\section{Introduction\label{sec1}}

Here we study some parabolic equations with eventual gradient absorption
terms. We are mainly concerned by the nonnegative solutions of the well known
viscous parabolic Hamilton-Jacobi equatio
\begin{equation}
u_{t}-\nu\Delta u+|\nabla u|^{q}=0 \label{un
\end{equation}
in $Q_{\Omega,T}=\Omega\times\left(  0,T\right)  ,$ $T\leqq\infty,$ where
$q>1,\nu\geqq0,$ and $\Omega=\mathbb{R}^{N},$ or $\Omega$ is a smooth bounded
domain of $\mathbb{R}^{N}$ and $u=0$ on $\partial\Omega\times\left(
0,T\right)  .$ We also consider the (signed) solutions of equations of the
type
\begin{equation}
u_{t}-\Delta_{p}u+\left\vert u\right\vert ^{\lambda-1}u|\nabla u|^{q}=0
\label{fru
\end{equation}
where $p>1$ and $\Delta_{p}$ is the $p$-Laplacian, or more generally involving
a quasilinear operator, nonnecessarily monotone,
\begin{equation}
u_{t}-\operatorname{div}(\mathrm{A}(x,t,u,\nabla u))+g(x,u,\nabla u)=0
\label{eqge
\end{equation}
with natural growth conditions on $\mathrm{A,}$ and nonnegativity conditions
\begin{equation}
\mathrm{A}(x,t,u,\eta).\eta\geqq\nu\left\vert \eta\right\vert ^{p},\qquad
g(x,u,\eta)u\geqq\gamma\left\vert u\right\vert ^{\lambda+1}|\nabla
u|^{q}\qquad\gamma\geqq0,\nu\geqq0, \label{alp
\end{equation}
where $\lambda\geqq0$. \medskip

We denote by$\mathcal{M}_{b}(\Omega)$ the set of bounded Radon measures in
$\Omega,$ and $\mathcal{M}_{b}^{+}(\Omega)$ the subset of nonnegative ones. We
set $Q_{\Omega,s,\tau}=\Omega\times\left(  s,\tau\right)  ,$ for any $0\leqq
s<\tau\leqq\infty,$ thus $Q_{\Omega,T}=Q_{\Omega,0,T}.$\medskip

We study the Cauchy problem with rough initial dat
\[
u(.,0)=u_{0},\qquad u_{0}\in L^{r}(\Omega),r\geqq1,\quad\text{or }u_{0
\in\mathcal{M}_{b}(\Omega).
\]
Our purpose is to give some decay estimates, and a regularizing effect
$L^{\infty}$ estimates, for the solutions, in terms of initial data $u_{0},$
and universal estimates when $\Omega$ is bounded, \textit{under very few
assumptions on the solutions}. In this problem two regularizing effects can
occur, the first one due to the gradient term $|\nabla u|^{q},$ when
$\gamma>0,$ the second to the operator when $\nu>0.$ A part of these estimates
are well known when the solutions can be approximated by smooth solutions, and
satisfy some conditions as $\left\vert x\right\vert \rightarrow$ $\infty$ when
$\Omega=\mathbb{R}^{N},$ of boundedness or integrability, for example
semi-group solutions. Our approach is different: our results are valid for all
the solutions of the equation \textit{in a weak sense}: in the sense of
distributions for the case of the Laplacian, in the renormalized sense in the
case of a general operator; and we make \textit{no assumption of existence or
uniqueness}. Moreover in the case of the Hamilton-Jacobi equation in
$\mathbb{R}^{N}$, we make \textit{no assumption as }$\left\vert x\right\vert
\rightarrow\infty$\textit{, all our assumptions are local. } As a consequence
we deduce \textit{new uniqueness results} for equation (\ref{un}) in
$\mathbb{R}^{N}$ or in bounded $\Omega.\bigskip$

In order to get regularizing properties, we give at Section \ref{sec2} an
iteration lemma based of Moser's method, inspired by the results of \cite{Ve},
and we compare it to results of \cite{Por} obtained from Stampacchia's method.
The Moser's method, based on the choice of test functions of the form
$\left\vert u\right\vert ^{\alpha-1}u,$ $\alpha>0,$ appears to be well adapted
to equations in a $L^{1}$ context. Since such functions are not always
admissible, we combine the method with a \textit{regularization} in case of
equation (\ref{un}) in $\mathbb{R}^{N}$, and a \textit{truncature} in the case
of the Dirichlet problem, for the same equation, and for the general equation
(\ref{eqge}). \bigskip

In Section \ref{Rn} we study the case of Hamilton-Jacobi equation (\ref{un})
in $\mathbb{R}^{N},$ for which there is a huge litterature. Among them we
quote only some significative contributions and refer to the references
therein: \cite{AmBA}, \cite{BeLa99}, \cite{BeBALa}, \cite{BASoWe},
\cite{SoZh}, see also \cite{BeBALa}, \cite{BeLaSc}, \cite{GaLa}. One of our
main results reads as follows:

\begin{theorem}
\label{main}Let $u\in L_{loc}^{1}(Q_{\mathbb{R}^{N},T}),$ with $|\nabla u|\in
L_{loc}^{q}(Q_{\mathbb{R}^{N},T}),$ be any nonnegative solution of equation
(\ref{un}) in$\hspace{0.05in}\mathcal{D}^{\prime}(Q_{\mathbb{R}^{N},T
).$\medskip

(i) Let $u_{0}\in L^{r}(\mathbb{R}^{N}),r\geqq1.$ Assume that $u\in C(\left[
0,T\right)  ;L_{loc}^{r}(\mathbb{R}^{N}))$ and $u(.,0)=u_{0}$. Then $u\in
C(\left[  0,T\right)  ;L^{r}(\mathbb{R}^{N}));$ and for any $t\in(0,T),$
$u(.,t)\in L^{\infty}(\mathbb{R}^{N})$ an
\begin{equation}
\Vert u(.,t)\Vert_{L^{r}(\mathbb{R}^{N})}\leqq\Vert u_{0}\Vert_{L^{r
(\mathbb{R}^{N})}, \label{pas
\end{equation
\[
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\sigma}\Vert
u_{0}\Vert_{L^{r}(\mathbb{R}^{N})}^{\varpi},\qquad C=C(N,q,r),
\]
where $\sigma,\varpi$ are given for $q<N$ by $\sigma=1/(rq/N+q-1)=N\varpi/rq;$
and if $\nu>0,$ $N>2,$
\[
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\frac{N}{2r}}\Vert
u_{0}\Vert_{L^{r}(\mathbb{R}^{N})},\qquad C=C(N,q,r,\nu).
\]


(ii) Let $u_{0}\in\mathcal{M}_{b}^{+}(\mathbb{R}^{N})$ and assume that
$u(.,t)$ converges weakly $^{\ast}$ to $u_{0}$ as $t\rightarrow0.$ Then $u\in
C((0,T);L^{1}(\mathbb{R}^{N})),$ and for any $t\in(0,T),$ $\Vert
u(.,t)\Vert_{L^{r}(\mathbb{R}^{N})}\leq
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0},$
\begin{align*}
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}  &  \leqq Ct^{-\sigma_{1,q}}
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0})^{\varpi_{1,q}},\qquad C=C(N,q),\\
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}  &  \leqq Ct^{-\frac{N}{2}
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0},\qquad C=C(N,q,\nu),\qquad\text{if }\nu>0.
\end{align*}

\end{theorem}

For any $q\leq2,$ we deduce estimates of the gradient, obtained from Bernstein
technique. As a consequence we improve some uniqueness results of
\cite{BeLa99} and \cite{BASoWe}:

\begin{theorem}
\label{uni1}(i) Let $1<q<(N+2)/(N+1),$ and $u_{0}\in\mathcal{M}_{b
^{+}(\mathbb{R}^{N}).$ Then there exists a unique nonnegative function $u\in
L_{loc}^{1}(Q_{\mathbb{R}^{N},T}),$ such that $|\nabla u|\in L_{loc
^{q}(Q_{\mathbb{R}^{N},T}),$ solution of equation (\ref{un}) in $\mathcal{D
^{\prime}(Q_{\mathbb{R}^{N},T})$ such that
\[
\lim_{t\rightarrow0}\int_{\mathbb{R}^{N}}u(.,t)\psi dx=\int_{\mathbb{R}^{N
}\psi du_{0},\qquad\forall\psi\in C_{c}(\mathbb{R}^{N}).
\]


(ii) Let $u_{0}\in L^{r}(\mathbb{R}^{N}),$ $r\geqq1$ and $1<q<(N+2r)/(N+r).$
Then there exists a unique nonnegative solution $u$ as above, such that $u\in
C\left(  \left[  0,T\right)  ;L_{loc}^{r}(\mathbb{R}^{N})\right)  $ and
$u(.,0)=u_{0}.$\bigskip
\end{theorem}

We also find again the existence result of \cite[Theorem 4.1]{BASoWe} for any
$u_{0}\in L^{r}(\mathbb{R}^{N}),$ $r\geqq1$, see Proposition \ref{pou}.
Finally we improve the estimate (\ref{pas}) when $q<(N+2r)/(N+r),$ see Theorem
\ref{esti}. \bigskip

In Section \ref{omega} we study the Dirichlet problem in a bounded domain
$\Omega$:
\begin{equation}
\left\{
\begin{array}
[c]{l
u_{t}-\nu\Delta u+|\nabla u|^{q}=0,\quad\text{in}\hspace{0.05in}Q_{\Omega
,T},\\
u=0,\quad\text{on}\hspace{0.05in}\partial\Omega\times(0,T),\\
u(x,0)=u_{0}\geqq0,
\end{array}
\right.  \label{1.1
\end{equation}
Here also the problem is the object of many works, such as \cite{CLS},
\cite{BeDa}, \cite{So}, \cite{BeDaLa}, \cite{Por}. We give decay properties
and regularizing effects valid for \textit{any weak solution} of the problem,
in particular the universal estimate
\[
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\frac{1}{q-1}}\quad\text{in
}\left(  0,T\right)  ,
\]
where $C=C(N,q),$ see Theorem \ref{deco}. And we improve the uniqueness
results of \cite{BeDa}:

\begin{theorem}
\label{uni2}Assume that $\Omega$ is bounded.\medskip

\noindent(i) Let $1<q<(N+2)/(N+1),$ and $u_{0}\in\mathcal{M}_{b}^{+}(\Omega).$
Then there exists a unique nonnegative function $u\in C((0,T);L^{1}\left(
\Omega\right)  )\cap L_{loc}^{1}((0,T);W_{0}^{1,1}\left(  \Omega\right)  ),$
such that $|\nabla u|^{q}\in L_{loc}^{1}((0,T);L^{1}\left(  \Omega\right)  )$
solution of equation (\ref{un}) in $\mathcal{D}^{\prime}(Q_{\Omega,T})$ such
that
\[
\lim_{t\rightarrow0}\int_{\Omega}u(.,t)\psi dx=\int_{\Omega}\psi du_{0
,\qquad\forall\psi\in C_{b}(\Omega).
\]


\noindent(ii) Let $u_{0}\in L^{r}(\Omega),$ $r\geqq1,$ and $1<q<(N+2r)/(N+r).$
Then there exists a unique nonnegative solution $u$ as above, such that $u\in
C\left(  \left[  0,T\right)  ;L^{r}(\Omega)\right)  $ and $u(.,0)=u_{0}.$
\end{theorem}

And we show the existence of solutions for any $u_{0}\in L^{r}(\Omega),$
$r\geqq1,$ see Proposition \ref{plu}.\bigskip

In Section \ref{quas} we extend some results of section \ref{omega} to the
case of the quasilinear equations (\ref{eqge}), with initial data $u_{0}\in
L^{r}\left(  \Omega\right)  $ or $u_{0}$ measure, and $u$ may be a signed
solution. In the case of equatio
\[
u_{t}-\Delta_{p}u=0,
\]
with rough initial data, several local or global $L^{\infty}$ estimates and
Harnack properties have been obtained in the last decades, see for example the
pioneer works of \cite{Ve}, \cite{DiBeHe1}, \cite{DiBeHe2}, \cite{Ju}, and
\cite{DiBeGV}, \cite{BonIaVa} and references therein. Regularizing properties
for equation (\ref{fru}) are given in \cite{Po} in an hilbertian context in
case $g=0$ or $p=2$. \bigskip

For this kind of problems, we combine our iteration method of Section
\ref{sec2} with a notion of \textit{renormalized solution}, developped by many
authors \cite{BlMu}, \cite{Po},\cite{Pr}, well adapted to our context of rough
initial data: we do not require that $u(.,t)\in L^{2}(\Omega),$ but we only
assume that the truncates $T_{k}(u)$ of $u$ by $k>0$ lie in $L^{p
((0,T);W^{1,p}(\Omega))$. We prove decay and $L^{\infty}$ estimates of the
following type: if $\gamma>0$, for any $r\geqq1,$ $p>1$ and for example
$q\in\left(  1,N\right)  ,$ the
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\sigma}\Vert u_{0
\Vert_{L^{r}(\Omega)}^{\varpi},\qquad\sigma=\frac{1}{\frac{rq}{N}+\lambda
+q-1}=\frac{N}{rq}\varpi, \label{ddd
\end{equation}
and we deduce a universal estimate as before. If $\nu>0,$ then for any
$r\geqq1,$ and any $p\in(1,N)$ such that $p>2N/(N+2),$
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\tilde{\sigma}}\Vert
u_{0}\Vert_{L^{r}(\Omega)}^{\tilde{\varpi}},\qquad\tilde{\sigma}=\frac
{1}{\frac{rp}{N}+p-2}=\frac{N}{rp}\tilde{\varpi}. \label{eee
\end{equation}
Such methods can also be extended to porous media equations, and doubly
nonlinear equations involving operators of the form $u\mapsto-\Delta
_{p}(\left\vert u\right\vert ^{m-1}u).$

\section{A Moser's type iteration lemma\label{sec2}}

We begin by a simple bootstrap property, used for example in \cite{Ve}: We
recall the proof for simplicity:

\begin{lemma}
\label{elem}Let $\omega\in\left(  0,1\right)  $ and $\sigma>0,$ and $K>0.$ Let
$y$ be any positive function on $(0,T)$ such that for any $0<s<t<T$
\[
y(t)\leqq K(t-s)^{-\sigma}y^{\omega}(s)
\]
and $y(t)\leqq Mt^{-\sigma}$ for some $M>0.$ Then $y$ satisfies an estimate
independent of $M:$ for any $t\in\left(  0,T\right)  $
\[
y(t)\leqq2^{\sigma(1-\omega)^{-2}}(Kt^{-\sigma})^{(1-\omega)^{-1}
\]

\end{lemma}

\begin{proof}
We get by induction
\[
\left\{
\begin{array}
[c]{c
y(t)\leqq K2^{\sigma}t^{-\sigma}y^{\omega}(t/2),\\
y^{\omega}(t/2)\leqq K^{\omega}2^{2\sigma\omega}t^{-\sigma\omega}y^{\omega
^{2}}(t/2^{2}),...\\
y^{\omega^{n-1}}(t/2^{n-1})\leqq K^{\omega^{n-1}}2^{n\sigma\omega^{n-1
}t^{-\sigma\omega^{n-1}}y^{\omega^{n}}(t/2^{n}),\\
y^{\omega^{n}}(t/2^{n})\leqq2^{n\sigma\omega^{n}}t^{-\sigma\omega^{n
}M^{\omega^{n}}.
\end{array}
\right.
\]
The
\[
y(t)\leqq K^{\sum_{k=0}^{n-1}\varpi^{k}}t^{-\sigma}{}^{\sum_{k=0}^{n
\varpi^{k}}2^{\sigma\sum_{k=0}^{n}(k+1)\varpi^{k})}M^{\omega^{n+1}
\]
and going to the limit as $n\rightarrow\infty,$ we get the conclusion, since
$\lim M^{\omega^{n+1}}=1.\medskip$
\end{proof}

In the sequel we use the following iteration property:

\begin{lemma}
\label{prod} Let $m>1,$ $\theta>1$ and $\lambda\in\mathbb{R}$ and $C_{0}>0$.
Let $v\in C(\left[  0,T\right)  ;L_{loc}^{1}(\Omega))$ be nonnegative, and
$v_{0}=v(x,0)\in L^{r}(\Omega)$ for some $r\geqq1$ such that
\begin{equation}
r>\frac{N}{m}(1-m-\lambda); \label{ele
\end{equation}
If $r>1$ we assume that for any $0\leqq s<t<T$ and any $\alpha\geqq r-1$ there
holds
\begin{equation}
\frac{1}{\alpha+1}\int_{\Omega}v^{\alpha+1}(.,t)dx+\frac{C_{0}}{\beta^{q}
\int_{s}^{t}(\int_{\Omega}v^{\beta m\theta}(.,\tau)dx)^{\frac{1}{\theta
})d\tau\leqq\frac{1}{\alpha+1}\int_{\Omega}v^{\alpha+1}(.,s)dx\leqq
\infty\label{mac
\end{equation}
where
\[
\beta=\beta(\alpha)=1+\frac{\alpha+\lambda}{m}.
\]
If $r=1$ we make one of the two following assumptions:\medskip

(H$_{1}$) (\ref{mac}) holds for any $\alpha\geqq0$,\medskip

(H$_{2}$) $\int_{\Omega}v(.,t)dx\leqq\int_{\Omega}v_{0}dx$ for any
$t\in\left(  0,T\right)  $, and $v_{0}\in L^{\rho}(\Omega)$ for some $\rho>1,$
and (\ref{mac}) holds for any $\alpha\geqq\rho-1.$\medskip

\noindent(i) Then there exists $C>0,$ depending on $N,m,r,\lambda,C_{0},$ and
eventually $\rho,$ such that for any $t\in(0,T),$
\begin{equation}
\Vert v(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\sigma_{r,m,\lambda,\theta
}\Vert v_{0}\Vert_{L^{r}(\Omega)}^{\varpi_{r,m,\lambda,\theta}}, \label{trap
\end{equation}
where
\begin{equation}
\sigma_{r,m,\lambda,\theta}=\frac{1}{\frac{r}{\theta^{\prime}}+\lambda
+m-1}=\frac{\theta^{\prime}}{r}\varpi_{r,m,\lambda,\theta}. \label{jam
\end{equation}
(ii) Moreover if $\lambda+m-1>0,$ and $\Omega$ is bounded, then a universal
estimate holds, with a constant $C$ depending on $N,m,r,\lambda,C_{0
,\left\vert \Omega\right\vert $ and eventually $\rho:$ for any $t\in(0,T),$
\[
\Vert v(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\frac{1}{m-1+\lambda}}.
\]

\end{lemma}

\begin{proof}
(i) $\bullet$ \textbf{Case }$r>1.$ Let $\alpha\geqq r-1$. From (\ref{mac}), It
implies the decay: $\int_{\Omega}v^{\alpha+1}(t)dx$ is decreasing for $t>s$.
And $\int_{\Omega}v^{\beta m\theta}(.,\xi)dx$ is finite for almost any $\xi
\in\left(  s,t\right)  .$ From assumption (\ref{ele}), and $\alpha\geqq r-1,$
there holds $\beta m\theta>\alpha+1.$ Replacing $\alpha$ by $\beta m\theta-1,$
taking $\xi_{n}\rightarrow s$ we have $\int_{\Omega}v^{\beta m\theta
(.,\xi)dx\leqq\int_{\Omega}u^{\beta m\theta}(.,\xi_{n})dx$ for any $\xi
>\xi_{n},$ then
\[
(t-\xi_{n})(\int_{\Omega}v^{\beta m\theta}(.,\xi_{n})dx)^{\frac{1}{\theta
}\leqq\int_{\xi_{n}}^{t}(\int_{\Omega}v^{\beta m\theta}(.,\xi)dx)^{\frac
{1}{\theta}}dt
\]
and also
\[
\int_{\Omega}v^{\alpha+1}(.,t)dx+\frac{C_{0}(\alpha+1)}{\beta^{q}}(\int
_{\xi_{n}}^{t}(\int_{\Omega}v^{\beta m\theta}(.,\xi)dx)^{\frac{1}{\theta
}dt\leqq\int_{\Omega}v^{\alpha+1}(.,\xi_{n})dx\leqq\int_{\Omega}v^{\alpha
+1}(.,s)dx.
\]
thus
\[
\int_{\Omega}u^{\alpha+1}(.,t)dx+\frac{C_{0}(\alpha+1)}{\beta^{q}}(t-\xi
_{n})(\int_{\Omega}v^{\beta m\theta}(.,\xi_{n})dx)^{^{\frac{1}{\theta}}
\leqq\int_{\Omega}v^{\alpha+1}(.,s)dx.
\]
Then going to the limit as $n\rightarrow\infty,$ since $v$ $\in C(\left[
0,T\right]  ;L_{loc}^{1}(\Omega))$ when $\xi_{n}\rightarrow s,$ $v(.,\xi
_{n})\rightarrow v(.,s)$ in $L^{1}(\Omega),$ and after extraction, a.e. in
$\Omega.$ Then from the Fatou lemma,
\[
\int_{\Omega}v^{\alpha+1}(.,t)dx+\frac{C_{0}(\alpha+1)}{\beta^{q}
(t-s)(\int_{\Omega}v^{\beta m\theta}(.,s)dx)^{^{\frac{1}{\theta}}}\leqq
\int_{\Omega}v^{\alpha+1}(.,s)dx.
\]
Hence
\begin{equation}
\Vert v(t)\Vert_{L^{\beta m\theta}(\Omega)}^{\beta m\theta}\leqq\left(
\frac{\beta^{m}}{C_{0}(\alpha+1)}\frac{1}{t-s}\Vert v(s)\Vert_{L^{\alpha
+1}(\Omega)}^{\alpha+1}\right)  ^{\theta}, \label{1.6
\end{equation}
We start from $s=0,$ we have $v_{0}\in L^{r}(\Omega).\medskip$

We take\textbf{ } $\alpha_{0}=r-1,$ thus $\int_{\Omega}v^{\alpha_{0}+1}(t)dx$
is finite, and set $\beta_{0}=1+\left(  r-1+\lambda\right)  /m.$ We define
increasing sequences $\left(  t_{n}\right)  ,\left(  \alpha_{n}\right)
,\left(  r_{n}\right)  ,\left(  \beta_{n}\right)  ,$ by $t_{0}=0,$ $r_{0}=r$
and for any $n\geqq1,$
\[
t_{n}=t(1-\frac{1}{2^{n}}),\qquad r_{n}=\alpha_{n}+1,\qquad\beta_{n
=1+\frac{\alpha_{n}+\lambda}{m},\qquad r_{n+1}=\beta_{n}m\theta=(r_{n
+\lambda+m-1)\theta.
\]
In (\ref{1.6}), we replace $s,t,\alpha,\beta m\theta,$ by $t_{n},t_{n+1
r_{n},r_{n+1},$ and get
\begin{equation}
\Vert v(t_{n+1})\Vert_{L^{r_{n+1}}(\Omega)}\leqq\left(  \frac{1}{C_{0
(m\theta)^{m}}\frac{r_{n+1}^{m}}{r_{n}}\frac{1}{t_{n+1}-t_{n}}\right)
^{\frac{\theta}{r_{n+1}}}\Vert v(t_{n})\Vert_{L^{r_{n}}(\Omega)}^{\frac
{\theta.r_{n}}{r_{n+1}}}. \label{1.7
\end{equation}
It follows that
\begin{equation}
\Vert v(t_{n+1})\Vert_{L^{r_{n+1}}(\Omega)}\leqq I_{n}J_{n}L_{n}\Vert
u_{0}\Vert_{L^{r}(\Omega)}^{\frac{\theta^{n+1}.r}{r_{n+1}}} \label{1.8
\end{equation}
where
\[
I_{n}
{\displaystyle\prod_{k=1}^{n+1}}
(\frac{r_{k}^{m}}{r_{k-1}})^{\frac{\theta^{n+2-k}}{r_{n+1}}},\quad J_{n}
{\displaystyle\prod_{k=1}^{n+1}}
\left(  \frac{1}{t_{k}-t_{k-1}}\right)  ^{\frac{\theta^{n+2-k}}{r^{n+1}
},\quad L_{n}=\left(  C_{0}(m\theta)^{q}\right)  ^{-\sum_{k=1}^{n+1
\frac{\theta^{n+2-k}}{r_{n+1}}}.
\]
Since $r_{n}=$ $\theta^{n}(r+(\lambda+m-1)\theta^{\prime}(1-\theta^{-n})),$ it
is clear that
\begin{equation}
\lim\frac{\theta^{n+1}r}{r_{n+1}}=\varpi_{r,m,\lambda,\theta},\qquad\lim
\frac{1}{r_{n+1}}\sum_{k=1}^{n+1}\theta^{n+2-k}=\sigma_{r,m,\lambda
,\qquad\lim\sum_{k=1}^{n+1}k\theta^{1-k}=\theta^{\prime2} \label{1.9
\end{equation}
Thus, it follows
\begin{equation}
\lim J_{n}=2^{-\frac{\varpi_{r,m,\lambda,\theta}}{r}\theta^{\prime2}
,\qquad\lim L_{n}=\left(  C_{0}(m\theta)^{q}\right)  ^{-\sigma_{r,m,\lambda
,\theta}}. \label{1.10
\end{equation}
And $I_{n}$ has a finite limit $\ell=\ell(N,m,r,\lambda)$ as $n\rightarrow
\infty.$ Indeed,
\[
\ln I_{n}=\frac{m}{r_{n+1}}\sum_{k=1}^{n+1}\theta^{n+2-k}\ln r_{k}-\frac
{1}{r_{n+1}}\sum_{k=0}^{n}\theta^{n+1-k}\ln r_{k}=\frac{\theta^{n+1}}{r_{n+1
}(m\theta\sum_{k=1}^{n+1}\theta^{-k}\ln r_{k}-\sum_{k=0}^{n}\theta^{-k}\ln
r_{k})
\]
and the sum $S=\sum_{k=0}^{n}\theta^{-k}\ln r_{k}$ is finite, since
$r_{k}\leqq\theta^{k}(r+\left\vert \lambda+m-1\right\vert \theta^{\prime})$.
Then $I_{n}$ has a finite limit $\ell=\ell(N,m,r,\lambda,\theta)=\exp
(r^{-1}\varpi_{r,m,\lambda}((m\theta-1)S-m\theta\ln r)).$ Thus we can go to
the limit in (\ref{1.8}), and the conclusion follows.$\medskip$

