On A Class of Degenerate And Singular Monge-Amp\`ere Equations

TL;DR

This paper proves existence, uniqueness, and regularity results for a class of degenerate and singular Monge-Ampère equations on convex domains, linking solution regularity to domain convexity.

Contribution

It establishes the existence, uniqueness, and Hölder continuity of solutions for a new class of degenerate and singular Monge-Ampère equations, with a novel relation to domain convexity.

Findings

Proved existence and uniqueness of solutions.

Established Hölder continuity of solutions.

Linked Hölder exponent to domain convexity.

Abstract

In this paper we shall prove the existence, uniqueness and global H$\ddot{o}$lder continuity for the Dirichlet problem of a class of Monge-Amp\`ere type equations which may be degenerate and singular on the boundary of convex domains. We will establish a relation of the H$\ddot{o}$lder exponent for the solutions with the convexity for the domains.

Equations

$$\begin{array}{rl}\det D^{2}u& =F(x,u)\text{in}\Omega ,\\ u& =0\text{on}\partial \Omega ,\end{array}$$

$$\begin{array}{r}F(x,t)\in C(\Omega \times (-\infty ,0))\text{ is non-decreasing in}t\text{ for any}x\in \Omega ;\end{array}$$

$$\begin{array}{rl}& \text{there are constants }A>0,\alpha \ge 0,\beta \ge n+1\text{ such that}\\ & 0<F(x,t)\le A{d}_x^{\beta -n-1}|t{|}^{-\alpha }\forall (x,t)\in \Omega \times (-\infty ,0),\end{array}$$

$$\gamma_{1}:=\left\{\begin{array}[]{cc}\frac{\beta-n+1}{n+\alpha},&{\rm if}\beta<\alpha+2n-1,\\ \text{any number in}(0,1),&{\rm if}\beta\geq\alpha+2n-1.\end{array}\right.$$

$$x=(x_1,x_2,\dots ,x_n)=(x^{\prime },x_n),x^{\prime }=(x_1,...,x_{n-1})$$

$$|x^{\prime }|=\sqrt{x_1^2+...+x_{n-1}^2}.$$

$$x_0=0\text{and}\Omega \subseteq \{x\in R^n|x_n\ge \eta |x^{\prime }{|}^a\}.$$

$$\gamma_{2}:=\left\{\begin{array}[]{cc}\frac{\beta-n+1}{n+\alpha}+\frac{2n-2}{a(n+\alpha)},&{\rm if}\beta<\alpha+2n-1-\frac{2n-2}{a},\\ \text{any number in}(0,1),&{\rm if}\beta\geq\alpha+2n-1-\frac{2n-2}{a}.\end{array}\right.$$

$$u\in C^{{\gamma }_2}(\overline{\Omega }).$$

$$\gamma_{3}:=\left\{\begin{array}[]{cc}\frac{\beta}{n+\alpha},&{\rm if}\beta<\alpha+n,\\ \text{any number in}(0,1),&{\rm if}\alpha+n\leq\beta<\alpha+n+1,\\ 1,&{\rm if}\beta\geq\alpha+n+1.\end{array}\right.$$

$$u\in C^{{\gamma }_3}(\overline{\Omega }).$$

$$A{d}_x^{\beta -n-1}|t{|}^{-\alpha }\le F(x,t),\forall (x,t)\in \Omega \times (-\infty ,0)$$

$$|u(y)|\ge C(d_y{)}^{{\gamma }_4},\forall y\in \Omega$$

$${\gamma }_4:=\frac{\beta }{n+\alpha }\in (0,1),$$

$$|u(x)|\le M{d_x}^{\gamma },\forall x\in \Omega ,$$

$$|u{|}_{C^{\gamma }}(\overline{\Omega })\le M\{1+[diam(\Omega ){]}^{\gamma }\}.$$

$$|u(x_2)-u(x_1)|\le \max \{|u(y_1+(x_2-x_1))-u(y_1)|,|u(y_2)-u(y_2-(x_2-x_1))|\}.$$

