\documentclass[twoside]{article} \usepackage{amssymb} % font used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil Singular solutions of doubly singular equations \hfil EJDE--2000/67} {EJDE--2000/67\hfil Yuanwei Qi \& Mingxin Wang \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol.~{\bf 2000}(2000), No.~67, pp. 1--22. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Singular solutions of doubly singular \\ parabolic equations with absorption % \thanks{ {\em Mathematics Subject Classifications:} 35K65, 35K15. \hfil\break\indent {\em Key words:} doubly singular parabolic equation, absorption, singular solutions. \hfil\break\indent \copyright 2000 Southwest Texas State University. \hfil\break\indent Submitted July 15, 2000. Published November 8, 2000. \hfil\break\indent Y.Q. was partially supported by HK RGC grant HKUST630/95P. \hfil\break\indent M.W. was partially supported by PRC grants NSFC-19771015 and 19831060, and \hfil\break\indent by HK RGC grant HKUST630/95P.} } \date{} % \author{ Yuanwei Qi \& Mingxin Wang } \maketitle \begin{abstract} In this paper we study a doubly singular parabolic equation with absorption, $$ u_t = \mathop{\rm div} ( |\nabla u^m|^{p-2}\nabla u^m ) -u^q $$ with $m>0$, $p>1$, $m(p-1)<1$, and $q>1$. We give a complete classification of solutions, which we call singular, that are non-negative, non-trivial, continuous in ${\mathbb R}^n \times [0, \infty)\backslash\{(0,0)\} $, and satisfy $u(x,0)=0$ for all $x\neq 0$. Applications of similar but simpler equations show that these solutions are very important in the study of intermediate asymptotic behavior of general solutions. \end{abstract} \newtheorem{theo}{Theorem}[section] \newtheorem{prop}{Proposition}[section] \newtheorem{lem}{Lemma}[section] \renewcommand{\theequation}{\thesection.\arabic{equation}} \catcode`@=11 \@addtoreset{equation}{section} \catcode`@=12 \section{Introduction} We are interested in the study of singular solutions to the doubly singular parabolic equation with absorption: \begin{eqnarray} u_t = \mathop{\rm div} ( |\nabla u^m|^{p-2}\nabla u^m ) -u^q \quad \mbox{in } {\mathbb R}^n\times(0, +\infty)\,, \label{a1} \end{eqnarray} where $m>0$, $p>1$, $m(p-1)<1$, and $q>1$. Here by a {\bf singular solution} we mean a non-negative and non-trivial solution which is continuous in ${\mathbb R}^n\times [0, +\infty)\backslash\{(0,0)\}$ and satisfies \begin{eqnarray} \lim_{t\searrow 0} \sup_{|x|>\varepsilon} u(x,t)=0 \quad\forall \varepsilon>0\,. \label{a2} \end{eqnarray} A singular solution is called a {\bf fundamental solution} (FS for short) if, for some $c>0$, \begin{eqnarray} \lim_{t\searrow 0} \int_{|x|\leq\varepsilon} u(x,t) dx =c \quad \forall \varepsilon>0\,. \label{t1} \end{eqnarray} A singular solution is called a {\bf very singular solution} (VSS for short) if \begin{eqnarray} \lim_{t\searrow 0} \int_{|x|\leq\varepsilon} u(x,t) dx =\infty \quad \forall ep>0\,. \label{a3} \end{eqnarray} By a {\bf self-similar solution} we mean a solution $u$ that has the form \begin{eqnarray} u(x,t)=\left (\frac{\alpha}{t} \right)^{\alpha}f \left (|x|\left(\frac{\alpha}{t} \right)^{\alpha \beta}\right), \quad\alpha:= \frac {1}{q-1}, \quad\beta:=\frac{q-m(p-1)}{p}, \label{a4} \end{eqnarray} where $f$ as a function of $r=|x|(\alpha /t)^{\alpha \beta}$ is defined on $[0, +\infty)$ and solves \begin{eqnarray} (|(f^m)'|^{p-2}(f^m)')'+\frac{n-1}{r} |(f^m)'|^{p-2}(f^m)'+\beta rf'+f-f^q=0 \quad \forall r>0. \label{a5} \end{eqnarray} Note that for $u$ given by (\ref{a4}), the condition (\ref{a2}) is equivalent to \begin{eqnarray} \lim_{r\to \infty} r^{1/\beta}f(r)=0. \label{a6} \end{eqnarray} Furthermore, if $q0,\ \ p>1$ satisfying $m(p-1)<1$ and $q>1.$ More importantly, we obtain the existence and uniqueness of both FS and VSS, self-similar or otherwise. Our main results read as follows: \begin{theo} Assume that $m>0$, $p>1$, $m(p-1)<1$ ,and $q>1$. Then the following statements hold: \begin{description} \item{\bf (i)} Every singular solution of (\ref{a1}) is either an FS or a VSS; \item{\bf (ii)} When $q\geq m(p-1)+p/n$, (\ref{a1}) does not have any singular solution; \item{\bf (iii)} When $q0$, a unique FS, $u_c$, with initial mass $c$. In addition, $u_{c_1}n$ and $k>0$, which is equivalent to $p>n(1+m)/(1+mn)$. \smallskip We note in passing that the present case of (\ref{a1}) is very different from the case of $m(p-1)>1$, which is similar to (\ref{a7}) with $m>1$ or (\ref{a8}) with $p>2$. In particular, when $m(p-1)>1$, there exist compact supported solutions and such solutions have finite speed of propagation. Whereas for our case, the propagation speed is infinite and any nontrivial, nonnegative solution has ${\mathbb R}^n$ as its support for $t>0$. As a matter of fact, the major effort is given to estimate the decay of singular solutions at $|x|=\infty$. Once we can do that, a lot of techniques in \cite{KP1}-\cite{KV1} which were developed for degenerate equations such as (\ref{a7}) with $m>1$ and (\ref{a8}) with $p>2$ can be adapted to study the present singular case. \section{Existence and Uniqueness of Very Singular Self-similar Solution} In this section we study (\ref{a5}) and prove the existence and uniqueness of very singular self-similar solution. Our proof of existence is different from the one given in \cite{L2}. In particular, through the classification of solutions in relation to their initial values, we prove the existence of self-similar VSS, rather than the shooting argument employed in \cite{L2}. We consider the solution of (\ref{a5}) with initial value \begin{eqnarray} f(0)=a, \quad f'(0)=0. \label{b1} \end{eqnarray} For each $a>0$, (\ref{a5}), (\ref{b1}) has a unique solution $f(r;a)$ and the solution is continuously differentiable in $a$ in a right neighbourhood of $ r > 0$ (see Proposition 1 in Appendix). Since $a\geq 1$ implies that $f'\geq 0$ in its existence interval, we need only consider the case $a\in(0, 1)$. For $a\in (0, 1)$, if we denote by $(0, R(a))$ the maximal existence interval where $f>0$, then $f'<0$ in $(0, R(a))$ and either (i) $R(a)=\infty$ and $\lim_{r\searrow \infty}f(r; a)=0$, or (ii) $R(a)<\infty$ and $f(R(a);a)=0$. The main results of this section read as follows. \begin{theo} Assume that $m>0,$ $p>1$, $m(p-1)<1$ and $q>1$. For each $a\in(0, 1)$, let $f(r; a)$ be the solution of (\ref{a5}), (\ref{b1}). Then the following