# On the Dirichlet problem for special Lagrangian curvature potential equation

Rongli Huang  School of Mathematics and Statistics, Guangxi Normal University,
Guilin, Guangxi 541004, People’s Republic of China
<EMAIL_ADDRESS>and  Yongmei Liang  School of Mathematics and
Statistics, Guangxi Normal University, Guilin, Guangxi 541004, People’s
Republic of China

###### Abstract.

In this paper, we study a class of special Lagrangian curvature potential
equations and obtain the existence of smooth solutions for Dirichlet problem.
The existence result is based on a priori estimates of global $C^{0}$, $C^{1}$
and $C^{2}$ norms of solutions under the assumption of existence of a
subsolution.

The authors were supported by National Science Foundation of
China(No.11771103) and Guangxi Natural Science
Foundation(No.2017GXNSFFA198017).

†† 2010 Mathematics Subject Classification. Primary 53C44; Secondary 53A10.
Keywords and phrases. Special Lagrangian curvature operator; Concavity;
Subsolution.

##  1\. Introduction

It is well known that the Dirichlet problem is well posed for elliptic
equations. The situation is complicated for fully nonlinear elliptic
equations; see for instance the seminal papers of L. Caffarelli. L. Nirenberg
and J. Spruck [1]-[4]. We will restrict our attention from now on to special
Lagrangian equation and some related problems. R. Harvey and H.B. Lawson
established the theory of calibrated geometry and introduced the special
Lagrangian equation in [5] back in 1982. They obtained that its solutions $u$
have the property that the graph $p=\nabla u$ in
$\mathbb{R}^{n}\times\mathbb{R}^{n}=\mathbb{C}^{n}$ being a Lagrangian
submanifold which is absolutely volume-minimizing and the linearization at any
solution being elliptic. Let $x=(x_{1},x_{2},\cdots,x_{n})$, $u=u(x)$ and
$\lambda(D^{2}u)=(\lambda_{1},\cdots,\lambda_{n})$ are the eigenvalues of
Hessian matrix $D^{2}u$ according to $x$. Yuan [6] proved that every classical
convex solution of special Lagrangian equation

$\sum_{i=1}^{n}\arctan\lambda_{i}=c,\quad\mathrm{in}\quad\mathbb{R}^{n},$

must be a quadratic polynomial. The second boundary value problem for special
Lagrangian equation was considered by Brendle and Warren [7]. They obtained
the existence and uniqueness of the solution by the elliptic method. Later
Brendle-Warren’ theorem was generalized by several authors (see, e.g.,
[8]-[12]).

Assume that

(A): $\Omega$ is $C^{4}$ bounded domain in $\mathbb{R}^{n}$,
$\varphi:\partial\Omega\longrightarrow\mathbb{R}$ be in
$C^{4}(\partial\Omega)$ and
$h:\bar{\Omega}\longrightarrow[(n-2)\frac{\pi}{2}+\delta,n\frac{\pi}{2})$ be
in $C^{2}(\bar{\Omega})$ where $\delta>0$ small enough.

In another case, Collins, Sebastien and Xuan [13] were concerned with
Lagrangian phase equation

(1.1) $\sum_{i=1}^{n}\arctan\lambda_{i}=h(x),\quad\mathrm{in}\quad\Omega,$

associated with Dirichlet problem

(1.2) $u=\varphi,\quad\mathrm{on}\quad\partial\Omega.$

Let $\alpha$ be any constant satisfying $0<\alpha<1$. They proved that

######  Theorem 1.1.

Suppose that $\Omega$, $\varphi$ and $h$ satisfy (A). If there exists a
$C^{4}(\bar{\Omega})$ subsolution of (1.1) and (1.2), i.e. there exists a
function $\underline{u}\in C^{4}(\bar{\Omega})$ such that

$\left\\{\begin{aligned} \sum_{i=1}^{n}\arctan\underline{\lambda}_{i}&\geq
h(x),&&\quad\mathrm{in}\quad\Omega,\\\
\underline{u}&=\varphi,&&\quad\mathrm{on}\quad\partial\Omega,\end{aligned}\right.$

where $\underline{\lambda}_{i}$ are the eigenvalues of $D^{2}\underline{u}$.
Then the Dirichlet problem (1.1) (1.2) admits a unique
$C^{3,\alpha}(\bar{\Omega})$ solution. In addition, if $\Omega,h,\varphi$ and
$\underline{u}$ are smooth, then the solution $u$ is smooth.

Here we noticed that they used the idea of B.Guan [14] that the curvature
assumption on $\partial\Omega$ is replaced by the assumption that there exists
an admissible subsolution of (1.1) and (1.2) to obtain the global $C^{2}$
estimates for the admissible solutions.

The aim of this paper is to study special Lagrangian curvature potential
equation

(1.3) $\sum_{i=1}^{n}\arctan\kappa_{i}=h(x),\quad\mathrm{in}\quad\Omega,$

in conjunction with Dirichlet condition

(1.4) $u=\varphi,\quad\mathrm{on}\quad\partial\Omega.$

where $(\kappa_{1},\cdots,\kappa_{n})$ are the principal curvatures of the
graph $\Gamma=\\{(x,u(x))\mid x\in\Omega\\}$. It shows that a very nice
geometric interpretation of special Lagrangian curvature operator
$F(\kappa_{1},\cdots,\kappa_{n})\triangleq\sum_{i=1}^{n}\arctan\kappa_{i}$ in
[15]. The background of the so-call special Lagrangian curvature potential
equation (1.3) can be seen in the literature [15] [16]. In addition, the first
author and Sitong Li considered (1.3) with the second boundary condition

(1.5) $Du(\Omega)=\tilde{\Omega},$

then they obtained the following result [17] for $h(x)$ being some constant.

######  Theorem 1.2.

Suppose that $\Omega$, $\tilde{\Omega}$ are uniformly convex bounded domains
with smooth boundary in $\mathbb{R}^{n}$. Then there exist a uniformly convex
solution $u\in C^{\infty}(\bar{\Omega})$ and a unique constant $c$ with
$h(x)=c$ solving (1.3) and (1.5), and $u$ is unique up to a constant.

It’s interested in considering Dirichlet problem (1.3) and (1.4) with the same
geometric conditions on $\Omega$ as the above theorem. The central result of
this paper is to find solution under the assumptions of the existence of the
subsolution, which originate from a series of Bo Guan’s paper (cf. [14] [18]
[19]).

Now we state our main result in the following which gives a generalization of
Theorem 1.1.

######  Theorem 1.3.

Suppose that $\Omega$, $\varphi$ and $h$ satisfy (A). If there exists a
$C^{4}(\bar{\Omega})$ subsolution of (1.3) and (1.4), i.e. there exists a
function $\underline{u}\in C^{4}(\bar{\Omega})$ such that

(1.6) $\left\\{\begin{aligned}
\sum_{i=1}^{n}\arctan\underline{\kappa}_{i}&\geq h(x),\ \
&&\quad\mathrm{in}\quad\Omega,\\\
\underline{u}&=\varphi,&&\quad\mathrm{on}\quad\partial\Omega,\end{aligned}\right.$

where $\underline{\kappa}_{i}$ are the principal curvatures of the graph
$\underline{\Gamma}=\\{(x,\underline{u}(x))\mid x\in\Omega\\}$. Then Dirichlet
problem (1.3) (1.4) admits a unique $C^{3,\alpha}(\bar{\Omega})$ solution. In
addition, if $\Omega,h,\varphi$ and $\underline{u}$ are smooth, then the
solution $u$ is smooth.

This article is dived into several sections. In section two, we state some
well known algebraic and inequalities according to the structure conditions
for the operator $F$. In section three, we can derive the zeroth and first
order estimates for the admissible solutions. In section four, we exhibit how
to estimate the second order derivatives of $u$ by the assumption of the
boundedness for them on $\partial\Omega$. In section five, we establish the
desired bound for the second order derivatives of $u$ on $\partial\Omega$
which is a key and difficult part in the paper. Finally we give the proof of
Theorem 1.3 by the continuity method.

##  2\. Preliminaries

In this section we collect various preliminary knowledge and auxiliary lemmas
that used in the proof of Theorem 1.3.

At the beginning, we recall some necessary geometric quantities associated
with the graph of a function $u\in C^{2}(\Omega)$. We use the Einstein
summation convention, if the indices are different from 1 and $n$.
$u_{i}=D_{i}u,u_{ij}=D_{ij}u,u_{ijk}=D_{ijk}u,\cdots$ Denote the all
derivatives of $u$ according to $x_{i},x_{j},x_{k},\cdots$. In the coordinate
system induced by the embedding $x\longmapsto(x,u(x))$ the metric of graph $u$
is given by

$g_{ij}=\delta_{ij}+u_{i}u_{j}$

which is called the first fundamental form and its inverse is

(2.1) $g^{ij}=\delta_{ij}-\frac{u_{i}u_{j}}{1+|Du|^{2}}.$

The second fundamental form is given by

(2.2) $h_{ij}=\frac{u_{ij}}{\sqrt{1+|Du|^{2}}}.$

Using (2.1) and (2.2), we obtain

$h_{ik}g^{kj}=\frac{u_{ik}}{\sqrt{1+|Du|^{2}}}\Big{(}\delta_{kj}-\frac{u_{k}u_{j}}{1+|Du|^{2}}\Big{)}.$

From [20] and [21], the principal curvatures of the graph of $u$ are the
eigenvalues of $[h_{ij}]$ relative to $[g_{ij}]$, thus they are the
eigenvalues of the generally nonsymmetric matrix $[h_{ik}g^{kj}]$.
Equivalently they are the eigenvalues of the symmetric matrix

$a_{ij}=b^{ik}h_{kl}b^{lj}=\frac{1}{w}b^{ik}u_{kl}b^{lj},$

where $w=\sqrt{1+|Du|^{2}}$ and $[b^{ij}]$ is the positive square root of
$[g^{ij}]$, i.e. $b^{ik}b^{kj}=g^{ij}$, given by

$b^{ij}=\delta_{ij}-\frac{u_{i}u_{j}}{w(1+w)}.$

Therefore

(2.3) $u_{ij}=wb_{ik}a_{kl}b_{lj},$

where $[b_{ij}]$ is the inverse of $[b^{ij}]$, given by

$b_{ij}=\delta_{ij}+\frac{u_{i}u_{j}}{1+w}.$

Explicitly we see

(2.4)
$a_{ij}=\frac{1}{w}\Big{\\{}u_{ij}-\frac{u_{i}u_{l}u_{jl}}{w(1+w)}-\frac{u_{j}u_{l}u_{il}}{w(1+w)}+\frac{u_{i}u_{j}u_{k}u_{l}u_{kl}}{w^{2}(1+w)^{2}}\Big{\\}}.$

Next we exhibit some properties of the function $F$, which can be found in
[22] and [21]. As mentioned in the introduction, the function $F$ was defined
by

(2.5) $F(\mathcal{A})=\sum_{i=1}^{n}\arctan\kappa_{i},$

where $(\kappa_{1},\cdots,\kappa_{n})$ are the eigenvalues of
$\mathcal{A}=[a_{ij}]$. The function $F$ satisfies

$F^{ij}(\mathcal{A})\eta_{i}\eta_{j}>0,\quad\eta\in\mathbb{R}^{n}-\\{0\\},$

where

$F^{ij}(\mathcal{A})=\frac{\partial F(\mathcal{A})}{\partial a_{ij}}.$

Note that $[F^{ij}]$ is symmetric because $\mathcal{A}$ is symmetric.
Furthermore, $[F^{ij}]$ is diagonal if $\mathcal{A}$ is diagonal.

As a matter of convenience, we use the notation

$\underline{a}_{ij}=\frac{1}{\underline{w}}\Big{\\{}\underline{u}_{ij}-\frac{\underline{u}_{i}\underline{u}_{l}\underline{u}_{jl}}{\underline{w}(1+\underline{w})}-\frac{\underline{u}_{j}\underline{u}_{l}\underline{u}_{il}}{\underline{w}(1+\underline{w})}+\frac{\underline{u}_{i}\underline{u}_{j}\underline{u}_{k}\underline{u}_{l}\underline{u}_{kl}}{\underline{w}^{2}(1+\underline{w})^{2}}\Big{\\}}.$

where $\underline{w}=\sqrt{1+|D\underline{u}|^{2}}$.

Since the technique in [13] that we will develop to prove Theorem 1.3 also
works for our setting, then we will use certain properties of $F$ at various
of points in the proof. These are summarized in the following lemmas, which
was proved in [13].

