An overdetermined eigenvalue problem
and the Critical Catenoid conjecture

¿0 & in  Ω,

ξ= 0  along  ∂Ω.

This problem is commonly known as a Dirichlet problem, and it typically has a
unique solution when one exists. Moreover, we introduce the following
additional condition to the problem:

$\frac{\partial\xi}{\partial\nu}=c\quad\text{along}\quad\partial\Omega,\quad
c\in\mathbb{R}^{+},$ (0.1)

where $\nu$ denotes the inner normal vector to $\partial\Omega$, it transforms
into an overdetermined elliptic problem (OEP) and solutions only exist for
specific domains $\Omega$. Consequently, a solution to the combined problem
(LABEL:DP) and (0.1) is represented as a pair $(\Omega,\xi)$, emphasizing that
not only the function $\xi$ but also the domain $\Omega$ are essential for
solving overdetermined problems. For instance, if we consider $\Omega$ as a
geodesic ball in a space form and $\xi$ as a function dependent solely on the
distance from the center of the ball that satisfies (LABEL:DP), it becomes
evident that $\xi$ satisfies condition (0.1). This naturally raises the
question of whether these are the only types of solutions to the OEP on
bounded domains. The interest in such problems traces back to Serrin’s seminal
paper [52], where he established the rigidity of problem (LABEL:DP) and (0.1)
when $\Omega$ is a bounded domain in Euclidean space and $f\equiv 1$. Serrin’s
work showed that if $(\Omega,\xi)$ is a $\mathcal{C}^{2}$ solution (i.e., both
the solution $\xi$ and the domain $\Omega$ exhibit $\mathcal{C}^{2}$
regularity) to the OEP (LABEL:DP) and (0.1), then $\Omega$ must be a round
ball and $\xi$ must be rotationally symmetric. Serrin employed the Alexandrov
reflection method and a modified maximum principle for domains with corners to
establish this result. Subsequently, Pucci and Serrin in [42] extended this
result to include a general $f$ with Lipschitz regularity. Serrin’s paper
marked the inception of OEPs in which the overdetermined condition (0.1)
implies the symmetry of the domain $\Omega$. However, beyond these well-known
examples, there exist more solutions to OEPs, apart from the previously
mentioned trivial ones. The first non-symmetric solutions to an OEP in the
Euclidean space were constructed by P. Sicbaldi in [55]. He showed the
existence of a family of solutions to (LABEL:DP) and (0.1) in
$\mathbb{R}^{n}$, $n\geq 3$, and $f(t)=\lambda t$ (a linear function). These
solutions were perturbations of a straight cylinder
$\mathbb{B}^{n-1}\times\mathbb{R}$ obtained by reformulating the
overdetermined problem in terms of a differential operator and applying the
Crandall-Rabinowicz bifurcation theorem. Later, many non-trivial solutions
have been discovered for various OEPs using bifurcation of symmetric domains
in more general Riemannian manifolds (cf. [13, 17, 39, 48, 50] and references
therein). Serrin’s seminal work [52] holds significance, not only for the
classification result itself, but for introducing the ‘moving plane method” in
the study of partial differential equations (PDEs) and, since then, this
method has been applied to prove similar results in other space forms, the
closed hemisphere $\textup{cl}(\mathbb{S}^{n}_{+})$ and hyperbolic space
$\mathbb{H}^{n}$, or extended to unbounded domains (cf. [3, 4, 14, 15, 16, 32,
36, 45] and references therein). Another widely-used method in the study of
OEPs is the one introduced by Weingberger [58]. Weingberger employed the
maximum principle with a specific function (a “P-function”) and a Pohozaev
identity to establish Serrin’s classification in an elementary manner. This
method, now known as the “P-function method”, has been successfully adapted to
different equations. For example, it has been employed to prove rigidity
statements in space forms (cf. [8, 43]), and product manifolds (cf. [18]).
While various analytical methods have been applied to study overdetermined
problems (e.g., Morse theory [40, 41] and complex analysis [28, 16]), in
recent years methods derived from differential geometry, particularly those
related to the theory of minimal and constant mean curvature (CMC)
hypersurfaces, have been successfully applied to OEPs in space forms (cf. [11,
16, 47]). Indeed, Serrin’s theorem in [52] can be seen as the OEP version of
Alexandrov’s well-known theorem regarding the uniqueness of compact embedded
CMC hypersurfaces in Euclidean space (cf. [2]). An interesting connection
between OEPs and the theory of minimal surfaces can be found in the paper by
Hélein et al. [23], where they established a link between a certain type of
minimal surfaces and solutions to the problem (LABEL:DP) and (0.1) in
$\Omega\subset\mathbb{R}^{n}$ with $f\equiv 0$, the “one phase problem”. In
particular, they constructed a Weierstrass-like representation of solutions to
the previous problem, which Traizet subsequently used in [57] to classify
solutions to the problem. Another connection, with greater relevance to the
objective of this paper in which the theory of minimal surfaces is employed to
classify solutions to an OEP, is presented in R. Souam’s paper [53] where he
investigates the overdetermined eigenvalue problem:

$\left\\{\begin{matrix}\Delta^{\hbox{\bbp S}^{2}}\xi+2\xi=0&\text{ in
}&\Omega,\\\\[5.69054pt] \xi=const.&\text{ along
}&\partial\Omega,\\\\[5.69054pt] \frac{\partial\xi}{\partial\nu}=const.&\text{
along }&\partial\Omega,\end{matrix}\right.$ (0.2)

being $\Omega$ a $\mathcal{C}^{2}$ domain in $(\mathbb{S}^{2},g_{\hbox{\bbp
S}^{2}})$, the unit sphere with the round metric. It is a known fact (cf. [22,
46]) that if $\xi$ solves the first equation of the previous system, then the
map

$X(z):=\nabla^{\hbox{\bbp S}^{2}}\xi(z)+\xi(z)\cdot z,\quad z\in\Omega$ (0.3)

produces a branched minimal surface. Conversely, the support function of any
branched minimal surface parameterized by its Gauss map provides a solution to
the equation $\Delta^{\hbox{\bbp S}^{2}}\xi+2\xi=0$. Using this correspondence
and Nitsche Theorem [38], Souam established that if $(\Omega,\xi)$ is a
solution to (0.2) and $\Omega$ is simply connected, then $\Omega$ must be a
geodesic disk and $\xi$ must exhibit rotational symmetry. Recall that, as said
above, the moving plane method in the sphere can only be applied for domains
contained in a closed hemisphere $\textup{cl}(\mathbb{S}^{n}_{+})$. Recently,
Espinar-Mazet [16] extended the classification of simply connected domains
(not necessarily contained in any closed hemisphere) in $\hbox{\bb S}^{2}$
supporting an overdetermined solution to $\Delta^{\hbox{\bbp
S}^{2}}\xi+f(\xi)=0$ as the rotationally symmetric ones for a wider class of
functions $f$, in particular, this method classify positive solutions to (0.2)
in simply connected domains. In fact, Serrin’s result is not fully
generalizable in the sphere $\hbox{\bb S}^{2}$, for there exist non-symmetric
solutions to Serrin’s overdetermined problem with equation
$-\epsilon\Delta^{\hbox{\bbp S}^{2}}\xi+\xi-\xi^{p}$ ($\epsilon>0,p>1$)
defined in simply connected domains (c.f. [49]). This paper focuses on
positive solutions to the eigenvalue problem:

$\left\\{\begin{matrix}\Delta^{\hbox{\bbp S}^{2}}\xi+2\xi=0&\text{ in
}&\Omega,\\\\[5.69054pt] \xi=0&\text{ along
}&\partial\Omega.\end{matrix}\right.$ (0.4)

Here, we assume that $\Omega\subset\hbox{\bb S}^{2}$ is a
$\mathcal{C}^{2}$-domain and that $\xi$ is of class $\mathcal{C}^{2}$ up to
the boundary, so it follows that $\partial\Omega$ is analytic (and so is the
function up to the boundary) because of the regularity results of [29]. Since
overdetermined positive solutions to (0.4) on simply connected domains are
rotationally symmetric (cf. [16, 53]), we consider “finite type domains”,
which are the complements of a finite number of disjoint simply connected
domains within $\mathbb{S}^{2}$. The concept at hand is to employ the
correlation established in [53], albeit in a distinct fashion, and the
methodologies recently advanced in [1] for the classification of solutions to
the Serrin equation within annular domains in $\mathbb{R}^{2}$. Such
techniques (cf. also [1, 5, 6, 7]) are particularly interesting for solutions
to OEPs in domains in the sphere where the moving plane method is not fully
available. We aim to classify rotationally symmetric solutions to (0.4) under
certain conditions, which, in turn, leads to the classification of minimal
surfaces with specific properties (see Section 6). In particular, we focus on
solutions to (0.4) that exhibit infinitely many maximum points. Our primary
classification result is as follows:

> Theorem A: Let $(\Omega,\xi)$ be a positive solution to (0.4), with $\Omega$
> being a topological annulus with $\mathcal{C}^{2}$ boundary. Suppose that
> $\xi$ has infinitely many maximum points and that the norm of its gradient
> is locally constant along the boundary, i.e.,
>
> $|\nabla^{\hbox{\bbp
> S}^{2}}\xi|=b_{\Gamma}>0,\quad\Gamma\in\pi_{0}(\partial\Omega),$ (0.5)
>
> where $b_{\Gamma}$ is a constant for each connected component $\Gamma$ of
> $\partial\Omega$. Then $\Omega$ is a rotationally symmetric neighborhood of
> an equator, and $\xi$ exhibits rotational symmetry with respect to the axis
> perpendicular to the plane defining this equator.

######  Remark 0.1.

It is worth noting that the constants considered in (0.5) could be distinct
for different connected components.

Using the correspondence established in Souam’s paper, we can derive an
intriguing consequence of Theorem A. Let $\Sigma\subset\mathbb{R}^{3}$ be an
open embedded minimal annulus with boundary, each boundary component
intersecting orthogonally a sphere centered at the origin (possibly of
different radius). Such surfaces are referred to as “minimal surfaces with
free boundaries” (see Definition 5.1). It is evident that if $\xi$ is a
solution to (0.2), and we consider the parametrization (0.3), we obtain a
surface of this type (a priori, not necessarily embedded). We arrive at the
following result:

> Theorem B: Let $\Sigma\subset\mathbb{B}^{3}$ be an embedded minimal annulus
> with free boundaries, and suppose its support function (or distance
> function) has infinitely many critical points. Then $\Sigma$ is a piece of a
> rotationally symmetric catenoid.

######  Remark 0.2.

It is not hard to see that, in the above situation, the set of critical points
of the support function coincides with the set of critical points of the
distance (to the origin) function.

Notably, if $\Sigma\subset\mathbb{B}^{3}$ (where $\mathbb{B}^{3}$ is the
Euclidean unit ball centered at the origin) and
$\partial\Sigma\subset\mathbb{S}^{2}$, then $\Sigma$ is a “free boundary
surface” within the unit ball. Such surfaces have been extensively studied in
recent years (see [9] or [33] for a survey of recent results). It is important
to note that there exists a unique embedded catenoid that intersects the unit
sphere $\mathbb{S}^{2}$ orthogonally, known as the “critical catenoid” (see
[12]). Consequently, an immediate consequence of the aforementioned result is:

> Corollary C: Let $\Sigma\subset\mathbb{B}^{3}$ be an embedded free boundary
> minimal annulus, and suppose its support function has infinitely many
> critical points. Then $\Sigma$ is the critical catenoid.

######  Remark 0.3.

To our knowledge, this is the first classification result of the critical
catenoid that does not use Schoen-Fraser classification result by first
Steklov eigenfunctions (c.f. [20, Theorem 1.2]).

The question of whether the critical catenoid is the only embedded free
boundary minimal annulus within the unit ball remains open. In recent years,
this question has garnered significant attention within the mathematical
community. The prevailing belief is that the answer is affirmative, and
numerous partial results have been obtained to substantiate this belief (see,
e.g., [12, 19, 31, 35]). This corollary can be regarded as a small
contribution to the proof of the so-called “critical catenoid conjecture,” an
outstanding open problem in the field.

######  Remark 0.4.

The tools employed in this paper have the potential for extension to positive
solutions of the general eigenvalue problem, specifically, positive solutions
to $\Delta\xi+\lambda\xi=0$ within the domain $\Omega$ and incorporating
overdetermined boundary conditions. However, our primary focus lies in
exploring the geometric applications associated with the particular case of
$\lambda=2$.

###  0.1 Organization of the paper

Our primary focus in Section 1 is to compute the rotationally symmetric
solutions to (0.4). We present these solutions as a one-parameter family of
normalized functions $(\Omega_{R},\xi_{R})$ that depends on the height
relative to the horizontal equator of $\mathbb{S}^{2}$, $R\in[0,1]$. We also
interpret this family of solutions as the support functions of a family of
vertical catenoids. In Section 2, we narrow our focus to a region
$\mathcal{U}$ within $\Omega$ without maximum points for a particular positive
solution $(\Omega,\xi)$ to (0.4). We introduce a parameter $R(\mathcal{U})$ in
a similar manner to [1]. We then establish that if $\Omega$ is not a
topological disk, then $R(\mathcal{U})=\bar{R}$ is well-defined and takes
values in $[0,1)$ (Theorem 2.3). This allows us to associate a model solution
$(\Omega_{\bar{R}},\xi_{\bar{R}})$ with a general solution within the region
$\mathcal{U}$. Section 3 introduces what is referred to as a “pseudo-radial
function” (see Definition 3.3), which we use to derive estimates for the norm
of the gradient of $\xi$ along its level sets (see Theorem 3.1) and for the
geodesic curvature of the level sets (see Theorem 3.2) within $\mathcal{U}$.
We also provide a relation between the length of the zero level set and the
top level set (Proposition 3.21). In Section 4, we examine positive solutions
to the OEP (0.4) and (0.5). We establish an estimate for the length of the
zero level sets of the region $\partial\mathcal{U}$ using the overdetermined
condition (Proposition 4.1). This estimate, combined with those obtained in
Section 4, enables us to conclude that either $\mathcal{U}\subset\Omega$ or
$\Omega\setminus\mathcal{U}$ is contained within an open hemisphere. The
statement of Theorem A follows from the moving plane method. In fact, we are
able to establish a more general result, as indicated in Theorem 4.1. In
Section 5, we delve into the geometric consequences of the classification
results obtained in Section 4. Through the study of minimal surfaces with free
boundaries, we ascertain that an embedded minimal annulus with each boundary
component intersecting a sphere (centered at the origin) orthogonally, must
possess an injective Gauss map and its support function must have a definite
sign. This immediately leads to the statement of Theorem B and Corollary C
using the correspondence from Souam’s paper.

##  1 The model solutions

In this section, we will describe the family of positive and bounded
rotationally symmetric solutions $(\Omega_{R},\xi_{R})$ to the Dirichlet
problem (0.4), depending on a parameter $R\in[0,1]$. In this family,
$\Omega_{0}$ is a symmetric tubular neighborhood of an equator and
$\Omega_{1}$ is an open hemisphere. Let $\hbox{\bb R}^{3}$ denote the usual
Euclidean space, where $(x,y,z)$ represent cartesian coordinates and
$\left\langle\cdot,\cdot\right\rangle$ is the Euclidean scalar product. We
parametrize the unit sphere in cylindrical coordinates:

$\hbox{\bb
S}^{2}=\left\\{(\sqrt{1-r^{2}}\cos\theta,\sqrt{1-r^{2}}\sin\theta,r)\,:\,r\in[-1,1],\,\theta\in[0,2\pi)\right\\},$

hence the induced metric in $\hbox{\bb S}^{2}$ in these coordinates is given
by

$g_{\hbox{\bbp S}^{2}}=\frac{1}{1-r^{2}}dr^{2}+(1-r^{2})d\theta^{2}.$

Define the orthonormal frame in $\hbox{\bb S}^{2}\setminus\left\\{{\bf s},{\bf
n}\right\\}$, where ${\bf s}:=(0,0,-1)$ and ${\bf n}:=(0,0,1)$ denote the
south and north pole respectively, given by

$n=\sqrt{1-r^{2}}\partial_{r}\quad\text{and}\quad
t=\frac{1}{\sqrt{1-r^{2}}}\partial_{\theta}.$

Then, the Christoffel symbols associated to this frame can be computed from

$\nabla_{n}n=0,\qquad\nabla_{n}t=0,\qquad\nabla_{t}n=-\frac{r}{\sqrt{1-r^{2}}}t,\qquad\nabla_{t}t=\frac{r}{\sqrt{1-r^{2}}}n.$

From now on, as we will be always considering differential equations on the
sphere, we denote by $\nabla$, $\Delta$ and $\nabla^{2}$ the gradient,
laplacian and hessian operators in $\mathbb{S}^{2}$.

###  1.1 Rotationally symmetric solutions

From the first equation of (0.4) using the cylindrical coordinates, assuming
that $\xi$ doesn’t depend on $\theta$, we obtain

$(1-r^{2})\frac{\partial^{2}\xi}{\partial r^{2}}-2r\frac{\partial\xi}{\partial
r}+2\xi=0.$

This is an ordinary differential equation of second order, and it has a two-
parameter family of solutions:

$\xi_{\alpha,\beta}(r)=\alpha(1-r\,\textup{arctanh}(r))+\beta r,\quad
r\in(-1,1),\quad\alpha,\beta\in\mathbb{R}.$ (1.1)

Hence, a function $\xi_{\alpha,\beta}$ will be a solution to (0.4) if there
exist $a,b\in[-1,1]$, $a<b$, depending on $\alpha$ and $\beta$ such that

$\xi_{\alpha,\beta}(a)=\xi(b)=0\quad\text{and}\quad
0<\left(\xi_{\alpha,\beta}\right)_{|(a,b)}<+\infty.$

  * •

Case 1: $\alpha=0$ and $\beta\neq 0$: We obtain the reescale height function
$\xi_{0,\beta}(r)=\beta r$. This is a monotone function that takes the zero
value only at the origin, and we obtain the solution
$(\Omega_{1},\xi_{0,\beta})$, where $\Omega_{1}$ is the open hemisphere
centered either at the north pole ${\bf n}$ if $\beta>0$ or at the south pole
${\bf s}$ if $\beta<0$.

  * •

Case 2: $\alpha\neq 0$ and $\beta=0$: We obtain the function

$\xi_{\alpha,0}(r)=\alpha(1-r\,\text{arctanh}(r)),\quad r\in(-1,1).$

If we assume that $\alpha>0$, then $\xi_{\alpha,0}(r)$ has a positive maximum
at the origin since $\xi_{\alpha,0}(-r)=\xi_{\alpha,0}(r)$, it is monotonic in
both intervals $(-1,0)$ and $(0,1)$ and

$\lim_{r\to 1^{-}}\xi_{\alpha,0}(r)=\lim_{r\to-1^{+}}\xi_{\alpha,0}=-\infty.$

Hence, it is clear that there exists $\bar{r}\in(0,1)$ such that
$\xi_{\alpha,0}(\bar{r})=\xi_{\alpha,0}(-\bar{r})=0$ and
$\left(\xi_{\alpha,0}\right)_{|(-\bar{r},\bar{r})}>0$. In this case, we can
easily compute that

$\frac{\partial\xi_{\alpha,0}}{\partial r}(0)=0\text{ and
}(\xi_{\alpha,0})_{{\rm max}}=\xi_{\alpha,0}(0)=\alpha.$

Therefore, $(\Omega_{0},\xi_{\alpha,0})$ is solution to (0.4) such that
$\Omega_{0}$ is a symmetric tubular neighborhood of the equator $\hbox{\bb
S}^{2}\cap\left\\{z=0\right\\}$. Finally, observe that
$\xi_{-\alpha,0}(r)=-\xi_{\alpha,0}(r)$ which means that it is negative in
$(-\bar{r},\bar{r})$ and positive in $(-1,1)\setminus[-\bar{r},\bar{r}]$.
However,

$\lim_{r\to
1^{-}}\xi_{-\alpha,0}(r)=\lim_{r\to-1^{+}}\xi_{-\alpha,0}=+\infty,$

that is, $\xi_{-\alpha,0}$ is not bounded.

