\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 24, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/24\hfil Dirichlet problems]
{Growth rate and existence of solutions to Dirichlet problems for
prescribed mean curvature equations on unbounded domains}

\author[Z. Jin\hfil EJDE-2008/24\hfilneg]
{Zhiren Jin}

\address{Zhiren Jin  \newline
 Department of Mathematics and Statistics \\
 Wichita State University \\
 Wichita, Kansas 67260-0033, USA}
\email{zhiren@math.wichita.edu}

\thanks{Submitted February 9, 2008. Published February 22, 2008.}
\subjclass[2000]{35J25, 35J60, 35J65}
\keywords{Elliptic boundary-value problem;
quasilinear elliptic equation; \hfill\break\indent
prescribed mean curvature equation;  unbounded domain;
Perron's method}

\begin{abstract}
 We prove growth rate estimates and existence of solutions to
 Dirichlet problems for prescribed mean curvature
 equation on unbounded domains inside the complement
 of a cone or a parabola like region in $\mathbb{R}^n$ ($n\geq 2$).
 The existence results are proved using a modified Perron's
 method by which a subsolution is a solution to the minimal
 surface equation, while the role played by a supersolution is
 replaced by estimates on the uniform $C^{0}$ bounds on
 the liftings of subfunctions on compact sets.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction and main results}

Let $\Omega$ be an unbounded domain with $C^{2,\gamma }$
($0<\gamma <1$) boundary in $\mathbb{R}^n$
($n\geq 2$), $\phi $ be a $C^{0}$ function on $\partial \Omega$, and
$\Lambda $ be a $C^{1}$ function on $\overline{\Omega}$,
we consider the Dirichlet problem
for the prescribed mean curvature
equation on $\Omega$ (here the summation convention is used):
\begin{gather}
((1+|Du|^{2}) \delta_{ij} - D_{i}u D_{j}u )D_{ij} u =
n \Lambda  (1+|Du|^{2})^{3/2} \quad\text{on } \Omega;
\label{eq:problem11} \\
u=\phi \quad\text{on } \partial\Omega .
\label{eq:problem12}
\end{gather}

In this paper, we  investigate the conditions from which we can
derive growth estimates and existence of solutions
$u$ for \eqref{eq:problem11}-\eqref{eq:problem12}.

When $\Omega$ is a bounded domain, Serrin proved in \cite{Serrin} that
\eqref{eq:problem11}-\eqref{eq:problem12} has a solution in
$C^{0}(\overline{\Omega})\cap C^{2}(\Omega )$ as long as one can get
$C^{0}$ estimates and the mean curvature $H'$ on the boundary
$\partial \Omega$ with respect to the inner normal satisfying
$H'\geq \frac{n}{n-1} |\Lambda |$
on $\partial \Omega$. Furthermore, a counterexample is given
\cite[page 480]{Serrin} to show that for some functions
$\Lambda $, \eqref{eq:problem11}-\eqref{eq:problem12}
do not have a $C^{2}$ solution (the only thing that did not work
out in the example is the
$C^{0}$ estimate).

When $n=2$, $\Omega$ is a strip and $\Lambda $ is a constant $H$,
there have been a lot of interest in investigating the solutions
of \eqref{eq:problem11}-\eqref{eq:problem12}. Finn \cite{Finn1}
showed that the solvability of \eqref{eq:problem11} in $\Omega$
implies that the width of $\Omega$ will be less than
$\frac{1}{|H|}$. When the width of a strip $\Omega$ is $1/|H|$,
the half cylinder of radius $1/(2|H|)$ is a graph with constant
mean curvature $H$ in the strip. Collin \cite{Collin} and Wang
\cite{Wang} showed independently that there are graphs with
constant mean curvature $H$ on the strip $\Omega$ with width
$1/|H|$ other than the half cylinder. When $\Lambda =H$ and
$\Omega $ is an unbounded convex domain on a plane, Lopez
\cite{Lopez2} proved that the necessary and sufficient condition
for \eqref{eq:problem11} to have solutions with zero boundary
value is that $\Omega$ is inside a strip of width $1/|H|$.


When $\Omega$ is a strip on the plane, the existence of constant
mean curvature graphs with prescribed boundary  was considered by
Lopez in \cite{Lopez4}. The approach used in \cite{Lopez4} is a
modified version of the classical Perron's method of super- sub-
solutions. The subsolution used in \cite{Lopez4} is a solution to
the minimal surface equation (i.e. a solution to
\eqref{eq:problem11}-\eqref{eq:problem12} with $\Lambda =0$),
while the role played by a supersolution is replaced by a family
of turned to side nodoids (the use of turned to side nodiods was
adopted from an idea used by Finn \cite{Finn}) that were used to
prove that liftings from subfunctions will be bounded uniformly on
any compact subset of $\Omega$.

When $\Omega$ is an unbounded domain inside a cone or cylinder, we
proved in \cite{Jin3} the existence of solutions to
\eqref{eq:problem11}-\eqref{eq:problem12} for certain class of
functions $\Lambda$. The approach used in \cite{Jin3} is also a
modified version of the classical Perron's method. There are new
difficulties in carrying out the Perron's method when $\Lambda $
is not a constant and $\Omega$ is not a slab. The main difficulty
is that the family of turned to side nodiods cannot be used
anymore. The difficulty was overcome in \cite{Jin3} by
constructing a family of auxiliary functions that were used to
prove that liftings from subfunctions will be bounded uniformly on
any compact subset of $\Omega$. However when $\Omega$ is outside a
cone (in the compliment of a cone) or inside a parabola-shaped
region, the family of auxiliary functions used in \cite{Jin3} can
no longer be used.In this paper, we construct a new family of
auxiliary functions so that we can use the Perron's method to
prove the existence of solutions to
\eqref{eq:problem11}-\eqref{eq:problem12}. As a by product, we can
also derive the growth estimates for solutions $u$ to
\eqref{eq:problem11}-\eqref{eq:problem12}.


For more historical notes and references on prescribed mean
curvature  equations, we refer readers to \cite{Collin},
\cite{Finn1}, \cite{Finn}, \cite{GT}, \cite{Lopez2},
\cite{Lopez4}, \cite{Wang}.

We will consider only those domains that are inside some special
regions. The first kind of regions is the compliment of a cone in
${\bf R}^n$ ($n\geq 2$) defined by (we use the notation
${\bf{x}}^{*}=(x_{1}, x_{2}, \cdot \cdot \cdot , x_{n-1})$)
$$
P(n) = \{  {\bf{x}} \in {\bf R}^n :   |x_{n}|<
\frac{1}{240n}|{\bf{x}}^{*}| \}.
$$
The second kind of regions is a parabola-shaped region defined by
$$
P(n, \alpha ,b) = \{ {\bf{x}} \in {\bf R}^n :   |x_{n}|<
b|{\bf{x}}^{*}|^{\alpha }   \}.
$$
for some fixed positive constants $\alpha $, $b$, $0<\alpha <1$.

For a general domain $\Omega$ inside $P(n)$, we can estimate the growth rate of
a solution.


\begin{theorem} \label{theorem:second}
Let $\Omega$ be a domain inside $P(n)$,  $|\Lambda ({\bf{x}})|$ satisfy
\begin{equation}
|\Lambda ({\bf{x}})| \leq \frac{15(n-1)}{14(n+1)} \frac{1}{|{\bf{x}}^{*}|}
\quad\text{on } \Omega ,
\label{eq:boundoflambda}
\end{equation}
then any $C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$
solution $u$ to \eqref{eq:problem11}-\eqref{eq:problem12}
satisfies that on $\Omega$,
\begin{equation}
|u({\bf{x}})|\leq \frac{1}{240n}|{\bf{x}}^{*}|+ \sup \{ |\phi
({\bf{p}},q)| : ({\bf{p}},q)\in \partial \Omega,  \frac{1}{2}|{\bf{x}}^{*}|\leq |{\bf{p}}|\leq 2 |{\bf{x}}^{*}| \} .
\label{eq:boundone}
\end{equation}
\end{theorem}

When $\Omega $ satisfies more geometric conditions, the existence
of  solutions to \eqref{eq:problem11}-\eqref{eq:problem12} can be
proved. First we list a set of conditions that will guarantee a
solution to the minimal surface equation with the same boundary
data on the same domain:
\begin{itemize}

\item[(A1)] There is a sequence of subdomains $\Omega_{j}$ such
that $\Omega_{j}\subset \Omega_{j+1} \subset \Omega $ for  all
$j\geq 1$, $\cup \Omega_{j}=\Omega$;

\item[(A2)] Each $\Omega_{j}$ is a $C^{2, \gamma }$ bounded
domain and has positive mean curvature on $\partial \Omega_{j}$
with respect to  the inner normal on $\partial \Omega_{j}$;

\item[(A3)] $dist ({\bf{0}},  \Omega \setminus \Omega_{j}) \to
\infty $  as $j\to \infty$.
\end{itemize}

The next condition on $\Omega$  will be used to prove the solution obtained by
Perron's method takes boundary
data $\phi $ continuously.

