\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 138, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/138\hfil 
Positive solutions for nonlocal singular elliptic problems]
{Existence, uniqueness and multiplicity of positive solutions
 for some nonlocal singular elliptic problems}

\author[B. Yan, Q. Ren \hfil EJDE-2017/138\hfilneg]
{Baoqiang Yan, Qianqian Ren}

\address{Baoqiang Yan (corresponding author) \newline
School of Mathematical Sciences,
Shandong Normal University,
Jinan 250014, China}
\email{yanbqcn@aliyun.com}

\address{Qianqian Ren \newline
School of Mathematical Sciences,
Shandong Normal University,
Jinan 250014, China}
\email{351191416@qq.com}

\dedicatory{Communicated by Claudianor Alves}

\thanks{Submitted January 10, 2017. Published May 24, 2017.}
\subjclass[2010]{35J60, 35J75, 47H10}
\keywords{Nonlocal elliptic equations; existence; uniqueness;
\hfill\break\indent Rabinowitz-type global bifurcation theory;
 multiplicity}

\begin{abstract}
 In this article, using the sub-supersolution method and Rabinowitz-type
 global bifurcation theory, we prove some results on existence,
 uniqueness and multiplicity of positive solutions for some singular
 nonlocal elliptic problems.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

In this article, we consider the nonlocal elliptic problems
\begin{equation}
\begin{gathered}
-a\Big(\int_{\Omega}|u(x)|^{\gamma}dx\Big)\Delta u=K(x)u^{-\mu},\quad x \text{ in }
 \Omega,\\
u(x)>0, \quad x \text{ in } \Omega,\\
u(x)=0,\quad x\text{ on }\partial\Omega
\end{gathered}. \label{e1.1}
\end{equation}
and
\begin{equation}
\begin{gathered}
-a\Big(\int_{\Omega}|u(x)|^{\gamma}dx\Big)\Delta u=\lambda(u^q+K(x)u^{-\mu}),\quad
 x \text{ in } \Omega,\\
u(x)>0, \quad x \text{ in } \Omega,\\
u(x)=0,\quad x\text{ on }\partial\Omega,
\end{gathered} \label{e1.2}
\end{equation}
where $\Omega\subseteq \mathbb{R}^N$ ($N\geq1$) is a sufficiently regularity domain,
$q>0$, $\lambda\geq 0$, $\mu>0$ and $\gamma\in(0,+\infty)$.

Obviously, if $a(t)\equiv 1$ for $t\in[0,+\infty)$, \eqref{e1.1} and
\eqref{e1.2} are singular elliptic boundary value problems and there
are many results on existence, uniqueness and multiplicity of positive solutions,
see \cite{c8,g1,h1,l1,r2,s1,s2,s3,z1} and their references.
Chipot and Lovat \cite{c2} considered the model problem
\begin{equation}
\begin{gathered}
 u_t - a\Big(\int_{\Omega}u(z,t)dz\Big)\Delta u=f, \quad \text{in }
\Omega\times (0,T),\\
u(x,t) = 0,\quad \text{on } \Gamma\times(0,T),\\
u(x,0)=u_0(x),\quad \text{on }\Omega.
\end{gathered}\label{e1.3}
\end{equation}
Here $\Omega$ is a bounded open subset in $\mathbb{R}^N$, $N\geq 1$ with smooth
boundary $\Gamma$, $T$ is some arbitrary time.
Notice that if $u(x,t)$ is independent from $t$, \eqref{e1.3} is a nonlocal
elliptic problems such as
\begin{equation}
\begin{gathered}
-a\Big(\int_{\Omega}|u(x)|^{\gamma}dx\Big)\Delta u=f(x,u),\quad x
 \text{ in } \Omega,\\
u(x)=0,\quad x\text{ on }\partial\Omega.
\end{gathered} \label{e1.4}
\end{equation}
And a more generalized problem of \eqref{e1.4} is
\begin{equation}
\begin{gathered}
-A(x,u)\Delta u=f(x,u),\quad x \text{ in } \Omega,\\
u(x)>0, \quad x \text{ in } \Omega,\\
u(x)=0,\quad x\text{ on }\partial\Omega,
\end{gathered} \label{e1.5}
\end{equation}
where $A:\Omega\times L^p(\Omega)\to R^+$ is a measurable function.

By establishing comparison principles, using the results on fixed point
index theory, sub-supersolution method, some authors obtained the existence
of at least one positive solutions for \eqref{e1.4} or \eqref{e1.5},
see \cite{c1,c3,c4,c5,c6,r3} and their references.
We notice that the nonlocal term $A(x,u)$ or
$a(\int_{\Omega}|u(x)|^{\gamma}dx)$ causes that the monotonic nondecreasing
of $f$ being necessary for using the sub-supersolution method.
Up to now, there are fewer results on the existence and multiplicity of
positive solutions for \eqref{e1.4} or \eqref{e1.5} when $f(x,u)$ is
singular at $u=0$. Very recently, an interesting result on the following problems
is obtained
\begin{equation}
\begin{gathered}
\begin{aligned}
-a\Big(\int_{\Omega}|u(x)|^{\gamma}dx\Big)\Delta u 
&=h_1(x,u)f\Big(\int_{\Omega}|u(x)|^{p}dx\Big)\\
&\quad +h_2(x,u)g\Big(\int_{\Omega}|u(x)|^rdx\Big),\quad 
 x\text{ in } \Omega,
\end{aligned}\\
u=0,\quad x\text{ on }\partial\Omega,
\end{gathered} \label{e1.6}
\end{equation}
where $\gamma,r,p\geq 1$ and in which Alves and Covei showed that the
 existence of solution for some classes of nonlocal problems without
of the monotonic nondecreasing of $h_1$ (see \cite{a4}) as
$h_1(x,u)=\frac{1}{u^{\alpha}}$, $\alpha\in(0,1)$. In \cite{o1}, applying the
change of variable and the theory of fixed point index on a cone,
 do \'{O} obtained the multiplicity of radial positive solutions for
some nonlocal and nonvariational elliptic systems when the nonlinearities
$f_i$ is nondecreasing in $u$ without singularity at $u=0$,
$i=1,2, \dots, n$ and $\Omega=\{x\in \mathbb{R}^N|0<r_1<|x|<r_2\}$.

In this article, we consider the existence, uniqueness and multiplicity
of positive solutions to \eqref{e1.1} and \eqref{e1.2} when $\mu>0$
is arbitrary.

This paper is organized as follows. In Section 2, according to the idea
in \cite{a4,c7}, we prove a new result on the existence of classical
solutions by using sub-supersolution method with maximum principle.
In section 3, using Theorem \ref{thm2.2}, the existence and uniqueness of positive
solution to \eqref{e1.1} are presented. In section 4,
by Rabinowitz-type global bifurcation theory, we discuss the global
results and obtain the multiplicity of positive solutions for \eqref{e1.2}.

\section{Sub-supersolution method}

 Now we consider a general problem
\begin{equation}
\begin{gathered}
-a\Big(\int_{\Omega}|u(x)|^{\gamma}dx\Big)\Delta u=F(x,u),\quad
 x\text{ in } \Omega,\\
u=0,\quad x\text{ on }\partial\Omega,
\end{gathered} \label{e2.1}
\end{equation}
where $\Omega\subseteq \mathbb{R}^N$ is a smooth bounded domain, $\gamma\in(0,+\infty)$
and $a:[0,+\infty)\to(0,+\infty)$ is continuous function with
\begin{equation}
\inf_{t\in[0,+\infty)}a(t)\geq a(0)=:a_0>0.\label{e2.2}
\end{equation}
Let
$C(\overline{\Omega})=\{u:\overline{\Omega}\to R| u$ be a continuous
function on $\overline{\Omega}\}$ with norm
$\|u\|=\max_{x\in\overline{\Omega}}|u(x)|$.

\begin{definition} \label{def2.1} \rm
 The pair functions $\alpha$ and $\beta$ with $\alpha$,
$\beta\in C(\overline{\Omega})\cap C^2(\Omega)$
are subsolution and supersolution of \eqref{e2.1} if $\alpha(x)\leq u\leq \beta(x)$
for $x\in\Omega$ and 
\begin{gather*}
-\Delta \alpha (x)\leq\frac{1}{b_0} F(x, \alpha(x)),\quad x\text{ in } \Omega,\\
\alpha\big|_{\partial \Omega}\leq 0
\end{gather*} 
and 
\begin{gather*}
-\Delta \beta(x)\geq \frac{1}{a_0}F(x, \beta(x)),\quad x\text{ in } \Omega,\\
\beta\big|_{\partial\Omega}\geq 0,
\end{gather*}
where $a_0=a(0)$ and 
\[
b_0=\sup_{t\in[0,\int_{\Omega}\max\{|\alpha(x)|,|\beta(x)|\}^{\gamma}dx]}a(t).
\]
\end{definition}


For a fixed $\lambda>0$, we state the problem
\begin{equation}
\begin{gathered}
-\Delta u+\lambda u(x)=h(x),\quad x\text{ in } \Omega,\\
u=0,\quad \text{on }\partial\Omega,
\end{gathered} \label{e2.3}
\end{equation}
where $\Omega\subseteq \mathbb{R}^N$ is a smooth bounded domain and
give the deformation of Agmon-Douglas-Nirenberg theorem for \eqref{e2.3}.



\begin{theorem}[Agmon-Douglas-Nirenberg \cite{a1}] \label{thm2.1}
If $h\in C^{\alpha}(\overline{\Omega})$, then \eqref{e2.3} has a unique solution
$u\in C^{2+\alpha}(\overline{\Omega})$ such that
$$
\|u\|_{2+\alpha}\leq C_1\|h|_{\infty};
$$
if $h\in L^p(\Omega)$$(p>1)$, then \eqref{e2.3} has a unique solution
$u\in W^2_p(\Omega)$ such that
$$
\|u\|_{2,p}\leq C_2\|h\|_{p},
$$
where $C_1$, $C_2$ ere independent from $u$, $h$.
\end{theorem}

 We define the unique solution $u=(-\Delta+\lambda)^{-1}h$ of \eqref{e2.3}.
Obviously $(-\Delta+\lambda)^{-1}$ is a linear operator.
To prove our theorem, we need the following Embedding theorem.

