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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 78, pp. 1--13.\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 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/78\hfil Multi-point problems on time scales]
{Positive solutions of four-point boundary-value problems for
four-order $p$-Laplacian dynamic equations on time scales}

\author[H. Su, B. Wang, Z. Wei\hfil EJDE-2006/78\hfilneg]
{Hua Su, Baohe Wang, Zhongli Wei}  % in alphabetical order

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

\address{Baohe Wang \newline
Shandong Administration Institute\\
 Jinan Shandong, 250014, China}

\address{Zhongli Wei \newline
School of  Mathematics and System Sciences\\
Shandong University, Jinan Shandong,  250100, China}
\email{jnwzl@yahoo.com.cn}

\date{}
\thanks{Submitted May 10, 2006. Published July 19, 2006.}
\subjclass[2000]{34B18}
\keywords{Time scales; four-point boundary-value problems; cone;
fixed points;\hfill\break\indent positive solutions}

\begin{abstract}
 We study  the existence  of positive solutions for the  
 nonlinear four-point singular  boundary-value problem with 
 $p$-Laplacian operator on time scales.
 By using the fixed-point index theory, the existence of positive
 solution and many positive solutions for nonlinear four-point
 singular boundary-value problem with $p$-Laplacian operator
 are obtained.
\end{abstract}

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

\section{Introduction}

In this paper,   we  study  the  existence of positive solutions for
the following nonlinear four-point singular boundary-value problem
with $p$-Laplacian operator on time scales
\begin{equation}
\begin{gathered}
(\phi_{p}(u^{\Delta}))^{\nabla}+g(t)f(u(t))=0,\quad t\in(0, T),\\
\alpha\phi_{p}(u(0))-\beta\phi_{p}(u^\Delta(\xi))=0,  \quad
\gamma\phi_{p}(u(T))+\delta\phi_{p}(u^\Delta(\eta))=0,
\end{gathered}\label{e1.1}
\end{equation}
where $\phi_{p}(s)$ is $p$-Laplacian operator,   i.e.
$\phi_{p}(s)=|s|^{p-2}s$,  $p>1$,
$\phi_{q}=\phi_{p}^{-1}$,
$\frac{1}{p}+\frac{1}{q}=1$,
$\xi,\eta\in(0,T)$ is prescribed and $\xi<\eta$,
$g:(0,1)\to[0,\infty)$, $\alpha>0$,  $\beta\geq0$,
$\gamma>0$, $\delta\geq0$.

In recent years, many authors have begun to pay attention to the
study of boundary-value problems on time scales. Here, two-point
boundary-value problems have been extensively studied; see
\cite{a1,a2,a3,e1,m1}
 and the references therein. However,  there are not many
concerning the $p$-Laplacian problems on time scales.

A time scale $\mathbf{T}$ is a nonempty closed subset of $R$. We make
the blanket assumption that $0, T$ are point in $\mathbf{T}$. By an
internal $(0, T)$, we always mean the intersection of the real
internal $(0, T)$ with the given time scale, that is $(0,
T)\cap\mathbf{T}$.

Sun and Li \cite{s1} considered the existence of positive solution
of the following dynamic equations on time scales:
\begin{gather*}
u^{\Delta\nabla}(t)+a(t)f(t, u(t))=0,\quad t\in(0, T), \\
\beta u(0)-\gamma u^{\Delta}(0)=0,\quad
 \alpha u(\eta)=u(T),
\end{gather*}
 they obtained the existence of
single and multiple positive solutions of the problem by using fixed
point theorem and Leggett-Williams fixed point theorem,
respectively.


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

\item[(H2)] $a(t)\in C_{ld}((0, T), [0, +\infty))$ and there exists
$t_{0}\in (0, T)$, such that
$$
a(t_{0})>0,\quad  0<\int_0^T a(s)  \nabla s<+\infty.
$$
\end{itemize}
In this paper,   by constructing one integral equation which is
equivalent to the problem \eqref{e1.1}, we research the existence of
positive solutions for nonlinear singular boundary-value problem
\eqref{e1.1} when  $g$ and $f$ satisfy some suitable conditions. Our main
tool of this paper is the following fixed point index theory.

\begin{theorem}[\cite{g1,g2}] \label{thm1.1}
Suppose $E$ is a real Banach space, $K\subset E$ is a cone, let
$\Omega_{r}=\{u\in K:\|u\|\leq r\}$. Let operator $T:$
$K:\Omega_{r}\to K$ be completely continuous and satisfy
$Tx\neq x$, for all $x\in \partial\Omega_{r}$. Then
\begin{itemize}
\item[(i)]  If $\|Tx\|\leq\|x\|,\ \forall x\in\partial\Omega_{r}$, then
$i(T,\Omega_{r},K)=1$;

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


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}

For convenience, we list here the following definitions which are
needed later.

A time scale $\mathbf{T}$ is an arbitrary nonempty closed subset of
real numbers $ R$. The operators $\sigma$ and $\rho$ from $\mathbf{T}$
to $\mathbf{T}$ which is defined in \cite{h1},
$$
\sigma(t)=\inf\{\tau\in\mathbf{T}:\tau> t\}\in\mathbf{T}, \quad
\rho(t)=\sup\{\tau\in\mathbf{T}:\tau< t\}\in\mathbf{T}.
$$
 are called
the  forward jump operator and  the backward jump operator,
respectively.

 The point $t\in\mathbf{T}$ is
 left-dense, left-scattered, right-dense, right-scattered if
$\rho (t)=t,\ \rho(t)<t,\ \sigma(t)=t,\ \sigma(t)>t$, respectively.
If $\mathbf{T}$ has a right scattered minimum $m$, define $\mathbf{T}_{k}=\mathbf{T}-\{m\}$; otherwise set $\mathbf{T}_{k}=\mathbf{T}$. If $\mathbf{T}$ has a left scattered maximum $M$, define $\mathbf{T}^{k}=\mathbf{T}-\{M\}$; otherwise set $\mathbf{T}^{k}=\mathbf{T}$.

