# Some rigidity results on compact hypersurfaces with capillary boundary in
the hyperbolic space

Yimin Chen  Department of Mathematics
Pusan National University
Busan 46241
Republic of Korea
<EMAIL_ADDRESS>and  Juncheol Pyo  Department of Mathematics
Pusan National University
Busan 46241
Republic of Korea
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we prove a Heintze-Karcher type inequality for capillary
hypersurfaces supported on various hypersurfaces in the hyperbolic space. The
equality case only occurs on capillary totally umbilical hypersurfaces. Then
we apply this result to prove the Alexandrov type theorem for embedded
capillary hypersurfaces in the hyperbolic space. In addition, we prove some
other rigidity results for capillary hypersurfaces supported on totally
geodesic plane in $\mathbb{B}^{n+1}_{+}$.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Capillary hypersurfaces and properties of the higher-order mean curvature $H_{r}$
    2. 2.2 Properties in the Poincaré ball model
    3. 2.3 Properties in the Poincaré half space model
  3. 3 Minkowski type formulae in the hyperbolic space
    1. 3.1 A Minkowski type formula for capillary hypersurfaces in Poincaré ball model
    2. 3.2 A Minkowski type formula for capillary hypersurfaces in Poincaré half space model
  4. 4 Proofs of Theorem 1 and Theorem 3
  5. 5 Proofs of Theorem 2 and Theorem 4
  6. 6 Capillary hypersurfaces in a geodesic ball
  7. 7 Other rigidity results for capillary hypersurfaces

##  1\. Introduction

Let $\mathbb{H}^{n+1}$ be the $(n+1)$-dimensional hyperbolic space. We will
use both the Poincaré ball model and the Poincaré half space model of
$\mathbb{H}^{n+1}$ throughout this paper. Let $\delta$ be the Euclidean metric
and $|\cdot|=\delta(\cdot,\cdot)^{1/2}$ be the Euclidean norm. The Poincaré
ball model is defined as follows:

$\displaystyle\left(\mathbb{B}^{n+1},\frac{4}{(1-|x|^{2})^{2}}\delta\right),$

where $\mathbb{B}^{n+1}$ is the unit Euclidean ball centered at the origin.

The Poincaré half space model is defined as follows:

$\displaystyle\left(\mathbb{R}^{n+1}_{+},\frac{1}{x_{n+1}^{2}}\delta\right),$

where $\mathbb{R}^{n+1}_{+}$ is the half Euclidean space on which the last
coordinate function $x_{n+1}$ is strictly positive.

Let $S$ be an umbilical hypersurface in $\mathbb{H}^{n+1}$ with its principal
curvature $\lambda$, and $\Sigma\hookrightarrow\mathbb{H}^{n+1}$ is an
immersed $n$-dimensional manifold into $\mathbb{H}^{n+1}$ such that
$x(\partial\Sigma)\subset S$. If $\Sigma$ intersects $S$ at a constant angle
$\theta$, $\Sigma$ is said to be a capillary hypersurface. In particular if
$\theta=\frac{\pi}{2}$, $\Sigma$ is said to be a free boundary hypersurface.
Furthermore, we call $S$ the supporting hypersurface of the capillary
hypersurface $\Sigma$. The study of capillary hypersurfaces has a long history
since the works by Young in [27] and Laplace in [12]. The notable result of
Fraser and Schoen in [4] reveals the relation between the Steklov eigenvalue
and the free boundary minimal hypersurfaces in the unit ball. Based on their
significant work, properties of free boundary hypersurfaces have been
established, including some geometric inequalities.

There are several results on geometric inequalities on capillary hypersurfaces
in a space form. In [22] and [26], Scheuer, Wang, Weng and Xia have
established a family of Alexandrov-Fenchel’s type inequalities on capillary
hypersurfaces in a geodesic ball. In [24] and [7], the authors give a complete
family of Alexandrov-Fenchel’s type inequalities on capillary hypersurfaces in
Euclidean half space. And in [3], Chen, Hu and Li proved Perez type inequality
on free boundary hypersurfaces in a geodesic ball.

In [21], Ros used Reilly’s formula (see [20]) to prove the following result.

######  Theorem A.

Let $\Omega$ be an $(n+1)$-dimensional Riemannian manifold with boundary and
non-negative Ricci curvature. Suppose the boundary $M=\partial\Omega$ is mean
convex (i.e. the mean curvature $H_{M}>0$), then

$\int_{M}\frac{1}{H_{M}}dA\geq(n+1)V(\Omega),$

where $V(\Omega)$ denotes the volume of $\Omega$. Moreover, the equality holds
if and only if $\Omega$ is isometric to a Euclidean ball.

There are several results on (weighted) Heintze-Karcher type inequalities for
embedded closed hypersurfaces in different Riemannian manifolds. In [2],
Brendle proved a weighted Heintze-Karcher type inequality for embedded, closed
mean convex hypersurfaces in a family of warped product spaces. In [19], Qiu
and Xia proved a weighted Heintze-Karcher type inequality for the case in
which the ambient space is a Riemannian manifold with a negative lower bound
on its sectional curvature. In [13], Li and Xia proved a weighted Heintze-
Karcher type inequality when the ambient space is a sub-static manifold.

For free boundary hypersurfaces, the second author proved a Heintze-Karcher
type inequality for those supported on totally geodesics hyperplane in a space
form in [18]. Guo and Xia proved the inequality for those supported on
horospheres and equidistant hypersurfaces in [6]. In [25], Wang and Xia prove
the case for those supported on geodesic balls.

Recently in the paper [10], Jia, Xia and Zhang proved a Heintze-Karcher type
inequality for capillary hypersurfaces in Euclidean half space. Furthermore,
they extended their results in Euclidean wedge and anisotropic half space (See
[8] and [9]). Inspired by their result, we will prove a Heintze-Karcher type
inequality for capillary hypersurfaces in $\mathbb{H}^{n+1}$.

In Poincaré ball model $(\mathbb{B}^{n+1},\frac{4}{(1-|x|^{2})^{2}}\delta)$,
let $\Sigma$ be a compact, embedded capillary hypersurface in

$\mathbb{B}^{n+1}_{+}=\\{x\in\mathbb{B}^{n+1}:\delta(x,E_{n+1})\geq 0\\}$

supported on a totally geodesic plane

$P=\\{x\in\mathbb{B}^{n+1}:\delta(x,E_{n+1})=0\\}.$

Denote $\Omega$ the region enclosed by $\Sigma$ and $P$, and $T\subset P$ the
region in $P$ with $\partial\Omega=\Sigma\cup T$. Denote
$V_{0}=\frac{1+|x|^{2}}{1-|x|^{2}}$, we have the following Heintze-Karcher
type inequality.

######  Theorem 1.

Let $\Sigma\subset\mathbb{B}^{n+1}_{+}$ be a compact, embedded capillary
hypersurface supported on $P$. Let $\Omega$ be a domain enclosed by $\Sigma$
and $T\subset P$. If the contact angle $\theta\in(0,\frac{\pi}{2}]$, and the
mean curvature $H_{1}>0$ on $\Sigma$, we have

(1)
$\displaystyle\int_{\Sigma}\frac{V_{0}}{H_{1}}dA\geq(n+1)\int_{\Omega}V_{0}d\Omega+n\cot\theta\frac{\left(\int_{T}V_{0}dA_{T}\right)^{2}}{\int_{\partial\Sigma}V_{0}ds}.$

In addition, the equality holds if and only if $\Sigma$ is umbilical.

Next, we consider Poincaré half space model
$(\mathbb{R}^{n+1}_{+},\frac{1}{x_{n+1}^{2}}\delta)$ of the hyperbolic space.
Denote the position vector in Poincaré half space model by $\widetilde{x}$.
Fixing $a$ a vector field in $\mathbb{H}^{n+1}$ which is constant with respect
to $(\mathbb{R}^{n+1},\delta_{+})$ and $0<\phi<\frac{\pi}{2}$, we consider a
umbilical hypersurface $L_{\phi,a}$ defined by

(2) $L_{\phi,a}=\\{\widetilde{x}\in\mathbb{R}^{n+1}_{+}:x_{n+1}-1=\tan\phi
x_{a}\\},$

where $x_{a}=\delta(\widetilde{x},a)$ and $a$ is a vector field in
$\mathbb{H}^{n+1}$ which is constant with respect to
$(\mathbb{R}^{n+1}_{+},\delta)$ and satisfies $\delta(a,E_{n+1})=0$ and
$\delta(a,a)=1$.It is well known that $L_{\phi,a}$ is a umbilical hypersurface
with principal curvature $0<\cos\phi\leq 1$. We define
$B^{\mathrm{int}}_{\phi,a}$ as follows:

$B^{\mathrm{int}}_{\phi,a}=\\{\widetilde{x}\in\mathbb{H}^{n+1}:x_{n+1}-1\geq\tan\phi
x_{a}\;\\}.$

In $B_{\phi}^{\mathrm{int}}$, we consider compact embedded capillary
hypersurfaces supported on $L_{\phi,a}$.

Analogously, denote $V_{n+1}=\frac{1}{x_{n+1}}$ and we have the following
Heintze-Karcher type inequality

######  Theorem 2.

Let $0\leq\phi<\frac{\pi}{2}$ and $\Sigma\subset B^{\mathrm{int}}_{\phi,a}$ be
a compact, embedded capillary hypersurface. The supporting hypersurface is
$L_{\phi}$. Let $\Omega$ be the domain bounded by $\Sigma$ and $T\subset
L_{\phi,a}$. If the contact angle $\theta\in(0,\frac{\pi}{2}]$, and the mean
curvature $H_{1}>0$ on $\Sigma$, we have

(3)
$\displaystyle\int_{\Sigma}\frac{V_{n+1}}{H_{1}}dA\geq(n+1)\int_{\Omega}V_{n+1}d\Omega+n\cos\theta\frac{\left(\int_{T}V_{n+1}dA_{T}\right)^{2}}{\int_{\partial\Sigma}\overline{g}(-\cos\phi\widetilde{x}+\sin\phi
a,\mu)ds}$

where $\mu$ is the outward unit conormal vector field of $\partial\Sigma$ in
$\Sigma$. In addition, the equality holds if and only if $\Sigma$ is
umbilical.

The classical Alexandrov type theorem says that the geodesic balls are the
only closed embedded hypersurfaces with constant mean curvature (CMC) in space
form. It has been proved using a method called moving planes (see [1]).

Another powerful tool to prove the Alexandrov type theorem is the method
provided by Ros. In 1986, Ros used a Heintze-Karcher type formula in [21] to
prove a higher-order version of an Alexandrov type theorem, which says that
embedded hypersurfaces with constant $k$-th mean curvature in Euclidean space
can only be a sphere. For the general ambient space, Brendle [2] proved that
any closed embedded CMC hypersurface in a family of space warped product
spaces can only be a slice.

As we mentioned above, there exists certain Heintze-Karcher type inequalities
on free boundary hypersurfaces on different supporting hypersurfaces, Wang and
Xia [25] proved the Alexandrov type theorem for free boundary hypersurface
supported on geodesic spheres in space forms. Moreover, for free boundary
hypersurfaces supported on horospheres and equidistant hypersurfaces, we refer
to Guo and Xia [6]. The second author [18] also studied the case of free
boundary hypersurfaces supported on totally geodesic hyperplane in space
forms. For capillary hypersurface, Jia, Xia and Zhang (see [10]) prove an
Alexandrov type theorem for capillary hypersurfaces in the Euclidean half
space and Euclidean ball. Jia, Wang, Xia and Zhang prove the capillary case of
the support hypersurface being Euclidean wedges and the geodesic plane in
anisotropic Euclidean half space (see [8] and [9]). In this paper, we will
prove the following Alexandrov type theorems for capillary hypersurfaces in
the hyperbolic space.

######  Theorem 3.

Let $\Sigma$ be a compact embedded CMC hypersurface in $\mathbb{B}^{n+1}_{+}$
supported on a the totally geodesic plane $P$, and the contact angle satisfies
$\theta\in(0,\frac{\pi}{2}]$. Then $\Sigma$ is totally umbilical except for
being totally geodesic.

######  Theorem 4.

Let $\Sigma$ be a compact embedded CMC capillary hypersurface in
$B^{\mathrm{int}}_{\phi,a}$ supported on $L_{\phi,a}$, and the contact angle
satisfies $\theta\in(0,\frac{\pi}{2}]$. In addition, we assume
$\phi+\theta>\frac{\pi}{2}$ for $\phi>0$ or $\theta=\frac{\pi}{2}$ for
$\phi=0$. Then $\Sigma$ is umbilical except for being totally geodesic.

######  Remark.

The assumption “ $\phi+\theta>\frac{\pi}{2}$ for $\phi>0$ or
$\theta=\frac{\pi}{2}$ for $\phi=0$” is only used for guaranteeing the
existence of the convex point, which can be substituted by assuming that the
constant mean curvature $H>0$, see Corollary 1.

