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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 223, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/223\hfil A sixth-order parabolic equation]
{A sixth-order parabolic equation describing continuum evolution of 
 film free surface}

\author[X. Zhao \hfil EJDE-2014/223\hfilneg]
{Xiaopeng Zhao}  % in alphabetical order

\address{Xiaopeng Zhao \newline
School of Science, Jiangnan University, Wuxi 214122, China}
\email{zhaoxiaopeng@sina.cn}

\thanks{Submitted June 24, 2014. Published October 21, 2014.}
\subjclass[2000]{35B65, 35K35, 35K55}
\keywords{Regularity; sixth-order parabolic equation; existence; 
\hfill\break\indent Campanato space}

\begin{abstract}
 In this article, we study the regularity of solutions for a sixth-order
 parabolic equation. Based on the Schauder type estimates and Campanato
 spaces, we prove the  existence of classical global solutions.
\end{abstract}

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

\section{Introduction} \label{sect1}

In the previous fifteen-twenty years, essentially sixth-order nonlinear parabolic
partial differential equationa, as models for applications in mechanics and physics,
have become more common in the literature on pure and applied PDEs.
Evans, Galaktionov and King\cite{Evans1,Evans2} studied the blow-up behavior and
global similarity patterns of solutions  for a  sixth order thin film equations
containing an unstable (backward parabolic) second-order term
$$
u_t=\nabla\cdot(|u|^n\nabla\Delta^2u)-\Delta(|u|^{p-1}u),\quad n>0,p>0,
$$
with bounded integrable initial data. J\"{u}ngel and Milisi\'{c}\cite{Junge1}
proved the global in time existence of weak nonnegative solutions to the
following initial value problem in one space dimension with periodic boundary
conditions:
\begin{gather*}
n_t=L[n]:=\Big[n\Big(\frac1n(n(\log n)_{xx})_{xx}+\frac12((\log n)_{xx})^2
\Big)_x\Big]_x,\quad x\in\mathbb{T}, t>0,
\\
n(x,0)=n_0(x),\quad x\in\mathbb{T}.
\end{gather*}
 In \cite{Korzec}, by an extension of the method of matched asymptotic expansions,
 Korzec, Evans, M\"{u}nch and Wangner derived the stationary solutions of a
1D driven sixth order Cahn-Hilliard equation which arises as a model for
epitaxially growing nano-structures. Li and Liu\cite{Liu} studied the radial
symmetric solutions for the following sixth order thin film equation:
$$
u_t=\nabla\cdot[|u|^n\nabla\Delta^2u],\quad x\text{ in the unit ball of }\mathbb{R}^2,
\; n>0.
$$
Recently, based on the Landau-Ginzburg theory, Pawlow and Zajaczkowski \cite{Paw}
proved that a 3D sixth order Cahn-Hilliard equation
 under consideration is well posed in the sense that it admits a unique global
smooth solution which depends continuously on the initial datum. We also refer
the solvability conditions in $H^6(\mathbb{R}^3)$ for sixth order linearized
Cahn-Hilliard problem is also studied in \cite{Vougalter}.

In the study of a thin, solid film grown on a solid substrate, in order to describe
the continuum evolution of the film free surface, there arise a classical
surface\-diffusion equation (see \cite{Golovin})
\begin{equation} \label{1-0}
v_n=\mathcal{D}\Delta_S\mu=\mathcal{D}\Delta_S(\mu_{\gamma}+\mu_{w})
=\mathcal{D}\Delta_S(\tilde{\gamma}_{\alpha\beta}C_{\alpha\beta}+\nu\Delta^2u+\mu_w),
\end{equation}
where $v_n$ is the normal surface velocity, $\mathcal{D}=D_SS_0\Omega_0V_0/(RT)^{23}$
($D_s$ is the surface diffusivity, $S_0$ is the number of atoms per unit area on
the surface, $\Omega_0$ is the atomic volume, $V_0$ is the molar volume of lattice
cites in the film, $R$ is the universal gas constant and $T$ is the absolute
temperature), $\Delta_S$
is the surface Laplace operator, $\nu$ is the regularization coefficient that
measures the energy of edges and corners, $C_{\alpha\beta}$ is the surface curvature
tensor and $\mu_w$ being an exponentially decaying function of $u$ that has a
singularity at $u\to 0$ (see \cite{Golovin}).

In the small-slop approximation, in the particular cases of high-symmetry
orientations of a crystal with cubic symmetry, then the evolution equation
\eqref{1-0} for the film thickness can be written in the following form
\begin{equation}\label{1-00}
\frac{\partial u}{\partial t}=D\left\{D^5u+D^3u-D[|Du|^2D^2u]+D[w_0(u)+w_2(u)|Du|^2
+w_3(u)D^2u]\right\},
\end{equation}
where $w_{0,2,3}(h)$ are smooth functions, respectively
$[w_3(h_0)=0,2w_2=\frac{d w_3}{dh}]$.

