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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 22, 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}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/22\hfil Positive solutions]
{Positive solutions for second-order multi-point boundary-value
problems at resonance in Banach spaces}

\author[W. Jiang, B. Wang\hfil EJDE-2011/22\hfilneg]
{Weihua Jiang, Bin Wang}  % in alphabetical order

\address{Weihua Jiang \newline
College of Sciences, Hebei University of Science and Technology\\
Shijiazhuang, 050018, Hebei, China}
\email{weihuajiang@hebust.edu.cn}

\address{Bin Wang \newline
Department of Basic Courses,
Hebei Professional and Technological College of Chemical and \\
Pharmaceutical Engineering, Shijiazhuang, 050026, Hebei,  China}
\email{wb@hebcpc.cn}

\thanks{Submitted July 14, 2010. Published February 9, 2011.}
\thanks{Supported by grants 10875094 from the Natural Science Foundation
of China, \hfill\break\indent A2009000664 from the Natural Science
Foundation of Hebei Province, 2008153 from the \hfill\break\indent
Foundation of Hebei Education Department, and XL200814 from the
Foundation of \hfill\break\indent Hebei University of Science and
Technology.} 
\subjclass[2000]{34B15} 
\keywords{Banach space; positive solution; strict set contraction; \hfill\break\indent
boundary value problem}

\begin{abstract}
 In this article, we study the existence and  multiplicity
 of positive solutions for a nonlinear second-order
 multi-point boundary-value problem at resonance in Banach spaces.
 The arguments are based upon a specially constructed equivalent
 equation and the fixed point theory in a cone for strict set
 contraction operators.
\end{abstract}

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

\section{Introduction}

 The theory of ordinary differential equations in Banach
spaces has become a new important branch (see, for example,
\cite{d1,g1,g2,l1} and references cited therein). In 1988, Guo and
Lakshmikantham \cite{g3} discussed the existence of multiple
solutions for two-point boundary value problem of ordinary
differential equations in Banach spaces. Since then, nonlinear
second-order multi-point boundary value problems at non-resonance
in Banach spaces have been studied by several authors (see, for
example, \cite{f1,l2,l3,z1} and references cited therein).
Recently, the existence of solutions for boundary value problems
at resonance have been studied by many papers, (see, for example
\cite{d2,d3,h1,k1,k2,l4,l5,z2}). Using the Krasnolsel'skii-Guo
fixed point theorem, Han \cite{h1} studied a second order
three-point BVP at resonance by rewriting the original BVP as an
equivalent one. Motivated by their results, in this paper, we will
discuss the existence of positive solutions for the second-order
$m$-point boundary value problem at resonance
\begin{gather}
y''(t)=f(t,y),\quad 0<t<1,\label{e1.1}\\
y'(0)=\theta,\quad y(1)=\sum_{i=1}^{m-2}k_i y(\xi_i ) \label{e1.2}
\end{gather}
in a real Banach space $E$, where $\theta$ is the zero element of
$E$, $0<\xi_1<\xi_2<\dots<\xi_{m-2}<1$,
 $k_i>0$, $i=1,2,\dots,m-2$, $\sum_{i=1}^{m-2}k_i =1$.

 The boundary value problem \eqref{e1.1}-\eqref{e1.2} is at
resonance when $\sum_{i=1}^{m-2}k_i =1$; that is, the corresponding
homogeneous boundary value problem
\begin{gather*}
y''(t)=0, \quad t\in[0,1],\\
y'(0)=0,\quad y(1)=\sum _{i=1}^{m-2}k_i y(\xi _i )
\end{gather*}
has nontrivial solutions.

To the best of our knowledge, no paper has
considered the existence of positive solutions for the boundary
value problems at resonance in Banach spaces. We shall fill this gap
in the literature. The organization of this paper is as follows.
We shall introduce a theorem and
some notations in the rest of this section. In Section 2, we provide
some necessary background. In particular, we state some properties
of Green's function associated with the equivalent problem of
\eqref{e1.1}-\eqref{e1.2}. In Section 3,
the main results will be stated and proved.

 \begin{theorem}[\cite{c1,p1}] \label{thm1.1}
 Let $K$ be a cone of the real Banach
 space $X$ and $K_{r,R}=\{x\in K |r\leq \|x\|\leq R\}$ with $R>r>0$.
 Assume that $A:~K_{r,R}\to K$ is a strict set contraction
 such that one of the following two conditions is satisfied
\begin{itemize}
\item[(i)] $Ax\not\leq x$ for all $x\in K$, $\|x\|=r$ and
$Ax\not\geq x$ for all $x\in K$, $\|x\|=R$.

