# Gauss curvature flow with shrinking obstacle

Ki-Ahm Lee  Department of Mathematical Sciences and Research Institute of
Mathematics, Seoul National University, Seoul 08826, Republic of Korea
<EMAIL_ADDRESS>and  Taehun Lee  School of Mathematics, Korea Institute for
Advanced Study, Seoul 02455, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

We consider a flow by powers of Gauss curvature under the obstruction that the
flow cannot penetrate a prescribed region, so called an obstacle. For all
dimensions and positive powers, we prove the optimal curvature bounds of
solutions and all time existence with its long time behavior. We also prove
the $C^{1}$ regularity of free boundaries under a uniform thickness
assumption.

###### Key words and phrases:

Gauss curvature flow, obstacle problem, free boundary problem

###### 2020 Mathematics Subject Classification:

53E10 (35R35, 35K96)

##  1\. Introduction

We study an evolution of hypersurfaces by powers of Gauss curvature under the
restriction that the hypersurface cannot enter a prescribed region. The
prescribed region is called an obstacle, and we assume that it shrinks slowly
with time.

Recall that for a given number $\alpha>0$, a one-parameter family of
immersions $X:M^{n}\times[0,T)\rightarrow{\mathbb{R}}^{n+1}$ defining complete
convex hypersurfaces $\Sigma_{t}=X(M^{n},t)$ is a solution of the
$\alpha$-Gauss curvature flow if $X$ satisfies

$\displaystyle\frac{\partial}{\partial t}X(p,t)=K^{\alpha}(p,t)\vec{n}(p,t).$

Here $K(p,t)$ and $\vec{n}(p,t)$ are the Gauss curvature and the inward unit
normal vector of $\Sigma_{t}$ at $X(p,t)$, respectively.

The classical Gauss curvature flow ($\alpha=1$) was first introduced by Firey
[14] to model the process of wearing stone on a beach. Later, Chow [9]
generalizes this flow to the $\alpha$-Gauss curvature flow ($\alpha>0$). Since
then, many authors have studied this flow, and most of them focused on the
analysis of singularities which was completely carried out when
$\alpha\geq\tfrac{1}{n+2}$, see [2, 3, 6, 5] and the references therein. See
also [4] for the case $0<\alpha<\tfrac{1}{n+2}$ and $n=1$. Note that the
evolution of any closed initial hypersurface under the $\alpha$-Gauss
curvature flow develops a singularity.

We consider an $\alpha$-Gauss curvature flow with an obstacle in
${\mathbb{R}}^{n+1}$, which will block the development of collapsing of the
hypersurface to a point. Precisely, given a strictly convex, closed
hypersurface $\Sigma_{0}$ in ${\mathbb{R}}^{n+1}$, we consider a time
dependent obstacle $\Phi=\Phi_{t}$ inside of $\Sigma_{0}$, which is a one-
parameter family of strictly convex, closed hypersurfaces. We then evolve the
hypersurface $\Sigma_{0}$ by the $\alpha$-Gauss curvature flow, $\alpha>0$, on
the condition that the evolving hypersurface $\Sigma_{t}$ cannot penetrate the
obstacle $\Phi_{t}$ for all time. Here the obstacle shrinks slowly with time,
and for example one can consider obstacles of the form

$\displaystyle\Phi_{t}=\tfrac{1+e^{-t}}{2}\Phi_{0}.$

We remark that if the obstacle shrinks quickly so that it disappears in a
finite time, then it cannot prevent the development of the collapse. The
precise definition of obstacles will be given in Definition 2.1 below.

Note that in general the evolutions of such hypersurfaces cannot be defined in
the classical sense since the speed $\tfrac{\partial}{\partial t}X$ has jump
discontinuities when the hypersurface $\Sigma_{t}$ touches the obstacle
$\Phi_{t}$ for the first time. Thus we need to consider a generalized concept
of solutions, that is, viscosity solutions.

A one-parameter family of immersions
$X:M^{n}\times[0,T)\rightarrow{\mathbb{R}}^{n+1}$ with $\Sigma_{t}=X(M^{n},t)$
is said to be a viscosity solution to the $\alpha$-Gauss curvature flow with a
shrinking obstacle if it satisfies

$\begin{split}\frac{\partial}{\partial
t}X(p,t)&=K^{\alpha}(p,t)\vec{n}(p,t)\quad\text{for all}\quad
X(p,t)\notin\Phi_{t},\\\
\mathrm{conv}(\Sigma_{t})&\supseteq\mathrm{conv}(\Phi_{t})\quad\text{for
all}\quad 0\leq t<T,\\\ \left\langle{\frac{\partial}{\partial
t}X},{\vec{n}}\right\rangle&\leq K^{\alpha}\quad\text{in}\quad
M^{n}\times[0,T),\end{split}$ (1.1)

in the viscosity sense (see Definition 2.3), where $\mathrm{conv(\Sigma)}$
denotes the convex hull of $\Sigma$.

As we discussed above, the solutions to (1.1) are at most Lipschitz continuous
in the time variable, which induces that solutions to (1.1) have at most
$C^{1,1}$ regularity. We will show that solutions to (1.1) have indeed
$C^{1,1}$ regularity, and therefore this is the optimal regularity.

######  Theorem 1.1.

Let $\Sigma_{0}$ be a closed strictly convex smooth hypersurface that encloses
a shrinking obstacle $\Phi$ (Definition 2.1). Then (1.1) starting from
$\Sigma_{0}$ has a unique viscosity solution $\\{\Sigma_{t}\\}_{t\geq 0}$ that
exists for all time, and the solution $\\{\Sigma_{t}\\}_{t\geq 0}$ has the
optimal regularity which is $C^{1,1}(\mathbb{S}^{n}\times[0,\infty))$. In
particular, principal curvatures $\lambda_{1},\ldots,\lambda_{n}$ of
$\Sigma_{t}$ are globally bounded away from zero and infinity. Moreover, there
exists a finite time $T^{*}=T^{*}(n,\alpha,\Sigma_{0},\Phi)$ such that
$\Sigma_{t}=\Phi_{t}$ for $t\geq T^{*}$.

It is worth noting that our result for the optimal regularity relies on
several curvature bounds. We first prove that the several curvatures are
bounded by some constant that depends on $T$. By showing that the final shape
of the solution becomes exactly the same shape as the obstacle after a finite
time $T^{*}$, we can remove the time dependence in the curvature estimates.

To prove curvature estimates, we employ the method of penalization, which
allows penetration of the obstacle up to order of $\delta>0$ and then recovers
the original obstacle problems by pushing the hypersurfaces out of the
obstacle as $\delta\rightarrow 0$. In the other words, we obtain smooth
approximate solutions, $u_{\delta}$, of ($\ast_{\delta}$) that will converge
to the solution of (1.1) as the penalty increases, i.e. $\delta\rightarrow 0$
.

A crucial ingredient in this paper is the smallest curvature estimate (Lemma
6.3) established by applying the maximum principle argument to the quantity

$\displaystyle\lambda_{\min}^{-1}e^{-\chi\beta_{\delta}(u-\varphi)},$

where $\lambda_{\min}=\min\\{\lambda_{1},\ldots,\lambda_{n}\\}$ and
$\beta_{\delta}$ is the penalty term defined in Section 2. The constant $\chi$
will be chosen in terms of the minimum value of Gauss curvature which we will
estimate in Corollary 4.2.

Let us review some related results for stationary obstacles,
$\Phi_{t}\equiv\Phi_{0}$. In [20], we studied the obstacle problem in
dimension $n=2$ with $0<\alpha\leq 1$. In that paper, we established the
optimal curvature bounds for principal curvatures,
$0\leq\lambda_{1},\lambda_{2}\leq C$. We point out that for stationary
obstacles the lowest principal curvature may become zero. For the mean
curvature flow, Rupflin and Schnürer [27] studied an obstacle problem by
considering graphical mean curvature flows in one dimension higher. They
proved local $C^{1,1}$ regularity for solutions in the graphical setting,
which implies the existence of solutions for all time in the original
dimension. See also [1, 15, 25].

Once the optimal regularity of solutions to obstacle problems is established,
the main interest will be a question of regularity of its free boundaries. In
obstacle problems, a free boundary of a solution is the interface between a
coincidence set and non-coincidence set of the solution and its obstacle.

To find a local regularity of free boundaries, we use the following local
graph representation of (1.1): for functions $w:Q_{1}\rightarrow{\mathbb{R}}$
and $\phi:Q_{1}\rightarrow{\mathbb{R}}$ such that $\Sigma_{t}$ and $\Phi_{t}$
are locally represented by the graphs of $w$ and $\phi$, respectively, we
consider

$\displaystyle\frac{\partial}{\partial t}w$
$\displaystyle=\frac{\det(D^{2}w)^{\alpha}}{(1+|Dw|^{2})^{\frac{(n+2)\alpha-1}{2}}}\quad$
$\displaystyle\text{in}\quad\Omega(w),$ (1.2) $\displaystyle w$
$\displaystyle\leq\phi\quad$ $\displaystyle\text{in}\quad Q_{1}$
$\displaystyle\frac{\partial}{\partial t}w$
$\displaystyle\leq\frac{\det(D^{2}w)^{\alpha}}{(1+|Dw|^{2})^{\frac{(n+2)\alpha-1}{2}}}\quad$
$\displaystyle\text{in}\quad Q_{1},$

where $Q_{r}=B_{r}\times(-r^{2},0)\subset{\mathbb{R}}^{n}\times{\mathbb{R}}$
is the parabolic ball of radius $r$ and $\Omega(w)=\\{(x,t)\in
Q_{1}:w(x,t)<\phi(x,t)\\}$ is the non-coincidence set. Note that the
coincidence set $\Lambda(w)$ and the free boundary $\Gamma(w)$ is then given
by $\Lambda(w)=Q_{1}\setminus\Omega(w)$ and $\Gamma(w)=\partial\Omega(w)$.

The regularity of free boundaries has been studied by many authors since the
seminal work of Caffarelli [7] which proves local $C^{1}$ regularity of free
boundaries in obstacle problems for a class of elliptic equations and the
Stefan problem. In the first named author’s thesis [19] and [23], the authors
analyzed the regularity of free boundaries in the obstacle problem for a fully
nonlinear operator. See also [21, 22] for double obstacle problems.

In the recent works [12, 13], Figalli and Shahgholian proved $C^{1}$
regularity of free boundaries for general fully nonlinear elliptic/parabolic
equations of the form $F(D^{2}w)=1$ or $F(D^{2}w)-\partial_{t}w=1$. Soon after
Indrei and Minne [17] extended the result to the equations $F(D^{2}w,x)=f(x)$
and $F(D^{2}w,x,t)-\partial_{t}w=f(x,t)$, where $F$ and $f$ are Lipschitz
continuous in both the space and time variable. We note that in these results
$F$ is assumed to be convex in the hessian variable.

However, our equation (1.2) does not fall into the known cases. This is
because our $F$ has $Dw$ dependence which is not Lipschitz continuous in the
time variable although it is Lipschitz in the space variable after
establishing the optimal $C^{1,1}$ regularity. We stress that most geometric
PDEs have similar gradient dependence on the operator, which requires some
extra work.

As indicated above, the theory for the free boundary regularity has been
developed for convex operators. Hence we assume $\alpha\leq\frac{1}{n}$ to
make the operator $-\det(\cdot)^{\alpha}$ convex. We will prove $C^{1}$
regularity of the free boundary $\Gamma(w)$ under the so called uniform
thickness assumption on the coincidence set $\Lambda(w)$.

To describe the thickness assumption, we first define the minimal diameter
(also known as the minimal width) $\mathrm{MD(E)}$ of a set $E$ by the minimum
distance between two parallel hyperplanes that contain the set $E$. Then the
thickness of $\Lambda(w)$ at $x_{0}$ in each time slice
$t\in[t_{0}-r^{2},t_{0}+r^{2}]$ is measured as

$\displaystyle\mathfrak{d}_{r}(w,x_{0},t_{0})=\inf_{|t-t_{0}|\leq
r^{2}}\frac{\mathrm{MD}(\Lambda(w)^{t}\cap B_{r}(x_{0}))}{r},$

where $\Lambda(w)^{t}=\\{x\in B_{1}:(x,t)\in\Lambda(w)\\}$ is the time slice
of $\Lambda(w)$ with respect to $t$. We say that $\Lambda(w)$ satisfies a
uniform thickness condition if there exists $\varepsilon>0$ such that for all
$0<r<\tfrac{1}{4}$ and $(x,t)\in\Gamma(w)\cap Q_{r}$, it holds
$\mathfrak{d}_{r}(w,x,t)>\varepsilon$.

######  Theorem 1.2.

Let $w$ be a solution to (1.2) with $\alpha\leq\frac{1}{n}$. If $\Lambda(w)$
satisfies a uniform thickness condition, then the free boundary $\Gamma(w)$ is
locally a $C^{1}$ graph in space-time.

The thickness condition is necessary for Theorem 1.2 since the free boundary
$\Gamma(w)$ may have arbitrary lower dimensional shapes such as a point when
solutions to (1.2) touch the obstacle for the first time.

We remark that there is another free boundary problem arising in the Gauss
curvature flow, the Gauss curvature flow with flat sides. Hamilton [16]
observed that if an initial hypersurface has flat sides, then its evolution by
the Gauss curvature flow also has flat sides for some time, in contrast with
the mean curvature flow whose flat sides disappear instantly. In the problem,
free boundaries are the interface between flat sides and non-flat sides.
Comparing with our obstacle problem that has growing contact sets, this free
boundary problem has shrinking flat sides. In dimension $n=2$, regularity for
the free boundary has been established by Daskalopoulos and Hamilton [10] for
short time and by Daskalopoulos and the first named author [11] for all time.
See also [8] for waiting time effects when an initial data is smooth enough.
Note that the same phenomena were observed for the $\alpha$-Gauss curvature
flow when $\frac{1}{2}<\alpha\leq 1$ and $n=2$, see [18].

The paper is organized as follows. In Section 2, we fix geometric notation and
define a shrinking obstacle. We also introduce a penalty term to approximate
(1.1) and present short time existence and evolution equations for the
approximate solutions. In Section 3, we prove that the penalty term is
bounded, which ensures uniform curvature estimates later and then the
convergence of approximating solutions to solutions of the obstacle problem.
In Section 4-6, we establish several curvature bounds possibly depending on
the existing time $T$. By estimating the speed function, we prove upper and
lower bounds for the Gauss curvature in Section 4 and 5, respectively. We then
obtain both the upper and lower bounds for principal curvatures in Section 6.
The proof of Theorem 1.1 and 1.2 are presented respectively in Section 7 and
8.

