\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 160, pp. 1--16.\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-2015/160\hfil Continuity of the free boundary] {Continuity of the free boundary in elliptic problems with Neuman boundary condition} \author[A. Saadi \hfil EJDE-2015/160\hfilneg] {Abderachid Saadi} \address{Abderachid Saadi \newline Ecole Normale Sup\'{e}rieure, 16050 Kouba, Algiers, Algeria.\newline Universit\'{e} de M'sila, B.P 166 ICHBELIA, M'sila, Algeria} \email{rachidsaadi81@gmail.com} \thanks{Submitted May 13, 2015. Published June 16, 2015.} \subjclass[2010]{35J15, 35R35} \keywords{Continuity; free boundary; Neuman boundary condition} \begin{abstract} We show the continuity of the free boundary in a class of two dimensional free boundary problems with Neuman boundary condition, which includes the aluminium electrolysis problem and the heterogeneous dam problem with leaky boundary condition. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{corollary}[theorem]{Corollary} \allowdisplaybreaks \section{Statement of the problem and preliminary results}\label{sec1} Let $\Omega$ be the open bounded domain of $\mathbb{R}^{2}$ defined by $$ \Omega=\{(x_1,x_2)\in \mathbb{R}^2:x_1\in(a_0,b_0),\, d_02$. Let $\Gamma=\{(x_1,\gamma(x_1)):x_1\in(a_0,b_0)\}$ and let $\beta(x,u)$ be a function defined on $\Gamma\times \mathbb{R}$ satisfying \begin{gather} \beta(x,.)\text{ is Lipschitz continuous for a.e. }x\in \Gamma \label{e1.6}\\ \beta(x,.)\text{ is non-decreasing for a.e. }x\in \Gamma. \label{e1.7} \end{gather} Let $\varphi$ be a Lipschitz continuous function on $\Gamma$, $e$ the vector $(0,1)$, and $\Upsilon=\partial\Omega\setminus\Gamma$. Then we consider the following problem \noindent \textbf{Problem (P1)} Find $(u, \chi) \in H^{1}(\Omega)\times L^\infty (\Omega)$ such that \begin{itemize} \item[(i)] $u\geq 0$, $0\leq \chi\leq 1$, $u(1-\chi) = 0$ a.e. in $\Omega$, \item[(ii)] \[ \int_\Omega \big( a(x) \nabla u + \chi h(x)e\big) \cdot\nabla\xi dx \leq \int_{\Gamma}\beta(x,\varphi-u)\xi d\sigma(x) \] for all $\xi \in H^{1}(\Omega)$ and $\xi \geq 0$ on $\Upsilon$. \end{itemize} This problem describes many free boundary problems including the aluminium electrolysis problem \cite{BMQ}, the heterogeneous dam problem with leaky boundary condition \cite{CC1,ChiL1,ChiL2,Ly1,Ly2,Ly3,Ly6}. For the problem with Dirichlet condition on $\Gamma$, we refer for example to \cite{A1} and \cite{Ly4} in the case of the heterogeneous dam problem, to \cite{AC} and \cite{BC} in the case of the lubrication problem, and to \cite{C,ChaL2,ChaL3} for a more general framework. Regarding the existence of a solution under suitable boundary conditions, we refer for example to \cite{CC2,CA,ChiL1,ChiL2,Ly1,Ly6}. In this paper, we shall be interested in studying the free boundary $\Gamma_f=\partial\{u>0\}\cap\Omega$ separating two different regions, which in the case of the dam and lubrication problems, separates the region that contains the fluid from the rest of the domain. In the case of the aluminium electrolysis problem, the free boundary separates the regions containing liquid and solid aluminium. The regularity of $\Gamma_f$ has been addressed in the case of Dirichlet boundary condition in \cite{C} and \cite{ChaL2}, where the authors have established that $\Gamma_f$ is a continuous curve $x_2=\Phi(x_1)$. This result was later on extended in \cite{ChaL3} to a more general framework and also in \cite{ChaL4} in the case of the $p-$Laplacian. \section{Preliminary results}\label{sec2} \begin{remark}\label{rmk1.2} \rm By Harnack's inequality \cite{GT}, we know that $u$ is locally bounded. Due to the local character of this study, we shall assume that there exists a positive constant $M$ such that \begin{equation}\label{e2.1} 0\leqslant u\leqslant M\quad\text{ a.e. in }\Omega. \end{equation} \end{remark} \begin{remark}\label{rmk1.1} \rm We have (see \cite[Remark 2.1]{ChaL2}) \begin{itemize} \item[(i)] $u\in C_{\rm loc}^{0,\alpha}( \Omega)$ for some $\alpha\in(0,1)$. As a consequence the set $\{u>0\}$ is open. \item[(ii)] If $a\in C_{\rm loc}^{0,\alpha}( \Omega)$ $(0<\alpha<1)$, then we have $u\in C_{\rm loc}^{1,\alpha} (\{u>0\})$. \end{itemize} \end{remark} The following three propositions were established in \cite{C} where the Dirichlet condition $u=0$ was imposed on $\Gamma$ instead of the Neuman boundary condition that we are considering in this work. The proofs are the same and will be omitted. \begin{proposition}\label{prop2.1} Let $(u,\chi)$ be a solution of {\rm (P1)}. We have \begin{equation}\label{e2.2} \chi_{x_2}\leq 0 \quad\text{in}\quad\mathcal{D}'(\Omega). \end{equation} \end{proposition} \begin{proposition} \label{prop2.2} Let $(u,\chi)$ be a solution of {\rm (P1)} and $x_0=(x_{01},x_{02})\in \Omega$. \begin{itemize} \item[(i)] If $u(x_0) > 0$, then there exists $\varepsilon >0$ such that $u(x_1,x_2) > 0$ for all $(x_1,x_2)\in C_\varepsilon(x_0)=B_\varepsilon(x_0)\cup \{ (x_1,x_2)\in \Omega: | x_2-x_{02}| <\varepsilon , \, x_2 < x_{02} \}$, where $B_\varepsilon(x_0)$ is the open ball of center $x_0$ and radius $r$. \item[(ii)] If $u(x_0)=0$, then $u(x_{01}, x_2) =0$ for all $x_2\geq x_{02}$. \end{itemize} \end{proposition} We then define the function $\Phi$ by \begin{equation}\label{e2.3} \Phi(x_1) =\begin{cases} d_0 \quad \text{if }\{x_2: (x_1,x_2)\in \Omega, \, u (x_1,x_2)> 0\}=\emptyset\\ \sup\{x_2: (x_1,x_2)\in \Omega, \, u(x_1,x_2) > 0\} \quad \text{otherwise.