\documentclass[reqno]{amsart}
\usepackage{epic} % For drawing figures

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 44, pp. 1--27.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2001/44\hfil Interface separating fresh and salt groundwater]
{On the behavior of the interface separating fresh
and salt groundwater in a heterogeneous coastal aquifer}

\author[S. Challal \& A. Lyaghfouri\hfil EJDE--2001/44\hfilneg]
{Samia Challal \&  Abdeslem Lyaghfouri}

\address{Samia Challal \hfill\break
King Fahd University of Petroleum and Minerals\\
P. O. Box 728, Dhahran 31261, Saudi Arabia}

\address{Abdeslem Lyaghfouri \hfill\break
King Fahd University of Petroleum and Minerals\\
P. O. Box 728, Dhahran 31261, Saudi Arabia}
\email{lyaghfo@kfupm.edu.sa}

\date{}
\thanks{Submitted December 17, 2002. Published April 17, 2003.}

\subjclass[2000]{35Q35, 35R35, 76S05} 
\keywords{Fresh-salt water, heterogeneous aquifer, nonlinear Darcy's law, 
\hfill\break\indent
monotone solution, comparison and uniqueness, continuity of the free boundary, 
limit behavior}


\begin{abstract}
  We consider a flow of fresh and salt groundwater in a
  two-dimensional heterogeneous horizontal aquifer. Assuming the
  flow governed by a nonlinear Darcy law and the permeability
  depending only on the vertical coordinate, we show the existence
  of a unique monotone solution that increases (resp. decreases)
  with respect to the salt (resp. fresh) water discharge. For this
  solution we prove that the free boundary is represented by the
  graph $x=g(z)$ of a continuous function. Finally we prove a limit
  behavior at the end points of the interval of definition of $g$.
\end{abstract}

\maketitle

\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}{Definition}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{corollary}{Corollary}
\newtheorem{remark}{Remark}
\allowdisplaybreaks

\section*{Introduction}\label{S:intro}

Fresh water and sea water are actually miscible fluids and
therefore the zone of contact between them takes the form of a
transition zone caused by hydrodynamics dispersion. Across this
zone the density of the mixed water varies from that of fresh
water to that of sea water.

 Under certain conditions the width of this zone is
relatively small (when compared with the thickness of the aquifer)
so that we assume that each liquid is confined to a well defined
portion of the flow domain with an abrupt interface separating the
two domains.

 We consider here a two-dimensional model for fresh-salt
water in a horizontally heterogeneous extended aquifer in the
$xz$-plane. We suppose that the scale of the problem is
sufficiently large so that the abrupt interface approximation is
applicable. Moreover we consider the model of a flow obeying to a
nonlinear Darcy law.

 In section 1, we indicate briefly how to obtain the weak
formulation and the existence and some properties of the
solutions, the definition of the free boundary $\Gamma=[x=g(z)]$
and the continuity of $g$ on its interval of definition $(-h^*,0)
$, $h^*>0$. All these results generalize previous works in
\cite{CL1} and \cite{CCL} (resp. \cite{CL2}) where the aquifer was
supposed homogeneous (resp. heterogeneous) and the flow governed
by a nonlinear (resp. linear) Darcy law.

 The aim of this paper is to study the behavior of the
free boundary when $z\to -h^*$ and $z\to 0$ under the assumption
that the permeability of the porous medium depends only on $z$
including the case of horizontal layers. Actually we establish in
section 3 that $\lim_{z\to -h^*}g(z)=+\infty$ by generalizing the
proof given in \cite{CL2}. We also give a second simple proof
which works only for a constant permeability. Then we prove that
$\lim_{z\to 0}g(z)=g(0-)$ exists. We recall that in \cite{AD} the
authors first proved (in the linear and homogeneous case) that
$\liminf_{z\to 0}g(z)>-\infty$ by using blow up arguments. They
also proved that $\limsup_{z\to 0}g(z)<0$ and used this result to
prove the existence of the limit $g(0-)$. Our proof does not
assume $\limsup_{z\to 0}g(z)<0$ and is valid in the general case.
Moreover we prove that $g(0-)$ is finite in more general cases.
Our proofs are systematically based on comparing the solution
locally or globally with explicit functions satisfying similar
equations. This method of comparison is developed in section 2 to
show that the solution increases with respect to the salt water
discharge $Q_s$ and decreases with respect to the fresh water
discharge $Q_f$. The uniqueness of the solution is obtained as a
corollary of this monotonicity result. We also deduce a limit
behavior of the solution when $Q_f$ or $Q_s$ goes to zero. Also by
a comparison argument, when the permeability is constant, we give
a simple proof of the fact that the set filled by fresh water is
star shaped with the origin and the free boundary is non
increasing in the region $[x>0]$. These two last results were
proved in the linear case in \cite{AD} by using blow up arguments.
Our proofs based on monotonicity arguments are much simpler.

\section{Description of the model}\label{1}

In this paper we are interested with the study of a stationary
flow of fresh and salt water in a heterogeneous coastal aquifer
$\Omega=\mathbb{R}\times(-h,0),\, h>0$, with permeability $A(X)$,
$X=(x,z)\in \mathbb{R}^2$. The velocity and the pressure of the
fluid are related by the following nonlinear Darcy law
\begin{equation}\label{1.1}
{\bf v}=-\left(\langle A(\nabla p+\gamma e_{z}),\nabla p+\gamma e_{z}
\rangle \right)^{\frac{r-2}{2}}A(\nabla p+\gamma e_{z})
\end{equation}
with $r>1$, $e_z=(0,1)$ and $\gamma$ is given by
\begin{equation}\label{1.2}
\gamma=\gamma_f\chi(\Omega_f)+\gamma_s\chi(\Omega_s)
\quad\mbox{with}\quad 0<\gamma_f<\gamma_s
\end{equation}
where $\gamma_f$ (resp. $\gamma_s$) represents the specific weight
of the fresh (resp. salt) water occupying the region $\Omega_f$
(resp. $\Omega_s$) of $\Omega$, $\chi(E)$ denotes the
characteristic function of the set $E$.

\begin{figure}[t]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(100,52)(0,0)
\put(0,35){\line(1,0){40}}
 \put(55,35){\line(1,0){45}}
\put(99,34.15){$\to$}
\put(0,0){\line(1,0){100}}
\put(40,35){\line(0,1){15}}
\put(39.1,50){$\uparrow$}
\put(35,50){$z$} \put(47,46){$Q_f$}
\put(47,40){$\Downarrow$}
\qbezier(20,35)(40,33)(60,25)
 \qbezier(60,25)(80,18)(100,15)
\put(20,40){$S$}
 \put(55,40){$A$}
 \put(35,40){$O$}
\put(100,37){$x$}
\put(0,20){Salt water}
 \put(30,20){$\Omega_s$}
\put(15,10){$Q_s \quad \Rightarrow$}
 \put(80,14){$\Gamma$}
\put(80,27){Fresh water}
\put(95,22){$\Omega_f$ }
\put(43.5,1.5){$z=-h$}
\end{picture}
\end{center}
\caption{}
\end{figure}

Fresh water is injected over the segment $[OA]$ ($A=(0,a)$ with
$a>0$) uniformly (see Figure 1) with a total amount of $Q_f$. From
infinity at the left of the aquifer, salt water arrives with a
total discharge of $Q_s$. We assume that the two fluids are
unmixed and separated by an interface $\Gamma$. The part of the
boundary $\partial\Omega\setminus[OA]$ is assumed to be impervious
and the flow incompressible. So the velocity satisfies
\begin{equation}\label{1.3}
\begin{gathered}\mathop{\rm div}
{\bf v} =0\quad \text{in } \Omega,\quad
{\bf v}=-\frac{Q_f}{a}e_z \quad\text{on } [OA],\\
{\bf v}\cdot\nu =0 \quad\text{on } \partial\Omega\setminus
[OA],\quad {\bf v_i}\cdot\nu =0 \quad \text{on }
\Gamma\quad(i=s,f),
\end{gathered}
\end{equation}
where ${\bf v_i}$ is the restriction of ${\bf v}$ to
$\Omega_i$ and $\nu$ is the outward unit normal to
$\partial\Omega$ or $\Gamma$. We deduce from (1.3) that there
exists a stream function $\psi$ satisfying
\begin{equation}\label{1.4}
\begin{gathered}
{\bf v} =\mathop{\rm Rot}\psi=\Big(-\frac{\partial\psi}
{\partial z},\frac{\partial\psi}{\partial x}\Big)\quad
\text{in }\Omega,\quad\ \psi =0 \quad\text{on}\quad\Gamma,\\
\psi(x,-h) =Q_s \quad\text{and}\quad \psi(x,0)=\phi_0
(x)\quad\text{for } x\in\mathbb{R}
\end{gathered}
\end{equation}
with
\begin{equation}\label{1.5}
\phi_0(x)=-Q_f\min\big(\frac{x^+}{a},1\big).
\end{equation}
We suppose the permeability matrix $A(X)$ such that
\begin{equation}\label{1.6}
\begin{gathered}
 A(X)\in\big[L^\infty(\Omega)\big]^4,\\
 \exists m_0>0 :\; \langle A(X)\xi,\xi\rangle \geq m_0\vert\xi\vert^2,\;
 \forall\xi\in\mathbb{R}^2,\quad\text{a.e. }
X\in\Omega,\\
^tA=A.\end{gathered}
\end{equation}
Then there exists a unique symmetric and strictly elliptic matrix
$\mathcal{A}$ satisfying $A(X)=\mathcal{A} ^t\mathcal{A}= \mathcal{A}^2$ (see
\cite{Ci}). Then (1.1) becomes
\begin{align*}
{\bf v} &=-\langle \mathcal{A}(X)(\nabla p+\gamma e_{z}),\mathcal{A}(X)
(\nabla p+\gamma e_{z} )\rangle ^{(r-2)/2}
\mathcal{A}(X)\mathcal{A}(X)(\nabla p+\gamma e_{z})\\
&=-\vert \mathcal{A}(X)(\nabla p+\gamma e_{z})\vert^{r-2}
\mathcal{A}(X)\mathcal{A}(X)(\nabla p+\gamma e_{z}).
\end{align*}
This leads to
\[
\vert\mathcal{A}^{-1}(X){\bf v}\vert
=\vert\mathcal{A}(X)(\nabla p+\gamma e_{z})\vert^{r-1}
\]
 and
\begin{equation}\label{1.7}
\nabla p+\gamma e_{z}=-\vert\mathcal{A}^{-1}(X){\bf v}\vert^{\frac{2-
r}{r-1}} \mathcal{A}^{-1}(X)(\mathcal{A}^{-1}(X){\bf v}).
\end{equation}
Now for $\zeta\in\mathcal{D}(\Omega)$ we get by \eqref{1.4} and
\eqref{1.7}
\[
\int_\Omega (\nabla p+\gamma e_z)\mathop{\rm Rot}\zeta=-\int_\Omega
\vert\mathcal{A}^{-1}(X)\mathop{\rm Rot}\psi\vert^{\frac{2-r}{r-1}}
\mathcal{A}^{-1}(X)(\mathcal{A}^{-1}(X)\mathop{\rm Rot}\psi)\cdot
\mathop{\rm Rot}\zeta.
\]
If we set
$\overline{A}(X)={\frac{1}{det A}A}$, then there
exists a unique symmetric and strictly elliptic matrix $B(X)$ such
that $\overline{A}(X)=B^2=B\cdot B$ and for which we have
\[
^t\mathcal{A}^{-1}(X)(\mathcal{A}^{-1}(X)\mathop{\rm Rot}\psi)\cdot
\mathop{\rm Rot}\zeta=\langle\overline{A}(X)
\nabla\psi,\nabla\zeta\rangle=B(X)\nabla\psi\cdot B(X)\nabla\zeta.
\]
Therefore
\begin{equation}\label{1.8}
\int_\Omega \vert B(X)\nabla\psi\vert^{\frac{2-r}{r-1}}B(X)
\nabla\psi\cdot B(X)\nabla\zeta\ +\gamma e_x\nabla\zeta=0
\quad\forall \zeta\in \mathcal{D}(\Omega).
\end{equation}
Setting $q=r/(r-1)$, we deduce from \eqref{1.8} by taking
$\zeta\in \mathcal{D} (\Omega_i)$ ($i=f,s$)
\begin{equation}\label{1.9}
\mathop{\rm div}\big(\mathcal{B}(X,\psi)\big)=0\quad\text{in }\mathcal{D}'
(\Omega_i)\quad i=f,s
\end{equation}
with
$$
\mathcal{B}(X,\xi)=\langle\overline{A}(X)\xi,\xi\rangle ^{\frac{q-2}{2}}
\overline{A}(X) \xi.
$$
Moreover if we assume that
\begin{equation}\label{1.10}
\lim_{x\to +\infty}\psi(x,z)\leq 0\quad\text{for }(x,z)\in
\Omega_f\quad\text{and} \quad
\lim_{x\to \pm\infty}\psi(x,z)\geq
0\quad\text{for }(x,z)\in \Omega_s
\end{equation}
and since
\begin{equation}\label{1.11}
\psi\leq 0\quad\text{on }\partial\Omega_f\quad\text{and}
\quad\psi\geq 0\quad\text{on }\partial\Omega_s,
\end{equation}
we deduce by \eqref{1.9}-\eqref{1.11} and the maximum
principle for $\mathcal{B}$-harmonic functions in unbounded domains
(see \cite{DL}, \cite{K}) that
\begin{equation}\label{1.12}
\psi<0\quad\text{in }\Omega_f\quad\text{and}\quad\psi>
0\quad\text{in }\Omega_s.
\end{equation}
 It follows from (1.2) and (1.12) that $\gamma\in
H(\psi)$, where $H$ is the maximal monotone graph defined by
\begin{equation*}
H(t)=\gamma_f \chi([t<0])+[ \gamma_f ,\gamma_s]\chi([t=0])
+\gamma_s\chi([t>0]),
\end{equation*}
$[ \gamma_f ,\gamma_s]$ being the closed interval of $\mathbb{R}$
with endpoints $\gamma_f$ and $\gamma_s$.

Then we are led to the following question:\\
\noindent\textbf{problem (P)}
 Find $(\psi,\gamma)\in W_{\rm loc}^{1,q}(\Omega)\times L^
{\infty}(\Omega)$  such that
\begin{itemize}
\item[(i)] $\int_\Omega\vert B(X)\nabla\psi\vert^{q-2}B(X)
\nabla\psi\cdot B(X)\nabla\zeta+\gamma e_x\cdot\nabla\zeta=0$
for all $\zeta\in W_0^{1,q}(\Omega)$  with
compact support in $\bar{\Omega}$

\item[(ii)] $\gamma\in H(\psi)$  a.e. in $\Omega$

\item[(iii)] $-Q_f\leq\psi\leq Q_s$  a.e. in $\Omega$

\item[(iv)] $\psi(x,-h)=Q_s$, $\psi(x,0)=\phi_0(x)$  for
all $x\in\mathbb{R}$.
\end{itemize} \smallskip

 Adapting technics in \cite{CCL}, \cite{CL1} and
\cite{CL2} we prove the following theorems.


\begin{theorem}\label{thm1.1}
\begin{itemize}
\item[(i)] There exists a solution $(\psi,\gamma)$ of $(P)$ that satisfies
$\psi\in C_{\rm loc}^{0,\alpha}(\bar{\Omega})$ for some
$\alpha\in(0,1)$. Moreover if $A(X)\in C_{\rm loc}^
{0,\sigma}(\Omega)$, then $\psi\in C_{\rm loc}^{1,\beta}
(\Omega\setminus[\psi=0])$, $\sigma, \beta\in (0,1)$.

