\documentclass[reqno]{amsart} 

\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol.~2000(2000), No.~68, pp.~1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu \quad ejde.math.unt.edu (login: ftp)}
\thanks{\copyright 2000 Southwest Texas State University.} 
\vspace{1cm}}

\begin{document} 

\title[\hfilneg EJDE--2000/68\hfil Regularity of the interface]
{Regularity of the interface for the porous medium equation} 

\author[Youngsang Ko\hfil EJDE--2000/68\hfilneg]
{Youngsang Ko}

\address{Youngsang Ko \hfill\break
Department of Mathematics, Kyonggi University, Suwon, Kyonggi-do, 
442-760, Korea}
\email{ysgo@kuic.kyonggi.ac.kr}

\date{}
\thanks{Submitted October 7, 2000. Published November 13, 2000.}
\thanks{Supported by a research grant form Kyonggi University}
\subjclass{35K65}
\keywords{porous medium, radial symmetry, interface, C-infinity regularity. }

\begin{abstract}
We establish the interface equation and 
prove the $C^{\infty}$ regularity of the interface for the 
porous medium equation whose solution is radial symmetry.
\end{abstract}

\maketitle

\newtheorem{thm}{Theorem}[section]
\newtheorem{lem}[thm]{Lemma}
\newtheorem{pro}[thm]{Proposition}


\makeatletter
\numberwithin{equation}{section}
\makeatother

\section{Introduction}
We consider the Cauchy problem of the form
\begin{gather} \label{eq:porous}
u_t = \Delta(u^m)\quad \text{in } 
 S= {\mathbb R}^N\times (0,\infty), \\  \label{eq:inporous}
u(x,0)=u_0 \quad \text{on } {\mathbb R}^N.
\end{gather}
Here we suppose that $m > 1$, and $u_0$ is a nonzero bounded nonnegative 
function with compact support. 

It is well known that  (\ref{eq:porous}) describes the evolution 
in time of various diffusion processes, in particular the flow of a gas 
through a porous medium. 
Here $u$ stands for the density, while $v=\frac{m}{m-1}u^{m-1}$ represents 
the pressure of the gas.
Then $v$ satisfies 
\begin{equation}\label{eq:pressure}
v_t = (m-1)v\Delta v + |\nabla v|^2 .
\end{equation}
If the solution is radial symmetry, then $v$ satisfies
\begin{equation}\label{eq:symm}
v_t = (m-1)v v_{rr} + \frac{\bar{m}}{r}vv_r + v_r^2,
\end{equation}
where $r=\sqrt{\sum_{i=1}^N x_i^2}$ and $\bar{m}=(m-1)(N-1)$. Since we are concerning about the regularity of 
the interface
we may assume $r\geq \epsilon_0$ for some positive number $\epsilon_0$.
In this paper we will show that, if the solution is radial symmetry then the interface
of (\ref{eq:symm}) can be represented by a $C^{\infty}$ function after a large time.

In the one-dimensional case
Aronson and V\'azuez \cite{AV} and independently H\"ollig and Kreiss \cite{HK} showed that the interfaces 
are smooth after the waiting time. Angenent \cite{A} showed that the interfaces are real analytic
after the waiting time.  For the dimensions $\geq 2$, 
Daskalopoulos and
Hamilton \cite{DH} showed that for $t\in (0, T)$, for some $T>0$, the interface can be described as a 
$C^{\infty}$ surface if the initial data $u_0$ satisfies some assumptions. On the other hand,
Caffarelli, Vazquez and Wolanski \cite{CVW} showed that after a large time the interface can be described 
as a Lipschitz surface if the $u_0$ satisfies some non-degeneracy conditions. Caffarelli and Wolanski 
\cite{CW} improved this result by showing that the interface can be described 
as a $C^{1,\alpha}$ surface under the same non-degeneracy conditions on the initial data. 
But many people believe that 
even after a large time, the interface can be described as a smooth surface.

In this paper, assuming the solution to (\ref{eq:porous}) is radial symmetry and the initial data
$u_0$ satisfies the same non-degeneracy conditions as in \cite{CW}, we obtain the following result :
\begin{thm}\label{thm:main}
If $v$ is a solution to (\ref{eq:symm}), then $v$ is a $C^{\infty}$ function
near the interface where $v>0$ and the interface is a $C^{\infty}$ function for $t>T$, for some $T>0$.
\end{thm}

This paper is divided into three parts :  In Part I, we obtain the interface equation. 
In Part II we obtain the upper and lower bound of $v_{rr}$ by
constructing a barrier function. In Part III, we obtain the upper and lower bound of 
$\left(\frac{\partial}{\partial r}\right)^j v\equiv v^{(j)}$ by constructing another barrier function.
In showing our results, we adapt the methods used in \cite{AV}. 

\section{Preliminaries}
In this section, we introduce some basic results which are necessary in showing the uniform boundedness
of the derivatives of the pressure $v$. First, since we are interested in the radial symmetry solution, let
$$P[u]=\{(r,t) \;|\; u(r,t)>0, r\geq \epsilon_0>0\}$$
for some $\epsilon_0>0$, be the positivity set. Then by \cite{CVW}, we can express the interface as
a nondecreasing and Lipschitz continuous 
$r=\zeta(t)=\sup\{r\geq \epsilon_0\;|\;u(r,t)>0\}$ and $r=\zeta(t)$
on $[T,\infty)$, for some $T>0$. 
In showing the interface is a $C^{\infty}$ function,
we need the following : 
\begin{thm}\label{thm:interface}
Assume that $u_0 >0$ on $I=(2\epsilon_0, a)$ and $u_0 =0$ on $[a,\infty)$. 
Let $v_0 = \frac{m}{m-1}u_{0}^{m-1}\in C^{1}(\bar{I})$. Then
$$\lim_{r\to  \zeta(t)}v_r (r,t) \equiv v_r (\zeta(t),t)$$
exists for all $t\geq 0$ and
$$v_r (\zeta(t),t)\left\{\zeta^{'}(t)+ v_r(\zeta(t),t)\right\}=0$$
for almost all $t\geq 0$.
\end{thm}

