\documentclass[twoside]{article} \usepackage{amssymb, amsmath} % used for R in Real numbers \pagestyle{myheadings} \markboth{Sets of admissible initial data } { Emmanuel Chasseigne \& Juan Luis Vazquez } \begin{document} \setcounter{page}{53} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent 2001-Luminy conference on Quasilinear Elliptic and Parabolic Equations and Systems,\newline Electronic Journal of Differential Equations, Conference 08, 2002, pp 53--83. \newline http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Sets of admissible initial data for porous-medium equations with absorption % \thanks{ {\em Mathematics Subject Classifications:} 35K55, 35K65. \hfil\break\indent {\em Key words:} Initial trace, Borel measures, initial projection, singular solution. \hfil\break\indent \copyright 2002 Southwest Texas State University. \hfil\break\indent Published October 21, 2002. } } \date{} \author{Emmanuel Chasseigne \& Juan Luis Vazquez} \maketitle \begin{abstract} In this article, we study a porous-medium equation with absorption in $\mathbb{R}^{N}\times (0,T)$ or in $\Omega \times (0,T)$: $$ u_{t}-\Delta u^{m}+u^{p}=0\,. $$ We give a rather complete qualitative picture of the initial trace problem in all the range $m>1$, $p\geqslant 0$. We consider nonnegative Borel measures as initial data (not necessarily locally bounded) and discuss whether or not the Cauchy problem admits a solution. In the case of non-admissible data we prove the existence of some projection operators which map any Borel measure to an admissible measure for this equation. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \numberwithin{equation}{section} \section{Introduction} \label{sec.intro} In this paper, we consider the equation \begin{equation} u_{t}-\Delta u^{m}+u^{p}=0, \label{eq1} \end{equation} where $m>1$ and $p>0$. In particular, we look for nonnegative weak solutions $u=u(x,t)$ defined in $Q_{T}=\mathbb{R}^{N}\times (0,T)$ for some $T\in (0,\infty ]$. We aim at describing the sets of nonnegative initial data for which there exists a solution of the Cauchy problem. We call these sets {\sl admissible sets}; they can be quite different depending on the value of the exponents. For convenience we think of $m$ as fixed and $p$ as a variable parameter, hence we denote the admissible set by $\mathcal{A}^{+}(p)$. \subsection{Measures in pure diffusion} Our equation can be seen as a perturbation of the heat equation $u_{t}=\Delta u$, and it will be convenient to review the situation for this equation in order to present the main ideas. It is well-known \cite{A} that any nonnegative distributional solution $u$ in $\mathbb{R}^{N}\times (0,T)$ of the heat equation has an initial trace $\mu$ which is a Radon measure (i.e., a locally finite measure). This means that for any $\varphi\in\mathrm{C}_0(\mathbb{R}^N)$, (i.e. continuous and compactly supported in $\mathbb{R}^N$) we have \begin{equation}\label{in.tr1} \int_{\mathbb{R}^N}u(x,t)\varphi(x)dx\to \int_{\mathbb{R}^N}\varphi(x)d\mu(x)\quad\mbox{as } t\to 0. \end{equation} Moreover, the initial trace satisfies the following growth property: \begin{equation} \int_{\mathbb{R}^{N}}e^{-|x|^{2}/4T}\,d\mu (x)<+\infty . \label{char1} \end{equation} On the other hand, given a nonnegative measure $\mu $ which satisfies the above condition, there exists a unique nonnegative solution, so that the nonnegative admissible data are measures which are characterized by (\ref {char1}). We have a similar result for the the porous-medium equation $u_{t}=\Delta u^{m}$, $m>1$ \cite{AC,BCP}, where the admissible non-negative data are non-negative Radon measures which satisfy \begin{equation} \sup_{R\geqslant 1}R^{-\frac{2}{m-1}} - \hskip-1.1em\int_{B_{R}}d\mu <+\infty , \label{char2} \end{equation} where $-\hskip-0.9em\int_{E}f$ denotes the average of $f$ on $E$: $-\hskip-0.9em\int_{E}f=|E|^{-1}\int_{E}f(x)dx$. On the other hand, the case $m<1$ usually called {\sl fast diffusion}, leads to difficulties that is better to postpone. \subsection{Complete equation and Borel measures} If we add an absorption term $u^{p}$, we get equation (\ref{eq1}) and then there is a big extension of the class of initial data. Namely, we may have Borel measures, $i.e.$, not necessarily locally finite measures as initial data of standard weak solutions, even locally bounded and continuous. A significant example of such a situation is well-known and goes back to the works of \cite{BPT} and \cite{GKS}. It concerns the heat equation with absorption \begin{equation} u_t-\Delta u + u^p=0,\quad p>1. \end{equation} For any $c>0$, this equation admits a solution $u_c$ with initial data $u_c(0)=c\delta_0$, so-called fundamental solution with mass $c>0$. Now, letting $c\to\infty$, one obtains a new kind of singular solution called {\sl Very Singular Solution} (VSS for short) which is continuous (and locally bounded in $Q_T$), although it takes on the initial data ``$+\infty\cdot\delta_0$''. Such a measure is of course not locally bounded and is a basic example of Borel measure, that we briefly recall below. Later on, Marcus and V\'eron \cite{MV} proved that any Borel measure is admissible as initial data for this equation, under some capacity condition if $p\geqslant 1+2/N$. We denote by $\mathcal{B}^{+}(\mathbb{R}^{N})$ the set of Borel measures. Let us recall \cite{MV,EC1} that any $\nu \in \mathcal{B}^{+}(% \mathbb{R}^{N})$ may be written as $\nu =(\mathcal{S},\mu )$, where $\mathcal{S}$ is a closed subset of $\mathbb{R}^{N}$ and $\mu \geqslant 0$ is a Radon measure on $\mathcal{R}=\mathbb{R}^{N}\setminus \mathcal{S}$. The set $\mathcal{S}$ is defined as follows: \begin{equation*} \mathcal{S}=\{x\in \mathbb{R}^{N} : \forall r>0,\quad \nu (B_{r}(x))=+\infty \}. \end{equation*} Thus, a Radon measure is a Borel measure with $\mathcal{S}=\emptyset$. Let us now define what kind of solutions we consider. \paragraph{Definition.} By a solution we mean a function $u\in \mathrm{C}^{0}(Q_{T})$, $u\geqslant 0 $ such that (\ref{eq1}) holds in the sense of distributions. We also consider $u\equiv +\infty $ as a solution to simplify the statements of our results. \smallskip It is important to notice that since $u$ is continuous in $Q_{T}$, then it is bounded on every compact subset of $Q_{T}$, and thus it is the uniquely determined on such compact sets \cite{EC1}. This proves that $u$ can be viewed as a limit solution, $i.e$. as the limit of a sequence of smooth solutions. We prove that the solutions we consider will take on initial data in the set of nonnegative Borel measures, $\mathcal{B}^+(\mathbb{R}^N)$. Now it means that there exists a measure $\nu\in\mathcal{B}^+(\mathbb{R}^N)$ such that for any $\varphi\in\mathrm{C}_0(\mathbb{R}^N)$, (i.e. continuous and compactly supported in $\mathbb{R}^N$), $\varphi\geqslant 0$, we have $$ \int_{\mathbb{R}^N}u(x,t)\varphi(x)dx\to \int_{\mathbb{R}^N}\varphi(x)d\nu(x)\quad\mbox{as } t\to 0. $$ Since $\nu=(\mathcal{S},\mu)$ may have nonempty singular set $\mathcal{S}$, the right-hand may be infinite (it is well-defined in $\mathbb{R}_+\cup\{+\infty\}$ since $\varphi$ is nonnegative). The measure $\nu$ will be called the {\sl initial trace} of $u$ at $t=0$, and we note $\mathrm{tr}_{\mathbb{R}^{N}}(u)=\nu $. \paragraph{Definition.