# Solutions of nonhomogeneous equations involving Hardy potentials with
singularities on the boundary

Huyuan Chen  Department of Mathematics, Jiangxi Normal University, Nanchang,
Jiangxi 330022, PR China<EMAIL_ADDRESS>,  Alexander Quaas  Departamento
de Matemática, Universidad Técnica Federico, Santa María Casilla V-110, Avda.
España 1680, Valparaíso, Chile<EMAIL_ADDRESS>and  Feng Zhou  Center
for PDEs, School of Mathematical Sciences, East China Normal University,
Shanghai Key Laboratory of PMMP, Shanghai, 200241, PR China
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we present a new distributional identity for the solutions of
elliptic equations involving Hardy potentials with singularities located on
the boundary of the domain. Then we use it to obtain the boundary isolated
singular solutions of nonhomogeneous problems.

###### Key words and phrases:

Distributional identity, Hardy potential, Boundary isolated singularity.

###### 2010 Mathematics Subject Classification:

35B40, 35J99.

##  1\. Introduction

The classical Hardy inequality is stated as following: For any smooth bounded
domain $\mathcal{O}$ in $\mathbb{R}^{N}$ containing the origin, there holds

(1.1) $\int_{\mathcal{O}}|\nabla u|^{2}dx\geq
c_{N}\int_{\mathcal{O}}|x|^{-2}|u|^{2}dx,\quad\forall\,u\in
H_{0}^{1}(\mathcal{O}),$

with the best constant $c_{N}=\frac{(N-2)^{2}}{4}$. The qualitative properties
of Hardy inequality and its improved versions have been studied extensively,
see for example [1, 4, 19, 21], motivated by great applications in the study
of stability of solutions to semilinear elliptic and parabolic equations (cf.
[5, 6, 13, 30, 31]). The isolated singular solutions of Hardy problem with
absorption nonlinearity have been studied in [11, 12, 23] and the one with
source nonlinearity has been done in [3, 16]. The related semilinear elliptic
problem involving the inverse square potential has been studied by variational
methods in [15, 14, 18] and the references therein. In a very recent work [9],
we established a new distributional identity with respect to a specific
weighted measure and we then classify the classical isolated singular
solutions of

$-\Delta u+\frac{\mu}{|x|^{2}}u=f\quad{\rm
in}\quad\mathcal{O}\setminus\\{0\\},$

subject to the homogeneous Dirichlet boundary condition with $\mu\geq-c_{N}$.
These results allow us to draw a complete picture of the existence, non-
existence and the singularities for classical solutions for the above problems
(cf. [10]).

It is of interest to consider the corresponding problem involving Hardy
potential with singularity on the boundary. While the sharp constant $c_{N}$
in Hardy inequality (1.1) could be replaced by $\frac{N^{2}}{4}$ when the
origin is addressed on the boundary of the domain, see [20, Corollary 2.4],
also [7, 8, 17].

Let $\Omega$ be a smooth bounded domain in $\mathbb{R}^{N}$ with
$0\in\partial\Omega$. We study boundary isolated singular solutions of
nonhomogeneous problems:

(1.2)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}u=f\quad{\rm
in}\;\;\Omega,\\\\[4.2679pt] \phantom{L_{\beta}}\displaystyle u=g\quad{\rm
on}\;\;\partial{\Omega}\setminus\\{0\\},\end{array}\right.$

where $f\in C^{\gamma}_{loc}(\bar{\Omega}\setminus\\{0\\})$ with
$\gamma\in(0,1)$, $g\in C(\partial\Omega\setminus\\{0\\})$ and
$\mathcal{L}_{\beta}:=-\Delta+\frac{\beta}{|x|^{2}}$ is the Hardy operator
which is singular at $0$ (with $N\geq 2$,
$\beta\geq\beta_{0}:=-\frac{N^{2}}{4}$). Recall that for $\beta\geq\beta_{0}$,
the problem

(1.3) $\left\\{\begin{array}[]{lll}\mathcal{L}_{\beta}u=0\quad{\rm
in}\;\;\mathbb{R}^{N}_{+},\\\\[2.84526pt]
\phantom{\mathcal{L}_{\beta}}u=0\quad{\rm
on}\;\;\partial\mathbb{R}^{N}_{+}\setminus\\{0\\}\end{array}\right.$

has two special solutions with the explicit formulas as

(1.4)
$\Lambda_{\beta}(x)=\left\\{\begin{array}[]{lll}x_{N}|x|^{\tau_{-}(\beta)}&{\rm
if}\;\;\beta>\beta_{0},\\\\[2.84526pt]
\phantom{}-x_{N}|x|^{\tau_{-}(\beta)}\ln|x|&{\rm
if}\;\;\beta=\beta_{0}\end{array}\right.\quad\;\;{\rm
and}\quad\lambda_{\beta}(x)=x_{N}|x|^{\tau_{+}(\beta)},$

where
$x=(x^{\prime},x_{N})\in\mathbb{R}^{N}_{+}:=\mathbb{R}^{N-1}\times(0,+\infty)$,
and

(1.5) $\tau_{-}(\beta)=-\frac{N}{2}-\sqrt{\beta-\beta_{0}}\quad{\rm
and}\quad\tau_{+}(\beta)=-\frac{N}{2}+\sqrt{\beta-\beta_{0}},$

are two roots of $\beta-\tau(\tau+N)=0$.

As in [10, 9], we first find a certain distributional identity which shows
that the singularity of solution $\Lambda_{\beta}$ for (1.3) is associated to
a Dirac mass. Let $C^{1.1}_{0}(\mathbb{R}^{N}_{+})$ be the set of functions in
$C^{1.1}(\overline{\mathbb{R}^{N}_{+}})$ vanishing on the boundary and having
compact support in $\overline{\mathbb{R}^{N}_{+}}$. Then we have

######  Theorem 1.1.

Let $d\gamma_{\beta}:=\lambda_{\beta}(x)dx$ and

(1.6)
$\mathcal{L}^{*}_{\beta}:=-\Delta-\frac{2\tau_{+}(\beta)}{|x|^{2}}\,x\cdot\nabla-\frac{2}{x_{N}}\frac{\partial}{\partial
x_{N}},\quad x=(x^{\prime},x_{N})\in\mathbb{R}^{N}_{+}.$

Then there holds

(1.7)
$\int_{\mathbb{R}^{N}_{+}}\Lambda_{\beta}\mathcal{L}^{*}_{\beta}(\frac{\zeta}{x_{N}})\,d\gamma_{\beta}=c_{\beta}\frac{\partial\zeta}{\partial
x_{N}}(0),\quad\forall\,\zeta\in C^{1.1}_{0}(\mathbb{R}^{N}_{+}),$

where

(1.8)
$c_{\beta}=\left\\{\begin{array}[]{lll}\sqrt{\beta-\beta_{0}}\,|\mathcal{S}^{N-1}|/N&{\rm
if}\;\;\beta>\beta_{0},\\\\[4.2679pt] \phantom{}|\mathcal{S}^{N-1}|/N&{\rm
if}\;\;\beta=\beta_{0},\end{array}\right.$

and $\mathcal{S}^{N-1}$ is the unit sphere of $\mathbb{R}^{N}$ and
$|\mathcal{S}^{N-1}|$ denotes its $(N-1)$-dimensional Hausdorff measure.

From the distributional identity (1.7), $\Lambda_{\beta}$ is called as a
fundamental solution of (1.3). We remark that when $\beta=0$,
$\mathcal{L}^{*}_{0}=-\Delta-\frac{2}{x_{N}}\frac{\partial}{\partial x_{N}}$,
$\lambda_{\beta}(x)=x_{N}$ and (1.7) could be reduced to

$\displaystyle c_{0}\frac{\partial\zeta}{\partial
x_{N}}(0)=\int_{\mathbb{R}^{N}_{+}}\Lambda_{0}\mathcal{L}^{*}_{0}(\frac{\zeta}{x_{N}})\,d\gamma_{\beta}=\int_{\mathbb{R}^{N}_{+}}\Lambda_{0}(-\Delta\zeta)\,dx,\quad\forall\;\zeta\in
C^{1.1}_{0}(\mathbb{R}^{N}_{+}),$

which coincides with the classical distributional identity proposed in [22].
On this classical subject, it has been vastly expanded in the works [2, 26,
27, 28, 29].

For simplicity, here and in the sequel, we always assume that $\Omega$ is a
bounded $C^{2}-$ domain satisfying that

(1.9) $B_{r_{0}}^{+}(0)\subset\Omega\subset B_{R_{0}}^{+}(0),$

for some $0<r_{0}<R_{0}<+\infty$ where
$B_{r}^{+}(0):=B_{r}(0)\cap\mathbb{R}^{N}_{+}$. Let
$d\omega_{\beta}(x):=|x|^{\tau_{+}(\beta)}d\omega(x)$, where $\omega$ is the
Hausdorff measure of $\partial\Omega$. We can state our main result as follows

######  Theorem 1.2.

Let $\mathcal{L}^{*}_{\beta}$ be given by (1.6), $f\in
C^{\theta}_{loc}(\bar{\Omega}\setminus\\{0\\})$ with $\theta\in(0,1)$, $g\in
C(\partial\Omega\setminus\\{0\\})$.

$(i)$ If

(1.10)
$\int_{\Omega}|f|\,d\gamma_{\beta}+\int_{\partial\Omega}|g|\,d\omega_{\beta}<+\infty,$

then for any $k\in\mathbb{R}$, problem (1.2) admits a unique solution
$u_{k}\in C^{2}(\Omega)\cap L^{1}(\Omega,|x|^{-1}d\gamma_{\beta})$ such that

(1.11)
$\int_{\Omega}u_{k}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=\int_{\Omega}\frac{f\xi}{x_{N}}\,d\gamma_{\beta}-\int_{\partial\Omega}g\frac{\partial\xi}{\partial\nu}d\omega_{\beta}+c_{\beta}k\frac{\partial\xi}{\partial
x_{N}}(0),\quad\forall\,\xi\in C^{1.1}_{0}(\Omega),$

where $\nu$ is the unit outward vector on $\partial\Omega$.

$(ii)$ If $f,g$ are nonnegative and

(1.12) $\lim_{r\to 0^{+}}\Big{(}\int_{\Omega\setminus
B_{r}(0)}f\,d\gamma_{\beta}+\int_{\partial\Omega\setminus
B_{r}(0)}g\,d\omega_{\beta}\Big{)}=+\infty,$

then problem (1.2) has no nonnegative solution.

When $g=0$ on $\partial\Omega$ and $f=0$ in $\Omega$, we prove in Proposition
3.2 in Section 3 that problem (1.2) admits an isolated singular solution
$\Lambda^{\Omega}_{\beta}$, which has the asymptotic behavior at the origin as
the fundamental function $\Lambda_{\beta}$. More precisely, we have

(1.13) $\lim_{t\to 0^{+}}\sup_{z\in
S^{N-1}_{+}}\Big{(}\frac{\Lambda^{\Omega}_{\beta}(tz)}{\Lambda_{\beta}(tz)}-1\Big{)}=0.$

When $g=0$ on $\partial\Omega$ and $f\in
C^{\theta}_{loc}(\bar{\Omega}\setminus\\{0\\})\cap
L^{1}(\Omega,d\gamma_{\beta})$, Theorem 4.1 in Section 4 shows that problem
(1.2) has a solution $u_{f}$ verifying the isolated singularity (see Remark
4.2)

(1.14) $\lim_{t\to 0^{+}}\inf_{z\in
S^{N-1}_{+}}\frac{u_{f}(tz)}{\Lambda_{\beta}(tz)}=0,$

which is less precise than (1.13) due to the lack of estimates of Green kernel
of Hardy operator with singularity on the boundary. However, when $f=0$ and
$g\not=0$, it is not convenient to use (1.14) to describe the singularity of
the solution $u_{g}$, so we may distinguish this by the distributional
identity

$\int_{\Omega}u_{g}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=-\int_{\partial\Omega}g\frac{\partial\xi}{\partial\nu}d\omega_{\beta},\quad\forall\,\xi\in
C^{1.1}_{0}(\Omega),$

All in all, the solution $u_{k}$ of (1.2) can be decomposed into three
components $k\Lambda^{\Omega}_{\beta}$, $u_{f}$ and $u_{g}$.

The method we use to prove the existence of solutions for problem (1.2) is
different from the classical method of the boundary data problem used by
Gmira-Véron in [22] due to the appearance of Hardy potential. They obtained
the very weak solutions by approximating the Dirac mass at boundary. Then they
considered the limit of the solutions to the corresponding problem where the
convergence is guaranteed by the Poisson kernel. In this paper, we prove the
existence of moderate singular solution by using the function
$\Lambda_{\beta}$ to construct suitable solutions of problem (1.2) with the
zero Dirichlet boundary condition. While for nonzero Dirichlet boundary
condition, we transform the boundary data into nonhomogeneous term. However,
for $\beta>0$, that transformation can not totally solve (1.2) with the
nonzero Dirichlet boundary condition, and our idea is to cut off the boundary
data and approximate the solutions.

The rest of the paper is organized as follows. In Section 2, we start from a
comparison principle for $\mathcal{L}_{\beta}$ and show the moderate singular
solution of (1.2) when $g=0$. Section 3 is devoted to prove the distributional
identity (1.7) for the fundamental solution $\Lambda_{\beta}$ in
$\mathbb{R}^{N}_{+}$, to consider its trace, the corresponding distributional
identity in bounded smooth domain. Section 4 is to study the qualitative
properties of the solutions for problem (1.2) when $g=0$ and then we give the
proof of Theorem 1.2 in the case of nonzero boundary data in Section 5. In
what follows, we denote by $c_{i}$ a generic positive constant in the proofs
of the results.

