\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 120, 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 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/120\hfil Multiple positive solutions]
{Multiple positive solutions to fourth-order singular boundary-value
problems in abstract spaces}

\author[Yansheng Liu\hfil EJDE-2004/120\hfilneg]
{Yansheng Liu}

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

\date{}
\thanks{Submitted September 22, 2004. Published October 14, 2004.}
\thanks{Supported by the Rewarded Foundation for Outstanding Middle and
 Young Scientist, \hfill\break\indent Shandong Province, China}
\subjclass[2000]{34G20, 34B15} 
\keywords{Banach space; singularity; positive solution; fourth order equation}

\begin{abstract}
 We prove the existence of multiple positive solutions
 to singular boundary-value problems for fourth-order equations in
 abstract spaces. Our results improve and extend that obtained in
 \cite{o1,r1,w1}, even in the scalar case.
\end{abstract}

\maketitle

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

\section {Introduction}

In this paper, we consider the following singular
boundary-value problem (BVP) for fourth-order differential
equations in a Banach  space $E$:
\begin{equation}
 x^{(4)}(t)=f(t, x(t)), \quad 0<t <1 \label{e1.1}
\end{equation}
subject to the boundary conditions
\begin{equation}
 x(0)=x(1)=x''(0)=x''(1)=0,\label{e1.2}
\end{equation}
where $f\in C[(0, 1)\times P\setminus\{\theta\}, P]$ which  may
 be singular at $t=0$, $t=1$, and $x=\theta$;
$P$ is a cone of Banach space $E$, which will be stated
in detail in section 2; $\theta$ is the zero element  of $E$.

Boundary-value problems arise from applied mathematical sciences,
and they have received a great deal of attention in the
literature. Problems of the form \eqref{e1.1} subject to
\eqref{e1.2}, for example, are used to model such phenomena as the
deflection of an elastic beam supported at the endpoints.  Most of
the available literature on fourth order boundary value problems,
for instance \cite{a1,a2,e1,g1,g2,j1,m1}, discuss the case where
$f$ is either continuous or a Caratheodory function. Recently some
papers such as \cite{o1,r1,w1}, by using approximating method or
upper-lower solution approach, investigate \eqref{e1.1} with
suitable boundary conditions in the case where $f$ may be singular
at $t=0$, $t=1$, or $x=0$. However, all of the above-mentioned
references consider \eqref{e1.1} only in scalar space; and
especially in the singular case, \cite{o1,r1,w1} concentrate on
the solvability of \eqref{e1.1} subject to some suitable boundary
conditions, not the existence of multiple solutions for such
problems. On the other hand, the theory of ordinary differential
equations (ODE) in abstract spaces is becoming an important branch
of mathematics in last thirty years because of its application in
partial differential equations and ODE's in appropriately infinite
dimensional spaces (see, for example \cite{d1,l1,m2}).  As a
result the goal of this paper is to fill up the gap in this area,
that is, to investigate the existence of multiple positive
solutions of \eqref{e1.1} with \eqref{e1.2} in a Banach space $E$.

 The technique used in this paper are the well-known Krein-Rutman
 theorem, a specially constructed cone, and the fixed point theorem of cone
 expansion and compression. It is remarkable that neither approximating method nor
 upper-lower solution approach is applied in present paper.

 This paper is organized as follows. Section 2 gives some
 preliminaries and some lemmas. Section 3 is  devoted to the main
 result and its proof. Some examples  are worked out in Section 4
 to indicate the application of  our main result.

\section{Preliminaries}

 Let $E$ be a real Banach space with norm $\|\cdot\|$, and $P\subset E$
 be a cone which defines a partial order relation  $\leq$  by $x\leq y$ if
 and only if $y-x\in P$, $x< y$ if
 and only if $x\leq y$ and $x\neq y$, where $x, y\in E$. Also suppose $P$ is a normal
 cone, that is, there exists $N>0$ such that $\|u\|\leq N\|v\|$ if
 $\theta\leq u\leq v$, where $\theta$ denotes the zero element of $E$.

Evidently, $C[I, E]$ is a Banach space with norm $\|x\|_c:=
\max_{t\in I}\|x(t)\|$. Moreover,
$$
C[I, P]:= \{x\in C[I, E]:\ x(t)\in P,\ t\in I\}
$$
is a normal cone of $C[I, E]$ with the
same normal constant $N$ as $P$ in $E$.

 A function $x$ is said to be a solution of \eqref{e1.1} subject to
  \eqref{e1.2} if $x\in C^2[I, E]\cap  C^4[(0, 1), E]$ satisfies \eqref{e1.1}
  and boundary conditions \eqref{e1.2}; in addition,  $x$ is said to
 be a positive solution  if $x(t)>\theta$ for $t\in (0, 1)$
 and $x$ is
 a solution of \eqref{e1.1} with \eqref{e1.2}.

Let $u: (0, 1]\to E$ be continuous. The abstract generalized
integral $\int_0^1 u(t)dt$ is said to be convergent if the limit
$\lim_{\epsilon\to 0^+}\int_{\epsilon}^1u(t)dt$ exists. The
convergency or divergency of other kinds of generalized integrals
can be defined similarly.

For a bounded subset $V$ of Banach space $E$, by $\alpha(V)$ we
denote the Kuratowskii measure of noncompactness of $V$(for
details, see \cite{d1,l1}). In this paper,
 the Kuratowskii measure of noncompactness of bounded set in $E$ and $C[I, E]$
  are denoted by $\alpha(\cdot)$ and $\alpha_c(\cdot)$, respectively.