  * •

Case 3: $\alpha\neq 0$ and $\beta\neq 0$: Consider $\alpha>0$. Since

$\xi_{\alpha,\beta}(r)=\alpha\,\xi_{1,\omega}(r)\text{ and
}\xi_{1,\omega}(-r)=\xi_{1,-\omega}(r)$

where $\omega=\beta/\alpha\neq 0$, we can write solutions to (1.1) in a more
interesting way

$\xi_{\alpha,\beta}(r)=\alpha\,\xi_{1,\omega}(r),\text{ where
}\xi_{1,\omega}(r)=1-r\,\textup{arctanh}(r)+\omega r.$

We first compute the zeros of $\xi_{\alpha,\beta}$. Since
$\xi_{1,\omega}(0)=1$ and $\lim_{r\to
1^{-}}\xi_{1,\omega}(r)=\lim_{r\to-1^{+}}\xi_{1,\omega}=-\infty$, then
$\xi_{1,\omega}$ there exist uniques $r_{-}\in(-1,0)$ and $r_{+}\in(0,1)$ such
that $\xi_{1,\omega}(r_{-})=\xi_{1,\omega}(r_{+})=0$. Moreover,

$-\frac{\partial\xi_{1,\omega}}{\partial
r}(r)=\text{arctanh}(r)+\frac{r}{1-r^{2}}-\omega,\quad r\in(r_{-},r_{+})$

being

$\frac{\partial\xi_{1,\omega}}{\partial
r}(r_{-})=-\frac{1}{r_{-}(1-r_{-}^{2})}>0\text{ and
}\frac{\partial\xi_{1,\omega}}{\partial
r}(r_{+})=-\frac{1}{r_{+}(1-r_{+}^{2})}<0,$

so we conclude that there exists a unique $R\in(r_{-},r_{+})$ such that
$\frac{\partial\xi_{1,\omega}}{\partial r}(R)=0$, which is a maximum. We have

$\omega=\text{arctanh}(R)+\frac{R}{1-R^{2}},$ (1.2)

and the value of the maximum of $\xi_{\alpha,\beta}$ is given by

$(\xi_{\alpha,\beta})_{\textup{max}}=\alpha\,\xi_{1,\omega}(R)=\frac{\alpha}{1-R^{2}}.$

Finally, we note that if $\alpha<0$ then $\xi_{\alpha,\beta}$ has a negative
sign in $(r_{-},r_{+})$ and a positive sign in $(-1,1)\setminus[r_{-},r_{+}]$;
but $\xi_{\alpha,\beta}$ is not bounded in $(-1,1)\setminus[r_{-},r_{+}]$.

Summarizing, we have obtained a two-parameter family of solutions to the
Dirichlet problem (0.4), where the parameters are a scale parameter $\alpha$
and a height parameter $R$, where $R$ indicates the parallel in $\hbox{\bb
S}^{2}$ where the curve of maximum points is located. In fact, we can describe
the solutions previously obtained as

$\xi_{\alpha,R}(r)=\alpha\left(1-r\textup{arctanh}(r)+\left(\textup{arctanh}(R)+\frac{R}{1-R^{2}}\right)r\right),\quad\forall
r\in[r_{-}(R),r_{+}(R)].$

However, note that the eigenvalue equation of the Laplace-Beltrami operator on
a general Riemannian manifold is invariant under a change of scale. Thus, the
parameter $\alpha$ is not important and for each $R$ we can fix a value of the
scale to obtain a 1-parameter family of rotationally symmetric solutions to
(0.4). Although the most natural way of obtaining the 1-parameter family is to
fix $\alpha=1$, we will use a different normalization for geometrical reasons.

Figure 1: The diagram shows the graphs of the functions that define the model
solutions $\xi_{R}$ with $R\in\\{0,0.25,0.5,0.75\\}$. In the blue part, the
function is increasing, while in the red part, the function is decreasing, and
the function takes its maximum values at the point $x=R$. With the
normalization at hand, the value of $(\xi_{R})_{\textup{max}}$ goes to zero as
$R$ approaches $1$, so $\xi_{R}\to 0$ in compact sets away from the north
pole. In order to embrace the positive solution in the disk-type domain
$\Omega_{1}$ we have to lose continuity at $R=1$ in the family $\xi_{R}$ (cf.
Proposition 1.1).

######  Proposition 1.1.

Let $(\Omega,\xi)$ be a positive bounded rotationally symmetric solution to
(0.4). Then, up to a rotation, a reflection with respect to the plane
$\\{z=0\\}$ and a dilation, $(\Omega,\xi)=(\Omega_{R},\xi_{R})$ for
$R\in[0,1]$ as described below:

  * •

If $R=1$, $(\Omega_{1},\xi_{1})$ is a solution to (0.4) in a disk-type domain,
where $\Omega_{1}$ is the open hemisphere centered at the north pole ${\bf n}$
and $\xi_{1}$ is the height function $\xi_{1}(r)=r$.

  * •

If $R\in[0,1)$, then $(\Omega_{R},\xi_{R})$ is a solution to (0.4) in a
annular-type domain given by

$\xi_{R}(r)=\alpha(R)\left(1-r\,\textup{arctanh}(r)+\left(\textup{arctanh}(R)+\frac{R}{1-R^{2}}\right)r\right),\quad\forall
r\in[r_{-}(R),r_{+}(R)],$

where $-1<r_{-}(R)<0<r_{+}(R)<1$ with $-r_{-}(0)=r_{+}(0)=\bar{r}$, being
$\alpha(R)=r_{+}(R)\sqrt{1-r_{+}(R)^{2}}$ and

$\Omega_{R}=\left\\{\left(\sqrt{1-r^{2}}\cos(\theta),\sqrt{1-r^{2}}\sin(\theta),r\right)\in\hbox{\bb
S}^{2}~{}\colon~{}r\in[r_{-}(R),r_{+}(R)],\,\theta\in[0,2\pi)\right\\}.$

Moreover,

$\max_{\textup{cl}(\Omega)}\xi_{R}:=(\xi_{R})_{\textup{max}}=\xi_{R}(R)=\frac{\alpha(R)}{1-R^{2}}=\frac{r_{+}(R)\sqrt{1-r_{+}(R)^{2}}}{1-R^{2}}$

and we denote the boundary components as

$\Gamma_{\pm}(R)=\left\\{\left(\sqrt{1-r_{\pm}(R)^{2}}\cos(\theta),\sqrt{1-r_{\pm}(R)^{2}}\sin(\theta),r_{\pm}(R)\right)\in\hbox{\bb
S}^{2}~{}\colon~{}\theta\in[0,2\pi)\right\\},$ (1.3)

where $\Gamma_{+}(R)$ and $\Gamma_{-}(R)$ are called the upper and lower
component of $\partial\Omega_{R}$ respectively (with respect to the equator
$\hbox{\bb S}^{2}\cap\\{z=0\\}$).

Observe that the constants $r_{-}(R)$ and $r_{+}(R)$ are smooth functions of
$R$ since they are implicit solutions to the equation

$1-r\,\textup{arctanh}(r)+\left(\textup{arctanh}(R)+\frac{R}{1-R^{2}}\right)r=0.$
(1.4)

Figure 2: The graphs of $r_{+}$ (above) and $r_{-}$ (below) as functions of
the parameter $R$. One can see that $-r_{-}(0)=r_{+}(0)$ and $r_{+}\to 1$ and
$r_{-}\to 0$ as $R$ goes to $1$.

Taking into account that $r_{\pm}(R)\neq 0$ for some $R\in[0,1]$ by (1.4), we
have that

$\frac{\partial r_{\pm}}{\partial
R}(R)=\frac{2r_{\pm}^{2}(1-r_{\pm}^{2})}{(1-R^{2})^{2}},\quad\forall
R\in[0,1),$ (1.5)

so both $r_{-}$ and $r_{+}$ are increasing functions of $R$. The following
properties of our family of solutions are straightforward.

######  Proposition 1.2.

Let $(\Omega_{R},\xi_{R})$, $R\in[0,1)$, be one of the model solutions
described in Proposition 1.1. Set $\alpha(R)=\alpha$. Then we have the
following identities:

  * •

The derivatives of the function up to second order:

$\frac{\partial\xi_{R}}{\partial
r}(r)=\frac{1}{r}\left(\xi_{R}-\frac{\alpha}{1-r^{2}}\right)\quad\textup{and}\quad\frac{\partial^{2}\xi_{R}}{\partial
r^{2}}(r)=-\frac{2\alpha}{(1-r^{2})^{2}}.$

  * •

The gradient and Hessian :

$\begin{split}&\nabla\xi_{R}(r)=\frac{\sqrt{1-r^{2}}}{r}\left(\xi_{R}-\frac{\alpha}{1-r^{2}}\right)n\quad\textup{and}\\\
&\nabla^{2}\xi_{R}(r)=\begin{pmatrix}-\xi_{R}-\frac{\alpha}{1-r^{2}}&0\\\
0&-\xi_{R}+\frac{\alpha}{1-r^{2}}\end{pmatrix}.\end{split}$

  * •

The norm of the gradient along the zero level sets:

$\left|\nabla\xi_{R}\right|=1\,\textup{along
}\Gamma_{+}(R)\quad\textup{and}\quad\left|\nabla\xi_{R}\right|=\frac{r_{+}\sqrt{1-r_{+}^{2}}}{r_{-}\sqrt{1-r_{-}^{2}}}\,\textup{
along }\Gamma_{-}(R).$

Moreover, the following identity follows away from the critical points of
$\xi_{R}$:

$\nabla^{2}\xi_{R}(r)=\left(\frac{r^{2}\alpha}{((1-r^{2})\xi_{R}-\alpha)^{2}}\left|\nabla\xi_{R}\right|-\xi\right)g_{\hbox{\bbp
S}^{2}}-\frac{2\alpha r^{2}}{((1-r^{2})\xi_{R}-\alpha)^{2}}d\xi_{R}\otimes
d\xi_{R}.$ (1.6)

###  1.2 Geometric interpretation

Let us consider the minimal catenoid $C_{\alpha,\omega}$, defined as the
unique rotationally symmetric minimal surface around the $z-$axis, symmetric
with respect to the plane $\left\\{z=-\omega\right\\}$ and necksize $\alpha$.
This catenoid can be parameterized by

$\psi_{\alpha,\omega}(r,\theta)=\alpha\left(\frac{\cos\theta}{\sqrt{1-r^{2}}},\frac{\sin\theta}{\sqrt{1-r^{2}}},{\rm
arctanh}(r)-\omega\right),\,\,r\in(-1,1)\text{ and }\theta\in[0,2\pi),$ (1.7)

with outward unit normal

$N(r,\theta)=\left(\sqrt{1-r^{2}}\,\cos\theta,\sqrt{1-r^{2}}\,\sin\theta,-r\right)\in\hbox{\bb
S}^{2}\setminus\left\\{{\bf s},{\bf n}\right\\}.$

The support function $\xi(r,\theta)=\langle\psi(r,\theta),N((r,\theta)\rangle$
is then given by

$\xi(r,\theta)=\alpha(1-r\,{\rm arctanh}(r)+\omega r).$

The coordinates $(r,\theta)\in(-1,1)\times[0,2\pi)$ can be seen as a
parametrization of $\hbox{\bb S}^{2}\setminus\left\\{{\bf s},{\bf n}\right\\}$
via the unit normal $N$. As we noted in the introduction, the support function
$\xi$ satisfies

$\Delta\xi+2\xi=0\text{ in }\hbox{\bb S}^{2}\setminus\left\\{{\bf s},{\bf
n}\right\\}.$

Thus, the rotationally symmetric solutions in annular domains described above
are nothing but the support function of a given catenoid in $\hbox{\bb R}^{3}$
(up to scaling and vertical translation) defined in the appropriate spherical
domain by the Gauss map. Now we can use this correspondence to derive some
geometric properties of the functions defined in the previous section. On the
one hand, note that the zero level set of $\xi$ gives us the points of
$C_{\alpha,\omega}$ where the position vector
$\psi(r,\theta)=|\psi(r,\theta)|$ is orthogonal to the normal vector
$N(r,\theta)$. Note also that $\xi$ depends only on $r$, so the image via
$\psi$ of the zero level set of $\xi$ on $C_{\alpha,\omega}$ are horizontal
circles contained in spheres (of different radii except for $\omega=0$)
centered at the origin, and $C_{\alpha,\omega}$ intersects these spheres
orthogonally. On the other hand, note that

$\lim_{r\to\pm 1}N(r,\theta)=\mp(0,0,1)\quad\textup{and}\quad\lim_{r\to\pm
1}\sqrt{1-r^{2}}\psi(r,\theta)=\alpha(\cos\theta,\sin\theta,0),$

and it is clear that the function $f(r):=\sqrt{1-r^{2}}\left({\rm
arctanh}(r)-\omega\right)$ satisfies $f((-1,1))=(-1,1)$. Thus, we can deduce
geometrically that there exist uniques $-1<r_{-}<0<r_{+}<1$ such that
$\xi(r_{+})=\xi(r_{-})=0$. Let us denote the image in $C_{\alpha,\omega}$ of
the two components of the zero level set of $\xi$ as
$\zeta_{\pm}=\\{\psi(r_{\pm},\theta)~{}\colon~{}\theta\in[0,2\pi)\\}$. Now,
observe that

$\left|\psi(r_{\pm},\theta)\right|^{2}=\alpha^{2}\left(\frac{1}{1-r_{\pm}^{2}}+\left(\textup{artanh}(r_{\pm})-\omega\right)^{2}\right)=\frac{\alpha^{2}}{r_{\pm}^{2}\left(1-r_{\pm}^{2}\right)},$
(1.8)

so taking $\alpha=r_{+}\sqrt{1-r_{+}^{2}}$ we obtain that $\zeta_{+}$ is
contained in the unit sphere $\mathbb{S}^{2}=\mathbb{S}^{2}(0,1)$. On the
other hand, it easy to check that the function
$f(R)=r_{+}^{2}(R)(1-r_{+}^{2}(R))/r_{-}^{2}(R)(1-r_{-}^{2}(R))$ with
$R\in[0,1)$ has derivative

$f^{\prime}(R)=\frac{4r_{+}^{2}(R)(1-r_{+}^{2}(R))r_{-}^{2}(R)(1-r_{-}^{2}(R))}{(1-R^{2})^{2}r_{-}^{4}(R)(1-r_{-}^{2}(R))^{2}}\left((r_{+}-2r_{+}^{3})-(r_{-}-2r_{-}^{3})\right),$

so taking into account that $\bar{r}>0.8$ it can be easily checked that
$f^{\prime}(R)<0$ for all $R\in[0,1)$. We deduce then that there exist
$\tilde{r}\leq 1$ such that $\zeta_{-}\in\mathbb{S}^{2}(\tilde{r})$, where
$\mathbb{S}^{2}(\tilde{r})$ denotes the euclidean sphere centered at the
origin and radius $\tilde{r}$. Thus, we conclude that solutions $\xi_{R}$
described in Proposition 1.1 correspond to support functions of pieces of
catenoids

$C_{R}:=\left\\{\psi_{\alpha(R),\omega(\alpha(R))}(r,\theta)~{}\colon~{}r\in(r_{-}(R),r_{+}(R)),\,\theta\in[0,2\pi)\right\\},$
(1.9)

where

$\alpha(R)=r_{+}(R)\sqrt{1-r_{+}^{2}(R)}\quad\textup{and}\quad\omega(R)=\textup{arctanh}(R)+\frac{R}{1-R^{2}},$

and $\psi_{\alpha(R),\omega(R)}$ is given by (1.7). It holds that
$C_{R}\subset\mathbb{B}^{3}$ for each $R\in[0,1)$. Also, since
$-r_{-}(0)=r_{+}(0)=\bar{r}$, it is clear that $C_{0}$ corresponds to the
critical catenoid, i.e. the unique free boundary minimal catenoid in the unit
sphere. Note that $N(C_{R})=\Omega_{-R}$ is the reflection with respect to the
plane $\\{z=0\\}$ of $\Omega_{R}$ for each $R\in[0,1)$.

Figure 3: Here, two model catenoids defined in (1.9). Left: model catenoid
with parameter $R=1/2$. It can be seen that $\zeta_{+}$ is contained in the
unit sphere and $\zeta_{-}$ is contained in a sphere of smaller radius. Right:
critical catenoid, which corresponds to $C_{0}$.

##  2 Normalized Wall Shear Stress

From now on, $\Omega$ will always denote a finite type region in the two-
sphere.

######  Definition 2.1.

Let $\Omega\subset\mathbb{S}^{2}$ be a domain with $\mathcal{C}^{2}$ boundary.
Then we say that $\Omega$ is of finite type if there exist a finite number of
disjoint simply connected domains $D_{1},\dots,D_{k}\subset\mathbb{S}^{2}$
such that

$\Omega=\mathbb{S}^{2}\setminus\left(\cup_{i=1}^{k}\textup{cl}(D_{i})\right).$
(2.1)

If $\Omega$ is a finite type domain with $k\geq 1$ boundary components, we
will always write $\partial\Omega=\Gamma_{1}\cup\dots\cup\Gamma_{k}$, being
$\Gamma_{i}=\partial D_{i}$ for $i\in\\{1,\dots,k\\}$. Following the ideas in
[1], in this section we will classify solutions $(\Omega,\xi)$ to the
eigenvalue problem

$\left\\{\begin{matrix}\Delta\xi+2\xi=0&\text{ in }&\Omega,\\\\[5.69054pt]
\xi>0&\text{ in }&\Omega,\\\\[5.69054pt] \xi=0&\text{ along
}&\partial\Omega,\end{matrix}\right.$ (2.2)

in terms of its normalized wall shear stress, a scale-invariant quantity. By
the maximum principle, a solution $(\Omega,\xi)$ to (2.2) does not have
interior minimums. Hence, the set of interior critical points

${\rm Crit}(\xi):=\left\\{p\in\Omega\,:\,\,\nabla\xi(p)=0\right\\}$

contains only saddle and maximum points. Given a solution $(\Omega,\xi)$ to
(2.2) we denote by

${\rm
Max}(\xi):=\left\\{p\in\Omega\,:\,\,\xi(p)=\xi_{\textup{max}}:=\max_{\textup{cl}(\Omega)}\xi\right\\},$

and it is clear that this set must be non-empty. We will refer to ${\rm
Max}(\xi)$ as the top level set of $\xi$.

######  Remark 2.1.

If a solution $(\Omega,\xi)$ to (2.2) has a connected component
$\mathcal{U}\subset\Omega\setminus\textup{Max}(\xi)$ such that
$\textup{cl}(\mathcal{U})\cap\partial\Omega=\emptyset$, then either
$\xi=\xi_{\textup{max}}$ in $\mathcal{U}$ or there exists an interior minimum
in $\mathcal{U}$, a contradiction in any case by the maximum principle.

We also observe a structure result for the top level set:

######  Lemma 2.1.

Let $(\Omega,\xi)$ be a solution to (2.2), $\Omega\subset\mathbb{S}^{2}$ of
finite type. Suppose that $\xi$ has infinite points of maximum, i.e.,
$\\#\textup{Max}(\xi)=+\infty$. Then

$\textup{Max}(\xi)=\gamma^{0}\cup\gamma^{1},$

where $\gamma^{0}$ is a finite set (could be empty) of points and $\gamma^{1}$
is a finite set of disjoint analytic closed curves. Moreover, for each
$\gamma\in\gamma^{1}$ it holds that
$\Omega\setminus\gamma=\Omega^{\gamma}_{1}\cup\Omega^{\gamma}_{2}$ with
$\partial\Omega^{\gamma}_{i}\cap\partial\Omega\neq\emptyset$ for $i=1,2$.

###### Proof.

We use the same arguments as in the proof of [1, Theorem D]. Solutions to
(2.2) are real analytic (cf. [29]), so by the Lojasiewicz structure theorem it
follows that $\textup{Max}(\xi)=\gamma^{0}\cup\gamma^{1}$, where $\gamma^{0}$
is a set of isolated points and $\gamma^{1}$ is a set of analytic curves. As
$\gamma^{0}\subset\mathbb{S}^{2}$ must be compact, it follows that
$\gamma^{0}$ is a finite set of isolated points, and by [7, Corollary 3.4.] we
get that $\gamma^{1}$ must be a finite set of analytic closed curves, possibly
intersecting at a finite set of points. Furthermore, the curves of
$\gamma^{1}$ are disjoint since, at the intersection points of two curves, the
hessian of $\xi$ must vanish, which contradicts (2.2). Finally, Remark 2.1
implies that $\gamma\in\gamma^{1}$ is not contractible in $\Omega$, that is,
$\Omega\setminus\gamma$ cannot contain a topological disk component $D$ such
that $\partial D\cap\partial\Omega=\emptyset$. Hence,
$\Omega\setminus\gamma=\Omega^{\gamma}_{1}\cup\Omega^{\gamma}_{2}$ with
$\partial\Omega^{\gamma}_{i}\cap\partial\Omega\neq\emptyset$ for $i=1,2$, as
claimed. ∎

The above lemma motivates the following

######  Definition 2.2.

Let $(\Omega,\xi)$ be a solution to (2.2) and assume that
$\gamma\subset\textup{Max}(\xi)$ is a closed embedded curve. Then,
$\Omega\setminus\gamma=\Omega_{1}^{\gamma}\cup\Omega_{2}^{\gamma}$ and we will
say that $\Omega_{i}^{\gamma}$, $i=1,2$, is the partition with respect to
$\gamma$.

