# Embeddings of weighted Sobolev spaces and a weighted fourth order elliptic
equation

Zongming Guo  Department of Mathematics, Henan Normal University, Xinxiang,
453007, P.R. China<EMAIL_ADDRESS>,  Fangshu Wan  Department of Mathematics,
Henan Normal University, Xinxiang, 453007, P.R. China<EMAIL_ADDRESS>and
Liping Wang  Department of Mathematics, Shanghai Key Laboratory of Pure
Mathematics and Mathematical Practice, East China Normal University, 200241,
P.R. China<EMAIL_ADDRESS>
###### Abstract.

New embeddings of weighted Sobolev spaces are established. Using such
embeddings, we obtain the existence and regularity of positive solutions with
Navier boundary value problems for a weighted fourth order elliptic equation.
We also obtain Liouville type results for the related equation. Some problems
are still open.

*: corresponding author.
The research of the first author is supported by NSFC 11571093 and the third
author is supported by NSFC 11671144.

Keywords: positive solutions, fourth order with weights, Liouville theorem

Mathematics Subject Classification 2000: 35B45; 35J40

##  1\. Introduction

We study structure of nonnegative solutions of the weighted fourth order
elliptic equation

$None$ $\Delta(|x|^{\alpha}\Delta u)=|x|^{l}u^{p}\;\;\mbox{in
$\mathbb{R}^{N}$},$

where $N\geq 5$, $p>1$, $\alpha,l\in\mathbb{R}$.

For a bounded smooth domain $\Omega\subset\mathbb{R}^{N}$ and $0\in\Omega$, we
also study the Navier boundary value problem

$None$ $\left\\{\begin{array}[]{ll}\Delta(|x|^{\alpha}\Delta
u)=|x|^{l}u^{p}&\mbox{in $\Omega$},\\\ u=\Delta u=0&\mbox{on
$\partial\Omega$}.\end{array}\right.$

By a positive solution of (P), we mean that $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$,
$|x|^{\alpha}\Delta u\in C^{0}(\mathbb{R}^{N})$, $u>0$ in $\mathbb{R}^{N}$ and
$u$ satisfies (P) in $\mathbb{R}^{N}\backslash\\{0\\}$. By a positive solution
of (Q), we mean that $u\in C^{4}(\Omega\backslash\\{0\\})\cap
C^{0}({\overline{\Omega}})$, $|x|^{\alpha}\Delta u\in
C^{0}({\overline{\Omega}})$, $u>0$ in $\Omega$ and $u$ satisfies (Q) in
$\Omega\backslash\\{0\\}$.

Throughout this paper, we assume

(1.1) $2N>N^{\prime}:=N+\alpha>4,\;\;\;\tau:=l-\alpha>-4,$ (1.2)
$p_{s}:=\frac{N^{\prime}+4+2\tau}{N^{\prime}-4}\;(>1).$

It is known from (1.1) that $\alpha\in(4-N,N)$.

Equation (P) and problem (Q) have been studied by many authors recently, in
particular, in the case of the pure biharmonic operator, see, for example, [2,
3, 5, 6, 8, 9, 10, 11, 13, 16, 18, 19, 20, 21, 23, 25] and the references
therein.

Under the assumptions in (1.1), the following Liouville theorem for (P) is
still open.

Open problem 1: Assume that $N\geq 5$ and (1.1) holds, $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$ and
$|x|^{\alpha}\Delta u\in C^{0}(\mathbb{R}^{N})$ is a nonnegative solution of
(P). Then $u\equiv 0$ in $\mathbb{R}^{N}$ provided $1<p<p_{s}$.

The positive answer to this open problem for $\alpha=l=0$ was given in [23]
via the moving plane argument. But it does not apply for our case here
(especially for $l>0$), since the weights do not match with the procedure of
this argument. In this paper, we will give a partial answer to this open
problem. More precisely, when $\alpha\in(4-N,N)$, we obtain the positive
answer if $1<p<\min\\{\frac{N+4}{N-4},p_{s}\\}$. Especially, when $\alpha=2$
we confirm the result for $1<p<p_{s}$. Namely, we have the following theorems.

######  Theorem 1.1.

Assume $N\geq 5$ and (1.1) holds. If $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$ and
$|x|^{\alpha}\Delta u\in C^{0}(\mathbb{R}^{N})$ is a nonnegative solution to
(P). Then $u\equiv 0$ in $\mathbb{R}^{N}$ provided
$1<p<\min\\{\frac{N+4}{N-4},p_{s}\\}$.

Note that $p_{s}\leq\frac{N+4}{N-4}$ is equal to
$\alpha\geq\frac{(N-4)}{4}\tau$ and $p_{s}>\frac{N+4}{N-4}$ provided
$\alpha<\frac{(N-4)}{4}\tau$. Hence Theorem 1.1 implies that if
$\alpha\geq\frac{(N-4)}{4}\tau$, the answer is positive for $1<p<p_{s}$ while
$\alpha<\frac{(N-4)}{4}\tau$, it only gives out the result for
$1<p<\frac{N+4}{N-4}$. So there is a gap $\frac{N+4}{N-4}\leq p<p_{s}$
provided $\alpha<\frac{(N-4)}{4}\tau$. But for the symmetric solutions or
$\alpha=2$, we can still get the result in Theorem 1.1.

######  Theorem 1.2.

Assume $N\geq 5$ and (1.1) holds. Assume also that $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$ with
$|x|^{\alpha}\Delta u\in C^{0}(\mathbb{R}^{N})$ is a nonnegative radial
solution to (P). Then $u\equiv 0$ in $\mathbb{R}^{N}$ provided $1<p<p_{s}$.

######  Theorem 1.3.

Assume that $N\geq 5$, $\alpha=2$ and $l>-2$. Assume also that $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$ and
$|x|^{2}\Delta u\in C^{0}(\mathbb{R}^{N})$ is a nonnegative solution to (P).
Then $u\equiv 0$ in $\mathbb{R}^{N}$ provided $1<p<p_{s}=\frac{N+2+2l}{N-2}$.

######  Remark 1.4.

There is no first order term in (3.24) if $\alpha=2$ and we can get the
monotonicity in the radial direction.

For a bounded domain $\Omega\subset\mathbb{R}^{N}$ of class $C^{2}$ with
$0\in\Omega$, we define $D^{1,\alpha}_{0}(\Omega)$ as the completion of
$C_{c}^{\infty}(\Omega\backslash\\{0\\})$ with the norm

$\|\phi\|_{1,\alpha}^{2}=\int_{\Omega}|x|^{\alpha}|\nabla\phi|^{2}dx$

and $D^{2,\alpha}_{0}(\Omega)$ as the completion of
$C_{c}^{\infty}(\Omega\backslash\\{0\\})$ with the norm

$\|\phi\|_{2,\alpha}^{2}=\int_{\Omega}|x|^{\alpha}|\Delta\phi|^{2}dx.$

Also we define $L_{l}^{q}(\Omega)\;(q\geq 1)$ the space of functions $\phi$
such that

$|x|^{\frac{l}{q}}|\phi|\in L^{q}(\Omega)$

with the norm

$\|\phi\|_{L_{l}^{q}(\Omega)}=\Big{(}\int_{\Omega}|x|^{l}|\phi|^{q}dx\Big{)}^{\frac{1}{q}}.$

We now establish the following embeddings:

######  Proposition 1.5.

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded domain of class $C^{2}$, but
not a ball, with $0\in\Omega$. Assume also that $N\geq 5$ and (1.1) holds with
$\alpha\geq\frac{(N-4)}{4}\tau$. Then, the embedding:

(1.3) $D^{2,\alpha}_{0}(\Omega)\hookrightarrow L_{l}^{q}(\Omega)$

is continuous for any $q\in[1,p_{s}+1]$ and this embedding is compact for any
$q\in[1,p_{s}+1)$.

Note that $p_{s}\leq\frac{N+4}{N-4}$ provided $\alpha\geq\frac{(N-4)}{4}\tau$.
The related embeddings from $D^{1,\alpha}_{0}(\Omega)$ to $L_{l}^{q}(\Omega)$
was obtained in [17]. Such embeddings may have been known already, but we can
not find a suitable reference.

######  Proposition 1.6.

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded domain of class $C^{2}$, but
not a ball, with $0\in\Omega$. Assume also that $N\geq 5$ and (1.1) holds with
$\alpha<\frac{(N-4)}{4}\tau$. Then, the embedding:

(1.4) $D^{2,\alpha}_{0}(\Omega)\hookrightarrow L_{l}^{q}(\Omega)$

is continuous for any $q\in[1,\frac{2N}{N-4}]$ and this embedding is compact
for any $q\in[1,\frac{2N}{N-4})$.

Note that $p_{s}>\frac{N+4}{N-4}$ provided $\alpha<\frac{(N-4)}{4}\tau$. So
the above two propositions establish the embedding theorem for
$1<p<\min\\{\frac{N+4}{N-4},p_{s}\\}$. Actually, for radial functions in the
ball, we can achieve better embedding theorem.

######  Proposition 1.7.

Let $B$ be a ball centered at 0. Assume also that $N\geq 5$ and (1.1) holds.
Then, the embedding:

$D^{2,\alpha}_{0,\mbox{{\tiny rad}}}(B)\hookrightarrow L_{l}^{q}(B)$

is continuous for any $q\in[1,p_{s}+1]$ and this embedding is compact for any
$q\in[1,p_{s}+1)$, where $D^{2,\alpha}_{0,\mbox{{\tiny rad}}}(B)=\\{\phi\in
D_{0}^{2,\alpha}(B):\;\phi(x)=\phi(|x|)\\}$.

Using the embeddings given in Propositions 1.5-1.7, we can obtain positive
solutions $u$ of (Q) or positive radial solutions $u$ of (Q) via variational
methods for $1<p<p_{s}$ or $1<p<\frac{N+4}{N-4}$. The existence of positive
radial solutions of the related weighted elliptic equations for second order
was obtained in [26]. The boot-strap argument implies that the obtained
solutions $u$ of (Q) satisfy $u\in C^{4}(\Omega\backslash\\{0\\})$ or $u\in
C^{4}(B\backslash\\{0\\})$. We will see from the theorems below that the
singular point 0 of $u$ is removable, i.e., $u\in
C^{4}(\Omega\backslash\\{0\\})\cap C^{0}({\overline{\Omega}})$ and
$|x|^{\alpha}\Delta u\in C^{0}({\overline{\Omega}})$ or $u\in
C^{4}(B\backslash\\{0\\})\cap C^{0}({\overline{B}})$ and $|x|^{\alpha}\Delta
u\in C^{0}({\overline{B}})$.

We know that $p_{s}>\frac{N+4}{N-4}$ provided $\alpha<\frac{N-4}{4}\tau$. The
embedding in Proposition 1.6 implies that we can only obtain the existence and
regularity of positive solutions of (Q) for $1<p<\frac{N+4}{N-4}$. The
following problem is still open.

Open problem 2. Assume that $\Omega$ is a bounded smooth domain in
$\mathbb{R}^{N}\;(N\geq 5)$ with $0\in\Omega$ but not a ball. Assume also that
(1.1) holds with $\alpha<\frac{N-4}{4}\tau$.

Does (Q) admit a positive solution $u\in C^{4}(\Omega\backslash\\{0\\})\cap
C^{0}({\overline{\Omega}})$ and $|x|^{\alpha}\Delta u\in
C^{0}({\overline{\Omega}})$ provided $\frac{N+4}{N-4}\leq p<p_{s}$?

In this paper, we use $C$ to denote a universal positive constant, which may
be changed from one line to another line.

##  2\. Embeddings of weighted Sobolev spaces and proof of Propositions
1.5-1.7

In this section, we present the proof of Propositions 1.5-1.7.

Proof of Proposition 1.5

Since $\alpha\geq\frac{N-4}{4}\tau$, we see that $p_{s}\leq\frac{N+4}{N-4}$.
It is known from Theorem 1.1 and Remark 2.2 of [6] that the embedding:

(2.1) $D^{2,\alpha}_{0}(\Omega)\hookrightarrow L_{l}^{p_{s}+1}(\Omega)$

is continuous. That is, there exists a constant $C>0$ independent of $u$ such
that

(2.2)
$\Big{(}\int_{\Omega}|x|^{l}|u|^{p_{s}+1}dx\Big{)}^{\frac{2}{p_{s}+1}}\leq
C\int_{\Omega}|x|^{\alpha}(\Delta u)^{2}dx.$

This inequality is corresponding to Caffarelli-Kohn-Nirenberg inequality for
the second order operator (see [4, 7]). Note that when $4-N<\alpha<N$,

$-\gamma_{\alpha}:=\Big{(}\frac{\alpha-2}{2}\Big{)}^{2}-\Big{(}\frac{N-2}{2}\Big{)}^{2}\not\in\Lambda(S^{N-1}),$

where $\Lambda(S^{N-1})$ is the Dirichlet spectrum of the Laplace-Beltrami
operator on $S^{N-1}$.

We now show that the embedding in (1.3) is continuous for any
$q\in[1,p_{s}+1)$. To see this, we just notice that, for $u\in
D^{2,\alpha}_{0}(\Omega)$ and $q\in[1,p_{s}+1)$,

(2.3) $\displaystyle\Big{(}\int_{\Omega}|x|^{l}|u|^{q}dx\Big{)}^{\frac{1}{q}}$
$\displaystyle\leq$
$\displaystyle\Big{(}\int_{\Omega}|x|^{l}|u|^{p_{s}+1}dx\Big{)}^{\frac{1}{p_{s}+1}}\Big{(}\int_{\Omega}|x|^{l}dx\Big{)}^{\frac{1}{q}-\frac{1}{p_{s}+1}}$
$\displaystyle\leq$ $\displaystyle C\Big{(}\int_{\Omega}|x|^{\alpha}(\Delta
u)^{2}dx\Big{)}^{\frac{1}{2}}\Big{(}\int_{\Omega}|x|^{l}dx\Big{)}^{\frac{1}{q}-\frac{1}{p_{s}+1}}\;\;\;\;\;(\mbox{by
\eqref{2.2}})$ $\displaystyle\leq$ $\displaystyle
C\Big{(}\int_{\Omega}|x|^{\alpha}(\Delta u)^{2}dx\Big{)}^{\frac{1}{2}}.$

This implies that the embedding is continuous. Note that
$\int_{\Omega}|x|^{l}dx\leq C$ provided $N+l=N^{\prime}+\tau>0$.

To prove the embedding is compact, we divide the proof into 2 steps.

Step 1. We show that if $\\{u_{m}\\}_{m=1}^{\infty}$ is a bounded sequence in
$D_{0}^{2,\alpha}(\Omega)$, there exists a subsequence (still denoted by
$\\{u_{m}\\}_{m=1}^{\infty}$) such that $u_{m}$ converges in
$L_{l}^{1}(\Omega)$.

Suppose that $\\{u_{m}\\}_{m=1}^{\infty}$ is a bounded sequence in
$D_{0}^{2,\alpha}(\Omega)$, then there exists $C>0$ independent of $m$ such
that $\|u_{m}\|_{D_{0}^{2,\alpha}(\Omega)}\leq C$. Setting
$v_{m}=|x|^{l}u_{m}$, we see from (2.3) that $v_{m}\in L^{1}(\Omega)$ and
$\|v_{m}\|_{L^{1}(\Omega)}\leq C$, where $C>0$ is independent of $m$. Let $B$
be a large ball such that $\Omega\subset\subset B$. If we denote

${\tilde{u}}_{m}(x)=\left\\{\begin{array}[]{ll}u_{m}(x),&x\in\Omega,\\\
0,&x\in B\backslash\Omega\end{array}\right.$

and ${\tilde{v}}_{m}(x)=|x|^{l}{\tilde{u}}_{m}(x)$, we see that
${\tilde{v}}_{m}\in L^{1}(B)$ and $\|{\tilde{v}}_{m}\|_{L^{1}(B)}\leq C$,
where $C>0$ is independent of $m$. Let

${\tilde{v}}_{m}^{\epsilon}(x)=\epsilon^{-N}\int_{B}\rho\Big{(}\frac{x-y}{\epsilon}\Big{)}{\tilde{v}}_{m}(y)dy,$

where $0<\epsilon<\delta:=$ dist$(\partial\Omega,\partial B)$ and $\rho\in
C^{\infty}(\mathbb{R}^{N})$, $\rho(x)=0$ for $|x|\geq 1$,
$\int_{\mathbb{R}^{N}}\rho(x)dx=1$. Obviously,
$\\{{\tilde{v}}_{m}^{\epsilon}\\}_{m=1}^{\infty}$ all have their supports in
$B$. We now claim that for each fixed $\epsilon>0$, the sequence
$\\{{\tilde{v}}_{m}^{\epsilon}\\}_{m=1}^{\infty}$ is uniformly bounded and
equi-continuous. Indeed, we see that, for $x\in B$,

$|{\tilde{v}}_{m}^{\epsilon}(x)|\leq\sup|\rho|\epsilon^{-N}\|{\tilde{v}}_{m}\|_{L^{1}(B)}\leq
C\epsilon^{-N},$
$|\nabla{\tilde{v}}_{m}^{\epsilon}(x)|\leq\sup|\nabla\rho|\epsilon^{-N-1}\|{\tilde{v}}_{m}\|_{L^{1}(B)}\leq
C\epsilon^{-N-1},$

where $C>0$ is independent of $m$. It follows from Arzela-Ascoli theorem that
$\\{{\tilde{v}}_{m}^{\epsilon}\\}_{m=1}^{\infty}$ is precompact in
$C^{0}({\overline{B}})$. Therefore, it is precompact in $L^{1}(B)$. We next
claim that

(2.4) ${\tilde{v}}_{m}^{\epsilon}\to{\tilde{v}}_{m}\;\;\;\mbox{in $L^{1}(B)$
as $\epsilon\to 0$ uniformly in $m$}.$