\begin{lemma}[\cite{a3}] \label{lem2.1}
Suppose $\Omega\subseteq \mathbb{R}^N$ is a bounded domain with smooth boundary
and $p>N$. Then there exists a $C(N,p,\Omega)>0$ such that
$$
|u|_{k+\alpha}\leq C(N,p,\Omega)\|u\|_{k+1,p},\quad \forall u\in W_p^{k+1}(\Omega),
$$
where $\alpha=1-\frac{N}{p}$.
\end{lemma}

Next we give our main theorem.

\begin{theorem} \label{thm2.2}
 Let $\Omega\subseteq \mathbb{R}^N (N\geq1)$ be a smooth bounded domain and
$\gamma\in(0,+\infty)$.
 Suppose that $F:\Omega\times R\to R$ is a continuous nonnegative function.
Assume $\alpha$ and $\beta$ are the subsolution and supersolution of \eqref{e2.1}
respectively.
Then problem \eqref{e2.1} has at least one solution $u$ such that, for all
$x\in\overline{\Omega}$,
$$
\alpha(x)\leq u(x)\leq\beta(x).
$$
\end{theorem}

\begin{proof}
Let
$$
\bar{F}(x,u)=\begin{cases}
F(x,\alpha(x)), &\text{if } u<\alpha(x);\\
F(x,u),&\text{if }\alpha(x)\leq u\leq \beta(x);\\
F(x,\beta(x)), &\text{if }u>\beta(x).
\end{cases}
$$
We will study the modified problem (for $\lambda>0$)
\begin{equation} \begin{gathered}
-\Delta u+\lambda u=\frac{\bar{F}(x, u)}{a(\int_{\Omega}|
\chi(x, u(x))|^{\gamma}dx)}+\lambda\chi(x, u), \quad x\in \Omega,\\
u|_{\partial\Omega}=0,
\end{gathered}\label{e2.4}
\end{equation}
here $\chi(x, u)=\alpha(x)+(u-\alpha(x))^+-(u-\beta(x))^+$.
\smallskip

\noindent\textbf{Step 1.} Every solution $u$ of \eqref{e2.4} is such that:
$\alpha(x)\leq u(x)\leq\beta(x)$, $x\in\overline{\Omega}$.
We prove that $\alpha(x)\leq u(x)$ on $\overline{\Omega}$. Obviously,
$|\chi(x, u(x))|\leq\max\{|\alpha(x)|,|\beta(x)|\}$, which implies that
$$
a_0\leq a(\int_{\Omega}|\chi(x,u(x))|^{\gamma}dx)\leq b_0.
$$
By contradiction, assume that $\max_{x\in\bar{\Omega}}(\alpha(x)-u(x))=M>0$. Note
that
$\alpha(x)-u(x)\not\equiv M$ on $ \bar{\Omega}$
($\alpha(x)-u(x)\leq 0$, $x\in \partial\Omega$).
If $x_0\in\Omega$ is such that $\alpha(x_0)-u(x_0)=M$, then
\begin{align*}
0&\leq -\Delta(\alpha(x_0)-u(x_0))\\
&\leq\frac{1}{b_0}F(x_0,\alpha(x_0))
 -\frac{1}{a(\int_{\Omega}|\chi(x, u(x))|^{\gamma}dx)}\bar{F}(x_0, u(x_0))
 -\lambda\chi(x_0, u(x_0))+\lambda u(x_0)\\
&\leq -\lambda(\alpha(x_0)-u(x_0))
<0.
\end{align*}
This is a contradiction.

Now we prove that $\beta(x)\geq u(x)$ on $\overline{\Omega}$.
By contradiction, assume $\min_{x\in\bar{\Omega}}(\beta(x)-u(x))=-m<0$. Note
that $\beta(x)-u(x)\not\equiv -m$ on $\bar{\Omega}$
($\beta(x)-u(x)\geq 0$, $x\in \partial\Omega$).
If $x_0\in\Omega$ is such that $\beta(x_0)-u(x_0)=-m$, then
\begin{align*}
0&\geq -\Delta(\beta(x_0)-u(x_0))\\
&\geq\frac{1}{a_0}F(x_0,\beta(x_0))
 -\frac{1}{a(\int_{\Omega}|\chi(x, u(x))|^{\gamma}dx)}\bar{F}(x_0, u(x_0))
-\lambda\chi(x_0, u(x_0))+\lambda u(x_0)\\
&\geq \lambda(u(x_0)-\beta(x_0))
>0.
\end{align*}
This is a contradiction.
Consequently,
$$
\alpha(x)\leq u(x)\leq \beta(x), \ \ x\in\overline{\Omega}.
$$
\smallskip

\noindent\textbf{Step 2.} Every solution of \eqref{e2.4} is a solution
 of \eqref{e2.1}.
Every solution of \eqref{e2.4} is such that :$\alpha(x)\leq u(x)\leq \beta(x)$.
By the definition of $\bar{F}$ and $\chi$, we have
$$
\bar{F}(x, u(x))=F(x, u(x)),\quad \chi(x, u(x))=u(x),\quad x\in\Omega
$$
and $u$ is a solution of \eqref{e2.1}.
\smallskip

\noindent\textbf{Step 3.} Problem \eqref{e2.4} has at least one solution.
Choose $p>N$, $\alpha=1-\frac{N}{p}$ and define an operator
$$
\overline{N}: C(\overline{\Omega})\to C(\overline{\Omega})
\subseteq L^p(\Omega); u\to \overline{F}(\cdot,u(\cdot)).
$$
Since $F$ is continuous, the definition of $\overline{F}$ implies that
$\overline{F}$ is continuous also, which guarantees
$\overline{N}: C(\overline{\Omega})\to C(\overline{\Omega})$ is well defined,
continuous and maps bounded sets to bounded sets. Since \eqref{e2.2} is true,
$a$ is continuous and
\[
\frac{1}{a(\int_{\Omega}|\chi(x, u(x))| ^{\gamma}dx)}\leq\frac{1}{a_0},
\]
the operator $\overline{N}_1u=\frac{1}{a(\int_{\Omega}|\chi(x,u(x))
|^{\gamma}dx)}\overline{N}u$ is continuous, and maps bounded sets to bounded sets.

For given $\lambda>0$, we define an operator
$\overline{A}: C(\overline{\Omega})\to C(\overline{\Omega})$ by
$$
\overline{A}(u)=(-\Delta+\lambda )^{-1}(\overline{N}_1u+\lambda\chi(\cdot, u)).
$$

Now we show that $\overline{A}: C(\overline{\Omega})\to C(\overline{\Omega})$
is completely continuous.

(1) By the construction of $\overline{F}$ and $\chi$, we have, for every
$u\in C(\overline{\Omega})$,
\begin{align*}
&\big|\frac{\overline{F}(x, u(x))}{a(\int_{\Omega}|\chi(x, u(x))|^{\gamma}dx)}
+\lambda \chi (x, u(x))\big| \\
&\leq \frac{1}{a_0}\max_{x\in\overline{\Omega},\alpha(x)\leq u\leq\beta(x)}
 F(x,u)+\lambda \max\{\|\alpha\|,\|\beta\|\},
\end{align*}
for all $x\in\overline{\Omega}$,
which guarantees that there exists a $K>0$ big enough such that
$N_1u+\lambda\chi(\cdot,u)\in B_{L^p}(0,K)$ for all $u\in C(\overline{\Omega})$,
where
$$
B_{L^p}(0,R)=\{u\in L_{p}(\Omega)|\|u\|_{p}\leq K\}.
$$
By Theorem \ref{thm2.1}, we have
\begin{equation}
\|\overline{A}(u)\|_{2,p}=\|(-\Delta+\lambda)^{-1}
(\overline{N}_1u+\lambda\chi(\cdot,u))\|_{2,p}
\leq C_2K,\quad \forall u\in C(\overline{\Omega}).\label{e2.5}
\end{equation}
Lemma \ref{lem2.1} implies that $\overline{A}(C(\overline{\Omega}))$ is bounded in
$C^{\alpha}(\overline{\Omega})$. Therefore,
$\overline{A}(C(\overline{\Omega}))$ is relatively compact in
$C(\overline{\Omega})$.

(2) For $u_1$, $u_2\in C(\overline{\Omega})$, by Theorem \ref{thm2.1}, one has
$$
\|\overline{A}(u_1)-\overline{A}(u_2)\|_{2,p}
\leq C_2\|\overline{N}_1u_1+\lambda\chi(\cdot,u_1)
-(\overline{N}_1u_2+\lambda\chi(\cdot,u_2))\|_{p}.
$$
Lemma \ref{lem2.1} and the continuity of the operator $N_1+\lambda\chi$ guarantee
 that $A:C(\overline{\Omega})\to C(\overline{\Omega})$ is continuous.
Consequently, $A:C(\overline{\Omega})\to C(\overline{\Omega})$
is completely continuous.

By \eqref{e2.5} and Lemma \ref{lem2.1}, there exists a $K_1>0$ big enough such that
$$
\overline{A}(C(\overline{\Omega}))\subseteq B_{C}(0,K_1),
$$
where $B_{C}(0,K_1)=\{u\in C(\overline{\Omega})|\|u\|\leq K_1\}$,
which implies
$$
\overline{A}(B_{C}(0,K_1))\subseteq B_{C}(0,K_1).
$$
The Schauder fixed point theorem guarantees that there exists a
$u\in B_{C}(0,K_1)$ such that
$$
u=\overline{A}u,
$$
i.e., $u$ is a solution of \eqref{e2.4}.

Consequently, steps 1 and 2 guarantee that $u$ in the step 3 is a solution
of \eqref{e2.1}.
The proof is complete.
\end{proof}

We remark that the difference between Theorem \ref{thm2.2} and \cite[Theorem 1]{a4}
 is that the solution $u$ is a classical solution and we use $\gamma>0$
instead of $\gamma\geq1$.
In the following sections, we assume that
$a(t):[0,+\infty)$ is continuous and increasing on $[0,+\infty)$
for convenience.

