1 Introduction

The large scale dynamics of an inviscid three-dimensional fluid subject to rapid background rotation and strong stratification can be described through the so-called quasi-geostrophic model. It is an asymptotic model derived from the Boussinesq system for vanishing Rossby and Froud numbers, for more details about its formal derivation we refer to [36]. Rigorous derivation can be found in [3, 10, 30].

We point out that this system is a pertinent model commonly used in the ocean and atmosphere circulations to describe the vortices and to track the emergence of long-lived structures. The quasi-geostrophic system is described by the potential vorticity q which is merely advected by the fluid,

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t q+u\partial _1 q+v\partial _2 q=0, &{} (t,x)\in [0,+\infty )\times {\mathbb {R}}^3, \\ \Delta \psi = q, &{} \\ u = -\partial _2 \psi , v = \partial _1 \psi , &{} \\ q(t=0,x) = q_0(x). &{} \end{array}\right. \end{aligned}$$
(1.1)

The second equation involving the standard Laplacian of \({{\,\mathrm{{\mathbb {R}}}\,}}^3\) can be formally inverted using Green’s function leading to the following representation of the stream function \(\psi \),

where dA denotes the usual Lebesgue measure. The velocity field (uv, 0) is solenoidal and can be recovered from q through the Biot–Savart law,

Notice that the velocity field is planar but its components depends on the all spatial variables and the potential vorticity is transported by the associated flow. The incompressibility of the velocity allows us to adapt without any difficulties the classical results known for 2D Euler equations. For instance, see [34], one may get global unique strong solutions when the initial data \(q_0\) belongs to Hölder class \({\mathscr {C}}^\alpha \), for \(\alpha \ge 0\). Yudovich theory [44] can also be implemented and one gets global unique solution when \(q_0\in L^1\cap L^\infty \). This latter context allows to deal with discontinuous vortices of the patch form, meaning a characteristic function of a bounded domain. This structure is preserved in time and the vortex patch problem consists of studying the regularity of the boundary and to analyze whether singularities can be formed in finite time on the boundary.

For the 2D Euler equations, the \({\mathscr {C}}^{1,\alpha }\) regularity of the boundary of the patch, with \(\alpha \in (0,1)\), is preserved in time, see [4, 11, 41]. The contour dynamics of the patch is in general hard to track and filamentation may occur. Therefore it is of important interest to look for ordered structure in turbulent flows like relative equilibria. It seems that only a few explicit examples are known in the literature in the patch form: the circular patches which are stationary and the elliptic ones which rotate uniformly with a constant angular velocity. This latter example is known as the Kirchhoff ellipses. However, a lot of implicit examples with higher symmetry have been constructed during the last decades and the first ones were discovered numerically by Deem and Zabusky [13]. Having this kind of V-state solutions in mind, Burbea [5] designed a rigorous approach to generate them close to Rankine vortices through complex analysis tools and bifurcation theory. Later this idea was fruitfully improved and extended in various directions, generating a lot of contributions dealing, for instance, with interesting topics like the regularity problem of the relative equilibria, their existence with different topological structure or for different active scalar equations and so forth. For more details about this active area we refer the reader to the works [6,7,8,9, 14,15,16, 19, 21,22,29] and the references therein.

Coming back to the 3D quasi-geostrophic system, it seems that stationary solutions in the patch form are more abundant than the planar case. Indeed, any domain with a revolution shape about the z-axis generates a stationary solution. The analogues to Kirchhoff ellipses still surprisingly survive in the 3d case. In [35] it is shown that a standing ellipsoid of arbitrary semi-axis lengths ab and c rotates steadily about the z-axis with the angular velocity

$$\begin{aligned} \Omega =\mu \frac{\lambda ^{-1} R_D(\mu ^2,\lambda ,\lambda ^{-1})-\lambda R_D(\mu ^2,\lambda ^{-1},\lambda )}{3(\lambda ^{-1}-\lambda )}, \end{aligned}$$

where \(\lambda =\frac{a}{b}\) is the horizontal aspect ratio, \(\mu := \frac{c}{\sqrt{ab}}\) the vertical aspect ratio and \(R_D\) denotes the elliptic integral of second order

$$\begin{aligned} R_D(x,y,z):=\frac{3}{2}\int _0^{+\infty }\frac{dt}{\sqrt{(t+x)(t+y)(t+z)}}\cdot \end{aligned}$$

For more details about the stability of those ellipsoids we refer to [17, 18, 20].

The main concern of this paper is to investigate the existence of non trivial relative equilibria close to the stationary revolution shapes. In our context, we mean by relative equilibria periodic solutions in the patch form, rotating uniformly about the vertical axis without any deformation. Very recently, Reinaud has explored numerically in [40] the existence and the linear stability of finite volume relative equilibria distributed around circular point vortex arrays. Similar analysis has been implemented in [39] for toroidal vortices. Apart from the numerical experiments, no analytical results had been yet developed and the main inquiry of this paper is to design some technical material allowing us to construct relative equilibria close to general smooth stationary revolution shapes. The basic tool is bifurcation theory but as we shall see its implementation is an involved task which requires refined and careful analysis. Let us explain more our strategy and how to proceed. First, we start with deriving the contour dynamic equation for rotating finite volume patches \({\mathbf{1}}_D\). To do so, we look for smooth domains D with the following parametrization,

$$\begin{aligned} D&= \left\{ (re^{i\theta },\cos (\phi ))\, :\, 0\le r\le r(\phi ,\theta ), 0\le \theta \le 2\pi , 0\le \phi \le \pi \right\} , \end{aligned}$$

where the shape is sufficiently close to a revolution shape domain, meaning that

$$\begin{aligned} r(\phi ,\theta )=r_0(\phi )+f(\phi ,\theta ), \end{aligned}$$

with small perturbation f. Since the domain is assumed to be smooth then we should prescribe the Dirichlet boundary conditions,

$$\begin{aligned} r_0(0)=r_0(\pi )=f(0,\theta )=f(\pi ,\theta )=0. \end{aligned}$$

Notice that without any perturbation, that is, \(f\equiv 0\), the initial data \(q_0=\mathbf{1}_D\) defines a stationary solution for (1.1), as we will prove in Lemma 2.1. Now a rotating solution about the vertical axis is a time-dependent solution taking the form,

$$\begin{aligned} q(t,x)=q_0(e^{-i\Omega t}x_h,x_3), \quad q_0=\mathbf{1}_D, \,x_h=(x_1,x_2). \end{aligned}$$

We shall see later that it is equivalent to check that

$$\begin{aligned} {F}(\Omega ,f)(\phi ,\theta ):=\psi _0(r(\phi ,\theta )e^{i\theta },\cos (\phi ))-\frac{\Omega }{2}r^2(\phi ,\theta )-m(\Omega ,f)(\phi )=0, \end{aligned}$$

for any \((\phi ,\theta )\in [0,\pi ]\times [0,2\pi ]\), where

where \(\psi _0\) stands for the stream function associated to \(q_0\). With this reformulation we visualize the smooth rotating surface as a collection of interacting stratified horizontal sections rotating with the same angular velocity but their size degenerates when we approach the north and south poles corresponding to \(\phi \in \{0,\pi \}\).

In order to apply a bifurcation argument, one has to deal with the linearized operator of F around \(f=0\). From Proposition 3.2 such linearized operator has a compact expression in terms of hypergeometric functions. Indeed, for \(h(\phi ,\theta )=\sum _{n\ge 1}h_n(\phi )\cos (n\theta )\), one gets

$$\begin{aligned} \partial _{f} F(\Omega ,0)h(\phi ,\theta )=r_0(\phi )\nu _\Omega (\phi )\sum _{n\ge 1}\cos (n\theta ){\mathcal {L}}_n(h_n)(\phi ), \end{aligned}$$

where

$$\begin{aligned} {\mathcal {L}}_n(h_n)(\phi )&= h_n(\phi )-{\mathcal {K}}_{n}^{\Omega }(h_n)(\phi )\\&:=h_n(\phi ) -\int _0^\pi \nu _\Omega ^{-1}(\phi )H_n(\phi ,\varphi )h_n(\varphi )d\varphi , \end{aligned}$$

with

$$\begin{aligned} \nu _\Omega (\phi ):=\int _0^\pi H_1(\phi ,\varphi )d\varphi -\Omega ,\quad R(\phi ,\varphi ):=(r_0(\phi )+r_0(\varphi ))^2+(\cos (\phi )-\cos (\varphi ))^2 \end{aligned}$$

and for \(n\ge 1\),

$$\begin{aligned} H_n(\phi ,\varphi )&:=\frac{2^{2n-1}\left( \frac{1}{2}\right) ^2_{n}}{(2n)!}\frac{\sin (\varphi )r_0^{n-1}(\phi )r_0^{n+1}(\varphi )}{\left[ R(\phi ,\varphi )\right] ^{n+\frac{1}{2}}} F_n\left( \frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right) . \end{aligned}$$

Here \(F_n\) denotes the hypergeometric function

$$\begin{aligned} F_n(x)=F\left( n+\frac{1}{2},n+\frac{1}{2},2n+1,x\right) , \quad x\in [0,1). \end{aligned}$$

An important observation is that the kernel study of \(\partial _{f} F(\Omega ,0)\) amounts to checking whether 1 is an eigenvalue for \({\mathcal {K}}_{n}^{\Omega }\). To do that we first start with symmetrizing this operator by working on suitable weighted Hilbert spaces. A natural candidate for that is the Hilbert space \(L^2_{\mu _\Omega }(0,\pi )\) of square integrable functions with respect to the measure

$$\begin{aligned} d\mu _{\Omega }(\varphi ):=\sin (\varphi )r_0^2(\varphi )\nu _{\Omega }(\varphi )d\varphi . \end{aligned}$$

In general this defines a signed measure and to get a positive one we should restrict the values of \(\Omega \) to the set \((-\infty ,\kappa )\), where \(\displaystyle {\kappa :=\inf _{\phi \in (0,\pi )}\int _0^\pi H_1(\phi ,\varphi )d\varphi }\).

In the next step we prove that for any \(n\ge 1\), the operator \({\mathcal {K}}_{n}^{\Omega }:L^2_{\mu _\Omega }\rightarrow L^2_{\mu _\Omega }\) acts as a compact self-adjoint integral operator. This gives us the structure of the eigenvalues which is a discrete set and we establish from the positivity of the kernel that the largest eigenvalue \(\lambda _n(\Omega )\) giving the spectral radius is positive and simple. For given integer \(n\ge 1\), we define the set

$$\begin{aligned} {\mathscr {S}}_n:=\Big \{\Omega \in (-\infty ,\kappa ) \quad \text {s.t.}\quad \lambda _n(\Omega )=1\Big \}, \end{aligned}$$

and in Proposition 4.2 we shall describe some basic properties of \(\lambda _n\) through precise study of the kernel. Those properties show in particular that the set \({\mathscr {S}}_n\) is formed by a single point denoted by \(\Omega _n\), see Proposition 4.3 for more details. In addition, we show that the sequence \(n\in N^\star \mapsto \Omega _n\) is strictly increasing which ensures that the kernel of the linearized operator is a one-dimensional vector space, see Proposition 4.6. Notice that the weighted space \(L^2_{\mu _\Omega }\) is chosen to be so weak in order to have it be stable under the nonlinear functional F. So we need to reinforce the regularity by selecting the standard Hölder spaces \({\mathscr {C}}^{1,\alpha }\) with Dirichlet boundary condition and \(\alpha \in (0,1)\). However this choice generates two delicate problems. The first one is to check that the eigenfunctions constructed in \(L^2_{\mu _\Omega }\) are sufficient smooth and belong to the new spaces. To reach this regularity we need to check that the function \(\nu _\Omega \) is \({\mathscr {C}}^{1,\alpha }\) and this requires more careful analysis due to the logarithmic singularity, see Proposition 4.1. Notice that the eigenfunctions satisfy the boundary condition provided that \(n\ge 2\) and which fails for \(n=1\). The second difficulty concerns the stability of the Hölder spaces by the nonlinear functional F, in fact not F but another modified functional \({\tilde{F}}\) deduced from the preceding one by removing the singularities coming from of the north and south poles, see (2.13). The deformation of the Euclidean kernel through the cylindrical coordinates generates singularities on the poles because the size of horizontal sections degenerates at those points. That is the central difficulty when we try to implement potential theory arguments to get the stability of the function spaces and will be discussed in Sect. 5.

Before stating our result, we need to make the following assumptions on the initial profile \(r_0\) and denoted throughout this paper by (H) :

(H1):

\(r_0\in {\mathscr {C}}^{2}([0,\pi ])\), with \(r_0(0)=r_0(\pi )=0\) and \(r_0(\phi )>0\) for \(\phi \in (0,\pi )\).

(H2):

There exists \(C>0\) such that

$$\begin{aligned} \forall \,\phi \in [0,\pi ],\quad C^{-1}\sin \phi \le r_0(\phi )\le C\sin (\phi ). \end{aligned}$$
(H3):

\(r_0\) is symmetric with respect to \(\phi =\frac{\pi }{2}\), i.e., \(r_0\left( \frac{\pi }{2}-\phi \right) =r_0\left( \frac{\pi }{2}+\phi \right) \), for any \(\phi \in [0,\frac{\pi }{2}]\).

Now we are ready to give a short version of the main result of this paper and the precise one is detailed in Theorem 6.1.

Theorem 1.1

Assume that \(r_0\) satisfies the assumptions (H). Then for any \(m\ge 2\), there exists a curve of non trivial rotating solutions with m-fold symmetry to the Eq. (1.1) bifurcating from the trivial revolution shape associated to \(r_0\) at the angular velocity \(\Omega _m\), the unique point of the set \({\mathscr {S}}_m\).

We specify that by m-fold shape symmetry of \({{\,\mathrm{{\mathbb {R}}}\,}}^3\), we mean a surface invariant under rotation with the vertical axis and angle \(\frac{2\pi }{m}\cdot \)

There is the particular case of \(r_0(\phi )=\sin (\phi )\) defining the unit sphere. Here, its associated stream function can be explicitly computed (see [32]) and it is quadratic inside the shape, that is,

$$\begin{aligned} \psi _0(x)=\frac{1}{6} (x_1^2+x_2^{2}+x_3^2-3). \end{aligned}$$

That gives us some interesting properties on the eigenvalues \(\Omega _m\) of the above Theorem 1.1. In particular, we achieve that the above eigenvalues \(\Omega _m\) belong to \((0,\frac{1}{3})\). The same properties occur also in the case of an ellipsoid of equal x and y axes defining a revolution shape around the z-axis. In this case, the associated stream function is also quadratic. See Sect. 6.1 for a more detailed discussion about those cases.

The paper is structured as follows. In Sect. 2, we provide different reformulations for the rotating patch problem and we introduce the appropriate function spaces. Section 3 is devoted to different useful expressions of the linearized operator around a stationary solution. The spectral study of the linearized operator will be developed in Sect. 4. In Sect. 5, we shall discuss the well-definition of the nonlinear functional and its regularity. In Sect. 6, we give the general statement of our result and provide its proof. We end this paper with three appendices concerning special functions, bifurcation theory and potential theory.

2 Vortex Patch Equations

Take an initial data uniformly distributed in a bounded domain of \({{\,\mathrm{{\mathbb {R}}}\,}}^3\), that is, \( q_0={\mathbf{1}}_{D}\). Then, this structure is preserved by the evolution and one gets for any time \(t\ge 0\)

$$\begin{aligned} q(t,x)={\mathbf{1}}_{D(t)}(x), \end{aligned}$$
(2.1)

for some bounded domain D(t). To track the dynamics of the boundary (which is a surface here) we can implement the contour dynamics method introduced by Deem and Zabusky for Euler equations [13]. Indeed, let \(\gamma _t: (\phi ,\theta )\in {{\,\mathrm{{\mathbb {T}}}\,}}^2\mapsto \gamma _t(\phi ,\theta )\in {{\,\mathrm{{\mathbb {R}}}\,}}^3\) be any parametrization of the boundary \(\partial D_t\). Since the boundary is transported by the flow then

$$\begin{aligned} \big (\partial _t\gamma _t-U(t,\gamma _t)\big )\cdot n(\gamma _t)=0, \end{aligned}$$
(2.2)

where \(U=(u,v,0)\) and \(n(\gamma _t)\) is a normal vector to the boundary at the point \(\gamma _t\). There is a special parametrization called Lagrangian parametrization given by

$$\begin{aligned} \partial _t\gamma _ t=U(t,\gamma _t), \end{aligned}$$

which is commonly used to follow the boundary motion. From the Biot–Savart law we deduce that

$$\begin{aligned} U(t,\gamma _t\big (\phi ,\theta )\big )=\frac{1}{4\pi }\int _{D_t}\frac{(\gamma _t\big (\phi ,\theta )-y)^\perp }{|\gamma _t\big (\phi ,\theta )-y|^3}dA(y) =\frac{1}{4\pi }\int _{\partial D_t}\frac{n^\perp (y)}{|\gamma _t\big (\phi ,\theta )-y|}d\sigma (y), \end{aligned}$$
(2.3)

where \(d\sigma \) denotes the Lebesgue surface measure of \(\partial D_t\). We have used the notation \(x^\perp =(-x_2,x_1,0)\in {{\,\mathrm{{\mathbb {R}}}\,}}^3\) for \(x=(x_1,x_2,x_3)\in {{\,\mathrm{{\mathbb {R}}}\,}}^3\).

2.1 Stationary patches

Our next goal is to check that any initial patch with revolution shape around the vertical axis generates a stationary solution. More precisely, we have the following result.

Lemma 2.1

Let \(r:[-1,1]\rightarrow {{\,\mathrm{{\mathbb {R}}}\,}}_+\) be a continuous function with \(r(-1)=r(1)=0\) and let D be the domain enclosed by the surface \(\left\{ (r(z)e^{i\theta },z),\ \theta \in [0,2\pi ], z\in [-1,1]\right\} \), then \(q(t,x)={\mathbf{1}}_{D}(x)\) defines a stationary solution for (1.1).

Proof

Recall from (2.3) that

(2.4)

Define

and let us prove that \(G\equiv 0\). Take \(\theta \in {{\,\mathrm{{\mathbb {R}}}\,}}\) and denote by \({\mathcal {R}}_\theta \) the rotation: \( x=(x_h,x_3)\mapsto (e^{i\theta } x_h, x_3)\). Since D is invariant by \({\mathcal {R}}_\theta \), changing variables leads to

$$\begin{aligned} G({\mathcal {R}}_\theta x)= G(x). \end{aligned}$$

Therefore \(G(x)=G(|x_h|,0,x_3)\), which means that

Since D is invariant by the reflexion: \(y\mapsto (y_1,-y_2,y_3)\) then a change of variables implies that \(G(x_1,x_2,x_3)=G(x_1,-x_2,x_ 3)=-G(x_1,x_2,x_3)\) and thus \(G(x)=0\). Consequently we get in particular that

$$\begin{aligned} U(x)\cdot x=0, \quad \forall x\in \partial D. \end{aligned}$$

On the other hand, we get from the revolution shape property of D that the horizontal component of the normal vector is \({n}_h(x)=(x_1,x_2)\), which implies

$$\begin{aligned} U(x)\cdot {n}(x)=(u,v)(x)\cdot {n}_h(x)=0, \quad \forall \, x\in \partial D. \end{aligned}$$
(2.5)

This implies that \({\mathbf{1}}_{D}\) is a stationary solution in the weak sense. \(\quad \square \)

2.2 Reformulations for periodic patches

In this section, we shall give two ways to write down rotating patches using respectively the velocity field and the stream function. Assume that we have a rotating patch around the \(x_3\) axis with constant angular velocity \(\Omega \in {{\,\mathrm{{\mathbb {R}}}\,}}\), that is \(D_t={\mathcal {R}}_{\Omega t}D\), with \({\mathcal {R}}_{\Omega t}\) being the rotation of angle \(\Omega t\) around the vertical axis. Inserting this expression into the Eq. (2.2) we get

$$\begin{aligned} \big (U(x)-\Omega x^\perp \big )\cdot {n}(x)=0, \quad \forall \, x\in \partial D. \end{aligned}$$

Since U is horizontal then this equation means also that each horizontal section \(D_{x_3}:=\{ y\in {{\,\mathrm{{\mathbb {R}}}\,}}^2,\, (y,x_3)\in D\}\) rotates with the same angular velocity \(\Omega \). Hence the horizontal sections satisfy the equation

$$\begin{aligned} \big (U(x)-\Omega x^\perp \big )\cdot {n}_{D_{x_3}}(x_h)=0, \quad x_h=(x_1,x_2)\in \partial D_{x_3}, \ x_3\in {{\,\mathrm{{\mathbb {R}}}\,}}, \end{aligned}$$

where \({n}_{D_{x_3}}\) denotes a normal vector to the planar curve \(\partial D_{x_3}\). Next we shall write down this equation in the particular case of simply connected domains that can be described through polar parametrization in the following way:

$$\begin{aligned} D&= \left\{ (re^{i\theta },\cos (\phi ))\, :\, 0\le r\le r(\phi ,\theta ), 0\le \theta \le 2\pi , 0\le \phi \le \pi \right\} . \end{aligned}$$
(2.6)

Notice that we have assumed in this description, and without any loss of generality, that the orthogonal projection onto the vertical axis is the segment \([-1,1]\). The horizontal sections are indexed by \(\phi \) and parametrized by the polar coordinates as \(\theta \mapsto r(\phi ,\theta )\) and it is obvious that

$$\begin{aligned} {n}_{\partial D_{x_3}}(r(\phi ,\theta )e^{i\theta })=\left( i\partial _{\theta }r(\phi ,\theta )-r(\phi ,\theta )\right) e^{i\theta }. \end{aligned}$$

Then, the equation of the sections reduces to

$$\begin{aligned}&\text {Re}\left[ \left\{ U_h(\phi ,\theta )-i\Omega r(\phi ,\theta )e^{i\theta }\right\} \left\{ \big [i\partial _{\theta }r(\phi ,\theta )+r(\phi ,\theta )\big ]e^{-i\theta }\right\} \right] =0,\quad \nonumber \\&\quad \forall (\phi ,\theta )\in [0,\pi ]\times [0,2\pi ], \end{aligned}$$
(2.7)

with, according to (2.3) and the change of variable \(y_3=\cos \varphi \),

(2.8)

We shall look for a rotating solution close to a stationary one described by a given revolution shape \((\theta ,\phi )\mapsto (r_0(\phi ) e^{i\theta },\cos (\phi ))\). This means that we are looking for a parametrization in the form

$$\begin{aligned} r(\phi ,\theta )=r_0(\phi )+f(\phi ,\theta ), \quad f(\phi ,\theta )=\sum _{n\ge 1}f_n(\phi )\cos (n\theta ). \end{aligned}$$
(2.9)

Implicitly, we have assumed that the domain D is symmetric with respect to the plane \(x_2=0\). In addition, we ask the following boundary conditions,

$$\begin{aligned} r_0(0)=r_0(\pi )=f(0,\theta )=f(\pi ,\theta )=0, \end{aligned}$$

meaning that the domain D intersects the vertical axis at the points \((0,0,-1)\) and (0, 0, 1).

Define the functionals

$$\begin{aligned}&F_\mathbf{v}(\Omega ,f)(\phi , \theta )\\&\quad :=\text {Re}\left[ \left\{ I_\mathbf{v}(f)(\phi ,\theta )-i\Omega r(\phi ,\theta )e^{i\theta }\right\} \left\{ i\partial _{\theta }r(\phi ,\theta )e^{-i\theta }+r(\phi ,\theta )e^{-i\theta }\right\} \right] , \end{aligned}$$

with

(2.10)

The subscript \(\mathbf{v}\) refers to the velocity formulation and we use it to compare it later to the stream function formulation. Hence, we need to study the equation:

$$\begin{aligned} F_\mathbf{v}(\Omega ,f)(\phi ,\theta )=0, \quad (\phi ,\theta )\in [0,\pi ]\times [0,2\pi ]. \end{aligned}$$

By Lemma 2.1, one has \(F_\mathbf{v}(\Omega ,0)(\phi , \theta )\equiv 0\), for any \(\Omega \in {{\,\mathrm{{\mathbb {R}}}\,}}\).

2.3 Stream function formulation

There is another way to characterize the rotating solutions described in the previous subsection by virtue of the stream function formulation.

For \(\phi \in [0,\pi ]\), let \( \theta \in [0,2\pi ]\mapsto \gamma _\phi (\theta ):=r(\phi ,\theta )e^{i\theta }, \) be the parametrization of \(\partial D_z\), where \(z=\cos (\phi )\). Then one can check without difficulties that (2.7) agrees with

$$\begin{aligned} \partial _\theta \left\{ \psi _0(\gamma _\phi (\theta ),\cos (\phi ))-\frac{\Omega }{2}|\gamma _\phi (\theta )|^2\right\} =0, \quad \forall (\phi ,\theta )\in [0,\pi ]\times [0,2\pi ]. \end{aligned}$$

Then, the equation can be integrated obtaining

$$\begin{aligned} \psi _0(\gamma _\phi (\theta ),\cos (\phi ))-\frac{\Omega }{2}|\gamma _\phi (\theta )|^2=m(\Omega ,f)(\phi ), \end{aligned}$$

where \(m(\Omega ,f)(\phi )\) is a function depending only on \(\phi \) and given by

(2.11)

Let us consider the functional

$$\begin{aligned} {F}_{\mathbf{s}}(\Omega ,f)(\phi ,\theta )&:=\psi _0(r(\phi ,\theta )e^{i\theta },\cos (\phi ))-\frac{\Omega }{2}r^2(\phi ,\theta )-m(\Omega ,f)(\phi ) \nonumber \\&= G(\Omega ,f)(\phi ,\theta )-\frac{1}{2\pi }\int _0^{2\pi } G(\Omega ,f)(\phi ,\eta )d\eta , \end{aligned}$$
(2.12)

where

$$\begin{aligned} G(\Omega ,f)(\phi ,\theta ):=\psi _0(r(\phi ,\theta )e^{i\theta },\cos (\phi ))-\frac{\Omega }{2}r^2(\phi ,\theta ), \end{aligned}$$

and the stream function is given by

Then, finding a rotating solution amounts to solving in f, for some specific angular velocity constant \(\Omega \), the equation

$$\begin{aligned} {F}_\mathbf{s}(\Omega ,f)(\phi ,\theta )=0, \quad \forall (\phi ,\theta )\in [0,\pi ]\times [0,2\pi ]. \end{aligned}$$

Remark that one may check directly from this reformulation that any revolution shape is a solution for any angular velocity \(\Omega \), meaning that, \({F}_\mathbf{s}(\Omega ,0)=0\), for any \(\Omega \). Motivated by the Sect. 3 on the structure of the linearized operator, we find it better to get rid of the singularities of the poles and work with the modified functional

$$\begin{aligned} {\tilde{F}}(\Omega ,f)(\phi ,\theta ):=\frac{F_\mathbf{s}(\Omega ,f)(\phi ,\theta )}{r_0(\phi )}\cdot \end{aligned}$$

Therefore, we get

$$\begin{aligned} {\tilde{F}}(\Omega ,f)(\phi ,\theta )=\frac{1}{r_0(\phi )}\left\{ I(f)(\phi ,\theta )-\frac{\Omega }{2}r(\phi ,\theta )^2-m(\Omega ,f)(\phi )\right\} , \end{aligned}$$
(2.13)

with

(2.14)

and

$$\begin{aligned} r(\phi ,\theta )=r_0(\phi )+f(\phi ,\theta ). \end{aligned}$$

2.4 Functions spaces

First we shall recall the Hölder spaces defined on an open nonempty set \({\mathscr {O}}\subset {{\,\mathrm{{\mathbb {R}}}\,}}^d\). Let \(\alpha \in (0,1)\) then

$$\begin{aligned} {\mathscr {C}}^{1,\alpha }({\mathscr {O}})=\Big \{ f:{\mathscr {O}}\mapsto {{\,\mathrm{{\mathbb {R}}}\,}}, \Vert f\Vert _{{\mathscr {C}}^{1,\alpha }}<\infty \Big \}, \end{aligned}$$

with

$$\begin{aligned} \Vert f\Vert _{{\mathscr {C}}^{1,\alpha }}=\Vert f\Vert _{L^\infty }+\Vert \nabla f\Vert _{L^\infty }+\sup _{x\ne y\in {\mathscr {O}}}\frac{|\nabla f(x)-\nabla f(y)|}{|x-y|^\alpha }\cdot \end{aligned}$$

It is known that \({\mathscr {C}}^{1,\alpha }({\mathscr {O}})\) is a Banach algebra, meaning a complete space satisfying

$$\begin{aligned} \Vert fg\Vert _{{\mathscr {C}}^{1,\alpha }}\le C \Vert f\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert g\Vert _{{\mathscr {C}}^{1,\alpha }}. \end{aligned}$$

Denote by \({{\,\mathrm{{\mathbb {T}}}\,}}\) the one-dimensional torus and we identify the space \({\mathscr {C}}^{1,\alpha }({{\,\mathrm{{\mathbb {T}}}\,}})\) with the space \({\mathscr {C}}_{2\pi }^{1,\alpha }({{\,\mathrm{{\mathbb {R}}}\,}})\) of \(2\pi \)-periodic functions that belongs to \({\mathscr {C}}^{1,\alpha }({{\,\mathrm{{\mathbb {R}}}\,}})\). Next, we shall introduce the function spaces that we use in a crucial way to study the stability of the functional \({\tilde{F}}\) defined in (2.13). For \(\alpha \in (0,1)\) and \(m\in {{\,\mathrm{{\mathbb {N}}}\,}}^\star \), set

$$\begin{aligned} X_m^\alpha :=\Big \{f\in {\mathscr {C}}^{1,\alpha }\big ((0,\pi )\times {{\,\mathrm{{\mathbb {T}}}\,}}\big ) , \, f\left( \phi ,\theta \right) =\sum _{n\ge 1}f_n(\phi )\cos (nm\theta )\Big \}, \end{aligned}$$
(2.15)

supplemented with the conditions

$$\begin{aligned}&\forall \theta \in [0,2\pi ]\quad f(0,\theta )=f(\pi ,\theta )= 0\quad \hbox {and}\quad \nonumber \\&\quad \forall (\phi ,\theta )\in [0,\pi ]\times [0,2\pi ] \quad f\left( \pi -\phi ,\theta \right) =f\left( \phi ,\theta \right) . \end{aligned}$$
(2.16)

This space is equipped with the same norm as \({\mathscr {C}}^{1,\alpha }((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}})\). The first assumption in (2.16) is a kind of partial Dirichlet condition and the second one is a symmetry property with respect to the equatorial \(\phi =\frac{\pi }{2}\). Notice that any function \(f\in {\mathscr {C}}^{1,\alpha }\big ((0,\pi )\times {{\,\mathrm{{\mathbb {T}}}\,}}\big ) \) admits a continuous extension up to the boundary, so the foregoing conditions are meaningful. Furthermore, the Dirichlet boundary conditions allow us to use Taylor’s formula to get a constant \(C>0\) such that for any \(f\in X_m^\alpha \)

$$\begin{aligned} |f(\varphi ,\eta )|&\le C\Vert f\Vert _{\text {Lip}}\sin \varphi ,\nonumber \\ \partial _\eta f(0,\eta )=\partial _\eta f(\pi ,\eta )=0&\quad \hbox {and}\quad |\partial _\eta f(\varphi ,\eta )|\le C\Vert f\Vert _{{\mathscr {C}}^{1,\alpha }}\sin ^\alpha (\varphi ). \end{aligned}$$
(2.17)

The notation \(B_{X_m^\alpha }(\varepsilon )\) means the ball in \(X_m^\alpha \) centered in 0 with radius \(\varepsilon \).

Next we shall discuss quickly some consequences needed for later purposes and following from the assumptions \({\mathbf{(H)}}\) on \(r_0\), given in the Introduction before our main statement.

  • From (H2) we have that \(r_0'(0)>0\) and by continuity of the derivative there exists \(\delta >0\) such that \(r_0'(\phi )>0\) for \(\phi \in [0,\delta ]\). Combining this with the mean value theorem, we deduce the arc-chord estimate: there exists \(C>0\) such that

    $$\begin{aligned} C^{-1}(\phi -\varphi )^2\le (r_0(\varphi )-r_0(\phi ))^2+(\cos (\phi )-\cos (\varphi ))^2\le C(\phi -\varphi )^2, \end{aligned}$$
    (2.18)

    for any \(\phi ,\varphi \in [0,\pi ]\).

  • We have that \(\frac{\sin (\cdot )}{r_0(\cdot )}\in {\mathscr {C}}^\alpha ([0,\pi ])\), and then \(\phi \in [0,\frac{\pi }{2}]\mapsto \frac{\phi }{r_0(\phi )}\) \(\in {\mathscr {C}}^\alpha ([0,\frac{\pi }{2}])\).

3 Linearized Operator

This section is devoted to show different expressions of the linearized operator around a revolution shape. We can find an useful one in terms of hypergeometric functions. See “Appendix A” for details about these special functions.

From now on, we will use the stream function formulation and then we omit the subscript \(\mathbf{s}\) from \(F_\mathbf{s}\) in order to alleviate the notation. The linearized operator of the velocity formulation is closely related to this one, see the previous section.

3.1 First representation

In the following, we provide the structure of the linearized operator of F around the trivial solution \((\Omega ,0)\).

Proposition 3.1

Let \({\tilde{F}}\) be as in (2.13) and \( (\phi ,\theta )\in [0,\pi ]\times [0,2\pi ]\mapsto h(\phi ,\theta )=\sum _{n\ge 1}h_n(\phi )\cos (n\theta )\) be a smooth function. Then,

(3.1)

Proof

First, note that

$$\begin{aligned}&|(re^{i\eta },\cos (\varphi ))-(r_0(\phi )e^{i\theta },\cos (\phi ))|^2\\&\quad =r^2+r_0^2(\phi )+(\cos (\phi )-\cos (\varphi ))^2-2rr_0(\phi )\cos (\theta -\eta ). \end{aligned}$$

The linearized operator at a state \(r_0\) is defined by Gateaux derivative,

$$\begin{aligned}&\partial _f {\tilde{F}}(\Omega ,0)h(\phi ,\theta ) :=\frac{d}{dt}{\tilde{F}}(\Omega ,th)\Big |_{t=0}(\phi ,\theta )\\&\quad = \frac{1}{r_0(\phi )}\left( \frac{d}{dt}G(\Omega ,th)\Big |_{t=0}(\phi ,\theta )-\frac{1}{2\pi }\int _0^{2\pi }\frac{d}{dt}G(\Omega ,th)\Big |_{t=0}(\phi ,\eta )d\eta \right) . \end{aligned}$$

Thus straightforward computations yield

with

$$\begin{aligned} A(\phi ,\theta ,\varphi ,\eta ):=r_0^2(\varphi )+r_0^2(\phi )+(\cos (\phi )-\cos (\varphi ))^2-2r_0(\varphi )r_0(\phi )\cos (\theta -\eta ). \end{aligned}$$

By expanding h in Fourier series we get

Let us analyze every term. For the first one, making the change of variable \(\theta -\eta \mapsto \eta \) we get using a symmetry argument,

Concerning the last integral term, we first use the identity

$$\begin{aligned}&\partial _r \frac{r}{(r^2+r_0^2(\phi )+(\cos (\phi )-\cos (\varphi ))^2-2rr_0(\phi )\cos (\eta ))^{\frac{1}{2}}}=\\&\quad \frac{{ 1}}{(r^2+r_0^2(\phi )+(\cos (\phi )-\cos (\varphi ))^2-2rr_0(\phi )\cos (\eta ))^{\frac{1}{2}}}\\&\quad -\frac{r(r-r_0(\phi )\cos \eta )}{(r^2+r_0^2(\phi )+(\cos (\phi )-\cos (\varphi ))^2-2rr_0(\phi )\cos (\eta ))^{\frac{3}{2}}}\cdot \end{aligned}$$

Consequently

Thus

Integrating by parts with respect to \(\eta \) gives

Putting together the preceding identities allows to get

Therefore we obtain

Now it is clear that

$$\begin{aligned} \frac{1}{2\pi }\int _0^{2\pi }\partial _f G(\Omega ,0)h(\phi ,\eta )d\eta =0, \end{aligned}$$

and so (3.1) is obtained. \(\quad \square \)

Remark 3.1

Notice that the local part of the linearized operator (3.1) can be directly related to the stream function \(\psi _0\) associated to the domain parametrized by \((\phi ,\theta )\mapsto (r_0(\phi )e^{i\theta },\cos (\phi ))\). Indeed, by differentiating the functional \({\widetilde{\psi }}_0:f\mapsto \psi _0((r_0(\phi )+f(\phi ,\theta ))e^{i\theta },\cos (\phi ))\) at \(f\equiv 0\) and in the direction h one gets

This form is useful later for spherical and ellipsoidal shapes where the stream functions admit explicit expressions inside these domains, see Sect. 6.1.

3.2 Second representation with hypergeometric functions

The main purpose of this subsection is to provide a suitable representation of the linearized operator. First we need to use some notations. For \(n\ge 1\), set

$$\begin{aligned} F_n(x):=F\left( n+\frac{1}{2},n+\frac{1}{2};2n+1;x\right) , \quad x\in [0,1), \end{aligned}$$

where the hypergeometric functions are defined in the “Appendix A”. Other useful notations are listed below,

$$\begin{aligned} R(\phi ,\varphi ):=(r_0(\phi )+r_0(\varphi ))^2+(\cos (\phi )-\cos (\varphi ))^2, \quad 0<\phi ,\varphi <\pi , \end{aligned}$$
(3.2)

and

$$\begin{aligned} H_n(\phi ,\varphi )&:=\frac{2^{2n-1}\left( \frac{1}{2}\right) ^2_{n}}{(2n)!}\frac{\sin (\varphi )r_0^{n-1}(\phi )r_0^{n+1}(\varphi )}{\left[ R(\phi ,\varphi )\right] ^{n+\frac{1}{2}}} F_n\left( \frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right) . \end{aligned}$$
(3.3)

Now we are ready to state the main result of this section.

Proposition 3.2

Let \({\tilde{F}}\) be as in (2.13) and \(h(\phi ,\theta )=\sum _{n\ge 1}h_n(\phi )\cos (n\theta ), (\phi ,\theta )\in [0,\pi ]\times [0,2\pi ]\), be a smooth function. Then,

$$\begin{aligned} \partial _{f} {\tilde{F}}(\Omega ,0)h(\phi ,\theta )=\sum _{n\ge 1}\cos (n\theta ){\mathcal {L}}_n^\Omega (h_n)(\phi ), \end{aligned}$$
(3.4)

where

Proof

With the help of Lemma A.1, we can simplify more the expression of the linearized operator given in Proposition 3.1. We shall first give another representation of the first integral of (3.1),

From Lemma A.1 we infer

Thus we deduce

Remark that the validity of Lemma A.1 is guaranteed since the inequality

$$\begin{aligned} {\frac{r_0^2(\phi )+r_0^2(\varphi )+(\cos (\phi )- \cos (\varphi ))^2}{2r_0(\phi )r_0(\varphi )}}>1, \end{aligned}$$

is satisfied provided that \(\phi \ne \varphi \) which leads to a negligible set. For the last integral in (3.1), we apply once again Lemma A.1,

It follows that

which gives the announced result. \(\quad \square \)

Remark 3.2

By virtue of Remark 3.1 and the previous expression one has that

$$\begin{aligned} \int _0^\pi H_1(\phi ,\varphi )d\varphi =\frac{1}{r_0(\phi )}\partial _R \psi _0(Re^{i\theta },\cos (\phi ))\left| _{R=r_0(\phi )}\right. , \end{aligned}$$
(3.5)

where \(\psi _0\) is the stream function associated to the domain parametrized by \((r_0(\phi )e^{i\theta } ,\cos (\phi ))\), for \((\phi ,\theta )\in [0,\pi ]\times [0,2\pi ]\).

3.3 Qualitative properties of some auxiliary functions

In the following lemma, we shall study some specific properties of the sequence of functions \(\{H_n\}_n\) introduced in (3.3). We shall study the monotonicity of the sequence \(n\mapsto H_n(\phi ,\varphi )\) which will be crucial later in the study of the monotonicity of the eigenvalues associated to the operators family \(\{{\mathcal {L}}_n, n\ge 1\}\). We will also study the decay rate of this sequence for large n.

Lemma 3.1

For any \(\varphi \ne \phi \in (0,\pi )\), the sequence \( n\in {{\,\mathrm{{\mathbb {N}}}\,}}^\star \mapsto H_n(\phi ,\varphi ) \) is strictly decreasing.

Moreover, if we assume that \(r_0\) satisfies (H2), then, for any \(0\le \alpha <\beta \le 1 \) there exists a constant \(C>0\) such that

$$\begin{aligned} |H_n(\phi ,\varphi )|\le C n^{-\alpha }\frac{\sin (\varphi ) r_0^{\frac{1}{2}}(\varphi )}{r_0^{\frac{3}{2}}(\phi )}|\phi -\varphi |^{-\beta }, \quad \forall n\ge 1, \phi \ne \varphi \in (0,\pi ). \end{aligned}$$
(3.6)

Proof

By virtue of (3.3) we may write

$$\begin{aligned} H_n(\phi ,\varphi )=\frac{2^{2n-1}\Gamma ^2\left( n+\frac{1}{2}\right) }{(2n)!\pi }\frac{\sin (\varphi )r_0(\varphi )^{\frac{1}{2}}}{4^{n+\frac{1}{2}}r_0(\phi )^{\frac{3}{2}}}x^{n+\frac{1}{2}}F_n(x), \end{aligned}$$

where \(x:=\frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\) belongs to [0, 1) provided that \(\varphi \ne \phi \). Now using the integral representation of hypergeometric functions (A.2) we obtain

(3.7)

with the notation

$$\begin{aligned} {\mathcal {H}}_n(x):= x^{n+\frac{1}{2}}\int _0^1 t^{n-\frac{1}{2}}(1-t)^{n-\frac{1}{2}}(1-tx)^{-n-\frac{1}{2}}dt. \end{aligned}$$

Therefore the desired result amounts to checking that \(n\mapsto {\mathcal {H}}_n(x)\) is strictly decreasing for any \(x\in (0,1)\). This follows from the fact that \(n\mapsto x^{n+\frac{1}{2}}\) is strictly decreasing combined with the identity

$$\begin{aligned} \int _0^1 t^{n-\frac{1}{2}}(1-t)^{n-\frac{1}{2}}(1-tx)^{-n-\frac{1}{2}}dt=\int _0^1 t^{-\frac{1}{2}}(1-t)^{-\frac{1}{2}}(1-tx)^{-\frac{1}{2}}\left( \frac{t(1-t)}{1-tx}\right) ^{n}dt, \end{aligned}$$

which shows the strict decreasing of this sequence since \(0<\frac{t(1-t)}{1-tx}<1\), for any \(t,x\in (0,1)\).

It follows that for any \(\phi \ne \varphi \), the sequence \(n\mapsto H_n(\phi ,\varphi )\) is strictly decreasing.

It remains to prove the decay estimate of \(H_n\) for large n. It is an immediate consequence of the following more precise estimate: for any \(\alpha \in [0,1]\), we get

$$\begin{aligned} |{\mathcal {H}}_n(x)|\le x^{n-\frac{1}{2}+\alpha }\frac{|\ln (1-x)|^{1-\alpha }}{n^\alpha (1-x)^\alpha }, \end{aligned}$$
(3.8)

for \(n\ge 1\) and \(0\le x<1.\) To see the connection with (3.6) recall first from (3.7) that

$$\begin{aligned} |H_n(\phi ,\varphi )|\lesssim \frac{\sin (\varphi )r_0(\varphi )^\frac{1}{2}}{r_0(\phi )^\frac{3}{2}} |{\mathcal {H}}_n(x)|. \end{aligned}$$

Since \(0\le x\le 1\) then we obtain from (3.8) that for any \(1\ge \beta >\alpha \ge 0\)

$$\begin{aligned} |{\mathcal {H}}_n(x)|\lesssim \frac{|\ln (1-x)|^{1-\alpha }}{n^\alpha (1-x)^\alpha }\lesssim {n^{-\alpha } (1-x)^{-\beta }}\cdot \end{aligned}$$

According to (4.10) we deduce that

$$\begin{aligned} |{\mathcal {H}}_n(x)|&\lesssim \frac{1}{n^\alpha |\phi -\varphi |^\beta }. \end{aligned}$$

which is the desired inequality. Let us now turn to the proof of (3.8). We write

where we have used that

$$\begin{aligned} 1-tx\ge 1-t, \end{aligned}$$

for any \(t\in [0,1]\) and \(0\le x<1\). Observe that we easily get the identity

(3.9)

which implies

and

By using interpolation, we obtain

which gives us

$$\begin{aligned} | {\mathcal {H}}_n(x)|\le x^{n+\frac{1}{2}}\frac{|\ln (1-x)|^{1-\alpha }}{x^{1-\alpha }(1-x)^\alpha }\frac{1}{n^\alpha }, \end{aligned}$$

for \(n\in {\mathbb {N}}^\star \) and \(0\le x<1\). \(\quad \square \)

4 Spectral Study

In this section, we aim to investigate some fundamental spectral properties of the linearized operator \(\partial _f {\widetilde{F}}(\Omega ,0)\) in order to apply the Crandall–Rabinowitz theorem. For this goal one must check that the kernel and the co-image of the linearized operator are one dimensional vector spaces. Noting that the study of the kernel agrees with the eigenvalue problem of a Hilbert–Schmidt operator, we achieve that the dimension is one. Moreover, we will study the Fredholm structure of the linearized operator, which will imply that the codimension of the image is one. At the end of the section, we characterize also the transversal condition.

4.1 Symmetrization of the linearized operator

The main strategy to explore some spectral properties of the linearized operator at each frequency level n is to construct a suitable Hilbert space, basically an \(L^2\) space with respect to a special Borel measure, on which it acts as a self-adjoint compact operator. Later we investigate the eigenspace associated with the largest eigenvalue and prove in particular that this space is one-dimensional.

Let us explain how to symmetrize the operator. Recall from (3.4) that for any smooth function \(\displaystyle {h(\phi ,\theta )=\sum _{n\ge 1}h_n(\phi ) \cos (n\theta )}\), we may write the operator \({\mathcal {L}}_n\) under the form

$$\begin{aligned} {\mathcal {L}}_n^\Omega (h_n)(\phi )=\nu _{\Omega }(\phi )\left\{ h_n(\phi )-\int _0^\pi K_n(\phi ,\varphi )h_n(\varphi )d\mu _{\Omega }(\varphi )\right\} , \quad \phi \in [0,\pi ], \end{aligned}$$
(4.1)

with

$$\begin{aligned}&K_n(\phi ,\varphi ):=\frac{H_n(\phi ,\varphi )}{\sin (\varphi )\nu _{\Omega }(\phi )\nu _{\Omega }(\varphi )r_0^2(\varphi )}, \end{aligned}$$
(4.2)
$$\begin{aligned}&\nu _{\Omega }(\phi ):=\int _0^\pi H_1(\phi ,\varphi )d\varphi -\Omega , \end{aligned}$$
(4.3)

and the signed measure

$$\begin{aligned} d\mu _{\Omega }(\varphi ):=\sin (\varphi )r_0^2(\varphi )\nu _{\Omega }(\varphi )d\varphi . \end{aligned}$$
(4.4)

Define the quantity

$$\begin{aligned} \kappa :=\inf _{\phi \in [0,\pi ]}\int _0^\pi H_1(\phi ,\varphi )d\varphi . \end{aligned}$$
(4.5)

We shall discuss in Proposition 4.1 below the existence of \(\kappa \) which allows to guarantee the positivity of the measure \(d\mu _\Omega \) provided that the parameter \(\Omega \) is restricted to lie in the interval \((-\infty ,\kappa )\). We shall also study the regularity of the function \(\nu _\Omega \) which is delicate and more involved. In particular, we prove that, under reasonable assumptions on the profile \(r_0\), this function is at least in the Hölder space \({\mathscr {C}}^{1,\alpha }\) for any \(\alpha \in (0,1)\).

Notice that the kernel \(K_ n\) is symmetric. Indeed, according to (3.3) we get the formula

$$\begin{aligned} K_n(\phi ,\varphi )&= \frac{2^{2n-1}\left( \frac{1}{2}\right) ^2_{n}}{(2n)!}\frac{r_0^{n-1}(\phi )r_0^{n-1}(\varphi )}{\nu _\Omega (\phi )\nu _\Omega (\varphi )\left[ R(\phi ,\varphi )\right] ^{n+\frac{1}{2}}} F_n\left( \frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right) , \end{aligned}$$
(4.6)

which gives the desired property in view of the symmetry of R, that is, \(R(\phi ,\varphi )=R(\varphi ,\phi )\).

We shall explore in Sect. 4.3 more spectral properties of the symmetric operator associated to the kernel \(K_n\).

4.2 Regularity of \(\nu _\Omega \)

This section is devoted to the study of the regularity of the function \(\nu _{\Omega }\) that arises in (4.1), which turns to be a very delicate problem. This is needed for getting enough regularity for the kernel elements that should belong to the function spaces where bifurcation will be applicable. For lower regularities than Lipschitz class, this can be implemented in a standard way using some boundary behavior of the hypergeometric functions. However for higher regularity of type \({\mathscr {C}}^{1,\alpha }\), the problem turns out to be more delicate due to some logarithmic singularity induced by \(H_1\). To get rid of this singularity we use some specific cancellation coming from the structure of the kernel. We shall also develop the local structure of \(\nu _\Omega \) near its minimum which appears to be crucial later especially in Proposition 4.2.

The main result of this section reads as follows.

Proposition 4.1

Let \(r_0\) be a profile satisfying (H1) and (H2). Then the following properties hold true.

  1. (1)

    The function \(\phi \in [0,\pi ]\mapsto \nu _\Omega (\phi )\) belongs to \({\mathscr {C}}^\beta ([0,\pi ])\), for all \(\beta \in [0,1)\).