Intriguingly, we also prove some other rigidity results on immersed capillary
hypersurfaces supported on a totally geodesic plane $P$ in the hyperbolic
space. Koh and Lee [11] used the Minkowski identity for closed hypersurfaces
in Euclidean space to classify the immersed hypersurfaces with constant ratio
of higher-order mean curvatures in Euclidean space. Since we will give a
Minkowski type formula on capillary hypersurfaces in the Section 3, we can
prove the following result:

######  Theorem 5.

Let $\Sigma\subset\mathbb{B}^{n+1}_{+}$ be an immersed capillary hypersurface
supported on $P$. Suppose $k>l\geq 1$. If there exists a constant $\alpha$
such that on $\Sigma$,

$\displaystyle\frac{H_{k}}{H_{l}}=\alpha,\quad H_{l}>0,$

we have that $\Sigma$ is totally umbilical except for being totally geodesic.

######  Remark.

We note that the second author proved the results in Theorem 1, Theorem 3 and
Theorem 5 for the free boundary case in [18]. Guo and Xia proved the result in
Theorem 2 and Theorem 4 for the free boundary case in [6].

Let $O\in\mathbb{H}^{n+1}$ be a point and $r(p)=\mbox{dist}(p,O)$ be the
distance function in the hyperbolic space. The metric in the hyperbolic space
can also be written as the following warped product form

$\overline{g}=dr^{2}+\sinh rg_{S^{n}}.$

It can be easily observed that the position vector field $x$ in Pioncaré ball
model satisfies $x=\sinh r\partial r$ if we choose $O$ be the origin of
$\mathbb{B}^{n+1}$. A hypersurface $M$ in $\mathbb{H}^{n+1}$ is called star-
shaped if $\overline{g}(x,\nu)=\overline{g}(\sinh r\partial r,\nu)>0$. It has
been proved that in some specific warped product manifolds (including the
hyperbolic space), an immersed, compact, closed, star-shaped hypersurface with
constant mean curvature must be a geodesic sphere (See [16, Corollary 5]). Now
we give a version of capillary hypersurfaces in the hyperbolic space.

######  Theorem 6.

Let $\Sigma\hookrightarrow\mathbb{B}^{n+1}_{+}$ be a compact, immersed CMC
capillary hypersurface supported on $P$. If $\Sigma$ is star-shaped, $\Sigma$
is totally umbilical.

Since $x$ is embedding immediately if it is star-shaped, Theorem 3 implies
Theorem 6 when $\theta\in(0,\frac{\pi}{2}]$.

The remainder of this paper is organized as follows. In Section 2, we collect
basic facts on the hyperbolic space. In Section 3, we prove Minkowski type
formulae. In Section 4, we give proofs of Theorem 1 and Theorem 3. In Section
5, we give proofs of Theorem 2 and Theorem 4. In Section 6, we consider case
of supporting hypersurfaces being geodesic spheres. Finally, in Section 7, we
prove two other rigidity results for the capillary hypersurfaces supported on
a totally geodesic plane in the hyperbolic space.

######  Acknowledgements.

This work is supported by the National Research Foundation of Korea (grant NRF
No.2021R1A4A1032418). We would also like to thank Prof. Haizhong Li from
Tsinghua University, Prof. Yingxiang Hu from Beihang University and Prof. Chao
Xia from Xiamen University for their useful comments and constant supports.

##  2\. Preliminaries

###  2.1. Capillary hypersurfaces and properties of the higher-order mean
curvature $H_{r}$

Let $\overline{g}$ and $\overline{\nabla}$ be the hyperbolic metric and the
Levi-Civita connection of $\mathbb{H}^{n+1}$, respectively. Let
$X:\Sigma\rightarrow\mathbb{\overline{}}B$ be an immersion of an
$n$-dimensional Riemannian manifold $\Sigma$ with boundary $\partial\Sigma$
into an $(n+1)$-dimensional Riemannian manifold $B$ with boundary $\partial
B$. $\Sigma$ is called a capillary hypersurface in $B$ if the immersion
satisfies the following

$\displaystyle X(\mbox{int}(\Sigma))\subset\mbox{int}(B),\quad
X(\partial\Sigma)\subset\partial B,$

and $\Sigma$ and $\partial B$ intersects with a constant contact angle
$\theta\in(0,\pi)$ along $\partial\Sigma$. On $\partial\Sigma$, there exists
four unit normal vector fields,
$\\{\nu,\,\mu,\,\overline{N},\,\overline{\nu}\\}$, on a 2-dimensional subspace
of $T_{p}\mathbb{H}^{n+1}$ as follows:

  1. (i)

$\nu$ is the outward unit normal vector field of isometric immersion
$X:\Sigma\rightarrow B$.

  2. (ii)

$\mu$ is the outward unit normal vector field of isometric immersion
$y:\partial\Sigma\rightarrow\Sigma$.

  3. (iii)

$\overline{N}$ is the outward unit normal vector field of isometric immersion
$z:\partial B\rightarrow B$.

  4. (iv)

$\overline{\nu}$ is the outward unit normal vector field of isometric
immersion $w:\partial\Sigma\rightarrow\partial B$.

The second fundamental form of $x:\Sigma\rightarrow\mathbb{H}^{n+1}$ denoted
by $h$ is defined as follows:

$\displaystyle h(Y,Z)=\overline{g}(\overline{\nabla}_{Y}\nu,Z),\mbox{ for
}Y,\;Z\in T(\Sigma),$

and the second fundamental form of $w:\partial\Sigma\rightarrow\partial B$
denoted by $h^{\partial\Sigma}$ is defined as follows:

$\displaystyle
h^{\partial\Sigma}(Y,Z)=\overline{g}(\overline{\nabla}_{Y}\overline{\nu},Z),\mbox{
for }Y,\;Z\in T(\partial\Sigma).$

$\displaystyle\nu$
$\mu$$\displaystyle\overline{N}$$\displaystyle\overline{\nu}$$\displaystyle\Sigma$$\displaystyle\partial
B$$\displaystyle B$$\displaystyle\theta$ Figure 1. Capillary hypersurface
supported on $\partial B$

Now we have the following relation of these vector fields,

(6)
$\displaystyle\left\\{\begin{array}[]{l}\mu=\sin\theta\overline{N}+\cos\theta\overline{\nu},\\\
\nu=-\cos\theta\overline{N}+\sin\theta\overline{\nu},\end{array}\right.$

where $\theta$ denotes the angle between $\overline{N}$ and $-\nu$. The
following lemma is well-known and fundamental.

######  Lemma 1.

Let $X:\Sigma\rightarrow B\subset\mathbb{H}^{n+1}$ be an isometric immersion
of a capillary hypersurface supported on the totally umbilical hypersurface
$\partial B$. Then $\mu$ is a principal direction of $\Sigma$, that is,

(7) $\displaystyle\overline{\nabla}_{\mu}\nu=h(\mu,\mu)\mu.$

###### Proof.

The proof can be found in [25, Proposition 2.1], we give a proof for
completeness. Without loss of generality, let $\\{e_{\alpha}\\}$ be the
orthonormal basis along $\partial\Sigma$ and $\alpha=1,2,\dots,n-1$. Then, we
have the following

$\displaystyle\overline{\nabla}_{\mu}\nu=$
$\displaystyle\overline{g}(\overline{\nabla}_{\mu}\nu,\mu)\mu+\sum_{\alpha=1}^{n-1}\overline{g}(\overline{\nabla}_{\mu}\nu,e_{\alpha})e_{\alpha}$
$\displaystyle=$ $\displaystyle
h(\mu,\mu)\mu-\sum_{\alpha=1}^{n-1}\overline{g}(\overline{\nabla}_{e_{\alpha}}(-\cos\theta\overline{N}+\sin\theta\overline{\nu}),\sin\theta\overline{N}+\cos\theta\overline{\nu})e_{\alpha}$
$\displaystyle=$ $\displaystyle
h(\mu,\mu)\mu-h^{\partial\Sigma}(e_{\alpha},\overline{\nu})e_{\alpha}$
$\displaystyle=$ $\displaystyle h(\mu,\mu)\mu,$

where the last equality holds because $\partial B$ is totally umbilical. ∎

Let $h_{ij}=h(e_{i},e_{j})$ and
$\lambda=(\lambda_{1},\lambda_{2},\dots,\lambda_{n})$ be the eigenvalues of
the matrix $(h_{ij})_{i,j=1}^{n}$, that is, the principal curvatures of
$\Sigma$. Let $k\in\\{1,2,\dots,n\\}$, the $k$-th mean curvature is defined by

$\displaystyle S_{k}=\frac{1}{k!}\sum\limits_{1\leq i_{1}<\cdots<i_{k}\leq
n}\lambda_{1}\lambda_{2}\dots\lambda_{i_{k}},$

and the normalized $k$-th mean curvature is defined by
$H_{k}=\binom{n}{k}^{-1}S_{k}$. The associated Newton transformation is
defined by induction,

$\displaystyle T_{0}=I,$ $\displaystyle T_{k}=S_{k}I-T_{k-1}\circ h.$

Using the induction formulae, we have the following properties.

$\displaystyle\mbox{tr}(T_{k})=(n-k)S_{k}=(n-k)\binom{n}{k}H_{k},$
$\displaystyle\mbox{tr}(T_{k}\circ
h)=\sum_{i,j=1}^{n}(T_{k})_{j}^{i}h_{i}^{j}=(k+1)S_{k+1}=(n-k)\binom{n}{k}H_{k+1}.$

Let $\Gamma^{+}_{k}=\\{\lambda\in\mathbb{R}^{n}:H_{i}>0,i=1,2,\dots,k\\}$.
Then, for $1\leq l<k\leq n$ and $\lambda\in\Gamma^{+}_{l}$, we have the
following classic Newton-Maclaurin inequality,

(8) $\displaystyle\frac{H_{l}}{H_{l-1}}\geq\frac{H_{k}}{H_{k-1}}.$

The equality in (8) holds if and only if $\lambda=c\mathcal{I}$, where $c>0$
is a constant and $\mathcal{I}=(1,1,\dots,1)\in\mathbb{R}^{n}$.

###  2.2. Properties in the Poincaré ball model

Let $x$ be the position vector in $\mathbb{B}^{n+1}$. The following fact is
well-known:

(9) $\displaystyle\overline{\nabla}x=V_{0}\overline{g},$

where $V_{0}=\frac{1+|x|^{2}}{1-|x|^{2}}$ is a smooth function in
$\mathbb{H}^{n+1}$ satisfying

$\overline{\nabla}^{2}V_{0}=V_{0}\overline{g}.$

A vector field $X$ is called a conformal Killing vector field in
$(\overline{M},\overline{g})$, if the Lie derivative $\mathcal{L}$ satisfies

$\displaystyle\mathcal{L}_{X}\overline{g}=f\overline{g}.$

In particular, $X$ is call a Killing vector if $f=0$. In this case, the key
Killing vector field $Y_{a}$ in Poincaré ball model we use is

$\displaystyle Y_{a}=\frac{1}{2}(1+|x|^{2})a-\delta(x,a)x,$

where $a$ is an arbitrary vector field in $\mathbb{H}^{n+1}$ which is constant
with respect to $(\mathbb{B}^{n+1},\delta)$. It is given by Wang and Xia in
[25].

Let $E_{1},E_{2},\dots,E_{n+1}$ form the coordinate frame in the conformal
Euclidean metric, and the corresponding normalized vectors
$\overline{E}_{1},\overline{E}_{2},\dots,\overline{E}_{n+1}$ form an
orthonormal frame on $(\mathbb{B}^{n+1},\overline{g})$. For
$A,B\in\\{1,2,\dots,n+1\\}$, it holds that

(10)
$\displaystyle\frac{1}{2}\left(\overline{g}(\overline{\nabla}_{\overline{E}_{A}}Y_{a},\overline{E}_{B})+\overline{g}(\overline{\nabla}_{\overline{E}_{B}}Y_{a},\overline{E}_{A})\right)=0.$

This can be verified by the following properties.

######  Proposition 1 (See [25]).

Let $a=\sum\limits_{i=1}^{n+1}a_{i}E_{i}$ be an arbitrary vector field
satisfying that $a_{i}$ are constant and $\delta(a,E_{n+1})=0$. Denote the
normalized vector of $a$ and $E_{n+1}$ in
$(\mathbb{B}^{n+1}_{+},\frac{4}{(1-|x|^{2})^{2}}\delta)$ by $\bar{a}$ and
$\bar{E}_{n+1}$ respectively, that is, $\bar{a}=\frac{1-|x|^{2}}{2}a$ and
$\bar{E}_{n+1}=\frac{1-|x|^{2}}{2}E_{n+1}$. Then, for any $Z\in
T\mathbb{H}^{n+1}$, we have the following:

  1. (a)

$\overline{\nabla}_{Z}a=\overline{g}(Z,x)\bar{a}+\overline{g}(\bar{a},x)Z-\overline{g}(Z,\bar{a})x$,

  2. (b)

$\overline{\nabla}_{Z}(\frac{1-|x|^{2}}{2}a)=\frac{1-|x|^{2}}{2}\left[\overline{g}(x,\bar{a})Z-\overline{g}(Z,\bar{a})x\right]$,

  3. (c)

$\overline{\nabla}_{Z}V_{0}=\overline{g}(x,Z)$,

  4. (d)

$\overline{\nabla}_{Z}Y_{a}=\overline{g}(x,Z)\bar{a}-\overline{g}(Z,\bar{a})x$.

