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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 42, pp. 1--11.\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-2006/42\hfil Multiple positive solutions]
{Multiple positive solutions for a class of nonresonant
 singular boundary-value problems}
\author[Y. Liu, B. Yan\hfil EJDE-2006/42\hfilneg]
{Yansheng Liu, Baoqiang Yan}

\address{Yansheng Liu \newline
Department of Mathematics, 
Shandong Normal University,
Jinan, 250014, China}
\email{ysliu6668@sohu.com}

\address{Baoqiang Yan \newline
Department of Mathematics, 
Shandong Normal University, 
Jinan, 250014, China} 
\email{bqyan819@beelink.com}


\date{}
\thanks{Submitted February 15, 2006. Published March 26, 2006.}
\thanks{Supported by grants 10571111 from the National Natural Science
 Foundation of China \hfill\break\indent
 and Y2005A07 from the Natural Science
 Foundation of Shandong Province}
\subjclass[2000]{34B15}
\keywords{Nonresonance; singular boundary-value problem; cone;
\hfill\break\indent positive solution}


\begin{abstract}
 Using a specially constructed cone
 and the fixed point index theory, this paper shows
 the existence of multiple positive solutions for a class of
 nonresonant singular boundary-value problem of second-order
 differential equations. The nonexistence of positive
 solution is also studied.
\end{abstract}


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

\section{Introduction }

 The theory of singular boundary-value problems (BVP, for short)
has become an important area of investigation in previous years; see
\cite{a1,h1,l1,l2,o1,t1,w1,w2,x1,z1}
 and references therein.
We consider the  nonresonant
singular boundary-value problem of second-order differential
equations
\begin{equation}
\begin{gathered}
-u''(t)+\rho p(t)u(t) = \lambda f(t, u(t)), \quad t\in (0,1); \\
u(0)=u(1) =0, \\
\end{gathered}  \label{e1.1}
\end{equation}
where $\rho>0$ and
\begin{gather*}
-u''(t)+\rho p(t)u(t) =0, \quad t\in (0,1); \\
 u(0)=u(1) =0
\end{gather*}
has only the trivial solution. Here the parameter $\lambda$ belongs
to $\mathbb{R}^+ = [0, +\infty)$, $p$ belongs to $C[(0, 1),
\mathbb{R}^+ ]$ with $\int_0^1  p(t)dt < +\infty$, and $f$ belongs
to  $C[(0,1)\times(0, +\infty), \mathbb{R}^+]$; that is, $f(t, u)$
may be singular at $t=0, 1$, and $u=0$.

In the special cases i) $p(t)=0$, $f(t,u)=p_1(t)u^{-\lambda_1}$,
$\lambda_1>0$, and ii) $p(t)=0$, $f(t,u)=p_1(t)u^{\lambda_1}$,
$0<\lambda_1<1$, where $p_1(t)>0$ for $t\in (0, 1)$, the existence
and uniqueness of positive solutions of \eqref{e1.1} as $\lambda=1$
have been studied completely by Taliaferro in \cite{t1} with the
shooting method and by Zhang \cite{z1} with the method of lower and
upper solutions, respectively. Also a sufficient condition for the
existence of $C[0, 1]$ solutions of the singular problem
\eqref{e1.1} with $\lambda=1$ was given by  O'Regan in \cite{o1} by
using a continuous theorem. In the special case iii): $f(t,u)$ is
quasi-homogeneous and sublinear in $u$, the existence of positive
solutions  \eqref{e1.1} as $\lambda=1$ have been studied by Wei and
Pang in \cite{w1,w2} with the method of lower and upper solutions.
In the special case iv): $p(t)=0$, $f(t,u)=p_1(t)g(u)$, $p_1(t)$ is
singular only at $t=0$ and $g(u)\geq e^u$, the existence of multiple
positive solutions of \eqref{e1.1} have been studied by Ha and Lee
in \cite{h1} with the method of lower and upper solutions.

In the present paper, we shall construct a special cone and use
fixed point index theory to investigate the existence of multiple
 positive solutions for \eqref{e1.1}, which is different from
 \cite{h1,o1,t1,w1,w2,z1}.
 Meanwhile, some results on nonexistence of positive solutions are
 given.

 The organization of this paper
is as follows. We shall introduce some lemmas and notations in the
rest of this section. The main result will be stated and proved in
Section 2. Finally  in Section 3 some examples are worked out to
demonstrate our main results.

 Now we present some lemmas and notation which will be
     used in Section 2. First from
\cite[Lemmas 2.3.1 and 2.3.3]{g1} we obtain the following lemma.

\begin{lemma} \label{lem1.1}
Let $P$ be a cone of real Banach space $E$, $\Omega$
     be a bounded open set of $E$, $\theta\in \Omega$,
$A: P\cap \overline{\Omega}\to P$ be
    completely continuous.
\begin{itemize}
\item[(i)] If $x\neq \mu Ax$ for $ x\in P\cap\partial\Omega$ and
 $\mu\in [0,1]$, then
     $i(A, P\cap\Omega, P)=1$.