Now, we are ready to introduce the definition of normalised wall shear stress
(NWSS):

######  Definition 2.3.

Let $(\Omega,\xi)$ be a solution to (2.2) and let
$\Gamma\in\pi_{0}(\partial\Omega)$ be a connected component of the boundary.
We define the normalised wall shear stress (NWSS) of $\Gamma$ as

$\overline{\tau}(\Gamma):=\frac{\max_{\Gamma}\left|\nabla\xi\right|}{\xi_{\textup{max}}}.$
(2.3)

If now $\mathcal{U}$ is a connected component of
$\Omega\setminus\textup{Max}(\xi)$,
$\partial\Omega\cap\textup{cl}(\mathcal{U})\neq\emptyset$, we define the NWSS
of $\mathcal{U}$ as

$\overline{\tau}(\mathcal{U}):=\max\left\\{\overline{\tau}(\Gamma)~{}\colon~{}\Gamma\in\pi_{0}(\partial\Omega\cap\textup{cl}(\mathcal{U}))\right\\}.$
(2.4)

Otherwise, we set $\overline{\tau}(\mathcal{U})=0$.

The rigidity results proved in the rest of the paper are derived by comparing
geometric quantities of a fixed solution to the eigenvalue problem
$(\Omega,\xi)$ with those of a certain model solution $(\Omega_{R},\xi_{R})$;
which motivates the following:

######  Definition 2.4.

We say that a solution $(\Omega,\xi)$ to (2.2) is equivalent to a model
solution $(\Omega_{R},\xi_{R})$, for some $R\in[0,1]$, if they differ up to a
rotation and a dilation. In such case, we will denote
$(\Omega,\xi)\equiv(\Omega_{R},\xi_{R})$.

In order to choose the appropriate model solution to compare with
$(\Omega,\xi)$, we will use the NWSS defined previously.

###  2.1 NWSS on the model solutions

The NWSS in our model solutions at each of the components of the boundary can
be computed in terms of the parameter $R$. In fact,

$\overline{\tau}_{\pm}(R):=\overline{\tau}\left(\Gamma_{\pm}(R)\right)=\pm\frac{1-R^{2}}{r_{\pm}(R)\sqrt{1-r_{\pm}(R)^{2}}},\quad
R\in[0,1)$ (2.5)

where $\Gamma_{+}(R)$ and $\Gamma_{-}(R)$ are defined in (1.3). Moreover, it
can be shown that $\overline{\tau}_{-}$ and $\overline{\tau}_{+}$ are
invertible functions. Set

$\overline{\tau}_{\pm}(0)=\frac{1}{\bar{r}\sqrt{1-\bar{r}^{2}}}=:\tau_{0}>1\quad\left(\textup{where}\quad
1-\bar{r}\,\textup{arctanh}(\bar{r})=0\right),$ (2.6)

where, recall, $r_{+}(0)=-r_{-}(0)=\bar{r}>0$.

######  Lemma 2.2.

The functions $\overline{\tau}_{+}:[0,1)\rightarrow[\tau_{0},+\infty)$ and
$\overline{\tau}_{-}:[0,1)\rightarrow(1,\tau_{0}]$ defined by (2.5) are
increasing and decreasing respectively.

###### Proof.

First, remember that in the previous section we proved

$\text{arctanh}(r_{\pm})-\frac{1}{r_{\pm}}=\text{arctanh}(R)+\frac{R}{1-R^{2}},\quad\forall
R\in[0,1),$ (2.7)

so we have that $r_{+}(R)\rightarrow 1$ and $r_{-}(R)\rightarrow 0$ as
$R\rightarrow 1$, and then by (1.5) we get

$r_{-}:[0,1)\rightarrow[-\bar{r},0)\quad\textup{and}\quad
r_{+}:[0,1)\rightarrow[\bar{r},1)$

are increasing functions of $R$.

> Claim A: $\overline{\tau}_{+}$ and $\overline{\tau}_{-}$ are monotonic.

###### Proof of Claim A.

We will see that the derivative of both functions cannot vanish. By
contradiction, suppose there exists $\bar{R}\in(0,1)$ such that
$\overline{\tau}_{\pm}^{\prime}(\bar{R})=0$. Then, by (2.5), we get

$2\bar{R}\,r_{\pm}(\bar{R})(1-r_{\pm}(\bar{R})^{2})+(1-2r_{\pm}(\bar{R})^{2})(1-\bar{R}^{2})r_{\pm}^{\prime}(\bar{R})=0$

and hence

$\frac{\partial r_{\pm}}{\partial
R}(\bar{R})=\frac{2(1-r_{\pm}(\bar{R})^{2})r_{\pm}(\bar{R})\bar{R}}{(1-\bar{R}^{2})(2r_{\pm}(\bar{R})^{2}-1)},$
(2.8)

so, comparing to (1.5), we get

$r_{\pm}(\bar{R})(2r_{\pm}(\bar{R})^{2}-1)+(1-\bar{R}^{2})\bar{R}=0.$ (2.9)

  * •

Assume first that $\bar{\tau}^{\prime}_{+}(\bar{R})=0$. Since
$r_{+}(R)\geq\bar{r}>0.8$ for all $R\in[0,1]$, then $2r_{+}^{2}(\bar{R})-1>0$,
which contradicts (2.9).

  * •

Second, assume that $\bar{\tau}^{\prime}_{-}(\bar{R})=0$. Then, (2.9) implies
that $2r_{-}(\bar{R})^{2}-1>0$ and, by (2.8), we get $\frac{\partial
r_{-}}{\partial R}(\bar{R})<0$ since $r_{-}(\bar{R})<0$, which contradicts
(1.5).

This proves Claim A. ∎

Next, we study the limit of $\overline{\tau}_{\pm}$ as $R$ goes to one. From
(2.7) we obtain

$\frac{1+r_{\pm}(R)}{1+R}(1-R)=(1-r_{\pm}(R))\textup{exp}\left(\frac{2}{r_{\pm}(R)}+\frac{2R}{1-R^{2}}\right),\quad\forall
R\in[0,1)$

that yields

$\lim_{R\to
1^{-}}(1-r_{\pm}(R))\textup{exp}\left(\frac{2}{r_{\pm}(R)}+\frac{2R}{1-R^{2}}\right)=0.$

Since $r_{+}(R)\to 1$ as $R\to 1$ and $e^{x}\geq x^{2}$ for all $x\geq 0$, we
get

$0=\lim_{R\to 1^{-}}\left|(1-r_{+}^{2}(R)){\rm
exp}\left(\frac{1}{1-R^{2}}\right)\right|\geq\lim_{R\to
1^{-}}\left|\frac{1-r_{+}^{2}(R)}{(1-R^{2})^{2}}\right|$

and we conclude that $\overline{\tau}_{+}\rightarrow+\infty$ as $R\rightarrow
1$. On the other hand, using again (2.7) and since $r_{-}(R)\to 0$ as $R\to 1$
it is easy to conclude that

$\begin{split}1&=\lim_{R\to
1^{-}}\frac{\textup{arctanh}(r_{-}(R))-\frac{1}{r_{-}(R)}}{\textup{arctanh}(R)+\frac{R}{1-R^{2}}}=\lim_{R\to
1^{-}}-\frac{\frac{1}{r_{-}(R)}}{\textup{arctanh}(R)+\frac{R}{1-R^{2}}}\\\
&=\lim_{R\to
1^{-}}-\frac{1-R^{2}}{r_{-}(R)\left((1-R^{2})\textup{arctanh}(R)+R\right)}=\lim_{R\to
1^{-}}-\frac{1-R^{2}}{r_{-}(R)},\end{split}$

and by the last equality we conclude that $\overline{\tau}_{-}(R)\to 1$ as
$R\to 1$. This finishes the proof. ∎

Looking at the previous lemma, one can figure out how to choose a model
solution associated to an arbitrary solution to the eigenvalue problem
$(\Omega,\xi)$: fixed $\mathcal{U}\subset\Omega\setminus\textup{Max}(\xi)$, if
we have that $\overline{\tau}(\mathcal{U})>1$ then there exists a unique model
solution $(\Omega_{R},\xi_{R})$ such that
$\overline{\tau}(\mathcal{U})=\overline{\tau}(\mathcal{V})$ for one of the
components $\mathcal{V}$ of $\Omega_{R}\setminus\textup{Max}(\xi_{R})$. This a
priori is not a good correspondence, as one could have a solution to the
eigenvalue problem with NWSS less or equal to one in a connected component of
$\Omega\setminus\textup{Max}(\xi)$. The key fact is that then one can prove
that the solution must be $(\Omega_{1},\xi_{1})$, so we can rule out this
case. To prove that we will need two technical results, which will be proven
using $P$-functions.

###  2.2 Associated $P$-function

We will find a $P$-function associated with (2.2), that is, a subharmonic
function associated with a solution to (2.2). The next result does not need to
impose boundary conditions.

######  Proposition 2.1.

Let $\Omega\subset\hbox{\bb S}^{2}$ be a smooth domain and
$\xi:\Omega\to\hbox{\bb R}$ be a smooth solution to $\Delta\xi+2\xi=0$. Then,
the $P$-function $\mathcal{P}:=|\nabla\xi|^{2}+\xi^{2}$ satisfies

$\Delta\mathcal{P}\geq 0\text{ in }\Omega.$ (2.10)

###### Proof.

By the Bochner-Weitzenböck formula, a solution $\xi$ to (2.2) satisfies

$\frac{1}{2}\Delta\left|\nabla\xi\right|^{2}=\left|\nabla^{2}\xi\right|^{2}-\left|\nabla\xi\right|^{2},$

and a standard computation shows

$\frac{1}{2}\Delta\xi^{2}=\xi\Delta\xi+\left|\nabla\xi\right|^{2}=-\frac{(\Delta\xi)^{2}}{2}+\left|\nabla\xi\right|^{2}.$

Thus, summing up the above two equations, and the Geometric-Arithmetic
Inequality, we obtain

$\frac{1}{2}\Delta\mathcal{P}=\left|\nabla^{2}\xi\right|^{2}-\frac{(\Delta\xi)^{2}}{2}\geq
0,$

that is, (2.10) holds. ∎

As a consequence of the above $P$-function, we can obtain a characterization
of disk-type solutions to (2.2) in terms of the normalized wall shear stress
(NWSS) of the boundary components. Specifically

######  Theorem 2.1.

Let $\Omega\subset\hbox{\bb S}^{2}$ be a smooth domain and
$\xi:\Omega\to\hbox{\bb R}$ be a non-vanishing smooth solution to
$\Delta\xi+2\xi=0$ in $\Omega$ and $\xi=0$ along $\partial\Omega$. Assume that

$\bar{\tau}(\Gamma)\leq 1\text{ for all }\Gamma\in\pi_{0}(\partial\Omega),$
(2.11)

then $\Omega$ is a geodesic disk and $\xi$ is rotationally symmetric.

###### Proof.

Observe that (2.11) and $\xi=0$ along $\partial\Omega$ imply that

${\rm max}_{\partial\Omega}\mathcal{P}\leq\xi_{\textup{max}}^{2}.$

Let $x\in\Omega$ be a point where $\xi^{2}$ attains its maximum, hence
$\xi(x)\nabla\xi(x)=0$ and $\xi(x)\neq 0$, thus $|\nabla\xi|(x)=0$ and
$\mathcal{P}(x)=\xi_{\textup{max}}^{2}$. Therefore, the strong maximum
principle implies that $\mathcal{P}=\xi_{\textup{max}}^{2}$ on
$\overline{\Omega}$ and $\nabla^{2}\xi=-\xi g_{\hbox{\bbp S}^{2}}$; now it
follows that $\Omega$ is a geodesic ball and $\xi$ is rotationally symmetric
using the same argument as in the proof of Lemma 2.4 of [8]. ∎

Now we will introduce an energy function along level sets of solutions to
(2.2) and we will show an important property of such energy function when
$\overline{\tau}(\mathcal{U})\leq 1$. Given
$\mathcal{U}\in\pi_{0}(\Omega\setminus\textup{Max}(\xi))$, we define the
energy function $E:[0,\xi_{\textup{max}})\to\hbox{\bb R}$ given by

$E(t)=\frac{1}{\xi_{\textup{max}}^{2}-t^{2}}\int_{\textup{cl}(\mathcal{U})\cap\left\\{\xi=t\right\\}}|\nabla\xi|.$
(2.12)

Note that $E$ is continuous in regular values of $[0,\xi_{\textup{max}})$
since the Lebesgue integral is absolutely continuous.

######  Lemma 2.3.

Let $(\Omega,\xi)$ be a solution to (2.2). Suppose that there exists
$\mathcal{U}\in\pi_{0}(\Omega\setminus\textup{Max}(\xi))$ such that
$\overline{\tau}(\mathcal{U})\leq 1$. Then $E$ is non-increasing.

###### Proof.

Given $\epsilon>0$, consider the inner region

$\mathcal{U}_{\epsilon}:=\left\\{p\in\mathcal{U}~{}\colon~{}\xi(p)<\xi_{\textup{max}}-\epsilon\right\\}\subset\mathcal{U}.$
(2.13)

Since the critical values of $\xi$ in $\mathcal{U}$ are isolated (cf. [54]),
then we can choose $\epsilon$ small enough so that
$\xi_{\textup{max}}-\epsilon$ is a regular value, and hence, the set
$\mathcal{U}\cap\\{\xi=\xi_{\textup{max}}-\epsilon\\}$ is the disjoint union
of a finite number of analytic curves by the Lojasiewicz structure theorem.
Using (2.10) and the maximum principle, we obtain that

$\max_{\mathcal{U}_{\epsilon}}\mathcal{P}\leq\max_{\partial\mathcal{U}_{\epsilon}}\mathcal{P}.$

On the one hand, $|\nabla\xi|\rightarrow 0$ as $\epsilon\rightarrow 0$ yields

$\lim_{\epsilon\to
0^{+}}\max_{\\{\xi=\xi_{\textup{max}}-\epsilon\\}}\mathcal{P}=\xi_{\textup{max}}^{2}.$

On the other hand, from Remark 2.1 and $\overline{\tau}(\mathcal{U})\leq 1$,
we get $|\nabla\xi|^{2}+\xi^{2}=|\nabla\xi|^{2}\leq\xi_{\textup{max}}^{2}$
along $\partial\Omega\cap\textup{cl}(\mathcal{U})$. Hence, we conclude that

$|\nabla\xi|^{2}+\xi^{2}=\mathcal{P}\leq\xi_{\textup{max}}^{2}\quad\textup{in}\quad\mathcal{U}.$

Hence, using $\Delta\xi+2\xi=0$ in $\Omega$, we obtain

$\begin{split}{\rm
div}\left(\frac{\nabla\xi}{\xi_{\textup{max}}^{2}-\xi^{2}}\right)=\frac{2\xi}{(\xi_{\textup{max}}^{2}-\xi^{2})^{2}}\left(\mathcal{P}-\xi_{\textup{max}}^{2}\right),\end{split}$

and, since $\xi\geq 0$ in $\Omega$ and $\mathcal{P}\leq\xi_{\textup{max}}^{2}$
in $\mathcal{U}$, we get

${\rm div}\left(\frac{\nabla\xi}{\xi_{\textup{max}}^{2}-\xi^{2}}\right)\leq
0\quad\textup{in}\quad\mathcal{U}.$

Now, we choose regular values $0\leq t_{1}<t_{2}<\xi_{\textup{max}}$ and we
integrate the above inequality along the finite perimeter set
$\\{t_{1}\leq\xi\leq t_{2}\\}$. Then, applying the Divergence Theorem:

$\begin{split}0\geq&\int_{\left\\{t_{1}\leq\xi\leq t_{2}\right\\}}{\rm
div}\left(\frac{\nabla\xi}{\xi_{\textup{max}}^{2}-\xi^{2}}\right)\\\
&=\int_{\left\\{u=t_{1}\right\\}}\langle\frac{\nabla\xi}{\xi_{\textup{max}}^{2}-\xi^{2}},n\rangle+\int_{\left\\{u=t_{2}\right\\}}\langle\frac{\nabla\xi}{\xi_{\textup{max}}^{2}-\xi^{2}},n\rangle\\\
&=-\frac{1}{\xi_{\textup{max}}^{2}-t_{1}^{2}}\int_{\left\\{u=t_{1}\right\\}}|\nabla\xi|+\frac{1}{\xi_{\textup{max}}^{2}-t_{2}^{2}}\int_{\left\\{u=t_{2}\right\\}}|\nabla\xi|\\\
&=-E(t_{1})+E(t_{2}),\end{split}$

where we have used that $\nu=-\nabla\xi/|\nabla\xi|$ along
$\left\\{u=t_{1}\right\\}$ and $\nu=\nabla\xi/|\nabla\xi|$ along
$\left\\{u=t_{2}\right\\}$ are the outer unit normal respectively. Therefore,

$E(t_{1})\geq E(t_{2})\text{ for }0\leq t_{1}<t_{2}<\xi_{\textup{max}}\text{
regular values}.$

Since $E$ is continuous and the set of critical values is finite, then $E$ is
a non-increasing function. ∎

###  2.3 Expected critical height

As said above, the NWSS on the model solutions will allow us to define an
expected value for any solution to (2.2). Specifically:

######  Definition 2.5.

Let $(\Omega,\xi)$ be a solution to (2.2) and
$\mathcal{U}\in\pi_{0}(\Omega\setminus\textup{Max}(\xi))$. Set $\tau_{0}$ as
in (2.6), if $\Omega$ is not a topological disk then we define the expected
critical height of $\mathcal{U}$ as follows:

  * •

if $\overline{\tau}(\mathcal{U})<\tau_{0}$, we set

$\displaystyle\bar{R}(\mathcal{U})=\overline{\tau}_{-}^{-1}\left(\overline{\tau}(\mathcal{U})\right),$
(2.14)

  * •

if $\overline{\tau}(\mathcal{U})\geq\tau_{0}$, we set

$\displaystyle\bar{R}(\mathcal{U})=\overline{\tau}_{+}^{-1}\left(\overline{\tau}(\mathcal{U})\right).$
(2.15)

If $\Omega$ is a topological disk, then we set $\bar{R}(\mathcal{U})=1$.

Observe that in (2.14) we are assuming implicitly that, if $\Omega$ is not a
topological disk, then $\overline{\tau}(\mathcal{U})>1$ for each connected
component $\mathcal{U}$ of $\Omega\setminus\textup{Max}(\xi)$. In order to
prove that Definition 2.5 is consistent we first show:

######  Theorem 2.2.

Let $(\Omega,\xi)$ be a solution to (2.2) and
$\mathcal{U}\in\pi_{0}(\Omega\setminus\textup{Max}(\xi))$. If
$\overline{\tau}(\mathcal{U})\leq 1$, then
$(\Omega,\xi)\equiv(\Omega_{1},\xi_{1})$.

