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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 15, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/15\hfil Viscosity solutions]
{Viscosity solutions to degenerate diffusion problems}

\author[Z. Chen, Y. Zhao\hfil EJDE-2007/15\hfilneg]
{Zu-Chi Chen, Yan-Yan Zhao}  % in alphabetical order

\address{Zu-Chi Chen \newline
 Department of mathematics\\
 University of Science and Technology of China\\
 Hefei 230026, China}
\email{chenzc@ustc.edu.cn}

\address{Yan-Yan Zhao \newline
Department of mathematics\\
 University of Science and Technology of China\\
 Hefei 230026, China}
\email{yyzhao@mail.ustc.edu.cn}

\thanks{Submitted November 4, 2006. Published January 23, 2007.}
\thanks{Supported by grant 10371116 from NNSF of China}
\subjclass[2000]{35K15, 35K55, 35K65}
\keywords{Viscosity solution; degenerate; H\"older regularity}

\begin{abstract}
 This paper concerns the weak solutions to a  Cauchy problem
 in $\mathbb{R}^N$  for a degenerate nonlinear parabolic equation.
 We obtain the  H\"older regularity  of the weak solutions
 to this problem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}


\section{Introduction}

We consider the Cauchy problem
\begin{equation} \label{e1}
\begin{gathered}
 u_t=\alpha_1u^{\beta_1}\Delta u+\alpha_2u^{\beta_2}w,\quad
w=\frac{1}{2}|\nabla u|^2, \quad (x,t)\in \mathbb{R}^N\times
\mathbb{R}^+ \\
u(x,0)=u_0(x),   \quad x\in \mathbb{R}^N
\end{gathered}
\end{equation}
where $\alpha_1,\alpha_2,\beta_1,\beta_2$ are constants and $u_0$
is a bounded continuous and nonnegative function on $R^N$, denote
$\Omega=\mathbb{R}^N\times \mathbb{R}^+$.

Problem \eqref{e1} degenerates at the points where $u$ vanishes.
Therefore, in general, it has no classical solutions and we have
to consider its weak solutions.  The weak solution is defined as
follows.

\begin{definition} \rm
A function $u\in L^\infty(\Omega)\cap L^2_{\rm
loc}([0,+\infty);H^1_{\rm loc}(\mathbb{R}^N))$ is called a weak
solution of  \eqref{e1} if $u\geq 0$ a.e. in $\Omega$ and for all
$T>0$,
$$
\int_{R^N}u_0\psi(0)dx+\int_{\mathbb{R}^N\times(0,T)} u\frac{\partial
\psi}{\partial t}-\alpha_1 \nabla u\cdot\nabla(u^{\beta_1}
\psi)+\alpha_2 u^{\beta_2}|\nabla u|^2\psi \,dx\,dt=0
$$
for all  $\psi\in C^{1,1}(\mathbb{R}^N\times[0,T])$ with the
compact support in $\mathbb{R}^N\times[0,T)$.
\end{definition}

 Let $u_\epsilon(x,t)\ge0$ be the classical solution of the
problem
\begin{equation} \label{e2}
\begin{gathered}
u_{\epsilon t}=\alpha_1u^{\beta_1}_{\epsilon}\Delta
u_\epsilon+\alpha_2u^{\beta_2}_{\epsilon}
w_{\epsilon},\quad w_\epsilon=\frac{1}{2}|\nabla u_\epsilon|^2, \quad
(x,t)\in \mathbb{R}^N \times \mathbb{R}^+ \\
u_\epsilon (x,0)=u_0(x)+\epsilon,   \quad x\in \mathbb{R}^N
\end{gathered}
\end{equation}
By the maximum principle $u_\epsilon(x,t)$ is decreasing with
respect to $\epsilon$, thus
$$
u(x,t)=\lim_{\epsilon\to 0}u_\epsilon(x,t)
$$
is well defined in $\bar\Omega$. The function $u$ is a weak solution of
 \eqref{e1}. Because $u_0$ is bounded, using the maximum
principle in problem \eqref{e2}, $u_\epsilon$ is bounded and
$\{u_\epsilon\}_{\epsilon\to 0}$  is uniformly bounded.

\begin{definition} \rm
The weak solution  defined above is called a viscosity solution
of  \eqref{e1}.
\end{definition}

As its special cases, Bertsh, Passo,  Ughi and Lu had considered
the equation $u_t=u\Delta u-\gamma|\nabla u|^2$ in [1-6]. When
$\alpha_1=m$, $\beta_1=m-1$, $\alpha_2=2m(m-1)$, $\beta_2=\beta_1-1 $,
problem \eqref{e1} is the porous medium equation, the well known
case.

