# Global smooth solutions for the inviscid SQG equation

Angel Castro, Diego Córdoba and Javier Gómez-Serrano

###### Abstract

In this paper, we show the existence of the first non trivial family of
classical global solutions of the inviscid surface quasi-geostrophic equation.

Keywords: global existence, surface quasi-geostrophic, incompressible,
computer-assisted

###### Contents

  1. 1 Introduction
  2. 2 The equations
  3. 3 Main theorem and Crandall-Rabinowitz (C-R) theorem
  4. 4 Checking the hypotheses of the C-R theorem for the equation (2.7)
    1. 4.1 Step 1. The functional setting and the hypothesis 1
    2. 4.2 Step 2. The partial derivatives of the functional $F$
    3. 4.3 Step 3. Analysis of the linear operator
      1. 4.3.1 The equation for the radial part
      2. 4.3.2 Existence of an element in the kernel of $\partial_{r}F[\rho,\lambda_{3}]$
      3. 4.3.3 One-dimensionality of the kernel of $\partial_{r}F[\rho,\lambda_{3}]$
    4. 4.4 Step 4. The transversality property 4
  5. A Asymptotics
  6. B Implementation of the computer-assisted part and rigorous numerical results
  7. C Finite projections

##  1 Introduction

We consider the initial value problem for the inviscid surface quasi-
geostrophic equation (SQG):

$\displaystyle\partial_{t}\theta(x,t)+u(x,t)\cdot\nabla\theta(x,t)$
$\displaystyle=0,\ \ (x,t)\in\mathbb{R}^{2}\times\mathbb{R}_{+}$ (1.1)
$\displaystyle u(x,t)$ $\displaystyle=(-R_{2}(\theta),R_{1}(\theta))(x,t)$
$\displaystyle\theta(x,0)$ $\displaystyle=\theta_{0}(x),$

where $R_{j}$ is the $j$-th Riesz transform:

$\displaystyle
R_{j}(\theta)(x)=\frac{1}{2\pi}P.V.\int_{\mathbb{R}^{2}}\frac{(x_{j}-y_{j})}{|x-y|^{3}}\theta(y)dy.$

This equation is derived considering small Rossby and Ekman numbers and
constant potential vorticity. It models the evolution of the temperature from
a general quasi-geostrophic system for atmospheric and oceanic flows (see [15,
33, 49, 44] for more details). The numerical and analytical study of the
equation was started by Constantin, Majda and Tabak in [15], since the SQG
system presents an analogy with the 3D Euler equations.

The aim of this paper is to address the main problem of whether its classical
solution corresponding to given initial data $\theta(x,0)=\theta_{0}(x)$ with
finite energy exists for all time or not. We remark that both the $L^{p}$
norms of theta $(1\leq p\leq\infty)$ and the $L^{2}$ norm of $u$ (the energy
of the system) are conserved quantities. Moreover, the $L^{p}$ norms of $u$
obey the following bounds:

$||u(\cdot,t)||_{L^{p}}\leq C_{p}||\theta_{0}||_{L^{p}}\quad 1<p<\infty.$

Local existence of solutions for (1.1) was first shown in [15] in Sobolev
spaces. By using different functional frameworks local existence has been also
addressed in several papers, see for example [9, 43, 56, 57].

Resnick, in his thesis [50], showed global existence of weak solutions in
$L^{2}$ using an extra cancellation due to the oddness of the Riesz transform.
Marchand [46] 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 in
$L^{2}$ is still a challenging open problem (see [37, 1, 52] and references
therein).

The problem of whether the SQG system presents finite time singularities or
there is global existence is open for the smooth case. Kiselev and Nazarov
[39] constructed solutions that started arbitrarily small but grew arbitrarily
big in finite time, and Friedlander and Shvydkoy [28] showed the existence of
unstable eigenvalues of the spectrum. Castro and Córdoba constructed singular
solutions with infinite energy in [5] and Dritschel [25] constructed global
solutions that have $C^{1/2}$ regularity.

The numerical simulations in [15] indicated a possible singularity in the form
of a hyperbolic saddle closing in finite time. Ohkitani and Yamada [48] and
Constantin et al [16] suggested that the growth was double exponential. The
question was settled by Cordoba [17] who bounded the growth by a quadruple
exponential and further improved by Cordoba and Fefferman [18] to a double
exponential (see also [24]). The same scenario was recomputed almost 20 years
with bigger computational power and improved algorithms by Constantin et al.
[14], yielding no evidence of blowup and the depletion of the hyperbolic
saddle past the previously computed times. In [45], Majda and Tabak compared
simulations for the SQG scenario with the Euler case. Scott, in [53], starting
from elliptical configurations, proposed a candidate that develops
filamentation and after a few cascades, blowup of $\nabla\theta$.

Several criteria of blowup have been found and blowup can only occur through
the blowup of either some geometric quantities or certain space-time norms.
For more details see [13, 10, 11, 12, 19, 36, 38, 4].

A different approach for the study of the formation of singularities for SQG
comes from the patch-problem, i.e., “sharp fronts” . In this problem one
considers that the scalar $\theta(x,t)$ is the characteristic function of some
compact and simple connected domain which depends on time and with smooth
boundary. Local existence for the patch was proven by Rodrigo [51] for a
$C^{\infty}$ boundary and by Gancedo [29] in Sobolev spaces. Córdoba et al.
found, in [21], strong numerical evidences of the formation of a singularity
in the boundary of the patch. For further numerical simulations addressing
formation of singularities see [54] and [53]. The possibility of a splash
singularity scenario (i.e. when the interface touches itself on a point but
the curve does not lose regularity) was ruled out by Gancedo and Strain [30].

Through a different motivation, Cordoba et al. [20], Fefferman and Rodrigo
[27] and Fefferman et al. [26] studied the existence of a special type of
solutions that are known as “almost sharp fronts” for SQG. These solutions can
be thought of as a regularization of a front, with a small strip around the
front in which the solution changes (reasonably) from one value of the front
to the other. These are strong solutions of the equation with large gradient
($\sim\text{(Width of the strip)}^{-1}$).

The main purpose of the paper is to show the following theorem:

###### Theorem 1.1

There is a nontrivial global smooth solution for the SQG equations that has
finite energy.

It is well known that radial functions are stationary solutions for (1.1) due
to the structure of the nonlinear term. The solutions that will be constructed
in this paper are a smooth perturbation in a suitable direction of a specific
radial function. The smooth profile we will perturb satisfies (in polar
coordinates)

$\displaystyle\theta(r)\equiv\left\\{\begin{array}[]{ccc}1&\text{for $0\leq
r\leq 1-a$}\\\ \text{smooth and decreasing}&\quad\text{for $1-a<r<1$}\\\
0&\text{for $1\leq r<\infty$}\end{array}\right.,$

where $a$ is a small number (below we will impose some more constraints in
this profile). In addition the dynamics of these solutions consist of global
rotating level sets with constant angular velocity. These level sets are a
perturbation of the circle. The limit case $a=0$ gives rise to the well known
V-state solution for SQG, i.e., a global rotating patch which solves weakly
(1.1). The existence of V-states with $C^{\infty}$ boundary for SQG was proven
in [7]. It was shown in [8] that the boundary of these solutions is actually
analytic.

The proofs of these results are motivated by the ones for 2D incompressible
Euler in the simply connected case; Burbea in [3] proved the existence of
V-states for Euler and $C^{\infty}$-regularity for its boundary was proved by
Hmidi at al. in [34] (see also [32]).

The paper is organized as follows: section 2 is devoted to the reformulation
of the equations (1.1) in new variables. In section 3 we state the main
theorem and present the Crandall-Rabinowitz (C-R) theorem which will be the
main tool in our proof. In section 4 we check that our equation satisfies the
hypotheses of the C-R theorem. This will be the main part of our work.

In particular section 4.3 is different from previous analysis. We stress the
following main differences:

  * •

The study of the linear problem is now reduced to a functional equation, as
opposed to a scalar equation (which was in the patch case). Even the existence
of nontrivial elements in the kernel of the linear part is not evident a
priori.

  * •

There is no algebraic formula for neither the eigenvalue nor the eigenvector,
not even in an implicit way (such as in [8]). This makes the proof of the
dimensionality of the kernel much harder since one needs to show that the
eigenvalue is simple and have some control of the rest of the eigenvalues.

###### Remark 1.2

In a forthcoming paper [6], by using the same techniques, we are able to
extend our construction to the 2D Incompressible Euler equations.

Finally, in Appendix A, we compute the asymptotics and bounds on the error
terms of some of the elliptic integrals that appear and Appendix B is devoted
to discuss the details of the computer-assisted code and its implementation,
as well as to show the rigorous numerical bounds used in the theorems.
Appendix C contains an explicit description of two big matrices used in the
proofs.

A major theme of our work is the interplay between rigorous computer
calculations and traditional mathematics. We use interval arithmetics as part
of a proof whenever they are needed.

Advances in computing power have made rigorous computer-assisted proofs
realizable. Naturally, floating-point operations can result in numerical
errors. In order to overcome these, we will employ interval arithmetics to
deal with this issue. The main paradigm is the following: instead of working
with arbitrary real numbers, we perform computations over intervals which have
representable numbers by the computer as endpoints in order to guarantee that
the true result at any point belongs to the interval by which is represented.
On these objects, an arithmetic is defined in such a way that we are
guaranteed that for every $x\in X,y\in Y$

$\displaystyle x\star y\in X\star Y,$

for any operation $\star$. For example,

$\displaystyle[\underline{x},\overline{x}]+[\underline{y},\overline{y}]$
$\displaystyle=[\underline{x}+\underline{y},\overline{x}+\overline{y}]$
$\displaystyle[\underline{x},\overline{x}]\times[\underline{y},\overline{y}]$
$\displaystyle=[\min\\{\underline{x}\underline{y},\underline{x}\overline{y},\overline{x}\underline{y},\overline{x}\overline{y}\\},\max\\{\underline{x}\underline{y},\underline{x}\overline{y},\overline{x}\underline{y},\overline{x}\overline{y}\\}].$

We can also define the interval version of a function $f(X)$ as an interval
$I$ that satisfies that for every $x\in X$, $f(x)\in I$. Rigorous computation
of integrals has been theoretically developed since the seminal works of Moore
and many others (see [2, 22, 40, 41, 47, 55] for just a small sample). In our
computations, all arithmetic will be double precision (64 bits).

##  2 The equations

In this section we describe the equation that a global rotating solution with
constant angular velocity of SQG must satisfy. We will look to the level sets
of this solution rather than the solution itself. Let’s assume that
$\theta(x,t)$ is a smooth solution of (1.1) with initial data $\theta_{0}(x)$.
On $\theta_{0}(x)$ we will assume that its level sets can be parameterized by
$z_{0}(\alpha,\rho)$ in such a way that

$\theta_{0}(z_{0}(\alpha,\rho))=f(\rho)$

for some smooth and even function $f\,:\,\mathbb{R}\to\mathbb{R}$. The
application $z_{0}(\alpha,\rho)$ satisfies:

  1. 1.

It is one to one from
$\\{(\alpha,\rho)\in\mathbb{R}^{2}\,:\,-\pi\leq\alpha<\pi,\,0<\rho<\infty\\}$
to $\mathbb{R}^{2}\setminus\\{0\\}$.

  2. 2.

For all $\alpha$, $z_{0}(\alpha,0)=(0,0)$.

  3. 3.

For fixed $\rho>0$, $z^{i}_{0}(\alpha,\rho)$, with $i=1,2$, are
$2\pi$-periodic and $z_{0}(\alpha,\rho)$, with $-\pi\leq\alpha<\pi$, is a
closed and $C^{1}$ curve in $\mathbb{R}^{2}$ satisfying the chord-arc
condition. Also we parametrize this curve counterclockwise with $\alpha$.

  4. 4.

It is differentiable with respect to $\alpha$ and $\rho$ in
$\mathbb{T}\times[0,\infty)$, $|z_{0\,\rho}(\alpha,\rho)|>c>0$ in
$\mathbb{T}\times[0,\infty)$, $|z_{0\,\alpha}(\alpha,\rho)|>0$ in
$\mathbb{T}\times(0,\infty)$ and $z_{0\,\alpha}^{\perp}(\alpha,\rho)\cdot
z_{0\,\rho}(\alpha,\rho)<0\quad\text{for $-\pi\leq\alpha<\pi$ and
$0<\rho<\infty$}.$

Because of the transport character of the equation (1.1) and by continuity we
can assume that the level sets of the solution $\theta(x,t)$ at time $t$ can
be parameterized by an application $z(\alpha,\rho,t)$ such that

$\displaystyle\theta(z(\alpha,\rho,t),t)=f(\rho),$ (2.1)

and it satisfies properties 1, 3 and 4.

Property 2 changes to

$z(\alpha,0,t)=c(t)\quad\text{for all $\alpha\in\mathbb{T}$},$

where, here, $c(t)$ is a vector in $\mathbb{R}^{2}$.

Differentiating (2.1) with respect to $\alpha$ and with respect to $\rho$ we
have that

$\displaystyle
z_{\alpha}(\alpha,\rho,t)\cdot\nabla\theta(z(\alpha,\rho,t),t)=0$
$\displaystyle
z_{\rho}(\alpha,\rho,t)\cdot\nabla\theta(z(\alpha,\rho,t),t)=f_{\rho}(\rho).$

Therefore

$\displaystyle\nabla\theta(z(\alpha,\rho,t))=\frac{f_{\rho}(\rho)}{z_{\alpha}^{\perp}\cdot
z_{\rho}(\alpha,\rho,t)}z_{\alpha}^{\perp}(\alpha,\rho,t),$ (2.2)

Taking a time derivative in (2.1) and using (1.1) and (2.2) yields

$\displaystyle 0=$
$\displaystyle\frac{d}{dt}\theta(z(\alpha,\rho,t),t)=\partial_{t}\theta(z(\alpha,\rho,t),t)+z_{t}(\alpha,\rho,t)\cdot\nabla\theta(z(\alpha,\rho,t),t)$
$\displaystyle=$
$\displaystyle(-u(z(\alpha,\rho,t),t)+z_{t}(\alpha,\rho,t))\cdot\nabla\theta(z(\alpha,\rho,t),t)$
$\displaystyle=$
$\displaystyle\left(-u(z(\alpha,\rho,t),t)+z_{t}(\alpha,\rho,t)\right)\cdot
z_{\alpha}^{\perp}(\alpha,\rho,t)\frac{f_{\rho}(\rho)}{z_{\alpha}^{\perp}\cdot
z_{\rho}(\alpha,\rho,t)}.$

This last expression is the equation that the level sets $z(\alpha,\rho,t)$ of
the solutions $\theta(x,t)$ satisfy.

Notice that since $u(x,t)=R^{\perp}\theta(x,t)$ we can also write
$u(x,t)=-\Lambda^{-1}\nabla^{\perp}\theta(x,t)$ and therefore

$\displaystyle u(z(\alpha,\rho,t),t)$
$\displaystyle=-\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\frac{1}{|z(\alpha,\rho,t)-y|}\nabla^{\perp}\theta(y,t)dy$
$\displaystyle=-\frac{1}{2\pi}\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{1}{|z(\alpha,\rho,t)-z(\alpha^{\prime},\rho^{\prime},t)|}f_{\rho}(\rho^{\prime})z_{\alpha}(\alpha^{\prime},\rho^{\prime})d\alpha^{\prime}d\rho^{\prime}.$

where we just did the change of variables
$y=z(\alpha^{\prime},\rho^{\prime},t)$ and used (2.2).

Conversely, if $z(\alpha,\rho,t)$ satisfies the equation

$\displaystyle\left(-u(z(\alpha,\rho,t),t)+z_{t}(\alpha,\rho,t)\right)\cdot
z_{\alpha}^{\perp}(\alpha,\rho,t)\frac{f_{\rho}(\rho)}{z_{\alpha}^{\perp}\cdot
z_{\rho}(\alpha,\rho,t)}=0$ (2.3) $\displaystyle
z(\alpha,\rho,0)=z_{0}(\alpha,\rho)$

with

$\displaystyle
u(z(\alpha,\rho,t),t)=-\frac{1}{2\pi}\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{1}{|z(\alpha,\rho,t)-z(\alpha^{\prime},\rho^{\prime},t)|}f_{\rho}(\rho^{\prime})z_{\alpha}(\alpha^{\prime},\rho^{\prime})d\alpha^{\prime}d\rho^{\prime},$
(2.4)

we can prove that the function
$\theta\,:\,\mathbb{R}^{2}\times\mathbb{R}\to\mathbb{R}$ defined by (2.1) is a
solution of the equation (1.1).

Let us assume now that ${\rm supp}\thinspace(f_{\rho})\subset[1-a,1]$, where
$0<a<1$ will be chosen later. Then, in order to find a solution for SQG, we
can solve the equation

$\displaystyle\left(z_{t}(\alpha,\rho,t)-u(z(\alpha,\rho,t),t)\right)\cdot
z_{\alpha}^{\perp}(\alpha,\rho,t)$ $\displaystyle=0,$

with $u(z(\alpha,\rho,t),t)$ given by (2.4), in the domain $\rho\in(1-a,1)$,
$\alpha\in\mathbb{T}$. After that we extend the solution $z(\alpha,\rho,t)$ to
the domain $0\leq\rho<\infty$, $-\pi\leq\alpha<\pi$ in a smooth way and
finally define $\theta(x,t)$ through equation (2.1).

In addition, we will assume that our solution rotates with angular velocity
$\lambda$, counterclockwise. Hence

$\displaystyle
z(\alpha,\rho,t)=\mathcal{O}(t)x(\alpha,\rho),\quad\mathcal{O}(t)=\left(\begin{array}[]{cc}\cos(\lambda
t)&-\sin(\lambda t)\\\ \sin(\lambda t)&\cos(\lambda t)\end{array}\right)$

This implies on one hand

$\displaystyle
z_{t}(\alpha,\rho,t)=\mathcal{O}_{t}(t)x(\alpha,\rho)=\lambda\left(\begin{array}[]{cc}-\sin(\lambda
t)&-\cos(\lambda t)\\\ \cos(\lambda t)&-\sin(\lambda
t)\end{array}\right)x(\alpha,\rho),\quad
z_{\alpha}(\rho,\alpha,t)=\mathcal{O}(t)x_{\alpha}(\alpha,\rho),$

and

$\displaystyle z_{t}\cdot
z_{\alpha}^{\perp}(\alpha,\rho,t)=\langle\mathcal{O}_{t}(t)x(\alpha,\rho),(\mathcal{O}(t)x_{\alpha}(\alpha,\rho))^{\perp}\rangle$
$\displaystyle=\langle\mathcal{O}_{t}(t)x(\alpha,\rho),\mathcal{O}(t)x_{\alpha}^{\perp}(\alpha,\rho)\rangle=\langle\mathcal{O}^{T}\mathcal{O}_{t}(t)x(\alpha,\rho),x_{\alpha}^{\perp}(\alpha,\rho)\rangle=-\lambda\langle
x(\alpha,\rho),x_{\alpha}(\alpha,\rho)\rangle.$

On the other hand from (2.4) we see that

$\displaystyle
u(z(\alpha,\rho,t),t)=-\mathcal{O}\frac{1}{2\pi}\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{1}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}f_{\rho}(\rho^{\prime})x_{\alpha}(\alpha^{\prime},\rho^{\prime})d\alpha^{\prime}d\rho^{\prime}\equiv\mathcal{O}\overline{u}(x(\alpha,\rho)).$

Thus

$\displaystyle\langle
u(z(\alpha,\rho,t),t),z_{\alpha}^{\perp}(\alpha,\rho,t)\rangle=\langle\mathcal{O}(t)\overline{u}(x(\alpha,\rho)),\mathcal{O}(t)x_{\alpha}^{\perp}(\alpha,\rho)\rangle=\langle\overline{u}(x(\alpha,\rho)),x_{\alpha}^{\perp}\rangle,$

which yields having to solve, for the pair $(x(\alpha,\rho),\lambda)$,

$\displaystyle-\lambda x(\alpha,\rho)\cdot
x_{\alpha}(\alpha,\rho)+\frac{1}{2\pi}x_{\alpha}^{\perp}(\alpha,\rho)\cdot\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{f_{\rho}(\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}x_{\alpha}(\alpha^{\prime},\rho^{\prime})d\alpha^{\prime}d\rho^{\prime}=0.$
(2.5)

We now write $x(\alpha,\rho)$ in polar coordinates

$\displaystyle x(\alpha,\rho)$
$\displaystyle=r(\alpha,\rho)(\cos(\alpha),\sin(\alpha)).$ (2.6)

This choice restricts the class of functions we are considering. However this
restriction will not be strong enough and we will be able to find a solution.