###### Proof.

By Lemma 2.1, $\textup{Max}(\xi)=\gamma^{0}\cup\gamma^{1}$, where $\gamma^{1}$
is the disjoint union of a finite number of analytic curves and $\gamma^{0}$
is a finite set of isolated points. We first show the result if
$\gamma^{1}=\emptyset$. In such a case, $\Omega\setminus\textup{Max}(\xi)$
would be connected and Theorem 2.1 implies that $(\Omega,\xi)$ is a
rotationally symmetric solution on a disk, so it must be
$(\Omega_{1},\xi_{1})$ after a rescaling. Hence, we can assume that
$\gamma^{1}$ is non-empty and is in contact with $\mathcal{U}$. Since
$\gamma^{1}\cap\textup{cl}(\mathcal{U})$ is a set of analytic closed curves,
by the Lojasiewicz inequality (see [7, Theorem 2.1]), there exists a
neighborhood $V$ of $\gamma^{1}\cap\textup{cl}(\mathcal{U})$,
$V\cap\gamma^{0}=\emptyset$, and two constants $c>0$ and $1/2\leq\theta<1$
such that

$\xi_{\textup{max}}-\xi<1\text{ and }\left|\nabla\xi\right|(x)\geq
c(\xi_{\textup{max}}-\xi(x))^{\theta},\quad\forall x\in V.$

Then, since $\\{\xi=t\\}\cap\textup{cl}(\mathcal{U})\subset V$ as
$t\rightarrow\xi_{\textup{max}}$, using the energy function (2.12) we get the
following chain of inequalities

$\begin{split}\frac{1}{\xi_{\textup{max}}^{2}}\int_{\textup{cl}(\mathcal{U})\cap\left\\{\xi=\xi_{\textup{max}}\right\\}}|\nabla\xi|=E(0)\geq^{\text{Lemma
}\ref{LemDecreasing}}E(t)&=\frac{1}{\xi_{\textup{max}}^{2}-t^{2}}\int_{\textup{cl}(\mathcal{U})\cap\left\\{\xi=t\right\\}}|\nabla\xi|\\\
&\geq\frac{1}{\xi_{\textup{max}}-t}\int_{\textup{cl}(\mathcal{U})\cap\left\\{\xi=t\right\\}}|\nabla\xi|\\\
&\geq\frac{c}{(\xi_{\textup{max}}-t)^{1-\theta}(\xi_{\textup{max}}+t)}\left|\\{\xi=t\\}\cap\textup{cl}(\mathcal{U})\right|,\end{split}$

We can assume that all $t$’s in the previous inequalities are regular values
of $\xi$, as we know by [54, Theorem 1] that if $Z$ is the set of critical
points of $\xi$ then $\xi(K\cap Z)$ is finite for any compact set
$K\subset\Omega$. Then
$\left|\\{\xi=t\\}\cap\textup{cl}(\mathcal{U})\right|>0$. Letting
$t\rightarrow\xi_{\textup{max}}$ we obtain that $E(0)=+\infty$, a
contradiction. ∎

Finally, we are ready to prove that Definition 2.5 is well-posed:

######  Theorem 2.3.

Let $(\Omega,\xi)$ be a solution to (2.2) and
$\mathcal{U}\in\pi_{0}(\Omega\setminus\textup{Max}(\xi))$. Then, the expected
critical height of $\mathcal{U}$, $\bar{R}(\mathcal{U})$, is well defined and
takes values in $[0,1]$. Moreover, $\bar{R}(\mathcal{U})=1$ if, and only if,
$(\Omega,\xi)\equiv(\Omega_{1},\xi_{1})$.

###### Proof.

First, from Remark 2.1, any
$\mathcal{U}\in\pi_{0}(\Omega\setminus\textup{Max}(\xi))$ must satisfy
$\partial\Omega\cap\overline{\mathcal{U}}\neq\emptyset$ and, from (2.3),
$\overline{\tau}(\mathcal{U})\geq 0$. Observe that, by definition,
$R(\mathcal{U})=1$ if, and only if, $\overline{\tau}(\mathcal{U})\in[0,1]$.
Then, Theorem 2.2 implies that $(\Omega,\xi)=(\Omega_{1},\xi_{1})$ up to a
rotation and a dilation. Therefore, there exists a bijective correspondence
between $\bar{R}(\mathcal{U})\in[0,1)$ and
$\bar{\tau}(\mathcal{U})\in(1,+\infty)$ by Lemma 2.2, which finishes the
proof. ∎

##  3 Comparison geometry

Now, using the expected critical height we can associate to each general
solution $(\Omega,\xi)$ one of the model solutions defined in Section 1.

######  Definition 3.1.

Let $(\Omega,\xi)$ be a solution to (2.2),
$\mathcal{U}\in\pi_{0}\left(\Omega\setminus\textup{Max}(\xi)\right)$ and set
$\bar{R}:=R(\mathcal{U})$ the expected critical height of $\mathcal{U}$. Then
we say that $(\Omega_{\bar{R}},\xi_{\bar{R}})$ is the associated model
solution to $\xi$ inside $\mathcal{U}$.

######  Remark 3.1.

From now on, we omit the parameter $\bar{R}$ and we only write
$(\bar{\Omega},\bar{\xi}):=(\Omega_{\bar{R}},\xi_{\bar{R}})$, we also denote
the extremal of the interval of definition and the scale parameter of
$\bar{\xi}$ as $\bar{r}_{\pm}:=r_{\pm}(\bar{R})$ and
$\bar{\alpha}=\alpha(\bar{R})$ respectively, since $\bar{R}$ is already fixed.

In this section, we will construct a function to compare the level sets of
$\xi$ with those of our model solution. To construct the comparison function,
we need our general function $\xi$ to be normalized.

######  Definition 3.2.

Let $(\Omega,\xi)$ be a solution to (2.2),
$\mathcal{U}\in\pi_{0}\left(\Omega\setminus\textup{Max}(\xi)\right)$ and set
$\bar{R}:=R(\mathcal{U})$ the expected critical height of $\mathcal{U}$. Then
we will say that the solution $\xi$ is normalized if
$\xi_{\textup{max}}=(\xi_{R})_{\textup{max}}$.

Let us suppose that $\xi$ is normalized, consider the function
$F:[0,\xi_{\textup{max}}]\times[\bar{r}_{-},\bar{r}_{+}]\to\hbox{\bb R}$ given
by

$F(\xi,r)=\xi-\bar{\alpha}(1-r\,{\rm arctanh}(r)+\omega r),$

where $\omega$ is given by (1.2) and $\bar{\alpha}$ is given in Proposition
1.1. Hence

$\bar{\alpha}^{-1}\frac{\partial F}{\partial r}(r)={\rm
arctanh}(r)+\frac{r}{1-r^{2}}-r.$

and therefore $\frac{\partial F}{\partial r}=0$ if, and only if, $r=\bar{R}$.
As a consequence of the Implicit Function Theorem, there exist two smooth
functions

$\chi_{-}:[0,\xi_{\max}]\to[\bar{r}_{-},\bar{R}]\quad\textup{and}\quad\chi_{+}:[0,\xi_{\max}]\to[\bar{R},\bar{r}_{+}]$

such that

$F(\xi,\chi_{\pm}(\xi))=0\quad\textup{for
all}\quad\xi\in[0,\xi_{\textup{max}}].$

Now we can define a pseudo-radial function following [1, Definition 3].

######  Definition 3.3.

Let $\mathcal{U}\subset\Omega\setminus\textup{Max}(\xi)$ be a connected
component and set $\tau_{0}$ as in (2.6). Then:

  * •

If $\overline{\tau}(\mathcal{U})\geq\tau_{0}$, we define the pseudo-radial
function $\Psi_{+}$ associated with the region $\mathcal{U}$ as

$\begin{split}\Psi_{+}:\,&\mathcal{U}\to[\bar{R},\bar{r}_{+}]\\\
&\,p\,\,\mapsto\Psi_{+}(p):=\chi_{+}\left(\xi(p)\right).\end{split}$ (3.1)

  * •

If $\overline{\tau}(\mathcal{U})<\tau_{0}$, we define the pseudo-radial
function $\Psi_{-}$ associated with the region $\mathcal{U}$ as

$\begin{split}\Psi_{-}:\,&\mathcal{U}\to[\bar{r}_{-},\bar{R}]\\\
&\,p\,\,\mapsto\Psi_{-}(p):=\chi_{-}\left(\xi(p)\right).\end{split}$ (3.2)

######  Remark 3.2.

Note that if $\mathcal{U}\subset\Omega\setminus\textup{Max}(\xi)$ is such that
$\overline{\tau}(\mathcal{U})=\tau_{0}$ then we can define the pseudo-radial
function associated to $\mathcal{U}$ as $\Psi_{+}:\mathcal{U}\to[0,\bar{r}]$
or as $\Psi_{-}:\mathcal{U}\to[-\bar{r},0]$. We can use both definitions
because the model solution $(\Omega_{0},\xi_{0})$ is symmetric with respect to
the meridian $\\{r=0\\}$ in cylindrical coordinates in the sphere, and both
components of
$\Omega_{0}\setminus\textup{Max}(\xi_{0})=\Omega\setminus\\{r=0\\}$ have the
same NWSS equal to $\tau_{0}$. To fix one, we choose to define the pseudo-
radial function as (3.1) when $\overline{\tau}(\mathcal{U})=\tau_{0}$.

######  Remark 3.3.

In order to perform some formal computations using the pseudo-radial function,
it is important to do previous considerations about the notation. First, we
will denote $\Psi_{\pm}=\Psi$ and $\chi_{\pm}=\chi$ and we will do the
computations considering both possibilities at the same time. On the other
hand, note that $\chi$ is a function of $\xi$ as a real parameter, while
$\Psi$ is a function defined on a domain in the sphere. Also, we denote the
derivatives with respect to $\xi$ with a dot. For example, $\dot{\chi}$ will
denote the derivative of $\chi$ with respect to the parameter $\xi$.

A straightforward computation gives

$0=\frac{dF}{d\xi}(\xi,\chi_{\pm}(\xi))=\frac{\partial
F}{\partial\xi}(\xi,\chi_{\pm}(\xi))+\frac{\partial F}{\partial
r}(\xi,\chi_{\pm}(\xi))\dot{\chi}_{\pm}(\xi)$

and

$0=\frac{d^{2}F}{d\xi^{2}}(\xi,\chi_{\pm}(\xi))=\frac{\partial^{2}F}{\partial
r^{2}}(\xi,\chi_{\pm}(\xi))\dot{\chi}_{\pm}(\xi)^{2}+\frac{\partial
F}{\partial r}(\xi,\chi_{\pm}(\xi))\ddot{\chi}_{\pm}(\xi)$

where we have used that $\frac{\partial
F}{\partial\xi}(\xi,\chi_{\pm}(\xi))=1$ and
$\frac{\partial^{2}F}{\partial\xi^{2}}(\xi,\chi_{\pm}(\xi))=0=\frac{\partial^{2}F}{\partial
r\partial\xi}(\xi,\chi_{\pm}(\xi))$. We get

$\dot{\chi}_{\pm}(\xi)=-\frac{\partial F}{\partial
r}(\xi,\chi_{\pm}(\xi))^{-1}=\mp\sqrt{1-\chi_{\pm}(\xi)^{2}}\left|\nabla\bar{\xi}\right|(\chi_{\pm}(\xi))^{-1}$
(3.3)

and

$\ddot{\chi}_{\pm}(\xi)=-\frac{\partial^{2}F}{\partial
r^{2}}(\xi,\chi_{\pm}(\xi))\frac{\partial F}{\partial
r}(\xi,\chi_{\pm}(\xi))^{-3}=\mp\frac{2\bar{\alpha}\left|\nabla\bar{\xi}\right|(\chi_{\pm}(\xi))^{-3}}{\sqrt{1-\chi_{\pm}(\xi)^{2}}}.$
(3.4)

We can also compute

$\nabla\Psi=\dot{\chi}\nabla\xi\,\text{ and
}\,\nabla^{2}\Psi=\dot{\chi}\nabla^{2}\xi+\ddot{\chi}d\xi\otimes d\xi.$

###  3.1 Gradient estimates

First, we will use the pseudo-radial function to compare the norm of the
gradient of the solution $\xi$ along level sets with that of the model
solution $\bar{\xi}$. Suppose that the function $\xi$ is normalized and
consider the comparison function

$W_{\bar{R}}=\left|\nabla\bar{\xi}\right|^{2}\circ\Psi=\frac{1-\Psi^{2}}{\Psi^{2}}\left(\xi-\frac{\bar{\alpha}}{1-\Psi^{2}}\right)^{2},$
(3.5)

and set

$W=\left|\nabla\xi\right|^{2}\text{ in }\mathcal{U}.$ (3.6)

Our aim is to prove the following result:

######  Theorem 3.1.

Let $(\Omega,\xi)$ be a solution to (2.2),
$\mathcal{U}\in\pi_{0}\left(\Omega\setminus\textup{Max}(\xi)\right)$ and set
$\bar{R}=R(\mathcal{U})\in[0,1)$ the expected critical height of
$\mathcal{U}$. Let $(\bar{\Omega},\bar{\xi})$ be its associated model solution
and assume that $\xi$ is normalized. Then, it holds

$W(p)\leq W_{\bar{R}}(p)\,\text{ for all }p\in\mathcal{U},$

moreover, if the equality holds at one single point of $\mathcal{U}$, then
$(\Omega,\xi)\equiv(\bar{\Omega},\bar{\xi})$.

To prove the previous theorem we will need to know how the previous functions
behaves near the top level set $\textup{Max}(\xi)$. The following lemma will
be useful later (cf. [1]).

######  Lemma 3.1.

Let $(\Omega,\xi)$, $(\bar{\Omega},\bar{\xi})$ and $\mathcal{U}$ be as in
Theorem 3.1 and consider $W_{\bar{R}}$ defined in (3.5). Then, if
$p\in\textup{Max}(\xi)$, it holds:

$\lim_{x\in\mathcal{U},x\to
p}\frac{W_{\bar{R}}}{\xi_{\textup{max}}-\xi}=4\xi_{\textup{max}}.$

###### Proof.

The proof of this lemma is a simple computation using the Taylor expansions of
$W_{\bar{R}}$ and $\xi_{\textup{max}}-\xi$ as functions of $\Psi$. Observe
that taking $z=\Psi-\bar{R}$ then it is clear that

$\begin{split}\xi_{\textup{max}}-\xi=\frac{\bar{\alpha}}{1-\bar{R}^{2}}-\xi&=\frac{\bar{\alpha}}{(1-\bar{R}^{2})^{2}}(\Psi-\bar{R})^{2}+O((\Psi-\bar{R})^{3})\\\
&=\frac{\bar{\alpha}}{(1-\bar{R}^{2})^{2}}z^{2}+O(z^{3}).\end{split}$ (3.7)

Note that we have done the Taylor expansion of $\xi\circ\Psi$ at
$\Psi=\bar{R}$, although $\Psi(p)$ is not defined in $\textup{Max}(\xi)$.
However, it is clear that we can extend $\Psi$ to $\textup{Max}(\xi)$ by
$\Psi(p)=\bar{R}$ for all $p\in\textup{Max}(\xi)$, so the previous expansion
makes sense. On the other hand, observe that

$W_{\bar{R}}(\Psi)=\frac{1-\Psi^{2}}{\Psi^{2}}\left(\xi-\frac{\bar{\alpha}}{1-\Psi^{2}}\right)^{2}=(1-\Psi^{2})\left(\frac{\partial\xi}{\partial\Psi}\right)^{2}.$

Hence

$\frac{\partial W_{\bar{R}}}{\partial\Psi}(\bar{R})=0\text{ and
}\frac{\partial^{2}W_{\bar{R}}}{\partial\Psi^{2}}(\bar{R})=2(1-\bar{R}^{2})\left(\frac{\partial^{2}\xi}{\partial\Psi^{2}}(\bar{R})\right)^{2}=\frac{8\bar{\alpha}^{2}}{(1-\bar{R}^{2})^{3}},$

so the Taylor expansion of $W_{\bar{R}}$ up to second order is

$W_{\bar{R}}=\frac{4\bar{\alpha}^{2}}{(1-\bar{R}^{2})^{3}}z^{2}+O(z^{3}).$
(3.8)

Then it is clear that

$\lim_{x\in\mathcal{U},x\to
p}\frac{W_{\bar{R}}}{\xi_{\textup{max}}-\xi}=\lim_{z\to
0}\frac{\frac{4\bar{\alpha}^{2}}{(1-\bar{R}^{2})^{3}}z^{2}+O(z^{3})}{\frac{\bar{\alpha}}{(1-\bar{R}^{2})^{2}}z^{2}+O(z^{3})}=\frac{4\bar{\alpha}}{1-\bar{R}^{2}}$

and the result is proved. ∎

Next, we shall establish some differential inequalities. Firstly, we want to
bound the norm of the hessian of $\xi$ in terms of the functions defined in
(3.5) and (3.6). From (1.6), we obtain the following expression:

$\displaystyle 0$ $\displaystyle\leq$
$\displaystyle\left|\nabla^{2}\xi+\frac{2\bar{\alpha}\Psi^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}d\xi\otimes
d\xi+\left(\xi-\frac{\Psi^{2}\bar{\alpha}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left|\nabla\xi\right|^{2}\right)g_{\hbox{\bbp
S}^{2}}\right|^{2}$ $\displaystyle=$
$\displaystyle\left|\nabla^{2}\xi\right|^{2}+\frac{4\bar{\alpha}^{2}\Psi^{4}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{4}}\left|d\xi\otimes
d\xi\right|^{2}+\left(\xi-\frac{\Psi^{2}\bar{\alpha}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left|\nabla\xi\right|^{2}\right)\left|g_{\hbox{\bbp
S}^{2}}\right|^{2}$
$\displaystyle+\frac{4\bar{\alpha}\Psi^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left(\xi-\frac{\Psi^{2}\bar{\alpha}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left|\nabla\xi\right|^{2}\right)\left\langle
d\xi\otimes d\xi,g_{\hbox{\bbp S}^{2}}\right\rangle$
$\displaystyle+\frac{4\bar{\alpha}\Psi^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left\langle\nabla^{2}\xi,d\xi\otimes
d\xi\right\rangle+2\left(\xi-\frac{\Psi^{2}\bar{\alpha}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left|\nabla\xi\right|^{2}\right)\left\langle\nabla^{2}\xi,g_{\hbox{\bbp
S}^{2}}\right\rangle.$

Using that

$\left|d\xi\otimes d\xi\right|^{2}=\left|\nabla\xi\right|^{4},\,\,\left\langle
d\xi\otimes d\xi,g_{\hbox{\bbp
S}^{2}}\right\rangle=\left|\nabla\xi\right|^{2}\,\,\textup{and}\,\,\left\langle\nabla^{2}\xi,g_{\hbox{\bbp
S}^{2}}\right\rangle=\Delta\xi,$

we get

$\displaystyle 0$ $\displaystyle\leq$
$\displaystyle\left|\nabla^{2}\xi\right|^{2}+\frac{4\bar{\alpha}\Psi^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left\langle\nabla^{2}\xi,d\xi\otimes
d\xi\right\rangle-2\xi^{2}+\frac{2\bar{\alpha}^{2}\Psi^{4}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{4}}\left|\nabla\xi\right|^{4}$
$\displaystyle+\frac{4\xi\bar{\alpha}\Psi^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left|\nabla\xi\right|^{2},$

and then we reach the inequality

$\begin{split}\left|\nabla^{2}\xi\right|^{2}\geq&-\frac{4\bar{\alpha}\Psi^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left\langle\nabla^{2}\xi,d\xi\otimes
d\xi\right\rangle+2\xi^{2}\\\
&-\frac{2\bar{\alpha}^{2}\Psi^{4}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{4}}\left|\nabla\xi\right|^{4}-\frac{4\xi\bar{\alpha}\Psi^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left|\nabla\xi\right|^{2}.\end{split}$
(3.9)

To compute the scalar product $\left\langle\nabla^{2}\xi,d\xi\otimes
d\xi\right\rangle$ we use the following lemma which is a standard computation
in geodesic coordinates.

