# Non planar free boundary minimal disks into ellipsoids

Romain Petrides  Romain Petrides, Université de Paris, Institut de
Mathématiques de Jussieu - Paris Rive Gauche, bâtiment Sophie Germain, 75205
PARIS Cedex 13, France<EMAIL_ADDRESS>
###### Abstract.

We prove the existence of embedded non planar free boundary minimal disks into
rotationally symmetric ellipsoids of $\mathbb{R}^{3}$. The construction relies
on the optimization of combinations of first and second Steklov eigenvalues
renormalized by the length of the boundary, among metrics on the disk. We also
prove that non planar free boundary harmonic maps from a disk into a ellipsoid
of $\mathbb{R}^{3}$ such that the coordinate functions are first or second
Steklov eigenfunctions with respect to the associated critical metric is a
minimal immersion (without branched points) and that if the critical metric is
even with respect to the coordinates of the disk, then the minimal immersion
is an embedding.

We consider the following 2-ellipsoid parametrized by
$\sigma:=(\sigma_{1},\sigma_{2},\sigma_{3})$:

$\mathcal{E}_{\sigma}:=\\{(x_{1},x_{2},x_{3})\in\mathbb{R}^{3};\sigma_{1}x_{1}^{2}+\sigma_{2}x_{2}^{2}+\sigma_{3}x_{3}^{2}=1\\}$

with semi-axes $\left(\sigma_{i}\right)^{-\frac{1}{2}}$ for $i=1,2,3$. The
equatorial curves $\\{x_{i}=0\\}\cap\mathcal{E}_{\sigma}$ for $i=1,2,3$ are
simple closed geodesic in $\mathcal{E}_{\sigma}$. By a result by Morse [Mor31]
(see also [Kli78]), they are also the only ones for ellipsoids close to the
sphere. In a celebrated result, proved by the combination of the min-max
method by Lusternik and Schnirelmann [LS47] and mean flow methods by Grayson
[Gra89], any closed Riemannian two-sphere contains at least 3 simple closed
geodesics. It was a refinement of the min-max method initiated by Birkhoff
[Bir17] to prove existence of simple closed geodesics in a Riemannian
topological sphere. Many generic ellipsoids $\mathcal{E}_{\sigma}$ realize
then the minimal number of simple closed geodesics $3$. Later, Viesel [Vie71]
proved existence of ellipsoids that contain an arbitrary number of simple
closed geodesics (see also [Kli95]).

In the same spirit, we ask for embedded free boundary minimal disks into
$\mathcal{E}_{\sigma}$. By definition, free boundary minimal disks into a
surface $\Sigma\subset\mathbb{R}^{3}$ are topological disks that are critical
points of the area functional among all disks $D$ such that $\partial
D\subset\Sigma$. They are exactly the minimal disks $D$ of $\mathbb{R}^{3}$
such that $\partial D\subset\Sigma$ and $D$ meets orthogonally $\Sigma$ on
$\partial D$. The planar disks $\\{x_{i}=0\\}\cap
co\left(\mathcal{E}_{\sigma}\right)$ are the first trivial examples of free
boundary minimal embedded disks with $\Sigma=\mathcal{E}_{\sigma}$. In the
current paper we are interested in the following question:

######  Question 1 (Dierkes, Hildebrandt, Küster, Wohlrab, 1993, [DHKW92]
p335).

Are there non-planar embedded free boundary minimal disks into ellipsoids of
$\mathbb{R}^{3}$ ?

One analogous question by Yau in 1987 [Yau87] was raised in the case of
2-spheres into 3-ellipsoids: are the only minimal 2-spheres in an ellipsoid
centered about the origin in $\mathbb{R}^{4}$ the planar ones? Haslhofer and
Ketover recently proved that the answer is no [HK19] for sufficiently
elongated ellipsoids, thanks to subtle min-max and mean curvature flow
methods. Since then, other constructions of non planar minimal spheres in
[BP22] and [Pet23] have been given by different methods. In [BP22], Bettiol
and Piccione build an arbitrary number of non planar minimal spheres for
sufficiently elongated rotationally symmetric ellipsoids by bifurcations
methods. In [Pet23], we use analogous techniques as in the current paper with
optimization of combination of eigenvalues of the Laplacian.

Notice that the free boundary minimal 2-disks into the boundary of a 3-d
manifold can purely be seen as a 1-d problem into a 2-d surface in a non local
setting: they are half-harmonic maps into the surface. More precisely, if it
is classical that $f:\mathbb{R}^{2}_{+}\to\mathbb{R}$ is a harmonic function,
then $-\partial_{y}f=\Delta^{\frac{1}{2}}f$ on $\mathbb{R}\times\\{0\\}$.
Following the definition by Da Lio and Rivière [DR11], if
$\Sigma\subset\mathbb{R}^{3}$ is a surface, we say that
$f:\mathbb{R}\to\Sigma$ is a half harmonic map into $\Sigma$ if it is a
critical for the energy
$\int_{\mathbb{R}}\left|\Delta^{\frac{1}{4}}f\right|^{2}$ under the constraint
$f\in\Sigma$ a.e. The Euler-Lagrange equation reads as

$\Delta^{\frac{1}{2}}f\perp T_{f}\Sigma\text{ on }\mathbb{R}\times\\{0\\}.$

We set $\Phi=\hat{f}\circ h:\mathbb{D}\to\mathbb{R}^{3}$, where
$h:\mathbb{D}\to\mathbb{R}^{2}_{+}$ is the biholomorphism such that $h(1)=0$,
$h(0)=1$ and $h(-1)=\infty$ and $\hat{f}:\mathbb{R}^{2}_{+}\to\mathbb{R}^{3}$
is the harmonic extension of $f$. Then $f$ is a half-harmonic map into
$\Sigma$ if and only if $\Phi$ is a free boundary harmonic map into $\Sigma$
(defined as critical maps of $\int_{\mathbb{D}}\left|\nabla\Psi\right|^{2}$
under the constraint $\Psi(\mathbb{S}^{1})\subset\Sigma$ a.e). In this case,
the equation reads as:

$\begin{cases}\Delta\Phi=0\text{ in }\mathbb{D}\\\
\partial_{r}\Phi\parallel\Delta^{\frac{1}{2}}f\perp T_{\Phi}\Sigma\text{ on
}\mathbb{S}^{1}.\end{cases}$

The restriction at the boundary then satisfies a non local version of the
equation of closed geodesics on $\Sigma$. By a classical argument on the Hopf
differential, if $\Phi$ is a free boundary harmonic map, then it is a branched
conformal immersion and then a free boundary minimal branched immersed disk
into $\Sigma$. If $\Sigma=\mathcal{E}_{\sigma}$, there is a natural weaker
question than the previous one:

######  Question 2.

Is there a non planar simple closed half harmonic maps
$\Phi:\mathbb{S}^{1}\to\mathcal{E}_{\sigma}$ ?

The answer of Question 1 and 2 is no if the ellipsoid is a round sphere
($\sigma_{1}=\sigma_{2}=\sigma_{3}$) by Nitsche [Nit85]: the only free
boundary minimal disks into spheres are the planar ones. This result is the
same as the classical one for 2-spheres into 3-spheres by Almgren [Alm66]. In
this context, free-boundary minimal disks are more rigid, because even if we
increase the dimension of the target sphere, the only free boundary branched
immersed minimal disks are the planar ones [FS15], while we have existence of
many non planar minimal spheres into higher dimensional spheres (leading to a
theory initiated by Calabi [Cal67] and Barbosa [Bar75]).

The construction of free boundary minimal disks has been investigated by
Struwe [Str84], Fraser [Fra00] (with methods inspired by Sacks and Uhlenbeck
[SU81]), Laurain, Petrides [LP19] and Lin, Sun, Zhou [LSZ20] (with min-max
methods inspired the replacement or shortening process by Birkhoff and Colding
and Minicozzi [CM08]), Gruter-Jost [GJ86] and Li-Zhou [LZ21] (with min-max
methods coming from geometric measure theory initiated by Almgren [Alm62]). In
all these constructions, the variational methods are not as refined as we need
to obtain new embedded free boundary minimal disks into 2-ellipsoids. As in
the closed case, the free boundary version [Wan20] by Wang of the proof of the
celebrated Yau conjecture by Song [Son23] is not helpful to build new disks,
since they do not give information on the topological type of their minimal
surfaces.

The submentioned methods in the closed case: [HK19] or [BP22] may still be
possible to address the free boundary case but involve new difficulties. For
instance in [BP22], the authors spectacularly use a rotational invariance
assumption to prove existence of an arbitrary number of embedded non-planar
rotationally invariant minimal spheres into sufficiently ellongated
rotationally symmetric 3-d ellipsoids with bifurcation methods. However, we
cannot build any embedded non-planar rotationally invariant free boundary
minimal disk into a rotationally symmetric 2-ellipsoid.

In the current paper, we shall use an indirect way: the shape optimization of
combinations of Steklov eigenvalues. Indeed, in [Pet21], (see also [Pet22] and
[PT23]) the author noticed that all possibly branched conformally immersed
minimal map $\Phi$ from a 2-surface with boundary $\Sigma$ into a n-ellipsoid
$\mathcal{E}$ can be seen as a shape associated to a critical metric of some
functional depending on combinations of Steklov eigenvalues with respect to
the metric. In addition, the $k$-th parameter of the ellipsoid is associated
to a Steklov eigenvalue involved in the functional at the critical metric, and
the $k$-th coordinate of the immersion is an eigenfunction associated to this
eigenvalue. This is a generalization of a result by Fraser and Schoen when
only one eigenvalue appears in the functional: the target manifold is then a
ball. Their discovery led to a method for the construction of free boundary
minimal surface of genus $0$ into 2-spheres [FS16] extended in [Pet19] for
higher genuses and higher eigenvalues. Applying the variational methods
developped for combinations of Steklov eigenvalues in [Pet21] and [Pet22], we
are in position to prove the following result:

######  Theorem 0.1.

For any $s\neq 0$, there is a one parameter family
$(p_{s,t})_{t>t_{\star}(s)}$ for some $t_{\star}(s)>0$ such that there is an
embedded free boundary non planar minimal topological disk $D_{s,t}$ into the
rotationally symmetric ellipsoid

$\mathcal{E}_{\sigma_{s,t}}:=\\{x\in\mathbb{R}^{3};p_{s,t}x_{0}^{2}+x_{1}^{2}+x_{2}^{2}=1\\}$

and such that the coordinates $x_{0},x_{2},x_{2}$ are first and second Steklov
eigenfunctions for an induced metric $g_{s,t}=e^{2v_{t}}\xi$ on $D_{s,t}$ such
that

$e^{v_{s,t}}=\frac{1}{\left(p_{s,t}^{2}x_{0}^{2}+x_{1}^{2}+x_{2}^{2}\right)^{\frac{1}{2}}}\text{
on }\partial{D_{s,t}}\text{ and }L_{s,t}=\int_{\partial D_{s,t}}dL_{g_{s,t}}$

where $\xi$ is the standard Euclidean metric. For any $s$ and $t_{1}<t_{2}$,
$D_{s,t_{1}}$ is not isometric to $D_{s,t_{2}}$. Moreover $t\mapsto
L_{s,t}p_{s,t}$ is decreasing and $t\mapsto L_{s,t}$ is increasing, and we
have that $p_{s,t}\to 0$ and $L_{s,t}\to 4\pi$ as $t\to+\infty$ and that
$D_{s,t}$ converges as varifolds to the disk $\\{0\\}\times\mathbb{D}^{2}$
with multiplicity $2$ as $t\to+\infty$.

This theorem gives a positive answer to Question 1 and Question 2. We obtain
$D_{s,t}=\Phi\left(\mathbb{D}\right)$ as the image of the minimal free
boundary immersion $\Phi:\mathbb{D}\to\mathcal{E}_{\sigma_{s,t}}$ associated
to a critical metric $g_{s,t}=e^{2v_{s,t}}(dx^{2}+dy^{2})$ for a minimization
problem with respect to the functionals

$F_{s,t}:g\mapsto\left(\bar{\sigma}_{1}(g)^{-s}+t\bar{\sigma}_{2}(g)^{-s}\right)^{\frac{1}{s}}$

for $t>0$ and $s\neq 0$, where for $k\geq 0$,
$\bar{\sigma}_{k}(g)=\sigma_{k}(g)L_{g}(\mathbb{S}^{1})$ denotes the
renormalized $k$-th Steklov eigenvalue $\sigma_{k}(g)$ by the length
$L_{g}(\mathbb{S}^{1})$ of the boundary of $\mathbb{D}$ with respect to $g$.
By convention, notice that $\sigma_{0}=0$ is the simple eigenvalue associated
to constant functions because $\mathbb{D}$ is connected and $\sigma_{1}(g)>0$
is the first non-zero eigenvalue and that we can have
$\sigma_{2}(g)=\sigma_{1}(g)$ if $\sigma_{1}(g)$ is multiple. Notice that we
can also try many other functionnals depending on $\bar{\sigma}_{1}$ and
$\bar{\sigma}_{2}$ than $F_{s,t}=:h_{s,t}(\bar{\sigma}_{1},\bar{\sigma}_{2})$.

With the notations of Theorem 0.1, we have
$\bar{\sigma}_{1}\left(D_{s,t},g_{s,t}\right)=p_{s,t}L_{s,t}$ and
$\bar{\sigma}_{2}\left(D_{s,t},g_{s,t}\right)=L_{s,t}$. Since
$\bar{\sigma}_{2}(g_{s,t})\to 4\pi$, 0.3 shows that $g_{s,t}$ is a particular
maximizing sequence for $\bar{\sigma}_{2}$ as $t\to+\infty$. Notice also that

$\begin{split}L_{s,t}=&\int_{\partial D_{s,t}}dL_{g_{s,t}}=\int_{\partial
D_{s,t}}(p_{s,t}x_{0}^{2}+x_{1}^{2}+x_{2}^{2})dL_{g_{s,t}}\\\
=&\int_{D_{s,t}}\left(\left|\nabla x_{0}\right|_{g_{s,t}}^{2}+\left|\nabla
x_{1}\right|_{g_{s,t}}^{2}+\left|\nabla
x_{2}\right|_{g_{s,t}}^{2}\right)dA_{g_{s,t}}=2A(D_{s,t})\end{split}$

where $A(D_{s,t})$ is the area of the minimal surface. As explained below, we
emphasize that $\Phi:\mathbb{D}\to\mathcal{E}_{\sigma_{s,t}}$ is an embedding
and its geometric properties (symmetries, intersection with equatorial planes)
are described in Claim 1.3.

Let’s explain the steps of proof for Theorem 0.1. It is not a priori clear
that the following properties hold true:

  * (1)

$F_{s,t}$ has a minimizer, and then there is a minimal (possibly branched)
free boundary immersion $\Phi$ from $\mathbb{D}$ into a $n$-ellipsoid
associated to this minimal metric.

  * (2)

Up to rearrangement of coordinates, $\Phi$ has at most $3$ coordinates and the
target of $\Phi$ is the convex hull of a $2$-ellipsoid
$\mathcal{E}_{\sigma_{s,t}}$ for
$\sigma_{s,t}=(\sigma_{1}(g_{s,t}),\sigma_{2}(g_{s,t}),\sigma_{2}(g_{s,t}))$.

  * (3)

$D_{s,t}=\Phi\left(\mathbb{D}\right)$ is non-planar.

  * (4)

$\Phi$ is an embedding.

Let’s explain every difficulty separately:

(1): For the choice $t=0$, we know by Weinstock inequality [Wei54]

(0.1) $\bar{\sigma}_{1}\leq 2\pi$

and that $2\pi$ is only realized for the flat disk. Weinstock arguments also
give that for $t=1$ and $s=1$,

(0.2) $\frac{1}{\bar{\sigma}_{1}}+\frac{1}{\bar{\sigma}_{2}}\geq\frac{1}{\pi}$

and the unique minimizer is the flat disk. For $t=\infty$: the only eigenvalue
which appears is $\bar{\sigma}_{2}$, then, by Hersch Payne and Schiffer
[HPS75]

(0.3) $\bar{\sigma}_{2}<4\pi$

and it is not realized (see [GP10]). $4\pi$ corresponds to the disjoint union
of two flat disks of same boundary length. In particular, there are maximizing
sequences approaching $4\pi$ which blow up and ”bubble converge” to a disjoint
union of two disks (see for instance Theorem 0.2 below). In [Pet21], we give
an optimal assumption on the functional to ensure that bubbling cannot occur
and then to ensure existence of a minimizer. In our context, setting
$h_{s,t}(\sigma_{1},\sigma_{2})=\left(\sigma_{1}^{-s}+t\sigma_{2}^{-s}\right)^{\frac{1}{s}}$,
the condition reads as

(0.4) $\inf_{g}F_{s,t}<h_{s,t}(0,4\pi).$

This condition is automatic if $s>0$, but needs to be discussed if $s<0$. In
section 3, we prove the following

######  Theorem 0.2.

There is a one parameter family of metrics
$h_{\varepsilon}=e^{2v_{\varepsilon}}(dx^{2}+dy^{2})$ such that

$\bar{\sigma}_{1}(h_{\varepsilon})=\frac{2\pi}{\ln\left(\frac{1}{\varepsilon}\right)}+O\left(\frac{1}{\ln\left(\frac{1}{\varepsilon}\right)^{2}}\right)\text{
and }\bar{\sigma}_{2}(g_{\varepsilon})=4\pi-16\pi\varepsilon+o(\varepsilon)$

as $\varepsilon\to 0$ and $e^{2v_{\varepsilon}}$ satisfies the following
symmetry properties

$\forall(x,y)\in\mathbb{D},e^{2v_{\varepsilon}}(x,y)=e^{2v_{\varepsilon}}(-x,y)=e^{2v_{\varepsilon}}(x,-y).$

This theorem ensures that a minimum also exists for $F_{s,t}$ with $s<0$ and
$t>0$.

(2): The coordinate functions of the minimal free boundary immersion
$\Phi:\mathbb{D}\to\mathbb{R}^{n+1}$ into a $n$-ellipsoid are associated to
first and second Steklov eigenvalues with respect to the minimal metric for
$F_{s,t}$. Up to rearrangement of coordinates, we can assume that they are
independent eigenfunctions. Knowing by Jammes [Jam16], that the multiplicity
of $\sigma_{1}$ and $\sigma_{2}$ on the disk is less than $2$, we obtain that
$n+1=2$ or $n+1=3$. In all the cases, the target surface is the convex hull of
$\mathcal{E}_{\sigma_{s,t}}$ with
$\sigma_{s,t}=(\bar{\sigma}_{1}(g_{s,t}),\bar{\sigma}_{2}(g_{s,t}),\bar{\sigma}_{2}(g_{s,t}))$
where if $n+1=2$, we set by convention $\phi_{3}=0$ and the minimal immersion
is planar.

(3): As noticed in (0.1) and (0.2), for $t=0$ or for $s=1$ and $t\leq 1$ the
only minimizers are flat disks, so that we do not build non planar minimal
disks into ellipsoids in these cases. Then, we have to choose carefully the
parameters $s$ and $t$ that ensure that minimizers cannot be planar ellipses.
In the current paper (see Proposition 1.2), we prove that any planar immersed
disk into the convex hull of an ellipse has to be a diffeomorphism. Then if
$\Phi\left(\mathbb{D}\right)$ is planar, there is only one candidate and the
associated critical metric $g_{c}$ satisfies

$\bar{\sigma}_{1}(g_{c})=2\pi\sqrt{\frac{p}{q}}\text{ and
}\bar{\sigma}_{2}(g_{c})=2\pi\sqrt{\frac{q}{p}}$

where up to a dilatation, $(\mathbb{D},g_{c})$ is isometric to
$E_{p,q}=\\{(x,y)\in\mathbb{R}^{2};px^{2}+qy^{2}=1\\}$ endowed with a metric
$e^{2v}(dx^{2}+dy^{2})$ such that the conformal factor satisfies

$\forall(x,y)\in\partial
E_{p,q},e^{v}(x,y)=\left(p^{2}x^{2}+q^{2}y^{2}\right)^{-\frac{1}{2}}.$

Notice that if $\sqrt{\frac{q}{p}}\geq 2$, then by 0.3, $\bar{\sigma}_{2}$
cannot correspond to a second eigenvalue and we obtain a contradiction. We
denote $\theta_{\star}$ the minimal value such that for any
$\theta:=\frac{q}{p}>\theta_{\star}$, $q$ is not a second Steklov eigenvalue
of $E_{p,q}$: we have $\theta_{\star}\leq 4$.

If in addition, $\Phi\left(\mathbb{D}\right)$ is not a flat disk (that is
$p<q$), we have an extra mass condition given in [Pet21] (see also [PT23] and
Proposition 2.1) coming from the choice of the combination $h_{s,t}$ in the
variational problem $F_{s,t}$, written as

$\frac{p}{q}=\frac{\partial_{2}h_{s,t}(p,q)}{\partial_{1}h_{s,t}(p,q)}$

so that $p^{-s}=tq^{-s}$ and

$\bar{\sigma}_{1}(g_{c})=2\pi t^{-\frac{1}{2s}}\text{ and
}\bar{\sigma}_{2}(g_{c})=2\pi t^{\frac{1}{2s}}$

It implies that if $s>0$ then $1<t\leq\theta_{\star}^{s}<4^{s}$ and that if
$s<0$ then $4^{s}<\theta_{\star}^{s}\leq t<1$.

Moreover, we can compare the value of the functional on the flat disk and on
$E_{p,q}$

$h_{s,t}(2\pi t^{-\frac{1}{2s}},2\pi
t^{\frac{1}{2s}})=\frac{\left(2\sqrt{t}\right)^{\frac{1}{s}}}{2\pi}\text{ and
}h_{s,t}(2\pi,2\pi)=\frac{(1+t)^{\frac{1}{s}}}{2\pi}$

so that if $s>0$ and $t>1$, then the flat disk $E_{1,1}$ is never a minimizer
and if $s<0$, then $E_{p,q}$ for $p<q$ is never a minimizer.