\item[(ii)] $Ax\not\geq x$ for all $x\in K$, $\|x\|=r$ and
$Ax\not\leq x$ for all $x\in K$, $\|x\|=R$.
\end{itemize}
Then $A$ has at least one fixed point $x\in K$ satisfying
$r<\|x\|<R$.
\end{theorem}

Let the real Banach space $E$ with norm $\|\cdot\|$ be partially
ordered by a normal cone $P$ of $E$; i.e., $x\leq y$ if and only if
$y-x\in P$, and $P^*$ denotes the dual cone of $P$; i.e.,
$P^*=\{\varphi\in E^*: \varphi(x)\geq0,\; x\in P\}$. Denote
the normal constant of $P$ by $N$; i.e., $\theta\leq x\leq y$
implies $\|x\|\leq N\|y\|$. Take $I=[0,1]$. For any $x\in C[I,E]$,
evidently, $(C[I,E],\|\cdot\|_c)$ is a Banach space with
$\|x\|_c=\max_{t\in I}\|x(t)\|$, and
$Q=\{x\in C[I,E]: x(t)\geq \theta \text{ for } t\in I\}$
is a cone of the Banach space $C[I,E]$.
A function $y\in C^2[I,E]$ is called a positive
solution of the boundary value problem \eqref{e1.1}-\eqref{e1.2}
if it satisfies \eqref{e1.1}-\eqref{e1.2} and
$y\in Q$, $y(t)\not\equiv \theta$.

 In this paper, we denote
$\alpha(\cdot)$ the Kuratowski measure of non-compactness of a
bounded set in $E$ and $C[I,E]$. The closed balls in spaces $E$ and
$C[I,E]$ are denoted by $T_r=\{x\in E:\|x\|\leq r\}(r>0)$ and
$B_r=\{y\in C[I,E]:\|y\|_c\leq r\}(r>0)$, respectively.

Define
$$
F(t,y):=f(t,y)+\beta^2 y,
$$
where $\beta\in(0,\frac{\pi}{2})$. Obviously, $y(t)$ is a solution
of the problem \eqref{e1.1}-\eqref{e1.2} if and only if it
is a solution of the problem
\begin{gather}
y''(t)+\beta^2 y(t)=F(t,y(t)), \quad 0<t<1,\label{e1.3}\\
y'(0)=\theta,\quad y(1)=\sum _{i=1}^{m-2}k_i y(\xi _i ),\label{e1.4}
\end{gather}
and the problem \eqref{e1.3}-\eqref{e1.4} is at non-resonance.

For convenience, we set
\begin{gather*}
a_0=\sum _{i=1}^{m-2}k_i \cos\beta\xi_i -\cos\beta,\quad
a_1=(a_0+1)\sin\beta+\sum_{i=1}^{m-2} k_i\sin\beta\xi_i, \\
a_2=1-\sum_{i=1}^{m-2}k_i\cos\beta(1-\xi_i).
\end{gather*}
In this paper, we assume the following conditions hold.
\begin{itemize}

\item[(H1)] $k_i>0$, $i=1,2,\dots,m-2$,
$0<\xi_1<\xi_2<\dots<\xi_{m-2}<1$, $\sum_{i=1}^{m-2}k_i =1$.

\item[(H2)] $P$ is a normal cone of $E$ and $N$ is the normal constant;
$F:I\times P\to P$, $F(t,\theta)=\theta$ for all $t\in I$; for
any $r>0$, $F(t,x)$ is uniformly continuous and bounded on
$I\times (P\cap T_r)$ and there exists a constant $L_r$ with
$0\leq L_r<(\beta a_0)/(2a_1)$ such that
$$
\alpha(F(I\times D))\leq L_r\alpha(D),\quad
\forall D\subset P\cap T_r.
$$
\end{itemize}