\item[(ii)] If $B(X)=B(z)$, then there exists a monotone
solution $(\psi,\gamma)$ of $(P)$ in the following sense:
\begin{equation}\label{1.13}
\partial_x\psi\leq 0\quad\text{and}
\quad\partial_x\gamma\leq 0\quad\text{in }\mathcal{D}'(\Omega).
\end{equation}
\end{itemize}
\end{theorem}

 For the rest of this article, we assume that
\begin{equation}\label{1.14}
B(X)=B(z)=\big(B_{ij}(z)\big)_{1\leq i,j\leq 2}\quad\hbox{a.e. in } \Omega
\end{equation}
and will consider only monotone solutions. Moreover we need to
introduce the following two functions defined for $z\in[- h,0]$.
\begin{equation}\label{1.15}
v_{+\infty}(z)=-Q_f+(Q_s+Q_f)\phi_1(z), \quad
v_{-\infty}(z)=Q_s\phi_1(z)
\end{equation}
where
\[
\phi_1(z)=\frac{{\int_z^0\frac{ds}{(B_{12}^2+B_{22}^2)^
{q'/2}(s)}}}
{{\int_{-h}^0\frac{ds}{(B_{12}^2+B_{22}^2)^{q'/2}(s)}}}.
\]
Then we have the following theorem.

\begin{theorem}\label{thm1.2}
Let $(\psi,\gamma)$ be a solution of (P) and let
$\Omega_{m,n}=(m,n)\times(-h,0)$ for $m,n\in \mathbb{R}$.
\begin{itemize}
\item[(i)]  For $r=\max(q,2)$, we have
$\lim_{R\to \pm\infty}\int_{\Omega_{R,R+1}}\vert\nabla(\psi-
v_{\pm\infty}\vert^r=0$.

\item[(ii)]  For all $z\in [-h,0]$,
$\psi(x,z)\to v_{\pm\infty}(z)$ as $x\to\pm\infty$.

\item[(iii)] $v_{+\infty}\leq\psi\leq v_{-\infty}$ in $\Omega$.

\item[(iv)] $\gamma(x+R,z)\rightharpoonup \gamma_{\pm\infty}(z)$
in $L^{q'}(\Omega_{0,1})$ as $R\to\pm\infty$, where
$\gamma_{\pm\infty}\in H(v_{\pm\infty})$.
\end{itemize}
\end{theorem}

\begin{remark}\label{rmk1.1}\rm
 From $(iii)$, one can see that the
strip $\mathbb{R} \times(-h,-h^*)$ is contained in $[\psi>0]$,
where $h^*\in (0,h)$ is defined by
$\phi_1(-h^*)=\frac{Q_f}{Q_s+Q_f}$. Therefore, the
free boundary $\Gamma=[\psi=0]$ is contained in $\mathbb{R}
\times[-h^*,0)$.
\end{remark}

 Arguing as in \cite{CCL} and \cite{CL2} we
can prove the continuity of the free boundary. The proof needs
Lemma \ref{lmA.3} (see Appendix) which requires the following regularity
of $B$: If $q\neq 2$, then
\begin{equation}
B(z)\in C^{0,1}_{\rm loc}(-h,0).
\end{equation}

\begin{remark}\label{rmk1.2} \rm
Under assumption (1.16), the critical points of any
$\mathcal{B}-$harmonic function in $\Omega$ are isolated (see \cite{AS}).
 For the rest of this article, we assume that (1.16) is satisfied.
\end{remark}

\begin{theorem}\label{thm1.3}
There exists a continuous function $g:(-h^*,0)\rightarrow \mathbb{R}$
such that
\[
[x=g(z)]\subset\Gamma\subset [x=g(z)]\cup[z=-h^*].\]
\end{theorem}


\begin{corollary}\label{coro1.1}
\begin{itemize}
\item[(i)] $\gamma=\gamma_s\chi([\psi>0])+\gamma_f\chi ([\psi<0])$ a.e. in
$\Omega$.
\item[(ii)] The sets $[\psi>0]$ and $[\psi<0]$ are connected by arcs.
\end{itemize}
\end{corollary}

\section{Comparison and Uniqueness}\label{2}

In this section we prove that solutions of $(P)$ increase with
respect to $Q_s$ and decrease with respect to $Q_f$. As a
consequence we obtain the uniqueness of the solution of $(P)$.
Let us denote by $(P(Q_s,Q_f))$ the problem $(P)$
corresponding to $Q_s$ and $Q_f$. Then we have the following
comparison result

\begin{theorem}\label{thm2.1}
Let $(\psi_1,\gamma_1)$ and $(\psi_2,\gamma_2)$ be solutions of
Problems $(P(Q_{s_1},Q_{f_1}))$ and $(P(Q_{s_2},Q_ {f_2}))$ respectively.
If $Q_{s_1}\leq Q_{s_2}$ and $Q_{f_2}\leq Q_ {f_1}$, then we have
$\psi_1\leq \psi_2$  and $\gamma_1\leq\gamma_2$
a.e. in $\Omega$.
\end{theorem}

  The proof of this theorem follows an idea
in \cite{CCL} and uses a recent result due to Alessandrini and
Sigalotti \cite {AS} regarding isolation of critical points of
$\mathcal{B}$-harmonic functions in planar domains. First we prove
the following lemma

\begin{lemma}\label{lm2.1}
Under the assumptions of Theorem \ref{thm2.1},
\[
T(\zeta)=\int_\Omega \Big(\mathcal{B}(z,\nabla\psi_1)
-\mathcal{B}(z,\nabla\psi_0)+(\gamma_1-\gamma_0)e_x\Big).\nabla\zeta=0
\quad\forall\zeta\in \mathcal{D}(\mathbb{R}^2),
\]
where
$\psi_0=\min(\psi_1,\psi_2)$ and $\gamma_0=\min
(\gamma_1,\gamma_2)$.
\end{lemma}

\begin{proof}
 Let $\zeta\in \mathcal{D}(\mathbb{R}^2)$ and let
$K=supp\zeta$, $M=\sup_{K}\vert\zeta\vert$. Then there exists
$R_0>a$ such that: $\forall R\geq R_0$, $K\subset (-
R,R)\times\mathbb{R}$. Consider $\zeta_R=M.\min\left(1,(-\vert
x\vert+R+1)^+\right)$ and set $\zeta_1=\zeta+\zeta_R$,
$\zeta_2=\zeta_R-\zeta$. Then for $\epsilon>0$ and $i=1,2$,
$\min(\zeta_i,\frac{\psi_1-\psi_0}{\epsilon})$ is a test function
and one has by the monotonicity of $\mathcal{B}(z,.)$
\begin{equation}\label{2.1}
\begin{aligned}
&\int_{[\psi_1-\psi_0\geq \epsilon\zeta_i]}\Big(\mathcal{B}
(z,\nabla\psi_1)-\mathcal{B}(z,\nabla\psi_0)\Big)
\nabla\zeta_i+\int_\Omega (\gamma_1-\gamma_0).\zeta_{ix}\\
&\leq \int_\Omega (\gamma_1-\gamma_0)\Bigl(\zeta_i-\frac
{\psi_1-\psi_0}{\epsilon}\Bigl)^+_x.
\end{aligned}
\end{equation}
Since $Q_{s_1}\leq Q_{s_2}$ and $Q_{f_2}\leq Q_{f_1}$ one has
\[
-h_1^*=\phi_1^{-1}\big(\frac{Q_{f_1}}{Q_{s_1}+Q_{f_1}}\big)
\leq \phi_1^{-1}\big(\frac{Q_{f_2}}{Q_{s_2}+Q_{f_2}}\big)
=-h_2^*
\]
and then if we denote by $I=\{z\in(-h_2^*,0) : g_2 (z)<g_1(z)\}$
\begin{align*}
\int_\Omega &(\gamma_1-\gamma_0)\big(\zeta_i-\frac{\psi_1-
\psi_0}{\epsilon}\big)^+_x\\
&=\int_{[\psi_0<0<\psi_1]}
(\gamma_s-\gamma_f)\big(\zeta_i-\frac{\psi_1-\psi_0}
{\epsilon}\big)^+_x\\
&=(\gamma_s-\gamma_f)\int_I\int_{g_2(z)}^{g_1(z)}
\big(\zeta_i-\frac{\psi_1-\psi_0}{\epsilon}\big)^+_x\\
&=(\gamma_s-\gamma_f)\int_I\big(\zeta_i+\frac{\psi_2}
{\epsilon}\big)^+(g_1(z),z)-
\big(\zeta_i-\frac{\psi_1}{\epsilon}\big)^+(g_2(z),z)dz
\end{align*}
which goes to zero when $\epsilon\to 0$. So from (2.1) we
get $T(\zeta_i)\leq 0$ for $i=1,2$. This leads to
\begin{equation}\label{2.2}
T(\zeta_R)\leq T(\zeta)\leq -T(\zeta_R).\end{equation} Moreover we
have
\begin{align*}
T(\zeta_R)&=M\int_{\Omega_{-R-1,-R}}\left(\mathcal{B}
(z,\nabla\psi_1)-\mathcal{B}(z,\nabla\psi_0)\right).e_x+M\int_
{\Omega_{-R-1,-R}}(\gamma_1-\gamma_0)\\
&\quad-M\int_{\Omega_{R,R+1}}\left(\mathcal{B}(z,\nabla\psi_1)
-\mathcal{B}(z,\nabla\psi_0)\right).e_x+M\int_{\Omega_{R,R+1}}(\gamma_1-
\gamma_0).
\end{align*}
Using Theorem \ref{thm1.2} and the fact that we have
either $v^1_ {+\infty}\equiv v^2_{+\infty}$ or $v^1_{+\infty}<v^2_
{+\infty}$ in $(-h,0)$, we deduce (see \cite{CL2,CCL})
that $\lim_{R\to +\infty}T(\zeta_R)=0$ which leads by (2.2) to
$T(\zeta)=0$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2.1}]
Let us denote by $D$ the domain $[\psi_1<0]$ (see Corollary
\ref{coro1.1}). First we remark from Lemma \ref{lm2.1} that
$(\psi_0,\gamma_0)$ is also a solution of $(P(Q_{s_0},Q_{f_0})) $,
that $\psi_0$ and $\psi_1$ are $\mathcal{B}$-Harmonic in $D$ and
that $\mathcal{B}(z,\nabla\psi_0).\nu=\mathcal{B}
(z,\nabla\psi_1).\nu$ on $(a,+\infty)\times\{0\}$.

 Since we have $\psi_0\leq\psi_1$ in $D$, $\psi_0=\psi_1$
on $(a,+\infty)\times\{0\}$ and $\psi_0$, $\psi_1\in C^1(D\cup
(a,+\infty)\times\{0\})$, it is enough according to Lemma
\ref{lmA.2}, to prove that $\nabla\psi_1$ does not vanish on some
part $\Gamma_0$ of $(a,+\infty)\times\{0\}$. Assume that for some
$x_1>a$ and ${0<r<\frac{x_1-a}{2}}$ we have
$\nabla\psi_1(x,0)=0\quad\forall x\in (x_1-r,x_1+r)$. Let
$B_r(x_1,0)$ be the ball of center $(x_1,0)$ and radius $r$. If we
extend $\psi_1$ by $-Q_{f_1}$ and $\mathcal{B}(z,\xi)$ by
$\mathcal{B} (0,\xi)$ to $B_r(x_1,0)\setminus\overline{D}$, it is
clear that $\psi_1\in W^{1,q}\big(B_r(x_1,0)\big)$ and is
$\mathcal{B}$-Harmonic in $B_r(x_1,0)$. Since the zeros of the
gradient of a nonconstant $\mathcal{B}$-Harmonic function are
isolated (see \cite{AS}) and $\nabla\psi_1=0$ in $B_r(x_1,0)\cap
[z>0] $, we deduce that $\psi_1=-Q_{f_1}$ in $B_r(x_1,0)$. By the
monotonicity of $\psi_1$ this leads to $\psi_1=-Q_{f_1}$ in the
strip $[x_1,+\infty)\times (-r,0)$ which contradicts the
asymptotic behavior of $\psi_1$ at $+\infty$. So there exists
$x_1'\in (x_1-r,x_1+r)$ such that $\nabla\psi_1 (x_1',0)\ne 0$.
Since $\psi_1\in C^1(D\cup(a,+\infty) \times\{0\})$ there exists
$r'>0$ such that $\nabla\psi_1(x,0)\ne 0$ $\forall x\in (x_1'-
r',x'_1+r')$. Therefore we get $\psi_0=\psi_1$ in $D$. In
particular $\psi_0=\psi_1$ in $[\psi_0<0]$. Similarly one can
prove that $\psi_0=\psi_1$ in $[\psi_0 >0]$ and then by continuity
$\psi_0=\psi_1$ in $\Omega$. Using Corollary \ref{coro1.1} we
deduce that $\gamma_0=\gamma_1$ in $\Omega$.
\end{proof}

As a direct consequence of Theorem \ref{thm2.1}, we have

\begin{corollary}\label{coro2.1}
The solution of problem $(P)$ is unique.
\end{corollary}

