\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 112, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/112\hfil Propagation of perturbations]
{Propagation of perturbations for a sixth-order thin film equation}

\author[Z. Li, C. Liu\hfil EJDE-2012/112\hfilneg]
{Zhenbang Li, Changchun Liu}  % in alphabetical order

\address{Zhenbang Li \newline
Department of Mathematics, Jilin University, Changchun 130012, China}
\email{jamesbom23@yahoo.com.cn}

\address{Changchun Liu \newline
Department of Mathematics, Jilin University, Changchun 130012, China}
\email{liucc@jlu.edu.cn}

\thanks{Submitted January 31, 2012. Published July 3, 2012.}
\thanks{Sponsored by the Scientific Research Foundation for the Returned Overseas
Chinese Scholars, \hfill\break\indent State Education Ministry}
\thanks{Changchun Liu is the corresponding author}
\subjclass[2000]{35D05, 35G25, 35Q99, 76A20}
\keywords{Sixth-order thin film equation; radial solution; existence; 
\hfill\break\indent finite  speed of propagation}

\begin{abstract}
 We consider an initial-boundary problem for a sixth-order thin film equation,
 which arises in the industrial application of the isolation oxidation of silicon.
 Relying on some necessary uniform estimates of the approximate solutions,
 we prove the existence of radial symmetric solutions to this problem in the
 two-dimensional space. The nonnegativity and the finite   speed of propagation
 of perturbations of solutions are also discussed.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

This article is devoted to the radial symmetric solutions for a sixth-order
 thin film equation
$$
\frac{\partial u}{\partial t}=\operatorname{div}[|u|^n\nabla\Delta^2 u],
\quad x\in B,
$$
with the boundary value conditions
$$
\frac{\partial u}{\partial\boldsymbol{\nu}}\big|_{\partial B}=
\frac {\partial\Delta u}{\partial\boldsymbol{\nu}}\big|_{\partial B}
=\frac {\partial\Delta^2 u}{\partial\boldsymbol{\nu}}\big|_{\partial B}=0,
$$
and the initial value condition
$$
u\big|_{t=0}=u_0(x),
$$
where $B$ is the unit ball in ${\mathbb R}^2$,  $n>0$ is a constant, and
$\boldsymbol{\nu}$ is the outward unit normal to $\partial B$.

The equation  is a typical higher order equation, which has a sharp physical
background and a rich theoretical connotation. It was first introduced in
\cite{King, King1} in the case $n = 3$
$$
\frac{\partial u}{\partial t}=\frac{\partial}{\partial x}
\Big(u^3\frac{\partial^5u}{\partial x^5}\Big).
$$
It describes the spreading of a thin viscous fluid under the driving force
of an elastica (or light plate).

During the past years,  only a few works have been devoted to
the sixth-order thin film equation \cite{bn3, FJC, Ga, King, King1}.
 Bernis and  Friedman \cite{bn3} have studied the initial boundary value
problems to the thin film equation
$$
\frac {\partial u}{\partial t}+(-1)^{m-1}\frac{\partial}{\partial x}
\Big(f(u)\frac{\partial^{2m+1}u}{\partial x^{2m+1}}\Big)=0,
$$
where $f(u)=|u|^nf_0(u)$, $f_0(u)>0$, $n\geq1$ and proved existence of weak
 solutions preserving
nonnegativity.  Barrett,  Langdon and Nuernberg \cite{JW} considered the
above equation with $m=2$.
 A finite element method is presented which proves to be well posed
and convergent. Numerical experiments illustrate the theory.

Recently,  J\"{u}ngel and Mili\v{s}i\'{c} \cite{AJ} studied the sixth-order nonlinear
parabolic equation
$$
\frac {\partial u}{\partial t}
=\Big[u\Big(\frac1u(u(\ln u)_{xx})_{xx}+\frac12((\ln u)_{xx})^2
\Big)_x\Big]_x.
$$
They proved the global-in-time existence of weak nonnegative
solutions in one space dimension with periodic boundary conditions.

Evans, Galaktionov and King
\cite{Ev, Ev2} considered the sixth-order thin film equation
containing an unstable (backward parabolic) second-order term
$$
\frac {\partial u}{\partial t}=\operatorname{div}\big[|u|^n\nabla\Delta^2 u\big]
-\Delta(|u|^{p-1}u), n>0,\quad  p>1.
$$
By a formal
matched expansion technique, they show that, for the first critical exponent
$p = p_0 = n + 1 +\frac4N$ for $ n \in( 0,\frac54)$,
where $N$ is the space dimension, the free-boundary problem  with
zero-height, zero-contact-angle, zero-moment, and zero-flux conditions at the
interface admits a countable set of continuous branches of radially symmetric
self-similar blow-up solutions $u_k(x,t)=(T-t)^{-\frac N{nN+6}}f_k(y)$, $k=1,2,\cdots$, $y=\frac x{(T-t)^{\frac1{nN+6}}}$,
where $T > 0$ is the blow-up time.

We also refer the following relevant equation
$$
\frac {\partial u}{\partial t}=-\frac{\partial}{\partial x}
\Big(u^n\frac{\partial^3u}{\partial x^3}\Big),
$$
which has been extensively studied.  Bernis and  Friedman \cite{bn3}
 studied the initial boundary value problems to the
thin film equation $n>0$ and proved existence of weak solutions preserving
nonnegativity (see also \cite{b2, L, L1, Y}). They proved that if $n\geq 2$
the support of the solutions $u(\cdot,t)$ is nondecreasing with respect to $t$.

Our purpose in this paper is to study the radial symmetric solutions for the
equation. We will study the problem in two-dimensional case, which has
particular physical derivation of modeling the oil film spreading
over a solid surface, see \cite{T1}. After introducing the radial variable
$r=|x|$, we see that the radial symmetric solution satisfies
\begin{gather}
\frac{\partial(ru)}{\partial t}
=\frac{\partial}{\partial r}\big\{r|u|^n\frac{\partial W}{\partial r}\big\},\quad
rW=\frac{\partial}{\partial r}\Big(r\frac{\partial V}{\partial r}\Big),\quad
rV=\frac{\partial}{\partial r}\Big(r\frac{\partial u}{\partial r}\Big),
\label{1-1}
\\
\frac{\partial u}{\partial r}\big|_{r=0}=\frac{\partial u}{\partial r}\big|_{r=1}=\frac{\partial V}{\partial r}\big|_{r=0}
=\frac{\partial  V}{\partial r}\big|_{r=1}=\frac{\partial  W}{\partial r}\big|_{r=0}=\frac{\partial  W}{\partial r}\big|_{r=1}=0,
\label{1-2}
\\
u\big|_{t=0}=u_0(r). \label{1-3}
\end{gather}
It should be noticed that the equation \eqref{1-1} is degenerate at the points where
$r=0$ or $u=0$, and hence the arguments for one-dimensional problem can not be
applied directly. Because of the degeneracy, the problem does not admit
classical solutions in general. So, we introduce the weak solutions in the
following sense

\begin{definition} \label{def1.1}\rm
 A function $u$ is said to be a weak solution of the
problem \eqref{1-1}--\eqref{1-3}, if the following conditions are fulfilled:
\begin{itemize}
\item[(1)] $ru(r,t)$ is continuous in $\overline Q_T$,
where $Q_T=(0,1)\times(0,T)$;

\item[(2)] $\sqrt{r}|u|^{n/2}\frac{\partial W}{\partial r}\in L^2(Q_T)$;

\item[(3)] For any $\varphi\in C^1(\overline Q_T)$, the following integral equality
holds
\begin{align*}
&\int_0^1ru(r,T)\varphi(r,T)\,dr-\int_0^1ru_0(r)\varphi(r,0)\,dr\\
&-\iint_{Q_T} ru\frac{\partial \varphi}{\partial t} \,dr\,dt
+\iint_{Q_T} r |u|^n\frac{\partial W}{\partial r}
 \frac{\partial \varphi}{\partial r} \,dr\,dt=0.
\end{align*}
\end{itemize}
\end{definition}

Our interest lies in the existence of weak solutions. Because of the degeneracy,
we will first consider the regularized problem.
Based on the uniform estimates for
the approximate solutions, we obtain the existence. Owing to the background, we are much interested in the nonnegativity of the
weak solutions and the solutions with the
property of finite speed of propagation of perturbations.
Using weighted Nirenberg's inequality and Hardy's inequality, we proved these properties.
 This paper is arranged as follows.
 We shall prove several preliminary lemmas and obtain some a priori estimates on the solutions of
 regularized problem in Section 2, and then establish the existence in Section 3.
Subsequently, we discuss the nonnegativity of weak solutions in Section 4 and
the finite speed of propagation in Section 5.