$$\begin{array}{rl}|u(y_1+(x_2-x_1))-u(y_1)|=& |u(y_1+(x_2-x_1))|\\ \le & M\{dist(y_1+x_2-x_1,\partial \Omega ){\}}^{\gamma }\\ \le & M|x_2-x_1{|}^{\gamma }.\end{array}$$

$$|u(y_2)-u(y_2-(x_2-x_1))|\le M|x_2-x_1{|}^{\gamma }.$$

$$|u{|}_{C^{{\gamma }_1}(\overline{\Omega })}\le C(\alpha ,\beta ,A,diam(\Omega ),n),$$

$$\beta <\alpha +2n-1.$$

$$0\in \overline{\Omega }\subseteq R_{+}^n.$$

$$W=-M{x}_n^{\gamma }\cdot \sqrt{N^2{l}^2-r^2}$$

$$\begin{array}{rl}{W}_i& =M{x}_n^{\gamma }\cdot \frac{x_i}{\sqrt{N^2{l}^2-r^2}},\\ W_{ij}& =M{x}_n^{\gamma }\cdot \frac{1}{\sqrt{N^2{l}^2-r^2}}({\delta }_{ij}+\frac{x_i{x}_j}{N^2{l}^2-r^2}),\\ W_n& =-M\gamma x_n^{\gamma -1}\cdot \sqrt{N^2{l}^2-r^2},\\ W_{in}& =M\gamma x_n^{\gamma -1}\cdot \frac{x_i}{\sqrt{N^2{l}^2-r^2}},\\ W_{nn}& =M\gamma (1-\gamma )x_n^{\gamma -2}\cdot \sqrt{N^2{l}^2-r^2}.\end{array}$$

$$D^{2}W:=\left(\begin{array}{cc}G& \xi \\ {\xi }^T& W_{nn}\end{array}\right)$$

$$det{D}^{2}W=detG\cdot (W_{nn}-{\xi }^T{G}^{-1}\xi ).$$

$$M{x}_n^{\gamma }\frac{1}{\sqrt{N^2{l}^2-r^2}},...,M{x}_n^{\gamma }\frac{1}{\sqrt{N^2{l}^2-r^2}},M{x}_n^{\gamma }\frac{N^2{l}^2}{(N^2{l}^2-r^2)\sqrt{N^2{l}^2-r^2}},$$

$$detG=M^{n-1}{N}^2{l}^2{x}_n^{(n-1)\gamma }\cdot (\frac{1}{\sqrt{N^2{l}^2-r^2}}{)}^{n+1}.$$

$$G\xi =\frac{N^2{l}^{2}M{x}_n^{\gamma }}{(N^2{l}^2-r^2{)}^{\frac{3}{2}}}\xi .$$

$$\begin{array}{rl}{\xi }^T{G}^{-1}\xi & =\frac{(N^2{l}^2-r^2{)}^{\frac{3}{2}}}{N^2{l}^{2}M{x}_n^{\gamma }}|\xi {|}^2\\ & =\frac{M{\gamma }^2}{N^2{l}^2}{x}_n^{\gamma -2}{r}^2\sqrt{N^2{l}^2-r^2}.\end{array}$$

Full text

On A Class of Degenerate And Singular Monge-Ampère Equations

This work was supported by NSFC 11771237

Huaiyu Jian     You Li      Xushan Tu

Department of Mathematics, Tsinghua University

Beijing 100084, China

Abstract: In this paper we shall prove the existence, uniqueness and global H$\ddot{o}$lder continuity for the Dirichlet problem of a class of Monge-Ampère type equations which may be degenerate and singular on the boundary of convex domains. We will establish a relation of the H$\ddot{o}$lder exponent for the solutions with the convexity for the domains.

Key Words: existence, uniqueness, global regularity, degenerate, singular, Monge-Ampère equation

AMS Mathematics Subject Classification: 35J60, 35J96, 53A15.

Running head: Degenerate And Singular Monge-Ampère Equations

On A Class of Degenerate And Singular Monge-Ampère Equations

Huaiyu Jian    You Li    Xushan Tu

1. Introduction

In this paper we study the Monge-Ampère type equation

[TABLE]

where $\Omega$ is a bounded convex domain in $R^{n}$ $(n\geq 2)$, and $F$ satisfies the following (1.2)-(1.3):

[TABLE]

[TABLE]

where $d_{x}=dist(x,\partial\Omega)$. Obviously, this problem is singular and degenerate at the boundary of the domain.