######  Lemma 3.2.

Let $(M,g)$ be a $n$-dimensional Riemannian manifold and
$f\in\mathcal{C}^{\infty}(M)$. Then:

$\left\langle\nabla\left|\nabla f\right|^{2},\nabla
f\right\rangle=2\nabla^{2}f(\nabla f,\nabla
f)=2\left\langle\nabla^{2}f,df\otimes df\right\rangle.$ (3.10)

Note that

$\nabla
W_{\bar{R}}=\frac{\partial}{\partial\Psi}\left(\frac{1-\Psi^{2}}{\Psi^{2}}\left(\xi-\frac{\bar{\alpha}}{1-\Psi^{2}}\right)^{2}\right)\dot{\chi}(\xi)\nabla\xi=-2\left(\xi+\frac{\bar{\alpha}}{1-\Psi^{2}}\right)\nabla\xi,$

and then using (3.10) we get

$\left\langle\nabla^{2}\xi,d\xi\otimes
d\xi\right\rangle=\frac{1}{2}\left\langle\nabla(W-W_{\bar{R}}),\nabla\xi\right\rangle-\left(\xi+\frac{\bar{\alpha}}{1-\Psi^{2}}\right)\left|\nabla\xi\right|^{2}.$

We use the previous equality and (3.9) to get

$\begin{split}\left|\nabla^{2}\xi\right|^{2}\geq&-\frac{2\bar{\alpha}\Psi^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left\langle\nabla(W-W_{\bar{R}}),\nabla\xi\right\rangle+2\xi^{2}\\\
&-\frac{2\bar{\alpha}^{2}\Psi^{4}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{4}}\left|\nabla\xi\right|^{4}+\frac{4\bar{\alpha}^{2}\Psi^{2}}{(1-\Psi^{2})\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left|\nabla\xi\right|^{2}.\end{split}$
(3.11)

Now, we want to apply the maximum principle to the function $W-W_{\bar{R}}$,
so we want this function to satisfy an elliptic inequality. With this in mind,
we compute the laplacian of this expression. Using the Bochner’s formula,

$\Delta(W-W_{\bar{R}})=2\left|\nabla^{2}\xi\right|^{2}-4\left(\xi+\frac{\bar{\alpha}}{1-\Psi^{2}}\right)+\frac{4\Psi^{2}\bar{\alpha}}{(1-\Psi^{2})\left((1-\Psi^{2})\xi-\bar{\alpha}\right)}W.$

From (3.11), we get

$\begin{split}\Delta(W-W_{\bar{R}})\geq&-\frac{4\bar{\alpha}\Psi^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left\langle\nabla(W-W_{\bar{R}}),\nabla\xi\right\rangle+\frac{8\bar{\alpha}^{2}\Psi^{2}}{(1-\Psi^{2})\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}W\\\
&-\frac{4\bar{\alpha}^{2}\Psi^{4}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{4}}W^{2}-\frac{4\xi\bar{\alpha}}{1-\Psi^{2}}+\frac{4\Psi^{2}\bar{\alpha}}{(1-\Psi^{2})\left((1-\Psi^{2})\xi-\bar{\alpha}\right)}W.\end{split}$

Then, using that

$W_{\bar{R}}=\frac{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}{\Psi^{2}(1-\Psi^{2})}$

and, after some simplifications, we get

$\begin{split}\Delta(W-W_{\bar{R}})\geq&-\frac{4\bar{\alpha}\Psi^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left\langle\nabla(W-W_{\bar{R}}),\nabla\xi\right\rangle\\\
&+\frac{4\Psi^{2}\bar{\alpha}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left(\xi-\frac{\Psi^{2}\bar{\alpha}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\left|\nabla\xi\right|^{2}\right)(W-W_{\bar{R}}).\end{split}$
(3.12)

Hence, we have arrived at a differential inequality that is not elliptic at
the quantity $W-W_{\bar{R}}$. To obtain an elliptic relation, we define the
function $F_{\beta}=\beta\cdot(W-W_{\bar{R}})$, where $\beta=\beta(\Psi)>0$ is
a function that we will determine later. First, we observe that

$\Delta
F_{\beta}=\beta\cdot\Delta(W-W_{\bar{R}})+2\left\langle\nabla\beta,\nabla(W-W_{\bar{R}})\right\rangle+(W-W_{\bar{R}})\Delta\beta.$
(3.13)

We compute the second and the third terms of the sum. Using that
$\nabla\beta=\beta^{\prime}(\Psi)\dot{\chi}(\xi)\nabla\xi$ (we use
$\beta^{\prime}$ to denote the derivative with respect to $\Psi$), we have
that

$2\left\langle\nabla\beta,\nabla(W-W_{\bar{R}})\right\rangle=2\dot{\chi}\frac{\beta^{\prime}}{\beta}\left\langle\beta\nabla(W-W_{\bar{R}}),\nabla\xi\right\rangle.$

It is clear that

$\begin{split}\left\langle\nabla
F_{\beta},\nabla\xi\right\rangle=\dot{\chi}\beta^{\prime}W(W-W_{\bar{R}})+\left\langle\beta\nabla(W-W_{\bar{R}}),\nabla\xi\right\rangle,\end{split}$

so the second term of the sum is

$2\left\langle\nabla\beta,\nabla(W-W_{\bar{R}})\right\rangle=2\dot{\chi}\frac{\beta^{\prime}}{\beta}\left\langle\nabla
F_{\beta},\nabla\xi\right\rangle-2\dot{\chi}^{2}\left(\frac{\beta^{\prime}}{\beta}\right)^{2}WF_{\beta}.$

For the third term of the sum, using the chain rule we get

$\Delta\beta=-2\xi\dot{\chi}\beta^{\prime}+\left(\ddot{\chi}\beta^{\prime}+\dot{\chi}^{2}\beta^{\prime\prime}\right)W,$

so the third term of (3.13) is

$(W-W_{\bar{R}})\Delta\beta=-2\xi\dot{\chi}\frac{\beta^{\prime}}{\beta}F_{\beta}+\left(\ddot{\chi}\frac{\beta^{\prime}}{\beta}+\dot{\chi}^{2}\frac{\beta^{\prime\prime}}{\beta}\right)WF_{\beta}.$

Finally observe that from (3.3) and (3.4) we obtain, using that
$\Psi(p)=\chi(\xi(p))$, that

$\dot{\chi}=\frac{\Psi(1-\Psi^{2})}{(1-\Psi^{2})\xi-\bar{\alpha}}\quad\textup{and}\quad\ddot{\chi}=\frac{2\bar{\alpha}\Psi^{3}(1-\Psi^{2})}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{3}},$

so combining (3.12) and (3.13), after some simplifications, it holds

$\begin{split}\Delta
F_{\beta}\geq&\frac{2\Psi(1-\Psi^{2})}{(1-\Psi^{2})\xi-\bar{\alpha}}\left(\frac{\beta^{\prime}}{\beta}-\frac{2\bar{\alpha}\Psi}{(1-\Psi^{2})\left((1-\Psi^{2})\xi-\bar{\alpha}\right)}\right)\left\langle\nabla
F_{\beta},\nabla\xi\right\rangle\\\
&-\frac{2\xi\Psi(1-\Psi^{2})}{(1-\Psi^{2})\xi-\bar{\alpha}}\left(\frac{\beta^{\prime}}{\beta}-\frac{2\Psi\bar{\alpha}}{(1-\Psi^{2})\left((1-\Psi^{2})\xi-\bar{\alpha}\right)}\right)F_{\beta}\\\
&+\frac{\Psi^{2}(1-\Psi^{2})^{2}\left|\nabla\xi\right|^{2}}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\Bigg{(}\left(\frac{\beta^{\prime}}{\beta}\right)^{\prime}-\left(\frac{\beta^{\prime}}{\beta}\right)^{2}+\frac{6\Psi\bar{\alpha}}{(1-\Psi^{2})\left((1-\Psi^{2})\xi-\bar{\alpha}\right)}\frac{\beta^{\prime}}{\beta}\\\
&-\frac{4\Psi^{2}\bar{\alpha}^{2}}{(1-\Psi^{2})^{2}\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}\Bigg{)}F_{\beta}.\end{split}$
(3.14)

Now we are ready to prove Theorem 3.1.

###### Proof of Theorem 3.1:.

We give the proof in two parts: first, we will prove that $F_{\beta}$
satisfies an elliptic inequality, hence $W\leq W_{\bar{R}}$ in $\mathcal{U}$
and $W\equiv W_{\bar{R}}$ in $\mathcal{U}$ if $W(p)=W_{\bar{R}}(p)$ at one
single point $p\in\mathcal{U}$ by the maximum principle. Second, we will show
the rigidity statement in the latter case. Define
$F_{\beta}=\beta\cdot(W-W_{\bar{R}})$, where

$\beta(\Psi):=\frac{\sqrt{1-\Psi^{2}}}{\sqrt{W_{\bar{R}}}}>0$

is a solution to the differential equation

$\frac{\beta^{\prime}}{\beta}-\frac{2\Psi\bar{\alpha}}{(1-\Psi^{2})\left((1-\Psi^{2})\xi-\bar{\alpha}\right)}=0.$

Take $\epsilon>0$ small enough and define $\mathcal{U}_{\epsilon}$ as in
(2.13), using (3.14) we get that $F_{\beta}$ satisfies the elliptic inequality

$\Delta
F_{\beta}-\frac{8\bar{\alpha}\Psi^{4}(1-\Psi^{2})\xi}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{4}}F_{\beta}\geq
0\text{ in }\mathcal{U}_{\epsilon}.$

Hence, using Lemma 3.1 and the Reverse Lojasiewicz Inequality [7, Theorem
2.2], it follows that

$\lim_{x\in\mathcal{U},x\to\textup{Max}(\xi)}\frac{W}{\sqrt{W_{\bar{R}}}}=\lim_{x\in\mathcal{U},x\to\textup{Max}(\xi)}\frac{\left|\nabla\xi\right|^{2}}{2\sqrt{\xi_{\textup{max}}}\sqrt{\xi_{\textup{max}}-\xi}}=0,$

so it is clear that

$\lim_{x\in\mathcal{U},x\to\textup{Max}(\xi)}F_{\beta}(p)=\lim_{x\in\mathcal{U},x\to\textup{Max}(\xi)}\sqrt{1-\Psi^{2}}\frac{W}{\sqrt{W_{\bar{R}}}}-\sqrt{1-\Psi^{2}}\sqrt{W_{\bar{R}}}=0.$

Therefore, applying the maximum principle to $F_{\beta}$ in
$\mathcal{U}_{\epsilon}$ and letting $\epsilon$ tend to zero (such that
$\xi_{\textup{max}}-\epsilon$ is always a regular value) we conclude the first
part of the proof. We continue with the second part of the proof. We proceed
as in [5, Theorem 4.2]. If we assume that $W(p)=W_{\bar{R}}(\Psi(\xi(p)))$ in
$\mathcal{U}$, then it follows that $W=\left|\nabla\xi\right|^{2}$ is a
positive function in $\mathcal{U}$ that only depends on $\xi$ and hence it is
constant along level sets. Therefore, there are no critical points of $\xi$ in
$\mathcal{U}$, so we can parametrize $\mathcal{U}$ by level sets with
coordinates $(\xi,\theta)$. In these coordinates, the metric on the sphere has
the form

$g_{\hbox{\bbp S}^{2}}=\frac{1}{W}d\xi^{2}+G(\xi,\theta)d\theta^{2},$

where $G$ is a positive function. It is easy to compute the hessian of $\xi$
on level sets in these coordinates:

$\nabla_{ij}^{2}\xi=\partial_{ij}^{2}\xi-\Gamma_{ij}^{k}\partial_{k}\xi=-\Gamma_{ij}^{\xi},$

and

$-\Gamma_{\xi\xi}^{\xi}=\frac{\dot{W}}{2W},\,-\Gamma_{\xi\theta}^{\xi}=-\Gamma_{\theta\xi}^{\xi}=0,\,\text{and}\,-\Gamma_{\theta\theta}^{\xi}=-\frac{1}{2}W\frac{\partial
G}{\partial\xi},$

where $\dot{W}$ denote the derivative of $W$ with respect to $\xi$. Then we
can compute the geodesic curvature of the level sets with respect to the
normal vector given by $\nabla\xi/\left|\nabla\xi\right|$ easily. Using the
formula given in [21], we have

$\kappa(\xi,\theta)=\frac{\nabla^{2}\xi(\nabla\xi,\nabla\xi)-\left|\nabla\xi\right|^{2}\Delta\xi}{\left|\nabla\xi\right|^{3}}=-\frac{\dot{W}}{2W^{5/2}}+\frac{2\xi}{W^{1/2}}.$
(3.15)

Thus, the geodesic curvature of the level sets only depends on $\xi$, so they
are curves of constant geodesic curvature. As the level set curves do not
cross, we have that they are intersections of parallel planes with
$\mathbb{S}^{2}$. Then, it follows that
$\Gamma=\partial\Omega\cap\overline{\mathcal{U}}$ is a circle on the sphere
and $\xi=\bar{\xi}$ in a neighborhood of $\Gamma$ in $\mathcal{U}$, so we
conclude that $\xi=\bar{\xi}$ in $\Omega$ since they are analytic functions.
This concludes the proof of the theorem. ∎

######  Remark 3.4.

Note that we are using the sign convention for the curvature such that the
geodesic curvature of $\partial\hbox{\bb D}({\bf n},s)$, where $\hbox{\bb
D}({\bf n},s)$ is the geodesic disk in $\hbox{\bb S}^{2}$ centered at ${\bf
n}$ of radius $s<\pi/2$, with respect to the inner normal is positive. It is
important to remark on this because in [1, 7] the authors used the opposite
sign convention.

###  3.2 Curvature estimates

Here, we are going to obtain curvature estimates of the level sets of our
solution $\xi$, taking advantage of the gradient estimates proved in Theorem
3.1. We recall that our model solution $\bar{\xi}$ satisfies:

$\nabla\bar{\xi}=\frac{\sqrt{1-r^{2}}}{r}\left(\bar{\xi}-\frac{\bar{\alpha}}{1-r^{2}}\right)n$

and

$\nabla^{2}\bar{\xi}=\begin{pmatrix}-\bar{\xi}-\frac{\bar{\alpha}}{1-r^{2}}&0\\\
0&-\bar{\xi}+\frac{\bar{\alpha}}{1-r^{2}}\end{pmatrix},$

so using (3.15), we have that

$\kappa(r)=\begin{cases}-\frac{r}{\sqrt{1-r^{2}}},&\mbox{if }r\geq 0,\\\
\frac{r}{\sqrt{1-r^{2}}},&\mbox{if }r<0,\end{cases}$

where the geodesic curvature is computed with respect to the inner orientation
to $\mathcal{U}$, i.e., using the conormal vector given by
$\nu=\nabla\xi/\left|\nabla\xi\right|$.

Figure 4: Here it appears the radial graph of the support function of the
critical catenoid, which is the graph
$\Sigma_{\xi_{0}}=\left\\{(1+\xi_{0}(r))(\sqrt{1-r^{2}}\cos\theta,\sqrt{1-r^{2}}\sin\theta,r)~{}\colon~{}r\in[-\bar{r},\bar{r}],\theta\in[0,2\pi)\right\\}$,
where $\xi_{0}$ is the model solution defined in Section 1. We see the zero-
level sets in blue, the top-level set in red, and also the level set of height
$r=0.5$ in purple. Here, the curvature of the level sets which are not in the
top stratum is computed with respect to the conormal vector
$\nu=\nabla\xi/\left|\nabla\xi\right|$, which points to the curve $\gamma$ in
each point of the domain.

In the next results, we estimate the geodesic curvature of the zero and top
level sets of $\xi$.

######  Proposition 3.1.

Let $(\Omega,\xi)$ be a solution to (2.2),
$\mathcal{U}\subset\Omega\setminus\textup{Max}(\xi)$ a connected component and
$\bar{R}=R(\mathcal{U})$ the expected critical height of the region. Let
$(\bar{\Omega},\bar{\xi})$ be its associated model solution in $\mathcal{U}$
and $p\in\partial\Omega$ such that

$\left|\nabla\xi\right|^{2}(p)=\max_{\partial\Omega\cap\textup{cl}(\mathcal{U})}\left|\nabla\xi\right|^{2}.$
(3.16)

Then, if $\kappa(p)$ denotes the curvature of $\partial\Omega$ at $p$ with
respect the inner orientation to $\mathcal{U}$, it holds

$\kappa(p)\leq-\frac{\bar{r}_{+}}{\sqrt{1-\bar{r}_{+}^{2}}}\quad\textup{if
}\overline{\tau}(\mathcal{U})\geq\tau_{0}\quad\textup{and}\quad\kappa(p)\leq\frac{\bar{r}_{-}}{\sqrt{1-\bar{r}_{-}^{2}}}\quad\textup{if
}\overline{\tau}(\mathcal{U})<\tau_{0},$

where $\bar{r}_{+}>0$ and $\bar{r}_{-}<0$ are defined in Remark 3.1.

###### Proof.

Since the geodesic curvature of the level sets of $\xi$ does not change after
a change of scale, we will assume that $\xi$ is normalized in the sense of
Definition 3.2. Consider the curve

$\sigma(s)=\text{exp}_{p}\left(\frac{\nabla\xi}{\left|\nabla\xi\right|}(p)s\right),\quad
s\in\mathbb{R},$

at some $p\in\textup{cl}(\mathcal{U})\cap\partial\Omega$, here the exponential
map is the one of $\hbox{\bb S}^{2}$. We compute the Taylor expansion of $W$
along this curve:

$W(\gamma(s))=W(p)+\left\langle\nabla
W,\nabla\xi/\left|\nabla\xi\right|\right\rangle(p)s+O(s^{2}),\quad
s\in(-\epsilon,\epsilon),$

for some $\epsilon>0$ small enough. Now, it is easy to check that

$\kappa(p)=\frac{\nabla^{2}\xi(\nabla\xi,\nabla\xi)-\left|\nabla\xi\right|^{2}\Delta\xi}{\left|\nabla\xi\right|^{3}}=\frac{\left\langle\nabla\left|\nabla\xi\right|^{2},\nabla\xi\right\rangle}{2\left|\nabla\xi\right|^{3}},$

so we have that

$W(\gamma(s))=W(p)+2\kappa(p)W(p)s+O(s^{2}),\quad s\in(-\epsilon,\epsilon).$

Next, we compute the Taylor expansion of $W_{\bar{R}}$ along the same curve:

$\displaystyle W_{\bar{R}}(\gamma(s))$ $\displaystyle=$ $\displaystyle
W_{\bar{R}}(p)+\left\langle\nabla
W_{\bar{R}}(p),\nabla\xi/\left|\nabla\xi\right|(p)\right\rangle s+O(s^{2})$
$\displaystyle=$ $\displaystyle
W_{\bar{R}}(p)+\left(\pm\frac{2\Psi}{\sqrt{1-\Psi^{2}}}\sqrt{W_{\bar{R}}}-\frac{4\bar{\alpha}}{1-\Psi^{2}}\right)\sqrt{W}(p)s+O(s^{2}),$

where $s\in[0,\delta)$ for some $0<\delta\leq\epsilon$. Since
$W_{\bar{R}}(p)=W(p)$ at $p\in\partial\Omega$ by (3.16) (this is clear using
the definition of $\overline{\tau}(\mathcal{U})$), Theorem 3.1 and the
expansions of $W$ and $W_{\bar{R}}$ near $p$ imply

$\kappa\leq\pm\frac{\Psi}{\sqrt{1-\Psi^{2}}}\frac{\sqrt{W_{\bar{R}}}}{\sqrt{W}}-\frac{2\bar{\alpha}}{(1-\Psi^{2})\sqrt{W}}\text{
at }p,$

where the sign is positive if $\overline{\tau}(\mathcal{U})\geq\tau_{0}$ and
negative if $\overline{\tau}(\mathcal{U})<\tau_{0}$. Finally, since

$\sqrt{W}(p)=\sqrt{W_{\bar{R}}}(p)=\pm\frac{\bar{\alpha}}{\bar{r}_{\pm}\sqrt{1-\bar{r}_{\pm}^{2}}}\text{
and }\Psi(p)=\bar{r}_{\pm},$

we conclude the following estimate for the curvature of the zero level set of
$\mathcal{U}$:

$\kappa(p)\leq\begin{cases}-\frac{\bar{r}_{+}}{\sqrt{1-\bar{r}_{+}^{2}}},&\mbox{if
}\overline{\tau}(\mathcal{U})\geq\tau_{0}\\\
\frac{\bar{r}_{-}}{\sqrt{1-\bar{r}_{-}^{2}}},&\mbox{if
}\overline{\tau}(\mathcal{U})<\tau_{0}.\end{cases}$

∎

######  Proposition 3.2.

Let $(\Omega,\xi)$ be a solution to (2.2),
$\mathcal{U}\subset\Omega\setminus\textup{Max}(\xi)$ a connected component and
$\bar{R}=R(\mathcal{U})$ the expected critical height of the region. Let
$(\bar{\Omega},\bar{\xi})$ be its associated model solution in $\mathcal{U}$.
Let $\gamma\subset\textup{cl}(\mathcal{U})\cap\textup{Max}(\xi)$ be an
analytic curve and $p\in\gamma$. Then if $\kappa(p)$ denotes the curvature of
$\gamma$ at $p$ with respect the inner orientation to $\mathcal{U}$, it holds