##  2\. Preliminaries

Throughout the paper we use the Einstein summation convention in which
repeated upper and lower indices are summed.

For a smooth hypersurface $\Sigma^{n}$ in ${\mathbb{R}}^{n+1}$, we denote the
induced metric by $g=\\{g_{ij}\\}$, the second fundamental form by
$h=\\{h_{ij}\\}$, and the Weingarten map by $W=\\{h_{i}^{j}\\}$. The principal
curvatures are the eigenvalues of $W$ denoted by
$\lambda_{1}\leq\cdots\leq\lambda_{n}$. We will use the mean curvature
$H=\lambda_{1}+\cdots+\lambda_{n}$ and the Gauss curvature
$K=\lambda_{1}\cdots\lambda_{n}$.

If $\Sigma^{n}$ is compact without boundary and strictly convex, then the
Gauss map $\nu:\Sigma^{n}\rightarrow\mathbb{S}^{n}$ given by the outward unit
normal vector to $\Sigma^{n}$ is a diffeomorphism. In the case we describe the
hypersurface $\Sigma^{n}$ as its support function
$u:\mathbb{S}^{n}\rightarrow{\mathbb{R}}$ defined by
$u(z)=\left\langle{\nu^{-1}(z)},{z}\right\rangle$. In other words, the
hypersurface $\Sigma^{n}$ is the image of the embedding
$\nu^{-1}(z)=\overline{\nabla}u(z)+u(z)z$, where $\overline{\nabla}$ is the
Levi-Civita connection on $\mathbb{S}^{n}$ induced by the standard metric
$\overline{g}=\\{\overline{g}_{ij}\\}$. It can be checked that the second
fundamental form $h$ is expressed as

$\displaystyle
h_{ij}=\overline{\nabla}_{i}\overline{\nabla}_{j}u+u\overline{g}_{ij},$ (2.1)

and that the eigenvalues of $h_{ij}$ with respect to $\overline{g}$ are the
reciprocal of the principal curvatures, i.e.,

$\displaystyle h_{ij}\overline{g}^{jk}=g_{ij}b^{jk}$ (2.2)

where $b=\\{b^{ij}\\}$ is the inverse of the second fundamental form $h$.

Initial data. Let $\Sigma_{0}$ be a strictly convex, closed initial
hypersurface embedded in ${\mathbb{R}}^{n+1}$ and
$u_{0}:\mathbb{S}^{n}\rightarrow{\mathbb{R}}$ be the support function of
$\Sigma_{0}$. Both $\Sigma_{0}$ and $u_{0}$ are used as initial data.
Geometric quantities associated to the initial data will be used with
subscript $0$, for example $H_{0}$, $K_{0}$, and so on.

Obstacle. Let $\\{\Omega(t)\\}_{t\geq 0}$ be a family of bounded open sets
that are strictly convex. We always assume that $\Omega(t)$ shrinks in $t$,
that is, $\Omega(t_{1})\supset\Omega(t_{2})$ for $t_{1}<t_{2}$, and that the
limit $\Omega_{\infty}:=\cap_{t\geq 0}\Omega(t)$ has an interior point. For
convenience, we assume that the limit obstacle $\Omega_{\infty}$ contains the
origin. We also assume that the initial obstacle $\Omega(0)$ is strictly
enclosed by the initial data, i.e.,
$\overline{\Omega(0)}\subset\operatorname{conv}(\Sigma_{0})$, where
$\operatorname{conv}(\Sigma_{0})$ denotes the convex hull of $\Sigma_{0}$. We
denote the boundary of $\Omega(t)$ by $\Phi_{t}$ and the support function of
$\Phi_{t}$ by $\varphi(\cdot,t):\mathbb{S}^{n}\rightarrow{\mathbb{R}}$. Denote
furthermore the principal curvatures of $\Phi_{t}$ by $\mu_{1},\cdots,\mu_{n}$
(not necessarily in increasing order). Other geometric quantities associated
to the obstacle will be used with subscript $\Phi$, for example $H_{\Phi}$,
$K_{\Phi}$, and so on.

######  Definition 2.1 (shrinking obstacles).

An obstacle $\Phi$ is said to be shrinking if its support function $\varphi$
is in $C^{3,1}(\mathbb{S}^{n}\times[0,\infty))$ and satisfies (i)
$\partial_{t}\varphi<0$, (ii) the speed $-\partial_{t}\varphi$ is non-
increasing, (iii) the final shape $\operatorname{conv}(\Phi_{\infty})$ has an
interior point, and (iv) the principal curvatures $\mu_{1},\ldots,\mu_{n}$ are
non-decreasing and bounded.

A natural example of the shrinking obstacle is any strictly convex
hypersurface that homothetically shrinks with decreasing speed, i.e., if
$A:[0,\infty)\rightarrow(0,1]$ with $A(0)=1$ and
$\lim_{t\rightarrow\infty}A(t)>0$ is decreasing and convex, then
$A(t)\varphi_{0}(z)$ is a shrinking obstacle. More generally, given an initial
shape $\Omega_{0}$ and a final shape $\Omega_{\infty}$ with their support
function $\varphi_{0}$ and $\varphi_{\infty}$ satisfying
$\varphi_{0}>\varphi_{\infty}>0$, the obstacle defined by

$\displaystyle\varphi(\cdot,t)=e^{-t}\varphi_{0}+(1-e^{-t})\varphi_{\infty}$

is clearly shrinking.

Note that for $a>0$, a rescaled obstacle
$\varphi^{a}(\cdot,t)=\varphi(\cdot,at)$ is also shrinking and has the same
initial and final shape with $\varphi$. Moreover, given an initial data
$\Sigma_{0}$, the following holds if we choose $a$ small enough:

$\displaystyle-\partial_{t}\varphi^{a}(\cdot,0)=-a\partial_{t}\varphi_{0}<\min_{\mathbb{S}^{n}}K_{0}^{\alpha}\quad\text{and}\quad-\partial_{t}\varphi^{a}(\cdot,0)<\min_{\mathbb{S}^{n}}K_{\Phi_{0}}^{\alpha}.$

Thus we additionally assume for simplicity that the shrinking obstacle
satisfies

$\displaystyle-\partial_{t}\varphi_{0}<\min_{\mathbb{S}^{n}}K_{0}^{\alpha}\quad\text{and}\quad-\partial_{t}\varphi_{0}<\min_{\mathbb{S}^{n}}K_{\Phi_{0}}^{\alpha}.$

######  Remark 2.2.

The condition (ii) and (iv) in Definition 2.1 ensure that the obstacle is a
supersolution of the $\alpha$-Gauss curvature flow. Indeed, since the
principal curvatures $\mu_{i}$ of $\Phi(\cdot,t)$ are non-decreasing and the
speed $-\partial_{t}\varphi$ is non-increasing, we obtain that
$-\partial_{t}\varphi\leq-\partial_{t}\varphi_{0}\leq
K_{\Phi_{0}}^{\alpha}\leq K_{\Phi_{t}}^{\alpha}$. We point out that if the
obstacle $\Phi$ is not a supersolution, then the solution to (1.1) might be
separated from the obstacle after it has the same shape with the obstacle.

Let $M^{n}$ be an $n$-dimensional smooth manifold, and let
$X(\cdot,t):M^{n}\rightarrow{\mathbb{R}}^{n+1}$ ($0\leq t<T$) be a one-
parameter family of smooth immersions for some $T\in(0,\infty]$ with an image
$\Sigma_{t}=X(M^{n},t)$ which is a strictly convex, closed hypersurface. For
the family $\\{\Sigma_{t}\\}$, we still denote by $g=\\{g_{ij}\\}$ the induced
metrics and by $h=\\{h_{ij}\\}$ the second fundamental forms, where they
depend on the variable $t$. We say that
$u:\mathbb{S}^{n}\times[0,T)\rightarrow{\mathbb{R}}$ is a support function of
the family $\\{\Sigma_{t}\\}$ if
$u(\cdot,t):\mathbb{S}^{n}\rightarrow{\mathbb{R}}$ is the support function of
$\Sigma_{t}$.

In terms of the support function, we rewrite the obstacle problem (1.1) as

$\displaystyle-\partial_{t}u$ $\displaystyle=K^{\alpha}\quad$
$\displaystyle\text{in}\quad\\{u>\varphi\\},$ (2.3) $\displaystyle u$
$\displaystyle\geq\varphi\quad$
$\displaystyle\text{in}\quad\mathbb{S}^{n}\times[0,T),$
$\displaystyle-\partial_{t}u$ $\displaystyle\leq K^{\alpha}\quad$
$\displaystyle\text{in}\quad\mathbb{S}^{n}\times[0,T).$

Here $u$ and $\varphi$ are the support functions of $\Sigma_{t}$ and
$\Phi_{t}$, respectively, and $K$ is the Gauss curvature of $\Sigma_{t}$.
Moreover, by (2.1) and (2.2), the Gauss curvature becomes

$\displaystyle
K=\det(g^{ij}h_{jk})=\det(b^{ij}\overline{g}_{jk})=\frac{\det(\overline{g}_{ij})}{\det(h_{ij})}=\frac{\det(\overline{g}_{ij})}{\det(\overline{\nabla}_{i}\overline{\nabla}_{j}u+u\overline{g}_{ij})}.$
(2.4)

Combining (2.3) and (2.4), we rewrite the obstacle problem (1.1) again as

$\displaystyle\min\left\\{\partial_{t}u+\left(\frac{\det(\overline{g}_{ij})}{\det(\overline{\nabla}_{i}\overline{\nabla}_{j}u+u\overline{g}_{ij})}\right)^{\alpha},u-\varphi\right\\}=0.$
(2.5)

We recall the notion of viscosity solutions to (2.5) or equivalently (1.1).

######  Definition 2.3 (viscosity solution).

A continuous function $u\in C(\mathbb{S}^{n}\times[0,T))$ is a viscosity
subsolution (supersolution) of (2.5) if for any point
$(x_{0},t_{0})\in\mathbb{S}^{n}\times(0,T)$, the left hand side of (2.5) is
nonpositive (nonnegative) for all test functions $\eta\in
C^{2}(\mathbb{S}^{n}\times[0,T))$ touching $u$ from above (below) at
$(x_{0},t_{0})$, i.e., $\eta(x_{0},t_{0})=u(x_{0},t_{0})$ and
$\eta(x,t)\geq(\leq)u(x,t)$. We say that $u$ is a viscosity solution of (2.5)
if it is both a subsolution and supersolution of (2.5).

Let $\beta:{\mathbb{R}}\rightarrow{\mathbb{R}}$ be a non-decreasing, concave,
smooth function such that

$\displaystyle\beta(x)=0\text{ for }x\geq
1,\quad\beta^{\prime\prime}(x)=0\text{ for }x<0,\quad\beta(0)=-1.$

Given a constant $\delta>0$ and an shrinking obstacle $\Phi$ in Definition
2.1, we define $\beta_{\delta}(x)=C_{0}\beta(x/\delta)$, where
$C_{0}=\left\lVert
K_{\Phi}\right\rVert_{L^{\infty}(\mathbb{S}^{n}\times[0,\infty))}^{\alpha}<\infty$.
For convenience, we choose $\delta<\min_{\mathbb{S}^{n}}(u_{0}-\varphi_{0})$
so that $\beta_{\delta}(u-\varphi)=0$ at the initial time.

To prove the existence and regularity results, we approximate (2.3) by the
following singular perturbation problem:

$\displaystyle-\partial_{t}u$
$\displaystyle=K^{\alpha}+\beta_{\delta}(u-\varphi)\quad$
$\displaystyle\text{in}\quad\mathbb{S}^{n}\times(0,T),$ ($\ast_{\delta}$)
$\displaystyle u(\cdot,0)$ $\displaystyle=u_{0}\quad$
$\displaystyle\text{on}\quad\mathbb{S}^{n}.$

We say that a one-parameter family $\\{\Sigma_{t}\\}_{0\leq t\leq T}$ of
hypersurfaces $\Sigma_{t}$ is a solution to ($\ast_{\delta}$) if the support
function of $\Sigma_{t}$ satisfies ($\ast_{\delta}$). We write
$\beta_{\delta}$ instead of $\beta_{\delta}(u-\varphi)$.

Let ${\mathcal{L}}$ denote the highest order terms of the linearized operator
of $-K^{\alpha}$. Then it follows from (2.4) that

$\displaystyle{\mathcal{L}}=\alpha
K^{\alpha}b^{ij}\overline{\nabla}_{i}\overline{\nabla}_{j}.$

The associated inner product
$\left\langle{\cdot},{\cdot}\right\rangle_{\mathcal{L}}$ and norm
$\left\lVert\cdot\right\rVert_{\mathcal{L}}$ are defined by
$\left\langle{\overline{\nabla}A},{\overline{\nabla}B}\right\rangle_{\mathcal{L}}=\alpha
K^{\alpha}b^{ij}\overline{\nabla}_{i}A\overline{\nabla}_{j}B$ and
$\left\lVert\overline{\nabla}A\right\rVert_{\mathcal{L}}^{2}=\left\langle{\overline{\nabla}A},{\overline{\nabla}A}\right\rangle_{\mathcal{L}}$,
respectively, for smooth functions $A$ and $B$ on $\mathbb{S}^{n}\times[0,T)$.

The short time existence follows from the standard argument as in [9] (see
also [20] for instance) that is based on the inverse function theorem. In
fact, since the linearized operator of $\beta_{\delta}(u-\varphi)$ does not
produce any second order term, the proof in [9] can be applied without
modifications.

######  Lemma 2.4 (Short time existence).

If $\Sigma_{0}=X_{0}(M^{n})$ is a strictly convex hypersurface, then
($\ast_{\delta}$) has a unique smooth solution $\\{\Sigma_{t}\\}_{0\leq t<T}$
for short time.

Under the flow ($\ast_{\delta}$), geometric quantities also evolve.