##  2\. Preliminary

###  2.1. Comparison principle

We start the analysis from a comparison principle for $\mathcal{L}_{\beta}$.
Let $\eta_{0}:[0,+\infty)\to[0,\,1]$ be a decreasing $C^{\infty}$ function
such that

(2.1) $\eta_{0}=1\quad{\rm in}\;\;[0,1]\quad{\rm and}\quad\eta_{0}=0\quad{\rm
in}\;\;[2,+\infty).$

######  Lemma 2.1.

Let $\Omega$ be a bounded open set in $\mathbb{R}^{N}_{+}$,
$L:\Omega\times[0,+\infty)\to[0,+\infty)$ be a continuous function satisfying
that for any $x\in\Omega$,

$L(x,s_{1})\geq L(x,s_{2})\quad{\rm if}\;\;s_{1}\geq s_{2},$

then $\mathcal{L}_{\beta}+L$ with $\beta\geq\beta_{0}$ verifies the comparison
principle, that is, if $u,\,v\in C^{1,1}(\Omega)\cap C(\bar{\Omega})$ verify
that

$\mathcal{L}_{\beta}u+L(x,u)\geq\mathcal{L}_{\beta}v+L(x,v)\quad{\rm
in}\;\;\Omega\quad{\rm and}\quad u\geq v\quad{\rm on}\;\;\partial\Omega,$

then $u\geq v\;{\rm in}\;\Omega.$

Proof. Let $w=u-v$ and then $w\geq 0$ on $\partial\Omega$. Denote
$w_{-}=\min\\{w,0\\}$, and we claim that $w_{-}\equiv 0$. Indeed, if
$\Omega_{-}:=\\{x\in\Omega:\,w(x)<0\\}$ is not empty, then it is a bounded
$C^{1,1}$ domain in $\Omega$ and $w_{-}=0$ on $\partial\Omega$. We observe
that $\Omega_{-}\subset\mathbb{R}^{N}_{+}$ and then from Hardy inequality [7,
(1.7)] (see also [25]), it holds that

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\int_{\Omega_{-}}(-\Delta
w_{-}+\frac{\beta}{|x|^{2}}w_{-})w_{-}dx+\int_{\Omega_{-}}[L(x,u)-L(x,v)]w_{-}\,dx$
$\displaystyle\geq$ $\displaystyle\int_{\Omega_{-}}\left(|\nabla
w_{-}|^{2}+\frac{\beta}{|x|^{2}}w_{-}^{2}\right)dx\geq
c_{1}\int_{\Omega_{-}}w_{-}^{2}dx,$

then $w_{-}=0$ in $\Omega_{-}$, by the continuity of $w_{-}$, which is
impossible with the definition of $\Omega_{-}$. $\Box$

######  Lemma 2.2.

Assume that $\beta\geq\beta_{0}$, $f_{1}$, $f_{2}$ are two functions in
$C^{\theta}_{loc}(\Omega)$ with $\theta\in(0,1)$, $g_{1}$, $g_{2}$ are two
continuous functions on $\partial\Omega\setminus\\{0\\}$, and

$f_{1}\geq f_{2}\quad{\rm in}\;\;\Omega\quad{\rm and}\quad g_{1}\geq
g_{2}\quad{\rm on}\;\;\partial\Omega\setminus\\{0\\}.$

Let $u_{i}$ ($i=1,2$) be the classical solutions of

$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}u=f_{i}\quad{\rm
in}\;\;{\Omega},\\\\[4.2679pt] \phantom{L_{\beta}}\displaystyle
u=g_{i}\quad{\rm on}\;\;\partial{\Omega}\setminus\\{0\\}.\end{array}\right.$

If

(2.2) $\lim_{r\to
0^{+}}\inf_{x\in\partial_{+}B_{r}(0)}[u_{1}(x)-u_{2}(x)]\Lambda_{\beta}^{-1}(x)\geq
0,$

where $\partial_{+}B_{r}(0)=\partial B_{r}(0)\cap\Omega$. Then $u_{1}\geq
u_{2}\quad{\rm in}\;\;\overline{\Omega}\setminus\\{0\\}.$

Proof. Let $w=u_{2}-u_{1}$, then $w$ satisfies

$\left\\{\begin{array}[]{lll}\displaystyle\qquad\mathcal{L}_{\beta}w\leq
0\quad{\rm in}\;\;{\Omega},\\\\[4.2679pt]
\phantom{L_{\beta}}\displaystyle\qquad w\leq 0\quad{\rm
on}\;\;\partial{\Omega}\setminus\\{0\\},\\\\[4.2679pt]
\phantom{}\displaystyle\lim_{r\to
0^{+}}\sup_{x\in\partial_{+}B_{r}(0)}w(x)\Lambda_{\beta}^{-1}(x)\leq
0.\end{array}\right.$

Thus for any $\epsilon>0$, there exists $r_{\epsilon}>0$ converging to zero as
$\epsilon\to 0$ such that

$w\leq\epsilon\Lambda_{\beta}\quad{\rm on}\quad\partial
B_{r_{\epsilon}}(0)\cap\Omega.$

We observe that $w\leq 0<\epsilon\Lambda_{\beta}\ {\rm on}\
\partial\Omega\setminus B_{r_{\epsilon}}(0),$ which implies by Lemma 2.1 that

$w\leq\epsilon\Lambda_{\beta}\quad{\rm
in}\quad\overline{\Omega}\setminus\\{0\\}.$

Therefore we obtain that $w\leq 0$ in $\overline{\Omega}\setminus\\{0\\}$
which ends the proof. $\Box$

For any $\varepsilon>0$, denote

(2.3)
$\mathcal{L}_{\beta,\varepsilon}=-\Delta+\frac{\beta}{|x|^{2}+\varepsilon}.$

We remark that $\mathcal{L}_{\beta,\varepsilon}$ is strictly elliptic operator
and we have the following existence result for related nonhomogeneous problem.

######  Lemma 2.3.

Assume that $\varepsilon\in(0,\,1)$, $\beta\geq\beta_{0}$,
$\mathcal{L}_{\beta,\varepsilon}$ is given by (2.3) and $f\in
C^{\theta}_{loc}(\Omega)\cap C(\bar{\Omega})$ with $\theta\in(0,1)$ and $g\in
C(\partial\Omega)$. Then the problem

(2.4)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta,\varepsilon}u=f&{\rm
in}\quad\Omega,\\\\[5.69054pt]
\phantom{\mathcal{L}_{\beta,\varepsilon}}u=g&{\rm
on}\quad\partial\Omega\end{array}\right.$

has a unique classical solution $u_{\varepsilon}\in C^{2}(\Omega)\cap
C(\bar{\Omega})$, which verifies that

(2.5)
$\int_{\Omega}u_{\varepsilon}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=\int_{\Omega}\frac{f\xi}{x_{N}}d\gamma_{\beta}-\int_{\partial\Omega}g\frac{\partial\xi}{\partial\nu}\,d\omega_{\beta}+\beta\varepsilon\int_{\Omega}\frac{u_{\varepsilon}\xi}{(|x|^{2}+\varepsilon)|x|^{2}x_{N}}\,d\gamma_{\beta},$

for any $\xi\in C^{1.1}_{0}(\Omega)$.

Assume more that $f\geq 0$ in $\Omega$ and $g\geq 0$ on $\partial\Omega$. Then
the mapping $\varepsilon\mapsto u_{\varepsilon}$ is decreasing if $\beta>0$,
and is increasing if $\beta_{0}\leq\beta<0$.

Proof. We first prove the existence of solution to problem (2.4). We introduce
Poisson kernel $P_{\Omega}$ of $-\Delta$ in $\Omega$, and denote Poisson
operator as

$\mathbb{P}_{\Omega}[g](x)=\int_{\partial\Omega}P_{\Omega}(x,y)g(y)dy.$

We observe that

$\mathcal{L}_{\beta,\varepsilon}\mathbb{P}_{\Omega}[g]=\frac{\beta}{|x|^{2}+\varepsilon}\mathbb{P}_{\Omega}[g]\in
C^{1}(\Omega)\cap C(\bar{\Omega}).$

Then the solution of (2.4) denoted by $u_{\varepsilon}$, could be reduced to
$u_{\varepsilon}=\mathbb{P}_{\Omega}[g]+u_{f}$, where $u_{f}$ is the solution
of

(2.6)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta,\varepsilon}u=f-\frac{\beta}{|x|^{2}+\varepsilon}\mathbb{P}_{\Omega}[g]&{\rm
in}\quad\Omega,\\\\[5.69054pt]
\phantom{\mathcal{L}_{\beta,\varepsilon}}u=0&{\rm
on}\quad\partial\Omega.\end{array}\right.$

For $\beta\geq\beta_{0}$, a solution $u_{f}$ in $H^{1}_{0}(\Omega)$ of (2.6)
could be derived by Ekeland’s variational methods as the critical point of the
functional

$I(u)=\int_{\Omega}|\nabla
u|^{2}dx+\beta\int_{\Omega}\frac{u^{2}}{|x|^{2}+\varepsilon}dx-\int_{\Omega}\Big{(}f-\frac{\beta}{|x|^{2}+\varepsilon}\mathbb{P}_{\Omega}[g]\Big{)}udx.$

That is well-defined in $H^{1}_{0}(\Omega)$ since $\beta\in(\beta_{0},0)$.
From the Hardy’s inequality in [17], we have that, for any $u\in
C_{0}^{2}(\Omega)$,

$\int_{\Omega}|\nabla
u|^{2}dx+\beta\int_{\Omega}\frac{u^{2}}{|x|^{2}+\varepsilon}dx\geq(\beta-\beta_{0})\int_{\Omega}|\nabla
u|^{2}dx,$

for $\beta=\beta_{0}$, from the improved Hardy inequality in [17], it holds

$\displaystyle c_{2}\int_{\Omega}u^{2}dx$ $\displaystyle\leq$
$\displaystyle\int_{\Omega}|\nabla
u|^{2}dx-|\beta_{0}|\int_{\Omega}\frac{u^{2}}{|x|^{2}}dx$ $\displaystyle<$
$\displaystyle\int_{\Omega}|\nabla
u|^{2}dx-|\beta_{0}|\int_{\Omega}\frac{u^{2}}{|x|^{2}+\varepsilon}dx.$

Finally it is trivial for the case $\beta\geq 0$.

By the standard regularity result (e.g. [24]), we have that $u_{f}$ is a
classical solution of (2.6). Then problem (2.4) admits a classical solution
and the uniqueness follows by comparison principle.

Finally, we prove (2.5). Multiple $\frac{\lambda_{\beta}\xi}{x_{N}}$ with
$\xi\in C^{1.1}_{0}(\Omega)$ and integrate over $\Omega$, we have that

$\displaystyle\int_{\Omega}\frac{\lambda_{\beta}\xi}{x_{N}}f\,dx=\int_{\Omega}\frac{\lambda_{\beta}\xi}{x_{N}}\mathcal{L}_{\beta,\varepsilon}u_{\varepsilon}\,dx$
$\displaystyle=$
$\displaystyle\int_{\Omega}u_{\varepsilon}(-\Delta\frac{\lambda_{\beta}\xi}{x_{N}})\,dx+\int_{\partial\Omega}g\frac{\partial(|x|^{\tau_{+}(\beta)}\xi)}{\partial\nu}\,d\omega(x)+\int_{\Omega}\frac{\beta}{|x|^{2}+\varepsilon}u_{\varepsilon}\lambda_{\beta}\xi\,dx$
$\displaystyle=$
$\displaystyle\int_{\Omega}u_{\varepsilon}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}+\int_{\partial\Omega}g\frac{\partial\xi}{\partial\nu}\,d\omega_{\beta}-\beta\varepsilon\int_{\Omega}\frac{u_{\varepsilon}\xi}{(|x|^{2}+\varepsilon)|x|^{2}x_{N}}\,d\gamma_{\beta}.$

Note that if $f\geq 0$ in $\Omega$ and $g\geq 0$ on $\partial\Omega$, then
$u_{\varepsilon}\geq 0$ in $\Omega$. Let $\varepsilon_{1}\geq\varepsilon_{2}$
and $u_{\varepsilon_{1}},\,u_{\varepsilon_{2}}$ be two solutions of (2.4)
respectively. If $\beta\geq\beta_{0}$, we observe that
$\mathcal{L}_{\beta,\varepsilon_{2}}u_{\varepsilon_{1}}\geq\mathcal{L}_{\beta,\varepsilon_{1}}u_{\varepsilon_{1}}=f,$
so $u_{\varepsilon_{1}}$ is a super solution of (2.4) with
$\varepsilon=\varepsilon_{2}$ and by comparison principle, it holds
$u_{\varepsilon_{1}}\geq u_{\varepsilon_{2}}\;\;{\rm in}\;\Omega.$ The proof
ends. $\Box$

Now we build the distributional identity for the classical solution of
nonhomogeneous problem with $g=0$ and moderate singularity at the origin, i.e.

(2.7) $\lim_{r\to
0^{+}}\sup_{x\in\partial_{+}B_{r}(0)}\frac{|u(x)|}{\Lambda_{\beta}(x)}=0.$

######  Proposition 2.4.

Let $\beta\geq\beta_{0}$, $N\geq 2$, $f\in C_{loc}^{\theta}(\bar{\Omega})$
with $\theta\in(0,1)$, then

(2.8)
$\left\\{\begin{array}[]{lll}\displaystyle\quad\mathcal{L}_{\beta}u=f\quad{\rm
in}\;\;{\Omega},\\\\[5.69054pt] \phantom{L_{\beta}}\quad\displaystyle
u=0\quad{\rm on}\;\;\partial{\Omega}\setminus\\{0\\},\end{array}\right.$

subjecting to (2.7), has a unique solution $u_{\beta}$, which satisfies the
distributional identity

(2.9)
$\int_{\Omega}u_{\beta}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=\int_{\Omega}\frac{f\xi}{x_{N}}\,d\gamma_{\beta},\quad\forall\,\xi\in
C^{1.1}_{0}(\Omega).$

Proof. The uniqueness follows by Lemma 2.2. Since $\mathcal{L}_{\beta}$ is a
linear operator, we only have to deal with the case that $f\geq 0$ in
$\Omega$.

Part 1: $\beta>0$. In this case, the mapping $\varepsilon\mapsto
u_{\varepsilon}$ is decreasing, where $u_{\varepsilon}>0$ is the solution of
(2.4) with $g=0$. Then $u_{\beta}:=\lim_{\varepsilon\to 0^{+}}u_{\varepsilon}$
exists, and by the standard regularity theory, we have that $u_{\beta}$ is a
classical solution of

(2.10)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}u=f\quad{\rm
in}\;\;{\Omega},\\\\[5.69054pt] \phantom{L_{\beta}}\displaystyle u=0\quad{\rm
on}\;\;\partial{\Omega}.\end{array}\right.$

Part 2: $\beta\in[\beta_{0},0)$. Without loss of generality, we assume that
$\Omega\subset B_{\frac{1}{2}}(0)$. Denote

$V_{t,s}(x):=\left\\{\begin{array}[]{lll}\displaystyle
tx_{N}|x|^{-\frac{N}{2}}-sx_{N}^{2}|x|^{\tau_{+}(\beta)}&{\rm
if}\;\;\beta\in(\beta_{0},0),\\\\[5.69054pt] \phantom{}\displaystyle
tx_{N}|x|^{-\frac{N}{2}}(-\ln|x|)^{\frac{1}{2}}-sx_{N}^{2}|x|^{-\frac{N}{2}}&{\rm
if}\;\;\beta=\beta_{0},\end{array}\right.$

where the parameters $s,t\geq 0$.