} Let $\nu \in \mathcal{B}^{+}(\mathbb{R}^{N})$. Then $\nu $ is said to be admissible and we note $\nu \in \mathcal{A}^{+}(p)$ if there exists a solution $u$ of (\ref{eq1}) such that $\mathrm{\rm tr}{}_{\mathbb{R}^{N}}(u)=\nu $. We will consider that $+\infty $ is allowed as initial data, associated to the special solution $u\equiv +\infty $. \smallskip For some values of $p$, the set of admissible initial data is known: \begin{itemize} \item The case $p=1$ is easily reduced to the pure diffusion case $u_{t}=\Delta u^{m}$ by the transformation $v(x,\tau )=e^{t}u(x,t)$, $d\tau /dt=e^{-(m-1)t}$, $\tau (0)=0$ so that $\mathcal{A}% ^{+}(1)$ is given by (\ref{char2}) for $m>1$, and (\ref{char1}) for $m=1$, with $+\infty $ allowed for $u\equiv +\infty $. \item By \cite{EC1} (and \cite{MV} in the case $m=1$), it is known that if $mm+2/N)$\newline \null\qquad 4.1 The operator $\mathbb{P}_{p}$\newline \null\qquad 4.2 Admissibility sets\newline \noindent\textbf{5.\quad Adaptations to the Case }$\Omega $ \textbf{Bounded} \section{Main Results}\label{sec.mr} Our results can be grouped in two areas: admissibility and the study of the projection operator. \paragraph{Admissibility.} In the case of very weak absorption, $0\leqslant p\leqslant 1$, we prove that the admissible initial data are exactly the same as in the case of the purely diffusive equation: \begin{equation*} \mathcal{A}^{+}(p)=\Big\{ \mu \in \mathcal{B}^{+}(\mathbb{R}^{N}) : \sup_{R\geqslant 1}R^{-\frac{2}{m-1}}% -\hskip-1.1em\int_{B_{R}}d\mu <+\infty \Big\} \cup \{+\infty \}, \end{equation*} where the measure $\nu =+\infty $ is associated to the ``special solution'' $u\equiv +\infty$. Moreover, we prove uniqueness of continuous weak solutions. The situation is different in the case of weak absorption $11$ and $\leqslant p1$, $p\geqslant 0$ and $u\geqslant 0$ be a solution of $u_{t}-\Delta u^{m}+u^{p}=0$. Then $u(t)$ has an initial trace when $t$ decreases to zero which is a Borel measure $\nu \geqslant 0$ in $\mathbb{R}% ^{N}$, in the following sense: for any continuous, compactly supported $% \varphi \in \mathrm{C}_{0}(\mathbb{R}^{N})$, $\varphi \geqslant 0,$% \begin{equation} \int_{\mathbb{R}^{N}}u(x,t)\varphi (x)dx\underset{t\rightarrow 0}{% \longrightarrow }\int_{\mathbb{R}^{N}}\varphi (x)d\nu (x), \label{convb} \end{equation} whether the last integral is finite or not. \end{theorem} \paragraph{Proof.} As was said, we already know the result if $10$. Then there exists sequences $t_n\to 0$ and $r_n\to 0$ such that $$ \forall n\in\mathbb{N},\quad\int_{B_{r_n}(y)}u(x,t_n)dx=c, $$ otherwise the integrals of $u$ would remain bounded near $y$, which could not be a singular point. Now by comparison, this implies that $u$ is not less than the solution $u_n$ in $(t_n,T)\times\mathbb{R}^N$ with initial data $u_n(t_n)=u(t_n)\chi_n$, where $\chi_n$ is the characteristic function of $B_{r_n}(y)$. Then by concentration, $u_n$ converges to the fundamental solution $v_{c,y}$ with initial data $c\delta_y$, so that $u\geqslant v_{c, \delta_y}$. Since $c$ is arbitrary, we let $c\to \infty$, thanks to Theorem \ref{thm.FS}, which yields that $u$ is either $+\infty$ everywhere or the flat solution whether $p$ is less or greater than $1$. \hfill$\square$ \section{Very Weak Absorption: $0\leqslant p\leqslant 1$} \label{sec.vwa} We consider equation (\ref{eq1}) in the range $0\leqslant p\leqslant 1$. We will see that in terms of initial data, the admissiblity condition is the same as for the diffusive case $u_{t}=\Delta u^{m}$ (see \cite{AC,BCP}). Recall that in this range, solutions are only local $a$ $priori$, as it is the case for the diffusive equation. A solution in $Q_{T}$ is thus understood to be defined up to $t=T$ (see remark after Theorem \ref{thmp<1}). \subsection{Harnack Inequality and Admissibility} We prove now that when $p<1$, $\mathcal{A}^{+}(p)=\mathcal{A}^{+}(1)$, in other terms, in this case, the absorption has no effect on admissibility of initial traces, compared with the purely diffusive case $u_{t}=\Delta u^{m}$ (we showed above that the case $p=1$ can be reduced to the diffusive equation with a suitable change of variables and functions). The following lemma reduces our study to the case $p=0$. \begin{lemma} \label{lem3}Let $\mu $ be a Radon measure. Then for every $p\in \lbrack 0,1], $ the following inclusion holds: \begin{equation*} \mathcal{A}^{+}(1)\subset \mathcal{A}^{+}(p)\subset \mathcal{A}^{+}(0). \end{equation*} \end{lemma} \paragraph{Proof.} The first inclusion is obvious since $\mathcal{A}^{+}(1)$ is related to equation $E_{1}:u_{t}-\Delta u^{m}+u=0$, and this equation has the same admissibility set than the purely diffusive equation $u_{t}=\Delta u^{m}$. Indeed, there exists a change of time variable which maps solutions of $u_{t}=\Delta u^{m}$ to $E_{1}$ and preserves the initial data. So if a trace is admissible for $E_{1}$, it is also for the diffusive equation, and then it is admissible if we add any absorption term. We are left to prove the second inclusion. Let $p\in \lbrack 0,1]$, and $\mu \in \mathcal{A}^{+}(p)$. Then we have $u^{p}\leqslant u+1$, and thus \begin{equation*} u_{t}-\Delta u^{m}+u+1\geqslant 0. \end{equation*} Now we use the same change of time variable (and of function) that maps $% E_{1}$ to $u_{t}=\Delta u^{m}:$ we get a function $w$ which has initial data $\mu $ and \begin{equation*} w_{t}-\Delta w^{m}+1\geqslant 0, \end{equation*} that is $w$ is a super-solution of $E_{0}$. Now it is easy to construct a solution with initial data $\mu :$ let $\mu _{n}$ be a sequence of bounded measures converging monotonically to $\mu $ and $v_{n}$ the sequence of minimal solutions of $E_{0}$ with initial data $\mu _{n}$. Then $v_{n}$ increases to some function $v$ and since $w$ is an upper bound for $v$, we know that $v$ has locally finite initial trace. Then by monotonicity, it is obvious that $\mathrm{\rm tr}{}_{\mathbb{R}^{N}}(v)=\mu $ (we have already proved this thing in the previous theorems). Thus $\mu \in \mathcal{A}^{+}(0)$ and the result is proved. \hfill$\square$ We now consider the case $p=0$, and more generally, we deal with nonnegative solutions of the equation \begin{equation} u_{t}=\Delta (u^{m})-a\chi (u>0),\qquad a\geqslant 0, \label{eq11} \end{equation} defined in $Q_{T}=\mathbb{R}^{N}\times (0,T)$. We obtain a Harnack inequality for this kind of equations, which includes in particular the porous medium equation when $a=0$. In this case, it was already obtained by Aronson and Caffarelli \cite{AC}, but our method is new for $a=0$ and quite simple. \begin{lemma} Let $u\in \mathrm{C}^{0}(\mathbb{R}^{N}\times \lbrack 0,T])$ be a nonnegative solution of (\ref{eq11}) in $p_{1}$ with $0\leqslant a\leqslant A $. Let \begin{equation} M=\int_{B_{1}}u(x,0)\,dx. \end{equation} There exist positive constants $M_{0}=M_{0}(N,m,A)$ and $k=k(N,m,A)$ such that for $M\geqslant M_{0}$ \begin{equation} u(0,1)\geqslant k\,M^{2\lambda },\quad \lambda =(N(m-1)+2)^{-1}. \end{equation} \end{lemma} \paragraph{Proof.