\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small {\em 
Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 23, pp. 1--7.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2005/23\hfil Extinction for fast diffusion equations] 
{Extinction for fast diffusion equations with nonlinear sources}

\author[Y. Li, J. Wu\hfil EJDE-2005/23\hfilneg]
{Yuxiang Li, Jichun Wu}

\address{Yuxiang Li \hfill\break
Department of Mathematics, Southeast University,
Nanjing 210096, China. \hfill\break
Department of Earth Sciences,
Nanjing University, Nanjing 210093, China}
\email{lieyuxiang@yahoo.com.cn}

\address{Jichun Wu\hfill\break
Department of Earth Sciences, Nanjing University, Nanjing 210093,
China} 
\email{jcwu@nju.edu.cn}

\date{}
\thanks{Partially supported by project40272106 from the National
 Science Foundation of China, \hfill\break\indent
 and by the Teaching and Research 
 Award Program for Outstanding Young Teacher \hfill\break\indent
  of the Ministry of Education, China.}

\thanks{Submitted September 15, 2004. Published February 20, 2005.}
\subjclass[2000]{35K20, 35K55} 
\keywords{Extinction; fast diffusion; first eigenvalue}


\begin{abstract}
 We establish conditions for the extinction of solutions,
 in finite time, of the fast diffusion problem
 $u_t=\Delta u^m+\lambda u^p$, $0<m<1$,
 in a bounded domain of $R^N$ with $N>2$.
 More precisely, we show that if $p>m$, the solution
 with small initial data vanishes in finite time,
 and if $p<m$,  the maximal solution is positive for all $t>0$.
 If $p=m$, then first eigenvalue of the Dirichlet problem plays
 a role.
\end{abstract}

\maketitle \numberwithin{equation}{section}
\newtheorem{thm}{Theorem}[section]
\newtheorem{defn}[thm]{Definition}
\newtheorem{rem}[thm]{Remark}


\section{Introduction}

In this paper we are concerned with the porous medium equation
\begin{equation}\label{e:main}
\begin{gathered}
u_t=\Delta u^m+\lambda u^p,\quad x\in\Omega,\;  t>0,\\
u=0,  \quad x\in\partial\Omega,\; t>0,\\
u(x,0)=u_0(x)\geq 0,  \quad x\in\Omega,
\end{gathered}
\end{equation}
with $0<m<1$ and $p,\ \lambda>0$, where $\Omega\subset R^N$,
$N>2$, is an open bounded domain with smooth boundary
$\partial\Omega$. We are interested in the extinction of the
nonnegative solution of (\ref{e:main}).

The phenomena of extinction have been studied extensively for
(\ref{e:main}) with $\lambda\leq 0$.  When $\lambda<0$, for the
case of slow diffusion, see \cite{CMM, GV1, GV2, GV3, Ka,Ker, RK}.
For $m=1$ we refer the reader to \cite{HV}.  And for the case of
fast diffusion, see \cite{BU, FV, Le, PZ1, PZ2}.  When $\lambda=0$
and $0<m<1$, we refer the reader to \cite{BC, BH, FK, King, GK,
GPV}.

For (\ref{e:main}) with $p>1$, it is well known that the solution
blows up in finite time for sufficiently large initial data; see
\cite{GV4, SGKM}.  In this paper we  show that the solution of
(\ref{e:main}) vanishes in finite time for sufficiently small
initial data. If $0<m<1$ and $p>m$, there is a maximal positive
solution of (\ref{e:main}). If $p<m$ and $p=m$, the first
eigenvalue $\lambda_1$ of the problem below plays a crucial role:
\begin{equation}\label{e:eigenvalue problem}
   -\Delta \psi(x)=\lambda\psi(x),\quad x\in\Omega;\quad
   \psi\big|_{\partial\Omega}=0.
\end{equation}

The existence and uniqueness for (\ref{e:main}) have been studied
in \cite{A1, A2}.  To state the definition of the weak solution,
we define class of nonnegative testing functions
\[
     \mathcal{F}=\{\xi: \xi_t, \Delta\xi, |\nabla\xi|\in
L^2(\Omega_T),\ \xi\geq
     0 \mbox{ and } \xi|_{(\partial\Omega)_T}=0\}.
\]

