Boundary blow-up solutions to fractional elliptic

equations in a measure framework

Huyuan<EMAIL_ADDRESS>Hichem<EMAIL_ADDRESS>Ying
<EMAIL_ADDRESS>

###### Abstract

Let $\alpha\in(0,1)$, $\Omega$ be a bounded open domain in $\mathbb{R}^{N}$
($N\geq 2$) with $C^{2}$ boundary $\partial\Omega$ and $\omega$ be the
Hausdorff measure on $\partial\Omega$. We denote by
$\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}$ a measure

$\langle\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}},f\rangle=\int_{\partial\Omega}\frac{\partial^{\alpha}f(x)}{\partial\vec{n}_{x}^{\alpha}}d\omega(x),\quad
f\in C^{1}(\bar{\Omega}),$

where $\vec{n}_{x}$ is the unit outward normal vector at point
$x\in\partial\Omega$. In this paper, we prove that problem

$\begin{array}[]{lll}(-\Delta)^{\alpha}u+g(u)=k\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}&{\rm
in}\quad\bar{\Omega},\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}+g(u)}u=0&{\rm
in}\quad\Omega^{c}\end{array}$ (0.1)

admits a unique weak solution $u_{k}$ under the hypotheses that $k>0$,
$(-\Delta)^{\alpha}$ denotes the fractional Laplacian with $\alpha\in(0,1)$
and $g$ is a nondecreasing function satisfying extra conditions. We prove that
the weak solution of (0.1) is a classical solution of

$\begin{array}[]{lll}\ \ \ (-\Delta)^{\alpha}u+g(u)=0&{\rm
in}\quad\Omega,\\\\[5.69054pt] \phantom{------\ }\ \ \ u=0&{\rm
in}\quad\mathbb{R}^{N}\setminus\bar{\Omega},\\\\[5.69054pt]
\phantom{}\lim_{x\in\Omega,x\to\partial\Omega}u(x)=+\infty.\end{array}$

Key Words: Fractional Laplacian, Green kernel, Boundary blow-up solution.

MSC2010: 35R11, 35J61, 35R06

##  1 Introduction

###  1.1 motivation

Let $\Omega$ be a bounded open domain in $\mathbb{R}^{N}$ ($N\geq 2$) with
$C^{2}$ boundary $\partial\Omega$. The pioneering works [13, 17] obtained that
the nonlinear reaction diffusion equation

$\begin{array}[]{lll}-\Delta u+h(u)=0&\mbox{in}&\Omega,\\\\[5.69054pt]
\phantom{-\Delta+h(u)}u=+\infty&\mbox{on}&\partial\Omega\end{array}$ (1.1)

admits a solution if $h$ is a locally Lipschitz continuous function which is
increasing and satisfies Keller-Osserman condition

$\int_{1}^{+\infty}\left[\int_{0}^{s}h(t)dt\right]^{-\frac{1}{2}}ds<+\infty.$

Great interests in existence, uniqueness and asymptotic behavior of boundary
blow-up solution to (1.1) have been taken, see [1, 9, 11, 14, 20, 22, 16]. It
is well known that when $h(s)=s^{p}$ with $p>1$, (1.1) has a unique solution
with boundary asymptotic behavior $\rho^{-\frac{2}{p-1}}(x)$, where
$\rho(x)={\rm dist}(x,\partial\Omega)$.

Comparing with the Laplacian case, a much richer structure for the solutions
set appears for the non-local case. Recently, the authors in [5] obtained very
different phenomena of the boundary blow-up solutions to elliptic equations
involving the fractional Laplacian, precisely,

$\displaystyle(-\Delta)^{\alpha}u+|u|^{p-1}u$ $\displaystyle=$ $\displaystyle
0\ \quad\hbox{in}\quad\Omega,$ $\displaystyle u$ $\displaystyle=$
$\displaystyle 0\quad\ \hbox{in}\quad\Omega^{c},$ (1.2)
$\displaystyle\lim_{x\in\Omega,x\to\partial\Omega}u(x)$ $\displaystyle=$
$\displaystyle+\infty,$

where $p>0$ and the fractional Laplacian $(-\Delta)^{\alpha}$ with
$\alpha\in(0,1)$ is defined by

$(-\Delta)^{\alpha}u(x)=\lim_{\epsilon\to
0^{+}}(-\Delta)_{\epsilon}^{\alpha}u(x),$

here for $\epsilon>0$,

$(-\Delta)_{\epsilon}^{\alpha}u(x)=-\int_{\mathbb{R}^{N}\setminus
B_{\epsilon}(x)}\frac{u(z)-u(x)}{|z-x|^{N+2\alpha}}dz$

The existence of boundary blow-up solution of (1.1) is derived by constructing
appropriate super and sub-solutions and this construction involves the one
dimensional truncated Laplacian of power functions given by

$C(\tau)=\int^{+\infty}_{0}\frac{\chi_{(0,1)}(t)|1-t|^{\tau}+(1+t)^{\tau}-2}{t^{1+2\alpha}}dt,$
(1.3)

where $\tau\in(-1,0)$ and $\chi_{(0,1)}$ is the characteristic function of the
interval $(0,1)$. It is known that there exists a unique zero point of (1.3)
in $(-1,0)$, denoting $\tau_{0}(\alpha)$. Then

###### Proposition 1.1

[5, Theorem 1.1] Assume that $\Omega$ is an open, bounded and connected domain
of class $C^{2}$ and $\alpha\in(0,1)$. Then we have:

Existence: Assume that

$1+2\alpha<p<1-\frac{2\alpha}{\tau_{0}(\alpha)},$

the equation (1.1) possesses at least one solution $u$ satisfying

$0<\liminf_{x\in\Omega,x\to\partial\Omega}u(x)d(x)^{\frac{2\alpha}{p-1}}\leq\limsup_{x\in\Omega,x\to\partial\Omega}u(x)d(x)^{\frac{2\alpha}{p-1}}<+\infty.$
(1.4)

Uniqueness: $u$ is the unique solution of (1.1) satisfying (1.4).

Nonexistence: In the following three cases:

  * i)

For any $\tau\in(-1,0)\setminus\\{-\frac{2\alpha}{p-1},\ \tau_{0}(\alpha)\\}$
and

$1+2\alpha<p<1-\frac{2\alpha}{\tau_{0}(\alpha)}\quad\mbox{ or}$

  * ii)

For any $\tau\in(-1,0)$ and

$p\geq 1-\frac{2\alpha}{\tau_{0}(\alpha)}\ \mbox{ or}$

  * iii)

For any $\tau\in(-1,0)\setminus\\{\tau_{0}(\alpha)\\}$ and

$1<p\leq 1+2\alpha,$

problem (1.1) does not have a solution $u$ satisfying

$0<\liminf_{x\in\Omega,x\to\partial\Omega}u(x)d(x)^{-\tau}\leq\limsup_{x\in\Omega,x\to\partial\Omega}u(x)d(x)^{-\tau}<+\infty.$
(1.5)

Special existence for $\tau=\tau_{0}(\alpha)$. Assume that

$\max\\{1-\frac{2\alpha}{\tau_{0}(\alpha)}+\frac{\tau_{0}(\alpha)+1}{\tau_{0}(\alpha)},1\\}<p<1-\frac{2\alpha}{\tau_{0}(\alpha)}.$

Then for any $t>0$, there is a positive solution $u$ of equation (1.1)
satisfying

$\lim_{x\in\Omega,x\to\partial\Omega}u(x)d(x)^{-\tau_{0}(\alpha)}=t.$

There are some challenging questions to ask:
_1\. Could $\tau_{0}(\alpha)$ be expressed explicitly?
2\. With what condition of general nonlinearity makes existence hold?
3\. The uniqueness and nonexistence restricts in the class functions (1.4) and
(1.5), so are there some solutions breaking the assumption (1.4)?
_

Our interest in this article is to introduce a new method to study the
boundary blow-up solutions of semilinear fractional elliptic equations and
answer above questions. The main idea is to find suitable type measure
concentrated on the whole boundary and then by making basic estimates to prove
that the corresponding weak solution solves (1.1). Our first result is stated
as follows:

###### Proposition 1.2

Let $\alpha\in(0,1)$ and $\tau_{0}(\alpha)$ is the zero point of $C(\cdot)$
when $C(\cdot)$ given by (1.3), then

$\tau_{0}(\alpha)=\alpha-1.$

We observe that the critical value $1-\frac{2\alpha}{\tau_{0}(\alpha)}$ in
Proposition 1.1 turns out to be $\frac{1+\alpha}{1-\alpha}$. In what follows,
we would like to show the details of our new method and answer the second and
third questions in the following.

###  1.2 A new method and main results

Let $\alpha\in(0,1)$ and $\omega$ be the Hausdorff measure on
$\partial\Omega$. We denote by
$\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}$ a measure

$\langle\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}},f\rangle=\int_{\partial\Omega}\frac{\partial^{\alpha}f(x)}{\partial\vec{n}_{x}^{\alpha}}d\omega(x),\qquad
f\in C^{\alpha}(\bar{\Omega}),$

where $\vec{n}_{x}$ is the unit inward normal vector of $\partial\Omega$ at
point $x$ and

$\frac{\partial^{\alpha}f(x)}{\partial\vec{n}_{x}^{\alpha}}=\lim_{t\to
0^{+}}\frac{f(x+t\vec{n}_{x})-f(x)}{t^{\alpha}}.$

In this paper, we are concerned with the existence and uniqueness of weak
solution to the semilinear fractional elliptic problem

$\begin{array}[]{lll}(-\Delta)^{\alpha}u+g(u)=k\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}&{\rm
in}\quad\bar{\Omega},\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}+g(u)}u=0&{\rm
in}\quad\bar{\Omega}^{c},\end{array}$ (1.6)

where $k>0$ and $g:\mathbb{R}_{+}\to\mathbb{R}_{+}$ is continuous.

In [6], the authors studied problem (1.6) replaced
$\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}$ by
$\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}$ where $\nu$ is a Radon
measure concentrated on boundary measure. They proved that such a problem has
a unique weak solution if $g$ is a continuous nondecreasing function
satisfying $g(0)\geq 0$ and

$\int_{1}^{\infty}g(s)s^{-1-\frac{N+\alpha}{N-\alpha}}ds<+\infty.$ (1.7)

Moreover, [6] analyzed the isolated singularity of weak solution of (1.6) in
the case that $\nu=\delta_{x_{0}}$ with $x_{0}\in\partial\Omega$. Our aim in
this article is to investigate how the Hausdorff measure on $\partial\Omega$
works on the weak solution of (1.6).

Before starting our main theorems we make precise the notion of weak solution
used in this note.

###### Definition 1.1

We say that $u$ is a weak solution of (1.6), if $u\in L^{1}(\Omega)$, $g(u)\in
L^{1}(\Omega,\rho^{\alpha}dx)$ and

$\int_{\Omega}[u(-\Delta)^{\alpha}\xi+g(u)\xi]dx=k\int_{\partial\Omega}\frac{\partial^{\alpha}\xi(x)}{\partial\vec{n}_{x}^{\alpha}}d\omega(x),\qquad\forall\xi\in\mathbb{X}_{\alpha}.$

where $\rho(x)={\rm dist}(x,\partial\Omega)$ and $\mathbb{X}_{\alpha}\subset
C(\mathbb{R}^{N})$ denotes the space of functions $\xi$ satisfying:

  *     1. $(i)$

${\rm supp}(\xi)\subset\bar{\Omega}$;

    1. $(ii)$

$(-\Delta)^{\alpha}\xi(x)$ exists for all $x\in\Omega$ and
$|(-\Delta)^{\alpha}\xi(x)|\leq C$ for some $C>0$;

    1. $(iii)$

there exist $\varphi\in L^{1}(\Omega,\rho^{\alpha}dx)$ and $\varepsilon_{0}>0$
such that $|(-\Delta)_{\varepsilon}^{\alpha}\xi|\leq\varphi$ a.e. in $\Omega$
for all $\varepsilon\in(0,\varepsilon_{0}]$.

