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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 05, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/05\hfil Four-point boundary-value problems]
{Positive solutions of four-point boundary-value problems for
higher-order with $p$-Laplacian operator}

\author[Y. Zhou, H. Su\hfil EJDE-2007/05\hfilneg]
{Yunming Zhou, Hua Su}  

\address{Yunming Zhou \newline
School of Mathematics and Information Science, 
Shandong University of Technology,
 Zibo Shandong, 255049, China}
\email{zym571219@126.com}

\address{Hua Su \newline
School of  Mathematics and System Sciences,  Shandong University,
Jinan Shandong,  250100, China \hfill\break
School of Mathematical Sciences, Qufu Normal University, Qufu
Shandong, 273165, China} 
\email{jnsuhua@163.com}

\thanks{Submitted September 12, 2006. Published January 2, 2007.}
\subjclass[2000]{34B18}
\keywords{Higher-order $p$-Laplacian operator;
four-point;  positive solutions; \hfill\break\indent
singular boundary-value; fixed-point index theory}

\begin{abstract}
 In this paper, we  study  the  existence of positive solutions for
 nonlinear four-point singular boundary-value problems for
 higher-order equation with the $p$-Laplacian operator.
 Using the fixed-point index theory, we find conditions
 for the existence of  one solution, and of multiple solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
%\allowdisplaybreaks

\section{Introduction}

In this paper, we study the quasi-linear equation, with $p$-Laplacian
operator,
\begin{equation}
(\phi_{p}(u^{(n-1)}))'+g(t)f(u(t),u'(t),
\dots, u^{(n-2)}(t))=0,\quad 0<t<1 , \label{e1.1}
\end{equation}
subject to the boundary conditions
\begin{equation}
\begin{gathered}
u^{(i)}(0)=0  \quad 0\leq i\leq n-3, \\
u^{(n-2)}(0)-B_0(u^{(n-1)}(\xi))=0 \quad n\geq 3,\\
u^{(n-2)}(1)+B_1(u^{(n-1)}(\eta))=0 \quad n\geq 3,
\end{gathered} \label{e1.2}
\end{equation}
where $\phi_{p}(s)$ is the $p$-Laplacian operator;   i.e.,
 $\phi_{p}(s)=|s|^{p-2}s$,  $p>1$,
$\phi_{q}=\phi_{p}^{-1}$, $\frac{1}{p}+\frac{1}{q}=1$.
$\xi,\eta\in(0,1)$ is prescribed and $\xi<\eta$,
$g:(0,1)\to[0,\infty)$, $B_0, B_1$ are both nondecreasing
continuous odd functions defined on $(-\infty,+\infty)$.

In recent years, the existence of positive solutions for nonlinear
boundary-value problems with $p$-Laplacian operator received wide
attention.
 Recently, for the existence of positive solutions of multi-points
boundary-value problems for second-order ordinary differential equation, some
 authors have obtained the existence results \cite{a3,a1,a2,e1,m1}. However,
 the multi-points boundary-value problems treated in the above
mentioned references do not discuss the problems with
 singularities and the higher-order $p$-Laplacian operator.
 For the singular case of multi-point boundary-value problems for
higher-order $p$-Laplacian operator, with the author's acknowledge,
no one has studied the
 existence of positive solutions in this case. Therefore
this paper mainly studies the existence of positive solutions for
nonlinear singular boundary-value problem
\eqref{e1.1}, \eqref{e1.2}.

In this paper, by constructing an integral equation which is
equivalent to the problem \eqref{e1.1}, \eqref{e1.2},
we research the existence of
positive solutions when  $g$ and $f$ satisfy some suitable conditions.

 For the rest of this paper, we make the following assumptions:
\begin{itemize}
\item[(H1)] $f\in C([0,+\infty)^{n-1},[0,+\infty))$;

\item[(H2)] $g:(0,1)\to[0,+\infty)$ and
$0<\int_{0}^{1}g(t)dt<\infty$;

\item[(H3)] $B_0,  B_1$ are both increasing, continuous, odd
functions defined on $(-\infty,+\infty)$ and at least one of them
satisfies the condition that there exists one $b>0$ such that
$$
0<B_i(v)\leq bv, \quad \forall  v\geq0,\; i=0,\text{ or } i=1.
$$
\end{itemize}
It is easy to check that condition (H2) implies
$$
0<\int_0^1\phi_{q}(\int_0^sg(s_1)ds_1)ds<+\infty.$$

This paper is organized as follows. In section 2, we present some
preliminaries and lemmas that will be used to prove our main
results. In section 3, we discuss the existence of single solution
of the systems \eqref{e1.1}. In section 4, we study the existence of at
least two solutions of the systems \eqref{e1.1}. In section 5, we give two
examples as an application.

\section{Preliminaries and Lemmas}

Let
$$
B=\big\{u\in C^{n-2}[0,1]: u^{(i)}(0)=0,\ 0\leq i\leq
n-3\big\}.
$$
Then $B$ is a Banach space with the norm
$\|u\|=\max_{t\in[0, 1]}|u^{(n-2)}(t)|$. And let
 $$
K=\big\{u\in B: u^{(n-2)}(t)\geq0,\, u^{(n-2)}(t)
 \hbox{ is concave function, } t\in[0,1]\big\}.
$$
Obviously, $K$ is a cone in $B$ and $0\leq u^{(i)}(t)\leq\|u\|$ on
$[0,1]$. Set $K_r=\{u\in K:\|u\|\leq r\}$.
 We can easily get the following
Lemmas.

\begin{lemma}\label{lem2.1}
Suppose condition $(H_2)$ holds.   Then there exists a constant
$\theta\in (0,1/2)$ that satisfies
$$
0<\int_{\theta}^{1-\theta}g(t)dt<\infty.
$$
Furthermore, the function
 $$
A(t)=\int_{\theta}^{t}\phi_{q}\big(\int_{s}^{t}g(s_1)ds_1\big)ds+
\int_{t}^{1-\theta}\phi_{q}\big(\int_{t}^{s}g(s_1)ds_1\big)ds,
\quad t\in[\theta, 1-\theta]
$$
is positive continuous function on
$[\theta, 1-\theta]$, therefore $A(t)$ has minimum on $[\theta,
1-\theta]$. Hence we suppose that there exists $L>0$ such that
$A\geq L$, $t\in[\theta, 1-\theta]$.
\end{lemma}


\begin{lemma}\label{lem2.2}
Let $u\in K$ and $\theta\in (0,1/2)$ in Lemma \ref{lem2.1}. Then
$$
u^{(n-2)}(t)\geq \theta\|u\|,  \quad t\in[\theta,1-\theta].
$$
\end{lemma}