The particular case of problem (1.1) includes a few geometric problems. When $F=|t|^{-(n+2)}$ and $u$ is a solution to problem (1.1), then the Legendre transform of $u$ is a complete affine hyperbolic sphere [4, 5, 7, 12, 14], and $(-u)^{-1}\sum u_{x_{i}x_{j}}dx_{i}dx_{j}$ gives the Hilbert metric (Poincare metric) in the convex domain $\Omega$ [19]. When $F=f(x)|t|^{-p},$ problem (1.1) may be obtained from $L_{p}$-Minkowski problem [20] and the Minkowski problem in centro-affine geometry[8, 13]. Also see p.440-441 in [15]. Generally, problem (1.1) can be applied to construct non-homogeneous complete Einstein-K$\ddot{a}$hler metrics on a tubular domain [5, 6].

Cheng and Yau in [5] proved that if $\Omega$ is a strictly convex $C^{2}$-domain and $F\in C^{k}$ ($k\geq 3$) satisfies (1.2)-(1.3), then problem (1.1) admits an unique convex generalized solution $u\in C(\bar{\Omega})$. Moreover, $u\in C^{k+1,\varepsilon}(\Omega)\bigcap C^{\gamma}(\bar{\Omega})$ for any $\varepsilon\in(0,1)$ and some $\gamma=C(\beta,\alpha,A,n,\partial\Omega)\in(0,1)$. We should emphasize that their methods need the strict convexity and the smoothness of $\Omega$, and the differentiability of $F$.

In this paper we find that the global H$\ddot{o}$lder regularity for problem (1.1) is independent of the smoothness of $\Omega$ and $F$, and the H$\ddot{o}$lder exponent depends only on the convexity of the domain. As a result, we can remove the smoothness of $\Omega$ as well as the differentiability of $F$ in [5]. Moreover, using the concept of $(a,\eta)$ type introduced in [12] to describe the convexity of the domain, we obtain a relation of the H$\ddot{o}$lder exponent for $u$ with the convexity for $\Omega$.

We have noticed that there are many papers on global regularity for equations of Monge-Ampère type. See, for example, [3, 9, 11, 18, 22, 23, 25] and the references therein. But, generally speaking, those results require that the domain $\Omega$ should be strictly convex and $\partial\Omega\in C^{1,1}$.

Our first result is stated as the following

Theorem 1.1**.**

Supposed that $\Omega$ is a bounded convex domain in $R^{n}$ and $F(x,t)$ satisfies (1.2)-(1.3). Let

[TABLE]

Then problem (1.1) admits an unique convex generalized solution $u\in C^{\gamma_{1}}(\overline{\Omega})$. Furthermore, $u\in C^{2,\gamma_{1}}(\Omega)$ if $F(x,t)\in C^{0,1}(\Omega\times(-\infty,0))$.

Here a generalized solution means the well-known Alexandrov solution. See, for example, [9, 10, 24] for the details.

To improve the regularity for the solution obtained in Theorem 1.1, we use the $(a,\eta)$ type in [12] to describe the convexity of $\Omega$. From now on, we denote

[TABLE]

and

[TABLE]

Definition 1.1. Supposed that $\Omega$ is a bounded convex domain in $R^{n}$, and $x_{0}\in\partial\Omega$. $x_{0}$ is called to be $(a,\eta)$ type if there are numbers $a\in[1,+\infty)$ and $\eta>0$, after translation and rotation transforms, we have

[TABLE]

$\Omega$ is called $(a,\eta)$ type domain if every point of $\partial\Omega$ is $(a,\eta)$ type.

Remark 1.1. The convexity requires that the number $a$ should be no less than 1. The less is $a$, the more convex is the domain. There is no $(a,\eta)$ type domain for $a\in[1,2)$, although part of $\partial\Omega$ may be $(a,\eta)$ type point for $a\in[1,2)$.