According to Theorem \ref{thm2.1} the solution of $(P)$ decreases
with respect to $Q_f$ and increases with respect to $Q_s$.
Intuitively one would expect that as $Q_f\to 0$ (resp. $Q_s\to 0)$
the aquifer would be saturated by salt (resp. fresh) water only.
More precisely we have

\begin{theorem}\label{thm2.2}
Let $(\psi_{Q_f},\gamma_{Q_f})$ be the solution of $(P(Q_
{s},Q_{f}))$. We have
\[
(\psi_{Q_f},\gamma_{Q_f})\to (v_{-
\infty},\gamma_s)\quad\hbox{in } W_{\rm loc}^{1,q}(\Omega)
\times L_{\rm loc}^{q'}(\Omega)\text{ as } Q_f\to 0,
\]
where $v_{- \infty}$ is given by (1.15).
\end{theorem}

\begin{proof} Using the monotonicity of $\psi_{Q_f}$ and
$\gamma_{Q_f}$ with respect to $Q_f$ and the fact that the two
functions are uniformly bounded, we deduce by Beppo-Levi's theorem
that there exists two functions $\psi$, $\gamma$ such that
\begin{gather*}
\psi_{Q_f}\to \psi \quad\hbox{in }L_{\rm loc}^r
(\Omega)\hbox{ a.e. in }\Omega\\
\gamma_{Q_f}\to \gamma \quad\hbox{in }L_{\rm loc}
^r(\Omega)\quad\forall r\geq 1.
\end{gather*}
 Now let $m>a$ and
$\eta\in W^{1,\infty}(\mathbb{R})$ such that $0\leq \eta\leq 1$,
$\eta=1$ in $(-m,m)$, $\eta=0$ for $\vert x\vert\geq m+1$ and
$\vert\eta'\vert\leq 1$. Let
$\Phi(x,z)=\phi_0(x)+\phi_1(z)(Q_s-\phi_0(x))$. Then
$\eta^q(\psi_{Q_f}-\Phi)$ is a test function for $(P(Q_
{s},Q_{f}))$ and we have
\begin{align*}
&\int_{\Omega_{m+1}}\eta^q \vert B(z)\nabla\psi_{Q_f} \vert^q\\
&=\int_{\Omega_{m+1}}\eta^q\vert B(z)\nabla\psi_{Q_f}
\vert^{q-2}B(z)\nabla\psi_{Q_f}\cdot B(z)\nabla\Phi\\
&\quad-\int_{\Omega_{m+1}}q\eta^{q-1}(\psi_{Q_f}-\Phi)\vert B(z)
\nabla\psi_{Q_f}\vert^{q-2}B(z)\nabla\psi_{Q_f}\cdot B(z)
\nabla\eta\\
&\quad-\int_{\Omega_{m+1}}\gamma_{Q_f}\eta^q\partial_x\psi_{Q_f}
+\int_{\Omega_{m+1}}\gamma_{Q_f}\eta^q\partial_x\Phi-\int_
{\Omega_{m+1}}\gamma_{Q_f}(\psi_{Q_f}-\Phi)q\eta^{q-1}
\eta'.
\end{align*}
Since $\psi_{Q_f},\Phi, \nabla\Phi,\eta,\eta',\gamma_ {Q_f}$
are uniformly bounded,
we deduce by using H\"{o}lder's inequality
\[
\vert\psi_{Q_f}\vert_{1,q,\Omega_m}\leq c_m
\]
where $c_m$ is a constant depending on $m$. Then by using a
diagonal process we get, up to a subsequence
\[
\psi_{Q_f}\rightharpoonup \psi \quad\hbox{in }W_{\rm loc}^
{1,q}(\Omega).
\]
Let us show that this convergence holds strongly.
Let $m>a$ and let $\rho\in\mathcal{D}(\Omega_m)$, $\rho\geq 0$ in
$\Omega_m$. Then $\rho^q(\psi_{Q_f}- \psi)$ is a test function for
$(P(Q_ {s},Q_{f}))$ and we have
\begin{align*}
&\int_{\Omega_{m}}\rho^q \vert B(z)\nabla\psi_{Q_f} \vert^q\\
&=\int_{\Omega_{m}}\rho^q\vert B(z)\nabla\psi_{Q_f}
\vert^{q-2}B(z)\nabla\psi_{Q_f}\cdot B(z)\nabla\psi\\
&\quad-\int_{\Omega_{m}}q\rho^{q-1}(\psi_{Q_f}-\psi)\vert B(z)
\nabla\psi_{Q_f}\vert^{q-2}B(z)\nabla\psi_{Q_f}\cdot B(z)
\nabla\rho\\
&\quad-\int_{\Omega_{m}}\rho^q\gamma_{Q_f}\partial_x(\psi_ {Q_f}-\psi)
-\int_{\Omega_{m}}\gamma_{Q_f}(\psi_{Q_f}-\psi)q\rho^{q-1}
\partial_x\rho.
\end{align*}
Applying H\"{o}lder's inequality and letting $Q_f\to 0 $, we get
\[
\limsup_{Q_f\to 0}\Big(\int_{\Omega_{m}}\rho^q \vert B
(z)\nabla\psi_{Q_f}\vert^q\Big)^{1/q}\leq
\Bigl(\int_{\Omega_{m}}\rho^q \vert B(z)
\nabla\psi\vert^q\Bigl)^{1/q}.
\]
Hence $\rho^q
B(z)\nabla\psi_{Q_f}\to \rho^q B(z) \nabla\psi$ in
$L^q(\Omega_{m})$ and in particular
\[
\nabla\psi_{Q_f}\to \nabla\psi\quad\hbox{in }
W_{\rm loc}^{1,q}(\Omega).
\]
Now using the monotonicity with respect
to $Q_f$ and the continuity of $\psi_{Q_f}$, it follows by Dini's
theorem
\[\lim_{Q_f\to 0}\big(\lim_{R\to \pm\infty}\psi_{Q_f}(x+R,z)
\big)= \lim_{R\to \pm\infty}\big(\lim_{Q_f\to 0}\psi_{Q_f}(x+R,z)
\big)
\]
and
\[\lim_{R\to \pm\infty}\psi(x+R,z)=\lim_{Q_f\to 0}v_
{\pm\infty}^{Q_f}(z)=v_{-\infty}(z)\quad\hbox{ for }(x,z)
\in\Omega_{0,1},
\]
where $v_{-\infty}^{Q_f}$ and
$v_{+\infty}^{Q_f}$ were defined by (1.15).
 Since $\partial_x\psi\leq 0$, we deduce that
$\psi(x,z)=v_{- \infty}(z)=Q_s\phi_1(z)$. Moreover $\gamma\in
H(\psi)$ and $\psi>0$ in $\Omega$, so $\gamma=\gamma_s$ a.e. in
$\Omega$.
 We conclude that when $Q_f\to 0$, the aquifer is
completely saturated by salt water only.
\end{proof}

\begin{theorem}\label{thm2.3}
Let $(\psi_{Q_s},\gamma_{Q_s})$ be the solution of $(P(Q_
{s},Q_{f}))$. We have
\[(\psi_{Q_s},\gamma_{Q_s})\to (\psi,\gamma_f)
\quad\hbox{in } W_{\rm loc}^{1,q}(\Omega)\times L_{\rm loc}^
{q'}(\Omega)\text{ as } Q_s\to 0,
\]
where $\psi$ is the solution of the following problem:
\end{theorem}
\noindent\textbf{Problem($P(Q_f)$):}
\begin{itemize}
\item[(i)] $\mathop{\rm div}\big(\mathcal{B}(z,\nabla\psi)\big)=0$
in $\mathcal{D}'(\Omega)$
\item[(ii)] $-Q_f<\psi<0$ in $\Omega$
\item[(iii)] $\psi(x,-h)=0$, $\psi(x,0)=\phi_0(x)$
for all $x\in \mathbb{R}$
\item[(iv)] $\lim_{x\to -\infty}\psi(x,z)=0$,
$\lim_{x\to +\infty}\psi(x,z)=Q_f(\phi_1(z)-1)$.
\end{itemize}

\begin{proof}[Proof of Theorem \ref{thm2.3}]
 Taking into account the
monotonicity of $\psi_ {Q_s}$ and $\gamma_{Q_s}$ with respect to
$Q_s$ and arguing as in the proof of Theorem \ref{thm2.2}, we deduce the
existence of two functions $\bar{\psi}$ and $\bar{\gamma}$ such
that
$$
\psi_{Q_s}\to \bar{\psi}\quad\mbox{in }W_{\rm loc}^{1,q}(\Omega) \quad
\mbox{and}\quad \gamma_{Q_s}\to \bar{\gamma}\quad\mbox{in }L_{\rm loc}^{q'}(\Omega).
$$
It follows that $(\bar{\psi},\bar{\gamma})$ satisfies:
\begin{itemize}
\item[(i)] $\mathop{\rm div}\big(\mathcal{B}(z,\nabla\bar{\psi})\big)=-
\partial_x\bar{\gamma}$ in $\mathcal{D}'(\Omega)$
\item[(ii)] $\bar{\gamma}\in H(\bar{\psi})$
\item[(iii)] $-Q_f\leq\bar{\psi}\leq 0$ in $\Omega$
\item[(iv)] $\bar{\psi}(x,-h)=0$, $\bar{\psi}(x,0)=\phi_0(x)$
for all $x\in \mathbb{R}$
\item[(v)] $\lim_{x\to -\infty}\bar{\psi}(x,z)=0$,
$\lim_{x\to +\infty}\bar{\psi}(x,z)=Q_f(\phi_1(z)-1)$
\item[(vi)] $\partial_x\bar{\psi}\leq 0$ and
$\partial_x\bar{\gamma}\leq 0$  in $\mathcal{D}'(\Omega)$.
\end{itemize}
By the weak maximum principle, we can compare $\bar{\psi}$ with
 the solution $\psi$ of $P(Q_f)$. This gives $\bar{\psi} \leq \psi$
in $\Omega$. Since $\psi$ satisfies by the strong maximum
principle $-Q_f<\psi<0$, we deduce that $\bar {\psi}<0$ in
$\Omega$ and then $\bar{\gamma}=\gamma_f$ a.e in $\Omega$.
Consequently $\bar{\psi}= \psi$ in $\Omega$. We conclude that when
$Q_s\to 0$, the aquifer is completely saturated by fresh water
only.\end{proof}

  The end of this section is devoted
to study the set $\Omega_f=[\psi<0]$. We would like to point out
that it was proved in \cite{AD} in the linear case that this set
is star shaped with the origin. The proof was based on blow-up
arguments. Here we propose a different proof based on a comparison
argument that works for the linear case as well as for the
nonlinear one with constant permeability. So we assume that
$\mathcal{B}$ does not depend on $z$.
 For any $r>0$, we consider
\[
\psi_r(x,z)=\frac{1}{r}\psi(rx,rz)\quad\text{and}\quad
\gamma_r(x,z)=\frac{1}{r}\gamma(rx,rz)
\]
defined on
$\Omega_r=\mathbb{R}\times(\frac{-h}{r},0)$.
It is easy to check that $(\psi_r,\gamma_r)$ is the solution of
the following problem:

\noindent\textbf{Problem $(P_r)$}
 Find $(\psi_r,\gamma_r)\in W_{\rm loc}^{1,q}(\Omega_r)\times
L^{\infty}(\Omega_r)$  such that:
\begin{itemize}
\item[(i)] ${\int}_{\Omega_r}\vert B\nabla\psi_r\vert^{q-2}B
\nabla\psi_r\cdot B\nabla\zeta+\gamma_r e_x.\nabla\zeta=0$
for all $\zeta\in W_0^{1,q}(\Omega_r)$ with
compact support in $\overline\Omega_r$

\item[(ii)] $\gamma_r\in H(\psi_r)$ a.e. in $\Omega_r$

\item[(iii)] $\psi_r(x,\frac{-h}{r})=\frac{Q_s}{r}$,
$\psi_r(x,0)=\frac{1}{r}\phi_0(rx)$  for all $x\in \mathbb{R}$

\item[(iv)] $\partial_x\psi_r\leq 0$  and
$\partial_x\gamma_r\leq 0$  in $\mathcal{D}'(\Omega_r)$.
\end{itemize}
>From the study of problem $(P)$, we know that problem $(P_r) $ has
a unique solution with a continuous free boundary $g_r$ and that
$\lim_{x\to-\infty}\psi_r(x,z)={\frac{1}{r}v_{-\infty}(rz)}$,
$\lim_{x\to+\infty}\psi_r(x,z)={\frac{1}{r} v_{+\infty}(rz)}$.
Moreover since we assume that $\mathcal{B}$ does not depend on
$z$, we have $v_{-\infty}(z)=-{\frac{Q_s} {h}z}$ and
$v_{+\infty}(z)=-{\frac{Q_f}{r}}-{\frac{Q_f+Q_s}{h}z}$.

\begin{theorem}\label{thm2.4}
For $0<r_1<r_2$, we have
$\psi_{r_1}\leq \psi_{r_2}$  in $\Omega_{r_2}
\subset\Omega_{r_1}$.
\end{theorem}

\begin{proof}
We remark that $(\psi_{r_1},\gamma_{r_1})$ and
$(\psi_{r_2},\gamma_{r_2})$ satisfy the same equation on
$\Omega_{r_2}$. Moreover one can check that
$$\psi_{r_1} \leq
\psi_{r_2}\quad\text{on }\partial\Omega_{r_2}\quad
\text{and}\quad\lim_{x\to\pm\infty}\psi_{r_1}(x,z)
\leq\lim_{x\to\pm\infty}\psi_{r_2}(x,z).
$$
Then we can derive a similar result as in Lemma \ref{lm2.1}.
Since
$\psi_{r_1}(x,0)=\psi_0(x,0)=-Q_f/r_1$
where $\psi_0=\min(\psi_{r_1},\psi_{r_2})$, one can argue as in
the proof of Theorem \ref{thm2.1} to prove that $\psi_{r_1}=\psi_0$ in
$[\psi_0<0]$.

 To prove that $\psi_{r_1}=\psi_0$ in $[\psi_0>0]$ it is
enough to verify that $\nabla\psi_{r_1}$ does not vanish on some
part of the left hand side of the line ${[z=-\frac{h}
{r_2}]}$. So assume that for some $x_0$, we have
${\nabla\psi_ {r_1}(x,-h/r_2)=0}$ for all
$x\leq x_0$. Then
${\psi_{r_1}(x_0,-h/r_2)}=C\in\mathbb{R}$ for all $x\leq x_0$.
By the asymptotic behavior
$C=Q_s/r_2>0$ and by continuity and
monotonicity, $\psi_{r_1}$ is positive in a strip
${D_\epsilon=(-\infty,x_0)\times(-\frac
{h}{r_2}-\epsilon,-\frac{h}{r_2}+\epsilon)}$ for some small
$\epsilon>0$. Therefore $\psi_{r_1}$ is $\mathcal{B}$-Harmonic in
$D_\epsilon$ and its gradient has non-isolated zeros. This means
that $\psi_{r_1}=C$ in $D_\epsilon$ which contradicts the
asymptotic behavior.
\end{proof}

\begin{corollary}\label{coro2.2}
$\Omega_f$ is star shaped with the origin.
\end{corollary}

 \begin{proof} Let $X_0\in \Omega_f$ and $t\in (0,1]$. We have by
Theorem \ref{thm2.4}, $\psi_t(X_0)\leq\psi_1(X_0)=\psi(X_0)<0 $. So
$\psi(tX_0)<0$, which means that $tX_0\in \Omega_f$.
\end{proof}

\begin{corollary}\label{coro2.3}
$(i)$ There exists $z_0\in(-h^*,0)$ such that $g(z)\geq 0$
for all $z\in (-h^*,z_0)$.\\
 $(ii)$ $g$ is non-increasing where it is nonnegative.
\end{corollary}

\begin{proof}  $(i)$ First note that the set
$[g>0]$ is nonempty. Indeed if $g(z)\leq 0$ $\forall z\in
(-h^*,0)$, then for all $(x,z)\in(0,+\infty)\times(-h^*,0)$ we
have $\psi (x,z)<0$. Since we also have $\psi(x,z)>0$ in
$(0,+\infty)\times(-h,-h^*)$, we get a contradiction with Lemma \ref{lmA.3}.
Let $z_0=\inf\{\,z\in (-h^*,0)\,/\,g(z)>0\}$ and take $z\in (-h^*,z_0)$.
Since ${r=\frac{z_0}{z}<1}$, we have
\begin{align*}
0&=\psi(g(z_0),z_0)=\frac{1}{r}\psi\Big(r\Big(\frac{g(z_0)}
{r}\Big),rz\Big)=\psi_r\Big(\frac{g(z_0)}{r},z\Big)\\
&\leq \psi_1\Big(\frac{g(z_0)}{r},z\Big)=\psi\Big(\frac{g
(z_0)}{r},z\Big)\leq\psi(g(z_0),z)
\end{align*}
since $g(z_0)\geq 0$ and ${\frac{1}{r}>1}$.
Thus $\psi(g (z_0),z)\geq 0$ and then $g(z)\geq g(z_0)\geq 0$.