######  Lemma 2.5.

Let $\\{\Sigma_{t}\\}$ be a solution to ($\ast_{\delta}$) and
$u:\mathbb{S}^{n}\times(0,T]\rightarrow{\mathbb{R}}$ be the support function
of $\\{\Sigma_{t}\\}$. Then the following holds:

$\displaystyle(\partial_{t}-\mathcal{L})u$ $\displaystyle=\alpha
K^{\alpha}Hu-(n\alpha+1)K^{\alpha}-\beta_{\delta}$ (2.6)
$\displaystyle(\partial_{t}-\mathcal{L})K^{\alpha}$ $\displaystyle=\alpha
K^{2\alpha}H+{\mathcal{L}}\beta_{\delta}+\alpha K^{\alpha}H\beta_{\delta}$
(2.7) $\displaystyle(\partial_{t}-\mathcal{L})h_{ij}$
$\displaystyle=-\mathcal{C}_{ij}+(n\alpha-1)K^{\alpha}\overline{g}_{ij}-\alpha
K^{\alpha}Hh_{ij}-\overline{\nabla}^{2}_{ij}\beta_{\delta}-\beta_{\delta}\overline{g}_{ij}$
(2.8)

where
$\mathcal{C}_{ij}=\alpha^{2}K^{\alpha}b^{kl}b^{pq}\overline{\nabla}_{i}h_{pq}\overline{\nabla}_{j}h_{kl}+\alpha
K^{\alpha}b^{kp}b^{lq}\overline{\nabla}_{i}h_{pq}\overline{\nabla}_{j}h_{kl}$.

###### Proof.

By (2.1) and (2.2), we have

$\displaystyle{\mathcal{L}}u=\alpha
K^{\alpha}b^{ij}\overline{\nabla}_{i}\overline{\nabla}_{j}u=\alpha
K^{\alpha}b^{ij}(h_{ij}-u\overline{g}_{ij})=n\alpha K^{\alpha}-\alpha
K^{\alpha}Hu.$

Then the equation (2.6) follows from
$\partial_{t}u=-K^{\alpha}-\beta_{\delta}$.

To prove (2.7), we recall (2.4) so that

$\displaystyle\partial_{t}K=-Kb^{ij}\partial_{t}h_{ij}=Kb^{ij}(\overline{\nabla}_{i}\overline{\nabla}_{j}(-\partial_{t}u)+(-\partial_{t}u)\overline{g}_{ij}).$

Using $-\partial_{t}u=K^{\alpha}+\beta_{\delta}$ and (2.2) again, we see that

$\displaystyle\partial_{t}K^{\alpha}=\alpha
K^{\alpha-1}\partial_{t}K={\mathcal{L}}(K^{\alpha}+\beta_{\delta})+(K^{\alpha}+\beta_{\delta})\alpha
K^{\alpha}H$

which gives (2.7).

For the last assertion, note first that

$\displaystyle\partial_{t}h_{ij}=-\overline{\nabla}_{i}\overline{\nabla}_{j}(K^{\alpha}+\beta_{\delta})-(K^{\alpha}+\beta_{\delta})\overline{g}_{ij}.$
(2.9)

On the other hand, since $\overline{\nabla}_{j}K^{\alpha}=-\alpha
K^{\alpha}b^{kl}\overline{\nabla}_{j}h_{kl}$ and
$\overline{\nabla}_{i}b^{kl}=-b^{kp}b^{lq}\overline{\nabla}_{i}h_{pq}$, we
have

$\begin{split}\overline{\nabla}_{i}\overline{\nabla}_{j}K^{\alpha}&=\alpha^{2}K^{\alpha}b^{kl}b^{pq}\overline{\nabla}_{i}h_{pq}\overline{\nabla}_{j}h_{kl}+\alpha
K^{\alpha}b^{kp}b^{lq}\overline{\nabla}_{i}h_{pq}\overline{\nabla}_{j}h_{kl}-\alpha
K^{\alpha}b^{kl}\overline{\nabla}_{i}\overline{\nabla}_{j}h_{kl}\\\
&=\mathcal{C}-\alpha
K^{\alpha}b^{kl}\overline{\nabla}_{i}\overline{\nabla}_{j}h_{kl}.\end{split}$
(2.10)

To proceed further we need the following result that we will prove in the next
lemma:

$\displaystyle\overline{\nabla}_{k}\overline{\nabla}_{l}h_{ij}=\overline{\nabla}_{i}\overline{\nabla}_{j}h_{kl}-h_{kj}\overline{g}_{il}+\overline{g}_{kl}h_{ij}-h_{kl}\overline{g}_{ij}+\overline{g}_{kj}h_{il}.$

Multiplying with $\alpha K^{\alpha}b^{kl}$ we get

$\displaystyle{\mathcal{L}}h_{ij}=\alpha
K^{\alpha}b^{kl}\overline{\nabla}_{i}\overline{\nabla}_{j}h_{kl}+\alpha
K^{\alpha}(-\overline{g}_{ij}+Hh_{ij}-n\overline{g}_{ij}+\overline{g}_{ij}),$

and substituting this into (2.10) we obtain

$\displaystyle\overline{\nabla}_{i}\overline{\nabla}_{j}K^{\alpha}=\mathcal{C}_{ij}-{\mathcal{L}}h_{ij}+\alpha
K^{\alpha}(Hh_{ij}-n\overline{g}_{ij}).$

This together with (2.9) gives the desired result. ∎

######  Lemma 2.6.

The second fundamental form $h_{ij}$ satisfies

  1. (i)

$\overline{\nabla}h$ is totally symmetric, i.e.,
$\overline{\nabla}_{k}h_{ij}=\overline{\nabla}_{i}h_{kj}=\overline{\nabla}_{j}h_{ki}$,

  2. (ii)

$\overline{\nabla}_{k}\overline{\nabla}_{l}h_{ij}=\overline{\nabla}_{i}\overline{\nabla}_{j}h_{kl}-h_{kj}\overline{g}_{il}+\overline{g}_{kl}h_{ij}-h_{kl}\overline{g}_{ij}+\overline{g}_{kj}h_{il}$.

###### Proof.

Since $\overline{\nabla}_{k}h_{ij}=\overline{\nabla}_{k}h_{ji}$, it suffices
to show that $\overline{\nabla}_{k}h_{ij}=\overline{\nabla}_{i}h_{kj}$. Recall
that $h_{ij}=\overline{\nabla}_{i}\overline{\nabla}_{j}u+u\overline{g}_{ij}$
and $\overline{\nabla}\overline{g}=0$. Then we compute

$\displaystyle\overline{\nabla}_{k}h_{ij}$
$\displaystyle=\overline{\nabla}_{k}(\overline{\nabla}_{i}\overline{\nabla}_{j}u+u\overline{g}_{ij})=\overline{\nabla}_{i}\overline{\nabla}_{k}\overline{\nabla}_{j}u-\overline{R}_{kipj}\overline{\nabla}^{p}u+(\overline{\nabla}_{k}u)\overline{g}_{ij},$

where $\overline{R}_{kipj}$ denotes the Riemannian curvature tensor on the
sphere. Since

$\displaystyle\overline{R}_{ijkl}=\overline{g}_{ik}\overline{g}_{jl}-\overline{g}_{il}\overline{g}_{jk},$
(2.11)

we find

$\displaystyle\overline{\nabla}_{k}h_{ij}=\overline{\nabla}_{i}(h_{kj}-u\overline{g}_{kj})-(\overline{g}_{kp}\overline{g}_{ij}-\overline{g}_{kj}\overline{g}_{ip})\overline{\nabla}^{p}u+(\overline{\nabla}_{k}u)\overline{g}_{ij}=\overline{\nabla}_{i}h_{kj},$

and the first assertion follows.

To prove the second assertion, note that we have shown
$\overline{\nabla}_{l}h_{ij}=\overline{\nabla}_{i}h_{lj}$. Again by (2.11), we
obtain that

$\displaystyle\overline{\nabla}_{k}\overline{\nabla}_{l}h_{ij}$
$\displaystyle=\overline{\nabla}_{k}\overline{\nabla}_{i}h_{lj}=\overline{\nabla}_{i}\overline{\nabla}_{k}h_{lj}-\overline{R}_{kipl}\overline{g}^{pr}h_{rj}-\overline{R}_{kipj}\overline{g}^{pr}h_{rl}$
$\displaystyle=\overline{\nabla}_{i}\overline{\nabla}_{j}h_{kl}-(\overline{g}_{kp}\overline{g}_{il}-\overline{g}_{kl}\overline{g}_{ip})\overline{g}^{pr}h_{rj}-(\overline{g}_{kp}\overline{g}_{ij}-\overline{g}_{kj}\overline{g}_{ip})\overline{g}^{pr}h_{rl}$
$\displaystyle=\overline{\nabla}_{i}\overline{\nabla}_{j}h_{kl}-h_{kj}\overline{g}_{il}+\overline{g}_{kl}h_{ij}-h_{kl}\overline{g}_{ij}+\overline{g}_{kj}h_{il}$

and the conclusion follows. ∎

##  3\. Uniform boundedness of the penalty term

To extract a solution from approximate solutions of ($\ast_{\delta}$), we need
several estimates that will be presented in this and the following three
sections. In this section we establish the uniform boundedness of the penalty
term $\beta_{\delta}(u-\varphi)$. Once we obtain the estimate, any possible
limit of subsequence of the approximate solutions can not pass through the
obstacle $\Phi$. Indeed, since $\beta_{\delta}(x)\rightarrow-\infty$ as
$\delta\rightarrow 0$ for each $x<0$, we conclude from the uniform boundedness
of $\beta_{\delta}(u-\varphi)$ that

$\displaystyle\liminf_{\delta\rightarrow 0}(u-\varphi)\geq 0.$

######  Lemma 3.1.

Let $u$ be a solution of ($\ast_{\delta}$) in $\mathbb{S}^{n}\times[0,T)$.
Then

$\displaystyle-C_{0}\leq\beta_{\delta}(u(z,t)-\varphi(z,t))\leq
0\quad\text{for}\quad(z,t)\in\mathbb{S}^{n}\times[0,T),$

where $C_{0}=\left\lVert
K_{\Phi}\right\rVert_{L^{\infty}(\mathbb{S}^{n}\times[0,\infty))}^{\alpha}$ is
independent of $\delta$.

###### Proof.

The inequality $\beta_{\delta}\leq 0$ follows from the definition of
$\beta_{\delta}$. To obtain the lower bound, assume that there exists a point
$(z_{0},t_{0})$ in $\mathbb{S}^{n}\times[0,T)$ such that $u=\varphi$ at
$(z_{0},t_{0})$ for the first time. Since the initial hypersurface
$\Sigma_{0}$ strictly encloses the initial obstacle $\Phi_{0}$, we see
$t_{0}>0$, and $u-\varphi$ attains an interior minimum at $(z_{0},t_{0})$ over
$\mathbb{S}^{n}\times[0,t_{0}]$. At the minimum point, we have

$\displaystyle\overline{\nabla}(u-\varphi)=0,\quad\overline{\nabla}^{2}(u-\varphi)\geq
0,\quad(u-\varphi)_{t}\leq 0,\quad u=\varphi,$

and it follows

$\displaystyle
K=\frac{\det\overline{g}_{ij}}{\det(\overline{\nabla}_{i}\overline{\nabla}_{j}u+u\overline{g}_{ij})}\leq\frac{\det\overline{g}_{ij}}{\det(\overline{\nabla}_{i}\overline{\nabla}_{j}\varphi+\varphi\overline{g}_{ij})}=K_{\Phi}.$

Then

$\displaystyle-\varphi_{t}\leq-u_{t}=K^{\alpha}+\beta_{\delta}(u-\varphi)\leq
K_{\Phi}^{\alpha}+\beta_{\delta}(0),$

and hence it is contradict to our choice of $\beta_{\delta}(0)=-\left\lVert
K_{\Phi}\right\rVert_{L^{\infty}(\mathbb{S}^{n}\times[0,\infty))}^{\alpha}$ in
the definition of the penalty term since $\varphi_{t}<0$. Therefore, we
conclude $u>\varphi$ for all $(z,t)\in\mathbb{S}^{n}\times[0,T)$. Hence, the
monotonicity of the function
$\beta_{\delta}:{\mathbb{R}}\rightarrow{\mathbb{R}}$ implies

$\beta_{\delta}(u-\varphi)\geq\beta_{\delta}(0)=-\left\lVert
K_{\Phi}\right\rVert_{L^{\infty}(\mathbb{S}^{n}\times[0,\infty))}^{\alpha}=-C_{0}$

for all $(x,t)\in\mathbb{S}^{n}\times[0,T)$. ∎

######  Remark 3.2.

In the proof, we have shown $u>\varphi$ for all
$(z,t)\in\mathbb{S}^{n}\times[0,T)$, which means that the evolving
hypersurface $\Sigma_{t}$ under ($\ast_{\delta}$) cannot penetrate or even
touch the obstacle. This is because $|\beta_{\delta}(0)|$ is chosen
sufficiently large. If one consider that the condition $|\beta_{\delta}(0)|$
has another uniform constant, for example $|\beta_{\delta}(0)|=1$, then one
can prove $u>\varphi-C(\delta)$ for some constant $C(\delta)$ with $C(0+)=0$,
in which case the hypersurface might penetrate the obstacle but the depth of
penetration is controlled.

##  4\. Speed estimates I: Uniform positive lower bounds

In this section we prove uniform positive lower bounds for the speed of the
solution to ($\ast_{\delta}$). As a direct corollary, we also obtain uniform
positive lower bounds on the Gauss curvature of $\Sigma_{t}$.

######  Lemma 4.1.

Let $\Sigma_{0}$ be an initial hypersurface and $\Phi$ be a shrinking
obstacle. If $u:\mathbb{S}^{n}\times[0,T)\rightarrow{\mathbb{R}}$ is a smooth
solution to ($\ast_{\delta}$), then

$\displaystyle\partial_{t}(u-\varphi)\leq 0\quad\text{in
}\mathbb{S}^{n}\times[0,T).$ (4.1)

###### Proof.

Since $\Sigma_{0}$ is strictly convex and
$\min_{\mathbb{S}^{n}}(u_{0}-\varphi(\cdot,0))\geq\delta>0$ by our choice of
$\delta$, we have at $t=0$,
$\partial_{t}u=-K^{\alpha}-\beta_{\delta}=-K_{0}^{\alpha}<0$. On the other
hand, from the definition of the shrinking obstacle $\Phi$, we have

$\displaystyle-\partial_{t}\varphi(\cdot,0)<\min_{\mathbb{S}^{n}}K_{0}^{\alpha}\leq
K_{0}^{\alpha}$

which implies (4.1) at $t=0$.