\noindent{\it{Serrin's condition}}: The mean curvature function
$H'$ on $\partial \Omega$ with respect to the inner normal
satisfies
\begin{equation}
H'> \frac{n}{n-1} |\Lambda ({\bf{x}})| \quad on\quad \partial \Omega .
\label{eq:meancurvature}
\end{equation}

\begin{remark} \label{rmk1.2} \rm
Conditions (A1)-(A3) and Serrin's condition
(\ref{eq:meancurvature}) are the same as those used in
\cite{Jin3}.
\end{remark}

Here is the first existence result.

\begin{theorem} \label{theorem:first}
Assume {\rm (A1)--(A3)}, Serrin's condition \eqref{eq:meancurvature} and
$\Omega$ is inside $P(n)$. If $\Lambda ({\bf{x}})$ satisfies
\eqref{eq:boundoflambda}, then
\eqref{eq:problem11}--\eqref{eq:problem12} has a solution
$u\in C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$.
\end{theorem}

When the domains are inside $P(n, \alpha ,b)$, we assume $\Omega$
is not very close to the origin:
\begin{equation}
|{\bf{x}}^{*}|\geq (120nb(\frac{3}{2})^{\alpha })^{\frac{1}{1-\alpha}} \quad {\text{for}} \ \  {\text{any}} \quad
{\bf{x}} \in \Omega .
\label{eq:boundaway0}
\end{equation}

\begin{remark} \label{rmk1.4} \rm
Condition \eqref{eq:boundaway0} is not absolutely necessary,
we use it here so that we can state
results more clearly. Without \eqref{eq:boundaway0}, the following
results are still true as long as
$\Lambda ({\bf{x}})$ is bounded appropriately where \eqref{eq:boundaway0}
does not hold.
\end{remark}

The growth estimate now is as follows.

\begin{theorem} \label{theorem:fourth}
Let $\Omega$ be a domain inside $P(n, \alpha ,b)$.
If $\Omega $ satisfies \eqref{eq:boundaway0} and
\begin{equation}
|\Lambda ({\bf{x}})| \leq \frac{(n-1)}{56n(n+1)}(\frac{1}{3})^{\alpha } \frac{1}{b|{\bf{x}}^{*}|^{\alpha }}
\quad\text{on } \Omega ,
\label{eq:boundoflambda20}
\end{equation}
then any $C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$
solution $u$ to \eqref{eq:problem11}-\eqref{eq:problem12}
satisfies that on $\Omega$
\begin{equation}
|u({\bf{x}})|\leq \frac{1}{2} (\frac{3}{2})^{\alpha}
|{\bf{x}}^{*}|^{\alpha} + \sup \{ |\phi ({\bf{p}},q)| :
({\bf{p}},q)\in \partial \Omega,  \frac{1}{2}|{\bf{x}}^{*}|\leq|{\bf{p}}|\leq 2 |{\bf{x}}^{*}| \}  .
\label{eq:boundtwo}
\end{equation}
\end{theorem}

Here is the existence results for domains in $P(n, \alpha, b)$.

\begin{theorem} \label{theorem:third}
Assume {\rm (A1)--(A3)}, Serrin's condition \eqref{eq:meancurvature} and $\Omega$ is
inside $P(n, \alpha ,b)$
satisfying \eqref{eq:boundaway0}. Then if $|\Lambda ({\bf{x}})|$
satisfies \eqref{eq:boundoflambda20},
\eqref{eq:problem11}-\eqref{eq:problem12} has a solution $u\in
C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$.
\end{theorem}


\section{A family of auxiliary functions and growth estimates}

In this section, we construct a family of auxiliary functions and
derive growth estimates for solutions of
\eqref{eq:problem11}-\eqref{eq:problem12}. The construction is
adapted from that in \cite{JL2} and \cite{JinKirk}  to fit our
needs here (in turn, the constructions in \cite{JL2} and
\cite{JinKirk} were inspired by \cite{Finn} and \cite{Serrin}).
Set
\begin{equation}
Q z \equiv \frac{((1+|Dz|^{2}) \delta_{ij} - D_{i}z D_{j}z )}{n+(n-1)|Dz|^{2}}D_{ij} z
\end{equation}
We first prove the existence of a family auxiliary functions that
will suit our needs
later.

\begin{lemma} \label{lemma:first}
For any numbers $M>0$, $H\geq 2$, and any point ${\bf{x}}^{*}_{0}\in R^{n-1}$,
there are positive decreasing functions  $\chi (t)$ (depending on $n$ only),
$h_{a}(t)$ (with the inverse $h_{a}^{-1}$) and a positive increasing function $A(t)$
(depending on $n$, $H$ and $M$ only) such that for any constant $\gamma$,
the function
\begin{equation}
z=z({\bf{x}})=\gamma +A(H)e^{\chi (H)} -\{ (h_{a}^{-1}(x_{n}+M))^2-|{\bf x}^{*}-{\bf x}^{*}_{0}|^2 \}^{1/2}
\label{eq:barrier}
\end{equation}
satisfies
\begin{equation}
Qz \leq
-\frac{n-1}{28(n+1) MH} \cdot \frac{(1+|Dz|^{2})^{3/2}}{n+(n-1)|Dz|^{2}}
\quad  in  \quad \Omega_{{\bf x}^{*}_{0},H,M}
\label{eq:estimates}
\end{equation}
where
\begin{equation}
\Omega_{{\bf x}^{*}_{0},H,M}
=\{{\bf x}:|x_{n}|< M,|{\bf x}^{*}-{\bf x}^{*}_{0}|< h_{a}^{-1}(x_{n}+M) \} .
\label{eq:domain}
\end{equation}
Furthermore
\begin{equation}
z({\bf{x}}^{*}_{0},x_{n})\leq \gamma +\frac{M}{H} \quad for \quad -M \leq x_{n} \leq M.
\label{eq:boundofz}
\end{equation}
\end{lemma}


\begin{proof}
 Set $E=\frac{1}{n-1}$, $G=\frac{1}{2n-1}$,
$c_{2}=\frac{2+E}{G}=4n+\frac{1}{n-1}$, and
$\Phi_{1}(\rho)=\rho^{-2}$  if $0<\rho<1$, $\Phi_{1}(\rho)=c_{2}$
if $\rho \geq 1$. We define a function $\chi$  by
$$
\chi(\alpha)=\int_{\alpha}^{\infty}
\frac{d\rho}{\rho^{3}\Phi_{1}(\rho)} \quad  \mbox{for } \alpha>0.
$$
It is clear that $\chi (\alpha )$ is a decreasing function with
range $(0,\infty).$  Let $\eta$  be the inverse of $\chi.$
Then $\eta$  is a positive, decreasing function with range $(0,\infty)$.