The proof of the above lemma is similar to the proof of  in
\cite[Lemma 2.2]{s1}, so we omit it.

\begin{lemma}\label{lem2.3}
Suppose that conditions (H1)--(H3) hold. Then
$u(t)\in K\cap C^{n-1}(0,1)$ is a solution of  boundary-value
problem \eqref{e1.1}, \eqref{e1.2}
if and only if $u(t)\in B$ is a solution of the integral
equation
$$
u(t)=\int_0^{t}\int_0^{s_1}\dots\int_0^{s_{n-3}}w(s_{n-2})
ds_{n-2}ds_{n-3}\dots ds_{1},
$$
where
\begin{equation}
w(t)=\begin{cases}
B_0\circ\phi_{q}\big(\int_{\xi}^{\delta}g(s)f(u(s),u'(s),\dots,
u^{(n-2)}(s))ds\big) \\
+\int_{0}^{t}\phi_{q}\big(\int_{s}^{\delta}g(r)f(u(r),u'(r),\dots,
u^{(n-2)}(r))dr\big)ds &0\leq t\leq\delta, \\[4pt]
B_1\circ\phi_{q}\big(\int_{\delta}^{\eta}g(s)f(u(s),u'(s),\dots,
u^{(n-2)}(s))ds\big) \\
+\int_{t}^{1}\phi_{q}\big(\int_{\delta}^{s}g(r)f(u(r),u'(r),\dots,
u^{(n-2)}(r))dr\big))ds &\delta\leq t\leq 1.
\end{cases}
 \label{e2.1}
\end{equation}
Here  $\delta$ is unique solution of the equation
$g_{1}(t)=g_{2}(t)$, where
\begin{align*}
g_{1}(t)&=B_0\circ\phi_{q}\big(\int_{\delta}^{\eta}g(s)f(u(s),u'(s),\dots,
u^{(n-2)}(s))ds\big) \\
&\quad +\int_{t}^{1}\phi_{q}\big(\int_{\delta}^{s}g(r)f(u(r),u'(r),\dots,
u^{(n-2)}(r))dr\big)ds,
\\
g_{2}(t)&=B_1\circ\phi_{q}(\int_{\delta}^{\eta}g(s)f(u(s),u'(s),\dots,
u^{(n-2)}(s))ds) \\
&\quad +\int_{t}^{1}\phi_{q}(\int_{\delta}^{s}g(r)f(u(r),u'(r),\dots,
u^{(n-2)}(r))dr)ds.
\end{align*}
The equation $g_{1}(t)=g_{2}(t)$
has unique solution in $(0,1)$ because $g_{1}(t)$ is strictly
increasing on $[0,1)$, and $g_{1}(0)=0$, while $g_{2}(t)$ is
strictly decreasing on $(0,1]$, and $g_{2}(1)=0$.
\end{lemma}

\begin{proof}
Necessity.  By the equation of the boundary
condition and (H3), we have $u^{(n-1)}(\xi)\geq 0$,
$u^{(n-1)}(\eta)\leq0$, then there exist a constant
$\delta\in [\xi,\eta]\subset(0,1)$ such that $u^{(n-1)}(\delta)=0$.
Firstly, by integrating the equation of the problems \eqref{e1.1}
on $(\delta,t)$,  we have
$$
\phi_{p}(u^{(n-1)}(t))=\phi_{p}(u^{(n-1)}(\delta))-\int_{\delta}^t
g(s)f\big(u(s),u'(s),\dots, u^{(n-2)}(s)\big)ds,
$$
then
\begin{equation}
u^{(n-1)}(t)=-\phi_{q}\Big(\int_{\delta}^t
g(s)f\big(u(s),u'(s),\dots,
u^{(n-2)}(s)\big)ds\Big),\label{e2.2}
\end{equation}
thus
\begin{equation}
u^{(n-2)}(t)=u^{(n-2)}(\delta)-\int_{\delta}^t\phi_{q}\Big(\int_{\delta}^s
g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds.\label{e2.3}
\end{equation}
 By $u^{(n-1)}(\delta)=0$ and condition \eqref{e1.2}, letting
$t=\eta$ on \eqref{e2.2}, we have
$$
u^{(n-2)}(1)=-B_1\big(u^{(n-1)}(\eta )\big)
=B_1\circ\phi_{q}\Big(\int_{\delta}^\eta
g(s)f\big(u(s),u'(s),\dots, u^{(n-2)}(s)\big)ds\Big).
$$
Then by \eqref{e2.3}, we have
\begin{equation}
\begin{aligned}
u^{(n-2)}(\delta)&=B_1\circ\phi_{q}\Big(\int_{\delta}^\eta
g(s)f\big(u(s),u'(s),\dots,
u^{(n-2)}(s)\big)ds\Big))\\
&\quad +\int_{\delta}^1\phi_{q}\Big(\int_{\delta}^s
g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds.
\end{aligned} \label{e2.4}
\end{equation}
Then
\begin{equation}
\begin{aligned}
u^{(n-2)}(t)&=B_1\circ\phi_{q}\Big(\int_{\delta}^\eta
g(s)f\big(u(s),u'(s),\dots,
u^{(n-2)}(s)\big)ds\Big)\\
&\quad +\int_{t}^1\phi_{q}\Big(\int_{\delta}^s
g(r)f\big(u(r),u'(r),\dots, u^{(n-2)}(r)\big)dr\Big)ds.
\end{aligned}\label{e2.5}
\end{equation}
Integrating \eqref{e2.5} for $n-2$ times on $(0,t)$, we have
\begin{align*}
u(t)&=\int_{0}^t\int_{0}^{s_1}\dots\int_{0}^{s_{n-3}}
B_1\circ\phi_{q}\Big(\int_{\delta}^\eta
g(s)f\big(u(s),u'(s),\\
&\quad \dots, u^{(n-2)}(s)\big)ds\Big)ds_{s_{n-2}}\dots
ds_2\,ds_1\\
&\quad +\int_{0}^t\int_{0}^{s_1}\dots\int_{0}^{s_{n-3}}
(\int_{s_{n-2}}^{1}\phi_{q}\Big(\int_{\delta}^s
g(r)f\big(u(r),u'(r),\\
&\quad \dots, u^{(n-2)}(r)\big)dr\Big)ds)ds_{s_{n-2}}\dots
ds_2\,ds_1.
\end{align*}
Similarly, for $t\in(0,\delta)$,
integrating  problems \eqref{e1.1} on $(0,\delta)$, we have
\begin{align*}
u(t)&=\int_{0}^t \int_{0}^{s_1}\dots\int_{0}^{s_{n-3}}
B_0\circ\phi_{q}\Big(\int_{\xi}^{\delta}
g(s)f\big(u(s),u'(s),\\
&\quad \dots, u^{(n-2)}(s)\big)ds\Big)ds_{s_{n-2}}\dots
ds_2\,ds_1\\
&\quad +\int_{0}^t\int_{0}^{s_1}\dots\int_{0}^{s_{n-3}}
(\int_{0}^{s_{n-2}}\phi_{q}\Big(\int_{s}^\delta
g(r)f\big(u(r),u'(r),\\
&\quad \dots, u^{(n-2)}(r)\big)dr\Big)ds)ds_{s_{n-2}}\dots
ds_2\,ds_1.
\end{align*}
Therefore,   for any $t\in [0,1]$, $u(t)$ can
be expressed as
$$
u(t)=\int_0^{t}\int_0^{s_1}\dots\int_0^{s_{n-3}}w(s_{n-2})
ds_{n-2}ds_{n-3}\dots ds_{1},
$$
 where $w(t)$ is expressed as \eqref{e2.1}.