\section{The existence and uniqueness of positive solution for \eqref{e1.1}}

 In this section, we consider the singular elliptic problems \eqref{e1.1},
where $K\in C^{\alpha}(\overline{\Omega})$ with $K(x)>0$ for
 $x\in\overline{\Omega}$, and $\mu>0$.
Let $\Phi_1$ is the eigenfunction corresponding to the principle eigenvalue
$\lambda_1$ of
\begin{equation}
\begin{gathered}
-\Delta u=\lambda u,\quad x\in\Omega\\
u|_{\partial\Omega}=0.
\end{gathered}\label{e3.1}
\end{equation}
It is found that $\lambda_1>0$, and
\begin{equation}
\Phi_1(x)>0,\quad |\nabla \Phi_1(x)|>0,\quad \forall x\in\partial\Omega.\label{e3.2}
\end{equation}

\begin{theorem} \label{thm3.1}
Let $\Omega\subseteq \mathbb{R}^N$, $N\geq 1$, be a bounded domain with
smooth boundary $\partial \Omega$ (of class $C^{2+\alpha}$, $0<\alpha<1$).
If $K\in C^{\alpha}(\overline{\Omega})$, $K(x)>0$ for all $x\in\overline{\Omega}$
and $\mu>0$, then there exists a unique function
$u\in C^{2+\alpha}(\Omega)\cap C(\overline{\Omega})$ such that
$u(x)>0$ for all $x\in\Omega$ and $u$ is a solution of \eqref{e1.1}.
If $\mu>1$, then there exist positive constants
$b_1$ and $b_2$ such that
$b_1\Phi_1(x)^{\frac{2}{1+\mu}}\leq u(x)\leq b_2\Phi_1(x)^{\frac{2}{1+\mu}}$,
$x\in\overline{\Omega}$.
\end{theorem}

\begin{proof}
The proof is based on Theorem \ref{thm2.2} and the construction of pairs of
 sub-supersolutions. The construction of supersolutions to \eqref{e1.1}
when $\mu>1$ is different from that when $0<\mu\leq 1$.

 (1) Assume first that $\mu>1$. In this case, let $t =2/(1+\mu)$ and let
$\Psi(x)=b\Phi_1(x)^t$
where $b>0$ is a constant. By \eqref{e3.1}, we deduce that
\begin{equation}
\Delta\Psi(x)+q(x, b)\Psi^{-\mu}(x) = 0, \quad x\in\Omega,\label{e3.3}
\end{equation}
where $q(x, b) = b^{1+\mu}[t(1-t)|\nabla\Phi_1(x)|^2+t\lambda_1\Phi_1(x)^2]$.
Inequality \eqref{e3.2} guarantees that
$\min_{x\in\overline{\Omega}}[t(1-t)|\nabla\Phi_1(x)|^2+t\lambda_1\Phi_1(x)^2]>0$,
which implies that there exists a positive constant $b$
such that
$$
\frac{1}{a_0}K(x)<q(x,b), \quad \forall x\in\Omega.
$$
Let $u(x) =b\Phi_1(x)^t$. Hence,
\begin{equation}
\Delta u(x) + \frac{1}{a_0}K(x)u(x)^{-\mu}
=\big[\frac{1}{a_0}K(x)-q(x, b)\big]u^{-\mu}(x)<0, \quad x\in\Omega.\label{e3.4}
\end{equation}

(2) Assume that $0<\mu\leq 1$. Let $s$ be chosen to satisfy the two inequalities
\begin{equation}
 0<s<1, s(1+\mu)<2\label{e3.5}
\end{equation}
and $u(x) =c\Phi_1(x)^s$, where $c$ is a large positive constant to be chosen. For
$x\in\Omega$, we have
\begin{align*}
&\Delta u(x) +\frac{1}{a_0}K(x)u(x)^{-\mu} \\
&=-\Phi_1(x)^{s-2}|\nabla\Phi_1(x)|^2cs(1-s)
 +\frac{1}{a_0}K(x)c^{-\mu}\Phi_1(x)^{-\mu s}
 -c\lambda_1s\Phi_1(x)^{s}\\
&=-\Phi_1(x)^{s-2}\big[|\nabla\Phi_1(x)|^2cs(1-s)
 -\frac{1}{a_0}K(x)c^{-\mu}\Phi_1(x)^{2-(1+\mu)s}\big]
 -c\lambda_1s\Phi_1(x)^{s}.
\end{align*}
From \eqref{e3.2}, there exists a open subset $\Omega'\subset\subset \Omega$
and a $\delta>0$ such that
$$
|\nabla\Phi_1(x)|>\delta,\quad \forall x\in \overline{\Omega}-\Omega',
$$
which together with $2-(1+\mu)s>0$ implies that there exists a $c_1>0$
big enough such that for all $c>c_1$,
$$
|\nabla\Phi_1(x)|^2cs(1-s)-\frac{1}{a_0}K(x)c^{-\mu}\Phi_1(x)^{2-(1+\mu)s}>0,\quad
 \forall x\in \overline{\Omega}-\Omega',
$$
i.e. for all $c>c_1$, $x\in \overline{\Omega}-\Omega'$
\begin{equation}
\begin{aligned}
&-\Phi_1(x)^{s-2}\big[|\nabla\Phi_1(x)|^2cs(1-s)
 -\frac{1}{a_0}K(x)c^{-\mu}\Phi_1(x)^{2-(1+\mu)s}\big]
 -c\lambda_1s\Phi_1(x)^{s}\\
&<0.
\end{aligned} \label{e3.6}
\end{equation}
Moreover, from $\min_{x\in\overline{\Omega'}}\Phi_1(x)>0$, there exists a
$c_2>0$ big enough such that for all $c>c_2$, one has
$$
\frac{1}{a_0}K(x)c^{-\mu}\Phi_1(x)^{-\mu s}-c\lambda_1s\Phi_1(x)^{s}<0,\quad
 \forall x\in \overline{\Omega}',
$$
i.e. for all $c>c_2$, $x\in\overline{\Omega}'$,
\begin{equation}
-\Phi_1(x)^{s-2}|\nabla\Phi_1(x)|^2cs(1-s)
+\frac{1}{a_0}K(x)c^{-\mu}\Phi_1(x)^{-\mu s}
-c\lambda_1s\Phi_1(x)^{s}<0.\label{e3.7}
\end{equation}
Now choose a $c>\max\{c_1,c_2\}$. Combining \eqref{e3.6} and \eqref{e3.7}, we have
\begin{equation}
\begin{aligned}
&\Delta u(x) +\frac{1}{a_0}K(x)u(x)^{-\mu}\\
&=-\Phi_1(x)^{s-2}\big[|\nabla\Phi_1(x)|^2cs(1-s)
-\frac{1}{a_0}K(x)c^{-\mu}\Phi_1(x)^{2-(1+\mu)s}\big]-c\lambda_1s\Phi_1(x)^{s}\\
&<0,\quad x\in\Omega.
\end{aligned} \label{e3.8}
\end{equation}
Choose $d=\max\{b,c\}$ and define
$$
u^*(x)=\begin{cases}
d\Phi_1^t(x), \quad x\in\overline{\Omega} &\text{if } \mu>1;\\
d\Phi_1^s(x), \quad x\in\overline{\Omega} &\text{if } 0<\mu\leq 1.
\end{cases}
$$
From \eqref{e3.4} and \eqref{e3.8}, we have
$$
\Delta u^*(x) +\frac{1}{a_0}K(x)u^*(x)^{-\mu}<0,\ \ \forall x\in\Omega.
$$
It follows that for each $n\in\mathbb{N}$,
\begin{equation}
\Delta u^*(x) +\frac{1}{a_0}K(x)\left(u^*(x)+\frac{1}{n}\right)^{-\mu}<\Delta u^*(x)
 +\frac{1}{a_0}K(x)u^*(x)^{-\mu}<0,\label{e3.9}
\end{equation}
for $x\in\Omega$.

Let $b_0=a(\int_{\Omega}|u^*(x)|^{\gamma}dx)$. Choose $\varepsilon>0$ small
enough such that
\begin{equation}
\frac{1}{b_0}K(x)2^{-\mu}-\varepsilon\lambda_1\Phi_1(x)>0,\quad
 \forall x\in\Omega,\label{e3.10}
\end{equation}
and
\begin{equation}
\varepsilon\Phi_1(x)< \min\{1,u^*(x)\}, \quad \forall x\in\Omega.\label{e3.11}
\end{equation}
From \eqref{e3.1}, \eqref{e3.10} and \eqref{e3.11}, one has that
for each $n\in\mathbb{N}$,
\begin{equation}
\Delta \varepsilon\Phi_1(x)+\frac{1}{b_0}K(x)\Big(\varepsilon\Phi_1(x)
+\frac{1}{n}\Big)^{-\mu}>\frac{1}{b_0}K(x)2^{-\mu}-\varepsilon\lambda_1\Phi_1(x)>0,
\label{e3.12}
\end{equation}
for $x\in\Omega$.

Let $u_*(x)=\varepsilon\Phi_1(x)$, $x\in\overline{\Omega}$.
By the definitions of $u_*$ and $u^*$, we have
$$
\max\{|u_*(x)|,|u^*(x)|\}^{\gamma}=u^*(x)^{\gamma}
$$
and so
$$
\sup_{t\in[0,\int_{\Omega}\max\{|u_*(x)|,|u^*(x)|\}^{\gamma}dx]}a(t)
=a\Big(\int_{\Omega}u^*(x)^{\gamma}dx\Big)=b_0.
$$
Then for $n\in\mathbb{N}$, from \eqref{e3.9} and \eqref{e3.12},
we have for each $n\in\mathbb{N}$,
\begin{gather*}
\Delta u^*(x) + \frac{1}{a_0}K(x)(u^*(x)+\frac{1}{n})^{-\mu}<0, \quad x\in\Omega,\\
u^*|_{\partial\Omega}=0
\end{gather*}
and
\begin{gather*}
\Delta u_*(x) + \frac{1}{b_0}K(x)(u_*(x)+\frac{1}{n})^{-\mu}>0, \quad x\in\Omega,\\
u^*|_{\partial\Omega}=0.
\end{gather*}
Now Theorem \ref{thm2.2} guarantees that for $n\in\mathbb{N}$, there exist
$\{u_n\}$ with $u_*(x)\leq u_n(x)\leq u^*(x)$ for all $x\in\overline{\Omega}$
such that
\begin{equation}
\begin{gathered}
 a\Big(\int_{\Omega}|u_n(x)|^{\gamma}dx\Big)\Delta u_n(x) +K(x)(u_n(x)
+\frac{1}{n})^{-\mu}=0, \quad x\in\Omega,\\
u_n|_{\partial\Omega}=0.
\end{gathered}
\label{e3.13}
\end{equation}