We study the sixth-order nonlinear parabolic equation
\begin{equation}\label{1-1}
\begin{aligned}
\frac{\partial u}{\partial t}
&=D\big\{m(u)\big[D^5u+D^3u-D(|Du|^{2}D^2u) \\
&\quad +D(w_0(u)+w_2(u)|Du|^2+w_3(u)D^2u)\big]\big\},
\end{aligned}
\end{equation}
where $(x,t)\in Q_T$, $Q_T\equiv (0,1)\times(0,T)$.
On the basis of physical consideration, Equation
\eqref{1-1} is supplemented by the following boundary conditions
\begin{equation}
\label{1-2}
Du(x,t)=D^3u(x,t)=D^5u(x,t)=0,\quad x=0,1,
\end{equation}
and initial condition
\begin{equation}
\label{1-3}
u(x,0)=u_0(x),\quad x\in[0,1].
\end{equation}

Our main purpose is to establish the  existence of classical global solutions under
much general assumptions. The main difficulties for treating the problem
\eqref{1-1}-\eqref{1-3} are caused by the nonlinearity of the principal part and
the lack of maximum principle.
Due to the nonlinearity of the principal part, there are more difficulties in
establishing the global existence of classical solutions. Our method for
investigating the regularity of solutions is based on uniform Schauder type
estimates for local in time solutions, which are relatively less used for
such kind of parabolic equations of sixth order. Our approach lies in the
combination of the energy techniques with
some methods based on the framework of Campanato spaces.
Now, we give the main results in this paper.

\begin{theorem} \label{thm1.1}
Assume that
\begin{itemize}
\item $u_0\in C^{6+\alpha}[0,1]$, $\alpha\in[0,1)$,
$D^iu_0(0)=D^iu_0(1)=0$ $(i=0,2,4)$;

 \item $m(s)\in C^{1+\alpha}(\mathbb{R})$, $\inf_{s\in\mathbb{R}} m(s)=m_0>0$;

\item $w_3(h_0)=0$, $2w_2(h)=w_3'(h)$, $W_0(s)=\int_0^sw_0(s)ds\geq\frac34[w_3(s)]^2$.
\end{itemize}
  Then  \eqref{1-1}-\eqref{1-3} admits a unique
classical solution $u(x,t)\in C^{6+\alpha,1+\frac{\alpha}6}(\bar{Q}_T)$.
\end{theorem}

\begin{remark} \rm
During the past few years, many authors studied the properties of solutions
(such as blow-up behavior and global
similarity patterns of solutions, weak nonnegative solutions, radial symmetric
solutions, stationary solutions, solvability conditions and so on) for
sixth-order parabolic equation, but only a few papers were devoted to
the existence of classical solution for sixth order parabolic equation.
In this article, based on the Schauder type estimates, Campanato
spaces and a result in \cite{Liu2}, we consider the existence of classical
solutions for a sixth-order parabolic equation which was introduced in \cite{Golovin}.
\end{remark}

\section{Proof of the main result}

Based on the classical approach, it is easy for us  to conclude that 
 problem \eqref{1-1}-\eqref{1-3} admits a unique classical
solution local in time. So, it is sufficient to make a priori estimates. 
First of all, we give the H\"{o}lder
norm estimate on the local in time solutions.

\begin{lemma} \label{lem2.1}
Assume that $u$ is a smooth solution of the problem \eqref{1-1}-\eqref{1-3}. 
Then there exists a constant $C$ depending only on the known quantities, 
such that for any $(x_1, t_1)$, $(x_2, t_2) \in Q_T$ and some $0 < \alpha < 1$,
\begin{gather*}
|u(x_1,t_1)-u(x_2,t_2)|\leq C(|t_1-t_2|^{\frac{\alpha}6}+|x_1-x_2|^{\alpha}),\\
|Du(x_1,t_1)-Du(x_2,t_2)|\leq C(|t_1-t_2|^{\frac1{12}}+|x_1-x_2|^{1/2}).
\end{gather*}
\end{lemma}