Definition 1.2. We say that a domain $\Omega$ in $R^{n}$ satisfies exterior (or interior) sphere condition with radius $R$ if for each $x_{0}\in\partial\Omega$, there is a $B_{R}(y_{0})\supseteq\Omega$ (or $B_{R}(y_{0})\subseteq\Omega$, respectively) such that $\partial B_{R}(y_{0})\bigcap\partial\Omega\ni x_{0}$.

In [12], we have proved that $(2,\eta)$ type domain is equivalent to the domain satisfies exterior sphere condition.

The following two theorems show the relation of the H$\ddot{o}$lder exponent for $u$ on $\bar{\Omega}$ with the convexity for $\Omega$.

Theorem 1.2**.**

Supposed that $\Omega$ is (a, $\eta$) type domain in $R^{n}$ with $a\in(2,+\infty)$, and $F$ satisfies (1.2)-(1.3). Let Let

[TABLE]

Then the convex generalized solution to problem (1.1)

[TABLE]

Furthermore $u\in C^{2,\gamma_{2}}(\Omega)$ if $F(x,t)\in C^{0,1}(\Omega\times(-\infty,0))$.

Theorem 1.3**.**

Let $\Omega$ be a bounded convex domain in $R^{n}$ and $u$ be a convex generalized solution to problem (1.1).

(i) Suppose that $\Omega$ satisfies exterior sphere condition and $F$ satisfies (1.2)-(1.3). Let

[TABLE]

Then

[TABLE]

Furthermore $u\in C^{2,\gamma_{3}}(\Omega)$ if $F(x,t)\in C^{0,1}(\Omega\times(-\infty,0)).$

(ii) If $\Omega$ satisfies interior sphere condition with radius $R$ and $F$ satisfies (1.2) and

[TABLE]

for some constants $A>0$, then

[TABLE]

for some constant $C=C(\beta,\alpha,A,n,R)>0$, where

[TABLE]

Remark 1.2. The H$\ddot{o}$lder regularity result of Theorem 1.1 can be viewed as the limit case of Theorem 1.2 as $a\to\infty$. Theorem 1.3 (i) shows that Theorem 1.2 is true for $a=2$, since a $(2,\eta)$ type domain is equivalent to that the domain satisfies exterior sphere condition.

In the following Sections 2, 3, and 4, we will prove Theorems 1.1, 1.2, and 1.3, respectively.

2. Proof of Theorem 1.1

We start at a primary result which is useful to proving that a convex function in $\Omega$ is H$\ddot{o}$lder continuous in $\bar{\Omega}$.

Lemma 2.1**.**

Let $\Omega$ be a bounded convex domain and $u\in C(\overline{\Omega})$ be a convex function in $\Omega$ with $u|_{\partial\Omega}=0$. If there are $\gamma\in(0,1]$ and $M>0$ such that

[TABLE]

then $u\in C^{\gamma}(\overline{\Omega})$ and

[TABLE]

Proof.

This was proved in [12]. Here we copy the arguments for the convenience.

For any two point $x_{1}$, $x_{2}\in\Omega$, consider the line determined by $x_{1}$ and $x_{2}$. The line will intersect $\partial\Omega$ at two points $y_{1}$ and $y_{2}$. Without loss generality we assume the four points are $y_{1}$, $x_{1}$, $x_{2}$, $y_{2}$ in order. By restricted onto the line, $u$ is one dimension convex function. By the monotonic proposition of convex functions, we have

[TABLE]

Moreover, since $y_{1}\in\partial\Omega$, by the assumption (2.1) we have

[TABLE]

Similarly,

[TABLE]

The above three inequalities, together with (2.1), implies the desired result. ∎

To prove Theorem 1.1, we need an a priori estimate result as follows, which holds without strictly convexity of $\Omega$ or any smoothness of $\Omega$ and of $F$.

Lemma 2.2**.**

Supposed that $\Omega$ is a bounded convex domain in $\mathbb{R}^{n}$ and $F(x,t)$ satisfies (1.2) and (1.3). If $u$ is a convex generalized solution to problem (1.1), then $u\in C^{\gamma_{1}}(\overline{\Omega})$ and

[TABLE]

where $\gamma_{1}$ is given by (1.4).

Proof.