}\end{cases} \end{equation} Then the function $\Phi$ is well defined. \begin{proposition}\label{prop1.3} $\Phi$ is lower semi-continuous on $(a_0,b_0)$ and $$ \{u>0\}=\{x_2<\Phi(x_1)\}. $$ \end{proposition} The following lemma is an extension of \cite[Lemma 3.4]{C}. \begin{lemma}\label{lem1.1} Let $(u,\chi)$ be a solution of {\rm (P1)}. Let $(x_{11},\underline{x}_2), (x_{12},\underline{x}_2)\in \Omega$ with $x_{11}0$ small enough, one sets: \[ \alpha_\epsilon(x_1) = \min\Big(1,\frac{(x_1-x_{11})^+}{\epsilon},\frac{(x_{12}-x_1)^+}{\epsilon}\Big). \] Note that \begin{equation}\label{e2.4} \alpha_\epsilon(x_1) = \begin{cases} \frac{x_1-x_{11}}{\epsilon} & \text{for } x_1\in (x_{11},x_{11}+\epsilon) \\ 1 & \text{for } x_1\in (x_{11}+\epsilon,x_{12}-\epsilon) \\ \frac{x_{12}-x_1}{\epsilon} & \text{for } x_1\in (x_{12}-\epsilon,x_{12}) \end{cases} \end{equation} Then $\chi(D)\alpha_\epsilon\zeta$ is a test function for (P), and we have: \begin{equation}\label{e2.5} {\int_D }\big( a(x) \nabla u + \chi h(x)e\big) \cdot\nabla(\alpha_\epsilon\zeta) dx \leq \int_{\partial D\cap\Gamma}\beta(x,\varphi-u)\alpha_\epsilon\zeta d\sigma(x) \end{equation} We set $\xi_\epsilon=(1-\alpha_\epsilon)\zeta$, and for $\delta>0$, we denote by $H_\delta$ the following approximation of the Heaviside function i.e. the function defined by \begin{equation}\label{e2.6} H_\delta(s) =\min\Big(1,\frac{s^+}{\delta}\Big) =\begin{cases} 1 & \text{for} s\geq\delta\\ s/\delta & \text{for } 0\leq s\leq\delta\\ 0 & \text{for } s\leq0 \end{cases} \end{equation} Then $\chi(D)H_\delta(u)\xi_\epsilon$ is a test function for (P1), and we have $$ \int_D \big( a(x) \nabla u + \chi h(x)e\big) \cdot\nabla(H_\delta(u)\xi_\epsilon) dx \leq \int_{\partial D\cap\Gamma}\beta(x,\varphi-u)H_\delta(u)\xi_\epsilon d\sigma(x) $$ or \begin{align*} & \int_D \big( H_\delta(u)a(x) \nabla u \nabla \xi_\epsilon + H'_\delta(u)a(x) \nabla u \nabla u + \chi h(x)(H_\delta(u)\xi_\epsilon)_{x_{2}}\big) dx \\ &\leq\int_{\partial D\cap\Gamma}\beta(x,\varphi-u)H_\delta(u)\xi_\epsilon d\sigma(x). \end{align*} which leads by \eqref{e1.2} and the monotonicity of $H_\delta$ to $$ \int_D \big( H_\delta(u)a(x) \nabla u \nabla \xi_\epsilon + \chi h(x)(H_\delta(u)\xi_\epsilon)_{x_{2}}\big) dx \leq \int_{\partial D\cap\Gamma}\beta(x,\varphi-u)H_\delta(u)\xi_\epsilon d\sigma(x) $$ Hence we have \begin{equation} \label{e2.7} \begin{aligned} &\int_D H_\delta(u)a(x) \nabla u \nabla \xi_\epsilon dx + \int_{D\cap\{u>0\}} \big((h.H_\delta(u)\xi_\epsilon)_{x_{2}} - h_{x_{2}} H_\delta(u)\xi_\epsilon\big)dx \\ &\leq \int_{\partial D\cap\Gamma}\beta(x,\varphi-u)H_\delta(u)\xi_\epsilon d\sigma(x). \end{aligned} \end{equation} Since $$ \int_{D\cap\{u>0\}}(h.H_\delta(u)\xi_\epsilon)_{x_2}dx = \int_{\partial D\cap \Gamma}\big(h.H_\delta(u)\xi_\epsilon\big) \nu_2d\sigma(x)\geq0, $$ it follows from \eqref{e2.7} that \begin{equation} \label{e2.8} \int_D H_\delta(u)a(x) \nabla u \nabla \xi_\epsilon dx - \int_{D\cap\{u>0\}} h_{x_{2}} H_\delta(u)\xi_\epsilon dx \leq \int_{\partial D\cap\Gamma}\beta(x,\varphi-u)H_\delta(u)\xi_\epsilon d\sigma(x). \end{equation} Letting $\delta$ go to $0$ in \eqref{e2.8}, we obtain \begin{equation} \label{e2.9} \begin{aligned} &\int_D a(x)\nabla u \nabla((1-\alpha_\epsilon)\zeta)dx\\ &\leq \int_{D\cap\{u>0\}} h_{x_{2}} (1-\alpha_\epsilon)\zeta dx + \int_{\partial D\cap\Gamma}\beta(x,\varphi-u)(1-\alpha_\epsilon)\zeta d\sigma(x). \end{aligned} \end{equation} Now, from \eqref{e2.5} and \eqref{e2.9}, we deduce that \begin{align*} &\int_D \big( a(x) \nabla u + \chi h(x)e\big)\cdot\nabla\zeta dx \\ &= {\int_D }\big( a(x) \nabla u + \chi h(x)e\big) \cdot\nabla(\alpha_\epsilon\zeta) dx \\ &\quad+\int_D \big( a(x) \nabla u + \chi h(x)e\big)\nabla((1-\alpha_\epsilon)\zeta) dx \\ &\leq \int_{\partial D\cap\Gamma}\beta(x,\varphi-u)(\alpha_\epsilon\zeta)d\sigma(x) +\int_D \chi h(x)(1-\alpha_\epsilon)\zeta_{x_{2}}dx\\ &\quad +\int_{D\cap\{u>0\}}h_{x_{2}} (1-\alpha_\epsilon)\zeta dx +\int_{\partial D\cap\Gamma}\beta(x,\varphi-u)(1-\alpha_\epsilon)\zeta d\sigma(x)\\ &\leq\int_D \chi h(x)(1-\alpha_\epsilon)\zeta_{x_{2}}dx +\int_{D\cap\{u>0\}}h_{x_{2}} (1-\alpha_\epsilon)\zeta dx\\ &\quad +\int_{\partial D\cap\Gamma}\beta(x,\varphi-u)\zeta d\sigma(x). \end{align*} Taking into account \eqref{e2.4}, the result follows by letting $\epsilon$ approach $0$. \end{proof} \begin{proposition} \label{prop2.4} Let $(u,\chi)$ be a solution of {\rm (P1)} and $B_r(x_0)\subset \Omega$. If $u=0$ in $B_r(x_0)$, then we have \[ \chi(x_1,x_2)=\frac{\beta((x_1,\gamma(x_1)),\varphi(x_1,\gamma(x_1)))} {h(x)\nu_2(x_1,\gamma(x_1))} \quad\text{for a.e. } (x_1,x_2)\in C_r(x_0). \] \end{proposition} \begin{proof} Since $u(x)=0$ in $B_r(x_0)$, we obtain by Proposition \ref{prop2.2} $$ u=0\quad \text{in } C_r(x_0). $$ Moreover, since we have in the distributional sense $$ \operatorname{div}\big(a(x)\nabla u+\chi h(x)e\big)=0\quad\text{in } C_r(x_0), $$ we obtain in particular \begin{equation}\label{e2.10} \big(\chi h(x)\big)_{x_2}=0\quad \text{in } \mathcal{D}'(C_r(x_0)). \end{equation} Let $\xi\in H^{1}(C_{r}(x_{0}))$ such that $\xi=0$ on $\partial C_{r}(x_{0})\cap\Omega$. Then $\pm\chi(C_{r}(x_{0}))\xi$ are test functions for {\rm (P1)}, and we have \begin{equation}\label{e2.11} \int_{C_r(x_0)} \chi h \xi_{x_2} dx = \int_{\partial C_r(x_0)\cap\Gamma}\beta(x,\varphi)\xi d\sigma(x). \end{equation} Integrating by parts and using \eqref{e2.10}, we obtain \begin{equation}\label{e2.12} \int_{C_r(x_0)} \chi h\xi_{x_2} dx =\int_{\partial C_r(x_0)\cap\Gamma}\chi h\nu_{2}\xi d\sigma(x). \end{equation} We deduce then from \eqref{e2.11} and \eqref{e2.12} that \[ \int_{\partial C_r(x_0)\cap\Gamma}\chi h\nu_{2}\xi d\sigma(x) = \int_{\partial C_r(x_0)\cap\Gamma}\beta(x,\varphi)\xi d\sigma(x), \] for