Then for $\beta\in(\beta_{0},0)$, we see that $V_{t,s}(x)>0$ for $x\in\Omega$
if $t\geq s$ and

$\mathcal{L}_{\beta}V_{t,s}(x)=tc_{\beta}(-N/2)x_{N}|x|^{-\frac{N}{2}-2}+2s|x|^{\tau_{+}(\beta)}+2s\tau_{+}(\beta)x_{N}^{2}|x|^{\tau_{+}(\beta)-2},$

where $c_{\beta}(-N/2)>0$ and $\tau_{+}(\beta)<0$. Since $f$ is bounded in
$\Omega$, let

$s_{0}=\frac{1}{2}\sup_{x\in\Omega}\frac{|f(x)|}{|x|^{\tau_{+}(\beta)}}$

and then we fix $t_{0}\geq s_{0}$ such that

$t_{0}c_{\beta}(-N/2)x_{N}|x|^{-\frac{N}{2}-2}+2s_{0}\tau_{+}(\beta)x_{N}^{2}|x|^{\tau_{+}(\beta)-2}\geq
0.$

So $V_{t_{0},s_{0}}$ is a positive supersolution of (2.8).

For $\beta=\beta_{0},\,\tau_{-}(\beta)=-\frac{N}{2}$, we have that

$\mathcal{L}_{\beta}V_{t,s}(x)=\frac{t}{4}x_{N}|x|^{-\frac{N}{2}-2}(-\ln|x|)^{-\frac{1}{2}}+2s|x|^{-\frac{N}{2}}-2sNx_{N}^{2}|x|^{-\frac{N}{2}-2}.$

We take $s_{0}$ as above where $\beta$ is replaced by $\beta_{0}$ and we fix
$t_{0}\geq s_{0}$ such that

$\frac{t_{0}}{4}x_{N}|x|^{-\frac{N}{2}-2}(-\ln|x|)^{-\frac{1}{2}}-2s_{0}Nx_{N}^{2}|x|^{-\frac{N}{2}-2}\geq
0.$

So $V_{t_{0},s_{0}}$ is also a positive supersolution of (2.8) in this case
which implies, by comparison principle, that we have

$u_{\varepsilon}(x)\leq V_{t_{0},s_{0}}(x),\quad\forall\,x\in\Omega.$

Proof of (2.9). We need to estimate
$\displaystyle\int_{\Omega}\frac{u_{\varepsilon}\xi}{(|x|^{2}+\varepsilon)|x|^{2}x_{N}}\,d\gamma_{\beta}$
for $0<\varepsilon<\varepsilon_{0}$ for some $\varepsilon_{0}>0$ fixed. we
first consider the case $\beta>0$. We observe that

$\displaystyle\varepsilon\int_{\Omega\setminus
B_{\sqrt{\varepsilon}}(0)}\frac{u_{\varepsilon}\xi\lambda_{\beta}(x)}{(|x|^{2}+\varepsilon)|x|^{2}x_{N}}\,dx$
$\displaystyle\leq$
$\displaystyle\varepsilon\|u_{\varepsilon_{0}}\|_{L^{\infty}(\Omega)}\|\xi/\rho\|_{L^{\infty}(\Omega)}\int_{\Omega\setminus
B_{\sqrt{\varepsilon}}(0)}\frac{|x|^{\tau_{+}(\beta)-2}}{|x|^{2}+\varepsilon}\,dx$
$\displaystyle\leq$
$\displaystyle\|u_{\varepsilon_{0}}\|_{L^{\infty}(\Omega)}\|\xi/\rho\|_{L^{\infty}(\Omega)}\varepsilon^{\frac{N-2+\tau_{+}(\beta)}{2}}\int_{B_{\frac{1}{2\sqrt{\varepsilon}}}(0)\setminus
B_{1}(0)}|y|^{\tau_{+}(\beta)-4}\,dy$ $\displaystyle\leq$ $\displaystyle
c_{3}\|u_{\varepsilon_{0}}\|_{L^{\infty}(\Omega)}\|\xi/\rho\|_{L^{\infty}(\Omega)}(2^{-\tau_{+}(\beta)+4-N}\varepsilon+\varepsilon^{\frac{N-2+\tau_{+}(\beta)}{2}})$
$\displaystyle\to$ $\displaystyle 0\ \quad{\rm as}\quad\ \varepsilon\to 0^{+}$

and

$\displaystyle\varepsilon\int_{B_{\sqrt{\varepsilon}}(0)}\frac{u_{\varepsilon}\xi\lambda_{\beta}(x)}{(|x|^{2}+\varepsilon)|x|^{2}x_{N}}\,\,dx$
$\displaystyle\leq$
$\displaystyle\|u_{\varepsilon_{0}}\|_{L^{\infty}(\Omega)}\|\xi/\rho\|_{L^{\infty}(\Omega)}\int_{B_{\sqrt{\varepsilon}}(0)}|x|^{\tau_{+}(\beta)-2}\,dx$
$\displaystyle\leq$ $\displaystyle
c_{4}\|u_{\varepsilon_{0}}\|_{L^{\infty}(\Omega)}\|\xi/\rho\|_{L^{\infty}(\Omega)}\varepsilon^{\frac{N-2+\tau_{+}(\beta)}{2}}$
$\displaystyle\to$ $\displaystyle 0\ \quad{\rm as}\quad\ \varepsilon\to
0^{+},$

where $\rho(x)={\rm dist}(x,\,\partial\Omega)$ and
$\frac{N-2+\tau_{+}(\beta)}{2}>0$. Therefore, passing to the limit of (2.5),
we obtain (2.9).

For $\beta\in(\beta_{0},\,0)$, from the increasing monotonicity and the upper
bound $V_{s_{0},t_{0}}$, we have that

$\lim_{\varepsilon\to
0^{+}}\int_{\Omega}u_{\varepsilon}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})d\gamma_{\beta}=\int_{\Omega}u_{\beta}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})d\gamma_{\beta}$

and

$\varepsilon\int_{\Omega}\frac{\xi
u_{\varepsilon}\lambda_{\beta}(x)}{(|x|^{2}+\varepsilon)|x|^{2}x_{N}}dx\leq
c_{5}\varepsilon\int_{\Omega}\frac{|x|^{-N+\sqrt{\beta-\beta_{0}}}}{|x|^{2}+\varepsilon}dx.$

By directly compute, we have that

$\displaystyle\varepsilon\int_{\Omega\setminus
B_{\sqrt{\varepsilon}}(0)}\frac{|x|^{-N+\sqrt{\beta-\beta_{0}}}}{|x|^{2}+\varepsilon}dx$
$\displaystyle\leq$ $\displaystyle
c_{6}\varepsilon^{\frac{\sqrt{\beta-\beta_{0}}}{2}}\int_{B_{\frac{1}{2\sqrt{\varepsilon}}}(0)\setminus
B_{1}(0)}|y|^{-N-2+\sqrt{\beta-\beta_{0}}}\,dy$ $\displaystyle\leq$
$\displaystyle
c_{7}(\varepsilon+\varepsilon^{\frac{\sqrt{\beta-\beta_{0}}}{2}})\to
0\quad{\rm as}\quad\varepsilon\to 0^{+}$

and

$\displaystyle\varepsilon\int_{B_{\sqrt{\varepsilon}}(0)}\frac{|x|^{-N+\sqrt{\beta-\beta_{0}}}}{|x|^{2}+\varepsilon}dx$
$\displaystyle\leq$
$\displaystyle\int_{B_{\sqrt{\varepsilon}}(0)}|x|^{-N+\sqrt{\beta-\beta_{0}}}\,dx$
$\displaystyle\leq$ $\displaystyle
c_{8}\varepsilon^{\frac{\sqrt{\beta-\beta_{0}}}{2}}\to 0\quad{\rm
as}\quad\varepsilon\to 0^{+},$

As a conclusion, passing to the limit in (2.5) as $\varepsilon\to 0^{+}$, we
have that $u_{\beta}$ satisfies that

(2.11)
$\int_{\Omega}u_{\beta}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=\int_{\Omega}\frac{f\xi}{x_{N}}\,d\gamma_{\beta},\quad\forall\,\xi\in
C^{1.1}_{0}(\Omega).$

Finally, we prove (2.9) with $\beta=\beta_{0}$, We claim that the mapping
$\beta\mapsto u_{\beta}$ with $\beta\in(\beta_{0},0)$ is decreasing. In fact,
if $\beta_{0}<\beta_{1}\leq\beta_{2}<0$, we know that

$\displaystyle f=\mathcal{L}_{\beta_{1}}u_{\beta_{1}}$ $\displaystyle=$
$\displaystyle-\Delta u_{\beta_{1}}+\frac{\beta_{1}}{|x|^{2}}u_{\beta_{1}}$
$\displaystyle\leq$ $\displaystyle-\Delta
u_{\beta_{1}}+\frac{\beta_{2}}{|x|^{2}}u_{\beta_{1}}=\mathcal{L}_{\beta_{2}}u_{\beta_{1}},$

by Lemma 2.2, which implies that $u_{\beta_{1}}\geq u_{\beta_{2}}.$

We know that $V_{s_{0},t_{0}}$ is a super solution of (2.8) with
$\beta\in(\beta_{0},0)$. So it follows by Lemma 2.2 that
$\\{u_{\beta}\\}_{\beta}$ is uniformly bounded by the upper bound
$V_{s_{0},t_{0}}\in L^{1}(\Omega,\frac{1}{x_{N}}d\gamma_{\beta})$.

For $\xi\in C^{1.1}_{0}(\Omega)$, we have that

$|\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})|\leq
c_{9}(\|\frac{\xi}{x_{N}}\|_{C^{1.1}(\Omega)}+\|\frac{\xi}{x_{N}}\|_{C^{1}(\Omega)}x_{N}^{-1}),$

where $c_{9}>0$ is independent of $\beta$.

From the dominate monotonicity convergence theorem and the uniqueness of the
solution, we have that

$u_{\beta}\to u_{\beta_{0}}\quad{\rm a.e.\ in}\ \Omega\;{\rm
as}\;\;\beta\to\beta^{+}_{0}\quad{\rm
and\;in}\;\;L^{1}(\Omega,\,x_{N}^{-1}d\gamma_{\beta})$

and $u_{\beta_{0}}$ is a classical solution of (2.8) with $\beta=\beta_{0}$.
Passing to the limit of (2.11) as $\beta\to\beta^{+}_{0}$ to obtain that

$\int_{\Omega}u_{\beta_{0}}\mathcal{L}^{*}_{\beta_{0}}(\frac{\xi}{x_{N}})\,d\gamma_{\beta_{0}}=\int_{\Omega}\frac{f\xi}{x_{N}}d\gamma_{\beta_{0}}.$

The proof ends. $\Box$

######  Remark 2.5.

We note that when $\beta\geq 0$ and $f$ is bounded, the moderate singular
solution of problem (2.8) is no longer singular, that means, it is a classical
solution of

(2.12)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}u=f\quad{\rm
in}\;\;{\Omega},\\\\[4.2679pt] \phantom{L_{\beta}}\displaystyle u=0\quad{\rm
on}\;\;\partial{\Omega}.\end{array}\right.$

Now we prove the following

######  Lemma 2.6.

$(i)$ The problem

(2.13)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}^{*}(\frac{u}{x_{N}})=1\quad{\rm
in}\;\;\Omega,\\\\[5.69054pt] \phantom{\mathcal{L}_{\beta}^{*}-\
\,\,\,}\displaystyle u=0\quad{\rm on}\;\;\partial{\Omega}\end{array}\right.$

has a unique positive solution $w_{1}\in C^{2}(\Omega)\cap
C^{0.1}_{0}(\Omega)$.

$(ii)$ The problem

(2.14)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}^{*}(\frac{u}{x_{N}})=\frac{1}{x_{N}}&{\rm
in}\;\;\Omega,\\\\[5.69054pt] \phantom{\mathcal{L}_{\beta}^{*}--}\displaystyle
u=0&{\rm on}\;\;\partial{\Omega}\end{array}\right.$

has a unique positive solution $w_{2}\in C^{2}(\Omega)\cap
C^{1}_{0}(\bar{\Omega}\setminus\\{0\\})\cap C^{0.1}_{0}(\Omega)$.

Proof. We first claim that problem (2.8) has a unique classical positive
solution $w_{\beta}$ under the constraint (2.7) when $f(x)=\lambda_{\beta}(x)$
or $f(x)=|x|^{\tau_{+}(\beta)}$.

In fact, let $f_{n}(x)=\lambda_{\beta}(x)\eta_{0}(n|x|)$, where
$\eta_{0}:[0,+\infty)\to[0,\,1]$ is a decreasing $C^{\infty}$ function
satisfying (2.1). Then $f_{n}\in C^{\theta}(\bar{\Omega})$ with
$\theta\in(0,1),f_{n}\leq f$, and by Proposition 2.4, let $w_{n}$ be the
solution of problem

(2.15)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}u=f_{n}&{\rm
in}\;\;{\Omega},\\\\[5.69054pt] \phantom{L_{\beta}}\displaystyle u=0&{\rm
on}\;\;\partial{\Omega}\setminus\\{0\\},\end{array}\right.$

subject to (2.7). We know that the mapping: $n\to w_{n}$ is increasing by the
increasing monotone of $\\{f_{n}\\}$. So we only construct a suitable upper
bound for $w_{n}$ in the cases that $f(x)=\lambda_{\beta}(x)$ and
$f(x)=|x|^{\tau_{+}(\beta)}$ respectively.