\begin{defn}\label{defn:weak solutions} \rm
A function $u(x,t)\in L^{\infty}(\Omega_T)$ is called a
subsolution (supersolution) of (\ref{e:main}) in $\Omega_T$ if the
following conditions hold:
\begin{enumerate}
  \item[(i)]    $u(x,0)\leq (\geq)\ u_0(x)$ in $\Omega$,
  \item[(ii)]   $u(x,t)\leq (\geq)\ 0$ on $(\partial\Omega)_T$,
  \item[(iii)]  For every $t\in (0,T)$ and every $\xi\in
       \mathcal{F}$,
\[
  \int_{\Omega}u(x,t)\xi(x,t)dx\
   \leq (\geq) \int_{\Omega}u_0(x)\xi(x,0)dx
    +\int_0^t\int_{\Omega}\{u\xi_t+u^m\Delta\xi+u^p\xi\}dx\,ds.
\]
\end{enumerate}
A function $u(x,t)$ is called a (local) solution of (\ref{e:main})
if it is both a subsolution and a supersolution for some $T>0$.
\end{defn}

According to \cite[Thm. 2.1]{A1} and \cite[Thm. 2.1, 2.2,
2.3]{A2}, if $p>m$ or $p=m$ and $\lambda\leq \lambda_1$, the
nonnegative solution of (\ref{e:main}) is unique.  Moreover, if
$u_0\geq v_0\geq 0$, then $u\geq v$.  If $p<m$ or $p=m$ and
$\lambda>\lambda_1$, then the maximal solution $U(x,t)$ of
(\ref{e:main}) with $u_0\equiv 0$ has $U(x,t)\neq 0$, and $U(x,t)$
satisfies a subsolution comparison theory.  Put
\begin{equation}\label{e:subsolution}
    v(x,t)=g(t)\psi^{1/m}(x),
\end{equation}
where  $\psi(x)$ is the first eigenfunction of (\ref{e:eigenvalue
problem}) with $\max\psi(x)=1$.  If $g(t)$ satisfies the ordinary
differential equation
\begin{gather*}
   g'(t)=(\lambda-\lambda_1)g^m(t),\quad g(0)=0,\\
   g(t)>0,\quad\mbox{for } t>0,
\end{gather*}
it can be  verified easily that $v(x,t)$ is a subsolution of
(\ref{e:main}) for $p=m$ and $\lambda>\lambda_1$.  If $p<m$, let
$g(t)$ in (\ref{e:subsolution}) be the solution of
\begin{gather*}
   g'(t)=-\lambda_1 g^m(t)+\lambda g^p(t),\quad g(0)=0,\\
   g(t)>0,\quad\mbox{for } t>0.
\end{gather*}
Then $v(x,t)$ is also a subsolution of (\ref{e:main}).  The fact
that $U(x,t)>0$ in $\Omega$ for all $t>0$ follows from the
subsolution comparison theory. From the above, we have the
following statement.

\begin{thm}
Assume that $p<m$ or $p=m$ and $\lambda>\lambda_1$.  Then for any
nonnegative initial data $u_0\in L^{\infty}(\Omega)$, the maximal
solution $U(x,t)$ of {\rm (\ref{e:main})} can't vanishes in finite
time.
\end{thm}

For the case $p=m$ and $\lambda=\lambda_1$, $k\psi(x)$, $k>0$, is
a steady state solution of (\ref{e:main}).  Then for any
nontrivial nonnegative initial data, the solution $u(x,t)$ of
(\ref{e:main}) satisfies that $u(x,t)>0$ in $\Omega$ for $t>0$ or
$u(x,t)$ is identically zero.

In the next section we consider the case $p>m$ or $p=m$ and
$\lambda\leq \lambda_1$.

\section{Extinction in finite time}

The regularities of the solution of (\ref{e:main}) can be found in
\cite{Sa}. Multiplying the first equation of (\ref{e:main}) by
$u^{s-1}$, $s>1$, and integrating over $\Omega$, we obtain
\begin{equation}\label{e:main integral}
   \frac{1}{s}\frac{d}{dt}\int_{\Omega}u^sdx+\frac{4m(s-1)}{(m+s-1)^2}
   \int_{\Omega}|\nabla u^{\frac{m+s-1}{2}}|^2dx
   =\lambda\int_{\Omega}u^{p+s-1}dx.
\end{equation}

\begin{thm}\label{thm:p large than m}
Assume that $0<m<1$ and $p>m$.  Then the unique solution of {\rm
(\ref{e:main})} vanishes in finite time for small initial data.
\end{thm}