Recall that
$K=\det\overline{g}_{ij}/\det(\overline{\nabla}_{i}\overline{\nabla}_{j}u+u\overline{g}_{ij})$.
If we differentiate $-\partial_{t}u=K^{\alpha}+\beta_{\delta}$ with respect to
$t$, we obtain

$\displaystyle\partial_{t}(-\partial_{t}u)=-\alpha
K^{\alpha}b^{ij}(\overline{\nabla}_{i}\overline{\nabla}_{j}\partial_{t}u+\partial_{t}u\overline{g}_{ij})+\beta_{\delta}^{\prime}(u-\varphi)_{t}.$
(4.2)

For $Z(x,t):=\partial_{t}(u(x,t)-\varphi(x,t))$, (4.2) can be rewritten as

$\displaystyle-Z_{t}-\partial^{2}_{t}\varphi=-\alpha
K^{\alpha}b^{ij}(\overline{\nabla}_{i}\overline{\nabla}_{j}(Z+\partial_{t}\varphi)+(Z+\partial_{t}\varphi)\overline{g}_{ij})+\beta_{\delta}^{\prime}Z.$

Thus we have

$\displaystyle Z_{t}={\mathcal{L}}Z+(\alpha
K^{\alpha}H-\beta_{\delta}^{\prime})Z+\alpha
K^{\alpha}b^{ij}(\overline{\nabla}_{i}\overline{\nabla}_{j}\partial_{t}\varphi+\partial_{t}\varphi\overline{g}_{ij})-\partial_{t}^{2}\varphi.$

To estimate the terms involving the obstacle $\varphi$, recall that the
shrinking obstacle satisfies $\partial_{t}^{2}\varphi\geq 0$ and
$\partial_{t}\mu_{i}(\cdot,t)\geq 0$, where
$\mu_{1}(\cdot,t),\cdots,\mu_{n}(\cdot,t)$ is the principal curvatures of
$\Phi_{t}$. Since the second fundamental form of the obstacle is given by

$\displaystyle
h_{ij}^{\varphi}=\overline{\nabla}_{i}\overline{\nabla}_{j}\varphi+\varphi\overline{g}_{ij},$

it follows from the strict convexity of the solution and the property
$\partial_{t}\mu_{i}\geq 0$ that

$\displaystyle\alpha
K^{\alpha}b^{ij}(\overline{\nabla}_{i}\overline{\nabla}_{j}\partial_{t}\varphi+\partial_{t}\varphi\overline{g}_{ij})=\alpha
K^{\alpha}b^{ij}\partial_{t}(h_{ij}^{\varphi})=\alpha
K^{\alpha}b^{ij}\overline{g}_{jk}\partial_{t}(\overline{g}^{kl}h_{lj}^{\varphi})\leq
0.$

Note that in the last inequality we have used that the eigenvalues of
$\overline{g}^{kl}h_{lj}^{\varphi}=(b^{\varphi})^{kl}g^{\varphi}_{lj}$ are
$1/\mu_{1},\cdots,1/\mu_{n}$ which are decrease in time. Hence,

$\displaystyle Z_{t}\leq{\mathcal{L}}Z+(\alpha
K^{\alpha}H-\beta_{\delta}^{\prime})Z.$ (4.3)

Assuming the contrary, we take a time $t_{0}$ such that
$\sup_{\mathbb{S}^{n}\times[0,t_{0}]}Z>0$. Set
$m=\sup_{\mathbb{S}^{n}\times[0,t_{0}]}(\alpha
K^{\alpha}H-\beta_{\delta}^{\prime})<\infty$ and let $\tilde{Z}=Ze^{-mt-t}$.
By (4.3), $\tilde{Z}$ satisfies

$\displaystyle\tilde{Z}_{t}=Z_{t}e^{-mt-t}-(m+1)Ze^{-mt-t}\leq{\mathcal{L}}\tilde{Z}-\tilde{Z}.$
(4.4)

For a small number $\varepsilon>0$, we take a point $(x_{1},t_{1})$ satisfying
$\tilde{Z}(x_{1},t_{1})=\varepsilon$ for the first time. Clearly, $t_{0}>0$.
Then we have that $\tilde{Z}\leq\varepsilon$ on
$\mathbb{S}^{n}\times[0,t_{1}]$ and $\tilde{Z}(z_{1},t_{1})=\varepsilon$, and
that

$\displaystyle\partial_{t}\tilde{Z}\geq
0,\quad\overline{\nabla}_{i}\tilde{Z}=0,\quad\overline{\nabla}_{i}\overline{\nabla}_{j}\tilde{Z}\leq
0\quad\text{at }(z_{1},t_{1}),$

which is contradict to (4.4). Thus $\sup_{\mathbb{S}^{n}\times[0,t]}Z\leq 0$
for all $t\in(0,T)$. ∎

Using the lemma above, we obtain the uniform positive lower bound for the
Gauss curvature.

######  Corollary 4.2.

Let $\Sigma_{0}$ be an initial hypersurface and $\Phi$ be a shrinking
obstacle. If $\\{\Sigma_{t}\\}_{0\leq t<T}$ is the solution to
($\ast_{\delta}$), then the Gauss curvature of $\Sigma_{t}$ has a uniform
positive lower bound, i.e.,

$\displaystyle\inf_{\mathbb{S}^{n}\times[0,T)}K\geq c_{T}>0,$

where $c_{T}=c(\alpha,T,\Phi)$ is a constant independent of $\delta$.

###### Proof.

By Lemma 4.1 and the non-positivity of $\beta_{\delta}$, we have

$\displaystyle
K^{\alpha}=-\partial_{t}u-\beta_{\delta}\geq-\partial_{t}u\geq-\partial_{t}\varphi\geq\min_{\mathbb{S}^{n}\times[0,T]}(-\partial_{t}\varphi)>0.$

Now the conclusion follows by choosing
$c_{T}=\min_{\mathbb{S}^{n}\times[0,T]}(-\partial_{t}\varphi)^{1/\alpha}$. ∎

We will remove the time dependence of the constant $c_{T}$ above after
analyzing the long time behavior of the solution $\\{\Sigma_{t}\\}$.

##  5\. Speed estimates II: Uniform upper bounds

In the previous section we proved the positive lower bound on the speed
$-\partial_{t}u$ and the Gauss curvature $K$. In this section we will obtain
the opposite bounds on the speed and the Gauss curvature.

######  Lemma 5.1.

Let $\Sigma_{0}$ be an initial hypersurface and $\Phi$ be a shrinking
obstacle. If $u:\mathbb{S}^{n}\times[0,T)\rightarrow{\mathbb{R}}$ is a smooth
solution of the penalized problem ($\ast_{\delta}$), then

$\displaystyle-\partial_{t}u\leq C\quad\text{in }\mathbb{S}^{n}\times[0,T),$

where $C=C(n,\alpha,\Sigma_{0},\Phi)$ is a constant independent of $\delta$.

###### Proof.

Let $\varphi$ be the support function of the obstacle and
$\rho_{0}=\frac{1}{2}\min_{\mathbb{S}^{n}}\varphi_{\infty}>0$. We consider an
auxiliary function on $\mathbb{S}^{n}\times[0,T)$

$\displaystyle
w=\frac{K^{\alpha}(z,t)+\beta_{\delta}(u(z,t)-\varphi(z,t))}{u(z,t)-\rho_{0}}.$

Notice that the denominator remains positive since
$u\geq\varphi\geq\varphi_{\infty}\geq 2\rho_{0}$ by Lemma 3.1 and the
definition of $\rho_{0}$.

Our first task is to derive the evolution equation for the quantity $w$. Since

$\displaystyle\overline{\nabla}_{j}w$
$\displaystyle=\frac{\overline{\nabla}_{j}(K^{\alpha}+\beta_{\delta})}{u-\rho_{0}}-\frac{K^{\alpha}+\beta_{\delta}}{(u-\rho_{0})^{2}}\overline{\nabla}_{j}(u-\rho_{0}),$
$\displaystyle{\mathcal{L}}w$
$\displaystyle=\frac{{\mathcal{L}}(K^{\alpha}+\beta_{\delta})}{u-\rho_{0}}-\frac{2\left\langle{\overline{\nabla}(K^{\alpha}+\beta_{\delta})},{\overline{\nabla}u}\right\rangle}{(u-\rho_{0})^{2}}-\frac{K^{\alpha}+\beta_{\delta}}{(u-\rho_{0})^{2}}{\mathcal{L}}u+\frac{2(K^{\alpha}+\beta_{\delta})}{(u-\rho_{0})^{3}}\left\lVert
u\right\rVert_{\mathcal{L}}^{2}$
$\displaystyle=\frac{{\mathcal{L}}(K^{\alpha}+\beta_{\delta})}{u-\rho_{0}}-\frac{K^{\alpha}+\beta_{\delta}}{(u-\rho_{0})^{2}}{\mathcal{L}}u+\frac{2}{u-\rho_{0}}\left\langle{\overline{\nabla}w},{\overline{\nabla}u}\right\rangle_{\mathcal{L}},$

we obtain

$\displaystyle\begin{split}(\partial_{t}-\mathcal{L})w&=\frac{(\partial_{t}-\mathcal{L})(K^{\alpha}+\beta_{\delta})}{u-\rho_{0}}-\frac{K^{\alpha}+\beta_{\delta}}{(u-\rho_{0})^{2}}(\partial_{t}-\mathcal{L})u-\frac{2}{u-\rho_{0}}\left\langle{\overline{\nabla}w},{\overline{\nabla}u}\right\rangle_{\mathcal{L}}.\end{split}$
(5.1)

It follows from (2.7) in Lemma 2.5 that

$\displaystyle(\partial_{t}-\mathcal{L})(K^{\alpha}+\beta_{\delta})=\alpha
K^{\alpha}H(K^{\alpha}+\beta_{\delta})+(u-\varphi)_{t}\beta_{\delta}^{\prime}.$

Plugging this and the evolution equation (2.6) in Lemma 2.5 into (5.1) gives
that

$\displaystyle(\partial_{t}-\mathcal{L})w$
$\displaystyle=\frac{(u-\varphi)_{t}\beta_{\delta}^{\prime}}{u-\rho_{0}}-\frac{K^{\alpha}+\beta_{\delta}}{(u-\rho_{0})^{2}}(\alpha
K^{\alpha}H\rho_{0}-(n\alpha+1)K^{\alpha}-\beta_{\delta})$
$\displaystyle\quad-\frac{2}{u-\rho_{0}}\left\langle{\overline{\nabla}w},{\overline{\nabla}u}\right\rangle_{\mathcal{L}}.$

Therefore, we arrive at

$\displaystyle(\partial_{t}-\mathcal{L})w$
$\displaystyle=-\frac{2}{u-\rho_{0}}\left\langle{\overline{\nabla}w},{\overline{\nabla}u}\right\rangle_{\mathcal{L}}$
$\displaystyle\quad+\left(-w-\frac{\varphi_{t}}{u-\rho_{0}}\right)\beta_{\delta}^{\prime}-\frac{w}{u-\rho_{0}}(\alpha
K^{\alpha}H\rho_{0}-(n\alpha+1)K^{\alpha}-\beta_{\delta}).$

Next, we apply the maximum principle argument to $w$. Fix a time
$T^{\prime}\in(0,T)$. Since $\mathbb{S}^{n}\times[0,T^{\prime}]$ is a compact
set and $\Sigma_{0}$ is strictly convex, $w$ attains positive maximum value at
some point $(z_{0},t_{0})\in\mathbb{S}^{n}\times[0,T^{\prime}]$. If $t_{0}=0$,
then

$\displaystyle w\leq\frac{\max_{\mathbb{S}^{n}}(K_{0})^{\alpha}}{\rho_{0}},$
(5.2)

where $K_{0}$ is the Gauss curvature of the initial hypersurface $\Sigma_{0}$.
Now we assume that $t_{0}>0$. Then $w$ has an interior maximum in
$\mathbb{S}^{n}\times[0,T^{\prime}]$ and thus $w$ satisfies at the interior
maximum point

$\displaystyle\overline{\nabla}w=0,\quad\overline{\nabla}^{2}w\leq
0,\quad\partial_{t}w\geq 0.$

This gives that

$\displaystyle
0\leq\left(-w-\frac{\varphi_{t}}{u-\rho_{0}}\right)\beta_{\delta}^{\prime}-\frac{w}{u-\rho_{0}}(\alpha
K^{\alpha}H\rho_{0}-(n\alpha+1)K^{\alpha}-\beta_{\delta})$

at $(z_{0},t_{0})$. If
$\left(-w-\frac{\varphi_{t}}{u-\rho_{0}}\right)(z_{0},t_{0})\geq 0$, then at
the same point

$\displaystyle
w\leq\frac{-\partial_{t}\varphi}{\rho_{0}}\leq\frac{-\partial_{t}\varphi_{0}}{\rho_{0}}<\frac{\min_{\mathbb{S}^{n}}K_{0}^{\alpha}}{\rho_{0}}$
(5.3)

since $\varphi$ is convex in time variable. Otherwise, it follows from
$\beta_{\delta}\leq 0$ and $\beta_{\delta}^{\prime}\geq 0$ that

$\displaystyle 0\leq\frac{w}{u-\rho_{0}}(-\alpha
K^{\alpha}H\rho_{0}+(n\alpha+1)K^{\alpha})$

which implies by the arithmetic-geometric mean and $K^{\alpha}>0$,

$\displaystyle nK^{\frac{1}{n}}\leq H\leq\frac{n\alpha+1}{\alpha\rho_{0}}$
(5.4)

at the point $(z_{0},t_{0})$. Hence, combining (5.2), (5.3) and (5.4), we
conclude that

$\displaystyle\max_{\mathbb{S}^{n}\times[0,T^{\prime}]}w\leq\frac{1}{\rho_{0}}\max\left\\{(\max_{\mathbb{S}^{n}}K_{0})^{\alpha},\left(\frac{n\alpha+1}{n\alpha\rho_{0}}\right)^{n\alpha}\right\\}=:C_{1}.$

Since $T^{\prime}$ is an arbitrary number in $(0,T)$ and the constant $C_{1}$
does not depends on $T^{\prime}$, we conclude that
$\max_{\mathbb{S}^{n}\times[0,T)}w\leq C_{1}$.

Finally, by Lemma 4.1, we observe $u(z,t)\leq
u_{0}(z)\leq\max_{\mathbb{S}^{n}}u_{0}$, which completes the proof since
$-\partial_{t}u\leq w(u-\rho_{0})\leq C_{1}\max_{\mathbb{S}^{n}}u_{0}=:C$. ∎

Using the uniform boundedness of the penalty term $\beta_{\delta}(u-\varphi)$
and upper bounds on the speed, we can obtain the following upper bound on the
Gauss curvature $K$.