We have the following relations:

$\displaystyle x_{\alpha}(\alpha,\rho)$
$\displaystyle=r_{\alpha}(\alpha,\rho)(\cos(\alpha),\sin(\alpha))+r(\alpha,\rho)(-\sin(\alpha),\cos(\alpha))=r_{\alpha}(\alpha,\rho)n(\alpha)+r(\alpha,\rho)t(\alpha)$
$\displaystyle x_{\alpha}^{\perp}(\alpha,\rho)$
$\displaystyle=r_{\alpha}(\alpha,\rho)n^{\perp}(\alpha)+r(\alpha,\rho)t^{\perp}(\alpha)=r_{\alpha}(\alpha,\rho)t(\alpha)-r(\alpha,\rho)n(\alpha)$
$\displaystyle x_{\alpha}(\alpha^{\prime},\rho^{\prime})\cdot
x_{\alpha}^{\perp}(\alpha,\rho)$
$\displaystyle=r_{\alpha}(\alpha,\rho)r_{\alpha}(\alpha^{\prime},\rho^{\prime})n(\alpha^{\prime})\cdot
t(\alpha)-r_{\alpha}(\alpha^{\prime},\rho^{\prime})r(\alpha,\rho)n(\alpha)\cdot
n(\alpha^{\prime})$
$\displaystyle+r_{\alpha}(\alpha,\rho)r(\alpha^{\prime},\rho^{\prime})t(\alpha^{\prime})\cdot
t(\alpha)-r(\alpha,\rho)r(\alpha^{\prime},\rho^{\prime})n(\alpha)\cdot
t(\alpha^{\prime}),$

where

$\displaystyle n(\alpha)\cdot n(\alpha^{\prime})$
$\displaystyle=\cos(\alpha-\alpha^{\prime})$ $\displaystyle t(\alpha)\cdot
t(\alpha^{\prime})$ $\displaystyle=\cos(\alpha-\alpha^{\prime})$
$\displaystyle n(\alpha^{\prime})\cdot t(\alpha)$
$\displaystyle=-\sin(\alpha-\alpha^{\prime})$ $\displaystyle n(\alpha)\cdot
t(\alpha^{\prime})$ $\displaystyle=\sin(\alpha-\alpha^{\prime}).$

Moreover, we have that

$\displaystyle x(\alpha,\rho)\cdot
x_{\alpha}(\alpha,\rho)=r(\alpha,\rho)n(\alpha)\cdot(r_{\alpha}(\alpha,\rho)n(\alpha)+r(\alpha,\rho)t(\alpha))=r(\alpha,\rho)r_{\alpha}(\alpha,\rho).$

Therefore, equation (2.5) reads

$\displaystyle F[r,\,\lambda]=0\quad\text{in $\rho\in(1-a,1)$,
$\alpha\in\mathbb{T}$},$ (2.7)

with

$\displaystyle F[r,\,\lambda]$ (2.8) $\displaystyle\equiv\lambda
r_{\alpha}(\alpha,\rho)-\frac{1}{2\pi}\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{f_{\rho}(\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}\cos(\alpha-\alpha^{\prime})(r_{\alpha}(\alpha^{\prime},\rho^{\prime})-r_{\alpha}(\alpha,\rho))d\alpha^{\prime}d\rho^{\prime}$
$\displaystyle+\frac{r_{\alpha}(\alpha,\rho)}{2\pi
r(\alpha,\rho)}\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{f_{\rho}(\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}\cos(\alpha-\alpha^{\prime})(r(\alpha^{\prime},\rho^{\prime})-r(\alpha,\rho))d\alpha^{\prime}d\rho^{\prime}$
$\displaystyle-\frac{1}{2\pi
r(\alpha,\rho)}\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{f_{\rho}(\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}\sin(\alpha-\alpha^{\prime})(r(\alpha,\rho)r(\alpha^{\prime},\rho^{\prime})+r_{\alpha}(\alpha,\rho)r_{\alpha}(\alpha^{\prime},\rho^{\prime}))d\alpha^{\prime}d\rho^{\prime},$

where $x(\alpha,\rho)$ is given by (2.6) and we have added and subtracted

$\frac{r_{\alpha}(\alpha,\rho)}{2\pi
r(\alpha,\rho)}\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{f_{\rho}(\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}\cos(\alpha-\alpha^{\prime})r(\alpha,\rho)d\alpha^{\prime}d\rho^{\prime}$

for cosmetic reasons.

The rest of the paper consists in finding a nontrivial solution
$(r(\alpha,\rho),\lambda)$ of (2.7) and with $r_{\rho}(\alpha,\rho)>c>0.$

##  3 Main theorem and Crandall-Rabinowitz (C-R) theorem

This section is devoted to state the main theorem of the paper. But firstly,
we will fix the function $f(\rho)$. The derivative of this function will be
given by the expression

$\displaystyle\frac{a}{2}f_{\rho}\left(\rho\right)$ (3.1)
$\displaystyle=\left\\{\scriptsize\begin{array}[]{cc}-\frac{1}{2\beta^{9}}(126\beta^{4}-420\beta^{3}(1+\tilde{\rho})+540\beta^{2}(1+\tilde{\rho})^{2}-315\beta(1+\tilde{\rho})^{3}+70(1+\tilde{\rho})^{4})(1+\tilde{\rho})^{5}&\text{
if }-1\leq\tilde{\rho}\leq-1+\beta\\\ -0.5&\text{ if
}-1+\beta\leq\tilde{\rho}\leq 1-\beta\\\
\frac{1}{2\beta^{9}}(126\beta^{4}+420\beta^{3}(-1+\tilde{\rho})+540\beta^{2}(-1+\tilde{\rho})^{2}+315\beta(-1+\tilde{\rho})^{3}+70(-1+\tilde{\rho})^{4})(-1+\tilde{\rho})^{5}&\text{
if }1-\beta\leq\tilde{\rho}\leq 1\\\ \end{array}\right.$ (3.5)

where $\beta$ will be chosen later and $\tilde{\rho}=\frac{2}{a}(\rho-1)+1$.
We take $f(1)=0$. This expression will be used in order to compute some
integrals where $f_{\rho}(\rho)$ arises. Next we describe the main properties
of this function:

  1. 1.

The function $f(\rho)$ defined in $(1-a,1)$ admits a $C^{4}$-extension (we
still call it $f$) to $\mathbb{R}$. This extension is given by

$\displaystyle f(\rho)=\left\\{\begin{array}[]{ccc}1&\rho\leq 1-a\\\
f(\rho)&1-a<\rho<1\\\ 0&\rho\geq 1\end{array}\right\\}.$

  2. 2.

It is strictly decreasing in $(1-a,1)$.

  3. 3.

The derivative $f_{\rho}(\rho)$ is constant for
$1-a+\frac{a\beta}{2}\leq\rho\leq 1-\frac{a\beta}{2}$.

###### Remark 3.1

Due to the computer-assisted nature of some parts of the proof, the choice of
$f(\rho)$ and all the parameters of the problem $(n,a,\beta)$ need to be
explicit. A similar strategy works for other $n$-fold symmetric solutions, and
more regular solutions can be obtained by choosing more regular (explicit)
profiles $f$ if the computer-assisted parts of the proof yield suitable
numbers.

###### Theorem 3.2

Let $a=0.05$ and $\beta=\frac{2}{512}=2^{-8}$ and consider the domain

$\Omega_{a}\equiv\\{(\alpha,\,\rho)\,:\,\alpha\in\mathbb{T},\,1-a<\rho<1\\}$

and $f\in C^{4}([1-a,1])$ as in (3.1). Then there exists a branch of
nontrivial smooth solutions, with $3-$fold symmetry, of equation (2.7), in
$H^{4,3}(\Omega_{a})$, bifurcating from $r(\alpha,\rho)=\rho$ and
$\lambda=\lambda_{3}$ for some $\lambda_{3}\in\mathbb{R}$.

###### Remark 3.3

Nontrivial solutions means that the function $r(\alpha,\rho)$ depends on
$\alpha$ in a nontrivial way. See Definition 4.5 for a precise definition of
the space $H^{k,l}({\Omega_{a}})$.

From section 2 is clear that Theorem 1.1 follows from Theorem 3.2.

The proof of Theorem 3.2 relies on the Crandall-Rabinowitz theorem. We recall
here the statement of this theorem from [23] for expository purposes.

######  Theorem 3.4 (Crandall-Rabinowitz)

Let $X$, $Y$ be Banach spaces, $V$ a neighborhood of 0 in $X$ and

$\displaystyle F\,:\,$ $\displaystyle V\times(-1,\,1)\rightarrow Y$
$\displaystyle\,\qquad(r,\,\mu)\,\,\,\,\rightarrow F[r,\,\mu]$

have the properties

$F[0,\,\mu]=0$ for any $|\mu|<1$. (3.6) The partial derivatives
$\partial_{\mu}F$, $\partial_{r}F$ and $\partial^{2}_{\mu r}F$ exist and are
continuous. (3.7) $\mathcal{N}(\partial_{r}F[0,0])$ and
$Y/\mathcal{R}(\partial_{r}F[0,0])$ are one-dimensional. (3.8)
$\partial^{2}_{\mu r}F[0,0]r_{0}\not\in\mathcal{R}(\partial_{r}F[0,0])$, where
$\mathcal{N}(\partial_{r}F[0,0])=\text{span }r_{0}.$ (3.9)

(Here $\mathcal{N}$ and $\mathcal{R}$ denote the kernel and range
respectively). If $Z$ is any complement of $\mathcal{N}(\partial_{r}F(0,0))$
in $X$, then there is a neighborhood $U$ of (0, 0) in $X\times\mathbb{R}$, an
interval $(-b,b)$, and continuous functions

$\displaystyle\phi\,:\,(-b,b)\rightarrow\mathbb{R}$
$\displaystyle\psi\,:\,(-b,b)\rightarrow Z$

such that $\phi(0)=0,\psi(0)=0$ and

$F^{-1}(0)\cap U=\\{(\xi
r_{0}+\xi\psi(\xi),\phi(\xi))\,:\,|\xi|<b\\}\cup\\{(0,t)\,:\,(0,t)\in U\\}.$

We will check in the following section the hypotheses of the C-R theorem.

##  4 Checking the hypotheses of the C-R theorem for the equation (2.7)

In this section we will check the hypotheses of the C-R theorem in suitable
Banach spaces $X$ and $Y$ in order to find a nontrivial branch of the solution
$(r(\alpha,\rho),\lambda)$ of (2.7). In order to be able to apply this theorem
in the way that it is written in section 3 we will define new variables:

$\displaystyle\overline{r}(\alpha,\rho)$ $\displaystyle\equiv
r(\alpha,\rho)-\rho$ (4.1) $\displaystyle\mu$
$\displaystyle\equiv\lambda-\lambda_{3}$ (4.2)
$\displaystyle\overline{F}[\overline{r},\mu]$ $\displaystyle\equiv
F[\overline{r}+\rho,\lambda_{3}+\mu]$ (4.3)

with $\lambda_{3}$ to be fixed later. Thus we understand (2.7) as an equation
for $(\overline{r},\mu)$ rather that for $(r,\lambda)$. In fact, we look for
solutions of

$\displaystyle\overline{F}[\overline{r},\mu]=0\quad\text{in $\Omega_{a}$}.$
(4.4)

Let us also define the spaces $H^{k,l}(\Omega_{a})$ for $k,\,l\in\mathbb{N}$
and $k\geq l$ as follows:

$\displaystyle\left\\{r\in
L^{2}(\Omega_{a})\,:\,||r||^{2}_{L^{2}(\Omega_{a})}+||\partial^{l}_{\rho}r||_{L^{2}(\Omega_{a})}^{2}+\sum_{j=0}^{l}||\partial^{k-j}_{\alpha}\partial_{\rho}^{j}r||^{2}_{L^{2}(\Omega_{a})}<\infty\right\\}.$
(4.5)

We will work in the space $H^{4,3}$ and notice that $H^{4,3}\subset C^{2}$.
This is shown below. The reason why we work with this space is because the
functional $\overline{F}$ takes 1 derivative in $\alpha$ but no derivatives in
$\rho$. Due to this anisotropy we would not be able to apply the C-R theorem
by solely using homogeneous spaces (see [42] for additional information).

###### Lemma 4.1

Let $\Omega\in[0,1]\times[-\pi,\pi]$ and $f:\Omega\rightarrow\mathbb{R}\in
H^{4,3}(\Omega)$. Then:

$\displaystyle\|f\|_{L^{\infty}(\Omega)}+\|\partial_{x}^{2}f\|_{L^{\infty}(\Omega)}+\|\partial_{\alpha}^{2}f\|_{L^{\infty}(\Omega)}\leq
C\|f\|_{H^{4,3}(\Omega)}$

###### Proof.

If $f\in H^{4,3}(\Omega)$, then:

$\displaystyle\|f\|_{L^{2}(\Omega)}^{2}+\|\partial_{x}^{3}f\|_{L^{2}(\Omega)}^{2}+\|\partial_{\alpha}^{4}f\|_{L^{2}(\Omega)}^{2}+\|\partial_{\alpha}^{3}\partial_{x}f\|_{L^{2}(\Omega)}^{2}+\|\partial_{\alpha}^{2}\partial_{x}^{2}f\|_{L^{2}(\Omega)}^{2}+\|\partial_{\alpha}\partial_{x}^{3}f\|_{L^{2}(\Omega)}^{2}<C$

On one hand, $f_{\alpha}\in H^{3}(\Omega)$, since

$\displaystyle\|f_{\alpha}\|_{H^{3}(\Omega)}^{2}$
$\displaystyle=\|f_{\alpha}\|_{L^{2}(\Omega)}^{2}+\|\partial_{x}^{3}f_{\alpha}\|_{L^{2}(\Omega)}^{2}+\|\partial_{\alpha}^{3}f_{\alpha}\|_{L^{2}(\Omega)}^{2}$
$\displaystyle\lesssim\|f\|_{L^{2}(\Omega)}^{2}+\|\partial_{\alpha}^{4}f\|_{L^{2}(\Omega)}^{2}+\|\partial_{x}^{3}\partial_{\alpha}f\|_{L^{2}(\Omega)}^{2}+\|\partial_{\alpha}^{4}f\|_{L^{2}(\Omega)}^{2}\lesssim\|f\|_{H^{4,3}(\Omega)}$

This implies that $f_{\alpha}\in C^{1+\gamma}(\overline{\Omega})$, yielding
$\|f_{\alpha\alpha}\|_{L^{\infty}(\overline{\Omega})}\leq C$ and $\|f_{\alpha
x}\|_{L^{\infty}(\overline{\Omega})}\leq C$ for some constant $C$.

On the other hand, if we define

$\displaystyle g(\alpha)=\int_{0}^{1}\partial_{x}^{2}f(x,\alpha)dx,$

we claim that $g\in H^{1}([-\pi,\pi])$. In order to see this, we can compute

$\displaystyle\|g\|_{L^{2}([-\pi,\pi])}^{2}=\int_{-\pi}^{\pi}\left(\int\partial_{x}^{2}f(x,\alpha)dx\right)^{2}d\alpha\leq\int_{-\pi}^{\pi}\int_{0}^{1}|\partial_{x}^{2}f(x,\alpha)|^{2}dxd\alpha\leq\|f\|_{H^{4,3}(\Omega)}.$

In addition

$\displaystyle\|g_{\alpha}\|_{L^{2}([-\pi,\pi])}^{2}=\int_{-\pi}^{\pi}\left(\int\partial_{x}^{2}\partial_{\alpha}f(x,\alpha)dx\right)^{2}d\alpha\leq\int_{-\pi}^{\pi}\int_{0}^{1}|\partial_{x}^{2}\partial_{\alpha}f(x,\alpha)|^{2}dxd\alpha\leq\|f\|_{H^{4,3}(\Omega)}.$

Therefore

$\displaystyle\|g\|_{L^{\infty}(\Omega)}\leq\|g\|_{L^{\infty}([-\pi,\pi])}\leq\|g\|_{H^{1}([-\pi,\pi])}\leq
C\|f\|_{H^{4,3}(\Omega)}.$

We have that

$\displaystyle\partial_{x}^{2}f(x,\alpha)=\partial_{x}^{2}f(x,\alpha)-\int_{0}^{1}\partial_{x}^{2}f(x^{\prime},\alpha)dx^{\prime}+g(\alpha),$

which we can bound in the following way:

$\displaystyle\|\partial_{x}^{2}f\|_{L^{\infty}(\Omega)}\leq\left\|\int_{0}^{1}\partial_{x}^{2}f(x,\alpha)-\partial_{x}^{2}f(x^{\prime},\alpha)dx^{\prime}\right\|_{L^{\infty}(\Omega)}+\|g\|_{L^{\infty}(\Omega)}.$

In addition:

$\displaystyle\int_{0}^{1}\partial_{x}^{2}f(x,\alpha)-\partial_{x}^{2}f(x^{\prime},\alpha)dx^{\prime}=\int_{0}^{1}\int_{x^{\prime}}^{x}\partial_{x}^{3}f(x^{\prime\prime},\alpha)dx^{\prime\prime}dx^{\prime}=h(\alpha,x)$

We now fix $x$ and we show that $\|g(\alpha,x)\|_{H^{1}([-\pi,\pi])}$ is
uniformly bounded. We achieve that by using the following estimate:

$\displaystyle\|h(\alpha,x)\|_{L^{2}([-\pi,\pi])}$
$\displaystyle\leq\int_{0}^{1}\int_{x^{\prime}}^{x}\left(\int_{-\pi}^{\pi}|\partial_{x}^{3}f(x^{\prime\prime},\alpha)|^{2}d\alpha\right)^{\frac{1}{2}}dx^{\prime\prime}dx^{\prime}$
$\displaystyle\leq\int_{0}^{1}|x-x^{\prime}|^{\frac{1}{2}}\left(\int_{x^{\prime}}^{x}\int_{-\pi}^{\pi}|\partial_{x}^{3}f(x^{\prime\prime},\alpha)|^{2}\text{sign}(x-x^{\prime})d\alpha
dx^{\prime\prime}\right)^{\frac{1}{2}}dx^{\prime}$ $\displaystyle\leq
C\left(\int_{0}^{1}\left(\int_{0}^{1}\int_{-\pi}^{\pi}|\partial_{x}^{3}f(x^{\prime\prime},\alpha)|^{2}d\alpha
dx^{\prime\prime}\right)^{\frac{1}{2}}dx^{\prime}\right)$ $\displaystyle\leq
C\|f\|_{H^{4,3}(\Omega)},$

where $C$ is independent of $x$. We can do the same procedure with
$\partial_{\alpha}h(\alpha,x)$, getting

$\displaystyle\|h(\alpha,x)\|_{L^{\infty}([-\pi,\pi])}\leq\|h(\alpha,x)\|_{H^{1}([-\pi,\pi])}\leq
C\|f\|_{H^{4,3}(\Omega)}$

Taking the supremum over $x$ yields the desired result.

∎

The theorem we will prove is the following:

###### Theorem 4.2

Let $f$ and $a$ be as in Theorem 3.2. Then there exist
$(\overline{r}_{0}(\alpha,\rho),\lambda_{3})\in
H^{4,3}(\Omega_{a})\times\mathbb{R}$, an interval $(-b,b)$, and continuous
functions

$\displaystyle\phi\,:\,(-b,b)\rightarrow\mathbb{R}$
$\displaystyle\psi\,:\,(-b,b)\rightarrow Z$

with $\phi(0)=0,\psi(0)=0$, such that, if $Z$ is any complement of
$\text{span}\\{\overline{r}_{0}\\}$ in $H^{4,3}(\Omega_{a})$,

$\overline{F}[\xi\overline{r}_{0}+\xi\psi(\xi),\phi(\xi)]=0,$

for $|\xi|<b$.

In addition these solutions will have $3-$fold symmetry.

Here it is important to remark that this theorem provides a nontrivial
solution $r(\alpha,\rho)=\rho+\xi\overline{r}_{0}+\xi\psi(\xi)$ of (2.7) with
$\lambda=\lambda_{3}+\phi(\xi)$ satisfying $|r_{\rho}(\alpha,\rho)|>c>0$ if we
take $\xi$ small enough. Theorem 3.2 follows from Theorem 4.2.

###  4.1 Step 1. The functional setting and the hypothesis 1

Our first step is to define the spaces we will work with in order to apply the
C-R theorem. The spaces $H^{k,l}_{3,\text{even}}(\Omega_{a})$ and
$H^{k,l}_{3,\text{odd}}(\Omega_{a})$ will be given by

$\displaystyle\left\\{\overline{r}\in
H^{k,l}(\Omega_{a})\,:\,\overline{r}(\alpha,\rho)=\sum_{m=1}^{\infty}\hat{r}_{m}(\rho)\cos(3m\alpha)\right\\},$

and

$\displaystyle\left\\{\overline{r}\in
H^{k,l}(\Omega_{a})\,:\,\overline{r}(\alpha,\rho)=\sum_{m=1}^{\infty}\hat{r}_{m}(\rho)\sin(3m\alpha)\right\\},$

respectively.

One of the purposes to introducing these spaces, which only represent
frequencies multiples of $3$, is to be able to show the $3-$fold symmetry of
the solution. Our starting space $X$ will be
$H^{4,3}_{3,\text{even}}(\Omega_{a})$.

The target space $Y$ will be $H^{3,3}_{3,\text{odd}}(\Omega_{a})$. Notice that
a function in $H^{4,3}(\Omega_{a})$ belongs to
$C^{2,2}(\overline{\Omega_{a}})=C^{1}(\overline{\Omega_{a}})$. Finally we take
the neighbourhood $V$ of $0$ in $X$ to be

$\displaystyle V\equiv\left\\{\overline{r}\in
H^{4,3}_{3,\text{even}}(\Omega_{a})\,:\,||\overline{r}||_{H^{4,3}_{3,\text{even}}(\Omega_{a})}<\delta\right\\}$

for $\delta>0$. The parameter $\delta$ will be fixed later (small enough).

Given these definitions we need to show the following lemma:

###### Lemma 4.3

Let $\overline{F}[\overline{r},\mu]$ be as in (4.3). Then, for fixed
$a\in(0,1)$, there exists $\delta>0$ small enough so that

$\displaystyle\overline{F}\,:\,V\times[-1,1]\to
H^{3,3}_{3,\,\emph{odd}}(\Omega_{a}).$

###### Proof.

Here we recall the definition of the functional

$\displaystyle F[r,\,\lambda]$ $\displaystyle\equiv\lambda
r_{\alpha}(\alpha,\rho)-\frac{1}{2\pi}\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{f_{\rho}(\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}\cos(\alpha-\alpha^{\prime})(r_{\alpha}(\alpha^{\prime},\rho^{\prime})-r_{\alpha}(\alpha,\rho))d\alpha^{\prime}d\rho^{\prime}$
$\displaystyle+\frac{r_{\alpha}(\alpha,\rho)}{2\pi
r(\alpha,\rho)}\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{f_{\rho}(\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}\cos(\alpha-\alpha^{\prime})(r(\alpha^{\prime},\rho^{\prime})-r(\alpha,\rho))d\alpha^{\prime}d\rho^{\prime}$
$\displaystyle-\frac{1}{2\pi
r(\alpha,\rho)}\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{f_{\rho}(\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}\sin(\alpha-\alpha^{\prime})(r(\alpha,\rho)r(\alpha^{\prime},\rho^{\prime})+r_{\alpha}(\alpha,\rho)r_{\alpha}(\alpha^{\prime},\rho^{\prime}))d\alpha^{\prime}d\rho^{\prime}$
$\displaystyle\equiv F_{1}[r,\lambda]+F_{2}[r]+F_{3}[r]+F_{4}[r],$

with $r=\rho+\overline{r}$ and $x=r(\cos(\alpha),\sin(\alpha))$. It is easy to
check that $F_{1}[r,\lambda]=\lambda\partial_{\alpha}r$ belongs to
$H^{3,3}_{3,\,\text{odd}}(\Omega_{a})$.

