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A Constrained Mean Curvature Flow On Capillary Hypersurface Supported on totally geodesic plane

Xiaoxiang Chai Department of Mathematics, POSTECH, Pohang, Gyeongbuk, South Korea [email protected], [email protected]  and  Yimin Chen Department of Mathematics, Pusan National University, Busan, South Korea [email protected]
Abstract.

We prove a new Minkowski type formula for capillary hypersurfaces supported on totally geodesic hyperplanes in hyperbolic space. It leads to a volume-preserving flow starting from a star-shaped initial hypersurface. We prove the long-time existence of the flow and its uniform convergence to a θ\theta-totally umbilical cap. Additionally, we establish that a θ\theta-totally umbilical cap is an energy minimizer for a given enclosed volume.

1. Introduction

Mean curvature flow has a rich history, dating back to significant works such as Huisken [13]. Huisken showed that a convex and closed hypersurface will flow to a sphere under the properly rescaled mean curvature flow. A constrained curvature flow is a flow which preserves some geometric quantities. In 2\mathbb{R}^{2}, Gage  [5] used a constrained curve shortening flow to prove an isoperimetric inequality. In higher dimensions, a constrained mean curvature flow was applied to prove the isoperimetric inequality by Huisken [14]. This constrained flow preserves the enclosed volume while decreasing the area of the hypersurface.

An alternative approach to create a flow that preserves the enclosed volume is by employing the Minkowski formula on the hypersurface. This has been explored in [6], which investigated such flows in space forms. Furthermore, this approach has been extended to warped product spaces in [7], where they considered the flow x:(×N,g)(M¯,g¯=dr2+ϕ2(r)gN)x:(\mathbb{R}\times N,g)\rightarrow(\bar{M},\bar{g}=dr^{2}+\phi^{2}(r)g_{N}) satisfying

xt=(nϕuH)ν.\frac{\partial x}{\partial t}=(n\phi^{\prime}-uH)\nu.

Under appropriate assumptions on the metric g¯\bar{g}, the flow is expected to converge to a level set of ϕ\phi. See for example [1], [2], [3], [9] and [10] for various types of fully nonlinear curvature flow and anisotropic curvature flow in different ambient spaces.

In recent years, there has been considerable interest in geometric flows of capillary hypersurfaces, for instance, inverse mean curvature flow with free boundary in the Euclidean unit ball [15].

Diving into the main topic of this paper, constrained curvature flow on capillary hypersurfaces has yielded significant results. These include: constrained inverse mean curvature type [19], [23] and mean curvature type flow [12], [20] for capillary hypersurfaces in the Euclidean unit ball; curvature flows [11], [16] and [21] in capillary hypersurfaces in Euclidean half space;  mean curvature type flow [17] in geodesic ball in space forms.

In this paper, we consider a new constrained mean curvature type flow for capillary hypersurfaces, which are supported on totally geodesic hyperplanes in hyperbolic space n+1\mathbb{H}^{n+1}. We use the well-known Poincare´\acute{\mathrm{e}} half space model (n+1,,)=(+n+1,1xn+12,δ)(\mathbb{H}^{n+1},\langle\cdot,\cdot\rangle)=\left(\mathbb{R}^{n+1}_{+},\frac{1}{x^{2}_{n+1}}\langle\cdot,\cdot\rangle_{\delta}\right), where xn+1x_{n+1} is the (n+1)(n+1)-th coordinate, ,δ\langle\cdot,\cdot\rangle_{\delta} is the Euclidean metric and +n+1:={xn+1:xn+1>0}\mathbb{R}^{n+1}_{+}:=\{x\in\mathbb{R}^{n+1}:x_{n+1}>0\}. Let P:={xn+1:x1=0}P:=\{x\in\mathbb{H}^{n+1}:x_{1}=0\} be a totally geodesic hyperplane. Denote by xx the position vector in +n+1\mathbb{R}^{n+1}_{+}and {Ei}i=1n+1\{E_{i}\}_{i=1}^{n+1} the coordinate basis of (n+1,,δ)(\mathbb{R}^{n+1},\langle\cdot,\cdot\rangle_{\delta}).

Throughout the paper, we consider x0:Mn+1x_{0}:M\rightarrow\mathbb{H}^{n+1} as an immersion of hypersurface in n+1\mathbb{H}^{n+1}. If Σ=x(M)\Sigma=x(M) satisfies that

  1. i.

    intΣ=x(intM)P+:={xn+1:x1>0}\operatorname{int}\Sigma=x(\operatorname{int}M)\subset P_{+}:=\{x\in\mathbb{H}^{n+1}:x_{1}>0\},

  2. ii.

    Σ=x(M){xn+1:x1=0}=P\partial\Sigma=x(\partial M)\subset\{x\in\mathbb{H}^{n+1}:x_{1}=0\}=P,

  3. iii.

    Σ\Sigma and PP contacts at a constant angle θ\theta on Σ=ΣP\partial\Sigma=\Sigma\cap P,

we call Σ\Sigma a θ\theta-capillary hypersurface supported on PP, and PP the supporting hypersurface.

On a θ\theta-capillary hypersurface supported on totally geodesic hyperplane PP, we introduce a novel condition called star-shapedness with respect to cEn+1cE_{n+1}.

Definition 1.

Let x:MP+n+1x:M\rightarrow P_{+}\subset\mathbb{H}^{n+1} be a θ\theta-capillary hypersurface supported on PP. We say Σ\Sigma is star-shaped with respect to cEn+1cE_{n+1} if it satisfies that

xcEn+1,ν>0.\langle x-cE_{n+1},\nu\rangle>0.

We consider a flow, defined as a family of embeddings x:M×[0,T)n+1x:M\times[0,T)\rightarrow\mathbb{H}^{n+1} with x(Σ,)Px(\partial\Sigma,\cdot)\subset P, such that

(1) (tx)=qcν,inM×[0,T);ν,N¯x=cosθ,onM×[0,T);x(,0)=x0(),onM,\begin{array}[]{rclccl}(\partial_{t}x)^{\bot}&=&q_{c}\nu,&&&\operatorname{in}M\times[0,T);\\ \langle\nu,\bar{N}\circ x\rangle&=&-\cos\theta,&&&\operatorname{on}\partial M\times[0,T);\\ x(\cdot,0)&=&x_{0}(\cdot),&&&\operatorname{on}M,\end{array}

where the normal velocity qcq_{c} is given by

qc=ncxn+1nccosθE1,νHxcEn+1,ν.q_{c}=\frac{nc}{x_{n+1}}-nc\cos\theta\langle E_{1},\nu\rangle-H\langle x-cE_{n+1},\nu\rangle.

We denote by κ\kappa the principal curvature of an umbilical hypersurface 𝒞\mathcal{C}. Umbilical hypersurfaces in hyperbolic space can be classified into three types depending on κ\kappa as depicted in the figure. In the case of κ=0\kappa=0, it is a totally geodesic hyperplane; in the case of κ>1\kappa>1, it is a geodesic sphere, and for 0<κ<10<\kappa<1, it is an equidistant hypersurface, and if κ=1\kappa=1, it is a horosphere. In the Poincaré half space model, 𝒞\mathcal{C} can be represented as a plane or sphere with respect to the Euclidean metric. It is easy to see that compact umbilical θ\theta-capillary hypersurface, can be part of a geodesic sphere, a horosphere and an equidistant hypersurface.

Refer to caption
Figure 1. 𝒞1\mathcal{C}_{1}, equidistant hypersurface; 𝒞2\mathcal{C}_{2}, geodesic ball; 𝒞3\mathcal{C}_{3}, horosphere.

To state our main theorem, we need the following definition.

Definition 2.

We define the θ\theta-umbilical cap 𝒞c,R,θ\mathcal{C}_{c,R,\theta} as follows

(2) 𝒞c,R,θ(a)={xn+1:|xRcosθE1acEn+1|δR,x10},\mathcal{C}_{c,R,\theta}(a)=\{x\in\mathbb{H}^{n+1}:|x-R\cos\theta E_{1}-a-cE_{n+1}|_{\delta}\leq R,x_{1}\geq 0\},

where aa is a constant vector perpendicular to both E1E_{1} and En+1E_{n+1}.

For a θ\theta-umbilical cap, we define a constant K0(c,R,θ)K_{0}(c,R,\theta) by

K0(c,R,θ)={cRsinθθπ/2cRθ>π/2.K_{0}(c,R,\theta)=\left\{\begin{array}[]{llll}c-R\sin\theta&&&\theta\leq\pi/2\\ c-R&&&\theta>\pi/2\end{array}\right..

Note that K0(c,R,θ)>0K_{0}(c,R,\theta)>0 if and only if the 𝒞c,R,θ(a)\mathcal{C}_{c,R,\theta}(a) is compact. Combining with Remark 2, we know that its principal curvature κ=cR>sinθ\kappa=\frac{c}{R}>\sin\theta.

Now we are ready to state our main theorem.

Theorem 1.

Let x0:MP+n+1x_{0}:M\rightarrow P_{+}\subset\mathbb{H}^{n+1} be an embedding of a compact capillary hypersurface Σ0=x0(M)\Sigma_{0}=x_{0}(M), supported on the totally geodesic plane PP with constant contact angle θ\theta. Suppose there exist constants c, R such that K0(c,R,θ)>c(n1)/4nK_{0}(c,R,\theta)>c(n-1)/4n, and  Σ0\Sigma_{0} is contained in the cap 𝒞c,R,θ(a)\mathcal{C}_{c,R,\theta}(a) and star-shaped with respect to cEn+1cE_{n+1}. We assume that in addition, θ\theta satisfies that

(3) |cosθ|<4nK0(c,R,θ)c(n1)4nK0(c,R,θ)+c(n1).|\cos\theta|<\frac{4nK_{0}(c,R,\theta)-c(n-1)}{4nK_{0}(c,R,\theta)+c(n-1)}.

Then the flow (1) exists globally with uniform CC^{\infty}-estimates. Moreover, x(,t)x(\cdot,t) uniformly converges to an umbilical cap in CC^{\infty} topology as tt\rightarrow\infty, with the same volume of its enclosed domain as Σ0\Sigma_{0}.

Remark 1.

In [17] and [20], the angle condition |cosθ|3n+15n1|\cos\theta|\leq\frac{3n+1}{5n-1} is required similar to our angle condition (3). By rearranging the terms, we find that the condition (3) is equivalent to

|cosθ|<3n+14nsinθR/c5n14nsinθR/c,ifθπ/2;|\cos\theta|<\frac{3n+1-4n\sin\theta R/c}{5n-1-4n\sin\theta R/c},\quad\operatorname{if}\hskip 11.99998pt\theta\leq\pi/2;

and

|cosθ|<3n+14nR/c5n14nR/c,ifθ>π/2.|\cos\theta|<\frac{3n+1-4nR/c}{5n-1-4nR/c},\quad\operatorname{if}\hskip 11.99998pt\theta>\pi/2.

Compared to the condition in [17] and [20], our condition is stricter due to R/cR/c.

The following remark is crucial, which shows that some specific umbilical caps are static along the flow (1).

Remark 2.

For any r>0r>0, qc,θq_{c,\theta} is identically zero on umbilical cap 𝒞c,r,θ(a)\mathcal{C}_{c,r,\theta}(a). It is well-known that the mean curvature by conformality can be written as

H\displaystyle H =\displaystyle= ew(Hδ+n~ν~w)\displaystyle e^{-w}(H_{\delta}+n\tilde{\nabla}_{\tilde{\nu}}w)
=\displaystyle= nxn+1(1r+~ν~log1xn+1)\displaystyle nx_{n+1}\left(\frac{1}{r}+\tilde{\nabla}_{\tilde{\nu}}\log\frac{1}{x_{n+1}}\right)
=\displaystyle= nxn+1(1rxn+11x+rcosθE1cEn+1r,En+1δ)\displaystyle nx_{n+1}\left(\frac{1}{r}-x^{-1}_{n+1}\langle\frac{x+r\cos\theta E_{1}-cE_{n+1}}{r},E_{n+1}\rangle_{\delta}\right)
=\displaystyle= ncr,\displaystyle n\frac{c}{r},

where ν~=r1(x+rcosθE1cEn+1)\tilde{\nu}=r^{-1}(x+r\cos\theta E_{1}-cE_{n+1}) and we use the fact that the mean curvature of 𝒞c,r,θ(0)\mathcal{C}_{c,r,\theta}(0) with respect to the metric δ\delta is Hδ=nrH_{\delta}=\frac{n}{r}. Hence, we get

qc\displaystyle q_{c} =\displaystyle= ncxn+1nE1,νccosθHxcEn+1,ν\displaystyle\frac{nc}{x_{n+1}}-n\langle E_{1},\nu\rangle c\cos\theta-H\langle x-cE_{n+1},\nu\rangle
=\displaystyle= ncxn+1ncxn+11ν¯1cosθncrrcosθE1+cEn+1+rν¯cEn+1,xn+1ν¯\displaystyle\tfrac{nc}{x_{n+1}}-ncx_{n+1}^{-1}\bar{\nu}_{1}\cos\theta-\tfrac{nc}{r}\langle-r\cos\theta E_{1}+cE_{n+1}+r\bar{\nu}-cE_{n+1},x_{n+1}\bar{\nu}\rangle
=\displaystyle= nxn+1(cν¯1cosθ+crcosθrν¯1)\displaystyle\tfrac{n}{x_{n+1}}(-c\bar{\nu}_{1}\cos\theta+c\tfrac{r\cos\theta}{r}\bar{\nu}_{1})
=\displaystyle= 0.\displaystyle 0.

Therefore umbilical caps are static along the flow (1).

Note that the principal curvature of an umbilical cap depends not only on the radius but also on the last coordinate of its center (in the Euclidean metric sense).

The paper is organized as follows:

In Section 2, we introduce basic notations and definitions of hypersurfaces. In Section 3, we prove a new Minkowski formula on capillary hypersurface supported on a totally geodesic hyperplane, comparing to the one in [4]. In Sections 4, 5 and 6, we follow the method in [20] and [16] to study the scalar equation of the flow (1), in particular, we prove the C0C^{0} and C1C^{1} estimates. In Section 7, we prove the uniform convergence of the flow.