######  Lemma 2.1.

Let $0<\delta<\frac{\pi}{2}$. Assume that
$\kappa_{1}\geq\kappa_{2}\geq\cdots\geq\kappa_{n}$ satisfy

$\sum_{i=1}^{n}\arctan\kappa_{i}\geq(n-2)\frac{\pi}{2}+\delta.$

Then there hold
(i) $\kappa_{1}\geq\kappa_{2}\geq\cdots\geq\kappa_{n-1}>0$,
$|\kappa_{n}|\leq\kappa_{n-1}$;
(ii) $\sum_{i=1}^{n}\kappa_{i}\geq 0$;
(iii) $\kappa_{n}\geq-\frac{1}{\tan(\delta)}$;
(iv) if $\kappa_{n}<0$, then
$\sum_{i=1}^{n}\frac{1}{\kappa_{i}}\leq-\tan(\delta)$;
(v) For any $\sigma\in((n-2)\frac{\pi}{2},n\frac{\pi}{2})$,

$\Gamma^{\sigma}=\\{\kappa\in\mathbb{R}^{n}:\sum_{i=1}^{n}\arctan\kappa_{i}>\sigma\\}.$

is a convex set, and $\partial\Gamma^{\sigma}$ is a smooth convex
hypersurface.

######  Lemma 2.2.

Let $\Omega$, $\varphi$ and $h$ satisfy (A), $u\in C^{3,\alpha}(\bar{\Omega})$
satisfy the Dirichlet problem (1.3) and (1.4). Suppose that there exists a
function $\underline{u}$ such that for each point $x\in\Omega$ and each index
$i$, there holds

(2.6) $\lim_{t\rightarrow\infty}F(\underline{\kappa}+te_{i})>h(x),$

where $\underline{\kappa}_{i}$ are the principal curvatures of the graph
$\underline{\Gamma}=\\{(x,\underline{u}(x))|x\in\Omega\\}$,
$\underline{\kappa}=(\underline{\kappa}_{1},\underline{\kappa}_{2},\cdots,\underline{\kappa}_{n})$,
$e_{i}$ is the $i$th standard basis vector and
$F(\kappa)=\sum_{i=1}^{n}\arctan\kappa_{i}$.

Then there are constants $R_{0}$, $C_{1}>0$ to arrive at the following
inequalities. If $|\kappa|\geq R_{0}$, we either have

(2.7)
$\sum_{i,j=1}^{n}F^{ij}(\mathcal{A})\big{[}\underline{a}_{ij}-a_{ij}\big{]}>C_{1}\sum_{p=1}^{n}F^{pp}(\mathcal{A}),$

or

(2.8) $F^{ii}(\mathcal{A})>C_{1}\sum_{p=1}^{n}F^{pp}(\mathcal{A})$

for each $i$.

######  Remark 2.3.

Without loss of generality, in the following we set $C_{1}$, $C_{2}$, $\ldots$
to be positive constants depending only on the known data.

Introducing Corollary 3.2 in [13], we infer that

######  Lemma 2.4.

Assume that $\Omega$, $\varphi$ and $h$ satisfy $(A)$, $\underline{u}$ is a
subsolution satisfying (1.6), $u\in C^{3,\alpha}(\bar{\Omega})$ is an
admissible solution to (1.3) and (1.4). As before, let
$\kappa=(\kappa_{1},\cdots,\kappa_{n})$ are the principal curvatures of the
graph $\Gamma=\\{(x,u(x))\mid x\in\Omega\\}$. Then there exsits $R_{0}$
depending only on $\underline{u}$ and $\delta$, such that for any
$|\kappa|\geq R_{0}$, we have

(2.9) $F^{ij}(\mathcal{A})[\underline{a}_{ij}-a_{ij}]\geq\tau>0,$

where $\tau$ is a constant depending on only $\underline{u}$ and $\delta$.

##  3\.  Zeroth and first order estimates

In this section, we will prove the zeroth and first order estimates for
Dirichlet problem (1.3) and (1.4). The two-sided bound for $u$ can be directly
inferred from the existence of the sub-solution and super-solution.

To see this, it is enough to consider $\overline{u}$ satisfying

(3.1) $\left\\{\begin{aligned} \triangle\overline{u}&=0,\ \
&&\quad\mathrm{in}\quad\Omega,\\\
\overline{u}&=\varphi,&&\quad\mathrm{on}\quad\partial\Omega.\end{aligned}\right.$

Review the identity: $u_{ij}=wb_{ik}a_{kl}b_{lj}$. Let $D^{2}u=[u_{ij}]$,
$\mathcal{B}=[b_{ij}]$, $\mathcal{A}=[a_{ij}]$, where
$\mathcal{B}^{T}=\mathcal{B}$. By Lemma 2.3 we see that
$\mathrm{tr}\mathcal{A}=\sum_{i=1}^{n}\kappa_{i}\geq 0$. Since
$D^{2}u=w\mathcal{B}^{T}\mathcal{A}\mathcal{B}$ where $w=\sqrt{1+|Du|^{2}}$,
then we arrive at

$\mathrm{tr}D^{2}u=\mathrm{tr}(w\mathcal{B}^{T}\mathcal{A}\mathcal{B})\geq 0.$

Next we have the following result.

######  Lemma 3.1.

Suppose that $\Omega$, $\varphi$ and $h$ satisfy (A), $\underline{u}$ is a
admissible subsolution defined by (1.6), $u$ satisfy (1.3) and (1.4). Let
$\overline{u}:\Omega\rightarrow\mathbb{R}$ be the function satisfying (3.1).
Then we have

$\underline{u}\leq u\leq\overline{u},\quad\mathrm{in}\quad\overline{\Omega}.$

###### Proof.

Note that $\Delta u=\mathrm{tr}D^{2}u\geq 0=\triangle\overline{u}$ and
$F(\mathcal{A})\leq F(\mathcal{\underline{A}})$ where
$\underline{\mathcal{A}}=[\underline{a}_{ij}]$. Therefore, the conclusion is
derived from the maximum principle. ∎

By Lemma 3.1, we have the following Hopf Lemma, due to Li-Wang[23].

######  Lemma 3.2.

Suppose that $\Omega$, $\varphi$ and $h$ satisfy (A), $\underline{u}$ is a
admissible subsolution defined by (1.6), $u$ satisfy (1.3) and (1.4). Suppose
that $\hat{x}\in\partial\Omega$, $\hat{s}\in\mathbb{R}$, then we have

$\liminf_{\hat{s}\rightarrow
0^{+}}\frac{(u-\underline{u})(\hat{x}+\hat{s}\nu(\hat{x}))}{\hat{s}}>0,$

where $\nu$ is the inner unit normal to $\partial\Omega$ at $\hat{x}$.

One of the obstacles in the present work is the lack of concavity of the
operator $F$ appearing in (2.5). It is important to note that one of the key
elements of this paper is to transform the equation (1.3) to find hidden
concavity properties when $h(x)>(n-2)\frac{\pi}{2}$, thus it allows us to
obtain the gradient estimate. It is worthy of mention that we use the argument
following from [13] and [22].

######  Lemma 3.3.

Let $F(\kappa)=\sum_{i=1}^{n}\arctan\kappa_{i}$ be defined on

$\\{\kappa\in\mathbb{R}^{n}:\sum_{i=1}^{n}\arctan\kappa_{i}\geq(n-2)\frac{\pi}{2}+\delta\\}.$

Then there exists a sufficiently large $A$ depending only on $\delta$ such
that $G(\kappa)=-e^{-AF(\kappa)}$ is a concave function.

Similarly, an consequence of Lemma 3.3 is

######  Lemma 3.4.

Let $\Gamma:=\\{\mathcal{M}\in
sym(n):F(\mathcal{M})\geq(n-2)\frac{\pi}{2}+\delta\\}$, then there exists
$A=A(\delta)$ such that the operator

$G(\mathcal{A})=-e^{-AF(\mathcal{A})}$

is elliptic and concave on $\Gamma$.

The readers can see the proof of the above two lemmas in [13]. Now we use the
concavity of the operator $G$ from Lemma 3.4. Without loss of generality, we
assume that $\mathcal{A}$ is diagonal. It is easy to verify that

$G^{ij}(\mathcal{A})=\frac{\partial G(\mathcal{A})}{\partial
a_{ij}}=\frac{\partial G(\mathcal{A})}{\partial a_{ii}}\delta_{ij},\quad
G^{ij,kl}(\mathcal{A})=\frac{\partial^{2}G(\mathcal{A})}{\partial
a_{ij}\partial a_{kl}},$

then we have

(3.2) $G^{ij,kl}M_{ij}M_{kl}\leq 0,$

for any real symmetric tensor $M_{ij}$. A direct computation shows that there
exists a positive constant $\widetilde{c}$ such that

(3.3) $\sum_{i=1}^{n}\kappa_{i}\frac{\partial
G(\kappa)}{\partial\kappa_{i}}=Ae^{-AF(\kappa)}\sum_{i}\frac{\kappa_{i}}{1+\kappa_{i}^{2}}\geq-\widetilde{c},\quad\mathrm{in}\quad\overline{\Gamma^{\sigma}}\setminus\\{0\\}.$

By the arguments in [13] one can conclude that there exists some constant
$C_{2}$ such that

(3.4) $|\kappa|^{2}\sum_{i=1}^{n}\frac{\partial
G(\kappa)}{\partial\kappa_{i}}\leq C_{2}\sum_{i=1}^{n}\frac{\partial
G(\kappa)}{\partial\kappa_{i}}\kappa_{i}^{2},\quad\mathrm{in}\quad\overline{\Gamma^{\sigma}}\setminus\\{0\\}.$

After introducing the new operator $G$, then (1.3) and (1.4) are equivalent to
the following Dirichlet problem

(3.5) $\left\\{\begin{aligned} G(\mathcal{A})&=-e^{-Ah(x)}:=\psi(x),\ \
&&\quad\mathrm{in}\quad\Omega,\\\
u&=\varphi,&&\quad\mathrm{on}\quad\partial\Omega.\end{aligned}\right.$

It can be followed from Lemma 2.1 that

(3.6) $\frac{\partial
G(\kappa)}{\partial\kappa_{n}}=Ae^{-AF(\kappa)}\frac{1}{1+\kappa_{n}^{2}}\geq
C_{3}>0.$

where $C_{3}$ is a positive constant depending only on $\delta$.

For the convenient of reader, we denote

$g_{\alpha}=\frac{\partial G(\kappa)}{\partial\kappa_{\alpha}}(1\leq\alpha\leq
n-1),\,\,g_{i}=\frac{\partial G(\kappa)}{\partial\kappa_{i}}(1\leq i\leq
n),\,\,g_{n}=\frac{\partial G(\kappa)}{\partial\kappa_{n}}.$

After the corresponding transformation, inspired by literatures in [3] and
[4], then we derive that

######  Lemma 3.5.

Suppose that $\Omega$, $\varphi$ and $h$ satisfy (A). If there exists a
$C^{2}$ admissible subsolution $\underline{u}$ satisfying (1.6), $u\in
C^{3}(\overline{\Omega})$ satisfy (1.3) and (1.4), then we have the following
estimate

(3.7) $\sup_{\overline{\Omega}}|Du|\leq C_{4},$

where $C_{4}$ is a positive constant that depends on $\overline{\Omega}$,
$\|\underline{u}\|_{C^{2}(\overline{\Omega})}$,
$\|h\|_{C^{2}(\overline{\Omega})}$, $\delta$.