To conclude this section, we list  three lemmas which will be used
in Section 3.

\begin{lemma}[\cite{g4}] \label{lem2.1}
If $V\subset C[J,E]$ is bounded and equicontinuous,
 then $\alpha(V(t))$ is continuous on $J$ and $\alpha_c(V)=\max\{\alpha(V(t))|
 t\in J\}$, where $V(t)=\{x(t)|x\in V\}$.
\end{lemma}

\begin{lemma}[\cite{n1}] \label{lem2.2}
Let $K$ be a cone of a real Banach space
$E$ and $B$: $K\to K$ a completely continuous operator. Assume
that $B$ is order-preserving and positively homogeneous of degree
$1$ and that there exist $v\in K\setminus \{\theta\}$, $\lambda>0$
such that $Bv\geq \lambda v$. Then $r(B)\geq\lambda$, where $r(B)$ denotes the
spectral radius of $B$.
\end{lemma}

\begin{lemma}[Fixed point theorem of cone
 expansion and compression \cite{g4}] \label{lem2.3}
 Let $P$ be a cone of a real Banach space $E$ and
$ P_{r,s}=\{x\in P: r\leq\|x\|\leq s\}$ with $s>r>0$. Suppose that
$A$: $P_{r, s}\to P$ is a strict contraction such that one of the
following two conditions is satisfied:
\begin{itemize}
\item[(i)] $Ax\not\leq x$ for $x\in P$, $\|x\|=r$ and
 $Ax\not\geq x$ for $x\in P$,  $\|x\|=s$.

\item[(ii)]  $Ax\not\geq x$ for $x\in P$, $\|x\|=r$ and $Ax\not\leq x$
for $x\in P$,   $\|x\|=s$.
\end{itemize}
     Then, the operator $A$ has a fixed point $x\in P$ such that $r<\|x\|<s$.
\end{lemma}

\section{Main Result}
 For convenience, we list the following assumptions:
\begin{itemize}
\item[(H0)]  $f\in C[(0, 1)\times P\setminus\{\theta\} , P]$ and satisfies
$$
0 <\int_0^1t(1-t)f_{r, R}(t)dt < +\infty,\quad \forall R \geq r>0,
$$ where for $t\in (0, 1)$, $z(t):= \min\{t, 1-t\}$ and
$$
f_{r, R}(t):= \sup\{\|f(t, x)\|:\ \frac{z(t)}{N}r\leq \|x\|\leq
R,\ x\in P\}.
$$

\item[(H1)] For every $[c, d]\subset (0, 1)$, and positive numbers
$R_2> R_1> 0$, $f(t, x)$ is uniformly continuous on $[c, d]\times\overline{P_{R_2}}\setminus P_{R_1}$ with
respect to $t$. Here $P_r=$: $\{x\in P:\|x\|< r\}$
for each $r>0$.

\item[(H2)]
For every $ t\in (0, 1)$ and  every bounded subset
$D\subset \overline{P_{R_2}}\setminus P_{R_1}$ $(R_2>R_1>0)$, we have
$\alpha(f(t, D))\leq l\alpha(D)$, where $l$ is a constant
with $l<15$.

\item[(H3)] There exist $\varphi\in L[I, R^+]$,
$[c, d]\subset [0, 1]$, and  $\varphi^*\in P^*$ (here $P^*$ denotes the dual
cone of $P$) with $\|\varphi^*\|=1$  such that
$$
\liminf_{x\to\theta,\; x\in P}\varphi^*(f(t, x)) \geq
\varphi (t)
$$
uniformly with respect to $t\in [c, d]$ and
$\int_c^ds(1-s)\varphi(s)ds > 0$.

\item[(H4)] There exist functions $a \in C[I, R^+]$ and
$b\in C[I, P]$ with $a(t)\not\equiv 0$  on every subinterval of $I$ such that
$$
f(t, x)\geq a(t)x- b(t)\quad {for }\ t\in (0, 1)\  { and }\ x\in
P\setminus\{\theta\}.
$$

\item[(H5)] There exists a positive number $R$ such that
$$
N\int_0^1s(1-s)f_{R,R}(s)ds < 8R,
$$
where $f_{R,R}(s)$ is the same as in (H0).\end{itemize}

Note that assumption (H3) is reasonable since $f(t, x)$ is
singular at $x=\theta$.

We assume that (H0) holds throughout the remainder of the
paper. To overcome the difficulties arising from singularities, we
define
\begin{equation}
Q:=\{x\in C[I, P]:\ x(t)\geq z(t)x(s)\geq \theta,\ \forall t, s\in
I\}.\label{e3.1}
\end{equation}
It is easy to see that $Q$ is a
nonempty (notice $t(1-t)\in Q$), convex, and closed subset of $C[I,E]$.
Furthermore, $Q$ is a cone of the Banach space $C[I, E]$ and
for every $x\in Q\setminus\{\theta\}$, by \eqref{e3.1} and the normality of the cone $P$,
we have
\begin{equation}
\|x(t)\|\geq \frac{z(t)}{N}\|x\|_c > 0\quad \quad\mbox{for }t\in (0, 1).\label{e3.2}
\end{equation}
Therefore, $x$ is a positive solution of \eqref{e1.1}-\eqref{e1.2} provided
that $x\in Q\setminus\{\theta\}$ is a solution of \eqref{e1.1}-\eqref{e1.2}.

