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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 111, 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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/111\hfil Positive solutions]
{Positive solutions for an $m$-point \\ boundary-value problem}

\author[L. X. Truong, L. T. P. Ngoc, N. T. Long\hfil EJDE-2008/111\hfilneg]
{Le Xuan Truong, Le Thi Phuong Ngoc, Nguyen Thanh Long}

\address{Le Xuan Truong  \newline
University of Technical Education in HoChiMinh City, 01 Vo Van
Ngan Str., Thu Duc Dist., HoChiMinh City, Vietnam}
\email{lxuantruong@gmail.com}

\address{Le Thi Phuong Ngoc \newline
Nhatrang Educational College, 01 Nguyen Chanh Str.,
Nhatrang City, Vietnam}
\email{ngoc1966@gmail.com}

\address{Nguyen Thanh Long \newline
Department of Mathematics and Computer Science,
University of Natural Science, Vietnam National University HoChiMinh City,
227 Nguyen Van Cu Str., Dist. 5, HoChiMinh City, Vietnam}
\email{longnt@hcmc.netnam.vn, longnt2@gmail.com}

\thanks{Submitted April 22, 2008. Published August 15, 2008.}
\subjclass[2000]{34B07, 34B10, 34B18, 34B27}
\keywords{Multi-point boundary; positive solution; \hfill\break\indent
Guo-Krasnoselskii fixed point theorem}

\begin{abstract}
In this paper, we obtain sufficient conditions for the existence of
a positive solution, and infinitely many positive solutions, of the
$m$-point boundary-value problem
\begin{gather*}
  x''(t) = f(t, x(t)), \quad 0 < t < 1,  \\
  x'(0) = 0, \quad x(1)=\sum_{i=1}^{m-2}\alpha _{i}x(\eta _{i})\,.
\end{gather*}
Our main tools are the  Guo-Krasnoselskii's fixed point theorem
and the monotone iterative technique.
We also show that the set of positive solutions is compact.
\end{abstract}

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

\section{Introduction}

The existence and multiplicity of positive solutions for
boundary-value problems have been extensively studied by many authors
using various techniques, such  fixed point theorem in cones, the
nonlinear alternative of Leray-Schauder,
the Leggett-William's fixed point theorem,
monotone iterative techniques.
We refer the reader to the references in this article and
the references therein for the
results of multi-point boundary-value problems.

Han \cite{h1}  studied the existence of positive solutions for the
three-point boundary-value problem at resonance
\begin{gather}\label{1.01}
    x''(t) = f(t, x(t)), \quad 0 < t < 1, \\
\label{1.02}
    x'(0) = 0, \quad x(\eta) = x(1),
\end{gather}
where $\eta \in (0, 1)$. The main tool is the fixed point theorem in cones.
By the same method, Long and Ngoc \cite{l1} have studied the equation
\eqref{1.01} together with the boundary conditions
\begin{equation}\label{1.03}
    x'(0) = 0, \quad x(1) = \alpha x(\eta),
\end{equation}
where $\alpha$ and $\eta$ in $(0,1)$.
The authors proved the existence of positive solutions and established
the compactness of  the set of positive solutions.

Based on the above works, we investigate the $m$-point boundary-value
problem consisting of the equation $\eqref{1.01}$ together with
the boundary conditions
\begin{equation}\label{1.04}
    x'(0) = 0, \quad x(1)=\sum_{i=1}^{m-2}\alpha _{i}x(\eta _{i}),
\end{equation}
where $m\geq 3$, $0 < \eta _{1} < \eta _{2} < \dots < \eta _{m-2} < 1$
and $\alpha_{i} \geq 0$, for all $i = 1, 2, \dots m -2$ such that
$\sum_{i=1}^{m-2}\alpha _{i} < 1$. We shall establish the existence
and multiplicity of positive solutions by applying well-know
Guo-Krasnoselskii's fixed point theorem and applying the monotone
iterative technique.