\noindent $(ii)$ Let $z_1, z_2\in (-h^*,z_0)$ such that $z_1<z_2$.
If $g(z_2)\geq 0$, we can argue as in $i)$ and obtain $g (z_1)\geq
g(z_2)$.\end{proof}

\section{Behavior of the free boundary near $z=-h^*$ }\label{3}

In \cite{CL2} we proved in the linear and heterogeneous case
($q=2$) that the free boundary has the line $[z=-h^*]$ as an
asymptote. Here we generalize the proof to the nonlinear case.
Before this, we give a second proof which is much simpler but
works only when the permeability is constant. \vskip 0.3cm

\begin{theorem}\label{thm3.1}
 $(i)$  The set $S=\{x\in \mathbb{R}:\psi(x,- h^*)=0\}$
  is empty and $\Gamma=[x=g(z)]$.\\
 $(ii)$ ${\lim_{z\to -h^*}g(z)=+\infty}$.
\end{theorem}

 \textbf{Case of a constant permeability:}
 $(ii)$ Since $g$ is non-increasing in $(-h^*,z_0)$ (see
Corollary \ref{coro2.3}), there exists a limit $L$ for $g$ as
$z\to - h^*$. Assume that $L$ is finite. By the monotonicity of
$g$ we get $g(z)\leq L$ $\forall z\in (-h^*,z_0)$ and then
\[
\forall x>L,\quad\forall z\in (-h^*,z_0), \quad\psi(x,z)<0.
\]
Since $\psi>0$ for $z<-h^*$ we deduce by continuity that we
have necessarily $\psi(x,-h^*)=0$ for all $x>L$. This leads to a
contradiction with Lemma \ref{lmA.3}. Thus $L=+\infty$.

\noindent $(i)$ Assume that $S\ne\emptyset$. Then there exists
$x_0\in \mathbb{R}$ such that $\psi(x_0,-h^*)=0$. For $A_0>x_0$
there exists $\delta>0$ by $(ii)$ such that: for all
$z \in(-h^*,-h^*+\delta)$, $g(z)>A_0$. So for
$(x,z)\in(-\infty,A_0)\times(-h^*,-h^*+\delta)$ we have $\psi(x,z)>0$.
By monotonicity of $\psi$, we deduce that
$\psi(x,-h^*)=0$ $\forall x\geq x_0$ since
one has $v_{+\infty}(-h^*)=0\leq \psi(x,-h^*)\leq \psi
(x_0,-h^*)=0$. Hence we have $\psi>0$ in $(x_0,A_0)\times\Big
((-h^*-\delta,-h^*+\delta)\setminus\{-h^*\}\Big)$ and
$\psi(x,-h^*) =0$ $\forall x\in( x_0,A_0)$ which contradicts Lemma
\ref{lmA.3}.

\noindent\textbf{The general case:}
 Since we follow the proof given in \cite{CL2} for the
linear case, we will only give an outline of it.

\noindent $(i)$ Assume that $S\ne\emptyset$. Then
$S=[\alpha,+\infty)$ with $\alpha=\inf S>-\infty$. We need some
lemmas.

\begin{lemma}\label{lm3.1}
There exists $u\in L^\infty(-h,0)$ such that $u>0$ for a.e
$z\in(-h,0) $
and
\[
E\big(u(z)\big)=\left(\overline{A}_{22}u^2(z)-2\overline{A}_{12}u(z)+
\overline{A}_{11}\right)^{\frac{q-2}{2}}
\left(\overline{A}_{22}u(z)-\overline{A}_{12}\right)-C_0=0
\]
for some constant
$C_0>-\big(\overline{A}_{11}^{\frac{q-2}{2}}\overline{A}_{12}
\big)(z)$ for a.e $z\in(-h,0)$. $\overline{A}_{ij}$ being the
entries of the matrix $\overline{A}$ introduced in Section 1.
\end{lemma}

\begin{proof}
 The function $E: \mathbb{R}\to \mathbb{R}$ is continuous and satisfies
 ${\lim_{t\to +\infty}E(t)=+\infty}$,
$E(0)=-\overline{A}_{11}^{\frac{q-2}{2}}\overline{A}_{12}-C_0<0$.
So we deduce that for a.e $z\in(-h,0)$, there exists $u(z)>0$ such
that $E\big(u(z)\big)=0$. This clearly defines a positive and
uniformly bounded function $u$ on $(-h,0)$.
\end{proof}

Set $\alpha'=\alpha+\frac{1}{2}$ and define
$f(z)=\int_{-h}^z u(s)ds+\alpha'$.
For $k>0$ define $v$ and $\theta$ by
\begin{gather*}
v(x,z)=\big(k(\gamma_s-\gamma_f)\big)^{1/(q-1)}
(f(z)-x)^+\\
\theta(x,z)=\gamma_s\chi\big([x<f(z)]\big)+\gamma_f\chi\big
([x>f(z)]\big)
\end{gather*}
for $(x,z)\in D(z_1)=(\alpha',+\infty)\times(- h^*,z_1)$
with $z_1\in (-h^*,0).$ Then we have the following lemma.

\begin{lemma}\label{lm3.2}
There exists $k>0$ such that
\[
\int_{D(z_1)}\Big(\mathcal{B}(z,\nabla v)+\theta e_x\Big)
\nabla\xi\geq 0\quad\forall \xi\in\mathcal{D}\big(D(z_1)
\big).
\]
\end{lemma}
To prove this lemma, one can
adapt the proof of \cite[Lemma 6.3]{CL2}.

\begin{lemma}\label{lm3.3}
Let $(\psi,\gamma)$ be the solution of $(P)$. Then there exists
$z_0\in (-h^*,0)$ such that
\[
\int_{D(z_0)}\big(\mathcal{B}(z,\nabla \psi^+)-\mathcal{B}
(z,\nabla v_0)+(\gamma-\theta_0) e_x\big)\nabla\zeta=0
\quad\forall \zeta\in\mathcal{D}\big(\mathbb{R}^2\big)
\]
where $v_0=\min(\psi^+,v)$ and $\theta_0=\min(\gamma,\theta) $.
\end{lemma}

To prove this lemma, one can adapt the proof in
\cite[Lemma 6.4]{CL2}.

\begin{proof}[completion of the proof for Theorem \ref{thm3.1}]
Let $z_*\in (-h^*,z_0)$. \\
$\bullet$ If $\psi^+(\alpha',z_*)=0$, then by monotonicity
we have $\psi^+(x,z_*)=0\,\,\forall x\geq \alpha'$. \\
$\bullet$ If $\psi^+(\alpha',z_*)>0$, then
since we can choose $k$ such that
$\psi^+(\alpha',z_*)<v(\alpha',z_*)$, we deduce by
continuity that there exists a small ball $B_r
(\alpha',z_*)$ in which one has $\psi^+>0$ and $\psi^+< v$.
Then $\psi^+= v_0$ in $B_r(\alpha',z_*) $. Let us denote by
$C_*$ the connected component of $D(z_0) \cap[\psi^+>0]$ which
contains $B_r(\alpha',z_*)\cap D(z_0)$.

 Let $D^+(z_0)=D(z_0)\cap [x<f(z)]$. By Lemma \ref{lm3.3} we have
\[
\mathop{\rm div}\Big(\mathcal{B}(z,\nabla \psi^+)-\mathcal{B}(z,\nabla v_0)
\Big)+(\gamma-\theta_0)_x=0\quad\hbox{in } \mathcal{D}
'\big(C_*\cap D^+(z_0)\big).
\]
But since in $C_*\cap D^+(z_0)$ we have $\psi^+>0$, $v>0$, it follows that
$\gamma=\theta=\theta_0=\gamma_s$ and then
\[
\mathop{\rm div}\big(\mathcal{B}(z,\nabla \psi^+)\big)=
\mathop{\rm div}\big(\mathcal{B}(z,\nabla v_0)\big)=0\quad\hbox{in }
\mathcal{D}'\big(C_*\cap D^+(z_0)\big).
\]
So we have
\begin{gather*}
\mathop{\rm div}\big(\mathcal{B}(z,\nabla \psi^+)\big)=0,\quad
\mathop{\rm div}\big (\mathcal{B}(z,\nabla v_0)=0\quad\hbox{in }
\mathcal{D}'\big(C_*\cap D^+(z_0)\big)\\
\psi^+\geq v_0 \quad \hbox{in }C_*\cap D^+(z_0)\\
\quad\psi^+= v_0 \quad \hbox{in }B_r(\alpha',z_*) \cap D^+(z_0)
\end{gather*}
which leads by Lemma \ref{lmA.1} (see Appendix) to $\psi^+= v_0$ in
$C_*\cap D^+(z_0)$. We then conclude
as in the end of the proof of Theorem 6.1 in \cite{CL2}.

\noindent $(ii)$ See Corollary 6.6 of \cite{CL2}.
\end{proof}

\section{Behavior of the free boundary near $z=0$ }\label{4}

 Now we study the free boundary near $z=0$. We first
prove the existence of a limit $g(0-)\leq 0$ as $z$ approaches
zero. Then we prove that this limit is finite in two cases.

\subsection{Existence of the limit of $g$ at $z=0$}\label{4.1}

\begin{theorem} \label{thm4.1}
The function $g$ has a limit when $z$ approaches
zero.
\end{theorem}

\begin{proof}
It suffices to show that ${\liminf_{z\to 0}g(z) =\limsup_{z\to 0}g(z)}$.\\
First, if ${\limsup_{z\to 0}g(z)=-\infty}$, then
$$\liminf_ {z\to 0}g(z)=\limsup_{z\to 0}g(z)=\lim_{z\to 0}g(z)=-\infty.
$$
 Now assume that $\limsup_{z\to 0}g(z)=a^+>-\infty$. Note
that $a^+\leq 0$. Indeed if $a^+>0$, then $\psi(a^+,0)<0$ and by
continuity of $\psi$ there exists $\epsilon>0$ such that $\psi<0$
in $(a^+-\epsilon,a^++\epsilon)\times(-\epsilon,0) $. So $g(z)\leq
a^+-\epsilon\quad\forall z\in(-\epsilon,0)$ and then $a^+\leq
a^+-\epsilon$ which is impossible.

Set $a^-=\liminf_{z\to 0}g(z)$ and assume that $a^-<a^+$. Let
$x_1\in(a^-,a^+)$, $x_2\in(x_1,a^+)$ and let $(z_n)_n$ be a
sequence that satisfies
$$
\lim_{n\to +\infty}z_n=0\quad\text{ and }\quad
\lim_{n\to +\infty}g(z_n)=a^-.
$$
So there exists $n_1\in \mathbb{N}$ such that
$g(z_n)\leq x_1$ for all $n\geq n_1$ and then
\begin{equation}
\psi^+(x_1,z_n)=0\quad\forall n\geq n_1.
\end{equation}
Arguing as in the proof of Lemma \ref{lm3.1}, we prove the
existence of a negative function $u\in L^\infty(-h,0)$ such that
for a.e $z\in(-h,0)$
\begin{equation}
E\big(u(z)\big)=\big(\overline{A}_{22}u^2(z)-2\overline{A}_{12}u(z)
+\overline{A}_{11}\big)^
{\frac{q-2}{2}}\big(\overline{A}_{22}u(z)-\overline{A}_{12}\big)-C_1=0
\end{equation}
 for some constant
$C_1<-\big(\overline{A}_{11}^{\frac{q-2}{2}}\overline{A}_{12}
\big)(z)$ for a.e $z\in(-h,0)$. We then set
\[
f(z)=\int_{0}^z u(s)ds+x_1'\quad\hbox{for }
x_1'\in(x_1,x_2).
\]
Since $f$ is continuous and
non-increasing there exists $n_2 \in\mathbb{N}$ such that
\begin{equation}
f(0)=x_1'<f(z)<x_2\quad\forall z\in(z_n,0) \; \forall
n\geq n_2.
\end{equation}
Now for $k>0$ we define $v$ and $\theta$
by
\begin{equation}
\begin{gathered}
v(x,z)=\big(k(\gamma_s-\gamma_f)\big)^{\frac{1}{q-1}}(f(z)-x)^+\\
\theta(x,z)=\gamma_s\chi\big([x<f(z)]\big)+\gamma_f\chi\big
([x>f(z)]\big)
\end{gathered}\end{equation}
for $(x,z)\in D(z_n)=(x_1,+\infty)\times(z_n,0)$ with $n\geq n_2$.
Then as in Lemma \ref{lm3.2}, one can deduce from (4.2) and(4.4) the existence
of $k>0$ such that
\[
\int_{D(z_n)}\Big(\mathcal{B}(z,\nabla v)+\theta e_x\Big)
\nabla\xi\geq 0\quad\forall \xi\in\mathcal{D}\big(D(z_n) \big).
\]
Using the fact that $\psi^+(x_1,0)=0$, the continuity of $\psi$,
the monotonicity of $f$, (4.1) and (4.3), there exists
$n\geq\sup(n_1,n_2)$ such that $\psi^+(x_1,z)\leq v
(x_1,z)\quad\forall z\in(z_n,0)$. Moreover one can check that
\[
\psi^+\leq v\quad\hbox{on } \partial D(z_n)\quad\hbox
{and }\quad \lim_{x\to+\infty}\psi^+(x,z)=\lim_
{x\to+\infty}v(x,z)=0.
\]
Arguing as in the proof of Theorem \ref{thm3.1}, we
obtain $\psi^+\leq v$ in $D(z_n)$ from which we deduce that
$\psi^+ (x,z)=0\quad\forall (x,z)\in(x_2,a^+)\times(z_n,0)$ and
then $g(z) \leq x_2\quad\forall z\in(z_n,0)$. So
${a^+=\limsup_{z\to 0}g (z)\leq x_2<a^+}$ and we get a
contradiction. Thus we have $a^+=a^-$ and ${\lim_{z\to 0}g(z)}$
exists.\end{proof}

 The remainder of this section is devoted to
show that $g(0- )$ is finite in two general cases.


\begin{figure}[t]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(100,62)(0,-12)
\put(-2,35){\line(1,0){42}}
\put(55,35){\line(1,0){45}}
\put(99,34.15){$\to$}
\put(0,0){\line(1,0){100}}
\put(40,35){\line(0,1){10}}
\put(39.1,45){$\uparrow$}
\put(35,45){$z$}
\put(-2,38){$-b$}
\put(37,37){0}
\put(18,37){$x_R$}
\dashline{2}(20,35)(20,30)
\put(55,38){$A$}
\put(100,37){$x$}
\put(0,-15){\line(0,1){52}}
\put(-5,30){$z_R$}
\put(-5,-1){$-h$}
\put(-6,-13){$-R$}
\put(38,12){$f(x)$}
\put(69,20){$f_d(x)$}
\drawline(0,-12)(20,30)(80,-5)
\drawline(40,35)(100,0)
\dashline{2}(0,30)(20,30)
\qbezier(0,35)(15,34)(20,30)
\end{picture}
\end{center}
\caption{}
\end{figure}