###  2.3. Properties in the Poincaré half space model

In Poincaré half space model
$\mathbb{H}^{n+1}=(\mathbb{R}^{n+1}_{+},\frac{1}{x_{n+1}^{2}}\delta)$, we
denote $\widetilde{x}$ the position vector field in $\mathbb{R}^{n+1}_{+}$.

Let $E_{1},E_{2},\dots,E_{n+1}$ form the coordinate frame in the Euclidean
metric, and the normalized vectors
$\overline{E}_{1},\overline{E}_{2},\dots,\overline{E}_{n+1}$ form an
orthonormal frame on $(\mathbb{R}^{n+1}_{+},\overline{g})$. A conformal
Killing vector field given by

$X_{n+1}:=\widetilde{x}-E_{n+1}$

plays an important role in [6] and [5]. For $A,B\in\\{1,2,\dots,n+1\\}$, it
holds that

(11)
$\displaystyle\frac{1}{2}\left(\overline{g}(\overline{\nabla}_{\overline{E}_{A}}X_{n+1},\overline{E}_{B})+\overline{g}(\overline{\nabla}_{\overline{E}_{B}}X_{n+1},\overline{E}_{A})\right)=V_{n+1}\overline{g},$

where $V_{n+1}=\frac{1}{x_{n+1}}$ is a smooth function satisfying

$\overline{\nabla}^{2}V_{n+1}=V_{n+1}\overline{g}.$

Similarly, in the Poincaré half space model, we have the following
proposition, the proof of which is omitted.

######  Proposition 2 (See [5]).

Let $a=\sum\limits_{i=1}^{n+1}a_{i}E_{i}$ be an arbitrary vector field
satisfying that $a_{i}$ are constant and $\delta(a,E_{n+1})=0$. Denote the
normalized vector of $a$ and $E_{n+1}$ in
$(\mathbb{R}^{n+1}_{+},\frac{1}{x_{n+1}^{2}}\delta)$ by $\bar{a}$ and
$\bar{E}_{n+1}$ respectively, that is, $\bar{a}=x_{n+1}a$ and
$\bar{E}_{n+1}=x_{n+1}E_{n+1}$. Then, for any $Y\in T\mathbb{H}^{n+1}$, we
have the following:

  1. (a)

$\overline{\nabla}_{Y}\widetilde{x}=-\overline{g}(Y,\overline{E}_{n+1})\widetilde{x}+\overline{g}(Y,\widetilde{x})\overline{E}_{n+1}$,

  2. (b)

$\overline{\nabla}_{Y}a=-\overline{g}(Y,\overline{E}_{n+1})\overline{a}+\overline{g}(Y,\overline{a})E_{n+1}$,

  3. (c)

$\overline{\nabla}_{Y}E_{n+1}=-\frac{1}{x_{n+1}}Y$,

  4. (d)

$\overline{\nabla}_{Y}\overline{E}_{n+1}=\overline{g}(\overline{E}_{n+1},Y)\overline{E}_{n+1}-Y$.

It is easy to see that (11) can be obtained from (a) and (c) in Proposition 2.
Moreover, from (a) and (b) in Proposition 2, we have the following:

(12)
$\displaystyle\frac{1}{2}\left(\overline{g}(\overline{\nabla}_{\overline{E}_{A}}a,\overline{E}_{B})+\overline{g}(\overline{\nabla}_{\overline{E}_{B}}a,\overline{E}_{A})\right)=0,$

and

(13)
$\displaystyle\frac{1}{2}\left(\overline{g}(\overline{\nabla}_{\overline{E}_{A}}\widetilde{x},\overline{E}_{B})+\overline{g}(\overline{\nabla}_{\overline{E}_{B}}\widetilde{x},\overline{E}_{A})\right)=0,$

that is, $a$ and $\widetilde{x}$ are Killing vector fields in
$(\mathbb{R}^{n+1}_{+},\frac{1}{x_{n+1}^{2}}\delta)$.

##  3\. Minkowski type formulae in the hyperbolic space

In [25], the authors give a Minkowski type formula for capillary hypersurface
$M$ in a geodesic ball of radius $R$ in a space form $\mathbb{M}^{n+1}(K)$
($K=0,\,-1,\,1$),

$\displaystyle\int_{M}\left(n(V_{a}+\mbox{sn}_{K}(R)\cos\theta\overline{g}(Y_{a},\nu))-S_{1}\overline{g}(X_{a},\nu)\right)dA=0,$

where

$\displaystyle\mbox{sn}_{K}(r)=\left\\{\begin{array}[]{ll}r,&\mbox{ when
}K=0;\\\ \sin r,&\mbox{ when }K=1;\\\ \sinh r,&\mbox{ when }K=-1,\\\
\end{array}\right.$

and $X_{a}$ is a conformal Killing vector field tangent to the geodesic sphere
of radius $R$, which is the supporting hypersurface of $M$. We introduce our
results on the case (supported on geodesic sphere in $\mathbb{H}^{n+1}$) in
the Section 6.

###  3.1. A Minkowski type formula for capillary hypersurfaces in Poincaré
ball model

In $\mathbb{B}^{n+1}_{+}$, the outer unit normal vector field along $P$ is
$\overline{N}=-\frac{1-|x|^{2}}{2}E_{n+1}$. We note that the position vector
field $x$ in the Poincaré ball model is tangent to $P$, therefore
$\overline{g}(x,\overline{N})=0$.

$\displaystyle\nu$$\displaystyle\overline{\nu}$$\displaystyle\overline{N}$$\displaystyle\mu$$\displaystyle
E_{n+1}$$\displaystyle\Sigma$$\displaystyle P$$\displaystyle O$ Figure 2.
Capillary hypersurface $\Sigma$ supported on a totally geodesic plane $P$.

Let $Y_{n+1}=Y_{E_{n+1}}=\frac{1}{2}(1+|x|^{2})E_{n+1}-\delta(x,E_{n+1})x$ and
$E_{n+1}$ be the unit constant vector with respect to $\delta$. Then, we can
derive a Minkowski type formula as follows.

######  Proposition 3.

Let $x:\Sigma\subset\mathbb{B}^{n+1}_{+}$ be a compact, immersed capillary
hypersurface supported on $P$, intersecting $P$ at a constant angle $\theta$.
For $k\in\\{1,2,\dots,n\\}$, we have

(14)
$\displaystyle\int_{\Sigma}\left[H_{k-1}(V_{0}-\cos\theta\overline{g}(Y_{n+1},\nu))-H_{k}\overline{g}(x,\nu)\right]dA=0.$

###### Proof.

Restricting (9) to $\Sigma$ and let $\\{e_{i}\\}_{i=1}^{n}$ be the orthonormal
frame on $\Sigma$, we have

(15)
$\displaystyle\overline{g}(\nabla_{e_{i}}x^{T},e_{j})=V_{0}\delta_{ij}-h_{ij}\overline{g}(x,\nu),$

where $\nabla$ is the Levi-Civita connection of $(\Sigma,g)$. Applying the
Newton transformation $T_{k-1}$ on both sides of (15), we get

(16)
$\displaystyle(n-k+1)\binom{n}{k-1}\int_{\Sigma}V_{0}H_{k-1}-H_{k}\overline{g}(x,\nu)dA$
$\displaystyle=$
$\displaystyle\int_{\Sigma}\mbox{div}_{\Sigma}(T_{k-1}(x^{T}))dA$
$\displaystyle=$ $\displaystyle\int_{\partial\Sigma}T_{k-1}(x^{T},\mu)ds$
$\displaystyle=$
$\displaystyle\cos\theta\int_{\partial\Sigma}S_{k-1;\mu}\overline{g}(x,\overline{\nu})ds.$

In the last equality, we use (6) and the fact that $\mu$ is a principal
direction of $\Sigma$ and $\overline{g}(x,\overline{N})=0$. Therefore,
$T_{k-1}(x^{T},\mu)=S_{k-1;\mu}\overline{g}(x,\mu)=\cos\theta
S_{k-1;\mu}\overline{g}(x,\overline{\nu})$ where $S_{k-1;\mu}$ is given by

$\displaystyle
S_{k-1;\mu}=\sum_{\begin{subarray}{c}1\leq\alpha_{1}<\dots<\alpha_{k-1}\leq
n-1\\\
\mu\notin\\{e_{\alpha_{1}},\dots,e_{\alpha_{k-1}}\\}\end{subarray}}\lambda_{\alpha_{1}}\lambda_{\alpha_{2}}\cdots\lambda_{\alpha_{k-1}}.$

On the other hand, let
$U_{n+1}=\overline{g}(\nu,e^{-u}E_{n+1})x-\overline{g}(x,\nu)e^{-u}E_{n+1}$ be
a tangent vector field on $\Sigma$. We can easily see that
$\overline{g}(U_{n+1},\nu)=0$ along $\Sigma$. Then using Proposition 1, we
have

(17) $\displaystyle\overline{g}(\nabla_{e_{i}}U_{n+1},e_{j})=$
$\displaystyle\overline{g}\left(\overline{\nabla}_{e_{i}}\left(\overline{g}(\nu,e^{-u}E_{n+1})x^{T}-\overline{g}(x,\nu)(e^{-u}E_{n+1})^{T}\right),e_{j}\right)$
$\displaystyle=$
$\displaystyle-e^{-u}\overline{g}(x,\nu)\overline{g}(e^{-u}E_{n+1},e_{i})\overline{g}(x,e_{j})+\overline{g}(e^{-u}E_{n+1},\nu)(V_{0}\delta_{ij}$
$\displaystyle-
h_{ij}\overline{g}(x,\nu))-\overline{g}(x,\nu)\big{[}e^{-u}(\delta_{ij}\overline{g}(x,e^{-u}E_{n+1})$
$\displaystyle-\overline{g}(e^{-u}E_{n+1},e_{i})\overline{g}(x,e_{j})-h_{ij}\overline{g}(\nu,e^{-u}E_{n+1})\big{]}$
$\displaystyle=$
$\displaystyle\left(\overline{g}(e^{-u}E_{n+1},\nu)V_{0}-e^{-u}\overline{g}(x,\nu)\overline{g}(x,e^{-u}E_{n+1})\right)\delta_{ij}$
$\displaystyle=$ $\displaystyle\overline{g}(Y_{n+1},\nu)\delta_{ij},$

where we use the fact that $e^{-u}V_{0}E_{n+1}-\langle x,E_{n+1}\rangle
x=Y_{n+1}$. Again, applying the Newton transformation $T_{k-1}$ on both sides
of (17) and integrating on $\Sigma$, we have

(18)
$\displaystyle(n-k+1)\binom{n}{k-1}\int_{\Sigma}H_{k-1}\overline{g}(Y_{n+1},\nu)dA$
$\displaystyle=$
$\displaystyle\int_{\Sigma}(T_{k-1})_{ij}\overline{g}(\nabla_{e_{i}}U_{n+1},e_{j})dA=\int_{\partial\Sigma}T_{k-1}(U_{n+1},\mu)ds$
$\displaystyle=$
$\displaystyle\int_{\partial\Sigma}S_{k-1;\mu}\overline{g}(U_{n+1},\mu)ds$
$\displaystyle=$
$\displaystyle\int_{\partial\Sigma}S_{k-1;\mu}\overline{g}(x,\overline{\nu})ds,$

where in the last equality, we use (6) on $\partial\Sigma$ to obtain

$\displaystyle\overline{g}(U_{n+1},\mu)=$
$\displaystyle\overline{g}(\nu,e^{-u}E_{n+1})\overline{g}(x,\mu)-\overline{g}(x,\nu)\overline{g}(e^{-u}E_{n+1},\mu)$
$\displaystyle=$
$\displaystyle\left(\cos^{2}\theta\overline{g}(x,\overline{\nu})+\sin^{2}\theta\overline{g}(x,\overline{\nu})\right)=\overline{g}(x,\overline{\nu}).$

Combining (16) and (18), we obtain the Minkowski type formula (14).

∎

###  3.2. A Minkowski type formula for capillary hypersurfaces in Poincaré
half space model

In Poincaré half space model, the inner normal vector field along $L_{\phi}$
is given by

(19) $\overline{N}=x_{n+1}(-\cos\phi E_{n+1}+\sin\phi a).$

Similarly, it can be easily to observe that the conformal Killing vector field
$X_{n+1}=\widetilde{x}-E_{n+1}$ is tangential to $L_{\phi}$. Indeed, we have

(20) $\displaystyle\overline{g}(X_{n+1},\overline{N})=$ $\displaystyle
x_{n+1}\overline{g}(\widetilde{x}-E_{n+1},-\cos\phi E_{n+1}+\sin\phi a)$
$\displaystyle=$
$\displaystyle-\cos\phi+\sin\phi\frac{x_{a}}{x_{n+1}}+\frac{\cos\phi}{x_{n+1}}$
$\displaystyle=$ $\displaystyle 0,$

where in the last equality we use (2).

$x_{n+1}$$x_{a}$$\phi$$B^{int}$$\displaystyle{\color[rgb]{0.72,0.91,0.53}\definecolor[named]{pgfstrokecolor}{rgb}{0.72,0.91,0.53}\Sigma}$$\overline{N}$$L_{\phi}$$\overline{\nu}$$\nu$$\mu$
Figure 3. Capillary hypersurface $\Sigma$ supported on an equidistant
hypersurface $L_{\phi}$.

Thus, we have the following Minkowski type formula analogously.