First, we may assume

[TABLE]

Since for the case $\beta\geq\alpha+2n-1$, we take a $\hat{\beta}<\alpha+2n-1$ such that $\frac{\hat{\beta}-n+1}{n+\alpha}$ can be any number in $(0,1)$. (Note $n\geq 2$). Obviously, (1.3) still holds with $\beta$ replaced by $\hat{\beta}$. Hence, this case is reduced to the case (2.3).

Next, we assume for the time being that

[TABLE]

Then we are going to construct a sub-solution to problem (1.1).

For brevity, write $l=diam(\Omega)$. Set

[TABLE]

where $r=\sqrt{x_{1}^{2}+...+x_{n-1}^{2}}$. We will choose positive constants $\gamma$, $M$, $N$ such that $W$ is an sub-solution to problem (1.1) under the assumptions (2.3) and (2.4).

For $i,j\in\{1,2,...,n-1\}$, write $W_{i}=\frac{\partial W}{\partial x_{i}},W_{ij}=\frac{\partial^{2}W}{\partial x_{i}\partial x_{j}}$. Then we have

[TABLE]

Denote

[TABLE]

where $\xi^{T}=(W_{n1},...,W_{n(n-1)})$, and $G$ is the $(n-1)$-order matrix. Then

[TABLE]

Since all the eigenvalues of $G$ are

[TABLE]

[TABLE]

It is direct to verify that

[TABLE]

It follows that

[TABLE]

Hence, we obtain that

[TABLE]

We want to prove

[TABLE]

Since (1.3) and (2.4) implies that

[TABLE]

we see that (2.6) can be deduced from

[TABLE]

which is equivalent to

[TABLE]

By (2.5), (2.8) is nothing but

[TABLE]

Now we choose $\gamma=\frac{\beta-n+1}{n+\alpha}$ such that

[TABLE]

Since $\gamma\in(0,1)$ by (2.3) and $r=|x^{\prime}|\leq diam(\Omega)=l$ in $\Omega$, we first take $N=C(\gamma)$ large enough such that

[TABLE]

Noting $N^{2}l^{2}-r^{2}\in[(N^{2}-1)l^{2},\ N^{2}l^{2}]$,  we then take $M=C(A,\alpha,\gamma,N,n,l)$ large enough such that

[TABLE]

we obtain (2.9) and thus have proved (2.6).

Finally, for any point $y\in\Omega$, letting $z\in\partial\Omega$ be the nearest boundary point to $y$, by some translations and rotations, we assume $z=0$, $\Omega\subseteq R_{+}^{n}$ and the line $yz$ is the $x_{n}-axis$. This is to say that (2.4) is satisfied. Therefore we have (2.6). Obviously, $W\leq 0$ on $\overline{\Omega}$. Hence, $W$ is a sub-solution to problem (1.1). By comparison principle for generalized solutions (see [9, 10, 24] for example), we have

[TABLE]

which, together with Lemma 2.1, implies the desired result (2.2).

Note that we have used the fact that problem (1.1) is invariant under translation and rotation transforms, since $detD^{2}u$ is invariant and $F(x,u)$ is transformed to the one satisfying the same condition as $F$. This fact will be again used a few times in the following. ∎

Proof of Theorem 1.1.  We prove the theorem by three steps.

Step 1.  Suppose that $\Omega$ is bounded convex but $F(x,t)\in C^{k}(\Omega\times(-\infty,0))$ $(k\geq 3)$ satisfies (1.2) and (1.3).

We choose a sequence of bounded and strictly convex domains $\{\Omega_{i}\}$ such that

[TABLE]

Then by Theorem 5 in [5], there exists a convex generalized solution $u_{i}$ to problem (1.1) in the domain $\Omega_{i}$ for each $i$. We assume $u_{i}(x)=0$ for all $x\in R^{n}\setminus\Omega_{i}$. By Lemma 2.2, We have the uniform estimations

[TABLE]

which implies that there is a subsequence, still denoted by itself, convergent to a $u$ in the space $C(\overline{\Omega})$. Moreover, by (2.11) again, we have

[TABLE]

By the well-known convergence result for convex generalized solutions (see Lemma 1.6.1 in [10] for example), we see that $u$ is a convex generalized solution to problem (1.1).

Step 2.  Drop the restriction on the smoothness for $F$.