\item[(ii)] If $\inf_{x\in P\cap\partial\Omega}\|Ax\| > 0$ and
$Ax\neq \mu  x$ for  $ x\in P\cap \partial\Omega$ and  $\mu\in(0,
1]$, then
     $i(A, P\cap\Omega, P)=0$.
\end{itemize}
\end{lemma}

Next noticing that $\int_0^1  p(t)dt < +\infty$, it is not difficult
from \cite{w1,w2} to obtain the following lemma.

\begin{lemma} \label{lem1.2}
 (i) The boundary-value problem
\begin{gather*}
-u''+ \rho p(t)u =0,\quad\text{for }  t\in (0, 1)\\
 u(0)=0,\quad    u'(0)=1
\end{gather*}
has an increasing positive solution
$e_1(t)= tw_1(t)\in C[0, 1]\cap C^1[0, 1)$,
where $w_1\in C[0, 1]$ is the unique solution of the
integral equation
\begin{equation}
 w_1(t)= 1 + \frac{\rho}{t}\int_0^t\int_0^s \tau p(\tau)w_1(\tau)d\tau ds. \label{e1.2}
\end{equation}
(ii) The boundary-value problem
\begin{gather*}
-u''+ \rho p(t)u =0,\quad\text{for }  t\in (0, 1)\\
 u(1)=0, \quad  u'(1)=-1
\end{gather*}
has a decreasing positive solution
$e_2(t)= (1-t)w_2(t)\in C[0, 1]\cap C^1(0, 1]$,
  where $w_2\in C[0, 1]$ is the unique solution of the integral equation
\begin{equation}
w_2(t)= 1 + \frac{\rho}{1-t}\int_t^1\int_s^1(1- \tau)
  p(\tau)w_2(\tau)d\tau ds. \label{e1.3}
\end{equation}
(iii) The Wronskian  $\omega=\det\begin{pmatrix}
e_1(t) & e_1'(t) \\ e_2(t) & e_2'(t)
\end{pmatrix}$ is a positive constant.
\end{lemma}

Let $J=[0, 1]$. The basic space used in this paper is
$E=C[J, \mathbb{R}]$.  It is well known that $C[J, \mathbb{R}]$ is a
Banach space
with norm $\|u\|=\max_{t\in J}|u(t)|$\ ($\forall u\in C[J, \mathbb{R}]$).
 From Lemma \ref{lem1.2}, it is easy to see
\begin{equation}
Q:= \{ u\in C[J,\mathbb{R}^+]: u(t)\geq q(t)u(s),\; \forall t, s\in J\}
\label{e1.4}
\end{equation}
is a cone of $C[J, \mathbb{R}]$, where
\begin{equation}
q(t):= \frac{e_1(t)e_2(t)}{e_1(1)e_2(0)}, \quad t\in J.
\label{e1.5}
\end{equation}
Moreover, by \eqref{e1.4}-\eqref{e1.5},  we have for all $u\in Q$,
\begin{equation}
u(t)\geq q(t)\|u\|,\quad \forall t\in J. \label{e1.6}
\end{equation}
A function $ u$ is said to be a solution of \eqref{e1.1} if
$\lambda\geq 0$ and $u$  satisfies \eqref{e1.1}. In addition, if
$\lambda >0$, $u(t)>0$ for $t\in (0, 1)$, then $u$ is said to be a
positive solution of \eqref{e1.1}. Obviously, if $u\in  Q\setminus\{\theta\}$ is a
solution of \eqref{e1.1}, then $u$ is a positive solution of
\eqref{e1.1}, where $\theta$ denotes the zero element of Banach
space $C[J, \mathbb{R}]$.

\section {Main Results}

For convenience, we list the following assumptions.
\begin{itemize}

\item[(H1)] $f\in C[(0, 1)\times (0, +\infty), \mathbb{R}^+]$ and for every pair
of positive numbers  $R$ and $r$ with $R>r>0$,
$$
\int_0^1  s(1-s)f_{r, R}(s)ds < +\infty,
$$
where
$f_{r, R}(s):= \max\{ f(s, u): u\in [rs(1-s), R]\}$,
for all $s\in (0,1)$.

\item[(H2)] For every $R>0$, there exists $\psi_R\in C[J, \mathbb{R}^+]$
($\psi\neq \theta$) such that
$f(t, u)\geq \psi_R(t)$ for
$t\in (0, 1)$ and $u\in (0, R]$.

\item[(H3)] There exists an interval $[a, b]\subset (0, 1)$ such that
$\lim_{x\to +\infty} f(s, u)/ u = +\infty$ uniformly with
respect to $s\in [a, b]$.
\end{itemize}


We remark that (H2) allows $f(t, u)$ being singular at $t=0, 1$,
and $u=0$.  Assumption (H3) shows
that $f$ is superlinear in $u$.
The following theorems are our main results of this paper.

\begin{theorem} \label{thm2.1}
Assume {\rm (H1)-(H3)} are satisfied.
Then there exist positive numbers $\lambda^*$ and
$\lambda^{**}$ with $\lambda^*<\lambda^{**}$ such that \eqref{e1.1}
has at least two positive solutions for $\lambda\in (0, \lambda^*)$
and no solution for
$\lambda > \lambda^{**}$.
\end{theorem}

