\documentclass{amsart} 
\begin{document} 
{\noindent\small {\em Electronic Journal of Differential Equations},
Vol.\ 1999(1999), No.~01, 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 1999 Southwest Texas State University  and 
University of North Texas.} 
\vspace{1.5cm}

\title[\hfilneg EJDE--1999/01\hfil C-infinity interfaces of solutions]
{$C$-infinity interfaces of solutions for one-dimensional parabolic 
$p$-Laplacian equations} 

\author[Yoonmi Ham \& Youngsang Ko\hfil EJDE--1999/01\hfilneg]
{Yoonmi Ham \& Youngsang Ko }

\address{ {\sc Yoonmi Ham}\hfill\break
Department of Mathematics\\
Kyonggi University \hfill\break
Suwon, Kyonggi-do, 442-760, Korea }
\email{ymham@@kuic.kyonggi.ac.kr}

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

\date{}
\thanks{Submitted November 11, 1998. Published January 5, 1999.}
\subjclass{35K65}
\keywords{p-Laplacian, free boundary, C-infinity regularity.}
\maketitle

\begin{abstract}
We study the regularity of a moving interface $x = \zeta (t)$ of the 
solutions for the initial value problem
$$ \displaylines{
u_t = \left(|u_x|^{p-2}u_x \right)_x \cr
u(x,0) =u_0 (x)\,,}
$$
where $u_0\in L^1({\mathbb R})$ and $p>2$. We prove that each side of the 
moving interface is $C^{\infty}$.
\end{abstract}

\newtheorem{thm}{Theorem}[section]
\newtheorem{pro}[thm]{Proposition}
\newcommand{\ol}{\overline}
\makeatletter
\numberwithin{equation}{section}
\makeatother

\section{\bf Introduction}
We consider the Cauchy problem of the form
\begin{equation}\label{eq:plap}  \begin{gathered}
u_t = \left(|u_x|^{p-2}u_x\right)_{x}\text{ in  }
     S := {\mathbb R}\times (0,\infty) \\
u(x,0) =u_0 (x)
\end{gathered} \end{equation}
where $p > 2$. This equation has application to many physical situations, 
and has been studied by many authors; see for example \cite{EV88} and
references therein.  In the study of this equation, the velocity of 
propagation, $V(x,t)$, is very important, and can be obtained in terms of
$u$ by writing (\ref{eq:plap}) as the conservation law
$$u_t + (uV)_x = 0\,.$$ 
In this way we obtain $V= -v_x |v_x |^{p-2}$,
where the nonlinear potential $v(x,t)$ is 
\begin{equation}\label{eq:defv}
v = \frac{p-1}{p-2}u^{(p-2)/(p-1)}.
\end{equation}
By a direct computation, we realize that 
\begin{equation}\label{eq:eqv}
v_t = (p-2)v|v_x|^{p-2}v_{xx} + |v_x|^p.
\end{equation} 
In \cite{EV88}, it is
shown that  $V$ satisfies $V_x \leq \frac{1}{2(p-1)t}$
which can also be written as 
\begin{equation}\label{eq:V}
(v_x |v_x|^{p-2})_x \geq -\frac{1}{2(p-1)t}\,.
\end{equation}
Without loss of generality, we assume that $u_0$ vanishes 
on ${\mathbb R}^{-}$ and that  $u_0$ is a continuous positive 
function on an interval $(0,a)$ with $a>0$. Let 
$$P[u] =\{(x,t) \in S:u(x,t) >0\}$$ 
be the positivity set of a solution $u$. Then $P[u]$ is bounded from the left 
in the $(x,t)$-plane by the left interface curve $x=\zeta(t)$, 
where
$$\zeta(t) = \inf\{x\in {\mathbb R} :u(x,t) >0\}\,.$$
Moreover, there is a time $t^{*}\in [0,\infty)$, called the waiting time,
such that $\zeta(t) =0$ for $0 \leq t \leq t^{*}$ and $\zeta(t) <0$ 
for $t>t^{*}$. It is shown in
\cite{EV88} that $t^*$ is finite (possibly zero) and $\zeta(t)$ is a 
non-increasing $C^{1}$ function on $(t^{*},\infty)$. Actually it is shown 
that $\zeta'(t) <0$ for 
every $t>t^*$, i.e., a moving interface never stops. 

On the other hand, D. G. Aronson and J. L. Vazquez \cite{AV87}
established Theorem~\ref{thm:1} below.

Let $D= \{(x,t): t>t^{*}, \zeta(t) \leq x \leq 0\}$, and let $v$ be the 
pressure for the solution of the porous medium equation
\begin{equation}\label{eq:porous}
u_t = (u^m)_{xx} \quad in\quad Q_T = {\mathbb R}\times (0,T).
\end{equation}

\begin{thm}\label{thm:1}
$v$ is a $C^{\infty}$ function on $D$, and $\zeta(t)$ is a $C^{\infty}$ function
on $(t^*, \infty)$.
\end{thm}

This theorem is proven by  finding bounds for $v^{(k)}$ with $k\geq 2$. \medskip

The purpose of this paper is to discuss the $C^{\infty}$ regularity of the  
moving part of the interface of the solution to (\ref{eq:plap}). 
To accomplish this end, we use some ideas from \cite{AV87}.

\section{\bf Upper and Lower Bounds for $v_{xx}$}

Let $q=(x_0, t_0)$ be a point on the left interface, so that
$x_0 = \zeta(t_0)$, $v(x,t_0) = 0$ for all $x\leq \zeta(t_0)$, and 
$v(x,t_0) >0$ for all sufficiently small $x>\zeta(t_0)$. We assume the 
left interface 
is moving at $q$. Thus $t_0 >t^{*}$. We shall use the notation
$$R_{\delta, \eta} =R_{\delta, \eta}(t_0) 
= \{(x,t) \in {\mathbb R}^2 :\zeta(t) <x\leq \zeta(t) + \delta, t_0 - \eta \leq t
\leq t_0 + \eta\}.$$ 

\begin{pro}\label{pro:1}
Let $q$ be the point as above. Then there exist positive constants $C$,
$\delta$, and
$\eta$ depending only on $p$, $q$, and $u$ such that
$$v_{xx} \geq C \quad\text{in } R_{\delta,\eta/2}.$$
\end{pro}