Finally, if $s<0$ and if $\Phi\left(\mathbb{D}\right)$ is a flat disk, we
obtain that

$\frac{(1+t)^{\frac{1}{s}}}{2\pi}=h_{s,t}(2\pi,2\pi)=\inf_{g}F_{s,t}<h_{s,t}(0,4\pi)=\frac{t^{\frac{1}{s}}}{4\pi}$

so that $t<\frac{2^{s}}{1-2^{s}}$.

As a conclusion, we obtain that if

$s>0\text{ and }t>\theta_{\star}^{s}\text{ or }s<0\text{ and
}t\geq\frac{1}{2^{-s}-1}$

then $\Phi\left(\mathbb{D}\right)$ has to be non planar.

(4): While we easily prove that planar possibly branched immersed free
boundary minimal disks into an ellipsoid by first and second eigenfunctions
have to be embedded (see Proposition 1.2), the non-planar case is less
obvious. In the current paper, we prove it under symmetry assumptions,

######  Theorem 0.3.

Let $\Phi:\mathbb{D}\to co\left(\mathcal{E}_{\sigma}\right)$ be a non-planar
possibly branched free boundary minimal immersion such that
$\sigma=(\sigma_{1},\sigma_{2},\sigma_{2})$ only contains first and second
Steklov eigenvalues with respect to the critical weight on $\mathbb{S}^{1}$

$e^{v}:=\frac{\left|\Phi_{\theta}\right|}{\left(\sigma_{1}^{2}\phi_{0}^{2}+\sigma_{2}^{2}\left(\phi_{1}^{2}+\phi_{2}^{2}\right)\right)^{\frac{1}{2}}}=\frac{\left|\Phi_{r}\right|}{\left(\sigma_{1}^{2}\phi_{0}^{2}+\sigma_{2}^{2}\left(\phi_{1}^{2}+\phi_{2}^{2}\right)\right)^{\frac{1}{2}}}$

then $\Phi$ does not have any branched point. Moreover, if

$\forall(x,y)\in\mathbb{S}^{1},e^{v}(x,y)=e^{v}(-x,y)=e^{v}(x,-y),$

then $\Phi$ is an embedding.

Theorem 0.3 is a similar result as the embeddedness of (possibly branched)
minimal free boundary immersions into balls by first Steklov eigenfunctions
for surfaces with boundary of genus $0$ [FS16] but the proof of Theorem 0.3
needs much more refinement (see Section 1.5). We do no know if it is possible
to remove the symmetry assumptions on $e^{v}$ and get the same conclusion, or
at least the weaker conclusion: $\Phi:\mathbb{S}^{1}\to\mathcal{E}_{\sigma}$
parametrizes a simple curve.

At that stage, we do not a priori know if a minimizer of $F_{s,t}$ has
symmetries. In order to obtain Theorem 0.1, we then perform a variational
method on combinations of eigenvalues under symmetry constraints on the
metrics (see Section 2). Thanks to the symmetry properties of the test
functions of Theorem 0.2, condition (0.4) is still realized for the infimum
among symmetric metrics. Then again, no blow up happens on symmetric
minimizing sequences of $F_{s,t}$ for $t\geq t_{\star}(s)$. This leads to the
existence of the expected embedded non planar free boundary minimal disk into
some ellipsoid. By other simple arguments, we deduce the complete proof of
Theorem 0.1 (see Section 2.4)

The current paper is fairly self-contained up to several small lemma already
written in [Pet22], and all the steps of our construction can also be
performed in the analogous closed case (see [Pet23]). In the end of Section
1.2, we also suggest ways to make the construction more explicit in the same
spirit as the bifurcation methods in the closed case [BP22]. New branches of
embedded free boundary minimal disks should appear as soon as the parameter of
the ellipsoid crosses values having multiple eigenvalues for a Steklov problem
on ellipses endowed with a metric conformal to the flat one. Such a
bifurcation method would even give an arbitrary number of embedded disks for
sufficiently elongated ellipsoids.

##  1\. Properties of the free boundary minimal disks into an ellipsoid

###  1.1. Free boundary minimal surfaces into ellipsoids and Steklov
eigenvalues

We start with the following remark: let $\mathcal{E}\subset\mathbb{R}^{n}$ be
an ellipsoid of parameters
$\sigma=diag\left(\sigma_{1},\cdots,\sigma_{n}\right)$, with $\sigma_{i}>0$,
defined by

$\mathcal{E}=\\{(x_{1},\cdots,x_{n})\in\mathcal{E};\sigma_{1}x_{1}^{2}+\cdots+\sigma_{n}x_{n}^{2}=1\\}\hskip
2.84544pt,$

endowed with the induced metric of the Euclidean metric $\xi$. We know that
$x$ is a harmonic map. We compute the outward normal derivative of $x$ on
$\mathcal{E}$:

$\partial_{\nu}x=\nu\hskip 2.84544pt,$

where the outward normal of the ellipsoid is denoted by $\nu=\frac{\sigma
x}{\left|\sigma x\right|}$ where $\left|\sigma
x\right|=\left(\sum_{i=1}^{n}\sigma_{i}^{2}x_{i}^{2}\right)^{\frac{1}{2}}$.
Therefore, if we endow $\mathcal{E}$ with the metric
$g=\frac{\xi}{\left|\sigma x\right|}$, $x_{i}$ is a Steklov eigenfunction on
$(\mathcal{E},g)$ associated to the eigenvalue $\sigma_{i}$.

Now, let $\Phi:(\Sigma,h)\to\mathbb{R}^{n}$ be such that
$\Phi(\partial\Sigma)\in\mathcal{E}$, a $n-1$ dimentional ellipsoid of
parameter $\sigma=\left(\sigma_{1},\cdots,\sigma_{n}\right)$. A well-known
characterisation of $\Phi:(\Sigma,h)\to\mathbb{R}^{n}$ to be minimal with free
boundary in $\mathcal{E}$ is free boundary harmonicity in $\mathcal{E}$ and
conformality. We recall that $\Phi$ is harmonic in $\mathcal{E}$ with free
boundary if it is a critical point of the energy

$E(\Phi)=\frac{1}{2}\int_{\Sigma}\left|\nabla\Phi\right|_{h}^{2}dA_{h}$

under the constraint $\Phi(\partial\Sigma)\subset\mathcal{E}$. The Euler-
Lagrange characterization is

$\Delta_{h}\Phi=0\hbox{ in }\Sigma\hbox{ and
}\partial_{\nu}\Phi\in\left(T_{\Phi}\mathcal{E}\right)^{\perp}\hbox{ on
}\partial\Sigma$

Then $\partial_{\nu}\Phi=f\nu$ for some function $f=\Phi.\partial_{\nu}\Phi$.
Conformality is characterized by the vanishing of

$0=\left|\nabla\Phi\right|_{h}^{2}\frac{h}{2}-d\Phi\otimes
d\Phi:=\sum_{i=1}^{n}\left(\left|\nabla\Phi_{i}\right|_{h}^{2}\frac{h}{2}-d\Phi_{i}\otimes
d\Phi_{i}\right)\hskip 2.84544pt.$

For a smooth positive function $e^{2u}$, such that $g=e^{2u}h$, we have

$\Delta_{g}f=e^{-2u}\Delta_{h}f\hbox{ and
}\partial_{\nu_{g}}f=e^{-u}\partial_{\nu_{h}}f\hskip 2.84544pt,$

and if $\Phi:(\Sigma,h)\to\mathbb{R}^{n}$ is a minimal isometric immersion
with free boundary in $\mathcal{E}$, setting $g=e^{2u}h$ for any function $u$
extending the following formula on the boundary,

$e^{u}=\Phi.\partial_{\nu}\Phi=\frac{1}{\left|\sigma\Phi\right|}\hbox{ on
}\partial\Sigma\hskip 2.84544pt,$

the coordinates of $\Phi$ are Steklov eigenfunctions on $(\Sigma,g)$ with
eigenvalues $\sigma_{1},\cdots,\sigma_{n}$.

###  1.2. The case of planar ellipses

Let’s consider the simplest example: we assume that
$\Phi=id:\mathcal{E}_{p}\to\mathcal{E}_{p}$ into a $2$-d ellipse
$\mathcal{E}_{p}$ of parameters $p=(p_{1},p_{2})$. Setting the metric
conformal to $\xi$

$g_{p}=e^{2u}\xi\hbox{ where }e^{u}=\frac{1}{\left|\sigma
x\right|}=\frac{1}{\sqrt{p_{1}^{2}x_{1}^{2}+p_{2}^{2}x_{2}^{2}}}\hbox{ on
}\mathcal{E}_{p}\hskip 2.84544pt,$

on $(\mathcal{E}_{p},g_{p})$ we then have

$\Delta_{g_{p}}x_{i}=0\hbox{ and }\partial_{\nu_{p}}x_{i}=p_{i}x_{i}$

for all $i=1,2$. This means that the coordinate functions are eigenfunctions
on $(\mathcal{E}_{p},g_{p})$ with eigenvalues $p_{1},p_{2}$. However, these
are not necessarily eigenfunctions associated to the first and second
eigenvalue of $(\mathcal{E}_{p},g_{p})$. Let’s compute their renormalized
eigenvalue. By invariance of the indices of eigenvalues by diltation of the
ellipse, we study $\mathcal{E}_{p}=\\{px^{2}+y^{2}=1\\}$, for some $0\leq
p\leq 1$, where the boundary is endowed with the conformal factor
$e^{u_{p}(x)}=\left(p^{2}x^{2}+y^{2}\right)^{-\frac{1}{2}}$. We know that
there are $k_{1}\leq k_{2}$ such that
$\sigma_{k_{1}}(\mathcal{E}_{p},e^{u_{p}})=p$ and
$\sigma_{k_{2}}(\mathcal{E}_{p},e^{u_{p}})=1$. By parametrisation of the
ellipse on the circle $(\frac{1}{\sqrt{p}}\cos\theta,\sin\theta)$, we have
that

$dL_{\mathcal{E}_{p}}=\sqrt{\frac{1}{p}\sin^{2}\theta+\cos^{2}\theta}d\theta=\frac{1}{\sqrt{p}}e^{-u}d\theta$

We obtain that the total length of the boundary is

$L_{\mathcal{E}_{p}}(e^{u_{p}}dL_{\mathcal{E}_{p}})=\int_{\mathcal{E}_{p}}e^{u_{p}}dL_{\mathcal{E}_{p}}=\int_{0}^{2\pi}\frac{d\theta}{\sqrt{p}}=\frac{2\pi}{\sqrt{p}}$

and the renormalized eigenvalues are

$\bar{\sigma}_{k_{1}}(\mathcal{E}_{p},e^{u_{p}})=p\sigma_{k_{1}}(\mathcal{E}_{p},e^{u_{p}})=pL=2\pi\sqrt{p}\text{
and }\bar{\sigma}_{k_{2}}(\mathcal{E}_{p},e^{u_{p}})=L=\frac{2\pi}{\sqrt{p}}$

Then, for degenerating ellipses, we have that

$\bar{\sigma}_{k_{1}}(\mathcal{E}_{p},e^{u_{p}})\to 0\text{ and
}\bar{\sigma}_{k_{2}}(\mathcal{E}_{p},e^{u_{p}})\to+\infty\hbox{ as }p\to
0\hskip 2.84544pt.$

We know that $\bar{\sigma}_{k}(\mathcal{E}_{p},e^{u_{p}})$ has to be bounded
by $2\pi k$ for any $k$. Since $\sigma_{k_{2}}(\mathcal{E}_{p},g_{p})=1$, we
have that $k_{2}\to+\infty$ as $p\to 0$. It would be interesting to know the
value of

$k_{1}(p)=\inf\\{k\in\mathbb{N};\sigma_{k}(\mathcal{E}_{p},g_{p})=p\\}\text{
and }k_{2}(p)=\inf\\{k\in\mathbb{N};\sigma_{k}(\mathcal{E}_{p},g_{p})=1\\}$

for $i=1,2$ on the Riemannian manifold $(\mathcal{E}_{p},g_{p})$ defined
above. For $p=1$, $k_{i}(p)=1$ since the ellipse is a circle. The points $p$
such that $k_{i}$ jumps are points of bifurcation of eigenvalues. Computing
precisely the set $\\{p\in\mathbb{R}^{2}\setminus\\{0\\};\forall
i\in\\{1,2\\},k_{i}(p)\leq 2\\}$ would give the minimal value $\theta_{\star}$
such that if $\frac{1}{p}>\theta_{\star}$, $k_{2}(p)\geq 3$. More generally,
we conjecture that $k_{1}(p)=1$ for any $0\leq p\leq 1$ and that a new
bifurcation branch of non planar free boundary minimal surfaces is created if
$\frac{1}{p}$ crosses

$\theta_{\star}^{k}:=\inf\left\\{\frac{1}{p};0\leq p\leq 1\text{ and
}k_{2}(p)=k\right\\}$

as $p$ decreases.

###  1.3. Minimal immersions into ellipsoids do not have branch points at the
boundary

Let $\Phi:\Sigma\to\mathbb{R}^{n}$ be some (possibly branched) conformal
minimal immersion into an ellipsoid

$\mathcal{E}=\\{(x_{1},\cdots,x_{n})\in\mathcal{E};\sigma_{1}x_{1}^{2}+\cdots+\sigma_{n}x_{n}^{2}=1\\}\hskip
2.84544pt,$

parametrized by $\sigma=diag(\sigma_{1},\cdots,\sigma_{n})$. We let
$e^{2u}\left(dx^{2}+dy^{2}\right)=\Phi^{\star}eucl$ be the pull-back metric of
the Euclidean one by $\Phi$. Notice that branched points correspond to
singularities $u(x)=-\infty$. We know that the coordinate functions of $\Phi$
are Steklov eigenfunctions with respect to the conformal facotor
$e^{v}=\frac{e^{u}}{\left|\sigma\Phi\right|}$ at the boundary. We first prove
the following

######  Claim 1.1.

$\Phi$ cannot have any branched points at the boundary. In other words $e^{u}$
and $e^{v}$ are positive everywhere.

###### Proof.

Let $x\in\partial\Sigma$. We set
$\psi=\sigma_{1}\phi_{1}(x)\phi_{1}+\cdots+\sigma_{n}\phi_{n}(x)\phi_{n}$. We
have that for any $y\in\mathbb{D}$,

$\psi(y)=\left\langle\sigma\phi(x),\phi(y)\right\rangle\leq\sqrt{\left\langle\sigma\phi(x),\phi(x)\right\rangle}\sqrt{\left\langle\sigma\phi(y),\phi(y)\right\rangle}=\sqrt{\left\langle\sigma\phi(y),\phi(y)\right\rangle}$

The function $f:y\mapsto\left\langle\sigma\phi(y),\phi(y)\right\rangle$ is
subharmonic since

$\Delta f=-\left\langle\sigma\nabla\phi,\nabla\phi\right\rangle\leq 0\hskip
2.84544pt,$

so that $f$ realizes its maximum at the boundary, and $f=1$ at the boundary
gives that

$\psi(y)\leq 1=\psi(x)\hskip 2.84544pt.$

$\psi$ is a harmonic function that realizes its maximum at
$x\in\mathbb{S}^{1}$. Since $\psi$ is harmonic, by the Hopf lemma, we must
have $\partial_{\nu}\psi(x)\neq 0$. And we have that

$e^{u(x)}=\left|\sigma\Phi(x)\right|^{2}e^{v(x)}=\left\langle\sigma\Phi(x),\partial_{\nu}\Phi(x)\right\rangle=\partial_{\nu}\psi(x)\neq
0\hskip 2.84544pt.$

∎

###  1.4. Free boundary minimal disks by first and second eigenfunctions do
not have branched points

The main property of the subsection relies on the following claim:

######  Claim 1.2.

Let $x\in\mathbb{D}$ and let $\psi$ a first or second Steklov eigenfunction on
the disk, associated to some positive weight $e^{v}$ on the boundary. Then

$\psi(x)=0\Rightarrow\nabla\psi(x)\neq 0\hskip 2.84544pt.$

###### Proof.

By the Courant nodal theorem, $\psi$ has at most three nodal domains.
Moreover, the nodal set is either a smooth curve having two ends on the
boundary or the disjoint union of two connected curves having two ends at the
boundary. Indeed, since eigenfunctions are non constant and harmonic, the
nodal set cannot contain a closed curve, and, the nodal set cannot contain a
singularity in the interior of the disk since otherwise, the eigenfunction
would have at least four nodal domains. Now let $x\in\mathbb{D}$ be such that
$\psi(x)=0$. Let $D$ be a nodal domain such that $x\in\partial D$. Then
$\partial D$ is smooth at $x$ and $x$ is an extremal point of the harmonic
function $\psi$ on $D$. By the Hopf lemma $\partial_{\nu}\psi(x)\neq 0$. This
ends the proof of the claim. ∎

It is clear that $\Phi$ has at least two coordinates since first and second
eigenfunctions cannot be constant. We know by [Jam16] that the multiplicity of
first eigenfunctions on the disk is less than 2 and that the multiplicity of
second eigenfunctions on the disk is less than 2. Therefore, $\Phi$ has at
most 3 coordinates. We consider the cases $n=2$ and $n=3$ separately.

######  Proposition 1.1.

We first assume
$\Phi=(\phi_{0},\phi_{1},\phi_{2}):\mathbb{D}\to\mathbb{R}^{3}$ is a possibly
branched free boundary minimal immersion into
$\mathcal{E}=\\{x\in\mathbb{R}^{2};\sigma_{1}x_{0}^{2}+\sigma_{2}(x_{1}^{2}+x_{2}^{2})=1\\}$,
where $\sigma_{1}<\sigma_{2}$ and $\phi_{0}$ is a first eigenfunction and
$\phi_{1}$ and $\phi_{2}$ are second eigenfunctions. Then $\Phi$ does not have
any branched point.

###### Proof.

Notice that by Claim 1.1, $\Phi$ does not have any branched point at the
boundary. It remains to prove that for $z\in\mathbb{D}$, we have that
$\nabla\Phi(z)\neq 0$. In fact, we will even prove that $\nabla\eta(z)\neq 0$,
where $\eta=(\phi_{1},\phi_{2})$. Let $v\in\mathbb{S}^{1}$ be such that
$\left\langle v,\eta(z)\right\rangle=0$. Then $\left\langle
v,\eta\right\rangle$ vanishes at $z$ and the previous claim implies that
$\langle v,\nabla\eta(z)\rangle=\nabla\left(\langle
v,\nabla\eta\rangle\right)(z)\neq 0$. ∎

In the following proposition, we even obtain that planar free boundary minimal
disks into ellipsoids by first and second eigenfunction are embeddings.

######  Proposition 1.2.

We assume that $\Phi=(\phi_{1},\phi_{2}):\mathbb{D}\to\mathbb{R}^{2}$ is a
possibly branched free boundary minimal immersion into
$\mathcal{E}=\\{x\in\mathbb{R}^{2};\sigma_{1}x_{1}^{2}+\sigma_{2}x_{2}^{2}=1\\}$,
where $\sigma_{1}\leq\sigma_{2}$ and $\phi_{1}$ is a first eigenfunction and
$\phi_{2}$ is either a first or second eigenfunction. Then $\Phi$ is a
diffeomorphism.

###### Proof.

We first prove that the curve
$\Phi_{|\mathbb{S}^{1}}:\mathbb{S}^{1}\to\mathcal{E}$ is an embedding.

Since $\Phi$ is conformal, we must have that
$\left|\partial_{r}\Phi\right|^{2}=\left|\partial_{\theta}\Phi\right|^{2}$ on
$\mathbb{S}^{1}$. Moreover, by Claim 1.1,
$\left|\partial_{r}\Phi\right|^{2}=\left|\sigma\Phi\right|^{2}e^{2v}$ never
vanishes. Therefore, $\Phi_{|\mathbb{S}^{1}}:\mathbb{S}^{1}\to\mathcal{E}$ is
an immersion. Then it is a covering map. Knowing that $\phi_{1}$ is a first
eigenfunction, $\phi_{1}$ has at most two nodal domains, and the nodal line
meets the boundary at two points. Therefore, the degree of
$\Phi_{|\mathbb{S}^{1}}$ must be equal to $1$ and $\phi_{|\mathbb{S}^{1}}$ is
an embedding.

As a conclusion, since the ellipse $\mathcal{E}$ encloses a convex curve and
$\Phi$ is harmonic in $\mathbb{D}$ a classical result by Kneser [Kne26] (see
also Choquet [Cho45] or a more general result by Alessandrini-Nesi [AN09])
gives that $\Phi$ is a diffeomorphism. ∎

In the following section, we also prove embeddedness of immersed non planar
free boundary disks into ellipsoids by first and second eigenfunctions with
symmetry assumptions.