Now we are ready to state our first result for problem (1.6).

###### Theorem 1.1

Assume that $k>0$, $\rho(x)={\rm dist}(x,\partial\Omega)$ and $g$ is a
continuous nondecreasing function satisfying $g(0)\geq 0$ and

$\int_{1}^{\infty}g(s)s^{-1-\frac{1+\alpha}{1-\alpha}}ds<+\infty.$ (1.8)

Then $(i)$ problem (1.6) admits a unique positive weak solution $u_{k}$;

$(ii)$ the mapping $k\to u_{k}$ is increasing and there exists $c_{1}\geq 1$
independent of $k$ such that

$\frac{k}{c_{1}}\rho(x)^{\alpha-1}\leq u_{k}(x)\leq
c_{1}k\rho(x)^{\alpha-1},\qquad\forall x\in\Omega;$ (1.9)

$(iii)$ if we assume additionally that $g$ is $C^{\beta}$ locally in
$\mathbb{R}$ with $\beta>0$, then $u_{k}$ is a classical solution of

$\begin{array}[]{lll}\ \ (-\Delta)^{\alpha}u+g(u)=0&{\rm
in}\quad\Omega,\\\\[5.69054pt] \phantom{------\ }\ \ u=0&{\rm
in}\quad\mathbb{R}^{N}\setminus\bar{\Omega},\\\\[5.69054pt]
\phantom{}\lim_{x\in\Omega,x\to\partial\Omega}u(x)=+\infty.\end{array}$ (1.10)

We remark that in Theorem 1.1 extends the special existence of boundary blow
up solutions to fractional elliptic equation (1.10) with general nonlinearity
$g$ in integral subcritical case with the critical exponents
$\frac{1+\alpha}{1-\alpha}$, which is larger than $\frac{N+\alpha}{N-\alpha}$.
Specially, letting $g\equiv 0$, there exists infinitely many boundary blow up
$\alpha-$harmonic functions.

Since $\alpha-1>-\frac{2\alpha}{p-1}$, so we may call the solutions of (1.10)
as the weak boundary blow-up solution from the asymptotic behavior (1.9). Our
second interest is to consider the limit of weak boundary blow-up solutions.

###### Theorem 1.2

Let $g(s)=s^{p}$ with $p\in(0,\frac{1+\alpha}{1-\alpha})$ and $u_{k}$ be the
weak solution of (1.6), then

$(i)$ if $p\in(1+2\alpha,\frac{1+\alpha}{1-\alpha})$, then the limit of
$\\{u_{k}\\}$ as $k\to\infty$ exists, denoting $u_{\infty}$, which is a
classical solution of (1.1). Moreover, $u_{\infty}$ satisfies

$\frac{1}{c_{2}}\rho(x)^{-\frac{2\alpha}{p-1}}\leq u_{\infty}(x)\leq
c_{2}\rho(x)^{-\frac{2\alpha}{p-1}},\qquad\forall x\in\Omega,$ (1.11)

where $c_{2}\geq 1$.

$(ii)$ if $p\in(0,1+2\alpha]$, then

$\lim_{k\to\infty}u_{k}(x)=+\infty,\qquad\forall x\in\Omega.$

We notice that the limit of weak boundary blow-up solutions is the solution of
(1.1) with behavior (1.4) when $p\in(1+2\alpha,\frac{1+\alpha}{1-\alpha})$
stated in Proposition 1.1. As a consequence of Theorem 1.2 $(ii)$,
$p\in(0,1+2\alpha]$, there is no solution $u$ of (1.1) such that

$\lim_{x\in\Omega,x\in\Omega}u(x)\rho^{1-\alpha}(x)=0\ \ {\rm or}\ \ +\infty.$

From [6] and Theorem 1.1, the Dirac mass and Hausdorff measure have different
contribution to the solution of

$(-\Delta)^{\alpha}u+g(u)=0\quad{\rm in}\quad\Omega.$

Our interest is to understand what singularity of the solution to

$\begin{array}[]{lll}(-\Delta)^{\alpha}u+g(u)=\frac{\partial^{\alpha}(\omega+\delta_{x_{0}})}{\partial\vec{n}^{\alpha}}&{\rm
in}\quad\bar{\Omega},\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}+g(u)}u=0&{\rm
in}\quad\bar{\Omega}^{c},\end{array}$ (1.12)

where $x_{0}\in\partial\Omega$ and $\delta_{x_{0}}$ is the Dirac mass
concentrated $x_{0}$ on the boundary. Inspired by Definition 1.1, it is
natural to give the definition of weak solution of (1.12) as following.

###### Definition 1.2

We say that $u$ is a weak solution of (1.12), if $u\in L^{1}(\Omega)$,
$g(u)\in L^{1}(\Omega,\rho^{\alpha}dx)$ and

$\int_{\Omega}[u(-\Delta)^{\alpha}\xi+g(u)\xi]dx=\int_{\partial\Omega}\frac{\partial^{\alpha}\xi(x)}{\partial\vec{n}_{x}^{\alpha}}d\omega(x)+\frac{\partial^{\alpha}\xi(x_{0})}{\partial\vec{n}_{x_{0}}},\qquad\forall\xi\in\mathbb{X}_{\alpha}.$

###### Theorem 1.3

Assume that $x_{0}\in\partial\Omega$, $g$ is a continuous nondecreasing
function satisfying $g(0)\geq 0$,

$\int_{1}^{\infty}g(s)s^{-1-\frac{N+\alpha}{N-\alpha}}ds<+\infty$ (1.13)

and for some $\lambda>0$,

$g(s+t)\leq\lambda[g(s)+g(t)],\quad\forall s,t>0.$ (1.14)

Then problem (1.6) admits a unique positive weak solution $v$ such that

$\frac{1}{c_{3}}\left[\rho(x)^{\alpha-1}+\frac{\rho(x)^{\alpha}}{|x-x_{0}|^{N}}\right]\leq
v(x)\leq
c_{3}\left[\rho(x)^{\alpha-1}+\frac{\rho(x)^{\alpha}}{|x-x_{0}|^{N}}\right],\qquad\forall
x\in\Omega.$ (1.15)

Moreover, if assume additionally that $g$ is $C^{\beta}$ locally in
$\mathbb{R}$ with $\beta>0$, then $v$ is a classical solution of (1.10).

From Theorem 1.3, we find out a classical solution of (1.10) with explosive
rate $\rho(x)^{\alpha-1}+\frac{\rho(x)^{\alpha}}{|x-x_{0}|^{N}}$, this answers
the question 3 in the first part of the introduction.

The boundary blow-up solutions of (1.10) could be searched for by making use
of measure type data on boundary and the main difficulty is to do the estimate
of
$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]$
and
$g(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])$.
Especially, it is dedicate to make the estimate of
$g(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])$
near the boundary when the nonlinearity $g$ is just integral-subcritical, i.e.
(1.8).

This article is organized as follows. In Section §2 we present some
preliminaries to the Marcinkiewicz type estimate for
$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]$
and present the existence and uniqueness of weak solution of (1.6) when $g$ is
bounded. Section §3, §4 are devoted to prove Theorem 1.1 and Theorem 1.2.
Finally, we obtain one typical solution that blows up along the boundary with
different power rate.

##  2 Preliminary

###  2.1 The Marcinkiewicz type estimate

In order to obtain the weak solution of (1.6) with integral subcritical
nonlinearity, we have to introduce the Marcinkiewicz space and recall some
related estimate.

###### Definition 2.1

Let $\Theta\subset\mathbb{R}^{N}$ be a domain and $\varpi$ be a positive Borel
measure in $\Theta$. For $\kappa>1$, $\kappa^{\prime}=\kappa/(\kappa-1)$ and
$u\in L^{1}_{loc}(\Theta,d\mu)$, we set

$\|u\|_{M^{\kappa}(\Theta,d\varpi)}=\inf\left\\{c\in[0,\infty]:\int_{E}|u|d\varpi\leq
c\left(\int_{E}d\varpi\right)^{\frac{1}{\kappa^{\prime}}},\ \forall
E\subset\Theta,\,E\ {\rm Borel}\right\\}$ (2.1)

and

$M^{\kappa}(\Theta,d\varpi)=\\{u\in
L_{loc}^{1}(\Theta,d\varpi):\|u\|_{M^{\kappa}(\Theta,d\varpi)}<+\infty\\}.$
(2.2)

The space $M^{\kappa}(\Theta,d\varpi)$ is called the Marcinkiewicz space of
exponent $\kappa$, or weak $L^{\kappa}$-space and
$\|.\|_{M^{\kappa}(\Theta,d\varpi)}$ is a quasi-norm.

###### Proposition 2.1

[2, 7] Assume that $1\leq q<\kappa<\infty$ and $u\in
L^{1}_{loc}(\Theta,d\varpi)$. Then there exists $c_{4}>0$ dependent of
$q,\kappa$ such that

$\int_{E}|u|^{q}d\varpi\leq
c_{4}\|u\|_{M^{\kappa}(\Theta,d\varpi)}\left(\int_{E}d\varpi\right)^{1-q/\kappa}$

for any Borel set $E$ of $\Theta$.

Denote by $G_{\alpha}$ the Green kernel of $(-\Delta)^{\alpha}$ in
$\Omega\times\Omega$ and by $\mathbb{G}_{\alpha}[\cdot]$ the Green operator
defined as

$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)=\lim_{t\to
0^{+}}\int_{\partial\Omega}G_{\alpha}(x,y+t\vec{n}_{y})t^{-\alpha}d\omega(y).$

Our purpose in this subsection is to do Marcinkiewicz type estimate for
$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]$.

###### Lemma 2.1

There exists $c_{5}\geq 1$ such that for any $x\in\Omega$,

$\frac{1}{c_{5}}\rho(x)^{\alpha-1}\leq\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)\leq
c_{5}\rho(x)^{\alpha-1}.$ (2.3)

Proof. Since $\partial\Omega$ is $C^{2}$, then there exists
$t_{0}\in(0,\frac{1}{2})$ such that for any
$x\in\Omega_{t}:=\\{z\in\Omega,\rho(x)<t\\}$ with $t<t_{0}$, there exists a
unique $x_{\partial}\in\partial\Omega$ such that

$|x-x_{\partial}|=\rho(x)$

and for $t\in(0,t_{0})$ letting

$\mathcal{C}_{t}=\\{x\in\Omega:\ \rho(x)=t\\},$

$\mathcal{C}_{t}$ is $C^{2}$ for $t\in(0,t_{0})$ and any Borel set $E_{t}$ in
$\mathcal{C}_{t}$, there exists unique set $E\subset\partial\Omega$ such that
for any $x_{t}\in E_{t}$, there exists a unique $x\in E$ such that

$|x_{t}-x|=t$ (2.4)

and for any $x\in E$, there exists a unique $x_{t}\in E_{t}$ satisfying (2.4).
Moreover, for $x\in\mathcal{C}_{t}$ with $t\in(0,t_{0})$, there exists a
unique $x_{\partial}\in\partial\Omega$ such that

$x=t\vec{n}_{x_{\partial}}+x_{\partial}\quad{\rm
and}\quad|x-x_{\partial}|=t=\rho(x).$

Denotes by $\omega_{t}$ a measure on $\mathcal{C}_{t}$ generated by $\omega$
such that for $t\in(0,t_{0})$,

$\omega_{t}(E_{t})=\omega(E)\qquad{\rm for\ any\ Borel\ set\ }\
E_{t}\subset\mathcal{C}_{t}.$ (2.5)

By compactness we only have to prove that (2.3) holds in a neighborhood of any
point $\bar{x}\in\partial\Omega$ and without loss of generality, we may assume
that