In proving Theorem \ref{thm:interface}, we need
\begin{lem}\label{lem:0}
Assume that $v_0$ has bounded derivatives of all orders, and that there exist positive constants
$\mu$, $M$ such that
$$\mu \leq v_0(r)\leq M \quad\text{on } [\epsilon_0, \infty).$$
Then the Cauchy problem (\ref{eq:symm}) has a unique classical solution $v$ in 
$S=[\epsilon_0, \infty)\times [0, \infty)$ such that
$$\mu \leq v(r,t)\leq M \quad\text{on } S $$
and $v \in C^{\infty}(S)$. 
\end{lem}

\noindent\textbf{Proof.} To show the existence of $v$, we need to have an a priori lower bound for $v$. To 
this end, let $\varphi=\varphi(s)$ denote a $C^{\infty}({\mathbb R}^1)$ function such that $\varphi(s)=s$
for $s \geq \mu$, $\varphi(s)=\mu/2$ for $s\leq 0$, and $\varphi$ increases from $\mu/2$ to $\mu$
as $s$ increases from $0$ to $\mu$. Now consider
\begin{equation} \label{eq:sym}
\begin{gathered} v_{t}=(m-1)\varphi(v)v_{rr} + \frac{\bar{m}}{r}vv_r + v_r^2
\quad\text{in } [\epsilon_0,\infty)\times (0,\infty),   \\  
v(x, 0) = v_{0}(x)\quad \text{on } [\epsilon_0,\infty).
\end{gathered}
\end{equation}
Then it is easily verified that equation (\ref{eq:sym}) satisfies all the hypotheses of Theorem 5.2 
in \cite{LSU}( pp. 564-565). Let $\tau\in(0,\infty)$ and $\beta\in(0,1)$ be arbitrary, and let 
$R_{\tau}=[\epsilon_0,\infty)\times [0,\tau]$. Then there exists a unique solution $v$ of equation 
(\ref{eq:sym})
such that $|v|\leq M$ in $R_{\tau}$ and $v \in H^{2+\beta, 1+\beta/2}(R_{\tau})$.In \cite{LSU}, 
the solution of (\ref{eq:sym}) is obtained as the limit as $n\to \infty$ of the solutions
$v^n$ of the sequence of the first boundary value problems
\begin{gather*}
v_t = (m-1)\varphi(v)v_{rr} + \frac{\bar{m}}{r}vv_r + v_r^2 
\quad\text{in } [\epsilon_0+1/n,n]\times (0, \tau),\\
v(r,0)=v_0(r)\quad\text{in } [\epsilon_0+ \frac{1}{n}, n], \\
v(n,t)=v_0(n)\quad\text{in } [0,\tau].
\end{gather*} 
Then by the method used in proving Theorem 2 in \cite{AR1}, we have $v=\lim_{n\to \infty}v^n$
belongs to $C^{\infty}(S)$ and $\mu\leq v\leq M$. \hfill $\Box$


Now let us prove Theorem \ref{thm:interface}.
Let $t>0$ and assume $r<r^{'} \in I_t =[2\epsilon_0, \zeta(t))$. By mean value theorem,
$$v_r(r^{'}, t) = v_r(r,t)+(r-r^{'})v_{rr}(\tilde{r},t)$$ 
for some $\tilde{r}\in (r,r^{'})$. Since $|\nabla v|=|v_r |\leq L$ \cite{CVW}, 
by the following famous result
$$\Delta v = v_{rr} + \frac{N-1}{r}v_r \geq -\frac{1}{(m-1+2/N)t}$$
established in \cite{AB}, together with the assumption that $r\geq \epsilon_0$, the lower bound for
$v_{rr}$ is obtained. Hence
$$ v_r(r^{'}, t) \geq v_r(r,t)-\alpha(r^{'}-r),$$
for some positive constant $\alpha$. Also, since $v_r$ is bounded above,
$$\sup_{I_t}|v_r(r,t)|\leq C.$$
Therefore the inferior and superior limits of $v_r(r,t)$ exist and are finite when $r\to  \zeta(t)$.
It follows that
$$\liminf_{r^{'}\to \zeta(t)}v_r(r^{'},t)\geq v_r(r,t)-\alpha\{\zeta(t)-r\}$$
for all $r\in I_t$, and 
$$\liminf_{r^{'}\to \zeta(t)}v_r(r^{'},t)\geq \limsup_{r\to \zeta(t)}v_r(r^{'},t).$$
Therefore
$$\lim_{r\to \zeta(t)}v_r(r,t)\equiv v_r(\zeta(t),t)$$
exists.