$\bullet$ \textbf{Case }$r=1.$ If (H$_{1}$) holds we can take $\alpha
_{0}=r-1=0$ and the proof is done. Next assume (H$_{2}$) Then we obtain, for
any $0\leqq s<t<T,$ and a constant $C$ as before,
\begin{align*}
\Vert v(.,t)\Vert_{L^{\infty}(\Omega)}  &  \leqq C(t-s)^{-\sigma
_{\rho,m,\lambda,\theta}}\Vert v(.,s)\Vert_{L^{\rho}(\Omega)}^{\varpi
_{\rho,m,\lambda,\theta}}\\
&  \leqq C(t-s)^{-\sigma_{\rho,m,\lambda}}\Vert v(.,s)\Vert_{L^{\infty
(\Omega)}^{\varpi_{\rho,m,\lambda,\theta}(\rho-1)/\rho}\Vert v(.,s)\Vert
_{L^{1}(\Omega)}^{\varpi_{\rho,m,\lambda,\theta}/\rho}\\
&  \leqq C\Vert v_{0}\Vert_{L^{1}(\Omega)}^{\varpi_{\rho,m,\lambda,\theta
}/\rho}(t-s)^{-\sigma_{\rho,m,\lambda}}\Vert v(.,s)\Vert_{L^{\infty}(\Omega
)}^{\varpi_{\rho,m,\lambda,\theta}(\rho-1)/\rho
\end{align*}
Let $y(t)=\Vert v(.,t)\Vert_{L^{\infty}(\Omega)}.$ We can apply Lemma
\ref{elem} to $y$, with
\[
\sigma=\sigma_{\rho,m,\lambda,\theta},\quad\quad\omega=\frac{\varpi
_{\rho,m,\lambda,\theta}}{\rho^{\prime}},\quad\quad K=C\Vert v_{0}\Vert
_{L^{1}(\Omega)}^{\varpi_{\rho,m,\lambda,\theta}/\rho},\quad\quad M=C\Vert
v_{0}\Vert_{L^{\rho}(\Omega)}^{\varpi_{\rho,m,\lambda,\theta/\rho}}.
\]
Indeed $\omega<1$ from assumption (\ref{ele}) with $r=1$. Then there holds
\[
\Vert v(.,t)\Vert_{L^{\infty}(\Omega)}\leqq2^{\sigma(1-\omega)^{-2
}(Kt^{-\sigma})^{(1-\omega)^{-1}}=2^{\sigma(1-\omega)^{-2}}C^{(1-\omega)^{-1
}t^{-\sigma(1-\omega)^{-1}}\Vert v_{0}\Vert_{L^{1}(\Omega)}^{\varpi
_{\rho,q,\lambda,\theta}/\rho((1-\omega))}.
\]
And we observe that $\sigma(1-\omega)^{-1}=\sigma_{1,m,\lambda,\theta}$ and
$\varpi_{\rho,m,\lambda,\theta}/\rho((1-\omega))=\varpi_{1,m,\lambda,\theta}$,
then with a new constant $C$, now depending on $\rho,$
\begin{equation}
\Vert v(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\sigma_{1,m,\lambda,\theta
}\Vert v_{0}\Vert_{L^{1}(\Omega)}^{\varpi_{1,m,\lambda,\theta}}. \label{blo
\end{equation}
\bigskip

\noindent(ii) Assume that $\Omega$ is bounded, thus $L^{r}\left(
\Omega\right)  \subset L^{1}\left(  \Omega\right)  .$ We use the result with
$r=1$ and obtain (\ref{blo}), and, for any $0<s<t<T,
\[
\Vert v(t)\Vert_{L^{\infty}(\Omega)}\leqq C(t-s)^{-\sigma_{1,m,\lambda,\theta
}}\Vert v(s)\Vert_{L^{1}(\Omega)}^{\varpi_{1,m,\lambda}}\leqq C(t-s)^{-\sigma
_{1,m,\lambda}}\left\vert \Omega\right\vert ^{\varpi_{1,m,\lambda,\theta
}\Vert v(s)\Vert_{L^{\infty}(\Omega)}^{\varpi_{1,m,\lambda,\theta}
\]
where $C=C(N,m,\lambda,C_{0})$ (or $C=C(N,m,\lambda,C_{0},\rho).$ And
$\varpi_{1,m,\lambda,\theta}<1,$ because $\lambda+m-1>0.$ Then we can apply
Lemma \ref{elem}, and we get
\[
\Vert v(\tau)\Vert_{L^{\infty}(\Omega)}\leqq2^{\sigma_{1,m,\lambda
(1-\varpi_{1,m,\lambda,\theta})^{-2}}(C\left\vert \Omega\right\vert
^{\varpi_{1,m,\lambda,\theta}}t^{-\sigma_{1,m,\lambda,\theta}})^{(1-\varpi
_{1,m,\lambda,\theta})^{-1}}=M\tau^{-\frac{1}{m-1+\lambda}},\forall\tau
\in(0,T).
\]
with $M=M(N,m,\lambda,C_{0},\left\vert \Omega\right\vert )$ (or
$M=M(N,m,\lambda,C_{0},\left\vert \Omega\right\vert ,\rho)$).
\end{proof}

\begin{remark}
This lemma can be compared with the result of \cite[Theorem 2.1]{Por} obtained
by the Stampacchia's method. In order to obtain decay estimates for the
solutions $u$ of a parabolic equation such as (\ref{un}) or (\ref{egge}), the
Moser's method consists to take as test functions powers $\left\vert
u\right\vert ^{\alpha-1}u$ of $u;$ the Stampacchia's method uses test
functions of the form $(u-k)^{+}\mathrm{sign}u.$ If one applies to
sufficiently smooth solutions, both techniques leed to decay estimates of the
same type. In the case of weaker solutions, the second method supposes that
the functions $(u-k)^{+}$ are admissible in the equation, which leads to
assume that $u(.,t)\in W^{1,2}(\Omega),$ see \cite{Por}. In the sequel we
combine Moser's method with regularization or truncature of $u$, in order to
admit powers as test functions.
\end{remark}

\section{The Hamilton-Jacobi equation in $\mathbb{R}^{N}$\label{Rn}}

\subsection{Different notions of solution\label{diff}}

The Hamilton-Jacobi equation was first studied with smooth initial data. Let
us recall the main results:\medskip

$\bullet$ For any nonnegative $u_{0}\in C_{b}^{2}\left(  \mathbb{R
^{N}\right)  ,$ from \cite{AmBA} there a unique global solution in
$C^{2,1}(\mathbb{R}^{N}\times\left[  0,\infty\right)  )$ such that
\[
\left\Vert u(.,t)\right\Vert _{L^{\infty}(\mathbb{R}^{N})}\leqq\left\Vert
u_{0}\right\Vert _{L^{\infty}(\mathbb{R}^{N})},\qquad\left\Vert \nabla
u(.,t)\right\Vert _{L^{\infty}(\mathbb{R}^{N})}\leqq\left\Vert \nabla
u_{0}\right\Vert _{L^{\infty}(\mathbb{R}^{N})}.
\]
Some estimates of the gradient, independant of $\nu,$ have been obtained
\textbf{for this solution}, by using the Bersnstein technique, which consists
in derivating the equation, and computing the equation satisfied by $|\nabla
u|^{2}:$ first from \cite{Li
\[
\left\Vert \nabla u(.,t)\right\Vert _{L^{\infty}(\mathbb{R}^{N})}^{q
\leqq\frac{\left\Vert u_{0}\right\Vert _{L^{\infty}(\mathbb{R}^{N})}}{t},
\]
then from \cite{BeLa99}
\begin{equation}
\Vert\nabla(u^{\frac{q-1}{q}})(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq
C_{q}t^{-1/2}\Vert u_{0}\Vert_{L^{\infty}(\mathbb{R}^{N})}^{\frac{q-1}{q}},
\label{cor
\end{equation
\begin{equation}
\left\Vert \nabla(u^{(q-1)/q}(.,t)\right\Vert _{L^{\infty}(\mathbb{R}^{N
)}\leqq\frac{q-1}{q}\frac{1}{t^{1/q}},\qquad\text{equivalently}\qquad
\left\vert \nabla u(.,t)\right\vert ^{q}\leqq\frac{u(.,t)}{t}\quad
a.e.\text{in }\mathbb{R}^{N} \label{bela
\end{equation}


$\bullet$ For any nonnegative $u_{0}\in C_{b}\left(  \mathbb{R}^{N}\right)  $,
from \cite{GiGuKe} there exists a unique solution such that $u\in
C^{2,1}(Q_{\mathbb{R}^{N},\infty})$ and $u\in C(\mathbb{R}^{N}\times\left[
0,\infty\right)  \cap L^{\infty}(\mathbb{R}^{N}\times\left(  0,\infty\right)
),$ and from \cite{BeBALa} estimates (\ref{cor}) and (\ref{bela}) are still
valid$.\medskip$

In case of rough initial data $u_{0}\in\mathcal{M}_{b}(\mathbb{R}^{N})$ or
$u\in L^{r}(\mathbb{R}^{N}),$ $r\geqq1$, existence results have been obtained
in \cite{BeLa99}, \cite{BASoWe} at section by using different formulations
involving the semi-group of heat equation. Here we consider the solutions in a
weaker sense, which does not use this formulation.

\begin{definition}
We say that a \textit{nonnegative} function $u$ is a \textbf{weak solution}
(resp. subsolution) of equation of (\ref{un}) in $Q_{\mathbb{R}^{N},T},$ if
$u\in L_{loc}^{1}(Q_{\mathbb{R}^{N},T}),$ and $|\nabla u|\in L_{loc
^{q}(Q_{\mathbb{R}^{N},T}),$ and
\begin{equation}
\int_{0}^{T}\int_{\Omega}(-u\varphi_{t}-u\Delta\varphi+|\nabla u|^{q
\varphi)dxdt=0,\quad(\text{resp.}\leqq),\quad\forall\varphi\in\mathcal{D
^{+}(Q_{\mathbb{R}^{N},T}). \label{for
\end{equation}

\end{definition}

\begin{remark}
Recall that from \cite{BiDao1}, any weak solution satisfies
\begin{equation}
u\in L_{loc}^{\infty}(Q_{\mathbb{R}^{N},T}),\qquad\nabla u\in L_{loc
^{2}(Q_{\mathbb{R}^{N},T}),\qquad u\in C((0,T);L_{loc}^{\rho}(\mathbb{R
^{N}))\quad\forall\rho\geqq1. \label{2.1
\end{equation}
Hence (\ref{for}) is equivalent to:
\begin{equation}
\int_{0}^{T}\int_{\Omega}(-u\varphi_{t}+\nabla u.\nabla\varphi+|\nabla
u|^{q}\varphi)dxdt=0,\forall\varphi\in\mathcal{D}(Q_{\mathbb{R}^{N},T}),
\label{fur
\end{equation}
and we have and there holds for any $s,\tau\in(0,T),$
\begin{equation}
\int_{\mathbb{R}^{N}}u(.,\tau)\varphi(.,\theta)dx-\int_{\mathbb{R}^{N
}u(.,s)\varphi(.,s)dx+\int_{s}^{\tau}\int_{\mathbb{R}^{N}}(-u\varphi
_{t}+\nabla u.\nabla\varphi+|\nabla u|^{q}\varphi)dxdt=0
\end{equation}
and then for any $\psi\in$ $C_{c}^{2}\left(  \mathbb{R}^{N}\right)  ,$
\begin{equation}
\int_{\mathbb{R}^{N}}u(.,\tau)\psi dx-\int_{\mathbb{R}^{N}}u(.,s)\psi
dx+\int_{s}^{\tau}\int_{\mathbb{R}^{N}}(\nabla u.\nabla\psi+|\nabla u|^{q}\psi
dxdt=0
\end{equation}

\end{remark}

\noindent In this section we study the Cauchy proble
\begin{equation}
\left\{
\begin{array}
[c]{l
u_{t}-\Delta u+|\nabla u|^{q}=0,\quad\text{in}\hspace{0.05in}Q_{\mathbb{R
^{N},T},\\
u(x,0)=u_{0}\geqq0\quad\text{in}\hspace{0.05in}\mathbb{R}^{N},
\end{array}
\right.  \label{cau
\end{equation}


\begin{definition}
Let $u_{0}\in L_{loc}^{r}\left(  \mathbb{R}^{N}\right)  ,r\geqq1.\medskip$

We say that $u$ is a \textbf{weak }$L_{loc}^{r}$\textbf{ solution }if $u$ is a
weak solution of (\ref{un}) and the extension of $u$ by $u_{0}$ at time $0$
satisfies $u\in C\left(  \left[  0,T\right)  ;L_{loc}^{r}(\mathbb{R
^{N}\right)  ).\medskip$

We say that $u$ is a \textbf{weak }$r$ \textbf{solution} of problem
(\ref{cau}) if it is a weak solution of equation (\ref{un}) such that
\begin{equation}
\lim_{t\rightarrow0}\int_{\mathbb{R}^{N}}u^{r}(.,t)\psi dx=\int_{\mathbb{R
^{N}}u_{0}^{r}\psi dx,\qquad\forall\psi\in C_{c}(\mathbb{R}^{N}). \label{int
\end{equation}

\end{definition}

\begin{definition}
Let $u_{0}$ be any nonnegative Radon measure in $\mathbb{R}^{N},$ we say that
$u$ is a \textbf{weak }$\mathcal{M}_{loc}$\textbf{ solution }of problem
(\ref{cau}) if it is a weak solution of (\ref{un}) such that \textbf{ }
\begin{equation}
\lim_{t\rightarrow0}\int_{\mathbb{R}^{N}}u(.,t)\psi dx=\int_{\mathbb{R}^{N
}\psi du_{0},\qquad\forall\psi\in C_{c}(\mathbb{R}^{N}), \label{mea
\end{equation}

\end{definition}

\textbf{ }

\begin{remark}
\label{comp}Obviously, any weak $L_{loc}^{r}$ solution is a weak $r$ solution.
When $r=1,$ the notions of weak $1$-solution and weak $\mathcal{M}_{loc}$
solution coincide\textbf{. }When $r>1,$ $u$ is a weak $L_{loc}^{r}$ solution
if and only if it is a weak $r$ solution and
\begin{equation}
\lim_{t\rightarrow0}\int_{\mathbb{R}^{N}}u(.,t)\psi dx=\int_{\mathbb{R}^{N
}u_{0}\psi dx,\qquad\forall\psi\in C_{c}(\mathbb{R}^{N}). \label{fai
\end{equation}
Indeed if $u(.,t)$ converges to $u_{0}$ in $L_{loc}^{r}(\mathbb{R}^{N})$ as
$t\rightarrow0,$ then it satisfies (\ref{int}) and (\ref{fai}). The converse
is true: let $u$ satisfying (\ref{int}) and (\ref{fai}), then $u(.,t)$ is
bounded in $L_{loc}^{r},$ there exists $t_{n}\rightarrow0$ such that
$u(.,t)\rightarrow v$ in $\mathcal{D}^{\prime}(\mathbb{R}^{N})$ with $v\in
L_{loc}^{r}.$ And $u(.,t_{n})\rightarrow u_{0}$ in $\mathcal{D}^{\prime
}(\mathbb{R}^{N}),$ then $v=u_{0}$, hence it is true for any $t\rightarrow0.$
Then for any nonnegative $\psi\in C_{c}(\mathbb{R}^{N}),$ $u(.,t)\psi
\rightarrow u_{0}\psi$ weakly and in norm, thus strongly in $L^{r
(\mathbb{R}^{N}),$ thus $u(.,t)$ converges to $u_{0}$ in $L_{loc
^{r}(\mathbb{R}^{N}).$
\end{remark}

Other types of solutions using the semigroup of the heat equation have been
introduced in (\cite{BASoWe}):

\begin{definition}
Let $u_{0}\in L^{r}\left(  \mathbb{R}^{N}\right)  .$ A function $u$ is called
\textbf{mild }$L^{r}$\textbf{ solution} of problem (\ref{cau}) if $u\in
C(\left[  0,T\right)  ;L^{r}\left(  \mathbb{R}^{N}\right)  )$, and $|\nabla
u|^{q}\in L_{loc}^{1}(\left[  0,T\right)  ;L^{r}\left(  \mathbb{R}^{N}\right)
)$ and
\[
u(.,t)=e^{t\Delta}u_{0}-\int_{0}^{t}e^{(t-s)\Delta}|\nabla u(.,s)|^{q
ds\qquad\text{in }L^{r}(\mathbb{R}^{N}).
\]
Here $e^{t\Delta}$ is the semi-group of the heat equation acting on
$L^{r}\left(  \mathbb{R}^{N}\right)  .$
\end{definition}

\begin{definition}
Let $u_{0}\in\mathcal{M}_{b}^{+}(\mathbb{R}^{N}).$ A function $u$ is called
\textbf{mild }$\mathcal{M}$ \textbf{solution} of (\ref{cau}) if $u\in
C_{b}((0,T);L^{1}\left(  \mathbb{R}^{N}\right)  )$ and $|\nabla u|^{q}\in
L_{loc}^{1}(\left[  0,T\right)  ;L^{1}\left(  \mathbb{R}^{N}\right)  )$ and
for any $0<t<T,$
\begin{equation}
u(.,t)=e^{t\Delta}u_{0}(.)-\int_{0}^{t}e^{(t-s)\Delta}|\nabla u(.,s)|^{q
ds\qquad\text{in }L^{1}(\mathbb{R}^{N}), \label{hc
\end{equation}
where $e^{t\Delta}$ is defined on $\mathcal{M}_{b}^{+}(\mathbb{R}^{N})$ as the
adjoint of the operator $e^{t\Delta}$ on $C_{0}(\mathbb{R}^{N}),$ the space of
continuous functions on $\mathbb{R}^{N}$ which tend to $0$ as $\left\vert
x\right\vert \rightarrow\infty.$
\end{definition}

\begin{remark}
Every mild $L^{r}$ solution is a weak $L_{loc}^{r}$ solution$.$
\end{remark}

\begin{remark}
Any mild $\mathcal{M}$ solution is a weak $\mathcal{M}_{loc}$ solution. Indeed
for any $0<\epsilon<t<T,$ we find
\[
u(.,t)=e^{(t-\epsilon)\Delta}u(.,\epsilon)-\int_{\epsilon}^{t}e^{(t-s)\Delta
}|\nabla u(.,s)|^{q}ds\qquad\text{in }L^{1}(\mathbb{R}^{N}),
\]
and $u(.,\epsilon)\in L^{1}(\mathbb{R}^{N}),$ then $u$ is a weak solution on
$\left(  \epsilon,T\right)  ,$ then on $(0,T).$ As $t\rightarrow0,$
$u(.,t)-e^{t\Delta}u_{0}(.)$ converges to $0$ in $L^{1}(\mathbb{R}^{N}),$ then
weakly *, and $e^{t\Delta}u_{0}(.)\rightarrow u_{0}$ weakly *, then
(\ref{mea}) holds.
\end{remark}

Another definition of solution with initial data measure was given in
(\cite{BeLa99}):

\begin{definition}
Let $u_{0}\in\mathcal{M}_{b}^{+}(\mathbb{R}^{N}).$ A function $u$ is called
\textbf{weak semi-group solution} if $u\in C((0,T);L^{1}\left(  \mathbb{R
^{N}\right)  )$ and $|\nabla u|^{q}\in L_{loc}^{1}(\left[  0,T\right)
;L^{1}\left(  \mathbb{R}^{N}\right)  )$ and for any $0<\epsilon<t<T,$
\begin{equation}
u(.,t)=e^{(t-\epsilon)\Delta}u(.,\epsilon)-\int_{\epsilon}^{t}e^{(t-s)\Delta
}|\nabla u(.,s)|^{q}ds\qquad\text{in }L^{1}(\mathbb{R}^{N}), \label{ha
\end{equation}
\begin{subequations}
\label{0
\begin{equation}
\lim_{t\rightarrow0}\int_{\mathbb{R}^{N}}u(.,t)\varphi dx=\int_{\mathbb{R
^{N}}\varphi du_{0},\qquad\forall\varphi\in C_{b}(\mathbb{R}^{N}), \label{hb
\end{equation}

\end{subequations}
\end{definition}

We first prove that the two definitions\textbf{ }coincide:

\begin{lemma}
\label{lem} Let $u_{0}\in\mathcal{M}_{b}^{+}(\mathbb{R}^{N}).$ Then
\[
u\text{ is a mild }\mathcal{M}\text{ solution of (\ref{cau})
\Longleftrightarrow u\text{ is a weak semi-group solution of (\ref{cau}).
\]

\end{lemma}

\begin{proof}
(i) Let $u$ be a mild $\mathcal{M}$ solution. Then clearly (\ref{ha}) holds.
Moreover for any $\psi\in C_{0}\left(  \mathbb{R}^{N}\right)  ,$ from the
assumption on the gradient,
\[
<e^{t\Delta}\mu_{0},\psi>=<\mu_{0},e^{t\Delta}\psi>=\int_{\mathbb{R}^{n
}e^{t\Delta}\psi d\mu_{0}=\int_{\mathbb{R}^{N}}(u(.,t)+\int_{0}^{t
e^{(t-s)\Delta}|\nabla u(.,s)|^{q}ds)\psi dx
\]
By approximation the relation extends to any $\varphi\in C_{b}\left(
\mathbb{R}^{N}\right)  :$
\begin{align*}
\int_{\mathbb{R}^{N}}e^{t\Delta}\varphi d\mu_{0}  &  =\int_{\mathbb{R}^{N
}u(.,t)\varphi dx+\int_{\mathbb{R}^{N}}(\int_{0}^{t}e^{(t-s)\Delta}|\nabla
u(.,s)|^{q}ds)\varphi dx\\
&  =\int_{\mathbb{R}^{N}}u(.,t)\varphi dx+\int_{0}^{t}\int_{\mathbb{R}^{N
}|\nabla u|^{q}\varphi dxds
\end{align*}
since the measure is bounded. And from the integrability of the gradient and
the Lebesgue theorem in $L^{1}(\mathbb{R}^{N},d\mu_{0}),$ we deduce
\[
\lim_{t\rightarrow0}\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla u|^{q}\varphi
dxds=0,\qquad\lim_{t\rightarrow0}\int_{\mathbb{R}^{N}}e^{t\Delta}\varphi
d\mu_{0}=\int_{\mathbb{R}^{N}}\varphi d\mu_{0},
\]
since $\left\Vert e^{t\Delta}\varphi\right\Vert _{L^{\infty}\left(
\mathbb{R}^{N}\right)  }\leq\left\Vert \varphi\right\Vert _{L^{\infty}\left(
\mathbb{R}^{N}\right)  }$ and $e^{t\Delta}\varphi$ converges to $\varphi$
everywhere as $t\rightarrow0;$ thus (\ref{hb}) holds.\medskip

(ii) Let $u$ be a weak semi-group solution. Then obviously $u\in
C_{b}((0,T);L^{1}\left(  \mathbb{R}^{N}\right)  ).$ As $\epsilon\rightarrow0,$
we have
\[
\lim_{\epsilon\rightarrow0}\int_{\epsilon}^{t}e^{(t-s)\Delta}|\nabla
u(.,s)|^{q}ds=\int_{0}^{t}e^{(t-s)\Delta}|\nabla u(.,s)|^{q}ds\qquad\text{in
}L^{1}(\mathbb{R}^{N}).
\]
Then
\[
\lim_{\epsilon\rightarrow0}e^{(t-\epsilon)\Delta}u(.,\epsilon)=u(.,t)+\int
_{0}^{t}e^{(t-s)\Delta}|\nabla u(.,s)|^{q}ds\qquad\text{in }L^{1
(\mathbb{R}^{N}).
\]
Moreover (\ref{hb}) entails that that $u(.,\epsilon)\rightarrow u_{0}$ in
$\mathcal{S}^{\prime}(\mathbb{R}^{N})$ and
\begin{equation}
\lim_{\epsilon\rightarrow0}e^{(t-\epsilon)\Delta}u(.,\epsilon)=e^{t\Delta
}u_{0}\qquad\text{in }\mathcal{S}^{\prime}(\mathbb{R}^{N}); \label{chic
\end{equation}
indeed for any $\varphi\in\mathcal{S}(\mathbb{R}^{N}),$
\begin{align*}
\left\vert <e^{(t-\epsilon)\Delta}u(.,\epsilon)-e^{t\Delta}u_{0
,\varphi>\right\vert  &  \leq\left\vert <e^{t\Delta}(u(.,\epsilon
)-u_{0}(.),\varphi>\right\vert +\left\vert \int_{\mathbb{R}^{N}
(u(x,\epsilon)((e^{(t-\epsilon)\Delta}-e^{t\Delta})\varphi)(x)dx\right\vert \\
&  \leq\left\vert <e^{t\Delta}(u(.,\epsilon)-u_{0}(.),\varphi>\right\vert \\
&  +\left\Vert u(.,\epsilon)\right\Vert _{L^{1}(\mathbb{R}^{N})}\left\Vert
(e^{(t-\epsilon)\Delta}-e^{t\Delta})\varphi\right\Vert _{L^{\infty
(\mathbb{R}^{N})})
\end{align*}
and $e^{t\Delta}is$ continuous on $\mathcal{S}(\mathbb{R}^{N})$. Then for any
$\varphi\in\mathcal{S}(\mathbb{R}^{N}),$ we have
\[
<e^{t\Delta}u_{0},\varphi>=\int_{\mathbb{R}^{n}}u(.,t)\varphi dx+\int
_{\mathbb{R}^{n}}(\int_{0}^{t}e^{(t-s)\Delta}|\nabla u(.,s)|^{q}ds)\varphi dx
\]
which extends to any $\varphi\in C_{0}(\mathbb{R}^{N})$ by density. Thus
(\ref{hc}) follows. \medskip
\end{proof}

Let us recall the main existence results using semi-groups: \medskip\ 

$\bullet$ If $1<q<(N+2)/(N+1),$ for any $u_{0}\in\mathcal{M}_{b
(\mathbb{R}^{N}),$ from \cite{BeLa99}, there exists a weak semi-group solution
$u$ of problem (\ref{cau}), obtained by approximation, and $u\in
C^{2,1}(Q_{\mathbb{R}^{N},\infty})$. The existence of a mild $\mathcal{M}$
solution is also proved in \cite{BASoWe} from the Banach fixed point theorem,
and the notions are equivalent from Lemma \ref{lem}. In any case uniqueness
results are obtained \textit{under additional conditions of punctual or
integral conditions on the gradient.}\medskip\ 