$x_{\partial}=0\quad{\rm and}\quad\vec{n}_{x_{\partial}}=e_{N}.$

From [6, Lemma 2.1], there exists $c_{6}>0$ such that

$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)\leq\int_{\partial\Omega}\frac{c_{6}}{|x-y|^{N-\alpha}}d\omega(y),\qquad\forall
x\in\Omega.$ (2.6)

Let $\phi:B^{\prime}_{t_{0}}(0)\to\mathbb{R}$ such that
$(y^{\prime},\phi(y^{\prime})\in\partial\Omega$, where $B^{\prime}_{t_{0}}(0)$
is the ball centered at origin with radium $t_{0}$ in $\mathbb{R}^{N-1}$. We
choose some $s_{0}\in(0,t_{0})$ small enough, there exists $c_{7}\geq 1$ such
that for any Borel set $E\subset B_{s_{0}}(0)\cap\partial\Omega$,

$\frac{1}{c_{7}}|E^{\prime}|\leq\omega(E)\leq c_{7}|E^{\prime}|,$

where

$E^{\prime}=\\{y^{\prime}\in\mathbb{R}^{N-1}:\quad(y^{\prime},\phi(y^{\prime}))\in
E\\}.$

For $s_{0}>0$ small, there exists $c_{8}>0$ such that for
$y=(y^{\prime},y_{N})\in B_{s_{0}}(0)\cap\partial\Omega$

$|te_{N}-y|\geq
c_{8}|te_{N}-(y^{\prime},0)|=c_{8}\sqrt{t^{2}+|y^{\prime}|^{2}}$

Therefore,

$\displaystyle\int_{B_{s_{0}}(0)\cap\partial\Omega}\frac{1}{|te_{N}-y|^{N-\alpha}}d\omega(y)$
$\displaystyle\leq$ $\displaystyle
c_{9}\int_{B_{s_{0}}(0)\cap\partial\Omega}\frac{1}{(t^{2}+|y^{\prime}|^{2})^{\frac{N-\alpha}{2}}}d\omega(y)$
$\displaystyle\leq$ $\displaystyle
c_{9}\int_{B^{\prime}_{s_{0}}(0)}\frac{1}{(t^{2}+|y^{\prime}|^{2})^{\frac{N-\alpha}{2}}}dy^{\prime}$
$\displaystyle=$ $\displaystyle
c_{10}\int_{0}^{s_{0}}\frac{s^{N-2}}{(t^{2}+s^{2})^{\frac{N-\alpha}{2}}}ds$
$\displaystyle=$ $\displaystyle
c_{10}t^{\alpha-1}\int_{0}^{\frac{s_{0}}{t}}\frac{s^{N-2}}{(1+s^{2})^{\frac{N-\alpha}{2}}}ds$
$\displaystyle\leq$ $\displaystyle c_{11}t^{\alpha-1},$

where $c_{9},c_{10}>0$ and
$c_{11}=\int_{0}^{+\infty}\frac{c_{10}}{(1+s^{2})^{\frac{2-\alpha}{2}}}ds<+\infty$
since $\frac{2-\alpha}{2}>\frac{1}{2}$. For $y\in\partial\Omega\setminus
B_{s_{0}}(0)$, there exists $c_{12}>0$ such that $|te_{N}-y|\geq c_{12}s_{0}$,
then

$\displaystyle\int_{\partial\Omega\setminus
B_{s_{0}}(0)}\frac{1}{|te_{N}-y|^{N-\alpha}}d\omega(y)\leq
c_{12}s_{0}^{\alpha-N}\int_{\partial\Omega\setminus
B_{s_{0}}(0)}d\omega(y)\leq c_{12}s_{0}^{\alpha-N}\omega(\partial\Omega).$

Therefore, for $t\in(0,t_{0})$,

$\int_{\partial\Omega}\frac{1}{|te_{N}-y|^{N-\alpha}}d\omega(y)\leq
c_{12}t^{\alpha-1}.$

For $x\in\Omega\setminus\Omega_{t_{0}}$ and $y\in\partial\Omega$, we observe
that $|x-y|\geq t_{0}$, then
$\int_{\partial\Omega}\frac{1}{|x-y|^{N-\alpha}}d\omega(y)$ is bounded by some
constant dependent of $t_{0}$ and the diameter of $\Omega$, thus, (2.3) holds.

We now prove that for $t\in(0,t_{0})$,

$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](te_{N})\geq\frac{1}{c_{6}}t^{\alpha-1}.$
(2.7)

For all $s\in(0,\frac{t}{8})$, we have that

$|te_{N}-y|>\frac{t}{2}\quad{\rm for}\quad y\in\mathcal{C}_{s}\cap
B_{\frac{t}{4}}(se_{N}).$

Therefore,

$|te_{N}-y|>\frac{t}{2}=\frac{1}{2}\max\\{\rho(y),\rho(te_{N})\\}$

and apply [4, Theorem 1.2] to derive that there exists $c_{13}>0$ such that
for all $s\in(0,\frac{t}{8})$

$G_{\alpha}(te_{N},y)\geq
c_{13}\frac{\rho^{\alpha}(y)\rho^{\alpha}(te_{N})}{|te_{N}-y|^{N}}=c_{13}\frac{t^{\alpha}s^{\alpha}}{|te_{N}-y|^{N}},\quad
y\in\mathcal{C}_{s}\cap B_{\frac{t}{4}}(se_{N}).$ (2.8)

Thus,

$\mathbb{G}_{\alpha}[s^{-\alpha}\omega_{s}](te_{N})\geq
c_{13}\int_{\mathcal{C}_{s}\cap
B_{\frac{t}{4}}(se_{N})}\frac{t^{\alpha}}{|te_{N}-y|^{N}}d\omega_{s}(y).$

Denote

$D_{t,s}=\mathcal{C}_{s}\cap B_{\frac{t}{4}}(se_{N})\quad{\rm and}\quad
D_{t}=\partial\Omega\cap B_{\frac{t}{4}}(0).$

We observe that

$|te_{N}-y|\leq c_{14}t,\quad\forall y\in D_{t,s}$

and

$\frac{1}{c_{14}}t^{N-1}\leq\omega_{s}(D_{t,s})\leq c_{14}t^{N-1},$

where $c_{14}>1$, then for any $s\in(0,\frac{t}{8})$

$\displaystyle\int_{\mathcal{C}_{s}}\frac{t^{\alpha}}{|te_{N}-y|^{N}}d\omega_{s}(y)$
$\displaystyle\geq$
$\displaystyle\int_{D_{t,s}}\frac{t^{\alpha}}{|x-y|^{N}}d\omega_{\mathcal{C}_{s}}(y)$
$\displaystyle\geq$ $\displaystyle
c_{15}t^{\alpha}t^{-N}\omega_{s}(D_{t,s})\geq c_{16}t^{\alpha-1},$

which implies (2.7) by passing the limit of $s\to 0^{+}$. $\Box$

###### Proposition 2.2

Let
$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]$
given by (2.6) and $p^{*}=\frac{1+\alpha}{1-\alpha}$. Then there exists
$c_{17}>0$ such that

$\|\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]\|_{M^{p^{*}}(\Omega,\rho^{\alpha}dx)}\leq
c_{17}.$ (2.9)

Proof. For any Borel set $E$ of $\Omega$ satisfying

$0<|E|<|\Omega_{t_{0}}|,$

where $\Omega_{r}=\\{x\in\Omega,\ \rho(x)<r\\}$ for $r>0$, there exists
$t\in(0,t_{0})$ such that

$|E|=|\Omega_{t}|.$

Then there exists $c_{18}>0$ such that

$|\Omega_{t}|=\int_{0}^{t}\omega_{t}(\mathcal{C}_{t})dt\leq c_{18}t.$

We observe that

$|E\setminus\Omega_{t}|=|E|-|E\cap\Omega_{t}|=|\Omega_{t}\setminus E|$

and

$\rho(y)\leq\rho(z),\quad\forall y\in E\setminus\Omega_{t},\ \forall
z\in\Omega_{t}\setminus E.$

Then

$\displaystyle\int_{E}\rho^{\alpha}dx$ $\displaystyle\geq$
$\displaystyle\int_{\Omega_{t}}\rho^{\alpha}(x)dx=c_{19}\int_{0}^{t}\int_{\mathcal{C}_{s}}s^{\alpha}d\omega_{s}ds\geq
c_{20}t^{\alpha+1}$

and together with (2.3), we deduce that

$\displaystyle\int_{E}\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)\rho^{\alpha}(x)dx$
$\displaystyle\leq$ $\displaystyle c_{21}\int_{E}\rho^{2\alpha-1}(x)dx\leq
c_{21}\int_{\Omega_{t}}\rho^{2\alpha-1}(x)dx$ $\displaystyle=$ $\displaystyle
c_{22}\int_{0}^{t}\int_{\mathcal{C}_{s}}s^{2\alpha-1}d\omega_{s}ds=c_{23}t^{2\alpha}$
$\displaystyle\leq$ $\displaystyle
c_{23}(\int_{E}\rho^{\alpha}dx)^{\frac{2\alpha}{1+\alpha}},$

where $c_{22},c_{23}>0$. Therefore,

$\int_{E}\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)\rho^{\alpha}(x)dx\leq
c_{23}(\int_{E}\rho^{\alpha}dx)^{\frac{2\alpha}{1+\alpha}}.$

Together with

$\frac{2\alpha}{1+\alpha}=\frac{p^{*}-1}{p^{*}},$

we derive (2.9). This completes the proof. $\Box$

###  2.2 Existence for bounded nonlinearity

We extend Hausdorff measure $\omega$ to $\bar{\Omega}$ by zero inside
$\Omega$, still denoting $\omega$. For bounded $C^{2}$ domain, it follows [19,
p 57] that $\omega$ is a Radon measure in $\bar{\Omega}$. In the approximating
to weak solution of (1.6), we consider a sequence $\\{g_{n}\\}$ of $C^{1}$
nonnegative functions defined on $\mathbb{R}_{+}$ such that $g_{n}(0)=g(0)$,

$g_{n}\leq g_{n+1}\leq g,\quad\sup_{s\in\mathbb{R}_{+}}g_{n}(s)=n\quad{\rm
and}\quad\lim_{n\to\infty}\|g_{n}-g\|_{L^{\infty}_{loc}(\mathbb{R}_{+})}=0.$
(2.10)

###### Proposition 2.3

Assume that $\\{g_{n}\\}_{n}$ is given by (2.10). Then

$\begin{array}[]{lll}(-\Delta)^{\alpha}u+g_{n}(u)=k\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}&{\rm
in}\quad\Omega,\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}+g_{n}(u)}u=0&{\rm
in}\quad\Omega^{c}\end{array}$ (2.11)

admits a unique positive weak solution $u_{k,n}$ satisfying

$(i)$ the mapping $k\to u_{k,n}$ is increasing, the mapping $n\to u_{k,n}$ is
decreasing

$k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)-k\mathbb{G}_{\alpha}[g_{n}(k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])](x)\leq
u_{k,n}(x)\leq
k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x),\quad\forall
x\in\Omega;$ (2.12)

$(ii)$ $u_{k,n}$ is a classical solution of

$\begin{array}[]{lll}\ (-\Delta)^{\alpha}u+g_{n}(u)=0&{\rm
in}\quad\Omega,\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}+g_{n}(u)}\ u=0&{\rm
in}\quad\mathbb{R}^{N}\setminus\bar{\Omega},\\\\[5.69054pt]
\phantom{}\lim_{x\in\Omega,x\to\partial\Omega}u(x)=+\infty.\end{array}$ (2.13)

Proof. Since $\omega$ is a Radon measure in $\bar{\Omega}$, we could apply [6,
Theorem 1.1] to obtain that problem (2.11) admits a unique weak solution
$u_{k,n}$ satisfying that $(i)$ and $u_{k,n}$ is a classical solution of

$(-\Delta)^{\alpha}u+g_{n}(u)=0\quad{\rm in}\quad\Omega,\qquad\ u=0\quad{\rm
in}\quad\mathbb{R}^{N}\setminus\bar{\Omega}.$

From Lemma 2.1 and (2.12), there exists $c_{24}\geq 1$ such that

$\frac{1}{c_{24}}\rho(x)^{\alpha-1}\leq u_{k,n}(x)\leq
c_{24}\rho(x)^{\alpha-1},\quad x\in\Omega.$ (2.14)

Therefore, $u_{k,n}$ is a classical solution of (2.13). $\Box$

In particular, let $g_{0}\equiv 0$, we have that

###### Corollary 2.1

$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]$
is a classical solution of

$\begin{array}[]{lll}\qquad\quad\ \ (-\Delta)^{\alpha}u=0&{\rm
in}\quad\Omega,\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}}\qquad\quad\ \
u=0&{\rm in}\quad\mathbb{R}^{N}\setminus\bar{\Omega},\\\\[5.69054pt]
\phantom{}\lim_{x\in\Omega,x\to\partial\Omega}u(x)=+\infty.\end{array}$ (2.15)

With the help of Corollary 2.1, we are in the position to prove Proposition
1.2.