Next, let $\psi$ be a $C(S)$ function which has compact support in ${\mathbb R}^1$ for each fixed $t\geq 0$.
Suppose further that $\psi_r \in C(S)$ and that $\psi$ possesses a weak derivative $\psi_t$
with respect to $t$ in $S$. Define
$$ 
\tilde{\psi}(r,t)=\begin{cases}
\psi(r,t)& \text{for } t\geq 0\\
\psi(r,0)& \text{for } t\leq 0\,.
\end{cases}
$$ 
Then it is shown \cite{AR2} that $\tilde{\psi}\in C({\mathbb R}^2)$, $\tilde{\psi}$ has compact support as
a function of $r$ for each fixed $t$, $\tilde{\psi}_r \in C({\mathbb R}^2)$, and $\tilde{\psi}$ is
weakly differentiable with respect to $t$ in ${\mathbb R}^2$. Moreover, $\tilde{\psi}_t$ coincides with
$\psi_t$ for $t>0$. Let
\[
\psi_n(r,t)=\iint_{{\mathbb R}^2}k_n(r-\xi,t-\tau)\tilde{\psi}(\xi,\tau)d\xi d\tau,
\]
where $k_n(r,t)$ denotes an averaging kernel with support in 
$(-\frac{1}{n}, \frac{1}{n})\times (-\frac{1}{n},\frac{1}{n})$ for each integer $n\geq 1$. 
Then $\psi_n$ satisfies
\begin{equation}\label{eq:wk} 
\begin{split}
&\int_{{\mathbb R}^1}\psi_n(r,t_2)v(r,t_2)dr 
+\int_{t_1}^{t_2}\int_{{\mathbb R}^1}\{(m-1+\frac{c}{r})vv_r\psi_{nr} +(m-2)v_r^2\psi_n-v\psi_{nt}\}\,dr\,dt\\
&=\int_{{\mathbb R}^{1}}\psi_{n} (r,t_1)v(r,t_1)dr.
\end{split}
\end{equation}
Recall that $v$ and its weak derivative $v_r$ are bounded in $S$. On the other hand, 
it is shown in \cite{AR2} that
$\psi_n \to  \tilde{\psi}$ and $\psi_{nr}\to  \tilde{\psi}_r$ uniformly
on any compact subsets of ${\mathbb R}^2$, while $\psi_{nt}\to  \tilde{\psi}_t$ strongly
in $L_{loc}^1({\mathbb R}^2)$. Therefore (\ref{eq:wk}) also holds for the limit of the sequence $\psi_n$, that is, 
for any set function which satisfies the conditions listed at the beginning of this paragraph. Now define
a function
$$ 
K(r)=\begin{cases}
C\cdot \exp\{-1/(1-r^2)\} &\text{for } |r|\leq 1\\
0 &\text{for } |r|\geq 1\,,
\end{cases} 
$$ 
where $C$ is chosen so that
$$\int_{{\mathbb R}^1}K(r)dr=1\,,$$
and set $k_n(r)=nK(nr)$
for each integer $n\geq 1$. Then $k_n(r)$ is an even averaging kernel and $k_n(\zeta(t)-r)$ belongs to $C(S)$
and has compact support in ${\mathbb R}^1$ for each $t\geq 0$. Also $\frac{d}{dr}k_n(\zeta(t)-r) \in C(S)$. 
Since $\zeta$ is Lipschitz continuous, $\zeta^{'}$ exists almost everywhere and is bounded above. 
Moreover $\zeta$ is weakly differentiable and its weak derivative can be represented by $\zeta^{'}$.
It follows that $k_n(\zeta(t)-r)$ is also weakly differentiable with respect to $t$ in $S$
with weak derivative given by $\zeta^{'}(t)k_n'(\zeta(t)-r)$ which belongs to $L^1({\mathbb R}^1\times (t_1,t_2)$
for any $0\leq t_1<t_2<\infty$. Thus $k_n(\zeta(t)-r)$ is an admissible test function in (\ref{eq:wk}).
In particular, if we set $\psi_n(r,t)=k_n(\zeta(t)-r)$ in (\ref{eq:wk}), we have
\begin{equation}\label{eq:wk*} 
\begin{split}
\int_{{\mathbb R}^1}k_n(\zeta(t_2)-r)v(r,t_2)dr 
+\int_{t_1}^{t_2}\int_{{\mathbb R}^1}k_n'(\zeta-r)v(r,t)\{-(m-1)v_r-\zeta'\}\,dr\,dt\\
+(m-2)\int_{t_1}^{t_2}\int_{{\mathbb R}^1}k_n v_r^2\,dr\,dt 
-\int_{t_1}^{t_2}\int_{{\mathbb R}^1}\frac{\bar{m}}{r}vv_rk_n(\zeta-r)\,dr\,dt\\
=\int_{{\mathbb R}^{1}}k_n(\zeta(t_1)-r)v(r,t_1)dr.\\
\end{split}
\end{equation}
 For fixed $t$, $v(r,t)$ is a continuous function of $r$ in ${\mathbb R}^1$ and $v(r,t)=0$ for
$r\in {\mathbb R}^1\setminus I_t$. Hence for any $t\geq 0$
\[
\lim_{n\to  \infty}\int_{{\mathbb R}^1}k_n(\zeta(t)-r)v(r,t)dx=v(\zeta(t),t)=0.
\] 
Similarly, since $|v_r|$ is bounded above, we have
\[
\lim_{n\to \infty}\int_{t_1}^{t_2}\int_{{\mathbb R}^1}\frac{\bar{m}}{r}vv_rk_n(\zeta-r)dr=0.
\]
Since $v_r(r,t)\to  v_r(\zeta(t),t) $ as $r\to  \zeta-$ and $v_r(r,t)=0$ on ${\mathbb R}^1\setminus I_t$,
we have
\[
\int_{{\mathbb R}^1}k_n(\zeta-r)v_r^2(r,t)dr=\int_{\zeta-\frac{1}{n}}^{\zeta}k_n(\zeta-r)v_r^2(r,t)dr.
\]
Since $k_n$ is even, we have
\[
\lim_{n\to  \infty}\int_{{\mathbb R}^1}k_n(\zeta(t)-r)v_r^2(r,t)dx=\frac{1}{2}v_r^2(\zeta,t)
\]
for each $t>0$. Moreover, since $|v_r|\leq L$,
\[\left|\int_{{\mathbb R}^1}k_n(\zeta-r)v_r^2(r,t)dr\right|\leq L^2.\]
Thus by the Lebesgue's dominated convergence theorem,
\[
\lim_{n\to  \infty}\int_{t_1}^{t_2}\int_{{\mathbb R}^1}k_n(\zeta(t)-r)v_r^2(r,t)dxdt
=\frac{1}{2}\int_{t_1}^{t_2}v_r^2(\zeta,t)dt.
\] 
Next define
$$
w(r,t)=
\begin{cases}
\dfrac{-v(r,t)}{\zeta-r} &\text{for } r<\zeta(t)\\ 
v_r(\zeta(t),t) &\text{for } r=\zeta(t)\\
0&\text{for } r>\zeta(t)\,.
\end{cases} $$ 
Note that for fixed $t$, $w$ is continuous on $[\epsilon_0,\zeta(t)]$ and $|w(x,t)|\leq C$.
Then the second integral on the left in (\ref{eq:wk*}) can be written in the form
\[
I_n=\int_{t_1}^{t_2}\int_{{\mathbb R}^1}(\zeta-r)k_n'(\zeta-r)w(r,t)\{(m-1)v_r+\zeta'\}\,dr\,dt.
\]
It is easily verified that the function $-rk_n^{'}(r)$ is also an even averaging kernel. Thus for each $t$,
$$\lim_{n\to  \infty}\int_{{\mathbb R}^1}(\zeta-r)k_n^{'}(\zeta-r)w\{(m-1)v_r+\zeta'\}dr
=-\frac{1}{2}v_r(\zeta,t)\{ (m-1)v_r+\zeta'\}$$ 
and 
\[
\left|\int_{{\mathbb R}^1}(\zeta-r)k_n^{'}(\zeta-r)w\{(m-1)v_r+\zeta'\}dr \right|
\leq (m-1)L^2 +L\zeta'\in L^1(t_1,t_2).
\]
It follows that
\[
\lim_{n\to \infty}I_n =-\frac{1}{2}\int_{t_1}^{t_2} v_r(\zeta,t)\{ (m-1)v_r+\zeta'\}dt.
\]
Hence if we let $n\to  \infty$ in (\ref{eq:wk*}), we obtain
\[
-\frac{1}{2}\int_{t_1}^{t_2} v_r(\zeta,t)( \zeta^{'}(t)+v_r(\zeta,t))dt=0
\]
for any $0\leq t_1<t_2<\infty $. Therefore
\[
v_r(\zeta,t)( \zeta^{'}(t)+v_r(\zeta,t))=0
\]
for almost all $t\geq 0$. \hfill $\Box$ 