Sufficiency. Suppose that
$u(t)=\int_0^{t}\int_0^{s_1}\dots\int_0^{s_{n-3}}w(s_{n-2})
ds_{n-2}ds_{n-3}\dots ds_{1}$. Then by \eqref{e2.1}, we have
\begin{equation}
u^{(n-1)}(t)=\begin{cases}
\phi_{q}\Big(\int_{t}^{\delta}g(s)f\big(u(s),u'(s),\dots,
u^{(n-2)}(s)\big)ds\Big)ds\geq0, &0\leq t<\delta, \\
 -\phi_{q}\Big(\int_{\delta}^{t}g(s)f\big(u(s),u'(s),\dots,
u^{(n-2)}(s)\big)ds\Big)ds\leq0,
   &\delta< t\leq1,
\end{cases}\label{e2.6}
\end{equation}
So that $(\phi_{p}(u^{(n-1)}))'+g(t)f(u(t),u'(t), \dots,
u^{(n-2)}(t))=0$,   $ 0<t<1$, $t\neq\delta $. These imply that
 \eqref{e1.1} holds. Furthermore, by letting $t=0$ and $t=1$ on
\eqref{e2.1} and \eqref{e2.6}, we  obtain the boundary-value equations of
\eqref{e1.2}. The proof is complete.
\end{proof}

Now, we define a mapping $T:K\to C^{n-1}[0,1]$ given by
$$
(Tu)(t)=\int_0^{t}\int_0^{s_1}\dots\int_0^{s_{n-3}}w(s_{n-2})
ds_{n-2}ds_{n-3}\dots ds_{1},
$$
where $w(t)$ is given by \eqref{e2.1}.

\begin{lemma}\label{lem2.4}
Suppose that conditions (H1), (H2) hold. Then the solution
$u(t)\in K$ of  \eqref{e1.1}, \eqref{e1.2} satisfies
$$
u(t)\leq u'(t)\leq \dots\leq u^{(n-3)}(t),\quad t\in[0,1],
$$
and for $\theta\in(0,1/2)$ in Lemma \ref{lem2.1}, we
have
$$
u^{(n-3)}(t)\leq \frac{1}{\theta}u^{(n-2)}(t),\quad t\in[\theta,1-\theta].
$$
\end{lemma}

 \begin{proof}
 If $u(t)$ is the solution of problem \eqref{e1.1}, \eqref{e1.2},
then $u^{(n-2)}(t)$ is concave function, and $u^{(i)}(t)\geq 0$,
$i=0,1,\dots,n-2$, $t\in[0,1]$, Thus we have
$$
u^{(i)}(t)=\int_0^t u^{(i+1)}(s)ds\leq tu^{(i+1)}(t)\leq u^{(i+1)}(t),
\quad i=0,1,\dots,n-4,
$$
i.e.,
$u(t)\leq u'(t)\leq \dots\leq u^{(n-3)}(t)$, $t\in[0,1]$.
Next, by Lemma \ref{lem2.2}, for $t\in[\theta,1-\theta]$, we have
$u^{(n-2)}(t)\geq \theta\|u^{(n-2)}\|$. Then from
$$
u^{(n-3)}(t)=\int_0^t u^{(n-2)}(s)ds\leq \|u^{(n-2)}\|,
$$
we have
$$
u^{(n-3)}(t)\leq \frac{1}{\theta}u^{(n-2)}(t),\quad t\in[\theta,1-\theta].
$$
The proof is complete.
\end{proof}

\begin{lemma}\label{lem2.5}
The operator $T:K\to K$ is completely continuous.
\end{lemma}

\begin{proof}
 Because
\begin{align*}
&(Tu)^{(n-1)}(t)\\
&=w'(t)=\begin{cases}
\phi_{q}\Big(\int_{t}^{\delta}g(s)f\big(u(s),u'(s),\dots,
u^{(n-2)}(s)\big)ds\Big)\geq0 &0\leq t\leq\delta, \\
 -\phi_{q}\Big(\int_{\delta}^{t}g(s)f\big(u(s),u'(s),\dots,
u^{(n-2)}(s)\big)ds\Big)\leq0   &\delta\leq t\leq1,
\end{cases}
\end{align*}
is continuous, decreasing on $[0,1]$ and satisfies
$(Tu)^{(n-1)}(\delta)=0$. Then, $Tu\in K$ for each $u\in K$ and
$(Tu)^{(n-2)}(\delta)=\max_{t\in[0,1]}(Tu)^{(n-2)}(t)$. This
shows that $TK\subset K$. Furthermore, it is easy to check by
Arzela-ascoli Theorem that $T:K\to K$ is completely
continuous.
\end{proof}

Obviously, we can obtain the following results,
$$
w(0)-B_0w'(\xi)=0,\quad w(1)+B_1w'(\eta)=0.
$$
Our main tool of this paper is the following fixed point index theorem.