Define the operator $A$ on $Q\setminus\{\theta\}$ by
\begin{equation}
(Ax)(t):= \int_0^1 J(t, \tau)f(\tau, x(\tau))d\tau, \quad\forall x\in Q\setminus\{\theta\},\label{e3.3}
\end{equation}
where $J(t, \tau)=\int_0^1 G(t,s)G(s, \tau)ds\quad$ and
\begin{equation}
G(t, s)=\begin{cases}
t(1-s),& 0\leq t\leq s\leq 1;\\
s(1-t),& 0\leq s\leq t\leq 1.
\end{cases}
\label{e3.4}
\end{equation}
Now we show the operator $A$ is well defined on $Q\setminus\{\theta\}$. First we
claim that for each $x\in Q\setminus\{\theta\}$, $\int_0^1 G(s, \tau)f(\tau, x(\tau))d\tau $  is
convergent. In fact, since $x\in Q\setminus\{\theta\}$, we can see by \eqref{e3.2}
 that $\|x\|_c\neq 0$ and
$$
\frac{z(t)}{N}\|x\|_c\leq \|x(\tau)\| \leq \|x\|_c\quad \mbox{for each }  \tau\in (0, 1).
$$
This together with $G(s, \tau)\leq \tau(1-\tau)$ for all $ s, \tau\in I$ and (H0)
implies that $\int_0^1 G(s, \tau)f(\tau, x(\tau))d\tau $ is convergent and
$$
\int_0^1 \tau(1-\tau)\|f(\tau, x(\tau))\|d\tau < +\infty\quad \mbox{for each } x\in Q\setminus\{\theta\}.
$$
The Lebesgue dominated convergence theorem yields that
for every $x\in Q\setminus\{\theta\}$, $\int_0^1 G(s, \tau)f(\tau, x(\tau))d\tau $ is
continuous in $s$ on $I$. Therefore, by \eqref{e3.3} we obtain that
$Ax\in C^2[I, P]$ and
\begin{equation}
\begin{gathered}
(Ax)^{(4)}(t)=f(t, x(t)), \quad 0<t <1;\\
(Ax)(0)=(Ax)(1)=(Ax)''(0)=(Ax)''(1)=0.
\end{gathered}\label{e3.5}
\end{equation}
Now we are in position to state the following lemma.

\begin{lemma} \label{lem3.1}
The boundary-value problem \eqref{e1.1}-\eqref{e1.2} has a positive solution
in $C^2[I, P]\cap  C^4[(0, 1), P]$ if
and only if $A$ has a fixed point $x$ in $Q\setminus\{\theta\}$.
\end{lemma}

\begin{proof} From the above, sufficiency is evident. In what
follows, we  prove only necessity.
Suppose $x$ is a positive solution of \eqref{e1.1}-\eqref{e1.2}, that is,
$x(t)>\theta$ for $t\in (0, 1)$ and satisfies \eqref{e1.1}-\eqref{e1.2}. Now we
show $x\in Q\setminus\{\theta\}$. To see this notice that
\begin{equation}
x(t)=  \int_0^1 G(t, s)\int_0^1 G(s, \tau) f(\tau, x(\tau))d\tau ds\quad \mbox{for }
 t\in I.\label{e3.6}
\end{equation}
This and
\begin{equation}
\frac{G(t, s)}{G(\xi, s)}=
\begin{cases}
\frac{t(1-s)}{\xi (1-s)}\geq t\geq z(t), & t, \xi\leq s;\\[3pt]
\frac{s(1-t)}{s(1-\xi)}\geq 1-t\geq z(t),& t, \xi\geq s;\\[3pt]
\frac{t(1-s)}{s(1-\xi)}\geq t\geq z(t), & t<s<\xi; \\[3pt]
\frac{s(1-t)}{\xi(1-s)}\geq 1-t \geq z(t), &  t>s>\xi
\end{cases} \label{e3.7}
\end{equation}
yield
$$
x(t)\geq z(t)\int_0^1 G(\xi,s)\int_0^1 G(s, \tau)f(\tau, x(\tau))d\tau\quad \mbox{for }  t,
\xi\in I,
$$
which implies $x\in Q\setminus\{\theta\}$.
On the other hand, it is easy to see by \eqref{e3.6} that $Ax=x$. This
completes the proof.
\end{proof}

Consequently, the existence of positive solution for \eqref{e1.2} is
equivalent to that of fixed point of $A$ in $Q\setminus\{\theta\}$. By \eqref{e3.5} and
the process similar to the proof of Lemma \ref{lem3.1}, we also obtain
the following Lemma.

\begin{lemma} \label{lem3.2} $A(Q\setminus\{\theta\})\subset Q$.
\end{lemma}

\begin{lemma} \label{lem3.3}
 For every pair of positive numbers  $R_2$ and $R_1$ with $R_2>R_1>0$,
 $A: \overline{Q_{R_2}}\setminus Q_{R_1}\to Q$ is a strict set contraction,
  where $Q_r:= \{x\in Q:\ \|x\|_c<r\}(r>0)$.
\end{lemma}

\begin{proof}
First, under the assumptions for $R_2$ and $R_1$,  (H0) guarantees that
for each $x\in\overline{Q_{R_2}}\setminus Q_{R_1}$,
\begin{equation}
\int_0^1 \tau(1-\tau)\|f(\tau, x(\tau))\|d\tau \leq \int_0^1\tau(1-\tau)f_{R_1, R_2}(\tau)d\tau < +\infty \label{e3.8}
\end{equation}
which implies $A$: $\overline{Q_{R_2}}\setminus Q_{R_1}\to Q$ is bounded.

Next we show $A$: $\overline{Q_{R_2}}\setminus Q_{R_1}\to Q$ is
continuous. To see this from \eqref{e3.3} it follows that for
$x\in\overline{Q_{R_2}}\setminus Q_{R_1}$ and $t_1, t_2\in I$,
\begin{equation}
\|(Ax)(t_1)-(Ax)(t_2)\|\leq  \int_0^1
 |G(t_1, s)-G(t_2, s)|ds\int_0^1 G(s, \tau)\|f(\tau, x(\tau))\|d\tau. \label{e3.9}
\end{equation}
This and \eqref{e3.8} yield that for every subset
$V\subset \overline{Q_{R_2}}\setminus Q_{R_1}$, $(AV)(t)$ is equicontinuous on $I$, where
$(AV)(t)= \{(Ax)(t): x\in V\}$, $t\in I$.