######  Lemma 5.2.

Let $\Sigma_{0}$ be an initial hypersurface and $\Phi$ be a shrinking
obstacle. If $\\{\Sigma_{t}\\}_{0\leq t<T}$ be a solution to
($\ast_{\delta}$), then the Gauss curvature of $\Sigma_{t}$ satisfies

$\displaystyle\max_{\mathbb{S}^{n}\times[0,T)}K\leq C,$

where $C=C(n,\alpha,\Sigma_{0},\Phi)$ is a constant independent of $\delta$.

###### Proof.

By Lemma 3.1 and Lemma 5.1, we have

$\displaystyle K^{\alpha}=-\partial_{t}u-\beta_{\delta}(u-\varphi)\leq C$

which completes the proof. ∎

What we have proved in this and the previous section is the following uniform
estimate on the Gauss curvature: if $\\{\Sigma_{t}\\}_{0\leq t<T}$ is a
solution of ($\ast_{\delta}$), then its Gauss curvature satisfies

$\displaystyle 0<c_{T}\leq K(z,t)\leq C\quad\text{for
all}\quad(z,t)\in\mathbb{S}^{n}\times[0,T),$ (5.5)

where $c_{T}$ and $C$ are constants independent of $\delta$. This, however,
does not give sufficient controls on each principal curvature. In the next
section we will obtain uniform bounds on each principal curvature, which
implies the optimal regularity for the solution of (2.3).

##  6\. Uniform bounds on principal curvatures

Here we establish uniform (independent of $\delta$) positive lower bounds on
the principal curvatures of the solution to ($\ast_{\delta}$) using the bounds
(5.5). We start with two lemmas that will be used in the proof of Lemma 6.3
below. In the following lemma and its proof, we will not use the Einstein
summation convention temporarily.

######  Lemma 6.1 (Euler type formula).

Let $\Sigma\subset{\mathbb{R}}^{n+1}$ be a smooth, strictly convex
hypersurface, and let $X:\mathbb{S}^{n}\rightarrow{\mathbb{R}}^{n+1}$ be an
immersion such that $\Sigma=X(\mathbb{S}^{n})$ parameterized through the
inverse of the Gauss map, i.e., $\nu(X(z))=z$ for any $z\in\mathbb{S}^{n}$.
Then for any $z\in\mathbb{S}^{n}$ and $1\leq i\leq n$,

$\displaystyle\frac{h_{ii}(z)}{\overline{g}_{ii}(z)}\leq\frac{1}{\lambda_{\min}(z)},$

where $\overline{g}_{ij}$ is the standard metric on $\mathbb{S}^{n}$ and
$h_{ij}$ is the second fundamental form of $\Sigma$.

###### Proof.

Fix a point $z\in\mathbb{S}^{n}$ and an orthonormal basis
$\\{E_{1},\cdots,E_{n}\\}$ of $T_{z}\Sigma$ such that
$L(E_{j})=\lambda_{j}E_{j}$ for $j=1,\cdots,n$, where $L$ is the Weingarten
map and $\lambda_{1},\cdots,\lambda_{n}$ are the principal curvatures of
$\Sigma$ at $z$. Write $\overline{\nabla}_{i}X=\sum_{j}a_{ij}E_{j}$ with
$a=(a_{ij})$, and denote by $c=\\{c_{ij}\\}$ the diagonal matrix
$\operatorname{diag}(\lambda_{1},\cdots,\lambda_{n})$. Since
$L(\overline{\nabla}_{i}X)=\sum_{j}h_{i}^{j}(\overline{\nabla}_{j}X)$, we get

$\displaystyle\sum_{j,k}a_{ij}c_{jk}E_{k}=L(\overline{\nabla}_{i}X)=h_{i}^{j}(\overline{\nabla}_{j}X)=\sum_{j,k}h_{i}^{j}a_{jk}E_{k}$

which implies

$\displaystyle(ac)_{ik}=\sum_{j}a_{ij}c_{jk}=\sum_{l,j}h_{il}g^{lj}a_{jk}=(hg^{-1}a)_{ik},$
(6.1)

where $h$ and $g$ are $n\times n$ matrices whose $(i,j)$-components are
$h_{ij}$ and $g_{ij}$, respectively, and $g^{-1}$ is the inverse matrix of
$g$. Observing that

$\displaystyle
g_{ij}=\left\langle{\overline{\nabla}_{i}X},{\overline{\nabla}_{j}X}\right\rangle=\sum_{k,l}a_{ik}\delta_{kl}a_{jl}=\sum_{k}a_{ik}a_{jk}=(aa^{T})_{ij},$

it follows from (6.1) that $h=aca^{-1}g=aca^{T}$.

On the other hand, from (2.2) we get

$\displaystyle\overline{g}=hg^{-1}h=(aca^{T})(aa^{T})^{-1}aca^{T}=aca^{T}(a^{T})^{-1}a^{-1}aca^{T}=ac^{2}a^{T}$

which implies

$\displaystyle\overline{g}_{ii}=\sum_{j,k,l}a_{ij}c_{jk}c_{kl}a_{il}\geq\lambda_{\min}\sum_{j,l,l}a_{ij}c_{jk}\delta_{kl}a_{il}=\lambda_{\min}\sum_{j,k}a_{ij}c_{jk}a_{ik}=\lambda_{\min}h_{ii}.$

This completes the proof. ∎

######  Lemma 6.2.

If $u:\mathbb{S}^{n}\times[0,T)\rightarrow{\mathbb{R}}$ is a smooth solution
to ($\ast_{\delta}$), then the evolution equation of the following quantity
involving $\beta_{\delta}=\beta_{\delta}(u-\varphi)$ is

$\displaystyle(\partial_{t}-\mathcal{L})e^{-\chi\beta_{\delta}}=e^{-\chi\beta_{\delta}}\left[-\chi\beta_{\delta}^{\prime}(\partial_{t}-\mathcal{L})(u-\varphi)+(-\chi^{2}(\beta_{\delta}^{\prime})^{2}+\chi\beta_{\delta}^{\prime\prime})\left\lVert\overline{\nabla}(u-\varphi)\right\rVert^{2}_{\mathcal{L}}\right],$

where $\chi$ is a given constant.

###### Proof.

The proof follows from direct computations. In fact, we have

$\displaystyle\partial_{t}e^{-\chi\beta_{\delta}}$
$\displaystyle=-\chi\beta_{\delta}^{\prime}e^{-\chi\beta_{\delta}}\partial_{t}(u-\varphi),$
$\displaystyle\overline{\nabla}_{j}e^{-\chi\beta_{\delta}}$
$\displaystyle=-\chi\beta_{\delta}^{\prime}e^{-\chi\beta_{\delta}}\overline{\nabla}_{j}(u-\varphi),$
$\displaystyle\overline{\nabla}_{i}\overline{\nabla}_{j}e^{-\chi\beta_{\delta}}$
$\displaystyle=-\chi\beta_{\delta}^{\prime}e^{-\chi\beta_{\delta}}\overline{\nabla}_{i}\overline{\nabla}_{j}(u-\varphi)+((\chi\beta_{\delta}^{\prime})^{2}-\chi\beta_{\delta}^{\prime\prime})e^{-\chi\beta_{\delta}}\overline{\nabla}_{i}(u-\varphi)\overline{\nabla}_{j}(u-\varphi)$

so that
${\mathcal{L}}e^{-\chi\beta_{\delta}}=-\chi\beta_{\delta}^{\prime}e^{-\chi\beta_{\delta}}{\mathcal{L}}(u-\varphi)+((\chi\beta_{\delta}^{\prime})^{2}-\chi\beta_{\delta}^{\prime\prime})e^{-\chi\beta_{\delta}}\left\lVert\overline{\nabla}(u-\varphi)\right\rVert_{\mathcal{L}}^{2}$
and the conclusion follows. ∎

Now we are ready to prove the main result of this section.

######  Lemma 6.3.

Let $\\{\Sigma_{t}\\}_{0\leq t<T}$ be a solution to ($\ast_{\delta}$) and
$\Phi$ be a shrinking obstacle. Then the principal curvatures
$\lambda_{1}(\cdot,t),\cdots,\lambda_{n}(\cdot,t)$ of $\Sigma_{t}$ satisfy

$\displaystyle\inf_{\mathbb{S}^{n}\times[0,T)}\lambda_{i}\geq c_{T}>0,$ (6.2)

where $c_{T}$ is a constant depending only on $n,\alpha,T,\Phi$, and
$\Sigma_{0}$. In particular, the constant $c_{T}$ depends on the minimum speed
of the obstacle, $\inf_{\mathbb{S}^{n}\times[0,T]}|\partial_{t}\Phi|$.

######  Remark 6.4.

If the obstacle is stationary, i.e. $\Phi_{t}\equiv\Phi_{0}$, then there is no
positive lower bounds for $\lambda_{i}$, $i=1,\ldots,n$, see [20].

###### Proof.

Take a time $T^{\prime}\in(0,T)$. To establish the lower bound (6.2), we
estimate an upper bound of a function

$\displaystyle\tilde{W}(z,t)=\lambda_{\min}^{-1}(z,t)e^{-\chi\beta_{\delta}(u-\varphi)},$

where $\displaystyle\lambda_{\min}(z,t)=\min_{i=1,\cdots,n}\lambda_{i}(z,t)$
and $\chi$ is a constant to be determined later. Assume that $\tilde{W}$
attains its maximum value over $\mathbb{S}^{n}\times[0,T^{\prime}]$ at an
interior point $(z_{0},t_{0})$ with $t_{0}>0$. Now we choose a coordinate
chart of $z_{0}$ such that

$\displaystyle\overline{g}_{ij}(z_{0},t_{0})=\delta_{ij}\quad\text{and}\quad
h_{ij}(z_{0},t_{0})=\lambda_{i}^{-1}(z_{0},t_{0})\delta_{ij}$ (6.3)

with $\lambda_{1}\leq\cdots\leq\lambda_{n}$. We then note that a function

$\displaystyle
W(z,t)=\frac{h_{11}}{\overline{g}_{11}}e^{-\chi\beta_{\delta}(u-\varphi)}$

also has the same maximum at the same point $(z_{0},t_{0})$ since, by Lemma
6.1, we see that

$\displaystyle W(z,t)$
$\displaystyle\leq\tilde{W}(z,t)\leq\tilde{W}(z_{0},t_{0}),$ $\displaystyle
W(z_{0},t_{0})$
$\displaystyle=\lambda_{1}^{-1}e^{-\chi\beta_{\delta}(u-\varphi)}(z_{0},t_{0})=\tilde{W}(z_{0},t_{0}).$

Thus we can obtain the upper bound for the function $\tilde{W}$ by estimating
the function $W$, and we have at the point $(z_{0},t_{0})$,

$\displaystyle\partial_{t}W\geq
0,\quad\overline{\nabla}W=0,\quad\text{and}\quad\overline{\nabla}^{2}W\leq 0.$
(6.4)

Our next task is to derive an evolution equation for $W$. We first observe
that

$\displaystyle\partial_{t}\overline{g}=0\quad\text{and}\quad\overline{\nabla}\overline{g}=0,$

and recall the evolution equations for $h_{ij}$ from (2.8):

$\displaystyle(\partial_{t}-{\mathcal{L}})h_{ij}=-\mathcal{C}_{ij}+(n\alpha-1)K^{\alpha}\overline{g}_{ij}-\alpha
K^{\alpha}Hh_{ij}-\overline{\nabla}_{ij}^{2}\beta_{\delta}-\beta_{\delta}\overline{g}_{ij}.$

From these, we have

$\displaystyle(\partial_{t}-{\mathcal{L}})\left(\frac{h_{11}}{\overline{g}_{11}}\right)=-\frac{\mathcal{C}_{11}}{\overline{g}_{11}}+(n\alpha-1)K^{\alpha}-\alpha
K^{\alpha}H\frac{h_{11}}{\overline{g}_{11}}-\frac{\overline{\nabla}_{11}^{2}\beta_{\delta}}{\overline{g}_{11}}-\beta_{\delta}.$

Observe that at the point $(z_{0},t_{0})$ we have
$n-Hh_{11}/\overline{g}_{11}=n-H/\lambda_{1}\leq 0$, and thus

$\displaystyle(n\alpha-1)K^{\alpha}-\alpha
K^{\alpha}H\frac{h_{11}}{\overline{g}_{11}}-\beta_{\delta}$
$\displaystyle=\alpha
K^{\alpha}\left(n-H\frac{h_{11}}{\overline{g}_{11}}\right)-K^{\alpha}-\beta_{\delta}<0$
(6.5)

since $K^{\alpha}+\beta_{\delta}=-\partial_{t}u>-\partial_{t}\varphi>0$ by
Lemma 4.1. However, the second derivative of $\beta_{\delta}$ produces bad
terms, which makes us to consider the auxiliary function
$e^{-\chi\beta_{\delta}}=e^{-\chi\beta_{\delta}(u(z,t)-\varphi(z,t))}$. By
Lemma 6.2, we have that

$\displaystyle\begin{split}(\partial_{t}-{\mathcal{L}})W&=(\partial_{t}-{\mathcal{L}})\left(\frac{h_{11}}{\overline{g}_{11}}\right)e^{-\chi\beta_{\delta}}+\frac{h_{11}}{\overline{g}_{11}}(\partial_{t}-{\mathcal{L}})\left(e^{-\chi\beta_{\delta}}\right)\\\
&\quad-2\left\langle{\overline{\nabla}\left(\frac{h_{11}}{\overline{g}_{11}}\right)},{\overline{\nabla}e^{-\chi\beta_{\delta}}}\right\rangle_{\mathcal{L}}\\\
&=\left(-\frac{\mathcal{C}_{11}}{\overline{g}_{11}}+(n\alpha-1)K^{\alpha}-\alpha
K^{\alpha}H\frac{h_{11}}{\overline{g}_{11}}-\frac{\overline{\nabla}_{11}^{2}\beta_{\delta}}{\overline{g}_{11}}-\beta_{\delta}\right)e^{-\chi\beta_{\delta}}\\\
&\quad+\left(-\chi\beta_{\delta}^{\prime}e^{-\chi\beta_{\delta}}(\partial_{t}-\mathcal{L})(u-\varphi)+(-\chi^{2}(\beta_{\delta}^{\prime})^{2}+\chi\beta_{\delta}^{\prime\prime})e^{-\chi\beta_{\delta}}\left\lVert\overline{\nabla}(u-\varphi)\right\rVert^{2}_{\mathcal{L}}\right)\frac{h_{11}}{\overline{g}_{11}}\\\
&\quad-2e^{\chi\beta_{\delta}}\left\langle{\overline{\nabla}W},{\overline{\nabla}e^{-\chi\beta_{\delta}}}\right\rangle_{\mathcal{L}}+2\frac{h_{11}}{\overline{g}_{11}}e^{\chi\beta_{\delta}}\left\lVert\overline{\nabla}e^{-\chi\beta_{\delta}}\right\rVert^{2}_{\mathcal{L}}.\end{split}$
(6.6)