Next we show that

$\displaystyle F_{i}[r]\in H^{3,3}(\Omega_{a}),\quad i=2,3,4.$ (4.6)

In order to do it we notice that, since
$||\cdot||_{C^{2}(\overline{\Omega_{a}})}\leq
C||\cdot||_{H^{4,3}(\Omega_{a})}$, we can choose, for fixed $a\in(0,1)$,
$\delta$ small enough to have that $r(\alpha,\rho)>c_{0}(a,\delta)>0$ and
$r_{\rho}(\alpha,\rho)>c_{1}(a,\delta)>0$, for every
$(\alpha,\rho)\in\Omega_{a}$. By comparing equations (2.5) and (2.7) we see
that

$\displaystyle\left(F_{2}[r]+F_{3}[r]+F_{4}[r]\right)(\alpha,\rho)=$
$\displaystyle-\frac{x_{\alpha}^{\perp}(\alpha,\rho)}{2\pi
r(\alpha,\rho)}\cdot\int_{0}^{\infty}\int_{-\pi}^{\pi}f_{\rho}(\rho^{\prime})\frac{x_{\alpha}(\alpha^{\prime},\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}d\alpha^{\prime}d\rho^{\prime}$
$\displaystyle=$ $\displaystyle\frac{x_{\alpha}^{\perp}(\alpha,\rho)}{2\pi
r(\alpha,\rho)}\cdot\int_{0}^{\infty}\int_{-\pi}^{\pi}f_{\rho}(\rho^{\prime})\frac{x_{\alpha}(\alpha,\rho)-x_{\alpha}(\alpha^{\prime},\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha^{\prime},\rho^{\prime})|}d\alpha^{\prime}d\rho^{\prime}.$

with $x(\alpha,\rho)=r(\alpha,\rho)(\cos(\alpha),\sin(\alpha))$ and
$r(\alpha,\rho)=\rho+\overline{r}(\alpha,\rho)$. We will extend the
restriction of the function $f_{\rho}$ to the positive real axis
$f_{\rho}|_{\mathbb{R}^{+}}$ to $\mathbb{R}$ by zero. We still call this
extension $f_{\rho}$. Thus, a change of variables yields,

$\displaystyle\left(F_{2}[r]+F_{3}[r]+F_{4}[r]\right)(\alpha,\rho)$
$\displaystyle=\underbrace{\frac{x_{\alpha}^{\perp}(\alpha,\rho)}{2\pi
r(\alpha,\rho)}}_{g(\alpha,\rho)}\cdot\underbrace{\int_{-\infty}^{\infty}\int_{-\pi}^{\pi}f_{\rho}(\rho-\rho^{\prime})\frac{x_{\alpha}(\alpha,\rho)-x_{\alpha}(\alpha-\alpha^{\prime},\rho-\rho^{\prime})}{|x(\alpha,\rho)-x(\alpha-\alpha^{\prime},\rho-\rho^{\prime})|}d\alpha^{\prime}d\rho^{\prime}}_{B[f,x](\alpha,\rho)}.$

We will use the following notation. For a general function $h(\alpha,\rho)$ we
define

$\displaystyle h=$ $\displaystyle h(\alpha,\rho)$ $\displaystyle h^{\prime}=$
$\displaystyle h(\alpha^{\prime},\rho^{\prime})$ $\displaystyle
h^{\prime\prime}=$ $\displaystyle
h(\alpha-\alpha^{\prime},\rho-\rho^{\prime})$ $\displaystyle\Delta h=$
$\displaystyle h-h^{\prime}$ $\displaystyle\Delta h^{\prime}=$ $\displaystyle
h-h^{\prime\prime}$

Thus we can write

$\displaystyle
B[f,x]=\int_{-\infty}^{\infty}\int_{-\pi}^{\pi}f_{\rho}^{\prime\prime}\frac{\Delta
x^{\prime}_{\alpha}}{|\Delta x^{\prime}|}d\alpha^{\prime}d\rho^{\prime}.$

Next we look at

$\displaystyle\sum_{j=0}^{3}\partial^{3-j}_{\alpha}\partial^{j}_{\rho}\left(g(\alpha,\rho)B[f,x](\alpha,\rho)\right).$
(4.7)

We will consider two groups of terms. Group 1 consists of the terms

$\displaystyle
g\partial^{3}_{\alpha}B,\,\,\,\,\partial_{\alpha}g\partial_{\alpha}^{2}B,\,\,\,\,\partial_{\alpha}g\partial^{2}_{\alpha\rho}B,\,\,\,\,\partial_{\rho}g\partial^{2}_{\alpha}B,\,\,\,\,g\partial^{2}_{\alpha}\partial_{\rho}B,\,\,\,\,\partial_{\alpha}g\partial^{2}_{\rho}B,\,\,\,\,\partial_{\rho}g\partial^{2}_{\alpha\rho}B,\,\,\,\,g\partial^{2}_{\rho}\partial_{\alpha}B,\,\,\,\,\partial_{\rho}g\partial^{2}_{\rho}B,\,\,\,\,g\partial^{3}_{\rho}B.$

Group 2 consists of the terms

$\displaystyle\partial^{3}_{\alpha}gB,\,\,\,\,\partial^{2}_{\alpha}g\partial_{\alpha}B,\,\,\,\,\partial^{2}_{\alpha}g\partial_{\rho}B,\,\,\,\,\partial^{2}_{\alpha\rho}g\partial_{\alpha}B,\,\,\,\,\partial^{2}_{\rho}\partial_{\alpha}gB,\,\,\,\,\partial^{2}_{\rho}g\partial_{\alpha}B,\,\,\,\,\partial^{2}_{\alpha\rho}g\partial_{\rho}B,\,\,\,\,\partial^{2}_{\rho}g\partial_{\rho}B,\,\,\,\,\partial^{3}_{\rho}gB.$

It is easy to check that expression (4.7) is given by a linear combination of
the terms in group 1 and group 2. On one hand, since in group 1 there is no
more than one derivative acting on $g$ and

$||g||_{L^{\infty}(\overline{\Omega_{a}})},\quad||\partial_{\alpha}g||_{L^{\infty}(\overline{\Omega_{a}})},\quad||\partial_{\rho}g||_{L^{\infty}(\overline{\Omega_{a}})}\leq
C(\delta,a)$

in order to bound its terms we need to estimate

$\displaystyle\partial^{3}_{\alpha}B,\,\,\,\,\partial_{\alpha}^{2}B,\,\,\,\,\partial^{2}_{\alpha\rho}B,\,\,\,\,\partial^{2}_{\alpha}\partial_{\rho}B,\,\,\,\,\partial^{2}_{\rho}B,\,\,\,\,\partial^{2}_{\rho}\partial_{\alpha}B,\,\,\,\,\partial^{3}_{\rho}B.$

On the other hand, since

$\displaystyle\partial^{2}_{\alpha}g,\quad\partial^{2}_{\rho}g,\quad\partial^{2}_{\alpha\rho}g,\quad\partial^{3}_{\alpha}g,\quad\partial^{3}_{\rho}g,\quad\partial^{2}_{\alpha}\partial_{\rho}g,\quad\partial^{2}_{\rho}\partial_{\alpha}g,$

have $L^{2}$-norms bounded by some constant depending on $\delta$ and $a$, in
order to estimate the terms in group 2 we just need to control the
$L^{\infty}$ norms of $B$, $\partial_{\alpha}B$, $\partial_{\rho}B$. These
norms are bounded by $||B||_{H^{3}(\overline{\Omega_{a}})}$.

In addition since the $L^{2}$-norm of $B[f,x]$ is easy to control we just have
to estimate the $L^{2}$-norms of the derivatives of order 3 of $B[f,x]$.

###### Lemma 4.4

Let $r=\rho+\overline{r}$, where $\overline{r}\in V$, and
$x=r(\cos(\alpha),\sin(\alpha))$, the derivatives
$\partial^{3}_{\sigma_{1},\sigma_{2},\sigma_{3}}B[f,x]$ where $\sigma_{i}$ is
either $\alpha$ or $\rho$, with $i=1,2,3$, are in $L^{2}$ with norm bounded by
a constant $C$ just depending on $\delta$, $a$ and $||f||_{C^{4}}$.

###### Proof.

This lemma will be proved by using the following lemma:

###### Lemma 4.5

Let $r=\rho+\overline{r}$, where $\overline{r}\in V$, and
$x=r(\cos(\alpha),\sin(\alpha))$, then there exists a constant $c(a,\delta)>0$
such that

$\displaystyle|x(\alpha,\rho)-x(\alpha-\alpha^{\prime},\rho-\rho^{\prime})|^{2}\geq
c(a,\delta)\left(\alpha^{\prime 2}+\rho^{\prime 2}\right).$

###### Proof.

Because of the definition we have that

$\displaystyle|\Delta x^{\prime}|^{2}=r^{2}+r^{\prime\prime
2}-2rr^{\prime\prime}\cos(\alpha^{\prime})=(\Delta
r^{\prime})^{2}+4rr^{\prime\prime}\sin^{2}\left(\frac{\alpha^{\prime}}{2}\right).$

Now we notice that $r=\rho+\overline{r}$ and then, for $\rho\in(1-a,1)$, there
exists $c_{0}(a,\delta)>0$ such that $r\geq c_{0}(a,\delta)$, where
$c_{0}(a,\delta)$ is increasing with $a$ and decreasing with $\delta$. Since
$\rho-\rho^{\prime}$ belongs to $(1-a,1)$, $r^{\prime\prime}\geq
c_{0}(a,\delta)$ too. Both inequalities together yield

$\displaystyle|\Delta
x^{\prime}|^{2}\geq(r-r^{\prime\prime})^{2}+c_{0}(a,\delta)\sin^{2}\left(\frac{\alpha^{\prime}}{2}\right).$

In addition, $r-r^{\prime}=\rho^{\prime}+\Delta\overline{r}^{\prime}$, and
$|\overline{r}^{\prime}|\leq C(\delta)(|\alpha^{\prime}|+|\rho^{\prime}|)$
where $C(\delta)\to 0$ when $\delta\to 0$. Then

$\displaystyle(r-r^{\prime\prime})^{2}\geq\rho^{\prime
2}-C(\delta)\left(\alpha^{\prime 2}+\rho^{\prime 2}\right)$

where $C(\delta)\to 0$ when $\delta\to 0$. Therefore we obtain that, by making
$\delta$ small enough

$\displaystyle|\Delta x^{\prime}|^{2}\geq c(a,\delta)\left(\alpha^{\prime
2}+\rho^{\prime 2}\right).$

∎

Let $\partial$ mean differentiation with respect to either $\alpha$ or $\rho$.
Then, the derivatives $\partial^{3}\left(f_{\rho}^{\prime\prime}\frac{\Delta
x^{\prime}_{\alpha}}{|\Delta x^{\prime}|}\right)$, consist of terms of the
form

$\displaystyle\partial^{3}f_{\rho}^{\prime\prime}\frac{\Delta
x^{\prime}_{\alpha}}{|\Delta x|},\quad
f^{\prime\prime}_{\rho}\partial^{3}\Delta x_{\alpha}^{\prime}\frac{1}{|\Delta
x^{\prime}|},\quad f_{\rho}^{\prime\prime}\Delta
x_{\alpha}^{\prime}\partial^{3}\frac{1}{|\Delta x^{\prime}|},$
$\displaystyle\partial^{2}f^{\prime\prime}_{\rho}\partial\Delta
x^{\prime}_{\alpha}\frac{1}{|\Delta
x^{\prime}|},\quad\partial^{2}f^{\prime\prime}_{\rho}\Delta
x_{\alpha}^{\prime}\partial\frac{1}{|\Delta x^{\prime}|},\quad\partial
f^{\prime\prime}_{\rho}\partial^{2}\Delta x^{\prime}_{\alpha}\frac{1}{|\Delta
x^{\prime}|},$ $\displaystyle f^{\prime\prime}_{\rho}\partial^{2}\Delta
x^{\prime}_{\alpha}\partial\frac{1}{|\Delta x|},\quad\partial
f^{\prime\prime}_{\rho}\Delta x^{\prime}_{\alpha}\partial^{2}\frac{1}{|\Delta
x^{\prime}|},\quad f^{\prime\prime}_{\rho}\partial\Delta
x^{\prime}_{\alpha}\partial^{2}\frac{1}{|\Delta x^{\prime}|}$
$\displaystyle\partial f^{\prime\prime}_{\rho}\partial\Delta
x^{\prime}_{\alpha}\partial\frac{1}{|\Delta x^{\prime}|}.$

Since $f\in C^{4}$, Lemma 4.5, the fact that

$\int_{\rho-1}^{\rho-(1-a)}\int_{-\pi}^{\pi}\frac{1}{\sqrt{\alpha^{\prime
2}+\rho^{\prime 2}}}d\alpha^{\prime}d\rho^{\prime}\leq C$

and that $||\overline{r}||_{C^{2}}\leq C(\delta)$ we obtain that the terms in
$\partial^{3}B$ coming from $\partial^{3}f^{\prime\prime}_{\rho}\frac{\Delta
x_{\alpha}^{\prime}}{|\Delta x^{\prime}|}$ and
$\partial^{2}f_{\rho}^{\prime\prime}\partial\Delta
x_{\alpha}^{\prime}\frac{1}{|\Delta x^{\prime}|}$ are in $L^{\infty}$ with
$L^{\infty}-$norm bounded by some constant
$C\left(a,\delta,||f||_{C^{4}(\overline{\Omega_{a}})}\right)$.

The terms in $\partial^{3}B$ coming from
$f_{\rho}^{\prime\prime}\partial^{3}\Delta x_{\alpha}^{\prime}\frac{1}{|\Delta
x^{\prime}|}$ and $\partial f_{\rho}^{\prime\prime}\partial^{2}\Delta
x^{\prime}_{\alpha}\frac{1}{|\Delta x^{\prime}|}$ all can be bounded in
$L^{2}$ in the following way. Again we will use the bound for $f$ in $C^{4}$
and Lemma 4.5. Let us focus on

$\displaystyle\int_{-\infty}^{\infty}\int_{-\pi}^{\pi}f^{\prime\prime}_{\rho}\frac{\partial^{3}\Delta
x_{\alpha}}{|\Delta
x^{\prime}|}d\alpha^{\prime}d\rho^{\prime}=\partial^{3}x_{\alpha}\int_{\rho-1}^{\rho-(1-a)}\int_{-\pi}^{\pi}f^{\prime\prime}_{\rho}\frac{1}{|\Delta
x^{\prime}|}d\alpha^{\prime}d\rho^{\prime}-\int_{-\infty}^{\infty}\int_{-\pi}^{\pi}f^{\prime\prime}_{\rho}\frac{\partial^{3}x^{\prime\prime}_{\alpha}}{|\Delta
x^{\prime}|}d\alpha^{\prime}d\rho^{\prime}\equiv I_{1}+I_{2}.$ (4.8)

Since

$\displaystyle\left|\int_{\rho-1}^{\rho-(1-a)}\int_{-\pi}^{\pi}f^{\prime\prime}_{\rho}\frac{1}{|\Delta
x^{\prime}|}d\alpha^{\prime}d\rho^{\prime}\right|\leq
C\left(a,\delta,||f||_{C^{4}}\right)$

we have that $||I_{1}||_{L^{2}(\overline{\Omega_{a}})}\leq
C\left(a,\delta,||f||_{C^{4}}\right)||\partial^{3}x_{\alpha}||_{L^{2}(\overline{\Omega_{a}})}$.

In order to bound $I_{2}$ we notice that, after a change of variables,

$\displaystyle|I_{2}|=\left|\int_{1-a}^{1}\int_{-\pi}^{\pi}\frac{f^{\prime}_{\rho}}{|\Delta
x|}\partial^{3}x_{\alpha}^{\prime}d\alpha^{\prime}d\rho^{\prime}\right|\leq
C(a,\delta)\int_{1-a}^{1}\int_{-\pi}^{\pi}\frac{|f^{\prime}_{\rho}|}{|\sqrt{(\alpha-\alpha^{\prime})^{2}+(\rho-\rho^{\prime})^{2}}|}|\partial^{3}x_{\alpha}^{\prime}|d\alpha^{\prime}d\rho^{\prime}$

and therefore Young’s inequality applies to yield

$||I_{2}||_{L^{2}(\overline{\Omega_{a}})}\leq
C\left(a,\delta,||f||_{C^{4}}\right)||\partial^{3}x_{\alpha}||_{L^{2}(\overline{\Omega_{a}})}.$

It remains to bound the terms with derivatives acting on the factor
$\frac{1}{|\Delta x^{\prime}|}$.

We first will deal with the terms in $\partial^{3}B$ with a factor
$\partial\frac{1}{|\Delta x^{\prime}|}$ i.e., the terms coming from
$\partial^{2}_{\rho}f^{\prime\prime}_{\rho}\Delta
x_{\alpha}^{\prime}\partial\frac{1}{|\Delta x^{\prime}|}$,
$f^{\prime\prime}_{\rho}\partial^{2}\Delta
x_{\alpha}^{\prime}\partial\frac{1}{|\Delta x^{\prime}|}$ and $\partial
f^{\prime\prime}_{\rho}\partial\Delta
x^{\prime}_{\alpha}\partial\frac{1}{|\Delta x^{\prime}|}$. Just a computation
shows that

$\displaystyle\partial\frac{1}{|\Delta x^{\prime}|}=-\frac{\partial\Delta
x^{\prime}\cdot\Delta x^{\prime}}{|\Delta x^{\prime}|^{3}}$

and therefore by Lemma 4.5 and because of $x\in C^{2}$ we have that

$\displaystyle\left|\partial\frac{1}{|\Delta x^{\prime}|}\right|\leq
C(a,\delta)\frac{1}{\sqrt{\alpha^{\prime 2}+\rho^{\prime 2}}}.$

Therefore the terms in $\partial^{3}B$ coming from
$\partial^{2}_{\rho}f^{\prime\prime}_{\rho}\Delta
x_{\alpha}^{\prime}\partial\frac{1}{|\Delta x^{\prime}|}$, and $\partial
f^{\prime\prime}_{\rho}\partial\Delta
x^{\prime}_{\alpha}\partial\frac{1}{|\Delta x^{\prime}|}$ are actually bounded
in $L^{\infty}$. The term coming from
$f^{\prime\prime}_{\rho}\partial^{2}\Delta
x_{\alpha}^{\prime}\partial\frac{1}{|\Delta x^{\prime}|}$ is bounded as we did
before for $I_{1}+I_{2}$ in (4.8).

The term with two derivatives of the factor $\frac{1}{|\Delta x^{\prime}|}$
which causes more difficulties is $f_{\rho}^{\prime\prime}\partial\Delta
x_{\alpha}\partial^{2}\frac{1}{|\Delta x^{\prime}|}$. We will use the
following embedding: since $x_{\alpha}\in
H^{3,3}(\Omega_{a})=H^{3}(\Omega_{a})$ we know that $x_{\alpha}\in
C^{1+\gamma}(\overline{\Omega_{a}})$ with
$C^{1+\gamma}(\overline{\Omega_{a}})-$norm bounded for some constant
$C(\delta)$. Then, since,

$\displaystyle\partial^{2}\frac{1}{|\Delta
x^{\prime}|}=-\frac{\partial^{2}\Delta x^{\prime}\cdot\Delta
x^{\prime}}{|\Delta x^{\prime}|^{3}}-\frac{\partial\Delta
x^{\prime}\cdot\partial\Delta x^{\prime}}{|\Delta
x^{\prime}|^{3}}+3\frac{\partial\Delta x^{\prime}\cdot\Delta
x^{\prime}\partial\Delta x^{\prime}\cdot\Delta x^{\prime}}{|\Delta
x^{\prime}|^{5}}$

we can use the previous embedding to estimate

$\displaystyle\left|\partial\Delta
x^{\prime}_{\alpha}\partial^{2}\frac{1}{|\Delta x^{\prime}|}\right|\leq
C(a,\delta)\left(\frac{1}{\left(\alpha^{\prime 2}+\rho^{\prime
2}\right)^{1-\frac{\alpha}{2}}}|\partial^{2}\Delta
x^{\prime}|+\frac{1}{\sqrt{\alpha^{\prime 2}+\rho^{\prime 2}}}|\partial\Delta
x^{\prime}_{\alpha}|\right).$

Therefore by using again Young’s inequality we bound the term

$\int_{-\infty}^{\infty}\int_{-\pi}^{\pi}f_{\rho}^{\prime}\Delta
x_{\alpha}^{\prime}\partial^{2}\frac{1}{|\Delta
x^{\prime}|}d\alpha^{\prime}d\rho^{\prime}\leq C(a,\delta,||f||_{C^{4}}).$

Finally we compute three derivatives of the factor $\frac{1}{|\Delta
x^{\prime}|}$. The terms arising from these derivatives are linear
combinations of terms with the structures

$\displaystyle\frac{\partial^{3}\Delta x^{\prime}\cdot\Delta
x^{\prime}}{|\Delta x^{\prime}|^{3}},\quad\frac{\partial^{2}\Delta
x^{\prime}\cdot\partial\Delta x^{\prime}}{|\Delta
x^{\prime}|^{3}},\frac{\partial^{2}\Delta x^{\prime}\cdot\Delta
x^{\prime}\partial\Delta x^{\prime}\cdot\Delta x^{\prime}}{|\Delta
x^{\prime}|^{5}},$ $\displaystyle\frac{\partial\Delta
x^{\prime}\cdot\partial\Delta x^{\prime}\partial\Delta x^{\prime}\cdot\Delta
x^{\prime}}{|\Delta x^{\prime}|^{5}},\quad\frac{\partial\Delta
x^{\prime}\cdot\Delta x^{\prime}\partial\Delta x^{\prime}\cdot\Delta
x^{\prime}\partial\Delta x^{\prime}\cdot\Delta x^{\prime}}{|\Delta
x^{\prime}|^{7}}$

and then, a similar analysis we did before helps us to prove that

$\displaystyle\int_{-\infty}^{\infty}\int_{-\pi}^{\pi}f_{\rho}^{\prime}\Delta
x^{\prime}_{\alpha}\partial^{3}\frac{1}{|\Delta
x^{\prime}|}d\alpha^{\prime}d\rho^{\prime},$

is bounded in $L^{2}$ for a constant $C(a,\delta,||f||_{C^{4}}$). This
concludes the proof of Lemma 4.4.∎

Thus we have proven that (4.6) holds. This finishes the proof of Lemma 4.3. ∎

Therefore, in order to prove that,

$\overline{F}\,:\,V\times(-1,1)\to H^{3,3}_{3,\text{odd}}(\Omega_{a})$

we just need to show that if

$\displaystyle\overline{r}(-\alpha,\rho)=$
$\displaystyle\overline{r}(\alpha,\rho)$

and

$\displaystyle\overline{r}\left(\alpha+\frac{2n\pi}{3},\rho\right)=$
$\displaystyle\overline{r}(\alpha,\rho)$

for $n\in\mathbb{N}$, then

$\displaystyle\overline{F}(-\alpha,\rho)=-$
$\displaystyle\overline{F}(\alpha,\rho)$

and

$\displaystyle\overline{F}\left(\alpha+\frac{2n\pi}{3},\rho\right)=$
$\displaystyle\overline{F}(\alpha,\rho)$

for $n\in\mathbb{N}$. These two properties are easy to check.