Let $f:\mathbf{T}\to  R$ and $t\in \mathbf{T}^{k}$ (assume $t$ is not
left-scattered if $t=\sup\mathbf{T}$), then the delta derivative of $f$
at the point $t$ is defined to be the number $f^{\Delta}(t)$
(provided it exists) with the property that for each $\epsilon>0$
there is a neighborhood $U$ of $t$ such that
$$
|f(\sigma(t))-f(s)-f^{\Delta}(t)(\sigma(t)-s) |\le | \sigma(t)-s|,
\quad \mbox{for all } s\in U.
$$
Similarly, for $t\in \mathbf{T}$ (assume $t$ is not right-scattered if
$t=\inf\mathbf{T}$),  the nabla derivative of $f$ at the point $t$ is
defined in \cite{a1} to be the number $f^{\nabla}(t)$ (provided it exists)
with the property that for each $\epsilon >0$ there is a
neighborhood $U$ of $t$ such that
$$
|f(\rho(t))-f(s)-f^{\nabla}(t)(\rho(t)-s) |\le | \rho(t)-s |, \quad
\mbox{for all } s\in U.
$$
If $\mathbf{T}=\mathbb{R}$, then $x^\Delta(t)=x^\nabla(t)=x'(t)$. If  $\mathbf{T}=Z$, then $x^\Delta(t)=x(t+1)-x(t)$ is the forward difference
operator while $x^\nabla(t)=x(t)-x(t-1)$ is the backward difference
operator.

 A function
$f$ is left-dense continuous (i.e., $ld$-continuous), if $f$ is
continuous at each left-dense point in $\mathbf{T}$ and its right-sided
limit exists at each right-dense point in $\mathbf{T}$. It is
well-known that if $f$ is $ld$-continuous.

If  $F^{\nabla}(t)=f(t)$, then we define the nabla integral by
$$
\int^b_a f(t)\nabla t=F(b)-F(a).
$$
If  $F^{\Delta}(t)=f(t)$, then we define the delta integral by
$$
\int^b_a f(t)\Delta t=F(b)-F(a).
$$
In the rest of this article, $\mathbf{T}$ is closed subset of $\mathbb{R}$ with
$0\in\mathbf{T}_k$, $T\in\mathbf{T}^k$. And let $E=C_{ld}([0, T],\ R)$ which
is a Banach space with the maximum norm $\|u\|=\max_{t\in[0,
T]}|u(t)|$. And define the cone $K\subset E$ by
$$
K=\big\{u\in E: u(t)\geq0,\quad u(t)
 \text{\ is concave function, } t\in[0,1]\big\}.
$$
We now state and prove several lemmas before stating our main
results.

\begin{lemma}\label{lem2.1}
Suppose condition (H2) holds,   then there exists a constant
$\theta\in (0,\frac{1}{2})$ satisfies
$$
0<\int_{\theta}^{T-\theta}a(t)\nabla t<\infty.
$$
Furthermore, the function
\begin{equation}
A(t)=\int_{\theta}^{t}\phi_{q}\Big(\int_{s}^{t}a(t)\nabla
t\Big)\Delta s+
\int_{t}^{T-\theta}\phi_{q}\Big(\int_{t}^{s}a(t)\nabla
t\Big)\Delta s, \quad t\in[\theta,   T-\theta] \label{e1.2}
\end{equation}
is positive continuous functions on $[\theta,T-\theta]$, therefore
$A(t)\ $ has minimum on $[\theta, 1-\theta]$, hence we suppose that
there exists $L>0$ such that $A(t)\geq L$, $t\in[\theta,T-\theta]$.
\end{lemma}


\begin{proof}
 At first, it is easily seen that $A(t)$ is continuous
on $[\theta,T-\theta]$. Nest, let
$$
A_1(t)=\int_{\theta}^{t}\phi_{q}\Big(\int_{s}^{t}a(s_1)\nabla s_1\Big)
\Delta s,\quad
A_2(t)=\int_{t}^{T-\theta}\phi_{q}\Big(\int_{t}^{s}a(s_1)\nabla
s_1\Big)\Delta s.
$$
 Then, from condition (H2), we have the
function $A_1(t)$ is strictly monotone nondecreasing on
$[\theta,1-\theta]$ and $A_1(\theta)=0$, the function $A_2(t)$ is
strictly monotone nonincreasing on $[\theta,T-\theta]$ and
$A_2(T-\theta)=0$, which implies
$L=\min_{t\in[\theta,T-\theta]}A(t)>0$. The proof is
complete.
\end{proof}


\begin{lemma}\label{lem2.2}
Let $u\in K$ and   $\theta$ of Lemma \ref{lem2.1}, then
\begin{equation}
u(t)\geq\theta\|u\|,  \quad t\in[\theta,   T-\theta]. \label{e1.3}
\end{equation}
\end{lemma}

\begin{proof}
Suppose $\tau=\inf\big\{\xi\in[0,T]:\sup_{t\in[0,T]}u(t)=u(\xi)\big\}$.
we shall discuss it from three perspectives.

(i)\ $\tau\in[0,\theta]$. It follows from the concavity of $u(t)$
that each point on chord between $(\tau,u(\tau))$ and $(T,u(T))$ is
below the graph of $u(t)$, thus
$$
u(t)\geq u(\tau)+\frac{u(T)-u(\tau)}{T-\tau}(t-\tau),\quad
 t\in[\theta,   T-\theta],
$$
then
\begin{align*}
u(t)&\geq \min_{t\in[\theta,   T-\theta]}
\big[u(\tau)+\frac{u(T)-u(\tau)}{T-\tau}(t-\tau)\big]\\
&=u(\tau)+\frac{u(T)-u(\tau)}{T-\tau}(T-\theta-\tau)\\
&=\frac{T-\theta-\tau}{T-\tau}u(T)+\frac{\theta}{T-\tau}u(\tau)
\geq\theta u(\tau),
\end{align*}
 this implies $u(t)\geq\theta\|u\|$, $t\in[\theta, T-\theta]$.