Acknowledgments .

We would like to thank Juncheol Pyo for helpful comments and hospitality. X. Chai has been partially supported by National Research Foundation of Korea grant No. 2022R1C1C1013511. Y. Chen has been supported by National Research Foundation of Korea grant No. RS-2023-00247299 and partially supported by National Research Foundation of Korea grant No. NRF-2020R1A01005698.

2. Preliminaries

Throughout this paper, we assume that x:MP+n+1x:M\rightarrow P_{+}\subset\mathbb{H}^{n+1} is an embedding of capillary hypersurface along PP. We denote the second fundamental form of the embedding by hijh_{ij}. Since ν\nu is the outer normal field on Σ\Sigma, hij=eiν,ejh_{ij}=\langle\nabla_{e_{i}}\nu,e_{j}\rangle. Let κ=(κ1,,κn)\kappa=(\kappa_{1},\cdots,\kappa_{n}) be the eigenvalues of (hij)(h_{ij}), i.e., the principal curvatures of Σ\Sigma. The kk-th mean curvature SkS_{k} is defined by

Sr=1r!1i1<<irnκ1κ2κir,S_{r}=\frac{1}{r!}\sum_{1\leq i_{1}<\cdots<i_{r}\leq n}\kappa_{1}\kappa_{2}\cdots\kappa_{i_{r}},

and the normalized mean curvature is defined by Hr=(nr)1SrH_{r}=\binom{n}{r}^{-1}S_{r}.

The following so-called Newton transformation defined on the tangent bundle is essential to our formula:

T0=id,T_{0}=\operatorname{id},
Tk=SkITk1h.T_{k}=S_{k}I-T_{k-1}\circ h.

The following properties are well-known,

tr(Tr)=(nr)Sr=(nr)(nr)Hr,\operatorname{tr}(T_{r})=(n-r)S_{r}=(n-r)\binom{n}{r}H_{r},
tr(Trh)=(r+1)Sr+1=(nk)(nr)Hr+1.\operatorname{tr}(T_{r}\circ h)=(r+1)S_{r+1}=(n-k)\binom{n}{r}H_{r+1}.

Let N¯\bar{N} denote the outer normal field of PP (P+P_{+} is the interior side), ν\nu denote the outer normal field of the immersed hypersurface Σ\Sigma, ν¯\bar{\nu} denote the outer normal field of ΣP\partial\Sigma\subset P and η\eta denote the outer conormal field along ΣΣ\partial\Sigma\subset\Sigma. Without loss of generality, assuming θ\theta as the angle between ν-\nu and N¯\bar{N}, we have the following relation

(4) {μ=sinθN¯+cosθν¯ν=cosθN¯+sinθν¯.\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..

Then the following lemma is widely recognized, and we refer to [22] for its proof.

Lemma 1.

Let x:ΣP+n+1x:\Sigma\rightarrow P_{+}\subset\mathbb{H}^{n+1} be an isometric immersion of a capillary hypersurface supported on PP. Then μ\mu is a principal direction of Σ\Sigma, that is,

(5) ¯μν=h(μ,μ)μ.\bar{\nabla}_{\mu}\nu=h(\mu,\mu)\mu.

This property of capillary hypersurfaces is essential in the proof of the Minkowski type formula in the next section.

3. Minkowski type formula

In this section, we introduce a new Minkowski type formula, which is based on the the following properties (see [8]).

Proposition 1.

It satisfies that

(6) ¯Zx=Z,E¯n+1x+Z,xE¯n+1,\bar{\nabla}_{Z}x=-\langle Z,\overline{E}_{n+1}\rangle x+\langle Z,x\rangle\overline{E}_{n+1},
(7) ¯Zx¯=x¯,ZE¯n+1,\bar{\nabla}_{Z}\bar{x}=\langle\bar{x},Z\rangle\bar{E}_{n+1},
(8) ¯ZE1=Z,E¯n+1E1+Z,E1E¯n+1,\bar{\nabla}_{Z}E_{1}=-\langle Z,\overline{E}_{n+1}\rangle E_{1}+\langle Z,E_{1}\rangle\overline{E}_{n+1},
(9) ¯Z(En+1)=1xn+1Z,\overline{\nabla}_{Z}(-E_{n+1})=\frac{1}{x_{n+1}}Z,

and

(10) ¯Z(E¯n+1)=ZE¯n+1,ZE¯n+1.\bar{\nabla}_{Z}(-\overline{E}_{n+1})=Z-\langle\overline{E}_{n+1},Z\rangle\overline{E}_{n+1}.

These can be directly calculated by the relation between Levi-Civita connections of the metrics conformal to each other, we refer the proof to [8]. From (6), (8), and (9), we can easily see that xx, E1E_{1}, and En+1E_{n+1} are conformal Killing vector fields, that is,

(11) (xg¯)(X,Y)=12(¯Yx,X+¯Xx,Y)=0,(\mathcal{L}_{x}\bar{g})(X,Y)=\frac{1}{2}(\langle\overline{\nabla}_{Y}x,X\rangle+\langle\overline{\nabla}_{X}x,Y\rangle)=0,
(12) (E1g¯)(X,Y)=12(¯YE1,X+¯XE1,Y)=0,(\mathcal{L}_{E_{1}}\bar{g})(X,Y)=\frac{1}{2}(\langle\overline{\nabla}_{Y}E_{1},X\rangle+\langle\overline{\nabla}_{X}E_{1},Y\rangle)=0,

and

(13) En+1X,Y=12(¯Y(En+1),X+¯X(En+1),Y)=1xn+1X,Y,\begin{array}[]{ccl}\mathcal{L}_{-E_{n+1}}\langle X,Y\rangle&=&\frac{1}{2}(\langle\overline{\nabla}_{Y}(-E_{n+1}),X\rangle+\langle\overline{\nabla}_{X}(-E_{n+1}),Y\rangle)\\ &=&\frac{1}{x_{n+1}}\langle X,Y\rangle,\end{array}

for any X,YTn+1X,Y\in T\mathbb{H}^{n+1}, where g¯\bar{g} denotes the metric ,\langle\cdot,\cdot\rangle.

Restricting the equation (13) to TΣT\Sigma, we have

(14) eiΣE1,ej=ei,E¯n+1E1,ej+ei,E1E¯n+1,ejhijE1,ν,\langle\nabla^{\Sigma}_{e_{i}}E_{1},e_{j}\rangle=-\langle e_{i},\overline{E}_{n+1}\rangle\langle E_{1},e_{j}\rangle+\langle e_{i},E_{1}\rangle\langle\overline{E}_{n+1},e_{j}\rangle-h_{ij}\langle E_{1}{,}\nu\rangle,
(15) eiΣx,ej=ei,E¯n+1x,ej+ei,xE¯n+1,ejhijx,ν,\langle\nabla^{\Sigma}_{e_{i}}x,e_{j}\rangle=-\langle e_{i},\overline{E}_{n+1}\rangle\langle x,e_{j}\rangle+\langle e_{i},x\rangle\langle\overline{E}_{n+1},e_{j}\rangle-h_{ij}\langle x{,}\nu\rangle,

and

(16) eiΣ(En+1),ej=1xn+1ei,ejhijEn+1,ν.\langle\nabla^{\Sigma}_{e_{i}}(-E_{n+1}),e_{j}\rangle=\frac{1}{x_{n+1}}\langle e_{i},e_{j}\rangle-h_{ij}\langle-E_{n+1}{,}\nu\rangle.

Let x:Σ(n+1,,)=(+n+1,1xn+12,δ)x:\Sigma\rightarrow(\mathbb{H}^{n+1},\langle\cdot,\cdot\rangle)=\left(\mathbb{R}^{n+1}_{+},\frac{1}{x^{2}_{n+1}}\langle\cdot,\cdot\rangle_{\delta}\right) be an immersion of θ\theta-capillary hypersurface supported on the hyperplane PP. By using the facts above, we can prove the following Minkowski type formula on Σ\Sigma.

Proposition 2.

For k=1,,n,k=1,\cdots,n,and c>0c>0, it satisfies that

(17) Σ[ncHk1(1xn+1cosθE1,ν)HkxcEn+1,ν]𝑑A=0,\int_{\Sigma}\left[ncH_{k-1}\left(\frac{1}{x_{n+1}}-\cos\theta\langle E_{1},\nu\rangle\right)-H_{k}\langle x-cE_{n+1},\nu\rangle\right]dA=0,

where HkH_{k} is the kk-th mean curvature of Σ\Sigma.

Proof.

Let TrT_{r} acts on the both side of (15)+c(16)\eqref{divx}+c\eqref{divE} and integrate. Using divergence theorem, we have

(nr+1)(nr1)Σ[Hr1(cxn+1)HrxcEn+1,ν]𝑑A\displaystyle(n-r+1)\binom{n}{r-1}\int_{\Sigma}\left[H_{r-1}\left(\frac{c}{x_{n+1}}\right)-H_{r}\langle x-cE_{n+1},\nu\rangle\right]dA
=\displaystyle= ΣdivΣ(Tr1(xcEn+1))𝑑A\displaystyle\int_{\Sigma}\operatorname{div}_{\Sigma}(T_{r-1}(x-cE_{n+1}))dA
=\displaystyle= ΣTr1(xcEn+1,μ)𝑑A\displaystyle\int_{\partial\Sigma}T_{r-1}(x-cE_{n+1},\mu)dA
=\displaystyle= cosθΣSr1;μxcEn+1,ν¯𝑑s,\displaystyle\cos\theta\int_{\partial\Sigma}S_{r-1;\mu}\langle x-cE_{n+1},\bar{\nu}\rangle ds,

where in the last equality we used (5), the angle relation (4) and the fact that N¯=xn+1E1\bar{N}=-x_{n+1}E_{1}.

Let Z1,n+1=E1,νE¯n+1E¯n+1,νE1Z_{1,n+1}=\langle E_{1},\nu\rangle\bar{E}_{n+1}-\langle\bar{E}_{n+1},\nu\rangle E_{1}, applying Proposition 1, we have

eiZ1,n+1,ej\displaystyle\langle\nabla_{e_{i}}Z_{1,n+1},e_{j}\rangle =\displaystyle= δijE1,ν+hik[E1,ekE¯n+1,ej\displaystyle-\delta_{ij}\langle E_{1},\nu\rangle+h_{ik}[\langle E_{1},e_{k}\rangle\langle\overline{E}_{n+1},e_{j}\rangle
E¯n+1,ekE1,ej].\displaystyle-\langle\overline{E}_{n+1},e_{k}\rangle\langle E_{1},e_{j}\rangle].

Let Tr1T_{r-1} act on both sides, we get

divΣ(Tr1(Z1,n+1))=(nr+1)(nr1)Hr1E1,ν.\operatorname{div}_{\Sigma}(T_{r-1}(Z_{1,n+1}))=-(n-r+1)\binom{n}{r-1}H_{r-1}\langle E_{1},\nu\rangle.

Considering the vector field Z1=g¯(E1,ν)x¯g¯(x¯,ν)E1Z_{1}=\overline{g}(E_{1},\nu)\bar{x}-\overline{g}(\bar{x},\nu)E_{1}, from Proposition 1 we have

¯eiZ1,ej\displaystyle\langle\bar{\nabla}_{e_{i}}Z_{1},e_{j}\rangle =\displaystyle= E¯n+1,ν[ei,E1x¯,ejei,x¯E1,ej]\displaystyle\langle\bar{E}_{n+1},\nu\rangle[\langle e_{i},E_{1}\rangle\langle\bar{x},e_{j}\rangle-\langle e_{i},\bar{x}\rangle\langle E_{1},e_{j}\rangle]
+E1,ν[ei,x¯E¯n+1,ejei,E¯n+1x¯,ej]\displaystyle+\langle E_{1},\nu\rangle[\langle e_{i},\bar{x}\rangle\langle\bar{E}_{n+1},e_{j}\rangle-\langle e_{i,}\bar{E}_{n+1}\rangle\langle\bar{x},e_{j}\rangle]
+hik[E1,ekx¯,ejx¯,ekE1,ej]\displaystyle+h_{ik}[\langle E_{1},e_{k}\rangle\langle\bar{x},e_{j}\rangle-\langle\bar{x},e_{k}\rangle\langle E_{1},e_{j}\rangle]
+x¯,ν[E¯n+1,eiE1,ejE1,eiE¯n+1,ej].\displaystyle+\langle\bar{x},\nu\rangle[\langle\bar{E}_{n+1},e_{i}\rangle\langle E_{1},e_{j}\rangle-\langle E_{1},e_{i}\rangle\langle\bar{E}_{n+1},e_{j}\rangle].

Similarly, we have

divΣ(Tr1(Z1))=0.\operatorname{div}_{\Sigma}(T_{r-1}(Z_{1}))=0.

Combining all the equations above, we have

(nk+1)(nk1)Σ(cHk1xn+1HkxcEn+1,ν)𝑑A\displaystyle(n-k+1)\binom{n}{k-1}\int_{\Sigma}\left(\frac{cH_{k-1}}{x_{n+1}}-H_{k}\langle x-cE_{n+1},\nu\rangle\right)dA
=\displaystyle= cosθΣSk1;μxcEn+1,ν¯𝑑s\displaystyle\cos\theta\int_{\partial\Sigma}S_{k-1;\mu}\langle x-cE_{n+1},\bar{\nu}\rangle ds
=\displaystyle= cosθΣSk1;μZ1cZ1,n+1,μ𝑑s\displaystyle\cos\theta\int_{\partial\Sigma}S_{k-1;\mu}\langle Z_{1}-cZ_{1,n+1},\mu\rangle ds
=\displaystyle= cosθΣTk1(Z1cZ1,n+1,μ)𝑑s\displaystyle\cos\theta\int_{\partial\Sigma}T_{k-1}(Z_{1}-cZ_{1,n+1},\mu)ds
=\displaystyle= cosθΣdivΣ(Tk1(Z1cZ1,n1))𝑑s\displaystyle\cos\theta\int_{\Sigma}\operatorname{div}_{{}_{\Sigma}}(T_{k-1}(Z_{1}-cZ_{1,n-1}))ds
=\displaystyle= (nk+1)(nk1)cosθΣHk1cE1,ν𝑑A.\displaystyle(n-k+1)\binom{n}{k-1}\cos\theta\int_{\Sigma}H_{k-1}\langle cE_{1},\nu\rangle dA.