\begin{theorem}[\cite{g1,g2}] \label{thm2.6}
 Suppose $E$ is a real Banach space and
$K\subset E$ is a cone. Let $\Omega_{r}=\{u\in K:\|u\|\leq r\}$, and
the operator $T:\Omega_{r}\to K$ be completely continuous
and satisfy $Tx\neq x$ for all $ x\in\partial\Omega_{r}$.
Then
\begin{itemize}
\item[(i)] If $\|Tx\|\leq\|x\|$ for all $x\in\partial\Omega_{r}$, then
$i(T,\Omega_{r},K)=1$;

\item[(ii)] If $\|Tx\|\geq\|x\|$ for all $x\in\partial\Omega_{r}$, then
$i(T,\Omega_{r},K)=0$.
\end{itemize}
\end{theorem}


For convenience,  we set
$$
\theta^*=\frac{2}{L},   \quad
\theta_*=\frac{1}{(b+1)\phi_{q}\big(\int_{0}^{1}g(r)dr\big)}.
$$
where $L$ is the constant in Lemma \ref{lem2.1}. By Lemma
\ref{lem2.4}, we can also
set
\begin{gather*}
f^0=\lim_{u_{n-1}\to0}
 \max_{0\leq u_1\leq\dots\leq  u_{n-2}\leq u_{n-1}/\theta }
 \frac{f(u_1,u_2,\dots,u_{n-1})}{(u_{n-1})^{p-1}}, \\
f_\infty=\lim_{u_{n-1}\to\infty}
 \min_{0\leq u_1\leq\dots\leq  u_{n-2}\leq u_{n-1}/\theta }
 \frac{f(u_1,u_2,\dots,u_{n-1})}{(u_{n-1})^{p-1}},\\
f_0=\lim_{u_{n-1}\to0}
 \min_{0\leq u_1\leq\dots\leq  u_{n-2}\leq u_{n-1}/\theta }
 \frac{f(u_1,u_2,\dots,u_{n-1})}{(u_{n-1})^{p-1}}, \\
f^\infty=\lim_{u_{n-1}\to\infty}
 \max_{0\leq u_1\leq\dots\leq u_{n-2}\leq u_{n-1}/\theta }
 \frac{f(u_1,u_2,\dots,u_{n-1})}{(u_{n-1})^{p-1}}.
\end{gather*}

\section{Existence of Positive Solutions}

 In this section, we present our main results.

\begin{theorem}\label{thm3.1}
Suppose that condition (H1)--(H3) hold.
Assume that $f$ also satisfies
\begin{itemize}
\item[(A1)] $f(u_1,u_2,\dots,u_{n-1})\geq(mr)^{p-1}$
for $\theta r\leq u_{n-1}\leq r$, $0\leq u_1\leq\dots\leq
 u_{n-2}\leq u_{n-1}/\theta $;

\item[(A2)] $f(u_1,u_2,\dots,u_{n-1})\leq(MR)^{p-1}$
for $0\leq u_{n-1}\leq R$,
$0\leq u_1\leq\dots\leq  u_{n-2}\leq u_{n-1}/\theta$,
where $m\in(\theta^*,\infty), M\in(0,\theta_*)$.
\end{itemize}
Then the boundary-value problem \eqref{e1.1}, \eqref{e1.2}
has a solution $u$ such that $\|u\|$ lies between $r$ and $R$.
\end{theorem}

\begin{theorem}\label{thm3.2}
 Suppose that condition (H1)--(H3) hold. Assume that $f$ also satisfies
\begin{itemize}
\item[(A3)] $f^0=\varphi\in[0,(\theta_*/4)^{p-1})$;

\item[(A4)] $f_\infty=\lambda\in (2\theta^*/\theta)^{p-1}, \infty)$.
\end{itemize}
Then   the boundary-value problem \eqref{e1.1},
\eqref{e1.2} has a solution $u$ which is bounded in the norm $\|\cdot\|$.
\end{theorem}

\begin{theorem}\label{thm3.3}
Suppose that condition (H1)--(H3) hold.
Assume that $f$ also satisfies
\begin{itemize}
\item[(A5)] $f^\infty=\lambda\in[0, (\theta_*/4)^{p-1})$;

\item[(A6)] $f_0=\varphi\in ((2\theta^*/\theta)^{p-1}, \infty)$.
\end{itemize}
Then   the boundary-value problem \eqref{e1.1},
\eqref{e1.2} has a solution $u$ which is bounded in  the norm $\|\cdot\|$.
\end{theorem}

\begin{proof}[Proof of Theorem \ref{thm3.1}]
 Without loss of
generality, we suppose that $r<R$ and $0<B_0(v)\leq bv$ for all
$v\geq0$. For any  $u\in K$, by Lemma \ref{lem2.2},  we have
\begin{equation}
u^{(n-2)}(t)\geq\theta\|u\|, \quad
 t\in[\theta, 1-\theta].
\label{e3.1}
\end{equation}
We define the following two open subset of $E$:
$$
\Omega_{1}=\{u\in K:\|u\|<r\}, \quad
\Omega_{2}=\{u\in K:\|u\|<R\}.
$$
For each $u\in\partial\Omega_{1}$, by \eqref{e3.1} we have
$$
r=\|u\|\geq u^{(n-2)}(t)\geq\theta\|u\|=\theta r, \quad
t \in[\theta, 1-\theta].
$$
For $t\in [\theta, 1-\theta]$ and
$u\in\partial\Omega_{1}$, we shall discuss it from three
perspectives.

