# Gauss curvature type flow and Alexandrov-Fenchel inequalities in the
hyperbolic space

Tianci Luo  School of Mathematical Sciences, University of Science and
Technology of China, Hefei 230026, P.R. China<EMAIL_ADDRESS>and
Rong Zhou<EMAIL_ADDRESS>
###### Abstract.

We consider the Gauss curvature type flow for uniformly convex hypersurfaces
in the hyperbolic space $\mathbb{H}^{n+1}\ (n\geqslant 2)$. We prove that if
the initial closed hypersurface is smooth and uniformly convex, then the
smooth solution exists for all positive time and converges smoothly and
exponentially to a geodesic sphere centered at the origin. The key step is to
prove the upper bound of Gauss curvature and that uniform convexity preserves
along the flow. As an application, we provide a new proof for an Alexandrov-
Fenchel inequality comparing $(n-2)$th quermassintegral and volume of convex
domains in $\mathbb{H}^{n+1}\ (n\geqslant 2)$.

###### Key words and phrases:

Gauss curvature flow, quermassintegrals, hyperbolic space

###### 2020 Mathematics Subject Classification:

53C42; 53E10

##  1\. Introduction

In this paper, we consider a Gauss curvature type flow for uniformly convex
hypersurfaces in the hyperbolic space $\mathbb{H}^{n+1}\ (n\geqslant 2)$ and
we also provide a new proof for a quermassintegral inequality for convex
domains.

Let $n\geqslant 2$. We view $\mathbb{H}^{n+1}=\mathbb{S}^{n}\times[0,+\infty)$
as a warped product manifold equipped with metric

$\overline{g}_{\mathbb{H}^{n+1}}=\mathrm{d}\rho^{2}+\phi^{2}(\rho)g_{\mathbb{S}^{n}},$

where $\phi(\rho)=\sinh\rho,\ \rho\in[0,+\infty)$. Define
$V=\phi(\rho)\partial_{\rho}$. It’s well known that $V$ is a conformal Killing
field. For a hypersurface $M\subset\mathbb{H}^{n+1}$, let $\nu$ be its unit
outward normal vector. We denote $u=\langle V,\nu\rangle$ be its support
function. Recall that, $M$ is star-shaped, if its support function is positive
everywhere on $M$; while $M$ is uniformly convex (resp. convex) if its
principal curvatures are all positive (resp. non-negative).

Consider a smooth family of embeddings
$X:\mathbb{S}^{n}\times[0,T)\to\mathbb{H}^{n+1}$ satisfying

$\begin{cases}\partial_{t}X(x,t)=\left(\phi^{\prime}(\rho)-uK^{\frac{1}{n}}\right)\nu(x,t),\\\
X(\cdot,0)=X_{0}(\cdot),\end{cases}$ (1.1)

where $K(\cdot,t)$ denotes the Gauss curvature of $M_{t}=X(\mathbb{S}^{n},t)$.
The volume of the enclosed domain $\Omega_{t}$ is non-decreasing while the
$(n-2)$th quermassintegral $\mathscr{A}_{n-2}(\Omega_{t})$ (see Section 2.2
for precise definition) is non-increasing along the flow (1.1). We will prove
the following long time existence and convergence results for the flow (1.1)
with uniformly convex initial hypersurfaces.

######  Theorem 1.1.

Let $M_{0}=X_{0}(\mathbb{S}^{n})$ be a smooth, closed, uniformly convex
hypersurface in hyperbolic space $\mathbb{H}^{n+1}$ containing the origin,
then the flow (1.1) has a unique smooth uniformly convex solution $M_{t}$ for
all time $t\in[0,+\infty)$. When $t\rightarrow+\infty$, $M_{t}$ converges
smoothly to a unique standard smooth geodesic sphere centered at the origin,
and the convergence of $M_{t}$ in any $C^{k}$-norm is exponential.

Gauss curvature type flows in hyperbolic space have been concentrated widely
in recent years, and they have many applications in geometric problems.
Andrews and Chen [4] considered Gauss curvature type flow in $\mathbb{H}^{3}$

$\partial_{t}X(x,t)=-(K-1)\nu(x,t).$

They proved that if the initial hypersurface has positive scalar curvature,
then the solution surface is uniformly convex and converges to a point in
finite time. After rescalings, the solution is asymptotic to a shrinking
sphere in finite time. Chen and Huang [11] considered the flow by $\alpha$-th
power of the Gauss curvature in $\mathbb{H}^{n+1}$

$\partial_{t}X(x,t)=-K^{\alpha}\nu(x,t).$

They proved that if the initial hypersurface is uniformly convex, then the
solution hypersurface exists and converges to a point in finite time for
$\alpha>0$, and converges to a geodesic sphere after rescaling for
$\alpha>\frac{1}{n+2}$. Their work generalized the corresponding results in
Euclidean space $\mathbb{R}^{n+1}$, see [1, 2, 6, 8]. In [25], Wei, Yang and
Zhou considered a non-local type volume preserving Gauss curvature flow in
$\mathbb{H}^{n+1}$

$\partial_{t}X(x,t)=(\phi(t)-K^{\alpha})\nu(x,t),$

where $\alpha>0$ and
$\phi(t)=\dfrac{1}{|M_{t}|}\int_{M_{t}}K^{\alpha}\mathrm{d}\mu_{t}$. They
confirmed that for a smooth closed uniformly convex initial hypersurface, the
solution exists for all time $t\in[0,+\infty)$ and converges smoothly and
exponentially to a geodesic sphere as $t\to+\infty$. Li and Zhang [19] studied
the prescribed Gaussian curvature problem in $\mathbb{H}^{n+1}$ by using the
Gauss curvature flow method.

Our flow (1.1) is a locally constrained curvature flow. Note that by the
standard theory of parabolic equation, the flow (1.1) exists for a short time
if initial hypersurface is uniformly convex. The proof of long time existence
and convergence require a priori estimates. $C^{0}$ estimate is obtained by
parametrizing the flow as a graph of radial function on $\mathbb{S}^{n}$ and
using standard maximum principle. By [13, Theorem 2.7.10], for a convex
solution to the flow (1.1), once we obtain the $C^{0}$ estimate, the $C^{1}$
estimate follows, and we conclude that the star-shapedness preserves along the
flow (1.1). Therefore, the main difficulty for the long time existence is the
$C^{2}$ estimate. Inspired by the method in [19, Section 4] and [21, Lemma
4.2], we will choose appropriate auxiliary functions to prove the upper bound
of Gauss curvature $K$ and that uniform convexity preserves along the flow
(1.1), from which we obtain the $C^{2}$ estimate. Then by the standard
procedure, we obtain the long time existence. Note that by the arithmetic mean
and geometric mean inequality, the volume of the enclosed domain is
monotonically non-decreasing along the flow (1.1). This property helps us to
verify that the limit hypersurface is a geodesic sphere centered at the origin
and to prove the smooth convergence. The exponential convergence follows by
estimating the gradient of radial function along the flow (1.1) inspired by
[18, Section 8.3].

Locally constrained curvature flows play an important role in proving
Alexandrov-Fenchel type inequalities, which ask whether

$\mathscr{A}_{k}(\Omega)\geqslant\xi_{k,l}\left(\mathscr{A}_{\ell}(\Omega)\right)\
\ (-1\leqslant\ell<k\leqslant n-1)$ (1.2)

holds for a smooth bounded domain $\Omega$ in space forms. Here
$\mathscr{A}_{k}(\Omega)$ are quermassintegrals for the smooth bounded domain
$\Omega$, see Section 2.2 for precise definition. $\xi_{k,l}(s)$ is the unique
function such that equality in (1.2) holds if and only if $\Omega$ is a
geodesic ball. Equality in (1.2) holds if and only if $\Omega$ is a geodesic
ball.

In [14], Guan and Li used inverse curvature flows to prove quermassintegral
inequalities (1.2) for $k$-convex (i.e. $\sigma_{j}(\kappa)\geqslant 0,\
j=1,2,\cdots,k$) and star-shaped hypersurfaces in $\mathbb{R}^{n+1}$ when
$-1\leqslant\ell<k\leqslant n-1$. In [15], Guan and Li studied the mean
curvature type flow

$\partial_{t}X=(n\phi^{\prime}(\rho)-uH)\nu$ (1.3)

in space forms, where $H$ is the mean curvature of the evolving hypersurfaces.
They proved that if the initial closed hypersurface is star-shaped, then the
solution of (1.3) exists for all positive time and converges smoothly and
exponentially to a geodesic sphere centered at the origin. They also proved
that the uniform convexity preserves along the flow (1.3) in
$\mathbb{R}^{n+1}$. Note that the volume preserves while area decreases along
the flow (1.3), the isoperimetric inequality for star-shaped hypersurfaces in
$\mathbb{R}^{n+1}$ follows. Later, Chen-Guan-Li-Scheuer [10] proved that the
uniform convexity preserves along the flow (1.3) in $\mathbb{S}^{n+1}$. Chen
and Sun [12] proved quermassintegral inequalities (1.2) for convex
hypersurfaces in $\mathbb{S}^{n+1}$ when $\ell=k-2,\ 1\leqslant k\leqslant
n-1$.

In $\mathbb{H}^{n+1}$, Li, Wei and Xiong [20] proved quermassintegral
inequalities (1.2) when $\ell=-1,\ k=2$ for star-shaped and two-convex
hypersurfaces. Wang and Xia [24] proved quermassintegral inequalities (1.2)
when $-1\leqslant\ell<k\leqslant n-1$ for h-convex (i.e. $\kappa_{i}\geqslant
1,\ i=1,\cdots,n$) domains. Later, Hu-Li-Wei [18] gave a new proof of (1.2)
for h-convex domains. Andrews, Chen and Wei [7] proved (1.2) when
$-1=\ell<k\leqslant n-1$ for domains whose boundary has positive sectional
curvatures. Brendle, Guan and Li [9, 16] proved quermassintegral inequalities
(1.2) when $-1\leqslant\ell<k=n-1$ for convex domains and $\ell=0,\ k=1$ for
star-shaped and mean convex domains. Hu and Li [17, Theorem 1.3] proved (1.2)
when $-1\leqslant\ell<k=n-2$ for uniformly convex domains.

As far as we know, quermassintegral inequalities (1.2) for convex domains in
$\mathbb{H}^{n+1}$ remain open in many cases. Note that $(n-2)$th
quermassintegral is non-increasing while the volume is non-decreasing along
our flow (1.1). As an application, we will use the convergence results of flow
(1.1) to provide a new proof for the Alexandrov-Fenchel inequality comparing
$(n-2)$th quermassintegral and the volume of convex domains in
$\mathbb{H}^{n+1}$.

######  Theorem 1.2.

Let $M$ be a closed, smooth, convex hypersurface in $\mathbb{H}^{n+1}\
(n\geqslant 2)$ enclosing a domain $\Omega$, then

$\mathscr{A}_{n-2}(\Omega)\geqslant\xi_{n-2,-1}\left(\mathscr{A}_{-1}(\Omega)\right),$
(1.4)

where $\xi_{n-2,-1}=\xi_{n-2}\circ\xi_{-1}^{-1}$,
$\xi_{k}(r)=\mathscr{A}_{k}(B_{r})$ is a monotone function, $B_{r}$ is a
geodesic ball of radius $r$ centered at the origin, $\xi_{k}^{-1}$ is the
inverse function of $\xi_{k}$. Equality holds if and only if $M$ is a geodesic
sphere.

We will first prove the inequality (1.4) for uniformly convex hypersurfaces by
using the convergence result Theorem 1.1. Then for a convex hypersurface,
Theorem 1.2 follows by an approximation argument as in [14] and [25].