\subsection{The Linear Nonhomogeneous Case}\label{4.2}

 We assume that  $q=2$ and $B_{21}'(z)\leq 0$ in $\mathcal{D}'(-h,0) $,
 then we have the following theorem.

\begin{theorem}\label{thm4.2}
$$\lim_{z\to 0}g(z)>-\infty.$$
\end{theorem}

The idea of the proof is to compare $\psi$
with a suitable function. First let $R, b>0$ and consider
\begin{equation*}
f(x)=
\begin{cases}
-R+\sqrt{R^2-(x+b)^{+2}} &\hbox{if }x\leq {x_R=\frac{R(a+b)}
{\sqrt{R^2+(a+b)^2}}-b},\\
z_R-{\frac{a+b}{R}}(x-x_R) &\hbox{if }x\geq x_R
\end{cases}
\end{equation*}
with ${z_R=\frac{R(a+b)}{\sqrt{R^2+(a+b)^2}}-R}$ (see Figure 2).
 Set
\[\omega(z)=h\frac{{\int}_0^z\frac{ds}{B_{22}(s)}}{
{\int}_{-h}^0
\frac{ds}{B_{22}(s)}}={\kappa\int_0^z\frac{ds}{B_{22}(s)}}
\quad\hbox{ for }z\in(-h,0).
\]
For $t>0$, consider
$$G(t)=\lambda[(t+1)^2-1]\quad \text{with}\quad
\lambda=\frac{Q_s}{h^2+2h}$$ and define $v_1$ by
\[
v_1(x,z)=G(f(x)-\omega(z))\quad\hbox{for }(x,z)\in
D_1=\{(x,z)\in \Omega/\,-h<z<\omega^{-1}(x)\}.
\]
For $d>0$, set
$f_d(x)=f(x)+d$ and  for $t>0$ consider
$$
K(t) =\mu \log(1+t)\quad \text{with}\quad
\mu=\frac{Q_f}{\log(1+d)}.
$$
Then we define $v_2$ by
\[ v_2(x,z)=-K(\omega(z)-f(x))\quad\hbox{for }
(x,z)\in D_2
\]
with $D_2=\{(x,z)\in \Omega :\omega^{-1}(x)
<z<\omega^{-1}of_d(x)\}$.
  Now set
\[
v=\chi(D_1)v_1+\chi(D_2)v_2\quad\hbox{and}\quad
\theta=\chi(D_1)\gamma_s+\chi(D_1)\gamma_f.
\]
Remark that $v\in H^1_{\rm loc}(D)$ since $v_1(x,\omega^{-1}(x))=G(0)=0$
and $v_2(x,\omega^{-1}(x))=-K(0)=0$. We will compare $(\psi,\gamma)$
with $(v,\theta)$ on the domain
$D=(\overline{D_1\cup D_2})\cap\Omega$.  The proof needs some preliminary
Lemmas.

 \begin{lemma}\label{lm4.1} There exists $\alpha_1>0$
such that
\begin{equation}
\forall\alpha\in(0,\alpha_1),\;\forall R\geq
\max\Big({\frac{2M^2}{\kappa^2},\frac{a}{\alpha}}\Big),\quad
\mathop{\rm div}\big(B(z)\nabla v_1\big)\geq 0\quad\hbox{ in }\mathcal{D}'
(D_1).
\end{equation}
\end{lemma}

\begin{proof}
 Indeed let $C_1$ be a constant
such that $B_{21}(z)+C_1>0$ for a.e $z\in(-h,0)$. Then we have
\begin{equation}
\begin{aligned}
&\mathop{\rm div}\big(B(z)\nabla v_1\big)\\
&=\frac{\partial}{\partial x}\big(B_{11}\frac{\partial v_1}{\partial x}
+ B_{12}\frac {\partial v_1}{\partial z}\big)
+\frac{\partial}{\partial z}\big(-C_1\frac{\partial v_1}{\partial x}
+ B_{22}\frac{\partial v_1}{\partial z}\big)
+\frac{\partial}{\partial z}\big((B_{21}+C_1)\frac
{\partial v_1}{\partial x}\big)\\
&=B_{11}\big(G''(f(x)-\omega(z))(f')^2+G'(f(x)-\omega(z))f''\big)\\
&\quad+(B_{12}-C_1)\big(\frac{-\kappa}{B_{22}}f' G''(f(x)-\omega(z))\big)
+\frac{\kappa^2}{B_{22}}G''(f(x)-\omega(z)) \\
&\quad -\frac{\kappa}{B_{22}}(B_{21}+C_1)G''(f(x)-
\omega(z))f'+B_{21}' G'(f(x)-\omega(z))f'\\
&\geq 2\lambda B_{11}\big((f')^2+\frac{\kappa^2}{B_
{22}B_{11}}-\frac{\kappa}{B_{22}B_{11}}(B_{12}-C_1)f'
+(f(x)-\omega(z)+1)f''\big)\\
&=2\lambda B_{11}I_1
\end{aligned}
\end{equation}
since $B_{21}+C_1>0,\,G''=2\lambda,\,f' (x)\leq
0,\,B_{21}'\leq 0\hbox{ in } \mathcal{D}'(- h,0)$ and
$G'(t)=2\lambda (t+1)>0$ for $t>0$.

 We shall distinguish three cases:
\begin{itemize}
\item  For  $x\leq -b$, we have
$f(x)=0$ and then
${I_1=\frac{\kappa^2}{B_{22}B_{11}}>0}$.

\item   For  $x\in(x_R,f^{-1}(-h))$, we have $f''(x)=0$,
${f'(x)=-\frac{a+b}{R}=- \alpha}$, $m\leq
B_{11},B_{22}\leq M$, $\vert B_{12}\vert\leq M$, where $m$ and $M$
are two positive constants such that
\begin{equation}
m\vert\xi\vert^2\leq \langle B(z) \xi,\xi\rangle
\leq M\vert\xi\vert^2\quad\forall\xi \in\mathbb{R}^2,\text{ a.e. }
z\in(-h,0).
\end{equation}
 So
\begin{align*}
I_1 &=\alpha^2+\frac{\kappa^2}{B_{22}B_{11}}
+\frac{\alpha\kappa}{B_{22}B_{11}}(B_{12}-C_1)\\
&\geq \alpha^2+\frac{\kappa^2}{M^2}-\frac{\alpha\kappa}{m^2}
(M+C_1)\to \frac{\kappa^2}{M^2}>0\quad\hbox{as }
\alpha\to 0.
\end{align*}
Therefore, there exists $\alpha_0>0$ such that
$\forall\alpha\in(0,\alpha_0)\quad I_1\geq 0$.

\item  For $x\in (-b,x_R)$, we have
\begin{gather*}
 f'(x)=\frac{-(x+b)}{\sqrt{R^2-(x+b)^2}}, \quad
 f''(x)=-\frac{R^2}{(R^2-(x+b)^2)^{3/2}},\\
1+{f'}^2(x)=\frac{R^2}{R^2-(x+b)^2}.
\end{gather*}
Then
\begin{align*}
I_1 &=\frac{R^2}{(R^2-(x+b)^2)^{3/2}}\Big(\sqrt{R^2-(x+b)^2}\\
&-f(x)+\frac{(R^2-(x+b)^2)^{3/2}}{R^2}\big(\frac{\kappa^2}
{B_{22}B_{11}}-1\big)\\
&+\frac{\kappa(B_{12}-C_1)}{B_{22}B_{11}}
\frac{(R^2-(x+b)^2)}{R^2}(x+b)+\omega(z)-1\Big).
\end{align*}
\end{itemize}
Note that
\begin{itemize}

\item $\sqrt{R^2-(x+b)^2}-f(x)=R$.

\item ${\frac{(R^2-(x+b)^2)^{3/2}}{R^2}\left(\frac{\kappa^2}{B_{22}
B_{11}}-1\right)\geq R\left(\frac{\kappa^2}{M^2}-1\right)}$
provided that $M>\kappa$.

\item ${\frac{\kappa(B_{12}-C_1)}{B_{22}B_{11}}\frac{(R^2-(x+b)
^2)}{R^2}(x+b)\geq-C_2\frac{\alpha R}{\sqrt{1+\alpha^2}}
\geq-C_2(x_R+b)}$
since \\
${\frac{\kappa(B_{12}-C_1)}{B_{22}B_{11}}>-C_2}$
for some positive constant $C_2$.

\item $\omega(z)\geq -h$.
\end{itemize}
Therefore
\begin{align*}
I_1 &\geq\frac{R^2}{(R^2-(x+b)^2)^{3/2}}\big(R\frac
{\kappa^2}{M^2}-\frac{C_2\alpha }{\sqrt{1+\alpha^2}}R-h-1\big)\\
&\geq\frac{R^3}{(R^2-(x+b)^2)^{3/2}}\Big(\big(\frac
{\kappa^2}{2M^2}-C_2\alpha \big)+\big(\frac{\kappa^2}{2M^2}-
\frac{h+1}{R}\big)\Big)
&\geq 0
\end{align*}
provided that $\alpha\leq \frac{\kappa^2}{2M^2 C_2}$
and $R>\frac{2M^2}{\kappa^2}(h+1)$.

Finally for
${\alpha\in\big(0,\alpha_1=\min\big(\alpha_0,\frac{\kappa^2}{2M^2C_2}\big)\big)}$
and ${R>\frac {2M^2}{\kappa^2}(h+1)}$, one has $I_1\geq
0$ and then by (4.6) we obtain (4.5).
\end{proof}

\begin{lemma}\label{lm4.2}
There exists $\alpha_2>0$ such that
\begin{equation}
\forall\alpha\in(0,\alpha_2),\; \forall R\geq
\max\big({\frac{2M^2}{\kappa^2},\frac{a}{\alpha}}\big),\quad
\mathop{\rm div}\left(B(z)\nabla v_2\right)\geq 0\quad\hbox{in }
\mathcal{D}'(D_2).
\end{equation}
\end{lemma}

\begin{proof}  Indeed we have
\begin{equation}
 \begin{aligned}
&\mathop{\rm div}\left(B(z)\nabla v_2\right)\\
&=\frac{\partial}{\partial x}
\big(B_{11}\frac{\partial v_2}{\partial x}+ B_{12}\frac {\partial
v_2}{\partial z}\big)+\frac{\partial}{\partial z}
\big(-C_1\frac{\partial v_2}{\partial x}+ B_{22}\frac
{\partial v_2}{\partial z}\big)
+\frac{\partial}{\partial z}\big((B_{21}+C_1)\frac
{\partial v_2}{\partial x}\big)\\
&=B_{11}\big(-K''(\omega(z)-f(x))(f')
^2+K'(\omega(z)-f(x))f''\big)\\
&\quad +(B_{12}-C_1)\frac{\kappa}{B_{22}}f'K''(\omega(z)-f(x))
-\frac{\kappa^2}{B_{22}}K''(\omega(z)-f(x))\\
&\quad +\frac{\kappa}{B_{22}}(B_{21}+C_1)K''(\omega(z)-f(x))f'
+B_{21}' K'(\omega(z)-f(x))f'\\
&\geq \frac{\mu B_{11}}{(1+\omega(z)-f(x))^2}\Big((f')^2
+\frac{\kappa^2}{B_{22}B_{11}}-\frac{\kappa(B_{12}-C_1)}{B_{22}B_{11}}f'
+(1+\omega(z)-f(x))f''\Big)\\
&=\frac{\mu B_{11}}{(1+\omega(z)-f(x))^2}I_2,
 \end{aligned}
\end{equation}
since
$B_{21}+C_1>0,\,K''(t)=-{\frac{\mu}{(1+t)
^2}<0,\,K'(t)=\frac{\mu}{1+t}>0,}$ $f'(x)\leq 0$,
$B_{21}'\leq 0$  in $\mathcal{D}'(-h,0)$.

 We distinguish two cases:

\noindent$\bullet$  For $x\in (x_R,f_d^{-1}(-h))$, we
have $f''(x)=0$,
${f'(x)=-\frac{a+b}{R}=- \alpha}$ and then
\begin{align*}
I_2 &=\alpha^2+\frac{\kappa^2}{B_{22}B_{11}}
+\frac{\alpha\kappa}{B_{22}B_{11}}(B_{12}-C_1)\\
&\geq \alpha^2+\frac{\kappa^2}{M^2}-\frac{\alpha\kappa}{m^2}
(M+C_1)\to \frac{\kappa^2}{M^2}>0\quad\hbox{as }
\alpha\to 0.
\end{align*}
So there exists $\alpha_0>0$ such that
$I_2\geq 0 \quad\forall\alpha\in(0,\alpha_0)$.

\noindent  $\bullet$  For $x\in(-b,x_R)$, we have
\begin{align*}
I_2
&=\frac{R^2}{(R^2-(x+b)^2)}+\frac{\kappa^2}{B_{22}B_{11}}-1
+\frac{\kappa(B_{12}-C_1)}{B_{22}B_{11}}\frac{(x+b)}{\sqrt{R^2-(x+b)^2}}\\
&\quad -\frac{R^2}{(R^2-(x+b)^2)^{3/2}}(1+\omega(z)-f(x))\\
&=\frac{R^2}{(R^2-(x+b)^2)^{3/2}}\Big(
\sqrt{R^2-(x+b)^2}+\frac{(R^2-(x+b)^2)^{3/2}}{R^2}
\big(\frac{\kappa^2}{B_{22}B_{11}}-1\big)\\
&\quad+\frac{\kappa(B_{12}-C_1)}{B_{22}B_{11}}(x+b)
\frac{(R^2-(x+b)^2)}{R^2}-(1+\omega(z)-f(x))\Big).
\end{align*}
Taking into account the fact that $x\in (-b,x_R]$ and (4.7), it follows that
\begin{align*}
&I_2 \\
&\geq \frac{R^2}{(R^2-(x+b)^2)^{3/2}}\Big(\!-R+2\sqrt{R^2-(x_R+b)^2}
+ R\big(\frac{\kappa^2}{M^2}-1\big)-C_2(x_R+b)-1\Big)\\
&=\frac{R^2}{(R^2-(x+b)^2)^{3/2}}\Big(-R+\frac{2R}{\sqrt{1+\alpha^2}}
+R\big(\frac{\kappa^2}{M^2}-1\big)-C_2\frac{\alpha}{\sqrt{1+\alpha^2}}R-1\Big)\\
&=\frac{R^3}{(R^2-(x+b)^2)^{3/2}}\Big(\frac{2}{\sqrt{1+\alpha^2}}-2
+\big(\frac{\kappa^2}{2M^2}-\frac{C_2\alpha}{\sqrt{1+\alpha^2}}\big)
+\frac{\kappa^2}{2M^2}-\frac{1}{R}\Big)\geq 0
\end{align*}
 provided that ${R>\frac{2M^2}{\kappa^2}}$
and $\alpha\in (0,\alpha'_0)$ for a small
$\alpha'_0>0$.
 Finally for
$\alpha\in(0,\alpha_2=\min(\alpha_0,{\alpha_0} '))$, we
have $I_2\geq 0$ and from (4.9) we obtain (4.8).
\end{proof}