\noindent (i) If $\delta\in[\theta, 1-\theta]$,    thus for
$u\in\partial\Omega_{1}$, by (A1) and Lemma \ref{lem2.3},  we have
\begin{align*}
2\|Tu\|
&=2(Tu)^{(n-2)}(\delta) \\
&\geq \int_{0}^{\delta}\phi_{q}\Big(\int_{s}^{\delta}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds  \\
&\quad +\int_{\delta}^{1}\phi_{q}
\Big(\int_{\delta}^{s}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds  \\
&\geq \int_{\theta}^{\delta}\phi_{q}\Big(\int_{s}^{\delta}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds  \\
&\quad +\int_{\delta}^{1-\theta}\phi_{q}
\Big(\int_{\delta}^{s}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds  \\
&\geq mrA(\delta)\geq mrL\\
&>2r=2\|u\|.
\end{align*}

\noindent(ii) If $\delta\in(1-\theta, 1]$,  thus for
$u\in\partial\Omega_{1}$, by (A1) and Lemma \ref{lem2.3},  we have
\begin{align*}
\|Tu\|&=(Tu)^{(n-2)}(\delta) \\
& \geq B_0\circ
\phi_{q}\Big(\int_{\xi}^{\delta}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)  \\
&\quad +\int_{0}^{\delta}\phi_{q}
(\int_{s}^{\delta}g(r)f(u(r),u'(r),\dots,
u^{(n-2)}(r))dr)ds \\
&\geq \int_{\theta}^{1-\theta}\phi_{q}
\Big(\int_{s}^{1-\theta}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds \\
&\geq mrA(1-\theta)\geq mrL\\
&>2r>r=\|u\|.
\end{align*}

\noindent (iii)
 If $\delta\in(0,\theta)$, thus for $u\in\partial\Omega_{1}$,
by (A1) and Lemma \ref{lem2.3},  we have
\begin{align*}
 \|Tu\|&=(Tu)^{(n-2)}(\delta) \\
& \geq B_1\circ \phi_{q}\Big(\int_{\delta}^{\eta}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)  \\
&\quad +\int_{\delta}^{1}\phi_{q}
\Big(\int_{\delta}^{s}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds \\
&\geq \int_{\theta}^{1-\theta}\phi_{q}
\Big(\int_{\theta}^{s}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds \\
&\geq mrA(\theta)\geq mrL\\
&>2r>r=\|u\|.
\end{align*}
Therefore,   under all  condition, we have
$\|Tu\|>\|u\|$ for all $u\in\partial\Omega_{1}$. Then by Theorem \ref{thm2.6},
\begin{equation}
i(T,\Omega_{1},K)=0.\label{e3.2}
\end{equation}
On the other hand,  for $u\in\partial\Omega_{2}$, we have
$u^{(n-2)}(t)\leq\|u\|=R$,   by (A2),
\begin{align*}
\|Tu\|&=(Tu)^{(n-2)}(\delta) \\
& \leq  B_0\circ \phi_{q}\Big(\int_{0}^{1}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)  \\
&\quad +\int_{0}^{1}\phi_{q}
\Big(\int_{s}^{\delta}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds \\
&\leq bMR\phi_{q} \big(\int_{0}^{1}g(r)dr\big)+
MR\phi_{q} \big(\int_{0}^{1}g(r)dr\big) \\
& =(b+1)MR\phi_{q} \big(\int_{0}^{1}g(r)dr\big)\\
&\leq R=\|u\|.
\end{align*}
Thus
$\|Tu\|<\|u\|$ for all $u\in\partial\Omega_{2}$.
Then by Theorem \ref{thm3.1}, we have
\begin{equation}
i(T,\Omega_{2},K)=1.\label{e3.3}
\end{equation}
Therefore, by \eqref{e3.2}, \eqref{e3.3}, $r<R$ we have
$$
i(T,\Omega_2\setminus\overline{\Omega}_{1},K)=1.
$$
Then operator $T$ has a fixed point
$u\in(\Omega_2\setminus\overline{\Omega}_{1})$ and $r\leq\|u\|\leq
R$.  This completes the proof
 \end{proof}

\begin{proof}[Proof of Theorem \ref{thm3.2}]
First,   from $f^0=\varphi\in[0,(\theta_*/4)^{p-1})$,
  for $\epsilon=(\theta_*/4)^{p-1}-\varphi$,
there exists an appropriately small positive number $\rho$,
 such that $0\leq u_{n-1}\leq \rho$. Since $u_{n-1}\neq0$,   we have
\begin{equation}
f(u_1,u_2,\dots,u_{n-1})\leq(\varphi+\epsilon)(u_{n-1})^{p-1}\leq
(\theta_*/4)^{p-1}\rho^{p-1}
=(\theta_*\rho/4)^{p-1}.\label{e3.4}
\end{equation}
Then let $R=\rho$, $M=\frac{\theta_*}{4}\in(0,\theta_*)$, thus by
 \eqref{e3.4}
\begin{gather*}
f(u_1,u_2,\dots,u_{n-1})\leq(MR)^{p-1},\\
0\leq u_{n-1}\leq R,\quad  0\leq u_1\leq\dots\leq
 u_{n-2}\leq u_{n-1}/\theta .
\end{gather*}
So condition (A2) holds.

Next,  by condition (A4), $f_\infty=
 \lambda\in((2\theta^*/\theta)^{p-1}, \infty)$,
then for $\epsilon=\lambda-(2\theta^*/\theta)^{p-1}$,
 there exists an adequately big positive number $r\neq R$,  such that
 $u_{n-1}\geq \theta r$, $0\leq u_1\leq\dots\leq  u_{n-2}\leq u_{n-1}/\theta $,
we have
\begin{equation}
f(u_1,u_2,\dots,u_{n-1})\geq(\lambda-\epsilon)(u_{n-1})^{p-1}
\geq(2\theta^*/\theta)^{p-1}(\theta
r)^{p-1} =(2\theta^*r)^{p-1},   \label{e3.5}
\end{equation}
 Let $m=2\theta^*>\theta^*$, thus by \eqref{e3.5},  condition (A1) holds.
Therefore by Theorem \ref{thm3.1} we know that the results of
Theorem \ref{thm3.2}
hold. The proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3.3}]
First,   by condition (A6),
$f_0=\varphi\in ((2\theta^*/\theta)^{p-1},\infty)$, then for
$\epsilon=\varphi-(2\theta^*/\theta)^{p-1}$,
there exists an appropriately small positive number $r$, such that
$0\leq u_{n-1}\leq r$,  $ u_{n-1}\neq0$,    we have
$$
f(u_1,u_2,\dots,u_{n-1})\geq(\varphi-\epsilon)(u_{n-1})^{p-1}
=(2\theta^*/\theta)^{p-1}(u_{n-1})^{p-1},
$$
thus when $\theta r\leq u_{n-1}\leq r$,   we have
\begin{equation}
f(u_1,u_2,\dots,u_{n-1})\geq(2\theta^*/\theta)^{p-1}(\theta r)^{p-1}
=(2\theta^*r)^{p-1}.\label{e3.6}
\end{equation}
Let $m=2\theta^*>\theta^*$,   so by
\eqref{e3.6}, condition (A1) holds.