For $\alpha >1$,  we have
\begin{equation}
\chi(\alpha)=\int_{\alpha}^{\infty}\frac{d\rho}{\rho^{3}\Phi_{1}(\rho)}
=\int_{\alpha}^{\infty}\frac{d\rho}{c_{2}\rho^{3}}
= \frac{1}{2c_{2}} \alpha^{-2} <1.
\label{eq:chi2}
\end{equation}
Thus
\begin{equation}
\eta (\beta ) = (2c_{2}\beta)^{-1/2}\quad \text{for } 0<\beta <(2c_{2})^{-1}.
\label{eq:beta1}
\end{equation}
For $H\geq 2$, since $\eta (\chi (H)) =H$ and $\eta $ is decreasing, we have
$\eta(\beta)> H$ and $\eta (\beta )= (2c_{2}\beta)^{-1/2}$ for $0<\beta< \chi(H)$.
We define a function $A(H)=A(H,M)$ by
\begin{equation}
A(H) = 2M (\int_{1}^{e^{\chi(H)}} \eta (\ln t)  dt)^{-1} .
\label{eq:ah}
\end{equation}
 For the rest of this article, we set $a=A(H)$ and define
\begin{equation}
h_{a}(r)=\int_{r}^{ae^{\chi(H)}}
\eta (\ln \frac{t}{a} ) dt \quad\text{for }
a\le r\le ae^{\chi(H)}.
\end{equation}
Then
\begin{equation}
h_{a}(ae^{\chi (H)})=0, \quad h_{a}(a) =h_{A(H)}(A(H))= 2M.
\label{eq:chi3}
\end{equation}
For $a<r\le ae^{\chi(H)}$,
\begin{equation}
h_{a}'(r)=-\eta(\ln \frac{r}{a} )<0,\quad
|h_{a}'(r)|>H, \quad
h_{a}''(r)=\frac{1}{r}(\eta(\ln \frac{r}{a} ))^{3}\Phi_{1}
(\eta(\ln \frac{r}{a} )).
\label{eq:derivativelarge}
\end{equation}
Thus for $a<r\le ae^{\chi(H)}$,
\begin{equation}
\frac{h_{a}''(r)}{(h_{a}'(r))^2}=-\frac{h_{a}'(r)}{r}\Phi_{1}(-h_{a}'(r)) .
\label{eq:hequ}
\end{equation}
Let $h_{a}^{-1}$ be the inverse of $h_{a}.$  Then $h_{a}^{-1}$ is decreasing and
\begin{equation}
h_{a}^{-1}(0)=A(H)e^{\chi (H)}, \quad
h_{a}^{-1}(2M) = A(H) .
\label{eq:valueofinverseh}
\end{equation}
Further for $-M\leq x_{n}\leq M$,
$$
(h_{a}^{-1})'(x_{n}+M) =\frac{1}{h'_{a}(h_{a}^{-1}(x_{n}+M))},
$$
\begin{align*}
(h_{a}^{-1})''(x_{n}+M)& =(\frac{1}{h'_{a}(h_{a}^{-1}(x_{n}+M))})'\\
&= - \frac{h''_{a}(h_{a}^{-1}(x_{n}+M))(h_{a}^{-1})'(y+M)}
{ (h'_{a}(h_{a}^{-1}(x_{n}+M)))^{2}} \\
&= - \frac{h''_{a}(h_{a}^{-1}(x_{n}+M))}{ (h'_{a}(h_{a}^{-1}(x_{n}+M)))^{3}}\\
&= \frac{1}{h_{a}^{-1}(x_{n}+M)} \Phi_{1} (-h'_{a}(h_{a}^{-1}(x_{n}+M))) .
\end{align*}
Thus for $-M <x_{n}<M$,
\begin{equation}
(h_{a}^{-1})''(y+M)h_{a}^{-1}(x_{n}+M)
=\Phi_{1} (-h'_{a}(h_{a}^{-1}(x_{n}+M))) .
\label{eq:equationofinverse}
\end{equation}
For $H\geq 2$, by (\ref{eq:chi2}), (\ref{eq:beta1}), we have
\begin{align*}
A(H)^{-1} &= (2M)^{-1} \int_{1}^{e^{\chi(H)}} \eta (\ln t) \, dt \\
&=(2M)^{-1} \int_{0}^{\chi(H)} \eta (m)e^{m} \, dm \\
&=(2M)^{-1} \int_{0}^{\chi(H)}\frac{e^{m}}{\sqrt{2c_{2}m}} \,  dm.
\end{align*}
 From
$$
\int_{0}^{\chi(H)}\frac{1}{\sqrt{2c_{2}m}}   dm \leq
\int_{0}^{\chi(H)}\frac{e^{m}}{\sqrt{2c_{2}m}}  dm
\leq
\frac{e^{\chi (H)}}{\sqrt{2c_{2}}}
\int_{0}^{\chi(H)} m^{-1/2}  dm,
$$
we have
$$
\frac{1}{c_{2}H}=
\frac{2\sqrt{\chi (H)}}{\sqrt{2c_{2}} } \leq
\int_{0}^{\chi(H)}\frac{e^{m}}{\sqrt{2c_{2}m}}   dm
\leq
\frac{2e^{\chi (H)}\sqrt{\chi (H)}}{\sqrt{2c_{2}}}
=\frac{e^{\frac{1}{2c_{2}H^{2}}}}{c_{2} H} .
$$
Thus
\begin{equation}
2Mc_{2}H\geq A(H)\geq 2Mc_{2}He^{-\chi (H)}=
2Mc_{2}He^{-\frac{1}{2c_{2}H^{2}}} .
\label{eq:ah2}
\end{equation}

For ${\bf x}_{0}\in {\bf R}^{n-1}$, $H\geq 2$ and $M>0$,
we define a domain $\Omega_{{\bf x}^{*}_{0},H,M}$
in ${\bf{x}}$ space by (\ref{eq:domain}) and
define a function $z=z_{{\bf x}^{*}_{0},H,M}({\bf{x}})$
by (\ref{eq:barrier}).  It is clear that the function
$z$
is well defined on $\Omega_{{\bf x}^{*}_{0},H,M}$.
Let
\begin{equation}
S=((h_{a}^{-1}(x_{n}+M))^2-|{\bf x}^{*}-{\bf x}_{0}^{*}|^2 )^{1/2}.
\label{eq:defofs}
\end{equation}
then for $1\le i\le n-1 $,  we have
\begin{equation}
\frac{\partial z}{\partial x_i}=
\frac{1}{S}(x_i-x_{0i}),\quad
\frac{\partial z}{\partial x_{n}}=
- \frac{1}{S}h_{a}^{-1} (h_{a}^{-1})'.
\label{eq:derivative}
\end{equation}
Since
$h_{a}^{-1}(r)$ and $\eta $ are decreasing functions,
for $H\geq 2$, $|y|\leq M$, we have
\begin{equation}
\begin{aligned}
0&< -(h_{a}^{-1})' =
\frac{-1}{h_{a}'(h_{a}^{-1}(x_{n}+M))}
=\frac{1}{ \eta(\ln(\frac{1}{a}h_{a}^{-1}(x_{n}+M))) } \\
&\leq
\frac{1}{\eta (\ln e^{\chi (H)})}
=\frac{1}{\eta(\chi(H))}= \frac{1}{H}
\quad {\rm{for}}  |x_{n}|\le -M.
\end{aligned} \label{eq:boundofeta}
\end{equation}
Then (\ref{eq:boundofz}) follows from the facts that
$z({\bf{x}}^{*}_{0},-M)=\gamma +A(H)e^{\chi (H)} -h_{a}^{-1}(0) =\gamma$
and
$$
\frac{\partial z}{\partial x_{n}} ({\bf{x}}^{*}_{0},x_{n})=
- \frac{1}{S}h_{a}^{-1} (h_{a}^{-1})'=- (h_{a}^{-1})'\leq \frac{1}{H}.
$$
Now if $|{\bf x}^{*}-{\bf x}^{*}_{0}|\geq \frac{1}{2}h_{a}^{-1}(x_{n}+M))$
and $H\geq 2$,
\begin{equation}
(\frac{\partial z}{\partial x_{n}})^{2}
=\frac{1}{S^{2}} (h_{a}^{-1})^{2} ((h_{a}^{-1})')^{2}
\leq \frac{1}{S^{2}} (\frac{1}{2}h_{a}^{-1})^{2} \frac{4}{H^{2}}
\leq \sum_{i=1}^{n-1} (\frac{\partial z}{\partial x_{i}})^{2} .
\label{eq:derivative2}
\end{equation}
If $|{\bf x}^{*}-{\bf x}^{*}_{0}|\leq \frac{1}{2}h_{a}^{-1}(x_{n}+M))$
and $H\geq 2$, then
$$
S^{2}=(h_{a}^{-1}(x_{n}+M)))^{2} -|{\bf x}^{*}-{\bf x}^{*}_{0}|^{2} \geq \frac{3}{4}(h_{a}^{-1}(x_{n}+M)))^{2},
$$
and
\begin{equation}
(\frac{\partial z}{\partial x_{n}})^{2}
=\frac{1}{S^{2}} (h_{a}^{-1})^{2} ((h_{a}^{-1})')^{2}
\leq \frac{1}{S^{2}} (h_{a}^{-1})^{2} \frac{1}{H^{2}}
\leq \frac{4}{3H^{2}}\leq 1.
\label{eq:derivative3}
\end{equation}
Therefore,
$$
(\frac{\partial z}{\partial x_{n}})^{2}\leq \sum_{i=1}^{n-1} (\frac{\partial z}{\partial x_{i}})^{2} +1.
$$
We set the notation
$$
a_{ij} =\frac{(1+|p|^{2})\delta_{ij} -p_{i}p_{j}}{n+(n-1)|p|^{2}},
  p_{i}=\frac{\partial z}{\partial x_{i}}
 \quad 1\leq i,  j \leq n.
$$
Then  $|p_{n}|^{2}\leq \sum_{i=1}^{n-1}p_{i}^{2} +1$ and
\begin{equation}
a_{nn}=\frac{1+\sum_{i=1}^{n-1}p_{i}^{2}}{n+(n-1)|p|^{2}}
\geq \frac{1+\sum_{i=1}^{i=n-1}p_{i}^{2}}{2n-1+2(n-1)\sum_{i=1}^{i=n-1}p_{i}^{2}}
\geq \frac{1}{2n-1}=G
\label{eq:derivative4}
\end{equation}
and
\begin{equation}
\sum_{i,j=1}^{n} a_{ij} \frac{\partial z}{\partial x_{i}} \frac{\partial z}{\partial x_{j}}
=\frac{|p|^{2}}{n+(n-1)|p|^{2}} \leq \frac{1}{n-1} =E.
\label{eq:derivative5}
\end{equation}
Thus on
$\Omega_{{\bf{x}}^{*}_{0}, H,M}$, we have
\begin{align*}
Qz&=\sum_{i,j=1}^{n} a_{ij}D_{ij}z \\
&= \frac{1}{S}\sum_{i=1}^{n-1} a_{ii}
+\frac{1}{S^{3}}\sum_{i,j=1}^{n-1} a_{ij}(x_{i}-x_{i}^{0})(x_{j}-x_{j}^{0})
-\frac{1}{S^{3}}\sum_{i=1}^{n-1} a_{in}(x_{i}-x_{i}^{0})h_{a}^{-1}
(h_{a}^{-1})' \\
&\quad - \frac{1}{S}a_{nn}((h_{a}^{-1})^{2} + h_{a}^{-1} (h_{a}^{-1})'')
+ \frac{1}{S^{3}}a_{nn}(h_{a}^{-1})^{2} ((h_{a}^{-1})')^{2}\\
&= \frac{1}{S} \big\{
1-a_{nn}+\sum_{i,j=1}^{n} a_{ij} \frac{\partial z}{\partial x_{i}}
\frac{\partial z}{\partial x_{j}}
-a_{nn} ((h_{a}^{-1})^{2} + h_{a}^{-1} (h_{a}^{-1})'') \big\} \\
& \leq \frac{1}{S}\big\{
1+\sum_{i,j=1}^{n} a_{ij} \frac{\partial z}{\partial x_{i}}
\frac{\partial z}{\partial x_{j}}
-a_{nn} h_{a}^{-1} (h_{a}^{-1})'' \big\} \\
&\leq \frac{1}{S}\{ 1+E - Gh_{a}^{-1} (h_{a}^{-1})'' \}=
\frac{-1}{S}
\end{align*}
(by (\ref{eq:equationofinverse})), (\ref{eq:derivative4}),
(\ref{eq:derivative5}) and the definition of $\Phi $).
Then (\ref{eq:estimates}) follows from the following inequality
\begin{equation}
\frac{n-1}{28(n+1)MH} \cdot \frac{(1+|Dz|^{2})^{3/2}}{n+(n-1)|Dz|^{2}}
\leq
\frac{1}{S}.
\label{eq:lastone}
\end{equation}
To prove (\ref{eq:lastone}), since $\chi (H)<1$ for $H\geq 2$,
we have
\begin{align*}
&\frac{(1+|Dz|^{2})^{3/2}}{n+(n-1)|Dz|^{2}}\\
&\leq \frac{1}{n-1}(1+|Dz|^{2})^{1/2} \\
&=\frac{1}{n-1} (1+\frac{1}{S^{2}} (|{\bf{x}}^{*}
 -{\bf{x}}_{0}^{*}|^{2} +
(h_{a}^{-1})^{2} ((h_{a}^{-1})')^{2}))^{1/2} \quad
\text{by  \eqref{eq:derivative}} \\
&= \frac{1}{(n-1)S} (S^{2}+|{\bf{x}}^{*} -{\bf{x}}^{*}_{0}|^{2} +
(h_{a}^{-1})^{2} ((h_{a}^{-1})')^{2})^{1/2} \\
&=\frac{1}{(n-1)S} ((h_{a}^{-1})^{2} +
(h_{a}^{-1})^{2} ((h_{a}^{-1})')^{2})^{1/2} \quad
\text{by  \eqref{eq:defofs}}\\
&=\frac{1}{(n-1)S} (h_{a}^{-1})(1+((h_{a}^{-1})')^{2})^{1/2} \\
&\leq \frac{1}{(n-1)S} (h_{a}^{-1})(1+\frac{1}{H^{2}})^{1/2}
\leq \frac{1}{(n-1)S} A(H)e^{\chi (H)}(1+\frac{1}{4})^{1/2} \quad
\text{by   \eqref{eq:valueofinverseh}, \eqref{eq:boundofeta}}\\
&\leq \frac{1}{(n-1)S} (\frac{5}{4})^{1/2} 2c_{2}e^{\chi (H)}MH
\leq \frac{1}{(n-1)S} c_{2}5^{1/2} e MH   \quad
\text{by  \eqref{eq:ah2}}\\
&=\frac{1}{(n-1)S} (4n+\frac{1}{n-1}) 5^{1/2} e MH
\leq \frac{28(n+1)}{n-1} MH \frac{1}{S}  \quad
\text{by    the   definition   of   $c_{2}$}.
\end{align*}
\end{proof}