Let $\Omega_k=\{x\in\Omega|u_*(x)>\frac{1}{k}\}$, $k\in\mathbb{N}$.
From \eqref{e3.13}, we have
$$
|\Delta u_n(x)|\leq \frac{1}{a_0}K(x)u_*(x)^{-\mu}\
leq \frac{1}{a_0}\max_{x\in\overline{\Omega}}K(x)
(\min_{x\in\overline{\Omega}_k}u_*(x))^{-\mu}, \quad x\in\overline{\Omega}_k,
$$
which implies that
$\{u_n(x)\}$ is equicontinous and uniformly bounded on
$\overline{\Omega}_k$, $ k\in\mathbb{N}$.
Therefore, $\{u_n(x)\}$ has a uniformly convergent subsequence on every
$\overline{\Omega}_k$. By Diagonal method, we can choose a subsequence of
$\{u_n(x)\}$ which converges a $u_0$ on every $\overline{\Omega}_k$ uniformly.
 Without loss of generality, assume that
$$
\lim_{n\to+\infty}u_n(x)=u_0(x), \quad \text{ uniformly on } \overline{\Omega}_k, \;
 k\in\mathbb{N}.
$$
Obviously,
$$
u_*(x)\leq u_0(x)\leq u^*(x), \quad x\in\Omega,
$$
which implies that
$$
\lim_{x\to y\in\partial\Omega}u_0(x)=0, \quad \forall y\in\partial\Omega.
$$
Hence, we define $u_0(x)=0$, for $x\in\partial\Omega$.
And the Dominated Convergence Theorem implies that
$$
\lim_{n\to+\infty}\int_{\Omega}|u_n(x)|^{\gamma}dx
=\int_{\Omega}|u_0(x)|^{\gamma}dx,
$$
which together with the continuity of $a(t)$ yields
$$
\lim_{n\to+\infty}a\Big(\int_{\Omega}|u_n(x)|^{\gamma}dx\Big)
=a\Big(\int_{\Omega}|u_0(x)|^{\gamma}dx\Big).
$$

Now we claim that $u_0\in C^{2+\alpha}(\Omega)$ and that
\begin{equation}
a\Big(\int_{\Omega}|u_0(x)|^{\gamma}dx\Big)\Delta u_0(x)
+ K(x)u_0(x)^{-\mu}= 0,\quad \forall x\in\Omega.\label{e3.14}
\end{equation}
Although the proof is similar as the standard arguments for the the theory
of the Elliptic problems (see \cite{l1}), we still give it in details.

Let $x_0\in\Omega$ and let $r>0$ be chosen so that
$\overline{B(x_0, r)}\subseteq \Omega$, where $B(x_0, r)$
denotes the open ball of radius $r$ centered at $x_0$.
Let $\Psi$ be a $C^{\infty}$ function
which is equal to $1$ on $\overline{B(x_0,r/2)}$ and equal to
$0$ off $\overline{B(x_0,r)}$. We have
$$
\Delta(\Psi(x) u_n(x))=\begin{cases}
 2\nabla\Psi(x)\cdot\nabla u_n(x)+u_n(x)\Delta\Psi(x)\\
+\Psi(x)\frac{1}{a(\int_{\Omega}|u_n(x)|^{\gamma}dx)}K(x)u_n^{-\mu}(x),
& \forall x\in \overline{B(x_0, r)},\\
 0,& \forall x\in \Omega-\overline{B(x_0, r)}.
\end{cases}
$$
Let
$$
p_n(x)=\begin{cases}
\Psi(x)\frac{1}{a(\int_{\Omega}|u_n(x)|^{\gamma}dx)}K(x)u_n^{-\mu}(x),
&\forall x\in \overline{B(x_0, r)},\\
0,&\forall x\in \Omega-\overline{B(x_0, r)}.
\end{cases}
$$
It is easy to see that $p_n$ is a term whose $L^{\infty}$ norm is bounded
independently of
$n$ (note $\inf_{t\in[0,+\infty)}a(t)\geq a(0)=a_0>0$).
Therefore, for $n>1$, we have
$$
\Psi(x)u_n(x)\Delta(\Psi(x)u_n(x))
=\sum_{j=1}^Nb_{n,j}\frac{\partial(\Psi(x)u_n(x))}{\partial x_j}+q_n,
$$
where $b_{n,j}$, $j=1,2,\dots, N$, $q_n$ are terms whose $L^{\infty}$
norm is bounded independently of $n$.
Integrating the above equation, we have that there exist constants
$c_3>0$, $c_4>0$, independent of $n$, such that
$$
\int_{B(x_0,r)}|\nabla(\Psi u_n)|^2dx
\leq c_3 (\int_{B(x_0,r)}|\nabla(\Psi u_n)|^2dx)^{\frac{1}{2}}+c_4.
$$

From this, it follows that the $L^2(B(x_0, r))$-norm of $|\nabla(\Psi u_n)|$
is bounded independently of $n$. Hence, $L^2(B(x_0,\frac{r}{2}))$-norm of
$|\nabla u_n|$ is bounded independently of $n$.
Let $\Psi_1$ be a $C^{\infty}$ function
which is equal to $1$ on $\overline{B(x_0,r/4)}$ and equal to $0$ off
 $\overline{B(x_0,\frac{r}{2})}$.
 We have $\Delta(\Psi_1(x) u_n(x))= 2\nabla\Psi_1(x)\cdot\nabla u_n(x)+p_{n,1}$,
 $p_{n,1}$ is a term whose $L^{\infty}(B(x_0,\frac{r}{2}))$ norm is bounded
independently of $n$.
From standard elliptic theory, the $W^{2,2}(B(x_0,\frac{r}{2}))$-norm
of $\Psi_1u_n$ is bounded independently of $n$ and hence, the
$W^{2,2}(B(x_0,\frac{r}{4}))$-norm of $u_n$ is bounded independently of $n$.
Since the $W^{1,2}(B(x_0,\frac{r}{4}))$-norms of the components of
 $\nabla u_n$ are bounded independently of $n$, it follows
from the Sobolev imbedding theorem that, if $q=2N/(N-2)>2$ if $N > 2$
and $q>2$ is arbitrary if $N\leq 2$, then the $L^q(B(x_0,\frac{r}{4}))$-norm
of $|\nabla u_n|$ is bounded independently of $n$.
If $\Psi_2$ is a $C^{\infty}$ function which is equal to $1$ on
$\overline{B(x_0, \frac{r}{8})}$ and equal to $0$ off
$\overline{B(x_0, \frac{r}{4})}$, then
$\Delta(\Psi_2(x) u_n(x))= 2\nabla\Psi_2(x)\cdot\nabla u_n(x)+p_{n,2}$,
$p_{n,2}$ is a term whose $L^{\infty}(B(x_0,\frac{r}{4}))$ norm is
bounded independently of $n$. Since the righthand
side of the above equation is bounded in $L^q(B(x_0,\frac{r}{4}))$, independently
of $n$, the $W^{2,q}(B(x_0,\frac{r}{4}))$-norm of $\Psi_2u_n$ is also bounded
independently of $n$. Hence, the $W^{2,q}(B(x_0, \frac{r}{8}))$-norm of
 $u_n$ is bounded independently of $n$.
Continuing the line of reasoning, after a finite number of steps,
we find a number $r_1 >0$ and $q_1 > N/(1-\alpha)$ such that the
$W^{2,q_1}(B(x_0,r_1))$-norm of $u_n$ is
bounded independently of $n$. Hence, there is a subsequence of $\{u_n\}$, which
we may assume is the sequence itself, which converges in
$C^{1+\alpha}(B(x_0, r_1))$. If
$\theta$ is a $C^{\infty}$ function which is equal to $1$ on
$\overline{B(x_0, \frac{r_1}{2})}$ and equal to $0$ off $B(x_0, r_1)$, then
$$
\Delta(\theta u_n)=\nabla \Psi\nabla u_n+\tilde{p}_n,
$$
 where $\tilde{p}_n=\theta\Delta u_n+u_n\Delta\theta$.
 The right-hand side of the above equation converges in $C^{\alpha}(B(x_0, r_1))$.
So, by Schauder theory, $\{\theta u_n\}$converges in $C^{2+\alpha}(B(x_0, r_1))$
and hence $\{u_n\}$ converges in $C^{2+\alpha}(B(x_0,\frac{r_1}{2}))$.
Since $x_0\in \Omega$ is arbitrary, this shows that
$u_0\in C^{2+\alpha}(\Omega)$. Clearly, \eqref{e3.14} holds.

Consequently, we have
\begin{gather*}
 a\Big(\int_{\Omega}|u_0(x)|^{\gamma}dx\Big)\Delta u_0(x) +K(x)u_0(x)^{-\mu}=0, \quad
 x\in\Omega,\\
u_0|_{\partial\Omega}=0.
\end{gather*}
By \cite[Theorem 1]{l1}, we have if $\mu>1$, there exist a $b_1>0$ and $b_2>0$
such that
$$
b_1\Phi_1(x)^{\frac{2}{1+\mu}}\leq u_0(x)\leq b_2\Phi_1(x)^{\frac{2}{1+\mu}},\quad
 \forall x\in\overline{\Omega}.
$$

Next we consider the uniqueness of positive solutions of \eqref{e3.1}.
Assume that $u_1$ and $u_2$ are two positive solutions.
Let $c_i=(a(\int_{\Omega}u_i(x)^{\gamma}dx))^{1/(\mu+1)}$ and
$v_i=c_iu_i$, $i=1,2$.
Then $v_i$ satisfies
\begin{gather*}
-\Delta v_i=K(x)v_i^{-\mu},\\
v_i|_{\partial\Omega}=0.
\end{gather*}
Now \cite{l1} guarantees that
\begin{gather*}
-\Delta v=K(x)v^{-\mu},\\
v|_{\partial\Omega}=0
\end{gather*}
has a unique positive solution, which implies $v_1=v_2$, i.e.,
\begin{equation}
\Big(a\Big(\int_{\Omega}u_1(x)^{\gamma}dx\Big)\Big)^{1/(\mu+1)}u_1(x)
=\Big(a\Big(\int_{\Omega}u_2(x)^{\gamma}dx\Big)\Big)^{1/(\mu+1)}u_2(x), \quad
\label{e3.15}
\end{equation}
for $x\in\overline{\Omega}$, and so
$$
\Big(a\Big(\int_{\Omega}u_1(x)^{\gamma}dx\Big)\Big)^{\gamma/(\mu+1)}u_1^{\gamma}(x)
=\Big(a\Big(\int_{\Omega}u_2(x)^{\gamma}dx\Big)\Big)^{\gamma/(\mu+1)}u_2^{\gamma}(x),
 \quad \forall x\in\overline{\Omega}.
$$
Integration on $\Omega$ yields
 $$
\Big(a\Big(\int_{\Omega}u_1(x)^{\gamma}dx\Big)\Big)^{\gamma/(\mu+1)}
\int_{\Omega}u_1^{\gamma}(x)dx
=\Big(a\Big(\int_{\Omega}u_2(x)^{\gamma}dx\Big)\Big)^{\gamma/(\mu+1)}
\int_{\Omega}u_2^{\gamma}(x)dx .
$$
 The monotonicity of $a$ implies that $(a(t))^{\gamma/(\mu+1)}t$ is increasing
on $[0,+\infty)$, which guarantees that
$$
\int_{\Omega}u_1(x)^{\gamma}dx=\int_{\Omega}u_2(x)^{\gamma}dx,
$$
and so
$$
\Big(a\Big(\int_{\Omega}u_1(x)^{\gamma}dx\Big)\Big)^{1/(\mu+1)}
=\Big(a\Big(\int_{\Omega}u_2(x)^{\gamma}dx\Big)\Big)^{1/(\mu+1)},
$$
which together with \eqref{e3.15} yields
$u_1(x)=u_2(x)$.
The proof is complete.
\end{proof}


