\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 229, pp. 1--5.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2015 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/229\hfil Porosity of the free boundary] {Porosity of the free boundary for singular $p$-parabolic obstacle problems} \author[A. Lyaghfouri \hfil EJDE-2015/229\hfilneg] {Abdeslem Lyaghfouri} \address{Abdeslem Lyaghfouri \newline American University of Ras Al Khaimah, Department of Mathematics and Natural Sciences, Ras Al Khaimah, UAE} \email{abdeslem.lyaghfouri@aurak.ac.ae} \thanks{Submitted June 2, 2015. Published September 10, 2015.} \subjclass[2010]{35K59, 35K67, 35K92, 35R35} \keywords{Singular $p$-parabolic obstacle problem; free boundary; porosity} \begin{abstract} In this article we establish the exact growth of the solution to the singular quasilinear p-parabolic obstacle problem near the free boundary from which follows its porosity. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction}\label{S:intro} Let $\Omega$ be an open bounded domain of $\mathbb{R}^n$, $n\geq 2$, $T>0$. We consider the problem: Find $u\in L^p(0,T; W^{1,p}(\Omega))$ such that: \begin{itemize} \item[(i)] $u\geq 0$ in $\Omega_T=\Omega\times(0,T)$, \item[(ii)] $L_p(u)=u_t-\Delta_p u=-f(x)$ in $\{u>0\}$, \item[(iii)] $u=g$ on $\partial_p\Omega_T=(\Omega\times\{0\})\cup(\partial\Omega\times(0,T))$, \end{itemize} where $p>1$, $\Delta_p$ is the $p$-Laplacian defined by $\Delta_p u=\operatorname{div}\big(|\nabla u|^{p-2}\nabla u\big)$, and $f$, $g$ are functions defined in $\Omega_T$ and satisfying for two positive constants $\lambda_0$ and $\Lambda_0 $ \begin{equation}\label{e1.1} \lambda_0 \leq f \leq \Lambda_0 \quad \text{a.e. in }\Omega_T. \end{equation} Moreover we assume that \begin{gather} f\text{ is non-increasing in }t. \label{e-1.2}\\ g(x,0)=0\quad \text{ a.e. in }\Omega.\label{e1.3}\\ g\text{ is non-decreasing in }t. \label{e1.4} \end{gather} The variational formulation of the above problem is: Find \[ u\in K_g=\{v\in V^{1,p}(\Omega_T)/v=g\text{ on } \partial_p\Omega_T,\quad v\geq0 \text{ a.e. in } \Omega_T\} \] such that for all $ h>0 $ and $tT-h\,. \] Let us recall the following existence and uniqueness theorem of the solution of the problem \eqref{eP} \cite{S}. \begin{theorem}\label{thm1.1} Assume that $f$ and $g$ satisfy \eqref{e1.1}--\eqref{e1.4}. Then there exists a unique solution $u$ of the problem \eqref{eP} which satisfies \begin{gather} 0\leq u \leq M=\| g\|_{\infty,\Omega_T} \quad \text{in }\Omega_T, \label{e1.5} \\ u_t \geq 0\quad \text{in } \Omega_T. \nonumber \\ f \chi_{\{u>0\}} \leq \Delta_p u-u_t\leq f \quad \text{a.e. in }\Omega_T. \label{e1.6} \end{gather} \end{theorem} \begin{remark}\label{rem1.1} \rm We deduce from \eqref{e1.5}--\eqref{e1.6} (see \cite[Theorems 7 and 8]{Ch}) that $u\in C_{loc}^{0,\alpha}(\Omega_T)\cap C_{x,loc}^{1,\alpha}(\Omega_T)$ for some $\alpha\in(0,1)$. \end{remark} The main result of this article is as follows. \begin{theorem}\label{thm1.3} Assume that $10\})\cap K\cap\{t=t_0\} $ is porous in $\mathbb{R}^{n}$ with porosity constant depending only on $n$, $p$, $\lambda_0$, $\Lambda_0$, $M$, and $\operatorname{dist}(K,\partial_p \Omega_T)$. \end{theorem} We recall that a set $E\subset \mathbb{R}^n$ is called porous with porosity $\delta$, if there is an $r_0>0$ such that for all $x\in E$ and all $r\in (0,r_0)$, there exists $y\in \mathbb{R}^n$ such that \[ B_{\delta r}(y)\subset B_{ r}(x)\setminus E. \] A porous set has Hausdorff dimension not exceeding $n-c\delta^n$, where $c=c(n) >0$ is a constant depending only on $n$. In particular a porous set has Lebesgue measure zero. Theorem \ref{thm1.3} extends the result established in \cite{S} in the quasilinear degenerate and linear cases $p\geq 2$. The proof is based on the exact growth of the solution of the problem \eqref{eP} near the free boundary which is given by the next theorem. \begin{theorem}\label{thm1.4} Assume that $10\})\cap K$, the following estimates hold \begin{equation}\label{e1.7} c_0 r^q\leq \sup_{B_r(x_0)} u(.,t_0)\leq C_0 r^q, \end{equation} where $q=p/(p-1)$ is the conjugate of $p$. \end{theorem} Since the proof of Theorem \ref{thm1.3} relies on the one of Theorem \ref{thm1.4}, it will be enough to prove the latter one. On the other hand we observe that the left hand side inequality in \eqref{e1.7} was established in \cite[Lemma 2.1]{S} for any $p>1$, while the right hand side inequality in \eqref{e1.7} was established only for $p\geq 2$. In the next section, we shall establish the second inequality for a class of functions in the singular case i.e. for $10\}\}$ for $(x,t)\in \{u>0\}$, and $d(x,t)=0$ otherwise, and where $Q_r(x,t)=B_r(x)\times(s-r^q,s+r^q)$. \end{theorem} To prove Theorem \ref{thm2.1}, we need to introduce some notation inspired from \cite{S}. For a nonnegative bounded function $u$, we define the quantities \[ Q_r^-=B_r\times(-r^q,0),\quad S(r,u) = \sup_{(x,t)\in Q_r^-} u(x,t). \] Also for $u\in\mathcal{F}$ define the set \begin{equation}\label{e2.5} \mathbb{M}(u) = \{ j\in \mathbb{N}\cup\{0\}: A S( 2^{-j-1},u)\geq S( 2^{-j},u) \} \end{equation} where $A= 2^q \max\big(1,1/ C_0\big) $ and $C_0$ is the constant in \eqref{e1.7}. As in \cite{S}, we first show a weaker version of the inequality. \begin{lemma}\label{lem2.2} There exists a constant $C_1=C_1(p,n,M,\Lambda_0)$ such that $$ S( 2^{-j-1},u) \leq C_1 2^{-qj} \quad \forall u\in \mathcal{F} , \; \forall j\in \mathbb{M}(u). $$ \end{lemma} \begin{proof} We argue by contradiction and assume that: for all $k\in \mathbb{N}$ there exist $u_k \in \mathcal{F}$ and $j_k\in \mathbb{M}(u_k)$ such that \begin{equation}\label{e2.6} S( 2^{-j_k-1},u_k) \geq k 2^{-qj_k}. \end{equation} Let $\alpha_k=2^{-pj_k}(S(2^{-j_k-1},u_k))^{2-p}$, and let \[ v_k(x,t)=\frac{u_k(2^{-j_k}x,\alpha_k t)}{S(2^{-j_k-1},u_k)} \] for $(x,t)\in Q_1$. First we observe that since $u(0,0)=0$ and $u$ is continuous, we have $\alpha_k\to 0$ as $k\to\infty$. Moreover, we have \begin{equation}\label{e2.7} \begin{gathered} \nabla v_k(x,t)=\frac{2^{-j_k}}{S( 2^{-j_k-1},u_k)} \nabla u_k(2^{-j_k}x,\alpha_k t) \\ v_{kt}(x,t)=\frac{\alpha_k}{S( 2^{-j_k-1},u_k)} u_{kt}(2^{-j_k}x,\alpha_k t) =\Big(\frac{2^{-qj_k}}{S(2^{-j_k-1},u_k)}\Big)^{p-1} u_{kt}(2^{-j_k}x,\alpha_k t) \end{gathered} \end{equation} and \begin{equation}\label{e2.8} \begin{aligned} &\Delta_p v_k(x,t) \\ &=\operatorname{div}\big(|\nabla