When $f(x)=\lambda_{\beta}(x)$, let
$V_{t,s}(x)=t\lambda_{\beta}(x)-sx_{N}|x|^{\tau_{+}(\beta)+2}$ for $s,t>0$. It
is know that

$\mathcal{L}_{\beta}V_{t,s}=-sc_{\tau_{+}(\beta)+2}\lambda_{\beta}(x),\quad
x\in\mathbb{R}^{N}_{+},$

for some $c_{\tau_{+}(\beta)+2}<0$. So fix $s=-1/c_{\tau_{+}(\beta)+2}$ and
then fix $t>0$ such that

$V_{t,s}(x)>0,\quad\forall x\in\Omega.$

The limit of $\\{w_{n}\\}_{n}$, denoting by $w_{\beta,1}$, is a solution of
(2.7) satisfying $w_{\beta,1}\leq V_{t,s}(x).$

When $f(x)=|x|^{\tau_{+}(\beta)}$, let

$W_{t,s,l}(x)=t\lambda_{\beta}(x)-s(x_{N}|x|^{\tau_{+}(\beta)+2}+lx_{N}^{2}|x|^{\tau_{+}(\beta)+2}),$

where $s,t,l>0$. We observe that

$\mathcal{L}_{\beta}W_{t,s,l}(x)=s[-c_{\tau_{+}(\beta)+2}\lambda_{\beta}(x)+2l|x|^{\tau_{+}(\beta)}+2l\tau_{+}(\beta)x_{N}^{2}|x|^{\tau_{+}(\beta)}],\;x\in\mathbb{R}^{N}_{+},$

with the same constant $c_{\tau_{+}(\beta)+2}<0$ as above. Then we choose
$l>0$ such that $-2c_{\tau_{+}(\beta)+2}l\tau_{+}(\beta)x_{N}>0$ for
$x\in\Omega$, $s=\frac{1}{2l}$ and we take $t>0$ such that $W_{t,s,l}>0$ in
$\Omega$ and

$\mathcal{L}_{\beta}W_{t,s,l}(x)\geq|x|^{\tau_{+}(\beta)}.$

Thus, the limit of $\\{w_{n}\\}_{n}$, denoting by $w_{\beta,2}$, is a solution
of (2.7) such that

$w_{\beta,2}(x)\leq W_{t,s,l}(x).$

As a conclusion, for $i=1,2$,

(2.16) $w_{\beta,i}\leq t\lambda_{\beta}\quad{\rm in}\quad\Omega.$

Denote $w_{i}=w_{\beta,i}x_{N}/\lambda_{\beta},$ we observe that

$\displaystyle
1=\lambda_{\beta}^{-1}\mathcal{L}_{\beta}w_{\beta,1}=\lambda_{\beta}^{-1}\mathcal{L}_{\beta}(\lambda_{\beta}w_{1}/x_{N})=\mathcal{L}_{\beta}^{*}(w_{1}/x_{N})$

and

$\displaystyle
1/x_{N}=\lambda_{\beta}^{-1}\mathcal{L}_{\beta}w_{\beta,2}=\lambda_{\beta}^{-1}\mathcal{L}_{\beta}(\lambda_{\beta}w_{2}/x_{N})=\mathcal{L}_{\beta}^{*}(w_{2}/x_{N}).$

Moreover, by (2.16), it follow that $w_{i}\leq tx_{N}.$ Then we have that
$w_{i}\in C^{2}(\Omega)\cap C^{0.1}_{0}(\Omega)$ for $i=1,2$. Away from the
origin, Hardy’s operator is uniform elliptic, thus $u\in
C^{1}_{0}(\bar{\Omega}\setminus\\{0\\})$ and then $u\in C^{2}(\Omega)\cap
C^{1}_{0}(\bar{\Omega}\setminus\\{0\\})\cap C^{0.1}_{0}(\Omega).$ $\Box$

Although $C^{2}(\Omega)\cap C^{1}_{0}(\bar{\Omega}\setminus\\{0\\})\cap
C^{0.1}_{0}(\Omega)$ is not suitable as test function space for problem (1.2),
$w_{1},\,w_{2}$ are still valid as test functions for formula (1.11) with
$k=0$ in the distributional sense.

For given $f\in C^{1}(\bar{\Omega})$, a direct consequence of Lemma 2.6 can be
stated as follows

######  Corollary 2.7.

Assume that $f\in C^{1}(\bar{\Omega}\setminus\\{0\\})$ satisfying for some
$c_{10}>0$

$|f(x)|\leq\frac{c_{10}}{x_{N}}.$

Then there exists a unique solution of $w_{f}\in C^{2}(\Omega)\cap
C^{0.1}_{0}(\Omega)$ of

(2.17)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}^{*}(\frac{u}{x_{N}})=f&{\rm
in}\;\;\Omega,\\\\[5.69054pt] \phantom{\mathcal{L}_{\beta}^{*}-\
\,\,\,}\displaystyle u=0&{\rm on}\;\;\partial{\Omega}.\end{array}\right.$

##  3\. Fundamental solution

###  3.1.  In half space

In this subsection, we give the proof of Theorem 1.1.

Proof of Theorem 1.1. For any $\xi\in C^{1.1}_{0}(\mathbb{R}^{N}_{+})$, we
know there exists a unique $\zeta\in C^{1.1}_{c}(\ \mathbb{R}^{N})$ such that
$\xi(x)=x_{N}\zeta(x)$ for $x\in\overline{\mathbb{R}^{N}_{+}}$. Moreover, we
have that $\frac{\partial\xi}{\partial x_{N}}(0)=\zeta(0)$.

Take $\zeta\in C^{1.1}_{c}(\mathbb{R}^{N})$, multiplying
$\lambda_{\beta}\zeta$ in (1.3) and integrating over
$\mathbb{R}^{N}_{+}\setminus\overline{B_{r}(0)}$, then we have that

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N}_{+}\setminus\overline{B_{r}(0)}}\mathcal{L}_{\beta}(\Lambda_{\beta})\lambda_{\beta}\zeta\,dx=\int_{\mathbb{R}^{N}_{+}\setminus\overline{B_{r}(0)}}\Lambda_{\beta}\mathcal{L}_{\beta}^{*}(\zeta)\,d\gamma_{\beta}$
$\displaystyle+\int_{\partial_{+}B_{r}(0)}\Big{(}-\nabla\Lambda_{\beta}\cdot\frac{x}{|x|}\lambda_{\beta}+\nabla\lambda_{\beta}\cdot\frac{x}{|x|}\Lambda_{\beta}\Big{)}\zeta\,d\omega$
$\displaystyle+\int_{\partial_{+}B_{r}(0)}\Lambda_{\beta}\lambda_{\beta}\Big{(}\nabla\zeta\cdot\frac{x}{|x|}\Big{)}\,d\omega,$

where $\partial_{+}B_{r}(0)=\partial B_{r}(0)\cap\mathbb{R}^{N}_{+}$. For
$\beta\geq\beta_{0}$, we see that for $r=|x|>0$ small,

$\displaystyle-\nabla\Lambda_{\beta}(x)\cdot\frac{x}{|x|}\lambda_{\beta}(x)+\nabla\lambda_{\beta}(x)\cdot\frac{x}{|x|}\Lambda_{\beta}(x)$
$\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{lll}2\sqrt{\beta-\beta_{0}}\,x_{N}^{2}r^{-N-1}&{\rm
if}\;\;\beta>\beta_{0},\\\\[4.2679pt] \phantom{}x_{N}^{2}r^{-N-1}&{\rm
if}\;\;\beta=\beta_{0}\end{array}\right.$

and

$|\zeta(x)-\zeta(0)|\leq c_{11}r,$

then

$\displaystyle\int_{\partial_{+}B_{r}(0)}\sqrt{\beta-\beta_{0}}\,x_{N}^{2}r^{-N-1}\zeta(0)x_{N}d\omega(x)$
$\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{lll}\sqrt{\beta-\beta_{0}}\,\displaystyle\int_{\partial_{+}B_{1}(0)}x_{N}^{2}d\omega(x)\,\zeta(0)&{\rm
if}\;\;\beta>\beta_{0},\\\\[5.69054pt]
\phantom{}\displaystyle\int_{\partial_{+}B_{1}(0)}x_{N}^{2}d\omega(x)\,\zeta(0)&{\rm
if}\;\;\beta=\beta_{0}\end{array}\right.$ $\displaystyle=$ $\displaystyle
c_{\beta}\zeta(0)$

and

$\displaystyle\Big{|}\int_{\partial_{+}B_{r}(0)}\Big{(}-\nabla\Lambda_{\beta}\cdot\frac{x}{|x|}\lambda_{\beta}+\nabla\lambda_{\beta}\cdot\frac{x}{|x|}\Lambda_{\beta}\Big{)}\zeta\,d\omega-
c_{\beta}\zeta(0)\Big{|}$ $\displaystyle\leq$ $\displaystyle
c_{12}(\sqrt{\beta-\beta_{0}}+1)\,r\int_{\partial_{+}B_{1}(0)}x_{N}^{2}d\omega(x)$
$\displaystyle\to$ $\displaystyle 0\quad{\rm as}\quad r\to 0^{+},$

that is,

$\lim_{r\to
0}\Big{(}\int_{\partial_{+}B_{r}(0)}-\nabla\Lambda_{\beta}\cdot\frac{x}{|x|}\lambda_{\beta}\zeta\,d\omega+\int_{\partial
B_{r}(0)}\nabla\lambda_{\beta}\cdot\frac{x}{|x|}\Lambda_{\beta}\zeta\,d\omega\Big{)}=c_{\beta}\zeta(0).$

Moreover, we see that

$\Big{|}\int_{\partial_{+}B_{r}(0)}\Lambda_{\beta}\lambda_{\beta}\Big{(}\nabla\zeta\cdot\frac{x}{|x|}\Big{)}\,d\omega\Big{|}\leq\|\zeta\|_{C^{1}}\,r\int_{\partial_{+}B_{1}(0)}x_{N}^{2}d\omega\to
0\quad{\rm as}\quad r\to 0^{+}.$

Therefore, we have that

$\lim_{r\to
0^{+}}\int_{\mathbb{R}^{N}\setminus\overline{B_{r}(0)}}\Lambda_{\beta}\mathcal{L}_{\beta}^{*}(\zeta)d\gamma_{\beta}=c_{\beta}\zeta(0),$

which implies (1.7). The proof ends. $\Box$

###  3.2.  Trace of $\Lambda_{\beta}$.

The following theorem shows the trace of $\Lambda_{\beta}$.

######  Theorem 3.1.

Let $d\omega_{\beta}(x^{\prime})=|x^{\prime}|^{\tau_{+}(\beta)}dx^{\prime}$
for $x^{\prime}\in\mathbb{R}^{N-1}$, then for any $\zeta\in
C_{c}(\mathbb{R}^{N-1})$,

(3.3) $\lim_{t\to
0^{+}}\int_{\mathbb{R}^{N-1}}\Lambda_{\beta}(x^{\prime},t)\zeta(x^{\prime})d\omega_{\beta}(x^{\prime})=b_{N}\zeta(0),$

where

$b_{N}=\int_{\mathbb{R}^{N-1}}(1+|y^{\prime}|^{2})^{-\frac{N}{2}}dy^{\prime}>0.$

This is to say that the trace of $\Lambda_{\beta}$ is $\delta_{0}$ in the
$d\gamma_{\beta}$-distributional sense.

Proof. For any $\zeta\in C_{c}(\mathbb{R}^{N-1})$, there exists $R>0$ such
that supp$\,\zeta\subset B_{R}^{\prime}(0)$, here and in the sequel, denoting
by $B_{R}^{\prime}(0)$ the ball in $\mathbb{R}^{N-1}$. By direct computations,
we have that

$\displaystyle\int_{\mathbb{R}^{N-1}}\Lambda_{\beta}(x^{\prime},t)\zeta(x^{\prime})\,d\omega_{\beta}(x^{\prime})$
$\displaystyle=$
$\displaystyle\int_{B_{R}^{\prime}(0)}\Lambda_{\beta}(x^{\prime},t)\zeta(x^{\prime})\,d\omega_{\beta}(x^{\prime})$
$\displaystyle=$
$\displaystyle\int_{B_{R/t}^{\prime}(0)}(|y^{\prime}|^{2}+1)^{\frac{\tau_{-}(\beta)}{2}}|y^{\prime}|^{\tau_{+}(\beta)}\zeta(ty^{\prime})dy^{\prime}.$

For any $\varepsilon>0$, there exists $R_{\varepsilon}>1$ such that

$\displaystyle\int_{B_{R/t}^{\prime}(0)\setminus
B^{\prime}_{R_{\varepsilon}}(0)}(|y^{\prime}|^{2}+1)^{\frac{\tau_{-}(\beta)}{2}}|y^{\prime}|^{\tau_{+}(\beta)}\zeta(ty^{\prime})dy^{\prime}$
$\displaystyle\leq$
$\displaystyle\|\zeta\|_{L^{\infty}(\mathbb{R}^{N-1})}\int_{\mathbb{R}^{N-1}\setminus
B^{\prime}_{R_{\varepsilon}}(0)}|y^{\prime}|^{-N}dy^{\prime}$
$\displaystyle\leq$
$\displaystyle\|\zeta\|_{L^{\infty}(\mathbb{R}^{N-1})}|\mathcal{S}^{N-2}|\varepsilon,$

where $R_{\varepsilon}\leq\frac{1}{\varepsilon}$. Let

$A:=\int_{B^{\prime}_{R_{\varepsilon}}(0)}(|y^{\prime}|^{2}+1)^{\frac{\tau_{-}(\beta)}{2}}|y^{\prime}|^{\tau_{+}(\beta)}\zeta(ty^{\prime})dy^{\prime}-\int_{\mathbb{R}^{N-1}}(|y^{\prime}|^{2}+1)^{\frac{\tau_{-}(\beta)}{2}}|y^{\prime}|^{\tau_{+}(\beta)}\zeta(0)dy^{\prime},$

we have that

$\displaystyle|A|$ $\displaystyle\leq$
$\displaystyle\int_{B^{\prime}_{R_{\varepsilon}}(0)}(|y^{\prime}|^{2}+1)^{\frac{\tau_{-}(\beta)}{2}}|y^{\prime}|^{\tau_{+}(\beta)}\left|\zeta(ty^{\prime})-\zeta(0)\right|\,dy^{\prime}+\varepsilon|\zeta(0)||\mathcal{S}^{N-2}|$
$\displaystyle\leq$ $\displaystyle
t\|\zeta\|_{C^{1}(\mathbb{R}^{N-1})}\int_{B^{\prime}_{R_{\varepsilon}}(0)}(|y^{\prime}|^{2}+1)^{\frac{\tau_{-}(\beta)}{2}}|y^{\prime}|^{\tau_{+}(\beta)}dy^{\prime}+\varepsilon|\zeta(0)||\mathcal{S}^{N-2}|$
$\displaystyle=$ $\displaystyle
R_{\varepsilon}t\|\zeta\|_{C^{1}(\mathbb{R}^{N-1})}+\varepsilon|\zeta(0)||\mathcal{S}^{N-2}|$
$\displaystyle\leq$
$\displaystyle\left(\|\zeta\|_{C^{1}(\mathbb{R}^{N-1})}+|\zeta(0)||\mathcal{S}^{N-2}|\right)\varepsilon,$

if we take $t=\varepsilon^{2}$. Passing to the limit as $\varepsilon\to 0$, we
derive (3.3). $\Box$

###  3.3. Fundamental solution in bounded domain

In this subsection, we do an approximation of the isolated singular solution.