conclusions hold: \begin{description} \item{\bf (i)} If $n\beta\geq 1$, then $f>0$ and $f'<0$ in $(0, \infty)$ and $\lim\inf_{r\to \infty} r^{1/\beta}f(r; a)>0$. \item{\bf (ii)} If $n\beta<1$, then there exists $a^*\in (0, 1)$ such that the following classification is valid: \\ {\bf (a)} If $a\in(0, a^*)$, then there exists $R(a)<\infty$ such that $f'<0$ in $(0, R(a)]$ and $f(R(a); a)=0$. \\ {\bf (b)} If $a\in(a^*, 1)$, then $f'<0, \ \ f>0, \ \ f_a:=\frac{d}{da}f>0$, and $(r^{\mu}f)'>0$ in $(0, \infty)$. In addition, $\lim_{r\to \infty}r^{1/\beta}f(r; a)$ has a finite limit $k(a)$ which, as a function of $a$ defined on $(a^*, 1)$, is positive, continuous and strictly increasing, and satisfies $\lim_{a\searrow a^*}k(a)=0$ and $\lim_{a\nearrow 1}k(a)= \infty$. \\ {\bf (c)} If $a=a^*$, then $\lim_{r\to \infty}r^{1/\beta}f(r; a)=0$, $\lim_{r\to\infty}r^{m\mu}f^m(r; a)=F^*$, where $$ F^{*}=\left\{\frac{(m\mu)^{p-1}(\mu-n) [1-m(p-1)]}{q-1}\right\}^{m/[1-m(p-1)]}$$ \end{description} \end{theo} \subsection*{Nonexistence Results} Now we prove nonexistence results of very singular self-similar solutions. We note the same result was proved in \cite{L2}. For completeness we give a simple proof here. \paragraph{Proof of Theorem 2.1(i).} Multiplying (\ref{a5}) by $r^{1/\beta-1}$ we have, for $r$ in $(0, R(a))$, $$ (r^{1/\beta-1}|(f^m)'|^{p-2}(f^m)' +\beta r^{1/\beta}f)'=(n-1/\beta)r^{1/\beta-2} |(f^m)'|^{p-1}+r^{1/\beta-1}f^q>0 $$ since $n\beta\geq 1$. Thus the function $g(r):=r^{1/\beta-1} |(f^m)'|^{p-2}(f^m)'+\beta r^{1/\beta}f$ is strictly increasing in $(0, R(a))$. Note that $\lim_{r\searrow 0}g(r)=0$, we get $g>0$ in $(0, R(a))$. Since $f'<0$ we conclude that $R(a)=\infty$ and $f\searrow 0$ as $r\nearrow \infty$. Since $g(r)$ is increasing, $$\lim_{r\to\infty}(r^{1/\beta-1} |(f^m)'|^{p-2}(f^m)'+\beta r^{1/\beta}f)=\lim_{r\to\infty}g(r)=g_{\infty} $$ exists, where $g_{\infty}$ is either a positive constant or $\infty$. Thus, $\lim\inf_{r\to\infty}r^{1/\beta}f>0$. This completes the proof. \hfill {\tiny {\small Q.E.D.}} \subsection*{A Monotonicity Lemma} Observe that $f=f(r;a)$ satisfies \begin{eqnarray} -|(f^m)'|^{p-2}(f^m)'=\beta rf+r^{1-n}\int_{0}^{r}s^{n-1} [1-n\beta-f^{q-1}]f ds. \label{b2} \end{eqnarray} Since $f'<0$, we have that $$ -(f^m)'=\left (\beta rf+r^{1-n}\int_{0}^{r}s^{n-1} [1-n\beta-f^{q-1}]f ds\right)^{1/(p-1)}. $$ Using $f(0)=a$ it follows that as $r\searrow 0$, \begin{eqnarray} f^m(r;a)=a^m-\frac{p-1}{p}(a-a^q)^{1/(p-1)}n^{-1/(p-1)}r^{p/(p-1)} (1+\circ (r))\,. \label{b3} \end{eqnarray} To study the behavior of the solution $f(r;a)$, we introduce a function $F= F(r;a)$ defined by \begin{eqnarray} F(r;a):=\{r^{\mu}f(r;a)\}^m, \quad\mbox{where} \quad \mu=\frac{p}{1-m(p-1)}>0. \label{b4} \end{eqnarray} Then we have $(f^m)'=r^{-m\mu-1}(rF'-m\mu F)$ and \begin{eqnarray*} \lefteqn{ (|(f^m)'|^{p-2}(f^m)')'}\\ &=&(m\mu+1)(p-1)r^{-(m\mu+1)(p-1)-1} |m\mu F-rF'|^{p-2}(m\mu F-rF')\\ &&-(p-1)r^{-(m\mu+1)(p-1)} |m\mu F-rF'|^{p-2}(m\mu F'-F'-rF''). \end{eqnarray*} Since $\mu=p/(1-m(p-1))$, substituting the above expressions into (\ref{a5}) gives \begin{eqnarray} (p-1)r^2F''+[n-1-2(p-1)m\mu]rF'+m\mu(\mu-n)F&& \label{b5}\\ +(m\mu F-rF')^{2-p}\{\frac{\beta}{m}rF'F^{(1-m)/m}+(1-\beta \mu) F^{1/m}-r^{\mu (1-q)}F^{q/m}\}&=&0\,. \nonumber \end{eqnarray} In addition, a differentiation in $a$ gives, for $F_a:=\frac{\partial F} {\partial a}$, \begin{eqnarray} {\cal L}(F_a)&:=&(p-1)r^2F_a''+[n-1-2(p-1)m\mu]rF_a' +m\mu(\mu-n)F_a \nonumber\\ &&+(2-p)(m\mu F-rF')^{1-p}(m\mu F_a-rF_a') \{\frac{\beta}{m}rF'F^{(1-m)/m}\nonumber\\ &&+(1-\beta \mu)F^{1/m} -r^{\mu (1-q)}F^{q/m}\} +(m\mu F-rF')^{2-p}\nonumber\\ &&\times\Big\{\frac{\beta}{m}rF_a'F^{(1-m)/m} +\frac{\beta(1-m)}{m^2}rF'F^{(1-2m)/m}F_a \nonumber\\ &&+\frac{1-\beta \mu}{m} F^{(1-m)/m}F_a-\frac{q}{m}r^{\mu (1-q)}F^{(q-m)/m}F_a\Big\}=0\,. \label{b6} \end{eqnarray} \begin{lem} If $F'>0$ in a finite interval $(0, r_1)$, then $\mu aF_a>rF'$ in $(0, r_1)$ and $F_a>0$ in $(0, r_1]$. \end{lem} \paragraph{Proof.} Applying the differential operator $r\frac{d}{dr}$ to (\ref{b5}) and using the identity $r[r^2F'']'= r^2[rF']''$, we get $$ {\cal L}(rF')=\mu(1-q)(m \mu F-rF')^{2-p}r^{\mu(1-q) }F^{q/m}<0 \quad\mbox{in} \ \ (0, R(a)). $$ In the interval $(0, r_1)$, write $F_a=C(r)rF'$. Using the expansion (\ref{b3}) we have that, for all $r$ sufficiently small, \begin{eqnarray*} C(r)&=&(\mu a)^{-1}\Big\{1+ \frac{q-1}{mp}n^{-1/(p-1)}a^{q-m}(a-a^q)^{(2-p)/(p-1)}r^{p/(p-1)}\\ &&\quad+O(r^{2p/(p-1)})\Big\}. \end{eqnarray*} It then follows that $C(0)=(\mu a)^{-1}$, and $C'(r)>0$ near the origin. Substituting $F_a=C(r)rF'$ into (\ref{b6}) we find $$ (p-1)r^2C''[rF']+ C'[\cdots]+C{\cal L}(rF')=0. $$ Because ${\cal L}(rF')<0$ and $rF'>0$ in $(0, r_1)$, we know that $C'(r)$ can not attain its first zero in $(0, r_1)$. Hence $C' (r)>0$ in $(0, r_1)$. Consequently, $F_a=C(r)rF'>(\mu a)^{-1}rF'>0 $ in $(0, r_1)$. It remains to show that $F_a>0$ at $r_1$. For later application, here we provide an elaborated proof. Let $r_0=\mbox{min}\{1, r_1/2\}$ and $\psi$ be the solution to ${\cal L}(\psi)=0$ in $(0, R(a))$ with the initial values $\psi(r_0) =0$ and $\psi'(r_0)=1$. Then $\psi>0$ in $(r_0, r_1]$ since between any two zeros of $\psi$ there is a zero of $F_a$. Set $k_0= C'rF'|_{r=r_0}>0$ and $c_0=C(r_0)$. We consider the function $\phi=F_a-k_0\psi$. It is obvious that ${\cal L}(\phi)=0$ in $(0, R(a))$. In addition, at $r=r_0, \quad\phi=F_a=c_0rF'$ and $\phi'=\{C' rF'+C(rF')'-k_0\psi'\}|_{r=r_0} =c_0(rF')'|_{r=r_0}.$ Writing $\phi=\overline{C}(r)rF'$, we have that $\overline{C}(r_0)=c_0, \quad\overline{C}' (r_0)=0$, and $\overline{C}$ satisfies the same equation as that for $C$. As $\overline{C} ''(r_0)>0$ (from the differential equation), we get that $\overline{C}'>0$ in $(r_0, r_1)$. Therefore $\phi=\overline{C}(r)rF'>0$ in $[r_0, r_1)$. Consequently, $F_a\geq k_0\psi>0$ in $(r_0, r_1]$. This completes the proof of the lemma. \hfill {\small Q.E.D.