(ii)\ $\tau\in[\theta,T-\theta]$. If $t\in[\theta, \tau]$,
similarly, we have
$$
u(t)\geq u(\tau)+\frac{u(\tau)-u(0)}{\tau}(t-\tau),\quad t\in[\theta,   \tau],
$$
then
\begin{align*}
u(t)&\geq \min_{t\in[\theta,   1-\theta]}
\big[u(\tau)+\frac{u(\tau)-u(0)}{\tau}(t-\tau)\big] \\
&=\frac{\theta}{\tau}u(\tau)+\frac{\tau-\theta}{\tau}u(0)
\geq\theta u(\tau),
\end{align*}
If $t\in[\tau, T-\theta]$, similarly, we have
$$
u(t)\geq u(\tau)+\frac{u(T)-u(\tau)}{T-\tau}(t-\tau),\quad t\in[\tau, T-\theta],
$$
then
\begin{align*}
u(t)&\geq \min_{t\in[\theta,   T-\theta]}
\big[u(\tau)+\frac{u(T)-u(\tau)}{T-\tau}(t-\tau)\big] \\
&=\frac{\theta}{T-\tau}u(\tau)+\frac{T-\theta-\tau}{T-\tau}u(T)
\geq\theta u(\tau),
\end{align*}
this implies $u(t)\geq\theta\|u\|$,
$t\in[\theta,   1-\theta]$.

(iii)\ $\tau\in[T-\theta,T]$. similarly, we have
$$
u(t)\geq u(\tau)+\frac{u(\tau)-u(0)}{\tau}(t-\tau),\quad t\in[\theta, T-\theta],
$$
then
\begin{align*}
u(t)\geq \min_{t\in[\theta,   1-\theta]}
\big[u(\tau)+\frac{u(\tau)-u(0)}{\tau}(t-\tau)\big]
=\frac{\theta}{\tau}u(\tau)+\frac{\tau-\theta}{\tau}u(0)
\geq\theta u(\tau),
\end{align*}
this implies $u(t)\geq\theta\|u\|$,
$t\in[\theta, T-\theta]$. From the above, we know
$u(t)\geq\theta\|u\|$, $t\in[\theta, T-\theta]$. The proof is
complete.
\end{proof}

\begin{lemma}\label{lem2.3}
Suppose that conditions (H1), (H2) hold. Then $u(t)$ is a
solution of  boundary-value problems \eqref{e1.1} if and only if
$u(t)\in E$ is a solution of the integral equation
\begin{equation}
u(t)=\begin{cases}
\phi_{q}\big(\frac{\beta}{\alpha}\int_{\xi}^{\sigma}a(r)f(u(r))\nabla
r\big)
+\int_{0}^{t}\phi_{q}\big(\int_{s}^{\sigma}a(r)f(u(r))\nabla
r\big)\Delta s, \\
\qquad \text{if } 0\leq t\leq\sigma, \\[3pt]
\phi_{q}\big(\frac{\delta}{\gamma}\int_{\theta}^{\eta}a(r)f(u(r))\nabla
r\big)
+\int_{t}^{T}\phi_{q}\big(\int_{\theta}^{s}a(r)f(u(r))\nabla
r\big)\Delta s, \\
\qquad \text{if } \sigma\leq t\leq T.
\end{cases}\label{e2.1}
\end{equation}
\end{lemma}

\begin{proof}
 \emph{Necessity}.  By the equation of the boundary
condition we have $u^\Delta(\xi)\geq 0$, $u^\Delta(\eta)\leq0$, then
there exist a constant $\sigma\in [\xi,\eta]\subset(0,1)$ such that
$u^\Delta(\sigma)=0$.

First, by integrating the equation of the problems \eqref{e1.1} on
$(\theta,1)$ we have,
\begin{equation}
\phi_{p}(u^\Delta(t))=\phi_{p}(u^\Delta(\sigma))-\int_{\sigma}^t
a(s)f(u)(s)\nabla s,\label{e2.2}
\end{equation}
then
$u^\Delta(t)=u^\Delta(\sigma)-\phi_{q}\Big(\int_{\sigma}^ta(s)f(u)(s)\nabla
s\Big)$, thus
\begin{equation}
u(t)=u(\sigma)+u^\Delta(\sigma)(t-\sigma)-
\int_{\sigma}^t\phi_{q}\Big(\int_{\sigma}^s
a(r)f(u)(r)\nabla r\Big)\Delta s,\label{e2.3}
\end{equation}
 by
$u^\Delta(\sigma)=0$, let $t=\eta$ on \eqref{e2.2}, we have
$\phi_{p}(u'(\eta))=-\int_{\sigma}^\eta
a(s)f(u)(s)\nabla s$. By the equation of the boundary condition
\eqref{e1.1}, we have
$\phi_{p}(u(T))=-\frac{\delta}{\gamma}\phi_{p}(u'(\eta))$,
then
\begin{equation}
u(T)=\phi_{q}\Big(\frac{\delta}{\gamma}\int_{\sigma}^\eta a(s)f(u)(s)
\nabla s\Big).\label{e2.4}
\end{equation}
by \eqref{e2.3}, \eqref{e2.4} and let let $t=T$ on \eqref{e2.3}, we have
\begin{equation}
u(\sigma)=\phi_{q}\Big(\frac{\delta}{\gamma}
\int_{\sigma}^\eta a(s)f(u)(s)\nabla s\Big)
+\int_{\sigma}^T\phi_{q}\Big(\int_{\sigma}^s
a(r)f(u)(r)\nabla r\Big)\Delta s,\label{e2.5}
\end{equation}
by \eqref{e2.3} and \eqref{e2.5},
for $t\in(\sigma,T)$ we know
$$
u(t)=\phi_{q}\Big(\frac{\delta}{\gamma}\int_{\sigma}^\eta
a(s)f(u)(s)\nabla s\Big)+\int_{t}^T\phi_{q}\Big(\int_{\sigma}^s
a(r)f(u)(r)\nabla r\Big)\Delta s.
$$
Similarly, for $t\in(0,\sigma)$, by integrating the equation of
problems \eqref{e1.1} on $(0,\sigma)$, we have
$$
u(t)=\phi_{q}\Big(\frac{\beta}{\alpha}\int_{\xi}^{\sigma}a(r)f((u)(r))
\nabla r\Big)
+\int_{0}^{t}\phi_{q}\Big(\int_{s}^{\sigma}a(r)f((u)(r))\nabla
r\Big)\Delta s.
$$
 Then \eqref{e2.1} holds.