\begin{lemma} \label{lemma:second}
Let $\phi $ be a continuous function defined on $\partial \Omega$.
For any ${\bf{x}}^{*}_{0} \in R^{n-1}$, we set
\begin{equation}
\gamma = \gamma({\bf{x}}^{*}_{0}) =\sup \{ |\phi ({\bf{x}})| :
{\bf{x}}\in \partial \Omega, \quad
\frac{1}{2}|{\bf{x}}^{*}_{0}|\leq |{\bf{x}}^{*}|\leq
\frac{3}{2}|{\bf{x}}^{*}_{0}| \} . \label{eq:definitionofgamma}
\end{equation}
For any ${\bf{x}}^{*}_{0} \in R^{n-1}$ such that $({\bf{x}}^{*}_{0},x_{n})\in \Omega$ for some $x_{n}$,
in the function $z=z_{{\bf{x}}^{*}_{0}} $ defined in
(\ref{eq:barrier}),
we set
\begin{equation}
\gamma=\gamma ({\bf{x}}^{*}_{0}), \quad
H=2, \quad
M=\frac{1}{120n}|{\bf{x}}^{*}_{0}|.
\label{eq:setofmh}
\end{equation}
Then $z=z_{{\bf{x}}^{*}_{0}} $ satisfies
\begin{equation}
Qz\leq -n \Lambda_{0} ({\bf{x}})
\frac{(1+|Dz|^{2})^{3/2}}{n+(n-1)|Dz|^{2}}
\quad  in  \quad \Omega_{{\bf x}_{0},H,M} \cap P(n)
\label{eq:supersolution}
\end{equation}
where
\begin{equation}
\Lambda_{0} ({\bf{x}}) = \frac{15(n-1)}{14(n+1)|{\bf{x}}^{*}|} .
\label{eq:boundoflambda0}
\end{equation}
Furthermore
\begin{itemize}
\item[(i)] $z({\bf{x}}^{*}_{0},x_{n})\leq \frac{1}{240n} |{\bf{x}}^{*}_{0}| +
\gamma({\bf{x}}^{*}_{0})$ for $|x_{n}|<M$;

\item[(ii)] $\partial \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n) \subset \{
{\bf{x}} : |x_{n}|< M \}$;

\item[(iii)] For the unit outer normal ${\bf{n}}$ on
$\partial \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n)$,
$$
\frac{\partial z}{\partial {\bf{n}}} =+\infty \quad\text{on }
\partial \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n) .
$$
\end{itemize}
\end{lemma}

\begin{proof}
  From the choices of $H$ and $M$ and the definition of
$\Omega_{{\bf{x}}^{*}_{0}, H, M}$, we have
that for any ${\bf{x}}\in \Omega_{{\bf{x}}^{*}_{0}, H, M}$,
 by (\ref{eq:ah2}),
$$
|{\bf{x}}^{*} -{\bf{x}}^{*}_{0}|\leq h_{a}^{-1}(x_{n}+M)\leq A(H)e^{\chi (H)}
\leq 2c_{2}eHM\leq 60n M=\frac{1}{2}|{\bf{x}}^{*}_{0}|.
$$
Thus for any ${\bf{x}}\in \Omega_{{\bf{x}}^{*}_{0}, H, M}$,
$\frac{1}{2} |{\bf{x}}^{*}_{0}|\leq |{\bf{x}}^{*}|
\leq \frac{3}{2} |{\bf{x}}^{*}_{0}|$.
Again from the choices of $M$ and $H$, we have
$$
\frac{n-1}{28(n+1)MH}=
\frac{n-1}{56(n+1)M}=
\frac{15n(n-1)}{7(n+1)|{\bf{x}}^{*}_{0}|}.
$$
Then if we set
$$
\Lambda_{0} ({\bf{x}}) = \frac{15(n-1)}{14(n+1)|{\bf{x}}^{*}|},
$$
we have that on $\Omega_{{\bf{x}}^{*}_{0}, H, M}$
$$
n\Lambda_{0} ({\bf{x}}) =\frac{15n(n-1)}{14(n+1)|{\bf{x}}^{*}|}
\leq \frac{15n(n-1)}{7(n+1)|{\bf{x}}^{*}_{0}|}
= \frac{n-1}{28(n+1)MH}.
$$
Now (\ref{eq:supersolution}) follows from Lemma \ref{lemma:first}.
\begin{itemize}
\item[(i)] is clear from (\ref{eq:boundofz}) and the definitions
of $M$ and $H$.

\item[(ii)] follows from $|{\bf{x}}^{*}|\leq \frac{3}{2} |{\bf{x}}^{*}_{0}|$
and $|x_{n}|\leq \frac{1}{240n}|{\bf{x}}^{*}|$
for all ${\bf{x}}\in \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n)$:
$$
|x_{n}|\leq \frac{1}{240n}|{\bf{x}}^{*}|\leq \frac{3}{480n}|{\bf{x}}^{*}_{0}|
<\frac{1}{120n}|{\bf{x}}^{*}_{0}|=M \quad for \quad
{\bf{x}}\in \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n) .
$$

\item[(iii)] is obvious since $|{\bf{x}}^{*}-{\bf{x}}^{*}_{0}|
=h_{a}^{-1}(x_{n}+M)$ on
 $\partial \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n)$.