$\bullet$ If $u_{0}\in L^{r}(\mathbb{R}^{N}),$ $r\geqq1,$ and
$r>N(q-1)/(2-q),$ which means $q<(N+2r)/(N+r),$ there exists a mild $L^{r}$
solution of (\ref{cau}), and uniqueness holds in the class of pointwise mild
solutions such that $u\in C(\left[  0,T\right)  ;L^{r}\left(  \mathbb{R
^{N}\right)  )\cap C((0,T);W^{1,qr}\left(  \mathbb{R}^{N}\right)  )$, from
\cite[Theorem 2.1]{BASoWe}. Moreover if $q\leq2$, there exists a pointwise
mildsolution of (\ref{cau}) for any $r\geqq1$ but uniqueness is not known for
$q<2,$ see \cite[Theorem 4.1]{BASoWe}. For $q>2,$ existence holds under the
restriction that $u_{0}$ is a limit of a monotone sequence of continuous
functions, and is not known in the general case.

\begin{remark}
\label{all}All the definitions of semi-group solutions assume an integrability
property of $|\nabla u|^{q}$, global in space$.$ Observe also that (\ref{hb})
is assumed for any $\varphi\in C_{b}(\mathbb{R}^{N})$. On the contrary, our
definitions of weak\textbf{ }solutions are \textrm{local in space}, they do
not require such global properties.
\end{remark}

Finally we mention another weaker form of semi-group solutions, given in
(\cite{BASoWe}), which will be used in the sequel:

\begin{definition}
Let $u_{0}\in\mathcal{M}_{b}^{+}(\mathbb{R}^{N}).$ Then $u$ is \textit{a
pointwise mild solution} of (\ref{cau}) if $u\in L_{loc}^{1}(Q_{\mathbb{R
^{N},T}),$ and $|\nabla u|^{q}\in L_{loc}^{1}(Q_{\mathbb{R}^{N},T}),$ and
\[
u(x,t)=(e^{t\Delta}u_{0})(x)-\int_{0}^{t}\int_{\mathbb{R}^{N}
g(x-y,t-s)|\nabla u(y,s)|^{q}dyds\qquad\text{for a.e. }(x,t)\in Q_{\mathbb{R
^{N},T},
\]
where $g$ is the heat kernel.
\end{definition}

\begin{remark}
For $r\geqq1,$ it is clear that every mild $L^{r}$ solution is a pointwise
mild solution. If $u_{0}\in L^{1}\left(  \mathbb{R}^{N}\right)  $ every
pointwise mild solution is a mild $L^{1}$ solution; if $u_{0}\in
\mathcal{M}_{b}^{+}(\mathbb{R}^{N})$, every pointwise mild solution, is a mild
$\mathcal{M}$ solution. see \cite[Proposition 1.1 and Remark 1.2]{BASoWe}.
\end{remark}

\subsection{ Decay of the norms}

\textbf{ }Next we show a \textit{decay result} for the solutions of Hamilton
Jacobi equations, which is valid for any $q>1,$ and for \textit{all the weak
solutions}, \textit{with no condition of boundedness at infinity}.\medskip

When $q\leqq2,$ any weak solution $u$ of equation (\ref{un}) is smooth: $u\in
C^{2,1}\left(  Q_{\mathbb{R}^{N},T}\right)  ,$ from \cite[Theorem
2.15]{BiDao1}. Since it may be false for $q>2,$ we regularize $u$ by
convolution, setting
\[
u_{\varepsilon}=u\ast\varrho_{\varepsilon},
\]
where $(\varrho_{\varepsilon})_{\varepsilon>0}$ is a sequence of mollifiers.
We recall that for given $0<s<\tau<T$, and $\varepsilon$ small enough,
$u_{\varepsilon}$ is a \textit{subsolution} of equation (\ref{un})$,$ see
\cite{BiDao1}:
\begin{equation}
(u_{\varepsilon})_{t}-\nu\Delta u_{\varepsilon}+|\nabla u_{\varepsilon
|^{q}\leqq0,\qquad\text{in }Q_{\mathbb{R}^{N},s,\tau} \label{epsi
\end{equation}


\begin{theorem}
\label{decay}Assume $q>1.$ Let $r\geqq1.$ Let $u_{0}\in L^{r}(\mathbb{R}^{N})$
be nonnegative. Let $u$ be any non-negative weak $r$ solution of problem
(\ref{cau}).

(i) Then $u(.,t)\in L^{r}\left(  \mathbb{R}^{N}\right)  $ for any $t\in\left(
0,T\right)  ,$ an
\begin{equation}
\int_{\mathbb{R}^{N}}u^{r}(.,t)dx\leqq\int_{\mathbb{R}^{N}}u_{0}^{r}dx.
\label{ghi
\end{equation}
(ii) Moreover \textbf{ }$u^{r-1}|\nabla u|^{q}\in L_{loc}^{1}(\left[
0,T\right)  ;L^{1}\left(  \mathbb{R}^{N}\right)  );$ and $u^{r-2}|\nabla
u|^{2}\in L_{loc}^{1}(\left[  0,T\right)  ;L^{1}\left(  \mathbb{R}^{N}\right)
)$ if $r>1$ and $\nu>0;$ and for any $t\in\left(  0,T\right)  ,$
\begin{equation}
\int_{\mathbb{R}^{N}}u^{r}(.,t)dx+r\int_{0}^{t}\int_{\mathbb{R}^{N}
u^{r-1}|\nabla u|^{q}dxdt+r(r-1)\nu\int_{0}^{t}\int_{\mathbb{R}^{N}
u^{r-2}|\nabla u|^{2}dxdt=\int_{\mathbb{R}^{N}}u_{0}^{r}dx,\qquad\text{if
}r>1, \label{egge
\end{equation
\begin{equation}
\int_{\mathbb{R}^{N}}u(.,t)dx+\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla
u|^{q}dxdt=\int_{\mathbb{R}^{N}}u_{0}dx,\qquad\text{if }r=1, \label{agga
\end{equation
\begin{equation}
\lim_{t\rightarrow0}\int_{\mathbb{R}^{N}}u^{r}(.,t)dx=\int_{\mathbb{R}^{N
}u_{0}^{r}dx. \label{limu
\end{equation}
(iii) $u^{q-1+r}\in L_{loc}^{1}((\left[  0,T\right)  ;W^{1,1}\left(
\mathbb{R}^{N}\right)  ),$ and if $\nu>0,$ then $u^{r/2}\in L_{loc
^{2}(\left[  0,T\right)  ;W^{1,2}\left(  \mathbb{R}^{N}\right)  ).$

\noindent(iv) If $u$ is a weak $L_{loc}^{r}$ solution, then $u\in C(\left[
0,T\right)  ;L^{r}\left(  \mathbb{R}^{N}\right)  ).$
\end{theorem}

\begin{proof}
(i) \textbf{ First step: case } $q^{\prime}>N/r.$ That means $r\geqq N$ or $q$
is small enough: $1<q<N/(N-r).$

Let $0<s<\tau<T$ be fixed and $\varepsilon>0$ small enough. Let $\delta>0,$
and $u_{\varepsilon,\delta}=u_{\varepsilon}+\delta.$ For any $R>0,$ we
consider $\xi(x)=\xi_{R}(x)=\psi(x/R),$where $\psi(x)\in\left[  0,1\right]
,\psi(x)=1$ for $\left\vert x\right\vert \leqq1,\psi(x)=0$ for $\left\vert
x\right\vert \geqq2.$ Then multiplying (\ref{epsi}) by $u_{\varepsilon,\delta
}^{r-1}\xi^{\lambda}$ where $\lambda>0$, we get
\begin{align*}
&  \frac{d}{dt}\left(  \frac{1}{r}\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta
}^{r}\xi^{\lambda}dx\right)  +(r-1)\nu\int_{\mathbb{R}^{N}}u_{\varepsilon
,\delta}^{r-2}\left\vert \nabla u_{\varepsilon,\delta}\right\vert ^{2
\xi^{\lambda}dx+\int_{\mathbb{R}^{N}}|\nabla u_{\varepsilon,\delta
|^{q}u_{\varepsilon,\delta}^{r-1}\xi^{\lambda}dx\\
&  \leqq-\lambda\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r-1}\xi
^{\lambda-1}\nabla u_{\varepsilon,\delta}.\nabla\xi dx,
\end{align*}
and from the H\"{o}lder inequality
\[
\lambda\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r-1}\xi^{\lambda
-1}\left\vert \nabla u_{\varepsilon,\delta}\right\vert \left\vert \nabla
\xi\right\vert dx\leqq\frac{1}{2}\int_{\mathbb{R}^{N}}|\nabla u_{\varepsilon
,\delta}|^{q}u_{\varepsilon,\delta}^{r-1}\xi^{\lambda}dx+C(q,\lambda
)\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r-1}\xi^{\lambda-q^{\prime
}\left\vert \nabla\xi\right\vert ^{q^{\prime}}dx,
\
\[
\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r-1}\xi^{\lambda-q^{\prime
}\left\vert \nabla\xi\right\vert ^{q^{\prime}}dx\leqq\left(  \int
_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r}\xi^{\lambda}dx\right)
^{1/r^{\prime}}\left(  \int_{\mathbb{R}^{N}}\xi^{\frac{\lambda}{r}-q^{\prime
}\left\vert \nabla\xi\right\vert ^{rq^{\prime}}dx\right)  ^{1/r}.
\]
Choosing $\lambda=rq^{\prime}$ we deduce
\[
\frac{d}{dt}\left(  \left(  \int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r
\xi^{\lambda}dx\right)  ^{1/r}\right)  \leqq C(q,\lambda)\left(
\int_{\mathbb{R}^{N}}\left\vert \nabla\xi\right\vert ^{rq^{\prime}}dx\right)
^{1/r}\leqq CR^{\frac{N}{r}-q^{\prime}
\]
where $C=C(N,q,\alpha,\psi).$ By integration, for any $0<s\leqq\sigma
<t\leqq\tau,$
\[
\left(  \int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r}(.,t)\xi^{\lambda
}dx\right)  ^{1/r}\leqq\left(  \int_{\mathbb{R}^{N}}u_{\varepsilon,\delta
^{r}(.,\sigma)\xi^{\lambda}dx\right)  ^{1/r}+C\tau R^{\frac{N}{r}-q^{\prime
}.
\]
with a new constant $C$ as above. Let $R_{0}>0$ be fixed and take $R>R_{0},$
thus
\[
\left(  \int_{B_{R_{0}}}u_{\varepsilon,\delta}^{r}(.,t)dx\right)  ^{1/r
\leqq\left(  \int_{B_{2R}}u_{\varepsilon,\delta}^{r}(.,\sigma)\xi^{\lambda
}dx\right)  ^{1/r}+C\tau R^{\frac{N}{r}-q^{\prime}
\]
Then we make successively $\delta\rightarrow0,$ and then $\varepsilon
\rightarrow0.$ From (\ref{int}), we deduce that
\begin{equation}
\left(  \int_{B_{R_{0}}}u(.,t)^{r}dx\right)  ^{1/r}\leqq\left(  \int
_{\mathbb{R}^{N}}u(.,\sigma)^{r}\xi^{\lambda}dx\right)  ^{1/r}+C\tau
R^{\frac{N}{r}-q^{\prime}} \label{fis
\end{equation}
and then from (\ref{int}) we can make $\sigma\rightarrow0$ and obtain
\[
\left(  \int_{B_{R_{0}}}u(.,t)^{r}dx\right)  ^{1/r}\leqq\left(  \int
_{\mathbb{R}^{N}}u(.,\sigma)^{r}\xi^{\lambda}dx\right)  ^{1/r}+C\tau
R^{\frac{N}{r}-q^{\prime}}\leqq\left(  \int_{\mathbb{R}^{N}}u_{0
^{r}dx\right)  ^{1/r}+C\tau R^{\frac{N}{r}-q^{\prime}
\]
and finally we make $R\rightarrow\infty$ and then $R_{0}\rightarrow
\infty.\medskip$

\textbf{Second step: case } $q^{\prime}<N/r.$ Then $r<N$ and $q\geqq
N/(N-r)>1.$ Then we fix some $k\in\left(  1,N/(N-r)\right)  .$ For any
$\eta\in\left(  0,1\right)  $, we have $\eta|\nabla u|^{k}\leqq\eta+|\nabla
u|^{q},$ hence the function
\[
w_{\eta}=\eta^{1/(k-1)}(u-\eta t)
\]
satisfies
\[
(w_{\eta})_{t}-\Delta w_{\eta}+|\nabla w_{\eta}|^{k}\leqq0
\]
in the weak sense. Thanks to Kato's inequality, see \cite[Lemma 1]{BrFr},
\cite{BaPi}, we deduce that
\begin{equation}
(w_{\eta}^{+})_{t}-\Delta w_{\eta}^{+}+|\nabla w_{\eta}^{+}|^{k}\leqq0,
\label{II1.36
\end{equation}
in $\hspace{0.05in}\mathcal{D}^{\prime}(Q_{\mathbb{R}^{N},T}).$ And $w_{\eta
}^{+}$ has the same regularity as $u,$ and moreover it satisfies an analogous
property to (\ref{int}):
\begin{equation}
\lim_{t\rightarrow0}\int_{\mathbb{R}^{N}}(w_{\eta}^{+})^{r}(.,t)\psi
dx=\int_{\mathbb{R}^{N}}(\eta^{1/(k-1)}u_{0})^{r}\psi dx,\qquad\forall\psi\in
C_{c}(\mathbb{R}^{N}), \label{mol
\end{equation}
Indeed
\begin{align*}
\left\vert \int_{\mathbb{R}^{N}}((u-\eta t)^{+})^{r}-u^{r}(.,t))\psi
dx\right\vert  &  \leqq\int_{\left\{  u\geqq\eta t\right\}  }\left\vert
u(.,t)-\eta t)^{r}-u^{r}(.,t)\right\vert \psi dx+\int_{\left\{  u\leqq\eta
t\right\}  }u^{r}(.,t))\psi dx\\
&  \leqq r\eta t\int_{\mathbb{R}^{N}}u^{r-1}(.,t)\psi dx+Ct^{r}\\
&  \leqq r\eta t(\int_{\mathbb{R}^{N}}u^{r}(.,t)dx)^{1/r^{\prime}
(\int_{\mathbb{R}^{N}}\psi^{r}dx)^{1/r}+Ct^{r
\end{align*}
then
\[
\lim_{t\rightarrow0}\int_{\mathbb{R}^{N}}((u-\eta t)^{+})^{r}-u^{r}(.,t))\psi
dx=0
\]
and (\ref{mol}) follows from (\ref{int}) applied to $\eta^{1/(k-1)}u.$ From
the first step we deduce that $w_{\eta}^{+}(t)\in L^{r}\left(  \mathbb{R
^{N}\right)  $ and
\[
\int_{\mathbb{R}^{N}}(w_{\eta}^{+})^{r}(.,t)dx\leqq\eta^{r/(k-1)
\int_{\mathbb{R}^{N}}u_{0}^{r}dx.
\]
Then $\left\Vert (u-\eta t)^{+}\right\Vert _{L^{r}\left(  \mathbb{R
^{N}\right)  }\leqq\left\Vert u_{0}\right\Vert _{L^{r}\left(  \mathbb{R
^{N}\right)  }.$ Then for any $R>0,$ since $u\leqq\eta t+(u-\eta t)^{+},$
\[
\left\Vert u(.,t)\right\Vert _{L^{r}\left(  B_{R}\right)  }\leqq\left\Vert
u_{0}\right\Vert _{L^{r}\left(  \mathbb{R}^{N}\right)  }+\eta t\left\vert
B_{R}\right\vert ^{r
\]
Going to the limit as $\eta\rightarrow0$ for fixed $R,$ we get $\left\Vert
u(.,t)\right\Vert _{L^{r}\left(  B_{R}\right)  }\leqq\left\Vert u_{0
\right\Vert _{L^{r}\left(  \mathbb{R}^{N}\right)  },$ then going to the limit
as $R\rightarrow\infty$ we deduce that $u(.,t)\in L^{r}\left(  \mathbb{R
^{N}\right)  $ and (\ref{ghi}) holds$.\medskip$

(ii) Considering again $u_{\varepsilon,\delta}$ as above, and setting
$F_{\varepsilon}=\left\vert \nabla u\right\vert ^{q}\ast\varrho_{\varepsilon
},$ there holds
\[
(u_{\varepsilon,\delta})_{t}-\nu\Delta u_{\varepsilon,\delta}+F_{\varepsilon
}=0,
\]
then
\begin{align*}
&  \frac{d}{dt}\left(  \int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r
\xi^{\lambda}dx\right)  +r(r-1)\nu\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta
}^{\alpha-1}\left\vert \nabla u_{\varepsilon,\delta}\right\vert ^{2
\xi^{\lambda}dx+r\int_{\mathbb{R}^{N}}F_{\varepsilon}u_{\varepsilon,\delta
}^{\alpha}\xi^{\lambda}dx\\
&  =-r\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r-1}\nabla u_{\varepsilon
,\delta}.\nabla(\xi^{\lambda})dx=\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta
}^{r}\Delta(\xi^{\lambda})dx
\end{align*}
then for any $0<\sigma<t<T,$
\begin{subequations}
\begin{align}
&  \int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r}(.,t)\xi^{\lambda
dx+r\int_{\sigma}^{t}\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{\alpha
}F_{\varepsilon}\xi^{\lambda}dxdt\nonumber\\
+r(r-1)\nu\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{\alpha-1}\left\vert
\nabla u_{\varepsilon,\delta}\right\vert ^{2}\xi^{\lambda}dx  &
=\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r}(.,\sigma)\xi^{\lambda
dx+\int_{\sigma}^{t}\int_{\mathbb{R}^{N}}u_{\varepsilon,\delta}^{r}\Delta
(\xi^{\lambda})dx\nonumber
\end{align}
First we can go to the limit as $\varepsilon\rightarrow0,$ because $u\in
L_{loc}^{\infty}(Q_{\mathbb{R}^{N},T}),$ and $\left\vert \nabla u\right\vert
^{2}$ $\in L_{loc}^{1}(Q_{\mathbb{R}^{N},T}),$ and $F_{\varepsilon
\rightarrow\left\vert \nabla u\right\vert ^{q}$ in $L_{loc}^{1}(Q_{\mathbb{R
^{N},T}).$ Setting $v_{\delta}=u+\delta,$ we obtain for almost any $\sigma,t,$
and in fact for any $\sigma,t$ by the continuity,
\end{subequations}
\begin{subequations}
\begin{align}
&  \int_{\mathbb{R}^{N}}v_{\delta}^{r}(.,t)\xi^{\lambda}dx+r\int_{\sigma
^{t}\int_{\mathbb{R}^{N}}v_{\delta}^{\alpha}\left\vert \nabla u\right\vert
^{q}\psi dxdt\nonumber\\
+r(r-1)\nu\int_{\mathbb{R}^{N}}v_{\delta}^{\alpha-1}\left\vert \nabla
u\right\vert ^{2}\xi^{\lambda}dx  &  =\int_{\mathbb{R}^{N}}v_{\delta
^{r}(.,\sigma)\xi^{\lambda}dx+\int_{\sigma}^{t}\int_{\mathbb{R}^{N}}v_{\delta
}^{r}\Delta(\xi^{\lambda})dx\nonumber
\end{align}
Next we go to the limit as $\delta\rightarrow0:$ from the Fatou Lemma we
deduce that $\int_{\sigma}^{t}\int_{\mathbb{R}^{N}}u^{r-1}\left\vert \nabla
u\right\vert ^{q}\psi dxdt$ and $(r-1)\nu\int_{\mathbb{R}^{N}}u^{\alpha
-1}\left\vert \nabla u\right\vert ^{2}\xi^{\lambda}dx$ are finite, and then
from Lebesgue we obtain the equalit
\end{subequations}
\begin{align*}
&  \int_{\mathbb{R}^{N}}u^{r}(.,t)\xi^{\lambda}dx+r\int_{\sigma}^{t
\int_{\mathbb{R}^{N}}u^{\alpha}\left\vert \nabla u\right\vert ^{q}\xi
^{\lambda}dxdt\\
+r(r-1)\nu\int_{\mathbb{R}^{N}}u^{\alpha-1}\left\vert \nabla u\right\vert
^{2}\xi^{\lambda}dx  &  =\int_{\mathbb{R}^{N}}u^{r}(.,\sigma)\xi^{\lambda
}dx+\int_{\sigma}^{t}\int_{\mathbb{R}^{N}}u^{r}\Delta(\xi^{\lambda})dx.
\end{align*}
Next we go to the limit as $\sigma\rightarrow0,$ from (\ref{int}). In the same
way we deduce that
\begin{align*}
&  \int_{\mathbb{R}^{N}}u^{r}(.,t)\xi^{\lambda}dx+r\int_{0}^{t}\int
_{\mathbb{R}^{N}}u^{\alpha}\left\vert \nabla u\right\vert ^{q}\xi^{\lambda
}dxdt\\
+\alpha r\int_{0}^{t}\int_{\mathbb{R}^{N}}u^{\alpha-1}\left\vert \nabla
u\right\vert ^{2}\xi^{\lambda}dx  &  =\int_{\mathbb{R}^{N}}u_{0}^{r
(.,\sigma)\xi^{\lambda}dx+\int_{\sigma}^{t}\int_{\mathbb{R}^{N}}u^{r
\Delta(\xi^{\lambda})dx
\end{align*}
Now $u(.,t)\in L^{r}\left(  \mathbb{R}^{N}\right)  $ for any $t\in\left[
s,\tau\right]  $, and
\[
\int_{\sigma}^{t}\int_{\mathbb{R}^{N}}u^{r}\Delta(\xi^{\lambda})dx\leqq
\frac{C}{R^{2}}\tau\int_{\mathbb{R}^{N}}u^{r}(\sigma)dx
\]
and we can make $R\rightarrow\infty$. We get that
\[
\int_{0}^{\tau}\int_{\mathbb{R}^{N}}u^{r-1}|\nabla u|^{q}dxdt+(r-1)\nu\int
_{0}^{\tau}\int_{\mathbb{R}^{N}}u^{r-2}|\nabla u|^{2}dxdt<\infty
\]
and, from the Lebesque theorem, we deduce
\begin{equation}
\int_{\mathbb{R}^{N}}u^{r}(.,t)dx+r\int_{0}^{t}\int_{\mathbb{R}^{N}}u^{\alpha
}|\nabla u|^{q}dxdt+r(r-1)\nu\int_{0}^{t}\int_{\mathbb{R}^{N}}u^{\alpha
-1}\left\vert \nabla u\right\vert ^{2}dx=\int_{\mathbb{R}^{N}}u_{0}^{r}dx
\label{lon
\end{equation}
Hence (\ref{egge}) follows, which implies directly (\ref{limu}).\medskip

(iii) Setting $v=u^{m}$ with $m=(q-1+r)/q<$ $r,$ we have $|\nabla v|^{q}\in
L_{loc}^{1}(\left[  0,T\right)  ;L^{1}(\mathbb{R}^{N})),$ and $v\in L^{\infty
}((0,T);L^{\frac{r}{m}}\left(  \mathbb{R}^{N}\right)  ).$ From the
Gagliardo-Nirenberg inequality, we deduce tha
\begin{equation}
\left\Vert v(.,t)\right\Vert _{L^{q}\left(  \mathbb{R}^{N}\right)  
\leqq\left\Vert v(.,t)\right\Vert _{L^{\frac{r}{m}}\left(  \mathbb{R
^{N}\right)  }^{1-k}\left\Vert \nabla v(.,t)\right\Vert _{L^{q}\left(
\mathbb{R}^{N}\right)  }^{k},\qquad\frac{1}{k}=1+\frac{rq^{\prime}}{N}.
\label{gal
\end{equation}
Then by integration, for any $0<\tau<T,$ using H\"{o}lder inequality,
\[
\int_{0}^{\tau}\int_{\mathbb{R}^{N}}v^{q}(.,t)dxdt=\int_{0}^{\tau
\int_{\mathbb{R}^{N}}u^{q-1+r}(.,t)dxdt\leqq C(\tau)\left\Vert v\right\Vert
_{L^{\infty}((0,\tau);L^{\frac{r}{m}}\left(  \mathbb{R}^{N}\right)  
^{(1-k)q}(\int_{0}^{\tau}\int_{\mathbb{R}^{N}}\left\vert \nabla v\right\vert
^{q}dxdt)^{k
\]
Then $u\in L^{q-1+r}(Q_{\mathbb{R}^{N},\tau}),$ and $v^{q}=u^{q-1+r}\in
L^{1}((0,\tau);W^{1,1}\left(  \mathbb{R}^{N}\right)  ),$ $v\in L^{q
((0,\tau);W^{1,q}\left(  \mathbb{R}^{N}\right)  )$. If $\nu>0,$ we also have
$u^{r-2}\left\vert \nabla u\right\vert ^{2}=\left\vert \nabla(u^{r/2
)\right\vert ^{2}\in L^{1}(Q_{\mathbb{R}^{N},\tau}),$ and $u^{r/2}\in
L^{2}(Q_{\mathbb{R}^{N},\tau}),$ then $u^{r/2}\in L^{2}((0,\tau);W^{1,2
\left(  \mathbb{R}^{N}\right)  ).$\medskip