The paper is organized as follows. In Section 2, we will gather some
preliminaries for our proof, including quermassintegrals, formulas for
hypersurfaces in the hyperbolic space, the general evolution equations for
curvature flows, and parametrization by radial graph. In Section 3, we will
prove $C^{0}$ and $C^{1}$ estimates. Especially, we conclude that star-
shapedness preserves along the flow (1.1). In Section 4, we will prove the
upper bound of Gauss curvature and the preservation of uniform convexity to
establish the $C^{2}$ estimate. Then the long time existence follows by the
standard procedure. In Section 5, we will obtain the smooth and exponential
convergence results and complete the proof of Theorem 1.1. In Section 6, we
will prove Theorem 1.2, the quermassintegral inequality for convex domains.

######  Acknowledgements.

The authors would like to thank Professor Yong Wei for his helpful discussions
and support. This work is supported by National Key Research and Development
Program of China 2021YFA1001800.

##  2\. Preliminaries

In this section, we will collect some preliminaries for our proof, including
quermassintegrals, formulas for hypersurfaces in hyperbolic space, general
evolution equations and parametrization by radial graph.

###  2.1. Hypersurfaces in hyperbolic space

We consider $\mathbb{H}^{n+1}=\mathbb{S}^{n}\times[0,\infty)$ as a warped
product manifold equipped with metric

$\overline{g}=\mathrm{d}\rho^{2}+\phi^{2}(\rho)g_{\mathbb{S}^{n}},$

where $\phi(\rho)=\sinh\rho$, $\rho\in[0,\infty)$, $g_{\mathbb{S}^{n}}$
denotes the standard metric on $\mathbb{S}^{n}$. Let

$\Phi(\rho)=\int_{0}^{\rho}\phi(s)\mathrm{d}s=\cosh\rho-1.$

It’s well known that $V=D\Phi=\phi(\rho)\partial_{\rho}$ is a conformal
Killing field on $\mathbb{H}^{n+1}$, that is

$DV=\phi^{\prime}(\rho)\overline{g}.$

Let $M$ be a smooth closed hypersurface in $\mathbb{H}^{n+1}\ (n\geqslant 2)$
with induced metric $g$ and unit outward normal vector $\nu$. Let
$\\{x^{1},x^{2},\cdots,x^{n}\\}$ be a local coordinate system on $M$, we
denote $g_{ij}=g(\partial_{i},\partial_{j})$ and its second fundamental form
$h_{ij}=\langle D_{\partial_{i}}\nu,\partial_{j}\rangle$. The Weingarten
matrix is denoted by $\mathcal{W}=(h^{i}_{j})$, where
$h^{i}_{j}=g^{ik}h_{kj}$, and $(g^{ij})$ is the inverse matrix of $(g_{ij})$.
The principal curvatures $\kappa=(\kappa_{1},\kappa_{2},\cdots,\kappa_{n})$
are eigenvalues of the Weingarten matrix $\mathcal{W}$. The connection on $M$
is denoted by $\nabla$.

The following formulas hold for smooth hypersurfaces in $\mathbb{H}^{n+1}$,
see [15, Lemmas 2.2 and 2.6].

######  Lemma 2.1.

Let $(M,g)$ be a smooth hypersurface in $\mathbb{H}^{n+1}$ with local frame
$\\{\partial_{1},\cdots,\partial_{n}\\}$, then $\Phi|_{M}$ satisfies

$\nabla_{i}\Phi=\langle V,e_{i}\rangle,\ \
\nabla_{j}\nabla_{i}\Phi=\phi^{\prime}g_{ij}-uh_{ij},$ (2.1)

and the support function $u=\langle V,\nu\rangle$ satisfies

$\nabla_{i}u=\langle V,e_{l}\rangle h_{i}^{l},\ \
\nabla_{j}\nabla_{i}u=\langle V,\nabla
h_{ij}\rangle+\phi^{\prime}h_{ij}-u(h^{2})_{ij},$ (2.2)

where $(h^{2})_{ij}=h_{ik}h^{k}_{j}$.

Then we have the first and second derivatives of the distance function $\rho$.

######  Corollary 2.2 ([19, Corollary 2.1]).

We have

$\nabla_{i}\rho=\dfrac{\langle V,e_{i}\rangle}{\phi},\ \
\nabla_{j}\nabla_{i}\rho=\dfrac{\phi^{\prime}}{\phi}(g_{ij}-\nabla_{i}\rho\nabla_{j}\rho)-\dfrac{uh_{ij}}{\phi}.$
(2.3)

###### Proof.

Observe that

$\nabla_{i}\Phi=\phi\nabla_{i}\rho,\quad\nabla_{j}\nabla_{i}\Phi=\phi\nabla_{j}\nabla_{i}\rho+\phi^{\prime}\nabla_{j}\rho\nabla_{i}\rho.$

Combining with (2.1), we get (2.3) by a direct calculation. ∎

The $k$th elementary symmetric polynomial of $\kappa$ is defined by

$\sigma_{k}(\kappa)=\sum\limits_{1\leqslant i_{1}<i_{2}<\cdots<i_{k}\leqslant
n}\kappa_{i_{1}}\kappa_{i_{2}}\cdots\kappa_{i_{k}}.$

We also have the convention that $\sigma_{0}=1$ and $\sigma_{k}=0$ for $k>n$.
This includes the mean curvature $H=\sigma_{1}(\kappa)$ and Gauss curvature
$K=\sigma_{n}(\kappa)$ as special cases. We have the following Minkowski
formulas.

######  Lemma 2.3 ([15, Proposition 2.5]).

Let $M$ be a smooth closed hypersurface in $\mathbb{H}^{n+1}$. Then

$(n-k)\int_{M}\phi^{\prime}\sigma_{k}(\kappa)\mathrm{d}\mu=(k+1)\int_{M}u\sigma_{k+1}(\kappa)\mathrm{d}\mu,\
\ k=0,1,\cdots,n-1,$ (2.4)

where $\sigma_{k}(\kappa)$ is the $k$th elementary symmetric polynomial of
$\kappa$.

###  2.2. Quermassintegrals

Recall that for a convex domain $\Omega$ in hyperbolic space
$\mathbb{H}^{n+1}$, the quermassintegrals of $\Omega$ are defined as follows
(see [23, Definition 2.1]):

$\mathscr{A}_{k}(\Omega)=(n-k){n\choose
k}\dfrac{\omega_{k}\cdots\omega_{0}}{\omega_{n-1}\cdots\omega_{n-k-1}}\int_{\mathcal{L}_{k+1}}\chi(L_{k+1}\cap\Omega)\mathrm{d}L_{k+1},\
k=0,1,\cdots,n-1,$

where $\omega_{k}=|\mathbb{S}^{k}|$ denotes the area of $k$-dimensional
sphere, $\mathcal{L}_{k+1}$ is the space of $k+1$-dimensional totally geodesic
subspaces $L_{k+1}$ in $\mathbb{H}^{n+1}$. The function $\chi$ is defined to
be 1 if $L_{k+1}\cap\Omega\neq\varnothing$ and to be 0 otherwise. Furthermore,
we set

$\mathscr{A}_{-1}(\Omega)=|\Omega|,\ \
\mathscr{A}_{0}(\Omega)=|\partial\Omega|,\ \
\mathscr{A}_{n}(\Omega)=|\mathbb{B}^{n+1}|=\dfrac{\omega_{n}}{n+1}.$

The quermassintegrals and the curvature integrals of a smooth domain $\Omega$
in $\mathbb{H}^{n+1}$ are related by the following equations:

$\displaystyle\mathscr{A}_{1}(\Omega)=\int_{\partial\Omega}\sigma_{1}(\kappa)\mathrm{d}\mu_{g}-n\mathscr{A}_{-1}(\Omega),$
$\displaystyle\mathscr{A}_{k}(\Omega)=\int_{\partial\Omega}\sigma_{k}(\kappa)\mathrm{d}\mu_{g}-\dfrac{n-k+1}{k-1}\mathscr{A}_{k-2}(\Omega),\
\ k=2,\cdots,n.$ (2.5)

The quermassintegrals for smooth domains satisfy a nice variation property:

$\displaystyle\dfrac{\mathrm{d}}{\mathrm{d}t}\mathscr{A}_{-1}(\Omega_{t})=\int_{M_{t}}f\mathrm{d}\mu_{t},$
(2.6)
$\displaystyle\dfrac{\mathrm{d}}{\mathrm{d}t}\mathscr{A}_{k}(\Omega_{t})=(k+1)\int_{M_{t}}f\sigma_{k+1}(\kappa)\mathrm{d}\mu_{t},\
\ k=0,1,\cdots,n-1$ (2.7)

along any outward normal variation with speed $f$.

###  2.3. Evolution equations

For convenience, we rewrite the flow (1.1) as

$\partial_{t}X(x,t)=(\phi^{\prime}-\Theta)\nu(x,t),$ (2.8)

where $\Theta=uK^{\frac{1}{n}}$. The following lemma can be found in [15,
Lemma 3.1] and [19, Lemma 2.2]. We include a proof here for reader’s
convenience.

######  Lemma 2.4.

Along the flow (2.8), we have the following evolution equations. The induced
metric evolves by

$\partial_{t}g_{ij}=-2\Theta h_{ij}+2\phi^{\prime}h_{ij},$ (2.9)

The support function evolves by

$\partial_{t}u=-\phi^{\prime}\Theta+(\phi^{\prime})^{2}-\phi^{2}|\nabla\rho|^{2}+\langle
V,\nabla\Theta\rangle,$ (2.10)

The Weingarten matrix evolves by

$\partial_{t}h_{i}^{j}=\nabla_{i}\nabla^{j}\Theta+\Theta
h_{i}^{k}h_{k}^{j}-\phi^{\prime}h_{i}^{k}h_{k}^{j}+uh_{i}^{j}-\Theta\delta_{i}^{j},$
(2.11)

where $\nabla$ is the Levi-Civita connection of the induced metric on $M_{t}$.