\emph{Sufficiency}. Suppose that \eqref{e2.1} holds. Then by \eqref{e2.1},
we have
\begin{equation}
u^\Delta(t)=\begin{cases}
\phi_{q}\Big(\int_{t}^{\sigma}a(r)f((u)(r))\nabla
r\Big)\geq0, & 0\leq t\leq\sigma, \\
 -\phi_{q}\Big(\int_{\sigma}^{t}a(r)f((u)(r))\nabla r\Big)\leq0,
  &\sigma\leq t\leq T,
\end{cases}\label{e2.6}
\end{equation}
So, $(\phi_{p}(u^\Delta))^\nabla+a(t)f(u(t))=0$,   $0<t<1 $.
These imply that the first equation of \eqref{e1.1} holds. Furthermore, by
letting $t=0$ and $t=T$ on \eqref{e2.1} and \eqref{e2.6}, we can obtain the
boundary value equations of \eqref{e1.1}. The proof of Lemma \ref{lem2.3} is
complete.
\end{proof}

Now,   we define a mapping $T:K\to E$ given by
$$
(T(u))(t)=\begin{cases}
\phi_{q}\big(\frac{\beta}{\alpha}\int_{\xi}^{\sigma}a(r)f(u(r))\nabla
r\big)
+\int_{0}^{t}\phi_{q}\big(\int_{s}^{\sigma}a(r)f(u(r))\nabla
r\big)\Delta s, \\
\qquad\text{if } 0\leq t\leq\sigma, \\[3pt]
\phi_{q}\big(\frac{\delta}{\gamma}\int_{\sigma}^{\eta}a(r)f(u(r))\nabla
r\big)
+\int_{t}^{T}\phi_{q}\big(\int_{\sigma}^{s}a(r)f(u(r))\nabla
r\big)\Delta s, \\
\qquad \text{if } \sigma\leq t\leq T.
\end{cases}
$$
Because of
$$
(T(u))'(t)=\begin{cases}
\phi_{q}\big(\int_{t}^{\sigma}a(r)f((u)(r))\nabla
r\big)\geq0,  &0\leq t\leq\sigma, \\
 -\phi_{q}\big(\int_{\sigma}^{t}a(r)f((u)(r))\nabla r\big)\leq0,
 &\sigma\leq t\leq T,
\end{cases}
$$
the operator $T$ is monotone decreasing continuous and
$(T(u)^\Delta)(\sigma)=0$,   and for any $u\in K$, we have
$$
(\phi_{q}(T(u))^\Delta)^\nabla(t)=-a(t)f((u)(t),
 \quad \text{a.e. }t\in(0,   1),
$$
and $(T(u))(\sigma)=\|T(u)\|$. Therefore,  $T(K)\subset K$.

\begin{lemma}\label{lem2.4}
$T:K\to K$ is completely continuous.
\end{lemma}


\begin{proof}
 Suppose $D\subset K$ is a bounded set, Let $M>0$ such
that $\|u\|\leq M$, $u\in D$. For any $u\in D$,  we have
\begin{align*}
\|Tu\|&=(Tu)(\sigma)\\
&=\phi_{q}\Big(\frac{\beta}{\alpha}\int_{\xi}^{\sigma}a(r)f(u(r))\nabla
r\Big)
+\int_{0}^{\sigma}\phi_{q}\Big(\int_{s}^{\sigma}a(r)f(u(r))\nabla
r\Big)\Delta s, \\
&\leq\Big[\phi_{q}\Big(\frac{\beta}{\alpha}
\int_{0}^{\sigma}a(r)\nabla r\Big)
+\int_{0}^{\sigma}\phi_{q}\Big(\int_{s}^{\sigma}a(r)\nabla r\Big)\Delta s
\Big] \phi_{q}\Big(\sup_{u\in D}f(u)\Big).
 \end{align*}
Then $T(D)$ is bounded.
Furthermore it is easy see by Arzela-ascoli Theorem and Lebesgue
dominated convergence Theorem that $T:K\to K$ is completely
continuous. The proof is complete.
\end{proof}

For convenience, we set $\theta^*=2/L$ and
$$
\theta_*=\frac{1}{\Big(T+\phi_{q}\big(\frac{\beta}{\alpha}\big)\Big)
\phi_{q}\Big(\int_{0}^{T}a(r)\nabla r\Big)}.
$$

\section{The Existence of Positive Solutions}

 The main results of this
part are the following three Theorems.

\begin{theorem}\label{thm3.1}
Suppose that conditions (H1), (H2) hold. Assume that $f$ also satisfies
\begin{itemize}
\item[(A1)] $f(u)\geq(mr)^{p-1}$,   $\theta r\leq u\leq r$;

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

\begin{theorem}\label{thm3.2}
 Suppose that conditions (H1), (H2)
hold. Assume that $f$ also satisfy
\begin{itemize}
\item[(A3)] $f_\infty=\lambda\in
\big((\frac{2\theta^*}{\theta})^{p-1}, \infty\big)$;

