\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol.~2000(2000), No.~68, pp.~1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu \quad ejde.math.unt.edu (login: ftp)} \thanks{\copyright 2000 Southwest Texas State University.} \vspace{1cm}} \begin{document} \title[\hfilneg EJDE--2000/68\hfil Regularity of the interface] {Regularity of the interface for the porous medium equation} \author[Youngsang Ko\hfil EJDE--2000/68\hfilneg] {Youngsang Ko} \address{Youngsang Ko \hfill\break Department of Mathematics, Kyonggi University, Suwon, Kyonggi-do, 442-760, Korea} \email{ysgo@kuic.kyonggi.ac.kr} \date{} \thanks{Submitted October 7, 2000. Published November 13, 2000.} \thanks{Supported by a research grant form Kyonggi University} \subjclass{35K65} \keywords{porous medium, radial symmetry, interface, C-infinity regularity. } \begin{abstract} We establish the interface equation and prove the $C^{\infty}$ regularity of the interface for the porous medium equation whose solution is radial symmetry. \end{abstract} \maketitle \newtheorem{thm}{Theorem}[section] \newtheorem{lem}[thm]{Lemma} \newtheorem{pro}[thm]{Proposition} \makeatletter \numberwithin{equation}{section} \makeatother \section{Introduction} We consider the Cauchy problem of the form \begin{gather} \label{eq:porous} u_t = \Delta(u^m)\quad \text{in } S= {\mathbb R}^N\times (0,\infty), \\ \label{eq:inporous} u(x,0)=u_0 \quad \text{on } {\mathbb R}^N. \end{gather} Here we suppose that $m > 1$, and $u_0$ is a nonzero bounded nonnegative function with compact support. It is well known that (\ref{eq:porous}) describes the evolution in time of various diffusion processes, in particular the flow of a gas through a porous medium. Here $u$ stands for the density, while $v=\frac{m}{m-1}u^{m-1}$ represents the pressure of the gas. Then $v$ satisfies $$\label{eq:pressure} v_t = (m-1)v\Delta v + |\nabla v|^2 .$$ If the solution is radial symmetry, then $v$ satisfies $$\label{eq:symm} v_t = (m-1)v v_{rr} + \frac{\bar{m}}{r}vv_r + v_r^2,$$ where $r=\sqrt{\sum_{i=1}^N x_i^2}$ and $\bar{m}=(m-1)(N-1)$. Since we are concerning about the regularity of the interface we may assume $r\geq \epsilon_0$ for some positive number $\epsilon_0$. In this paper we will show that, if the solution is radial symmetry then the interface of (\ref{eq:symm}) can be represented by a $C^{\infty}$ function after a large time. In the one-dimensional case Aronson and V\'azuez \cite{AV} and independently H\"ollig and Kreiss \cite{HK} showed that the interfaces are smooth after the waiting time. Angenent \cite{A} showed that the interfaces are real analytic after the waiting time. For the dimensions $\geq 2$, Daskalopoulos and Hamilton \cite{DH} showed that for $t\in (0, T)$, for some $T>0$, the interface can be described as a $C^{\infty}$ surface if the initial data $u_0$ satisfies some assumptions. On the other hand, Caffarelli, Vazquez and Wolanski \cite{CVW} showed that after a large time the interface can be described as a Lipschitz surface if the $u_0$ satisfies some non-degeneracy conditions. Caffarelli and Wolanski \cite{CW} improved this result by showing that the interface can be described as a $C^{1,\alpha}$ surface under the same non-degeneracy conditions on the initial data. But many people believe that even after a large time, the interface can be described as a smooth surface. In this paper, assuming the solution to (\ref{eq:porous}) is radial symmetry and the initial data $u_0$ satisfies the same non-degeneracy conditions as in \cite{CW}, we obtain the following result : \begin{thm}\label{thm:main} If $v$ is a solution to (\ref{eq:symm}), then $v$ is a $C^{\infty}$ function near the interface where $v>0$ and the interface is a $C^{\infty}$ function for $t>T$, for some $T>0$. \end{thm} This paper is divided into three parts : In Part I, we obtain the interface equation. In Part II we obtain the upper and lower bound of $v_{rr}$ by constructing a barrier function. In Part III, we obtain the upper and lower bound of $\left(\frac{\partial}{\partial r}\right)^j v\equiv v^{(j)}$ by constructing another barrier function. In showing our results, we adapt the methods used in \cite{AV}. \section{Preliminaries} In this section, we introduce some basic results which are necessary in showing the uniform boundedness of the derivatives of the pressure $v$. First, since we are interested in the radial symmetry solution, let $$P[u]=\{(r,t) \;|\; u(r,t)>0, r\geq \epsilon_0>0\}$$ for some $\epsilon_0>0$, be the positivity set. Then by \cite{CVW}, we can express the interface as a nondecreasing and Lipschitz continuous $r=\zeta(t)=\sup\{r\geq \epsilon_0\;|\;u(r,t)>0\}$ and $r=\zeta(t)$ on $[T,\infty)$, for some $T>0$. In showing the interface is a $C^{\infty}$ function, we need the following : \begin{thm}\label{thm:interface} Assume that $u_0 >0$ on $I=(2\epsilon_0, a)$ and $u_0 =0$ on $[a,\infty)$. Let $v_0 = \frac{m}{m-1}u_{0}^{m-1}\in C^{1}(\bar{I})$. Then $$\lim_{r\to \zeta(t)}v_r (r,t) \equiv v_r (\zeta(t),t)$$ exists for all $t\geq 0$ and $$v_r (\zeta(t),t)\left\{\zeta^{'}(t)+ v_r(\zeta(t),t)\right\}=0$$ for almost all $t\geq 0$. \end{thm} In proving Theorem \ref{thm:interface}, we need \begin{lem}\label{lem:0} Assume that $v_0$ has bounded derivatives of all orders, and that there exist positive constants $\mu$, $M$ such that $$\mu \leq v_0(r)\leq M \quad\text{on } [\epsilon_0, \infty).