Next,  by condition (A5): $f^\infty=\lambda\in[0,(\theta_*/4)^{p-1})$,   then
for
$\epsilon=(\theta_*/4)^{p-1}-\lambda$,
there exists an suitably big positive number $\rho\neq r$,  such that
$u_{n-1}\geq\rho$,    we have
\begin{equation}
f(u_1,u_2,\dots,u_{n-1})\leq(\lambda+\epsilon)(u_{n-1})^{p-1}\leq
(\theta_*/4)^{p-1}(u_{n-1})^{p-1}.\label{e3.7}
\end{equation}
If $f$ is unbounded,   by the continuation of $f$ on $[0,
\infty)^{n-1}$, then exists constant
$R\geq\rho$, $R\neq r$, and
a point $(u_{01},u_{02},\dots,u_{0(n-1)})\in[0, \infty)^{n-1}$ such
that $$\rho\leq u_{0(n-1)}\leq R$$ and
$$
f(u_1,u_2,\dots,u_{n-1})\leq f(u_{01},u_{02},\dots,u_{0(n-1)}),  \quad
0\leq u_{n-1}\leq R.
$$
Thus,   by $\rho\leq u_{0(n-1)}\leq R, $  we know
\begin{align*}
f(u_1,u_2,\dots,u_{n-1})
&\leq f(u_{01},u_{02},\dots,u_{0(n-1)})\\
&\leq(\theta_*/4)^{p-1}(u_{0(n-1)})^{p-1}\\
&\leq(\theta_*R/4)^{p-1}.
\end{align*}
Choose
$M=\frac{\theta_*}{4}\in(0,   \theta_*)$. Then,   we
have
\begin{gather*}
f(u_1,u_2,\dots,u_{n-1})\leq (MR)^{p-1},  \\
 0\leq u_{n-1}\leq R,\quad  0\leq u_1\leq\dots\leq
 u_{n-2}\leq u_{n-1}/\theta .
\end{gather*}
If $f$ is bounded, we suppose
$f(u_1,u_2,\dots,u_{n-1})\leq \overline{M}^{p-1}$,
$u_{n-1}\in[0, \infty)$,
$\overline{M}^{p-1} \in R_+$,   there exists an adequately
 big positive number $R>4\overline{M}/\theta_*$, then choose
$M=\theta_*/4\in(0,  \theta_*)$,
for $0\leq u_1\leq\dots\leq  u_{n-2}\leq u_{n-1}/\theta $, we
have
$$
f(u_1,u_2,\dots,u_{n-1})\leq
\overline{M}^{p-1}\leq(\theta_*R/4)^{p-1}=(MR)^{p-1},
\quad 0\leq u_{n-1}\leq R
$$
Therefore,   condition
(A2) holds. Therefore,  by Theorem \ref{thm3.1},  we know that the
results of Theorem \ref{thm3.3} holds. The proof is complete.
\end{proof}

 \section{Existence of Many Positive Solutions  }

Next, we discuss the existence of  many positive solutions.

\begin{theorem}\label{thm4.1}
 Suppose that conditions (H1)--(H3) and (A2) hold. Assume that
$f$ also satisfies
\begin{itemize}
\item[(A7)] $f_0=+\infty$;
\item[(A8)] $f_\infty=+\infty$.
\end{itemize}
Then   the boundary-value problem \eqref{e1.1},   \eqref{e1.2} has
at least two solutions $u_1, u_2$ such that
$$0<\|u_1\|<R<\|u_2\|.
$$
\end{theorem}

\begin{proof}
  First, by condition (A7), for
any $M>\frac{2}{\theta L}$, there exists a constant
$\rho_*\in(0,R)$ such that
\begin{equation}
\begin{gathered}
f(u_1,u_2,\dots,u_{n-1})\geq(Mu_{n-1})^{p-1},\\
 0<u_{n-1}\leq\rho_*,\quad 0\leq u_1\leq\dots\leq
 u_{n-2}\leq u_{n-1}/\theta .
\end{gathered} \label{e4.1}
\end{equation}
Set $\Omega_{\rho_*}=\{u\in K:\|u\|<\rho_*\}$, for any
$u\in\partial\Omega_{\rho_*}$. By \eqref{e4.1} and Lemma \ref{lem2.2}, similar to
the  proof of Theorem \ref{thm3.1}, we  have from the three
perspectives,
$$
\|Tu\|\geq\|u\|,\ \forall\ u\in\partial\Omega_{\rho_*}.
$$
Then by Theorem \ref{thm2.6}, we have
\begin{equation}
i(T,\Omega_{\rho_*},K)=0.\label{e4.2}
\end{equation}
Next, by condition (A8), for
any $\overline{M}>\frac{2}{\theta L}$, there exists a
constant $\rho_0>0$ such that
\begin{equation}
\begin{gathered}
f(u_1,u_2,\dots,u_{n-1})\geq(\overline{M}u_{n-1})^{p-1},\\
 u_{n-1}>\rho_0,\quad 0\leq u_1\leq\dots\leq
 u_{n-2}\leq u_{n-1}/\theta .
\end{gathered}\label{e4.3}
\end{equation}
We choose a constant $\rho^*>\max\{R, \frac{\rho_0}{\theta}\}$,
obviously $\rho_*<R<\rho^*$. Set $\Omega_{\rho^*}=\{u\in K:\|u\|<\rho^*\}$.
For any $u\in\partial\Omega_{\rho^*}$, by Lemma \ref{lem2.2}, we have
$$
u^{(n-2)}(t)\geq\theta\|u\|=\theta \rho^*>\rho_0,\quad t\in[\theta,1-\theta].
$$
Then by \eqref{e4.3} and also similar to the proof of Theorem \ref{thm3.1},
we  have from the three perspectives,
$$
\|Tu\|\geq\|u\| \quad \forall u\in\partial\Omega_{\rho^*}.
$$
Then by Theorem \ref{thm2.6}, we have
\begin{equation}
i(T,\Omega_{\rho^*},K)=0.\label{e4.4}
\end{equation}
Finally, set $\Omega_{R}=\{u\in K:\|u\|<R\}$, For each
$u\in\partial\Omega_{R}$, by (A2), Lemma \ref{lem2.2} and also similar to
the proof of Theorem \ref{thm3.1}, we can also have
$$
\|Tu\|\leq\|u\|\quad \forall u\in\partial\Omega_{R}.
$$
Then by Theorem \ref{thm2.6}, we have
\begin{equation}
i(T,\Omega_{R},K)=1.\label{e4.5}
\end{equation}
Therefore, by \eqref{e4.2}, \eqref{e4.4}, \eqref{e4.5}, $\rho_*<R<\rho^*$
we have
$$
i(T,\Omega_R\setminus\overline{\Omega}_{\rho_*},K)=1,\quad
i(T,\Omega_{\rho^*}\setminus\overline{\Omega}_{R},K)=-1.
$$
Then $T$ have fixed point
$u_1\in\Omega_R\setminus\overline{\Omega}_{\rho_*}$, and fixed point
$u_2\in\Omega_{\rho^*}\setminus\overline{\Omega}_{R}$. Obviously,
$u_1,\ u_2$ are all positive solutions of problem \eqref{e1.1},\eqref{e1.2} and
$0<\|u_1\|<R<\|u_2\|$. The proof is complete.
\end{proof}