\item[(A4)] $f_0=\varphi\in [0,(\frac{\theta_*}{4})^{p-1})$.
Then,   the boundary-value problem \eqref{e1.1} has a positive
solution $u$  such that $\|u\|$ is between $r$ and $R$.
\end{itemize}
\end{theorem}

\begin{theorem}\label{thm3.3}
Suppose that conditions (H1), (H2)
hold. Assume that $f$ also satisfy
\begin{itemize}
\item[(A5)]
$f_0=\varphi\in\big((\frac{2\theta^*}{\theta})^{p-1},
\infty\big)$;

\item[(A6)] $f_\infty=\lambda\in \big[0, (\frac{\theta_*}{4})^{p-1}\big)$.
\end{itemize}
Then,   the boundary-value problem \eqref{e1.1} has a positive
solution $u$  such that $\|u\|$ is between $r$ and $R$.
\end{theorem}


\begin{proof}[Proof of Theorem \ref{thm3.1}].
 Without loss of generality, we suppose
that $r<R$. For any  $u\in K$, by Lemma \ref{lem2.2},  we have
\begin{equation}
u(t)\geq\theta\|u\|,  \quad  t\in[\theta, T-\theta].\label{e3.1}
\end{equation}
we define two open subset  $\Omega_{1}$ and $\Omega_{2}$ of $E$,
$$
\Omega_{1}=\{u\in K:\|u\|<r\},\quad  \Omega_{2}=\{u\in K:\|u\|<R\}
$$
For $u\in\partial\Omega_{1}$,  by \eqref{e3.1},  we have
$$
r=\|u\|\geq u(t)\geq\theta\|u\|=\theta r, \quad
 t \in[\theta,    T-\theta].
$$
  For $t\in[\theta, T-\theta]$, if
(A1) hold, we shall discuss it from three perspectives.

(i)\  If $\theta\in[\theta, T-\theta]$,    thus for
$u\in\partial\Omega_{1}$,   by (A1) and Lemma \ref{lem2.1},  we have
\begin{align*}
2\|T(u)\|&=2(T(u))(\sigma) \\
&\geq\int_{0}^{\theta}\phi_{q}\Big(\int_{s}^{\theta}
a(r)f((u)(r))\nabla r\Big)\Delta s
+\int_{\theta}^{T}\phi_{q}
\Big(\int_{\theta}^{s}a(r)f((u)(r))\nabla
r\Big)\Delta s \\
&\geq(mr)\Big(\int_{\theta}^{\theta}\phi_{q}
\Big(\int_{s}^{\theta}a(r)\nabla r\Big)\Delta s\Big)
+(mr)\Big(\int_{\theta}^{T-\theta}\phi_{q}\Big(\int_{\theta}^{s}a(r)\nabla
r \Big)\Delta s \Big)\\
&\geq mrA(\theta)\geq mrL\\
&>2r=2\|u\|.
\end{align*}

(ii)\  If $\theta\in(T-\theta,    T]$,    thus for
$u\in\partial\Omega_{1}$,    by (A1) and Lemma \ref{lem2.1},  we have
\begin{align*}
\|T(u)\|&=(T(u))(\theta)\\
&\geq\int_{0}^{\theta}\phi_{q}
\Big(\int_{s}^{\theta}a(r)f((u)(r))\nabla r\Big)\Delta s \\
&\geq\int_{\theta}^{T-\theta}\phi_{q}
\Big(\int_{s}^{T-\theta}a(r)f((u)(r))\nabla
r\Big)\Delta s\\
&\geq mr\int_{\theta}^{T-\theta}\phi_{q}
\Big(\int_{s}^{T-\theta}a(r)\nabla r\Big)\Delta s\\
&=mrA(T-\theta)\geq mrL\\
&>2r>r=\|u\|.
\end{align*}

(iii)\   If $\theta\in[0,    \theta)$,     thus for
$u\in\partial\Omega_{1}$,   by ($A_{1}$) and Lemma \ref{lem2.1},  we have
\begin{align*}
\|T(u)\|&=(T(u))(\theta)\\
&\geq\int_{\theta}^{T}\phi_{q}
\Big(\int_{\theta}^{s}a(r)f((u)(r))\nabla r\Big)\Delta s \\
&\geq\int_{\theta}^{T-\theta}\phi_{q}
\Big(\int_{\theta}^{s}a(r)f((u)(r))\nabla r\Big)\Delta s\\
&\geq mr\int_{\theta}^{T-\theta}\phi_{q}
\Big(\int_{\theta}^{s}a(r)\nabla r\Big)\Delta s\\
&=mr A(\theta)\geq mrL\\
&>2r>r =\|u\|.
\end{align*}
Therefore,
$\|Tu\|\geq\|u\|$,  for all $u\in\partial\Omega_{1}$.
Then by Theorem \ref{thm3.1},
\begin{equation}
i(T,\Omega_{1},K)=0.\label{e3.2}
\end{equation}
On the other hand,  since $u\in\partial\Omega_{2}$, we have
$u(t)\leq\|u\|=R$.   By (A2),  we know
\begin{align*}
\|T(u)\|&=(T(u))(\theta)\\
&=\phi_{q}\Big(\frac{\beta}{\alpha}\int_{\xi}^{\theta}a(r)f((u)(r))\nabla
r\Big)
+\int_{0}^{t}\phi_{q}\Big(\int_{0}^{t}a(r)f((u)(r))\nabla
r\Big)\Delta s\\
&\leq\phi_{q}\Big(\frac{\beta}{\alpha}\int_{0}^{T}a(r)f((u)(r))\nabla
r\Big)
+\int_{0}^{T}\phi_{q}\Big(\int_{0}^{T}a(r)f((u)(r))\nabla
r\Big)\Delta s\\
&=\big[T+\phi_{q}(\frac{\beta}{\alpha})\big]
\phi_{q}\Big(\int_{0}^{T}a(r)f((u)(r))\nabla r\Big) \\
&\leq\big[T+\phi_{q}(\frac{\beta}{\alpha})\big]MR
\phi_{q}\Big(\int_{0}^{T}a(r)\nabla r\Big)\\
&=\big[T+\phi_{q}(\frac{\beta}{\alpha})\big]MR
\phi_{q}\Big(\int_{0}^{T}a(r)\nabla r\Big) \\
&\leq R=\|u\|.
\end{align*}
Therefore,
$\|T(u)\|\leq\|u\|$, for all  $u\in\partial\Omega_{2}$.
Then by Theorem \ref{thm3.1},
\begin{equation}
i(T,\Omega_{2},K)=1.\label{e3.3}
\end{equation}
Therefore, by \eqref{e3.2}, \eqref{e3.3} and $r<R$,   we have
$$
i(T,\Omega_2\setminus\overline{\Omega}_{1},K)=1.
$$
 Then $T$ has a fixed
point $u\in(\Omega_{2}\setminus\overline{\Omega}_{1})$. Obviously,
$u$ is positive solution of problem \eqref{e1.1} and $r<\|u\|<R$. The proof
of Theorem \ref{thm3.1} is complete.
 \end{proof}