\end{itemize}
\end{proof}

\begin{lemma} \label{lemma:third2}
 Let $\phi $ be a continuous function defined on $\partial \Omega$.
For any ${\bf{x}}^{*}_{0} \in R^{n-1}$ such that $|{\bf{x}}^{*}_{0}|\geq
(120bn(\frac{3}{2})^{\alpha })^{\frac{1}{1-\alpha }}$
and $({\bf{x}}^{*}_{0},x_{n})\in \Omega$ for some $x_{n}$,
in the function $z=z_{{\bf{x}}^{*}_{0}} $ defined in
(\ref{eq:barrier}),
we set ($\gamma ({\bf{x}}^{*}_{0})$ is defined in (\ref{eq:definitionofgamma}))
\begin{equation}
\gamma=\gamma ({\bf{x}}^{*}_{0}), \quad
H=2, \quad
M=(\frac{3}{2})^{\alpha} b|{\bf{x}}^{*}_{0}|^{\alpha }.
\label{eq:setofmh2}
\end{equation}
Then $z=z_{{\bf{x}}^{*}_{0}} $ satisfies
\begin{equation}
Qz\leq -n \Lambda_{1} ({\bf{x}})
\frac{(1+|Dz|^{2})^{3/2}}{n+(n-1)|Dz|^{2}}
\quad  in  \quad \Omega_{{\bf x}_{0},H,M} \cap P(n, \alpha ,b)
\label{eq:supersolution2}
\end{equation}
where
\begin{equation}
\Lambda_{1} ({\bf{x}}) = \frac{(n-1)}{56n(n+1)b}(\frac{2}{3})^{\alpha } \frac{1}{|{\bf{x}}^{*}_{0}|^{\alpha }}.
\label{eq:boundoflambda1}
\end{equation}
Furthermore
\begin{itemize}
\item[(i)] $z({\bf{x}}^{*}_{0},x_{n})\leq \frac{1}{2}(\frac{3}{2})^{\alpha} b|{\bf{x}}^{*}_{0}|^{\alpha } +
\gamma({\bf{x}}^{*}_{0})$ for $|x_{n}|<M$;

\item[(ii)] $\partial \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n, \alpha , b)
\subset \{ {\bf{x}} : |x_{n}|< M \}$;

\item[(iii)] For the unit outer normal ${\bf{n}}$ on $\partial \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n, \alpha , b)$,
$$
\frac{\partial z}{\partial {\bf{n}}} =+\infty \quad\text{on }
\partial \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n, \alpha ,b) .
$$
\end{itemize}
\end{lemma}

\begin{proof}
  From the choices of $H$ and $M$ and the definition of
$\Omega_{{\bf{x}}^{*}_{0}, H, M}$, we have
that for any ${\bf{x}}\in \Omega_{{\bf{x}}^{*}_{0}, H, M}$, by
(\ref{eq:ah2}),
$$
|{\bf{x}}^{*} -{\bf{x}}^{*}_{0}|\leq h_{a}^{-1}(x_{n}+M)
\leq A(H)e^{\chi (H)}
\leq 2c_{2}eHM\leq 60n M<60nb (\frac{3}{2})^{\alpha }
|{\bf{x}}^{*}_{0}|^{\alpha }.
$$
Thus when $|{\bf{x}}^{*}_{0}|
\geq (120nb (\frac{3}{2})^{\alpha })^{\frac{1}{1-\alpha }}$,
for any ${\bf{x}}\in \Omega_{{\bf{x}}^{*}_{0}, H, M}$,
$\frac{1}{2} |{\bf{x}}^{*}_{0}|\leq |{\bf{x}}^{*}|
\leq \frac{3}{2} |{\bf{x}}^{*}_{0}|$.
Again from the choices of $M$ and $H$, we have
$$
\frac{n-1}{28(n+1)MH}=
\frac{n-1}{56(n+1)M}=
\frac{(n-1)}{56(n+1)}(\frac{2}{3})^{\alpha }
\frac{1}{b|{\bf{x}}^{*}_{0}|^{\alpha }}.
$$
Then if we set
$$
\Lambda_{1} ({\bf{x}}) =
\frac{(n-1)}{56n(n+1)}(\frac{1}{3})^{\alpha } \frac{1}{b|{\bf{x}}^{*}|^{\alpha }},
$$
we have that on $\Omega_{{\bf{x}}^{*}_{0}, H, M}$
$$
n\Lambda_{1} ({\bf{x}})
\leq \frac{n-1}{28(n+1)MH}.
$$
Now (\ref{eq:supersolution2}) follows from Lemma \ref{lemma:first}.
(i) and (iii) are proved in the same way as in that of
Lemma \ref{lemma:second}.
(ii) follows from $|{\bf{x}}^{*}|< \frac{3}{2} |{\bf{x}}^{*}_{0}|$ and $|x_{n}|\leq b|{\bf{x}}^{*}|^{\alpha }$
for all ${\bf{x}}\in \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n, \alpha ,b)$:
$$
|x_{n}|\leq b|{\bf{x}}^{*}|^{\alpha }\leq (\frac{3}{2})^{\alpha } b |{\bf{x}}^{*}_{0}|^{\alpha } =M
  for   {\bf{x}}\in  \Omega_{{\bf{x}}^{*}_{0}, M,H}\cap P(n, \alpha , b) .
$$
\end{proof}

Now we are ready to prove growth estimates for solutions.

\begin{lemma} \label{lemma:growth1}
Let $\Omega$ be a domain inside $P(n)$
and  $\Lambda ({\bf{x}})$ satisfy \eqref{eq:boundoflambda},
then any $C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$
solution $v$ to \eqref{eq:problem11}-\eqref{eq:problem12}
satisfies
\begin{equation}
|v| \leq z_{{\bf{x}}^{*}_{0}}\quad\text{on } \Omega_{{\bf{x}}^{*}_{0},M,H} \cap \Omega
\label{eq:growth1}
\end{equation}
where $z_{{\bf{x}}^{*}_{0}}$ is defined in Lemma \ref{lemma:second}.
\end{lemma}

\begin{proof}
  Since $\Lambda_{0}$ defined in \eqref{eq:boundoflambda} is positive,
(\ref{eq:supersolution}) implies that
$z_{{\bf{x}}^{*}_{0}}$ is
also a supersolution to \eqref{eq:problem11} on
$\Omega_{{\bf{x}}^{*}_{0},M,H}\cap \Omega$. Furthermore,
the definition of $z_{{\bf{x}}^{*}_{0}}$ implies that
$z_{{\bf{x}}^{*}_{0}}\geq |\phi | $ on
$\Omega_{{\bf{x}}^{*}_{0},M,H}\cap \partial \Omega$. Now
$v- z_{{\bf{x}}^{*}_{0}}$ cannot achieve its maximum value in
$\overline{\Omega_{{\bf{x}}^{*}_{0},M,H}\cap  \Omega}$
on $\partial \Omega_{{\bf{x}}^{*}_{0},M,H} \cap \Omega$ since the
directional derivative of $z_{{\bf{x}}^{*}_{0}}$ with respect
to outer normal is $+\infty$ on
$\partial \Omega_{{\bf{x}}^{*}_{0},M,H}\cap \Omega $
by (iii) in Lemma \ref{lemma:second}.
A comparison argument also concludes that
$v- z_{{\bf{x}}^{*}_{0}}$ cannot achieve a local maximum inside
$\Omega_{{\bf{x}}^{*}_{0},M,H}\cap \Omega$. Thus
$v -z_{{\bf{x}}^{*}_{0}}$ achieves its maximum value
in
$\overline{\Omega_{{\bf{x}}^{*}_{0},M,H}\cap  \Omega }$
on $\Omega_{{\bf{x}}^{*}_{0},M,H}\cap \partial \Omega$.
Then
$z_{{\bf{x}}^{*}_{0}}\geq |\phi | $ on
$\Omega_{{\bf{x}}^{*}_{0},M,H}\cap \partial \Omega$.
implies $v- z_{{\bf{x}}^{*}_{0}}\leq 0$ on
$\Omega_{{\bf{x}}^{*}_{0},M,H}\cap  \Omega$.
Now we apply the same argument with
$z_{{\bf{x}}^{*}_{0}}$ and $-v$ ($-v$ satisfies \eqref{eq:problem11}
with $\Lambda $ replaced by $-\Lambda $ and boundary data $-\phi $),
we can conclude that $-v\leq z_{{\bf{x}}^{*}_{0}}$.
\end{proof}

\begin{lemma} \label{lemma:growth2}
Let $\Omega$ be a domain inside $P(n, \alpha, b)$ satisfying
\eqref{eq:boundaway0}
and  $\Lambda ({\bf{x}})$ satisfy \eqref{eq:boundoflambda20},
then any $C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$
solution $v$ to \eqref{eq:problem11}-\eqref{eq:problem12}
satisfies
\begin{equation}
|v| \leq z_{{\bf{x}}^{*}_{0}}\quad\text{on }
\Omega_{{\bf{x}}^{*}_{0},M,H} \cap \Omega \label{eq:growth2}
\end{equation}
where $z_{{\bf{x}}^{*}_{0}}$ is defined in Lemma \ref{lemma:third2}.
\end{lemma}

The proof of the above lemma is completely parallel to that of
Lemma \ref{lemma:growth1} with Lemma \ref{lemma:second} replaced by
\ref{lemma:third2}.