$$ Then the Cauchy problem (\ref{eq:symm}) has a unique classical solution $v$ in $S=[\epsilon_0, \infty)\times [0, \infty)$ such that $$\mu \leq v(r,t)\leq M \quad\text{on } S$$ and $v \in C^{\infty}(S)$. \end{lem} \noindent\textbf{Proof.} To show the existence of $v$, we need to have an a priori lower bound for $v$. To this end, let $\varphi=\varphi(s)$ denote a $C^{\infty}({\mathbb R}^1)$ function such that $\varphi(s)=s$ for $s \geq \mu$, $\varphi(s)=\mu/2$ for $s\leq 0$, and $\varphi$ increases from $\mu/2$ to $\mu$ as $s$ increases from $0$ to $\mu$. Now consider $$\label{eq:sym} \begin{gathered} v_{t}=(m-1)\varphi(v)v_{rr} + \frac{\bar{m}}{r}vv_r + v_r^2 \quad\text{in } [\epsilon_0,\infty)\times (0,\infty), \\ v(x, 0) = v_{0}(x)\quad \text{on } [\epsilon_0,\infty). \end{gathered}$$ Then it is easily verified that equation (\ref{eq:sym}) satisfies all the hypotheses of Theorem 5.2 in \cite{LSU}( pp. 564-565). Let $\tau\in(0,\infty)$ and $\beta\in(0,1)$ be arbitrary, and let $R_{\tau}=[\epsilon_0,\infty)\times [0,\tau]$. Then there exists a unique solution $v$ of equation (\ref{eq:sym}) such that $|v|\leq M$ in $R_{\tau}$ and $v \in H^{2+\beta, 1+\beta/2}(R_{\tau})$.In \cite{LSU}, the solution of (\ref{eq:sym}) is obtained as the limit as $n\to \infty$ of the solutions $v^n$ of the sequence of the first boundary value problems \begin{gather*} v_t = (m-1)\varphi(v)v_{rr} + \frac{\bar{m}}{r}vv_r + v_r^2 \quad\text{in } [\epsilon_0+1/n,n]\times (0, \tau),\\ v(r,0)=v_0(r)\quad\text{in } [\epsilon_0+ \frac{1}{n}, n], \\ v(n,t)=v_0(n)\quad\text{in } [0,\tau]. \end{gather*} Then by the method used in proving Theorem 2 in \cite{AR1}, we have $v=\lim_{n\to \infty}v^n$ belongs to $C^{\infty}(S)$ and $\mu\leq v\leq M$. \hfill $\Box$ Now let us prove Theorem \ref{thm:interface}. Let $t>0$ and assume $r0$. Let $\psi_n(r,t)=\iint_{{\mathbb R}^2}k_n(r-\xi,t-\tau)\tilde{\psi}(\xi,\tau)d\xi d\tau,$ where $k_n(r,t)$ denotes an averaging kernel with support in $(-\frac{1}{n}, \frac{1}{n})\times (-\frac{1}{n},\frac{1}{n})$ for each integer $n\geq 1$. Then $\psi_n$ satisfies $$\label{eq:wk} \begin{split} &\int_{{\mathbb R}^1}\psi_n(r,t_2)v(r,t_2)dr +\int_{t_1}^{t_2}\int_{{\mathbb R}^1}\{(m-1+\frac{c}{r})vv_r\psi_{nr} +(m-2)v_r^2\psi_n-v\psi_{nt}\}\,dr\,dt\\ &=\int_{{\mathbb R}^{1}}\psi_{n} (r,t_1)v(r,t_1)dr. \end{split}$$ Recall that $v$ and its weak derivative $v_r$ are bounded in $S$. On the other hand, it is shown in \cite{AR2} that $\psi_n \to \tilde{\psi}$ and $\psi_{nr}\to \tilde{\psi}_r$ uniformly on any compact subsets of ${\mathbb R}^2$, while $\psi_{nt}\to \tilde{\psi}_t$ strongly in $L_{loc}^1({\mathbb R}^2)$. Therefore (\ref{eq:wk}) also holds for the limit of the sequence $\psi_n$, that is, for any set function which satisfies the conditions listed at