\begin{proof}[Proof of Theorem \ref{thm3.2}]
 Firstly,   by
$f_0=\lim_{u\to0}\frac{f(u)}{u^{p-1}}=\varphi$,
for
$\epsilon=(\frac{\theta_*}{4})^{p-1}-\varphi$,
there exists an adequate small positive number $\rho$. Since
 $0\leq u\leq \rho$,  $u\neq0$,   we have
\begin{equation}
f(u)\leq(\varphi+\epsilon)u^{p-1}\leq
(\frac{\theta_*}{4})^{p-1}(2\rho)^{p-1}
=\big(\frac{\theta_*}{2}\rho\big)^{p-1}.\label{e3.4}
\end{equation}
Let $R=\rho$, $M=\frac{\theta_*}{2}\in(0,\theta_*)$,
 thus by \eqref{e3.4}, we have
$$
f(u)\leq(MR)^{p-1},  \quad 0\leq u\leq R.
$$
 Then condition (A2) holds.
Next,   by condition (A3),
$$
f_\infty=\lim_{u\to0}\frac{f(u)}{u^{p-1}}=\lambda\in
\big((\frac{2\theta^*}{\theta})^{p-1}.
\infty\big)
$$
Then for $\epsilon=\lambda-(\frac{2\theta^*}{\theta})^{p-1}$,
 there exists an adequate big positive number $r\neq R$. Since
 $u\geq \theta r$,    we have
\begin{equation}
 f(u)\geq(\lambda-\epsilon)u^{p-1}
\geq\big(\frac{2\theta^*}{\theta}\big)^{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}
holds. The proof of Theorem \ref{thm3.2} is complete.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm3.3}]
 Firstly,  by condition $f_0=\varphi$, then for
$\epsilon=\varphi-(\frac{2\theta^*}{\theta})^{p-1}$,
there exists an adequate small positive number $r$.
Since  $0\leq u\leq r, u\neq0$,    we have
$$
f(u)\geq(\varphi-\epsilon)u^{p-1}
=\big(\frac{2\theta^*}{\theta}\big)^{p-1}u^{p-1}.
$$
Thus when $\theta r\leq u\leq r$,   we have
\begin{equation}
f(u)\geq\big(\frac{2\theta^*}{\theta}\big)^{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 (A6): $f_\infty=\lambda$,   then for
$\epsilon=(\frac{\theta_*}{4})^{p-1}-\lambda$,
there exists an adequate big positive number $\rho\neq r$.   Since
$u\geq\rho$, we have
\begin{equation}
f(u)\leq(\lambda+\epsilon)u^{p-1}\leq
\big(\frac{\theta_*}{4}\big)^{p-1}u^{p-1}.\label{e3.7}
\end{equation}
If $f$ is non-boundary,   by the continuation of $f$ on
$[0,\infty)$, then exists constant $R(\neq r)\geq\rho$,   and a point
$u_0\in[0, \infty)$ such that $\rho\leq u\leq R$ and $f(u)\leq
f(u_0),  \ 0\leq u\leq R$. Thus,   by $\rho\leq u_0\leq R$, we know
$$
f(u)\leq f(u_0)\leq\big(\frac{\theta_*}{4}\big)^{p-1}u_0^{p-1}
\leq\big(\frac{\theta_*}{4}R\big)^{p-1}.
$$
 Let
$M=\theta_*/4 \in(0,   \theta_*)$,    we have
$$
f(u)\leq (MR)^{p-1},  \quad   0\leq u\leq R.
$$
If $f$ is boundary,   we suppose $f(u)\leq \overline{M}^{p-1}$,
$u\in[0,   \infty)$.   There exists an adequate big positive number
$R>\frac{4}{\theta_*}\overline{M}$, then let
$M=\frac{\theta_*}{4}\in(0,   \theta_*)$,   we have
$$
f(u)\leq \overline{M}^{p-1}\leq\big(\frac{\theta_*}{4}R\big)^{p-1}=
(MR)^{p-1},  \quad  0\leq u\leq R.
$$
 Therefore,   condition (A2) holds. Therefore, by Theorem \ref{thm3.1},
we know that the results of Theorem \ref{thm3.3} holds. The proof of
Theorem \ref{thm3.3} is complete.
\end{proof}

 \section{The Existence of Many Positive Solutions}

Now, we will discuss the existence of  many positive solutions.

\begin{theorem}\label{thm4.1}
 Suppose that conditions (H1), (H2) and (A2) in Theorem \ref{thm3.1} hold.
Assume that $f$ also satisfy
\begin{itemize}
\item[(A7)]  $f_0=+\infty$;