###### Proof.

By a direct calculation, we have

$\displaystyle\partial_{t}g_{ij}=\partial_{t}\langle\partial_{i}X,\partial_{j}X\rangle$
$\displaystyle=\langle
D_{i}((\phi^{\prime}-\Theta)\nu(x,t)),\partial_{j}X\rangle+\langle\partial_{i}X,D_{j}((\phi^{\prime}-\Theta)\nu(x,t))\rangle$
$\displaystyle=(\phi^{\prime}-\Theta)(\langle
D_{i}\nu,\partial_{j}X\rangle+\langle\partial_{i}X,D_{j}\nu\rangle)=-2\Theta
h_{ij}+2\phi^{\prime}h_{ij}.$

Since $\partial_{t}\nu$ is tangential,

$\displaystyle\partial_{t}\nu=\langle\partial_{t}\nu,\partial_{i}X\rangle
g^{il}\partial_{l}X$
$\displaystyle=-\langle\nu,\partial_{i}((\phi^{\prime}-\Theta)\nu)\rangle
g^{il}\partial_{l}X$ (2.12)
$\displaystyle=\partial_{i}(-\phi^{\prime}+\Theta)g^{il}\partial_{l}X$
$\displaystyle=\nabla(-\phi^{\prime}+\Theta)$
$\displaystyle=\nabla\Theta-\phi\nabla\rho.$

Using (2.12), we obtain the evolution of the support function $u$ as follows:

$\displaystyle\partial_{t}u=\partial_{t}\langle V,\nu\rangle$
$\displaystyle=\phi^{\prime}\langle(\phi^{\prime}-\Theta)\nu,\nu\rangle+\langle
V,\nabla\Theta-\phi\nabla\rho\rangle$
$\displaystyle=-\phi^{\prime}\Theta+(\phi^{\prime})^{2}-\phi^{2}|\nabla\rho|^{2}+\langle
V,\nabla\Theta\rangle.$

Now we calculate the evolution of $h_{ij}$ as

$\displaystyle\partial_{t}h_{ij}$ $\displaystyle=-\partial_{t}\langle
D_{\partial_{i}X}\partial_{j}X,\nu\rangle$ $\displaystyle=-\langle
D_{\partial_{i}X}D_{\partial_{j}X}((\phi^{\prime}-\Theta)\nu),\nu\rangle-R^{\mathbb{H}^{n+1}}(\partial_{i}X,\partial_{t}X,\partial_{j}X,\nu)-\langle
D_{\partial_{i}X}\partial_{j}X,\nabla\Theta-\nabla\phi^{\prime}\rangle$
$\displaystyle=\partial_{i}\partial_{j}(\Theta-\phi^{\prime})-(\Theta-\phi^{\prime})(h^{2})_{ij}+(\phi^{\prime}-\Theta)g_{ij}-\langle
D_{\partial_{i}X}\partial_{j}X,\nabla\Theta-\nabla\phi^{\prime}\rangle$
$\displaystyle=\nabla_{i}\nabla_{j}(\Theta-\phi^{\prime})-(\Theta-\phi^{\prime})h_{i}^{k}h_{kj}+(\phi^{\prime}-\Theta)g_{ij}$
$\displaystyle=\nabla_{i}\nabla_{j}\Theta-\Theta
h_{i}^{k}h_{kj}+\phi^{\prime}h_{i}^{k}h_{kj}-\Theta g_{ij}+uh_{ij},$

where we used (2.1) in the last equality. From (2.9), we have

$\partial_{t}g^{ij}=-g^{il}(\partial_{t}g_{lm})g^{mj}=2\Theta
g^{il}h_{l}^{j}-2\phi^{\prime}g^{il}h_{l}^{j}.$

Thus

$\partial_{t}h_{i}^{j}=\partial_{t}h_{il}g^{lj}+h_{il}\partial_{t}g^{lj}=\nabla_{i}\nabla^{j}\Theta+\Theta
h_{i}^{k}h_{k}^{j}-\phi^{\prime}h_{i}^{k}h_{k}^{j}+uh_{i}^{j}-\Theta\delta_{i}^{j}.$

∎

###  2.4. Parametrization by radial graph

For a closed star-shaped hypersurface $M\subset\mathbb{H}^{n+1}$, we can
parametrize it as a graph of the radial function
$\rho(\theta):\mathbb{S}^{n}\rightarrow\mathbb{R}^{+}$, i.e.,

$M^{n}=\\{(\rho(\theta),\theta)\mid\rho:\mathbb{S}^{n}\rightarrow\mathbb{R}^{+},\theta\in\mathbb{S}^{n}\\},$

where $\theta=(\theta_{1},\cdots,\theta_{n})$ is a local normal coordinate
system of $\mathbb{S}^{n}$ and $\rho$ is a smooth function on
$\mathbb{S}^{n}$. Let $f_{i}=\overline{\nabla}_{i}f$,
$f_{ij}=\overline{\nabla}^{2}_{ij}f$, where $\overline{\nabla}$ is the Levi-
Civita connection on $\mathbb{S}^{n}$ with respect to the standard metric
$g_{\mathbb{S}^{n}}$.

The tangent space of $M^{n}$ is spanned by
$X_{i}=\rho_{i}\partial_{\rho}+\partial_{\theta_{i}}$ (see [18, Section 2.3])
and the unit outward normal vector is

$\nu=\dfrac{\partial_{\rho}-\frac{\rho^{i}\partial_{\theta_{i}}}{\phi^{2}}}{w},$

where we set

$w=\sqrt{1+\frac{|\overline{\nabla}\rho|^{2}}{\phi^{2}}}.$ (2.13)

Then the support function and the induced metric can be expressed as

$u=\dfrac{\phi^{2}}{\sqrt{\phi^{2}+|\overline{\nabla}\rho|^{2}}}=\dfrac{\phi}{w},$
(2.14) $g_{ij}=\phi^{2}\delta_{ij}+\rho_{i}\rho_{j},\quad
g^{ij}=\dfrac{1}{\phi^{2}}(\delta^{ij}-\dfrac{\rho_{i}\rho_{j}}{\phi^{2}+|\overline{\nabla}\rho|^{2}}).$
(2.15)

The second fundamental form is given by

$h_{ij}=\dfrac{-\phi\rho_{ij}+2\phi^{\prime}\rho_{i}\rho_{j}+\phi^{2}\phi^{\prime}\delta_{ij}}{\sqrt{\phi^{2}+|\overline{\nabla}\rho|^{2}}},$
(2.16)

and we have the Weingarten matrix

$h_{i}^{j}=\dfrac{1}{\phi^{2}\sqrt{\phi^{2}+|\overline{\nabla}\rho|^{2}}}(\delta^{jk}-\dfrac{\rho_{j}\rho_{k}}{\phi^{2}+|\overline{\nabla}\rho|^{2}})(-\phi\rho_{ki}+2\phi^{\prime}\rho_{k}\rho_{i}+\phi^{2}\phi^{\prime}\delta_{ki}).$
(2.17)

Similar to [15], the flow (2.8) can be written as a scalar parabolic PDE for
the radial function $\rho$:

$\begin{cases}\partial_{t}\rho(\theta,t)=-\phi
K^{\frac{1}{n}}+\phi^{\prime}w,\quad\mbox{for}\
(\theta,t)\in\mathbb{S}^{n}\times[0,+\infty),\\\
\rho(\cdot,0)=\rho_{0}(\cdot),\end{cases}$ (2.18)

where $w$ is the function defined in (2.13).

Let $\gamma(\rho)$ satisfy

$\dfrac{\mathrm{d}\gamma}{\mathrm{d}\rho}=\dfrac{1}{\phi(\rho)},$ (2.19)

then the flow (2.8) can be equivalent to the scalar parabolic equation for
$\gamma$ on $\mathbb{S}^{n}$:

$\partial_{t}\gamma=\dfrac{\partial_{t}\rho}{\phi(\rho)}=\dfrac{\phi^{\prime}}{\phi}w-K^{\frac{1}{n}},\
\ (\theta,t)\in\mathbb{S}^{n}\times[0,+\infty).$ (2.20)

##  3\. $C^{0}$ and $C^{1}$ estimates

In this section, we will establish the $C^{0}$ and $C^{1}$ estimates of the
flow (1.1) for the proof of Theorem 1.1. Especially, we show that the flow
hypersurface $M_{t}$ preserves star-shapedness along (1.1).

###  3.1. $C^{0}$ Estimate

We first use the maximum principle to show that the radial function $\rho$ of
(2.18) has uniform bounds.

######  Lemma 3.1.

Let $\rho(\cdot,t)$ be a smooth, positive, uniformly convex solution to (2.18)
on $\mathbb{S}^{n}\times[0,T)$. Then

$\underset{\mathbb{S}^{n}}{\min}\rho(\cdot,0)\leq\rho(\cdot,t)\leq\underset{\mathbb{S}^{n}}{\max}\rho(\cdot,0),\qquad\forall\
t\in[0,T).$ (3.1)

###### Proof.

Fix time $t$ and suppose that $\rho$ attains its spatial maximum at point
$(p_{0},t)$. At $(p_{0},t)$, we have $|\overline{\nabla}\rho|=0$ and
$(\rho_{ij})\leq 0$. From (2.17),

$h_{i}^{j}=\dfrac{\phi^{\prime}}{\phi}\delta_{i}^{j}-\dfrac{1}{\phi^{2}}\rho_{i}^{j}\geq\dfrac{\phi^{\prime}}{\phi}\delta_{i}^{j}.$
(3.2)

Hence we have $K^{\frac{1}{n}}\geq\dfrac{\phi^{\prime}}{\phi}$. Substituting
it into (2.18), we obtain

$\partial_{t}\rho_{\max}\leq-\phi\dfrac{\phi^{\prime}}{\phi}+\phi^{\prime}=0.$
(3.3)

This implies that $\underset{\mathbb{S}^{n}}{\max}\rho(\cdot,t)$ is non-
increasing. In particular,

$\rho(\cdot,t)\leq\underset{\mathbb{S}^{n}}{\max}\rho(\cdot,0),\qquad\forall\
t\in[0,T).$

Similarly, $\underset{\mathbb{S}^{n}}{\min}\rho(\cdot,t)$ is non-decreasing,
and

$\rho(\cdot,t)\geq\underset{\mathbb{S}^{n}}{\min}\rho(\cdot,0),\qquad\forall\
t\in[0,T).$

This proves the lemma. ∎

###  3.2. $C^{1}$ Estimate

We now give the uniform upper bound of the gradient of $\rho$. It is obtained
by combining [13, Theorem 2.7.10] and the $C^{0}$ estimate (Lemma 3.1).

######  Lemma 3.2.

Let $\rho(\cdot,t)$ be a smooth, positive, uniformly convex solution to (2.18)
on $\mathbb{S}^{n}\times[0,T)$. Then

$|\overline{\nabla}\rho|\leq C,$ (3.4)

where $C$ depends only on the initial hypersurface.

###### Proof.

Since $\rho(\cdot,t)$ is a smooth uniformly convex solution to (2.18) on
$\mathbb{S}^{n}\times[0,T)$, we have
$|\overline{\nabla}\rho|(\cdot,t)\leqslant
C\left(\max_{\mathbb{S}^{n}}\rho(\cdot,t)-\min_{\mathbb{S}^{n}}\rho(\cdot,t)\right)$
by [13, Theorem 2.7.10]. Combining the $C^{0}$ estimate (3.1), we obtain
(3.4), where the positive constant $C$ depends only on the initial
hypersurface. ∎

Along the flow (2.8), star-shapedness preservation follows by combining Lemmas
3.1 and 3.2.

######  Corollary 3.3.

Along the flow (2.8), the uniformly convex hypersurface $M_{t}$ preserves
star-shapedness and the support function $u$ satisfies

$\frac{1}{C}\leq u\leq C$ (3.5)

for all $t\in[0,T)$ for some constant $C>0$, where $C$ only depends on the
initial hypersurface.

###### Proof.

Recall (2.14),

$u=\dfrac{\phi}{\sqrt{1+\frac{|\overline{\nabla}\rho|^{2}}{\phi^{2}}}}.$

The upper and lower bounds of $u$ follow from Lemmas 3.1 and 3.2. Besides, we
have

$\langle\partial_{\rho},\nu\rangle=\dfrac{u}{\phi}=\dfrac{1}{w}=\dfrac{1}{\sqrt{1+\frac{|\overline{\nabla}\rho|^{2}}{\phi^{2}}}}\geq\dfrac{1}{C^{{}^{\prime}}}$

for some $C^{\prime}>0$ depends on
$\max_{\mathbb{S}^{n}\times[0,T)}|\overline{\nabla}\rho|$ and
$\min_{\mathbb{S}^{n}\times[0,T)}\rho$. Thus the hypersurface $M_{t}$
preserves star-shapedness along the flow (2.8). ∎

##  4\. $C^{2}$ estimate

In this section, we will prove the $C^{2}$ estimate of the flow (2.8).

###  4.1. The upper bound of Gauss curvature $K$

We show that $K$ is bounded above along (2.8).