} It is the combination of several steps. The letter $C$ will denote different positive constants that depend only on $N$ and $m$. \noindent $\bullet $\quad By comparison we may assume that $u_{0}$ is supported in the unit ball $B_{1}$. Indeed, for general $u_{0}$, then $u_{0}$ is greater than $u_{0}\eta $, $\eta $ being a suitable cut-off function compactly supported in $B_{1}$ and less than one. Thus if $v$ is the solution with initial data $u_{0}\eta $ (existence and uniqueness are well-known in this case), we obtain \begin{equation*} \int_{B_{1}}u(x,0)dx\geqslant \int_{B_{1}}u_{0}\eta =M, \end{equation*} and if the lemma holds true for $v$, then \begin{equation*} u(0,1)\geqslant v(0,1)\geqslant kM^{2\lambda }. \end{equation*} We may then take the domain of definition as $p=\mathbb{R}^{N}\times (0,\infty )$. \noindent $\bullet \quad $By comparison with the porous medium equation without absorption, we know the a priori estimate for the solution \cite[p. 54]{BCP} \begin{equation} 0\leqslant u(x,t)\leqslant C\,M^{2\lambda }\,t^{-N\lambda } \end{equation} and the a priori estimate for the support at time $t$, \begin{equation} \text{supp\thinspace }u(\cdot ,t)\subset B_{R}(t),\,\quad R(t)=C\,M^{(m-1)\lambda }\,t^{\lambda }. \end{equation} Hence, if $M$ is large this radius is much larger than 1 at $t=1$. \noindent $\bullet \quad $The reflection argument of Aleksandrov used in Lemma 2.2 of \cite{AC} means that for $|x|\geqslant 2$ we have \begin{equation} u(0,t)\geqslant u(x,t). \end{equation} \noindent $\bullet \quad $Let us now estimate the mass at time $t$ \begin{equation} \int u(x,t)\,dx=\int u_{0}(x)\,dx-a\int_{0}^{t}\int \chi (u>0)\,dxdt. \end{equation} The last term is bounded above by $C\,a\,t\,R(t)^{N}$, while the first member can be split into the integrals \begin{equation*} \int_{|x|\geqslant 2}u(x,t)\,dx+\int_{|x|\leqslant 2}u(x,t)\,dx, \end{equation*} and the last term can be estimated by \begin{equation*} C\,M^{2\lambda }\,t^{-N\lambda }2^{N}. \end{equation*} We conclude that \begin{equation*} Cu(0,t)\,(R(t)^{N}-2^{N})\geqslant \int_{|x|\geq 2}u(x,t)\,dx\geqslant M-C\,a\,t\,R(t)^{N}-C\,M^{2\lambda }\,t^{-N\lambda }2^{N}, \end{equation*} hence for $t=1$, \begin{equation*} C\,u(0,1)\,(M^{N(m-1)\lambda }-2^{N})\geqslant M-C\,a\,M^{N(m-1)\lambda }-C\,M^{2\lambda }, \end{equation*} so that there are three constants $c_{1},c_{2},c_{3}(m,N)$ such that \begin{equation*} u(0,1)\geqslant c_{1}M^{2\lambda }-ac_{2}-c_{3}M^{\gamma \lambda },\quad \gamma =2-(m-1)N. \end{equation*} Since $\gamma <2$, there exists some constants $M_{0}$ and $k$ such that \begin{equation*} c_{1}M^{2\lambda }-ac_{2}-c_{3}M^{\gamma \lambda }\geqslant kM^{2\lambda } \end{equation*} holds for every $M\geqslant M_{0}$, and this proves the Lemma. \hfill$\square$ Now we give the Harnack-type inequality. \begin{lemma} \label{lemHar}The estimate of the form \begin{equation} \int_{B_{r}(x_{0})}u(x,t)\,dx\leqslant C\left( r^{1/\lambda (m-1)}T^{-1/(m-1)}+T^{N/2}u^{1/2\lambda }(x_{0},T)\right) \label{eq12} \end{equation} holds for all nonnegative solutions if $r\geqslant T^{m/2}$ and $t\leqslant T$. \end{lemma} \paragraph{Proof.} We can use the previous lemma on $(t,T)$ since $u\in \mathrm{C}^{0}(Q_{T})$, perform the transformation \begin{equation} u^{\ast }(x,t)=r^{-2/(m-1)}T^{1/(m-1)}u(rx,Tt) \end{equation} as in \cite[p. 361]{AC}, and look at the equation satisfied by $u^{\ast }$. Now it is the same (\ref{eq11}) but for the constant $a$ which becomes \begin{equation} a'=a\,r^{-2/(m-1)}\,T^{m/(m-1)}. \end{equation} Hence, in order to apply the previous lemma we need to impose the condition $r\geqslant T^{m/2}$. \hfill$\square$ \paragraph{Remark.} The previous lemma gives a direct proof of the existence of the initial trace for $E_{0}$, which is a Radon measure. \begin{theorem} \label{thmp<1} For every $p\in \lbrack 0,1]$, we have that following characterization: \begin{equation*} \mu \in \mathcal{A}^{+}(p)\Leftrightarrow \sup_{R\geqslant 1}R^{-\frac{2}{m-1% }-N}\int_{B_{R}}d\mu (x)<\infty . \end{equation*} In other words, the admissibility condition is the same as in the purely diffusive equation. \end{theorem} \paragraph{Proof.} By Lemma \ref{lem3}, we have only to prove the converse inclusion \begin{equation*} \mathcal{A}^{+}(0)\subset \mathcal{A}^{+}(1). \end{equation*} If $\mu $ is admissible for $E_{0}$, there exists a minimal solution $u$. Hence by (\ref{eq12}), we get \begin{equation} r^{1/\lambda (m-1)}\int_{B_{r}(x_{0})}d\mu \leqslant C\left( T^{-1/(m-1)}+r^{-1/\lambda (m-1)}T^{N/2}u^{1/2\lambda }(x_{0},T)\right) , \label{eq.har} \end{equation} and since $(\lambda (m-1))^{-1}=N+2/(m-1)$, this implies that \begin{equation*} \sup_{R\geqslant 1}R^{-\frac{2}{m-1}}% -\hskip-1.1em\int_{B_{R}}d\mu <\infty , \end{equation*} thus $\mu $ is admissible for $E_{1}$, and the theorem is proved. \hfill$\square$ \paragraph{Remark.} The existence of a solution of $E_{p}$ with initial data $\mu $ is only valid up a time $T_{p}(\mu )$ in this range. It is obvious that $T_{p}(\mu )$ is not less that the blow-up time $T(\mu )$ in the case of the purely diffusive equation \cite{BCP}: \begin{equation*} T_{p}(\mu )\geqslant T(\mu )\geqslant C(m,N)/\ell (\mu )^{m-1}, \end{equation*} where \begin{equation*} \ell (\mu )=\lim_{r\rightarrow \infty }\sup_{R\geqslant r}R^{-\frac{2}{m-1}}% -\hskip-1.1em\int_{B_{R}}d\mu . \end{equation*} In fact, in the case $p=1$, the blow-up time can be computed thanks to the exponential change of time variable, and we find: \begin{equation*} T_{1}(\mu )\geqslant \frac{1}{m-1}\exp \big( (m-1)\frac{C(m,N)}{\ell (\mu )} \big) , \end{equation*} which is greater than $T(\mu )$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Optimal Uniqueness} We have seen that in the case $0\leqslant p\leqslant 1$, the admissibility condition for $E_{p}$ was the same as for the case $u_{t}=\Delta u^{m}$, and since uniqueness holds for this diffusive equation with no growth restriction \cite{DK}, one can reasonably think that the same holds for $% E_{p}$. We prove here that it is indeed the case. The following Lemma shows some $a$ $priori$ estimates for solutions of $% E_{p}$, similar to the one satisfied by the solutions of the diffusive equation. For every $\alpha >0$, we note \begin{equation*} \rho _{\alpha }(x)=[1+|x|^{2}]^{\alpha }. \end{equation*} \begin{lemma} \label{lemap}Let $u$ be a solution of $E_{p}$ with initial data $\mu \in \mathcal{A}^{+}(p)$, $i.e.