We finally estimate (6.6) at the point $(z_{0},t_{0})$. Since
$\left\lVert\overline{\nabla}e^{-\chi\beta_{\delta}}\right\rVert^{2}_{\mathcal{L}}=(\chi\beta_{\delta}^{\prime})^{2}e^{-2\chi\beta_{\delta}}\left\lVert\overline{\nabla}(u-\varphi)\right\rVert^{2}_{\mathcal{L}}$,
by dividing by $e^{-\chi\beta_{\delta}}$ in (6.6), we see that (6.4) and (6.5)
implies at the point $(z_{0},t_{0})$,

$\displaystyle 0$
$\displaystyle<-\frac{\mathcal{C}_{11}}{\overline{g}_{11}}-\frac{\overline{\nabla}_{11}^{2}\beta_{\delta}}{\overline{g}_{11}}+\left(-\chi\beta_{\delta}^{\prime}(\partial_{t}-\mathcal{L})(u-\varphi)+(\chi^{2}(\beta_{\delta}^{\prime})^{2}+\chi\beta_{\delta}^{\prime\prime})\left\lVert\overline{\nabla}(u-\varphi)\right\rVert^{2}_{\mathcal{L}}\right)\frac{h_{11}}{\overline{g}_{11}}.$

By a direct computation, we have

$\displaystyle\overline{\nabla}^{2}_{11}\beta_{\delta}=\beta_{\delta}^{\prime}\overline{\nabla}^{2}_{11}(u-\varphi)+\beta_{\delta}^{\prime\prime}|\overline{\nabla}_{1}(u-\varphi)|^{2}.$

We then deduce from (6.3) that

$\displaystyle\begin{split}0&<-\mathcal{C}_{11}-\beta_{\delta}^{\prime\prime}|\overline{\nabla}_{1}(u-\varphi)|^{2}+(\chi^{2}(\beta_{\delta}^{\prime})^{2}+\chi\beta_{\delta}^{\prime\prime})\lambda_{1}^{-1}\left\lVert\overline{\nabla}(u-\varphi)\right\rVert^{2}_{\mathcal{L}}\\\
&\quad-\beta_{\delta}^{\prime}\overline{\nabla}^{2}_{11}(u-\varphi)-\chi\beta_{\delta}^{\prime}\lambda_{1}^{-1}(\partial_{t}-\mathcal{L})(u-\varphi).\end{split}$
(6.7)

On the other hand, using (6.3) again to simplify the quantity
$\mathcal{C}_{11}$, we see

$\displaystyle\mathcal{C}_{11}$
$\displaystyle=\alpha^{2}K^{\alpha}\left(\sum_{k=1}^{n}\lambda_{k}\overline{\nabla}_{1}h_{kk}\right)^{2}+\alpha
K^{\alpha}\sum_{k,l=1}^{n}\lambda_{k}\lambda_{l}(\overline{\nabla}_{1}h_{kl})^{2}$
$\displaystyle\geq\alpha
K^{\alpha}\sum_{i=1}^{n}\lambda_{i}\lambda_{1}(\overline{\nabla}_{1}h_{i1})^{2}=\alpha
K^{\alpha}(\chi\beta_{\delta}^{\prime})^{2}\sum_{i=1}^{n}\frac{\lambda_{i}}{\lambda_{1}}|\overline{\nabla}_{i}(u-\varphi)|^{2},$

where we have used
$\lambda_{1}\overline{\nabla}h_{11}=\chi\beta_{\delta}^{\prime}\overline{\nabla}(u-\varphi)$
which follows from $\overline{\nabla}W=0$ at $(z_{0},t_{0})$ and the Codazzi
equation (i) in Lemma 2.6. Thus we obtain

$\displaystyle\frac{(\chi^{2}(\beta_{\delta}^{\prime})^{2}+\chi\beta_{\delta}^{\prime\prime})}{\lambda_{1}}\left\lVert\overline{\nabla}(u-\varphi)\right\rVert^{2}_{\mathcal{L}}$
$\displaystyle\leq\mathcal{C}_{11}+\alpha
K^{\alpha}\chi\beta_{\delta}^{\prime\prime}|\overline{\nabla}_{1}(u-\varphi)|^{2}$

since $\left\lVert\cdot\right\rVert^{2}_{\mathcal{L}}=\alpha
K^{\alpha}\sum_{i=1}^{n}\lambda_{i}|\overline{\nabla}_{i}\cdot|^{2}$.
Moreover, using Corollary 4.2, we can take $\chi=\chi(\alpha,T,\Phi)>0$
satisfying $\alpha K^{\alpha}\chi\geq 1$ which implies

$\displaystyle(\alpha
K^{\alpha}\chi-1)\beta_{\delta}^{\prime\prime}|\overline{\nabla}_{1}(u-\varphi)|^{2}\leq
0$

since $\beta_{\delta}^{\prime\prime}\leq 0$. Combining these facts together,
therefore, the inequality (6.7) becomes, after dividing
$\beta_{\delta}^{\prime}$,

$\displaystyle
0<-\overline{\nabla}^{2}_{11}(u-\varphi)-\chi\lambda_{1}^{-1}(\partial_{t}-\mathcal{L})(u-\varphi).$
(6.8)

To finish the proof, we observe that at the point $(z_{0},t_{0})$,

$\displaystyle\overline{\nabla}^{2}_{ii}u=h_{ii}-u=\frac{1}{\lambda_{i}}-u\quad\text{and}\quad\overline{\nabla}^{2}_{ii}\varphi=h_{ii}^{\varphi}-\varphi=\frac{1}{\mu_{i}}-\varphi,$

where $\\{\mu_{i}\\}_{i=1,\cdots,n}$ is principal curvatures of $\Phi$ and
$h^{\varphi}$ is the second fundamental form of $\Phi$. Using the evolution
equation of $u$ in Lemma (2.5), we have

$\displaystyle(\partial_{t}-\mathcal{L})(u-\varphi)$ $\displaystyle=\alpha
K^{\alpha}Hu-(n\alpha+1)K^{\alpha}-\beta_{\delta}-\varphi^{\prime}+\alpha
K^{\alpha}\sum_{i=1}^{n}\lambda_{i}\left(\frac{1}{\mu_{i}}-\varphi\right)$
$\displaystyle=\alpha
K^{\alpha}H(u-\varphi)-(n\alpha+1)K^{\alpha}-\beta_{\delta}-\varphi^{\prime}+\alpha
K^{\alpha}\sum_{i=1}^{n}\frac{\lambda_{i}}{\mu_{i}}$
$\displaystyle\geq-(n\alpha+1)K^{\alpha}+\alpha
K^{\alpha}\sum_{i=1}^{n}\frac{\lambda_{i}}{\mu_{i}}$

since $H=\lambda_{1}+\cdots+\lambda_{n}>0$, $\beta_{\delta}\leq 0$,
$\varphi^{\prime}<0$, and $u-\varphi>0$. Hence, (6.8) becomes

$\displaystyle
0<-\frac{1}{\lambda_{1}}+\frac{1}{\mu_{\min}}+(u-\varphi)+\frac{K^{\alpha}\chi}{\lambda_{1}}\left(n\alpha+1-\alpha\frac{\lambda_{n}}{\mu_{\max}}\right).$

Thus we conclude that

$\displaystyle-\frac{1}{\lambda_{1}}+\frac{1}{\mu_{\min}}+(u-\varphi)>0\quad\text{or}\quad
n\alpha+1-\alpha\frac{\lambda_{n}}{\mu_{\max}}>0,$

which is equivalent to

$\displaystyle\frac{1}{\lambda_{1}}<\frac{1}{\mu_{\min}}+u-\varphi\quad\text{or}\quad\lambda_{n}<\left(n+\frac{1}{\alpha}\right)\mu_{\max}.$
(6.9)

For the latter inequality, we relate the largest eigenvalue $\lambda_{n}$ with
the smallest eigenvalue $\lambda_{1}$ by using Corollary 4.2. In fact, we have

$\displaystyle\frac{1}{\lambda_{1}}=\frac{\lambda_{2}\cdots\lambda_{n}}{K}\leq\frac{\lambda_{n}^{n-1}}{c},$
(6.10)

where $c$ is the constant in Corollary 4.2. From (6.9) and (6.10), there
exists a positive constant $C=C(n,\alpha,T,\Phi,\Sigma_{0})$ such that

$\displaystyle\frac{1}{\lambda_{1}}\leq C.$

Using this and Lemma 3.1, we finally conclude that

$\displaystyle\frac{1}{\lambda_{\min}}\leq\max_{\mathbb{S}^{n}\times[0,T^{\prime}]}\tilde{W}=\tilde{W}(z_{0},t_{0})=W(z_{0},t_{0})=\frac{e^{-\chi\beta_{\delta}(u-\varphi)}}{\lambda_{1}}\leq
Ce^{\chi C_{0}}.$ (6.11)

Since $T^{\prime}$ is an arbitrary number in $(0,T)$ and the upper bound in
(6.11) does not depend on $T^{\prime}$, we obtain the conclusion by taking
$T^{\prime}\rightarrow T$. ∎

The lemma above automatically gives uniform upper bounds on principal
curvatures. Indeed, the largest principal eigenvalue satisfies

$\displaystyle\lambda_{n}=\frac{K}{\lambda_{1}\cdots\lambda_{n-1}}\leq\frac{K}{\lambda_{1}^{n-1}}$

which is bounded by Lemma 5.2 and Lemma 6.3. In summary, all principal
curvatures of the solution $\\{\Sigma_{t}\\}$ to ($\ast_{\delta}$) over
$\mathbb{S}^{n}\times[0,T)$ satisfy the following uniform estimates:

$\displaystyle 0<c_{T}\leq\lambda_{i}(z,t)\leq C_{T}\quad\text{for
all}\quad(z,t)\in\mathbb{S}^{n}\times[0,T),1\leq i\leq n,$ (6.12)

where $c_{T}$ and $C_{T}$ are constants independent of $\delta$.

##  7\. Proof of Theorem 1.1

The proof of Theorem 1.1 consists of two parts. First, we provide an existence
result for the long time solution that have the optimal $C^{1,1}$ regularity.
In this part we will use the uniform estimates obtained in the previous
sections. Second, we show that the motion of the solution is identically equal
to that of the obstacle after some time.

###### proof of Theorem 1.1.

From Lemma 2.4, approximate solutions to ($\ast_{\delta}$) exist at least for
short time. Let $T>0$ be the maximal time for which the solutions exist. We
claim that $T=\infty$. If not, we apply Lemma 5.2 and Lemma 6.3 to the
solutions so that (6.12) holds over $\mathbb{S}^{n}\times[0,T)$. Then the
linearized operator ${\mathcal{L}}$ is uniformly parabolic, i.e.,

$\displaystyle 0<\frac{1}{C}|\xi|^{2}\leq\alpha
K^{\alpha}b^{ij}\xi_{i}\xi_{j}\leq C|\xi|^{2}\quad\text{on
}\mathbb{S}^{n}\times[0,T)$

for all $\xi=(\xi_{i})\in{\mathbb{R}}^{n}$, where
$C=C(n,\alpha,\Sigma_{0},\Phi,T)$ is a positive constant. By applying the
standard argument in parabolic theory [24], the solutions exist beyond $T$,
which is contradict to the maximality of $T$. Hence, the solutions to
($\ast_{\delta}$) exist for all time.

Next we prove the uniqueness of solutions to (2.3). Let $u_{1}$ and $u_{2}$ be
two viscosity solutions of (2.3) with the same initial data. Assume that
$u_{1}<u_{2}$ at some point $(z_{0},t_{0})$. Since $u_{1}=u_{2}$ at the
initial time, we have $t_{0}>0$. Observing that $\varphi\leq u_{1}<u_{2}$ in
the set

$\displaystyle G=\\{\mathbb{S}^{n}\times(0,t_{0}]:u_{1}<u_{2}\\},$

we have $-\partial_{t}u_{2}=K_{2}^{\alpha}$, where $K_{2}$ is the Gauss
curvature of the hypersurface produced by $u_{2}$. On the other hand, $u_{1}$
satisfies $-\partial_{t}u_{1}\leq K_{1}^{\alpha}$, where $K_{1}$ is the Gauss
curvature with respect to $u_{1}$, similarly. In other words, $u_{1}$ is a
supersolution and $u_{2}$ is a subsolution of the same equation
$-\partial_{t}u=K^{\alpha}$. Since $u_{1}=u_{2}$ on the parabolic boundary of
$G$, it follows from the comparison principle that $u_{2}\leq u_{1}$ in $G$
which is a contradiction. Therefore, $u_{1}\equiv u_{2}$, and the solution to
(2.3) is unique.

Fix a time $T\in(0,\infty)$, and let $u^{\delta}$ be a solution of
($\ast_{\delta}$) over $\mathbb{S}^{n}\times[0,T)$. By the estimates on the
principal curvatures (6.12), the family $\\{u^{\delta}\\}_{\delta>0}$ of
solutions is uniformly bounded in $C^{1,1}$ so that there exists a function
$u\in C^{1,1}$ such that $u^{\delta}$ converges over a subsequence
$\delta=\delta_{j}\rightarrow 0$ to $u$ in $C^{1,\beta}$ for any $0<\beta<1$.
Moreover, it follows from the uniform boundedness of the penalty term
$\beta_{\delta}$ obtained in Lemma 3.1 that $u$ satisfies $u\geq\varphi$.
Since $T$ is arbitrary, we have the viscosity solution $u$ of (2.3) that
exists for all time with controlled principal curvatures

$\displaystyle\frac{1}{C_{T}}\leq\lambda_{i}\leq C_{T},\quad\text{for
}i=1,\cdots,n,$ (7.1)

in $\mathbb{S}^{n}\times[0,T)$, where $C_{T}$ is a constant depending on $T$.
We will remove the dependence of $T$ in the constant $C_{T}$ after analyzing
the long time behavior of the solution.