(iv) Here we assume that $u\in C(\left[  0,T\right)  ;L_{loc}^{r}\left(
\mathbb{R}^{N}\right)  )$. First assume $r>1.$ Then from a diagonal procedure,
there exists $t_{n}\rightarrow0$ such that $u(.,t_{n})\rightarrow u_{0}$ a.e.
in $\mathbb{R}^{N},$ and $\left\Vert u(.,t_{n})\right\Vert _{L^{r}\left(
\mathbb{R}^{N}\right)  }\rightarrow\left\Vert u_{0}\right\Vert _{L^{r}\left(
\mathbb{R}^{N}\right)  }$, and $u(.,t_{n})\rightarrow u_{0}$ weakly in
$L^{r}\left(  \mathbb{R}^{N}\right)  .$ Then it holds from any sequence, and
$u\in C(\left[  0,T\right)  ;L^{r}\left(  \mathbb{R}^{N}\right)  )$. Next
assume $r=1;$ let $t_{n}\rightarrow t\in\left[  0,T\right)  .$ We have for any
$p>0,$
\begin{align*}
\int_{\mathbb{R}^{N}}\left\vert u(t_{n})-u_{0}\right\vert dx  &  \leqq
\int_{B_{p}}\left\vert u(t_{n})-u_{0}\right\vert dx+\int_{\mathbb{R
^{N}\backslash B_{p}}\left\vert u(t_{n})-u_{0}\right\vert dx\\
&  \leqq\int_{B_{p}}\left\vert u(t_{n})-u_{0}\right\vert dx+\int
_{\mathbb{R}^{N}\backslash B_{p}}u(t_{n})dx+\int_{\mathbb{R}^{N}\backslash
B_{p}}u_{0}dx\\
&  =\int_{B_{p}}\left\vert u(t_{n})-u_{0}\right\vert dx+\int_{\mathbb{R}^{N
}u(t_{n})dx-\int_{B_{p}}u_{0}dx+\int_{B_{p}}(u(t)-u_{0})dx+\int_{\mathbb{R
^{N}\backslash B_{p}}u_{0}dx\\
&  \leqq2\int_{B_{p}}\left\vert u(t_{n})-u_{0}\right\vert dx+\int
_{\mathbb{R}^{N}}u(t_{n})dx-\int_{\mathbb{R}^{N}}u_{0}dx+2\int_{\mathbb{R
^{N}\backslash B_{p}}u_{0}dx
\end{align*}
And the result follows because $\int_{\mathbb{R}^{N}\backslash B_{p}
u_{0}dx\rightarrow0$ as $p\rightarrow\infty,$ since $u_{0}\in L^{1}\left(
\mathbb{R}^{N}\right)  .\medskip$
\end{proof}

The decay result is also available for initial data measures, where we do not
assume that $q<(N+2)/(N+1):$

\begin{theorem}
\label{inmea}Assume $q>1.$ Let $u_{0}\in\mathcal{M}_{b}^{+}(\mathbb{R}^{N})$
and $u$ be any non-negative weak $\mathcal{M}_{loc}$ solution of equation
(\ref{cau}) in$\hspace{0.05in}Q_{\mathbb{R}^{N},T}.$ Then $u(.,t)\in
L^{1}\left(  \mathbb{R}^{N}\right)  $ for any $t>0,$ an
\begin{equation}
\int_{\mathbb{R}^{N}}u(.,t)dx\leq
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0}. \label{mis
\end{equation}
Moreover $u\in C((0,T);L^{1}\left(  \mathbb{R}^{N}\right)  ),$ $|\nabla
u|^{q}\in L_{loc}^{1}(\left[  0,T\right)  ;L^{1}\left(  \mathbb{R}^{N}\right)
)$ and
\begin{equation}
\int_{\mathbb{R}^{N}}u(.,t)dx+\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla
u|^{q}dxdt=\
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0}, \label{his
\end{equation}
and
\begin{equation}
\lim_{t\rightarrow0}\int_{\mathbb{R}^{N}}u(.,t)\varphi dx
{\displaystyle\int_{\mathbb{R}^{N}}}
\varphi du_{0},\qquad\forall\varphi\in C_{b}(\mathbb{R}^{N}) \label{ris
\end{equation}

\end{theorem}

\begin{proof}
We obtain in the same way, as in (\ref{fis}),
\[
\int_{B_{R_{0}}}u(.,t)dx\leqq\int_{\mathbb{R}^{N}}u(.,t)\xi^{\lambda
dx\leqq\int_{\mathbb{R}^{N}}u(.,\sigma)\xi^{\lambda}dx+C\tau R^{N-q^{\prime}
\]
and we can go to the limit as $\sigma\rightarrow0$ from (\ref{mea}), then
\[
\int_{B_{R_{0}}}u(.,t)dx\leq
{\displaystyle\int_{\mathbb{R}^{N}}}
\xi^{\lambda}du_{0}+C\tau R^{N-q^{\prime}}\leq
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0}+C\tau R^{N-q^{\prime}
\]
Then going to the limit as $R\rightarrow\infty,$ and then as $R_{0
\rightarrow\infty,$ we deduce that (\ref{mis}) holds, and we still obtain
(\ref{his}) holds. And $u\in C((0,T);L^{1}\left(  \mathbb{R}^{N}\right)  ),$
from the Lebesgue theorem, because $u\in C((0,T);L_{loc}^{1}\left(
\mathbb{R}^{N}\right)  ),$ and $u\in L^{\infty}((0,T);L^{1}\left(
\mathbb{R}^{N}\right)  )$

Let us show (\ref{ris}): let $\varphi\in C_{b}(\mathbb{R}^{N})$ be
nonnegative, we can assume that $\varphi$ takes its values in $\left[
0,1\right]  .$ Let $t_{n}\rightarrow0.$ We know that $\lim\int_{\mathbb{R
^{N}}u(.,t_{n})dx
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0}.$ Let $\psi_{p}\in\mathcal{D}(\mathbb{R}^{N})$ with values in $\left[
0,1\right]  ,$ $\psi_{p}(x)=1$ if $\left\vert x\right\vert \leqq p,$ $0$ if
$\left\vert x\right\vert \geqq2p.$ Then for fixed $p,$ $\lim\int
_{\mathbb{R}^{N}}u(.,t_{n})\varphi\psi_{p}dx
{\displaystyle\int_{\mathbb{R}^{N}}}
\varphi\psi_{p}du_{0}.$ Let $\eta>0.$
\begin{align*}
&  \left\vert \int_{\mathbb{R}^{N}}u(.,t_{n})\varphi dx
{\displaystyle\int_{\mathbb{R}^{N}}}
\varphi du_{0}\right\vert \\
&  \leqq\left\vert \int_{\mathbb{R}^{N}}u(.,t_{n})\varphi\psi_{p}dx
{\displaystyle\int_{\mathbb{R}^{N}}}
\varphi\psi_{p}du_{0}\right\vert 
{\displaystyle\int_{\mathbb{R}^{N}}}
\varphi(1-\psi_{p})du_{0}+\int_{\mathbb{R}^{N}}u(.,t_{n})\varphi(1-\psi_{p})dx
\end{align*}
and 
{\displaystyle\int_{\mathbb{R}^{N}}}
(1-\psi_{p})du_{0}\rightarrow0$ as $p\rightarrow\infty$ from the Lebesgue
Theorem, then for some $p_{\eta}$ we have 
{\displaystyle\int_{\mathbb{R}^{N}}}
(1-\psi_{p})du_{0}\leqq\eta.$ As $n\rightarrow\infty,$
\[
\int_{\mathbb{R}^{N}}u(.,t_{n})(1-\psi_{p_{\eta}})dx\rightarro
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0}
{\displaystyle\int_{\mathbb{R}^{N}}}
\psi_{p_{\eta}}du_{0}
{\displaystyle\int_{\mathbb{R}^{N}}}
(1-\psi_{p_{\eta}})du_{0
\]
Then $\int_{\mathbb{R}^{N}}u(.,t_{n})\varphi(1-\psi_{p_{\eta}})dx\leqq
2p_{\eta}$ for large $n,$ and $\left[  \int_{\mathbb{R}^{N}}u(.,t_{n
)\varphi\psi_{p_{\eta}}dx
{\displaystyle\int_{\mathbb{R}^{N}}}
\varphi\psi_{p_{\eta}}du_{0}\right]  \leqq p_{\eta}$ for large $n,$ hence we
majorizate by $4\eta,$ hence the result.\medskip
\end{proof}

\subsection{Regularizing effects}

Here we deduce of the decay estimates a regularizing effect \textit{without
any condition at}\textbf{ }$\infty,$ achieving the proof of Theorem \ref{main}.

\begin{theorem}
\label{bound}Let $q>1.$ Let $r\geqq1$ and $u_{0}\in L^{r}(\mathbb{R}^{N})$.
Let $u$ be any non-negative weak $L_{loc}^{r}$ solution of problem (\ref{cau})
in$\hspace{0.05in}Q_{\mathbb{R}^{N},T}$ (\ref{int}). Then $u(.,t)\in
L^{\infty}(\mathbb{R}^{N})$ for any $t\in$ $(0,T)$ and
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\sigma_{rq,N}}\Vert
u_{0}\Vert_{L^{r}(\mathbb{R}^{N})}^{\varpi_{r,q,N}}, \label{uinf
\end{equation}
where $C=C(N,q,r)$ and $\sigma_{r,q,N},\varpi_{r,q,N}$ are given by
\begin{equation}
\sigma_{r,q,N}=\frac{1}{\frac{rq}{N}+q-1},\qquad\varpi_{r,q,N}=\frac{rq
{N}\sigma_{r,q},\qquad\text{if }q<N, \label{sigr
\end{equation}
and
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\frac{N}{2r}}\Vert
u_{0}\Vert_{L^{r}(\mathbb{R}^{N})},\qquad\text{if }\nu>0,\;2<N. \label{use
\end{equation}
where $C=C(N,q,r,\nu).$
\end{theorem}

\begin{proof}
Since $u$ is a weak $L_{loc}^{r}$ solution, then $u\in C(\left[  0,T\right)
;L^{r}\left(  \mathbb{R}^{N}\right)  ),$ from Theorem \ref{decay}, thus for
any $0\leqq s<T,$ $u$ is a weak $r$ solution in $Q_{\mathbb{R}^{N},s,T}$; and
$\int_{\mathbb{R}^{N}}u^{r}(s)dx<\infty$ with $r\geqq1$; for any $s\leqq t<T,$
and for any $\alpha\geqq0$ such that $\int_{\mathbb{R}^{N}}u^{\alpha
+1}(s)dx<\infty,$ applying Theorem \ref{decay} to $u$ starting at point $s,$
denoting $\beta=1+\alpha/q,$ we have
\begin{equation}
\frac{1}{\alpha+1}\int_{\mathbb{R}^{N}}u^{\alpha+1}(.,t)dx+\int_{s}^{t
\int_{\mathbb{R}^{N}}|\nabla(u^{\beta})|^{q}dxdt\leqq\frac{1}{\alpha+1
\int_{\mathbb{R}^{N}}u^{\alpha+1}(.,s)dx \label{miss
\end{equation}
and $u^{\beta}(.,t)\in L^{q}(Q_{\mathbb{R}^{N},s,\tau})$ for a.e.$t$. \medskip

(i) Proof of (\ref{uinf}).\ First suppose $q<N.$ Then from the Sobolev
injection of $W^{1,q}\left(  \mathbb{R}^{N}\right)  $ into $L^{q^{\ast
}\left(  \mathbb{R}^{N}\right)  $,
\[
\frac{1}{\alpha+1}\int_{\mathbb{R}^{N}}u^{\alpha+1}(.,t)dx+\frac{C_{N,q
}{\beta^{q}}\int_{s}^{t}(\int_{\mathbb{R}^{N}}u^{\beta q^{\ast}
(.,t)dx)^{\frac{q}{q^{\ast}}})dt\leqq\frac{1}{\alpha+1}\int_{\mathbb{R}^{N
}u^{\alpha+1}(.,s)dx
\]
so that we can apply Lemma \ref{prod} with $m=q$ and $\theta=N/(N-q)$ and
deduce (\ref{uinf}). If $q\geqq N$ we still obtain (\ref{uinf}), with
$\sigma_{r,q,N}=1/(q+r-1)=\varpi_{r,q,N}/r$ if $q>N,$ and $\sigma
_{r,q,N}=1/(N(1-\delta)+r-1)=\varpi_{r,q,N}/r(1-\delta)$ if $q=N,$ where
$\delta\in\left(  0,1\right)  $ is arbitrary$.$ Indeed, if $q=N,$then
$W^{1,q}\left(  \mathbb{R}^{N}\right)  \subset$ $L^{q\theta}\left(
\mathbb{R}^{N}\right)  $ for any $\theta>0,$ and Lemma \ref{prod} applies. If
$q>N,$ $W^{1,q}\left(  \mathbb{R}^{N}\right)  \subset$ $L^{\infty}\left(
\mathbb{R}^{N}\right)  ,$ and then $t\mapsto\left\Vert u(.,t)\right\Vert
_{L^{\infty}\left(  \mathbb{R}^{N}\right)  }^{\beta q}=\left\Vert
u(.,t)\right\Vert _{L^{\infty}\left(  \mathbb{R}^{N}\right)  }^{q+r-1}$ is
nonincreasing, thus for any $r\geqq1,$ from (\ref{miss}),
\[
\int_{\mathbb{R}^{N}}u^{r}(.,t)dx+C_{N,N}rt\left\Vert u(.,t)\right\Vert
_{L^{\infty}\left(  \mathbb{R}^{N}\right)  }^{q+r-1}\leqq\int_{\mathbb{R}^{N
}u_{0}^{r}dx.
\]


(ii) Proof of (\ref{use}). Assume $\nu>0,N>2.$ For any $\alpha>0$ such that
$\int_{\mathbb{R}^{N}}u^{\alpha+1}(s)dx<\infty$
\[
\frac{1}{\alpha+1}\int_{\mathbb{R}^{N}}u^{\alpha+1}(t)dx+\frac{\alpha
{\tilde{\beta}^{2}}\nu\int_{\sigma}^{t}\int_{\mathbb{R}^{N}}\left\vert
\nabla(u^{\tilde{\beta}})\right\vert ^{2}dx\leqq\frac{1}{\alpha+1
\int_{\mathbb{R}^{N}}u^{\alpha+1}(\sigma)dx
\]
where $\tilde{\beta}=(\alpha+1)/2;$ and $u^{\tilde{\beta}}\in L_{loc
^{2}((0,\tau);W^{1,2}\left(  \mathbb{R}^{N}\right)  ).$ From the Sobolev
injection of $W^{1,2}\left(  \mathbb{R}^{N}\right)  $ into $L^{2^{\ast
}\left(  \mathbb{R}^{N}\right)  $, we get
\[
\frac{1}{\alpha+1}\int_{\mathbb{R}^{N}}u^{\alpha+1}(t)dx+\frac{\alpha C_{N
}{\tilde{\beta}^{2}}\nu\int_{s}^{t}(\int_{\mathbb{R}^{N}}u^{\tilde{\beta
}2^{\ast}})^{2^{\ast}/2}dx\leqq\frac{1}{\alpha+1}\int_{\mathbb{R}^{N
}u^{\alpha+1}(s)dx.
\]
First suppose $r>1.$ Then we can apply Lemma \ref{prod} with $C_{0
=(r-1)C_{N}\nu,$ $q=2,\theta=N/(N-2)$ and $\lambda=-1,$ since $\tilde{\beta
}=1+(\alpha-1)/2$, and $r>N(1-2+1)/2,$ and obtain (\ref{use}). Next assume
$r=1.$ Then $u\in C(\left[  0,T\right)  ;L^{1}(\mathbb{R}^{N}))\cap
L_{loc}^{\infty}((0,T);L^{\infty}(\mathbb{R}^{N}))$ \textit{because of
estimate} (\ref{uinf}), then $C(\left[  0,T\right)  ;L^{\rho}(\mathbb{R
^{N}))$ for any $\rho>1,$ for example with $\rho=2,$ and $\Vert u(.,t)\Vert
_{L^{1}(\mathbb{R}^{N})}$ is nonincreasing, from Theorem \ref{decay}, hence we
can still apply Lemma \ref{prod} on $\left(  \epsilon,t\right)  $ for
$0<\epsilon<t<T$
\[
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq C(t-\epsilon)^{-\frac
{N}{2}}\Vert u(.,\epsilon)\Vert_{L^{1}(\mathbb{R}^{N})}\leqq C(t-\epsilon
)^{-\frac{N}{2}}\Vert u_{0}\Vert_{L^{1}(\mathbb{R}^{N})
\]
Then we still obtain (\ref{use}) with $C=C(N,q,r,\nu).$
\end{proof}

\begin{remark}
\label{parti} If $N\leqq2$ we obtain similarly that $\Vert u(.,t)\Vert
_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\tilde{\sigma}}\Vert u_{0}\Vert
_{L^{r}(\mathbb{R}^{N})}^{\tilde{\varpi}}$ with $\tilde{\sigma}=1/r=\tilde
{\varpi}/r$ if $N=1,$ and $\tilde{\sigma}=1/(r-2\delta)$=$\tilde{\varpi
}/r(1-\delta)$ if $N=2.$
\end{remark}

\begin{remark}
As a consequence, for any $k\geqq1,q<N,
\begin{equation}
\Vert u(.,t)\Vert_{L^{kr}(\mathbb{R}^{N})}\leqq Ct^{-\frac{\sigma_{r,q
}{k^{\prime}}}\Vert u_{0}\Vert_{L^{r}(\mathbb{R}^{N})}^{\frac{\varpi_{r,q
}{k^{\prime}}+\frac{1}{k}}, \label{uk
\end{equation
\begin{equation}
\Vert u(.,t)\Vert_{L^{kr}(\mathbb{R}^{N})}\leqq Ct^{-\frac{N}{2rk^{\prime}
}\Vert u_{0}\Vert_{L^{r}(\mathbb{R}^{N})},\qquad\text{if }\nu>0. \label{uke
\end{equation}
Indeed it follows from (\ref{ghi}) and (\ref{uinf}), (\ref{use}) by
interpolation
\[
\Vert u(.,t)\Vert_{L^{kr}(\mathbb{R}^{N})}\leqq\Vert u(.,t)\Vert_{_{L^{\infty
}(\mathbb{R}^{N})}}^{1/k^{\prime}}\Vert u(.,t)\Vert_{L^{r}(\mathbb{R}^{N
)}^{1/k}.
\]

\end{remark}

\begin{remark}
\label{q2}If $q\leqq2,$ then $u\in C^{2,1}\left(  Q_{\mathbb{R}^{N},T}\right)
,$ from the regularity result of \cite[Theorem 2.12]{BiDao1}. In this case
\textbf{we do not need to introduce the regularization by }$u_{\varepsilon};$
we only need to introduce $u+\delta,$ when $r>1$ and make $\delta
\rightarrow0.$
\end{remark}

In case of initial data measures, we obtain in the same way:

\begin{theorem}
\label{bmea}Assume $q>1.$ Let $u_{0}\in\mathcal{M}_{b}^{+}(\mathbb{R}^{N})$
and $u$ be any non-negative weak $\mathcal{M}_{loc}$ solution of equation
(\ref{cau}) in$\hspace{0.05in}Q_{\mathbb{R}^{N},T}.$Then
\[
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\sigma_{1,q}}
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0})^{\varpi_{1,q}},
\]
where $\sigma_{1,q},\varpi_{1,q}$ are given at (\ref{sigr}), and $C=C(N,q)$.
Moreover if $\nu>0,$ with $C=C(N,q,\nu),$
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\frac{N}{2}
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0}. \label{mas
\end{equation}

\end{theorem}

\begin{proof}
Taking $\epsilon>0,$ we have for any $t\geqq\epsilon,
\[
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq C(t-\epsilon)^{-\sigma
_{1}}\Vert u(.,t)\Vert_{L^{1}(\mathbb{R}^{N})}^{\sigma_{2}}\leqq
C(t-\epsilon)^{-\sigma_{1}}
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0})^{\sigma_{2}
\]
and then we make $\varepsilon\rightarrow0$ and deduce the estimate. If
$\nu>0,$ we also obtain for $t>s>0$
\[
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq C(t-s)^{-\frac{N}{2}
\int_{\mathbb{R}^{N}}u(.,s)dx
\]
and going to the limit as $s\rightarrow0,$ we deduce (\ref{mas}).
\end{proof}

\begin{remark}
Up to now, the decay estimate (\ref{ghi}) and the $L^{\infty}$ estimate
(\ref{uinf}) of $u$ were proved in case $u_{0}\in C_{b}\left(  \mathbb{R
^{N}\right)  \cap L^{r}\left(  \mathbb{R}^{N}\right)  $ and for the unique
bounded solution $u$ of problem (\ref{cau}), and based on the estimate
(\ref{bela}) given in \cite[Theorem 5.6]{BASoWe}; indeed (\ref{uinf}) follows
from the Gagliardo-Nirenberg estimate:
\[
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq\Vert\nabla u(.,t)\Vert
_{L^{\infty}(\mathbb{R}^{N})}^{\frac{N}{N+r}}\Vert u(.,t)\Vert_{L^{r
(\mathbb{R}^{N})}^{\frac{r}{N+r}}\leqq C(q,r)\Vert u(.,t)\Vert_{L^{\infty
}(\mathbb{R}^{N})}^{\frac{N}{q(N+r)}}\Vert u_{0}\Vert_{L^{r}(\mathbb{R}^{N
)}^{\frac{r}{N+r}}.
\]

\end{remark}

\subsection{Further estimates and convergence results for $q\leqq2.$}

Here we consider the case $1<q\leqq2.$ From the $L^{\infty}$ estimates above,
and the interior regularity of $u,$ we deduce new local estimates and
convergence results:

\begin{corollary}
\label{pass}Assume $1<q\leqq2.\medskip$

(i) Any nonnegative weak $L_{loc}^{r}$ solution (resp. $\mathcal{M}_{loc}$
solution) $u$ of problem (\ref{cau}) with initial data $u_{0}\in
L^{r}(\mathbb{R}^{N}),$ $r\geqq1$ (resp. $u_{0}\in\mathcal{M}_{b
^{+}(\mathbb{R}^{N})$) satisfies $u\in C^{2,1}(Q_{\mathbb{R}^{N},T})\cap
L_{loc}^{\infty}(\left(  0,T\right)  ;C_{b}(\mathbb{R}^{N})).\medskip$

(ii) Let $(u_{0,n})$ be any bounded sequence in $L^{r}(\mathbb{R}^{N}),$
$r\geqq1$ (resp. in $\mathcal{M}_{b}^{+}(\mathbb{R}^{N})).$ For any
$n\in\mathbb{N},$ let $u_{n}$ be any nonnegative weak $L_{loc}^{r}$ solution
(resp. $\mathcal{M}_{loc}$ solution) of problem (\ref{cau}) with initial data
$u_{0,n}.$ Then one can extract a subsequence converging in $C_{loc
^{2,1}(Q_{\mathbb{R}^{N},T})$ to a weak solution $u$ of (\ref{un}) in
$Q_{\mathbb{R}^{N},T}$.
\end{corollary}

\begin{proof}
From \cite[Theorem 2.16]{BiDao1} there there exists $\gamma\in\left(
0,1\right)  $ such that for any nonnegative weak solution of equation
(\ref{un}) $u$ in $Q_{\mathbb{R}^{N},T}$ and any ball $B_{R}\subset
\mathbb{R}^{N},$ and $0<s<\tau<T,
\[
\left\Vert u\right\Vert _{C^{2+\gamma,1+\gamma/2}(Q_{B_{R},s,\tau})}\leqq
C\Phi(\left\Vert u\right\Vert _{L^{\infty}(Q_{B_{2R},s/2,\tau})}).
\]
where $C=C(N,q,R,s,\tau)$ and $\Phi$ is a continuous increasing function. From
estimates (\ref{uinf}), (\ref{mas}), we deduce that $u\in L^{\infty}(\left(
0,T\right)  ;C_{b}(\mathbb{R}^{N}))$ and
\begin{equation}
\left\Vert u\right\Vert _{C^{2+\gamma,1+\gamma/2}(Q_{B_{R},s,\tau})}\leqq
C\Phi(\left\Vert u_{0}\right\Vert _{L^{r}(\mathbb{R}^{N})}),\qquad\text{(resp.
}\left\Vert u\right\Vert _{C^{2+\gamma,1+\gamma/2}(Q_{B_{R},s,\tau})}\leqq
C\Phi(\int_{\mathbb{R}^{N}}du_{0}) \label{gff
\end{equation}
and the conclusions follow.$\medskip$
\end{proof}

We also deduce global gradient estimates in $\mathbb{R}^{N}:$

\begin{corollary}
\label{grad}Assume $1<q\leqq2$ (i) Let $u_{0}\in L^{r}(\mathbb{R}^{N}),$
$r\geqq1.$ Then any weak $L_{loc}^{r}$ solution $u$ of problem (\ref{cau})
satisfies
\begin{equation}
\Vert\nabla u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\vartheta
_{r,q}}\Vert u_{0}\Vert_{L^{r}(\mathbb{R}^{N})}^{\varkappa_{_{r,q}}},
\label{sic
\end{equation
\[
\vartheta_{r,q}=\frac{N+r}{rq+N(q-1)},\qquad\varkappa_{_{r,q}}=\frac
{r}{rq+N(q-1)
\]
and $\left\vert \nabla u\right\vert ^{q}\in L_{loc}^{\infty}((0,T);L^{r
(\mathbb{R}^{N})),$ and
\begin{equation}
\int_{\mathbb{R}^{N}}\left\vert \nabla u(.,t)\right\vert ^{qr}dx\leqq
C_{q}t^{-r(\frac{q}{2}+\sigma_{r,q}(q-1))}\Vert u_{0}\Vert_{L^{r
(\mathbb{R}^{N})}^{(1+\varpi_{r,q}(q-1))r} \label{vou
\end{equation}
And for $\nu>0,$
\begin{equation}
\Vert\nabla u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\frac{1
{q}(\frac{N}{2r}+1)}\Vert u_{0}\Vert_{L^{r}(\mathbb{R}^{N})}^{\frac{1}{q}};
\label{sac
\end{equation
\begin{equation}
\int_{\mathbb{R}^{N}}\left\vert \nabla u(.,t)\right\vert ^{qr}dx\leqq
C_{q}t^{-r(\frac{q}{2}+\frac{N}{2r}(q-1))}\Vert u_{0}\Vert_{L^{r
(\mathbb{R}^{N})}^{qr}, \label{voi
\end{equation}
moreover if $q<2,$ $u$ is a pointwise mild solution.\medskip

(ii) Let $u_{0}\in\mathcal{M}_{b}^{+}(\mathbb{R}^{N}).$ Then any weak
$\mathcal{M}_{loc}$ solution of (\ref{cau}) satisfies
\[
\Vert\nabla u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\vartheta
_{1,q}}
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0})^{\varkappa_{_{1,q}}},
\]
and $\left\vert \nabla u\right\vert \in L_{loc}^{\infty}((0,T);L^{q
(\mathbb{R}^{N});$ and for $\nu>0,$
\[
\Vert\nabla u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq Ct^{-\frac{1
{q}(\frac{N}{2}+1)}
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0})^{1/q}.
\]