\section{Regularized problem} \label{sec2}

  Bernis and Friedman \cite{bn3} obtained
several uniform estimations for the regularized solutions of fourth order
thin film equation with the initial boundary value problems.
 To discuss the existence of weak solutions of problem \eqref{1-1}-\eqref{1-3},
 we adopt the method of parabolic regularization,
namely, the desired solution will be obtained as the limit of some subsequence
of solutions of the following regularized problem
\begin{gather}
\frac{\partial(r_\varepsilon u)}{\partial t}=\frac{\partial}{\partial r}
\big\{r_\varepsilon m_\varepsilon(u)\frac {\partial W}{\partial r}\big\},\quad
 r_\varepsilon W=\frac{\partial}{\partial r}
 \Big(r_\varepsilon\frac{\partial V}{\partial r}\Big),\quad
 r_\varepsilon V=\frac{\partial}{\partial r}
\Big(r_\varepsilon\frac {\partial u} {\partial r}\Big), \label{2-1}
\\
\frac {\partial u}{\partial r}\big|_{r=0}=\frac {\partial u}{\partial r}\big|_{r=1}=
\frac{\partial V}{\partial r}\big|_{r=0}=\frac{\partial V}{\partial r}\big|_{r=1}=\frac{\partial W}{\partial r}\big|_{r=0}
=\frac{\partial W}{\partial r}\big|_{r=1}=0,
\label{2-2}
\\
u\big|_{t=0}=u_{0\varepsilon}(r),\label{2-3}
\end{gather}
where $r_\varepsilon=r+\varepsilon$,
$m_\varepsilon(u)=(|u|^2+\varepsilon)^{n/2}$ and
$u_{0\varepsilon}(r)$ is a smooth approximation of the initial data $u_0(r)$.

From the classical approach \cite{bn3}, it is not difficult to conclude that
the problem \eqref{2-1}--\eqref{2-3} admits a global classical solution.
 We need some uniform estimates on the classical solutions.

We first introduce some notation. Let $I=(0,1)$ and for any fixed
$\varepsilon\geq0$ denote by $W^{1,2}_{*,\varepsilon}(I)$ the class
of all functions satisfying
$$
\|u\|_{*,\varepsilon}=\Big(\int_0^1(r+\varepsilon)|u'(r)|^2dr\Big)^{1/2}
+\Big(\int_0^1(r+\varepsilon)|u(r)|^2dr\Big)^{1/2}<+\infty.
$$
It is obvious that $W^{1,2}(I)\subset W^{1,2}_{*,0}(I)$, but the
class $W^{1,2}_{*,0}(I)$ is quite different from $W^{1,2}(I)$. In
particular, we notice that the functions in $W^{1,2}_{*,0}(I)$ may
not be bounded. For $\varepsilon>0$ the spaces $W^{1,2}(I)$ and
$W^{1,2}_{*,\varepsilon}(I)$ coincide. However, it is not difficult
to prove that for
$u\in W^{1,2}_{*,\varepsilon}(I)$, the following properties hold:

\begin{lemma}\label{lem2.1}
 If $0<\alpha\leq1$, and $u\in W^{1,2}_{*,\varepsilon}(I)$. Then
$$
\sup_{0<r\leq1}((r+\varepsilon)^\alpha|u(r)|)\leq C\|u\|_{*,\varepsilon},
$$
where $C$ is a constant depending only on $\alpha$;
\end{lemma}

\begin{proof}
First we consider the case that $0<r<1/2$. From the mean value theorem,
we obtain immediately
$$
\int_r^1u(x)dx=u(\xi)(1-r)
$$
for some $\xi\in [r,1]$. Thus
\begin{align*}
|u(r)|&\leq |u(r)-u(\xi)|+|u(\xi)|
\leq\Big|\int_r^\xi u'(x)dx\Big|+\frac1{1-r}\Big|\int_r^1 u(x)dx\Big|
\\
&\leq \int_r^1 |u'(x)|dx+2\int_r^1 |u(x)|dx.
\end{align*}
%%
It follows that
%%
\begin{align*}
(r+\varepsilon)^\alpha|u(r)|
&\leq \int_r^1(r+\varepsilon)^\alpha |u'(x)|dx
+2\int_r^1(r+\varepsilon)^\alpha |u(x)|dx
\\
&\leq \int_0^1(r+\varepsilon)^\alpha |u'(r)|dr
+2\int_0^1(r+\varepsilon)^\alpha |u(r)|dr
\\
&\leq \Big(\int_0^1(r+\varepsilon) |u'(r)|^2dr\Big)^{1/2}
\Big(\int_0^1(r+\varepsilon)^{2\alpha-1}dr\Big)^{1/2}
\\
&\quad +2\Big(\int_0^1(r+\varepsilon) |u(r)|^2dr\Big)^{1/2}
\Big(\int_0^1(r+\varepsilon)^{2\alpha-1}dr\Big)^{1/2}
\\
&\leq C(\alpha)\|u\|_{*,\varepsilon}.
\end{align*}
Finally, we discuss the case that $1/2\leq r\leq1$, we have
\begin{align*}
(r+\varepsilon)^\alpha|u(r)|
&\leq(1+\varepsilon)^\alpha |u(r)|
\\
&\leq (1+\varepsilon)^\alpha\Big[\int_{1/2}^1 |u'(x)|dx
+2\int_{1/2}^1 |u(y)|dy\Big]
\\
&\leq (1+\varepsilon)^\alpha\Big[
\Big(\int_0^1(r+\varepsilon) |u'(r)|^2dr\Big)^{1/2}
+2\Big(\int_0^1(r+\varepsilon) |u(r)|^2dr\Big)^{1/2}
\Big]
\\
&\leq C(\alpha)\|u\|_{*,\varepsilon}.
\end{align*}
The proof is complete.
\end{proof}

\begin{lemma}\label{lem2.2}
If $0<\alpha\leq 1/2$, and $u\in W^{1,2}_{*,\varepsilon}(I)$.
Then for any $\beta<\alpha$,
$$
|(r_1+\varepsilon)^\alpha u(r_1)-(r_2+\varepsilon)^\alpha u(r_2)|
\leq C|r_1-r_2|^\beta\|u\|_{*,\varepsilon},
$$
where $C$ is a constant depending only on $\alpha$ and $\beta$.
\end{lemma}

\begin{proof}
For fixed $0<r_2<r_1<1$, we have
$$
|u(r_2)-u(r_1)|\leq\int_{r_2}^{r_1}|u'(t)|dt.
$$
It follows that
%%
\begin{align*}
(r_2+\varepsilon)^\alpha|u(r_1)-u(r_2)|
&\leq(r_2+\varepsilon)^\alpha\int_{r_2}^{r_1}|u'(t)|dt
\leq\int_{r_2}^{r_1}(t+\varepsilon)^\alpha|u'(t)|dt
\\
&\leq \Big(\int_{r_2}^{r_1}(t+\varepsilon) |u'(t)|^2dt\Big)^{1/2}
\Big(\int_{r_2}^{r_1}(t+\varepsilon)^{2\alpha-1}dt\Big)^{1/2}
\\
&\leq  C(\alpha)\|u\|_{*,\varepsilon}
((r_1+\varepsilon)^{2\alpha}-(r_2+\varepsilon)^{2\alpha})^{1/2}
\\
&\leq  C(\alpha)\|u\|_{*,\varepsilon}|r_1-r_2|^\alpha.
\end{align*}
On the other hand, from Lemma \ref{lem2.1}, we have
\begin{align*}
|((r_1+\varepsilon)^\alpha-(r_2+\varepsilon)^\alpha)u(r_1)|
&\leq2|r_1-r_2|^\alpha|u(r_1)|
\\
&\leq 2|r_1-r_2|^\beta(r_1+\varepsilon)^{\alpha-\beta}|u(r_1)|\\
&\leq C|r_1-r_2|^\beta\|u\|_{*,\varepsilon}.
\end{align*}
Therefore,
\begin{align*}
&|(r_1+\varepsilon)^\alpha u(r_1)-(r_2+\varepsilon)^\alpha u(r_2)|
\\
&\leq |((r_1+\varepsilon)^\alpha-(r_2+\varepsilon)^\alpha)u(r_1)|+
(r_2+\varepsilon)^{\alpha}|u(r_2)-u(r_1)|
\\
&\leq C|r_1-r_2|^\beta\|u\|_{*,\varepsilon}
\end{align*}
with $C$ depending only on $\alpha$ and $\beta$. The proof is complete.
\end{proof}

\begin{remark} \label{rmk2.1} \rm
Let $0<\beta<1/2$ and $u\in W^{1,2}_{*,\varepsilon}(I)$. Then
$$
|(r_1+\varepsilon) u(r_1)-(r_2+\varepsilon) u(r_2)|
\leq C(\beta)|r_1-r_2|^\beta\|u\|_{*,\varepsilon},
$$
where $C(\beta)$ is a constant depending only on $\beta$.
\end{remark}

\begin{lemma}\label{lem2.3}
Let $u$ be a smooth solution of problem \eqref{2-1}--\eqref{2-3}
and for any $\alpha\in(0,1/2]$ and $\beta<\alpha$, there
is a constant $M$ independent of $\varepsilon$ such that
\begin{gather*}
|r_\varepsilon^\alpha u(r,t)-s_\varepsilon^\alpha u(s,t)|\leq M|r-s|^\beta,\\
|r_\varepsilon u(r,t)-s_\varepsilon u(s,t)|\leq M|r-s|^\beta
\end{gather*}
for all $r,s\in (0,1)$, where $s_\varepsilon=s+\varepsilon$.
\end{lemma}