  2. (2)

    We have \(\kappa >0\) and for any \( \Omega \in (-\infty ,\kappa )\) we get

    $$\begin{aligned} \forall \phi \in [0,\pi ],\quad \nu _{\Omega }(\phi )\ge \kappa -\Omega >0. \end{aligned}$$
  3. (3)

    The function \(\nu _\Omega \) belongs to \({\mathscr {C}}^{1,\alpha }([0,\pi ])\), for any \(\alpha \in (0,1)\), with

    $$\begin{aligned} \nu _\Omega ^\prime (0)=\nu _\Omega ^\prime (\pi )=0. \end{aligned}$$
  4. (4)

    Let \(\Omega \in (-\infty ,\kappa ]\) and assume that \(\nu _\Omega \) reaches its minimum at a point \(\phi _0\in [0,\pi ]\) then there exists \(C>0\) independent of \(\Omega \) such that,

    $$\begin{aligned} \forall \phi \in [0,\pi ],\quad 0\le \nu _\Omega \big (\phi \big )-\nu _\Omega \big (\phi _0\big )\le C|\phi -\phi _0|^{1+\alpha }. \end{aligned}$$

    Moreover, for \(\Omega =\kappa \) this result becomes

    $$\begin{aligned} \forall \phi \in [0,\pi ],\quad 0\le \nu _\kappa \big (\phi \big )\le C|\phi -\phi _0|^{1+\alpha }. \end{aligned}$$

Proof

\({\mathbf{(1)}}\) To start, notice that according to (3.3)

$$\begin{aligned} H_1(\phi ,\varphi )&= \frac{1}{4}\frac{\sin (\varphi )r_0^{2}(\varphi )}{\left[ R(\phi ,\varphi )\right] ^{\frac{3}{2}}} F_1\left( \frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right) , \quad \forall \, \varphi \ne \phi \in (0,\pi ) \nonumber \\&= {\mathscr {K}}_1(\phi ,\varphi ){\mathscr {K}}_2(\phi ,\varphi ), \end{aligned}$$
(4.7)

where

$$\begin{aligned} {\mathscr {K}}_1(\phi ,\varphi ):=\frac{1}{4}\frac{\sin (\varphi )r_0^{2}(\varphi )}{R^{\frac{3}{2}}(\phi ,\varphi )}, \end{aligned}$$
(4.8)

and

$$\begin{aligned} {\mathscr {K}}_2(\phi ,\varphi ):=F_1\left( \frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right) . \end{aligned}$$
(4.9)

Therefore we may write

$$\begin{aligned} \forall \phi \in (0,\pi ),\quad \nu _\Omega (\phi )=\int _0^\pi {\mathscr {K}}_1(\phi ,\varphi ) {\mathscr {K}}_2(\phi ,\varphi )d\varphi -\Omega . \end{aligned}$$

Using the boundary behavior of hypergeometric functions stated in Proposition A.1 we deduce that

$$\begin{aligned} 1\le {\mathscr {K}}_2(\phi ,\varphi )&\le C+C\left| \ln \left( 1-\frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right) \right| \\&\le C+C \left| \ln \left( \frac{(r_0(\phi )+r_0(\varphi ))^2+(\cos \phi -\cos \varphi )^2}{(r_0(\phi )-r_0(\varphi ))^2+(\cos \phi -\cos \varphi )^2}\right) \right| . \end{aligned}$$

From the assumption (H2) on \(r_0\) we can write, using the mean value theorem

$$\begin{aligned} (r_0(\phi )+r_0(\varphi ))^2+(\cos \phi -\cos \varphi )^2\le C(\sin \phi +\sin \varphi )^2+ (\phi -\varphi )^2. \end{aligned}$$

In view of (2.18) we get for all \(\phi \ne \varphi \in [0,\pi ]\),

$$\begin{aligned} 1\le \frac{(r_0(\phi )+r_0(\varphi ))^2+(\cos \phi -\cos \varphi )^2}{(r_0(\phi )-r_0(\varphi ))^2+(\cos \phi -\cos \varphi )^2}\le C\frac{(\sin \phi +\sin \varphi )^2}{(\phi -\varphi )^2}+C. \end{aligned}$$
(4.10)

Consequently, we get

$$\begin{aligned} 1\le {\mathscr {K}}_2(\phi ,\varphi )\le C+C \left| \ln \left( \frac{\sin (\phi )+\sin (\varphi )}{|\phi -\varphi |}\right) \right| . \end{aligned}$$
(4.11)

On the other hand, it is obvious using the assumption (H2) on \(r_0\) that

$$\begin{aligned} 0\le {\mathscr {K}}_1(\phi ,\varphi )\le \frac{\sin (\varphi )r_0^{2}(\varphi )}{\left[ R(\phi ,\varphi )\right] ^{\frac{3}{2}}}&\le \frac{\sin \varphi }{r_0(\varphi )}\le C, \quad \forall \varphi ,\phi \in (0,\pi ). \end{aligned}$$
(4.12)

It follows that

$$\begin{aligned} \sup _{\phi \in (0,\pi )}|\nu _\Omega (\phi )|&\le C+C\sup _{\phi \in (0,\pi )}\int _0^\pi \ln \left( \frac{\sin (\phi )+\sin (\varphi )}{|\phi -\varphi |}\right) d\varphi \le C, \end{aligned}$$

which ensures that \(\nu _\Omega \) is bounded.

Now let us check the Hölder continuity by estimating

$$\begin{aligned} |\nu _\Omega (\phi _1)-\nu _\Omega (\phi _2)|&\le \int _0^\pi |{\mathscr {K}}_2(\phi _1,\varphi )||{\mathscr {K}}_1(\phi _1,\varphi )-{\mathscr {K}}_1(\phi _2,\varphi )|d\varphi \\&\quad +\int _0^\pi |{\mathscr {K}}_1(\phi _2,\varphi )||{\mathscr {K}}_2(\phi _1,\varphi )-{\mathscr {K}}_2(\phi _2,\varphi )|d\varphi \\&=: {\mathscr {H}}_1(\phi _1,\phi _2)+{\mathscr {H}}_2(\phi _1,\phi _2). \end{aligned}$$

Let us begin with \({\mathscr {H}}_1\). Notice that

$$\begin{aligned} \partial _\phi R(\phi ,\varphi )=2r_0^\prime (\phi )(r_0(\phi )+r_0(\varphi ))+2\sin \phi (\cos \varphi -\cos \phi ), \end{aligned}$$
(4.13)

which implies that

$$\begin{aligned} |\partial _\phi R(\phi ,\varphi )|\le CR^{\frac{1}{2}}(\phi ,\varphi ). \end{aligned}$$

It follows that

$$\begin{aligned} |\partial _\phi R^{-\frac{3}{2}}(\phi ,\varphi )|&\lesssim R^{-2}(\phi ,\varphi )\lesssim r_0^{-4}(\varphi ). \end{aligned}$$

Differentiating \({\mathscr {K}}_1\) with respect to \(\phi \) yields

$$\begin{aligned} 4\pi \partial _\phi {\mathscr {K}}_1(\phi ,\varphi )=\partial _\phi (R^{-\frac{3}{2}})(\phi ,\varphi )\sin \varphi \,r_0^2(\varphi ). \end{aligned}$$

Hence using (H2) we deduce that

$$\begin{aligned} \sup _{\phi \in (0,\pi )}|\partial _\phi {\mathscr {K}}_1(\phi ,\varphi )|&\le C\frac{\sin \varphi }{r_0^2(\varphi )}\lesssim \frac{1}{\sin \varphi }\cdot \end{aligned}$$
(4.14)

From an interpolation argument using the boundedness of \({\mathscr {K}}_1\) we find, according to the mean value theorem,

$$\begin{aligned} |{\mathscr {K}}_1(\phi _1,\varphi )-{\mathscr {K}}_1(\phi _2,\varphi )|&= |{\mathscr {K}}_1(\phi _1,\varphi )-{\mathscr {K}}_1(\phi _2,\varphi )|^{1-\beta }|{\mathscr {K}}_1(\phi _1,\varphi )-{\mathscr {K}}_1(\phi _2,\varphi )|^\beta \nonumber \\ \le&\big (2\Vert {\mathscr {K}}_1\Vert _{L^\infty }\big )^{1-\beta }\Vert {\mathscr {K}}_1\Vert _{\text {Lip}}^{\beta } |\phi _1-\phi _2|^\beta \nonumber \\&\lesssim \frac{|\phi _1-\phi _2|^\beta }{\sin ^\beta \varphi }\cdot \end{aligned}$$
(4.15)

Using (4.11), we obtain

$$\begin{aligned} {\mathscr {H}}_1(\phi _1,\phi _2)\le & {} C|\phi _1-\phi _2|^\beta \int _0^\pi \frac{1}{\sin ^\beta \varphi }\left\{ 1+\ln \left( \frac{\sin (\phi _1)+\sin (\varphi )}{|\phi _1-\varphi |}\right) \right\} \nonumber \\ d\varphi\le & {} C|\phi _1-\phi _2|^\beta , \end{aligned}$$
(4.16)

for any \(\beta \in (0,1)\).

Next we shall proceed to the estimate \({\mathscr {H}}_2\). From (4.12), one finds that

$$\begin{aligned} {\mathscr {H}}_2(\phi _1,\phi _2)\le C\int _0^\pi |{\mathscr {K}}_2(\phi _1,\varphi )-{\mathscr {K}}_2(\phi _2,\varphi )|d\varphi . \end{aligned}$$

We separate the last integral as follows:

where \(d=|\phi _1-\phi _2|\), \(B_{\phi }(r)=\{\varphi \in [0,\pi ]: |\varphi -\phi |<r\}\) and \(B^c_{\phi }(r)\) denotes its complement set. For the first term, \({\mathscr {H}}_{2,1}\), we simply use (4.11)

Since for \(\varphi \in B_{\phi _1}(d)\cup B_{\phi _2}(d)\) one has \(|\phi _1-\varphi |\le 2 |\phi _1-\phi _2|\) and \(|\phi _2-\varphi |\le 2 |\phi _1-\phi _2|\), then one achieves

(4.17)

for any \(\beta \in (0,1)\). For the second term of \({\mathscr {H}}_2\), we observe that for any \({\varphi \in B_{\phi _1}^c(d)\cap B_{\phi _2}^c(d)}\) one has

$$\begin{aligned} \frac{1}{2}|\phi _1-\varphi |\leqslant |\phi _2-\varphi |\leqslant 2|\phi _1-\varphi |. \end{aligned}$$
(4.18)

On the other hand, direct computations yield

$$\begin{aligned} \partial _\phi {\mathscr {K}}_2(\phi ,\varphi )=4r_0(\varphi )\frac{r_0^\prime (\phi )R(\phi ,\varphi )-r_0(\phi )\partial _\phi R(\phi ,\varphi )}{R^2(\phi ,\varphi )}F_1^\prime \left( \frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right) . \end{aligned}$$

We know that

$$\begin{aligned} \forall x\in (-1,1),\quad F_1^\prime (x)=\frac{3}{4}F(5/2,5/2;4;x), \end{aligned}$$

and by virtue of the boundary behavior stated in Proposition A.1 we get

$$\begin{aligned} \forall x\in [0,1),\quad |F_1^\prime (x)|\lesssim (1-x)^{-1}. \end{aligned}$$

It follows that, using (2.18),

$$\begin{aligned}&\forall \phi \ne \varphi \in (0,\pi )\quad \Big |F_1^\prime \left( \frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right) \Big |\nonumber \\&\quad \lesssim \frac{(r_0(\phi )+r_0(\varphi ))^2+(\cos \phi -\cos \varphi )^2}{(r_0(\phi )-r_0(\varphi ))^2+(\cos \phi -\cos \varphi )^2}\lesssim \frac{R(\phi ,\varphi )}{|\phi -\varphi |^2}\cdot \end{aligned}$$
(4.19)

By explicit calculation using (4.13) we get

$$\begin{aligned} r_0^\prime (\phi )R(\phi ,\varphi )-r_0(\phi )\partial _\phi R(\phi ,\varphi )&= r_0^\prime (\phi )\left( r_0^2(\varphi )-r_0^2(\phi )+(\cos \phi -\cos \varphi )^2\right) \\&\quad +2r_0(\phi )\sin \phi (\cos \phi -\cos \varphi ). \end{aligned}$$

Then using the mean value theorem we find

$$\begin{aligned} |r_0^\prime (\phi )R(\phi ,\varphi )-r_0(\phi )\partial _\phi R(\phi ,\varphi )|&\lesssim |\phi -\varphi |\big (r_0(\varphi )+r_0(\phi )+|\cos \phi -\cos \varphi |\big )\\&\quad +r_0(\phi )\sin \phi |\cos \phi -\cos \varphi |\\&\lesssim |\phi -\varphi | R^{\frac{1}{2}}(\phi ,\varphi ). \end{aligned}$$

Putting together the preceding estimates we find

$$\begin{aligned} |\partial _\phi {\mathscr {K}}_2(\phi ,\varphi )|&\lesssim r_0(\varphi )|\phi -\varphi | R^{-\frac{3}{2}}(\phi ,\varphi )\Big |F_1^\prime \left( \frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right) \Big |\lesssim |\phi -\varphi |^{-1}. \end{aligned}$$

Then applying once again the mean value theorem, we get for any \({\varphi \in B_{\phi _1}^c(d)\cap B_{\phi _2}^c(d)}\) a value \(\phi \in (\phi _1,\phi _2)\) such that

$$\begin{aligned} |{\mathscr {K}}_2(\phi _1,\varphi )-{\mathscr {K}}_2(\phi _2,\varphi )|&\lesssim {|\phi _1-\phi _2|}{|\phi -\varphi |^{-1}}\\&\lesssim {|\phi _1-\phi _2|}{|\phi _1-\varphi |^{-1}}. \end{aligned}$$

Combining this estimate with (4.11) and (4.18), and using an interpolation argument we get for any \(\varepsilon >0\),

$$\begin{aligned} |{\mathscr {K}}_2(\phi _1,\varphi )-{\mathscr {K}}_2(\phi _2,\varphi )|&\lesssim |\phi _1-\phi _2|^\beta {|\phi _1-\varphi |^{-\beta }}\nonumber \\&\times \left( C+\left| \ln \left( \frac{\sin \phi _1+\sin \varphi }{|\phi _1-\varphi |}\right) \right| +\left| \ln \left( \frac{\sin \phi _2+\sin \varphi }{|\phi _2-\varphi |}\right) \right| \right) ^{1-\beta }\nonumber \\&\lesssim |\phi _1-\phi _2|^\beta \left( 1+\big |\ln (\sin \varphi )\big |^{1-\beta }\right) \left( {|\phi _1-\varphi |^{-\beta -\varepsilon }}\right) . \end{aligned}$$
(4.20)

Hence, we find that

(4.21)

for any \(\beta \in (0,1)\). Finally, putting together (4.17)–(4.21) we get

$$\begin{aligned} {\mathscr {H}}_2(\phi _1,\phi _2)\le C|\phi _1-\phi _2|^\beta , \end{aligned}$$

for any \(\beta \in (0,1)\). Then, we find that \(\nu _\Omega \in {\mathscr {C}}^\beta \), for any \(\beta \in (0,1)\).

\({\mathbf{(2)}}\) The function \(\phi \mapsto \Omega +\nu _\Omega (\phi )\) is continuous over the compact set \([0,\pi ]\) then it reaches its minimum at some point \(\phi _0\in [0,\pi ]\). Thus from the definition of \(\kappa \) in (4.5) we deduce that

$$\begin{aligned} \kappa =\inf _{\phi \in (0,\pi )}\int _0^\pi H_1(\phi ,\varphi )d\varphi =\int _0^\pi H_1(\phi _0,\varphi )d\varphi >0, \end{aligned}$$

which implies that

$$\begin{aligned} \forall \,\phi \in [0,\pi ],\quad \nu _\Omega (\phi )&\ge \int _0^\pi H_1(\phi _0,\varphi )d\varphi -\Omega \ge \kappa -\Omega . \end{aligned}$$

Hence we infer that for any \(\Omega \in (-\infty ,\kappa )\)

$$\begin{aligned} \forall \,\phi \in [0,\pi ],\quad \nu _\Omega (\phi )\ge \kappa -\Omega >0. \end{aligned}$$

\({\mathbf{(3)}}\) The proof is long and technical and for clarity of presentation it will be divided into two steps. In the first one we prove that \(\nu _\Omega \) is \({\mathscr {C}}^1\) in the full closed interval \([0,\pi ]\). This is mainly based on two principal ingredients. The first one is an important algebraic cancellation in the integrals allowing us to get rid of the logarithmic singularity coming from the boundary and the second one is the boundary behavior of the hypergeometric functions allowing us to deal with the diagonal singularity lying inside the domain of integration. Notice that in order to apply Lebesgue theorem and recover the continuity of the derivative up to the boundary we use a rescaling argument. This rescaling argument shows in addition a surprising effect concerning the derivative at the boundary points \(\nu _\Omega ^\prime (0)\) and \(\nu _\Omega ^\prime (\pi )\): they are independent of the global structure of the profile \(r_0\) and they depend only on the derivative \(r_0^\prime (0)\). This property allows us to compute \(\nu _\Omega '(0)\) in the special case of \(r_0(\phi )=\sin (\phi )\) by using the special geometry of the sphere and observe that this derivative vanishes. As to the second step, it is devoted to the proof of \(\nu _\Omega ^\prime \in {\mathscr {C}}^\alpha (0,\pi )\) which is involved and requires more refined analysis.

\(\bullet \) Step 1: \(\nu _\Omega \in {\mathscr {C}}^1([0,\pi ])\). The first step is to check that \(\nu _\Omega \) is \({\mathscr {C}}^1\) on \([0,\pi ]\). Define

$$\begin{aligned} \varrho (\phi ,\varphi ):=\frac{4 r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}, \end{aligned}$$

then we can check that

$$\begin{aligned}&\partial _\phi H_1(\phi ,\varphi )\\&\quad = {\mathscr {K}}_1(\phi ,\varphi )\left( -\frac{3}{2} R^{-1}(\phi ,\varphi ) \partial _\phi R(\phi ,\varphi )F_1(\varrho (\phi ,\varphi ))+F_1^\prime (\varrho (\phi ,\varphi ))\partial _\phi \varrho (\phi ,\varrho )\right) , \end{aligned}$$

which implies after simple manipulations that

$$\begin{aligned} \partial _\phi H_1&= {\mathscr {K}}_1\left( -\frac{3}{2} \frac{\partial _\phi R}{R}+\frac{3}{4}\partial _\phi \rho -\frac{3}{2}\frac{\partial _\phi R}{R}\big (F_1(\rho )-1\big )+\big (F_1^\prime (\rho )-3/4\big )\big [\partial _\phi \rho +\partial _\varphi \rho \big ]\right) \\&\quad -{\mathscr {K}}_1\big (F_1^\prime (\rho )-3/4\big )\partial _\varphi \rho . \end{aligned}$$

In addition using the identity

$$\begin{aligned} {\mathscr {K}}_1\big (F_1^\prime (\rho )-3/4\big )\partial _\varphi \rho&= {\mathscr {K}}_1\partial _\varphi \big [F(\rho )-3/4\rho -1\big ]\\&= \partial _\varphi \left( {\mathscr {K}}_1\big [F(\rho )-3/4\rho -1\big ]\right) -(\partial _\varphi {\mathscr {K}}_1)\big [F(\rho )-3/4\rho -1\big ], \end{aligned}$$

we find

$$\begin{aligned} \partial _\phi H_1=\varkappa _0+\varkappa _1\big (F_1(\varrho )-1\big )+\varkappa _2\big (F_1^\prime (\varrho )-3/4\big )-\partial _\varphi \left( {\mathscr {K}}_1\big [F(\rho )-3/4\rho -1\big ]\right) , \end{aligned}$$

with

$$\begin{aligned} \varkappa _0&:={\mathscr {K}}_1\left( -\frac{3}{2} \frac{\partial _\phi R}{R}+\frac{3}{4}\partial _\phi \rho \right) -\frac{3}{4}\rho \partial _\varphi {\mathscr {K}}_1,\nonumber \\ \varkappa _1&:=-\frac{3}{2}\frac{\partial _\phi R}{R}{\mathscr {K}}_1+\partial _\varphi {\mathscr {K}}_1,\nonumber \\ \varkappa _2&:={\mathscr {K}}_1\left( \partial _\phi \rho +\partial _\varphi \rho \right) . \end{aligned}$$
(4.22)

Notice that \(F_1(0)=1,\,F_1^\prime (0)=\frac{3}{4}\). Assuming that the following functions are well-defined and using the boundary conditions then we can write

$$\begin{aligned} \nu _\Omega ^\prime (\phi )&= \int _0^\pi \Big (\varkappa _0(\phi ,\varphi )+\varkappa _1(\phi ,\varphi )\big [F_1(\varrho (\phi ,\varphi ))-1\big ]\nonumber \\&\quad +\varkappa _2(\phi ,\varphi )\big [F_1^\prime (\varrho (\phi ,\varphi ))-3/4\big ]\Big )d\varphi \nonumber \\&:=\zeta _1(\phi )+\zeta _2(\phi )+\zeta _3(\phi ), \end{aligned}$$
(4.23)

with

$$\begin{aligned} \zeta _1(\phi ):=\int _0^\pi \varkappa _0(\phi ,\varphi )d\varphi ,\quad \zeta _2(\phi ):=\int _0^\pi \varkappa _1(\phi ,\varphi )\big [F_1(\varrho (\phi ,\varphi ))-1\big ]d\varphi \end{aligned}$$

and

$$\begin{aligned} \zeta _3(\phi )=\int _0^{\pi } \varkappa _2(\phi ,\varphi )\big [F_1^\prime (\varrho (\phi ,\varphi ))-3/4\big ]d\varphi . \end{aligned}$$

Direct computations show that

$$\begin{aligned} \partial _\phi \rho (\phi ,\varphi )&= \frac{4 r_0(\varphi )r_0^\prime (\phi )}{R^2(\phi ,\varphi )}\Big (R(\phi ,\varphi )-2r_0(\phi )\big (r_0(\varphi )+r_0(\phi )\big )\Big )\\&\quad -\frac{8 r_0(\phi )r_0(\varphi )}{R^2(\phi ,\varphi )}\sin \phi (\cos \varphi -\cos \phi ). \end{aligned}$$

According to (4.13) and using some cancellation, it implies that

$$\begin{aligned} \varkappa _0(\phi ,\varphi )&= -3r_0^\prime (\phi )\frac{r_0(\phi ){\mathscr {K}}_1(\phi ,\varphi )}{R(\phi ,\varphi )}\left( 1+2\frac{r_0(\varphi )(r_0(\phi )+r_0(\varphi ))}{R(\phi ,\varphi )}\right) \nonumber \\&\quad +6\frac{{\mathscr {K}}_1(\phi ,\varphi )}{R^2(\phi ,\varphi )}\sin (\phi )\, r_0(\phi )\,r_0(\varphi )\big (\cos \phi -\cos \varphi \big )\nonumber \\&\quad +3\frac{{\mathscr {K}}_1(\phi ,\varphi )}{R(\phi ,\varphi )}\sin (\phi )\big (\cos \phi -\cos \varphi \big ) -3\partial _\varphi {\mathscr {K}}_1\frac{r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}. \end{aligned}$$
(4.24)

We point out that this simplification is crucial and allows to get rid of the logarithmic singularity.

Now we shall start with the regularity of the function

$$\begin{aligned} \zeta _1:\phi \in (0,\pi )\mapsto \int _0^{\pi }\varkappa _0(\phi ,\varphi )d\varphi , \end{aligned}$$

and prove first that it is continuous in \([0,\pi ]\). It is obvious from (4.24) that \(\varkappa _0\) is \({\mathscr {C}}^1\) over any compact set contained in \((0,\pi )\times [0,\pi ]\) and therefore \(\zeta _1\) is \({\mathscr {C}}^1\) over any compact set contained in \((0,\pi )\). Thus it remains to check that this function is continuous at the points 0 and \(\pi \). The proofs for both cases are quite similar and we shall only check the continuity at the origin. For this purpose it is enough to check that \(\zeta _1\) admits a limit at zero. Before that let us check that \(\zeta _1\) is bounded in \((0,\pi )\). From the definition of R stated in (3.2) and using elementary inequalities it is easy to verify the following estimates: for any \((\phi ,\varphi )\in (0,\pi )^2\)

$$\begin{aligned} \frac{r_0(\phi )(r_0(\phi )+r_0(\varphi ))}{R(\phi ,\varphi )}&\le 1,\\ \frac{r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}&\le {\frac{1}{2}},\\ r_0(\phi )r_0(\varphi )\big |\cos \phi -\cos \varphi \big |&\le R(\phi ,\varphi ). \end{aligned}$$

In addition, the assumption (H2) implies that

$$\begin{aligned} \sup _{\phi ,\varphi \in (0,\pi )}{\mathscr {K}}_1(\phi ,\varphi )<\infty . \end{aligned}$$

Thus we find according to (4.14) and (H2)

$$\begin{aligned} \forall (\phi ,\varphi )\in (0,\pi )^2,\,|\varkappa _0(\phi ,\varphi )|&\lesssim \frac{\sin (\phi )}{R(\phi ,\varphi )}+|\partial _\varphi {\mathscr {K}}_1(\phi ,\varphi )|\frac{r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\lesssim \frac{\sin (\phi )}{R(\phi ,\varphi )}\cdot \end{aligned}$$
(4.25)

Hence we deduce that

$$\begin{aligned} \forall \phi \in (0,\pi ),\quad |\zeta _1(\phi )|&\lesssim \int _0^\pi \frac{\sin (\phi )}{R(\phi ,\varphi )} d\varphi \\&\lesssim \int _0^{\frac{\pi }{2}} \frac{\sin \phi }{(\sin \phi +\sin \varphi )^2} d\varphi . \end{aligned}$$

Making the change of variables \(\sin \varphi =x\) we get

$$\begin{aligned} \int _0^{\frac{\pi }{2}} \frac{\sin \phi }{(\sin \phi +\sin \varphi )^2} d\varphi&= \sin \phi \int _0^1\frac{1}{(\sin \phi +x)^2}\frac{dx}{\sqrt{1-x^2}}\\&\lesssim \sin \phi \int _0^{\frac{1}{2}}\frac{1}{(\sin \phi +x)^2}{dx}+\sin \phi \lesssim 1. \end{aligned}$$

Thus

$$\begin{aligned} \sup _{\phi \in (0,\pi )}|\zeta _1(\phi )|<\infty .\end{aligned}$$
(4.26)

Let us now prove that \(\zeta _1\) admits a limit at the origin and compute its value. For this goal, take \(0<\delta<<1\) small enough and write

$$\begin{aligned} \zeta _1(\phi )=\int _0^{\delta }\varkappa _0(\phi ,\varphi )d\varphi +\int _{\delta }^\pi \varkappa _0(\phi ,\varphi )d\varphi . \end{aligned}$$

The assumption \(\mathbf{(H2) }\) combined with standard trigonometric formula allow to get the estimate

$$\begin{aligned} R(\phi ,\varphi )&\gtrsim (\sin \phi +\sin \varphi )^2+(\cos \phi -\cos \varphi )^2=2\big (1-\cos (\phi +\varphi )\big )\nonumber \\&\gtrsim 1-\cos (\phi +\varphi ). \end{aligned}$$
(4.27)

From this we infer that

$$\begin{aligned} \forall \,\phi \in [0,\pi /2],\forall \varphi \in (\delta ,\pi ),\quad R(\phi ,\varphi )\gtrsim 1-\cos \delta . \end{aligned}$$
(4.28)

Thus we get from (4.25)

$$\begin{aligned} \left| \int _{\delta }^\pi \varkappa _0(\phi ,\varphi )d\varphi \right|&\lesssim \int _\delta ^\pi \frac{\phi }{(1-\cos \delta )}d\varphi \lesssim \phi \delta ^{-2}. \end{aligned}$$

This implies that for given small parameter \(\delta \) one has

$$\begin{aligned} \lim _{\phi \rightarrow 0}\int _{\delta }^\pi \varkappa _0(\phi ,\varphi )d\varphi =0. \end{aligned}$$

Therefore

$$\begin{aligned} \lim _{\phi \rightarrow 0}\zeta _1(\phi )=\lim _{\phi \rightarrow 0}\int _0^{\delta }\varkappa _0(\phi ,\varphi )d\varphi . \end{aligned}$$

Making the change of variables \(\varphi =\phi \theta \) we get

$$\begin{aligned} \int _0^{\delta }\varkappa _0(\phi ,\varphi )d\varphi =\int _0^{\frac{\delta }{\phi }}\phi \varkappa _0(\phi ,\phi \theta )\,d\theta . \end{aligned}$$

From (4.25) and (H2) one may write

$$\begin{aligned} \forall (\phi ,\varphi )\in (0,\delta )^2,\quad |\varkappa _0(\phi ,\varphi )|\lesssim \frac{\phi }{(\phi +\varphi )^2}, \end{aligned}$$

which yields after simplification to the uniform bound on \(\phi \),

$$\begin{aligned} \forall \theta \in [0,\delta /\phi ],\quad \phi \varkappa _0(\phi ,\phi \theta )\lesssim \frac{1}{(1+\theta )^2}\cdot \end{aligned}$$

This gives a domination which is integrable over \((0,+\infty )\). In order to apply classical dominated Lebesgue theorem, it remains to check the convergence almost everywhere in \(\theta \) as \(\phi \) goes to zero. This can be done through the first-order Taylor expansion around zero. First one has the expansion

$$\begin{aligned} r_0(\phi \theta )=c_0\phi \theta +\phi \theta \epsilon (\phi \theta );\quad R(\phi , \phi \theta )=c_0^2\phi ^2\big (1+\theta +\theta \epsilon (\phi \theta )\big )^2, \end{aligned}$$

with \(c_0=r_0^\prime (0)\) and \(\displaystyle {\lim _{x\rightarrow 0}\epsilon (x)=0.}\) Thus, from the definitions (3.2) and (4.8) it is straightforward that

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}{\mathscr {K}}_1(\phi ,\phi \theta )=\frac{{{c_0^{-1}}}\theta ^3}{(1+\theta )^3},\quad \lim _{\phi \rightarrow 0}\frac{\phi r_0(\phi )}{R(\phi ,\phi \theta )}=\frac{c_0^{-1}}{(1+\theta )^2}\cdot \end{aligned}$$
(4.29)

Hence

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0} r_0^\prime (\phi )\frac{\phi r_0(\phi ){\mathscr {K}}_1(\phi ,\phi \theta )}{R(\phi ,\phi \theta )} \left( 1+2\frac{r_0(\phi \theta )(r_0(\phi )+r_0(\phi \theta ))}{R(\phi ,\phi \theta )}\right) =\frac{{{c_0^{-1}}}\theta ^3(1+3\theta )}{(1+\theta )^6}\cdot \nonumber \\ \end{aligned}$$
(4.30)

Similarly we get

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}\frac{{\mathscr {K}}_1(\phi ,\phi \theta )}{R^2(\phi ,\phi \theta )}\sin \phi \, r_0(\phi )r_0(\phi \theta )\big (\cos \phi -\cos (\phi \theta )\big )=0, \end{aligned}$$
(4.31)

and

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}\frac{{\mathscr {K}}_1(\phi ,\phi \theta )}{R(\phi ,\phi \theta )}\sin (\phi )\big (\cos \phi -\cos (\phi \theta )\big )=0. \end{aligned}$$
(4.32)

Standard computations yield

$$\begin{aligned}&4\pi \partial _\varphi {\mathscr {K}}_1(\phi ,\varphi )\nonumber \\&\quad = - 3 R^{-\frac{5}{2}}(\phi ,\varphi )\sin (\varphi )\, r_0^2(\varphi )\Big (r_0^\prime (\varphi )(r_0(\phi )+r_0(\varphi ))+\sin (\varphi )\,(\cos \phi -\cos \varphi )\Big )\nonumber \\&\qquad + \frac{\cos (\varphi )\,r_0^2(\varphi )+2\sin (\varphi ) r_0(\varphi )\, r_0^\prime (\varphi )}{R^\frac{3}{2}(\phi ,\varphi )}. \end{aligned}$$
(4.33)

Thus

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}\phi \partial _\varphi {\mathscr {K}}_1(\phi ,\phi \theta )&= c_0^{-1}\left( -3\frac{\theta ^3}{(1+\theta )^4}+{\frac{3\theta ^2}{(1+\theta )^3}}\right) ={3 c_0^{-1}\frac{\theta ^2}{(1+\theta )^4}}. \end{aligned}$$
(4.34)

Therefore

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}\phi \partial _\varphi {\mathscr {K}}_1(\phi ,\phi \theta )\frac{r_0(\phi )r_0({\phi \theta })}{R(\phi ,\phi \theta )}= {3c_0^{-1}\frac{\theta ^3}{(1+\theta )^6}}\cdot \end{aligned}$$
(4.35)

Plugging (4.30), (4.31), (4.32) and (4.35) into (4.24)

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}\phi \varkappa _0(\phi ,\phi \theta )&= -\frac{3 c_0^{-1}\theta ^3}{(1+\theta )^6}\Big ({4}+3\theta \Big ). \end{aligned}$$

Using Lebesgue dominated theorem we deduce that

Computing the integrals we finally get

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}\zeta _1(\phi )=-{\frac{4}{5}} c_0^{-1}. \end{aligned}$$
(4.36)

Let us now move to the regularity of the function \(\zeta _2\) defined in (4.23) through

$$\begin{aligned} \phi \in (0,\pi ),\quad \zeta _2(\phi )=\int _0^\pi \varkappa _1(\phi ,\varphi )\big (F_1(\rho (\phi ,\varphi ))-1\big )d\varphi , \end{aligned}$$

where \(\varkappa _1\) is defined in (4.22). From direct computations using \(|\partial _\phi R|\lesssim R^\frac{1}{2}\), the boundedness of \({\mathscr {K}}_1\), the assumption (H2) and (4.33) one can check that

$$\begin{aligned} |\varkappa _1(\phi ,\varphi )|&\lesssim \frac{|\partial _\phi R(\phi ,\varphi )|{\mathscr {K}}_1(\phi ,\varphi )}{R(\phi ,\varphi )}+|\partial _\varphi {\mathscr {K}}_1(\phi ,\varphi )|\\&\lesssim {R^{-\frac{1}{2}}(\phi ,\varphi )}. \end{aligned}$$

Using Proposition A.1 we get

$$\begin{aligned} |F_1(\varrho (\phi ,\varphi ))-1|&\lesssim \varrho (\phi ,\varphi )\left( 1+|\ln (1-\varrho (\phi ,\varphi ))|\right) \nonumber \\&\lesssim \frac{r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )} \left( 1+\ln \left( \frac{(r_0(\phi )+r_0(\varphi ))^2+(\cos (\phi )-\cos (\varphi ))^2}{(r_0(\phi )-r_0(\varphi ))^2+(\cos (\phi )-\cos (\varphi ))^2}\right) \right) . \end{aligned}$$
(4.37)

Thus

$$\begin{aligned}&|\varkappa _1(\phi ,\varphi )[F_1(\varrho (\phi ,\varphi ))-1]|\\&\quad \lesssim \frac{r_0(\phi )r_0(\varphi )}{R^{\frac{3}{2}}(\phi ,\varphi )} \left( 1+\ln \left( \frac{(r_0(\phi )+r_0(\varphi ))^2+(\cos (\phi )-\cos (\varphi ))^2}{(r_0(\phi )-r_0(\varphi ))^2+(\cos (\phi )-\cos (\varphi ))^2}\right) \right) . \end{aligned}$$

Hence using the arc-chord property (2.18) we find

$$\begin{aligned} |\varkappa _1(\phi ,\varphi )[F_1(\varrho (\phi ,\varphi ))-1]|\lesssim \frac{r_0(\phi )r_0(\varphi )}{R^{\frac{3}{2}}(\phi ,\varphi )} \left( C+\ln \left( \frac{R(\phi ,\varphi )}{|\phi -\varphi |^2}\right) \right) . \end{aligned}$$
(4.38)

for some constant \(C>0\). In addition, using (4.27) we get

$$\begin{aligned} \inf _{\begin{array}{c} \phi \in [0,{\frac{\pi }{2}}],\\ \varphi \in [{\frac{\pi }{2}},\pi ] \end{array}}R(\phi ,\varphi )> 0, \end{aligned}$$
(4.39)

which leads to

$$\begin{aligned} \forall \,\phi \in [0,{\pi /2}],\,\varphi \in [{\pi /2},\pi ],\quad |\varkappa _1(\phi ,\varphi )[F_1(\varrho (\phi ,\varphi ))-1]|\lesssim 1+|\ln |\phi -\varphi ||. \end{aligned}$$

This implies that

$$\begin{aligned}&\sup _{\phi \in [0,\pi /2]}\int _{\frac{\pi }{2}}^\pi |\varkappa _1(\phi ,\varphi )[F_1(\varrho (\phi ,\varphi ))-1]|d\varphi \nonumber \\&\quad \lesssim 1+\sup _{\phi \in [0,\pi /2]}\int _0^\pi |\ln |\phi -\varphi ||d\varphi <\infty . \end{aligned}$$
(4.40)

Now in the region \(\phi ,\varphi \in [0,\pi /2], \) we use the estimate (H2) leading to

$$\begin{aligned} (\phi +\varphi )^2\lesssim R(\phi ,\varphi )\lesssim {(\phi +\varphi )^2}. \end{aligned}$$

Plugging this into (4.38) we find

Making the change of variables \(\varphi =\phi \theta \) we obtain

which implies that

$$\begin{aligned} \sup _{\phi \in [0,\pi /2]}\int _0^{\frac{\pi }{2}}|\varkappa _1(\phi ,\varphi )[F_1(\varrho (\phi ,\varphi ))-1]|d\varphi <\infty . \end{aligned}$$

Therefore we obtain by virtue of (4.40)

$$\begin{aligned} \sup _{\phi \in [0,\pi /2]}\int _0^{\pi }|\varkappa _1(\phi ,\varphi )[F_1(\varrho (\phi ,\varphi ))-1]|d\varphi <\infty . \end{aligned}$$

By symmetry we get similar estimate for \(\phi \in [\frac{\pi }{2},\pi ]\) and hence

$$\begin{aligned} \sup _{\phi \in (0,\pi )}|\zeta _2(\phi )|<\infty .\end{aligned}$$
(4.41)

Let us now calculate the limit when \(\phi \) goes to 0 of \(\zeta _2\). We shall proceed in a similar way to \(\zeta _1\). Let \(0<\delta \ll 1\) be enough small, then using (4.38) combined with (4.28) we obtain

$$\begin{aligned} \lim _{\phi \rightarrow 0}\int _\delta ^\pi |\varkappa _1(\phi ,\varphi )||F_1(\rho (\phi ,\varphi ))-1|d\varphi =0. \end{aligned}$$

Hence

$$\begin{aligned} \lim _{\phi \rightarrow 0}\zeta _2(\phi )=\lim _{\phi \rightarrow 0}\int _0^\delta \varkappa _1(\phi ,\varphi )\big (F_1(\rho (\phi ,\varphi ))-1\big )d\varphi . \end{aligned}$$

Now we make the change of variables \(\varphi =\phi \theta \) and then

$$\begin{aligned} \lim _{\phi \rightarrow 0}\zeta _2(\phi )=\lim _{\phi \rightarrow 0}\int _0^{\frac{\delta }{\phi }}\phi \varkappa _1(\phi ,\phi \theta )\big (F_1(\rho (\phi ,\phi \theta ))-1\big ){d\theta }. \end{aligned}$$

According to (4.22) one has

$$\begin{aligned} \phi \varkappa _1(\phi ,\phi \theta )=-\frac{3}{2}\frac{\phi \partial _\phi R(\phi ,\phi \theta )}{R(\phi ,\phi \theta )}{\mathscr {K}}_1(\phi ,\phi \theta )+\phi \partial _\varphi {\mathscr {K}}_1(\phi ,\phi \theta ). \end{aligned}$$

From the differentiating of the expression of R stated in (3.2) we get

$$\begin{aligned} \frac{\phi \partial _\phi R(\phi ,\phi \theta )}{R(\phi ,\phi \theta )}&= 2\frac{\phi r_0^\prime (\phi )(r_0(\phi )+r_0(\phi \theta ))+ \phi \sin \phi (\cos (\phi \theta )-\cos \phi )}{(r_0(\phi )+ r_0(\phi \theta ))^2 +(\cos \phi -\cos (\phi \theta ))^2}\cdot \end{aligned}$$

Taking Taylor expansion to first order we deduce the pointwise convergence,

$$\begin{aligned} \lim _{\phi \rightarrow 0}\frac{\phi \partial _\phi R(\phi ,\phi \theta )}{R(\phi ,\phi \theta )}&= \frac{2}{1+\theta }\cdot \end{aligned}$$

Combined with (4.29) it implies that

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}-\frac{3}{2}\frac{\phi \partial _\phi R(\phi ,\phi \theta )}{R(\phi ,\phi \theta )}{\mathscr {K}}_1(\phi ,\phi \theta )=-\frac{3{c_0^{-1}}\theta ^3}{(1+\theta )^4}\cdot \end{aligned}$$

Plugging (4.34) and the preceding estimates into the expression of \(\varkappa _1\) given by (4.22) we find

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}\phi \varkappa _1(\phi ,\phi \theta )&= -\frac{3{c_0^{-1}}\theta ^3}{(1+\theta )^4}+{3c_0^{-1}\frac{\theta ^2}{(1+\theta )^4}}\\&= {3c_0^{-1}\frac{\theta ^2(1-\theta )}{(1+\theta )^4}}\cdot \end{aligned}$$

From the result

$$\begin{aligned} \lim _{\phi \rightarrow 0} \frac{r_0(\phi )r_0(\phi \theta )}{R(\phi ,\phi \theta )}=\frac{\theta }{(1+\theta )^2}, \end{aligned}$$

we deduce the point-wise convergence

$$\begin{aligned} \lim _{\phi \rightarrow 0}F_1(\rho (\phi ,\phi \theta ))=F_1\left( \frac{4\theta }{(1+\theta )^2}\right) . \end{aligned}$$

Consequently,

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}\phi \varkappa _1(\phi ,\phi \theta )\Big [F_1(\rho (\phi ,\phi \theta ))-1\Big ]= {3c_0^{-1}\frac{\theta ^2(1-\theta )}{(1+\theta )^4}} \left( F_1\left( \frac{4\theta }{(1+\theta )^2}\right) -1\right) \cdot \end{aligned}$$

Therefore,

(4.42)

Note that we can apply the dominated convergence theorem in the previous integral since

$$\begin{aligned} |\phi \varkappa _1(\phi ,\phi \theta )[F_1(\rho (\phi ,\phi \theta ))-1]|\le C\frac{\theta }{(1+\theta )^3}\left( 1+\ln \left( \frac{1+\theta }{|1-\theta |}\right) \right) , \end{aligned}$$

which is integrable. Next, we shall implement a similar study for \(\zeta _3\) defined in (4.23). Straightforward computations yield

$$\begin{aligned} \partial _\phi \varrho (\phi ,\varphi )+\partial _\varphi \varrho (\phi ,\varphi )=\varrho _1(\phi ,\varphi )+\varrho _2(\phi ,\varphi ), \end{aligned}$$
(4.43)

with

$$\begin{aligned} \varrho _1(\phi ,\varphi ):=4\frac{r_0^2(\varphi )-r_0^2(\phi )}{R^2(\phi ,\varphi )}r_0^\prime (\phi )\big (r_0(\varphi )-r_0(\phi )\big ), \end{aligned}$$
(4.44)

and

$$\begin{aligned} \varrho _2(\phi ,\varphi )&:=4\frac{r_0^2(\varphi )-r_0^2(\phi )}{R^2(\phi ,\varphi )}r_0(\phi )\Big (r_0^\prime (\phi )-r_0^\prime (\varphi )\Big )\nonumber \\&\quad +8\frac{r_0(\varphi )r_0(\phi )}{R^2(\phi ,\varphi )}\big (\cos \phi -\cos \varphi )\big (\sin \phi -\sin \varphi \big )\nonumber \\&\quad +4\frac{(\cos \phi -\cos \varphi )^2}{R^2(\phi ,\varphi )}\Big (r_0(\varphi )r_0^\prime (\phi )+r_0(\phi )r_0^\prime (\varphi )\Big ).\end{aligned}$$
(4.45)

Since \(r_0^\prime \) is Lipschitz then using the mean value theorem we get

$$\begin{aligned} \forall \,\phi ,\varphi \in (0,\pi ),\quad |\varrho _1(\phi ,\varphi )|+|\varrho _2(\phi ,\varphi )|\lesssim \frac{(\phi -\varphi )^2}{R^{\frac{3}{2}}(\phi ,\varphi )}\cdot \end{aligned}$$
(4.46)

From Proposition A.1 combined with (2.18) we get

$$\begin{aligned} |F_1^\prime (\varrho (\phi ,\varphi ))-3/4|&= \frac{3}{4}|F(5/2,5/2;4;\rho (\phi ,\varphi ))-1|\nonumber \\&\lesssim \frac{\varrho (\phi ,\varphi ) R(\phi ,\varphi )}{(r_0(\phi )-r_0(\varphi )^2+(\cos \phi -\cos \varphi )^2}\nonumber \\&\lesssim \frac{r_0(\phi ) r_0(\varphi )}{(\phi -\varphi )^2}. \end{aligned}$$
(4.47)

In addition

$$\begin{aligned} |\varkappa _2(\phi ,\varphi ))|&= |{\mathscr {K}}_1(\phi ,\varphi )||\varrho _1(\phi ,\varphi )+\varrho _2(\phi ,\varphi )|\\&\lesssim \frac{\sin \varphi \, r_0^2(\varphi )}{R^{\frac{3}{2}}(\phi ,\varphi )}\frac{(\phi -\varphi )^2}{R^{\frac{3}{2}}(\phi ,\varphi )}\\&\lesssim \frac{\sin \varphi \, (\varphi -\phi )^2}{R^{2}(\phi ,\varphi )}\cdot \end{aligned}$$

Consequently, we obtain in view of (H2)

$$\begin{aligned} |\varkappa _2(\phi ,\varphi )[F_1^\prime (\varrho (\phi ,\varphi ))-3/4]|&\lesssim \frac{\sin (\varphi )r_0(\varphi )r_0(\phi )}{R^{2}(\phi ,\varphi )}\lesssim \frac{ \sin \phi }{R(\phi ,\varphi )}\cdot \end{aligned}$$
(4.48)

As before, we can assume without any loss of generality that \(\phi \in [0,\,\pi /2]\), then by (H2)

$$\begin{aligned} \int _0^\pi |\varkappa _2(\phi ,\varphi )[F_1^\prime (\varrho (\phi ,\varphi ))-3/4]|d\varphi&\lesssim \int _0^{\frac{\pi }{2}}\frac{ \sin \phi }{(\sin \phi +\sin \varphi )^2}d\varphi \\&\lesssim \int _0^{\frac{\pi }{2}}\frac{\phi }{\big (\phi +\varphi \big )^2}d\varphi . \end{aligned}$$

By the change of variables \(\varphi =\phi \theta \) we get

$$\begin{aligned} \int _0^{\frac{\pi }{2}}\frac{\phi }{\big (\phi +\varphi \big )^2}d\varphi&= \int _0^{\frac{\pi }{2\phi }}\frac{1}{\big (1+\theta \big )^2}d\theta \le \int _0^{+\infty }\frac{1}{\big (1+\theta \big )^2}d\theta <\infty . \end{aligned}$$

Therefore

$$\begin{aligned} \sup _{\phi \in [0,\pi /2]}\int _0^\pi |\varkappa _2(\phi ,\varphi )[F_1^\prime (\varrho (\phi ,\varphi ))-3/4]|d\varphi <\infty . \end{aligned}$$

Consequently

$$\begin{aligned} \sup _{\phi \in [0,\pi ]}|\zeta _3(\phi )|<\infty . \end{aligned}$$
(4.49)

Now, we shall calculate the limit of \(\zeta _3\) at the origin. Let \(0<\delta \ll 1\) be enough small, then using (4.48) combined with (4.28) and (H2) we obtain

$$\begin{aligned} \lim _{\phi \rightarrow 0}\int _\delta ^\pi \varkappa _2(\phi ,\varphi ) [F_1^\prime (\varrho (\phi ,\varphi ))-3/4]|d\varphi =0. \end{aligned}$$

It follows that

$$\begin{aligned} \lim _{\phi \rightarrow 0}\zeta _3(\phi )=\lim _{\phi \rightarrow 0}\int _0^\delta \varkappa _2(\phi ,\varphi ) [F_1^\prime (\varrho (\phi ,\varphi ))-3/4]|d\varphi . \end{aligned}$$

Making the change of variables \(\varphi =\phi \theta \) yields

$$\begin{aligned} \lim _{\phi \rightarrow 0}\zeta _3(\phi )=\lim _{\phi \rightarrow 0}\int _0^{\frac{\delta }{\phi }}\phi \varkappa _2(\phi ,\phi \theta )\big [F_1^\prime (\rho (\phi ,\phi \theta ))-3/4\big ]{d\theta }. \end{aligned}$$

Using Taylor expansion to first order one in (4.44) and (4.45) we can check that

$$\begin{aligned} \lim _{\phi \rightarrow 0}\phi \varrho _1(\phi ,\phi \theta )=4\frac{(\theta -1)^2}{(1+\theta )^3},\quad \lim _{\phi \rightarrow 0}\phi \varrho _2(\phi ,\phi \theta )=0. \end{aligned}$$

Hence we get in view of the definition of \(\varkappa _2\) and (4.29) the point-wise limit

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}\phi \varkappa _2(\phi ,\phi \theta )&= 4\pi \lim _{\phi \rightarrow 0}\phi {\mathscr {K}}_1(\phi ,\phi \theta )\Big (\varrho _1(\phi ,\phi \theta )+\varrho _2(\phi ,\phi \theta )\Big )=4{{c_0^{-1}}}\frac{\theta ^3(\theta -1)^2}{(1+\theta )^6}\cdot \end{aligned}$$

It follows that

$$\begin{aligned} 4\pi \lim _{\phi \rightarrow 0}\phi \varkappa _2(\phi ,\phi \theta )\Big (F_1^\prime (\rho (\phi ,\phi \theta ))-3/4\Big )&=4{{c_0^{-1}}}\frac{\theta ^3(\theta -1)^2}{(1+\theta )^6}\left( F_1^\prime \Big (\frac{4\theta }{(1+\theta )^2}\Big )-3/4\right) . \end{aligned}$$

Moreover, by (4.48)

$$\begin{aligned} |\phi \varkappa _2(\phi ,\phi \theta )\Big (F_1^\prime (\rho (\phi ,\phi \theta ))-3/4\Big )|\le C\frac{\theta ^2}{(1+\theta )^4}, \end{aligned}$$

so we can apply dominated convergence theorem obtaining

(4.50)

Putting together (4.23), (4.36), (4.42) and (4.50) we find

Notice that the real number \(\eta \) is well-defined since all the integrals converge. This shows the existence of the derivative of \(\nu _\Omega \) at the origin. It is important to emphasize that number \(\eta \) is independent of the profile \(r_0\) and we claim that the number \(\eta \) is zero. It is slightly difficult to check this result directly from the integral representation of \( \eta \), however we shall check it in a different way by calculating its value for the unit ball

$$\begin{aligned} \big \{ (r e^{i\theta },z),\,r^2+z^2\le 1, \theta \in {{\,\mathrm{{\mathbb {R}}}\,}}\big \} \end{aligned}$$

whose boundary can be parametrized by \((\phi ,\theta )\mapsto (r_0(\phi ) e^{i\theta },\cos \phi )\) with \(r_0(\phi )=\sin \phi \). Now according to the identity (3.5) one has

$$\begin{aligned} \int _0^\pi H_1(\phi ,\varphi )d\varphi =\frac{1}{r_0(\phi )}\partial _r \psi _0(re^{i\theta },\cos (\phi ))\left| _{r=r_0(\phi )}\right. . \end{aligned}$$

However it is known [32] that the stream function \(\psi _0\) is radial and quadratic inside the domain taking the form

$$\begin{aligned} 0\le r\le \sin \phi ,\quad \psi _0(re^{i\theta },\cos (\phi ))=\frac{1}{6}\big ( r^2+\cos ^2\phi \big ). \end{aligned}$$

Consequently, with this special geometry the function \(\displaystyle {\nu _\Omega }\) is constant and therefore

$$\begin{aligned} \nu _\Omega ^\prime (0)=\nu _\Omega ^\prime (\pi )=0. \end{aligned}$$

\(\bullet \) Step 2: \(\nu _\Omega ^\prime \in {\mathscr {C}}^\alpha (0,\pi )\). We shall prove that \(\nu _\Omega ^\prime \) is \({\mathscr {C}}^\alpha (0,\pi )\) and for this purpose we start with the first term in (4.23), i.e., \(\zeta _1\). According to (4.24) it can be split into several terms and to fix the ideas let us describe how to proceed with the first term given by

and check that it belongs to \({\mathscr {C}}^\alpha (0,\pi )\). The remaining terms of \(\zeta _1\) can be treated in a similar way and to alleviate the discussion we leave them to the reader.