Let $x_n, x\in \overline{Q_{R_2}}\setminus Q_{R_1}$ with $\|x_n-x\|_c\to 0 $ as $n\to +\infty$ . This
implies
$$
\|x_n(t)-x(t)\|\to 0\quad \mbox{as } n\to +\infty \quad\mbox{for } t\in I.
$$
 From Lebesgue dominated convergence theorem and \eqref{e3.8}, it follows
that
$$ \|(Ax_n)(t)-(Ax)(t)\|\to 0\quad \mbox{as } n\to +\infty.
$$
 Thus,  $\{(Ax_n)(t)\}$ is
 relatively compact for every $t\in I$. From this and the equicontinuity
  of $\{Ax_n(t)\}$ by the  Ascoli-Arzela theorem, we obtain that $\{Ax_n\}$
  is a relatively compact subset of $Q$.

 Now it remains to show $\|Ax_n-Ax\|_c\to 0$ as $n\to +\infty$.
In fact, if this is not true, then there is a constant $\epsilon_0>0$ and a
 subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that
 $\|Ax_{n_i}-Ax\|_c\geq \epsilon_0$ ($i=1,2,\dots$). However, the relative
compactness   of $\{Ax_n\}$ implies that $\{Ax_{n_i}\}$ contains a subsequence
which  converges in $C[I, P]$. Without loss of generality, we may assume that
$\{Ax_{n_i}\}$ itself converges to $y$, that is,
$\|Ax_{n_i}-y\|_c\to 0 $  as  $i\to +\infty $.
So we have $y=Ax$. This is a contradiction.
Therefore, $A$ is continuous.

Finally, we show $A$: $\overline{Q_{R_2}}\setminus Q_{R_1}\to Q$ is a strict set contraction,
that is, there exists $k\in (0, 1)$ such that
$\alpha_c(AV)\leq k\alpha_c(V)\quad$ for each $V\subset\overline{Q_{R_2}}\setminus Q_{R_1}$.
Fix $V\subset\overline{Q_{R_2}}\setminus Q_{R_1}$. Let
\begin{equation}
(A_nx)(t):= \int_{1/n}^{1-(1/n)} J(t, s)f(s, x(s))ds\quad \mbox{for }
 x\in V.\label{e3.10}
\end{equation}
By \eqref{e3.8} we know
\begin{equation}
(A_nx)(t)\to   (Ax)(t)\quad \mbox{as } n\to +\infty\quad \mbox{for each }
 x\in V \mbox{ and } t\in I.\label{e3.11}
\end{equation}
  This implies
$$
d_H((A_nV)(t), (AV)(t))\to 0\quad \mbox{as } n\to +\infty \quad
\mbox{for each } t\in I,
$$
where $d_H(\cdot, \cdot)$ denotes the Hausdorff
metric. Thus, by the property of noncompactness measure,
\begin{equation}
\alpha((A_nV)(t))\to\alpha((AV)(t))\quad \quad\mbox{for }t\in I.\label{e3.12}
\end{equation}
In what follows, we estimate $\alpha((A_nV)(t))$ for each $t\in
I$. Note that $x(s)\geq z(s)x(\tau)\geq \theta\quad$ for $s, \tau
\in I$ and $x\in V.$ Thus,
$$
\frac{R_1}{nN}\leq \frac{\|x\|_c}{nN}\leq \|x(s)\|\leq R_2\quad \quad\mbox{for }
s\in[\frac{1}{n}, 1-\frac{1}{n}].
$$
By the definition of integration and \eqref{e3.4}, respectively, we have
$$
\int_{1/n}^{1-(1/n)} J(t, s)f(s,
x(s))ds\in(1-\frac{2}{n})\overline{co}(\{J(t,s)f(s, x(s)):
s\in[\frac{1}{n}, 1-\frac{1}{n}]\}
$$
and
$$
J(t, s)=\int_0^1 G(t, \tau)G(\tau, s)d\tau\leq \int_0^1 \tau^2(1-\tau)^2d\tau
=\frac{1}{30}\quad \mbox{for all } t, s\in I.
$$
These, (H1), (H2), and Lemma \ref{lem2.1} guarantee that
\begin{equation}
\begin{aligned}
\alpha((A_nV)(t))
&=\alpha (\{\int_{1/n}^{1-(1/n)} J(t, s)f(s, x(s))ds |\ x\in V\})\\
&\leq (1-\frac{2}{n})\alpha (\overline{co}\{J(t, s)f(s, x(s)) |\
s\in [\frac{1}{n},\ 1-\frac{1}{n}],\
x\in V\})\\
&\leq  \alpha(\{J(t, s)f(s, x(s))|\ s\in [\frac{1}{n},\
1-\frac{1}{n}],\ x\in
V\})\\
&\leq \frac{1}{30}\alpha(\{f(s, x(s))|\ s\in [\frac{1}{n},\ 1-\frac{1}{n}],\ x\in V\})\\
&\leq \frac{1}{30}\alpha(f(I_n\times V(I_n))\leq
\frac{1}{30}l\cdot\alpha V(I_n))\leq \frac{1}{15}l\alpha_c(V),
\end{aligned}\label{e3.13}
\end{equation}
 where $I_n= [\frac{1}{n}, 1-\frac{1}{n}]$ and
$V(I_n)= \{x(s)$: $x\in V,\ s\in I_n\}$. Combining  Lemma
\ref{lem2.1}  with \eqref{e3.9}, \eqref{e3.12}, and \eqref{e3.13}
again, one can obtain
$$\alpha_c(AV)= \max_{t\in I}\alpha ((AV)(t))\leq \frac{1}{15}l\cdot \alpha_c(V).
$$
This implies $A$ is a strict set contraction with
$k=\frac{1}{15}l< 1$ from $\overline{Q_{R_2}}\setminus Q_{R_1}$ to $Q$.
\end{proof}

Using (H4), we define an operator $L$ on $C[I, R]$ by
$$
(Lu)(t):= \int_0^1 J(s, t)a(s)u(s)ds=\int_0^1 G(\tau, t)\int_0^1 G(s,
\tau)a(s)u(s)dsd\tau
$$
for $u\in C[I, R]$ where $J$ is given
by \eqref{e3.4}, and $a$ is the same as in  (H4).