\begin{proof}
Multiplying  \eqref{2-1} by $W$ and integrating with respect to
$r$ over $(0,1)$, we obtain
\begin{align*}
0&=\int_0^1\big\{\frac{\partial r_\varepsilon u}{\partial t} W
-\frac{\partial}{\partial r}\big[r_\varepsilon  m_\varepsilon(u)
\frac{\partial W}{\partial r}\big]W\big\}dr
\\
&=\int_0^1\big\{\frac12\frac{\partial r_\varepsilon V^2}{\partial t}
+r_\varepsilon  m_\varepsilon(u)\big(\frac{\partial W}{\partial r}\big)^2\big\}dr.
\end{align*}
Hence, integrating also with respect to $t$, we hve
\begin{gather}
\int_0^1r_\varepsilon V^2dr\leq C, \label{2-4} \\
\iint_{Q_T}r_\varepsilon m_\varepsilon(u)
\big|\frac{\partial W}{\partial r}\big|^2dr\leq C.
\label{2-5}
\end{gather}
It is easy to see that
$$
\int_0^1r_\varepsilon V^2dr=\int_0^1\frac{\partial}{\partial r}
\big(r_\varepsilon\frac{\partial u}{\partial r}\big)\cdot
\frac1{r_\varepsilon}\frac{\partial}{\partial r}
\big(r_\varepsilon\frac{\partial u}{\partial r}\big)\,dr\leq C.
$$
A simple calculation shows that
$$
\int_0^1r_\varepsilon\Big(\frac{\partial^2u}{\partial r^2}\Big)^2dr
+\int_0^1\frac1{r_\varepsilon}\Big(\frac{\partial u}{\partial r}\Big)^2dr
+2\int_0^1\frac{\partial^2u}{\partial r^2}\frac{\partial u}{\partial r}dr\leq C.
$$
Using boundary condition \eqref{2-2}, we have
$$
\int_0^1\frac{\partial^2u}{\partial r^2}\frac{\partial u}{\partial r}dr=0.
$$
Hence
\begin{gather}
\int_0^1r_\varepsilon\Big(\frac{\partial u}{\partial r}\Big)^2dr
\leq\int_0^1\frac1{r_\varepsilon}\Big(\frac{\partial u}{\partial r}\Big)^2dr\leq C.
\label{2-6} \\
\int_0^1r_\varepsilon\Big(\frac{\partial^2u}{\partial r^2}\Big)^2dr\leq C,
\label{2-7}
\end{gather}
On the other hand, integrating the equation (2.1) on $Q_t=(0,1)\times(0,t)$, we have
\begin{equation}
\int_0^1r_\varepsilon u(r,t)\,dr=\int_0^1r_\varepsilon u_{0\varepsilon}(r)\,dr.
\label{2-8}
\end{equation}
Note that, for any $\rho\in (0,1)$,
\begin{align*}
&\frac{1+2\varepsilon}2u(\rho,t)-\int_0^1s_\varepsilon u(s,t)ds\\
&=\int_0^1s_\varepsilon[u(\rho,t)-u(s,t)]ds
\\
&= \int_0^1\int_s^\rho s_\varepsilon\frac{\partial u}{\partial r}(r,t)\,dr\,ds
=\int_0^\rho\int_s^\rho s_\varepsilon\frac{\partial u}{\partial r}(r,t)\,dr\,ds+\int_\rho^1\int_s^\rho
s_\varepsilon\frac{\partial u}{\partial r}(r,t)\,dr\,ds
\\
&= \int_0^\rho\int_0^r s_\varepsilon\frac{\partial u}{\partial r}(r,t)dsdr+\int_\rho^1\int_r^1
s_\varepsilon\frac{\partial u}{\partial r}(r,t)dsdr
\\
&= \int_0^\rho\Big(\frac{r^2}2+\varepsilon r\Big)\frac{\partial u}{\partial r}(r,t)\,dr
+\int_\rho^1\Big[\frac12(1-r^2)+\varepsilon(1-r)\Big]\frac{\partial u}{\partial r}(r,t)\,dr
\\
&\leq \int_0^\rho r_\varepsilon\big|
 \frac{\partial u}{\partial r}(r,t)\big|dr+2\int_\rho^1
\big|\frac{\partial u}{\partial r}(r,t)\big|dr.
\end{align*}
Setting $\rho_\varepsilon=\rho+\varepsilon$ and multiplying the above
inequality by $2\rho_\varepsilon^{1/2}$,
we obtain
\begin{equation}
\begin{aligned}
&\big|(1+2\varepsilon)\rho_\varepsilon^{1/2}u(\rho,t)
-2\rho_\varepsilon^{1/2}\int_0^1s_\varepsilon u(s,t)ds\big|\\
&\leq 2\rho_\varepsilon^{1/2}\int_0^\rho r_\varepsilon\big|\frac{\partial u}{\partial r}(r,t)\big|dr
+4\rho_\varepsilon^{1/2}\int_\rho^1\big|\frac{\partial u}{\partial r}(r,t)\big|dr
\\
&\leq 2\rho_\varepsilon^{1/2}\int_0^\rho r_\varepsilon\big|\frac{\partial u}{\partial r}(r,t)\big|dr
+4\int_\rho^1r^{1/2}_\varepsilon\big|\frac{\partial u}{\partial r}(r,t)\big|dr
\\
&\leq C\left(\int_0^1 r_\varepsilon\big|\frac{\partial u}{\partial r}(r,t)\big|^2dr\right)^{1/2}.
\end{aligned} \label{2-9}
\end{equation}
From \eqref{2-6}, \eqref{2-8} and \eqref{2-9}, we see that
$r_\varepsilon^{1/2}u(r,t)$ is
uniformly bounded on $Q_T$. Furthermore
$u(\cdot,t)\in W^{1,2}_{*,\varepsilon}(I)$ for any fixed $t\in (0,T)$, with
$\|u(\cdot,t)\|_{*,\varepsilon}$ bounded by a constant $C$ independent of
$\varepsilon$. The desired estimates then follow from the properties of
$W^{1,2}_{*,\varepsilon}(I)$ mentioned above.

From the above results and using the Remark, we conclude the second inequality.
The proof is complete.
\end{proof}

\begin{lemma}\label{lem2.4}
For any $\alpha>0$, there is a constant $M$ independent of $\varepsilon$
such that
\begin{gather}
r_\varepsilon^\alpha|u(r,t)|\leq M,\quad
\|u\|_{*,\varepsilon}\leq M , \label{2-10}\\
|r_\varepsilon u(r,t_2)-r_\varepsilon u(r,t_1)|\leq M|t_2-t_1|^{1/{16}}
\label{2-11}
\end{gather}
for all $r\in (0,1)$, $t_1,t_2\in(0,T)$.
\end{lemma}