 To overcome difficulties arising from singularity we first
 consider the  approximate problem
\begin{equation}
\begin{aligned}
-u''(t)+\rho p(t)u(t) = \lambda f_n(t, u(t)), &\quad t\in (0,1); \\
u(0)=u(1) =0,&
\end{aligned}
\label{e2.1}
\end{equation}
where $ f_n(t, u)=$: $f(t, \max\{\frac{1}{n},u\})$, $n\in \mathbb{N}$.
Define an operator $ A_n^\lambda $ on $Q$ by
\begin{equation}
( A_n^\lambda u)(t):= \lambda\int_0^1  G(t, s)f_n(s, u(s))ds,\label{e2.2}
\end{equation}
where
\begin{equation}
G(t, s)=\begin{cases}
\frac{1}{\omega}e_1(t)e_2(s), & 0\leq t\leq s\leq 1; \\
\frac{1}{\omega}e_1(s)e_2(t), & 0\leq s\leq t\leq 1,\\
\end{cases} \label{e2.3}
\end{equation}
where $e_1(t)$, $e_2(t)$ and $\omega$ are defined
as in Lemma \ref{lem1.2}.

Obviously, $u= A_n^\lambda u$ is the corresponding integral equation of
\eqref{e2.1}. Therefore,
$u\in C[J, \mathbb{R}^+] \bigcap C^2[(0, 1), \mathbb{R}^+]$
is a solution of \eqref{e2.1} if $u\in C[J, \mathbb{R}^+]$ is a fixe point
of $ A_n^\lambda$.
Furthermore, $u$ is a positive solution of \eqref{e2.1} if $u\in  Q\setminus\{\theta\}$ is
a  fixed point of $ A_n^\lambda$.

By \eqref{e2.1}-\eqref{e2.3}, it is easy to see that $ A_n^\lambda$ is well
defined on $Q$ for each $n\in \mathbb{N}$ if  condition (H1)  holds.
For the sake of  proving our main results we first prove some lemmas.

\begin{lemma} \label{lem2.1}
Under condition (H1), $ A_n^\lambda:Q\to Q$ is  completely continuous.
\end{lemma}

\begin{proof} First we show $ A_n^\lambda Q\subset Q$ for each $n\in \mathbb{N}$
and $\lambda> 0$.
 From \eqref{e2.3} and the monotonicity of $e_1(t)$ and $e_2(t)$ obtained
from Lemma \ref{lem1.2}, it follows that
$$
\frac{G(t, s)}{G(\tau, s)}=
\begin{cases}
\frac{e_1(t)e_2(s)}{e_1(\tau)e_2(s)}\geq\frac{e_1(t)}{e_1(1)}
\geq\frac{e_1(t)e_2(t)}{e_1(1)e_2(0)}, &
0< t, \tau\leq s< 1; \\[3pt]
\frac{e_1(s)e_2(t)}{e_1(s)e_2(t)}\geq\frac{e_2(t)}{e_2(0)}
\geq\frac{e_1(t)e_2(t)}{e_1(1)e_2(0)},
& 0< s\leq t, \tau< 1;\\[3pt]
\frac{e_1(t)e_2(s)}{e_1(s)e_2(\tau)}\geq\frac{e_1(t)}{e_1(s)}
\geq\frac{e_1(t)e_2(t)}{e_1(1)e_2(0)},
& 0< t\leq s\leq \tau< 1;\\[3pt]
\frac{e_1(s)e_2(t)}{e_1(\tau)e_2(s)}\geq\frac{e_2(t)}{e_2(s)}
\geq\frac{e_1(t)e_2(t)}{e_1(1)e_2(0)},
& 0< \tau\leq s\leq t< 1;
\end{cases}
$$
This equality and \eqref{e2.2} guarantee
that
\begin{align*}
 ( A_n^\lambda u)(t)& =  \lambda\int_0^1  G(t, s)f_n(s, u(s))ds\\
 &\geq \frac{e_1(t)e_2(t)}{e_1(1)e_2(0)}\cdot \lambda \int_0^1
 G(\tau, s)f_n(s,u(s))ds\\
&= \frac{e_1(t)e_2(t)}{e_1(1)e_2(0)}\cdot ( A_n^\lambda u)(\tau),\quad
\forall t, \tau\in J, u\in Q.
\end{align*}
Therefore, $ A_n^\lambda Q\subset Q$ for each $n\in \mathbb{N}$ and $\lambda> 0$.

Next by standard methods and Ascoli-Arzela theorem one can prove
$ A_n^\lambda:Q\to Q$ is  completely continuous. So it is
omitted.
\end{proof}

\begin{lemma} \label{lem2.2}
Suppose conditions (H1) and (H2) hold.  Then for each $r>0$,
there exists a positive number $\lambda(r)$ such that
$$
i( A_n^\lambda, Q_r, Q)=1
$$
for $\lambda\in (0, \lambda(r))$ and $ n$
sufficiently large, where $ Q_r=\{u\in Q:\|u\|<r\}$.
\end{lemma}