We see

(2.5) $\displaystyle\|{\tilde{v}}_{m}^{\epsilon}-{\tilde{v}}_{m}\|_{L^{1}(B)}$
$\displaystyle=$
$\displaystyle\|{\tilde{v}}_{m}^{\epsilon}-{\tilde{v}}_{m}\|_{L^{1}(B_{2\epsilon})}+\|{\tilde{v}}_{m}^{\epsilon}-{\tilde{v}}_{m}\|_{L^{1}(B\backslash
B_{2\epsilon})}$ $\displaystyle\leq$ $\displaystyle
2\|{\tilde{v}}_{m}\|_{L^{1}(B_{3\epsilon})}+\|{\tilde{v}}_{m}^{\epsilon}-{\tilde{v}}_{m}\|_{L^{1}(B\backslash
B_{2\epsilon})}$ $\displaystyle\leq$ $\displaystyle
C\epsilon^{\eta}+\|{\tilde{v}}_{m}^{\epsilon}-{\tilde{v}}_{m}\|_{L^{1}(B\backslash
B_{2\epsilon})}$

where $B_{2\epsilon}=\\{x:\;|x|<2\epsilon\\}$,
$\eta=\Big{[}1-\frac{1}{p_{s}+1}\Big{]}(N^{\prime}+\tau)>0$. Note that, for
$q\in[1,p_{s}+1)$ and $v\in D_{0}^{2,\alpha}(\Omega)$,

$\displaystyle\Big{(}\int_{B_{3\epsilon}}|x|^{l}|v|^{q}dx\Big{)}^{\frac{1}{q}}$
$\displaystyle\leq$
$\displaystyle\Big{(}\int_{B_{3\epsilon}}|x|^{l}|v|^{p_{s}+1}dx\Big{)}^{\frac{1}{p_{s}+1}}\Big{(}\int_{B_{3\epsilon}}|x|^{l}dx\Big{)}^{\frac{1}{q}-\frac{1}{p_{s}+1}}$
$\displaystyle\leq$ $\displaystyle C\Big{(}\int_{\Omega}|x|^{\alpha}(\Delta
v)^{2}dx\Big{)}^{\frac{1}{2}}\epsilon^{(\frac{1}{q}-\frac{1}{p_{s}+1})[N^{\prime}+\tau]}.$

We also see that

$\displaystyle\int_{B\backslash
B_{2\epsilon}}|{\tilde{v}}_{m}^{\epsilon}(x)-{\tilde{v}}_{m}(x)|dx$
$\displaystyle\leq$ $\displaystyle\int_{|z|\leq
1}\rho(z)\int_{0}^{1}\int_{B\backslash
B_{2\epsilon}}\Big{|}\nabla{\tilde{v}}_{m}(x-\epsilon tz)\cdot\epsilon
z\Big{|}dxdtdz$ $\displaystyle\leq$ $\displaystyle\epsilon\int_{B\backslash
B_{\epsilon}}|\nabla{\tilde{v}}_{m}(z)|dz$ $\displaystyle\leq$ $\displaystyle
C\epsilon\int_{B\backslash
B_{\epsilon}}|x|^{l-1}|{\tilde{u}}_{m}(x)dx+\epsilon\int_{B\backslash
B_{\epsilon}}|x|^{l}|\nabla{\tilde{u}}_{m}(x)|dx$ $\displaystyle\leq$
$\displaystyle C\epsilon\Big{(}\int_{B\backslash
B_{\epsilon}}|x|^{l}|{\tilde{u}}_{m}|^{p_{s}+1}dx\Big{)}^{\frac{1}{p_{s}+1}}\Big{(}\int_{B\backslash
B_{\epsilon}}|x|^{l-\frac{2(N^{\prime}+\tau)}{N^{\prime}+4+2\tau}}\Big{)}^{\frac{N^{\prime}+4+2\tau}{2(N^{\prime}+\tau)}}$
$\displaystyle+\epsilon\Big{(}\int_{B\backslash
B_{\epsilon}}|x|^{\alpha-2}|\nabla{\tilde{u}}_{m}(x)|^{2}dx\Big{)}^{\frac{1}{2}}\Big{(}\int_{B\backslash
B_{\epsilon}}|x|^{l+2+\tau}dx\Big{)}^{\frac{1}{2}}$ $\displaystyle\leq$
$\displaystyle
C\epsilon\Big{(}\int_{B}|x|^{\alpha}(\Delta{\tilde{u}}_{m}(x))^{2}dx\Big{)}^{\frac{1}{2}}\Big{(}\int_{B\backslash
B_{\epsilon}}|x|^{l-\frac{2(N^{\prime}+\tau)}{N^{\prime}+4+2\tau}}\Big{)}^{\frac{N^{\prime}+4+2\tau}{2(N^{\prime}+\tau)}}$
$\displaystyle+\epsilon\Big{(}\int_{B}|x|^{\alpha}(\Delta{\tilde{u}}_{m}(x))^{2}dx\Big{)}^{\frac{1}{2}}\Big{(}\int_{B\backslash
B_{\epsilon}}|x|^{l+2+\tau}dx\Big{)}^{\frac{1}{2}}$ $\displaystyle\leq$
$\displaystyle
C\epsilon\|{\tilde{u}}_{m}\|_{D_{0}^{2,\alpha}(B)}\Big{[}\Big{(}\int_{B\backslash
B_{\epsilon}}|x|^{l-\frac{2(N^{\prime}+\tau)}{N^{\prime}+4+2\tau}}dx\Big{)}^{\frac{N^{\prime}+4+2\tau}{2(N^{\prime}+\tau)}}+\Big{(}\int_{B\backslash
B_{\epsilon}}|x|^{l+2+\tau}dx\Big{)}^{\frac{1}{2}}\Big{]},$

where we have used that the fact (see [14, 15, 24])

(2.6) $\int_{B}|x|^{\alpha-2}|\nabla{\tilde{u}}_{m}|^{2}dx\leq
C\int_{B}|x|^{\alpha}(\Delta{\tilde{u}}_{m})^{2}dx,$

where $C>0$ is independent of $m$. We consider two cases here: (i)
$N^{\prime}+2\tau\geq 0$, (ii) $N^{\prime}+2\tau<0$.

For the case (i), we have that

$\int_{B\backslash
B_{\epsilon}}|x|^{l-\frac{2(N^{\prime}+\tau)}{N^{\prime}+4+2\tau}}dx\leq C,$
$\int_{B\backslash B_{\epsilon}}|x|^{l+2+\tau}dx\leq C,$

where $C>0$ is independent of $\epsilon$, and hence

$\int_{B\backslash
B_{2\epsilon}}|{\tilde{v}}_{m}^{\epsilon}(x)-{\tilde{v}}_{m}(x)|dx\leq
C\epsilon.$

For the case (ii), we have that

$\epsilon\Big{(}\int_{B\backslash
B_{\epsilon}}|x|^{l-\frac{2(N^{\prime}+\tau)}{N^{\prime}+4+2\tau}}dx\Big{)}^{\frac{N^{\prime}+4+2\tau}{2(N^{\prime}+\tau)}}\leq
C\epsilon^{\frac{N^{\prime}+2\tau+4}{2}},$ $\epsilon\Big{(}\int_{B\backslash
B_{\epsilon}}|x|^{l+2+\tau}dx\Big{)}^{\frac{1}{2}}\leq
C\epsilon^{\frac{N^{\prime}+2\tau+4}{2}},$

where $C>0$ is independent of $\epsilon$, and hence

$\int_{B\backslash
B_{2\epsilon}}|{\tilde{v}}_{m}^{\epsilon}(x)-{\tilde{v}}_{m}(x)|dx\leq
2C\epsilon^{\eta_{2}}\|{\tilde{u}}_{\epsilon}\|_{D_{0}^{2,\alpha}(B)},$

where

$\eta_{2}:=\frac{N^{\prime}+2\tau+4}{2}>0.$

Therefore, our claim (2.4) holds.

Now fix $\sigma>0$, we claim that there exists a subsequence
$\\{{\tilde{v}}_{m_{i}}\\}_{i=1}^{\infty}$ of
$\\{{\tilde{v}}_{m}\\}_{m=1}^{\infty}$ such that

(2.7)
$\lim_{i,j\to\infty}\sup\|{\tilde{v}}_{m_{i}}-{\tilde{v}}_{m_{j}}\|_{L^{1}(B)}\leq\sigma.$

To show this, we choose $\epsilon>0$ so small that

$\|{\tilde{v}}_{m}^{\epsilon}-{\tilde{v}}_{m}\|_{L^{1}(B)}\leq\frac{\sigma}{2},\;\;\;\mbox{for
$m=1,2,\ldots$.}$

Since $\\{{\tilde{v}}_{m}^{\epsilon}\\}_{m=1}^{\infty}$ is precompact in
$L^{1}(B)$, there exists a subsequence
$\\{{\tilde{v}}_{m_{i}}^{\epsilon}\\}_{i=1}^{\infty}$ of
$\\{{\tilde{v}}_{m}^{\epsilon}\\}_{m=1}^{\infty}$ that converges uniformly in
$L^{1}(B)$. Therefore,
$\lim_{i,j\to\infty}\sup\|{\tilde{v}}_{m_{i}}^{\epsilon}-{\tilde{v}}_{m_{j}}^{\epsilon}\|_{L^{1}(B)}=0$.
Then

$\displaystyle\int_{B}|{\tilde{v}}_{m_{i}}-{\tilde{v}}_{m_{j}}|dx$
$\displaystyle\;\;\;\;\;\;\leq\int_{B}|{\tilde{v}}_{m_{i}}-{\tilde{v}}_{m_{i}}^{\epsilon}|dx+\int_{B}|{\tilde{v}}_{m_{i}}^{\epsilon}-{\tilde{v}}_{m_{j}}^{\epsilon}|dx+\int_{B}|{\tilde{v}}_{m_{j}}^{\epsilon}-{\tilde{v}}_{m_{j}}|dx$
$\displaystyle\;\;\;\;\;\;\leq\frac{\sigma}{2}+\frac{\sigma}{2}\;\;\;\mbox{as
$i,j\to\infty$}.$

The claim (2.7) holds. We next employ assertion (2.7) with
$\sigma=1,\frac{1}{2},\ldots,\frac{1}{i},\ldots$ and a standard diagonal
argument to extract a subsequence $\\{{\tilde{v}}_{i}\\}_{i=1}^{\infty}$ of
$\\{{\tilde{v}}_{m}\\}_{m=1}^{\infty}$ satisfying

$\lim_{i,j\to\infty}\sup\|{\tilde{v}}_{i}-{\tilde{v}}_{j}\|_{L^{1}(B)}=0$

and hence ${\tilde{v}}_{i}$ converges to ${\tilde{v}}\in L^{1}(B)$ as
$i\to\infty$, i.e.,

(2.8) $\int_{B}||x|^{l}{\tilde{u}}_{i}-{\tilde{v}}|dx\to 0\;\;\mbox{as
$i\to\infty$}.$

Let ${\tilde{u}}=|x|^{-l}{\tilde{v}}$. We easily see from (2.8) that
${\tilde{u}}\in L_{l}^{1}(B)$ and

(2.9) $\int_{B}|x|^{l}|{\tilde{u}}_{i}-{\tilde{u}}|dx\to 0\;\;\mbox{as
$i\to\infty$}.$

This also implies that $u_{i}$ converges to $u$ in $L_{l}^{1}(\Omega)$ and the
proof of step 1 is complete.

Step 2. We show that if $\\{u_{m}\\}_{m=1}^{\infty}$ is a bounded sequence in
$D_{0}^{2,\alpha}(\Omega)$, there exists a subsequence (still denoted by
$\\{u_{m}\\}_{m=1}^{\infty}$) such that $u_{m}$ converges in
$L^{q}_{l}(\Omega)$ for $q\in(1,p_{s}+1)$.

We see that for $q\in(1,p_{s}+1)$, there exists $\lambda\in(0,1)$ such that
$\frac{1}{q}=\lambda+\frac{1-\lambda}{p_{s}+1}$. Then for any $u_{m}\in
D_{0}^{2,\alpha}(\Omega)$,

$\displaystyle\Big{(}\int_{\Omega}|x|^{l}|u_{m}|^{q}dx\Big{)}^{\frac{1}{q}}$
$\displaystyle\leq$
$\displaystyle\Big{(}\int_{\Omega}|x|^{l}|u_{m}|dx\Big{)}^{\lambda}\Big{(}\int_{\Omega}|x|^{l}|u_{m}|^{p_{s}+1}dx\Big{)}^{\frac{1-\lambda}{p_{s}+1}}$
$\displaystyle\leq$ $\displaystyle
C\Big{(}\int_{\Omega}|x|^{l}|u_{m}|dx\Big{)}^{\lambda}\Big{(}\int_{\Omega}|x|^{\alpha}(\Delta
u_{m})^{2}dx\Big{)}^{\frac{1-\lambda}{2}}$ $\displaystyle\leq$ $\displaystyle
C\Big{(}\int_{\Omega}|x|^{l}|u_{m}|dx\Big{)}^{\lambda}.$

It follows from the proof of step 1 that there is a subsequence
$\\{u_{i}\\}_{i=1}^{\infty}$ of $\\{u_{m}\\}_{m=1}^{\infty}$ such that
$\|u_{i}-u_{j}\|_{L_{l}^{1}(\Omega)}\to 0$ as $i,j\to\infty$. Then

(2.10)
$\Big{(}\int_{\Omega}||x|^{\frac{l}{q}}u_{i}-|x|^{\frac{l}{q}}u_{j}|^{q}dx\Big{)}^{\frac{1}{q}}\leq
C\Big{(}\int_{\Omega}|x|^{l}|u_{i}-u_{j}|dx\Big{)}^{\lambda}\to 0\;\;\mbox{as
$i,j\to\infty$}.$

There exists $v\in L^{q}(\Omega)$ such that

(2.11) $\int_{\Omega}||x|^{\frac{l}{q}}u_{i}-v|^{q}dx\to 0\;\;\mbox{as
$i\to\infty$}.$

Set $u=|x|^{-\frac{l}{q}}v$. We see from (2.11) that $u\in L_{l}^{q}(\Omega)$
and $u_{i}$ converges to $u$ in $L^{q}_{l}(\Omega)$ as $i\to\infty$. This
completes the proof of step 2 and the proof of this proposition. ∎

Proof of Proposition 1.6

Since $\alpha<\frac{N-4}{4}\tau$, we can choose $l_{1}$ such that
$\tau_{1}:=l_{1}-\alpha>-4$, $\frac{N-4}{4}\tau_{1}=\alpha$. This implies that
$l_{1}<l$ and
$p_{s}^{1}:=\frac{N^{\prime}+4+2\tau_{1}}{N^{\prime}-4}=\frac{N+4}{N-4}$. It
follows from Proposition 1.5 that the embedding:

(2.12) $D^{2,\alpha}_{0}(\Omega)\hookrightarrow L_{l_{1}}^{q}(\Omega)$

is continuous for any $q\in[1,p_{s}^{1}+1]\;(:=[1,\frac{2N}{N-4}])$ and this
embedding is continuous and compact for any $q\in[1,\frac{2N}{N-4})$.
Moreover,

$\int_{\Omega}|x|^{l}|u|^{q}dx=\int_{\Omega}|x|^{l_{1}}|x|^{l-l_{1}}|u|^{q}dx\leq
C\int_{\Omega}|x|^{l_{1}}|u|^{q}dx\;\;(\mbox{note that $l_{1}<l$}).$

This implies that

(2.13) $D^{2,\alpha}_{0}(\Omega)\hookrightarrow L_{l}^{q}(\Omega)$

for any $q\in[1,\frac{2N}{N-4}]$ is continuous and compact for any
$q\in[1,\frac{2N}{N-4})$. ∎

Proof of Proposition 1.7

The conclusions of the special case of $\alpha=0$ and $l>0$ have been obtained
in Theorem 1.1 and Corollary 1.2 of [12]. We prove this proposition by
different arguments.

For any $u\in C_{c}^{2}(B\backslash\\{0\\})$ and $u(x)=u(r)$ with $r=|x|$, we
make the transformation:

$u(r)=r^{-\frac{N^{\prime}-4}{2}}w(t),\;\;\;t=-\ln r.$

Then, we have $w\in H^{2}_{0}(-\ln R,\infty)$ and

(2.14) $\int_{B}|x|^{\alpha}(\Delta u)^{2}dx=\omega_{N}\int_{-\ln
R}^{\infty}\Big{(}|w^{\prime\prime}|^{2}+2{\tilde{\delta}}_{\alpha}|w^{\prime}|^{2}+\delta_{\alpha}|w|^{2}\Big{)}dt$

where $\omega_{N}=|S^{N-1}|$,

${\tilde{\delta}}_{\alpha}=\Big{(}\frac{N-2}{2}\Big{)}^{2}+\Big{(}\frac{\alpha-2}{2}\Big{)}^{2},\;\;\;\delta_{\alpha}=\Big{[}\Big{(}\frac{N-2}{2}\Big{)}^{2}-\Big{(}\frac{\alpha-2}{2}\Big{)}^{2}\Big{]}^{2}.$

We see that $\delta_{a}\neq 0$ provided $4-N<\alpha<N$ and $N\geq 5$.
Moreover,

(2.15) $\int_{B}|x|^{l}|u|^{p_{s}+1}dx=\omega_{N}\int_{-\ln
R}^{\infty}|w|^{p_{s}+1}dt.$

Since $H^{2}_{0}(-\ln R,\infty)\hookrightarrow L^{p_{s}+1}(-\ln R,\infty)$ by
the Sobolev embedding theorem, we obtain from (2.14) and (2.15) that the
embedding:

$D^{2,\alpha}_{0,\mbox{{\tiny rad}}}(B)\hookrightarrow L_{l}^{p_{s}+1}(B)$

is continuous. The embedding:

$D^{2,\alpha}_{0,\mbox{{\tiny rad}}}(B)\hookrightarrow L_{l}^{q}(B)$

is continuous and compact for any $q\in[1,p_{s}+1)$ can be obtained from the
interpolations arguments similar to those in the proof of Proposition 1.5. ∎

We now obtain the existence of positive solutions of (Q). To simplify
notation, $H_{\alpha}(\Omega)$ stands for $W^{2,2}_{\alpha}(\Omega)\cap
D_{0}^{1,\alpha}(\Omega)$, where $W^{2,2}_{\alpha}(\Omega)$ is the completion
of $C^{\infty}(\Omega\backslash\\{0\\})$ with respect to the norm

$\|\phi\|_{W^{2,2}_{\alpha}(\Omega)}^{2}=\int_{\Omega}|x|^{\alpha}(|u|^{2}+|\nabla
u|^{2}+|D^{2}u|^{2})dx.$

We see that the norm $[\int_{\Omega}|x|^{\alpha}(|u|^{2}+|\nabla
u|^{2}+|D^{2}u|^{2})dx]^{1/2}$ of $W^{2,2}_{\alpha}(\Omega)$ is equivalent to
$(\int_{\Omega}|x|^{\alpha}(\Delta u)^{2}dx)^{1/2}$ on $H_{\alpha}(\Omega)$.
We consider $H_{\alpha}(\Omega)$ endowed with the norm

$\|u\|_{H_{\alpha}(\Omega)}=\Big{(}\int_{\Omega}|x|^{\alpha}(\Delta
u)^{2}dx\Big{)}^{\frac{1}{2}},\;\;\;u\in H_{\alpha}(\Omega).$

######  Remark 2.1.

It is known from Remarks 2.2 and 2.3 of [6] that the functions

$u\mapsto\max_{i,j=1,\ldots,N}\int_{\Omega}|x|^{\alpha}|\partial_{i,j}u|^{2}dx,\;\;\;u\mapsto\int_{\Omega}|x|^{\alpha}(\Delta
u)^{2}dx$

define two equivalent norms in $C_{c}^{\infty}(\Omega\backslash\\{0\\})$
provided $4-N<\alpha<N$. Moreover, we see from [14, 15, 24] that

$\int_{\Omega}|x|^{\alpha}|\nabla u|^{2}dx\leq
C\int_{\Omega}|x|^{\alpha-2}|\nabla u|^{2}dx\leq
C\int_{\Omega}|x|^{\alpha}(\Delta u)^{2}dx$

for any $u\in D^{2,\alpha}_{0}(\Omega)$. Note that $|x|^{\alpha}\leq
C|x|^{\alpha-2}$ for $x\in\Omega$.