\begin{theorem} \label{thm3.2}
 The solution $u$ of Theorem \ref{thm3.1} is in $W^{1,2}$ if and only if $\mu<3$. 
If $\mu>1$, then $u$ is not in $C^1(\overline{\Omega})$.
\end{theorem}

\begin{proof}
Suppose $u$ is a positive solution in Theorem \ref{thm3.1}.
Let 
\[
p(x)=\frac{K(x)}{a\big(\int_{\Omega}|u(x)|^{\gamma}dx\big)}.
\]
Then $p\in C(\overline{\Omega})$, $p(x)>0$ for all $x\in\overline{\Omega}$
and $u(x)$ satisfies that
\begin{equation}
\begin{gathered}
-\Delta u=p(x)u^{-\mu},\\
u|_{\partial\Omega}=0.
\end{gathered}\label{e3.16}
\end{equation}
By \cite[Theorem 2]{l1}, $u$ is in $W^{1,2}$ if and only if $\mu<3$. If
$\mu>1$, then $u$ is not in $C^1(\overline{\Omega})$.The proof is complete.
\end{proof}

 The monotonicity of $a(t)$ on $[0,+\infty)$ is very important for the
uniqueness of positive solution to \eqref{e1.1}.
For example, assume that $c=\int_{\Omega}|u_1(x)|dx$, where $u_1$ is the
unique positive solution of the following problem (see \cite[Theorem 1]{l1}
\begin{equation}
\begin{gathered}
-\Delta u=u^{-\mu},\\
u|_{\partial\Omega}=0.
\end{gathered} \label{e3.17}
\end{equation}
Let
$$
a(t)=\begin{cases}
3, &t=0;\\
2+((\frac{t}{c})^{-(1+\mu)}-2)|\sin \frac{t}{c}|^{1+\mu},&t>0.
\end{cases}
$$
It is easy to see that $a(t)$ is not monotone on $[0,+\infty)$.
Let $\lambda_k=2k\pi+\frac{\pi}{2}$. Then
\begin{equation}
a(\lambda_k c)=2+((\lambda_k)^{-(1+\mu)}-2)|\sin \lambda_k|^{1+\mu}
=(\lambda_k)^{-(1+\mu)},\ \ k\in\mathbb{N}.\label{e3.18}
\end{equation}
Let $u_k(x)=\lambda_ku_1(x)$, $x\in\overline{\Omega}$.
 Then, from \eqref{e3.17} and \eqref{e3.18}, we have
 $$
\Delta u_k(x)=\lambda_k\Delta u_1(x)=-\lambda_ku_1^{-\mu}(x), \quad x\in\Omega,
$$
 and
\begin{align*}
 \frac{1}{a(\int_{\Omega}|u_k(x)|dx)}u_k(x)^{-\mu}
&=\frac{1}{a(\int_{\Omega}\lambda_k|u_1(x)|dx)}u_k(x)^{-\mu}\\
 &=\frac{1}{a(\lambda_kc)}(\lambda_ku_1(x))^{-\mu} \\
 &=\lambda_k^{1+\mu}\lambda_k^{-\mu}u_1(x)^{-\mu}
=\lambda_k u_1(x)^{-\mu}
 \end{align*}
Hence,
\begin{gather*}
 \Delta u_k(x)+\frac{1}{a(\int_{\Omega}|u_k(x)|dx)}u_k(x)^{-\mu}=0,\quad x\in\Omega,\\
 u_k|_{\partial\Omega}=0,
 \end{gather*}
i.e.,
\begin{gather*}
 a\Big(\int_{\Omega}|u(x)|dx\Big) \Delta u(x)+u(x)^{-\mu}=0,\quad x\in\Omega,\\
 u|_{\partial\Omega}=0
 \end{gather*}
has at  infinitely many positive solutions.

\section{Global structure of positive solutions for \eqref{e1.2}}

 In this section, we consider the singular nonlocal elliptic problems \eqref{e1.2},
where $q\in(0,+\infty)$, $\mu>0$, $K\in C^{\alpha}(\overline{\Omega})$ 
with $K(x)>0$ for all $x\in\overline{\Omega}$.

To sutudy equation \eqref{e1.2}, for each $n\in \mathbb{N}$, we study the 
equations
\begin{equation}
\begin{gathered}
 a\Big(\int_{\Omega}|u(x)|^{\gamma}dx\Big)\Delta u(x)
+\lambda\big[u^q+K(x)\big(u(x)+\frac{1}{n}\big)^{-\mu}\big]=0,\quad x\in\Omega,\\
 u|_{\partial\Omega}=0.
 \end{gathered} \label{e4.1}
\end{equation}
Let $u$ denote the inward normal derivative of $u$ on $\partial\Omega$ and define
$$
P=\{u\in C^{1,\alpha}(\overline{\Omega}): u(x)>0\; \forall x\in\Omega, \;
 u(x)=0 \text{ on } \partial\Omega \text{ and }\frac{\partial u}{\partial v}>0
\text{ on } \partial\Omega\},
$$
where $\alpha\in(0,1)$. It follows from \cite[Theorem 3.7]{r1} that for
$n\in\mathbb{N}$ there is a set $C_n$ of solutions of \eqref{e4.1}
which is a connected and unbounded subset of
$\mathbb{R}^+\times (P\cup \{(0,0)\})$ (in the topology of
$\mathbb{R}\times C^{1,\alpha}(\overline{\Omega})$) and contains $(0,0)$.
Obviously,
$$
\|u\|\leq \|u\|_{1+\alpha},\quad \forall u\in C_n,
$$
which guarantees that
\begin{equation}
\begin{gathered}
\|u\|\to +\infty \text{ implies that }\|u\|_{1+\alpha}\to+\infty,
\forall u\in C_n,\\
\|u-u_0\|_{1+\alpha}\to0 \hbox { implies that }\|u-u_0\|\to 0.
\end{gathered}\label{e4.2}
\end{equation}
On the other hand, by Lemma \ref{lem2.1} and Theorem \ref{thm2.1}, for $u\in C_n$, one has
\begin{align*}
\|u\|_{1+\alpha}
&\leq C(n,p,\Omega)\|u\|_{2,p}\\
&\leq C(n,p,\Omega)\lambda\frac{1}{a(\int_{\Omega}|u(x)|^{\gamma}dx)}
\Big(\int_{\Omega}\big[u^q+K(x)\big(u(x)+\frac{1}{n}\big)^{-\mu}\big]^pdx
 \Big)^{1/p}\\
&\leq C(n,p,\Omega)\lambda\frac{1}{a_0}
\Big(\int_{\Omega}\big[u^q+K(x)\big(u(x)+\frac{1}{n}\big)^{-\mu}\big]^pdx
 \Big)^{1/p}\\
&\leq C(n,p,\Omega)\lambda\frac{1}{a_0}|\Omega|^{1/p}[\|u\|^q+n\|K\|],\quad
 \forall u\in C_n
\end{align*}
and
\begin{align*}
\|u-u_0\|_{1+\alpha}
&\leq C(n,p,\Omega)\|u-u_0\|_{2,p}\\
 &\leq C(n,p,\Omega)\lambda\Big(\int_{\Omega}|\Psi_n(u)(x)
-\Psi_n(u_0)(x)|^pdx\Big)^{1/p},\quad \forall u, u_0\in C_n,
\end{align*}
where
$$
\Psi_n(u)(x)=\frac{1}{a(\int_{\Omega}|u(x)|^{\gamma}dx)}[u^q(x)
+\frac{1}{(u(x)+\frac{1}{n})^{\mu}}],
$$
which guarantees that
\begin{equation} \begin{gathered}
\|u\|_{1+\alpha}\to +\infty \text{ implies that }\|u\|\to+\infty, \forall u\in C_n,\\
\|u-u_0\|\to0\hbox { implies that }\|u-u_0\|_{1+\alpha}\to 0.
\end{gathered}\label{e4.3}
\end{equation}
Combining \eqref{e4.2} and \eqref{e4.3}, we know that $C_n$ is connected and
unbounded in $\mathbb{R}\times C(\overline{\Omega})$.

Let $\phi\in C^{2,\alpha}(\overline{\Omega})$ defined by
\begin{equation}
-\Delta \phi=1,\quad x\in\Omega; \phi(x)=0,\quad  x\in\partial\Omega.\label{e4.4}
\end{equation}

\begin{lemma} \label{lem4.1} 
 Let $M>0$ and $(\lambda_n, u_n)\in (0,+\infty)\times P$ be a solution of
\eqref{e4.1} satisfying $\lambda_n\leq M$ and $\|u_n\|\leq M$. 
There is a number $\overline{\varepsilon}>0$ and a pair of functions 
$\overline{\Gamma}(M)>0$, $\overline{K}(\beta, M)>0$ such that if $\phi$
is given by \eqref{e4.3} and $0<\frac{1}{n}<\overline{\varepsilon}$, then
\begin{equation}
\lambda_n\overline{\Gamma}(M)\phi(x)\leq u_n(x)
\leq\beta+\lambda_n\overline{K}(\beta, M)\phi(x), \quad x\in\Omega\label{e4.5}
\end{equation}
for $\beta\in (0, M]$.
\end{lemma}

 \begin{proof} Set
\begin{equation}
\overline{K}(\beta,M)=\max\{\frac{1}{a_0}(r^q+K(x)r^{-\mu})
:(x,r)\in\overline{\Omega}\times[\beta,1+M]\}.\label{e4.6}
\end{equation}
Let $(\lambda_n, u_n)$ be as in the Lemma \ref{lem4.1}, $0 <\frac{1}{n}<1$
and $\beta\in(0,M]$. Set
$A_{\beta}=\{x\in\Omega|u_n(x)>\beta\}$. By \eqref{e4.4} and \eqref{e4.6},
one has
\begin{align*}
&-\Delta(\beta+\lambda_n\overline{K}(\beta,M)\phi-u_n) \\
&=\lambda_n\overline{K}(\beta,M)-\lambda_n\frac{1}
{a\big(\int_{\Omega}u_n(x)^{\gamma}dx\big)}[u_n^q+K(x)(u_n+\frac{1}{n})^{-\mu}]\\
&\geq \lambda_n\overline{K}(\beta,M)-\lambda_n\frac{1}{a_0}[u_n^q+K(x)(u_n)^{-\mu}]
\geq 0,\quad x\in A_{\beta},
\end{align*}
and
$$
u_n(x)=\beta,\quad x\in\partial A_{\beta}.
$$
Thus $\beta+\lambda_n\overline{K}(\beta,M)\phi(x)\geq u_n(x)$ on
$\overline{A}_{\beta}$ by the maximum principle and the
right-hand inequality of \eqref{e4.5} is established.