\section{Main Result}

\begin{theorem}
If $\alpha_1>0,\beta_2=\beta_1-1$, there exists a constant $s$
such that
$$
2\alpha_2\beta_1-2\alpha_2-s\alpha_2+2s(s+1)\alpha_1+N\alpha_1\beta^2_1\le 0
$$
and
$$
|\nabla({u_0^{1+\frac{s}{2}}})|\le M
$$
for a nonnegative constant $M$.
Then the viscosity solution $u$ of
 \eqref{e1} satisfies $|\nabla({u^{1+\frac{s}{2}}})|\le M$
in $\bar\Omega$.
\label{thm1}
\end{theorem}

\begin{proof}
 In the definition of the viscosity solution, we let
$u_\epsilon>0$ be the classical solution of  \eqref{e2}. Then
$$
u(x,t)=\lim_{\epsilon\to 0}u_\epsilon(x,t)
$$
is the viscosity solution of \eqref{e1}.
In the following we use the notation $u_{\epsilon,.}$ to
denote the derivative of function $u_\epsilon $ with respect to
its independent variables.
  At first, we have
\begin{align*}
w_{\epsilon,  t} & =(\frac{1}{2}|\nabla
       u_\epsilon|^2)_t=\sum_{i=1}^Nu_{\epsilon,  x_i}(u_{\epsilon, x_i})_t \\
    & =\sum_{i=1}^Nu_{\epsilon, x_i}(\alpha_1u^{\beta_1}_{\epsilon}\Delta
       u_\epsilon+\alpha_2u^{\beta_2}_{\epsilon}w_{\epsilon, x_i} \\
    & =\sum_{i=1}^Nu_{\epsilon, x_i}(\alpha_1\beta_1u_{\epsilon, x_i}
    u^{\beta_1-1}_{\epsilon}\Delta
       u_\epsilon+\alpha_1u^{\beta_1}_{\epsilon}\Delta
       u_{\epsilon, x_i}\\
    &\quad +\alpha_2\beta_2u_{\epsilon, x_i}u^{\beta_2-1}_{\epsilon}w_\epsilon
       +\alpha_2u^{\beta_2}_{\epsilon}w_{\epsilon, x_i})\\
    & =2\alpha_1\beta_1u^{\beta_1-1}_{\epsilon}w_\epsilon\Delta
       u_\epsilon+\alpha_1u^{\beta_1}_{\epsilon}\sum_{i=1}^Nu_{\epsilon, x_i}
       \Delta u_{\epsilon, x_i}\\
    &\quad +2\alpha_2\beta_2u^{\beta_2-1}_{\epsilon}w^2_{\epsilon}
       +\alpha_2u^{\beta_2}_{\epsilon}\sum_{i=1}^Nu_{\epsilon, x_i}w_{\epsilon, x_i}\\
    & =2\alpha_1\beta_1u^{\beta_1-1}_{\epsilon}w_\epsilon\Delta
       u_\epsilon+\alpha_1u^{\beta_1}_{\epsilon}\Delta
       w_\epsilon-\alpha_1u^{\beta_1}_{\epsilon}\sum_{i,j=1}^Nu_{\epsilon,
       x_ix_j}^2 \\
    &\quad +2\alpha_2\beta_2u^{\beta_2-1}_{\epsilon}w^2_{\epsilon}
        +\alpha_2u^{\beta_2}_{\epsilon}
       \sum_{i=1}^Nu_{\epsilon, x_i}w_{\epsilon, x_i}.
\end{align*}
Let
\begin{equation} \label{e3}
 z_\epsilon=u^s_\epsilon w_\epsilon,
\end{equation}
then
\begin{equation} \label{e4}
\begin{aligned}
z_{\epsilon, t}
&=u^s_{\epsilon, t}w_\epsilon+u^s_{\epsilon}w_{\epsilon, t}  \\
 &=s\alpha_1u^{s+\beta_1-1}_\epsilon w_\epsilon\Delta u_\epsilon+s\alpha_2u^{s+\beta_2-1}_\epsilon w^2_\epsilon
    +2\alpha_1\beta_1u^{s+\beta_1-1}_\epsilon w_\epsilon\Delta
    u_\epsilon +\alpha_1u^{s+\beta_1}_\epsilon\Delta w_\epsilon \\
 & \quad -\alpha_1u^{s+\beta_1}_\epsilon\sum_{i,j=1}^Nu_{\epsilon, x_ix_j}^2
   +2\alpha_2\beta_2u^{s+\beta_2-1}_\epsilon
    w^2_\epsilon+\alpha_2u^{s+\beta_2}_\epsilon\sum_{i=1}^N
    u_{\epsilon, x_i}w_{\epsilon, x_i}.
\end{aligned}
\end{equation}
From \eqref{e3} and \eqref{e4},