\begin{proof}[Proof of Theorem \ref{theorem:second}]
 From Lemma \ref{lemma:growth1}, for any
$({\bf{x}}^{*}_{0},x_{n})\in \Omega$,
$$
|v({\bf{x}})| \leq z_{{\bf{x}}^{*}_{0}}({\bf{x}})\quad\text{on }
 \Omega_{{\bf{x}}^{*}_{0},M,H} \cap \Omega.
$$
In particular,
$$
|v({\bf{x}}^{*}_{0},x_{n})| \leq z_{{\bf{x}}^{*}_{0}}
({\bf{x}}^{*}_{0},x_{n})\quad\text{on }
\Omega_{{\bf{x}}^{*}_{0},M,H} \cap \Omega .
$$
Then Theorem \ref{theorem:second} follows from (i) in
Lemma \ref{lemma:second}, the definition of $\gamma ({\bf{x}}^{*}_{0})$
and the fact that
$({\bf{x}}^{*}_{0},x_{n})$ can be an arbitrary point in $\Omega$.
\end{proof}


\begin{proof}[Proof of Theorem \ref{theorem:fourth}]
The proof is completely parallel to that of Theorem \ref{theorem:second}
with Lemma
\ref{lemma:growth1} replaced by Lemma \ref{lemma:growth2}.
\end{proof}

\section{Proofs of Theorems \ref{theorem:first} and \ref{theorem:third}}

We will give only the proof for Theorem \ref{theorem:first}.
The proof of Theorem \ref{theorem:third} is
completely parallel to that of Theorem \ref{theorem:first}
with Lemma \ref{lemma:second} replaced by
Lemma \ref{lemma:third2} and Lemma \ref{lemma:growth1} replaced
by Lemma \ref{lemma:growth2}.

Once we have proved the lemmas in the previous section, the
proof of Theorem \ref{theorem:first}  is very similar to the
\cite[Theorems 1.2 and 1.3]{Jin3} that in turn is similar
that of \cite[Theorem 1]{Lopez4}. For reader's convenience,
we still carry out the details here.

We define $\Pi$, a family of open subsets of $\Omega $, as follows:
If ${\bf{x}}_{1}\in \Omega$, we choose a small ball $O$ centered
${\bf{x}}_{1}$ such that $O\subset \Omega$ and the mean curvature
function $H'$ on $\partial O$ with respect to inner normal satisfies
\begin{equation}
H'\geq \frac{n-1}{n} |\Lambda ({\bf{x}})| .
\label{eq:meancurvatureofo}
\end{equation}
If ${\bf{x}}_{1}\in \partial \Omega$, we choose a domain $O$ such that
$O\subset \Omega$, $O$ has $C^{2, \mu}$ boundary and
$\partial O \cap\partial \Omega$
is a neighborhood of ${\bf{x}}_{1}$ in $\partial \Omega$.
Furthermore, the mean curvature function $H'$ on $\partial O$
with respect to inner normal satisfies (\ref{eq:meancurvatureofo}).
The existence of such domains can
be proved under the assumption that $\Omega$ satisfies Serrin's
condition (\ref{eq:meancurvature}), for details of a proof, one may
see \cite[Lemma A.3]{Jin3}.

Let $v>0$ be a continuous function on $\overline{\Omega}$,
for each open set $O\in \Pi$,
we define a new function $M_{O}(v)$, called the
lifting of $v$ over $O$ as follows:
$$
M_{O}(v)({\bf{x}})=v({\bf{x}})  \quad\text{if }   {\bf{x}}\in
\Omega\setminus O, \quad
M_{O}(v)({\bf{x}},y)=w({\bf{x}})  \quad\text{if }   {\bf{x}}\in O
$$
where $w({\bf{x}})$ is the solution of the boundary-value problem
\begin{gather}
((1+|Dw|^{2}) \delta_{ij} - D_{i}w D_{j}w )D_{ij} w =
n \Lambda ({\bf{x}}) (1+|Dw|^{2})^{3/2}
 \quad\text{in } O, \label{eq:lift1} \\
w=v  \quad\text{on }   \partial O \,. \label{eq:lift2}
\end{gather}

\begin{remark} \rm
By (\ref{eq:meancurvatureofo}), Lemma \ref{lemma:growth1} and
\cite{Serrin} or \cite[Theorem 16.9]{GT}, there is a unique solution
$w\in C^{2}(O)\cap C^{0}(\overline{\Omega})$ to
(\ref{eq:lift1})-(\ref{eq:lift2}). Thus $M_{O}(v)$ is well defined.
\end{remark}

We define a class $\Xi $ of functions $v$, called subfunctions,
such that:
\begin{enumerate}
\item  $v\in C^{0}(\overline{\Omega })$ and $v\leq \phi $
on $\partial \Omega$;

\item For any $O \in \Pi $, $v\leq M_{O}(v)$;

\item $v\leq z_{{\bf{x}}^{*}_{0}} $ on
$\Omega \cap \Omega_{{\bf{x}}^{*}_{0}, M, H} $ for
any ${\bf{x}}^{*}_{0} \in R^{n-1}$ such that
$({\bf{x}}^{*}_{0},x_{n})\in \Omega$ for some
$x_{n}$, where $z_{{\bf{x}}^{*}_{0}}$ are those functions
defined in Lemma \ref{lemma:second}.

\end{enumerate}

Now we prove some properties for subfunctions in the class $\Xi$.

\begin{lemma}\label{lemma:fourth}
 If $v_{1}\leq v_{2}$, then $M_{O}(v_{1})\leq M_{O}(v_{2})$
for any $O\in \Pi $.
\end{lemma}

\begin{proof} Let $w_{1}$, $w_{2}$ be the
solutions of the following two problems, respectively:
\begin{gather*}
((1+|Dw_{k}|^{2}) \delta_{ij} - D_{i}w_{k} D_{j}w_{k} )
D_{ij} w_{k} =
n \Lambda ({\bf{x}}) (1+|Dw_{k}|^{2})^{3/2}
   {\rm{in}}   O, \\
w_{k}=v_{k} \quad \text{on }   \partial O, \quad k=1,2.
\end{gather*}
Since $w_{1}=v_{1}\leq v_{2}=w_{2}$ on $\partial O$,
by a comparison principle for quasilinear
elliptic equations (e.g. see \cite[Theorem 10.1]{GT}), we have
$w_{1}\leq w_{2}$ on $O$. On $\Omega \setminus  O$,
$M_{O}(v_{1})=v_{1}$, $M_{O}(v_{2})=v_{2}$.
Thus $M_{O}(v_{1})\leq M_{O}(v_{2})$.
\end{proof}

\begin{lemma} \label{lemma:fifth}
 If $v_{1}\in \Xi$, $v_{2}\in \Xi$, then $\max \{ v_{1}, v_{2} \} \in \Xi$.
\end{lemma}

\begin{proof} If $v_{1}\in \Xi$, $v_{2}\in \Xi$, then
$\max \{ v_{1}, v_{2} \}\in C^{0}(\overline{\Omega })$,
 and $\max \{ v_{1}, v_{2} \}\leq \phi $ on $\partial \Omega$.
It is also clear that
$\max \{ v_{1}, v_{2} \}\leq z_{{\bf{x}}^{*}_{0}}$ on
$\Omega_{{\bf{x}}^{*}_{0}, M, H } \cap \Omega$.

Since $v_{1}\leq \max \{ v_{1}, v_{2} \}$,
$v_{2} \leq \max \{ v_{1}, v_{2}\}$,
we have (by Lemma \ref{lemma:fourth}) that for any $O\in \Pi $,
$$
M_{O}(v_{1}) \leq
M_{O}(\max \{ v_{1}, v_{2}\} ),\quad
M_{O}(v_{2}) \leq
M_{O}(\max \{ v_{1}, v_{2}\} ).
$$
Since $v_{1}\in \Xi $ and $v_{2}\in \Xi $ imply
$v_{1} \leq M_{O}(v_{1})$, $v_{2} \leq M_{O}(v_{2})$,
we have
$\max \{ v_{1}, v_{2} \}\leq M_{O}(\max \{ v_{1}, v_{2}\})$.
Thus $\max \{ v_{1}, v_{2} \}\in \Xi$.
\end{proof}

\begin{lemma} \label{lemma:sixth}
If $v\in \Xi $, then $M_{O}(v)\in \Xi$
for any $O\in \Pi$.
\end{lemma}