\noindent\textbf{Proof.} From (\ref{eq:V}) we have, $v_{xx} \geq 
-\displaystyle\frac{1}{2(p-1)^{2}|v_{x}|^{p-2}t}$.
However, from Lemma 4.4 in \cite{EV88}, $v_{x}$ is bounded away and above 
from zero near $q$, where $u(x,t) >0$. \hfill $\Box$

\begin{pro}\label{pro:2}
Let $q=(x_0,t_0)$ be as above. Then there exist positive 
constants $C_{2}$, $\delta$, and
$\eta$ depending only on $p$, $q$, and $u$ such that
$$v_{xx} \leq C_{2} \quad\text{in } R_{\delta,\eta/2}\,.$$
\end{pro}

\noindent\textbf{Proof.} From Theorem 2 and Lemma 4.4 in \cite{EV88} we have 
\begin{equation}\label{eq:ev1}
\zeta'(t_0)=-v_x|v_x|^{p-2}=-v_x^{p-1}=-a 
\end{equation}
and
\begin{equation}\label{eq:ev2}
v_t = |v_x |^p 
\end{equation} 
on the moving part of the interface $\{x=\zeta(t), t>t^*\}$. 
Choose $\epsilon>0$ such that
\begin{equation}\label{eq:eeps}
(p-1)a-5p\epsilon\geq 4[(p-2)^2 + (p-1)^2](a+\epsilon)\epsilon.
\end{equation}
Then by Theorem 2 in \cite{EV88}, there exists a $\delta=\delta(\epsilon)>0$ 
and $\eta=\eta(\epsilon)\in(0,t_0 -t^*)$
such that $R_{\delta,\eta}\subset P[u]$,  
\begin{equation}\label{eq:1}
(a-\epsilon)^{\frac{1}{p-1}}<v_x < (a+\epsilon)^{\frac{1}{p-1}}
\end{equation}
and
\begin{equation}\label{eq:ev3}
vv_{xx} \leq (a-\epsilon)^{\frac{2}{p-1}}\epsilon
\end{equation}
in $R_{\delta,\eta}$. Then from (\ref{eq:1}) we have
\begin{equation}\label{eq:ev4}
(a-\epsilon)^{\frac{1}{p-1}}(x-\zeta)<v(x,t) 
< (a+\epsilon)^{\frac{1}{p-1}}(x-\zeta)
\end{equation}
in $R_{\delta,\eta}$ and 
\begin{equation}\label{eq:ev5}
-(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:z*}
\zeta^{*}(t) = \zeta(t_1) - b(t-t_1)
\end{equation}
where $b=a + 2\epsilon$. Then clearly $\zeta(t) > \zeta^{*}(t)$ in 
$(t_1,t_2]$.
On $P[u]$, $w\equiv v_{xx}$ satisfies
\begin{eqnarray*}
L(w)&=&w_t-(p-2)v|v_x|^{p-2}w_{xx}-(3p-4)|v_x|^{p-2}v_x w_x \\
&&- [(p-2)^2+2(p-1)^2]|v_x|^{p-2}w^2\\
    && - 3(p-2)^2 v|v_x|^{p-4}v_x ww_x - (p-2)^2(p-3)v|v_x|^{p-4}w^3\\
&=&0\,.
\end{eqnarray*} 
We shall construct a barrier for $w$ in $R_{\delta,\eta}$ of the form
$$
\phi(x,t) \equiv \frac{\alpha}{x-\zeta (t)}+\frac{\beta}{x-\zeta^{*}(t)},
$$
where $\alpha$ and $\beta$ will be decided later.

By a direct computation we have
\begin{eqnarray*}
L(\phi)
&=&\frac{\alpha}{(x-\zeta)^2}\{\zeta' - (p-2)v|v_x|^{p-2}\frac{2}{x-\zeta}
+ (3p-4)|v_x|^{p-2}v_x \}\\
&& + \frac{\beta}{(x-\zeta^*)^2}\{\zeta^{*'} 
- (p-2)v|v_x|^{p-2}\frac{2}{x-\zeta^*}
+ (3p-4)|v_x|^{p-2}v_x \}\\ 
&& - [(p-2)^2 + 2(p-1)^2]|v_x|^{p-2}\phi^2 + \bar{G}
\end{eqnarray*} 
where 
\begin{eqnarray*}
\lefteqn{\bar{G} }\\
 &=& -3(p-2)^2 vv_x |v_x|^{p-4}\phi \phi_{x} 
- (p-2)^2 (p-3)v|v_x|^{p-4}\phi^{3}\\
  &=& (p-2)^2 v|v_x|^{p-4}\phi\left(3v_x[\frac{\alpha}{(x-\zeta)^2}
+\frac{\beta}{(x-\zeta^*)^2}]
-(p-3)[\frac{\alpha}{x-\zeta} + \frac{\beta}{x-\zeta*}]^2\right).
\end{eqnarray*}
If we choose $\alpha$ and $\beta$ satisfying
$$v_x \geq|p-3|\max(\alpha,\beta),$$
then $\bar{G} \geq 0$ in $R_{\delta, \eta}$. Now set  
$\bar{A}= \frac{\alpha}{(x-\zeta)^2}$ and $\bar{B}=
\frac{\beta}{(x-\zeta^*)^2}$. Then  we have
\begin{eqnarray*}
\lefteqn{ L(\phi)  } \\
&\geq& \bar{A}\left\{\zeta' +
|v_{x}|^{p-2}\{-(p-2)v\frac{2}{x-\zeta} + (3p-4)v_x -2[(p-2)^2 +
2(p-1)^2 ]\alpha\}\right\}\\ &&+ \bar{B}\left\{\zeta^{*'} +
|v_{x}|^{p-2}\{-(p-2)v\frac{2}{x-\zeta^*} + (3p-4)v_x -2[(p-2)^2 +
2(p-1)^2 ]\beta\}\right\}\\ &\geq& \bar{A}\left\{(p-1)a -
(5p-7)\epsilon -2[(p-2)^2
+2(p-1)^2](a+\epsilon)^{\frac{p-2}{p-1}}\alpha\right\}\\ &&+ \bar{B}
\left\{(p-1)a - (5p-6)\epsilon  -2[(p-2)^2
+2(p-1)^2](a+\epsilon)^{\frac{p-2}{p-1}}\beta\right\}.
\end{eqnarray*}
Set
$$0< \alpha\leq 
\frac{(p-1)a -(5p-7)\epsilon}
{2[(p-2)^2 + 2(p-1)^2](a+\epsilon)^{\frac{p-2}{p-1}}}=\alpha_0$$
and
\begin{equation}\label{eq:beta}
\beta =\frac{(p-1)a-(5p-6\epsilon)}
{2[(p-2)^2 + 2(p-1)^2](a+\epsilon)^{\frac{p-2}{p-1}}}\,.
\end{equation}
Then from (\ref{eq:eeps}), $\beta >0$ and $L(\phi)\geq 0$ in
$R_{\delta,\eta}$ for all $\alpha\in (0, \alpha_0]$ and $\beta$.