$% \begin{equation*} \sup_{R\geqslant 1}R^{-\frac{2}{m-1}}% -\hskip-1.1em\int_{B_{R}}d\mu <\infty . \end{equation*} Then the following bound holds: \begin{equation*} u(x,t)\leqslant C(t)\rho _{\frac{1}{m-1}}(x)\quad \text{in\quad }% \{|x|\geqslant 1\}\times (0,T), \end{equation*} where $C(\cdot )\in L_{\rm loc}^{\infty }(0,T)$. Moreover, \begin{equation} \int_{\mathbb{R}^{N}}u(x,t)\rho _{\alpha }dx\underset{t\rightarrow 0}{% \longrightarrow }\int_{\mathbb{R}^{N}}\rho _{\alpha }(x)d\mu (x), \label{eqconv} \end{equation} for every $\alpha >1+\frac{N}{2}+\frac{1}{m-1}$. \end{lemma} \paragraph{Proof.} Let $s>0$. Since $u$ is a solution of $E_{p}$ on $(s,T)$, necessarily, \begin{equation*} \sup_{R\geqslant 1}R^{-\frac{2}{m-1}}% -\hskip-1.1em\int_{B_{R}}u(s)<\infty , \end{equation*} and thus $u(s)$ is also an admissible initial data for $u_{t}-\Delta u^{m}=0. $ We call $v_{s}$ the solution associated with $v_{s}(0)=u(s)$ for the diffusive equation ($v_{s}$ is unique by the results of \cite{DK}). Moreover, since $u$ is the limit of a sequence of smooth solutions on $% (s,T), $ we can compare $u$ with $v_{s}$: \begin{equation*} u(x,t)\leqslant v_{s}(x,t)\quad \text{in\quad }\mathbb{R}^{N}\times (s,t). \end{equation*} By the estimates on $v_{s}$ \cite[Rem. 3]{BCP}, we know that \begin{equation*} v_{s}(x,t)\leqslant \frac{c(s)}{t^{\lambda }}[1+|x|^{2}]^{\frac{1}{m-1}% }\quad \text{in\quad }\{|x|\geqslant 1\}\times (s,T), \end{equation*} where $\lambda =\frac{N}{N(m-1)+2}$ and \begin{equation*} c(s)\leqslant c(N,m)\Big[ \sup_{R\geqslant 1}R^{-\frac{2}{m-1}} -\hskip-1.1em\int_{B_{R}}u(s)\Big] ^{2\lambda /N}. \end{equation*} The function $c(s)$ remains bounded when $s$ decreases to zero (this is a consequence of the Harnack inequality (\ref{eq12})), we find that for some function $C(\cdot )\in L_{\rm loc}^{\infty }(0,T),$% \begin{equation*} u(x,t)\leqslant C(t)[1+|x|^{2}]^{\frac{1}{m-1}}\quad \text{in\quad }\mathbb{% \{}|x|\geqslant 1\}\times (0,T). \end{equation*} Moreover, these techniques show that \begin{equation*} u(x,t)\leqslant v(x,t)\quad \text{in\quad }Q_{T}, \end{equation*} where $v$ is the unique solution of the diffusive equation $u_{t}=\Delta u^{m}$ with initial data $\mu $. But the convergence property (\ref{eqconv}) holds for $v$ (see \cite[p. 81]{BCP}), so that it also holds for $u$. Indeed, (all integrals are taken over $\mathbb{R}^{N}$) \begin{equation*} \int u(t)\rho _{\alpha }-\int \rho _{a}d\mu =\int \underset{\leqslant 0}{% \underbrace{(u-v)}}(t)\rho _{a}+\int v(t)\rho _{\alpha }-\int \rho _{\alpha }d\mu , \end{equation*} so that \begin{equation*} \underset{t\rightarrow 0}{\sup \lim }\int u(t)\rho _{\alpha }\leqslant \int \rho _{a}d\mu , \end{equation*} and since $u(t)\rightarrow \mu $ weakly in measure, clearly \begin{equation*} \int u(t)\rho _{\alpha }\underset{t\rightarrow 0}{\longrightarrow }\int \rho _{a}d\mu . \end{equation*} \hfill$\square$ \begin{theorem} Let $0\leqslant p\leqslant 1\;$and $\nu \in \mathcal{A}^{+}(p)$, $i.e.$, $% \nu $ satisfies \begin{equation*} \sup_{R\geqslant 1}R^{-\frac{2}{m-1}}% -\hskip-1.1em\int_{B_{R}}d\nu <\infty . \end{equation*} Then there exists a unique solution $u$ to(\ref{eq1}) such that $\mathrm{\rm tr}{}_{\mathbb{R}^{N}}(u)=\nu $. \end{theorem} \paragraph{Proof.} Thanks to the previous a priori estimate, we can use the same techniques as in \cite[Prop. 2.1]{BCP}, which consists in solving the dual problem. Before this, we need to construct a minimal solution, which can be obtained as in \cite[Sec. 4.2]{EC1}: if $u$ is any solution, let $u_{R,\tau }$ be the unique solution of the problem \begin{gather*} \partial _{t}u_{R,\tau }-\Delta u_{R,\tau }^{m}+u_{R,\tau }^{p}=0 \quad \text{in }B_{R}\times (\tau ,T), \\ u_{R,\tau }(x,t)=0 \quad \text{on }\partial B_{R}\times (\tau ,T), \\ u_{R,\tau }(\tau )=u(\tau ) \quad \text{in }B_{R}. \end{gather*} By comparison in this set, since both solutions are bounded, \begin{equation*} u_{R,\tau }(x,t)\leqslant u(x,t)\quad \text{in\quad }B_{R}\times (\tau ,T). \end{equation*} Then if we let $\tau $ decrease to zero, we see that $u_{R,\tau }$ converges locally uniformly to a solution $u_{R}$ with initial data $\mu /_{B_{R}}$ and zero lateral data on $\partial B_{R}\times (0,T)$. Moreover, $u_{R}$ is uniquely determined, as was proved in \cite[Theorem 6.2]{EC1}, so that it can be constructed independently of any solution, and in the limit, \begin{equation*} u_{R}(x,t)\leqslant u(x,t)\quad \text{in\quad }B_{R}\times (0,T). \end{equation*} Finally, when $R$ increase to $+\infty $, $u_{R}$ increases to some solution $\underline{u}$ with initial data $\mu $, which is the (unique) minimal solution since $\underline{u}(x,t)\leqslant u(x,t)$ in $Q_{T}$. Now if $u$ is any solution with initial data $\mu $, we will prove that $u\equiv \underline{u}$, hence uniqueness since $\underline{u}$ can be constructed independently of any solution. Let us first fix $s>0$ and $t\in (s,T)$. Since $u\geqslant \underline{u}$, we have \begin{equation*} (u-\underline{u})_{t}-\Delta (u^{m}-\underline{u})=\underline{u}% ^{p}-u^{p}\leqslant 0, \end{equation*} and thanks to the $a$ $priori$ estimate given by Lemma \ref{lemap}, \begin{equation*} u\rho _{\frac{1}{m-1}},\underline{u}\rho _{\frac{1}{m-1}}\in L^{\infty }(% \mathbb{R}^{N}\times (s,t)). \end{equation*} Then the techniques of \cite[Prop. 2.1]{BCP} apply $verbatim$ and give \begin{equation*} \int_{\mathbb{R}^{N}}(u-\underline{u})(t)\theta \leqslant \int_{\mathbb{R}% ^{N}}|u-u\underline{u}|(s)\rho _{\beta }, \end{equation*} where $\theta \in \mathrm{C}_{0}^{\infty }(\mathbb{R}^{N})$, $0\leqslant \theta \leqslant 1$ is arbitrary and $\beta >\frac{N-1}{2}+\frac{m}{m-1}$ can also be chosen freely. In \cite{BCP}, since both solutions have the same value at $s=0$ in $L_{\rm loc}^{1}$, the conclusion is that the solutions coincide everywhere. But here we have to let $s$ decrease to zero, so we use that fact that $u[\nu ]$ is minimal: \begin{equation*} \int |u-\underline{u}|(s)\rho _{\beta }=\int u(s)\rho _{\beta }-\int \underline{u}(s)\rho _{\beta }, \end{equation*} which both converge to the same value when $s$ goes to zero, thanks to (\ref {eqconv}) with a suitable $\beta $. Hence in the limit, \begin{equation*} \int_{\mathbb{R}^{N}}(u-\underline{u})(t)\theta \leqslant 0, \end{equation*} which proves that $u\equiv \underline{u}$ since $\theta \geqslant 0$ and $% t>0 $ are arbitrary. \hfill$\square$ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Weak Absorption: $1\leqslant p\leqslant m$} \label{sec.wa} \subsection{The Projection Operator} We now construct a projection operator $\mathbb{P}_{p}:\mathcal{B}% ^{+}(\Omega )\rightarrow \mathcal{B}^{+}(\Omega )$. Actually, the construction remains valid in the range $0\leqslant p\leqslant 1$ previously studied, so that