||I_{2}||_{Y^{\sigma,\delta}}$
$\displaystyle\leq$ $\displaystyle
C||\Omega||_{X_{2}^{p}}\left\\{\left\|\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f\right\\}^{-\frac{1}{2\mu}}\partial_{\overline{\beta\varphi}}f\right\|_{\mathcal{X}_{\sigma,\alpha}}+\left\|\left[\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f\right\\}^{-\frac{1}{2\mu}}-\left\\{-\frac{\alpha-1}{1+\alpha}\right\\}^{-\frac{1}{2\mu}}\right]\right\|_{\mathcal{X}_{\sigma,\alpha}}\right\\}$
$\displaystyle\leq$ $\displaystyle
C||\Omega||_{X_{2}^{p}}||f||_{X_{1}^{\sigma,\alpha}}.$

Moreover, Proposition 3.9 gives us that $I_{1}$ and $I_{2}$ are $C^{1}$ in
$f$. It remains to check the well-posedness and $C^{1}$ regularity of $I_{3}$
and $I_{4}$. Let us start with $I_{4}$.

Note that

$\displaystyle I_{4}(f,\Omega)(\beta,\phi)=$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\Omega(\Phi)-1)dbd\Phi}{\left\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle=$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}\beta^{\frac{\gamma}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-2}{1+\alpha}}(\Omega(\Phi)-1)dbd\Phi}{\left\\{\beta^{\frac{2}{1+\alpha}}+b^{\frac{2}{1+\alpha}}-2\beta^{\frac{1}{1+\alpha}}b^{\frac{1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle=$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}\beta^{\frac{\gamma}{1+\alpha}}I_{4,1}(f,\Omega)(\beta,\phi),$

where $C$ is some constant which is not important in this proof. Let us first
check the $L^{\infty}_{\sigma}$ norm. Take first $\beta\in[0,\beta_{0}]$.
Since we have in the numerator $\Omega-1$, we get an extra cancellation at
$\beta=0$. This comes from the expression of $\Omega-1$:

$\Omega(\phi)-1=\sum_{0\neq k\in\operatorname{\mathbb{Z}}}g_{k}e^{ik\phi},$

where the mode $k=0$ vanishes implying that

$\int_{0}^{2\pi}(\Omega(\Phi)-1)d\Phi=0.$

In that way, we obtain that

$\displaystyle I_{4,1}(f,\Omega)(0,\phi)=$
$\displaystyle\lim_{\varepsilon\rightarrow
0}\int_{0}^{2\pi}\int_{\varepsilon}^{1/\varepsilon}\frac{b^{\frac{-2}{1+\alpha}}(\Omega(\Phi)-1)dbd\Phi}{\left\\{b^{\frac{2}{1+\alpha}}\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle=$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\int_{\varepsilon}^{1/\varepsilon}\frac{b^{\frac{-2}{1+\alpha}}db}{\left\\{b^{\frac{2}{1+\alpha}}\right\\}^{\frac{\gamma}{2}}}\int_{0}^{2\pi}(\Omega(\Phi)-1)d\Phi$
$\displaystyle=$ $\displaystyle 0.$

Then:

$\displaystyle\sup_{\beta\in[0,\beta_{0}]}\frac{(1+\beta)^{2\sigma}}{\beta^{\sigma}}I_{4}(f,\Omega)(\beta,\phi)\leq\sup_{\beta\in[0,\beta_{0}]}\frac{\beta^{-\frac{\alpha-1}{1+\alpha}}\beta^{\frac{\gamma}{1+\alpha}}}{\beta^{\sigma}}<C,$

since $\sigma<\frac{\gamma+1-\alpha}{1+\alpha}$ due to (56). Now take
$\beta\in[\beta_{0},\infty)$ and note that $I_{4}$ can be also written as

$\displaystyle I_{4}(f,\Omega)(\beta,\phi)=$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\Omega(\Phi)-1)dbd\Phi}{\left\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}^{\frac{\gamma}{2}}}$
(81) $\displaystyle=$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}I_{4,2}(f,\Omega)(\beta,\phi),$

where

$\lim_{\beta\rightarrow+\infty}I_{4,2}(f,\Omega)(\beta,\phi)=0,$

since $\Omega-1$ has a vanishing $0$ Fourier mode as before. Since
$\sigma<\frac{\alpha-1}{1+\alpha}$ by (56) we get