######  Lemma 4.1.

Let $X(\cdot,t)$ be a uniformly convex solution to the flow (2.8) which
encloses the origin for $t\in[0,T)$. Then there is a positive constant C
depending only on n, $M_{0}$ and the uniform bounds of $\rho$ such that

$K(\cdot,t)\leq C,\qquad\forall\ t\in[0,T).$ (4.1)

###### Proof.

Consider the auxiliary function

$Q=\log\Theta-\log(u-a),$

where $\Theta=uK^{\frac{1}{n}}$ and $a=\frac{1}{2}\inf_{M\times[0,T)}u>0$.

Using (2.10) and (2.11), we then have

$\displaystyle\partial_{t}\Theta=$
$\displaystyle\partial_{t}uK^{\frac{1}{n}}+\frac{1}{n}\dfrac{\Theta}{K}\dfrac{\partial
K}{\partial h_{i}^{j}}\left(\nabla_{i}\nabla^{j}\Theta+\Theta
h_{i}^{k}h_{k}^{j}-\phi^{\prime}h_{i}^{k}h_{k}^{j}+uh_{i}^{j}-\Theta\delta_{i}^{j}\right)$
(4.2) $\displaystyle=$
$\displaystyle[(\phi^{\prime})^{2}-\phi^{2}|\nabla\rho|^{2}-\phi^{\prime}\Theta+\langle
V,\nabla\Theta\rangle]K^{\frac{1}{n}}$
$\displaystyle+\frac{1}{n}\dfrac{\Theta}{K}\dot{K}^{ij}\Theta_{ij}+\dfrac{\Theta^{2}}{n}H-\dfrac{\Theta\phi^{\prime}}{n}H+u\Theta-\dfrac{\Theta^{2}}{n}\dfrac{\sigma_{n-1}}{K},$

where we used

$\dfrac{\partial K}{\partial
h_{i}^{j}}\delta_{i}^{j}=\sigma_{n-1}\quad\mbox{and}\quad\dfrac{\partial
K}{\partial h_{i}^{j}}h_{i}^{k}h_{k}^{j}=KH$

in the last equality and denote $\dot{K}^{ij}=\dfrac{\partial K}{\partial
h_{i}^{j}}$. Recall (2.10),

$\displaystyle\partial_{t}u=$
$\displaystyle(\phi^{\prime})^{2}-\phi^{2}|\nabla\rho|^{2}-\phi^{\prime}\Theta+\langle
V,\nabla\Theta\rangle$ $\displaystyle=$
$\displaystyle(\phi^{\prime})^{2}-\phi^{2}|\nabla\rho|^{2}-\phi^{\prime}\Theta+K^{\frac{1}{n}}\langle
V,\nabla u\rangle+\dfrac{uK^{\frac{1}{n}-1}}{n}\langle V,\nabla K\rangle$
$\displaystyle=$
$\displaystyle(\phi^{\prime})^{2}-\phi^{2}|\nabla\rho|^{2}-\phi^{\prime}\Theta+K^{\frac{1}{n}}\langle
V,\nabla u\rangle+\dfrac{\Theta}{nK}\dot{K}^{ij}\langle V,\nabla
h_{ij}\rangle.$

Combining with (2.2), we obtain the evolution equation of $u$:

$\partial_{t}u=(\phi^{\prime})^{2}-\phi^{2}|\nabla\rho|^{2}-2\phi^{\prime}\Theta+K^{\frac{1}{n}}\langle
V,\nabla u\rangle+\dfrac{u\Theta}{n}H+\dfrac{\Theta}{nK}\dot{K}^{ij}u_{ij}.$
(4.3)

At the spatial maximum point $(p_{0},t)$ of $Q$ on $M_{t}$, we have

$\dfrac{\nabla\Theta}{\Theta}=\dfrac{\nabla u}{u-a}\quad\mbox{and}\quad
Q_{ij}=\dfrac{\Theta_{ij}}{\Theta}-\dfrac{u_{ij}}{u-a}\leq 0.$ (4.4)

Combining with (4.2), (4.3) and (4.4), we obtain

$\displaystyle\partial_{t}Q=$
$\displaystyle\dfrac{\partial_{t}\Theta}{\Theta}-\dfrac{\partial_{t}u}{u-a}$
$\displaystyle=$
$\displaystyle\dfrac{((\phi^{\prime})^{2}-\phi^{2}|\nabla\rho|^{2})}{u}-\dfrac{\phi^{\prime}\Theta}{u}+\dfrac{\langle
V,\nabla\Theta\rangle}{u}+\dfrac{1}{nK}\dot{K}^{ij}\Theta_{ij}+\dfrac{1}{n}\Theta
H-\dfrac{\phi^{\prime}}{n}H+u-\dfrac{\Theta}{n}\dfrac{\sigma_{n-1}}{K}$
$\displaystyle-\dfrac{((\phi^{\prime})^{2}-\phi^{2}|\nabla\rho|^{2})}{u-a}+2\dfrac{\phi^{\prime}\Theta}{u-a}-\dfrac{u}{n(u-a)}\Theta
H-\dfrac{\Theta}{n(u-a)K}\dot{K}^{ij}u_{ij}-\dfrac{\Theta}{u(u-a)}\langle
V,\nabla u\rangle$ $\displaystyle\leq$
$\displaystyle\dfrac{\Theta}{nK}Q_{ij}-\dfrac{a}{u(u-a)}((\phi^{\prime})^{2}-\phi^{2}|\nabla\rho|^{2})+u+\left(\dfrac{2\phi^{\prime}}{u-a}-\dfrac{\phi^{\prime}}{u}\right)\Theta-\dfrac{a}{n(u-a)}\Theta
H,$

where we used (4.4) in the last inequality. Then we use the arithmetic mean
and geometric mean inequality $H\geq nK^{\frac{1}{n}}$ and the $C^{0}$
estimate to obtain

$\partial_{t}Q\leq C_{1}+C_{2}\Theta-C_{3}\Theta^{2}$

at the spatial maximum point $(p_{0},t)$ for some $C_{1},C_{2},C_{3}>0$
depending only on $n$, $M_{0}$ and the uniform bounds of $\rho$. Besides,
there is a constant $C_{4}$ depending on the uniform upper and lower bounds of
$u$ such that

$\dfrac{1}{C_{4}}\mathrm{e}^{Q}\leq\Theta=(u-a)\mathrm{e}^{Q}\leq
C_{4}\mathrm{e}^{Q}.$

We finally obtain that

$\partial_{t}Q\leq
C_{1}+C_{2}C_{4}\mathrm{e}^{Q}-C_{3}(C_{4})^{-2}\mathrm{e}^{2Q}$

holds at the spatial maximum point $(p_{0},t)$. Therefore,
$Q\leq\max\\{C_{5},Q_{\max}(0)\\}$, where $C_{5}$ is a positive constant
depending on $C_{1},C_{2},C_{3}$ and $C_{4}$. Hence
$K=(\frac{u-a}{u}\mathrm{e}^{Q})^{n}$ is bounded from above by some positive
constant $C$ depending only on $n$, $M_{0}$ and the uniform bounds of $\rho$.
∎

###  4.2. The bound of principal curvatures

We show that the principal curvatures of $M_{t}$ are uniformly bounded along
the flow (2.8).

######  Lemma 4.2.

Let $X(\cdot,t)$ be a uniformly convex solution to the flow (2.8) which
encloses the origin for $t\in[0,T)$. Then there is a positive constant $C$
depending only on n, $M_{0}$ and the uniform bounds of $\rho$ such that the
principal curvatures satisfy

$\dfrac{1}{C}\leq\kappa_{i}\leq C$ (4.5)

for all $t\in[0,T)$ and $i=1,2,\dots,n$.

###### Proof.

Denote by $\lambda(x,t)$ the maximal principal radii at $X(x,t)$. Let $A$ and
$B$ be the positive constants to be determined later. Let
$\tilde{Q}=\log\lambda(x,t)-Au+B\rho$. Fix an arbitrary time $T_{0}\in[0,T)$.
Assume that $\tilde{Q}$ attains its maximum at $(x_{0},t_{0})$ provided
$t\in[0,T_{0}]$. We then introduce a normal coordinate system
$\\{\partial_{i}\\}$ around $(x_{0},t_{0})$ such that
$\nabla_{\partial_{i}X}\partial_{j}X(x_{0},t_{0})=0$ for all $i,j=1,2,\dots,n$
and $h_{ij}(x_{0},t_{0})=\kappa_{i}(x_{0},t_{0})\delta_{ij}$. Furthermore, we
can choose $\partial_{1}(x_{0},t_{0})$ as the eigenvector with respect to
$\lambda(x_{0},t_{0})$, i.e., $\lambda(x_{0},t_{0})=h^{11}(x_{0},t_{0})$.
Assume that $\\{h^{ij}\\}$ is the inverse matrix of $\\{h_{ij}\\}$. Clearly,

$\lambda(x,t)=\max\\{h^{ij}(x,t)\xi_{i}\xi_{j}\,|\,g^{ij}(x,t)\xi_{i}\xi_{j}=1\\}.$
(4.6)

For the continuity, using this coordinate system, we consider the auxiliary
function

$Q=\log v-Au+B\rho,$ (4.7)

where $v=\dfrac{h^{11}}{g^{11}}$. Note that $v(x,t)\leq\lambda(x,t)$ from
(4.6) and $v(x_{0},t_{0})=\lambda(x_{0},t_{0})$. Thus
$Q(x,t)\leq\tilde{Q}(x_{0},t_{0})$ for all $t\in[0,T_{0}]$.

Now we calculate the derivatives of $v$ at $(x_{0},t_{0})$ as follows:

$\displaystyle\partial_{t}v$
$\displaystyle=-(h^{11})^{2}\partial_{t}h_{11}+h^{11}\partial_{t}g_{11}=-(h^{11})^{2}\partial_{t}h_{1}^{1},$
(4.8) $\displaystyle\nabla_{i}v$
$\displaystyle=\dfrac{1}{g^{11}}\dfrac{\partial h^{11}}{\partial
h_{pq}}\nabla_{i}h_{pq}=-\dfrac{1}{g^{11}}h^{1p}h^{1q}\nabla_{i}h_{pq},\quad\nabla_{i}v(x_{0},t_{0})=-(h^{11})^{2}\nabla_{i}h_{11}.$

Then

$\displaystyle\nabla_{j}\nabla_{i}v(x_{0},t_{0})$
$\displaystyle=\nabla_{j}(-\dfrac{1}{g^{11}}h^{1p}h^{1q}\nabla_{i}h_{pq})$
$\displaystyle=-h^{1q}\nabla_{j}h^{1p}\nabla_{i}h_{pq}-h^{1p}\nabla_{j}h^{1q}\nabla_{i}h_{pq}-h^{1p}h^{1q}\nabla_{j}\nabla_{i}h_{pq}$
$\displaystyle=h^{1q}h^{1r}h^{ps}\nabla_{j}h_{rs}\nabla_{i}h_{pq}+h^{1p}h^{1r}h^{qs}\nabla_{j}h_{rs}\nabla_{i}h_{pq}-h^{1p}h^{1q}\nabla_{j}\nabla_{i}h_{pq}$
$\displaystyle=-(h^{11})^{2}\nabla_{j}\nabla_{i}h_{11}+2(h^{11})^{2}h^{pp}\nabla_{i}h_{1p}\nabla_{j}h_{1p}.$