######  Proposition 4.

Let $x:\Sigma\subset B^{\mathrm{int}}_{\phi,a}$ be a compact, immersed
capillary hypersurface supported on $L_{\phi,a}$, intersecting $L_{\phi,a}$ at
a constant angle $\theta$. For $k\in\\{1,2,\dots,n\\}$, we have

(21)
$\int_{\Sigma}[H_{k-1}(V_{n+1}-\cos\theta\sec\phi\overline{g}(\widetilde{x},\nu))-H_{k}\overline{g}(X_{n+1},\nu)]dA=0.$

###### Proof.

As the proof of Proposition 3, restricting (11) to $\Sigma$ and using the
divergence theorem, we have

(22)
$\displaystyle(n-k+1)\binom{n}{k-1}\int_{\Sigma}V_{n+1}H_{k-1}-H_{k}\overline{g}(X_{n+1},\nu)dA$
$\displaystyle=$ $\displaystyle\int_{\partial\Sigma}T_{k-1}(X_{n+1},\mu)ds$
$\displaystyle=$
$\displaystyle\cos\theta\int_{\partial\Sigma}S_{k-1;\mu}\overline{g}(X_{n+1},\overline{\nu})ds,$

where in the last equality we use (6), (20) and the fact that $\mu$ is a
principal direction of $\Sigma$ along $\partial\Sigma$.

On the other hand, consider a vector field
$Z_{n+1}=\overline{g}(x,\nu)\overline{E}_{n+1}-\overline{g}(\overline{E}_{n+1},\nu)x$.
By a direct computation, we have

$\displaystyle\overline{g}(\nabla_{e_{i}}Z_{n+1},e_{j})$ $\displaystyle=$
$\displaystyle-\overline{g}(e_{i},\overline{E}_{n+1})\overline{g}(\widetilde{x},\nu)\overline{g}(\overline{E}_{n+1},e_{j})+\overline{g}(\widetilde{x},e_{i})\overline{g}(\overline{E}_{n+1},\nu)\overline{g}(\overline{E}_{n+1},e_{j})$
$\displaystyle+h_{ik}\overline{g}(\widetilde{x},e_{k})\overline{g}(\overline{E}_{n+1},e_{j})+\overline{g}(\widetilde{x},\nu)(\overline{g}(\overline{E}_{n+1},e_{i})\overline{g}(\overline{E}_{n+1},e_{j})-\delta_{ij})$
$\displaystyle-\overline{g}(\overline{E}_{n+1},e_{i})\overline{g}(\overline{E}_{n+1},\nu)\overline{g}(\widetilde{x},e_{j})-h_{ik}\overline{g}(\overline{E}_{n+1},e_{k})\overline{g}(\widetilde{x},e_{j})$
$\displaystyle-\overline{g}(\overline{E}_{n+1},\nu)\left[-\overline{g}(e_{i},\overline{E}_{n+1})\overline{g}(\widetilde{x},e_{j})+\overline{g}(\widetilde{x},e_{j})\overline{g}(\overline{E}_{n+1},e_{i})\right]$
$\displaystyle=$
$\displaystyle-\delta_{ij}\overline{g}(\widetilde{x},\nu)+\overline{g}(\overline{E}_{n+1},\nu)\left(\overline{g}(\widetilde{x},e_{i})\overline{g}(\overline{E}_{n+1},e_{j})-\overline{g}(\overline{E}_{n+1},e_{i})\overline{g}(\widetilde{x},e_{j})\right).$

Therefore,

(23)
$\displaystyle\frac{1}{2}\left(\overline{g}(\nabla_{e_{i}}Z_{n+1},e_{j})+\overline{g}(\nabla_{e_{j}}Z_{n+1},e_{i})\right)=-\overline{g}(\widetilde{x},\nu)\delta_{ij}.$

Using the divergence Theorem, (6), (7) and (19), we have

(24)
$\displaystyle-(n-k+1)\binom{n}{k-1}\int_{\Sigma}H_{k-1}\overline{g}(\widetilde{x},\nu)dA$
$\displaystyle=$ $\displaystyle\int_{\partial\Sigma}T_{k-1}(Z_{n+1},\mu)ds$
$\displaystyle=$
$\displaystyle\int_{\partial\Sigma}S_{k-1;\mu}\overline{g}(Z_{n+1},\mu)ds$
$\displaystyle=$
$\displaystyle\int_{\partial\Sigma}S_{k-1;\mu}\left[\overline{g}(\widetilde{x},-\cos\theta\overline{N}+\sin\theta\overline{\nu})\overline{g}(\overline{E}_{n+1},\sin\theta\overline{N}+\cos\theta\overline{\nu})\right.$
$\displaystyle-\left.\overline{g}(\overline{E}_{n+1},-\cos\theta\overline{N}+\sin\theta\overline{\nu})\overline{g}(\widetilde{x},\sin\theta\overline{N}+\cos\theta\overline{\nu})\right]ds$
$\displaystyle=$
$\displaystyle-\int_{\partial\Sigma}S_{k-1;\mu}\left[(-\cos\phi+\sin\phi\frac{x_{a}}{x_{n+1}})\overline{g}(\overline{E}_{n+1},\overline{\nu})+\cos\phi\overline{g}(\widetilde{x},\overline{\nu})\right]ds$
$\displaystyle=$ $\displaystyle-\int_{\partial\Sigma}\cos\phi
S_{k-1;\mu}\overline{g}(X_{n+1},\overline{\nu})ds,$

where the last equality we use the fact from (2) that on
$\partial\Sigma\subset L_{\phi,a}$, $x_{n+1}=\tan\phi x_{a}+1$. As
consequence, the proof follows by combining (22) and (24). ∎

##  4\. Proofs of Theorem 1 and Theorem 3

Reilly established an important integral formula for general Riemannian
manifolds in [20]. Reilly’s formula is very powerful to obtain some properties
on the hypersurfaces in Riemannian manifolds with non-negative Ricci
curvatures. For instance, Ros used it to establish a Heintze-Karcher type
inequality in [21]. Later on, Li and Xia [13] generalized it to a weighted
Reilly-type formula to obtain a broader range of results. Let
$(\overline{\nabla},\overline{\Delta},\overline{\nabla}^{2})$ and
$(\nabla,\Delta,\nabla^{2})$ be the triples of Levi-Civita connection,
Laplacian and Hessian operator on $(\overline{\Omega},\overline{g})$ and
$(\partial\Omega,g)$ respectively. Then their Reilly-type formula is as
follows:

######  Theorem 7 (See [13]).

Let $(\overline{\Omega},\overline{g})$ be a compact Riemannian manifold with
piecewise smooth boundary $\partial\Omega$, $V\in
C^{\infty}(\overline{\Omega})$ be a positive smooth function. Suppose
$\frac{\overline{\nabla}^{2}V}{V}$ is continuous up to $\partial\Omega$. Then
for any smooth function $f\in C^{\infty}(\overline{\Omega})$, the following
identity holds.

(25)
$\displaystyle\int_{\Omega}V\left(\left(\overline{\Delta}f-\frac{\overline{\Delta}V}{V}f\right)^{2}-\left|\overline{\nabla}^{2}f-\frac{\overline{\nabla}^{2}V}{V}f\right|^{2}\right)d\Omega$
$\displaystyle=$
$\displaystyle\int_{\Omega}\left(\overline{\Delta}V\overline{g}-\overline{\nabla}^{2}V+V\overline{\mbox{Ric}}\right)\left(\overline{\nabla}f-\frac{\overline{\nabla}V}{V}f,\overline{\nabla}f-\frac{\overline{\nabla}V}{V}f\right)d\Omega$
$\displaystyle+\int_{\partial\Omega}V\left(f_{\nu_{\partial\Omega}}-\frac{V_{\nu_{\partial\Omega}}}{V}f\right)\left(\Delta
f-\frac{\Delta V}{V}f\right)dA$
$\displaystyle-\int_{\partial\Omega}Vg\left(\nabla\left(f_{\nu_{\partial\Omega}}-\frac{V_{\nu_{\partial\Omega}}}{V}f\right),\nabla
f-\frac{\nabla V}{V}f\right)dA$
$\displaystyle+\int_{\partial\Omega}nVH_{1}\left(f_{\nu_{\partial\Omega}}-\frac{V_{\nu_{\partial\Omega}}}{V}f\right)^{2}dA$
$\displaystyle+\int_{\partial\Omega}\left(h-\frac{V_{\nu_{\partial\Omega}}}{V}g\right)\left(\nabla
f-\frac{\nabla V}{V}f,\nabla f-\frac{\nabla V}{V}f\right)dA,$

where $\nu_{\partial\Omega}$ is the outer normal vector field on
$\partial\Omega$, $h(\cdot,\cdot)$ and $H_{1}$ are the second fundamental form
and mean curvature on $\partial\Omega$ respectively, and $\mathrm{Ric}$ is the
Ricci curvature in $\Omega$.

Let $\Omega$ be a domain in $\mathbb{B}_{+}^{n+1}$ enclosed by $\Sigma$ and
$T$, where $T$ is a domain on $P$ enclosed by $\partial\Sigma$. Inspired by
the idea in [10], we will give a proof of Theorem 1 now.

###### Proof of Theorem 1.

For $\theta=\frac{\pi}{2}$, we refer to [18]. Therefore we will deal with the
case of $\theta\in(0,\frac{\pi}{2})$. We consider the following mixed boundary
problem,

(29)
$\displaystyle\left\\{\begin{array}[]{ll}\overline{\Delta}f-(n+1)f=1,&\mbox{
in }\Omega;\\\ f=0,&\mbox{ on }\Sigma;\\\ f_{\overline{N}}=c_{0},&\mbox{ on
}T,\end{array}\right.$

where
$c_{0}=-\frac{n}{n+1}\cot\theta\frac{\int_{T}V_{0}dA_{T}}{\int_{\partial\Sigma}V_{0}ds}$.
Since the idea about the existence and regularity is well established in [10],
we will leave details of the following conclusion in the appendix: The
solution $f$ of (29) satisfies $f\in
C^{\infty}(\overline{\Omega}\setminus\partial\Sigma)\cup C^{2}(\Omega\cup
T)\cup C^{1,\alpha}(\overline{\Omega})$ and
$|\overline{\nabla}^{2}f|_{\overline{g}}\in L^{1}(T)$. This makes $f$
applicable in the generalized Reilly’s formula (25) and by letting $V=V_{0}$.
Using (25), we have

(30) $\displaystyle\frac{n}{n+1}\int_{\Omega}V_{0}d\Omega=$
$\displaystyle\frac{n}{n+1}\int_{\Omega}V_{0}\left(\overline{\Delta}f-\frac{\overline{\Delta}V_{0}}{V_{0}}f\right)^{2}d\Omega$
$\displaystyle\geq$
$\displaystyle\int_{\Omega}V_{0}\left(\left(\overline{\Delta}f-\frac{\overline{\Delta}V_{0}}{V_{0}}f\right)^{2}-\left|\overline{\nabla}^{2}f-\frac{\overline{\nabla}^{2}V_{0}}{V_{0}}f\right|^{2}\right)d\Omega$
$\displaystyle=$
$\displaystyle\int_{\Sigma}nV_{0}H_{1}f_{\nu}^{2}dA+c_{0}\int_{T}(V_{0}\Delta
f-f\Delta V_{0})dA_{T}$ $\displaystyle=$
$\displaystyle\int_{\Sigma}nV_{0}H_{1}f_{\nu}^{2}dA+c_{0}\int_{\partial\Sigma}V_{0}g(\nabla^{T}f,\overline{\nu})ds.$

In the second equality we use the fact that
$\overline{\nabla}^{2}V_{0}=V_{0}\overline{g}$,
$\overline{\mbox{Ric}}=n\overline{g}$ and
$(V_{0})_{\overline{N}}=\overline{g}(\widetilde{x},\overline{N})=0$ on $T$,
and in the third equality we apply the Green’s formula (See [17]).