Therefore, we obtain the Minkowski type formula (17). ∎

Let k=1k=1, the Minkowski formula (17) becomes

(18) Σ[nc(1xn+1cosθE1,ν)HxcEn+1,ν]𝑑A=0,\int_{\Sigma}\left[nc\left(\frac{1}{x_{n+1}}-\cos\theta\langle E_{1},\nu\rangle\right)-H\langle x-cE_{n+1},\nu\rangle\right]dA=0,

which holds for any capillary hypersurfaces Σ\Sigma supported on totally geodesic hyperplane PP. Here H=nH1H=nH_{1} is the mean curvature of Σ\Sigma.

Remark 3.

The second author and Juncheol Pyo [4] gave another version of Minkowski type formula on capillary hypersurfaces supported on a totally geodesic plane, which is presented in Poincaré ball model (n+1,g¯)=(𝔹n+1,4(1|x|2)2δ)(\mathbb{H}^{n+1},\bar{g})=(\mathbb{B}^{n+1},\frac{4}{(1-|x|^{2})^{2}}\delta) as follows (𝔹n+1\mathbb{B}^{n+1} is an Euclidean unit ball and δ\delta is the Euclidean metric in 𝔹n+1\mathbb{B}^{n+1}).

(19) Σ[nV0ncosθg¯(Yn+1,ν)Hg¯(x,ν)]𝑑A=0,\int_{\Sigma}[nV_{0}-n\cos\theta\bar{g}(Y_{n+1},\nu)-H\bar{g}(x,\nu)]dA=0,

where Σ\Sigma is a θ\theta-capillary hypersurface supported on a totally geodesic hypersurface P={xn+1,g¯(x,En+1)=0}P^{\prime}=\{x\in\mathbb{H}^{n+1},\bar{g}(x,E_{n+1})=0\} (in Poincaré ball model), V0=(1+|x|2)/(1|x|2)V_{0}=(1+|x|^{2})/(1-|x|^{2}), xx is the position vector and Yn+1=δ(x,En+1)x12(1+|x|2)En+1Y_{n+1}=\delta(x,E_{n+1})x-\frac{1}{2}(1+|x|^{2})E_{n+1}. But unfortunately, we cannot find any umbilical θ\theta-capillary hypersurfaces where the integrand in (19) is identically zero.

Let Σ^\hat{\Sigma} denote the domain enclosed by Σ\Sigma and PP, and |Σ^||\widehat{\partial\Sigma}| be the domain enclosed by Σ\partial\Sigma on PP. The energy functional defined by

𝒬(Σ)=|Σ|cosθ|Σ^|\mathcal{Q}(\Sigma)=|\Sigma|-\cos\theta|\widehat{\partial\Sigma}|

is well-known since the critical hypersurface of this functional under any volume preserving variation is a θ\theta-capillary hypersurface with constant mean curvature, see [18] and [22]. Under a flow Σt=xt(M)\Sigma_{t}=x_{t}(M) with with the given normal velocity ff and capillary boundary condition as in (1), the following variation formula is well-known:

ddt|Σ^t|=Σtf𝑑At\frac{d}{dt}|\hat{\Sigma}_{t}|=\int_{\Sigma_{t}}fdA_{t}

and

ddt𝒬(Σt)=ΣtHfdAt.\frac{d}{dt}\mathcal{Q}(\Sigma_{t})=\text{$\int_{\Sigma_{t}}HfdA_{t}.$}

From the Minkowski formula (18)\eqref{Mink}, it is evident that the flow described in (1) is a volume preserving flow.

4. Scalar equation of the flow

In this section, we express the flow (1) by a scalar equation of the radius function.

Let xx be the position vector defined on MM which is represented by

x=cEn+1+ρ(z)z,zΩ𝕊¯+n,x=cE_{n+1}+\rho(z)z,z\in\Omega\subset\bar{\mathbb{S}}^{n}_{+},

where ρ\rho defined on 𝕊+n\mathbb{S}^{n}_{+} is the distance between xx and cEn+1cE_{n+1} in the Euclidean metric. Since Σ\Sigma is smooth, star-shaped, the function ρ\rho is well-defined and smooth on Ω\Omega. Let u=logρu=\log\rho, then uC2(Ω)C0(Ω¯)u\in C^{2}(\Omega)\cap C^{0}(\bar{\Omega}). Define v=1+|u|2v=\sqrt{1+|\nabla u|^{2}}, where u\nabla u is the gradient of uu with respect to the ordinary metric on 𝕊+n\mathbb{S}^{n}_{+}. From the basic facts for radial function, it is well-known that

ν~:=xn+11ν=ζρ1uv,\tilde{\nu}:=x^{-1}_{n+1}\nu=\frac{\zeta-\rho^{-1}\nabla u}{v},

where ζ=ρ\zeta=\partial_{\rho}.

Throughout the paper, we let the indices i,j,ki,j,k range from 1 to nn and we will apply the Einstein convention.

We use polar coordinates (ρ,β,γ,ξ)[0,+)×[0,π2]×[0,2π]×𝕊n2(\rho,\beta,\gamma,\xi)\in[0,+\infty)\times\left[0,\frac{\pi}{2}\right]\times[0,2\pi]\times\mathbb{S}^{n-2}, where ξ\xi is the spherical coordinate on 𝕊n2\mathbb{S}^{n-2}, and the star-shaped hypersurface Σ:=x(M)\Sigma:=x(M) can be written as

(20) xcEn+1=ρ(z)z=ρ(β,γ,ξ)z,wherez:=(β,γ,ξ)𝕊¯+n,x-cE_{n+1}=\rho(z)z=\rho(\beta,\gamma,\xi)z,\hskip 18.00005pt\operatorname{where}z:=(\beta,\gamma,\xi)\in\bar{\mathbb{S}}_{+}^{n},

where x1=ρcosβx_{1}=\rho\cos\beta and xn+1=ρsinβcosγ+c=ewx_{n+1}=\rho\sin\beta\cos\gamma+c=e^{-w}. Then the Euclidean metric ds2=,δds^{2}=\langle\cdot,\cdot\rangle_{\delta} can be written as

ds2=dρ2+ρ2σ𝕊+n=dρ2+ρ2dβ2+ρ2sin2βdγ2+ρ2sin2βsin2γσ𝕊n2,ds^{2}=d\rho^{2}+\rho^{2}\sigma_{\mathbb{S}^{n}_{+}}=d\rho^{2}+\rho^{2}d\beta^{2}+\rho^{2}\sin^{2}\beta d\gamma^{2}+\rho^{2}\sin^{2}\beta\sin^{2}\gamma\sigma_{\mathbb{S}^{n-2}},

and we have

ρ2=x12+i=2nxi2+(xn+1c)2.\rho^{2}=x_{1}^{2}+\sum_{i=2}^{n}x_{i}^{2}+(x_{n+1}-c)^{2}.

Now we can represent E1E_{1}, En+1E_{n+1} by the coordinate (20). Indeed, since

ρx1=x1ρ=cosβ,\frac{\partial\rho}{\partial x_{1}}=\frac{x_{1}}{\rho}=\cos\beta,

and

sinββx1=(cosβ)x1=(x1/ρ)x1=sin2βρ,-\sin\beta\frac{\partial\beta}{\partial x_{1}}=\frac{\partial(\cos\beta)}{\partial x_{1}}=\frac{\partial(x_{1}/\rho)}{\partial x_{1}}=\frac{\sin^{2}\beta}{\rho},

we can represent E1E_{1} by the polar coordinate defined above,

(21) E1=x1=ρx1ρ+βx1β=cosβρsinβρβ.E_{1}=\frac{\partial}{\partial x_{1}}=\frac{\partial\rho}{\partial x_{1}}\partial_{\rho}+\frac{\partial\beta}{\partial x_{1}}\partial_{\beta}=\cos\beta\partial_{\rho}-\frac{\sin\beta}{\rho}\partial_{\beta}.

Similarly, since

ρxn+1=xn+1cρ=sinβcosγ,\frac{\partial\rho}{\partial x_{n+1}}=\frac{x_{n+1}-c}{\rho}=\sin\beta\cos\gamma,
sinββxn+1=(cosβ)xn+1=(x1/ρ)xn+1=1ρsinβcosβcosγ-\sin\beta\frac{\partial\beta}{\partial x_{n+1}}=\frac{\partial(\cos\beta)}{\partial x_{n+1}}=\frac{\partial(x_{1}/\rho)}{\partial x_{n+1}}=-\frac{1}{\rho}\sin\beta\cos\beta\cos\gamma

and

sinγγxn+1=(cosγ)xn+1=((xn+1c)/(ρsinβ))xn+1=sin2γρsinβ,-\sin\gamma\frac{\partial\gamma}{\partial x_{n+1}}=\frac{\partial(\cos\gamma)}{\partial x_{n+1}}=\frac{\partial((x_{n+1}-c)/(\rho\sin\beta))}{\partial x_{n+1}}=\frac{\sin^{2}\gamma}{\rho\sin\beta},

we have

(22) En+1=xn+1=sinβcosγρ+1ρcosβcosγβsinγρsinβγ.E_{n+1}=\frac{\partial}{\partial x_{n+1}}=\sin\beta\cos\gamma\partial_{\rho}+\frac{1}{\rho}\cos\beta\cos\gamma\partial_{\beta}-\frac{\sin\gamma}{\rho\sin\beta}\partial_{\gamma}.

Denoting uβ=σ(u,β)u_{\beta}=\sigma(\nabla u,\partial_{\beta}) and uγ=σ(u,γ)u_{\gamma}=\sigma(\nabla u,\partial_{\gamma}), from (21) and (22) we have

(23) E1,ν=e2wcosβρsinβρβ,ewζρ1uvδ=ewcosβsinβuβv,\langle E_{1},\nu\rangle=e^{2w}\langle\cos\beta\partial_{\rho}-\frac{\sin\beta}{\rho}\partial_{\beta},e^{-w}\frac{\zeta-\rho^{-1}\nabla u}{v}\rangle_{\delta}=e^{w}\frac{\cos\beta-\sin\beta u_{\beta}}{v},

and

(24) En+1,ν~δ=\displaystyle\langle E_{n+1},\tilde{\nu}\rangle_{\delta}= sinβcosγρ+1ρcosβcosγβsinγρsinβγ,ζρ1uvδ\displaystyle\langle\sin\beta\cos\gamma\partial_{\rho}+\frac{1}{\rho}\cos\beta\cos\gamma\partial_{\beta}-\frac{\sin\gamma}{\rho\sin\beta}\partial_{\gamma},\frac{\zeta-\rho^{-1}\nabla u}{v}\rangle_{\delta}
=\displaystyle= 1v(sinβcosγcosβcosγuβ+sinβsinγuγ).\displaystyle\frac{1}{v}(\sin\beta\cos\gamma-\cos\beta\cos\gamma u_{\beta}+\sin\beta\sin\gamma u_{\gamma}).

By definition, we have

(25) xcEn+1,ν=e2wρζ,ewζρ1uv=ρewv.\langle x-cE_{n+1},\nu\rangle=e^{2w}\langle\rho\zeta,e^{-w}\frac{\zeta-\rho^{-1}\nabla u}{v}\rangle=\frac{\rho e^{w}}{v}.

Moreover, by the conformality of mean curvature,

(26) H=\displaystyle H= ew[nρv1ρv(σijuiujv2)uij]nDν~ew\displaystyle e^{-w}\left[\frac{n}{\rho v}-\frac{1}{\rho v}\left(\sigma^{ij}-\frac{u^{i}u^{j}}{v^{2}}\right)u_{ij}\right]-nD_{\tilde{\nu}}e^{-w}
=\displaystyle= ew[nρv1ρv(σijuiujv2)uij]nEn+1,ν~δ\displaystyle e^{-w}\left[\frac{n}{\rho v}-\frac{1}{\rho v}\left(\sigma^{ij}-\frac{u^{i}u^{j}}{v^{2}}\right)u_{ij}\right]-n\langle E_{n+1},\tilde{\nu}\rangle_{\delta}
=\displaystyle= 1ρvew(σijuiujv2)uij+1v(cosβcosγuβsinβsinγuγ)+ncρv,\displaystyle-\frac{1}{\rho ve^{w}}\left(\sigma^{ij}-\frac{u^{i}u^{j}}{v^{2}}\right)u_{ij}+\frac{1}{v}(\cos\beta\cos\gamma u_{\beta}-\sin\beta\sin\gamma u_{\gamma})+\frac{nc}{\rho v},

where σij\sigma^{ij} corresponds to the inverse of the metric on 𝕊+n\mathbb{S}^{n}_{+}, ui=σijuju^{i}=\sigma^{ij}u_{j} is the ii-th component of u\nabla u (the gradient of uu on 𝕊+n\mathbb{S}^{n}_{+}) and uiju_{ij} is the (i,j)(i,j)-th component of the Hessian 2u\nabla^{2}u on 𝕊+n\mathbb{S}^{n}_{+}. Then by calculation,

qc\displaystyle q_{c} =\displaystyle= nc(1xn+1cosθE1,ν)HxcEn+1,ν\displaystyle nc\left(\frac{1}{x_{n+1}}-\cos\theta\langle E_{1},\nu\rangle\right)-H\langle x-cE_{n+1},\nu\rangle
=\displaystyle= ncewnccosθv1ew(cosθsinβuβ)\displaystyle nce^{w}-nc\cos\theta v^{-1}e^{w}(\cos\theta-\sin\beta u_{\beta})
(1ρvew(σijuiujv2)uij+1v(cosβcosγuβsinβsinγuγ)+ncρv)ρewv.\displaystyle-\left(-\frac{1}{\rho ve^{w}}\left(\sigma^{ij}-\frac{u^{i}u^{j}}{v^{2}}\right)u_{ij}+\frac{1}{v}(\cos\beta\cos\gamma u_{\beta}-\sin\beta\sin\gamma u_{\gamma})+\frac{nc}{\rho v}\right)\frac{\rho e^{w}}{v}.