Let us now compare $w$ and $\phi$ on the parabolic boundary of 
$R_{\delta,\eta}$.
In view of (\ref{eq:ev3}) and (\ref{eq:ev4}) we have
$$v_{xx}\leq\frac{\epsilon(a-\epsilon)^{\frac{1}{p-1}}}{x-\zeta}
\quad\text{in } R_{\delta,\eta}$$
and in particular
$$v_{xx}(\zeta(t) + \delta, t) 
\leq \frac{\epsilon(a-\epsilon)^{\frac{1}{p-1}}}{\delta} 
\text{ in  } [t_1, t_2 ]\,.$$
By the Mean Value Theorem and (\ref{eq:ev5}), we have 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
$$\eta = \min\{\eta(\epsilon), \delta(\epsilon)/6\epsilon \}.$$
Since $\epsilon$ satisfies (\ref{eq:eeps}) and $\beta$ is given 
by (\ref{eq:beta}) it follows that
$$\phi(\zeta + \delta, t) 
\geq \frac{\beta}{2\delta} 
\geq \frac{(p-1)a-(5p-6\epsilon)}
{4[(p-2)^2 +2(p-1)^2](a+\epsilon)^{\frac{p-2}{p-1}}\delta}
\geq\frac{(a+\epsilon)^{\frac{1}{p-1}}}{\delta}\epsilon\geq v_{xx}\,,
$$  
on $[t_1,t_2]$. 
Moreover from (\ref{eq:eps}) and (\ref{eq:beta})
\[\phi(x,t_1) \geq \frac{\beta}{x-\zeta(t_1)}
>\frac{\epsilon(a-\epsilon)^{\frac{1}{p-1}}}
{x-\zeta(t_{1})}>v_{xx}(x,t_{1})\text{ on  } (\zeta(t_1), \zeta(t_1) 
+\delta]\,.\]
Let $\Gamma = \{(x,t) \in {\mathbb R}^2 : x=\zeta(t), t_1 \leq t\leq t_2\}$. 
Clearly $\Gamma$ is a 
compact subset of ${\mathbb R}^2$. Fix $\alpha \in (0,\alpha_0)$. For each point 
$s\in \Gamma$ there 
is an open ball $B_s $ centered at $s$ such that
$$(vv_{xx})(x,t) \leq \alpha(a-\epsilon)^{\frac{1}{p-1}}\quad\text{in }
 B_s \cap P[u]\,.$$ 
In view of (\ref{eq:ev4}) we have
$$\phi(x,t) \geq \frac{\alpha}{x-\zeta}
\geq v_{xx}(x,t)\quad\text{in } B_s \cap P[u]\,.$$
Since $\Gamma$ can be covered by a finite number of these balls it follows 
that there is a 
$\gamma=\gamma(\alpha) \in (0,\delta)$ such that
$$\phi(x,t) \geq w(x,t)\quad\text{in } R_{\delta,\eta}.$$
Thus for every $\alpha \in (0, \alpha_0)$, $\phi$ is a barrier for
$w$  in $R_{\delta,\eta}$.
By the comparison principle for parabolic equations \cite{LSU} we conclude 
that
$$v_{xx}(x,t) 
\leq \frac{\alpha}{x-\zeta(t)} + \frac{\beta}{x-\zeta^{*}(t) } 
\quad\text{in } R_{\delta,\eta}\,,$$ 
where $\beta$ is given by (\ref{eq:beta}) and $\alpha \in(0,\alpha_0)$ is 
arbitrary. Now as $\alpha$ approaches zero, we obtain
$$
v_{xx}(x,t) \leq \frac{\beta}{x-\zeta^*} 
\leq \frac{2\beta}{\epsilon\eta} \quad\text{in } {\mathbb R}.
$$


\section{\bf Bounds for $\left(\frac{\partial}{\partial{x}}\right)^{3}v$}

In this section we find the
estimates of the  derivatives of the form
$$
v^{(3)} \equiv \left(\frac{\partial}{\partial x}\right)^{3}v.
$$
By a direct computation we have,
\begin{eqnarray}\label{eq:**}
L_{3}(v^{(3)})&=&v^{(3)}_t-(p-2)vv_{x}^{p-2}v^{(3)}_{xx} - (A+B)v_{x}^{(3)} 
-Cv^{(3)} -D(v^{(3)})^2\\
&&-Ev_{x}^{p-3}v_{xx}^{3} 
- (p-2)^{2}(p-3)(p-4)vv_{x}^{p-5}v_{xx}^{4}=0\,,\nonumber 
\end{eqnarray}                             
where 
\begin{eqnarray*}
A &=& (p-2)v_{x}^{p-1} + (p-2)^{2}vv_{x}^{p-3}v_{xx}\,,\\ 
B &=& (3p-4)v_{x}^{p-1} + 3(p-2)^{2}vv_{x}^{p-3}v_{xx}\,,\\
C &=& v_{xx}v_{x}^{p-2}\{(3p-4)(p-1) + 2[(p-2)^{2} \\
  && + 2(p-1)^{2}] + 6(p-2)^{2}(p-3)vv_{x}^{-2}v_{xx} + 3(p-2)^2 \}\,,\\
D &=& 3(p-2)^2 vv_{x}^{p-3}\,, \\
E &=&[(p-2)^2 +2(p-1)^2 ](p-2) + (p-2)^{2}(p-3)\,.
\end{eqnarray*} 