###  1.5. Non planar free boundary minimal disks with symmetry properties are
embedded

From now to the end of the subsection, we assume that
$\Phi=(\phi_{0},\phi_{1},\phi_{2}):\mathbb{D}\to\mathbb{R}^{3}$ is a free
boundary minimal immersion into
$\mathcal{E}=\\{x\in\mathbb{R}^{2};\sigma_{1}x_{0}^{2}+\sigma_{2}\left(x_{1}^{2}+x_{2}^{2}\right)=1\\}$,
where $\sigma_{1}<\sigma_{2}$ and $\phi_{0}$ is a first eigenfunction and
$\phi_{1}$ and $\phi_{2}$ are second eigenfunctions with respect to the
positive function $e^{v}=\Phi.\partial_{r}\Phi$ on $\mathbb{S}^{1}$ (positive
thanks to Claim 1.1). We also assume that $e^{v}$ satisfies the following
symmetry assumtions:

$\forall(x,y)\in\mathbb{S}^{1},e^{v(x,y)}=e^{v(-x,y)}=e^{v(x,-y)}$

######  Claim 1.3.

Up to reparametrization and rotation of $\Phi$, we must have that

$\forall x,y\in\mathbb{D},\phi_{0}(x,-y)=-\phi_{0}(x,y)\text{ and
}\phi_{0}(-x,y)=\phi_{0}(x,y)$ $\forall
x,y\in\mathbb{D},\phi_{1}(x,-y)=\phi_{1}(x,y)\text{ and
}\phi_{1}(-x,y)=-\phi_{1}(x,y)$ $\forall
x,y\in\mathbb{D},\phi_{2}(x,-y)=\phi_{2}(x,y)\text{ and
}\phi_{2}(-x,y)=\phi_{2}(x,y)$

where $\phi_{0}$ and $\phi_{1}$ have exactly $2$ nodal domains and $\phi_{2}$
has exactly $3$ nodal domains. Moreover, $\phi_{2}$ does not vanish on
$[-1,1]\times\\{0\\}\cup\\{(0,\pm 1)\\}$.

###### Proof.

The proof the claim is based on the 3 following simple facts

Fact 1: For any Steklov eigenfunction $\phi:\mathbb{D}\to\mathbb{R}$, if for
all $(x,y)\in\mathbb{D}$, $\phi(x,y)=\phi(-x,y)=\phi(x,-y)$ then $\phi$ has at
least $3$ nodal domains and if for all $(x,y)\in\mathbb{D}$,
$\phi(x,y)=-\phi(-x,y)=-\phi(x,-y)$ then $\phi$ has at least $4$ nodal
domains.

Fact 2: If $\phi$ is a Steklov eigenfunction on $\mathbb{D}$ associated to the
symmetric weight $e^{v}:\mathbb{S}^{1}\to\mathbb{R}$, then $\phi(-x,y)$ and
$\phi(x,-y)$ are also Steklov eigenfunctions.

Fact 3: A second eigenfunction $\phi$ vanishes on at most 4 points at the
boundary. If $\phi$ vanishes at 2 points $p_{1},p_{2}\in\partial{D}$, then it
has exactly 2 nodal domains and the nodal set is a smooth curve ending at
$p_{1}$ and $p_{2}$. If $\phi$ vanishes at 3 points
$p_{0},p_{1},p_{2}\in\partial\mathbb{D}$, then (up to a permutation of the
indices of $p_{i}$) it has exactly 3 nodal domains and the nodal set is the
union of two smooth curves: one ending at $p_{0}$ and $p_{1}$, the other one
ending at $p_{0}$ and $p_{2}$ and they only intersect at $p_{0}$. If $\phi$
vanishes at 4 points $p_{1},p_{2},p_{3},p_{4}\in\partial{D}$, then (up to a
permutation of the indices of $p_{i}$) it has exactly 3 nodal domains and the
nodal set is a disjoint union of a smooth curve ending at $p_{1}$ and $p_{2}$
and a smooth curve ending at $p_{3}$ and $p_{4}$.

Step 1 Up to a quarter rotation in the set of parametrizations, $\phi_{0}$ has
the expected symmetries, and there are orthogonal second eigenfunctions
$\eta_{1}$ and $\eta_{2}$ such that

$\forall(x,y)\in\mathbb{D},\eta_{i}(x,y)=\varepsilon_{i}^{1}\eta_{i}(-x,y)=\varepsilon_{i}^{2}\eta_{i}(x,-y).$

Proof of Step 1

From fact 2, since the first eigenvalue is simple, we must have that

$\phi_{0}(x,y)=\pm\phi_{0}(-x,y)=\pm\phi_{0}(x,-y).$

From fact 1 and the Courant nodal theorem ($\phi_{0}$ have at most two nodal
domains), we deduce that, up to a quarter rotation on the set of
parametrization, we have the expected symmetries on $\phi_{0}$.

From fact 2, up to replace $\phi_{i}$ by
$\frac{\phi_{i}(x,y)+\phi_{i}(-x,y)+\phi_{i}(x,-y)+\phi_{i}(-x,-y)}{4}$ for
$i=1,2$, we can find two Steklov eigenfunctions $\eta_{1}$ and $\eta_{2}$
associated to $\sigma_{2}(e^{v})$ such that for $i=1,2$, $j=1,2$, there are
$\varepsilon_{i}^{j}\in\\{\pm 1\\}$ such that

$\forall(x,y)\in\mathbb{D},\eta_{i}(x,y)=\varepsilon_{i}^{1}\eta_{i}(-x,y)=\varepsilon_{i}^{2}\eta_{i}(x,-y).$

Step 2 Up to exchange the indices of $\eta_{1}$ and $\eta_{2}$, we have that
$\varepsilon_{1}^{2}=\varepsilon_{2}^{2}=1$, that $\varepsilon_{1}^{1}=-1$ and
that $\varepsilon_{2}^{1}=1$:

$\forall x,y\in\mathbb{D},\eta_{1}(x,-y)=\eta_{1}(x,y)\text{ and
}\eta_{1}(-x,y)=-\eta_{1}(x,y)$ $\forall
x,y\in\mathbb{D},\eta_{2}(x,-y)=\eta_{2}(x,y)\text{ and
}\eta_{2}(-x,y)=\eta_{2}(x,y)$

Proof of Step 2:

Let $i\in\\{1,2\\}$. From fact 1 and the Courant nodal theorem, we cannot have
$\varepsilon_{i}^{1}=\varepsilon_{i}^{2}=-1$.

More precisely, let’s prove that $\varepsilon_{1}^{2}=\varepsilon_{2}^{2}=1$.
We assume by contradiction that for some $i\in\\{1,2\\}$,
$\varepsilon_{i}^{2}=-1$. Then $\varepsilon_{i}^{1}=1$, and knowing that the
function $\eta_{i}$ is orthogonal to $\phi_{0}$, $\eta_{i}$ must vanish at
some point $(x_{0},y_{0})\in\mathbb{S}^{1}$ with $y_{0}\neq 0$. By symmetries
this implies that $\eta_{i}$ vanishes at $(x_{0},-y_{0})\in\mathbb{S}^{1}$.
Since $\\{y=0\\}$ is a nodal set, there are at least two nodal domains in
$\mathbb{D}_{+}$ and two other nodal domains in $\mathbb{D}_{-}$. Then
$\eta_{i}$ has at leats four nodal domains: this contradicts the Courant nodal
theorem. Then $\varepsilon_{1}^{2}=\varepsilon_{2}^{2}=1$.

By a similar argument, we can prove that
$1\in\\{\varepsilon_{1}^{1},\varepsilon_{2}^{1}\\}$. Indeed, if not,
$\eta_{1}$ and $\eta_{2}$ vanish on $\\{x=0\\}$. Since $\eta_{1}$ is
orthogonal to $\eta_{2}$, symmetries and the Courant nodal theorem give a
contradiction.

Now, we set $\eta=(\eta_{1},\eta_{2})$: we have that
$\forall(x,y)\in\mathbb{D},\eta(x,-y)=\eta(x,y)$. By contradiction, we assume
that we also have $\varepsilon_{1}^{1}=\varepsilon_{2}^{1}=1$, that is
$\forall(x,y)\in\mathbb{D},\eta(-x,y)=\eta(x,y)$. By fact $1$, it follows that
for any $v\in\mathbb{S}^{1}$, $\left\langle\eta,v\right\rangle$ is a second
eigenfunction having at least 3 nodal domains and therefore exactly 3 nodal
domains by the Courant nodal theorem. By continuity, the map
$N:v\in\mathbb{S}^{1}\mapsto$ the number of positive nodal domains of
$\left\langle\eta,v\right\rangle$ is a constant map. We obtain a contradiction
because for any $v\in\mathbb{S}^{1}$, $\\{N(v),N(-v)\\}=\\{1,2\\}$. Then, up
to exchange the indices of $\eta_{1}$ and $\eta_{2}$, we can assume that
$\varepsilon_{1}^{1}=-1$ and that $\varepsilon_{2}^{1}=1$.

Step 3: $\eta_{1}$ has exactly 2 nodal domains and
$\eta_{1}^{-1}(\\{0\\})=\\{0\\}\times[-1,1]$. $\eta_{2}$ has exactly $3$ nodal
domains and $\eta_{2}$ cannot vanish on $[-1,1]\times\\{0\\}\cup\\{(0,\pm
1)\\}$.

Proof of Step 3:

First of all, we assume by contradiction that $\eta_{2}$ vanishes on $\\{(\pm
1,0),(0,\pm 1)\\}$, for instance at $(1,0)$. By symmetry $(x,y)\mapsto(x,-y)$,
$(1,0)$ has to be the ending point of at least two smooth vanishing curves. By
symmetry $(x,y)\mapsto(-x,y)$, $(-1,0)$ is also a zero of $\eta_{2}$ and the
ending point of at least two smooth curves. This contradicts Fact 3. Then,
$\eta_{2}$ vanishes at $z\in\mathbb{S}^{1}\setminus\\{(\pm 1,0),(0,\pm 1)\\}$
and by symmetries at $z,\bar{z},-z,-\bar{z}$.

By Fact 3, $\\{z,\bar{z},-z,-\bar{z}\\}$ is the nodal set at the boundary and
the two disjoint smooth nodal curves $C_{1}$ and $C_{2}$ must not connect $z$
and $-z$ or $\bar{z}$ and $-\bar{z}$. Let’s prove that up to change the
indices of $C_{i}$, $C_{1}$ connects $z$ to $-\bar{z}$ and $C_{2}$ connects
$-z$ to $\bar{z}$. We assume by contradiction that $C_{1}$ connects $z$ to
$\bar{z}$ and $-z$ to $-\bar{z}$. Then, let’s study the nodal set of the
second eigenfunctions $\left\langle\eta,v\right\rangle$ as $v$ describes
$\mathbb{S}^{1}$. In particular, following Fact 3, we can write

$\mathbb{S}^{1}=A_{2}\sqcup A_{3}\sqcup A_{4}$

where $A_{k}$ is the set of $v\in\mathbb{S}^{1}$ such that
$\left\langle\eta,v\right\rangle$ vanishes at exactly $k$ points at the
boundary. $A_{2}$ and $A_{4}$ are open sets and $A_{3}$ is a closed set. For
instance, $(0,1)\in A_{4}$ and $(1,0)\in A_{2}$. Let $v_{0}\in A_{3}$ be a
point at the boundary of the connected component of $A_{4}$ containing
$(0,1)$. Then by symmetries, the point $z_{0}$ intersecting the two smooth
branches of the nodal set of $\left\langle\eta,v_{0}\right\rangle$ has to
belong to $\\{(\pm 1,0);(0,\pm 1)\\}$. If $z_{0}=\pm(0,1)$, then by symmetry
of $\left\langle\eta,v_{0}\right\rangle$ with respect to $(x,y)\mapsto(x,-y)$,
$-z_{0}$ is also such a singular vanishing point. This contradicts fact 3.
Then $z_{0}=\pm(1,0)$. However, following the nodal set of
$\left\langle\eta,v\right\rangle$ for $v\in](0,1),v_{0}]\subset A_{4}$ we
obtain that the nodal set of $\left\langle\eta,v_{0}\right\rangle$ contains a
smooth curve having two ends at $z_{0}$ : it does not satisfy fact 3. It is a
contradiction.

Therefore, $C_{1}$ connects $z$ to $-\bar{z}$ and $C_{2}$ connects $-z$ to
$\bar{z}$. Finally, for $i=1,2$, $C_{i}\cap[-1,1]\times\\{0\\}=\emptyset$
because if we assume $C_{i}\cap[-1,1]\times\\{0\\}\neq\emptyset$, then the set
contains at least two points and by symmetries, $\eta_{2}$ would have a nodal
domain included in the interior $\mathbb{D}$: this is absurd.

Conclusion:

Up to replace $\Phi$ by
$(\phi_{0},\cos\theta\phi_{1}+\sin\theta\phi_{2},-\sin\theta\phi_{1}+\cos\theta\phi_{2})$,
we can assume that $\phi_{1}=\alpha\eta_{1}$ for some $\alpha\in\mathbb{R}$
and $\phi_{1}$ have the same symmetries as $\eta_{1}$. Knowing that

$\sigma_{1}\phi_{0}^{2}+\sigma_{2}\left(\phi_{1}^{2}+\phi_{2}^{2}\right)=1\text{
on }\mathbb{S}^{1}$

we obtain that
$\forall(x,y)\in\mathbb{S}^{1},\phi_{2}(x,y)^{2}=\phi_{2}(x,-y)^{2}=\phi_{2}(-x,y)^{2}$.
By harmonic extension, it is also true in $\mathbb{D}$ and $\phi_{2}$ must
have the same symmetries as $\eta_{2}$. ∎

######  Proposition 1.3.

$\Phi$ is an embedding

###### Proof.

We aim at proving that a projection of the map $\Phi$, the map
$\eta:=(\phi_{1},\phi_{2}):\mathbb{D}_{+}\to\eta(\mathbb{D}_{+})$ is a
diffomorphism. By symmetries, it will prove that
$\Phi:\mathbb{D}\to\mathbb{R}^{3}$ is injective and since it is an immersion,
we will obtain that $\Phi$ is an embedding.

Step 1: $0\notin\eta\left(\partial\left({\mathbb{D}_{+}}\right)\right)$ and
$\frac{\eta}{\left|\eta\right|}:\partial\left({\mathbb{D}_{+}}\right)\to\mathbb{S}^{1}$
is a homeomorphism. In other words,
$\eta:\partial\left({\mathbb{D}_{+}}\right)\to\mathbb{R}^{2}$ is an injective
closed curve and up to change the orientation, $\eta\wedge\eta_{\theta}$ is
non-negative on $\mathbb{S}^{1}_{+}$ and $\eta\wedge\eta_{x}$ is non negative
on $[-1,1]\times\\{0\\}$. In particular,
$\eta\left(\partial\left({\mathbb{D}_{+}}\right)\right)$ encloses a star-
shaped domain with respect to $0$.

Proof of Step 1: $\phi_{1}^{-1}(\\{0\\})=\\{0\\}\times(-1,1)$ so that
$\phi_{1}^{-1}(\\{0\\})\cap\partial\left({\mathbb{D}_{+}}\right)=\\{(0,0),(0,1)\\}$
and $\phi_{2}$ does not vanish on this set. Therefore, we can consider
$\frac{\eta}{\left|\eta\right|}:\partial\left({\mathbb{D}_{+}}\right)\to\mathbb{S}^{1}$
and prove that it is monotone. In other words, let’s prove that for any
$v\in\mathbb{S}^{1}$, and every $x\in\partial\left({\mathbb{D}_{+}}\right)$
such that $\eta(x)\in D_{v}$, there is an arc
$(a,b)\subset\partial\left({\mathbb{D}_{+}}\right)$ such that $x\in(a,b)$ and

$\forall y\in(a,x),\eta(y)\in H_{v}^{+}\text{ and }\forall
y\in(x,b),\eta(y)\in H_{v}^{-}$

where

$D_{v}=\\{z\in\mathbb{R}^{2};\left\langle
z,v\right\rangle=0\\},H_{v}^{+}=\\{z\in\mathbb{R}^{2};\left\langle
z,v\right\rangle>0\\},H_{v}^{-}=\\{z\in\mathbb{R}^{2};\left\langle
z,v\right\rangle<0\\}.$

We assume that it is not true for some
$x\in\partial\left({\mathbb{D}_{+}}\right)$ and $v\in\mathbb{S}^{1}$.

If $x\neq(\pm 1,0)$, $\eta$ is analytic at $x$ and we deduce that there is an
arc $(a,b)\subset\mathbb{S}^{1}$ such that $x\in(a,b)$ and either for all
$y\in(a,b)$, $\eta(y)\in D_{v}$ or for all $y\in(a,b)\setminus\\{x\\}$,
$\eta(y)\in H_{v}^{+}$ or for all $y\in(a,b)\setminus\\{x\\}$, $\eta(y)\in
H_{v}^{-}$. Since $\left\langle\eta,v\right\rangle$ cannot vanish on an arc of
$\mathbb{S}^{1}$, nor on an arc of $[-1,1]\times\\{0\\}$ by symmetries, up to
take $-v$, we assume that for all $y\in(a,b)\setminus\\{x\\}$, $\eta(y)\in
H_{v}^{+}$. Then, knowing that the nodal set is smooth in the interior of
$\mathbb{D}$, $\left\langle\eta,v\right\rangle$ has at least 3 distinct nodal
domains at the neighborhood of $x$. By symmetries,
$\left\langle\eta,v\right\rangle$ has at least four nodal domains and it is a
contradiction.

Now we assume that $x=(\pm 1,0)$. We assume that there is an arc
$(a,b)\subset\mathbb{S}^{1}$ such that $x\in(a,b)$ and either for all
$y\in(a,b)\setminus\\{x\\}$, $\eta(y)\in H_{v}^{+}$ or for all
$y\in(a,b)\setminus\\{x\\}$, $\eta(y)\in H_{v}^{-}$. Then, knowing that the
nodal set is smooth in the interior of $\mathbb{D}$,
$\left\langle\eta,v\right\rangle$ has at least $3$ nodal domains at the
neighborhood of $x$ in $\mathbb{D}_{+}$. By symmetries,
$\left\langle\eta,v\right\rangle$ has at least $5$ nodal domains and it is a
contradiction. Therefore, such an arc $(a,b)$ does not exist, but by
analyticity of $\eta$ on $\partial\left(\mathbb{D}_{+}\right)\setminus\\{(\pm
1,0)\\}$, such an arc $(a,b)$ would exist at the neighborhood of a point
$y\in\partial\left(\mathbb{D}_{+}\right)\setminus\\{(\pm 1,0)\\}$ and it is
not possible.

In order to conclude step 1, we just notice that $\phi_{1}$ and $\phi_{2}$
vanish only twice on $\partial\left(\mathbb{D}_{+}\right)$, so that the degree
of
$\frac{\eta}{\left|\eta\right|}:\partial\left(\mathbb{D}_{+}\right)\to\mathbb{S}^{1}$
is $1$ and it is a homeomorphism.

Step 2: $\eta_{r}\wedge\eta_{\theta}=\eta_{x}\wedge\eta_{y}$ does not vanish
on $\mathbb{S}^{1}_{+}$.

Proof of Step 2: We denote $n=e^{-2u}\Phi_{x}\wedge\Phi_{y}$ a normal vector
of the free-boundary minimal surface. Then
$n_{0}=e^{-2u}\eta_{x}\wedge\eta_{y}$ and $n$ satisfies the equation of
harmonic maps into $\mathbb{S}^{2}$. Notice that

$\begin{cases}n_{x}=-ee^{-2u}\Phi_{x}-fe^{-2u}\Phi_{y}\\\
n_{y}=-fe^{-2u}\Phi_{x}+ee^{-2u}\Phi_{y}\end{cases}$

Let $z\in\mathbb{S}^{1}_{+}$. Notice in particular that $z\neq(\pm 1,0)$ and
$\phi_{0}(z)\neq 0$. We assume by contradiction that $n_{0}(z)=0$. By step 1,
$n_{0}$ has a constant sign at the neighborhood of $z$. Therefore, we must
have that $n_{0,\theta}(z)=0$. Up to a rotation in the set of parametrization,
we assume that for any function $f:\mathbb{D}\to\mathbb{R}$, we have
$f_{r}=f_{x}$ and $f_{\theta}=f_{y}$ on $\mathbb{S}^{1}$. By a Taylor
expansion,

$\begin{split}\Phi(z+h)=&\Phi(z)+h_{1}\Phi_{x}(z)+h_{2}\Phi_{y}(z)\\\
&+\frac{1}{2}\left((h_{1})^{2}\Phi_{xx}(z)+2h_{1}h_{2}\Phi_{xy}(z)+(h_{2})^{2}\Phi_{yy}(z)\right)+o(\left|h\right|^{2})\end{split}$

Since $n_{0}(z)=0$, we have that
$\left\langle\nabla\eta(z),\tilde{n}(z)\right\rangle=\left\langle\nabla\Phi(z),n(z)\right\rangle=0$
and by the Steklov eigenfunction equation,
$\sigma_{2}e^{v(z)}\left\langle\eta(z),\tilde{n}(z)\right\rangle=\left\langle\eta(z),\tilde{n}(z)\right\rangle=0$
so that

$\left\langle\eta(z+h),\tilde{n}(z)\right\rangle=\frac{1}{2}\left(\left((h_{1})^{2}-(h_{2})^{2}\right)e+2h_{1}h_{2}f\right)+o(\left|h\right|^{2})$

Since the nodal set of the eigenfunction
$\left\langle\eta,\tilde{n}(z)\right\rangle$ has only one branch starting from
$z$ (see Step 1), then by the Morse lemma, we must have $e=0$. Since
$n_{0,y}(z)=0$, we obtain that $-f\phi_{0,x}(z)=0$. Since by the Steklov
eigenfunction equation $\phi_{0,x}(z)=\sigma_{1}e^{v(z)}\phi_{0}(z)$, we
obtain that either $\phi_{0}(z)=0$, either $f=0$. Since $\phi_{0}(z)\neq 0$ we
obtain $f=0$. Now, since $\left\langle\eta,\tilde{n}(z)\right\rangle$ is
harmonic and smooth until the boundary of $\mathbb{D}$, it has a harmonic
extension at the neighborhood of $z$. Since $e=f=0$ at $z$, the nodal set of
$\left\langle\eta,\tilde{n}(z)\right\rangle$ has a singularity of multiplicity
higher than $3$ at $z$, so that the nodal set
$\left\langle\eta,\tilde{n}(z)\right\rangle$ has at least two branches
starting from $z$ in $\mathbb{D}$, and it is a contradiction.