###### Proof.

Our first goal is to estimate $|Du|$ in $\Omega$. Let

$z=|Du|e^{\widetilde{A}u},$

where $\widetilde{A}=\big{[}\frac{2}{C_{3}}\max|D\psi(x)|\big{]}^{1/2}$,
$C_{3}$ is the positive constant from (3.6). We assume that it achieves its
maximum at a point $x_{0}$ in $\Omega$. At this point we may suppose
$|Du|=u_{n}>0$ and $u_{\alpha}(x_{0})$=0 for $1\leq\alpha\leq n-1$, then $\log
u_{n}+\widetilde{A}u$ reaches its maximum. It is obvious that

(3.8) $\frac{u_{ni}}{u_{n}}+\widetilde{A}u_{i}=0.$

Thus for $1\leq\alpha\leq n-1$ we have

(3.9) $u_{n\alpha}(x_{0})=-\widetilde{A}u_{\alpha}(x_{0})u_{n}(x_{0})=0,\quad
u_{nn}(x_{0})=-\widetilde{A}u_{n}^{2}(x_{0}).$

By rotation, we may assume that $u_{ij}(x_{0})$ is diagonal. For any $i$ we
also get

(3.10)
$\Big{(}\frac{u_{nii}}{u_{n}}-\frac{u_{ni}^{2}}{u_{n}^{2}}-\widetilde{A}u_{ii}\Big{)}(x_{0})\leq
0.$

From (2.1) we can find $a_{ij}(x_{0})$ is also diagonal and

(3.11) $\displaystyle a_{nn}$
$\displaystyle=\frac{u_{nn}}{w}\Big{(}1-\frac{2u_{n}^{2}}{w(1+w)}+\frac{u_{n}^{4}}{w^{2}(1+w)^{2}}\Big{)},$
$\displaystyle a_{\alpha\alpha}$
$\displaystyle=\frac{u_{\alpha\alpha}}{w},\quad 1\leq\alpha\leq n-1.$

Differentiating both sides of (3.5) with respect to $x_{n}$, one can deduced
that

(3.12) $\psi_{n}=G^{ij}(\mathcal{A})\frac{\partial a_{ij}}{\partial x_{n}}.$

It follows from (3.11) that

$\displaystyle a_{\alpha\alpha,n}$
$\displaystyle=(\frac{1}{w})_{n}u_{\alpha\alpha}+\frac{1}{w}u_{\alpha\alpha
n}=-\frac{u_{n}u_{nn}}{w^{2}}a_{\alpha\alpha}+\frac{u_{\alpha\alpha n}}{w},$
$\displaystyle a_{nn,n}$
$\displaystyle=(\frac{u_{nn}}{w})_{n}\Big{(}1-\frac{2u_{n}^{2}}{w(1+w)}+\frac{u_{n}^{4}}{w^{2}(1+w)^{2}}\Big{)}+\frac{u_{nn}}{w}\Big{(}1-\frac{2u_{n}^{2}}{w(1+w)}+\frac{u_{n}^{4}}{w^{2}(1+w)^{2}}\Big{)}_{n}$
$\displaystyle=\frac{u_{nnn}}{w^{3}}-\frac{3u_{n}u_{nn}^{2}}{w^{5}}.$

Thus we obtain

$\displaystyle\psi_{n}$
$\displaystyle=\frac{1}{w^{3}}g_{n}u_{nnn}-\frac{3g_{n}u_{n}u_{nn}^{2}}{w^{5}}-\frac{u_{n}u_{nn}}{w^{2}}\sum_{1\leq\alpha\leq
n-1}g_{\alpha}a_{\alpha\alpha}+\frac{1}{w}\sum_{1\leq\alpha\leq
n-1}g_{\alpha}u_{n\alpha\alpha}$
$\displaystyle=\frac{1}{w^{3}}g_{n}u_{nnn}-\frac{3g_{n}u_{n}u_{nn}^{2}}{w^{5}}+\frac{u_{n}u_{nn}}{w^{2}}g_{n}a_{nn}-\frac{u_{n}u_{nn}}{w^{2}}g_{n}a_{nn}$
$\displaystyle~{}~{}~{}~{}-\frac{u_{n}u_{nn}}{w^{2}}\sum_{1\leq\alpha\leq
n-1}g_{\alpha}a_{\alpha\alpha}+\frac{1}{w}\sum_{1\leq\alpha\leq
n-1}g_{\alpha}u_{n\alpha\alpha}$
$\displaystyle=\frac{1}{w^{3}}g_{n}u_{nnn}-\frac{2g_{n}u_{n}u_{nn}^{2}}{w^{5}}-\frac{u_{n}u_{nn}}{w^{2}}g_{i}a_{ii}+\frac{1}{w}\sum_{1\leq\alpha\leq
n-1}g_{\alpha}u_{n\alpha\alpha}.$

As a result, we have

(3.13)
$\frac{2g_{n}u_{n}u_{nn}^{2}}{w^{5}}+\frac{u_{n}u_{nn}}{w^{2}}g_{i}a_{ii}=-\psi_{n}+\frac{1}{w^{3}}g_{n}u_{nnn}+\frac{1}{w}\sum_{1\leq\alpha\leq
n-1}g_{\alpha}u_{n\alpha\alpha}.$

By substituting (3.10) into (3.13), after some computation, we obtain

(3.14)
$\frac{2g_{n}u_{n}u_{nn}^{2}}{w^{5}}+\frac{u_{n}u_{nn}}{w^{2}}g_{i}a_{ii}-\frac{g_{n}u_{nn}^{2}}{w^{3}u_{n}}+\widetilde{A}u_{n}g_{i}a_{ii}\leq-\psi_{n}.$

Arranging (3.14), we get

$\frac{g_{n}u_{nn}^{2}}{u_{n}w^{5}}(w^{2}-2)+\frac{u_{n}u_{nn}}{w^{2}}g_{i}a_{ii}+\widetilde{A}u_{n}g_{i}a_{ii}\leq-\psi_{n}.$

Combining with (3.8), we can get that

(3.15)
$\frac{g_{n}\widetilde{A}^{2}u_{n}^{3}}{w^{5}}(w^{2}-2)+\frac{\widetilde{A}}{w^{2}}u_{n}g_{i}a_{ii}\leq-\psi_{n}.$

It’s obvious that $w=\sqrt{1+u_{n}^{2}}$ at $x_{0}$. By Lemma 2.1, here we
noted that there exists a positive constant $\tau>0$ such that

$\sum_{1\leq\alpha\leq n-1}g_{\alpha}a_{\alpha\alpha}=\sum_{1\leq\alpha\leq
n-1}g_{\alpha}\kappa_{\alpha}=Ae^{-F(\kappa)}\sum_{1\leq\alpha\leq
n-1}\frac{\kappa_{\alpha}}{1+\kappa_{\alpha}^{2}}\geq\tau>0,$

therefore

$\frac{g_{n}\widetilde{A}^{2}u_{n}^{3}}{w^{5}}(u_{n}^{2}-1)+\frac{\widetilde{A}}{w^{2}}u_{n}g_{n}a_{nn}\leq\max|D\psi|.$

From (3.9) and (3.11) we obtain

$\frac{g_{n}\widetilde{A}^{2}u_{n}^{3}}{w^{5}}(u_{n}^{2}-1)-\frac{g_{n}\widetilde{A}^{2}u_{n}^{3}}{w^{5}}=\frac{g_{n}\widetilde{A}^{2}u_{n}^{3}}{w^{5}}(u_{n}^{2}-2)\leq\max|D\psi|.$

Using (3.6) we derive that

$\frac{u_{n}^{3}(u_{n}^{2}-2)}{w^{5}}=\frac{u_{n}^{3}(u_{n}^{2}-2)}{(1+u_{n}^{2})^{5/2}}\leq\frac{\max|D\psi|}{C_{3}\widetilde{A}^{2}}=\frac{1}{2}.$

Hence we conclude that $\sup_{\Omega}|Du|\leq C_{4}$.

Finally, we assume that $x_{0}\in\partial\Omega$ and $\nu$ is the inner unit
normal to $\partial\Omega$ at $x_{0}$ such that

$D_{\nu}u(x_{0})=\sup_{\overline{\Omega}}|Du|.$

We choose $\widetilde{p}>0$ such that line segment
$\\{x_{0}+\widetilde{p}\nu:0\leq\widetilde{p}\leq\widetilde{p}_{0}\\}$ is
contained in $\Omega$. Since $\underline{u}=u=\overline{u}$ on
$\partial\Omega$, it follows from Lemma 3.1 that we obtain

$\frac{\underline{u}(x_{0}+\widetilde{p}\nu)-\underline{u}(x_{0})}{\widetilde{p}}\leq\frac{u(x_{0}+\widetilde{p}\nu)-u(x_{0})}{\widetilde{p}}\leq\frac{\overline{u}(x_{0}+\widetilde{p}\nu)-\overline{u}(x_{0})}{\widetilde{p}}.$

Therefore $D_{\nu}\underline{u}(x_{0})\leq D_{\nu}u(x_{0})\leq
D_{\nu}\overline{u}(x_{0})$. Thus this completes the proof of Lemma 3.5. ∎

From (3.5), we define the equation

(3.16) $\widetilde{G}(D^{2}u,Du)\triangleq G(\mathcal{A})=\psi(x).$

We will use the notation

$\widetilde{G}_{ij}=\frac{\partial\widetilde{G}}{\partial
r_{ij}},\quad\widetilde{G}_{i}=\frac{\partial\widetilde{G}}{\partial p_{i}},$

where $r$ represents for the second derivative and $p$ represents for gradient
variables. By the property of the operator $G$ and the boundedness of $|Du|$,
the operator $\widetilde{G}(\cdot,p)$ satisfies the structure conditions as
same as Lemma 2.14 in [24]. Then another version of Lemma 2.4 in the following
holds.

######  Corollary 3.6.

Assume that $\Omega$, $\varphi$ and $h$ satisfy $(A)$, $\underline{u}$ is an
admissible subsolution satisfying (1.6), $u\in C^{3,\alpha}(\bar{\Omega})$ is
an admissible solution to (1.3) and (1.4). As before, let
$\lambda(D^{2}u)=(\lambda_{1},\cdots,\lambda_{n})$ are the eigenvalues of
Hessian matrix $D^{2}u$ according to $x$. Then there exsits $R_{0}$ depending
only on $\underline{u}$ and $\delta$, such that for any $|\lambda|\geq R_{0}$,
we have

(3.17) $\widetilde{G}_{ij}[\underline{u}_{ij}-u_{ij}]\geq\tau>0,$

where $\tau$ is a constant depending on only $\underline{u}$ and $\delta$.

##  4\. Estimates for Second Derivatives from Bounds on the Boundary

In this section we will show that how to estimate the second derivatives of
$u$ in $\Omega$ if we know bounds for them on $\partial\Omega$. Let us assume
that

######  Lemma 4.1.

Suppose that $\Omega$, $\varphi$ and $h$ satisfy (A). If there exists a
$C^{4}$ admissible subsolution $\underline{u}$ of (1.6), $u\in
C^{3}(\overline{\Omega})$ satisfy (1.3) and (1.4), then we have the following
estimate

(4.1) $\sup_{\partial\Omega}|D^{2}u|\leq C_{5}.$

where $C_{5}$ is a constant depending on $\partial\Omega$, $\underline{u}$,
$h$, $\delta$ and $n$.

The proof of this statements will be deferred until the section 5. If (4.1)
holds then we have

######  Lemma 4.2.

Suppose that $\Omega$, $\varphi$ and $h$ satisfy (A). If there exists a
$C^{4}$ admissible subsolution $\underline{u}$ satisfying (1.6), $u\in
C^{3}(\overline{\Omega})$ satisfy (1.3) and (1.4), then we have the following
estimate

(4.2) $\sup_{\overline{\Omega}}|D^{2}u|\leq C_{6},$

where $C_{6}$ is a constant depending on $\overline{\Omega}$,
$\|\underline{u}\|_{C^{4}(\overline{\Omega})}$,
$\|h\|_{C^{2}(\overline{\Omega})}$, $\delta$.

We present useful knowledge for the proof of Lemma 4.2.