\section{Upper  and Lower Bounds for $v_{rr}$}
Let $v=v(x,t)$ be the pressure corresponding to a solution $u=u(x,t)$
of equation (\ref{eq:porous}). If $u$ is radial symmetry then $v=v(r,t)$ satisfies (\ref{eq:symm})
in the positivity set $P[u]=\{(r,t)\;|\; u(r,t)>0, r\geq \epsilon_0\}$. Assume that
$v_0(r)$ has compact support containing $[2\epsilon_0,a]$ for some $a>0$ and satisfies 
the non-degeneracy condition (1.4) of the Theorem 1 in \cite{CW}.
Let $q=(r_0, t_0)$ be a point on the interface $r=\zeta(t)$ so that $r_0 = \zeta(t_0)$, 
$v(r,t_0)=0$ for all $r\geq \zeta(t_0)$, and $v(r,t_0)>0$ for all $0<r<\zeta(t_0)$.
Since the lower bound for
$v_{rr}$ is obtained already, we need to show $v_{rr}$ is bounded above. In showing this we adopt
the methods used in \cite{AV}. 

Now let $T>0$ be the positive constant established in \cite{CVW}. 
Assume $t_0 >T$ so that the interface is moving at $q$. 
Then from Theorem \ref{thm:interface}, we have
\begin{equation}\label{eq:21}
\zeta^{'}(t_0) = -v_r(\zeta,t)=a>0,
\end{equation}
and on the moving interface we have
\begin{equation}\label{eq:2}
v_t = v_r^2 .
\end{equation}
As in \cite{AV}, we use the notation
$$R_{\delta, \eta} 
= R_{\delta, \eta}(t_0) \equiv \{(r,t) \in {\mathbb R}^2 \;|\; \zeta(t)-\delta < r \leq \zeta(t), 
\;t_0 -\eta \leq t\leq t_0 + \eta\}.$$ 
\begin{lem}\label{lem:21}
Let $q$ be a point on the interface and assume (\ref{eq:21}) holds. Then there exist positive
constants $C$, $\delta$ and $\eta$ depending only on $N$, $\epsilon_0$, $m$, $q$ and $u$ such that
$$ |v_{rr}| \leq C \qquad \text{in } R_{\delta, \eta/2} .$$
\end{lem}

\noindent\textbf{Proof}.
It is well known that $v_t$, $v_r$ and $vv_{rr}$ are continuous in a closed neighborhood in 
$\overline{P}[u]$ of any point on the interface if $t>T$, and that
$$(vv_{rr})(r,t)=\frac{1}{m-1}\left(v_t -\frac{\bar{m}}{r}vv_r - v_r^2 \right)\to  0\;
as\; P[u]\ni (r,t) \to  (\zeta(t),t)$$
for any $t>T$. Choose now an $\epsilon >0$ such that
\begin{equation}\label{eq:24}
(a-(8m-3)\epsilon)(a-\epsilon) \geq 4(m+1)\epsilon >0.
\end{equation}
Then there exist $\frac{\epsilon_0}{2\bar{m}}\leq\delta=\delta(\epsilon)>0$ 
and $\eta=\eta_{1} (\epsilon)\in (0,t_0-T)$
such that $R_{\delta,\eta}\subset P[u]$,
\begin{equation}\label{eq:25}
-a-\epsilon <v_r <-a+\epsilon,
\end{equation}
and 
\begin{equation}\label{eq:26}
vv_{rr} \leq \epsilon
\end{equation}
in $R_{\delta,\eta}$. In view of (\ref{eq:25}) we have
\begin{equation}\label{eq:27}
(a-\epsilon)(\zeta(t)-r)<v(r,t)<(a+\epsilon)(\zeta(t)-r) \quad\text{in } R_{\delta,\eta}
\end{equation}
and
\begin{equation}\label{eq:28}
a-\epsilon<\zeta^{'}(t)<a+\epsilon\quad\text{in } [t_1,t_2]
\end{equation}
where $t_1 =t_0-\eta$ and $t_2 = t_0 +\eta$. We set                    
\begin{equation}\label{eq:29}
\zeta^* (t) \equiv \zeta_1 + b(t-t_1),
\end{equation}
where $\zeta_1 = \zeta(t_1)$ and $b= a+2\epsilon$. Clearly $\zeta(t)<\zeta^* (t)$ in $(t_1,t_2]$.