}\medskip For convenience, we denote \begin{eqnarray*} {\cal A}&=&\big\{a\in(0, 1): \mbox{there exists $R_1(a)\in(0, R(a))$ such that $F'(R_1(a); a)=0$}\big\}\\ {\cal B}&=&\big\{a\in(0, 1): F'(\cdot; a)>0 \quad\mbox{in } (0, \infty), \quad\lim_{r\to\infty}F(r; a)<\infty\big\}\\ {\cal C}&=&\{a\in(0, 1): F'(\cdot; a)>0 \quad\mbox{in } (0, \infty), \quad\lim_{r\to\infty}F(r; a)=\infty\big\}. \end{eqnarray*} Since $F'(r; a)>0$ near the origin, then $F'(r; a)>0$ in $(0, R(a))$ if $a\in (0, 1)$ is not in ${\cal A}$, this implies that $R(a)=\infty$, so that $a\in {\cal B}\bigcup{\cal C}$. Thus, ${\cal A}, {\cal B}$ and ${\cal C}$ are disjoint with each other and ${\cal A}\bigcup{\cal B}\bigcup{\cal C}=(0, 1)$. \subsection*{Characterization of the set ${\cal A}$} \begin{lem} Let $a\in (0, 1)$. Then the following statements are equivalent: \begin{description} \item{\bf (i)} $a\in {\cal A}$; \item{\bf (ii)} there exists $R_1\in(0, R(a))$ such that $F'(r; a)>0$ in $(0, R_1(a))$,\\ $F''(R_1(a); a)<0$, and $F'(r; a)<0$ in $(R_1(a), R(a))$; \item{\bf (iii)} {\tt sup}$_{r\in (0, R(a))}F(r; a)0$; \item{\bf (v)} $R(a)<\infty$ and $(f^m)'(R(a); a)<0$; \item{\bf (vi)} $R(a)<\infty$. \end{description} \end{lem} \paragraph{Proof.} {\bf (i)} $\Rightarrow$ {\bf (ii)}. Let $(0, R_1(a))$ be the maximal interval where $F'>0$. Since $a\in{\cal A}$, $R_1(a)0$ in $(0, R_1(a))$. Therefore, $F'' (R_1(a); a)<0$. Next we show that $F'(r; a)<0$ in $(R_1(a), R(a))$. In fact, if this is not true, then there exists $R_2(a)\in(R_1(a), R(a))$ such that $F'(R_2(a); a) =0$ and $F'(r; a)<0$ in $(R_1(a), R_2(a))$. Evaluating (\ref{b5}) at $r=R_1(a)$ with \\ $F'(R_1(a); a)=0$ and $F''(R_1(a); a)<0$, and at $r=R_2(a)$ with $F'(R_2(a); a)=0$ and $F''(R_2(a); a)\geq 0$, and using the definition $F=r^{m\mu}f^m$, we obtain \begin{eqnarray} \lefteqn{ \{\frac{q-1}{1-m(p-1)}+f^{q-1}\}F^{(1-m(p-1))/m}|_{r=R_1(a)} }\nonumber\\ &<& (m\mu)^{p-1}(\mu-n) \label{b7}\\ &\leq& \{\frac{q-1}{1-m(p-1)}+f^{q-1}\}F^{(1-m(p-1))/m}|_{r=R_2(a)}.\nonumber \end{eqnarray} However, this is impossible since $F(R_1(a); a)>F(R_2(a); a)$ and $f(R_1(a); a)> f(R_2(a); a)$. Hence $F'(r; a)<0$ in $(R_1(a), R(a))$.\medskip {\bf (ii)} $\Rightarrow$ {\bf (iii)}. Note that the maximum of $F$ is obtained at $r=R_1(a)$, so the assertion follows from the first inequality of (\ref{b7}). \medskip {\bf (iii)} $\Rightarrow$ {\bf (iv)}. Assume for the contrary that $\int_{0}^{r}s^{n-1}(1- n\beta-f^{q-1})f ds\leq 0$ for all $r\in (0, R(a))$. Then from (\ref{b2}) we have that $-|(f^m)'|^{p-2}(f^m)'\leq \beta rf$, i.e. $-(f^m)'\leq (\beta rf)^{1/(p-1)}$ for all $r\in(0, R(a))$. Upon integrating this inequality over $(0, r)$ we have $$ f(r; a) \geq \left(a^{[m(p-1)-1]/(p-1)}+\frac{1-m(p-1)}{mp}\beta^{1/(p-1)} r^{p/(p-1)}\right)^{(p-1)/[m(p-1)-1]} $$ for all $r\in(0, R(a))$. Then it follows that $R(a)=\infty$ and, using (\ref{b4}), $\hat{F}: =\liminf_{r\to \infty}F>0$. Note that either $F'>0$ in $(0, \infty)$, or if $F'$ changes sign, then $a\in {\cal A}$ and hence $F'<0$ in $(R_1(a), \infty)$. In either case we have $\lim_{r\to \infty}F=\hat{F}$ and $\liminf_{r\to \infty} |rF'|=0$. Let $\{r_j\}_{j=1}^{\infty}$ be a sequence with $\lim_{j\to \infty} r_j=\infty$ and $\lim_{j\to \infty}(rF')|_{r=r_j}=0$. We claim that $\{r_j\}$ can be chosen such that in addition $\lim_{j\to \infty} (r^2F'')|_{r=r_j}=0$. In fact, if $|rF'|$, which is positive for all large $r$, oscillates infinitely many times, then one can choose $\{r_j\}$ to be the local minimum points of $|rF'|$ so that $0=(rF')'=rF'' +F'$ on $\{r_j\}$. That is, \ $$\lim_{j\to \infty}(r^2F'')|_{r=r_j}=-\lim_{j\to \infty}(rF')|_{r=r_j}=0\,. $$ If $|rF'|$ does not oscillate infinitely many times, then $|rF'|$ eventually monotonically decreases to zero. So that, one can choose $\{r_j\}$ along which $r(|rF'|)'$ approaches zero, namely, $r^2F''=r(rF')'-rF'$ approaches zero along the sequence $\{r_j\}$. Now evaluating (\ref{b5}) at $r_j$ and sending $j\to \infty$ we obtain $\hat{F}=F^{*}$, which is a contradiction to the assumption $\sup_{r\in (0, R(a))}F0$ we have that $z>0$ for all $r\in[r_1, R(a))$. It then follows that for some $\delta >0, \ \ \int_{0}^{r}s^{n-1} (1-n\beta-f^{q-1})f ds\geq \delta$ in $[r_1, R(a))$. From (\ref{b2}) we have \begin{eqnarray} -|(f^m)'|^{p-2}(f^m)'\geq \beta rf+\delta r^{1-n} \quad \forall r\in[r_1, R(a)). \label{b8} \end{eqnarray} Since $10$, one can choose $\varepsilon: 1-m(p-1)<\varepsilon<\min\{1, p/n\}$. Therefore, $0<\varepsilon <1$ and satisfies \begin{eqnarray} m-(1-\varepsilon)/(p-1)>0, \quad 1+(1-n\varepsilon)/(p-1)>0. \label{b9} \end{eqnarray} Using the inequality $\beta rf+\delta r^{1-n} \geq (\beta rf)^{1-\varepsilon}(\delta r^{1-n})^{\varepsilon} =\beta ^{1-\varepsilon}\delta^{\varepsilon}f^{1-\varepsilon}r^{1-n\varepsilon}$, we obtain from (\ref{b8}) that $$ |(f^m)'|^{p-1} \geq \beta ^{1-\varepsilon} \delta^{\varepsilon}f^{1-\varepsilon}r^{1-n\varepsilon} \quad\mbox{for all } r\in(r_1, R(a)). $$ Due to $(f^m)'<0$, we have $$ -mf^{m-1-(1-\varepsilon)/(p-1)}f'\geq Cr^{(1-n\varepsilon)/(p-1)}. $$ Integrating this inequality over $[r_1, r)$, $rF^*\,.$$ \end{lem} \paragraph{Proof.} The only if part follows from the definition of ${\cal C}$. \\ If $\sup_{r\in(0, R(a))}F(r; a)>F^{*}$, then by Lemma 2.2 (iii), $a\not\in{\cal A}$, and so $a\in {\cal B}\bigcup {\cal C}$. However, if $\hat{F}:= \lim_{r\to\infty}F(r; a)$ is finite, then a sequence $\{r_j\}$ can be found along which $rF'$ and $r^2F''$ approach zero. It follows from (\ref{b5}) that $\hat{F}=F^{*}$. This contradicts the assumption that $\sup_{r\in(0, R(a))}F(r; a)>F^{*}$. \hfill {\small Q.E.D.} \begin{lem} There exists $a^{*}\in(0, 1)$ such that ${\cal C}=(a^{*}, 1)$. In addition, for every $a\in{\cal C}$ there exists $k(a)>0$ such that $$ \lim_{r\to \infty}r^{1/{\beta}}f(r; a)=k(a). $$ Furthermore, $k(a),$ as a function of $a\in(a^{*}, 1)$, is positive, continuous, strictly increasing, and $$ \lim_{a\searrow a^{*}}k(a)=0, \quad\lim_{a\nearrow 1}k(a)=\infty. $$ \end{lem} \paragraph{Proof. Step 1:} We first prove that ${\cal C}$ is open and non-empty. Since, $a\in{\cal C}$ if and only if $\sup_{r\in(0, R(a))}F(r; a)>F^{*}$, by the continuous dependence of initial data, ${\cal C}$ is open. In view of $\lim_{a\nearrow 1}f(r; a)=f(r; 1) \equiv 1$ uniformly in any compact subset of $[0, \infty)$ and\\ $\lim_{a\nearrow 1}F((2F^{*})^{1/(m\mu)}; a)=2F^{*}$, we have $(1-\varepsilon, 1)\subset {\cal C}$ for some sufficiently small positive $\varepsilon $. Because ${\cal A}=(0, a_{*}), \ \ [a_{*}, 1)\subset {\cal B}\bigcup {\cal C}$, we know that $F'(r; a)>0$ for all $r\in(0, \infty)$ and all $a\in [a_{*}, 1)$. Consequently, by Lemma 2.1, $F_a(r; a)>0$ for all $r\in(0, \infty)$ and all $a\in [a_{*}, 1)$. This implies that ${\cal C}=(a^{*}, 1)$ where $a^*=\inf \{a\geq a_* |\lim_{r\to \infty}F(r; a)>F^*\}$. As a by-product, ${\cal B}=[a_*, a^*]=\{a| R(a)=\infty, \mbox{and} \ \ F(r; a) \nearrow F^* \quad\mbox{as} \ \ r\to \infty \}$. \medskip \noindent{\bf Step 2.