Proof of Proposition 1.2. We first prove that $\tau_{0}(\alpha)\leq\alpha-1$.
Inversely, if $\tau_{0}(\alpha)>\alpha-1$, then we have that

$1-\frac{2\alpha}{\tau_{0}(\alpha)}>\frac{1+\alpha}{1-\alpha}>1+2\alpha.$

On the one hand, it follows by [5, Theorem 1.1] that for
$p=\frac{1+\alpha}{1-\alpha}$, problem

$\begin{array}[]{lll}(-\Delta)^{\alpha}u+|u|^{p-1}u=0&{\rm
in}\quad\Omega,\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}+|u|^{p-1}u}u=0&{\rm
in}\quad\mathbb{R}^{N}\setminus\bar{\Omega},\\\\[5.69054pt] \phantom{\
}\lim_{x\in\Omega,x\to\partial\Omega}u(x)=+\infty\end{array}$ (2.16)

admits a solution $w$ such that

$\frac{1}{c_{25}}\leq w(x)\rho^{1-\alpha}(x)\leq c_{25},\quad x\in\Omega,$

where $c_{25}>1$ and

$-\frac{2\alpha}{\frac{1+\alpha}{1-\alpha}-1}=\alpha-1.$

On the other hand, form Corollary 2.1 we know that for any $\mu>0$,
$\mu\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]$
is a super solution of problem (2.16). Furthermore, from Lemma 2.1,

$\frac{1}{c_{5}}\leq\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)\rho^{1-\alpha}\leq
c_{5}.$

Now choosing $\mu_{1}=\frac{1}{2c_{25}c_{5}}$ and $\mu_{2}=2c_{25}c_{5}$, we
derive that

$\limsup_{x\in\Omega,\
x\to\partial\Omega}\mu_{1}\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)\rho(x)^{1-\alpha}<\frac{1}{c_{25}},\
\ \liminf_{x\in\Omega,\
x\to\partial\Omega}\mu_{2}\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)\rho(x)^{1-\alpha}>c_{25}.$

thus, from [5, Proposition 6.1], there is no solution $u$ such that

$\frac{1}{c_{25}}\leq\liminf_{x\in\Omega,\
x\to\partial\Omega}u(x)\rho(x)^{1-\alpha}\leq\limsup_{x\in\Omega,\
x\to\partial\Omega}u(x)\rho(x)^{1-\alpha}\leq c_{25}.$

The contradiction is obvious.

We finally prove that $\tau_{0}(\alpha)\geq\alpha-1$. Inversely, if
$\tau_{0}(\alpha)<\alpha-1$, then we have that

$1-\frac{2\alpha}{\tau_{0}(\alpha)}<\frac{1+\alpha}{1-\alpha}.$

From [5, Theorem 1.1] nonexistence $(ii)$, there is no solution $u$ of problem
(2.16) with

$\max\\{1-\frac{2\alpha}{\tau_{0}(\alpha)},\frac{2\alpha}{1-\alpha}\\}<p<\frac{1+\alpha}{1-\alpha}$
(2.17)

such that

$\displaystyle 0<\liminf_{x\in\Omega,\
x\to\partial\Omega}u(x)\rho(x)^{1-\alpha}\leq\limsup_{x\in\Omega,\
x\to\partial\Omega}u(x)\rho(x)^{1-\alpha}<+\infty.$ (2.18)

Let $\bar{\tau}=2\alpha-(1-\alpha)p$, then

$\bar{\tau}>2\alpha-(1-\alpha)\frac{1+\alpha}{1-\alpha}=\alpha-1$

and

$\bar{\tau}<2\alpha-(1-\alpha)\frac{2\alpha}{1-\alpha}=0.$

For $t_{0}>0$ small, $\Omega_{t_{0}}=\\{x\in\Omega,\rho(x)<t_{0}\\}$ is
$C^{2}$ and define

$V_{1}(x)=\left\\{\begin{array}[]{lll}d(x)^{\bar{\tau}},&x\in\Omega_{t_{0}},\\\\[5.69054pt]
l(x),&x\in\Omega\setminus\Omega_{t_{0}},\\\\[5.69054pt]
0,&x\in\Omega^{c},\end{array}\right.$ (2.19)

where the function $l$ is positive such that $V_{1}$ is $C^{2}$ in $\Omega$.
From [5, Proposition 3.2 $(ii)$], there exists $\delta_{1}\in(0,t_{0}]$ and
$c_{26}>1$ such that

$\frac{1}{c_{26}}\rho(x)^{\bar{\tau}-2\alpha}\leq(-\Delta)^{\alpha}V_{1}(x)\leq
c_{26}\rho(x)^{\bar{\tau}-2\alpha},\ \ \forall x\in\Omega_{\delta_{1}}.$

We observe that
$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]$
is a super solution of (2.16) with $p$ in (2.17). Now we define

$W_{\mu}(x)=\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]-\mu
V_{\bar{\tau}}(x)-\mu^{2}\mathbb{G}_{\alpha}[1],$

where $\mathbb{G}_{\alpha}[1]$ is the solution of

$\begin{array}[]{lll}(-\Delta)^{\alpha}u=1&{\rm in}\quad\Omega,\\\\[5.69054pt]
\phantom{(-\Delta)^{\alpha}}u=0&{\rm
in}\quad\mathbb{R}^{N}\setminus\Omega.\end{array}$

We see that $\bar{\tau}-2\alpha=(\alpha-1)p$, there exists $\mu_{1}>0$ such
that for $\mu\geq\mu_{1}$ and $x\in\Omega_{\delta_{1}}$,

$\displaystyle(-\Delta)^{\alpha}W_{\mu}(x)+|W_{\mu}|^{p-1}W_{\mu}(x)$
$\displaystyle\leq$ $\displaystyle-
c_{26}\mu\rho(x)^{\bar{\tau}-2\alpha}+c_{5}^{p}\rho(x)^{(\alpha-1)p}\leq 0$

and there exists $\mu_{2}>0$ such that for $\mu\geq\mu_{2}$ and
$x\in\Omega\setminus\Omega_{\delta_{1}}$,

$\displaystyle(-\Delta)^{\alpha}W_{\mu}(x)+|W_{\mu}|^{p-1}W_{\mu}(x)$
$\displaystyle\leq$ $\displaystyle
c_{26}\mu\max_{\Omega\setminus\Omega_{\delta_{1}}}|(-\Delta)^{\alpha}V_{1}|-\mu^{2}+\left(\min_{\Omega\setminus\Omega_{\delta_{1}}}\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]\right)^{p}\leq
0,$

Therefore, for $\mu=\max\\{\mu_{1},\mu_{2}\\}$, $W_{\mu}$ is a sub solution of
(2.16) with $p$ in (2.17) and

$c_{5}\rho^{\alpha-1}\geq\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]\geq
W_{\mu}\quad{\rm and}\quad\liminf_{x\in\Omega,\
x\to\partial\Omega}W_{\mu}(x)\rho^{1-\alpha}(x)\geq\frac{1}{c_{5}}.$

By [5, Theorem 2.6], there exists a solution $u$ of (2.16) with $p$ in (2.17)
satisfying (2.18). A contradiction is obtained and the proof is complete.
$\Box$

##  3 Proof of Theorem 1.1

###### Lemma 3.1

$(i)$ Assume that $g$ is a continuous nondecreasing function satisfying
$g(0)\geq 0$ and (1.8). Then

$\lim_{\rho(x)\to
0^{+}}\mathbb{G}_{\alpha}[g(\mathbb{G}_{\alpha}[k\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])](x)\rho(x)^{1-\alpha}=0.$
(3.1)

$(ii)$ Assume that $p\in(0,\frac{1+\alpha}{1-\alpha})$, then there exists
$c_{27}>0$ such that for any $x\in\Omega_{t}$ with $t\in(0,t_{0})$,

$\mathbb{G}_{\alpha}[(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])^{p}](x)\leq
c_{27}\rho(x)^{2\alpha-(1-\alpha)p}+c_{33}$ (3.2)

Proof. $(i)$ Without loss of generality, we may assume that

$0\in\partial\Omega,\quad\vec{n}_{0}=e_{N},\quad x_{s}=se_{N}$

and we just need prove (3.2) and (3.1) for $x_{s}$ with $s\in(0,t_{0})$. It
follows by Lemma 2.1 that

$\frac{1}{c_{5}}\rho(x)^{\alpha-1}\leq\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)\leq
c_{5}\rho(x)^{\alpha-1},\qquad\forall x\in\Omega.$ (3.3)

Combining with monotonicity of $g$, we have that

$\displaystyle\mathbb{G}_{\alpha}[g(\mathbb{G}_{\alpha}[k\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])](x_{s})s^{1-\alpha}$
$\displaystyle\leq$
$\displaystyle\int_{\Omega}G_{\alpha}(x_{s},z)g(c_{5}k\rho(z)^{\alpha-1})dzs^{1-\alpha}$
$\displaystyle\leq$
$\displaystyle\int_{\Omega}\frac{c_{5}\rho^{\alpha}(z)}{|x_{s}-z|^{N-\alpha}}g(c_{5}k\rho(z)^{\alpha-1})dzs^{1-\alpha}$
$\displaystyle=$ $\displaystyle
c_{5}\left[\int_{B_{t_{0}}(0)\cap\Omega}\frac{s^{1-\alpha}\rho^{\alpha}(z)}{|x_{s}-z|^{N-\alpha}}g(c_{5}k\rho(z)^{\alpha-1})dz\right.$
$\displaystyle\left.+\int_{\Omega\setminus
B_{t_{0}}(0)}\frac{s^{1-\alpha}\rho^{\alpha}(z)}{|x_{s}-z|^{N-\alpha}}g(c_{5}k\rho(z)^{\alpha-1})dz\right]$
$\displaystyle:=$ $\displaystyle A_{1}(s)+A_{2}(s).$

Let $B^{\prime}_{t_{0}}(0)$ be the ball with radium $t_{0}$ and centered at
the origin in $\mathbb{R}^{N-1}$ and since $\partial\Omega$ is $C^{2}$, there
exists a $C^{2}$ function $\psi:B^{\prime}_{\eta}(0)\to\mathbb{R}$ such that

$(z^{\prime},\psi(z^{\prime})\in\partial\Omega\quad{\rm for\ any}\
z^{\prime}\in B^{\prime}_{\eta}(0),$

where $\eta>0$. Denote

$\Psi(y)=(y^{\prime},\psi(y^{\prime}))+y_{N}\vec{n}_{(y^{\prime},\psi(y^{\prime}))},\quad\forall
y=(y^{\prime},y_{N})\in Q_{\eta},$

where

$Q_{\eta}=\\{z=(z^{\prime},z_{N})\in\mathbb{R}^{N-1}\times\mathbb{R},\
|z^{\prime}|<\eta,\ 0<z_{N}<\eta\\}.$