\section{Preliminary lemmas}

\begin{lemma} \label{lem2.1}
Assume $\sum_{i=1}^{m-2}k_i=1$, then
for $h(t)\in C[I,E]$, the problem
\begin{gather}
y''(t)+\beta^2y(t)=h(t),\quad 0<t<1, \label{e2.1}\\
y'(0)=\theta,\quad y(1)=\sum_{i=1}^{m-2}k_i y(\xi_i ) \label{e2.2}
\end{gather}
has a unique solution
\begin{equation}
\begin{aligned}
y(t)&=\frac{1}{\beta}\int_0^t\sin
\beta(t-s)h(s)ds +\frac{\cos\beta t}{\beta a_0}[\int_0^1\sin
\beta(1-s)h(s)ds\\
&\quad -\sum_{i=1}^{m-2} k_i\int_0^{\xi_i}\sin
\beta(\xi_i-s)h(s)ds]\\
&=:\int_{0}^{1}G(t,s)h(s)ds,
\end{aligned} \label{e2.3}
\end{equation}
 where
$$
G(t,s)=\begin{cases}
\frac{1}{\beta}\sin\beta(t-s)+
\frac{\cos\beta t}{\beta a_0}[\sin\beta(1-s)
 -\sum _{j=i}^{m-2}k_{j}\sin\beta(\xi_{j}-s)],\\
 \quad \text{if }\xi_{i-1}\leq s\leq \min\{t,\xi_i \},; i=1,2,\dots,m-1;\\[4pt]
\frac{\cos\beta t}{\beta a_0}[\sin\beta(1-s)-\sum
_{j=i}^{m-2}k_{j}\sin\beta(\xi_{j}-s)],\\
\quad \text{if }\max\{\xi_{i-1},t\}\leq s\leq \xi_i ,\; i=1,2,\dots,m-1.
\end{cases}
$$
\end{lemma}

 The proof of the above lemma is easy, so we omit it.

\begin{lemma} \label{lem2.2}
There exist $c_1$, $c_2>0$ such that
 $$
c_1(1-s)\leq G(t,s)\leq c_2(1-s),\quad t,s\in[0,1].
$$
\end{lemma}

\begin{proof}
Take $H(t,s)=c(1-s)-G(t,s)$. We will prove that
$H(t,s)\geq 0$, $t,s\in[0,1]$, when $c$ is sufficiently large.
For $t,s\in[0,1]$, we have
\begin{align*}
H(t,s)
&\geq c(1-s)-\frac{1}{\beta}\sin\beta(t-s)-
 \frac{\cos\beta t}{\beta a_0}\sin\beta(1-s)\\
&\geq c(1-s)-\frac{1}{\beta}\sin\beta(1-s)-
 \frac{\sin\beta(1-s)}{\beta a_0}\\
&=c(1-s)-\frac{1}{\beta}[1+\frac{1}{a_0}]\sin\beta(1-s)\\
&\geq(c-1-\frac{1}{a_0})(1-s).
\end{align*}
Take $c_2\geq1+\frac{1}{a_0}$, then $H(t,s)\geq 0$, $t,s\in[0,1]$.

 Now, we prove $H(t,s)\leq 0$, $t,s\in[0,1]$, when $c$ is
sufficiently small.
For $t\in[0,1]$, $s\in(\xi_1,1]$, we have
\begin{align*}
H(t,s)
&\leq c(1-s)- \frac{\cos\beta t}{\beta a_0}[\sin\beta(1-s)-\sum
_{j=i}^{m-2}k_{j}\sin\beta(\xi_{j}-s)]\\
&\leq c(1-s)-\frac{\cos\beta}{\beta a_0}[\sin\beta(1-s)-\sum
_{j=2}^{m-2}k_{j}\sin\beta(\xi_{j}-s)]\\
&\leq c(1-s)-\frac{k_1\cos\beta}{\beta a_0}\sin\beta(1-s).
\end{align*}
Since
$$
g(x)=\begin{cases}
\frac{\sin x}{x}, & 0<x\leq \pi/2,\\
1, & x=0
\end{cases}
$$
is continuous on $[0,\pi/2]$. So, we obtain
$$
\min_{x\in[0,\pi/2]}g(x):=m_0>0;
$$
i.e., $\sin x\geq m_0x$, $x\in[0,\pi/2]$.
Therefore,
\begin{align*}
H(t,s)
&\leq c(1-s)-\frac{m_0k_1\cos\beta}{a_0}(1-s)\\
&=(c-\frac{m_0k_1\cos\beta}{a_0})(1-s).
\end{align*}
 For $t\in[0,1]$, $s\in[0,\xi_1]$, we obtain
\begin{align*}
H(t,s)
&\leq c(1-s)- \frac{\cos\beta t}{\beta a_0}[\sin\beta(1-s)-\sum
_{j=1}^{m-2}k_{j}\sin\beta(\xi_{j}-s)]\\
&\leq c(1-s)-\frac{\cos\beta}{\beta a_0}2\sum
_{j=1}^{m-2}k_{j}\cos\frac{\beta(1+\xi_{j}-2s)}{2}\sin\frac{\beta(1-\xi_{j})}{2}\\
&\leq c-\frac{2\cos\beta}{\beta a_0}\sum
_{i=1}^{m-2}k_i \cos\frac{\beta(1+\xi_i )}{2}\sin\frac{\beta(1-\xi_i )}{2}.\\
\end{align*}
Take
$$
0<c_1\leq\min\big\{\frac{m_0k_1\cos\beta}{a_0},
\frac{2\cos\beta(\sum_{i=1}^{m-2}k_i 
\cos\frac{\beta(1+\xi_i )}{2}\sin\frac{\beta(1-\xi_i )}{2})}{\beta a_0}
\big\}.
$$
Then we have $c_1(1-s)\leq G(t,s)$, $t,s\in[0,1]$.
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.3}
Assume {\rm (H1)} holds. If $h\in Q$, then the unique solution $y$
of  \eqref{e2.1}-\eqref{e2.2} satisfies $y(t)\geq \theta$, $t\in
I$ and $y(t)\geq \gamma y(s)$ for all $t,s\in I$, where
$\gamma={c_1}/{c_2}$.
\end{lemma}