\begin{proof} 
Now, we set
$$
F(t)=\int_0^1\Big(\frac12|D^2u|^2-\frac12|Du|^2+\frac1{12}|Du|^4+W_0(u)
-\frac12w_3(u)|Du|^2\Big)dx.
$$
Integrating by parts, from the boundary value condition \eqref{1-2}, we deduce that
\begin{align*}
\frac {d}{dt}F(t)
&=\int_0^1[D^2uD^2u_t-DuDu_t+\frac13|Du|^2DuDu_t+w_0(u)u_t\\
&\quad -w_3(u)DuDu_t-\frac12w_3'(u)|Du|^2u_t]dx\\
&=\int_0^1[D^2uD^2u_t-DuDu_t+\frac13|Du|^2DuDu_t+w_0(u)u_t\\
&\quad +w_3(u)D^2uu_t+\frac12w_3'(u)|Du|^2u_t]dx\\
&=\int_0^1\Big[D^4u+D^2u-\frac13D(|Du|^2Du)+w_0(u)\\
&\quad +w_2(u)|Du|^2+w_3(u)D^2u\Big]u_tdx \\
&=-\int_0^1m(u)\Big|D\Big[D^4u+D^2u-\frac13D(|Du|^2Du)+w_0(u) \\
&\quad +w_2(u)|Du|^2+w_3(u)D^2u\Big]\Big|^2dx 
\leq 0.
\end{align*}
Hence $F(t)\leq F(0)$, that is
\begin{equation}\label{2-3}
\begin{aligned}
&\int_0^1(\frac12|D^2u|^2+\frac1{12}|Du|^4+W_0(u))dx\\
&\leq F(0)+\frac12\int_0^1(|Du|^2+w_3(u)|Du|^2)dx.
\end{aligned}
\end{equation}
It then from Poincar\'{e}'s inequality and the boundary value
 condition \eqref{1-2} follows that
\begin{equation} \label{yue-1}
\int_0^1|Du|^2dx\leq\frac1{\pi^2}\int_0^1|D^2u|^2dx.
\end{equation}
On the other hand, we have
\begin{equation}\label{yue-2}
\int_0^1w_3(u)|Du|^2dx\leq\frac16\int_0^1|Du|^4dx+\frac32\int_0^1[w_3(u)]^2dx.
\end{equation}
Adding \eqref{2-3}, \eqref{yue-1} and \eqref{yue-2}, noticing that 
$W_0(u)\geq\frac34[w_3(u)]^2$, we obtain
\begin{equation} \label{2-8}
\sup_{0<t<T}\int_0^1|D^2u|^2dx\leq C.
\end{equation}
Combing \eqref{yue-1} and \eqref{2-8}  gives
\begin{equation} \label{2-7}
\sup_{0<t<T}\int_0^1|Du|^2dx\leq C,
\end{equation}
The integration of \eqref{1-1} over the interval $(0, 1)$ yields
 $\int_0^1\frac{\partial u}{\partial t}dx=0$, hence we obtain
$$
\int_0^1u(x,t)dx=\int_0^1u_0(x)dx.
$$
Applying the mean value theorem, we see that for some $x_t^*\in(0,1)$
$$
u(x_t^*,t)=\int_0^1u_0(x)dx=M.
$$
Then
$$
|u(x,t)|\leq|u(x,t)-u(x_t^*,t)|+|u(x_t^*,t)|\leq|\int_{x_t^*}^xDu(t,y)dy|+M.
$$
Taking this into account, we deduce that
\begin{equation} \label{2-9}
\sup_{Q_T}|u(x,t)|\leq C,
\end{equation}
On the other hand, a simple calculation shows that
\begin{equation} \label{wuxi-1}
\int_0^1u^2dx\leq \sup_{Q_T}|u(x,t)|^2\leq C.
\end{equation}
Combing \eqref{wuxi-1}, \eqref{2-7} and \eqref{2-8} together, 
using Sobolev's embedding theorem, we derive that
\begin{equation}\label{wuxi-2}
\sup_{Q_T}|Du(x,t)|\leq C\Big(\int_0^1u^2dx+\int_0^1|Du|^2dx
+\int_0^1|D^2u|^2dx\Big)^{1/2}\leq C.
\end{equation}
Multiplying both sides of  \eqref{1-1} by $D^4u$, integrating the resulting 
relation with respect to $ x$
over $(0, 1)$, integrating by parts, we have
\begin{equation}
\begin{aligned}
&\frac12\frac d{dt}\int_0^1|D^2u|^2dx+\int_0^1m(u)|D^5u|^2dx\\
&=-\int_0^1m(u)D^3uD^5u\,dx+\int_0^1m(u)D(|Du|^2D^2u)D^5u\,dx\\
&\quad -\int_0^1m(u)Dw_0(u)D^5u\,dx -\frac12\int_0^1m(u)D(w_3'(u)|Du|^2)D^5du\,dx\\
&\quad -\int_0^1m(u)D(w_3(u)D^2u)D^5u\,dx
\\
&=-\int_0^1m(u)D^3uD^5u\,dx+\int_0^1m(u)|Du|^2D^3uD^5u\,dx\\
&\quad +2\int_0^1m(u)Du|D^2u|^2D^5u\,dx
-\int_0^1m(u)w_0'(u)DuD^5u\,dx\\
&\quad -\frac12\int_0^1m(u)w_3'' (u)|Du|^3D^5u\,dx
 -2\int_0^1m(u)w_3'(u)DuD^2uD^5u\,dx\\
&\quad -\int_0^1m(u)w_3(u)D^3uD^5u\,dx \\
&=:I_1+I_2+I_3+I_4+I_5+I_6+I_7.
\end{aligned}\label{yue-3}
\end{equation}
By Nirenberg's inequality, we derive that
\begin{align*}
&\int_0^1|D^3u|^2dx\\
&\leq  \Big(C'\Big(\int_0^1|D^5u|^2dx\Big)^{1/6}
\Big(\int_0^1|D^2u|^2dx\Big)^{1/3}+C''
\Big(\int_0^1|D^2u|^2dx\Big)^{1/2}\Big)^2 \\
&\leq \varepsilon\int_0^1|D^5u|^2dx+C_{\varepsilon}.
\end{align*}
and
\begin{align*}
&\int_0^1|D^2u|^4dx\\
&\leq \Big(C'\Big(\int_0^1|D^5u|^2dx\Big)^{1/24}
 \Big(\int_0^1|D^2u|^2dx\Big)^{11/24} +C''
\Big(\int_0^1|D^2u|^2dx\Big)^{1/2}\Big)^4\\