Now we prove that the solution coincides with the shrinking obstacle after
some time $T_{*}$. Fix a point
$(z_{1},t_{1})\in\mathbb{S}^{n}\times[0,\infty)$ and consider the point
$\Phi(z_{1},t_{1})$ on the obstacle $\Phi(\cdot,t_{1})$. Since the principal
curvatures of the shrinking obstacle satisfy

$\displaystyle\mu_{i}(\cdot,t)\geq\mu_{i}(\cdot,0)>0,\quad\text{for all
}i=1,\cdots,n,$ (7.2)

we can take a ball touching at the point $\Phi(z_{1},t_{1})$ and containing
the obstacle $\Phi(\cdot,t_{1})$ at $t_{1}$. Denoting by $a(z_{1},t_{1})$ and
$r(z_{1},t_{1})$ the center and the radius of the ball, respectively, we write
the ball as

$\displaystyle a(z_{1},t_{1})+B_{r(z_{1},t_{1})}.$ (7.3)

By (7.2) and
$\operatorname{conv}\Phi(\cdot,t)\subset\operatorname{conv}\Phi(\cdot,0)$ for
all $t\geq 0$, we may assume that $r(z_{1},t_{1})\leq\rho$ for some constant
$\rho$ independent of $(z_{1},t_{1})$.

We will construct a barrier for the point $\Phi(z_{1},t_{1})$ with

$\displaystyle t_{1}>\frac{1}{n\alpha+1}(\left\lVert
u_{0}\right\rVert_{\infty}+\rho)^{n\alpha+1}.$ (7.4)

Recall that the solution of the $\alpha$-Gauss curvature flow of the ball
centered at $a_{0}$ with radius $R_{0}$ at time $t$ is given by
$a_{0}+\partial B_{R(t)}$ where
$R(t)=(R_{0}^{n\alpha+1}-(n\alpha+1)t))^{1/(n\alpha+1)}$. Thus the
$\alpha$-Gauss curvature flow whose shape at $t_{1}$ is equal to the boundary
of (7.3) can be written as

$\displaystyle a(z_{1},t_{1})+\partial
B_{R(t)}=\\{x\in{\mathbb{R}}^{n+1}:|x-a(z_{1},t_{1})|=R(t)\\}$ (7.5)

where
$R(t)=(r(z_{1},t_{1})^{n\alpha+1}+(n\alpha+1)(t_{1}-t))^{1/(n\alpha+1)}$. Then
the support function $u_{p}$ of the ball (7.5) is

$\displaystyle u_{p}(z,t)=\left\langle{a(z_{1},t_{1})},{z}\right\rangle+R(t)$

Note that $u_{p}$ is concave in $t$ so that $u_{p}-\varphi$ is also concave in
$t$, and $u_{p}-\varphi$ is nonnegative in $\mathbb{S}^{n}\times\\{t_{1}\\}$
since the ball $a(z_{1},t_{1})+B_{R(z_{1},t_{1})}$ contains the obstacle
$\Phi(\cdot,t_{1})$. Furthermore, from the fact that the origin is contained
in the obstacle $\Phi(\cdot,t_{1})$ which is again contained in the ball
$a(z_{1},t_{1})+B_{R(t)}$, we have

$\displaystyle|a(z_{1},t_{1})|=|0-a(z_{1},t_{1})|\leq
R(t_{1})=r(z_{1},t_{1}).$

This gives that at $t=0$,

$\displaystyle u_{p}(z,0)-\varphi(z,0)$
$\displaystyle=R(0)+\left\langle{a(z_{1},t_{1})},{z}\right\rangle-\varphi_{0}(z)$
$\displaystyle\geq(r(z_{1},t_{1})^{n+1}+(n\alpha+1)t_{1})^{1/(n\alpha+1)}-|a(z_{1},t_{1})|-\left\lVert\varphi_{0}\right\rVert_{\infty}$
$\displaystyle\geq((n\alpha+1)t_{1})^{1/(n\alpha+1)}-\rho-\left\lVert
u_{0}\right\rVert_{\infty},$

where we have used $|a(z_{1},t_{1})|\leq r(z_{1},t_{1})\leq\rho$ and
$\varphi_{0}\leq u_{0}$. Then it follows from (7.4) that $u_{p}-\varphi>0$ at
$t=0$. This, together with the concavity of $u_{p}-\varphi$ and
$u_{p}-\varphi\geq 0$ at $t=t_{1}$, concludes that $u_{p}>\varphi$ in
$\mathbb{S}^{n}\times[0,t_{1})$, and therefore, the solution $u_{p}$ of the
$\alpha$-Gauss curvature flow is also the solution of (2.3), the
$\alpha$-Gauss curvature flow with the shrinking obstacle. Moreover, at the
initial time $t=0$, from (7.4), we have

$\displaystyle
u_{p}(z,0)-u_{0}\geq((n\alpha+1)t_{1})^{1/(n\alpha+1)}-\rho-\left\lVert
u_{0}\right\rVert_{\infty}>0.$

In other words, both
$u_{p}:\mathbb{S}^{n}\times[0,t_{1})\rightarrow{\mathbb{R}}$ and
$u:\mathbb{S}^{n}\times[0,t_{1})\rightarrow{\mathbb{R}}$ are the solution to
the $\alpha$-Gauss curvature flow with the shrinking obstacle, and the initial
data of $u_{p}$ encloses the initial data $u_{0}$, which imply $u_{p}\geq u$
in $\mathbb{S}^{n}\times[0,t_{1})$ by the comparison principle. Since
$u_{p}(z_{1},t_{1})=\varphi(z_{1},t_{1})$ and $u_{p}\geq u\geq\varphi$, we
conclude that $u(z_{1},t_{1})=\varphi(z_{1},t_{1})$. Hence, setting
$T_{*}=(\left\lVert u_{0}\right\rVert_{\infty}+\rho)^{n\alpha+1}/(n\alpha+1)$
yields $u\equiv\varphi$ in $\mathbb{S}^{n}\times[T_{*},\infty)$ since
$(z_{1},t_{1})$ was an arbitrary point in
$\mathbb{S}^{n}\times(T_{*},\infty)$.

It remains to prove that the principal curvatures
$\lambda_{1},\cdots,\lambda_{n}$ are globally bounded from above and below in
$\mathbb{S}^{n}\times[0,\infty)$. Taking $T=T_{*}$ in (7.1), we get
$C_{T_{*}}^{-1}\leq\lambda_{i}\leq C_{T_{*}}$ on the time interval
$[0,T_{*})$. On another time interval $[T_{*},\infty)$, since $u$ is
identically equal to $\varphi$ and the principal curvatures
$\mu_{1},\cdots,\mu_{n}$ of the obstacle is globally bounded from above and
below in $\mathbb{S}^{n}\times[0,\infty)$, we can obtain the desired bounds.
This completes the proof. ∎

##  8\. Proof of Theorem 1.2

To prove that the free boundary is locally $C^{1}$ graph, we need several
ingredients: non-degeneracy (Lemma 8.1), classification of blowups (8.4),
continuity of speed (8.5), and directional monotonicity (Lemma 8.2 and (8.7)).

Let $w$ be a solution to the local graph representation of the $\alpha$-Gauss
curvature flow with a shrinking obstacle, (1.2). Set $v=\phi-w$. Then
$v:Q_{1}\rightarrow{\mathbb{R}}$ satisfies

$\displaystyle F(D^{2}v,Dv,x,t)-\frac{\partial}{\partial t}v$
$\displaystyle=f(Dv,x,t)\quad$
$\displaystyle\text{in}\quad\Omega(v)=\\{v>0\\},$ (8.1) $\displaystyle v$
$\displaystyle\geq 0\quad$ $\displaystyle\text{in}\quad Q_{1},$
$\displaystyle|D^{2}v|+|\partial_{t}v|$ $\displaystyle\leq C\quad$
$\displaystyle\text{in}\quad Q_{1},$

where

$\displaystyle F(M,p,x,t)$ $\displaystyle=\frac{\det
D^{2}\phi(x,t)^{\alpha}-\det(D^{2}\phi(x,t)-M)^{\alpha}}{(1+|D\phi(x,t)-p|^{2})^{\frac{(n+2)\alpha-1}{2}}},$
$\displaystyle f(p,x,t)$ $\displaystyle=-\frac{\partial}{\partial
t}\phi(x,t)+\frac{\det
D^{2}\phi(x,t)^{\alpha}}{(1+|D\phi(x,t)-p|^{2})^{\frac{(n+2)\alpha-1}{2}}}.$

Observe that $F(0,Dv,x,t)=0$, $f\geq c>0$ for some constant $c$, and
$F(\cdot,Dv)$ is convex if $\alpha\leq\frac{1}{n}$. Moreover, the operator $F$
is uniformly elliptic since the eigenvalues of $D^{2}w$ are bounded above and
below by positive constants; there exists a constant $\theta>0$ such that
$\theta^{-1}|\xi|^{2}\leq F^{ij}\xi_{i}\xi_{j}\leq\theta|\xi|^{2}$, where

$\displaystyle F^{ij}:=\frac{\partial F}{\partial
M_{ij}}(D^{2}v,Dv,x,t)=\frac{\alpha\det(D^{2}w)^{\alpha}[(D^{2}w)^{-1}]^{ij}}{(1+|D\phi-
Dv|^{2})^{\frac{(n+2)\alpha-1}{2}}}.$

Let us fix some notations. For a point $X=(x,t)$, let
$Q_{r}(X)=B_{r}(x)\times(t-r^{2},t)$. We denote by $\partial_{p}Q_{r}(X)$ the
parabolic boundary of $Q_{r}(X)$.

In Theorem 1.1, we proved the optimal curvature bounds for the solution, which
implies the optimal $C^{1,1}$ regularity of the solution. In particular, the
solution has at most quadratic growth especially near the free boundary. The
following lemma says the quadratic growth is actually achieved near the free
boundary.

######  Lemma 8.1 (non-degeneracy).

Let $v$ be a solution of (8.1) with $\Omega=\Omega(v)$. Then for any
$X_{0}=(x_{0},t_{0})\in\overline{\Omega}\cap Q_{1/2}$, we have the inequality

$\displaystyle\sup_{\partial_{p}Q_{r}(X_{0})}v\geq
v(X_{0})+\frac{cr^{2}}{2n\theta+1}\quad\text{for all}\quad 0<r<1/4.$ (8.2)

###### Proof.

By an approximation argument, it suffices to prove (8.2) for
$X_{0}\in\Omega\cap Q_{1/2}$. For a point $X_{0}=(x_{0},t_{0})\in\Omega\cap
Q_{1/2}$, we define a function $\overline{v}:Q_{1}\rightarrow{\mathbb{R}}$ by

$\displaystyle\overline{v}(x,t)=v(x,t)-\frac{c(|x-x_{0}|^{2}-(t-t_{0}))}{2n\theta+1}.$

We claim that $\overline{v}$ satisfies
$F(D^{2}\overline{v},Dv,x,t)-\partial_{t}\overline{v}\geq 0$ in $\Omega\cap
Q_{1}$. If the claim holds, then it follows from
$\overline{v}(X_{0})=v(X_{0})\geq 0$ and
$\overline{v}=-\frac{c(|x-x_{0}|^{2}-(t-t_{0}))}{2n\theta+1}<0$ on
$\partial\Omega$ that

$\displaystyle 0\leq\sup_{\Omega\cap
Q_{r}(X_{0})}\overline{v}\leq\sup_{\partial_{p}Q_{r}(X_{0})}\overline{v}\leq\sup_{\partial_{p}Q_{r}(X_{0})}v-\frac{cr^{2}}{2n\theta+1},$

by the maximum principle, which proves (8.2).

We now prove the claim. Note that
$D^{2}\overline{v}=D^{2}v-\frac{2c}{2n\theta+1}I$, where $I$ denotes the
$n\times n$ identity matrix. Then by the uniformly ellipticity, we have

$\displaystyle F(D^{2}\overline{v},Dv,x,t)-\partial_{t}\overline{v}$
$\displaystyle=F(D^{2}v-\frac{2c}{2n\theta+1}I,Dv,x,t)-\partial_{t}v-\frac{c}{2n\theta+1}$
$\displaystyle\geq F(D^{2}v,Dv,x,t)-\frac{2n\theta
c}{2n\theta+1}-\partial_{t}v-\frac{c}{2n\theta+1}$
$\displaystyle=f(Dv,x,t)-c\geq 0\quad\text{in }\Omega.$

Therefore, the claim holds. ∎

Take a free boundary point $X_{0}=(x_{0},t_{0})\in\partial\Omega\cap Q_{1/8}$
and consider the rescaled function $v_{r}$ ($r>0$) of $v$ around $X_{0}$
defined by

$\displaystyle
v_{r}(y,s)=\frac{v(x_{0}+ry,t_{0}+r^{2}s)-v(x_{0},t_{0})}{r^{2}},\quad(y,s)\in
Q_{\tfrac{1}{2r}}.$

By Theorem 1.1 and the scaling properties
$D^{2}_{y}v_{r}(y,s)=D^{2}_{x}v(x_{0}+ry,t_{0}+r^{2}s)$ and
$\partial_{s}v_{r}(y,s)=\partial_{t}v(x_{0}+ry,t_{0}+r^{2}s)$, the rescaled
functions $\\{v_{r}\\}_{r>0}$ have uniform $C^{1,1}$ estimates. Then we can
extract a converging subsequence $v_{r_{j}}\rightarrow v_{0}$ in
$C_{\text{loc}}^{1,\gamma}({\mathbb{R}}^{n}\times{\mathbb{R}})$ for any
$0<\gamma<1$, where $v_{0}\in
C^{1,1}_{\text{loc}}({\mathbb{R}}^{n}\times{\mathbb{R}})$.