\begin{theorem}\label{thm4.2}
 Suppose that conditions (H1)--(H3)and (A1) hold. Assume that
$f$ also satisfies
\begin{itemize}
\item[(A9)] $f^0=0$;

\item[(A10)] $f^\infty=0$.
\end{itemize}
Then  the boundary-value problem \eqref{e1.1},  \eqref{e1.2} has at least two
solutions $u_1, u_2$ such that $0<\|u_1\|<r<\|u_2\|$.
\end{theorem}

\begin{proof}
 First, from $f^0=0$, for
$\eta_1\in(0,\theta_*)$, there exists a constant $\rho_*\in(0,r)$
such that
\begin{equation}
\begin{gathered}
f(u_1,u_2,\dots,u_{n-1})\leq(\eta_1u_{n-1})^{p-1}, \\
0<u_{n-1}\leq\rho_*,\quad 0\leq u_1\leq\dots\leq
 u_{n-2}\leq u_{n-1}/\theta .
\end{gathered}\label{e4.6}
\end{equation}
Set $\Omega_{\rho_*}=\{u\in K:\|u\|<\rho_*\}$, for each
$u\in\partial\Omega_{\rho_*}$, by \eqref{e4.6},
we have
\begin{align*}
\|Tu\|&=(Tu)^{(n-2)}(\delta) \\
&\leq B_0\circ \phi_{q}\Big(\int_{0}^{1}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)  \\
&\quad +\int_{0}^{1}\phi_{q}
\Big(\int_{s}^{\delta}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds  \\
& \leq B_0\circ
\phi_{q}\Big(\int_{0}^{1}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)  \\
&\quad +\phi_{q} \Big(\int_{0}^{1}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)\\
&\leq (b+1)\eta_1\rho_*\phi_{q}
\Big(\int_{0}^{1}g(r)dr\Big)\\
&\leq \rho_*=\|u\|.
\end{align*}
i.e.,
$\|Tu\|\leq\|u\|$ for all $u\in\partial\Omega_{\rho_*}$.
Then by Theorem \ref{thm2.6}, we have
\begin{equation}
i(T,\Omega_{\rho_*},K)=1.\label{e4.7}
\end{equation}
Next, let $f^*(x)=\max_{0\leq u_{n-1}\leq x,0\leq u_1\leq\dots\leq
 u_{n-2}\leq u_{n-1}/\theta }f(u_1,u_2,\dots,u_{n-1})$,
note that $f^*(x)$ is monotone
increasing with respect to $x\geq0$. Then  from $f^\infty=0$, it is
easy to see that
$$
\lim_{x\to\infty}\frac{f^*(x)}{x^{p-1}}=0.
$$
Therefore, for any $\eta_2\in(0,\theta_*)$, there exists a constant
$\rho^*>r$ such that
\begin{equation}
f^*(x)\leq(\eta_2 x)^{p-1},\quad x\geq\rho^*.\label{e4.8}
\end{equation}
Set $\Omega_{\rho^*}=\{u\in K:\|u\|<\rho^*\}$, for each
$u\in\partial\Omega_{\rho^*}$, by \eqref{e4.8},  we have
\begin{align*}
\|Tu\|&=(Tu)^{(n-2)}(\delta) \\
&\leq B_0\circ \phi_{q}\Big(\int_{0}^{1}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)  \\
&\quad +\int_{0}^{1}\phi_{q}
\Big(\int_{s}^{\delta}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)ds \\
&\leq B_0\circ \phi_{q}\Big(\int_{0}^{1}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big)  \\
&\quad +\phi_{q} \Big(\int_{0}^{1}g(r)f\big(u(r),u'(r),\dots,
u^{(n-2)}(r)\big)dr\Big) \\
&\leq (b+1)\phi_{q}
\Big(\int_{0}^{1}g(r)f^*(\rho^*)dr\Big) \\
&\leq (b+1)\eta_2\rho^*\phi_{q}
\Big(\int_{0}^{1}g(r)dr\Big)\\
&\leq \rho^*=\|u\|.
\end{align*}
i.e.,
$\|Tu\|\leq\|u\|$ for all $u\in\partial\Omega_{\rho^*}$.
Then by Theorem \ref{thm2.6}, we have
\begin{equation}
i(T,\Omega_{\rho^*},K)=1.\label{e4.9}
\end{equation}
Finally, set $\Omega_{r}=\{u\in K:\|u\|<r\}$, For any
$u\in\partial\Omega_{r}$, by $(A_1)$, Lemma \ref{lem2.2} and also similar to
the previous proof of Theorem \ref{thm3.1}, we can also have
$\|Tu\|\geq\|u\|$ for all $u\in\partial\Omega_{r}$.
Then by Theorem \ref{thm2.6}, we have
\begin{equation}
i(T,\Omega_{r},K)=0.\label{e4.10}
\end{equation}
Therefore, by \eqref{e4.7}, \eqref{e4.9}, \eqref{e4.10}, $\rho_*<r<\rho^*$,
 we have
$$
i(T,\Omega_r\setminus\overline{\Omega}_{\rho_*},K)=-1,\quad
i(T,\Omega_{\rho^*}\setminus\overline{\Omega}_{r},K)=1.
$$
Then $T$ has a fixed points
$u_1\in\Omega_r\setminus\overline{\Omega}_{\rho_*}$, and
$u_2\in\Omega_{\rho_*}\setminus\overline{\Omega}_{r}$. Obviously,
$u_1, u_2$ are all positive solutions of problem \eqref{e1.1},\eqref{e1.2} and
$0<\|u_1\|<r<\|u_2\|$. The proof  is complete.
\end{proof}