\begin{proof}
Obviously, $y(t)\geq \theta$ for all $t\in I$.
By Lemma \ref{lem2.2}, we obtain
$$
y(t)\geq c_1 \int_{0}^{1}(1-s)h(s)ds
=\frac{c_1}{c_2} \int_{0}^{1}c_2(1-s)h(s)ds
\geq\gamma y(r),\quad \forall t,~r\in I.
$$
The proof is complete.
\end{proof}

Define an operator $A:Q\to C[I,E]$ as follows
\begin{equation}
\begin{aligned}
A(y(t))&:=\frac{1}{\beta}\int_0^t\sin
\beta(t-s)F(s,y(s))ds +\frac{\cos\beta t}{\beta a_0}[\int_0^1\sin
\beta(1-s)F(s,y(s))ds\\
&\quad -\sum_{i=1}^{m-2} k_i\int_0^{\xi_i}\sin
\beta(\xi_i-s)F(s,y(s))ds].
\end{aligned} \label{e2.4}
\end{equation}
By Lemmas \ref{lem2.1} and  \ref{lem2.3}, we obtain that
$A:Q\to C^2[I,E]\cap Q$, and $y(t)$ is a positive solution
of  \eqref{e1.1}-\eqref{e1.2} if
and only if $y(t)\in C^2[I,E]\cap Q$ and $y(t)\not\equiv \theta$ is
a fixed point of the operator $A$.

\begin{lemma} \label{lem2.4}
Assume {\rm (H1), (H2)} hold. Then, for any
$r>0$, the operator $A$ is a strict set contraction on $Q\cap
B_r$.
\end{lemma}

\begin{proof}
 Since $F(t,x)$ is uniformly continuous and bounded on
$I\times (P\cap T_r)$, we see from \eqref{e2.4} that $A$
is continuous and bounded on $Q\cap B_r$.
For any $S\subset Q\cap B_r$, by \eqref{e2.4}, we
can easily get that functions $A(S)=\{Ay|y\in S\}$ are uniformly
bounded and equicontinuous.
By \cite{l1}, we have
\begin{equation}
\alpha(A(S))=\sup_{t\in I}\alpha(A(S(t))),\label{e2.5}
\end{equation}
where $A(S(t))=\{Ay(t):y\in S,\; t\in I\text{ is fixed}\}$.
For any
$y\in C[I,E]$, $g\in C[I,I]$, by $\int_0^tg(s)y(s)ds\in
\overline{\rm co}(\{g(t)y(t)|t\in I\}\cup\{\theta\})
\subset\overline{\rm co}(\{y(t)|t\in I\}\cup\{\theta\})$, we obtain
\begin{align*}
&\alpha(A(S(t)))\\
&=\alpha(\{\frac{1}{\beta}\int_0^t\sin
\beta(t-s)F(s,y(s))ds +\frac{\cos\beta t}{\beta a_0}\Big[\int_0^1\sin
\beta(1-s)F(s,y(s))ds\\
&\quad -\sum_{i=1}^{m-2} k_i\int_0^{\xi_i}\sin
\beta(\xi_i-s)F(s,y(s))ds\Big]:y \in S\})\\
&\leq\frac{\sin\beta}{\beta}\alpha(\overline{\rm co}(\{F(s,y(s)):s\in
I,~y\in S\}\cup\{\theta\}))\\
&\quad +\frac{\sin\beta}{\beta
a_0}\alpha(\overline{\rm co}(\{F(s,y(s)):s\in I,\;y\in
S\}\cup\{\theta\}))\\
&\quad +\frac{\sum_{i=1}^{m-2}
k_i\sin\beta\xi_i}{\beta a_0}\alpha(\overline{\rm co}(\{F(s,y(s)):s\in
I,~y\in S\}\cup\{\theta\}))\\
&=\frac{a_1}{\beta a_0}\alpha(\{F(s,y(s)):s\in I,\; y\in S\})\\
&\leq\frac{a_1}{\beta a_0}\alpha(F(I\times B)),
\end{align*}
where $B=\{y(s):s\in I,~y\in S\}\subset P\cap T_r$.