The last part of this section will be to check that the hypothesis 1 in the
C-R theorem holds. This fact is a consequence of radial functions being
stationary solutions of the SQG equation but let us check it on (2.8). If we
take $\overline{r}=0$, i.e., $r=\rho$, the only term in (2.8) that is not
trivially zero is the last integral. In order to check that this integral is
zero we just notice that the integrand is odd in $\alpha$.

###  4.2 Step 2. The partial derivatives of the functional $F$

We need to prove the existence and the continuity of the Gateaux derivatives
$\partial_{\overline{r}}\overline{F}[\overline{r},\lambda]$,$\partial_{\lambda}\overline{F}[\overline{r},\lambda]$
and $\partial^{2}_{\overline{r}\lambda}\overline{F}[\overline{r},\lambda]$. We
have the following lemma

###### Lemma 4.6

For all $\overline{r}\in V^{\delta}$ and $\mu\in\mathbb{R}$ the partial
derivatives
$\partial_{\overline{r}}\overline{F}[\overline{r},\lambda]$,$\partial_{\lambda}\overline{F}[\overline{r},\lambda]$
and $\partial^{2}_{\overline{r}\lambda}\overline{F}[\overline{r},\lambda]$
exist and are continuous. In addition

$\displaystyle\partial_{r}\overline{F}[0,\mu]\tilde{r}(\alpha,\rho)=\partial_{r}F[\rho,\lambda]\tilde{r}(\alpha,\rho)$
$\displaystyle=\lambda\tilde{r}_{\alpha}(\alpha,\rho)-\frac{1}{2\pi}\int\int\frac{f_{\rho}(\rho^{\prime})\cos(\alpha-\alpha^{\prime})(\tilde{r}_{\alpha}(\alpha^{\prime},\rho^{\prime})-\tilde{r}_{\alpha}(\alpha,\rho))}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(\alpha-\alpha^{\prime})}}d\alpha^{\prime}d\rho^{\prime}$
$\displaystyle+\frac{\tilde{r}_{\alpha}(\alpha,\rho)}{2\pi}\int\int\frac{f_{\rho}(\rho^{\prime})\cos(\alpha-\alpha^{\prime})(\rho^{\prime}-\rho)}{\rho\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(\alpha-\alpha^{\prime})}}d\alpha^{\prime}d\rho^{\prime}$
$\displaystyle-\frac{1}{2\pi}\int\int\frac{f_{\rho}(\rho^{\prime})(\rho-\rho^{\prime}\cos(\alpha-\alpha^{\prime}))\sin(\alpha-\alpha^{\prime})(\rho\tilde{r}(\alpha^{\prime},\rho^{\prime})-\rho^{\prime}\tilde{r}(\alpha,\rho))}{(\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(\alpha-\alpha^{\prime}))^{\frac{3}{2}}}d\alpha^{\prime}d\rho^{\prime}$
$\displaystyle\equiv
I_{0}[\tilde{r},\lambda](\alpha,\rho)+I_{1}[\tilde{r}](\alpha,\rho)+I_{2}[\tilde{r}](\alpha,\rho)+I_{3}[\tilde{r}](\alpha,\rho).$

###### Proof.

The lemma is trivial for the derivatives involving $\lambda$. The continuity
of the derivative with respect to $\overline{r}$ also follows since $f\in
C^{4}$ and is compactly supported. ∎

###  4.3 Step 3. Analysis of the linear operator

Now we have to study the dimension of both the kernel and image of the
operator $\partial_{r}F[\rho,\lambda]$. We will first show that for a certain
value of $\lambda$ that we will call $\lambda_{3}$ the kernel of the operator
$\partial_{r}F[\rho,\lambda_{3}]$ is one dimensional. After that we will show
that the codimension of the image of $\partial_{r}F[\rho,\lambda_{3}]$ is also
one dimensional. This will finish the checking of the hypothesis 3 in the C-R
theorem. Propositions 4.7 and 4.35 are the main results of this section.

###### Proposition 4.7

There exists a pair $(\tilde{r}_{0},\lambda_{3})\in
H^{4,3}_{3,\text{even}}(\Omega_{a})\times\mathbb{R}$, with $\tilde{r}_{0}$ not
identically zero, such that

$\displaystyle\partial_{r}F[\rho,\lambda_{3}]\tilde{r}_{0}(\alpha,\rho)=0.$
(4.9)

Moreover $\tilde{r}_{0}$ is unique modulo multiplication by constant.

###### Proof.

The proof of Proposition 4.7 consists of the following steps:

  1. 1.

The equation for the radial part. We introduce in (4.9) the $m$-fold ansatz:

$\displaystyle\tilde{r}(\alpha,\rho)=\rho B(\rho)\cos(3m\alpha)$

and we obtain an equation for the pair $(B(\rho),\lambda)$, which we will
write in the following form:

$\displaystyle\Theta^{m}B-\lambda B=0\quad\text{in $(1-a,1)$}.$ (4.10)

(See the equation (4.14) below).

  2. 2.

Existence of solutions of equation (4.9). We solve the equation (4.10) for
$m=3$ and find a solution $(B^{3},\lambda_{3})\in
H^{3}((1-a,1))\times\mathbb{R}$ of (4.10). Therefore $\tilde{r}_{0}=\rho
B^{3}(\rho)\cos(3\alpha)\in H^{4,3}_{3,\text{even}}(\Omega_{a})$ satisfies
(4.9).

  3. 3.

Uniqueness for the equation (4.9). We notice that we still need to show
uniqueness for (4.9), since, until now, we have that, given $\lambda_{3}$
there is a unique $B^{3}$ such that (4.10) holds. But this fact does not imply
that there is only one solution (modulo multiplication by constants),
$\tilde{r}_{0}$, to (4.9). Indeed, we need to show that the equation

$\partial_{r}F[\rho,\lambda_{3}]\left(b^{3m}(\rho)\cos(3m\alpha)\right)=0\quad\text{for
$m>1$}$

implies $b^{3m}(\rho)=0$ for $m>1$.

####  4.3.1 The equation for the radial part

Taking $\tilde{r}(\alpha,\rho)=\rho B(\rho)\cos(m\alpha)$ we have that
$\partial_{r}F[\rho,\lambda]\tilde{r}(\alpha,\rho)$ is given by the following
terms:

$\displaystyle I_{0}[\tilde{r},\lambda](\alpha,\rho)$ $\displaystyle=-\lambda
m\rho B(\rho)\sin(m\alpha)$ $\displaystyle I_{2}[\tilde{r}](\alpha,\rho)$
$\displaystyle=-\frac{m}{2\pi}B(\rho)\sin(m\alpha)\int_{-\infty}^{\infty}\int_{-\pi}^{\pi}\frac{f_{\rho}(\rho^{\prime})\cos(\alpha-\alpha^{\prime})(\rho^{\prime}-\rho)}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(\alpha-\alpha^{\prime})}}d\alpha^{\prime}d\rho^{\prime}$
$\displaystyle=-\frac{m}{2\pi}B(\rho)\sin(m\alpha)\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})(\rho^{\prime}-\rho)\left(\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x)}}dx\right)d\rho^{\prime}$

We move on to $I_{1}[\tilde{r},\lambda](\alpha,\rho)$. We have that

$\displaystyle\tilde{r}_{\alpha}(\alpha^{\prime},\rho^{\prime})-\tilde{r}_{\alpha}(\alpha,\rho)$
$\displaystyle=-m(\rho^{\prime}B(\rho^{\prime})\sin(m\alpha^{\prime})-\rho
B(\rho)\sin(m\alpha))$
$\displaystyle=-m(\rho^{\prime}B(\rho^{\prime})\sin(m\alpha)\cos(m(\alpha-\alpha^{\prime}))$
$\displaystyle-\underbrace{\rho^{\prime}B(\rho^{\prime})\cos(m\alpha)\sin(m(\alpha-\alpha^{\prime}))}_{\text{will
integrate to }0}-\rho B(\rho)\sin(m\alpha)).$

Therefore

$\displaystyle I_{1}[\tilde{r}](\alpha,\rho)$
$\displaystyle=-\frac{m}{2\pi}\sin(m\alpha)\int_{-\infty}^{\infty}\int_{-\pi}^{\pi}f_{\rho}(\rho^{\prime})\frac{(\rho
B(\rho)-\rho^{\prime}B(\rho^{\prime})\cos(mx))\cos(x)}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x)}}dxd\rho^{\prime}$

Finally, we develop $I_{3}[\tilde{r}](\alpha,\rho)$. Using:

$\displaystyle\rho\tilde{r}(\alpha^{\prime},\rho^{\prime})-\rho^{\prime}\tilde{r}(\alpha,\rho)$
$\displaystyle=\rho\rho^{\prime}(B(\rho^{\prime})\cos(m\alpha^{\prime})-B(\rho)\cos(m\alpha))$
$\displaystyle=\rho\rho^{\prime}(B(\rho^{\prime})\sin(m(\alpha-\alpha^{\prime}))\sin(m\alpha)$
$\displaystyle+\underbrace{B(\rho^{\prime})\cos(m(\alpha-\alpha^{\prime}))\cos(m\alpha)-B(\rho)\cos(m\alpha)}_{\text{will
integrate to }0})$

This implies that

$\displaystyle I_{3}[\tilde{r}](\alpha,\rho)$
$\displaystyle=-\frac{1}{2\pi}\sin(m\alpha)\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\sin(mx)(\rho-\rho^{\prime}\cos(x))\rho\rho^{\prime}\sin(x)}{(\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x))^{\frac{3}{2}}}dx\right)d\rho^{\prime}$

Integrating by parts, using that

$\displaystyle\frac{\rho\rho^{\prime}\sin(x)}{(\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x))^{\frac{3}{2}}}=-\partial_{x}\left(\frac{1}{(\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x))^{\frac{1}{2}}}\right)$

we get that $I_{3}[\tilde{r}](\alpha,\rho)$ is given by:

$\displaystyle I_{3}[\tilde{r}](\alpha,\rho)$
$\displaystyle=-\frac{m}{2\pi}\sin(m\alpha)\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\cos(mx)(\rho-\rho^{\prime}\cos(x))}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x)}}dx\right)d\rho^{\prime}$
$\displaystyle-\frac{1}{2\pi}\sin(m\alpha)\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\rho^{\prime}B(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\sin(mx)\sin(x)}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x)}}dx\right)d\rho^{\prime}$

Putting all the pieces together and dividing by $\sin(m\alpha)$, the equation
we want to solve is:

$\displaystyle B(\rho)\left(-\lambda
m\rho-\frac{m}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\rho^{\prime}\left(\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x)}}dx\right)d\rho^{\prime}\right)$
(4.11)
$\displaystyle+\frac{2m}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\rho^{\prime}B(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\cos(mx)\cos(x)}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x)}}dx\right)d\rho^{\prime}$
$\displaystyle-\frac{m}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\rho
B(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\cos(mx)}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x)}}dx\right)d\rho^{\prime}$
$\displaystyle-\frac{1}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\rho^{\prime}B(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\sin(mx)\sin(x)}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x)}}dx\right)d\rho^{\prime}=0.$

The inner integrals can be explicitly calculated in terms of EllipticE and
EllipticK functions for any $m$. We can simplify the equation (4.11) in the
following way. Letting $s(\rho,\rho^{\prime})=\frac{\rho}{\rho^{\prime}}$, we
obtain

$\displaystyle\frac{1}{\sqrt{\rho^{2}+(\rho^{\prime})^{2}-2\rho\rho^{\prime}\cos(x)}}=\frac{1}{\rho^{\prime}}\frac{1}{\sqrt{1+\left(\frac{\rho}{\rho^{\prime}}\right)^{2}-2\frac{\rho}{\rho^{\prime}}\cos(x)}}=\frac{1}{\rho^{\prime}}\frac{1}{\sqrt{1+s^{2}-2s\cos(x)}},$

thus equation (4.11) reads:

$\displaystyle
B(\rho)\left(-\lambda\rho-\frac{1}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{1+s^{2}-2s\cos(x)}}dx\right)d\rho^{\prime}\right)$
(4.12)
$\displaystyle+\frac{2}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\cos(mx)\cos(x)}{\sqrt{1+s^{2}-2s\cos(x)}}dx\right)d\rho^{\prime}$
$\displaystyle-\frac{1}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\frac{\rho}{\rho^{\prime}}B(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\cos(mx)}{\sqrt{1+s^{2}-2s\cos(x)}}dx\right)d\rho^{\prime}$
$\displaystyle-\frac{1}{2m\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\sin(mx)\sin(x)}{\sqrt{1+s^{2}-2s\cos(x)}}dx\right)d\rho^{\prime}=0.$

We focus on the term

$\displaystyle\int_{-\pi}^{\pi}\frac{2\cos(mx)\cos(x)-s\cos(mx)-\frac{1}{m}\sin(mx)\sin(x)}{\sqrt{1+s^{2}-2s\cos(x)}}dx\equiv
T(s).$

We remark that $\cos(mx)=\frac{1}{m}\partial_{x}\sin(mx)$. This implies, on
the one hand:

$\displaystyle\int_{-\pi}^{\pi}\frac{2\cos(mx)\cos(x)}{\sqrt{1+s^{2}-2s\cos(x)}}dx$
$\displaystyle=\frac{1}{m}\int_{-\pi}^{\pi}\frac{2\cos(x)\partial_{x}\sin(mx)}{\sqrt{1+s^{2}-2s\cos(x)}}dx$
$\displaystyle=\frac{1}{m}\int_{-\pi}^{\pi}\frac{2\sin(x)\sin(mx)}{\sqrt{1+s^{2}-2s\cos(x)}}dx+\frac{1}{m}\int_{-\pi}^{\pi}\frac{2s\cos(x)\sin(mx)\sin(x)}{(1+s^{2}-2s\cos(x))^{3/2}}dx$

On the other

$\displaystyle
s\int_{-\pi}^{\pi}\frac{\cos(mx)}{\sqrt{1+s^{2}-2s\cos(x)}}dx=\frac{s^{2}}{m}\int_{-\pi}^{\pi}\frac{\sin(mx)\sin(x)}{(1+s^{2}-2s\cos(x))^{3/2}}dx$

Therefore, $T(s)$ can be transformed into

$\displaystyle T(s)$
$\displaystyle=\frac{1}{m}\int_{-\pi}^{\pi}\frac{\sin(x)\sin(mx)}{\sqrt{1+s^{2}-2s\cos(x)}}dx+\frac{1}{m}\int_{-\pi}^{\pi}\frac{2s\cos(x)\sin(mx)\sin(x)}{(1+s^{2}-2s\cos(x))^{3/2}}dx$
$\displaystyle-\frac{s^{2}}{m}\int_{-\pi}^{\pi}\frac{\sin(mx)\sin(x)}{(1+s^{2}-2s\cos(x))^{3/2}}dx$
$\displaystyle=\frac{1}{m}\int_{-\pi}^{\pi}\frac{\sin(mx)\sin(x)}{(1+s^{2}-2s\cos(x))^{3/2}}dx=\frac{1}{s}\int_{-\pi}^{\pi}\frac{\cos(mx)}{\sqrt{1+s^{2}-2s\cos(x)}}dx.$

Substituting into (4.12), we have to solve:

$\displaystyle
B(\rho)\left(-\lambda\rho-\frac{1}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{1+s^{2}-2s\cos(x)}}dx\right)d\rho^{\prime}\right)$
(4.13)
$\displaystyle+\frac{1}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})\frac{\rho^{\prime}}{\rho}\int_{-\pi}^{\pi}\frac{\cos(mx)}{\sqrt{1+s^{2}-2s\cos(x)}}dxd\rho^{\prime}=0.$

From now on, we will call

$\displaystyle I(\rho)$
$\displaystyle=-\frac{1}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\left(\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{1+\left(\frac{\rho}{\rho^{\prime}}\right)^{2}-2\left(\frac{\rho}{\rho^{\prime}}\right)\cos(x)}}dx\right)d\rho^{\prime}$
$\displaystyle T^{m}B(\rho)$
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})\frac{\rho^{\prime}}{\rho}\int_{-\pi}^{\pi}\frac{\cos(mx)}{\sqrt{1+\left(\frac{\rho}{\rho^{\prime}}\right)^{2}-2\left(\frac{\rho}{\rho^{\prime}}\right)\cos(x)}}dxd\rho^{\prime}$

We will also define

$\displaystyle K^{m}(s)$ $\displaystyle=\frac{1}{2\pi
s}\int_{-\pi}^{\pi}\frac{\cos(mx)}{\sqrt{1+s^{2}-2s\cos(x)}}dx$ $\displaystyle
T^{m}B(\rho)$
$\displaystyle=\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}$

This allows us to write (4.13) as:

$\displaystyle\tilde{I}(\rho)B(\rho)+\tilde{T}^{m}B(\rho)=\lambda
B(\rho),\quad\text{in $(1-a,1)$,}$ (4.14)

where $\tilde{T}^{m}=\frac{1}{\rho}T^{m}$ and
$\tilde{I}=\frac{1}{\rho}I(\rho)$. Thus, using the notation of (4.10),

$\Theta^{m}B(\rho)\equiv\tilde{I}(\rho)B(\rho)+\tilde{T}^{m}B(\rho).$

####  4.3.2 Existence of an element in the kernel of
$\partial_{r}F[\rho,\lambda_{3}]$

In this part we will study the equation (4.14) in order to obtain an element
in the kernel or $\partial_{r}F[\rho,\lambda_{3}]$ for some value
$\lambda_{3}\in\mathbb{R}$ . We shall show the following proposition:

###### Proposition 4.8

There exists a solution $(B^{3},\lambda_{3})\in
H^{3}((1-a),1)\times\mathbb{R}$ to the equation (4.14). In addition,
$\lambda_{3}$ is simple.

We remark that this proposition yields the next corollary:

###### Corollary 4.9

The function $\tilde{r}_{0}(\alpha,\rho)=\rho B^{3}(\rho)\cos(3\alpha)$
belongs to $H^{4,3}_{3,\text{even}}(\Omega_{a})$ and solves (4.9).

###### Proof.

The proof of Proposition 4.8 is divided in two parts. In the first one we deal
with the operator $\tilde{T}$ in (4.14). In the second one we show existence
of pair $(B^{3},\lambda_{3})\in H^{3}((1-a,1))\times\mathbb{R}$ solving (4.14)
and that $\lambda_{3}$ is simple.

  1. 1.

Study of the operator $\tilde{T}^{m}$. This part is devoted to studying the
operator $\tilde{T}^{m}$ and its derivatives until order 3. Here we recall its
definition:

$\displaystyle\tilde{T}^{m}B(\rho)=\frac{1}{\rho}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}$

with

$\displaystyle K^{m}(s)$ $\displaystyle=\frac{1}{2\pi
s}\int_{-\pi}^{\pi}\frac{\cos(mx)}{\sqrt{1+s^{2}-2s\cos(x)}}dx.$

The main results here are Corollary 4.15 and Lemma 4.18 that state that the
operator $\tilde{T}^{m}$ and its adjoint $\tilde{T}^{m*}$ are compact
operators acting from $H^{k}$ to $H^{k+1}$ for $k=0,1,2$.

Let’s compute the derivatives of $\tilde{T}^{3m}$.