\begin{proof} First we show for each $u\in Q\setminus\{\theta\}$, there exists $l>0$
such that
\begin{equation}
u(t)\geq lt(1-t)\|u\|,\quad \forall t\in J. \label{e2.4}
\end{equation}
In fact, by \eqref{e1.2}-\eqref{e1.3}, one can obtain $w_1(0)=1$
and $w_2(1)=1$.
This together with the monotonicity of $e_1(t)$ and $e_2(t)$
guarantees that $\min_{t\in J}w_1(t)>0$ and
$\min_{t\in J}w_2(t)>0$. Therefore, choose
$$
 l:= \frac{1}{w_1(1)w_2(0)}\big(\min_{t\in
J}w_1(t)\big)\cdot\big(\min_{t\in J}w_2(t)\big)>0.
$$
Immediately from \eqref{e1.4}-\eqref{e1.6}, \eqref{e2.4} follows.
Next for each $r>0$ and $ n>\frac{1}{r}$, let
$$
\lambda(r):=  r[\int_0^1  G(s, s)f_{rl, r}(s)ds]^{-1}. \label{e2.5}
$$
We assert $\| A_n^\lambda u\|<\|u\|$ for each $\lambda\in (0,
\lambda(r))$ and $u\in \partial Q_r$. In fact, using \eqref{e2.4}
and $G(t, s)\leq G(s, s)$ for $t, s\in J$ one can obtain
\begin{align*}
\| A_n^\lambda u\| &\leq  \lambda\int_0^1  G(s, s)f_n(s, u(s))ds\\
&\leq  \lambda\int_0^1  G(s, s)f_{rl, r}(s)ds\\
&<  r=\|u\|,\quad \text{for $\lambda\in (0, \lambda(r))$ and $u\in \partial Q_r$}.
\end{align*}
Therefore, by Lemma \ref{lem1.1} we have $i( A_n^\lambda, Q_r, Q)=1$ for
$ \lambda\in (0, \lambda(r))$.
\end{proof}

\begin{lemma} \label{lem2.3}
Suppose conditions (H1) and (H2) hold. Then for any given
$\lambda\in (0, \lambda(r))$, there exists
$r'\in (0, r)$ such that
$$
i( A_n^\lambda, Q_{r'}, Q)=0
$$
for  $n$ sufficiently large, where $r$ and $\lambda(r)$ are the same
as in Lemma \ref{lem2.2}.
\end{lemma}

\begin{proof}
Choose a positive number $r'$ with
$ r'< \min\big\{r,  \lambda\max_{t\in J}\int_0^1  G(t, s)\psi_r(s)ds\big\}$,
where $\psi_r(s)$ is defined as in (H2).
 Now, we claim that
\begin{equation}
 A_n^\lambda u\neq \mu  u\quad \text{ for }  u\in \partial Q_{r'}
\text{ and }\mu\in(0, 1].
\label{e2.6}
\end{equation}
 for  $n>1/r'$. Suppose, on the contrary, there exists
$u_0\in \partial Q_{r'}$ and
$\mu_0\in(0, 1]$ such that $ A_n^\lambda u_0=\mu_0 u_0$, namely,
$$
u_0(t)\geq ( A_n^\lambda u_0)(t)=\lambda\int_0^1  G(t, s)f_n(s, u_0(s))ds,\quad
\forall t\in J.
$$
Notice that $|u_0(s)|\leq r'<r$ and  $ n>\frac{1}{r'}$
implies
$f_n(s, u_0(s))\geq \psi_r(s)$ for $s\in (0, 1)$. Therefore,
$$
u_0(t)\geq ( A_n^\lambda u_0)(t)=\lambda\int_0^1  G(t, s)\psi_r(s)ds,
$$
that is,
$$
 r'\geq \lambda\max_{t\in J}\int_0^1  G(t,s)\psi_r(s)ds,
$$
which is in contradiction with the
selection of $r'$. This means \eqref{e2.6}
holds. Thus, by Lemma \ref{lem1.1} we have
$i( A_n^\lambda, Q_{r'}, Q)=0$ for $ n>\frac{1}{r'}$.
\end{proof}

\begin{lemma} \label{lem2.4}
Suppose condition (H3) holds. Then for
every $\lambda\in (0, \lambda(r))$, there exists $R> r$ such that
$$
i( A_n^\lambda, Q_R, Q)=0
$$
for all $n\in \mathbb{N}$, where $\lambda(r)$ is the same as in
Lemma \ref{lem2.2}.
\end{lemma}

\begin{proof} By (H3) we know there exists
 $R'>\max\{r, 1\}$ such that
\begin{equation}
\frac{f(t, u)}{u}> L:=  [la(1-b)
\big(\lambda\min_{t\in [a,b]}\int_{a}^{b}G(t, s)ds\big)]^{-1}
\quad\text{for } u>R'. \label{e2.7}
\end{equation}
Let $ R:=  1+\frac{R'}{la (1-b)}$. Then for $u\in\partial Q_R$, by
\eqref{e2.4} we have  $u(t)\geq la(1-b)\|u\|>R' $ as $t\in [a, b]$. Now we
show that
\begin{equation} \label{e2.8}
A_n^\lambda u\neq \mu u \quad\text{for }u\in \partial Q_R,\quad
\text{and}\quad \mu\in(0, 1].
\end{equation}
Suppose, on the contrary, there exists $u_0\in\partial Q_R$ and
$\mu_0\in(0, 1]$ such that $ A_n^\lambda u_0=\mu_0 u_0$, that is,
$$
u_0(t)\geq ( A_n^\lambda u_0)(t)=\lambda\int_0^1  G(t,
s)f_n(s, u_0(s))ds,\quad \forall t\in J.
$$
Furthermore,
\begin{align*}
u_0(t)&\geq ( A_n^\lambda u_0)(t)> \lambda\Big(\int_{a}^{b}
G(t, s)\cdot Lu_0(s)ds\Big)\\
&> \Big(\lambda\min_{t\in [a, b]}\int_{a}^{b}G(t, s)ds\Big)Lla(1-b)R=R
\end{align*}
for $t\in [a, b]$.
This is in contradiction with $\|u_0\|=R$. This means that \eqref{e2.8}
holds. Therefore, by Lemma \ref{lem1.1} we have
$i( A_n^\lambda, Q_R, Q)=0$ for $n\in \mathbb{N}$.
\end{proof}