It is easy to see  $L$: $C[I, R]\to C[I, R]$ is a completely
continuous positive operator. Note that if  $v(t)=t(1-t)$ on
$I$, then $\|v\|_c=\frac{1}{4}$. By $G(t, \tau)\geq t\tau (1-t)(1-\tau)$
for $t, \tau\in I$, we know
$$
(Lv)(t)\geq \int_0^1\tau ^2(1-\tau)^2d\tau\int_0^1 s^2(1-s)^2a(s)ds\cdot
v(t)=\delta_0 v(t)\quad\quad\mbox{for }t\in I,
$$
where $ \delta_0 =\frac{1}{30}\int_0^1 s^2(1-s)^2a(s)ds > 0$.  From Lemma \ref{lem2.2}
 it follows
that the spectral radius $r(L)\geq \delta_0>0$. So the well-known
 Krein-Rutman theorem \cite{n1} guarantees that there exists an $p\in C[I, R^+]$
with $p(t)\not\equiv 0$ on $I$ such that
\begin{equation}
(Lp)(t)=\int_0^1J(s,t)a(s)p(s)ds=r(L)p(t)\quad\quad\mbox{for }t\in I.
\label{e3.14}
\end{equation}
 From \eqref{e3.7} one deduces that
\begin{align*}
p(t) &=\frac{1}{r(L)}\int_0^1(\int_0^1
G(s, \tau)G(\tau, t)d\tau)a(s)p(s)ds\\
&\geq \frac{1}{r(L)}\int_0^1 z(s)\int_0^1
G(\xi, \tau)G(\tau, t)d\tau)a(s)p(s)ds\\
&\geq \frac{1}{r(L)}\int_0^1 z(s)a(s)p(s)ds\cdot J(\xi, t)\quad\mbox{for all }
 t, \xi\in I.
\end{align*}
Therefore,
\begin{equation}
p(t)\geq \delta J(\xi, t)\quad{\rm for\ all}\ t, \xi\in I, \label{e3.15}
\end{equation}
where $\delta:= \frac{1}{r(L)}\int_0^1 z(s)a(s)p(s)ds$. This and
\eqref{e3.14} guarantees that
\begin{equation}
\int_0^1 p(t)a(t)dt\geq \delta r(L).\label{e3.16}
\end{equation}
Now we are ready to give the main result of the present paper.

\begin{theorem} \label{thm3.1}
Assume that (H0)--(H5) hold and $r(L)>1$.
Then \eqref{e1.1} subject to \eqref{e1.2} has at least two positive solutions.
\end{theorem}

\begin{proof}
Set
\begin{equation}
K:= \{x\in Q:\quad \int_0^1p(t)a(t)x(t)dt\geq \delta r(L)x(s),\quad \forall s\in I\},\label{e3.17}
\end{equation}
where $Q$ is given by \eqref{e3.1}, $a(t)$ is given by
(H4), $p(t)$, $r(L)$, and $\delta$ are given by \eqref{e3.14} and \eqref{e3.15}.

By \eqref{e3.16} it is easy to see $K\setminus\{\theta\}\neq\emptyset$ and $K$ is also a cone of
$C[I, E]$. We now prove that the operator $A$ defined by \eqref{e3.3}
maps $Q\setminus\{\theta\}$ into $K$. In fact, for $x\in Q\setminus\{\theta\}$, it follows from
\eqref{e3.3}, \eqref{e3.15}, and \eqref{e3.2} that
\begin{align*}
\int_0^1p(t)a(t)(Ax)(t)dt
&=\int_0^1p(t)a(t)\int_0^1 J(t, s)f(s, x(s))dsdt\\
&=\lim_{n\to +\infty} \int_0^1
p(t)a(t)\int_{1/n}^{1-(1/n)} J(t, s)f(s, x(s))dsdt\\
&= \lim_{n\to +\infty} \int_{1/n}^{1-(1/n)} f(s, x(s))ds\int_0^1 J(t, s)a(t)p(t)dt\\
&= r(L)\lim_{n\to +\infty} \int_{1/n}^{1-(1/n)} p(s)f(s, x(s))ds \\
&\geq \delta r(L)\lim_{n\to +\infty}  \int_{1/n}^{1-(1/n)} J(\tau, s)f(s, x(s))ds\\
&= \delta r(L)\lim_{n\to +\infty} \int_0^1 J(\tau, s)f(s, x(s))ds \\
&= \delta r(L)(Ax)(\tau)\quad \mbox{for all } \tau\in I,
\end{align*}
which implies $A(Q\setminus\{\theta\})\subset K$.
Consequently,  we obtain by Lemma \ref{lem3.2} and Lemma \ref{lem3.3} that $A:
\overline{K_{R_2}}\setminus K_{R_1}\to K$ is a strict set contraction for every pair of positive
numbers  $R_2$ and $R_1$ with $R_2> R_1>0$, where $K_{R_1}= \{x\in
K: \|x\|_c< R_1\}$.