\begin{lemma}\label{lm4.3}
There exists $\alpha_*>0$ and $R_*>0$ such that
\begin{equation}
\forall\alpha\in(0,\alpha_*),\;\forall R\geq R_*,\quad
\mathop{\rm div}\big(B(z)\nabla v+\theta e_x\big)\geq 0\quad
\hbox{in } \mathcal{D}'(D).
\end{equation}
\end{lemma}

\begin{proof} Let $\xi\in \mathcal{D}(D)$, $\xi\geq 0$. Then by (4.5) and
(4.8),
\begin{align*}
I(\xi) &=\int_D \big(B(z)\nabla v+\theta e_x\big)\nabla\xi\\
&=\int_{D_1} \big(B(z)\nabla v_1+\gamma_s e_x\big)\nabla\xi
+\int_{D_2} \big(B(z)\nabla v_2+\gamma_f e_x\big)\nabla\xi\\
&=\langle-\mathop{\rm div}\big(B(z)\nabla v_1\big),\xi\rangle
+\langle -\mathop{\rm div}\big(B(z)\nabla
v_2\big),\xi\rangle\\
&\quad +\int_{[z=\omega^{-1}(x)]\cap\Omega}\big(B(z)(\nabla v_1-
\nabla v_2).\nu+(\gamma_s-\gamma_f)\nu_x\big)\xi\\
&\leq \int_{[z=\omega^{-1}(x)]\cap\Omega}\big(B(z)(\nabla
v_1-\nabla v_2).\nu+(\gamma_s-\gamma_f)\nu_x\big)\xi
\end{align*}
provided that $\alpha\in(0,\min(\alpha_1,\alpha_2))$ and
${R>\frac{2M^2}{\kappa^2}(h+1)}$.  Moreover
one has
\begin{align*}
\nabla v_1(x,\omega^{-1}(x))
&=G'(0) ^t\big(f'(x),-\frac{\kappa}{B_{22}(\omega^{-1}(x))}\big)\\
&=\frac{2Q_s}{h^2+2h} ^t\big(f'(x),-\frac{\kappa}{B_
{22}(\omega^{-1}(x))}\big)
\end{align*}
\begin{align*}
\nabla v_2(x,\omega^{-1}(x))&=K'(0) ^t\big(f'
(x),-\frac{\kappa}{B_{22}(\omega^{-1}(x))}\big)\\
&=\frac{Q_f}{\log(1+d)} ^t\big(f'(x),-\frac{\kappa}{B_
{22}(\omega^{-1}(x))}\big)
\end{align*}
\begin{align*}
\nu(x,\omega^{-1}(x))
&=\frac{1}{\sqrt{1+{(\omega^{-1}of)'}^2(x)}} ^t
\big(-(\omega^{-1}of)'(x),1\big)\\
&=\frac{1}{\sqrt{\big({\frac{\kappa}{B_{22}(\omega^{-1}of
(x))}}\big)^2+{f'}^2(x)}} ^t\big(-f'(x),\frac
{\kappa}{B_{22}(\omega^{-1}(x))}\big).
\end{align*}
Then
\begin{align*}
J_1&=B(z)(\nabla v_1-\nabla v_2).\nu+(\gamma_s-\gamma_f)
\nu_x\\
&=\frac{-1}{\sqrt{{\big({\frac{\kappa}{B_{22}(\omega^{-1}of
(x))}}\big)}^2+{f'}^2(x)}}{\Big(\frac{2Q_s}{h^2+2h}-\frac
{Q_f}{\log(1+d)}\Big)}\\
& \times B(z) ^t\Big(f'(x),\frac{-\kappa}{B_{22}(\omega^{-1}
(x))}\Big)^t{\big(f'(x),\frac{-\kappa}{B_{22}(\omega^
{-1}(x))}\big)}\\
&-\frac{f'(x)}{\sqrt{{\big({\frac{\kappa}{B_{22}(\omega^{-1}of
(x))}}\big)}^2+{f'}^2(x)}}(\gamma_s-\gamma_f)\\
&=\frac{-1}{\sqrt{{\big({\frac{\kappa}{B_{22}(\omega^{-1}of
(x))}}\big)}^2+{f'}^2(x)}}{\Big(\frac{Q_s}{h^2+2h}-\frac
{Q_f}{\log(1+d)}\Big)}\\
& \times B(z)
^t{\Big(f'(x),\frac{-\kappa}{B_{22}(\omega^{-1}
(x))}\Big)}^t{\big(f'(x),\frac{-\kappa}{B_{22}(\omega^
{-1}(x))}\big)}\\
&-\frac{1}{\sqrt{{\big({\frac{\kappa}{B_{22}(\omega^{-1}of
(x))}}\big)}^2+{f'}^2(x)}}\Big(\frac{Q_s}{h^2+2h} B(z)
^t{\big(f'(x),\frac{-\kappa}{B_{22}(\omega^{-1}of
(x))}\big)}\\
&\times {}^t{\big(f'(x),\frac{-\kappa}{B_{22}(\omega^{-1}of
(x))}\big)}+(\gamma_s-\gamma_f)f'(x)\Big).
\end{align*}
Using (4.7), we obtain
\begin{align*}
&J_2\\
&=-\frac{1}{\sqrt{\big({\frac{\kappa}{B_{22}(\omega^{-1}of
(x))}}\big)^2+{f'}^2(x)}}\Big(\frac{Q_s}{h^2+2h} B(z)
^t\big(f'(x),\frac{-\kappa}{B_{22}(\omega^{-1}of
(x))}\big)\\
& \times{} ^t\big(f'(x),\frac{-\kappa}{B_{22}(\omega^{-1}of
(x))}\big)+(\gamma_s-\gamma_f)f'(x)\Big)\\
&\leq \sqrt{\big({\frac{\kappa}{B_{22}(\omega^{-1}of
(x))}}\big)^2+{f'}^2(x)}\Big(\frac{-mQ_s}{h^2+2h}+
(\gamma_s-\gamma_f)\frac{-f'(x)}{{f'}^2(x)+
\big({\frac{\kappa}{B_{22}(\omega^{-1}of
(x))}}\Big)^2}\big).
\end{align*}
Note that
$${F(x)=\frac{-f'(x)}{{f'}^2(x)
+\big({\frac{\kappa}{B_{22}(\omega^{-1}of
(x))}}\big)^2}\leq -\frac { M^2}{\kappa^2}f'(x)}
$$
and then
\begin{itemize}
\item  If $x\in (x_R,f_d^{-1}(-h))$ we have $f'(x)=-\alpha$ and
${F(x)\leq\frac{ \alpha M^2} {\kappa^2}}$.

\item  If $x\in(-b,x_R)$
\begin{align*}
F(x)&\leq-\frac{ M^2}{\kappa^2}f'(x)=\frac{ M^2}
{\kappa^2}\frac{x+b}{\sqrt{(R^2-(x+b)^2)}} \\
&\leq \frac{
M^2}{\kappa^2}\frac{x_R+b}{R}=\frac{ M^2} {\kappa^2}\frac{\alpha
}{\sqrt{1+\alpha^2}}\leq \alpha\frac { M^2}{\kappa^2}.
\end{align*}
 So if
${\alpha<\frac{mQ_s}{h^2+2h}\frac{1}{\gamma_s-\gamma_f}
\frac{\kappa^2}{ M^2}=\alpha_3}$, we get $J_2\leq 0$ and then
\begin{align*}
J_1\leq & \frac{1} {\sqrt{\big({\frac{\kappa}{B_{22}(\omega^{-1}of
(x))}}\big)^2+{f'}^2(x)} } \Big(\frac{Q_f}{\log(1+d)}-
\frac{Q_s}{h^2+2h}\Big)\\
& \times B(z){} ^t\Big(f'(x),\frac{-\kappa}{B_{22}(\omega^{-1}
(x))}\Big)^t\Big(f'(x),\frac{-\kappa}{B_{22}(\omega^
{-1}(x))}\Big).
\end{align*}
Note that
\[
\frac{Q_f}{\log(1+d)}<\frac{Q_s}{h^2+2h}\Leftrightarrow
R>\frac{e^{\frac{Q_f}{Q_s}(h^2+2h)}-1+a\alpha}{\sqrt
{1+\alpha^2}\big(\sqrt{1+\alpha^2}-1\big)}.
\] Thus if
$\alpha\in(0,\alpha_*=\min (\alpha_1,\alpha_2,\alpha_3))$
and
\[
R\geq R_*(\alpha)=\max\Big(\frac{a}{\alpha},\frac{ 2M^2}
{\kappa^2}(h+1),\frac{e^{\frac{Q_f}{Q_s}(h^2+2h)}-1+a\alpha}
{\sqrt{1+\alpha^2}\big(\sqrt{1+\alpha^2}-1\big)}\Big),
\]
 we get $I(\xi)\leq 0$  for all $\xi\in \mathcal{D}(D)$,
$\xi\geq 0$. Therefore (4.10) holds.
\end{itemize}
\end{proof}

\begin{lemma}\label{lm4.4}
For $R>\max\big(\frac{a}{\alpha},\frac{a\alpha}
{\sqrt{1+\alpha^2}(\sqrt{1+\alpha^2}-1)}\big)$,
\begin{equation}
 v\leq\psi\quad\hbox{on }\partial D\quad\hbox{and}
\quad\lim_{x\to-\infty}v(x,z)\leq \lim_{x\to-\infty}\psi
(x,z).
\end{equation}
\end{lemma}

\begin{proof} Indeed we have
\begin{itemize}
\item $v(x,-h)=G(f(x)-\omega(-h))=G(f(x)+h)\leq
    G(h)=Q_s=\psi(x,-h)$ for $x<f^{-1}(-h)$.
\item $v(x,-h)=-K(\omega(-h)-f(x))=-K(-h-f(x))
\leq 0<Q_s=\psi(x,-h)$ for $x\in(f^{-1}(-h),f_d^{-1}(-h))$.
\item $v(x,0)=G(f(x)-\omega(0))=G(0)=0=\psi(x,0)$ for $x\leq-b$.
\item $v(x,0)=-K(\omega(0)-f(x))=-K(-f(x))\leq 0=\psi(x,0)$
for $x\in (-b,x_d)$,
where $x_d$ is defined by  $f_d(x_d)=0$ and is to be chosen such
that $x_d=0$. In fact $f_d(x_d)=0$ is equivalent to
\begin{align*}
x_d &=(z_R+d)\frac{R}{a+b}+x_R \\
&=\frac{R}{a+b}\Bigl( \frac{R}{a+b}
\frac{R}{\sqrt{1+\big(\frac{R}{a+b}\big)^2}}
-R+d+\frac{a+b}{\sqrt{1+\big(\frac{R}{a+b}\big)^2}}\\
&\quad-\frac{(a+b)^2}{R}+\frac{a(a+b)}{R}\Bigl)
=\frac{1}{\alpha}\big(R\sqrt{1+\alpha^2}-(1+\alpha^2)
R+d+a\alpha\big).
\end{align*}
Then we can choose $x_d$=0 if $d=-a\alpha+R\sqrt{1+\alpha^2}
(\sqrt{1+\alpha^2}-1)>0$ which holds if
\[R>\frac{a\alpha}{\sqrt{1+\alpha^2}(\sqrt{1+\alpha^2}-
1)}.
\]
 \item $v(x,\omega^{-1}(x))=-K(f_d(x)-f(x))
=-K(d)=-Q_f\leq\psi(x,\omega^{-1}(x))$

 \item $\lim_{x\to-\infty}v(x,z)=\lim_{x\to-\infty}v_1(x,z)=\lim_
{x\to-\infty}G(f(x)-\omega(z))\\
=G(-\omega(z))\leq
{\frac{Q_s}{h}\omega(z)}=v_{-\infty}
(z)=\lim_{x\to-\infty}\psi(x,z).$
\end{itemize}
This (4.11) holds.
\end{proof}

\begin{lemma}\label{lm4.5}
For all $\alpha\in(0,\alpha_*)$, $R>\max\big(R_*
(\alpha),\frac{a\alpha}{\sqrt{1+\alpha^2}(\sqrt{1+\alpha^2}-1)}\big)$,
one has
\begin{equation}
\int_D \left(B(z)\nabla(v-\psi_0)+(\theta-\gamma_0)e_x\right)
\nabla\zeta=0\quad\forall\zeta\in\mathcal{D}(\mathbb{R}^2)
\end{equation}
 where $\psi_0=\min(\psi,v)$ and
$\gamma_0=\min(\gamma,\theta)$.
\end{lemma}

\begin{proof} Let $\zeta\in\mathcal{D}(\mathbb{R}^2)$ and
consider for $i=1,2$ the function $\zeta_i$ defined as in the
proof of Lemma \ref{lm2.1}. Then for $\epsilon>0$,
${\min\big(\zeta_i,\frac{v- \psi_0}{\epsilon}\big)\in
H^1_0(D)}$ by (4.11) and is nonnegative with compact support. By
(4.10) and $(P)i)$, we get
\begin{align*}
&\int_{D\cap[v-\psi_0\geq\epsilon\zeta_i]} B(z)\nabla(v-
\psi_0)\nabla\zeta_i+\int_D (\theta-\gamma_0)\zeta_{ix}\\
&\leq \int_D
(\theta-\gamma_0)\big(\zeta_i-\frac{v-\psi_0}
{\epsilon}\big)_x^+\\
&\leq(\gamma_s-\gamma_f)\int_{I_0}\big(\zeta_i+\frac{\psi(f^
{-1}o\omega(z),z)}{\epsilon}\big)^+
-\big(\zeta_i-\frac{v(g(z),z)}{\epsilon}\big)^+
\end{align*}
where $I_0=\{z\in(-h^*,0)/\,g(z)<f^{-1}o\omega(z)\}$. Letting
$\epsilon\to 0$, we get
\[
\int_D \big(B(z)\nabla(v-\psi_0)+(\theta-\gamma_0)e_x\big)
\nabla\zeta_i\leq 0.
\]
To obtain (4.12), we argue as in the proof
of Lemma \ref{lm2.1}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm4.2}]
 Let $u=(v-\psi_0)\chi(D)$ and $\zeta\in\mathcal{D}([z>\omega^{-
1}(x)]\cap\Omega)$. Note that $\theta=\gamma_f=\gamma_0$ in
$D_2$. We deduce from (4.12)
\[
\int_{[z>\omega^{-1}(x)]\cap D} B(z)\nabla u.\nabla\zeta=0.
\]
Since $u$ vanishes on $[z=\omega^{-1}of_d(x)]$,
then $u\in H_{\rm loc}^1(\Omega)$ and
\[
\int_{[z>\omega^{-1}(x)]\cap \Omega} B(z)\nabla
u.\nabla\zeta=0.
\]
Moreover $u\geq 0$ and $u=0$ in
$\Omega\setminus D$. So by the strong maximum principle we get
$u=0$ in $[z>\omega^{-1} (x)]\cap \Omega$ which leads to $v\leq
\psi$ in $D_2$. In particular we obtain
$\psi(f^{-1}o\omega(z),z)\geq v(f^{-1}
o\omega(z),z)=v(x,\omega^{-1}(x))=0$. So $g(z)\geq f^{-1}
o\omega(z)\quad\forall z\in(-h^*,0)$ and then
\[
g(0-)\geq f^{-1}o\omega(0)=f^{-1}(0)=-b>-\infty.
\]
\end{proof}