From the assumptions (H) on \(r_0\) we have \(r_0^\prime ,\,\phi \mapsto \frac{r_0(\phi )}{\sin \phi }\in {\mathscr {C}}^\alpha (0,\pi )\) , then using the fact that \({\mathscr {C}}^\alpha \) is an algebra, it suffices to verify that

This function is locally \({\mathscr {C}}^1\) in \((0,\pi )\) and so the problem reduces to check the regularity close to the boundary \(\{0,\pi \}\). By symmetry it suffices to check the regularity near the origin. Decompose the integral as follows

Since we are considering \(\phi \in (0,\pi /2)\), it is easy to check that the last integral term defines a \({\mathscr {C}}^1\) function in \([0,\pi /2]\). Since \(\sin (\phi )/\phi \) is \({\mathscr {C}}^\alpha (0,\pi /2)\), then the problem amounts to checking that the function

is \({\mathscr {C}}^\alpha \) close to zero. Making the change of variables \(\varphi =\phi \theta \) we get

Let us now define the following functions

and

(4.51)

We will show that \(T_{1,\phi }\in {\mathscr {C}}^\alpha [\phi ,\pi /2]\) and \(T_{2,\phi }\in {\mathscr {C}}^\alpha (0,\phi ]\) uniformly in \(\phi \in (0,\frac{\pi }{2})\). Thus we get in particular a constant \(C>0\) such that for any \(0<\phi _1\le \phi _2< \frac{\pi }{2}\),

$$\begin{aligned} |T_{1,\phi _1}(\phi _1)-T_{1,\phi _1}(\phi _2)|\le C|\phi _1-\phi _2|^\alpha \end{aligned}$$
(4.52)

and

$$\begin{aligned} {|T_{2,\phi _2}(\phi _1)-T_{2,\phi _2}(\phi _2)|\le C|\phi _1-\phi _2|^\alpha .} \end{aligned}$$
(4.53)

By combining (4.52) and (4.53), since \(T_{1,\phi _1}(\phi _ 2)= T_{2,\phi _2}(\phi _1)\) we get

$$\begin{aligned}&|\zeta _{1,1}(\phi _1)-\zeta _{1,1}(\phi _2)|\\&\quad \le |T_{1,\phi _1}(\phi _1)-T_{1,\phi _1}(\phi _2)|+{{|T_{2,\phi _2}(\phi _1)-T_{2,\phi _2}(\phi _2)|}}\le C|\phi _1-\phi _2|^\alpha . \end{aligned}$$

This ensures that \(\zeta _{1,1}\in {\mathscr {C}}^\alpha (0,\pi /2)\).

It remains to show that \(T_{1,\phi }\in {\mathscr {C}}^\alpha ([\phi ,\pi /2])\) and \(T_{2,\phi }\in {\mathscr {C}}^\alpha ((0,\phi ])\) uniformly in \(\phi \in (0,\frac{\pi }{2})\). We start with the term \(T_{1,\phi }\). Then straightforward computations imply

$$\begin{aligned} \forall \, s\in [\phi ,\pi /2],\quad |\partial _sT_{1,\phi }(s)|&= \frac{\pi }{2s^2}\frac{\phi ^2\sin (\phi \pi /2s)r_0^2(\phi \pi /2s)}{R^\frac{5}{2}(\phi ,\phi \pi /2s)}\\&\le C\frac{1}{s^5}\frac{\phi ^5}{\phi ^5(1+\frac{\pi }{2s})^5} \le C, \end{aligned}$$

for any \(\phi ,s\in (0,\pi /2]\). Notice that we have used in the last line the following inequalities which follow from the assumptions (H2),

$$\begin{aligned} \phi \lesssim r_0(\phi )\lesssim \phi , \quad \forall \phi \in [0,\pi /2] \end{aligned}$$

and

$$\begin{aligned} \theta \lesssim \frac{r_0(\phi \theta )}{\phi }\lesssim \theta , \quad \forall \theta \in [0,\pi /2\phi ]. \end{aligned}$$
(4.54)

Hence \(T_{1,\phi }\in \text {Lip}([\phi ,\pi /2])\), uniformly with respect to \(\phi \in (0,\pi /2)\).

Let us move to the term \(T_{2,\phi }\). First, we write

$$\begin{aligned} \frac{\sin (\phi \theta )}{\phi }=\theta \int _0^1\cos (\phi \theta \tau )d\tau , \end{aligned}$$

and taking the derivative with respect to \(\phi \) we obtain

$$\begin{aligned} \partial _\phi \left( \frac{\sin (\phi \theta )}{\phi } \right) =-\theta ^2\int _0^1\sin (\phi \theta \tau )\tau \,d\tau . \end{aligned}$$

Hence,

$$\begin{aligned} \Big |\partial _\phi \left( \frac{\sin (\phi \theta )}{\phi } \right) \Big |\le \theta ^2. \end{aligned}$$
(4.55)

By the mean value theorem we infer

$$\begin{aligned} \left| \frac{\sin (s_1\theta )}{s_1}-\frac{\sin (s_2\theta )}{s_2}\right| \le |s_1-s_2|\theta ^2 \end{aligned}$$
(4.56)

Interpolating between (4.54), which is also true for \(r_0=\sin \), and (4.56) we obtain

$$\begin{aligned} \left| \frac{\sin (s_1\theta )}{s_1}-\frac{\sin (s_2\theta )}{s_2}\right| \le C|s_1-s_2|^\alpha \theta ^{1-\alpha }\theta ^{2\alpha }=C|s_1-s_2|^\alpha \theta ^{1+\alpha }. \end{aligned}$$
(4.57)

Using Taylor’s formula

$$\begin{aligned} r_0(\phi \theta )=\phi \theta \int _0^1r_0'(\tau \phi \theta )d\tau , \end{aligned}$$

one finds that if \(0\le \phi \theta \le \pi /2\) then

$$\begin{aligned} \left| \partial _\phi \left( \frac{r_0(\phi \theta )}{\phi }\right) \right| \le C\theta ^2. \end{aligned}$$
(4.58)

As before, one gets that if \(0\le s_1 \theta ,\, s_2\theta \le \pi /2\) hence

$$\begin{aligned} \left| \frac{r_0(s_1\theta )}{s_1}-\frac{r_0(s_2\theta )}{s_2}\right| \le C|s_1-s_2|^\alpha \theta ^{1+\alpha }. \end{aligned}$$
(4.59)

Now, let us check that \(T_{2,\phi }\) is \({\mathscr {C}}^\alpha (0,\phi ]\) uniformly in \(\phi \in (0,\pi /2)\). Let \(s_1,s_2\in (0,\phi ]\), then using the estimates (4.56) and (4.54), we achieve for any \(s\in (0,\phi ]\),

for \(\alpha \in (0,1)\). In the same way

To analyze the difference of the denominator in \(T_{2,\phi }\) we first write that for any \(0\le s\theta \le \frac{\pi }{2}\),

$$\begin{aligned} s^5R^{-\frac{5}{2}}(s,s\theta )=\left( \left( \frac{r_0(s)+r_0(s\theta )}{s}\right) ^2+\left( \frac{\cos (s)-\cos (s\theta )}{s}\right) ^2\right) ^{-\frac{5}{2}}\lesssim (1+\theta )^{-5}. \end{aligned}$$

Using an slight variant of the argument in (4.55) one gets

$$\begin{aligned} |1-\cos (\phi )|\le |\phi | \end{aligned}$$
(4.60)

and

$$\begin{aligned} \Big |\partial _\phi \left( \frac{\cos (\phi \theta )-1}{\phi } \right) \Big |\le \theta ^2. \end{aligned}$$
(4.61)

By differentiation, and using (4.61) and (4.58), we find that if \( 0\le s\theta \le \frac{\pi }{2}\) hence

$$\begin{aligned} \left| \partial _s\left( s^5R^{-\frac{5}{2}}(s,s\theta )\right) \right| \lesssim (1+\theta )^{-4}. \end{aligned}$$

This implies in view of the mean value theorem

$$\begin{aligned} \forall \, 0\le s_1\theta , s_2\theta \le \frac{\pi }{2},\quad \left| s_1^5R^{-\frac{5}{2}}(s_1,s_1\theta )-s_2^5R^{-\frac{5}{2}}(s_2,s_2\theta )\right| \lesssim (1+\theta )^{-4}|s_1-s_2|. \end{aligned}$$

Moreover

$$\begin{aligned} \forall \, 0\le s_1\theta , s_2\theta \le \frac{\pi }{2},\quad \left| s_1^5R^{-\frac{5}{2}}(s_1,s_1\theta )-s_2^5R^{-\frac{5}{2}}(s_2,s_2\theta )\right| \lesssim (1+\theta )^{-5}. \end{aligned}$$

Then by interpolation we get

$$\begin{aligned} \forall \, 0\le s_1\theta , s_2\theta \le \frac{\pi }{2},\quad \left| s_1^5R^{-\frac{5}{2}}(s_1,s_1\theta )-s_2^5R^{-\frac{5}{2}}(s_2,s_2\theta )\right| \lesssim (1+\theta )^{\alpha -5}|s_1-s_2|^\alpha .\nonumber \\ \end{aligned}$$
(4.62)

Therefore we obtain

which converges since \(\alpha \in (0,1)\).

Combining the preceding estimates one deduces that

$$\begin{aligned} \forall \, s_1, s_0\in (0,\phi ],\quad |T_{2,\phi }(s_1)-T_{2,\phi }(s_2)|\le C|s_1-s_2|^\alpha , \end{aligned}$$

uniformly in \(\phi \in (0,\pi /2)\). Hence, we conclude that \(\zeta _{1,1}\) is \({\mathscr {C}}^\alpha (0,\pi /2)\), for any \(\alpha \in (0,1)\).

The argument used to prove that the other terms in (4.24) are in \({\mathscr {C}}^\alpha (0,\pi /2)\), for any \(\alpha \in (0,1)\), are quite similar, but for the reader convenience we will sketch some details. The second term in the sum is

Let us consider the functions

and

To prove that this second term is in \({\mathscr {C}}^\alpha (0,\pi /2)\) we will see that \(T_{1,\phi }\in \text {Lip}([\phi ,\pi /2])\) and \(T_{2, \phi }\) is in \({\mathscr {C}}^\alpha (0,\phi )\). Using the mean value theorem and the estimate

$$\begin{aligned} \forall \, s\in [\phi ,\pi /2],\quad |\partial _sT_{1,\phi }(s)| \le C\frac{1}{s^7}\frac{\phi ^7}{\phi ^7(1+\frac{\pi }{2s})^7} \le C, \end{aligned}$$

we can easily see that \(T_{1,\phi }\) is in the desired space. Now we will check that \(T_{2,\phi }\) is \({\mathscr {C}}^\alpha (0,\phi ]\) uniformly in \(\phi \in (0,\pi /2)\). Let \(s_1,s_2\in (0,\phi ]\), then using the estimates (4.54) and (4.56), we achieve for any \(s\in (0,\phi ]\),

for \(\alpha \in (0,1)\). In the same way

Now by differentiation, and using (4.58) and (4.61) we find that if \( 0\le s\theta \le \frac{\pi }{2}\) then

$$\begin{aligned} \left| \partial _s\left( s^7R^{-\frac{7}{2}}(s,s\theta )\right) \right| \lesssim (1+\theta )^{-6}. \end{aligned}$$
(4.63)

Therefore, using interpolation argument we obtain

and the last integral converges since \(\alpha \in (0,1)\).

Another term to consider in (4.24) is given by

which will be treated exactly in the same way as the previous one. For this reason we will not repeat again the arguments. The next term in (4.24) can be written as

As in the previous cases we can consider the auxiliary functions

and

For this term it is enough to check that the auxiliary functions are in the \({\mathscr {C}}^\alpha \) space. For \(T_{1,\phi }\) we will use the mean value theorem. Thus the inequality

$$\begin{aligned} \forall \, s\in [\phi ,\pi /2],\quad |\partial _sT_{1,\phi }(s)| \le C\frac{1}{s^6}\frac{\phi ^7}{\phi ^7(1+\frac{\pi }{2s})^6} \le C, \end{aligned}$$

will be enough. To estimate \(T_{2,\phi }\) we will follow the same arguments developed in the previous cases. Hence, using inequalities (4.58), (4.59), (4.55), (4.56), (4.54), (4.61) and (4.60) one gets the following estimates for \(\alpha \in (0,1)\).

and

On the other hand, using (4.61), (4.60) and interpolation, one obtains

$$\begin{aligned} \left| \frac{1-\cos (s_1\theta )}{s_1}-\frac{1-\cos (s_2\theta )}{s_2}\right| \le C\theta ^{1+\alpha }|s_1-s_2|^{\alpha }. \end{aligned}$$
(4.64)

Then, by (4.64) we obtain

To estimate the last term of \(T_{2,\phi }\), using (4.63)

The next function to analyze in (4.24) is

We will not repeat the arguments for this function because they are quite similar to the preceding case. The last function to consider in (4.24) is given by

$$\begin{aligned} \partial _{\varphi }{\mathscr {K}}_1(\phi ,\varphi )\frac{r_0(\phi )r_0(\varphi )}{R(\phi , \varphi )}&= 2\frac{r_0(\phi )r_0^2(\varphi )r_0'(\varphi )\sin (\varphi )}{(R(\phi , \varphi ))^{5/2}}+ \frac{r_0(\phi )r_0^3(\varphi )\cos (\phi )}{(R(\phi , \varphi ))^{5/2}}\\&\quad -3\frac{r_0(\phi )r_0^3(\varphi ) \sin (\varphi )}{(R(\phi , \varphi ))^{7/2}}\\&\quad \big ((r_0(\phi )+r_0(\varphi ))r_0'(\varphi )+(\cos (\phi )-\cos (\varphi ))\sin (\varphi )\big ) \end{aligned}$$

This term generates several functions. Some of them are similar to the functions estimated in the previous cases and the others are similar between them. For this reason we will only check the first one. Let us prove that the function

is in \({\mathscr {C}}^\alpha (0,\pi /2)\), for any \(\alpha \in (0,1)\). Since the integral in the interval \([\frac{\pi }{2},\pi ]\) provides a function in \({\mathscr {C}}^\alpha \), as in the above cases we can reduce the integral to the interval \([0,\pi /2]. \) Now the strategy is again to consider the auxilary functions

and

Since

$$\begin{aligned} \forall \, s\in [\phi ,\pi /2],\quad |\partial _sT_{1,\phi }(s)| \le C\frac{1}{s^5}\frac{\phi ^5}{\phi ^5(1+\frac{\pi }{2s})^5} \le C, \end{aligned}$$

the function \(T_{1,\phi }\) is in \({\mathscr {C}}^\alpha \), for any \(\alpha \in (0,1)\). To establish that \(T_{2,\phi }\) is in the same space we need the following estimates.

in the first inequality we have used that \(r_0'\) is a bounded function. The next estimate is

Using the inequality \(|r_0'(s_1\theta )-r_0'(s_2\theta )|\le C\theta ^{\alpha }|s_1-s_2|^{\alpha }\) we get

The last term to estimate is

Hence, we obtain the announced result. The remaining terms can be studied using the same inequalities and for this reason we will avoid them.

Let us now move to the regularity of \(\zeta _2\) defined in (4.23) which takes the form

and

The first function can be split into two parts as follows

As before, by evoking the symmetry property of r we can restrict the study to \(\phi \in [0,\frac{\pi }{2}]\). The second term is the easiest one and we claim that \(I_2\in W^{1,\infty }\). Indeed,

It can be transformed into

Integrating by parts yields

Notice that the last term is bounded uniformly on \(\phi \in [0,\pi /2]\). In fact, one has from the definition of R in (3.2)

$$\begin{aligned} \forall \phi \in [0,\pi /2],\quad \frac{1}{R(\phi ,\frac{\pi }{2})}\le {\frac{1}{r_0^2(\pi /2)}}\cdot \end{aligned}$$

Using (4.13) we get

$$\begin{aligned} \partial _\phi R(\phi ,\pi /2)=2r_0^\prime (\phi )(r_0(\phi )+r_0(\pi /2))-2\sin \phi \cos \phi . \end{aligned}$$

Moreover, since \(r_0\) is symmetric with respect to \(\pi /2\) then we get \(r_0^\prime \left( \frac{\pi }{2}\right) =0\), which implies that \(\partial _\phi R(\pi /2,\pi /2)=0\) and by the mean value theorem,

$$\begin{aligned} \forall \phi \in (0,\pi ),\quad \Big |(\partial _\phi R)\left( \phi ,\pi /2\right) \Big |\lesssim \left| \phi -\frac{\pi }{2}\right| . \end{aligned}$$

Hence, combining (4.37) and (4.10) we find

$$\begin{aligned} \forall \phi \in \left( 0,\frac{\pi }{2}\right) ,\quad \left| F_1\left( \rho \left( \phi ,\frac{\pi }{2}\right) \right) -1\right|&\lesssim \rho \left( \phi ,\frac{\pi }{2}\right) \Big (1{+}\ln \Big [1-\rho \left( \phi ,\frac{\pi }{2}\right) \Big ]\Big )\\&\lesssim { 1+}\left| \ln \left( \frac{\pi }{2}-\phi \right) \right| . \end{aligned}$$

Consequently

$$\begin{aligned} \forall \phi \in \left( 0,\frac{\pi }{2}\right) ,\quad&\frac{\Big |\partial _\phi R(\phi ,\frac{\pi }{2})\Big |}{R^\frac{5}{2}(\phi ,\frac{\pi }{2})}r_0^2\left( \frac{\pi }{2}\right) \Big |F_1\left( \rho \left( \phi ,\frac{\pi }{2}\right) \right) -1\Big |\nonumber \\&\quad \lesssim \big (\frac{\pi }{2}-\phi \big )\left| {1+}\ln \left( \frac{\pi }{2}-\phi \right) \right| , \end{aligned}$$
(4.65)

which ensures that this quantity is bounded in the interval \((0,\frac{\pi }{2})\).

Next, let us check the boundedness of the integral terms of \(I_2^\prime \). Inequality (4.39) allows to get

$$\begin{aligned} \sup _{\begin{array}{c} \phi \in [0,\pi /2]\\ \varphi \in [\pi /2,\pi ] \end{array}} \left| \partial _\phi \left( \frac{\partial _\phi R(\phi ,\varphi )}{R(\phi ,\varphi )^\frac{5}{2}}\right) \right| +\left| \frac{\partial _\phi R(\phi ,\varphi )}{R(\phi ,\varphi )^\frac{5}{2}}\right| +\left| \partial _\varphi \left( \frac{\partial _\phi R(\phi ,\varphi )}{R(\phi ,\varphi )^\frac{5}{2}}\sin (\varphi )r_0(\varphi )^2\right) \right| <\infty , \end{aligned}$$

which implies

Therefore, (4.10) combined with (4.19) and (4.46) yield

Let us move to \(I_1\). First, we do the change of variables \(\varphi =\phi \theta \) leading to

We will check that \(I_1\) is \({\mathscr {C}}^\alpha (0,\pi /2)\), for any \(\alpha \in (0,1)\). Indeed, take \(\phi _1\le \phi _2\in (0,\frac{\pi }{2})\), then

(4.66)

where

We follow the ideas done for \(\zeta _1\). In order to estimate \(I_{1,1}\), define

Then

$$\begin{aligned} \forall \, s\in [\phi ,\pi /2),\quad \partial _sG_{1,\phi }(s)= & {} -\frac{\pi }{2s^2}\frac{\phi (\partial _\phi R)(\phi ,\frac{\pi \phi }{2s})}{R^\frac{5}{2}(\phi ,\frac{\pi \phi }{2s}) }\sin \left( \frac{\pi \phi }{2s}\right) r_0^2\left( \frac{\pi \phi }{2s}\right) \\&\left[ F_1\left( \rho \left( \phi ,\frac{\pi \phi }{2s}\right) \right) -1\right] , \end{aligned}$$

for \(\phi \in (0,\pi /2)\). Taking the derivative in \(\phi \) of the function R

$$\begin{aligned} (\partial _\phi R)(\phi ,\phi \theta )=2r_0'(\phi )(r_0(\phi )+r_0(\phi \theta ))+2\sin (\phi )(\cos (\phi \theta )-\cos (\phi )), \end{aligned}$$

we get

$$\begin{aligned} {|(\partial _\phi R)(\phi ,\phi \theta )|}\le C(\phi (1+\theta )+\phi |1-\theta |). \end{aligned}$$
(4.67)

Moreover, proceeding as before in (4.37) combined with the assumptions (H) we find

$$\begin{aligned} \big |F_1(\rho (\phi ,\phi \theta ))-1\big |\lesssim \frac{1}{1+\theta }\Big (1+\ln \left| \frac{1+\theta }{1-\theta }\right| \Big ). \end{aligned}$$
(4.68)

Putting together the preceding estimates allows to get

$$\begin{aligned} |\partial _sG_{1,\phi }(s)|&\lesssim \frac{1}{s^2}\frac{\phi \left\{ \phi (1+\frac{\pi }{2s})+\phi |1-\frac{\pi }{2s}|\right\} }{\phi ^5(1+\frac{\pi }{2s})^5}\frac{\phi ^3}{s^3}\frac{1}{1+\frac{\pi }{2s}}\left( 1+\ln \left| \frac{1+\frac{\pi }{2s}}{1-\frac{\pi }{2s}}\right| \right) \\&\lesssim \frac{(s+\frac{\pi }{2})+|s-\frac{\pi }{2}|}{(s+\frac{\pi }{2})^6}\left( 1+\ln \left| \frac{1+\frac{\pi }{2s}}{1-\frac{\pi }{2s}}\right| \right) . \end{aligned}$$

It follows that

$$\begin{aligned} \forall s\in (0,\pi /2),\quad \sup _{\phi \in (0,s]}\left| \partial _sG_{1,\phi }(s)\right|&\lesssim 1+\left| \ln \left( \frac{\pi }{2}-s\right) \right| . \end{aligned}$$

Now using this estimate combined with the mean value theorem we get for \(0<\phi _1\le \phi _2<\frac{\pi }{2}\)

$$\begin{aligned} |I_{1,1}|\le&\int _{\phi _1}^{\phi _2}\left| \partial _sG_{1,\phi _1}(s)\right| ds\\ \le&C|\phi _1-\phi _2|+\int _{\phi _1}^{\phi _2}\left| \ln \left( \frac{\pi }{2}-s\right) \right| ds. \end{aligned}$$

Using Hölder inequality yields for any \(\alpha \in (0,1)\),

$$\begin{aligned} \int _{\phi _1}^{\phi _2}\left| \ln \left( \frac{\pi }{2}-s\right) \right| ds\le&|\phi _1-\phi _2|^\alpha \left( \int _{0}^{\frac{\pi }{2}}\left| \ln \left( \frac{\pi }{2}-s\right) \right| ^{\frac{1}{1-\alpha }}ds\right) ^{1-\alpha }\le C_\alpha |\phi _1-\phi _2|^\alpha .\end{aligned}$$

Notice that the constant \(C_\alpha \) blows up when \(\alpha \) approaches 1. Thus

$$\begin{aligned} \forall \, \phi _1,\phi _2\in (0,\pi /2),\quad |I_{1,1}|\le C_\alpha |\phi _1-\phi _2|^\alpha . \end{aligned}$$

Next, let us move to the estimate of \(I_{1,2}\). Using (4.19) we arrive at

$$\begin{aligned} |F_1'(\rho (\phi ,\phi \theta ))|\le C\frac{R(\phi ,\phi \theta )}{\phi ^2(1-\theta )^2}\le C\frac{(1+\theta )^2}{(1-\theta )^2}\cdot \end{aligned}$$
(4.69)

Set

$$\begin{aligned} {\mathscr {R}}(\theta ,\phi ):=\rho (\phi ,\phi \theta )=\frac{4r_0(\phi ) r_0(\phi \theta )}{R(\phi ,\phi \theta )}, \end{aligned}$$

then differentiating with respect to \(\theta \) we get

$$\begin{aligned}&\partial _\theta {\mathscr {R}}(\theta ,\phi )\\&\quad =-\frac{8r_0(\phi )r_0(\phi \theta )}{R(\phi ,\phi \theta )^2}\Big ((r_0(\phi )+r_0(\phi \theta ))\phi r_0'(\phi \theta )+(\cos (\phi )-\cos (\phi \theta ))\phi \sin (\phi \theta )\Big )\\&\qquad +\frac{4r_0(\phi ) \phi r_0^\prime (\phi \theta )}{R(\phi ,\phi \theta )}\cdot \end{aligned}$$

Using the assumption (H2) we may check that

$$\begin{aligned} \forall \, 0\le \phi \theta \le \pi /2,\quad |\partial _\theta {\mathscr {R}}(\theta ,\phi )|\le \frac{C}{(1+\theta )^2}, \end{aligned}$$

where C depends only on \(\Vert r_0^\prime \Vert _{L^\infty }\). Now by rewriting

$$\begin{aligned} \partial _\theta {\mathscr {R}}(\theta ,\phi )&= \frac{\frac{4r_0(\phi )}{\phi } r_0^\prime (\phi \theta )}{\left( \frac{r_0(\phi )+r_0(\phi \theta )}{\phi }\right) ^2+\left( \frac{\cos (\phi )-\cos (\phi \theta )}{\phi }\right) ^2}\\&\quad -\frac{8\frac{r_0(\phi )}{\phi }\frac{r_0(\phi \theta )}{\phi }}{\left\{ \left( \frac{r_0(\phi )+r_0(\phi \theta )}{\phi }\right) ^2+\left( \frac{\cos (\phi )-\cos (\phi \theta )}{\phi }\right) ^2\right\} ^2}\\&\quad \times \left[ \frac{(r_0(\phi )+r_0(\phi \theta ))}{\phi } r_0'(\phi \theta )+{(\cos (\phi )-\cos (\phi \theta ))}\frac{\sin (\phi \theta )}{\phi }\right] , \end{aligned}$$

and differentiating in \(\phi \) we get the estimate

$$\begin{aligned} \forall \, 0\le \phi \theta \le \pi /2,\quad |\partial _\phi \partial _\theta {\mathscr {R}}(\theta ,\phi )|\le \frac{C}{(1+\theta )}, \end{aligned}$$

where C depends only on \(\Vert r_0\Vert _{C^2}\). Taylor’s formula

$$\begin{aligned} {\mathscr {R}}(\theta ,\phi )={\mathscr {R}}(1,\phi )+\int _1^\theta \partial _\theta {\mathscr {R}}(\tau ,\phi )d\tau \end{aligned}$$

combined with \({\mathscr {R}}(1,\phi )=1\) yields

$$\begin{aligned} \partial _\phi {\mathscr {R}}(\theta ,\phi )=\int _1^\theta \partial _\phi \partial _\theta {\mathscr {R}}(\tau ,\phi )d\tau . \end{aligned}$$

This implies in turn that

$$\begin{aligned} \sup _{\phi \in (0,\frac{\pi }{2\theta })}|\partial _\phi {\mathscr {R}}(\theta ,\phi )|\le C\left| \ln \left( \frac{1+\theta }{2}\right) \right| . \end{aligned}$$
(4.70)

Combining this estimate with (4.69) we deduce that

$$\begin{aligned} \sup _{\phi \in (0,\frac{\pi }{2\theta })}\big |\partial _\phi \big [F_1(\rho (\phi ,\phi \theta ))\big ]\big |\le C\frac{(1+\theta )^2}{(1-\theta )^2}\left| \ln \left( \frac{1+\theta }{2}\right) \right| . \end{aligned}$$

Following an interpolation argument combining the preceding estimate with (4.68) yields \(\hbox { for any} \alpha \in [0,1]\) and for \(0<\phi _1\le \phi _2\le \frac{\pi }{2\theta }\)

$$\begin{aligned}&\big |F_1(\rho (\phi _1,\phi _1\theta ))-F_1(\rho (\phi _2,\phi _2\theta ))\big |\nonumber \\&\quad \le C|\phi _1-\phi _2|^\alpha \frac{(1+\theta )^{3\alpha -1}}{|1-\theta |^{2\alpha }}\left| \ln \left( \frac{1+\theta }{2}\right) \right| ^\alpha \left( 1+\ln \left| \frac{1+\theta }{1-\theta }\right| \right) ^{1-\alpha }. \end{aligned}$$
(4.71)

Plugging this estimate into the definition of \(I_{1,2}\) given in (4.66) implies

This integral converges, close to 1 and at \(\infty \), provided that \(0\le \alpha <1\). We mention that to get the integrability close to 1 we use the approximation

$$\begin{aligned} \ln \left( \frac{1+\theta }{2}\right) \overset{1}{ \sim } \frac{\theta -1}{2}\cdot \end{aligned}$$

As to the estimate of the term \(I_{1,3}\) described in (4.66) we roughly implement similar ideas. For that purpose, we introduce the function

Then combining (4.56), (4.67) and (4.68), we deduce that

provided that \(\alpha \in (0,1)\). Implementing the same analysis for the remaining terms and using (4.59) and (4.62) as for \(\zeta _{1}\), we find

$$\begin{aligned} \forall \, 0\le s_1,s_2\le \phi ,\quad |G_{2,\phi }(s_1)-G_{2,\phi }(s_2)|\le C|s_1-s_2|^\alpha , \end{aligned}$$

uniformly for \(\phi \in (0,\pi /2)\). Therefore from the definition (4.66) we obtain for any \(0\le \phi _1\le \phi _2\le \frac{\pi }{2}\),

$$\begin{aligned} |I_{1,3}|&= \big |G_{2,\phi _2}(\phi _1)- G_{2,\phi _2}(\phi _2)\big |\le C|\phi _1-\phi _2|^\alpha . \end{aligned}$$

Now let us consider the next term in \(\zeta _2(\phi )\)

The first function that we intend to study is

Since the function \(\frac{r_0(\phi )}{\phi }\) has bounded derivatives, it is enough to estimate the function I. Thus,

The arguments to estimate the terms \(I_1(\phi )\) and \(I_2(\phi )\) are similar to the case of the function \(\zeta _1\), but we will repeat them for the reader convenience. First we will prove that \(I_2'(\phi )\) is a bounded function. By direct computations we infer

The estimate of \(I_{2,2}\) can be done using (A.10), (4.39) and (4.12), leading to

$$\begin{aligned} |I_{2,2}|\leqslant C+\int _{\pi /2}^{\pi }\ln \big |\frac{\phi +\varphi }{\phi -\varphi }\big |d\varphi \le C. \end{aligned}$$

For the term \(I_{2,1}\) we may apply (4.46), (2.18) and (4.39) in order to get

$$\begin{aligned} |I_{2,1}|\leqslant \int _{\pi /2}^{\pi }\frac{R(\phi ,\varphi )|\phi -\varphi |^2}{|\phi -\varphi |^2R(\phi ,\varphi )^4}d\varphi \le C. \end{aligned}$$

To estimate the term \(I_{2,3}\) we decompose it in different terms.

Since \(r_0'(\pi /2)=r_0(\pi )=0\) then integration by parts allows to get

We can then proceed similarly to the terms \(I_{2,1}\) and \(I_{2,2}\) in order to get that \(I_{2,2}\) is bounded. The next step will be to check that the function \(I_1\) is in \(\mathscr {C}^{\alpha }(0,\pi /2)\), for all \(\alpha \in (0,1)\). If we do the change of variable \(\varphi =\phi \theta \) we find

$$\begin{aligned} I_1(\phi )=\int _0^{\frac{\pi }{2\phi }} \frac{\phi ^2 r_0^2(\theta \phi )r_0'(\theta \phi )\sin (\theta \phi )}{R(\phi ,\theta \phi )^{\frac{5}{2}}} (F_1(\rho (\phi ,\theta \phi ))-1)d\theta . \end{aligned}$$

If \(\phi _1\) and \(\phi _2\) are in \((0,\pi /2)\), then

To estimate the function \(I_{1,1}\), let us consider the auxiliary function

First we may write for \(0<\phi _1\leqslant \phi _2 <\pi /2\),

$$\begin{aligned} |I_{1,1}|=|G_{1,\phi _1}(\phi _2)-G_{1,\phi _1}(\phi _1)|\le \int _{\phi _1}^{\phi _2}|\partial _sG_{1,\phi _1}(s)|ds. \end{aligned}$$

Hence, it is enough to obtain an appropriate bound for \(\partial _sG_{1,\phi _1}\). Taking the derivative in s, applying (4.68) and some standard estimates used before we have

$$\begin{aligned} |\partial _sG_{1,\phi }(s)|\le C\left( 1+\Big |\ln \left( \frac{\pi }{2}-s\right) \Big |\right) , \end{aligned}$$

and so by Hölder inequality

for all \(\alpha \in (0,1)\). Let us now move to the estimate of the term \(I_{1,2}\). By (4.71) and some standard estimates used along the work we obtain, for any \(\alpha \in (0,1)\)

where in the last inequality we have used that \(\ln (\frac{1+\theta }{2})\simeq \frac{\theta -1}{2}\) if \(\theta \) is enough close to 1.

To estimate the term \(I_{1,3}, \) let us take the function

Now, \(|I_{2,3}|=|G_{2,\phi _2}(\phi _1)-G_{2,\phi _2}(\phi _2)|. \) As in the case of the function \(\zeta _1\) we will get the desired estimate through decomposing the integral in several terms. The first term to consider is

Using (4.59) and (4.68) one can see that this term is bounded by

For the next adding term, using (4.56) and (4.68) we obtain

For the next term we will use the regularityof \(r_0\) and (4.68), obtaining

Let us move to the last term in \(I_{1,3}\). By (4.68) and (4.62)

The two remaining terms in \(\zeta _2\) are left to the reader because they are very similar to the first one. It remains to estimate the term \(\zeta _3\) defined by (4.23). It can be split as follows,

Recall that we restrict to check the regularity for \(\phi \in (0,\pi /2)\) without loss of generality and later we extend it to \(\phi \in (0,\pi )\). Then, we will show that \(I_3\) belongs to \({\mathscr {C}}^\alpha (0,\frac{\pi }{2})\). We will skip here the details for the regularity of \(I_4\) since this term is less singular and the same procedure works, see the estimates for \(I_2\) previously done. To estimate \(I_3\) we proceed as before through the use of the change of variables \(\varphi =\phi \theta \),

Define the functions

In order to check that \(I_3\) belongs to \({\mathscr {C}}^\alpha (0,\frac{\pi }{2})\), it suffices to prove that each function \(H_{i,\phi }\) is in \({\mathscr {C}}^\alpha (0,\frac{\pi }{2})\) uniformly in \(\phi \in (0,\pi /2)\), for any \(i=1,\dots , 4\). Let us start with \(H_{1,\phi }\) showing that its derivative is bounded.

From straightforward calculus it is easy to check that for any \(0\le \phi \le s<\frac{\pi }{2}\),

$$\begin{aligned} |H_{1,\phi }'(s)|&\le \frac{\pi }{2s^2}\frac{\phi \sin \left( \frac{\phi \pi }{2s}\right) r_0^2\left( \frac{\phi \pi }{2s}\right) }{R^\frac{3}{2}\left( \phi ,\frac{\phi \pi }{2s}\right) } \Big |(\partial _\phi \rho )\left( \phi ,{\phi \pi }/{2s}\right) +(\partial _\varphi \rho ) \left( \phi ,{\phi \pi }/{2s}\right) \Big |\\&\quad \left| F_1'\left( \rho \big (\phi ,{\phi \pi }/{2s}\big )\right) -\frac{3}{4}\right| . \end{aligned}$$

Hence, we obtain

$$\begin{aligned} |H_{1,\phi }'(s)|&\lesssim \frac{1}{s^5}\frac{\phi ^4}{\phi ^3(1+\frac{\pi }{2s})^3}\Big |(\partial _\phi \rho ) \left( \phi ,{\phi \pi }/{2s}\right) \\&+(\partial _\varphi \rho )\left( \phi ,{\phi \pi }/{2s}\right) \Big |\left| F_1'\left( \rho \big (\phi ,{\phi \pi }/{2s}\big )\right) -\frac{3}{4}\right| . \end{aligned}$$

Using (4.43)–(4.46)–(4.47) allows to get

$$\begin{aligned} |H_{1,\phi }'(s)|&\lesssim \frac{1}{s^5}\frac{\phi ^4}{\phi ^3(1+\frac{\pi }{2s})^3}\frac{\phi ^2\left( 1-\frac{\pi }{2s}\right) ^2}{\phi ^3\left( 1+\frac{\pi }{2s}\right) ^3}\frac{\phi ^2\frac{\pi }{2s}}{\phi ^2\left( 1-\frac{\pi }{2s}\right) ^2}\lesssim 1, \end{aligned}$$

which is uniformly bounded on \(0\le \phi \le s<\frac{\pi }{2}\). We shall skip the details for \(H_{2,\phi }\) which can be analyzed following the same lines of the term \(T_{2,\phi }\) introduced in (4.51).

Let us now focus on the estimate of \(H_{3,\phi }\). Set

$$\begin{aligned} {\mathscr {T}}(\theta ,s):=R^\frac{1}{2}(s,s\theta )\big ((\partial _\phi \rho )(s, s\theta )+(\partial _\varphi \rho )(s,s\theta )\big ), \end{aligned}$$

then using (4.46), we deduce

$$\begin{aligned} |{\mathscr {T}}(\theta ,s)|\le C\frac{(1-\theta )^2}{(1+\theta )^2}\cdot \end{aligned}$$
(4.72)

By ranging the expression of \({\mathscr {T}}\) as follows

$$\begin{aligned} {\mathscr {T}}(\theta ,s)&= 4\frac{\frac{r_0^2(s\theta )}{s^2}-\frac{r_0^2(s)}{s^2}}{\frac{R^\frac{3}{2}(s,s\theta )}{s^{3}}}r_0^\prime (s)\left( \frac{r_0(s\theta )}{s}-\frac{r_0(s)}{s}\right) \\&\quad +4\frac{\frac{r_0^2(s\theta )}{s^2}-\frac{r_0^2(s)}{s^2}}{\frac{R^\frac{3}{2}(s,s\theta )}{s^3}}\frac{r_0(s)}{s}\Big (r_0^\prime (s)-r_0^\prime (s\theta )\Big ) \\&\quad {+8s\frac{\frac{r_0(s\theta )}{s}\frac{r_0(s)}{s}}{\frac{R^\frac{3}{2}(s,s\theta )}{s^3}}\frac{\big (\cos s-\cos s\theta )}{s}\left( \frac{\sin s}{s}-\frac{\sin s\theta }{s}\right) }\\&\quad +4\frac{\frac{(\cos s-\cos s\theta )^2}{s^2}}{\frac{R^\frac{3}{2}(s,s\theta )}{s^3}}\left( \frac{r_0(s\theta )}{s}r_0^\prime (s)+\frac{r_0(s)}{s}{r_0^\prime (s\theta )}\right) , \end{aligned}$$

and differentiating with respect to s we find

$$\begin{aligned} |\partial _s {\mathscr {T}}(\theta ,s)|\le C\frac{|1-\theta |}{1+\theta }. \end{aligned}$$
(4.73)

We will not give the full details for this estimate because the computations are long and tedious, but to get a more precise idea how this works we shall just explain the estimate of the first term in \(\partial _s {\mathscr {T}}\) given by

$$\begin{aligned} \forall \,\,0<\theta s< \frac{\pi }{2}{, s<\frac{\pi }{2}},\quad {\mathscr {T}}_1(\theta ,s) =: 4\frac{\partial _s \left( \frac{r_0(s\theta )-r_0(s)}{s}\right) \frac{r_0(s\theta )+r_0(s)}{s}}{\frac{R^\frac{3}{2}(s,s\theta )}{s^3}}r_0^\prime (s)\frac{r_0(s\theta )-r_0(s)}{s}\cdot \end{aligned}$$

Note that the other terms can be treated similarly and we use similar estimates but with \(\sin \), \(\cos \) and \(r_0'\) instead of \(r_0\). In particular, here we use \(r_0\in {\mathscr {C}}^2\) in order to bound \(r_0''\). Define

$$\begin{aligned} \forall \,\,0<\theta s< \frac{\pi }{2},{s<\frac{\pi }{2}}\quad g(\theta ,s):=\frac{r_0(s\theta )-r_0(s)}{s}\cdot \end{aligned}$$

Then, one has \(\partial _\theta g(\theta ,s)=r_0'(s\theta )\) and then

$$\begin{aligned} |\partial _s\partial _\theta g(\theta ,s)|=|\theta r_0''(s\theta )|\le C\theta .\end{aligned}$$
(4.74)

Since \(g(1,s)=0\), we can write by Taylor’s formula

$$\begin{aligned} g(\theta ,s)=\int _1^\theta \partial _\theta g(\tau ,s)d\tau , \end{aligned}$$

and hence

$$\begin{aligned} \partial _s g(\theta ,s)=\int _1^\theta \partial _s \partial _\theta g(\tau ,s)d\tau . \end{aligned}$$

Using (4.74), we achieve

$$\begin{aligned} \forall \,\,0<\theta s< \frac{\pi }{2},\quad |\partial _s g(\theta ,s)|\le C|1-\theta |{(1+\theta )}. \end{aligned}$$

Plugging this into the the definition of \({\mathscr {T}}_1\) and using the mean value theorem yields to the estimate

$$\begin{aligned} |{\mathscr {T}}_1|\le C\frac{|1-\theta |{(1+\theta )^2}|1-\theta |}{(1+\theta )^3}\le C\frac{(1-\theta )^2}{(1+\theta )}\cdot \end{aligned}$$

Now, interpolating between (4.72) and (4.73), we find that for any \(\alpha \in (0,1)\)

$$\begin{aligned} |{\mathscr {T}}(\theta ,s_1)-{\mathscr {T}}(\theta ,s_2)|\le C|s_1-s_2|^\alpha \frac{|1-\theta |^{2-\alpha }}{(1+\theta )^{2-\alpha }}. \end{aligned}$$
(4.75)

Using (4.47) we get

$$\begin{aligned} \forall \, 0\le \phi \theta \le \frac{\pi }{2},\quad \left| F_1'(\rho (\phi ,\phi \theta ))-\frac{3}{4}\right| \le C\frac{\theta }{(1-\theta )^2}. \end{aligned}$$
(4.76)

Combining this estimate with (4.75) and (4.47), we conclude that for any \(0\le s_1,s_2\le \phi \le \frac{\pi }{2}\)

$$\begin{aligned} |H_{3,\phi }(s_1)-H_{3,\phi }(s_2)|&\le C|s_1-s_2|^\alpha \int _0^{+\infty }\frac{\phi ^4\theta ^3}{\phi ^4(1+\theta )^4}\frac{(1-\theta )^{2-\alpha }}{(1+\theta )^{2-\alpha }}\frac{\theta }{(1-\theta )^2} d\theta \nonumber \\&\le C|s_1-s_2|^\alpha , \end{aligned}$$
(4.77)

for any \(\alpha \in (0,1)\). Let us finish working with \(H_{4,\phi }\). Using (4.54) and the standard inequality

$$\begin{aligned} \Big |\frac{\cos (\theta \phi )-1}{\phi }\Big |\le \theta ,\end{aligned}$$

one gets

$$\begin{aligned} \Big |\frac{1}{1-\rho (\phi \theta ,\phi )}\Big |\le \frac{(1+\theta )^2}{(1-\theta )^2}. \end{aligned}$$
(4.78)

As a consequence of (A.9), (4.70) and (4.78) one has

$$\begin{aligned} \left| \partial _s \left( F_1'(\rho (s,s\theta ))-\frac{3}{4}\right) \right|\le & {} C|F_1''(\rho (s,s\theta ))||\partial _s(\rho (s,s\theta ))| \nonumber \\\le & {} C\frac{(1+\theta )^4}{(1-\theta )^4} \left| \ln \left( \frac{1+\theta }{2}\right) \right| . \end{aligned}$$
(4.79)

Interpolating between (4.76) and (4.79) we achieve

$$\begin{aligned}&\left| F_1'(\rho (s_1,s_1\theta ))-F_1'(\rho (s_2,s_2\theta ))\right| \nonumber \\&\quad \le C|s_1-s_2|^\alpha \frac{(1+\theta )^{4\alpha }}{(1-\theta )^{4\alpha }}\left| \ln \left( \frac{1+\theta }{2}\right) \right| ^\alpha \frac{\theta ^{1-\alpha }}{(1-\theta )^{2(1-\alpha )}}\nonumber \\&\quad \le C|s_1-s_2|^\alpha \frac{(1+\theta )^{1+3\alpha }}{(1-\theta )^{2+2\alpha }}\left| \ln \left( \frac{1+\theta }{2}\right) \right| ^\alpha . \end{aligned}$$
(4.80)

Finally, using (4.72) and (4.80) we obtain for any \(0\le s_1,s_2\le \phi \)

the convergence of the integral is guaranteed pro

$$\begin{aligned} \Big |\partial _\phi \left( \frac{\sin (\phi \theta )}{\phi } \right) \Big |\le \theta ^2. \end{aligned}$$
(4.81)

vided that \(\alpha \in (0,1)\). This achieves the proof of \(\nu _\Omega \in {\mathscr {C}}^{1,\alpha }(0,\pi )\) for any \(\alpha \in (0,1)\).