we give it in its full generality below. It is based on the following Lemma: \begin{lemma} \label{lem1}Let $0\leqslant p\leqslant m$, and $\nu \in \mathcal{B}^{+}(\mathbb{% R}^{N})$. Let $\mu _{n}\geqslant 0$ be a sequence of compactly supported measures converging monotonically to $\nu $ and $u_{n}$ the associated sequence of solutions. Then $u_{n}$ converges locally uniformly to the minimal solution $u[\nu ]$ with initial data $\nu $. Moreover, if $\mathrm{tr% }_{\mathbb{R}^{N}}(u[\nu ])$ is not locally finite, then \begin{equation} \begin{aligned} 0 \leqslant p\leqslant 1 &\Rightarrow u[\nu ]\equiv +\infty,\\ 1p$ and that $\mu \in \mathcal{A}^{+}(p)$. Then we shall show that $\mu $ is also an admissible data for $E_{p'}$, that is, $\mu \in \mathcal{A}^{+}(p')$. In fact, we will first construct a super-solution for equation $E_{p'}$ on some small interval $(0,t_{0})$, with initial data $\mu $, and then we will easily show that there exists a solution with initial data $\mu $ on $(0,T)$. Let $u$ be the minimal solution associated with $\mu $, and let \begin{equation*} v(x,t)=(1+ct)\cdot u(x,t),\quad (x,t)\in \mathbb{R}^{N}\times (0,t_{0}), \end{equation*} for some parameters $c,t_{0}>0$ which will be specified later on. Then a straightforward calculation gives \begin{equation*} v_{t}=(1+ct)^{1-m}\Delta v^{m}+c(1+ct)^{-1}v-(1+ct)^{1-p}v^{p}, \end{equation*} and changing the time variable in $\tau \in (0,\tau _{0}(t_{0}))$ such that \begin{equation*} \frac{d\tau }{dt}=(1+ct)^{1-m}, \end{equation*} we get, setting $v(x,t)=w(x,\tau ),$% \begin{equation*} w_{\tau }=\Delta w^{m}-w^{p}(1+ct)^{m-p}+cw(1+ct)^{m-2}. \end{equation*} Note that if we work with classical solutions, \begin{equation*} w(x,0)=v(x,0)=u(x,0), \end{equation*} and this is again the case if the initial data is a measure, the above equality being understood in the sense of initial traces. Moreover, for a good choice of $t_{0}$ and $c$, we show that $w$ is a super-solution of equation $E_{p'}$ with initial data $\mu :$ let $c$ be a free parameter for the moment and put $t_{0}=1/c$, then for every $t\in (0,t_{0}), $% \begin{equation} \min \{1;2^{m-2}\}<(1+ct)^{m-p},(1+ct)^{m-2}<2^{m}. \label{eq2} \end{equation} Now since $p'>p$, there exists a constant $k(m,p,p'),$such that for $w\geqslant k$, \begin{equation*} w^{p'}\geqslant 2^{m}w^{p}, \end{equation*} and for $w\leqslant k$ (fixed above), thanks to (\ref{eq2}) there exists some $c(m,p,p')$ (maybe big but finite), such that \begin{equation*} cw(1+ct)^{m-2}\geqslant w^{p}(1+ct)^{m-p}. \end{equation*} Thus, we have obtained that whatever the value of $w,$% \begin{equation*} w^{p'}+cw(1+ct)^{m-2}\geqslant (1+ct)^{m-p}w^{p},\quad \text{on}% \quad (0,t_{0}), \end{equation*} so that there exists some $\tau _{0}>0$ (only depending on $m$, $p$ and $% p'$ through $c$ and $k$) such that for $\tau \in (0,\tau _{0})$% \begin{equation*} w_{\tau }\geqslant \Delta w^{m}-w^{p'}, \end{equation*} that is, $w$ is a super-solution of $E_{p'}$ with initial data $\mu $.\smallskip Now we can construct a solution of $E_{p'}$ with initial data $\mu :$ let $\mu _{n}$ be a sequence of bounded measures converging to $% \mu $ monotonically and $u_{n}$ be the associated sequence of minimal solutions. Then $u_{n}$ also increases to some distributional solution $u\in \mathrm{C}^{0}(\mathbb{R}^{N}\times (0,T))$, and $u_{n}$ has an initial trace $\nu \in \mathcal{B}^{+}(\mathbb{R}^{N})$. We also construct the associated sequence of functions $w_{n}$ on $(0,\tau _{0})$ by the same process as above, taking the initial data $\mu _{n}$. By construction, we have on $(0,\tau _{0})$, \begin{equation*} u_{n}(t)\leqslant w_{n}(t)\leqslant w(t), \end{equation*} and since $u\in L^{m}(0,T;L_{\rm loc}^{m}(\mathbb{R}^{N}))$, then so is $w$, so that the $\{u_{n}\}$ remain uniformly bounded by $w$ which is in $% L^{m}(0,T;L_{\rm loc}^{m}(\mathbb{R}^{N}))$, and in $L_{\rm loc}^{1}(\mathbb{R}^{N})$ for every $\tau \in (0,\tau _{0})$. As we already seen, this argument allows us to pass to the limit in the following equation, where $\varphi \in \mathrm{C}_{0}^{2}(\mathbb{R}^{N})$: \begin{equation*} \int_{\mathbb{R}^{N}}u_{n}(t)\varphi -\int_{0}^{t}\int_{\mathbb{R}% ^{N}}u_{n}^{m}\Delta \varphi +\int_{0}^{t}\int_{\mathbb{R}% ^{N}}u_{n}^{p}\varphi =\int_{\mathbb{R}^{N}}\varphi d\mu _{n}, \end{equation*} and thus we get \begin{equation*} \int_{\mathbb{R}^{N}}u(t)\varphi -\int_{0}^{t}\int_{\mathbb{R}% ^{N}}u^{m}\Delta \varphi +\int_{0}^{t}\int_{\mathbb{R}^{N}}u^{p}\varphi =\int_{\mathbb{R}^{N}}\varphi d\mu , \end{equation*} hence initial trace of $u$ is $\mu $. Thus $\mu $ is an admissible data for $% E_{p'}$, and the theorem is proved.\newline Finally, for $pm$ and $\{u_{n}\}_{n\in \mathbb{N}}$ be a sequence of solution of (% \ref{eq1}) in $\mathbb{R}^{N}\times (0,T)$ with initial trace $\nu _{n}$ such that $u_{n}$ is increasing and converges locally uniformly to some $u$. Then $u$ is a solution of (\ref{eq1}) with initial trace \begin{equation*} \mathrm{\rm tr}{}_{\Omega }(u)=\lim \nu _{n}. \end{equation*} \end{lemma} \paragraph{Proof.} Let us first notice that $\nu _{n}=\mathrm{\rm tr}_{\Omega }(u_{n})$ is well-defined since $u_{n}$ is a weak solution, as well as $\mathrm{\rm \{tr}_{\Omega }(u)=(\mathcal{S},\mu )$. Let us call $\nu =\lim \nu_{n}$, which is well-defined by monotonicity. Now for any $\varphi \in \mathrm{C}_{0}(\mathcal{R})$, and $n$ big enough, we may write \begin{equation} \int_{\mathbb{R}^{N}}u_{n}(t)\varphi (t)+\int_{0}^{t} \int_{\mathbb{R}^{N}}\{-u_{n}\varphi _{t}-u_{n}^{m}\Delta \varphi +u_{n}^{p}\varphi \}=\int_{\mathbb{R}^{N}}\varphi d\nu _{n}<\infty . \label{eqlim} \end{equation} Indeed, by monotonicity of $u_{n}$, the sequence of Borel measures $\nu _{n}=(\mathcal{S}^{n},\mu _{n})$ is also monotone so that $\varphi $ has compact support in $\mathbb{R}^{N}\setminus \mathcal{S}^{n}$, for $n$ sufficiently. To get some bounds, let us take $\varphi (x,t)=\varphi (x)\in \mathrm{C}_{0}^{\infty }(\mathbb{R}^{N})$, $0\leqslant \varphi \leqslant 1$, with support in some ball $B_{r}\subset \mathcal{R}$, and let us define \begin{equation*} X_{n}(t)=\int_{0}^{t}\int_{\mathbb{R}^{N}}u_{n}\varphi ,\quad Y_{n}(t)=\int_{0}^{t}\int_{\mathbb{R}^{N}}u_{n}^{p}\varphi . \end{equation*} Then for a good choice of $\varphi $ (see \cite{EC1}), we have \begin{equation*} \int_{0}^{t}\int_{\mathbb{R}^{N}}u_{n}^{m}|\Delta \varphi |\leqslant c(\varphi )Y_{n}(t)^{m/p}, \end{equation*} so that we arrive at the following differential inequality, where $C=\nu (B_{r})<\infty :$% \begin{equation*} \frac{dX_{n}(t)}{dt}-c(\varphi )Y_{n}(t)^{m/p}+Y_{n}(t)\leqslant C. \end{equation*} Then clearly, we get a uniform bound for $dX_{n}/dt=\int_{\mathbb{R}% ^{N}}u_{n}(t)\varphi (t)$, so that now we can use Lemma \ref{lemZL}: it yields that $u_{n}$ is uniformly bounded in $L^{q}(0,T;L_{\rm loc}^{q}(\mathcal{R% }))$, for