\begin{proof}
We  consider first the case $p\leq 1$.  For $\frac{N-2}{N+2}\leq
m<1$, let $s=1+m$ in (\ref{e:main integral}). By the H\"{o}lder
inequality and the embedding theorem, we have
\begin{align*}
  \| u(\cdot,t)\|^m_{1+m,\Omega}
  &\leq |\Omega|^{\frac{m}{1+m}-\frac{N-2}{2N}}
    \| u^m(\cdot,t)\|_{\frac{2N}{N-2},\Omega}\\
 &\leq \gamma\mid\Omega\mid^{\frac{m}{1+m}-\frac{N-2}{2N}}
     \| \nabla u^m(\cdot,t)\|_{2, \Omega}.
\end{align*}
where $\gamma$ is the embedding constant.  This remarks in
(\ref{e:main integral}) yields the differential inequality
\[
  \frac{d}{dt}\| u(\cdot,t)\|_{1+m,\Omega}
  + \gamma^{-2}|\Omega|^{\frac{N-2}{N}-\frac{2m}{1+m}}
    \| u(\cdot,t)\|^m_{1+m,\Omega}
 \leq  \lambda|\Omega|^{1-\frac{p+m}{1+m}}
     \| u(\cdot,t)\|^p_{1+m,\Omega}.
\]
Choose
\[
     \|
     u_0\|^{p-m}_{1+m,\Omega}<\lambda^{-1}\gamma^{-2}
      |\Omega|^{\frac{p-m}{1+m}-\frac{2}{N}}\,.
\]
Then
\begin{equation}\label{e:s 1 m}
  \frac{d}{dt}\| u(\cdot,t)\|_{1+m,\Omega}
  +c_1\| u(\cdot,t)\|^m_{1+m,\Omega}\leq 0,
\end{equation}
where
\[
     c_1=\gamma^{-2}|\Omega|^{\frac{N-2}{N}-\frac{2m}{1+m}}
      -\lambda|\Omega|^{1-\frac{p+m}{1+m}}
     \| u_0\|^{p-m}_{1+m,\Omega}.
\]
Integrating (\ref{e:s 1 m}) gives
\[
     \| u(\cdot,t)\|^{1-m}_{1+m,\Omega}\leq
     \| u_0\|^{1-m}_{1+m,\Omega}-(1-m)c_1t,
\]
as long as the right side is nonnegative.  From this,
\[
     \| u(\cdot,t)\|_{1+m,\Omega}\leq
     \| u_0\|_{1+m,\Omega}\big\{1-\frac{(1-m)c_1t}
     {\| u_0\|^{1-m}_{1+m,\Omega}}\big\}_+^{\frac{1}{1-m}}.
\]
Next we take $m$ in such that $0<m<(N-2)/N$. In (\ref{e:main
integral}), let
\[
    s=\frac{N}{2}(1-m)>1.
\]
By the embedding theorem and the specific choice of $s$, we obtain
\[
  \|u(\cdot,t)\|^{\frac{m+s-1}{2}}_{s,\Omega}
  =\|u^{\frac{m+s-1}{2}}(\cdot,t)\|_{\frac{2N}{N-2},\Omega}
  \leq \gamma\| \nabla u^{\frac{m+s-1}{2}}(\cdot,t)\|_{2,\Omega}.
\]
We conclude that
\[
  \frac{d}{dt}\| u(\cdot,t)\|_{s,\Omega}
  + \gamma^{-2}\frac{4m(s-1)}{(m+s-1)^2}
    \| u(\cdot,t)\|^m_{s,\Omega}
  \leq \lambda|\Omega|^{1-\frac{p+s-1}{s}}
     \| u(\cdot,t)\|^p_{s,\Omega}.
\]
Choose
\[
\|u_0\|^{p-m}_{s,\Omega}<\lambda^{-1}\gamma^{-2}\frac{4m(s-1)}{(m+s-1)^2}
      |\Omega|^{\frac{p+s-1}{s}-1}\,.
\]
Then
\[
  \frac{d}{dt}\| u(\cdot,t)\|_{s,\Omega}
  +c_2\| u(\cdot,t)\|^m_{s,\Omega}\leq 0,
\]
where
\[
     c_2=\gamma^{-2}\frac{4m(s-1)}{(m+s-1)^2}
    -\lambda|\Omega|^{1-\frac{p+s-1}{s}}
     \| u_0\|^{p-m}_{s,\Omega}.
\]
By integration, we have
\[
     \| u(\cdot,t)\|_{s,\Omega}\leq
     \| u_0\|_{s,\Omega}\big\{1-\frac{(1-m)c_2t}
     {\| u_0\|^{1-m}_{s,\Omega}}\big\}_+^{\frac{1}{1-m}}.
\]
For the case $p>1$, for sufficiently small $k>0$, it can be easily
verified that $k\psi^{1/m}(x)$ is a supersolution of
(\ref{e:main}), where $\psi(x)$ is the first eigenfunction of
(\ref{e:eigenvalue problem}) with $\max\psi(x)=1$.  Then
\[
    u(x,t)\leq k\psi^{1/m}(x), \quad  t>0,
\]
by the comparison principle if $u_0(x)\leq k\psi^{1/m}(x)$ in
$\Omega$.  From this (\ref{e:main integral}) can be rewritten as
\[
   \frac{1}{s}\frac{d}{dt}\int_{\Omega}u^sdx+\frac{4m(s-1)}{(m+s-1)^2}
   \int_{\Omega}|\nabla u^{\frac{m+s-1}{2}}|^2dx
          \leq\lambda k^{p-1}\int_{\Omega}u^sdx,
\]
to which the above argument can be applied. The proof is
completed.
\end{proof}