\begin{equation} \label{e5}
\begin{aligned}
z_{\epsilon, t}&= \alpha_1u^{\beta_1}_\epsilon\Delta
z_\epsilon+(\alpha_2u^{\beta_2}_\epsilon-2s\alpha_1u^{\beta_1-1}_\epsilon)\sum_{i=1}^Nu_{\epsilon, x_i}z_{\epsilon,
x_i}\\
&\quad +[(2\alpha_2\beta_2-s\alpha_2)u^{\beta_2-s-1}_\epsilon+2s(s+1)\alpha_1u^{\beta_1-s-2}_\epsilon]z^2_\epsilon
\\
&\quad +2\alpha_1\beta_1u^{\beta_1-1}_\epsilon z_\epsilon\Delta
u_\epsilon-\alpha_1u^{s+\beta_1}_\epsilon\sum_{i,j=1}^Nu_{\epsilon, x_ix_j}^2 .
\end{aligned}
\end{equation}
If $\beta_2=\beta_1-1,\alpha_1>0$,  then
\begin{align*}
z_{\epsilon, t}&= \alpha_1u^{\beta_1}_\epsilon\Delta
z_\epsilon+(\alpha_2-2s\alpha_1)u^{\beta_1-1}_\epsilon\sum_{i=1}^Nu_{\epsilon, x_i}z_{\epsilon, x_i}\\
&\quad +[(2\alpha_2\beta_2-s\alpha_2)
+2s(s+1)\alpha_1]u^{\beta_1-s-2}_\epsilon
z^2_\epsilon\\
&\quad +2\alpha_1\beta_1u^{\beta_1-1}_\epsilon z_\epsilon\Delta
u_\epsilon-\alpha_1u^{s+\beta_1}_\epsilon\sum_{i,j=1}^Nu_{\epsilon, x_ix_j}^2.
\end{align*}
Since
$$
\sum_{i,j=1}^Nu_{\epsilon, x_ix_j}^2\ge\frac{1}{N}(\Delta u_\epsilon)^2,
$$
it follows that
\begin{equation} \label{e6}
\begin{aligned}
z_{\epsilon, t} \le&\alpha_1u^{\beta_1}_\epsilon\Delta
  z_\epsilon+(\alpha_2-2s\alpha_1)u^{\beta_1-1}_\epsilon\sum_{i=1}^Nu_{\epsilon, x_i}z_{\epsilon, x_i}\\
  &\quad +[(2\alpha_2\beta_1-2\alpha_2-s\alpha_2)+2s(s+1)\alpha_1]u^{\beta_1-s-2}_\epsilon z^2_\epsilon \\
&\quad +2\alpha_1\beta_1u^{\beta_1-1}_\epsilon z_\epsilon\Delta
u_\epsilon-\frac{\alpha_1}{N}u^{s+\beta_1}_\epsilon
(\Delta u)^2_\epsilon \\
&= \alpha_1u^{\beta_1}_\epsilon\Delta
z_\epsilon+(\alpha_2-2s\alpha_1)u^{\beta_1-1}_\epsilon\sum_{i=1}^N
u_{\epsilon, x_i}z_{\epsilon, x_i}\\
  &\quad -(\sqrt\frac{\alpha_1}{N}u_{\epsilon}^{\frac{s+\beta_1}
  {2}}\Delta u_\epsilon-\beta_1\sqrt{N\alpha_1}u_{\epsilon}^{\frac{\beta_1-s-2}{2}}z_\epsilon)^2\\
&\quad +[(2\alpha_2\beta_1-2\alpha_2-s\alpha_2)+2s(s+1)\alpha_1+N\alpha_1\beta_1^2]u_{\epsilon}^{\beta_1-s-2}z_{\epsilon}^2
\end{aligned}
\end{equation}
By the condition
$$
2\alpha_2\beta_1-2\alpha_2-s\alpha_2+2s(s+1)\alpha_1+N\alpha_1\beta^2_1\le0
$$
and \eqref{e6}, we obtain
$$
z_{\epsilon, t}\leq \alpha_1 u_{\epsilon}^{\beta_1}\Delta z_\epsilon
+(\alpha_2-2s\alpha_1) u_{\epsilon}^{\beta_1-1}
\sum_{i=1}^Nu_{\epsilon, x_i}z_{\epsilon,x_i}.
$$
Using the maximum principle, we obtain
$$
\|z_\epsilon\|_{\infty}\leq \|z_0\|_{\infty}.
$$
Because $z_\epsilon=u^s_\epsilon w_\epsilon
=\frac{1}{2}u^s_\epsilon |\nabla u_\epsilon |^2$, thus
$$
\|u^s_\epsilon|\nabla u_\epsilon|^2\|_{\infty}
\leq \|u^s_0|\nabla u_0|^2\|_{\infty}\leq M+\epsilon.
$$
Since $\nabla(u_{\epsilon}^{1+\frac{s}{2}})$ is continuous,
\begin{equation}
|\nabla(u_\epsilon ^{1+\frac{s}{2}})|\leq M+\epsilon.
\end{equation}
Because $u(x,t)=\lim_{\epsilon\to 0}u_\epsilon(x,t)$, then
\begin{equation}
|\nabla(u^{1+\frac{s}{2}})|\leq M.
\end{equation}
\end{proof}