Let $\beta \in (0, \frac{\pi}{2})$. Obviously, problem \eqref{1.01},
\eqref{1.04} is equivalent to the problem
\begin{gather}\label{2.01}
    x''(t) + \beta^{2}x(t) = g(t, x(t)), \\
\label{2.02}
    x'(0) = 0, \quad x(1)=\sum_{i=1}^{m-2}\alpha _{i}x(\eta _{i}),
\end{gather}
where
\begin{equation}\label{2.03}
    g(t, x) = f(t, x) + \beta^{2}x.
\end{equation}

In this paper, we sue the following assumptions:
\begin{itemize}
\item [(H1)] $\sum_{i=1}^{m-2}\alpha_{i}\cos{\beta\eta_{i}} - \cos{\beta} > 0$;

\item [(H2)] $f: [0, 1] \times [0, +\infty) \to \mathbb{R}$ is a continuous
function such that
\begin{equation}\label{2.04}
    f(t, x) \geq - \beta^{2}x, \forall t\in [0,1], x \in [0, +\infty);
\end{equation}

\item [(H2')] The function $f(t,x)$ is nondecreasing in $x$ and satisfy
(H2)
\end{itemize}

We put:
\begin{gather*}
K_{m} = \frac{1}{\sum_{i=1}^{m-2}\alpha_{i}\cos{\beta\eta_{i}}
 - \cos{\beta}};\\
 M = \frac{\sin{\beta}}{\beta}(1+K_{m});\\
 M_{0} = \frac{K_{m}\cos{\beta}}{\beta}
 \Big(1-\sum_{i=1}^{m-2}\alpha_{i}\Big)\sin{\beta(1- \eta_{m-2})}.
\end{gather*}
The main results for the existence and multiplicity of positive solutions
are the following theorems, in which the operator $T$ and constant $c$,
 $ 0<c<1$ will be defined in next section.
Applying well-know Guo-Krasnoselskii's fixed point theorem, we obtain
the following result.

\begin{theorem} \label{thm1.1}
    Let {\rm (H1)--(H2)} hold. If there exist two constants
$R_{1}, R_{2}$ such that $R_{1} < cR_{2}$ and one the following
two conditions is satisfied:
\begin{equation} \label{a1}
\begin{gathered}
 f(t,x)+\beta^{2}x \leq \frac{R_1}{M}, \quad \forall (t,x)
 \in [0,1]\times [cR_{1},R_{1}],\\
 f(t,x)+\beta^{2}x \geq \frac{R_{2}}{M_{0}\eta_{m-2}}, \quad
  \forall (t,x) \in [0,1]\times [cR_{2},R_{2}],
\end{gathered}
\end{equation}
or
\begin{equation}\label{a2}
\begin{gathered}
 f(t,x)+\beta^{2}x \geq \frac{R_1}{M_{0}\eta_{m-2}}, \quad
  \forall (t,x) \in [0,1]\times [cR_{1},R_{1}],\\
 f(t,x)+\beta^{2}x \leq \frac{R_{2}}{M}, \quad
 \forall (t,x) \in [0,1]\times [cR_{2},R_{2}].
\end{gathered}
\end{equation}
Then the boundary-value problem \eqref{2.01}-\eqref{2.03}
has a positive solution.
\end{theorem}

Using the monotone iterative technique, we have the following result.

\begin{theorem}  \label{thm1.2}
    Let {\rm (H1),  (H2')} hold. Suppose there exist two positive
numbers $R_{1} < R_{2}$ such that
\begin{equation}
    \sup_{t\in [0,1]}g(t, R_{2}) \leq \frac{R_{2}}{M}, \quad
 \inf_{t\in [0,1]}g(t,cR_{1}) \geq \frac{R_{1}}{M_{0}\eta_{m-2}}.
\end{equation}
Then  problem \eqref{2.01}-\eqref{2.03} has positive solutions
$x_{1}^{*}$, $x_{2}^{*}$,
$x_{1}^{*}$ and $x_{2}^{*}$ may coincide with
\[
    R_{1}\leq \|x_{1}^{*}\| \leq R_{2} \quad \text{and}\quad
\lim_{n\to +\infty}T^{n}x_{0} = x_{1}^{*}, \quad
\text{where }  x_{0}(t) = R_{2}, t\in [0,1],
\]
and
\[
    R_{1}\leq \|x_{2}^{*}\| \leq R_{2} \quad \text{and}\quad
 \lim_{n\to +\infty}T^{n}\widehat{x}_{0} = x_{2}^{*}, \quad \text{where }
 \widehat{x}_{0}(t) = R_{1}, t\in [0,1].
\]
\end{theorem}