Now we calculate the first term in (2.11) as

$\displaystyle\nabla_{j}\nabla_{i}\Theta$
$\displaystyle=u\nabla_{j}\nabla_{i}K^{\frac{1}{n}}+K^{\frac{1}{n}}\nabla_{j}\nabla_{i}u+\nabla_{j}u\nabla_{i}K^{\frac{1}{n}}+\nabla_{j}K^{\frac{1}{n}}\nabla_{i}u$
$\displaystyle=u\nabla_{j}(\dfrac{1}{n}K^{\frac{1}{n}-1}\dot{K}^{pq}\nabla_{i}h_{pq})+K^{\frac{1}{n}}\nabla_{j}\nabla_{i}u+\dfrac{1}{n}K^{\frac{1}{n}-1}\dot{K}^{pq}\nabla_{i}h_{pq}\nabla_{j}u+\dfrac{1}{n}K^{\frac{1}{n}-1}\dot{K}^{pq}\nabla_{j}h_{pq}\nabla_{i}u$
$\displaystyle=\dfrac{\Theta}{nK}\dot{K}^{pq}\nabla_{j}\nabla_{i}h_{pq}+\dfrac{1}{n}(\dfrac{1}{n}-1)\dfrac{\Theta}{K^{2}}\dot{K}^{pq}\dot{K}^{rs}\nabla_{j}h_{rs}\nabla_{i}h_{pq}+\dfrac{\Theta}{nK}\ddot{K}^{pq,rs}\nabla_{j}h_{rs}\nabla_{i}h_{pq}$
$\displaystyle\qquad+\dfrac{\Theta}{u}\nabla_{j}\nabla_{i}u+\dfrac{\Theta}{nuK}\dot{K}^{pq}\nabla_{i}h_{pq}\nabla_{j}u+\dfrac{\Theta}{nuK}\dot{K}^{pq}\nabla_{j}h_{pq}\nabla_{i}u.$

Due to Codazzi equation and Ricci identity, we have

$\displaystyle\dot{K}^{pq}\nabla_{j}\nabla_{i}h_{pq}$
$\displaystyle=\dot{K}^{pq}\nabla_{j}\nabla_{q}h_{ip}=\dot{K}^{pq}(\nabla_{q}\nabla_{j}h_{ip}+h_{lp}R_{liqj}+h_{li}R_{lpqj})$
$\displaystyle=\dot{K}^{pq}(\nabla_{q}\nabla_{p}h_{ij}+h_{lp}h_{lq}h_{ij}-h_{lp}h_{lj}h_{iq}+h_{li}h_{lq}h_{pj}-h_{li}h_{lj}h_{pq}$
$\displaystyle\qquad\qquad-
h_{lp}\delta_{lq}\delta_{ij}+h_{lp}\delta_{lj}\delta_{iq}-h_{li}\delta_{lq}\delta_{pj}+h_{li}\delta_{lj}\delta_{pq})$
$\displaystyle=\dot{K}^{pq}\nabla_{q}\nabla_{p}h_{ij}+KHh_{ij}-nKh_{il}h_{jl}-nK\delta_{ij}+\dot{K}^{pp}h_{ij}.$

It is direct to calculate that

$\ddot{K}^{pq,rs}\nabla_{j}h_{rs}\nabla_{i}h_{pq}=\dfrac{\partial\left(Kh^{pq}\right)}{\partial
h_{rs}}\nabla_{j}h_{rs}\nabla_{i}h_{pq}=\dfrac{\nabla_{j}K\nabla_{i}K}{K}-Kh^{pp}h^{qq}\nabla_{j}h_{pq}\nabla_{i}h_{pq},$

where we used $\dot{K}^{pq}=Kh^{pq}$ and $\dfrac{\partial h^{pq}}{\partial
h_{rs}}=-h^{pr}h^{qs}$. Hence, at $(x_{0},t_{0})$,

$\displaystyle\nabla_{j}\nabla_{i}\Theta=$
$\displaystyle\dfrac{\Theta}{nK}\dot{K}^{pq}\nabla_{q}\nabla_{p}h_{ij}-\dfrac{\Theta}{n}h^{pp}h^{qq}\nabla_{i}h_{pq}\nabla_{j}h_{pq}$
(4.9) $\displaystyle+\dfrac{\Theta}{n}Hh_{ij}-\Theta
h_{il}h_{jl}-\Theta\delta_{ij}+\dfrac{\Theta}{n}h^{pp}h_{ij}+\dfrac{\Theta}{u}\nabla_{j}\nabla_{i}u$
$\displaystyle+\dfrac{\Theta}{n^{2}}\nabla_{i}\log K\nabla_{j}\log
K+\dfrac{\Theta}{nu}\nabla_{j}\log
K\nabla_{i}u+\dfrac{\Theta}{nu}\nabla_{i}\log K\nabla_{j}u.$

Direct computation gives

$\displaystyle\nabla_{j}\nabla_{i}Q$
$\displaystyle=\dfrac{\nabla_{j}\nabla_{i}v}{v}-\dfrac{\nabla_{j}v\nabla_{i}v}{v^{2}}-A\nabla_{j}\nabla_{i}u+B\nabla_{j}\nabla_{i}\rho$
(4.10)
$\displaystyle=-h^{11}\nabla_{j}\nabla_{i}h_{11}+2h^{11}h^{pp}\nabla_{j}h_{1p}\nabla_{i}h_{1p}-(h^{11})^{2}\nabla_{j}h_{11}\nabla_{i}h_{11}$
$\displaystyle\ \ \ \ -A\nabla_{j}\nabla_{i}u+B\nabla_{j}\nabla_{i}\rho.$

Recall (2.11). By (4.8) and (4.9), at $(x_{0},t_{0})$ we obtain

$\displaystyle\partial_{t}Q$
$\displaystyle=\dfrac{\partial_{t}v}{v}-A\partial_{t}u+B\partial_{t}\rho$
(4.11)
$\displaystyle=-h^{11}(\nabla_{1}\nabla_{1}\Theta+\Theta(h_{11})^{2}-\phi^{\prime}(h_{11})^{2}+uh_{11}-\Theta)-A\partial_{t}u+B\partial_{t}\rho$
$\displaystyle=-\dfrac{\Theta}{nK}\dot{K}^{pq}h^{11}\nabla_{q}\nabla_{p}h_{11}+\dfrac{\Theta}{n}h^{11}h^{pp}h^{qq}\nabla_{1}h_{pq}\nabla_{1}h_{pq}-\dfrac{\Theta}{n}H+\Theta
h_{11}+\Theta h^{11}-\dfrac{\Theta}{n}h^{pp}$
$\displaystyle\quad\quad-\dfrac{\Theta}{u}h^{11}\nabla_{1}\nabla_{1}u-\dfrac{\Theta}{n^{2}}(\nabla_{1}\log
K)^{2}h^{11}-2\dfrac{\Theta}{nu}h^{11}\nabla_{1}\log K\nabla_{1}u-\Theta
h_{11}+\phi^{\prime}h_{11}$ $\displaystyle\quad\quad-u+\Theta
h^{11}-A\partial_{t}u+B\partial_{t}\rho$
$\displaystyle\leq-\dfrac{\Theta}{nK}\dot{K}^{pq}h^{11}\nabla_{q}\nabla_{p}h_{11}+\dfrac{\Theta}{n}h^{11}h^{pp}h^{qq}\nabla_{1}h_{pq}\nabla_{1}h_{pq}+2\Theta
h^{11}-\dfrac{\Theta}{u}h^{11}\nabla_{1}\nabla_{1}u$
$\displaystyle\quad\quad-\dfrac{\Theta}{n^{2}}(\nabla_{1}\log
K)^{2}h^{11}-2\dfrac{\Theta}{nu}h^{11}\nabla_{1}\log
K\nabla_{1}u+\phi^{\prime}h_{11}-A\partial_{t}u+B\partial_{t}\rho.$

Substituting (4.10) into (4.11) and using Cauchy-Schwarz inequality, we get

$\displaystyle\partial_{t}Q$
$\displaystyle\leq\dfrac{\Theta}{nK}\dot{K}^{ij}\nabla_{j}\nabla_{i}Q-2\dfrac{\Theta}{n}h^{11}h^{ij}h^{pp}\nabla_{j}h_{1p}\nabla_{i}h_{1p}+\dfrac{\Theta}{n}(h^{11})^{2}h^{ij}\nabla_{j}h_{11}\nabla_{i}h_{11}$
(4.12)
$\displaystyle\quad\quad+A\dfrac{\Theta}{nK}\dot{K}^{ij}\nabla_{j}\nabla_{i}u-B\dfrac{\Theta}{n}h^{ij}\nabla_{j}\nabla_{i}\rho+\dfrac{\Theta}{n}h^{11}h^{pp}h^{qq}\nabla_{1}h_{pq}\nabla_{1}h_{pq}+2\Theta
h^{11}$
$\displaystyle\quad\quad-\dfrac{\Theta}{u}h^{11}\nabla_{1}\nabla_{1}u+\dfrac{\Theta}{u^{2}}|\nabla_{1}u|^{2}h^{11}+\phi^{\prime}h_{11}-A\partial_{t}u+B\partial_{t}\rho$
$\displaystyle\leq-B\dfrac{\Theta}{n}h^{ij}\nabla_{j}\nabla_{i}\rho+2\Theta
h^{11}+\dfrac{\Theta}{u^{2}}|\nabla_{1}u|^{2}h^{11}+\phi^{\prime}h_{11}+B\partial_{t}\rho$
$\displaystyle\quad\quad+A\dfrac{\Theta}{nK}\dot{K}^{ij}\nabla_{j}\nabla_{i}u-\dfrac{\Theta}{u}h^{11}\nabla_{1}\nabla_{1}u-A\partial_{t}u.$

Recall (2.2), we have $\nabla_{1}u=\phi\nabla_{1}\rho h_{11}$, hence
$|\nabla_{1}u|^{2}\leq\phi^{2}h_{11}^{2}$. Meanwhile, noting that
$\frac{1}{K^{\frac{1}{n}}}\leq h^{11}$, we divide (4.12) by
$\Theta=uK^{\frac{1}{n}}$ on both sides to obtain

$\dfrac{\partial_{t}Q}{\Theta}\leq-\dfrac{B}{n}h^{ij}\nabla_{j}\nabla_{i}\rho+2h^{11}+\dfrac{\phi^{2}}{u^{2}}h_{11}+\dfrac{\phi^{\prime}}{u}+B\dfrac{\partial_{t}\rho}{\Theta}+\dfrac{A}{nK}\dot{K}^{ij}\nabla_{j}\nabla_{i}u-\dfrac{\nabla_{1}\nabla_{1}u}{u}h^{11}-A\dfrac{\partial_{t}u}{\Theta}.$
(4.13)

Recall (2.2) again, we have

$\dfrac{\nabla_{1}\nabla_{1}u}{u}h^{11}=\dfrac{h^{11}}{u}\langle V,\nabla
h_{11}\rangle+\dfrac{\phi^{\prime}}{u}-h_{11}.$

Since $\nabla Q(x_{0},t_{0})=0$, we also have

$\dfrac{\nabla v}{v}=A\nabla u-B\nabla\rho\quad\Rightarrow\quad h^{11}\nabla
h_{11}=-A\nabla u+B\nabla\rho.$

Hence we have

$\dfrac{\nabla_{1}\nabla_{1}u}{u}h^{11}=-\dfrac{A}{u}\langle V,\nabla
u\rangle+\dfrac{B}{u}\langle
V,\nabla\rho\rangle+\dfrac{\phi^{\prime}}{u}-h_{11}.$ (4.14)