$\sup_{\beta\in[\beta_{0},+\infty)}I_{4}(f,\Omega)(\beta,\phi)<C.$

Now make the change of variables $b=\beta z$ in the expression of $I_{4}$ to
get

$\displaystyle I_{4}(f,\Omega)(\beta,\phi)=$ $\displaystyle
C\int_{0}^{2\pi}\int_{0}^{\infty}\frac{z^{\frac{-2}{1+\alpha}}(\Omega(\Phi)-1)dbd\Phi}{\left\\{1+z^{\frac{2}{1+\alpha}}-2z^{\frac{1}{1+\alpha}}\cos(\phi+\beta-\Phi-z\beta)\right\\}^{\frac{\gamma}{2}}},$
(82)

where we can prove easily the $C^{\delta}$ regularity in $\beta$ and the Lip
regularity in $\phi$ using such expression and the ideas developed in
Proposition 3.9. Taking $\partial_{\varphi}$ to (81) we find

$\displaystyle\partial_{\varphi}I_{4}(f,\Omega)(\beta,\phi)=C\beta^{-1}I_{4}(f,\Omega)(\beta,\phi)$
$\displaystyle+C\beta^{-1}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\Omega(\Phi)-1)\left\\{\beta^{\frac{-2}{1+\alpha}}-\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}dbd\Phi}{\left\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}^{\frac{2+\gamma}{2}}}.$

Let us check the $L^{\infty}_{\sigma}$ norm of $\beta\partial_{\varphi}I_{4}$.
The first term is clearly in $L^{\infty}_{\sigma}$ and the second term can be
written as

$\displaystyle\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\Omega(\Phi)-1)\left\\{\beta^{\frac{-2}{1+\alpha}}-\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}dbd\Phi}{\left\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}^{\frac{2+\gamma}{2}}}$
$\displaystyle=$
$\displaystyle\beta^{-\frac{\alpha-1}{1+\alpha}}I_{4,3}(\beta,\phi)$
$\displaystyle=$
$\displaystyle\beta^{-\frac{\alpha-1}{1+\alpha}}\beta^{\frac{\gamma}{1+\alpha}}I_{4,4}(\beta,\phi)$

with $I_{4,4}(0,\phi)=0$ due to the structure of $\Omega-1$, and

$\lim_{\beta\rightarrow+\infty}I_{4,3}(\beta,\phi)=0.$

Arguing as before for $I_{4}$ working with $I_{4,4}$ for
$\beta\in[0,\beta_{0}]$ and with $I_{4,3}$ for $\beta\in[\beta_{0},+\infty)$
we find that $\beta\partial_{\varphi}I_{4}\in L^{\infty}_{\sigma}$. In the
same way we can prove the Hölder continuity of such term and the $C^{1}$
regularity in $f$.

Let us work with $I_{3}$. A priori, it can not be written in terms of
$\mathcal{H}$, however it has the same scaling and the estimates works
similarly. Let us just gave the first ideas to estimate it. Recall

$\displaystyle I_{3}(f,\Omega)=$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\Omega(\Phi)}{D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}}$
$\displaystyle\times\left\\{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}-D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}\right\\}dbd\Phi.$

Notice that

$\displaystyle|D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}-D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}|\leq$
$\displaystyle
C\frac{|D(0)(\beta,\phi,b,\Phi)-D(f)(\beta,\phi,b,\Phi)|}{|D(0)(\beta,\phi,b,\Phi)^{1-\frac{\gamma}{2}}+D(f)(\beta,\phi,b,\Phi)^{1-\frac{\gamma}{2}}|}.$

Via Proposition 3.7 and Remark 3.8, we have that $D(f)$ and $D(0)$ are
comparable: $D(f)(\beta,\phi,b,\Phi)\geq D(0)(\beta,\phi,b,\Phi)$ and then

$|I_{3}(f,\Omega)|\leq
C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}|\Omega(\Phi)|}{D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}}\frac{|D(0)(\beta,\phi,b,\Phi)-D(f)(\beta,\phi,b,\Phi)|}{D(0)(\beta,\phi,b,\Phi)}dbd\Phi.$
(83)