Now we are in position to prove Theorem \ref{thm2.1}.

\begin{proof}[Proof of Theorem \ref{thm2.1}]
For each $r>0$, by Lemma \ref{lem2.2}-\ref{lem2.4}, there exist
three positive numbers $\lambda(r), r'$, and $R$ with $r'<r<R$ such
that
\begin{equation}
i( A_n^\lambda, Q_{r'}, Q)=0,\quad i( A_n^\lambda, Q_r, Q)=1,\quad i( A_n^\lambda, Q_R, Q)=0
\label{e2.9}
\end{equation}
 for $n$ sufficiently large.
Without loss of generality, suppose \eqref{e2.9} holds for $n\geq n_0$. By
virtue of the excision property of the fixed point index, we get
$$
i( A_n^\lambda, Q_r\setminus\overline{Q_{r'}}, Q)=1,\quad i( A_n^\lambda, Q_R\setminus\overline{Q_{r}},
Q)=-1
$$
 for $n\geq n_0$. Therefore, using the solution property of
 the fixed point index, there exist $u_n\in
Q_r\setminus\overline{Q_{r'}}$ and $v_n\in Q_R\setminus\overline{Q_{r}}$
satisfying $ A_n^\lambda u_n=u_n$ and $ A_n^\lambda v_n=v_n$ as $n\geq n_0$. By the
proof of Lemma \ref{lem2.2} we know that there is no positive fixed point
on $\partial Q_r$. Thus, $u_n\neq v_n$. Moreover, from \eqref{e2.4} it follows
that
\begin{equation}
lr't(1-t)\leq u_n(t)< r\quad \text{and}\quad lrt(1-t)< v_n(t)\leq R,\quad
{\rm for}\ t\in J.\label{e2.10}
\end{equation}

 In the following we show $\{u_n(t)\}_{n\geq n_0}$
are equicontinuous on $J$. To see this we need to prove only that
$\lim_{t\to 0+}u_n(t)=0$ and $\lim_{t\to 1-}u_n(t)=0$ both uniformly
with respect to $n\in \{n_0, n_0+1, n_0+2,\dots \}$ and
$\{u_n(t)\}_{n\geq n_0}$ are equicontinuous on any subinterval of
$(0, 1)$. We first claim that $\lim_{t\to 0+}u_n(t)=0$ uniformly
with respect to $n\in \{n_0, n_0+1, n_0+2,\dots \}$. According
to \eqref{e2.3} and Lemma \ref{lem1.2}, it is easy to see
\begin{equation}
G(t, s)\leq \begin{cases}
\bar l s(1-t),& 0\leq s\leq t\leq 1;\\
\bar l t(1-s),& 0\leq t\leq s\leq 1,
\end{cases} \label{e2.11}
\end{equation}
where $\bar l=\max_{t\in  J}w_1(t)\cdot \max_{t\in J}w_2(t)$, $w_1(t)$
and $w_2(t)$  are stated in Lemma \ref{lem1.2}. On the other hand,
for arbitrary $\varepsilon>0$, by (H1) there exists $\bar\delta>0$ such that
\begin{equation}
\lambda\bar l\int_0^{\bar\delta}s(1-s)f_{r'l, r}(s)ds
 \leq\frac{\varepsilon}{3}.\label{e2.12}
\end{equation}
Choose $\delta\in (0, \bar\delta)$ sufficiently small such that
\begin{equation}
\lambda\bar l\delta{\bar\delta}^{-1}\int_0^1  s(1-s)f_{r'l, r}(s)ds<\frac{\varepsilon}{3}.
\label{e2.13}
\end{equation}
 Therefore, by \eqref{e2.11}-\eqref{e2.13}, we know for $t\in (0, \delta)$
and $\forall n\geq n_0$ that
 \begin{align*}
 u_n(t) &= \lambda \int_0^1  G(t, s)f_n(s, u_n(s))ds\\
 &\leq   \lambda\bar l\int_0^ts(1-t)f_{r'l, r}(s)ds
+ \lambda\bar l\int_t^1t(1-s)f_{r'l, r}(s)ds\\
&\leq
 \lambda\bar l\int_0^ts(1-s)f_{r'l, r}(s)ds
 + \lambda\bar l\Big(\int_t^{\bar\delta}+
 \int_{\bar\delta}^1\Big)t(1-s)f_{r'l, r}(s)ds\\
&\leq 2\lambda\bar l\int_0^{\bar\delta}s(1-s)f_{r'l, r}(s)ds
 +\frac{\lambda\bar l t}{\bar\delta}\int_{\bar\delta}^1 s(1-s)f_{r'l, r}(s)ds\\
&\leq
 2\lambda\bar l\int_0^{\bar\delta}s(1-s)f_{r'l, r}(s)ds
 +\frac{\lambda\bar l t}{\bar\delta}\int_0^1 s(1-s)f_{r'l, r}(s)ds\\
&\leq
 \frac{2\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon.
\end{align*}
This implies $\lim_{t\to 0+}u_n(t)=0$ uniformly with respect to
$n\in \{n_0, n_0+1, n_0+2,\dots \}$.
Similarly, one can show that $\lim_{t\to 1-}u_n(t)=0$ uniformly
with respect to $n\in \{n_0, n_0+1, n_0+2,\dots \}$ also.