(4) Since the function \(\nu _\Omega \) reaches its minimum at a point \(\phi _0\in [0,\pi ]\), we have that if this point belongs to the open set \((0,\pi )\) then necessary \(\nu _\Omega ^\prime (\phi _0)=0\). However when \(\phi _0\in \{0,\pi \}\) then from the point (3) of Proposition 4.1 we deduce also that the derivative is vanishing at \(\phi _0\). Using the mean value theorem, we obtain for any \(\phi \in [0,\pi ]\)

$$\begin{aligned} \nu _\Omega (\phi )&= \nu _\Omega (\phi _0)+\nu _\Omega ^\prime \big ({{\overline{\phi }}}\big )(\phi -\phi _0)=\nu _\Omega (\phi _0)+\big (\nu _\Omega ^\prime \big ({{\overline{\phi }}}\big )-\nu _\Omega ^\prime \big (\phi _0\big )\big )(\phi -\phi _0), \end{aligned}$$

for some \({{\overline{\phi }}}\in (\phi _0,\phi )\). Since \(\nu _\Omega ^\prime \in {\mathscr {C}}^\alpha \) then

$$\begin{aligned} \Big |\nu _\Omega ^\prime \big ({{\overline{\phi }}}\big )-\nu _\Omega ^\prime \big (\phi _0\big )\Big |\le \Vert \nu _\Omega ^\prime \Vert _{{\mathscr {C}}^\alpha }|\phi -\phi _0|^\alpha . \end{aligned}$$

Notice that \(\Vert \nu _\Omega ^\prime \Vert _{{\mathscr {C}}^\alpha }\) is independent of \(\Omega \). Consequently

$$\begin{aligned} \forall \phi \in [0,\pi ],\quad 0\le \nu _\Omega \big (\phi \big )-\nu _\Omega \big (\phi _0\big )\le C|\phi -\phi _0|^{1+\alpha }, \end{aligned}$$

for some absolute constant C. In the particular case \(\Omega =\kappa \) we get from the definition (4.5) that \(\nu _\kappa (\phi _0)=0\) and therefore the preceding result becomes

$$\begin{aligned} \forall \phi \in [0,\pi ],\quad 0\le \nu _\kappa \big (\phi \big )\le C|\phi -\phi _0|^{1+\alpha },\quad \nu _\kappa \big (\phi _0\big )=0. \end{aligned}$$

\(\square \)

4.3 Eigenvalue problem

In Sect. 4.1 we have checked that the operator \({\mathcal {L}}_n^\Omega \) defined in (4.1) is of integral type. Then studying the kernel of this operator reduces to solving the integral equation

$$\begin{aligned} {\mathcal {K}}_n^{{\Omega }} h_n(\phi ):=\int _0^\pi K_n(\phi ,\varphi )h_n(\varphi )d\mu _{\Omega }(\varphi )=h_n(\phi ),\quad \forall \phi \in [0,\pi ], \end{aligned}$$
(4.82)

where the kernel \(K_n\) and the measure \(d\mu _\Omega \) are defined successively in (4.2) and (4.4). The parameter \(\Omega \) ranges over the interval \((-\infty ,\kappa )\). This latter condition is imposed to guarantee the positivity of the measure \(d\mu _\Omega \) through the positivity of \(\nu _\Omega \) according to Lemma 4.1. We point out that studying the kernel of \({\mathcal {L}}_n^\Omega \) amounts to finding the values of \(\Omega \) such that 1 is an eigenvalue of \({\mathcal {K}}_n^{{\Omega }}\). To investigate the spectral study of \({\mathcal {K}}_n^{{\Omega }}\) we need to introduce the Hilbert space \(L^2_{\mu _\Omega }\) of measurable functions \(f:[0,\pi ]\rightarrow {\mathbb {R}}\) such that

$$\begin{aligned} \Vert f\Vert _{\mu _\Omega }:=\left( \int _0^\pi |f(\varphi )|^2d\mu _{\Omega }(\varphi )\right) ^{\frac{1}{2}}<\infty . \end{aligned}$$
(4.83)

Notice that the space \(L^2_{\mu _\Omega }\) is equipped with the usual inner product:

$$\begin{aligned} \langle f,g \rangle _{{\Omega }} =\int _{0}^{\pi }f(\varphi )g(\varphi )d\mu _{\Omega }(\varphi ), \quad \forall \, f,g\in L^2_{\mu _\Omega }. \end{aligned}$$
(4.84)

Remarks 4.1

  1. (1)

    Since \(d\mu _\Omega \) is a nonnegative bounded Borel measure for any \(\Omega \in (-\infty ,\kappa )\), then the Hilbert space \(L^2_{\mu _\Omega }\) is separable.

  2. (2)

    For any \(\Omega \in (-\infty ,\kappa )\), the space \(L^2_{\mu _\Omega }\) is isomorphic to the space \(L^2_{\mu }\) where

    $$\begin{aligned} d\mu (\varphi )=\sin (\varphi )\, r_0^2(\varphi )\, d\varphi . \end{aligned}$$

    This follows from Proposition 4.1-(2) which ensures that \(\nu _\Omega \) is nowhere vanishing. However this property fails for the critical value \(\Omega =\kappa \) because \(\nu _\kappa \) is vanishing at some points.

The next proposition deals with some basic properties of the operator \({\mathcal {K}}_n^{{\Omega }}\).

Proposition 4.2

Let \(\Omega \in (-\infty ,\kappa )\) and \(r_0\) satisfies the assumptions (H1) and (H2). Then, the following assertions hold true.

  1. (1)

    For any \(n\ge 1\), the operator \({\mathcal {K}}_n^{{\Omega }}:L^2_{\mu _\Omega }\rightarrow L^2_{\mu _\Omega }\) is Hilbert–Schmidt and self-adjoint.

  2. (2)

    For any \(n\ge 1\), the eigenvalues of \({\mathcal {K}}_n^{{\Omega }}\) form a countable family of real numbers. Let \(\lambda _n(\Omega )\) be the largest eigenvalue, then it is strictly positive and satisfies

    for any function \(\varrho \) such that \(\displaystyle {\int _0^\pi \varrho ^2(\varphi )d\varphi =1.}\)

  3. (3)

    We have the following decay: for any \(\alpha \in [0,1)\) there exists \(C>0\) such that

  4. (4)

    The eigenvalue \(\lambda _n(\Omega )\) is simple and the associated nonzero eigenfunctions do not vanish in \((0,\pi )\).

  5. (5)

    For any \(\Omega \in (-\infty ,\kappa )\), the sequence \(n\in {{\,\mathrm{{\mathbb {N}}}\,}}^{\star }\mapsto \lambda _n(\Omega )\) is strictly decreasing.

  6. (6)

    For any \(n\ge 1\) the map \(\Omega \in (-\infty ,\kappa )\mapsto \lambda _n(\Omega )\) is differentiable and strictly increasing.

Proof

(1) In order to check that \({\mathcal {K}}_n^{{\Omega }}\) is a Hilbert–Schmidt operator, we need to verify that the kernel \(K_n\) satisfies the integrability condition

$$\begin{aligned} \Vert {\mathcal {K}}_n^{{\Omega }}\Vert _{\mu _{\Omega }}:=\left( \int _0^\pi \int _0^\pi |K_n(\phi ,\varphi )|^2d\mu _{\Omega }(\varphi )d\mu _{\Omega }(\phi )\right) ^{\frac{1}{2}}<+\infty . \end{aligned}$$

Indeed, by (4.2) and (3.3), one gets

for some constant \(C_n\) and R was defined in (3.2). Remark that

$$\begin{aligned} \left| \frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right| \le 1. \end{aligned}$$

Moreover, according to Proposition 4.1 the function \(\nu _{\Omega }(\varphi )\) is not vanishing in the interval \([0,\pi ]\) provided that \(\Omega <\kappa \). Therefore we get

By (A.7) and the assumption (H2) we deduce that

It suffices now to use the inequality (4.10) to get

This concludes that the operator \({\mathcal {K}}_n^{{\Omega }}\) is bounded and is of Hilbert–Schmidt type. As a consequence from the general theory this operator is necessarily compact.

On the other hand, as we have mentioned before the kernel \(K_n\) is symmetric in view of the formula (4.6) and the symmetry of R defined in (3.2). Therefore we deduce that \({\mathcal {K}}_n^{{\Omega }}\) is a self-adjoint operator

(2) From the spectral theorem on self-adjoint compact operators, we know that the eigenvalues of \({\mathcal {K}}_n^{{\Omega }}\) form a countable family of real numbers. Define the real numbers

$$\begin{aligned} m=\inf _{\Vert h\Vert _{\mu _{\Omega }}=1} \langle {\mathcal {K}}_n^{{\Omega }} h, h\rangle _{{\Omega }} \quad \text {and} \quad M=\sup _{\Vert h\Vert _{\mu _{\Omega }}=1} \langle {\mathcal {K}}_n^{{\Omega }} h, h\rangle _{{\Omega }}. \end{aligned}$$

Since \({\mathcal {K}}\) is self-adjoint, we obtain \(\sigma ({\mathcal {K}}_n^{{\Omega }})\subset [m,M]\), with \(m\in \sigma ({\mathcal {K}}_n^{{\Omega }})\) and \(M\in \sigma ({\mathcal {K}}_n^{{\Omega }})\), where the set \(\sigma ({\mathcal {K}}_n^{{\Omega }})\) denotes the spectrum of \({\mathcal {K}}_n^{{\Omega }}\). Since \(\lambda _n(\Omega )\) is the largest eigenvalue, then

$$\begin{aligned} \lambda _n(\Omega )=M=\sup _{\Vert h\Vert _{\mu _{\Omega }}=1} \langle {\mathcal {K}}_n^{{\Omega }} h, h\rangle _{{\Omega }}. \end{aligned}$$
(4.85)

We shall prove that \(M>0\) and \(|m|\le M\). Indeed, for any \(h\in L^2_{\mu _\Omega }\), the positive function |h| belongs also to \( L^2_{\mu _\Omega }\) with the same norm and using the positivity of the kernel \(K_n\) we obtain

$$\begin{aligned} \sup _{\Vert h\Vert _{\mu _{\Omega }}=1} \langle {\mathcal {K}}_n^{{\Omega }} h, h\rangle _{{\Omega }}=\sup _{h\ge 0, \Vert h\Vert _{\mu _{\Omega }}=1} \langle {\mathcal {K}}_n^{{\Omega }} h, h\rangle _{{\Omega }}. \end{aligned}$$

Using once again the positivity of the kernel one deduces that

$$\begin{aligned} \forall h\ge 0, \Vert h\Vert _{\mu _{\Omega }}=1\Longrightarrow \langle {\mathcal {K}}_n^{{\Omega }} h, h\rangle _{{\Omega }}>0. \end{aligned}$$

Consequently, we obtain that \(M>0\). In order to prove that \(|m|\le M\), we shall proceed as follows. Using the positivity of the kernel, we achieve

$$\begin{aligned} |m| \le \langle {\mathcal {K}}_n^{{\Omega }} |h|, |h|\rangle _{{\Omega }}\le M,\quad \forall \,\Vert h\Vert _{\mu _\Omega }=1. \end{aligned}$$

This implies that M is nothing but the spectral radius of the operator \({\mathcal {K}}_n^{{\Omega }}\), that is,

$$\begin{aligned} M=\Vert {\mathcal {K}}_n^{{\Omega }}\Vert _{{\mathcal {L}}(L^2_{\mu _\Omega })}. \end{aligned}$$

From the Cauchy–Schwarz inequality, one deduces that

$$\begin{aligned} \Vert {\mathcal {K}}_n^{{\Omega }}\Vert _{{\mathcal {L}}(L^2_{\mu _\Omega })}^2 \le \int _0^\pi \int _0^\pi |K_n(\phi ,\varphi )|^2d\mu _{\Omega }(\phi )d\mu _{\Omega }(\varphi ), \end{aligned}$$

which implies that

$$\begin{aligned} \lambda _n^2(\Omega )\le \int _0^\pi \int _0^\pi |K_n(\phi ,\varphi )|^2d\mu _{\Omega }(\phi )d\mu _{\Omega }(\varphi ). \end{aligned}$$

For the lower bound, we shall work with the special function

$$\begin{aligned} f(\varphi )=\frac{\varrho (\varphi )}{\sin (\varphi )^\frac{1}{2} r_0(\varphi )\nu _{\Omega }(\varphi )^{\frac{1}{2}}},\quad \varphi \in (0,\pi ), \end{aligned}$$

with the normalized condition \(\Vert f\Vert _{\mu _{\Omega }}=1\) which is equivalent to

$$\begin{aligned} \int _0^\pi \varrho ^2(\varphi )d\varphi =1 \end{aligned}$$

and

This gives the announced lower bound for the largest eigenvalue.

(3) From the expression of \(K_n\) given by (4.2) we easily get

Using the definition (4.5) of \(\kappa \) we infer

$$\begin{aligned} \forall \, \phi \in [0,\pi ],\quad \nu _\Omega (\phi )\ge \kappa -\Omega \end{aligned}$$

and we obtain

Applying Lemma 3.1 combined with the assumption (H2) yields for any \(0\le \alpha <\beta \le 1\)

By taking \(\beta <\frac{1}{2}\) we get the convergence of the integral and consequently we obtain the desired result,

$$\begin{aligned} {\Vert {\mathcal {K}}_n^{{\Omega }}\Vert _{\mu _\Omega }^2\lesssim (\kappa -\Omega )^{-2} n^{-2\alpha }.} \end{aligned}$$
(4.86)

(4) First, let us check that any nonzero eigenfunction associated to the largest eigenvalue \(\lambda _n(\Omega )\) should be with a constant sign. Indeed, let f be a nonzero normalized eigenfunction and assume that it changes the sign over a non negligible set. From the strict positivity of the kernel in the interval \((0,\pi )\), we deduce that

$$\begin{aligned} {\mathcal {K}}_n^{{\Omega }} f(\phi )<{\mathcal {K}}_n^{{\Omega }} |f|(\phi ),\quad \forall \phi \in (0,\pi ). \end{aligned}$$

First, by the assumption on f we get

$$\begin{aligned} \int _0^\pi {\mathcal {K}}_n^{{\Omega }}(f)(\phi )f(\phi )d\mu _{\Omega }(\phi )=\lambda _n(\Omega )\int _0^\pi f^2(\phi ) d\mu _{\Omega }(\phi )=\lambda _n(\Omega ). \end{aligned}$$

Second, from (4.85) we have that

$$\begin{aligned} \int _0^\pi {\mathcal {K}}_n^{{\Omega }}(|f|)(\phi )|f(\phi )|d\mu _{\Omega }(\phi )\le \lambda _n(\Omega ). \end{aligned}$$

Consequently,

$$\begin{aligned} \lambda _n(\Omega )=\int _0^\pi {\mathcal {K}}_n^{{\Omega }}(f)(\phi )f(\phi )d\mu _{\Omega }(\phi )< \int _0^\pi {\mathcal {K}}_n^{{\Omega }}(|f|)(\phi )|f(\phi )|d\mu _{\Omega }(\phi )\le \lambda _n(\Omega ), \end{aligned}$$

achieving a contradiction. Hence, any nonzero eigenfunction of \(\lambda _n(\Omega )\) must have a constant sign. Now let us check that f is not vanishing in \((0,\pi )\). First we write

$$\begin{aligned} f(\phi )=\frac{1}{\lambda _n(\Omega )}{\mathcal {K}}_n^{{\Omega }} f(\phi )=\frac{1}{\lambda _n(\Omega )\nu _\Omega (\phi )}\int _0^\pi H_n(\phi ,\varphi ) f(\varphi ) d\varphi . \end{aligned}$$

From (3.3) and Proposition 4.1 we get

$$\begin{aligned} \forall \phi ,\varphi \in (0,\pi ),\quad H_n(\phi ,\varphi )>0, \quad \nu _\Omega (\phi )>0. \end{aligned}$$

The first assertion follows from the strict positivity of the associated hypergeometric function. Combined with the positivity of f we deduce that

$$\begin{aligned} \forall \phi \in (0,\pi ),\quad f(\phi )>0. \end{aligned}$$

Finally, we shall check that the subspace generated by the eigenfunctions associated to \(\lambda _n(\Omega )\) is one-dimensional. Assume that we have two independent eigenfunctions \(f_0\) and \(f_1\), which are necessarily with constant sign, then there exists \(a, b\in {{\,\mathrm{{\mathbb {R}}}\,}}\) such that the eigenfunction \(af_0+bf_1\) changes its sign. This is a contradiction.

(5) Using (4.2) combined with Lemma 3.1, we get that \(n\in {{\,\mathrm{{\mathbb {N}}}\,}}^{\star }\mapsto K_n(\phi ,\varphi )\) is strictly decreasing for any \(\varphi \ne \phi \in (0,\pi )\). Then, for any \(\Omega \in (-\infty ,\kappa )\) and for any nonnegative function f, we get

$$\begin{aligned} \forall \, \phi \in (0,\pi ),\quad {\mathcal {K}}_n^{{\Omega }} f(\phi )>{\mathcal {K}}_{n+1}^{{\Omega }}f(\phi ), \end{aligned}$$

which implies in turn that

$$\begin{aligned} \int _0^\pi {\mathcal {K}}_n^{{\Omega }}(f)(\phi )f(\phi )d\mu _\Omega (\phi )>\int _0^\pi {\mathcal {K}}_{n+1}^{{\Omega }}(f)(\phi )f(\phi )d\mu _\Omega (\phi ). \end{aligned}$$

Since the largest eigenvalue \(\lambda _{n+1}(\Omega )\) is reached at some positive normalized function \(f_{n+1}\ge 0\), then

$$\begin{aligned} \lambda _{n+1}(\Omega )&=\int _0^\pi {\mathcal {K}}_{n+1}^{{\Omega }}(f_{n+1})(\phi ) f_{n+1}(\phi )d\mu _{\Omega }(\phi )\\&<\int _0^\pi {\mathcal {K}}_n^{{\Omega }}(f_{n+1})(\phi ) f_{n+1}(\phi )d\mu _{\Omega }(\phi )\\&<\sup _{\Vert f\Vert _{\mu _\Omega }=1}\int _0^\pi {\mathcal {K}}_n^{{\Omega }}(f)(\phi ) f(\phi )d\mu _{\Omega }(\phi )\\&<\lambda _{n}(\Omega ). \end{aligned}$$

This provides the announced result.

(6) Fix \(\Omega _0\in (-\infty ,\kappa )\) and denote by \(f_n^\Omega \) the positive normalized eigenfunction associated to the eigenvalue \(\lambda _n(\Omega )\). Using the definition of the eigenfunction yields

$$\begin{aligned} \lambda _n(\Omega )=\frac{\langle {\mathcal {K}}_n^{{\Omega }} f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}{\langle f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}, \quad \Vert f_n^\Omega \Vert _{\mu _{\Omega _0}}=1. \end{aligned}$$
(4.87)

The regularity follows from the general theory using the fact that this eigenvalue is simple. However we can in our special case give a direct proof for its differentiability in the following way. From the decomposition

$$\begin{aligned} \frac{1}{\nu _\Omega (\phi )}=\frac{1}{\nu _{\Omega _0}(\phi )}+\frac{\Omega -\Omega _0}{\nu _\Omega (\phi )\nu _{\Omega _0}(\phi )}, \end{aligned}$$

we get according to the expression of \({\mathcal {K}}_n^{{\Omega }}\)

(4.88)

with

(4.89)

Therefore we obtain

$$\begin{aligned} \lambda _n(\Omega )=\frac{\langle {\mathcal {K}}_n^{{\Omega _0}} f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}{\langle f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}+(\Omega -\Omega _0)\frac{\langle {\mathscr {R}}_n^{\Omega _0,\Omega }f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}{\langle f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}. \end{aligned}$$

As \({\mathcal {K}}_n^{{\Omega _0}}\) is self-adjoint on the Hilbert space \(L^2_{\mu _{\Omega _0}}\) then

$$\begin{aligned} \frac{\langle {\mathcal {K}}_n^{{\Omega _0}} f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}{\langle f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}=\frac{\langle f_n^\Omega , {\mathcal {K}}_n^{{\Omega _0}}f_n^{\Omega _0}\rangle _{{\Omega _0}}}{\langle f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}=\lambda _n(\Omega _0). \end{aligned}$$

Let us assume for a while that

$$\begin{aligned} {\lim _{\Omega \rightarrow \Omega _0}\frac{\langle {\mathscr {R}}_n^{\Omega _0,\Omega }f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}{\langle f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}}={\langle {\mathscr {R}}_n^{\Omega _0,\Omega _0}f_n^{\Omega _0}, f_n^{\Omega _0}}\rangle _{{\Omega _0}}}. \end{aligned}$$
(4.90)

Then we deduce that \(\Omega \mapsto \lambda _n(\Omega )\) is differentiable at \(\Omega _0\) with

Since

$$\begin{aligned} \forall \varphi , \phi \in (0,\pi ), \quad H_n(\phi , \varphi )>0, \,\,f_n^{\Omega _0}(\phi )>0, \,\,\nu _{\Omega _0}(\phi )>0, \end{aligned}$$

we find that \( \lambda _n^\prime (\Omega _0)>0\), which achieves the proof of the suitable result.

It remains to prove (4.90). First, for the numerator of the left hand side we first make the splitting

$$\begin{aligned} \langle {\mathscr {R}}_n^{\Omega _0,\Omega }f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}&=\langle {\mathscr {R}}_n^{\Omega _0,\Omega _0}f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}+\langle ( {\mathscr {R}}_n^{\Omega _0,\Omega }-{\mathscr {R}}_n^{\Omega _0,\Omega _0})f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}} \nonumber \\&:= {\mathcal {I}}_1(\Omega )+{\mathcal {I}}_2(\Omega ). \end{aligned}$$
(4.91)

To estimate the second term \({\mathcal {I}}_2\) we use the identities (4.88) and (4.89) leading to

$$\begin{aligned} {\mathscr {R}}_n^{\Omega _0,\Omega }-{\mathscr {R}}_n^{\Omega _0,\Omega _0}=\frac{\Omega -\Omega _0}{\nu _{\Omega _0}^2(\phi )}{\mathcal {K}}_n^{{\Omega }}. \end{aligned}$$

It follows that

$$\begin{aligned} {\mathcal {I}}_2(\Omega )=(\Omega -\Omega _0)\langle {\nu ^{-2}_{\Omega _0}} {\mathcal {K}}_n^{\Omega }f_n^\Omega ,f_n^{\Omega _0}\rangle _{{\Omega _0}}. \end{aligned}$$

Hence, applying Proposition 4.1-(2) combined with Cauchy-Schwarz inequality and the normalization assumption of the eigenfunctions in (4.87) we infer

$$\begin{aligned} \big |{\mathcal {I}}_2(\Omega )\big |\lesssim |\Omega -\Omega _0| \Vert {\mathcal {K}}_n^{\Omega }f_n^\Omega \Vert _{\mu _{\Omega _0}}. \end{aligned}$$
(4.92)

From Remarks 4.1-(2), (4.86) and (4.87) we may write for \(\Omega \) close to \(\Omega _0\)

$$\begin{aligned} \big |{\mathcal {I}}_2(\Omega )\big |&\lesssim |\Omega -\Omega _0| \Vert {\mathcal {K}}_n^{\Omega }f_n^\Omega \Vert _{\mu _{\Omega }}\\&\lesssim |\Omega -\Omega _0|. \end{aligned}$$

This obviously gives

$$\begin{aligned} \lim _{\Omega \rightarrow \Omega _0} {\mathcal {I}}_2(\Omega )=0. \end{aligned}$$
(4.93)

Let us move to the term \({\mathcal {I}}_1\) introduced in (4.91). Then combining (4.89) with the fact that \({\mathcal {K}}_n^{{\Omega _0}}\) is self-adjoint on the Hilbert space \(L^2_{\mu _{\Omega _0}}\) allows to get

$$\begin{aligned} {\mathcal {I}}_1(\Omega )&=\langle f_n^\Omega , g\rangle _{{\Omega _0}}\quad \hbox {with}\quad g:={\mathcal {K}}_n^{{\Omega _0}}\big (\nu ^{-1}_{\Omega _0}f_n^{\Omega _0}\big ). \end{aligned}$$
(4.94)

Applying Proposition 4.1-(2) we easily get that \(g\in L^2_{\mu _{\Omega _0}}\). Now we claim that

$$\begin{aligned} \lim _{\Omega \rightarrow \Omega _0}\Vert f_n^\Omega -f_n^{\Omega _0}\Vert _{\mu _{\Omega _0}}=0. \end{aligned}$$
(4.95)

Before giving its proof, let us see how to conclude. It is easy to check from (4.94) and (4.95) that

$$\begin{aligned} \lim _{\Omega \rightarrow \Omega _0} {\mathcal {I}}_1(\Omega )={\mathcal {I}}_1(\Omega _0)\quad \hbox {and}\quad \lim _{\Omega \rightarrow \Omega _0}\langle f_n^\Omega , f_n^{\Omega _0}\rangle _{{\Omega _0}}=1. \end{aligned}$$

Thus combining this result with (4.91) and (4.93) yields (4.90). It remains to check (4.95) which is a consequence of classical results on perturbation theory. One can use for instance [42] or [38, Chapter XII], where the analytic dependence of the eigenvalues and the associated eigenfunctions is analyzed. Let us briefly discuss the main arguments used to get the continuity of the eigenfunctions with respect to the parameter \(\Omega \). First we set

$$\begin{aligned} A(\Omega ):={\mathcal {K}}_n^{{\Omega }}. \end{aligned}$$

Then using (4.3) we finds

$$\begin{aligned} \frac{1}{\nu _{\Omega }(\phi )}=\sum _{m\in {{\,\mathrm{{\mathbb {N}}}\,}}}\frac{(\Omega -\Omega _0)^m}{\nu ^{m+1}_{\Omega _0}(\phi )}\cdot \end{aligned}$$

Then similarly to (4.88) we obtain the decomposition

$$\begin{aligned} A(\Omega )=\sum _{m\in {{\,\mathrm{{\mathbb {N}}}\,}}}(\Omega -\Omega _0)^mA_m,\quad \quad \hbox {with}\quad A_m:=\nu ^{-m}_{\Omega _0}(\phi ){{\mathcal {K}}_n^{{\Omega _0}}}. \end{aligned}$$
(4.96)

By applying the lower bound of Proposition 4.1-(2) we get that \(A_m\) is bounded with

$$\begin{aligned} \Vert A_m\Vert _{{\mathcal {L}}(L^2_{\mu _{\Omega _0}})}\lesssim (\kappa -\Omega _0)^{-m} \end{aligned}$$

This shows that \(A(\Omega )\) is analytic for \(\Omega \) close enough to \(\Omega _0\). Now define the sets

$$\begin{aligned} \rho (A(\Omega )):= & {} \Big \{z\in {\mathbb {C}}, A(\Omega )-z\text {Id}\, \hbox { is invertible }\Big \}\quad \hbox {and}\quad \\ \Gamma:= & {} \Big \{(\Omega , z)\in {\mathbb {C}}^2, z\in \rho (A(\Omega ))\Big \}. \end{aligned}$$

Then it is known that the resolvent set \(A(\Omega )\) and \(\Gamma \) are open, see Theorem XII.7 in [38]. Now, since \(\lambda _n(\Omega _0)\) is an isolated simple eigenvalue and \(\Gamma \) is open then we may find \(\delta >0\) such that the oriented circle \(\gamma :=\big \{z, |z- \lambda _n(\Omega _0)|=\delta \big \}\) is contained in \(\rho (A(\Omega ))\) for all \(|\Omega -\Omega _0|\leqslant \delta \). Thus, for \(|\Omega -\Omega _0|\leqslant \delta \) the operator

$$\begin{aligned} P_n(\Omega )=-\frac{1}{2i \pi }\int _{\gamma }\big (A(\Omega )-z\text {Id}\big )^{-1} dz \end{aligned}$$

is well-defined and it is analytic in view of (4.96). Notice that one may get the estimate

$$\begin{aligned} \Vert P_n(\Omega )-P_n(\Omega _0)\Vert _{{\mathcal {L}}(L^2_{\mu _{\Omega _0}})}\lesssim |\Omega -\Omega _0|. \end{aligned}$$
(4.97)

Then from classical results on spectral theory, see for instance Theorems XII.5-XII.6-XII.8 in [38], \(P_n(\Omega _0)\) is a projection on the one dimensional eigenspace associate to \(\lambda _n(\Omega _0)\). In addition, \(A_n(\Omega )\) admits only one eigenvalue inside the circle \(\Gamma \) which necessary coincides with \(\lambda _n(\Omega )\). Notice that this latter claim can be proved using the continuity in \(\Omega \) of the largest eigenvalue which can be checked from (4.85). Furthermore, \(P_n(\Omega )\) is still a one dimensional projection on the eigenspace associated to the \(\lambda _n(\Omega )\). As a consequence, if \(f_n^{\Omega _0}\) is a normalized eigenfunction of the operator \(A(\Omega _0)={\mathcal {K}}_n^{{\Omega _0}}\) associated to \(\lambda _n(\Omega _0)\), then \( P_n(\Omega )f_n^{\Omega _0} \) is an eigenfunction of \(A(\Omega )\) associated to \(\lambda _n(\Omega )\). Applying (4.97) yields

$$\begin{aligned} \lim _{\Omega \rightarrow \Omega _0}\Vert P_n(\Omega )f_n^{\Omega _0}-f_n^{\Omega _0}\Vert _{{\mu _{\Omega _0}}}=0, \end{aligned}$$

where we have used the fact that \(P_n(\Omega _0)f_n^{\Omega _0}=f_n^{\Omega _0}\). Now, by taking

$$\begin{aligned} f_n^{\Omega }:=\frac{P_n(\Omega )f_n^{\Omega _0}}{\Vert P_n(\Omega )f_n^{\Omega _0}\Vert }_{{\mu _{\Omega _0}}} \end{aligned}$$

we get a normalized eigenfunction in the sense of (4.87) and the family \(\Omega \mapsto f_n^\Omega \in L^2_{\mu _{\Omega _0}}\) is continuous at \(\Omega _0\), which ensures (4.95). This ends the proof of the desired result. \(\quad \square \)

Next we shall establish the following result.

Proposition 4.3

Let \(n\ge 1\) and \(r_0\) satisfies the assumptions (H1) and (H2). Set

$$\begin{aligned} {\mathscr {S}}_n:=\Big \{\Omega \in (-\infty ,\kappa ) \quad \text {s.t.}\quad \lambda _n(\Omega )=1\Big \}. \end{aligned}$$
(4.98)

Then the following holds true

  1. (1)

    The set \({\mathscr {S}}_n\) is formed by a single point denoted by \(\Omega _n\) .

  2. (2)

    The sequence \((\Omega _n)_{n\ge 1}\) is strictly increasing and satisfies

    $$\begin{aligned} \lim _{n\rightarrow \infty }\Omega _n=\kappa . \end{aligned}$$

Proof

(1) To check that the set \({\mathscr {S}}_n\) is non empty we shall use the mean value theorem. From the upper bound in Proposition 4.2-(2) and (4.2) we find that

Thus by taking the limit as \(\Omega \rightarrow -\infty \) we deduce that

$$\begin{aligned} \lim _{\Omega \rightarrow -\infty } \lambda _n(\Omega )=0. \end{aligned}$$
(4.99)

Next, we intend to show that

$$\begin{aligned} \lim _{\Omega \rightarrow \kappa }\lambda _n(\Omega )=\infty . \end{aligned}$$
(4.100)

Using the lower bound of \(\lambda _n(\Omega )\) in Proposition 4.2-(2), we find by virtue of Fatou Lemma

for any nonnegative \(\varrho \) satisfying \(\displaystyle {\int _0^\pi \varrho ^2(\phi )d\phi =1}\). According to Proposition 4.1-(4), the function \(\nu _\kappa \) reaches its minimum at a point \(\phi _0\in [0,\pi ]\) and

$$\begin{aligned} \forall \, \phi \in [0,\pi ],\quad 0\le \nu _\kappa (\phi )\le C|\phi -\phi _0|^{1+\alpha }. \end{aligned}$$

There are two possibilities: \(\phi _0\in (0,\pi )\) or \(\phi _0\in \{0,\pi \}\). Let us start with the first case and we shall take \(\varrho \) as follows

$$\begin{aligned} \varrho (\phi )=\frac{c_\beta }{|\phi -\phi _0|^\beta }, \end{aligned}$$

with \(\beta <\frac{1}{2}\) and the constant \(c_\beta \) is chosen such that \(\varrho \) is normalized. Hence using the preceding estimates we get

(4.101)

Let \(\varepsilon >0\) such that \([\phi _0-\varepsilon ,\phi _0+\varepsilon ]\subset (0,\pi )\). According to (3.3) the function \(H_n\) is strictly positive in the domain \((0,\pi )^2\), hence there exists \(\delta >0\) such

$$\begin{aligned} \forall \, (\phi ,\varphi )\in [\phi _0-\varepsilon ,\phi _0+\varepsilon ]^2, \quad \frac{H_n(\phi ,\varphi )\sin ^\frac{1}{2}(\phi ) r_0(\phi )}{\sin ^\frac{1}{2}(\varphi ) r_0(\varphi )}\ge \delta . \end{aligned}$$

Thus we obtain

By taking \(\frac{1+\alpha }{2}+\beta >1\), which is an admissible configuration, we find

$$\begin{aligned} \lim _{\Omega \rightarrow \kappa } \lambda _n(\Omega )=+\infty . \end{aligned}$$

Now let us move to the second possibility where \(\phi _0\in \{0,\pi \}\) and without any loss of generality we can only deal with the case \(\phi _0=0\). From (3.3) and using the inequality

$$\begin{aligned} \forall \, x\in [0,1),\quad F_n(x)\ge 1, \end{aligned}$$

we obtain

$$\begin{aligned} \forall \phi ,\varphi \in (0,\pi ),\quad H_n(\phi ,\varphi )&\ge c_n\frac{\sin (\varphi )r_0^{n-1}(\phi )r_0^{n+1}(\varphi )}{\left[ R(\phi ,\varphi )\right] ^{n+\frac{1}{2}}}. \end{aligned}$$

Combined with the assumption \({\mathbf{(H2)}}\), it implies

$$\begin{aligned} \forall \phi ,\varphi \in (0,\pi ),\quad H_n(\phi ,\varphi )&\ge c_n\frac{\sin ^{n+2}(\varphi )\sin ^{n-1}(\phi )}{\left[ R(\phi ,\varphi )\right] ^{n+\frac{1}{2}}}. \end{aligned}$$

Plugging this into (4.101) we find

Let \(\varepsilon >0\) sufficiently small, then using Taylor expansion we get according to (3.2)

$$\begin{aligned} 0\le \phi ,\varphi \le \varepsilon \Longrightarrow R(\phi ,\varphi )\le C(\phi +\varphi )^2. \end{aligned}$$

Thus

which gives after simplification

Making the change of variables \(\varphi =\phi \theta \) we obtain

This integral diverges provided that \(\alpha +2\beta >1\) and thus under this assumption

$$\begin{aligned} \lim _{\Omega \rightarrow \kappa } \lambda _n(\Omega )=+\infty . \end{aligned}$$

Hence we obtain (4.100). By the intermediate mean value, we achieve the existence of at least one solution for the equation

$$\begin{aligned} \lambda _n(\Omega )=1. \end{aligned}$$

Consequently, using Proposition 4.2 we deduce by the mean value theorem that the set \({\mathscr {S}}_n\) contains only one element.

(2) Since \(\Omega _n\) satisfies the equation

$$\begin{aligned} \lambda _n(\Omega _n)=1. \end{aligned}$$

According to Proposition 4.2-(5) the sequence \(k\mapsto \lambda _k(\Omega _n)\) is strictly decreasing. It implies in particular that

$$\begin{aligned} \lambda _{n+1}(\Omega _n)<\lambda _n(\Omega _n)=1. \end{aligned}$$

Hence by (4.100) one may apply the mean value theorem and find an element of the set \({\mathscr {S}}_{n+1}\) in the interval \((\Omega _n,\kappa )\). This means that \(\Omega _{n+1}>\Omega _n\) and thus this sequence is strictly increasing. It remains to prove that this sequence is converging to \(\kappa \). The convergence of this sequence to some element \({\overline{\Omega }}\le \kappa \) is clear. To prove that \({\overline{\Omega }}=\kappa \) we shall argue by contradiction by assuming that \({\overline{\Omega }}<\kappa \). By the construction of \(\Omega _n\) one has necessarily

$$\begin{aligned} \forall n\ge 1,\quad \lambda _n({\overline{\Omega }})>1. \end{aligned}$$

Using the upper-bound estimate stated in Proposition 4.2-(2) combined with the point (3) we obtain for any \(\alpha \in (0,1)\)

$$\begin{aligned} \forall \, n\ge 1,\quad 0< \lambda _n({\overline{\Omega }})\lesssim {(\kappa -{\overline{\Omega }})^{-1} n^{-\alpha }}. \end{aligned}$$
(4.102)

By taking the limit as \(n\rightarrow +\infty \) we find

$$\begin{aligned} \lim _{n\rightarrow \infty }\lambda _n({\overline{\Omega }})=0. \end{aligned}$$

This contradicts (4.102) which achieves the proof. \(\quad \square \)

4.4 Eigenfunctions regularity

This section is devoted to the strong regularity of the eigenfunctions associated to the operator \({\mathcal {K}}_n^{{\Omega }}\) and constructed in Proposition 4.2. We have already seen that these eigenfunctions belong to a weak function space \(L^2_{\mu _\Omega }\). Here we shall show first their continuity and later their Hölder regularity.

4.4.1 Continuity

The main result of this section reads as follows.

Proposition 4.4

Let \(\Omega \in (-\infty ,\kappa )\), \(n\ge 1\), \(r_0\) satisfies the assumptions (H1) and (H2), and f be an eigenfunction for \({\mathcal {K}}_n^{{\Omega }}\) associated to a non-vanishing eigenvalue. Then f is continuous \(\hbox { over}\ [0,\pi ]\), and for \(n\ge 2\) it satisfies the boundary condition \(f(0)=f(\pi )=0\). However this boundary condition fails for \(n=1\) at least with the eigenfunctions associated to the largest eigenvalue \(\lambda _1(\Omega )\).

Proof

Let \(f\in L^2_{\mu _{\Omega }}\) be any non trivial eigenfunction of the operator \({\mathcal {K}}_n^{{\Omega }}\) defined in (4.82) and associated to an eigenvalue \(\lambda \ne 0\), then

$$\begin{aligned} f(\phi )=\frac{1}{\lambda \, \nu _\Omega (\phi )}\int _0^\pi H_n(\phi ,\varphi ) f(\varphi )d\varphi , \quad \forall \phi \in (0,\pi )\,\text {a.e.} \end{aligned}$$
(4.103)

Since \(f\in L^2_{\mu _\Omega }\), then the function \(g: \varphi \in [0,\pi ]\mapsto r_0^{\frac{3}{2}}(\varphi )f(\varphi )\)belongs to \( L^2((0,\pi );d\varphi )\). Therefore the equation (4.103) can be written in terms of g as follows

Coming back to the definition of \(H_n\) in (3.3) we obtain for some constant \(c_n\) the formula

$$\begin{aligned} r_0^{-\frac{3}{2}}(\varphi ){r_0^{\frac{3}{2}}(\phi )}H_n(\phi ,\varphi )&= c_n\frac{\sin (\varphi ) r_0^{n-\frac{1}{2}}(\varphi )r_0^{n+\frac{1}{2}}(\phi )}{\left[ R(\phi ,\varphi )\right] ^{n+\frac{1}{2}}} F_n\left( \frac{4r_0(\phi )r_0(\varphi )}{R(\phi ,\varphi )}\right) . \end{aligned}$$

Using (A.7) and the assumption (H2) yields

$$\begin{aligned} r_0^{-\frac{3}{2}}(\varphi ){r_0^{\frac{3}{2}}(\phi )}H_n(\phi ,\varphi )&\lesssim \frac{r_0^{n+\frac{1}{2}} (\varphi )r_0^{n+\frac{1}{2}}(\phi )}{R^{n+\frac{1}{2}}(\phi ,\varphi )}\left( 1+\ln \left( \frac{\sin (\phi )+\sin (\varphi )}{|\phi -\varphi |}\right) \right) \nonumber \\&\lesssim 1+\ln \left( \frac{\sin (\phi )+\sin (\varphi )}{|\phi -\varphi |}\right) . \end{aligned}$$
(4.104)

This implies, using Cauchy–Schwarz inequality and the fact that \(\nu _\Omega \) is bounded away from zero

It follows that g is bounded. Now inserting this estimate into (4.103) allows to get

(4.105)

Using once again estimat-1 and the assumption (H2) we deduce that

By symmetry we may restrict the analysis to \(\phi \in [0,\frac{\pi }{2}]\). Thus, splitting the integral given in (4.105) and using that

$$\begin{aligned} \inf _{\begin{array}{c} \varphi \in [\pi /2,\pi ]\\ \phi \in [0,\pi /2] \end{array}}R(\phi ,\varphi )>0, \end{aligned}$$

we obtain

It follows that

Using the change of variables \(\varphi =\phi \theta \) we get

Consequently we find

$$\begin{aligned} \sup _{\phi \in (0,\pi )}r_0^{\frac{1}{2}}(\phi )| f(\phi )|\lesssim \Vert f\Vert _{\mu _\Omega }. \end{aligned}$$

Inserting this estimate into (4.103) and using (4.104) yields

As before we can restrict \(\phi \in [0,\frac{\pi }{2}]\) and by using the fact

$$\begin{aligned} \inf _{\begin{array}{c} \varphi \in [\pi /2,\pi ]\\ \phi \in [0,\pi /2] \end{array}}R(\phi ,\varphi )>0, \end{aligned}$$

we deduce after splitting the integral

Making the change of variables \(\varphi =\phi \theta \) leads to

Consequently we get

$$\begin{aligned} \forall \phi \in (0,\pi ),\quad |f(\phi )|&\lesssim \Vert f\Vert _{\mu _\Omega }(r_0^{n-1}(\phi )+{r_0^{\frac{1}{2}}(\phi )}). \end{aligned}$$

This shows that f is bounded over \((0,\pi )\) and by the dominated convergence theorem one can show that f is in fact continuous on \([0,\pi ]\) and satisfies for \(n\ge 2\) the boundary condition

$$\begin{aligned} f(0)=f(\pi )=0. \end{aligned}$$

Last, we shall check that this boundary condition fails for \(n=1\) with the largest eigenvalue \(\lambda _1(\Omega )\). Indeed, according to (4.103) we have

$$\begin{aligned} f(0)=\frac{1}{\lambda \nu _\Omega (0)}\int _0^\pi H_1(0,\varphi ) f(\varphi )d\varphi . \end{aligned}$$

However, from (3.3) we get

$$\begin{aligned} \forall \, \varphi \in (0,\pi ),\quad H_1(0,\varphi )=c_1 \frac{r_0^2(\varphi )\sin (\varphi )}{R^\frac{3}{2}(0,\varphi )}>0. \end{aligned}$$

Combining this with the fact that f does not change the sign allows to get that \(f(0)\ne 0\). \(\quad \square \)

4.4.2 Hölder continuity

The main goal of this section is to prove the Hölder regularity of the eigenfunctions.

Proposition 4.5

Assume that \(r_0\) satisfies the conditions (H) and let \(\Omega \in (-\infty ,\kappa )\), then any solution h of the equation

$$\begin{aligned} h(\phi )=\frac{1}{\lambda \,\nu _\Omega (\phi )}\int _0^\pi H_n(\varphi ,\phi )h(\varphi )d\varphi ,\quad \forall \phi \in (0,\pi ), \end{aligned}$$
(4.106)

with \(\lambda \ne 0\), belongs to \({\mathscr {C}}^{1,\alpha }(0,\pi )\), for any \(n\ge 2\). The functions involved in the above expression can be found in (3.3)–(4.2)–(4.3).

Proof

From the initial expression of the linearized operator (3.1) in Proposition 3.1 and combining it with Proposition 3.2, one has

$$\begin{aligned} {\mathcal {F}}_n(h)(\phi )&:=\int _0^\pi H_n(\varphi ,\phi )h(\varphi )d\varphi {=} \frac{1}{4\pi }\frac{1}{r_0(\phi )} \int _0^\pi \int _0^{2\pi }{\mathcal {H}}_n(\phi ,\varphi ,\eta )h(\varphi )d\eta d\varphi , \end{aligned}$$
(4.107)

with

$$\begin{aligned} {\widehat{R}}(\phi ,\varphi ,\eta )&:=(r_0(\phi )-r_0(\varphi ))^2+ 2r_0(\phi )r_0(\varphi )(1-\cos \eta )+(\cos \phi -\cos \varphi )^2,\\ {\mathcal {H}}_n(\phi ,\varphi ,\eta )&:=\frac{\sin (\varphi )r_0(\varphi )\cos (n\eta )}{{\widehat{R}}^{\frac{1}{2}}(\phi ,\varphi ,\eta )}. \end{aligned}$$

It is clear that any solution h of (4.106) is equivalent to a solution of

$$\begin{aligned} \forall \, \phi \in (0,\pi ),\quad h(\phi )=\frac{{\mathcal {F}}_n(h)(\phi )}{\lambda \,\nu _\Omega (\phi )}\cdot \end{aligned}$$

From Proposition 4.1 we know that \(\nu _\Omega \in {\mathscr {C}}^{1,\alpha }(0,\pi )\) and does not vanish when \(\Omega \in (-\infty ,\kappa )\). Therefore to check the regularity \(h\in {\mathscr {C}}^{1,\alpha }(0,\pi )\) it is enough to establish that \({\mathcal {F}}_n(h)\in {\mathscr {C}}^{1,\alpha }(0,\pi )\), due to the fact that \({\mathscr {C}}^\alpha \) is an algebra. Since h is symmetric with respect to \(\phi =\frac{\pi }{2}\), then one can verify that \({\mathcal {F}}_n(h)\) preserves this symmetry and hence we shall only study the regularity in the interval \([0,\frac{\pi }{2}]\) and check that the left and right derivative at \(\pi /2\) coincide. Notice that Proposition 4.4 tells us that h is continuous in \([0,\pi ]\), for any \(n\ge 1\).

In order to prove such regularity, let us first check that

$$\begin{aligned} {\widehat{R}}(\phi ,\varphi ,\eta )\ge & {} C\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2 (\eta /2)\right\} , \nonumber \\&\forall \,\phi ,\varphi \in (0,\pi ), \eta \in (0,2\pi ), \end{aligned}$$
(4.108)

which is the key point in this proof. In order to do so, recall first from (2.18) that

$$\begin{aligned} {\widehat{R}}(\phi ,\varphi ,\eta )\ge C(\phi -\varphi )^2. \end{aligned}$$
(4.109)

On the other hand, define the function

$$\begin{aligned} g_1(x)=x^2+r_0(\varphi )^2-2xr_0(\varphi )\cos (\eta )+(\cos (\varphi )-\cos (\phi ))^2, \end{aligned}$$

which obviously verifies \(g_1(r_0(\phi ))={\widehat{R}}(\phi ,\varphi ,\eta )\). Such function has a minimum located at

$$\begin{aligned} x_c=r_0(\varphi )\cos (\eta ). \end{aligned}$$

Now we shall distinguish two cases: \(\cos \eta \in [0,1]\) and \(\cos \eta \in [-1,0]\). In the first case we get

$$\begin{aligned} g_1(x)&\ge g_1(x_c)=r_0^2(\varphi )\sin ^2(\eta )+(\cos (\varphi )-\cos (\phi ))^2\\&\ge r_0^2(\varphi )\sin ^2(\eta ). \end{aligned}$$

From elementary trigonometric relations we deduce that

$$\begin{aligned} \sin ^2\eta = 2\sin ^2(\eta /2)\big (1+\cos (\eta )\big )\ge 2\sin ^2(\eta /2). \end{aligned}$$

This implies in particular that, for \(\cos \eta \in [0,1]\)

$$\begin{aligned} {\widehat{R}}(\phi ,\varphi ,\eta )\ge 2 r_0^2(\varphi )\sin ^2(\eta /2). \end{aligned}$$

As to the second case \(\cos \eta \in [-1,0]\), we simply notice that the critical point \(x_c\) is negative and therefore the second degree polynomial \(g_1\) is strictly increasing in \({{\,\mathrm{{\mathbb {R}}}\,}}_+\). This implies that

$$\begin{aligned} {\widehat{R}}(\phi ,\varphi ,\eta )=g_1(r_0(\phi ))&\ge g_1(0) \ge r_0^2(\varphi ) \ge r_0^2(\varphi )\sin ^2(\eta /2). \end{aligned}$$

Therefore we get in both cases

$$\begin{aligned} {\widehat{R}}(\phi ,\varphi ,\eta )\ge r_0^2(\varphi )\sin ^2(\eta /2). \end{aligned}$$
(4.110)

By the symmetry property \({\widehat{R}}(\phi ,\varphi ,\eta )={\widehat{R}}(\varphi ,\phi ,\eta )\) we also get

$$\begin{aligned} {\widehat{R}}(\phi ,\varphi ,\eta )\ge r_0^2(\phi )\sin ^2(\eta /2). \end{aligned}$$
(4.111)

Adding together (4.109)–(4.110)–(4.111), we achieve

$$\begin{aligned} 3{\widehat{R}}(\phi ,\varphi ,\eta )\ge C(\phi -\varphi )^2+(r_0^2(\phi )+r_0^2(\varphi ))\sin ^2(\eta /2). \end{aligned}$$
(4.112)

It suffices now to combine this inequality with the assumption (H2) on \(r_0\) in order to get the desired estimate (4.108).

Let us now prove that \({\mathcal {F}}_n(h)\in {\mathscr {C}}^{1,\alpha }\) and for this aim we shall proceed in four steps.

\(\bullet \) Step 1: If \(h\in L^\infty \) then \({\mathcal {F}}_n(h)\in {\mathscr {C}}^\alpha (0,\pi )\).