&\leq \varepsilon\int_0^1|D^5u|^2dx+C_{\varepsilon}.
\end{align*}
Hence
\begin{gather}\label{2-11}
\begin{aligned}
I_1 &\leq \sup_{Q_T}|m(u)|\int_0^1|D^3uD^5u|dx\leq C\int_0^1|D^3uD^5u|dx\\
&\leq \varepsilon\int_0^1|D^5u|^2dx+C_{\varepsilon}\int_0^1|D^3u|^2dx\\
&\leq 2\varepsilon\int_0^1|D^5u|^2dx+C.
\end{aligned}\\
\label{yue-4}
\begin{aligned}
I_2&\leq \sup_{Q_T}|m(u)(Du)^2|\int_0^1|D^3uD^5u|dx\\
&\leq C\int_0^1|D^3uD^5u|dx\\
&\leq  2\varepsilon\int_0^1|D^5u|^2dx+C.
\end{aligned} \\
\label{yue-5}
\begin{aligned}
I_3&\leq 2\sup_{Q_T}|m(u)Du|\int_0^1|(D^2u)^2D^5u|dx\\
& \leq\varepsilon\int_0^1|D^5u|^2dx+C_{\varepsilon}\int_0^1|D^2u|^4dx
\leq  2\varepsilon\int_0^1|D^5u|^2dx+C.
\end{aligned} \\
 \label{yue-6}
\begin{aligned}
I_4
&\leq \sup_{Q_T}|m(u)w_0'(u)|\int_0^1|DuD^5u|dx\leq C\int_0^1|DuD^5u|dx\\
&\leq  \varepsilon\int_0^1|D^5u|^2dx+C_{\varepsilon}\int_0^1|Du|^2dx
\leq \varepsilon\int_0^1|D^5u|^2dx+C.
\end{aligned} \\
\label{yue-7}
\begin{aligned}
I_5 &\leq \sup_{Q_T}|m(u)w_3'' (u)|Du|^2|\int_0^1|DuD^5u|dx\leq C\int_0^1|DuD^5u|dx\\
&\leq  \varepsilon\int_0^1|D^5u|^2dx+C_{\varepsilon}\int_0^1|Du|^2dx
 \leq \varepsilon\int_0^1|D^5u|^2dx+C.
\end{aligned} \\
\label{yue-8}
\begin{aligned}
I_6&\leq 2\sup_{Q_T}|m(u)w_3{'}(u)Du|\int_0^1|D^2uD^5u|dx
 \leq C\int_0^1|D^2uD^5u|dx \\
&\leq  \varepsilon\int_0^1|D^5u|^2dx+C_{\varepsilon}\int_0^1|D^2u|^2dx
\leq \varepsilon\int_0^1|D^5u|^2dx+C.
\end{aligned} \\
\label{2-11b}
\begin{aligned}
I_7 &\leq  \sup_{Q_T}|m(u)w_3(u)|\int_0^1|D^3uD^5u|dx
 \leq C\int_0^1|D^3uD^5u|dx \\
&\leq \varepsilon\int_0^1|D^5u|^2dx+C_{\varepsilon}\int_0^1|D^3u|^2dx
\leq 2\varepsilon\int_0^1|D^5u|^2dx+C.
\end{aligned}
\end{gather}
Summing up, noticing that $m(s)\geq m_0>0$, we obtain
$$
\frac d{dt}\int_0^1|D^2u|^2dx+(2m_0-22\varepsilon)\int_0^1|D^5u|^2dx\leq C,
$$
where $\varepsilon$ is small enough, it satisfies $2m_0-10\varepsilon>0$. 
Therefore,
\begin{equation} \label{2-13}
\int   \int_{Q_T}|D^5u|^2dxdt\leq C.
\end{equation}
Multiplying both sides of the equation \eqref{1-1} by $D^6u$, integrating 
the resulting relation with respect to $ x$
over $(0, 1)$, after integrating by parts, and using the boundary value conditions, 
we have
\begin{align*} %\begin{aligned}
&\frac12\frac d{dt}\int_0^1|D^3u|^2dx
+\int_0^1D(m(u)D^5u)D^6u\,dx+\int_0^1D(m(u)D^3u)D^6u\,dx
\\
&=\int_0^1D[m(u)D(|Du|^2D^2u)]D^6u\,dx\\
&\quad-\int_0^1D[m(u)D(w_0(u)+w_2(u)|Du|^2+w_3(u)D^2u)]D^6u\,dx.
\end{align*}
Simple calculations show that
%\label{yueyue-1}
\begin{align*}
&\frac12\frac d{dt}\int_0^1|D^3u|^2dx+\int_0^1m(u)|D^6u|^2dx
\\
&=-\int_0^1m'(u)DuD^5uD^6u\,dx-\int_0^1m(u)D^4uD^6u\,dx
 -\int_0^1m'(u)DuD^3uD^6u\,dx \\
&\quad +\int_0^1m(u)(|Du|^2D^4u\,dx+6DuD^2uD^3u+2|D^2u|^2D^2u)D^6u\,dx\\
&\quad +\int_0^1m'(u)Du(|Du|^2D^3u+2Du|D^2u|^2)D^6u\,dx\\
&\quad -\int_0^1m(u)(w_0'(u)D^2u+w_0'' (u)|Du|^2)D^6u\,dx
 -\int_0^1m{'}(u)w_0'(u)|Du|^2D^6u\,dx\\
&\quad -\int_0^1m'(u)Du(w_2'(u)|Du|^2Du+2w_2(u)DuD^2u)D^6u\,dx\\
&\quad -\int_0^1m(u)[w_2''(u)|Du|^4+5w_2'(u)|Du|^2D^2u+2w_2(u)|D^2u|^2\\
&\quad +2w_2(u)DuD^3u]D^6u\,dx\\
&\quad -\int_0^1m(u)[w_3'' (u)DuD^2u+2w_3'(u)D^3u+w_3'(u)D^4u]D^6u\,dx\\
&\quad -\int_0^1m'(u)Du(w_3'(u)D^2u+w_3(u)D^3u)D^6u\,dx\\
&=:I_8+I_9+I_{10}+I_{11}+I_{12}+I_{13}+I_{14}+I_{15}+I_{16}+I_{17}+I_{18}.
\end{align*}
By Nirenberg's inequality, we deduce that
\begin{align*}
&\int_0^1|D^5u|^2dx\\
&\leq  \Big(C'\Big(\int_0^1|D^6u|^2dx\Big)^{\frac3{8}}
\Big(\int_0^1|D^2u|^2dx\Big)^{\frac{1}{8}}+C'' 
\Big(\int_0^1|D^2u|^2dx\Big)^{1/2}\Big)^2\\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}.
\end{align*}
\begin{align*}
&\int_0^1|D^4u|^2dx\\
&\leq  \Big(C'\Big(\int_0^1|D^6u|^2dx\Big)^{\frac1{4}}