\begin{theorem}
Suppose $\alpha_1$, $\alpha_2$, $\beta_1$, $\beta_2$,$u_0$ are as
in Theorem 2.1, if there exists a nonpositive constant $s\neq-2$
satisfying
$$
2\alpha_2\beta_1-2\alpha_2-s\alpha_2+2s(s+1)\alpha_1+N\alpha_1\beta^2_1\le0 ,
$$
then the viscosity solution $u(x,t)$ of problem \eqref{e1} is
Lipschitz continuous in $x$ and H\"{o}lder continuous in $t$ with
exponent $1/2$ in $\overline\Omega$. \label{thm2}
\end{theorem}

\begin{proof}
 Because $\{u_\epsilon\}_{\epsilon\to 0}$
is uniformly bounded, $u(x,t)=\lim_{\epsilon\to 0} u_\epsilon$,
so there exists a constant $M_1$ such that $|u|<M_1$.
By Theorem \ref{thm1},
$|\nabla({u^{1+\frac{s}{2}}})|\le M$, then
$$
|\nabla u|\le |1+\frac{s}{2}|^{-1}M|u^{\frac{-s}{2}}|\le
|1+\frac{s}{2}|^{-1}M{M_1}^{\frac{-s}{2}}.
$$
Therefore, $u$ is Lipschitz continuous with respect to $x$. Hence,
we get directly from [7] that $u$ is H\"{o}lder continuous in $t$
with exponent $1/2$ in $\overline\Omega$.
\end{proof}

\section{Examples}

\begin{example}  \rm
Consider the problem
\begin{equation} \label{e8}
\begin{gathered}
u_t=u\Delta u-\gamma|\nabla u|^2, \quad (x,t)\in \Omega\\
u(x,0)=u_0(x), \quad   x\in \mathbb{R}^N
\end{gathered}
\end{equation}
If $\gamma\geq \sqrt{N-1}$ $(N\not=10)$, and there are
constants
\begin{gather*}
\tau=\frac{3-\sqrt{N-1}}{2},\\
s=\frac{1-\gamma-2\tau}{2\tau}+\frac{\sqrt{2\gamma^2-2N
+2-[2\tau-(3-\gamma)]^2}}{2\tau}
\end{gather*}
satisfying $|\nabla u^{\tau(1+\frac{s}{2})}_{0}|\leq M$, then the
viscosity solution of \eqref{e8} is Lipschitz continuous in $x$
and H\"{o}lder continuous  in $t$ with exponent $1/2$ in
$\overline\Omega$. \label{exa1}
\end{example}