\item[(A8)] $f_\infty=+\infty$.
\end{itemize}
Then,   the boundary-value problem
\eqref{e1.1} has at least two solutions $u_1$, $u_2$ such that
$$
0<\|u_1\|<R<\|u_2\|.
$$
\end{theorem}

\begin{proof}
 Firstly, by condition (A7), for any $M>\frac{2}{L}$, there exists a constant
$\rho_*\in(0,R)$ such that
\begin{equation}
f(u)\geq(Mu)^{p-1},\quad 0<u\leq\rho_*,\quad u\neq 0. \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 previous proof of Theorem \ref{thm3.1}, we can have from three
perspectives
$$
\|Tu\|\geq\|u\|,\quad \forall u\in\partial\Omega_{\rho_*}.
$$
Then by Theorem \ref{thm3.1}, we have
\begin{equation}
i(T,\Omega_{\rho_*},K)=0.\label{e4.2}
\end{equation}
 Next, by condition ($A_{8}$), for
any $\overline{M}>\frac{2}{L}$, there exists a constant
$\rho_0>0$ such that
\begin{equation}
f(u)\geq(\overline{M}u)^{p-1},\quad u>\rho_0.
\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
$$
u(t)\geq\theta\|u\|=\theta \rho^*>\rho_0,\quad t\in[\theta,1-\theta].
$$
Then by \eqref{e4.3} and also similar to the previous proof of
Theorem \ref{thm3.1},
we can also have from three perspectives
$$
\|Tu\|\geq\|u\|,\quad \forall  u\in\partial\Omega_{\rho^*}.
$$
Then by Theorem \ref{thm3.1}, 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 any
$u\in\partial\Omega_{R}$, by (A2), Lemma \ref{lem2.2} and also similar to
the latter proof of Theorem \ref{thm3.1}, we can also have
$$
\|Tu\|\leq\|u\|,\quad \forall u\in\partial\Omega_{R}.
$$
Then by Theorem \ref{thm3.1},
\begin{equation}
i(T,\Omega_{R},K)=1.\label{e4.5}
\end{equation}
Therefore, by \eqref{e4.2}, \eqref{e4.4}, \eqref{e4.5} and  $\rho_*<R<\rho^*$,  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  fixed points
$u_1\in\Omega_R\setminus\overline{\Omega}_{\rho_*}$, 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 of Theorem \ref{thm4.1} is complete.
\end{proof}

\begin{theorem}\label{thm4.2}
 Suppose that conditions (H1), (H2)
and (A1) in Theorem \ref{thm3.1} hold. Assume that $f$ also satisfy
\begin{itemize}
\item[(A9)] $f_0=0$;

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

\begin{proof}
 Firstly, by $f_0=0$, for
$\epsilon_1\in(0,\theta_*)$, there exists a constant
$\rho_*\in(0,r)$ such that
\begin{equation}
f(u)\leq\left(\epsilon_1u\right)^{p-1},\quad
 0<u\leq\rho_*.\label{e4.6}
\end{equation}
 Set $\Omega_{\rho_*}=\{u\in
K:\|u\|<\rho_*\}$, for any $u\in\partial\Omega_{\rho_*}$, by \eqref{e4.6},
we have
\begin{align*}
\|Tu\|&=(Tu)(\delta)\\
&=\phi_{q}\Big(\frac{\beta}{\alpha}\int_{\xi}^{\sigma}a(r)f(u(r))\nabla
r\Big)
+\int_{0}^{\delta}\phi_{q}\Big(\int_{s}^{\sigma}a(r)f(u(r))\nabla
r\Big)\Delta s \\
&\leq \phi_{q}\Big(\frac{\beta}{\alpha}\int_{0}^{T}a(r)f(u(r))\nabla
r\Big) +T\phi_{q}\Big(\int_{0}^{T}a(r)f(u(r))\nabla
r\Big)\\
&\leq \big(\phi_{q}(\frac{\beta}{\alpha})+T\big)\epsilon_1\rho_*\phi_{q}
\Big(\int_{0}^{T}a(r)\nabla r\Big)\\
&\leq \rho_*=\|u\|.
\end{align*}
i.e.,
$\|Tu\|\leq\|u\|$, for all $u\in\partial\Omega_{\rho_*}$.
Then by Theorem \ref{thm3.1}, we have
\begin{equation}
i(T,\Omega_{\rho_*},K)=1.\label{e4.7}
\end{equation}
Next, let $f^*(x)=\max_{0\leq u\leq x}f(u)$, 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 $\epsilon_2\in(0,\theta_*)$, there exists a
constant $\rho^*>r$ such that
\begin{equation}
f^*(x)\leq(\epsilon_2 x)^{p-1},\quad x\geq\rho^*.\label{e4.8}
\end{equation}
Set $\Omega_{\rho^*}=\{u\in K:\|u\|<\rho^*\}$. For any
$u\in\partial\Omega_{\rho^*}$, by \eqref{e4.8},  we have
\begin{align*}
\|Tu\|&=(Tu)(\delta)\\
&=\phi_{q}\Big(\frac{\beta}{\alpha}\int_{\xi}^{\sigma}a(r)f(u(r))\nabla
r\Big)
+\int_{0}^{\delta}\phi_{q}\Big(\int_{s}^{\sigma}a(r)f(u(r))\nabla
r\Big)\Delta s \\
&\leq \phi_{q}\Big(\frac{\beta}{\alpha}\int_{0}^{T}a(r)f(u(r))\nabla
r\Big) +T\phi_{q}\Big(\int_{0}^{T}a(r)f(u(r))\nabla r\Big)\\
&\leq \phi_{q}\Big(\frac{\beta}{\alpha}\int_{0}^{T}a(r)f^*(\rho^*))\nabla
r\Big) +T\phi_{q}\Big(\int_{0}^{T}a(r)f^*(\rho^*))\nabla
r\Big)\\
&\leq \big(\phi_{q}(\frac{\beta}{\alpha})+T\big)\epsilon_\rho^*\phi_{q}
\Big(\int_{0}^{T}a(r)\nabla r\Big)\leq
\rho^*=\|u\|.
\end{align*}
i.e.,
$\|Tu\|\leq\|u\|$, for all $u\in\partial\Omega_{\rho^*}$.
Then by Theorem \ref{thm3.1}, 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 (A1),  Lemma \ref{lem2.2} and also similar to
the previous proof of Theorem \ref{thm3.1}, we  have
$$
\|Tu\|\geq\|u\|,\quad \forall u\in\partial\Omega_{r}.
$$
Then by Theorem \ref{thm3.1}, 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 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 of Theorem \ref{thm4.2} is complete.
\end{proof}