Thus, $\Psi$ is a $C^{2}$ diffeomorphism mapping such that

$\Psi(y)=y,\quad\forall y=te_{N},\ t\in(0,\eta).$

Therefore, if $t_{0}>0$ is chosen small enough, we have that $\Omega\cap
B_{t_{0}}(0)\subset\Psi(Q_{\eta})$ and there exists $c_{28}>1$ such that for
$z=\Psi(y)\in\Psi(Q_{\eta})$,

$\rho(z)=y_{N}\quad{\rm and}\quad\frac{1}{c_{28}}|se_{N}-y|\leq|se_{N}-z|\leq
c_{28}|se_{N}-y|.$ (3.4)

Then we have that

$\displaystyle A_{1}(s)$ $\displaystyle=$
$\displaystyle\int_{B_{t_{0}}(0)\cap\Omega}\frac{s^{1-\alpha}\rho^{\alpha}(z)}{|x_{s}-z|^{N-\alpha}}g(c_{5}k\rho(z)^{\alpha-1})dz$
$\displaystyle\leq$ $\displaystyle
c_{28}\int_{B_{t_{0}}(0)\cap\Omega}\frac{s^{1-\alpha}\rho^{\alpha}(z)}{|x_{s}-\Psi^{-1}(z)|^{N-\alpha}}g(c_{5}k\rho(z)^{\alpha-1})dz$
$\displaystyle\leq$ $\displaystyle
c_{28}\int_{Q_{\eta}}\frac{s^{1-\alpha}y_{N}^{\alpha}}{|x_{s}-y|^{N-\alpha}}g(c_{5}ky_{N}^{\alpha-1})dy.$

For $s\in(0,\frac{1}{8}\eta)$, we decompose $Q_{\eta}$ as following

$Q_{i,0}=\\{z=(z^{\prime},z_{N})\in Q_{t_{0}}:\
\frac{is}{2}\leq|z^{\prime}|<\frac{(i+1)s}{2},\ 0<z_{N}<\frac{s}{2}\\}$

and

$Q_{i,j}=\\{z=(z^{\prime},z_{N})\in Q_{t_{0}}:\
\frac{is}{2}\leq|z^{\prime}|<\frac{(i+1)s}{2},\ (j+\frac{1}{2})s\leq
z_{N}<(j+\frac{3}{2})s\\},$

where $i=0,1,\cdots N_{s}$, $j=1,\cdots N_{s}$ and $N_{s}$ is the largest
integer number such that $N_{s}\leq\frac{\eta}{8s}$.

For $y\in Q_{0,1}$, we have that $\frac{s}{2}\leq y_{N}\leq\frac{3s}{2}$ and

$\displaystyle\int_{Q_{0,1}}\frac{s^{1-\alpha}y_{N}^{\alpha}}{|x_{s}-y|^{N-\alpha}}g(c_{5}ky_{N}^{\alpha-1})dy$
$\displaystyle\leq$ $\displaystyle c_{5}s^{1+\alpha}g\
(c_{29}ks^{\alpha-1})\int_{B_{2/3}(e_{N})}\frac{1}{|e_{N}-z|^{N-\alpha}}dz$
$\displaystyle=$ $\displaystyle
c_{5}r^{-\frac{1+\alpha}{1-\alpha}}g(c_{29}kr)\int_{B_{2/3}(e_{N})}\frac{1}{|\vec{n}_{0}-z|^{N-\alpha}}dz$
$\displaystyle\leq$ $\displaystyle
c_{30}r^{-\frac{1+\alpha}{1-\alpha}}g(c_{29}kr),$

where $r=s^{\alpha-1}$ and $c_{29},c_{30}>0$.

For $y\in Q_{i,0}$ with $i=0,\cdots,N_{s}$, we obtain that
$|x_{s}-y|\geq\frac{i+1}{4}s$ and

$\displaystyle\int_{Q_{i,0}}\frac{s^{1-\alpha}y_{N}^{\alpha}}{|x_{s}-y|^{N-\alpha}}g(c_{5}ky_{N}^{\alpha-1})dy$
$\displaystyle\leq$
$\displaystyle\frac{c_{31}s^{1-N}}{(1+i)^{N-\alpha}}\int_{Q_{i,0}}y_{N}^{\alpha}g(c_{5}ky_{N}^{\alpha-1})dy$
$\displaystyle=$
$\displaystyle\frac{c_{32}}{(1+i)^{N-\alpha}}\int_{0}^{\frac{s}{2}}t^{\alpha}g(c_{5}kt^{\alpha-1})dt$
$\displaystyle=$
$\displaystyle\frac{c_{32}(1-\alpha)^{-1}}{(1+i)^{N-\alpha}}\int_{s^{-\frac{1}{N-\alpha}}}^{\infty}\tau^{-1-\frac{1+\alpha}{1-\alpha}}g\left(c_{5}k\tau\right)d\tau,$

where $r=s^{\alpha-1}$ and $c_{31},c_{32}>0$.

For $y\in Q_{i,j}$ with $i=0,\cdots,N_{s}$, $j=1,\cdots,N_{s}$ and
$(i,j)\not=(0,1)$, we derive that $|x_{s}-y|\geq\frac{i+j}{4}s$ and
$(j+\frac{1}{2})s\leq y_{N}<(j+\frac{3}{2})s$

$\displaystyle\int_{Q_{i,j}}\frac{s^{1-\alpha}y_{N}^{\alpha}}{|x_{s}-y|^{N-\alpha}}g(c_{5}ky_{N}^{\alpha-1})dy$
$\displaystyle\leq$
$\displaystyle\frac{c_{33}s^{1-N}}{(i+j)^{N-\alpha}}(js)^{\alpha}g\left(c_{5}k(js)^{\alpha-1}\right)|Q_{i,j}|$
(3.7) $\displaystyle=$
$\displaystyle\frac{c_{34}j^{\alpha}}{(i+j)^{N-\alpha}}r^{-\frac{1+\alpha}{1-\alpha}}g(c_{5}kj^{\alpha-1}r),$

where $r=s^{\alpha-1}$ and $c_{33},c_{34}>0$.

Therefore, there exist $c_{35},c_{36}>0$ such that

$A_{1}(s)\leq\sum_{i=0,j=0}^{N_{s}}\frac{c_{35}j^{\alpha}r^{-\frac{1+\alpha}{1-\alpha}}}{(i+j+1)^{N-\alpha}}g(c_{29}kj^{\alpha-1}r)+\sum_{i=0}^{N_{s}}\frac{c_{36}}{(1+i)^{N-\alpha}}\int_{s^{-\frac{1}{N-\alpha}}}^{\infty}\tau^{-1-\frac{1+\alpha}{1-\alpha}}g\left(c_{5}k\tau\right)d\tau.$

Since

$\sum_{i=0}^{N_{s}}\frac{c_{36}}{(1+i)^{N-\alpha}}\int_{s^{-\frac{1}{N-\alpha}}}^{\infty}\tau^{-1-\frac{1+\alpha}{1-\alpha}}g\left(c_{5}k\tau\right)d\tau\leq
c_{37}\int_{s^{-\frac{1}{N-\alpha}}}^{\infty}\tau^{-1-\frac{1+\alpha}{1-\alpha}}g\left(c_{5}k\tau\right)d\tau$

which tends to 0 as $s\to 0^{+}$ by hypothesis (1.8).

For any $\epsilon>0$, there exists $n_{\epsilon}>1$ such that

$\sum_{i,j=n_{\epsilon}}^{N_{s}}\frac{c_{35}}{(i+j+1)^{N-\alpha}}\leq\epsilon,$

and since
$\\{(j^{\alpha-1}r)^{-\frac{1+\alpha}{1-\alpha}}g(c_{29}kj^{\alpha-1}r)\\}$ is
uniformly bounded, we imply that

$\displaystyle\sum_{i,j=n_{\epsilon}}^{N_{s}}\frac{c_{35}j^{\alpha}}{(i+j+1)^{N-\alpha}}r^{-\frac{1+\alpha}{1-\alpha}}g(c_{29}kj^{\alpha-1}r)$
$\displaystyle=$
$\displaystyle\sum_{i,j=n_{\epsilon}}^{N_{s}}\frac{c_{35}j^{-1}}{(i+j+1)^{N-\alpha}}(j^{\alpha-1}r)^{-\frac{1+\alpha}{1-\alpha}}g(c_{29}kj^{\alpha-1}r)$
$\displaystyle\leq$ $\displaystyle c_{36}\epsilon.$

For $i,j<n_{\epsilon}$, there exists $s_{\epsilon}\in(0,\eta)$ such that

$j^{\alpha}r^{-\frac{1+\alpha}{1-\alpha}}g(c_{29}kj^{1-\alpha}r)\leq\epsilon,\quad{\rm
for}\ r\geq s_{\epsilon}^{\alpha-1},$

thus, for any $\epsilon>0$, there exists $s_{\epsilon}$ such that for
$s\in(0,s_{\epsilon})$

$\displaystyle\sum_{i=0,j=0}^{n_{\epsilon}}\frac{c_{35}j^{\alpha}}{(i+j+1)^{N-\alpha}}r^{-\frac{1+\alpha}{1-\alpha}}g(c_{29}kj^{\alpha-1}r)$
$\displaystyle\leq$
$\displaystyle\epsilon\sum_{i=0,j=0}^{N_{s}}\frac{c_{35}j^{-1}}{(i+j+1)^{N-\alpha}}$
$\displaystyle\leq$ $\displaystyle c_{37}\epsilon.$

Then we deduce that

$\lim_{s\to 0^{+}}A_{1}(s)=0.$

Therefore,

$\lim_{s\to
0^{+}}\mathbb{G}_{\alpha}[g(\mathbb{G}_{\alpha}[k\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])](x_{s})s^{1-\alpha}=0.$

Since $|x_{s}-z|\geq c_{38}t_{0}$ for $\Omega\setminus B_{t_{0}}(0)$,
therefore,

$\displaystyle A_{2}$ $\displaystyle\leq$ $\displaystyle
c_{39}s^{1-\alpha}\int_{\Omega\setminus
B_{t_{0}}(0)}\rho^{\alpha}(z)g(c_{5}k\rho(z)^{\alpha-1})dz$
$\displaystyle\leq$ $\displaystyle
c_{39}s^{1-\alpha}\omega(\partial\Omega)\int_{0}^{d_{0}}t^{\alpha}g(c_{5}kt^{\alpha-1})dt$
$\displaystyle\leq$ $\displaystyle c_{40}s^{1-\alpha},$

where $c_{39},c_{40}>0$ and $d_{0}=\max_{x\in\Omega}\rho(x)$. Thus, (3.1)
holds.