} We are now in a position to study the behavior of the solution $f(\cdot; a)$ for $a\in {\cal C}$. For simplicity, we write $f(r; a)$ and $F(r; a)$ as $f(r)$ and $F(r)$ respectively. It is convenient to use the variable $s=\ln r$. Because $f$ is positive, we can write $f(e^s)=f(1)\exp(-\int_0^s G(\sigma)d\sigma)$. Since $f'<0$ and $F'=r^{m\mu-1}[r(f^m)' +m\mu f^m]>0$ for all $r>0$, we get $00$ there exists $s_{\varepsilon}>0$ such that $H(G, s)>0$ for all $G\in(0, (1-\varepsilon)/\beta]$ and $s>s_{\varepsilon}$; $H(G, s)<0$ for all $G\in((1+\varepsilon)/\beta, \mu]$ and $s>s_{\varepsilon}$. It then follows from an invariant region argument that $$ \lim_{s\to \infty}G(s)=1/{\beta}. $$ \noindent{\bf Step 3.} We show that, as $s\to \infty, \ \ G(s)$ approaches $1/{\beta}$ exponentially fast, with an exponent at least $\nu=\frac{1}{2}\min\{\frac{q-1}{\beta}, (1-m(p-1))(\mu-\frac{1}{\beta})\}$. Considering the function $G^{-}(s)=\frac{1}{\beta}[1-\frac{1}{2}e^{\nu (S-s)}]$ defined on $[S, \infty)$. We want to prove that $G^{-}(s)$ is a sub-solution of the equation $(p-1)\dot{G}=H(G, s)$ in $[S, \infty)$ provided that $S$ is sufficiently large. For this purpose, first, we let $S$ be large enough such that $f^{q-1}(e^S) <\frac{1}{4}$ and $G(s)>1/(2\beta)$ for all $s>S$. Then $$ f^{q-1}(e^s)= f^{q-1}(e^S)\exp\{-(q-1)\int_S^sG(\sigma)d\sigma\}<\frac{1}{4}e^{\nu (S-s)} \quad\mbox{for all } s\geq S. $$ Next, by taking a larger $S$ if necessary, we assume that $G(s) \leq 1/{\beta}+\frac{1}{2}(\mu-1/{\beta})$ for all $s\geq S$. Then \begin{eqnarray*} F^{(1-m(p-1))/m}(e^s) &=&F^{(1-m(p-1))/m}(e^S)\exp\Big\{[1-m(p-1)]\int_S^s(\mu-G)d\sigma\Big\} \\ &\geq& F^{(1-m(p-1))/m}(e^S)e^{\nu (s-S)} \quad\forall s\geq S. \end{eqnarray*} Hence, \begin{eqnarray*} \lefteqn{ \{1-\beta G^{-}(s)-f^{q-1}(e^s)\}F^{(1-m(p-1))/m}(e^s) }\\ &\geq& [\frac{1}{2}e^{\nu (S-s)}-f^{q-1}(e^s)]F^{(1-m(p-1))/m}(e^s)\\ &\geq& \frac{1}{4}e^{\nu (S-s)}F^{(1-m(p-1))/m}(e^s) \\ &\geq& \frac{1}{4}F^{(1-m(p-1))/m}(e^S) \quad\forall s\geq S\,. \end{eqnarray*} Using the fact that $1/(2\beta)S$ with $S$ large enough, since $F(e^S)\to \infty$ as $S\to \infty$. Here $$ \lambda= \left\{\begin{array}{ll} (2\beta)^{p-2}, & {\rm if } p\leq 2,\\ {\beta}^{p-2}, & {\rm if } p>2. \end{array} \right. $$ Comparing $G(s)$ to $G^{-}(s)$ in $[S, \infty)$ we obtain that $G(s)\geq G^{-}(s) =\frac{1}{\beta}[1-\frac{1}{2}e^{\nu (S-s)}]$ in $[S, \infty)$. In a similar manner we can prove that $G(s)\leq G^{+}(s)= \frac{1}{\beta}[1+\frac{1}{2}(\mu-1/{\beta})e^{\nu (S-s)}]$. Therefore, $|G-1/{\beta}|\leq \frac{1+\mu}{2\beta}e^{\nu(S-s)}$. Consequently, as $r\to \infty$, \begin{eqnarray} r^{1/{\beta}}f(r) & = & f(1)\exp\Big\{-\int_0^{\ln r}(G(\sigma)-1/{\beta})d\sigma \Big\} \nonumber \\ & \to & f(1)\exp\Big\{-\int_0^{\infty}(G(\sigma) -1/{\beta})d\sigma\Big\}=:k(a). \nonumber \end{eqnarray} Since $(f^m)_a=r^{-m\mu}F_a>0$ in $(0, \infty)$, we know that $k(\cdot)$ is positive, continuous, and non-decreasing in $(a^*, 1)$. As $G(s)$ approaches $1/{\beta}$ exponentially fast, we have that $r(r^{1/{\beta}}f)'=O(r^{-\nu})$ as $r\to \infty$. Recall from Lemma 2.1 that $F_a \geq \frac{1}{a\mu}rF'$, which implies that $$ f_a\geq \frac{1}{a\mu}[\mu f+rf'], $$ i.e. $$ (r^{1/{\beta}}f)_a\geq \frac{1}{a\mu}[(\mu-1/{\beta})r^{1/{\beta}}f+r(r^{1/{\beta}}f)']. $$ Hence, for any $a^*1/{\beta}$ and $k(a)>0$, the above inequality show that $k(\cdot)$ is strictly increasing. Now if $\lim_{a\searrow a^*}k(a)>0$, it can be derived that $\sup_{r>0}r^{m\mu} f^m(r; a^*)>F^*$ because $\mu>1/{\beta}$, which would imply that $a^*\in {\cal C}$. It contradicts to the definition of $a^*$. Therefore, $\lim_{a\searrow a^*}k(a)=0$. Finally, if $K=r^{1/{\beta}}f$ achieves a local maximum, say, at $r=r_1$, which is the first one, then at $r=r_1, \ \ K'=0,$ and $K''\leq 0$, \ \ i.e. $\beta rf'+f=0$ and $r^2f''\leq\frac{1+\beta}{\beta ^2}f$. Substituting these two relations into (\ref{a5}) then yields $$ K(r_1; a)0$ for all $r>0$, and $F(r; a)\nearrow F^*$ as $r\nearrow \infty$. It remains to show that $a_*=a^*$. We claim that if $a\in {\cal B}$, then $\lim_{r\to \infty}F_a(r; a) =\infty$. To this aim we use the independent variable $s=\ln r$. Note that $rF'=\dot{F}$ vanishes as $s\to \infty$. The linear operator ${\cal L}$ in (\ref{b6}) takes the form, for $s$ sufficiently large, $$ {\cal L}(\phi)=(p-1)\ddot{\phi}+[b+\circ (1)]\dot{\phi}-[c+\circ(1)]\phi $$ where $\circ(1)\to 0$ as $s\to \infty, \ \ b$ is a certain constant, and $c=\mu(\mu-n)(1+m-mp)>0$ because $10$, it is easy to prove that the solution to ${\cal L}(\phi _1)=0$ in $(S, \infty)$ with initial value $\phi_1(S)=0, \quad\dot{\phi}_1(S)=1$, with $S$ large enough, will have the property that $\phi_1 \to \infty$ exponentially fast as $s\to \infty$. Note that the function $\psi$, constructed in the proof of Lemma 2.1, is positive in $(r_0, \infty)$. As $F_a$ and $\psi$ are linearly independent, one of them will be unbounded. Since $F_a\geq k_0\psi$, we have that $F_a\to \infty$ as $r\to \infty$. Finally we prove that $a^*=a_*$. In fact, if $a^*>a_*$, then by Fatou's lemma, \begin{eqnarray*} 0=\lim_{r\to \infty}(F(r; a^*)-F(r; a_*)) &=&\lim_{r\to \infty}\int _{a_*}^{a^*}F_a(r; a)da\\ &\geq&\int_{a_*}^{a^*}\liminf_{r\to \infty} F_a(r; a)da=\infty, \end{eqnarray*} which is impossible. This completes the proof of the lemma.\hfill {\small Q.E.D.