any $qFrom this we deduce that on $\mathbb{R}^{N}\setminus \mathcal{S}$, $\mu =\nu $. It remains to show that the blow-up set of $\nu $ is exactly $\mathcal{S}, $ which is easy: for any $y$ in the blow-up set of $\nu $, and any $r>0,$% \begin{equation*} \int_{B_{r}(y)}u(x,t)dx\geqslant \int_{B_{r}(y)}u_{n}(x,t)dx, \end{equation*} so that as $t\rightarrow 0,$% \begin{equation*} \lim_{t\rightarrow 0}\int_{B_{r}(y)}u(x,t)dx\geqslant \nu _{n}(B_{r}(y)). \end{equation*} Thus when $n\rightarrow \infty $, we find \begin{equation*} \lim_{t\rightarrow 0}\int_{B_{r}(y)}u(x,t)dx=+\infty , \end{equation*} which means that $y\in \mathcal{S}$. On the other hand, if we assume that $% y\in \mathcal{S}$, there exists some $r>0$ such that $\nu (B_{r}(y))<\infty . $ This implies that for any $n\in \mathbb{N}$, \begin{equation*} \nu _{n}(B_{r}(y))\leqslant \nu (B_{r}(y))<\infty . \end{equation*} Thus by Lemma \ref{lemZL}, we have a uniform bound for $u_{n}^{m}$ in $% L^{1}((0,T)\times B_{r}(y))$, and thus also for $u_{n}^{p}$ (this derives from (\ref{eqlim})), so that in the limit, $u^{m}$ and $u^{p}$ will be locally integrable near $y$. This proves that $y\notin \mathcal{S}$, and we reach a contradiction. \hfill$\square$ \begin{lemma} \label{lem-op-st-rn}Let $1p$, and $v$ be the solution associated with equation $E_{p'}$. We would like to show that there exists a solution $u$ of equation $E_{p}.$\newline Let us first remark that using the same technique as in Theorem \ref {thmcomp1}, we can easily show that there exist some constants $t_{0},c>0$ such that the function \begin{equation*} w(x,\tau )=(1-ct)\cdot v(x,t),\quad \frac{d\tau }{dt}=(1-ct)^{1-m}, \end{equation*} satisfies \begin{equation*} w_{\tau }-\Delta w^{m}+w^{p}\leqslant 0\quad \text{in\quad }\mathbb{R}% ^{N}\times (0,\tau _{0}), \end{equation*} where $\tau _{0}>0$ depends on $t_{0}$ by the change of variable. Hence $w$ is a sub-solution of equation $E_{p}$ in $\mathbb{R}^{N}\times (0,\tau _{0}), $ and remembering that the constants $t_{0}$ and $c$ depend only on $% m,p,p'$, we take $n$ big enough so that $1/n<\tau _{0}.$\newline Now let $u_{n}$ be the solution of the following problem: \begin{gather*} \partial _{t}u_{n}-\Delta u_{n}^{m}+u_{n}^{p}=0 \quad \text{in }Q_{T}, \\ u_{n}(1/n)=w(1/n) \quad \text{in }\mathbb{R}^{N}. \end{gather*} It is well-known that $u_{n}$ exists and is unique since $w(1/n)$ is continuous in $\mathbb{R}^{N}$ (see \cite{VW} or \cite{EC1}). Using the same arguments as in the previous Subsection, it is clear that the $u_{n}$ will converge (up to extraction) to a solution $u$ such that \begin{equation*} \mathrm{\rm tr}{}_{\mathbb{R}^{N}}(u)\leqslant \nu . \end{equation*} But both $u_{n}$ and $v$ (thus $w$) are limits of regular solutions in $% \mathbb{R}^{N}\times (1/n,T)$, and thus they can be compared in $\mathbb{R}% ^{N}\times (1/n,\tau _{0})$, which gives that \begin{equation*} u_{n}(x,\tau )\geqslant w(x,\tau )\quad \text{in\quad }\mathbb{R}^{N}\times (1/n,\tau _{0}). \end{equation*} Thus in the limit, \begin{equation*} u(x,\tau )\geqslant w(x,\tau )\quad \text{in\quad }\mathbb{R}^{N}\times (0,\tau _{0}), \end{equation*} and this proves that the initial trace of $u$ is controlled (from below) by $% w$. Hence we have necessarily \begin{equation*} \mathrm{\rm tr}{}_{\mathbb{R}^{N}}(u)=\nu , \end{equation*} so that $\nu \in \mathcal{A}^{+}(p)$, and the result is proved. \hfill$\square$ We now give some qualitative properties for the sets $\mathcal{A}^{+}(p)$ in the super-critical range. The first result is a consequence Lemma \ref {lemconv1}, concerning monotone convergence. \begin{lemma} \label{lemconv2} Let $p\geqslant m+2/N$, and $\nu _{n}$ be an increasing sequence of admissible Borel measures such that $\mu _{n}$ converges in the sense of Borel measures to a measure $\nu \in \mathcal{B}^{+}(\mathbb{R}^{N})$. Then $\nu \in \mathcal{A}^{+}(p)$, $i.e.$, $\nu $ is also admissible. \end{lemma} \paragraph{Proof.} Let $\nu =(\mathcal{S},\mu )$ and $u_{n}$ be the sequence associated with $% \nu _{n}$. Then $u_{n}$ converges monotonically and locally uniformly to a solution $u$ of equation $E_{p}$. By Lemma \ref{lemconv1}, we know that \begin{equation*} \mathrm{\rm tr}{}_{\mathbb{R}^{N}}(u)\leqslant \nu , \end{equation*} but here we can use monotone convergence instead of Fatou's lemma in the proof of Lemma \ref{lemconv1}, which gives that on $\mathbb{R}^{N}\setminus \mathcal{S}$, the trace of $u$ is exactly $\mu $. It remains to prove that $% u $ blows up on $\mathcal{S}$. This is easy since for every open set $U$ such that $U\cap \mathcal{S}\neq \emptyset ,$% \begin{equation*} \int_{U}u(x,t)dx\geqslant \int_{U}u_{n}(x,t)dx, \end{equation*} and thus for every $n\in \mathbb{N},$% \begin{equation*} \underset{t\rightarrow 0}{\inf \lim }\int_{U}u(x,t)dx\geqslant \nu _{n}(U\cap \mathcal{S}), \end{equation*} which blows up when $n$ goes to infinity, because $\nu _{n}$ converges to $% +\infty $ on $\mathcal{S}$. Thus we have obtained that $\mathrm{\rm tr}_{\mathbb{R}^{N}}(u)=\nu$, hence $\nu \in \mathcal{A}^{+}(p)$. \hfill$\square$ Let us give now some basic examples of admissible measures, and admissible singular sets (see definition just below). \paragraph{Definition.} A closed set $\mathcal{S}\subset \mathbb{R}^{N}$ is said to be admissible as a singular set if the measure $\nu =(\mathcal{S},0)$ is admissible as initial data. \begin{proposition} \label{propball}Every closed ball of positive radius is admissible as a singular set. \end{proposition} \paragraph{Proof.} Let $\overline{B}_{r}$ be such a ball, and $\chi $ be the characteristic function of this ball. Let $u_{n}$ be the unique solution of equation $E_{p}$ with initial data \begin{equation*} u_{n}(0)=n\cdot \chi . \end{equation*} Then $u_{n}$ is not zero since $\chi $ is not zero almost everywhere, and it is increasing, and by a direct application of Lemma \ref{lemconv2}, we obtain the result: $u_{n}$ converges to a solution $u$ with initial trace $% +\infty \cdot \chi $, which is thus an admissible data. Of course, $+\infty \cdot \chi $ is the Borel measure which is represented by \begin{equation*} \nu =(\overline{B}_{r},0), \end{equation*} hence the closed balls are admissible sets. \hfill$\square$ \paragraph{Remarks.} \textbf{1. }If we want to do the same with a set of zero Lebesgue measure, this method cannot work since $u_{n}(0)=0$ almost everywhere, and then $% u_{n}\equiv 0$. In fact it is clear that in the super-critical case, the singular set has to be dense enough to insure existence of a non trivial solution, since for instance a point is not admissible as a singular set (non existence of the $V.S.S.$). Obviously, balls are dense enough.