v_k|^{p-2}\nabla v_k\big) \\ &=\Big(\frac{2^{-j_k}}{S(2^{-j_k-1},u_k)}\Big)^{p-1}\operatorname{div} \big(|\nabla u_k(2^{-j_k}x,\alpha_k t)|^{p-2}\nabla u_k(2^{-j_k}x,\alpha_k t)\big) \\ &=2^{-j_k}\Big(\frac{2^{-j_k}}{S(2^{-j_k-1},u_k)}\Big)^{p-1} \Delta_p u_k(2^{-j_k}x,\alpha_k t) \\ &=\Big(\frac{2^{-qj_k}}{S(2^{-j_k-1}},u_k)\Big)^{p-1} \Delta_p u_k(2^{-j_k}x,\alpha_k t). \end{aligned} \end{equation} We deduce from \eqref{e2.7}--\eqref{e2.8} that \begin{equation}\label{e2.9} v_{kt}-\Delta_p v_k(x,t) = \Big(\frac{2^{-qj_k}}{S(2^{-j_k-1},u_k)}\Big)^{p-1}(u_{kt} -\Delta_p u_k)(2^{-j_k}x,\alpha_k t). \end{equation} Combining \eqref{e2.1}--\eqref{e2.6} and \eqref{e2.9}, we obtain \begin{gather} 0 \leq -v_{kt}+\Delta_p v_k\leq \frac{\Lambda_0}{k^{p-1}} \quad \text{in } Q_1, \label{e2.10} \\ 0\leq v_k\leq\frac {S( 2^{-j_k},u_k)}{S( 2^{-j_k-1},u_k)} \leq A\quad \text{in } Q_1^-, \label{e2.11}\\ v_{kt}\geq 0\quad \text{in } Q_1^-, \label{e2.12}\\ {\sup_{Q_{1/2}^-} } v_k=1, \label{e2.13}\\ v_k(0,t)=0\quad\forall t\in(-1,0). \label{e2.14} \end{gather} Taking into account \eqref{e2.10}--\eqref{e2.11}, and using \cite[Theorem 1.1]{L} and \cite[Theorem 1]{Ch}, we deduce that $v_k$ is locally uniformly bounded in $L^\infty(Q_1)$ independently of $k$. Therefore we obtain from \cite[Theorems 7 and 8]{Ch}, that $v_k$ is uniformly bounded in $C^{0,\alpha}(\overline{Q_{3/4}})$ and in $C_x^{1,\alpha}(\overline{Q_{3/4}})$ independently of $k$, for a constant $\alpha=\alpha(n, p, A, \Lambda_0)\in(0,1)$. It follows then from Ascoli-Arzella's theorem that there exists a subsequence, still denoted by $v_k$, and a function $v\in C^{0,\alpha}(\overline{Q_{3/4}})\cap C_x^{1,\alpha}(\overline{Q_{3/4}})$ such that $ v_k \to v$ and $ \nabla v_k \to \nabla v$ uniformly in $\overline{Q_{3/4}}$. Moreover, using \eqref{e2.10}--\eqref{e2.14}, we see that $v$ satisfies \begin{gather*} v_t-\Delta_p v =0 \quad \text{in } Q_{3/4}^-,\quad v, v_t\geq 0\quad \text{in } Q_{3/4}^-,\\ \sup_{x\in Q_{1/2}^-} v(x,t)=1,\quad v(0,t)=0\quad\forall t\in(-3/4,0). \end{gather*} We discuss two cases: \smallskip \noindent\textbf{Case 1:} for all $(x,t)\in Q_{3/4}^-$ $v(x,t)=0$. In particular we have $v\equiv 0$ in $Q_{1/2}^-$ which contradicts the fact that ${\sup_{x\in Q_{1/2}^-} v(x)=1}$. \smallskip \noindent\textbf{Case 2:} There exists $(x_0,t_0)\in Q_{3/4}^-$ such that $v(x_0,t_0)>0$. Since $v(.,t_0)$ is not identically zero and $v(0,t_0/2)=0$, we get from the strong maximum principle (see \cite{N}) that $v(x,t_0/2)=0$ for all $x\in B_{3/4}$. By the monotonicity of $v$ with respect to $t$ and the fact that $v$ is nonnegative, we have necessarily $v(x,t)=0$ for all $(x,t)\in B_{3/4}\times(-3/4,t_0/2)$, which is in contradiction with the fact that $v(x_0,t_0)>0$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2.1}] Using Lemma \ref{lem2.2}, the proof follows exactly as the one of \cite[Theorem 2.2]{S}. \end{proof} \begin{thebibliography}{0} \bibitem{CL1} S. Challal, A. Lyaghfouri; \emph{Porosity of Free Boundaries in $A$-Obstacle Problems}, Nonlinear Analysis: Theory, Methods \& Applications, Vol. 70, No. 7, 2772-2778 (2009). \bibitem{CL2} S. 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