Now we are in position to show $\{u_n(t)\}_{n\geq n_0}$ are
equicontinuous on any subinterval $[a, b]$ of $(0, 1)$. Notice
that
$$
u_n(t)=\frac{\lambda}{\omega}\Big(\int_0^te_1(s)e_2(t)f_n(s, u_n(s))ds
+ \int_t^1e_1(t)e_2(s)f_n(s, u_n(s))ds\Big),
$$
for all $t\in (0, 1)$.
Thus, by Lemma \ref{lem1.2}, for $t\in [a, b]$, we have
\begin{align*}
 &|u_n'(t)|\\
&=\frac{\lambda}{\omega}\Big|\int_0^te_1(s)e'_2(t)f_n(s, u_n(s))ds +
\int_t^1e'_1(t)e_2(s)f_n(s, u_n(s))ds\Big|
\\
&\leq \frac{\lambda}{\omega}\Big(\int_0^te_1(s)|e'_2(t)|f_{r'l, r}(s)ds +
\int_t^1|e'_1(t)|e_2(s)f_{r'l, r}(s)ds \Big)
\\
&\leq \frac{\lambda}{\omega}\max_{t\in [a, b]}\{|e'_1(t)|,
|e'_2(t)|\}\Big(\int_0^t\frac{e_1(s)e_2(t)}{e_2(t)}f_{r'l, r}(s)ds
 + \int_t^1\frac{e_1(t)e_2(s)}{e_1(t)}f_{r'l, r}(s)ds\Big)\\
&\leq \lambda \bar l\max_{t\in [a, b]}\{|e'_1(t)|,
|e'_2(t)|\}\cdot\max\Big\{\frac{1}{e_1(a)}, \frac{1}{e_2(b)}\Big\}\int_0^1
s(1-s)f_{r'l, r}(s)ds
< +\infty,
\end{align*}
which implies $\{u_n(t)\}_{n\geq n_0}$ are equicontinuous on  $[a, b]$.
Similarly as above, we can get $\{v_n(t)\}_{n\geq n_0}$ are
equicontinuous on  $[0, 1]$.

Then, the Ascoli-Arzela theorem guarantees 
the existence of  $u, v\in C[J, \mathbb{R}^+]$ and  two
subsequences $\{u_{n_i}\}$ of
$\{u_n\}$ and $\{v_{n_i}\}$ of $\{v_n\}$ such that
$\lim_{i\to +\infty}u_{n_i}(t)=u(t)$ and
$\lim_{i\to +\infty}v_{n_i}(t)=v(t)$ both uniformly with
respect to $t\in J$.
Moreover, by (H1), \eqref{e2.10},  and Lebesgue dominated convergence
theorem, we obtain
$$
u(t)=\lambda\int_0^1  G(t, s)f(s, u(s))ds,\quad
v(t)=\lambda\int_0^1  G(t, s)f(s, v(s))ds,\quad \forall t\in J
$$
with $r'\leq \|u\|\leq r\leq \|v\|\leq R$. On the other hand,
similar to the proof of Lemma \ref{lem2.2}, it is easy to see
 $\|u\|< r< \|v\|$.

Choose $r=1$. From above we know there exists $\lambda(1)>0$ such that
for each $\lambda\in (0, \lambda(1))$, \eqref{e1.1} has at least two positive
solutions $u_{\lambda}$ and $v_{\lambda}$ with
$0<\|u_{\lambda}\|<1<\|v_{\lambda}\|$. Let
$$
\lambda^*:= \sup\{\bar \lambda>0: \eqref{e1.1} \mbox{ has at least
two positive solutions as } \lambda\in (0,\bar\lambda)\}.
$$
So we get the existence of $\lambda^*$ satisfying that
\eqref{e1.1} has multiple positive solutions as
$\lambda\in (0, \lambda^*)$.

Now we are in position to prove the existence of $\lambda^{**}$.
As above, still choose $r=1$ and corresponding $\lambda(1)$, $R$,
$r'$. Here we show \eqref{e1.1} has no positive solution as $\lambda$
sufficiently large.