\newline \textbf{2. }The same result holds true if we consider the Borel measure $\nu =(\overline{B}_{r},\mu )$, where $\mu $ is an admissible Radon measure. The modification is the following: let $v$ be the solution of $E_{p}$ with initial data $\mu $. Then if we put \begin{equation*} u_{n}(0)=n\cdot \chi +v(1/n), \end{equation*} it is obvious that $u_{n}$ will converge locally uniformly to a solution $u$ with initial trace $+\infty $ on $\overline{B}_{r}$, and to check that the initial trace is not lost on the complement of $\overline{B}_{r}$, we use the fact that $u\geqslant v$ (by construction), so that finally, $\mathrm{\rm tr}_{\mathbb{R}^{N}}(u)=\nu $. In the same spirit, one can easily prove the following results: \begin{proposition} Let $\mathcal{S}$ be a closed set in $\mathbb{R}^{N}$ such that $\mathcal{S}$ is equal to the closure of its interior. Then $\mathcal{S}$ is admissible as a singular set. \end{proposition} \paragraph{Proof.} Let $u_{n}$ be defined as in the case of a ball as the solution of $E_{p}$ with initial data \begin{equation*} u_{n}(0)=n\cdot \chi , \end{equation*} where $\chi $ is the indicator function of $\mathrm{int}(\mathcal{S})$. In particular, for every $y\in \mathrm{int}(\mathcal{S})$, there exists a ball $% B_{y}$ of positive Radius centered at $y$ such that $B_{y}\subset \mathrm{int% }(\mathcal{S})$, and if $\chi _{y}$ is the indicator function of $B_{y},$% \begin{equation*} u_{n}(0)\geqslant n\cdot \chi _{y}. \end{equation*} Thus when $n$ increases, $u_{n}$ also increases to a limit solution $u$ with initial trace $\nu $, and clearly for every $y\in \mathrm{int}(\mathcal{S}),$% $\nu =+\infty$ on $B_{y}$. Then it is obvious that the singular set of $\nu $ contains the closure of $% \mathrm{int}(\mathcal{S})$, $i.e$. $\mathcal{S}$ itself. But on the complement of $\mathcal{S}$, it is also obvious that $\nu =0$, by using exactly the same method as in Proposition \ref{propball} (or using directly Lemma \ref{lemconv1}). \hfill$\square$ \begin{theorem} Assume that $\nu =(\mathcal{S},\mu )$, where $\mu \in \mathcal{A}^{+}(p)$ and $\mathcal{S}$ is admissible as a singular set, $i.e.$, there exists a solution $V_{\mathcal{S}}$ of $E_{p}$ such that \begin{equation*} \mathrm{\rm tr}{}_{\mathbb{R}^{N}}(V_{\mathcal{S}})=(\mathcal{S},0). \end{equation*} Then $\nu \in \mathcal{A}^{+}(p)$. \end{theorem} \paragraph{Proof.} Since $\mu $ is admissible, there exists a solution $v$ with initial data $% \mu $, and if we take the solution $u_{n}$ of $E_{p}$ with initial data \begin{equation*} u_{n}(0)=V_{\mathcal{S}}(1/n)+v(1/n), \end{equation*} then obviously, $u_{n}\geqslant \max \{V_{\mathcal{S}};v\}$ on $\mathbb{R}% ^{N}\times (1/n,T)$, so that $u_{n}$ converges to a limit solution $u$ locally uniformly, and $\mathrm{\rm tr}_{\mathbb{R}^{N}}(u)$ is exactly $(\mathcal{S},\mu )$. Indeed, it is obvious that $u$ will blow-up on $\mathcal{S}$ since it is greater than $V_{\mathcal{S}}$, and on the regular set, there is no loss of initial trace because $u\geqslant v$, which takes on the initial data $\mu $. \hfill$\square$ \paragraph{Remark.} The Radon measure $\mu $ is defined on the complement of $\mathcal{S}$, hence it may not be a Radon measure in $\mathbb{% R}^{N}$. Indeed, one may have $\mu (\partial \mathcal{S})=+\infty $. Hence in this result, we say that $\mu \in \mathcal{A}^{+}(p)$ if $\mu $, extended by zero on $\mathcal{S}$ is a Borel measure in $\mathbb{R}^{N}$ which is admissible. Note also that by Lemma \ref{lemconv2}, we know that if $\mu /_{K}$ is admissible for every compact set $K\subset \mathbb{R}^{N}\setminus \mathcal{S}$, then $\mu $ is admissible. \begin{proposition} Let $p\geqslant m+2/N$ and $\mu \in \mathcal{A}^{+}(p)$, locally finite. Then for every $\mu \in \mathcal{B}^{+}(\mathbb{R}^{N}),$% \begin{equation*} 0\leqslant \mu '\leqslant \mu \Rightarrow \mu '\in \mathcal{A}^{+}(p). \end{equation*} \end{proposition} \paragraph{Proof.} Let $\mu $ and $\mu '$ as above ($\mu '\leqslant \mu $ implies that $\mu '$ is also locally finite). Since $\mu \in \mathcal{A}^{+}(p)$, here exists a solution $u$ of $E_{p}$ such that $\mathrm{\rm tr}_{\mathbb{R}^{N}}(u)=\mu $, and by the uniqueness result in \cite[Sec. 4.2]{EC1}, recalled at the beginning of this section, $u$ is unique. Thus the solution $u$ can be viewed as the limit of the $u_{n}$, where $u_{n}$ is the unique solution of $E_{p}$ with initial data \begin{equation*} \mu _{n}=\rho _{n}\star \mu , \end{equation*} $\rho _{n}$ being a convolution kernel in $\mathbb{R}^{N}$. And if we call $v_{n}$ the solution with initial data \begin{equation*} \mu _{n}'=\rho _{n}\star \mu ', \end{equation*} then by construction, $v_{n}\leqslant u_{n}$ in $Q_{T}$, and $v_{n}$ will converge to a solution $v$ of $E_{p}$. The problem is to check that $v$ takes on the initial data $\mu '$, and to achieve this, it is sufficient to prove that the sequence $v_{n}$ converges in $% L^{p}(0,T;L_{\rm loc}^{p}(\mathbb{R}^{N}))$. Indeed, in this case, we can pass to the limit in the following weak formulation: \begin{equation*} \int_{\mathbb{R}^{N}}v_{n}(t)\varphi -\int_{0}^{t}\int_{\mathbb{R}% ^{N}}v_{n}^{m}\Delta \varphi +\int_{0}^{t}\int_{\mathbb{R}% ^{N}}v_{n}^{p}\varphi =\int_{\mathbb{R}^{N}}\varphi d\mu _{n}', \end{equation*} which gives \begin{equation*} \int_{\mathbb{R}^{N}}v(t)\varphi -\int_{0}^{t}\int_{\mathbb{R}% ^{N}}v^{m}\Delta \varphi +\int_{0}^{t}\int_{\mathbb{R}^{N}}v^{p}\varphi =\int_{\mathbb{R}^{N}}\varphi d\mu ', \end{equation*} hence $\mathrm{\rm tr}_{\mathbb{R}^{N}}(v)=\mu '$. But since $v_{n}\leqslant u_{n}$, it is sufficient to prove that the $u_{n}$ converge in $L^{p}(0,T;L_{\rm loc}^{p}(\mathbb{R}^{N}))$. So let us use the weak formulation for the $u_{n}$, where $\varphi \in \mathrm{C}^{2}(\mathbb{R}% ^{N})$, $\mathrm{\mathrm{supp}}(\varphi (t))\subset K$ fixed: \begin{equation*} \int_{\mathbb{R}^{N}}u_{n}(t)\varphi -\int_{0}^{t}\int_{\mathbb{R}% ^{N}}u_{n}\varphi _{t}-\int_{0}^{t}\int_{\mathbb{R}^{N}}u_{n}^{m}\Delta \varphi +\int_{0}^{t}\int_{\mathbb{R}^{N}}u_{n}^{p}\varphi =\int_{\mathbb{R}% ^{N}}\varphi d\mu _{n}, \end{equation*} and in the limit, $u$ is a solution with initial trace $\mu :$% \begin{equation*} \int_{\mathbb{R}^{N}}u(t)\varphi -\int_{0}^{t}\int_{\mathbb{R}^{N}}u\varphi _{t}-\int_{0}^{t}\int_{\mathbb{R}^{N}}u^{m}\Delta \varphi +\int_{0}^{t}\int_{% \mathbb{R}^{N}}u^{p}\varphi =\int_{\mathbb{R}^{N}}\varphi d\mu . \end{equation*} Since $u_{n}$, $u_{n}^{m}$ converge in $L^{1}(0,T;L_{\rm loc}^{1}(\mathbb{R}% ^{N}))$ (from equi-integrability given by Lemma \ref{lemZL}), then \begin{equation} \int_{0}^{t}\int_{\mathbb{R}^{N}}u_{n}^{p}\varphi \underset{n\rightarrow \infty }{\longrightarrow }\int_{0}^{t}\int_{\mathbb{R}^{N}}u^{p}\varphi . \label{eqconv2} \end{equation} Then it is easy to see that $u_{n}$ converges in $L^{p}(0,T;L_{\rm loc}^{p}(% \mathbb{R}^{N})):$ if we take $\varphi \geqslant 0,$% \begin{equation*} \int_{0}^{t}\int_{\mathbb{R}^{N}}(u^{p}-u_{n}^{p})_{+}\varphi \underset{% n\rightarrow \infty }{\longrightarrow }0 \end{equation*} by dominated convergence, and combining this with (\ref{eqconv2}), we obtain the same for $(u^{p}-u_{n}^{p})_{-}$, which in turn implies the strong convergence: \begin{equation*} \int_{0}^{t}\int_{\mathbb{R}^{N}}|u^{p}-u_{n}^{p}|\varphi \underset{% n\rightarrow \infty }{\longrightarrow }0. \end{equation*} Thus $v$ is indeed a solution of $E_{p}$ with initial data $\mu '$ (and in fact it is unique), hence $\mu '\in \mathcal{A}^{+}(p)$. \hfill$\square$ \paragraph{Remark.