Suppose that $q=(x_0,t_0)$ is a point on the left 
interface for which (\ref{eq:ev1}) holds. Fix $\epsilon \in (0,a)$ and 
take $\delta_0 = \delta_0(\epsilon) >0$
and $\eta_0=\eta(\epsilon)\in (0,t_0-t^{*})$ such that 
$R_0\equiv R_{\delta_0,\eta_0}(t_0)\subset P[u]$ and
(\ref{eq:ev3})  holds. Thus we also have (\ref{eq:ev4}) and
(\ref{eq:ev5}) in $R_0$.  Then by rescaling and interior estimate
we have 
\begin{pro}\label{pro:3}
There are constants $K\in {\mathbb R}^{+}$, $\delta\in (0,\delta_0)$, 
and $\eta \in (0,\eta_0)$ depending only on $p$, $q$, and $C_{2}$ such
that
$$|v^{(3)}(x,t)|\leq \frac{K}{x-\zeta(t)}\quad\text{in } R_{\delta,\eta}\,.$$
\end{pro}

\noindent\textbf{Proof.} Set 
$$\delta = \min\{\frac{2\delta_0}{3},2s\eta_0\},\quad 
\eta = \eta_0-\frac{\delta}{4s}\,,$$
and define 
\[ R(\ol{x},\ol{t})\equiv \left\{(x,t) \in {\mathbb R}^2 : |x-\ol{x}| 
< \frac{\lambda}{2}, 
\ol{t} -\frac{\lambda}{4s} <t\leq \ol{t}\right\}\]
for $(\ol{x},\ol{t})\in R_{\delta,\eta}$, where $s=a+\epsilon$ and 
$\lambda = \ol{x}-\zeta(\ol{t})$. Then $(\ol{x},\ol{t})\in R_{\delta,\eta}$ 
implies that 
$R(\ol{x},\ol{t})\subset R_0$. 
Since $\delta_0 \geq \frac{3\delta}{2}$, $\lambda < \delta$ and $\zeta$ 
is non-increasing, we have
\begin{eqnarray*}
&t_0-\eta_0 = t_0-\eta - \frac{\lambda}{4s} < t < t_0 
+ \eta < t_0 + \eta_0\,,&\\
&\ol{x}-\frac{\lambda}{2} = \ol{x}-\frac{\ol{x} + \zeta(\ol{t})}{2} 
= \frac{ \ol{x} + \zeta(\ol{t})}{2} > \zeta(t_0+ \eta_0)\,,&\\
&\zeta(t_0-\eta) + \delta + \frac{\lambda}{2} <\zeta(t_0
-\eta_0)\,.&
\end{eqnarray*}
Also observe that for each 
$(\ol{x},\ol{t})\in R_{\delta,\eta}$, $R(\ol{x},\ol{t})$ lies to the right 
of the line 
$x=\zeta(\ol{t}) + s(\ol{t} - t)$.
Next
set $x=\lambda\xi + \ol{x}$ and  $t=\lambda\tau + \ol{t}$. The function
$$W(\xi,\tau)\equiv v_{xx}(\lambda\xi + \ol{x},\lambda\tau + \ol{t})
=v_{xx}(x,t)$$
satisfies the equation
\begin{eqnarray}\label{eq:*}
W_{\tau}&=& \left\{(p-2)\frac{v}{\lambda}v_{x}^{p-2}W_{\xi} 
+ (3p-4)v_{x}^{p-1}W\right\}_{\xi}
\nonumber\\
&&+ [2(p-2)^{2}vv_{x}^{p-3}v_{xx}-(p-2)v_{x}^{p-1}]W_{\xi}\\
&&+\lambda [(p-2)^{2}(p-3)vv_{x}^{p-4}(v_{xx})^3 
- (p-2)v_{x}^{p-2}(v_{xx})^2 ]\nonumber
\end{eqnarray} 
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 $|W|\leq C_{2}$ in $B$. In view of (\ref{eq:ev4}) and (\ref{eq:ev5})
$$(a-\epsilon)^{\frac{1}{p-1}}\frac{x-\zeta(t)}{\lambda}
\leq\frac{v(x,t)}{\lambda}
\leq (a+\epsilon)^{\frac{1}{p-1}}\frac{x-\zeta(t)}{\lambda}$$
and
$$\zeta(\ol{t})\leq\zeta(t)\leq\zeta(\ol{t})+s(\ol{t}-t)
\leq\zeta(\ol{t})+\frac{\lambda}{4}\,.$$
Therefore,
$$\frac{\lambda}{4} = \ol{x}-\frac{\lambda}{2}-\zeta(\ol{t})
-\frac{\lambda}{4}
\leq x - \zeta(t) \leq \ol{x} + \frac{\lambda}{2} - \zeta(\ol{t}) 
= \frac{3\lambda}{2}$$
which implies
$$\frac{(a-\epsilon)^{\frac{1}{p-1}}}{4}\leq \frac{v}{\lambda}
\leq \frac{3(a+\epsilon)^{\frac{1}{p-1}}}{2}\,.$$
Hence by (\ref{eq:1}) equation (\ref{eq:*}) is uniformly parabolic in $B$. 
Moreover, it follows
from Proposition \ref{pro:2} that $W$ satisfies all of the hypotheses of 
Theorem 5.3.1 of \cite{LSU}. Thus we conclude that there exists a constant 
$K=K(a,p,C_{2})>0$ such that
$$\left|\frac{\partial}{\partial\xi}W(0,0)\right| \leq K ;$$
that is,
$$|v^{(3)}(\ol{x},\ol{t})|\leq \frac{K}{\lambda}\,.$$
Since $(\ol{x},\ol{t})\in R_{\delta,\eta}$ is arbitrary, this proves the 
proposition.\hfill$\Box$\medskip