\Big(\int_0^1|D^2u|^2dx\Big)^{\frac{1}{4}}
+C'' \Big(\int_0^1|D^2u|^2dx\Big)^{1/2}\Big)^2 \\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}.
\end{align*}
\begin{align*}
& \int_0^1|D^3u|^2dx\\
&\leq  \Big(C'\Big(\int_0^1|D^6u|^2dx\Big)^{1/8}
\Big(\int_0^1|D^2u|^2dx\Big)^{3/8}+C'' \Big(\int_0^1|D^2u|^2dx\Big)^{1/2}\Big)^2
\\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}.
\end{align*}
\begin{align*}
&\int_0^1|D^2u|^4dx\\
&\leq 
\Big(C'\Big(\int_0^1|D^6u|^2dx\Big)^{\frac1{32}}
\Big(\int_0^1|D^2u|^2dx\Big)^{\frac{15}{32}}
+C'' \Big(\int_0^1|D^2u|^2dx\Big)^{1/2}\Big)^4\\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}.
\end{align*}
\begin{align*}
& \int_0^1|D^2u|^6dx\\
&\leq \Big(C'\Big(\int_0^1|D^6u|^2dx\Big)^{\frac1{24}}
\Big(\int_0^1|D^2u|^2dx\Big)^{\frac{11}{24}}+C'' \Big(\int_0^1|D^2u|^2dx\Big)^{1/2}
\Big)^4\\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}.
\end{align*}
and
\begin{align*}
& \int_0^1|D^3u|^4dx\\
&\leq \Big(C'\Big(\int_0^1|D^6u|^2dx\Big)^{\frac5{32}}
\Big(\int_0^1|D^2u|^2dx\Big)^{\frac{11}{32}}
+C'' \Big(\int_0^1|D^2u|^2dx\Big)^{1/2}\Big)^4\\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}.
\end{align*}
Therefore,
\begin{align*}
I_8
&\leq \sup_{Q_T}|m'(u)Du|\int_0^1|D^5uD^6u|dx
\leq C\int_0^1|D^5uD^6u|dx\\
&\leq \varepsilon \int_0^1|D^6u|^2dx+C_{\varepsilon}\int_0^1|D^5u|^2dx
\leq 2\varepsilon \int_0^1|D^6u|^2dx+C'_{\varepsilon}.
\end{align*}
\begin{align*}
I_9&\leq \sup_{Q_T}|m(u)|\int_0^1|D^4uD^6u|dx\leq C\int_0^1|D^4uD^6u|dx
\\
&\leq \varepsilon \int_0^1|D^6u|^2dx+C_{\varepsilon}\int_0^1|D^4u|^2dx\leq 2\varepsilon \int_0^1|D^6u|^2dx+C'_{\varepsilon}.
\end{align*}
\begin{align*}
I_{10}
&\leq \sup_{Q_T}|m'(u)Du|\int_0^1|D^3uD^6u|dx
\leq C\int_0^1|D^3uD^6u|dx\\
&\leq \varepsilon \int_0^1|D^6u|^2dx+C_{\varepsilon}\int_0^1|D^3u|^2dx
\leq 2\varepsilon \int_0^1|D^6u|^2dx+C'_{\varepsilon}.
\end{align*}
\begin{align*}
I_{11}
&\leq \sup_{Q_T}|m(u)(Du)^2|\int_0^1|D^4uD^6u|dx
+6\sup_{Q_T}|m(u)Du|\int_0^1|D^2uD^3uD^6u|dx\\
&\quad +2\sup_{Q_T}|m(u)|\int_0^1|D^2u|^3D^6u\,dx\\
&\leq  C\int_0^1|D^3uD^6u|dx+C\int_0^1|D^2uD^3uD^6u|dx
 +C\int_0^1|D^2u|^3D^6u\,dx\\
&\leq  \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}(\int_0^1|D^3u|^2dx
 +\int_0^1|D^2u|^4dx+\int_0^1|D^3u|^4dx\\
&\quad +\int_0^1|D^2u|^6dx\\
&\leq  2\varepsilon \int_0^1|D^6u|^2dx+C.
\end{align*}
\begin{align*}
I_{12}
&\leq \sup_{Q_T}|m'(u)(Du)^3|\int_0^1|D^3uD^6u|dx
 +2\sup_{Q_T}|m'(u)(Du)^2|\int_0^1|(D^2u)^2D^6u|dx\\
&\leq C\int_0^1|D^3uD^6u|dx+C\int_0^1|(D^2u)^2D^6u|dx\\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}
\Big(\int_0^1|D^3u|^2dx+\int_0^1|D^2u|^4dx\Big)\\
&\leq  2\varepsilon \int_0^1|D^6u|^2dx+C.
\end{align*}
\begin{align*}
I_{13}&\leq \sup_{Q_T}|m(u)w_0'(u)|\int_0^1|D^2uD^6u|dx
+\sup_{Q_T}|m(u)w_0'' (u)Du|\int_0^1|DuD^6u|dx \\
&\leq C\int_0^1|D^2uD^6u|dx+C\int_0^1|DuD^6u|dx\\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}
\Big(\int_0^1|D^2u|^2dx+\int_0^1|Du|^2dx\Big)\\
&\leq  \varepsilon \int_0^1|D^6u|^2dx+C.
\end{align*}
\[
I_{14}\leq \sup_{Q_T}|m'(u)w_0'(u)Du|\int_0^1|DuD^6u|dx
\leq C\int_0^1DuD^6u\,dx\leq \varepsilon \int_0^1|D^6u|^2dx+C.
\]
\begin{align*}
I_{15}
&\leq \sup_{Q_T}|m'(u)w_2'(u)(Du)^3|\int_0^1|DuD^6u|dx\\
&\quad +2\sup_{Q_T}|m'(u)w_2(u)(Du)^2|\int_0^1|D^2uD^6u|dx \\
&\leq C\int_0^1|DuD^6u|dx+C\int_0^1|D^2uD^6u|dx\\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}(\int_0^1|Du|^2dx
+\int_0^1|D^2u|^2dx)\\
&\leq  \varepsilon \int_0^1|D^6u|^2dx+C.
\end{align*}
\begin{align*}
I_{16}
&\leq \sup_{Q_T}|m(u)w_2'' (u)(Du)^3|\int_0^1|DuD^6u|dx\\
&\quad +5\sup_{Q_T}|m(u)w_2'(u)(Du)^2|\int_0^1|D^2uD^6u|dx
\\
&\quad +2\sup_{Q_T}|m(u)w_2(u)|\int_0^1|(D^2u)^2D^6u|dx\\