On $P[u]$, $p\equiv v_{rr}$ satisfies
\begin{eqnarray*}
L(p)&=&p_t- (m-1)vp_{rr} - \left(2mv_r + \frac{\bar{m}}{r}v\right)p_r\\
    &&-(m+1)p^2 +\left(\frac{2\bar{m}}{r^2}v-\frac{3\bar{m}}{r}v_r\right)p 
-\frac{2\bar{m}}{r^3}(vv_r-rv_r^2 )=0
\end{eqnarray*}
where $\bar{m}=(m-1)(N-1)$.  
As in \cite{AV} we construct a barrier function for $p$ of the form
\begin{equation}\label{eq:barrier1}
\phi (r,t)\equiv \frac{\alpha}{\zeta(t)-r} + \frac{\beta}{\zeta^* (t)-r},
\end{equation}
where $\alpha$ and $\beta$ are positive constants and will be decided later.
\begin{eqnarray*}
L(\phi) &=& \frac{\alpha}{(\zeta-r)^2}\left\{-\zeta^{'} -2(m-1)\frac{v}{\zeta-r} 
- 2mv_r - \frac{\bar{m}}{r}v\right\}\\
&&+ \frac{\beta}{(\zeta^*-r)^2}\left\{-\zeta^{*'} -2(m-1)\frac{v}{\zeta^{*}-r} 
- 2mv_r - \frac{\bar{m}}{r}v\right\}\\ 
&& -(m+1)\phi^2 + (\frac{2\bar{m}}{r^2}v-\frac{3\bar{m}}{r}v_r)\phi -\frac{2\bar{m}}{r^3}(vv_r-rv_r^2)\\
&&\geq \frac{\alpha}{(\zeta-r)^2}\left\{-\zeta^{'} -2(m-1)\frac{v}{\zeta-r} 
- 2mv_r -\frac{\bar{m}}{r}v- 2(m+1)\alpha \right\}\\
&&+ \frac{\beta}{(\zeta^{*}-r)^2}\left\{-\zeta^{*'} -2(m-1)\frac{v}{\zeta^{*}-r} 
- 2mv_r -\frac{\bar{m}}{r}- 2(m+1)\beta \right\},
\end{eqnarray*}
since $v_r<0$.
>From the choice of $\delta$ and the estimates (\ref{eq:25}), (\ref{eq:27}), (\ref{eq:28}) 
and the definition (\ref{eq:29}) of
$\zeta^{*}$ we conclude that
\begin{eqnarray*}
L(\phi) &\geq& \frac{\alpha}{2(\zeta-r)^2} \{a-(8m-5)\epsilon-4(m+1)\alpha\}\\
&&+ \frac{\beta}{2(\zeta^*-r)^2}\{a-(8m-3)\epsilon -4(m+1)\beta\}.
\end{eqnarray*}
Now set 
\begin{equation}\label{eq:212}
\beta=\frac{a-(8m-3)\epsilon}{8(m+1)}
\end{equation}
and note that (\ref{eq:24}) implies that $\beta>0$. Then $L(\phi)\geq 0$ in $R_{\delta,\eta}$ 
for all $\alpha\in(0,\alpha_0]$, where $\alpha_0 =\{a-(8m-5)\epsilon\}/4(m+1)$.

Let us compare $p$ and $\phi$ on the parabolic boundary of $R_{\delta,\eta}$. In view of 
(\ref{eq:26}) and (\ref{eq:27}) we have 
$$ v_{rr} < \frac{\epsilon}{(a-\epsilon)(\zeta-r)}\quad\text{in } R_{\delta,\eta},$$
so that, in particular,
$$v_{rr}(\zeta(t)-\delta,t)\leq \frac{\epsilon}{(a-\epsilon)\delta}\quad\text{in } [t_1,t_2].$$ 
By the mean value theorem and (\ref{eq:28}) it follows that for some $\tau\in (t_1,t_2)$
\begin{eqnarray*}
\zeta^*(t)+\delta - \zeta(t) &=& \delta + (a+2\epsilon)(t-t_1)-\zeta^{'}(\tau)(t-t_1)\\
                               &\leq&\delta + 3\epsilon(t-t_1)\leq \delta + 6\epsilon\eta.
\end{eqnarray*}
Now set
$$\eta\equiv \min\{\eta_1 (\epsilon),\delta(\epsilon)/6\epsilon\}.$$
Since $\epsilon$ satisfies (\ref{eq:24}) and $\beta$ is given by (\ref{eq:212}) it follows that
$$\phi(\zeta +\delta,t)\geq \frac{\beta}{2\delta}\geq \frac{\epsilon}{(a-\epsilon)\delta}
\geq v_{rr}(\zeta+\delta,t)\quad\text{on } [t_1,t_2].$$ 
Moreover,
$$\phi(r,t_1)\geq \frac{\beta}{\zeta_1-r}>\frac{\epsilon}{(a-\epsilon)(\zeta_1-r)}>v_{rr}(r,t_1)
\quad\text{on } [\zeta_1-\delta, \zeta_1).$$
Let $\Gamma=\{(r,t)\in {\mathbb R}^2 \;:\;r=\zeta(t), t_1 \leq t\leq t_2\}$. Then $\Gamma$ is a compact subset
of ${\mathbb R}^2$. Now fix $\alpha\in(0,\alpha_0)$. For each point $s\in\Gamma$ there is an open ball $B_s$
centered at $s$ such that
$$(vv_{rr})(r,t)\leq \alpha(a-\epsilon)\quad\text{in } B_s \cap P[u],$$
In view of (\ref{eq:27}) we have
$$\phi(r,t)\geq \frac{\alpha}{\zeta-r}\geq v_{rr}(r,t)\quad\text{in }  B_s \cap P[u].$$ 
Since $\Gamma$ is compact, finite number of these balls can cover $\Gamma$ and hence there is
a $\gamma=\gamma(\alpha)\in(0,\delta)$ such that
$$\phi(r,t)\geq p(r,t)\quad\text{in } R_{\gamma,\eta}.$$
Thus for every $\alpha \in (0,\alpha_0)$, $\phi$ is a barrier for $p$ in $R_{\delta,\eta}$.
Hence by the comparison principle we conclude 
$$v_{rr}(r,t) \leq \frac{\alpha}{\zeta-r} +\frac{\beta}{\zeta^*(t)-r}\quad\text{in } R_{\delta,\eta},$$
where $\beta$ is given by (\ref{eq:212}) and $\alpha \in (0,\alpha_0)$ is arbitrary. 
Now let $\alpha\downarrow 0$ to obtain
$$v_{rr}(r,t) \leq \frac{\beta}{\zeta^*(t)-r}\leq \frac{2\beta}{\epsilon\eta}
\quad\text{in } R_{\delta,\eta/2},$$  \hfill $\Box$