We now turn to the barrier construction. If $\gamma \in (0,\delta)$ we will 
use the notation 
$$R_{\delta,\eta}^{\gamma}=R_{\delta,\eta}^{\gamma}(t_0)\equiv
\{(x,t)\in {\mathbb R}^2 :\zeta(t)+\gamma\leq x\leq\zeta(t)+\delta, t_0-\eta
\leq t\leq t_0 + \eta\}.$$

\begin{pro}\label{pro:4}
Let $R_{\delta_{1},\eta_{1}}$ be the region constructed in the proof of 
Proposition \ref{pro:2} with
\begin{equation}\label{eq:delta1} 
0<\delta_1 < \frac{(p-1)a^{\frac{1}{p-1}}}{12(p-2)^{2}K}\,.
\end{equation} 
For $(x,t)\in R_{\delta_{1},\eta_{1}}^{\gamma}$,
let
\begin{equation}\label{eq:ba2}
\phi_{\gamma}(x,t) \equiv \frac{\alpha}{x-\zeta(t) -\gamma/3} 
+ \frac{\beta}{x-\zeta^{*}(t)}\,,
\end{equation}
where $\zeta^{*}$ is given by (\ref{eq:z*}), and $\alpha$ and
$\beta$ are positive constant less than $K/2$. Then there exist $\delta\in(0,\delta_1 )$ and
$\eta\in(0,\eta_1 )$ depending  only on $a$, $p$ and $C_{2}$ such that
$$L_{3}(\phi_{\gamma})\geq 0 \quad\text{in } R_{\delta,\eta}^{\gamma}$$
for all $\gamma\in (0,\delta)$.
\end{pro}                      

\noindent\textbf{Proof.} Choose $\epsilon$ such that
\begin{equation}\label{eq:eps}
0<\epsilon < \frac{(p-1)a}{13p-23}\,.
\end{equation}
There exist $\delta_{2}\in (0,\delta_{1})$ and $\eta\in (0,\eta_{1})$ such 
that (\ref{eq:1}),
(\ref{eq:ev4}) and (\ref{eq:ev5}) hold in $R_{\delta_{2},\eta}$. 
Fix $\gamma\in(0,\delta_{2})$.
For $(x,t) \in R_{\delta_{2},\eta}^{\gamma}$, we have
\begin{eqnarray*}
L_{3}(\phi_{3})&=&\frac{\alpha}{(x-\zeta-\gamma/3)^2}\left\{\zeta^{'}
-\frac{2(p-2)vv_{x}^{p-2}}{x-\zeta-\gamma/3} + A+B\right\}\\ 
&& +\frac{\alpha}{(x-\zeta^{*})^2}\left\{\zeta^{*'}
-\frac{2(p-2)vv_{x}^{p-2}}{x-\zeta^{*}} + A+B\right\}\\  
&&-C\phi_{3} -D(\phi_{3})^2 -Ev_{x}^{p-3}v_{xx}^{3} 
- (p-2)^{2}(p-3)(p-4)vv_{x}^{p-5}v_{xx}^4\,
\end{eqnarray*}
where $A$, $B$, $C$, $D$, and $E$ are as above.

>From (\ref{eq:ev4}), together with the fact that $x-\zeta^{*}\geq x-\zeta-\gamma/3$ 
we have
$$\frac{v}{x-\zeta^{*}}\leq \frac{v}{x-\zeta-\gamma/3}
\leq (a+\epsilon)^{\frac{1}{p-1}}\frac{x-\zeta}{x-\zeta-\gamma/3} 
\leq (a+\epsilon)^{\frac{1}{p-1}}\frac{\gamma}{\gamma-\gamma/3} 
= \frac{3}{2}(a+\epsilon)^{\frac{1}{p-1}}\,.$$
>From (\ref{eq:delta1}), we have 
\begin{equation}\label{eq:ab}
D\alpha, D\beta <
1/2DK<DK\leq\frac{(p-1)a}{4} +\frac{(p-1)\epsilon}{4} \,.
\end{equation}
Then since
$|C|$ is bounded and from (\ref{eq:1}) and (\ref{eq:ev4}), we have  
\begin{eqnarray*}\label{eq:****}
\lefteqn{ L_{3}(\phi_{3})  }\\
&\geq&
\frac{\alpha}{Y^{2}}
\left\{(p-1)a-(7p-11)\epsilon-|C|Y-2D\alpha-\ol{E}\frac{Y^{2}}{\alpha}
\right\} \\  &&+
\frac{\beta}{(x-\zeta^{*})^2}\left\{(p-1)a-(7p-10)\epsilon-|C|(x-\zeta
^{*})-2D\beta-\ol{E}\frac{(x-\zeta^{*})^2}{\beta} \right\} \\
&\geq&
\frac{\alpha}{Y^2}\left\{\frac{(p-1)a}{2}-\frac{13p-23}{2}\epsilon-
\delta_{2}(|C|-\ol{E}\frac{Y}{\alpha} )\right\}\\ 
&&+
\frac{\beta}{(x-\zeta^{*})^2 } \left\{\frac{(p-1)a}{2}- 
\frac{13p-21}{2}\epsilon-\delta_{2}(|C|-\ol{E}\frac{x-\zeta^*}{\beta})
\right\}
\end{eqnarray*} 
where $Y= x-\zeta-\gamma/3$ and 
$\ol{E}=|E|v_{x}^{p-3}v_{xx}^{3}$.
Since $\epsilon$ satisfies (\ref{eq:eps}) we can choose 
$\delta=\delta_{2}(\epsilon, p, a, C_{2})>0$ so small that 
$L_{3}(\phi_{3}) \geq 0$
in $R_{\delta,\eta}^{\gamma}$. \hfill $\Box$ \medskip

\noindent{\bf Remark 3.1.} 
>From (\ref{eq:ab}) the Proposition \ref{pro:4} will be true for any
$\alpha,\beta \in (0, K)$.


\begin{pro}  \label{pro:5}
(Barrier Transformation). Let $\delta$ and $\eta$ be as in Proposition 
\ref{pro:4} with the additional restriction that
\begin{equation}\label{eq:sepis}
\eta <\frac{\delta}{6\epsilon},
\end{equation}
where $\epsilon$ is as in Proposition \ref{pro:4}. Suppose that for
some nonnegative constant $\beta$
\begin{equation}\label{eq:semi}
v^{(3)}(x,t) \leq \frac{\alpha}{x-\zeta(t)} 
+ \frac{\beta}{x-\zeta^{*}(t)} \quad \text{in  } R_{\delta,\eta}.
\end{equation}
Then $v^{(3)}$ also satisfies
\begin{equation}\label{eq:final}
v^{(3)}(x,t) \leq \frac{2\alpha/3}{x-\zeta(t)} 
+ \frac{\beta + 2\alpha/3}{x-\zeta^{*}(t)}  
\quad\text{in } R_{\delta,\eta}.
\end{equation}
\end{pro}