######  Proposition 3.2.

Let $\Omega$ be a $C^{2}$ domain verifying (1.9). Then the problem

(3.4)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}u=0\quad{\rm
in}\;\;{\Omega},\\\\[5.69054pt] \phantom{L_{\beta}}\displaystyle u=0\quad{\rm
on}\;\;\partial{\Omega}\setminus\\{0\\},\\\\[5.69054pt]
\phantom{}\displaystyle\lim_{r\to 0^{+}}\sup_{x\in
B_{r}^{+}(0)}\frac{|u(x)-\Lambda_{\beta}(x)|}{\Lambda_{\beta}(x)}=0\end{array}\right.$

admits a unique solution $\Lambda^{\Omega}_{\beta}$ satisfying the following
distributional identity:

(3.5)
$\int_{\Omega}\Lambda^{\Omega}_{\beta}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=c_{\beta}\frac{\partial\xi}{\partial
x_{N}}(0),\quad\forall\,\xi\in C^{1.1}_{0}(\Omega).$

Proof. Let $\eta_{r_{0}}(t)=\eta_{0}(\frac{2}{r_{0}}t)$, which satisfies that

(3.6) $\eta_{r_{0}}=1\quad{\rm in}\quad[0,r_{0}/2]\quad{\rm
and}\quad\eta_{r_{0}}=0\quad{\rm in}\quad[r_{0},+\infty).$

For $i=1,2$ the problem

(3.7)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}w_{i}=-\nabla\eta_{r_{0}}\cdot\nabla\Lambda_{\beta}-\Lambda_{\beta}\Delta\eta_{r_{0}}\quad{\rm
in}\;\;{\Omega},\\\\[5.69054pt] \phantom{L_{\beta}}\displaystyle
w_{i}=0\quad{\rm on}\;\;\partial{\Omega}\setminus\\{0\\},\\\\[5.69054pt]
\phantom{}\displaystyle\lim_{e\in\mathcal{S}^{N}_{+},\,t\to
0^{+}}w_{i}(te)\Lambda_{\beta}^{-1}(te)=2-i,\end{array}\right.$

admits a unique solutions $w_{1}$ and $w_{2}$ respectively. Obviously,

$w_{1}=\Lambda_{\beta}\eta_{r_{0}}$

and
$-\nabla\eta_{r_{0}}\cdot\nabla\Lambda_{\beta}-\Lambda_{\beta}\Delta\eta_{r_{0}}$
has compact set in $\Omega\cap(\overline{B_{r_{0}}(0)\setminus
B_{\frac{r_{0}}{2}}(0)})$ and then
$-\nabla\eta_{r_{0}}\cdot\nabla\Lambda_{\beta}-\Lambda_{\beta}\Delta\eta_{r_{0}}$
is smooth and bounded, it follows by the proof of Proposition 2.4 that there
exist $s_{0},t_{0}>0$ such that $|w_{2}|\leq V_{s_{0},t_{0}}$.

For $i=1$, following the proof of Theorem 1.1, we get then for any $\xi\in
C^{1.1}_{0}(\Omega)$,

(3.8)
$\int_{\Omega}w_{1}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=\int_{\Omega}\Big{(}-\nabla\eta_{r_{0}}\cdot\nabla\Lambda_{\beta}-\Lambda_{\beta}\Delta\eta_{r_{0}}\Big{)}\frac{\xi}{x_{N}}\,d\gamma_{\beta}+c_{\beta}\frac{\partial\xi}{\partial
x_{N}}(0).$

For $i=2$, it follows by Proposition 2.4 that for any $\xi\in
C^{1.1}_{0}(\Omega)$,

(3.9)
$\int_{\Omega}w_{2}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})d\gamma_{\beta}=\int_{\Omega}\Big{(}-\nabla\eta_{r_{0}}\cdot\nabla\Lambda_{\beta}-\Lambda_{\beta}\Delta\eta_{r_{0}}\Big{)}\frac{\xi}{x_{N}}\,d\gamma_{\beta}.$

Let $\Lambda^{\Omega}_{\beta}=\Lambda_{\beta}\eta_{r_{0}}-w_{2},$ it follows
by (3.8) and (3.9) that

$\int_{\Omega}\Lambda^{\Omega}_{\beta}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=c_{\beta}\frac{\partial\xi}{\partial
x_{N}}(0),\quad\forall\,\xi\in C^{1.1}_{0}(\Omega).$

Finally, it’s clear that if $u_{1}$ and $u_{2}$ are two solutions of (3.4),
then $w:=u_{1}-u_{2}$ satisfies

$\lim_{r\to 0^{+}}\sup_{x\in
B_{r}^{+}(0)}\frac{|w(x)|}{\Lambda_{\beta}(x)}=0.$

Combining with the fact that

$\mathcal{L}_{\beta}w=0\;\;{\rm in}\;\;{\Omega}\quad{\rm and}\quad w=0\;\;{\rm
on}\;\;\partial{\Omega}\setminus\\{0\\},$

and Lemma 2.2, we have that $w\equiv 0$. Thus the uniqueness is proved. $\Box$

##  4\. Existence

###  4.1. Zero Dirichlet boundary

Our purpose in this section is to clarify the isolated singularities of the
nonhomogeneous problem

(4.1)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}u=f\quad{\rm
in}\;\;\Omega,\\\\[4.2679pt] \phantom{L_{\beta}}\displaystyle u=0\quad{\rm
on}\;\;\partial{\Omega}\setminus\\{0\\},\end{array}\right.$

where $f\in C^{\theta}_{loc}(\bar{\Omega}\setminus\\{0\\})$ with
$\theta\in(0,1)$. Recall that $\mathcal{L}^{*}_{\beta}$ is given by (1.6) and
$d\gamma_{\beta}(x)=\lambda_{\beta}(x)dx$. We prove the following

######  Theorem 4.1.

$(i)$ Assume that $f\in L^{1}(\Omega,\,d\gamma_{\beta})$ and $u\in
L^{1}(\Omega,\frac{1}{|x|}d\gamma_{\beta})$ is a classical solution of problem
(4.1), then there exists some $k\in\mathbb{R}$ such that there holds

(4.2)
$\int_{\Omega}u\,\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=\int_{\Omega}\frac{f\xi}{x_{N}}\,d\gamma_{\beta}+k\frac{\partial\xi}{\partial
x_{N}}(0),\quad\forall\,\xi\in C^{1.1}_{0}(\Omega).$

$(ii)$ Inversely, assume that $f\in L^{1}(\Omega,\,d\gamma_{\beta})$, then for
any $k\in\mathbb{R}$, problem (4.1) has a unique solution $u_{k}\in
L^{1}(\Omega,\frac{1}{|x|}d\gamma_{\beta})$ verifying (4.2) with such $k.$

Proof. $(i)$ Let $\tilde{\Omega}$ be the interior set of
$\bar{\Omega}\cup\overline{\\{(x^{\prime},-x_{N}):(x^{\prime},x_{N})\in\Omega\\}}$
and extend $u$ (resp. $f$) by the $x_{N}$-odd extension to $\tilde{u}$ (resp.
$\tilde{f}$) in $\tilde{\Omega}$, then
$\mathcal{L}_{\beta}\tilde{u}=\tilde{f}$. Our aim is to see the distributional
property at the origin. Denote by $L$ the operator related to
$\mathcal{L}_{\beta}\tilde{u}-\tilde{f}$ in the distribution sense, i.e.

(4.3)
$L(\zeta)=\int_{\tilde{\Omega}}\Big{(}\tilde{u}\mathcal{L}_{\beta}^{*}(\zeta)-\tilde{f}\zeta\Big{)}|x_{N}||x|^{\tau_{+}(\beta)}\,dx,\quad\forall\zeta\in
C^{\infty}_{c}(\tilde{\Omega}).$

For any $\zeta\in C^{\infty}_{c}(\tilde{\Omega}\setminus\\{0\\})$, we have
that $L(\zeta)=0.$ In fact, there exists $\varepsilon>0$ such that
supp$(\zeta)\subset\tilde{\Omega}\setminus B_{\varepsilon}(0)$ and then

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
2\int_{\Omega}\zeta(\mathcal{L}_{\beta}u-f)\,d\gamma_{\beta}=\int_{\tilde{\Omega}}\zeta(\mathcal{L}_{\beta}\tilde{u}-\tilde{f})\,d\tilde{\gamma}_{\beta}$
$\displaystyle=$
$\displaystyle-\int_{\tilde{\Omega}}\tilde{f}\zeta\,d\tilde{\gamma}_{\beta}+\int_{\Omega\setminus
B_{\varepsilon}(0)}u\mathcal{L}_{\beta}^{*}\zeta
d\gamma_{\beta}+\int_{\partial(\Omega\setminus
B_{\varepsilon}(0))\cap(\mathbb{R}^{N-1}\times\\{0\\})}\frac{\partial
u}{\partial x_{N}}\zeta d\omega_{\beta}$
$\displaystyle+\int_{(-\Omega)\setminus
B_{\varepsilon}(0)}(-u)\mathcal{L}_{\beta}^{*}\zeta
d\tilde{\gamma}_{\beta}+\int_{\partial(-\Omega\setminus
B_{\varepsilon}(0))\cap(\mathbb{R}^{N-1}\times\\{0\\})}\frac{\partial\tilde{u}}{\partial(-x_{N})}\zeta
d\omega_{\beta}$ $\displaystyle=$ $\displaystyle\int_{\tilde{\Omega}\setminus
B_{\varepsilon}(0)}(\tilde{u}\mathcal{L}_{\beta}^{*}\zeta-\tilde{f}\zeta)\,d\tilde{\gamma}_{\beta}$
$\displaystyle=$
$\displaystyle\int_{\tilde{\Omega}}(\tilde{u}\mathcal{L}_{\beta}^{*}\zeta-\tilde{f}\zeta)\,d\tilde{\gamma}_{\beta},$

where $d\tilde{\gamma}_{\beta}=|\tilde{\lambda}_{\beta}(x)|dx$,
$\tilde{\lambda}_{\beta}$ is the odd extension of $\lambda_{\beta}$ and

$\int_{\partial(\Omega\setminus
B_{\varepsilon}(0))\cap(\mathbb{R}^{N-1}\times\\{0\\})}\frac{\partial
u}{\partial x_{N}}\zeta d\omega_{\beta}=-\int_{\partial(-\Omega\setminus
B_{\varepsilon}(0))\cap(\mathbb{R}^{N-1}\times\\{0\\})}\frac{\partial\tilde{u}}{\partial(-x_{N})}\zeta
d\omega_{\beta}.$

By Theorem XXXV in [33] (see also Theorem 6.25 in [32]), it implies that

(4.4) $L=\sum_{|a|=0}^{p}k_{a}D^{a}\delta_{0},$

where $p\in\mathbb{N}$, $a=(a_{1},\cdots,a_{N})$ is a multiple index with
$a_{i}\in\mathbb{N}$, $|a|=\sum_{i=1}^{N}a_{i}$ and in particular,
$D^{0}\delta_{0}=\delta_{0}$. Then we have that

(4.5)
$L(\zeta)=\int_{\tilde{\Omega}}\Big{(}\tilde{u}\mathcal{L}_{\beta}^{*}\zeta-f\zeta\Big{)}\,d\tilde{\gamma}_{\beta}=\sum_{|a|=0}^{\infty}k_{a}D^{a}\zeta(0),\quad\
\ \forall\zeta\in C^{\infty}_{c}(\tilde{\Omega}).$

For any multiple index $a=(a_{1},\cdots,a_{N})$, let $\zeta_{a}$ be a
$C^{\infty}$ function such that

(4.6) ${\rm supp}(\zeta_{a})\subset\overline{B_{2}(0)}\quad{\rm
and}\quad\zeta_{a}(x)=k_{a}\prod_{i=1}^{N}x_{i}^{a_{i}}\quad{\rm for}\ \ x\in
B_{1}(0).$

Now we use the test function
$\zeta_{\varepsilon,a}(x):=\zeta_{a}(\varepsilon^{-1}x)$ for
$x\in\tilde{\Omega}$ in (4.5), we have that

$\sum_{|a|\leq
q}k_{a}D^{a}\zeta_{\varepsilon,a}(0)=\frac{k_{a}^{2}}{\varepsilon^{|a|}}\prod^{N}_{i=1}a_{i}!,$

where $a_{i}!=a_{i}\cdot(a_{i}-1)\cdots 1>0$ and $a_{i}!=1$ if $a_{i}=0$.