Furthermore, since $f\in C^{1}(\overline{\Omega})$ and $f=0$ on $\Sigma$, we
have $f_{\mu}=0$. From the relation (6), we have the following:

$\displaystyle\overline{\nu}=$ $\displaystyle\cos\theta\mu+\sin\theta\nu$
$\displaystyle=$
$\displaystyle\cos\theta\mu+\sin\theta(-\cos\theta\overline{N}+\sin\theta\overline{\nu}),$
$\displaystyle=$
$\displaystyle\cos\theta\mu-\sin\theta\cos\theta\overline{N}+\sin^{2}\theta\overline{\nu}.$

Since $\overline{\nu}=\frac{1}{\cos\theta}\mu-\tan\theta\overline{N}$, we have

$\displaystyle c_{0}\int_{\partial\Sigma}V_{0}f_{\overline{\nu}}ds=$
$\displaystyle
c_{0}\int_{\partial\Sigma}V_{0}\overline{g}(\overline{\nabla}f-f_{\overline{N}}\overline{N},\overline{\nu})ds$
$\displaystyle=$ $\displaystyle
c_{0}\int_{\partial\Sigma}V_{0}\overline{g}(\nabla^{T}f,\frac{1}{\cos\theta}\mu-\tan\theta\overline{N})ds$
$\displaystyle=$ $\displaystyle-
c_{0}^{2}\tan\theta\int_{\partial\Sigma}V_{0}ds$ $\displaystyle=$
$\displaystyle-\left(\frac{n}{n+1}\right)^{2}\cot\theta\frac{\left(\int_{T}V_{0}dA_{T}\right)^{2}}{\int_{\partial\Sigma}V_{0}ds}.$

Therefore, (30) is equivalent to the following inequality,

(31)
$\displaystyle\frac{n}{n+1}\left(\int_{\Omega}V_{0}d\Omega+\frac{n}{n+1}\cot\theta\frac{\left(\int_{T}V_{0}dA_{T}\right)^{2}}{\int_{\partial\Sigma}V_{0}ds}\right)\geq\int_{\Sigma}nV_{0}H_{1}f_{\nu}^{2}dA.$

On the other hand, by Green’s formula to the equation above, we have

$\displaystyle\int_{\Omega}V_{0}d\Omega=$
$\displaystyle\int_{\Omega}V_{0}\left(\overline{\Delta}f-\frac{\overline{\Delta}V_{0}}{V_{0}}f\right)d\Omega$
$\displaystyle=$
$\displaystyle\int_{\Sigma}(V_{0}f_{\nu}-f(V_{0})_{\nu})dA+\int_{T}(V_{0}f_{\overline{N}}-f(V_{0})_{\overline{N}})dA_{T}$
$\displaystyle=$
$\displaystyle\int_{\Sigma}V_{0}f_{\nu}dA+c_{0}\int_{T}V_{0}dA_{T}$
$\displaystyle=$
$\displaystyle\int_{\Sigma}V_{0}f_{\nu}dA-\frac{n}{n+1}\cot\theta\frac{\left(\int_{T}V_{0}dA_{T}\right)^{2}}{\int_{\partial\Sigma}V_{0}ds}.$

Applying Hölder’s inequality, we obtain

(32)
$\displaystyle\int_{\Omega}V_{0}d\Omega+\frac{n}{n+1}\cot\theta\frac{\left(\int_{T}V_{0}d\nu_{T}\right)^{2}}{\int_{\partial\Sigma}V_{0}ds}$
$\displaystyle=$
$\displaystyle\int_{\Sigma}V_{0}f_{\mu}dA\leq\left(\int_{\Sigma}\frac{V_{0}}{nH_{1}}dA\right)^{\frac{1}{2}}\left(\int_{\Sigma}nV_{0}H_{1}f_{\nu}^{2}dA\right)^{\frac{1}{2}}.$

Combining (31) and (32), we have

(33)
$\displaystyle\int_{\Sigma}\frac{V_{0}}{H_{1}}dA\geq(n+1)\int_{\Omega}V_{0}d\Omega+n\cot\theta\frac{\left(\int_{T}V_{0}dA_{T}\right)^{2}}{\int_{\partial\Sigma}V_{0}ds}.$

If the equality in (33) holds, all the inequalities above become equalities.
In particular from (30) we have,

$\displaystyle\overline{\nabla}^{2}f-f\overline{g}=\overline{\nabla}^{2}f-\frac{\overline{\nabla}^{2}V_{0}}{V_{0}}f=\frac{1}{n+1}\left(\overline{\Delta}f-\frac{\overline{\Delta}V_{0}}{V_{0}}f\right)\overline{g}=\frac{1}{n+1}\overline{g},$

which is equivalent to

(34)
$\displaystyle\overline{\nabla}^{2}\left(f+\frac{1}{n+1}\right)=\left(f+\frac{1}{n+1}\right)\overline{g}$

Restricting (34) to $\Sigma$ where $f=0$, we can see that

$f_{\nu}h_{ij}=\frac{1}{n+1}\overline{g}_{ij}.$

This implies that $\Sigma$ is totally umbilical. ∎

######  Remark.

In the case of $\theta=\frac{\pi}{2}$, we note that the last term of the right
hand side in (30) vanishes, therefore there is no need for the Green’s formula
used in the second equality in (30), the theory in [14] is sufficient. See the
Appendix for detail.

Before giving a proof of Theorem 3, we need the following identities.

######  Proposition 5.

On the embedded capillary hypersurface $\Sigma\subset\mathbb{B}^{n+1}_{+}$
supported on $P$, the following identities holds,

(35)
$\displaystyle\int_{T}V_{0}dA_{T}=\int_{\Sigma}\overline{g}(Y_{n+1},\nu)dA,$

and

(36)
$\displaystyle\sin\theta\int_{\partial\Sigma}V_{0}ds=\int_{\Sigma}nH_{1}\overline{g}(Y_{n+1},\nu)dA.$

###### Proof.

Since $Y_{n+1}$ is a Killing vector field, then
$\overline{\mathrm{div}}(Y_{n+1})=0$ on $\Omega$. Integrate on $\Omega$, we
have

$\displaystyle 0=$
$\displaystyle\int_{\Omega}\overline{\mathrm{div}}(Y_{n+1})d\Omega$
$\displaystyle=$
$\displaystyle\int_{\Sigma}\overline{g}(Y_{n+1},\nu)dA+\int_{T}\overline{g}(Y_{n+1},\overline{N})dA_{T}.$

Since $\overline{g}(Y_{n+1},\overline{N})=-V_{0}$, we have

$\displaystyle\int_{T}V_{0}dA_{T}=\int_{\Sigma}\overline{g}(Y_{n+1},\nu)dA.$

Restricting (10) to $\Sigma$ and using the divergence theorem, we have

$\displaystyle-\int_{\Sigma}nH_{1}\overline{g}(Y_{n+1},\nu)dA=\int_{\Sigma}\mathrm{div}_{\Sigma}(P_{\Sigma}Y_{n+1})dA=\int_{\partial\Sigma}\overline{g}(Y_{n+1},\mu)ds=-\sin\theta\int_{\partial\Sigma}V_{0}ds,$

where $P_{\Sigma}$ denotes the projection on $T\Sigma$. ∎

We are now ready to prove Theorem 3.

###### Proof of the Theorem 3.

Since $\Sigma$ is compact, there exists a $t\in(0,\pi)$ such that $\Sigma$ is
contained in a domain enclosed by $P$ and consider 1-parameter family fo
equidistant hypersurfaces $S_{t}$.

$\displaystyle S_{t}=\left\\{x\in\mathbb{H}^{n+1}:|x+\tan
tE_{n+1}|^{2}=\frac{1}{\cos^{2}t},\;t\in(0,\frac{\pi}{2})\right\\}.$

When $t\rightarrow 0+$, $S_{t}$ tends to $\partial_{\infty}\mathbb{H}^{n+1}$.
On the other hand, when $t\rightarrow\frac{\pi}{2}-$, $S_{t}$ tends to $P$.
Given that $\partial\Sigma$ is compact, and $\partial S_{t}=\partial
P\subset\partial_{\infty}\mathbb{H}^{n+1}$, therefore when
$t\rightarrow\frac{\pi}{2}-$, $S_{t}$ would only touch the interior of
$\Sigma$ other than its boundary. As $t\rightarrow\frac{\pi}{2}$, there must
exist a $t_{0}$ such that $S_{t_{0}}$ touches $\mbox{int}(\Sigma)$ at first
and tangent to $\Sigma$ at a point $p\in\mbox{int}(\Sigma$). Hence, we have
$h(p)\geq\lambda_{S}(p)>0$. Then $H_{1}\equiv H_{1}(p)>0$. See the Figure 4.

$\displaystyle{\color[rgb]{0.72,0.91,0.53}\definecolor[named]{pgfstrokecolor}{rgb}{0.72,0.91,0.53}\Sigma}$$S_{t}$$\displaystyle
P$ Figure 4. Existence of the convex point on capillary hypersurface $\Sigma$
supported on $P$

Now applying (35) and (36) to the inequality (1), we can see that (1) is
equivalent to the following:

(37)
$\displaystyle\int_{\Sigma}\frac{V_{0}}{H_{1}}dA\geq(n+1)\int_{\Omega}V_{0}d\Omega+\cos\theta\frac{\left(\int_{\Sigma}\overline{g}(Y_{n+1},\nu)dA\right)^{2}}{\int_{\Sigma}H_{1}\overline{g}(Y_{n+1},\nu)dA}.$

Since $H_{1}$ is constant, (37) can be written as

(38)
$\displaystyle\int_{\Sigma}\frac{V_{0}-\cos\theta\overline{g}(Y_{n+1},\nu)}{H_{1}}dA\geq$
$\displaystyle(n+1)\int_{\Omega}V_{0}d\Omega$ $\displaystyle=$
$\displaystyle\int_{\Omega}\overline{\mathrm{div}}(x)d\Omega$ $\displaystyle=$
$\displaystyle\int_{\Sigma}\overline{g}(x,\nu)dA.$

The last equality holds because $\overline{g}(x,\overline{N})=0$ on $T$. Since
$H_{1}$ is constant, from Proposition 3 ($k=1$) we can see that the equality
in (38) holds. According to the condition in Theorem 1, we obtain that
$\Sigma$ is umbilical, but it is not totally geodesic since $H_{1}>0$. ∎

######  Remark.

We consider the equality case of Theorem 1. From [25, Remark 4.2], we know
that any function in $\\{V\in
C^{\infty}(\mathbb{H}^{n+1}):\overline{\nabla}^{2}V=V\overline{g}\\}$ can be
represented by the linear combination of the $n+2$ functions
$\\{V_{0},V_{1},\dots,V_{n+1}\\}$, where $V_{i}=\frac{2\langle
x,E_{i}\rangle}{1-|x|^{2}}$. From (34) we can see that there exists constants
$B,\,c_{i}\in\mathbb{R}$ and a constant vector (in the Euclidean sense)
$a\in\mathbb{R}^{n+1}$ such that

$\displaystyle f+\frac{1}{n+1}=$ $\displaystyle
BV_{0}+\sum\limits_{i=1}^{n+1}c_{i}V_{i}$ $\displaystyle=$ $\displaystyle
B\frac{1+|x|^{2}}{1-|x|^{2}}+\frac{2\langle x,C\rangle}{1-|x|^{2}}$
$\displaystyle=$ $\displaystyle\frac{B(1+|x|^{2})+2\langle
x,C\rangle}{1-|x|^{2}},$

where $C=\sum\limits_{i=1}^{n+1}c_{i}E_{i}$.

  1. (1)

When $B\neq-\frac{1}{n+1}$, $\Sigma$ lies in a hypersurface defined by

$\displaystyle\left\\{x\in\mathbb{H}^{n+1}:\left|x+\frac{(n+1)}{1+(n+1)B}C\right|^{2}=\frac{1-(n+1)B}{1+(n+1)B}+\left(\frac{(n+1)|C|}{1+(n+1)B}\right)^{2}\right\\}.$

  2. (2)

When $B=-\frac{1}{n+1}$ and $C\neq 0$, $\Sigma$ lies in a hypersurface defined
by

$\displaystyle\left\\{x\in\mathbb{H}^{n+1}:\langle x,a\rangle=1/|C|\right\\}.$

Therefore, $\Sigma$ is a totally umbilical hypersurface in $\mathbb{H}^{n+1}$
and can be a geodesic ball, a horosphere or an equidistant hypersurfaces, and
$B,\,C\in\mathbb{R},a\in\mathbb{R}^{n+1}$ can be chosen such that $\Sigma$ is
a part of one of the illustrated above and compact (compactness excludes the
totally geodesic case). For example, a horosphere can be seen Figure 5.

$\displaystyle a$EOF$\displaystyle P$$\displaystyle\Sigma$ Figure 5. Compact
capillary horosphere

From the Figure 5, we can see

$\displaystyle|OF|=\sqrt{\left|\frac{1-(n+1)B}{1+(n+1)B}\right|},\quad|OE|=\left|\frac{(n+1)}{1+(n+1)B}C\right|.$

##  5\. Proofs of Theorem 2 and Theorem 4

In this section, we consider the case when the supporting hypersurface is
either an equidistant hypersurface or a horospheres. Before we give a proof of
Theorem 2, we need the following lemma.

######  Proposition 6.

On the embedded capillary hypersurfaces
$\Sigma\hookrightarrow\mathbb{R}^{n+1}_{+}$ supported on $L_{\phi,a}$ in the
Poincaré half space model, the following identities holds

(39)
$\int_{\Sigma}\overline{g}(\widetilde{x},\nu)dA=\cos\phi\int_{T}V_{n+1}dA_{T};$
(40) $\int_{\Sigma}\overline{g}(a,\nu)dA=-\sin\phi\int_{T}V_{n+1}dA_{T};$ (41)
$\int_{\Sigma}(-nH_{1}\overline{g}(\widetilde{x},\nu))dA=\int_{\partial\Sigma}\overline{g}(\widetilde{x},\mu)ds;$
(42)
$\int_{\Sigma}(-nH_{1}\overline{g}(a,\nu))dA=\int_{\partial\Sigma}\overline{g}(a,\mu)ds;$

and

(43)
$\int_{\Sigma}n\cos\theta\overline{g}(\widetilde{x},\nu)dA=\int_{\partial\Sigma}\cos\phi\overline{g}(X_{n+1},\mu)ds.$

###### Proof.

To prove (39) and (41), we know from (13) that
$\overline{\mathrm{div}}\widetilde{x}=0$. Let $\Omega$ be a domain enclosed by
$\Sigma$ and $T$. Using the divergence Theorem, we have

$\displaystyle 0=$
$\displaystyle\int_{\Omega}\overline{\mathrm{div}}\widetilde{x}d\Omega$
$\displaystyle=$
$\displaystyle\int_{\Sigma}\overline{g}(\widetilde{x},\nu)dA+\int_{T}\overline{g}(\overline{N},\widetilde{x})dA_{T}$
$\displaystyle=$
$\displaystyle\int_{\Sigma}\overline{g}(\widetilde{x},\nu)dA+\int_{T}\frac{1}{x_{n+1}}(-\cos\phi
x_{n+1}+\sin\phi x_{a})dA_{T}$ $\displaystyle=$
$\displaystyle\int_{\Sigma}\overline{g}(\widetilde{x},\nu)dA-\int_{T}\cos\phi
V_{n+1}dA_{T}.$

In the last equality, we use (2).