Step 3: $\eta\wedge\eta_{x}$ does not vanish on $(-1,1)\times\\{0\\}$.

Proof of Step 3: We assume that for some $z\in(-1,1)\times\\{0\\}$,
$\eta(z)\wedge\eta_{x}(z)=0$. By symmetries, we know that $\phi_{0,x}(z)=0$
and $\eta_{y}(z)=0$. Therefore, if we denote $n=(n_{0},\tilde{n})$, and
$\varphi=\left\langle\eta,\tilde{n}(z)\right\rangle$, we have obtain that
$\nabla\varphi(z)=0$ and $\varphi(z)=0$ so that the nodal set of $\varphi$ is
singular at $z$: it is a contradiction.

Step 4: Setting
$e=\left\langle\Phi_{xx},n\right\rangle=-\left\langle\Phi_{yy},n\right\rangle$
and
$\tilde{e}=\left\langle\Phi_{xxx},n\right\rangle=-\left\langle\Phi_{yyx},n\right\rangle$,
where we denote $n=e^{-2u}\Phi_{x}\wedge\Phi_{y}$, then $e$ or $\tilde{e}$
never vanish in $\\{(\pm 1,0)\\}$

Proof of Step 4: We have by a Taylor extension that for $z\in\\{(\pm 1,0)\\}$

$\eta(z+h)=\sum_{k=0}^{+\infty}\frac{1}{k!}\cos(k\theta)r^{k}\eta^{(k)_{x}}(z)$

for $h=r(\cos\theta,\sin\theta)$, because $\eta$ is harmonic and satisfies
$\eta(x,-y)=\eta(x,y)$. Now, thanks to the Steklov equation, $\eta_{x}(z)$ and
$\eta(z)$ are parallel vectors and are orthogonal to $\tilde{n}(z)$, where
$n(z)=(n_{0}(z),\tilde{n}(z))$, since $n_{0}(z)=0$. Then, if $e=\tilde{e}=0$,

$\left\langle\eta(z+h),\tilde{n}(z)\right\rangle=c\cos(k\theta)r^{k}+o(r^{k})$

for a constant $c\neq 0$ and $k\geq 4$ and it is impossible because of the
structure of the nodal set of the second eigenfunction
$\left\langle\eta,\tilde{n}(z)\right\rangle$.

Step 5: We still denote
$\Phi=(\phi_{0},\eta):\mathbb{D}_{r_{0}}\to\mathbb{R}^{2}$ a harmonic
extension of $\Phi$ where $r_{0}>1$. There is $\varepsilon_{0}>0$ such that
for $\varepsilon\leq\varepsilon_{0}$, there is a smooth function
$\eta_{\varepsilon}:\mathbb{D}_{r_{0}}\to\mathbb{R}^{2}$ for $r_{0}>1$,
satisfying that

  * •

$\eta_{\varepsilon}:\partial\left(\mathbb{D}_{+}\right)\to\mathbb{R}^{2}$
parametrizes a Jordan curve,

  * •

$\eta_{\varepsilon,x}\wedge\eta_{\varepsilon,y}$ is positive on
$\partial\left(\mathbb{D}_{+}\right)$,

  * •

there is a positive smooth function $\sigma_{\varepsilon}$ such that

$-div\left(\sigma_{\varepsilon}\nabla\eta_{\varepsilon}\right)=0\text{ in
}\mathbb{D}_{+}$

  * •

$\eta_{\varepsilon}$ converges to $\eta$ in
$\mathcal{C}^{2}\left(\mathbb{D}_{r_{0}}\right)$ as $\varepsilon\to 0$.

Proof of Step 5: By Step 1 and Step 2,
$\eta_{x}\wedge\eta_{y}=\eta_{r}\wedge\eta_{\theta}$ is positive on
$\mathbb{S}^{1}_{+}$. By a straightforword computation, using the Steklov
eigenvalue equation

$\tilde{\eta}_{x}\wedge\tilde{\eta}_{y}=\tilde{\eta}_{r}\wedge\tilde{\eta}_{\theta}=\frac{1+\varepsilon\left(1-\frac{\sigma_{1}}{\sigma_{2}}\right)\phi_{0}}{\left(1+\varepsilon\phi_{0}\right)^{3}}\eta_{r}\wedge\left(\eta_{\theta}-\frac{\varepsilon\phi_{0,\theta}}{1+\varepsilon\phi_{0}}\eta\right)=\frac{1+\varepsilon\left(1-\frac{\sigma_{1}}{\sigma_{2}}\right)\phi_{0}}{\left(1+\varepsilon\phi_{0}\right)^{3}}\eta_{r}\wedge\eta_{\theta}>0$

on $\mathbb{S}^{1}_{+}$ for $\varepsilon>0$ small enough. Now, we compute
$\tilde{\eta}_{x}\wedge\tilde{\eta}_{y}$ on $[-1,1]\times\\{0\\}$. By the
symmetries, we know that $\eta_{y}=0$ and $\phi_{0}=\phi_{0,x}=0$. Then

$\tilde{\eta}_{x}\wedge\tilde{\eta}_{y}=\eta_{x}\wedge\left(-\varepsilon\phi_{0,y}\eta\right)=\varepsilon\phi_{0,y}\left(\eta\wedge\eta_{x}\right)$

on $[-1,1]\times\\{0\\}$. Up to choose $-\phi_{0}$, we assume that $\phi_{0}$
is positve on $\mathbb{D}_{+}$ so that by the Hopf lemma, $\phi_{0,y}>0$. By
step 1 and step 3, $\tilde{\eta}_{x}\wedge\tilde{\eta}_{y}$ is then positive
on $(-1,1)\times\\{0\\}$.

Now, we assume that $e(0,1)\neq 0$ and we set
$\eta_{\varepsilon}=\frac{\eta+\alpha_{\varepsilon}\eta_{y}}{1+\varepsilon\phi_{0}}$
where $\alpha_{\varepsilon}>0$ and $\alpha_{\varepsilon}\to 0$ as
$\varepsilon\to 0$. By similar computations as for $\tilde{\eta}$,

$\eta_{\varepsilon,x}\wedge\eta_{\varepsilon,y}\frac{1+\varepsilon\left(1-\frac{\sigma_{1}}{\sigma_{2}}\right)\phi_{0}}{\left(1+\varepsilon\phi_{0}\right)^{3}}\eta_{r}\wedge\eta_{\theta}+\alpha_{\varepsilon}\left(\eta_{xy}\wedge\eta_{y}+\eta_{x}\wedge\eta_{yy}\right)+O\left(\alpha_{\varepsilon}^{2}+\varepsilon^{2}\right)$

on $\mathbb{S}^{1}_{+}$ and

$\eta_{\varepsilon,x}\wedge\eta_{\varepsilon,y}=\eta_{x}\wedge\left(-\varepsilon\phi_{0,y}\eta\right)=\varepsilon\phi_{0,y}\left(\eta\wedge\eta_{x}\right)+\alpha_{\varepsilon}\eta_{x}\wedge\eta_{yy}+O\left(\alpha_{\varepsilon}^{2}+\varepsilon^{2}\right)$

on $[-1,1]\times\\{0\\}$, knowing that $\eta_{xy}=0$ on $[-1,1]\times\\{0\\}$.
Notice that in both formula,
$O\left(\alpha_{\varepsilon}^{2}+\varepsilon^{2}\right)$ is uniform in
$\mathbb{D}$. Moreover, we have by the symmetries
$\phi_{1}(-x,y)=-\phi_{1}(x,y)$ and $\phi_{2}(-x,y)=\phi_{2}(x,y)$ that

$\left(\eta_{x}\wedge\eta_{yy}\right)(-1,0)=\left(\eta_{x}\wedge\eta_{yy}\right)(1,0)$

and that
$\eta_{x}\wedge\eta_{yy}(1,0)=-e(1,0)\eta_{x}(1,0)\wedge\tilde{n}(1,0)\neq 0$.
Then, up to take $-\phi_{1}$ instead of $\phi_{1}$ and choosing
$\alpha_{\varepsilon}$ such that $\alpha_{\varepsilon}=o(\varepsilon)$ and
$\varepsilon^{2}=o(\alpha_{\varepsilon})$, then there is $\varepsilon_{0}>0$
such that for any $\varepsilon\leq\varepsilon_{0}$,
$\eta_{\varepsilon}:\mathbb{D}_{r_{0}}\to\mathbb{R}^{2}$ is a local
diffeomorphism at the neighbourhood of any point of $\partial{\mathbb{D}_{+}}$
and $\eta_{\varepsilon}:\partial\left(\mathbb{D}_{+}\right)\to\mathbb{R}^{2}$
parametrizes a Jordan curve. Finally, setting
$\sigma_{\varepsilon}=(1+\varepsilon\phi_{0})^{2}$, we have

$\begin{split}-div\left(\sigma_{\varepsilon}\nabla\eta_{\varepsilon}\right)=&-div\left((1+\varepsilon\phi_{0})\nabla\left(\eta+\alpha_{\varepsilon}\eta_{y}\right)-\left(\eta+\alpha_{\varepsilon}\eta_{y}\right)\nabla\left((1+\varepsilon\phi_{0})\right)\right)\\\
=&(1+\varepsilon\phi_{0})\Delta\left(\eta+\alpha_{\varepsilon}\eta_{y}\right)-\left(\eta+\alpha_{\varepsilon}\eta_{y}\right)\Delta(1+\varepsilon\phi_{0})=0\end{split}$

and the expected equation is satisfied. Notice that if $e=0$, then by Step 4,
$\tilde{e}\neq 0$ and the same argument occurs by setting
$\eta_{\varepsilon}=\frac{\eta+\alpha_{\varepsilon}\eta_{xy}}{1+\varepsilon\phi_{0}}$.

Conclusion: By Alessandrini and Nesi [AN21], we obtain that
$\eta_{\varepsilon}:\Omega_{\varepsilon}\to\eta_{\varepsilon}(\Omega_{\varepsilon})$
is a diffeomorphism, where $\Omega_{\varepsilon}$ is chosen as a smooth domain
such that $\mathbb{D}_{+}\subset\Omega_{\varepsilon}\subset\mathbb{D}_{r_{0}}$
and $\eta_{\varepsilon}:\partial\Omega_{\varepsilon}\to\mathbb{R}^{2}$ is a
parametrization of a Jordan curve. Then
$\eta_{\varepsilon}:\mathbb{D}_{+}\to\eta(\mathbb{D}_{+})$ is a
diffeomorphism. Letting $\varepsilon\to 0$, we obtain that
$\eta_{x}\wedge\eta_{y}\geq 0$ on $\mathbb{D}_{+}$. The zero set of
$\eta_{x}\wedge\eta_{y}$ is the same as the zero set of the first coordinate
$n_{0}$ of the normal vector to the surface, which satisfies an eigenfunction
equation $\Delta n_{0}=\left|\nabla n\right|^{2}n_{0}$ (because $n$ is a
harmonic map into the sphere). $n_{0}$ has only one nodal domain. Then
$\eta_{x}\wedge\eta_{y}>0$ on $\mathbb{D}_{+}$. Then $\eta_{\varepsilon}^{-1}$
converges in $\mathcal{C}^{2}(K)$ for any compact subset $K$ of
$\eta\left(\mathbb{D}_{+}\right)$ to a function
$\psi:\eta\left(\mathbb{D}_{+}\right)\to\mathbb{D}_{+}$ that is the inverse
function of $\eta$. In particular, $\eta$ is injective on $\mathbb{D}_{+}$. By
symmetries, $\Phi$ is injective on $\mathbb{D}$ and we obtain the expected
result. ∎

##  2\. Optimization of combinations of Steklov eigenvalues

In the current section, we prove Theorem 0.1. In particular, we gather the
previous techniques for maximization of combinations of Steklov eigenvalues
[Pet21] and [Pet22], and adapt them for the first time on a restriction of the
admissible conformal factors at the boundary $e^{v}$ to symmetric ones on a
surface $\Sigma$. For simplicity, we choose to write it in our case
$\Sigma=\mathbb{D}$ and

$\forall x,y\in\mathbb{S}^{1},e^{v(x,y)}=e^{v(-x,y)}=e^{v(x,-y)}$

but it appears that such a method for symmetry constraints can be setted in a
general framework.

###  2.1. Critical metrics

We denote $\sigma_{k}(f)$ is the $k$-th non zero Steklov eigenvalue with
respect to the density $f:\mathbb{S}^{1}\to\mathbb{R}^{*}_{+}$ on
$\mathbb{D}$, that is the restriction at the boundary of the density of the
measure associated to some metric conformal to the flat one on the disk. We
set $\bar{\sigma_{k}}(f)=\sigma_{k}(f)L_{f}(\mathbb{S}^{1})$, where
$L_{f}\left(\mathbb{S}^{1}\right)=\int_{\mathbb{S}^{1}}fd\theta$ is the length
of the boundary with respect to $f$. We set

$V=\mathcal{C}^{0}\left(\mathbb{S}^{1}\right)\text{ and
}X=\\{f\in\mathcal{C}^{0}(\mathbb{S}^{1}),f>0\text{ and }\forall
x,y\in\mathbb{S}^{1},f(x,y)=f(-x,y)=f(x,-y)\\}$

and for $f\in X$ we set

$E(f):=F\left(\bar{\sigma}_{1}(f),\cdots,\bar{\sigma}_{m}(f)\right)$

where $F:\mathbb{R}^{m}\to\mathbb{R}_{+}$ is a $\mathcal{C}^{1}$ function such
that for all $i\in\\{1,\cdots,m\\}$, $\partial_{i}F\leq 0$ everywhere. Notice
that in the current paper, $F=h_{s,t}$ only depends on first and second
Steklov eigenfunctions. We denote $\partial E(f)$ the subdifferential of $E$
at $f$.

We proved in [PT23]

(2.1) $\begin{split}\partial
E(f)\subset\overline{co}_{(\phi_{1},\cdots,\phi_{m})\in\mathcal{O}_{E(f)}}\left\\{\sum_{i=1}^{m}d_{i}\bar{\sigma}_{i}(f)\left(\frac{1}{\int_{\mathbb{S}^{1}}f}-\left(\phi_{i}\right)^{2}\right)\right\\}\end{split},$

where $(\phi_{1},\cdots,\phi_{m})$ lies in the set $\mathcal{O}_{E(f)}$ of
$L^{2}(\partial\Sigma,fd\theta)$-orthonormal families where $\phi_{i}\in
E_{i}(f)$ and
$d_{i}=\partial_{i}F\left(\bar{\sigma}_{1}(f),\cdots,\bar{\sigma}_{m}(f)\right)\leq
0$. Notice that in this case the subdifferential is a space of functions on
$\partial\mathbb{S}^{1}$, while it is defined as a subspace of $V^{\star}$,
the set of Radon measures on $\mathbb{S}^{1}$: here, we identify a function
$\psi$ to the measure $\psi d\theta$ on $\mathbb{S}^{1}$.

From this (see also [Pet21]) and the multiplicity results by [Jam16] for the
first and second eigenvalues of the disk (and Proposition 1.2 and Proposition
1.3 for the embedding part), we obtain in our case $F=h_{s,t}$ that the
critical densities $e^{u}$ (densities such that $0\in\partial E(e^{u})$)
satisfy:

######  Proposition 2.1.

If $e^{u}$ is a critical density of
$h_{s,t}(\bar{\sigma}_{1},\bar{\sigma}_{2})=\left(\bar{\sigma}_{1}^{-s}+t\bar{\sigma}_{2}^{-s}\right)^{\frac{1}{s}}$,
then there is a free boundary minimal immersion
$\Phi=(\phi_{0},\phi_{1},\phi_{2}):\mathbb{D}\to co(\mathcal{E}_{\sigma})$,
where
$\sigma=(\bar{\sigma}_{1}(e^{u}),\bar{\sigma}_{2}(e^{u}),\bar{\sigma}_{2}(e^{u}))$,
$\phi_{0}$ is a Steklov eigenfunction associated to $\sigma_{1}(e^{u})$ and
$\phi_{1}$ and $\phi_{2}$ are associated to $\sigma_{2}(e^{u})$ such that
$(\phi_{0},\phi_{1},\phi_{2})$ are independent function in
$L^{2}(\mathbb{S}^{1},e^{u}d\theta)$ or $\phi_{2}=0$ and $(\phi_{0},\phi_{1})$
are independent. Moreover,

$\int_{\mathbb{S}^{1}}\phi_{0}^{2}e^{u}d\theta=\frac{\bar{\sigma}_{1}(e^{u})^{-s-1}}{f_{s,t}(\bar{\sigma}_{1}(e^{u}),\bar{\sigma}_{2}(e^{u}))}\int_{\mathbb{S}^{1}}e^{u}d\theta$

and

$\int_{\mathbb{S}^{1}}\left(\phi_{1}^{2}+\phi_{2}^{2}\right)e^{u}d\theta=\frac{t\bar{\sigma}_{2}(e^{u})^{-s-1}}{f_{s,t}(\bar{\sigma}_{1}(e^{u}),\bar{\sigma}_{2}(e^{u}))}\int_{\mathbb{S}^{1}}e^{u}d\theta.$

If in addition $f$ is symmetric with respect to the two axes of
$\mathbb{S}^{1}$, then $\Phi$ is an embedding.

###  2.2. Construction of Palais-Smale sequences

We set

$\left|\partial E(f)\right|=-\inf_{h\in Y}\max_{\psi\in\partial
E(f)}\int_{\mathbb{S}^{1}}h\psi d\theta$

where $Y=\\{h\in\mathcal{C}^{0}(\mathbb{S}^{1});h\geq
0,\int_{\mathbb{S}^{1}}hd\theta=1,\forall
x,y\in\mathbb{S}^{1},h(x,y)=h(-x,y)=h(x,-y)\\}$. We proved in [Pet22] the
following deformation lemma:

######  Proposition 2.2.

If there is $\varepsilon_{0}>0$ and $\delta>0$ such that

$\forall f\in
X;\forall\varepsilon\in(0,\varepsilon_{0}),\left|E(f)-c\right|\leq\varepsilon\Rightarrow\left|\partial
E(f)\right|\geq\delta\hskip 2.84544pt,$

Then $\forall\varepsilon\in(0,\varepsilon_{0})$, there is $\eta:X\to X$ such
that

  * •

$\eta(f)=f$ for any $f\in\\{E\geq c+\varepsilon_{0}\\}\cup\\{E\leq
c-\varepsilon_{0}\\}$

  * •

$\forall f\in X,E(\eta(f))\leq E(f)$

  * •

$\eta(\\{E\leq c+\varepsilon\\})\subset\\{E\leq c-\varepsilon\\}$

Thanks to this lemma, we have existence of a Palais-Smale minimizing sequence
$f_{\varepsilon}\in X$

$\left|E(f_{\varepsilon})-c\right|\leq\varepsilon\text{ and
}\delta_{\varepsilon}=\left|\partial E(f_{\varepsilon})\right|\to 0\text{ as
}\varepsilon\to 0.$

We rewrite $\delta_{\varepsilon}=\left|\partial E(f_{\varepsilon})\right|$ as

$\forall\tau\in\bar{Y},\exists\psi\in\partial
E(f_{\varepsilon});-\int_{\mathbb{S}^{1}}\psi d\tau\leq\delta_{\varepsilon}$

where
$\bar{Y}=\\{\tau\in\mathcal{P}(\mathbb{S}^{1});\tau=s_{1}^{\star}\tau=s_{2}^{\star}\tau\\}$
is a subset of the set of probability measures $\mathcal{P}(\mathbb{S}^{1})$
with symmetries $s_{1}(-x,y)=s_{1}(x,y)$ and $s_{2}(x,-y)=s_{2}(x,y)$. It can
be rewritten as

$\forall\tau\in\bar{Y},\exists\psi\in\partial
E(f_{\varepsilon});-\int_{\mathbb{S}^{1}}(\psi+\delta_{\varepsilon})d\tau\leq
0$

Notice that since for any eigenfunction $\phi$, $\phi\circ s_{1}$ and
$\phi\circ s_{2}$ are eigenfunctions with the same mass:

$\psi\in\partial E(f_{\varepsilon})\Rightarrow\psi\circ s_{1}\in\partial
E(f_{\varepsilon})\text{ and }\psi\circ s_{2}\in\partial E(f_{\varepsilon})$

and we can deduce that

(2.2) $-\left(\partial
E(f_{\varepsilon})+\\{\delta_{\varepsilon}\\}\right)\cap\\{a\in\mathcal{C}^{0};a\leq
0\\}\neq\emptyset.$

Indeed, if not, we use the classical Hahn-Banach theorem to separate these two
spaces (the first one is compact, the second one is closed in
$\mathcal{C}^{0}$) by $\tau\in\left(\mathcal{C}^{0}\right)^{\star}$ satisfying

$\begin{split}\forall\psi\in\partial
E(f_{\varepsilon}),\int_{\mathbb{S}^{1}}-(\psi+\delta_{\varepsilon})d\tau>0\end{split}$

and

$\forall a\in\mathcal{C}^{0};a\leq 0,\left\langle\tau,a\right\rangle\leq 0.$

The second condition implies that $\tau$ is a non-negative Radon measure. Up
to a renormalization, and up to a symmetrization, we assume that
$\tau\in\bar{Y}$ and we obtain a contradiction. Then the Palais-Smale
condition (PS) can be written as the assumption of Proposition 2.3.