Let $r>0$, we obtain that

$\displaystyle\Big{|}\int_{\tilde{\Omega}}\tilde{u}\mathcal{L}_{\beta}^{*}\zeta_{\varepsilon}\,d\tilde{\gamma}_{\beta}\Big{|}$
$\displaystyle=$
$\displaystyle\Big{|}\int_{B_{2\varepsilon}(0)}\tilde{u}\mathcal{L}_{\beta}^{*}\zeta_{\varepsilon}\,d\tilde{\gamma}_{\beta}\Big{|}$
$\displaystyle\leq$
$\displaystyle\frac{1}{\varepsilon^{2}}\Big{|}\int_{B_{2\varepsilon}(0)}\tilde{u}(x)(-\Delta)\zeta_{a}(\varepsilon^{-1}x)\,d\tilde{\gamma}_{\beta}\Big{|}$
$\displaystyle+\frac{2|\tau_{+}(\beta)|}{\varepsilon}\,\Big{|}\int_{B_{2\varepsilon}(0)}\tilde{u}(x)\frac{x}{|x|^{2}}\cdot\nabla\zeta_{a}(\varepsilon^{-1}x)\,d\tilde{\gamma}_{\beta}\Big{|}$
$\displaystyle\leq$ $\displaystyle
c_{13}\left[\frac{1}{\varepsilon^{2}}\int_{B_{2\varepsilon}(0)}|\tilde{u}(x)|\,d\tilde{\gamma}_{\beta}+\frac{1}{\varepsilon}\,\int_{B_{2\varepsilon}(0)}\frac{|\tilde{u}(x)|}{|x|}\,d\tilde{\gamma}_{\beta}\right]$
$\displaystyle\leq$
$\displaystyle\frac{c_{14}}{\varepsilon}\,\int_{B_{2\varepsilon}(0)}\frac{|\tilde{u}(x)|}{|x|}\,d\tilde{\gamma}_{\beta},$

then, by the fact that $u\in L^{1}(\Omega,\frac{1}{|x|}d\gamma_{\beta})$, it
follows that

(4.7) $\lim_{\varepsilon\to
0^{+}}\displaystyle\int_{B_{2\varepsilon}(0)}\frac{|\tilde{u}(x)|}{|x|}\,d\tilde{\gamma}_{\beta}=0\quad{\rm
and}\quad\lim_{\varepsilon\to
0^{+}}\varepsilon\Big{|}\int_{\tilde{\Omega}}\tilde{u}\mathcal{L}_{\beta}^{*}\zeta_{\varepsilon}\,d\tilde{\gamma}_{\beta}\Big{|}=0.$

For $|a|\geq 1$, we have that

$k_{a}^{2}\leq
c_{15}\varepsilon^{|a|-1}\Big{|}\int_{\tilde{\Omega}}\tilde{u}\mathcal{L}_{\beta}^{*}\zeta_{\varepsilon}\,d\tilde{\gamma}_{\beta}\Big{|}\to
0\quad{\rm as}\quad\varepsilon\to 0,$

then we have $k_{a}=0$ by arbitrary of $\varepsilon>0$ in (4.5) with $|a|\geq
1$, thus,

(4.8)
$L(\zeta)=\int_{\tilde{\Omega}}\left[\tilde{u}\mathcal{L}_{\beta}^{*}\zeta-\tilde{f}\zeta\right]\,d\tilde{\gamma}_{\beta}=k_{0}\zeta(0),\quad\
\forall\,\xi\in C^{\infty}_{c}(\tilde{\Omega}).$

For any $\zeta\in C^{1.1}_{c}(\tilde{\Omega})$, by taking a sequence
$\zeta_{n}\in C^{\infty}_{c}(\tilde{\Omega})$ converging to $\zeta$, we obtain
that (4.8) holds for any $\zeta\in C^{1.1}_{c}(\tilde{\Omega})$.

Now we fix $\xi\in C^{1.1}_{0}(\Omega)$ with compact support in
$\Omega\cup\\{(x^{\prime},0)\in\mathbb{R}^{N-1}\times\mathbb{R}:|x^{\prime}|<r_{0}\\}$,
then $\xi/x_{N}\in C^{1.1}(\bar{\Omega})$ and we may do $x_{N}$-even extension
of $\xi/x_{N}$ in $\tilde{\Omega}$, denoting by $\tilde{\xi}$, then
$\tilde{\xi}\in C^{1.1}_{c}(\tilde{\Omega})$, by the $x_{N}$-even extension,
we have that

$\tilde{\xi}(0)=\frac{\partial\xi}{\partial x_{N}}(0).$

So it follows from (4.8) that

(4.9)
$\int_{\Omega}\Big{(}\tilde{u}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})-\tilde{f}\frac{\xi}{x_{N}}\Big{)}\,d\gamma_{\beta}=k_{0}\frac{\partial\xi}{\partial
x_{N}}(0),\quad\ \forall\,\xi\in C^{\infty}_{c}(\tilde{\Omega}),$

so (4.2) holds.

$(ii)$ By the linearity of $\mathcal{L}_{\beta}$, we may assume that $f\geq
0$. Let $f_{n}=f\eta_{n}$, where $\eta_{n}(r)=1-\eta_{0}(nr)$ for $r\geq 0$,
where $\eta_{0}$ satisfies (2.1) and let $v_{n}$ be solution of (2.8) where
$f$ is replaced by $f_{n}$. We see that $f_{n}$ is bounded and for any $\xi\in
C^{1.1}_{0}(\Omega)$,

(4.10)
$\int_{\Omega}v_{n}\,\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=\int_{\Omega}f_{n}\frac{\xi}{x_{N}}\,d\gamma_{\beta}.$

Then taking $\xi=w_{2}$ in Lemma 2.6, we have that $v_{n}$ is uniformly
bounded in $L^{1}(\Omega,\,d\gamma_{\beta})$ and in
$L^{1}(\Omega,\,x_{N}^{-1}d\gamma_{\beta})$, that is,

$\|v_{n}\|_{L^{1}(\Omega,\,x_{N}^{-1}d\gamma_{\beta})}\leq\|\frac{\xi}{x_{N}}\|_{L^{\infty}(\Omega)}\|f_{n}\|_{L^{1}(\Omega,d\gamma_{\beta})}\leq\|\frac{\xi}{x_{N}}\|_{L^{\infty}(\Omega)}\|f\|_{L^{1}(\Omega,d\gamma_{\beta})}.$

Moreover, $\\{v_{n}\\}$ is increasing, and then there exists $v_{f}$ such that

$v_{n}\to v_{f}\quad{\rm a.e.\ in}\ \ \Omega\quad{\rm and\ in}\ \
L^{1}(\Omega,\,x_{N}^{-1}d\gamma_{\beta}).$

Then we have that

$\int_{\Omega}v_{f}\mathcal{L}_{\beta}^{*}(\xi)\,d\gamma_{\beta}=\int_{\Omega}f\xi\,d\gamma_{\beta},\quad\forall\,\xi\in
C^{1.1}_{0}(\Omega).$

Since $f\in C^{\gamma}(\overline{\Omega}\setminus\\{0\\})$, then it follows by
the standard regularity theory that $v_{f}\in C^{2}(\Omega)$.

We claim that $v_{f}$ is a classical solution of (4.1). From Corollary 2.8 in
[31] with $L^{*}=\mathcal{L}_{\beta}^{*}$, which is strictly elliptic in
$\Omega\setminus B_{r}(0)$, we have that for $q<\frac{N}{N-1}$,

(4.11) $\displaystyle\|v_{n}\lambda_{\beta}\|_{W^{1,q}(\Omega_{2r})}$
$\displaystyle\leq$ $\displaystyle
c_{16}\|f\lambda_{\beta}\|_{L^{1}(\Omega\setminus
B_{r}(0))}+c_{16}\|v_{n}\lambda_{\beta}\|_{L^{1}(\Omega\setminus B_{r}(0))}$
$\displaystyle\leq$ $\displaystyle
c_{17}\|f\|_{L^{1}(\Omega,\,d\gamma_{\beta})},$

where $\Omega_{2r}=\\{x\in\Omega\setminus B_{2r}(0):\,\rho(x)>2r\\}.$ We see
that

$\displaystyle-\Delta v_{n}=-\frac{\beta}{|x|^{2}}v_{n}+f.$

For any compact set $K$ in $\Omega$, it is standard to improve the regularity
$v_{n}$

$\|v_{n}\|_{C^{2,\lambda}(K)}\leq
c_{18}[\|f\|_{L^{1}(\Omega,\,d\gamma_{\beta})}+\|f\|_{C^{\lambda}(K)}]$

where $c_{18}>0$ is independent of $n$. Then $v_{f}$ is a classical solution
of (4.1) verifying the identity

(4.12)
$\int_{\Omega}v_{f}\,\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=\int_{\Omega}\frac{f\xi}{x_{N}}\,d\gamma_{\beta},\quad\forall\,\xi\in
C^{1.1}_{0}(\Omega).$

Together with the fact that $u_{k,f}=k\Lambda^{\Omega}_{\beta}+v_{f}$, we
conclude that the function $u_{k,f}$ is a solution of (4.1), verifying the
identity (4.2) by (4.12).

Finally, we prove the uniqueness. In fact, let $w_{k,f}$ be a solution of
(4.1) verifying the identity (4.2).

$\int_{\Omega}(u_{k,f}-w_{k,f})\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=0.$

For any Borel subset $O$ of $\Omega$, Corollary 2.7 implies that problem

(4.13)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}^{*}(\frac{u}{x_{N}})=\zeta_{n}&{\rm
in}\;\;\Omega,\\\\[2.84526pt] \phantom{\mathcal{L}_{\mu}^{*}}\displaystyle
u=0&{\rm on}\;\;\partial{\Omega},\end{array}\right.$

has a solution $\eta_{\omega,n}\in C^{2}(\Omega)\cap C^{0.1}_{0}(\Omega)$,
where $\zeta_{n}:\bar{\Omega}\mapsto[0,1]$ is a $C^{1}(\bar{\Omega})$ function
such that $\zeta_{n}\to\chi_{O}\;{\rm{in}}\ L^{\infty}(\Omega)\;{\rm{as}}\
n\to\infty.$ Therefore by passing to the limit as $n\to\infty$, we have that

$\displaystyle\int_{O}(u_{k,f}-w_{k,f})d\gamma_{\beta}=0,$

which implies that $u_{k,f}=w_{k,f}$ a.e. in $\Omega$ and then the uniqueness
holds true. $\Box$

######  Remark 4.2.

Let $u_{f}$ be the solution of (4.1) verifying the identity (4.2) with $k=0$,
then $u_{f}$ satisfies the isolated singular behavior (1.14). In fact, letting
$f\geq 0$, then $u_{f}\geq 0$ in $\Omega$. So if (1.14) fails, it implies by
the positivity of $u_{f}$, that $\liminf_{t\to 0^{+}}\inf_{z\in
S^{N-1}_{+}}\frac{u_{f}(tz)}{\Lambda_{\beta}(tz)}=l_{0}>0$ and
$\tilde{u}_{f}:=u_{f}-l_{0}\Lambda_{\beta}^{\Omega}$ is a solution of (4.1).
By Lemma 2.2, we have that $\tilde{u}_{f}\geq 0$ in $\Omega$, By the
approximating procedure, $\tilde{u}_{f}$ verifies the identity (4.2) with
$k=0$, which is impossible with the fact that
$u_{f}-\tilde{u}_{f}=l_{0}\Lambda_{\beta}^{\Omega}$, which satisfies

$\int_{\Omega}(u_{f}-\tilde{u}_{f})\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=l_{0}c_{\beta}\frac{\partial\xi}{\partial
x_{N}}(0),\quad\forall\,\xi\in C^{1.1}_{0}(\Omega).$

###  4.2.  Nonzero Dirichlet boundary

Recall that $P_{\Omega}$ is Poisson’s Kernel of $-\Delta$ in $\Omega$ and
$\mathbb{P}_{\Omega}[g](x)=\displaystyle\int_{\partial\Omega}P_{\Omega}(x,y)g(y)d\omega(y).$
It is known that if $g$ is continuous, $\mathbb{P}_{\Omega}[g]$ is a solution
of

(4.14) $\left\\{\begin{array}[]{lll}\displaystyle-\Delta u=0\quad{\rm
in}\;\;\Omega,\\\\[2.84526pt] \phantom{-\Delta}\displaystyle u=g\quad{\rm
on}\;\;\partial{\Omega}.\end{array}\right.$

Multiply $\frac{\xi\lambda_{\beta}}{x_{N}}$ where $\xi\in C^{1.1}_{0}(\Omega)$
and integrate over $\Omega$, then we have that

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\int_{\Omega}(-\Delta\mathbb{P}_{\Omega}[g])\frac{\xi\lambda_{\beta}}{x_{N}}dx$
$\displaystyle=$
$\displaystyle\int_{\partial\Omega}\mathbb{P}_{\Omega}[g]\nabla(\frac{\xi\lambda_{\beta}}{x_{N}})\cdot\nu
d\omega+\int_{\Omega}\mathbb{P}_{\Omega}[g]\Big{(}-\Delta(\frac{\xi\lambda_{\beta}}{x_{N}})\Big{)}dx$
$\displaystyle=$
$\displaystyle\int_{\partial\Omega}g\frac{\partial\xi}{\partial\nu}d\omega_{\beta}+\int_{\Omega}\mathbb{P}_{\Omega}[g]\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}-\beta\int_{\Omega}\frac{\mathbb{P}_{\Omega}[g]}{|x|^{2}}\frac{\xi}{x_{N}}d\gamma_{\beta},$

that is, for any $\xi\in C^{1.1}_{0}(\Omega)$, there holds

(4.15)
$\int_{\Omega}\mathbb{P}_{\Omega}[g]\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=\beta\int_{\Omega}\frac{\mathbb{P}_{\Omega}[g]}{|x|^{2}}\frac{\xi}{x_{N}}d\gamma_{\beta}-\int_{\partial\Omega}g\frac{\partial\xi}{\partial\nu}d\omega_{\beta}.$

######  Lemma 4.3.

Let $\beta\in[\beta_{0},\,+\infty)\setminus\\{0\\}$,
$d\tilde{\omega}_{\beta}=(1+|x|^{\tau_{+}(\beta)})d\omega(x)$ and $g\geq 0$.
We have that

$(i)$ If $g\in C(\partial\Omega\setminus\\{0\\})\cap
L^{1}(\partial\Omega,\,d\tilde{\omega}_{\beta})$, then
$\frac{1}{|\cdot\;|^{2}}\mathbb{P}_{\Omega}[g]\in
L^{1}(\Omega,\,d\gamma_{\beta}).$

$(ii)$ If $g\in C(\partial\Omega\setminus\\{0\\})$ and

(4.16) $\lim_{r\to 0^{+}}\int_{\partial\Omega\setminus
B_{r}(0)}g\,d\tilde{\omega}_{\beta}=+\infty,$

then

$\lim_{r\to 0^{+}}\int_{\Omega\setminus
B_{r}(0)}\frac{1}{|x|^{2}}\mathbb{P}_{\Omega}[g](x)d\gamma_{\beta}=+\infty.$