Writing the flow as a function uu of tt, from the flow (1), we see that

qc=tx,ν=ρtewe2wζ,ζρ1u1+|u|2δ=ρtew1+|u|2=utρewv.q_{c}=\langle\partial_{t}x,\nu\rangle=\rho_{t}e^{-w}e^{2w}\langle\zeta,\frac{\zeta-\rho^{-1}\nabla u}{\sqrt{1+|\nabla u|^{2}}}\rangle_{\delta}=\frac{\rho_{t}e^{w}}{\sqrt{1+|\nabla u|^{2}}}=\frac{u_{t}\rho e^{w}}{v}.

Therefore, we obtain the evolution equation of uu as follows,

(27) ut=\displaystyle u_{t}= vρewqc\displaystyle\frac{v}{\rho e^{w}}q_{c}
=\displaystyle= ncvρnccosθρ(cosβsinβuβ)+1ρewv(σijuiujv2)uij\displaystyle\frac{ncv}{\rho}-\frac{nc\cos\theta}{\rho}(\cos\beta-\sin\beta u_{\beta})+\frac{1}{\rho e^{w}v}\left(\sigma^{ij}-\frac{u^{i}u^{j}}{v^{2}}\right)u_{ij}
+nv(cosβcosγuβ+sinβsinγuγ)ncρv\displaystyle+\frac{n}{v}(-\cos\beta\cos\gamma u_{\beta}+\sin\beta\sin\gamma u_{\gamma})-\frac{nc}{\rho v}
=\displaystyle= ncρ|u|2vnccosθρ(cosβsinβuβ)\displaystyle\frac{nc}{\rho}\frac{|\nabla u|^{2}}{v}-\frac{nc\cos\theta}{\rho}(\cos\beta-\sin\beta u_{\beta})
+1ρewv(σijuiujv2)uij+nv(cosβcosγuβ+sinβsinγuγ)\displaystyle+\frac{1}{\rho e^{w}v}\left(\sigma^{ij}-\frac{u^{i}u^{j}}{v^{2}}\right)u_{ij}+\frac{n}{v}(-\cos\beta\cos\gamma u_{\beta}+\sin\beta\sin\gamma u_{\gamma})
=:\displaystyle=: Qc(2u,u,ρ,β,γ).\displaystyle Q_{c}(\nabla^{2}u,\nabla u,\rho,\beta,\gamma).

As for the boundary condition in (1),

cosθ\displaystyle-\cos\theta =ν,N¯\displaystyle=\langle\nu,\bar{N}\rangle
=e2wewν~,ewE1δ\displaystyle=e^{2w}\langle e^{-w}\tilde{\nu},-e^{-w}E_{1}\rangle_{\delta}
=ζρ1uv,1ρβδ\displaystyle=\langle\frac{\zeta-\rho^{-1}\nabla u}{v},\frac{1}{\rho}\partial_{\beta}\rangle_{\delta}
=1vuβ.\displaystyle=-\frac{1}{v}u_{\beta}.

Therefore we have

(28) uβ=ρvcosθ=1+|u|2cosθ.u_{\beta}=\rho v\cos\theta=\sqrt{1+|\nabla u|^{2}}\cos\theta.

Combining (27) and (28), we know that the flow (1) can be written as the parabolic equation of the scalar function uu as follows,

(29) {ut=Qc(2u,u,ρ,β,γ),in𝕊+n×[0,T);uβ=1+|u|2cosθ,on𝕊+n×[0,T);u(,0)=u0(.),on𝕊+n,\left\{\begin{array}[]{lllll}u_{t}=Q_{c}(\nabla^{2}u,\nabla u,\rho,\beta,\gamma),&&&\operatorname{in}&\mathbb{S}^{n}_{+}\times[0,T);\\ u_{\beta}=\sqrt{1+|\nabla u|^{2}}\cos\theta,&&&\operatorname{on}&\partial\mathbb{S}^{n}_{+}\times[0,T);\\ u(\cdot,0)=u_{0}(.),&&&\operatorname{on}&\mathbb{S}^{n}_{+},\end{array}\right.

where u0=logρ0u_{0}=\log\rho_{0} and ρ0=|x0cEn+1|δ\rho_{0}=|x_{0}-cE_{n+1}|_{\delta} is the radial function with respect to cEn+1cE_{n+1} of the initial hypersurface Σ0=x0(M)\Sigma_{0}=x_{0}(M).

The short time existence of the flow can be guaranteed by applying the standard PDE theory to (29), due to the assumption on the star-shapedness of Σ0=x0(M)\Sigma_{0}=x_{0}(M). In the following section, we will show the uniform C0C^{0} and C1C^{1}-estimates for the equation.

Before we start, we need the following calculation.

(30) Qcij:=Qc(λ,ψ,ρ,β,γ)λij|λ=2u,ψ=u=1ρvew(σijuiujv2),Q_{c}^{ij}:=\left.\frac{\partial Q_{c}(\lambda,\psi,\rho,\beta,\gamma)}{\partial\lambda_{ij}}\right|_{\lambda=\nabla^{2}u,\psi=\nabla u}=\frac{1}{\rho ve^{w}}\left(\sigma^{ij}-\frac{u^{i}u^{j}}{v^{2}}\right),
(31) Qc,ψi\displaystyle Q_{c,\psi_{i}} :=Qc(λ,ψ,ρ,β,γ)ψi|λ=2u,ψ=u\displaystyle:=\left.\frac{\partial Q_{c}(\lambda,\psi,\rho,\beta,\gamma)}{\partial\psi_{i}}\right|_{\lambda=\nabla^{2}u,\psi=\nabla u}
=ncρ(2uiv|u|2uiv3)uiρewv3aklukl2ρewv3ailukluk\displaystyle=\frac{nc}{\rho}\left(\frac{2u_{i}}{v}-\frac{|\nabla u|^{2}u_{i}}{v^{3}}\right)-\frac{u_{i}}{\rho e^{w}v^{3}}a^{kl}u_{kl}-\frac{2}{\rho e^{w}v^{3}}a^{il}u_{kl}u_{k}
+nccosθsinβρσ(β,ei)+nuiv3(cosβcosγuβ+sinβsinγuγ)\displaystyle\quad+\frac{nc\cos\theta\sin\beta}{\rho}\sigma(\partial_{\beta},e_{i})+\frac{nu_{i}}{v^{3}}(-\cos\beta\cos\gamma u_{\beta}+\sin\beta\sin\gamma u_{\gamma})
nv(cosβcosγσ(β,ei)sinβsinγσ(γ,ei)),\displaystyle\quad-\frac{n}{v}(\cos\beta\cos\gamma\sigma(\partial_{\beta},e_{i})-\sin\beta\sin\gamma\sigma(\partial_{\gamma},e_{i})),
(32) Qc,ρ\displaystyle Q_{c,\rho} :=Qc(λ,ψ,ρ,β,γ)ρ|λ=2u,ψ=u\displaystyle:=\left.\frac{\partial Q_{c}(\lambda,\psi,\rho,\beta,\gamma)}{\partial\rho}\right|_{\lambda=\nabla^{2}u,\psi=\nabla u}
=ncρ2|u|2v+nccosθρ2(cosβsinβuβ)1ρ2vaijuij,\displaystyle=\frac{nc}{\rho^{2}}\frac{|\nabla u|^{2}}{v}+\frac{nc\cos\theta}{\rho^{2}}(\cos\beta-\sin\beta u_{\beta})-\frac{1}{\rho^{2}v}a^{ij}u_{ij},
(33) Qc,β\displaystyle Q_{c,\beta} :=Qc(λ,ψ,ρ,β,γ)β|λ=2u,ψ=u\displaystyle:=\left.\frac{\partial Q_{c}(\lambda,\psi,\rho,\beta,\gamma)}{\partial\beta}\right|_{\lambda=\nabla^{2}u,\psi=\nabla u}
=1vcosβcosγaijuij+nccosθ(sinβρ+cosβρuβ)\displaystyle=\frac{1}{v}\cos\beta\cos\gamma a^{ij}u_{ij}+nc\cos\theta\left(\frac{\sin\beta}{\rho}+\frac{\cos\beta}{\rho}u_{\beta}\right)
+nv(sinβcosγuβ+cosβsinγuγ),\displaystyle\quad+\frac{n}{v}(\sin\beta\cos\gamma u_{\beta}+\cos\beta\sin\gamma u_{\gamma}),
(34) Qc,γ\displaystyle Q_{c,\gamma} :=Qc(λ,ψ,ρ,β,γ)γ|λ=2u,ψ=u\displaystyle:=\left.\frac{\partial Q_{c}(\lambda,\psi,\rho,\beta,\gamma)}{\partial\gamma}\right|_{\lambda=\nabla^{2}u,\psi=\nabla u}
=1vsinβsinγaijuij+nv(cosβsinγuβ+sinβcosγuγ),\displaystyle=-\frac{1}{v}\sin\beta\sin\gamma a^{ij}u_{ij}+\frac{n}{v}(\cos\beta\sin\gamma u_{\beta}+\sin\beta\cos\gamma u_{\gamma}),

where aij=σijuiujv2a^{ij}=\sigma^{ij}-\frac{u^{i}u^{j}}{v^{2}}. Now we are ready to prove the C0C^{0} and C1C^{1}-estimates.

5. C0C^{0} estimate

Wang and Weng [20] proved a C0C^{0}-estimate for a similar constrained mean curvature type flow of capillary hypersurfaces in the Euclidean unit ball. More specifically, if the initial surface is bounded by two spherical caps that remains stationary under a specified constrained mean curvature flow, the flow will remain bounded by the same two spherical caps. Therefore, a certain pair of spherical caps can be used as barriers of their designed constrained mean curvature flow. A similar property holds in the case of geodesic ball in space forms, see [17].

According to the discussion in Remark 2, the umbilical caps defined by (2) can be regarded as barriers of the flow (1). Then we have a similar corresponding C0C^{0}-estimate.

Proposition 3.

Let x0(M)x_{0}(M) be an initial star-shaped hypersurface with respect to cEn+1cE_{n+1} which satisfies that

x0(M)𝒞c,R1,θ\𝒞c,R2,θx_{0}(M)\subset\mathcal{C}_{c,R_{1},\theta}\backslash\mathcal{C}_{c,R_{2},\theta}

for some R1>R2>0R_{1}>R_{2}>0, where 𝒞c,R,θ\mathcal{C}_{c,R,\theta} is defined in (2). Then it holds that

x(M,t)𝒞c,R1,θ\𝒞c,R2,θx(M,t)\subset\mathcal{C}_{c,R_{1},\theta}\backslash\mathcal{C}_{c,R_{2},\theta}

for all tt along the flow (1). In particular, if u(x,t)u(x,t) solves the initial boundary value problem (29) on [0,)[0,\infty), then for any T>0T>0,

uC0(𝕊+n×[0,T])C,\|u\|_{C^{0}(\mathbb{S}^{n}_{+}\times[0,T])}\leq C,

where C=C(u0,u0,2u0)C=C(u_{0},\nabla u_{0},\nabla^{2}u_{0}).

Proof.

Let φ\varphi be the defining logarithmic radial function of Cc,R1,θC_{c,R_{1},\theta}. Then, φ\varphi is a static solution of the equation of (29), we have

t(uφ)\displaystyle\partial_{t}(u-\varphi) =\displaystyle= Qc(2u,u,eu,β,γ)Qc(2φ,φ,eφ,β,γ)\displaystyle Q_{c}(\nabla^{2}u,\nabla u,e^{u},\beta,\gamma)-Q_{c}(\nabla^{2}\varphi,\nabla\varphi,e^{\varphi},\beta,\gamma)
=\displaystyle= Aijij(uφ)+bj(uφ)j+B(uφ),\displaystyle A^{ij}\nabla_{ij}(u-\varphi)+b^{j}\cdot(u-\varphi)_{j}+B(u-\varphi),

where

Aij=01Qcij(2(su+(1s)φ),(su+(1s)φ),esu+(1s)φ,β,γ)𝑑s,A^{ij}=\int_{0}^{1}Q_{c}^{ij}(\nabla^{2}(su+(1-s)\varphi),\nabla(su+(1-s)\varphi),e^{su+(1-s)\varphi},\beta,\gamma)ds,
bj=01Qc,pj(2(su+(1s)φ),(su+(1s)φ),esu+(1s)ψ,β,γ)𝑑sb_{j}=\int_{0}^{1}Q_{c,p_{j}}(\nabla^{2}(su+(1-s)\varphi),\nabla(su+(1-s)\varphi),e^{su+(1-s)\psi},\beta,\gamma)ds

and

B=01Qc,ρ(2(su+(1s)φ),(su+(1s)φ),esu+(1s)ψ,β,γ)esu+(1s)ψ𝑑s.B=\int_{0}^{1}Q_{c,\rho}(\nabla^{2}(su+(1-s)\varphi),\nabla(su+(1-s)\varphi),e^{su+(1-s)\psi},\beta,\gamma)e^{su+(1-s)\psi}ds.

Since Qc,ρQ_{c,\rho} has no singular point if ρ0\rho\neq 0, BB is bounded and we can denote λ=sup𝕊+n×[0,T]|B|\lambda=-\sup_{\mathbb{S}^{n}_{+}\times[0,T]}|B|. We get

t(eλt(uφ))=eλt(Aijij(uφ)+bj(uφ)j+(B+λ)(uφ)).\partial_{t}(e^{\lambda t}(u-\varphi))=e^{\lambda t}(A^{ij}\nabla_{ij}(u-\varphi)+b^{j}\cdot(u-\varphi)_{j}+(B+\lambda)(u-\varphi)).

Let (y0,t0)(y_{0},t_{0}) be the point where eλt(uφ)e^{\lambda t}(u-\varphi) attains  its nonnegative maximum value. By maximum principle, (y0,t0)(y_{0},t_{0}) can only be located on the parabolic boundary, say (y0,t0)(y_{0},t_{0}). That is,

eλt(u(x,t)φ(x))sup𝕊+n×[0,T)𝕊+n×{0}{0,eλt(u(x,t)φ(x))}.e^{\lambda t}(u(x,t)-\varphi(x))\leq\sup_{\partial\mathbb{S}^{n}_{+}\times[0,T)\cup\mathbb{S}^{n}_{+}\times\{0\}}\{0,e^{\lambda t}(u(x,t)-\varphi(x))\}.