######  Definition 1.

We say that $u\in H_{\alpha}(\Omega)$ is a weak solution of

(2.16) $\left\\{\begin{array}[]{ll}\Delta(|x|^{\alpha}\Delta
u)=|x|^{l}|u|^{p-1}u&\mbox{in $\Omega$},\\\ u=\Delta u=0,&\mbox{on
$\partial\Omega$},\end{array}\right.$

if $u$ is a critical point of the $C^{1}(H_{\alpha}(\Omega),\mathbb{R})$
functional

$J(u)=\frac{1}{2}\int_{\Omega}|x|^{\alpha}(\Delta
u)^{2}dx-\frac{1}{p+1}\int_{\Omega}|x|^{l}|u|^{p+1}dx,\;\;\;u\in
H_{\alpha}(\Omega),$

that is, $u\in H_{\alpha}(\Omega)$ and satisfies

$\int_{\Omega}|x|^{\alpha}\Delta u\Delta
vdx=\int_{\Omega}|x|^{l}|u|^{p-1}uvdx,\;\;\;\forall v\in H_{\alpha}(\Omega).$

Now, we set

$m_{\alpha,l}=\inf_{u\in H_{\alpha}(\Omega),u\neq
0}\frac{\int_{\Omega}|x|^{\alpha}(\Delta
u)^{2}dx}{(\int_{\Omega}|x|^{l}|u|^{p+1}dx)^{2/(p+1)}}.$

We see that under the assumptions of Proposition 1.5, $m_{\alpha,l}$ is
attained by a function $u\in H_{\alpha}(\Omega)$ provided $1<p<p_{s}$, since
the embedding: $H_{\alpha}(\Omega)\hookrightarrow L^{q}_{l}(\Omega)$ is
continuous and compact for any $q\in[1,p_{s}+1)$. In addition, by standard
arguments, a suitable multiple of $u$ turns to be a weak solution, as defined
above, of (2.16).

######  Theorem 2.2.

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded smooth domain, but not a ball,
with $0\in\Omega$. Assume also that $N\geq 5$ and (1.1) holds with
$\alpha\geq\frac{(N-4)}{4}\tau$. Let $u\in H_{\alpha}(\Omega)$, $u\not\equiv
0$ be a minimizer of $m_{\alpha,l}$ for $1<p<p_{s}$. Then $u\in
C^{4}(\Omega\backslash\\{0\\})\cap C^{0}({\overline{\Omega}})$,
$|x|^{\alpha}\Delta u\in C^{0}({\overline{\Omega}})$ and
$u>0,-|x|^{\alpha}\Delta u>0$ in $\Omega$. Therefore, (Q) admits a positive
solution $u$.

Proof. Let $u\in H_{\alpha}(\Omega)$ be a minimizer for $m_{\alpha,l}$. The
conclusions of this theorem are two parts:

(i) $u>0$ and $-|x|^{\alpha}\Delta u>0$ in $\Omega$,

(ii) $u\in C^{4}(\Omega\backslash\\{0\\})\cap C^{0}({\overline{\Omega}})$,
$|x|^{\alpha}\Delta u\in C^{0}({\overline{\Omega}})$.
We prove the first part here. The proof of the second part will be obtained in
the next section. In fact the proof of the second part relies on the
assumptions $u>0$ and $-|x|^{\alpha}\Delta u>0$ in $\Omega\backslash\\{0\\}$.

To see the conclusions in (i), it suffices to prove that $\Delta u$ does not
change sign in $\Omega$. Indeed, suppose by contradiction that it does.
Consider $w$ be the solution of

$-\Delta w=|\Delta u|\;\;\mbox{in $\Omega$},\;\;\;w=0\;\;\mbox{on
$\partial\Omega$}.$

Since

$\int_{\Omega}|x|^{\alpha}(\Delta w)^{2}dx=\int_{\Omega}|x|^{\alpha}(\Delta
u)^{2}dx,$

we see that $w\in H_{\alpha}(\Omega)$. Regularity of $-\Delta$ and
$\Delta^{2}$ and the boot-strap argument imply that $u\in
C^{2}(\Omega\backslash\\{0\\})$ and $w\in C^{2}(\Omega\backslash\\{0\\})$.
(Note that $p_{s}\leq\frac{N+4}{N-4}$, for any
$\Omega^{\prime}\subset\subset\Omega\backslash\\{0\\}$, we can show $u\in
C^{4}(\Omega^{\prime})$ by the boot-strap argument.) Observe that
$-\Delta(w\pm u)\geq 0$ in $\Omega$. Then, the strong maximum principle
implies $w>|u|$ in $\Omega\backslash\\{0\\}$. Using $w$ in the quotient that
defines $m_{\alpha,l}$ we get a contradiction. Since $(u,v)$ satisfies the
following system of equations:

(2.17) $\left\\{\begin{array}[]{ll}-\Delta u=|x|^{-\alpha}v&\mbox{in
$\Omega$},\\\ -\Delta v=|x|^{l}|u|^{p-1}u&\mbox{in $\Omega$},\\\
u=v=0&\mbox{on $\partial\Omega$},\end{array}\right.$

regularity of $-\Delta$ and the boot-strap argument imply that $u\in
C^{2}(\Omega\backslash\\{0\\})$ and $v\in C^{2}(\Omega\backslash\\{0\\})$. The
strong maximum principle guarantees $u$, $v>0$ in $\Omega\backslash\\{0\\}$ or
$u$, $v<0$ in $\Omega\backslash\\{0\\}$. Without loss of generality, we assume
the first case occurs. The proof in the next section implies that the limits
$\lim_{|x|\to 0}u(x)$ and $\lim_{|x|\to 0}v(x)$ exist and to be positive.
Therefore, the conclusions in (i) hold. If the second case occurs, we easily
see that $-u$, $-v>0$ in $\Omega$. In the next section, we will see that $u\in
C^{4}(\Omega\backslash\\{0\\})\cap C^{0}({\overline{\Omega}})$,
$|x|^{\alpha}\Delta u\in C^{0}({\overline{\Omega}})$ provided $u>0$,
$-|x|^{\alpha}\Delta u>0$ in $\Omega\backslash\\{0\\}$. The proof of this
theorem is complete. ∎

Similarly, we have the following theorems.

######  Theorem 2.3.

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded smooth domain, but not a ball,
with $0\in\Omega$. Assume also that $N\geq 5$ and (1.1) holds with
$\alpha<\frac{(N-4)}{4}\tau$. Let $u\in H_{\alpha}(\Omega)$, $u\not\equiv 0$
be a minimizer of $m_{\alpha,l}$ for $1<p<\frac{N+4}{N-4}$. Then $u\in
C^{4}(\Omega\backslash\\{0\\})\cap C^{0}({\overline{\Omega}})$,
$|x|^{\alpha}\Delta u\in C^{0}({\overline{\Omega}})$ and $u>0$,
$-|x|^{\alpha}\Delta u>0$ in $\Omega$. Therefore, (Q) admits a positive
solution $u$.

######  Theorem 2.4.

Let $B$ be a ball centered at 0. Assume also that $N\geq 5$ and (1.1) holds.
Let $u\in H_{\alpha,\mbox{{\tiny rad}}}(B)$, $u\not\equiv 0$ be a minimizer of
$m_{\alpha,l}$ for $1<p<p_{s}$. Then $u\in C^{4}(B\backslash\\{0\\})\cap
C^{0}({\overline{B}})$, $|x|^{\alpha}\Delta u\in C^{0}({\overline{B}})$ and
$u>0$, $-|x|^{\alpha}\Delta u>0$ in $B$. Therefore, (Q) admits a positive
radial solution $u$.

##  3\. Removable singularities of solutions of (P)

In this section, we obtain some removable singularity results for solutions to
the equation (P). Such results can be seen as regularity results of solutions
to the equation (P), which have important applications in the boundary value
problems.

######  Lemma 3.1.

Assume that $1<p<p_{s}$ and (1.1) holds. Let $u(x)=u(|x|)\in
H_{\alpha,\mbox{{\tiny rad}}}(B)$ be a weak radial solution to the problem (Q)
and $u>0$, $-|x|^{\alpha}\Delta u>0$ in $B\backslash\\{0\\}$. Then $u\in
C^{4}(B\backslash\\{0\\})\cap C^{0}({\overline{B}})$, $|x|^{\alpha}\Delta u\in
C^{0}({\overline{B}})$.

Proof. Define $v(r)=-r^{\alpha}\Delta u(r)$. We see that $(u,v)$ satisfies the
system of equations

(3.1)
$\left\\{\begin{array}[]{ll}-(r^{N-1}u^{\prime}(r))^{\prime}=r^{N-\alpha-1}v(r)&\mbox{in
$(0,R)$},\\\
-(r^{N-1}v^{\prime}(r))^{\prime}=r^{N^{\prime}+\tau-1}u^{p}(r)&\mbox{in
$(0,R)$},\\\ u(R)=v(R)=0.\end{array}\right.$

Moreover, we easily see from the equations in (3.1) that
$r^{N-1}u^{\prime}(r)$ and $r^{N-1}v^{\prime}(r)$ are decreasing functions for
$r\in(0,R)$ and thus both $\lim_{r\to 0}r^{N-1}u^{\prime}(r)$ and $\lim_{r\to
0}r^{N-1}v^{\prime}(r)$ exist (maybe $\infty$). The facts $u\in
H_{\alpha,\mbox{{\tiny rad}}}(B)$ and the embedding $H_{\alpha,\mbox{{\tiny
rad}}}(B)\hookrightarrow L^{q}_{l}(B)$ for $1\leq q\leq p_{s}+1$ imply that

(3.2)
$\int_{B}|x|^{l}u^{p}dx<\infty,\;\;\;\int_{B}|x|^{-\alpha}v^{2}dx<\infty,$

and

(3.3)
$\int_{B}|x|^{-\alpha}|v|dx\leq\Big{(}\int_{B}|x|^{-\alpha}v^{2}dx\Big{)}^{\frac{1}{2}}\Big{(}\int_{B}|x|^{-\alpha}dx\Big{)}^{\frac{1}{2}}<\infty,$

since $\alpha<N$. Therefore, for any $\epsilon>0$,

$-\int_{B_{\epsilon}}|x|^{l}u^{p}dx=\int_{B_{\epsilon}}\Delta
vdx=C\epsilon^{N-1}v^{\prime}(\epsilon),$
$-\int_{B_{\epsilon}}|x|^{-\alpha}vdx=\int_{B_{\epsilon}}\Delta
udx=C\epsilon^{N-1}u^{\prime}(\epsilon).$

These, (3.2) and (3.3) imply that $r^{N-1}u^{\prime}(r)\to 0$,
$r^{N-1}v^{\prime}(r)\to 0$ as $r\to 0$. Integrations of the equations in
(3.1) give that $u\in C^{2}(0,R)$, $v\in C^{2}(0,R)$.

To prove this lemma, we only need to show the following claim:

(3.4) $\lim_{r\to 0}u(r)=\tau_{u}\in(0,\infty),\;\;\;\lim_{r\to
0}v(r)=\tau_{v}\in(0,\infty),$

where $\tau_{u}$ and $\tau_{v}$ are constants.

Since $u\in H_{\alpha,\mbox{{\tiny rad}}}(B)$ and (1.1) holds, we have

(3.5) $\int_{0}^{R}r^{N^{\prime}-1-2\alpha}v^{2}(r)dr<\infty,$

note that $\Delta u=-|x|^{-\alpha}v$. We also obtain from the embedding:

$H_{\alpha,\mbox{{\tiny rad}}}(B)\hookrightarrow L^{p_{s}+1}_{l}(B)$

that

(3.6)
$\int_{0}^{R}r^{N^{\prime}+\tau-1}u^{\frac{2(N^{\prime}+\tau)}{N^{\prime}-4}}(r)dr<\infty.$

This implies that, for $r$ near 0,

(3.7)
$u(r)=o\Big{(}r^{-\frac{N^{\prime}-4}{2}}\Big{)},\;\;\;v(r)=o\Big{(}r^{-\frac{N-\alpha}{2}}\Big{)}.$

We only show $\eqref{3.6}_{1}$ by using (3.6). The proof of $\eqref{3.6}_{2}$
is similar to that of $\eqref{3.6}_{1}$ by using (3.5). We easily see from
(3.6) that

$\int_{0}^{r}s^{N^{\prime}+\tau-1}u^{\frac{2(N^{\prime}+\tau)}{N^{\prime}-4}}(s)ds=o(1)\;\;\mbox{for
$r$ near 0}.$

Using the fact that $u(r)$ is decreasing, we have that

$\int_{0}^{r}s^{N^{\prime}+\tau-1}u^{\frac{2(N^{\prime}+\tau)}{N^{\prime}-4}}(s)ds\geq\frac{r^{N^{\prime}+\tau}}{N^{\prime}+\tau}u^{\frac{2(N^{\prime}+\tau)}{N^{\prime}-4}}(r).$

This implies that

$r^{N^{\prime}+\tau}u^{\frac{2(N^{\prime}+\tau)}{N^{\prime}-4}}(r)=o(1)$

and

$u(r)=o\Big{(}r^{-\frac{N^{\prime}-4}{2}}\Big{)}.$

Let

$w(t)=r^{\frac{N^{\prime}-4}{2}}u(r),\;\;z(t)=r^{\frac{N-\alpha}{2}}v(r),\;\;\;t=-\ln
r.$

We see that $(w(t),z(t))$ satisfies the problem

(3.8)
$\left\\{\begin{array}[]{ll}w_{tt}+(\alpha-2)w_{t}-\frac{(N-\alpha)(N^{\prime}-4)}{4}w+z=0,&t\in(-\ln
R,\infty),\\\
z_{tt}+(2-\alpha)z_{t}-\frac{(N-\alpha)(N^{\prime}-4)}{4}z+e^{-p_{*}t}w^{p}=0,&t\in(-\ln
R,\infty),\\\ w(-\ln R)=z(-\ln R)=0,\end{array}\right.$

where and in the following

$p_{*}=\frac{[N^{\prime}+4+2\tau]-[N^{\prime}-4]p}{2}.$

Note that $p_{*}>0$ provided $1<p<p_{s}$. We know from (3.7) and $1<p<p_{s}$
that $w(t)\to 0$, $z(t)\to 0$ and $e^{-p_{*}t}\to 0$ as $t\to\infty$. It
follows from the ODE theory on perturbation of linear systems (see [22]) that
there exist $\tau_{u}>0$, $\tau_{v}>0$ such that

(3.9)
$w(t)=(\tau_{u}+o(1))e^{-\frac{N^{\prime}-4}{2}t},\;\;\;z(t)=(\tau_{v}+o(1))e^{-\frac{N-\alpha}{2}t}\;\;\mbox{as
$t\to\infty$}.$

Note that the system (3.8) can be written to the following system

(3.10) $\left\\{\begin{array}[]{l}\dot{w_{1}}=w_{2},\\\
\dot{w_{2}}=\frac{(N-\alpha)(N^{\prime}-4)}{4}w_{1}-(\alpha-2)w_{2}-w_{3},\\\
\dot{w_{3}}=w_{4},\\\
\dot{w_{4}}=\frac{(N-\alpha)(N^{\prime}-4)}{4}w_{3}+(\alpha-2)w_{4}-e^{-p_{*}t}w_{1}^{p},\end{array}\right.$

where $(w_{1},w_{2},w_{3},w_{4})=(w,\dot{w},z,\dot{z})$. Since
$(w_{1}(t),w_{2}(t),w_{3}(t),w_{4}(t))\to(0,0,0,0)$ as $t\to\infty$, we
linearize the system (3.10) at (0,0,0,0) and obtain the system

(3.11) $\left\\{\begin{array}[]{l}\dot{h_{1}}=h_{2},\\\
\dot{h_{2}}=\frac{(N-\alpha)(N^{\prime}-4)}{4}h_{1}-(\alpha-2)h_{2}-h_{3},\\\
\dot{h_{3}}=h_{4},\\\
\dot{h_{4}}=\frac{(N-\alpha)(N^{\prime}-4)}{4}h_{3}+(\alpha-2)h_{4}.\end{array}\right.$

The matrix of (3.11) is

(3.12) ${\mathcal{A}}=\left(\begin{array}[]{llll}0&1&0&0\\\
\frac{(N-\alpha)(N^{\prime}-4)}{4}&2-\alpha&-1&0\\\ 0&0&0&1\\\
0&0&\frac{(N-\alpha)(N^{\prime}-4)}{4}&\alpha-2\end{array}\right).$

By simple calculations, we see that the four eigenvalues of $\mathcal{A}$ are:
$\lambda_{1,2}=\mp\frac{N^{\prime}-4}{2}$,
$\lambda_{3,4}=\mp\frac{N-\alpha}{2}$. We only choose
$\lambda_{1}=-\frac{N^{\prime}-4}{2}$ and $\lambda_{3}=-\frac{N-\alpha}{2}$,
since $\lambda_{2}=\frac{N^{\prime}-4}{2}>0$ and
$\lambda_{4}=\frac{N-\alpha}{2}>0$ which do not meet our requirement. We also
see that there are eigenvectors for $\lambda_{1}$ and $\lambda_{2}$
respectively: $e^{\lambda_{1}t}(1,-\frac{N^{\prime}-4}{2},0,0)$ and
$e^{\lambda_{2}t}(0,0,1,-\frac{N-\alpha}{2})$. These imply that (3.9) holds
and hence our claim (3.4) holds. The proof of this lemma is complete. ∎

######  Lemma 3.2.

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded smooth domain, but not a ball,
with $0\in\Omega$. Assume also that $N\geq 5$ and (1.1) holds with
$\alpha\geq\frac{(N-4)}{4}\tau$. Let $u\in H_{\alpha}(\Omega)$ be a weak
solution of (Q) for $1<p<p_{s}$ and $u>0$, $-|x|^{\alpha}\Delta u>0$ in
$\Omega\backslash\\{0\\}$. Then $u\in C^{4}(\Omega\backslash\\{0\\})\cap
C^{0}({\overline{\Omega}})$, $|x|^{\alpha}\Delta u\in
C^{0}({\overline{\Omega}})$.