$\kappa(p)\leq\frac{\bar{R}}{\sqrt{1-\bar{R}^{2}}}\quad\textup{if
}\overline{\tau}(\mathcal{U})\geq\tau_{0}\quad\textup{and}\quad\kappa(p)\leq-\frac{\bar{R}}{\sqrt{1-\bar{R}^{2}}}\quad\textup{if
}\overline{\tau}(\mathcal{U})<\tau_{0}.$

###### Proof.

As in the proof of Proposition 3.1, we assume again that the solution $\xi$ is
normalized. To estimate the geodesic curvature of $\gamma$ we use [7, Theorem
3.1]. If we denote

$\rho(x)=\textup{dist}(x,\gamma),\quad x\in\mathcal{U},$

then

$\xi=\xi_{\textup{max}}-\frac{\varphi}{2}\rho^{2}-\frac{\varphi(\xi_{\textup{max}})}{6}\kappa\,\rho^{3}+O(\rho^{4}),$

where $\varphi$ is such that $\Delta\xi=-\varphi(\xi)$. Replacing the explicit
terms in the previous expansion, we obtain:

$\xi=\frac{\bar{\alpha}}{1-\bar{R}^{2}}-\frac{\bar{\alpha}}{1-\bar{R}^{2}}\rho^{2}-\frac{\bar{\alpha}}{3(1-\bar{R}^{2})}\kappa\,\rho^{3}+O(\rho^{4})$

in a certain neighborhood of $\gamma$ inside $\mathcal{U}$. Computing the
gradient of the previous expression, we get

$\nabla\xi=-\frac{\bar{\alpha}}{1-\bar{R}^{2}}\rho\left(2+\kappa\,\rho\right)\partial_{\rho}+O(\rho^{3}),$

and then we conclude that

$W=\frac{4\bar{\alpha}^{2}\rho^{2}}{(1-\bar{R}^{2})^{2}}\left(1+\kappa\,\rho\right)+O(\rho^{4}).$
(3.17)

We want to obtain an expansion of $W_{\bar{R}}$ as the previous one. To
achieve this, we use the Taylor expansions obtained in the proof of Lemma 3.1.
Set $z=\Psi-\bar{R}$. Observe that from (3.7) it is easy to check that

$\frac{\xi_{\textup{max}}-\xi}{z^{2}}=\frac{\xi_{\textup{max}}-\xi}{(\Psi-\bar{R})^{2}}\rightarrow\textup{constant,
as }\Psi\rightarrow\bar{R},$

so we conclude that

$z=\pm\frac{1-\bar{R}^{2}}{\sqrt{\bar{\alpha}}}\sqrt{\xi_{\textup{max}}-\xi}+O\left(\xi_{\textup{max}}-\xi\right)^{3/2},$
(3.18)

where the sign of the first term is positive if
$\overline{\tau}(\mathcal{U})\geq\tau_{0}$ and negative otherwise. We will
need the third order term in the expansion (3.8), so we compute the third
derivative of $W_{\bar{R}}$. A straightforward computation shows

$\frac{\partial^{3}W_{\bar{R}}}{\partial\Psi^{3}}(\bar{R})=-12\bar{R}\frac{4\bar{\alpha}^{2}}{(1-\bar{R}^{2})^{4}}+6(1-\bar{R}^{2})\left(\frac{2\bar{\alpha}}{(1-\bar{R}^{2})^{2}}\frac{8\bar{\alpha}\bar{R}}{(1-\bar{R}^{2})^{3}}\right)=\frac{48\bar{R}\bar{\alpha}^{2}}{(1-\bar{R}^{2})^{4}},$

so the expansion of $W_{\bar{R}}(\Psi)$ at $\Psi=\bar{R}$ up to third order is
given by

$W_{\bar{R}}=\frac{4\bar{\alpha}^{2}}{(1-\bar{R}^{2})^{3}}z^{2}+\frac{8\bar{R}\bar{\alpha}^{2}}{(1-\bar{R}^{2})^{4}}z^{3}+O(z^{4}).$

Substituting (3.18) in the previous expansion we get

$W_{\bar{R}}=\frac{4\bar{\alpha}}{1-\bar{R}^{2}}(\xi_{\textup{max}}-\xi)\pm\frac{8\bar{R}\bar{\alpha}^{1/2}}{1-\bar{R}^{2}}(\xi_{\textup{max}}-\xi)^{3/2}+O(\xi_{\textup{max}}-\xi)^{2},$

using again [7, Theorem 3.1] and simplifying we conclude that

$W_{\bar{R}}=\frac{4\bar{\alpha}^{2}}{(1-\bar{R}^{2})^{2}}\rho^{2}\left(1+\left(\frac{1}{3}\kappa\pm\frac{2}{3}\frac{\bar{R}}{\sqrt{1-\bar{R}^{2}}}\right)\rho\right)+O(\rho^{7/2}).$
(3.19)

Finally, we proceed as at the end of the proof of Proposition 3.1. By Theorem
3.1 we have that $W\leq W_{\bar{R}}$ in $\mathcal{U}$, comparing the
expansions near $\gamma$, (3.17) and (3.19), we have that

$\kappa(p)\leq\frac{\kappa(p)}{3}\pm\frac{2}{3}\frac{\bar{R}}{\sqrt{1-\bar{R}^{2}}},$

that is,

$\kappa(p)\leq\begin{cases}\frac{\bar{R}}{\sqrt{1-\bar{R}^{2}}},&\mbox{if
}\overline{\tau}(\mathcal{U})\geq\tau_{0}\\\
-\frac{\bar{R}}{\sqrt{1-\bar{R}^{2}}},&\mbox{if
}\overline{\tau}(\mathcal{U})<\tau_{0}.\end{cases}$

∎

At this point, we can announce the following result:

######  Theorem 3.2.

Let $(\Omega,\xi)$ be a solution to (2.2). Assume that there exists an
analytic closed curve $\gamma\subset\textup{Max}(\xi)$, let
$\Omega^{\gamma}_{1},\Omega^{\gamma}_{2}\subset\Omega$ be the partition with
respect to $\gamma$ and set $R_{1}=R(\Omega^{\gamma}_{1})$ and
$R_{2}=R(\Omega^{\gamma}_{2})$ the expected critical heights associated to
that regions.

  * •

If $\overline{\tau}(\Omega^{\gamma}_{1})\leq\tau_{0}$, then
$\overline{\tau}(\Omega^{\gamma}_{2})\geq\tau_{0}$ and for each $p\in\gamma$
it holds

$\frac{R_{1}}{\sqrt{1-R_{1}^{2}}}\leq\kappa(p)\leq\frac{R_{2}}{\sqrt{1-R_{2}^{2}}},$
(3.20)

where $\kappa(p)$ is the geodesic curvature of $\gamma$ in $p$ with respect to
the normal vector pointing to the region $\Omega^{\gamma}_{2}$.

  * •

If
$\overline{\tau}(\Omega^{\gamma}_{1})=\overline{\tau}(\Omega^{\gamma}_{2})=\tau_{0}$,
then $\kappa=0$ along $\gamma$.

In particular, $R_{2}\geq R_{1}$ and equality holds if, and only if,
$(\Omega,\xi)\equiv(\bar{\Omega},\bar{\xi})$, where $(\bar{\Omega},\bar{\xi})$
is the associated model solution given in Definition 3.1.

###### Proof.

First, since we are assuming that
$\overline{\tau}(\Omega^{\gamma}_{1})\leq\tau_{0}$, by Proposition 3.2 we
conclude that

$0\leq\frac{R_{1}}{\sqrt{1-R_{1}^{2}}}\leq\kappa(p)\quad\forall p\in\gamma,$

so, in particular, $\kappa\geq 0$ with respect to the normal orientation
pointing to $\Omega^{\gamma}_{2}$. Then
$\overline{\tau}(\Omega^{\gamma}_{2})\geq\tau_{0}$, because otherwise, using
Proposition 3.2 with this region, we would conclude that $\kappa<0$ along
$\gamma$, a contradiction. It follows also that it must be $R_{1}\geq R_{2}$,
and thus we obtain the chain of inequalities (3.20). Assume now that
$R_{1}=R_{2}=\bar{R}$, then $\gamma$ must have constant geodesic curvature, so
it must be a parallel at height $\bar{R}$. Let us suppose that $\xi$ is
normalized. Then, we get that $(\Omega,\xi)$ is a model solution as follows:
take $\mathcal{N}$ a tubular neighborhood of $\gamma$ of radius
$\epsilon<\min{(-\bar{r}_{-},\bar{r}_{+})}$, then $\xi$ is a solution to

$\left\\{\begin{matrix}\Delta\xi+2\xi=0&\text{ in
}&\mathcal{N},\\\\[5.69054pt] \xi=\xi_{\textup{max}}&\text{ along
}&\gamma,\\\\[5.69054pt] \frac{\partial\xi}{\partial\nu}=0&\text{ along
}&\gamma.\end{matrix}\right.$

Since $\gamma$ is non-characteristic, it follows from the Cauchy-Kovalevskaya
theorem (cf. [30]) that $\xi$ is the unique solution to the previous problem.
But the associated model solution $(\bar{\Omega},\bar{\xi})$ is also a
solution to the previous problem, so we conclude that $\xi\equiv\bar{\xi}$ in
$\mathcal{N}$ (up to a rotation if necessary). It is clear then that it must
be $(\Omega,\xi)\equiv(\bar{\Omega},\bar{\xi})$ by analyticity. Finally,
consider the case
$\overline{\tau}(\Omega^{\gamma}_{1})=\overline{\tau}(\Omega^{\gamma}_{2})=\tau_{0}$.
It follows from Proposition 3.2 that $\kappa\leq 0$ with respect to the inner
and to the outer normal to $\Omega^{\gamma}_{2}$ simultaneously, so it must be
$\kappa=0$ along $\gamma$. Then we get that it must be
$(\Omega,\xi)\equiv(\Omega_{0},\xi_{0})$ using the same argument as in the
previous case. ∎

###  3.3 Length estimates

We show here a relation between the length of the zero and top level sets:

######  Proposition 3.3.

Let $(\Omega,\xi)$ be a solution to (2.2),
$\mathcal{U}\subset\Omega\setminus\textup{Max}(\xi)$ a connected component and
$\bar{R}=R(\mathcal{U})$ the expected critical height of the region. Suppose
that $\textup{cl}(\mathcal{U})\cap\textup{Max}(\xi)=\gamma^{\mathcal{U}}$ and
$\textup{cl}(\mathcal{U})\cap\partial\Omega=\Gamma^{\mathcal{U}}$ are sets of
analytic closed curves. Then

$\frac{\left|\gamma^{\mathcal{U}}\right|}{\sqrt{1-\bar{R}^{2}}}\leq\frac{\left|\Gamma^{\mathcal{U}}\right|}{\sqrt{1-r_{+}^{2}}}\quad\textup{if
}\overline{\tau}(\mathcal{U})\geq\tau_{0}\quad\textup{and}\quad\frac{\left|\gamma^{\mathcal{U}}\right|}{\sqrt{1-\bar{R}^{2}}}\leq\frac{\left|\Gamma^{\mathcal{U}}\right|}{\sqrt{1-r_{-}^{2}}}\quad\textup{if
}\overline{\tau}(\mathcal{U})<\tau_{0},$ (3.21)

where $\left|\gamma^{\mathcal{U}}\right|$ and
$\left|\Gamma^{\mathcal{U}}\right|$ denotes the sum of the lengths of each set
of curves.

###### Proof.

We follow the proof of [1, Proposition 5.5]. Assume that $\xi$ is normalized
in the sense of Definition 3.2. Given $\epsilon>0$, set
$\mathcal{U}_{\epsilon}$ as in (2.13) (as always, we suppose that
$\xi_{\textup{max}}-\epsilon$ is regular). Then, by the divergence theorem, it
holds:

$\begin{split}\int_{\mathcal{U}_{\epsilon}}\frac{-2\xi}{(1-\Psi^{2})\xi-\bar{\alpha}}\,dA&=\int_{\mathcal{U}_{\epsilon}}\frac{\Delta\xi}{(1-\Psi^{2})\xi-\bar{\alpha}}\,dA\\\
&=\int_{\mathcal{U}_{\epsilon}}\left\langle\frac{\nabla\xi}{(1-\Psi^{2})\xi-\bar{\alpha}},\nabla\xi\right\rangle\,dA+\int_{\partial\mathcal{U}_{\epsilon}}\frac{\left\langle\nabla\xi,\nu\right\rangle}{(1-\Psi^{2})\xi-\bar{\alpha}}\,ds,\end{split}$

where $\Psi$ is the pseudo-radial function given in Definition 3.3,
$\bar{\alpha}=\alpha(\bar{R})$ as in Proposition 1.1 and $\nu$ is the inner
unit normal to $\mathcal{U}_{\epsilon}$. Thus, on the one hand

$\nabla\frac{\Psi}{(1-\Psi^{2})\xi-\bar{\alpha}}=\frac{2\Psi^{3}(1-\Psi^{2})\xi}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{3}}\nabla\xi$

and on the other hand

$\frac{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{3}}{2\Psi^{3}(1-\Psi^{2})\xi}\cdot\frac{2\Psi\xi}{(1-\Psi^{2})\xi-\bar{\alpha}}=\frac{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{2}}{\Psi^{2}(1-\Psi^{2})}=W_{\bar{R}},$

so we obtain the identity

$\int_{\mathcal{U}_{\epsilon}}\frac{2\Psi^{3}(1-\Psi^{2})\xi}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{3}}(W-W_{\bar{R}})\,dA=-\int_{\Gamma_{\mathcal{U}}}\frac{\left|\nabla\xi\right|\Psi}{(1-\Psi^{2})\xi-\bar{\alpha}}\,ds+\int_{\gamma^{\mathcal{U}}_{\epsilon}}\frac{\left|\nabla\xi\right|\Psi}{(1-\Psi^{2})\xi-\bar{\alpha}}\,ds,$
(3.22)

where
$\gamma^{\mathcal{U}}_{\epsilon}=\\{p\in\mathcal{U}~{}\colon~{}\xi(p)+\epsilon=\xi_{\textup{max}}\\}$.
It is clear that for $\epsilon>0$ small enough,
$\gamma^{\mathcal{U}}_{\epsilon}$ is a set of analytic curves, and
$\gamma_{\epsilon}^{\mathcal{U}}\to\gamma^{\mathcal{U}}$ when $\epsilon\to 0$.
Next, we analyze both sides of the previous identity. We begin with the right-
hand side. Since $\Psi=r_{\pm}$ and
$\max_{\Gamma^{\mathcal{U}}}\left|\nabla\xi\right|=\pm\bar{\alpha}/r_{\pm}\sqrt{1-r_{\pm}^{2}}$,
it is clear that

$\int_{\Gamma^{\mathcal{U}}}\frac{\left|\nabla\xi\right|\Psi}{(1-\Psi^{2})\xi-\bar{\alpha}}\,ds=\mp\frac{1}{\sqrt{1-r_{\pm}^{2}}}\int_{\Gamma^{\mathcal{U}}}\frac{\left|\nabla\xi\right|}{\max_{\Gamma^{\mathcal{U}}}\left|\nabla\xi\right|}\left\\{\begin{matrix}\geq-\frac{\left|\Gamma^{\mathcal{U}}\right|}{\sqrt{1-r_{+}^{2}}}&\text{
if }&\overline{\tau}(\mathcal{U})\geq\tau_{0},\\\\[5.69054pt]
\leq\frac{\left|\Gamma^{\mathcal{U}}\right|}{\sqrt{1-r_{-}^{2}}}&\text{ if
}&\overline{\tau}(\mathcal{U})<\tau_{0}.\end{matrix}\right.$ (3.23)

Also, note that by the Taylor expansions given in (3.17) and (3.19) we have

$\lim_{p\to\gamma^{\mathcal{U}}}\frac{W}{W_{\bar{R}}}=1.$

Since

$\frac{\left|\nabla\xi\right|\Psi}{(1-\Psi^{2})\xi-\bar{\alpha}}=\mp\frac{1}{\sqrt{1-\Psi^{2}}}\sqrt{\frac{W}{W_{\bar{R}}}},$

(as always, the upper sign corresponds to the case
$\overline{\tau}(\mathcal{U})\geq\tau_{0}$ and the lower sign otherwise) we
conclude that

$\lim_{\epsilon\to
0^{+}}\int_{\gamma^{\mathcal{U}}_{\epsilon}}\frac{\left|\nabla\xi\right|\Psi}{(1-\Psi^{2})\xi-\bar{\alpha}}\,ds=\mp\frac{\left|\gamma^{\mathcal{U}}\right|}{1-\bar{R}^{2}}.$
(3.24)

Now, we look at the left-hand side of (3.22). Note that

$\frac{\Psi^{3}(1-\Psi^{2})\xi}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{3}}=\mp\frac{1}{W_{\bar{R}}^{3/2}\sqrt{1-\Psi^{2}}},$

therefore, using Theorem 3.1, we conclude that

$\int_{\mathcal{U}_{\epsilon}}\frac{2\Psi^{3}(1-\Psi^{2})\xi}{\left((1-\Psi^{2})\xi-\bar{\alpha}\right)^{3}}(W-W_{\bar{R}})\,dA\left\\{\begin{matrix}\geq
0&\text{ if }&\overline{\tau}(\mathcal{U})\geq\tau_{0},\\\\[5.69054pt] \leq
0&\text{ if }&\overline{\tau}(\mathcal{U})<\tau_{0}.\end{matrix}\right.$
(3.25)

Finally, combining (3.23), (3.24) and (3.25) we obtain the inequalities
(3.21). ∎

##  4 Overdetermined solutions

Let $\Omega\subset\hbox{\bb S}^{2}$ be a domain of finite type with $k\geq 2$
boundary components and consider the overdetermined elliptic problem

$\left\\{\begin{matrix}\Delta\xi+2\xi=0&\text{ in }&\Omega,\\\\[5.69054pt]
\xi>0&\text{ in }&\Omega,\\\\[5.69054pt] \xi=0&\text{ along
}&\partial\Omega,\\\\[5.69054pt] \left|\nabla\xi\right|=b_{i}>0&\text{ along
}&\Gamma_{i},&\text{ for }i\in\\{1,\dots,k\\},\end{matrix}\right.$ (4.1)

where we have that $\partial\Omega=\bigcup_{i=1}^{k}\Gamma_{i}$ using the
notation introduced in Definition 2.1.

######  Remark 4.1.

When $\Omega$ is a domain of finite type with $k=1$, i.e., a topological disk,
a solution $\xi$ to (4.1) must be rotationally symmetric and $\Omega$ a
geodesic disk by [16, 53], that is, $(\xi,\Omega)\equiv(\xi_{1},\Omega_{1})$.

In this section, we will classify the solutions $(\Omega,\xi)$ to (4.1) which
have infinitely many maximum points. To do so, we will use the estimates
computed in Section 3 and a length estimate that uses the overdetermined
condition and Proposition 3.1.