If y0𝕊+n,y_{0}\in\partial\mathbb{S}^{n}_{+}, by Hopf Lemma we have

𝕊+n(uφ)(y0,t0)=0;n(uφ)(y0,t0)<0,\nabla^{\partial\mathbb{S}^{n}_{+}}(u-\varphi)(y_{0},t_{0})=0;\quad\nabla_{n}(u-\varphi)(y_{0},t_{0})<0,

where 𝕊+n\nabla^{\partial\mathbb{S}^{n}_{+}} denotes the gradient of a function on 𝕊+n\partial\mathbb{S}^{n}_{+}, and n\nabla_{n} is the normal derivative on 𝕊+n\partial\mathbb{S}^{n}_{+}. Then we have

|𝕊+nu(y0,t0)|=|𝕊+nφ(y0,t0)|:=k, and nu(y0,t0)<nφ(y0,t0).|\nabla^{\partial\mathbb{S}^{n}_{+}}u(y_{0},t_{0})|=|\nabla^{\partial\mathbb{S}^{n}_{+}}\varphi(y_{0},t_{0})|:=k,\text{ and }\nabla_{n}u(y_{0},t_{0})<\nabla_{n}\varphi(y_{0},t_{0}).

From the boundary condition in (29),

nu1+k2+|nu|2=cosθ=nφ1+k2+|nφ|2,\frac{\nabla_{n}u}{\sqrt{1+k^{2}+|\nabla_{n}u|^{2}}}=-\cos\theta=\frac{\nabla_{n}\varphi}{\sqrt{1+k^{2}+|\nabla_{n}\varphi|^{2}}},

which is a contradiction to nu(y0,t0)<nφ(y0,t0)\nabla_{n}u(y_{0},t_{0})<\nabla_{n}\varphi(y_{0},t_{0}) by monotonicity of the function g(x)=x/1+k2+x2.g(x)=x/\sqrt{1+k^{2}+x^{2}}.

Therefore we can only have t0=0t_{0}=0, that is,

eλt(u(y,t)φ(y,t))u0(y0)φ(y0)0,(y,t)𝕊+n×[0,T],e^{\lambda t}(u(y,t)-\varphi(y,t))\leq u_{0}(y_{0})-\varphi(y_{0})\leq 0,\quad\forall\;(y,t)\in\mathbb{S}^{n}_{+}\times[0,T],

which gives uu an upper bound by the umbilical cap 𝒞c,R2,θ\mathcal{C}_{c,R_{2},\theta}’s defining function φ\varphi. Similarly, the desired lower bound can be obtained. We have finished the proof of Proposition 3. ∎

6. C1C^{1}-estimate

In this section, we prove a C1C^{1}-estimate of the flow (29). Inspired by [20], we use the similar auxiliary function. Let d:neigh(𝕊+n)d:\operatorname{neigh}(\partial\mathbb{S}^{n}_{+})\rightarrow\mathbb{R} be a nonnegative smooth function on neigh(𝕊+n)𝕊+n\operatorname{neigh}(\partial\mathbb{S}^{n}_{+})\subset\mathbb{S}^{n}_{+} defined by

d(x):=distσ(x,𝕊+n).d(x):=\operatorname{dist}_{\sigma}(x,\partial\mathbb{S}^{n}_{+}).

Indeed, the function is well defined on a neighborhood of 𝕊+n\partial\mathbb{S}^{n}_{+} but it can be extended to the whole 𝕊+n\mathbb{S}^{n}_{+} and satisfies that

d0,|d|1,in𝕊¯+n.d\geq 0,\hskip 11.99998pt|\nabla d|\leq 1,\;\operatorname{in}\hskip 11.99998pt\bar{\mathbb{S}}^{n}_{+}.\hskip 11.99998pt

The following C1C^{1}-estimate is essential for the flow (1).

Proposition 4.

If θ\theta satisfies that |cosθ|<4nK0(c,R,θ)c(n1)4nK0(c,R,θ)+c(n1)|\cos\theta|<\frac{4nK_{0}(c,R,\theta)-c(n-1)}{4nK_{0}(c,R,\theta)+c(n-1)}, for any (x,t)𝕊+n×[0,T](x,t)\in\mathbb{S}^{n}_{+}\times[0,T] (T<T)(T<T^{\ast}), we have

|u|(x,t)C,|\nabla u|(x,t)\leq C,

for some positive constant C=C(2u0,u0,u0)C=C(\nabla^{2}u_{0},\nabla u_{0},u_{0}).

Proof.

Define an auxiliary function inspired by [20],

Ψ:=(1+Kd)v+cosθσ(u,d)\Psi:=(1+Kd)v+\cos\theta\sigma(\nabla u,\nabla d)

where KK is a constant to be determined later. Let (y0,t0)𝕊¯+n×[0,t](y_{0},t_{0})\in\bar{\mathbb{S}}^{n}_{+}\times[0,t] be the point where Ψ\Psi attains its maximum. We will discuss case by case to prove the theorem.

Case 1: (y0,t0)𝕊+n×[0,T](y_{0},t_{0})\in\partial\mathbb{S}^{n}_{+}\times[0,T]. At the point x0𝕊+nx_{0}\in\partial\mathbb{S}^{n}_{+}, we choose a local coordinate {y1,,yn}\{y_{1},\cdots,y_{n}\} near y0y_{0}, such that y1=β\frac{\partial}{\partial y_{1}}=-\partial_{\beta} is an inner normal vector of 𝕊+n\partial\mathbb{S}^{n}_{+}, and {yi}i=2n+1\{y_{i}\}_{i=2}^{n+1} be the geodesic coordinate near y0𝕊+ny_{0}\in\partial\mathbb{S}^{n}_{+} along y1=ty_{1}=t, (0<t<ε)(0<t<\varepsilon) in the neighborhood of x0x_{0}. Under this coordinate, y1=d\frac{\partial}{\partial y_{1}}=\nabla d on 𝕊+n\partial\mathbb{S}^{n}_{+}.

First of all, from the boundary condition of (29), we know that u1:=βu=cosθvu_{1}:=\nabla_{-\partial_{\beta}}u=-\cos\theta v, denote u=uu1\nabla^{\prime}u=\nabla u-u_{1}, then from 1+|u|2=v2u12=sinθv21+|\nabla^{\prime}u|^{2}=v^{2}-u_{1}^{2}=\sin\theta v^{2} we have,

Ψ|𝕊+n=vvcos2θ=1+|u|2+u12sin2θ=|sinθ|1+|u|2.\Psi|_{\partial\mathbb{S}^{n}_{+}}=v-v\cos^{2}\theta=\sqrt{1+|\nabla^{\prime}u|^{2}+u_{1}^{2}}\sin^{2}\theta=|\sin\theta|\sqrt{1+|\nabla^{\prime}u|^{2}}.

Applying the Gauss-Weingarten equation, we have

1v\displaystyle\nabla_{1}v =kuk1uv=i=2niu1iuvcosθ11u\displaystyle=\frac{\nabla_{k}u\nabla_{k1}u}{v}=\frac{\sum_{i=2}^{n}\nabla_{i}u\nabla_{1i}u}{v}-\cos\theta\nabla_{11}u
=1vi=2n(uiui1+j=2nuihij𝕊+nuj)cosθ11u\displaystyle=\frac{1}{v}\sum_{i=2}^{n}\left(u_{i}u_{i1}+\sum_{j=2}^{n}u_{i}h^{\partial\mathbb{S}^{n}_{+}}_{ij}u_{j}\right)-\cos\theta\nabla_{11}u
=i=2nuiui1vcosθ11u,\displaystyle=\sum_{i=2}^{n}\frac{u_{i}u_{i1}}{v}-\cos\theta\nabla_{11}u,

where we have used the fact that hij𝕊+n=0h^{\partial\mathbb{S}^{n}_{+}}_{ij}=0 since 𝕊+n\partial\mathbb{S}^{n}_{+} is totally geodesic in 𝕊+n\mathbb{S}^{n}_{+}. Then, from Hopf Lemma we have,

(35) 0\displaystyle 0 1Ψ(x0)=1v+Kv1d+1(ukdk)cosθ\displaystyle\geq\nabla_{1}\Psi(x_{0})=\nabla_{1}v+Kv\nabla_{1}d+\nabla_{1}(u_{k}d_{k})\cos\theta
=1v+Kv1d+k1udkcosθ+ukk1dcosθ\displaystyle=\nabla_{1}v+Kv\nabla_{1}d+\nabla_{k1}ud_{k}\cos\theta+u_{k}\nabla_{k1}d\cos\theta
=1vi=2nuiu1i+Kv+ukk1dcosθ,\displaystyle=\frac{1}{v}\sum_{i=2}^{n}u_{i}u_{1i}+Kv+u_{k}\nabla_{k1}d\cos\theta,

and for i2i\geq 2, we have

iΨ(x0)=vi+u1icosθ.\nabla_{i}^{\prime}\Psi(x_{0})=v_{i}+u_{1i}\cos\theta.

Differentiating the boundary condition (29), together with the fact above we have

u1i=i(cosθv)=cos2θu1i,u_{1i}=-\nabla_{i}^{\prime}(\cos\theta v)=\cos^{2}\theta u_{1i},

therefore we know

(36) u1i=0,1in1.u_{1i}=0,\quad\forall 1\leq i\leq n-1.

Then applying (36) to (35), we have

0Kv+ukk1dcosθΨ(Ksin2θC1),0\geq Kv+u_{k}\nabla_{k1}d\cos\theta\geq\Psi\left(\frac{K}{\sin^{2}\theta}-C_{1}\right),

where C1C_{1} is a universal constant which satisfies ||u|cosθ|v|sin2θ|C1\frac{||\nabla u|\cos\theta|}{v|\sin^{2}\theta|}\leq C_{1}. Then KK can be chosen large enough so that it comes to a contradiction.

Case 2: If (x0,t0)𝕊¯+n×{0}(x_{0},t_{0})\in\bar{\mathbb{S}}^{n}_{+}\times\{0\}, then

Ψ(x,t)Ψ(x0,0)=(1+Kd)1+|u0|2+cosθσ(u0,d)C2.\Psi(x,t)\leq\Psi(x_{0},0)=(1+Kd)\sqrt{1+|\nabla u_{0}|^{2}}+\cos\theta\sigma(\nabla u_{0},\nabla d)\leq C_{2}.

Therefore v(x,t)C,v(x,t)\leq C, for any (x,t)𝕊¯×[0,T](x,t)\in\bar{\mathbb{S}}\times[0,T].

Case 3: In the case of (x0,t0)int(𝕊+n)(x_{0},t_{0})\in\operatorname{int}(\mathbb{S}^{n}_{+}), we have

(37) 0=\displaystyle 0= iΨ(x0,t0)=(1+Kd)vi+Kdiv+cosθ(ukdk)i\displaystyle\nabla_{i}\Psi(x_{0},t_{0})=(1+Kd)v_{i}+Kd_{i}v+\cos\theta(u_{k}d_{k})_{i}
=\displaystyle= ((1+Kd)kuv+kdcosθ)iku+kuikdcosθ+Kdiv.\displaystyle\left((1+Kd)\frac{\nabla_{k}u}{v}+\nabla_{k}d\cos\theta\right)\nabla_{ik}u+\nabla_{k}u\nabla_{ik}d\cos\theta+Kd_{i}v.

Choose a geodesic coordinate, up to a rotation of the one in Case 1, such that

u1=|u|>0,and{αβu}2α,βnisdiagonal.u_{1}=|\nabla u|>0,\quad\operatorname{and}\quad\{\nabla_{\alpha\beta}u\}_{2\leq\alpha,\beta\leq n}\,\operatorname{is}\,\operatorname{diagonal}.

Assume u1=|u|u_{1}=|\nabla u| is large enough such that u1u_{1}, v=1+u12v=\sqrt{1+u_{1}^{2}} and Ψ=(1+Kd)v+u1d1cosθ\Psi=(1+Kd)v+u_{1}d_{1}\cos\theta are equivalent. Otherwise, we could obtain the desire C1C^{1} estimate for uu. Here we denote d1=σ(d,e1)=u11σ(d,u)d_{1}=\sigma(\nabla d,e_{1})=u_{1}^{-1}\sigma(\nabla d,\nabla u).

Firstly, from (37) by letting i=αi=\alpha, we have

(38) [(1+Kd)u1v+cosθd1]u1α=cosθuααdαcosθu1d1αKdαv,\left[(1+Kd)\frac{u_{1}}{v}+\cos\theta d_{1}\right]u_{1\alpha}=-\cos\theta u_{\alpha\alpha}d_{\alpha}-\cos\theta u_{1}d_{1\alpha}-Kd_{\alpha}v,

and letting i=1i=1, we have

(39) [(1+Kd)u1v+cosθd1]u11=cosθuα1dαcosθu1d11Kd1v,\left[(1+Kd)\frac{u_{1}}{v}+\cos\theta d_{1}\right]u_{11}=-\cos\theta u_{\alpha 1}d_{\alpha}-\cos\theta u_{1}d_{11}-Kd_{1}v,

We can see from the (38) that

(40) u1α=S1cosθuααdαS1(cosθu1d1α+Kdαv),u_{1\alpha}=-S^{-1}\cos\theta u_{\alpha\alpha}d_{\alpha}-S^{-1}(\cos\theta u_{1}d_{1\alpha}+Kd_{\alpha}v),

where S=(1+Kd)u1v+cosθd1S=(1+Kd)\frac{u_{1}}{v}+\cos\theta d_{1}, and it is clear that 2+πKSC(δ,θ)2+\pi K\geq S\geq C(\delta,\theta), if there exists some positive δ>0\delta>0 such that u1(x0,t0)δu_{1}(x_{0},t_{0})\geq\delta. Indeed, if not, we would have already proved the theorem.