Proof. We know that $p_{s}\leq\frac{N+4}{N-4}$ provided that (1.1) holds with
$\alpha\geq\frac{N-4}{4}\tau$. Then the facts $u\in H_{\alpha}(\Omega)$ and
$1<p<p_{s}$ and the boot-strap argument imply that $u\in
C^{4}(\Omega\backslash\\{0\\})$. Note that for any subset
$\Omega^{\prime}\subset\Omega\backslash\\{0\\}$, $u\in
W^{2,2}(\Omega^{\prime})$. The boot-strap argument (note that
$1<p<\frac{N+4}{N-4}$ if $1<p<p_{s}$) implies that $u\in
C^{4}(\Omega^{\prime})$.

Since $0\in\Omega$, we can choose a $0<\sigma<\frac{1}{10}$ such that
$B_{\sigma}\subset\subset\Omega$, where
$B_{\sigma}=\\{x\in\Omega:\;|x|<\sigma\\}$. The proof is divided to 2 steps:

(1) We show that there exists $C>0$ such that

(3.13) $u(x)\leq C|x|^{-\frac{4+\tau}{p-1}},\;\;\;v(x)\leq
C|x|^{\alpha-\frac{2(p+1)+\tau}{p-1}},\;\;\;0<|x|<\sigma/2,$

where $v(x)=-|x|^{\alpha}\Delta u(x)$. Note that $u>0$, $v>0$ in
$B_{\sigma/2}$.

(2) Using the estimates obtained in (3.13) and Harnack’s inequality of
systems, we obtain our conclusion.

Step 1. For any $x_{0}\in B_{\sigma}$ with $0<|x_{0}|<\sigma/2$, we denote

$R=\frac{|x_{0}|}{2}$

and observe that, for all $y\in B_{1}$,
$\frac{|x_{0}|}{2}<|x_{0}+Ry|<\frac{3|x_{0}|}{2}$, so that $x_{0}+Ry\in
B_{\sigma}$. Let us define

$U(y)=R^{\frac{4+\tau}{p-1}}u(x_{0}+Ry),\;\;\;V(y)=R^{\frac{2(p+1)+\tau}{p-1}-\alpha}v(x_{0}+Ry).$

Then $(U,V)$ satisfies the system

(3.14) $\left\\{\begin{array}[]{ll}-\Delta
U=|y+\frac{x_{0}}{R}|^{-\alpha}V&\mbox{in $B_{1}$},\\\ -\Delta
V=|y+\frac{x_{0}}{R}|^{l}U^{p}&\mbox{in $B_{1}$}.\end{array}\right.$

Note that $|y+\frac{x_{0}}{R}|\in[1,3]$ for all $y\in{\overline{B_{1}}}$.

We now claim that there exists $C>0$ depending only on $p,N^{\prime},\tau$,
independent of $x_{0}$, such that

(3.15) $U^{\frac{p-1}{4}}(y)+V^{\frac{p-1}{2(p+1)}}(y)\leq
C(1+\mbox{dist}^{-1}(y,\partial B_{1})),\;\;y\in B_{1}.$

Arguing by contradiction, we suppose that there exist sequences $x_{k}\in
B_{\sigma/2}\backslash\\{0\\}$, points $z_{k}\in B_{1}$ and $(U_{k},V_{k})$
satisfying

(3.16) $\left\\{\begin{array}[]{ll}-\Delta
U_{k}=|y+\frac{x_{k}}{R_{k}}|^{-\alpha}V_{k}&\mbox{in $B_{1}$},\\\ -\Delta
V_{k}=|y+\frac{x_{k}}{R_{k}}|^{l}U_{k}^{p}&\mbox{in
$B_{1}$},\end{array}\right.$

where $R_{k}=\frac{1}{2}|x_{k}|$ such that the functions

$M_{k}=|U_{k}|^{\frac{p-1}{4}}+|V_{k}|^{\frac{p-1}{2(p+1)}}$

satisfy

$M_{k}(z_{k})>2k(1+\mbox{dist}^{-1}(z_{k},\partial
B_{1}))>2k\mbox{dist}^{-1}(z_{k},\partial B_{1}).$

By doubling lemma in [27], there exists $y_{k}\in B_{1}$ such that

$M_{k}(y_{k})\geq
M_{k}(z_{k}),\;\;M_{k}(y_{k})>2k\mbox{dist}^{-1}(y_{k},\partial B_{1}),$

and

(3.17) $M_{k}(y)\leq 2M_{k}(y_{k}),\;\;\mbox{for all $y$ such that
$|y-y_{k}|\leq kM_{k}^{-1}(y_{k})$}.$

We have

(3.18) $\lambda_{k}:=M_{k}^{-1}(y_{k})\to 0\;\;\mbox{as $k\to\infty$},$

due to $M_{k}(y_{k})\geq M_{k}(z_{k})>2k$.

Next we let

${\tilde{U}}_{k}(z)=\lambda_{k}^{\frac{4}{p-1}}U_{k}(y_{k}+\lambda_{k}z),\;\;{\tilde{V}}_{k}(z)=\lambda_{k}^{\frac{2(p+1)}{p-1}}V_{k}(y_{k}+\lambda_{k}z).$

Note that
$|{\tilde{U}}_{k}|^{\frac{p-1}{4}}(0)+|{\tilde{V}}_{k}|^{\frac{p-1}{2(p+1)}}(0)=1$,

(3.19)
$\Big{[}|{\tilde{U}}_{k}|^{\frac{p-1}{4}}+|{\tilde{V}}_{k}|^{\frac{p-1}{2(p+1)}}\Big{]}(z)\leq
2,\;\;|z|\leq k,$

due to (3.17) and we see that $({\tilde{U}}_{k},{\tilde{V}}_{k})$ satisfies

(3.20)
$\left\\{\begin{array}[]{ll}-\Delta{\tilde{U}}_{k}=|y_{k}+\lambda_{k}z+\frac{x_{k}}{R_{k}}|^{-\alpha}{\tilde{V}}_{k},&|z|\leq
k,\\\
-\Delta{\tilde{V}}_{k}=|y_{k}+\lambda_{k}z+\frac{x_{k}}{R_{k}}|^{l}{\tilde{U}}_{k}^{p},&|z|\leq
k.\end{array}\right.$

Now, for each $A>0$ and $1<q<\infty$, by (3.19), (3.20) and interior elliptic
$L^{q}$ estimates, the sequence $({\tilde{U}}_{k},{\tilde{V}}_{k})$ is
uniformly bounded in $W^{2,q}(B_{A})\times W^{2,q}(B_{A})$. Note that

$\Big{|}y_{k}+\lambda_{k}z+\frac{x_{k}}{R_{k}}\Big{|}\in[1,3]\;\;\mbox{for all
$k$}.$

Using standard embeddings and interior elliptic Schauder estimates, after
extracting subsequences (still denoted by $\\{y_{k}\\}$,
$\\{\frac{x_{k}}{R_{k}}\\}$ and $\\{({\tilde{U}}_{k},{\tilde{V}}_{k})\\}$), we
may assume $y_{k}\to y_{0}\in{\overline{B_{1}}}$,
$\frac{x_{k}}{R_{k}}\to{\tilde{x}}\in\partial B_{2}$,
$({\tilde{U}}_{k},{\tilde{V}}_{k})\to({\tilde{U}},{\tilde{V}})$ in
$C^{2}_{loc}(\mathbb{R}^{N})\times C^{2}_{loc}(\mathbb{R}^{N})$. It follows
that $({\tilde{U}},{\tilde{V}})\geq 0$ is a classical solution of

(3.21)
$\left\\{\begin{array}[]{ll}-\Delta{\tilde{U}}=|y_{0}+{\tilde{x}}|^{-\alpha}{\tilde{V}},&z\in\mathbb{R}^{N},\\\
-\Delta{\tilde{V}}=|y_{0}+{\tilde{x}}|^{l}{\tilde{U}}^{p},&z\in\mathbb{R}^{N}.\end{array}\right.$

(3.21) implies that ${\tilde{U}}\in C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap
C^{2}(\mathbb{R}^{N})$ satisfies

(3.22) $\Delta^{2}{\tilde{U}}=C{\tilde{U}}^{p}\;\;\mbox{in
$\mathbb{R}^{N}\backslash\\{0\\}$},$

where $0<C=|y_{0}+{\tilde{x}}|^{\tau}<\infty$ and
${\tilde{U}}^{\frac{p-1}{4}}(0)+{\tilde{V}}^{\frac{(p-1)}{2(p+1)}}(0)=1$.
Since $1<p<p_{s}\leq\frac{N+4}{N-4}$, this contradicts the Liouville-type
result in [23] and concludes our claim (3.15). It also implies that
$U(0)+V(0)\leq C$, hence

$u(x_{0})\leq C|x_{0}|^{-\frac{4+\tau}{p-1}},\;\;v(x_{0})\leq
C|x_{0}|^{\alpha-\frac{2(p+1)+\tau}{p-1}},$

which completes the proof of step 1.

Step 2. Let $\sigma_{0}=\sigma/4$. We make the transform:

(3.23)
$w(t,\omega)=r^{\frac{N^{\prime}-4}{2}}u(r\omega),\;\;z(t,\omega)=r^{\frac{N-\alpha}{2}}v(r\omega),\;\;\;t=-\ln
r.$

Note that $u>0$ and $v>0$ in $B_{\sigma_{0}}\subset\subset\Omega$. Then
$(w(t,\omega),z(t,\omega))$ satisfies

(3.24)
$\left\\{\begin{array}[]{ll}w_{tt}+\Delta_{S^{N-1}}w+(\alpha-2)w_{t}-\frac{(N-\alpha)(N^{\prime}-4)}{4}w+z=0,&(t,\omega)\in
I\times S^{N-1},\\\
z_{tt}+\Delta_{S^{N-1}}z+(2-\alpha)z_{t}-\frac{(N-\alpha)(N^{\prime}-4)}{4}z+e^{-p_{*}t}w^{p}=0,&(t,\omega)\in
I\times S^{N-1},\end{array}\right.$

where $I=(-\ln\sigma_{0},\infty)$ since $B_{\sigma_{0}}\subset\subset\Omega$.
We can write (3.24) in the form:

(3.25)
$\left\\{\begin{array}[]{ll}w_{tt}+\Delta_{S^{N-1}}w+(\alpha-2)w_{t}-\frac{(N-\alpha)(N^{\prime}-4)}{4}w+z=0,&(t,\omega)\in
I\times S^{N-1},\\\
z_{tt}+\Delta_{S^{N-1}}z+(2-\alpha)z_{t}-\frac{(N-\alpha)(N^{\prime}-4)}{4}z+a(t,\omega)w=0,&(t,\omega)\in
I\times S^{N-1},\end{array}\right.$

with $a(t,\omega)=e^{-p_{*}t}w^{p-1}$. It is known from (3.13) that

$w(t,\omega)\leq
Ce^{(\frac{4+\tau}{p-1}-\frac{N^{\prime}-4}{2})t}\;\;\;\mbox{for
$(t,\omega)\in I\times S^{N-1}$}.$

Therefore,

$e^{-p_{*}t}w^{p-1}\leq
Ce^{[(\frac{4+\tau}{p-1}-\frac{N^{\prime}-4}{2})(p-1)-p_{*}]t}=C\;\;\;\mbox{for
$(t,\omega)\in I\times S^{N-1}$}.$

Note that

$\Big{(}\frac{4+\tau}{p-1}-\frac{N^{\prime}-4}{2}\Big{)}(p-1)-p_{*}=0.$

Since $|a(t,\omega)|\leq C$ for $t\in(-\ln\sigma_{0},\infty)\times S^{N-1}$,
by Harnack’s inequality (see Theorem 2.1 of [1]) of systems over
$[L-1,L]\times S^{N-1}$ with $L-1>-\ln\sigma_{0}$, there exists $C>0$
independent of $t\in[-\ln\sigma_{0}+1,\infty)$ such that

(3.26) $\max_{S^{N-1}}\Big{\\{}w(t,\cdot),z(t,\cdot)\Big{\\}}\leq
C\min_{S^{N-1}}\Big{\\{}w(t,\cdot),z(t,\cdot)\Big{\\}}\;\;\;\mbox{for
$t\geq-\ln\sigma_{0}+1$}.$

Let ${\overline{w}}(t)=\frac{1}{|S^{N-1}|}\int_{S^{N-1}}w(t,\omega)d\omega$
and ${\overline{z}}(t)=\frac{1}{|S^{N-1}|}\int_{S^{N-1}}z(t,\omega)d\omega$.
It follows from (3.26) that there exist $0<c_{1}<c_{2}$ and $0<c_{3}<c_{4}$
such that

$c_{1}{\overline{w}}\leq w\leq c_{2}{\overline{w}}\;\;\mbox{for
$(t,\omega)\in(-\ln\sigma_{0}+1,\infty)\times S^{N-1}$},$
$c_{3}{\overline{z}}\leq z\leq c_{4}{\overline{z}}\;\;\mbox{for
$(t,\omega)\in(-\ln\sigma_{0}+1,\infty)\times S^{N-1}$}.$

These imply that

(3.27) $c_{1}{\overline{u}}\leq u\leq
c_{2}{\overline{u}},\;\;\;c_{3}{\overline{v}}\leq v\leq
c_{4}{\overline{v}}\;\;\;\mbox{for $x\in
B_{e^{-1}\sigma_{0}}\backslash\\{0\\}$},$

where ${\overline{u}}(r):=\frac{\int_{\partial
B_{r}}u(r,\omega)d\omega}{|\partial B_{r}|}$,
${\overline{v}}(r):=\frac{\int_{\partial B_{r}}v(r,\omega)d\omega}{|\partial
B_{r}|}$.

On the other hand, it follows from $u\in H_{\alpha}(\Omega)$ and the embedding
given in Proposition 1.5 that

(3.28)
$\int_{B_{e^{-1}\sigma_{0}}}|x|^{l}|u|^{\frac{2(N^{\prime}+\tau)}{N^{\prime}-4}}dx<\infty,\;\;\;\int_{B_{e^{-1}\sigma_{0}}}|x|^{-\alpha}v^{2}dx<\infty$

provided that (1.1) holds with $\alpha\geq\frac{N-4}{4}\tau$. This and (3.27)
imply

(3.29)
$\int_{B_{e^{-1}\sigma_{0}}}|x|^{l}{\overline{u}}^{\frac{2(N^{\prime}+\tau)}{N^{\prime}-4}}dx<\infty,\;\;\;\int_{B_{e^{-1}\sigma_{0}}}|x|^{-\alpha}{\overline{v}}^{2}dx<\infty.$

Then, arguments similar to those in the proof of (3.7) imply that there exists
$0<\sigma_{1}<e^{-1}\sigma_{0}$ such that

(3.30)
${\overline{u}}(r)=o\Big{(}r^{-\frac{N^{\prime}-4}{2}}\Big{)},\;\;\;{\overline{v}}(r)=o\Big{(}r^{-\frac{N-\alpha}{2}}\Big{)}\;\;\;\forall
r\in(0,\sigma_{1}).$

Note that

${\overline{w}}(t)=r^{\frac{N^{\prime}-4}{2}}{\overline{u}}(r),\;\;\;{\overline{z}}(t)=r^{\frac{N-\alpha}{2}}{\overline{v}}(r),\;\;\;t=-\ln
r.$

(3.30) implies that

(3.31) ${\overline{w}}(t)\to 0,\;\;\;{\overline{z}}(t)\to 0\;\;\;\mbox{as
$t\to\infty$}.$

On the other hand, $({\overline{w}},{\overline{z}})$ satisfies

(3.32)
$\left\\{\begin{array}[]{ll}{\overline{w}}_{tt}+(\alpha-2){\overline{w}}_{t}-\frac{(N-\alpha)(N^{\prime}-4)}{4}{\overline{w}}+{\overline{z}}=0,&t\in
I,\\\
{\overline{z}}_{tt}+(2-\alpha){\overline{z}}_{t}-\frac{(N-\alpha)(N^{\prime}-4)}{4}{\overline{z}}+e^{-p_{*}t}{\overline{w^{p}}}=0,&t\in
I,\end{array}\right.$

and due to (3.26),

$c_{5}{\overline{w}}^{p}\leq{\overline{w^{p}}}\leq
c_{6}{\overline{w}}^{p}\;\;\mbox{for
$(t,\omega)\in(-\ln\sigma_{0}+1,\infty)\times S^{N-1}$}.$

By (3.31), it follows from the ODE theory on perturbation of linear systems
(see [22]) that, there exist ${\tilde{\tau}}_{1}>0$ and ${\tilde{\tau}}_{2}>0$
such that

(3.33)
${\overline{w}}(t)=({\tilde{\tau}}_{1}+o(1))e^{-\frac{N^{\prime}-4}{2}t}\;\;\;{\overline{z}}(t)=({\tilde{\tau}}_{2}+o(1))e^{-\frac{N-\alpha}{2}t}\;\;\;\mbox{as
$t\to\infty$}.$

In view of (3.26), (3.33) implies

$w(t,\omega)=O(e^{-\frac{N^{\prime}-4}{2}t}),\;\;\;z(t,\omega)=O(e^{-\frac{N-\alpha}{2}t})\;\;\mbox{as
$t\to\infty$ uniformly for $\omega\in S^{N-1}$}.$

These imply that there exist $\tau_{1}>0$, $\tau_{2}>0$ such that

(3.34) $u(x)\to\tau_{1},\;\;\;v(x)\to\tau_{2}\;\;\mbox{as $|x|\to 0$}.$

The proof of this lemma is completed. ∎

######  Lemma 3.3.

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded smooth domain, but not a ball,
with $0\in\Omega$. Assume also that $N\geq 5$ and (1.1) holds with
$\alpha<\frac{(N-4)}{4}\tau$. Let $u\in H_{\alpha}(\Omega)$ be a weak solution
of (Q) for $1<p<\frac{N+4}{N-4}$ and $u>0$, $-|x|^{\alpha}\Delta u>0$ in
$\Omega\backslash\\{0\\}$. Then $u\in C^{4}(\Omega\backslash\\{0\\})\cap
C^{0}({\overline{\Omega}})$, $|x|^{\alpha}\Delta u\in
C^{0}({\overline{\Omega}})$.