###### Lemma 4.10

Let $B\in C^{3}$ then the following equalities hold:

$\displaystyle\partial_{\rho}\tilde{T}^{m}B(\rho)=$
$\displaystyle\tilde{T}^{m}_{1}\partial_{\rho}B(\rho)+\frac{1}{\rho}\int_{-\infty}^{\infty}\partial^{2}_{\rho}f(\rho^{\prime})B(\rho^{\prime})\frac{\rho^{\prime}}{\rho}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime},$
$\displaystyle\partial^{2}_{\rho}\tilde{T}^{m}B(\rho)=$ $\displaystyle
T^{m}_{2}\partial^{2}_{\rho}B(\rho)+\frac{1}{\rho}\sum_{j=1}^{2}\int_{-\infty}^{\infty}\partial^{j+1}_{\rho}f(\rho^{\prime})\partial^{2-j}_{\rho}B(\rho^{\prime})\left(\frac{\rho^{\prime}}{\rho}\right)^{2}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}$
$\displaystyle\partial^{3}_{\rho}\tilde{T}^{m}B(\rho)=$ $\displaystyle
T^{m}_{3}\partial^{3}_{\rho}B(\rho)+\sum_{j=1}^{3}\int_{-\infty}^{\infty}\partial^{j+1}_{\rho}f(\rho^{\prime})\partial^{3-j}_{\rho}B(\rho^{\prime})\left(\frac{\rho^{\prime}}{\rho}\right)^{3}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}$

where

$\displaystyle\tilde{T}^{m}_{1}B=$
$\displaystyle\frac{1}{\rho}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})\frac{\rho^{\prime}}{\rho}K^{m}\left(\frac{1}{\gamma}\right)d\rho^{\prime}$
(4.15) $\displaystyle\tilde{T}^{m}_{2}B(\rho)=$
$\displaystyle\frac{1}{\rho}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})\left(\frac{\rho^{\prime}}{\rho}\right)^{2}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}$
(4.16) $\displaystyle\tilde{T}^{m}_{3}B(\rho)=$
$\displaystyle\frac{1}{\rho}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})B(\rho^{\prime})\left(\frac{\rho^{\prime}}{\rho}\right)^{3}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}.$
(4.17)

###### Proof.

We notice that after the change of variable
$\gamma=\frac{\rho^{\prime}}{\rho}$ we have that

$\displaystyle\tilde{T}^{m}B(\rho)=\int_{-\infty}^{\infty}g(\rho\gamma)K^{m}\left(\frac{1}{\gamma}\right)d\gamma$

where $g=f_{\rho}B$. Taking one derivative we have that

$\displaystyle\partial_{\rho}\tilde{T}^{m}B(\rho)=$
$\displaystyle\int_{-\infty}^{\infty}(\partial_{\rho}g)(\gamma\rho)\gamma
K^{m}\left(\frac{1}{\gamma}\right)d\gamma$ $\displaystyle=$
$\displaystyle\frac{1}{\rho}\int_{-\infty}^{\infty}\partial_{\rho}g(\rho^{\prime})\frac{\rho^{\prime}}{\rho}K^{m}\left(\frac{1}{\gamma}\right)d\rho^{\prime}$
$\displaystyle=$
$\displaystyle\frac{1}{\rho}\int_{-\infty}^{\infty}\partial^{2}_{\rho}f(\rho^{\prime})B(\rho^{\prime})\frac{\rho^{\prime}}{\rho}K^{m}\left(\frac{1}{\gamma}\right)d\rho+\frac{1}{\rho}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\partial_{\rho}B(\rho^{\prime})\frac{\rho^{\prime}}{\rho}K^{m}\left(\frac{1}{\gamma}\right)d\rho^{\prime},$

so that

$\displaystyle\partial_{\rho}\tilde{T}^{m}B(\rho)=\tilde{T}^{m}_{1}\partial_{\rho}B(\rho)+\frac{1}{\rho}\int_{-\infty}^{\infty}\partial^{2}_{\rho}f(\rho^{\prime})B(\rho^{\prime})\frac{\rho^{\prime}}{\rho}K^{m}\left(\frac{1}{\gamma}\right)d\rho^{\prime}.$

Computing in a similar way, by taking two derivatives, we have that

$\displaystyle\partial_{\rho}^{2}\tilde{T}^{m}B(\rho)=$
$\displaystyle\frac{1}{\rho}\int_{-\infty}^{\infty}\partial_{\rho}^{2}g(\rho^{\prime})\left(\frac{\rho^{\prime}}{\rho}\right)^{2}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}$
$\displaystyle\partial_{\rho}^{3}\tilde{T}^{m}B(\rho)=$
$\displaystyle\frac{1}{\rho}\int_{-\infty}^{\infty}\partial_{\rho}^{3}g(\rho^{\prime})\left(\frac{\rho^{\prime}}{\rho}\right)^{3}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}.$

And we can write

$\displaystyle\partial^{2}_{\rho}\tilde{T}^{m}B(\rho)=$
$\displaystyle\tilde{T}^{m}_{2}\partial^{2}_{\rho}B(\rho)+\frac{1}{\rho}\sum_{j=1}^{2}\int_{-\infty}^{\infty}\partial^{j+1}_{\rho}f(\rho^{\prime})\partial^{2-j}_{\rho}B(\rho^{\prime})\left(\frac{\rho^{\prime}}{\rho}\right)^{2}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}$
$\displaystyle\partial^{3}_{\rho}\tilde{T}^{m}B(\rho)=$
$\displaystyle\tilde{T}^{m}_{3}\partial^{3}_{\rho}B(\rho)+\sum_{j=1}^{3}\int_{-\infty}^{\infty}\partial^{j+1}_{\rho}f(\rho^{\prime})\partial^{3-j}_{\rho}B(\rho^{\prime})\left(\frac{\rho^{\prime}}{\rho}\right)^{3}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}.$

∎

Some of the properties of the operator $\tilde{T}^{m}$ come from the sign of
its kernel. We study this sign in the following lemma.

###### Lemma 4.11

Let $T^{m}\left(\frac{\rho}{\rho^{\prime}}\right)$ be defined as:

$\displaystyle
T^{m}\left(\frac{\rho}{\rho^{\prime}}\right)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\cos(mx)}{\sqrt{1+\left(\frac{\rho}{\rho^{\prime}}\right)^{2}-2\left(\frac{\rho}{\rho^{\prime}}\right)\cos(x)}}dx$

Then, we have that, for every
$(\rho,\rho^{\prime})\in\mathbb{R}^{2},\rho\neq\rho^{\prime}$

    1. (a)

$T^{m}\left(\frac{\rho}{\rho^{\prime}}\right)>0$

    2. (b)

$T^{m+1}\left(\frac{\rho}{\rho^{\prime}}\right)<T^{m}\left(\frac{\rho}{\rho^{\prime}}\right)$.

###### Proof.

Let $r=\frac{\rho}{\rho^{\prime}}$. We have that

$\displaystyle
T^{m}\left(\frac{\rho}{\rho^{\prime}}\right)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\cos(mx)}{(1+r^{2}-2r\cos(x))^{\frac{1}{2}}}dx$

We do first the $r<1$ case. The $r>1$ case follows from the property

$\displaystyle
T^{m}\left(\frac{\rho}{\rho^{\prime}}\right)=\left(\frac{\rho}{\rho^{\prime}}\right)^{3}T^{m}\left(\frac{\rho^{\prime}}{\rho}\right)$

$\displaystyle T^{m}\left(\frac{\rho}{\rho^{\prime}}\right)$
$\displaystyle=\frac{1}{1+r}\int_{-\pi}^{\pi}\frac{\cos(mx)}{\left(1-\frac{4r}{(1+r)^{2}}\cos^{2}\left(\frac{x}{2}\right)\right)^{\frac{1}{2}}}=\frac{1}{2\pi}\frac{2}{1+r}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos(2mx)}{\left(1-\frac{4r}{(1+r)^{2}}\cos^{2}\left(x\right)\right)^{\frac{1}{2}}}$
$\displaystyle=\frac{1}{2\pi}\frac{4}{1+r}\sum_{k=0}^{\infty}\int_{0}^{\frac{\pi}{2}}\cos(2mx)\left(\frac{4r}{(1+r)^{2}}\right)^{k}\cos^{2k}\left(x\right)\frac{1}{k!}\left(1/2\right)_{k}dx$
$\displaystyle=\frac{1}{2\pi}\frac{4}{1+r}\sum_{k=0}^{\infty}\left(\frac{4r}{(1+r)^{2}}\right)^{k}\frac{1}{k!}\left(1/2\right)_{k}\int_{0}^{\frac{\pi}{2}}\cos(2mx)\cos^{2k}\left(x\right)dx$
$\displaystyle=\frac{1}{2\pi}\frac{4}{1+r}\sum_{k=m}^{\infty}\left(\frac{4r}{(1+r)^{2}}\right)^{k}\frac{1}{k!}\left(1/2\right)_{k}\frac{\pi}{2^{2k+1}}\frac{\Gamma(2k+1)}{\Gamma(1+k+m)\Gamma(1+k-m)}$
$\displaystyle=\sum_{k=m}^{\infty}A_{k}\frac{1}{\Gamma(1+k+m)\Gamma(1+k-m)}.$

Since $A_{k}>0$ for every $k$, this shows the first item. Next, we compute for
$r<1$:

$\displaystyle
T^{m}\left(\frac{\rho}{\rho^{\prime}}\right)-T^{m+1}\left(\frac{\rho}{\rho^{\prime}}\right)$
$\displaystyle=\frac{A_{m}}{\Gamma(1+2m)}$
$\displaystyle+\sum_{k=m+1}^{\infty}A_{k}\left(\frac{1}{\Gamma(1+k+m)\Gamma(1+k-m)}-\frac{1}{\Gamma(2+k+m)\Gamma(k-m)}\right)$
$\displaystyle=\frac{A_{m}}{\Gamma(1+2m)}+\sum_{k=m+1}^{\infty}\frac{A_{k}}{\Gamma(1+k+m)\Gamma(1+k-m)}\left(\frac{1+2m}{1+k+m}\right)>0.$

This completes the proof of the lemma. ∎

In order to prove the compactness of the operator $\tilde{T}^{m}$ we will use
the following decomposition:

###### Lemma 4.12

The function $T^{m}(s)$ satisfies

$T^{m}(s)=-\frac{2}{\sqrt{s}}\log(|1-s|)+E^{m}(s)$

where $E^{m}(s)$ is a $C^{1}-$function with

$||E^{m}||_{C^{1}}\leq C(m).$

###### Proof.

We will split $T^{m}(s)$ in two parts

$\displaystyle T^{m}(s)=$
$\displaystyle\int_{-\pi}^{\pi}\frac{1}{\sqrt{(1-s)^{2}+4s\sin^{2}\left(\frac{x}{2}\right)}}dx+\int_{-\pi}^{\pi}\frac{\cos(mx)-1}{(1-s)^{2}+4s\sin^{2}\left(\frac{x}{2}\right)}dx$
$\displaystyle\equiv$ $\displaystyle T_{1}(s)+T^{m}_{2}(s).$

We now focus on the term $T_{1}(s)$. By making the change
$y=\sin\left(\frac{x}{2}\right)$ yields

$\displaystyle T_{1}(s)=$ $\displaystyle
4\int_{0}^{1}\frac{1}{\sqrt{(1-s)^{2}+4s\sin^{2}\left(\frac{x}{2}\right)}}\frac{dy}{\sqrt{1-y^{2}}}$
$\displaystyle=\frac{2}{\sqrt{s}}\int_{0}^{\frac{1}{\varepsilon}}\frac{1}{\sqrt{1+z^{2}}}\frac{dz}{\sqrt{1-\varepsilon^{2}z^{2}}},$

with $\varepsilon=\frac{|1-s|}{2\sqrt{s}}.$ We will break $T_{1}(s)$ in two
parts

$\displaystyle T_{1}(s)=$
$\displaystyle\frac{2}{\sqrt{s}}\int_{0}^{\frac{1}{\varepsilon}}\frac{1}{\sqrt{1+z^{2}}}\left(\frac{1}{\sqrt{1-\varepsilon^{2}z^{2}}}-1\right)dz$
$\displaystyle+$
$\displaystyle\frac{2}{\sqrt{s}}\int_{0}^{\frac{1}{\varepsilon}}\frac{1}{\sqrt{1+z^{2}}}dz$
$\displaystyle\equiv$ $\displaystyle T_{11}(s)+T_{12}(s).$

The integral in the term $T_{12}(s)$ can be computed exactly. We obtain that

$\displaystyle
T_{12}(s)=\frac{2}{\sqrt{s}}\operatorname{arcsinh}\left(\frac{1}{\varepsilon}\right),$
(4.18)

where we recall that

$\displaystyle\operatorname{arcsinh}(x)=\log\left(x+\sqrt{x^{2}+1}\right)$

and then,

$\displaystyle\operatorname{arcsinh}\left(\frac{1}{\varepsilon}\right)=\log\left(\frac{2\sqrt{s}}{|1-s|}+\sqrt{\frac{4s}{|1-s|^{2}}+1}\right)=-\log\left(|1-s|\right)+2\log(1+\sqrt{s}).$

Finally,

$\displaystyle T_{12}(s)=$
$\displaystyle-\frac{2}{\sqrt{s}}\log(|1-s|)+\frac{4}{\sqrt{s}}\log(1+\sqrt{s})=-\frac{2}{\sqrt{s}}\log(|1-s|)+S(s).$

where $S(s)$ is an smooth function. Next we will show that the first
derivative of the function $T_{11}(s)$ is continuous. We notice that undoing
the change of variable we can write

$\displaystyle T_{11}(s)$
$\displaystyle=4\int_{0}^{1}\frac{1}{\sqrt{(1-s)^{2}+4sy^{2}}}\left(\frac{1}{\sqrt{1-y^{2}}}-1\right)dy$
$\displaystyle=4\int_{0}^{1}\frac{1}{\sqrt{(1-s)^{2}+4sy^{2}}}\left(\frac{y^{2}}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}\right)dy$

Since the function

$\frac{y^{2}}{\sqrt{(s-1)^{2}+4sy^{2}}}$

is in $L^{\infty}$, by the dominated convergence theorem (DCT), $T_{11}(s)$ is
continuous at $s=1$. In addition, for $s\neq 1$, we can differentiate to get
that in a weak sense

$\displaystyle\partial_{s}T_{11}(s)=$
$\displaystyle-4\int_{0}^{1}\frac{(s-1)+2y^{2}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{3}{2}}}\left(\frac{y^{2}}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}\right)dy$
$\displaystyle=T_{111}(s)+T_{112}(s)$

where

$\displaystyle
T_{111}(s)=-4\int_{0}^{1}\frac{(s-1)}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{3}{2}}}\left(\frac{y^{2}}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}\right)dy$

and

$\displaystyle
T_{112}(s)=-8\int_{0}^{1}\frac{y^{4}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{3}{2}}}\left(\frac{1}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}\right)dy$

where we can prove that $T_{112}(s)$ is a continuous function at $s=1$ by DCT.
To analyze $T_{111}(s)$ we split this term into two parts

$\displaystyle T_{111}(s)=$
$\displaystyle-2\int_{0}^{1}\frac{(s-1)y^{2}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{3}{2}}}dy$
$\displaystyle-4\int_{0}^{1}\frac{(s-1)y^{2}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{3}{2}}}\left(\frac{1}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}-\frac{1}{2}\right)dy.$

The second integral is continuous at $s=1$ again by DCT. The first integral
can be computed analytically. We obtain that

$\displaystyle-2\int_{0}^{1}\frac{(s-1)y^{2}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{3}{2}}}dy=\frac{1-s}{3s}\left(-\frac{2\sqrt{s}}{1+s}+\operatorname{arcsinh}\left(\frac{2\sqrt{s}}{|1-s|}\right)\right),$

to show that $T_{111}(s)$ is also a continuous function at $s=1$. Therefore we
have that

$T_{1}(s)=-\frac{2}{\sqrt{s}}\log(|1-s|)+E(s)$

where $E(s)$ is a $C^{1}$ function.

For $T^{m}_{2}(s)$ we use Taylor’s integral remainder formula to show that

$\cos(mx)-1=-\frac{m^{2}x^{2}}{2}+m^{4}x^{4}\int_{0}^{1}\gamma^{3}\int_{0}^{1}\mu^{2}\int_{0}^{1}\nu\int_{0}^{1}\cos(m\gamma\mu\nu\tau
x)d\nu d\mu d\gamma d\tau\equiv\frac{m^{2}x^{2}}{2}+m^{4}x^{4}R^{m}(x),$

and that

$\displaystyle
2\sin^{2}\left(\frac{x}{2}\right)=(1-\cos(x))=\frac{x^{2}}{2}-x^{4}R^{1}(x)$

Therefore

$\displaystyle\cos(mx)-1=2m^{2}\sin^{2}\left(\frac{x}{2}\right)+x^{4}\tilde{R}^{m}(x).$

where $\tilde{R}^{m}(x)$ is a bounded function. We now can write
$T^{m}_{2}(s)$ as follows

$\displaystyle T^{m}_{2}(s)=$
$\displaystyle\int_{-\pi}^{\pi}\frac{2m^{2}\sin^{2}\left(\frac{x}{2}\right)}{\sqrt{(1-s)^{2}+4s\sin^{2}\left(\frac{x}{2}\right)}}dx+\int_{-\pi}^{\pi}\frac{x^{4}\tilde{R}^{m}(x)}{\sqrt{(1-s)^{2}+4s\sin^{2}\left(\frac{x}{2}\right)}}dx$

By DCT $T^{m}_{2}(s)$ is continuous at $s=1$. For $s\neq 1$ we differentiate
to get

$\displaystyle\partial_{s}T^{m}_{2}(s)=$
$\displaystyle-\int_{-\pi}^{\pi}\frac{(s-1)+2\sin^{2}\left(\frac{x}{2}\right)}{\left((1-s)^{2}+4s\sin^{2}\left(\frac{x}{2}\right)\right)^{\frac{3}{2}}}\sin^{2}\left(\frac{x}{2}\right)dx$
$\displaystyle-\int_{-\pi}^{\pi}\frac{(s-1)+2\sin^{2}\left(\frac{x}{2}\right)}{\left((1-s)^{2}+4s\sin^{2}\left(\frac{x}{2}\right)\right)^{\frac{3}{2}}}x^{4}\tilde{R}^{m}(x)dx.$

The second integral on the right hand side is a continuous function. In the
first one we make the change of variables $y=\sin\left(\frac{x}{2}\right)$ to
get that

$\displaystyle\int_{-\pi}^{\pi}\frac{(s-1)+2\sin^{2}\left(\frac{x}{2}\right)}{\left((1-s)^{2}+4s\sin^{2}\left(\frac{x}{2}\right)\right)^{\frac{3}{2}}}\sin^{2}\left(\frac{x}{2}\right)dx=4\int_{0}^{1}\frac{(s-1)+2y^{2}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{3}{2}}}\frac{y^{2}}{\sqrt{1-y^{2}}}dy.$

Here, the difficulty to show continuity comes from the integral

$\displaystyle\int_{0}^{1}\frac{(s-1)}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{3}{2}}}\frac{y^{2}}{\sqrt{1-y^{2}}}dy$

but it has been already proven that this integral is continuous in the
analysis of $T_{111}(s)$. This concludes the proof of the lemma. ∎

We also need to study the derivatives of the function $\tilde{T}^{3*}B(\rho)$.
We start with the following lemma:

###### Lemma 4.13

The function $(1-s)^{3}T^{3}(s)$ is $C^{2}$ in a neighbourhood of $s=1$.

###### Proof.

In order to prove the lemma we will split $T^{3}(s)$ in different parts and we
will deal with every one separately. We first notice that we can write

$\displaystyle T^{3}(s)$
$\displaystyle=\int_{-\pi}^{\pi}\frac{\cos(3x)}{\sqrt{(1-s)^{2}+4s\sin\left(\frac{x}{2}\right)^{2}}}dx$

We will use the formula

$\displaystyle\cos(3x)=1-18\sin^{2}\left(\frac{x}{2}\right)+48\sin^{4}\left(\frac{x}{2}\right)-32\sin^{6}\left(\frac{x}{2}\right)$

and make the change $y=\sin\left(\frac{x}{2}\right)$ to get

$\displaystyle T^{3}(s)=$ $\displaystyle
2\int_{-1}^{1}\frac{1-18y^{2}+48y^{4}-32y^{6}}{\sqrt{(1-s)^{2}+4sy^{2}}}\frac{dy}{\sqrt{1-y^{2}}}$
$\displaystyle=$ $\displaystyle
4\int_{0}^{1}\frac{1-18y^{2}+48y^{4}-32y^{6}}{\sqrt{(1-s)^{2}+4sy^{2}}}\frac{dy}{\sqrt{1-y^{2}}}$

In addition we change again of variable by making $z=\frac{2\sqrt{s}}{|1-s|}y$
so that

$\displaystyle T^{3}(s)=$
$\displaystyle\frac{2}{\sqrt{s}}\int_{0}^{\frac{1}{\varepsilon}}\frac{1-18\varepsilon^{2}z^{2}+48\varepsilon^{4}z^{4}-32\varepsilon^{6}z^{6}}{\sqrt{1+z^{2}}}\frac{dz}{\sqrt{1-\varepsilon^{2}z^{2}}},$

where $\varepsilon=\frac{|1-s|}{2\sqrt{s}}$. Now we define

$\displaystyle T_{1}(s)=$
$\displaystyle\frac{2}{\sqrt{s}}\int_{0}^{\frac{1}{\varepsilon}}\frac{1}{\sqrt{1+z^{2}}}\frac{dz}{\sqrt{1-\varepsilon^{2}z^{2}}}$
$\displaystyle T_{2}^{3}(s)=$
$\displaystyle\frac{2}{\sqrt{s}}\int_{0}^{\frac{1}{\varepsilon}}\frac{-18\varepsilon^{2}z^{2}+48\varepsilon^{4}z^{4}-32\varepsilon^{6}z^{6}}{\sqrt{1+z^{2}}}\frac{dz}{\sqrt{1-\varepsilon^{2}z^{2}}}.$

Thus $T^{3}(s)=T_{1}(s)+T^{3}_{2}(s)$ and $T_{1}(s)$ is the same function that
in the proof of Lemma 4.12. Then we know that $T_{1}(s)=T_{11}(s)+T_{12}(s)$,
where

$T_{12}(s)=-\frac{2}{\sqrt{s}}\log(|1-s|)+S(s),$

with $S(s)$ a smooth function. Therefore $(1-s)^{3}T_{12}(s)$ is a
$C^{2}-$function. Also we know that $T_{11}(s)$ is a $C^{1}-$ function. In
order to analyze two derivatives of $T_{11}(s)$ we differentiate for $s\neq 1$
to get

$\displaystyle\partial^{2}_{s}T_{11}(s)=$
$\displaystyle-4\int_{0}^{1}\frac{1}{\left((1-s)+4sy^{2}\right)^{\frac{3}{2}}}\left(\frac{y^{2}}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}\right)dy$
$\displaystyle+12\int_{0}^{1}\frac{\left((s-1)^{2}+4y^{2}\right)^{2}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{5}{2}}}\left(\frac{y^{2}}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}\right)dy.$

Multiplying by $(1-s)$ we have that

$\displaystyle(1-s)\partial^{2}_{s}T^{3}_{11}(s)=-T^{3}_{111}(s)+6\int_{0}^{1}(1-s)\frac{\left((s-1)+4y^{2}\right)^{2}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{5}{2}}}\left(\frac{y^{2}}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}\right)dy,$

where we have already checked that $T^{3}_{111}(s)$ is continuous. Next we
check that this function is also continuous. This will be a consequence of the
continuity of the following terms

$\displaystyle\int_{0}^{1}(1-s)\frac{(s-1)^{2}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{5}{2}}}\left(\frac{y^{2}}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}\right)dy$
$\displaystyle\int_{0}^{1}(1-s)\frac{(s-1)y^{2}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{5}{2}}}\left(\frac{y^{2}}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}\right)dy$
$\displaystyle\int_{0}^{1}(1-s)\frac{y^{4}}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{5}{2}}}\left(\frac{y^{2}}{\sqrt{1-y^{2}}\left(1+\sqrt{1-y^{2}}\right)}\right)dy.$

The last term is continuous just by applying DCT. The other two terms can be
treated in a similar way we did before to show continuity.