} In this result, the restriction to Radon measures is essential. Indeed, we have for instance ($\chi $ being the characteristic function of the ball $B_{1}(0)$): \begin{equation*} 0\in \mathcal{A}^{+}(p),\quad \delta _{0}\not\in \mathcal{A}^{+}(p),\quad +\infty \cdot \chi \in \mathcal{A}^{+}(p), \end{equation*} although these measures are ordered. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Adaptations to the Case $\Omega $ Bounded} \label{sec.bdd} We give below the adapations of our results to the case when equation (\ref {eq1}) is considered in $Q_{T}=\Omega \times (0,T)$, with $\Omega $ open, bounded and regular. Actually, the main difference is the existence of the so-called \textsl{Friendly Giant}, which was studied first by Dahlberg and Kenig \cite {DK3} for the purely diffusive equation. We will use the same definition of weak solution with zero boundary data: $u\in \mathrm{C}^{0}(0,T;\mathrm{C}% ^{0}(\overline{\Omega }))\cap L_{\rm loc}^{1}(0,T;L^{1}(\Omega ,\delta ))$, where $\delta (x)=\mathrm{\rm dist}(x,\partial \Omega )$, and for any $\varphi \in \mathrm{C}^{0}((0,T)\times \overline{\Omega })$, with compact support in some time interval $I\subset (0,T)$, \begin{equation*} \iint\nolimits_{Q_{T}}\{-u\varphi _{t}-u^{m}\Delta \varphi +u^{p}\varphi \}=0. \end{equation*} The limit of fundamental solutions in the range $0\leqslant p\leqslant m$ is thus the Friendly Giant: \begin{theorem} Let $\Omega \subset \mathbb{R}^{N}$ be open, bounded, regular and $% 0\leqslant p\leqslant m$. Then the following limit holds: \begin{equation*} v_{c}\rightarrow \mathcal{V}_{p}\quad \text{locally uniformly\quad in\quad }% \overline{\Omega }\times (0,T), \end{equation*} where $\mathcal{V}_{p}$ is the so-called ``Friendly Giant'', which has the following properties: $\mathcal{V}_{p}$ takes on the initial trace $\nu =+\infty $ in $\Omega $, $\mathcal{V}_{p}\in \mathrm{C}^{0}((0,T)\times \overline{\Omega })$, $\mathcal{V}_{p}=0$ on $\partial \Omega \times (0,T)$. \end{theorem} \paragraph{Proof.} It is clear by uniqueness of the $v_{c}$'s we have comparison with the Friendly Giant for $u_{t}=\Delta u^{m}$, that is: \begin{equation*} v_{c}(x,t)\leqslant t^{-\frac{1}{m-1}}f(x), \end{equation*} where $f$ satisfies $-\Delta f-1/(m-1)f=0$ with zero boundary conditions on $% \partial \Omega $. This gives a universal bound up to the boundary, hence by equi-continuity results \cite{DiB}, $v_{c}$ will converge in $\mathrm{C}% ^{0}((0,T)\times \overline{\Omega })$ to some $\mathcal{V}_{p}$. Now, we are left to show that the initial trace of $\mathcal{V}_{p}$ is $+\infty $. We argue as in \cite{EC2}, with a little modification. For $k>1$, let $T_{k}$ be the transformation \begin{equation*} T_{k}(u)(x,t)=k^{1/(m-1)}u(x,kt). \end{equation*} The set $\Omega $ is invariant under $T_{k}$, and $T_{k}$ maps $v_{c}$ into a sub-solution with initial data $ck^{1/(m-1)}\delta _{y}$. Indeed, since $% p\leqslant m$ and $k>1$, we have \begin{equation*} \frac{\partial }{\partial t}T_{k}(v_{c})-\Delta T_{k}(v_{c})^{m}+T_{k}(v_{c})^{p}=k^{m/(m-1)}u^{p}-k^{p/(m-1)}u^{p}\leqslant 0. \end{equation*} By uniqueness results in $\Omega $ and zero lateral data, we know that $v_{ck^{1/(m-1)}}\geqslant T_{k}(v_{c})$, and thus in the limit as $c\rightarrow \infty $, we obtain \begin{equation*} \mathcal{V}_{p}(x,t)\geqslant k^{1(m-1)}\mathcal{V}_{p}(x,kt). \end{equation*} So taking $k=t^{-1}$, which is greater than $1$ for small $t$, it turns out that \begin{equation*} \mathcal{V}_{p}(x,t)\geqslant t^{-1/(m-1)}\mathcal{V}_{p}(x,1). \end{equation*} In fact, it is clear by positivity properties that $\mathcal{V}_{p}(x,1)>0$ at least on some open set $B_{\eta }(0)\subset \Omega $, which proves that the singular set of the initial trace of $\mathcal{V}_{p}$ contains $B_{\eta }(0)$. Now for any $z\in B_{\eta }(0)$, we can easily prove by comparison that for any $c>0,$% \begin{equation*} \mathcal{V}_{p}(x,t)\geqslant v_{c}(x-y,t), \end{equation*} so that $\mathcal{V}_{p}(x,t)\geqslant \mathcal{V}_{p}(x-y,t)$. Hence, the singular set of $\mathrm{\rm tr}_{\Omega }(\mathcal{V}_{p})$ contains also $B_{2\eta }(0)\cap \Omega $, and by induction, we deduce that finally $\mathrm{\rm tr}_{\Omega }(\mathcal{V}_{p})=\Omega $. \hfill$\square$ The trace that we defined in $\mathbb{R}^{N}$ is based on purely local arguments, so that it is also valid in $\Omega $ bounded. Our solutions are clearly comparable with the solutions of the diffusive equation $% u_{t}=\Delta u^{m}$, so that by the results of \cite{DK3}, the following result holds: \begin{theorem} If $0\leqslant p\leqslant m$, the initial trace of any solution $u\not\equiv \mathcal{V}_{p}$ in $\Omega $ is a Radon measure $\mu $ in $\Omega $ which satisfies \begin{equation*} \int_{\Omega }\mathrm{\rm dist}(x,\partial \Omega )\,d\mu (x)<\infty \end{equation*} \end{theorem} For $0\leqslant p\leqslant m$, the operator $\mathbb{P}_{p}$ is constructed as in $\Omega =\mathbb{R}^{N}$. \begin{theorem} Let $0\leqslant p\leqslant m$. Then the operator $\mathbb{P}_{p}$ is a projection in $\mathcal{B}^{+}(\Omega )$, $i.e$. $\mathbb{P}_{p}\circ \mathbb{P}_{p}=\mathbb{P}_{p}$. Moreover, \begin{itemize} \item $\mathbb{P}_{p}(\nu ) =\nu$ if and only if $\nu$ is admissible, and then $u[\nu ]$ is the unique solution with trace $\nu$, \\ \item $\mathbb{P}_{p}(\nu ) =+\infty $ if $\nu$ is not admissible and then $u[\nu ]=\mathcal{V}_{p}$ (the Friendly Giant). \end{itemize} \end{theorem} Therefore, for bounded $\Omega$'s, we do not find any case of projection by complete blow-up. The situation for $p>m$ is similar to the case $\Omega =\mathbb{R}^{N}$. Actually, the Friendly Giant gives a bound for $t>0$, up to the boundary which allows to pass to the limit up to the boundary. Thus the limits are still weak solutions and the problem of the initial trace is exactly the same as for $\Omega =\mathbb{R}^{N}$. Thus, all the qualitative aspects of the sets $\mathcal{A}^{+}(p)$ for $p\geqslant m+2/N$ remain valid in $\Omega $. It is important to notice that contrary to what happens in the fast diffusion case $(m<1)$ \cite{CV2}, the initial Borel measure can have singular points at the boundary. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Some open problems and extensions} \label{sec.ope} We list some of the questions that have not been solved in the preceding discussion. \begin{itemize} \item To characterize the admissibility set for $1