\begin{rem} \label{rmk2.2}\rm
The method of the above proof is a modification of the argument in
\cite[Prop. 10]{BC} and \cite[Prop. VII. 2.1]{D}.
\end{rem}

\begin{thm} \label{thm2.3}
Assume that $0<m=p<1$ and $\lambda<\lambda_1$.  Then for any
nonnegative initial data, the solution of {\rm (\ref{e:main})}
vanishes in finite time.
\end{thm}

\begin{proof}
First we apply the argument in the above theorem to get some
results.  We consider two cases: $\frac{N-2}{N+2}\leq m<1$ and
$m<\frac{N-2}{N+2}$.  In the first case, let $s=1+m$ in
(\ref{e:main integral}).  Noticing that
\[
    \lambda_1=\inf_{v\in H_0^1(\Omega), v\neq 0}
          \frac{\int_{\Omega}|\nabla
          v|^2dx}{\int_{\Omega}|v|^2dx},
\]
we obtain
\[
  \frac{1}{1+m}\frac{d}{dt}\| u(\cdot,t)\|^{1+m}_{1+m,\Omega}
  + \big(1-\frac{\lambda}{\lambda_1}\big)\| \nabla
  u^m(\cdot,t)\|^2_{2,\Omega}\leq 0.
\]
Since $\lambda<\lambda_1$, as in the above proof of
Theorem~\ref{thm:p large than m}, there exists $T^*(u_0)<\infty$
such that $u(x,t)\equiv 0$ for all $t\geq T^*(u_0)$.  In the
second case, let $s=\frac{N}{2}(1-m)>1$ in (\ref{e:main
integral}).  Then we have
\[
  \frac{1}{s}\frac{d}{dt}\| u(\cdot,t)\|^s_{s,\Omega}
  + \big(\frac{4m(s-1)}{(m+s-1)^2}-\frac{\lambda}{\lambda_1}\big)\|
\nabla
  u^{\frac{m+s-1}{2}}(\cdot,t)\|^2_{2,\Omega}\leq 0.
\]
Set
\[
      \lambda^*=\frac{(m+s-1)^2}{4m(s-1)}\lambda>\lambda.
\]
Then if $\lambda_1>\lambda^*$, $u(x,t)$ with any initial data
vanishes in finite time.

To fill the gap where $m<\frac{N-2}{N+2}$ and
$\lambda<\lambda_1<\lambda^*$, we apply a supersolution argument.
In fact this supersolution argument can apply to all the case of
$0<m<1$ and $\lambda<\lambda_1$.  Denote by $\psi(x)$ the first
eigenfunction of (\ref{e:eigenvalue problem}) with
$\max_{x\in\Omega}\psi(x)=1$.  Let $g(t)$ be the solution of the
differential equation
\begin{gather*}
   g'(t)=-(\lambda_1-\lambda)g^m(t),\\
   g(0)=\theta,
\end{gather*}
where $\theta$ is chosen that $u_0(x)\leq \theta(\psi)^{1/m}(x)$
in $\Omega$.  Thus $v(x,t)=g(t)(\psi)^{1/m}(x)$ is a supersolution
of (\ref{e:main}).  Since $0<m<1$, $g(t)$ vanishes in finite time.
Then the theorem follows from the comparison principle.
\end{proof}