\subsection{The Nonlinear Homogeneous Case}
Assume that $\mathcal{B}
(z,\xi)=\vert\xi\vert^{q-2}\xi$. Then we have the following theorem.

\begin{theorem}\label{thm4.3}
$g(0-)=\lim_{z\to-\infty}g(z)>-\infty$.
\end{theorem}

We follow the proof of Theorem \ref{thm4.2} and we use
the same notation with $\omega(z)=z$. We will need the following three
Lemmas.

\begin{lemma}\label{lm4.6} We have
\begin{equation}
\Delta_q v_1>0\quad\text{in}\quad D_1,\quad
\forall R>R_1=\frac{h+1} {q-1}(\vert
q-2\vert+q-1).
\end{equation}
\end{lemma}

\begin{proof} Indeed we have
\begin{align*}
\Delta_q v_1 &=\mathop{\rm div}(\vert\nabla v_1\vert^{q-2}\nabla v_1)\\
&=\frac{\partial}{\partial x}\big(\vert\nabla v_1\vert^{q-2}
\frac{\partial v_1}{\partial x}\big)+\frac{\partial} {\partial
z}\big(\vert\nabla v_1\vert^{q-2}\frac{\partial
v_1}{\partial z}\big)\\
&=(G'(f(x)-z))^{q-2}(1+{f'}^2(x))^{\frac{q-2}{2}}\\
&\times\Big((q-1)G''(f(x)-z)(1+{f'}^2(x))
+G'(f(x)-z)f''(x)\frac{1+(q-1){f'}^2(x)}{1+{f'}^2(x)}\Big).
\end{align*}
We distinguish two cases:

\noindent $\bullet$ If $x<-b$ or $x>x_R$, we have
$f''(x)=0$ and then $\Delta_q v_1>0\hbox{ in }D_1.$

\noindent $\bullet$ If $-b<x\leq x_R$, we have
\begin{align*}
\Delta_q v_1 &=(2\lambda(1+f(x)-z))^{q-2}(1+{f'}^2(x))
^{\frac{q-2}{2}}\Big(2\lambda\frac{(R^2-(x+b)^2)^{3/2}}{R^2}
\Big)\\
&\quad\times\Big((q-1)(R+z-1)+(q-2)(f(x)-z+1)\frac{(R^2-(x+b)^2)}{R^2}
\Big)\\
&\geq
\left(2\lambda(1+f(x)-z)\right)^{q-2}\left(1+{f'}^2(x)\right)^{\frac
{q-2}{2}}\Big(2\lambda\frac{(R^2-(x+b)^2)^{3/2}}{R^2}\Big)\\
&\quad\times\Big((q-1)(R-h-1)-\vert q-2\vert(h+1)\Big)
\end{align*}
since ${\big\vert(q-2)(f(x)-z+1)\frac{(R^2-(x+b)^2)} {R^2}\big\vert\leq
 \vert q-2\vert(h+1)}$.
 So for $R>R_1$ we get $\Delta_q v_1>0$ in $D_1$
\end{proof}

\begin{lemma}\label{lm4.7} For all
$$R>R_2(\alpha)
=\max\Big(\frac{a}{\alpha},\frac{(q-1)+\vert q-2\vert(h+1)}
{(q-1)\Big(\frac{2}{\sqrt{1+\alpha^2}}-1\Big)}\Big),
$$
we have
\begin{equation}
\Delta_q v_2> 0\quad\text{in}\quad D_2,\quad
\forall\alpha\in(0,1).
\end{equation}
\end{lemma}

\begin{proof} We have
\begin{align*}
\Delta_q v_2 &=(K'(z-f(x)))^{q-2}(1+{f'}^2(x))^
{\frac{q-2}{2}}\Big(-(q-1)K''(z-f(x))(1+{f'}^2(x))\\
&+K'(z-f(x))f''(x)\big(q-1-\frac{q-2}{1+{f'}^2(x)}\big)\Big).
\end{align*}
We distinguish two cases:

\noindent $\bullet$ For $x_R<x<f_d^{-1}(-h)$, we have
$f''(x)=0$ and then $\Delta_q v_2> 0\hbox{ in }D_2.$

\noindent $\bullet$ If $-b<x\leq x_R$, we have
\begin{align*}
&\Delta_q v_2 \\
&=\mu(K'(z-f(x)))^{q-2}(1+{f'}^2
(x))^{\frac{q-2}{2}}\Big(\frac{q-1}{(1+z-f(x))^2}\frac{R^2}
{R^2-(x+b)^2}\\
&\quad+\frac{1}{1+z-f(x)}\big(\frac{-R^2}{(R^2-(x+b)^2)^{3/2}}
\big){\big((q-1)-(q-2)
\frac{(R^2-(x+b)^2)}{R^2}\big)}\Big)\\
&=\mu(K'(z-f(x)))^{q-2}(1+{f'}^2(x))^{\frac{q-2}
{2}}\frac{R^2}{(R^2-(x+b)^2)^{3/2}}\frac{1}{(1+z-f(x))^2}
\\
&\quad \times\Big((q-1)(2\sqrt{R^2-(x+b)^2}-R)-(q-1)(1+z)\\
&\quad +(q-2)(1+z-f(x))\frac{(R^2-(x+b)^2)}{R^2}\Big)\\
&\geq \mu(K'(z-f(x)))^{q-2}(1+{f'}^2(x))^{\frac
{q-2}{2}}\frac{R^2}{(R^2-(x+b)^2)^{3/2}}\frac{1}{(1+z-f(x))
^2}\\
&\quad\times \Big((q-1)\Big(\frac{2}{\sqrt{1+\alpha^2}}-1\Big)R-(q-1)-
\vert q-2\vert(h+1)\Big)
\end{align*}
 since ${2\sqrt{R^2-(x+b)^2}-R\geq
2\sqrt{R^2-(x_R+b)^2}-
R=R\Big(\frac{2}{\sqrt{1+\alpha^2}}-1\Big)}$.

 If we choose $\alpha\in(0,1)$ such that
${\frac{2}{\sqrt {1+\alpha^2}}-1>0}$ and
${R\geq\max(\frac{a}{\alpha},R_2(\alpha))}$, we get
(4.14).
\end{proof}

\begin{lemma}\label{lm4.8}
There exists $\alpha_*>0$ and $R_*>0$ such that
\begin{equation}
\forall\alpha\in(0,\alpha_*),\quad\forall R>R_*,\, \hbox{
one has }\,\Delta_q v +\theta_x\geq 0\hbox{ in }\mathcal{D}'(D).
\end{equation}
\end{lemma}

 \begin{proof}
Let $\xi\in \mathcal{D}(D)$, $\xi\geq 0$. For $\alpha\in(0,1)$
and $R\geq\max\big(R_1, R_2(\alpha),\frac{a}{\alpha}\big)$,
we have by (4.14)-(4.15)
\begin{align*}
I(\xi)&=\int_D(\vert\nabla v\vert^{q-2}\nabla v+\theta e_x)
\nabla\xi\\
&=\int_{D_1}(\vert\nabla v_1\vert^{q-2}\nabla v_1+\gamma_s
e_x)\nabla\xi+\int_{D_2}(\vert\nabla v_2\vert^{q-2}\nabla
v_2+\gamma_f e_x)\nabla\xi\\
&=\langle-\Delta_q v_1,\xi>+<-\Delta_q v_2,\xi\rangle\\
&\quad+\int_{[z=f(x)]\cap D}\big((\vert\nabla v_1\vert^{q-2} \nabla
v_1-\vert\nabla v_2\vert^{q-2}\nabla v_2).\nu+
(\gamma_s-\gamma_f)\nu_x\big)\xi\\
&\leq \int_{[z=f(x)]\cap D}\big((\vert\nabla v_1\vert^{q-2} \nabla
v_1-\vert\nabla v_2\vert^{q-2}\nabla v_2).\nu+
(\gamma_s-\gamma_f)\nu_x\big)\xi\\
&=\int_{[z=f(x)]\cap D}J_1\xi.
\end{align*}
 Recall that
\begin{eqnarray*}
&&\nabla v_1(x,f(x))=\frac{2Q_s}{h^2+2h}(f'(x),-1)\cr
&&\vert\nabla v_1\vert^{q-2}\nabla
v_1=\Big(\frac{2Q_s}{h^2+2h}\Big)^{q-1}\left(1+{f'}^2(x)\right)^{\frac{q-2}{2}}(f'(x),-1)\cr
&&\nabla v_2(x,f(x))=\frac{Q_f}{\log(1+d)}(f'(x),-1)\cr
&&\vert\nabla v_1\vert^{q-2}\nabla
v_1=\Big(\frac{Q_f}{\log(1+d)}\Big)^{q-1}\left(1+{f'}^2(x)\right)^{\frac{q-2}{2}}(f'(x),-1)\cr
&&\nu(x,f(x))=\frac{1}{\sqrt{1+{f'}^2(x)}}(-f'(x),1).
\end{eqnarray*}
 Then
\begin{align*}
J_1&=(1+{f'}^2(x))^{\frac{q-1}{2}}\Big(\big(\frac
{Q_f}{\log(1+d)}\big)^{q-1}-\big(\frac{2Q_s}{h^2+2h}\big)^{q-1}
\Big)\\
&\quad +(\gamma_s-\gamma_f)\frac{-f'(x)}{\sqrt{1+{f'}^2(x)}}\\
&=\left(1+{f'}^2(x)\right)^{\frac{q-1}{2}}\Big(\big(\frac{Q_f}
{\log(1+d)}\big)^{q-1}-\frac{1}{2}\big(\frac{2Q_s}{h^2+2h}
\big)^{q-1}\Big)\\
&\quad+\left(1+{f'}^2(x)\right)^{\frac{q-1}{2}}\Big((\gamma_s-
\gamma_f)\frac{-f'(x)}{(1+{f'}^2(x))^{q/2}}-
\frac{1}{2}{\big(\frac{2Q_s}{h^2+h}\big)}^{q-1}\Big).
\end{align*}
Moreover we have
\[
\frac{-f'(x)}{(1+{f'}^2(x))^{q/2}} \leq -f'(x)\leq
\alpha\quad\forall x\in(-b,f_d^{-1}(- h))
\]
 and for
\[
R>R_3(\alpha)=
\frac{ e^{ \frac{Q_f(h^2+2h)}{2^{\frac{q-2}{q-1}}Q_s} } -1+a\alpha
} {\sqrt{1+\alpha^2}(\sqrt{1+\alpha^2}-1)}
\]
one has
\[
\left(\frac{Q_f}{\log(1+d)}\right)^{q-1}<\frac{1}{2}\left(\frac
{2Q_s}{h^2+2h}\right)^{q-1}=\Big(\frac{Q_s}{2^{\frac{1}
{q-1}}(h^2+2h)}\Big)^{q-1}.
\]
Thus if
\[
\alpha \in\Big(0,\alpha_*=\min\big(1,\frac{1}{2
(\gamma_s-\gamma_f)}\big(\frac{2Q_s}{h^2+2h}\big)^{q-1}\big)\Big)
\]
 and $R\geq R_*=\max\left(R_1,R_2(\alpha),R_3
(\alpha),\frac{a}{\alpha}\right)$,
 we get $J_1\leq 0$ and then
 $I(\xi)\leq 0$  for all $\xi\in \mathcal{D}(D)$,
 $\xi\geq 0$.
 Hence (4.15) holds.\end{proof}

\begin{proof}[Proof of Theorem \ref{thm4.3}]
Clearly  for ${R>\max\big(\frac{a\alpha}{\sqrt{1+\alpha^2}(\sqrt
{1+\alpha^2}-1)},\frac{a}{\alpha}\big)}$. Then  we have
(see Lemma \ref{lm4.4})
\[
v\leq\psi\quad\hbox{on }\partial D\quad\hbox{and}
\quad\lim_{x\to-\infty}v(x,z)\leq \lim_{x\to-\infty}\psi (x,z).
\]
Using (4.15) and arguing as in the proof of Lemma \ref{lm2.1}, we can establish
that for all $\alpha\in (0,\alpha_*)$ and ${R>\max\big(R_*,\frac{a
\alpha}{\sqrt {1+\alpha^2}(\sqrt{1+\alpha^2}-1)}\big)}$,
\[
(\Delta_q v-\Delta_q\psi_0)+(\gamma-\theta_0)_x=0
\quad\text{in}\quad \mathcal{D}'(D).
\]
 Assume that $D_1\cap [\psi<0]\ne\emptyset$ and note that we have
$\gamma=\gamma_f=\gamma_0$ and $\theta=\gamma_s$ in
$D_1\cap[\psi<0]$. We also have $\psi=\psi_0$ in $D_1\cap[\psi<0]$
since $\psi<0<v$. So we deduce that
\[
\Delta_q v=\Delta_q\psi_0-(\gamma-\theta_0)_x
=\Delta_q\psi-(\gamma_f-\gamma_s)_x=0
\quad\text{in }\mathcal{D}'({D_1\cap[\psi<0]}).
\]
This leads to a contradiction since $\Delta_q v=\Delta_q
v_1>0$ in $D_1$. Therefore
$D_1\cap[\psi<0]=\emptyset$ and $\psi\geq 0$ in $D_1$. In
particular we have $g(z)\geq f^{- 1}(z)\quad\forall z\in(-h^*,0)$
and then $g(0-)\geq f^{-1} (0)=-b>-\infty$.
\end{proof}

\subsection{The Linear Homogeneous Case}
Under the assumption $\mathcal{B} (z,\xi)=\xi$
we show that $g(0-)<0$. This result was announced in \cite{AD}
without an explicit proof. In that paper the authors indicated
that this result can be obtained by using hodograph techniques to
get a semi-explicit expression for $\bar{\psi}$ which is the
unique harmonic function satisfying the same boundary and limit
conditions as $\psi$. However an explicit proof was not given.
That is why we propose here a proof of this result.