all $\xi\in H^{1}(C_{r}(x_{0}))$, $\xi=0$ on $\partial C_{r}(x_{0})\cap\Omega$, which leads to $\chi h\nu_{2}=\beta(x,\varphi)$, or \[ \chi(x)=\frac{\beta((x_1,\gamma(x_1)),\varphi(x_1,\gamma(x_1)))} {h(x)\nu_2(x_1,\gamma(x_1))}\quad\text{a.e. in } C_r(x_0). \] \end{proof} \begin{proposition} \label{prop2.5} Let $(u,\chi)$ be a solution of {\rm (P1)}. If the Lebesgue measure of the free boundary is zero, then we have \begin{align*} \chi=\chi_{\{u>0\}}+\frac{\beta(x,\varphi)}{h\nu_2} \chi_{\{u=0\}} \quad\text{for a.e. } x\in \Omega. \end{align*} \end{proposition} \begin{proof} From (P1)(i), we know that \begin{equation}\label{e2.13} \chi=1\quad \text{a.e. in }\{u>0\}. \end{equation} Let $x\in Int(\{u=0\})$. Then there exists a ball $B_r(x)$ of center $x$ and radius $r$ such that $B_r(x)\subset \{u=0\}$, i.e. $u=0$ in $B_r(x)$. By Proposition \ref{prop2.4}, we have $\chi=\frac{\beta(x,\varphi)}{h\nu_2}$ a.e. in $B_r(x)$. Therefore \begin{equation}\label{e2.14} \chi=\frac{\beta(x,\varphi)}{h\nu_2} \quad \text{a.e. in } \operatorname{Int}(\{u=0\}). \end{equation} Since $\partial\{u>0\}\cap\Omega$ is of measure zero, from \eqref{e2.13}-\eqref{e2.14} we obtain \begin{align*} \chi=\chi_{\{u>0\}}+\frac{\beta(x,\varphi)} {h\cdot\nu_2}\chi_{\{u=0\}} \quad\text{a.e. in }\Omega. \end{align*} \end{proof} \begin{proposition}\label{prop2.6} Let $(u, \chi)$ be a solution of {\rm (P1)}. If $x_0=(x_{01},x_{02})\in\Omega$ and $r>0$ are such that $B_r(x_0)\subset \Omega $, then we cannot have the following situations in $B_r(x_0)$: \begin{itemize} \item[(i)] $u(x)=0$ for $x_1=x_{01}$ and $u(x)>0$ for $x_1\neq x_{01}$. \item[(ii)] $u(x)=0$ for $x_1\geq x_{01}$ and $u(x)>0$ for $x_10$ for $ x_1>x_{01}$ and $u (x)=0$ for $x_1\leq x_{01}$. \end{itemize} \end{proposition} \begin{proof} Let $\xi\in\mathcal{D}(B_r), \xi\geq0$. Since $\pm\xi$ are test functions for (P1), we have \begin{equation}\label{e2.15} \int_{B_r(x_0)}\big(a(x)\nabla u+\chi h(x)e \big)\nabla\xi dx=0. \end{equation} (i) Under assumption (i), we have $\chi=1$ a.e. in $B_r(x_0)$. Taking into account \eqref{e1.5}, from \eqref{e2.15} we obtain \begin{equation*} \int_{B_r(x_0)}a(x)\nabla u\nabla\xi dx = -\int_{B_r(x_0)}h(x) \xi_{x_{2}}\geq0 \end{equation*} Using the maximum principle, we deduce that either $u>0$ in $B_r(x_0)$ or $u=0$ in $B_r(x_0)$, which both contradict $(i)$. (ii) From (P1)(i) and Proposition \ref{prop2.5}, we have under the assumption (ii), $\chi=1$ a.e. in $B_r^-=B_r(x_0)\cap\{x_1x_{01}\}$. Then from \eqref{e2.15}, it follows that \begin{equation} \label{e2.16} \begin{aligned} \int_{B_r^-}a(x)\nabla u\nabla\xi dx &=-\int_{B_r^-}\chi h(x)\xi_{x_{2}}dx -\int_{B_r^+}\chi h(x)\xi_{x_{2}}dx \\ &= - \int_{B_r^-}h(x)\xi_{x_{2}}dx-\int_{B_r^+} \frac{\beta(x,\varphi)}{\nu_2}\xi_{x_{2}}dx. \end{aligned} \end{equation} Integrating by parts, we have \begin{equation}\label{e2.17} \int_{B_r^+}\frac{\beta(x,\varphi)}{\nu_2}\xi_{x_{2}}dx = \int_{B_r^+}\big(\frac{\beta(x,\varphi)}{\nu_2}\xi\big)_{x_{2}}dx=0. \end{equation} It follows from \eqref{e2.16}-\eqref{e2.17} and taking into account \eqref{e1.5} that \begin{equation*} \int_{B_r(x_0)}a(x)\nabla u\nabla\xi dx = -\int_{B_r^-}h\xi_{x_{2}}=\int_{B_r^-}h_{x_2}\xi\geq0. \end{equation*} We deduce from the maximum principle that either $u>0$ in $B_r(x_0)$ or $u=0$ in $B_r(x_0)$, which both contradict (ii). The proof (iii) is similar to the proof of (ii), an it is omitted. \end{proof} \begin{theorem}\label{thm2.1} Let $(u, \chi)$ be a solution of {\rm (P1)}. Then we have $$ \chi\geq \kappa=\min\Big(1, \frac{\beta(x,\varphi)}{h\nu_2}\Big)\quad \text{a.e. in }\Omega. $$ \end{theorem} To prove the above theorem we need the following lemma. \begin{lemma}\label{lem2.2} Let $(a,b)\subset(a_0,b_0)$, $y_0$ such that $(a,b)\times\{y_0\}\subset \Omega$ and let $D=((a,b)\times(y_0,\infty))\cap\Omega$. Then we have \[ \int_{D} h(\kappa - \chi)^{+}\xi_{x_2} dx\leq 0,\quad \forall \xi\in H^1(D),\; \xi\geq 0,\; \xi=0\text{ on }(\partial D)\cap\Omega. \] \end{lemma} \begin{proof} Let $\Gamma'=\{(x_{1},\gamma(x_{1})): x_1\in(a,b)\}$, and $\xi\in H^1(D)$ such that $\xi\geq0$ and $\xi=0$ on $(\partial D)\cap\Omega$. Using $\pm\chi(D)(H_\delta(u)-1)\xi$ as test functions for {\rm (P1)}, we obtain \[ \int_D \big(a(x)\nabla u+\chi he\big) \nabla\big((H_\delta(u)-1)\xi\big)dx= \int_{\Gamma'} \beta(x,\varphi-u)(H_\delta(u)-1)\xi d\sigma(x) \] which can be written as \begin{align*} &\int_D H'_\delta(u)a(x) \nabla u \cdot\nabla u .\xi dx + \int_D (H_\delta(u)-1)a(x) \nabla u \nabla \xi dx \\ &+\int_D \chi h[H_\delta(u)\xi]_{x_2} dx-\int_D\chi h(x)\xi_{x_2} dx \\ &\leq \int_{\Gamma'} \beta(x,\varphi-u)(H_\delta(u)-1)\xi d\sigma(x)\,. \end{align*} By taking into account the monotonicity of $H_\delta$, integrating by parts, and using the ellipticity of $a(x)$, we have \begin{equation}\label{e2.18} \begin{aligned} -\int_D \chi h\xi_{x_2} dx &\leq\int_D (1-H_\delta(u))a(x) \nabla u \nabla \xi dx -\int_D h[H_\delta(u)\xi]_{x_2} dx \\ &\quad+\int_{\Gamma'} \beta(x,\varphi-u)(H_\delta(u)-1)\xi d\sigma(x) \\ &=\int_D (1-H_\delta(u))a(x) \nabla u\cdot\nabla \xi dx +\int_D H_\delta(u) h_{x_2}\xi dx \\ &\quad+\int_{\Gamma'}(H_\delta(u)-1) [\beta(x,\varphi-u)-h\nu_2]\xi d\sigma(x) -\int_{\Gamma'} h\nu_2 \xi d\sigma(x). \end{aligned} \end{equation} Next, integrating by parts again, we have \begin{equation}\label{e2.19} \int_D\kappa h\xi_{x_2} dx=\int_{\Gamma'}\kappa h\nu_2 \xi d\sigma(x) -\int_D(\kappa h)_{x_2}\xi dx. \end{equation} Adding \eqref{e2.18} and \eqref{e2.19}, and using the fact that $(\kappa h)_{x_2}=h_{x_2}\chi_{\{\beta(x,\varphi)\geq