######  Proposition 4.1.

Let $(\Omega,\xi)$ be a solution to (4.1) and
$\mathcal{U}\subset\Omega\setminus\textup{Max}(\xi)$ an annular connected
component, i.e., homeomorphic to an annulus, such that
$\partial\mathcal{U}=\Gamma\cup\gamma$, $\Gamma\subset\partial\Omega$ and
$\gamma\subset{\rm Max}(\xi)$. Then, it holds

$\left|\Gamma\right|\leq 2\pi\sqrt{1-r_{+}^{2}}\quad\textup{if
}\overline{\tau}(\mathcal{U})\geq\tau_{0}\quad\textup{and}\quad\left|\Gamma\right|\leq
2\pi\sqrt{1-r_{-}^{2}}\quad\textup{if }\overline{\tau}(\mathcal{U})<\tau_{0}.$

###### Proof.

First, observe that Proposition 3.1 and the overdetermined condition $\xi=0$
and $\left|\nabla\xi\right|=b_{i}>0$ along each
$\Gamma_{i}\subset\partial\Omega$ connected component imply that

$\left|\kappa(p)\right|\geq\left|\frac{\bar{r}_{\pm}}{\sqrt{1-\bar{r}_{\pm}}}\right|\text{
for all }p\in\Gamma$

where $\bar{r}_{+}>0$ and $\bar{r}_{-}<0$ are defined in Remark 3.1 and
$\kappa$ is the geodesic curvature of $\Gamma$. Then, by Blaschke turning
theorem (generalized to space forms in [27]) we conclude that there exists
$p\in\mathbb{S}^{2}$ such that $\Gamma$ is contained in the geodesic disk
$D:=\mathbb{D}(p,|r_{\pm}|)$. Now, given a closed convex curve
$\sigma:I\to\mathbb{S}^{2}$, Crofton’s formula (cf. [37]) tells us that

$\textup{Length}(\sigma)=\frac{1}{4}\int_{\mathbb{S}^{2}}n_{\sigma}(a)\,da,$

where $n_{\sigma}(a)$ measures the number of points in which the plane
$P_{a}:=<\left\\{a\right\\}>^{\perp}$, $a\in\mathbb{S}^{2}$, and the curve
$\sigma$ intersects. Since any $P_{a}$ intersects an embedded strictly convex
closed curve in the sphere at most at two points and $\Gamma\subset D$, the
result follows. Recall that $\Gamma$ is strictly convex with the correct
orientation. ∎

Now, we have the ingredients to prove our main classification result:

Figure 5: Here, a finite domain in the unit sphere. The domain is
$\Omega=\mathbb{S}^{2}\setminus\bigcup_{i=1}^{3}D_{i}$, where $D_{1},D_{2}$
and $D_{3}$ are the blue regions. We have that
$\Omega\setminus\gamma=\Omega^{\gamma}_{1}\cup\Omega^{\gamma}_{2}$ is a
partition of $\Omega$, and the region $\Omega^{\gamma}_{1}$ has the topology
of an annulus. If $\xi$ solves (4.1) and $\gamma\in\textup{Max}(\xi)$ then
$(\Omega,\xi)$ satisfies the hypothesis of Theorem 4.1.

######  Theorem 4.1.

Let $(\Omega,\xi)$ be a solution to the overdetermined problem (4.1), where
$\Omega$ is a $\mathcal{C}^{2}$-domain of finite type with $k\geq 2$ boundary
components. Suppose that $\textup{Max}(\xi)$ contains a closed curve $\gamma$
such that the partition with respect to $\gamma$,
$\Omega\setminus\gamma=\Omega_{1}^{\gamma}\cup\Omega_{2}^{\gamma}$ contains a
component, say $\Omega_{1}^{\gamma}$, that is a topological annulus. Then,
$k=2$ and $\Omega$ is rotationally symmetric. In particular, there exists an
$R\in[0,1)$ such that $(\Omega,\xi)=(\Omega_{R},\xi_{R})$ up to a rotation and
a dilation.

###### Proof.

By hypothesis, $\Omega^{\gamma}_{1}$ is a topological annulus and, up to
rearranging the indexes if necessary,
$\partial\Omega_{1}^{\gamma}=\Gamma_{1}\cup\gamma$, where
$\Gamma_{1}=\partial\Omega\cap\partial\Omega_{1}^{\gamma}$. It follows from
Proposition 3.21 and Proposition 4.1 that

$\left|\gamma_{1}\right|\leq
2\pi\sqrt{1-R_{1}^{2}}\frac{\left|\Gamma_{1}\right|}{\sqrt{1-r_{\pm}^{2}}}\leq
2\pi\sqrt{1-R_{1}^{2}},$

where $R_{1}=R(\Omega_{1}^{\gamma})\in[0,1)$ is the expected critical height
of the region $\Omega_{1}^{\gamma}$. Then, [37, Proposition 9.54] implies that
$\gamma_{1}$ is either an equator or it is contained in an open hemisphere,
denoted by $\hbox{\bb S}^{2}_{+}$. In the first case, we conclude that
$(\Omega,\xi)=(\Omega_{0},\xi_{0})$ up to a rotation and a dilation applying
the Cauchy-Kovalevskaya theorem. In the second case, since $\gamma$ is not
contractible in $\Omega$ by Lemma 2.1, either
$\Omega^{\gamma}_{1}\subset\hbox{\bb S}^{2}_{+}$ or
$\Omega_{2}^{\gamma}\subset\hbox{\bb S}^{2}_{+}$. In any case, we can apply
the moving plane method in $\hbox{\bb S}^{2}_{+}$ (a slight modification of
the arguments in [32, 43]) to the component
$\Omega_{i}^{\gamma}\subset\hbox{\bb S}^{2}_{+}$ to conclude that $\Omega$ and
$\xi$ are rotationally symmetric. Since the model solutions are the only
solutions to (2.2) with this property, this proves the result. ∎

######  Remark 4.2.

Let $\Omega$ be a finite domain with $k\geq 2$ and $\xi$ a solution to (4.1),
a simple topological argument and Lemma 2.1 imply that, if the top level set
of $\xi$ contains $k-1$ curves $\gamma_{i}$, $i=1,\ldots,k$, then
$\Omega\setminus\bigcup_{i=1}^{k-1}\gamma_{i}$ contains an annular component
in the conditions of Theorem 4.1. Hence, $(\Omega,\xi)=(\Omega_{R},\xi_{R})$
up to a rotation and a dilation.

In the case that $\Omega$ is a topological annulus, Theorem 4.1 and Remark 4.2
imply:

> Theorem A: Let $(\Omega,\xi)$ be a solution to (4.1), $\Omega$ an annular
> domain (i.e., $k=2$). Suppose that $\xi$ has infinitely many maximum points
> inside $\Omega$. Then, there exist $R\in[0,1)$ such that
> $(\Omega,\xi)\equiv(\Omega_{R},\xi_{R})$.

We conclude this section by noting that we can change the topological
condition on $\Omega\setminus\gamma$ in Theorem 4.1 by a restriction in the
NWSS of the solution. In fact, using Theorem 3.2 we can obtain the following
result:

######  Theorem 4.2.

Let $(\Omega,\xi)$ be a solution to the overdetermined problem (4.1), where
$\Omega$ is of finite type with $k\geq 2$. Suppose that $\xi$ has infinitely
many maximum points and that at least one of the connected components of
$\Omega\setminus\textup{Max}(\xi)$ has NWSS less or equal to $\tau_{0}$. Then,
$\Omega$ is rotationally symmetric. In particular, there exists $R\in[0,1)$
such that $(\Omega,\xi)\equiv(\Omega_{R},\xi_{R})$.

###### Proof.

Let $\mathcal{U}\subset\Omega\setminus\textup{Max}(\xi)$ such that
$\overline{\tau}(\mathcal{U})\leq\tau_{0}$ by hypothesis and set
$\partial\mathcal{U}\cap{\rm Max}(\xi)=\bigcup_{j=1}^{m}\gamma_{j}$, the
boundary components of $\mathcal{U}$ in the top level set. Theorem 3.2 implies
that the geodesic curvature of $\gamma_{j}$, with respect to the inner
orientation to $\mathcal{U}$, is non-positive. Therefore, each $\gamma_{j}$ is
contained in a closed hemisphere and hence, there exists, at least, a
connected component $\gamma\subset\partial\mathcal{U}\cap{\rm Max}(\xi)$ such
that one of the components,
$\Omega\setminus\gamma=\Omega^{\gamma}_{1}\cup\Omega_{2}^{\gamma}$, of the
partition with respect to $\gamma$, say $\Omega_{1}^{\gamma}$, is contained in
a closed hemisphere, denoted by $\textup{cl}(\mathbb{S}^{2}_{+})$. We
distinguish two cases. First, assume that the geodesic curvature of $\gamma$
vanishes identically. In this case, Cauchy-Kovalevskaya theorem implies that
$(\xi,\Omega)=(\xi_{0},\Omega_{0})$, up to a rotation and scaling. Second, if
the geodesic curvature does not vanish identically, we can move slightly
$\textup{cl}(\mathbb{S}^{2}_{+})$ to obtain an open hemisphere, still denoted
by $\hbox{\bb S}^{2}_{+}$, such that
$\textup{cl}(\Omega_{1}^{\gamma})\subset\hbox{\bb S}_{2}^{+}$. Hence, again by
the moving plane method using the overdetermined condition, we infer that
$(\xi,\Omega)\equiv(\xi_{R},\Omega_{R})$ for some $R\in(0,1)$. ∎

##  5 Geometric application

In this section, we will exploit the correspondence of solutions to (4.1) and
free boundary minimal surfaces explained in the introduction. Let
$\mathbb{B}^{3}(R)$ denote the open Euclidean ball of radius $R>0$ centered at
the origin, $\partial\mathbb{B}^{3}(R)=\mathbb{S}^{2}(R)$. For simplicity, we
also set $\mathbb{B}^{3}=\mathbb{B}^{3}(1)$ and $\hbox{\bb
S}^{2}\equiv\hbox{\bb S}^{2}(1)$.

######  Definition 5.1.

Let $\Sigma\subset\mathbb{R}^{3}$ be an open immersed minimal surface with
boundary. We will say that $\Sigma$ has free boundaries if each boundary
component of $\Sigma$ meets orthogonally a sphere centered at the origin,
possibly of different radii. If $\Sigma\subset\mathbb{B}^{3}$ and
$\partial\Sigma\subset\mathbb{S}^{2}$ then $\Sigma$ is said to be a free
boundary minimal surface in the unit ball.

We recall the correspondence between minimal surfaces with free boundaries and
the solutions to (4.1) (cf. [53, Proposition 2.1]).

######  Proposition 5.1.

Let $\Omega\subset\mathbb{S}^{2}$ be a $\mathcal{C}^{2}$-domain in the sphere.
Suppose that $\Omega$ is not a topological disk and there exists a nonzero
function
$\xi\in\mathcal{C}^{2}(\Omega)\cap\mathcal{C}^{1}(\textup{cl}(\Omega))$
solution to the problem

$\left\\{\begin{matrix}\Delta\xi+2\xi=0&\text{ in }&\Omega,\\\\[5.69054pt]
\xi=0&\text{ along }&\partial\Omega,\\\\[5.69054pt]
\left|\nabla\xi\right|^{2}=b_{i}^{2}&\text{ along
}&\Gamma_{i}\in\pi_{0}(\partial\Omega),\,i\in\left\\{1,\ldots,k\right\\},\end{matrix}\right.$
(5.1)

where $\nu$ is the exterior unit normal to $\Omega$ and $b_{i}$ is a constant
for each $i\in\left\\{1,\ldots,k\right\\}$. Then, the map:

$\begin{split}X_{\xi}:\,&\Omega\longrightarrow\mathbb{R}^{3}\\\
&\,z\,\,\mapsto X_{\xi}(z):=\nabla^{\hbox{\bbp S}^{2}}\xi(z)+\xi(z)\cdot
z.\end{split}$

defines a branched minimal surface in $\mathbb{R}^{3}$ with free boundaries,
where each boundary component $X_{\xi}(\Gamma_{i})$ lies in the sphere
$\mathbb{S}^{2}\left(\left|b_{i}\right|\right)$. Conversely, let $\Sigma$ be a
minimal surface with free boundaries,
$\partial\Sigma=\bigcup_{i=1}^{k}\zeta_{i}$, and injective Gauss map
$N:\Sigma\to\mathbb{S}^{2}$. Set $u(p):=\langle p,N(p)\rangle$ its support
function. Then

$\xi(z):=(u\circ N^{-1})(z)=\left\langle N^{-1}(z),z\right\rangle,\quad\forall
z\in N(\Sigma)=\Omega$ (5.2)

satisfies (5.1), where $\partial\Omega=\bigcup_{i=1}^{k}\Gamma_{i}$,
$N(\zeta_{i})=\Gamma_{i}$, and $\left|b_{i}\right|$ is the radius of the
sphere in which $\zeta_{i}\in\pi_{0}(\partial\Sigma)$ is contained.

From now on, up to a dilation in $\hbox{\bb R}^{3}$,
$\Sigma\subset\mathbb{B}^{3}$ will denote an embedded minimal annulus with
free boundaries, $\partial\Sigma=\zeta_{1}\cup\zeta_{2}$, such that
$\zeta_{1}\subset\mathbb{S}^{2}$ and
$\zeta_{2}\subset\mathbb{S}^{2}(\tilde{r})$ for some $0<\tilde{r}\leq 1$. We
will always assume that the boundaries of $\Sigma$ intersect the spheres from
the inside, i.e., let $\nu:\partial\Sigma\to\mathbb{S}^{2}$ the exterior
conormal to $\Sigma$ then, $\left\langle p,\nu(p)\right\rangle>0$ for all
$p\in\partial\Sigma$. Now, we are ready to prove Theorem B. We rewrite it here
using the notation introduced so far.

> Theorem B: Let $\Sigma\subset\mathbb{B}^{3}$ be an embedded minimal annulus
> with free boundaries, $\partial\Sigma=\zeta_{1}\cup\zeta_{2}$, such that
> $\zeta_{1}\subset\mathbb{S}^{2}$ and
> $\zeta_{2}\subset\mathbb{S}^{2}(\tilde{r})$ for some $0<\tilde{r}\leq 1$.
> Suppose that its support function has infinitely many critical points. Then,
> there exists $R\in[0,1)$ such that $\Sigma=C_{R}$, up to a rotation around
> the origin, where $C_{R}$ is one of the model catenoids defined in (1.9).

###### Proof of Theorem B.

First, since $\Sigma$ is minimal, the Hopf differential is holomorphic and so
the interior zeros are isolated and of negative index. Moreover, since
$\textup{cl}(\Sigma)$ intersects some sphere at a constant angle,
Joachimstahl’s theorem [10, p. 152] implies that each boundary component is a
line of curvature, and so the zeros at the boundary are also isolated and of
negative index. Since the zeros of the Hopf differential correspond to umbilic
points we have that the index at any umbilical point is negative. Using that
$\Sigma$ is a topological annulus, the Poincaré-Hopf index theorem asserts
that $\Sigma$ is either totally umbilical, which is not possible, or there are
no umbilic points (cf. [24, 38] for details). Next, since there are no umbilic
points, the Gaussian curvature $K$ is strictly negative in ${\rm cl}(\Sigma)$
and the boundary curves $\zeta_{1}\subset\hbox{\bb S}^{2}$ and
$\zeta_{2}\subset\hbox{\bb S}^{2}(\tilde{r})$ are strictly convex spherical
curves. This also implies that the Gauss map
$N:\textup{cl}(\Sigma)\to\hbox{\bb S}^{2}$ is a local diffeormorphism.
Moreover, $N:\zeta_{i}\to N(\zeta_{i})=\Gamma_{i}$, $i=1,2$, is one to one.
Consider the projection map $\varrho:\hbox{\bb S}^{2}(\tilde{r})\to\hbox{\bb
S}^{2}$ given by $\varrho(p)=p/|p|$. We next show:

> Claim A: Let $\Sigma$ be an embedded minimal annulus with free boundaries.
> Then $\mathbb{S}^{2}\setminus(\zeta_{1}\cup\varrho(\zeta_{2}))=A\cup
> B_{1}\cup\varrho(B_{2})$, where $A$ is a topological annulus and
> $B_{1}\subset\hbox{\bb S}^{2}$ and $\varrho(B_{2})$, $B_{2}\subset\hbox{\bb
> S}^{2}(\tilde{r})$, are disjoint topological disks with boundaries
> $\zeta_{1}$ and $\varrho(\zeta_{2})$ respectively. In addition, $B_{1}$ and
> $\varrho(B_{2})$ are contained in two different closed hemispheres of
> $\mathbb{S}^{2}$.

###### Proof.

Let $F_{i}=\int_{\zeta_{i}}\nu$ be the flux along $\zeta_{i}$ (for $i=1,2$).
Then, since $\Sigma$ is a minimal annulus, we have $F_{1}=-F_{2}$. But
$\Sigma$ has free boundaries, then it follows that $\nu(p)=p/|p|$ for each
$p\in\zeta_{i}$, $i=1,2$, so we obtain that $\zeta_{1}$ and
$\varrho(\zeta_{2})$ are convex curves contained in $\mathbb{S}^{2}$ with
opposite center mass. On the other hand, since the convex hull of $\zeta_{1}$
(resp. $\varrho(\zeta_{2})$), ${\rm conv}(\zeta_{1})$ (resp. ${\rm
conv}(\varrho(\zeta_{2}))$), is given by the intersection (see [51]) of all
closed halfspaces containing $\zeta_{1}$ (resp. $\varrho(\zeta_{2})$), if $H$
is a closed halfspace of equation $\langle q,v_{0}\rangle\leq r_{0}$, for
$v_{0}\in\hbox{\bb R}^{3}\setminus\\{0\\},r_{0}\in\hbox{\bb R}$, with
$\zeta_{1}\subset H$ (resp. $\varrho(\zeta_{2})\subset H$) then it is clear
that $F_{1}\in H$ (resp. $F_{2}\in H$). Moreover, $F_{i}\in\partial H$ if and
only if $\zeta_{i}\subset\partial H$. Therefore, $F_{1}$ (resp. $F_{2}$) is a
non zero vector with $F_{1}\in{\rm conv}(\zeta_{1})\setminus\zeta_{1}$ (resp.
$F_{2}\in{\rm conv}(\varrho(\zeta_{2}))\setminus\varrho(\zeta_{2})$). This
fact proves that both $\zeta_{1}$ and $\varrho(\zeta_{2})$ cannot be contained
in a common closed hemisphere of $\hbox{\bb S}^{2}$, otherwise
$F_{1}+F_{2}\neq 0$. Thus the result is proven. ∎

Let $B_{i}$ be the convex inner domain inside $\mathbb{S}^{2}(r_{i})$,
$r_{1}=1$ and $r_{2}=\tilde{r}$, bounded by the boundary component
$\zeta_{i}$, $i\in\\{1,2\\}$. Consider the closed cones of $\hbox{\bb R}^{3}$
given by

${\cal C}_{i}=\\{\lambda p:\ \lambda\geq 0,\ p\in\zeta_{i}\cup B_{i}\\},\quad
i\in\\{1,2\\}.$ (5.3)

From now on we fix the outward orientation $N$, that is, $N$ points towards
$\hbox{\bb S}^{2}(r_{i})\setminus\textup{cl}(B_{i})$ along each boundary
component $\zeta_{i}$. For every support plane $P$ to ${\cal C}_{i}$ consider
its exterior unit normal $v_{P}$. The set ${\cal N}_{i}$ of exterior unit
normals $v_{P}$ to ${\cal C}_{i}$ is a convex set of the sphere (see [51]).
Moreover, since $\Sigma$ meets $\hbox{\bb S}^{2}$ and
$\mathbb{S}^{2}(\tilde{r})$ at constant angles, the boundary of ${\cal N}_{i}$
is determined by the curve $N(\zeta_{i})$.

> Claim B: $\mathcal{N}_{1}\cap\mathcal{N}_{2}=\emptyset$.

###### Proof of Claim B.

First, by contradiction, assume that $N(\zeta_{1})\cap
N(\zeta_{2})\neq\emptyset$. Then, there would exist $N_{0}\in N(\zeta_{1})\cap
N(\zeta_{2})$, and so the plane $P$ with equation $\langle q,N_{0}\rangle=0$
satisfies that $\zeta_{1}\cup\zeta_{2}$ is contained in a closed halfspace
determined by $P$. But, this contradicts that $\zeta_{1}$ and
$\varrho(\zeta_{2})$ cannot be contained in a common closed hemisphere of
$\hbox{\bb S}^{2}$ by Claim A. Second, by contradiction again, assume
$\mathcal{N}_{1}\subset\mathcal{N}_{2}$. An argument as above shows that
$\zeta_{1}$ and $\varrho(\zeta_{2})$ are contained in the same hemisphere.
Which is a contradiction. ∎

Therefore, by Claim B, $\hbox{\bb S}^{2}\setminus(N(\zeta_{1})\cup
N(\zeta_{2}))$ has three connected components: two open disks whose closure
are ${\cal N}_{1}$ and ${\cal N}_{2}$ and an open annulus $\Omega$ whose
boundary curves are $N(\zeta_{1})\cup N(\zeta_{2})$. Hence, as the degree of a
differential map between compact manifolds is the same in all regular values
(see [25, Chapter 5, Lemma 1.4]) an elementary topological argument shows that
$N$ is a global diffeomorphism from $\textup{cl}(\Sigma)$ onto the closure of
the annulus $\Omega:=N(\Sigma)\subset\hbox{\bb S}^{2}$. Thus, we can
pushforward the support function $u(p):=\langle p,N(p)\rangle$,
$p\in\textup{cl}(\Sigma)$, to $\textup{cl}(\Omega):=N(\textup{cl}(\Sigma))$,
$\xi$ defined in (5.2), and $\xi$ satisfies (5.1) and has infinitely many
critical points. As a consequence of the injectivity of the Gauss map, we will
deduce that the support function of $\Sigma$ must have a constant sign.