First suppose $\lambda\geq \lambda^*$. If \eqref{e1.1} has a positive
solution $u$ for some $\lambda\geq \lambda^*$, then by corresponding
integral equation
\begin{equation}
u(t)= \lambda\int_0^1  G(t, s)f(s, u(s))ds  \label{e2.14}
\end{equation}
and a process similar to one in the proof of Lemmas \ref{lem2.3}
and \ref{lem2.4} (replacing $\lambda$ in \eqref{e2.7} with $\lambda(1)$),
we obtain $r'<\|u\|< R$. This together with condition
(H2) and \eqref{e2.14} guarantees that
$ u(t)\geq \lambda\int_0^1  G(t, s)\psi_R(s)ds$, that is,
$ R>\|u\|\geq\lambda\cdot\max_{t\in J}\int_0^1  G(t, s)\psi_R(s)ds$,
which implies
$\lambda<\Big(\max_{t\in J}\int_0^1  G(t, s)\psi_R(s)ds\Big)^{-1}R$.
Therefore, we obtain the existence of
$\lambda^{**}$.
The proof of Theorem \ref{thm2.1} is complete.
\end{proof}

 If $f(t, u)$ is not singular at $u=0$, we have the
following result, under the hypothesis
\begin{itemize}
\item[(H4)] $f\in C[(0, 1)\times [0, +\infty), \mathbb{R}^+]$ and for every
positive number  $R$,
$$
\int_0^1  s(1-s)f_{0, R}(s)ds < +\infty,
$$
 where $f_{0, R}(s)=\max\{ f(s, u): u\in [0, R]\}$, for all
$s\in (0,1)$.
\end{itemize}

\begin{theorem} \label{thm2.2}
Assume that conditions (H2)-(H4) hold. Then there exist two positive
numbers $\lambda^*$ and $\lambda^{***}$ with
$\lambda^*\leq\lambda^{***}$ such that
\begin{itemize}
\item[(i)] \eqref{e1.1} has at least two positive solutions for
$\lambda\in (0, \lambda^*)$;

\item[(ii)] \eqref{e1.1} has at least one positive solution for
$\lambda\in (0, \lambda^{***}]$;

\item[(iii)] \eqref{e1.1} has no solutions for $\lambda > \lambda^{***}$
\end{itemize}
\end{theorem}

\begin{proof}  Notice that condition (H4) implies (H1).
Therefore, the existence of $\lambda^*$ can be obtained as in
Theorem \ref{thm2.1}. Now we claim that the condition
\begin{equation}
\lambda^{***}:=  \sup\{\lambda\in \mathbb{R}^+:\eqref{e1.1}\ \mbox{has at least one
positive solution in }C[0, 1]\bigcap C^2(0, 1)\} \label{e2.15}
\end{equation}
is required.
First from the proof of Theorem \ref{thm2.1}, we
know $\lambda^{***}\leq \lambda^{**}$. In the following we prove that
\eqref{e1.1} with $\lambda=\lambda^{***}$
 has a positive solution $u^*\in C[0, 1]\bigcap
C^2(0, 1)\bigcap Q$.

By \eqref{e2.15}, there exist two sequences $\{\lambda_n\}$ and
$\{u_n\}\subset  Q\setminus\{\theta\}$ such that
$\{u_n\}$ is a positive solution of \eqref{e1.1} with
$\lambda=\lambda_n$ and  $\lambda_1<\lambda_2<\dots <\lambda_n\to
\lambda^{***}$. Without loss of generality, suppose
$\lambda_n\geq \lambda^*/2$ for each $n\in \mathbb{N}$. Similar to the proof of
Lemmas \ref{lem2.3}, \ref{lem2.4} and Theorem \ref{thm2.1}, we can obtain
that there exists two
positive numbers $r_1$ and $R_1$ satisfying
$r_1\leq\|u_n\|\leq R_1$
for each $n\in \mathbb{N}$ and $\{u_n\}$ has a subsequence $\{u_{n_k}\}$
which convergence to a function
$u^*\in \overline Q_{R_1}\setminus Q_{r_1}$ uniformly as
$t\in J$. Notice that
$$
u_{n_k}(t)=\lambda_{n_k}\int_0^1  G(t, s)f(s, u_{n_k}(s))ds,\quad\forall t\in J.
$$
Letting $k\to +\infty$, by condition (H4) and Lebesgue
dominated convergence theorem, we get
$$
u^*(t)=\lambda^{***}\int_0^1  G(t, s)f(s, u^*(s))ds,\quad \forall t\in J.
$$
This implies that $u^*(t)$ is a positive solution of
\eqref{e1.1} with $\lambda=\lambda^{***}$.

Now we are in position to prove that BVP \eqref{e1.1} has at least
one positive solution $u_\lambda(t)$ for each $\lambda\in (0,
\lambda^{***})$. Notice that for $\lambda\in (0, \lambda^{***})$,
\begin{gather*}
-{u^*}''(t)+\rho p(t)u^*(t) = \lambda^{***} f(t, u^*(t))\geq
\lambda f(t, u^*(t)), \quad t\in (0,1); \\
u^*(0)=u^*(1) =0.
\end{gather*}
This implies $u^*(t)$ is an upper solution
of \eqref{e1.1}. On the other hand, $u(t)\equiv 0$ is a lower solution
for BVP \eqref{e1.1}. Applying  \cite[Theorem 3.2]{h2} and the method used in
\cite{z1,w1}, one can obtain that BVP \eqref{e1.1} has at least one positive
solution $u_\lambda(t)\in [0, u^*(t)]\ (t\in J)$ for each
$\lambda\in (0, \lambda^{***})$.
\end{proof}