h\cdot\nu_2\}}\geq0$, we obtain \begin{equation} \label{e2.20} \begin{aligned} \int_D(\kappa-\chi) h\xi_{x_2} dx &\leq\int_D (1-H_\delta(u))a(x) \nabla u\cdot\nabla \xi \, dx +\int_D H_\delta(u)h_{x_2}(x)\xi dx\\ &\quad +\int_{\Gamma'}(H_\delta(u)-1) [\beta(x,\varphi-u)-h\nu_2]\xi d\sigma(x)\\ &\quad +\int_{\Gamma'} ( \kappa-1 )h\nu_2 \xi d\sigma(x). \end{aligned} \end{equation} Letting $\delta$ go to $0$ in \eqref{e2.20}, we obtain \begin{align*} \int_D h(\kappa-\chi)\xi_{x_2} dx &\leq -\int_{\Gamma'\cap \{u=0\}}(\beta(x,\varphi)-h\nu_2]\xi d\sigma(x) \\ &\quad +\int_D\chi_{\{u>0\}}h_{x_2}\xi dx +\int_{\Gamma'}( \kappa-1 )h\nu_2 \xi d\sigma(x) \end{align*} which leads to \begin{align*} &\int_D h(\kappa-\chi)\xi_{x_2} dx \\ &\leq \int_D \chi_{\{u>0\}}h_{x_2}\xi dx +\int_{\Gamma'\cap \{u=0\}} \Big(\kappa-\frac{\beta(x,\varphi)}{h\nu_2} \Big)h\nu_2 \xi d\sigma(x) \\ &\quad +\int_{\Gamma'\cap \{u>0\}}( \kappa-1 )h\nu_2 \xi d\sigma(x) \\ &\leq \int_D\chi_{\{u>0\}}h_{x_2}\xi dx \quad \forall \xi\in H^{1}(D), \quad \xi\geq0,\; \xi=0\text{ on }\partial D\cap\Omega. \end{align*} Hence we arrive at \begin{equation} \label{e2.21} \int_D(h\kappa-h\chi)\xi_{x_2} dx \leq \int_D\chi_{\{u>0\}}h_{x_2}\xi dx, \end{equation} for all $\xi\in H^1(D)$, $\xi\geq0$, $\xi=0$ on $\partial D\cap\Omega$. Now we extend the functions $h\kappa$, $h\chi$, and $\chi_{\{u>0\}}h_{x_2}$ by $0$ and denote the extensions respectively by $\overline{\kappa}$, $\overline{\chi}$, and $\theta$. In particular \eqref{e2.21} holds for any $\xi\in C^{0,1}(\mathbb{R}^2), \xi\geq0$, with $\xi$ having a compact support in $D\cup\Gamma'$. Let $\xi$ be such a function and let \[ \epsilon_0 = \operatorname{dist}(\operatorname{supp}(\xi_{|\overline{D}}), \partial D\cap\Omega). \] Then for each $y\in B_{\epsilon}(O)$, and $\epsilon\in (0,\epsilon_0/2)$, the function $x\to \xi(x+y)$ is nonnegative, belongs to $C^{0,1}(\mathbb{R}^2)$, and has a compact support in $D\cup\Gamma'$. Therefore we obtain from \eqref{e2.21} $$ \int_{\mathbb{R}^2} (\overline{\kappa}-\overline{\chi})\xi_{x_2}(x+y) dx \leq \int_{\mathbb{R}^2}\theta(x)\xi(x+y) dx $$ which leads to $$ \int_{\mathbb{R}^2}\rho_{\epsilon}(y)\Big(\int_{\mathbb{R}^2} (\overline{\kappa}-\overline{\chi}).\xi_{x_2}(x+y) dx \Big)dy \leq \int_{\mathbb{R}^2}\rho_{\epsilon}(y) \Big(\int_{\mathbb{R}^2}\theta(x)\xi(x+y) dx\Big)dy $$ where $\rho_{\epsilon}$ is a smooth function satisfying $\rho_{\epsilon}\geq 0, supp\rho_{\epsilon}\subset B_{\epsilon}(O)$ and ${\int_{\mathbb{R}^2}\rho_{\epsilon}=1}$. Writing $f_{\epsilon}=\rho_{\epsilon}*f$ for a function $f$, we obtain \[ \int_{\mathbb{R}^2}(\overline{\kappa}_{\epsilon}(x) -\overline{\chi}_{\epsilon}(x))\xi_{x_2}dx \leq\int_{\mathbb{R}^2}\theta_{\epsilon}(x)\xi dx\quad \forall\xi\in C^{0,1}(\mathbb{R}^2),\; \xi\geq0,; \operatorname{supp}(\xi)\subset D\cup\Gamma'. \] In particular, we obtain for the function $\xi=\min\big(1,\frac{(\overline{\kappa}_{\epsilon}-\overline{ \chi}_{ \epsilon})^+}{\delta}\big)\zeta$, with $\delta>0$, $\zeta\in C^{0,1}(\mathbb{R}^2)$, $\zeta\geq0$, $\operatorname{supp}(\zeta)\subset D\cup\Gamma'$, \begin{equation} \label{e2.22} \begin{aligned} &\int_{\mathbb{R}^2}(\overline{\kappa}_{\varepsilon} -\overline{\chi}_{\varepsilon}) \Big(\min\Big(1,\frac{(\overline{\kappa}_{\varepsilon} -\overline{\chi}_{\varepsilon})^+}{\delta}\Big) \zeta\Big)_{x_2}dx \\ &\leq\int_{\mathbb{R}^2} \theta_{\varepsilon}(x) \min\Big(1,\frac{(\overline{\kappa}_{\varepsilon} -\overline{\chi}_{\varepsilon})^+}{\delta}\Big)\zeta dx\\ &\leq\int_{\mathbb{R}^2} \theta_{\varepsilon}(x)\zeta dx. \end{aligned} \end{equation} Note that \begin{equation} \label{e2.23} \begin{aligned} &\int_{\mathbb{R}^2}(\overline{\kappa}_{\varepsilon} -\overline{\chi}_{\varepsilon})\Big(\min\Big(1,\frac{(\overline{\kappa}_{\varepsilon} -\overline{\chi}_{\varepsilon})^+}{\delta}\Big)\zeta\Big)_{x_2}dx \\ &=\int_{\mathbb{R}^2}\min\Big(1,\frac{(\overline{\kappa}_{\epsilon} -\overline{\chi}_{\epsilon})^+}{\delta}\Big) (\overline{\kappa}_{\epsilon}-\overline{\chi}_{\epsilon})\zeta_{x_2}dx \\ &\quad+\int_{\mathbb{R}^2} (\overline{\kappa}_{\epsilon} -\overline{\chi}_{\epsilon}) \Big(\min\Big(1,\frac{(\overline{\kappa}_{\epsilon} -\overline{\chi}_{\epsilon})^+}{\delta}\Big)\Big)_{x_2}\zeta dx. \end{aligned} \end{equation} As $\delta\to0$, we have \begin{equation} \label{e2.24} \begin{aligned} \int_{\mathbb{R}^2}\min\Big(1,\frac{(\overline{\kappa}_{\epsilon} -\overline{\chi}_{\epsilon})^+}{\delta}\Big) (\overline{\kappa}_{\epsilon}-\overline{\chi}_{\epsilon})\zeta_{x_2}dx \to \int_{\mathbb{R}^2} (\overline{\kappa}_{\epsilon}-\overline{\chi}_{\epsilon})^+ \zeta_{x_2} dx \end{aligned} \end{equation} and \begin{equation}\label{e2.25} \begin{aligned} &\int_{\mathbb{R}^2} (\overline{\kappa}_{\epsilon}-\overline{\chi}_{\epsilon}) \Big(\min\Big(1,\frac{(\overline{\kappa}_{\epsilon} -\overline{\chi}_{\epsilon})^+}{\delta}\Big)\Big)_{x_2}\zeta dx \\ &=\int_{\mathbb{R}^2 \cap\{0<\overline{\kappa}_{\epsilon} -\overline{\chi}_{\epsilon}<\delta\}} \frac{\zeta}{\delta} (\overline{\kappa}_{\epsilon}-\overline{\chi}_{\epsilon})^+ (\overline{\kappa}_{\varepsilon}-\overline{\chi}_{\varepsilon})_{x_2}dx \\ &=\int_{\mathbb{R}^2} \frac{\zeta}{2\delta} ((\min(\delta,(\overline{\kappa}_{\epsilon} -\overline{\chi}_{\epsilon})^+))^2)_{x_2}dx \\ &=-\int_{\mathbb{R}^2} (\min(\delta,(\overline{\kappa}_{\epsilon} -\overline{\chi}_{\epsilon})^+))^2\frac{\zeta_{x_2}}{2\delta}dx \to 0. \end{aligned} \end{equation} It follows from \eqref{e2.22}-\eqref{e2.25} that $$ \int_{\mathbb{R}^2} (\overline{\kappa}_{\epsilon} -\overline{\chi}_{\epsilon})^+ \zeta_{x_2} dx \leq \int_{\mathbb{R}^2} \theta_{\epsilon}\zeta dx $$ which by letting $\epsilon\to0$ leads to \begin{equation}\label{e2.26} \int_D (\kappa-\chi)^+h\zeta_{x_2} dx \leq \int_D \chi_{\{u>0\}} h_{x_2}\zeta dx \end{equation} Taking $\zeta=(1-H_\delta(u))\zeta$ with $\zeta\in H^1(\mathbb{R}^2)$, $\zeta\geq0$, $\operatorname{supp}(\zeta)\subset D\cup\Gamma'$ in \eqref{e2.26}, we obtain \begin{equation}\label{e2.27} \begin{aligned} &\int_D h(\kappa-\chi)^+\zeta_{x_2} dx\\ &\leq \int_D (1-H_\delta(u)) \chi_{\{u>0\}}h_{x_2}\zeta dx +\int_D \zeta h(\kappa-\chi)^+(H_\delta(u))_{x_2} dx \end{aligned} \end{equation} Since $\chi=1$ a.e. in $\{u>0\}$ and $\kappa\leq 1$, we have \begin{equation}\label{e2.28} \int_D \zeta h(\kappa-\chi)^+(H_\delta(u))_{x_2}dx =\int_{D\cap\{u>0\}} \zeta h(\kappa-1)^+(H_\delta(u))_{x_2} dx=0. \end{equation} Then we deduce from \eqref{e2.27}--\eqref{e2.28} that \begin{equation}\label{e2.29} \int_D h(\kappa-\chi)^+\zeta_{x_2} dx \leq \int_D (1-H_\delta(u)) \chi_{\{u>0\}}h_{x_2}\zeta dx. \end{equation} Letting $\delta$ go to $0$ in \eqref{e2.29}, we obtain \[ \int_D h(\kappa-\chi)^+\zeta_{x_2} dx\leq 0\quad \forall \zeta \in H^1(D),\; \zeta\geq 0,\; \zeta=0 \text{ on }\partial D\cap \Omega \] which completes the proof of the lemma. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2.1}] Let $D$ be a domain below $\Gamma$ given by $D= ((a,b))\times(y_0,\infty) )\cap \Omega$, with $\{y_0\}\times(a,b) \subset \Omega$. By Lemma \ref{lem2.2}, we have for $\zeta= l(x_1) (y-y_0) \chi (D)$, where $ l(x_1)= (x_1-a) (b-x_1)$ \begin{equation*} \int_D h(\kappa-\chi)^+(x_1-a)(b-x_1)dx \leq 0 \end{equation*} which leads to $(\kappa-\chi)^+=0$ a.e. in $D$ and $\kappa\leq\chi$ a.e. in $D$. Since this holds for all such domains $D$, we obtain $\chi\geq\kappa$ a.e in $\Omega$. \end{proof} \begin{corollary}\label{coro2.1} Let $(u, \chi)$ be a solution of {\rm (P1)} and let $\Gamma'=\{(x_{1},\gamma(x_{1})),x_1\in(a,b)\}\subset\Gamma$, with $a_0 1\quad \text{ a.e. in }\Gamma'. $$ Then we have $u>0$ and $\chi=1$ below $\Gamma'$. \end{corollary} \begin{proof} Let $D= ((a,b))\times(y_0,\infty) )\cap \Omega$ such that $\{y_0\}\times(a,b) \subset \Omega$. We deduce from Theorem \ref{thm2.1} and the assumption that $\chi=1$ a.e. in $D$. Using (P1)(ii) and \eqref{e1.5}, we obtain $\operatorname{div}(a(x)\nabla u)=-h_{x_2}\leq 0$ in $D$, which leads by the maximum principle to either $u=0$ in $D$ or $u>0$ in $D$. Now, assume that $u=0$ in $D$ and let $\zeta\in H^1(D)$ with $\zeta\geq 0$ in $D$ and $\zeta=0$ on $\partial D\cap\Omega$. Since $\pm\chi(D)\xi$ are a test functions for {\rm (P1)}, we have \[ \int_D h\zeta_{x_{2}}=\int_{\Gamma'}\beta(x,\varphi)\zeta d\sigma(x) \] which becomes after integrating by parts and taking into account \eqref{e1.5} \[ \int_{\Gamma'}(\beta(x,\varphi)-h\nu_2)\zeta d\sigma(x) =-\int_D \zeta h_{x_2}\leq 0. \] Since $\zeta$ is arbitrary, we obtain $\beta(x,\varphi)-h\nu_2\leq 0$ a.e. in $\Gamma'$, which contradicts the assumption. \end{proof} \section{A barrier function}\label{3} For the rest of this article, we assume that \begin{gather} \label{e3.1} a\in C^{0,\alpha}_{\rm loc}( \Omega),\quad \alpha\in(0,1),\\ h_{x_2} \leq c_0 \quad \text{in } \mathcal{D}'(\Omega). \label{e3.2} \end{gather} In this section, we construct a barrier function that will be used in Section 4 to establish the continuity of the function $\phi$. \begin{theorem}\label{thm3.1} Let $(u, \chi)$ be a solution of {\rm (P1)}. Then $ u\in C^{0,1}_{\rm loc}(\Omega)$. \end{theorem} \begin{proof} We refer to \cite{ChaL2} when $H(x)=h(x)e$, and to \cite{ChaL3} or \cite{Ly5} for more general situations. Let $(x_{11},\underline{x}_2)$, $(x_{12}, \underline{x}_2)\in \Omega$ such that $x_{11}0\})+\frac{\beta(x,\varphi)}{h\nu_2} \chi (\{v=0\})$. The main result of this section is the following Lemma. \begin{lemma}\label{lem3.1} Assume that for some positive number $\mu$ we have \begin{equation}\label{e3.5} h(x)-\frac{\beta(x,\varphi(x))}{\nu_2} >\mu\quad\text{on }T. \end{equation} Then for $\epsilon>0$ small enough we have \begin{equation}\label{e3.6} \begin{gathered} \int_{D}\Big(a(x) \nabla v + \theta h(x)e\Big)\cdot\nabla\zeta \geq \int_{\Gamma}\beta(x,\varphi)d\sigma(x) \\ \forall \zeta\in H^1(D),\; \zeta\geq 0,\; \zeta=0 \text{ on }\partial D\setminus\Gamma. \end{gathered} \end{equation} \end{lemma} \begin{proof} Let $\nu$ be the outward unit normal vector to $D$. First we have by \eqref{e1.1}, \eqref{e3.4} and \eqref{e3.5} for $\epsilon $ small enough \begin{equation}\label{e3.7} a(x) \nabla v. \nu +h(x)-\frac{\beta(x,\varphi)}{\nu_2} \geq -\Lambda C .\epsilon^{(1-\frac{2}{p})} +\mu\geq 0 \quad\text{on } T. \end{equation} Next, for $\zeta\in H^1(D)$, $\zeta\geq 0$, $\zeta=0$ on $\partial D\setminus\Gamma$, by integrating by parts and using \eqref{e3.5} and \eqref{e3.7} we obtain \begin{align*} &\int_D\Big(a(x) \nabla v +\theta h(x)e\Big)\cdot\nabla\zeta dx\\ &=\int_{D\cap [v>0]}\Big(a(x) \nabla v + \theta h(x)e\Big)\cdot\nabla\zeta dx +\int_{D\cap [v=0]}\frac{\beta(x,\varphi)}{\nu_2} \zeta_{x_2} dx\\ &=- \int_{D\cap [v>0]}\Big(div (a(x) \nabla v) + h_{x_2}(x)\Big)\zeta dx +\int_{\Gamma}\beta(x,\varphi)\zeta d\sigma(x)\\ &\quad +\int_T\Big(a(x) \nabla v\cdot \nu + h(x)-\frac{\beta(x,\varphi)}{\nu_2} \Big)\zeta d\sigma\\ &\geq \int_{\Gamma}\beta(x,\varphi)\zeta d\sigma(x). \end{align*} \end{proof} \section{Continuity of the free boundary}\label{4} This last section is devoted to the upper semi-continuity of $\phi$. We assume that \begin{gather} \label{e4.1} \gamma\in C_{\rm loc}^{1,\alpha}(a_0,b_0), \\ a\in C_{\rm