Suppose $F_{j}\in C^{k}(\Omega\times(-\infty,0))$ $(k\geq 3)$ satisfy the same assumption as $F$ in the Step 1 and $F_{j}$ locally uniform convergence to $F$ in as $j\to\infty$. (For example we can take $F_{j}=F*\eta_{\varepsilon_{j}}$, $\varepsilon_{j}$ convergence to 0 as $j$ tend to $+\infty$.) Then by the result of Step 1, for each $j$, there exists a convex generalized solution $u_{j}\in C^{\frac{\beta-n+1}{n+\alpha}}(\overline{\Omega})$ to problem (1.1) with $F$ replaced by $F_{j}$. Moreover, we have

[TABLE]

for all $j$. Using this estimate, Lemma 1.6.1 in [10], and the same argument as in Step 1, we obtain a solution $u$ to problem (1.1), which is the limit of a subsequence of $u_{j}$ in the space space $C(\overline{\Omega})$. Furthermore, we have $u\in C^{\frac{\beta-n+1}{n+\alpha}}(\overline{\Omega})$ by (2.12). The uniqueness for (1.1) is directly from the comparison principle (see [9, 10, 24] for example).

Step 3.   We are going to prove $u\in C^{2,\ \frac{\beta-n+1}{n+\alpha}}(\Omega)$ if $F(x,t)\in C^{0,1}(\Omega\times(-\infty,0))$.

It is enough to prove

[TABLE]

for any convex $\Omega_{1}\subset\subset\Omega$.

Taking a convex $\Omega^{\prime}$ such that $\Omega_{1}\subset\subset\Omega^{\prime}\subset\subset\Omega$, if there exists $z\in\overline{\Omega^{\prime}}\subset\Omega$ such that $u(z)=0$, then $u\equiv 0$ in $\Omega$ by convexity and the boundary condition $u|_{\partial\Omega}=0$. Hence we obtain (2.13). Otherwise, $u(x)<0$  for all $x\in\overline{\Omega^{\prime}}$. Then $F(x,u(x))\in C^{\frac{\beta-n+1}{n+\alpha}}(\overline{\Omega^{\prime}})$ and is positive on $\overline{\Omega^{\prime}}$. By the Caffarelli’s local $C^{2,\alpha}$ regularity in [2] (also see [16] for another proof), we obtain (2.13), too.

3. Proof of Theorem 1.2

In this section we establish the relation between the H$\ddot{o}$lder exponent and the convexity of the domain $\Omega$ and thus prove Theorem 1.2.

Assume that $\Omega$ is a $(a,\eta)$ type domain with $a\in(2,\infty)$, $F$ satisfies (1.2)-(1.3), and $u$ is the unique solution to problem (1.1) as in Theorem 1.1. To prove Theorem 1.2, it is sufficient to prove (1.6). See the Step 3 in the proof of Theorem 1.1.

As (2.3) we may assume

[TABLE]

Hence, in the following we have

[TABLE]

By Lemma 2.1, (1.6) can be deduced from

[TABLE]

for some positive constant$C=C(a,n,\alpha,\eta,A,diam\Omega)$.

We are going to prove (3.2). For any $y\in\Omega$, we can find $z\in\partial\Omega$, such that $|y-z|=d_{y}.$ Since the domain $\Omega$ is $(a,\eta)$ type and the problem (1.1) is invariant under translation and rotation transforms, we may assume $z=0$, and take the line determined by $z$ and $y$ as the $x_{n}-axis$ such that

[TABLE]

We will prove (3.2) by three steps.

Step 1. Let

[TABLE]

where $r=|x^{\prime}|=\sqrt{x_{1}^{2}+\cdots,x_{n-1}^{2}}$, $b$ and $\varepsilon$ are positive constants to be determined. We want to find a sufficient condition for which $W$ is a sub-solution to problem (1.1).