\begin{proof}
The first two estimates have already been seen from the arguments in
Lemma \ref{lem2.1}.
Now, we begin to show \eqref{2-11}. Without loss of generality, we assume that
$t_1<t_2$ and set $\Delta t=t_2-t_1$. Integrating both sides of the equation
\eqref{2-1} over $(t_1,t_2)\times(y,y+(\Delta t)^\alpha)$ and then integrating the
resulting relation with respect to $y$ over $(x,x+(\Delta t)^\alpha)$, we obtain
%%
\begin{align*}
&(\Delta t)^\alpha\int_x^{x+(\Delta t)^\alpha}
\int_0^1(y+\theta(\Delta t)^\alpha+\varepsilon)
\Big[u(y+\theta(\Delta t)^\alpha,t_2)
-u(y+\theta(\Delta t)^\alpha,t_1)\Big]\,d\theta\,dy
\\
&= \int_x^{x+(\Delta t)^\alpha}\int_y^{y+(\Delta t)^\alpha}\int_{t_1}^{t_2}
\frac{\partial}{\partial r}\left\{r_\varepsilon m_\varepsilon(u)
\frac{\partial W}{\partial r}\right\}\,d\tau \,dr\,dy
\\
&= \int_x^{x+(\Delta t)^\alpha}\int_{t_1}^{t_2}
\Big[(y+(\Delta t)^\alpha+\varepsilon)m_\varepsilon(u(y+(\Delta t)^\alpha))
\\
&\quad\times\frac{\partial}{\partial y} W(y+(\Delta t)^\alpha,\tau)
-(y+\varepsilon) m_\varepsilon(u(y))
\frac{\partial}{\partial y} W(y,\tau)\Big]\,d\tau\, dy.
\end{align*}
By the mean value theorem, there exists $x^*=y^*+\theta^*(\Delta t)^\alpha$,
$y^*\in(x,x+(\Delta t)^\alpha),\theta^*\in(0,1)$ such that the left hand side
of the above equality can be expressed by
\begin{align*}
&(\Delta t)^\alpha\int_x^{x+(\Delta t)^\alpha}
\int_0^1(y+\theta(\Delta t)^\alpha+\varepsilon)
\left[u(y+\theta(\Delta t)^\alpha,t_2)-u(y+\theta(\Delta t)^\alpha,t_1)
\right]\,d\theta\,dy
\\
&= (\Delta t)^{2\alpha}(y^*+\theta^*(\Delta t)^\alpha+\varepsilon)
\Big[u(y^*+\theta^*(\Delta t)^\alpha,t_2)-u(y^*+\theta^*(\Delta t)^\alpha,t_1)\Big].
\end{align*}
For the right hand side, we have
\begin{align*}
&\int_x^{x+(\Delta t)^\alpha}\int_{t_1}^{t_2}
\Big[(y+(\Delta t)^\alpha+\varepsilon)m_\varepsilon(u(y+(\Delta t)^\alpha))
\\
&\quad\times\frac{\partial}{\partial y} W(y+(\Delta t)^\alpha,\tau)
-(y+\varepsilon) m_\varepsilon(u(y))
\frac{\partial}{\partial y} W(y,\tau)\Big]\,d\tau\,dy
\\
&= \int_{x+(\Delta t)^\alpha}^{x+2(\Delta t)^\alpha}\int_{t_1}^{t_2}
(r+\varepsilon) m_\varepsilon(u(r))
\frac{\partial}{\partial r} W(r,\tau)\,d\tau\,dr
\\
&\quad -\int_x^{x+(\Delta t)^\alpha}(r+\varepsilon) m_\varepsilon(u(r))
\frac{\partial}{\partial r} W(r,\tau)\,d\tau\,dr
\\
&\leq \int_{x+(\Delta t)^\alpha}^{x+2(\Delta t)^\alpha}\int_{t_1}^{t_2}
r_\varepsilon m_\varepsilon(u(r))
\big|\frac{\partial}{\partial r} W(r,\tau)\big|\,d\tau\,dr.
\end{align*}
By \eqref{2-5}, we see that
$$
|x^*_\varepsilon u(x^*,t_2)-x^*_\varepsilon u(x^*,t_1)|
\leq C(\Delta t)^{\frac{1-3\alpha}2},
$$
which implies, by setting $\alpha=1/4$ and using the properties of the
functions in $W^{1,2}_{*,\varepsilon}(I)$, that
$$
|r_\varepsilon u(r,t_2)-r_\varepsilon u(r,t_1)|\leq C(\Delta t)^{1/16}.
$$
The proof is complete.
\end{proof}

\section{Existence}\label{sec3}

After the discussion of the regularized problem, we can now turn to the
investigation of the existence of weak solutions of the problem
\eqref{1-1}--\eqref{1-3}.
The main existence result is the following

\begin{theorem}\label{th3.1}
If $u_0(r) \in H^2(I)$, then problem \eqref{1-1}--\eqref{1-3} admits at least
one weak solution.
\end{theorem}

\begin{proof}
Let $u_\varepsilon$ be the approximate solution of  \eqref{2-1}--\eqref{2-3}
constructed in the previous section. Using the estimates in Lemma \ref{lem2.3} and \ref{lem2.4},
for any $\beta<\frac12$, and $(r_1,t_2),(r_2,t_1)\in Q_T$, we have
$$
|r_{1\varepsilon} u_\varepsilon(r_1,t_2)-r_{2\varepsilon} u_\varepsilon(r_2,t_1)|
\leq C(|r_1-r_2|^\beta+|t_1-t_2|^{\beta/4})
$$
with constant $C$ independent of $\varepsilon$. So, we may extract a
subsequence from $\{r_\varepsilon u_\varepsilon\}$, denoted also by
$\{r_\varepsilon u_\varepsilon\}$, such that
$$
r_\varepsilon u_\varepsilon(r,t)\to ru(r,t)\hbox{ uniformly in }\overline Q_T,
$$
and the limiting function $ru\in C^{1/4,1/{16}}(\overline Q_T)$.

Now we prove that $u(r,t)$ is a weak solution of problem \eqref{1-1}-\eqref{1-3},
let $\delta>0$ be fixed and set
$P_\delta=\{(r,t):r|u|^n>\delta\}$.
We choose $\varepsilon_0(\delta)>0$, such that
\begin{equation}
r_\varepsilon (|u_\varepsilon|^2+\varepsilon)^{n/2}
\geq \frac{\delta}2,\quad (r,t)\in
P_\delta,\;0<\varepsilon<\varepsilon_0(\delta).
\label{3-1}
\end{equation}
Then from \eqref{2-5}
\begin{equation}
\iint_{P_\delta}\Big(\frac{\partial W_\varepsilon}{\partial r}\Big)^2 \,dr\,dt
\leq \frac C\delta.
\label{3-2}
\end{equation}
To prove the integral equality in the definition of solutions, it suffices to pass
the limit as $\varepsilon\to0$ in
\begin{align*}
&\int_0^1r_\varepsilon u_\varepsilon(r,T)\varphi(r,T)\,dr
-\int_0^1r_\varepsilon u_{0\varepsilon}\varphi(r,0)\,dr
-\iint_{Q_T}r_\varepsilon u_\varepsilon\frac{\partial \varphi}{\partial t}\,dr\,dt
\\
&+\iint_{Q_T} r_\varepsilon
(u_\varepsilon^2+\varepsilon)^{n/2}\frac{\partial W_\varepsilon}{\partial r}\frac{\partial\varphi}{\partial r}\,dr\,dt=0.
\end{align*}
The limits
\begin{gather*}
\lim_{\varepsilon\to0}\int_0^1r_\varepsilon
u_\varepsilon(r,T)\varphi(r,T)=\int_0^1ru(r,T)\varphi(r,T)\,dr,
\\
\lim_{\varepsilon\to0}\int_0^1r_\varepsilon
u_{0\varepsilon}(r)\varphi(r,0)\,dr=\int_0^1u_0(r)\varphi(0,r)\,dr,
\\
\lim_{\varepsilon\to0}\iint_{Q_T}
r_\varepsilon u_\varepsilon\frac{\partial\varphi}{\partial t}\,dr\,dt=\iint_{Q_T}ru\frac{\partial\varphi}{\partial t}\,dr\,dt,
\end{gather*}
are obvious. It remains to show that
\begin{equation}
\lim_{\varepsilon\to0}\iint_{Q_T} r_\varepsilon
(u_\varepsilon^2+\varepsilon)^{n/2}\frac{\partial W_\varepsilon}{\partial r}\frac{\partial \varphi}{\partial r}\,dr\,dt
=\iint_{Q_T} r |u|^n\frac{\partial W}{\partial r}\frac{\partial \varphi}{\partial r}\,dr\,dt.
\label{3-3}
\end{equation}
In fact, for any fixed $\delta>0$,
\begin{align*}
&\big|\iint_{Q_T} r_\varepsilon
(u_\varepsilon^2+\varepsilon)^{n/2}\frac{\partial W_\varepsilon}{\partial r}\frac{\partial \varphi}{\partial r}\,dr\,dt
-\iint_{Q_T} r |u|^n\frac{\partial W}{\partial r}\frac{\partial \varphi}{\partial r}\,dr\,dt\big|
\\
&\leq \big|\iint_{P_\delta} r_\varepsilon
(u_\varepsilon^2+\varepsilon)^{n/2}\frac{\partial W_\varepsilon}{\partial r}\frac{\partial \varphi}{\partial r}\,dr\,dt
-\iint_{P_\delta} r|u|^n\frac{\partial W}{\partial r}\frac{\partial \varphi}{\partial r}\,dr\,dt\big|
\\
&\quad +\big|\iint_{{Q_T}\backslash{P_\delta}} r_\varepsilon
(u_\varepsilon^2+\varepsilon)^{n/2}\frac{\partial W_\varepsilon}{\partial r}\frac{\partial \varphi}{\partial r}\,dr\,dt\big|
+\big|\iint_{{Q_T}\backslash{P_\delta}} r|u|^n\frac{\partial W}{\partial r}
\frac{\partial \varphi}{\partial r}\,dr\,dt\big|.
\end{align*}
From the estimates \eqref{2-5}, we have
\begin{gather*}
\Big|\iint_{{Q_T}\backslash {P_\delta}} r_\varepsilon
(u_\varepsilon^2+\varepsilon)^{n/2}\frac{\partial W_\varepsilon}{\partial r}
\frac{\partial \varphi}{\partial r}\,dr\,dt\Big|
\leq C\delta\sup\big|\frac{\partial \varphi}{\partial r}\big|,\quad 0<\varepsilon<\varepsilon_0(\delta),
\\
\Big|\iint_{{Q_T}\backslash {P_\delta}} r|u|^n\frac{\partial W}{\partial r}
\frac{\partial \varphi}{\partial r}\,dr\,dt\Big|
\leq C\delta\sup\big|\frac{\partial \varphi}{\partial r}\big|,
\\
\begin{aligned}
&\Big|\iint_{P_\delta} r_\varepsilon
(u_\varepsilon^2+\varepsilon)^{n/2}\frac{\partial W_\varepsilon}{\partial r}\frac{\partial \varphi}{\partial r}\,dr\,dt
-\iint_{P_\delta} r|u|^n\frac{\partial W}{\partial r}
\frac{\partial \varphi}{\partial r}\,dr\,dt\Big|
\\
&\leq \iint_{P_\delta}
\big|r_\varepsilon(u_\varepsilon^2+\varepsilon)^{n/2} -r|u|^n\big|
\big|\frac{\partial W_\varepsilon}{\partial r}\big|\big|
\frac{\partial \varphi}{\partial r}\big|\,dr\,dt\\
 &\quad +\Big|\iint_{P_\delta} r|u|^n
\Big(\frac{\partial W_\varepsilon}{\partial r}-\frac{\partial W}{\partial r}\Big)
 \frac{\partial \varphi}{\partial r}\,dr\,dt\Big|
\\
&\leq \sup|r_\varepsilon(u_\varepsilon^2+\varepsilon)^{n/2} -r|u|^n|
\big|\frac{\partial \varphi}{\partial r}\big|\frac C{\sqrt{\delta}}
+\Big|\iint_{P_\delta}r|u|^n\Big(\frac{\partial W_\varepsilon}{\partial r}
-\frac{\partial W}{\partial r}\Big)
\frac{\partial \varphi}{\partial r}\,dr\,dt\Big|
\end{aligned}
\end{gather*}
and hence
$$
\limsup_{\varepsilon\to0}
\Big|\iint_{Q_T} r_\varepsilon
\frac{\partial W_\varepsilon}{\partial r}\frac{\partial \varphi}{\partial r}\,dr\,dt
-\iint_{Q_T} r\frac{\partial W}{\partial r}\frac{\partial \varphi}{\partial r}\,dr\,dt
\Big|
\leq C\delta\sup\big|\frac{\partial \varphi}{\partial r}\big|.
$$
By the arbitrariness of $\delta$, we see that the limit \eqref{3-3} holds.
 The proof is complete.
\end{proof}