Moreover, using the expression of $D(f)$ in (51) we find

$\displaystyle|D(0)(\beta,\phi,b,\Phi)-D(f)(\beta,\phi,b,\Phi)|\leq
C|D(0)(\beta,\phi,b,\Phi)^{\frac{1}{2}}+D(f)(\beta,\phi,b,\Phi)^{\frac{1}{2}}|$
$\displaystyle\times\Bigg{[}\left|\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-\beta^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}\right\\}^{\frac{1}{2}}\right|$
$\displaystyle+\left|b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}-b^{\frac{-1}{1+\alpha}}\left\\{\frac{\alpha-1}{1+\alpha}\right\\}^{\frac{1}{2}}\right|$
$\displaystyle+\beta^{\frac{-1}{1+\alpha}}\left|\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-\left\\{\frac{\alpha-1}{1+\alpha}\right\\}^{\frac{1}{2}}\right|\left|\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\phi)\right|^{\frac{1}{2}}||\sin((\beta+\phi-b-\Phi)/2)|$
$\displaystyle+b^{\frac{-1}{1+\alpha}}\left|\frac{\alpha-1}{1+\alpha}\right|^{\frac{1}{2}}\left|\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}-\left\\{\frac{\alpha-1}{1+\alpha}\right\\}^{\frac{1}{2}}\right||\sin((\beta+\phi-b-\Phi)/2)|\Bigg{]}.$

Since $\partial_{b}f(\beta,\Phi)$ is bounded in $L^{\infty}$ we get

$\displaystyle|D(0)(\beta,\phi,b,\Phi)-D(f)(\beta,\phi,b,\Phi)|\leq
C|D(0)(\beta,\phi,b,\Phi)^{\frac{1}{2}}+D(f)(\beta,\phi,b,\Phi)^{\frac{1}{2}}|$
$\displaystyle\times\Bigg{[}\beta^{\frac{-1}{1+\alpha}}\left|\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\beta}}f(\beta,\phi)\right\\}^{\frac{1}{2}}-\left\\{\frac{\alpha-1}{1+\alpha}\right\\}^{\frac{1}{2}}\right|+b^{\frac{-1}{1+\alpha}}\left|\left\\{\frac{\alpha-1}{1+\alpha}+\partial_{\overline{b}}f(b,\Phi)\right\\}^{\frac{1}{2}}-\left\\{\frac{\alpha-1}{1+\alpha}\right\\}^{\frac{1}{2}}\right|\Bigg{]}.$

Now, we wish to take into account the bound for $f$ and hence using that the
norm of $f$ is small, we can write it as

$\displaystyle|D(0)(\beta,\phi,b,\Phi)-D(f)(\beta,\phi,b,\Phi)|\leq$
$\displaystyle
C|D(0)(\beta,\phi,b,\Phi)^{\frac{1}{2}}+D(f)(\beta,\phi,b,\Phi)^{\frac{1}{2}}|$
$\displaystyle\times\Bigg{[}\beta^{\frac{-1}{1+\alpha}}\left|\partial_{\overline{\beta}}f(\beta,\phi)\right|+b^{\frac{-1}{1+\alpha}}\left|\partial_{\overline{b}}f(b,\Phi)\right|\Bigg{]}.$

In this way, we separate $I_{3}$ in two integrals and use again Remark 3.8 to
compare $D(f)$ with $D(0)$:

$\displaystyle|I_{3}(f,\Omega)|\leq$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{\frac{-\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}|\Omega(\Phi)|}{D(f)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}}\frac{\beta^{\frac{-1}{1+\alpha}}|\partial_{\overline{\beta}}f(\beta,\phi)|+b^{\frac{-1}{1+\alpha}}|\partial_{\overline{b}}f(b,\Phi)|}{D(0)(\beta,\phi,b,\Phi)^{\frac{1}{2}}}dbd\Phi.$

By using similar arguments than in Proposition 3.9, we finally achieve that

$||I_{3}||_{L^{\infty}_{\sigma}}\leq
C||\Omega||_{L^{p}}||\partial_{\overline{\beta}}f||_{L^{\infty}_{\sigma}},$

for any $p>1$.