Here we check that \({\mathcal {F}}_n(h)\in {\mathscr {C}}^\alpha (0,\pi )\) for any \(n\ge 1\). In order to avoid the singularity in the denominator coming from \(r_0\), we integrate by parts in the variable \(\eta \)

Introduce

$$\begin{aligned} K_1(\phi ,\varphi ,\eta ):=\frac{\sin (\varphi )r_0^2(\varphi )\sin (n\eta )\sin (\eta ) h(\varphi )}{{\widehat{R}}^{\frac{3}{2}}(\phi ,\varphi ,\eta )}, \end{aligned}$$

and according to Chebyshev polynomials we know that

$$\begin{aligned} \sin (n\eta )=\sin (\eta ) \,U_{n-1}(\cos \eta ), \end{aligned}$$
(4.113)

with \(U_{n}\) being a polynomial of degree n. Thus

$$\begin{aligned} K_1(\phi ,\varphi ,\eta )=\frac{\sin (\varphi )r_0^2(\varphi )U_{n-1}(\cos \eta )\sin ^2(\eta ) h(\varphi )}{{\widehat{R}}^{\frac{3}{2}}(\phi ,\varphi ,\eta )}\cdot \end{aligned}$$

Using the assumption (H2) combined with the estimate (4.108) for the denominator \({\widehat{R}}(\phi ,\varphi ,\eta )\), we achieve

$$\begin{aligned} |K_1(\phi ,\varphi ,\eta )|&\lesssim \frac{\Vert h\Vert _{L^\infty }\sin ^3(\varphi ) \sin ^2(\eta /2)}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2 (\eta /2)\right) ^\frac{3}{2}}\\&\lesssim \frac{\sin (\varphi )}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2 (\eta /2)\right) ^\frac{1}{2}}\cdot \end{aligned}$$

Interpolating between the two inequalities

$$\begin{aligned} \frac{\sin (\varphi )}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2 (\eta /2)\right) ^\frac{1}{2}}\le |\varphi -\phi |^{-1} \end{aligned}$$

and

$$\begin{aligned} \frac{\sin (\varphi )}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2 (\eta /2)\right) ^\frac{1}{2}}\le \sin ^{-1} (\eta /2), \end{aligned}$$

we deduce that for any \(\beta \in [0,1]\)

$$\begin{aligned} \frac{\sin (\varphi )}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2 (\eta /2)\right) ^\frac{1}{2}}\lesssim |\varphi -\phi |^{-(1-\beta )}\sin ^{-\beta } (\eta /2).\qquad \quad \end{aligned}$$
(4.114)

Then,

$$\begin{aligned} |K_1(\phi ,\varphi ,\eta )|&\lesssim \frac{1}{|\phi -\varphi |^{1-\beta }\sin ^\beta (\eta /2)}\cdot \end{aligned}$$

Let us now bound the derivative \(\partial _\phi K_1(\phi ,\varphi ,\eta )\). For this purpose, let us first show that

$$\begin{aligned} {|\partial _\phi {\widehat{R}}(\phi ,\varphi ,\eta )|}\lesssim {\widehat{R}}^{\frac{1}{2}}(\phi ,\varphi ,\eta ). \end{aligned}$$
(4.115)

Indeed

$$\begin{aligned}&\frac{\partial _\phi {\widehat{R}}(\phi ,\varphi ,\eta )}{{\widehat{R}}^\frac{1}{2}(\phi ,\varphi ,\eta )}\\&\quad =\frac{2r_0'(\phi )(r_0(\phi )-r_0(\varphi ))+2r_0'(\phi )r_0(\varphi )(1-\cos (\eta ))+2\sin (\phi )(\cos (\varphi )-\cos (\phi ))}{{\widehat{R}}^\frac{1}{2}(\phi ,\varphi ,\eta )}\cdot \end{aligned}$$

Using the identity \(1-\cos (\eta )=2 \sin ^2(\eta /2)\) and (4.108), we get a constant C such that

$$\begin{aligned} \frac{\big |\partial _\phi {\widehat{R}}(\phi ,\varphi ,\eta )\big |}{{\widehat{R}}^\frac{1}{2}(\phi ,\varphi ,\eta )}\lesssim \frac{|\phi -\varphi |+\sin (\varphi )\sin ^2(\eta /2)}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right) ^\frac{1}{2}}\le C, \end{aligned}$$

achieving (4.115). Therefore, taking the derivative in \(\phi \) of \(K_1\) yields

$$\begin{aligned} \left| \partial _\phi K_1(\phi ,\varphi ,\eta )\right|&\le C\Vert h\Vert _{L^\infty }\frac{\sin ^3(\varphi )\sin ^2(\eta /2)}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right) ^2}\\&\lesssim \frac{\sin (\varphi )}{(\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)}\\&\lesssim \frac{|\phi -\varphi |^{-1}\sin (\varphi )}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right) ^{\frac{1}{2}}}\cdot \end{aligned}$$

Hence (4.114) allows to get or any \(\beta \in (0,1)\),

$$\begin{aligned} \left| \partial _\phi K_1(\phi ,\varphi ,\eta )\right|&\lesssim \frac{1}{|\phi -\varphi |^{2-\beta }\sin ^{\beta }(\eta /2)}\cdot \end{aligned}$$

Here, we can use Proposition C.1 to the case where the operator \({\mathcal {K}}\) depends only on one variable by taking \(g_1(\theta ,\eta )=g_3(\theta ,\eta )=\sin ^{-\beta }(\eta /2)\). Then, we infer \({\mathcal {F}}_n(h)\in {\mathscr {C}}^\beta (0,\pi )\) for any \(\beta \in (0,1)\).

\(\bullet \) Step 2: For \(n\ge 2\), if h is bounded then \(h(0)=h(\pi )=0\).

Notice that this property was shown in Proposition 4.4 and we give here an alternative proof. Since \(\nu _\Omega \) is not vanishing then this amounts to checking that \({\mathcal {F}}_n(h)(0)=0\). By continuity, it is clear by Fubini that

which is vanishing if \(n\ge 2\). Hence, \(h(0)=0\), for any \(n\ge 2\). In the same way, we achieve \(h(\pi )=0\).

\(\bullet \) Step 3: If \(h\in {\mathscr {C}}^\alpha (0,\pi )\) and \(h(0)=h(\pi )=0\), then \({\mathcal {F}}_n(h)\in W^{1,\infty }(0,\pi )\).

We have shown before that \({\mathcal {F}}_n(h)\in {\mathscr {C}}^\beta (0,\pi )\) for any \(\beta \in (0,1)\). Then it is enough to check that \({\mathcal {F}}_n(h)^\prime \in L^\infty (0,\pi )\). For this aim, we write

(4.116)

Adding and subtracting some appropriate terms, we find

(4.117)

Let us bound each term separately. Using (4.108), (4.113) and (4.115) we achieve

We write in view of (4.114)

$$\begin{aligned}&\frac{\sin (\varphi )}{(\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)}\\&\quad \le \frac{|\phi -\varphi |^{-1}\sin (\varphi )}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right) ^\frac{1}{2}}\\&\quad \lesssim \frac{1}{|\phi -\varphi |^{2-\beta }\sin ^\beta (\eta /2)}\cdot \end{aligned}$$

Therefore by imposing \(1-\alpha<\beta <1\) we get

which implies immediately that \(I_1\in L^\infty \). Now let us move to the boundedness of \(I_2\). From direct computations we get

$$\begin{aligned} \Big |\big (\partial _\phi +\partial _\varphi \big ){\widehat{R}}(\phi ,\varphi ,\eta )\Big |&= \Big |2(r_0(\phi )-r_0(\varphi ))(r_0^\prime (\phi )-r_0^\prime (\varphi ))\nonumber \\&\quad +2(\cos \phi -\cos \varphi )(\sin \varphi -\sin \phi )\nonumber \\&\quad +2(1-\cos \eta )\big (r_0^\prime (\phi )r_0(\varphi )+r_0^\prime (\varphi ) r_0(\phi )\big )\Big |. \end{aligned}$$
(4.118)

Combining the assumption (H2) with \(r_0'\in W^{1,\infty }\) and the mean value theorem yields

$$\begin{aligned} \Big |\big (\partial _\phi +\partial _\varphi \big ){\widehat{R}}(\phi ,\varphi ,\eta )\Big |&\le C\Big ( |\phi -\varphi |^2+(\sin \varphi +\sin \phi )\sin ^2(\eta /2) \Big ). \end{aligned}$$
(4.119)

Hence, using (4.108) and (4.113) we obtain

Interpolation inequalities imply

$$\begin{aligned} \frac{1}{\big ((\phi -\varphi )^2+(\sin ^2(\varphi )+\sin ^2(\phi ))\sin ^2(\eta /2)\big )^{\frac{1}{2}}}\le |\phi -\varphi |^{\alpha -1}\sin ^{-\alpha }(\phi )\sin ^{-\alpha }(\eta /2).\nonumber \\ \end{aligned}$$
(4.120)

Therefore we get for any \(\phi \in (0,\pi /2), \varphi \in (0,\pi )\) and \(\eta \in (0,2\pi )\),

$$\begin{aligned} \frac{\phi ^{\alpha }}{\big ((\phi -\varphi )^2+(\sin (\phi )+\sin (\varphi ))^2\sin ^2(\eta /2)\big )^{\frac{1}{2}}} \lesssim |\phi -\varphi |^{\alpha -1}\sin ^{-\alpha }(\eta /2). \end{aligned}$$
(4.121)

It follows that

which gives the boundedness of \(I_2\). It remains to bound the last term \(I_3\). Then integrating by parts we infer

To make the previous integration by parts rigorously, we should split the integral in \(\varphi \in (\varepsilon ,\phi -\varepsilon )\) and \(\varphi \in (\pi +\varepsilon ,\pi -\varepsilon )\) and later taking the limit as \(\varepsilon \rightarrow 0\).

Then, since \(h(0)=0\) and \(h\in {\mathscr {C}}^\alpha \) we find according to the assumptions (H), (4.108) and (4.113)

Applying (4.121) yields

That implies that \(h\in W^{1,\infty }(0,\pi /2)\).

Moreover, since \(r_0'(\pi /2)=0\) (this comes from the symmetry of \(r_0\) with respect to \(\pi /2\)) and using (4.116) we find that \(h'(\pi /2)=0\), which justifies why we can check the regularity only on \(\phi \in (0,\pi /2)\). Finally, we get the desired result, that is, \(h\in W^{1,\infty }(0,\pi )\).

\(\bullet \) Step 4: If \(h^\prime \in L^\infty (0,\pi )\) and \(h(0)=h(\pi )=0\), then \({\mathcal {F}}_n(h)^\prime \in {\mathscr {C}}^\beta (0,\pi )\) for any \(\beta \in (0,1)\).

Coming back to (4.117) and integrating by parts in the last integral we deduce

(4.122)

with

$$\begin{aligned} T_2(\phi ,\varphi ,\eta )=\frac{\partial _\varphi \big (\sin (\varphi ) r_0(\varphi )^2h(\varphi )\big )\sin (n\eta )\sin (\eta ) }{{\widehat{R}}^{\frac{3}{2}}(\phi ,\varphi ,\eta )}\cdot \end{aligned}$$

As in the previous step, integration by parts can be justified by splitting the integral in \(\varphi \in (\varepsilon ,\phi -\varepsilon )\) and \(\varepsilon \in (\phi +\varepsilon ,\pi -\varepsilon )\) and later taking limits as \(\varepsilon \rightarrow 0\).

We want to apply Proposition C.1 to each of those terms. First, for \(T_1\) we use (H) and (4.119) combined with (4.108), we arrive at

$$\begin{aligned} |T_1(\phi ,\varphi ,\eta )|&\lesssim \frac{\sin ^3(\varphi )|h(\varphi )|\sin ^2(\eta )\big (|\phi -\varphi |^2+(\sin \phi +\sin \varphi )\sin ^2({\eta /2})\big )}{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^\frac{5}{2}}\\&\lesssim \frac{|h(\varphi )|}{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^\frac{1}{2}}\cdot \end{aligned}$$

Since \(h'\in L^\infty \) and \(h(0)=h(\pi )=0\) then we can write \(h(\varphi )=\sin (\varphi ){\widehat{h}}(\varphi )\), with \({\widehat{h}}\in L^\infty (0,\pi )\).

Consequently,

$$\begin{aligned} |T_1(\phi ,\varphi ,\eta )|&\lesssim ||{\widehat{h}}||_{L^\infty }\frac{\sin (\varphi )}{\left\{ (\phi -\varphi )^2+\sin ^2(\varphi )\sin ^2(\eta /2)\right\} ^\frac{1}{2}}\\&\lesssim \frac{1}{\left\{ (\phi -\varphi )^2+\sin ^2(\eta /2)\right\} ^\frac{1}{2}}\cdot \end{aligned}$$

Interpolating again, we find that for any \(\beta \in [0,1]\)

$$\begin{aligned} |T_1(\phi ,\varphi ,\eta )|\lesssim \frac{1}{|\phi -\varphi |^{1-\beta }\sin ^\beta (\eta /2)}\cdot \end{aligned}$$

Let us mention that we have proven that

$$\begin{aligned} \frac{|h(\varphi )|}{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^\frac{1}{2}}\le C\frac{1}{|\phi -\varphi |^{1-\beta }\sin ^\beta (\eta /2)}, \end{aligned}$$
(4.123)

for any \(\beta \in (0,1)\), \(\phi \in (0,\pi /2), \varphi \in (0,\pi )\) and \(\eta \in (0,2\pi )\), which will be useful later.

Now we shall estimate the derivative of \(T_1\) with respect to \(\phi \). We start with

$$\begin{aligned} \Big |\partial _\phi \left\{ \big (\partial _\phi +\partial _\varphi \big ){\widehat{R}}(\phi ,\varphi ,\eta )\right\} \Big |&= \Big |2r_0'(\phi )(r_0^\prime (\phi )-r_0^\prime (\varphi ))+2(r_0(\phi )-r_0(\varphi ))r_0''(\phi )\nonumber \\&\quad -2\sin (\phi )(\sin \varphi -\sin \phi )-2(\cos \phi -\cos \varphi )\cos (\phi )\nonumber \\&\quad +2(1-\cos \eta )\big (r_0''(\phi )r_0(\varphi )+r_0^\prime (\varphi ) r_0'(\phi )\big )\Big |. \end{aligned}$$
(4.124)

Using that \(r_0''\in L^\infty \), we find

$$\begin{aligned} \Big |\partial _\phi \left\{ \big (\partial _\phi +\partial _\varphi \big ){\widehat{R}}(\phi ,\varphi ,\eta )\right\} \Big |&\le C\left( |\phi -\varphi |+\sin ^2(\eta /2)\right) . \end{aligned}$$
(4.125)

Thus,

$$\begin{aligned} |\partial _\phi T_1(\phi ,\varphi ,\eta )|&\lesssim \frac{\sin ^3(\varphi )|h(\varphi )|\sin ^2(\eta )\Big |\partial _\phi \left\{ \big (\partial _\phi +\partial _\varphi \big ){\widehat{R}}(\phi ,\varphi ,\eta )\right\} \Big |}{{\widehat{R}}^\frac{5}{2}(\phi ,\varphi ,\eta )}\\&\quad +\frac{\sin ^3(\varphi )|h(\varphi )|\sin ^2(\eta )\left| \big (\partial _\phi +\partial _\varphi \big ){\widehat{R}}(\phi ,\varphi ,\eta )\right| \left| \partial _\phi {\widehat{R}}(\phi ,\varphi ,\eta )\right| }{{\widehat{R}}^\frac{7}{2}(\phi ,\varphi ,\eta )}\cdot \end{aligned}$$

Using (4.108)–(4.119)–(4.125), we find

$$\begin{aligned} |\partial _\phi T_1(\phi ,\varphi ,\eta )|&\lesssim \frac{\sin ^3(\varphi )|h(\varphi )|\sin ^2(\eta )\left\{ |\phi -\varphi |+\sin ^2(\eta /2)\right\} }{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^\frac{5}{2}}\\&\quad +\frac{\sin ^3(\varphi )|h(\varphi )|\sin ^2(\eta )\left\{ |\phi -\varphi |^2+(\sin (\varphi )+\sin (\phi ))\sin ^2(\eta /2)\right\} }{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^3}\cdot \end{aligned}$$

It follows that

$$\begin{aligned} |\partial _\phi T_1(\phi ,\varphi ,\eta )|&\lesssim \frac{\sin (\varphi )|h(\varphi )|\left\{ |\phi -\varphi |+\sin ^2(\eta /2)\right\} }{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^\frac{3}{2}}\\&\quad +\frac{\sin (\varphi )|h(\varphi )|\left\{ |\phi -\varphi |^2+(\sin (\varphi )+\sin (\phi ))\sin ^2(\eta /2)\right\} }{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^2}\\&\lesssim \frac{|h(\varphi )|}{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} }\cdot \end{aligned}$$

Putting together this estimate with (4.123) we infer

$$\begin{aligned} |\partial _\phi T_1(\phi ,\varphi ,\eta )|&\lesssim \frac{|\phi -\varphi |^{-1}|h(\varphi )|}{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^{\frac{1}{2}}}\\&\lesssim C|\phi -\varphi |^{-(2-\beta )}\sin ^{-\beta }(\eta /2), \end{aligned}$$

for any \(\beta \in (0,1)\).

Concerning the estimate of the term \(T_2\), we first make appeal to (4.108) and (4.113) leading to

$$\begin{aligned} |T_2(\phi ,\varphi ,\eta )|&\lesssim \frac{(\sin ^3(\varphi )+\sin ^2(\varphi )|h(\varphi )|)\sin ^2(\eta /2)}{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^\frac{3}{2}}\\&\lesssim \frac{(\sin (\varphi )+|h(\varphi )|)}{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^\frac{1}{2}}\cdot \end{aligned}$$

Applying (4.114) and (4.123), one finds

$$\begin{aligned} |T_2(\phi ,\varphi ,\eta )|\le C |\phi -\varphi |^{\beta -1}\sin ^{-\beta }(\eta /2), \end{aligned}$$

for any \(\beta \in (0,1)\). The next stage is devoted to the estimate of \(\partial _\phi T_2\) and one gets from direct computations

$$\begin{aligned} |\partial _\phi T_2(\phi ,\varphi ,\eta )|&\lesssim \frac{(\sin ^3(\varphi )+\sin ^2(\varphi )|h(\varphi )|)\sin ^2(\eta /2)\left| \partial _\phi {\widehat{R}}(\phi ,\varphi ,\eta )\right| }{{\widehat{R}}^\frac{5}{2}(\phi ,\varphi ,\eta )}\cdot \end{aligned}$$

Using (4.115) and (4.108), it implies

$$\begin{aligned} |\partial _\phi T_2(\phi ,\varphi ,\eta )|&\lesssim \frac{(\sin ^3(\varphi )+\sin ^2(\varphi )|h(\varphi )|)\sin ^2(\eta /2)}{\left\{ (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)\right\} ^2}\\&\lesssim \frac{(\sin (\varphi )+|h(\varphi )|)}{(\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2(\eta /2)}\cdot \end{aligned}$$

Therefore (4.114) and (4.123) allows to get

$$\begin{aligned} |\partial _\phi T_2(\phi ,\varphi ,\eta )|\lesssim |\phi -\varphi |^{-(2-\beta )}\sin ^\beta (\eta /2), \end{aligned}$$

for any \(\beta \in (0,1)\). Hence, by Proposition C.1, adapted to a one variable function, we achieve that \({\mathcal {F}}_n(h)'\in {\mathscr {C}}^\beta \), for any \(\beta \in (0,1)\), which achieves the proof of the announced result. \(\quad \square \)

4.5 Fredholm structure

In this section we shall be concerned with the Fredholm structure of the linearized operator \(\partial _f {\tilde{F}}(\Omega ,0)\) defined through (3.4) and (4.1). Our main result reads as follows.

Proposition 4.6

Let \(m\ge 2\), \(\alpha \in (0,1)\) and \(\Omega \in (-\infty ,\kappa )\), then \(\partial _f {\tilde{F}}(\Omega ,0):X_m^\alpha \rightarrow X_m^\alpha \) is a well-defined Fredholm operator with zero index. In addition, for \(\Omega =\Omega _m\), the kernel of \(\partial _f {\tilde{F}}(\Omega _m,0)\) is one-dimensional and its range is closed and of co-dimension one.

Recall that the spaces \(X_m^\alpha \) have been introduced in (2.15) and \(\Omega _m\) in Proposition 4.3.

Proof

We shall first prove the second part, assuming the first one. The structure of the linearized operator is detailed in (3.4) and one has for \(h(\phi ,\theta )=\sum _{n\ge 1} h_n(\phi )\cos (n\theta )\)

$$\begin{aligned} \partial _{f} {\tilde{F}}(\Omega ,0)h(\phi ,\theta )=\sum _{n\ge 1}\cos (n\theta ){\mathcal {L}}_n^\Omega (h_n)(\phi ), \end{aligned}$$

where

$$\begin{aligned} {\mathcal {L}}_n^\Omega (h_n)(\phi )&= \nu _\Omega (\phi )h_n(\phi )-\int _0^\pi H_n(\phi ,\varphi )h_n(\varphi )d\varphi , \quad \phi \in [0,\pi ]. \end{aligned}$$

In view of (4.1) and (4.82), \({\mathcal {L}}_n^\Omega (h)=0\) can be written in the form

$$\begin{aligned} {\mathcal {K}}_n^\Omega h=h. \end{aligned}$$

We define the dispersion set by

$$\begin{aligned} {\mathcal {S}}=\big \{\Omega \in (-\infty ,\kappa ),\quad \text {Ker} \partial _{f} {\tilde{F}}(\Omega ,0)\ne \{0\}\big \}. \end{aligned}$$

Hence \(\Omega \in {\mathcal {S}}\) if and only if there exists \(m\ge 1\) such that the equation

$$\begin{aligned} \forall \phi \in [0,\pi ],\quad {\mathcal {K}}_m^\Omega (h_m)(\phi )=h_m(\phi ), \end{aligned}$$

admits a nontrivial solution satisfying the regularity \(h_m\in {\mathscr {C}}^{1,\alpha }(0,\pi )\) and the boundary condition \(h_m(0)=h_m(\pi )=0\). By virtue of Propositions 4.4 and 4.5 the foregoing conditions are satisfied for any eigenvalue provided that \(m\ge 2\). On the other hand, we have shown in Proposition 4.2-(4) that for \(\Omega =\Omega _m\) the kernel of \({\mathcal {L}}_m\) is one-dimensional. Moreover, Proposition 4.2-(5) ensures that for any \(n>m\) we have \(\lambda _n(\Omega _m)<\lambda _m(\Omega _m)=1\). Since by construction \(\lambda _n(\Omega _m)\) is the largest eigenvalue of \({\mathcal {K}}_n^{\Omega _m}\), then 1 could not be an eigenvalue of this operator and the equation

$$\begin{aligned} {\mathcal {K}}_n^{\Omega _m} h=h, \end{aligned}$$

admits only the trivial solution. Thus the kernel of the restricted operator \(\partial _f {\tilde{F}}(\Omega _m,0):X_m^\alpha \rightarrow X_m^\alpha \) is one-dimensional and is generated by the eigenfunction

$$\begin{aligned} (\phi ,\theta )\mapsto h_m(\phi ) \cos ( m\theta ). \end{aligned}$$

We emphasize that this element belongs to the space \(X_m^\alpha \) because it belongs to the function space \({\mathscr {C}}^{1,\alpha }((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}})\) since \(\phi \mapsto h_m(\phi )\in {\mathscr {C}}^{1,\alpha }(0,\pi )\). That the range of \(\partial _f {\tilde{F}}(\Omega _m,0) \) is closed and of co-dimension one follows from the fact this operator is Fredholm of zero index.

Next, let us show that \(\partial _f {\tilde{F}}(\Omega ,0) \) is Fredholm of zero index. By virtue of the computations developed in Proposition 3.1 and the expression of (4.1), we assert that

$$\begin{aligned} \partial _f {\tilde{F}}(\Omega ,0)h(\phi ,\theta )=\nu _\Omega (\phi )h(\phi ,\theta )-\frac{1}{4\pi }G(h)(\phi ,\theta ), \end{aligned}$$

with

(4.126)

Since \(\Omega \in (-\infty ,\kappa )\), the function \(\nu _\Omega \) is not vanishing. Moreover, by Proposition 4.1 one has that \(\nu _\Omega \in {\mathscr {C}}^{1,\beta }\), for any \(\beta \in (0,1)\).

Define the linear operator \(\nu _\Omega \text {Id}: X_m^\alpha \rightarrow X_m^\alpha \) by

$$\begin{aligned} (\nu _\Omega \text {Id})(h)(\phi ,\theta )=\nu _\Omega (\phi )h(\phi ,\theta )\cdot \end{aligned}$$

We shall check that it defines an isomorphism. The continuity of this operator follows from the regularity \(\nu _\Omega \in {\mathscr {C}}^{1,\alpha }(0,\pi )\) combined with the fact that \( {\mathscr {C}}^{1,\alpha }((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}})\) is an algebra. The Dirichlet boundary condition, the m-fold symmetry and the absence of the frequency zero are immediate for the product \(\nu _\Omega h\), which finally belongs to \(X_m^\alpha \). Moreover, since \(\nu _\Omega \) is not vanishing, one has that \(\nu _\Omega \text {Id}\) is injective. In order to check that such an operator is an isomorphism, it is enough to check that it is surjective, as a consequence of the Banach theorem. Take \(k\in X_m^\alpha \), and we will find \(h\in X_m^\alpha \) such that \((\nu _\Omega \text {Id})(h)=k\). Indeed, h is given by

$$\begin{aligned} h(\phi ,\theta )=\frac{d(\phi ,\theta )}{\nu _\Omega (\phi )}\cdot \end{aligned}$$

Using the regularity of \(\nu _\Omega \) and the fact that it is not vanishing, it is easy to check that its inverse \(\frac{1}{\nu _\Omega }\) still belongs to \({\mathscr {C}}^{1,\alpha }(0,\pi )\). Similar arguments as before allow to get \(h\in X_m^\alpha \). Hence \(\nu _\Omega \text {Id}\) is an isomorphism, and thus it is a Fredholm operator of zero index. From classical results on index theory, it is known that to get \(\partial _f {\tilde{F}}(\Omega ,0)\) is Fredholm of zero index, it is enough to establish that the perturbation \({G}: X_m^\alpha \rightarrow X_m^\alpha \) is compact. To do so, we prove that for any \(\beta \in (\alpha ,1)\) one has the smoothing effect

$$\begin{aligned} \forall \, h\in X_{m}^\alpha ,\quad \Vert G(h)\Vert _{{\mathscr {C}}^{1,\beta }}\le C\Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}, \end{aligned}$$

that we combine with the compact embedding \({\mathscr {C}}^{1,\beta }\big ((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}}\big )\hookrightarrow {\mathscr {C}}^{1,\alpha }\big ((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}}\big )\).

Take \(h\in X_m^\alpha \) and let us show that \(G(h)\in {\mathscr {C}}^{1,\beta }\big ((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}}\big )\), for any \(\beta \in (0,1)\). We shall first deal with a preliminary fact. Define the following function

$$\begin{aligned} \forall \, \varphi \in [0,\pi ],\, \theta ,\eta \in {{\,\mathrm{{\mathbb {R}}}\,}},\quad g_\theta (\varphi ,\eta )=\int _\theta ^\eta h(\varphi ,\tau )d\tau . \end{aligned}$$
(4.127)

By (2.17) we infer

$$\begin{aligned} |g_\theta (\varphi ,\eta )|\le C\Vert h\Vert _{\text {Lip}}|\theta -\eta |\sin (\varphi ), \end{aligned}$$

According to the definition of the space \(X_m^\alpha \), the partial function \(\tau \mapsto h(\varphi ,\tau )\) is \(2\pi \)-periodic and with zero average, that is, \(\displaystyle {\int _{0}^{2\pi }h(\varphi ,\tau )d\tau =0}\). This allows to get that \(\eta \mapsto g_\theta (\varphi ,\eta )\) is also \(2\pi \)-periodic, and from elementary arguments we find

$$\begin{aligned} |g_\theta (\varphi ,\eta )|\le C\Vert h\Vert _{\text {Lip}}\,|\sin ((\theta -\eta )/2)|\sin (\varphi ), \end{aligned}$$
(4.128)

for any \(\varphi \in [0,\pi ]\) and \(\theta ,\eta \in [0,2\pi ]\). In addition, it is immediate that \(g_\theta \in {\mathscr {C}}^{1,\alpha }\big ((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}}\big )\) and

$$\begin{aligned} \partial _\varphi g_\theta (\varphi ,\eta )=\int _\theta ^\eta \partial _\varphi h(\varphi ,\tau )d\tau . \end{aligned}$$

The same arguments as before show that the partial function \(\tau \mapsto \partial _\varphi h(\varphi ,\tau )\) is \(2\pi \)-periodic and with zero average. Moreover, \(\eta \mapsto \partial _\varphi g_\theta (\varphi ,\eta )\) is also \(2\pi \)-periodic and

$$\begin{aligned} |\partial _\varphi g_\theta (\varphi ,\eta )|\le C\Vert h\Vert _{\text {Lip}}\,|\sin ((\theta -\eta )/2)|, \end{aligned}$$
(4.129)

for any \(\varphi \in [0,\pi ]\) and \(\theta ,\eta \in [0,2\pi ]\). Using the auxiliary function \(g_\theta \), one can integrate by parts in G(h) in the variable \(\eta \) obtaining

(4.130)

We can justify the integration by parts by splitting the integral in \(\eta \in (0,\theta -\varepsilon )\) and \(\eta \in (\theta +\varepsilon ,2\pi )\) and later taking limits as \(\varepsilon \rightarrow 0\). The boundary term in the above integration by parts is vanishing due to the periodicity in \(\eta \) of the involved functions. It follows from (4.108),

$$\begin{aligned} A(\phi ,\theta ,\varphi ,\eta )\gtrsim (\phi -\varphi )^2 +(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2), \end{aligned}$$
(4.131)

for any \(\phi ,\varphi \in (0,\pi )\) and \(\theta ,\eta \in (0,2\pi )\), and this estimate is crucial in the proof.

The boundedness of G(h) can be implemented by using (4.128) and (4.131). Indeed, we write

Therefore, we obtain

From the assumption \(\mathbf{(H2)}\) on \(r_0\) combined with (4.114) we get for any \(\beta \in (0,1)\), and then

Therefore

$$\begin{aligned} \Vert G(h)\Vert _{L^\infty }\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}. \end{aligned}$$
(4.132)

The next step is to check now that \(\partial _\phi G(h)\in {\mathscr {C}}^\beta \) by making appeal to Proposition C.1. From direct computations using (4.130) we find

Note that we can insert the derivative inside the integral, to make this rigorous, cut off the integral in \(\eta \) away from \(\theta \) and take a limit. Adding and subtracting in the numerator \(\partial _\varphi A(\phi ,\theta ,\varphi ,\eta )\), it can be written in the form

Integrating by parts in \(\varphi \) in the last term yields

The goal is to check the kernel assumptions for Proposition C.1 in order to prove that \({\mathscr {G}}_1\) and \({\mathscr {G}}_2\) belong to \({\mathscr {C}}^\beta \), for any \(\beta \in (0,1)\). For this aim, we define the kernels

$$\begin{aligned} {K}_1(\phi ,\theta ,\varphi ,\eta ):=\frac{\sin (\varphi )r_0^2(\varphi )\sin (\theta -\eta )g_\theta (\varphi ,\eta )(\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta )}{A(\phi ,\theta ,\varphi ,\eta )^\frac{5}{2}}, \end{aligned}$$

and

$$\begin{aligned} K_2(\phi ,\theta ,\varphi ,\eta ):=\frac{\partial _\varphi \left( \sin (\varphi )r_0^2(\varphi )g_\theta (\varphi ,\eta )\right) \sin (\theta -\eta )}{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}\cdot \end{aligned}$$

Let us start with \(K_1\) and show that it satisfies the hypothesis of Proposition C.1. From straightforward calculus we obtain in view of the assumptions \({\mathbf{(H)}}\) and the mean value theorem

$$\begin{aligned} |(\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta )|&= |2(r_0'(\phi )-r_0'(\varphi ))(r_0(\phi )-r_0(\varphi ))\nonumber \\&\quad +2(r_0(\phi )r_0'(\varphi )+r_0(\varphi )r_0'(\phi ))(1-\cos (\theta -\eta ))\nonumber \\&\quad -2(\sin (\phi )-\sin (\varphi ))(\cos (\phi )-\cos (\varphi ))|\nonumber \\&\lesssim \,(\phi -\varphi )^2+\big (\sin \varphi +\sin \phi \big )\sin ^2((\theta -\eta )/2). \end{aligned}$$
(4.133)

Using the inequality \(|ab|\le \frac{1}{2}(a^2+b^2)\) allows to get

$$\begin{aligned} \sin \varphi |(\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta )|&\lesssim \,(\phi -\varphi )^2+\big (\sin ^2\varphi +\sin ^2\phi \big )\sin ^2((\theta -\eta )/2). \end{aligned}$$

Thus, applying (4.131) we deduce that

$$\begin{aligned} \sin \varphi \big |(\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta )\big |&\lesssim |A(\phi ,\theta ,\varphi ,\eta )|. \end{aligned}$$
(4.134)

Then, putting together (4.128), \({\mathbf{(H2)}}\), (4.131) and (4.134) we find

$$\begin{aligned} |K_1(\phi ,\theta ,\varphi ,\eta )|&\lesssim \Vert h\Vert _{\text {Lip}}\frac{\sin (\varphi )r_0^2(\varphi )\sin ^2((\theta -\eta )/2)}{\left( (\phi -\varphi )^2 +(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\right) ^\frac{3}{2}}\\&\lesssim \Vert h\Vert _{\text {Lip}}\frac{\sin (\varphi )}{\left( (\phi -\varphi )^2 +(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\right) ^\frac{1}{2}}\cdot \end{aligned}$$

As a consequence of (4.114), we immediately get

$$\begin{aligned} |K_1(\phi ,\theta ,\varphi ,\eta )|\lesssim \Vert h\Vert _{\text {Lip}} |\phi -\varphi |^{\beta -1}|\sin ((\theta -\eta )/2)|^{-\beta }, \end{aligned}$$
(4.135)

for any \(\beta \in (0,1)\). Let us compute the derivative with respect to \(\phi \) of \(K_1\),

$$\begin{aligned}&\partial _\phi K_1(\phi ,\theta ,\varphi ,\eta )\\&\quad = \frac{\sin (\varphi )r_0^2(\varphi )\sin (\theta -\eta )g_\theta (\varphi ,\eta )\partial _\phi ((\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta ))}{A(\phi ,\theta ,\varphi ,\eta )^\frac{5}{2}}\\&\qquad -\frac{5}{2} \frac{\sin (\varphi )r_0^2(\varphi )\sin (\theta -\eta )g_\theta (\varphi ,\eta )((\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta )) \partial _\phi A(\phi ,\theta ,\varphi ,\eta ) }{A(\phi ,\theta ,\varphi ,\eta )^\frac{7}{2}}\cdot \end{aligned}$$

From direct computations, we easily get

$$\begin{aligned} |\partial _\phi ((\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta ))|\lesssim |\phi -\varphi |+\sin ^2((\theta -\eta )/2) \end{aligned}$$
(4.136)

and

$$\begin{aligned} |\partial _\phi A(\phi ,\theta ,\varphi ,\eta )|\lesssim |\phi -\varphi |+\sin (\varphi )\sin ^2((\theta -\eta )/2). \end{aligned}$$
(4.137)

Then, it is clear from (4.131) that

$$\begin{aligned} |\partial _\phi A(\phi ,\theta ,\varphi ,\eta )|\lesssim A^\frac{1}{2} (\phi ,\theta ,\varphi ,\eta ). \end{aligned}$$
(4.138)

In addition, one may check that

$$\begin{aligned} |(\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta )|&\lesssim \,(\phi -\varphi )^2+\big (\sin \varphi +\sin \phi \big )\sin ^2((\theta -\eta )/2)\nonumber \\&\lesssim A^{\frac{1}{2}}(\phi ,\theta ,\varphi ,\eta )\big (|\varphi -\phi |+|\sin ((\theta -\eta )/2)|\big ). \end{aligned}$$
(4.139)

By using (4.128), \({\mathbf{(H2)}}\), (4.131), (4.136), (4.138) and (4.139), one achieves

$$\begin{aligned} |\partial _\phi K_1(\phi ,\theta ,\varphi ,\eta )|&\lesssim \Vert h\Vert _{\text {Lip}}\frac{\sin ^2(\varphi )\left( |\phi -\varphi |+\sin ^2((\theta -\eta )/2)\right) }{\left( (\phi -\varphi )^2 +(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\right) ^\frac{3}{2}}\\&\quad +\Vert h\Vert _{\text {Lip}} \frac{\sin ^2(\varphi )\big (|\varphi -\phi |+|\sin ((\theta -\eta )/2)|\big )}{\big ((\phi -\varphi )^2 +(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\big )^{\frac{3}{2}}}\\&\lesssim \Vert h\Vert _{\text {Lip}}\frac{\sin ^2(\varphi )\left( |\phi -\varphi |+|\sin ((\theta -\eta )/2)|\right) }{\left( (\phi -\varphi )^2 +(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\right) ^\frac{3}{2}}\cdot \end{aligned}$$

Therefore, using some elementary inequalities allow to get

$$\begin{aligned} |\partial _\phi K_1(\phi ,\theta ,\varphi ,\eta )|&\lesssim \Vert h\Vert _{\text {Lip}} \frac{\sin \varphi }{(\phi -\varphi )^2 +(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)}\\&\lesssim \Vert h\Vert _{\text {Lip}}\frac{|\phi -\varphi |^{-1}\,\sin \varphi }{\left( (\phi -\varphi )^2 +(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\right) ^{\frac{1}{2}}}\cdot \end{aligned}$$

Applying (4.114) implies for any \(\beta \in (0,1)\)

$$\begin{aligned} |\partial _\phi K_1(\phi ,\theta ,\varphi ,\eta )|\le C \Vert h\Vert _{\text {Lip}}|\phi -\varphi |^{-(2-\beta )}|\sin ((\theta -\eta )/2)|^{-\beta }. \end{aligned}$$

Now, let us move to the estimate of the partial derivative \(\partial _\theta K_1\), given by

$$\begin{aligned}&\partial _\theta K_1(\phi ,\theta ,\varphi ,\eta )\\&\quad = \frac{\sin (\varphi )r_0^2(\varphi )\partial _\theta (\sin (\theta -\eta )g_\theta (\varphi ,\eta ))(\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta )}{A(\phi ,\theta ,\varphi ,\eta )^\frac{5}{2}}\\&\qquad +\frac{\sin (\varphi )r_0^2(\varphi )\sin (\theta -\eta )g_\theta (\varphi ,\eta )\partial _\theta \left\{ (\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta )\right\} }{A(\phi ,\theta ,\varphi ,\eta )^\frac{5}{2}}\\&\qquad -\frac{5}{2} \frac{\sin (\varphi )r_0^2(\varphi )\sin (\theta -\eta )g_\theta (\varphi ,\eta )((\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta ))(\partial _\theta A(\phi ,\theta ,\varphi ,\eta ))}{A(\phi ,\theta ,\varphi ,\eta )^\frac{7}{2}}\cdot \end{aligned}$$

By definition of \(g_\theta \) in (4.127) and (4.128), one concludes in view of (2.17) and (4.131) that

$$\begin{aligned} |\partial _\theta (\sin (\theta -\eta )g_\theta (\varphi ,\eta ))|&\lesssim \Vert h\Vert _{\text {Lip}}\sin (\varphi )|\sin ((\theta -\eta )/2)|\lesssim \Vert h\Vert _{\text {Lip}} A^{\frac{1}{2}}(\phi ,\theta ,\varphi ,\eta ). \end{aligned}$$
(4.140)

Moreover, one gets

$$\begin{aligned} |\partial _\theta \left\{ (\partial _\phi +\partial _\varphi )A(\phi ,\theta ,\varphi ,\eta )\right\} |&\lesssim (\sin (\phi )+\sin (\varphi ))|\sin (\theta -\eta )|\lesssim A^{\frac{1}{2}}(\phi ,\theta ,\varphi ,\eta ). \end{aligned}$$
(4.141)

Using also the definition of A, we obtain

$$\begin{aligned} |\partial _\theta A(\phi ,\theta ,\varphi ,\eta )|\le C\sin (\phi )\sin (\varphi )|\sin (\theta -\eta )|\lesssim \sin (\varphi )A^\frac{1}{2}(\phi ,\theta ,\varphi ,\eta ). \end{aligned}$$
(4.142)

Then, with the help of (4.128), (4.131), (4.133) (4.140), (4.141) and (4.142), we can estimate \(\partial _\theta K_1\) as

$$\begin{aligned} |\partial _\theta K_1(\phi ,\theta ,\varphi ,\eta )|&\lesssim \Vert h\Vert _{\text {Lip}}\frac{\sin ^3(\varphi )\left( (\phi -\varphi )^2+(\sin (\varphi )+\sin (\phi ))\sin ^2((\theta -\eta )/2)\right) }{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\right) ^2}\\&\quad +\Vert h\Vert _{\text {Lip}}\frac{\sin ^2(\varphi )}{(\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)}\\&\quad + \Vert h\Vert _{\text {Lip}}\frac{\sin ^3(\varphi )\left( (\phi -\varphi )^2+(\sin (\varphi )+\sin (\phi ))\sin ^2((\theta -\eta )/2)\right) }{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\right) ^2}\cdot \end{aligned}$$

Consequently we get

$$\begin{aligned} |\partial _\theta K_1(\phi ,\theta ,\varphi ,\eta )|&\lesssim \frac{\Vert h\Vert _{\text {Lip}}\,\sin ^2(\varphi )}{(\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)}\\&\lesssim \frac{\Vert h\Vert _{\text {Lip}}\,|\sin ((\theta -\eta )/2)|^{-1}\,\sin (\varphi )}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\right) ^{\frac{1}{2}}}\cdot \end{aligned}$$

Therefore we obtain by virtue of (4.114)

$$\begin{aligned} |\partial _\theta K_1(\phi ,\theta ,\varphi ,\eta )|\le C\Vert h\Vert _{\text {Lip}}|\phi -\varphi |^{-\beta }|\sin ((\theta -\eta )/2|^{-(2-\beta )}, \end{aligned}$$
(4.143)

for any \(\beta \in (0,1)\). Hence, all the hypothesis of Proposition C.1 are satisfied and therefore we deduce that \({\mathscr {G}}_1\) belongs to \({\mathscr {C}}^\beta \big ((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}}\big )\), for any \(\beta \in (0,1)\). The estimates of the kernel \(K_2\) we are quite similar to those of \(K_1\) modulo some slight adaptations. We shall not develop all the estimates which are straightforward and tedious. We will restrict this discussion to the analogous estimate to (4.135) and (4.143). First note that thanks to (4.128) and (4.129) one gets

$$\begin{aligned} \big |\partial _\varphi (\sin (\varphi )\, r_0^2(\varphi )g_\theta (\varphi ,\theta ))\big |\lesssim \Vert h\Vert _{\text {Lip}}\sin ^3(\varphi )|\sin ((\theta -\eta )/2)|. \end{aligned}$$

This implies that

$$\begin{aligned} \big |K_2(\phi ,\theta ,\varphi ,\eta )\big |&\lesssim \Vert h\Vert _{\text {Lip}}\frac{\sin ^3(\varphi )\sin ^2((\theta -\eta )/2)}{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}\\&\lesssim \Vert h\Vert _{\text {Lip}}\frac{\sin (\varphi )}{A(\phi ,\theta ,\varphi ,\eta )^\frac{1}{2}}\cdot \end{aligned}$$

It follows from (4.131) and (4.114) that

$$\begin{aligned} | K_2(\phi ,\theta ,\varphi ,\eta )|\lesssim \Vert h\Vert _{\text {Lip}}|\phi -\varphi |^{-\beta }|\sin ((\theta -\eta )/2|^{-(1-\beta )}, \end{aligned}$$
(4.144)

which is the announced estimate. As to the estimate of \(\partial _\theta K_2\) we first write

$$\begin{aligned} \partial _\theta K_2(\phi ,\theta ,\varphi ,\eta )&= -\frac{\partial _\varphi \left( \sin (\varphi )r_0^2(\varphi )h(\varphi ,\theta )\right) \sin (\theta -\eta )}{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}\nonumber \\&\quad +\frac{\partial _\varphi \left( \sin (\varphi )r_0^2(\varphi ) g_\theta (\varphi ,\eta )\right) \cos (\theta -\eta )}{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}\nonumber \\&\quad - \frac{3}{2} K_2(\phi ,\theta ,\varphi ,\eta )\frac{\partial _\theta A(\phi ,\theta ,\varphi ,\eta )}{A(\phi ,\theta ,\varphi ,\eta )} \end{aligned}$$
(4.145)

Straightforward calculations using \({\mathbf{(H2)}}\) and (2.17) show that

$$\begin{aligned} \frac{\left| \partial _\varphi \left( \sin (\varphi )r_0^2(\varphi )h(\varphi ,\theta )\right) \sin (\theta -\eta )\right| }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}&\lesssim \Vert h\Vert _{\text {Lip}} \frac{\sin ^3(\varphi )|\sin ((\theta -\eta )/2)|}{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}\\&\lesssim \Vert h\Vert _{\text {Lip}} \frac{\sin ^2(\varphi )}{A(\phi ,\theta ,\varphi ,\eta )}\cdot \end{aligned}$$

Putting together (4.131) and (4.114) implies

$$\begin{aligned} \frac{\sin ^2(\varphi ) }{A(\phi ,\theta ,\varphi ,\eta )}&\lesssim |\sin ((\theta -\eta )/2)|^{-1} \frac{\sin (\varphi ) }{A^{\frac{1}{2}}(\phi ,\theta ,\varphi ,\eta )}\nonumber \\&\lesssim \frac{1}{|\phi -\varphi |^{\beta }|\sin ((\theta -\eta )/2)|^{2-\beta }}\cdot \end{aligned}$$
(4.146)

Therefore we find

$$\begin{aligned} \frac{\left| \partial _\varphi \left( \sin (\varphi )r_0^2(\varphi )h(\varphi ,\theta )\right) \sin (\theta -\eta )\right| }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}&\lesssim \frac{\Vert h\Vert _{\text {Lip}} }{|\phi -\varphi |^{\beta }|\sin ((\theta -\eta )/2)|^{2-\beta }}\cdot \end{aligned}$$

As to the second term of the right-hand side of (4.145), we get in view of \({\mathbf{(H2)}}\), (4.128) and (4.129)

$$\begin{aligned} \frac{\left| \partial _\varphi \left( \sin (\varphi )r_0^2(\varphi ) g_\theta (\varphi ,\eta )\right) \cos (\theta -\eta )\right| }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}&\lesssim \Vert h\Vert _{\text {Lip}} \frac{\sin ^3(\varphi )|\sin ((\theta -\eta )/2)|}{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}\\&\lesssim \Vert h\Vert _{\text {Lip}} \frac{\sin ^2(\varphi )}{A(\phi ,\theta ,\varphi ,\eta )}\cdot \end{aligned}$$

It follows from (4.146) that

$$\begin{aligned} \frac{\left| \partial _\varphi \left( \sin (\varphi )r_0^2(\varphi ) g_\theta (\varphi ,\eta )\right) \cos (\theta -\eta )\right| }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}&\lesssim \Vert h\Vert _{\text {Lip}} \frac{1}{|\phi -\varphi |^{\beta }|\sin ((\theta -\eta )/2)|^{2-\beta }}\cdot \end{aligned}$$

Concerning the last term of (4.145), we put together (4.142), (4.144), (4.131) and (4.114) that

$$\begin{aligned}&|K_2(\phi ,\theta ,\varphi ,\eta )|\frac{|\partial _\theta A(\phi ,\theta ,\varphi ,\eta )|}{A(\phi ,\theta ,\varphi ,\eta )}\\&\quad \lesssim \Vert h\Vert _{\text {Lip}}|\phi -\varphi |^{-\beta }|\sin ((\theta -\eta )/2)|^{-(1-\beta )}\frac{\sin \varphi }{A^{\frac{1}{2}}(\phi ,\theta ,\varphi ,\eta )}\\&\quad \lesssim \Vert h\Vert _{\text {Lip}}|\phi -\varphi |^{-\beta }|\sin ((\theta -\eta )/2)|^{-(2-\beta )}. \end{aligned}$$

Therefore collecting the preceding estimates allows to get the suitable estimate for \(\partial _\theta K_2\),

$$\begin{aligned} |\partial _\theta K_2(\phi ,\theta ,\varphi ,\eta )|\lesssim \Vert h\Vert _{\text {Lip}}|\phi -\varphi |^{-\beta }|\sin ((\theta -\eta )/2)|^{-(2-\beta )}. \end{aligned}$$

The estimate for \(\partial _\phi K_2\) can be done similarly in a straightforward way. Consequently the assumptions of Proposition C.1 hold true and one deduces that \({\mathscr {G}}_1\in {\mathscr {C}}^\beta \big ((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}}\big )\). Hence we obtain \(\partial _\phi G\in {\mathscr {C}}^\beta \big ((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}}\big )\), with the estimate

$$\begin{aligned} \Vert \partial _\phi G(h)\Vert _{{\mathscr {C}}^{\beta }}\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}. \end{aligned}$$
(4.147)

The next stage is to show that \(\partial _\theta G(h)\in {\mathscr {C}}^\beta \big ((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}}\big )\) following the same strategy as before. From (4.126), we get

Direct computations show that

$$\begin{aligned} \partial _\theta A(\phi ,\theta ,\varphi ,\eta )&= 2r_0(\phi ) r_0(\varphi )\sin (\theta -\eta )=-\partial _\eta A(\phi ,\theta ,\varphi ,\eta ). \end{aligned}$$

It follows

On the other hand, integration by parts in \(\eta \) (this can be done by cutting off the integral in \(\eta \) away from \(\theta \) and taking a limit) yields

Thus we deduce by subtraction

(4.148)

Since \(h\in {\mathscr {C}}^{1,\alpha }\), then the mean value theorem implies

$$\begin{aligned} |h(\varphi ,\theta )-h(\varphi ,\eta )|\lesssim \Vert h \Vert _{\text {Lip}}|\theta -\eta |. \end{aligned}$$

Moreover, by the \(2\pi \)-periodicity of h in \(\eta \) one obtains

$$\begin{aligned} |h(\varphi ,\theta )-h(\varphi ,\eta )|\lesssim \Vert h \Vert _{{\mathscr {C}}^{1,\alpha }}\big |\sin \big ((\theta -\eta )/2\big )\big |. \end{aligned}$$
(4.149)

Define the kernel

$$\begin{aligned} K_3(\phi ,\theta ,\varphi ,\eta ):=\frac{\sin (\varphi )r_0^2(\varphi )\sin (\eta -\theta )\big (h(\varphi ,\eta )-h(\varphi ,\theta )\big ) }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}, \end{aligned}$$

and let us check the hypothesis of Proposition C.1. First using (4.131), \({\mathbf{(H2)}}\) and (4.149) we obtain

$$\begin{aligned} |K_3(\phi ,\theta ,\varphi ,\eta )|&\lesssim \Vert h \Vert _{{\mathscr {C}}^{1,\alpha }} \frac{\sin ^{3}(\varphi )\sin ^2((\theta -\eta )/2)}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\right) ^\frac{3}{2}}\\&\lesssim \Vert h \Vert _{{\mathscr {C}}^{1,\alpha }}\frac{\sin (\varphi )}{\left( (\phi -\varphi )^2+(\sin ^2(\phi )+\sin ^2(\varphi ))\sin ^2((\theta -\eta )/2)\right) ^\frac{1}{2}}. \end{aligned}$$