\section{Nonnegativity}\label{sec4}

Just as mentioned by several authors, it is much interesting to discuss
the physical solutions. For the two-dimensional problem \eqref{1-1}--\eqref{1-3},
a very typical example is the modeling of oil films spreading over a
solid surface, where the unknown function $u$ denotes the height from the
surface of the oil film to the solid surface. Motivated by this idea, we
devote this section to the discussion of the nonnegativity of solutions.

\begin{theorem}\label{th4.1}
The weak solution $u$ obtained in Section 3 satisfies $u(r,t)\geq0$,
if $u_0(r)\geq0$.
\end{theorem}

\begin{proof}
Suppose the contrary, that is, the set
\begin{equation}
E=\{(r,t)\in\overline Q_T; u(r,t)<0\}
\label{4-1}
\end{equation}
is nonempty.
For any fixed $\delta>0$, choose a $C^\infty$ function $H_\delta(s)$ such that
$H_\delta(s)=-\delta$ for $s\geq-\delta$, $H_\delta(s)=-1$, for $s\leq-2\delta$ 
and that $H_\delta(s)$ is nondecreasing for $-2\delta<s<-\delta$. Also, we extend the
function $u(r,t)$ to be defined in the whole plane ${\mathbb R}^2$ such that the
extension $\bar{u}(r,t)=0$ for $t\geq T+1$ and $t\leq -1$. Let $\alpha(s)$
be the kernel of mollifier in one-dimension, that is, $\alpha(s)\in
C^\infty({\mathbb R})$, $\hbox{supp}\alpha=[-1,1]$, $\alpha(s)>0$ in $(-1,1)$, and
$\int_{-1}^1\alpha(s)ds=1$. For any fixed $k>0,\delta>0$, define
\begin{gather*}
u^h(r,t)=\int_{\mathbb R}\bar{u}(s,r)\alpha_h(t-s)ds, \\
\beta_\delta(t)=\int_t^{+\infty}\alpha\Big(\frac{s-\frac T2}{\frac
T2-\delta}\Big)\frac1{\frac T2-\delta}ds,
\end{gather*}
where $\alpha_h(s)=\frac1h\alpha(s/h)$.

The function
$$
\varphi_\delta^h(r,t)\equiv[\beta_\delta(t)H_\delta(u^h)]^h
$$
is clearly an admissible test function, that is, the following integral
equality holds
\begin{equation}
\begin{aligned}
&\int_0^1ru(r,T)\varphi^h_\delta(T,r)\,dr-\int_0^1ru_0(r)\varphi^h_\delta(r,0)\,dr\\
&-\iint_{Q_T} ru\frac {\partial\varphi^h_\delta} {\partial t} \,dr\,dt
+\iint_{Q_T} r|u|^n\frac{\partial W}{\partial r}
\frac {\partial\varphi^h_\delta}{\partial r} \,dr\,dt=0.
\end{aligned}\label{4-2}
\end{equation}
To proceed further, we give an analysis on the properties of the test
function $\varphi^h_\delta(r,t)$. The definition of $\beta_\delta(t)$ implies that
\begin{equation}
\varphi_\delta^h(r,t)=0,\quad t\geq T-\frac \delta2, \quad  h<\frac\delta2.
\label{4-3}
\end{equation}
Since $\bar{u}(r,t)$ is continuous, for fixed $\delta$, there exists
$\eta_1(\delta)>0$, such that
\begin{equation}
u^h(r,t)\geq-\frac\delta2,\quad t\leq \eta_1(\delta),\quad
0\leq r\leq1,\quad h<\eta_1(\delta),
\label{4-4}
\end{equation}
which together with the definition of $\beta_\delta(t)$ and $H_\delta(s)$ imply
\begin{equation}
H_\delta(u^h(r,t))=-\delta,\quad t\leq \eta_1(\delta),\quad
0\leq r\leq1,\quad h<\eta_1(\delta) \label{4-5}
\end{equation}
and hence
\begin{equation}
\varphi_\delta^h=-\delta,\quad t\leq \frac12\eta_1(\delta),\quad 0\leq r\leq1,\quad
h<\frac12\eta_1(\delta).
\label{4-6}
\end{equation}
We note also that for any functions $f(t),g(t)\in L^2(R)$,
\begin{align*}
\int_{\mathbb R}f(t)g^h(t)dt&=\int_{\mathbb R}f(t)dt\int_{\mathbb R}g(s)\alpha_n(t-s)ds
=\int_{\mathbb R}f(t)\int_{\mathbb R}g(s)\alpha_n(s-t)ds
\\
&=\int_{\mathbb R}g(s)ds\int_{\mathbb R}f(t)\alpha_n(s-t)dt=\int_{\mathbb R}f^h(t)g(t)dt.
\end{align*}
Taking this into account and using \eqref{4-3}, \eqref{4-5}, \eqref{4-6}, we have
\begin{align*}
\iint_{Q_T}ru\frac{\partial}{\partial t}\varphi_\delta^h\,dr\,dt
&=\int_{-\infty}^{+\infty}dt\int_0^1ru\big[\frac{\partial}{\partial t}\big(\beta_\delta(t)
H_\delta(u^h)\big)\big]^hdr
\\
&= \iint_{Q_T}(ru)^h\frac{\partial}{\partial t}\big(\beta_\delta(t)H_\delta(u^h)\big)\,dr\,dt
\end{align*}
and hence by integrating by parts
\begin{align*}
&\iint_{Q_T}(ru)^h\frac{\partial}{\partial t}\left(\beta_\delta(t)H_\delta(u^h)\right)\,dr\,dt
\\
&= \int_0^1(ru)^h(r,T)\beta_\delta(T)H_\delta(u^h(r,T))\,dr
-\int_0^1(ru)^h(r,0)\beta_\delta(0)H_\delta(u^h(r,0))\,dr
\\
&\quad -\iint_{Q_T}\beta_\delta(t)H_\delta(u^h)\frac{\partial(ru)^h}{\partial t}\,dr\,dt
\\
&= \delta\int_0^1(ru)^h(r,0)\,dr
-\iint_{Q_T}r\beta_\delta(t)\frac{\partial}{\partial t}F_\delta(u^h)\,dr\,dt,
\end{align*}
where $F_\delta(s)=\int_0^sH_\delta(\sigma)d\sigma$.