Proof. Since $\alpha<\frac{N-4}{2}\tau$, we see that $\frac{N+4}{N-4}<p_{s}$.
The conclusions can be obtained by arguments exactly the same as those in the
proof of Lemma 3.2. The only thing we need to show is

(3.35)
${\overline{u}}(r)=o\Big{(}r^{-\frac{N^{\prime}-4}{2}}\Big{)}\;\;\mbox{for $r$
near 0}.$

To see (3.35), we choose $l_{1}$ and $\tau_{1}$ as in the proof of Proposition
1.6, then

$\int_{B_{\sigma}}|x|^{l_{1}}u^{\frac{2(N^{\prime}+\tau_{1})}{N^{\prime}-4}}dx\;\Big{(}:=\int_{B_{\sigma}}|x|^{l_{1}}u^{\frac{2N}{N-4}}dx\Big{)}<\infty.$

By Harnack’s inequality as in the proof of Lemma 3.2, we have

$\int_{B_{\sigma}}|x|^{l_{1}}{\overline{u}}^{\frac{2(N^{\prime}+\tau_{1})}{N^{\prime}-4}}dx<\infty.$

Now, (3.35) can be obtained by similar arguments to those in the proof of
(3.7). This completes the proof of this lemma. ∎

######  Corollary 3.4.

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded smooth domain with
$0\in\Omega$. Assume that (1.1) holds and $\varphi_{1}\in H_{\alpha}(\Omega)$
is the first eigenfunction corresponding to the first eigenvalue $\lambda_{1}$
of the problem

(3.36)
$\left\\{\begin{array}[]{rl}\Delta(|x|^{\alpha}\Delta\phi(x))=\lambda|x|^{l}\phi(x),&x\in\Omega,\\\
\phi(x)=0,\;\;\;\Delta\phi(x)=0,&x\in\partial\Omega.\end{array}\right.$

Then $\varphi_{1}\in C^{4}(\Omega\backslash\\{0\\})\cap
C^{0}({\overline{\Omega}})$, $|x|^{\alpha}\Delta\varphi_{1}\in
C^{0}({\overline{\Omega}})$.

Proof. The existence of a nontrivial $\varphi_{1}\in H_{\alpha}(\Omega)$ as a
minimizer of

$\inf_{\phi\in
H_{\alpha}(\Omega)\backslash\\{0\\}}\frac{\int_{\Omega}|x|^{\alpha}(\Delta\phi)^{2}}{\int_{\Omega}|x|^{l}|\phi|^{2}dx}$

can be obtained from the compact embedding: $H_{\alpha}\hookrightarrow
L_{l}^{2}(\Omega)$. We need to consider two cases respectively: (1.1) holds
with $\alpha\geq\frac{N-4}{4}\tau$ and (1.1) holds with
$\alpha<\frac{N-4}{4}\tau$. Similar arguments to those in the proof of
Theorems 2.2 and 2.3 imply that $\varphi_{1}>0$,
$-|x|^{\alpha}\Delta\varphi_{1}>0$ in $\Omega$. Other properties can be also
obtained. Note that we can use Harnack’s inequality here, since in our case
here $p_{*}=4+\tau>0$ (note that $p=1$) and

$a(t,\omega)=e^{-(4+\tau)t},\;\;\;(t,\omega)\in I\times S^{N-1},$

which implies that $|a(t,\omega)|\leq C$ for $(t,\omega)\in I\times S^{N-1}$.
∎

######  Corollary 3.5.

Let $B=\\{x\in\mathbb{R}^{N}:\;|x|<R\\}$ be a ball in $\mathbb{R}^{N}$
centered at 0. Assume that (1.1) holds. Then the first eigenfunction
$\varphi_{1}$ corresponding to the first eigenvalue $\lambda_{1}$ of the
eigenvalue problem

(3.37)
$\left\\{\begin{array}[]{rl}\Delta(|x|^{\alpha}\Delta\phi(x))=\lambda|x|^{l}\phi(x),&x\in
B,\\\ \phi(x)=0,\;\;\Delta\phi(x)=0,&x\in\partial B\end{array}\right.$

satisfies that $\varphi_{1}\in C^{4}(B\backslash\\{0\\})\cap
C^{0}({\overline{B}})$, $|x|^{\alpha}\Delta\varphi_{1}\in
C^{0}({\overline{B}})$, $\varphi_{1}(x)=\varphi_{1}(r)$, $\varphi_{1}(r)>0$
and $\varphi_{1}^{\prime}(r)<0$ for $r\in(0,R)$.

Proof. The existence of a nontrivial nonnegative $\varphi_{1}\in
H_{\alpha,\mbox{{\tiny rad}}}(B)$ as a minimizer of

$\inf_{\phi\in H_{\alpha,\mbox{{\tiny
rad}}}(B)\backslash\\{0\\}}\frac{\int_{B}|x|^{\alpha}(\Delta\phi)^{2}}{\int_{B}|x|^{l}|\phi|^{2}dx}$

can be obtained from the compact embedding: $H_{\alpha,\mbox{{\tiny
rad}}}(B)\hookrightarrow L_{l}^{2}(B)$. Other properties can be obtained by
similar arguments to those in the proof of Theorem 2.4 and Lemma 3.1. ∎

##  4\. Liouville type results: Proof of Theorems 1.1 and 1.3

In this section, we present the proof of Theorems 1.1 and 1.3. We first obtain
the following result.

######  Lemma 4.1.

Assume that $N\geq 5$, (1.1) holds and $1<p<\min\\{\frac{N+4}{N-4},p_{s}\\}$.
Then, there exists a constant $C=C(N,p,\alpha,\tau)>0$ such that the following
holds

(i) Any nonnegative solution $u\in C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap
C^{0}(\mathbb{R}^{N})$ with $|x|^{\alpha}\Delta u\in C^{0}(\mathbb{R}^{N})$ of
(P) in $\Omega=\\{x\in\mathbb{R}^{N}:\;0<|x|<\rho\\}\;(\rho>0)$ satisfies

(4.1) $u(x)\leq C|x|^{-\frac{4+\tau}{p-1}}\;\;\mbox{and}\;\;|\nabla u(x)|\leq
C|x|^{-\frac{p+\tau+3}{p-1}},\;\;\;0<|x|<\frac{\rho}{2},$ (4.2) $|v(x)|\leq
C|x|^{\alpha-\frac{2(p+1)+\tau}{p-1}}\;\;\mbox{and}\;\;|\nabla v(x)|\leq
C|x|^{\alpha-1-\frac{2(p+1)+\tau}{p-1}},\;\;\;0<|x|<\frac{\rho}{2},$

where $v(x):=-|x|^{\alpha}\Delta u(x)$.

(ii) Any nonnegative solution $u\in C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap
C^{0}(\mathbb{R}^{N})$ with $|x|^{\alpha}\Delta u\in C^{0}(\mathbb{R}^{N})$ of
(P) in $\Omega=\\{x\in\mathbb{R}^{N}:\;|x|>\rho\\}\;(\rho>0)$ satisfies

(4.3) $u(x)\leq C|x|^{-\frac{4+\tau}{p-1}}\;\;\mbox{and}\;\;|\nabla u(x)|\leq
C|x|^{-\frac{p+\tau+3}{p-1}},\;\;\;|x|>2\rho,$ (4.4) $|v(x)|\leq
C|x|^{\alpha-\frac{2(p+1)+\tau}{p-1}}\;\;\mbox{and}\;\;|\nabla v(x)|\leq
C|x|^{\alpha-1-\frac{2(p+1)+\tau}{p-1}},\;\;\;|x|>2\rho.$

Proof. The proof of this lemma is similar to the proof of the first step of
Lemma 3.2. For any $x_{0}\in\Omega=\\{x\in\mathbb{R}^{N}:\;0<|x|<\rho\\}$ and
$0<|x_{0}|<\rho/2$, or $x_{0}\in\Omega=\\{x\in\mathbb{R}^{N}:\;|x|>\rho\\}$
and $|x_{0}|>2\rho$. We denote $R=\frac{1}{2}|x_{0}|$ and observe that, for
all $y\in B_{1}$, $\frac{|x_{0}|}{2}<|x_{0}+Ry|<\frac{3|x_{0}|}{2}$, so that
$x_{0}+Ry\in\Omega$. Let us define

$U(y)=R^{\frac{4+\tau}{p-1}}u(x_{0}+Ry),\;\;\;V(y)=R^{\frac{2(p+1)+\tau}{p-1}-\alpha}v(x_{0}+Ry).$

Then $(U,V)$ satisfies the system

(4.5) $\left\\{\begin{array}[]{ll}-\Delta
U=|y+\frac{x_{0}}{R}|^{-\alpha}V&\mbox{in $B_{1}$},\\\ -\Delta
V=|y+\frac{x_{0}}{R}|^{l}U^{p}&\mbox{in $B_{1}$}.\end{array}\right.$

Note that $|y+\frac{x_{0}}{R}|\in[1,3]$ for all $y\in{\overline{B_{1}}}$.
Using the fact $1<p<\min\\{\frac{N+4}{N-4},p_{s}\\}$ and the blow-up argument
as in the proof of the first step of Lemma 3.2, we obtain that there exists
$C>0$ depending only on $N,p,\alpha,l$, independent of $x_{0}$, such that

(4.6) $U^{\frac{p-1}{4}}(y)+|\nabla
U(y)|^{\frac{p-1}{p+3}}+|V(y)|^{\frac{p-1}{2(p+1)}}+|\nabla
V(y)|^{\frac{p-1}{3p+1}}\leq C(1+\mbox{dist}^{-1}(y,\partial B_{1})),\;\;y\in
B_{1}.$

This implies that $U(0)+|\nabla U(0)|+|V(0)|+|\nabla V(0)|\leq C$. This
implies that our conclusions in (4.1), (4.2), (4.3) and (4.4) hold. ∎

Proof of Theorem 1.1

We obtain from Lemma 5.1 in Appendix that, if $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$,
$v(x):=-|x|^{\alpha}\Delta u\in C^{0}(\mathbb{R}^{N})$, is a nonnegative
solution to (P), the following Rellich-Pohozaev identity holds:

$\displaystyle\Bigl{[}\frac{N^{\prime}+\tau}{p+1}-\frac{N^{\prime}-4}{2}\Bigr{]}\int_{B_{R}}|x|^{l}u^{p+1}dx=\int_{\partial
B_{R}}\bigg{[}R^{l+1}\frac{u^{p+1}}{p+1}$
$\displaystyle+R^{1-\alpha}\frac{v^{2}}{2}+2Ru^{\prime}v^{\prime}-R\nabla
u\cdot\nabla
v+\frac{N-\alpha}{2}vu^{\prime}+\frac{N^{\prime}-4}{2}uv^{\prime}\bigg{]}d\sigma_{R}$

for all $R>0$. Define

(4.7) $F(R)=\int_{B_{R}}|x|^{l}u^{p+1}dx.$

By Rellich-Pohozaev identity, we have

(4.8) $F(R)\leq C\left(G_{1}(R)+G_{2}(R)\right),$

where

(4.9) $G_{1}(R)=R^{N^{\prime}+\tau}\int_{S^{N-1}}u^{p+1}(R,\theta)d\theta$

and

(4.10) $\displaystyle G_{2}(R)=$ $\displaystyle
R^{N^{\prime}}\int_{S^{N-1}}\bigg{[}(\Delta_{x}u(R,\theta))^{2}+R^{-\alpha}\left|\nabla
u(R,\theta)\right|\left|\nabla v(R,\theta)\right|$
$\displaystyle+R^{-(1+\alpha)}\Big{(}|v(R,\theta)||\nabla
u(R,\theta)|+u(R,\theta)|\nabla v(R,\theta)|\Big{)}\bigg{]}d\theta.$

Now, by (4.1)-(4.4) in Lemma 4.1, we have

(4.11) $u(x)\leq C|x|^{-\frac{4+\tau}{p-1}}\;\;\mbox{and}\;\;|\Delta u(x)|\leq
C|x|^{-\frac{2p+2+\tau}{p-1}},\;\;\;x\neq 0,$ (4.12) $|\nabla u(x)|\leq
C|x|^{-\frac{p+\tau+3}{p-1}},\;\;\;|\nabla v(x)|\leq
C|x|^{\alpha-1-\frac{2(p+1)+\tau}{p-1}},\;\;\;x\neq 0,$ (4.13) $|v(x)|\leq
C|x|^{\alpha-\frac{2(p+1)+\tau}{p-1}},\;\;\;x\neq 0.$

Due to $p<p_{s}$, it follows that

$G_{1}(R)+G_{2}(R)\leq
CR^{\frac{(N^{\prime}-4)p-(N^{\prime}+4+2\tau)}{p-1}}\longrightarrow
0,\qquad\quad\mbox{as}\ R\to\infty.$

Therefore, $u\equiv 0$ by (4.8). This completes the proof of Theorem 1.1. ∎

Proof of Theorem 1.2

A little variant of the proof of Lemma 4.3 below implies that

$v(r)\geq 0\;\;\;\mbox{for $r>0$}$

provided that $u(r)\geq 0$ for $r>0$. The strong maximum principle implies
that $u(r)>0$ and $v(r)>0$ for $r\in[0,\infty)$ provided that $u(r)$ is
nontrivial.

We first show that

(4.14) $ru^{\prime}(r)+(N-2)u(r)\geq 0,\;\;\;rv^{\prime}(r)+(N-2)v(r)\geq
0,\;\;\;\forall r>0.$

Since $\Delta u(r)\leq 0$ and $\Delta v(r)\leq 0$, we see that

(4.15) $(ru^{\prime}(r)+(N-2)u(r))^{\prime}\leq
0,\;\;\;(rv^{\prime}(r)+(N-2)v(r))^{\prime}\leq 0,\;\;\;\forall r>0.$

Suppose that there exists $r_{0}>0$ such that
$M_{0}:=r_{0}u^{\prime}(r_{0})+(N-2)u(r_{0})<0$, we obtain from
$\eqref{a-4.13}_{1}$ that

(4.16) $u^{\prime}(r)\leq r^{-1}[ru^{\prime}(r)+(N-2)u(r)]\leq
M_{0}r^{-1}<0\;\;\;\forall r\geq r_{0}.$

Integrating (4.16) in $(r_{0},r)$ and sending $r$ to $\infty$, we derive a
contradiction. The proof of $\eqref{a-4.12}_{2}$ is similar.

We now claim that

(4.17) $u(r)\leq Cr^{-\frac{4+\tau}{p-1}},\;\;\;v(r)\leq
Cr^{\alpha-\frac{2(p+1)+\tau}{p-1}},\;\;\;\forall r>0,$ (4.18)
$|u(r)v^{\prime}(r)|+|u^{\prime}(r)v(r)|+|u^{\prime}(r)v^{\prime}(r)|\leq
Cr^{\alpha-1-\frac{2(p+3+\tau)}{p-1}},\;\;\;\forall r>0,$

where $C$ is a positive constant depending only on $N,p,\alpha,l$.

Since $(u(r),v(r))$ satisfies the equations:

(4.19)
$-(r^{N-1}u^{\prime}(r))^{\prime}=r^{N-\alpha-1}v(r),\;\;\;-(r^{N-1}v^{\prime}(r))^{\prime}=r^{N+l-1}u^{p}(r),$

with $ru^{\prime}(r)\to 0$, $rv^{\prime}(r)\to 0$ as $r\to 0^{+}$ (we know
that $u$ and $v$ are continuous at $r=0$), we obtain that

(4.20)
$-r^{N-1}u^{\prime}(r)\geq\frac{r^{N-\alpha}}{N-\alpha}v(r),\;\;\;-r^{N-1}v^{\prime}(r)\geq\frac{r^{N+l}}{N+l}u^{p}(r),\;\;\;\forall
r>0,$

by integrating both the equations in (4.19) in $(0,r)$ and using the facts
$u^{\prime}(r)<0$ and $v^{\prime}(r)<0$. Using (4.14), we have

(4.21) $0\geq-
ru^{\prime}(r)-(N-2)u(r)\geq\frac{r^{2-\alpha}}{N-\alpha}v(r)-(N-2)u(r),$
(4.22) $0\geq-
rv^{\prime}(r)-(N-2)v(r)\geq\frac{r^{2+l}}{N+l}u^{p}(r)-(N-2)v(r).$

These two inequalities imply that the two inequalities in (4.17) hold. Using
the two inequalities in (4.14) and (4.17), we obtain (4.18). Now the proof is
finished by using Rellich-Pohozaev identity given in Lemma 5.1 and the same
arguments as those in the proof of Theorem 1.1. ∎

To prove Theorem 1.3, we first obtain the following proposition.

######  Proposition 4.2.

Assume that $N\geq 5$, $\alpha=2$, $l>-2$ and $1<p<\frac{N+2+2l}{N-2}$. Let
$u\in C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$ and
$|x|^{2}\Delta u\in C^{0}(\mathbb{R}^{N})$ be a positive solution of (P) in
$\mathbb{R}^{N}$. Then $|x|^{\frac{N-2}{2}}u(x)$ and $|x|^{\frac{N-2}{2}}v(x)$
are strictly increasing in the radius $|x|$, where $v(x)=-|x|^{2}\Delta u(x)$.