\begin{proof}  By the definition of $M_{O}( v)$, it is clear that
$M_{O}( v)\in C^{0}(\overline{\Omega} )$ and
$M_{O}(v) \leq \phi $ on $\partial \Omega $.
First we show that for any $O_{1}\in \Pi$,
\begin{equation}
M_{O}( v)({\bf{x}})\leq M_{O_{1}}(M_{O}(v))({\bf{x}}).
\label{eq:mv}
\end{equation}
We  need to prove only that (\ref{eq:mv}) is true for
${\bf{x}}\in O_{1}$.
Since $v\leq M_{O}( v)$ on $\Omega$,
we have (by Lemma \ref{lemma:fourth})
$M_{O_{1}}( v)\leq M_{O_{1}}(M_{O}( v))$.
Combining this with $v\leq M_{O_{1}}( v)$,
we have $v\leq M_{O_{1}}(M_{O}(v))$.
Thus for ${\bf{x}}\in
O_{1} \setminus O$,
\begin{equation}
M_{O}( v)({\bf{x}})=v({\bf{x}})
\leq M_{O_{1}}(M_{O}(v))({\bf{x}}).
\label{eq:boundary1}
\end{equation}
That is, (\ref{eq:mv}) is true on $O_{1} \setminus O$,
Now for $\Omega_{1}=O_{1} \cap O$,
if we set
$$
M_{O}( v)=w_{1}, \quad
M_{O_{1}}(M_{O}(v))=w_{2},
$$
we have  that on $\Omega_{1}$, $k=1,2$,
$$
((1+|Dw_{k}|^{2}) \delta_{ij} - D_{i}w_{k} D_{j}w_{k} )D_{ij} w_{k} =
n \Lambda ({\bf{x}}) (1+|Dw_{k}|^{2})^{3/2} .
$$
On $\partial \Omega_{1}$,
$w_{1}\leq w_{2}$ on $O_{1}\cap \partial O$
by (\ref{eq:boundary1}) and
$w_{1}\leq w_{2}$ on $\partial O_{1}\cap O$  since
(\ref{eq:mv}) is true on $\Omega\setminus O_{1}$.
Then a comparison argument implies $w_{1}\leq w_{2}$ on $\Omega_{1}$.
Thus (\ref{eq:mv}) is true on $O_{1} \cap O$ and on
$O_{1}$.
\end{proof}


Now we prove that $M_{O}(v)\leq z_{{\bf{x}}^{*}_{0}}$ on
$\Omega_{{\bf{x}}^{*}_{0},M,H}\cap \Omega $.
Since $v\in \Xi$, $v\leq z_{{\bf{x}}^{*}_{0}}$ on
$\Omega_{{\bf{x}}^{*}_{0},M,H}\cap \Omega $.
Thus by the definition of $M_{O}(v)$, we only need to show
$M_{O}(v)\leq z_{{\bf{x}}^{*}_{0}}$ on
$\Omega_{{\bf{x}}^{*}_{0},M,H}\cap O$.
If $O$ does not intersect with
$\Omega_{{\bf{x}}^{*}_{0},M,H}$, the conclusion is trivial.
In the case that $O$ is at least partly covered
by $\Omega_{{\bf{x}}^{*}_{0},M,H}$. $M_{O}(v)- z_{{\bf{x}}^{*}_{0}}$
cannot achieve its maximum value in
$\overline{\Omega_{{\bf{x}}^{*}_{0},M,H}\cap  O}$ on
$\partial \Omega_{{\bf{x}}^{*}_{0},M,H} \cap O$ since the directional
derivative of $z_{{\bf{x}}^{*}_{0}}$ with respect
to outer normal is $+\infty$ on
$\partial \Omega_{{\bf{x}}^{*}_{0},M,H}\cap O $ by (iii)
in Lemma \ref{lemma:second}.
Furthermore since
$z_{{\bf{x}}^{*}_{0}}$ satisfies (\ref{eq:supersolution}) and
$|\Lambda ({\bf{x}})|\leq
\Lambda_{0}({\bf{x}})$ by \eqref{eq:boundoflambda} and
(\ref{eq:boundoflambda0}),
a comparison argument concludes that
$M_{O}(v)- z_{{\bf{x}}^{*}_{0}}$ cannot achieve a local maximum
inside $\Omega_{{\bf{x}}^{*}_{0},M,H}\cap O$. Thus
$M_{O}(v) -z_{{\bf{x}}^{*}_{0}}$ achieves its maximum value in
$\overline{\Omega_{{\bf{x}}^{*}_{0},M,H}\cap  O}$
on $\Omega_{{\bf{x}}^{*}_{0},M,H}\cap \partial O$.
Then $M_{O}(v) -z_{{\bf{x}}^{*}_{0}}\leq 0$ on
$\Omega_{{\bf{x}}^{*}_{0},M,H}\cap  O$ follows from
$M_{O}(v) -z_{{\bf{x}}^{*}_{0}}=v-z_{{\bf{x}}^{*}_{0}}\leq 0$ on
$\Omega_{{\bf{x}}^{*}_{0},M,H}\cap \partial O$.
%\end{proof}

Now we will show that $\Xi$ is not empty by proving the existence of
a solution to the minimal surface equation with the same boundary-value
and on the same domain.

\begin{lemma} \label{lemma:seventh}
 If $v\in C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$ is
a solution of the problem
\begin{equation}
((1+|Dv|^{2}) \delta_{ij} - D_{i}v D_{j}v )D_{ij} v = 0
   {\rm{in}}  \Omega ,
v=\phi   on  \partial \Omega .
\label{eq:subsolution221}
\end{equation}
Then for any $({\bf{x}}^{*}_{0},x_{n})\in \Omega$,
\begin{equation}
|v| \leq z_{{\bf{x}}^{*}_{0}}\quad\text{on } \Omega_{{\bf{x}}^{*}_{0},M,H} \cap \Omega
\label{eq:boundofminimal}
\end{equation}
\end{lemma}

The proof of the above lemma is just a special case of
Lemma \ref{lemma:growth1} with $\Lambda ({\bf{x}})=0$.


\begin{lemma} \label{lemma:eigth}
Assume {\rm (A1)--(A3)}. Then the boundary-value problem
\begin{equation}
((1+|Dv|^{2}) \delta_{ij} - D_{i}v D_{j}v )D_{ij} v = 0
  \quad\text{in }    \Omega ,\quad
v=\phi \quad\text{on }   \partial \Omega .
\label{eq:subsolution1}
\end{equation}
has a solution $u\in C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$.
\end{lemma}


\begin{proof}
 This is \cite[Lemma 4.5 ]{Jin3}
(Though a slight difference should be noted there. That
is, the bound for solutions of the minimal surface equation
is given by (\ref{eq:boundofminimal}) which will play
the same role as the Lemma 4.4 in \cite{Jin3} (By the way,
the Lemma 3.1 quoted in the proof of Lemma 4.5 in
\cite{Jin3} should be Lemma 4.4 in \cite{Jin3}).
\end{proof}