Inserting (40) into (39),

(41) u11=\displaystyle u_{11}= S1cosθuα1dαS1(cosθu1d11+Kd1v)\displaystyle-S^{-1}\cos\theta u_{\alpha 1}d_{\alpha}-S^{-1}(\cos\theta u_{1}d_{11}+Kd_{1}v)
=\displaystyle= cos2θS2α=2ndα2uαα+[α=2ncosθdαS2(cosθu1d1α+Kdαv)\displaystyle\frac{\cos^{2}\theta}{S^{2}}\sum_{\alpha=2}^{n}d_{\alpha}^{2}u_{\alpha\alpha}+\left[\sum_{\alpha=2}^{n}\frac{\cos\theta d_{\alpha}}{S^{2}}(\cos\theta u_{1}d_{1\alpha}+Kd_{\alpha}v)\right.
S1(cosθu1d11+Kd1v)]\displaystyle-S^{-1}(\cos\theta u_{1}d_{11}+Kd_{1}v)\Bigg{]}
=\displaystyle= cos2θS2α=2ndα2uαα+O(v).\displaystyle\frac{\cos^{2}\theta}{S^{2}}\sum_{\alpha=2}^{n}d_{\alpha}^{2}u_{\alpha\alpha}+O(v).

Applying the condition of second derivative on (x0,t0)(x_{0},t_{0}), we have

(42) 0\displaystyle 0 (tQcijijQc,pii)Φ\displaystyle\leq\left(\partial_{t}-Q_{c}^{ij}\nabla_{ij}{-Q_{c,p_{i}}}\nabla_{i}\right)\Phi
=(1+Kd)vuk(uktQcijukijQc,piuki)\displaystyle=\frac{(1+Kd)}{v}u_{k}(u_{kt}-Q_{c}^{ij}u_{kij}-Q_{c,p_{\mathit{i}}}u_{ki})
+dkcosθ(uktQcijukijQc,piuki)\displaystyle\quad+d_{k}\cos\theta(u_{kt}-Q_{c}^{ij}u_{kij}-Q_{c,p_{i}}u_{ki})
+(1+Kd)(Qcijululiukukjv3Qcijuliuljv)\displaystyle\quad+(1+Kd)\left(\frac{Q_{c}^{ij}u_{l}u_{li}u_{k}u_{kj}}{v^{3}}-\frac{Q_{c}^{ij}u_{li}u_{lj}}{v}\right)
(2Qcijukidkjcosθ+2KQcijdivj)\displaystyle\quad-(2Q_{c}^{ij}u_{ki}d_{kj}\cos\theta+2KQ_{c}^{ij}d_{i}v_{j})
(Qcijukdkijcosθ+KQcijdijv)\displaystyle\quad-(Q_{c}^{ij}u_{k}d_{kij}\cos\theta+KQ_{c}^{ij}d_{ij}v)
Qc,ψi(Kdiv+cosθukdki)\displaystyle\quad-Q_{c,\psi_{i}}(Kd_{i}v+\cos\theta u_{k}d_{ki})
:=L1+L2+L3+L4+L5+L6.\displaystyle:=L_{1}+L_{2}+L_{3}+L_{4}+L_{5}+L_{6}.

Let us examine the term L1L_{1} and L2L_{2} at first. Differentiating the parabolic equation (29) with respect to tt, we have

(43) utk=Qcijuijk+Qc,piuik+Qc,ρρuk+Qc,βσ(β,ek)+Qc,γσ(γ,ek).u_{tk}=Q_{c}^{ij}u_{ijk}+Q_{c,p_{i}}u_{ik}+Q_{c,\rho}\rho u_{k}+Q_{c,\beta}\sigma(\partial_{\beta},e_{k})+Q_{c,\gamma}\sigma(\partial_{\gamma},e_{k}).

Ricci identity on 𝕊+n\mathbb{S}^{n}_{+} gives that

(44) uijk=ukij+ujσilulσij.u_{ijk}=u_{kij}+u_{j}\sigma_{il}-u_{l}\sigma_{ij}.

And by definition, we have

(45) aijuij\displaystyle a^{ij}u_{ij} =(σijuiujv2)uij=u11+α=2nuαα|u|2v2u11\displaystyle=\left(\sigma^{ij}-\frac{u^{i}u^{j}}{v^{2}}\right)u_{ij}=u_{11}+\sum_{\alpha=2}^{n}u_{\alpha\alpha}-\frac{|\nabla u|^{2}}{v^{2}}u_{11}
=1v2u11+α=2nuαα.\displaystyle=\frac{1}{v^{2}}u_{11}+\sum_{\alpha=2}^{n}u_{\alpha\alpha}.

Applying (43), (44) and (45) to the terms L1L_{1} and L2L_{2} gives,

(46) L1\displaystyle L_{1} =1+Kdvuk(uktQijukijQc,piuki)\displaystyle=\frac{1+Kd}{v}u_{k}(u_{kt}-Q^{ij}u_{kij}-Q_{c,p_{i}}u_{ki})
=1+Kdvuk(Qcij(ujσikukσij)+Qc,ρρuk+Qc,βσ(β,ek)\displaystyle=\frac{1+Kd}{v}u_{k}(Q_{c}^{ij}(u_{j}\sigma_{ik}-u_{k}\sigma_{ij})+Q_{c,\rho}\rho u_{k}+Q_{c,\beta}\sigma(\partial_{\beta},e_{k})
+Qc,γσ(γ,ek))\displaystyle\quad+Q_{c,\gamma}\sigma(\partial_{\gamma},e_{k}))
=1+Kdv4(cosβcosγuβsinβsinγuγc|u|2ρ)u11\displaystyle=\frac{1+Kd}{v^{4}}\left(\cos\beta\cos\gamma u_{\beta}-\sin\beta\sin\gamma u_{\gamma}-c\frac{|\nabla u|^{2}}{\rho}\right)u_{11}
c1+Kdρv2|u|2(α=2nuαα)\displaystyle\quad-c\frac{1+Kd}{\rho v^{2}}|\nabla u|^{2}\left(\sum_{\alpha=2}^{n}u_{\alpha\alpha}\right)
nc1+Kdv|u|2(|u|2ρv+cosθsinβρuβ)\displaystyle\quad-nc\frac{1+Kd}{v}|\nabla u|^{2}\left(\frac{|\nabla u|^{2}}{\rho v}+\frac{\cos\theta\sin\beta}{\rho}u_{\beta}\right)
+(1+Kd){1v2(cosβcosγuβsinβsinγuγ)α=2nuαα\displaystyle\quad+(1+Kd)\{\frac{1}{v^{2}}(\cos\beta\cos\gamma u_{\beta}-\sin\beta\sin\gamma u_{\gamma})\sum_{\alpha=2}^{n}u_{\alpha\alpha}
+ncosθcosβvρ|u|2+nv2(sinβcosγuβ2+2cosβsinγuγuβ\displaystyle\quad+\frac{n\cos\theta\cos\beta}{v\rho}|\nabla u|^{2}+\frac{n}{v^{2}}(\sin\beta\cos\gamma u_{\beta}^{2}+2\cos\beta\sin\gamma u_{\gamma}u_{\beta}
+sinβcosγuγ2)+(1n)(1+Kd)|u|2ρewv2}\displaystyle\quad+\sin\beta\cos\gamma u^{2}_{\gamma})+\frac{(1-n)(1+Kd)|\nabla u|^{2}}{\rho e^{w}v^{2}}\}
:=L11+L12+L13+L14.\displaystyle:=L_{11}+L_{12}+L_{13}+L_{14}.

It is obvious that L14=O(v1)α=2nuαα+O(v)L_{14}=O(v^{-1})\sum_{\alpha=2}^{n}u_{\alpha\alpha}+O(v). And using (41) we have

(47) L11\displaystyle L_{11} =1+Kdv4(cosβcosγuβsinβsinγuγc|u|2ρ)(cos2θS2α=2ndα2uαα+O(u1))\displaystyle=\frac{1+Kd}{v^{4}}\left(\cos\beta\cos\gamma u_{\beta}-\sin\beta\sin\gamma u_{\gamma}-c\frac{|\nabla u|^{2}}{\rho}\right)\left(\frac{\cos^{2}\theta}{S^{2}}\sum_{\alpha=2}^{n}d_{\alpha}^{2}u_{\alpha\alpha}+O(u_{1})\right)
=O(v2)α=2n|uαα|+O(v1).\displaystyle=O(v^{-2})\sum_{\alpha=2}^{n}|u_{\alpha\alpha}|+O(v^{-1}).

Similarly, the term L2L_{2} be written as follows,

(48) L2\displaystyle L_{2} =dkcosθ(uktQcijukijQc,piuki)\displaystyle=d_{k}\cos\theta(u_{kt}-Q_{c}^{ij}u_{kij}-Q_{c,p_{i}}u_{ki})
=cosθv3(cosβcosγdβsinβsinγdγcσ(u,d)ρ)u11\displaystyle=\frac{\cos\theta}{v^{3}}\left(\cos\beta\cos\gamma d_{\beta}-\sin\beta\sin\gamma d_{\gamma}-c\frac{\sigma(\nabla u,\nabla d)}{\rho}\right)u_{11}
ccosθρvσ(u,d)α=2nuαα\displaystyle\quad-c\frac{\cos\theta}{\rho v}\sigma(\nabla u,\nabla d)\sum_{\alpha=2}^{n}u_{\alpha\alpha}
ncσ(u,d)(|u|2ρv+cosθsinβρuβ)\displaystyle\quad-nc\sigma(\nabla u,\nabla d)\left(\frac{|\nabla u|^{2}}{\rho v}+\frac{\cos\theta\sin\beta}{\rho}u_{\beta}\right)
+cosθ[1v(cosβcosγdβsinβsinγdγ)α=2nuαα\displaystyle\quad+\cos\theta\left[\frac{1}{v}(\cos\beta\cos\gamma d_{\beta}-\sin\beta\sin\gamma d_{\gamma})\sum_{\alpha=2}^{n}u_{\alpha\alpha}\right.
+ncosθcosβρσ(u,d)+nv(sinβcosγuβ+cosβsinγuγ)dβ\displaystyle\quad\quad+\frac{n\cos\theta\cos\beta}{\rho}\sigma(\nabla u,\nabla d)+\frac{n}{v}(\sin\beta\cos\gamma u_{\beta}+\cos\beta\sin\gamma u_{\gamma})d_{\beta}
+nv(sinβcosγuβ+cosβsinγuγ)dα+(1n)1ρew(u,d)]\displaystyle\quad\quad+\frac{n}{v}(\sin\beta\cos\gamma u_{\beta}+\cos\beta\sin\gamma u_{\gamma})d_{\alpha}+(1-n)\frac{1}{{\rho e^{w}}}(\nabla u,\nabla d)]
:=L21+L22+L23+L24.\displaystyle:=L_{21}+L_{22}+L_{23}+L_{24}.

And for the same reason, we have L21=O(v2)α=2n|uαα|+O(v1)L_{21}=O(v^{-2})\sum_{\alpha=2}^{n}|u_{\alpha\alpha}|+O(v^{-1}) and L24=O(1v)α=2nuαα+O(v)L_{24}=O\left(\frac{1}{v}\right)\sum_{\alpha=2}^{n}u_{\alpha\alpha}+O(v).

For the term L3L_{3}, we have

(49) L3\displaystyle L_{3} =(1+Kd)(Qcijululiukukjv3Qcijuliuljv)\displaystyle=(1+Kd)\left(\frac{Q_{c}^{ij}u_{l}u_{li}u_{k}u_{kj}}{v^{3}}-\frac{Q_{c}^{ij}u_{li}u_{lj}}{v}\right)
=1+Kdρvew(σijuiujv2)(ululiukukjv3uliuljv)\displaystyle=\frac{1+Kd}{\rho ve^{w}}\left(\sigma^{ij}-\frac{u_{i}u_{j}}{v^{2}}\right)\left(\frac{u_{l}u_{li}u_{k}u_{kj}}{v^{3}}-\frac{u_{li}u_{lj}}{v}\right)
=1+Kdρvew(1v5u1122v3α=2nu1α21vα=2nuαα2)\displaystyle=\frac{1+Kd}{\rho ve^{w}}\left(-\frac{1}{v^{5}}u_{11}^{2}-\frac{2}{v^{3}}\sum_{\alpha=2}^{n}u_{1\alpha}^{2}-\frac{1}{v}\sum_{\alpha=2}^{n}u_{\alpha\alpha}^{2}\right)
=1+Kdρvew(1v5u1122v3α=2nu1α2)(1ε)1+Kdρv2ewα=2nuαα2\displaystyle=\frac{1+Kd}{\rho ve^{w}}\left(-\frac{1}{v^{5}}u_{11}^{2}-\frac{2}{v^{3}}\sum_{\alpha=2}^{n}u_{1\alpha}^{2}\right)-(1-\varepsilon)\frac{1+Kd}{\rho v^{2}e^{w}}\sum_{\alpha=2}^{n}u_{\alpha\alpha}^{2}
ε1+Kdρv2ewα=2nuαα2\displaystyle\quad-\varepsilon\frac{1+Kd}{\rho v^{2}e^{w}}\sum_{\alpha=2}^{n}u_{\alpha\alpha}^{2}
:=L31+L32+L33.\displaystyle:=L_{31}+L_{32}+L_{33}.

Then we compile the following three terms

(50) L12+L22+L32\displaystyle L_{12}+L_{22}+L_{32} =cρv(1+Kdv|u|2+cosθσ(u,d))α=2nuαα\displaystyle=-\frac{c}{\rho v}\left(\frac{1+Kd}{v}|\nabla u|^{2}+\cos\theta\sigma(\nabla u,\nabla d)\right)\sum_{\alpha=2}^{n}u_{\alpha\alpha}
(1ε)1+Kdρv2ewα=2nuαα2\displaystyle\quad-(1-\varepsilon)\frac{1+Kd}{\rho v^{2}e^{w}}\sum_{\alpha=2}^{n}u_{\alpha\alpha}^{2}
=cρvS|u|α=2nuαα(1ε)1+Kdρv2ewα=2nuαα2\displaystyle=-\frac{c}{\rho v}S|\nabla u|\sum_{\alpha=2}^{n}u_{\alpha\alpha}-(1-\varepsilon)\frac{1+Kd}{\rho v^{2}e^{w}}\sum_{\alpha=2}^{n}u_{\alpha\alpha}^{2}
c2(n1)ewS2|u|24ρ(1ε)(1+Kd)\displaystyle\leq\frac{c^{2}(n-1)e^{w}S^{2}|\nabla u|^{2}}{4\rho(1-\varepsilon)(1+Kd)}
c2(n1)(1+|cosθ|)Sew4ρ(1ε)|u|2.\displaystyle\leq\frac{c^{2}(n-1)(1+|\cos\theta|)Se^{w}}{4\rho(1-\varepsilon)}|\nabla u|^{2}.