Choose a positive number $R_0$ with $R_0 > R$ such that
$$
R_0> \frac{N\|b\|_c}{30\delta(r(L) - 1)}\int_0^1 a(t)p(t)dt.
$$
We proceed to prove
\begin{equation}
Ax\not\leq x\quad\mbox{for all }  x\in\partial K_{R_0}.
\label{e3.18}
\end{equation}
Suppose, on the contrary, there exists an $x_0\in\partial K_{R_0}$ such
that $Ax_0\leq x_0$. Therefore,
$$
x_0(t)\geq (Ax_0)(t) = \int_0^1 J(t,s)f(s, x_0(s))ds\quad\mbox{for all }  t\in I.
$$
Multiply by $p(t)a(t)$ and integrate from $0$ to $1$ to obtain
\begin{align*}
& \int_0^1 p(t)a(t)x_0(t)dt \geq \int_0^1 p(t)a(t)\int_0^1 J(t, s)f(s, x_0(s))dsdt\\
&\geq  \int_0^1 p(t)a(t)\int_0^1 J(t, s)a(s)x_0(s)dsdt - \int_0^1\int_0^1 J(t,
s)a(t)p(t)b(s)dsdt\\
&= \int_0^1(\int_0^1 J(t, s)a(t)p(t)dt)a(s)x_0(s)ds
-\int_0^1\int_0^1 J(t, s)a(t)p(t)b(s)dsdt\\
&=  r(L)\int_0^1 p(s)a(s)x_0(s)ds
- \int_0^1\int_0^1 J(t, s)a(t)p(t)b(s)dsdt.
\end{align*}
 This and \eqref{e3.17} yield
\begin{align*}
\int_0^1\int_0^1 J(t, s)a(t)p(t)b(s)dsdt
&\geq (r(L) -1)\int_0^1 p(s)a(s)x_0(s)ds\\
&\geq (r(L)-1)\delta x_0(\tau)\geq \theta\quad\mbox{for all } \tau\in I.
\end{align*}
The normality of the cone $P$ and $|J(t, s)|\leq \frac{1}{30}$ for all
 $t, s\in I$ guarantee that
$$
\frac{N\|b\|_c}{30}\int_0^1 a(t)p(t)dt\geq \delta(r(L) - 1)R_0.
 $$
This is a contradiction with the selection of $R_0$.
Consequently, \eqref{e3.18} holds.

 In what follows we show
\begin{equation}
 Ax\not\geq x\quad\mbox{for all }  x\in\partial  K_R. \label{e3.19}
\end{equation}
If this is false, then there exists an  $x_1\in\partial K_R$ such that $x_1\leq Ax_1$,
that is,
$$
\theta\leq x_1(t)\leq  (Ax_1)(t)\quad\quad\mbox{for }t\in I.
$$
Since $x_1\in K\subset Q$, we get
$$
x_1(t)\geq z(t)x_1(\tau)\geq \theta\quad\quad\mbox{for }t, \tau\in I.
$$
As a  result,
$$\frac{z(t)}{N}R = \frac{z(t)}{N}\|x_1\|_c \leq \|x_1(t)\| \leq R\quad\quad\mbox{for }t\in I.
$$
This implies
$$
\|f(t, x_1(t))\| \leq f_{R, R}(t)\quad\quad\mbox{for }t\in (0, 1).
$$
Combining the above with (H5) we know
\begin{align*}
\|x_1(t)\|&\leq  N\|(Ax_1)(t)\|\leq N\|\int_0^1 J(t, s)f(s, x_1(s))ds\|\\
&\leq  N \int_0^1 J(t, s)\|f(s, x_1(s))\|ds\\
&\leq N\int_0^1 J(t, s)f_{R, R}(s)ds\\
&\leq  N\int_0^1 G(t, \tau)(\int_0^1 s(1-s)f_{R, R}(s)ds)d\tau\\
&\leq \frac{N}{8}\int_0^1  s(1-s)f_{R, R}(s)ds
<  R\quad\quad\mbox{for }t\in I.
\end{align*}
 This is a contradiction. Then  \eqref{e3.19} follows.
Finally, we prove that there exists a positive number $R'$ with
$R' < R$ such that
\begin{equation}
Ax\not\leq x\quad\quad\mbox{for }x\in \partial K_{R'}.\label{e3.20}
\end{equation}
In fact, by (H3), given $\epsilon\in (0, \int_c^d J(\frac{1}{2},s)\phi(s)ds)$,
there exists an $R''> 0$ such that
\begin{equation}
\phi^*(f(t, x(t))\geq \phi(t) -\epsilon\quad\quad\mbox{for }t\in [c, d] \mbox{ and }
x\in P_{R''}\setminus\{\theta\}.\label{e3.21}
\end{equation}
Choose
\begin{equation}
R':= \min\big\{\frac{R}{2},\ R'', \ \int_c^d J(\frac{1}{2},
s)\phi(s)ds- \epsilon\big\}. \label{e3.22}
\end{equation}
Now we are ready to prove \eqref{e3.20} holds. Suppose, on the contrary,
there exists an $x_2\in \partial K_{R'}$ such that $Ax_2\leq x_2$. Then by
\eqref{e3.2} we know
$$
\frac{z(t)}{N}R'\leq \|x_2(t)\| \leq  R'\leq R''\quad\quad\mbox{for }t\in (0, 1).
$$
This and \eqref{e3.21} guarantee that
 \begin{align*}
 \phi^*(x_2(t))&\geq \int_0^1 J(t, s)\phi^*(f(s, x_2(s))ds\\
 &\geq  \int_c^d J(t, s)[\phi(s) - \epsilon]ds \\
 &>  \int_c^d J(t, s)\phi(s)ds -
 \frac{\epsilon}{2}\quad\quad\mbox{for }t\in (0, 1).
\end{align*}
Consequently,
$$
\|x_2\|_c\geq \phi^*(x_2(\frac{1}{2}))\geq  \int_c^d J(\frac{1}{2}, s)\phi(s)ds
- \frac{\epsilon}{2} > R'.
$$
This is a  contradiction with $x_2\in \partial K_{R'}$.
Then the conclusion follows from Lemma \ref{lem2.3}.
\end{proof}