By (H2), we obtain
\begin{equation}
\alpha(A(S(t)))\leq\frac{a_1}{\beta a_0}L_r\alpha(B).\label{e2.6}
\end{equation}
For any given $\varepsilon>0$, there exists a partition
$S=\cup_{j=1}^{l}S_j$ such that
\begin{equation}
\operatorname{diam}(S_j)<\alpha(S)+\frac{\varepsilon}{3},
\quad j=1,2,\dots,l.\label{e2.7}
\end{equation}
Now, choose $y_j\in S_j$, $j=1,2,\dots,l$ and a partition
$0=t_0<t_1<\dots<t_k=1$ such that
\begin{equation}
\|y_j(t)-y_j(\overline{t})\|<\frac{\varepsilon}{3},\quad
\forall t,\; \overline{t}\in[t_{i-1},t_i],\;
j=1,2,\dots,l,\; i=1,2,\dots,k.\label{e2.8}
\end{equation}
 Obviously,
$B=\cup_{j=1}^{l}\cup_{i=1}^{k}B_{ij}$, where
$B_{ij}=\{y(t):y\in S_j,t\in[t_{i-1},~t_i]\}$. For any
$y(t),\overline{y}(\overline{t})\in B_{ij}$, by \eqref{e2.7}
and \eqref{e2.8}, we obtain
\begin{align*}
\|y(t)-\overline{y}(\overline{t})\|
&\leq\|y(t)-y_j(t)\|+\|y_j(t)-y_j(\overline{t})\|
+\|y_j(\overline{t})-\overline{y}(\overline{t})\|\\
&\leq\|y-y_j\|_c+\frac{\varepsilon}{3}+\|y_j-\overline{y}\|_c\\
&\leq2\operatorname{diam}(S_j)+\frac{\varepsilon}{3}
<2\alpha(S)+\varepsilon,
\end{align*}
which implies $\operatorname{diam}(B_{ij})\leq 2\alpha(S)+\varepsilon$,
and so, $\alpha(B)\leq 2\alpha(S)+\varepsilon$. Since $\varepsilon$ is
arbitrary, we obtain
\begin{equation}
\alpha(B)\leq 2\alpha(S).\label{e2.9}
\end{equation}
It follows from \eqref{e2.5}, \eqref{e2.6} and \eqref{e2.9} that
$$
\alpha(A(S))\leq \frac{2a_1}{\beta a_0}L_r\alpha(S),\quad
\forall S\subset Q\cap B_r.
$$
 By (H2), we obtain that $A$ is a strict set contraction on
$Q\cap B_r$.
\end{proof}

\section{Main results}

 Let $K=\{y\in Q:y(t)\geq \gamma y(s),\; \forall t,\;s\in I\}$.
Clearly, $K\subset Q$ is a cone of $C[I,E]$. By
 Lemmas \ref{lem2.1} and \ref{lem2.3}, we obtain $AQ\subset K$. 
 So, $AK\subset K$.

For convenience, for any $x\in P$ and $\varphi\in P^*$, we set
\begin{gather*}
F^{0}=\limsup_{\|x\|\to 0}\sup_{t\in I}\frac{\|F(t,x)\|}{\|x\|},\quad
 F^{\infty}=\limsup_{\|x\|\to\infty}
\sup_{t\in I}\frac{\|F(t,x)\|}{\|x\|},\\
F^\varphi_{0}=\liminf_{\|x\|\to 0}\inf_{t\in
I}\frac{\varphi(F(t,x))}{\varphi(x)},\quad
F^\varphi_{\infty}=\liminf_{\|x\|\to
\infty}\inf_{t\in I}\frac{\varphi(F(t,x))}{\varphi(x)}
\end{gather*}
and list the following assumptions:
\begin{itemize}
\item[(H3)] There exists $\varphi\in P^*$ such that $\varphi(x)>0$ for
any $x>\theta$ and
$F^\varphi_{0}>\frac{\beta^2a_0}{\gamma a_2}$.

\item[(H4)] There exists $\varphi\in P^*$ such that $\varphi(x)>0$ for
any $x>\theta$ and
$F^\varphi_{\infty}>\frac{\beta^2a_0}{\gamma a_2}$.

\item[(H5)] $F^0<\frac{\beta^2a_0}{N(1+a_0)(1-\cos\beta)}$.

\item[(H6)] $F^\infty<\frac{\beta^2a_0}{N(1+a_0)(1-\cos\beta)}$.