Combining (4.3) with (4.14), we obtain

$\displaystyle\quad\quad\dfrac{A}{nK}\dot{K}^{ij}\nabla_{j}\nabla_{i}u-\dfrac{\nabla_{1}\nabla_{1}u}{u}h^{11}-A\dfrac{\partial_{t}u}{\Theta}$
(4.15)
$\displaystyle=-B\dfrac{\phi}{u}|\nabla\rho|^{2}-\dfrac{\phi^{\prime}}{u}+h_{11}-A\dfrac{{\phi^{\prime}}^{2}-\phi^{2}|\nabla\rho|^{2}}{\Theta}+2A\phi^{\prime}-A\dfrac{u}{n}H$
$\displaystyle\leq h_{11}-\dfrac{A}{\Theta}+2A\phi^{\prime},$

where we used
${\phi^{\prime}}^{2}-\phi^{2}|\nabla\rho|^{2}\geq{\phi^{\prime}}^{2}-\phi^{2}=1$.
Differentiating $\langle V,V\rangle=\phi^{2}$, we obtain

$\phi\phi^{\prime}\partial_{t}\rho=\langle\partial_{t}V,V\rangle=\phi^{\prime}\langle\partial_{t}X,V\rangle.$

By (2.8), we have

$\partial_{t}\rho=-\dfrac{\Theta}{w}+\dfrac{\phi^{\prime}}{w}\leq\dfrac{\phi^{\prime}}{w}.$
(4.16)

Substituting (4.15) and (4.16) into (4.13), we get

$\displaystyle\dfrac{\partial_{t}Q}{\Theta}$
$\displaystyle\leq-\dfrac{B}{n}h^{ij}\nabla_{j}\nabla_{i}\rho+2h^{11}+\dfrac{\phi^{2}}{u^{2}}h_{11}+B\dfrac{\phi^{\prime}}{w\Theta}+h_{11}-\dfrac{A}{\Theta}+2A\phi^{\prime}$
(4.17)
$\displaystyle\leq-\dfrac{B}{n}h^{ij}\nabla_{j}\nabla_{i}\rho+C_{1}h_{11}+C_{2}h^{11}+AC_{3}$

for some $C_{1},C_{2},C_{3}>0$ depending only on $n$, $M_{0}$ and the uniform
bounds of $\rho$ and provided $A=BC_{5}\geq
2B\sup_{\mathbb{S}^{n}\times[0,T)}\dfrac{\phi^{\prime}}{w}$, where $C_{5}$
depends only on the uniform upper and lower bounds of $\rho$.

Finally, we wipe off some nonpositive terms. Recall (2.3), at $(x_{0},t_{0})$,
we have

$|\nabla\rho|^{2}=\dfrac{\sum\limits_{i=0}^{n}\langle
X_{i},V\rangle^{2}}{\phi^{2}}=\dfrac{|V|^{2}}{\phi^{2}}-\dfrac{\langle\nu,V\rangle}{\phi^{2}}=1-\left(\dfrac{u}{\phi}\right)^{2}\leq
1-C_{6},$ (4.18)

where $C_{6}$ depends on the lower bound of $u$ and the upper bound of $\rho$.
Now substituting (2.3) into (4.17) and using (4.18), we obtain that at
$(x_{0},t_{0})$,

$\displaystyle 0\leq\dfrac{\partial_{t}Q}{\Theta}$
$\displaystyle\leq-\dfrac{B}{n}h^{ii}\left(\dfrac{\phi^{\prime}}{\phi}\left(1-\left(\nabla_{i}\rho\right)^{2}\right)-\dfrac{u}{\phi}h_{ii}\right)+C_{1}h_{11}+C_{2}h^{11}+AC_{3}$
(4.19) $\displaystyle\leq-
BC_{0}h^{11}+BC_{4}+C_{1}\dfrac{1}{h^{11}}+C_{2}h^{11}+AC_{3}$

holds for some positive constants $C_{0},C_{1},C_{2},C_{3},C_{4}$ depending on
the uniform upper and lower bounds of $\rho$ and $n$. We deduce from (4.19)
that if we choose $B=2\dfrac{C_{1}}{C_{0}}$ and $A=BC_{5}$,
$\lambda(x_{0},t_{0})$ cannot be too large, i.e., $\lambda(x_{0},t_{0})$ has a
uniform upper bound $C(C_{0},C_{1},C_{2},C_{3},C_{4},C_{5})$ independent of
time $T_{0}$. Thus, $Q(x_{0},t_{0})\leqslant C$, which implies that

$Q(x,t)\leqslant Q(x_{0},t_{0})\leqslant C.$

Therefore, $\lambda(x,t)\leqslant C$ for all $(x,t)\in M_{t}$, $t\in[0,T)$. It
implies that $\kappa_{i}\geqslant c_{0}\ (i=1,2,\cdots,n)$.

Meanwhile, by Lemma 4.1, we obtain that $C\geq K\geq c_{0}^{n-1}\kappa_{\max}$
for some constant $C$ from Lemma 4.1. Hence the principal curvatures are
bounded from above. This completes the proof of Lemma 4.2. ∎

By Lemma 4.2, equation (2.18) is uniformly parabolic. Combining the $C^{0}$
estimate and the gradient estimate in Lemmas 3.1 and 3.2, one obtains the
Hölder continuity of $\overline{\nabla}^{2}\rho$ and $\partial_{t}\rho$ by [3,
Theorem 6]. Estimates for higher derivatives then follow by the standard
regularity theory for uniformly parabolic equations. Hence we obtain the long
time existence and regularity of solutions for the flow (1.1).

######  Theorem 4.3.

Let $M_{0}$ be a smooth, closed, uniformly convex hypersurface in
$\mathbb{H}^{n+1}$ which encloses the origin. Then the flow (1.1) has a unique
smooth, uniformly convex solution $M_{t}$ for all time $t\in[0,+\infty)$.

Moreover, we have a priori estimates

$\|\rho\|_{C^{k,m}_{(\theta,t)}(\mathbb{S}^{n}\times[0,+\infty))}\leq
C_{k,m},$

where $C_{k,m}>0$ depends only on $k$, $m$, and the geometry of $M_{0}$.

##  5\. convergence

In this section, we will prove the smooth convergence and exponential
convergence of the flow (1.1).

###  5.1. Smooth Convergence

We first consider subsequential convergence.

######  Lemma 5.1.

There exists a sequence $\\{t_{j}\\}$ such that when $t_{j}\to\infty$,
$Q(t_{j})$ monotonically decreasing converges to 0, where

$Q(t)=\int_{M_{t}}\left(\phi^{\prime}(\rho)-uK^{\frac{1}{n}}\right)\mathrm{d}\mu_{t},$
(5.1)

and $M_{t_{j}}$ converges smoothly to a geodesic sphere centered at the
origin.

###### Proof.

Note that $\dfrac{\mathrm{d}}{\mathrm{d}t}\mathscr{A}_{-1}(\Omega_{t})=Q(t)$,
then

$\int_{0}^{t}Q(\tau)\mathrm{d}\tau=\mathscr{A}_{-1}(\Omega_{t})-\mathscr{A}_{-1}(\Omega).$
(5.2)

By $K^{\frac{1}{n}}\leqslant\dfrac{H}{n}$ and Minkowski formulas (2.4) when
$k=0$, we know

$Q(t)\geqslant\dfrac{1}{n}\int_{M_{t}}\left(n\phi^{\prime}-uH\right)\mathrm{d}\mu_{t}=0.$

By $C^{0}$ estimate,
$\mathscr{A}_{-1}(\Omega_{t})\leqslant|\mathbb{S}^{n}|\Phi(\max\limits_{\mathbb{S}^{n}}\rho_{0})$.
Thus $Q(t)$ is a nonnegative $L^{1}(0,\infty)$ function. Then by Arzelà-Ascoli
theorem and uniform regularity estimates, there exists $\\{t_{j}\\}$ such that
$M_{t_{j}}$ converges smoothly to $M_{\infty}$ and the following holds:

$\int_{M_{\infty}}\left(\phi^{\prime}(\rho)-uK^{\frac{1}{n}}\right)\mathrm{d}\mu_{\infty}=0.$

Thus, we have

$\int_{M_{\infty}}u\left(K^{\frac{1}{n}}-\dfrac{H}{n}\right)\mathrm{d}\mu_{\infty}=0.$
(5.3)

Since $\frac{1}{C}\leqslant u\leqslant C$ by Corollary 3.3, we have
$K^{\frac{1}{n}}=\dfrac{H}{n}$ on $M_{\infty}$. Therefore, $M_{\infty}$ is a
geodesic sphere.

We now prove that $M_{\infty}$ is a geodesic sphere centered at the origin.
Suppose the radius of $M_{\infty}$ is $\rho_{\infty}$, its principal
curvatures are
$\kappa_{i}=\dfrac{\phi^{\prime}(\rho_{\infty})}{\phi(\rho_{\infty})}\
(i=1,\cdots,n)$. We argue by contradiction. Suppose $M_{\infty}$ is not
centered at the origin, then at the point where $\rho$ attains its spatial
maximum of $\theta\in\mathbb{S}^{n}$,

$\dfrac{\partial\rho}{\partial
t}=\phi^{\prime}(\rho)-\phi(\rho)\dfrac{\phi^{\prime}(\rho_{\infty})}{\phi(\rho_{\infty})}=\dfrac{\phi(\rho_{\infty}-\rho)}{\phi(\rho_{\infty})}<0,$
(5.4)

which contradicts that $M_{\infty}$ is the stationary solution to the flow
(1.1). ∎

Then we get smooth convergence.

######  Proposition 5.2.

There exists $\rho_{\infty}>0$ such that along the flow (1.1), when
$t\to\infty$, there holds
$\|\rho(\cdot,t)-\rho_{\infty}\|_{C^{k}(\mathbb{S}^{n})}\to 0$, $\forall\
k\in\mathbb{N}$.

###### Proof.

We have proved that there exist $t_{j}\to\infty$ and $\rho_{\infty}$ such that
$\|\rho(\cdot,t_{j})-\rho_{\infty}\|_{C^{k}(\mathbb{S}^{n})}\to 0,\ \forall\
k\in\mathbb{N}$. Since
$\rho_{\max}(t)=\max\limits_{\mathbb{S}^{n}}\rho(\theta,t)$ is non-increasing,
$\rho_{\min}(t)=\min\limits_{\mathbb{S}^{n}}\rho(\theta,t)$ is non-decreasing,
we have $\|\rho(\cdot,t)-\rho_{\infty}\|_{C^{0}(\mathbb{S}^{n})}\to 0\
(t\to\infty)$.

Recall the special case of Gagliardo-Nirenberg interpolation inequality (see
[22]): For any positive integer $j$ and $k$, where $j<k$, and let
$\delta=\frac{j}{k}$, there exists a constant $C(j,k)$, such that for any
$f\in W^{k,2}(\mathbb{S}^{n})$,

$\left\|\dfrac{\partial^{j}}{\partial\theta^{j}}f\right\|_{L^{2}(\mathbb{S}^{n})}\leqslant\left\|\dfrac{\partial^{k}}{\partial\theta^{k}}f\right\|_{L^{2}(\mathbb{S}^{n})}^{\delta}\|f\|_{L^{2}(\mathbb{S}^{n})}^{1-\delta}.$
(5.5)

Choose $f=\rho(\theta,t)-\rho_{\infty}$, we have

$\left\|\dfrac{\partial^{j}}{\partial\theta^{j}}\rho(\theta,t)\right\|_{L^{2}(\mathbb{S}^{n})}\leqslant\left\|\dfrac{\partial^{k}}{\partial\theta^{k}}\rho(\theta,t)\right\|_{L^{2}(\mathbb{S}^{n})}^{\delta}\|\rho(\theta,t)-\rho_{\infty}\|_{L^{2}(\mathbb{S}^{n})}^{1-\delta}.$
(5.6)

For any positive integer $j$, and any $\varepsilon>0$, for a large enough time
$t$, by the boundedness of $\|\rho\|_{C^{k}(\mathbb{S}^{n})}$ and
$\|\rho(\cdot,t)-\rho_{\infty}\|_{C^{0}(\mathbb{S}^{n})}<\varepsilon$, we know
that
$\left\|\dfrac{\partial^{j}}{\partial\theta^{j}}\rho(\theta,t)\right\|_{L^{2}(\mathbb{S}^{n})}<C\varepsilon$.
Then Sobolev embedding theorem implies that
$\|\rho(\cdot,t)-\rho_{\infty}\|_{C^{j-1}(\mathbb{S}^{n})}<C\varepsilon$.
Hence we prove that $\rho(\cdot,t)$ converges smoothly to $\rho_{\infty}$ as
$t\to\infty$. ∎

###  5.2. Exponential convergence

We now prove the exponential convergence of the flow (1.1).