To obtain the left-hand inequality, choose $R>0$ so that
\[
\frac{1}{a(\int_{\Omega}(\beta+M\overline{K}(\beta, M)\phi(x))^{\gamma}dx)}K(x)
r^{-\mu}>1
\]
 if $0<r<R$. Define
$\overline{\Gamma}(M)=\min\{1, R/(2M\|\phi\|)\}$. 
Then, for $\frac{1}{n}\in(0,R/2]$, $\eta\in(0,\overline{\Gamma}(M)]$ and
$\lambda_n\in(0,M]$, from the
right-hand inequality of \eqref{e4.5} and the monotonicity of $a(t)$, one has
\begin{equation}
\begin{aligned}
-\Delta(\lambda_n\eta\phi(x))
&=\lambda_n\eta\\
&<\lambda_n\frac{1}{a(\int_{\Omega}(\beta+M\overline{K}
 (\beta, M)\phi(x))^{\gamma}dx)}K(x)(\lambda_n\eta\phi+\frac{1}{n})^{-\mu}\\
&\leq\lambda_n\frac{1}{a(\int_{\Omega}u_n(x)^{\gamma}dx)}
 [(\lambda_n\eta\phi)^q+K(x)(\lambda_n\eta\phi+\frac{1}{n})^{-\mu}].
\end{aligned}\label{e4.7}
\end{equation}
From this we will deduce that $\lambda_n\overline{\Gamma}(M)\phi(x)<u_n(x)$,
 $x\in\Omega$. Since $\frac{\partial u_n}{\partial v}|_{\partial\Omega}>0$,
$u_n(x)>0$ for $x\in\Omega$, there exists a $\Omega'\subset\subset\Omega$
and $m>0$ such that $\frac{\partial u_n}{\partial v}|_{\partial\Omega}\geq m>0$
for all $x\in \overline{\Omega}-\Omega'$ and $u_n(x)\geq m>0$ for all
$x\in\overline{\Omega}'$, which implies that there exists a $s>0$ such that
$$
u_n-\tau\lambda_n\phi\in P,\ \ \forall \tau\in[0,s].
$$
Since $\lim_{s\to+\infty}\|s\lambda_n\phi\|=+\infty$, there exists a $s'>0$
such that $u_n-s'\lambda_n\phi\not\in P$.
Define
$$
\eta^*=\sup\{s>0|u_n-\tau\lambda_n\phi\in P,\quad \forall \tau\in[0,s]\}.
$$
It is easy to see that $0<\eta^*\leq s'$ and
$u_{n}-\eta \lambda_n\phi\in P$ for $0<\eta<\eta^*$ and
$u_{n}-\eta^* \lambda_n\phi\not\in P$. It suffices to
show $\eta^*>\overline{\Gamma}(M)$. If $\eta^*\leq\overline{\Gamma}(M)$,
let $w=u_n-\lambda_n\eta^*\phi\geq 0$ in $\overline{\Omega}$ and, by \eqref{e4.7}
for $C>0$, we have
\begin{align*}
-\Delta w+ Cw
&=Cw+\lambda_n\frac{1}{a(\int_{\Omega}u_n(x)^{\gamma}dx)}
 [u_n(x)^q+K(x)(u_n(x)+\frac{1}{n})^{-\mu}]-\lambda_n\eta^*\\
&>Cw+\lambda_n\frac{1}{a(\int_{\Omega}u_n(x)^{\gamma}dx)}
 \Big([u_n(x)^q+K(x)(u_n(x)+\frac{1}{n})^{-\mu}] \\
&\quad -[(\lambda_n\eta^*\phi)^q+K(x)(\lambda_n\eta^*\phi+\frac{1}{n})^{-\mu}]\Big).
\end{align*}
By the Mean Value Theorem we have
 $$
[u_n(x)^q+K(x)(u_n(x)+\frac{1}{n})^{-\mu}]-[(\lambda_n\eta^*\phi(x))^q+K(x)
(\lambda_n\eta^*\phi(x)+\frac{1}{n})^{-\mu}]\geq C_0w,
$$
where
$$
C_0=\min_{x\in\overline{\Omega}}\inf_{r\in[\frac{1}{n},
\frac{1}{n}+\|u_n\|+\lambda_n\eta^*\|\phi\|]}K(x)(-\mu)r^{-(1+\mu)}.
$$
Choose
$$
C+\lambda_n\frac{1}{a(\int_{\Omega}u_n(x)^{\gamma}dx)}C_0>0.
$$
Then
$$
-\Delta w+Cw>0,
$$
which means that
$w\in P$. This is a contradiction. Consequently,
 $\eta^*>\overline{\Gamma}(M)$ and so
 $\lambda_n\overline{\Gamma}(M)\phi(x)<u_n(x)$, $x\in\Omega$.
The proof is complete.
\end{proof}

\begin{theorem} \label{thm4.1} 
There is a set $C$ of solutions
of \eqref{e1.2} satisfying the following:
\begin{itemize}
\item[(i)] $C$ is connected in $\mathbb{R}\times C(\overline{\Omega})$;

\item[(ii)] $C$ is unbounded in $\mathbb{R}\times C(\overline{\Omega})$;

\item[(iii)] $(0,0)$ lies in the closure of $C$ in
 $\mathbb{R}\times C(\overline{\Omega})$.
\end{itemize}
\end{theorem}


\begin{proof} For $M>0$, define
$$
B((0,0),M)=\{(\lambda,u)\in \mathbb{R}\times C(\overline{\Omega})|\lambda^2
+\|u\|^2<M^2\}.
$$
Let $(\lambda_{n}, u_n)\in \partial B((0,0),M)\cap (0,+\infty)\times P$
be solutions of \eqref{e4.1} as above, $n\to+\infty$ and $\lambda_n\to \lambda$. 
If $\lambda=0$, we deduce from \eqref{e4.5} that
$$
0< \limsup_{n\to+\infty} \sup_{x\in\overline{\Omega}}u_n(x)\leq \beta, \quad
 \forall \beta\in(0,M]
$$
and hence that $u_n\to 0$ in $C(\overline{\Omega})$. Then 
$(\lambda_n,u_n)\to (0,0)$ as $n\to+\infty$ in 
$\mathbb{R}\times C(\overline{\Omega})$. Since 
$(\lambda_n,u_n)\in \partial B((0,0),M)$, this is impossible. Then $\lambda>0$.

From \eqref{e4.5} and $\lambda> 0$, we see that $u_n$ is bounded from below
by a function which is positive in $\Omega$ and from above by a constant.
Arguing as in the proof of Theorem \ref{thm3.1}, without loss of generality, passing to
the limit in \eqref{e4.5},
there is a $u_0\in C(\overline{\Omega})$ such that
\begin{equation}
\lim_{n\to+\infty}u_n(x)=u_0(x), \quad
 \text{ uniformly } x\in\overline{\Omega}_0\subset\Omega,\label{e4.8}
\end{equation}
where $\Omega_0$ is arbitrary sub-domain in $\Omega$ and
\begin{equation}
\lambda\overline{\Gamma}(M)\phi(x)\leq u(x)
\leq\beta+\lambda\overline{K}(\beta, M)\phi(x), \quad x\in\Omega\label{e4.9}
\end{equation}
for $\beta\in (0,M]$.
From \eqref{e4.5} and \eqref{e4.9} we have
$$
\lim_{x\to\partial \Omega}u_0(x)=0
$$
and
\begin{equation}
\lim_{x\to\partial \Omega}u_n(x)=0, \quad\text{uniformly for } n\in \mathbb{N}.
\label{e4.10}
\end{equation}
Now \eqref{e4.8} and \eqref{e4.10} imply that $u_n\to u_0$ as $n\to+\infty$.
It follows that $(\lambda_n,u_n)\to(\lambda, u_0)$ in
$\mathbb{R}\times C(\overline{\Omega})$ and hence
$(\lambda,u_0)\in \partial B((0,0),M)$.

A standard argument as the proof of Theorem \ref{thm3.1} shows that
$u_0$ satisfies 
\begin{gather*}
 a\Big(\int_{\Omega}|u_0(x)|^{\gamma}dx\Big)\Delta u_0(x) 
+ \lambda(u_0(x)^q+K(x)u_0(x)^{-\mu})=0, \quad x\in\Omega,\\
u_0|_{\partial\Omega}=0.
\end{gather*}
Wee omit the proof.

At this point we have shown that if $B((0,0),M)$ is a bounded neighborhood
of $(0,0)$ in $\mathbb{R}\times C(\overline{\Omega})$, then there is 
a solution $(\lambda, u_0)\in \partial B((0,0),M))$ of
\eqref{e1.2}. Since $M$ is arbitrary, 
$C=\{(\lambda,u_{\lambda})\in B((0,0),M)| u_{\lambda}$ is a positive 
solution for \eqref{e1.2}. The proof is complete.
\end{proof}

\begin{corollary} \label{coro4.1}
 If $q<1$, then $\lambda\in(0,+\infty)$. In particular,
\eqref{e1.2} with $\lambda=1$ has a solution.
\end{corollary}

\begin{proof} 
Suppose $C$ is the connected and unbounded set of positive solutions 
for \eqref{e1.2} in Theorem \ref{thm4.1}. Now we show that $\lambda\in(0,+\infty)$.