Again by \eqref{4-5}
\begin{align*}
F_\delta(u^h(r,0))&=\int_0^{u^h(r,0)}H_\delta(\sigma)d\sigma
=\int_0^1H_\delta(\lambda u^h(r,0))d\lambda\cdot u^h(r,0)\\
&=-\delta u^h(r,0)
\end{align*}
and hence
\begin{equation}
\begin{aligned}
&\iint_{Q_T}(ru)^h\frac{\partial}{\partial t}(\beta_\delta(t)H_\delta(u^h))\,dr\,dt
\\
&= \delta\int_0^1(ru)^h(r,0)\,dr
+\int_0^1r\beta_\delta(0)F_\delta(u^h(r,0))\,dr
+\iint_{Q_T}rF_\delta(u^h)\beta'_\delta(t)\,dr\,dt
\\
&= -\frac1{\frac T2-\delta}\iint_{Q_T}rF_\delta(u^h)\alpha
\Big(\frac{t-\frac T2}{\frac T2-\delta}\Big)\,dr\,dt.
\end{aligned} \label{4-7}
\end{equation}
From \eqref{4-3}, \eqref{4-6} it is clear that
\begin{gather}
\int_0^1ru(r,T)\varphi^h_\delta(T,r)\,dr=0,\quad 0<h<\frac12\eta_1(\delta),
\label{4-8} \\
-\int_0^1ru_0(r)\varphi^h_\delta(r,0)\,dr=\delta\int_0^1ru_0(r)\,dr.
\label{4-9}
\end{gather}
Substituting \eqref{4-7}, \eqref{4-8} and \eqref{4-9} into \eqref{4-2}, we have
\begin{equation}
\begin{split}
&\frac2{ T-2\delta}\iint_{Q_T}rF_\delta(u^h)\alpha
\Big(\frac{t-\frac T2}{\frac T2-\delta}\Big)\,dr\,dt+\delta\int_0^1ru_0(r)\,dr\\
&+\iint_P r|u|^n\frac{\partial W}{\partial r}
\frac {\partial\varphi^h_\delta}{\partial r} \,dr\,dt=0.
\end{split}\label{4-10}
\end{equation}
By the uniform continuity of $u(r,t)$ in $\overline Q_T$, there exists
$\eta_2(\delta)>0$, such that
\begin{equation}
u(r,t)\geq-\frac\delta2\quad\forall (r,t)\in P^\delta, \label{4-11}
\end{equation}
where $P^\delta=\{(r,t);\operatorname{dist}((r,t),P)<\eta_2(\delta)\}$.
Here we have used the fact that $u(r,t)>0$ in $P$. Thus
$$
H_\delta(u^h(r,t))=-\delta,\quad\forall (r,t)\in P^{\delta/2},\quad
0<h<\frac12\eta_2(\delta)
$$
where $P^{\delta/2}=\{(r,t);\operatorname{dist}((r,t),P)<\frac12\eta_2(\delta)\}$.

By this and the definition of $u^h, H_\delta(s)$ shows that the function
$\varphi_\delta^h(r,t)$ is only a function of $t$ in $P$, whenever
$h<\frac12\eta_2(\delta)$. Therefore
\begin{equation}
D\varphi_\delta^h(r,t)=0,\quad \forall (r,t)\in P,\quad 0<h<\frac12\eta_2(\delta)
\label{4-12}
\end{equation}
and so \eqref{4-10} becomes
\begin{equation}
\delta\int_0^1ru_0(r)\,dr+\frac2{ T-2\delta}\iint_{Q_T}rF_\delta(u^h)\alpha
\Big(\frac{2t- T}{T-2\delta}\Big)\,dr\,dt=0,
\label{4-13}
\end{equation}
where $\eta(\delta)=\hbox{min}(\eta_1(\delta),\eta_2(\delta))$. Letting $h$ tend 
to zero, we have
\begin{equation}
\delta\int_0^1ru_0(r)\,dr+\frac2{ T-2\delta}\iint_{Q_T}rF_\delta(u)\alpha
\Big(\frac{2t- T} {T-2\delta}\Big)\,dr\,dt=0.
\label{4-14}
\end{equation}
From the definition of $F_\delta(s)$ and $H_\delta(s)$, it is easily seen that
$$
F_\delta(u(r,t))\to-\chi_E(r,t)u(r,t)\quad (\delta\to0)
$$
and so by letting $\delta$ tend to zero in \eqref{4-14}, we have
$$
\iint_E|u(r,t)|\alpha\Big(\frac{2t- T} T\Big)\,dr\,dt=0,
$$
which contradicts the fact that $\alpha\big(\frac{2t- T}T\big)>0$ for $0<t<T$. 
We have thus proved the theorem.
\end{proof}

\begin{lemma}\label{lem4.1}
Let $u$ be the limit function of the approximate solutions obtained
above. Then the following integral
inequality holds
$$
\int_0^1ru^{2-n}dr+\iint_{Q_t} r\Big(\frac{\partial V}{\partial r}\Big)^2\,dr\,ds
\leq\int_0^1ru_0^{2-n}dr.
$$
\end{lemma}

\begin{proof}
Let $u_\varepsilon$ be the solution of the problem \eqref{2-1}-\eqref{2-3}.
 Denote
$$
g_\varepsilon(u)=\int_0^u\frac{dr}{(|r|^2+\varepsilon)^{n/2}},\quad
G_\varepsilon(u)=\int_0^ug_\varepsilon(r)\,dr.
$$
Multiplying both sides of the equation \eqref{2-1} by $g_\varepsilon(u_\varepsilon)$,
 and then integrating over $Q_t$, we obtain
\begin{equation}
\int_0^1(r+\varepsilon)G_\varepsilon(u_\varepsilon(r,t))\,dr
+\iint_{Q_t}(r+\varepsilon)\frac{\partial W}{\partial r}\frac {\partial u_\varepsilon}{\partial r}\,dr\,ds
=\int_0^1(r+\varepsilon)G_\varepsilon(u_{0\varepsilon}(r))\,dr.
\label{5-1}
\end{equation}
Integrating by parts, we obtain
$$
\int_0^1(r+\varepsilon)G_\varepsilon(u_\varepsilon(r,t))\,dr
+\iint_{Q_t}(r+\varepsilon)\big(\frac{\partial V}{\partial r}\big)^2\,dr\,ds
=\int_0^1G_\varepsilon(u_{0\varepsilon}(r))\,dr.
$$
Letting $\varepsilon\to0$ and using the fact that
$G_\varepsilon(u_\varepsilon)\to u^{2-n}/(1-n)(2-n)$ and $u_\varepsilon\to u$
pointwise and the lower semi-continuity of the integrals, we immediately
get the conclusion of the lemma. The proof is complete.
\end{proof}

\begin{theorem}\label{th4.2}
Suppose that $u_0(r)>0$ and $n\geq 6$, then the weak solution $u$ satisfies 
$u(r,t)>0$.
\end{theorem}

\begin{proof} 
Since we have proved that $u(r,t)\geq0$, if the conclusion were false, then there
would exist a point $(r_0,t_0)\in Q_T$, such that $u(r_0, t_0)=0$. 
From the H\"{o}lder continuity of $ru$,
we see that
$$
ru(r, t_0)\leq C|r-r_0|^{1/4}.
$$
Since $n\geq 6$, we have
$$
\int_0^1ru(r,t_0)^{2-n}dr\geq C\int_0^1|r-r_0|^{(2-n)/4}dr=\infty.
$$
On the other hand, by Lemma \ref{lem4.1},
$$
\int_0^1ru(r,t_0)^{2-n}dr\leq C,
$$
which is a contradiction. The proof is complete.
\end{proof}

\section{Finite speed of propagation of perturbations}

As is well known, one of the important properties of solutions of the porous medium 
equation is the finite speed of propagation of perturbations. 
So from the point of view of physical background, it seems to be natural to 
investigate this property for thin film equation.  Bernis and Friedman \cite{bn3},
 Bernis \cite{bn4} considered this property for thin film
equation. On the other hand, the mathematical description of this property is that if
$\operatorname{supp} u_0$ is bounded, then for any $t>0$, 
$\operatorname{supp} u(\cdot,t)$ is also bounded.
So from the point of view of mathematics, this problem seems to be quite interesting.
We adopt the weighted energy method and the main technical tools are
weighted  Nirenberg's inequality and Hardy's inequality.

\begin{theorem}\label{th5.1}
Assume $0<n<1$, $u_0\in H^1_0(I)\cap H^2(I)$, $u_0\geq0$,
$\operatorname{supp}u_0\subset[r_1,r_2]$, $0<r_1<r_2<1$, and 
$u$ is the weak solution of the problem \eqref{1-1}-\eqref{1-3}, then for any
fixed $t>0$, we have
$$
\operatorname{supp}u(x,\cdot)\subset[r_1(t),r_2(t)]\cap[0,1],
$$
where $r_1(t)=r_1-C_1t^\gamma$, $r_2(t)=r_2+C_2t^\gamma$,
$C_1, C_2, \gamma>0$.
\end{theorem}

We need some uniform estimates on such approximate solutions $u_\varepsilon$.