######  Lemma 4.3.

There exists a positive constant $\Lambda$ only depending on $\delta$ such
that

$\sum_{i}g_{i}=Ae^{-AF(\kappa)}\sum_{i}\frac{1}{1+\kappa_{i}^{2}}\geq\Lambda>0.$

###### Proof.

Using Lemma 2.1 and $\sup_{\overline{\Omega}}F<n\frac{\pi}{2}$, we have an
estimate for the smallest eigenvalue

$|\kappa_{n}|\leq C_{\delta},$

where $C_{\delta}$ is a positive constant depending on $\delta$. Therefore
there exist two positive constants $\Lambda_{1}$ and $\Lambda$ only depending
on $\delta$ such that

$\sum_{i}g_{i}=Ae^{-AF(\kappa)}\sum_{i}\frac{1}{1+\kappa_{i}^{2}}\geq\Lambda_{1}\frac{1}{1+\kappa_{n}^{2}}\geq\Lambda_{1}\frac{1}{1+C_{\delta}^{2}}\geq\Lambda>0.$

∎

In order to prove Lemma 4.2, we need to use the knowledge of [3] as follows.
From the proof of Lemma 3.4, we have a bound for

(4.3) $C_{7}=2\max_{\overline{\Omega}}w,$

where $w=\sqrt{1+|Du|^{2}}$. Let

(4.4) $\widetilde{\theta}\triangleq\frac{1}{w},$ (4.5)
$a=\frac{1}{k}\triangleq\frac{1}{2}\min_{\overline{\Omega}}\widetilde{\theta}.$

Obviously we have

(4.6) $\frac{1}{\widetilde{\theta}-a}\leq\frac{1}{a}=C_{7}.$

For any $x^{0}\in\Omega$, it is convenient to use the new coordinates by
defining the surface by $\zeta(y)$, where $y$ are tangential coordinates to
the surface at the point $(x^{0},u(x^{0}))$. We choose a local orthonormal
frame $\\{e_{i}\\}$ for $1\leq i\leq n$ in neighborhood of $(x^{0},u(x^{0}))$
in $\zeta(y)$, and $e_{n+1}$ is the normal. Introduce the new orthonormal
vectors

$\varepsilon_{1},\cdots,\varepsilon_{n},\varepsilon_{n+1},$

where $\varepsilon_{n+1}$ being the normal vector at $x^{0}$ with
$\varepsilon_{n+1}=\frac{1}{w}(-u_{1},\cdots,-u_{n},1)$ and $\varepsilon_{1}$
corresponding to the tangential direction at $x^{0}$ with largest principal
curvature. After proper rotation, we represent the surface near
$(x^{0},u(x^{0}))$ by tangential coordinates $y_{1},\cdots,y_{n}$ and
$\zeta(y)$ (summation is of the from 1 to $n$)

$x_{j}e_{j}+u(x)e_{n+1}=x_{j}^{0}e_{j}+u(x^{0})e_{n+1}+y_{j}\varepsilon_{j}+\zeta(y)\varepsilon_{n+1},$

so $\nabla\zeta(0)=0$ and $x=x^{0}$ if and only if $y=0$. In this case we set
that $\Upsilon=\sqrt{1+|\nabla\zeta|^{2}}$. Following [20], the first
fundamental form of the graph of $\zeta(y)$ is

$\widetilde{g}_{ij}=\delta_{ij}+\zeta_{i}\zeta_{j}.$

The second fundamental form is given by

$\widetilde{h}_{ij}=\frac{\zeta_{ij}}{\sqrt{1+|\nabla\zeta|^{2}}}.$

Thus the normal curvature in the $\varepsilon_{1}$ direction is

$\kappa=\frac{\widetilde{h}_{11}}{\widetilde{g}_{11}}=\frac{\zeta_{11}}{(1+\zeta_{1}^{2})\Upsilon}.$

In the $y$-coordinate we have the normal vector

(4.7)
$N=-\frac{1}{\Upsilon}\zeta_{j}\varepsilon_{j}+\frac{1}{\Upsilon}\varepsilon_{n+1},$
(4.8) $\widetilde{\theta}=\frac{1}{w}=N\cdot e_{n+1}=\frac{1}{\Upsilon
w}-\frac{1}{\Upsilon}\sum a_{j}\zeta_{j},$

where $a_{j}=\varepsilon_{j}\cdot e_{n+1}$, thus $\sum a_{j}^{2}\leq 1$. And
the equation (3.5) locally reads as

(4.9) $G(\mathcal{A})=\widetilde{\psi}(y),$

where $\widetilde{\psi}(y)\triangleq\psi(x).$ By $\nabla\zeta(0)=0$, it
follows from (4.8) that at $y=0$ we have

(4.10) $\widetilde{\theta}_{i}=-a_{i}\zeta_{ii},\quad i=1,\cdots,n.$ (4.11)
$\widetilde{\theta}_{ii}=-a_{j}\zeta_{jii}-\frac{\zeta_{ii}^{2}}{w},\quad
i=1,\cdots,n.$

######  Lemma 4.4.

We claim that $\zeta_{1j}(x_{0})=0$ for $j>1$ at $x_{0}\in\partial\Omega$.

###### Proof.

Let $e_{\theta}=e_{1}\cos\theta+e_{j}\sin\theta$. Then

$\zeta_{e_{\theta}e_{\theta}}=\zeta_{11}\cos^{2}\theta+2\zeta_{1j}\cos\theta\sin\theta+\zeta_{jj}\sin^{2}\theta$

has a maximum at $\theta=0$. It follows that

$\frac{d}{d\theta}\big{(}\zeta_{e_{\theta}e_{\theta}}(x_{0})\big{)}\Big{|}_{\theta=0}=0.$

This gives $\zeta_{1j}(x_{0})=0$. ∎

Using the knowledge of [3] as above, then we show the following

Proof of Lemma 4.2.

Set

$M=\max_{x\in\overline{\Omega},1\leq i\leq
n}\frac{1}{\widetilde{\theta}-a}\kappa_{i}(x),$

where the maximum is taken over all principal curvatures $\kappa_{i}$. We
assume that the maximum is attained at some point $x^{0}\in\Omega$ and $M>0$.
At this point corresponding to $y=0$, the function

$\frac{1}{\widetilde{\theta}-a}\frac{\zeta_{11}}{(1+\zeta_{1}^{2})\Upsilon}$

takes its maximum value equal to $M$ by the previous assumption. At the point
$y=0$, because $y_{1}$ direction is a direction of principal curvature, then
by Lemma 4.4, we get that $\zeta_{1j}=0$ for $j>1$. By rotating the
$\varepsilon_{2},\cdots,\varepsilon_{n}$, without loss of generality we may
have that $\zeta_{ij}(0)$ is diagonal. At $y=0$, the function
$\ln\frac{1}{\widetilde{\theta}-a}\frac{\zeta_{11}}{(1+\zeta_{1}^{2})\Upsilon}$
reaches its maximum value, and taking the derivative twice in a row can be
concluded as follows

(4.12)
$\frac{\zeta_{11i}}{\zeta_{11}}-\frac{\widetilde{\theta}_{i}}{\widetilde{\theta}-a}-\frac{2\zeta_{1}\zeta_{1i}}{1+\zeta_{1}^{2}}-\frac{\Upsilon_{i}}{\Upsilon}=0,\quad
i=1,\cdots n,$ (4.13)
$0\geq\frac{\zeta_{11ii}}{\zeta_{11}}-\frac{\zeta_{11i}^{2}}{\zeta_{11}^{2}}-\Big{(}\frac{\widetilde{\theta}_{i}}{\widetilde{\theta}-a}\Big{)}_{i}-2\zeta_{1i}^{2}-\zeta_{ii}^{2},\quad
i=1,\cdots n.$

Observing (2.4), at the origin we also see that $a_{ij}=\zeta_{ij}$ is
diagonal and

(4.14) $\displaystyle\frac{\partial a_{ij}}{\partial y_{k}}$
$\displaystyle=a_{ij,k}=\zeta_{ijk},$
$\displaystyle\frac{\partial^{2}a_{ij}}{\partial^{2}y_{1}}$
$\displaystyle=a_{ij,11}=\zeta_{ij11}-\zeta_{11}^{2}\Big{(}\zeta_{ij}+\delta_{i1}\zeta_{1j}+\delta_{j1}\zeta_{1i}\Big{)}.$

It should be noted that at a matrix $\mathcal{A}=[a_{ij}]$ which is diagonal,

(4.15) $\frac{\partial G}{\partial a_{ij}}=\frac{\partial
G}{\partial\kappa_{i}}\delta_{ij}=g_{i}\delta_{ij}.$

We proceed the differentiate the equation

$G(\mathcal{A})=\widetilde{\psi}(y),$

then there holds

$\frac{\partial G}{\partial a_{ij}}\frac{\partial a_{ij}}{\partial
y_{k}}=\widetilde{\psi}_{k}(y).$

Using (4.14) and (4.15), we find that at $y=0$,

$g_{i}\zeta_{iik}=\widetilde{\psi}_{k}$

for all $k$. Using the concavity of $G$ and (4.14), one can see that at $y=0$,

$\widetilde{\psi}_{11}=\frac{\partial^{2}G}{\partial a_{ij}\partial
a_{kl}}\frac{\partial a_{ij}}{\partial y_{1}}\frac{\partial a_{kl}}{\partial
y_{1}}+\frac{\partial G}{\partial a_{ij}}a_{ij,11}\leq\frac{\partial
G}{\partial a_{ij}}a_{ij,11}.$

Thus we have, at $y=0$,

(4.16) $\displaystyle\widetilde{\psi}_{11}$ $\displaystyle\leq\frac{\partial
G}{\partial
a_{ij}}a_{ij,11}=g_{i}(\zeta_{ii11}-\zeta_{ii}\xi_{11}^{2})-2g_{1}\zeta_{11}^{3}$
$\displaystyle\leq
g_{i}\Big{(}\frac{\zeta_{i11}^{2}}{\zeta_{11}}+\zeta_{11}(\frac{\widetilde{\theta}_{i}}{\widetilde{\theta}-a})_{i}+2\zeta_{11}\zeta_{1i}^{2}+\zeta_{11}\zeta_{ii}^{2}-\zeta_{ii}\zeta_{11}^{2}\Big{)}-2g_{1}\zeta_{11}^{3}$
$\displaystyle=\zeta_{11}g_{i}\Big{(}\frac{\zeta_{i11}^{2}}{\zeta_{11}^{2}}+(\frac{\widetilde{\theta}_{i}}{\widetilde{\theta}-a})_{i}+\zeta_{ii}^{2}-\zeta_{11}\zeta_{ii}\Big{)}$
$\displaystyle=\zeta_{11}g_{i}\Big{[}(\frac{\widetilde{\theta}_{i}}{\widetilde{\theta}-a})^{2}+(\frac{\widetilde{\theta}_{i}}{\widetilde{\theta}-a})_{i}+\zeta_{ii}^{2}-\zeta_{11}\zeta_{ii}\Big{]}$
$\displaystyle=\zeta_{11}g_{i}\Big{[}\frac{\widetilde{\theta}_{ii}}{\widetilde{\theta}-a}+\zeta_{ii}^{2}-\zeta_{11}\zeta_{ii}\Big{]}$
$\displaystyle=\zeta_{11}g_{i}\Big{[}\frac{1}{\widetilde{\theta}-a}(-a_{j}\zeta_{jii}-\frac{\zeta_{ii}^{2}}{w})+\zeta_{ii}^{2}-\zeta_{11}\zeta_{ii}\Big{]}$
$\displaystyle=-\zeta_{11}a_{j}g_{i}\frac{\zeta_{jii}}{\widetilde{\theta}-a}+\zeta_{11}g_{i}\Big{[}(1-\frac{1}{\widetilde{\theta}-a}\frac{1}{w})\zeta_{ii}^{2}-\zeta_{11}\zeta_{ii}\Big{]}$
$\displaystyle=-\zeta_{11}a_{j}\widetilde{\psi}_{j}\frac{1}{\widetilde{\theta}-a}+\zeta_{11}g_{i}\Big{[}(1-\frac{\widetilde{\theta}}{\widetilde{\theta}-a})\zeta_{ii}^{2}-\zeta_{11}\zeta_{ii}\Big{]}$
$\displaystyle=-\zeta_{11}a_{j}\widetilde{\psi}_{j}\frac{1}{\widetilde{\theta}-a}-\zeta_{11}g_{i}\zeta_{ii}^{2}\frac{a}{\widetilde{\theta}-a}-\zeta_{11}^{2}g_{i}\zeta_{ii}.$

Here the first step follows from (4.14) while the second step comes from
(4.13). Using (4.16) we have

(4.17)
$\zeta_{11}g_{i}\zeta_{ii}^{2}\frac{a}{\widetilde{\theta}-a}+\zeta_{11}^{2}g_{i}\zeta_{ii}\leq-\zeta_{11}a_{j}\widetilde{\psi}_{j}\frac{1}{\widetilde{\theta}-a}-\widetilde{\psi}_{11}.$

Recall that $\widetilde{\psi}(y)=-e^{-Ah(y)}$, a routine computation gives
rise to

$\widetilde{\psi}_{11}=-A^{2}e^{-Ah(y)}(\frac{\partial h}{\partial
y_{1}})^{2}+Ae^{-Ah(y)}\frac{\partial^{2}h}{\partial^{2}y_{1}}.$

Since $\sum a_{j}^{2}<1$, we now choose a sufficiently large constant $A$ so
that

$-\zeta_{11}a_{j}\widetilde{\psi}_{j}\frac{1}{\widetilde{\theta}-a}-\widetilde{\psi}_{11}\leq
C_{8}(1+\zeta_{11}),$

where $C_{8}$ is a positive constant depending on $\overline{\Omega}$,
$\widetilde{\psi}$, $\underline{u}$ and $\widetilde{\theta}$. Thus

$\zeta_{11}g_{i}\zeta_{ii}^{2}\frac{a}{\widetilde{\theta}-a}+\zeta_{11}^{2}g_{i}\zeta_{ii}\leq
C_{8}(1+\zeta_{11}).$

Since (4.17) and the definition of $M$, we have

(4.18)
$Ma\sum_{i}g_{i}\zeta_{ii}^{2}+M^{2}(\widetilde{\theta}-a)^{2}\sum_{i}g_{i}\zeta_{ii}\leq
C_{9}(1+M).$

where $C_{9}$ is a positive constant depending only on the known data. With
the aid of (3.3), we can see that
$\sum_{i}g_{i}\zeta_{ii}=\sum_{i}g_{i}\kappa_{i}\geq-\widetilde{c}$. It
follows that from (4.18) that

$Ma\sum_{i}g_{i}\zeta_{ii}^{2}\leq\widetilde{c}M^{2}(\widetilde{\theta}-a)^{2}+C_{9}(1+M).$

Using (3.4) and the definition of $M$, there exists a positive constant
$C_{10}$ such that

$C_{10}M^{3}\sum_{i}g_{i}\leq\widetilde{c}M^{2}(\widetilde{\theta}-a)^{2}+C_{9}(1+M).$

By Lemma 4.3, we obtain that $\sum_{i}g_{i}\geq\Lambda>0$, then

$M\leq C_{6},$

where a suitable constant $C_{6}$ depends on $\overline{\Omega}$,
$\|\underline{u}\|_{C^{4}(\overline{\Omega})}$,
$\|h\|_{C^{2}(\overline{\Omega})}$, $\delta$. The proof is completed.