Let’s first derive the evolution equation of
$\dfrac{|\overline{\nabla}\gamma|^{2}}{2}$ along (2.20).

######  Lemma 5.3.

Let $\xi=\dfrac{|\overline{\nabla}\gamma|^{2}}{2}$, then along the flow
(2.20), there holds

$\displaystyle\mathcal{L}\xi$ $\displaystyle=$
$\displaystyle-\dfrac{1}{n}K^{\frac{1}{n}-1}\dot{K}^{ij}\left[\dfrac{1}{\phi
w^{3}}\xi^{i}\xi_{j}-\dfrac{3}{\phi
w^{5}}\gamma^{k}\xi_{k}\gamma^{i}\xi_{j}-\dfrac{2\phi^{\prime}}{\phi
w^{3}}\xi\gamma^{i}\xi_{j}-\dfrac{\phi^{\prime}}{\phi
w^{3}}\gamma^{k}\xi_{k}\delta^{i}_{j}+\dfrac{1}{\phi
w^{3}}\gamma^{k}\xi_{k}\gamma^{i}_{j}\right]$ (5.7)
$\displaystyle+\dfrac{\phi^{\prime}}{\phi\omega}\gamma^{k}\xi_{k}-\dfrac{1}{n}K^{\frac{1}{n}-1}\dot{K}^{ij}\left[\dfrac{1}{\phi
w}\gamma^{i}_{k}\gamma^{k}_{j}-\dfrac{1}{\phi
w}\gamma^{i}\gamma_{j}+\dfrac{2\phi^{\prime}}{\phi
w}\xi\gamma^{i}_{j}\right]-\dfrac{2w\xi}{\phi},$

where the operator $\mathcal{L}$ is defined by

$\mathcal{L}\xi=\partial_{t}\xi-\dfrac{1}{n\phi
w^{3}}K^{\frac{1}{n}-1}\dot{K}^{ij}(w^{2}e^{ik}-\gamma^{i}\gamma^{k})\overline{\nabla}_{k}\overline{\nabla}_{j}\xi,$
(5.8)

and we denote $\dot{K}^{ij}=\dfrac{\partial K}{\partial h^{i}_{j}}$,
$\gamma^{i}_{j}=e^{ik}\gamma_{kj}$. Here $e_{ij}$ is the standard metric on
$\mathbb{S}^{n}$.

###### Proof.

Recall that

$w=\sqrt{1+|\overline{\nabla}\gamma|^{2}},\ \
\overline{\nabla}w=\dfrac{\overline{\nabla}\xi}{w}.$ (5.9)

By (2.20), we have

$\displaystyle\partial_{t}\left(\dfrac{|\overline{\nabla}\gamma|^{2}}{2}\right)$
$\displaystyle=$
$\displaystyle\overline{\nabla}\gamma\cdot\overline{\nabla}\gamma_{t}$ (5.10)
$\displaystyle=$
$\displaystyle\overline{\nabla}\gamma\cdot\overline{\nabla}\left(\dfrac{\phi^{\prime}}{\phi}w-K^{\frac{1}{n}}\right)$
$\displaystyle=$
$\displaystyle\overline{\nabla}\gamma\cdot\left[\dfrac{w\overline{\nabla}\phi^{\prime}+\phi^{\prime}\overline{\nabla}w}{\phi}-\dfrac{\phi^{\prime}w\overline{\nabla}\phi}{\phi^{2}}-\overline{\nabla}K^{\frac{1}{n}}\right].$

Note that,

$\displaystyle\overline{\nabla}\phi^{\prime}=\phi^{\prime\prime}\phi\overline{\nabla}\gamma,\
\ \overline{\nabla}\phi=\phi^{\prime}\phi\overline{\nabla}\gamma,$
$\displaystyle\overline{\nabla}K^{\frac{1}{n}}=\dfrac{1}{n}K^{\frac{1}{n}-1}\dot{K}^{ij}\overline{\nabla}h^{i}_{j},$
(5.11)

and combining $(\phi^{\prime})^{2}-\phi^{\prime\prime}\phi=1$, we have

$\displaystyle\partial_{t}\xi=-\dfrac{2w\xi}{\phi}+\dfrac{\phi^{\prime}}{\phi
w}\overline{\nabla}\gamma\cdot\overline{\nabla}\xi-\dfrac{1}{n}K^{\frac{1}{n}-1}\dot{K}^{ij}\overline{\nabla}\gamma\cdot\overline{\nabla}h^{i}_{j}.$
(5.12)

Recall that by (2.17) and (2.19),

$h^{i}_{j}=\dfrac{1}{\phi
w}\left(\phi^{\prime}\delta^{i}_{j}-\gamma^{i}_{j}+\dfrac{1}{w^{2}}\gamma^{i}\xi_{j}\right).$
(5.13)

By a direct computation,

$\displaystyle\overline{\nabla}_{k}h^{i}_{j}$ $\displaystyle=$
$\displaystyle\dfrac{1}{\phi
w}\left(\phi^{\prime\prime}\phi\gamma_{k}\delta^{i}_{j}-\overline{\nabla}_{k}\gamma^{i}_{j}+\dfrac{1}{w^{2}}\gamma^{i}_{k}\xi_{j}+\dfrac{1}{w^{2}}\gamma^{i}\xi_{jk}-\dfrac{2}{w^{4}}\xi_{k}\gamma^{i}\xi_{j}\right)$
(5.14)
$\displaystyle-\dfrac{w^{2}\phi^{\prime}\phi\gamma_{k}+\phi\xi_{k}}{\phi^{2}w^{3}}\left(\phi^{\prime}\delta^{i}_{j}-\gamma^{i}_{j}+\dfrac{1}{w^{2}}\gamma^{i}\xi_{j}\right).$

Using Ricci identity to commute the covariant derivatives on the sphere
$\mathbb{S}^{n}$, we obtain

$\displaystyle\gamma^{k}\overline{\nabla}_{k}\gamma^{i}_{j}$ $\displaystyle=$
$\displaystyle\gamma^{k}\left(\overline{\nabla}_{j}\gamma^{i}_{k}+\gamma_{j}\delta^{i}_{k}-\gamma_{k}\delta^{i}_{j}\right)$
(5.15) $\displaystyle=$
$\displaystyle\xi^{i}_{j}-\gamma^{i}_{k}\gamma^{k}_{j}+\gamma^{i}\gamma_{j}-2\xi\delta^{i}_{j}.$

Hence

$\displaystyle\gamma^{k}\overline{\nabla}_{k}h^{i}_{j}$ $\displaystyle=$
$\displaystyle\dfrac{1}{\phi
w^{3}}\left(\gamma^{i}\gamma^{k}-w^{2}e^{ik}\right)\xi_{kj}+\dfrac{1}{\phi
w^{3}}\xi^{i}\xi_{j}-\dfrac{3}{\phi
w^{5}}\gamma^{k}\xi_{k}\gamma^{i}\xi_{j}-\dfrac{2\phi^{\prime}}{\phi
w^{3}}\xi\gamma^{i}\xi_{j}$ (5.16) $\displaystyle-\dfrac{\phi^{\prime}}{\phi
w^{3}}\gamma^{k}\xi_{k}\delta^{i}_{j}+\dfrac{1}{\phi
w^{3}}\gamma^{k}\xi_{k}\gamma^{i}_{j}+\dfrac{1}{\phi
w}\gamma^{i}_{k}\gamma^{k}_{j}-\dfrac{1}{\phi
w}\gamma^{i}\gamma_{j}+\dfrac{2\phi^{\prime}}{\phi w}\xi\gamma^{i}_{j}.$

Substituting (5.16) into (5.12), we obtain the result. ∎

######  Proposition 5.4.

Let $\gamma(\theta,t),\ (\theta,t)\in\mathbb{S}^{n}\times[0,+\infty)$ be a
smooth solution of initial value problem (2.20), and $\rho_{\infty}$ be the
radius of limit geodesic sphere. Then for any positive constant $\alpha$
satisfying

$\alpha<\dfrac{n-1}{n\phi(\rho_{\infty})}$

which depends only on the initial hypersurface $M_{0}$, we have

$|\overline{\nabla}\gamma|^{2}(\theta,t)\leqslant C\mathrm{e}^{-\alpha t},\ \
\forall\ (\theta,t)\in\mathbb{S}^{n}\times[0,+\infty),$

where $C$ is a positive constant depending only on $M_{0}$ and $\alpha$.

###### Proof.

Since $M_{t}$ converges smoothly to a geodesic sphere as $t\to\infty$, there
exists $T_{0}>0$ such that for any $t>T_{0}$,

$\max\limits_{1\leqslant i<j\leqslant
n}|\kappa_{i}-\kappa_{j}|\leqslant\varepsilon$ (5.17)

holds on $M_{t}$ by the regularity estimates of $M_{t}$. We only need to
obtain the estimate of $\xi=\frac{|\overline{\nabla}\gamma|^{2}}{2}$ for large
enough time $t>T_{0}$. We also assume that

$\xi\leqslant\varepsilon<1\ \ \ \mbox{and}\ \ \
\dfrac{1}{n}K^{\frac{1}{n}-1}\dot{K}^{ii}\in(\frac{1}{n}-\varepsilon,\frac{1}{n}+\varepsilon)$

hold on $M_{t}$ for each time $t>T_{0}$.