On the other hand, restricting (13) on $\Sigma$, we have

$\displaystyle\int_{\Sigma}\left(-nH_{1}\overline{g}(\widetilde{x},\nu)\right)dA=\int_{\Sigma}\mathrm{div}_{\Sigma}(P_{\Sigma}\widetilde{x})dA=\int_{\partial\Sigma}\overline{g}(\widetilde{x},\mu)ds,$

where $P_{\Sigma}$ denotes the projection on $T\Sigma$. Then we obtain (39)
and (41). Using (12), we can prove (42) and (40). Moreover, (43) can be noted
from in the proof of Proposition 4. ∎

###### Proof of Theorem 2.

Consider the following mixed boundary problem in the Poincaré half space
model,

(44) $\left\\{\begin{array}[]{ll}\overline{\Delta}f-(n+1)f=1,&\mbox{ in
}\Omega;\\\ f=0&\mbox{ on }\Sigma;\\\ f_{\overline{N}}-\cos\phi f=c_{0}&\mbox{
on }T,\end{array}\right.$

where

$c_{0}=-\frac{n}{n+1}\cos\theta\frac{\int_{T}V_{n+1}dA_{T}}{\int_{\partial\Sigma}\overline{g}(-\cos\phi\widetilde{x}+\sin\phi
a,\mu)ds}.$

According to the Lieberman’s theory (see the appendix), there exists a
solution $f$ such that $f\in
C^{\infty}(\overline{\Omega}\setminus\partial\Sigma)\cup C^{2}(\Omega\cup
T)\cup C^{1,\alpha}(\overline{\Omega})$ and
$|\overline{\nabla}^{2}f|_{\overline{g}}\in L^{1}(T)$. Therefore, for the same
reason as the proof of Theorem 1, we can insert the solution $f$ to the
generalized Reilly’s type formula (25).

$\displaystyle\frac{n}{n+1}\int_{\Omega}V_{n+1}d\Omega=$
$\displaystyle\frac{n}{n+1}\int_{\Omega}V_{n+1}\left(\overline{\Delta}f-\frac{\overline{\Delta}V_{n+1}}{V_{n+1}}f\right)^{2}d\Omega$
$\displaystyle\geq$
$\displaystyle\int_{\Omega}V_{n+1}\left(\left(\overline{\Delta}f-\frac{\overline{\Delta}V_{n+1}}{V_{n+1}}f\right)^{2}-\left|\overline{\nabla}^{2}f-\frac{\overline{\nabla}^{2}V_{n+1}}{V_{n+1}}f\right|^{2}\right)d\Omega$
$\displaystyle=$
$\displaystyle\int_{\Sigma}nV_{n+1}H_{1}f_{\nu}^{2}dA+c_{0}\int_{T}(V_{n+1}\Delta
f-f\Delta V_{n+1})dA_{T}$
$\displaystyle+nc_{0}^{2}\cos\phi\int_{T}V_{n+1}dA_{T}$ $\displaystyle=$
$\displaystyle\int_{\Sigma}nV_{n+1}H_{1}f^{2}_{\nu}dA+c_{0}\int_{\partial\Sigma}V_{n+1}\overline{g}(\nabla^{T}f,\overline{\nu})ds+nc_{0}^{2}\int_{\Sigma}\overline{g}(\widetilde{x},\nu)dA$
$\displaystyle=$
$\displaystyle\int_{\Sigma}nV_{n+1}H_{1}f^{2}_{\nu}dA+c_{0}\int_{\partial\Sigma}V_{n+1}\overline{g}(\nabla^{T}f,\overline{\nu})ds$
$\displaystyle+\frac{c_{0}^{2}\cos\phi}{\cos\theta}\int_{\partial\Sigma}\overline{g}(X_{n+1},\mu)ds,$

where we use the fact that
$(V_{n+1})_{\overline{N}}=-\frac{1}{x_{n+1}^{2}}\langle
E_{n+1},\overline{N}\rangle=\cos\phi V_{n+1}$ and (43) in the last equality.
On $\partial\Sigma$, using (6) and the boundary condition in (44), we have

$\displaystyle\overline{g}(\nabla^{T}f,\overline{\nu})=$
$\displaystyle\overline{g}(\overline{\nabla}f-\overline{g}(\overline{\nabla}f,\overline{N})\overline{N},\frac{1}{\cos\theta}\mu)$
$\displaystyle=$
$\displaystyle-\frac{c_{0}}{\cos\theta}\overline{g}(\overline{N},\mu).$

Then we have

$\displaystyle\frac{n}{n+1}\int_{\Omega}V_{n+1}d\Omega\geq$
$\displaystyle\int_{\Sigma}nV_{n+1}H_{1}f_{\nu}^{2}dA-\frac{c_{0}^{2}}{\cos\theta}\int_{\partial\Sigma}V_{n+1}\overline{g}(\overline{N},\mu)ds$
$\displaystyle+\frac{c_{0}^{2}\cos\phi}{\cos\theta}\int_{\partial\Sigma}\overline{g}(X_{n+1},\mu)ds$
$\displaystyle=$
$\displaystyle\int_{\Sigma}nV_{n+1}H_{1}f_{\nu}^{2}dA-\frac{c_{0}^{2}}{\cos\theta}\int_{\partial\Sigma}\overline{g}(-\cos\phi\widetilde{x}+\sin\phi
a,\mu)ds$ $\displaystyle=$
$\displaystyle\int_{\Sigma}nV_{n+1}H_{1}f_{\nu}^{2}dA-\left(\frac{n}{n+1}\right)^{2}\cos\theta\frac{\left(\int_{T}V_{n+1}dA_{T}\right)^{2}}{\int_{\partial\Sigma}\overline{g}(-\cos\phi\widetilde{x}+\sin\theta
a,\mu)ds}.$

Therefore,

(45)
$\displaystyle\frac{n}{n+1}\left(\int_{\Omega}V_{n+1}d\Omega+\frac{n}{n+1}\cos\theta\frac{\left(\int_{T}V_{n+1}dA_{T}\right)^{2}}{\int_{\partial\Sigma}\overline{g}(-\cos\phi\widetilde{x}+\sin\theta
a,\mu)ds}\right)$ $\displaystyle\geq$
$\displaystyle\int_{\Sigma}nV_{n+1}H_{1}f_{\nu}^{2}dA.$

On the other hand, using the divergence theorem we have

(46) $\displaystyle\int_{\Omega}V_{n+1}d\Omega=$
$\displaystyle\int_{\Omega}V_{n+1}\left(\overline{\Delta}f-\frac{\overline{\Delta}V_{n+1}}{V_{n+1}}f\right)d\Omega$
$\displaystyle=$
$\displaystyle\int_{\Sigma}(V_{n+1}f_{\nu}-f(V_{n+1})_{\nu})dA+\int_{T}(V_{n+1}f_{\overline{N}}-f(V_{n+1})_{\overline{N}})dA_{T}$
$\displaystyle=$
$\displaystyle\int_{\Sigma}V_{n+1}f_{\nu}dA+c_{0}\int_{T}V_{n+1}dA_{T}$
$\displaystyle=$
$\displaystyle\int_{\Sigma}V_{n+1}f_{\nu}dA-\frac{n}{n+1}\cos\theta\frac{\left(\int_{T}V_{n+1}dA_{T}\right)^{2}}{\int_{\partial\Sigma}\overline{g}(-\cos\phi\widetilde{x}+\sin\phi
a,\mu)ds}.$

Combining (45) and (46) and using the similar argument as the proof of Theorem
1, we prove the inequality in Theorem 2. The equality case is also the same as
Theorem 1. ∎

At this point, we are ready to prove the Alexandrov type theorem for the
capillary hypersurface in $B^{\mathrm{int}}_{\phi,a}$.

###### Proof of the Theorem 4.

In order to prove the existence of convex points on $\Sigma$, we consider both
the case that $\phi>0$ and the case that $\phi=0$ independently.
Case 1 ($\phi>0$): Now $L_{\phi,a}$ is an equidistant hypersurface. Let
$\eta\in(0,\phi)$, which will be determined later. We consider a set

$P_{\eta}=\\{\widetilde{x}\in\mathbb{H}^{n+1}:x_{n+1}=\tan\eta(x_{a}+1)\\}.$

Let $p\in P_{\eta}$ and $S_{p}$ be a family of equidistant hypersurface
defined as

$S_{p}=\\{\widetilde{x}\in\mathbb{H}^{n+1}:|\widetilde{x}-p|=|p-\mathrm{dist}_{\mathbb{R}}(p,\partial
P_{\eta})|\mbox{ and }p\in P_{\eta}\\},$

where $\mathrm{dist}_{\mathbb{R}}$ is the Euclidean distance in the Poincaré
half space model. Since $\Sigma$ is a compact capillary hypersurface supported
on $L_{\phi,a}$, we can find a proper $p_{0}\in P_{\eta}$ sufficiently far (in
the Euclidean sense) from $\partial P_{\eta}$, such that $\Sigma$ is contained
in the domain enclosed by $S_{p_{0}}$ and $L_{\phi,a}$ and $\eta>0$ guarantees
that $\Sigma$ stays in the interior side of $S_{p_{0}}$. Then for any $p\in P$
the contact angle $\theta_{0}$ between $L_{\phi,a}$ and $S_{p_{0}}$ is
constant satisfying $\theta_{0}=\frac{\pi}{2}-\phi+\eta$.

Since $\theta+\phi>\frac{\pi}{2}$, there exists an $\eta\in(0,\phi)$ such that
$\theta>\theta_{0}=\frac{\pi}{2}-\phi+\eta$.

Now moving $p_{0}$ toward $\partial P_{\eta}$ such that
$\mathrm{dist}_{\mathbb{R}}(p_{0},\partial P_{\eta})$ goes to $0$, we can see
that there exists $p_{1}$ such that $S_{p_{1}}$ firstly touches $\Sigma$ at
some $q\in\mathrm{int}(\Sigma)$ since $\theta>\theta_{0}$. Meanwhile, since
$\eta>0$, $\Sigma$ remains within the interior side of $S_{p_{1}}$. See Figure
6.

$x_{n+1}$$x_{a}$$\phi$$B^{int}$$\displaystyle{\color[rgb]{0.72,0.91,0.53}\definecolor[named]{pgfstrokecolor}{rgb}{0.72,0.91,0.53}\Sigma}$$\eta$$\overline{N}$$\theta_{0}$$P_{\eta}$$L_{\phi}$$p_{0}$$S_{p_{0}}$
Figure 6. Existence of the convex point for hypersurfaces supported on a
equidistant hypersurface

Then we can see that the principal curvature
$\lambda(q)\geq\lambda_{S_{p_{1}}}(q)>0$ of $\Sigma$, then $H_{1}\equiv
H_{1}(q)>0$.
Case 2 ($\phi=0$): Now $L_{\phi,a}$ is a horosphere and $\theta=\frac{\pi}{2}$
by assumption. Consider a set

$P_{r_{0}}=\\{x\in\mathbb{H}^{n+1}:x_{n+1}=r_{0}\\},$

for $0<r_{0}<1$. Let $S_{R,p}$ be the hypersurface defined by

$S_{R,p}=\\{x\in\mathbb{H}^{n+1}:|\widetilde{x}-p|=R,p\in P_{r_{0}}\\}.$

Since $\Sigma$ is compact, there must exist $p_{0}\in P_{r_{0}}$ and a
positive $R_{0}$ big enough such that $\Sigma$ is contained in the domain
enclosed by $S_{R_{0},p_{0}}$ and $L_{\phi,a}$. Now fix $p_{0}$ and shrinking
the radius $R_{0}$ of $S_{R_{0},p_{0}}$ in the Euclidean sense until it
touches $\Sigma$ at $p$ for the first time when the radius is $R_{1}$. Since
the contact angle of $S_{R_{1},p_{0}}$ is less than $\frac{\pi}{2}$, the
firstly touching point $q$ must be the interior point on $\Sigma$. Since the
contact angle of $S_{R_{1},p_{0}}$ and
$\partial_{\infty}\mathbb{H}^{n+1}:=\\{x_{n+1}=0\\}$ is greater than
$\frac{\pi}{2}$, $\Sigma$ is staying in the interior side of
$S_{R_{1},p_{0}}$. See Figure 7.