\section{Bound for $\left(\frac{\partial}{\partial r}\right)^j v$}
As in \cite{AV}, if we can show 
$$|v^{(j)}\equiv \left(\frac{\partial}{\partial x}\right)^j v|\leq C_j,$$
for each $j\geq 2$, then Theorem \ref{thm:main} follows. First by a direct computation for
$j\geq 3$, $ v^{(j)}$ satisfy the following equation
\begin{align}\label{eq:31}
L_j v^{(j)}&\equiv  v_t^{(j)} - 
(m-1)vv_{rr}^{(j)}- (2+j(m-1))v_rv_r^{(j)} -\frac{\bar{m}}{r}vv_r^{(j)}-F_j(r,t)v^{(j)}\notag\\
&-G_j(r,t),
\end{align}
where $F_j(r,t)$ and $G_j(r,t)$ are functions of $r, v$ and derivatives of $v$ of order $<j$ only.
Then our result is
\begin{pro}\label{pro:31}
Let $q=(r_0,t_0)$ be a point on the interface for which (\ref{eq:21}) holds. For each integer $j\geq 2$
there exist constants $C_j$, $\delta$ and $\eta$ depending only on $N$, $\epsilon_0$,
$m$, $j$, $q$ and $u$ such that
\begin{equation}\label{eq:32}
\left|\left(\frac{\partial}{\partial r}\right)^j v\right|\leq C_j\quad\text{in } R_{\delta,\eta/2}.
\end{equation}
\end{pro}
The proof proceeds as in \cite{AV} by induction on $j$. Suppose that $q$ is a point on the interface
for which (\ref{eq:21}) holds. Fix $\epsilon\in (0,a)$ and
take $\delta_0 =\delta_0(\epsilon)>0$ and $\eta_0=\eta_0(\epsilon)\in(0,t_0-T)$
such that $R_0 \equiv R_{\delta_0\eta_0}(t_0)\subset P[u]$ and (\ref{eq:25}) holds. Thus we also have
(\ref{eq:27}) and (\ref{eq:28}) in $R_0$. Assume that there are constants $C_k\in {\mathbb R}^+$
for $k=2, 3, \ldots, j-1$ such that
\begin{equation}\label{eq:33}
|v^{(k)}|\leq C_k\quad\text{on } R_0\quad \text{for}\;k=2, \ldots,j-1.
\end{equation}
Observe that, by Lemma \ref{lem:21}, the estimate (\ref{eq:33}) holds for $k=2$. As in \cite{AV}
by rescaling and using interior estimates we obtain the following estimate near $\zeta$.
\begin{lem}\label{lem:32}
There are constants $K\in {\mathbb R}^+$, $\delta\in(0,\delta_0)$ and $\eta\in(0,\eta_0)$, depending
only on $q$, $m$ and the $C_k$ for $k\in [2,j-1]$ with $j\geq 3$, such that
\[ |v^{(j)}(r,t)|\leq \frac{K}{\zeta(t)-r}\quad\text{in } R_{\delta,\eta}.\]
\end{lem}
\noindent\textbf{Proof}. Set
$$\delta=\min\{\frac{2\delta_0}{3},2s\eta_0\},$$
$$\eta=\eta_0-\frac{\delta}{4s},$$
and define
$$
R(\bar{r},\bar{t})\equiv
\left\{(r,t)\in{\mathbb R}^2 \;:\;|r-\bar{r}|<\frac{\lambda}{2},
\;\bar{t}-\frac{\lambda}{4s}<t\leq\bar{t}\right\}
$$ 
for $(\bar{r},\bar{t})\in R_{\delta,\eta}$, where $s=a+\epsilon$ and 
$\lambda=\zeta(\bar{t})-\bar{r}$. Then $(\bar{r},\bar{t})\in R_{\delta,\eta}$
implies that $ R_{\bar{\delta},\bar{\eta}}\subset R_0$. Also observe that for each
$(\bar{r},\bar{t})\in R_{\delta,\eta}$,  $ R(\bar{r},\bar{t})$ lies to the left of the line 
$r=\zeta(\bar{t})+ s(t-\bar{t})$. Now set $r=\lambda\xi+\bar{r}$ and $t=\lambda\tau + \bar{t}$. 
Then the function 
$$V^{(j-1)}(\xi,\tau)\equiv v^{(j-1)}(\lambda\xi+\bar{r},\lambda\tau + \bar{t})=v^{(j-1)}(r,t)$$
satisfies the equation
\begin{align}\label{eq:34}
V_{\tau}^{(j-1)} &=\left\{(m-1)\frac{v}{\lambda}V_{\xi}^{(j-1)} 
+ (2+(j-1)(m-1))v_rV^{(j-1)}\right\}_{\xi}\notag\\
&-(m-1)v_rV_{\xi}^{(j-1)} +\frac{\bar{m}}{r}vV_{\xi}^{(j-1)}\\
&+\lambda (F_{j-1}(r,t)-(2+(j-1)(m-1))v_{rr})V^{(j-1)}+\lambda G_{j-1}(r,t)\notag
\end{align}
in the region
$$B\equiv\left\{(\xi,\tau)\in {\mathbb R}^2\;:\;|\xi|\leq\frac{1}{2},\; -\frac{1}{4s}<\tau\leq 0\right\},$$
and $|V^{(j-1)}|\leq C_{j-1}$ in $B$. In view of (\ref{eq:27}) and (\ref{eq:28})
we have
$$(a-\epsilon)\frac{\zeta(t)-r}{\lambda}\leq \frac{v(r,t)}{\lambda}\leq (a+\epsilon)\frac{\zeta(t)-r}{\lambda}
$$
and 
$$\zeta(t)\leq\zeta(\bar{t})\leq\zeta(t) + s(\bar{t}-t)\leq\zeta(t)+\frac{\lambda}{4}.$$
Therefore
$$
\frac{\lambda}{4}=\zeta(\bar{t})-\frac{\lambda}{4}-\bar{r}-\frac{\lambda}{2}
\leq\zeta(t)-r\leq\zeta(\bar{t})-\bar{r}+\frac{\lambda}{2}=\frac{3\lambda}{2}$$
which implies
$$\frac{a-\epsilon}{4}\leq \frac{v}{\lambda}\leq\frac{3(a+\epsilon)}{2}.$$
that is, equation (\ref{eq:34}) is uniformly parabolic in $B$. Moreover, it follows from
(\ref{eq:25}) and (\ref{eq:33}) that $V^{(j-1)}$ satisfies all of the hypotheses of the Theorem 5.3.1 of 
\cite{LSU}. Thus we conclude that there is a constant $K=K(a, m, C_1, \ldots, C_{j-1})>0$
such that
$$\left|\frac{\partial}{\partial\xi}V^{(j-1)}(0,0)\right|\leq K,$$
that is 
$$|v^{(j-1)}(\bar{r},\bar{t})|\leq K/\lambda.$$
Since $( \bar{r},\bar{t})\in R_{\delta,\eta}$ is arbitrary, this proves the lemma. \hfill $\Box$