Applying (4.114) yields

$$\begin{aligned} |K_3(\phi ,\theta ,\varphi ,\eta )|\lesssim \Vert h \Vert _{{\mathscr {C}}^{1,\alpha }}|\phi -\varphi |^{-(1-\beta )}|\sin ((\theta -\eta )/2)|^{-\beta }, \end{aligned}$$
(4.150)

for any \(\beta \in (0,1)\). Let us estimate \(\partial _\phi K_3\) which is explicitly given by,

$$\begin{aligned} \partial _\phi K_3(\phi ,\theta ,\varphi ,\eta )&= -\frac{3}{2} \frac{\sin (\varphi )r_0^2(\varphi )\sin (\eta )\big (h(\varphi ,\eta )-h(\varphi ,\theta )\big )\partial _\phi A(\phi ,\theta ,\varphi ,\eta ) }{A(\phi ,\theta ,\varphi ,\eta )^\frac{5}{2}}\\&= -\frac{3}{2} K_3(\phi ,\theta ,\varphi ,\eta )\frac{\partial _\phi A(\phi ,\theta ,\varphi ,\eta )}{A(\phi ,\theta ,\varphi ,\eta )}\cdot \end{aligned}$$

By virtue of (4.138) and (4.150), we achieve

$$\begin{aligned} |\partial _\phi K_3(\phi ,\theta ,\varphi ,\eta )|&\lesssim \Vert h \Vert _{{\mathscr {C}}^{1,\alpha }}|\phi -\varphi |^{-(1-\beta )}\big |\sin \big ((\theta -\eta )/2\big )\big |^{-\beta } A^{-\frac{1}{2}}(\phi ,\theta ,\varphi ,\eta ),\\&\lesssim \frac{\Vert h \Vert _{{\mathscr {C}}^{1,\alpha }}}{|\phi -\varphi |^{2-\beta }|\sin ((\theta -\eta )/2)|^{\beta }}, \end{aligned}$$

for any \(\beta \in (0,1)\). It remains to establish the suitable estimates for \(\partial _\theta K_3\). First we have

$$\begin{aligned} \partial _\theta K_3(\phi ,\theta ,\varphi ,\eta )&= -\frac{3}{2} K_3(\phi ,\theta ,\varphi ,\eta )\frac{\partial _\theta A(\phi ,\theta ,\varphi ,\eta )}{A(\phi ,\theta ,\varphi ,\eta )}\\&\quad - \frac{\sin (\varphi )r_0^2(\varphi )\sin (\eta -\theta )\partial _\theta h(\varphi ,\theta ) }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}} \\&\quad -\frac{\sin (\varphi )r_0^2(\varphi )\cos (\eta -\theta )\big (h(\varphi ,\eta )-h(\varphi ,\theta )\big ) }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}\cdot \end{aligned}$$

Using (4.142) and (4.150) (where we exchange \(\beta \) by \(1-\beta \)) we get

$$\begin{aligned} \big |K_3(\phi ,\theta ,\varphi ,\eta )\big |\frac{\big |\partial _\theta A(\phi ,\theta ,\varphi ,\eta )\big |}{A(\phi ,\theta ,\varphi ,\eta )}&\lesssim \frac{\Vert h \Vert _{{\mathscr {C}}^{1,\alpha }}}{|\phi -\varphi |^{\beta }|\sin ((\theta -\eta )/2)|^{1-\beta }}\frac{\sin \varphi }{A^{\frac{1}{2}}(\phi ,\theta ,\varphi ,\eta )}\\&\lesssim \frac{\Vert h \Vert _{{\mathscr {C}}^{1,\alpha }}}{|\phi -\varphi |^{\beta }|\sin ((\theta -\eta )/2)|^{2-\beta }}\cdot \end{aligned}$$

For the second term of the right-hand side of \(\partial _\theta K_3\) we write in view of \({\mathbf{(H2)}}\)

$$\begin{aligned} \frac{\sin (\varphi )r_0^2(\varphi )|\sin (\eta -\theta )||\partial _\theta h(\varphi ,\theta )| }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}&\lesssim \Vert h\Vert _{\text {Lip}} \frac{\sin ^3(\varphi )|\sin ((\theta -\eta )/2)| }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}\\&\lesssim \Vert h\Vert _{\text {Lip}} \frac{\sin ^2(\varphi ) }{A(\phi ,\theta ,\varphi ,\eta )}\cdot \end{aligned}$$

Applying (4.146) yields

$$\begin{aligned} \frac{\sin (\varphi )r_0^2(\varphi )|\sin (\eta -\theta )||\partial _\theta h(\varphi ,\theta )| }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}&\lesssim \frac{\Vert h\Vert _{\text {Lip}} }{|\phi -\varphi |^{\beta }|\sin ((\theta -\eta )/2)|^{2-\beta }}\cdot \end{aligned}$$

Concerning the last term of the right-hand side of \(\partial _\theta K_3\), it is similar to the foregoing one. Indeed, using (4.149) and \({\mathbf{(H2)}}\) we get

$$\begin{aligned} \frac{\sin (\varphi )r_0^2(\varphi )|\cos (\eta -\theta )|\big |h(\varphi ,\eta )-h(\varphi ,\theta )\big | }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}&\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }} \frac{\sin ^3(\varphi )|\sin ((\eta -\theta )/2)|}{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}\\&\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }} \frac{\sin ^2(\varphi )}{A(\phi ,\theta ,\varphi ,\eta )}\cdot \end{aligned}$$

It suffices to use (4.146) to obtain

$$\begin{aligned} \frac{\sin (\varphi )r_0^2(\varphi )|\cos (\eta -\theta )|\big |h(\varphi ,\eta )-h(\varphi ,\theta )\big | }{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}&\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }} \frac{\sin ^3(\varphi )|\sin ((\eta -\theta )/2)|}{A(\phi ,\theta ,\varphi ,\eta )^\frac{3}{2}}\\&\lesssim \frac{\Vert h\Vert _{{\mathscr {C}}^{1,\alpha }} }{|\phi -\varphi |^{\beta }|\sin ((\theta -\eta )/2)|^{2-\beta }}\cdot \end{aligned}$$

Therefore we get from the preceding estimates

$$\begin{aligned} |\partial _\theta K_3(\phi ,\theta ,\varphi ,\eta )|&\lesssim \frac{\Vert h\Vert _{{\mathscr {C}}^{1,\alpha }} }{|\phi -\varphi |^{\beta }|\sin ((\theta -\eta )/2)|^{2-\beta }}\cdot \end{aligned}$$

Consequently, all the assumptions of Proposition C.1 are verified by the kernel \(K_3\) and thus we deduce that \(\partial _\theta G(h)\in {\mathscr {C}}^\beta \big ((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}}\big )\) for any \(\beta \in (0,1)\), with the estimate

$$\begin{aligned} \Vert \partial _\theta G(h)\Vert _{{\mathscr {C}}^{\beta }}\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}. \end{aligned}$$

Putting together this estimate with (4.147) and (4.132) yields

$$\begin{aligned} \Vert G(h)\Vert _{{\mathscr {C}}^{1,\beta }}\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}, \end{aligned}$$

and this achieves the proof of the proposition. \(\quad \square \)

4.6 Transversality

We have shown in Proposition 4.6 that when \(\Omega \) belongs to the discrete set \(\{\Omega _m, m\ge 2\}\) then the linearized operator \(\partial _f {\tilde{F}}(\Omega ,0)\) is of Fredholm type with one-dimensional kernel. This property is not enough to bifurcate to nontrivial solutions for the nonlinear problem. A sufficient condition for that, according to Theorem B.1, is the the transversal assumption which amounts to checking

$$\begin{aligned} \partial ^2_{\Omega , f} {\tilde{F}}(\Omega _m,0)f_m^\star \notin \text {Im}(\partial _f {\tilde{F}}(\Omega _m,0)), \end{aligned}$$

where \(f_m^\star \) is a generator of the kernel of \(\partial _f {\tilde{F}}(\Omega _m,0)\). Note that as a consequence of (4.1) and (4.2), for a function \(h:(\phi ,\theta )\mapsto \sum _{n\ge 1}h_n(\phi ) \cos (n\theta ) \in X_m^\alpha \), we get

$$\begin{aligned} \partial _{f} {\tilde{F}}(\Omega ,0)h(\phi ,\theta )=\sum _{n\ge 1}{\mathcal {L}}_n^\Omega h_n(\phi ) \cos (n\theta ), \end{aligned}$$

with

where \({\mathcal {K}}_n^\Omega \) is defined in (4.82). Hence, the second mixed derivative takes the form,

$$\begin{aligned} \partial ^2_{\Omega ,f} {\tilde{F}}(\Omega ,0)h(\phi ,\theta )=-h(\phi ,\theta ). \end{aligned}$$

Our main result of this section reads as follows.

Proposition 4.7

Let \(m\ge 2\), then the transversal condition holds true, that is,

$$\begin{aligned} \partial ^2_{\Omega , f} {\tilde{F}}(\Omega _m,0)f_m^\star \notin \text {Im}(\partial _f {\tilde{F}}(\Omega _m,0)), \end{aligned}$$

where \(f_m^\star \) is a generator of the kernel of \(\partial _f {\tilde{F}}(\Omega _m,0)\).

Proof

Recall from the proof of Proposition 4.6 that the function \(f_m^\star \) has the form

$$\begin{aligned} f_m^\star (\phi ,\theta )=h_m^\star (\phi )\cos (m\theta ) \end{aligned}$$

and \(h_m^\star \) is a nonzero solution to the equation

$$\begin{aligned} {\mathcal {K}}_m^{\Omega _m} h_m^\star (\phi )=h_m^\star (\phi ). \end{aligned}$$

It follows that

$$\begin{aligned} \partial ^2_{\Omega ,f} {\tilde{F}}(\Omega _m,0)f_m^\star (\phi ,\theta )=-h_m^\star (\phi )\cos (m\theta ). \end{aligned}$$

Assume that this element belongs to the range of \(\partial _{f} {\tilde{F}}(\Omega _m,0)\). Then we can find \(h_m\) such that

$$\begin{aligned} h_m^\star (\phi )=\nu _{\Omega _m}(\phi )\big ( h_m(\phi )-{\mathcal {K}}_m^{\Omega _m} h_m(\phi )\big ). \end{aligned}$$

Dividing this equality by \(\nu _{\Omega _m}\) and taking the inner product with \(h_m^{\star }\), with respect to \(\langle \cdot ,\cdot \rangle _{\Omega _m}\) defined in (4.84) yields by the symmetry of \( {\mathcal {K}}_m^{\Omega _m}\)

$$\begin{aligned} \Big \langle \frac{h_m^{\star }}{\nu _{\Omega _m}},h_m^{\star }\Big \rangle _{\Omega _m}&= \Big \langle h_m,h_m^{\star }\Big \rangle _{\Omega _m}-\Big \langle {\mathcal {K}}_m^{\Omega _m} h_m,h_m^{\star }\Big \rangle _{\Omega _m}\\&= \Big \langle h_m,h_m^{\star }\Big \rangle _{\Omega _m}-\Big \langle h_m, {\mathcal {K}}_m^{\Omega _m}h_m^{\star }\Big \rangle _{\Omega _m}\\&= \Big \langle h_m,h_m^{\star }-{\mathcal {K}}_m^{\Omega _m}h_m^{\star }\Big \rangle _{\Omega _m}\\&= 0. \end{aligned}$$

Coming back to the definition of the inner product (4.84) and (4.4), we find

From the assumption \({\mathbf{(H)}}\) we know that \(r_0\) does not vanish in \((0,\pi )\). Then we get from the continuity of \(h_m^\star \) that this latter function should vanish everywhere in \((0,\pi )\), which is a contradiction. Hence, we deduce that \(f_m^\star \) does not belong to the range of \(\partial _f {\tilde{F}}(\Omega _m,0)\) and then the transversal condition is satisfied. \(\quad \square \)

5 Nonlinear Action

This section is devoted to the regularity study of the nonlinear functional \({\tilde{F}}\) defined in (2.13) that we recall for the convenience of the reader,

$$\begin{aligned} {\tilde{F}}(\Omega ,f)(\phi ,\theta )=\frac{1}{r_0(\phi )}\left\{ I(f)(\phi ,\theta )-\frac{\Omega }{2}r^2(\phi ,\theta )-m(\Omega ,f)(\phi )\right\} , \end{aligned}$$

for any \((\phi ,\theta )\in (0,\pi )\times (0,2\pi )\) and where

the mean m is defined in (2.11) and

$$\begin{aligned} r(\phi ,\theta )=r_0(\phi )+f(\phi ,\theta ). \end{aligned}$$

We would like in particular to analyze the symmetry/regularity persistence of the function spaces \(X_m^\alpha \) defined in (2.15) and (2.16) through the action of the nonlinear functional \({\tilde{F}}\).

5.1 Symmetry persistence

The main task here is to check the symmetry persistence of the function spaces \(X_m^\alpha \) defined in (2.15) through the nonlinear action of \({\tilde{F}}\). Notice that at this level, we do not raise the problem of whether or not this functional is well-defined and this target is postponed later in Sect. 5. First recall that

$$\begin{aligned} X_m^\alpha&=\Big \{f:[0,\pi ]\times [0,2\pi ]\rightarrow {{\,\mathrm{{\mathbb {R}}}\,}}\, :\, \, f\in {\mathscr {C}}^{1,\alpha }, f(0,\theta )=f(\pi ,\theta )\equiv 0,\,\\&\quad f\left( \frac{\pi }{2}-\phi ,\theta \right) =f\left( \frac{\pi }{2}+\phi ,\theta \right) , \, f\left( \phi ,\theta \right) \\&\quad =\sum _{n\ge 1}f_n(\phi )\cos (nm\theta )\Big \}. \end{aligned}$$

Proposition 5.1

Let \(\Omega \in {{\,\mathrm{{\mathbb {R}}}\,}}\), \(f\in X_m^\alpha \) with \(m\ge 1\) and assume that \(r_0\) satisfies the conditions \({\mathbf{(H)}}\). Then the following assertions hold true.

  1. (1)

    The equatorial symmetry:

    $$\begin{aligned}{\tilde{F}}(\Omega ,f)\left( \pi -\phi ,\theta \right) ={\tilde{F}}(\Omega ,f)\left( \phi ,\theta \right) ,\quad \forall \, (\phi ,\theta )\in [0,\pi ]\times {{\,\mathrm{{\mathbb {R}}}\,}}. \end{aligned}$$
  2. (2)

    We get the algebraic structure,

    $$\begin{aligned} {\tilde{F}}(\Omega ,f)(\phi ,\theta )=\sum _{n\ge 1} f_n(\phi )\cos (n\theta ), \end{aligned}$$

    for some functions \(f_n\) and for any \((\phi ,\theta )\in [0,\pi ]\times [0,2\pi ]\).

  3. (3)

    The m-fold symmetry: \({\tilde{F}}(\Omega ,f)(\phi ,\theta +\frac{2\pi }{m})={\tilde{F}}(\Omega ,f)(\phi ,\theta )\), for any \((\phi ,\theta )\in [0,\pi ]\times {{\,\mathrm{{\mathbb {R}}}\,}}\).

Proof

\({\mathbf{(1)}}\) From the expression of \({\tilde{F}}\) in (2.13), it is suffices to check the property for I(f). One can easily verify using the symmetry of the functions \(\cos \) and r combined with the change of variables \(\varphi \mapsto \pi -\varphi \)

\({\mathbf{(2)}}\) In order to get the desired structure, it suffices to check the following symmetry

$$\begin{aligned} I(f)(\phi ,-\theta )=I(f)(\phi ,\theta ),\quad \forall \, (\phi ,\theta )\in [0,\pi ]\times {{\,\mathrm{{\mathbb {R}}}\,}}. \end{aligned}$$

To do that, we use the symmetry of r, that is \(r(\varphi ,-\theta )=r(\varphi ,\theta )\), combined with the change of variables \(\eta \mapsto -\eta \) allowing one to get

\({\mathbf{(3)}}\) First, since r belongs to \(X_m^\alpha \) then it satisfies \(r(\varphi , \theta +\frac{2\pi }{m})=r(\varphi ,\theta )\). Thus we get by the change of variables \(\eta \mapsto \eta +\frac{2\pi }{m}\)

Notice that we have used the fact that the Euclidean distance in \({\mathbb {C}}\) is invariant by the rotation action \(z\mapsto e^{i\frac{2\pi }{m}} z\). \(\quad \square \)

The next discussion is devoted to the symmetry effects of the surface of the vortices on the velocity structure. We shall show the following.

Lemma 5.1

If \(r_0\) satisfies \({\mathbf{(H)}}\) and \(f\in X_m^\alpha \), with \(m\ge 2\), then

As a consequence, the velocity field defined in (2.4) is vanishing at the vertical axis, that is,

$$\begin{aligned} U(0,0,z)=0, \end{aligned}$$

for any \(z\in {{\,\mathrm{{\mathbb {R}}}\,}}\).

Proof

Set for any \(z\in {{\,\mathrm{{\mathbb {R}}}\,}}\),

Observe that from the periodicity in \(\eta \) we may write

Since \(f\in X_m^\alpha \), then \(r(\varphi ,-\eta )=r(\varphi ,\eta )\) and so \((\partial _\eta r)(\varphi ,-\eta )=-(\partial _\eta r)(\varphi ,\eta )\). Therefore making the change of variables \(\eta \mapsto -\eta \) allows to get \(I_1(z)=0\).

To check \(I_2(z)=0\) we shall use the m-fold symmetry of r. In fact by the change of variables \(\eta \mapsto \eta +\frac{2\pi }{m}\) and using the \(2\pi \)-periodicity in \(\eta \) and some elementary trigonometric identity, we find

Since \(m\ge 2\) and \(I_1(z)=0\) then we get \(I_2(z)=0\).

Coming back to (2.4) and following the change of variables giving (2.8) we easily get

$$\begin{aligned} U(0,0,z)=\big (I_1(z), I_2(z),0\big ), \end{aligned}$$

which gives the announced result. \(\quad \square \)

5.2 Deformation of the Euclidean norm

The spherical change of coordinates used to recover both the velocity and the stream function from the surface geometry of the patch yields a deformation of the Green function. Notice that in the usual Cartesian coordinates the Green kernel is radial and thus it is isotropic with respect to all the variables. In the new coordinates we lose this property and the Green kernel becomes anisotropic and the north and south poles are degenerating points. To deal with these defects one needs refined treatments in the behavior of the kernel or also the adaptation of the function spaces which are of Dirichlet type. The following lemma is crucial to deal with the anisotropy of the kernel.

Lemma 5.2

Let \(m\ge 1, \alpha \in (0,1)\), \(r_0\) satisfies \({\mathbf{(H)}}\), \(f\in X_m^\alpha \) such that \(\Vert f\Vert _{X_m^\alpha }\le \varepsilon \) with \(\varepsilon \) small enough and set \(r=r_0+f\). Define for any \(\phi \in [0,\frac{\pi }{2}]\), \(\varphi \in [0,\pi ]\), \(\theta ,\eta \in [0,2\pi ]\) and \(s\in [0,1]\)

$$\begin{aligned} J_s(\phi ,\theta ,\varphi ,\eta ):= & {} (r(\varphi ,\eta )-sr(\phi ,\theta ))^2\\&+2sr(\phi ,\theta )r(\varphi ,\eta ) (1-\cos (\theta -\eta ))+(\cos (\phi )-\cos (\varphi ))^2. \end{aligned}$$

Then

$$\begin{aligned} |J_0(\phi ,\theta ,\varphi ,\eta )|&\ge C\sin ^2(\varphi ),\end{aligned}$$
(5.1)
$$\begin{aligned} |J_s(\phi ,\theta ,\varphi ,\eta )|&\ge C\Big (\big (\sin ^2(\varphi )+s^2\phi ^2\big ){\sin ^2((\theta -\eta )/2)}+(\varphi +\phi )^2(\phi -\varphi )^2\Big ), \end{aligned}$$
(5.2)

with C an absolute constant. Remark that we have restricted \(\phi \) to \(\in [0,\pi /2]\) instead of \([0,\pi ]\) because of the symmetry of r with respect to \(\frac{\pi }{2}\).

Proof

Since \(f\in B_{X_m^\alpha }(\varepsilon )\), for some \(\varepsilon <1\), and \(r_0\) verifies (H2) then

$$\begin{aligned} r(\varphi ,\eta )&= r_0(\varphi )+f(\varphi ,\eta )\ge 2C \sin \varphi -|f(\varphi ,\eta )|. \end{aligned}$$

In addition f satisfies (2.17) and in particular

$$\begin{aligned} \frac{|f(\varphi ,\eta )|}{\sin (\varphi )}\le C_1\Vert f\Vert _{\text {Lip}}. \end{aligned}$$

It follows that,

$$\begin{aligned} r(\varphi ,\eta )\ge \left( 2C -C_1\Vert f\Vert _{\text {Lip}}\right) \sin (\varphi ). \end{aligned}$$

By imposing \(\Vert f\Vert _{\text {Lip}}\le \varepsilon = \frac{C}{C_1}\), we infer

$$\begin{aligned} r(\varphi ,\eta )\ge C\sin (\varphi ). \end{aligned}$$
(5.3)

Consequently, we obtain

$$\begin{aligned} J_0(\phi ,\theta ,\varphi ,\eta )&= r^2(\varphi ,\eta )+(\cos (\varphi )-\cos (\phi ))^2\ge C\sin ^2(\varphi ), \end{aligned}$$

which gives the estimate (5.1). Let us now check the validity of (5.2). First, we remark that

$$\begin{aligned} J_s(\phi ,\theta ,\varphi ,\eta )= & {} r^2(\varphi ,\eta )+s^2r^2(\phi ,\theta )-2sr(\varphi ,\eta )r(\phi ,\theta )\cos (\theta -\eta )\\&+(\cos (\varphi )-\cos (\phi ))^2. \end{aligned}$$

Denote

$$\begin{aligned} g_1(x):=r^2(\varphi ,\eta )+x^2-2xr(\varphi ,\eta )\cos (\theta -\eta )+(\cos (\varphi )-\cos (\phi ))^2, \end{aligned}$$

and therefore we get the relation \(g_1(sr(\phi ,\theta ))=J_s(\phi ,\theta ,\varphi ,\eta )\). From variation arguments we infer that the function \(g_1\) reaches its global minimum at the point

$$\begin{aligned} x_c=r(\varphi ,\eta )\cos (\theta -\eta ). \end{aligned}$$

Let us distinguish the cases \(\cos (\theta -\eta )\in [0,1]\) and \(\cos (\theta -\eta )\in [-1,0]\). In the first case, one has according to (5.3)

$$\begin{aligned} J_s(\phi ,\theta ,\varphi ,\eta )&= g_1(sr(\phi ,\theta ))\\&\ge g_1(x_c) =r^2(\varphi ,\eta )\sin ^2(\theta -\eta )+(\cos (\varphi )-\cos (\phi ))^2 \\&\ge C(\sin ^2(\varphi ){\sin ^2(\theta -\eta )}+(\cos (\varphi )-\cos (\phi ))^2). \end{aligned}$$

Using that \(\cos (\theta -\eta )\in [0,1]\), one gets

$$\begin{aligned}\sin ^2(\theta -\eta )=2\sin ^2((\theta -\eta )/2)(1+\cos (\theta -\eta ))\ge 2\sin ^2((\theta -\eta )/2).\end{aligned}$$

Moreover, since \(\phi \in [0,\frac{\pi }{2}]\) and \(\varphi \in [0,\pi ]\), we obtain

$$\begin{aligned} |\cos (\varphi )-\cos (\phi )|&= |(1-\cos (\phi ))-(1-\cos (\varphi ))|\nonumber \\&= 2|\sin ^2(\phi /2)-\sin ^2(\varphi /2)|\nonumber \\&= 2|\sin (\phi /2)-\sin (\varphi /2)|(\sin (\phi /2)+\sin (\varphi /2))\nonumber \\&\ge C|\phi -\varphi ||\phi +\varphi |. \end{aligned}$$
(5.4)

Hence

$$\begin{aligned} J_s(\phi ,\theta ,\varphi ,\eta )&\ge C\Big (\sin ^2(\varphi ){\sin ^2((\theta -\eta )/2)}+(\phi +\varphi )^2(\phi -\varphi )^2\Big ). \end{aligned}$$
(5.5)

In the second case where \(\cos (\theta -\eta )\in [-1,0]\), the critical point is negative, \(x_c\le 0\), and one has from the variations of \(g_1\), the estimate (5.3) and (5.4)

$$\begin{aligned} J_s(\phi ,\theta ,\varphi ,\eta )&= g_1(sr(\phi ,\theta ))\nonumber \\&\ge g_1(0) =r(\varphi ,\eta )^2+(\cos (\varphi )-\cos (\phi ))^2\nonumber \\&\ge C(\sin ^2(\varphi ){\sin ^2((\theta -\eta )/2)}+(\phi +\varphi )^2(\phi -\varphi )^2). \end{aligned}$$
(5.6)

Putting together (5.5) and (5.6), one deduces that

$$\begin{aligned} J_s(\phi ,\theta ,\varphi ,\eta )&\ge C\Big (\sin ^2(\varphi ){\sin ^2((\theta -\eta )/2)}+(\phi +\varphi )^2(\phi -\varphi )^2\Big ), \end{aligned}$$
(5.7)

for any \(\phi \in [0,\pi /2]\), \(\varphi \in [0,\pi ]\) and \(\theta ,\eta \in [0,2\pi ]\).

Following the same ideas, we introduce the function

$$\begin{aligned} g_2(x):=x^2+s^2{r^2(\phi ,\theta )}-2sxr(\phi ,\theta )\cos (\theta -\eta )+(\cos (\varphi )-\cos (\phi ))^2, \end{aligned}$$

which satisfies \(g_2(r(\varphi ,\eta ))=J_s(\phi ,\theta ,\varphi ,\eta )\). Then as before we can check easily that the function \(g_2\) reaches its minimum at the point \( {\tilde{x}}_c=sr(\phi ,\theta )\cos (\theta -\eta ). \) Similarly we distinguishing between two cases \(\cos (\theta -\eta )\in [0,1]\) and \(\cos (\theta -\eta )\in [-1,0]\). For the first case, using (5.4), we have

$$\begin{aligned} J_s(\phi ,\theta ,\varphi ,\eta )&\ge C(s^2\sin ^2(\phi ){\sin ^2((\theta -\eta )/2)}+(\phi +\varphi )^2(\phi -\varphi )^2). \end{aligned}$$

Since \(\phi \in [0,\pi /2]\), we have that \(\sin (\phi )\ge \frac{2}{\pi }\phi \), and then

$$\begin{aligned} J_s(\phi ,\theta ,\varphi ,\eta )&\ge C(s^2\phi ^2{\sin ^2((\theta -\eta )/2)}+(\phi +\varphi )^2(\phi -\varphi )^2). \end{aligned}$$
(5.8)

In the other case, i.e. \(\cos (\theta -\eta )\in [-1,0]\), one has that \({\tilde{x}}_c<0\) and then

$$\begin{aligned} J_s(\phi ,\theta ,\varphi ,\eta )=g_2(r(\varphi ,\eta ))\ge g_2(0)&= s^2r(\phi ,\theta )^2+(\cos (\varphi )-\cos (\phi ))^2 \nonumber \\&\ge s^2\sin ^2(\phi )\sin ^2((\theta -\eta )/2)+(\phi +\varphi )^2(\phi -\varphi )^2. \end{aligned}$$
(5.9)

By summing up (5.7)–(5.8)–(5.9) we achieve (5.2). \(\quad \square \)

5.3 Regularity persistence

In this section we shall investigate the regularity of the function \({\tilde{F}}\) introduced in (2.13). The main result reads as follows.

Proposition 5.2

Let \(m\ge 2, \alpha \in (0,1)\) and \(r_0\) satisfy \({\mathbf{(H)}}\). There exists \(\varepsilon \in (0,1)\) small enough such that the functional

$$\begin{aligned}{\tilde{F}}:{{\,\mathrm{{\mathbb {R}}}\,}}\times B_{X_m^\alpha }(\varepsilon )\rightarrow X_m^\alpha \end{aligned}$$

is well-defined and of class \({\mathscr {C}}^1\). The function spaces \(X_m^\alpha \) are defined in (2.15) and (2.16).

Proof

First we shall split the functional \({\tilde{F}}\) into two pieces

with

$$\begin{aligned} F_1(f)(\phi ,\theta )&= \frac{I(f)(\phi ,\theta )}{r_0(\phi )},\\ F_2(f)(\phi ,\theta )&= 2f(\phi ,\theta )+\frac{f^2(\phi ,\theta )}{r_0(\phi )}\cdot \end{aligned}$$

Define also

Note that I(f) is defined (2.14) and it is nothing but the stream function \(\psi _0\) associated to the domain parametrized by

$$\begin{aligned} (\phi ,\theta )\in [0,\pi ]\times [0,2\pi ]\mapsto \left( \big ( r_0(\phi )+f(\phi ,\theta )\big )e^{i\theta }, \cos \phi \right) \cdot \end{aligned}$$

Thus

$$\begin{aligned} F_1(f)(\phi ,\theta )=\frac{\psi _0\left( \big (r_0(\phi )+f(\phi ,\theta )\big )e^{i\theta }, \cos \phi \right) }{r_0(\phi )}\cdot \end{aligned}$$
(5.10)

We point out that according to the general potential theory the stream function \(\psi _0\) belongs at least to the space \({\mathscr {C}}^{1,\alpha }({{\,\mathrm{{\mathbb {R}}}\,}}^3)\). The proof will be divided into three steps.

Step 1: \(f\mapsto {\mathscr {F}}_2(f)\) is \({\mathscr {C}}^1\). In this step, we check that \({\mathscr {F}}_2\) is well-defined and of class \({\mathscr {C}}^1\). Note that checking the regularity for \({\mathscr {F}}_2\) is equivalent to do it for \(F_2\). The first term of \(F_2\) is trivial to check. As to the second one, it is clear by Taylor’s formula using the boundary conditions and \({\mathbf{(H2)}}\) that the function \(\frac{f^2}{r_0}\) is bounded and vanishes at the points \(\phi =0,\pi \). For the regularity, we differentiate with respect to \(\phi \),

$$\begin{aligned} \partial _\phi \left( \frac{f^2(\phi ,\theta )}{{ r_0(\phi )}}\right) =-r_0'(\phi ) \left( \frac{f(\phi ,\theta )}{r_0(\phi )}\right) ^2+2\frac{f(\phi ,\theta )}{r_0(\phi )}\partial _\phi f(\phi ,\theta ). \end{aligned}$$

Using again Taylor’s formula and the assumptions \({\mathbf{(H)}}\) on \(r_0\) we deduce that the functions \((\phi ,\theta )\mapsto \frac{f(\phi ,\theta )}{\sin \phi }\) and \((\phi ,\theta )\mapsto \frac{\sin \phi }{r_0(\phi )}\) belongs to \({\mathscr {C}}^{\alpha }\). Thus using the algebra structure of this latter space we infer that \((\phi ,\theta )\mapsto \frac{f(\phi ,\theta )}{r_0(\phi )}\) belongs also to \({\mathscr {C}}^\alpha \). The same algebra structure allows to get \(\partial _\phi \left( \frac{f^2}{{ r_0}}\right) \in {\mathscr {C}}^\alpha \). Following the same argument we obtain \(\partial _\theta F_2\) belongs to \({\mathscr {C}}^\alpha \). Concerning the symmetry; it can be derived from Proposition 5.1 combined with the fact that frequency \(n=0\) is eliminated in the definition of \({\mathscr {F}}_2\) by subtracting the mean value in \(\theta \).

Now let us check the \({\mathscr {C}}^1\) dependence in f of \(F_2\). First we can check that its Frechet derivative takes the form

$$\begin{aligned} \partial _f F_2(f)h(\phi ,\theta )=2h(\phi ,\theta )+2\frac{f(\phi ,\theta )h(\phi ,\theta )}{r_0(\phi )}\cdot \end{aligned}$$

Using similar ideas as before, we can easily get that

$$\begin{aligned} \Vert \partial _f {{\mathscr {F}}_2(f_1)h-\partial _f {\mathscr {F}}_2(f_2)h}\Vert _{X_m^\alpha }\le C\Vert f_1-f_2\Vert _{X_m^\alpha }\Vert h\Vert _{X_m^\alpha }. \end{aligned}$$

This implies that \(f\mapsto \partial _f {\mathscr {F}}_2(f)\) is continuous and therefore \({\mathscr {F}}_2\) is of class \({\mathscr {C}}^1\).

Step 2: \(f\mapsto {\mathscr {F}}_1(f)\) is well-defined. This is more involved than \({\mathscr {F}}_2\). According to Proposition 5.1 the functional \({\mathscr {F}}_1\) is symmetric with respect to \(\phi =\frac{\pi }{2}\) and therefore it suffices to check the desired regularity in the range \(\phi \in (0,\pi /2)\) and check that the derivative is not discontinuity at \(\pi /2\). Let us emphasize that we need to check the regularity not for \(F_1\) but for \({\mathscr {F}}_1\) First, we shall check that \({\mathscr {F}}_1\) is bounded and satisfies the boundary condition \({\mathscr {F}}_1(0,\theta )=0\), for any \(\theta \in (0,2\pi )\). The remaining boundary condition \({\mathscr {F}}_1(\pi ,\theta )=0\) follows from the symmetry with respect to the equatorial. For this purpose, we write by virtue of Taylor’s formula

$$\begin{aligned} \forall x_h\in {{\,\mathrm{{\mathbb {R}}}\,}}^2,\quad \psi _0(x_h,\cos \phi )&= \psi _0(0,0,\cos \phi )+x_h\cdot \int _0^1\nabla _h\psi _0\big (\tau x_h,\cos \phi \big )d\tau . \end{aligned}$$
(5.11)

Making the substitution \(x_h=(r_0(\phi )+f(\phi ,\theta ))e^{i\theta }\) and using (5.10) we infer

$$\begin{aligned} F_1(f)(\phi ,\theta )&= \frac{\psi (0,0,\cos \phi )}{r_0(\phi )}+\left( 1+\frac{f(\phi ,\theta )}{r_0(\phi )}\right) e^{i\theta }\\&\quad \cdot \int _0^1\nabla _h\psi _0\left( \tau (r_0(\phi )+f(\phi ,\theta ))e^{i\theta },\cos \phi \right) d\tau \\&=: \frac{\psi (0,0,\cos \phi )}{r_0(\phi )}+{\mathscr {F}}_{1,1}(\phi ,\theta ). \end{aligned}$$

We observe that the \(\cdot \) denotes the usual Euclidean inner product of \({{\,\mathrm{{\mathbb {R}}}\,}}^2\). Consequently, we obtain

$$\begin{aligned} {\mathscr {F}}_1(\phi ,\theta )={\mathscr {F}}_{1,1}(\phi ,\theta )-\langle {\mathscr {F}}_{1,1}\rangle _\theta . \end{aligned}$$
(5.12)

Let us analyze the term \( {\mathscr {F}}_{1,1}\) and check its continuity and the Dirichlet boundary condition. First we observe from the assumption \({\mathbf{(H2)}}\) that 0 is a simple zero for \(r_0\) and we know that \(f(0,\theta )=0\), then one may easily obtain the bound

$$\begin{aligned} |{\mathscr {F}}_{1,1}(\phi ,\theta )|\le C(1+\Vert \partial _\phi f\Vert _{L^\infty })\Vert \nabla _h\psi _0\Vert _{L^\infty ({{\,\mathrm{{\mathbb {R}}}\,}}^3)}. \end{aligned}$$

Furthermore, according to Lebesgue dominated convergence theorem we infer

$$\begin{aligned} \lim _{\phi \rightarrow 0}{\mathscr {F}}_{1,1}(\phi ,\theta )=\left( 1+\frac{\partial _\phi f(0,\theta )}{r_0^\prime (0)}\right) e^{i\theta }\cdot \nabla _h\psi _0\left( 0,0,1\right) , \end{aligned}$$

and this convergence is uniform in \(\theta \in (0,2\pi )\). Notice that the same tool gives the continuity of \( {\mathscr {F}}_{1,1}\) in \([0;\pi /2]\times [0;2\pi ]\).

Now, applying Lemma 5.1 we get \(\nabla _h\psi _0\left( 0,0, 1\right) =0\), and therefore

$$\begin{aligned} \forall \,\theta \in (0,2\pi ),\quad \lim _{\phi \rightarrow 0}{\mathscr {F}}_{1,1}(\phi ,\theta )=\lim _{\phi \rightarrow 0}\langle {\mathscr {F}}_{1,1}\rangle _\theta =0. \end{aligned}$$

This implies that \({\mathscr {F}}_1\) is continuous in \([0,\pi ]\times [0,2\pi ]\) and it satisfies the required Dirichlet boundary condition \({\mathscr {F}}_1(0,\theta )={\mathscr {F}}_1(\pi ,\theta )=0\).

The next step is to establish that \(\partial _\theta {\mathscr {F}}_1\) and \(\partial _\phi {\mathscr {F}}_1\) are \({\mathscr {C}}^\alpha \). We will relate such derivatives to the two-components velocity field \(U=\nabla _h^\perp \psi _0\). Differentiating (5.10) with respect to \(\theta \) leads to

$$\begin{aligned}&\partial _\theta {\mathscr {F}}_1(\phi ,\theta )\nonumber \\&\quad = \partial _\theta F_1(f)(\phi ,\theta ) \nonumber \\&\quad = r_0^{-1}(\phi )\,\nabla _h \psi _0(r(\phi ,\theta )e^{i\theta }, \cos (\phi ))\cdot \left( r(\phi ,\theta )ie^{i\theta }+\partial _\theta r(\phi ,\theta )e^{i\theta }\right) \nonumber \\&\quad = {-}\frac{r(\phi ,\theta )}{r_0(\phi )} U(r(\phi ,\theta )e^{i\theta }, \cos (\phi ))\cdot e^{i\theta }{+} \partial _\theta r(\phi ,\theta ) \frac{U(r(\phi ,\theta )e^{i\theta },\cos (\phi ))}{r_0(\phi )}\cdot ie^{i\theta }, \end{aligned}$$
(5.13)

where \(r(\phi ,\theta )=r_0(\phi )+f(\phi ,\theta )\) and recall that \(\cdot \) is the usual Euclidean inner product of \({{\,\mathrm{{\mathbb {R}}}\,}}^2\).

Concerning the regularity of the partial derivative in \(\phi \), we achieve

$$\begin{aligned}&\partial _\phi F_1(f)(\phi ,\theta )\nonumber \\&\quad = -\frac{r_0'(\phi )}{r_0^2(\phi )}\psi _0(r(\phi ,\theta )e^{i\theta }, \cos (\phi ))+ \frac{\partial _\phi r(\phi ,\theta )}{r_0(\phi )}\nabla _h \psi _0(r(\phi ,\theta )e^{i\theta },\cos (\phi ))\cdot e^{i\theta }\nonumber \\&\qquad -\frac{\sin (\phi )}{r_0(\phi )}\partial _z\psi _0(r(\phi ,\theta )e^{i\theta },\cos (\phi ))\nonumber \\&\quad = -\frac{r_0'(\phi )}{r_0^2(\phi )}\psi _0(r(\phi ,\theta )e^{i\theta }, \cos (\phi ))- \partial _\phi r(\phi ,\theta )\frac{U(r(\phi ,\theta )e^{i\theta },\cos (\phi ))}{r_0(\phi )} \cdot ie^{i\theta }\nonumber \\&\qquad -\frac{\sin (\phi )}{r_0(\phi )}\partial _z\psi _0(r(\phi ,\theta )e^{i\theta },\cos (\phi )). \end{aligned}$$
(5.14)

Define

$$\begin{aligned} {\mathscr {J}}_{1}(\phi ,\theta )=\frac{r_0'(\phi )}{r_0^2(\phi )}\psi _0(r(\phi ,\theta )e^{i\theta }, \cos (\phi )), \end{aligned}$$

and

$$\begin{aligned} {\mathscr {J}}_{2}(\phi ,\theta )=\partial _\phi r(\phi ,\theta )\frac{U(r(\phi ,\theta )e^{i\theta },\cos (\phi ))}{r_0(\phi )} \cdot ie^{i\theta }. \end{aligned}$$

Then from (5.14) we may write

$$\begin{aligned} \partial _\phi F_1(f)(\phi ,\theta )=-{\mathscr {J}}_{1}(\phi ,\theta )-{\mathscr {J}}_{2}(\phi ,\theta )-\frac{\sin (\phi )}{r_0(\phi )}\partial _z\psi _0(r(\phi ,\theta )e^{i\theta },\cos (\phi )). \end{aligned}$$

Let us justify why we can restrict ourselves to prove the regularity for \(\phi \in [0,\pi /2]\) by symmetry. Indeed, since \(r(\pi -\phi ,\theta )=r(\phi ,\theta )\) and \(r_0(\pi -\phi )=r_0(\phi )\), then we obtain that

$$\begin{aligned} r_0'(\pi -\phi )=-r_0'(\phi ), \quad \partial _\phi r(\pi -\phi ,\theta )=\partial _\phi r(\phi ,\theta ). \end{aligned}$$

That implies that \(r_0'(\pi /2)=0\) and \(\partial _\phi r(\pi /2,\theta )=0\). By this way, it is easy to check that \({\mathscr {J}}_1(\pi /2,\theta )={\mathscr {J}}_2(\pi /2,\theta )=0\) yielding that there is not any jump at \(\pi /2\). Moreover, note that the last term belongs to \({\mathscr {C}}^\alpha \) for any \(\phi \in [0,\pi ]\) (here we do not need to resctrict ourselves to \(\phi \in [0,\pi /2]\)) and then, by symmetry, we can extend it to \((0,\pi )\). Indeed, as \((\phi ,\theta )\mapsto \big (r(\phi ,\theta )e^{i\theta },\cos (\phi )\big )\) belongs to \({\mathscr {C}}^{1,\alpha }\) and \(\partial _z\psi _0\in {\mathscr {C}}^{\alpha }({{\,\mathrm{{\mathbb {R}}}\,}}^3)\) then by composition we infer \((\phi ,\theta )\mapsto \partial _z\psi _0\big (r(\phi ,\theta )e^{i\theta },\cos (\phi )\big )\) is in \({\mathscr {C}}^{\alpha }((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}}\big )\). On the other hand, the function \(\frac{\sin }{r_0}\) belongs to \( {\mathscr {C}}^\alpha \) and thus by the algebra structure of \({\mathscr {C}}^\alpha \) we obtain the desired result.

Concerning the term \({\mathscr {J}}_{1}\), we use Taylor’s formula for the stream function \(\psi _0\) as in (5.11) finding that

We observe that the first term is singular and depends only on \(\phi \) and therefore it does not contribute in \({\mathscr {J}}_{1}-\langle {\mathscr {J}}_{1}\rangle _\theta \). Since \((\phi ,\theta )\mapsto \frac{r(\phi ,\theta )}{r_0(\phi )}\) belongs to \({\mathscr {C}}^\alpha \) then to get \({\mathscr {J}}_{1}-\langle {\mathscr {J}}_{1}\rangle _\theta \in {\mathscr {C}}^\alpha \) it suffices to prove that

(5.15)

On the other hand to obtain \({\mathscr {J}}_2\in {\mathscr {C}}^\alpha \) it is enough to get

$$\begin{aligned} (\phi ,\theta )\mapsto \frac{U(r(\phi ,\theta )r^{i\theta }, \cos (\phi ))}{r_0(\phi )}\cdot i\,e^{i\theta }\in {\mathscr {C}}^\alpha . \end{aligned}$$
(5.16)

From (5.13) we get that \(\partial _\theta F_1(f)\in {\mathscr {C}}^\alpha \) provided that (5.15) and (5.16) are satisfied together with

$$\begin{aligned} (\phi ,\theta )\mapsto {U(r(\phi ,\theta )r^{i\theta }, \cos (\phi ))}\cdot e^{i\theta }\in {\mathscr {C}}^\alpha . \end{aligned}$$
(5.17)

By virtue of (2.10) and the fact that \(U=\nabla _h^\perp \psi \), we find that

(5.18)

Next we intend to prove (5.15), (5.16) and (5.17).