To estimate L13+L23L_{13}+L_{23}, we need the following arguments. Let c0c_{0} be a constant which satisfies c0(|cosθ|,4nτK0(c,R,θ)c(n1)4nτK0(c,R,θ)+c(n1))c_{0}\in\left(|\cos\theta|,\frac{4n\tau K_{0}(c,R,\theta)-c(n-1)}{4n\tau K_{0}(c,R,\theta)+c(n-1)}\right), for some 0<τ<10<\tau<1. Then we can assume that

(51) |u|2v+cosθsinβuβ(1c0)|u|.\frac{|\nabla u|^{2}}{v}+\cos\theta\sin\beta u_{\beta}\geq(1-c_{0})|\nabla u|.

Otherwise,

1c0\displaystyle 1-c_{0} >\displaystyle> |u|v+cosθsinβuβ|u|\displaystyle\frac{|\nabla u|}{v}+\cos\theta\sin\beta\frac{u_{\beta}}{|\nabla u|}
\displaystyle\geq |u|v|cosθ|\displaystyle\frac{|\nabla u|}{v}-|\cos\theta|

would give an upper bound for |u||\nabla u|, the required estimate is obtained.

Considering L13+L23L_{13}+L_{23} and using (51), we have

(52) L13+L23\displaystyle L_{13}+L_{23} =nc1+Kdv|u|2(|u|2ρv+cosθsinβuβρ)\displaystyle=-nc\frac{1+Kd}{v}|\nabla u|^{2}\left(\frac{|\nabla u|^{2}}{\rho v}+\frac{\cos\theta\sin\beta u_{\beta}}{\rho}\right)
nccosθσ(u,d)(|u|2ρv+cosθsinβuβρ)\displaystyle\quad-nc\cos\theta\sigma(\nabla u,\nabla d)\left(\frac{|\nabla u|^{2}}{\rho v}+\frac{\cos\theta\sin\beta u_{\beta}}{\rho}\right)
=nc|u|ρ1(|u|2v+cosθsinβuβ)S\displaystyle=-nc|\nabla u|\rho^{-1}\left(\frac{|\nabla u|^{2}}{v}+\cos\theta\sin\beta u_{\beta}\right)S
nc(1c0)ρ1S|u|2.\displaystyle\leq-nc(1-c_{0})\rho^{-1}S|\nabla u|^{2}.

Combining (50) and (52), we have

L13+L23+L12+L22+L32\displaystyle L_{13}+L_{23}+L_{12}+L_{22}+L_{32}
\displaystyle\leq nc(1c0)ρ1S|u|2+c2(n1)(1+|cosθ|)S4(1ε)ρew|u|2\displaystyle-nc(1-c_{0})\rho^{-1}S|\nabla u|^{2}+\frac{c^{2}(n-1)(1+|\cos\theta|)S}{4(1-\varepsilon)\rho}e^{w}|\nabla u|^{2}
\displaystyle\leq (nc(1c0)ρ1+c2(n1)(1+c0)4(1ε)ρxn+1)S|u|2.\displaystyle\left(-nc(1-c_{0})\rho^{-1}+\frac{c^{2}(n-1)(1+c_{0})}{4(1-\varepsilon)\rho x_{n+1}}\right)S|\nabla u|^{2}.

For the last inequality, Proposition 3 shows that Σt𝒞^c,R,θ\Sigma_{t}\subset\hat{\mathcal{C}}_{c,R,\theta}, and hence ew=xn+1>K0(c,R,θ)e^{-w}=x_{n+1}>K_{0}(c,R,\theta). Therefore

nc(1c0)+c2(n1)(1+c0)4(1ε)xn+1\displaystyle-nc(1-c_{0})+\frac{c^{2}(n-1)(1+c_{0})}{4(1-\varepsilon)x_{n+1}}
\displaystyle\leq 2c2n(n1)4τcnK0(c,R,θ)+(n1)+2c2τnK0(c,R,θ)(n1)xn+1(1ε)(4τcnK0(c,R,θ)+(n1))\displaystyle-\frac{2c^{2}n(n-1)}{4\tau cnK_{0}(c,R,\theta)+(n-1)}+\frac{2c^{2}\tau nK_{0}(c,R,\theta)(n-1)}{x_{n+1}(1-\varepsilon)(4\tau cnK_{0}(c,R,\theta)+(n-1))}
\displaystyle\leq 2c2n(n1)4τcnK0(c,R,θ)+(n1)(1τ1ε).\displaystyle-\frac{2c^{2}n(n-1)}{4\tau cnK_{0}(c,R,\theta)+(n-1)}\left(1-\frac{\tau}{1-\varepsilon}\right).

Let ε=(1τ)/2(0,1)\varepsilon=(1-\tau)/2\in(0,1), from the C0C^{0}-estimate in Proposition 3, there is an uniform lower bound ρ0\rho_{0} on ρ\rho. Hence letting a0=2c2n(n1)(1τ)(4τcnK0(c,R,θ)+(n1))(1+τ)ρ0Sa_{0}=\frac{2c^{2}n(n-1)(1-\tau)}{(4\tau cnK_{0}(c,R,\theta)+(n-1))(1+\tau)}\rho_{0}S, we have

(53) L13+L23+L12+L22+L32<a0|u|2.L_{13}+L_{23}+L_{12}+L_{22}+L_{32}<-a_{0}|\nabla u|^{2}.

Now we consider L4L_{4} and L6L_{6}. Using (41), (42) and (45), we obtain

L4+L6\displaystyle L_{4}+L_{6}
=\displaystyle= (2Qcijukidkjcosθ+2KQcijdivj)Qc,ψi(Kdiv+cosθukdki)\displaystyle-(2Q_{c}^{ij}u_{ki}d_{kj}\cos\theta+2KQ_{c}^{ij}d_{i}v_{j})-Q_{c,\psi_{i}}(Kd_{i}v+\cos\theta u_{k}d_{ki})
=\displaystyle= O(1v)α=2n|uαα|+O(1),\displaystyle O\left(\frac{1}{v}\right)\sum_{\alpha=2}^{n}|u_{\alpha\alpha}|+O(1),

and it is easy to see L5=(Qcijukdkijcosθ+KQcijdijv)=O(1)L_{5}=-(Q_{c}^{ij}u_{k}d_{kij}\cos\theta+KQ_{c}^{ij}d_{ij}v)=O(1).

Then adding all the terms back to (42), we have

0\displaystyle 0 1+Kdρewv(1v5u1122v3α=2n|u1α|)ε01+Kd2ρewv2α=2n|uαα|α0u12\displaystyle\leq\frac{1+Kd}{\rho e^{w}v}\left(-\frac{1}{v^{5}}u^{2}_{11}-\frac{2}{v^{3}}\sum_{\alpha=2}^{n}|u_{1\alpha}|\right)-\varepsilon_{0}\frac{1+Kd}{2\rho e^{w}v^{2}}\sum_{\alpha=2}^{n}|u_{\alpha\alpha}|-\alpha_{0}u_{1}^{2}
+O(1v)α=2n|uαα|+O(v)\displaystyle\quad+O\left(\frac{1}{v}\right)\sum_{\alpha=2}^{n}|u_{\alpha\alpha}|+O(v)
ε01+Kd2ρewv2α=2nuαα2+C2vα=2n|uαα|a0u12+C1v\displaystyle\leq-\varepsilon_{0}\frac{1+Kd}{2\rho e^{w}v^{2}}\sum_{\alpha=2}^{n}u^{2}_{\alpha\alpha}+\frac{C_{2}}{v}\sum_{\alpha=2}^{n}|u_{\alpha\alpha}|-a_{0}u_{1}^{2}+C_{1}v
a0u12+C1v+C22ρew2ε0(1+Kd),\displaystyle\leq-a_{0}u_{1}^{2}+C_{1}v+\frac{C^{2}_{2}\rho e^{w}}{2\varepsilon_{0}(1+Kd)},

This gives an upper bound for u1u_{1}. ∎

7. Convergence of the flow

The higher order a-priori estimates of uu follow from the uniform C0C^{0} and C1C^{1} estimates. The same argument as in [20] gives the following result.

Proposition 5.

If u(,t)u(\cdot,t) solves the boundary value problem (29) on the interval [0,T)[0,T^{\ast}), and the initial hypersurfaces Σ0\Sigma_{0} and the contact angle θ\theta satisfies the same condition in Theorem 1. Then for any 0<T<T0<T<T^{\ast}, we have

u(,t)CkC,0<t<T,\|u(\cdot,t)\|_{C^{k}}\leq C,\quad 0<t<T,

where C=C(k,u0,u0,2u0)>0C=C(k,u_{0},\nabla u_{0},\nabla^{2}u_{0})>0. In addition, it follows that T=T^{\ast}=\infty.

Before we prove Theorem 1, we need the following lemma.

Lemma 2.

Let 𝒞c1,R1,θ(a1)\mathcal{C}_{c_{1},R_{1},\theta}(a_{1}) and 𝒞c2,R2,θ(a2)\mathcal{C}_{c_{2},R_{2},\theta}(a_{2}) defined by (2). If their enclosed volume are equal, that is, |𝒞^c1,R1,θ(a1)|=|𝒞^c2,R2,θ(a2)||\hat{\mathcal{C}}_{c_{1},R_{1}{,}\theta}(a_{1})|=|\hat{\mathcal{C}}_{c_{2},R_{2}{,}\theta}(a_{2})|, it holds that

c1R1=c2R2,\frac{c_{1}}{R_{1}}=\frac{c_{2}}{R_{2}},

that is, the umbilical θ\theta-caps with the same enclosed volume have the same principal curvature.

Proof.

Let χ=χ2χ1\chi=\chi_{2}\circ\chi_{1} be an isometry composed by two isometries in n+1\mathbb{H}^{n+1}, where χ1\chi_{1} is a translation along the hyperbolic geodesic γt=(0,,t)+n+1\gamma_{t}=(0,\cdots,t)\in\mathbb{R}_{+}^{n+1}, such that,

χ1(x1,,xn+1)=c2c1(x1,,xn+1),\chi_{1}(x_{1},\cdots,x_{n+1})=\frac{c_{2}}{c_{1}}(x_{1},\cdots,x_{n+1}),

and χ2\chi_{2} is a translation defined by

χ2(x~,xn+1)=(x~+a2a1,xn+1).\chi_{2}(\tilde{x},x_{n+1})=(\tilde{x}+a_{2}-a_{1},x_{n+1}).

They are both isometric transformation. Hence

|𝒞^c2,R2,θ(a2)|=|𝒞^R1,c1,θ(a1)|=|χ(𝒞^R1,c1,θ(a1))|=|𝒞^c2,c2R1/c1,θ(a2)|.|\hat{\mathcal{C}}_{c_{2},R_{2}{,}\theta}(a_{2})|=|\hat{\mathcal{C}}_{R_{1},c_{1},\theta}(a_{1})|=|\chi(\hat{\mathcal{C}}_{R_{1},c_{1},\theta}(a_{1}))|=|\hat{\mathcal{C}}_{c_{2},c_{2}R_{1}/c_{1},\theta}(a_{2})|.

Since |𝒞^c2,R,θ(θ)||\hat{\mathcal{C}}_{c_{2},R,\theta}(\theta)| is monotonically decreasing as RR decreasing, so c2R1/c1=R2c_{2}R_{1}/c_{1}=R_{2}. ∎

Remark 4.

We can see from the proof that, the umbilical caps 𝒞1\mathcal{C}_{1} and 𝒞2\mathcal{C}_{2} on totally geodesic hyperplane with the same enclosed volume is equivalent to the same area and the same wetting area, which is the area of the domain enclosed by 𝒞1\partial\mathcal{C}_{1} and 𝒞2\partial\mathcal{C}_{2}.

Now we are ready to prove Theorem 1.

Proof of the Theorem 1.

Let r=2r=2 in the Minkowski type formula (17), we have

Σt[H1(cxn+1ccosθE1,ν)H2xcEn+1,ν]𝑑A=0.\int_{\Sigma_{t}}\left[H_{1}\left(\frac{c}{x_{n+1}}-c\cos\theta\langle E_{1},\nu\rangle\right)-H_{2}\langle x-cE_{n+1},\nu\rangle\right]dA=0.