By the standard argument for blowups (see Lemma 18 in [17] or Proposition 3.17
in [26]), together with the non-degeneracy (Lemma 8.1), $v_{0}$ satisfies

$\displaystyle
F(D^{2}v_{0}(y,s),0,x_{0},t_{0})-\partial_{s}v_{0}(y,s)=f(0,x_{0},t_{0})\quad\text{in}\quad\Omega(v_{0}).$
(8.3)

Indeed, it follows from the equation

$\displaystyle
F(D^{2}v_{r_{j}},r_{j}Dv_{r_{j}},x_{0}+r_{j}y,t_{0}+r_{j}^{2}s)-\partial_{s}v_{r_{j}}$
$\displaystyle=f(r_{j}Dv_{r_{j}},x_{0}+r_{j}y,t_{0}+r_{j}^{2}s)$

in $\Omega(v_{r_{j}})$ by taking $j\rightarrow\infty$, where we used again the
scaling properties $D_{y}v_{r}=D_{x}v/r$, $D^{2}_{y}v_{r}=D^{2}v$, and
$D_{s}v_{r}=D_{t}v$.

Recall that $F$ is a convex operator in $D^{2}v$ variable since $\alpha\leq
1/n$, that $F(0,0,x_{0},t_{0})=0$, and that $F(\cdot,0,x_{0},t_{0})$ is
uniformly elliptic. By the work of Figalli and Shahgholian [13, Proposition
3.2] on the classification of global solutions to (8.3), it follows from the
uniform thickness condition in Theorem 1.2 that, after a rotation,

$\displaystyle v_{0}(y,s)=\tfrac{\gamma}{2}[(x_{1})_{+}]^{2},$ (8.4)

where $\gamma\in(1/\theta,\theta)$ such that $F(\gamma e_{1}\otimes
e_{1},0,x_{0},t_{0})=f(0,x_{0},t_{0})$. In particular, $v_{0}$ is time-
independent.

We now claim the continuity of the speed,

$\displaystyle\lim_{X\rightarrow\partial\Omega}\partial_{t}v(X)=0,$ (8.5)

under the uniform thickness assumption. We prove the claim by contradiction.
If there exists a sequence $X_{j}\in\Omega(v)$ such that
$X_{j}\rightarrow\partial\Omega(v)$ and
$|\partial_{t}v(X_{j})|\geq\varepsilon$ for some $\varepsilon>0$, we define
$d_{j}=\text{dist}(X_{j},\partial\Omega(v))$ and consider

$\displaystyle
v_{d_{j}}(y,s)=\frac{v(x_{j}+d_{j}y,t_{j}+d_{j}^{2}s)-v(x_{j},t_{j})}{d_{j}^{2}}.$

Then it can be verified that $v_{d_{j}}$ converges to a global solution
$\tilde{v}_{0}$ of (8.3) with $|\partial_{s}\tilde{v}(0)|>0$, which
contradicts the fact that the global solution $\tilde{v}_{0}$ is time-
independent.

Our next lemma is the key ingredient of the directional monotonicity for
solutions to

$\displaystyle F(D^{2}v,rDv,x_{0}+ry,t_{0}+r^{2}s)-\partial_{s}v$
$\displaystyle=f(rDv,x_{0}+ry,t_{0}+r^{2}s).$ (8.6)

Note that the rescaled function $v_{r}$ solves (8.6) in $\Omega(v_{r})$. For
simplicity, we assume for a moment $(x_{0},t_{0})=(0,0)$.

######  Lemma 8.2.

Assume that $v$ satisfies (8.6) in $\Omega(v)$. For a number $\kappa\in(0,1)$,
let $e=(e_{x},e_{t})\in\mathbb{S}^{n}$ be any space-time direction satisfying
$e\cdot(e_{1},0)\geq\kappa>0$. If
$C_{\kappa}\partial_{e}v-v\geq-\varepsilon_{0}$ in $Q_{1}$ for some constants
$C_{\kappa}$ and $\varepsilon_{0}$, then $C_{\kappa}\partial_{e}v-v\geq 0$ in
$Q_{1/2}$ provided that $\varepsilon_{0}<\frac{c}{32(2n\theta+1)}$ and
$0<r\leq\frac{c}{2C}$, where $C$ is a constant depending only on $\left\lVert
f\right\rVert_{C^{1}},\left\lVert F\right\rVert_{C^{1}},\left\lVert
v\right\rVert_{C^{1}}$ and $\kappa$.

###### Proof.

By convexity of $F(\cdot,rDv,ry,r^{2}s)$ and $F(0,rDv,ry,r^{2}s)=0$, we have

$\displaystyle
F(D^{2}v,rDv,ry,r^{2}s)-F^{ij}(D^{2}v,rDv,ry,r^{2}s)\partial_{ij}v\leq
F(0,rDv,ry,r^{2}s)=0.$

Thus we obtain $F^{ij}\partial_{ij}v-\partial_{s}v\geq F-\partial_{s}v=f\geq
c>0$. On the other hand, by differentiating (8.6) with respect to a direction
$e\in\mathbb{S}^{n}$, we get

$\displaystyle
Lv_{e}:=(F^{ij}\partial_{ij}-\partial_{s}+r(F_{p_{i}}-f_{p_{i}})\partial_{i})v_{e}=r(f_{x}-F_{x})\cdot
e_{x}+r^{2}(f_{t}-F_{t})e_{t},$

where $\partial_{i}=\tfrac{\partial}{\partial y_{i}}$ and $v_{e}=e_{x}\cdot
D_{y}v+e_{t}\partial_{s}v$. Since $f_{x}$, $F_{x}$, $f_{t}$, and $F_{t}$ are
bounded, the right hand side is as small as we want, provided $r>0$ is
sufficiently small.

Assume by contradiction that there is $(y_{0},s_{0})\in
Q_{\frac{1}{2}}\cap\Omega(v)$ such that
$C_{\kappa}\partial_{e}v(y_{0},s_{0})-v(y_{0},s_{0})<0$. If we consider

$\displaystyle\overline{v}(x,t)=C_{\kappa}\partial_{e}v-v+\frac{c(|y-y_{0}|^{2}-(s-s_{0}))}{2(2n\theta+1)},$

where $c$ is the lower bound of $f$ and $\theta$ is the ellipticity constant,
then

$\displaystyle L\overline{v}$ $\displaystyle\leq
C_{\kappa}(r(f_{x}-F_{x})\cdot e_{x}+r^{2}(f_{t}-F_{t})e_{t})$
$\displaystyle\quad-
f-r((F_{p_{i}}-f_{p_{i}})\partial_{i}v+\frac{c}{2}+r(F_{p_{i}}-f_{p_{i}})(y-y_{0})_{i}$
$\displaystyle\leq rC(\left\lVert f\right\rVert_{C^{1}},\left\lVert
F\right\rVert_{C^{1}},\left\lVert v\right\rVert_{C^{1}},\kappa)-\frac{c}{2}.$

Hence, for sufficiently small $r$, we have $L\overline{v}\leq 0$ in
$Q_{1/4}(y_{0},s_{0})\cap\Omega(v)$. By the minimum principle, the minimum is
achieved on the boundary of $Q_{1/4}(y_{0},s_{0})\cap\Omega(v)$. However,
since $\overline{v}>0$ on $\Omega(v)$ and $\overline{v}(y_{0},s_{0})<0$, the
function $\overline{v}$ attains its minimum on
$\partial_{p}Q_{1/4}(y_{0},s_{0})$. Therefore, we conclude that

$\displaystyle
0>\min_{\partial_{p}(Q_{1/4}(y_{0},s_{0})\cap\Omega(v))}\overline{v}=-\varepsilon_{0}+\frac{c}{32(2n\theta+1)},$

which is a contradiction. ∎

To finish the proof of Theorem 1.2, we take any number $\kappa\in(0,1)$ and
consider any direction $e=(e_{x},e_{t})\in\mathbb{S}^{n}\subset R^{n+1}$ such
that $e\cdot(e_{1},0)=e_{x}\cdot e_{1}\geq\kappa>0$. Recall that the limit
$v_{0}$ of $v_{r_{j}}$ has the form $\frac{\gamma}{2}[(x_{1})_{+}]^{2}$. Then
for the constant $C_{\kappa}=2/\kappa$, we have
$C_{\kappa}\partial_{e}v_{0}-v_{0}\geq 0$ in $Q_{1}$. Using the $C^{1}$
convergence $v_{r_{j}}$ in the space variable and the continuity of the speed
(8.5), we induce
$C_{\kappa}\partial_{e}v_{r_{j}}-v_{r_{j}}\geq-\varepsilon_{0}$ in $Q_{1}$ for
sufficiently large $j\geq j_{0}$, where $\varepsilon_{0}$ is the constant
given in Lemma 8.2. Then by Lemma 8.2, we obtain the improved inequality

$\displaystyle C_{\kappa}\partial_{e}v_{r_{j}}-v_{r_{j}}\geq
0\quad\text{in}\quad Q_{1/2}$ (8.7)

for $j\geq j_{0}$. Since $v_{r_{j}}\geq 0$, this implies
$\partial_{e}v_{r_{j}}\geq 0$ in $Q_{1/2}$. Scaling back to $v$, we conclude
that there exists $r=r(\kappa)>0$ such that $\partial_{e}v\geq 0$ in
$Q_{r}(X_{0})$ for all directions $e=(e_{x},e_{t})\in\mathbb{S}^{n}$
satisfying $e\cdot(e_{1},0)\geq\kappa>0$. This together with a simple
compactness argument implies that the free boundary $\Gamma(v)$ is
$\kappa$-Lipschitz graph for any $\kappa\in(0,1)$. Then the $C^{1}$ graphness
of the free boundary $\Gamma(v)$ follows by the standard argument, see [26]
for instance.

Acknowledgements. We are grateful to the referee for helpful comments. Ki-Ahm
Lee was supported by NRF grant NRF-2020R1A2C1A01006256 funded by the Korean
government (MSIP). Taehun Lee was supported by the NRF grant RS-2023-00211258
and KIAS Individual Grant MG079501.

Conflict of interest. On behalf of all authors, the corresponding author
states that there is no conflict of interest.

## References

  * [1] Almeida, L., Chambolle, A., and Novaga, M. Mean curvature flow with obstacles.  Ann. Inst. H. Poincaré Anal. Non Linéaire 29, 5 (2012), 667–681.
  * [2] Andrews, B. Contraction of convex hypersurfaces by their affine normal.  J. Differential Geom. 43, 2 (1996), 207–230.
  * [3] Andrews, B. Gauss curvature flow: the fate of the rolling stones.  Invent. Math. 138, 1 (1999), 151–161.
  * [4] Andrews, B. Classification of limiting shapes for isotropic curve flows.  J. Amer. Math. Soc. 16, 2 (2003), 443–459.
  * [5] Andrews, B., Guan, P., and Ni, L. Flow by powers of the Gauss curvature.  Adv. Math. 299 (2016), 174–201.
  * [6] Brendle, S., Choi, K., and Daskalopoulos, P. Asymptotic behavior of flows by powers of the Gaussian curvature.  Acta Math. 219, 1 (2017), 1–16.
  * [7] Caffarelli, L. A. The regularity of free boundaries in higher dimensions.  Acta Math. 139, 3-4 (1977), 155–184.
  * [8] Chopp, D., Evans, L. C., and Ishii, H. Waiting time effects for Gauss curvature flows.  Indiana Univ. Math. J. 48, 1 (1999), 311–334.
  * [9] Chow, B. Deforming convex hypersurfaces by the $n$th root of the Gaussian curvature.  J. Differential Geom. 22, 1 (1985), 117–138.
  * [10] Daskalopoulos, P., and Hamilton, R. The free boundary in the Gauss curvature flow with flat sides.  J. Reine Angew. Math. 510 (1999), 187–227.
  * [11] Daskalopoulos, P., and Lee, K.-A. Worn stones with flat sides all time regularity of the interface.  Invent. Math. 156, 3 (2004), 445–493.
  * [12] Figalli, A., and Shahgholian, H. A general class of free boundary problems for fully nonlinear elliptic equations.  Arch. Ration. Mech. Anal. 213, 1 (2014), 269–286.
  * [13] Figalli, A., and Shahgholian, H. A general class of free boundary problems for fully nonlinear parabolic equations.  Ann. Mat. Pura Appl. (4) 194, 4 (2015), 1123–1134.
  * [14] Firey, W. J. Shapes of worn stones.  Mathematika 21 (1974), 1–11.
  * [15] Giga, Y., Tran, H. V., and Zhang, L. On obstacle problem for mean curvature flow with driving force.  Geom. Flows 4, 1 (2019), 9–29.
  * [16] Hamilton, R. S. Worn stones with flat sides.  In A tribute to Ilya Bakelman (College Station, TX, 1993), vol. 3 of Discourses Math. Appl. Texas A & M Univ., College Station, TX, 1994, pp. 69–78.
  * [17] Indrei, E., and Minne, A. Regularity of solutions to fully nonlinear elliptic and parabolic free boundary problems.  Ann. Inst. H. Poincaré Anal. Non Linéaire 33, 5 (2016), 1259–1277.
  * [18] Kim, L., Lee, K.-a., and Rhee, E. $\alpha$-Gauss curvature flows with flat sides.  J. Differential Equations 254, 3 (2013), 1172–1192.
  * [19] Lee, K.-A. Obstacle problems for the fully nonlinear elliptic operators.  ProQuest LLC, Ann Arbor, MI, 1998.  Thesis (Ph.D.)–New York University.
  * [20] Lee, K.-A., and Lee, T. Gauss curvature flow with an obstacle.  Calc. Var. Partial Differential Equations 60, 5 (2021), Paper No. 166.
  * [21] Lee, K.-A., and Park, J. The regularity theory for the parabolic double obstacle problem.  Math. Ann. 381, 1-2 (2021), 685–728.
  * [22] Lee, K.-A., Park, J., and Shahgholian, H. The regularity theory for the double obstacle problem.  Calc. Var. Partial Differential Equations 58, 3 (2019), Paper No. 104, 19.
  * [23] Lee, K.-A., and Shahgholian, H. Regularity of a free boundary for viscosity solutions of nonlinear elliptic equations.  Comm. Pure Appl. Math. 54, 1 (2001), 43–56.
  * [24] Lieberman, G. M. Second order parabolic differential equations.  World Scientific Publishing Co., Inc., River Edge, NJ, 1996.
  * [25] Mercier, G., and Novaga, M. Mean curvature flow with obstacles: existence, uniqueness and regularity of solutions.  Interfaces Free Bound. 17, 3 (2015), 399–426.
  * [26] Petrosyan, A., Shahgholian, H., and Uraltseva, N. Regularity of free boundaries in obstacle-type problems, vol. 136 of Graduate Studies in Mathematics.  American Mathematical Society, Providence, RI, 2012.
  * [27] Rupflin, M., and Schnürer, O. C. Weak solutions to mean curvature flow respecting obstacles.  Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 20, 4 (2020), 1429–1467.