In fact, suppose set $\{\lambda|(\lambda,u)\in C\}$ is finite and let
 $\Lambda_0=\{\lambda>0|(\lambda,u)\in C\}$. The unboundedness of $C$ 
means that there exist $\{(\lambda_n,u_n)\}$ such that
$$
\lim_{n\to+\infty}\|u_n\|=+\infty.
$$
Set
$A_{1}=\{x\in\Omega|u_n(x)>1\}$ and
\begin{equation}
\overline{K}_n=\frac{1}{a_0}(\|u_n\|^q
+\max_{x\in\overline{\Omega}}K(x)).\label{e4.11}
\end{equation}
It follows from \eqref{e4.4} and \eqref{e4.11} that
\begin{align*}
 -\Delta(1+\lambda_n\overline{K}_n\phi-u_n)
&=\lambda_n\overline{K}_n-\lambda_n\frac{1}{a(\int_{\Omega}u_n(x)^{\gamma}dx)}
 [u_n^q+K(x)(u_n)^{-\mu}]\\
&\geq \lambda_n\overline{K}_n-\lambda_n\frac{1}{a_0}[\|u_n\|^q
 +\max_{x\in\overline{\Omega}}K(x)]
\geq 0,\quad x\in A_{1},
\end{align*}
and
$$
u_n(x)=1,\quad x\in\partial A_{1}.
$$
Thus
$1+\lambda_n\overline{K}_n\phi(x)\geq u_n(x)$ on $\overline{A}_{1}$
by the maximum principle and so
$$
u_n(x)\leq 1+\lambda_n\overline{K}_n\phi(x),\quad
 \forall x\in\overline{\Omega},
$$
which implies
$$
\|u_n\|\leq 1+\Lambda_0(\|u_n\|^q+\max_{x\in\overline{\Omega}}K(x))
\max_{x\in\overline{\Omega}}\phi(x).
$$
By $q<1$, one has
$$
1\leq \lim_{n\to+\infty}\Big[\frac{1}{\|u_n\|}+\Lambda_0(\|u_n\|^{q-1}
+\max_{x\in\overline{\Omega}}K(x)/\|u_n\|)\max_{x\in\overline{\Omega}}\phi(x)\Big]=0.
$$
This is a contradiction. Therefore, $\Lambda_0=+\infty$.
The proof is complete.
\end{proof}

Now we consider the case $q>1$. Let $K(x)=K(|x|)$ and we consider the problem 
\eqref{e1.2} when
 $\Omega=\{x\in \mathbb{R}^N|0<r_1<|x|<r_2\}$ and $N\geq 3$ and discuss the radial 
positive solutions for \eqref{e1.2}, i.e., \eqref{e1.2} is equivalent to 
the  problem
\begin{equation}
\begin{gathered}
\begin{aligned}
&-a\left(N\omega_N\int_{r_1}^{r_2}r^{N-1}|u(r)|^{\gamma}dr\right)(u''_{rr}
+\frac{N-1}{r}u_r) \\
&=\lambda [u(r)^{q}+K(|r|)u^{-\mu}(r))],\quad r\text{in } (r_1,r_2),
\end{aligned}\\
u(r)>0, \quad t\in(r_1,r_2),\\
u(r_1)=0,\quad u(r_2)=0,
\end{gathered}\label{e4.12}
\end{equation}
where $\omega_N$ denotes the area of unit sphere in $\mathbb{R}^N$.

By \cite{o1}, applying the change of variable $t=l(r)$ and $u(r)=z(t)$ with
$$
t=l(r)=-\frac{A}{r^{N-2}}+B\Longleftrightarrow r=(\frac{A}{B-t})^{\frac{1}{N-2}},
$$
where
$$
A=\frac{(r_1r_2)^{N-2}}{r_2^{N-2}-r_1^{N-2}},\quad
B=\frac{r_2^{N-2}}{r_2^{N-2}-r_1^{N-2}},
$$
we obtain
\begin{align*}
&N\omega_N\int_{r_1}^{r_2}r^{N-1}|u(r)|^{\gamma}dr \\
&=N\omega_N\int_0^1(\frac{A}{B-s})^{\frac{N-1}{N-2}}{A}^{\frac{1}{N-2}}
 \frac{1}{N-2}(B-s)^{-\frac{N-1}{N-2}}|z(s)|^{\gamma}ds\\
&=A_N\int_0^1B_N(s)|z(s)|^{\gamma}ds\\
\end{align*}
where
$$
A_N=N\frac{\omega_N}{N-2}A^{\frac{N}{N-2}},\quad
B_N(s)=(B-s)^{\frac{2(N-1)}{2-N}},
$$
and
\begin{gather*}
u'_r=z'_tt'_r=z'_t(-A)(2-N)r^{1-N}, \\
 u''_{rr}=z''_{tt}((-A)(2-N)r^{1-N})^2+z'_t(-A)(2-N)(1-N)r^{-N},
\end{gather*}
which implies 
$$
u''_{rr}+\frac{N-1}{r}u_r=((-A)(2-N)r^{1-N})^2z''_{tt}.
$$
And then \eqref{e4.12} is equivalent to  the  problem
\begin{equation}
\begin{gathered}
\begin{aligned}
&-a\Big(A_N\int_{0}^{1}B_N(s)|z(s)|^{\gamma}ds\Big)z''(t)\\
&=\lambda d(t)[z(t)^q+K((\frac{A}{B-t})^{1/(N-2)})z^{-\mu}(t))],\quad
 t\text{ in } (0,1),
\end{aligned}\\
z(t)>0, \quad t\in(0,1),\\
z(0)=0, \quad z(1)=0,
\end{gathered}\label{e4.13}
\end{equation}
where
$$
d(t)=\frac{A^{2/(2-N)}}{(N-2)^2(B-t)^{2(N-1)/(N-2)}},\quad t\in[0,1]
$$
and the related integral equation is
\begin{equation}
\begin{aligned}
z(t)&=\lambda\frac{1}{a\big(A_N\int_{0}^{1}B_N(s)|z(s)|^{\gamma}ds\big)}
\int_0^1G(t,s)d(s)\\
&\quad\times \big[z(s)^q+K((\frac{A}{B-s})^{1/(N-2)})z^{-\mu}(s)\big]ds,
\end{aligned}\label{e4.14}
\end{equation}
for $t\in(0,1)$, where
$$
G(t,s)=\begin{cases}
s(1-t),& 0\leq s\leq t\leq 1;\\
t(1-s),& 0\leq t\leq s\leq 1.
\end{cases}
$$

\begin{lemma}[{see \cite[page 18]{a2}}] \label{lem4.2} 
 Suppose $z\in C[0,1]$ is concave on $[0,1]$ with $z(t)\geq 0$ for all $t\in[0,1]$. 
Then 
 $z(t)\geq \|z\|t(1-t)$ for $t\in [0,1]$
\end{lemma}

\begin{corollary} \label{coro4.2}
 If $\lim_{t\to+\infty}\frac{t^{q-1}}{a(t^{\gamma})}=+\infty$, then
$C$ in Theorem \ref{thm4.1} satisfies:
\begin{itemize}
\item[(i)] there exists $\Lambda_0>$ satisfying 
$C\cap((\Lambda_0,+\infty)\times C_0[0,1])=\emptyset$;

\item[(ii)] for every $\lambda\in(0,\Lambda_0]$, 
$C\cap([0, \lambda]\times C_0[0,1])$ is unbounded;

\item[(iii)] there exists $\lambda_0\leq \Lambda_0$ such that for every 
$\lambda \in (0,\lambda_0)$, \eqref{e4.10} has at least two
positive solutions
$z_{1,\lambda}$ and $z_{2,\lambda}$ with
$$
\lim_{\lambda\to 0, (\lambda,z_{1,\lambda})\in C}\|z_{1,\lambda}\|=0,\quad
\lim_{\lambda\to 0, (\lambda,z_{2,\lambda})\in C}\|z_{2,\lambda}\|=+\infty.
$$
\end{itemize}
\end{corollary}

\begin{proof} 
(i) Suppose that $(\lambda,z_{\lambda})\in C$. 
Since $z_{\lambda}''(t)\leq 0$ and $z_{\lambda}(0)=z_{\lambda}(1)=0$, 
we have $z$ is concave on $[0,1]$ with $z(t)\geq 0$ for all $t\in[0,1]$. 
Now  Lemma \ref{lem4.2} implies
$$
z_{\lambda}(t)\geq t(1-t)\|z_{\lambda}\|,\quad \forall t\in [0,1].
$$
If $\|z_{\lambda}\|\leq 1$, it follows from \eqref{e4.14}
\begin{align*}
1&\geq\|z_{\lambda}\|\\
&=\lambda\frac{1}{a(A_N\int_{0}^{1}B_N(s)|z_{\lambda}(s)|^{\gamma}ds)}
\max_{t\in[0,1]}\int_0^1G(t,s)d(s) \\
&\quad\times \big[z_{\lambda}(s)^q+K((\frac{A}{B-s})^{1/(N-2)})z_{\lambda}^{-\mu}(s)
 \big]ds\\
&>\lambda\frac{1}{a(A_N\int_{0}^{1}B_N(s)ds)}\max_{t\in[0,1]}
\int_0^1G(t,s)d(s)K((\frac{A}{B-s})^{1/(N-2)})ds,
\end{align*}
and so
\begin{equation}
\lambda\leq \frac{a(A_N\int_{0}^{1}B_N(s)ds)}{\max_{t\in[0,1]}
\int_0^1G(t,s)d(s)K((\frac{A}{B-s})^{1/(N-2)})ds}. \label{e4.15}
\end{equation}
Since
$$
\lim_{t\to+\infty}\frac{t^{q-1}}{a(t^{\gamma})}=+\infty,
$$
one has
\begin{equation}
\lim_{t\to+\infty}\frac{t^{q-1}}{a(t^{\gamma}A_N\int_{0}^{1}B_N(s)ds)}
=\lim_{s\to+\infty}\frac{s^{q-1}(A_N\int_{0}^{1}B_N(s)ds)^{-(q-1)/\gamma}}
{a(s^{\gamma})}=+\infty,\label{e4.16}
\end{equation}
which implies that there is an $M_0>0$ such that
\begin{equation}
\frac{a(t^{\gamma}A_N\int_{0}^{1}B_N(s)ds)}{t^{q-1}}\leq M_0,\quad
 \forall t\in[1,+\infty).\label{e4.17}
\end{equation}
If $\|z_{\lambda}\|\geq 1$, from \eqref{e4.14} and \eqref{e4.17}, one has
\begin{align*}
\|z_{\lambda}\|
&\geq \lambda\frac{1}{a\big(\|z\|^{\gamma}A_N\int_{0}^{1}B_N(s)ds\big)}
\max_{t\in[0,1]}\int_{0}^{1}G(t,s)d(s)[z_{\lambda}(s)^q]ds\\
&\geq \lambda\frac{\|z_{\lambda}\|^q}
{a\big(\|z_{\lambda}\|^{\gamma}A_N\int_{0}^{1}B_N(s)ds\big)}
\max_{t\in[0,1]}\int_{0}^{1}G(t,s)d(s)[s(1-s)]^qds,\\
\end{align*}
and so
\begin{equation}
\begin{aligned}
\lambda
&\leq \frac{a\big(\|z_{\lambda}\|^{\gamma}A_N\int_{0}^{1}B_N(s)ds\big)}
{\|z\|^{q-1}}\frac{1}{\max_{t\in[0,1]}\int_{0}^{1}G(t,s)d(s)[s(1-s)]^qds}\\
&\leq M_0\frac{1}{\max_{t\in[0,1]}\int_{0}^{1}G(t,s)d(s)[s(1-s)]^qds}.
\end{aligned}\label{e4.18}
\end{equation}
It follows from \eqref{e4.15} and \eqref{e4.18} that
\begin{gather*}
\Lambda_0=\sup\{\lambda|(\lambda,z_{\lambda})\in C\}<+\infty, \\
C\cap((\Lambda_0,+\infty)\times C_0[0,1])=\emptyset.
\end{gather*}