For $i,j\in\{1,2,...,n-1\}$, by direct computation we have

[TABLE]

Let

[TABLE]

where $\xi^{T}=(W_{n1},...,W_{n(n-1)})$, and $G$ is the matrix of $n-1$ order all of which eigenvalues are

[TABLE]

and one of which eigenvector with respect to the eigenvalue $W_{rr}$ is $\xi$. As obtaining (2.5), we have

[TABLE]

Obviously, $W\leq 0$ on $\partial\Omega$. Therefore we conclude that $W$ is a sub-solution to problem (1.1) if and only if

[TABLE]

We use the expression of $W$ to compute

[TABLE]

Using the expression of $W$ again we have

[TABLE]

[TABLE]

Hence,

[TABLE]

To estimate $I_{1}$, $I_{2}$ and $I_{3}$,  we will choose a small $\delta=C(a,\alpha,\beta,n)>0$. Now for this $\delta$, we choose a small $\varepsilon=C(\delta,a,\eta)>0$ such that

[TABLE]

Then we have

[TABLE]

By (3.8) we have

[TABLE]

Since $a>2$, we have two case: $a\geq\frac{2\alpha+2}{\beta-n+1}$ and $a<\frac{2\alpha+2}{\beta-n+1}$ if $\frac{2\alpha+2}{\beta-n+1}>2$.

Step 2. Assume that $\frac{2\alpha+2}{\beta-n+1}>2$ and $2<a<\frac{2\alpha+2}{\beta-n+1}$. We want to find $b>1$ and $\varepsilon>0$ such that (3.4) is satisfied, by which we will prove (3.2).

Since $a>2$ and $b>1$, $I_{1},I_{2}$ and $I_{3}$ in (3.6) are all positive.

[TABLE]

Observe that $d_{x}\leq x_{n}$ in $\Omega$. Hence, by (1.3), (3.4) and (3.5) we obtain

[TABLE]

it follows from (3.9) that

[TABLE]

Therefore, we arrive at

[TABLE]

Now, we set

[TABLE]

which is equivalent to

[TABLE]

Since $a\in(2,\frac{2\alpha+2}{\beta-n+1}),$ we see that $b>1$ by (3.1). Observing that $\beta-n+1>0$, we can choose $\varepsilon=C(a,\eta,A,\alpha,\beta,n)>0$ small enough again, such that $H[W]\geq 1$. This proves (3.4), which is to say that $W$ is a sub-solution to problem (1.1). By comparison principle, we have

[TABLE]

Restricting this inequality onto the $x_{n}$ axis, we obtain

[TABLE]

which is (3.2) exactly.

**Step 3. ** Assume that $a\geq\frac{2\alpha+2}{\beta-n+1}$. Note that $a>2$ by the assumption of the theorem. We will find $b\in(0,1)$ and $\varepsilon>0$ such that the function $W$ is a sub-solution to problem (1.1), and thus prove (3.2).

By (3.9) we have

[TABLE]

Since $a>2$, $b\in(0,1)$ and (3.9) yields

[TABLE]

we obtain

[TABLE]

Again by (3.9), we have

[TABLE]

Hence, we have

[TABLE]

Therefore, we obtain

[TABLE]

where

[TABLE]

Using above estimates, together with (1.3) and (3.9) we have

[TABLE]

Now, we set

[TABLE]

which ie equivalent to

[TABLE]

Since $a\geq\frac{2\alpha+2}{\beta-n+1}$, we see that $b\in(0,1]$. Of course, we also need

[TABLE]

which is equivalent to

[TABLE]

Since $\gamma_{2}=\frac{\beta-n+1}{n+\alpha}+\frac{2n-2}{a(n+\alpha)}\in(0,1)$ by (3.1), we see that

[TABLE]

Using this and taking $\delta=C(a,\alpha,\beta,n)>0$ small enough, we obtain (3.12) and thus (3.11).

Finally, choosing a positive

[TABLE]

smaller if necessary, by (3.10) and (3.11) we obtain that $H[W]\geq 1$ in $\Omega$, which implies $W$ is an sub-solution to problem (1.1) by (3.4). As in the end of Step 2, we have proved (3.2).

4. Proof of Theorem 1.3

As the proof of Theorem 1.2, the proof of (i) of Theorem 1.3 follows directly from

[TABLE]

for some positive constant$C=C(a,n,\alpha,\eta,A,diam\Omega)$.