$(ii)$ When $g(s)=s^{p}$ with $p\in(0,\frac{1+\alpha}{1-\alpha})$, we observe
that (3) becomes that

$\displaystyle\int_{Q_{0,1}}\frac{s^{1-\alpha}y_{N}^{\alpha}}{|x_{s}-y|^{N-\alpha}}(c_{5}y_{N}^{\alpha-1})^{p}dy\leq
c_{5}^{p}s^{1+\alpha+(\alpha-1)p},$

(3) turns out to

$\displaystyle\int_{Q_{i,0}}\frac{s^{1-\alpha}y_{N}^{\alpha}}{|x_{s}-y|^{N-\alpha}}(c_{5}y_{N}^{\alpha-1})^{p}dy$
$\displaystyle=$
$\displaystyle\frac{c_{41}}{(1+i)^{N-\alpha}}\int_{0}^{\frac{s}{2}}t^{\alpha+(\alpha-1)p}dt$
$\displaystyle\leq$
$\displaystyle\frac{c_{42}}{(1+i)^{N-\alpha}}s^{1+\alpha+(\alpha-1)p}$

and (3.7) becomes that

$\displaystyle\int_{Q_{i,j}}\frac{s^{1-\alpha}y_{N}^{\alpha}}{|x_{s}-y|^{N-\alpha}}(c_{5}y_{N}^{\alpha-1})^{p}dy$
$\displaystyle\leq$
$\displaystyle\frac{c_{43}}{(1+i+j)^{N-\alpha}}s^{1+\alpha+(\alpha-1)p}.$

Therefore, we have that

$\int_{B_{t_{0}}(0)\cap\Omega}\frac{\rho^{\alpha}(y)}{|x_{s}-y|^{N-\alpha}}\rho(y)^{(\alpha-1)p}dy\leq
c_{44}s^{2\alpha+(\alpha-1)p}.$

which, combining (3), implies (3.2). $\Box$

Proof of Theorem 1.1. To prove the existence of weak solution. Take
$\\{g_{n}\\}$ a sequence of $C^{1}$ nondecreasing functions defined on
$\mathbb{R}$ satisfying $g_{n}(0)=g(0)$ and (2.10). By Proposition 2.3,
problem (2.11) admits a unique weak solution $u_{k,n}$ such that

$0<u_{k,n}\leq\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]\quad{\rm
in}\quad\Omega$

and

$\int_{\Omega}[u_{k,n}(-\Delta)^{\alpha}\xi+g_{n}(u_{k,n})\xi]dx=k\int_{\partial\Omega}\frac{\partial^{\alpha}\xi(x)}{\partial\vec{n}_{x}^{\alpha}}d\omega(x),\quad\forall\xi\in\mathbb{X}_{\alpha}.$
(3.9)

For any compact set $\mathcal{K}\subset\Omega$, we observe from [6, Lemma 3.2]
that for some $\beta\in(0,\alpha)$,

$\|u_{k,n}\|_{C^{\beta}(\mathcal{K})}\leq c_{45}k.$

Therefore, up to some subsequence, there exists $u_{k}$ such that

$\lim_{n\to\infty}u_{k,n}=u_{k}\quad{\rm\ in}\ \Omega.$

Then $g_{n}(u_{k,n})$ converge to $g(u_{k})$ in $\Omega$ as $n\to\infty$. By
Proposition 2.2 and (3.19) in [6], we have that

$u_{k,n}\to u_{k}\ {\rm in}\
L^{1}(\Omega),\quad\|g_{n}(u_{n})\|_{L^{1}(\Omega,\rho^{\alpha}dx)}\leq
c_{46}\|\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}|\nu|}{\partial\vec{n}^{\alpha}}]\|_{L^{1}(\Omega)}$

and

$m(\lambda)\leq c_{47}\lambda^{-\frac{1+\alpha}{1-\alpha}}\ \quad{\rm for}\ \
\ \lambda>\lambda_{0},$

where

$m(\lambda)=\int_{S_{\lambda}}\rho_{\partial\Omega}^{\alpha}(x)dx\quad{\rm
with}\quad
S_{\lambda}=\\{x\in\Omega:\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}|\nu|}{\partial\vec{n}^{\alpha}}]>\lambda\\}.$

For any Borel set $E\subset\Omega$, we have that

$\displaystyle\begin{array}[]{lll}\displaystyle\int_{E}|g_{n}(u_{n})|\rho_{\partial\Omega}^{\alpha}(x)dx\leq\int_{E\cap\tilde{S}^{c}_{\frac{\lambda}{k}}}g\left(k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}|\nu|}{\partial\vec{n}^{\alpha}}]\right)\rho_{\partial\Omega}^{\alpha}(x)dx+\int_{E\cap\tilde{S}_{\frac{\lambda}{k}}}g\left(k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}|\nu|}{\partial\vec{n}^{\alpha}}]\right)\rho_{\partial\Omega}^{\alpha}(x)dx\\\\[11.38109pt]
\phantom{\int_{E}|g(u_{t})|\rho^{\alpha}_{\partial\Omega}(x)dx}\displaystyle\leq\tilde{g}\left(\frac{\lambda}{k}\right)\int_{E}\rho^{\alpha}_{\partial\Omega}(x)dx+\int_{\tilde{S}_{\frac{\lambda}{k}}}\tilde{g}\left(k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}|\nu|}{\partial\vec{n}^{\alpha}}]\right)\rho^{\alpha}_{\partial\Omega}(x)dx\\\\[11.38109pt]
\phantom{\int_{E}|g(u_{t})|\rho^{\alpha}_{\partial\Omega}(x)dx}\displaystyle\leq\tilde{g}\left(\frac{\lambda}{k}\right)\int_{E}\rho^{\alpha}_{\partial\Omega}(x)dx+\tilde{m}\left(\frac{\lambda}{k}\right)\tilde{g}\left(\frac{\lambda}{k}\right)+\int_{\frac{\lambda}{k}}^{\infty}\tilde{m}(s)d\tilde{g}(s),\end{array}$

where $\tilde{g}(r)=g(|r|)-g(-|r|)$.

On the other hand,

$\int_{\frac{\lambda}{k}}^{\infty}\tilde{g}(s)d\tilde{m}(s)=\lim_{T\to\infty}\int_{\frac{\lambda}{k}}^{T}\tilde{g}(s)d\tilde{m}(s).$

Thus,

$\displaystyle\begin{array}[]{lll}\displaystyle\tilde{m}\left(\frac{\lambda}{k}\right)\tilde{g}\left(\frac{\lambda}{k}\right)+\int_{\frac{\lambda}{k}}^{T}\tilde{m}(s)d\tilde{g}(s)\leq
c_{47}\tilde{g}\left(\frac{\lambda}{k}\right)\left(\frac{\lambda}{k}\right)^{-\frac{1+\alpha}{1-\alpha}}+c_{47}\int_{\frac{\lambda}{k}}^{T}s^{-\frac{1+\alpha}{1-\alpha}}d\tilde{g}(s)\\\\[11.38109pt]
\phantom{-----\ \int_{\lambda}^{T}\tilde{g}(s)d\omega(s)}\displaystyle\leq
c_{48}T^{-\frac{1+\alpha}{1-\alpha}}\tilde{g}(T)+\frac{c_{49}}{\frac{1+\alpha}{1-\alpha}+1}\int_{\frac{\lambda}{k}}^{T}s^{-1-\frac{1+\alpha}{1-\alpha}}\tilde{g}(s)ds.\end{array}$

By assumption (1.8) and [6, Lemma 3.4] with $p=\frac{1+\alpha}{1-\alpha}$,
$T^{-\frac{N+\alpha}{N-\alpha}}\tilde{g}(T)\to 0$ when $T\to\infty$,
therefore,

$\tilde{m}\left(\frac{\lambda}{k}\right)\tilde{g}\left(\frac{\lambda}{k}\right)+\int_{\frac{\lambda}{k}}^{\infty}\tilde{m}(s)\
d\tilde{g}(s)\leq\frac{c_{49}}{\frac{1+\alpha}{1-\alpha}+1}\int_{\frac{\lambda}{k}}^{\infty}s^{-1-\frac{1+\alpha}{1-\alpha}}\tilde{g}(s)ds.$

Notice that the above quantity on the right-hand side tends to $0$ when
$\lambda\to\infty$. The conclusion follows: for any $\epsilon>0$ there exists
$\lambda>0$ such that

$\frac{c_{49}}{\frac{1+\alpha}{1-\alpha}+1}\int_{\frac{\lambda}{k}}^{\infty}s^{-1-\frac{1+\alpha}{1-\alpha}}\tilde{g}(s)ds\leq\frac{\epsilon}{2}.$

For $\lambda$ fixed, there exists $\delta>0$ such that

$\int_{E}\rho_{\partial\Omega}^{\alpha}(x)dx\leq\delta\Longrightarrow\tilde{g}\left(\frac{\lambda}{k}\right)\int_{E}\rho_{\partial\Omega}^{\alpha}(x)dx\leq\frac{\epsilon}{2},$

which implies that $\\{g_{n}\circ u_{n}\\}$ is uniformly integrable in
$L^{1}(\Omega,\rho_{\partial\Omega}^{\alpha}dx)$. Then $g_{n}\circ u_{n}\to
g\circ u_{\nu}$ in $L^{1}(\Omega,\rho_{\partial\Omega}^{\alpha}dx)$ by Vitali
convergence theorem.

Passing to the limit as $n\to+\infty$ in the identity (3.9), it implies that

$\int_{\Omega}[u_{k}(-\Delta)^{\alpha}\xi+g(u_{k})\xi]dx=k\int_{\partial\Omega}\frac{\partial^{\alpha}\xi(x)}{\partial\vec{n}^{\alpha}_{x}}d\omega(x),\quad\forall\xi\in\mathbb{X}_{\alpha}.$

Then $u_{k}$ is a weak solution of (1.6). Moreover, it follows by the fact

$k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]-g(k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])\leq
u_{k}\leq
k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]\quad{\rm
in}\ \Omega.$ (3.10)

Uniqueness of weak solution. Let $u_{1},u_{2}$ be two weak solutions of (1.6)
and $w=u_{1}-u_{2}$. Then $(-\Delta)^{\alpha}w=g_{n}(u_{2})-g_{n}(u_{1})$ and
$g_{n}(u_{2})-g_{n}(u_{1})\in L^{1}(\Omega,\rho^{\alpha}dx)$. By Kato’s
inequatlity, see [8, Proposition 2.4], for $\xi\in\mathbb{X}_{\alpha}$,
$\xi\geq 0$, we have that

$\displaystyle\int_{\Omega}|w|(-\Delta)^{\alpha}\xi
dx+\int_{\Omega}[g_{n}(u_{1})-g_{n}(u_{2})]{\rm sign}(w)\xi dx\leq 0.$

Combining with $\int_{\Omega}[g_{n}(u_{1})-g_{n}(u_{2})]{\rm sign}(w)\xi
dx\geq 0$, then we have

$w=0\quad{\rm a.e.\ in}\ \ \Omega.$

Regularity of $u_{k,n}$ and $u_{k}$. Since $g_{n}$ is $C^{1}$ in $\mathbb{R}$,
then by [6, Lemma 3.2], we have

$\|u_{k,n}\|_{C^{2\alpha+\beta}(\mathcal{K})}\leq c_{50}k,$ (3.11)

for any compact set $\mathcal{K}$ and some $\beta\in(0,\alpha)$. Then
$u_{k,n}$ is $C^{2\alpha+\beta}$ locally in $\Omega$. Together with the fact
that $u_{n,k}$ is classical solution of (2.13), we derive by Theorem 2.2 in
[5] that $u_{k}$ is a classical solution of (2.13).

To prove (1.9.)  Plugging (3.1) and (3.3) into (3.10), we obtain that (1.9).
$\Box$

##  4 Proof of Theorem 1.2

###  4.1 Strong singularity for $p\in(1+2\alpha,\frac{1+\alpha}{1-\alpha})$

In this subsection, we consider the limit of $\\{u_{k}\\}$ as $k\to\infty$,
where $u_{k}$ is the weak solution of

$\begin{array}[]{lll}(-\Delta)^{\alpha}u+u^{p}=k\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}&{\rm
in}\quad\ \ \bar{\Omega},\\\\[8.53581pt] \phantom{-----\ }u=0&{\rm in}\quad\ \
\bar{\Omega}^{c},\end{array}$

here $p\in(1+2\alpha,\frac{1+\alpha}{1-\alpha})$. From Theorem 1.1, we know
that $k\mapsto u_{k}$ is increasing and $u_{k}$ is a classical solution of
(1.1).