the beginning of this paragraph. Now define a function $$K(r)=\begin{cases} C\cdot \exp\{-1/(1-r^2)\} &\text{for } |r|\leq 1\\ 0 &\text{for } |r|\geq 1\,, \end{cases}$$ where $C$ is chosen so that $$\int_{{\mathbb R}^1}K(r)dr=1\,,$$ and set $k_n(r)=nK(nr)$ for each integer $n\geq 1$. Then $k_n(r)$ is an even averaging kernel and $k_n(\zeta(t)-r)$ belongs to $C(S)$ and has compact support in ${\mathbb R}^1$ for each $t\geq 0$. Also $\frac{d}{dr}k_n(\zeta(t)-r) \in C(S)$. Since $\zeta$ is Lipschitz continuous, $\zeta^{'}$ exists almost everywhere and is bounded above. Moreover $\zeta$ is weakly differentiable and its weak derivative can be represented by $\zeta^{'}$. It follows that $k_n(\zeta(t)-r)$ is also weakly differentiable with respect to $t$ in $S$ with weak derivative given by $\zeta^{'}(t)k_n'(\zeta(t)-r)$ which belongs to $L^1({\mathbb R}^1\times (t_1,t_2)$ for any $0\leq t_10$. Moreover, since $|v_r|\leq L$, $\left|\int_{{\mathbb R}^1}k_n(\zeta-r)v_r^2(r,t)dr\right|\leq L^2.$ Thus by the Lebesgue's dominated convergence theorem, $\lim_{n\to \infty}\int_{t_1}^{t_2}\int_{{\mathbb R}^1}k_n(\zeta(t)-r)v_r^2(r,t)dxdt =\frac{1}{2}\int_{t_1}^{t_2}v_r^2(\zeta,t)dt.$ Next define $$w(r,t)= \begin{cases} \dfrac{-v(r,t)}{\zeta-r} &\text{for } r<\zeta(t)\\ v_r(\zeta(t),t) &\text{for } r=\zeta(t)\\ 0&\text{for } r>\zeta(t)\,. \end{cases}$$ Note that for fixed $t$, $w$ is continuous on $[\epsilon_0,\zeta(t)]$ and $|w(x,t)|\leq C$. Then the second integral on the left in (\ref{eq:wk*}) can be written in the form $I_n=\int_{t_1}^{t_2}\int_{{\mathbb R}^1}(\zeta-r)k_n'(\zeta-r)w(r,t)\{(m-1)v_r+\zeta'\}\,dr\,dt.$ It is easily verified that the function $-rk_n^{'}(r)$ is also an even averaging kernel. Thus for each $t$, $$\lim_{n\to \infty}\int_{{\mathbb R}^1}(\zeta-r)k_n^{'}(\zeta-r)w\{(m-1)v_r+\zeta'\}dr =-\frac{1}{2}v_r(\zeta,t)\{ (m-1)v_r+\zeta'\}$$ and $\left|\int_{{\mathbb R}^1}(\zeta-r)k_n^{'}(\zeta-r)w\{(m-1)v_r+\zeta'\}dr \right| \leq (m-1)L^2 +L\zeta'\in L^1(t_1,t_2).$ It follows that $\lim_{n\to \infty}I_n =-\frac{1}{2}\int_{t_1}^{t_2} v_r(\zeta,t)\{ (m-1)v_r+\zeta'\}dt.$ Hence if we let $n\to \infty$ in (\ref{eq:wk*}), we obtain $-\frac{1}{2}\int_{t_1}^{t_2} v_r(\zeta,t)( \zeta^{'}(t)+v_r(\zeta,t))dt=0$ for any $0\leq t_10, r\geq \epsilon_0\}$. Assume that $v_0(r)$ has compact support containing $[2\epsilon_0,a]$ for some $a>0$ and satisfies the non-degeneracy condition (1.4) of the Theorem 1 in \cite{CW}. Let $q=(r_0, t_0)$ be a point on the interface $r=\zeta(t)$ so that $r_0 = \zeta(t_0)$, $v(r,t_0)=0$ for all $r\geq \zeta(t_0)$, and $v(r,t_0)>0$ for all $00$ be the positive constant established in \cite{CVW}. Assume $t_0 >T$ so that the interface is moving at $q$. Then from Theorem \ref{thm:interface}, we have $$\label{eq:21} \zeta^{'}(t_0) = -v_r(\zeta,t)=a>0,$$ and on the moving interface we have $$\label{eq:2} v_t = v_r^2 .