###  2.3. Bubble convergence of Palais-Smale sequences

In [Pet22] we stated the following proposition (in our case
$\Sigma=\mathbb{D}$, endowed with the Euclidean metric)

######  Proposition 2.3.

Let $e^{u_{\varepsilon}}$, $\sigma_{\varepsilon}$,
$\Phi_{\varepsilon}:\Sigma\to\mathbb{R}^{m}$ be a smooth sequence of maps
satisfying the Palais-Smale assumption $(PS)$ as $\varepsilon\to 0$, that is

  * •

$\sigma_{\varepsilon}=diag(\sigma_{k_{1}^{\varepsilon}}^{\varepsilon},\cdots,\sigma_{k_{m}^{\varepsilon}}^{\varepsilon})$
where the diagonal terms are eigenvalues associated to $e^{u_{\varepsilon}}$
with uniformly bounded spectral indices $k_{i}^{\varepsilon}$, and
$\sigma_{\varepsilon}\to\sigma$ as $\varepsilon\to 0$.

  * •

$\Delta_{g}\Phi_{\varepsilon}=0$ in $\Sigma$ and
$\partial_{\nu}\Phi_{\varepsilon}=e^{u_{\varepsilon}}\sigma_{\varepsilon}\Phi_{\varepsilon}$
on $\partial\Sigma$

  * •

$\int_{\partial\Sigma}e^{u_{\varepsilon}}dL_{g}=\int_{\partial\Sigma}\left|\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}e^{u_{\varepsilon}}dL_{g}=\int_{\Sigma}\left|\nabla\Phi_{\varepsilon}\right|_{g}^{2}dA_{g}=1$

  * •

$\left|\Phi\right|_{\sigma_{\varepsilon}}^{2}\geq 1-\delta_{\varepsilon}$
uniformly on $\partial\Sigma$, where $\delta_{\varepsilon}\to 0$ as
$\varepsilon\to 0$.

Then, up to a subsequence $\Phi_{\varepsilon}$ bubble tree converges in
$W^{1,2}$ to $\Phi_{0}:\Sigma\to co\left(\mathcal{E}_{\sigma}\right)$, and
$\Phi_{j}:\mathbb{D}\to co\left(\mathcal{E}_{\sigma}\right)$ for
$j=1,\cdots,l$ ($l\geq 0$) with an energy identity:

$1=\int_{\Sigma}\left|\nabla\Phi_{0}\right|_{g}^{2}dA_{g}+\sum_{j=1}^{l}\int_{\mathbb{D}}\left|\nabla\Phi_{j}\right|_{h}^{2}dA_{h}$

Moreover, $\Phi_{j}$ are smooth harmonic maps with free boundary for
$j=0,\cdots,l$ and their $i$-th coordinates are eigenfunctions associated to
$\lambda_{k_{i}}$ on the surface $\Sigma\cup\bigcup_{1\leq i\leq l}\mathbb{D}$
with respect to the metrics $e^{2u}g$ on $\Sigma$ such that
$e^{u}=\partial_{\nu}\Phi_{0}.\Phi_{0}$ on $\partial\Sigma$ and
$e^{2v_{j}}\xi$ on $\mathbb{D}$ such that
$e^{v_{j}}=\partial_{\nu}\Phi_{j}.\Phi_{j}$ on $\mathbb{S}^{1}$.

We proved it in [Pet22] under the assumption that
$\sigma_{k_{j}^{\varepsilon}}^{\varepsilon}$ does not converge to $0$, leaving
to the reader the remaining case, that is analogous to the Laplacian case
proved in all generality in [Pet22]. In the current paper, for the sake of
completeness, we shall add the missing arguments for convergence of Palais
Smale sequences in the specific case $\Sigma=\mathbb{D}$, and
$e^{u_{\varepsilon}}$ symmetric with respect to the two axes,
$\sigma_{\varepsilon}=(\sigma_{1}^{\varepsilon},\sigma_{2}^{\varepsilon},\sigma_{2}^{\varepsilon})$,
$\Phi_{\varepsilon}=(\phi_{0}^{\varepsilon},\phi_{1}^{\varepsilon},\phi_{2}^{\varepsilon})=(\phi_{0}^{\varepsilon},\eta_{\varepsilon})$,
where $\phi_{0}^{\varepsilon}$ is a eigenfunction, $\phi_{1}^{\varepsilon}$
and $\phi_{2}^{\varepsilon}$ are independent second eigenfunctions (or
possibly $\phi_{2}^{\varepsilon}=0$), $\sigma_{1}^{\varepsilon}\to 0$ and
$\sigma_{2}^{\varepsilon}$ uniformly lower bounded as $\varepsilon\to 0$.

Following [Pet22] (with a slight improvement in the current paper) we first
have:

######  Claim 2.1.

Let $\omega_{\varepsilon}$ be the harmonic extension of
$\left|\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}$ in $\mathbb{D}$ and
$\widetilde{\Phi_{\varepsilon}}=\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}$.
Then

(2.3)
$\int_{\mathbb{D}}\left|\nabla\omega_{\varepsilon}\right|^{2}+\int_{\mathbb{D}}\left|\nabla\left(\Phi_{\varepsilon}-\widetilde{\Phi_{\varepsilon}}\right)\right|_{\sigma_{\varepsilon}}^{2}=O\left(\delta_{\varepsilon}\right)\text{
as }\varepsilon\to 0.$

###### Proof.

We have

$\begin{split}\int_{\mathbb{D}}\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}-\int_{\mathbb{D}}\left|\nabla\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}-\int_{\mathbb{D}}\left|\nabla\left(\Phi_{\varepsilon}-\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right)\right|_{\sigma_{\varepsilon}}^{2}\\\
=2\int_{\mathbb{D}}\left\langle\nabla\Phi_{\varepsilon},\nabla\left(\Phi_{\varepsilon}-\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right)\right\rangle_{\sigma_{\varepsilon}}\\\
=2\int_{\mathbb{D}}\Delta\Phi_{\varepsilon}{\sigma_{\varepsilon}}.\left(\Phi_{\varepsilon}-\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right)+2\int_{\mathbb{S}^{1}}\partial_{r}\Phi_{\varepsilon}.{\sigma_{\varepsilon}}\left(\Phi_{\varepsilon}-\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right)\\\
=2\int_{\mathbb{S}^{1}}e^{u_{\varepsilon}}\sigma_{\varepsilon}.\Phi_{\varepsilon}.{\sigma_{\varepsilon}}.\left(\Phi_{\varepsilon}-\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right)\\\
=2\int_{\mathbb{S}^{1}}e^{u_{\varepsilon}}\frac{\left|\sigma_{\varepsilon}\Phi_{\varepsilon}\right|^{2}}{\omega_{\varepsilon}^{2}}\left(\omega_{\varepsilon}^{2}-\omega_{\varepsilon}\right)=O\left(\delta_{\varepsilon}\right)\end{split}$

where we integrated $\Delta\Phi_{\varepsilon}=0$ in $\mathbb{D}$ and
$\partial_{r}\Phi_{\varepsilon}=\sigma_{\varepsilon}e^{u_{\varepsilon}}\Phi_{\varepsilon}$
in $\mathbb{S}^{1}$ against
$\sigma_{\varepsilon}.\left(\Phi_{\varepsilon}-\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right)$,
and we used that
$\omega_{\varepsilon}^{2}=\left|\Phi_{\varepsilon}\right|^{2}\geq
1-\delta_{\varepsilon}$ and that
$\int\left(\omega_{\varepsilon}^{2}-1\right)e^{u_{\varepsilon}}=0$. In
particular, we have

$0\leq\int_{\mathbb{D}}\left|\nabla\left(\Phi_{\varepsilon}-\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right)\right|_{\sigma_{\varepsilon}}^{2}\leq\int_{\mathbb{D}}\left(\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}-\left|\nabla\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}\right)+O(\delta_{\varepsilon})$

and knowing that with the straightforward computations we have

$\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}-\left|\nabla\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}=\left(1-\omega_{\varepsilon}^{2}\right)\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}+\left|\nabla\omega_{\varepsilon}\right|^{2}\frac{\left|\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}}{\omega_{\varepsilon}^{2}}-\nabla\ln\omega_{\varepsilon}\nabla\left|\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}$

and knowing that

$\begin{split}\int_{\mathbb{D}}\nabla\ln\omega_{\varepsilon}\nabla\left|\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}=&\int_{\mathbb{D}}\Delta\ln\omega_{\varepsilon}\left|\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}+\int_{\mathbb{S}^{1}}\omega_{\varepsilon}^{2}\partial_{r}\ln\omega_{\varepsilon}\\\
=&\int_{\mathbb{D}}\left|\nabla\omega_{\varepsilon}\right|^{2}\frac{\left|\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}}{\omega_{\varepsilon}^{2}}+\int_{\mathbb{D}}\Delta\frac{\omega_{\varepsilon}^{2}}{2}\\\
=&\int_{\mathbb{D}}\left|\nabla\omega_{\varepsilon}\right|^{2}\frac{\left|\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}}{\omega_{\varepsilon}^{2}}+\int_{\mathbb{D}}\left|\nabla\omega_{\varepsilon}\right|^{2}\end{split}$

we obtain that

$0\leq\int_{\mathbb{D}}\left|\nabla\omega_{\varepsilon}\right|^{2}+\int_{\mathbb{D}}\left|\nabla\left(\Phi_{\varepsilon}-\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right)\right|_{\sigma_{\varepsilon}}^{2}\leq\int_{\mathbb{D}}\left(1-\omega_{\varepsilon}^{2}\right)\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}+O(\delta_{\varepsilon})\leq
O\left(\delta_{\varepsilon}\right)$

as $\varepsilon\to 0$. ∎

Following [Pet22] we define concentration points as points
$p\in\mathbb{S}^{1}$ such that

$\lim_{r\to 0}\liminf_{\varepsilon\to
0}\int_{\mathbb{S}^{1}\setminus\mathbb{D}_{r}(p)}e^{u_{\varepsilon}}d\theta>0$

and bad points as points $p\in\mathbb{S}^{1}$ such that there are sequences
$p_{\varepsilon}\in\mathbb{S}^{1}$ and $r_{\varepsilon}>0$ with
$p_{\varepsilon}\to p$ and $r_{\varepsilon}\to 0$ as $\varepsilon\to 0$ such
that the eigenvalue

$\sigma_{\star}\left(p_{\varepsilon},r_{\varepsilon},e^{u_{\varepsilon}}d\theta\right):=\inf_{\varphi\in\mathcal{C}^{\infty}_{c}(\mathbb{D}_{r_{\varepsilon}(p_{\varepsilon})}\cap\overline{\mathbb{D}})\setminus\\{0\\}}\frac{\int_{\mathbb{D}_{r_{\varepsilon}(p_{\varepsilon})}}\left|\nabla\varphi\right|^{2}}{\int_{\mathbb{D}_{r_{\varepsilon}(p_{\varepsilon})}\cap\mathbb{S}^{1}}\varphi^{2}e^{u_{\varepsilon}}d\theta}$

satisfies

$\sigma_{\star}\left(p_{\varepsilon},r_{\varepsilon},e^{u_{\varepsilon}}d\theta\right)\leq\sigma_{2}^{\varepsilon}$

and good points are the other points of $\mathbb{S}^{1}$.

######  Claim 2.2.

Let $p\in\mathbb{S}^{1}$ be a good point. Then

$\begin{split}\lim_{r\to 0}\limsup_{\varepsilon\to
0}&\int_{\mathbb{S}^{1}\cap\mathbb{D}_{r}(p)}e^{u_{\varepsilon}}d\theta=\lim_{r\to
0}\limsup_{\varepsilon\to
0}\int_{\mathbb{D}\cap\mathbb{D}_{r}(p)}\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}\\\
=&\lim_{r\to 0}\limsup_{\varepsilon\to
0}\int_{\mathbb{D}\cap\mathbb{D}_{r}(p)}\left|\nabla\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}=\lim_{r\to
0}\limsup_{\varepsilon\to
0}\int_{\mathbb{D}\cap\mathbb{D}_{r}(p)}\left|\nabla\eta_{\varepsilon}\right|^{2}=0\end{split}$

###### Proof.

We choose $r$ small enough such that
$\sigma_{\star}\left(p,\sqrt{r},e^{u_{\varepsilon}}d\theta\right)\geq\sigma_{2}^{\varepsilon}$.
Let $\eta\in\mathcal{C}_{c}^{\infty}\left(\mathbb{D}_{\sqrt{r}}(p)\right)$
such that $0\leq\eta\leq 1$, $\eta=1$ in $\mathbb{D}_{r}(p)$ and
$\int_{\mathbb{R}^{2}}\left|\nabla\eta\right|^{2}\leq\frac{C}{\ln\frac{1}{r}}$

$\int_{\mathbb{S}^{1}}\eta
e^{u_{\varepsilon}}\leq\left(\int_{\mathbb{S}^{1}}\eta^{2}e^{u_{\varepsilon}}\right)^{\frac{1}{2}}\leq\left(\frac{1}{\sigma_{\star}\left(p,\sqrt{r},e^{u_{\varepsilon}}d\theta\right)}\int_{\mathbb{D}_{\sqrt{r}}(p)}\left|\nabla\eta\right|^{2}\right)^{\frac{1}{2}}\leq\left(\frac{1}{\sigma_{2}^{\varepsilon}}\frac{C}{\ln\frac{1}{r}}\right)^{\frac{1}{2}}$

so that $p$ is not a concentration point. Then we integrate the equation
$\Delta\Phi_{\varepsilon}=0$ and
$\partial_{r}\Phi_{\varepsilon}=\sigma_{\varepsilon}e^{u_{\varepsilon}}\Phi_{\varepsilon}$
against
$\eta\sigma_{\varepsilon}\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}^{2}}$
and we obtain

$\int_{\mathbb{D}}\eta\left\langle\nabla{\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}^{2}}},\nabla\Phi_{\varepsilon}\right\rangle_{\sigma_{\varepsilon}}+\int_{\Sigma}\nabla\eta\nabla{\sigma_{\varepsilon}\Phi_{\varepsilon}}\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}^{2}}=\int_{\mathbb{S}^{1}}\eta\frac{\left|\sigma_{\varepsilon}\Phi_{\varepsilon}\right|^{2}}{\omega_{\varepsilon}^{2}}e^{u_{\varepsilon}}d\theta$

and since we have that

$\begin{split}\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}=&\left\langle\frac{\nabla\Phi_{\varepsilon}}{\omega_{\varepsilon}},\omega_{\varepsilon}\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}^{2}}+\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}^{2}}\textpm\nabla\omega_{\varepsilon}\right\rangle_{\sigma_{\varepsilon}}-\left\langle\Phi_{\varepsilon}\frac{\nabla\omega_{\varepsilon}}{\omega_{\varepsilon}^{2}},\frac{\nabla\Phi_{\varepsilon}}{\omega_{\varepsilon}}-\Phi_{\varepsilon}\frac{\nabla\omega_{\varepsilon}}{\omega_{\varepsilon}^{2}}\right\rangle_{\sigma_{\varepsilon}}\\\
=&\left\langle\nabla{\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}^{2}}},\nabla\Phi_{\varepsilon}\right\rangle_{\sigma_{\varepsilon}}+\frac{\left|\nabla\omega_{\varepsilon}\right|^{2}\left|\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}}{\omega_{\varepsilon}^{4}}\end{split}$

we obtain

$\int_{\mathbb{D}}\eta\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}\leq\sigma_{2}^{\varepsilon}\int_{\mathbb{S}^{1}}\eta
e^{u_{\varepsilon}}+\left(\frac{C}{\ln\frac{1}{r}}\int_{\mathbb{D}}\left|\nabla\Phi_{\varepsilon}\right|^{2}\right)^{\frac{1}{2}}\sup_{\mathbb{D}}\frac{\left|\sigma_{\varepsilon}\Phi_{\varepsilon}\right|}{\omega_{\varepsilon}^{2}}+\int_{\mathbb{D}}\eta\frac{\left|\nabla\omega_{\varepsilon}\right|^{2}\left|\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}}{\omega_{\varepsilon}^{4}}$

so that
$\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}$
does not concentrate at $p$. And

$\begin{split}\int_{\mathbb{D}}\eta\sigma_{2}^{\varepsilon}\left|\nabla\eta_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}\leq\int_{\mathbb{D}}\eta\left|\nabla\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}\leq&\int_{\mathbb{D}}\eta\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}+\int_{\mathbb{D}}\left|\nabla\left(\Phi_{\varepsilon}-\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right)\right|_{\sigma_{\varepsilon}}\left|\nabla\left(\Phi_{\varepsilon}+\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right)\right|_{\sigma_{\varepsilon}}\\\
=&\int_{\mathbb{D}}\eta\left|\nabla\frac{\Phi_{\varepsilon}}{\omega_{\varepsilon}}\right|_{\sigma_{\varepsilon}}^{2}+O\left(\delta_{\varepsilon}^{\frac{1}{2}}\right)\end{split}$

completes the proof of the claim. ∎

We deduce thanks to test function arguments (see [Pet22]) that since
$\sigma_{2}^{\varepsilon}$ is uniformly lower bounded,
$e^{u_{\varepsilon}}d\theta$ has at most two concentration points and that

  * •

If $e^{u_{\varepsilon}}d\theta$ does not have concentration points, then in
addition, by symmetries, any $p\in\mathbb{S}^{1}$ has to be a good point (see
test function arguments in [Pet22]).

  * •

If $e^{u_{\varepsilon}}d\theta$ has one concentration point, then by symmetry
and up to a quarter rotation there are two concentration points
$(1,0),(-1,0)$. Then, up take half of the density
$1_{x_{1}>0}.e^{u_{\varepsilon}}$ and to rescale by the Mobius group of
$\mathbb{D}$, we have thanks to the symmetry that

$\lim_{r\to 0}\liminf_{\varepsilon\to
0}\int_{\mathbb{S}^{1}\setminus\mathbb{D}_{r}((-1,0))}e^{u_{\varepsilon}}d\theta=\frac{1}{2}$

and that $e^{u_{\varepsilon}}d\theta$ does not concentrate on
$\mathbb{S}^{1}\setminus\\{(-1,0)\\}$. In addition, any
$p\in\mathbb{S}^{1}\setminus\\{(-1,0)\\}$ has to be a good point (see test
function arguments in [Pet22]).

Without loss of generality, we assume that the second case occurs (we recall
this assumption in Proposition 2.4)

######  Claim 2.3.

For any $r>0$, there is $C_{r}>0$ such that for $j=0,1,2$,

(2.4)
$\int_{\mathbb{S}^{1}\cap\mathbb{D}_{r}((-1,0))}\left(\phi_{\varepsilon}^{j}-m_{\varepsilon,r}^{j}\right)^{2}d\theta\leq
C_{r}\int_{\mathbb{D}}\left|\nabla\phi_{\varepsilon}^{j}\right|^{2},$

where $M_{\varepsilon,r}=(m_{\varepsilon,r}^{j})_{j=0,1,2}$ satisfies

$M_{\varepsilon,r}=\frac{\int_{\mathbb{S}^{1}\setminus\mathbb{D}_{r}((-1,0))}\Phi_{\varepsilon}e^{u_{\varepsilon}}d\theta\textdiv}{\int_{\mathbb{S}^{1}\setminus\mathbb{D}_{r}((-1,0))}e^{u_{\varepsilon}}d\theta}$

and satisfies $\left|M_{\varepsilon,r}\right|_{\sigma_{\varepsilon}}\leq 1$.
In particular, $\eta_{\varepsilon}$ is bounded in
$H^{1}_{loc}\left(\mathbb{D}\setminus\\{(-1,0)\\}\right)$ and
$\sqrt{\sigma_{1}^{\varepsilon}}\phi_{0}^{\varepsilon}$ converges to a
constant function in
$H^{1}_{loc}\left(\mathbb{D}\setminus\\{(-1,0)\\}\right)$.

This Claim is the consequence of a Poincaré inequality given from a uniform
$H^{-\frac{1}{2}}$ control of
$\frac{e^{u_{\varepsilon}}\textdiv}{\int_{\mathbb{S}^{1}\setminus\mathbb{D}_{r}((-1,0))}e^{u_{\varepsilon}}d\theta}$
on $\mathbb{S}^{1}\setminus\mathbb{D}_{r}((-1,0))$ as $\varepsilon\to 0$
because all the points of $\mathbb{S}^{1}\setminus\\{(-1,0)\\}$ are good
points. It is left to the reader (follow the proof of Claim 2.6 in [Pet22])
Thanks to the previous claim, we have the assumptions of Proposition 2.4.

######  Proposition 2.4.

We assume that (up to some rescaling by the Mobius group of the disk)
$\left(\sigma_{1}^{\varepsilon}\right)^{\frac{1}{2}}\phi_{0}^{\varepsilon}$
converges to a constant function $c$ in
$H_{loc}^{1}(\mathbb{D}^{1}\setminus\\{(-1,0)\\})$, then either
$\left|c\right|=1$ and $\eta_{\varepsilon}$ bubble tree converges to two
constant functions in $W^{1,2}$ or $\left|c\right|<1$ and up to a subsequence
as $\varepsilon\to 0$,
$\frac{\sigma_{2}^{\varepsilon}}{\sqrt{1-c^{2}}}\eta_{\varepsilon}$ bubble
tree converges in $W^{1,2}$ to two identity maps of $\mathbb{D}$.

Notice that we a posteriori have that for maximizing sequences,
$\eta_{\varepsilon}$ cannot bubble tree converge to constant functions since
$\sigma_{2}^{\varepsilon}$ does not converge to $0$. We do not prove the case
$c=1$ because it is not possible in the current paper with the choice
$h_{s,t}$ because $c=0$ in these cases (see Proposition 2.1). In the proof of
Proposition 2.4, we will use the following

######  Claim 2.4.

[Sch06], lemma 3.1 Let $\Phi:\mathbb{D}^{+}\to\mathbb{R}^{n}$ be a Euclidean
harmonic map. We assume that $\left|\Phi\right|\geq 1$ on
$[-1,1]\times\\{0\\}$. Then for any $\alpha>0$ there is
$\varepsilon_{\alpha}>0$ such that if

$\int_{\mathbb{D}_{+}}\left|\nabla\Phi\right|^{2}\leq\varepsilon_{\alpha}$

then $\left|\Phi\right|^{2}\geq 1-\alpha$ on $\mathbb{D}_{\frac{1}{2}}^{+}$.