} The proof of Theorem 2.1: follows directly from Lemmas 2.3, 2.5 and 2.6. \section {Existence and Uniqueness of Singular Solutions} In this section we prove the existence and uniqueness of singular solutions of (\ref{a1}), and discuss their properties as well as those of the following equation \begin{equation} u_t = \mathop{\rm div}(|\nabla u^m|^{p-2}\nabla u^m ) \quad \mbox{in } {\mathbb R}^n\times(0, +\infty). \label{d1} \end{equation} \subsection*{Properties of Singular Solutions and Non-existence Results} \begin{lem} Assume that $u$ is a singular solution of (\ref{a1}), or (\ref{d1}). Then either (\ref{t1}) or (\ref{a3}) holds. That is, every singular solution is either an FS or a VSS. \end{lem} Its proof is similar to that of Lemma 2.1 in \cite{CQW3}. We omit the details here. \begin{lem} \begin{description} \item{\bf (i)} If $u$ is singular solution of (\ref{a1}), then for $A:= (\frac{1}{q-1})^{1/(q-1)}$, \begin{equation} u(x, t)\leq At^{-1/(q-1)} \quad \quad\mbox{in } {\mathbb R}^{n}\times(0, \infty). \label{d2} \end{equation} \item{\bf (ii)} If $u$ is a singular solution to (\ref{a1}) or (\ref{d1}), then for $B=\big\{nk(m\mu)^{p-1}\big\}^{\mu/p}$ which is equal to $\big\{[p-n(1-m(p-1))](mp/[1-m(p-1)])^{p-1}\big\}^{1/[1-m(p-1)]}$, we have \begin{equation} u(x, t)\leq B(t^{1/p}|x|^{-1})^{\mu}=Bt^{1/[1-m(p-1)]}|x|^{-p/[1-m(p-1)]} \quad\mbox{in } {\mathbb R}^n\times(0, \infty). \label{d3} \end{equation} \end{description}\end{lem} \paragraph{Proof. (i)} The proof is obvious since $At^{-1/(q-1)}$ is a solution of (\ref{a1}) with initial value $\infty$ in ${\mathbb R}^n$. \medskip \noindent{\bf (ii)} Direct calculation shows that for any $\varepsilon>0$, the function $B(t+\varepsilon)^{\mu/p}(|x|-\varepsilon)^{-\mu}$ is a solution to $w_t= \mathop{\rm div}(|\nabla w^m|^{p-2}\nabla w^m )$ in $\{(x, t)| \ \ |x|>\varepsilon, \ \ t\geq 0\}$. Comparing this function with $u$ in the domain $\{(x, t)| \ \ |x|>\varepsilon, \ \ t\geq 0\}$ then gives $u(x, t)\leq B(t+\varepsilon)^{\mu/p}(|x|-\varepsilon)^{-\mu}$ for all $|x|>\varepsilon, \ \ t>0$. Let $\varepsilon\searrow 0$ then yields $u(x, t)\leq Bt^{\mu/p}|x|^{-\mu}$. The desired results are proved. \begin{lem} If (\ref{a1}) has a singular solution, then it must have a maximal singular solution $u^{*}$ having the following properties: \begin{description} \item{\bf (i)} Every singular solution of (\ref{a1}) is no bigger than $u^{*}$. \item{\bf (ii)} $u^{*}$ is self-similar; namely, there exists a smooth function $f(\cdot):[0, \infty)\to [0, \infty)$ such that $u^{*}=(\alpha/t)^{\alpha}f(|x|(\alpha/t)^{\alpha\beta})$ and $f$ solves (\ref{a5}). \end{description}\end{lem} \paragraph{Proof.} For any $\tau>0$, let $u_{\tau}(x, t)$ be the solution of (\ref{a1}) in ${\mathbb R}^n\times(\tau, \infty)$ with initial value $$u_{\tau}(x,\tau)=\min\{A\tau^{-1/(q-1)}, \ \ B(\tau^{1/p}|x|^{-1})^{\mu}\} \ \quad\mbox{on } {\mathbb R}^n\times\{t=\tau\}. $$ By comparison principle we have \begin{eqnarray} u_{\tau}(x,t)\leq\min\{At^{-1/(q-1)}, \ \ B(t^{1/p}|x|^{-1})^{\mu}\} \ \quad\mbox{on } {\mathbb R}^n\times[t, \infty). \label{s1} \end{eqnarray} Consequently, for any $\tau_1>\tau_2>0$, $u_{\tau_1}(\cdot, \tau_1)\geq u_{\tau_2}(\cdot, \tau_1)$, so that by comparison, $u_{\tau_1}\geq u_{\tau_2}$ in ${\mathbb R}^n\times[\tau_1, \infty)$. Hence, $\lim_{\tau\searrow 0}u_{\tau}$ exists for all $(x, t)\in {\mathbb R}^n\times(0, \infty)$. We denote this limit by $u^{*}$, which is necessarily a solution of (\ref{a1}). Since each $u_{\tau}$ satisfies (\ref{s1}), it follows that $u^{*}(x, t)\leq \min\{At^{-1/(q-1)}, \ \ B(t^{1/p}|x|^{-1})^{\mu}\}$. It then yields that $u^{*}$ satisfies (\ref{a2}). To show that $u^{*}$ is non-trivial, we only need to show that $u^{*}$ is no lees than any singular solution of (\ref{a1}). In fact, if $u$ is a singular solution of (\ref{a1}), then from Lemma 3.2 and comparison principle, $u\leq u_{\tau}$ in ${\mathbb R}^n\times[s, \infty)$ for any $0<\tau\leq s$. Thus, $u\leq u^{*}$ in ${\mathbb R}^n\times (0, \infty)$. Consequently, $u^{*}$ is non-trivial, and is the maximal singular solution of (\ref{a1}) if (\ref{a1}) has one. Due to the symmetry and the scaling invariance $u\to u_{h}(x, t)$ of the equation (\ref{a1}), here $u_h(x,t)=h^{1/(q-1)} u(h^{(q+m-pm)/[p(q-1)]}x, ht)$, we know that the maximal singular solution $u^{*}$ must be self-similar and has the form (\ref{a4}). \hfill {\small Q.E.D.} \begin{theo} \begin{description} \item{\bf (i)} If $q\geq m(p-1)+p/n$, then (\ref{a1}) does not have any singular solution. \item{\bf (ii)} If $p< n(1+m)/(1+mn)$, then neither (\ref{a1}) nor (\ref{d1}) has any singular solution. \end{description}\end{theo} \paragraph{Proof. (i)} By the results of \S 2 (see also \cite{L2}), we know that if $q\geq m(p-1)+p/n$ then (\ref{a5}), (\ref{a6}) has no positive solution. Using the Lemma 3.3 we know that assertion holds. \noindent{\bf (ii)} $p0$, applying Lemma 3.1 we have $$ \int_{|x|\leq 1}u(x, t)dx\leq \int_{|x|\leq 1}Bt^{\mu/p}|x|^{-\mu}dx \leq \frac{1}{n-\mu}B\omega_nt^{\mu/p}\to 0 \quad\mbox{as} \ \ t\searrow 0,$$ where $\omega_n$ is the area of unit sphere in ${\mathbb R}^n$. This contradicts Lemma 3.1. \hfill {\small Q.E.D.} \subsection*{Singular Solutions of (\ref{d1})} \begin{theo} Assume that $p>n(1+m)/(1+mn)$. Then for any $c>0$, (\ref{d1}) has a unique FS with initial mass $c$. It is given by \begin{eqnarray} E_c(x, t):=t^{-nk}\{a+b(|x|t^{-k})^{p/(p-1)}\}^{-\theta}, \label{d4} \end{eqnarray} where $b=k^{1/(p-1)}[1-m(p-1)]/(mp)$, $\theta=(p-1)/[1-m(p-1)]$ and $a= a(c)>0$ is the unique constant such that $\int_{{\mathbb R}^n}(a+b|y|^{p/(p-1)})^{-\theta}dy=c$. \end{theo} \paragraph{Proof.} It is clear that $E_c(x, t)$ is an FS of (\ref{d1}) with initial mass $c$. We need only to prove the uniqueness. Assume that $u$ is any FS of (\ref{d1}) with initial mass $c$, we shall show that $u=E_c$. The proof is divided into three steps. \noindent{\bf Step 1.} Consider the sequence $\{u^h\}_{h>0}$, where $u^h(x, t)=h^{nk}u(h^kx, ht)$. Direct calculation shows that $u^h$ is a solution of (\ref{d1}), and $$ \int_{{\mathbb R}^n}u^h(x, t)dx=\int_{{\mathbb R}^n}u(y, ht)dx=c \quad\forall h>0,\; t>0\,.$$ In view of (\ref{d3}) we have $u^h(x, t)=h^{nk}u(h^kx, ht)\leq Bt^{\mu/p}|x|^{-\mu}$. By the regularity results (see \cite{Zhou}) we know that $\{u^h(\cdot, 1)\}$ is equi-continuous in any bounded domain of ${\mathbb R}^n$, so there exists a subsequence of $\{u^h\}$, denote also by $\{u^h\}$, and a function $u_0$ such that $u^h(\cdot, 1)\to u_0(\cdot)$ as $h\searrow 0$ uniformly in any compact subset of ${\mathbb R}^n$. Since $\mu>n$ and $u^h(x, 1) \leq B|x|^{-\mu}$, the Lebesgue's dominated convergence theorem then gives $u^h(\cdot, 1)\to u_0$ in $L^1({\mathbb R}^n)$. Let $v(x,t)$ be the solution of (\ref{d1}) in ${\mathbb R}^n\times(1, \infty)$ with initial data $v(\cdot, 1)=u_0$. Then the contraction principle yields, for all $t>1$, \begin{eqnarray} \int_{{\mathbb R}^n}|u^h(\cdot, t)-v(\cdot, t)|\leq \int_{{\mathbb R}^n}|u^h(\cdot, 1)-v(\cdot, 1)|\to 0 \quad\mbox{as} \ \ h\to 0. \label{w1} \end{eqnarray} \noindent{\bf Step 2.