&\quad +2\sup_{Q_T}|m(u)w_2(u)Du|\int_0^1|D^3uD^6u|dx
\\
&\leq C\int_0^1|DuD^6u|dx+C\int_0^1|D^2uD^6u|dx
+C\int_0^1|(D^2u)^2D^6u|dx\\
&\quad +C\int_0^1|D^3uD^6u|dx
\\
&\leq \varepsilon\int_0^1|D^6u|^2dx
+C_{\varepsilon}\Big(\int_0^1|Du|^2dx+\int_0^1|D^2u|^2dx+\int_0^1|D^2u|^4dx\\
&\quad +\int_0^1|D^3u|^2dx\Big)
\\
&\leq 2 \varepsilon \int_0^1|D^6u|^2dx+C.
\end{align*}
\begin{align*}
I_{17}
&\leq \sup_{Q_T}|m(u)w_3'' (u)Du|\int_0^1|D^2uD^6u|dx
+2\sup_{Q_T}|m(u)w_3'(u)|\int_0^1|D^3uD^6u|dx\\
&\quad +\sup_{Q_T}|m(u)w_3(u)|\int_0^1|D^4uD^6u|dx
\\
&\leq C\int_0^1|D^2uD^6u|dx+C\int_0^1|D^3uD^6u|dx+C\int_0^1|D^4uD^6u|dx
\\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}(\int_0^1|D^2u|^2dx
+\int_0^1|D^3u|^2dx+\int_0^1|D^4u|^2dx)\\
&\leq 2 \varepsilon \int_0^1|D^6u|^2dx+C.
\end{align*}
\begin{align*}
I_{18}
&\leq \sup_{Q_T}|m'(u)w_3'(u)Du|\int_0^1|D^2uD^6u|dx
+\sup_{Q_T}|m'(u)w_3(u)Du|\int_0^1|D^3uD^6u|dx \\
&\leq C\int_0^1|D^2uD^6u|dx+C\int_0^1|D^3uD^6u|dx\\
&\leq \varepsilon\int_0^1|D^6u|^2dx+C_{\varepsilon}
\Big(\int_0^1|D^2u|^2dx+\int_0^1|D^3u|^2dx\Big)
\\
&\leq 2 \varepsilon \int_0^1|D^6u|^2dx+C.
\end{align*}
Summing up, noticing that $m(s)\geq m_0>0$, we obtain
$$
\frac  d{dt}\int_0^1|D^3u|^2dx+(2m_0-38\varepsilon)\int_0^1|D^6u|^2dx\leq C,
$$
where $\varepsilon$ is small enough, it satisfies $2m_0-38\varepsilon>0$. Hence
\begin{equation}
\label{jiangnan}
\sup_{0<t<T}\int_0^1|D^3u|^2dx\leq C.
\end{equation}
Combing \eqref{wuxi-1}, \eqref{2-7}, \eqref{2-8} and \eqref{jiangnan} together, 
using Sobolev's embedding theorem, we derive that
\begin{equation}\label{wuxi-jiangnan}
\sup_{Q_T}|D^2u(x,t)|\leq C\Big(\int_0^1[u^2dx+|Du|^2|D^2u|^2|D^3u|^2]dx\Big)^{1/2}
\leq C.
\end{equation}
By \eqref{2-7} and \eqref{2-9}, we deduce that
\begin{equation} \label{2-14}
|u(x_1,t)-u(x_2,t)|\leq C|x_1-x_2|^{\alpha},\quad 0\leq\alpha<1.
\end{equation}
Integrating the equation \eqref{1-1} with respect to $x$ over 
$(y,y+(\Delta t)^{1/6})\times(t_1,t_2)$, where $0<t_1<t_2<T$,
$\Delta t=t_2-t_1$, we deduce that
\begin{align*}
&\int_y^{y+(\Delta t)^{1/6}}[u(z,t_2)-u(z,t_1)]dz
\\
&=\int_{t_1}^{t_2}\Big[m(u(y',s))\Big(D^5u(y',s)+D^3u(y',s)
-D(|Du(y',s)|^{2}D^2u(y',s))
\\
&\quad +D(w_0(u(y',s))+w_2(u(y',s))|Du(y',s)|^2+w_3(u(y',s))D^2u(y',s))\Big)
\\
&\quad -m(u(y,s)) \Big(D^5u(y,s)+D^3u(y,s)-D(|Du(y,s)|^{2}D^2u(y,s))
\\
&\quad +D(w_0(u(y,s))+w_2(u(y,s))|Du(y,s)|^2+w_3(u(y,s))D^2u(y,s))\Big)\Big]ds.
\end{align*}
Set
\begin{align*}
N(s,y)
&=m(u(y',s))\Big(D^5u(y',s)+D^3u(y',s)-D(|Du(y',s)|^{2}D^2u(y',s))
\\
&\quad +D(w_0(u(y',s))+w_2(u(y',s))|Du(y',s)|^2+w_3(u(y',s))D^2u(y',s))\Big)
\\
&-m(u(y,s))\Big(D^5u(y,s)+D^3u(y,s)-D(|Du(y,s)|^{2}D^2u(y,s))
\\
&+D(w_0(u(y,s))+w_2(u(y,s))|Du(y,s)|^2+w_3(u(y,s))D^2u(y,s))\Big),
\end{align*}
where $y'=y+(\Delta t)^{1/6}$. Then, the above equality is converted into
$$
(\Delta t)^{1/6}\int_0^1[u(y+\theta(\Delta t)^{1/6},t_2)
-u(y+\theta(\Delta t)^{1/6},t_1)]d\theta=\int_{t_1}^{t_2}N(s,y)ds.
$$
Integrating above equality with respect to $y$ over $(x,x+(\Delta t)^{1/6})$, 
we immediately obtain
$$
(\Delta t)^{1/3}(u(x^*,t_2)-u(x^*,t_1))
=\int_{t_1}^{t_2}   \int_x^{x+(\Delta t)^{1/6}}N(s,y)dyds.
$$
Here, we have used the mean value theorem, where 
$x^*=y^*+\theta^*(\Delta t)^{1/6}$, $y^*\in(x,x+(\Delta t)^{1/6})$, 
$\theta\in(0,1)$. Then, by H\"{o}lder's inequality and \eqref{2-8}, \eqref{2-9}, 
\eqref{2-13}, we obtain
$$
|u(x^*,t_2)-u(x^*,t_1)|\leq C(\Delta t)^{\frac{\alpha}6},~0<\alpha<1.
$$
Similar to the above discussion, we have
\begin{equation} \label{2-16}
|Du(x_1,t_1)-Du(x_2,t_2)|\leq C(|x_1-x_2|^{1/2}+|t_1-t_2|^{\frac1{12}}).
\end{equation}
The proof is complete.
\end{proof}