\begin{remark} \label{rmk3.2}\rm
 If $r(L)>1$ is replaced with $\int_0^1 s^2(1-s)^2a(s)ds > 30$ in Theorem \ref{thm3.1},
the conclusion of Theorem \ref{thm3.1} also holds. In fact, by $v(t)=t(1-t)\in Q$
and $G(t, s)\geq ts(1-t)(1-s)$ for $t, s\in I$, and using the definition of the
operator $L$, one can obtain
$$
Lv\geq \frac{1}{30}\int_0^1 s^2(1-s)^2a(s)ds\cdot v.
$$
Then by Lemma \ref{lem2.2}, $r(L)>1$ follows.
\end{remark}
For the next theorem we replace (H4) by
\begin{itemize}
\item[(H'4)]
 There exists an $\psi^*\in P^*$ with $\|\psi^*\| = 1$
 and a subinterval $[c', d']\subset (0, 1)$ such that
$$
\lim_{\|x\|\to  +\infty,\; x\in P}\frac{\psi^*(f(t, x))}{\|x\|} > \mu
$$
uniformly  with respect to $t\in [c', d']$, where
$$
\mu= N\Big(\min\{c',  1-d'\}\cdot\int_{c'}^{d'} J(\frac{1}{2}, s)ds\Big)^{-1}.
$$
\end{itemize}

\begin{theorem} \label{thm3.2}
Assume that (H0)-(H3), (H'4), and (H5) hold. Then \eqref{e1.1}
subject to \eqref{e1.2} has at least two positive solutions.
\end{theorem}

\begin{proof} Consider the operator $A$ in $Q\setminus\{\theta\}$, where $Q$ is
defined by \eqref{e3.1}. From the proof of Theorem \ref{thm3.1}, it is not
difficult to see that \eqref{e3.19} and \eqref{e3.20} hold for $x\in\partial Q_{R'}$
and $x\in\partial Q_{R}$, respectively; where $R'$ is given by \eqref{e3.22}
and $R$ is given by (H5).

It remains to show that there exists a positive number $R_0$
with $R_0> R$ such that \eqref{e3.18} holds for $x\in\partial Q_{R_0}$.
By (H'4), there exist an $\epsilon>0$ and an $R_1>0$ such that
\begin{equation}
\psi^*(f(t, x))\geq (\mu+\epsilon)\|x\|\quad\quad\mbox{for }t\in[c', d'],\ x\in P,
\mbox{ and }\|x\|\geq R_1.\label{e3.23}
\end{equation}
Choose
$$
R_0\geq \max\big\{R+1,\ \frac{NR_1}{\min\{c', 1-d'\}}\big\}.
$$
We proceed to prove \eqref{e3.18} holds for $x\in\partial Q_{R_0}$. If this is
false, then there exists an $x_0\in\partial Q_{R_0}$ such that $Ax_0\leq
x_0$. By \eqref{e3.1} we know
$$
x_0(t)\geq z(t)x_0(s)\geq \min\{c',1-d'\}\cdot x_0(s)\geq \theta\quad\quad
\mbox{for }t\in[c', d'] \mbox{ and } s\in I.
$$
>From the normality of the cone $P$ it follows that
$$
N\|x_0(t)\| \geq \min\{c',\ 1-d'\}\cdot\|x_0\|_c = \min\{c',\
1-d'\}\cdot R_0\quad\quad\mbox{for }t\in[c',\ d'],
$$
that is, $\|x_0(t)\|\geq R_1$ for $t\in [c,\ d']$. This,
\eqref{e3.23}, and \eqref{e3.3} guarantee that
\begin{align*}
 R_0 &\geq \psi^*(x_0(\frac{1}{2}))\\
&\geq \int_0^1 J(\frac{1}{2}, s)\psi^*(f(s, x_0(s))ds\\
&\geq \int_{c'}^{d'} J(\frac{1}{2}, s)(\mu+\epsilon)\|x_0(s)\|ds \\
&\geq \frac{\min\{c', 1-d'\}}{N}\int_{c'}^{d'} J(\frac{1}{2}, s)(\mu+\epsilon)R_0ds
> R_0,
\end{align*}
which is a contradiction. Then the statement of Theorem \ref{thm3.2} follows.
\end{proof}

\begin{corollary} \label{coro0}
Suppose (H0), (H1), (H2), and
one of the following conditions are satisfied:
\begin{itemize}
\item[(i)]  (H3) and (H5).

\item[(ii)] (H4), (H5), and $r(L)>1$.

\item[(iii)] (H'4) and (H5).
\end{itemize}
Then \eqref{e1.1} subject to \eqref{e1.2} has at least one positive solution.
\end{corollary}

\begin{remark} \label{rmk3.3} \rm
By the same method used above,
we can study the existence of multiple positive solutions of
second order nonlinear singular boundary-value problems in scalar
or in abstract space.
\end{remark}

\section{Examples}
\begin{example} \label{ex1} \rm
 Consider the boundary-value problem consisting of a finite system
of fourth-order scalar nonlinear differential equations.
\begin{equation}
\begin{gathered}
 x_n^{(4)}(t)=\frac{1}{\sqrt{t(1-t)}}\big(x_n^{\frac{3}{2}} +
\frac{1}{\max_{1\leq i\leq m}|x_i|}\big), \quad 0<t <1;\\
x_n(0)=x_n(1)=x_n''(0)=x_n''(1)=0,\quad(n=1, 2,\dots
m).\end{gathered}\label{e4.1}
\end{equation}
\textbf{Claim:} \eqref{e4.1} has at least two positive solutions
$x^*(t)= (x_1^*(t), x_2^*(t),\dots x_m^*(t))$ and
$x^{**}(t)= (x_1^{**}(t), x_2^{**}(t),\dots x_m^{**}(t))$
such that
$$
0< \max_{1\leq i\leq m,\; t\in[0, 1]}|x_i^*(t)| < 1 <
\max_{1\leq i\leq m,\; t\in[0, 1]} |x_i^{**}(t)|.
$$
\end{example}