\item[(H7)] There exists $r_0>0$ such that
$$
\sup_{\stackrel {t\in I,~x\in P}{ \gamma r_0/N\leq\|x\|\leq
r_0}}\|F(t,~x)\|<\frac{\beta^2a_0}{N(1+a_0)(1-\cos\beta)}r_0.
$$

\item[(H8)] There exist $R_0>0$ and $\varphi\in P^*$ with $\varphi(x)>0$
for any $x>\theta$ such that
$$
\inf_{\stackrel {t\in I, x\in P}{\gamma R_0/N\leq\|x\|\leq R_0}}
\frac{\varphi(F(t,x))}{\varphi(x)}>
\frac{\beta^2a_0}{\gamma a_2}.
$$
\end{itemize}

\begin{theorem} \label{thm3.1}
Assume {\rm (H1), (H2)} hold. If one of the
following conditions is satisfied:
\begin{itemize}
\item[(i)] {\rm (H4)} and {\rm (H5)} hold.

\item[(ii)] {\rm (H3)} and {\rm (H6)} hold.
\end{itemize}
Then the problem \eqref{e1.1}-\eqref{e1.2} has at least
one positive solution.
\end{theorem}

\begin{proof}
(i) By (H4), we obtain that there exist constants
$M>\frac{\beta^2a_0}{\gamma a_2}$ and $r_1>0$ such that
\begin{equation}
\varphi(F(t,x))\geq M\varphi(x),\quad
\forall t\in I, x\in P,\quad \|x\|>r_1.\label{e3.1}
\end{equation}
For any $R>N r_1/\gamma$, we will show that
\begin{equation}
Ay\not\leq y,\quad \forall y\in K,\; \|y\|_c=R.\label{e3.2}
\end{equation}
In fact, if not, there exists $y_0\in K$, $\|y_0\|_c=R$ such that
$Ay_0\leq y_0$. By
\begin{equation}
y_0(t)\geq \gamma y_0(s)\geq \theta,\quad \forall t,\;
s\in I,\label{e3.3}
\end{equation}
we have
\begin{equation}
\|y_0(t)\|\geq \frac{\gamma}{N}\|y_0\|_c>r_1,\quad
 \forall t\in I.\label{e3.4}
\end{equation}
By \eqref{e2.4}, for any $t\in I$, we have
\begin{align*}
A(y_0(t))
&=\frac{1}{\beta}\int_0^t\sin
\beta(t-s)F(s,y_0(s))ds +\frac{\cos\beta t}{\beta a_0}\Big[\int_0^1\sin
\beta(1-s)F(s,y_0(s))ds\\
&\quad -\sum_{i=1}^{m-2} k_i\int_0^{\xi_i}\sin
\beta(\xi_i-s)F(s,y_0(s))ds\Big]\\
&\geq\frac{\cos\beta t}{\beta
a_0}\sum_{i=1}^{m-2} k_i\int_{\xi_i}^1\sin
\beta(1-s)F(s,y_0(s))ds.
\end{align*}
This inequality, \eqref{e3.1}, \eqref{e3.3} and \eqref{e3.4}, imply
\begin{align*}
\varphi(Ay_0(0))
&\geq\frac{1}{\beta a_0}\sum_{i=1}^{m-2} k_i\int_{\xi_i}^1
\sin \beta(1-s)M\gamma \varphi(y_0(0))ds\\
&=\frac{a_2}{\beta^2a_0}M\gamma \varphi(y_0(0)).
\end{align*}
Considering $Ay_0\leq y_0$, we obtain
\begin{equation}
\varphi(y_0(0))\geq\frac{\gamma
a_2}{\beta^2a_0}M\varphi(y_0(0)).\label{e3.5}
\end{equation}
It is easy to see that $\varphi(y_0(0))>0$ (In fact, if
$\varphi(y_0(0))=0$, by \eqref{e3.3}, we obtain
$\varphi(y_0(0))\geq \gamma\varphi(y_0(s))\geq 0$ for all $s\in I$.
So, we have $\varphi(y_0(s))\equiv 0$ for all $s\in I$. That is,
$y_0(s)\equiv\theta$. This is a contradiction with $\|y_0\|_c=R$).
So, \eqref{e3.5} contradicts with $M>\frac{\beta^2a_0}{\gamma
a_2}$. Therefore, \eqref{e3.2} is true.