> Claim C: Let $\Sigma$ be an embedded minimal annulus with free boundaries.
> Then,
>
> $\Sigma\cap{\cal C}_{i}=\emptyset,\quad i\in\\{1,2\\},$
>
> where ${\cal C}_{i}$ are the closed convex cones defined by (5.3).

######  Remark 5.1.

This result can be easily thought of if we replace the cones ${\cal C}_{i}$
with a convex body whose boundary is a smooth surface and agrees with the cone
except in a small neighborhood of the vertex. And also replace support planes
with tangent planes.

###### Proof of Claim C.

Let $v_{i}\in B_{i}\subset{\cal C}_{i}$ a fixed vector contained in the
interior of ${\cal C}_{i}$. We first observe that the cone $\mu v_{i}+{\cal
C}_{i}$ given as the translation of ${\cal C}_{i}$ using the vector $\mu
v_{i}$ is included in the interior of ${\cal C}_{i}$ for each $\mu>0$. Then,
by Claim A, $(\mu v_{i}+{\cal C}_{i})\cap\textup{cl}(\Sigma)=\emptyset$ for
$\mu$ large enough. Now, decrease $\mu$ until $(\mu v_{i}+{\cal
C}_{i})\cap\overline{\Sigma}\neq\emptyset$ for the first time. Denote by
$\mu_{0}$ that $\mu$. If $\mu_{0}>0$ then $\mu_{0}v_{i}+{\cal C}_{i}$
intersects $\textup{cl}(\Sigma)$ at an interior point $p_{0}\in\Sigma$. So,
the tangent plane of $\Sigma$ at $p_{0}$ is a support plane of
$\mu_{0}v_{i}+{\cal C}_{i}$ (or equivalently, a support plane of ${\cal
C}_{i}$), with the same interior unit normal $N(p_{0})$. This is a
contradiction because the image of the interior unit normal of $\Sigma$ and
${\cal C}_{i}$ are two disjoint sets of the sphere by Claim B. If $\mu_{0}=0$
then $\zeta_{i}\subset{\cal C}_{i}\cap\textup{cl}(\Sigma)$. But it is
impossible the existence of another point $p_{0}\in\Sigma$ at ${\cal
C}_{i}\cap\textup{cl}(\Sigma)$ using the same reasoning as in the case
$\mu_{0}>0$. ∎

It is important to recall that we are considering the outward orientation.
Then, Claim C implies that the support function $u(p)=\langle p,N(p)\rangle$
is positive at some point on $\Sigma$.

> Claim D: Let $\Sigma$ be an embedded minimal annulus with free boundaries.
> Then, its support function $u(p)$ is strictly positive in $\Sigma$ and
> vanishes along $\partial\Sigma$. Here, $N$ is the outward orientation.

###### Proof of Claim D.

First, we will show that $\Sigma\cap L({\bf o},p)=\left\\{p\right\\}$ for all
$p\in\Sigma$, where $L({\bf o},p)$ is the half-line starting at the origin
${\bf o}$ passing through $p\in\Sigma$. Assume this was not the case. Consider
the dilation $\Sigma_{\lambda}:=\lambda\Sigma$, $\lambda\geq 1$. Observe that
$\partial\Sigma_{\lambda}\subset\mathcal{C}_{1}\cup\mathcal{C}_{2}$, where
$\mathcal{C}_{i}$, $i=1,2$, are the closed cones defined in (5.3). Then, for
$\lambda>1$ big enough, $\Sigma\cap\Sigma_{\lambda}=\emptyset$. If there were
a point $q\in\Sigma$ so that $\left|\Sigma\cap L({\bf o},p)\right|\geq 2$,
then there exists $\bar{\lambda}>1$ so that
$\Sigma\cap\Sigma_{\bar{\lambda}}\neq\emptyset$ and
$\Sigma\cap\Sigma_{\lambda}=\emptyset$ for all $\bar{\lambda}<\lambda$,
observe that the intersection points must be at the interior of the surfaces.
This implies that the tangent planes at the intersection points
$\Sigma\cap\Sigma_{\bar{\lambda}}$ must be equal and, also, the normals must
be equal. But this contradicts that the Gauss map is a global diffeomorphism.
Therefore, $u(p)=\langle p,N(p)\rangle\geq 0$ (with the outward orientation)
on $\textup{cl}(\Sigma)$. Since $u=0$ along $\partial\Sigma$ and $u$ is a
Jacobi function, it is standard that either $u\equiv 0$, which is impossible,
or $u>0$ on the interior. This finishes the proof. ∎

######  Remark 5.2.

When $\Sigma\subset\hbox{\bb B}^{3}$ is an embedded free boundary minimal
surface, Kusner and McGrath in [31, Corollary 4.2] proved that $\Sigma$ is a
radial graph. However, their proof depends strongly on the two-piece property
for embedded free boundary minimal surfaces in $\hbox{\bb B}^{3}$ (see [34]),
a property that is not clear to hold in our case.

Observe that Claim D implies that $\textup{cl}(\Sigma)$ is a radial graph.
Finally, we need to prove:

> Claim E: If $\\#{\rm Crit}(u)=+\infty$, then $\tilde{\gamma}={\rm
> Max}(u)={\rm Crit}(u)$ is a closed simple curve.

###### Proof of Claim E.

If $\\#{\rm Crit}(u)=+\infty$, by analyticity, we already know that there
exists, at least, one simple curve $\tilde{\gamma}$ contained in ${\rm
Crit}(u)$; we will see that this is the only one. Consider
$\varrho:\textup{cl}(\Sigma)\to\hbox{\bb S}^{2}$ the radial projection
$\varrho(p)=p/|p|$ and $N:\textup{cl}(\Sigma)\to\hbox{\bb S}^{2}$ the outward
orientation. Is it easy to observe that $p\in{\rm Crit}(u)$ if, and only if,
$\varrho(p)=N(p)$ and hence
$\gamma=\varrho(\tilde{\gamma})=N(\tilde{\gamma})$. Without loss of
generality, up to a rotation, we can assume that the flux along the boundary
components of $\Sigma$ is vertical. Thus, since $\textup{cl}(\Sigma)$ is a
radial graph, $\gamma$ divides $\hbox{\bb S}^{2}$ into two components
$S^{+}\cup S^{-}=\hbox{\bb S}^{2}\setminus\gamma$ where $S^{+}$ contains the
north pole and $S^{-}$ contains the south pole. Moreover,
$\gamma\subset\varrho(\textup{cl}(\Sigma))\cap N(\textup{cl}(\Sigma))$ divides
both set into two components, that is,

$\mathcal{F}^{\pm}=\varrho(\textup{cl}(\Sigma))\cap S^{\pm}\text{ and
}\Omega^{\pm}=\ N(\textup{cl}(\Sigma))\cap S^{\pm}$

Since both, $\varrho$ and $N$, are one-to-one, they map a component
$\textup{cl}(\Sigma)\setminus\tilde{\gamma}$ into a component of their image.
Let $\Sigma^{+}$ be the connected component of $\Sigma\setminus\tilde{\gamma}$
whose boundary component $\zeta^{+}=\partial\Sigma^{+}\setminus\tilde{\gamma}$
has flux in the positive vertical axis. Thus, by the free boundary condition
$N(\Sigma^{+})=\Omega^{-}$ and $\varrho(\Sigma^{+})=\mathcal{F}^{+}$, which
implies that $N(\Sigma^{+})\cap F(\Sigma^{+})=\emptyset$. The same happens on
the other component $\Sigma^{-}$ of $\Sigma\setminus\tilde{\gamma}$ and
boundary component $\zeta^{-}=\partial\Sigma^{-}\setminus\tilde{\gamma}$.
Therefore

$N(\Sigma\setminus\tilde{\gamma})\cap\varrho(\Sigma\setminus\tilde{\gamma})=\emptyset,$

which means that the only critical points of $u$ are those of
$\tilde{\gamma}$. So far, we have proven that $\tilde{\gamma}={\rm
Crit}(u)={\rm Crit}(d)$, where $d(p)=|p|^{2}$ is the square distance function.
Since $\Sigma$ is minimal, the only possibility is that $\tilde{\gamma}$ is a
curve of absolute minimums of $d$, that is, there exists $\tilde{R}\in(0,1)$
so that $\tilde{\gamma}\subset\hbox{\bb S}^{2}(\tilde{R})$, $\Sigma$ is
tangential to $\hbox{\bb S}^{2}(\tilde{R})$ along $\tilde{\gamma}$, and
$\tilde{\gamma}$ is line of curvature of $\Sigma$ by Joachimstahl’s theorem
[10, p. 152]. This proves Claim E. ∎

Then, we conclude that $(\Omega:=N(\Sigma),\xi:=u\circ N^{-1})$ is a solution
to (4.1) satisfying the conditions of Theorem A since
$\gamma=N(\tilde{\gamma})={\rm Max}(\xi)$ by Claim E. Hence
$(\xi,\Omega)\equiv(\xi_{R},\Omega_{R})$ for some $R\in[0,1)$, this finishes
the proof of Theorem B. ∎

Finally, Corollary C follows as an immediate consequence of Theorem B, taking
into account that $C_{0}$ is the only free boundary catenoid of the family in
$\hbox{\bb B}^{3}$.

## Acknowledgments

The first author is partially supported by Spanish MIC Grant
PID2020-118137GB-I00 and MIC-NextGenerationEU Grant 30.RP.23.00.04
CONSOLIDACION2022. The second author is partially supported by the Maria de
Maeztu Excellence Unit IMAG, reference CEX2020-001105-M, funded by
MCINN/AEI/10.13039/ 501100011033/CEX2020-001105-M.

## References

  * [1] V. Agostiniani, S. Borghini, L. Mazzieri, On the Serrin problem for ring-shaped domains, to appear on J. Eur. Math. Soc.
  * [2] A. D. Alexandrov, _Uniqueness theorems for surfaces in the large,_ Amer. Math. Soc. Transl., Vol. 21 no. 2 (1962), 412-416.
  * [3] H. Berestycki, L. Caffarelli, L. Nirenberg, _Monotonicity for elliptic equations in unbounded Lipschitz domains_ , Communications on Pure and Applied Mathematics, Vol. 50 (1997), 1089-1111.
  * [4] H. Berestycki, L. Nirenberg, _On the method of moving planes and the sliding method_. Bol. Soc. Bras. Mat., Vol. 22 (1991), 1-37.
  * [5] S. Borghini, Static Black Hole Uniqueness for nonpositive masses, Nonlinear Analysis, Vol. 220 (2022), paper no. 112843, 1-12.
  * [6] S. Borghini, _Symmetry results for Serrin-type problems in doubly connected domains_ , Math. Eng., Vol. 5 no. 2 (2023), paper no. 027, 1-16.
  * [7] S. Borghini, P. T. Chrusciel, L. Mazzieri, On the uniqueness of Schwarzschild-de Sitter spacetime, to appear on Archive for Rational Mechanics and Analysis, (2023).
  * [8] G. Ciraolo, L. Vezzoni, On Serrin’s overdetermined problem in space forms, Manuscripta Math., Vol. 159 (2019), 445-452.
  * [9] A. Carlotto, _Free boundary minimal surfaces: a survey of recent results_ , Rendiconto dell’Accademia delle Scienze Fisiche e Matematiche, Serie 4, Vol. 86 no. 1 (2019), 103-121.
  * [10] M. P. do Carmo, Differential Geometry of Curves and Surfaces. Prentice-Hall, Englewood Cliffs, New Jersey (1976).
  * [11] M. del Pino, F. Pacard, J. Wei, _Serrin’s overdetermined problem and constant mean curvature surfaces_ , Duke Math. J., Vol. 164 (2015), 2643-2722.
  * [12] B. Devyver, Index of the critical catenoid Geom. Dedicata, Vol. 199 (2019), 355-371.
  * [13] M. Domínguez-Vázquez, A. Enciso, D. Peralta-Salas, _Overdetermined boundary problems with nonconstant Dirichlet and Neumann data_ , Anal. P.D.E., in press.
  * [14] J.M. Espinar, A. Farina, L. Mazet, _f-extremal domains in hyperbolic space_ , Calc. Var., Vol. 60 (2021), paper no. 112, 1-19.
  * [15] J.M. Espinar, J. Mao, _Extremal domains on Hadamard manifolds_ , Journal of Differential Equations, Vol. 265 (2018), 2671-2707.
  * [16] J. M. Espinar, L. Mazet, _Characterization of f-extremal disks_ , Journal of Differential Equations, Vol. 266 (2019), 2052-2077.
  * [17] M. M. Fall, I. A. Minlend, T. Weth, Serrin’s overdetermined problem on the sphere Calc. Var., Vol. 57 no. 3 (2018), paper no. 3, 1-24.
  * [18] A. Farina, A. Roncoroni, Serrin’s type problems in warped product manifolds, Communications in Contemporary Mathematics, Vol. 24 no. 4 (2022), paper no. 2150020, 1-21.
  * [19] A. Fraser, R. Schoen, _The first Steklov eigenvalue, conformal geometry, and minimal surfaces_ , Advances in Mathematics, Vol. 226 no. 5 (2011), 4011-4030.
  * [20] A. Fraser, R. Schoen, _Sharp eigenvalue bounds and minimal surfaces in the ball_ , Invent. Math., Vol. 203 no. 3 (2016), 823-890.
  * [21] R. Goldman, Curvature formulas for implicit curves and surfaces, Computer Aided Geometric Design, Vol. 22 no. 7 (2005), 632-658.
  * [22] P. Hartman, A. Winter, _On the Third Fundamental Form of a Surface_ , American Journal of Mathematics, Vol. 75 no. 2 (1953), 298-334.
  * [23] F. Hélein, L. Hauswirth, F. Pacard, _A note on some overdetermined elliptic problem_ , hal-00444828 (2010).
  * [24] H. Hopf, Differential Geometry in the large, Springer, Berlin, Lecture Notes in Math., Vol. 1000 (1983).
  * [25] M. W. Hirsch, _Differential Topology_ , Graduate Texts in Mathematics, Vol. 33 (1994).
  * [26] N. Kamburov, L. Sciaraffia , Nontrivial solutions to Serrin’s problem in annular domains, Ann. I. H. Poincaré AN, Vol. 38 no. 1 (2021), 1-22.
  * [27] H. Karcher, Umkreise und Inkreise konvexer Kurven in der spharischen und der hyperbolischen Geometrie, Math. Ann., Vol. 177 (1968), 122-132.
  * [28] D. Khavinson, A.Y. Solynin, D. Vassilev, _Overdetermined Boundary Value Problems, Quadrature Domains and Applications_ , Computational Methods and Function Theory, Vol. 5 no. 1 (2005), 19-48.
  * [29] D. Kinderlherer, L. Nierenberg, Regularity in free boundary value problems, Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4e series, Vol. 4 no. 2 (1977), 373-391.
  * [30] S. G. Krantz, H. R. Parks, A primer of real analytic functions. Birkhäuser Advanced Texts: BaslerLehrbücher. [Birkhäuser Advanced Texts: Basel Textbooks]. Birkhäuser Boston, Inc., Boston, MA, second edition (2002).
  * [31] R. Kusner and P. McGrath, _On Steklov Eigenspaces for Free Boundary Minimal Surfaces in the Unit Ball_ , arXiv:2011.06884.
  * [32] S. Kumaresan, J. Prajapat, Serrin’s result for hyperbolic space and sphere Duke Mathematical Journal, Vol. 91 no. 1 (1998), 17-28.
  * [33] M. Li, _Free boundary minimal surfaces in the unit ball: recent advances and open questions_ , Proceedings of the International Consortium of Chinese Mathematicians, 2017 (First Annual Meeting), 401-436.
  * [34] V. Lima, A. Menezes, _A two-piece property for free boundary minimal surfaces in the ball_ , Transactions of the American Mathematical Society, Vol. 374 no. 3 (2021), 1661-1686.
  * [35] P. McGrath, _A Characterization of the Critical Catenoid_ , Indiana University Mathematics Journal, Vol. 67 no. 2 (2018), 889-897.
  * [36] P. Mira, _Overdetermined elliptic problems in topological disks_ , Journal of Differential Equations, Vol 264 (2018), 6994-7005,
  * [37] S. Montiel, A. Ros, _Cuves and Surfaces_ , Graduate Studies in Mathematics, Vol. 69 (2009).
  * [38] J. C. C. Nitsche, Stationary partitioning of convex bodies, Arch. Rational Mech. Anal., Vol. 89 no. 1 (1985), 1-19.
  * [39] F. Pacard, P. Sicbaldi, _Extremal domains for the first eigenvalue of the Laplace-Beltrami operator_ , Ann. Inst. Fourier, Grenoble, Vol. 59 no. 2 (2009) 515-542.
  * [40] F. Pacella, _Symmetry Results for Solutions of Semilinear Elliptic Equations with Convex Nonlinearities_ , Journal of Functional Analysis, Vol. 192 (2002), 271-282.
  * [41] F. Pacella, T. Weth, _Symmetry of solutions to semilinear elliptic equations via Morse index_ , Proceedings of the American Mathematical Society, Vol. 135 no. 6 (2007), 1753-1762.
  * [42] P. Pucci, J. Serrin, _The Maximum Principle_ , Progress in Nonlinear Differential Equations and Their Applications, Vol. 73 (2007).
  * [43] G. Qiu, C. Xia, Overdetermined Boundary Value Problems in $\mathbb{S}^{n}$, J. Math.Study, Vol. 50 no. 2 (2017), 165-173.
  * [44] W. Reichel, _Radial symmetry by moving planes for semilinear elliptic BVPs on annuli and other non-convex domains_ , Elliptic and Parabolic equations (1994), 164-182.
  * [45] W. Reichel, _Radial Symmetry for Elliptic Boundary-Value Problems on Exterior Domains_ , Arch. Rational Mech. Anal., Vol. 137 no. 4 (1997), 381-394.
  * [46] A. G. Reznikov, Linearization and explicit solutions of the minimal surface equation, Publicacions Matematiques, Vol. 36 (1992), 39-46.
  * [47] A. Ros, P. Sicbaldi, _Geometry and topology of some overdetermined elliptic problems_ , Journal of Differential Equations, Vol. 255 no. 5 (2013), 951-977.
  * [48] D. Ruiz, _Nonsymmetric sign-changing solutions to overdetermined elliptic problems in bounded domains_ , arXiv:2211.14014.
  * [49] D. Ruiz, P. Sicbaldi, J. Wu, _Overdetermined elliptic problems in nontrivial contractible domains of the sphere_ , to appear in J. Math. Pures Appliquées.
  * [50] F. Schlenk, P. Sicbaldi, _Bifurcating extremal domains for the first eigenvalue of the Laplacian_ , Advances in Mathematics, Vol. 229 (2012), 602-632.
  * [51] R. Schneider, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and its Applications, Vol. 44 (1993).
  * [52] J. Serrin, A symmetry problem in potential theory, Arch. Rational Mech. Anal., Vol. 43 (1971), 304-318.
  * [53] R. Souam, Schiffer’s problem and an isoperimetric inequality for the first buckling eigenvalue of domains on $\hbox{\bb S}^{2}$, Ann. Global Anal. Geom., Vol. 27 no. 4 (2005), 341-354.
  * [54] J. Soucek, V. Soucek, _Morse-Sard theorem for real-analytic functions_ , Commentationes Mathematicae Universitatis Carolinae, Vol. 13 no. 1 (1972), 45-51.
  * [55] P. Sicbaldi, _New extremal domains for the first eigenvalue of the Laplacian in flat tori_ , Calc. Var., Vol. 37 (2010), 329-344.
  * [56] B. Sirakov, _Symmetry for exterior elliptic problems and two conjectures in potential theory_ , Ann. Inst. Henri Poincaré, Anal. non linéaire, Vol. 18 no. 2 (2001), 135-156.
  * [57] M. Traizet, _Classification of the solutions to an overdetermined elliptic problem in the plane_ , Geom. Funct. Anal. Vol. 24 (2014), 690-720.
  * [58] H. F. Weinberger, Remark on the preceding paper of Serrin. Arch. Rational Mech. Anal., Vol. 43 (1971), 319-320.