\begin{lemma}\label{lem5.1}
Let $u$ be the limit function of the approximate solutions obtained above.
Then for any $y\in\mathbb R^+$,
the following integral inequality holds
\begin{align*}
&\int_0^1r(r-y)_+^\alpha u^{2-n}dr
+\frac 12\iint_{Q_t}r(r-y)_+^\alpha\Big(\frac{\partial^3 u}{\partial r^3}\Big)^2\,dr\,ds
\\
&\leq C\iint_{Q_t}r(r-y)_+^{\alpha-4}\Big(\frac{\partial u}{\partial r}\Big)^2\,dr\,ds
+C\iint_{Q_t}r(x-y)_+^{\alpha-2}\Big(\frac{\partial^2 u}{\partial r^2}\Big)^2\,dr\,ds
\\
&\quad+C\int_0^1r(r-y)_+^\alpha|u_0|^{2-n}dr
 +C\Big(\iint_{Q_t}r|u_0|^{2-n}\,dr\,ds\Big)^{1/2},
\end{align*}
where $C$ depends only on $n,u_0$ and $\alpha\ge 2p-1$, where $(r-y)_+$ denotes the
positive part of $r-y$.
\end{lemma}

\begin{proof}
Let $g_\varepsilon(u)$ and $G_\varepsilon(u)$ be defined as in the proof
of Lemma \ref{lem4.1}. Let $u_\varepsilon$ be the approximate solutions derived from
the problem \eqref{2-1}--\eqref{2-3}. Then, using the equation \eqref{2-1} and 
integrating by parts, we obtain
\begin{align*}
&\int_0^1r(r-y)_+^\alpha
G_\varepsilon(u_\varepsilon)\,dr
-\int_0^1r(r-y)_+^\alpha G_\varepsilon(u_0)\,dr
\\
&= -\iint_{Q_t}r_\varepsilon(|u_\varepsilon|^2+\varepsilon)^{n/2}
\frac{\partial W}{\partial r} \frac{\partial}{\partial r}\left[(r-y)_+^\alpha g_\varepsilon(u_\varepsilon)\right]\,dr\,ds
\\
&= -\iint_{Q_t}r_\varepsilon
\frac{\partial W}{\partial r} (r-y)_+^\alpha \frac{\partial u_\varepsilon}{\partial r} \,dr\,ds
\\
&\quad-\iint_{Q_t}r_\varepsilon(|u_\varepsilon|^2+\varepsilon)^{n/2}
\frac{\partial W}{\partial r}
\left[\alpha(r-y)_+^{\alpha-1} g_\varepsilon(u_\varepsilon)\right]\,dr\,ds
\\
&\equiv I_1+I_2.
\end{align*}
As for $I_1$, integrating by parts, we have
%%
\begin{align*}
&I_1\\
&= -\iint_{Q_t}r_\varepsilon
\frac{\partial W}{\partial r} (r-y)_+^\alpha \frac{\partial u_\varepsilon}{\partial r} \,dr\,ds
\\
&= \iint_{Q_t}Wr_\varepsilon\frac{\partial u_\varepsilon}{\partial r}\alpha(r-y)_+^{\alpha-1}\,dr\,ds
+\iint_{Q_t}W\frac{\partial}{\partial r}
 \Big(r_\varepsilon\frac{\partial u_\varepsilon}{\partial r}\Big)(r-y)_+^\alpha \,dr\,ds
\\
&= -\iint_{Q_t}r_\varepsilon\frac{\partial V}{\partial r}\frac{\partial^2 u_\varepsilon}{\partial r^2}\alpha(r-y)_+^{\alpha-1}\,dr\,ds
-\iint_{Q_t}r_\varepsilon\frac{\partial V}{\partial r}\frac{\partial u_\varepsilon}{\partial r}\alpha(\alpha-1)(r-y)_+^{\alpha-2}\,dr\,ds
\\
&\quad -\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
 \Big(\frac{\partial V}{\partial r}\Big)^2\,dr\,ds
-\iint_{Q_t}r_\varepsilon\alpha(r-y)_+^{\alpha-1}\frac{\partial V}{\partial r}V \,dr\,ds.
\end{align*}
Therefore,
\begin{align*}
&\int_0^1r_\varepsilon(r-y)_+^\alpha
G_\varepsilon(u_\varepsilon)\,dr
-\int_0^1r_\varepsilon(r-y)_+^\alpha G_\varepsilon(u_0)\,dr
+\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha\Big(\frac{\partial V}{\partial r}\Big)^2\,dr\,ds
\\
&= -\iint_{Q_t}r_\varepsilon\frac{\partial V}{\partial r}\frac{\partial^2 u_\varepsilon}{\partial r^2}\alpha(r-y)_+^{\alpha-1}\,dr\,ds
-\iint_{Q_t}r_\varepsilon\frac{\partial V}{\partial r}\frac{\partial u_\varepsilon}{\partial r}\alpha(\alpha-1)(r-y)_+^{\alpha-2}\,dr\,ds
\\
&\quad -\iint_{Q_t}r_\varepsilon\alpha(r-y)_+^{\alpha-1}\frac{\partial V}{\partial r} V\,dr\,ds
\\
&\quad -\iint_{Q_t}r_\varepsilon(|u_\varepsilon|^2+\varepsilon)^{n/2}
\frac{\partial W}{\partial r}
\left[\alpha(r-y)_+^{\alpha-1} g_\varepsilon(u_\varepsilon)\right]\,dr\,ds
\\
&\equiv I_a+I_b+I_c+I_d.
\end{align*}
H\"{o}lder's inequality yields
\begin{gather*}
|I_a|\leq
\frac 18\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
\Big(\frac{\partial V}{\partial r}\Big)^2\,dr\,ds
+C\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-2}
 \Big(\frac{\partial^2 u_\varepsilon}{\partial r^2}\Big)^2\,dr\,ds,
\\
|I_b| \leq \frac 18\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha\left(\frac{\partial V}{\partial r}\right)^2\,dr\,ds
+C_1\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-4}
\Big(\frac{\partial u_\varepsilon}{\partial r}\Big)^2\,dr\,ds,
\\
|I_c| \leq\frac 14\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
\Big(\frac{\partial V}{\partial r}\Big)^2\,dr\,ds
+C_2\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-2}V^2\,dr\,ds.
\end{gather*}
Noticing that
$$
(|u_\varepsilon|^2+\varepsilon)^{n/2}|g_\varepsilon(u_\varepsilon)|
\leq\frac2{1-n}|u_\varepsilon|,
$$
using \eqref{2-5}, we have
\begin{align*}
|I_d|&\leq  C\Big(\iint_{Q_t}r_\varepsilon(|u_\varepsilon|^2+\varepsilon)^{n/2}
\Big(\frac{\partial W}{\partial r}\Big)^2\,dr\,ds\Big)^{1/2}\\
&\quad\times \Big(\iint_{Q_t}r_\varepsilon(|u_\varepsilon|^2
 +\varepsilon)^{n/2}g_\varepsilon(u_\varepsilon)^2\,dr\,ds\Big)^{1/2}
\\
&\leq C\Big(\iint_{Q_t}r_\varepsilon|u_\varepsilon|^{2-n}\,dr\,ds\Big)^{1/2}.
\end{align*}
Summing up, we have
\begin{align*}
&\int_0^1r_\varepsilon(r-y)_+^\alpha
G_\varepsilon(u_\varepsilon)\,dr
-\int_0^1r_\varepsilon(r-y)_+^\alpha G_\varepsilon(u_0)\,dr\\
&+\frac12\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
\Big(\frac{\partial V}{\partial r}\Big)^2\,dr\,ds
\\
&\leq C\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-2}
\Big(\frac{\partial^2 u_\varepsilon}{\partial r^2}\Big)^2\,dr\,ds
+C_1\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-4}
\Big(\frac{\partial u_\varepsilon}{\partial r}\Big)^2\,dr\,ds
\\
&\quad +C_2\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-2}V^2\,dr\,ds
+C_3\Big(\iint_{Q_t}r_\varepsilon|u_\varepsilon|^{2-n}\,dr\,ds\Big)^{1/2}.
\end{align*}
A simple calculation shows that
\begin{align*}
&\frac12\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
 \Big(\frac{\partial V}{\partial r}\Big)^2\,dr\,ds
\\
&= \frac12\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
\bigg(\Big(\frac{\partial^3u_\varepsilon}{\partial r^3}\Big)^2+
2\frac{\partial^3u_\varepsilon}{\partial r^3}\frac{\partial}{\partial r}
\Big(\frac1{r_\varepsilon}\frac{\partial u_\varepsilon}{\partial r}\Big)
+\Big(\frac{\partial}{\partial r}\Big(\frac1{r_\varepsilon}
 \frac{\partial u_\varepsilon}{\partial r}\Big)\Big)^2\bigg)\,dr\,ds
\\
&= \frac12\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
 \Big(\frac{\partial^3u_\varepsilon}{\partial r^3}\Big)^2\,dr\,ds
+\frac12\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
\Big(\frac{\partial}{\partial r}\Big(\frac1{r_\varepsilon}
 \frac{\partial u_\varepsilon}{\partial r}\Big)\Big)^2\,dr\,ds
\\
&\quad +\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
\frac{\partial^3u_\varepsilon}{\partial r^3}\frac{\partial}{\partial r}
\Big(\frac1{r_\varepsilon}\frac{\partial u_\varepsilon}{\partial r}\Big)\,dr\,ds.
\end{align*}
Note that
\begin{align*}
&\Big|\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
 \frac{\partial^3u_\varepsilon}{\partial r^3}\frac{\partial}{\partial r}
\Big(\frac1{r_\varepsilon}\frac{\partial u_\varepsilon}{\partial r}\Big)\,dr\,ds\Big|
\\
&= \Big|\iint_{Q_t}(r-y)_+^\alpha\frac{\partial^3u_\varepsilon}{\partial r^3}\frac{\partial^2u_\varepsilon}{\partial r^2}\,dr\,ds
-\iint_{Q_t}\frac1{r_\varepsilon}(r-y)_+^\alpha
 \frac{\partial^3u_\varepsilon}{\partial r^3}
\frac{\partial u_\varepsilon}{\partial r}\,dr\,ds\Big|
\\
&\leq \frac18\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
\Big(\frac{\partial^3u_\varepsilon}{\partial r^3}\Big)^2\,dr\,ds
+C\iint_{Q_t}\frac{r_\varepsilon}{r_\varepsilon^2}(r-y)_+^\alpha
\Big(\frac{\partial^2u_\varepsilon}{\partial r^2}\Big)^2\,dr\,ds
\\
&\quad +\frac18\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
\Big(\frac{\partial^3u_\varepsilon}{\partial r^3}\Big)^2\,dr\,ds
+C\iint_{Q_t}\frac{r_\varepsilon^2}{r_\varepsilon^4}(r-y)_+^\alpha
\Big(\frac{\partial u_\varepsilon}{\partial r}\Big)^2\,dr\,ds
\\
&\leq \frac14\iint_{Q_t}r_\varepsilon(r-y)_+^\alpha
\Big(\frac{\partial^3u_\varepsilon}{\partial r^3}\Big)^2\,dr\,ds
+C\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-2}
\Big(\frac{\partial^2u_\varepsilon}{\partial r^2}\Big)^2\,dr\,ds
\\
&\quad +C\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-4}
\Big(\frac{\partial u_\varepsilon}{\partial r}\Big)^2\,dr\,ds.
\end{align*}
On the other hand, we have
\begin{align*}
&\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-2}V^2\,dr\,ds
\\
&\leq 2\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-2}
 \Big(\frac{\partial^2u_\varepsilon}{\partial r^2}\Big)^2\,dr\,ds
+2\iint_{Q_t}r_\varepsilon(r-y)_+^{\alpha-4}
\Big(\frac{\partial u_\varepsilon}{\partial r}\Big)^2\,dr\,ds.
\end{align*}
Letting $\varepsilon\to0$,
we immediately get the desired conclusion and complete the proof of the
lemma.
\end{proof}