\noindent\textbf{Remark.} It is interesting to investigate  what
conditions should guarantee  $\lambda^*= \lambda^{***}$.  Ha and Lee
\cite{h1} obtained $\lambda^*= \lambda^{***}$  when  $p(t)=0$,
$f(t,u)=p_1(t)g(u)$, $p_1(t)>0$ is singular only at $t=0$ and $g(u)$
is nondecreasing with $g(u)\geq e^u$.  Xu \cite{x1}, also obtained
$\lambda^*= \lambda^{***}$ when $p(t)=0$, $f(t,u)=p_1(t)g(u)$,
$p_1(t)>0$ is singular  at $t=0, 1$ and $g(u)$ is nondecreasing with
$g(u)\geq \delta x^m$\ $(m\geq 2)$. To the best of our knowledge, it
seems an open problem when $f(t, u)$ is singular at $u=0$.

\section {Examples}

\begin{example} \label{ex3.1} \rm
Consider the  nonresonant singular boundary-value problem
\begin{equation}
\begin{gathered}
-u''(t)+\frac{3}{\sqrt{t}}u(t)
= \lambda \Big[\frac{1}{\sqrt{t(1-t)}}\Big( u^{-4/3}
+ x^{\alpha}\sin^2 t\Big)\Big], \quad t\in (0,1); \\
u(0)=u(1) =0,
\end{gathered} \label{e3.1}
\end{equation}
 where $\alpha>1$. Then there exist positive
numbers $\lambda^*$ and $\lambda^{**}$ with $\lambda^*<\lambda^{**}$ such that
BVP \eqref{e3.1} has at least two positive solutions for
$\lambda\in (0, \lambda^*)$ and no solution for $\lambda > \lambda^{**}$.
\end{example}

\begin{proof} BVP \eqref{e3.1} can be regarded as an BVP of the form
\eqref{e1.1}, where $\rho=3$, $p(t)= \frac{1}{\sqrt{t}}$, and
$$
f(t, u)= \frac{1}{\sqrt{t(1-t)}}\Big( u^{-4/3} +
x^{\alpha}\sin^2 t\Big).
$$
First we notice that $\int_0^1  \frac{1}{\sqrt{t}}dt< +\infty$, and
$ \int_0^1  t(1-t) \frac{1}{\sqrt{t}}dt =\frac{4}{15}$. By Banach
fixed point theorem, it is easy to see that
\begin{gather*}
 -u''(t)+\frac{3}{\sqrt{t}}u(t)
= 0, \quad t\in (0,1); \\
u(0)=u(1) =0,
\end{gather*}
has only the trivial solution.
Next we prove that $f(t, u)$ satisfies conditions (H1)-(H3).
For each pair of positive numbers $R$ and $r$ with $R>r>0$, we
know
$$
f_{r, R}(t)\leq \frac{1}{\sqrt{t(1-t)}}\Big(
\big(rt(1-t)\big)^{-4/3} + R^{\alpha}\Big).
$$
Then
$$\int_0^1  t(1-t)f_{r, R}(t)dt\leq \int_0^1  \sqrt{t(1-t)}\Big(
\big(rt(1-t)\big)^{-4/3} + R^{\alpha}\Big)dt <+\infty.
$$
This means condition (H1) is satisfied.
To see that (H2) holds, we notice that for each $R>0$, one can choose
$\psi_R(t)=R^{-4/3}/\sqrt{t(1-t)}$, which satisfies
$\psi_R\neq \theta$ and $f(t, u)\geq \psi_R(t)$ for $t\in (0, 1)$ and
$u\in (0, R]$.
Finally it is easy to see (H3) is satisfied since we can choose
any subinterval of $[a, b]\subset (0, 1)$ satisfying
$\lim_{x\to +\infty}f(s, u)/u = +\infty$ uniformly with
respect to $s\in [a, b]$.
By Theorem \ref{thm2.1}, the conclusion follows.
\end{proof}

Analogously, using Theorem \ref{thm2.2}, we can prove that the following
statement holds.

\begin{example} \label{ex3.2} \rm
Consider the  nonresonant singular boundary-value problem
\begin{equation}
\begin{gathered}
-u''(t)+\frac{1}{2}t^{-3/2}u(t)
= \lambda \big(t(1-t)\big)^{-3/2} \big( 2+ \sin u
+ x^{\alpha}\cos t\big), \quad t\in (0,1); \\
u(0)=u(1) =0,
\end{gathered} \label{e3.2}
\end{equation}
where $\alpha>1$. Then there exist two positive
numbers $\lambda^*$ and $\lambda^{***}$ with $\lambda^*\leq\lambda^{***}$ such that:
\begin{itemize}
\item[(i)] BVP \eqref{e3.2} has at least two positive solutions for
$\lambda\in (0, \lambda^*)$;

\item[(ii)]  BVP \eqref{e3.2} has at least one positive solution for
$\lambda\in (0, \lambda^{***}]$;

\item[(iii)] BVP \eqref{e3.2} has no solution for $\lambda > \lambda^{***}$
\end{itemize}
\end{example}


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\end{document}