Similar to Theorem \ref{thm3.1}, we obtain the following Theorems.

\begin{theorem}\label{thm4.3}
 Suppose that conditions (H1)--(H3), (A2), (A4), and (A6)hold.
Then  the boundary-value problem \eqref{e1.1},  \eqref{e1.2} has at
least two solutions $u_1, u_2$ such that $0<\|u_1\|<R<\|u_2\|$.
\end{theorem}

\begin{theorem}\label{thm4.4}
Suppose that conditions (H1)--(H3),  (A1),  (A3) and (A5) hold.
Then  the boundary-value problem \eqref{e1.1},  \eqref{e1.2} has at
least two solutions $u_1, u_2$ such
that $0<\|u_1\|<r<\|u_2\|$.
\end{theorem}

 \section{Applications}

\begin{example} \label{exa5.1} \rm
Consider the following third-order singular
boundary-value problem (SBVP), with $p$-Laplacian,
\begin{equation}
\begin{gathered}
(\phi_{p}(u''))'+\frac{\sqrt{3}}{36}t^{-\frac{1}{2}}(u')^{1/2}
[\frac{1}{5}+\frac{\frac{94}{5}e^{2u'}}{120u+7e^{u'}+e^{2u'}}]=0 \quad
 0<t<1 , \\
u(0)=0,  \\
u'(0)-u''(1/4)=0,\quad u'(1)+3u''(1/2))=0,
\end{gathered}\label{e5.1}
\end{equation}
where
\begin{gather*}
p=\frac{3}{2},\quad \xi=\frac{1}{4},\quad
\eta=\frac{1}{2},\quad b=2,\quad \theta=\frac{1}{4}, \\
g(t)=\frac{\sqrt{3}}{36}t^{-\frac{1}{2}},  \quad
f(u_1,\ u_2)=(u_2)^{1/2}
\big[\frac{1}{5}+\frac{94e^{2u_2}/5}{120u_1+7e^{u_2}+e^{2u_2}}\big].
\end{gather*}
Then obviously,
\begin{gather*}
q=3, \quad f^0=\varphi=\lim_{u_2\to0^+}\max_{0\leq u_1\leq 4u_2}
 \frac{f(u_1,\ u_2)}{u_2^{p-1}}=\frac{51}{20},\\
 f_\infty=l=\lim_{u_2\to\infty}\min_{0\leq u_1\leq
4u_2}
 \frac{f(u_1,\ u_2)}{u_2^{p-1}}=l=\frac{95}{5}, \quad
 \int_0^1  g(t)dt=\frac{\sqrt{3}}{18},
\\
B_0(v)=v<2v=bv,\quad B_1(v)=3v,\quad  \forall v\geq0,
\end{gather*}
 so conditions (H1)--(H3) hold.
Next,
$$
\theta_*= \frac{1}{(b+1)\phi_{q}
\big(\int_{0}^{1}g(r)dr\big)}=36,
$$
then $(\theta_*/4)^{p-1}=3>\frac{51}{20}$,
i.e. $\varphi\in[0, (\theta_*/4)^{p-1})$, so
conditions (A3) holds.
For $\theta=1/4$,  by calculating, it is easy see that
$$
L=\min_{t\in[\theta,1-\theta]}A(t)=\frac{1}{16}
(\frac{7}{36}+\frac{\sqrt{3}}{3}).
$$
Because
$$
(2\theta^*/\theta)^{p-1}=96\times(\frac{1}{7+12\sqrt{3}})^{1/2}
<\frac{95}{5},$$
we have
 $$
l\in ((2\theta^*/\theta)^{p-1},
\infty),
$$
 so conditions (A4) holds. Then by Theorem \ref{thm3.2},  \eqref{e5.1} has at
 least a positive solution.
\end{example}

\begin{example} \label{exam5.2} \rm
Consider the following third-order singular
boundary-value problem, with $p$-Laplacian,
\begin{equation}
\begin{gathered}
(\phi_{p}(u''))'+\frac{1}{64\pi^4}t^{-\frac{1}{2}}(1-t)
[u+(u')^{2}+(u')^{4}]=0 \quad  0<t<1 , \\
u(0)=0,   \\
u'(0)-u''(1/4)=0,\quad u'(1)+5u''(1/3))=0,
\end{gathered} \label{e5.2}
\end{equation}
where
\begin{gather*}
p=4,\quad \xi=\frac{1}{4},\quad \eta=\frac{1}{3},\quad
 \theta=\frac{1}{4}, \\
g(t)=\frac{1}{64\pi^4}t^{-\frac{1}{2}}(1-t),\quad
f(u_1, u_2)=u_1+u_2^{2}+u_2^{4}.
\end{gather*}
Then obviously,
\begin{gather*}
 q=\frac{4}{3},\quad
 \int_0^1g(t)dt=\frac{1}{64\pi^3},\quad
 f_\infty=+\infty,\quad f_0=+\infty, \\
 B_0(v)=v<2v=bv,\quad B_1(v)=5v\quad  \forall v\geq0,
\end{gather*}
 conditions (H1)--(H3), (A7), (A8) hold.
Next,
$$
\phi_{q}\big(\int_0^1g(t)dt\big)=\frac{1}{4\pi},\quad
\theta_*=\frac{4\pi}{3},
$$
we choose $R=3$, $M=2$ and for $\theta=\frac{1}{4}$,
because of the  monotone increasing of $f(u_1, u_2)$ on
$[0,\infty)\times[0,\infty)$, then
 $$
f(u_1,\ u_2)\leq f(12,\ 3)=12+90=102,\quad
  0\leq u_2\leq 3,\quad 0\leq u_1  \leq 4u_2.
$$
Therefore, by
$$
M\in  (0,\theta_*),\quad  (MR)^{p-1}=(6)^3=216,
 $$
we know that
 $$
f(u_1, u_2)\leq(MR)^{p-1},\quad 0\leq u_2\leq 3,\; 0\leq u_1
 \leq 4u_2,
$$
so conditions (A2) holds.
 Then by Theorem \ref{thm4.1},  \eqref{e5.2} has at
 least two positive solutions $v_1, v_2$ and $0<\|v_1\|<3<\|v_2\|$.
\end{example}

\subsection*{Acknowledgments}
The authors are very grateful to the anonymous referee
for his/her valuable suggestions and comments.

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