For any time $t_{0}>T_{0}$, let $\theta_{t_{0}}\in\mathbb{S}^{n}$ be the
spatial maximum point of $\xi$ at $t=t_{0}$. Then at $(\theta_{t_{0}},t_{0})$,
we have

$\overline{\nabla}\xi=0,\ \ \left(\overline{\nabla}^{2}\xi\right)\leqslant 0.$
(5.18)

Note that $\dot{K}^{ij}$ and $(w^{2}e^{ik}-\gamma^{i}\gamma^{k})$ are positive
definite, we get that at $(\theta_{t_{0}},t_{0})$,

$\dfrac{\partial}{\partial t}\xi\leqslant\dfrac{1}{n\phi
w}K^{\frac{1}{n}-1}\dot{K}^{ij}\gamma^{i}\gamma_{j}-\dfrac{2w\xi}{\phi}-\dfrac{2\phi^{\prime}}{\phi
w}\dfrac{1}{n}K^{\frac{1}{n}-1}\dot{K}^{ij}\gamma^{i}_{j}\xi.$ (5.19)

At this point, we choose orthonormal frame such that

$\gamma_{1}=|\overline{\nabla}\gamma|,\ \ \gamma_{j}=0,\ \ 2\leqslant
j\leqslant n,$

and $(\gamma_{ij})$ is diagonal with $\gamma_{1j}=0$ for all $j=1,2,\cdots,n$.
Then the Weingarten matrix

$h^{i}_{j}=\dfrac{1}{\phi
w}\left(-\gamma_{ij}+\phi^{\prime}\delta^{i}_{j}\right)$ (5.20)

is diagonalized at $(\theta_{t_{0}},t_{0})$, and $\gamma_{11}=0$,
$|\gamma_{ii}|=|\gamma_{ii}-\gamma_{11}|\leqslant\varepsilon\phi w$ for
$2\leqslant i\leqslant n$ by (5.17). Now inequality (5.19) becomes

$\displaystyle\dfrac{\partial}{\partial t}\xi$ $\displaystyle\leqslant$
$\displaystyle\dfrac{2}{n\phi
w}K^{\frac{1}{n}-1}\dot{K}^{11}\xi-\dfrac{2w\xi}{\phi}-\dfrac{2\phi^{\prime}}{\phi
w}\left(\dfrac{1}{n}K^{\frac{1}{n}-1}\sum\limits_{i=2}^{n}\dot{K}^{ii}\gamma_{ii}\right)\xi$
(5.21) $\displaystyle\leqslant$
$\displaystyle-\left(\dfrac{n-1}{n\phi}-\tilde{\varepsilon}\right)\xi,$

where $\tilde{\varepsilon}\leqslant C\varepsilon$ is a small constant. Let

$\alpha_{0}=\dfrac{n-1}{n\phi(\rho_{\infty})}.$

Then for any $\alpha-\alpha_{0}$, there exists a time $T_{0}$ depending on
$M_{0}$ and $\alpha_{0}-\alpha$ such that for all $t>T_{0}$, there holds

$\dfrac{\partial}{\partial t}\xi\leqslant-\alpha\xi,\ \ t\in[T_{0},+\infty)$

at the spatial maximum of $\xi$. Integrating the above inequality, we obtain

$\xi(t)\leqslant\xi(T_{0})\mathrm{e}^{-\alpha(t-T_{0})}$

for all time $t\geqslant T_{0}$. This completes the proof. ∎

By Gagliardo-Nirenberg interpolation inequality again, we get the exponential
convergence of the flow (1.1) in $C^{k}$ norm ($k\in\mathbb{N}$). Hence, we
complete the proof of Theorem 1.1.

##  6\. Alexandrov-Fenchel inequalities

As an application, we will compare quermassintegrals
$\mathscr{A}_{n-2}(\Omega)$ and $\mathscr{A}_{-1}(\Omega)$ for a smooth convex
domain $\Omega\subset\mathbb{H}^{n+1}\ (n\geqslant 2)$. Now let us prove
Theorem 1.2.

###### Proof of Theorem 1.2.

We divide the proof into two steps.

Step 1. Assume that $M$ is a smooth uniformly convex hypersurface enclosing a
domain $\Omega$. We evolve $M$ along the flow (1.1) and obtain a family of
uniformly convex hypersurface $M_{t}$ for all time by Theorem 1.1. As
$t\to\infty$, $M_{t}$ converges smoothly and exponentially to a geodesic
sphere $M_{\infty}$ centered at the origin with radius $r_{\infty}$. Let
$\Omega_{\infty}$ be the domain enclosed by $M_{\infty}$. On the one hand,
along the flow (1.1),

$\displaystyle\dfrac{\mathrm{d}}{\mathrm{d}t}\mathscr{A}_{n-2}(\Omega_{t})$
$\displaystyle=$
$\displaystyle(n-1)\int_{M_{t}}\left(\phi^{\prime}(\rho)-uK^{\frac{1}{n}}\right)\sigma_{n-1}(\kappa)\mathrm{d}\mu_{t}$
(6.1) $\displaystyle\leqslant$
$\displaystyle(n-1)\int_{M_{t}}\left(\phi^{\prime}(\rho)\sigma_{n-1}(\kappa)-nu\sigma_{n}(\kappa)\right)\mathrm{d}\mu_{t}=0,$

where we used $\sigma_{n-1}(\kappa)K^{\frac{1}{n}}\geqslant
n\sigma_{n}(\kappa)$ and Minkowski formulas (2.4) with $k=n-1$. Equality holds
in (6.1) if and only if $\Omega_{t}$ is a geodesic ball. On the other hand,

$\displaystyle\dfrac{\mathrm{d}}{\mathrm{d}t}\mathscr{A}_{-1}(\Omega_{t})$
$\displaystyle=$
$\displaystyle\int_{M_{t}}\left(\phi^{\prime}(\rho)-uK^{\frac{1}{n}}\right)\mathrm{d}\mu_{t}$
(6.2) $\displaystyle\geqslant$
$\displaystyle\dfrac{1}{n}\int_{M_{t}}(n\phi^{\prime}(\rho)-uH)\mathrm{d}\mu_{t}=0,$

where we used $K^{\frac{1}{n}}\leqslant\dfrac{H}{n}$ and Minkowski formulas
(2.4) with $k=0$. Equality holds in (6.2) if and only if $\Omega_{t}$ is a
geodesic ball. Thus,

$\mathscr{A}_{n-2}(\Omega)\geqslant\mathscr{A}_{n-2}(\Omega_{\infty})=\xi_{n-2}(r_{\infty}),$
(6.3)

and

$\mathscr{A}_{-1}(\Omega)\leqslant\mathscr{A}_{-1}(\Omega_{\infty})=\xi_{-1}(r_{\infty}).$
(6.4)

Since $\xi_{-1}$ is strictly increasing, its inverse function $\xi_{-1}^{-1}$
is well-defined. Let $\xi_{n-2,-1}=\xi_{n-2}\circ\xi_{-1}^{-1}$, then
inequality (1.4) for uniformly convex $M$ holds. Equality in (1.4) holds if
and only if

$\dfrac{\mathrm{d}}{\mathrm{d}t}\mathscr{A}_{n-2}(\Omega_{t})=\dfrac{\mathrm{d}}{\mathrm{d}t}\mathscr{A}_{-1}(\Omega_{t})=0$

holds for all time, which is equivalent to that $\Omega$ is a geodesic ball.

Step 2. Let $M$ be a smooth convex hypersurface enclosing the domain $\Omega$,
then there exists a sequence of uniformly convex hypersurfaces
$M_{\varepsilon}$ such that $M_{\varepsilon}\to M$ as $\varepsilon\to 0$. In
fact, we can project $\Omega$ into $B_{1}(O)\subset\mathbb{R}^{n+1}$ by Klein
model (see [5, Section 5] and [25, Section 3]).

For convenience, we review the Klein model here. Denote $\mathbb{R}^{1,n+1}$
be the Minkowski space time, which is the vector space $\mathbb{R}^{n+2}$
equipped with Minkowski spacetime metric $\langle\cdot,\cdot\rangle$: For any
vector $X=(X_{0},X_{1},X_{2},\cdots,X_{n+1})\in\mathbb{R}^{n+2}$,

$\langle X,X\rangle=-X_{0}^{2}+\sum\limits_{i=1}^{n+1}X_{i}^{2}.$

The hyperbolic space $\mathbb{H}^{n+1}$ is then viewed as a hypersurface in
$\mathbb{R}^{1,n+1}$,

$\mathbb{H}^{n+1}=\left\\{X\in\mathbb{R}^{1,n+1}:\langle X,X\rangle=-1,\
X_{0}>0\right\\}.$ (6.5)

The Klein model parametrizes the hyperbolic space by using the unit disc,
which induces a projection from an embedding $X:M^{n}\to\mathbb{H}^{n+1}$ to
an embedding $Y:M^{n}\to B_{1}(O)\subset\mathbb{R}^{n+1}$ by

$X=\dfrac{(1,Y)}{\sqrt{1-|Y|^{2}}}.$ (6.6)

Let $\nu\in T\mathbb{H}^{n+1}$, $h_{ij}^{X}$ and $N\in\mathbb{R}^{n+1}$,
$h_{ij}^{Y}$ denote the unit normal vector, the second fundamental form of
$X(M)\subset\mathbb{H}^{n+1}$ and $Y(M)\subset\mathbb{R}^{n+1}$ respectively.
We have the relation:

$h_{ij}^{X}=\dfrac{h_{ij}^{Y}}{\sqrt{(1-|Y|^{2})(1-\langle N,Y\rangle^{2})}}.$
(6.7)

By (6.7) we know the image $\hat{\Omega}$ under projection is also convex. By
mean curvature flow, there exists a sequence of uniformly convex domains
$\hat{\Omega}_{\varepsilon}$ that approximate $\hat{\Omega}$. Since the
projection is a diffeomorphism, we find a family of uniformly convex domains
$\Omega_{\varepsilon}\subset\mathbb{H}^{n+1}$ that approximate $\Omega$.
Inequality (1.4) for convex $M$ follows by letting $\varepsilon\to 0$ in the
inequality (1.4) for the uniformly convex $M_{\varepsilon}$.

Now assume $M$ is convex and equality in (1.4) holds. Let

$M_{+}=\\{p\in M:\kappa_{i}(p)>0,\ 1\leqslant i\leqslant n\\}.$ (6.8)

Since a compact embedded hypersurface in $\mathbb{H}^{n+1}$ contains at least
one elliptic point, $M_{+}$ is a nonempty open set. We claim that $M_{+}$ is
also closed. Following [14], we consider a normal variation of $M$

$\partial_{t}X=-\varphi\nu,$ (6.9)

where $\varphi\in C_{c}^{\infty}(M_{+})$ is a smooth function with compact
support in $M_{+}$. For sufficiently small time
$t\in(-\varepsilon,\varepsilon)$, $M_{t}$ is convex. Set

$Q(t)=\mathscr{A}_{n-2}(\Omega_{t})-\xi_{n-2,-1}\left(\mathscr{A}_{-1}(\Omega_{t})\right),$
(6.10)

then

$Q(0)=0,\ \ Q(t)\geqslant 0,\ \forall\ t\in(-\varepsilon,\varepsilon),$ (6.11)

which implies that $Q(t)$ attains its minimum at $t=0$. Thus,

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\left.\dfrac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\left[\mathscr{A}_{n-2}(\Omega_{t})-\xi_{n-2,-1}\left(\mathscr{A}_{-1}(\Omega_{t})\right)\right]$
(6.12) $\displaystyle=$
$\displaystyle(n-1)\int_{M}\sigma_{n-1}(\kappa)(-\varphi)\mathrm{d}\mu-\xi_{n-2,-1}^{\prime}\left(\mathscr{A}_{-1}(\Omega)\right)\int_{M}(-\varphi)\mathrm{d}\mu$
$\displaystyle=$
$\displaystyle-\int_{M}\left[(n-1)\sigma_{n-1}(\kappa)-\xi_{n-2,-1}^{\prime}\left(\mathscr{A}_{-1}(\Omega)\right)\right]\varphi\mathrm{d}\mu,\
\ \forall\ \varphi\in C_{c}^{\infty}(M_{+}).$

In particular,

$M_{+}=\left\\{p\in
M:(n-1)\sigma_{n-1}(\kappa)=\xi_{n-2,-1}^{\prime}\left(\mathscr{A}_{-1}(\Omega)\right)\right\\}.$
(6.13)

Therefore, $M_{+}$ is closed in $M$ and the connectedness of $M$ implies
$M_{+}=M$. We conclude that $M$ is uniformly convex. By the equality case in
Step 1, we obtain that $\Omega$ is a geodesic ball. Hence we complete the
proof of Theorem 1.2. ∎

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