\begin{theorem}\label{thm4.4}
\begin{equation}
\lim_{z\to 0^-}g(z)=g(0-)\leq -\frac{h}{\pi}\log
\Big(\frac {e^{\frac{a\pi(Q_s+Q_f)}{h Q_f}}-1}{e^{\frac{a\pi(Q_s+Q_f)} {h
Q_f}}-e^{\frac{a\pi}{h }}}\Big)<0.
\end{equation}
\end{theorem}

\begin{proof}
Since we have $\Delta\psi\geq 0$ and $\Delta \bar{\psi}=0$ in $\Omega$,
we get by taking into account the boundary conditions and the limit behavior
at infinity
that $\psi\leq \bar{\psi}$ in $\Omega$. Moreover if we denote by
$\bar{g}$ the function defined by $\bar{\psi}(\bar{g}(z),z)=0$
$\forall z\in(-h^*,0)$, we get
$g(z)\leq \bar{g}(z)$ for all $z\in(-h^*,0)$ which leads to
\begin{equation}
g(0-)\leq \bar{g}(0-).
\end{equation}
Next, we will prove that
\begin{equation}\bar{g}(0-)=-\frac{h}{\pi}\log\left(\frac{e^{\frac{a\pi
(Q_s+Q_f)}{h Q_f}}-1}{e^{\frac{a\pi(Q_s+Q_f)}{h Q_f}}-e^
{\frac{a\pi}{h }}}\right)<0.
\end{equation}
For simplicity we introduce the function
\[
\psi_0(x,z)=\bar{\psi}(hx,h(z-1))\quad\hbox{for }(x,z)
\in \mathbb{R}\times[0,1]
\]
which is harmonic in $\mathbb{R}\times(0,1)$ and satisfies
$\psi_0(x,0)=Q_s$, $\psi_0(x,1)=\phi_0(x)$ and
$\lim_{x\to \pm\infty}\psi_0(x,z)=v_{\pm\infty}(h(z-1))$.
 In \cite{LC}, one can find that
\begin{align*}
\psi_0(x,z)&=\mathop{\rm Re}\left( \frac{i}{2} \int_{-\infty}^{+\infty}Q_s
\coth\frac{\pi(t-x-iz)}{2}+\phi_0
(ht)th\frac{\pi(t-x-iz)}{2}dt\right)\\
&=-\frac{Q_s}{2}\int_{-\infty}^{+\infty}\frac{ch\pi(t-x) +\cos\pi
z}{sh^2\pi(t-x)+\sin^2\pi z}\sin\pi z dt\\
&\quad+\frac{1}{2}\int_{-\infty}^{+\infty}\phi_0(ht)\frac{\sin\pi
z}{ch\pi(t-x)+\cos\pi z }dt\\
&=-\frac{Q_s}{2}\int_{-\infty}^{+\infty}\frac{ch\pi s+\cos\pi
z}{sh^2\pi s+\sin^2\pi z}\sin\pi z ds+\frac{1}{2}I (x,z)
\end{align*}
where
\begin{align*}
I(x,z)&=\int_0^{\frac{a}{h}}-\frac{Q_f}{a}ht\frac{\sin\pi z}
{ch\pi(t-x)+\cos\pi z}dt
+\int_{\frac{a}{h}}^{+\infty}-Q_f\frac{\sin\pi z}{ch\pi(t-x)
+\cos\pi z}dt\\
&=-\frac{Q_f}{a}h\int_{-x}^{\frac{a}{h}-x}t\frac{\sin\pi z} {ch\pi
t+\cos\pi z}dt
-\frac{Q_f}{a}hx\int_{-x}^{\frac{a}{h}-x}\frac{\sin\pi z}
{ch\pi t+\cos\pi z}dt\\
&\quad-Q_f\int_{\frac{a}{h}-x}^{+\infty}\frac{\sin\pi z}{ch\pi t+\cos\pi
z}dt.
\end{align*}
Then we deduce that
\begin{align*}
\frac{\partial \psi_0}{\partial x}&=\frac{1}{2}\frac {\partial
I}{\partial x}=-\frac{Q_f}{2a}h\int_{-x}^{\frac{a}
{h}-x}\frac{\sin\pi z}{ch\pi t+\cos\pi z}dt\\
&=-\frac{hQ_f }{a\pi}\Big[\mathop{\rm Arctan}\Big(\frac{e^{(\frac{a}
{h}-x)\pi}+\cos\pi z} {\sin \pi
z}\Big)-\mathop{\rm Arctan}\left(\frac{e^{-\pi x}+\cos\pi z}{\sin \pi
z}\right)\Big]
\end{align*}
and for $(x,z)\in\Omega$
\begin{multline*}
\bar{\psi}(x,z)=v_{-\infty}(z)\\
-\frac{hQ_f }{a\pi}\int_{-
\infty}^{\frac{x}{h}} \Big(\mathop{\rm Arctan}\left(\frac{e^{-\pi
t}-\cos\frac{\pi z}{h}}{\sin \frac{\pi
z}{h}}\right)-\mathop{\rm Arctan}\Big(\frac{e^{\big(\frac{a}{h}-t\big)\pi}-\cos\frac
{\pi z}{h}}{\sin \frac{\pi z}{h}}\Big)\Big)dt.
\end{multline*}
Using this formula we give an asymptotic behavior of $\bar {\psi}$
near $z=0$. Note that for $x\leq 0$ and ${t<\frac{x} {h}}$, we
have $e^{-\pi t}>1$ and ${e^{\big({\frac{a}{h}}-t\big) \pi}>1}$.
Therefore
\begin{gather*}
\mathop{\rm Arctan}\left(\frac{e^{-\pi t}-\cos\frac{\pi z}{h}}{\sin \frac {\pi
z}{h}}\right)=-\frac{\pi}{2}+\frac{\pi}{h}\frac{1}{1-e^{-
\pi t}}z+\frac{z^2}{2}p(z,t)\\
\mathop{\rm Arctan}\Big(\frac{e^{\big(\frac{a}{h}-t\big)\pi}-\cos\frac {\pi
z}{h}}{\sin \frac{\pi z}{h}}\Big)=-\frac{\pi}{2}+\frac
{\pi}{h}\frac{1}{1-e^{\big(\frac{a}{h}-t\big)\pi}}z+\frac
{z^2}{2}q(z,t)
\end{gather*}
with $p(z,t), q(z,t)\approx e^{\pi t}$ as $t\to -\infty$.
It follows that
\begin{align*}
0=\bar{\psi}(\bar{g}(z),z)&=-\frac{Q_s}{h}z-\frac{Q_f}{a}
z\int_{-\infty}^{\bar{g}(z)}\Big(\frac{1}{1-e^{-\pi t}}-
\frac{1}{1-e^{\big(\frac{a}{h}-t\big)\pi}}\Big)dt\\
&\quad-\frac{hQ_f}{2\pi a}z^2\int_{-\infty}^{\bar{g}(z)}(p(z,t)-q
(z,t))dt-\frac{Q_s}{h}\\
&\quad-\frac{Q_f}{a}\int_{-\infty}^{\bar{g}(z)}\left
(\frac{1}{1-e^{-\pi t}}-\frac{1}{1-e^{\big(\frac{a}{h}-
t\big)\pi}}\right)dt\\
&=z\int_{-\infty}^{\bar{g}(z)}(p(z,t)-q(z,t))dt.
\end{align*}
Letting $z\to 0$, we obtain
\[
\frac{Q_f}{a}\int_{-\infty}^{\bar{g}(0)}\big(\frac{1}{1-e^
{-\pi t}}-\frac{1}{1-e^{\big(\frac{a}{h}-t\big)\pi}}\big)
dt=\frac{Q_s}{h}
\]
and by evaluating the last integral we get
(4.17). Taking into account (4.17) and (4.18), we obtain
(4.16).\end{proof}

\begin{remark}\label{rmk4.1} \rm
As a consequence of the above theorem, we have also $g(0-)<0 $
when
\[
{\mathcal{B}(z,\xi)=\left(\frac{C^2}{B(z)}\xi_1,B(z)\xi_2\right)
\quad\hbox{with }
C={\frac{-h}{{\int_{-h}^0}{\frac{ds}{B (s)}}}.}  }
\]
\end{remark}

Let $v(x,z)=\bar{\psi}(x,\omega(z))$ with
${\omega(z) =C{\int_z^0\frac{ds}{B(s)}}}$.
We remark that
\[
\mathop{\rm div}(\mathcal{B}(z,\nabla v))=\frac{C^2}{B(z)}\Delta\bar{\psi}
(x,\omega(z))=0.
\]
Moreover $v=\psi$ on $\partial\Omega$ and $\lim_
{x\to\pm\infty}v=\lim_{x\to\pm\infty}\psi$. Then we deduce that
$\psi\leq v$ in $\Omega$ which leads to
\[
\psi(\bar{g}(\omega(z)),z)\leq v (\bar{g}(\omega(z)),z)
=\bar{\psi}(\bar{g}(\omega(z)),\omega(z))=0\quad\forall z\in
(-h^*,0).
\]
Therefore $g(z)\leq \bar{g}(\omega(z))$ for all
$z\in(-h^*,0)$ and then $g(0-)\leq \bar{g}(0-)<0$.

\section*{Appendix}\label{A:Append}

This section is devoted to some technical Lemmas.

\begin{lemma}[Strong maximum principle] \label{lmA.1}
Let $\Omega$ be a domain of $\mathbb{R}^2$ and let $\mathcal{B}$ defined
on $\Omega\times\mathbb{R}^2$ by $\mathcal{B}(X,\xi)=\vert
B(X)\xi\vert^{q-2}B^2(X) \xi$, where $B(X)$ is a locally Lipschitz
continuous symmetric and strictly elliptic matrix.
 Assume that $u_1$, $u_2 \in W_{\rm loc}^{1,q}(\Omega)$ are such that
\begin{gather}
\mathop{\rm div}(\mathcal{B}(X,\nabla u_1))
=\mathop{\rm div}(\mathcal{B}(X,\nabla u_2))=0
\quad\hbox{in }\mathcal{D}'(\Omega),\label{A.1} \\
 u_1\leq u_2 \quad\hbox{in }\Omega.\label{A.2}
\end{gather}
Then we have either $u_1=u_2$ in $\Omega$ or $u_1<u_2$
in $\Omega$.
\end{lemma}

For the proof of this lemma see \cite{D}.

\begin{lemma} \label{lmA.2}
Under the same hypothesis of Lemma \ref{lmA.1} we assume that there exists
$\Gamma_0\subset\partial\Omega$ of class $C^{1,\alpha}$ such that
\begin{gather}
u_1=u_2\quad\hbox{on }\Gamma_0,\quad u_1, u_2\in C^1
(\Omega\cup\Gamma_0),\label{A.3} \\
\mathcal{B}(X,\nabla u_1).\nu=\mathcal{B}(X,\nabla u_2).\nu\quad
\hbox{ on }\Gamma_0,\label{A.4} \\
\nabla u_1(X)\ne 0 \quad \forall X\in\Gamma_0
\quad\hbox{ or }\quad \nabla u_2(X)\ne 0
\quad \forall X\in\Gamma_0.\label{A.5}
\end{gather}
Then $u_1=u_2\hbox{ in }\Omega$.
\end{lemma}

\begin{proof} We first prove that $u_1=u_2$ in a small open set
near $\Gamma_0$. Then we conclude by Lemma \ref{lmA.1}.
 Assume for example that we have $\nabla u_1(X)\ne 0$
for all $X\in\Gamma_0$. Since $u_1\in C^1(\Omega\cup\Gamma_0)$,
there exists a small ball $B(X_0,\epsilon_0)$ centered at some
point $X_0$ of $\Gamma_0$ such that
\[
\nabla u_1(X)\ne 0 \quad\forall X\in B(X_0,\epsilon_0)\cap
(\Omega\cup\Gamma_0)
\]
and then there exists two positive
constants $c_0$ and $c_1 $ such that
\begin{equation}
c_0\leq\vert\nabla u_1(X)\vert\leq c_1\quad\forall X\in
K_1=\overline{B(X_0,\epsilon_0)\cap
(\Omega\cup\Gamma_0)}.\label{A.6}
\end{equation}
Let $u=u_1-u_2$ and $u_t=tu_2+(1-t)u_1$ for any $t\in[0,1] $. Then
we deduce from \eqref{A.1} and \eqref{A.4}
\begin{equation}
\int_{B(X_0,\epsilon_0)\cap\Omega}(\mathcal{B}(X,\nabla u_2)
- \mathcal{B}(X,\nabla u_1)).\nabla\zeta=0\quad
\forall\zeta\in \mathcal{D}(B(X_0,\epsilon_0)).\label{A.7}
\end{equation}
Remark that
\[
\mathcal{B}(X,\nabla u_2)-\mathcal{B}(X,\nabla u_1)=\int_0^1
\frac{d}{dt}(\mathcal{B}(X,\nabla u_t))dt,
\]
then \eqref{A.7} becomes
\begin{equation}
\int_{B(X_0,\epsilon_0)\cap\Omega}\mathcal{C}(X)\nabla
u.\nabla\zeta=0\quad\forall\zeta\in\mathcal{D}(B
(X_0,\epsilon_0))\label{A.8}\end{equation} where
\begin{align*}
\mathcal{C}(X)&=\int_0^1 \langle\mathcal{B}(X,\nabla u_t),\nabla
u_t\rangle^{\frac{q-2}{2}} \\
&\quad\times \left(\mathcal{B}(X,\nabla u_t)
+(q-2)\frac{\mathcal{B}(X,\nabla u_t).^t(\mathcal{B}(X,\nabla u_t))}
{\langle \mathcal{B}(X,\nabla u_t),\nabla u_t\rangle }\right)dt
\end{align*}
from which we deduce
\begin{equation}
c_2\vert Y\vert^2\lambda(X)\leq \mathcal{C}(X).Y.Y\leq c_3 \vert
Y\vert^2\lambda(X)\quad\forall(X,Y)\in(B(X_0,\epsilon_0)
\cap\Omega)\times\mathbb{R}^2\label{A.9}
\end{equation}
where $c_2,c_3$ are two positive constants and ${\lambda(X)
=\int_0^1 \vert \nabla u_t\vert^{q-2}dt}$.  Using \eqref{A.6}
and arguing as in \cite{L}, we can prove that $\lambda(X)$ is
bounded from both sides by two positive constants
$\lambda_0,\lambda_1$ in $B(X_0,\epsilon_0') \cap\Omega$ for
some $\epsilon_0'\in(0,\epsilon_0)$. So $\mathcal{C}(X)$ is
strictly elliptic in $B (X_0,\epsilon_0')\cap\Omega$.
 By \eqref{A.3} we can extend $u$ by $0$ to $B
(X_0,\epsilon_0')\setminus\Omega$ so that $u\in W^
{1,q}(B(X_0,\epsilon_0'))$. One can also extend $\mathcal{C}(X)$ by
$c_2\lambda_0 I_2$ to $B(X_0,\epsilon_0')\setminus\Omega$ so that it remains
strictly elliptic in $B(X_0,\epsilon_0')$. Then from \eqref{A.8},
\begin{equation}
\int_{B(X_0,\epsilon_0')}\mathcal{C}(X)\nabla
u.\nabla\zeta=0\quad\forall\zeta\in\mathcal{D}(B
(X_0,\epsilon_0')).\label{A.10}
\end{equation}
Now since $u\geq 0$ and $u=0$ in $B(X_0,\epsilon_0') \setminus\Omega$,
we deduce from \eqref{A.10} and the strong maximum principle for
linear elliptic equations that $u=0$ in $B
(X_0,\epsilon_0')$ which means that $u_1=u_2$ in $B
(X_0,\epsilon_0')\cap\Omega$.
\end{proof}

\begin{lemma}[Non-oscillation Lemma] \label{lmA.3}
Let $z_0\in(- h^*,0)$, $x_0\in\mathbb{R}$, $r>0$ and assume that
$S_{eg}=\{(x,z_0)/\,\vert x-x_0\vert\leq r\}\subset\Gamma$, then we
cannot have
\[
\forall(x,z)\in B_r(x_0,z_0)\setminus S_{eg}\quad\psi(x,z)
\ne 0
\]
where $B_r(x_0,z_0)$ is the open ball of center
$(x_0,z_0)$ and radius $r$.
\end{lemma}

The proof follows as in \cite[Lemma 5.1]{CCL} and uses
Lemma \ref{lmA.1}.

\subsection*{Acknowledgments}
We would like to thank KFUPM for the facilities provided and for
its support.

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