In addition the analysis of $T^{3}_{2}(s)$ does not introduce any new
difficulty and we will not give the details here. This concludes the proof of
the lemma. ∎

###### Lemma 4.14

The function $(1-s)T^{m}(s)$ satisfies

$\displaystyle\partial_{s}\left((1-s)T^{m}(s)\right)=-T^{m}(s)+E^{m}(s)$

where $E^{m}(s)$ is a continuous function with $L^{\infty}-$norm independent
on $m$.

###### Proof.

By Lemma 4.12 $(1-s)T^{m}(s)$ is a continuous function. For $s\neq 1$ we can
differentiate to get

$\displaystyle\partial_{s}\left((1-s)T^{m}(s)\right)=-T^{m}(s)+E^{m}(s)$

with

$\displaystyle
E^{m}(s)=\int_{-\pi}^{\pi}\frac{-(1-s)^{2}+2(1-s)\sin^{2}\left(\frac{x}{2}\right)}{\left((1-s)^{2}+2\sin^{2}\left(\frac{x}{2}\right)\right)^{\frac{3}{2}}}\cos(mx)dx.$

In addition we have that

$\displaystyle\left|E^{m}(s)\right|\leq\int_{-\pi}^{\pi}\frac{(1-s)^{2}+2|1-s|\sin^{2}\left(\frac{x}{2}\right)}{\left((1-s)^{2}+s\sin^{2}\left(\frac{x}{2}\right)\right)^{\frac{3}{2}}}dx.$

For the first integral we have that

$\displaystyle\int_{-\pi}^{\pi}$
$\displaystyle\frac{(1-s)^{2}}{\left((1-s)^{2}+4s\sin^{2}\left(\frac{x}{2}\right)\right)^{\frac{3}{2}}}dx=\frac{2}{\sqrt{s}}\int_{0}^{\frac{2\sqrt{s}}{|1-s|}}\frac{1}{\left(1+y^{2}\right)^{\frac{3}{2}}}\frac{dy}{\sqrt{1-\frac{|1-s|^{2}}{2s}y^{2}}}$
$\displaystyle=\frac{2}{\sqrt{s}}\int_{0}^{\frac{2\sqrt{s}}{|1-s|}}\frac{1}{\left(1+y^{2}\right)^{\frac{3}{2}}}dy+\frac{2}{\sqrt{s}}\int_{0}^{\frac{2\sqrt{s}}{|1-s|}}\frac{1}{\left(1+y^{2}\right)^{\frac{3}{2}}}\left(\frac{1}{\sqrt{1-\frac{|1-s|^{2}}{2s}y^{2}}}-1\right)dy$
$\displaystyle=\frac{2}{\sqrt{s}}\int_{0}^{\frac{2\sqrt{s}}{|1-s|}}\frac{1}{\left(1+y^{2}\right)^{\frac{3}{2}}}dy+4\int_{0}^{1}\frac{|1-s|}{\left((1-s)^{2}+4sy^{2}\right)^{\frac{3}{2}}}\left(\frac{1}{\sqrt{1-y^{2}}}-1\right)dy.$
(4.19)

In addition, the second integral, can be written

$\displaystyle\int_{-\pi}^{\pi}\frac{2|1-s|\sin^{2}\left(\frac{x}{2}\right)}{\left((1-s)^{2}+2\sin^{2}\left(\frac{x}{2}\right)\right)^{\frac{3}{2}}}dx=8\int_{0}^{1}\frac{|1-s|y^{2}}{\left(|1-s|^{2}+4sy^{2}\right)^{\frac{3}{2}}}\frac{dy}{\sqrt{1-y^{2}}}$
$\displaystyle=\frac{|1-s|}{s^{\frac{3}{2}}}\int_{0}^{\frac{2\sqrt{s}}{|1-s|}}\frac{y^{2}}{(1+y^{2})^{\frac{3}{2}}}\frac{dy}{\sqrt{1-\frac{|1-s|^{2}}{4s}y^{2}}}=\frac{|1-s|}{s^{\frac{3}{2}}}\int_{0}^{\frac{2\sqrt{s}}{|1-s|}}\frac{y^{2}}{(1+y^{2})^{\frac{3}{2}}}dy$
$\displaystyle+\frac{|1-s|}{s^{\frac{3}{2}}}\int_{0}^{\frac{2\sqrt{s}}{|1-s|}}\frac{y^{2}}{(1+y^{2})^{\frac{3}{2}}}\left(\frac{1}{\sqrt{1-\frac{|1-s|^{2}}{4s}y^{2}}}-1\right)dy$
$\displaystyle\frac{|1-s|}{s^{\frac{3}{2}}}\int_{0}^{\frac{2\sqrt{s}}{|1-s|}}\frac{y^{2}}{(1+y^{2})^{\frac{3}{2}}}dy+8\int_{0}^{1}\frac{|1-s|y^{2}}{\left(|1-s|^{2}+4sy^{2}\right)^{\frac{3}{2}}}\left(\frac{1}{\sqrt{1-y^{2}}}-1\right)dy.$
(4.20)

From expression (1), (1) and DCT it is easy to achieve the conclusion of the
lemma. ∎

Now we can state and prove the main lemmas of this section.

###### Lemma 4.15

Let $f\in C^{4}(\mathbb{R}^{+})$ and $m\geq 1$. Then, $\tilde{T}^{m}$ and
$\tilde{T}^{m}_{i}$ , with $i=1,2,3$ are a compact operators, acting between
$L^{2}$ and $H^{1}$, with

$\displaystyle||\tilde{T}_{i}^{m}v||_{L^{2}}\leq C||v||_{L^{2}},$
$\displaystyle\|\tilde{T}_{i}^{m}v\|_{H^{1}}\leq C(m)\|v\|_{L^{2}},$

where the constant $C$ in the first inequality only depends on $\|f\|_{C^{4}}$
and $\Omega_{a}$ and the constant $C(m)$ in the second one also depends on
$m$.

###### Proof.

Because of Lemma 4.12 $\tilde{T}^{3}_{i}$ is bounded from $L^{2}$ to $L^{2}$.
Then, by Lemma 4.11 and the monotonicity of $f(\rho)$, every
$\tilde{T}^{m}_{i}$ is also bounded from $L^{2}$ to $L^{2}$ by the same
constant as $\tilde{T}^{3}_{i}$. In order to show the boundedness from $L^{2}$
to $H^{1}$ we first show the estimate for smooth functions. This is done just
using Lemma 4.12 and the $L^{2}-$boundedness of the Hilbert transform. Finally
we proceed by a density argument. ∎

###### Corollary 4.16

The operator $\tilde{T}^{m}$ is bounded from $H^{k}$ to $H^{k+1}$, for
$k=0,1,2$, with norm depending only on $m$, $||f||_{C^{4}}$ and $a$.

###### Proof.

We first prove the bound for smooth functions. This can be done by using Lemma
4.10 and 4.15. Then we proceed by a density argument. ∎

Finally, we will study the adjoint operator of $\tilde{T}^{3}$ given by the
expression

$\displaystyle\tilde{T}^{3*}B(\rho)=f_{\rho}(\rho)\int_{1-a}^{1}B(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho^{\prime}}.$

###### Lemma 4.17

Let $B\in C^{3}$ then the following equalities hold:

    1. (a)

$\displaystyle\partial_{\rho}\tilde{T}^{3*}B(\rho)$
$\displaystyle=\partial^{2}_{\rho}f(\rho)\int_{1-a}^{1}K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho^{\prime}}-\frac{1}{\rho^{\prime}}V.P.\int_{1-a}^{1}B(\rho^{\prime}),\partial_{\rho^{\prime}}\left(K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\right)d\rho^{\prime}$

    2. (b)

$\displaystyle\partial^{2}_{\rho}\tilde{T}^{3*}B(\rho)$
$\displaystyle=\partial^{3}_{\rho}f(\rho)\int_{1-a}^{1}B(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho^{\prime}}$
$\displaystyle-\frac{\partial^{2}_{\rho}f(\rho)}{\rho}P.V.\int_{1-a}^{1}B(\rho^{\prime})\partial_{\rho^{\prime}}K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)d\rho^{\prime}-\partial_{\rho}\left(B(1)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1}{\rho}\right)}{\rho}-B(1-a)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1-a}{\rho}\right)}{\rho}\right)$
$\displaystyle-\frac{\partial^{2}_{\rho}f(\rho)}{\rho}\int_{1-a}^{1}B^{\prime}(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)+\frac{f_{\rho}(\rho)}{\rho}P.V.\int_{1-a}^{1}B^{\prime}(\rho^{\prime})\partial_{\rho^{\prime}}\left(\rho^{\prime}K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\right)d\rho^{\prime},$

    3. (c)

$\displaystyle\partial^{3}_{\rho}\tilde{T}^{3*}(s)$
$\displaystyle=\partial^{4}_{\rho}f(\rho)\int_{1-a}^{1}B(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho^{\prime}}+\frac{\partial^{3}_{\rho}f(\rho)}{\rho}\left(-B(1)K^{3}\left(\frac{1}{\rho}\right)+B(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle+\partial^{3}_{\rho}f(\rho)\int_{1-a}^{1}B^{\prime}(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho}$
$\displaystyle+\partial_{\rho}\left(\frac{\partial^{2}_{\rho}f(\rho)}{\rho}\left(-B(1)K^{3}\left(\frac{1}{\rho}\right)+B(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)\right)$
$\displaystyle+2\partial^{3}_{\rho}f(\rho)\int_{1-a}^{1}B^{\prime\prime}(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho}++2\frac{\partial^{2}_{\rho}f(\rho)}{\rho^{2}}\left(-B^{\prime}(1)K^{3}\left(\frac{1}{\rho}\right)+(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle+2\partial^{2}_{\rho}f(\rho)\int_{1-a}^{1}B^{\prime\prime}(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{\rho^{\prime}}{\rho}\frac{d\rho^{\prime}}{\rho}-\partial^{2}_{\rho}\left(B(1)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1}{\rho}\right)}{\rho}-B(1-a)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1-a}{\rho}\right)}{\rho}\right)$
$\displaystyle\partial_{\rho}^{2}f(\rho)\int_{1-a}^{1}B^{\prime\prime}(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{\rho^{\prime}}{\rho}\frac{d\rho^{\prime}}{\rho}-f_{\rho}(\rho)P.V.\int_{1-a}^{1}\partial_{\rho^{\prime}}\left(\rho^{\prime
2}K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\right)\frac{d\rho^{\prime}}{\rho}.$

###### Proof.

First we notice that, after a change of variables,

$\displaystyle\tilde{T}^{3*}B(\rho)=f_{\rho}(\rho)\int_{\frac{{1-a}}{\rho}}^{\frac{1}{\rho}}B(s\rho)K^{3}\left(s\right)\frac{ds}{s}.$

Then, taking a derivative yields,

$\displaystyle\partial_{\rho}\tilde{T}^{3*}B(\rho)=\partial^{2}_{\rho}f(\rho)\int_{\frac{{1-a}}{\rho}}^{\frac{1}{\rho}}B(\rho
s)K^{3}(s)\frac{ds}{s}+\frac{1}{\rho}f_{\rho}(\rho)\left(B(1)K^{3}\left(\frac{1}{\rho}\right)-B(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle+f_{\rho}(\rho)\int_{\frac{{1-a}}{\rho}}^{\frac{1}{\rho}}K^{3}(s)B^{\prime}(\rho
s)ds$ (4.21)

and changing variable we have that

$\displaystyle\partial_{\rho}\tilde{T}^{3*}B(\rho)=\partial^{2}_{\rho}f(\rho)\int_{1-a}^{1}B(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho^{\prime}}+\frac{1}{\rho}f_{\rho}(\rho)\left(B(1)K^{3}\left(\frac{1}{\rho}\right)-B(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle-
f_{\rho}(\rho)\int_{1-a}^{1}K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)B^{\prime}(\rho^{\prime})\frac{d\rho^{\prime}}{\rho}.$

And integration by parts in the last integral yields

$\displaystyle\partial_{\rho}\tilde{T}^{3*}B(\rho)=\partial^{2}_{\rho}f(\rho)\int_{1-a}^{1}B(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho^{\prime}}+f_{\rho}(\rho)V.P.\int_{1-a}^{1}\partial_{\rho^{\prime}}K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)B(\rho^{\prime})\frac{d\rho^{\prime}}{\rho}$

Taking a derivative on the equation (1) we obtain

$\displaystyle\partial^{2}_{\rho}\tilde{T}^{3*}B(\rho)=\partial^{3}_{\rho}f(\rho)\int_{\frac{{1-a}}{\rho}}^{\frac{1}{\rho}}B(\rho
s)K^{3}(s)\frac{ds}{s}+\partial^{2}_{\rho}f(\rho)\frac{1}{\rho}\left(-B(1)K^{3}\left(\frac{1}{\rho}\right)+B(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle+\partial^{2}_{\rho}f(\rho)\int_{\frac{{1-a}}{\rho}}^{\frac{1}{\rho}}B^{\prime}(\rho
s)K^{3}(s)ds-\partial_{\rho}\left(B(1)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1}{\rho}\right)}{\rho}-B(1-a)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1-a}{\rho}\right)}{\rho}\right)$
$\displaystyle+\partial^{2}_{\rho}f(\rho)\int_{\frac{1-a}{\rho}}^{\frac{1}{\rho}}B^{\prime}(\rho
s)K^{3}(s)ds+\frac{f_{\rho}(\rho)}{\rho^{2}}\left(B^{\prime}(1)K^{3}\left(\frac{1}{\rho}\right)-B^{\prime}(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle+f_{\rho}(\rho)\int_{\frac{{1-a}}{\rho}}^{\frac{1}{\rho}}K^{3}(s)B^{\prime\prime}(\rho
s)sds$ (4.22)

A change of variable and integration by parts then yield

$\displaystyle\partial^{2}_{\rho}\tilde{T}^{3*}B(\rho)=\partial^{2}_{\rho}f(\rho)\int_{1-a}^{1}B(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho^{\prime}}$
$\displaystyle-\frac{\partial^{2}_{\rho}f(\rho)}{\rho}P.V.\int_{1-a}^{1}B(\rho^{\prime})\partial_{\rho^{\prime}}K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)d\rho^{\prime}-\partial_{\rho}\left(B(1)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1}{\rho}\right)}{\rho}-B(1-a)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1-a}{\rho}\right)}{\rho}\right)$
$\displaystyle+\frac{\partial^{2}_{\rho}f(\rho)}{\rho}\int_{1-a}^{1}B^{\prime}(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)+\frac{f_{\rho}(\rho)}{\rho}P.V.\int_{1-a}^{1}B^{\prime}(\rho^{\prime})\partial_{\rho^{\prime}}\left(\rho^{\prime}K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\right)d\rho^{\prime}.$

Finally we take a derivative of the equation 1. We have that

$\displaystyle\partial^{3}_{\rho}\tilde{T}^{3*}=\partial^{4}_{\rho}f(\rho)\int_{\frac{1-a}{\rho}}^{\frac{1}{\rho}}B(\rho
s)K^{3}(s)\frac{ds}{s}+\frac{\partial^{3}_{\rho}f(\rho)}{\rho}\left(-B(1)K^{3}\left(\frac{1}{\rho}\right)+B(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle+\partial^{3}_{\rho}f(\rho)\int_{\frac{1-a}{\rho}}^{\frac{1}{\rho}}B^{\prime}(\rho
s)K^{3}(s)ds$
$\displaystyle+\partial_{\rho}\left(\frac{\partial^{2}_{\rho}f(\rho)}{\rho}\left(-B(1)K^{3}\left(\frac{1}{\rho}\right)+B(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)\right)$
$\displaystyle+2\partial^{3}_{\rho}f(\rho)\int_{\frac{1-a}{\rho}}^{\frac{1}{\rho}}B^{\prime}(\rho
s)K^{3}(s)ds+2\frac{\partial^{2}_{\rho}f(\rho)}{\rho^{2}}\left(-B^{\prime}(1)K^{3}\left(\frac{1}{\rho}\right)+(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle+2\partial^{2}_{\rho}f(\rho)\int_{\frac{1-a}{\rho}}^{\frac{1}{\rho}}B^{\prime\prime}(\rho
s)K^{3}(s)sds-\partial^{2}_{\rho}\left(B(1)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1}{\rho}\right)}{\rho}-B(1-a)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1-a}{\rho}\right)}{\rho}\right)$
$\displaystyle+\partial^{2}_{\rho}f(\rho)\int_{\frac{1-a}{\rho}}^{\frac{1}{\rho}}K^{3}(s)B^{\prime\prime}(\rho
s)sds-
f_{\rho}(\rho)\left(-\frac{1}{\rho^{3}}B^{\prime\prime}(1)K^{3}\left(\frac{1}{\rho}\right)+\frac{(1-a)^{2}}{\rho^{3}}B^{\prime\prime}(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle
f_{\rho}(\rho)\int_{\frac{1-a}{\rho}}^{\frac{1}{\rho}}K^{3}(s)B^{\prime\prime\prime}(\rho
s)s^{2}ds.$

And a change of variable and an integration by parts yield

$\displaystyle\partial^{3}_{\rho}\tilde{T}^{3*}(s)=\partial^{4}_{\rho}f(\rho)\int_{1-a}^{1}B(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho^{\prime}}+\frac{\partial^{3}_{\rho}f(\rho)}{\rho}\left(-B(1)K^{3}\left(\frac{1}{\rho}\right)+B(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle+\partial^{3}_{\rho}f(\rho)\int_{1-a}^{1}B^{\prime}(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho}$
$\displaystyle+\partial_{\rho}\left(\frac{\partial^{2}_{\rho}f(\rho)}{\rho}\left(-B(1)K^{3}\left(\frac{1}{\rho}\right)+B(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)\right)$
$\displaystyle+2\partial^{3}_{\rho}f(\rho)\int_{1-a}^{1}B^{\prime\prime}(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{d\rho^{\prime}}{\rho}++2\frac{\partial^{2}_{\rho}f(\rho)}{\rho^{2}}\left(-B^{\prime}(1)K^{3}\left(\frac{1}{\rho}\right)+(1-a)K^{3}\left(\frac{1-a}{\rho}\right)\right)$
$\displaystyle+2\partial^{2}_{\rho}f(\rho)\int_{1-a}^{1}B^{\prime\prime}(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{\rho^{\prime}}{\rho}\frac{d\rho^{\prime}}{\rho}-\partial^{2}_{\rho}\left(B(1)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1}{\rho}\right)}{\rho}-B(1-a)\frac{f_{\rho}(\rho)K^{3}\left(\frac{1-a}{\rho}\right)}{\rho}\right)$
$\displaystyle\partial_{\rho}^{2}f(\rho)\int_{1-a}^{1}B^{\prime\prime}(\rho^{\prime})K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\frac{\rho^{\prime}}{\rho}\frac{d\rho^{\prime}}{\rho}-f_{\rho}(\rho)P.V.\int_{1-a}^{1}\partial_{\rho^{\prime}}\left(\rho^{\prime
2}K^{3}\left(\frac{\rho^{\prime}}{\rho}\right)\right)\frac{d\rho^{\prime}}{\rho}.$

∎

###### Lemma 4.18

The operator $\tilde{T}^{3*}$ is bounded from $H^{k}$ to $H^{k+1}$, for
$k=0,1,2$, with norm depending only on $||f||_{C^{4}}$ and $a$.

###### Proof.

In order to prove this lemma we prove the estimates for smooth functions. This
can be done by using Lemma 4.17. Indeed, in order to control the evaluations
of $B$ and its derivatives we use the standard Sobolev embedding $H^{k}\subset
C^{l}$, for $k>l+\frac{1}{2}$ in one dimension ($H^{3}\subset C^{2}$). In
addition we use Lemmas 4.12, 4.13, together with the $L^{2}$-boundedness of
the Hilbert transform to control the terms involving the derivative of the
kernel $K^{3}$. Finally we proceed by a density argument. ∎

Finally, we finish this section by studying the regularity of
$\tilde{I}(\rho)$.

###### Lemma 4.19

Let $f\in C^{4}(\mathbb{R}^{+})$. Then, $\tilde{I}(\rho)\in C^{3}(1-a,1)$.

###### Proof.

The proof follows easily by the fact that $f\in C^{4}$ and Lemma 4.12. ∎

  2. 2.

Existence of a solution for the equation (4.14). We will regard our operator
$\Theta^{3}$ as a perturbation of its symmetric part $\Theta_{S}^{3}$ (which
we can do by taking $a$ and $\beta$ small enough). Then we estimate the first
and second eigenvalues (and the first eigenfunction) and see that there is a
gap. If the antisymmetric perturbation is small enough, then there is still a
gap and we are good. The strategy is to use a computer-assisted proof for the
estimation of the eigenvalue-eigenvector pair and the norms of the different
operators that appear.

Let $(B_{sj},\lambda^{*})$ be the approximate eigenvector-eigenvalue pair for
the symmetric operator $\Theta^{3}_{S}$, satisfying

$\displaystyle\Theta_{S}^{3}B_{sj}=\lambda^{*}B_{sj}+e,$ (4.23)

where $e$ is small (see the next lemma for an explicit bound on $e$).

###### Lemma 4.20

Let
$B_{sj}=\frac{1}{\sqrt{a-a\beta}}{1}_{[1-a+\frac{a\beta}{2},1-\frac{a\beta}{2}]}$
and $\lambda^{*}=0.3482$. Then:

$\displaystyle\|e\|_{L^{2}}<0.0905$ $\displaystyle|\langle
e,B_{sj}\rangle|<0.0101$

###### Proof.

The proof is computer-assisted and the code can be found in the supplementary
material. We refer to the appendix for details about the implementation. ∎

The symbol $B_{sj}^{\perp}$ will denote the orthogonal space to $B_{sj}$ in
$L^{2}((1-a,1))$, i.e.