Let Σ^\widehat{\partial\Sigma} be the domain enclosed by Σ\partial\Sigma on PP. The first variation formula of the energy 𝒬(Σ^t)=1n[|Σt|cosθ|Σ^|]\mathcal{Q}(\hat{\Sigma}_{t})=\frac{1}{n}[|\Sigma_{t}|-\cos\theta|\widehat{\partial\Sigma}|] gives that

ddt𝒬(t)\displaystyle\frac{d}{dt}\mathcal{Q}(t) =ΣtH1(cxn+1ccosθE1,νH1xcEn+1,ν)𝑑A\displaystyle=\int_{\Sigma_{t}}H_{1}\left(\frac{c}{x_{n+1}}-c\cos\theta\langle E_{1},\nu\rangle-H_{1}\langle x-cE_{n+1},\nu\rangle\right)dA
=ΣtH1(cxn+1ccosθE1,νH1xcEn+1,ν)𝑑A\displaystyle=\int_{\Sigma_{t}}H_{1}\left(\frac{c}{x_{n+1}}-c\cos\theta\langle E_{1},\nu\rangle-H_{1}\langle x-cE_{n+1},\nu\rangle\right)dA
Σt[H1(cxn+1ccosθE1,ν)H2xcEn+1,ν]𝑑A\displaystyle\quad-\int_{\Sigma_{t}}\left[H_{1}\left(\frac{c}{x_{n+1}}-c\cos\theta\langle E_{1},\nu\rangle\right)-H_{2}\langle x-cE_{n+1},\nu\rangle\right]dA
=Σt(H12H2)xcEn+1,ν𝑑A\displaystyle=\int_{\Sigma_{t}}(H_{1}^{2}-H_{2})\langle x-cE_{n+1},\nu\rangle dA
=1n2(n1)i<jΣt(κjκi)2xcEn+1,ν𝑑A\displaystyle=-\frac{1}{n^{2}(n-1)}\sum_{i<j}\int_{\Sigma_{t}}(\kappa_{j}-\kappa_{i})^{2}\langle x-cE_{n+1},\nu\rangle dA
=1n2(n1)i<j𝕊+n(κi(y)κj(y))2𝑑g(y),\displaystyle=-\frac{1}{n^{2}(n-1)}\sum_{i<j}\int_{\mathbb{S}^{n}_{+}}(\kappa_{i}(y)-\kappa_{j}(y))^{2}dg(y),

where dg(y)=ρewvdydg(y)=\frac{\rho e^{w}}{v}dy and κi\kappa_{i}’s are the principle curvatures at y𝕊+ny\in\mathbb{S}_{+}^{n} identified as a point on Σt\Sigma_{t}. Therefore, the energy 𝒬(Σ)=Area(Σ)cosθ𝒲(Σ)\mathcal{Q}(\Sigma)=\operatorname{Area}(\Sigma)-\cos\theta\mathcal{W}(\partial\Sigma) is monotonically decreasing from Proposition 3, we know that the energy is bounded from above and below. Then integrating both sides of the equation above on [0,+)[0,+\infty), we have

i<j0𝕊+n(κi(y,t)κj(y,t))2𝑑g(y)𝑑tC.\sum_{i<j}\int_{0}^{\infty}\int_{\mathbb{S}^{n}_{+}}(\kappa_{i}(y,t)-\kappa_{j}(y,t))^{2}dg(y)dt\leq C.

From the uniform CkC^{k} estimate in Proposition 5, we have

limt|κiκj|2=0.\lim_{t\rightarrow\infty}|\kappa_{i}-\kappa_{j}|^{2}=0.

Therefore any convergent subsequence of x(,t)x(\cdot,t) must converge to an umbilical cap 𝒞c,R,θ(a)\mathcal{C}_{c,R_{\infty},\theta}(a_{\infty}) as tt\rightarrow\infty. Hence from the boundary condition in (1), we know that they are given by

𝒞c,R,θ(a)={xn+1:|x+RcosθE1acEn+1|=R},\mathcal{C}_{c^{\prime},R^{\prime},\theta}(a_{\infty})=\{x\in\mathbb{H}^{n+1}:|x+R^{\prime}\cos\theta E_{1}-a_{\infty}-c^{\prime}E_{n+1}|=R^{\prime}\},

where aa_{\infty} is a constant vector perpendicular to both E1E_{1} and En+1E_{n+1}.

It remains to show that the limit umbilical cap is unique. We follow the proof in [17] and [21]. Let R>0R_{\infty}>0 be the unique number such that |Σ^|=|𝒞^c,R,θ||\hat{\Sigma}|=|\hat{\mathcal{C}}_{c,R_{\infty},\theta}|.

Denote by R(,t)R(\cdot,t) the radius of the unique umbilical cap

𝒞c,R(,t),θ={xn+1:|xR(,t)cosθE1cEn+1|=R(,t)},\mathcal{C}_{c,R(\cdot,t),\theta}=\{x\in\mathbb{H}^{n+1}:|x-R(\cdot,t)\cos\theta E_{1}-cE_{n+1}|=R(\cdot,t)\},

passing through the point x(,t)x(\cdot,t). Let R(t)=maxxMR(x,t)R^{\ast}(t)=\max_{x\in M}R(x,t) and there exists a point ξt\xi_{t} attaining R(t)R^{\ast}(t) by compactness. It then follows that R(t)R^{\ast}(t) is non-increasing since the 𝒞c,R(,t),θ\mathcal{C}_{c,R^{\ast}(\cdot,t),\theta} is a barrier of x(,t)x(\cdot,t) by Proposition 3. We claim that

(54) limtR(t)=R.\lim_{t\rightarrow\infty}R^{\ast}(t)=R_{\infty}.

We suppose otherwise, then there exist ε>0\varepsilon>0 and tt large enough, such that

(55) R(t)>R+ε,R^{\ast}(t)>R_{\infty}+\varepsilon,

from the definition of R(,t)R(\cdot,t), we have

|x+R(,t)cosθE1cEn+1|δ2=R2(,t),|x+R(\cdot,t)\cos\theta E_{1}-cE_{n+1}|_{\delta}^{2}=R^{2}(\cdot,t),

or equivalently

|x|δ2+c22cx,En+1δ2R(,t)cosθx,E1δ=R2(,t)sin2θ.|x|_{\delta}^{2}+c^{2}-2c\langle x,E_{n+1}\rangle_{\delta}-2R(\cdot,t)\cos\theta\langle x,E_{1}\rangle_{\delta}=R^{2}(\cdot,t)\sin^{2}\theta.

Taking the derivative of the equation above with respect to tt, we have

(56) xt,xcEn+1+R(,t)cosθE1δ=Rt(,t)(R(,t)sin2θcosθx,E1δ).\langle x_{t},x-cE_{n+1}+R(\cdot,t)\cos\theta E_{1}\rangle_{\delta}=R_{t}(\cdot,t)(R(\cdot,t)\sin^{2}\theta-\cos\theta\langle x,E_{1}\rangle_{\delta}).

Now we evaluate at the point (ξt,t)(\xi_{t},t), since Σt\Sigma_{t} is tangential to 𝒞c,ρ(,t),θ\mathcal{C}_{c,\rho^{\ast}(\cdot,t),\theta} at the point, then the normal vector at (ξt,t)(\xi_{t},t) is

ν~(ξt,t)=ν𝒞c,ρ(t),θ(x(ξt,t))=x+R(t)cosθE1cEn+1R(t).\tilde{\nu}(\xi_{t},t)=\nu_{\mathcal{C}_{c,\rho^{\ast}(t),\theta}}(x(\xi_{t},t))=\frac{x+R^{\ast}(t)\cos\theta E_{1}-cE_{n+1}}{R^{\ast}(t)}.

Inserting the flow equation (1) to (56), we have

(57) (R(t)sin2θcosθx,E1δ)tR(t)\displaystyle(R^{\ast}(t)\sin^{2}\theta-\cos\theta\langle x,E_{1}\rangle_{\delta})\partial_{t}R^{\ast}(t)
=\displaystyle= xn+1|x+R(t)cosθE1cEn+1|δ2R(t)qc\displaystyle x_{n+1}\frac{|x+R^{\ast}(t)\cos\theta E_{1}-cE_{n+1}|_{\delta}^{2}}{R^{\ast}(t)}q_{c}
=\displaystyle= xn+1R(t)(ncxn+1nccosθE1,νH(ξt,t)xcEn+1,ν).\displaystyle x_{n+1}R^{\ast}(t)\left(\frac{nc}{x_{n+1}}-nc\cos\theta\langle E_{1},\nu\rangle-H(\xi_{t},t)\langle x-cE_{n+1},\nu\rangle\right).

From the calculation in Remark 2, we have

(58) H𝒞c,ρ(t),θ=ncR(t),H_{\mathcal{C}_{c,\rho^{\ast}(t),\theta}}=\frac{nc}{R^{\ast}(t)},

then

(59) ncxn+1nccosθE1,νxcEn+1,ν|(ξt,t)=ncxn+1nccosθE1,νxcEn+1,ν|𝒞c,ρ(t),θ=ncR(t).\left.\frac{\frac{nc}{x_{n+1}}-nc\cos\theta\langle E_{1},\nu\rangle}{\langle x-cE_{n+1},\nu\rangle}\right|_{(\xi_{t},t)}=\left.\frac{\frac{nc}{x_{n+1}}-nc\cos\theta\langle E_{1},\nu\rangle}{\langle x-cE_{n+1},\nu\rangle}\right|_{\mathcal{C}_{c,\rho^{\ast}(t),\theta}}=\frac{nc}{R^{\ast}(t)}.
  

From (25), (58) and (59), we can see that for ε>0\varepsilon>0 small enough, there exist a T>0T>0 such that for any t>Tt>T, the following

(60) xn+1R(t)(ncxn+1nccosθE1,νH(ξt,t)xcEn+1,ν)\displaystyle x_{n+1}R^{\ast}(t)\left(\frac{nc}{x_{n+1}}-nc\cos\theta\langle E_{1},\nu\rangle-H(\xi_{t},t)\langle x-cE_{n+1},\nu\rangle\right)
=\displaystyle= xn+1(ncH(ξt,t)R(t))xcEn+1,ν\displaystyle x_{n+1}(nc-H(\xi_{t},t)R^{\ast}(t))\langle x-cE_{n+1},\nu\rangle

holds at (ξt,t)(\xi_{t},t).

On the other hand, since x(,t)x(\cdot,t) converge to 𝒞c,ρ,θ(a)\mathcal{C}_{c,\rho_{\infty},\theta}(a_{\infty}) and ρ\rho_{\infty} is uniquely determined, we get

limtH(ξt,t)=ncR.\lim_{t\rightarrow\infty}H(\xi_{t},t)=\frac{nc^{\prime}}{{R}^{\prime}}.

Therefore, for ε0=nc2R\varepsilon_{0}=\frac{nc^{\prime}}{2R^{\prime}}, there exists a T>0T>0, such that for any t>Tt>T,

(61) H(ξt,t)>ncRε0>0.H(\xi_{t},t)>\frac{nc^{\prime}}{R^{\prime}}-\varepsilon_{0}>0.

Then from (55), (60) and (61), we get

(62) xn+1R(t)(ncxn+1nccosθE1,νH(ξt,t)xcEn+1,ν)\displaystyle\;x_{n+1}R^{\ast}(t)\left(\frac{nc}{x_{n+1}}-nc\cos\theta\langle E_{1},\nu\rangle-H(\xi_{t},t)\langle x-cE_{n+1},\nu\rangle\right)
\displaystyle\leq 12xn+1(nc(ncRε0)(R+ε))xcEn+1,ν\displaystyle\;\frac{1}{2}x_{n+1}\left(nc-\left(\frac{nc^{\prime}}{R^{\prime}}-\varepsilon_{0}\right)(R_{\infty}+\varepsilon)\right)\langle x-cE_{n+1},\nu\rangle
\displaystyle\leq 12xn+1(ε0Rε(ncRε0))xcEn+1,ν\displaystyle\;\frac{1}{2}x_{n+1}\left(-\varepsilon_{0}R_{\infty}-\varepsilon\left(\frac{nc^{\prime}}{R^{\prime}}-\varepsilon_{0}\right)\right)\langle x-cE_{n+1},\nu\rangle
<\displaystyle<  0\displaystyle\;0

In the second inequality we have used Proposition 5. On the other hand, since 0x,E1δ1ρ(t)cosθ0\leq\langle x,E_{1}\rangle_{\delta}\leq 1-\rho^{\ast}(t)\cos\theta, we have

(63) R(t)sin2θcosθx,E1R(t)(1cosθ)>0,R^{\ast}(t)\sin^{2}\theta-\cos\theta\langle x,E_{1}\rangle\geq R^{\ast}(t)(1-\cos\theta)>0,

then combining (57), (62) and (63), we have

tR(t)<0.\partial_{t}R^{\ast}(t)<0.

This leads to a contradiction to that limt0ddtR(t)=0\lim_{t\rightarrow 0}\frac{d}{dt}R^{\ast}(t)=0. Hence (54) holds. Similarly, we can show

limtR(t)=R,\lim_{t\rightarrow\infty}R_{\ast}(t)=R_{\infty},

where R(t)R_{\ast}(t) is defined by R(t)=minxMR(x,t)=R(χt,t)R_{\ast}(t)=\min_{x\in M}R(x,t)=R(\chi_{t},t) and χt\chi_{t} is the point achieving R(t)R_{\ast}(t). Therefore 𝒞c,R,θ(a)=𝒞c,R,θ\mathcal{C}_{c^{\prime},R^{\prime},\theta}(a_{\infty})=\mathcal{C}_{c,R_{\infty},\theta}, and we obtain the uniqueness of the limit. ∎

Remark 5.

In Poincaré half space model of hyperbolic space, the volume of an umbilical cap is determined not only by its Euclidean radius but also by the location of its center, particularly the (n+1)(n+1)-th coordinate. As indicated by Remark 2, the location of its center also influences the principal curvatures of the cap. Therefore, we cannot identity the radius by the volume of the domain bounded by initial hypersurfaces Σ0\Sigma_{0}.

Note that all the isometries appearing in the proof of Lemma 2 on Poincare´\acute{\mathrm{e}} half-space model will keep the ratio c/Rc/R of an umbilical cap 𝒞c,R,θ\mathcal{C}_{c,R{,}\theta}.

From the monotonicity of the energy 𝒬\mathcal{Q}, we have the following corollary.

Corollary 1.

Let x:Mn+1x:M\rightarrow\mathbb{H}^{n+1} be a θ\theta-capillary hypersurface supported on the totally geodesic hyperplane PP. If

  1. (1)

    Σ=x(M)\Sigma=x(M) is contained in an umbilical cap 𝒞c,R,θ(a)\mathcal{C}_{c,R,\theta}(a) which satisfies that K(c,R,θ)>c(n1)/4nK(c,R,\theta)>c(n-1)/4n.

  2. (2)

    the contacting angle θ\theta satisfies

    |cosθ|<4nK0(c,R,θ)c(n1)4nK0(c,R,θ)+c(n1),|\cos\theta|<\frac{4nK_{0}(c,R,\theta)-c(n-1)}{4nK_{0}(c,R,\theta)+c(n-1)},

then θ\theta-umbilical caps with the same enclosed volume with Σ=x(M)\Sigma=x(M) are the only minimizers of the energy 𝒬\mathcal{Q}.

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