\begin{proof} Set $v_\epsilon=u_\epsilon^\tau$. From problem
\eqref{e8},  we obtain
\begin{align*}
v_{\epsilon, t} & =\tau v\Delta u_\epsilon-\tau\gamma u_{\epsilon}^{\tau-1}|\nabla u_\epsilon|^2\\
    & =\tau v_\epsilon\sum^{N}_{i=1}(\frac{1}{\tau}v_{\epsilon}^{\frac{1}{\tau}-1}v_{\epsilon, x_i})_{x_i}-
       \tau\gamma v_{\epsilon}^{\frac{\tau-1}{\tau}}|\frac{1}{\tau}v_{\epsilon}^{\frac{1}{\tau}-1}\nabla v_\epsilon|^2\\
    & =v_{\epsilon}^{\frac{1}{\tau}}\Delta v_\epsilon + v_\epsilon\sum^{N}_{i=1}(\frac{1}{\tau}-1)v_{\epsilon}^{
    \frac{1}{\tau}-2}v^2_{\epsilon, x_i}-
       \frac{\gamma}{\tau}v_{\epsilon, }^{\frac{1}{\tau}-1}|\nabla v_\epsilon|^2\\
    & =v_{\epsilon}^{\frac{1}{\tau}}\Delta v_\epsilon+\frac{1-\gamma-\tau}{\tau}v_{\epsilon}^{\frac{1}{\tau}-1}
    |\nabla v_\epsilon|^2 .
\end{align*}
In problem \eqref{e1}, with
$\alpha_1=1$, $\beta_1=\frac{1}{\tau}$,
$\alpha_2=\frac{2-2\gamma -2\tau}{\tau}$, $\beta_2=\beta_1-1$,
we have
\begin{align*}
 & 2\alpha_2\beta_1-2\alpha_2-s\alpha_2+2s(s+1)\alpha_1+N\alpha_1\beta_1^2\\
&=  \frac{4(1-\gamma-\tau)}{\tau^2}-\frac{4(1-\gamma-\tau)}{\tau}
 -\frac{2s(1-\gamma-\tau)}{\tau}+2s(s+1)
+\frac{N}{\tau^2}\\
&=  2(s-\frac{1-\gamma-2\tau}{2\tau})^2-\frac{(1-\gamma-2\tau)^2}{2\tau^2}
 +\frac{4(1-\gamma-\tau)}{\tau^2}-
 \frac{4(1-\gamma-\tau)}{\tau}+\frac{N}{\tau^2}\\
&=  2(s-\frac{1-\gamma-2\tau}{2\tau})^2+\frac{1}{2\tau^2}[-\gamma^2
 +(4\tau-6)\gamma+4\tau^2-12\tau+2N+7]\\
&=  0.
\end{align*}
From Theorem \ref{thm1} we get
$|\nabla({u^{\tau(1+\frac{s}{2})}})|\le M$. Because
$\tau(1+\frac{s}{2})-1\leq 0$, we have
 $$
|\nabla u|\le |\tau(1+\frac{s}{2})|^{-1}M|u^{-\tau(1+\frac{s}{2})+1}|\le M_2.
$$
 We get the H\"{o}lder continuity of $u$ with respect to $t$ from [7]
directly.
\end{proof}

\begin{remark} \rm
For the case $N=10$, we take $\tau$ as a positive number, say
$\delta$, then similar to the above arguments we can get the
result that when $\gamma\geq2\delta-3+\sqrt{2(2\delta-3)^2+2N-2}$
and $|\nabla u^{\delta(1+\frac{s}{2})}_{0}|\leq M$, then $u $ is
Lipschitz continuous in $x$ and H\"{o}lder continuous in $t$ with
exponent $1/2$ in $\overline\Omega$. Since $\delta$ is any
positive number which can be taken small enough so our conclusion
for the parameter $\gamma$ is that when $\gamma>\sqrt{N-1}$ the
solution is Lipschitz continuous in $x$ and  H\"older
continuous in $t$ with exponent $1/2$ in
$\overline\Omega$. It is a improved result of the one in
\cite{zhao-chen}.
\end{remark}

\begin{remark} \rm
Let $\tau=1$ in Example \ref{exa1}, we could get the result as
$\gamma\geq\sqrt{2N}-1$. It is the main result in \cite{lu-qian}.
\end{remark}

\begin{example} \label{exa2}
The initial problem for the porous medium equation
\begin{equation} \label{e9}
\begin{gathered}
u_t=\Delta(u^m), \quad (x,t)\in \Omega\\u(x,0)=u_0(x),
\quad x\in \mathbb{R}^N
\end{gathered}
\end{equation}
If $m>0$, $\frac{10+2N-\sqrt{16+2N}} {7+2N}\leq m\leq
\frac{10+2N+\sqrt{16+2N}} {7+2N}$, and
$|\nabla({u_0^{\frac{m+2}{4}}})|\le M $, then the viscosity
solution $u(x,t)$ is Lipschitz continuous in $x$ and H\"{o}lder
continuous in $t$ with exponent $1/2$ in
$\overline\Omega$.
\end{example}