As a consequence, in any case $u$ is defined on $\left(  0,\infty\right)
$.\bigskip
\end{corollary}

\begin{proof}
(i) Let $u_{0}\in L^{r}(\mathbb{R}^{N}),$ $r\geqq1.$ Then for any
$\epsilon>0,$ $u(.,\epsilon)\in C_{b}(\mathbb{R}^{N}),$ from Corollary
\ref{pass}. From \cite{GiGuKe}, $u$ is the unique solution $v$ such that $v\in
C^{2,1}\left(  \mathbb{R}^{N}\times\left(  \epsilon,T\right)  \right)  \cap
C_{b}\left(  \mathbb{R}^{N}\times\left[  \epsilon,T\right)  \right)  $, and
$v(.,\epsilon)=u(.,\epsilon)$; since $v\in C_{b}^{2}\left(  \mathbb{R
^{N}\times(\epsilon,T\right)  ),$ we deduce that $u\in C_{b}^{2}\left(
\mathbb{R}^{N}\times(0,T\right)  );$ and for any $\epsilon\leqq t<T,$
\[
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq\Vert u(.,\epsilon
)\Vert_{L^{\infty}(\mathbb{R}^{N})},\qquad\Vert\nabla u(.,t)\Vert_{L^{\infty
}(\mathbb{R}^{N})}\leqq\Vert\nabla u(.,\epsilon)\Vert_{L^{\infty
(\mathbb{R}^{N})},
\]
and
\begin{equation}
\left\vert \nabla u(.,t)\right\vert ^{q}\leqq C_{q}\frac{u(.,t)}{t-\epsilon
},\qquad\text{a.e. in }\mathbb{R}^{N}. \label{lis
\end{equation}
From the decay estimates, we also have $\Vert u(.,\epsilon)\Vert
_{L^{r}(\mathbb{R}^{N})}\leqq\Vert u_{0}\Vert_{L^{r}(\mathbb{R}^{N})}.$ And
$u(.,\epsilon)\in L^{\tilde{r}}(\mathbb{R}^{N})$ for any $\tilde{r}\in\left[
r,\infty\right]  ,$ and $u\in C(\left[  \epsilon,T\right)  ;L^{\tilde{r
}(\mathbb{R}^{N})).$ Going to the limit in (\ref{lis}) as $\epsilon
\rightarrow0,$ we deduce (\ref{sic}) from (\ref{uinf}), and (\ref{sac}) from
(\ref{use}). Moreover $\left\vert \nabla u\right\vert ^{q}\in L_{loc}^{\infty
}((0,T);L^{r}(\mathbb{R}^{N})),$ since
\[
\Vert\nabla u(.,t)\Vert_{L^{qr}(\mathbb{R}^{N})}\leqq Ct^{-1/q}\Vert
u_{0}\Vert_{L^{r}(\mathbb{R}^{N})}^{1/q}.
\]
More precisely we get from estimate (\ref{cor}),
\[
\Vert\nabla(u^{\frac{q-1}{q}}(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq
C_{q}(t-\epsilon)^{-1/2}\Vert u(.,\epsilon)\Vert_{L^{\infty}(\mathbb{R}^{N
)}^{\frac{q-1}{q}
\]
then from estimate (\ref{use}), for any $t\in\left(  0,T\right)  ,$ with other
constants $C_{q},
\[
\Vert\nabla(u^{\frac{q-1}{q}}(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq
C_{q}t^{-1/2}\Vert u(.,\frac{t}{2})\Vert_{L^{\infty}(\mathbb{R}^{N})
^{\frac{q-1}{q}
\
\[
\left\vert \nabla u(.,t)\right\vert ^{q}\leqq C_{q}t^{-q/2}\Vert u(.,\frac
{t}{2})\Vert_{L^{\infty}(\mathbb{R}^{N})}^{q-1}u(.,t),
\]
then from estimate (\ref{uinf}) we get
\[
\int_{\mathbb{R}^{N}}\left\vert \nabla u(.,t)\right\vert ^{qr}dx\leqq
C_{q}\Vert u_{0}\Vert_{L^{r}(\mathbb{R}^{N})}^{\varpi_{r,q}(q-1)r
t^{-r(\frac{q}{2}+\sigma_{r,q}(q-1))}\int_{\mathbb{R}^{N}}u(.,t)^{r}dx
\]
then (\ref{vou}) follows.$\medskip$

Assume that $\nu>0$; then (\ref{voi}) follows from (\ref{use}). Moreover, from
\cite[Theorem 6]{GiGuKe}, $u(.,t)\in C_{b}^{2}(\mathbb{R}^{N})$ for any
$t\in\left(  \epsilon,T\right)  ,$ in particular $u(.,2\epsilon)\in C_{b
^{2}(\mathbb{R}^{N}),$ then for any $t\geqq\epsilon,$ and any $x\in
\mathbb{R}^{N},$
\begin{equation}
u(x,t)=e^{(t-2\epsilon)\Delta}u(x,2\epsilon)-\int_{2\epsilon}^{t
\int_{\mathbb{R}^{N}}g(x-y,t-s)|\nabla u(y,s)|^{q}dyds, \label{brou
\end{equation}
see for example \cite[Proposition 4.2 ]{BeBALa}. But $u(x,2\epsilon)$
converges to $u_{0}$ in $L^{r}(\mathbb{R}^{N}),$ and then $e^{(t-2\epsilon
)\Delta}u(.,\epsilon)$ converges to $e^{t\Delta}u_{0}$ in $L^{r
(\mathbb{R}^{N}).$ Then we can go to the limit as $\epsilon\rightarrow0$ in
(\ref{brou}), for a.e. $x\in\mathbb{R}^{N}:$ the integral is convergent, then
$u$ is a pointwise mild solution.\medskip

(ii) For Theorem \ref{inmea}, we have $u(.,t)\in L^{1}(\mathbb{R}^{N})$ for
$t\geqq\epsilon>0,$ which gives from (i)
\[
\Vert u(.,t)\Vert_{L^{\infty}(\mathbb{R}^{N})}\leqq C(t-\varepsilon
)^{-\sigma_{1,q}}\Vert u(.,\epsilon)\Vert_{L^{1}(\mathbb{R}^{N})
^{\varpi_{1,q}}\leqq C(t-\varepsilon)^{-\sigma_{1,q}}
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0})^{\varpi_{1,q}
\]
and then we go to the limit as $\epsilon\rightarrow0.$ And $\left\vert \nabla
u\right\vert \in L_{loc}^{\infty}((0,T);L^{q}(\mathbb{R}^{N}),$ since
\[
\Vert\nabla u(.,t)\Vert_{L^{q}(\mathbb{R}^{N})}\leqq Ct^{-1/q}
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0})^{1/q}.
\]
And the estimates (\ref{vou}) and (\ref{voi}) hold with $r=1$ and $\Vert
u_{0}\Vert_{L^{1}(\mathbb{R}^{N})}$ replaced by 
{\displaystyle\int_{\mathbb{R}^{N}}}
du_{0}.$
\end{proof}

\subsection{Existence and uniqueness results for $q\leq2$}

Let $u_{0}\in L^{r}(\mathbb{R}^{N}),$ $r\geqq1.$ We first consider the
subcritical case $q<(N+2r)/(N+r)$, equivalently $q<2$ and $r>N(q-1)/(2-q).$

\begin{theorem}
\label{val}Let $u_{0}\in L^{r}(\mathbb{R}^{N}),$ $r\geqq1.$ Suppose
$1<q<(N+2r)/(N+r)$. Then any weak $L_{loc}^{r}$ solution $u$ of problem
(\ref{cau}) satisfies
\begin{equation}
\left\vert \nabla u\right\vert ^{q}\in L_{loc}^{1}(\left[  0,T\right)
;L^{r}(\mathbb{R}^{N})). \label{dim
\end{equation}
And
\[
u\text{ is a weak }L_{loc}^{r}\text{ solution}\Longleftrightarrow u\text{ is a
mild }L^{r}\text{ solution.
\]

\end{theorem}

\begin{proof}
Let $u$ be any weak $L_{loc}^{r}$ solution. Then from (\ref{vou}),
\[
\int_{0}^{\tau}\left\Vert \nabla u(.,t)\right\Vert _{L^{qr}(\mathbb{R}^{N
)}^{q}dt=\int_{0}^{\tau}(\int_{\mathbb{R}^{N}}\left\vert \nabla
u(.,t)\right\vert ^{qr}dx)^{\frac{1}{r}}dt\leqq C\int_{0}^{\tau}t^{-(\frac
{q}{2}+\sigma_{r,q}(q-1))}dt
\]
with $C=C_{q}\Vert u_{0}\Vert_{L^{r}(\mathbb{R}^{N})}^{(1+\varpi_{r,q
(q-1))r},$ and $q/2+\sigma_{r,q}(q-1)<1$ is equivalent to $q<(N+2r)/(N+r)$; if
$\nu>0$, the estimate (\ref{voi}) leads to the same conclusion, since
$q/2+(q-1)N/2r<1$ is still equivalent to $q<(N+2r)/(N+r).$ Then (\ref{dim})
holds. Moreover from Corollary \ref{grad}, $u$ is a mild pointwise solution:
\begin{equation}
u(.,t)=e^{t\Delta}u_{0}(.)-\int_{0}^{t}\int_{\mathbb{R}^{N}}g(x-y,t-s)|\nabla
u(y,s)|^{q}dyds; \label{frou
\end{equation}
and $u\in C(\left[  0,T\right)  ;L^{r}\left(  \mathbb{R}^{N}\right)  )$ from
Theorem \ref{decay}, and $f=\left\vert \nabla u\right\vert ^{q}\in L_{loc
^{1}(\left[  0,T\right)  ;L^{r}(\mathbb{R}^{N})),$ thus the relation
(\ref{frou}) holds in $L^{r}(\mathbb{R}^{N}),$
\begin{equation}
u(.,t)=(e^{t\Delta}u_{0})-\int_{0}^{t}e^{(t-s)\Delta}\left\vert \nabla
u(.,s)\right\vert ^{q}(s)ds\qquad\text{in }L^{r}(\mathbb{R}^{N}), \label{truc
\end{equation}
that means $u$ is a mild $L^{r}$solution. Conversely it is clear that any mild
$L^{r}$solution is a weak $L_{loc}^{r}$ solution$.\medskip$
\end{proof}

Next we deduce the uniqueness results of Theorem \ref{uni1}.

\begin{theorem}
\label{unir} Let $u_{0}\in L^{r}(\mathbb{R}^{N}).$ Assume $1<q<(N+2r)/(N+r),$
or $q=2.$ Then there exists a unique weak $L_{loc}^{r}$ solution $u$ of
problem (\ref{cau}). In the first case, $u\in C((0,T);W^{1,qr}(\mathbb{R
^{N})).$
\end{theorem}

\begin{proof}
(i) Case $1<q<(N+2r)/(N+r).$ From \cite[Theorem 2.1]{BASoWe}, there exists a
mild $L^{r}$ solution, then it is a $L_{loc}^{r}$ solution. Let us show the
uniqueness. Let $u$ be any weak $L_{loc}^{r}$ solution, thus $u$ is a mild
$L^{r}$ solution, from Theorem \ref{val}. And $u\in L^{\infty}((0,T);L^{r
(\mathbb{R}^{N}))$ from Theorem \ref{decay}, and $u\in L_{loc}^{\infty
}((0,T);W^{1,qr}\left(  \mathbb{R}^{N}\right)  )$, since $\left\vert \nabla
u\right\vert \in L_{loc}^{\infty}((0,T);L^{qr}\left(  \mathbb{R}^{N}\right)
)$ from Theorem \ref{val} and $u\in L_{loc}^{\infty}((0,T);L^{qr}\left(
\mathbb{R}^{N}\right)  )$ by interpolation. . Then we enter in the class of
uniqueness $u\in L_{loc}^{\infty}((0,T);W^{1,qr}\left(  \mathbb{R}^{N}\right)
)$ required in \cite[Lemma 2.2 and Remark 2.5]{BASoWe}. Thus $u$ is unique,
and satisfies $u\in C((0,T);W^{1,qr}(\mathbb{R}^{N})),$ from \cite[Theorem
2.1]{BASoWe}.\medskip

(ii) Case $q=2.$ From \cite[Theorem 4.2]{BASoWe} there exists a unique
solution $u$ such that $u\in C(\left[  0,T\right)  ;L^{r}\left(
\mathbb{R}^{N}\right)  )\cap u\in C^{2,1}(\left(  Q_{\mathbb{R}^{N},\infty
}\right)  $ solution of (\ref{un}) at each point. Then it is a weak
$L_{loc}^{r}$ solution. Reciprocally any weak $L_{loc}^{r}$ solution $u$
satisfies the conditions above, from Theorem \ref{decay} and \cite[Theorem
2.16]{BiDao1}.
\end{proof}

\begin{theorem}
\label{souc} Assume that $1<q<(N+2)/(N+1).$ Let $u_{0}\in\mathcal{M}_{b
^{+}(\mathbb{R}^{N}).$ Then there exists a unique weak $\mathcal{M}_{loc}$
solution of problem (\ref{cau}).
\end{theorem}

\begin{proof}
The existence of a weak semi-group solution was obtained in \cite{BeLa99} by
approximation. The existence of a mild $\mathcal{M}$ solution was proved in
\cite[Theorem 2.2]{BASoWe}, and the two notions are equivalent from Lemma
\ref{lem}. In any case the solution is a weak $\mathcal{M}_{loc}$ solution.
Next consider any solution $\mathcal{M}_{loc}$ solution $u$. Then $u(.,t)\in
L^{\infty}(\mathbb{R}^{N})$ for any $t\in\left(  \epsilon,T\right)  $ by
applying theorem \ref{bound} from $\epsilon/2$. Then again we deduce
$u(.,\epsilon)\in C_{b}(\mathbb{R}^{N}),$ then (\ref{lis}). From Theorem
\ref{decay} we obtain again that $u\in L_{loc}^{\infty}((0,T);W^{1,q}\left(
\mathbb{R}^{N}\right)  ).$ And moreover from the uniqueness after $\epsilon,$
we have $u\in C((\epsilon,T);W^{1,q}\left(  \mathbb{R}^{N}\right)  )$ from
Theorem \ref{unir}. Then $u\in C((0,T);W^{1,q}\left(  \mathbb{R}^{N}\right)
).$ And $u$ satisfies (\ref{ris}) from Theorem \ref{inmea}. Then $u$ is a weak
semi-group solution, then a mild $\mathcal{M}$ solution from Lemma \ref{lem}.
Then we enter the class of uniqueness of \cite[Theorem 2.2]{BASoWe}. We can
also prove the uniqueness directly: if $u_{1},u_{2}$ are two solutions, since
they are mild $\mathcal{M}$ solutions, we have
\[
(u_{1}-u_{2})(.,t)=\int_{0}^{t}e^{(t-s)\Delta}(\left\vert \nabla
u_{1}(.,s)\right\vert ^{q}-\left\vert \nabla u_{2}(.,s)\right\vert ^{q})ds
\]
and we know that $\left\vert \nabla u_{j}\right\vert ^{q}\in C((0,T);L^{r
(\mathbb{R}^{N})),$ henc
\begin{align*}
\left\Vert \nabla(u_{1}-u_{2})(.,t)\right\Vert _{L^{qr}\left(  \mathbb{R
^{N}\right)  }  &  \leqq\int_{0}^{t}\left\Vert \nabla(e^{(t-s)\Delta
})\right\Vert _{L^{1}\left(  \mathbb{R}^{N}\right)  }\left\Vert \left\vert
\nabla u_{1}(.,s)\right\vert ^{q}-\left\vert \nabla u_{2}(.,s)\right\vert
^{q}\right\Vert _{L^{qr}\left(  \mathbb{R}^{N}\right)  }ds\\
&  \leqq C\int_{0}^{t}(t-s)^{-1/2}\max_{j=1,2}\left\Vert \nabla u_{j
(.,s)\right\Vert _{L^{\infty}\left(  \mathbb{R}^{N}\right)  }^{q-1}\left\Vert
\nabla(u_{1}-u_{2})(.,s)\right\Vert _{L^{qr}\left(  \mathbb{R}^{N}\right)
}ds\\
&  \leqq C\int_{0}^{t}(t-s)^{-1/2}s^{-(q-1)\vartheta_{1,q}}\left\Vert
\nabla(u_{1}-u_{2})(.,s)\right\Vert _{L^{qr}\left(  \mathbb{R}^{N}\right)  }ds
\end{align*}
and we can apply the singular Gronwall lemma when $2<(q-1)\vartheta_{1,q},$
which means precisely $q<\frac{N+2}{N+1}.$ Then $\nabla(u_{1}-u_{2})(.,t)=0$
in $L^{qr}\left(  \mathbb{R}^{N}\right)  ,$ hence $u_{1}=u_{2}.\medskip$
\end{proof}

Finally we give a short proof of the existence result of \cite[Theorem
4.1]{BASoWe}.

\begin{proposition}
\label{pou}Let $1<q<2$. For any nonnegative $u_{0}\in L^{r}(\mathbb{R
^{N}),r\geq1,$ there exists a mild pointwise solution $u$ of problem
(\ref{cau}), and $u\in C(\left[  0,T\right)  ;L^{r}\left(  \mathbb{R
^{N}\right)  )$.
\end{proposition}

\begin{proof}
Let $u_{0,n}=\min(u_{0},n).$ Then $u_{0,n}\in L^{\rho}(\mathbb{R}^{N})$ for
any $\rho\geq r.\ $Choosing $\rho>N(q-1)/(2-q),$ that means $q<q_{\rho},$ from
\cite[Theorem 2.1]{BASoWe}, there exists a mild $L^{\rho}$ solution $u_{n}$
with initial data $u_{0,n},$ and $u_{n}\in C((0,T);C_{b}^{2}(\mathbb{R
^{N}))\cap C^{2,1}(Q_{\mathbb{R}^{N},T}).$ Then $\left(  u_{n}\right)  $ is
nondecreasing from the comparison principle, and $u_{n}(.,t)\leq e^{t\Delta
}u_{0}\leq Ct^{-N/2r}\left\Vert u_{0}\right\Vert _{L^{r}(\mathbb{R}^{N})}.$
From Corollary \ref{pass}, $\left(  u_{n}\right)  $ converges in
$C_{loc}^{2,1}(Q_{\mathbb{R}^{N},T})$ to a weak solution $u$ of (\ref{un}) in
$Q_{\mathbb{R}^{N},T}$, and $u(.,t)\leq e^{t\Delta}u_{0}.$ Moreoever
$(\left\vert \nabla u_{n}\right\vert ^{q})$ is bounded in $L_{loc}^{1}\left(
\left[  0,T\right)  ;L_{loc}^{1}(\mathbb{R}^{N})\right)  :$ indeed for any
$\xi\in\mathcal{D}^{+}(\mathbb{R}^{N}),$ with values in $\left[  0,1\right]
,$ and any $0<s<t<T,
\begin{align*}
&  \int_{\mathbb{R}^{N}}u_{n}(t,.)\xi^{q^{\prime}}dx+\int_{s}^{t
\int_{\mathbb{R}^{N}}|\nabla u_{n}|^{q}\xi^{q^{\prime}}dx\leqq-q^{\prime
\int_{s}^{t}\int_{\mathbb{R}^{N}}\xi^{1/(q-1)}\nabla u_{n}.\nabla\xi
dx+\int_{\mathbb{R}^{N}}u_{n}(s,.)\xi^{q^{\prime}}dx\\
&  \leqq\frac{1}{2}\int_{s}^{t}\int_{\mathbb{R}^{N}}|\nabla u_{n}|^{q
\xi^{q^{\prime}}dx+Ct\int_{\mathbb{R}^{N}}|\nabla\xi|^{q^{\prime}
dx+\int_{\mathbb{R}^{N}}u_{n}(s,.)\xi^{q^{\prime}}dx
\end{align*}
and $u_{n}\in C(\left[  0,T\right)  ;L^{\rho}\left(  \mathbb{R}^{N}\right)
),$ thus we can go to the limit as $s\rightarrow0:$
\[
\int_{\mathbb{R}^{N}}u_{n}(t,.)\xi^{q^{\prime}}dx+\frac{1}{2}\int_{s}^{t
\int_{\mathbb{R}^{N}}|\nabla u_{n}|^{q}\xi^{q^{\prime}}dx\leqq Ct\int
_{\mathbb{R}^{N}}|\nabla\xi|^{q^{\prime}}dx+\int_{\mathbb{R}^{N}}u_{0
\xi^{q^{\prime}}dx.
\]
Thus \textbf{ }$\left\vert \nabla u\right\vert ^{q}\in L_{loc}^{1}\left(
\left[  0,T\right)  ;L_{loc}^{1}(\mathbb{R}^{N})\right)  ,$ hence, from
\cite[Proposition 2.11]{BiDao1}, $u$ admits a trace as $t\rightarrow0:$ there
exists a Radon measure $\mu_{0}$ in $\mathbb{R}^{N},$ such that $u(.,t)$
converges weakly* to $\mu_{0}.$ And $e^{t\Delta}u_{0}$ converges to $u_{0}$ in
$L^{r}(\mathbb{R}^{N}),$ thus $\mu_{0}\in L_{loc}^{1}(\mathbb{R}^{N})$ and
$0\leq\mu_{0}\leq u_{0}$; and $u_{n}\leq u,$ thus $u_{0,n}\leq\mu_{0},$ hence
$\mu_{0}=u_{0}.$ Also there exists a function $g\in L^{r}(\mathbb{R}^{N})$
such that $u(.,t)\leqq g$ for small $t.$ Then the nonnegative function
$e^{t\Delta}u_{0}-u(.,t)$ converges weakly* to $0,$ and then in $L_{loc
^{1}(\mathbb{R}^{N}).$ Hence $u(.,t)$ converges to $u_{0}$ in $L_{loc
^{1}(\mathbb{R}^{N}),$ then in $L^{r}(\Omega)$ from the Lebesgue theorem. Thus
$u\in C(\left[  0,T\right)  ;L^{r}\left(  \mathbb{R}^{N}\right)  )$. In
particular $u$ is a weak $L_{loc}^{r}$ solution, then a pointwise mild
solution, from Corollary \ref{grad}.
\end{proof}

\begin{remark}
The uniqueness of the solution is still an open problem when $u_{0}\in
L^{r}(\mathbb{R}^{N})$ and $q\geqq(N+2r)/(N+r).$
\end{remark}

\subsection{More decay estimates for $q<(N+2r)/(N+r)$}

Here, we exploit theorem \ref{decay} to obtain a better decay estimate of the
$L^{r}$ norm when $u_{0}\in L^{r}(\mathbb{R}^{N})$ and $q<(N+2r)/(N+r),$ which
appears to be new for $r>1.$ In case $r=1$ we find again the result of
\cite{AnTeU}, proved \textit{under the assumption that the energy relation
(\ref{his}) holds.}

\begin{theorem}
\label{esti}Let $r\geqq1$ and $1<q<(N+2r)/(N+r).$ Let $u$ be any non-negative
weak $r$ solution of problem (\ref{cau}) with $u_{0}\in L^{r}(\mathbb{R}^{N
)$. Then there exists $C=C(N,q,r)$ such that for any $t\in\left(  0,T\right)
$
\begin{equation}
\int_{\mathbb{R}^{N}}u^{r}(.,t)dx\leq C(\int_{\{|x|>\sqrt{t}\}}u_{0
^{r}(x)dx+t^{-\frac{ar-N}{2}},\qquad a=\frac{2-q}{q-1}. \label{alb
\end{equation}
As a consequence, $\lim_{t\longrightarrow\infty}\Vert u(t)\Vert_{L^{r
(\mathbb{R}^{N})}=0$ and
\[
r\int_{0}^{\infty}\int_{\mathbb{R}^{N}}u^{r-1}|\nabla u|^{q}dxdt+r(r-1)\nu
\int_{0}^{\infty}\int_{\mathbb{R}^{N}}u^{r-2}|\nabla u|^{2}dxdt=\int
_{\mathbb{R}^{N}}u_{0}^{r}dx.
\]