$$ As in \cite{AV}, we use the notation $$R_{\delta, \eta} = R_{\delta, \eta}(t_0) \equiv \{(r,t) \in {\mathbb R}^2 \;|\; \zeta(t)-\delta < r \leq \zeta(t), \;t_0 -\eta \leq t\leq t_0 + \eta\}.$$ \begin{lem}\label{lem:21} Let $q$ be a point on the interface and assume (\ref{eq:21}) holds. Then there exist positive constants $C$, $\delta$ and $\eta$ depending only on $N$, $\epsilon_0$, $m$, $q$ and $u$ such that $$|v_{rr}| \leq C \qquad \text{in } R_{\delta, \eta/2} .$$ \end{lem} \noindent\textbf{Proof}. It is well known that $v_t$, $v_r$ and $vv_{rr}$ are continuous in a closed neighborhood in $\overline{P}[u]$ of any point on the interface if $t>T$, and that $$(vv_{rr})(r,t)=\frac{1}{m-1}\left(v_t -\frac{\bar{m}}{r}vv_r - v_r^2 \right)\to 0\; as\; P[u]\ni (r,t) \to (\zeta(t),t)$$ for any $t>T$. Choose now an $\epsilon >0$ such that $$\label{eq:24} (a-(8m-3)\epsilon)(a-\epsilon) \geq 4(m+1)\epsilon >0.$$ Then there exist $\frac{\epsilon_0}{2\bar{m}}\leq\delta=\delta(\epsilon)>0$ and $\eta=\eta_{1} (\epsilon)\in (0,t_0-T)$ such that $R_{\delta,\eta}\subset P[u]$, \label{eq:25} -a-\epsilon From the choice of $\delta$ and the estimates (\ref{eq:25}), (\ref{eq:27}), (\ref{eq:28}) and the definition (\ref{eq:29}) of $\zeta^{*}$ we conclude that \begin{eqnarray*} L(\phi) &\geq& \frac{\alpha}{2(\zeta-r)^2} \{a-(8m-5)\epsilon-4(m+1)\alpha\}\\ &&+ \frac{\beta}{2(\zeta^*-r)^2}\{a-(8m-3)\epsilon -4(m+1)\beta\}. \end{eqnarray*} Now set $$\label{eq:212} \beta=\frac{a-(8m-3)\epsilon}{8(m+1)}$$ and note that (\ref{eq:24}) implies that $\beta>0$. Then $L(\phi)\geq 0$ in $R_{\delta,\eta}$ for all $\alpha\in(0,\alpha_0]$, where $\alpha_0 =\{a-(8m-5)\epsilon\}/4(m+1)$. Let us compare $p$ and $\phi$ on the parabolic boundary of $R_{\delta,\eta}$. In view of (\ref{eq:26}) and (\ref{eq:27}) we have $$v_{rr} < \frac{\epsilon}{(a-\epsilon)(\zeta-r)}\quad\text{in } R_{\delta,\eta},$$ so that, in particular, $$v_{rr}(\zeta(t)-\delta,t)\leq \frac{\epsilon}{(a-\epsilon)\delta}\quad\text{in } [t_1,t_2].$$ By the mean value theorem and (\ref{eq:28}) it follows that for some $\tau\in (t_1,t_2)$ \begin{eqnarray*} \zeta^*(t)+\delta - \zeta(t) &=& \delta + (a+2\epsilon)(t-t_1)-\zeta^{'}(\tau)(t-t_1)\\ &\leq&\delta + 3\epsilon(t-t_1)\leq \delta + 6\epsilon\eta. \end{eqnarray*} Now set $$\eta\equiv \min\{\eta_1 (\epsilon),\delta(\epsilon)/6\epsilon\}.$$ Since $\epsilon$ satisfies (\ref{eq:24}) and $\beta$ is given by (\ref{eq:212}) it follows that $$\phi(\zeta +\delta,t)\geq \frac{\beta}{2\delta}\geq \frac{\epsilon}{(a-\epsilon)\delta} \geq v_{rr}(\zeta+\delta,t)\quad\text{on } [t_1,t_2].