For any $y\in\Omega$, we can find $z\in\partial\Omega$, such that $|y-z|=d_{y}.$ Since the domain $\Omega$ satisfies exterior sphere condition with radius $R$ and the problem (1.1) is invariant under translation and rotation transforms, we may assume

[TABLE]

Since $z=\mathbf{0}$ satisfies $|y-z|=d_{y}$, the tangent plane of $\Omega$ at $z=\mathbf{0}$ is unique. And it is easy to check $y$ is on the line dertermined by $\mathbf{0}$ and $y_{0}$. Hence $d_{y}=|y|=|y_{0}|-|y_{0}-y|=R-|y_{0}-y|$.

Consider the function

[TABLE]

where $r=|x-y_{0}|$, $M$ and $b$ are positive constants to be determined later. As (3.3), we obtain that

[TABLE]

But

[TABLE]

[TABLE]

Hence

[TABLE]

Observing that $W\leq 0$ on $\partial\Omega$, we see that $W$ is a sub-solution to problem (1.1) if and only if

[TABLE]

for all $x\in\Omega$ and $r=|x-y_{0}|$.

First, we consider the case

[TABLE]

As (2.3), we need only to consider the case $\beta<n+\alpha.$ We take

[TABLE]

Then in this case $b=\gamma_{3}\in(0,1)$ and $|2b-1|<1$. Hence,

[TABLE]

It follows from (4.2) that

[TABLE]

Therefore, by (1.3), (4.5), (4.8) and (4.9) that

[TABLE]

Note that

[TABLE]

by (4.7). Hence, by (4.10) and (4.11) we can choose a large $M=C(A,b,R,\alpha,n,\beta)$ such that

[TABLE]

Next, we consider the case

[TABLE]

In this case, we take

[TABLE]

Then , by (1.3) and (4.4) we have

[TABLE]

Therefore, (4.12) still holds true.

To sum up, we have obtained (4.5). By comparison principle, we see that

[TABLE]

In particular, we obtain that

[TABLE]

This is desired (4.1) and hence we have proved the (i) of Theorem 1.3.

To prove (ii) of Theorem 1.3, we notice that $u\in C(\bar{\Omega})$ and $u<0$ in $\Omega$ and $u=0$ on $\partial\Omega$. By comparing the graph of the convex function $u$ with the cone whose vortex is $(x_{0},u(x_{0}))$ and whose upper bottom is $\bar{\Omega}$, where $u(x_{0})=\min_{\bar{\Omega}}u$, we see easily that (1.10) is true for $\gamma_{4}\geq 1$. Hence, we need only to consider that case $\gamma_{4}<1$ in the following, which implies that $\beta<n+1$.

Since (1.10) holds naturally for all $y\in\{x\in\Omega:d_{x}\geq\frac{R}{2}\}$, where $R$ is the radius of the interior sphere for the $\Omega$. Hence, it is sufficient to prove

[TABLE]

Take such a $y$. We can find $z\in\partial\Omega$, such that $|y-z|=d_{y}.$ we may assume

[TABLE]

Since the tangent plane of $\Omega$ at $z=\mathbf{0}$ is unique. And it is easy to check $y$ is on the line determined by $\mathbf{0}$ and $y_{0}$. Hence $d_{y}=|y|=|y_{0}|-|y_{0}-y|=R-|y_{0}-y|$.

Observing that in this case, instead of (4.8) we have

[TABLE]

First, we require $b\in(0,1)$, which implies $2b-1\in(-1,1)$. Similarly to the arguments of (i),  by (4.16) we find that the function $W$, given by (4.3), satisfies

[TABLE]

Taking $b=\frac{\beta}{n+\alpha}=\gamma_{4}\in(0,1)$ we have

[TABLE]

Using (4.17)-(4.18), we see that $W$ is a super-solution to problem (1.1) in the domain $B_{R}(y_{0})$ for sufficiently small $M=C(A,b,R,\alpha,n,\beta)>0$. Since $u$ is a solution on $\Omega$ and $u|_{\partial B_{R}(y_{0})}\leq 0$, thus $u$ is a sub-solution on $B_{R}(y_{0})$. Therefore, we have

[TABLE]

Which is the desired (4.14) exactly. In this way, the proof of Theorem 1.3 has been completed.