Proof. From Proposition 2.1 in [2] that

(4.17) $c_{19}\rho(x)|x-y|^{-N}\leq P_{\Omega}(x,y)\leq
c_{20}\rho(x)|x-y|^{-N},\quad x\in\Omega,\ y\in\partial\Omega,$

where $\rho(x)={\rm dist}(x,\partial\Omega)$. Since $g$ is continuous in
$\partial\Omega\setminus\\{0\\}$ and $\Omega$ is flat near the origin, we can
only consider the integrability of
$\frac{1}{|\cdot\;|^{2}}\mathbb{P}_{\Omega}[g]$ near the origin. Fix
$r=r_{0}/2$, let
$B^{\prime}_{r}(0)=\\{x^{\prime}\in\mathbb{R}^{N-1}:|x^{\prime}|<r\\}$ and
$e_{(y^{\prime},0)}=(\frac{y^{\prime}}{|y^{\prime}|},\,0)$ for
$y^{\prime}\not=0$, then

$\displaystyle\int_{B_{r}^{+}(0)}\frac{1}{|x|^{2}}\mathbb{P}_{\Omega}[g]d\gamma_{\beta}$
$\displaystyle\geq$ $\displaystyle
c_{21}\int_{B_{r}^{+}(0)}\int_{B^{\prime}_{r}(0)\setminus\\{0\\}}g(y^{\prime})|x-(y^{\prime},0)|^{-N}\frac{x_{N}^{2}}{|x|^{2}}|x|^{\tau_{+}(\beta)}\,dy^{\prime}dx$
$\displaystyle=$ $\displaystyle
c_{22}\int_{B_{r}^{\prime}(0)\setminus\\{0\\}}g(y^{\prime})|y^{\prime}|^{\tau_{+}(\beta)}\int_{B_{r/|y^{\prime}|}^{+}(0)}|z-e_{(y^{\prime},0)}|^{-N}\frac{z_{N}^{2}}{|z|^{2}}|z|^{\tau_{+}(\beta)}dzdy^{\prime}$

and

$\displaystyle\int_{B_{r}^{+}(0)}\frac{1}{|x|^{2}}\mathbb{P}_{\Omega}[g]d\gamma_{\beta}$
$\displaystyle\leq$ $\displaystyle
c_{23}\int_{B_{r}^{+}(0)}\int_{B^{\prime}_{r}(0)\setminus\\{0\\}}g(y^{\prime})|x-(y^{\prime},0)|^{-N}\frac{x_{N}^{2}}{|x|^{2}}|x|^{\tau_{+}(\beta)}\,dy^{\prime}dx$
$\displaystyle=$ $\displaystyle
c_{24}\int_{B_{r}^{\prime}(0)\setminus\\{0\\}}g(y^{\prime})|y^{\prime}|^{\tau_{+}(\beta)}\int_{B_{r/|y^{\prime}|}^{+}(0)}|z-e_{(y^{\prime},0)}|^{-N}\frac{z_{N}^{2}}{|z|^{2}}|z|^{\tau_{+}(\beta)}dzdy^{\prime}.$

Now we do estimates for

$\int_{B_{r/|y^{\prime}|}^{+}(0)}I(z)dz:=\int_{B_{r/|y^{\prime}|}^{+}(0)}|z-e_{(y^{\prime},0)}|^{-N}\frac{z_{N}^{2}}{|z|^{2}}|z|^{\tau_{+}(\beta)}dz,$

we have

$\displaystyle 0<\int_{B_{\frac{1}{2}}^{+}(0)}I(z)\,dz\leq
2^{N}\int_{B_{\frac{1}{2}}^{+}(0)}|z|^{\tau_{+}(\beta)}dz,$ $\displaystyle
0<\int_{B_{\frac{1}{2}}^{+}(e_{(y^{\prime},0)})}I(z)\,dz$ $\displaystyle\leq$
$\displaystyle
2^{|\tau_{+}(\beta)|+2}\int_{B_{\frac{1}{2}}^{+}(e_{(y^{\prime},0)})}|z-e_{(y^{\prime},0)}|^{-N}z_{N}^{2}dz$
$\displaystyle\leq$ $\displaystyle
2^{|\tau_{+}(\beta)|+2}\int_{B_{\frac{1}{2}}^{+}(0)}|z|^{2-N}dz,$

and

$\displaystyle\int_{B_{r/|y^{\prime}|}^{+}(0)\setminus\left(B_{\frac{1}{2}}^{+}(0)\cup
B_{\frac{1}{2}}^{+}(e_{(y^{\prime},0)})\right)}I(z)dz$ $\displaystyle\leq$
$\displaystyle c_{25}\int_{B_{r/|y^{\prime}|}^{+}(0)\setminus
B_{\frac{1}{2}}^{+}(0)}|z|^{-N+\tau_{+}(\beta)}\,dz$ $\displaystyle\leq$
$\displaystyle\left\\{\begin{array}[]{lll}\displaystyle
c_{26}\int_{\mathbb{R}^{N}\setminus
B_{\frac{1}{2}}(0)}|z|^{-N+\tau_{+}(\beta)}\,dz&{\rm
if}\quad\beta<0,\\\\[5.69054pt] \phantom{}\displaystyle
c_{26}|y^{\prime}|^{-\tau_{+}(\beta)}&{\rm if}\quad\beta>0\end{array}\right.$
$\displaystyle\leq$ $\displaystyle c_{27}(1+|y^{\prime}|^{-\tau_{+}(\beta)})$

and

$\displaystyle\int_{B_{r/|y^{\prime}|}^{+}(0)\setminus\left(B_{\frac{1}{2}}^{+}(0)\cup
B_{\frac{1}{2}}^{+}(e_{(y^{\prime},0)})\right)}I(z)\,dz$ $\displaystyle\geq$
$\displaystyle c_{28}\int_{B_{r/|y^{\prime}|}^{+}(0)\setminus
B_{\frac{1}{2}}^{+}(0)}|z|^{-N+\tau_{+}(\beta)}\,dz$ $\displaystyle\geq$
$\displaystyle c_{29}(1+|y^{\prime}|^{-\tau_{+}(\beta)}).$

Thus, we have that

(4.19)
$c_{30}\int_{B_{r}^{\prime}(0)\setminus\\{0\\}}g(y^{\prime})d\tilde{\omega}(y^{\prime})\leq\int_{B_{r}^{+}(0)}\frac{1}{|x|^{2}}\mathbb{P}_{\Omega}[g]d\gamma_{\beta}\leq
c_{31}\int_{B_{r}^{\prime}(0)\setminus\\{0\\}}g(y^{\prime})d\tilde{\omega}(y^{\prime}),$

which, together with the fact that $\mathbb{P}_{\Omega}[g]$ is nonnegative and
bounded in $\Omega\setminus B_{r}^{+}(0)$, proves Lemma 4.3. $\Box$

We remark that Lemma 4.3 provides estimates for transforming the boundary data
into the nonhomogeneous term. Now we are ready to prove Theorem 1.2 part $(i)$
where we distinguish two cases $\beta\in[\beta_{0},\,0]$ and $\beta>0$.

Proof of Theorem 1.2. Part $(i)$. The existence for $g\in
L^{1}(\partial\Omega,\,d\tilde{\omega}_{\beta})$.  Let
$\bar{f}=f-\frac{\beta}{|\cdot\;|^{2}}\mathbb{P}_{\Omega}[g].$ Then it follows
from Lemma 4.3 part $(i)$ that $\bar{f}\in L^{1}(\Omega,\,d\gamma_{\beta})$
and applying Theorem 4.1 part $(i)$, problem (4.1) verifying (4.2) for
$k\in\mathbb{R}$ and replaced $f$ by $\bar{f}$ admits a unique solution of
$u_{f}$. Denote $u_{f,g}:=u_{f}+\mathbb{P}_{\Omega}[g]$, then

$\mathcal{L}_{\beta}u_{f,g}=f\quad{\rm and}\quad u_{f,g}=g\quad{\rm on}\ \
\partial\Omega\setminus\\{0\\}.$

Together with (4.2) and (4.15), we have that $u_{f,g}$ verifies (1.11) and it
is the unique solution of problem (4.1) verifying (4.2) for that $k$.

Case of $\beta\in[\beta_{0},\,0]$. Then $d\tilde{\omega}_{\beta}$ is
equivalent to $d\omega_{\beta}$, so
$L^{1}(\partial\Omega,\,d\tilde{\omega}_{\beta})=L^{1}(\partial\Omega,\,d\omega_{\beta})$
and we are done.

Case of $\beta>0$.  We note that

$L^{1}(\partial\Omega,\,d\tilde{\omega}_{\beta})\subsetneqq
L^{1}(\partial\Omega,\,d\omega_{\beta}).$

So for $g\in L^{1}(\partial\Omega,\,d\omega_{\beta})\setminus
L^{1}(\partial\Omega,\,d\tilde{\omega}_{\beta})$, we may assume $g\geq 0$ by
linearity of $\mathcal{L}_{\beta}$. Let

(4.20) $\eta_{n}(s)=1-\eta_{0}(ns)\quad{\rm and}\quad
g_{n}(x)=g(x)\eta_{n}(|x|),$

where $\eta_{0}$ is defined in (2.1). Then $\\{g_{n}\\}_{n}\subset
L^{1}(\partial\Omega,\,d\tilde{\omega}_{\beta})$ is an increasing sequence of
functions. For simplicity, we assume that $f=0$. Then the problem

(4.21) $\left\\{\begin{array}[]{lll}\mathcal{L}_{\beta}^{*}u=0&{\rm
in}\;\;\Omega,\\\\[2.84526pt] \phantom{\mathcal{L}_{\beta}}u=g_{n}&{\rm
on}\;\;\partial\Omega\setminus\\{0\\}\end{array}\right.$

has a unique solution of $u_{n}$ verifying the identify

(4.22)
$\int_{\Omega}u_{n}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=-\int_{\partial\Omega}g_{n}\frac{\partial\xi}{\partial\nu}d\omega_{\beta},\quad\forall\,\xi\in
C^{1.1}_{0}(\Omega).$

Since $0\leq g_{n}\leq g$ and $g\in L^{1}(\partial\Omega,\,d\omega_{\beta})$,
we may expand the text function space including $w_{1}$, $w_{2}$, which are
the solutions of (2.13) and (2.14) respectively. Taking $\xi=w_{1}$ and then
$w_{2}$ , we derive that

$\|u_{n}\|_{L^{1}(\Omega)}\leq
c_{32}\|g_{n}\|_{L^{1}(\partial\Omega,\,d\omega_{\beta})}\leq
c_{33}\|g\|_{L^{1}(\partial\Omega,\,d\omega_{\beta})}$

and

$\|u_{n}\|_{L^{1}(\Omega,\,x_{N}^{-1}d\gamma_{\beta})}\leq
c_{34}\|g\|_{L^{1}(\partial\Omega,\,d\omega_{\beta})}.$

We notice that $u_{n}\geq 0$ and the mapping $n\mapsto u_{n}$ is increasing,
then by the monotone converge theorem, we have that there exists $u$ such that
$u_{n}$ converging to $u$ in $L^{1}(\Omega,\frac{1}{x_{N}}d\gamma_{\beta})$.
Since $\xi\in C_{0}^{1.1}(\Omega)$, we have that
$|\mathcal{L}_{\beta}^{*}(\xi/x_{N})|\leq cx_{N}^{-1}.$ Pass to the limit of
(4.22), we have that $u$ verifies that

(4.23)
$\int_{\Omega}u\mathcal{L}_{\beta}^{*}(\xi/x_{N})\,d\gamma_{\beta}=-\int_{\partial\Omega}g\frac{\partial\xi}{\partial\nu}d\omega_{\beta},\quad\forall\,\xi\in
C^{1.1}_{0}(\Omega).$

From standard interior regularity, we have that $u$ is a classical solution

$\left\\{\begin{array}[]{lll}\mathcal{L}_{\beta}^{*}u=0\quad{\rm
in}\;\;\Omega,\\\\[2.84526pt] \phantom{\mathcal{L}_{\beta}}u=g\quad{\rm
on}\;\;\partial\Omega\setminus\\{0\\},\end{array}\right.$

which ends the proof. $\Box$

##  5\. Nonexistence

In this subsection, we establish the approximation of the fundamental solution
$G_{\mu}$.

######  Lemma 5.1.

$(i)$ Let $\\{\delta_{n}\\}_{n}$ be a sequence of nonnegative
$L^{\infty}$-functions defined in $\Omega$ such that ${\rm
supp}\,\delta_{n}\subset B_{r_{n}}(0)\cap\Omega,$ where $r_{n}\to 0$ as
$n\to+\infty$ and

$\int_{\Omega}\delta_{n}\xi dx\to\frac{\partial\xi(0)}{\partial
x_{N}}\quad{\rm as}\quad n\to+\infty,\quad\forall\xi\in C_{0}^{1}(\Omega).$

For any $n$, let $w_{n}$ be the unique solution of the problem in the
$d\gamma_{\beta}$-distributional sense

(5.1)
$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}u=\delta_{n}/\lambda_{\beta}&{\rm
in}\;\;{\Omega}\setminus\\{0\\},\\\\[5.69054pt]
\phantom{L_{\beta}}\displaystyle u=0&{\rm
on}\;\;\partial{\Omega},\\\\[5.69054pt] \phantom{}\displaystyle\lim_{r\to
0^{+}}\sup_{x\in\partial_{+}B_{r}(0)}\frac{|u(x)|}{\Lambda_{\beta}(x)}=0.\end{array}\right.$

Then

$\lim_{n\to+\infty}w_{n}(x)=\frac{1}{c_{\beta}}\Lambda_{\beta}^{\Omega}(x),\quad\forall\,x\in\Omega\setminus\\{0\\}$

and for any compact set $K\subset\Omega\setminus\\{0\\}$,

(5.2) $w_{n}\to\frac{1}{c_{\beta}}\Lambda_{\beta}^{\Omega}\quad{\rm as}\quad
n\to+\infty\ \ {\rm in}\ \ C^{2}(K).$

$(ii)$ Let $\\{\sigma_{n}\\}_{n}$ be a sequence of nonnegative $L^{\infty}$
functions defined on $\partial\Omega$ such that ${\rm
supp}\,\sigma_{n}\subset\partial\Omega\cap B_{r_{n}}(0),$ where $r_{n}\to 0$
as $n\to+\infty$ and

$\int_{\partial\Omega}\sigma_{n}\zeta d\omega(x)\to\zeta(0)\quad{\rm as}\quad
n\to+\infty,\quad\forall\zeta\in C^{1}(\partial\Omega).$

For any $n$, let $v_{n}$ be the unique solution of the problem

(5.3) $\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\beta}u=0&{\rm
in}\;\;{\Omega}\setminus\\{0\\},\\\\[5.69054pt] \phantom{}\displaystyle
u=\frac{\sigma_{n}}{|\cdot|^{\tau_{+}(\beta)}}&{\rm
on}\;\;\partial{\Omega}\setminus\\{0\\}\end{array}\right.$

subject to

$\int_{\Omega}v_{n}\mathcal{L}_{\beta}^{*}(\xi/x_{N})\,d\gamma_{\beta}=-\int_{\partial\Omega}\sigma_{n}\frac{\partial\xi}{\partial\nu}d\omega,\quad\forall\,\xi\in
C^{1.1}_{0}(\Omega).$

Then

$\lim_{n\to+\infty}v_{n}(x)=\frac{1}{c_{\beta}}\Lambda_{\beta}^{\Omega}(x),\quad\forall\,x\in\Omega\setminus\\{0\\}$

and for any compact set $K\subset\Omega\setminus\\{0\\}$, (5.2) holds true.