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

\begin{theorem}\label{thm4.3}
 Suppose that conditions (H1), (H2)
and (A2) in Theorem \ref{thm3.1}, (A4) in Theorem \ref{thm3.2} and (A6) in
Theorem \ref{thm3.3} hold. Then   the boundary-value problem \eqref{e1.1}
 has at last 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), (H2)
and (A1) in Theorem \ref{thm3.1}, (A3) in Theorem \ref{thm3.2} and (A5) in
Theorem \ref{thm3.3} hold. Then   the boundary-value problem \eqref{e1.1} has at
last 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 singular boundary-value problem (SBVP)
with $p$-Laplacian:
\begin{equation}
\begin{gathered}
(\phi_{p}(u^\Delta))^\nabla
+\frac{1}{4}t^{-\frac{1}{2}}u^{1/2}
\big[\frac{1}{3}+\frac{64e^{2u}}{120+7e^{u}+e^{2u}}\big]=0, \quad
0<t<\frac{3}{2} , \\
4\phi_{p}(u(0))-\phi_{p}(u^\Delta(\frac{1}{4}))=0,  \quad
\phi_{p}(u(\frac{3}{2}))+\delta\phi_{p}(u^\Delta(\frac{1}{2}))=0,
\end{gathered}\label{e5.1}
\end{equation}
where
$\beta=\gamma=1$, $\alpha=4$, $p=\frac{3}{2}$,
$\delta\geq0$, $\xi=\frac{1}{4}$,
$\eta=\frac{1}{2}$, $T=\frac{3}{2}$,
$$
a(t)=\frac{1}{4}t^{-1/2},  \quad
 f(u)=u^{1/2} \big[\frac{1}{3}+\frac{64e^{2u}}{120+7e^{u}
+e^{2u}}\big].
$$
Then obviously,
\begin{gather*}
q=3, \quad
f_0=\varepsilon=\lim_{u\to0^+}  \frac{f(u)}{u^{p-1}}=\frac{5}{6},\\
f_\infty=\lim_{u\to\infty}
 \frac{f(u)}{u^{p-1}}=64+\frac{1}{3}, \quad
\int_0^T  a(t)\nabla t=\frac{\sqrt{6}}{4},
\end{gather*}
  so conditions (H1), (H2) hold.
Next,
$$
\theta_*=\frac{1}{\big(T+\phi_{q}(\frac{\beta}{\alpha})\big)\phi_{q}
\big(\int_{0}^{T}a(r)\nabla
r\big)}=\frac{32\sqrt{6}}{75},
$$
 then
$\varepsilon\in[0,(\frac{\theta_*}{4})^{p-1})=[0,1.97)$,
so conditions (A4) holds.
We choose $\theta=1/4$, then it is easy see by
calculating that
$$
L=\min_{t\in[\theta,1-\theta]}A(t)=\frac{1}{16}
\big(\frac{7}{36}+\frac{\sqrt{3}}{3}\big).
$$
Because of
$$
\big(\frac{2\theta^*}{\theta}\big)^{p-1}=
96\times\big(\frac{1}{7+12\sqrt{3}}\big)^{1/2}<64+\frac{1}{3},
$$
then
$f_\infty=\lambda\in \big((\frac{2\theta^*}{\theta})^{p-1}, \infty\big)$,
 so conditions (A3) holds. Then by Theorem \ref{thm3.2}, SBVP \eqref{e5.1}
has at  least a positive solution.
\end{example}

\begin{example} \label{exa5.2}\rm
 Consider the following singular boundary-value problem (SBVP) with
$p$-Laplacian
\begin{equation}
\begin{gathered}
(\phi_{p}(u'))'+\frac{3}{256\pi^3}t^{-\frac{1}{2}}(1-t)
[u^{2}+u^{4}]=0,\quad 0<t<1 , \\
2\phi_{p}(u(0))-\phi_{p}(u'(\frac{1}{4}))=0, \
\phi_{p}(u(1))+\delta\phi_{p}(u'(\frac{1}{2}))=0,
\end{gathered}\label{e5.2}
\end{equation}
where
$\beta=\gamma=1$, $\alpha=2$,  $p=4$, $\delta\geq0$,
$\xi=\frac{1}{4}$, $\eta=\frac{1}{2}$, $T=1$,
$$
a(t)=\frac{3}{256\pi^3}t^{-1/2}(1-t),
$$
 and $f(u)=u^{2}+u^{4}$.
  Then obviously,
$$
q=\frac{4}{3},\quad
\int_0^Ta(t)\nabla t=\frac{1}{64\pi^3},\quad
f_\infty=+\infty,\quad  f_0=+\infty,
$$
so conditions  (H1), (H2), (A7), (A8) hold.
 Next,
$$
\phi_{q}\Big(\int_0^Ta(t)\nabla t\Big)=\frac{1}{4\pi},\quad
 \theta_*=\frac{4\pi}{1+\sqrt[3]{2}},
$$
 we choose $R=3$, $M=2$. Because $f(u)$ is  monotone increasing
on $[0,\infty)$, we have
$f(u)\leq f(3)=90$, for $0\leq u\leq 3$.
  Therefore, because $M\in (0,\theta_*)$, $(MR)^{p-1}=(6)^3=216$.
Also we know that
$$
f(u)\leq(MR)^{p-1},\quad 0\leq u\leq 3,
$$
so conditions (A2) holds.
 Then by Theorem \ref{thm4.1}, SBVP \eqref{e5.2} has at
 least two positive solutions $u_1$, $u_2$ and $0<\|u_1\|<3<\|u_2\|$.
\end{example}

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\end{document}