loc}^{0,\alpha}(\Omega\cup\Gamma), \label{e4.2}\\ \beta(x,\varphi)-h\nu_2\in C_{\rm loc}^{0}(\Gamma). \label{e4.3} \end{gather} The main result is the following theorem. \begin{theorem}\label{thm4.1} Let $x_{01}\in (a_0,b_0)$ such that $(x_{01},\Phi(x_{01}))\in\Omega$ and \begin{equation}\label{e4.4} \frac{\beta(x_{01},\gamma(x_{01}),\varphi(x_{01},\gamma(x_{01}) ) )} {\nu_2(x_{01},\gamma(x_{01})) } x_{01}$ such that $u(x_1,\gamma(x_1))=0$ for all $x_1\in (x_{01},x_{12})$. \end{itemize} \end{lemma} \begin{proof} Let $\mu>0$ small enough and $Z_\mu=\big((x_{01}-\mu,x_{01}+\mu)\times(x_{02},+\infty)\big)\cap\Omega$. We denote by $v$ the unique solution in $H^1(Z_\mu)$ of \begin{equation}\label{e4.5} \begin{gathered} \operatorname{div}( a(x) \nabla v)= - h_{x_2} \quad \text{in }Z_\mu \\ v= \varphi(x)= \gamma(x_1)-x_2 \quad \text{on }\partial Z_\mu. \end{gathered} \end{equation} We have (see \cite[p. 212]{GT}) $v\in C_{\rm loc}^{1, \alpha}\big(Z_\mu\cup \Gamma_\mu)$, where $\Gamma_\mu=\{(x_1,\gamma(x_1))~/~x_1\in(x_{01}-\mu,x_{01}+\mu)~\}$. Moreover, given the sign of the functions $\gamma(x_1)-x_2$ and $h_{x_2}$, we have by the maximum principle $v>0$ in $Z_\mu$. Since $\beta(x,\varphi)x_{01}$ and $u(x_{12},x_{22})=0$. \end{itemize} Let us assume for example that (i) holds. Set $\underline{x}_2=\max(x_{02},x_{21})$, $Z=\big((x_{11},x_{01})\times(\underline{x}_2,+\infty)\big)\cap\Omega$, $v_\epsilon=\lambda_\epsilon v$ and assume that $\mu_1$ is small enough. Since $(x_{11},x_{01})\times\{\underline{x}_2\}\subset \overline{B}_{\mu_1}(x_0)$, we have by \eqref{e4.7} \begin{equation}\label{e4.8} u(x_1,\underline{x}_2)\leq v_\epsilon(x_1,\underline{x}_2),\quad \forall x_1\in(x_{11},x_{01}) \end{equation} Moreover, since $u(x_{11},\underline{x}_2)=u(x_{01},\underline{x}_2)=0$, we obtain by Proposition \ref{prop2.2} (ii) that \begin{equation}\label{e4.9} u(x_{11},x_2)=u(x_{01},x_2)=0,\quad \forall x_2\geq \underline{x}_{2} \end{equation} Taking into account \eqref{e4.8}-\eqref{e4.9}, we have $u\leq v_\epsilon$ on $(\partial Z)\cap\Omega$. Therefore we obtain by using $(u-v_\epsilon)^+\chi_Z\in H^1(\Omega)$ as a test function in (P1) (ii) and \eqref{e4.5} \begin{gather}\label{e4.10} \begin{aligned} &\int_{Z} (a(x)\nabla u+\chi h(x)e)\nabla(u-v_\epsilon)^+ dx\\ &=\int_{(\partial Z)\cap\Gamma}\beta(x,\varphi-u)(u-v_\epsilon)^+ d\sigma(x). \end{aligned}\\ \begin{aligned} &\int_{Z} (a(x)\nabla u+h(x)e)\nabla(u-v_\epsilon)^+ dx \\ &=\int_{(\partial Z)\cap\Gamma}(\lambda_\epsilon a(x)(\nabla v) \cdot\nu+h(x)\nu_2)(u-v_\epsilon)^+ d\sigma(x). \end{aligned} \label{e4.11} \end{gather} Subtracting \eqref{e4.11} from \eqref{e4.10} and using \eqref{e4.6} and the fact that $\chi h(x)e\nabla(u-v_\epsilon)^+=h(x)e\nabla(u-v_\epsilon)^+$ a.e. in $Z$, we obtain \begin{equation}\label{e4.12} \begin{aligned} &\int_{Z} a(x)\nabla (u-v_\epsilon)^+\cdot\nabla(u-v_\epsilon)^+ dx \\ &=\int_{(\partial Z)\cap\Gamma}\big(\beta(x,\varphi-u) -\lambda_\epsilon a(x)(\nabla v)\cdot\nu-h(x)\nu_2\big)(u-v_\epsilon)^+ d\sigma(x) \\ &\leq\int_{(\partial Z)\cap\Gamma}\big(\beta(x,\varphi) -\lambda_\epsilon a(x)(\nabla v)\cdot\nu-h(x)\nu_2\big)(u-v_\epsilon)^+ d\sigma(x) \leq0. \end{aligned} \end{equation} We deduce from \eqref{e4.12} and \eqref{e1.2} that $\nabla(u-v_\epsilon)^+=0$ a.e. in $Z$. Since $(u-v_\epsilon)^+=0$ on $(\partial Z)\cap\Omega$, we obtain $u\leq v_\epsilon$ in $Z$ and in particular $u(x_1,\gamma(x_1))=0\quad\forall x_1\in (x_{11},x_{01})$. If (ii) holds, in a similar way we obtain that $u(x_1,\gamma(x_1))=0\quad\forall x_1\in (x_{01},x_{12})$. \end{proof} \begin{lemma}\label{lem4.2} Let $\epsilon>0$ small enough and let $v$ be the barrier function defined by \eqref{e3.3}. Assume that $(u, \chi)$ is a solution of {\rm (P1)} such that: \begin{gather}\label{e4.13} u(x_1,\underline{x}_2) \leq v(x_1,\underline{x}_2) \forall x_1\in (x_{11},x_{12}) \\ u(x_{i1},\underline{x}_2)=0 \quad i=1,2 \label{e4.14} \\ u(x_1,\gamma(x_1))=0\quad\forall x_1\in (x_{11},x_{12}). \label{e4.15} \end{gather} Then for $\delta>0$ and $D_\delta= D\cap \{v>0\} \cap\{0< u-v< \delta\}$ we have $$ \lim_{\delta\to 0} \frac{1}{\delta} \int_{D_\delta } a(x) \nabla (u- v)^+. \nabla (u- v)^+ dx = 0. $$ \end{lemma} \begin{proof} Let $\delta, \eta>0$. We consider the function $d_\eta(x_2)=H_\eta(x_2-\overline{x}_2),$ where $H_\eta(s)$ was defined in \eqref{e2.6} and $\overline{x}_2=\underline{x}_2+\varepsilon$. Then we introduce the non-negative function $\zeta=H_\delta(u-v)+ d_\eta(1-H_\delta(u))$ which belongs to $H^1(D)\cap L^\infty(D)$. Using the fact that $d_\eta(\underline{x}_2)=0$ and \eqref{e4.13}, we see that $\zeta$ vanishes on $\{x=\underline{x}_2\}$. Therefore by Lemma \ref{lem2.2} we have \begin{equation}\label{e4.16} \begin{aligned} &\int_D (a(x)\nabla u+\chi h(x)e)\nabla(H_\delta(u-v))dx \\ &\leq\int_\Gamma \beta (x,\varphi-u)(H_\delta(u-v)+d_\eta(1-H_\delta(u)))d\sigma(x) \\ &\quad-\int_D (a(x)\nabla u+\chi h(x)e)\nabla(d_\eta(1-H_\delta(u)))dx. \end{aligned} \end{equation} Given that $u(x_{1i},\underline{x}_2)=0$ for $i=1,2$, from Proposition \ref{prop2.2} (ii) we deduce that $u(x_{1i},x_2)=0, \forall x_2\geq \underline{x}_2, i=1,2$. This leads to \begin{equation}\label{e4.17} u(x_{1i},x_2)\leq v(x_{1i},x_2),\quad \forall x_2\geq \underline{x}_2,\; i=1,2. \end{equation} Combining \eqref{e4.13} and \eqref{e4.17}, we obtain $u \leq v$ on $\partial D\cap\Omega$, and therefore since $H_\delta(s)=0$ for $s\leq0$, we obtain $H_\delta(u-v)=0$ on $\partial D\cap\Omega= \partial D\backslash\Gamma$. So using $H_\delta(u-v)$ as a test function in \eqref{e3.6} we obtain \begin{equation}\label{e4.18} -\int_D (a(x)\nabla v+\theta h(x)e)\nabla(H_\delta(u-v))dx \leq - \int_\Gamma \beta (x,\varphi)H_\delta(u-v)d\sigma(x)=0. \end{equation} Adding \eqref{e4.16} and \eqref{e4.18}, we obtain by taking into account the fact that $u=0$ on $\Gamma\cap\partial D$, \begin{align*} &\int_D a(x)\nabla (u-v)\nabla(H_\delta(u-v))dx \\ &\leq \int_D (\theta-\chi)h(x)(H_\delta(u-v))_{x_2}dx\\ &\quad -\int_D (a(x)\nabla u+\chi h(x)e)\nabla(d_\eta(1-H_\delta(u-v)))dx \\ &\quad+\int_\Gamma \beta (x,\varphi)d_\eta(1-H_\delta(u))d\sigma(x)\,. \end{align*} Taking into account that $u=0$ on $\Gamma\cap\partial D$ and $d_\eta=0$ in $\{v>0\}$ the above inequality becomes \begin{equation}\label{e4.19} \begin{aligned} &\int_{D\cap\{v>0\}} H'_\delta(u) a(x)\nabla (u-v)\nabla(u-v)dx \\ &\leq -\int_{D\cap\{v=0\}} H'_\delta(u) a(x)\nabla u\nabla u -\int_{D\cap\{v=0\}}\chi h(x)(H_\delta(u))_{x_2}dx \\ &\quad +\int_{D\cap\{v=0\}} \frac{\beta(x,\varphi)}{\nu_2}(H_\delta(u))_{x_2}dx \\ &\quad +\int_{D\cap\{v=0\}} (a(x)\nabla u+\chi h(x)e)\nabla((1-d_\eta)(1-H_\delta(u)))dx \\ &\quad+\int_{D\cap\{v=0\}} (a(x)\nabla u+\chi h(x)e)\nabla(H_\delta(u))dx \\ &=I_1^\delta+ I_2^\delta+ I_3^\delta+ I_4^\delta+ I_5^\delta. \end{aligned} \end{equation} We observe that $I_1^\delta+ I_2^\delta+ I_5^\delta=0$. Moreover, integrating by parts, we have since $u=0$ on $\Gamma\cap\partial D$ \begin{equation}\label{e4.20} \begin{aligned} I_3^\delta &= \int_\Gamma \beta(x,\varphi)H_\delta(u) d\sigma(x)-\int_{\{x_2=\overline{x}_2\}} \frac{\beta(x,\varphi)}{\nu_2}(H_\delta(u))dx_1 \\ &= -\int_{\{x_2=\overline{x}_2\}} \frac{\beta(x,\varphi)}{\nu_2}(H_\delta(u))dx_1 \leq 0. \end{aligned} \end{equation} From \eqref{e4.19}-\eqref{e4.20} we obtain \begin{align*} &\int_{D\cap\{v>0\}} H'_\delta(u) a(x)\nabla (u-v)\nabla(u-v)dx\\ &\leq\int_{D\cap\{v=0\}} (a(x)\nabla u+\chi h(x)e) \nabla((1-d_\eta)(1-H_\delta(u)))dx. \end{align*} At this point the proof follows step by step the one of \cite[Lemma 5.1]{ChaL2}. \end{proof} \begin{proof}[Proof of Theorem \ref{thm4.1}] Let $\epsilon >0$ be small enough. Let $x_{01}\in (a_0,b_0)$. Set $x_0= (x_{01},\phi(x_{01}))=(x_{01},x_{02})$ and assume that $x_0\in \Omega$. Using the continuity of $\beta(x,\varphi)-h(x)\nu_2$ at $\overline{x}_0=(x_{01},\gamma(x_{01}))$, there exists for $\epsilon$ small enough a positive number $\mu_\epsilon$ such that \begin{equation}\label{e4.21} h(x)\nu_2-\beta(x,\varphi)>\mu_\epsilon\quad\text{on } \Gamma_\epsilon=\{(x_1,\gamma(x_1)):x_1\in(x_{01}-\epsilon,x_{01}+\epsilon)\}. \end{equation} Since $u(x_0)=0$ and $u$ is continuous, there exists $\eta_1 \in (0, \epsilon)$ such that \begin{equation}\label{e4.22} u(x_1,x_2) \leq \epsilon^2 \quad \forall (x_1,x_2)\in B_{\eta_1} (x_0). \end{equation} By Proposition \ref{prop2.6}, one of the following situations holds \begin{itemize} \item[(i)] There exists $x_n=(x_{n1},x_{n2}) \in B_{\eta_1} (x_0)$ such that $ x_{n1}x_{01}$, $u(x_{n1},x_{n2})=0$ and $\lim_{n\to\infty}x_n=x_0$. \end{itemize} Let us assume that $i)$ holds. Then there exists by Lemma \ref{lem4.1} a positive integer $n$ large enough such that \begin{equation}\label{e4.23} u(x_1,\gamma(x_1))=0\quad\forall x_1\in (x_{n1},x_{01}). \end{equation} Set $\underline{x}_2=\max (\phi(x_{01}),x_{n2})$ and assume that $\epsilon$ is small enough so that $$ (x_{n1}-\epsilon, x_{01}+\epsilon)\times(\underline{x}_2-2\epsilon, \underline{x}_2 + 2\epsilon) \Subset \Omega. $$ Let $v_1$ be the barrier function defined by \eqref{e3.3} in the set $Z_1= (x_{n1}-\epsilon, x_{01}+\epsilon)\times(\underline{x}_2,\underline{x}_2 + 2\epsilon)$. We consider the extension by $0$ of $v_1$ to $D_1= \big((x_{n1},x_{01}) \times (\underline{x}_2, + \infty)\big) \cap \Omega$. Taking into account \eqref{e4.4}, we see that $v_1$ satisfies \eqref{e3.6}. Now since $(x_{n1},x_{01}) \times \{\underline{x}_2\}\subset B_{\eta_1} (x_0)$, by \eqref{e4.22} we have \begin{equation}\label{e4.24} u(x_1,\underline{x}_2) \leq \epsilon^2=v_1(x_1,\underline{x}_2) \quad \forall x_1\in (x_{n1},x_{01}). \end{equation} Moreover since $u(x_{n1},\underline{x}_2)=u(x_{01},\underline{x}_2)=0$, by Proposition \ref{prop2.2} (ii) we obtain \begin{equation}\label{e4.25} u(x_{n1},x_2)=u(x_{01},x_2)=0 \quad \forall x_2\geq \underline{x}_2. \end{equation} Combining \eqref{e4.23}-\eqref{e4.25}, we see that Lemma \ref{lem4.2} holds for $D_1=(x_{n1},x_{01}) \times (\underline{x}_2,\underline{x}_2+\epsilon)$. Then we can argue as in \cite{ChaL2} to obtain for $\Delta_1=(x_{n1},x_{01}) \times (\underline{x}_2-\epsilon,\underline{x}_2+\epsilon)$ $$ \int_{\Delta_1} a(x) \nabla (u-v_1)^+. \nabla\zeta dx\, = \,0 \quad \forall\zeta \in \mathcal{D}(\Delta_1) $$ which by \eqref{e4.24} and the strong maximum principle leads to $(u-v_1)^+\equiv 0$ in $\Delta_1$. Consequently we have $u\leq v_1$ in $D_1$ and in particular $u(x_1,\underline{x}_2+\epsilon) =0$ $\quad \forall x_1\in (x_{n1},x_{01})$. Therefore $$ u(x_1,x_2) =0 \quad \forall x_2\geq \underline{x}_2+\epsilon = \bar{x}_2,\quad \forall x_1\in [x_{n1},x_{01}]. $$ Now, by continuity of $u$ there exists $\eta_2 \in (0,x_{01}-x_{n1})$ such that \begin{equation*} u(x_1,x_2) \leq \epsilon^2 \quad \forall (x_1,x_2) \in B_{\eta_2}(x_{01},\bar{x}_2). \end{equation*} By Proposition \ref{prop2.6}, there exists $(x_{m1},x_{m2})\in B_{\eta_2}(x_{01},\bar{x}_2)$ such that \begin{equation*}%\label{1} x_{m2}> \bar{x}_2 , \quad x_{m1}> x_{01},\quad u(x_{m1},x_{m2})=0. \end{equation*} Set $\underline{x}_2'=x_{m2}$ and assume that $\epsilon$ is small enough so that $$ (x_{n1}- \epsilon,x_{m1}+ \epsilon)\times(\underline{x}_2', \underline{x}_2' + 