$B_{sj}^{\perp}\equiv\\{v\in L^{2}((1-a,1)),\,\>\,\langle
B_{sj},v\rangle_{L^{2}((1-a,1))}=0\\}.$

Let us find a solution $(B^{3},\lambda)$ such that

$\displaystyle\Theta^{3}B^{3}=\lambda B^{3},\quad B=B_{sj}+v,\quad v\in
B_{sj}^{\perp}.$ (4.24)

We expect $v$ to be small if $(B_{sj},\lambda^{*})$ is an accurate enough
approximation to the true eigenpair. Plugging this ansatz for $B^{3}$ into the
previous equation and using the equation (4.23), we obtain

$\displaystyle\Theta^{3}v$ $\displaystyle=\Theta^{3}B^{3}-\Theta^{3}B_{sj}$
$\displaystyle=\lambda B_{sj}+\lambda
v-\lambda^{*}B_{sj}-e-\Theta^{3}_{A}B_{sj}$
$\displaystyle=(\lambda-\lambda^{*})B_{sj}+\lambda v-e-\Theta^{3}_{A}B_{sj},$

where $\Theta^{3}_{A}=\Theta^{3}-\Theta^{3}_{S}$ is the antisymmetric part of
$\Theta^{3}$. From now on, the pairing $\langle\cdot,\cdot\rangle$ is assumed
to be taken in $L^{2}((1-a,1))$. Let us consider the functional equation

$\displaystyle\langle\Theta^{3}v-\lambda v,u\rangle=-\langle
e+\Theta^{3}_{A}B_{sj},u\rangle\quad\forall u\in B_{sj}^{\perp},$

and let $c^{*}$ be defined by

$\displaystyle c^{*}=\inf_{v\in
B_{sj}^{\perp}}\frac{\langle\Theta^{3}v,v\rangle}{\|v\|_{L^{2}}^{2}}.$

A rigorous bound for $c^{*}$ will be given in the next lemma:

###### Lemma 4.21

Let
$B_{sj}=\frac{1}{\sqrt{a-a\beta}}{1}_{[1-a+\frac{a\beta}{2},1-\frac{a\beta}{2}]}$.
Then:

$\displaystyle c^{*}\geq 0.8526$

###### Proof.

The proof is computer-assisted and the code can be found in the supplementary
material. We refer to the appendix for details about the implementation. ∎

Then, if $\lambda<c^{*}$ the operator $\Theta^{3}-\lambda$ is coercive in
$B_{sj}^{\perp}$. By Lax-Milgram, for every $\lambda<c^{*}$ there exists a
function $v^{\lambda}\in B_{sj}^{\perp}$ such that

$\displaystyle\langle\Theta^{3}v^{\lambda}-\lambda
v^{\lambda},u\rangle=-\langle e+\Theta^{3}_{A}B_{sj},u\rangle\quad\forall u\in
B_{sj}^{\perp}.$ (4.25)

Thus, there exists a function $d(\lambda)$ such that

$\displaystyle\Theta^{3}v^{\lambda}$
$\displaystyle=(\lambda-\lambda^{*})B_{sj}+\lambda
v^{\lambda}-e-\Theta^{3}_{A}B_{sj}+d(\lambda)B_{sj}.$ (4.26)

By computing the scalar product of (4.26) with $v^{\lambda}$ we obtain

$\displaystyle(c^{*}-\lambda)\|v^{\lambda}\|_{L^{2}}^{2}\leq\langle(\Theta^{3}-\lambda)v^{\lambda},v^{\lambda}\rangle=-\langle\Theta^{3}_{A}B_{sj},v^{\lambda}\rangle-\langle
e,v^{\lambda}\rangle\leq|\langle\Theta^{3}_{A}B_{sj}+e,v^{\lambda}\rangle|\leq\|\Theta^{3}_{A}B_{sj}+e\|_{L^{2}}\|v^{\lambda}\|_{L^{2}}.$

This inequality implies

$\displaystyle\|v^{\lambda}\|_{L^{2}}\leq\frac{\|\Theta^{3}_{A}B_{sj}+e\|_{L^{2}}}{c^{*}-\lambda}.$
(4.27)

In addition, using that

$\displaystyle\langle\Theta^{3}v^{\lambda},B_{sj}\rangle$
$\displaystyle=\langle\Theta^{3}_{S}v^{\lambda},B_{sj}\rangle+\langle\Theta^{3}_{A}v^{\lambda},B_{sj}\rangle=\langle
v^{\lambda},\Theta^{3}_{S}B_{sj}\rangle-\langle
v^{\lambda},\Theta^{3}_{A}B_{sj}\rangle=\langle v^{\lambda},e\rangle-\langle
v^{\lambda},\Theta^{3}_{A}B_{sj}\rangle,$

taking the scalar product with $B_{sj}$ of the equation (4.26) yields

$\displaystyle d(\lambda)\|B_{sj}\|_{L^{2}}^{2}$
$\displaystyle=\langle\Theta^{3}v^{\lambda},B_{sj}\rangle+(\lambda^{*}-\lambda)\|B_{sj}\|_{L^{2}}^{2}+\langle
e,B_{sj}\rangle$ $\displaystyle=\langle v^{\lambda},e\rangle-\langle
v^{\lambda},\Theta^{3}_{A}B_{sj}\rangle+(\lambda^{*}-\lambda)\|B_{sj}\|_{L^{2}}^{2}+\langle
e,B_{sj}\rangle.$

An important fact, that will be used, is the continuity of this function
$d(\lambda)$:

###### Lemma 4.22

The function $d(\lambda)$ defined in (4.26) is continuous.

###### Proof.

Let $\lambda,\gamma<c^{*}$. By definition of $d(\lambda)$ we have that

$\displaystyle|d(\lambda)-d(\gamma)|\leq
C\left(||B_{sj}||_{L^{2}},||e||_{L^{2}},||\Theta^{3}||_{L^{2}\to
L^{2}}\right)||v^{\lambda}-v^{\gamma}||_{L^{2}}+||B_{sj}||_{L^{2}}^{2}|\lambda-\gamma|.$

For every $u\in B^{\perp}_{sj}$ we know from (4.25) that

$\displaystyle\langle\Theta^{3}v^{\lambda}-\lambda v^{\lambda},u\rangle=$
$\displaystyle\langle f,u\rangle$
$\displaystyle\langle\Theta^{3}v^{\gamma}-\gamma v^{\gamma},u\rangle=$
$\displaystyle\langle f,u\rangle,$

where $f\in L^{2}$. And then

$\displaystyle\langle\Theta^{3}(v^{\lambda}-v^{\gamma})-\lambda(v^{\lambda}-v^{\gamma}),u\rangle=(\gamma-\lambda)\langle
v^{\gamma},u\rangle.$

Taking $u=v^{\lambda}-v^{\gamma}\in B^{\perp}_{sj}$ yields

$\displaystyle(c^{*}-\lambda)||v^{\lambda}-v^{\gamma}||_{L^{2}}^{2}\leq|\gamma-\lambda|\,||v^{\gamma}||_{L^{2}}\,||v^{\lambda}-v^{\gamma}||_{L^{2}}.$

Thus

$\displaystyle||v^{\lambda}-v^{\gamma}||_{L^{2}}\leq\frac{C}{c^{*}-\lambda}|\gamma-\lambda|\,||v^{\gamma}||_{L^{2}}.$

We achieve the conclusion of the lemma from the inequality (4.27). ∎

Our purpose is now to show that there exists a value of
$\lambda=\lambda^{**}<c^{*}$ such that $d(\lambda^{**})=0$. We have the
following upper bound for $d(\lambda)$:

$\displaystyle d(\lambda)$
$\displaystyle\geq-\frac{\|v^{\lambda}\|_{L^{2}}\|e-\Theta^{3}_{A}B_{sj}\|_{L^{2}}}{\|B_{sj}\|_{L^{2}}^{2}}+(\lambda^{*}-\lambda)-\frac{|\langle
e,B_{sj}\rangle|}{\|B_{sj}\|_{L^{2}}^{2}}$ (4.28)
$\displaystyle\geq-\frac{\|\Theta^{3}_{A}B_{sj}+e\|_{L^{2}}}{c^{*}-\lambda}\frac{\|e-\Theta^{3}_{A}B_{sj}\|_{L^{2}}}{\|B_{sj}\|_{L^{2}}^{2}}+(\lambda^{*}-\lambda)-\frac{|\langle
e,B_{sj}\rangle|}{\|B_{sj}\|_{L^{2}}^{2}}.$

We observe that $\lim_{\lambda\to-\infty}d(\lambda)=+\infty$. Hence, there
exists a $\lambda$ sufficiently small for which $d(\lambda)>0$. Similarly:

$\displaystyle d(\lambda)$
$\displaystyle\leq\frac{\|\Theta^{3}_{A}B_{sj}+e\|_{L^{2}}}{c^{*}-\lambda}\frac{\|e-\Theta^{3}_{A}B_{sj}\|_{L^{2}}}{\|B_{sj}\|_{L^{2}}^{2}}+(\lambda^{*}-\lambda)+\frac{|\langle
e,B_{sj}\rangle|}{\|B_{sj}\|_{L^{2}}^{2}}$ (4.29)

The objective is to use the previous inequality to show that there exist a
value of $\lambda$ smaller than $c^{*}$, such that $d(\lambda)<0$. We will use
the following lemma.

###### Lemma 4.23

Let $h(x)=\mathcal{A}-x+\frac{\mathcal{B}}{\mathcal{C}-x}$, where
$\mathcal{A},\mathcal{B},\mathcal{C}>0$. If

$\displaystyle\mathcal{C}>\mathcal{A}+2\sqrt{\mathcal{B}},$

then there exists an $x<\mathcal{C}$ such that $h(x)<0$.

###### Proof.

It is easy to notice that $h(x)$ is concave up since $h^{\prime\prime}(x)>0$
for $x<\mathcal{C}$. Thus, it is enough to check that the minimum of $h$ is
negative. We calculate the point $x_{m}$ where the minimum is attained.

$\displaystyle
h^{\prime}(x_{m})=0\Leftrightarrow(\mathcal{C}-x_{m})^{2}=\mathcal{B}\Leftrightarrow
x_{m}=\mathcal{C}-\sqrt{\mathcal{B}}.$

Evaluating at $x=x_{m}$:

$\displaystyle h(x_{m})=\mathcal{A}-\mathcal{C}+2\sqrt{\mathcal{B}}<0,$

by hypothesis. ∎

###### Lemma 4.24

Let
$B_{sj}=\frac{1}{\sqrt{a-a\beta}}{1}_{[1-a+\frac{a\beta}{2},1-\frac{a\beta}{2}]}$,
$\lambda^{*}=0.3482$ and let us define:

$\displaystyle\mathcal{A}=\lambda^{*}+\frac{|\langle
e,B_{sj}\rangle|}{\|B_{sj}\|_{L^{2}}^{2}},\quad\mathcal{B}=\frac{\|\Theta^{3}_{A}B_{sj}+e\|_{L^{2}}\|e-\Theta^{3}_{A}B_{sj}\|_{L^{2}}}{\|B_{sj}\|_{L^{2}}^{2}}$

Then, we have the following bounds:

$\displaystyle\mathcal{A}<0.3583$ $\displaystyle\sqrt{\mathcal{B}}<0.1534$

###### Proof.

The proof is computer-assisted and the code can be found in the supplementary
material. We refer to the appendix for details about the implementation. ∎

Finally, by applying Lemma 4.23 to the right hand side of (4.29) with

$\displaystyle\mathcal{A}=\lambda^{*}+\frac{|\langle
e,B_{sj}\rangle|}{\|B_{sj}\|_{L^{2}}^{2}},\quad\mathcal{B}=\frac{\|\Theta^{3}_{A}B_{sj}+e\|_{L^{2}}\|e-\Theta^{3}_{A}B_{sj}\|_{L^{2}}}{\|B_{sj}\|_{L^{2}}^{2}},\quad\mathcal{C}=c^{*}$

and by the bounds given by Lemma 4.24, we obtain that there exists a
$\lambda_{0}$ for which $d(\lambda_{0})<0$. More precisely, we have that

$\displaystyle\lambda_{0}$
$\displaystyle\leq\frac{\mathcal{C}+\mathcal{A}}{2}-\sqrt{\left(\frac{\mathcal{C}-\mathcal{A}}{2}\right)^{2}-\mathcal{B}}=\frac{\mathcal{A}\mathcal{C}+\mathcal{B}}{\frac{\mathcal{A}+\mathcal{C}}{2}+\sqrt{\left(\frac{\mathcal{A}-\mathcal{C}}{2}\right)^{2}-\mathcal{B}}}$
$\displaystyle=\frac{c^{*}+\lambda^{*}+\frac{|\langle
e,B_{sj}\rangle|}{\|B_{sj}\|_{L^{2}}^{2}}}{2}-\sqrt{\left(\frac{c^{*}-\lambda^{*}-\frac{|\langle
e,B_{sj}\rangle|}{\|B_{sj}\|_{L^{2}}^{2}}}{2}\right)^{2}-\frac{\|\Theta^{3}_{A}B_{sj}+e\|_{L^{2}}\|e-\Theta^{3}_{A}B_{sj}\|_{L^{2}}}{\|B_{sj}\|_{L^{2}}^{2}}}$

By the continuity of $d(\lambda)$ proved in Lemma 4.22, there has to be a
$\lambda^{**}\leq\lambda_{0}<c^{*}$ for which $d(\lambda^{**})=0$ and
therefore

$\displaystyle\Theta v^{\lambda^{**}}$
$\displaystyle=(\lambda^{**}-\lambda^{*})B_{sj}+\lambda^{**}v^{\lambda^{**}}-e-\Theta_{A}B_{sj}.$

which means that $(B_{sj}+v^{\lambda^{**}},\lambda^{**})$ is the eigenvalue-
eigenvector pair, $(B^{3},\lambda_{3})\in L^{2}((1-a,1))\times\mathbb{R}$, we
were looking for.

###### Corollary 4.25

We have just shown that $\lambda_{3}\leq\lambda_{0}<0.4117$.

###### Proof.

Follows from the numerical bounds obtained in Lemma 4.24 and Lemma 4.23. ∎

Next we show that $B^{3}$ is unique modulo multiplication by constants. In
order to prove it, let us assume that $w\in L^{2}((1-a,1))$ satisfies

$\displaystyle\Theta^{3}w=\lambda_{3}w.$

Now we write $w=B_{sj}+(w-B_{sj})$ and we notice we can decompose
$w-B_{sj}=\alpha B_{sj}+v$ where $\alpha\in\mathbb{R}$ and $v\in
B_{sj}^{\perp}$. Then $w=(1+\alpha)B_{sj}+v$ and by linearity we have that
$\frac{w}{1+\alpha}=B_{sj}+u$, with $u=\frac{v}{1+\alpha}\in B_{sj}^{\perp}$,
is also a solution. Then the uniqueness of the solutions in the Lax-Milgram
theorem implies $\frac{w}{1+\alpha}=B^{3}$. If $\alpha=-1$ the previous
argument fails. But, in this case, $w\in B_{sj}^{\perp}$ and then
$c^{*}\leq\lambda_{3}$ which has already been proven to be false.

In conclusion, we have shown that
dim($\mathcal{N}(\Theta^{3}-\lambda^{3}))=1$.

###### Lemma 4.26

We have that $\tilde{I}(\rho)-\lambda_{3}>0$. In particular, we have the
following bounds:

$\displaystyle\tilde{I}(\rho)-\lambda_{3}>0.8526$

###### Proof.

The proof is computer-assisted and the code can be found in the supplementary
material. We refer to the appendix for details about the implementation. ∎

Then it remains to prove the regularity of $B^{3}$, the solution of equation
(4.24). To do this we will bootstrap using Lemma 4.15. Since

$\displaystyle\tilde{I}(\rho)B^{3}(\rho)-\lambda_{3}B^{3}(\rho)=-\tilde{T}^{3}B^{3}(\rho)$
(4.30)

by Lemma 4.15 the function $(\tilde{I}(\rho)-\lambda_{3})B^{3}(\rho)\in
H^{1}((1-a,1))$. Since $\tilde{I}\in C^{3}$ and Lemma 4.26 we have that
$B^{3}(\rho)$ is in $H^{1}$. Let’s take two derivatives on equation (4.30), by
Lemma 4.10 we have that

$\displaystyle\tilde{I}(\rho)\partial^{2}_{\rho}B^{3}(\rho)-\lambda_{3}\partial^{2}_{\rho}B^{3}(\rho)=$
$\displaystyle-\sum_{j=1}^{2}\partial_{\rho}^{j)}\tilde{I}(\rho)\partial^{2-j)}_{\rho}B^{3}(\rho)-\partial_{\rho}T^{3}_{1}\partial_{\rho}B^{3}(\rho)$
$\displaystyle-\partial_{\rho}\left(\frac{1}{\rho}\int_{-\infty}^{\infty}\partial^{2}_{\rho}f(\rho^{\prime})B(\rho^{\prime})\frac{\rho^{\prime}}{\rho}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}\right)$

Then using that $\tilde{I}(\rho)\in C^{3}$, $f\in C^{4}$,
$\tilde{I}-\lambda_{3}>0$, Lemma 4.15 and $B^{3}\in H^{1}$ we have that
$B^{3}\in H^{2}$. Finally, taking three derivatives yields

$\displaystyle\tilde{I}(\rho)\partial^{3}_{\rho}B^{3}(\rho)-\lambda_{3}\partial^{3}_{\rho}B^{3}(\rho)=$
$\displaystyle-\sum_{j=1}^{3}\partial_{\rho}^{j)}\tilde{I}(\rho)\partial^{3-j)}_{\rho}B^{3}(\rho)-\partial_{\rho}T^{3}_{2}\partial^{2}_{\rho}B^{3}(\rho)$
$\displaystyle-\partial_{\rho}\left(\sum_{j=1}^{2}\frac{1}{\rho}\int_{-\infty}^{\infty}\partial^{j+1)}_{\rho}f(\rho^{\prime})\partial^{2-j)}_{\rho}B(\rho^{\prime})\left(\frac{\rho^{\prime}}{\rho}\right)^{2}K^{m}\left(\frac{\rho}{\rho^{\prime}}\right)d\rho^{\prime}\right)$

Then using, again, that $\tilde{I}(\rho)\in C^{3}$, $f\in C^{4}$,
$\tilde{I}-\lambda_{3}>0$, Lemma 4.15 and $B^{3}\in H^{2}$ we have that
$B^{3}\in H^{3}.$

This concludes the proof of Proposition 4.8.

∎

A similar proof as in Proposition 4.8 also works to show:

###### Proposition 4.27

There exists a solution $(B^{3}_{*},\lambda_{3})\in
H^{3}((1-a),1)\times\mathbb{R}$ to the equation

$\displaystyle\Theta^{3*}B^{3}_{*}=\lambda_{3}B^{3}_{*},$

where $\lambda_{3}$ is the same eigenvalue as in Proposition 4.8 and
$\Theta^{3*}$ is the adjoint operator of $\Theta^{3}$. In addition,
$\lambda_{3}$ is simple and we can decompose

$B_{*}^{3}=B_{sj}+v_{*}^{\lambda_{3}},$

where $v_{*}^{\lambda_{3}}\in\left(B_{sj}\right)^{\perp}$

and

$\displaystyle||v_{*}^{\lambda_{3}}||_{L^{2}}\leq\frac{||-\Theta^{3}_{A}B_{sj}+e||_{L^{2}}}{c^{*}-\lambda_{3}}$

###### Proof.

The proof that there exists a pair $(B^{3}_{*},\lambda_{3}^{*})\in
L^{2}\times\mathbb{R}$ satisfying

$\displaystyle(\Theta^{3*}-\lambda_{3}^{*})B^{3}_{*}=\lambda_{3}^{*}B^{3}_{*}$

runs the same steps than the proof of Proposition 4.8. The only modification
is the change $\Theta^{3}_{A}\to-\Theta^{3}_{A}$. In order to check that, in
fact, $\lambda_{3}^{*}=\lambda_{3}$ we notice that

$\displaystyle\lambda_{3}\langle
B^{3},B^{3}_{*}\rangle=\langle\Theta^{3}B^{3},B^{3}_{*}\rangle=\langle
B^{3},\Theta^{3*}B^{*}_{3}\rangle=\lambda_{3}^{*}\langle
B^{3},B^{3}_{*}\rangle.$

Therefore it is enough to show that $\langle B^{3},B^{3}_{*}\rangle\neq 0$. We
prove this result in the following lemma:

###### Lemma 4.28

The following inequality holds:

$\displaystyle\langle B^{3},B^{3}_{*}\rangle>0.$

###### Proof.

We can decompose $B^{3}$ and $B^{3}_{*}$ in the following way

$\displaystyle B^{3}=$ $\displaystyle B_{sj}+v^{\lambda_{3}}$ $\displaystyle
B^{3}_{*}=$ $\displaystyle B_{sj}+v^{\lambda_{3}}_{*}$

where $v^{\lambda_{3}},\,v^{\lambda_{3}}_{*}\in B_{sj}^{\perp}$. Thus

$\displaystyle\langle B^{3},B^{3}_{*}\rangle=||B_{sj}||_{L^{2}}^{2}+\langle
v^{\lambda_{3}},v^{\lambda_{3}}_{*}\rangle\geq||B_{sj}||_{L^{2}}^{2}-\frac{||-\Theta_{A}^{3}B_{sj}+e||_{L^{2}}||\Theta_{A}^{3}B_{sj}+e||_{L^{2}}}{(c^{*}-\lambda_{3})^{2}}.$
(4.31)

Using the notation from Lemma 4.23, we can bound

$\displaystyle c^{*}-\lambda_{3}\geq
c^{*}-\lambda_{0}=\frac{C-A}{2}+\sqrt{\left(\frac{C-A}{2}\right)^{2}-B}.$

This implies that the RHS of (4.31) is bounded below by

$\displaystyle||B_{sj}||_{L^{2}}^{2}\left(1-\frac{B}{\left(\frac{C-A}{2}+\sqrt{\left(\frac{C-A}{2}\right)^{2}-B}\right)^{2}}\right)>0,$

where in the last inequality we have used the already checked condition that

$\displaystyle\frac{C-A}{2}>\sqrt{B}.$

∎

Finally we prove that $B^{3}_{*}$ is in $H^{3}$ in the same way we did for
$B^{3}$ by using Lemma 4.18. ∎

####  4.3.3 One-dimensionality of the kernel of
$\partial_{r}F[\rho,\lambda_{3}]$

Until now, it has already been proven that there exists an element in the
kernel of the operator $\partial_{r}F[\rho,\lambda_{3}]$. In this section we
will prove that this kernel is the span of this element. As commented in point
3 at the beginning of the proof of Proposition 4.7 it is enough to prove that
the equation

$\Theta^{3n}u=\lambda_{3}u$

implies that $u=0$ for $n>1$.