\begin{proof}
In Theorem \ref{thm2}, let
$\alpha_1=m$, $\beta_1=m-1$, $\alpha_2=2m(m-1)$, $\beta_2=\beta_1-1$,
$s=\frac{m-2}{2}$.  Then the result follows.
\end{proof}

\begin{example} \label{exa3}
Consider the initial-value problem for the singular equation
\begin{equation} \label{e10}
\begin{gathered}
u_t=\Delta u+\frac{|\nabla u|^2}{u^m}, \quad (x,t)\in
\Omega\\
u(x,0)=u_0(x) \quad  x\in \mathbb{R}^N
\end{gathered}
\end{equation}
where $m\ge 0$. If $|\nabla u_0|\leq M$, then $u(x,t)$ is
Lipschitz continuous in $x$ and H\"{o}lder continuous in $t$ with
exponent $1/2$ in $\overline\Omega$.
\end{example}

\begin{proof} As the proof in Theorem \ref{thm1}, In problem
\eqref{e1}, we take $\alpha_1=1$, $\beta_1=0$, $\alpha_2=2$, $\beta_2=-m$.
From \eqref{e5},
$$
z_{\epsilon, t}\leq \Delta z_\epsilon +(2u_{\epsilon}^{-m}-2su_{\epsilon}^{-1})
\sum_{i=1}^Nu_{\epsilon, x_i}z_{\epsilon,
x_i}+[(-4m-2s)u_{\epsilon}^{-m+1}+2s(s+1)]z_{\epsilon}^2u_{\epsilon}^{-s-2}.
$$
Let $s=0$, then
 $$
z_{\epsilon, t}\leq \Delta z_\epsilon +2u_{\epsilon}^{-m}
\sum_{i=1}^Nu_{\epsilon, x_i}z_{\epsilon,  x_i}.
$$
 Thus
 $\|z_\epsilon\|_{\infty}\leq \|z_0\|_{\infty}$
 and so $ |\nabla u|\leq M$.
 As in the proof in Theorem \ref{thm2}, $u$ is Lipschitz continuous in
 $x$ and H\"{o}lder continuous  in $t$ with exponent $1/2$ in
$\overline\Omega$.
\end{proof}


\subsection*{Acknowledgements} The authors want to thank the
anonymous referee for his/her valuable comments and suggestions.


\begin{thebibliography}{00}

\bibitem{bertsch-ughi}
M. Bertsch and M. Ughi; \emph{Positivity properties of viscosity
solutions of a degenerate parabolic equation}, Nonlinear Anul.
TMA. 14 (1990), 7, 571-592.

\bibitem{bertsch-passo}
M. Bertsch, R. D. Passo and M. Ughi; \emph{Discontinuous viscosity
Solution of A degenerate parbolic equation}, Trans. AMS 320 (1990),
2, 779-798.

\bibitem{lu-qian}
Yungguang Lu and Liwen Qian; \emph{Regularity of viscosity solutions of
degenerate parabolic equation}, Proc. AMS. vol.130, No.4, 999-1004.

\bibitem{ughi}
M. Ughi; \emph{A dengenerate parabolic equation modelling the spread of
an epidemic}, Ann. Mat. Pura Appl. 143 (1986), 385-400.

\bibitem{passo-luckhans-ughi}
D. Passo and S. Luckhans; \emph{A degenarate diffusion problem not in
divergence form}. J. Diff. Eqns 69, 1-14 (1987).

\bibitem{bertsch-passo-ughi}
M. Bertsch, D. Passo and M. Ughi; \emph{Nonuniquenss of solutions of a
degenarate parabolic equation}, Ann. Mat. Pura Appl. (4) 161
(1992), 57-81.

\bibitem{gilding}
B. H. Gilding; \emph{H\"{o}lder continuity of solutions of parabolic
equations}, J. London Math. Soc. (2), 13 (1976), 103-106.

\bibitem{lu-willi}
Yunguang Lu and Willi J\"{a}ger; \emph{On solutions to nonliner
Reaction-diffusion-convection equations with dengenerate
diffusion}, J. Diff. Eqns 170, 1-21 (2001).

\bibitem{zhao-chen}
Yan-Yan Zhao and Zu-Chi Chen; \emph{Regularity of viscosity solutions to
a degenerate nonlinear cauchy problem}, to appear in Nonlinear
Analysis TMA.

\end{thebibliography}

\end{document}