###### Proof of Proposition 2.4.

We assume $c<1$. We can extend (the rescaling by the Mobius group of the disk
of) $\Phi_{\varepsilon}$ to a map still denoted
$\Phi_{\varepsilon}=(\phi_{0,\varepsilon},\eta_{\varepsilon}):\mathbb{R}^{2}\to\mathbb{R}^{3}$
by
$\Phi_{\varepsilon}(z)=\Phi_{\varepsilon}\left(\frac{z}{\left|z\right|^{2}}\right)$
for $\left|z\right|\geq 1$.

Step 1: We define a suitable replacement of $\eta_{\varepsilon}$, denoted by
$\psi_{\varepsilon}$

Let $p\in\mathbb{S}^{1}$ be a good point, $0<\delta$ be chosen later and $r>0$
be such that for any $\varepsilon>0$,

$\int_{\mathbb{D}_{2\sqrt{r}}(p)}\left|\nabla\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}\leq\delta.$

By the Courant-Lebesque lemma, we can find $r_{\varepsilon}\in(r,\sqrt{r})$
such that

(2.5)
$\int_{\partial\mathbb{D}_{r_{\varepsilon}}(p)}\left|\partial_{\theta}\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}d\theta\leq\frac{1}{\ln
2}\int_{\mathbb{D}_{\sqrt{r}}(p)}\left|\nabla\Phi_{\varepsilon}\right|_{\sigma_{\varepsilon}}^{2}\leq\frac{1}{\ln
2}\delta$

and as a consequence

(2.6) $\forall
q,q^{\prime}\in\partial\mathbb{D}_{r_{\varepsilon}}(p);\sigma_{1}^{\varepsilon}(\phi_{0,\varepsilon}(q)-\phi_{0,\varepsilon}(q^{\prime}))^{2}+\sigma_{2}^{\varepsilon}\left|\eta_{\varepsilon}(q)-\eta_{\varepsilon}(q^{\prime})\right|^{2}\leq\frac{\pi}{\ln
2}\delta.$

By the classical trace $L^{2}$ embedding into $H^{1}$ and (2.4), we have a
sequence $m_{\varepsilon,\rho}^{0}$ such that

$\int_{\partial\mathbb{D}_{r_{\varepsilon}}(p)}\left(\sqrt{\sigma_{1}^{\varepsilon}}\phi_{0,\varepsilon}-\sqrt{\sigma_{1}^{\varepsilon}}m_{\varepsilon,\rho}^{0}\right)^{2}=O\left(\sigma_{1}^{\varepsilon}\right)$

as $\varepsilon\to 0$ and using this and (2.6), we obtain for $\varepsilon$
small enough:

$\forall
q\in\partial\mathbb{D}_{r_{\varepsilon}}(p);\sigma_{1}^{\varepsilon}\left|\phi_{0,\varepsilon}(q)-m_{\varepsilon,\rho}^{0}\right|^{2}\leq\frac{\pi}{\ln
2}\delta+O\left(\sqrt{\sigma_{1}^{\varepsilon}}\right).$

We also have that

$\sigma_{1}^{\varepsilon}\left(m_{\varepsilon,\rho}^{0}\right)^{2}=c^{2}+o(1)$

Then, given $0<\alpha$ applying the claim 2.4 where up to reduce $r$, we can
reduce $\delta$ so that $\delta\leq\varepsilon_{\frac{\alpha}{4}}$ and
$2\sqrt{\frac{\pi\delta}{\ln 2}}\left|c\right|+\frac{\pi}{\ln
2}\delta\leq\frac{\alpha}{4}$, we obtain on
$\partial\mathbb{D}_{r_{\varepsilon}}(p)$,

$\begin{split}\sigma_{2}^{\varepsilon}\left|\eta_{\varepsilon}\right|^{2}=\left|\Phi_{\varepsilon}\right|^{2}_{\sigma_{\varepsilon}}-\sigma_{1}^{\varepsilon}\left(\phi_{0}^{\varepsilon}\right)^{2}&\geq
1-O(\delta_{\varepsilon})-\left(\sqrt{\sigma_{1}^{\varepsilon}}m_{\varepsilon,\rho}^{0}+\sqrt{\frac{\pi}{\ln
2}\delta}\right)^{2}\\\ &\geq 1-c^{2}-\alpha\end{split}$

for $\varepsilon$ small enough, so that we can define
$\frac{\eta_{\varepsilon}}{\left|\eta_{\varepsilon}\right|}$ on
$\partial\mathbb{D}_{r_{\varepsilon}}(p)$. Let
$\tilde{\eta}_{\varepsilon}:\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}\to\mathbb{D}$
be a free boundary harmonic (with respect to
$\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{S}^{1}$) extension of
$\frac{\eta_{\varepsilon}}{\left|\eta_{\varepsilon}\right|}:\partial\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}$.
Such a map is unique, as soon as its energy is small enough by the energy
convexity result (see [LP19]), that we apply to any map
$\varphi_{\varepsilon}$ such that
$\varphi_{\varepsilon}=\frac{\eta_{\varepsilon}}{\left|\eta_{\varepsilon}\right|}$
on $\partial\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}$ and
$\left|\varphi_{\varepsilon}\right|=1$ on
$\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{S}^{1}$:

(2.7)
$\frac{1}{2}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\left(\varphi_{\varepsilon}-\tilde{\eta}_{\varepsilon}\right)\right|^{2}\leq\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\varphi_{\varepsilon}\right|^{2}-\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\tilde{\eta}_{\varepsilon}\right|^{2}$

Before using this, we have to check that the energy of
$\tilde{\eta}_{\varepsilon}$ is small enough. Let $h_{\varepsilon}$ be the
harmonic extension of
$\eta_{\varepsilon}:\partial\mathbb{D}_{r_{\varepsilon}}(p)\to\mathbb{R}^{2}$
in $\mathbb{D}_{r_{\varepsilon}}(p)$, By the maximum principle and by (2.6),
we obtain that $\sigma_{2}^{\varepsilon}\left|h_{\varepsilon}\right|^{2}\geq
1-c^{2}-2\alpha$ so that

$\begin{split}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\tilde{\eta}_{\varepsilon}\right|^{2}\leq\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\frac{h_{\varepsilon}}{\left|h_{\varepsilon}\right|}\right|^{2}\leq\int_{\mathbb{D}_{r_{\varepsilon}}(p)}\frac{\left|\nabla
h_{\varepsilon}\right|^{2}}{\left|h_{\varepsilon}\right|^{2}}\\\
\leq\frac{\sigma_{2}^{\varepsilon}}{1-c^{2}-2\alpha}\int_{\mathbb{D}_{2\sqrt{r}}(p)}\left|\nabla
h_{\varepsilon}\right|^{2}\leq\frac{\sigma_{2}^{\varepsilon}}{1-c^{2}-2\alpha}\delta\end{split}$

and up to reduce $r$ again, the energy of $\tilde{\eta}_{\varepsilon}$ is
small enough to have (2.7). We set

$\tilde{\omega}_{\varepsilon}=\sqrt{\frac{\omega_{\varepsilon}^{2}-\sigma_{1}^{\varepsilon}\varphi_{\varepsilon}^{2}}{\sigma_{2}^{\varepsilon}}},$

where $\varphi_{\varepsilon}$ is the harmonic extension of
$\phi_{0}^{\varepsilon}:\partial\mathbb{D}_{r_{\varepsilon}}(p)\to\mathbb{R}$
in $\mathbb{D}_{r_{\varepsilon}}(p)$. We will use
$\psi_{\varepsilon}=\tilde{\omega}_{\varepsilon}\tilde{\eta}_{\varepsilon}$ as
the suitable replacement of $\eta_{\varepsilon}$ to prove the $H^{1}$-bubble
convergence.

Step 2: We test
$\frac{\psi_{\varepsilon}-\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}^{2}}$
against the equation satisfied by $\eta_{\varepsilon}$ and we obtain (2.8)

$\begin{split}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}&\left\langle\nabla\eta_{\varepsilon},\nabla\frac{\eta_{\varepsilon}-\psi_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}^{2}}\right\rangle=\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{S}^{1}}e^{u_{\varepsilon}}\left\langle\sigma_{2}^{\varepsilon}\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}},\frac{\eta_{\varepsilon}-\psi_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right\rangle\\\
=&\frac{1}{2}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}e^{u_{\varepsilon}}\sigma_{2}^{\varepsilon}\left|\nabla\left(\frac{\eta_{\varepsilon}-\psi_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right)\right|^{2}-\frac{1}{2}\sigma_{1}^{\varepsilon}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{S}^{1}}\frac{e^{u_{\varepsilon}}}{\widetilde{\omega_{\varepsilon}}^{2}}\left(\left(\phi_{0,\varepsilon}\right)^{2}-\left(\varphi_{\varepsilon}\right)^{2}\right)\\\
\leq&\frac{1}{2}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\left(\frac{\eta_{\varepsilon}-\psi_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right)\right|^{2}\\\
&+\frac{1}{c_{0}}\sigma_{1}^{\varepsilon}\left(\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{S}^{1}}\left(\phi_{0,\varepsilon}+\varphi_{\varepsilon}\right)^{2}e^{u_{\varepsilon}}\right)^{\frac{1}{2}}\left(\frac{1}{\sigma_{2}^{\varepsilon}}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\left(\phi_{0,\varepsilon}-\varphi_{\varepsilon}\right)\right|^{2}\right)^{\frac{1}{2}}\\\
\leq&\frac{1}{2}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\left(\frac{\eta_{\varepsilon}-\psi_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right)\right|^{2}+\frac{1}{c_{0}}\frac{\sigma_{1}^{\varepsilon}}{\sigma_{2}^{\varepsilon}}\left(\left|m_{\varepsilon,\rho}^{0}\right|+O\left(\sigma_{1}^{\varepsilon}\right)\right)\left(\sigma_{1}^{\varepsilon}\int_{\mathbb{S}^{1}}e^{u_{\varepsilon}}\left(\phi_{0}^{\varepsilon}\right)^{2}\right)^{\frac{1}{2}}\\\
\leq&\frac{1}{2}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\left(\frac{\eta_{\varepsilon}-\psi_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right)\right|^{2}+O\left(\left(\sigma_{1}^{\varepsilon}\right)^{\frac{1}{2}}\right)\end{split}$

which gives

(2.8)
$\begin{split}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right|^{2}-&\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\widetilde{\eta_{\varepsilon}}\right|^{2}\leq
O\left(\left(\sigma_{1}^{\varepsilon}\right)^{\frac{1}{2}}\right)\\\
&+\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\frac{\nabla\widetilde{\omega_{\varepsilon}}}{\widetilde{\omega_{\varepsilon}}}\left(\frac{\widetilde{\Phi_{\varepsilon}}}{\widetilde{\omega_{\varepsilon}}}\nabla\left(\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}-\widetilde{\eta_{\varepsilon}}\right)-\left(\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}-\widetilde{\eta_{\varepsilon}}\right)\frac{\nabla\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right)\\\
\leq&O\left(\left(\sigma_{1}^{\varepsilon}\right)^{\frac{1}{2}}+\delta_{\varepsilon}^{\frac{1}{2}}\right)\end{split}$

Step 3: We prove the $H_{loc}^{1}\left(\mathbb{S}^{1}\setminus\\{p\\}\right)$
convergence of $\eta_{\varepsilon}$ to a free boundary harmonic map in
$\mathbb{S}^{1}$

Now, we test
$\widetilde{\eta_{\varepsilon}}-\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}$
against the harmonic map equation satisfied by
$\widetilde{\eta_{\varepsilon}}$. We obtain in a similar way

$\begin{split}I:=2\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}&\left\langle\nabla\left(\widetilde{\eta_{\varepsilon}}-\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right),\nabla\widetilde{\eta_{\varepsilon}}\right\rangle=2\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{S}^{1}}\widetilde{\eta_{\varepsilon}}.\partial_{r}\widetilde{\eta_{\varepsilon}}\left\langle\sigma_{2}^{\varepsilon}\widetilde{\eta_{\varepsilon}},\widetilde{\eta_{\varepsilon}}-\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right\rangle\\\
=&\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{S}^{1}}\widetilde{\eta_{\varepsilon}}.\partial_{r}\widetilde{\eta_{\varepsilon}}\left(\sigma_{2}^{\varepsilon}\left|\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}-\widetilde{\eta_{\varepsilon}}\right|^{2}-\frac{\sigma_{1}^{\varepsilon}}{\widetilde{\omega_{\varepsilon}}^{2}}\left(\left(\varphi_{\varepsilon}\right)^{2}-\left(\phi_{0,\varepsilon}\right)^{2}\right)\right)\\\
\end{split}$

and using the consequence of the $\varepsilon$-regularity result on free
boundary harmonic maps
$\left|\nabla\widetilde{\eta_{\varepsilon}}\right|^{2}\leq\frac{C\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\widetilde{\eta_{\varepsilon}}\right|^{2}}{\left(1-\left|x\right|\right)^{2}}$
and a trace Hardy inequality (this is the way we prove energy convexity result
(2.7) in [LP19]), we obtain:

$\begin{split}I\leq&C\delta\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\left(\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}-\widetilde{\eta_{\varepsilon}}\right)\right|^{2}+C\frac{1}{c_{0}}\sigma_{1}^{\varepsilon}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\frac{\left(\varphi_{\varepsilon}\right)^{2}-\left(\phi_{0,\varepsilon}\right)^{2}}{\widetilde{\omega_{\varepsilon}}^{2}}\right|^{2}\\\
=:&C\delta\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\left(\eta_{\varepsilon}-\frac{\widetilde{\eta_{\varepsilon}}}{\widetilde{\omega_{\varepsilon}}}\right)\right|^{2}+\frac{C}{c_{0}}II\end{split}$

and we prove that $II=o(1)$ as $\varepsilon\to 0$, using that
$\sigma_{1}^{\varepsilon}\left(\varphi_{\varepsilon}^{2}+\phi_{0,\varepsilon}^{2}\right)\leq
C\tilde{\omega_{\varepsilon}}^{2}$ and several integration by parts

$\begin{split}II\leq&2\left(\sigma_{1}^{\varepsilon}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\frac{\left|\nabla\left(\varphi_{\varepsilon}\right)^{2}-\left(\phi_{0,\varepsilon}\right)^{2}\right|^{2}}{\widetilde{\omega_{\varepsilon}}^{2}}+C\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\widetilde{\omega_{\varepsilon}}\right|^{2}\right)\\\
\leq&2\left(C\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\Delta\left(\left(\varphi_{\varepsilon}\right)^{2}-\left(\phi_{0,\varepsilon}\right)^{2}\right)\right|+\sigma_{1}^{\varepsilon}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\widetilde{\omega_{\varepsilon}}\right|\left|\nabla\left(\varphi_{\varepsilon}-\phi_{0}^{\varepsilon}\right)\right|\frac{\left|\varphi_{\varepsilon}-\phi_{0}^{\varepsilon}\right|}{\widetilde{\omega_{\varepsilon}}}\right)\\\
&+2C\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\widetilde{\omega_{\varepsilon}}\right|^{2}\\\
=&4C\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\left|\nabla\varphi_{\varepsilon}\right|^{2}-\left|\nabla\phi_{0}^{\varepsilon}\right|^{2}\right|+O\left(\left(\sigma_{1}^{\varepsilon}\right)^{\frac{1}{2}}+\delta_{\varepsilon}^{\frac{1}{2}}\right)=o(1)\end{split}$

as $\varepsilon\to 0$ since $\varphi_{\varepsilon}-\phi_{0}^{\varepsilon}$
converges to $0$ in $H^{1}$. Letting $\delta$ be small enough such that
$1-C\delta\geq\frac{1}{2}$ we obtain that

$\begin{split}\frac{1}{2}\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}&\left|\nabla\left(\eta_{\varepsilon}-\frac{\widetilde{\eta_{\varepsilon}}}{\widetilde{\omega_{\varepsilon}}}\right)\right|^{2}\leq\left(1-C\delta\right)\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\left(\eta_{\varepsilon}-\frac{\widetilde{\eta_{\varepsilon}}}{\widetilde{\omega_{\varepsilon}}}\right)\right|^{2}\\\
\leq&\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\left(\eta_{\varepsilon}-\frac{\widetilde{\eta_{\varepsilon}}}{\widetilde{\omega_{\varepsilon}}}\right)\right|^{2}-2\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left\langle\nabla\left(\widetilde{\eta_{\varepsilon}}-\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right),\nabla\widetilde{\eta_{\varepsilon}}\right\rangle+o(1)\\\
\leq&\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right|^{2}-\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\widetilde{\eta_{\varepsilon}}\right|^{2}+o\left(1\right).\end{split}$

From (2.8) proved in Step 2, we obtain

$\int_{\mathbb{D}_{r_{\varepsilon}}(p)\cap\mathbb{D}}\left|\nabla\left(\widetilde{\eta_{\varepsilon}}-\frac{\eta_{\varepsilon}}{\widetilde{\omega_{\varepsilon}}}\right)\right|^{2}\to
0$

as $\varepsilon\to 0$. We obtain the expected local $H^{1}$ comparison of
$\frac{\eta_{\varepsilon}}{\tilde{\omega}_{\varepsilon}}$ to a free boundary
harmonic map into $\mathbb{S}^{1}$. Since $\tilde{\eta}_{\varepsilon}$
converges in
$\mathcal{C}^{1}\left(\mathbb{D}_{\frac{r}{2}}(p)\cap\mathbb{D}\right)$ we
obtain the expected local $H^{1}$ convergence of $\eta_{\varepsilon}$ to a
free boundary harmonic map $\eta:\mathbb{D}\setminus\\{(0,1)\\}\to\mathbb{D}$
at the neighborhood of any point $p\in\mathbb{S}^{1}\setminus\\{(0,1)$. By a
standard regularity and point removability argument [LP17] we have a smooth
free boundary harmonic map $\eta:\mathbb{D}\to\mathbb{D}$ at the limit.

Conclusion

By symmetries, we obtain a $H^{1}$ bubble convergence to two free boundary
harmonic maps $\eta:\mathbb{D}\to\mathbb{D}$. Up to a reparametrization by the
Möbius group, they have to be $z\mapsto z^{n}$. Since they came from second
eigenfunctions there energy is less than $2\pi$: we must have $n=1$ and this
completes the proof of the Proposition. ∎

In order to complete the proof of Proposition 2.3 (in our special case),
(again up to a Möbius reparametrization), we need to prove that the
weak-$\star$ limit $\nu$ of $e^{u_{\varepsilon}}d\theta$ is absolutely
continuous with respect to $d\theta$ with a smooth positive density. Let
$\zeta\in\mathcal{C}_{c}^{\infty}\left(\mathbb{D}_{r}(p)\cap\overline{\mathbb{D}}\right)$
where $p\in\mathbb{S}^{1}\setminus\\{(-1,0)\\}$. We have that

$\displaystyle\int_{\mathbb{S}^{1}}\zeta\left(\sigma_{2}^{\varepsilon}\eta_{\varepsilon}e^{u_{\varepsilon}}d\theta-\sigma\eta
d\nu\right)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{S}^{1}}\zeta\left(\sigma_{2}^{\varepsilon}\eta_{\varepsilon}-\sigma\eta\right)e^{u_{\varepsilon}}d\theta$
$\displaystyle\quad+\int_{\mathbb{S}^{1}}\zeta\sigma\eta\left(e^{u_{\varepsilon}}d\theta-d\nu\right)\hskip
2.84544pt.$

Then on the first right-hand term, we have that

$\displaystyle\int_{\mathbb{S}^{1}}\zeta\left(\sigma_{2}^{\varepsilon}\eta_{\varepsilon}-\sigma\eta\right)e^{u_{\varepsilon}}d\theta$
$\displaystyle\leq$
$\displaystyle\left(\int_{D_{r}(p)\cap\mathbb{S}^{1}}\zeta^{2}\left|\sigma_{2}^{\varepsilon}\eta_{\varepsilon}-\sigma\eta\right|^{2}e^{u_{\varepsilon}}\theta\right)^{\frac{1}{2}}$
$\displaystyle\leq$
$\displaystyle\left(\frac{1}{\sigma_{\star}(p,r,e^{u_{\varepsilon}}d\theta)}\int_{\mathbb{D}_{r}(p)\cap\mathbb{D}}\left|\nabla\left(\zeta\left|\sigma_{2}^{\varepsilon}\eta_{\varepsilon}-\sigma\eta\right|\right)\right|^{2}\right)^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle
C\left(\int_{\mathbb{D}_{r}(p)\cap\mathbb{D}}\left|\nabla\left(\eta_{\varepsilon}-\eta\right)\right|^{2}\right)^{\frac{1}{2}}$

for some constant $C$ independent of $\varepsilon$. Letting $\varepsilon\to 0$
in a weak sense to the eigenvalue equation $\Delta\eta_{\varepsilon}=0$ in
$\mathbb{D}$ and
$\partial_{r}\eta_{\varepsilon}=\sigma_{2}^{\varepsilon}\eta_{\varepsilon}e^{u_{\varepsilon}}$
on $\mathbb{S}^{1}$, we get by Proposition 2.4:

$\Delta\eta=0\text{ in }\mathbb{D}\text{ and
}\partial_{r}\eta=\sigma\Phi\nu\text{ on }\mathbb{S}^{1}$

and since $\eta$ is the identity by Proposition 2.4, we obtain that
$\nu=d\theta$: we obtain a smooth density for $\nu$.