} Denote, for each $h>0$, \begin{eqnarray} e^h(t)=\int_{{\mathbb R}^n}|u^h(\cdot, t)-E_c(\cdot, t)|. \label{w2} \end{eqnarray} The contraction principle implies that $e^h(t)$ is non-increasing. Because $E_c=E_c^h$, we have \begin{eqnarray*} e^h(t)&=&\int_{{\mathbb R}^n}|u^h(\cdot, t)-E_c^h(\cdot, t)|=h^{nk}\int_{{\mathbb R}^n}|u(h^kx, ht)-E(h^kx, ht)|dx\\ &=&\int_{{\mathbb R}^n}|u(x, ht)-E(x, ht)|dx=e^1(ht). \end{eqnarray*} Thus $e^h(t)$ is non-increasing in both $t$ and $h$. Since the initial mass of $u$ and $E_c$ is $c$, $e^h(t)$ is bounded by $2c$. It then follows that $\lim_{h\searrow 0}e^h(t)$ exists, and $$ \lim_{h\searrow 0}e^h(1)=\lim_{h\searrow 0}e^1(h)=\lim_{h\searrow 0}e^1(2h) =\lim_{h\searrow 0}e^h(2). $$Denote this limit by $e^0$. Then, in view of (\ref{w1}) and (\ref{w2}) we obtain \begin{eqnarray} e^0&=&\lim_{h\to 0}e^h(1)=\lim_{h\to 0} \int_{{\mathbb R}^n}|u^h(\cdot, 1)-E_c(\cdot, 1)|\nonumber\\ &=&\int_{{\mathbb R}^n}|v(\cdot,1)-E_c(\cdot, 1)|\\ &=&\lim_{h\to 0}e^h(2)=\int_{{\mathbb R}^n}|v(\cdot, 2)-E_c(\cdot, 2)|.\nonumber \end{eqnarray} \noindent{\bf Step 3.} We first show that $e^0=0$. Suppose for the contrary that $e^0>0$. We define $\overline{u}$ and $\underline{u}$ as the solution of (\ref{d1}) in ${\mathbb R}^n\times(1, \infty)$ with initial data $$\overline{u}(\cdot, 1):=\max\{v(\cdot, 1),\ E_c(\cdot, 1)\}, \quad \underline{u}(\cdot, 1):=\min\{v(\cdot, 1), \ E_c(\cdot, )\}. $$ Then the comparison principle gives $\overline{u}\geq\max\{v, E_c\}\geq\min\{v, E_c\}\geq\underline{u}$ in ${\mathbb R}^n\times[1, \infty).$ Since $v(\cdot,2)\not\equiv E_c(\cdot,2)$ and $\int_{{\mathbb R}^n}E_c(\cdot,2)=\int_{{\mathbb R}^n}v(\cdot,2)=c$, it follows that, \begin{eqnarray*} \int_{{\mathbb R}^n}[\overline{u}(\cdot, 2)-\underline{u}(\cdot, 2)] &>&\int_{{\mathbb R}^n}[\max\{v(\cdot,2), \ E_c(\cdot, 2)\}-\min\{v(\cdot, 2), \ E_c(\cdot, 2)\}]\\ &=&\int_{{\mathbb R}^n}|v(\cdot, 2)-E_c(\cdot, 2)|=e^0\,. \end{eqnarray*} On the other hand, by the contraction principle, $$ \int_{{\mathbb R}^n}|\overline{u}(\cdot, 2)-\underline{u}(\cdot, 2)|\leq\int_{{\mathbb R}^n}|\overline{u}(\cdot, 1)-\underline{u}(\cdot, 1)|=\int_{{\mathbb R}^n}|v(\cdot, 1)-E_c(\cdot, 1)|=e^0. $$ Here we obtain a contradiction. Therefore, $e^0=0$. Because $e^1(t)$ is non-increasing in $t$, then $0=e^0=\lim_{t\searrow 0}e^1(t)$ implies that $e^1(t)\equiv 0$. Consequently, $u\equiv E_c$. The proof is completed. \hfill {\small Q.E.D.} \subsection*{Singular Solutions of (\ref{a1})} \begin{theo} Assume that $1n(1+m)/(1+mn)$. Then for any $c>0$, (\ref{a1}) has a unique FS, denoted as $u_c$, with initial mass $c$. Moreover, $u_c$ is monotone increasing in $c$ and $u_c\to u_{\infty}$ as $c\to\infty$, and $u_{\infty}$ is a VSS of (\ref{a1}). \end{theo} \paragraph{Proof. Step 1: Existence.} Let $E_c(x, t)$ be given by (\ref{d4}), and $\phi_l(x)=E_c(x, 1/l)$. Then $$ \int_{{\mathbb R}^n}\phi_l(x)dx=c, \quad\mbox{and}\quad\lim_{l\to\infty}\phi_l(x)=0 \quad\forall\ \ x\not=0.$$ That is, $\{\phi_l(x)\}$ is a $\delta -$sequence. Let $u_l(x, t)$ and $w_l(x, t)$ be the solution of (\ref{a1}) and (\ref{d1}) with initial data $\phi_l(x)$ respectively. Because $E_c(x, t+1/l)$ satisfies (\ref{d1}) and has initial data $\phi_l(x)$, by the uniqueness we have $w_l(x, t)=E_c(x, t+1/l)$. From comparison it yields $u_l(x, t)\leq E_c(x, t+1/l)$. This shows that for any $\varepsilon>0$, $\{u_l(x, t)\}$ are uniformly bounded in ${\mathbb R}^n\times[\varepsilon, \infty)$. Consequently, by the regularity results (see \cite{Zhou}) it follows that $\{u_l\}$ is equi-continuous in any compact subset of ${\mathbb R}^n\times (0, \infty)\backslash \{(0, 0)\}$. Hence, there exist a function $u$ and a subsequence, denote still by $\{u_l\}$, such that $u_l\to u$ uniformly in any compact subset of ${\mathbb R}^n\times (0, \infty)\backslash \{(0, 0)\}$. The limit function $u$ is necessarily a (weak) solution of (\ref{a1}) in ${\mathbb R}^n\times(0, \infty)$. Now we show that $u$ is an FS of (\ref{a1}) with initial mass $c$. First, by (\ref{d3}) we have \begin{equation} u_l(x, t)\leq E_c(x, t+1/l)\leq B[(t+1/l)^{1/p}|x|^{-1}]^{\mu}. \label{d6} \end{equation} It follows that $u(x, t)\leq E_c(x, t)$. Therefore, $u$ satisfies (\ref{a2}). Next, by Fatou's lemma we have $$ \int_{{\mathbb R}^n}u(x, t)dx\leq\lim\inf_{l\to\infty}\int_{{\mathbb R}^n}u_l(x, t)dx\leq c \quad \forall \ \ t>0.$$ Now, we prove that for any $\delta>0$, \begin{equation} \lim_{t\searrow 0}\int_{|x|<\delta}u(x, t)dx=c. \label{d5} \end{equation} >From the differential equation of (\ref{a1}) we obtain $$\int_{{\mathbb R}^n}u_l(x, t)dx=\int_{{\mathbb R}^n}\phi_l(x)dx-\int_{0}^{t}\int_{{\mathbb R}^n}u_l^q(x, t)\,dx\,dt= c-\int_{0}^{t}\int_{{\mathbb R}^n}u_l^q(x, t)\,dx\,dt.$$ >From (\ref{d6}) it follows that \begin{eqnarray*} \lefteqn{\int_{0}^{t}\int_{{\mathbb R}^n}u_l^q(x, t)\,dx\,dt}\\ &\leq& \int_{0}^{t}\int_{{\mathbb R}^n} (t+\frac 1l)^{-qnk}\{a+b(|x|(t+\frac 1l)^{-k}) ^{p/(p-1)}\}^{-q\theta}\,dx\,dt\\ &=&\int_{0}^{t}\int_{{\mathbb R}^n} (t+\frac 1l)^{(1-q)nk}\{a+b|x|^{p/(p-1)}\}^{-q\theta}\,dx\,dt\\ &=&C[(t+\frac 1l)^{1-(q-1)nk}-(\frac 1l)^{1-(q-1)nk}], \end{eqnarray*} and $$\int_{|x|>\delta}u_ldx\leq \int_{|x|>\delta}B(t+1/l)^{\mu/p}|x|^{-\mu}dx= C_1\delta^{n-\mu}(t+1/l)^{\mu/p} \quad\mbox{since}\quad\mu>n.$$ Therefore, \begin{eqnarray*} \lefteqn{ \int_{|x|\leq\delta}u_l(x, t)dx}\\ &=&\int_{{\mathbb R}^n}u_l(x, t)dx -\int_{|x|>\delta}u_l(x, t)dx \\ &\geq&\left\{c-C[(t+1/l)^{1-(q-1)nk} -(1/l)^{1-(q-1)nk}]\right\}-C_1\delta^{n-\mu}(t+\frac 1l)^{\mu/p}. \end{eqnarray*} Since $nk(q-1)<1$, as $l\to \infty$ we obtain $$ \int_{|x|\leq\delta}u_l(x, t)dx\geq c-Ct^{1-(q-1)nk} -C_1\delta^{n-\mu}t^{\mu/p}.$$ As $t\searrow 0$ we have (\ref{d5}). Hence, $u$ is an FS with initial mass $c$. \medskip \noindent{\bf Step 2: Uniqueness.