$$ Moreover, $$\phi(r,t_1)\geq \frac{\beta}{\zeta_1-r}>\frac{\epsilon}{(a-\epsilon)(\zeta_1-r)}>v_{rr}(r,t_1) \quad\text{on } [\zeta_1-\delta, \zeta_1).$$ Let $\Gamma=\{(r,t)\in {\mathbb R}^2 \;:\;r=\zeta(t), t_1 \leq t\leq t_2\}$. Then $\Gamma$ is a compact subset of ${\mathbb R}^2$. Now fix $\alpha\in(0,\alpha_0)$. For each point $s\in\Gamma$ there is an open ball $B_s$ centered at $s$ such that $$(vv_{rr})(r,t)\leq \alpha(a-\epsilon)\quad\text{in } B_s \cap P[u],$$ In view of (\ref{eq:27}) we have $$\phi(r,t)\geq \frac{\alpha}{\zeta-r}\geq v_{rr}(r,t)\quad\text{in } B_s \cap P[u].$$ Since $\Gamma$ is compact, finite number of these balls can cover $\Gamma$ and hence there is a $\gamma=\gamma(\alpha)\in(0,\delta)$ such that $$\phi(r,t)\geq p(r,t)\quad\text{in } R_{\gamma,\eta}.$$ Thus for every $\alpha \in (0,\alpha_0)$, $\phi$ is a barrier for $p$ in $R_{\delta,\eta}$. Hence by the comparison principle we conclude $$v_{rr}(r,t) \leq \frac{\alpha}{\zeta-r} +\frac{\beta}{\zeta^*(t)-r}\quad\text{in } R_{\delta,\eta},$$ where $\beta$ is given by (\ref{eq:212}) and $\alpha \in (0,\alpha_0)$ is arbitrary. Now let $\alpha\downarrow 0$ to obtain $$v_{rr}(r,t) \leq \frac{\beta}{\zeta^*(t)-r}\leq \frac{2\beta}{\epsilon\eta} \quad\text{in } R_{\delta,\eta/2},$$ \hfill $\Box$ \section{Bound for $\left(\frac{\partial}{\partial r}\right)^j v$} As in \cite{AV}, if we can show $$|v^{(j)}\equiv \left(\frac{\partial}{\partial x}\right)^j v|\leq C_j,$$ for each $j\geq 2$, then Theorem \ref{thm:main} follows. First by a direct computation for $j\geq 3$, $v^{(j)}$ satisfy the following equation \begin{align}\label{eq:31} L_j v^{(j)}&\equiv v_t^{(j)} - (m-1)vv_{rr}^{(j)}- (2+j(m-1))v_rv_r^{(j)} -\frac{\bar{m}}{r}vv_r^{(j)}-F_j(r,t)v^{(j)}\notag\\ &-G_j(r,t), \end{align} where $F_j(r,t)$ and $G_j(r,t)$ are functions of $r, v$ and derivatives of $v$ of order $0$ and $\eta_0=\eta_0(\epsilon)\in(0,t_0-T)$ such that $R_0 \equiv R_{\delta_0\eta_0}(t_0)\subset P[u]$ and (\ref{eq:25}) holds. Thus we also have (\ref{eq:27}) and (\ref{eq:28}) in $R_0$. Assume that there are constants $C_k\in {\mathbb R}^+$ for $k=2, 3, \ldots, j-1$ such that $$\label{eq:33} |v^{(k)}|\leq C_k\quad\text{on } R_0\quad \text{for}\;k=2, \ldots,j-1.$$ Observe that, by Lemma \ref{lem:21}, the estimate (\ref{eq:33}) holds for $k=2$. As in \cite{AV} by rescaling and using interior estimates we obtain the following estimate near $\zeta$. \begin{lem}\label{lem:32} There are constants $K\in {\mathbb R}^+$, $\delta\in(0,\delta_0)$ and $\eta\in(0,\eta_0)$, depending only on $q$, $m$ and the $C_k$ for $k\in [2,j-1]$ with $j\geq 3$, such that $|v^{(j)}(r,t)|\leq \frac{K}{\zeta(t)-r}\quad\text{in } R_{\delta,\eta}.