\noindent\textbf{Proof.} By Remark 3.1, for any $\gamma \in (0,\delta)$
since  $\beta +2\alpha/3 \leq K$
the function
$$
\phi_{3}(x,t) 
= \frac{2\alpha/3}{x-\zeta -\gamma/3} + \frac{\beta +
2\alpha/3}{x-\zeta^{*}}
$$
satisfies $L_{3}(\phi_{3}) \geq 0$ in $R_{\delta,\eta}^{\gamma}$. On the 
other hand, on the parabolic boundary of $R_{\delta,\eta}^{\gamma}$ we have 
$\phi_{3} \geq v^{(3)}$. In fact, 
for $t = t_{1}$ and $\zeta_{1} + \gamma \leq x \leq \zeta_1 + \delta$,
with $\zeta_{1}=\zeta(t_1)$, we  have
$$
\phi_{3}(x,t_1)
=\frac{2\alpha}{x-\zeta_1 -\gamma/3} + \frac{\beta +
2\alpha/3}{x-\zeta_1} > \frac{4\alpha/3}{x-\zeta_1}
+\frac{\beta}{x-\zeta_1} > v^{(3)}(x,t_1)$$ while for $x=\zeta
+\delta$ and $t_{1}\leq t \leq t_2 $ we get,  in view of
(\ref{eq:sepis}),
\begin{eqnarray*}
\phi_{3}(\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{\delta}{\zeta + \delta - \zeta^{*}} 
       + \frac{\alpha/3}{\delta}\geq v^{(3)}(\zeta +\delta,t).
\end{eqnarray*}
Finally, for $x=\zeta + \gamma$, $t_1 \leq t \leq t_2 $ we have
$$\phi_{3}(\zeta + \delta, t)=\frac{2\alpha/3}{\gamma-\gamma/3} 
+ \frac{\beta + 2\alpha/3}{\zeta + \gamma - \zeta^{*}} 
\geq \frac{\alpha}{\gamma}+\frac{\beta }{\zeta+\gamma-\zeta^{*}}
\geq v^{(3)}(\zeta+\gamma,t).$$
By the comparison principle we get
$$\phi_{3} \geq v^{(3)} \quad\text{in } R_{\delta.\eta}^{\gamma}$$
for any $\gamma \in (0,\delta)$, and (\ref{eq:final}) follows by letting 
$\gamma \downarrow 0$.
\hfill $\Box$

\begin{pro}\label{pro:6}
Let $q=(x_0,t_0)$ be a point on the interface for which (\ref{eq:ev1}) holds. 
Then there exist constants $C_{3}$, $\delta$ and $\eta$ depending only on 
$p$, $q$ and $u$ such that
$$\left|\left(\frac{\partial}{\partial x}\right)^{3}v\right|\leq C_{3}  
\quad\text{in } R_{\delta,\eta/2}.$$
\end{pro}
 
\noindent\textbf{Proof.} By Proposition \ref{pro:3} we have, by letting
$\alpha=0$,
$$ v^{(3)}(x,t) \leq \frac{\beta}{x-\zeta^{*}}\leq
\frac{2\beta}{\epsilon\eta} \quad\text{in } R_{\delta,\eta/2}. $$
Even though the equation (\ref{eq:**}) is not linear for $v^{(3)}$, a
lower bound can be obtained in a similar way.
\hfill $\Box$ 

\section{\bf Main Result}

In this section we prove the interface is a $C^{\infty}$ function in 
$(t^{*},\infty)$. We follow the methods in \cite{AV87}. First we find the
estimates of the  derivatives of the form
$$
v^{(j)} \equiv \left(\frac{\partial}{\partial x}\right)^{j}v
$$
for $j\geq 4$.
For the porous medium equation, we have \cite{AV87} the following equation:
\begin{eqnarray*}
L_{j}v^{(j)} &\equiv& v_{t}^{(j)} - (m-1)vv_{xx}^{(j)} 
- (2+j(m-1))v_{x}v_{x}^{(j)} -c_{mj}v_{xx}v^{(j)}\\
             && - \sum_{l=3}^{j^{*}}d_{mj}^{l}v^{(l)}v^{(j+2-l)}=0
\end{eqnarray*}
for $j\geq 3$ in $P[u]$, where $j^{*} = [j/2] + 1$, and the $c_{mj}$ and 
$d_{mj}^{l}$ are 
constants which depend only on their indices, but whose precise values are 
irrelevant. Note that
$L_{j}$ is linear in $v^{(j)}$. On the other hand for the p-Laplacian 
equation by a direct computation we have the 
following equation for $j\geq 4$,
\begin{eqnarray}\label{eq:j4}
L_{j}v^{(j)} &=& v_{t}^{(j)}- (p-2)vv_{x}^{p-2}v_{xx}^{(j)} 
- ((j-2)A+B)v_{x}^{(j)} 
- C_{pj}v^{(j)}\\
&& - F(v, v_{x}, \ldots, v^{(j-1)})=0\nonumber 
\end{eqnarray}
where $A$ and $B$ are as before, and $C_{pj}$ involves only
$v$ and derivatives of order $< j$. Note that equation (\ref{eq:j4})
is linear in $v^{(j)}$. We also follow the method in \cite{AV87}.
Hence our result is

\begin{pro}\label{pro:7}
Let $q=(x_0,t_0)$ be a point on the interface for which (\ref{eq:ev1}) holds. 
For each integer $j \geq 2$ there exist constants $C_{j}$, $\delta$ and
$\eta$ depending only on $p$, $j$, $q$ and $u$ such that
$$\left|\left(\frac{\partial}{\partial x}\right)^{j}v\right|\leq C_{j} \quad 
\text{in } R_{\delta,\eta/2}.$$
\end{pro}

The proof is done by induction on $j$. 
Suppose that $q=(x_0,t_0)$ is a point on the left 
interface for which (\ref{eq:ev1}) holds. Fix $\epsilon \in (0,a)$ and 
take $\delta_0 = \delta_0(\epsilon) >0$
and $\eta_0=\eta(\epsilon)\in (0,t_0-t^{*})$ such that 
$R_0\equiv R_{\delta_0,\eta_0}(t_0)\subset P[u]$ and
(\ref{eq:ev3})  holds. Thus we also have (\ref{eq:ev4}) and
(\ref{eq:ev5}) in $R_0$. Assume that there are constants $C_k \in {\mathbb R}^{+}$ for
$k =3, \ldots, j-1$ such that
\begin{equation}\label{eq:ck}
|v^{(k)}| \leq C_k \quad\text{ on} R_0\quad for
\quad k=3,\ldots, j-1.
\end{equation}
Observe that (\ref{eq:ck}) hold for $k=3$ by Proposition \ref{pro:6}. 