##  5\. Proof of Lemma 4.1

The proof of Lemma 4.1 follows from the following steps.

First, we need to estimate the pure tangential second derivatives. Suppose
that the point is the origin and the $x_{n}-$axis is the inner normal there.
We may assume that the boundary near 0 is represented by

(5.1)
$x_{n}=\rho(x^{\prime})=\frac{1}{2}\sum_{\alpha,\beta<n}\rho_{\alpha\beta}(0)x_{\alpha}x_{\beta}+\mathrm{O}(|x^{\prime}|^{3}),$

where $x^{\prime}=(x_{1},x_{2},\cdots,x_{n-1})$. Using the Dirichlet boundary
condition, we get

(5.2) $(u-\underline{u})(x^{\prime},\rho(x^{\prime}))=0.$

Taking two derivatives for both sides of (5.2), we obtain

$(u-\underline{u})_{\alpha\beta}(0)=-(u-\underline{u})_{n}(0)\rho_{\alpha\beta}(0),\quad
1\leq\alpha,\beta\leq n-1.$

From the boundary gradient estimate it follows that

$|u_{\alpha\beta}(0)|\leq C_{11},\quad\alpha,\beta<n.$

The next thing to do in the proof is to estimate the mixed normal-tangential
derivative $u_{\alpha n}(0)$ for $\alpha<n$. Using the pre-knowledge in the
second section, we can obtain

(5.3) $\widetilde{G}_{ij}=G^{kl}\frac{\partial a_{kl}}{\partial
r_{ij}}=\frac{1}{w}b^{ik}G^{kl}b^{lj},$

and

$\widetilde{G}_{i}=G^{kl}\frac{\partial a_{kl}}{\partial
p_{i}}=G^{kl}\frac{\partial}{\partial
p_{i}}\Big{(}\frac{1}{w}b^{km}b^{ql}\Big{)}u_{mq}.$

Using a simple calculation, we have

(5.4) $\displaystyle\widetilde{G}_{i}$
$\displaystyle=-\frac{u_{i}}{w^{2}}G^{kl}a_{kl}-\frac{2}{w}G^{kl}a_{lm}b^{ik}u_{m}$
$\displaystyle=-\frac{u_{i}}{w^{2}}\sum_{j}g_{j}\kappa_{j}-\frac{2}{w}G^{kl}a_{lm}b^{ik}u_{m}.$

Recalling (3.7), we see that $|Du|$ is bounded and the eigenvalues of
$[b^{ij}]$ is bounded between two controlled positive constants. Since (5.4),
we see that in fact

(5.5) $\sum_{i}|\widetilde{G}_{i}|\leq C_{12}\sum g_{i}|\kappa_{i}|.$

Moreover, according to Lemma 3.5, (2.3) and (5.3), there exist two positive
constants $C_{13}$, $C_{14}$ such that

(5.6) $C_{13}\sum_{i}G^{ii}\leq\sum_{i}\widetilde{G}_{ii}\leq
C_{14}\sum_{i}G^{ii}.$

We also have

(5.7)
$\sum_{i}\frac{\partial\widetilde{G}}{\partial\lambda_{i}}\lambda_{i}=\widetilde{G}_{ij}u_{ij}=\frac{1}{w}b^{ik}G^{kl}b^{lj}u_{ij}=G^{kl}a_{kl}=\sum
g_{i}\kappa_{i}.$

where $(\lambda_{1},\cdots,\lambda_{n})$ are the eigenvalues of Hessian matrix
$D^{2}u$ of $x\in\Omega$. From Lemma 2.1, we have some positive constants
$s_{1}$ , $s_{2}$ and $s_{3}$ such that

(5.8) $0\leq\sum_{i}\frac{\partial G}{\partial\kappa_{i}}\leq s_{1},$

and

$s_{2}\leq\sum_{i}\frac{\partial
G}{\partial\kappa_{i}}\kappa_{i}^{2}=Ae^{-AF(\kappa)}\sum_{i}\frac{\kappa_{i}^{2}}{1+\kappa_{i}^{2}}=Ae^{-AF(\kappa)}\sum_{i}\Big{(}1-\frac{1}{1+\kappa_{i}^{2}}\Big{)}\leq
s_{3}.$

By taking (5.6) and (5.8) into consideration, then there exists a positive
constant $C_{15}$ such that

(5.9) $\sum\widetilde{G}_{ii}\leq C_{15}.$

We consider the operator

$L=\widetilde{G}_{ij}D_{i}D_{j}+\widetilde{G}_{i}D_{i}.$

It is necessary to obtain a version of differential inequality for the later
construction of auxiliary functions.

By the argument as before in this section, one can conclude that for any point
$x_{0}\in\partial\Omega$, we rotate the graph such that at $x_{0}$,
$D(u-\underline{u})=0$. But the principle curvatures are invariant through
such transformation. Then for any sufficiently small positive constant
$\varepsilon>0$, there exists $\delta^{\prime}>0$ depending only on
$\varepsilon$, $\Omega$, $\underline{u}$ and $h$ such that in $\Omega\cap
B_{\delta^{\prime}}(x_{0})$, we have

(5.10) $|D(u-\underline{u})|\leq\varepsilon.$

Based on the above argument, we obtain the following useful differential
inequality.

######  Corollary 5.1.

There exists uniform positive constant $C_{16}$ such that for $|\lambda|\geq
R_{0}$, we have

(5.11) $L(u-\underline{u})\leq-C_{16},\quad\mathrm{in}\quad\Omega\cap
B_{\delta^{\prime}}(x_{0}).$

where $R_{0}$ is a positive constant depending on $\tau$.

###### Proof.

Using (5.10) and (5.5), we can see that there exists sufficiently small
positive constant $\varepsilon_{1}$ with $\varepsilon_{1}\leq\frac{\tau}{2}$
such that in $\Omega\cap B_{\delta^{\prime}}(0)$, we have

(5.12) $\widetilde{G}_{i}(u-\underline{u})_{i}\leq\varepsilon_{1}.$

From the Corollary 3.6 and (5.12), for $|\lambda|\geq R_{0}$ we have

(5.13) $\displaystyle L(u-\underline{u})$
$\displaystyle=\widetilde{G}_{ij}(u-\underline{u})_{ij}+\widetilde{G}_{i}(u-\underline{u})_{i}$
$\displaystyle\leq-\tau+\varepsilon_{1}$ $\displaystyle\leq-C_{16}$

where $C_{16}=\frac{\tau}{2}$. ∎

######  Lemma 5.2.

There exist some uniform positive constants $\delta^{\prime}$ sufficiently
small depending only on $\underline{u},h,\Omega$ and choose positive constants
$t_{1},t_{2},\varepsilon_{2}$ depending on $\delta^{\prime}$ such that the
function

$v=t_{1}(u-\underline{u})+t_{2}d-d^{2},$

satisfy

$\displaystyle Lv$ $\displaystyle\leq-\varepsilon_{2},$
$\displaystyle\quad\mathrm{in}\quad\Omega\cap B_{\delta^{\prime}}(0),$
$\displaystyle v$ $\displaystyle\geq 0,$
$\displaystyle\quad\mathrm{on}\quad\partial(\Omega\cap
B_{\delta^{\prime}}(0)).$

where $d(x)=d(x,\partial\Omega)$ is the distance function.

###### Proof.

Note that $v=t_{1}(u-\underline{u})+(t_{2}-d)d$ and let
$t_{2}=2\delta^{\prime}>d$, it is obvious that $v\geq 0$ on
$\partial(\Omega\cap B_{\delta^{\prime}}(0))$. Since

$\widetilde{G}_{ij}D_{i}D_{j}d^{2}=2d\widetilde{G}_{ij}D_{i}D_{j}d+2\widetilde{G}_{ij}D_{i}dD_{j}d,\quad\quad\mathrm{in}\quad\Omega\cap
B_{\delta^{\prime}}(0),$

then we have

(5.14) $\displaystyle Lv$
$\displaystyle=t_{1}\Big{[}\widetilde{G}_{ij}(u_{ij}-\underline{u}_{ij})+\widetilde{G}_{i}(u_{i}-\underline{u}_{i})\Big{]}+t_{2}\widetilde{G}_{ij}D_{i}D_{j}d+t_{2}\widetilde{G}_{i}D_{i}d$
$\displaystyle~{}~{}-2d\widetilde{G}_{ij}D_{i}D_{j}d-2\widetilde{G}_{ij}D_{i}dD_{j}d-2d\widetilde{G}_{i}D_{i}d.$
$\displaystyle=t_{1}\Big{[}\widetilde{G}_{ij}(u_{ij}-\underline{u}_{ij})+\widetilde{G}_{i}(u_{i}-\underline{u}_{i})\Big{]}+(t_{2}-2d)\widetilde{G}_{ij}D_{i}D_{j}d$
$\displaystyle~{}~{}+(t_{2}-2d)\widetilde{G}_{i}D_{i}d-2\widetilde{G}_{ij}D_{i}dD_{j}d.$

Since $\partial\Omega$ is a $C^{4}$ bounded hypersurface, we can assume that
$d(x)$ is $C^{4}$ bounded in $\Omega\cap B_{\delta^{\prime}}(0)$. It can be
verified using a result of G-T (see [26] , Lemma 14.17) that $d(x)$ satisfy

$\displaystyle Dd(x)$ $\displaystyle=(0,\cdots,0,1),$ $\displaystyle
D^{2}d(x)$
$\displaystyle=diag(-\frac{\kappa_{1}}{1-\kappa_{1}d},\cdots,-\frac{\kappa_{n-1}}{1-\kappa_{n-1}d},0),$

in $\Omega\cap B_{\delta^{\prime}}(0)$, where $\kappa_{1},\cdots,\kappa_{n-1}$
are the principle curvature on $\partial\Omega$. Combining the boundedness of
$\widetilde{G}_{i}$ and (5.9), we obtain

$\displaystyle|\widetilde{G}_{ij}D_{i}D_{j}d|\leq
C_{17},\quad|\widetilde{G}_{i}D_{i}d|\leq C_{18},$

where $C_{17}$ and $C_{18}$ are positive constants depending on $\Omega$,
$\delta^{\prime}$, $\delta$ and $h$. There are two cases to consider: Firstly,
we consider that $|\lambda|\leq R_{0}$, where $R_{0}$ is a constant from
Corollary 3.6. Then there holds $|\kappa|\leq\tilde{R}_{0}$ for some constant
$\tilde{R}_{0}$ depending only on $R_{0}$ and the boundedness of $|Du|$.
Without loss of generality, we assume that $[a_{ij}]$ is diagonal. By making
use of (5.3) and the boundedness of $|Du|$, it is also easy to see that