We now turn to the barrier construction. If $\gamma\in (0,\delta)$ we will use the notation
$$R_{\delta,\eta}^{\gamma}= R_{\delta,\eta}^{\gamma}(t_0)
\equiv\{(r,t)\in {\mathbb R}^2\;:\;\zeta(t)-\delta\leq r\leq \zeta-\gamma, t_0-\eta\leq t \leq t_0 +\eta\}.$$
Then we have
\begin{lem}\label{lem:33}
Let $R_{\delta_1,\eta_1}$ be the region constructed in the proof of Lemma \ref{lem:21}. For 
$j\geq 3$ and $(r,t)\in R_{\delta_1,\eta_1}^{\gamma} $, let
%\begin{equation}\lebel{eq:35}
$$\phi_{j}(r,t)\equiv \frac{\alpha}{\zeta(t)-r-\gamma/3} +\frac{\beta}{\zeta^{*}(t)-r}
$$
%\end{equation}
where $\zeta^{*}$ is given by (\ref{eq:29}), and $\alpha$ and $\beta$ are positive constants. 
There exist $\delta\in (0,\delta_1)$ and $\eta\in (0,\eta_1)$ depending only on 
$a$, $m$, $C_1, \ldots, C_{j-1}$ such that
$$L_j (\phi_j)\geq 0 \quad\text{in } R_{\delta,\eta}^{\gamma}$$
for all $\gamma\in(0,\delta)$.
\end{lem}
\noindent\textbf{Proof}.
Choose $\epsilon$ such that
\begin{equation}\label{eq:36}
0<\epsilon<\frac{a}{2(4+2j(m-1))}.
\end{equation}
There exist $\delta_2\in (0,\delta_1)$ and $\eta\in (0,\eta_1)$ such that
(\ref{eq:25}), (\ref{eq:27}) and (\ref{eq:28}) hold in $R_{\delta_2,\eta}$.
Fix $\gamma\in (0,\delta_2)$. For $(r,t)\in R_{\delta_2,\eta}^{\gamma} $, we have
\begin{eqnarray*}
L_j(\phi_j) &=& \frac{\alpha}{(\zeta-r-\gamma/3)^2}\Big [-\zeta^{'} -\frac{2(m-1)v}{\zeta-r-\gamma/3} 
- (2+j(m-1))v_r-\frac{\bar{m}}{r}v \\ 
&&-F_j(r,t)(\zeta-r-\gamma/3)-\frac{(\zeta-r-\gamma/3)^2}{\alpha}G_j(r,t)\Big ]\\
&&+ \frac{\beta}{(\zeta^*-r)^2}\Big[-\zeta^{*'} -\frac{2(m-1)v}{\zeta^{*}-r} 
-  (2+j(m-1))v_r - \frac{\bar{m}}{r}v\\
&&-F_j(r,t)(\zeta^{*}-r)-\frac{(\zeta^{*}-r)^2}{\beta}G_j(r,t)\Big]. 
\end{eqnarray*} 
>From (\ref{eq:27}) and  by the fact that $\zeta^*-r \geq \zeta-r-\gamma/3$ we have
$$\frac{v}{\zeta^*-r}\leq \frac{v}{\zeta-r-\gamma/3}
\leq (a+\epsilon)\frac{\gamma}{\gamma-\gamma/3}=\frac{3}{2}(a+\epsilon).$$
%Let ax>1 or 1/x<a.
%Then1/(x-\gamma/3)<a\gamma/(\gamma-\gamma/3) if \gamma is very small
Thus it follows from (\ref{eq:25}), (\ref{eq:28}) and (\ref{eq:33}) that
\begin{eqnarray*}
L_j(\phi_j)&\geq & \frac{\alpha}{(r-\zeta-\gamma/3)^2}\Big\{a/2- (3+2j(m-1))\epsilon-\delta_2
\left (F_j +\frac{\delta_2}{\alpha}G_j\right )\Big\}\\
&& + \frac{\beta}{(r-\zeta^*)^2}\Big\{ a/2-(4+2j(m-1))\epsilon-\delta_2
\left (F_j +\frac{\delta_2}{\beta}G_j\right )\Big\}.
\end{eqnarray*}
Since $\epsilon$ satisfies (\ref{eq:36}) we can choose $\delta=\delta(\epsilon, m, C_2, \ldots, C_{j-1})>0$
so small that $L_j (\phi_j)\geq 0$ in $R_{\delta,\eta}^{\gamma}$. \hfill $\Box$
\begin{lem}\label{lem:34}(Barrier Transformation).
Let $\delta$ and $\eta$ be as in Lemma \ref{lem:33} with the additional restriction that
\begin{equation}\label{eq:37}
\eta<\frac{\delta}{6\epsilon},
\end{equation}
where $\epsilon$ satisfies (\ref{eq:36}). Suppose that for some nonnegative constants $\alpha$ and $\beta$
\begin{equation}\label{eq:38}
v^{(j)}(r,t)\leq \frac{\alpha}{\zeta(t)-r}+\frac{\beta}{\zeta^*(t)-r}\quad \text{in } R_{\delta,\eta},
\end{equation}
Then $v^{(j)}$ also satisfies
\begin{equation}\label{eq:39}
v^{(j)}(r,t)\leq \frac{2\alpha/3}{\zeta(t)-r}+\frac{\beta+2\alpha/3}{\zeta^*(t)-r}
\quad \text{in } R_{\delta,\eta}.
\end{equation}
\end{lem}
\noindent\textbf{Proof}. By Lemma \ref{lem:33}, for any $\gamma\in (0,\delta)$ the function
$$\phi_j(r,t)=\frac{2\alpha/3}{\zeta(t)-r-\gamma/3}+\frac{\beta+2\alpha/3}{\zeta^*(t)-r}$$
satisfies $L_j (\phi_j)\geq 0$ in $R_{\delta,\eta}^{\gamma}$. On the other hand, on the parabolic boundary of 
 $R_{\delta,\eta}^{\gamma}$ we have $\phi_j \geq v^{(j)}$. In fact, for $t=t_1$ and
$\zeta_1 -\delta\leq r\leq \zeta_1 -\gamma$, with $\zeta_1=\zeta(t_1)$, we have
$$\phi_j(r,t_1)=\frac{2\alpha/3}{\zeta_1-r-\gamma/3}+\frac{\beta+2\alpha/3}{\zeta_1-r}
>\frac{4\alpha/3}{\zeta_1-r}+\frac{\beta}{\zeta_1-r}>v^{(j)}(r,t_1),$$
while for $r=\zeta-\delta$ and $t_1 \leq t\leq t_2$ we have, in view of (\ref{eq:37}),
\begin{eqnarray*}
\phi_j(\zeta-\delta,t)&\geq &\frac{2\alpha/3}{\delta-\gamma/3}+\frac{\beta}{ \zeta^*+\delta-\zeta}
+\frac{2\alpha/3}{\delta+6\epsilon\eta}\\
&\geq&\frac{2\alpha/3}{\delta}+  \frac{\beta}{ \zeta^*+\delta-\zeta}
+\frac{\alpha/3}{\delta} \geq v^{(j)}( \zeta-\delta,t).
\end{eqnarray*}
Finally, for $r=\zeta-\gamma$, $t_1 \leq t\leq t_2$ we have
$$ \phi_j(\zeta-\gamma,t)=\frac{2\alpha/3}{\gamma-\gamma/3}+\frac{\beta+2\alpha/3}{ \zeta^*+\gamma-\zeta}
\geq \frac{\alpha}{\gamma}+\frac{\beta}{ \zeta^*+\gamma-\zeta}\geq v^{(j)}(\zeta-\gamma,t).$$
By the comparison principle we have
$$\phi_j \geq v^{(j)}\quad\text{in } R_{\delta,\eta}^{\gamma}$$
for any $\gamma\in (0,\delta)$, and (\ref{eq:39}) follows by letting $\gamma\downarrow 0$. \hfill $\Box$