(ii) For every $\lambda\in(0,\Lambda_0]$, we show that 
$C\cap([\lambda,\Lambda_0]\times C_0[0,1])$ is bounded.
In fact, if $C\cap([\lambda,\Lambda_0]\times C_0[0,1])$ is unbounded, 
there is $\{(\lambda_n,z_n)\}\subseteq C\cap([\lambda,\Lambda_0]\times C_0[0,1])$ 
such that
$$
\lambda_n^2+\|z_n\|^2\to+\infty, \quad \text{ as }n\to+\infty.
$$
Since $\{\lambda_n\}\subseteq [\lambda,\Lambda_0]$ is bounded, 
without loss of generality, we assume that $\lambda_n\to\lambda'>0$ as $n\to+\infty$.
It implies that 
$$
\|z_n\|^2\to+\infty, \quad \text{ as }n\to+\infty.
$$
From \eqref{e4.14}, one has
\begin{align*}
\|z_n\|
&\geq \lambda_n\frac{1}{a(\|z_n\|^{\gamma}A_N\int_{0}^{1}B_N(s)ds)}
 \max_{t\in[0,1]}\int_{0}^{1}G(t,s)d(s)[z_n(s)^q]ds\\
&\geq \lambda_n\frac{\|z_n\|^q}{a(\|z_n\|^{\gamma}A_N\int_{0}^{1}B_N(s)ds)}
 \max_{t\in[0,1]}\int_{0}^{1}G(t,s)d(s)[s(1-s)]^qds,\\
\end{align*}
and so
$$
1\geq \lambda\frac{\|z_n\|^{q-1}}{a(\|z_n\|^{\gamma}A_N\int_{0}^{1}B_N(s)ds)}
\max_{t\in[0,1]}\int_{0}^{1}G(t,s)d(s)[s(1-s)]^qds.
$$
From \eqref{e4.16}, letting $n\to+\infty$, one has
$1\geq +\infty$. This is a contradiction.
Hence, $C\cap([\lambda,\Lambda_0]\times C_0[0,1])$ is bounded for any 
$\lambda\in(0,\Lambda_0]$.


(iii) Choose $R>1>r>0$. Suppose $(\lambda,z_{\lambda})\in C$ with
 $r\leq \|z_{\lambda}\|\leq R$. By
$$
z^q+K(x)z^{-\mu}\geq z^q+\min_{x\in\overline{\Omega}}K(|x|) z^{-\mu},
$$
there is a $c_0>0$ such that
\begin{equation}
z^q+K(x)z^{-\mu}\geq c_0,\quad \forall z\in (0,+\infty), x\in\overline{\Omega}.
\label{e4.19}
\end{equation}
From \eqref{e4.14} and \eqref{e4.19} it follows that
\begin{align*}
z_{\lambda}(t)
&=\lambda\frac{1}{a(A_N\int_{0}^{1}B_N(s)|z_{\lambda}(s)|^{\gamma}ds)}
 \int_0^1G(t,s)d(s) \\
&\quad\times \big[z_{\lambda}(s)^q+K((\frac{A}{B-s})^{1/(N-2)})
 z_{\lambda}^{-\mu}(s)\big]ds\\
&\geq \lambda\frac{1}{a(R^{\gamma}A_N\int_{0}^{1}B_N(s)ds)}\int_0^1G(t,s)d(s)c_0ds,
\end{align*}
and so
$$
\|z_{\lambda}\|\geq \lambda\frac{1}{a(R^{\gamma}A_N\int_{0}^{1}B_N(s)ds)}
\max_{t\in[0,1]}\int_0^1G(t,s)d(s)c_0ds,
$$
which guarantees that
\begin{equation}
\lambda\leq \frac{R a(R^{\gamma}A_N\int_{0}^{1}B_N(s)ds)}
{\max_{t\in[0,1]}\int_0^1G(t,s)d(s)ds c_0}=:\lambda_{R}.\label{e4.20}
\end{equation}
One the other hand, since
\begin{gather*}
 z_{\lambda}''+\lambda\frac{1}{a(A_N\int_{0}^{1}B_N(s)
|z_{\lambda}(s)|^{\gamma}ds)}d(t)
[z_{\lambda}^q(t)+K((\frac{A}{B-t})^{1/(N-2)})z_{\lambda}^{-\mu}(t)]=0,\\
0<t<1,\\
 z_{\lambda}(0)=z_{\lambda}(1)=0,
\end{gather*}
there exists $t_{\lambda}\in (0,1)$
with $z_{\lambda}'(t) \geq 0$ on $(0,t_{\lambda})$ and $z_{\lambda}'(t) \leq 0$
on $(t_{\lambda},1)$. For $t\in(0,t_{\lambda})$ we have
\begin{gather*}
\begin{aligned}
-z_{\lambda}''(t)
& \leq \lambda \frac{1}{a_0}z_{\lambda}^{-\mu}(t)d(t)
\Big\{ \max_{t\in[0,1]}K((\frac{A}{B-t})^{1/(N-2)})+z_{\lambda}^{\mu+q}(t)\Big\}\\
&\leq \lambda \frac{1}{a_0}z_{\lambda}^{-\mu}(t)\max_{t\in[0,1]}d(t)
 \Big\{ \max_{t\in[0,1]}K((\frac{A}{B-t})^{1/(N-2)})+R^{\mu+q}\Big\}\\
&= \lambda \frac{1}{a_0}z_{\lambda}^{-\mu}(t)d_1,
\end{aligned}\\
 d_1:= \max_{t\in[0,1]}d(t)\Big\{ \max_{t\in[0,1]}K((\frac{A}{B-t})^{1/(N-2)})
+R^{\mu+q}\Big\}.
\end{gather*}
Integrate from  $t$ $(t\leq t_{\lambda})$ to $t_{\lambda}$
(note $z_{\lambda}(s)$ is increasing on $[t,t_{\lambda}]$) to obtain
\[
z_{\lambda}'(t)
 \leq \lambda \frac{1}{a_0}\int_t^{t_{\lambda}}z_{\lambda}^{-\mu}(s)ds d_1
 \leq \lambda \frac{1}{a_0}\int_t^{t_{\lambda}}z_{\lambda}^{-\mu}(t)ds d_1
 \leq \lambda \frac{1}{a_0}d_1z_{\lambda}^{-\mu}(t),
\]
i.e.
\begin{equation}
z_{\lambda}^{\mu}(t)z_{\lambda}'(t)\leq\lambda \frac{1}{a_0}d_1, \label{e4.21}
\end{equation}
and then integrate \eqref{e4.21} from $0$ to $t_{\lambda}$ to obtain
\[
\frac{1}{\mu+1}r^{\mu+1}\leq \int^{t_{\lambda}}_{0} z_{\lambda}^{\mu}(t)
d z_{\lambda}(t) \leq \lambda \frac{1}{a_0}d_1.
\]
Consequently
\begin{equation}
\lambda\geq \frac{r^{\mu+1}a_0}{(\mu+1)d_1}=:\lambda_{r}.\label{e4.22}
\end{equation}
It follows from \eqref{e4.20} and \eqref{e4.22} that
$(\lambda,u_{\lambda})\in [\lambda_r,\lambda_R]
\times (\{z|r\leq \|z\|\leq R\}\cap P)$ for all
$(\lambda,z_{\lambda})\in C$ with $r\leq \|z_{\lambda}\|\leq R$.
 Since $C$ comes from $(0,0)$, $C$ is connected and
$C\cap ((0,\lambda_r)\times C_0[0,1])$ is unbounded,
if $\lambda\in(0,\lambda_r)$, there exist at least two $x_{1,\lambda}$ and
$x_{2,\lambda}$ with $\|x_{1,\lambda}\|<r$ and $\|x_{2,\lambda}\|>R$.

Let
$$
\lambda_0=\sup\{\lambda_r: \eqref{e1.2} \text{ has at least two positive 
solutions for all }\lambda\in(0,\lambda_r)\}.
$$
Obviously, $\lambda_0\leq \Lambda_0$ and
\eqref{e1.2} has at least two positive solutions for all $\lambda\in(0,\lambda_r)$ 
and has at least one positive solution for all $\lambda\in[\lambda_0,\Lambda_0]$.
Since $R$ and $r$ are arbitrary, it follows that (iii) is true. 
The proof is complete.
\end{proof}

 If $N=1$, we can consider the problem
\begin{gather*}
-a\Big(\int_{0}^{1}|z(s)|^{\gamma}ds\Big)z''(t)
=\lambda [z(t)^{p}+K(t)z^{-\mu}(t))],\quad t\text{ in } (0,1),\\
z(t)>0, \quad t\in(0,1),\\
z(0)=0,\quad z(1)=0,
\end{gather*}
and obtain the similar results as Corollary \ref{coro4.2} for the above problem.


\subsection*{Acknowledgments}
This research is supported by the NSFC of China (61603226)
 and by the Fund of Science and Technology Plan of Shandong Province \\ (2014GGH201010).




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\end{document}