$\widetilde{G}_{ij}D_{i}dD_{j}d=\frac{1}{1+\kappa_{l}^{2}}\delta_{kl}\frac{1}{w}b^{ik}b^{jl}D_{i}dD_{j}d\geq\frac{C_{19}}{1+\tilde{R}_{0}^{2}}.$

For $|\lambda|\leq R_{0}$, we have

$\displaystyle|\widetilde{G}_{ij}(u_{ij}-\underline{u}_{ij})|$
$\displaystyle\leq
C_{20},\quad|\widetilde{G}_{i}(u_{i}-\underline{u}_{i})|\leq C_{21}.$

It follows from (5.14) that

$\displaystyle Lv$ $\displaystyle\leq
t_{1}(C_{20}+C_{21})+(t_{2}-2d)C_{17}+(t_{2}-2d)C_{18}-\frac{2C_{19}}{1+\tilde{R}_{0}^{2}}$
$\displaystyle\leq
t_{1}C_{22}+(t_{2}-2d)C_{23}-\frac{2C_{19}}{1+\tilde{R}_{0}^{2}}$
$\displaystyle\leq
t_{1}C_{22}+t_{2}C_{23}-\frac{2C_{19}}{1+\tilde{R}_{0}^{2}}.$

Secondly, if $|\lambda|\geq R_{0}$, by using Corollary 5.1 we can obtain

$\displaystyle Lv$ $\displaystyle\leq-
C_{16}t_{1}+(t_{2}-2d)C_{17}+(t_{2}-2d)C_{18}-2\widetilde{G}_{ij}D_{i}dD_{j}d$
$\displaystyle\leq-
C_{16}t_{1}+(t_{2}-2d)C_{23}-2\widetilde{G}_{ij}D_{i}dD_{j}d$
$\displaystyle\leq-C_{16}t_{1}+t_{2}C_{23}.$

We may fix $t_{1}$ and $t_{2}$ such that $C_{16}t_{1}=2t_{2}C_{23}$. Then for
$|\lambda|\leq R_{0}$, we have

$\displaystyle Lv$ $\displaystyle\leq
t_{2}C_{24}-\frac{2C_{19}}{1+\tilde{R}_{0}^{2}}=2\delta^{\prime}-\frac{2C_{19}}{1+\tilde{R}_{0}^{2}},$

and for $|\lambda|\geq R_{0}$, we get

$\displaystyle Lv$ $\displaystyle\leq-t_{2}C_{23}=-2\delta^{\prime}C_{23}.$

Let

$\delta^{\prime}\leq\frac{C_{19}}{2(1+\tilde{R}_{0}^{2})}.$

Then the proof is completed. ∎

By making use of Lemma 2.1 (i), we find that it can’t be all negative for
$\kappa$. Therefore, we can present

######  Remark 5.3.

By the assumption, $\lambda=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$
satisfy $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{n}$ and
$\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{n-1}\geq 0$. We claim that
$\lambda_{n}<-R_{0}$ is not satisfied.

In order to estimate $|u_{\alpha n}(0)|$, we introduce the motivation for the
vector field $T_{\alpha}$ by

$T_{\alpha}=\frac{\partial}{\partial
x_{\alpha}}+\sum_{\beta<n}\rho_{\alpha\beta}(0)\big{(}x_{\beta}\frac{\partial}{\partial
x_{n}}-x_{n}\frac{\partial}{\partial
x_{\beta}}\big{)},\quad\alpha\in\\{1,2,\cdots,k-1\\}.$

Furthermore, it follows immediately from (5.1) that for $\alpha<n$, on
$\partial\Omega$ near $0$ we have see

$T_{\alpha}=[\frac{\partial}{\partial x_{\alpha}}+\frac{\partial\rho}{\partial
x_{\alpha}}\frac{\partial}{\partial
x_{n}}]+\mathrm{O}(|x^{\prime}|^{2})\frac{\partial}{\partial
x_{n}}-\sum_{\beta<n}\rho_{\alpha\beta}(0)\rho(x^{\prime})\frac{\partial}{\partial
x_{\beta}}.$

Now, we can show that

######  Lemma 5.4.

The function $u-\underline{u}$ satisfies the following estimates

(5.15) $|T_{\alpha}(u-\underline{u})|\leq
C_{24}|x^{\prime}|^{2},\quad\mathrm{on}\quad\partial(\Omega\cap
B_{\delta^{\prime}}(0)).$ (5.16) $|LT_{\alpha}(u-\underline{u})|\leq
C_{25}(1+\sum_{i=1}^{n}\widetilde{G}_{ii})\leq
C_{26},\quad\mathrm{in}\quad\Omega\cap B_{\delta^{\prime}}(0).$

###### Proof.

Since $u=\underline{u}$ on $\partial\Omega$, then the first inequality follows
directly from $C^{1}$ estimate of Lemma 3.5. In the following we prove the
last inequality. Since $G(\kappa)$ is invariant under rotation, it follows
that for the operator

$X_{\beta}=x_{\beta}\frac{\partial}{\partial
x_{n}}-x_{n}\frac{\partial}{\partial x_{\beta}},$

which is the infinitesimal generator of a rotation. We can apply the operator
$X_{\beta}$ to the equation $\widetilde{G}(D^{2}u,Du)=\psi$. Consequently we
have

$X_{\beta}\psi=\widetilde{G}_{ij}D_{i}D_{j}(X_{\beta}u)+\widetilde{G}_{i}D_{i}(X_{\beta}u),$

from which it follows that

$T_{\alpha}\psi=L(T_{\alpha}u).$

Since $\widetilde{G}_{ii}$ is bounded, we have

$\displaystyle|LT_{\alpha}(u-\underline{u})|$
$\displaystyle\leq|LT_{\alpha}u|+|LT_{\alpha}\underline{u}|$
$\displaystyle=|T_{\alpha}\psi|+|LT_{\alpha}\underline{u}|$ $\displaystyle\leq
C_{25}\Big{(}1+\sum_{i}\widetilde{G}_{ii}\Big{)}$ $\displaystyle\leq
C_{26},\quad\mathrm{in}\quad\Omega\cap B_{\delta^{\prime}}(0).$

∎

The function $\wp$ introduced in the next Lemma will be the main part of a
barrier function which we shall construct in the following.

######  Lemma 5.5.

There exist two positive constants $\widehat{A}\gg\widehat{B}\gg 1$ such that

$\wp=\widehat{A}v+\widehat{B}|x|^{2}\pm T_{\alpha}(u-\underline{u})$

satisfies the inequalities

$\displaystyle L\wp$ $\displaystyle\leq 0,\quad\mathrm{in}\quad\Omega\cap
B_{\delta^{\prime}}(0),$ $\displaystyle\wp$ $\displaystyle\geq
0,\quad\mathrm{on}\quad\partial(\Omega\cap B_{\delta^{\prime}}(0)),$

where $\delta^{\prime}$ is a constant from Corollary 5.1.

###### Proof.

First the condition $\wp\geq 0$ on $\partial(\Omega\cap
B_{\delta^{\prime}}(0))$ follows if

$\widehat{B}|x|^{2}\pm T_{\alpha}(u-\underline{u})\geq
0,\quad\mathrm{on}\quad\partial(\Omega\cap B_{\delta^{\prime}}(0)).$

In view of Lemma 5.4, this can be arranged by choosing $\widehat{B}$
sufficiently large. By making use of Lemma 5.2 and Lemma 5.4, we can see that
the property $L\wp\leq 0$ now follows from the inequality

$\displaystyle L\wp$ $\displaystyle=\widehat{A}Lv+\widehat{B}L(|x|^{2})\pm
LT_{\alpha}(u-\underline{u})$
$\displaystyle\leq-\widehat{A}\varepsilon_{2}+\widehat{B}(2\widetilde{G}_{ii}+2\widetilde{G}_{i}x_{i})+C_{26}$
$\displaystyle\leq-\widehat{A}\varepsilon_{2}+\widehat{B}C_{27}+C_{26}$

with holds if $\widehat{A}$ is sufficiently large. ∎

The maximum principle applied to Lemma 5.5 shows that $\wp\geq 0$ in
$\Omega\cap B_{\delta^{\prime}}(0)$. Since
$\Big{(}\widehat{A}v+\widehat{B}|x|^{2}\pm
T_{\alpha}(u-\underline{u})\Big{)}(0)=0$, it follows that

$\frac{\partial}{\partial x_{n}}\big{(}\widehat{A}v+\widehat{B}|x|^{2}\pm
T_{\alpha}(u-\underline{u})\big{)}(0)\geq 0.$

Thus we obtain

$|\frac{\partial}{\partial
x_{n}}(T_{\alpha}(u-\underline{u}))(0)|\leq|\frac{\partial}{\partial
x_{n}}(\widehat{A}v+\widehat{B}|x|^{2})(0)|,$

and this gives

$|u_{\alpha n}(0)-\underline{u}_{\alpha
n}(0)|\leq|\widehat{A}u_{n}(0)|+|\sum_{\beta<n}\rho_{\alpha\beta}(0)(u-\underline{u})_{\beta}(0)|\leq
C_{28}.$

Hence we can obtain the mixed second-order derivative bound $|u_{\alpha
n}(0)|\leq C_{29}$ for all $\alpha<n$.

Finally, estimating the remaining second derivative $|u_{nn}|$ of the function
$u$ is somewhat complicated. We find that the principal curvatures admit a
uniform lower bound, i.e. $\mathcal{A}=[a_{ij}]\geq-C_{30}$, since the Lemma
2.1 and $F(\kappa)=\sum_{i}\arctan\kappa_{i}\geq(n-2)\frac{\pi}{2}+\delta$ in
$\overline{\Omega}$. Recall that $a_{ij}=\frac{1}{w}b^{ik}u_{kl}b^{lj}$ and
the gradient estimate, we can obtain that $D^{2}u\geq-C_{31}$. It suffices to
derive an upper bound

$u_{nn}\leq C_{32},\quad\mathrm{on}\quad\partial\Omega.$

For discussion purposes, we need to give some definition of the operator as
follows. Assume that the $(n-1)\times(n-1)$ unper block matrix
$u_{\alpha\beta}$ is diagonal. Before we deal with $D^{2}u$, using the
boundedness of $Du$, we consider $\widetilde{G}$ and $\widehat{G}$ as
functions through taking $D^{2}u$ as variable all alone. In other word, we can
define that

$\widetilde{G}[u_{ij}]:=\widetilde{G}(D^{2}u,p),$

and

$\widehat{G}[u_{\alpha\beta}]=\lim_{\mathrm{t}\rightarrow\infty}\widetilde{G}(D^{2}u+\mathrm{t}q\otimes
q,p),$

where $q=(0,\ldots,0,1)$ and $p$ represents for gradient variables.

Similar to Lemma 3.1, we can deduce that

######  Corollary 5.6.

The operator $\widehat{G}[u_{\alpha\beta}]$ is concave on the set
$u_{\alpha\beta}(\partial\Omega)$.

We need to consider

$\Theta=\min_{x\in\partial\Omega}\widehat{G}[u_{\alpha\beta}(x)]-\psi(x)$

and

$\mathbf{c}=\min_{x\in\partial\Omega}(\widehat{G}[\underline{u}_{\alpha\beta}]-\widetilde{G}[\underline{u}_{ij}])$

that demonstrate $u_{nn}\leq C_{32}$ on $\partial\Omega$. It’s obvious that
$\mathbf{c}>0$. According to an idea of Trudinger [25], we use the following
Lemma that implies the boundary $C^{2}$ estimate.