$L_{0}$$P_{r_{0}}{}$$S_{p_{0},R_{0}}$$\Sigma$$p_{0}$$x_{n+1}$$x_{a}$ Figure 7.
Existence of the convex point for hypersurfaces supported on a horosphere

Then it holds that $\lambda(q)\geq\lambda_{S_{R_{1},p_{0}}}>0$ and
$H_{1}\equiv H_{1}(q)>0$. Therefore the Heintze-Karcher inequality in Theorem
2 can be applied. And from (39), (39), (42) and (40), we are able to rewrite
(3) as

$\displaystyle\int_{\Sigma}\frac{V_{n+1}}{H_{1}}dA\geq$
$\displaystyle(n+1)\int_{\Omega}V_{n+1}d\Omega+n\cos\theta\frac{\left(\int_{T}V_{n+1}dA_{T}\right)^{2}}{\int_{\partial\Sigma}\overline{g}(-\cos\phi\widetilde{x}+\sin\phi
a,\mu)ds}$ $\displaystyle=$
$\displaystyle(n+1)\int_{\Omega}V_{n+1}d\Omega+\cos\theta\frac{\left(\int_{T}V_{n+1}dA_{T}\right)^{2}}{H_{1}\int_{\Sigma}\overline{g}(\cos\phi\widetilde{x}-\sin\phi
a,\nu)dA}$ $\displaystyle=$
$\displaystyle(n+1)\int_{\Omega}V_{n+1}d\Omega+\frac{\sec\theta\cos\phi}{H_{1}}\int_{\Sigma}\overline{g}(\widetilde{x},\nu)dA.$

From (11), we have

$\displaystyle(n+1)V_{n+1}d\Omega=$
$\displaystyle\int_{\Omega}\overline{\mathrm{div}}X_{n+1}d\Omega$
$\displaystyle=$
$\displaystyle\int_{\Sigma}\overline{g}(X_{n+1},\nu)dA+\int_{T}\overline{g}(X_{n+1},\overline{N})dA_{T}$
$\displaystyle=$ $\displaystyle\int_{\Sigma}\overline{g}(X_{n+1},\nu)dA.$

Now the rest is the same as the prove of Theorem 3. Together with Minkowski
type formula (21), we prove that $\Sigma$ is umbilical. ∎

######  Remark.

Indeed, the following example indicate that there exists a totally umbilical
capillary hypersurface in $B^{\mathrm{int}}_{\phi,a}$ without any convex
point. Consider a totally geodesic hypersurface is given as follows

$\Sigma=\\{\widetilde{x}\in\mathbb{H}_{n+1}:\delta(\widetilde{x},\widetilde{x})=2\\},$

and the supporting hypersurface is $L_{\frac{\pi}{6}}$. Then the totally
geodesic $\Sigma$ is a capillary hypersurface supported on $L_{\frac{\pi}{6}}$
and the contact angle $\theta=\arccos\frac{\sqrt{6}}{4}\approx 0.912$, but it
is apparent to have not any convex point. Also we can see that
$\theta+\phi\approx 1.435<\frac{\pi}{2}$, which is not satisfying the
condition in Theorem 4. See Figure 8.

Figure 8. Capillary hypersurfaces in $B^{\mathrm{int}}_{\phi,a}$ without any
convex point

We can see from the proof of Theorem 4 that the assumption
“$\phi+\theta>\frac{\pi}{2}$ for $\phi>0$ or $\theta=\frac{\pi}{2}$ for
$\phi=0$” can be replaced by $H_{1}>0$. Then, we have the following corollary.

######  Corollary 1.

Let $\Sigma$ be a compactly embedded CMC capillary hypersurface contained in
$B^{\mathrm{int}}_{\phi,a}$ supported on $L_{\phi,a}$, and the contact angle
satisfies $\theta\in(0,\frac{\pi}{2}]$. Assume $H_{1}>0$, then $\Sigma$ is
umbilical except for being totally geodesic.

##  6\. Capillary hypersurfaces in a geodesic ball

In this section, we give an Alexandrov type theorem for capillary
hypersurfaces in a geodesic ball. Consider a geodesic ball $B_{R}$ centered at
the origin in the Poincaré ball model, where $R$ the hyperbolic radius. Then
$B_{R}$ can be given by

$B_{R}=\\{x\in\mathbb{H}^{n+1}:g(x,x)\leq
R^{2}\\}=\\{x\in\mathbb{H}^{n+1}:|x|^{2}\leq R_{\delta}^{2}\\},$

where $R_{\delta}$ is the Euclidean radius of $B_{R}$. Thus, the following may
be easily noted:

$\cosh R=\frac{1+R_{\delta}^{2}}{1-R_{\delta}^{2}}\quad\mbox{ and }\quad\sinh
R=\frac{2R_{\delta}}{1-R_{\delta}^{2}}.$

Then the unit outward normal $\overline{N}$ of $B_{R}$ satisfies that

(47) $\overline{N}=\frac{1}{\sinh R}x.$

In the paper of Wang and Xia (See [25]), the authors give a family of
conformal Killing vector fields $X_{a}$. It is given by

$X_{a}=\frac{2}{1-R_{\delta}^{2}}\left[\delta(x,a)x-\frac{1}{2}(|x|^{2}+R_{\delta}^{2})a\right],$

where $a$ is a constant vector with respect to the Euclidean metric
$(\mathbb{B}^{n+1},\delta)$ satisfying that

$\frac{1}{2}\left(\overline{g}(\overline{\nabla}_{\overline{E}_{A}}X_{a},\overline{E}_{B})+\overline{g}(\overline{\nabla}_{\overline{E}_{B}}X_{a},\overline{E}_{A})\right)=V_{a}\overline{g},$

where $V_{a}=\frac{2\delta(x,a)}{1-|x|^{2}}$ satisfying
$\overline{\nabla}^{2}V_{a}=V_{a}\overline{g}$. A direct computation shows
that

(48) $\cosh RX_{a}=V_{a}x-\sinh^{2}RY_{a}.$

$\displaystyle B_{R}$$\displaystyle
O$$\overline{N}$$\mu$$\nu$$\overline{\nu}$$\Sigma$ Figure 9. Capillary
hypersurface $\Sigma$ supported on a geodesic ball $B_{R}$

We have the following properties analogously,

######  Proposition 7.

Let $\Sigma$ be an embedded capillary hypersurface supported on the geodesic
ball $B_{R}$. Let $\Omega$ be the domain enclosed by $\Sigma$ and $\partial
B_{R}$ and $\partial\Omega=\Sigma\cup T$, we have

(49) $\int_{T}V_{a}dA_{T}=-\sinh R\int_{\Sigma}\overline{g}(Y_{a},\nu)dA,$
(50)
$\int_{\partial\Sigma}\overline{g}(Y_{a},\mu)ds=-\int_{\Sigma}nH_{1}\overline{g}(Y_{a},\nu)dA,$
(51) $\int_{\Sigma}\left(V_{a}+\cos\theta\sinh
R\overline{g}(Y_{a},\nu)-H_{1}\overline{g}(X_{a},\nu)\right)dA=0,$

and

(52) $\int_{\Sigma}n\sinh
R\cos\theta\overline{g}(Y_{a},\nu)dA=\int_{\partial\Sigma}\overline{g}(X_{a},\mu)ds.$

###### Sketch of the proof.

The proof of (49) and (50) can be seen easily from the proof of other
supporting hypersurfaces cases, using the fact that $Y_{a}$ is a Killing
vector field, and (51) and (52) are given in the proof of [25, Prop 4.4]. ∎

Consider the following mixed boundary problem

(53) $\left\\{\begin{array}[]{ll}\overline{\Delta}f-(n+1)f=1&\mbox{ in
}\Omega;\\\ f=0&\mbox{ on }\Sigma;\\\ f_{\overline{N}}-\coth Rf=c_{0}&\mbox{
on }T.\end{array}\right.$

Where
$c_{0}=-\frac{n}{n+1}\cos\theta\frac{\int_{T}V_{a}dA_{T}}{\int_{\partial\Sigma}\sinh
R\overline{g}(Y_{a},\mu)ds}$ is chosen. The existence and regularity of the
solution to the problem (53) can be obtain from Lieberman’s theory. See the
Appendix.

######  Theorem 8.

Let $B_{R,a+}=\\{x\in B_{R}:V_{a}>0\\}$ be a half geodesic ball where $V_{a}$
is positive. Let $\Sigma\subset B_{R,a+}$ be a compact, embedded capillary
hypersurface in $B_{R,a+}$, and the supporting hypersurface is $\partial
B_{R}$. Let $\Omega$ be the domain enclosed by $\Sigma$ and $L_{\phi,a}$. If
the contact angle $\theta\in(0,\frac{\pi}{2}]$, and the mean curvature
$H_{1}>0$ on $\Sigma$, we have

(54)
$\displaystyle\int_{\Sigma}\frac{V_{a}}{H_{1}}dA\geq(n+1)\int_{\Omega}V_{a}d\Omega+n\cos\theta\frac{\left(\int_{T}V_{a}dA_{T}\right)^{2}}{\int_{\partial\Sigma}\sinh
R\overline{g}(Y_{a},\mu)ds},$

where $\mu$ is the outward unit conormal vector field of $\partial\Sigma$ in
$\Sigma$. In addition, the equality holds if and only if $\Sigma$ is
umbilical.

###### Proof.

Let $V=V_{a}$ and $f$ be the solution of (53) in (25). Using the relations
(47), (48) and Proposition 7 and noting that $(V_{a})_{\overline{N}}=\coth
RV_{a}$, we have

$\displaystyle\frac{n}{n+1}\int_{\Omega}V_{a}d\Omega\geq$
$\displaystyle\int_{\Sigma}nV_{a}H_{1}f_{\nu}^{2}dA+c_{0}\int_{T}(V_{a}\triangle
f-f\triangle V_{a})dA_{T}+c_{0}^{2}\int_{T}n\coth RV_{a}dA_{T}$
$\displaystyle=$
$\displaystyle\int_{\Sigma}V_{a}H_{1}f_{\nu}^{2}dA+c_{0}\int_{\partial\Sigma}V_{a}f_{\overline{\nu}}ds+c_{0}^{2}\int_{T}n\coth
RV_{a}dA_{T}$ $\displaystyle=$
$\displaystyle\int_{\Sigma}nV_{a}H_{1}f_{\nu}^{2}dA-\frac{c_{0}^{2}}{\cos\theta}\int_{\partial\Sigma}V_{a}\overline{g}(\mu,\frac{1}{\sinh
R}x)ds$ $\displaystyle-c_{0}^{2}\cosh
R\int_{\Sigma}n\overline{g}(Y_{a},\nu)dA$ $\displaystyle=$
$\displaystyle\int_{\Sigma}nV_{a}H_{1}f_{\nu}^{2}dA-\frac{c_{0}^{2}}{\cos\theta}\int_{\partial\Sigma}\overline{g}(\mu,\coth
RX_{a}+\sinh RY_{a})ds$
$\displaystyle+\frac{c_{0}^{2}}{\cos\theta}\int_{\partial\Sigma}\coth
R\overline{g}(Y_{a},\mu)ds$ $\displaystyle=$
$\displaystyle\int_{\Sigma}nV_{a}H_{1}f_{\nu}^{2}dA-\frac{c_{0}^{2}}{\cos\theta}\int_{\partial\Sigma}\sinh
R\overline{g}(\mu,Y_{a})ds.$

Therefore,

(55)
$\frac{n}{n+1}\left(\int_{\Omega}V_{a}d\Omega+\frac{n}{n+1}\cos\theta\frac{\left(\int_{T}V_{a}dA_{T}\right)^{2}}{\int_{\partial\Sigma}\sinh
R\overline{g}(Y_{a},\mu)ds}\right)\geq\int_{\Sigma}nV_{a}H_{1}f_{\nu}^{2}dA.$

On the other hand,

$\displaystyle\int_{\Omega}V_{a}d\Omega=\int_{\Omega}(V_{a}\overline{\triangle}f-f\overline{\triangle}V_{a})d\Omega=\int_{\Sigma}V_{a}f_{\nu}dA+c_{0}\int_{T}V_{a}dA_{T}.$

We have

(56) $\displaystyle\int_{\Sigma}V_{a}f_{\nu}dA=$
$\displaystyle\int_{\Omega}V_{a}d\Omega-c_{0}\int_{T}V_{a}dA_{T}$
$\displaystyle=$
$\displaystyle\int_{\Omega}V_{a}d\Omega+\frac{n}{n+1}\cos\theta\frac{\left(\int_{T}V_{a}dA_{T}\right)^{2}}{\int_{\partial\Sigma}\sinh
R\overline{g}(Y_{a},\mu)ds}.$

Hence, using (55) and (56) and the similar argument as the proof of Theorem 1,
we prove the inequality in Theorem 8. The equality case is also the same as
Theorem 1. ∎

Using (49), (50), (51) in Proposition 7 and the fact that
$\overline{g}(X_{a},\overline{N})=0$, we can prove the following Alexandrov
type theorem.

######  Theorem 9.

Let $\Sigma$ be an embedded CMC capillary hypersurface contained in $B_{R,a+}$
supported on $\partial B_{R}$. If $\theta\in(0,\frac{\pi}{2}]$ and $H_{1}>0$,
then $\Sigma$ is umbilical except for being totally geodesic.

######  Remark.

Since we do not have the existence of convex points on $\Sigma$, we assume
that $H_{1}>0$.