In order to control the limit of $\\{u_{k}\\}$ as $k\to\infty$, we have to
obtain barrier function, i.e. a suitable super solution of (1.1). To this end,
we consider $C^{2}$ function $w_{p}$ satisfying

$w_{p}(x)=\left\\{\begin{array}[]{lll}\rho(x)^{-\frac{2\alpha}{p-1}},&{\rm
for}\ \ x\in\Omega,\\\\[8.53581pt] \phantom{}0,&{\rm for}\ \
x\in\Omega^{c}.\end{array}\right.$ (4.1)

We see that $w_{p}\in L^{1}(\Omega)$ if $\frac{2\alpha}{p-1}<1$, i.e.
$p>1+2\alpha$.

###### Lemma 4.1

Assume that $p\in(1+2\alpha,\frac{1+\alpha}{1-\alpha})$ and $w_{p}$ is defined
in (4.1). Then there exists $\lambda_{0}>0$ such that $\lambda_{0}w_{p}$ is a
super solution of (1.1).

Proof. For $p\in(1+2\alpha,\frac{1+\alpha}{1-\alpha})$, we have that
$-\frac{2\alpha}{p-1}\in(-1,0)$ and from [5, Proposition 3.2], it shows that
there exists $c(p)<0$ such that

$(-\Delta)^{\alpha}w_{p}(x)\geq
c(p)\rho(x)^{-\frac{2\alpha}{p-1}-2\alpha},\quad x\in\Omega.$

Thus, taking $\lambda_{0}=|c(p)|^{\frac{1}{p-1}}$, we derive that

$(-\Delta)^{\alpha}(\lambda_{0}w_{p})+(\lambda_{0}w_{p})^{p}\geq 0\quad{\rm
in}\quad\Omega.$

Together with $\lambda_{0}w_{p}=0$ in $\Omega^{c}$, $\lambda_{0}w_{p}$ is a
super solution of (1.1). The proof ends. $\Box$

We observe that the super solution $\lambda_{0}w_{p}$ constructed in Lemma 4.1
provide a upper bound for $u_{\infty}$.

Proof of Theorem 1.2 $(i)$. For $p\in(1+2\alpha,\frac{1+\alpha}{1-\alpha})$,
we have that

$-\frac{2\alpha}{p-1}\in(-1,-1+\alpha)$

and it follows by (3.3) that

$u_{k}(x)\leq
k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)\leq\frac{c_{5}k}{|x|^{1-\alpha}},\quad
x\in\Omega.$

Then $\lim_{x\in\Omega,\rho(x)\to 0}\frac{u_{k}(x)}{w_{p}(x)}=0$ and we claim
that

$u_{k}\leq\lambda_{0}w_{p}\quad{\rm in}\quad\Omega.$

In fact, if it fails, then there exists $z_{0}\in\Omega$ such that

$(u_{k}-\lambda_{0}w_{p})(z_{0})=\inf_{\Omega}(u_{k}-\lambda_{0}w_{p})={\rm
ess}\inf_{\mathbb{R}^{N}}(u_{k}-\lambda_{0}w_{p})<0.$

Then we have $(-\Delta)^{\alpha}(u_{k}-\lambda_{0}w_{p})(z_{0})<0$, which
contradicts the fact that

$(-\Delta)^{\alpha}(u_{k}-\lambda_{0}w_{p})(z_{0})=\lambda_{0}w_{p}^{p}(z_{0})-u_{k}^{p}(z_{0})>0.$

By monotonicity of the mapping $k\to u_{k}$, there holds

$u_{\infty}(x):=\lim_{k\to\infty}u_{k}(x),\quad
x\in\mathbb{R}^{N}\setminus\\{0\\},$

which is a classical solution of (1.1) and

$u_{\infty}(x)\leq\lambda_{0}w_{p}(x)=\lambda_{0}\rho(x)^{-\frac{2\alpha}{p-1}},\quad\forall
x\in\Omega.$

By applying Stability Theorem [5, Theorem 2.4], we obtain that $u_{\infty}$ is
a classical solution of (1.1).

Finally, we claim that there exists $c_{51}>0$ such that for $t\in(0,t_{0})$,

$u_{\infty}(x)\geq c_{51}t^{-\frac{2\alpha}{p-1}},\quad\forall
x\in\mathcal{C}_{t}.$ (4.2)

Indeed, let $t_{k}=(\sigma^{-1}k)^{\frac{p-1}{(1-\alpha)p-1-\alpha}}$, where
$\sigma>0$ will be chosen later, then $k=\sigma
t_{k}^{\frac{(1-\alpha)p-1-\alpha}{p-1}}$ and for $x\in\mathcal{C}_{t}$, we
apply Lemma 3.1 with $p\in(1+2\alpha,\frac{1+\alpha}{1-\alpha})$ that

$\displaystyle u_{k}(x)$ $\displaystyle\geq$ $\displaystyle
k\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)-k^{p}\mathbb{G}_{\alpha}[(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])^{p}](x)$
$\displaystyle\geq$ $\displaystyle
c_{5}kt^{\alpha-N}[1-c_{52}k^{p-1}t^{(\alpha-1)p+\alpha+1}]$
$\displaystyle\geq$ $\displaystyle c_{5}\sigma
t_{k}^{-\frac{2\alpha}{p-1}}[1-c_{52}\sigma^{p-1}t_{k}^{p-1}(t_{k}/2)^{(\alpha-1)p+\alpha+1}]$
$\displaystyle\geq$ $\displaystyle c_{5}\sigma
t_{k}^{-\frac{2\alpha}{p-1}}[1-c_{52}\sigma^{p-1}2^{(1-\alpha)p-\alpha-1}]$
$\displaystyle\geq$
$\displaystyle\frac{c_{5}\sigma}{2}\rho(x)^{-\frac{2\alpha}{p-1}},$

where we choose $\sigma$ such that
$c_{52}\sigma^{p-1}2^{(N-\alpha)p-\alpha-N}=\frac{1}{2}$. Then for any
$x\in\Omega$, there exists $k>0$ such that $x\in\Omega$ and then

$u_{\infty}(x)\geq
u_{k}(x)\geq\frac{c_{5}\sigma}{2}\rho(x)^{-\frac{2\alpha}{p-1}},\qquad\forall
x\in\Omega.$

This ends the proof. $\Box$

###  4.2 The limit of $\\{u_{k}\\}$ blows up when $p\in(0,1+2\alpha]$

In this subsection, we derive the blow-up behavior of the limit of
$\\{u_{k}\\}$ when $p\in(0,1+2\alpha]$. To this end, we have to do more
estimate for $u_{k}$.

###### Lemma 4.2

Assume that $g(s)=s^{p}$ with $p\in(1,1+2\alpha]$ and $u_{k}$ is the solution
of (1.6) obtained by Theorem 1.1. Then there exist $c_{52}>0$,
$r_{0}\in(0,\frac{1}{4})$ and $\\{r_{k}\\}_{k}\subset(0,r_{0})$ satisfying
$r_{k}\to 0$ as $k\to\infty$ such that

$u_{k}(x)\geq\frac{c_{52}}{\rho(x)},\qquad\forall x\in\Omega_{t_{0}}.$ (4.3)

Proof. We divide $\alpha,p$ into 4 cases:
Case I: $1<\frac{2\alpha}{1-\alpha}<1+2\alpha$ and
$p\in[\frac{2\alpha}{1-\alpha},1+2\alpha]$; Case II:
$1<\frac{2\alpha}{1-\alpha}<1+2\alpha$ and $p\in[1,\frac{2\alpha}{1-\alpha})$;
Case III: $\frac{2\alpha}{1-\alpha}\leq 1$ and $p\in(1,1+2\alpha]$; Case IV:
$\frac{2\alpha}{1-\alpha}\geq 1+2\alpha$ and $p\in(1,1+2\alpha]$.

To prove (4.3) in Case I and Case III. Let $r_{j}=j^{-\frac{1}{\alpha}}$ with
$j\in(k_{0},k)$, then $j=r_{j}^{-\alpha}$. Applying Lemma 3.1 $(ii)$, for
$p\geq\frac{2\alpha}{1-\alpha}$, we have that for
$x\in\Omega_{r_{j}}\setminus\Omega_{\frac{r_{j}}{2}}$,

$\displaystyle u_{j}(x)$ $\displaystyle\geq$ $\displaystyle
j\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)-j^{p}\mathbb{G}_{\alpha}[(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])^{p}](x)$
$\displaystyle\geq$ $\displaystyle
c_{5}^{-1}jr_{j}^{\alpha-1}-c_{27}j^{p}\rho(x)^{(\alpha-1)p+2\alpha}$
$\displaystyle\geq$ $\displaystyle c_{5}^{-1}r_{j}^{-1}-c_{27}r_{j}^{-\alpha
p-(1-\alpha)p+2\alpha}$ $\displaystyle\geq$
$\displaystyle\frac{1}{2c_{5}}\rho(x)^{-1},$

where the last inequality holds since $-\alpha p-(1-\alpha)p+2\alpha>-1$ and
$r_{j}\to 0$ as $j\to\infty$. Then for any $x\in\Omega_{t_{0}}$, there exists
$j\in(k_{0},k)$ such that
$x\in\Omega_{t_{j}}\setminus\Omega_{\frac{t_{j}}{2}}$ and then

$u_{k}(x)\geq u_{j}(x)\geq\frac{1}{2c_{5}}\rho^{-1}(x),\qquad\forall
x\in\Omega_{t_{0}}.$

To prove (4.3) in Case II and Case IV. Let $r_{j}=j^{-\frac{1}{\alpha}}$ with
$j\in(k_{0},k)$, then $j=r_{j}^{-\alpha}$ and for
$x\in\Omega_{r_{j}}\setminus\Omega_{\frac{r_{j}}{2}}$, we have that

$\displaystyle u_{j}(x)$ $\displaystyle\geq$ $\displaystyle
j\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}](x)-j^{p}\mathbb{G}_{\alpha}[(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])^{p}](x)$
$\displaystyle\geq$ $\displaystyle c_{5}^{-1}j\rho(x)^{\alpha-1}-c_{27}j^{p}$
$\displaystyle\geq$ $\displaystyle c_{5}^{-1}r_{j}^{-1}-c_{27}r_{j}^{-\alpha
p}$ $\displaystyle\geq$ $\displaystyle\frac{1}{2c_{5}}\rho(x)^{-1},$

where the last inequality holds since $-\alpha p>-1$ and $r_{j}\to 0$ as
$j\to\infty$. For any $x\in\Omega_{t_{0}}$, there exists $j\in(k_{0},k)$ such
that $x\in\Omega_{r_{j}}\setminus\Omega_{\frac{r_{j}}{2}}$ and then

$u_{k}(x)\geq u_{j}(x)\geq\frac{1}{2c_{5}}\rho(x)^{-1},\qquad\forall
x\in\Omega_{t_{0}}.$

The proof ends. $\Box$

Proof of Theorem 1.2 $(ii)$. It derives by Lemma 4.2 that

$\pi_{k}:=\int_{\Omega_{t_{0}}\setminus\Omega_{\frac{t_{k}}{2}}}u_{k}(x)\geq
c_{52}\int_{\Omega_{t_{0}}\setminus\Omega_{\frac{t_{k}}{2}}}\rho^{-1}(x)dx\to\infty\quad{\rm
as}\ k\to\infty.$ (4.4)

Then

$\begin{array}[]{lll}(-\Delta)^{\alpha}u+u^{p}=0&{\rm
in}\quad\Omega\setminus\Omega_{t_{0}},\\\\[5.69054pt]
\phantom{(-\Delta)^{\alpha}+u^{p}}u=0&{\rm
in}\quad\mathbb{R}^{N}\setminus\Omega,\\\\[5.69054pt]
\phantom{(-\Delta)^{\alpha}+u^{p}}u=u_{k}&{\rm
in}\quad\Omega_{t_{0}}\end{array}$ (4.5)

admits a unique solution $w_{k}$. By Comparison Principle,

$u_{k}\geq w_{k}\quad{\rm in}\quad B_{\varrho_{0}}(y_{0}).$ (4.6)