###  2.4. Proof of Theorem 0.1

Thanks to the previous section, we have existence of a maybe disconnected
minimizer for $g\mapsto F_{s,t}(g)$ under symmetry constraints. Since the
second eigenvalue must not converge to $0$, there are at most two connected
components at the limit. Thanks to Theorem 0.2, involving symmetric test
functions, there is in fact at most one connected component at the limit by
comparison of the energy between one disk and two disjoint disks. By symmetry
assumptions, if there is one concentration point, we must have that
$e^{u_{\varepsilon}}d\theta$ weak $\star$ converges to a sum
$\frac{\delta_{1}+\delta_{-1}}{2}$ or $\frac{\delta_{i}+\delta_{-i}}{2}$ which
is impossible. Then the $H^{1}$ bubble convergence of Proposition 2.3 is in
fact a strong $H^{1}$ convergence. We obtain the existence of smooth
minimizers on the disk for the symmetric problem. By Theorem 0.3, we obtain
embedded free boundary minimal disks into ellipsoids. The remaining arguments
to obtain non planar ones are in the introduction. Let’s verify the remaining
properties of Theorem 0.1

####  2.4.1. Monotonicity

We prove that $L_{s,t}p_{s,t}$ is decreasing and that $L_{s,t}$ is increasing.
Let $t_{1}<t_{2}$, we set for $i=1,2$: $a_{i}=L_{s,t_{i}}p_{s,t_{i}}$ and
$b_{i}=L_{s,t_{i}}$ the first and second Steklov eigenvalues of the spectral
problem. We have

$h_{s,t_{1}}(a_{1},b_{1})\leq h_{s,t_{1}}(a_{2},b_{2})\text{ and
}h_{s,t_{2}}(a_{2},b_{2})\leq h_{s,t_{2}}(a_{1},b_{1})$

so that if $s<0$,

$a_{1}^{-s}+t_{1}b_{1}^{-s}\geq a_{2}^{-s}+t_{1}b_{2}^{-s}\text{ and
}a_{2}^{-s}+t_{2}b_{2}^{-s}\geq a_{1}^{-s}+t_{2}b_{1}^{-s}$

and by sum,

$(t_{2}-t_{1})(b_{2}^{-s}-b_{1}^{-s})\geq 0$

and dividing the equations by $t_{i}$, and a sum,

$\left(\frac{1}{t_{1}}-\frac{1}{t_{2}}\right)\left(a_{1}^{-s}-a_{2}^{-s}\right)\geq
0$

We argue the same way if $s>0$.

####  2.4.2. Energy estimates

By Theorem 0.2, we have that for any $s\neq 0$ $t>0$ and $\varepsilon>0$,
$h_{s,t}(D_{s,t},g_{s,t})\leq h_{s,t}(\mathbb{D},h_{\varepsilon})$ and a
straightforward computation gives that

$\bar{\sigma}_{1}\left(D_{s,t},g_{s,t}\right)\geq\left(a_{\varepsilon}^{-s}+t\left(b_{\varepsilon}^{-s}-\bar{\sigma}_{2}\left(D_{s,t},g_{s,t}\right)^{-s}\right)\right)^{-\frac{1}{s}}\geq\left(a_{\varepsilon}^{-s}+t\left(b_{\varepsilon}^{-s}-\left(4\pi\right)^{-s}\right)\right)^{-\frac{1}{s}},$

where
$a_{\varepsilon}=\frac{2\pi}{\ln\left(\frac{1}{\varepsilon}\right)}+O\left(\frac{1}{\ln\left(\frac{1}{\varepsilon}\right)^{2}}\right)$
and $b_{\varepsilon}=4\pi-16\pi\varepsilon+o(\varepsilon)$ as $\varepsilon\to
0$, so that

$\frac{\bar{\sigma}_{1}\left(D_{s,t},g_{s,t}\right)}{2\pi}\geq\ln\left(\frac{1}{\varepsilon}\right)^{-1}\left(1+O\left(\frac{1}{\ln\left(\frac{1}{\varepsilon}\right)}\right)+t\varepsilon\ln\left(\frac{1}{\varepsilon}\right)^{-s}s(4\pi)^{-s}16\pi(1+o(1))\right)^{-\frac{1}{s}}$

as $\varepsilon\to 0$, so that choosing
$t=\varepsilon^{-1}\ln\left(\frac{1}{\varepsilon}\right)^{s-1}$ we have

(2.9) $\bar{\sigma}_{1}\left(D_{s,t},g_{s,t}\right)\geq\frac{2\pi}{\ln
t}\left(1+O\left(\frac{1}{\ln t}\right)\right)$

as $t\to+\infty$. In addition, we have that for any $s\neq 0$

$\bar{\sigma}_{2}\left(D_{s,t},g_{s,t}\right)\geq\begin{cases}\left(\frac{-\bar{\sigma}_{1}\left(D_{s,t},g_{s,t}\right)^{-s}}{t}+(4\pi)^{-s}\right)^{-\frac{1}{s}}\text{
if }s<0\\\
\left(\frac{a_{\varepsilon}^{-s}}{t}+b_{\varepsilon}^{-s}\right)^{-\frac{1}{s}}\text{
if }s>0\end{cases},$

so that if $s<0$, we obtain $\bar{\sigma}_{2}\left(D_{s,t},g_{s,t}\right)\geq
4\pi-O\left(\frac{1}{t}\right)$ as $t\to+\infty$ and if $s>0$,

$\bar{\sigma}_{2}\left(D_{s,t},g_{s,t}\right)\geq
4\pi\left(\frac{\ln\left(\frac{1}{\varepsilon}\right)^{s}2^{s}}{t}\left(1+O\left(\frac{1}{\ln\left(\frac{1}{\varepsilon}\right)}\right)\right)+1+4s\varepsilon(1+o(1))\right)^{-\frac{1}{s}}$

as $\varepsilon\to 0$, so that setting $t=\varepsilon^{-1}$,

(2.10) $\bar{\sigma}_{2}\left(D_{s,t},g_{s,t}\right)\geq
4\pi-O\left(\frac{\left(\ln t\right)^{\max\\{s,0\\}}}{t}\right)$

as $t\to+\infty$.

Knowing that the critical metrics of $F_{s,t}$ satisfy the mass conditions of
Proposition 2.1 coming from the choice of the combination $h_{s,t}$, we obtain

$t=\frac{\int_{\partial
D_{s,t}}(x_{1}^{2}+x_{2}^{2})dL_{g_{s,t}}}{\int_{\partial
D_{s,t}}x_{0}^{2}dL_{g_{s,t}}}\frac{1}{p_{s,t}^{s+1}}$

for any $s$, and since $L_{s,t}=p_{s,t}\int_{\partial
D_{t}}x_{0}^{2}dL_{g_{s,t}}+\int_{\partial
D_{s,t}}(x_{1}^{2}+x_{2}^{2})dL_{g_{t}}$, we obtain

$\int_{\partial
D_{s,t}}x_{0}^{2}dL_{g_{s,t}}=\frac{L_{s,t}}{tp_{s,t}^{s+1}+p_{s,t}}\to
0\text{ as }t\to+\infty,$

since $\bar{\sigma}_{1}\left(D_{s,t},g_{s,t}\right)=p_{s,t}L_{s,t}$ satisfies
(2.9) and $\bar{\sigma}_{2}\left(D_{s,t},g_{s,t}\right)=L_{s,t}\to 4\pi$ by
(2.10). Then $x_{0}$ converges to $0$ in $H^{1}\left(D_{s,t}\right)$. Then
using again the arguments of Proposition 2.4, we obtain the remaining
convergence results as $t\to+\infty$ since $H^{1}$ bubble tree convergence
implies varifold convergence.

##  3\. Spectral gaps

###  3.1. Notations and preliminaries

On the Euclidean disk $(\mathbb{D},\xi)$, we set the following competitor
$g_{\varepsilon}=e^{2\omega_{\varepsilon}}\xi$, where for
$z=(z_{1},z_{2})\in\mathbb{D}^{2}$ we set

$e^{\omega_{\varepsilon}(z)}=\frac{\beta_{\varepsilon}^{2}-1}{\left|\beta_{\varepsilon}-z\right|^{2}}+\frac{\beta_{\varepsilon}^{2}-1}{\left|\beta_{\varepsilon}+z\right|^{2}}\hskip
2.84544pt,$

for

$\beta_{\varepsilon}=\frac{1+\varepsilon}{1-\varepsilon}>1\hskip 2.84544pt.$

Let’s describe the surface $(\mathbb{D},g_{\varepsilon})$ in conformal
coordinates. We often prefer to be in the chart of the $Im(x)\geq 0$ half
plane $\mathbb{H}_{+}$ via the holomorphic maps
$f_{\pm}:\mathbb{D}\to\mathbb{H}_{+}$ defined by

$f_{\pm}(z)=i\frac{1\mp z}{1\pm z}\hbox{ and
}f_{\pm}^{-1}(x)=\pm\frac{1+ix}{1-ix}\hskip 2.84544pt.$

Notice that

$\left(f_{\pm}^{-1}\right)^{\star}\xi=\frac{4}{\left|1-ix\right|^{4}}\xi=\frac{4}{\left(\left(1+x_{2}\right)^{2}+x_{1}^{2}\right)^{2}}\xi=e^{2u}\xi$

where the function

$u(x_{1},x_{2})=\ln\left(\frac{2}{\left(1+x_{2}\right)^{2}+x_{1}^{2}}\right)$

satisfies the Liouville equation

$\begin{cases}\Delta u=0\text{ in }\mathbb{H}_{+}\\\
-\partial_{y}u=e^{u}\text{ on }\mathbb{R}\times\\{0\\}\hskip
2.84544pt.\end{cases}$

Notice that the first Steklov eigenfunctions on the flat disk $z_{1}$ and
$z_{2}$ become on $\mathbb{H}_{+}$ endowed with the metric $e^{2u}\xi$:

(3.1)
$Z_{1}(x)=Re(f_{+}^{-1}(x))=\frac{1-\left|x\right|^{2}}{(1+x_{2})^{2}+(x_{1})^{2}}\text{
and }Z_{2}(x)=Im(f_{+}^{-1}(x))=\frac{2x_{1}}{(1+x_{2})^{2}+(x_{1})^{2}}\hskip
2.84544pt.$

The function $\omega_{\varepsilon}:\mathbb{D}\to\mathbb{R}$ was defined such
that $\omega_{\varepsilon}:=\tilde{\omega}_{\varepsilon}\circ f_{+}-u\circ
f_{+}$ where

$\tilde{\omega}_{\varepsilon}=\ln\left(e^{\tilde{u}_{\varepsilon}}+e^{\tilde{v}_{\varepsilon}}\right),\hskip
8.53581pte^{\tilde{u}_{\varepsilon}(x)}=\frac{1}{\varepsilon}e^{u\left(\frac{x}{\varepsilon}\right)}\text{
and }e^{\tilde{u}_{\varepsilon}(x)}=\varepsilon e^{u\left(\varepsilon
x\right)}$

where we notice that $u_{\varepsilon}:=\tilde{u}_{\varepsilon}\circ
f_{+}-u\circ f_{+}$ and $v_{\varepsilon}:=\tilde{v}_{\varepsilon}\circ
f_{+}-u\circ f_{+}$ satisfy

$e^{2u_{\varepsilon}}=\frac{\beta_{\varepsilon}^{2}-1}{\left|\beta_{\varepsilon}-z\right|^{2}}\text{
and
}e^{2v_{\varepsilon}}=\frac{\beta_{\varepsilon}^{2}-1}{\left|\beta_{\varepsilon}+z\right|^{2}}\hskip
2.84544pt.$

We denote by

$e^{\hat{\omega}_{\varepsilon}(x)}=\varepsilon
e^{\tilde{\omega}_{\varepsilon}(\varepsilon x)}$

the rescaled potential close to the point $(1,0)$ at the scale $\varepsilon$.
We see that the metric $e^{2\hat{\omega}_{\varepsilon}}\xi$ converges to
$e^{2u}\xi$ in any compact set of $\mathbb{H}_{+}$. By symmetry, the rescaled
potential close to the point $(-1,0)$ at the scale $\varepsilon$ also
converges to $e^{2u}\xi$ on any compact set of $\mathbb{H}_{+}$. The surface
$(\mathbb{D},g_{\varepsilon})$ then represents two flat disks attached by a
thin strip.

Now, except for the potentials $e^{\tilde{\omega}_{\varepsilon}}$, and
$e^{\hat{\omega}_{\varepsilon}(x)}$ just defined before, we denote for any
function $\varphi:\mathbb{D}\to\mathbb{R}$:

$\tilde{\varphi}(x)=\varphi\circ f_{+}^{-1}(x)\text{ and
}\hat{\varphi}(x)=\varphi\circ f_{+}^{-1}(\varepsilon
x)=\tilde{\varphi}(\varepsilon x)$

Notice that all the functions we consider in the following satisfy that
$\varphi\circ f_{+}^{-1}(x)=\pm\varphi\circ f_{-}^{-1}(x)$ for any $x$ so that
the analysis occuring for eigenfunctions at the neighbourhood of $(1,0)$ is
the same as the one occuring at the neighbourhood of $(-1,0)$.

We notice that $L_{g_{\varepsilon}}(\partial\mathbb{D})=4\pi$ as a sum of
potentials isometric to radius one flat disks. We set
$\sigma_{\varepsilon,1}:=\sigma_{1}(\mathbb{D},g_{\varepsilon})$ and
$\sigma_{\varepsilon,2}:=\sigma_{2}(\mathbb{D},g_{\varepsilon})$ the first and
second eigenvalues associated to $(\mathbb{D},g_{\varepsilon})$. We aim at
computing the asymptotic expansion of these quantities as $\varepsilon\to 0$.

###  3.2. Upper bounds for first and second eigenvalue

In this part, we aim at proving that

(3.2)
$\sigma_{\varepsilon,1}\leq\frac{1}{2\ln\frac{1}{\varepsilon}}+O\left(\frac{1}{\left(\ln\frac{1}{\varepsilon}\right)^{2}}\right)$
(3.3) $\sigma_{\varepsilon,2}\leq 1+O(\varepsilon)$

as $\varepsilon\to 0$. We set $f_{1}:\mathbb{D}\to\mathbb{R}$ and
$f_{2}:\mathbb{D}\to\mathbb{R}$ such that

(3.4) $f_{1}\circ f_{+}^{-1}(x)=\begin{cases}1&\hbox{ on
}\mathbb{D}_{\varepsilon}^{+}\\\
\frac{\ln\left(\left|x\right|\right)}{\ln\varepsilon}&\hbox{ on
}\mathbb{D}_{\frac{1}{\varepsilon}}^{+}\setminus\mathbb{D}_{\varepsilon}^{+}\\\
-1&\hbox{ on
}\mathbb{R}^{2}\setminus\mathbb{D}_{\frac{1}{\varepsilon}}^{+}\hskip
2.84544pt.\end{cases}$

and

(3.5)
$f_{2}(z)=\sqrt{\beta_{\varepsilon}^{2}-1}\frac{z_{2}}{\left|\beta_{\varepsilon}-z\right|^{2}}\hskip
2.84544pt,$

defined in order to have that

(3.6) $f_{2}\circ f_{+}^{-1}(\varepsilon
x)=\frac{2x_{1}}{(1+x_{2})^{2}+x_{1}^{2}}\hskip 2.84544pt,$

represents the first Steklov eigenfunction $z_{2}$ of the flat disk. By
symmetry, we have that

$\int_{\mathbb{S}^{1}}f_{1}d\theta=\int_{\mathbb{S}^{1}}f_{2}d\theta=\int_{\mathbb{S}^{1}}f_{1}f_{2}d\theta=0\hskip
2.84544pt.$

By conformal invariance of the Dirichlet energy, we have that

$\int_{\mathbb{D}}\left|\nabla
f_{1}\right|_{g_{\varepsilon}}^{2}dA_{g_{\varepsilon}}=\int_{\mathbb{D}}\left|\nabla
f_{1}\right|^{2}=\int_{\mathbb{R}^{2}_{+}}\left|\nabla f_{1}\circ
f_{+}^{-1}\right|^{2}=\pi\int_{\varepsilon}^{\frac{1}{\varepsilon}}\frac{dr}{r\left(\ln\varepsilon\right)^{2}}=\frac{2\pi}{\ln\frac{1}{\varepsilon}}$

and we compute the $L^{2}$ norm on the boundary as

$\int_{\mathbb{S}^{1}}\left(f_{1}\right)^{2}dL_{g_{\varepsilon}}=\int_{\mathbb{R}\times\\{0\\}}\left(\tilde{f}_{1}\right)^{2}e^{\tilde{\omega}_{\varepsilon}}dx=2\int_{[-1,1]\times\\{0\\}}\left(\tilde{f}_{1}\right)^{2}e^{\tilde{\omega}_{\varepsilon}}=2\int_{-\frac{1}{\varepsilon}}^{\frac{1}{\varepsilon}}\left(\hat{f}_{1}\right)^{2}e^{\hat{\omega}_{\varepsilon}}$

by symmetry and we get

$\int_{\mathbb{S}^{1}}\left(f_{1}\right)^{2}dL_{g_{\varepsilon}}=2\int_{-\frac{1}{\varepsilon}}^{\frac{1}{\varepsilon}}\left(1-\frac{\ln\left|x_{1}\right|}{\ln\varepsilon}\right)^{2}\frac{2}{1+\left(x_{1}\right)^{2}}dx_{1}+O(\varepsilon)=4\pi+O\left(\frac{1}{\ln\frac{1}{\varepsilon}}\right)\hskip
2.84544pt.$

Therefore the Rayleigh quotient gives (3.2).

Again, the conformal invariance of the Dirichlet energy gives that

$\int_{\mathbb{D}}\left|\nabla
f_{2}\right|_{g_{\varepsilon}}^{2}dA_{g_{\varepsilon}}=\int_{\mathbb{D}}\left|\nabla
x_{2}\right|^{2}=\pi$

and we have by symmetry and neglecting small terms in
$e^{\hat{\omega_{\varepsilon}}}$:

$\int_{\mathbb{S}^{1}}\left(f_{2}\right)^{2}dL_{g_{\varepsilon}}=2\int_{-\frac{1}{\varepsilon}}^{\frac{1}{\varepsilon}}\left(\hat{f}_{2}\right)^{2}e^{\hat{\omega}_{\varepsilon}}=2\int_{-\frac{1}{\varepsilon}}^{\frac{1}{\varepsilon}}\left(\frac{2x_{1}}{1+\left(x_{1}\right)^{2}}\right)^{2}\frac{2}{1+\left(x_{1}\right)^{2}}dx_{1}+O\left(\varepsilon\right)$

as $\varepsilon\to 0$. Easy computations by integration by parts finally give

$\int_{\mathbb{S}^{1}}\left(f_{2}\right)^{2}dL_{g_{\varepsilon}}=\pi+O\left(\varepsilon\right)$

as $\varepsilon\to 0$, so that taking the Rayleigh quotient, we get (3.3).

###  3.3. Convergence of the first eigenfunction and the first eigenvalue

In this subsection, we aim at proving that

(3.7)
$\sigma_{\varepsilon,1}=\frac{1}{2\ln\frac{1}{\varepsilon}}+O\left(\frac{1}{\left(\ln\varepsilon\right)^{2}}\right)$

as $\varepsilon\to 0$. Let $\varphi_{\varepsilon,1}$ be a first eigenfunction
associated to $\sigma_{1,\varepsilon}$. We have the following equation

(3.8) $\begin{cases}\Delta\varphi_{\varepsilon,1}=0&\text{ in }\mathbb{D}\\\
\partial_{r}\varphi_{\varepsilon,1}=\sigma_{\varepsilon,1}e^{\omega_{\varepsilon}}\varphi_{\varepsilon,1}&\text{
in }\mathbb{S}^{1}\end{cases}\text{ and
}\int_{\mathbb{S}^{1}}\left(\varphi_{\varepsilon,1}\right)^{2}dL_{g_{\varepsilon}}=4\pi\hskip
2.84544pt.$

Up to symmetrize or antisymmetrize in case of multiplicity (that a posteriori
do not happen), we can assume that

(3.9)
$\varphi_{\varepsilon,1}(z_{1},z_{2})^{2}=\varphi_{\varepsilon,1}(-z_{1},z_{2})^{2}=\varphi_{\varepsilon,1}(z_{1},-z_{2})^{2}$

for any $z=(z_{1},z_{2})\in\mathbb{D}$. Then $\hat{\varphi}_{\varepsilon,1}$
satisfies the following equation in $\mathbb{R}^{2}_{+}$:

(3.10) $\begin{cases}\Delta\hat{\varphi}_{\varepsilon,1}=0&\text{ in
}\mathbb{D}\\\
-\partial_{x_{2}}\hat{\varphi}_{\varepsilon,1}=\sigma_{\varepsilon,1}\frac{2}{1+(x_{1})^{2}}\hat{\varphi}_{\varepsilon,1}+\sigma_{\varepsilon,1}\frac{2\varepsilon^{2}}{1+(x_{1})^{2}\varepsilon^{4}}\hat{\varphi}_{\varepsilon,1}&\text{
on }\mathbb{R}\times\\{0\\}\end{cases}$

and

$\int_{\mathbb{R}\times\\{0\\}}\left(\hat{\varphi}_{\varepsilon,1}\right)^{2}\left(\frac{2}{1+(x_{1})^{2}}+\frac{2\varepsilon^{2}}{1+(x_{1})^{2}\varepsilon^{4}}\right)dx_{1}=4\pi\hskip
2.84544pt.$

In particular,
$\int_{\mathbb{R}\times\\{0\\}}\left(\hat{\varphi}_{\varepsilon,1}\right)^{2}\frac{2}{1+(x_{1})^{2}}dx_{1}\leq
4\pi$ and by standard Elliptic theory, since $\sigma_{\varepsilon,1}\to 0$ as
$\varepsilon\to 0$, we obtain that

$\hat{\varphi}_{\varepsilon,1}\to\varphi_{\star,1}^{+}\text{ in
}\mathcal{C}^{2}\left(\mathbb{D}_{\rho}\right)$

as $\varepsilon\to 0$, for any $\rho>0$. Letting (3.10) pass to the limit as
$\varepsilon\to 0$, we obtain that

(3.11) $\begin{cases}\Delta\hat{\varphi}_{\star,1}^{+}=0&\text{ in
}\mathbb{R}^{2}_{+}\\\ -\partial_{x_{2}}\hat{\varphi}_{\star,1}^{+}=0&\text{
on }\mathbb{R}\times\\{0\\}\end{cases}$

Since
$\int_{\mathbb{R}\times\\{0\\}}\left(\hat{\varphi}_{\varepsilon,1}\right)^{2}\frac{2}{1+(x_{1})^{2}}dx_{1}\leq
4\pi$, we get that $\varphi_{\star}^{+}$ is a constant function. Up to take
$-\varphi_{\varepsilon,1}$ instead of $\varphi_{\varepsilon,1}$, we can assume
that $\varphi_{\star,1}^{+}\geq 0$. Looking at
$\varphi_{\varepsilon,1}\circ\left(f_{-}\right)^{-1}(\varepsilon x)$ instead
of
$\hat{\varphi}_{\varepsilon,1}(x):=\varphi_{\varepsilon,1}\circ\left(f_{+}\right)^{-1}(\varepsilon
x)$, we also get a constant function $\varphi_{\star,1}^{-}$ at the limit and
by symmetry,
$\left(\varphi_{\star,1}^{+}\right)^{2}=\left(\varphi_{\star,1}^{-}\right)^{2}$.