\(\bullet \) Proof of (5.17). Using (5.18), we deduce that

Using the notation of Lemma 5.2 we find that

$$\begin{aligned} |(r(\phi , \theta )e^{i\theta },\cos (\phi ))-(r(\varphi ,\eta )e^{i\eta },\cos (\varphi ))|=J_1^{\frac{1}{2}}(\phi ,\theta ,\varphi ,\eta ), \end{aligned}$$

and therefore we may write

This can be split into two integral terms

(5.19)

Next, we shall prove that \({{\mathcal {I}}}_1\) is \({\mathscr {C}}^\alpha \). Notice that the second term \({\mathcal {I}}_2\) is easier to deal with than \({\mathcal {I}}_1\) because its kernel is more regular on the diagonal than that of \({\mathcal {I}}_1\). To get \({\mathcal {I}}_2\in {\mathscr {C}}^\alpha \) it suffices to use in a standard way Proposition C.1. We shall skip this part and focus our attention on the proof to the delicate part \({\mathcal {I}}_1\). For this aim let us define the kernel

$$\begin{aligned} {\mathscr {K}}_1(\phi ,\theta ,\varphi ,\eta )=\frac{\sin (\varphi )\partial _\eta r(\varphi ,\eta )\cos (\eta -\theta )}{J_1^\frac{1}{2}(\phi ,\theta ,\varphi ,\eta )}\cdot \end{aligned}$$

We shall start with checking that \({\mathscr {K}}_1\) is bounded. For this goal we use Lemma 5.2 which implies

$$\begin{aligned} |{\mathscr {K}}_1(\phi ,\theta ,\varphi ,\eta )|&\le \frac{C\,\sin (\varphi )|\partial _\eta r(\varphi ,\eta )|}{\{(\varphi +\phi )^2(\phi -\varphi )^2+(\sin ^2(\varphi )+\phi ^2)\sin ^2((\theta -\eta )/2)\}^{\frac{1}{2}}}\cdot \end{aligned}$$

It is easy to check the inequality

$$\begin{aligned}&\frac{\sin (\varphi )}{\Big ((\varphi +\phi )^2(\phi -\varphi )^2+(\sin ^2(\varphi )+\phi ^2)\sin ^2((\theta -\eta )/2)\Big )^{\frac{1}{2}}}\nonumber \\&\quad \le \frac{1}{\Big ((\phi -\varphi )^2+\sin ^2((\theta -\eta )/2)\Big )^\frac{1}{2}}\cdot \end{aligned}$$
(5.20)

By interpolating between

$$\begin{aligned} \Big ((\phi -\varphi )^2+\sin ^2((\theta -\eta )/2)\Big )^\frac{1}{2}\ge |\phi -\varphi |, \end{aligned}$$

and

$$\begin{aligned} \Big ((\phi -\varphi )^2+\sin ^2((\theta -\eta )/2)\Big )^\frac{1}{2}\ge |\sin ((\theta -\eta )/2)|, \end{aligned}$$

one finds that for any \(\beta \in [0,1]\) and \(\phi \le \pi /2\)

$$\begin{aligned}&\frac{\sin (\varphi )}{\Big ((\varphi +\phi )^2(\phi -\varphi )^2+(\sin ^2(\varphi )+\phi ^2)\sin ^2((\theta -\eta )/2)\Big )^{\frac{1}{2}}}\nonumber \\&\quad \le \frac{1}{|\phi -\varphi |^{1-\beta }|\sin ((\theta -\eta )/2)|^\beta }, \end{aligned}$$
(5.21)

implying that

$$\begin{aligned} |{\mathscr {K}}_1(\phi ,\theta ,\varphi ,\eta )|\le \frac{C}{|\phi -\varphi |^{1-\beta }|\sin ((\theta -\eta )/2)|^\beta }\cdot \end{aligned}$$
(5.22)

Therefore, we easily achieve that \({\mathcal {I}}_1\in L^\infty \). To establish that \({\mathcal {I}}_1\in {\mathscr {C}}^\alpha \), we proceed in a direct way using the definition. Before that we remark that to get the \({\mathscr {C}}^\alpha \) regularity in both variables \((\phi ,\theta )\) it is enough to check the \({\mathscr {C}}^\alpha \)-regularity separately in the partial variables. Thus we shall check that \(\phi \mapsto {\mathcal {I}}_1(\phi ,\theta )\) is \({\mathscr {C}}^\alpha (0,\pi )\) uniformly in \(\theta \in [0,2\pi ]\). Take \(\phi _1,\phi _2\in [0,\frac{\pi }{2}]\) with \(0<\phi _1<\phi _2\), then it is easy to check from some algebraic considerations that

Coming back to the definition of \(J_1\) seen in Lemma 5.2, we can check that

$$\begin{aligned}&J_1(\phi _1,\theta ,\varphi ,\eta )-J_1(\phi _2,\theta ,\varphi ,\eta )\\&\quad = \big (r(\phi _1,\theta )-r(\phi _2,\theta )\big )\big (r(\phi _1,\theta )-r(\varphi ,\eta )+r(\phi _2,\theta )-r(\varphi ,\eta )\big )\\&\qquad +2r(\varphi ,\eta )\big (r(\phi _1,\theta )-r(\phi _2,\theta )\big )\big (1-\cos (\theta -\eta )\big )\\&\qquad {-}\big (\cos \varphi -\cos \phi _1+\cos \varphi -\cos \phi _2\big )\big (\cos \phi _1-\cos \phi _2\big ). \end{aligned}$$

Since \(r\in \text {Lip}\) we infer by interpolation

$$\begin{aligned}&\big |r(\phi _1,\theta )-r(\phi _2,\theta )\big |\nonumber \\&\quad \le C|\phi _1-\phi _2|^\alpha \Big (|r(\phi _1,\theta )-r(\varphi ,\eta )|^{1-\alpha }+|r(\phi _2,\theta )-r(\varphi ,\eta )|^{1-\alpha }\Big ), \end{aligned}$$

and

$$\begin{aligned} \big |r(\phi _1,\theta )-r(\phi _2,\theta )\big |\le C|\phi _1-\phi _2|^\alpha \Big (r^{1-\alpha }(\phi _1,\theta )+r^{1-\alpha }(\phi _2,\theta )\Big ). \end{aligned}$$

Consequently we find

$$\begin{aligned}&|J_1(\phi _1,\theta ,\varphi ,\eta ) - J_1(\phi _2,\theta ,\varphi ,\eta )|\\&\quad \le C|\phi _1-\phi _2|^\alpha \Big (|r(\phi _1,\theta )-r(\varphi ,\eta )|^{2-\alpha }+|r(\phi _2,\theta )-r(\varphi ,\eta )|^{2-\alpha }\\&\qquad +r(\varphi ,\eta )\big (r^{1-\alpha }(\phi _1,\theta )+r^{1-\alpha }(\phi _2,\theta )\big )\big (1-\cos (\eta -\theta )\big )\\&\qquad +|\cos (\varphi )-\cos (\phi _1)|^{2-\alpha }+|\cos (\varphi )-\cos (\phi _2)|^{2-\alpha }\Big ). \end{aligned}$$

From straightforward calculus we observe that

$$\begin{aligned} \frac{|r(\phi _i,\theta )-r(\varphi ,\eta )|^{2-\alpha }+r(\varphi ,\eta )r(\phi _i,\theta )^{1-\alpha }\big (1-\cos (\eta -\theta )\big )+|(\cos (\varphi )-\cos (\phi _i)|^{2-\alpha }}{J_1(\phi _i,\theta ,\varphi ,\eta )^\frac{2-\alpha }{2}}\le C, \end{aligned}$$

and then we find

$$\begin{aligned}&|J_1(\phi _1,\theta ,\varphi ,\eta ) - J_1(\phi _2,\theta ,\varphi ,\eta )|\\&\quad \le C|\phi _1-\phi _2|^\alpha \Big (J_1^{\frac{2-\alpha }{2}}(\phi _1,\theta ,\varphi ,\eta )+J_1^{\frac{2-\alpha }{2}}(\phi _2,\theta ,\varphi ,\eta )\Big ).\end{aligned}$$

It follows that

$$\begin{aligned}&\frac{|J_1(\phi _1,\theta ,\varphi ,\eta )-J_1(\phi _2,\theta ,\varphi ,\eta )|}{J_1^\frac{1}{2}(\phi _2,\theta ,\varphi ,\eta )J_1(\phi _1,\theta ,\varphi ,\eta )^\frac{1}{2}(J_1(\phi _1,\theta ,\varphi ,\eta )^\frac{1}{2}+J_1^\frac{1}{2}(\phi _2,\theta ,\varphi ,\eta ))}\nonumber \\&\lesssim \frac{|\phi _1-\phi _2|^\alpha }{J_1^\frac{\alpha }{2}(\phi _2,\theta ,\varphi ,\eta )J_1^\frac{1}{2}(\phi _1,\theta ,\varphi ,\eta )}+\frac{|\phi _1-\phi _2|^\alpha }{J_1^\frac{1}{2}(\phi _2,\theta ,\varphi ,\eta )J_1^\frac{\alpha }{2}(\phi _1,\theta ,\varphi ,\eta )}\cdot \end{aligned}$$
(5.23)

Using (5.23), one finds

By virtue of (5.21), for any \(\beta \in (0,1)\) we obtain

Hence, we get in view of Lemma 5.2 and (2.17)

$$\begin{aligned} \frac{|\partial _\eta r(\varphi ,\eta )|}{J_1(\phi _i,\theta ,\varphi ,\eta )^\frac{\alpha }{2}}&\lesssim \frac{\varphi ^\alpha }{\left( (\varphi +\phi _i)^2(\phi _i-\varphi )^2+(\sin ^2(\varphi )+\phi _i^2)\sin ^2((\theta -\eta )/2)\right) ^\frac{\alpha }{2}}\\&\lesssim {|\phi _i-\varphi |^{-\alpha }}. \end{aligned}$$

for any \(i=1,2\). This implies that

It is classical that

provided that \(0\le \alpha<\beta <1\). Similarly we prove under the same condition that

Putting together the preceding estimates yields for any \(0<\alpha<\beta <1\)

$$\begin{aligned} \forall \theta \in (0,2\pi ),{\phi _1,\phi _2\in (0,\pi /2)},\quad |{\mathcal {I}}_1(\phi _2,\theta )-{\mathcal {I}}_1(\phi _1,\theta )|\le C|\phi _1-\phi _2|^\alpha , \end{aligned}$$

where the constant C is independent of \(\theta , \phi _1\) and \(\phi _2\). Let us move on the regularity of \({\mathcal {I}}_1\) in \(\theta \). Here we shall use the estimate later proved (5.34) for \(s=1\), that is, for \(\theta _1,\theta _2\in [0,2\pi ]\) we have

$$\begin{aligned}&\frac{|J_1(\phi ,\theta _2,\varphi ,\eta )-J_1(\phi ,\theta _1,\varphi ,\eta )|}{J_1^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+J_1^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )}\nonumber \\&\quad \lesssim |\theta _1-\theta _2|^\alpha \phi ^{\alpha ^2}(J_1^{\frac{1-\alpha }{2}}(\phi ,\theta _1,\varphi ,\eta )+J_1^{\frac{1-\alpha }{2}}(\phi ,\theta _2,\varphi ,\eta )). \end{aligned}$$
(5.24)

Note that

$$\begin{aligned}&{\mathcal {I}}_1(\phi ,\theta _1)-{\mathcal {I}}_1(\phi ,\theta _2) = \int _0^\pi \int _0^{2\pi }\frac{\sin (\varphi )\partial _\eta r(\varphi ,\eta )(\cos (\eta -\theta _1)-\cos (\eta -\theta _2))}{J_1^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )}d\eta d\varphi \\&\quad +\int _0^\pi \int _0^{2\pi }\frac{\sin (\varphi )\partial _\eta r(\varphi ,\eta )\cos (\eta -\theta _2)(J_1(\phi ,\theta _2,\varphi ,\eta )-J_1(\phi ,\theta _1,\varphi ,\eta ))}{J_1^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )J_1^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )(J_1^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )+J_1^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta ))}d\eta d\varphi \\&\quad =: \int _0^\pi \int _0^{2\pi }({\mathscr {K}}_{1,1}+{\mathscr {K}}_{1,2})(\phi ,\theta ,\varphi ,\eta )d\eta d\varphi \end{aligned}$$

On the one hand, using (5.21) we easily get

$$\begin{aligned} |{\mathscr {K}}_{1,1}(\phi ,\theta ,\varphi ,\eta )|\le \frac{C|\theta _1-\theta _2|}{|\phi -\varphi |^{1-\beta }|\sin ((\theta -\eta )/2)|^\beta }, \end{aligned}$$

for any \(\beta \in (0,1)\). On the other hand, we use (5.24) to estimate the second term:

$$\begin{aligned} |{\mathscr {K}}_{1,2}(\phi ,\theta ,\varphi ,\eta )|&\le C|\theta _1-\theta _2|^\alpha \frac{\sin (\varphi )^{1+\alpha }\phi ^{\alpha ^2}(J_1^{\frac{1-\alpha }{2}}(\phi ,\theta _1,\varphi ,\eta )+J_1^{\frac{1-\alpha }{2}}(\phi ,\theta _2,\varphi ,\eta ))}{J_1^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )J_1^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )}\\&\le C|\theta _1-\theta _2|^\alpha \frac{\sin (\varphi )^{1+\alpha }\phi ^{\alpha ^2}}{J_1^\frac{\alpha }{2}(\phi ,\theta _1,\varphi ,\eta )J_1^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )}\\&\quad +C|\theta _1-\theta _2|^\alpha \frac{\sin (\varphi )^{1+\alpha }\phi ^{\alpha ^2}}{J_1^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )J_1^\frac{\alpha }{2}(\phi ,\theta _1,\varphi ,\eta )}. \end{aligned}$$

By virtue of Lemma 5.2, we arrive at

$$\begin{aligned}&|{\mathscr {K}}_{1,2}(\phi ,\theta ,\varphi ,\eta )|\\&\quad \le C|\theta _1-\theta _2|^\alpha \frac{\sin (\varphi )^{1+\alpha }}{\sin (\varphi )^{1+\alpha }(|\phi -\varphi |+|\sin ((\theta _1-\eta )/2)|)^{\alpha }(|\phi -\varphi |+|\sin ((\theta _2-\eta )/2)|)}\\&\quad \quad +C|\theta _1-\theta _2|^\alpha \frac{\sin (\varphi )^{1+\alpha }}{\sin (\varphi )^{1+\alpha }(|\phi -\varphi |+|\sin ((\theta _2-\eta )/2)|)^{\alpha }(|\phi -\varphi |+|\sin ((\theta _1-\eta )/2)|)}\\&\quad \le C|\theta _1-\theta _2|^\alpha \frac{1}{\sin (\varphi )^{\gamma _1}|\sin ((\theta _1-\eta )/2)|^{\gamma _2}|\sin ((\theta _1-\eta )/2)|^{\gamma _3}}, \end{aligned}$$

for \(\gamma _1,\gamma _2,\gamma _3\in (0,1)\) and \(\gamma _2+\gamma _3<1\).

At the end, we can find that

$$\begin{aligned} |{\mathcal {I}}_1(\phi ,\theta _2)-{\mathcal {I}}_1(\phi ,\theta _1)|\le C|\theta _1-\theta _2|^\alpha . \end{aligned}$$

Finally, this allows to get that \({\mathcal {I}}_1\) is \({\mathscr {C}}^\alpha ((0,\pi )\times {{{\,\mathrm{{\mathbb {T}}}\,}}})\).

\(\bullet \) Proof of (5.16). In fact we shall establish a more refined result:

$$\begin{aligned} (\phi ,\theta )\mapsto \frac{U\big (sr(\phi ,\theta )e^{i\theta },\cos (\phi )\big )}{r_0(\phi )}\cdot ie^{i\theta }\in {\mathscr {C}}^\alpha ([0,\pi ]\times [0,2\pi ]). \end{aligned}$$

uniformly with respect to \(s\in [0,1]\). This allows to get the results (5.16) and (5.15).

Coming back to (5.18) and using \(J_s\) introduced in Lemma 5.2 we find the expression

(5.25)

Moreover, by Lemma 5.1 we have

and then we can subtract this vanishing term obtaining

Notice that we eliminate .the variable \(\theta \) from the definition of \(J_0\) because it is independent of this parameter. Since \((\phi ,\theta )\mapsto \frac{r(\phi ,\theta )}{r_0(\phi )}\) is \({\mathscr {C}}^\alpha \), then to get the desired regularity it is enough to check it for the the integral term. Denote

$$\begin{aligned} {\mathscr {K}}_2(s,\phi ,\theta ,\varphi ,\eta )=\frac{\sin (\varphi )\partial _\eta \big (r(\varphi ,\eta )\sin (\eta -\theta )\big )s\big (sr(\phi ,\theta )-2r(\varphi ,\eta )\cos (\theta -\eta )\big )}{J_s^\frac{1}{2}(\phi ,\theta ,\varphi ,\eta )J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta ,\varphi ,\eta )+J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big )},\nonumber \\ \end{aligned}$$
(5.26)

and let us show first that

belongs to \(L^\infty \). It is plain that

$$\begin{aligned} |{\mathscr {K}}_2(s,\phi ,\theta ,\varphi ,\eta )|&\lesssim \frac{\sin (\varphi )\big (r(\varphi ,\eta )+|\partial _\eta r(\varphi ,\eta )||\sin (\eta -\theta )|\big )s\big (sr(\phi ,\theta )+r(\varphi ,\eta )\big )}{J_s^\frac{1}{2}(\phi ,\theta ,\varphi ,\eta )J_0(\phi ,\varphi ,\eta )}\cdot \end{aligned}$$

Combined with the estimate (5.1), it yields

$$\begin{aligned} |{\mathscr {K}}_2(s,\phi ,\theta ,\varphi ,\eta )|&\lesssim \frac{(r(\varphi ,\eta )+|\partial _\eta r(\varphi ,\eta )||\sin (\eta -\theta )|)s(sr(\phi ,\theta )+r(\varphi ,\eta ))}{\sin (\varphi )J_s^\frac{1}{2}(\phi ,\theta ,\varphi ,\eta )}\cdot \end{aligned}$$

As we have mentioned before at different stages, the symmetry allows us to restrict the discussion to the interval \(\phi \in (0,\pi /2)\). Then using Lemma 5.2 and (2.17) we achieve that

$$\begin{aligned} \frac{sr(\phi ,\theta )+r(\varphi ,\eta )}{J_s^\frac{1}{2}(\phi ,\theta ,\varphi ,\eta )}&\lesssim \frac{s\phi +\sin \varphi }{\left\{ (\varphi +\phi )^2(\phi -\varphi )^2+(\sin ^2(\varphi )+s^2\phi ^2)\sin ^2((\theta -\eta )/2)\right\} ^\frac{1}{2}}\nonumber \\&\lesssim \frac{1}{\big ((\phi -\varphi )^2+\sin ^2((\theta -\eta )/2)\big )^\frac{1}{2}}\cdot \end{aligned}$$
(5.27)

Hence, the estimate of (2.17) allows to get

$$\begin{aligned} |{\mathscr {K}}_2(s,\phi ,\theta ,\varphi ,\eta )|&\lesssim \frac{\sin (\varphi )+\sin ^\alpha (\varphi )|\sin (\theta -\eta )|}{\sin (\varphi )\big ((\phi -\varphi )^2+\sin ^2((\theta -\eta )/2)\big )^\frac{1}{2}}\\&\lesssim \frac{1}{\big ((\phi -\varphi )^2+\sin ^2((\theta -\eta )/2)\big )^\frac{1}{2}}+\frac{1}{\sin ^{1-\alpha }(\varphi )}\cdot \end{aligned}$$

By interpolation we deduce for any \(\beta \in (0,1)\),

$$\begin{aligned} |{\mathscr {K}}_2(s,\phi ,\theta ,\varphi ,\eta )|&\lesssim \frac{1}{|\phi -\varphi |^{1-\beta }|\sin ((\theta -\eta )/2)|^\beta }+\sin ^{\alpha -1}(\varphi ). \end{aligned}$$

It follows that

Let us move to the \({\mathscr {C}}^\alpha \)-regularity of this latter function. This amounts to checking the partial regularity separately in \(\phi \) and \(\theta \). The strategy is the same for both of them and to alleviate the discussion, we shall establish the regularity in the variable \(\theta \), contrary to the preceding section where it was established for \({\mathcal {I}}_1\) in the direction of \(\phi \). The goal is to get a convenient estimate for the difference

where \(0\le \theta _1<\theta _2\le 2\pi \). Coming back to the definition of the kernel \({\mathscr {K}}_2\) in (5.26) one deduces through straightforward algebraic computations that

$$\begin{aligned} {\mathscr {K}}_2(s,\phi ,\theta _1,\varphi ,\eta )-{\mathscr {K}}_2(s,\phi ,\theta _2,\varphi ,\eta )={\mathcal {I}}_3+{\mathcal {I}}_4+{\mathcal {I}}_5+{\mathcal {I}}_6. \end{aligned}$$

with

$$\begin{aligned} {\mathcal {I}}_3=\frac{\sin (\varphi )\partial _\eta \big [r(\varphi ,\eta )\big (\sin (\eta -\theta _1)-\sin (\eta -\theta _2)\big )\big ]s\big [sr(\phi ,\theta _1)-2r(\varphi ,\eta )\cos (\theta _1-\eta )\big ]}{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )+J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big )}, \end{aligned}$$
$$\begin{aligned} {\mathcal {I}}_4&= \frac{\sin (\varphi )\partial _\eta \big [r(\varphi ,\eta )\sin (\eta -\theta _2)\big ]s\big [sr(\phi ,\theta _1)-2r(\varphi ,\eta )\cos (\theta _1-\eta )\big ]}{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )+J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big )}\\&\times \frac{J_s(\phi ,\theta _2,\varphi ,\eta )-J_s(\phi ,\theta _1,\varphi ,\eta )}{\big [J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big ]\big [J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )+J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )\big ]}, \end{aligned}$$
$$\begin{aligned} {\mathcal {I}}_5=\frac{\sin (\varphi )\partial _\eta \big [r(\varphi ,\eta )\sin (\eta -\theta _2)\big ]s\big [sr(\phi ,\theta _1)-sr(\phi ,\theta _2)-2r(\varphi ,\eta )\big (\cos (\theta _1-\eta )-\cos (\theta _2-\eta )\big )\big ]}{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big )} \end{aligned}$$

and

$$\begin{aligned} {\mathcal {I}}_6&= \frac{\sin (\varphi )\partial _\eta \big (r(\varphi ,\eta )\sin (\eta -\theta _2)\big )s\big [sr(\phi ,\theta _2)-2r(\varphi ,\eta )\cos (\theta _2-\eta )\big ]}{J_s(\phi ,\theta _1,\varphi ,\eta )^\frac{1}{2}J_0(\phi ,\varphi ,\eta )^\frac{1}{2}(J_s(\phi ,\theta _2,\varphi ,\eta )^\frac{1}{2}+J_0(\phi ,\varphi ,\eta )^\frac{1}{2})}\\&\times \frac{J_s(\phi ,\theta _2,\varphi ,\eta )-J_s(\phi ,\theta _1,\varphi ,\eta )}{J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )\big )}\cdot \end{aligned}$$

We shall estimate independently each one of those terms. Concerning the term \({\mathcal {I}}_3\) it can be estimated using (5.1)

$$\begin{aligned} |{\mathcal {I}}_3|&\lesssim |\theta _1-\theta _2|\frac{\sin (\varphi )\big (r(\varphi ,\eta )+|\partial _\eta r(\varphi ,\eta )|\big )s\big (sr(\phi ,\theta _1)+r(\varphi ,\eta )\big )}{\sin ^2(\varphi )J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )}\\&\lesssim |\theta _1-\theta _2|\frac{\big (r(\varphi ,\eta )+|\partial _\eta r(\varphi ,\eta )|\big )s\big (sr(\phi ,\theta _1)+r(\varphi ,\eta )\big )}{\sin (\varphi )J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )}\cdot \end{aligned}$$

Then by virtue of (5.27) and (2.17), we find

$$\begin{aligned} |{\mathcal {I}}_3|&\lesssim \frac{ |\theta _1-\theta _2|\big (\sin (\varphi )+\sin ^\alpha (\varphi )\big )}{\sin (\varphi )\left( (\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\right) ^\frac{1}{2}}\\&\lesssim \frac{ |\theta _1-\theta _2|\sin ^{\alpha -1}(\varphi )}{\left( (\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\right) ^\frac{1}{2}}\cdot \end{aligned}$$

Combining this estimate with the interpolation inequality: for any \(\beta \in [0,1]\)

$$\begin{aligned} \frac{1}{\left\{ (\phi -\varphi )^2+\sin ^2((\theta _1-\eta )^2)\right\} ^\frac{1}{2}}\le \frac{1}{|\varphi -\phi |^\beta |\sin ((\eta -\theta _1)/2)|^{1-\beta }}, \end{aligned}$$

we infer

$$\begin{aligned} |{\mathcal {I}}_3|&\lesssim \frac{|\theta _1-\theta _2|}{\sin ^{1-\alpha }(\varphi )|\varphi -\phi |^\beta |\sin ((\eta -\theta _1)/2)|^{1-\beta }}\cdot \end{aligned}$$

Thus

uniformly in \(\phi \in (0,\pi /2)\) and \(\theta _1,\theta _2\in (0,2\pi )\), provided that \(0<\beta <\alpha \le 1\).

Concerning the term \({\mathcal {I}}_4\), we first use the definition of \(J_s\) in Lemma 5.2 and one may check

$$\begin{aligned}&|J_s(\phi ,\theta _2,\varphi ,\eta )-J_s(\phi ,\theta _1,\varphi ,\eta )|\nonumber \\&\quad \le |sr(\phi ,\theta _1)-sr(\phi ,\theta _2)|(|sr(\phi ,\theta _1)-r(\varphi ,\eta )|\nonumber \\&\quad \quad +|sr(\phi ,\theta _2)-r(\varphi ,\eta )|)\nonumber \\&\quad \quad +2(sr(\phi ,\theta _1)-sr(\phi ,\theta _2))r(\varphi ,\eta )(1-\cos (\theta _1-\eta ))\nonumber \\&\quad \quad +2sr(\phi ,\theta _2)r(\varphi ,\eta )\big |\cos (\theta _2-\eta )-\cos (\theta _1-\eta )\big |. \end{aligned}$$
(5.28)

Using the trigonometric identity

$$\begin{aligned} 1-\cos (\theta -\eta )=2\sin ^2((\theta -\eta )/2), \end{aligned}$$
(5.29)

we get

$$\begin{aligned}&|J_s(\phi ,\theta _2,\varphi ,\eta )-J_s(\phi ,\theta _1,\varphi ,\eta )|\nonumber \\&\quad \lesssim |sr(\phi ,\theta _1)-sr(\phi ,\theta _2)|\Big (J_s^{\frac{1}{2}}(\phi ,\theta _2,\varphi ,\eta )+J_s^{\frac{1}{2}}(\phi ,\theta _1,\varphi ,\eta )\Big )\nonumber \\&\qquad +2sr(\phi ,\theta _2)r(\varphi ,\eta )\big |\cos (\theta _2-\eta )-\cos (\theta _1-\eta )\big |. \end{aligned}$$
(5.30)

From (2.17) combined with Taylor’s formula we find

$$\begin{aligned} |r(\phi ,\theta _2)-r(\phi ,\theta _1)|&\le \left| \int _{\theta _1}^{\theta _2}\partial _\eta r(\phi ,\eta )d\eta \right| \\&\lesssim |\theta _2-\theta _1| \sin ^\alpha (\phi ). \end{aligned}$$

Therefore we get by interpolation inequality

$$\begin{aligned} |sr(\phi ,\theta _1)-sr(\phi ,\theta _2)|\lesssim |\theta _1-\theta _2|^\alpha s^\alpha \phi ^{\alpha ^2}&\Big [|sr(\phi ,\theta _1)-r(\varphi ,\eta )|^{1-\alpha } \\&\quad +|sr(\phi ,\theta _2)-r(\varphi ,\eta )|^{1-\alpha }\Big ]. \end{aligned}$$

Hence

$$\begin{aligned} |sr(\phi ,\theta _1)-sr(\phi ,\theta _2)|\lesssim |\theta _1-\theta _2|^\alpha s^\alpha \phi ^{\alpha ^2}&\Big [J_s^{\frac{1-\alpha }{2}}(\phi ,\theta _1,\varphi ,\eta )+J_s^{\frac{1-\alpha }{2}}(\phi ,\theta _2,\varphi ,\eta )\Big ]. \end{aligned}$$
(5.31)

Combining (5.31) together with (5.30) implies

$$\begin{aligned}&|J_s(\phi ,\theta _2,\varphi ,\eta )-J_s(\phi ,\theta _1,\varphi ,\eta )|\nonumber \\&\quad \lesssim |\theta _1-\theta _2|^\alpha s^\alpha \phi ^{\alpha ^2}\Big (J_s^{1-\frac{\alpha }{2}}(\phi ,\theta _2,\varphi ,\eta )+J_s^{1-\frac{\alpha }{2}}(\phi ,\theta _1,\varphi ,\eta )\Big )\nonumber \\&\qquad +2sr(\phi ,\theta _2)r(\varphi ,\eta )\big |\cos (\theta _2-\eta )-\cos (\theta _1-\eta )\big |. \end{aligned}$$
(5.32)

Define

$$\begin{aligned} \xi :=|\cos (\theta _2-\eta )-\cos (\theta _1-\eta )|. \end{aligned}$$

Using once again (5.29) we get

$$\begin{aligned} \xi&\lesssim \Big |\sqrt{1-\cos (\theta _2-\eta )}-\sqrt{1-\cos (\theta _1-\eta )}\Big |\Big |\sqrt{1-\cos (\theta _2-\eta )}+\sqrt{1-\cos (\theta _1-\eta )}\Big |\\&\lesssim \Big (|\sin ((\theta _1-\eta )/2)|-|\sin ((\theta _2-\eta )/2)|\Big )\Big (|\sin ((\theta _1-\eta )/2)|+|\sin ((\theta _2-\eta )/2)|\Big ). \end{aligned}$$

Note that using the inequality

$$\begin{aligned} ||\sin x|-|\sin y||\le |x-y|, \end{aligned}$$

and interpolating, one achieves

$$\begin{aligned}&\Big (|\sin ((\theta _1-\eta )/2)|-|\sin ((\theta _2-\eta )/2)|\Big )\\&\quad \le C|\theta _1-\theta _2|^\alpha \Big (|\sin ((\theta _1-\eta )/2)|+|\sin ((\theta _2-\eta )/2)|\Big )^{1-\alpha }, \end{aligned}$$

for any \(\alpha \in (0,1)\). Putting everything together, one finds

$$\begin{aligned} \xi&\lesssim {|\theta _1-\theta _2|^\alpha \Big (|\sin ((\theta _1-\eta )/2)|^{2-\alpha }+|\sin ((\theta _2-\eta )/2)|^{2-\alpha }\Big ).} \end{aligned}$$
(5.33)

Now using the assumption (2.17) and (5.2) we find

$$\begin{aligned} sr(\phi ,\theta _2)r(\varphi ,\eta )\xi&\lesssim s\phi r_0(\varphi ){|\theta _1-\theta _2|^\alpha \Big (|\sin ((\theta _1-\eta )/2)|^{2-\alpha }+|\sin ((\theta _2-\eta )/2)|^{2-\alpha }\Big )}\\&\lesssim s^\alpha \phi ^\alpha {|\theta _1-\theta _2|^\alpha }\Big [J_s^{\frac{1-\alpha }{2}}(\phi ,\theta _1,\varphi ,\eta )+J_s^{\frac{1-\alpha }{2}}(\phi ,\theta _2,\varphi ,\eta )\Big ]. \end{aligned}$$

Inserting this inequality into (5.32) implies

$$\begin{aligned}&|J_s(\phi ,\theta _2,\varphi ,\eta )-J_s(\phi ,\theta _1,\varphi ,\eta )|\\&\quad \lesssim |\theta _1-\theta _2|^\alpha s^\alpha \phi ^{\alpha ^2}\Big (J_s^{1-\frac{\alpha }{2}}(\phi ,\theta _2,\varphi ,\eta )+J_s^{1-\frac{\alpha }{2}}(\phi ,\theta _1,\varphi ,\eta )\Big ). \end{aligned}$$

It follows that

$$\begin{aligned}&\frac{|J_s(\phi ,\theta _2,\varphi ,\eta )-J_s(\phi ,\theta _1,\varphi ,\eta )|}{J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )}\nonumber \\&\quad \lesssim |\theta _1-\theta _2|^\alpha s^\alpha \phi ^{\alpha ^2}\Big (J_s^{\frac{1-\alpha }{2}}(\phi ,\theta _2,\varphi ,\eta )+J_s^{\frac{1-\alpha }{2}}(\phi ,\theta _1,\varphi ,\eta )\Big ). \end{aligned}$$
(5.34)

Thus we get

$$\begin{aligned} |{\mathcal {I}}_4|&\lesssim |\theta _1-\theta _2|^\alpha \frac{\big (r(\varphi ,\eta )+\partial _\eta r(\varphi ,\eta )|\sin (\theta _2-\eta )|\big )s(sr(\phi ,\theta _1)+r(\varphi ,\eta ))s^\alpha \phi ^{\alpha ^2}}{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )\big (\sin (\varphi )+J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )\big )^\alpha \big (\sin (\varphi )+J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )\big )}\\&\quad +|\theta _1-\theta _2|^\alpha \frac{(r(\varphi ,\eta )+\partial _\eta r(\varphi ,\eta )|\sin (\theta _2-\eta )|)s(sr(\phi ,\theta _1)+r(\varphi ,\eta ))s^\alpha \phi ^{\alpha ^2}}{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )\big (\sin (\varphi )+J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )\big )\big (\sin (\varphi )+J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )\big )^\alpha }\cdot \end{aligned}$$

Applying (5.27) combined with (2.17) we arrive at

$$\begin{aligned} |{\mathcal {I}}_4|&\lesssim \frac{|\theta _1-\theta _2|^\alpha \big (\sin (\varphi )+\sin ^\alpha (\varphi )|\sin (\theta _2-\eta )|\big )ss^\alpha \phi ^{\alpha ^2}}{\left\{ (\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\right\} ^\frac{1}{2}(\sin (\varphi )+J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta ))^\alpha (\sin (\varphi )+J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta ))}\nonumber \\&\quad +\frac{|\theta _1-\theta _2|^\alpha \big (\sin (\varphi )+\sin ^\alpha (\varphi )|\sin (\theta _2-\eta )|\big )ss^\alpha \phi ^{\alpha ^2}}{\left\{ (\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\right\} ^\frac{1}{2}(\sin (\varphi )+J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta ))(\sin (\varphi )+J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta ))^\alpha }\nonumber \\&\lesssim \quad {\mathcal {J}}_1+{\mathcal {J}}_2. \end{aligned}$$
(5.35)

The right hand side terms \({\mathcal {J}}_1\) and \({\mathcal {J}}_2\) are treated similarly and we shall only focus on the first one. We find that

$$\begin{aligned}&|{\mathcal {J}}_1|\lesssim \frac{|\theta _1-\theta _2|^\alpha }{\left\{ (\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\right\} ^\frac{1}{2}\sin ^\alpha (\varphi )} \\&\qquad \qquad +\frac{|\theta _1-\theta _2|^\alpha |\sin (\theta _2-\eta )|\,s^{\alpha }\phi ^{\alpha ^2}}{\left\{ (\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\right\} ^\frac{1}{2}(\sin (\varphi )+J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta ))}\cdot \end{aligned}$$

Using (5.2) we deduce, since \(\alpha \in (0,1)\), that

$$\begin{aligned} |\sin (\theta _2-\eta )|\,s^{\alpha }\phi ^{\alpha ^2}&\le |\sin (\theta _2-\eta )|^{\alpha ^2}\,s^{\alpha ^2}\phi ^{\alpha ^2}\nonumber \\&\lesssim J_s^{\frac{\alpha ^2}{2}}(\phi ,\theta _2,\varphi ,\eta ). \end{aligned}$$
(5.36)

Thus

$$\begin{aligned}&|{\mathcal {J}}_1|\lesssim \frac{|\theta _1-\theta _2|^\alpha }{\left\{ (\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\right\} ^\frac{1}{2}\sin ^\alpha (\varphi )} \\&\quad +\frac{|\theta _1-\theta _2|^\alpha }{\left\{ (\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\right\} ^\frac{1}{2}\sin ^{1-\alpha ^2}(\varphi )}\cdot \end{aligned}$$

The same estimate holds true for \({\mathcal {J}}_2\). Therefore we find

$$\begin{aligned}&|{\mathcal {I}}_4|\lesssim |\theta _1-\theta _2|^\alpha \frac{\sin ^{-\alpha }(\varphi )+\sin ^{\alpha ^2-1}(\varphi )}{\left\{ (\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\right\} ^\frac{1}{2}}\cdot \end{aligned}$$

Hence by interpolation inequality we get for all \(\gamma \in (0,1)\)

$$\begin{aligned}&|{\mathcal {I}}_4|\lesssim |\theta _1-\theta _2|^\alpha \frac{\sin ^{-\alpha }(\varphi )+\sin ^{\alpha ^2-1}(\varphi )}{|\phi -\varphi |^\gamma |\sin ((\theta _1-\eta )/2)|^{1-\gamma }}\cdot \end{aligned}$$

It follows that

uniformly in \(\phi \in (0,\pi )\) and \(\theta _1,\theta _2\in (0,2\pi )\), provided that \(0<\gamma <\min (1-\alpha , \alpha ^2)\).

Let us focus on the estimate of the term \({\mathcal {I}}_5\). First we make the decomposition

$$\begin{aligned} {\mathcal {I}}_5={\mathcal {J}}_3+{\mathcal {J}}_4, \end{aligned}$$

with

$$\begin{aligned} {\mathcal {J}}_3=\frac{\sin (\varphi )\partial _\eta \big [r(\varphi ,\eta )\sin (\eta -\theta _2)\big ]s\big [sr(\phi ,\theta _1)-sr(\phi ,\theta _2))\big ]}{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big )}, \end{aligned}$$

and

$$\begin{aligned} {\mathcal {J}}_4=-2\frac{\sin (\varphi )\partial _\eta \big [r(\varphi ,\eta )\sin (\eta -\theta _2)\big ]sr(\varphi ,\eta )\big [\cos (\theta _1-\eta )-\cos (\theta _2-\eta )\big ]}{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+J_0^\frac{1}{2}(\phi ,\varphi ,\eta )\big )}\cdot \end{aligned}$$

Let us start with the last term \({\mathcal {J}}_4\). Using Lemma 5.2 combined with (5.33) and (2.17) yields

$$\begin{aligned} |{\mathcal {J}}_4|&\lesssim |\theta _1-\theta _2|^\alpha \frac{\big [\sin \varphi +\sin ^\alpha (\varphi )|\sin \big ((\eta -\theta _2)/2\big )|\big ]\sin \varphi }{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+\sin \varphi \big )}\cdot \end{aligned}$$

From (5.2) we infer

$$\begin{aligned} |{\mathcal {J}}_4|&\lesssim |\theta _1-\theta _2|^\alpha \frac{\sin ^2(\varphi )+\sin ^\alpha (\varphi )J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )|}{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+\sin \varphi \big )}\\&\lesssim |\theta _1-\theta _2|^\alpha \frac{\sin ^\alpha (\varphi )}{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )}\cdot \end{aligned}$$

Applying (5.27) leads to

$$\begin{aligned} |{\mathcal {J}}_4|&\lesssim |\theta _1-\theta _2|^\alpha \frac{\varphi ^{\alpha -1}}{\big ((\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\big )^\frac{1}{2}}\\&\lesssim |\theta _1-\theta _2|^\alpha \frac{\varphi ^{\alpha -1}}{|\phi -\varphi |^\beta |\sin ((\theta _1-\eta )/2)|^{1-\beta }}\cdot \end{aligned}$$

Therefore

uniformly in \(\phi \in (0,\pi )\) and \(\theta _1,\theta _2\in (0,2\pi )\) provided that \(0<\beta<\alpha <1\). Next we shall deal with the term \({\mathcal {J}}_5\). Then by virtue of (5.31) combined with (2.17) we may write

$$\begin{aligned} |{\mathcal {J}}_3|&\lesssim |\theta _1-\theta _2|^\alpha \frac{\sin (\varphi )+\sin ^\alpha (\varphi )|\sin (\theta _2-\eta )|}{J_s(\phi ,\theta _1,\varphi ,\eta )^\frac{1}{2}(J_s(\phi ,\theta _2,\varphi ,\eta )^\frac{1}{2} +\sin \varphi )} \\&\times (s^{1+\alpha }\phi ^{\alpha ^2}\Big [J_s^{\frac{1-\alpha }{2}}(\phi ,\theta _1,\varphi ,\eta )+J_s^{\frac{1-\alpha }{2}}(\phi ,\theta _2,\varphi ,\eta )\Big ]. \end{aligned}$$

It follows that

$$\begin{aligned} |{\mathcal {J}}_3|&\lesssim |\theta _1-\theta _2|^\alpha \frac{\big (\sin (\varphi )+\sin ^\alpha (\varphi )|\sin (\theta _2-\eta )|\big )s^{1+\alpha }\phi ^{\alpha ^2}}{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+\sin \varphi \big )^\alpha } \\&\quad +|\theta _1-\theta _2|^\alpha \frac{\big (\sin (\varphi )+\sin ^\alpha (\varphi )|\sin (\theta _2-\eta )|\big )s^{1+\alpha }\phi ^{\alpha ^2}}{J_s^\frac{\alpha }{2}(\phi ,\theta _1,\varphi ,\eta ) \big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+\sin \varphi \big )}\\&\lesssim |\theta _1-\theta _2|^\alpha {\mathcal {J}}_{3,1}+ |\theta _1-\theta _2|^\alpha {\mathcal {J}}_{3,2}. \end{aligned}$$

According to (5.36) and since \(\alpha \in (0,1)\) we find that

$$\begin{aligned} |{\mathcal {J}}_{3,1}|&\lesssim \frac{\sin (\varphi )+\sin ^\alpha (\varphi )|\sin (\theta _2-\eta )|^{\alpha ^2} s^{\alpha ^2}\phi ^{\alpha ^2} }{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+\sin \varphi \big )^\alpha }\\&\lesssim \frac{\sin (\varphi )+\sin ^\alpha (\varphi )J_s^\frac{\alpha ^2}{2}(\phi ,\theta _2,\varphi ,\eta ) }{J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )\big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+\sin \varphi \big )^\alpha } \end{aligned}$$

and similarly we obtain

$$\begin{aligned} |{\mathcal {J}}_{3,2}|&\lesssim \frac{\sin (\varphi )+\sin ^\alpha (\varphi )J_s^\frac{\alpha ^2}{2}(\phi ,\theta _2,\varphi ,\eta )}{J_s^\frac{\alpha }{2}(\phi ,\theta _1,\varphi ,\eta ) \big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+\sin \varphi \big )}\cdot \end{aligned}$$

Consequently, we get from (5.27)

$$\begin{aligned} |{\mathcal {J}}_{3,1}|&\lesssim \frac{\sin ^{-\alpha }(\varphi )+\sin ^{\alpha ^2-1}(\varphi )}{\big ((\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\big )^\frac{1}{2}}\\&\lesssim \frac{\sin ^{-\alpha }(\varphi )+\sin ^{\alpha ^2-1}(\varphi )}{|\phi -\varphi |^\gamma |\sin ((\theta _1-\eta )/2)|^{1-\gamma }}\cdot \end{aligned}$$

By integration, we obtain

uniformly in \(\phi \in (0,\pi )\) and \(\theta _1\in (0,2\pi )\) provided that \(0<\gamma <\min (\alpha ^2,1-\alpha )\).

Following the same ideas as before and using (5.27) we get

$$\begin{aligned} |{\mathcal {J}}_{3,2}|&\lesssim \left( \frac{\sin \varphi }{J_s^{\frac{1}{2}}(\phi ,\theta _1,\varphi ,\eta )}\right) ^\alpha \left( \sin ^{-\alpha }(\varphi )+ \big (J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )+\sin \varphi \big )^{\alpha ^2-1}\right) \\&\lesssim \frac{\sin ^{-\alpha }(\varphi )+\sin ^{\alpha ^2-1}(\varphi )}{\big ((\phi -\varphi )^2+\sin ^2((\theta _1-\eta )/2)\big )^\frac{\alpha }{2}}\cdot \end{aligned}$$

It follows that

$$\begin{aligned} |{\mathcal {J}}_{3,2}|&\lesssim \frac{\sin ^{-\alpha }(\varphi )+\sin ^{\alpha ^2-1}(\varphi )}{|\sin ((\theta _1-\eta )/2)|^{\alpha }}\cdot \end{aligned}$$

By integration, we deduce that

uniformly in \(\phi \in (0,\pi )\) and \(\theta _1\in (0,2\pi )\) provided that \(0<\alpha <1\). Putting together the preceding estimates allows to get

$$\begin{aligned} \int _0^{\pi }\int _0^{2\pi }|{\mathcal {I}}_{5}|d\eta d\varphi&\lesssim |\theta _1-\theta _2|^\alpha . \end{aligned}$$

Let us finish estimating the term \({\mathcal {I}}_6\), which is quite similar to \({\mathcal {I}}_4\). Applying (5.34), Lemma 5.2 and the ideas done for \({\mathcal {I}}_4\) we obtain

$$\begin{aligned} |{\mathcal {I}}_6|&\lesssim |\theta _1-\theta _2|^\alpha \frac{\sin (\varphi )s^\alpha \phi ^{\alpha ^2}(\sin (\varphi )+\sin (\varphi )^\alpha |\sin (\eta -\theta _2)|)(s\phi +\sin (\varphi ))}{\sin (\varphi )^2 J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )}\\&\times (J_s^\frac{1-\alpha }{2}(\phi ,\theta _1,\varphi ,\eta )+J_s^\frac{1-\alpha }{2}(\phi ,\theta _2,\varphi ,\eta ))\\&\lesssim |\theta _1-\theta _2|^\alpha \frac{s^\alpha \phi ^{\alpha ^2}(\sin (\varphi )+\sin (\varphi )^\alpha |\sin (\eta -\theta _2)|)(s\phi +\sin (\varphi ))}{\sin (\varphi ) J_s^\frac{\alpha }{2}(\phi ,\theta _1,\varphi ,\eta )J_s^\frac{1}{2}(\phi ,\theta _2,\varphi ,\eta )} \\&\quad + |\theta _1-\theta _2|^\alpha \frac{s^\alpha \phi ^{\alpha ^2}(\sin (\varphi )+\sin (\varphi )^\alpha |\sin (\eta -\theta _2)|)(s\phi +\sin (\varphi ))}{\sin (\varphi ) J_s^\frac{1}{2}(\phi ,\theta _1,\varphi ,\eta )J_s^\frac{\alpha }{2}(\phi ,\theta _2,\varphi ,\eta )}. \end{aligned}$$

Using again Lemma 5.2 we achieve

$$\begin{aligned} |{\mathcal {I}}_6|&\lesssim |\theta _1-\theta _2|^\alpha \frac{s^\alpha \phi ^{\alpha ^2}(\sin (\varphi )+\sin (\varphi )^\alpha |\sin (\eta -\theta _2)|)(s\phi +\sin (\varphi ))}{\sin (\varphi ) (\sin (\varphi )+s\phi )^{1+\alpha }(|\phi -\varphi |+|\sin ((\theta _1-\eta )/2)|)^\alpha (|\phi -\varphi |+|\sin ((\theta _2-\eta )/2)|)} \\&\quad + |\theta _1-\theta _2|^\alpha \frac{s^\alpha \phi ^{\alpha ^2}(\sin (\varphi )+\sin (\varphi )^\alpha |\sin (\eta -\theta _2)|)(s\phi +\sin (\varphi ))}{\sin (\varphi ) (\sin (\varphi )+s\phi )^{1+\alpha }(|\phi -\varphi |+|\sin ((\theta _2-\eta )/2)|)^\alpha (|\phi -\varphi |+|\sin ((\theta _1-\eta )/2)|)}\\&\lesssim |\theta _1-\theta _2|^\alpha \frac{1}{\sin (\varphi )^{\alpha -\alpha ^2} (|\phi -\varphi |+|\sin ((\theta _1-\eta )/2)|)^\alpha (|\phi -\varphi |+|\sin ((\theta _2-\eta )/2)|)}\\&\quad +|\theta _1-\theta _2|^\alpha \frac{1}{\sin (\varphi )^{1-\alpha ^2}(|\phi -\varphi |+|\sin ((\theta _1-\eta )/2)|)^\alpha }\\&\quad +|\theta _1-\theta _2|^\alpha \frac{1}{\sin (\varphi )^{\alpha -\alpha ^2} (|\phi -\varphi |+|\sin ((\theta _2-\eta )/2)|)^\alpha (|\phi -\varphi |+|\sin ((\theta _1-\eta )/2)|)}\\&\quad +|\theta _1-\theta _2|^\alpha \frac{1}{\sin (\varphi )^{1-\alpha ^2}(|\phi -\varphi |+|\sin ((\theta _2-\eta )/2)|)}\\&\lesssim |\theta _1-\theta _2|^\alpha \frac{1}{\sin (\varphi )^{\gamma _1}|\phi -\varphi |^{\gamma _2}|\sin ((\theta _2-\eta )/2)|^{\gamma _3}|\sin ((\theta _2-\eta )/2)|^{\gamma _4}}, \end{aligned}$$

for some \(\gamma _1,\gamma _2,\gamma _3,\gamma _4\in (0,1)\) with \(\gamma _1+\gamma _2<1\) and \(\gamma _3+\gamma _4<1\). Notice that this is possible since \(1+2\alpha -\alpha ^2<2\) for any \(\alpha \in (0,1)\).

Therefore, we obtain

$$\begin{aligned} \left| \int _0^{\pi }\int _0^{2\pi }({\mathscr {K}}_2(s,\phi ,\theta _1,\varphi ,\eta )-{\mathscr {K}}_2(s,\phi ,\theta _2,\varphi ,\eta ))d\eta d\varphi \right| \le C|\theta _1-\theta _2|^\alpha , \end{aligned}$$

uniformly in \(\phi \in (0,\pi )\). This concludes the proof of the stability of the function spaces by \({\tilde{F}}\).

Step 3: \(F_1\) is \({\mathscr {C}}^1\). In this last step, we check that \(F_1\) is \({\mathscr {C}}^1\).

More precisely, we intend to prove the following

$$\begin{aligned} \Vert \partial _f F_1(f_1)h-\partial _f F_1(f_2)h\Vert _{L^\infty }\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert f_1-f_2\Vert ^\gamma _{{\mathscr {C}}^{1,\alpha }}, \end{aligned}$$
(5.37)
$$\begin{aligned} \Vert \partial _\theta \big (\partial _f F_1(f_1)h-\partial _f F_1(f_2)h\big )\Vert _{{\mathscr {C}}^\alpha }\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert f_1-f_2\Vert ^\gamma _{{\mathscr {C}}^{1,\alpha }}, \end{aligned}$$
(5.38)

and

$$\begin{aligned} \Vert \partial _\phi \big (\partial _f F_1(f_1)h-\partial _f F_1(f_2)h\big )\Vert _{{\mathscr {C}}^\alpha }&\lesssim \Vert h|\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert f_1-f_2\Vert ^\gamma _{{\mathscr {C}}^{1,\alpha }}, \end{aligned}$$
(5.39)

for some \(\gamma >0\), and where \(f_1,f_2\in B_{X_m^\alpha }(\varepsilon )\), for some \(\varepsilon <1\).

Notice that (5.37)–(5.38)–(5.39) will imply the \({\mathscr {C}}^1\)-regularity of \(F_1\). Denote \(r_i(\phi ,\theta )=r_0(\phi )+f_i(\phi ,\theta )\), for \(i=1,2\). We will check directly the estimates for the derivatives, i.e., (5.38)–(5.39) and leave the first estimate which is less delicate. From the expressions (5.13) and (5.14), it is enough to check the estimates for the terms: \(U\cdot e^{i\theta }\) and \(\frac{U}{r_0}\cdot ie^{i\theta }\).

As we can guess the computations are very long, tedious and share lot of similarities. For this reason we shall focus only on one significant term given by (5.19) to illustrate how the estimates work, and restrict the discussion to some terms of \({\mathcal {I}}_2\). Notice that the Frechet derivative of each of the previous terms will correspond again to a singular integral where the kernel has the same order of singularity and then the estimates work similarly, even if the computations are longer.