\begin{proof}[Proof of Theorem \ref{th5.1}]
For any $y\geq r_2$, Lemma \ref{lem5.1} and Hardy's inequality \cite{HG} 
imply that for any $t\in [0,T]$,
\begin{equation}
\begin{aligned}
&\int_0^1r(r-y)_+^\alpha u^{2-n}dr+\frac 12\iint_{Q_t}r(r-y)_+^\alpha
\big|\frac{\partial^3u}{\partial r^3}\big|^2\,dr\,ds
\\
&\leq C\iint_{Q_t}r(r-y)_+^{\alpha-4}\big|\frac{\partial u}{\partial r}\big|^2\,dr\,ds
+C\iint_{Q_t}r(r-y)_+^{\alpha-2}\big|\frac{\partial^2u}{\partial r^2}\big|^2\,dr\,ds
\\
&\leq  C\iint_{Q_t}(r-y)_+^{\alpha-2}\big|\frac{\partial^2u}{\partial r^2}\big|^2\,dr\,ds.
\end{aligned}\label{5-2}
\end{equation}
For any positive number $m$, define
\[
f_m(y)=\int_0^t\int_0^1r(r-y)_+^m\big|\frac{\partial^3u}{\partial r^3}\big|^2\,dr\,ds,
\quad
f_0(y)=\int_0^t\int_y^1r\big|\frac{\partial^3u}{\partial r^3}\big|^2\,dr\,ds.
\]
By $y>r_2>0$, then, weighted Nirenberg's inequality \cite{BF} and
estimate \eqref{5-2} imply that
\begin{align*}
&f_{2p+1}(y)\\
&\leq  C\iint_{Q_t}(r-y)_+^{2p-1}\big|\frac{\partial^2u}{\partial r^2}\big|^2\,dr\,ds
\\
&\leq C_1\int_0^t
\Big(\int_0^1(r-y)_+^{2p-1}\big|\frac{\partial^3u}{\partial r^3}\big|^2dr\Big)^a
\Big(\int_0^1 (r-y)_+^{2p-1}| u|^qdr\Big)^{2(1-a)/q}ds
\\
&\leq C\sup_{0<t<T}\Big(\int_0^1r(r-y)_+^{2p-1}| u|^qdr\Big)^{2(1-a)/q}t^{1-a}
\Big(\iint_{Q_t}(r-y)_+^{2p-1}\big|\frac{\partial^3u}{\partial r^3}\big|^2\,dr\,ds
\Big)^a.
\end{align*}
Using \eqref{5-2} and Hardy's inequality, we have
$$
\sup_{0<t<T}\int_0^1r(r-y)_+^{2p-1}|u|^qdr
\leq C\iint_{Q_t}(r-y)_+^{2p-1}\big|\frac{\partial^3u}{\partial r^3}\big|^2\,dr\,ds
$$
and hence
$$
f_{2p+1}(y)
\leq C t^{1-a}\Big(\iint_{Q_t}(r-y)_+^{2p-1}\big|\frac{\partial^3u}{\partial r^3}
\big|^2\,dr\,ds\Big)^{a+2(1-a)/q},
$$
where $q=2-n$ and
$$
a=\frac{\frac12-\frac1p-\frac1q}{\frac12-\frac3{2p}-\frac1q}.
$$
Denote $\lambda=1-a,\mu=a+2(1-a)/q$, then $\lambda>0$, $1<\mu$. Applying
H\"{o}lder's inequality, we have
%%
\begin{align*}
f_{2p+1}(y)
&\leq C t^\lambda\Big[\iint_{Q_t}(r-y)_+^{2p-1}
\big|\frac{\partial^3u}{\partial r^3}\big|^2\,dr\,ds\Big]^\mu
\\
&\leq C t^\lambda\Big[\iint_{Q_t}r(r-y)_+^{2p-2}
\big|\frac{\partial^3u}{\partial r^3}\big|^2\,dr\,ds\Big]^\mu
\\
&\leq C t^\lambda\Big[\iint_{Q_t}r(r-y)_+^{2p+1}\big|
\frac{\partial^3u}{\partial r^3}\big|^2\,dr\,ds\Big]^{(2p-2)\mu/(2p+1)}\\
&\quad\times
\Big[\int_0^t\int_y^1r\big|\frac{\partial^3u}{\partial r^3}\big|^2\,dr\,ds
 \Big]^{3\mu/(2p+1)}
\\
&\leq C t^\lambda[f_{2p+1}(y)]^{(2p-2)\mu/(2p+1)}
[f_0(y)]^{3\mu/(2p+1)}.
\end{align*}
Therefore,
$$
f_{2p+1}(y)\leq Ct^{\lambda/\sigma}[f_0(y)]^{3\mu/(2p+1)\sigma},
\quad
\sigma=1-\frac{2p-2}{2p+1}\mu>0.
$$
Using H\"{o}lder's inequality again, we obtain
$$
f_1(y)\leq[f_0(y)]^{2p/2p+1}[f_{2p+1}(y)]^{1/2p+1}\leq Ct^\gamma[f_0(y)]^{1+\theta},
$$
where
$$
\gamma=\frac\lambda{(2p+1)\sigma},\quad
\theta=\frac{2\mu}{(2p+1)^2\sigma}-\frac1{2p+1}>0.
$$
Noticing that $f_1'(y)=-f_0(y)$, we obtain
$$
f_1'(y)\leq -Ct^{-\gamma/(\theta+1)}[f_1(y)]^{1/(\theta+1)}.
$$
If $f_1(r_2)=0$, then $\hbox{supp}u\subset[0,r_2]$. If $f_1(r_2)>0$, then there
exists a maximal interval $(r_2,r_2^*)$ in which $f_1(y)>0$ and
$$
\left[f_1(y)^{\theta/(\theta+1)}\right]'=\frac\theta{\theta+1}\frac{f_1'(y)}
{[f_1(y)]^{1/(\theta+1)}}\leq-Ct^{-\gamma/(\theta+1)}.
$$
Integrating the above inequality over $(r_2,r_2^*)$, we have
$$
f_1(r_2^*)^{\theta/(\theta+1)}-f_1(r_2)^{\theta/(\theta+1)}
\leq -Ct^{-\gamma/(\theta+1)}(r_2^*-r_2),
$$
which implies
$$
r_2^*\leq r_2+Ct^\gamma(f_0(r_2))^\theta.
$$
Therefore.
$$
\sup \operatorname{supp} u(\cdot,t)\le r_2+Ct^\gamma\equiv r_2(t).
$$
We have thus completed the proof of Theorem \ref{th5.1}.
\end{proof}

\subsection*{Acknowledgment}
The authors would like to express their gratitude to the anonymous  referees
for  their valuable suggestions which lead to improvements in this article.


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\end{document}