Let $\tilde{w}_{k}=w_{k}-u_{k}\chi_{\Omega\setminus\Omega_{t_{0}}},$ then
$\tilde{w}_{k}=w_{k}$ in $\Omega\setminus\Omega_{t_{0}}$ and for
$x\in\Omega\setminus\Omega_{t_{0}}$,

$\begin{array}[]{lll}(-\Delta)^{\alpha}\tilde{w}_{k}(x)=-\lim_{\epsilon\to
0^{+}}\int_{B_{\varrho_{0}}(y_{0})\setminus
B_{\epsilon}(x)}\frac{w_{k}(z)-w_{k}(x)}{|z-x|^{N+2\alpha}}dz\\\\[8.53581pt]
\phantom{--------}+\lim_{\epsilon\to
0^{+}}\int_{B_{\varrho_{0}}^{c}(y_{0})\setminus
B_{\epsilon}(x)}\frac{w_{k}(x)}{|z-x|^{N+2\alpha}}dz\\\\[8.53581pt]
\phantom{------}=-\lim_{\epsilon\to 0^{+}}\int_{\mathbb{R}^{N}\setminus
B_{\epsilon}(x)}\frac{w_{k}(z)-w_{k}(x)}{|z-x|^{N+2\alpha}}dz+\int_{B_{r_{0}}(0)}\frac{u_{k}(z)}{|z-x|^{N+2\alpha}}dz\\\\[8.53581pt]
\phantom{------}\geq(-\Delta)^{\alpha}w_{k}(x)+c_{53}\pi_{k},\end{array}$

where $c_{53}=(|y_{0}|+r_{0})^{-N-2\alpha}$ and the last inequality follows by
the fact of

$|z-x|\leq|x|+|z|\leq|y_{0}|+r_{0}\quad{\rm for}\ \forall z\in
B_{\frac{1}{4}}(0),\ \ \forall x\in\Omega\setminus\Omega_{t_{0}}.$

Therefore,

$\displaystyle(-\Delta)^{\alpha}\tilde{w}_{k}(x)+\tilde{w}_{k}^{p}(x)$
$\displaystyle\geq$
$\displaystyle(-\Delta)^{\alpha}w_{k}(x)+w_{k}^{p}(x)+c_{53}\pi_{k}$
$\displaystyle=$ $\displaystyle c_{53}\pi_{k},\qquad\forall
x\in\Omega\setminus\Omega_{t_{0}},$

that is, $\tilde{w}_{k}$ is a super solution of

$\begin{array}[]{lll}\displaystyle(-\Delta)^{\alpha}u+u^{p}=c_{53}\pi_{k}&{\rm
in}\quad\Omega\setminus\Omega_{t_{0}},\\\\[5.69054pt]
\phantom{(-\Delta)^{\alpha}+u^{p,}}u=0&{\rm
in}\quad(\Omega\setminus\Omega_{t_{0}})^{c}.\end{array}$ (4.7)

Let $\eta_{1}$ be the solution of

$\begin{array}[]{lll}(-\Delta)^{\alpha}u=1&{\rm
in}\quad\Omega\setminus\Omega_{t_{0}},\\\\[5.69054pt]
\phantom{(-\Delta)^{\alpha}}u=0&{\rm
in}\quad(\Omega\setminus\Omega_{t_{0}})^{c}.\end{array}$

Then
$(c_{53}\pi_{k})^{\frac{1}{p}}\frac{\eta_{1}}{2\max_{\mathbb{R}^{N}}\eta_{1}}$
is sub solution of (4.7) for $k$ large enough. By Comparison Principle, we
have that

$\tilde{w}_{k}(x)\geq(c_{53}\pi_{k})^{\frac{1}{p}}\frac{\eta_{1}(x)}{2\max_{\mathbb{R}^{N}}\eta_{1}},\quad\forall
x\in B_{\varrho_{0}}(y_{0}),$

which implies that

$w_{k}(y)\geq c_{54}(c_{53}\pi_{k})^{\frac{1}{p}},\qquad\forall
y\in\Omega\setminus\Omega_{t_{0}},$

where $c_{54}=\min_{x\in
B_{\varrho_{0}}(y_{0})}\frac{\eta_{1}(x)}{2\max_{\mathbb{R}^{N}}\eta_{1}}$.
Therefore, (4.6) and (4.4) imply that

$\lim_{k\to\infty}u_{k}(y)\geq\lim_{k\to\infty}w_{k}(y)=\infty,\qquad\forall\Omega\setminus\Omega_{t_{0}}.$

Similarly, we can prove

$\lim_{k\to\infty}u_{k}(y)\geq\lim_{k\to\infty}w_{k}(y)=\infty,\qquad\forall
y\in\Omega.$

The proof ends. $\Box$

##  5  Proof of Theorem 1.3

In this section, we are devoted to consider the solution of (1.10) with
different blow-up speeds. Without loss of generality, we assume that

$x_{0}=0\in\partial\Omega.$

For $k,j\in\mathbb{N}$, donte

$\nu=\omega+\delta_{0},$ (5.1)

and then

$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}]=\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}]+\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}],$

which is a $\alpha-$harmonic function of (2.15), since
$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]$
is a $\alpha-$harmonic function of (2.15) and
$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}]$
is $\alpha-$ harmonic.

###### Proposition 5.1

Let
$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}]$
given by (5.1) and $p^{*}_{N}=\frac{N+\alpha}{N-\alpha}$. Then there exists
$c_{55}>0$ such that

$\|\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}]\|_{M^{p^{*}_{N}}(\Omega,\rho^{\alpha}dx)}\leq
c_{55}.$ (5.2)

Proof. Since

$\|\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}]\|_{M^{p^{*}_{N}}(\Omega,\rho^{\alpha}dx)}\leq
k\|\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}]\|_{M^{p^{*}_{N}}(\Omega,\rho^{\alpha}dx)}+j\|\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}]\|_{M^{p^{*}_{N}}(\Omega,\rho^{\alpha}dx)}.$
(5.3)

From Proposition 2.2 and
$p*=\frac{1+\alpha}{1-\alpha}>\frac{N+\alpha}{N-\alpha}$, on the one hand, we
have that

$\displaystyle\|\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}]\|_{M^{p^{*}_{N}}(\Omega,\rho^{\alpha}dx)}$
$\displaystyle\leq$ $\displaystyle
c_{56}\|\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}]\|_{M^{p^{*}}(\Omega,\rho^{\alpha}dx)}$
(5.4) $\displaystyle\leq$ $\displaystyle c_{56}c_{55},$

On the other hand, for $\lambda>0$, denote

$S_{\lambda}=\\{x\in\Omega:\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}](x)>\lambda\\}\quad{\rm
and}\quad m(\lambda)=\int_{S_{\lambda}}\rho_{\partial\Omega}^{\alpha}(x)dx$

and from [6, Lemma 3.3] with $\nu=\delta_{0}$, there exist $\lambda_{0}>1$ and
$c_{57}>0$ such that for any $\lambda\geq\lambda_{0}$,

$\tilde{m}(\lambda)\leq c_{57}\lambda^{-\frac{N+\alpha}{N-\alpha}}.$

which implies that

$\|\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}]\|_{M^{p^{*}_{N}}(\Omega,\rho^{\alpha}dx)}\leq
c_{55},$ (5.5)

Thus, (5.2) follows by (5.4) and (5.5). Combining [6, Theorem 1.1] and the
proof of Proposition 2.3, we have following result.

###### Proposition 5.2

Assume that $\\{g_{n}\\}_{n}$ is given by (2.10). Then

$\begin{array}[]{lll}(-\Delta)^{\alpha}u+g_{n}(u)=k\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}&{\rm
in}\quad\Omega,\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}+g_{n}(u)}u=0&{\rm
in}\quad\Omega^{c}\end{array}$ (5.6)

admits a unique positive weak solution $u_{k,j,n}$ satisfying

$(i)$ the mappings $k\to u_{k,j,n}$, $j\to u_{k,j,n}$ are increasing, the
mapping $n\to u_{k,j,n}$ is decreasing

$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}](x)-\mathbb{G}_{\alpha}[g_{n}(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}])](x)\leq
u_{k,n}(x)\leq\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}](x),\quad\forall
x\in\Omega;$ (5.7)

$(ii)$ $u_{k,n}$ is a classical solution of (2.13).

###### Lemma 5.1

There exists $c_{58}>1$ such that

$\frac{1}{c_{58}}\rho^{\alpha}(x)|x|^{-N}\leq\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}]\leq
c_{58}\rho^{\alpha}(x)|x|^{-N},\qquad\forall x\in\Omega.$ (5.8)

Proof. For any $x\in\Omega$, there exists $s>0$ such that
$x\in\mathcal{C}_{s}$, and then letting $y_{t}=t\vec{n}_{0}$ with
$t\in(0,s/2)$, we have that

$|y_{t}-x|=s-t>\frac{s}{2}=\frac{1}{2}\max\\{\rho_{\partial\Omega}(y_{t}),\rho_{\partial\Omega}(x)\\}.$

Thus, it follows by apply [4, Theorem 1.1,Theorem 1.2] that

$\frac{1}{c_{59}}\frac{\rho^{\alpha}_{\partial\Omega}(y_{t})\rho^{\alpha}_{\partial\Omega}(x)}{|x-y_{t}|^{N}}\leq
G_{\alpha}(x,y_{t})\leq
c_{59}\frac{\rho^{\alpha}_{\partial\Omega}(y_{t})\rho^{\alpha}_{\partial\Omega}(x)}{|x-y_{t}|^{N}}.$

From

$\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}](x)=\lim_{t\to
0^{+}}G_{\alpha}(x,y_{t}),\qquad\forall x\in\Omega,$ (5.9)

we deduce that

$\frac{1}{c_{60}}\rho^{\alpha}(x)|x|^{-N}\leq\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}]\leq
c_{60}\rho^{\alpha}(x)|x|^{-N},\qquad\forall x\in\Omega.$

###### Lemma 5.2

Assume that $g$ is a continuous nondecreasing function satisfying $g(0)\geq
0$, (1.13) and (1.14). Then

$\lim_{\rho(x)\to
0^{+}}\frac{\mathbb{G}_{\alpha}[g(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}])](x)}{\rho(x)^{\alpha-1}+\rho^{\alpha}(x)|x|^{-N}}=0.$
(5.10)

Proof. From Lemma 3.1 and (1.13),

$\lim_{\rho(x)\to
0^{+}}\mathbb{G}_{\alpha}[g(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])](x)\rho^{1-\alpha}(x)=0.$
(5.11)

From [6, Lemma 4.1], we have that

$\lim_{s\to
0^{+}}\mathbb{G}_{\alpha}[g(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}])](s\vec{n}_{0})s^{N-\alpha}=0.$
(5.12)

By (1.14),

$\mathbb{G}_{\alpha}[g(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\nu}{\partial\vec{n}^{\alpha}}])](x)\leq\lambda\left[\mathbb{G}_{\alpha}[g(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\omega}{\partial\vec{n}^{\alpha}}])](x)+\mathbb{G}_{\alpha}[g(\mathbb{G}_{\alpha}[\frac{\partial^{\alpha}\delta_{0}}{\partial\vec{n}^{\alpha}}])](x)\right],$

together with (5.11) and (5.12), we implies (5.10). $\Box$

Proof of Theorem 1.3. The existence of weak solution just follows the
procedure of the proof of Theorem 1.1 by using Proposition 5.1 and Proposition
5.2. It is the same to prove the uniqueness and regularity of weak solution.
Finally, plugging (2.3), (5.8) and (5.10) into (5.7) replaced $g_{n}$ by $g$,
we obtain (1.15). $\Box$

Acknowledge: H. Chen is supported by NSFC, No: 11401270.

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