On the other hand, by (H5) and $F(t,\theta)=\theta$, we obtain that
there exist constants
$0<\varepsilon<\frac{\beta^2a_0}{N(1+a_0)(1-\cos\beta)}$
and  $0<r_2<R$ such that
\begin{equation}
\|F(t,x)\|\leq\varepsilon\|x\|,\quad \forall t\in I,\;
x\in P,\; \|x\|<r_2. \label{e3.6}
\end{equation}
For any $0<r<r_2$, we now prove that
\begin{equation}
Ay\not\geq y,\quad \forall y\in K,\; \|y\|_c=r.\label{e3.7}
\end{equation}
In fact, if not, there exists $y_0\in K$, $\|y_0\|_c=r$ such that
$Ay_0\geq y_0$. Since \eqref{e2.4} implies
\begin{equation}
Ay_0(t)\leq \frac{a_0+\cos\beta t}{\beta
a_0}\int_0^1\sin\beta(1-s)F(s,y_0(s))ds,~~\forall t\in
I.\label{e3.8}
\end{equation}
So, we have
$$
\theta\leq y_0(t)\leq\frac{a_0+\cos\beta t}{\beta
a_0}\int_0^1\sin\beta(1-s)F(s,y_0(s))ds,\quad \forall t\in I.
$$
This, together with \eqref{e3.6}, imply
$$
\|y_0(t)\|\leq\frac{N(1+a_0)\varepsilon}{\beta a_0}\int_0^1\sin\beta(1-s)\|y_0(s)\|ds
=\frac{N(1+a_0)(1-\cos\beta)\varepsilon\|y_0\|_c}{\beta^2a_0},
$$
for all $t\in I$. Therefore, we obtain $\varepsilon\geq
\beta^2a_0/N(1+a_0)(1-\cos\beta)$. This is a contradiction. So,
\eqref{e3.7} is true.

By \eqref{e3.2}, \eqref{e3.7}, Lemma \ref{lem2.4}
 and Theorem \ref{thm1.1}, we obtain that the operator
$A$ has at least one fixed point $y\in K$ satisfying
$r<\|y\|_c<R$.

(ii)  By (H3), in the same way as establishing \eqref{e3.2}
we can assert that there exists $r_2>0$ such that for any $0<r<r_2$,
\begin{equation}
Ay\not\leq y,\quad \forall y\in K,\; \|y\|_c=r.\label{e3.9}
\end{equation}
On the other hand, by (H6), we obtain that there exist constants
$r_1>0$ and $\varepsilon$, with
$0<\varepsilon<\frac{\beta^2a_0}{N(1+a_0)(1-\cos\beta)}$, such
that
$$
\|F(t,x)\|\leq\varepsilon\|x\|,\quad \forall t\in I,\;
x\in P,\; \|x\|>r_1.
$$
By (H2), we obtain
$$
\sup_{t\in I,\, x\in P\cap T_{r_1}} \|F(t,x)\|=:b<\infty.
$$
So, we have
\begin{equation}
\|F(t,x)\|\leq\varepsilon\|x\|+b,\quad \forall t\in I,\;
x\in P.\label{e3.10}
\end{equation}
Take
$$
R>\max\big\{r_2,\;
\frac{Nb(1+a_0)(1-\cos\beta)}{\beta^2a_0-N\varepsilon(1+a_0)
(1-\cos\beta)}\big\},
$$
we will prove that
\begin{equation}
Ay\not\geq y,\quad \forall y\in K,\; \|y\|_c=R. \label{e3.11}
\end{equation}
In fact, if there exists $y_0\in K$, $\|y_0\|_c=R$ such that
$Ay_0\geq y_0$. Then, by \eqref{e3.8} and \eqref{e3.10}, we obtain
\begin{align*}
\|y_0(t)\|&\leq\frac{N(a_0+\cos\beta t)}{\beta
a_0}\int_0^1\sin\beta(1-s)(\varepsilon\|y_0(s)\|+b)ds\\
&\leq\frac{N(1+a_0)(1-\cos\beta)}{\beta^2a_0}
(\varepsilon\|y_0\|_c+b),\quad \forall t\in I.
\end{align*}
So, we have
$$
\|y_0\|_c\leq\frac{Nb(1+a_0)(1-\cos\beta)}
{\beta^2a_0-N\varepsilon(1+a_0)(1-\cos\beta)}<R.
$$
A contradiction. Therefore, \eqref{e3.11} holds.

By \eqref{e3.9}, \eqref{e3.11}, Lemma \ref{lem2.4}
 and Theorem \ref{thm1.1}, we obtain that the
operator $A$ has at least one fixed point $y\in K$ satisfying
$r<\|y\|_c<R$. The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.2}
Assume {\rm (H1), (H2)} hold. If one of the
following conditions is satisfied:
\begin{itemize}
\item[(i)] {\rm (H3), (H4), (H7)} hold.