To prove Theorem \ref{thm1.1}, the key estimate is the H\"{o}lder estimate
 for $D^2u$. Now, we give the following lemma which can be seen in \cite{Liu2}.

\begin{lemma}\label{lem2.2} 
Assume that $\sup|f|<+\infty,~a(x,t)\in C^{\alpha,\frac{\alpha}6}(\bar{Q}_T)$,
$0<\alpha<1$, and there exist
two constants $a_0,b_0,A_0,B_0$ such that $0<a_0\leq a(x,t)\leq A_0$, 
$0<b_0\leq b(x,t)\leq B_0$ for all $(x,t)\in Q_T$. If $u$ is a smooth 
solution for the linear problem 
\begin{gather*}
\frac{\partial u}{\partial t}-D^3(a(x,t)D^3u)+D^3(b(x,t)Du)
=D^3f,\quad (x,t)\in Q_T,\\
Du(x,t)|_{x=0,1}=D^3u(x,t)|_{x=0,1}=D^5u(x,t)|_{x=0,1}=0,\quad t\in[0,T],\\
u(x,0)=u_0(x),\quad x\in[0,1],
\end{gather*}
then, for any $\delta\in (0,\frac12)$, there is a constant $K$
depending on $a_0$, $b_0$, $A_0$, $B_0$, $\delta$, $T$, 
$\iint_{Q_T}u^2dxdt$ and
$\iint_{Q_T}|D^3u|^2dxdt$, such that
$$
|u(x_1,t_1)-u(x_2,t_2)|\leq
K(1+\sup|f|)(|x_1-x_2|^{\delta}+|t_1-t_2|^{\frac{\delta}6}).
$$
\end{lemma}