Proof. From Lemma 2.3, problems (5.1) and (5.3) have unique solutions
$w_{n},v_{n}\geq 0$ respectively and satisfying that

(5.4)
$\int_{\Omega}w_{n}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=\int_{\Omega}\delta_{n}\xi\,dx,\quad\forall\,\xi\in
C^{1,1}_{0}(\Omega)$

and

(5.5)
$\int_{\Omega}v_{n}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=-\int_{\partial\Omega}\sigma_{n}\frac{\partial\xi}{\partial\nu}d\omega_{\beta},\quad\forall\,\xi\in
C^{1,1}_{0}(\Omega).$

By taking $\xi=\xi_{0}$, the solution of (2.13), we obtain that

$\|w_{n}\|_{L^{1}(\Omega,\,d\gamma_{\beta})}\leq\|\xi_{0}\|_{L^{\infty}(\Omega)}\|\delta_{n}\|_{L^{1}(\Omega)}=\|\xi_{0}\|_{L^{\infty}(\Omega)}.$

For any $r>0$, take $\xi$ with the support in $\Omega\setminus B_{r}(0)$, then
$\xi\in C^{1.1}_{c}(\overline{\Omega\setminus B_{r}(0)})$,

$\int_{\Omega\setminus
B_{r}(0)}w_{n}\mathcal{L}_{\mu}^{*}(\xi)\,d\gamma_{\beta}=0.$

Take $\xi$ the solution of (2.17) with $f(x)=\frac{1}{|x|}$, we have that

(5.6)
$\int_{\Omega}w_{n}|x|^{-1}\,d\gamma_{\beta}=\int_{\Omega}\delta_{n}\xi\,dx\leq\|\xi_{0}\|_{L^{\infty}(\Omega)}$

and

(5.7)
$\int_{\Omega}v_{n}|x|^{-1}\,d\gamma_{\beta}=-\int_{\partial\Omega}\sigma_{n}\frac{\partial\xi}{\partial\nu}d\omega_{\beta}\leq\|\nabla\xi_{0}\|_{L^{\infty}(\Omega)}.$

So $w_{n},v_{n}$ are uniform bounded in
$L^{1}(\Omega,\,|x|^{-1}d\gamma_{\beta})$.

From Corollary 2.8 in [31] with $L^{*}=\mathcal{L}_{\mu}^{*}$, which is
strictly elliptic in $\Omega\setminus B_{r}(0)$, we have that for
$q<\frac{N}{N-1}$,

$\displaystyle\|w_{n}\lambda_{\beta}\|_{W^{1,q}(\Omega_{2r})}\leq
c_{35}\|\delta_{n}\|_{L^{1}(\Omega\setminus
B_{r}(0))}+c_{36}\|w_{n}\|_{L^{1}(\Omega\setminus
B_{r}(0),\,d\gamma_{\beta})}\leq c_{37}$

and

$\displaystyle\|v_{n}\lambda_{\beta}\|_{W^{1,q}(\Omega_{2r})}$
$\displaystyle\leq$ $\displaystyle
c_{38}\|\sigma_{n}\|_{L^{1}(\partial\Omega\setminus
B_{r}(0))}+c_{39}\|v_{n}\|_{L^{1}(\Omega\setminus
B_{r}(0),\,d\gamma_{\beta})}\leq c_{40},$

where $\Omega_{2r}=\\{x\in\Omega\setminus B_{2r}(0):\,\rho(x)>2r\\}.$ By the
compact embedding $W^{1,q}(\Omega_{2r})\hookrightarrow L^{1}(\Omega_{2r}),$ up
to some subsequence, there exists $w_{\infty},\,v_{\infty}\in
W^{1,q}_{loc}(\Omega)\cap L^{1}(\Omega,\,d\gamma_{\beta})$ such that

$w_{n}\to w_{\infty}\quad{\rm as}\quad n\to+\infty\quad{\rm a.e.\ \ in}\ \
\Omega\ \ {\rm and\ in}\quad L^{1}(\Omega,\,d\gamma_{\beta})$

and it follows by (5.4) and (5.5) that for $\xi\in C^{1.1}_{0}(\Omega)$,

$\int_{\Omega}w_{\infty}\mathcal{L}_{\beta}^{*}(\xi)\,d\gamma_{\beta}=\int_{\Omega}v_{\infty}\mathcal{L}_{\beta}^{*}(\xi)\,d\gamma_{\beta}=\frac{\partial\xi}{\partial
x_{N}}(0).$

Furthermore,

$\int_{\Omega}(w_{\infty}-\frac{1}{c_{\beta}}G_{\beta})\mathcal{L}_{\beta}^{*}(\xi)\,d\gamma_{\beta}=0.$

From the Kato’s inequality, we deduce that

$w_{\infty}=v_{\infty}=\frac{1}{c_{\beta}}\Lambda_{\beta}^{\Omega}\quad{a.e.}\;\;\Omega.$

Proof of (5.2). For any $x_{0}\in\Omega\setminus\\{0\\}$, let
$r_{0}=\frac{1}{4}\\{|x_{0}|,\,\rho(x_{0})\\}$ and $\mu_{n}=w_{n}\eta,$ where
$\eta(x)=\eta_{0}(\frac{|x-x_{0}|}{r_{0}})$. There exists $n_{0}>0$ such that
for $n\geq n_{0}$, ${\rm supp}\mu_{n}\cap B_{r_{n}}(0)=\emptyset.$ Then

$\displaystyle-\Delta\mu_{n}(x)$ $\displaystyle=$ $\displaystyle-\Delta
w_{n}(x)\eta(x)-2\nabla w_{n}\cdot\nabla\eta-w_{n}\Delta\eta$ $\displaystyle=$
$\displaystyle-2\nabla w_{n}\cdot\nabla\eta-w_{n}\Delta\eta,$

where $\nabla\eta$ and $\Delta\eta$ are smooth.

We observe that $w_{n}\in W^{1,q}(B_{2r_{0}}(x_{0}))$ and $-2\nabla
w_{n}\cdot\nabla\eta-w_{n}\Delta\eta\in L^{q}(B_{2r_{0}}(x_{0})),$ then we
have that

$\|\mu_{n}\|_{W^{2,q}(B_{r_{0}}(x_{0}))}\leq
c\|w_{n}\|_{L^{1}(\Omega,\,d\gamma_{\beta})},$

where $c>0$ is independent of $n$. Thus, $-2\nabla w_{n}\cdot\nabla\eta-
w_{n}\Delta\eta\in W^{1,q}(B_{r_{0}}(x_{0})),$ repeat above process $N_{0}$
steps, for $N_{0}$ large enough, we deduce that

$\|w_{n}\|_{C^{2,\gamma}(B_{\frac{r_{0}}{2^{N_{0}}}}(x_{0}))}\leq
c\|w_{n}\|_{L^{1}(\Omega,\,d\gamma_{\beta})},$

where $\gamma\in(0,1)$ and $c>0$ is independent of $n$. As a conclusion, (5.2)
follows by Arzelà-Ascola theorem and Heine-Borel theorem. The above process
also holds for $v_{n}$. This ends the proof. $\Box$

Proof of Theorem 1.2. Part $(ii)$. From (1.12), one of the following two cases
holds true,

${\rm case}\ 1:\lim_{r\to 0^{+}}\int_{\Omega\setminus
B_{r}(0)}f\,d\gamma_{\beta}=+\infty,\;\;{\rm or\ case}\ 2:\;\;\lim_{r\to
0^{+}}\int_{\partial\Omega\setminus B_{r}(0)}g\,d\omega_{\beta}=+\infty.$

Case 1. We argue by contradiction. Assume that problem (1.2) has a nonnegative
solution of $u_{f}$. Let $\\{r_{n}\\}_{n}$ be a sequence of strictly
decreasing positive numbers converging to $0$. From the fact $f\in
C_{loc}^{\gamma}(\overline{\Omega}\setminus\\{0\\})$, for any $r_{n}$ fixed,
we have that

$\lim_{r\to 0^{+}}\int_{(B_{r_{n}}(0)\setminus
B_{r}(0))\cap\Omega}f(x)d\gamma_{\beta}=+\infty,$

then there exists $R_{n}\in(0,r_{n})$ such that

$\int_{(B_{r_{n}}(0)\setminus B_{R_{n}}(0))\cap\Omega}fd\gamma_{\beta}=n.$

Let $\delta_{n}=\frac{1}{n}\lambda_{\beta}f\chi_{B_{r_{n}}(0)\setminus
B_{R_{n}}(0)}$, then the problem

$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\mu}u\cdot\lambda_{\beta}=\delta_{n}\qquad{\rm
in}\quad{\Omega}\setminus\\{0\\},\\\\[2.84526pt]
\phantom{L_{\mu}--}\displaystyle u=0\qquad{\rm
on}\quad\partial{\Omega},\\\\[2.84526pt] \phantom{}\displaystyle\lim_{x\to
0}u(x)\Phi_{\mu}^{-1}(x)=0\end{array}\right.$

has a unique positive solution $w_{n}$ satisfying (in the usual sense)

$\int_{\Omega}w_{n}\mathcal{L}_{\mu}(\lambda_{\beta}\xi)dx=\int_{\Omega}\delta_{n}\xi
dx,\quad\forall\,\xi\in C^{1.1}_{0}(\Omega).$

For any $\xi\in C^{1.1}_{0}(\Omega)$, we have that

$\int_{\Omega}w_{n}\mathcal{L}_{\mu}^{*}(\xi)\,d\gamma_{\beta}=\int_{\Omega}\delta_{n}\xi\,dx\to\frac{\partial\xi}{\partial
x_{N}}(0)\quad{\rm as}\quad n\to+\infty.$

Therefore, by Lemma 5.1 for any compact set
$\mathcal{K}\subset\Omega\setminus\\{0\\}$

$\|w_{n}-\Lambda_{\beta}^{\Omega}\|_{C^{1}(\mathcal{K})}\to 0\quad{\rm
as}\quad{n\to+\infty}.$

We fix a point $x_{0}\in\Omega$ and let
$r_{0}=\frac{1}{2}\min\\{|x_{0}|,\,\rho(x_{0})\\}$ and
$\mathcal{K}=\overline{B_{r_{0}}(x_{0})}$, then there exists $n_{0}>0$ such
that for $n\geq n_{0}$,

(5.8) $w_{n}\geq\frac{1}{2}G_{\mu}\quad{\rm in}\quad\mathcal{K}.$

Let $u_{n}$ be the solution (in the usual sense) of

$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\mu}u\cdot\lambda_{\beta}=n\delta_{n}\quad{\rm
in}\;\;{\Omega}\setminus\\{0\\},\\\\[2.84526pt]
\phantom{L_{\mu}--}\displaystyle u=0\quad\ \ {\rm
on}\;\;\partial{\Omega},\\\\[2.84526pt] \phantom{}\displaystyle\lim_{r\to
0^{+}}\sup_{x\in\partial_{+}B_{r}(0)}\frac{|u(x)|}{\Lambda_{\beta}(x)}=0,\end{array}\right.$

then we have that $u_{n}\geq nw_{n}\;\;{\rm in}\;\;\Omega.$ Together with
(5.8), we derive that

$u_{n}\geq\frac{n}{2}\Lambda_{\mu}^{\Omega}\quad{\rm in}\;\;\mathcal{K}.$

Then by comparison principle, we have that $u_{f}(x_{0})\geq
u_{n}(x_{0})\to+\infty\quad{\rm as}\;\;n\to+\infty,$ which contradicts to the
fact that $u_{f}$ is classical solution of (4.1).

Case 2. Similarly for any $n\in\mathbb{N}$, we can take $r_{n}>R_{n}>0$ such
that $r_{n}\to 0$ as $n\to+\infty$ and

$\int_{(B_{r_{n}}(0)\setminus
B_{R_{n}}(0))\cap\partial\Omega}gd\omega_{\beta}=n.$

Let $\sigma_{n}=\frac{1}{n}g\chi_{B_{r_{n}}(0)\setminus B_{R_{n}}(0)}$,
$w_{n}$ be the solution of

$\left\\{\begin{array}[]{lll}\displaystyle\mathcal{L}_{\mu}u=0\quad{\rm
in}\;\;{\Omega}\setminus\\{0\\},\\\\[2.84526pt] \phantom{L_{\mu}}\displaystyle
u=\sigma_{n}/|\cdot|^{\tau^{+}(\beta)}\quad\ \ {\rm
on}\;\;\partial{\Omega},\end{array}\right.$

subject to

$\int_{\Omega}w_{n}\mathcal{L}_{\beta}^{*}(\frac{\xi}{x_{N}})\,d\gamma_{\beta}=-\int_{\partial\Omega}\sigma_{n}\frac{\partial\xi}{\partial\nu}d\omega,\quad\forall\,\xi\in
C^{1.1}_{0}(\Omega).$

Repeat the procedure in Case 1, we get a contradiction which completes the
proof. $\Box$

Acknowledgements: H. Chen is supported by the Natural Science Foundation of
China [Nos. 11661045, 11726614]. A. Quaas is partially supported by Fondecyt
Grant No. 1151180, Programa Basal, CMM in U. de Chile and Millennium Nucleus
Center for Analysis of PDE NC130017. F. Zhou is partially supported by NSFC
[Nos. 11726613, 11431005]; and STCSM [No. 18dz2271000].

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