\item[(ii)] {\rm (H5), (H6), (H8)} hold.
\end{itemize}
Then  \eqref{e1.1}-\eqref{e1.2} has at least two positive solutions.
\end{theorem}

\begin{proof}
(i)  By (H3), (H4) and the proof of
Theorem \ref{thm3.1}, we obtain that there exist $r,R$ with $0<r<r_0<R$ such
that
\begin{gather*}
Ay\not\leq y,\quad \forall y\in K,\; \|y\|_c=r,\label{e3.12}\\
Ay\not\leq y,\quad \forall y\in K,\; \|y\|_c=R.\label{e3.13}
\end{gather*}
Now, we  prove that
\begin{equation}
Ay\not\geq y,\quad \forall y\in K,\; \|y\|_c=r_0.\label{e3.14}
\end{equation}
In fact, if there exists $y_0\in K$, $\|y_0\|_c=r_0$ such that
$Ay_0\geq y_0$. By \eqref{e3.8} and (H7), we obtain
$$
\|y_0\|_c<\frac{N(1+a_0)}{\beta
a_0}\int_0^1\sin\beta(1-s)\frac{\beta^2a_0}{N(1+a_0)
(1-\cos\beta)}r_0ds=r_0.
$$
A contradiction. So, \eqref{e3.14} is true. By Lemma \ref{lem2.4}
and Theorem \ref{thm1.1}, we obtain that the operator $A$ has at
least two fixed points $y_1,~y_2\in K$ satisfying
$r<\|y_1\|_c<r_0<\|y_2\|_c<R$.

(ii) By (H5), (H6) and the proof of Theorem \ref{thm3.1},
we obtain that there exist $r,R$ with $0<r<R_0<R$ such that
\begin{gather*}
Ay\not\geq y,\quad \forall y\in K,\; \|y\|_c=r,\label{e3.15}\\
Ay\not\geq y,\quad \forall y\in K,\; \|y\|_c=R.\label{e3.16}
\end{gather*}
On the other hand, by (H8) and the same way as used in the proof
of \eqref{e3.2}, we can prove that
\begin{equation}
Ay\not\leq y,\quad \forall y\in K,\; \|y\|_c=R_0.\label{e3.17}
\end{equation}
By Lemma \ref{lem2.4} and Theorem \ref{thm1.1}, we obtain that the
operator $A$ has at least two fixed points $y_1,y_2\in K$
satisfying $r<\|y_1\|_c<R_0<\|y_2\|_c<R$. The proof is complete.
\end{proof}

Similar to the proofs of Theorem \ref{thm3.1} and
Theorem \ref{thm3.2}, we can
easily get the following corollaries.

\begin{corollary} \label{coro3.1}
Assume {\rm (H1), (H2)} hold. If one of the
following conditions is satisfied:
\begin{itemize}
\item[(i)] {\rm (H4), (H5), (H7), (H8)} hold with
$R_0<\gamma r_0/N$.

\item[(ii)] {\rm (H3), (H6), (H7), (H8)} hold with $r_0<\gamma R_0/N$.
\end{itemize}
Then  \eqref{e1.1}-\eqref{e1.2} has at least three positive
solutions.
\end{corollary}

\begin{corollary} \label{coro3.2}
Assume {\rm (H1), (H2)} hold. If one of the
following conditions is satisfied:
\begin{itemize}
\item[(i)] {\rm (H5)--(H7)} hold, and there exist
$R_i>0$, $\varphi_i\in P^*$ with $\varphi_i(x)>0$ for
$x>\theta$, $i=1,2$ such that
$$
\inf_{t\in I,\, x\in P,\, \gamma R_i/N\leq\|x\|\leq R_i}
\frac{\varphi_i(F(t,x))}{\varphi_i(x)}>
\frac{\beta^2a_0}{\gamma a_2},\quad i=1,2,
$$
where $R_1<\gamma r_0/N$, $r_0<\gamma R_2/N$.

\item[(ii)] {\rm (H3), (H4), (H8)} hold, and there exist
$r_1, r_2>0$ such that
$$
\sup_{t\in I,\,x\in P,\, \gamma r_i/N\leq\|x\|\leq r_i}
\|F(t,~x)\|<\frac{\beta^2a_0}{N(1+a_0)(1-\cos\beta)}r_i,\quad i=1,2,
$$
where $r_1<\gamma R_0/N$, $R_0<\gamma r_2/N$.
\end{itemize}
Then \eqref{e1.1}-\eqref{e1.2} has at least four positive solutions.
\end{corollary}

We can prove easily  the existence of multiple positive
solutions for \eqref{e1.1}-\eqref{e1.2}.

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\end{document}