Now we prove the Theorem \ref{theorem:first}. We set
$$
u({\bf{x}})=\sup \{ v({\bf{x}}) : v\in \Xi  \},
{\bf{x}}\in \overline{\Omega}.
$$
We first consider the case that $\Lambda ({\bf{x}})\leq 0$ on $\Omega $.
For such a choice of $\Lambda ({\bf{x}})$,
we will show that $u$ is in $C^{0}(\overline{\Omega})\cap C^{2}(\Omega)$
satisfying \eqref{eq:problem11}-\eqref{eq:problem12}.
It is well known and standard (for example, see \cite{CH}) that by
Perron's method, we can prove that $u$ is in $C^{2}(\Omega )$
and satisfies \eqref{eq:problem11}. Indeed, let ${\bf{x}}_{1}\in \Omega $.
By the definition of $u({\bf{x}}_{1})$, there is a sequence of functions
$v_{i}$ in $\Xi $ such that
$$
u({\bf{x}}_{1})=\lim_{i\to \infty } v_{i}({\bf{x}}_{1}).
$$
Let $v_{0} $ be a solution of (\ref{eq:subsolution1}).
Since $\Lambda ({\bf{x}})\leq 0$ on $\Omega $, by
Lemma \ref{lemma:seventh}, it is easy to check that
$v_{0}\in \Xi$.  By Lemma \ref{lemma:fifth} and  replacing $v_{i}$
by $\max \{ v_{i}, v_{0} \}$, we may assume
that $v_{i}\geq v_{0} $ on $\Omega$. Let $O$ be an open set
in $\Pi$ such that ${\bf{x}}_{1}\in O$.
We replace $v_{i}$ by $M_{O}(v_{i})$. Then we have a sequence of
functions $z_{i}$ defined on $O$ satisfying
\begin{gather*}
u({\bf{x}}_{1})=\lim_{i\to \infty } z_{i}({\bf{x}}_{1}) , \\
((1+|Dz_{i}|^{2}) \delta_{pq} - D_{p}z_{i} D_{q}z_{i})
D_{pq} z_{i} =
n \Lambda ({\bf{x}}) (1+|Dz_{i}|^{2})^{3/2}  \quad\text{on }    O, \\
z_{i}=v_{i}  \quad\text{on }    \partial O.
\end{gather*}
Since for all $i$, if $O\cap \Omega_{{\bf{x}}^{*}_{0},M,H}$ is not empty,
$$
v_{0} \leq v_{i}\leq z_{i} \leq z_{{\bf{x}}^{*}_{0}} \quad\text{on }
O\cap \Omega_{{\bf{x}}^{*}_{0},M,H},
$$
and we can cover $O$ by finitely many such domains
$\Omega_{{\bf{x}}^{*}_{0},M,H}$, thus
there is a number $K_{3}$ independent of
$i$, such that for all $i$,
$$
v_{0} \leq z_{i}\leq K_{3}  \quad\text{in } O.
$$
By \cite[Corollarys 16.6, 16.7]{GT},
there is a subsequence of $z_{i}$, for convenience still
denoted by $z_{i}$, converges to a $C^{2}(O)$
function $z(x)$ in $C^{2}(O)$. Thus $z(x)$ satisfies
$$
((1+|Dz|^{2}) \delta_{pq} - D_{p}z D_{q}z )D_{pq} z=
n \Lambda ({\bf{x}}) (1+|Dz|^{2})^{3/2}
\quad\text{on } O.
$$
Note that $u({\bf{x}}_{1})=z({\bf{x}}_{1})$ and
$u({\bf{x}})\geq z({\bf{x}})$ on $O$.
We claim that $u=z$ on $O$. Indeed, if there is another point
${\bf{x}}_{2} \in O$ such that
$u({\bf{x}}_{2})$ is not equal to $z({\bf{x}}_{2})$, we must have
$u({\bf{x}}_{2})>z({\bf{x}}_{2})$. Then there is a function
$u_{0}\in \Xi $, such that
$$
z({\bf{x}}_{2})< u_{0}({\bf{x}}_{2})\leq
u({\bf{x}}_{2}).
$$
Now the sequence $\max \{ u_{0}, M_{O}(v_{i}) \}$ satisfying
$$
v_{i} \leq \max \{ u_{0}, M_{O}(v_{i}) \} \leq u .
$$
Then similar to the way we obtain $z$,
$M_{O}(\max \{ u_{0}, M_{O}(v_{i}) \})$
will produce a $C^{2}$ function
$z_{1}$ satisfying
\begin{gather*}
((1+|Dz_{1}|^{2}) \delta_{pq} - D_{p}z_{1} D_{q}z_{1} )
D_{pq} z_{1} =
n \Lambda ({\bf{x}}) (1+|Dz_{1}|^{2})^{3/2} \quad\text{on } O,\\
z\leq z_{1} \quad\text{on } O, \quad
z({\bf{x}}_{2})<u_{0}({\bf{x}}_{2})\leq  z_{1}({\bf{x}}_{2}), \\
z({\bf{x}}_{1})=z_{1}({\bf{x}}_{1})=u({\bf{x}}_{1}) .
\end{gather*}
That is,  $z_{1}({\bf{x}})-z({\bf{x}})$ is non-negative,
not identically zero on $O$ and
achieves its minimum value zero inside $O$.
However, from the equations satisfied by $z$ and $z_{1}$, we have
that on $O$,
\begin{align*}
&((1+|Dz_{1}|^{2}) \delta_{pq} - D_{p}z_{1} D_{q}z_{1} )
D_{pq} (z_{1}-z) \\
&= E({\bf{x}}, z_{1}, z, Dz, Dz_{1}, D^{2}z, D^{2}z_{1})
(Dz_{1}-Dz)\quad\text{on } O
\end{align*}
for some continuous function $E$.
Then by the standard maximum principle (for example, see
\cite[Theorem 3.5]{GT}), we get a contradiction.
Thus $u=z$ on $O$. Therefore $u\in C^{2}(\Omega )$ and
$$
((1+|Du|^{2}) \delta_{ij} - D_{i}u D_{j}u )
D_{ij} u = n \Lambda ({\bf{x}}) (1+|Du|^{2})^{3/2}
\quad\text{on } \Omega.
$$
Since $v_{0}=\phi $ on $\partial \Omega$, from the definition of $u$ we
see that $u=\phi $ on $\partial \Omega$.
We still need to prove that $u\in C^{0}(\overline{\Omega})$.

Since $\Omega$ satisfies Serrin's condition (\ref{eq:meancurvature}),
we can find a $C^{2, \gamma}$ domain
$\Omega_{1}\subset \Omega$, $\Omega_{1}\in \Pi$, (for the existence of
$\Omega_{1}$, see Lemma A.3 in Appendix in \cite{Jin3}), such that
$\partial \Omega_{1} \cap \partial \Omega$ is
an open neighborhood of
${\bf{x}}_{1}$ in $\partial \Omega$ and
on $\partial \Omega_{1}$,  the mean curvature
$H'$ with respect to inner normal of $\partial \Omega_{1}$ satisfies
\begin{equation}
H'> \frac{n}{n-1} |\Lambda ({\bf{x}})| \quad\text{on } \partial \Omega_{1}.
\label{eq:boundaryofsmall}
\end{equation}
Since $\Omega_{1}$ can be covered by finitely many
$\Omega_{{\bf{x}}^{*}_{0}, M, H}$, there is
a number $K_{4}>0$, such that for all $v\in \Xi$,
\begin{equation}
v\leq K_{4} \quad\text{on } {\overline{\Omega}}_{1} .
\label{eq:boundofsmall}
\end{equation}
Now on $\partial \Omega_{1}$, we choose a smooth function $\phi^{*}$
as follows.
$\phi^{*}=K_{4}$ on $\partial \Omega_{1}\cap \Omega$. $\phi^{*}=\phi $
in a neighborhood of ${\bf{x}}_{1}$ in
$\partial \Omega_{1}$ and $\phi^{*}\geq \phi $ on the rest of
$\partial \Omega_{1}$
(since (\ref{eq:boundofsmall})
implies $\phi \leq K_{4}$ on $\partial \Omega_{1}\cap \partial \Omega$,
this is possible).
Now we consider the boundary-value problem
\begin{equation}
((1+|Du|^{2}) \delta_{ij} - D_{i}u D_{j}u )
D_{ij} u =
n \Lambda ({\bf{x}}) (1+|Du|^{2})^{3/2}
\quad\text{on }
\Omega_{1},
\label{eq:small}
\end{equation}
\begin{equation}
u=\phi^{*} \quad\text{on } \partial \Omega_{1} .
\label{eq:small22}
\end{equation}
 From (\ref{eq:boundaryofsmall}), Lemma \ref{lemma:growth1} and
\cite{Serrin} or \cite[Theorem 16.9]{GT},
(\ref{eq:small})-(\ref{eq:small22}) has a
solution $u_{1}\in C^{2}(\Omega_{1})\cap C^{0}({\overline{\Omega}}_{1})$.
From the definition of $u_{1}$,
(\ref{eq:boundofsmall}) and the fact that
$v=\phi $ on $\partial \Omega$ for any $v\in \Xi$,
a comparison argument shows that for any $v\in \Xi$,
$$
M_{\Omega_{1}}(v) \leq u_{1} \quad\text{on } \Omega_{1}\quad {\text{for  any}} \quad v\in \Xi.
$$
Therefore,
\begin{equation}
u\leq u_{1} \quad\text{on } \Omega_{1}.
\label{eq:upperhalf}
\end{equation}
Since we always have
$$
u\geq v_{0} \quad\text{on } \Omega
$$
for the solution $v_{0}$ of (\ref{eq:subsolution1}), we have
\begin{equation}
v_{0}\leq u\leq u_{1} \quad\text{on } \Omega_{1} .
\label{eq:whole}
\end{equation}
Then the continuity of $u$ at ${\bf{x}}_{1}$ follows from
the fact that $v_{0}=u_{1}=\phi $ on a neighborhood of ${\bf{x}}_{1}$
in $\partial \Omega$ and both $v_{0}$ and $u_{1}$ are continuous
in a neighborhood of ${\bf{x}}_{1}$ in $\overline{\Omega}$.
Since ${\bf{x}}_{1}\in \partial \Omega$ can be arbitrary, we
have $u\in C^{0}(\overline{\Omega})$.
Thus under the additional assumption that $\Lambda ({\bf{x}})\leq 0$ on
$\Omega$, we have proved Theorem \ref{theorem:first}.

In the case that $\Lambda ({\bf{x}})\geq 0$ on $\Omega$,
repeating above proof, we can find a function
$u\in C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$ satisfying
\begin{gather*}
((1+|Du|^{2}) \delta_{ij} - D_{i}u D_{j}u )
D_{ij} u = -n\Lambda ({\bf{x}}) (1+|Du|^{2})^{3/2}
\quad\text{on } \Omega , \\
u=-\phi \quad\text{on } \partial \Omega .
\end{gather*}
Then $-u$ will satisfy \eqref{eq:problem11}-\eqref{eq:problem12}.

In the general case of $\Lambda ({\bf{x}})$, we first find a function
$u_{0}\in C^{1}(\Omega)\cap C^{0}(\overline{\Omega})$ satisfying
\begin{gather*}
((1+|Du|^{2}) \delta_{ij} - D_{i}u D_{j}u )
D_{ij} u = n|\Lambda ({\bf{x}})| (1+|Du|^{2})^{3/2}
\quad\text{on } \Omega , \\
u=\phi \quad\text{on } \partial \Omega .
\end{gather*}
In the proof for the case that $\Lambda\leq 0$, we
replace $v_{0}$ (the solution of (\ref{eq:subsolution1}))
 by $u_{0}$, without changing
the rest of the proof, now we will obtain a function
$u\in C^{1}(\Omega)\cap C^{0}(\overline{\Omega})$ satisfies
\eqref{eq:problem11}-\eqref{eq:problem12}.
This completes the proof for Theorems \ref{theorem:first}.


\subsection*{Acknowledgements}
The author would like to thank Professor J. Serrin for his valuable
suggestions.

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\end{document}