$ \end{lem} \noindent\textbf{Proof}. Set $$\delta=\min\{\frac{2\delta_0}{3},2s\eta_0\},$$ $$\eta=\eta_0-\frac{\delta}{4s},$$ and define $$R(\bar{r},\bar{t})\equiv \left\{(r,t)\in{\mathbb R}^2 \;:\;|r-\bar{r}|<\frac{\lambda}{2}, \;\bar{t}-\frac{\lambda}{4s}0 such that$$\left|\frac{\partial}{\partial\xi}V^{(j-1)}(0,0)\right|\leq K,$$that is$$|v^{(j-1)}(\bar{r},\bar{t})|\leq K/\lambda.$$Since ( \bar{r},\bar{t})\in R_{\delta,\eta} is arbitrary, this proves the lemma. \hfill \Box We now turn to the barrier construction. If \gamma\in (0,\delta) we will use the notation$$R_{\delta,\eta}^{\gamma}= R_{\delta,\eta}^{\gamma}(t_0) \equiv\{(r,t)\in {\mathbb R}^2\;:\;\zeta(t)-\delta\leq r\leq \zeta-\gamma, t_0-\eta\leq t \leq t_0 +\eta\}.$$Then we have \begin{lem}\label{lem:33} Let R_{\delta_1,\eta_1} be the region constructed in the proof of Lemma \ref{lem:21}. For j\geq 3 and (r,t)\in R_{\delta_1,\eta_1}^{\gamma} , let %$$\lebel{eq:35} \phi_{j}(r,t)\equiv \frac{\alpha}{\zeta(t)-r-\gamma/3} +\frac{\beta}{\zeta^{*}(t)-r} %$$ where \zeta^{*} is given by (\ref{eq:29}), and \alpha and \beta are positive constants. There exist \delta\in (0,\delta_1) and \eta\in (0,\eta_1) depending only on a, m, C_1, \ldots, C_{j-1} such that$$L_j (\phi_j)\geq 0 \quad\text{in } R_{\delta,\eta}^{\gamma}$$for all \gamma\in(0,\delta). \end{lem} \noindent\textbf{Proof}. Choose \epsilon such that $$\label{eq:36} 0<\epsilon<\frac{a}{2(4+2j(m-1))}.$$ There exist \delta_2\in (0,\delta_1) and \eta\in (0,\eta_1) such that (\ref{eq:25}), (\ref{eq:27}) and (\ref{eq:28}) hold in R_{\delta_2,\eta}. Fix \gamma\in (0,\delta_2). For (r,t)\in R_{\delta_2,\eta}^{\gamma} , we have \begin{eqnarray*} L_j(\phi_j) &=& \frac{\alpha}{(\zeta-r-\gamma/3)^2}\Big [-\zeta^{'} -\frac{2(m-1)v}{\zeta-r-\gamma/3} - (2+j(m-1))v_r-\frac{\bar{m}}{r}v \\ &&-F_j(r,t)(\zeta-r-\gamma/3)-\frac{(\zeta-r-\gamma/3)^2}{\alpha}G_j(r,t)\Big ]\\ &&+ \frac{\beta}{(\zeta^*-r)^2}\Big[-\zeta^{*'} -\frac{2(m-1)v}{\zeta^{*}-r} - (2+j(m-1))v_r - \frac{\bar{m}}{r}v\\ &&-F_j(r,t)(\zeta^{*}-r)-\frac{(\zeta^{*}-r)^2}{\beta}G_j(r,t)\Big]. \end{eqnarray*} >From (\ref{eq:27}) and by the fact that \zeta^*-r \geq \zeta-r-\gamma/3 we have$$\frac{v}{\zeta^*-r}\leq \frac{v}{\zeta-r-\gamma/3} \leq (a+\epsilon)\frac{\gamma}{\gamma-\gamma/3}=\frac{3}{2}(a+\epsilon).$$%Let ax>1 or 1/x0 so small that L_j (\phi_j)\geq 0 in R_{\delta,\eta}^{\gamma}. \hfill \Box \begin{lem}\label{lem:34}(Barrier Transformation). Let \delta and \eta be as in Lemma \ref{lem:33} with the additional restriction that $$\label{eq:37} \eta<\frac{\delta}{6\epsilon},$$ where \epsilon satisfies (\ref{eq:36}). Suppose that for some nonnegative constants \alpha and \beta $$\label{eq:38} v^{(j)}(r,t)\leq \frac{\alpha}{\zeta(t)-r}+\frac{\beta}{\zeta^*(t)-r}\quad \text{in } R_{\delta,\eta},$$ Then v^{(j)} also satisfies $$\label{eq:39} v^{(j)}(r,t)\leq \frac{2\alpha/3}{\zeta(t)-r}+\frac{\beta+2\alpha/3}{\zeta^*(t)-r} \quad \text{in } R_{\delta,\eta}.