\begin{proof} Let $E$ be the $m$-dimensional Euclidean space
$R^m=\{x=(x_1, x_2, \dots x_m)\}$ with norm
$\|x\| =\max_{1\leq i\leq m}|x_i|$ and
$$
P=\{x=(x_1, x_2, \dots x_m): x_n\geq 0 \mbox{ for }n=1, 2, \dots m\}.
$$
Then $P$ is a normal cone in $E$, $P^*=P$ and the normal constant
is $N=1$. System \eqref{e4.1} can be regarded as a boundary-value problem
of form \eqref{e1.1} subject to\eqref{e1.2},
where $x=(x_1, x_2, \dots x_m)$,
$$
f(t, x)= (f_1(t, x_1, \dots
x_m),\dots, f_n(t, x_1, \dots x_m),\dots,f_m(t, x_1, \dots x_m))
$$
with
$$
f_n(t, x_1, \dots x_m):=\frac{1}{\sqrt{t(1-t)}}\big(x_n^{3/2} + \frac{1}{\max_
{1\leq i\leq m}|x_i|}\big).
$$
Evidently, $f\in C[ (0,1)\times P\setminus\{\theta\}, P]$ and is singular at $t=0$, $t=1$,
and $ x=\theta =(0,0,\dots, 0)$. Notice that
$$
f_{r, R}(t)\leq \frac{1}{\sqrt{t(1-t)}}\big( R^{\frac{3}{2}} +
\frac{1}{rt(1-t)}\big)\quad\quad\mbox{for }t\in (0, 1)\ \mbox{and }R\geq r>0$$
and
$$
\int_0^1 t(1-t)f_{1, 1}(t)dt\leq \int_0^1[\sqrt{t(1-t)} +
\frac{1}{\sqrt{t(1-t)}}]dt < (\frac{1}{8} + \pi) < 8.
$$
Therefore, by Theorem \ref{thm3.1} or Theorem \ref{thm3.2}, our conclusion follows.
\end{proof}

\begin{example} \label{ex2} \rm
Consider the boundary-value problem consisting of an infinite system of fourth
order scalar nonlinear differential equations.
\begin{equation}
\begin{gathered}
x_n^{(4)}(t)=f_n(t, x(t)), \quad t\in (0, 1); \\
x_n(0)=x_n(1)=x_n''(0)=x_n''(1)=0,\quad (n=1, 2,\dots ).
\end{gathered} \label{e4.2}
\end{equation}
where
\begin{gather*}
f_1(t,x)=\frac{1}{\sqrt{t(1-t)}}\big(x_1 + \frac{x_2}{2} +
(\sum_{i\geq 1}|x_i|)^2 + b_1\big),\\
f_2(t,x)=\frac{1}{\sqrt{t(1-t)}}\big( \frac{x_2}{2} + \frac{x_3}{3}+
\frac{1}{\sum_{i\geq 1}|x_i|} + b_2\big),\\
f_n(t, x)= \frac{1}{\sqrt{t(1-t)}}\big(\frac{x_n}{n} +
\frac{x_{n+1}}{n+1} + b_n\big),\quad n= 3, 4, \dots;\\
x=(x_1, x_2, \dots),\quad b_i\geq 0 \quad(i=1, 2, \dots),\quad \sum_{i\geq 1}b_i\leq 1.
\end{gather*}
\textbf{Claim:} System \eqref{e4.2} has at least two positive
solutions $x^*(t)= (x_1^*(t), x_2^*(t),\dots )$ and
 $x^{**}(t)= (x_1^{**}(t), x_2^{**}(t),\dots )$ such that
$$
0< \sum_{1\leq i,\; t\in[0, 1]}|x_i^*(t)| < 1 <
\sum_{1\leq i,\;t\in[0, 1]}|x_i^{**}(t)|.
$$
\end{example}

\begin{proof}
Let $E= l^{1}$ with norm $\|x\|= \sum_{i\geq 1}|x_i|$ and
$$
P=\{x=(x_1, x_2, \dots ): x_n\geq 0 \quad\mbox{for }n=1, 2, \dots \}.
$$
Then
$P$ is a normal cone in $E$ and the normal constant is $N=1$.
System \eqref{e4.2} can be regarded as a boundary-value problem form \eqref{e1.1}
with \eqref{e1.2}, where $x=(x_1, x_2, \dots )$,
$$
f(t, x)= (f_1(t, x_1, \dots ),\dots, f_n(t, x_1, \dots ),\dots,).
$$
Evidently, $f\in C[ (0,1)\times P\setminus\{\theta\}, P]$ and is singular at $t=0$, $t=1$, and
$ x=\theta =(0,0,\dots, )$. Note that
for $t\in (0, 1)$ and $R\geq r>0$,
$$
f_{r, R}(t)\leq \frac{1}{\sqrt{t(1-t)}}\big( 2R + R^2 + \frac{1}{rt(1-t)} +
\|b\|\big)
$$
So, (H0) is satisfied. In addition, (H1) is obvious. As in
\cite[Example 2.1.2]{g4}, one can see (H2) is satisfied with $l=0$.
Choosing $\phi^*=\psi^*=(1, 1, 0,\dots, 0, \dots)$, we know that
(H3) and  (H'4) holds for \eqref{e4.1}. From
$$
\int_0^1 s(1-s)f_{1,1}(s)ds\leq \int_0^1 ((3+\|b\|)\sqrt{s(1-s)} +
\frac{1}{\sqrt{s(1-s)}})ds \leq (\frac{4}{8} +\pi) < 8,
$$
it follows that (H5) is satisfied. By
Theorem \ref{thm3.2}, our conclusion follows.
\end{proof}

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