To prove this proposition, we first show the following lemma.

######  Lemma 4.3.

Assume that $N\geq 5$ and (1.1) holds with $1<p<p_{s}$. Assume also that $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$ satisfying
$|x|^{\alpha}\Delta u\in C^{0}(\mathbb{R}^{N})$ is a solution of (P) in
$\mathbb{R}^{N}\backslash\\{0\\}$ with $u(x)>0$ for $x\in\mathbb{R}^{N}$. Then
$v(x):=-|x|^{\alpha}\Delta u(x)>0$ for $x\in\mathbb{R}^{N}$.

###### Proof.

The proof of this lemma is quite elementary. We see that $(u,v)$ satisfies the
system

(4.23) $\left\\{\begin{array}[]{ll}-\Delta u=|x|^{-\alpha}v&\mbox{in
$\mathbb{R}^{N}$},\\\ -\Delta v=|x|^{l}u^{p}&\mbox{in
$\mathbb{R}^{N}$}.\end{array}\right.$

We consider three cases for $\alpha$: (a) $\alpha=2$, (b) $4-N<\alpha<2$, (c)
$2<\alpha<N$.

We give the proof of case (a). The proofs of case (b) and (c) are similar to
that of case (a).

Step 1. We show that $v(0)>0$.

The main idea of the proof is similar to that of Theorem 3.1 of [29]. Suppose
not, $v(0)\leq 0$. We introduce the average of a function

${\overline{f}}(r)=\frac{1}{|\partial B_{r}(0)|}\int_{\partial
B_{r}(0)}fd\sigma.$

Then we have by Jensen’s inequality

(4.24)
$\left\\{\begin{array}[]{l}\Delta{\overline{u}}+r^{-2}{\overline{v}}=0,\\\
\Delta{\overline{v}}+r^{l}{\overline{u}}^{p}\leq 0,\end{array}\right.$

where $r=|x|>0$. Since ${\overline{v}}(0)=v(0)\leq 0$ and
${\overline{v}}^{\prime}(r)<0$ for $r>0$, we have

(4.25) ${\overline{v}}(r)<{\overline{v}}(0)\leq 0\;\;\mbox{for all $r>0$}.$

Hence

$\Delta{\overline{u}}>0\;\;\;\mbox{for $r>0$}.$

This implies that

(4.26) ${\overline{u}}^{\prime}(r)>0\;\;\mbox{for $r>0$}.$

On the other hand, we see from (4.25) that for a fixed $\kappa_{0}>0$,

${\overline{v}}(r)<{\overline{v}}(\kappa_{0})<0\;\;\mbox{for $r>\kappa_{0}$}.$

Then

$\Delta{\overline{u}}\geq r^{-2}(-{\overline{v}}(\kappa_{0}))\;\;\mbox{for
$r\geq\kappa_{0}$}$

and

$r^{N-1}{\overline{u}}^{\prime}(r)-\kappa_{0}^{N-1}{\overline{u}}^{\prime}(\kappa_{0})\geq\frac{(-{\overline{v}}(\kappa_{0}))}{N-2}\Big{[}r^{N-2}-\kappa_{0}^{N-2}\Big{]}.$

It follows from (4.26) that

$r^{N-1}{\overline{u}}^{\prime}(r)\geq\frac{(-{\overline{v}}(\kappa_{0}))}{N-2}\Big{[}r^{N-2}-\kappa_{0}^{N-2}\Big{]}$

and

${\overline{u}}(r)\geq\frac{(-{\overline{v}}(\kappa_{0}))}{N-2}\ln\frac{r}{\kappa_{0}}-\frac{(-{\overline{v}}(\kappa_{0}))}{(N-2)^{2}}\geq
C_{0}>1\;\;\mbox{for $r\geq r_{0}>\kappa_{0}>0$},$

where $r_{0}>\kappa_{0}$ is a suitably large constant.

Suppose now that

${\overline{u}}(r)\geq\frac{(C_{0})^{p^{k}}}{A^{b_{k}}}r^{\sigma_{k}}\;\;\mbox{for
$r\geq r_{k}>r_{0}$},$

where

$A=[N+l+2pM]^{\frac{8}{2+l}},\;\;\sigma_{0}=b_{0}=0$

and $M>1$, $\sigma_{k}\geq 0$, $b_{k}\geq 0$ for $k\geq 1$ are determined
later. We consider two cases: (i) $l+p\sigma_{k}\neq-1$ for all
$k=0,1,\ldots$, (ii) there is an integer $k_{0}\geq 0$ such that
$l+p\sigma_{k_{0}}=-1$.

For the case (i), we have

$\begin{split}r^{N-1}{\overline{v}}^{\prime}(r)&\leq
r_{k}^{N-1}{\overline{v}}^{\prime}(r_{k})-\int_{r_{k}}^{r}s^{N+l-1}{\overline{u}}^{p}(s)ds\\\
&\leq-\frac{(C_{0})^{p^{k+1}}}{A^{b_{k}p}(N+l+p\sigma_{k})}\Big{[}r^{N+l+p\sigma_{k}}-r_{k}^{N+l+p\sigma_{k}}\Big{]}\\\
&\leq-\frac{(C_{0})^{p^{k+1}}}{2A^{b_{k}p}(N+l+p\sigma_{k})}r^{N+l+p\sigma_{k}}.\end{split}$

Hence

${\overline{v}}^{\prime}(r)\leq-\frac{(C_{0})^{p^{k+1}}}{2A^{b_{k}p}(N+l+p\sigma_{k})}r^{1+l+p\sigma_{k}}$

for $r\geq 2^{\frac{1}{N+l+p\sigma_{k}}}r_{k}$. Similarly

${\overline{v}}(r)\leq-\frac{(C_{0})^{p^{k+1}}}{4A^{b_{k}p}(N+l+p\sigma_{k})(2+l+p\sigma_{k})}r^{2+l+p\sigma_{k}}$

for $r\geq 2^{\frac{1}{2+l+p\sigma_{k}}}2^{\frac{1}{N+l+p\sigma_{k}}}r_{k}$.
Hence

${\overline{v}}(r)\leq-\frac{(C_{0})^{p^{k+1}}}{A^{b_{k}p}4(N+l+p\sigma_{k})^{2}}r^{2+l+p\sigma_{k}}.$

Note that

$2+l+p\sigma_{k}>0$

since $l>-2$. Therefore,

${\overline{u}}(r)\geq\frac{(C_{0})^{p^{k+1}}}{A^{b_{k}p}16(N+l+p\sigma_{k})^{4}}r^{2+l+p\sigma_{k}}$

for $r\geq 2^{\frac{4}{2+l+p\sigma_{k}}}r_{k}>r_{k}$.

Set

$\begin{array}[]{l}\sigma_{0}=0,\\\ \sigma_{k+1}=2+l+p\sigma_{k},\\\
r_{k+1}=2^{\frac{4}{2+l+p\sigma_{k}}}r_{k}.\end{array}$

First of all, by mathematical induction, it is easy to see that

$2^{4}(N+l+p\sigma_{k})^{4}\leq A^{(2+l)(k+1)}$

if we choose $M>1$ sufficiently large. Note that

(4.27) $\sigma_{k}=\frac{2+l}{p-1}\Big{(}p^{k}-1\Big{)}.$

Hence we can set

$b_{0}=0,\;\;\;b_{k+1}=pb_{k}+(2+l)(k+1).$

Then we have

(4.28)
${\overline{u}}(r)\geq\frac{(C_{0})^{p^{k+1}}}{A^{b_{k+1}}}r^{\sigma_{k+1}}\;\;\mbox{for
$r\geq r_{k+1}$}.$

Notice that

(4.29) $r_{k+1}\leq cr_{0},$

where $c$ can be chosen to be
$2^{\Sigma_{k=0}^{\infty}\frac{4}{2+l+p\sigma_{k}}}$. Also notice that, by
using the iteration formulas above, we have

$b_{k}=\frac{(2+l)}{(p-1)^{2}}\Big{[}p^{k+1}-(k+1)p+k\Big{]}.$

Hence, if we take $M>1$ large enough so that

$A^{\frac{p}{p-1}}>2cr_{0}$

and then take ${\tilde{r}}_{0}=A^{\frac{p}{p-1}}$, we see

${\overline{u}}({\tilde{r}}_{0})\geq(C_{0})^{p^{k+1}}A^{\frac{(2+l)}{p-1}(k+1)}\to\infty\;\;\mbox{as
$k\to\infty$}.$

Since ${\tilde{r}}_{0}$ is independent of $k$, a contradiction is reached.

For the case (ii), we see that $l+p\sigma_{k}\neq-1$ for $k\neq k_{0}$. We
only need to deal with $k=k_{0}$, since $1+l+p\sigma_{k_{0}}=0$. Arguments
similar to those in the case (i) imply that

$r^{N-1}{\overline{v}}^{\prime}(r)\leq-\frac{C_{0}^{p^{k_{0}+1}}}{2(N-1)A^{b_{k_{0}}p}}r^{N-1}\;\;\mbox{for
$r\geq 2^{\frac{1}{N-1}}r_{k_{0}}$}.$

Similarly,

${\overline{v}}(r)\leq-\frac{(C_{0})^{p^{k_{0}+1}}}{4(N-1)A^{b_{k_{0}}p}}r\;\;\mbox{for
$r\geq 2\times 2^{\frac{1}{N-1}}r_{k_{0}}$},$
${\overline{u}}(r)\geq\frac{(C_{0})^{p^{k_{0}+1}}}{2^{4}(N-1)^{2}A^{b_{k_{0}}p}}r\;\;\mbox{for
$r\geq 2^{2}\times 2^{\frac{2}{N-1}}r_{k_{0}}$}.$

This implies that

${\overline{u}}(r)\geq\frac{(C_{0})^{p^{k_{0}+1}}}{2^{4}(N-1)^{4}A^{b_{k_{0}}p}}r\;\;\mbox{for
$r\geq 2^{4}\times r_{k_{0}}$}.$

Now, we only need to make the change $r_{k_{0}+1}=2^{4}\times r_{k_{0}}$ and
choose the constant $c$ in (4.29) to be $2^{\Sigma_{k\neq
k_{0}}\frac{4}{2+l+p\sigma_{k}}+4}$, we can derive a contradiction by
arguments exactly the same as those in the case (i). Note that
$N+l+p\sigma_{k_{0}}=N-1$, $2+l+p\sigma_{k_{0}}=1$. This completes the proof
of Step 1.

Step 2. We claim

(4.30) $v(x)>0\;\;\mbox{for $x\in\mathbb{R}^{N}$}.$

Suppose not, there is $x_{0}\neq 0$ such that $v(x_{0})\leq 0$. Set

$w(y)=u(x),\;\;z(y)=v(x),\;\;y=x-x_{0}.$

We have that $(w,z)$ satisfies the system

(4.31) $\left\\{\begin{array}[]{ll}\Delta w+|y+x_{0}|^{-2}z=0&\mbox{in
$\mathbb{R}^{N}$},\\\ \Delta z+|y+x_{0}|^{l}w^{p}=0&\mbox{in
$\mathbb{R}^{N}$},\end{array}\right.$

and

(4.32)
$\left\\{\begin{array}[]{l}\Delta{\overline{w}}+{\overline{|y+x_{0}|^{-2}z}}=0,\\\
\Delta{\overline{z}}+{\overline{|y+x_{0}|^{l}w^{p}}}=0.\end{array}\right.$

It follows from the second equation of (4.32) that

(4.33)
${\overline{z}}^{\prime}(\rho)<0,\;\;{\overline{z}}(\rho)<{\overline{z}}(0)\leq
0\;\;\mbox{for $\rho>0$},$

where $\rho=|y|$. These imply that there is $-\infty\leq\vartheta<0$ such that

${\overline{z}}(\rho)\to\vartheta\;\;\mbox{as $\rho\to\infty$}.$

For $\rho>10|x_{0}|$ and sufficiently large, it is known from the first
equation of (4.32) that

(4.34)
$-\Delta{\overline{w}}=\rho^{-2}{\overline{\Big{|}\frac{y}{\rho}+\frac{x_{0}}{\rho}\Big{|}^{-2}z}}.$

Since $\Big{|}\frac{y}{\rho}+\frac{x_{0}}{\rho}\Big{|}\to 1$ as
$\rho\to\infty$, we see that

${\overline{\Big{|}\frac{y}{\rho}+\frac{x_{0}}{\rho}\Big{|}^{-2}z}}(\rho)\to\vartheta\;\;\mbox{as
$\rho\to\infty$}.$

This implies that there are
$\vartheta<{\tilde{\vartheta}}<\frac{1}{2}\vartheta$ and
${\overline{\kappa}}_{0}>0$ such that

(4.35) $\Delta{\overline{w}}\geq(-{\tilde{\vartheta}})\rho^{-2}\;\;\mbox{for
$\rho\geq{\overline{\kappa}}_{0}$}.$

Arguments similar to those in the proof of step 1 imply that we can reach a
contradiction. Note that it is seen from the second equation of (4.32) that

(4.36)
$-\Delta{\overline{z}}=\rho^{l}{\overline{\Big{|}\frac{y}{\rho}+\frac{x_{0}}{\rho}\Big{|}^{l}w^{p}}}\geq\frac{1}{2}\rho^{l}{\overline{w}}^{p}$

for $\rho\geq{\overline{\kappa}}_{0}$, if we choose ${\overline{\kappa}}_{0}$
large enough. By arguments exactly same as those in the proof of step 1 and by
choosing $A$ as in the proof of step 1 such that

$2^{5}(N+l+p\sigma_{k})^{4}\leq A^{(2+l)(k+1)},$

we can reach a contradiction. This implies that $v(x_{0})\leq 0$ does not
exist and our claim (4.30) holds. This completes the proof of case (a).

For the case (b), arguments similar to those in the proof of the case (a)
implies that

${\overline{u}}(r)\geq Cr^{2-\alpha}\geq C_{0}>1\;\;\mbox{for $r\geq
r_{0}>0$},$

where $r_{0}>0$ is a suitably large constant. By arguments similar to those in
the proof of the case (a), we obtain that

${\overline{u}}(r)\geq\frac{C_{0}^{p^{k}}}{A^{b_{k}}}r^{\sigma_{k}}\;\;\mbox{for
$r\geq r_{k}>r_{0}$},$

where

$A=[N+2+\tau+2pM]^{\frac{8}{4+\tau}},\;\;\sigma_{0}=b_{0}=0$

and $M>1$ with

$\begin{array}[]{l}\sigma_{0}=0,\\\ \sigma_{k+1}=4+\tau+p\sigma_{k},\\\
r_{k+1}=2^{\frac{4}{3+\tau+p\sigma_{k}}}r_{k}.\end{array}$

Therefore, we have

$\sigma_{k}=\frac{4+\tau}{p-1}\Big{(}p^{k}-1\Big{)}.$

(Note that $4+\tau>0$.) Hence we can set

$b_{0}=0,\;\;\;b_{k+1}=pb_{k}+(4+\tau)(k+1)$

and

$b_{k}=\frac{(4+\tau)}{(p-1)^{2}}\Big{[}p^{k+1}-(k+1)p+k\Big{]}.$

Moreover, we also need to consider $3+\tau+p\sigma_{k}=0$ or not. The main
idea is similar to that of the proof of the case (a). This completes the proof
of the case (b).

The proof of the case (c) is exactly same as that of the case (b). Note that
for $\alpha>2$, we see that $2+l>0$ since $l>\alpha-4$. Arguments similar to
those in the proof of case (a) imply

(4.37) ${\overline{u}}(r)\geq
C(-{\overline{v}}(\kappa_{0}))\kappa_{0}^{2-\alpha}\;\;\mbox{for $r\geq
r_{0}>\kappa_{0}>0$},$

where $r_{0}>\kappa_{0}>0$ is a suitably large constant. By a simple
calculation, we easily know that $-{\overline{v}}(\kappa_{0})\geq
C\kappa_{0}^{2+l}$ if we choose a sufficiently large $\kappa_{0}>0$ (note that
$2+l>0$). This and (4.37) imply that

(4.38) ${\overline{u}}(r)\geq C\kappa_{0}^{4+\tau}\geq C_{0}>1\;\;\mbox{for
$r\geq r_{0}>\kappa_{0}>0$}.$

Choosing $A$, $\sigma_{k}$, $b_{k}$ exactly the same as that in the case (b)
and using the same arguments, we complete the proof of the case (c). ∎

Proof of Proposition 4.2

We introduce the following transform

$w(t,\omega)=r^{\frac{N-2}{2}}u(r,\omega),\;\;\;z(t,\omega)=r^{\frac{N-2}{2}}v(r,\omega),\;\;t=\ln
r$

with $r=|x|$. Then $(w,z)$ satisfies the system of equations

(4.39)
$\left\\{\begin{array}[]{ll}w_{tt}+\Delta_{S^{N-1}}w-\frac{(N-2)^{2}}{4}w+z=0,&(t,\omega)\in(-\infty,\infty)\times
S^{N-1},\\\
z_{tt}+\Delta_{S^{N-1}}z-\frac{(N-2)^{2}}{4}z+e^{p^{*}t}w^{p}=0,&(t,\omega)\in(-\infty,\infty)\times
S^{N-1},\end{array}\right.$

where

$p^{*}=\frac{1}{2}\Big{[}(N+2+2l)-(N-2)p\Big{]}.$

Note that $p^{*}>0$ for
$1<p<\frac{N+2+2l}{N-2}\;(=\frac{N^{\prime}+4+2\tau}{N^{\prime}-4}:=p_{s})$.
Moreover, we have

(4.40)
$\lim_{t\to-\infty}e^{-\frac{N-2}{2}t}w(t,\omega)=u(0),\;\;\lim_{t\to-\infty}e^{-\frac{N-2}{2}t}z(t,\omega)=v(0)$

uniformly on $S^{N-1}$. Note that $0<u(0)<\infty$, $0<v(0)<\infty$ since $u$
and $v$ are continuous at $x=0$.