###### Lemma 4.29

Let $m>1$ and

$\displaystyle\lambda_{3m}^{s}=\inf_{\begin{subarray}{c}u\in L^{2}\\\
||u||_{L^{2}}=1\end{subarray}}\langle\Theta^{3m}u,u\rangle$

Then, if the pair $(u_{3m},\lambda_{3m})\in L^{2}\times\mathbb{R}$ (with
$u_{3m}$ not identically zero) satisfies

$\displaystyle\Theta^{3m}u_{3m}=\lambda_{3m},$

we have that

$\displaystyle\lambda_{3m}^{s}\leq\lambda_{3m}$

###### Proof.

Since we can take $u_{3m}$ with norm 1 we have that

$\displaystyle\lambda_{3m}=\langle\Theta^{3m}u_{3m},u_{3m}\rangle\geq\lambda_{3m}^{s},$

by definition of $\lambda_{3m}^{s}$. ∎

###### Lemma 4.30

Let $m>n$ and let $\lambda_{3m}^{s},\lambda_{3n}^{s}$ be given by

$\displaystyle\lambda_{3m}^{s}=\inf_{\begin{subarray}{c}u\in L^{2}\\\
||u||_{L^{2}}=1\end{subarray}}\langle\Theta^{3m}u,u\rangle\quad\text{and}\quad\lambda_{3n}^{s}=\inf_{\begin{subarray}{c}u\in
L^{2}\\\ ||u||_{L^{2}}=1\end{subarray}}\langle\Theta^{3n}u,u\rangle.$

Then

$\displaystyle\lambda_{3m}^{s}\geq\lambda_{3n}^{s}$

###### Proof.

First we will show that for every $m\geq 1$ and $u\in L^{2}$ the following
inequality holds

$\langle\Theta^{3m}u,u\rangle\geq\langle\Theta^{3m}|u|,|u|\rangle.$

By using the positivity of $T^{3m}(\rho,\rho^{\prime})$ proved in Lemma 4.11
we have that

$\displaystyle\langle\Theta^{3m}|u|,|u|\rangle$
$\displaystyle=\int\tilde{I}(\rho)|u(\rho)|^{2}d\rho+\int\int\frac{1}{\rho}\left(\frac{\rho^{\prime}}{\rho}\right)T^{3m}(\rho,\rho^{\prime})f_{\rho}(\rho^{\prime})|u(\rho)||u(\rho^{\prime})|d\rho
d\rho^{\prime}$ $\displaystyle=\int
I(\rho)|u(\rho)|^{2}d\rho+\int\int\frac{1}{\rho}\left(\frac{\rho^{\prime}}{\rho}\right)T^{3m}(\rho,\rho^{\prime})f_{\rho}(\rho^{\prime})u^{+}(\rho)u^{+}(\rho^{\prime})d\rho
d\rho^{\prime}$
$\displaystyle+\int\int\frac{1}{\rho}\left(\frac{\rho^{\prime}}{\rho}\right)T^{3m}(\rho,\rho^{\prime})f_{\rho}(\rho^{\prime})u^{-}(\rho)u^{-}(\rho^{\prime})d\rho
d\rho^{\prime}$
$\displaystyle+\int\int\frac{1}{\rho}\left(\frac{\rho^{\prime}}{\rho}\right)T^{3m}(\rho,\rho^{\prime})f_{\rho}(\rho^{\prime})u^{+}(\rho)u^{-}(\rho^{\prime})d\rho
d\rho^{\prime}$
$\displaystyle+\int\int\frac{1}{\rho}\left(\frac{\rho^{\prime}}{\rho}\right)T^{3m}(\rho,\rho^{\prime})f_{\rho}(\rho^{\prime})u^{-}(\rho)u^{+}(\rho^{\prime})d\rho
d\rho^{\prime}$ $\displaystyle\leq\int
I(\rho)|u(\rho)|^{2}d\rho+\int\int\frac{1}{\rho}\left(\frac{\rho^{\prime}}{\rho}\right)T^{3m}(\rho,\rho^{\prime})f_{\rho}(\rho^{\prime})u^{+}(\rho)u^{+}(\rho^{\prime})d\rho
d\rho^{\prime}$
$\displaystyle+\int\int\frac{1}{\rho}\left(\frac{\rho^{\prime}}{\rho}\right)T^{3m}(\rho,\rho^{\prime})f_{\rho}(\rho^{\prime})u^{-}(\rho)u^{-}(\rho^{\prime})d\rho
d\rho^{\prime}$
$\displaystyle-\int\int\frac{1}{\rho}\left(\frac{\rho^{\prime}}{\rho}\right)T^{3m}(\rho,\rho^{\prime})f_{\rho}(\rho^{\prime})u^{+}(\rho)u^{-}(\rho^{\prime})d\rho
d\rho^{\prime}$
$\displaystyle-\int\int\frac{1}{\rho}\left(\frac{\rho^{\prime}}{\rho}\right)T^{3m}(\rho,\rho^{\prime})f_{\rho}(\rho^{\prime})u^{-}(\rho)u^{+}(\rho^{\prime})d\rho
d\rho^{\prime}$ $\displaystyle=\int
I(\rho)u(\rho)^{2}d\rho+\int\int\frac{1}{\rho}\left(\frac{\rho^{\prime}}{\rho}\right)T^{3m}(\rho,\rho^{\prime})f_{\rho}(\rho^{\prime})u(\rho)u(\rho^{\prime})d\rho
d\rho^{\prime}$ $\displaystyle=\langle\Theta^{3m}u,u\rangle$

Then

$\inf_{u\in L^{2}\atop||u||_{L^{2}}=1}\langle\Theta^{3m}u,u\rangle=\inf_{u\in
L^{2}\atop||u||_{L^{2}}=1,\,u\geq 0}\langle\Theta^{3m}u,u\rangle,$

Now, for a positive $u\in L^{2}$, we write

$\displaystyle\langle\Theta^{3m}u,u\rangle=\langle\left(\Theta^{3m}-\Theta^{3n}\right)u,u\rangle+\langle\Theta^{3m}u,u\rangle.$

where

$\displaystyle\left(\Theta^{3m}-\Theta^{3n}\right)u=\frac{1}{\rho}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\left(K^{3m}-K^{3n}\right)\left(\frac{\rho}{\rho^{\prime}}\right)u(\rho^{\prime})d\rho^{\prime}$

and notice that

$\displaystyle\frac{1}{\rho}\int_{-\infty}^{\infty}f_{\rho}(\rho^{\prime})\left(K^{3m}-K^{3n}\right)\left(\frac{\rho}{\rho^{\prime}}\right)u(\rho^{\prime})d\rho^{\prime}\geq
0$

by Lemma 4.11.

We then obtain that

$\displaystyle\langle\Theta^{3m}u,u\rangle\geq\langle\Theta^{3n}u,u\rangle,$

for every positive $u\in L^{2}$. This concludes the proof of the lemma. ∎

Now our purpose is to find a bound from below for the number $\lambda_{s}^{6}$
defined by

$\lambda_{s}^{6}=\inf_{u\in
L^{2}\atop||u||_{L^{2}}=1}\langle\Theta^{6}u,u\rangle.$

In order to do this we will need Lemmas 4.31 and 4.32 below.

###### Lemma 4.31

Let
$B^{6\text{aprox}}_{s}=\frac{1}{\sqrt{a-a\beta}}{1}_{[1-a+\frac{a\beta}{2},1-\frac{a\beta}{2}]}$
and $\lambda^{s}_{6\text{aprox}}=0.573$. Then

$\displaystyle\Theta^{6}_{S}B^{6\text{aprox}}_{s}-\lambda^{s}_{6\text{aprox}}B^{6\text{aprox}}_{s}=e^{6}$

with

$||e^{6}||_{L^{2}}<0.0893.$

###### Proof.

The proof is computer-assisted and the codes can be found in the supplementary
material. We refer to the appendix for the implementation. ∎

###### Lemma 4.32

Let $B^{6\text{aprox}}_{s}$ be the approximation in Lemma 4.31. Then, if we
define the number $c^{6*}$ by

$\displaystyle
c^{6*}=\inf_{v\in\left(B^{6\text{aprox}}_{s}\right)^{\perp}\atop||v||_{L^{2}}=1}\langle\Theta^{6}v,v\rangle,$

the following bound holds

$\displaystyle c^{6*}>0.8355.$

###### Proof.

The proof is computer-assisted and the codes can be founded in the
supplementary material. We refer to the appendix for details on the
implementation. ∎

###### Lemma 4.33

Let $\lambda_{6}^{s}\in\mathbb{R}$ given by

$\displaystyle\lambda_{6}^{s}=\inf_{u\in
L^{2}\atop||u||_{L^{2}}=1}\langle\Theta^{6}u,u\rangle.$

Then, we have the following bound:

$\displaystyle\lambda_{6}^{s}>0.4837.$

###### Proof.

For a generic $B\in L^{2}$ with $||B||_{L^{2}}=1$ we can decompose $B=\alpha
B^{6\text{aprox}}_{s}+\beta v$, where
$v\in\left(B^{6\text{aprox}}_{s}\right)^{\perp}$, $||v||_{L^{2}}=1$ and
$\alpha^{2}+\beta^{2}=1$. Therefore

$\displaystyle\langle\Theta^{6}_{S}B,B\rangle$
$\displaystyle=\alpha^{2}\lambda^{s}_{6\text{aprox}}+\beta^{2}\langle\Theta^{6}_{S}v,v\rangle+2\alpha\beta\langle
e^{6},v\rangle$
$\displaystyle\geq\alpha^{2}\lambda^{s}_{6\text{aprox}}+\beta^{2}\langle\Theta^{6}_{S}v,v\rangle-2\alpha\sqrt{1-\alpha^{2}}\langle
e^{6},v\rangle.$

Finally we obtain that

$\displaystyle\langle\Theta^{6}_{S}B,B\rangle\geq\lambda^{s}_{6\text{aprox}}-||e^{6}||_{L^{2}},$

since

$\inf_{v\in\left(B^{6\text{aprox}}_{s}\right)^{\perp}\atop||v||_{L^{2}}=1}\langle\Theta^{6}_{S}v,v\rangle\geq
c^{6*}>\lambda^{s}_{6\text{aprox}}$

by Lemmas 4.31 and 4.32. ∎

Using Lemmas 4.29, 4.30 and 4.33 we can prove the following proposition:

###### Proposition 4.34

The bilinear forms $\langle(\Theta^{3m}-\lambda_{3})v,v\rangle$ are coercive
in $L^{2}$ for $m>1$. In addition

$\displaystyle\langle(\Theta^{3m}-\lambda_{3})v,v\rangle\geq\frac{1}{\lambda^{s}_{6}-\lambda_{3}}||v||_{L^{2}}^{2}.$

Therefore, if $u\in L^{2}$ satisfies

$\displaystyle\Theta^{3m}u=\lambda_{3}u,$

for $m>1$ then $u=0$.

Proposition 4.7 is then proven.

∎

To finish this section we study the codimension of the image of the operator
$\partial_{r}F[\rho,\lambda_{3}]$.

###### Proposition 4.35

The space
$H^{3,3}_{3,\text{odd}}(\Omega_{a})/\mathcal{R}\left(\partial_{r}F[\rho,\lambda_{3}]\right)$
has dimension one.

###### Proof.

In order to prove this proposition we will study the range of
$\partial_{r}F[\rho,\lambda_{3}]$. Let $G(\alpha,\rho)\in
H^{3,3}_{3,\text{odd}}(\Omega_{a})$. We shall try to find $g(\alpha,\rho)\in
H^{4,3}_{3,\text{even}}(\Omega_{a})$ such that

$\displaystyle\partial_{r}F[\rho,\lambda_{3}]g(\alpha,\rho)=G(\alpha,\rho)$
(4.32)

By using the expansions

$\displaystyle g(\alpha,\rho)=$ $\displaystyle\sum_{m=1}^{\infty}\rho
g^{3m}(\rho)\cos(3m\alpha)$ $\displaystyle G(\alpha,\rho)=$
$\displaystyle\sum_{m=1}^{\infty}\rho G^{3m}(\rho)\sin(3m\alpha)$

in (4.32) we have that

$\displaystyle\sum_{m=1}^{\infty}3m\rho\left(\Theta^{3m}g^{3m}(\rho)-\lambda_{3}g^{3m}(\rho)\right)\sin(3m\alpha)=\sum_{m=1}^{\infty}\rho
G^{3m}(\rho)\sin(3m\alpha).$

Taking the projection onto the $3m-$mode yields

$\displaystyle
3m\left(\Theta^{3m}g^{3m}-\lambda_{3}g^{3m}(\rho)\right)=G^{3m}(\rho)\quad\text{for
$m=1,2,3,...$}$ (4.33)

Next we shall study the existence of solutions for the equation (4.33) in
$L^{2}$ and after that the $H^{3}$-regularity.

  1. 1.

Existence in $L^{2}$. We will deal separately with two cases: in Lemma 4.36 we
take $m>1$ and in Lemma 4.37 $m=1$.

###### Lemma 4.36

For $m>1$ there exists an inverse operator

$\displaystyle\left(\Theta^{3m}-\lambda_{3}\right)^{-1}\,:\,L^{2}\to L^{2}$

with norm bounded independently of $m$.

###### Proof.

By Lemmas 4.29, 4.30 and the explicit bound for $\lambda^{s}_{6}$ in Lemma
4.33 we have that the bilinear form
$\langle(\Theta^{3m}-\lambda_{3})v,v\rangle$ is coercive in $L^{2}$ with

$\langle(\Theta^{3m}-\lambda_{3})v,v\rangle\geq\frac{1}{\lambda_{6}^{s}-\lambda_{3}}||v||_{L^{2}}^{2}\quad\text{for
$m>1$}.$

Since $G^{3m}\in H^{3}\subset L^{2}$ we can apply Lax-Milgram theorem in order
to obtain the existence of an inverse operator

$(\Theta^{3m}-\lambda_{3})^{-1}\,:\,L^{2}\to L^{2}.$

∎

###### Lemma 4.37

Let $B^{3}_{*}$ be defined by Proposition 4.27. Then, if
$G^{3}\in(B^{3}_{*})^{\perp}$, there exists a solution, $g^{3}\in L^{2}$, to
the equation (4.33) with $m=1$. However, if
$G^{3}\in\text{span}\left\\{B_{*}^{3}\right\\}$ there is no function in
$L^{2}$ satisfying the equation (4.33) with $m=1$.

###### Proof.

First we notice that we have already checked that
$\langle(\Theta^{3}-\lambda_{3})v,v\rangle$ is coercive in the space
$B_{sj}^{\perp}$. In the next lemma we prove that it is also coercive in
$(B^{3}_{*})^{\perp}.$

###### Lemma 4.38

There exists a constant $c>\lambda_{3}$ such that

$\displaystyle\inf_{v\in(B^{3})^{\perp}\atop||v||_{L_{2}}=1}\langle(\Theta^{3}-\lambda_{3})v,v\rangle\geq
c-\lambda_{3}.$

###### Proof.

We take $v\in(B_{*}^{3})^{\perp}$, with $L^{2}-$norm equal to 1. We can
decompose $v$ in the following form $v=\alpha B_{sj}+\beta h$ where $h\in
B^{\perp}_{sj}$ and $||h||_{L^{2}}=1$. Since
$B^{3}_{*}=B_{sj}+v^{\lambda_{3}}_{*}$ we have that

$\displaystyle\alpha||B_{sj}||^{2}_{L^{2}}+\beta\langle
h,v^{\lambda_{3}}_{*}\rangle=$ $\displaystyle 0$
$\displaystyle\alpha^{2}||B_{sj}||_{L^{2}}^{2}+\beta^{2}=1.$

Therefore, we can write,

$\displaystyle\langle(\Theta^{3}-\lambda_{3})v,v\rangle=\langle(\Theta^{3}_{S}-\lambda_{3})v,v\rangle,$

using that $\Theta^{3}_{S}B_{sj}=\lambda^{*}B_{sj}+e$,

$\displaystyle\langle(\Theta^{3}_{S}-\lambda_{3})v,v\rangle=\alpha^{2}\langle(\Theta^{3}_{S}-\lambda_{3})B_{sj},B_{sj}\rangle+2\alpha\beta\langle(\Theta^{3}_{S}-\lambda_{3})B_{sj},h\rangle+\beta^{2}\langle(\Theta^{3}_{S}-\lambda_{3})h,h\rangle$
$\displaystyle=\alpha^{2}\langle(\lambda^{*}-\lambda_{3})B_{sj}+e,B_{sj}\rangle+2\alpha\beta\langle(\lambda^{*}-\lambda_{3})B_{sj}+e,h\rangle+\beta^{2}\langle(\Theta^{3}_{S}-\lambda_{3})h,h\rangle.$

And substituting

$\displaystyle\alpha=-\beta\frac{\langle
h,v_{*}^{\lambda_{3}}\rangle}{||B_{sj}||_{L^{2}}^{2}}$

yields

$\displaystyle\langle(\Theta^{3}_{S}-\lambda_{3})v,v\rangle$
$\displaystyle=\beta^{2}\left(\frac{\langle
h,v_{*}^{\lambda_{3}}\rangle^{2}}{||B_{sj}||_{L^{2}}^{4}}\left((\lambda^{*}-\lambda_{3})\|B_{sj}\|_{L^{2}}^{2}+\langle
e,B_{sj}\rangle\right)+\langle(\Theta^{3}_{S}-\lambda_{3})h,h\rangle-2\frac{\langle
h,v_{*}^{\lambda_{3}}\rangle}{||B_{sj}||_{L^{2}}^{2}}\left\langle
e,h\right\rangle\right).$ (4.34)

Here we recall that

$\displaystyle||v_{*}^{\lambda_{3}}||_{L^{2}}\leq$
$\displaystyle\frac{||-\Theta^{3}_{A}B_{sj}+e||_{L^{2}}}{c^{*}-\lambda_{3}}$
$\displaystyle\langle(\Theta_{S}^{3}-\lambda_{3})h,h\rangle\geq$
$\displaystyle(c^{*}-\lambda_{3})$

and notice that

$\displaystyle\beta^{2}\geq\frac{1}{1+\frac{||-\Theta_{A}^{3}B_{sj}+e||^{2}_{L^{2}}}{(c^{*}-\lambda_{3})^{2}||B_{sj}||_{L^{2}}^{2}}}>0$

Therefore, it is enough to show that

$\displaystyle\left(\frac{\langle
h,v_{*}^{\lambda_{3}}\rangle^{2}}{||B_{sj}||_{L^{2}}^{4}}\left((\lambda^{*}-\lambda_{3})\|B_{sj}\|_{L^{2}}^{2}+\langle
e,B_{sj}\rangle\right)+\langle(\Theta^{3}_{S}-\lambda_{3})h,h\rangle-2\frac{\langle
h,v_{*}^{\lambda_{3}}\rangle}{||B_{sj}||_{L^{2}}^{2}}\left\langle
e,h\right\rangle\right)>0.$

However, the LHS can be bounded from below by

$\displaystyle
c^{*}-\lambda_{0}+\left(\frac{||-\Theta^{3}_{A}B_{sj}+e||_{L^{2}}}{\|B_{sj}\|_{L^{2}}^{2}(c^{*}-\lambda_{0})}\right)^{2}\left(((\lambda^{*}-\lambda_{0})\|B_{sj}\|_{L^{2}}^{2}-|\langle
e,B_{sj}\rangle|\right)-2\frac{||-\Theta^{3}_{A}B_{sj}+e||_{L^{2}}}{\|B_{sj}\|_{L^{2}}^{2}(c^{*}-\lambda_{0})}\|e\|_{L^{2}},$

since

$\displaystyle\langle(\Theta_{S}^{3}-\lambda_{3})h,h\rangle$
$\displaystyle\geq(c^{*}-\lambda_{3})\geq c^{*}-\lambda_{0}$
$\displaystyle\frac{\langle
h,v_{*}^{\lambda_{3}}\rangle^{2}}{||B_{sj}||_{L^{2}}^{4}}(\lambda^{*}-\lambda_{3})\|B_{sj}\|_{L^{2}}^{2}$
$\displaystyle\geq\left(\frac{||-\Theta^{3}_{A}B_{sj}+e||_{L^{2}}}{(c^{*}-\lambda_{3})\|B_{sj}\|_{L^{2}}^{2}}\right)^{2}(\lambda^{*}-\lambda_{3})\|B_{sj}\|_{L^{2}}^{2}$
$\displaystyle\geq\left(\frac{||-\Theta^{3}_{A}B_{sj}+e||_{L^{2}}}{(c^{*}-\lambda_{0})\|B_{sj}\|_{L^{2}}^{2}}\right)^{2}(\lambda^{*}-\lambda_{0})\|B_{sj}\|_{L^{2}}^{2}$
$\displaystyle\frac{\langle
h,v_{*}^{\lambda_{3}}\rangle^{2}}{||B_{sj}||_{L^{2}}^{4}}\langle
e,B_{sj}\rangle$
$\displaystyle\geq-\left(\frac{||-\Theta^{3}_{A}B_{sj}+e||_{L^{2}}}{(c^{*}-\lambda_{3})\|B_{sj}\|_{L^{2}}^{2}}\right)^{2}\left|\langle
e,B_{sj}\rangle\right|$
$\displaystyle\geq-\left(\frac{||-\Theta^{3}_{A}B_{sj}+e||_{L^{2}}}{(c^{*}-\lambda_{0})\|B_{sj}\|_{L^{2}}^{2}}\right)^{2}\left|\langle
e,B_{sj}\rangle\right|$ $\displaystyle-2\frac{\langle
h,v_{*}^{\lambda_{3}}\rangle}{||B_{sj}||_{L^{2}}^{2}}\left\langle
e,h\right\rangle$
$\displaystyle\geq-2\frac{||-\Theta^{3}_{A}B_{sj}+e||_{L^{2}}}{\|B_{sj}\|_{L^{2}}^{2}(c^{*}-\lambda_{3})}\|e\|_{L^{2}}\geq-2\frac{||-\Theta^{3}_{A}B_{sj}+e||_{L^{2}}}{\|B_{sj}\|_{L^{2}}^{2}(c^{*}-\lambda_{0})}\|e\|_{L^{2}}$