By rescaling and interior estimates, we have

\begin{pro}\label{pro:8}
There are constants $K\in {\mathbb R}^{+}$, $\delta\in (0,\delta_0)$, 
and $\eta \in (0,\eta_0)$ depending only on $p$,$q$ and $C_{k}$ for
$k\in[2, j-1]$ with $j\geq 4$ such that
$$|v^{(j)}(x,t)|\leq \frac{K}{x-\zeta(t)}\quad\text{in } R_{\delta,\eta}.$$
\end{pro}

\noindent\textbf{Proof.} Set 
$$\delta = \min\{\frac{2\delta_0}{3},2s\eta_0\}\,, \quad
\eta = \eta_0-\frac{\delta}{4s}\,,$$
and define 
\[ R(\ol{x},\ol{t})\equiv \left\{(x,t) \in {\mathbb R}^2 : |x-\ol{x}| 
< \frac{\lambda}{2}, 
\ol{t} -\frac{\lambda}{4s} <t\leq \ol{t}\right\}\]
for $(\ol{x},\ol{t})\in R_{\delta,\eta}$, where $s=a+\epsilon$ and 
$\lambda = \ol{x}-\zeta(\ol{t})$. Then $(\ol{x},\ol{t})\in R_{\delta,\eta}$ 
implies that 
$R(\ol{x},\ol{t})\subset R_0$. 
Since $\delta_0 \geq \frac{3\delta}{2}$, $\lambda < \delta$ and $\zeta$ 
is non-increasing, we have
\begin{eqnarray*} 
&t_0-\eta_0 = t_0-\eta - \frac{\lambda}{4s} < t < t_0 
+ \eta < t_0 + \eta_0 \,,&\\
& \ol{x}-\frac{\lambda}{2} = \ol{x}-\frac{\ol{x} + \zeta(\ol{t})}{2} 
= \frac{ \ol{x} + \zeta(\ol{t})}{2} > \zeta(t_0+ \eta_0)\,,&\\
&\zeta(t_0-\eta) + \delta + \frac{\lambda}{2} <\zeta(t_0
-\eta_0)\,. &
\end{eqnarray*}
Also observe that for each 
$(\ol{x},\ol{t})\in R_{\delta,\eta}$, $R(\ol{x},\ol{t})$ lies to the right 
of the line 
$x=\zeta(\ol{t}) + s(\ol{t} - t)$.
Next
set $x=\lambda\xi + \ol{x}$ and  $t=\lambda\tau + \ol{t}$. The function
$$V^{(j-1)}(\xi,\tau)\equiv v^{(j-1)}(\lambda\xi + \ol{x},\lambda\tau + \ol{t})
=v^{(j-1)}(x,t)$$
satisfies the equation
\begin{eqnarray}\label{eq:11}
V^{(j-1)}_{\tau}&=& \left\{(p-2)\frac{v}{\lambda}v_{x}^{p-2}V^{(j-1)}_{\xi} 
+[(j-2)A+B]v_{x}^{p-1}V^{(j-1)}\right\}_{\xi}
\nonumber\\
&&-[(p-2)v^{p-1}_{x}+(p-2)^2 vv^{p-3}_x
    v_{xx}+(j-2)A+B]V^{(j-1)}_{\xi}\\ 
&&+\lambda[C_{pj}-((j-2)A_x+B_x)]V^{(j-1)}+\lambda
F(v,\ldots,v^{(j-2)} \nonumber
\end{eqnarray} 
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 $|W|\leq C_{2}$ in $B$. In view of (\ref{eq:ev4}) and (\ref{eq:ev5})
$$(a-\epsilon)^{\frac{1}{p-1}}\frac{x-\zeta(t)}{\lambda}
\leq\frac{v(x,t)}{\lambda}
\leq (a+\epsilon)^{\frac{1}{p-1}}\frac{x-\zeta(t)}{\lambda}$$
and
$$\zeta(\ol{t})\leq\zeta(t)\leq\zeta(\ol{t})+s(\ol{t}-t)
\leq\zeta(\ol{t})+\frac{\lambda}{4}.$$
Therefore
$$\frac{\lambda}{4} = \ol{x}-\frac{\lambda}{2}-\zeta(\ol{t})
-\frac{\lambda}{4}
\leq x - \zeta(t) \leq \ol{x} + \frac{\lambda}{2} - \zeta(\ol{t}) 
= \frac{3\lambda}{2}$$
which implies
$$\frac{(a-\epsilon)^{\frac{1}{p-1}}}{4}\leq \frac{v}{\lambda}
\leq \frac{3(a+\epsilon)^{\frac{1}{p-1}}}{2}\,.$$
Hence by (\ref{eq:1}) equation (\ref{eq:*}) is uniformly parabolic in $B$. 
Moreover, it follows
from Proposition \ref{pro:2} that $W$ satisfies all of the hypotheses of 
Theorem 5.3.1 of \cite{LSU}. Thus we conclude that there exists a constant 
$K=K(a,p,C_{1},\ldots,C_{j-1})>0$ such that
$$\left|\frac{\partial}{\partial\xi}V^{(j-1)}(0,0)\right| \leq K ;$$
that is,
$$|v^{(j)}(\ol{x},\ol{t})|\leq \frac{K}{\lambda}\,.$$
Since $(\ol{x},\ol{t})\in R_{\delta,\eta}$ is arbitrary, this proves the 
proposition.\hfill$\Box$ \medskip