Let

$\Sigma=\\{(t,\omega):\;-\infty<t<\infty,\;\omega\in S^{N-1}\\}.$

For $T\in\mathbb{R}$ we define

$\Sigma_{T}=(-\infty,T)\times S^{N-1},\;\;S_{T}=\\{T\\}\times S^{N-1}.$

Further, we let $(t,\omega)_{T}$ be the reflection of $(t,\omega)$ with
respect to $\\{T\\}\times S^{N-1}$, namely,

$(t,\omega)_{T}=(2T-t,\omega).$

Then the functions

$w_{T}(t,\omega)=w(2T-t,\omega),\qquad z_{T}(t,\omega)=z(2T-t,\omega),$

and

${\tilde{w}}_{T}=w-w_{T},\qquad{\tilde{z}}_{T}=z-z_{T}$

are well defined in $\Sigma_{T}$ for $T>-\infty$. Moreover,
$({\tilde{w}}_{T},{\tilde{z}}_{T})$ satisfies the following system in
$\Sigma_{T}$

(4.41)
$\left\\{\begin{array}[]{l}({\tilde{w}}_{T})_{tt}+\Delta_{S^{N-1}}{\tilde{w}}_{T}-\frac{(N-2)^{2}}{4}{\tilde{w}}_{T}+{\tilde{z}}_{T}=0,\\\
({\tilde{z}}_{T})_{tt}+\Delta_{S^{N-1}}{\tilde{z}}_{T}-\frac{(N-2)^{2}}{4}{\tilde{z}}_{T}+e^{p^{*}t}w^{p}-e^{p^{*}(2T-t)}(w_{T})^{p}=0.\end{array}\right.$

For fixed $\omega\in S^{N-1}$, we have that if $1<p<p_{s}$,

$e^{p_{*}(2T-t)}(w_{T})^{p}\geq e^{p_{*}t}(w_{T})^{p}\;\;\mbox{for $t\leq
T$}.$

In turn, we have

(4.42)
$\left\\{\begin{array}[]{l}({\tilde{w}}_{T})_{tt}+\Delta_{S^{N-1}}{\tilde{w}}_{T}-\frac{(N-2)^{2}}{4}{\tilde{w}}_{T}+{\tilde{z}}_{T}=0,\\\
({\tilde{z}}_{T})_{tt}+\Delta_{S^{N-1}}{\tilde{z}}_{T}-\frac{(N-2)^{2}}{4}{\tilde{z}}_{T}+e^{p^{*}t}[w^{p}-(w_{T})^{p}]_{+}\geq
0\end{array}\right.$

in $\Sigma_{T}$, where we write $[h(t,\omega)]_{+}=\max\\{0,h(t,\omega)\\}$.
Therefore, $({\tilde{w}}_{T},{\tilde{z}}_{T})$ satisfies

(4.43)
$\left\\{\begin{array}[]{ll}({\tilde{w}}_{T})_{tt}+\Delta_{S^{N-1}}{\tilde{w}}_{T}-\frac{(N-2)^{2}}{4}{\tilde{w}}_{T}+{\tilde{z}}_{T}=0,&\mbox{in
$\Sigma_{T}$},\\\
({\tilde{z}}_{T})_{tt}+\Delta_{S^{N-1}}{\tilde{z}}_{T}-\frac{(N-2)^{2}}{4}{\tilde{z}}_{T}+c(t,\omega){\tilde{w}}_{T}\geq
0,&\mbox{in $\Sigma_{T}$},\\\ {\tilde{w}}_{T}\leq 0,\;\;{\tilde{z}}_{T}\leq
0,&\mbox{on $\partial\Sigma_{T}$},\end{array}\right.$

where

$c(t,\omega)=e^{p_{*}t}\frac{[w^{p}-(w_{T})^{p}]_{+}}{w-(w_{T})}.$

The fact that $({\tilde{w}}_{T},{\tilde{z}}_{T})\leq 0$ on
$\partial\Sigma_{T}$ can be obtained from $(w(t,\omega),z(t,\omega))\to(0,0)$
as $t\to-\infty$ uniformly for $\omega\in S^{N-1}$ and
$(w(t,\omega),z(t,\omega))>0$ for $(t,\omega)\in(-\infty,\infty)\times
S^{N-1}$. Note first that by (4.40) the functions $e^{-\frac{N-2}{2}t}w$ and
$e^{-\frac{N-2}{2}t}z$ are bounded in $\Sigma_{T}$ uniformly in
$T<{\overline{T}}$, where ${\overline{T}}$ is a sufficiently negative number.
We see that there exists $L=L({\overline{T}})>0$ such that, if we define
$\Sigma_{T}^{+}=\\{(t,\omega)\in\Sigma_{T}:{\tilde{w}}_{T}(t,\omega)>0\\}$,

(4.44) $0\leq c(t,\omega)\leq pLe^{(2+l)t}\;\;\mbox{on $\Sigma_{T}^{+}$ and
$T<{\overline{T}}$}.$

Therefore, we can choose a sufficiently negative number ${\hat{T}}\ll-1$ such
that

$c(t,\omega)<\frac{(N-2)^{2}}{4}$

uniformly for $(t,\omega)\in\Sigma_{T}^{+}$ and $T\leq{\hat{T}}$. This implies
that

(4.45) $c(t,\omega)-\frac{(N-2)^{2}}{4}<0\;\;\mbox{for
$(t,\omega)\in\Sigma_{T}$ and $T\leq{\hat{T}}$}.$

By (4.40), one has

$\lim_{t\to-\infty}\sup{\tilde{w}}_{T}(t,\omega)\leq
0,\;\;\;\lim_{t\to-\infty}\sup{\tilde{z}}_{T}(t,\omega)\leq 0,$

and

${\tilde{w}}_{T}\equiv 0,\;\;\;{\tilde{z}}_{T}\equiv 0\;\;\mbox{on $S_{T}$}.$

By the maximum principle of the cooperative systems (see Lemma 11 of [28]), in
view of (4.43) and (4.45), it follows that $({\tilde{w}}_{T},{\tilde{z}}_{T})$
must be non-positive, provided $T\leq{\hat{T}}$. Note that
$1-\frac{(N-2)^{2}}{4}<0$. Clearly

(4.46) $\frac{\partial{\tilde{w}}_{T}}{\partial t}(T,\omega)=2\frac{\partial
w}{\partial t}(T,\omega)\geq 0,\;\;\frac{\partial{\tilde{z}}_{T}}{\partial
t}(T,\omega)=2\frac{\partial z}{\partial t}(T,\omega)\geq
0,\;\;\;T\leq{\hat{T}}.$

Moreover, for any $T<{\hat{T}}$, either

(4.47) ${\tilde{w}}_{T}(t,\omega)<0\;\;\mbox{in
$\Sigma_{T}$},\;\;\;\frac{\partial{\tilde{w}}_{T}}{\partial
t}(t,\omega)>0\;\;\mbox{on $S_{T}$},$

and

(4.48) ${\tilde{z}}_{T}(t,\omega)<0\;\;\mbox{in
$\Sigma_{T}$},\;\;\;\frac{\partial{\tilde{z}}_{T}}{\partial
t}(t,\omega)>0\;\;\mbox{on $S_{T}$},$

by the strong maximum principle and the boundary lemma or one of two cases
occurs: ${\tilde{w}}_{T}\equiv 0,{\tilde{z}}_{T}\equiv 0$ in
$\Sigma_{\hat{T}}$. One readily sees that (4.47) and (4.48) hold for $T<0$
negative enough. For otherwise, one deduces that $w\equiv c_{1}$ or $z\equiv
c_{2}$ for some constants $c_{1}$ and $c_{2}$ and $T<0$ sufficiently negative,
thanks to the monotonicity (4.46). This is clearly impossible. Hence (4.47)
and (4.48) hold for sufficiently negative $T$.

Define

(4.49) $T_{0}=\sup\\{T_{1}>-\infty:\;\mbox{both \eqref{4.21} and \eqref{4.22}
hold for $T\leq T_{1}$}\\}.$

Arguments as the above imply that $T_{0}$ is well defined.

Obviously we only need to show $T_{0}=\infty$. Suppose for contradiction that
$T_{0}<\infty$. We want to show that

(4.50) ${\tilde{w}}_{T_{0}}(t,\omega)\equiv
0\;\;\mbox{or}\;\;{\tilde{z}}_{T_{0}}(t,\omega)\equiv 0,\;\;\mbox{for
$(t,\omega)\in\Sigma_{T_{0}}$}$

and for all $T\in(-\infty,T_{0})$,

(4.51) ${\tilde{w}}_{T}(t,\omega)<0\;\;\mbox{for
$(t,\omega)\in\Sigma_{T}$},\;\;\frac{\partial{\tilde{w}}_{T}}{\partial
t}(t,\omega)>0\;\;\mbox{for $(t,\omega)\in S_{T}$},$

and

(4.52) ${\tilde{z}}_{T}(t,\omega)<0\;\;\mbox{for
$(t,\omega)\in\Sigma_{T}$},\;\;\frac{\partial{\tilde{z}}_{T}}{\partial
t}(t,\omega)>0\;\;\mbox{for $(t,\omega)\in S_{T}$}.$

It is easily seen that (4.50)-(4.52) are impossible since the equation of
$(w,z)$ contains the term $e^{p_{*}t}$, which is strictly increasing with
respect to $t$. This implies that $T_{0}=\infty$, if we have shown
(4.50)-(4.52).

We also use contradiction arguments to show (4.50)-(4.52). Suppose that (4.50)
does not hold. Then by the strong maximum principle and the boundary lemma,

(4.53) ${\tilde{w}}_{T_{0}}<0\;\;\mbox{in
$\Sigma_{T_{0}}$},\;\;\frac{\partial{\tilde{w}}_{T_{0}}}{\partial
t}>0\;\;\mbox{on $S_{T_{0}}$},$

and

(4.54) ${\tilde{z}}_{T_{0}}<0\;\;\mbox{in
$\Sigma_{T_{0}}$},\;\;\frac{\partial{\tilde{z}}_{T_{0}}}{\partial
t}>0\;\;\mbox{on $S_{T_{0}}$}.$

These and the definition of $T_{0}$ imply that

$\frac{\partial w}{\partial t}(t,\omega)>0,\;\;\frac{\partial z}{\partial
t}(t,\omega)>0,\;\;\mbox{for $(t,\omega)\in(-\infty,T_{0}]\times S^{N-1}$}.$

In particular, by the compactness of $S^{N-1}$ and continuity, there exists
$\epsilon_{0}>0$ such that

$\frac{\partial w}{\partial t}(t,\omega)>0,\;\;\frac{\partial z}{\partial
t}(t,\omega)>0,\;\;\mbox{for $(t,\omega)\in(-\infty,T_{0}+\epsilon_{0}]\times
S^{N-1}$}.$

Next we choose from (4.44) a sufficiently negative value $T_{1}<T_{0}$ such
that for all $T\leq T_{0}+\epsilon_{0}$, we have

$c(t,\omega)-\frac{(N-2)^{2}}{4}<0\;\;\mbox{in
$\Sigma_{T}^{+}\cap\\{t<T_{1}\\}$}.$

Furthermore, thanks to the fact that $S^{N-1}$ has no boundary and by
continuity again, there exists $0<\epsilon_{1}<\epsilon_{0}$ such that

${\tilde{w}}_{T}(t,\omega)<0,\;\;{\tilde{z}}_{T}(t,\omega)<0,\;\;\mbox{in
$\Sigma_{T}\cap\\{T_{1}\leq t<T\\}$ for all $T\leq T_{0}+\epsilon_{1}$}.$

In turn, it follows that

$c(t,\omega)\equiv 0\;\;\mbox{in $\Sigma_{T}\cap\\{T_{1}\leq t<T\\}$ for all
$T\leq T_{0}+\epsilon_{1}$}.$

In particular, for all $T\leq T_{0}+\epsilon_{1}$, (4.43) holds with

$c(t,\omega)-\frac{(N-2)^{2}}{4}<0\;\;\mbox{in $\Sigma_{T}$}.$

Thus the maximum principle implies again that
$({\tilde{w}}_{T},{\tilde{z}}_{T})$ can not have a positive maximum in
$\Sigma_{T}$, provided $T\leq T_{0}+\epsilon_{1}$. Hence both (4.51) and
(4.52) hold for $T<T_{0}+\epsilon_{1}$ by the strong maximum principle and the
boundary lemma. This is again a contradiction to the definition of $T_{0}$ and
therefore (4.50) must hold. Clearly (4.51) and (4.52) are a direct consequence
of (4.50) and this finishes the proof. ∎

Proof of Theorem 1.3

Suppose that (P) admits a nontrivial nonnegative solution $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$ and
$v(x):=-|x|^{2}\Delta u\in C^{0}(\mathbb{R}^{N})$. Arguments similar to those
in the proof of Lemma 4.3 imply that $v\geq 0$ in $\mathbb{R}^{N}$. The
continuity of $u$ and $v$ at $x=0$ and the strong maximum principle implies
that $u>0$ and $v>0$ on $\mathbb{R}^{N}$.

Let $R>1$ and $\mathcal{F}_{R}=\\{x\in\mathbb{R}^{N}:\;|x|\geq R\\}$. We claim
that for any $R_{0}>1$ and $x\in{\mathcal{F}}_{R_{0}}$,

(4.55) $u(x)\geq
u_{0}\Big{(}\frac{R_{0}}{|x|}\Big{)}^{\frac{N-2}{2}},\;\;\;v(x)\geq
v_{0}\Big{(}\frac{R_{0}}{|x|}\Big{)}^{\frac{N-2}{2}},$

where

$u_{0}=\min_{|x|=R_{0}}u(x)>0,\;\;\;v_{0}=\min_{|x|=R_{0}}v(x)>0.$

Obviously, by Proposition 4.2, we have for $t>-\infty$,

$0<\frac{\partial w}{\partial
t}(t,\omega)=e^{\frac{N}{2}t}\Big{(}\frac{N-2}{2}\frac{u}{r}+\frac{\partial
u}{\partial r}(r,\omega)\Big{)},\;\;(t,\omega)\in S_{t}$

and

$0<\frac{\partial z}{\partial
t}(t,\omega)=e^{\frac{N}{2}t}\Big{(}\frac{N-2}{2}\frac{v}{r}+\frac{\partial
v}{\partial r}(r,\omega)\Big{)},\;\;(t,\omega)\in S_{t}$

where $r=|x|$. In turn

(4.56) $\frac{N-2}{2r}>-u^{-1}\frac{\partial u}{\partial
r}(r,\omega),\;\;\frac{N-2}{2r}>-v^{-1}\frac{\partial v}{\partial
r}(r,\omega),\;\;x\in\mathbb{R}^{N}\backslash\\{0\\}.$

Integrating (4.56) from $R_{0}$ to $r>R_{0}$ along the radius immediately
yields our claim (4.55).

For any $R>R_{0}$, set $\Omega_{R}=\\{x\in\mathbb{R}^{N}:\;||x|-3R|<R\\}$. We
see that $\Omega_{R}\subset\mathcal{F}_{R_{0}}$ and

(4.57) $|x|^{l}u^{p}\geq DR^{l}u^{p},\;\;|x|^{-2}v\geq DR^{-2}v,\;\;\;\mbox{in
$\Omega_{R}$},$

where $D=4^{-2}>0$.

For any $0<\delta\leq 1$, consider the problem

(4.58)
$\left\\{\begin{array}[]{ll}h_{ss}+\frac{N-1}{s+\frac{3}{\delta}}h_{s}+Dk=0&\mbox{in
$(-1,1)$},\\\ k_{ss}+\frac{N-1}{s+\frac{3}{\delta}}k_{s}+Dh^{p}=0&\mbox{in
$(-1,1)$},\\\ h(-1)=h(1)=k(-1)=k(1)=0.\end{array}\right.$

Direct calculation yields that, for suitably large $a>0$ and $K>0$ independent
of $\delta$,
$({\tilde{h}}(s),{\tilde{k}}(s))\equiv(K(1-s^{2})^{a},K^{\varrho}(1-s^{2})^{a})$
with $1<\varrho<p$ is a subsolution to (4.58), i.e., for any $0<\delta\leq 1$,

(4.59)
$\left\\{\begin{array}[]{ll}{\tilde{h}}_{ss}+\frac{N-1}{s+\frac{3}{\delta}}{\tilde{h}}_{s}+D{\tilde{k}}\geq
0&\mbox{in $(-1,1)$},\\\
{\tilde{k}}_{ss}+\frac{N-1}{s+\frac{3}{\delta}}{\tilde{k}}_{s}+D{\tilde{h}}^{p}\geq
0&\mbox{in $(-1,1)$},\\\
{\tilde{h}}(-1)={\tilde{h}}(1)={\tilde{k}}(-1)={\tilde{k}}(1)=0.\end{array}\right.$

Define

$u_{R}(x)=u_{R}(|x|):=R^{-\beta}{\tilde{h}}\Big{(}\frac{|x|-3R}{R}\Big{)},\;\;\;\;x\in\Omega_{R}=\\{x:\;2R<|x|<4R\\},$
$v_{R}(x)=v_{R}(|x|):=R^{-\beta}{\tilde{k}}\Big{(}\frac{|x|-3R}{R}\Big{)},\;\;\;\;x\in\Omega_{R}=\\{x:\;2R<|x|<4R\\},$

where

(4.60) $\beta:=\frac{2+l}{p-1}>\frac{N-2}{2}.$

In view of (4.55) and (4.60), we immediately deduce that there exists a
(sufficiently large) value $R_{1}\gg 1$ such that

(4.61) $u_{R_{1}}<u,\;\;v_{R_{1}}<v\;\;\mbox{in $\Omega_{R_{1}}$}.$

By (4.57) and the fact that $\beta(p-1)-2=l$, we obtain the following two
inequalities

(4.62) $\left\\{\begin{array}[]{ll}\Delta u+DR_{1}^{-2}v\leq 0\leq\Delta
u_{R_{1}}+DR_{1}^{-2}v_{R_{1}}&\mbox{in $\Omega_{R_{1}}$},\\\ \Delta
v+DR_{1}^{l}u^{p}\leq 0\leq\Delta v_{R_{1}}+DR_{1}^{l}u_{R_{1}}^{p}&\mbox{in
$\Omega_{R_{1}}$}.\end{array}\right.$

Next, for $0<\delta\leq 1$, we define the functions

$w_{\delta}(x)=w_{\delta}(|x|):=\delta^{-\gamma}R_{1}^{-\beta}{\tilde{h}}\Big{(}\frac{|x|-3R_{1}}{\delta
R_{1}}\Big{)},$
$z_{\delta}(x)=z_{\delta}(|x|):=\delta^{-(\gamma+2)}R_{1}^{-\beta}{\tilde{k}}\Big{(}\frac{|x|-3R_{1}}{\delta
R_{1}}\Big{)}$

in ${\tilde{\Omega}}_{\delta R_{1}}=\\{x:\;||x|-3R_{1}|<R_{1}\delta\\}$, where

$\gamma=\frac{4}{p-1}>0.$

A straight-forward computation implies

(4.63) $\left\\{\begin{array}[]{ll}\Delta w_{\delta}+DR_{1}^{-2}z_{\delta}\geq
0&\mbox{in ${\tilde{\Omega}}_{\delta R_{1}}$},\\\ \Delta
z_{\delta}+DR_{1}^{l}w_{\delta}^{p}\geq 0&\mbox{in ${\tilde{\Omega}}_{\delta
R_{1}}$},\\\ w_{\delta}=0,\;\;z_{\delta}=0&\mbox{on
$\partial{\tilde{\Omega}}_{\delta R_{1}}$}.\end{array}\right.$

Notice that $(w_{1},z_{1})\equiv(u_{R_{1}},v_{R_{1}})$.