We note that the unique solution of the problem
\begin{equation}\label{e:homogeneous problem}
\begin{gathered}
     u_t=\Delta u^m,\quad x\in\Omega,\; t>0, \\
     u=0, \quad x\in\partial\Omega,\; t>0,\\
     u(x,0)=u_0(x)\geq 0,\quad x\in\Omega,
\end{gathered}
\end{equation}
$0<m<1$, is a subsolution of (\ref{e:main}).  Since the solution
of the problem (\ref{e:homogeneous problem}) is positive
everywhere in $\Omega$ unless it is identically zero, by
comparison we conclude that, denoting by $T^*<\infty$ the
extinction time of the solution of (\ref{e:main}), we have
\[
     u(x,T^*)\equiv 0, \quad\mbox{and}\quad  u(x,t)>0\quad
     \mbox{in } \Omega,\; 0<t<T^*.
\]
In the following we consider the solution of (\ref{e:main}) with
negative initial energy.  Define
\begin{gather*}
   \mathcal{E}(u(t))=\frac{1}{2}\int_{\Omega}|\nabla
          u^m|^2dx-\frac{\lambda m}{p+m}\int_{\Omega}u^{p+m}dx\\
   \mathcal{H}(u(t))=\frac{1}{1+m}\int_{\Omega}u^{1+m}dx.
\end{gather*}
Differentiating $\mathcal{E}(u(t))$ and $\mathcal{H}(u(t))$, we
obtain
\[
   \frac{d}{dt}\mathcal{E}(u(t))=-\int_{\Omega}u_t(u^m)_tdx
   =-\frac{4m}{(1+m)^2}
   \int_{\Omega}\big[(u^{\frac{1+m}{2}})_t\big]^2dx,
\]
and
\begin{align*}
   \frac{d}{dt}\mathcal{H}(u(t))
   &=\int_{\Omega}u^mu_tdx\\
   &= -\int_{\Omega}|\nabla
          u^m|^2dx+\lambda\int_{\Omega}u^{p+m}dx\\
   &=-2\mathcal{E}(u(t))+\lambda\big(1-\frac{2m}{p+m}\big)
       \int_{\Omega}u^{p+m}dx.
\end{align*}
 From this, $\mathcal{E}(u(t))\leq 0$ provided that
$\mathcal{E}(u_0)\leq 0$.  Hence, if $p>m$, we have
\begin{equation}\label{e:differential inequality}
   \frac{d}{dt}\mathcal{H}(u(t))\geq\lambda\big(1-\frac{2m}{p+m}\big)
       \int_{\Omega}u^{p+m}dx.
\end{equation}
By the H\"{o}lder inequality, for $p\geq 1$,
\[
    \frac{d}{dt}\mathcal{H}(u(t))\geq
    c_3\mathcal{H}^{\frac{p+m}{1+m}}(u(t)),
\]
where
\[
     c_3=\lambda\big(1-\frac{2m}{p+m}\big)(1+m)^{\frac{p+m}{1+m}}
     |\Omega|^{1-\frac{p+m}{1+m}}.
\]
By integration, if $p>1$, there exists $T^*<\infty$ such that
\[
     \lim_{t\to T^*}\mathcal{H}(u(t))=\infty,
\]
provided that $\mathcal{H}(u_0)>0$.  When $p=1$, we have
\[
     \lim_{t\to \infty}\mathcal{H}(u(t))=\infty,
\]
if $\mathcal{H}(u_0)>0$.  For $m<p<1$, integrating
(\ref{e:differential inequality}) over $(0,t)$ gives
\[
    \mathcal{H}(u(t))\geq
\mathcal{H}(u_0)+\lambda\big(1-\frac{2m}{p+m}\big)
       \int_0^t\int_{\Omega}u^{p+m}dxds.
\]
Suppose on the contrary that $\|u(\cdot,t)\|_{\infty,\Omega}\leq
M<\infty$ for all $t>0$. Then,
\[
     \frac{M^{1-p}}{1+m}\int_{\Omega}u^{p+m}dx\geq
      \mathcal{H}(u_0)+\lambda\big(1-\frac{2m}{p+m}\big)
       \int_0^t\int_{\Omega}u^{p+m}dx\,ds\,.
\]
The Gronwall inequality implies that
\[
    \lim_{t\to \infty}\int_{\Omega}u^{p+m}dx=\infty,
\]
which is a contradiction.  Therefore, we have the following
statement.

\begin{thm}
Assume that $0<m<1$ and $p>m$.  If $u^m_0\in H_0^1(\Omega)$
satisfies
\[
    \mathcal{E}(u_0)\leq 0,\ \ \ \mathcal{H}(u_0)>0,
\]
then there exists $T^*\leq \infty$ such that
\[
    \lim_{t\to T^*}\|
u(\cdot,t)\|_{\infty,\Omega}=\infty.
\]
\end{thm}

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\end{document}