2\epsilon) \Subset \Omega. $$ Let $v_2$ be the barrier function defined by \eqref{e3.2} in the set $Z_2=(x_{n1}-\epsilon,x_{m1}+\epsilon)\times (\underline{x}_2',\underline{x}_2' + \epsilon)$. Clearly the extension by $0$ of $v_2$ to $D_2= \big( (x_{01},x_{m1})\times(\underline{x}_2', +\infty)\big) \cap \Omega$ satisfies \eqref{e3.6}. Then, since $(x_{01},x_{m1})\times \{\underline{x}_2'\}\subset B_{\eta_2}(x_{01},\bar{x}_2)$, we have \begin{equation*} u(x_1,\underline{x}_2') \leq \epsilon^2=v_2(x_1,\underline{x}_2') \quad \forall x_1\in (x_{01},x_{m1}). \end{equation*} Arguing as above, we show that $(u-v_2)^+\equiv 0$ in $D_2\cap [v_2>0]$, which leads to $$ u(x_1,x_2)\equiv 0\quad \forall x_2 \geq \underline{x}_2'+\epsilon, \quad \forall x_1\in [x_{01},x_{m1}]. $$ Hence we have $$ u(x_1,x_2)\equiv 0\quad \forall x_2 \geq \underline{x}_2'+\epsilon, \quad \forall x_1\in [x_{n1},x_{m1}]. $$ Note that if (ii) holds, we argue similarly to obtain the same conclusion. We have proved that for all $x_2\in (x_{n1},x_{m1})$, $$ \phi (x_1) \leq \underline{x}_2' + \epsilon < \bar{x}_2 + \eta_2 +\epsilon = \underline{x}_2 +\epsilon + \eta_2 +\epsilon < x_{02} +\eta_1 + \eta_2 +2\epsilon < \phi(x_{01}) +4\epsilon $$ which is the upper semi-continuity of $\phi$ at $x_{01}$. \end{proof} \begin{thebibliography}{00} \bibitem{A1} H. W. Alt; \emph{Str\"omungen durch inhomogene por\"ose Medien mit freiem Rand}. Journal f\"ur die Reine und Angewandte Mathematik 305, (1979), 89-115. \bibitem{A2} H. W. Alt; \emph{The fluid flow Through Porous Media. Regularity of the free Surface.} Manuscripta Math. 21, (1977), 255-272. \bibitem{AC} S. J. Alvarez, J. Carrillo; \emph{A free boundary problem in theory of lubrication}. Comm. Partial Differential Equations 19 (1994), no. 11-12, 1743–1761. \bibitem{BC} G. Bayada, M. Chambat; \emph{Existence and uniqueness for a lubrication problem with nonregular conditions on the free boundary}. Boll. Un. Mat. Ital. B (6) 3 (1984), no. 2, 543-557. \bibitem{BMQ} A. Berm\'udez, M. C. Mu$\tilde{n}$iz, P. Quintela; \emph{Existence and uniqueness for a free boundary problem in aluminum electrolysis}. J. Math. Anal. Appl. 191, No. 3, 497-527, (1995). \bibitem{CA} J. Carrillo, A. Alonso; \emph{A unified formulation for the boundary conditions in some convection-diffusion problem}. Elliptic and parabolic problems (Pont-\`{a}-Mousson, 1994), 51–63, Pitman Res. Notes Math. Ser., 325, Longman Sci. Tech., Harlow, 1995. \bibitem{CC1} J. Carrillo, M. Chipot; \emph{On the Dam Problem}. J. Differential Equations 45, (1982), 234 - 271. \bibitem{CC2} J. Carrillo, M. Chipot; \emph{The Dam Problem with Leaky Boundary conditions}. Applied Mathematics and Optimisation 28, (1993), 57-85. \bibitem{ChiL1} M. Chipot, A. Lyaghfouri; \emph{The dam problem with linear Darcy's law and nonlinear leaky boundary conditions}. Advances in Differential Equations Vol. 3, No. 1, 1-50, (1998). \bibitem{ChiL2} M. Chipot, A. Lyaghfouri; \emph{The dam problem with nonlinear Darcy's law and leaky boundary conditions}. Mathematical Methods in the Applied Sciences Vol. 20, No. 12, 1045-1068, (1997). \bibitem{C} M. Chipot; \emph{On the Continuity of the Free Boundary in some Class of Dimensional Problems}. Interfaces and Free Boundaries. Vol. 3, No. 1, 81-99, (2001). \bibitem{ChaL1} M. Challal, A. Lyaghfouri; \emph{A Filtration Problem through a Heterogeneous Porous Medium}. Interfaces and Free Boundaries 6, 55-79 (2004). \bibitem{ChaL2} S. Challal, A. Lyaghfouri; \emph{On the Continuity of the Free Boundary in Problems of type $\operatorname{div}(a(x) \nabla u) = -(\chi(u)h(x))_{x_1}$}. Nonlinear Analysis : Theory, Methods \& Applications, Vol. 62, No. 2, 283-300 (2005). \bibitem{ChaL3} S. Challal, A. Lyaghfouri; \emph{On a class of Free Boundary Problems of type $\operatorname{div}(a(X)\nabla u) = -\operatorname{div}(H(X)\chi(u))$}. Differential and Integral Equations, Vol. 19, No. 5, 481-516 (2006). \bibitem{ChaL4} S. Challal, A. Lyaghfouri; \emph{Continuity of the Free Boundary in the Problem $\Delta_p u = -(h(x,y)\chi(u))_x$ with $h_x\geqslant 0$}. Applicable Analysis, Vol. 86, No. 9, 1177-1184 (2007). \bibitem{ChaL5} S. Challal, A. Lyaghfouri; \emph{The Heterogeneous Dam problem with Leaky Boundary Condition}. Communications in Pure and Applied Analysis. Vol. 10, No. 1, 93-125 (2011). \bibitem{GT} D. Gilbarg, N.S. Trudinger; \emph{Elliptic Partial Differential Equations of Second Order}. Springer-Verlag 1983. \bibitem{Ly1} A. Lyaghfouri; \emph{A unified formulation for the dam problem}. Rivista di Matematica della Universit\`a di Parma. (6) 1, 113-148, (1998). \bibitem{Ly2} A. Lyaghfouri; \emph{On the uniqueness of the solution of a nonlinear filtration problem through a porous medium}. Calculus of Variations and Partial Differential Equations Vol. 6, No. 1, 67-94, (1998). \bibitem{Ly3} A. Lyaghfouri; \emph{A free boundary problem for a fluid flow in a heterogeneous porous medium}. Annali dell' Universita di Ferrara-Sez. VII-Sc. Mat., Vol. IL, 209-262 (2003). \bibitem{Ly4} A. Lyaghfouri; \emph{The Inhomogeneous Dam Problem with Linear Darcy's Law and Dirichlet Boundary Conditions}. Mathematical Models and Methods in Applied Sciences 8(6), (1996), 1051-1077. \bibitem{Ly5} A. Lyaghfouri; \emph{On the Lipschitz Continuity of the Solutions of a Class of Elliptic Free Boundary Problems}. Journal of Applied Analysis, Vol. 14, No. 2, 165-181 (2008). \bibitem{Ly6} A. Lyaghfouri; \emph{The dam Problem}. Handbook of Differential Equations, Stationary Partial Differential Equations, Vol. 3, ch. 06, 465-552 (2006). \end{thebibliography} \end{document}