##  7\. Other rigidity results for capillary hypersurfaces

In this section, we will prove the Theorem 5 and Theorem 6. The proof of
Theorem 5 is due to the Minkowski type formula in Proposition 3.

###### The proof of Theorem 5.

Since $(H_{k}/H_{l})(p)=\alpha>0$, from Newton-Maclaurin inequality, we have

$\displaystyle H_{k-1}H_{l}\geq H_{k}H_{l-1}=\alpha H_{l}H_{l-1}.$

and since $H_{l}>0$ are positive on $\Sigma$, we have $H_{k-1}-\alpha
H_{l-1}\geq 0$. On the other hand, applying Proposition 3, we have

(57) $\displaystyle 0=$
$\displaystyle\int_{\Sigma}\left[H_{k-1}(V_{0}-\cos\theta\overline{g}(Y_{n+1},\nu))-H_{k}\overline{g}(x,\nu)\right]dA$
$\displaystyle=$
$\displaystyle\int_{\Sigma}\left[H_{k-1}(V_{0}-\cos\theta\overline{g}(Y_{n+1},\nu))-\alpha
H_{l}\overline{g}(x,\nu)\right]dA$ $\displaystyle=$
$\displaystyle\int_{\Sigma}(H_{k-1}-\alpha
H_{l-1})(V_{0}-\cos\theta\overline{g}(Y_{n+1},\nu))dA.$

Now, we consider the term $V_{0}-\cos\theta\overline{g}(Y_{n+1},\nu)$. We see
that

$\displaystyle\overline{g}(Y_{n+1},Y_{n+1})=$
$\displaystyle\frac{1}{4}(1+|x|^{2})^{2}\overline{g}(E_{n+1},E_{n+1})+\langle
x,E_{n+1}\rangle^{2}\overline{g}(x,x)$ $\displaystyle-\langle
x,E_{n+1}\rangle(1+|x|^{2})\overline{g}(x,E_{n+1})$ $\displaystyle=$
$\displaystyle\frac{(1+|x|^{2})^{2}-4\langle
x,E_{n+1}\rangle^{2}}{(1-|x|^{2})^{2}}.$

Since $\langle x,E_{n+1}\rangle>0$ on $\mbox{int}(\Sigma)$, we have

$|\overline{g}(Y_{n+1},\nu)|\leq\sqrt{\overline{g}(Y_{n+1},Y_{n+1})}<V_{0},$

and therefore

(58) $\displaystyle V_{0}-\cos\theta\overline{g}(Y_{n+1},\nu)>0.$

Hence, combining (57) and (58), we have $H_{k-1}-\alpha H_{l-1}=0$. This
results in

(59) $\displaystyle H_{k-1}H_{l}=H_{k}H_{l-1}.$

Hence, examining the equality condition of Newton-Maclaurin inequality, we
obtain that $\Sigma$ is totally umbilical, completing the proof of Theorem 5.
∎

Now we give a proof of Theorem 6.

###### Proof of Theorem 6.

Let $G$ be a smooth function defined by $G=nH_{1}V_{0}-n\overline{g}(x,\nu)$.
Integrating the Laplacian of $G$, we have

$\displaystyle\int_{\Sigma}\Delta GdA=$
$\displaystyle\int_{\partial\Sigma}\nabla_{\mu}Gds$ $\displaystyle=$
$\displaystyle\int_{\partial\Sigma}(nH_{1}-nh(\mu,\mu))\overline{g}(x,\mu))ds$
$\displaystyle=$
$\displaystyle\cos\theta\int_{\partial\Sigma}(nH_{1}-nh(\mu,\mu))\overline{g}(x,\overline{\nu})ds.$

For any $e\in T(\partial\Sigma)$, we can see from(6) that

(60) $\displaystyle
h(e,e)=\overline{g}(\overline{\nabla}_{e}\nu,e)=\overline{g}(\overline{\nabla}_{e}(\cos\theta\overline{N}+\sin\theta\overline{\nu}),e)=\sin\theta
h^{\partial\Sigma}(e,e).$

By letting $k=1$ in (18), we have

$\displaystyle\int_{\Sigma}n\overline{g}(Y_{n+1},\nu)dA=\int_{\partial\Sigma}\overline{g}(x,\overline{\nu})ds.$

Then we have

(61) $\displaystyle\int_{\Sigma}\Delta GdA=$
$\displaystyle\cos\theta\int_{\partial\Sigma}(nH_{1}-nh(\mu,\mu))\overline{g}(x,\overline{\nu})ds$
$\displaystyle=$
$\displaystyle\cos\theta\int_{\partial\Sigma}(n(nH_{1}-h(\mu,\mu))-n(n-1)H_{1})\overline{g}(x,\overline{\nu})ds$
$\displaystyle=$
$\displaystyle\cos\theta\int_{\partial\Sigma}(n(n-1)\sin\theta
H^{\partial\Sigma}_{1}-n(n-1)H_{1})\overline{g}(x,\overline{\nu})ds$
$\displaystyle=$
$\displaystyle\cos\theta\left[\int_{\partial\Sigma}n(n-1)\sin\theta
H^{\partial\Sigma}_{1}\overline{g}(x,\overline{\nu})ds-n^{2}(n-1)H_{1}\right.$
$\displaystyle\quad\left.\times\int_{\Sigma}\overline{g}(Y_{n+1},\nu)dA\right]$
$\displaystyle=$
$\displaystyle\cos\theta\int_{\partial\Sigma}(n(n-1)\sin\theta
H^{\partial\Sigma}_{1}\overline{g}(x,\overline{\nu})-n(n-1)\sin\theta
V_{0})ds$ $\displaystyle=$ $\displaystyle 0.$

We use (36) in the fifth equality. In the last equality, we use a Minkowski
type formula for closed hypersurface in $P$ (See [2]), which is the
$(n-1)$-dimensional hyperbolic space.

On the other hand, it is well known that in hypersurface

$\displaystyle\Delta V_{0}=nV_{0}-nH_{1}\overline{\nabla}_{\nu}V_{0},$

and

$\displaystyle\Delta\overline{g}(x,\nu)=nH_{1}V_{0}-|h|^{2}\overline{g}(X,\nu)-n\overline{\nabla}_{\nu}V+n\overline{g}(X,\nu).$

Together with (61), we have

$\displaystyle 0=\int_{\Sigma}\Delta
GdA=\int_{\Sigma}n(|h|^{2}-nH_{1}^{2})\overline{g}(x,\nu)dA.$

Therefore, from the star-shapedness of $\Sigma$, we can see that $\Sigma$ is
umbilical. In addition, the totally geodesic case is excluded by the
compactness assumption on $\Sigma$. ∎

######  Remark.

All of our results require that $\Sigma$ is compact. Because $\Sigma$ is
totally umbilical and contained in $\mathbb{B}^{n+1}_{+}$, it can be compact
and a part of a horosphere or an equidistant hypersurface simultaneously. If
so, we notice that the contact angle $\theta$ must lie in the open interval
$(0,\frac{\pi}{2})$. See the Figure 10 and Figure 11.

$\displaystyle\mathbb{H}^{n+1}$$\displaystyle
P$$\displaystyle\Sigma$$\displaystyle O$ Figure 10. $\Sigma$ is a horosphere
$\displaystyle\mathbb{H}^{n+1}$$\displaystyle
P$$\displaystyle\Sigma$$\displaystyle O$ Figure 11. $\Sigma$ is an equidistant
hypersurface

######  Remark.

Since the star-shapedness of $\Sigma$ implies that $\Sigma$ is an embedded
hypersurface, Theorem 6 can be obtained immediately from Theorem 3 when the
contact angle $\theta\in(0,\frac{\pi}{2}]$.

## Appendix

To obtain the existence and sufficient regularity of the solution of the
mixed-boundary problem (29), (44) and (53), we write it in the conformal
Euclidean space as

$\displaystyle\left\\{\begin{array}[]{ll}\mathcal{L}f:=e^{-2u}(\overline{\Delta}_{\delta}f+(n-1)du(f))-(n+1)f=1&\mbox{
in }\Omega;\\\ f=0&\mbox{ on }\Sigma;\\\ f_{\overline{N}}-\gamma
f=c_{0}&\mbox{ on }T,\end{array}\right.$

where $\overline{\nabla}_{\delta}$ and $\overline{\Delta}_{\delta}$ are the
Levi-Civita connection and Laplacian with respect to the Euclidean metric
$\delta$ respectively. Since the conformal map preserves the angle, the
condition of $\theta$ is the same under both Euclidean metric and hyperbolic
metric.

We notice that the coefficients
$a_{ij}=e^{-2u}\delta_{ij},\,b^{i}=(n-1)e^{-2u}u^{i},\;c=-(n+1)$ satisfy the
conditions in [14], page 435, and $\gamma\geq 0$ satisfies the condition in
[23, Lemma 4.1]. From the theory of Lieberman and Theorem A.3 in [10], there
exists a solution $f\in
C^{\infty}(\overline{\Omega}\setminus\partial\Sigma)\cup
C^{2}(\mathrm{int}(\Omega)\cup T)$ with respect to the conformal metric
$\delta$.

Let $d_{\partial\Sigma}$ be the distance function from $\partial\Sigma$ and
define
$\Omega_{\varepsilon}=\\{x\in\Omega:d_{\partial\Sigma}(x)>\varepsilon\\}$. To
apply the result of Lieberman, we need to introduce the norm

$\displaystyle|f|^{b}_{a}=\sup_{\varepsilon>0}\varepsilon^{a+b}|f|_{a,\Omega_{\varepsilon}},$

where
$|f|_{a,\Omega_{\varepsilon}}=\sum\limits_{i=1}^{k}\sup_{x\in\Omega_{\varepsilon}}|\nabla_{\delta}^{k}f|+\sup_{x,y\in\Omega_{\varepsilon}}\frac{|D^{k}f(x)-D^{k}f(y)|}{|x-y|^{\alpha}}$
for $a>0$ and $a=k+\alpha$ for an integer $k$ and $0\leq\alpha<1$.

Let $f$ be the solution of the following boundary value problem

$\displaystyle\left\\{\begin{array}[]{ll}\mathcal{L}f=g&\mbox{ in }\Omega;\\\
f=0&\mbox{ on }\Sigma;\\\
\langle\overline{\nabla}_{\delta}f,\overline{N}\rangle=h&\mbox{ on
}T.\end{array}\right.$

From the theory by Lieberman (See Theorem 4 in [15]) and Lemma A.1 in [10], we
have the following estimate

$\displaystyle|f|_{a}^{-\lambda}\leq
C(|g|_{a-2}^{2-\lambda}+|h|_{a-1}^{1-\lambda}+|g|_{0}+|h|_{0}).$

Here we notice that the maximum principle (see [23, Lemma 4.1]) applied in
[10, Lemma A.1] holds for general operator $\mathcal{L}$ with $c=-(n+1)<0$ and
$\gamma\geq 0$. Furthermore, from the proof of Theorem 4 in [15], the
construction of the key Miller’s type barrier requires
$\lambda<\frac{\pi}{2\theta}$ (See [15, Lemma 4.1]).

From the discussion by Jia-Wang-Xia (See [10], Page 8-9 and Lemma A.1), we can
see that $|\overline{\nabla}_{\delta}^{2}f|_{\delta}\in L^{2}(\Omega)$
requires $|\overline{\nabla}^{2}_{\delta}f|_{\delta}\leq
Cd_{\partial\Sigma}^{-\beta}$ for $\beta\in(0,1)$ and $a>2$. Then, we can also
obtain that $|\nabla_{\delta}^{2}f|_{\delta}\in L^{1}(T)$.

Meanwhile in the proof of the Theorem 1, we need the regularity of $f$
satisfying $f\in C^{1,\alpha}(\overline{\Omega})$. From the definition and the
monotonicity of the norm $|\cdot|_{a}^{b}$, we have
$|f|_{\lambda}=|f|_{\lambda}^{-\lambda}\leq|f|_{a}^{-\lambda}$. Therefore it
is sufficient for $f\in C^{1,\alpha}(\overline{\Omega})$ when $\lambda>1$.

In conclusion, all the condition for the applicable regularity will be
satisfied if we let $1<\lambda<\mbox{min}\\{3,\frac{\pi}{2\theta}\\}$,
$a=\frac{\lambda+3}{2}>2$ and $\beta=a-\lambda=\frac{3-\lambda}{2}\in(0,1)$.
Hence, the condition $\theta<\frac{\pi}{2}$ is sufficient. From a well-known
fact, under conformal transformations of metrics we have

$\displaystyle\overline{\nabla}f=e^{-2u}\overline{\nabla}_{\delta}f,$

and

$\displaystyle\overline{\nabla}^{2}_{ij}f=\overline{\nabla}^{2}_{\delta,\;ij}f-u_{i}f_{j}-u_{j}f_{i}-\delta(\overline{\nabla}_{\delta}f,\overline{\nabla}_{\delta}u)\delta_{ij}.$

Then all the estimates above hold with respect to the hyperbolic metric
$\overline{g}$, that is, $f\in
C^{\infty}(\overline{\Omega}\setminus\partial\Sigma)\cup C^{2}(\Omega\cup
T)\cup C^{1,\alpha}(\overline{\Omega})$ and
$|\overline{\nabla}^{2}f|_{\overline{g}}\in L^{1}(T)$. Now we obtain the
existence and the regularity of the solution of (29).

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