Now we are ready to prove our main proposition.

\textbf{Completion of proof of Proposition \ref{pro:31}.} By Lemma \ref{lem:32}, we have an estimate for 
$v^{(j)}$ of the form (\ref{eq:38}) with $\alpha=K$ and $\beta=0$. Iterating this estimate by the 
Barrier Transformation Lemma we obtain the sequence of estimates
$$v^{(j)}(r,t)\leq \frac{\alpha_n}{\zeta(t)-r} +\frac{\beta_n}{\zeta^*-r} $$
with $\alpha_n = (2/3)^n K$ and $\beta_n=\{(\frac{2}{3}+\ldots +(\frac{2}{3} )^n\}K$. Thus if
we let $n\to  \infty$ we obtain an upper bound for $v^{(j)}$ of the form
\begin{equation}\label{eq:310}
v^{(j)}(r,t)\leq \frac{2K}{\zeta^*-r}\quad \text{in } R_{\delta,\eta}.
\end{equation}
As in the proof of Lemma \ref{lem:21}, this implies that $v^{(j)}$ is bounded above in$R_{\delta,\eta/2}$.

Since the equation (\ref{eq:31}) for $v^{(j)}$ is linear, a similar lower bound can be obtained in the same way 
and the induction step is complete. Therefore as we mentioned in the beginning of this section, 
Theorem \ref{thm:main} is proved. 

\vspace{0.5cm}

{\it Acknowledgement}. The author would like to thank Professor Zhengfang Zhou for his encouragement
and helpful discussions during preparation of this paper.

\vspace{0.5cm}

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\end{document}