######  Lemma 5.7.

There exist some positive constant $\omega_{0}>0$ depending only on the known
data, such that

(5.17) $\Theta\geq\omega_{0}>0.$

###### Proof.

Choose proper coordinates in $\mathbb{R}^{n}$ such that $\Theta$ is achieved
at $0\in\partial\Omega$. By the Corollary 5.6, there exists a symmetric
$\\{\widehat{G}_{0}^{\alpha\beta}\\}$ such that

(5.18)
$\widehat{G}_{0}^{\alpha\beta}\Big{(}u_{\alpha\beta}(x)-u_{\alpha\beta}(0)\Big{)}\geq\widehat{G}[u_{\alpha\beta}(x)]-\widehat{G}[u_{\alpha\beta}(0)],$

where $\widehat{G}_{0}^{\alpha\beta}=\frac{\partial\widehat{G}}{\partial
u_{\alpha\beta}}(u_{\alpha\beta}(0))$. Since (5.1) and (5.2), by calculating
we obtain

$(u-\underline{u})_{\alpha}+(u-\underline{u})_{n}\rho_{\alpha}=0,$
$(u-\underline{u})_{\alpha\beta}+(u-\underline{u})_{\alpha
n}\rho_{\beta}+(u-\underline{u})_{n\beta}\rho_{\alpha}+(u-\underline{u})_{n}\rho_{\alpha\beta}=0.$

Since $\rho_{\alpha}=\rho_{\beta}=0$ on $\partial\Omega$, we have

(5.19)
$u_{\alpha\beta}-\underline{u}_{\alpha\beta}=-(u-\underline{u})_{n}\rho_{\alpha\beta},\quad\mathrm{on}\quad\partial\Omega.$

It follows that

(5.20)
$\displaystyle(u-\underline{u})_{n}(0)\widehat{G}_{0}^{\alpha\beta}\rho_{\alpha\beta}(0)$
$\displaystyle=\widehat{G}_{0}^{\alpha\beta}\Big{(}\underline{u}_{\alpha\beta}(0)-u_{\alpha\beta}(0)\Big{)}$
$\displaystyle\geq\widehat{G}[\underline{u}_{\alpha\beta}(0)]-\widehat{G}[u_{\alpha\beta}(0)]$
$\displaystyle=\widehat{G}[\underline{u}_{\alpha\beta}(0)]-\psi(0)-\Theta$
$\displaystyle\geq\mathbf{c}-\Theta.$

Suppose now that

$(u-\underline{u})_{n}(0)\widehat{G}_{0}^{\alpha\beta}\rho_{\alpha\beta}(0)>\frac{1}{2}\mathbf{c}.$

Otherwise, taking (5.20) into consideration, we have
$\Theta\geq\frac{1}{2}\mathbf{c}$. Hence we see that (5.17) holds.

Let $\eta:=\widehat{G}_{0}^{\alpha\beta}\rho_{\alpha\beta}$. By Lemma 3.2 we
obtain $(u-\underline{u})_{n}>0$. Taking (3.7) and
$0<(u-\underline{u})_{n}\leq C_{\iota}$ into consideration, there exists a
universal constant $c_{\tau}>0$ depending on the $C^{1}$ estimate and
$\underline{u}$ such that

(5.21) $\eta(0)\geq\frac{1}{2}\mathbf{c}\frac{1}{(u-\underline{u})_{n}(0)}\geq
2c_{\tau}>0,$

where

$c_{\iota}=\frac{1}{4}\frac{\mathbf{c}}{C_{\iota}}.$

So we may assume that $\eta\geq c_{\tau}$ on $\Omega\cap
B_{\delta^{\prime}}(0)$ by choosing $\delta^{\prime}$ as in Lemma 5.2.

Consider the function

$\Phi=-(u-\underline{u})_{n}+\frac{1}{\eta}\widehat{G}_{0}^{\alpha\beta}\Big{(}\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0)\Big{)}-\frac{\psi(x)-\psi(0)}{\eta},\quad\mathrm{in}\quad\Omega\cap
B_{\delta^{\prime}}(0).$

We have $\Phi(0)=0$. Using (5.18) and (5.19), we see that

(5.22) $\displaystyle\Phi$
$\displaystyle=\frac{1}{\eta}\Big{\\{}-(u-\underline{u})_{n}\widehat{G}_{0}^{\alpha\beta}\rho_{\alpha\beta}+\widehat{G}_{0}^{\alpha\beta}\big{(}\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0))\Big{\\}}-\frac{1}{\eta}\Big{(}\psi(x)-\psi(0)\Big{)}$
$\displaystyle=\frac{1}{\eta}\Big{\\{}\widehat{G}_{0}^{\alpha\beta}(u_{\alpha\beta}(x)-\underline{u}_{\alpha\beta}(x)+\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0)\big{)}\Big{\\}}-\frac{1}{\eta}\Big{(}\psi(x)-\psi(0)\Big{)}$
$\displaystyle=\frac{1}{\eta}\Big{\\{}\widehat{G}_{0}^{\alpha\beta}(u_{\alpha\beta}(x)-u_{\alpha\beta}(0))\Big{\\}}-\frac{1}{\eta}\Big{(}\psi(x)-\psi(0)\Big{)}$
$\displaystyle\geq\frac{1}{\eta}\Big{\\{}\widehat{G}[u_{\alpha\beta}(x)]-\widehat{G}[u_{\alpha\beta}(0)]-(\psi(x)-\psi(0))\Big{\\}}$
$\displaystyle=\frac{1}{\eta}\Big{\\{}\widehat{G}[u_{\alpha\beta}(x)]-\psi(x)-\Theta\Big{\\}}$
$\displaystyle\geq 0,\quad\mathrm{on}\quad\partial\Omega.$

After direct computation, we have

$\displaystyle
D_{i}\Big{[}\frac{1}{\eta}\widehat{G}_{0}^{\alpha\beta}(\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0))\Big{]}$
$\displaystyle=\frac{1}{\eta}\widehat{G}_{0}^{\alpha\beta}\underline{u}_{\alpha\beta
i}-\frac{\eta_{i}}{\eta^{2}}\widehat{G}_{0}^{\alpha\beta}(\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0)),$
$\displaystyle
D_{i}D_{j}\Big{[}\frac{1}{\eta}\widehat{G}_{0}^{\alpha\beta}\big{(}\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0)\big{)}\Big{]}$
$\displaystyle=\frac{1}{\eta}\widehat{G}_{0}^{\alpha\beta}\underline{u}_{\alpha\beta
ij}+D_{i}D_{j}(\frac{1}{\eta})\widehat{G}_{0}^{\alpha\beta}(\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0))-\frac{2}{\eta^{2}}\widehat{G}_{0}^{\alpha\beta}\underline{u}_{\alpha\beta
i}\eta_{j}.$

Since $\widetilde{G}_{i}$ is bounded and $\psi$ is a bounded function, we can
obtain that

$\displaystyle L(\Phi)$
$\displaystyle=\widetilde{G}_{ij}D_{i}D_{j}\Phi+\widetilde{G}_{i}D_{i}\Phi$
$\displaystyle=-\widetilde{G}_{ij}D_{i}D_{j}(u-\underline{u})_{n}-\widetilde{G}_{i}D_{i}(u-\underline{u})_{n}+\widetilde{G}_{ij}D_{i}D_{j}\Big{[}\frac{1}{\eta}\widehat{G}_{0}^{\alpha\beta}\big{(}\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0)\big{)}\Big{]}$
$\displaystyle~{}+\widetilde{G}_{i}D_{i}\Big{[}\frac{1}{\eta}\widehat{G}_{0}^{\alpha\beta}(\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0))\Big{]}+\widetilde{G}_{ij}D_{i}D_{j}\frac{\psi(x)-\psi(0)}{\eta}+\widetilde{G}_{i}D_{i}\frac{\psi(x)-\psi(0)}{\eta}$
$\displaystyle=-\psi_{n}+\widetilde{G}_{ij}\underline{u}_{nij}+\widetilde{G}_{i}\underline{u}_{ni}+\frac{1}{\eta}\widetilde{G}_{ij}\widehat{G}_{0}^{\alpha\beta}\underline{u}_{\alpha\beta
ij}+D_{i}D_{j}(\frac{1}{\eta})\widetilde{G}_{ij}\widehat{G}_{0}^{\alpha\beta}(\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0))$
$\displaystyle~{}-\frac{2}{\eta^{2}}\widetilde{G}_{ij}\widehat{G}_{0}^{\alpha\beta}\underline{u}_{\alpha\beta
i}\eta_{j}+\frac{1}{\eta}\widetilde{G}_{i}\widehat{G}_{0}^{\alpha\beta}\underline{u}_{\alpha\beta
i}-\frac{\eta_{i}}{\eta^{2}}\widetilde{G}_{i}\widehat{G}_{0}^{\alpha\beta}(\underline{u}_{\alpha\beta}(x)-u_{\alpha\beta}(0))$
$\displaystyle~{}+\widetilde{G}_{ij}D_{i}D_{j}\frac{\psi(x)-\psi(0)}{\eta}+\widetilde{G}_{i}D_{i}\frac{\psi(x)-\psi(0)}{\eta}$
$\displaystyle\leq C_{33}(1+\sum_{i}\widetilde{G}_{ii})$ $\displaystyle\leq
C_{34}(1+\sum_{i}G^{ii}).$

We can again use Lemma 5.2, Lemma 5.5 and take $\widehat{A}\gg\widehat{B}\gg
1$ large enough to get

(5.23) $\displaystyle L(\widehat{A}v+\widehat{B}|x|^{2}+\Phi)$
$\displaystyle\leq 0,\quad\mathrm{in}\quad\Omega\cap B_{\delta^{\prime}}(0),$
$\displaystyle\widehat{A}v+\widehat{B}|x|^{2}+\Phi$ $\displaystyle\geq
0,\quad\mathrm{on}\quad\partial(\Omega\cap B_{\delta^{\prime}}(0)).$

By the maximum principle, we see that

$\widehat{A}v+\widehat{B}|x|^{2}+\Phi\geq 0,\quad\mathrm{in}\quad\Omega\cap
B_{\delta^{\prime}}(0).$

We have

$\widehat{A}v_{n}(0)+\Phi_{n}(0)=\frac{\partial}{\partial
x_{n}}(\widehat{A}v+\widehat{B}|x|^{2}+\Phi)(0)\geq 0,$

since $\widehat{A}v+\widehat{B}|x|^{2}+\Phi=0$ at the origin. Thus one can
deduce that $\Phi_{n}(0)\geq-\widehat{A}v_{n}(0)\geq-C_{35}$. Consequently we
can conclude that

$u_{nn}(0)\leq C_{32}.$

So that we obtain $|D^{2}u(0)|\leq C_{36}$ for some positive constant. By
letting $\mathrm{t}$ large enough, it yields that $[D^{2}u+\mathrm{t}q\otimes
q]>[D^{2}u]$ uniformly. Thus we see that

$\widehat{G}[u_{\alpha\beta}(0)]>\psi(0).$

As a result, (5.17) holds. ∎

Now we need the following Lemma from [3].

######  Lemma 5.8.

Consider the $n\times n$ symmetric matrix

$M=\left(\begin{matrix}d_{1}&0&\cdots&0&a_{1}\\\ 0&d_{2}&\cdots&0&a_{2}\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\ 0&0&\cdots&d_{n-1}&a_{n-1}\\\
a_{1}&a_{2}&\cdots&a_{n-1}&a\end{matrix}\right)$

with $d_{1},\ldots,d_{n-1}$ fixed, $a$ tending to infinity and
$|a_{\alpha}|\leq C_{37}$ for $1\leq\alpha\leq n-1$. Then the eigenvalues
$\lambda_{i}$ of $M$ behave like

$\displaystyle\lambda_{\alpha}$ $\displaystyle=d_{\alpha}+o(1),\quad
1\leq\alpha\leq n-1,$ $\displaystyle\lambda_{n}$
$\displaystyle=a\Big{(}1+O(\frac{1}{a})\Big{)},$

for $1\leq i\leq n$, where the $o(1)$ and $O(\frac{1}{a})$ are
uniform–depending only on $d_{1},\ldots,d_{n-1}$ and $C_{37}$.

Therefore, combining Lemma 5.7 with Lemma 5.8 we obtain that

######  Lemma 5.9.

There exist suitable constants $R_{0},\omega_{0}>0$ such that for all
$\mathrm{t}\geq R_{0}$ we have

$G(\kappa^{\prime}(a_{\alpha\beta}),\mathrm{t})>\psi(x)+\omega_{0}.$

where $\kappa^{\prime}(a_{\alpha\beta})$ is the eigenvalue of the
$(n-1)\times(n-1)$ matrix $a_{\alpha\beta}$.