} To prove the uniqueness, we first show that for any FS $u$ of (\ref{a1}) with initial mass $c$, $u(x, t)\leq E_c(x, t)$. Let $w(x, t)=B(t^{1/p}|x|^{-1})^{\mu}$, then (\ref{d3}) implies $u(x, t)\leq w(x, t)$. For any $\tau>0$, let $u_{\tau}$ be the solution to (\ref{d1}) for $t>\tau$ with initial value $u_{\tau}=u$ on $\{t=\tau\}$. Then by comparison, $u_{\tau}\geq u$ for all $t>\tau$. Therefore, when $\tau_1 \leq \tau_2$, $u_{\tau_{1}}\geq u_{\tau_2}$ for all $t>\tau_2$, i.e., $\{u_{\tau}\}_{\tau>0}$ is monotone decreasing in $\tau$. Consequently, the limiting function $v=\lim_{\tau\searrow 0}u_{\tau}$ exists. Since $u_{\tau}(x, \tau)=u(x, \tau)\leq w(x, \tau)$ and $w(x, t+\tau)$ satisfies (\ref{d1}). By comparison we have $u_{\tau}(x, t)\leq w(x, t+\tau)$. In view of the regularity of solutions of (\ref{d1}) we conclude that for any $t>0$, $u_{\tau}(\cdot, t)\to v(\cdot, t)$ as $\tau \searrow 0$, uniformly in any compact subset of ${\mathbb R}^n\times(0, \infty)\backslash\{(0, 0)\}$ and in $L^1({\mathbb R}^n)$. Since $\int_{{\mathbb R}^n}u_{\tau}(x, t)dx=\int_{{\mathbb R}^n}u(x, \tau)dx\to c$ as $\tau\to 0$, we assert that $\int_{{\mathbb R}^n}v(x, t)dx=c$ for all $t>0$. Thus, $v$ is an FS of (\ref{d1}) with initial mass $c$. By uniqueness of FS of (\ref{d1}), $v=E_c$. Consequently, $u\leq \lim_{\tau\searrow 0}u_{\tau}=v=E_c$. Let $u_1$ and $u_2$ be any two FSs of (\ref{a1}) with initial mass $c$. Then $u_i\leq E_c$ for $i=1, 2$. In view of contraction principle, for $t>s>0$, \begin{eqnarray*} \lefteqn{\int_{{\mathbb R}^n}|u_1(x, t)-u_2(x, t)|dx}\\ &\leq&\int_{{\mathbb R}^n}|u_1(x, s)-u_2(x, s)|dx\\ &\leq& \int_{{\mathbb R}^n}\{|u_1(x, s)-E_c(x, s)|+|E_c(x, s)-u_2(x, s)|\}dx\\ &=&\int_{{\mathbb R}^n}\{[E_c(x, s)-u_1(x, s)]+[E_c(x, s)-u_2(x, s)]\}dx. \end{eqnarray*} As $s\searrow 0$ we get that $u_1(x, t)=u_2(x, t)$. \medskip \noindent{\bf Step 3.} From the proof of Step 1 and the results of Step 2 we know that $u_c$ is monotone increasing in $c$. \medskip \noindent{\bf Step 4.} By Lemma 3.2 we have $$u_c(x, t)\leq At^{-1/(q-1)}+B(t^{1/p}|x|^{-1})^{\mu} \quad\forall c>0. $$ Similar to the arguments of Step 1 we have that the limit $\lim_{c\to\infty}u_c=u_{\infty}$ exists and $u_{\infty}$ is a VSS of (\ref{a1}). This completes the proof of the theorem. \hfill {\small Q.E.D.} \begin{theo} Assume that $10$ we define the truncated VSS $$ U_{\sigma}(x,t)=\left\{\begin{array}{ll} U(x,t) \quad &\mbox{if } U(x,t)<\sigma, \\ \sigma \quad &\mbox{if } U(x,t)\geq \sigma. \end{array} \right. $$ Because $n\beta<1$ and $U$ is a VSS, we conclude that there exist a sequence $\{\tau (l)\}$ with $\tau (l)\searrow 0$ and the corresponding $\{\sigma(l)\}$ such that $$ \int_{{\mathbb R}^n}U_{\sigma(l)}(x, \tau(l))dx=c. $$ Define $\psi_l(x)=U_{\sigma(l)}(x, \tau(l))$, and let $v_l$ be the solution of (\ref{a1}) with initial value $\psi_l(x)$. Because $\psi_l(x)\leq U(x, \tau(l))$, by the comparison principle we have $v_l(x, t)\leq U(x, t+\tau(l))\leq A(t+\tau(l))^{-1/(q-1)}+B\{(t+\tau(l))^{1/p} |x|^{-1}\}^{\mu}$. Similar to the argument of Step 1 of the proof of Theorem 3.3 we have that limit $\lim_{l\to\infty}v_l=v$ exists and $v$ is an FS of (\ref{a1}) with initial mass $c$, and $v(x, t) \leq U(x, t)$. By the uniqueness of FS of (\ref{a1}) it follows $u_c=v\leq U$. \noindent{\bf Step 2.} By Step 1 we know that the VSS $u_{\infty}$ obtained by Theorem 3.3 is the minimal VSS. Similar to the proof of Lemma 3.3 we have that $u_{\infty}$ is self-similar and has the form (\ref{a4}), and the corresponding function $f$ solves (\ref{a5}). Theorem 2.1 shows that $u_{\infty}=u^{*}$. Therefore, VSS is unique. \hfill {\small Q.E.D.} \section*{Appendix} \setcounter{equation}{0} %\addtocounter{equation}{-10} \renewcommand{\theequation}{A.\arabic{equation}} In this appendix, we show that the initial value problem (1.6), (2.1) has a unique solution in a right neighbourhood of $ r = 0$. \begin{prop} Suppose $ q > 1$, $ m > 0$, $ p > 1$ and $ m(p-1) < 1$. For each $ a \in R$, the following problem \begin{eqnarray} \label{eqa1} ( | ( |f|^{m-1} f)'|^{p-2} ( |f|^{m-1} f)' )' + \frac{n-1}{r} | ( |f|^{m-1} f)'|^{p-2} ( |f|^{m-1} f)' &&\nonumber\\ + \beta r f' + f - |f|^{q-1} f &=& 0 \nonumber \\ ( |f|^{m-1} f)'(0) = 0, \quad f(0) = a\hspace{2cm} && \end{eqnarray} has a unique solution with $ |f|^{m-1} f$ in $C^1$ and $ | ( |f|^{m-1} f)'|^{p-2} ( |f|^{m-1} f)' $ H\"{o}lder continuous in a right neighbourhood of $ r = 0$. Furthermore, $ f $ is continuously differentiable in $ a$ for $ a > 0$ and $ r $ sufficiently small and the following hold: \begin{enumerate} \item If $ a = 0$, then $f \equiv 0$; \item If $ a = 1$, then $ f \equiv 1$; \item If $ a > 1$, then $ f' > 0$. \end{enumerate} \end{prop} \paragraph{Proof:} First we derive an integral equation which is equivalent to the initial value problem (\ref{eqa1}). Integrating (\ref{eqa1}) over $(0, r)$ multiplied by $ r^{n-1}$, we find \begin{eqnarray} \label{eqaa1} - ( | ( |f|^{m-1} f)'|^{p-2} ( |f|^{m-1} f)' ) &= & \beta r f + \frac{1}{r^{n-1}} \int_0^r \rho^{n-1} [ 1 - n \beta - |f|^{q-1}] f d \rho \nonumber\\ & =: & { \cal G}[f](r) \end{eqnarray} Integrating the $ 1/(p-1)-$th power of both sides then gives \begin{equation} \label{eqa2} |f|^{m-1} f= |a|^{m-1} a - \int_0^r | { \cal G}[f](\rho) |^{( 2-p)/(p-1)} { \cal G}[f](\rho) d \rho=: { \cal H}[f](r) \end{equation} or equivalently, $$ % \label{eqa3} f= | { \cal H}[f](r) | ^{(1-m)/m} { \cal H}[f](r). $$ Now we proceed to prove that (\ref{eqa1}) has a uniques solution with the desired smoothness. The first case we consider is $ 1/(p-1) > 1$ and $ m \leq 1$. It is clear that $ | { \cal G}[f](\rho) |^{( 2-p)/(p-1)} { \cal G}[f](\rho) $ is continuously differentiable in $ f$, hence the right hand side of (\ref{eqa2}) is continuously differentiable in $ |f|^{m-1} f$. The existence and uniqueness of solution follows from classical Picard iteration and Gronwall's inequality. Next, we consider the case of $ 1/(p-1) > 1$ and $ m > 1$. For any two continuous functions $ f_1 $ and $ f_2$, \begin{eqnarray*} \lefteqn{ \Big| | { \cal H}[f_1](r) | ^{(1-m)/m} { \cal H}[f_1](r) - | { \cal H}[f_2](r) | ^{(1-m)/m} { \cal H}[f_2](r) \Big| }\\ & \leq& C( \|f_1\|_{ \infty}, \|f_2\|_{ \infty}) | { \cal H}[f_1](r) | ^{p-2} { \cal H}[f_1](r) - | { \cal H}[f_2](r) | ^{p-2} { \cal H}[f_2](r) | \\ &\leq& C( \|f_1\|_{ \infty}, \|f_2\|_{ \infty}) ( \int_0^r | { \cal G}[f_1]( \rho) - { \cal G}[f_2]( \rho) |^{1/(p-1)} d \rho )^{p-1} \end{eqnarray*} by triangle inequality and our assumption that $ m(p-1) < 1$. Again, we have the existence and uniqueness. At last, we deal with the case of $ 1/ (p-1) < 1$ and $ m < 1$. We show $ | { \cal G}[f](r) |^{( 2-p)/(p-1)} { \cal G}[f](r) $ is a Lipschitz continuous function of $ |f|^{m-1} f $. It is clear that we can think of $ { \cal G}[f](\rho) = \int_0^r F d \mu $ with certain measure $ \mu$, where $ F(r)$ is continuously