We now turn to the barrier construction. If $\gamma \in (0,\delta)$ we will 
use the notation 
$$R_{\delta,\eta}^{\gamma}=R_{\delta,\eta}^{\gamma}(t_0)\equiv
\{(x,t)\in {\mathbb R}^2 :\zeta(t)+\gamma\leq x\leq\zeta(t)+\delta, t_0-\eta
\leq t\leq t_0 + \eta\}\,.$$

\begin{pro}\label{pro:9}
Let $R_{\delta_{1},\eta_{1}}$ be the region constructed in the proof of 
Proposition \ref{pro:2} with
For $j\geq 4$ and $(x,t)\in R_{\delta_{1},\eta_{1}}^{\gamma}$,
let
\begin{equation}\label{eq:ba3}
\phi_{j}(x,t) \equiv \frac{\alpha}{x-\zeta(t) -\gamma/3} 
+ \frac{\beta}{x-\zeta^{*}(t)}
\end{equation}
where $\zeta^{*}$ is given by (\ref{eq:z*}), and $\alpha$ and
$\beta$ are positive constant. Then there exist $\delta\in(0,\delta_1 )$ and
$\eta\in(0,\eta_1 )$ depending  only on $a$, $p$, $C_{1}, \ldots,
C_{j-1}$ such that
$$L_{j}(\phi_{j})\geq 0 \quad\text{in } R_{\delta,\eta}^{\gamma}$$
for all $\gamma\in (0,\delta)$.
\end{pro}
                      
\noindent\textbf{Proof.} Choose $\epsilon$ such that
\begin{equation}\label{eq:2eps}
0<\epsilon < \frac{a}{(j-2)(p-2)+6p-8}\,.
\end{equation}
There exist $\delta_{2}\in (0,\delta_{1})$ and $\eta\in (0,\eta_{1})$ such 
that (\ref{eq:1}),
(\ref{eq:ev4}) and (\ref{eq:ev5}) hold in $R_{\delta_{2},\eta}$. 
Fix $\gamma\in(0,\delta_{2})$.
For $(x,t) \in R_{\delta_{2},\eta}^{\gamma}$, we have
\begin{eqnarray*}
L_{j}(\phi_{j})
&=&\frac{\alpha}{(x-\zeta-\gamma/3  )^2}\left\{\zeta^{'}
-\frac{2(p-2)vv_{x}^{p-2}}{x-\zeta-\gamma/3 } +(j-2)A+B\right\}\\
&&-\frac{\alpha}{(x-\zeta-\gamma/3  )^2}\left\{C_{pj}(x-\zeta-\gamma/3)
-\frac{(x-\zeta-\gamma/3)^{2}}{\alpha}F
\right\}\\ 
&&+ \frac{\beta}{(x-\zeta^*)^2}\left\{\zeta^{*'}
-\frac{2(p-2)vv_{x}^{p-2}}{x-\zeta^{*}} +(j-2) A+B
-C_{pj}(x-\zeta^*) \right\}
\end{eqnarray*}
where $A$, $B$, $C_{pj}$ and $F$ are as before.
>From (\ref{eq:ev4}), together with the fact that $x-\zeta^{*}\geq x-\zeta-\gamma/3$ 
we have
$$\frac{v}{x-\zeta^{*}}\leq \frac{v}{x-\zeta-\gamma/3}
\leq (a+\epsilon)^{\frac{1}{p-1}}\frac{x-\zeta}{x-\zeta-\gamma/3} 
\leq (a+\epsilon)^{\frac{1}{p-1}}\frac{\gamma}{\gamma-\gamma/3} 
= \frac{3}{2}(a+\epsilon)^{\frac{1}{p-1}}\,.$$
Then from (\ref{eq:1}), (\ref{eq:ev4}) and (\ref{eq:ck}), we have  
\begin{eqnarray}\label{eq:***}
L_{j}(\phi_{j}) 
&\geq&
\frac{\alpha}{(x-\zeta-\gamma/3)^{2}}
\left\{a-((j-2)(p-2)+6p-9)\epsilon-\delta_2
(|C_{pj}|+\frac{\delta}{\alpha}|F|\right\}\nonumber\\ 
&&+\frac{\beta}{(x-\zeta^{*})^2}\left\{a-((j-2)(p-2)+6p-8)-
\delta_2(|C_{pj}|\right\}\nonumber
\end{eqnarray} 
Since $\epsilon$ satisfies (\ref{eq:2eps}) we can choose 
$\delta=\delta_{2}(\epsilon, p, a, C_{2})>0$ so small that 
$L_{3}(\phi_{3}) \geq 0$
in $R_{\delta,\eta}^{\gamma}$. \hfill $\Box$ \medskip

Hence as in we have the following proposition whose proof can be found in 
\cite{AV87}.

\begin{pro}\label{pro:10}
(Barrier Transformation). Let $\delta$ and $\eta$ be as in Proposition 
\ref{pro:9} with the additional restriction that
\begin{equation}\label{eq:epis}
\eta <\frac{\delta}{6\epsilon}\,,
\end{equation}
where $\epsilon$ is as in Proposition \ref{pro:9}. Suppose that for
some nonnegative constant $\beta$
\begin{equation}\label{eq:ssemi}
v^{(j)}(x,t) \leq \frac{\alpha}{x-\zeta(t)} 
+ \frac{\beta}{x-\zeta^{*}(t)} \quad in 
\quad R_{\delta,\eta}\,.
\end{equation}
Then $v^{(j)}$ also satisfies
\begin{equation}\label{eq:ffinal}
v^{(j)}(x,t) \leq \frac{2\alpha/3}{x-\zeta(t)} 
+ \frac{\beta + 2\alpha/3}{x-\zeta^{*}(t)}  
\quad\text{in } R_{\delta,\eta}\,.
\end{equation}
\end{pro}

Then as in \cite{AV87}, we can prove the $C^{\infty}$ regularity of
the  interface.


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D. G. Aronson and J. L. Vazquez, Eventual $C^{\infty}$-regularity and 
concavity for flows in one-dimensional porous media, 
{\it Arch. Rational Mech. Anal.} {\bf 99} (1987),no.4, 329-348.

\bibitem{EV88}
J. R. Esteban and J. L. Vazquez, Homogeneous diffusion in R with 
power-like nonlinear diffusivity, {\it Arch. Rational Mech. Anal.}
{\bf 103}(1988) 39-80.


\bibitem{CF80}
L. A. Caffarelli and A. Friedman, 
Regularity of the free boundary of a gas flow in an 
n-dimensional porous medium, {\it Ind. Univ. Math. J.} {\bf 29} (1980), 
361-381.

\bibitem{LSU}
O. A. Ladyzhenskaya, N.A. Solonnikov and N.N. Uraltzeva, Linear 
and quasilinear equations of parabolic type, {\it Trans. Math. Monographs,} 
{\bf 23}, Amer. Math. Soc., Providence, R. I., 1968. 
\end{thebibliography} 


\end{document}