$$ \end{lem} \noindent\textbf{Proof}. By Lemma \ref{lem:33}, for any \gamma\in (0,\delta) the function$$\phi_j(r,t)=\frac{2\alpha/3}{\zeta(t)-r-\gamma/3}+\frac{\beta+2\alpha/3}{\zeta^*(t)-r}$$satisfies L_j (\phi_j)\geq 0 in R_{\delta,\eta}^{\gamma}. On the other hand, on the parabolic boundary of R_{\delta,\eta}^{\gamma} we have \phi_j \geq v^{(j)}. In fact, for t=t_1 and \zeta_1 -\delta\leq r\leq \zeta_1 -\gamma, with \zeta_1=\zeta(t_1), we have$$\phi_j(r,t_1)=\frac{2\alpha/3}{\zeta_1-r-\gamma/3}+\frac{\beta+2\alpha/3}{\zeta_1-r} >\frac{4\alpha/3}{\zeta_1-r}+\frac{\beta}{\zeta_1-r}>v^{(j)}(r,t_1),$$while for r=\zeta-\delta and t_1 \leq t\leq t_2 we have, in view of (\ref{eq:37}), \begin{eqnarray*} \phi_j(\zeta-\delta,t)&\geq &\frac{2\alpha/3}{\delta-\gamma/3}+\frac{\beta}{ \zeta^*+\delta-\zeta} +\frac{2\alpha/3}{\delta+6\epsilon\eta}\\ &\geq&\frac{2\alpha/3}{\delta}+ \frac{\beta}{ \zeta^*+\delta-\zeta} +\frac{\alpha/3}{\delta} \geq v^{(j)}( \zeta-\delta,t). \end{eqnarray*} Finally, for r=\zeta-\gamma, t_1 \leq t\leq t_2 we have$$ \phi_j(\zeta-\gamma,t)=\frac{2\alpha/3}{\gamma-\gamma/3}+\frac{\beta+2\alpha/3}{ \zeta^*+\gamma-\zeta} \geq \frac{\alpha}{\gamma}+\frac{\beta}{ \zeta^*+\gamma-\zeta}\geq v^{(j)}(\zeta-\gamma,t).$$By the comparison principle we have$$\phi_j \geq v^{(j)}\quad\text{in } R_{\delta,\eta}^{\gamma}$$for any \gamma\in (0,\delta), and (\ref{eq:39}) follows by letting \gamma\downarrow 0. \hfill \Box Now we are ready to prove our main proposition. \textbf{Completion of proof of Proposition \ref{pro:31}.} By Lemma \ref{lem:32}, we have an estimate for v^{(j)} of the form (\ref{eq:38}) with \alpha=K and \beta=0. Iterating this estimate by the Barrier Transformation Lemma we obtain the sequence of estimates$$v^{(j)}(r,t)\leq \frac{\alpha_n}{\zeta(t)-r} +\frac{\beta_n}{\zeta^*-r}  with $\alpha_n = (2/3)^n K$ and $\beta_n=\{(\frac{2}{3}+\ldots +(\frac{2}{3} )^n\}K$. Thus if we let $n\to \infty$ we obtain an upper bound for $v^{(j)}$ of the form $$\label{eq:310} v^{(j)}(r,t)\leq \frac{2K}{\zeta^*-r}\quad \text{in } R_{\delta,\eta}.$$ As in the proof of Lemma \ref{lem:21}, this implies that $v^{(j)}$ is bounded above in$R_{\delta,\eta/2}$. Since the equation (\ref{eq:31}) for $v^{(j)}$ is linear, a similar lower bound can be obtained in the same way and the induction step is complete. Therefore as we mentioned in the beginning of this section, Theorem \ref{thm:main} is proved. \vspace{0.5cm} {\it Acknowledgement}. 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