differentiable in $f$. In addition, for any two continuous functions $ f_1 $ and $ f_2$, the corresponding $F_1 $ and $F_2$ satisfy \begin{eqnarray*} \lefteqn{ \| \ | F_1|^{(2-p)/(p-1)} F_1 - | F_2|^{(2-p)/(p-1)} F_2 \|_{\infty} }\\ & \leq& C( \|f_1\|_{ \infty}, \|f_2\|_{ \infty}) \| \ | f_1|^{(2-p)/(p-1)} f_1 - | f_2|^{(2-p)/(p-1)} f_2 \|_{\infty} \\ & \leq& C( \|f_1\|_{ \infty}, \|f_2\|_{ \infty}) \| \ | f_1|^{m-1} f_1 - | f_2|^{m-1} f_2 \|_{\infty}, \end{eqnarray*} again by triangle inequality and our assumption that $ m(p-1) < 1$. If $ F_1, F_2 \geq 0$, then \begin{eqnarray*} \lefteqn{\Big| \ | { \cal G}[f_1](r) |^{( 2-p)/(p-1)} { \cal G}[f_1](r) - | { \cal G}[f_2](r) |^{( 2-p)/(p-1)} { \cal G}[f_2](r) \Big| }\\ & = & | ( \int_0^r F_1 d \mu )^{1/(p-1)}- ( \int_0^r F_2 d \mu )^{1/(p-1)} | \\ & \leq& ( \int_0^r | F_1^{1/(p-1)} - F_2 ^{1/(p-1)} |^{p-1} d \mu )^{1/(p-1)} \\ & \leq& C( \|f_1\|_{ \infty}, \|f_2\|_{ \infty}) \| \ | f_1|^{m-1} f_1 - | f_2|^{m-1} f_2 \|_{\infty}. \end{eqnarray*} In general, if $ a_1 =: \int_0^r F_1^+ d \mu \geq a_2 =: \int_0^r F_1^- d \mu$ and $ b_1=: \int_0^r F_2^+ d \mu \geq a_2 =: \int_0^r F_2^- d \mu$, $ | ( \int_0^r F_1 d \mu )^{1/p-1}- ( \int_0^r F_2 d \mu )^{1/p-1} | = | ( a_1 - a_2)^{1/(p-1)} - ( b_1 - b_2)^{1/(p-1)} | $, by the elementary inequality, \begin{eqnarray*} \lefteqn{ | ( a_1 - a_2)^{1/(p-1)} - ( b_1 - b_2)^{1/(p-1)}| }\\ &\leq& C \left( | a_1^{1/(p-1)} - b_1^{1/(p-1)}| + | a_2^{1/(p-1)} - b_2^{1/(p-1)} | \right), \end{eqnarray*} we reduce this case to the case of $ F_1, F_2 \geq 0$. Finally, if $ a_1 \geq a_2 $ but $ b_2 \geq b_1$, $$ | ( \int_0^r F_1 d \mu )^{1/p-1}- ( \int_0^r F_2 d \mu )^{1/p-1} | = | ( a_1 - a_2)^{1/(p-1)} + ( b_2 - b_1)^{1/(p-1)} |, $$ by the elementary inequality, \begin{eqnarray*} \lefteqn{| ( a_1 - a_2)^{1/(p-1)} + ( b_2 - b_1)^{1/(p-1)}| }\\ &\leq& C \left( | a_1^{1/(p-1)} - b_1^{1/(p-1)}| + | a_2^{1/(p-1)} - b_2^{1/(p-1)} | \right), \end{eqnarray*} we again reduce to the case of $ F_1, F_2 \geq 0$. Therefore, $ | { \cal G}[f](r) |^{( 2-p)/(p-1)} { \cal G}[f](r) $ is a Lipschitz continuous function of $ |f|^{m-1} f $. This complete the part of existence and uniqueness of solution with desired smoothness. The continuous differentiability of $f$ to $a$ when $ a > 0$ and $ r$ sufficiently small follows directly from the positivity of $ f$ and the continuous differentiability of $ |f|^{m-1} f $ to $a$. The assertions (1) and (2) follows from the uniqueness of solution. (3) can be seen directly from (\ref{eqaa1}). This completes the proof of the proposition. \begin{thebibliography}{00} \bibitem{BF} H.Brezis \& A.Friedman, \newblock{\em Nonlinear parabolic equations involving measures as initial conditions}, {J. Math. Pures Appl.} {\bf 62} (1983), 73-97. \bibitem{BPT} H.Brezis, L.A.Peletier \& D. Terman, \newblock{\em A very singular solution of the heat equation with absorption}, { Arch. Rat. Mech. Anal.} {\bf 96} (1986), 185-209. \bibitem{CQW1} X. Chen, Y. Qi \& M. Wang, \newblock{\em Self-similar very singular solutions of the parabolic p-Laplacian}, preprint. \bibitem{CQW2} X. Chen, Y. Qi \& M. Wang, \newblock{\em Existence and uniqueness of singular solutions of a fast diffusion porous medium equation}, preprint. \bibitem{CQW3} X. Chen, Y. Qi \& M. Wang, \newblock{\em Singular solutions of parabolic p-Laplacian with absorption}, preprint. \bibitem{DH} E. DiBenedetto \& M. Herrero, \newblock{\em Non-negative solutions of the evolution p-Laplacian equation. Initial traces and Cauchy problem when $ 1 < p < 2$}, { Arch. Rat. Mech. Anal.} {\bf 111} (1990), 225-290. \bibitem{EK} M. Escobedo \& O. Kavian, \newblock{\em Variational problems related to self-similar solutions of the heat equation,} { Nonlinear Analysis TMA} {\bf 11} (1987), 1103-1133. \bibitem{KP1} S. Kamin \& L. A. Peletier, \newblock{\em Singular solutions of the heat equation with absorption,} { Proc. Amer. Math. Soc.} {\bf 95} (1985), 205-210. \bibitem{KP2} S. Kamin \& L. A. Peletier, \newblock{\em Source-type solutions of degenerate diffusion equations with absorption,} { Isr. J. Math.} {\bf 50} (1985), 219-230. \bibitem{KPV} S. Kamin, L. A. Peletier \& J. Vasquez, \newblock{\em Classification of singular solutions of a nonlinear heat equation,} { Duke Math. J.,} {\bf 58} (1989), 601-615. \bibitem{KV1} S. Kamin \& J. Vasquez, \newblock{\em Singular solutions of some nonlinear parabolic equations,} { J. Analyse Math.} {\bf 59} (1992), 51-74. \bibitem{KV} S. Kamin \& L. Veron, \newblock{\em Existence and uniqueness of the very singular solution for the porous media equation with absorption,} { J. Analyse Math.} {\bf 51} (1988), 245-258. \bibitem{L1} G. Leoni, \newblock{\em A very singular solution for the porous media equation $u_t = \Delta ( u^m) - u^p$ when $ 0 < m < 1$,} { J. Diff. Eqns.} {\bf 132} (1996), 353-376. \bibitem{L2} G. Leoni, \newblock{\em On the existence of fast-decay solutions for a quasilinear elliptic equation with a gradient term,} preprint. \bibitem{O} L. Oswald, \newblock{\em Isolated, positive singularities for a nonlinear heat equation ,} { Houston J. Math.,} {\bf 14} (1988), 543-572. \bibitem{PT} L. A. Peletier \& D. Terman, \newblock{\em A very singular solution of the porous media equation with absorption,} { J. Diff. Eqns.} {\bf 65} (1986), 396-410. \bibitem{PW} L. A. Peletier \& Junyu Wang, \newblock{\em A very singular solution of a quasilinear degenerate diffusion equation with absorption}, Trans. Amer. Math. Soc. {\bf 307} (1988), 813-826. \bibitem{PZ1} L. A. Peletier \& Junning Zhao, \newblock{\em Large time behaviour of solution with absorption: The fast diffusion case}, Nonlinear Anal. TMA, {\bf 17} (1991), 991-1009. \bibitem{PZ2} Junning Zhao, \newblock{\em The asymptotic behaviour of solutions of a quasilinear degenerate parabolic equation}, J. Diff. Eqns., {\bf 102} (1993), 33-52. \bibitem{Zhou} S. L. Zhou, \newblock{\em H\"{o}lder continuity of solutions to second order quasilinear singular parabolic equations}, Ph. D Thesis, Beijing University, 1991. \end{thebibliography} \noindent{\sc Yuanwei Qi }\\ Department of Mathematics\\ Hong Kong University of Science \& Technology \\ Hong Kong,\\ email: maqi@uxmail.ust.hk \smallskip \noindent{\sc Mingxin Wang }\\ Department of Applied Mathematics, Southeast University,\\ Nanjing 210018, P. R. China \\ email: mxwang@seu.edu.cn \end{document}