###### Proof.

Since (5.17) holds, we have

$\lim_{t\rightarrow\infty}\widetilde{G}(r+\mathrm{t}q\otimes
q,p)\geq\psi(x)+\omega_{0}.$

Writing $\mathcal{A}(r,p)=[a_{ij}]$ and $\mathcal{A}(r+\mathrm{t}q\otimes
q,p)=[\widetilde{a}_{ij}]$, we have

$\widetilde{a}_{ij}=a_{ij}+\mathrm{t}b^{in}b^{nj},$

where $b^{ij}=\delta_{ij}-\frac{D_{i}uD_{j}u}{w(1+w)}$. After an orthonormal
transformation we may assume

$\widetilde{a}_{ij}=a_{ij}+\frac{\mathrm{t}}{\sqrt{1+|Du|^{2}}}\delta_{in}\delta_{nj}.$

By Lemma 5.8, the eigenvalues of $\mathcal{A}(r+\mathrm{t}q\otimes
q,p)=[\widetilde{a}_{ij}]$ are given by

$\kappa^{\prime}_{\alpha}=\kappa_{\alpha}+o(1),\quad\tilde{\kappa}_{n}=\mathrm{t}+\kappa_{n}+O(1),$

as $\mathrm{t}\rightarrow\infty$, where $\kappa^{\prime}_{\alpha}$ are the
eigenvalues of $[\tilde{a}_{\alpha\beta}]_{1\leq\alpha,\beta\leq n-1}$.
Through our observation, we have

$\widetilde{G}(r+\mathrm{t}q\otimes
q,p)=G(\kappa(\widetilde{a}_{ij}))=G(\kappa_{\alpha}+o(1),\mathrm{t}+\kappa_{n}+O(1)),$

where $\kappa_{n}\geq-\tan\delta$. Then there holds

$\lim_{t\rightarrow+\infty}G(\kappa^{\prime}(a_{\alpha\beta}),\mathrm{t})=\hat{G}(u_{\alpha\beta}).$

Applying Lemma 5.7, if $\mathrm{t}\rightarrow\infty$, we obtain

$G(\kappa^{\prime}(a_{\alpha\beta}),\mathrm{t})>\psi(x)+\omega_{0}.$

∎

Recall that

$G(\kappa)=-e^{-A\sum\arctan\kappa_{i}},\quad\psi(x)=-e^{-Ah(x)}.$

Suppose that $\kappa_{1},\ldots,\kappa_{n}$ are the eigenvalues of $a_{ij}(0)$
for $x\in\partial\Omega$. Applying Lemma 5.8, we see that the eigenvalues
$\kappa_{1},\kappa_{2},\ldots,\kappa_{n}$ behave like

(5.24) $\displaystyle\kappa_{\alpha}$
$\displaystyle=\frac{1}{w}\lambda_{\alpha}+o(1),$ $\displaystyle\kappa_{n}$
$\displaystyle=\frac{1}{w^{3}}u_{nn}(0)\Big{(}1+O(\frac{1}{|u_{nn}(0)|})\Big{)}.$

as $|u_{nn}(0)|\rightarrow\infty$, where $\lambda_{1},\ldots,\lambda_{n-1}$
are the eigenvalues of $u_{\alpha\beta}(0)$.

Next we prove the following conclusion:

######  Lemma 5.10.

We show that

$\max_{x\in\partial\Omega}|u_{nn}(x)|\leq C_{38}$

for all $x\in\partial\Omega$, where $C_{38}$ is positive constant depending on
$\partial\Omega,n,\delta,h$ and $\underline{u}$.

###### Proof.

Suppose that $|u_{nn}(x)|$ achieves its maximum at a point $x_{p}$ on
$\partial\Omega$. We assume that $u_{nn}(x_{p})\geq K$ where $K$ is a large
constant to be chosen. Denote
$\kappa_{\alpha}^{\prime}=(\kappa^{\prime}_{1},\ldots,\kappa^{\prime}_{n-1})$
eigenvalues of $a_{\alpha\beta}(x_{p})$. By (5.24), for every $\delta_{1}>0$,
there exists $R(\delta_{1})\gg 1$ such that if $u_{nn}(x_{p})\geq
R(\delta_{1})$, then the eigenvalues of $a_{ij}(x_{p})$ satisfy

$\kappa_{n}\geq R_{2},$

and

(5.25)
$|(\kappa_{1},\ldots,\kappa_{n-1},\kappa_{n})-(\kappa_{1}^{\prime},\ldots,\kappa^{\prime}_{n-1},\kappa_{n})|<\delta_{1}.$

Consequently, we find that there exists a positive constant $\delta_{1}>0$
depending on $\omega_{0}$ such that if $u_{nn}(x_{p})\geq
K\triangleq\max\\{R(\delta_{1}),R_{2}\\}$, then

$\displaystyle\psi(x_{p})=G(\kappa)(x_{p})$ $\displaystyle\geq
G(\kappa_{\alpha}^{\prime},\kappa_{n})(x_{p})-\frac{\omega_{0}}{2}$
$\displaystyle\geq G(\kappa_{\alpha}^{\prime},R_{2})-\frac{\omega_{0}}{2}$
$\displaystyle>\psi(x_{p})+\frac{\omega_{0}}{2},$

where the first step follows from the continuity of $G$, the second step
follows from monotonicity of $G$, while the last comes from the fact that
Lemma 5.9. So we obtain a contradictory result. Thus we complete the proof of
Lemma 5.10. ∎

Hence the boundary $C^{2}$ estimate is complete. Then this completes the proof
of Lemma 4.1.

##  6\. Proof of Theorem 1.3

The goal of this section is to prove Theorem 1.3. Since Lemma 4.1 and Lemma
4.2 have been established, by Evans-Krylov theorem and Schauder theory in
[26], we get the priori bound for the $C^{3,\alpha}$ norm of $u$ for any
$\alpha\in(0,1)$ to Dirichlet problem (3.5), i.e.

(6.1) $\|u\|_{C^{3,\alpha}(\overline{\Omega})}\leq C_{39},$

where $C_{39}$ is a positive constant depending on
$\Omega,\varphi,\|\underline{u}\|_{C^{4}(\overline{\Omega})},\|h\|_{C^{2}(\overline{\Omega})}$
and $\delta$. In this section we explain how to make use of (6.1) and
continuity methold to prove the existence of solution to (1.3) and (1.4).

Following the same proof as Lemma 5.2 in [10], we can obtain

######  Lemma 6.1.

If $\Omega,h,\varphi$ and $\underline{u}$ are smooth satisfying (A), then the
solution $u$ is smooth in $\bar{\Omega}$.

To solve the equation, we use the continuity method.

Proof of Theorem 1.3.

It’s enough for us to consider Dirichlet problem (3.5) as follows.

$\left\\{\begin{aligned} \widetilde{G}(D^{2}u,Du)&=\psi(x),\ \
&&\quad\mathrm{in}\quad\Omega,\\\
u&=\varphi,&&\quad\mathrm{on}\quad\partial\Omega.\end{aligned}\right.$

where $\widetilde{G}(D^{2}u,Du)=G(\mathcal{A})$ and $\psi(x)=-e^{-Ah(x)}$. One
can use the pre-knowledge in previous section. Assume that $\underline{u}$ is
$C^{4}(\overline{\Omega})$ subsolution satisfying (1.6). It yields that the
classical solution to (1.3) and (1.4) is unique by maximum principle. For each
$t\in[0,1]$, set

$J^{t}(D^{2}u,Du)=\widetilde{G}(D^{2}u,tDu).$

Consider

(6.2) $\left\\{\begin{aligned} J^{t}(D^{2}u,Du)&=\psi(x),\ \
&&\mathrm{in}\quad\Omega,\\\
u&=\varphi,&&\mathrm{on}\quad\partial\Omega.\end{aligned}\right.$

Define the set

$S:=\left\\{t\in[0,1]:\text{(\ref{e6.2}) has at least one admissible
solution}\right\\}.$

By the main results in [13], (6.2) is solvable if $t=0$. So $S$ is not empty.
We claim that $S=[0,1]$, which is equivalent to the fact that $S$ is not only
open, but also closed.

Define

$X:=\\{u\in C^{3,\alpha}(\bar{\Omega}):u\mid_{\partial\Omega}=\varphi\\}$

and

$Y:=C^{\alpha}(\bar{\Omega})\times C^{2,\alpha}(\partial\Omega).$

Then $X$ is the closed convex subset in Banach space
$C^{3,\alpha}(\bar{\Omega})$ and $Y$ is a Banach space. Define a map from
$X\times[0,1]$ to $Y$ as

$\mathfrak{F}(u,t):=\left(J^{t}(D^{2}u,Du)-\psi(x),u\right).$

Given $(u_{0},t_{0})\in X\times[0,1]$. For any $w\in
C^{3,\alpha}(\overline{\Omega})$ with $w\mid_{\partial\Omega}=0$, we have

(6.3) $\displaystyle\frac{d}{d\varepsilon}\mathfrak{F}(u_{0}+\varepsilon
w,t_{0}+\varepsilon)|_{\varepsilon=0}$ $\displaystyle=$
$\displaystyle\left(\widetilde{G}_{ij}(D^{2}u_{0},t_{0}Du_{0})\partial_{ij}w+\widetilde{G}_{i}(D^{2}u_{0},t_{0}Du_{0})t_{0}\partial_{i}w+\widetilde{G}_{i}(D^{2}u_{0},t_{0}Du_{0})\partial_{i}u_{0},0\right).$

Then the linearized operator of $\mathfrak{F}(u,t)$ at $(u_{0},t_{0})$ is
given by

$D\mathfrak{F}(u_{0},t_{0})=\left(\widetilde{G}_{ij}(D^{2}u_{0},t_{0}Du_{0})\partial_{ij}w+\widetilde{G}_{i}(D^{2}u_{0},t_{0}Du_{0})t_{0}\partial_{i}w+\widetilde{G}_{i}(D^{2}u_{0},t_{0}Du_{0})\partial_{i}u_{0},0\right).$

One can see that
$\widetilde{G}_{ij}(D^{2}u_{0},t_{0}Du_{0}),\widetilde{G}_{i}(D^{2}u_{0},t_{0}Du_{0})\in
C^{1,\alpha}(\overline{\Omega})$ and

$\mathfrak{L}\triangleq\widetilde{G}_{ij}(D^{2}u_{0},t_{0}Du_{0})\partial_{ij}+\widetilde{G}_{i}(D^{2}u_{0},t_{0}Du_{0})t_{0}\partial_{i}$

is a uniformly linear elliptic operator by calculation. Then making use of
Schauder theory in [26], $D\mathfrak{F}(u_{0},t_{0})$ is invertible for any
$u_{0}\in X$ and $t_{0}\in[0,1]$ being the solution to (6.2). Then the fact
that S is open follows from the invertibility of the linearized operator and
the implicit function theorem [26].

To finish the proof, we need to prove the fact that $S$ is a closed subset of
$[0,1]$. One can show that $S$ is closed is equivalent to the fact that for
any sequence $\\{t_{k}\\}\subset S$, if
$\lim_{k\rightarrow\infty}t_{k}=t_{0}$, then $t_{0}\in S$. For $t_{k}$,
$u_{k}$ is the solution of

(6.4) $\left\\{\begin{aligned} J^{t_{k}}(D^{2}u_{k},Du_{k})&=\psi(x),\ \
&&\mathrm{in}\quad\Omega,\\\
u_{k}&=\varphi,&&\mathrm{on}\quad\partial\Omega.\end{aligned}\right.$

The operator $J^{t}$ comes from $\widetilde{G}$. By the proof of the main
theorem in [17], the structure conditions of $J^{t}$ is as same as
$\widetilde{G}$. Then for any admissible solution of (6.2), the estimate (6.2)
also holds. So any solution $u_{k}$ satisfies

$\|u_{k}\|_{C^{3,\alpha}(\overline{\Omega})}\leq C_{40},$

where $C_{40}$ depends only on the known data and also is independent to
$t_{k}$. Using Arzela-Ascoli Theorem, we know that there exist $\hat{u}\in
C^{3,\alpha}(\bar{\Omega})$ and a subsequence of $\\{t_{k}\\}$, which is still
denoted as $\\{t_{k}\\}$, such that letting $k\rightarrow\infty$,

$\left\|u_{k}-\hat{u}\right\|_{C^{3}(\bar{\Omega})}\rightarrow 0.$

For equation (6.4), letting $k\rightarrow\infty$, we have

$\left\\{\begin{aligned} J^{t_{0}}(D^{2}\hat{u},D\hat{u})&=\psi(x),\ \
&&\mathrm{in}\quad\Omega,\\\
\hat{u}&=\varphi,&&\mathrm{on}\quad\partial\Omega.\end{aligned}\right.$

Therefore, $t_{0}\in S$, and thus $S$ is closed. Consequently, $S=[0,1]$.

It follows Lemma 6.1 that there exists a smooth solution $u\in
C^{\infty}(\overline{\Omega})$ to the Dirichlet problem (1.3) and (1.4) if all
data is smooth. Thus Theorem 1.3 is established. ∎

Acknowledgments: The authors thank Professor Hengyu Zhou for helpful
discussions and suggestions related to this work. The authors would also like
to thank the referees for useful comments that have improved the paper.

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