Now, we  prove the main result.

\begin{proof}[Proof of Theorem \ref{thm1.1}]
Suppose that $w=D^2u-D^2u_0$. Then $w$ satisfies the problem
\begin{align*}
\frac{\partial w}{\partial t}-D^3(a(x,t)D^3w)+D^3(b(x,t)Dw)=D^3f,\\
w(x,t)=D^2w(x,t)=D^4w(x,t)=0,\quad x=0,1,\\
w(x,0)=0,\quad x\in[0,1],
\end{align*}
where $a(x,t)=m(u)$, $b(x,t)=m(u)$ and 
$$
f(x,t)=m(u)[-D^5u_0-D^3u_0+D(|Du|^{2}D^2u)-D(w_0(u)+w_2(u)|Du|^2+w_3(u)D^2u)].
$$ 
It then follows from \eqref{2-8}-\eqref{wuxi-jiangnan} and Lemma \ref{lem2.2} that
$$
|D^2u(x_1,t_1)-D^2u(x_2,t_2)|\leq C(|x_1-x_2|^{\alpha/2}
+|t_1-t_2|^{\alpha/12}).
$$

The conclusion follows immediately from the classical theory, since we 
can transform the equation \eqref{1-1} into the form
\begin{align*}
&\frac{\partial u}{\partial t}+a_1(x,t)D^6u+a_2(x,t)D^5u+a_3(x,t)D^4u(x,t)
\\
&+a_4(x,t)D^3u(x,t)+a_5(x,t)D^2u(x,t)+a_6(x,t)Du(x,t)=0.
\end{align*}
with the H\"{o}lder norms on
\begin{align*}
a_1(x,t)&=-m(u(x,t)),\quad a_2(x,t)=-m'(u(x,t))Du(x,t),\\
a_3(x,t)&=m(u(x,t))(|Du(x,t)|^{2}+w_3(u(x,t))-1),\\
a_4(x,t)&=m'(u(x,t))[|Du(x,t)|^2Du(x,t)-Du(x,t)]\\
&\quad +m(u(x,t))[6Du(x,t)D^2u(x,t)+2w_2(u(x,t))Du(x,t)\\
&\quad +2w_3'(u)Du(x,t)],
\\
a_5(x,t)&=m(u(x,t))[2|D^2u(x,t)|^2+w_0'(u(x,t))+5w_2'(u(x,t))|Du(x,t)|^2
 \\
&\quad +2w_2(u(x,t))D^2u(x,t)+w_3'' (u(x,t))|Du(x,t)|^2+w_3'(u(x,t))D^2u(x,t)] \\
&\quad +m'(u(x,t))[2|Du(x,t)|^2D^2u(x,t)+2w_2(u(x,t))|Du(x,t)|^2\\
&\quad +w_3'(u(x,t))|Du(x,t)|^2+w_3(u(x,t))Du(x,t)]
\\
a_6(x,t) &=m'(u(x,t))[w_0'(u(x,t))Du(x,t)+w_2'(u(x,t))|Du(x,t)|^2Du(x,t)]\\
&\quad +m(u(x,t))[w_0'(u(x,t))Du(x,t)+w_2'' (u(x,t))|Du(x,t)|^2Du(x,t)].
\end{align*}
have been estimated in the above discussion. Then, the proof is complete
\end{proof}

\subsection*{Acknowledgments}
This research was supported by  the
Natural Science Foundation of China for Young Scholar (No. 11401258), and by
the  Natural Science Foundation of Jiangsu Province for Young Scholar (No. BK20140130).


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\end{document}






%\subsection*{Acknowledgements}