By working with $D(f)$ and using the ideas of Proposition 3.9, we can check
that the others norms in $X_{1}^{\sigma,\delta}$ are also bounded. ∎

##  4\. Linearized operator

This section aims to give different expression of the linear operator of
$\tilde{F}$ at the trivial solution $(0,1)$. First, we will show that there is
an extra cancellation that help us to prove that the linear operator is a
compact perturbation of a isomorphism. Hence, we can conclude that the
linearized operator is Fredholm of zero index, which will be useful in
checking that it is a isomorphism in the following section. Moreover, at the
end of this section, we give its expression in Fourier series.

###  4.1. Linearization at the equilibrium

First, let us give the expression for the linear operator of $\tilde{F}$ at
$(0,1)$.

######  Proposition 4.1.

The linearized operator of $\tilde{F}$ at the trivial solution $(0,1)$ reads
as follows:

$\displaystyle\partial_{f}\tilde{F}(0,1)h(\beta,\phi)=$
$\displaystyle\frac{(1+\alpha)}{2}\partial_{\beta}(\beta h(\beta,\phi))$
$\displaystyle-\frac{C_{\gamma}}{4\pi
C_{0}}\frac{1+\alpha}{\alpha-1}\left(\frac{\alpha-1}{1+\alpha}\right)^{\frac{\gamma}{2}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta^{-\frac{\gamma}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}\partial_{\overline{b\varPsi}}h(b,\Phi)dbd\Phi}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}},$
$\displaystyle-\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\frac{1}{\alpha-1}\left(\frac{\alpha-1}{1+\alpha}\right)^{\frac{\gamma}{2}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta^{-\frac{\gamma}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}\partial_{\overline{\varPsi}}h(b,\Phi)dbd\Phi}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}},$
$\displaystyle+\frac{\gamma}{2}\frac{C_{\gamma}}{2\pi
C_{0}}\frac{1}{1+\alpha}\left(\frac{\alpha-1}{1+\alpha}\right)^{\frac{\gamma}{2}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta^{-\frac{\gamma}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}+1}}$
$\displaystyle\times\left\\{b^{\frac{-2}{1+\alpha}}-\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}\partial_{\overline{b}}h(b,\Phi)dbd\Phi$
$\displaystyle=:$ $\displaystyle\frac{(1+\alpha)}{2}\partial_{\beta}(\beta
h(\beta,\phi))+I_{1}+I_{2}+I_{3}.$ (84)

###### Proof.

Note that

$\displaystyle\partial_{f}\tilde{F}(0,1)h(\beta,\phi)=h(\beta,\phi)-L(h)(\beta,\phi)$
$\displaystyle-\frac{C_{\gamma}(\alpha-1)^{\frac{\alpha-1}{2}}}{4\pi
C_{0}(1+\alpha)^{\frac{\alpha-1}{2}}}\left\\{\frac{\alpha-1}{1+\alpha}\right\\}^{-\frac{1}{2\mu}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta^{-\frac{\gamma}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}\partial_{\overline{b\varPsi}}h(b,\Phi)dbd\Phi}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}},$
$\displaystyle-\frac{1}{2\mu}\frac{C_{\gamma}(\alpha-1)^{\frac{\alpha-1}{2}}}{4\pi
C_{0}(1+\alpha)^{\frac{\alpha-1}{2}}}\left\\{\frac{\alpha-1}{1+\alpha}\right\\}^{-\frac{1}{2\mu}-1}\left\\{\frac{2(\alpha-1)}{(1+\alpha)^{2}}\right\\}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta^{-\frac{\gamma}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}\partial_{\overline{\varPsi}}h(b,\Phi)dbd\Phi}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}},$
$\displaystyle+\frac{\gamma}{2}\frac{C_{\gamma}(\alpha-1)^{\frac{\alpha-1}{2}}}{4\pi
C_{0}(1+\alpha)^{\frac{\alpha-1}{2}}}\left\\{\frac{\alpha-1}{1+\alpha}\right\\}^{-\frac{1}{2\mu}}\left\\{\frac{2(\alpha-1)}{(1+\alpha)^{2}}\right\\}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta^{-\frac{\gamma}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}+1}}$