On the other hand, clearly $w_{\delta}(3R_{1})\to\infty$,
$z_{\delta}(3R_{1})\to\infty$ as $\delta\to 0$. Therefore, there exists
$\delta\in(0,1)$ and a point ${\overline{x}}\in{\tilde{\Omega}}_{\delta
R_{1}}$ such that

(4.64) $u\geq w_{\delta},\;\;v\geq z_{\delta}\;\;\;\mbox{in
${\tilde{\Omega}}_{\delta R_{1}}$ and
$u({\overline{x}})=w_{\delta}({\overline{x}})$}$

or

(4.65) $u\geq w_{\delta},\;\;v\geq z_{\delta}\;\;\;\mbox{in
${\tilde{\Omega}}_{\delta R_{1}}$ and
$v({\overline{x}})=z_{\delta}({\overline{x}})$}.$

In view of (4.63), recall that $(u,v)$ is a supersolution and
$(w_{\delta},z_{\delta})$ is a subsolution to the problem:

$\left\\{\begin{array}[]{ll}\Delta f+DR_{1}^{-2}g=0&\mbox{in
${\tilde{\Omega}}_{\delta R_{1}}$},\\\ \Delta g+DR_{1}^{l}f^{p}=0&\mbox{in
${\tilde{\Omega}}_{\delta R_{1}}$},\\\ f=g=0&\mbox{on
$\partial{\tilde{\Omega}}_{\delta R_{1}}$}.\end{array}\right.$

The strong maximum principle implies $u\equiv w_{\delta}$ or $v\equiv
z_{\delta}$ in ${\overline{{\tilde{\Omega}}_{\delta R_{1}}}}$, which is
impossible (since $(u,v)>0$ on ${\overline{{\tilde{\Omega}}_{\delta R_{1}}}}$,
but $(w_{\delta},z_{\delta})\equiv 0$ on $\partial{\tilde{\Omega}}_{\delta
R_{1}}$). This contradiction completes the proof of our Theorem 1.3. ∎

##  5\. Appendix

We present Rellich-Pohozaev identity for a nonnegative solution $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$ and
$|x|^{\alpha}\Delta u\in C^{0}(\mathbb{R}^{N})$ of (P).

######  Lemma 5.1.

Assume $N\geq 5$, $p>1$ and (1.1) holds. Assume also that $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$ with
$v(x):=-|x|^{\alpha}\Delta u\in C^{0}(\mathbb{R}^{N})$ is a nonnegative
solution to (P), the following Rellich-Pohozaev identity holds:

$\displaystyle\Big{[}\frac{N^{\prime}+\tau}{p+1}-\frac{N^{\prime}-4}{2}\Big{]}\int_{B_{R}}|x|^{l}u^{p+1}dx=\int_{|x|=R}\Big{[}R^{l+1}\frac{u^{p+1}}{p+1}$
$\displaystyle+R^{1-\alpha}\frac{v^{2}}{2}+2Ru^{\prime}v^{\prime}-R\nabla
u\cdot\nabla
v+\frac{N-\alpha}{2}vu^{\prime}+\frac{N^{\prime}-4}{2}uv^{\prime}\Big{]}d\sigma_{R}$

for all $R>0$, where $B_{R}=\\{x\in\mathbb{R}^{N}:\;|x|<R\\}$,
$u^{\prime}=\nabla u\cdot\frac{x}{|x|}$, $v^{\prime}=\nabla
v\cdot\frac{x}{|x|}$.

###### Proof.

We know that $(u,v)$ satisfies the system:

(5.2) $\left\\{\begin{array}[]{ll}-\Delta u=|x|^{-\alpha}v&\mbox{in
$\mathbb{R}^{N}\backslash\\{0\\}$},\\\ -\Delta v=|x|^{l}u^{p}&\mbox{in
$\mathbb{R}^{N}\backslash\\{0\\}$}.\end{array}\right.$

For any $R>0$, we claim:

(5.3) $\int_{B_{R}}|\nabla
u|^{2}dx=\int_{B_{R}}|x|^{-\alpha}vudx+\int_{|x|=R}uu^{\prime}d\sigma_{R}<\infty,$
(5.4) $\int_{B_{R}}|\nabla
v|^{2}dx=\int_{B_{R}}|x|^{l}u^{p}vdx+\int_{|x|=R}vv^{\prime}d\sigma_{R}<\infty.$

In particular, there exists a sequence $\epsilon_{i}\to 0^{+}$ such that

(5.5) $\epsilon_{i}\int_{|x|=\epsilon_{i}}|\nabla
u|^{2}d\sigma_{\epsilon_{i}}\to
0,\;\;\;\epsilon_{i}\int_{|x|=\epsilon_{i}}|\nabla
v|^{2}d\sigma_{\epsilon_{i}}\to 0.$

We only show (5.3) and $\eqref{5.4}_{1}$, the proof of (5.4) and
$\eqref{5.4}_{2}$ is similar. For any $0<\rho<R$, we have

(5.6) $\displaystyle\int_{B_{R}\backslash B_{\rho}}|\nabla u|^{2}dx$
$\displaystyle=$ $\displaystyle-\int_{B_{R}\backslash B_{\rho}}u\Delta
udx+\int_{|x|=R}uu^{\prime}d\sigma_{R}-\int_{|x|=\rho}uu^{\prime}d\sigma_{\rho}$
$\displaystyle=$ $\displaystyle\int_{B_{R}\backslash
B_{\rho}}|x|^{-\alpha}vudx+\int_{|x|=R}uu^{\prime}d\sigma_{R}-\int_{|x|=\rho}uu^{\prime}d\sigma_{\rho}.$

On the other hand, we have

$\int_{|x|=\rho}uu^{\prime}d\sigma_{\rho}=\rho^{N-1}f^{\prime}(\rho),\;\;\mbox{where
$f(\rho):=\frac{1}{2}\int_{S^{N-1}}u^{2}(\rho,\theta)d\theta$}.$

Since $f\in C^{1}((0,R])\cap C^{0}([0,R])$ due to $u\in
C^{4}(\mathbb{R}^{N}\backslash\\{0\\})\cap C^{0}(\mathbb{R}^{N})$, we infer
the existence of a sequence $\rho_{i}\to 0^{+}$ such that
$\lim_{i\to\infty}\rho_{i}^{N-1}f^{\prime}(\rho_{i})=0$. Since $N\geq 5$,
$\alpha<N$ and $v\in C^{0}(\mathbb{R}^{N})$, passing to the limit in (5.6)
with $\rho=\rho_{i}$, we obtain (5.3). Note that
$\int_{B_{\rho_{i}}}|x|^{-\alpha}|vu|dx\leq\frac{C}{N-\alpha}\rho_{i}^{N-\alpha}\to
0$ as $i\to\infty$. Hence the right-hand side of (5.3) is finite. Since

$\int_{0}^{R}\int_{|x|=\epsilon}|\nabla
u|^{2}d\sigma_{\epsilon}d\epsilon=\int_{B_{R}}|\nabla u|^{2}dx,$

our claim $\eqref{5.4}_{1}$ follows. Moreover,

(5.7) $\int_{|x|=\epsilon_{i}}|\nabla u|d\sigma_{\epsilon_{i}}\leq
C\Big{(}\epsilon_{i}^{N-1}\int_{|x|=\epsilon_{i}}|\nabla
u|^{2}d\sigma_{\epsilon_{i}}\Big{)}^{\frac{1}{2}}\to 0\;\;\mbox{as
$i\to\infty$},$ (5.8) $\int_{|x|=\epsilon_{i}}|\nabla
v|d\sigma_{\epsilon_{i}}\leq
C\Big{(}\epsilon_{i}^{N-1}\int_{|x|=\epsilon_{i}}|\nabla
v|^{2}d\sigma_{\epsilon_{i}}\Big{)}^{\frac{1}{2}}\to 0\;\;\mbox{as
$i\to\infty$}.$

Define $w(x)=|x|^{\alpha}\Delta u(x)$. Then $w(x)=-v(x)$ and

$(x\cdot\nabla u)\Delta w=(x\cdot\nabla
u)|x|^{l}u^{p}=\mbox{div}\Big{(}x|x|^{l}\frac{u^{p+1}}{p+1}\Big{)}-\frac{N^{\prime}+\tau}{p+1}|x|^{l}u^{p+1}.$

Thus, for $0<\epsilon<R$, we have

$\int_{B_{R}\backslash B_{\epsilon}}(x\cdot\nabla u)\Delta
wdx=R^{l+1}\int_{|x|=R}\frac{u^{p+1}}{p+1}d\sigma_{R}-\epsilon^{l+1}\int_{|x|=\epsilon}\frac{u^{p+1}}{p+1}d\sigma_{\epsilon}-\frac{N^{\prime}+\tau}{p+1}\int_{B_{R}\backslash
B_{\epsilon}}|x|^{l}u^{p+1}dx.$

Letting $\epsilon\to 0$, using the continuity of $u$ and
$N+l=N^{\prime}+\tau>0$, we obtain

(5.9) $\int_{B_{R}}(x\cdot\nabla u)\Delta
wdx=R^{l+1}\int_{|x|=R}\frac{u^{p+1}}{p+1}d\sigma_{R}-\frac{N^{\prime}+\tau}{p+1}\int_{B_{R}}|x|^{l}u^{p+1}dx.$

Next, by direct calculations, we have the following identity

(5.10) $\mbox{div}((x\cdot\nabla u)\nabla w-x\nabla u\cdot\nabla
w)=(x\cdot\nabla u)\Delta w-(N-2)\nabla u\cdot\nabla w-\nabla(x\cdot\nabla
w)\cdot\nabla u.$

It follows, for $0<\epsilon<R$,

$\displaystyle\int_{B_{R}\backslash B_{\epsilon}}[(x\cdot\nabla u)\Delta
w-(N-2)\nabla u\cdot\nabla w-\nabla(x\cdot\nabla w)\cdot\nabla u]dx$
$\displaystyle=$ $\displaystyle\int_{|x|=R}((x\cdot\nabla u)\nabla w-x\nabla
u\cdot\nabla w)\frac{x}{|x|}d\sigma_{R}-\int_{|x|=\epsilon}((x\cdot\nabla
u)\nabla w-x\nabla u\cdot\nabla w)\frac{x}{|x|}d\sigma_{\epsilon}.$

Moreover,

$\displaystyle\int_{B_{R}\backslash B_{\epsilon}}\nabla u\cdot\nabla wdx$
$\displaystyle=$ $\displaystyle\int_{|x|=R}u\nabla
w\cdot\frac{x}{|x|}d\sigma_{R}-\int_{|x|=\epsilon}u\nabla
w\cdot\frac{x}{|x|}d\sigma_{\epsilon}-\int_{B_{R}\backslash
B_{\epsilon}}u\Delta wdx$ (5.12) $\displaystyle=$
$\displaystyle\int_{|x|=R}u\nabla
w\cdot\frac{x}{|x|}d\sigma_{R}-\int_{|x|=\epsilon}u\nabla
w\cdot\frac{x}{|x|}d\sigma_{\epsilon}-\int_{B_{R}\backslash
B_{\epsilon}}|x|^{l}u^{p+1}dx.$ $\displaystyle\int_{B_{R}\backslash
B_{\epsilon}}\nabla(x\cdot\nabla w)\cdot\nabla udx$ $\displaystyle=$
$\displaystyle\int_{|x|=R}(x\cdot\nabla w)\nabla
u\cdot\frac{x}{|x|}d\sigma_{R}-\int_{|x|=\epsilon}(x\cdot\nabla w)\nabla
u\cdot\frac{x}{|x|}d\sigma_{\epsilon}-\int_{B_{R}\backslash
B_{\epsilon}}(x\cdot\nabla w)\Delta udx$ $\displaystyle=$
$\displaystyle\int_{|x|=R}(x\cdot\nabla w)\nabla
u\cdot\frac{x}{|x|}d\sigma_{R}-\int_{|x|=\epsilon}(x\cdot\nabla w)\nabla
u\cdot\frac{x}{|x|}d\sigma_{\epsilon}-\int_{B_{R}\backslash
B_{\epsilon}}(x\cdot\nabla w)|x|^{-\alpha}wdx$ $\displaystyle=$
$\displaystyle\int_{|x|=R}(x\cdot\nabla w)\nabla
u\cdot\frac{x}{|x|}d\sigma_{R}-\int_{|x|=\epsilon}(x\cdot\nabla w)\nabla
u\cdot\frac{x}{|x|}d\sigma_{\epsilon}$
$\displaystyle-\int_{|x|=R}R^{1-\alpha}\frac{w^{2}}{2}d\sigma_{R}+\int_{|x|=\epsilon}\epsilon^{1-\alpha}\frac{w^{2}}{2}d\sigma_{\epsilon}+\frac{N-\alpha}{2}\int_{B_{R}\backslash
B_{\epsilon}}|x|^{\alpha}(\Delta u)^{2}dx$ $\displaystyle=$
$\displaystyle\int_{|x|=R}(x\cdot\nabla w)\nabla
u\cdot\frac{x}{|x|}d\sigma_{R}-\int_{|x|=\epsilon}(x\cdot\nabla w)\nabla
u\cdot\frac{x}{|x|}d\sigma_{\epsilon}+\int_{|x|=\epsilon}\epsilon^{1-\alpha}\frac{w^{2}}{2}d\sigma_{\epsilon}$
$\displaystyle-\int_{|x|=R}R^{1-\alpha}\frac{w^{2}}{2}d\sigma_{R}+\frac{N-\alpha}{2}\int_{|x|=R}\Big{[}w\nabla
u\cdot\frac{x}{|x|}-u\nabla w\cdot\frac{x}{|x|}\Big{]}d\sigma_{R}$
$\displaystyle-\frac{N-\alpha}{2}\int_{|x|=\epsilon}\Big{[}w\nabla
u\cdot\frac{x}{|x|}-u\nabla
w\cdot\frac{x}{|x|}\Big{]}d\sigma_{\epsilon}+\frac{N-\alpha}{2}\int_{B_{R}\backslash
B_{\epsilon}}|x|^{l}u^{p+1}dx.$

The last identity is obtained by multiplying $u$ on both the sides of (P) and
integrating it on $B_{R}\backslash B_{\epsilon}$. Letting
$\epsilon=\epsilon_{i}\to 0$, where $\epsilon_{i}$ is given in (5.5), (5.7)
and (5.8), we obtain from (5.5), (5.7), (5.8), (5), (5) and (5) that

$\displaystyle\int_{B_{R}}(x\cdot\nabla u)\Delta wdx$ $\displaystyle=$
$\displaystyle\int_{|x|=R}\Big{[}-R^{1-\alpha}\frac{w^{2}}{2}+\Big{(}(x\cdot\nabla
u)\nabla w-x\nabla u\cdot\nabla w+(N-2)u\nabla w+(x\cdot\nabla w)\nabla u$
$\displaystyle+\frac{N-\alpha}{2}(w\nabla u-u\nabla
w)\Big{)}\cdot\frac{x}{|x|}\Big{]}d\sigma_{R}+\Big{[}\frac{N-\alpha}{2}-(N-2)\Big{]}\int_{B_{R}}|x|^{l}u^{p+1}dx.$

Note that

$\displaystyle\Big{|}\int_{|x|=\epsilon}(x\cdot\nabla w)\nabla
u\cdot\frac{x}{|x|}d\sigma_{\epsilon}\Big{|}$ $\displaystyle\leq$
$\displaystyle\epsilon\int_{|x|=\epsilon}|w^{\prime}||u^{\prime}|d\sigma_{\epsilon}$
$\displaystyle\leq$ $\displaystyle\Big{(}\epsilon\int_{|x|=\epsilon}|\nabla
w|^{2}d\sigma_{\epsilon}\Big{)}^{\frac{1}{2}}\Big{(}\epsilon\int_{|x|=\epsilon}|\nabla
u|^{2}d\sigma_{\epsilon}\Big{)}^{\frac{1}{2}},$

and

$\epsilon^{1-\alpha}\int_{|x|=\epsilon}\frac{w^{2}}{2}d\sigma_{\epsilon}=\frac{\epsilon^{N-\alpha}}{2}\int_{S^{N-1}}w^{2}(\epsilon,\theta)d\theta.$

(We know that $N-\alpha>0$ and $w$ is continuous in $B_{\epsilon}$.)

Combining (5.9) and (5), we obtain

$\displaystyle\Big{[}\frac{N^{\prime}+\tau}{p+1}-\frac{N^{\prime}-4}{2}\Big{]}\int_{B_{R}}|x|^{l}u^{p+1}dx=\int_{|x|=R}\Big{[}R^{l+1}\frac{u^{p+1}}{p+1}$
$\displaystyle+R^{1-\alpha}\frac{v^{2}}{2}+2Ru^{\prime}v^{\prime}-R\nabla
u\cdot\nabla
v+\frac{N-\alpha}{2}vu^{\prime}+\frac{N^{\prime}-4}{2}uv^{\prime}\Big{]}d\sigma_{R}.$

Note that $v=-w$. ∎

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