\end{theorem}

\begin{proof}
We still consider $v=u^{m}$ with $m=(q-1+r)/q<$ $r.$ Let $E(s)=\int
_{\mathbb{R}^{N}}u^{r}(.,s)dx,$ thus from the energy relation (\ref{egge}) of
theorem \ref{decay}, $E\in W^{1,1}((0,T))$ and for almost any $s\in(0,T),
\[
E^{\prime}(s)=-r(r-1)\int_{\mathbb{R}^{N}}|\nabla u|^{2}u^{r-2}(.,s)dx-\int
_{\mathbb{R}^{N}}|\nabla u|^{q}u^{r-1}(.,s)dx\leq0.
\]
Next, we set $E=E_{1}+E_{2}$ with $E_{1}(s)=\int_{\{|x|<2R\}}u^{r}(x,s)dx,$
$E_{2}(s)=\int_{\{|x|\geq2R\}}u^{r}(x,s)dx.$ From the Gagliardo-Nirenberg
inequality (\ref{gal}), we obtain successively
\begin{align*}
E_{1}(s)  &  =\int_{\{|x|<2R\}}v^{\frac{r}{m}}(x,s)dx\leq\left(
\int_{\{|x|<2R\}}v^{q}(x,s)dx\right)  ^{\frac{r}{mq}}(2R)^{1-\frac{r}{mq}}\\
&  \leqq C\Vert\nabla v(s)\Vert_{L^{q}(\mathbb{R}^{N})}^{\frac{kr}{m}}\Vert
v(s)\Vert_{L^{r/m}(\mathbb{R}^{N})}^{\frac{(1-k)r}{m}}R^{N(1-\frac{r}{mq})}\\
&  \leqq\frac{1}{2}\Vert v(s)\Vert_{L^{r/m}(\mathbb{R}^{N})}^{\frac{r}{m
}+C(N,q,r)\Vert\nabla v(s)\Vert_{L^{q}(\mathbb{R}^{N})}^{\frac{kr}{m}
R^{\frac{N}{k}(1-\frac{r}{mq})},
\end{align*}
thus
\begin{equation}
E(s)\leqq C(\Vert\nabla v(s)\Vert_{L^{q}(\mathbb{R}^{N})}^{\frac{r}{m
}R^{\frac{N}{k}(1-\frac{r}{mq})}+2E_{2}(s)). \label{3.35a
\end{equation}
Consider two smooth cut-off functions $\varphi$, $\eta$ with values in
$\left[  0,1\right]  ,$ such that $\varphi=1$ in $B_{1},$ with support in
$\overline{B_{2}}$, and $\eta=1-\varphi$, and put $\varphi_{l}(x)=\varphi
(\frac{x}{l}),\hspace{0.05in}\eta_{R}(x)=\eta(\frac{x}{R})$. As in the first
step of theorem \ref{decay}, we obtain for any $0<\sigma<s<t<T$, and $l>2R$,
\begin{equation}
\left(  \int_{\mathbb{R}^{N}}u^{r}(.,s)\varphi_{l}^{\lambda}\eta_{R}^{\lambda
}dx\right)  ^{\frac{1}{r}}\leq\left(  \int_{\mathbb{R}^{N}}u^{r
(.,\sigma)\varphi_{l}^{\lambda}\eta_{R}^{\lambda}dx\right)  ^{\frac{1}{r
}+C(s-\sigma)(R^{\frac{N}{r}-q^{\prime}}+l^{\frac{N}{r}-q^{\prime}}),
\label{3.36a
\end{equation}
with $\lambda=rq^{\prime}$ Noting that our assumption on $q$ implies
$N<rq^{\prime}$. As $\sigma\rightarrow0$ and $l\rightarrow\infty$. we deduce
\[
\left(  \int_{\mathbb{R}^{N}}u^{r}(x,s)\eta_{R}dx\right)  ^{\frac{1}{r}
\leq\left(  \int_{\mathbb{R}^{N}}u_{0}^{r}(x)\eta_{R}dx\right)  ^{\frac{1}{r
}+CsR^{\frac{N}{r}-q^{\prime}},
\]
hence, taking $R=\sqrt{t}$, and setting
\[
\rho=r+\frac{N-rq^{\prime}}{2}=\frac{(N+2r)-q(N+r)}{2(q-1)}=\frac{ar-N}{2},
\]
we find
\[
E_{2}(s)\leq A(t)=C\left(  \int_{\{|x|>\sqrt{t}\}}u_{0}^{r}(x)dx+t^{-\rho
}\right)  ,
\]
where with a new constant $C.$ Next, we set $F(s)=E(s)-2A(t).$ $\hspace
{0.05in}$Either there exists $t_{0}\in(0,t)$ such that $F(t_{0})\leq0$, then
$F(s)\leq0,\hspace{0.05in}\forall s\in(t_{0},t),$ thus by continuity,
$E(t)\leq2A(t),$ hence (\ref{alb}) holds. Or $F(s)>0$, \hspace{0.05in}$\forall
s\in(0,t).$ Since
\begin{equation}
-F^{\prime}(s)\geq\int_{\mathbb{R}^{N}}|\nabla u|^{q}u^{r-1}(x,s)dx=\int
_{\mathbb{R}^{N}}|\nabla v(x,s)|^{q}dx \label{3.39a
\end{equation}
it follows from (\ref{3.35a}) that $F(s)\leqq C(-F^{\prime}(s))^{\frac{r}{mq
}t^{\frac{N}{2k}(1-\frac{r}{mq})}.$ Thus by integration
\[
C(t-s)t^{-\frac{N}{2k}(1-\frac{r}{mq})}\leqq F(t)^{-\frac{q-1}{r}
-F(s)^{\frac{q-1}{r}}.
\]
Then as $s\longrightarrow0$ we get $F(t)\leqq Ct^{-\rho},$ since
$\rho=r/(q-1)-N/2k$, and (\ref{alb}) still holds.
\end{proof}

\begin{remark}
The case $r=1$ has been the object of many works, assuming that $u_{0}\in
L^{1}(\mathbb{R}^{N})\cap W^{1,\infty}(\mathbb{R}^{N}).$ There holds
$\lim_{t\longrightarrow\infty}\Vert u(t)\Vert_{L^{1}(\mathbb{R}^{N})}=0$ if
and only if $q\leq(N+2)/(N+1),$ see \cite{AmBA}, \cite{BeLa99}, \cite{BAKo},
\cite{GaLa}. When $q<(N+2)/(N+1)$, the absorption plays a role in the
asymptotics. From \cite{BeKaLa}, if lim$_{\left\vert x\right\vert
\rightarrow\infty}\left\vert x\right\vert ^{a}$ $u_{0}(x)=0,$ then $u(.,t)$
converges as $t\rightarrow\infty$ to the very singular solution constructed in
\cite{QW}, \cite{BeLa01}. In that case $\int_{\mathbb{R}^{N}}u(.,t)dx$ behaves
like $t^{-(a-N)/2}$ for large $t,$ and estimate (\ref{alb}) is sharp$.$ If
$q>(N+2)/(N+1),$ and $u_{0}\in L^{1}(\mathbb{R}^{N}),$ then $u(.,t)$ behaves
as the fundamental solution of heat equation, see \cite{BeKaLa}.

Our result is new when $u_{0}\in L^{r}(\mathbb{R}^{N}),$ $r>1$ and
$u_{0}\not \in L^{1}(\mathbb{R}^{N}).$ When $q>(N+2)/(N+1),$ and $u_{0}$ is
bounded and behaves like $\left\vert x\right\vert ^{-b}$ as $\left\vert
x\right\vert \rightarrow\infty$ with $b\in(a,N),$ it has been shown that
$u(.,t)$ behaves as the selfsimilar solution of the heat equation with initial
data $\left\vert x\right\vert ^{-b},$ see \cite{BiGuKa}. In that case
$u_{0}\in L^{r}(\mathbb{R}^{N})$ for any $r>N/b$ and $\int_{\mathbb{R}^{N
}u^{r}(.,t)dx$ behaves like $t^{-(br-N)/2}$. Thus (\ref{alb}) is sharp as
$b\rightarrow a.$
\end{remark}

\section{The Dirichlet problem in $Q_{\Omega,T}$\label{omega}}

Here we study equation (\ref{un}) in case of a regular bounded domain
$\Omega,$ with Dirichlet conditions on $\partial\Omega\times(0,T),$ with
$\nu=1:$
\begin{equation}
(D_{\Omega,T})\left\{
\begin{array}
[c]{l
u_{t}-\Delta u+|\nabla u|^{q}=0,\quad\text{in}\hspace{0.05in}Q_{\Omega,T},\\
u=0\quad\text{on}\hspace{0.05in}\partial\Omega\times(0,T),
\end{array}
\right.  \label{ome
\end{equation}


Let us recall some well-known results in case of smooth initial data. For any
nonnegative $u_{0}\in C_{0}^{1}\left(  \overline{\Omega}\right)  $, there
exists a unique solution $u\in C^{2,1}\left(  \Omega\times(0,\infty)\right)
\cap$ $C\left(  \overline{\Omega}\times\left[  0,\infty\right)  \right)  ,$
such that $|\nabla u|\in C\left(  \overline{\Omega}\times\left[
0,\infty\right)  \right)  .$ Universal a priori estimates are given in
\cite{CLS}: there exists a constant $C>0$ and a function $D\in C((0,\infty)$
such that
\begin{equation}
u(.,t)\leqq C(1+t^{-1/(q-1)})d(x,\partial\Omega),\qquad|\nabla u(.,t)|\leqq
D(t). \label{gd
\end{equation}
The estimate on $u$ is based on the construction of supersolutions, and the
estimate of the gradient is deduced from the first one by the Bernstein
technique.\medskip

As in section \ref{Rn}, we study the problem with rough initial data, and
introduce different notions of solutions.\medskip

\subsection{Solutions of the heat equation with $L^{1}$ data}

In the following, since the regularization used at Section does not provide
estimates up to the boundary, thus we use another argument: the notion of
\textit{ entropy solution, }introduce din \cite{Pr}, for the problem\textit{
}
\begin{equation}
\left\{
\begin{array}
[c]{l
u_{t}-\Delta u=f,\quad\text{in}\hspace{0.05in}Q_{\Omega,s,\tau},\\
u=0\quad\text{on}\hspace{0.05in}\partial\Omega\times(s,\tau),\\
u(.,s)=u_{s}\geqq0
\end{array}
\right.  \label{ggs
\end{equation}
when $f$ and $u_{s}$ are integrable, that we recall now. For any $k>0$ and
$\theta\in\mathbb{R},$ we define as usual the trucation fonction $T_{k}$ and a
primitive $\Theta_{k}$ by
\begin{equation}
T_{k}(\theta)=\max(-k,\min(k,\theta)),\qquad\Theta_{k}(s)=\int_{0}^{r
T_{k}(\theta)d\theta. \label{tro
\end{equation}


\begin{definition}
Let $s,\tau\in\mathbb{R}$ with $s<\tau,$ and $f\in L^{1}(Q_{\Omega,s,\tau})$
and $u_{s}\in L^{1}(\Omega).$ A function $u\in C(\left[  s,\tau\right]
;L^{1}(\Omega))$ is an entropy solution of the problem (\ref{ggs}) if and for
any $k>0,T_{k}(u)\in L^{2}(\left(  s,\tau\right)  ;W_{0}^{1,2}(\Omega))$ and
\begin{equation
\begin{array}
[c]{c
\int_{\Omega}\Theta_{k}(u-\varphi)(.,\tau)dx-\int_{\Omega}\Theta_{k
(u_{s}-\varphi(.,s)dx+\int_{s}^{\tau}\langle\varphi_{t},T_{k}(u-\varphi
)\rangle dt\\
+\int_{s}^{\tau}\int_{\Omega}(\nabla u.\nabla T_{k}(u-\varphi)-fT_{k
(u-\varphi)dxdt\leq0
\end{array}
\label{nnn
\end{equation}
for any $\varphi\in L^{2}((s,\tau);W^{1,2}(\Omega))\cap L^{\infty}\left(
Q_{\Omega,\tau}\right)  $ such that $\varphi_{t}\in L^{2}((s,\tau
);W^{-1,2}(\Omega))$.
\end{definition}

Other notions of solutions have been used for this problem, see \cite{BeDa},
recalled below. In fact they are equivalent: here $e^{t\Delta}$ denotes the
semi-group of the heat equation with Dirichlet conditions acting on
$L^{1}\left(  \Omega\right)  ,$

\begin{lemma}
\label{pare} Let -$\infty<s<\tau<\infty,$ $f\in L^{1}(Q_{\Omega,s,\tau})$,
$u_{s}\in L^{1}(\Omega)$ and $u\in C(\left[  s,\tau\right]  ;L^{1}(\Omega)),$
$u(.,s)=u_{s}.$ Then denoting the three properties are equivalent:\medskip

(i) u is a weak solution of problem (\ref{ggs}) in $Q_{\Omega,s,\tau},$ that
means $u\in$ $L_{loc}^{1}((s,\tau);W_{0}^{1,1}\left(  \Omega\right)  )$ and
\begin{equation}
u_{t}-\Delta u=f,\quad\text{in}\hspace{0.05in}\mathcal{D}^{\prime
(Q_{\Omega,s,\tau}); \label{fil
\end{equation}


(ii) $u$ is a mild solution of (\ref{ggs}), that means, for any $t\in\left[
s,\tau\right]  ,$
\begin{equation}
u(.,t)=e^{(t-s)\Delta}u_{s}+\int_{s}^{t}e^{(t-\sigma)\Delta}f(\sigma
)d\sigma\qquad\text{in }L^{1}\left(  \Omega\right)  ; \label{seg
\end{equation}


(iii) $u$ is an entropy solution of (\ref{ggs}).\medskip

\noindent Such a solution exists, is unique, and will be called weak solution
of (\ref{ggs}).
\end{lemma}

\begin{proof}
It follows from the existence and uniqueness of the solutions of (i) from
\cite[Lemma 3.4]{BaPi}, as noticed in \cite{BeDa}, and of the entropy
solutions, see \cite{BlMu}.$\medskip$
\end{proof}

As a consequence, when $u$ is bounded, we can admit test functions of the form
$u^{\alpha}:$

\begin{lemma}
\label{dil}Let $s,\tau\in\mathbb{R}$ with $s<\tau,$ and $f\in L^{1
(Q_{\Omega,s,\tau})$ and $u$ be any nonnegative \textbf{bounded }weak solution
in $Q_{\Omega,s,\tau}$ of (\ref{ggs}). \medskip

Then for any $\alpha>0$, we have $u^{\alpha-1}\left\vert \nabla u\right\vert
^{2}\in L^{1}(Q_{\Omega,s,\tau})$ and
\begin{equation}
\frac{1}{\alpha+1}\int_{\Omega}u^{\alpha+1}(.,\tau))dx+\alph
{\displaystyle\int}
\int_{Q_{\Omega,s,\tau}}u^{\alpha-1}\left\vert \nabla u\right\vert
^{2}dxdt=\frac{1}{\alpha+1}\int_{\Omega}u^{\alpha+1}(.,s))dx+\int_{s}^{\tau
}\int_{\Omega}fu^{\alpha}dxdt. \label{hol
\end{equation}

\end{lemma}

\begin{proof}
There holds $u\in L^{2}((s,\tau);W_{0}^{1,2}(\Omega))\cap L^{\infty}\left(
Q_{\Omega,s,\tau}\right)  ,$ and $u_{t}\in L^{2}((s,\tau);W^{-1,2
(\Omega))+L^{1}\left(  Q_{\Omega,s,\tau}\right)  ,$ then any function
$\varphi\in L^{2}((s,\tau);W_{0}^{1,2}(\Omega))\cap L^{\infty}\left(
Q_{\Omega,s,\tau}\right)  $ is admissible in equation (\ref{fil}). In
particular for any $\alpha>0$, we can take $\varphi=M_{\alpha,\delta
}(u)=(u+\delta)^{\alpha}-\delta^{\alpha},$ with $\delta>0.$ Integrating on
$\left[  s,\tau\right]  $ we deduce that
\[
\int_{s}^{\tau}<u_{t},\varphi>+\alph
{\displaystyle\int}
\int_{Q_{\Omega,s,\tau}}(u+\delta)^{\alpha-1}\left\vert \nabla u\right\vert
^{2}dxdt=\int_{s}^{\tau}\int_{\Omega}fM_{\alpha,\delta}(u)dxdt.
\]
Let $k>0$ such that $\sup_{Q_{\Omega,s,\tau}}u\leqq k,$ thus $u=T_{k}(u).$
Moreover the function $\theta\mapsto M(\theta)=(T_{k}(\theta)+\delta)^{\alpha
}-\delta^{\alpha}$ is continuous on $\mathbb{R}^{+}$and piecewise $C^{1}$ such
that $M(0)=0$ and $M^{\prime}$ has a compact support. Denoting $\mathcal{M
_{\alpha,\delta}(r)=(u+\delta)^{\alpha+1}/(\alpha+1)-\delta^{\alpha}u,$ we can
integrate by parts from \cite[Lemma 7.1]{DrPr}, and deduce that
\[
\int_{\Omega}\mathcal{M}_{\alpha,\delta}(u)(.,\tau))dx-\int_{\Omega
}\mathcal{M}_{\alpha,\delta}(u)(.,s))dx+\alph
{\displaystyle\int}
\int_{Q_{\Omega,s,\tau}}(u+\delta)^{\alpha-1}\left\vert \nabla u\right\vert
^{2}dxdt=\int_{s}^{\tau}\int_{\Omega}fM_{\alpha,\delta}(u)dxdt
\]
and then we go to the limit as $\delta\rightarrow0$ from the Fatou Lemma and
then from the Lebesgue theorem. Thus (\ref{hol}) holds for $\alpha>0.$
\end{proof}

\begin{remark}
From \cite{DrPr}, the notion of entropy solution of (\ref{ggs}) is also
equivalent to the notion of renormalized solution, that we develop in Section
\ref{quas}. Lemma \ref{dil} is a special case of a much more general property
of the truncates when $u$ is not necessarily bounded, see Lemma \ref{admi}.
\end{remark}

\subsection{Different notions of solutions of problem $(D_{\Omega,T})$}

\begin{definition}
We say thay $u$ is a \textbf{weak } \textbf{solution} of the problem
$(D_{\Omega,T})$ if $u\in C((0,T);L^{1}\left(  \Omega\right)  )\cap
L_{loc}^{1}((0,T);W_{0}^{1,1}\left(  \Omega\right)  ),$ such that $|\nabla
u|^{q}\in L_{loc}^{1}((0,T);L^{1}\left(  \Omega\right)  )$ and $u$ satisfies
\begin{equation}
u_{t}-\Delta u+|\nabla u|^{q}=0,\quad\text{in}\hspace{0.05in}\mathcal{D
^{\prime}(Q_{\Omega,T}). \label{dis
\end{equation}

\end{definition}

Next we study the Cauchy proble
\begin{equation}
\left\{
\begin{array}
[c]{l
u_{t}-\Delta u+|\nabla u|^{q}=0,\quad\text{in}\hspace{0.05in}Q_{\Omega,T},\\
u=0\quad\text{on}\hspace{0.05in}\partial\Omega\times(0,T),\\
u(x,0)=u_{0}\geqq0
\end{array}
\right.  \label{def
\end{equation}
with $u_{0}\in L^{r}\left(  \Omega\right)  ,$ $r\geqq1,$ or only $u_{0
\in\mathcal{M}_{b}^{+}(\Omega).$ Here in any case $u_{0}\in\mathcal{M}_{b
^{+}(\Omega).$

\begin{definition}
If $u_{0}\in L^{r}(\Omega),r\geqq1,$ we say that $u$ is a \textbf{weak 
$L^{r}$\textbf{ solution } of problem (\ref{def}) if it is a weak solution of
$(D_{\Omega,T})$, such that the extension of $u$ by $u_{0}$ at time $0$
satisfies $u\in C\left(  \left[  0,T\right)  ;L^{r}(\Omega\right)  ).$
\end{definition}

\begin{definition}
For any $u_{0}\in\mathcal{M}_{b}^{+}(\Omega),$ we say that $u$ is a weak
$\mathcal{M}$ solution of problem (\ref{def}) if it is a weak solution of
$(D_{\Omega,T})$, such that
\begin{equation}
\lim_{t\rightarrow0}\int_{\Omega}u(.,t)\psi dx=\int_{\Omega}\psi du_{0
,\qquad\forall\psi\in C_{b}(\Omega). \label{jkl
\end{equation}

\end{definition}

Some semi-group notions of solutions have been introduced in \cite{BeDa}, for
any nonnegative $u_{0}\in\mathcal{M}_{b}^{+}(\Omega).$ Here $e^{t\Delta
u_{0}=\int_{\Omega}g_{\Omega}(.,y,t)du_{0}(y),$ where $g_{\Omega}$ is the heat
kernel with Dirichlet conditions on $\partial\Omega.$

\begin{definition}
For any $u_{0}\in\mathcal{M}_{b}^{+}(\Omega),$ a function $u$ is a
\textbf{mild } \textbf{solution }of problem (\ref{def}) if $u\in
C((0,T);L^{1}\left(  \Omega\right)  )$, and $|\nabla u|^{q}\in L_{loc
^{1}(\left[  0,T\right)  ;L^{1}\left(  \Omega\right)  )$ and
\begin{equation}
u(.,t)=e^{t\Delta}u_{0}(.)-\int_{0}^{t}e^{(t-s)\Delta}|\nabla u(.,s)|^{q
ds\qquad\text{in }L^{1}\left(  \Omega\right)  , \label{abc
\end{equation}

\end{definition}

\begin{remark}
As it was shown in \cite[p.1420]{BeDa}, from Lemma \ref{pare}, u is a
mild\textbf{ } solution if and only if $u$ is a weak $\mathcal{M}$ solution
such that $|\nabla u|^{q}\in L_{loc}^{1}(\left[  0,T\right)  ;L^{1}\left(
\Omega\right)  );$ and then $u\in L_{loc}^{1}(\left[  0,T\right)  ;W_{0
^{1,1}\left(  \Omega\right)  ).$
\end{remark}

\begin{remark}
\label{few}As in Remark \ref{all}, the definition of mild solution requires an
integrability property of the gradient up to time $0$, namely $|\nabla
u|^{q}\in L_{loc}^{1}(\left[  0,T\right)  ;L^{1}\left(  \Omega\right)  ).$ The
definition of weak solution only assumes that $|\nabla u|^{q}\in L_{loc
^{1}((0,T);L^{1}\left(  \Omega\right)  ).$
\end{remark}

\subsection{Decay and regularizing effect}

Here $\Omega$ is bounded, then the situation is simpler than in $\mathbb{R
^{N}$, because we take benefit of the regularizing effect of the semi-group
$e^{t\Delta}$ associated with the first eigenvalue $\lambda_{1}$ of the
Laplacian, and also since $L^{r}(\Omega)\subset$ $L^{1}(\Omega).$

\begin{lemma}
\label{sim}Let $q>1,$ and $u_{0}\in L^{r}(\Omega),$ $r\geqq1.$ 1) Let $u$ be
any non-negative weak $L^{r}$\textbf{ }-solution of problem (\ref{def
).$\medskip$

(i) Then $u(.,t)\in L^{\infty}(\Omega)$ for any $t>0,$ and
\begin{equation}
\Vert u(.,t)\Vert_{L^{r}(\Omega)}\leqq Ce^{-\lambda_{1}t}\Vert u_{0
\Vert_{L^{r}(\Omega)},\qquad\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq
Ct^{-\frac{N}{2r}}e^{-\lambda_{1}t}\Vert u_{0}\Vert_{L^{r}(\Omega)}.
\label{clo
\end{equation}
(ii) Moreover $|\nabla u|^{q}\in L_{loc}^{1}(\left[  0,T\right)  ;L^{1}\left(
\Omega\right)  ),$ and
\begin{equation}
\int_{\Omega}u(.,t)dx+\int_{0}^{t}\int_{\Omega}|\nabla u|^{q}dxdt\leqq
\int_{\Omega}u_{0}dx. \label{ple
\end{equation}
If $r>1,$ then \textbf{ }$u^{r-1}|\nabla u|^{q}\in L_{loc}^{1}(\left[
0,T\right)  ;L^{1}\left(  \Omega\right)  );$ we have $u^{r-2}|\nabla u|^{2}\in
L_{loc}^{1}(\left[  0,T\right)  ;L^{1}\left(  \Omega\right)  )$ and
\begin{equation}
\frac{1}{r}\int_{\Omega}u^{r}(.,t)dx+\int_{0}^{t}\int_{\Omega}u^{r-1}|\nabla
u|^{q}dxdt+(r-1)\int_{0}^{t}\int_{\Omega}u^{r-2}|\nabla u|^{2}dxdt=\frac{1
{r}\int_{\Omega}u_{0}^{r}dx, \label{plo
\end{equation}
As a consequence, $u^{q-1+r}\in L_{loc}^{1}((\left[  0,T\right)  ;W_{0
^{1,1}\left(  \Omega\right)  ).\medskip$

\noindent2) Let $u_{0}\in\mathcal{M}_{b}^{+}(\Omega)$ and $u$ be any
non-negative weak $\mathcal{M}$ solution of problem (\ref{def}). Then
(\ref{clo}) and (\ref{ple}) still hold as in case $u_{0}\in L^{1}(\Omega),$
where the norm $\Vert u_{0}\Vert_{L^{1}(\Omega)}$ is replaced by 
{\displaystyle\int_{\Omega}}
du_{0}.$ In particular $u$ is a mild solution.
\end{lemma}

\begin{proof}
1) (i) Let $0<\epsilon<\tau<T.$ Since $u$ is a weak solution of $(D_{\Omega
,T})$, we can apply Lemma \ref{pare} with $f=-|\nabla u|^{q}$ in
$Q_{\Omega,\epsilon,\tau}.$ Thus $u$ is a mild solution of the problem in
$Q_{\Omega,\epsilon,\tau}:$ for any $t\in\left[  \epsilon,\tau\right]  ,$
\[
u(.,t)=e^{(t-\epsilon)\Delta}u(.,\epsilon)-\int_{\epsilon}^{t}e^{(t-\sigma
)\Delta}|\nabla u|^{q}d\sigma\qquad\text{in }L^{1}\left(  \Omega\right)  .
\]
thus $u(.,t)\leqq e^{(t-\epsilon)\Delta}u(.,\epsilon).$ From our assumptions
$u\in C\left(  \left[  0,T\right)  ;L^{r}(\Omega\right)  ),$ thus we deduce
$u(.,t)\leqq e^{t\Delta}u_{0}$ as $\epsilon\rightarrow0.$ Then (\ref{clo})
follows.$\medskip$

(ii) The function $u$ is bounded in $Q_{\Omega,s,\tau},$ thus from Lemma
\ref{dil}, for any $\rho>1,$
\begin{equation}
\frac{1}{\rho}\int_{\Omega}u^{\rho}(.,t)dx+\int_{\epsilon}^{t}\int_{\Omega
}u^{\rho-1}|\nabla u|^{q}dxdt+(\rho-1)\int_{\epsilon}^{t}\int_{\Omega
u^{\rho-2}|\nabla u|^{2}dxdt=\frac{1}{\rho}\int_{\Omega}u^{\rho
(.,\epsilon)dx. \label{plm
\end{equation}
and we make $\rho\rightarrow1.$ From Fatou Lemma we deduce that $|\nabla
u|^{q}\in L^{1}\left(  Q_{\Omega,\epsilon,\tau}\right)  $ and
\[
\int_{\Omega}u(.,t)dx+\int_{\epsilon}^{t}\int_{\Omega}|\nabla u|^{q
dxdt\leqq\int_{\Omega}u(.,\epsilon)dx.
\]
As $\epsilon\rightarrow0$ we deduce that $|\nabla u|^{q}\in L^{1}\left(
Q_{\Omega,\tau}\right)  $ and (\ref{ple}) holds. If $r>1,$ we can take
$\rho=r$ in (\ref{plm}) and obtain (\ref{plo}) as $\epsilon\rightarrow0$. Then
$u^{q-1+r}\in L_{loc}^{1}((\left[  0,T\right)  ;W_{0}^{1,1}\left(
\Omega\right)  )$ as in the case of $\mathbb{R}^{N}.$

\noindent2) The same estimates hold because $\lim_{\epsilon\rightarrow0}\Vert
u(.,\epsilon)\Vert_{L^{1}(\Omega)}
{\displaystyle\int_{\Omega}}
du_{0}.$\medskip
\end{proof}

\begin{theorem}
\label{deco}Let $q>1$ and $u_{0}\in L^{r}(\Omega),$ $r\geqq1.$ 1) Let $u$ be
any non-negative weak $L^{r}$\textbf{ }-solution of problem (\ref{def}). Then
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\sigma_{r,q}}\Vert u_{0
\Vert_{L^{r}(\Omega)}^{\varpi_{r,q}}, \label{yac
\end{equation}
where $\sigma_{r,q},\varpi_{r,q}$ are given at (\ref{sigr}).$\medskip$

\noindent2) Any non-negative weak solution $u$ of $(D_{\Omega,T})$ satisfies
the universal estimate, where $C=C(N,q)>0,
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\frac{1}{q-1}}. \label{wad
\end{equation}