Then, let us show the idea for \({\mathcal {I}}_2\) and note that

where

$$\begin{aligned} |(r(\phi , \theta )e^{i\theta },\cos (\phi ))-(r(\varphi ,\eta )e^{i\eta },\cos (\varphi ))|^2=J_1(f)(\phi ,\theta ,\varphi ,\eta ),\quad r=r_0+f, \end{aligned}$$

and

$$\begin{aligned} \frac{1}{2}\partial _f J_1(f)h(\phi ,\theta ,\varphi ,\eta )&= (r(\varphi ,\eta )-r(\phi ,\theta ))(h(\varphi ,\eta )-h(\phi ,\theta )) \nonumber \\&\quad +(r(\varphi ,\eta )h(\phi ,\theta )+h(\varphi ,\eta )r(\phi ,\theta ))(1-\cos (\eta -\theta )). \end{aligned}$$
(5.40)

Let us exhibit the main ideas of the term \({\mathcal {T}}_1\). The goal is to check

$$\begin{aligned} \Vert {\mathcal {T}}_1(f_1)h-{\mathcal {T}}_1(f_2)h\Vert _{{{\mathscr {C}}^{\alpha }}}&\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert f_1-f_2\Vert ^\gamma _{{\mathscr {C}}^{1,\alpha }}, \end{aligned}$$

for some \(\gamma >0\). We observe that

$$\begin{aligned} (J_1(f_1)-J_1(f_2))(\phi ,\theta ,\varphi ,\eta )&= (r_1(\varphi ,\eta )-r_1(\phi ,\theta ))^2-(r_2(\varphi ,\eta )-r_2(\phi ,\theta ))^2\nonumber \\&\quad +2(r_1-r_2)(\phi ,\theta )r_1(\varphi ,\eta )(1-\cos (\theta -\eta ))\nonumber \\&\quad +2r_2(\phi ,\theta )(r_1-r_2)(\varphi ,\eta )(1-\cos (\theta -\eta )). \end{aligned}$$
(5.41)

Now we write for any \(\phi ,\varphi \in (0,\pi )\) and \(\theta ,\eta \in (0,2\pi )\)

$$\begin{aligned} |r(\phi ,\theta )-r(\varphi ,\eta )|\le |r(\phi ,\theta )-r(\varphi ,\theta )|+|r(\varphi ,\theta )-r(\varphi ,\eta )|, \end{aligned}$$

By the \({\mathscr {C}}^{1,\alpha }\) regularity of r one has

$$\begin{aligned} |r(\phi ,\theta )-r(\varphi ,\theta )|\lesssim |\phi -\varphi | \Vert r\Vert _{\text {Lip}}. \end{aligned}$$

In addition, we claim that

$$\begin{aligned} |r(\varphi ,\theta )-r(\varphi ,\eta )|\lesssim |\sin ((\theta -\eta )/2)| \Vert r\Vert _{\text {Lip}}. \end{aligned}$$
(5.42)

Indeed, and without any restriction to the generality we can impose that \(0\le \eta \le \theta \le 2\pi \). We shall discuss two cases: \(0\le {\theta -\eta }\le \pi \) and \(\pi \le {\theta -\eta }\le 2\pi \). In the first case, we simply write

$$\begin{aligned} \frac{|r(\varphi ,\theta )-r(\varphi ,\eta )|}{|\sin ((\theta -\eta )/2)|}= \frac{|r(\varphi ,\theta )-r(\varphi ,\eta )|}{|\theta -\eta |}\frac{|\theta -\eta |}{|\sin ((\theta -\eta )/2)|} \le C\Vert r\Vert _{\text {Lip}}, \end{aligned}$$

with C a constant. As to the second case \(\pi \le {\theta -\eta }\le 2\pi \), by setting \({\widehat{\eta }}=\eta +2\pi \) we get

$$\begin{aligned} {\widehat{\eta }}-\theta \in [0,\pi ],\quad \sin ((\theta -\eta )/2)=-\sin ((\theta -{\widehat{\eta }})/2). \end{aligned}$$

Since \(\eta \mapsto r(\varphi ,\eta )\) is \(2\pi \)-periodic then using the result of the first case yields

$$\begin{aligned} \frac{|r(\varphi ,\theta )-r(\varphi ,\eta )|}{|\sin ((\theta -\eta )/2)|}&= \frac{|r(\varphi ,\theta )-r(\varphi ,{{\widehat{\eta }}})|}{|\sin ((\theta -{{\widehat{\eta }}})/2)|}\\&\le C\Vert r\Vert _{\text {Lip}}. \end{aligned}$$

This achieves the proof of the (5.42). Consequently we find

$$\begin{aligned} |r(\phi ,\theta )-r(\varphi ,\eta )|\lesssim \Vert r\Vert _{\text {Lip}}\big (|\phi -\varphi |+|\sin ((\theta -\eta )/2)|\big ). \end{aligned}$$
(5.43)

From algebraic calculus we easily get

$$\begin{aligned}&|(r_1(\varphi ,\eta )-r_1(\phi ,\theta ))^2-(r_2(\varphi ,\eta )-r_2(\phi ,\theta ))^2|\\&\quad = \big |((r_1-r_2)(\varphi ,\eta )-(r_1-r_2)(\phi ,\theta )\big |\\&\qquad \times \big |((r_1+r_2)(\varphi ,\eta )-(r_1+r_2)(\phi ,\theta ))\big |. \end{aligned}$$

Therefore we deduce successively from (5.43)

$$\begin{aligned}&|(r_1(\varphi ,\eta )-r_1(\phi ,\theta ))^2-(r_2(\varphi ,\eta )-r_2(\phi ,\theta ))^2|\\&\quad \lesssim \Vert r_1-r_2\Vert _{\text {Lip}}\big (|\phi -\varphi |+|\sin ((\theta -\eta )/2)|\big )\\&\qquad \times \Big (|r_1(\varphi ,\eta )-r_1(\phi ,\theta )|+|r_2(\varphi ,\eta )-r_2(\phi ,\theta )|\Big ), \end{aligned}$$

and

$$\begin{aligned}&|(r_1(\varphi ,\eta )-r_1(\phi ,\theta ))^2-(r_2(\varphi ,\eta )-r_2(\phi ,\theta ))^2|\\&\quad \lesssim |r_1(\varphi ,\eta )-r_1(\phi ,\theta )|^2+|r_2(\varphi ,\eta )-r_2(\phi ,\theta )|^2. \end{aligned}$$

By interpolation, we infer for any \(\gamma \in [0,1]\),

$$\begin{aligned}&\Big |(r_1(\varphi ,\eta )-r_1(\phi ,\theta ))^2 - (r_2(\varphi ,\eta )-r_2(\phi ,\theta ))^2\Big |\nonumber \\&\quad \lesssim \Vert r_1-r_2\Vert _{\text {Lip}}^{\gamma }\Big (|\varphi -\phi |^\gamma +|\sin ((\theta -{\eta })/2)|^\gamma |\Big )\nonumber \\&\qquad \times \Big (|r_1(\varphi ,\eta )-r_1(\phi ,\theta )|^{2-\gamma }+|r_2(\varphi ,\eta )-r_2(\phi ,\theta )|^{2-\gamma }\Big ), \end{aligned}$$
(5.44)

On the other hand, coming back to the definition of \(J_1\) we get

$$\begin{aligned} J_1(f_1)(\phi ,\theta ,\varphi ,\eta )\ge |r_1(\phi ,\theta )-r_1(\varphi ,\eta )|^2. \end{aligned}$$

Thus, putting together this inequality with (5.44) and (5.41) yield

$$\begin{aligned}&\frac{|(J_1(f_1)-J_1(f_2))(\phi ,\theta ,\varphi ,\eta )|}{J_1^\frac{1}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )+J_1^\frac{1}{2}(f_2)(\phi ,\theta ,\varphi ,\eta )}\nonumber \\&\quad \lesssim \Vert r_1-r_2\Vert ^\gamma _{{\mathscr {C}}^1}\Big \{\big (|\varphi -\phi |^\gamma +|\sin ((\theta -{\eta })/2)|^\gamma \big )\nonumber \\&\qquad \times \big [J_1^{\frac{1-\gamma }{2}}(f_1)(\phi ,\theta ,\varphi ,\eta )+J_1^{\frac{1-\gamma }{2}}(f_2)(\phi ,\theta ,\varphi ,\eta )\big ]+\phi |\sin ((\theta -\eta )/2)|\Big \}. \end{aligned}$$
(5.45)

Now, we shall give an estimate of \({\mathcal {T}}_1(f_1)-{\mathcal {T}}_1(f_2)\) in \( L^\infty \). For this purpose, define the quantity

$$\begin{aligned} {\mathscr {K}}_3(f)(\phi ,\theta ,\varphi ,\eta )=\frac{\sin (\varphi )h(\varphi ,\eta )\sin (\eta -\theta )}{J_1^\frac{1}{2}(f)(\phi ,\theta ,\varphi ,\eta )}, \end{aligned}$$

then one can easily check that

$$\begin{aligned} {\mathcal {I}}_7(\phi ,\theta ,\varphi ,\eta )&:={\mathscr {K}}_3(f_1)(\phi ,\theta ,\varphi ,\eta )-{\mathscr {K}}_3(f_2)(\phi ,\theta ,\varphi ,\eta ) = \frac{\sin (\varphi ) h(\varphi ,\eta )\sin (\eta -\theta )}{J_1^\frac{1}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )J_1^\frac{1}{2}(f_2)(\phi ,\theta ,\varphi ,\eta )}\nonumber \\&\quad \times \frac{J_1(f_2)(\phi ,\theta ,\varphi ,\eta )-J_1(f_1)(\phi ,\theta ,\varphi ,\eta )}{J_1^\frac{1}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )+J_1^\frac{1}{2}(f_2)(\phi ,\theta ,\varphi ,\eta )}\cdot \end{aligned}$$
(5.46)

From this definition, it follows that

$$\begin{aligned} \big ({\mathcal {T}}_1(f_1)-{\mathcal {T}}_1(f_2)\big )(\phi ,\theta )=\int _0^\pi \int _0^{2\pi }{\mathcal {I}}_7(\phi ,\theta ,\varphi ,\eta )d\varphi d\eta . \end{aligned}$$

According to (5.2) we get

$$\begin{aligned} \phi |\sin ((\theta -\eta )/2)|&\lesssim |\sin ((\theta -\eta )/2)|^{\gamma } \phi ^{1-\gamma }|\sin ((\theta -\eta )/2)|^{1-\gamma }\\&\lesssim |\sin ((\theta -\eta )/2)|^{\gamma }J_1^{\frac{1-\gamma }{2}}(f_1)(\phi ,\theta ,\varphi ,\eta ). \end{aligned}$$

Combining this inequality with (5.45) and (5.46) leads to

$$\begin{aligned}&|{\mathcal {I}}_7(\phi ,\theta ,\varphi ,\eta )|\nonumber \\&\quad \lesssim \Vert r_1-r_2\Vert _{{\mathscr {C}}^{1,\alpha }}^\gamma \frac{\sin (\varphi )|h(\varphi ,\eta )|\big (|\varphi -\phi )|^\gamma +|\sin ((\theta -\eta )/2)|^\gamma \big )}{J_1^\frac{1}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )J_1^\frac{1}{2}(f_2)(\phi ,\theta ,\varphi ,\eta )}\nonumber \\&\qquad \times \Big (J_1^{\frac{1-\gamma }{2}}(f_1)(\phi ,\theta ,\varphi ,\eta )+J_1^{\frac{1-\gamma }{2}}(f_2)(\phi ,\theta ,\varphi ,\eta )\Big ). \end{aligned}$$
(5.47)

Applying Lemma 5.2, we infer

$$\begin{aligned}&|{\mathcal {I}}_7(\phi ,\theta ,\varphi ,\eta )|\\&\quad \lesssim \Vert r_1-r_2\Vert _{{\mathscr {C}}^{1,\alpha }}^\gamma \frac{\sin (\varphi )|h(\varphi ,\eta )|(|\varphi -\phi )|^\gamma +|\sin ((\theta -\eta )/2)|^\gamma )}{J_1^\frac{1}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )J_1^\frac{\gamma }{2}(f_2)(\phi ,\theta ,\varphi ,\eta )}\\&\qquad +\Vert r_1-r_2\Vert _{{\mathscr {C}}^{1,\alpha }}^\gamma \frac{\sin (\varphi )| h(\varphi ,\eta )|(|\varphi -\phi )|^\gamma +|\sin ((\theta -\eta )/2)|^\gamma )}{J_1^\frac{\gamma }{2}(f_1)(\phi ,\theta ,\varphi ,\eta )J_1^\frac{1}{2}(f_2)(\phi ,\theta ,\varphi ,\eta )}\\&\quad \lesssim \Vert r_1-r_2\Vert _{{\mathscr {C}}^{1,\alpha }}^\gamma \frac{\sin (\varphi )|h(\varphi ,\eta )|(|\varphi -\phi )|^\gamma +|\sin ((\theta -\eta )/2)|^\gamma )}{\left\{ (\varphi +\phi )^2(\phi -\varphi )^2+(\sin ^2(\varphi )+\phi ^2)\sin ^2((\theta -\eta )/2)\right\} ^\frac{1+\gamma }{2}}\cdot \end{aligned}$$

Using the inequality \(\varphi ^2\ge \sin ^2(\varphi )\) for any \(\varphi \in {{\,\mathrm{{\mathbb {R}}}\,}}\), one achieves

$$\begin{aligned} |{\mathcal {I}}_7(\phi ,\theta ,\varphi ,\eta )|&\lesssim \Vert r_1-r_2\Vert _{{\mathscr {C}}^{1,\alpha }}^\gamma \frac{\sin (\varphi )|h(\varphi ,\eta )|(|\varphi -\phi )|^\gamma +|\sin ((\theta -\eta )/2)|^\gamma )}{(\sin (\varphi )+\phi )^{1+\gamma }\left\{ (\phi -\varphi )^2+\sin ^2((\theta -\eta )/2)\right\} ^\frac{1+\gamma }{2}}\\&\lesssim \frac{|h(\varphi ,\eta )|\Vert r_1-r_2\Vert _{{\mathscr {C}}^{1,\alpha }}^\gamma }{\sin ^\gamma (\varphi )\left\{ (\phi -\varphi )^2+\sin ^2((\theta -\eta )/2)\right\} ^\frac{1}{2}}\cdot \end{aligned}$$

The boundary conditions \(h(0,\eta )=h(\pi ,\eta )=0\) allow to cancel the singularity and one gets

$$\begin{aligned} |{\mathcal {I}}_7(\phi ,\theta ,\varphi ,\eta )|&\lesssim \frac{\Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert r_1-r_2\Vert _{{\mathscr {C}}^{1,\alpha }}^\gamma }{\left\{ (\phi -\varphi )^2+\sin ^2((\theta -\eta )/2)\right\} ^\frac{1}{2}}\cdot \end{aligned}$$

Interpolating we find that for any \(\beta \in (0,1)\),

$$\begin{aligned} |{\mathcal {I}}_7(\phi ,\theta ,\varphi ,\eta )|&\lesssim \frac{\Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert r_1-r_2\Vert _{{\mathscr {C}}^{1,\alpha }}^\gamma }{|\phi -\varphi |^{1-\beta }|\sin ((\eta -\theta )/2)|^\beta }\cdot \end{aligned}$$

Thus, we have that \({\mathcal {I}}_7\) is integrable in the variable \((\varphi ,\eta )\) uniformly in \((\phi ,\theta )\), and then

$$\begin{aligned} \Vert {\mathcal {T}}_1(f_1)h-{\mathcal {T}}_1(f_2)h\Vert _{L^\infty }&\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert f_1-f_2\Vert ^\gamma _{{\mathscr {C}}^{1,\alpha }}. \end{aligned}$$

The next purpose is to establish the partial \({\mathscr {C}}^\alpha \)-regularity in \(\phi \), and the partial regularity in \(\theta \) can be done similarly. We want to prove the following

$$\begin{aligned}&|({\mathcal {T}}_1(f_1)-{\mathcal {T}}_1(f_2))h(\phi _1,\theta )-(T_1(f_1)-T_1(f_2))h(\phi _2,\theta )|\nonumber \\&\quad \lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert f_1-f_2\Vert ^\gamma _{{\mathscr {C}}^{1,\alpha }}|\phi _1-\phi _2|^\alpha . \end{aligned}$$
(5.48)

For this goal we need to study the kernel \( |{\mathcal {I}}_7(\phi _1)-{\mathcal {I}}_7(\phi _2)|. \) To alleviate the notation we simply denote \({\mathcal {I}}_7(\phi , \theta ,\varphi ,\eta )\) by \({\mathcal {I}}_7(\phi )\) and \(J_1(f_i)(\phi _i,\theta ,\varphi ,\eta )\) by \(J_1(f_i)(\phi _i)\). Adding and subtracting some appropriate terms, one finds

$$\begin{aligned} |{\mathcal {I}}_7(\phi _1)-{\mathcal {I}}_7(\phi _2)|\lesssim {\mathcal {I}}_8+{\mathcal {I}}_9+{\mathcal {I}}_{10}+{\mathcal {I}}_{11}+{\mathcal {I}}_{12} \end{aligned}$$

with

$$\begin{aligned} {\mathcal {I}}_8&= \frac{\sin (\varphi )|h(\varphi ,\eta )|}{J_1^\frac{1}{2}(f_1)(\phi _1)J_1^\frac{1}{2}(f_1)(\phi _2)J_1^\frac{1}{2}(f_2)(\phi _1)}\frac{|J_1(f_2)(\phi _1)-J_1(f_1)(\phi _1)|}{J_1^\frac{1}{2}(f_1)(\phi _1)+J_1^\frac{1}{2}(f_2)(\phi _1)}\\&\times \frac{|J_1(f_1)(\phi _1)-J_1(f_1)(\phi _2)|}{J_1^\frac{1}{2}(f_1)(\phi _1)+J_1^\frac{1}{2}(f_1)(\phi _2)},\\ {\mathcal {I}}_9&= \frac{\sin (\varphi )|h(\varphi ,\eta )|}{J_1(f_1)(\phi _2)^\frac{1}{2}J_1(f_2)(\phi _1)^\frac{1}{2}}\frac{|J_1(f_2)(\phi _1)-J_1(f_1)(\phi _1)|}{(J_1^\frac{1}{2}(f_1)(\phi _1)+J_1^\frac{1}{2}(f_2)(\phi _1))(J_1(f_1)(\phi _1)^\frac{1}{2}+J_1(f_2)(\phi _2)^\frac{1}{2})}\\&\times \frac{|J_1(f_2)(\phi _1)-J_1(f_2)(\phi _2)|}{J_1^\frac{1}{2}(f_2)(\phi _1)+J_1^\frac{1}{2}(f_2)(\phi _2)},\end{aligned}$$
$$\begin{aligned} {\mathcal {I}}_{10}&= \frac{\sin (\varphi )|h(\varphi ,\eta )|}{J_1^\frac{1}{2}(f_1)(\phi _2)J_1^\frac{1}{2}(f_2)(\phi _1)}\frac{|(J_1(f_2)-J_1(f_1))(\phi _1)-(J_1(f_2)-J_1(f_1))(\phi _2)|}{J_1^\frac{1}{2}(f_1)(\phi _1)+J_1^\frac{1}{2}(f_2)(\phi _2)}, \\ {\mathcal {I}}_{11}&= \frac{\sin (\varphi )|h(\varphi ,\eta )|}{J_1^\frac{1}{2}(f_1)(\phi _2)J_1^\frac{1}{2}(f_2)(\phi _1)}\frac{|J_1(f_2)(\phi _2)-J_1(f_1)(\phi _2)|}{(J_1^\frac{1}{2}(f_1)(\phi _2)+J_1^\frac{1}{2}(f_2)(\phi _2))(J_1^\frac{1}{2}(f_1)(\phi _1)+J_1^\frac{1}{2}(f_2)(\phi _2))}\\&\times \frac{|J_1(f_1)(\phi _1)-J_1(f_1)(\phi _2)|}{J_1^\frac{1}{2}(f_1)(\phi _1)+J_1^\frac{1}{2}(f_1)(\phi _2)}\\ \end{aligned}$$

and

$$\begin{aligned} {\mathcal {I}}_{12}&= \frac{\sin (\varphi )|h(\varphi ,\eta )|}{J_1^\frac{1}{2}(f_1)(\phi _2)J_1^\frac{1}{2}(f_2)(\phi _1)J_1^\frac{1}{2}(f_2)(\phi _2)}\frac{|J_1(f_2)(\phi _2)-J_1(f_1)(\phi _2)|}{J_1^\frac{1}{2}(f_1)(\phi _2)+J_1^\frac{1}{2}(f_2)(\phi _2)}\\&\times \frac{|J_1(f_2)(\phi _1)-J_1(f_2)(\phi _2)|}{J_1^\frac{1}{2}(f_2)(\phi _1)+J_1^\frac{1}{2}(f_2)(\phi _2)}. \end{aligned}$$

The estimate of those terms are quite similar and we shall restrict the discussion to the term \({\mathcal {I}}_{10}\) which involves more computations. The analysis is straightforward and we will just give the basic ideas. First one should give a suitable estimate for the quantity

$$\begin{aligned} |(J_1(f_2)-J_1(f_1))(\phi _1)-(J_1(f_2)-J_1(f_1))(\phi _2)|. \end{aligned}$$

By using (5.41)–(5.44), one finds

$$\begin{aligned}&|(J_1(f_2)-J_1(f_1))(\phi _1)-(J_1(f_2)-J_1(f_1))(\phi _2)|\\&\quad \lesssim |(r_1-r_2)(\phi _1,\theta )-(r_1-r_2)(\phi _2,\theta )||(r_1+r_2)(\phi _1,\theta )-(r_1+r_2)(\varphi ,\eta )|\\&\qquad +|(r_1-r_2)(\phi _2,\theta )-(r_1-r_2)(\varphi ,\eta )||(r_1+r_2)(\phi _1,\theta )-(r_1+r_2)(\phi _2,\theta )|\\&\qquad +|(r_1-r_2)(\phi _1,\theta )-(r_1-r_2)(\phi _2,\theta )|r_1(\varphi ,\eta )\sin ^2((\theta -\eta )/2)\\&\qquad +|r_2(\phi _1,\theta )-r_2(\phi _2,\theta )||(r_1-r_2)(\varphi ,\eta )|\sin ^2((\theta -\eta )/2). \end{aligned}$$

Moreover,

$$\begin{aligned}&|(r_1-r_2)(\phi _1,\theta )-(r_1-r_2)(\phi _2,\theta )|\lesssim \Vert r_1-r_2\Vert ^\alpha |\phi _1-\phi _2|^\alpha \left( |r_1(\phi _1,\theta )-r_1(\varphi ,\eta )|^{1-\alpha }\right. \\&\left. +|r_2(\phi _1,\theta )-r_2(\varphi ,\eta )|^{1-\alpha }+|r_1(\phi _2,\theta )-r_1(\varphi ,\eta )|^{1-\alpha }+|r_2(\phi _2,\theta )-r_2(\varphi ,\eta )|^{1-\alpha }\right) , \end{aligned}$$

and

$$\begin{aligned} |(r_1+r_2)(\phi _1,\theta )&- (r_1+r_2)(\phi _2,\theta )|\lesssim |\phi _1-\phi _2|^\alpha (|r_1(\phi _1,\theta )-r_1(\varphi ,\eta )|^{1-\alpha }\\&\quad +|r_2(\phi _1,\theta )-r_2(\varphi ,\eta )|^{1-\alpha }+|r_1(\phi _2,\theta )-r_1(\varphi ,\eta )|^{1-\alpha }\\&\quad +|r_2(\phi _2,\theta )-r_2(\varphi ,\eta )|^{1-\alpha }). \end{aligned}$$

In a similar way, we deduce first by triangular inequality

$$\begin{aligned} |(r_1+r_2)(\phi _1,\theta )-(r_1+r_2)(\varphi ,\eta )|&\le |r_1(\phi _1,\theta )-r_1(\varphi ,\eta )|+|r_2(\phi _1,\theta )-r_1(\varphi ,\eta )| \end{aligned}$$

and second from (5.44)

$$\begin{aligned}&|(r_1-r_2)(\phi _2,\theta )-(r_1-r_2)(\varphi ,\eta )|\\&\quad \lesssim \Vert r_1-r_2|\Vert ^\gamma (|\phi _2-\varphi |^ \gamma +|\sin ((\theta -\eta )/2)|^\gamma )\\&\qquad \times \left( |r_1(\phi _2,\theta )-r_1(\varphi ,\eta )|^{1-\gamma }+|r_2(\phi _2,\theta )-r_2(\varphi ,\eta )|^{1-\gamma }\right) . \end{aligned}$$

Combining the preceding estimate we achieve

$$\begin{aligned}&|(J_1(f_2)-J_1(f_1))(\phi _1)-(J_1(f_2)-J_1(f_1))(\phi _2)|\\&\quad \lesssim |\phi _1-\phi _2|^\alpha \Vert f_1-f_2\Vert ^\gamma \left( |r_1(\phi _1,\theta )-r_1(\varphi ,\eta )|^{2-\alpha }+|r_2(\phi _1,\theta )-r_2(\varphi ,\eta )|^{2-\alpha }\right. \\&\qquad \left. +|r_1(\phi _2,\theta )-r_1(\varphi ,\eta )|^{2-\alpha }+|r_2(\phi _2,\theta )-r_2(\varphi ,\eta )|^{2-\alpha }\right) \\&\quad \lesssim |\phi _1-\phi _2|^\alpha \Vert f_1-f_2\Vert ^\gamma ({\mathscr {E}}_{1,1}^{2-\alpha }+{\mathscr {E}}_{2,1}^{2-\alpha }+{\mathscr {E}}_{1,2}^{2-\alpha }+{\mathscr {E}}_{2,2}^{2-\alpha }). \end{aligned}$$

where we use the notation

$$\begin{aligned} {\mathscr {E}}_{i,j}=|r_i(\phi _j,\theta )-r_i(\varphi ,\eta )|; \, i,j\in \{1,2\} \end{aligned}$$

Hence,

$$\begin{aligned} |{\mathcal {I}}_{10}|&\lesssim |\phi _1-\phi _2|^\alpha \Vert f_1-f_2\Vert ^\gamma \frac{\sin (\varphi )| h(\varphi ,\eta )|}{J_1^\frac{1}{2}(f_1)(\phi _2)J_1^\frac{1}{2}(f_2)(\phi _1)}\frac{\sum _{i,j=1}^2{\mathscr {E}}_{i,j}^{2-\alpha }}{J_1^\frac{1}{2}(f_1)(\phi _1)+J_1^\frac{1}{2}(f_2)(\phi _2)}\cdot \end{aligned}$$

By using the definition of \(J_1\) in Lemma 5.2, we immediately get

$$\begin{aligned} {\mathscr {E}}_{i,j}\le J_1^\frac{1}{2}(f_i)(\phi _j), \end{aligned}$$

that we combine with (5.43) in order to get

$$\begin{aligned} {\mathscr {E}}_{i,j}\lesssim |\phi _j-\varphi |+|\sin ((\theta -\eta )/2)|. \end{aligned}$$

We shall analyze the term associated to \({\mathscr {E}}_{1,1}\) and the treatment of the other ones are quite similar. First we note

$$\begin{aligned}&\frac{{\mathscr {E}}_{1,1}^{2-\alpha }}{J_1^\frac{1}{2}(f_1)(\phi _2)J_1^\frac{1}{2}(f_2)(\phi _1)\big (J_1^\frac{1}{2}(f_1)(\phi _1)+J_1^\frac{1}{2}(f_2)(\phi _2)\big )}\\&\quad \lesssim \frac{|\phi _1-\varphi |^{1-\alpha }+|\sin ((\theta -\eta )/2)|^{1-\alpha }}{J_1^\frac{1}{2}(f_1)(\phi _2)J_1^\frac{1}{2}(f_2)(\phi _1)}\cdot \end{aligned}$$

Making appeal to (5.27) and (2.17), we infer

$$\begin{aligned}&\frac{\sin (\varphi )|h(\varphi )|}{J_1^\frac{1}{2}(f_1)(\phi _2)J_1^\frac{1}{2}(f_2)(\phi _1)}\\&\quad \lesssim \frac{\Vert h\Vert _{\text {Lip}}}{\left\{ (\phi _1-\varphi )^2+\sin ^2((\theta -\eta )/2)\right\} ^\frac{1}{2}\left\{ (\phi _2-\varphi )^2+\sin ^2((\theta -\eta )/2)\right\} ^\frac{1}{2}}\cdot \end{aligned}$$

By interpolation we obtain for any \(\gamma , \beta \in [0,1]\),

$$\begin{aligned} \frac{\sin (\varphi )|h(\varphi )|}{J_1^\frac{1}{2}(f_1)(\phi _2)J_1^\frac{1}{2}(f_2)(\phi _1)}\lesssim \Vert h\Vert _{\text {Lip}}\frac{|\phi _1-\varphi |^{-\gamma }|\phi _2-\varphi |^{-\beta }}{|\sin ((\theta -\eta )/2)|^{2-\gamma -\beta }}\cdot \end{aligned}$$

Combining the preceding inequalities gives for any \(\gamma _1, \gamma _2, \beta _1,\beta _2\in [0,1]\)

$$\begin{aligned}&\frac{\sin (\varphi )|h(\varphi ,\eta )|}{J_1(f_1)(\phi _2)^\frac{1}{2}J_1(f_2)(\phi _1)^\frac{1}{2}}\frac{{\mathscr {E}}_{1,1}^{2-\alpha }}{J_1(f_1)(\phi _1)^\frac{1}{2}+J_1(f_2)(\phi _2)^\frac{1}{2}}\\&\quad \lesssim \Vert h\Vert _{\text {Lip}}\frac{|\phi _1-\varphi |^{1-\alpha -\gamma _1}|\phi _2-\varphi |^{-\beta _1}}{|\sin ((\theta -\eta )/2)|^{2-\gamma _1-\beta _1}}\\&\qquad + \Vert h\Vert _{\text {Lip}}\frac{|\phi _1-\varphi |^{-\gamma _2}|\phi _2-\varphi |^{-\beta _2}}{|\sin ((\theta -\eta )/2)|^{1+\alpha -\gamma _2-\beta _2}}\cdot \end{aligned}$$

The majorant functions are integrable in the variable \((\varphi ,\eta )\) uniformly in \(\phi _1,\phi _2,\theta \) provided that

$$\begin{aligned} 1<\gamma _1+\beta _1<2-\alpha \quad \hbox {and}\quad \alpha<\gamma _2+\beta _2<1, \end{aligned}$$

and under these constraints one can find admissible parameters. Consequently,

$$\begin{aligned} \int _0^\pi \int _{0}^{2\pi } {\mathcal {I}}_{10} d\varphi d\eta \lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert f_1-f_2\Vert ^\gamma _{{\mathscr {C}}^{1,\alpha }}|\phi _1-\phi _2|^\alpha . \end{aligned}$$

They, we are able to find that

$$\begin{aligned} |{\mathcal {I}}_7(\phi _1)-{\mathcal {I}}_7(\phi _2)|\lesssim ||h||_{{\mathscr {C}}^{1,\alpha }}||f_1-f_2||^\gamma _{{\mathscr {C}}^{1,\alpha }}|\phi _1-\phi _2|^\alpha , \end{aligned}$$

for some \(\gamma \in (0,1)\).

Let us now move on \({\mathcal {T}}_2(f)h(\phi ,\theta )\) and show the main ideas. Define

$$\begin{aligned} {\mathcal {K}}_4(f)(\phi ,\theta ,\varphi ,\eta ):=\frac{\sin (\varphi )r(\varphi ,\eta )\sin (\eta -\theta )}{J_1^\frac{3}{2}(\phi ,\theta ,\varphi ,\eta )}\partial _f J_1(f)h(\phi ,\theta ,\varphi ,\eta ), \end{aligned}$$

and then

$$\begin{aligned}&{\mathcal {K}}_4(f_1)(\phi ,\theta ,\varphi ,\eta ) - {\mathcal {K}}_4(f_2)(\phi ,\theta ,\varphi ,\eta ) =\frac{\sin (\varphi )(r_1-r_2)(\varphi ,\eta )\sin (\eta -\theta )}{J_1^\frac{3}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )}\partial _f J_1(f_1)h(\phi ,\theta ,\varphi ,\eta )\nonumber \\&\qquad +\sin (\varphi )r_2(\varphi ,\eta )\sin (\eta -\theta )\left( \frac{\partial _f J_1(f_1)h(\phi ,\theta ,\varphi ,\eta )}{J_1(f_1)^\frac{3}{2}(\phi ,\theta ,\varphi ,\eta )}-\frac{\partial _f J_1(f_2)h(\phi ,\theta ,\varphi ,\eta )}{J_2(f_1)^\frac{3}{2}(\phi ,\theta ,\varphi ,\eta )}\right) \nonumber \\&\quad =: ({\mathcal {I}}_{13}+{\mathcal {I}}_{14})(\phi ,\theta ,\varphi ,\vartheta ). \end{aligned}$$
(5.49)

Let us analyze \({\mathcal {I}}_{13}\) and note that \({\mathcal {I}}_{14}\) is more involved (it includes more computations) but has same order of singularity. Moreover, recall the expression of \(\partial _f J_1\) in (5.40). Define also

$$\begin{aligned} {\mathcal {T}}_{2,1}(\phi ,\theta ):=\int _0^\pi \int _0^{2\pi }{\mathcal {I}}_{13}(\phi ,\theta ,\varphi ,\eta )d\varphi d\eta . \end{aligned}$$

Let us begin studying the \(L^\infty \) norm of \({\mathcal {T}}_{2,1}\). Indeed, by (5.40) and (5.43) we are able to find that

$$\begin{aligned}&\frac{|\partial _f J_1(f)h(\phi ,\theta ,\varphi ,\eta )|}{J_1^\frac{3}{2}(f)(\phi ,\theta ,\varphi ,\eta )}\le C ||h||_{\text {Lip}}\nonumber \\&\times \frac{|r_1(\varphi ,\eta )-r_1(\phi ,\theta )|(|\phi -\varphi |+|\sin ((\theta -\eta )/2)|)+\sin (\phi )\sin (\varphi )\sin ^2((\theta -\eta )/2))}{J_1^\frac{3}{2}(f)(\phi ,\theta ,\varphi ,\eta )}\nonumber \\&\le C\frac{ ||h||_{\text {Lip}}}{|\sin \varphi +\phi |J_1^\frac{1}{2}(f)(\phi ,\theta ,\varphi ,\eta )}. \end{aligned}$$
(5.50)

Then, using Lemma 5.2 we achieve

$$\begin{aligned} |{\mathcal {I}}_{13}(\phi ,\theta ,\varphi ,\eta )|&\le C||h||_{\text {Lip}}||r_1-r_2||_{\text {Lip}}\frac{\sin ^2(\varphi )|\sin (\eta -\theta )|}{(\sin \varphi +\phi )J_1^\frac{1}{2}(\phi ,\theta ,\varphi ,\eta )\nonumber }\\&\le C||h||_{\text {Lip}}||r_1-r_2||_{\text {Lip}}\frac{\sin ^2(\varphi )|\sin (\eta -\theta )|}{(\sin \varphi +\phi )^2(|\phi -\varphi |+|\sin ((\theta -\eta )/2)|)}\nonumber \\&\le C ||h||_{\text {Lip}}||r_1-r_2||_{\text {Lip}}, \end{aligned}$$
(5.51)

finding that \({\mathcal {T}}_{2,1}\) is bounded.

Since we showed the estimates in \(\phi \) of \({\mathcal {T}}_1\) (see (5.48)), let us work here with the variable \(\theta \). Indeed, our purpose will be showing

$$\begin{aligned}&|({\mathcal {T}}_2(f_1)-{\mathcal {T}}_2(f_2))h(\phi ,\theta _1)-(T_2(f_1)-T_2(f_2))h(\phi ,\theta _2)|\nonumber \\&\quad \lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert f_1-f_2\Vert ^\gamma _{{\mathscr {C}}^{1,\alpha }}|\theta _1-\theta _2|^\alpha , \end{aligned}$$
(5.52)

for any \(\theta _1,\theta _2\in [0,2\pi ]\) with \(\theta _1<\theta _2\). Since we have decomposed \({\mathcal {T}}_2(f_1)-{\mathcal {T}}_2(f_2)\) in two terms in (5.49), let us work with \({\mathcal {T}}_{2,1}\) and show

$$\begin{aligned} |{\mathcal {T}}_{2,1}(\phi ,\theta _1)-T_{2,1}(\phi ,\theta _2)|&\lesssim \Vert h\Vert _{{\mathscr {C}}^{1,\alpha }}\Vert f_1-f_2\Vert ^\gamma _{{\mathscr {C}}^{1,\alpha }}|\theta _1-\theta _2|^\alpha . \end{aligned}$$
(5.53)

Here, we will use Proposition C.1 by fixing \(\phi \). The kernel of \({\mathcal {T}}_{2,1}\), i.e. \({\mathcal {I}}_{13}\), has been already bounded in (5.51). That gives us hypothesis (C.2) of such proposition and it remains to estimate \(\partial _\theta {\mathcal {I}}_{13}\) (see hypothesis (C.4)). By using the expression of \({\mathcal {I}}_{13}\) we get

$$\begin{aligned}&|\partial _\theta {\mathcal {I}}_{13}(\phi ,\theta ,\varphi ,\eta )|\nonumber \\&\quad \le C||r_1-r_2||_{\text {Lip}}\frac{\sin ^2(\varphi )|\partial _f J_1(f_1)h(\phi ,\theta ,\varphi ,\eta )|}{J_1^\frac{3}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )}\nonumber \\&\qquad +C||r_1-r_2||_{\text {Lip}}\frac{\sin ^2(\varphi )|\sin (\eta -\theta )||\partial _\theta \partial _f J_1(f_1)h(\phi ,\theta ,\varphi ,\eta )|}{J_1^\frac{3}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )}\nonumber \\&\qquad +C||r_1-r_2||_{\text {Lip}}\frac{\sin ^2(\varphi )|\sin (\eta -\theta )||\partial _f J_1(f_1)h(\phi ,\theta ,\varphi ,\eta )||\partial _\theta J_1(f_1)h(\phi ,\theta ,\varphi ,\eta )|}{J_1^\frac{5}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )}\nonumber \\&\quad =: C||r_1-r_2||_{\text {Lip}}({\mathcal {I}}_{13,1}+{\mathcal {I}}_{13,2}+{\mathcal {I}}_{13,3}). \end{aligned}$$
(5.54)

For \({\mathcal {I}}_{13,1}\) we use (5.50) and Lemma 5.2 finding

$$\begin{aligned} {\mathcal {I}}_{13,1}&\le C||h||_{\text {Lip}}\frac{\sin ^2(\varphi )}{|\sin \varphi +\phi |J_1^\frac{1}{2}(f)(\phi ,\theta ,\varphi ,\eta )}\\&\le C ||h||_{\text {Lip}}\frac{1}{(|\phi -\varphi |+|\sin ((\theta -\eta )/2)|)}\\ \le&C ||h||_{\text {Lip}}\frac{1}{|\sin ((\theta -\eta )/2)|}. \end{aligned}$$

That gives us hypothesis (C.4) for the first term \({\mathcal {I}}_{13,1}\). In order to work with \({\mathcal {I}}_{13,2}\), using (5.43) note that

$$\begin{aligned}&\frac{|\partial _\theta \partial _f J_1(f_1)h(\phi ,\theta ,\varphi ,\eta )|}{J_1^\frac{3}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )}\le C||h||_{\text {Lip}}\\&\times \frac{|\sin (\phi )|^\alpha (|\phi -\varphi |+|\sin ((\theta -\eta )/2)|))+\sin (\varphi )\phi ^\alpha \sin ^2((\theta -\eta )/2)+\sin (\varphi )\phi |\sin ((\theta -\eta )/2)|}{ J_1^\frac{3}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )}\\&\le C\frac{||h||_{\text {Lip}}}{(\phi +\sin \varphi )^{1-\alpha }J_1(f_1)(\phi ,\theta ,\varphi ,\eta )}, \end{aligned}$$

and then \({\mathcal {I}}_{13,2}\) follows as

$$\begin{aligned} {\mathcal {I}}_{13,2}&\le C ||h||_{\text {Lip}}||r_1-r_2||_{\text {Lip}}\frac{\sin ^2(\varphi )|\sin (\eta -\theta )|}{(\phi +\sin \varphi )^{3-\alpha }(|\phi -\varphi |+|\sin ((\theta -\eta )/2)|)^2}\\&\le C ||h||_{\text {Lip}}||r_1-r_2||_{\text {Lip}}\frac{1}{(\phi +\sin \varphi )^{1-\alpha }(|\phi -\varphi |+|\sin ((\theta -\eta )/2)|)}\\&\le C ||h||_{\text {Lip}}||r_1-r_2||_{\text {Lip}}\frac{1}{(\phi +\sin \varphi )^{1-\alpha }|\sin ((\theta -\eta )/2)|}. \end{aligned}$$

Similarly, we can work with \({\mathcal {I}}_{13,3}\). First note that

$$\begin{aligned} \frac{|\partial _\theta J_1(f_1)h(\phi ,\theta ,\varphi ,\eta )|}{J_1(f_1)(\phi ,\theta ,\varphi ,\eta )}\le C\frac{\phi ^\alpha }{J_1^\frac{1}{2}(f_1)(\phi ,\theta ,\varphi ,\eta )}, \end{aligned}$$

which, together with (5.50) implies

$$\begin{aligned} {\mathcal {I}}_{13,3}&\le C ||h||_{\text {Lip}}||r_1-r_2||_{\text {Lip}}\frac{\sin ^2(\varphi )|\sin (\eta -\theta )|\phi ^\alpha }{|\sin (\varphi )+\phi |J_1(f_1)(\phi ,\theta ,\varphi ,\eta )}\\&\le C ||h||_{\text {Lip}}||r_1-r_2||_{\text {Lip}}\frac{1}{|\sin ((\theta -\eta )/2)|}. \end{aligned}$$

Putting everything together we achieves that \(\partial _\theta {\mathcal {I}}_{13}\) satisfies hypothesis (C.4) of Proposition C.1. Then, such proposition can be applied to find (5.53) concluding the proof. \(\quad \square \)

6 Main Result

In this section we shall provide a general statement that precise Theorem 1.1 and give its proof using all the previous results. Recall that the search of rotating solutions in the patch form to the equation (1.1), that is, solutions in the form

$$\begin{aligned} q(t,x)=q_0(e^{-i\Omega t}(x_1,x_2), x_3), \quad q_0=\mathbf{1}_{D}, \end{aligned}$$

where D is a bounded simply-connected domain surrounded by a surface parametrized by

$$\begin{aligned} (\phi , \theta )\in [0,\pi ]\times [0,2\pi ]\mapsto ((r_0(\phi )+f(\phi ,\theta ))e^{i\theta }, \cos (\phi )), \end{aligned}$$

reduces to solving the following infinite-dimensional equation

$$\begin{aligned}{\tilde{F}}(\Omega , f)=0\end{aligned}$$

with f in a small neighborhood of the origin in the Banach space \(X_m^\alpha \) and \({\tilde{F}}\) is introduced in (2.13). Notice that a solution is nontrivial means that the associated shape is not invariant by rotation along the vertical axis. Looking to the structure of the elements of space \(X_m^\alpha \) one can easily see that a nonzero element guarantees a nontrivial shape. Our result stated below asserts that solutions to this functional equation do exist and are organized in a countable family of one-dimensional curves bifurcating from the trivial solution at the largest eigenvalues of the linearized operator at the origin. More precisely, we have the following.

Theorem 6.1

Let \(m\ge 2\) be a fixed integer and \(r_0:[0,\pi ]\rightarrow {{\,\mathrm{{\mathbb {R}}}\,}}\) satisfies the conditions:

  1. (H1)

    \(r_0\in {\mathscr {C}}^{2}([0,\pi ])\), with \(r_0(0)=r_0(\pi )=0\) and \(r_0(\phi )>0\) for \(\phi \in (0,\pi )\).

  2. (H2)

    There exists \(C>0\) such that

    $$\begin{aligned} \forall \, \phi \in [0,\pi ],\quad C^{-1}\sin \phi \le r_0(\phi )\le C\sin (\phi ). \end{aligned}$$
  3. (H3)

    \(r_0\) is symmetric with respect to \(\phi =\frac{\pi }{2}\), i.e., \(r_0\left( \frac{\pi }{2}-\phi \right) =r_0\left( \frac{\pi }{2}+\phi \right) \), for any \(\phi \in [0,\frac{\pi }{2}]\).

Then there exist \(\delta >0\) and two one-dimensonal \({\mathscr {C}}^1\)-curves \(s\in (-\delta ,\delta )\mapsto f_m(s)\in X_m^\alpha \) and \( s\in (-\delta ,\delta )\mapsto \Omega _m(s)\in {{\,\mathrm{{\mathbb {R}}}\,}}, \) with

$$\begin{aligned}f_m(0)=0, \quad f_m(s)\ne 0, \, \forall \, s\ne 0\quad \hbox {and}\quad \Omega _m(0)=\Omega _m, \end{aligned}$$

where \(\Omega _m\) is defined in Proposition 4.3, such that

$$\begin{aligned} \forall \, s\in (-\delta ,\delta ),\quad {\tilde{F}}\big (\Omega _m(s),f_m(s)\big )=0. \end{aligned}$$

Proof

The main material to prove this result is Crandall–Rabinowitz theorem, recalled in Theorem B.1. First the well-possednes and the regularity of \({\tilde{F}}:X_m^\alpha \rightarrow X_m^\alpha \) were discussed in Proposition 5.2. Thus it remains to check the suitable spectral properties of the linearized operator at the origin. The expression of this operator is detailed in Proposition 3.2 and it is a of Fredholm type of zero index according to Proposition 4.6. In addition for \(\Omega =\Omega _m\) the kernel is a one-dimensional vector space. Finally, the transversal condition is satisfied by virtue of Proposition 4.7. \(\quad \square \)

6.1 Special case: sphere and ellipsoid

In this section we aim to show the particular case of bifurcating from spherical or ellipsoidal shapes. The main particularity of these shapes is that their associated stream function is well-known in the literature, see [32]. More specifically, let \({\mathscr {E}}\) be an ellipsoid inside the region

$$\begin{aligned} \frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}+\frac{x_3^2}{c^2}=1. \end{aligned}$$

The associated stream function given by

$$\begin{aligned} \psi _0(x)=-\frac{1}{4\pi }\int _{{\mathscr {E}}}\frac{dA(y)}{|x-y|}, \end{aligned}$$

can be computed inside the ellipsoid as

$$\begin{aligned} \psi _0(x)=\frac{abc}{4}\int _0^\infty \left\{ \frac{x_1^2}{a^2+s}+\frac{x_2^2}{b^2+s}+\frac{x_3^2}{c^2+s}-1\right\} \frac{ds}{\sqrt{(a^2+s)(b^2+s)(c^2+s)}}\cdot \end{aligned}$$

In the case that \(a=b\) we have that the ellipsoid is invariant under rotations about the z-axis and then it defines a stationary patch, see Lemma 2.1. Moreover and without loss of generality we can take \(c=1\). Note that in this case

$$\begin{aligned} \psi _0(x)=\alpha _1(a) (x_1^2+x_2^3)+\alpha _2 (a)x_3^2+\alpha _3(a), \end{aligned}$$

where

$$\begin{aligned} \alpha _1(a):=\frac{a^2}{4}\int _0^\infty \frac{ds}{(a^2+s)^2\sqrt{1+s}}, \\ \alpha _2(a):=\frac{a^2}{4}\int _0^\infty \frac{ds}{(a^2+s)\sqrt{(1+s)^3}}, \end{aligned}$$

and

$$\begin{aligned} \alpha _3(a):=-\frac{a^2}{4}\int _0^\infty \frac{ds}{(a^2+s)^2\sqrt{1+s}}. \end{aligned}$$

The sphere coincides with the case \(a=1\) having \(\alpha _1(1)=\alpha _2(1)=\frac{1}{6}\) and \(\alpha _3(1)=\frac{1}{2}\).

The above expression of the stream function together with Remark 3.2 gives us that

$$\begin{aligned} \int _0^\pi H_1(\phi ,\varphi )d\varphi =2\alpha _1(a), \end{aligned}$$

for any \(\phi \in [0,\pi ]\). Recall that \(H_n\) is defined in (3.3). Now, by virtue of Proposition 3.2 one has

$$\begin{aligned} \partial _{f} {\tilde{F}}(\Omega ,0)h(\phi ,\theta )=\sum _{n\ge 1}\cos (n\theta ){\mathcal {L}}_n^\Omega (h_n)(\phi ), \end{aligned}$$

where

Moreover, the function \(\nu _\Omega \) used in the spectral study and defined in (4.3) agrees with

$$\begin{aligned} \nu _\Omega (\phi )=2\alpha _1(a)-\Omega , \end{aligned}$$

which now is constant on \(\phi \). Also the constant \(\kappa \) in (4.5) equals now to \(2\alpha _1(a)\). Hence, the key point in Sect. 4.1 is the symmetrization of the above operator. For that reason, we have defined the signed measure \(d{\mu _\Omega }\) as

$$\begin{aligned} d\mu _\Omega (\varphi )=\sin (\varphi )r_0^2(\varphi )\nu _\Omega (\varphi )d\varphi , \end{aligned}$$

in (4.4) and the operator \({\mathcal {K}}_n^\Omega \) in (4.82). However, since in this case \(\nu _\Omega (\varphi )\) is constant on \(\varphi \), there is no need to introduce it in the measure with the goal of symmetryzing the operator. Following the ideas developed above, we deduce that the kernel study of the linearized operator agrees in this case with the following eigenvalue problem

$$\begin{aligned} \tilde{{\mathcal {K}}}_n(\phi )=(2\alpha _1(a)-\Omega )h(\phi ). \end{aligned}$$

Here, we define

$$\begin{aligned} \tilde{{\mathcal {K}}}_n(\phi ):=(2\alpha _1(a)-\Omega ){\mathcal {K}}_n^\Omega (\phi ), \end{aligned}$$

which does not depend now on \(\Omega \) by definition of \({\mathcal {K}}_n^\Omega \). Note that both operators have similar properties. Hence \(\tilde{{\mathcal {K}}}_n\) sets the properties given in Proposition 4.2 taking the Lebesgue space \(L^2_{{\tilde{\mu }}_\Omega }\) with

$$\begin{aligned}d{\tilde{\mu }}_\Omega (\varphi )=\sin (\varphi )r_0^2(\varphi )d\varphi .\end{aligned}$$

Denote by \(\beta _{n,i}\) the eigenvalues of \(\tilde{{\mathcal {K}}}_n\) (for each n we have a family of eigenvalues). Then, we have necessary that

$$\begin{aligned} \Omega _n=2\alpha _1(a)-\beta _{n,i}. \end{aligned}$$

In Theorem 6.1, bifurcation occurs from \(\Omega _n^\star \) given by

$$\begin{aligned} \Omega _n^\star =2\alpha _1(a)-\beta _{n}^\star , \end{aligned}$$

with

$$\begin{aligned} \beta _n^\star =\max _{i}\beta _{n,i}. \end{aligned}$$

Moreover, we know that \(\beta _n^\star \) is positive and then \(\Omega _n^\star <2\alpha _1(a)\). In particular, by Proposition 4.3, we have that \(\Omega _n^\star \) tends to \(\kappa =2\alpha _1(a)\). Furthermore, \(\Omega _n^\star \) increases in n and then we can bound it below by \(\Omega _1^\star \). Using the equation for \(\beta _1^\star \), that is

$$\begin{aligned} \int _0^\pi H_1(\varphi ,\phi )h(\varphi )d\varphi =\beta _1^\star h(\phi ), \end{aligned}$$

one finds that \(\beta _1^\star \le 2\alpha _1(a)\) and then \(\Omega _1^\star \) is positive. This implies that \(\Omega _n^\star \) is positive for any n. Then, in Theorem 6.1 bifurcation holds at some \(\Omega _n^\star \in (0,2\alpha _1(a))\). Let us remark that in the case of the sphere, meaning \(a=1\), one has \(2\alpha _1(a)=\frac{1}{3}\).

There is an interesting open problem concerning, first the spectral distribution of the eigenvalues \(\beta _{n,i}\) (whether or not they are finite, simple or multiple), and second if bifurcation occurs at the eigenvalues \(\Omega _n=2\alpha _1(a)-\beta _{n,i}\) (which is shown to happen only for the largest eigenvalue \(\beta _n^\star \)). Notice that the simplicity and the monotonicity of the eigenvalues is a delicate problem and could be related to the geometry of the revolution shape. Finally we observe that since \(\beta _{n,i}<\beta _n^\star \) then \(\Omega _n=2\alpha _1(a)-\beta _{n,i}>\frac{1}{3}-\beta _n^\star >0\).