We denote by

$m_{\varepsilon,1}(r)=\frac{1}{\pi}\int_{0}^{\pi}\tilde{\varphi}_{\varepsilon,1}(r\cos\theta,r\sin\theta)d\theta$

the mean value of $\tilde{\varphi}_{\varepsilon,1}$ on half circle
(corresponding to the mean value of $\varphi_{\varepsilon,1}$ on the lines
$\left|\frac{1-z}{1+z}\right|=constant$). It is a radial function. We also set

$\psi_{\varepsilon,1}(x)=\tilde{\varphi}_{\varepsilon,1}(x)-m_{\varepsilon,1}(\left|x\right|)\hskip
2.84544pt.$

We have

(3.12) $\begin{cases}\Delta\psi_{\varepsilon,1}=-\Delta
m_{\varepsilon,1}=-\frac{\sigma_{\varepsilon,1}}{\pi
r}e^{\tilde{\omega}_{\varepsilon}}\left(\tilde{\varphi}_{\varepsilon,1}(r,0)+\tilde{\varphi}_{\varepsilon,1}(-r,0)\right)&\text{
in }\mathbb{R}^{2}_{+}\\\
-\partial_{x_{2}}\psi_{\varepsilon,1}=\sigma_{\varepsilon,1}e^{\tilde{\omega}_{\varepsilon}}\tilde{\varphi}_{\varepsilon,1}&\text{
on }\mathbb{R}\times\\{0\\}\hskip 2.84544pt.\end{cases}$

Let $x_{\varepsilon}\in\mathbb{D}_{+}$ be such that
$\psi_{\varepsilon}(x_{\varepsilon})=\left\|\psi_{\varepsilon,1}\right\|_{\infty}$.
We aim at proving that

(3.13) $\left\|\psi_{\varepsilon,1}\right\|_{\infty}\leq
C\sigma_{\varepsilon,1}$

for some positive constant $C$. We have two cases:

If $\left|x_{\varepsilon}\right|=O\left(\varepsilon\right)$, then
$\hat{\psi}_{\varepsilon,1}$ satisfies

(3.14)
$\begin{cases}\Delta\hat{\psi}_{\varepsilon,1}=-\frac{\sigma_{\varepsilon,1}}{\pi
r}\left(\frac{2}{1+r^{2}}+\frac{2\varepsilon^{2}}{1+\varepsilon^{4}r^{2}}\right)\left(\hat{\varphi}_{\varepsilon,1}(r,0)+\hat{\varphi}_{\varepsilon,1}(-r,0)\right)&\text{
in }\mathbb{R}^{2}_{+}\\\
-\partial_{x_{2}}\hat{\psi}_{\varepsilon,1}=\sigma_{\varepsilon,1}\frac{2}{1+(x_{1})^{2}}\hat{\varphi}_{\varepsilon,1}+\sigma_{\varepsilon,1}\frac{2\varepsilon^{2}}{1+\varepsilon^{4}(x_{1})^{2}}\hat{\varphi}_{\varepsilon,1}&\text{
on }\mathbb{R}\times\\{0\\}\hskip 2.84544pt.\end{cases}$

We also know by assumption that for any $\rho>0$ we have
$\int_{\mathbb{D}_{\rho}^{+}}\psi_{\varepsilon,1}=0$. We also know that
$\psi_{\varepsilon,1}$ has a bounded energy. By asumption, the right-hand side
term of the second equation is bounded in
$L^{2}\left(\mathbb{R}^{2}_{+}\right)$. Moreover, by Hölder inequalities, the
right-hand side term in the first equation is bounded in $L^{p}$ for any
$p<2$. By standard elliptic theory (3.13) holds true.

If $\varepsilon=o\left(\left|x_{\varepsilon}\right|\right)$, then we set
$\phi_{\varepsilon}(x)=\psi_{\varepsilon,1}(\left|x_{\varepsilon}\right|x)$
and we have the equation

(3.15)
$\begin{cases}\Delta\phi_{\varepsilon}=&-\frac{\sigma_{\varepsilon,1}}{\pi
r}\left(\frac{2\frac{\varepsilon}{\left|x_{\varepsilon}\right|}}{\frac{\varepsilon^{2}}{\left|x_{\varepsilon}\right|^{2}}+r^{2}}+\frac{2\varepsilon^{2}\left|x_{\varepsilon}\right|}{1+\varepsilon^{4}\left|x_{\varepsilon}\right|^{2}r^{2}}\right)\left(\tilde{\varphi}_{\varepsilon,1}(r\left|x_{\varepsilon}\right|,0)+\tilde{\varphi}_{\varepsilon,1}(-r\left|x_{\varepsilon}\right|,0)\right)\text{
in }\mathbb{R}^{2}_{+}\\\
-\partial_{x_{2}}\phi_{\varepsilon}=&\sigma_{\varepsilon,1}\frac{2\frac{\varepsilon}{\left|x_{\varepsilon}\right|}}{\frac{\varepsilon^{2}}{\left|x_{\varepsilon}\right|^{2}}+(x_{1})^{2}}\tilde{\varphi}_{\varepsilon,1}(-x_{1}\left|x_{\varepsilon}\right|,0)\\\
&+\sigma_{\varepsilon,1}\frac{2\varepsilon^{2}\left|x_{\varepsilon}\right|}{1+\varepsilon^{4}\left|x_{\varepsilon}\right|^{2}(x_{1})^{2}}\tilde{\varphi}_{\varepsilon,1}(x_{1}\left|x_{\varepsilon}\right|,0)\text{
on }\mathbb{R}\times\\{0\\}\hskip 2.84544pt.\end{cases}$

so that by standard elliptic estimates in
$A=\mathbb{D}_{2}^{+}\setminus\mathbb{D}_{\frac{1}{2}}^{+}$, knowing that
$\int_{A}\phi_{\varepsilon}=0$, we obtain thanks to $L^{p}$ boundedness in the
left-hand side terms of the equation for $p<2$ that
$\left\|\psi_{\varepsilon}\right\|_{\infty}\leq C\sigma_{1}^{\varepsilon}$ for
some positive constant $C$ independent from $\varepsilon$ and (3.13) holds
true.

We deduce from (3.13) that

(3.16)
$4\pi=\int_{\mathbb{S}^{1}}\left(\varphi_{\varepsilon,1}\right)^{2}dL_{g_{\varepsilon}}=\int_{\mathbb{S}^{1}}\left(m_{\varepsilon,1}\right)^{2}dL_{g_{\varepsilon}}+O\left(\sigma_{\varepsilon,1}\right)$

as $\varepsilon\to 0$.

By the Courant nodal theorem, and symmetry assumptions 3.9, the only
possibilities for nodal sets are $\\{z_{1}=0\\}$ or $\\{z_{2}=0\\}$. But if
$\\{z_{2}=0\\}$ is a nodal set, we obtain that
$\varphi_{\varepsilon,1}(z_{1},z_{2})=\varphi_{\varepsilon,1}(z_{1},-z_{2})$
by the Courant nodal theorem and we would get $m_{\varepsilon,1}=0$. However,
since $\sigma_{\varepsilon,1}\to 0$ as $\varepsilon\to 0$, (3.16) gives a
contradiction. Therefore, $\\{z_{1}=0\\}$ is the nodal line of
$\varphi_{\varepsilon,1}$ and $\\{\left|x\right|=1\\}$ is the zero set for
$m_{\varepsilon,1}$. Moreover $m_{\varepsilon,1}$ and
$\tilde{\varphi}_{\varepsilon,1}$ are positive in $\mathbb{D}$ and negative in
$\mathbb{R}^{2}\setminus\mathbb{D}$. $m_{\varepsilon,1}$ satisfies:

$\Delta
m_{\varepsilon,1}:=-\frac{1}{r}\partial_{r}\left(rm_{\varepsilon,1}^{\prime}\right)=\frac{\sigma_{\varepsilon,1}}{\pi
r}e^{\tilde{\omega}_{\varepsilon}}\left(\tilde{\varphi}_{\varepsilon,1}(r,0)+\tilde{\varphi}_{\varepsilon,1}(-r,0)\right)\hskip
2.84544pt.$

Integrating between $0$ and $r\leq 1$, we get

(3.17)
$m_{\varepsilon,1}(r)-m_{\varepsilon,1}(0)=-\int_{0}^{r}\frac{1}{s}\left(\int_{[-s,s]\times\\{0\\}}\sigma_{\varepsilon,1}e^{\tilde{\omega_{\varepsilon}}}\tilde{\varphi}_{\varepsilon,1}\right)\leq
0$

Then, $m_{\varepsilon,1}$ realizes its maximum at $0$. Then, we have that for
any $\rho>0$,

$\int_{\rho\varepsilon}^{1}\left(\tilde{\varphi}_{\varepsilon,1}\right)^{2}e^{\tilde{\omega}_{\varepsilon}}dx_{1}\leq\left(\varphi_{\varepsilon,1}(0)\right)^{2}\int_{\rho\varepsilon}^{1}e^{\tilde{\omega}_{\varepsilon}}dx_{1}\leq\frac{C}{\rho}$

for a positive constant $C$ independent from $\rho$ and $\varepsilon$. Letting
$\varepsilon\to 0$, and then $\rho\to 0$ in (3.16), we obtain that
$\left(\varphi_{\star,1}^{+}\right)^{2}+\left(\varphi_{\star,1}^{-}\right)^{2}=2$.
Then, since we know that
$\varphi_{\varepsilon,1}(z_{1},z_{2})=-\varphi(-z_{1},z_{2})$, and
$\varphi_{\star,1}^{+}\geq 0$, we have that $\varphi_{\star,1}^{+}=1$ and
$\varphi_{\star,1}^{-}=-1$.

Taking $r=1$ in (3.17), we obtain

$\displaystyle m_{\varepsilon,1}(0)$ $\displaystyle=$
$\displaystyle\frac{\sigma_{\varepsilon,1}}{\pi}\int_{0}^{1}\frac{1}{s}\left(\int_{[-s,s]\times\\{0\\}}e^{\tilde{\omega}_{\varepsilon}}\tilde{\varphi}_{\varepsilon,1}\right)$
$\displaystyle\leq$
$\displaystyle\frac{\sigma_{\varepsilon,1}}{\pi}\tilde{\varphi}_{\varepsilon,1}(0)\int_{0}^{1}\frac{1}{s}\int_{-s}^{s}\frac{2\varepsilon}{\varepsilon^{2}+u^{2}}du+O\left(\varepsilon\right)$
$\displaystyle\leq$
$\displaystyle\frac{\sigma_{\varepsilon,1}}{\pi}\tilde{\varphi}_{\varepsilon,1}(0)\int_{0}^{\frac{1}{\varepsilon}}\frac{2}{1+r^{2}}\left(\ln\frac{1}{r\varepsilon}\right)dr+O\left(\varepsilon\right)$
$\displaystyle\leq$
$\displaystyle\frac{\sigma_{\varepsilon,1}}{\pi}\tilde{\varphi}_{\varepsilon,1}(0)\left(\ln\frac{1}{\varepsilon}\right)\int_{0}^{+\infty}\frac{2}{1+r^{2}}dr+O\left(\sigma_{\varepsilon,1}\right)+O\left(\varepsilon\right)$

and knowing that $\tilde{\varphi}_{\varepsilon,1}(0)\to 1$ as $\varepsilon\to
0$, we get

$\sigma_{\varepsilon,1}\geq\frac{1}{2\ln\frac{1}{\varepsilon}}+O\left(\frac{\sigma_{\varepsilon,1}}{\ln\frac{1}{\varepsilon}}\right)$

so that with (3.2), we obtain (3.7).

###  3.4. The second eigenfunction and second eigenvalue

In this section, we aim at proving that

(3.18) $\sigma_{\varepsilon,2}=1-4\varepsilon+o(\varepsilon)$

as $\varepsilon\to 0$.

We focus on the equation satisfied by $\varphi_{\varepsilon,2}$, an
eigenfunction associated to the second non-zero eigenvalue
$\sigma_{\varepsilon,2}$, satisfying

(3.19)
$\int_{\mathbb{S}^{1}}\left(\varphi_{\varepsilon,2}\right)^{2}e^{\omega_{\varepsilon}}dz=\pi\hbox{
and
}\int_{\mathbb{S}^{1}}\varphi_{\varepsilon,2}e^{\omega_{\varepsilon}}dz=\int_{\mathbb{S}^{1}}\varphi_{\varepsilon,1}\varphi_{\varepsilon,2}e^{\omega_{\varepsilon}}dz=0\hskip
2.84544pt,$

where $\varphi_{\varepsilon,1}$ is the radial first eigenfunction of the
previous section. Up to symmetrization again, in case of multiplicity, we
assume in addition that

(3.20)
$\varphi_{\varepsilon,2}(z_{1},z_{2})^{2}=\varphi_{\varepsilon,2}(-z_{1},z_{2})^{2}=\varphi_{\varepsilon,2}(z_{1},-z_{2})^{2}$

In fact, we must have

(3.21)
$\varphi_{\varepsilon,2}(z_{1},z_{2})=\varphi_{\varepsilon,2}(-z_{1},z_{2})\hskip
2.84544pt.$

Indeed, if not, we would have antisymmetry and in particular that
$\varphi_{\varepsilon,2}(0,z_{2})=0$. By orthogonality with
$\varphi_{\varepsilon,1}$ which is anti-symmetric, $\varphi_{\varepsilon,2}$
must vanish elsewhere: $\varphi_{\varepsilon,2}(z_{1},z_{2})=0$ for $z_{1}>0$.
The nodal line containing $z$ is a line connecting two points of
$\left(\mathbb{S}^{1}\cap\\{z_{1}>0\\}\right)\cup\\{z_{1}=0\\}$. Then
$\varphi_{\varepsilon,2}$ has at least two nodal domains in $\\{z_{1}>0\\}$.
By symmetry, $\varphi_{\varepsilon,2}$ has at least four nodal domains,
contradicting the Courant nodal theorem. We also deduce from the orthogonality
with the constants that

(3.22)
$\int_{\mathbb{S}^{1}\cap\\{z_{1}>0\\}}\varphi_{\varepsilon,2}e^{\omega_{\varepsilon}}dz=\int_{\mathbb{S}^{1}\cap\\{z_{1}<0\\}}\varphi_{\varepsilon,2}e^{\omega_{\varepsilon}}dz=0\hskip
2.84544pt.$

Rescaling at the neighbourhood of $(1,0)$, $\hat{\varphi}_{\varepsilon,2}$
satisfies the equation

(3.23) $\begin{cases}\Delta\hat{\varphi}_{\varepsilon,2}=0&\text{ in
}\mathbb{D}\\\
-\partial_{x_{2}}\hat{\varphi}_{\varepsilon,2}=\sigma_{\varepsilon,2}\frac{2}{1+(x_{1})^{2}}\hat{\varphi}_{\varepsilon,2}+\sigma_{\varepsilon,2}\frac{2\varepsilon^{2}}{1+(x_{1})^{2}\varepsilon^{4}}\hat{\varphi}_{\varepsilon,2}&\text{
on }\mathbb{R}\times\\{0\\}\end{cases}$

with

$\int_{\mathbb{R}\times\\{0\\}}\left(\frac{2}{\left(1+(x_{1})^{2}\right)^{2}}+\frac{2\varepsilon^{2}}{1+\varepsilon^{4}(x_{1})^{2}}\right)\left(\hat{\varphi}_{\varepsilon,2}\right)^{2}dx=\pi\hskip
2.84544pt.$

In particular,
$\int_{\mathbb{R}\times\\{0\\}}\frac{2}{\left(1+(x_{1})^{2}\right)^{2}}\left(\hat{\varphi}_{\varepsilon,2}\right)^{2}dx\leq\pi$.
$\sigma_{\varepsilon,2}$ is bounded by (3.3) and converges up to the
extraction of a subsequence to $\sigma_{\star,2}$. By standard elliptic
estimates, up to the extraction of a subsequence,

(3.24) $\hat{\varphi}_{\varepsilon,2}\to\varphi_{\star,2}^{+}\hbox{ in
}\mathcal{C}^{2}\left(\mathbb{D}_{\rho}^{+}\right)$

for any $\rho>0$ and

(3.25) $\begin{cases}\Delta\varphi_{\star,2}^{+}=0&\text{ in
}\mathbb{R}^{2}_{+}\\\
-\partial_{x_{2}}\varphi_{\star,2}^{+}=\sigma_{\star,2}e^{u}\varphi_{\star,2}^{+}&\text{
on }\mathbb{R}\times\\{0\\}\end{cases}$

By boundedness of the energy, $\varphi_{\star,2}^{+}\circ f_{+}$ has to be a
Steklov eigenfunction associated to $\sigma_{\star,2}$ on the disk. Since
$\sigma_{\star,2}\leq 1$ by (3.3), we have either $\sigma_{\star,2}=0$ or
$\sigma_{\star,2}=1$. We aim at proving that $\sigma_{\star,2}=1$ and at
getting an estimate on $\delta_{\varepsilon}:=\sigma_{\varepsilon,2}-1$ as
$\varepsilon\to 0$.

Let’s deal with the second symmetry in (3.20). In the following subsections we
separate two cases:

(3.26)
$\varphi_{\varepsilon,2}(z_{1},z_{2})=-\varphi_{\varepsilon,2}(z_{1},-z_{2})$

is called the antisymmetric case and is handled in subsection 3.4.1 and in
this case, we assume in addition that $\varphi_{\varepsilon,2}(0,1)\geq 0$ up
to take $-\varphi_{\varepsilon,2}$.

(3.27)
$\varphi_{\varepsilon,2}(z_{1},z_{2})=\varphi_{\varepsilon,2}(z_{1},-z_{2})$

is called the symmetric case and is handled in subsection 3.4.2, and in this
case, we assume in addition that $\varphi_{\varepsilon,2}(1,0)\geq 0$ up to
take $-\varphi_{\varepsilon,2}$. We a posteriori prove that this case cannot
occur for the second eigenfunction.

####  3.4.1. The antisymmetric case

By antisymmetry (3.26), we must have that for any $r>0$,

$\int_{0}^{\pi}\tilde{\varphi}_{\varepsilon,2}(r\cos\theta,r\sin\theta)d\theta=0$

so that we can handle elliptic estimates in $\tilde{\varphi}_{\varepsilon,2}$
at any scales. In particular we have that $\tilde{\varphi}_{\varepsilon,2}$ is
uniformly bounded. We then obtain that

(3.28)
$\int_{\rho\varepsilon}^{1}\tilde{\varphi}_{\varepsilon,2}e^{\tilde{\omega}_{\varepsilon}}dx_{1}\leq\frac{C}{\rho}$

for a constant $C>0$ independant from $\rho>0$ and $\varepsilon>0$. By (3.19),
(3.22) and (3.24) we obtain that

$\int_{\mathbb{R}\times\\{0\\}}\left(\varphi_{\star,2}^{+}\right)^{2}e^{u}dx_{1}=\frac{\pi}{2}\text{
and }\int_{\mathbb{R}\times\\{0\\}}\varphi_{\star,2}^{+}e^{u}dx_{1}=0$

By (3.25), and remarks below, we must have $\sigma_{\star,2}=1$ and
$\varphi_{\star,2}^{+}\circ f_{+}$ has to be a first Steklov eigenfunction on
the disk, that is

$\varphi_{\star,2}^{+}=a_{\star,1}Z_{1}+a_{\star,2}Z_{2}\hskip 2.84544pt,$

where $Z_{1}=Re(f_{+}^{-1})$ and $Z_{2}=Im(f_{+}^{-1})$ and
$\left(a_{\star,1}\right)^{2}+\left(a_{\star,2}\right)^{2}=1$. By antisymmetry
(3.26), $a_{\star,1}=0$ and $a_{\star,2}=1$.

Now, let’s work on $\delta_{\varepsilon}=\sigma_{\varepsilon,2}-1$.

We define $R_{\varepsilon}:\mathbb{D}\to\mathbb{R}$ such that the function