\end{theorem}

\begin{proof}
1) For any $\alpha>0,$ setting $\rho=1+\alpha,$ and $0<\epsilon\leqq s<t<T,$
setting $\beta=1+\alpha/q,$ we have from (\ref{plm}),
\[
\frac{1}{\alpha+1}\int_{\Omega}u^{\alpha+1}(.,t)dx+\int_{s}^{t}\int_{\Omega
}\left\vert \nabla(u^{\beta})\right\vert ^{q}dxdt\leqq\frac{1}{\alpha+1
\int_{\Omega}u^{\alpha+1}(.,s)dx.
\]
And $u^{\beta}(.,t)\in L^{\infty}(Q_{\Omega,s,\tau})$ for a.e. $t>0,$ then
$u^{\beta}(.,t)\in W^{1,q}\left(  \Omega\right)  $; and $u(.,t)\in W_{0
^{1,1}\left(  \Omega\right)  )$ hence $u^{\beta}(.,t)\in W_{0}^{1,q}\left(
\Omega\right)  ,$ then from the Sobolev injection of $W_{0}^{1,q}\left(
\Omega\right)  $ into $L^{q^{\ast}}\left(  \Omega\right)  $, for any $s<t,$
\[
\frac{1}{\alpha+1}\int_{\Omega}u^{\alpha+1}(.,t)dx+\frac{C_{N,q}}{\beta^{q
}\int_{s}^{t}(\int_{\Omega}u^{\beta q^{\ast}}(.,\sigma)dx)^{\frac{q}{q^{\ast
}}dt\leqq\frac{1}{\alpha+1}\int_{\Omega}u^{\alpha+1}(.,s)dx.
\]
Then we can apply Lemma \ref{prod} on $\left[  \epsilon,T\right)  $, and
deduce estimates for $\epsilon<t<T,$
\[
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq C(t-\epsilon)^{-\sigma_{r,q}}\Vert
u(.,\epsilon)\Vert_{L^{r}(\Omega)}^{\varpi_{r,q}},
\
\[
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq C(t-\epsilon)^{-\frac{1}{q-1}}.
\]
and we deduce (\ref{yac}) and (\ref{wad}) as $\epsilon\rightarrow0$.\medskip

\noindent2) Let $u$ be any weak solution of $(D_{\Omega,T}).$ Let
$\epsilon>0.$ Since $u\in C(\left[  \epsilon,T\right)  ;L^{1}(\Omega))$ we
find, for any $t\in\left[  \epsilon,T\right)  ,$
\[
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq C(t-\epsilon)^{-\frac{1}{q-1}
\]
with $C=C(N,q),$ and deduce (\ref{wad}) for any $t\in\left(  0,T\right)  $ by
letting $\epsilon$ tend to $0.$
\end{proof}

\begin{remark}
The same decay estimates where shown in \cite{Por} in case $q<2$, for any weak
$L^{r}$ solution $u$ such that $u\in C((0,T);L^{2}\left(  \Omega\right)  )\cap
L^{2}((0,T);W_{0}^{1,2}\left(  \Omega\right)  ),$ and $(u-k)^{+}$ is
admissible as a test function in the equation; this implies integrability
properties of $u|\nabla u|^{q}.$ Our result is valid without any of these conditions.
\end{remark}

\subsection{Existence and uniqueness results for $q\leqq2$}

Here we consider the case $1<q\leqq2.$ From the universal a priori estimate
(\ref{wad}), we deduce new convergence results:

\begin{corollary}
\label{patt}Assume $1<q\leqq2.$ Then \medskip

(i) any weak solution $u$ of problem $(D_{\Omega,T})$ satisfies $u\in
C^{2,1}\left(  Q_{\mathbb{R}^{N},T}\right)  \cap C^{1}\left(  \overline
{\Omega}\times\left(  0,T\right)  \right)  ;$\medskip

(ii) for any sequence of weak solutions $\left(  u_{n}\right)  $ of
$(D_{\Omega,T}),$ one can extract a subsequence converging in $C_{loc
^{2,1}(Q_{\mathbb{R}^{N},T})\cap C^{1}\left(  \overline{\Omega}\times\left(
0,T\right)  \right)  $ to a weak solution $u$ of $(D_{\Omega,T})$.
\end{corollary}

\begin{proof}
(i) From \cite[Theorem 2.17]{BiDao1}, any weak solution $u$ of $(D_{\Omega
,T})$ such that $u\in L_{loc}^{\infty}(\left(  0,T\right)  ;L^{\infty
(\Omega))$ satisfies $u\in C^{2,1}\left(  Q_{\mathbb{R}^{N},T}\right)  \cap
C^{1}\left(  \overline{\Omega}\times\left(  0,T\right)  \right)  .$ And any
weak solution $u\in L_{loc}^{\infty}(\left(  0,T\right)  ;L^{\infty
(\Omega)),$ from Theorem \ref{deco},3.\medskip

(ii) Moreover $(u_{n})$ is uniformly bounded in $L_{loc}^{\infty}\left(
0,T);L^{\infty}\left(  \Omega\right)  \right)  $. From \cite[Theorem
2.13]{BiDao1}, there exists $\upsilon\in\left(  0,1\right)  $ such that, for
any $0<s<\tau<T,$
\begin{equation}
\left\Vert u_{n}\right\Vert _{C(\overline{\Omega}\times\left[  s,\tau\right]
)}+\left\Vert \nabla u_{n}\right\Vert _{C^{\upsilon,\upsilon/2}(\overline
{\Omega}\times\left[  s,\tau\right]  )}\leqq C\Phi(\left\Vert u_{n}\right\Vert
_{L^{\infty}(Q_{\Omega,s/2,\tau})}) \label{gou
\end{equation}
where $C=C((N,q,\Omega,s,\tau,\upsilon),$ and $\Phi$ is an increasing
function. The conclusion follows.
\end{proof}

\begin{theorem}
\label{unic} Suppose $1<q<(N+2)/(N+1).$ For any $u_{0}\in\mathcal{M}_{b
^{+}(\Omega),$ problem (\ref{def}) admits a unique weak $\mathcal{M}$ solution.
\end{theorem}

\begin{proof}
From Theorem \ref{deco}, $u$ is a mild $\mathcal{M}$ solution, and then it is
the unique mild $\mathcal{M}$ solution, from \cite[Theorem 3.2]{BeDa
.$\medskip$
\end{proof}

Next assume that $u_{0}\in L^{r}(\Omega)$ and $q<(N+2r)/(N+r).$ In
\cite[Theorem 3.3]{BeDa}, it is proved that there exists a weak $L^{r}$
solution such that $u\in L_{loc}^{q}(\left[  0,T\right)  ;W_{0}^{1,qr}\left(
\Omega\right)  ),$ and \textit{it is unique in this space.} The local
existence in an interval $\left(  0,T_{1}\right)  $ is obtained by the Banach
fixed point theorem in a ball of radius $K_{1}$ of the space
\[
X_{K_{1}}(T_{1})=\left\{  u\in C(\left(  0,T_{1}\right]  ,W_{0}^{1,qr}\left(
\Omega\right)  :\sup_{\left(  0,t_{1}\right]  }t^{\theta}(\left\Vert
u(.,t)\right\Vert _{L^{qr}\left(  \Omega\right)  }+t^{\frac{1}{2}}\left\Vert
\nabla u(.,t)\right\Vert _{L^{qr}\left(  \Omega\right)  }<\infty\right\}
\]
where $\theta=r/2Nq^{\prime},$ under the condition
\begin{equation}
\left\Vert u_{0}\right\Vert _{L^{r}\left(  \Omega\right)  }+K_{1}^{q
T_{1}^{\gamma}\leqq CK_{1},\qquad\text{where }\gamma=1-q(\theta+1/2)\quad
\text{and }C=C(N,q,r,\Omega). \label{fac
\end{equation}
We prove the uniqueness \textit{with no condition of integrability}:

\begin{theorem}
\label{nouveau}Assume that $u_{0}\in L^{r}(\Omega)$ and $1<q<(N+2r)/(N+r).$
Then problem (\ref{def}) admits a unique weak $L^{r}$ solution.
\end{theorem}

\begin{proof}
Let $\epsilon>0.$ From Theorem \ref{deco}, $u$ is bounded on $(\epsilon,T)$
for any $\epsilon\in\left(  0,T\right)  $. Then $u\in C^{2,1}(Q_{\Omega
,T})\cap C^{1,0}(\overline{\Omega}\times\left(  0,T\right)  )$ because $q<2,$
from \cite[Theorem 2.16]{BiDao1}. From (\ref{gd}), there exists a function
$D\in C((0,\infty)$ such that for any $\epsilon>0$ and for $t\geqq\epsilon$
\[
\Vert\nabla u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq D(t-\epsilon).
\]
Then $\left\vert \nabla u\right\vert $ is bounded in $Q_{\epsilon,T,\Omega}$
for any $\epsilon>0.$ Thus $u\in C((0,T),W_{0}^{1,qr}\left(  \Omega\right)
).$ The problem with initial data $u(.,\epsilon)$ at time $0$ has a unique
solution $v_{\epsilon}$ such that $v_{\epsilon}\in C((0,T-\epsilon
),W_{0}^{1,qr}\left(  \Omega\right)  ),$ then $v_{\epsilon
(.,t)=u(.,t+\epsilon).$ Let $K_{1}$ and $T_{1}$ such that (\ref{fac}) holds.
Since $\left\Vert u(.,\epsilon)\right\Vert _{L^{r}\left(  \Omega\right)
}\leqq$ $\left\Vert u_{0}\right\Vert _{L^{r}\left(  \Omega\right)  }$, we also
have $\left\Vert v_{\epsilon}(0)\right\Vert _{L^{r}\left(  \Omega\right)
}+K_{1}^{q}T_{1}^{\gamma}\leqq CK_{1},$ thus for any $t\in(0,T_{1})
\[
t^{\theta}(\left\Vert v_{\epsilon}(.,t)\right\Vert _{L^{qr}\left(
\Omega\right)  }+t^{\frac{1}{2}}\left\Vert \nabla v_{\epsilon}(.,t)\right\Vert
_{L^{qr}\left(  \Omega\right)  }\leqq K_{1
\]
Going to the limit as $\epsilon\rightarrow0$ from the Fatou Lemma, we obtain
\[
t^{\theta}(\left\Vert u(.,t)\right\Vert _{L^{qr}\left(  \Omega\right)
}+t^{\frac{1}{2}}\left\Vert \nabla u(.,t)\right\Vert _{L^{qr}\left(
\Omega\right)  }\leqq K_{1
\]
Hence we enter the class of uniqueness. Then $u$ is the unique solution
constructed in \cite{BeDa}.\medskip
\end{proof}

Finally we give existence results for any $u_{0}\in L^{r}(\Omega),r\geq1$,
extending the results of \cite[Theorem 3.4]{BeDa} for $u_{0}\in L^{1
(\Omega),$ also proved for more general operators in \cite{Po}. We proceed as
in Proposition \ref{pou}.

\begin{proposition}
\label{plu} Let $1<q\leq2$. For any nonnegative $u_{0}\in L^{r}(\Omega
),r\geq1,$ there exists a weak $L^{r}$ solution of problem (\ref{def}). And it
is unique if $q=2.$
\end{proposition}

\begin{proof}
(i) Case $q<2.$ Let $u_{0,n}=\min(u_{0},n).$ Then for $\rho>N(q-1)/(2-q),$
from \cite[Theorem 3.3]{BeDa}, there exists a mild solution $u_{n}$ with
initial data $u_{0,n},$ and $u_{n}\in C(\left[  0,T\right)  ;L^{\rho
(\Omega))\cap L^{q}((0,T);W_{0}^{1,q\rho}(\Omega)\cap C^{2,1}\left(
Q_{\Omega,T}\right)  .$ Then $u_{n}(.,t)\leq e^{t\Delta}u_{0},$ and $\left(
u_{n}\right)  $ is nondecreasing and $\left\vert \nabla u_{n}\right\vert ^{q}$
is bounded in $L_{loc}^{1}\left(  \left[  0,T\right)  ;L^{1}(\Omega)\right)  $
from (\ref{ple}). From Corollary \ref{pass}, $\left(  u_{n}\right)  $
converges in $C_{loc}^{2,1}(Q_{\Omega,T})$ to a weak solution $u$ of
(\ref{un}) in $Q_{\Omega,T}$, and then $u(.,t)\leq e^{t\Delta}u_{0}$ and
$\left\vert \nabla u\right\vert ^{q}\in$ $L_{loc}^{1}\left(  \left[
0,T\right)  ;L^{1}(\Omega)\right)  .$ Thus from \cite[Proposition
2.11]{BiDao1}, $u(.,t)$ converges weakly* to some Radon measure $\mu_{0}$ on
$\Omega$. And $e^{t\Delta}u_{0}$ converges to $u_{0}$ in $L^{r}(\Omega),$ thus
$\mu_{0}\in L_{loc}^{1}(\Omega)$ and $0\leq\mu_{0}\leq u_{0}$. Since
$u_{n}\leq u,$ there holds $u_{0,n}\leq\mu_{0},$ hence $\mu_{0}=u_{0}\in
L^{r}(\Omega).$ Also there exists a function $g\in L^{r}(\Omega)$ such that
$u(.,t)\leqq g$ for small $t.$ Then the nonnegative function $e^{t\Delta
u_{0}-u(.,t)$ converges weakly* to $0,$ and then in $L_{loc}^{1}(\Omega).$
Hence $u(.,t)$ converges to $u_{0}$ in $L_{loc}^{1}(\Omega),$ then in
$L^{r}(\Omega)$ from the Lebesgue theorem. Thus $u\in C(\left[  0,T\right)
;L^{r}\left(  \Omega\right)  ).\medskip$

(ii) Case $q=2.$ As in \cite[Theorem 4.2]{BASoWe}, using the classical
transformation $v=1-e^{-u},$ it can be shown that there exists a unique
solution $u$ such that $u\in C(\left[  0,T\right)  ;L^{r}\left(
\Omega\right)  )\cap C^{2,1}\left(  Q_{\Omega,T}\right)  \cap C^{1}\left(
\overline{\Omega}\times\left(  0,T\right)  \right)  $. Then it is a weak
$L^{r}$ solution. Reciprocally any weak $L^{r}$ solution $u$ satisfies the
conditions above, from Corollary \ref{patt} and \cite[Theorem 2.17]{BiDao1}.
\end{proof}

\section{Regularizing effects for quasilinear Dirichlet problems\label{quas}}

Next we extend some results of section \ref{omega} to a general quasilinear
problem, where $u$ may be a signed solution. In this section, we suppose
$\Omega$ is a smooth bounded domain in $\mathbb{R}^{N}.$ Let $p,q>1.$ Let
$\mathrm{A}$ be a Caratheodory function on $Q_{\Omega,\infty}\times
\mathbb{R}\times\mathbb{R}^{N}$ such that for any $(u,\eta)\in\mathbb{R
\times\mathbb{R}^{N},$ and a.e. $(x,t)\in Q_{\Omega,\infty},$
\begin{equation}
\left\vert \mathrm{A}(x,t,u,\eta)\right\vert \leqq C(\left\vert \eta
\right\vert ^{p-1}+b(x,t)),\qquad C>0,\quad b\in L^{p^{\prime}}(Q_{\Omega
,\infty}), \label{hypa
\end{equation}
and $\mathrm{A}$ is nonnegative operator:
\begin{equation}
\mathrm{A}(x,t,u,\eta).\eta\geqq\nu\left\vert \eta\right\vert ^{p}\qquad
\nu\geqq0, \label{coe
\end{equation}
with no monotonicity assumption.\medskip

Let $g$ be a Caratheodory function on $Q_{\Omega,\infty}\times\mathbb{R
^{+}\times\mathbb{R}^{N}$, such tha
\begin{equation}
g(x,t,u,\eta)u\geqq\gamma\left\vert u\right\vert ^{\lambda+1}\left\vert
\eta\right\vert ^{q},\text{ }\qquad\text{ }\lambda\geqq0,\quad\gamma\geqq0.
\label{mino
\end{equation}


\begin{definition}
We say that $A$ is coercive if (\ref{coe}) holds with $\nu>0,$ and $g$ is
coercive if (\ref{mino}) holds with $\gamma>0.$
\end{definition}

We consider the solutions of the Dirichlet proble
\begin{equation}
(P_{\Omega,T})\left\{
\begin{array}
[c]{l
u_{t}-\operatorname{div}(\mathrm{A}(x,t,u,\nabla u))+g(x,t,u,\nabla
u)=0,\quad\text{in}\hspace{0.05in}Q_{\Omega,T},\\
u=0,\quad\text{on}\hspace{0.05in}\partial\Omega\times(0,T),\\
u(x,0)=u_{0
\end{array}
\right.  \label{gen
\end{equation}
where $u_{0}\in L^{r}\left(  \Omega\right)  ,$ $r\geqq1$ or only $u_{0
\in\mathcal{M}_{b}(\Omega).$

\subsection{Solutions of quasilinear heat equation with $L^{1}$ data}

Here we consider the problem in $Q_{\Omega,s,\tau}$
\begin{equation}
\left\{
\begin{array}
[c]{l
u_{t}-\operatorname{div}(\mathrm{A}(x,t,u,\nabla u))=f,\quad\text{in
\hspace{0.05in}Q_{\Omega,s,\tau},\\
u=0,\quad\text{on}\hspace{0.05in}\partial\Omega\times(s,\tau),\\
u(x,s)=u_{s
\end{array}
\right.  \label{cru
\end{equation}
First we recall the notion of renormalized solution introduced in \cite{BlMu}
for this problem with $L^{1}$ data:

\begin{definition}
\label{renor}Let $s,\tau\in\mathbb{R}$ with $s<\tau,$ and $f\in L^{1
(Q_{\Omega,s,\tau})$ and $u_{s}\in L^{1}(\Omega).$ A function $u\in L^{\infty
}((s,\tau);L^{1}(\Omega))$ is a renormalized solution in $Q_{\Omega,s,\tau}$
of (\ref{cru}) if $T_{k}(u)\in L^{p}((s,\tau);W_{0}^{1,p}(\Omega))$ for any
$k\geqq0,$ and for any $S\in W^{2,\infty}(\mathbb{R})$ such that $S^{\prime}$
has a compact support,
\begin{equation}
(S(u))_{t}-\operatorname{div}(\mathrm{A}(x,t,u,\nabla u)S^{\prime
}(u))+S^{\prime\prime}(u)(\mathrm{A}(x,t,u,\nabla u).\nabla u-S^{\prime
}(u)f=0\quad\text{in}\hspace{0.05in}\mathcal{D}^{\prime}(Q_{\Omega,s,\tau}),
\label{quo
\end{equation}
and $u(s)=u_{s},$ and
\begin{equation}
\lim_{n\rightarrow\infty
{\displaystyle\int}
\int_{Q_{\Omega,s,\tau}\cap\left\{  n\leqq u\leqq n+1\right\}  }|\nabla
u|^{p}dxdt=0, \label{ruo
\end{equation}

\end{definition}

\begin{remark}
\label{zzz} The initial condition takes sense from \cite{BlMu}, because $S(u)$
lies in the set
\begin{equation}
E=\left\{  \varphi\in L^{p}((0,T);W_{0}^{1,p}(\Omega)):\varphi_{t}\in
L^{p^{\prime}}((0,T);W^{-1,p^{\prime}}(\Omega))+L^{1}\left(  Q_{\Omega
,T}\right)  \right\}  \label{ens
\end{equation}
and $E\subset C(\left[  0,T\right]  ;L^{1}(\Omega));$ and any function
$\varphi\in L^{p}((0,T);W_{0}^{1,p}(\Omega))\cap L^{\infty}\left(
Q_{\Omega,T}\right)  $ is admissible in equation (\ref{quo}). Moreover from
\cite[Lemma 7.1]{DrPr}, $v=S(u)$ satisfies for any $\psi\in C^{\infty}(\left[
s,\tau\right]  \times\bar{\Omega})$ the integration formula
\begin{equation}
\int_{s}^{\tau}<v_{t},M(v)\psi>=\int_{\Omega}\mathcal{M}(v(.,\tau))\psi
(.,\tau)dx-\int_{\Omega}\mathcal{M}(v(.,s))\psi(.,s)dx-\int_{s}^{\tau
\int_{\Omega}\psi_{t}\mathcal{M}(v)dxdt, \label{form
\end{equation}
for any function $M$ continuous and piecewise $C^{1}$ such that $M(0)=0$ and
$M^{\prime}$ has a compact support, where $\mathcal{M}(r)=\int_{0}^{r
M(\theta)d\theta.$
\end{remark}

A main point in the sequel is the choice of test functions: here we
approximate $\left\vert u\right\vert ^{\alpha-1}u$ for $\alpha>0$ by
truncation. In the following lemma, we solve some technical difficulties
arising because the truncates are not smooth enough to apply the integration
formula, and moreover we do not assume $\alpha\geqq1.$

\begin{lemma}
\label{admi}Let $s,\tau\in\mathbb{R}$ with $s<\tau,$ and $f\in L^{1
(Q_{\Omega,s,\tau})$ and $u\in C(\left[  s,\tau\right]  ;L^{1}(\Omega))$ be
any nonnegative renormalized solution in $Q_{\Omega,s,\tau}$ of (\ref{cru}),
with $u_{s}=u(.,s)$. For any $\alpha>0$ and $k>0,$ we set
\[
\mathcal{T}_{k,\alpha}(r)=\int_{0}^{r}\left\vert T_{k}(\theta)\right\vert
^{\alpha-1}T_{k}(\theta)d\theta.
\]
Then $\left\vert T_{k}(u)\right\vert ^{\alpha-1}\mathrm{A}(x,t,u,\nabla
u).\nabla(T_{k}(u))\in L^{1}(Q_{\Omega,s,\tau})$ and
\begin{align}
&  \int_{\Omega}\mathcal{T}_{k,\alpha}(u)(.,\tau))dx+\alph
{\displaystyle\int}
\int_{Q_{\Omega,s,\tau}}\left\vert T_{k}(u)\right\vert ^{\alpha-1
\mathrm{A}(x,t,u,\nabla u).\nabla(T_{k}(u))dxdt\nonumber\\
&  =\int_{\Omega}\mathcal{T}_{k,\alpha}(u)(.,s))dx+\int_{s}^{\tau}\int
_{\Omega}f\left\vert T_{k}(u)\right\vert ^{\alpha-1}T_{k}(u)dxdt. \label{rela
\end{align}

\end{lemma}

\begin{proof}
Let $\alpha>0,k>0$ be fixed, and for any $n\geqq2,$ and $\theta\in\mathbb{R},$
\textbf{ 
\[
S_{n}(\theta)=\int_{0}^{\theta}(1-\left\vert T_{1}(s-T_{n}(s)\right\vert
)ds,\qquad n\geqq2.
\]
This function, introduced in \cite{BlMu}, is a smoothing of the truncate
$T_{n+1},$ such that $0\leqq S_{n}(\theta)\theta\leqq T_{n+1}(\theta)\theta,$
supp $S_{n}^{\prime}\subset\left[  -(n+1),n+1\right]  $, and $S_{n
(\theta)=S_{n}(T_{k}(\theta))$ for any $n>k.$ Let $\delta\in\left(
0,k\right)  ,$ and $n>k.$ Setting
\[
T_{\delta,k,\alpha}(\theta)=((T_{k}(\left\vert \theta\right\vert
)+\delta))^{\alpha}-\delta^{\alpha})\mathrm{sign}\theta,\qquad\mathcal{T
_{\delta,k,\alpha}(r)=\int_{0}^{r}T_{\delta,k,\alpha}(\theta)d\theta
\]
Then we can take in (\ref{quo}) $S=S_{n}$ and $\varphi=T_{\delta,k,\alpha
}(u)=T_{\delta,k,\alpha}(S_{n}(u)).$ We obtai
\begin{align*}
\int_{s}^{t}  &  <(S_{n}(u))_{t},\varphi>+\int_{s}^{t}\int_{\Omega
S_{n}^{\prime}(u)\mathrm{A}(x,t,u,\nabla u).\nabla\varphi dxdt\\
&  =\int_{s}^{t}\int_{\Omega}S_{n}^{\prime}(u)f\varphi dxdt-\int_{s}^{t
\int_{\Omega}S_{n}^{\prime\prime}(u)(\mathrm{A}(x,t,u,\nabla u).\nabla
u)\varphi dxdt.
\end{align*}
then from (\ref{form}), we deduce
\begin{align*}
&  \int_{\Omega}\mathcal{T}_{\delta,k,\alpha}(S_{n}(u)(.,\tau))dx+\alph
{\displaystyle\int}
\int_{Q_{\Omega,s,\tau}}(T_{k}(\left\vert u\right\vert )+\delta)^{\alpha
-1}\mathrm{A}(x,t,u,\nabla u).\nabla(T_{k}(u))dxdt\\
&  =\int_{\Omega}\mathcal{T}_{\delta,k,\alpha}(S_{n}(u)(.,s))dx+\int_{s
^{\tau}\int_{\Omega}S_{n}^{\prime}(u)f\varphi dxdt-\int_{s}^{t}\int_{\Omega
}S_{n}^{\prime\prime}(u)(\mathrm{A}(x,t,u,\nabla u).\nabla u)\varphi dxdt
\end{align*}
First we make $\delta\rightarrow0.$ We have $\left\vert \mathcal{T
_{\delta,k,\alpha}(\theta)\right\vert \leqq k^{\alpha}\left\vert
\theta\right\vert $ for any $\theta\in\mathbb{R},$ and $S_{n}(u)\in C(\left[
0,T\right]  ;L^{1}(\Omega)),$ and $S_{n}^{\prime}$ is bounded, thus we can go
to the limit in the right hand side$.$ In the left hand side, From the
positivity of $A,$ and the Fatou Lemma we deduce that $T_{k}(\left\vert
u\right\vert )^{\alpha-1}\mathrm{A}(x,u,\nabla u).\nabla T_{k}(u)$ $\in
L^{1}(Q_{\Omega,s,\tau}).$ Then we can apply Lebesgue theorem: indeed
$\mathrm{A}(x,u,\nabla u).\nabla T_{k}(u)$ $\in L^{1}(Q_{\Omega,s,\tau})$ from
(\ref{hypa}), since $T_{k}(u)\in L^{p}((s,\tau);W_{0}^{1,p}(\Omega)),$ and
$(T_{k}(\left\vert u\right\vert )+\delta)^{\alpha-1}\leqq$ $\max(T_{k
^{\alpha-1}(\left\vert u\right\vert ),(k+1)^{\alpha-1}).$ Then the same
relation holds with $\delta=0$, with $T_{0,k,\alpha}(r)=T_{k}^{\alpha
-1}(\left\vert u\right\vert )T_{k}(u):$
\begin{align*}
&  \int_{\Omega}\mathcal{T}_{k,\alpha}(S_{n}(u)(.,\tau))dx-\int_{\Omega
}\mathcal{T}_{k,\alpha}(S_{n}(u)(.,s))dx+\alpha\int_{s}^{t}\int_{\Omega
T_{k}^{\alpha-1}(\left\vert u\right\vert )\mathrm{A}(x,u,\nabla u).\nabla
(T_{k}(u))dxdt\\
&  =\int_{s}^{\tau}\int_{\Omega}S_{n}^{\prime}(u)fT_{0,k,,\alpha
(u)dxdt-\int_{s}^{t}\int_{\Omega}S_{n}^{\prime\prime}(u)(\mathrm{A
(x,t,u,\nabla u).\nabla u)T_{0,k,,\alpha}(u)dxdt.
\end{align*}
Then we make $n\rightarrow\infty.$ Since $u\in C(\left[  0,T\right]
;L^{1}(\Omega)),$ for any $t\in\left[  s,\tau\right]  $
\[
\lim_{n\rightarrow\infty}\int_{\Omega}\mathcal{T}_{k,\alpha}(S_{n
(u)(.,t))dx=\int_{\Omega}\mathcal{T}_{k,\alpha}(u(.,t))dx,
\]
moreover
\[
\lim_{n\rightarrow\infty}\int_{s}^{t}\int_{\Omega}S_{n}^{\prime\prime
}(u)(\mathrm{A}(x,t,u,\nabla u).\nabla u)\;T_{0,k,\alpha}(u)dxdt=0
\]
from (\ref{ruo}), (\ref{hypa}), since $S_{n}^{\prime\prime}=-1_{\left[
n,n+1\right]  }+1_{\left[  -n,-n-1\right]  }.$ Moreove
\[
\lim_{n\rightarrow\infty}\int_{s}^{\tau}\int_{\Omega}S_{n}^{\prime
}(u)fT_{0,k,,\alpha}(u)dxdt=\int_{s}^{\tau}\int_{\Omega}fT_{0,k,,\alpha
}(u)dxdt
\]
since $S_{n}^{\prime}(u)$ $\rightarrow1$ a.e. and is uniformly bounded. Then
(\ref{rela}) follows.
\end{proof}

\subsection{Notion of solutions of problem $(P_{\Omega,T})$}

\begin{definition}
We say that $u$ is a \textbf{renormalized solution} of problem $(P_{\Omega
,T})$ if:\medskip

(i) $u\in C((0,T);L^{1}(\Omega)),$ $T_{k}(u)\in L_{loc}^{p}((0,T);W_{0
^{1,p}(\Omega))$ for any $k\geqq0,$ and $g(x,u,\nabla u)\in L_{loc
^{1}((0,T);L^{1}(\Omega)),$\medskip

(i) for any $0<s<\tau<T,$ $u$ is a renormalized solution of proble
\[
\left\{
\begin{array}
[c]{l
u_{t}-\operatorname{div}(\mathrm{A}(x,t,u,\nabla u))+g(x,t,u,\nabla
u)=0,\quad\text{in}\hspace{0.05in}Q_{\Omega,s,\tau},\\
u=0,\quad\text{on}\hspace{0.05in}\partial\Omega\times(0,T),
\end{array}
\right.
\]
with initial data $u(.,s);$\medskip

(ii) for $u_{0}\in L^{r}(\Omega),$ the extension of $u$ by $u_{0}$ at time $0$
belongs to $C(\left[  0,T\right)  ;L^{r}(\Omega));$ for $u_{0}\in
\mathcal{M}_{b}(\Omega),$ there hold
\begin{equation}
\lim_{t\rightarrow0}\int_{\Omega}u(.,t)\psi dx=\int_{\Omega}\psi du_{0
,\qquad\forall\psi\in C_{b}(\Omega). \label{wou
\end{equation}
\medskip
\end{definition}

\begin{remark}
Recall that $\nabla u$ is defined by $\nabla u=\nabla(T_{k}(u))$ on the set
$\left\vert u\right\vert \leq k.$ The assumption on $g$ means that, for any
$0<s<\tau<T,$
\[
\int_{Q_{\Omega,s,\tau}}\left\vert g(.,u,\nabla u)\right\vert dxdt=\sum
_{k=1}^{\infty}\int_{Q_{\Omega,s,\tau}\cap\left\{  k-1\leqq\left\vert
u\right\vert \leqq k\right\}  }\left\vert g(.,u,\nabla(T_{k}(u))\right\vert
dxdt<\infty.
\]

\end{remark}

We first prove decay properties of the solutions.

\begin{theorem}
\label{dec}Let $p,q>1,$ and $\Omega$ be a regular bounded domain of
$\mathbb{R}^{N}.$ Let $A$ and $g$ satisfy (\ref{hypa}) (\ref{coe}) and
(\ref{mino}).\medskip

1) Let $u_{0}\in L^{r}(\Omega),r\geqq1$ and $u$ be any renormalized solution
of $(P_{\Omega,T}).$ Then for any $t\in\left[  0,T\right)  ,$
\begin{equation}
\int_{\Omega}\left\vert u\right\vert ^{r}(.,t)dx\leqq\int_{\Omega}\left\vert
u_{0}\right\vert ^{r}dx. \label{des
\end{equation}
Moreover if $r>1,$ or if $g$ is coercive, then $\gamma\left\vert u\right\vert
^{\lambda+r-1}|\nabla u|^{q}+\nu\left\vert u\right\vert ^{r-2}|\nabla
u|^{p}\in L_{loc}^{1}(\left[  0,T\right)  ;L^{1}\left(  \Omega\right)  ),$
and
\begin{equation}
\int_{\Omega}\left\vert u\right\vert ^{r}(.,t)dx+r\gamma\int_{0}^{t
\int_{\Omega}\left\vert u\right\vert ^{\lambda+r-1}|\nabla u|^{q
dxdt+r(r-1)\nu\int_{0}^{t}\int_{\Omega}\left\vert u\right\vert ^{r-2}|\nabla
u|^{p}dxdt\leqq\int_{\Omega}\left\vert u_{0}\right\vert ^{r}dx. \label{bac
\end{equation}


2) Let $u_{0}\in\mathcal{M}_{b}^{+}(\Omega)$ and $u$ be any nonnegative
renormalized solution of $(P_{\Omega,T})$ of problem (\ref{def}). Then the
same conclusions hold as in case $u_{0}\in L^{1}(\Omega),$ where the norm
$\Vert u_{0}\Vert_{L^{1}(\Omega)}$ is replaced by 
{\displaystyle\int_{\Omega}}
du_{0}.$
\end{theorem}

\begin{proof}
\textbf{ } 1) Let $0<s<t<T,$ then we have for any $\alpha>0,$ any $k>0,$ from
Lemma \ref{admi},
\begin{align*}
&  \text{ }\int_{\Omega}\mathcal{T}_{k,\alpha}(u)(.,\tau))dx+\alpha\int
_{s}^{t}\int_{\Omega}\left\vert T_{k}(u)\right\vert ^{\alpha-1}\mathrm{A
(x,t,u,\nabla u).\nabla(T_{k}(u))dxdt\\
&  =\int_{\Omega}\mathcal{T}_{k,\alpha}(u)(.,s))dx-\int_{s}^{\tau}\int
_{\Omega}\left\vert T_{k}(u)\right\vert ^{\alpha-1}T_{k}(u)g(.,u,\nabla u)dxdt
\end{align*}
And $\left\vert T_{k}(u)\right\vert ^{\alpha-1}T_{k}(u)g(.,u,\nabla
u)\geq\gamma\left\vert T_{k}(u)\right\vert ^{\alpha+\lambda}\left\vert \nabla
T_{k}(u)\right\vert ^{q}$ from (\ref{mino}). Then $\int_{\Omega
\mathcal{T}_{k,\alpha}(u)(.,t))$ is decreasing for any $k,\alpha>0,$ and
\begin{align}
&  \int_{\Omega}\mathcal{T}_{k,\alpha}(u)(.,\tau))dx+\gamma\int_{s}^{t
\int_{\Omega}\left\vert T_{k}(u)\right\vert ^{\alpha+\lambda}\left\vert \nabla
T_{k}(u)\right\vert ^{q}dxdt+\alpha\nu\int_{s}^{t}\int_{\Omega}\left\vert
T_{k}(u)\right\vert ^{\alpha-1}\left\vert \nabla T_{k}(u)\right\vert
^{p}dxdt\nonumber\\
&  \leqq\int_{\Omega}\mathcal{T}_{k,\alpha}(u)(.,s))dx \label{arc
\end{align}
If $r>1,$ we can take $\alpha=r-1>0$ in (\ref{arc}) and get
\begin{align}
&  \int_{\Omega}\mathcal{T}_{k,r-1}(u)(.,t))dx+\gamma\int_{s}^{t}\int_{\Omega
}\left\vert T_{k}(u)\right\vert ^{r-1+\lambda}\left\vert \nabla T_{k
(u)\right\vert ^{q}dxdt+\alpha\nu\int_{s}^{t}\int_{\Omega}\left\vert
T_{k}(u)\right\vert ^{r-2}\left\vert \nabla T_{k}(u)\right\vert ^{p
dxdt\nonumber\\
&  \leqq\int_{\Omega}\mathcal{T}_{k,r-1}(u)(.,s))dx\leqq\frac{1}{r
\int_{\Omega}\left\vert u\right\vert ^{r}(.,s)dx \label{koc
\end{align}
Since $u\in C(\left[  0,T\right)  ;L^{r}(\Omega))$ we can go to the limit as
$k\rightarrow\infty,$ and $s\rightarrow0,$ and deduce that $\gamma\left\vert
u\right\vert ^{r-1+\lambda}\left\vert \nabla u\right\vert ^{q}$ and $\alpha
\nu\left\vert u\right\vert ^{r-2}\left\vert \nabla u\right\vert ^{p}$ belong
to $L_{loc}^{1}\left(  \left[  0,T\right)  ;L^{1}(\Omega)\right)  $ and for
any $t\in\left(  0,T\right)  ,$
\[
\int_{\Omega}\left\vert u\right\vert ^{r}(.,t)dx+r\gamma\int_{0}^{t
\int_{\Omega}\left\vert u\right\vert ^{r-1+\lambda}\left\vert \nabla
u\right\vert ^{q}dxdt+r(r-1)\nu\int_{0}^{t}\int_{\Omega}\left\vert
u\right\vert ^{r-2}\left\vert \nabla u\right\vert ^{p}dxdt\leqq\int_{\Omega
}\left\vert u_{0}\right\vert ^{r}dx.
\]
If $r=1,$ we take any $\alpha>0$ in (\ref{arc}) and observe that for any
$\theta>0,$
\begin{equation}
\frac{\left\vert T_{k}(\theta)\right\vert ^{\alpha+1}}{\alpha+1
\leqq\mathcal{T}_{k,\alpha}(\theta)\leqq k^{\alpha}\left\vert \theta
\right\vert \label{per
\end{equation}
Then
\[
\int_{\Omega}\left\vert T_{k}(u)\right\vert ^{\alpha+1}(.,t))dx+(\alpha
+1)\gamma\int_{s}^{t}\int_{\Omega}\left\vert T_{k}(u)\right\vert
^{\alpha+\lambda}\left\vert \nabla T_{k}(u)\right\vert ^{q}dxdt\leqq
(\alpha+1)k^{\alpha}\int_{\Omega}\left\vert u\right\vert (.,s)dx
\]
Then we go to the limit as $\alpha\rightarrow0,$ we deduce
\begin{equation}
\int_{\Omega}\left\vert T_{k}(u)\right\vert (.,t))dx+\gamma\int_{s}^{t
\int_{\Omega}\left\vert T_{k}(u)\right\vert ^{\lambda}\left\vert \nabla
T_{k}(u)\right\vert ^{q}dxdt\leqq\int_{\Omega}\left\vert u\right\vert (.,s)dx
\label{rou
\end{equation}
and then as $s\rightarrow0$ we find
\begin{equation}
\int_{\Omega}T_{k}(u)(.,t))dx+\gamma\int_{s}^{t}\int_{\Omega}\left\vert
T_{k}(u)\right\vert ^{\lambda}\left\vert \nabla T_{k}(u)\right\vert
^{q}dxdt\leqq\int_{\Omega}\left\vert u_{0}\right\vert dx \label{loi
\end{equation}
and finally $k\rightarrow\infty,$ and deduce that $\int_{\Omega}\left\vert
u\right\vert (.,t)dx\leqq\int_{\Omega}\left\vert u_{0}\right\vert dx.$
Moreover if $\gamma>0,$ we find
\[
\int_{\Omega}\left\vert u\right\vert (.,t)dx+\gamma\int_{0}^{t}\int_{\Omega
}\left\vert u\right\vert ^{\lambda}|\nabla u|^{q}dxdt\leqq\int_{\Omega
}\left\vert u_{0}\right\vert dx,
\]
thus (\ref{bac}) still holds with $r=1.\medskip$

\noindent2) We still find (\ref{rou}). And $\lim_{s\rightarrow0}\int_{\Omega
}u(.,s)dx
{\displaystyle\int_{\Omega}}
du_{0}$ from (\ref{wou}), hence the conclusion.\medskip
\end{proof}

Next we deduce $L^{\infty}$ estimates, in particular a universal one.

\begin{theorem}
\label{Bound2}Let $p>1,$ $1<q<N,$ and $\Omega$ be a regular bounded domain of
$\mathbb{R}^{N}.$ Assume (\ref{hypa}) (\ref{coe}) and (\ref{mino}). Let
$u_{0}\in L^{r}(\Omega),r\geqq1,$ and $u$ be any renormalized solution of
$(P_{\Omega,T}).$\medskip

(i) If $g$ is coercive, there exists $C=C(N,q,\lambda,\gamma,\Omega)$
independent of $\mathrm{A,}$ such that
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\sigma_{r,q,\lambda}}\Vert
u_{0}\Vert_{L^{r}(\Omega)}^{\varpi_{r,q,\lambda}}, \label{xac
\end{equation}
where
\[
\sigma_{r,q,\lambda}=\frac{1}{\frac{rq}{N}+\lambda+q-1},\qquad\varpi
_{r,q,\lambda}=\frac{rq}{N}\sigma_{r,q,\lambda
\]
and there exists $C=C(N,q,\lambda,\left\vert \Omega\right\vert )$ such that
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\frac{1}{q-1+\lambda}}.
\label{niv
\end{equation}


(ii) If $\mathrm{A}$ is coercive and $r>(2-p)N/p,$ (in particular if
$p>2N/(N+1)$), and $p<N,$ then
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\sigma_{r,p,-1}}\Vert
u_{0}\Vert_{L^{r}(\Omega)}^{\varpi_{r,p,-1}}, \label{yoc
\end{equation}
where
\[
\sigma_{r,p,-1}=\frac{1}{\frac{rp}{N}+p-2},\qquad\varpi_{r,p,-1}=\frac
{1}{1+\frac{N}{rp}(p-2)
\]
and if $p>2,$ there exists $C=C(N,p,\left\vert \Omega\right\vert )$ such that
\begin{equation}
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq Ct^{-\frac{1}{p-2}}. \label{pni
\end{equation}
(iii) The same conclusions hold if $u$ is nonnegative and $u_{0}\in
\mathcal{M}_{b}^{+}(\Omega),$ as in case $u_{0}\in L^{1}(\Omega),$ where the
norm $\Vert u_{0}\Vert_{L^{1}(\Omega)}$ is replaced by 
{\displaystyle\int_{\Omega}}
du_{0}.$
\end{theorem}

\begin{proof}
(i) Let $0<s<t<T$. Since $g$ is coercive, from Theorem \ref{dec}, for any
$\alpha\geqq0$ such that\textbf{ }$\left\vert u\right\vert ^{\alpha+1}(.,s)\in
L^{1}(\Omega),$ we get from (\ref{bac}
\[
\int_{\Omega}\left\vert u\right\vert ^{\alpha+1}(.,t)dx+(\alpha+1)\gamma
\int_{s}^{t}\int_{\Omega}\left\vert u\right\vert ^{\lambda+\alpha}|\nabla
u|^{q}dxdt\leqq\int_{\Omega}\left\vert u\right\vert ^{\alpha+1}(.,s)dx,
\]
and in particular
\[
\int_{\Omega}\left\vert T_{k}(u)\right\vert ^{\alpha+1}(.,t)dx+(\alpha
+1)\gamma\int_{s}^{t}\int_{\Omega}(T_{k}(u))^{\lambda+\alpha}\left\vert \nabla
T_{k}(u)\right\vert ^{q}dxdt\leqq\int_{\Omega}\left\vert u\right\vert
^{\alpha+1}(.,s)dx
\]
And $\left\vert u\right\vert ^{\lambda+\alpha}|\nabla u|^{q}=|\nabla
(\left\vert u\right\vert ^{\beta-1}u)|^{q}$ with $\beta=1+(\alpha
+\lambda)/q\geqq1.$ Then $|\nabla((\left\vert u\right\vert ^{\beta
-1}u)(.,t))|$, and also $|\nabla((\left\vert T_{k}(u)\right\vert ^{\beta
-1}T_{k}(u))(.,t))|$ belong to $L^{q}(\Omega)$ for a.e. $t>0.$ Since
$\left\vert T_{k}(u)\right\vert ^{\beta-1}T_{k}(u)(.,t)\in L^{\infty}(\Omega)$
it follows that $\left\vert T_{k}(u)\right\vert ^{\beta-1}T_{k}(u)(.,t)\in
W^{1,q}\left(  \Omega\right)  $. Moreover $T_{k}(u)(.,t)\in W_{0}^{1,p}\left(
\Omega\right)  ),$ hence $\left\vert T_{k}(u)\right\vert ^{\beta-1
T_{k}(u)(.,t)\in W_{0}^{1,q}\left(  \Omega\right)  .$ Then from the Sobolev
injection of $W_{0}^{1,q}\left(  \Omega\right)  $ into $L^{q^{\ast}}\left(
\Omega\right)  $,
\[
\int_{\Omega}\left\vert T_{k}(u)\right\vert ^{\alpha+1}(.,t)dx+\gamma
\frac{C_{0}(\alpha+1)}{\beta^{q}}\int_{s}^{t}(\int_{\Omega}\left\vert
T_{k}(u)\right\vert ^{\beta q^{\ast}}(.,\sigma)dx)^{\frac{q}{q^{\ast}}
d\sigma\leqq\int_{\Omega}\left\vert u\right\vert ^{\alpha+1}(.,s)dx.
\]
Going to the limit as $k\rightarrow\infty,$ we find
\[
\int_{\Omega}\left\vert u\right\vert ^{\alpha+1}(.,t)dx+\gamma\frac
{C_{0}(\alpha+1)}{\beta^{q}}\int_{s}^{t}(\int_{\Omega}\left\vert u\right\vert
^{\beta q^{\ast}}(.,\sigma)dx)^{\frac{q}{q^{\ast}}}d\sigma\leqq\frac{1
{\alpha+1}\int_{\Omega}\left\vert u\right\vert ^{\alpha+1}(.,s)dx.
\]
Then we can apply Lemma \ref{prod} on $\left[  \epsilon,T\right)  $, with
$m=q$ and $\theta=N/(N-q)$; we deduce the estimate for $\left[  \epsilon
,T\right)  ,$
\[
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq C(t-\epsilon)^{-\sigma
_{r,q,\lambda}}\Vert u(.,\epsilon)\Vert_{L^{r}(\Omega)}^{\varpi_{r,q,\lambda
},
\]
with $C=C(N,q,\lambda,\gamma,\Omega).$ Finally we go to the limit as
$\epsilon\rightarrow0,$ and get (\ref{xac}) for $u_{0}\in L^{r}(\Omega),$ and
the analogous when $u$ is nonnegative and $u_{0}\in\mathcal{M}_{b}^{+
(\Omega),$ and also (\ref{niv}).\medskip

(ii) Assume that $\mathrm{A}$ is coercive. Then for any $\alpha>0$,
\begin{equation}
\int_{\Omega}\mathcal{T}_{k,\alpha}(u)(.,t))dx+\alpha\nu\int_{s}^{t
\int_{\Omega}\left\vert T_{k}(u)\right\vert ^{\alpha-1}\left\vert \nabla
T_{k}(u)\right\vert ^{p}dxdt\leqq\int_{\Omega}\mathcal{T}_{k,\alpha
}(u)(.,s))dx\nonumber
\end{equation}
from (\ref{arc}). From the Sobolev injection of $W_{0}^{1,p}\left(
\Omega\right)  $ into $L^{p^{\ast}}\left(  \Omega\right)  $, since $p<N,$ we
deduce
\[
\frac{1}{\alpha+1}\int_{\Omega}u^{\alpha+1}(.,t)dx+\alpha\frac{C_{0}}{k^{p
}\int_{s}^{t}(\int_{\Omega}u^{kp^{\ast}}(.,\sigma)dx)^{\frac{q}{q^{\ast}
}dt\leqq\frac{1}{\alpha+1}\int_{\Omega}u^{\alpha+1}(.,s)dx.
\]
with $k=1+(\alpha-1)/p.$\medskip\ 

$\bullet$ First suppose $r>1;$ then we start from $\alpha_{0}=r-1>0$, and we
can apply Lemma \ref{prod} with $m=p,$ $\theta=N/(N-q)$ and $\lambda=-1.$ The
condition (\ref{ele}) is satisfied, since $r>N(2-p)/p$.\medskip

$\bullet$ Next suppose $r=1.$ Then $1>(2-p)N/p,$ thus $p-1+p/N>1.$ For any
$\alpha>0,$
\[
\int_{\Omega}\left\vert T_{k}(u)\right\vert ^{\alpha+1}(.,t))dx+\alpha
(\alpha+1)\nu\int_{s}^{t}\int_{\Omega}\left\vert T_{k}(u)\right\vert
^{\alpha-1}\left\vert \nabla T_{k}(u)\right\vert ^{p}dxdt\leqq(\alpha
+1)k^{\alpha}\int_{\Omega}\left\vert u\right\vert (.,s)dx
\]
Taking $\alpha=1,$ we get from (\ref{des}),
\[
\nu\int_{s}^{t}\int_{\Omega}\left\vert \nabla T_{k}(u)\right\vert
^{p}dxdt\leqq k\int_{\Omega}\left\vert u\right\vert (.,s)dx\leqq k\int
_{\Omega}\left\vert u_{0}\right\vert dx.
\]
And from (\ref{des}), $u\in L^{\infty}\left(  (s,T);L^{1}\left(
\Omega\right)  \right)  ,$ then from standard estimates, there holds $u\in
L^{\rho}(Q_{\Omega,s,t})$ for any $\rho\in\left(  1,p-1+p/N\right)  ,$ see
\cite{BoGa}. Then $\left\vert u\right\vert ^{\rho}(.,t)$ $\in L^{1}\left(
\Omega\right)  $ for almost any $t\in\left(  0,T\right)  ,$ hence we can apply
Lemma \ref{prod} on $\left[  \epsilon,T\right)  $ for $\epsilon>0,$ with the
same parameters, after fixing such a $\rho=\rho_{p,N}.$ We obtain that
\[
\Vert u(.,t)\Vert_{L^{\infty}(\Omega)}\leqq C(t-\epsilon)^{-\sigma_{1,p,-1
}\Vert u(.,\epsilon)\Vert_{L^{1}(\Omega)}^{\varpi_{1,p,-1}},
\]
where $C=C(N,p,\lambda,\rho_{p,N})=C(N,p,\lambda);$ finally we go to the limit
as $\epsilon\rightarrow0$ because $u\in C(\left[  0,T\right]  ;L^{1
(\Omega)).$ Estimate (\ref{pni}) follows, since $-1+p-1>0.\medskip$

(iii) Similarly, if $u_{0}\in\mathcal{M}_{b}^{+}(\Omega)$ and $u$ is
nonnegative, we are lead to the same conclusions, where $\left\Vert
u_{0}\right\Vert _{L^{1}(\Omega)}$ is replaced by $\int_{\Omega}du_{0}.$ In
particular (\ref{yoc}) holds for $p>2N/(N+1),$ and (\ref{pni}) for $p>2.$
\end{proof}

\begin{remark}
As at Remark \ref{parti}, we still obtain $L^{\infty}$ estimates for $q\geq N$
or $p\geq N.$ If $g$ is coercive, we get (\ref{xac}) with $\sigma
_{r,q,N}=1/(q+r-1+\lambda)=\varpi_{r,q,N}/r$ if $q>N,$ and $\sigma
_{r,N,N}=1/(N(1-\delta)+r-1+\lambda)=\varpi_{r,N,N}/r(1-\delta)$ where
$\delta\in\left(  0,1\right)  $ is arbitrary$.$ If $\mathrm{A}$ is coercive we
get (\ref{yoc}) with $\sigma_{r,p,-1}=1/(r+p-2)=\varpi_{r,p,-1}/r$ if $p>N,$
and $\sigma_{r,N,-1}=1/(N(1-\delta)+p-2)=\varpi_{r,N,-1}/r(1-\delta).$
\end{remark}

\begin{remark}
Our results apply in particular to the problem
\[
\left\{
\begin{array}
[c]{l
u_{t}-\operatorname{div}(\mathrm{A}(x,t,u,\nabla u))=0,\quad\text{in
\hspace{0.05in}Q_{\Omega,T},\\
u=0,\quad\text{on}\hspace{0.05in}\partial\Omega\times(0,T),\\
u(x,0)=u_{0
\end{array}
\right.
\]
Thus we find again the estimates of \cite[Theorem 5.3]{Por}, with less
regularity on the solutions: those estimates were proved for solutions $u\in
C(\left[  0,T\right)  ;L^{r}(\Omega))$ such that $u\in L^{p}(\left(
0,T\right)  ;W_{0}^{1,p}(\Omega))\cap C(\left[  0,T\right)  ;L^{2}(\Omega))$.
The notion of renormalized solutions, equivalent to the notion of entropy
solutions of \cite{Pr} (see \cite{DrPr}), is weaker.\bigskip
\end{remark}

\begin{remark}
The extension of results of section \ref{Rn} to the case of equation of type
(\ref{fru}) in the case $\Omega=\mathbb{R}^{N}$ will be treated a further article.
\end{remark}

\begin{acknowledgement}
We thank Professor F. Weissler for helpfull discussions during the preparation
of this article.
\end{acknowledgement}