Clearly, in the above theorem, we not only obtain the existence as
in Theorem \ref{thm1.1}, but also we establish a sequence which converges to a
solution of problem \eqref{2.01}-\eqref{2.03}.

This paper consists of five sections. In Section 2, we present the
lemmas that will be used to prove the  existence results.
The proofs and two corollaries of  Theorems \ref{thm1.1}, \ref{thm1.2}
 will be given
in Section 3. In Section 4, we give sufficient conditions for
existence of infinitely many positive solutions, furthermore,
an example is also given here. Finally, in section 5, we show that
the set of positive solutions is compact.

\section{Preliminaries}

Consider the Banach spaces $C[0,1]$ and $C^2[0,1]$ equipped with the norms
\begin{gather*}
    \|x\|=\max\{|x(t)|: 0\leq t\leq 1\},\\
    \|x\|_{2}=\max\{\|x\|, \|x'\|, \|x''\|\},
\end{gather*}
respectively. We define a linear operator
$L: D(L) \subset C^{2}[0,1]] \to C[0, 1]$ by setting
\begin{equation}\label{2.05}
    Lx:= x'' + \beta^{2}x,
\end{equation}
in which $D(L) = \{x \in C^{2}[0, 1]: x'(0) =0, \;
  x(1)=\sum_{i=1}^{m-2}\alpha _{i}x(\eta _{i})\}$.
We shall proceed with some properties of the inverse operator of $L$.

\begin{lemma} \label{lem2.1}
    Let $\beta \in (0, \frac{\pi}{2})$. Then for each
$h \in C[0, 1]$, there a unique function $x = A(h) \in D(L)$ such
that $ Lx = h$ in $(0, 1)$. The function $A(h)$ is defined by
\begin{equation}\label{2.06}
    Ah(t) = \int_{0}^{1}G(t, s)h(s)ds,
\end{equation}
where
\begin{equation}\label{2.07}
\begin{aligned}
 G(t,s) &= \begin{cases}
  \frac{1}{\beta}\sin{\beta(t-s)}, & 0\leq s\leq t\leq 1,  \\
   0, & 0\leq t \leq s\leq 1
  \end{cases} \\
  &\quad+\frac{K_{m}}{\beta}\cos{\beta t}
\begin{cases}
        \sin{\beta(1-s)} - \sum_{i=1}^{m-2}\alpha_{i}\sin{\beta(\eta_{i}-s)},
& 0\leq s\leq \eta_{1}, \\
        \sin{\beta(1-s)} - \sum_{i=2}^{m-2}\alpha_{i}\sin{\beta(\eta_{i}-s)},
& \eta_{1}\leq s\leq \eta_{2}, \\
        \sin{\beta(1-s)} - \sum_{i=3}^{m-2}\alpha_{i}\sin{\beta(\eta_{i}-s)},
& \eta_{2}\leq s\leq \eta_{3}, \\
        \dots \\
        \sin{\beta(1-s)} - \sum_{i=k}^{m-2}\alpha_{i}\sin{\beta(\eta_{i}-s)},
& \eta_{k-1}\leq s\leq \eta_{k}, \\
        \dots \\
        \sin{\beta(1-s)},& \eta_{m-2}\leq s\leq 1.
\end{cases}
\end{aligned}
\end{equation}
\end{lemma}

\begin{lemma} \label{lem2.2}
Let $\beta \in (0, \frac{\pi}{2})$. We have
    \begin{itemize}
\item[(i)] The operator $A: C[0, 1] \to C[0, 1]$ is completely continuous
linear operator.
\item[(ii)] For any positive function $h\in C[0,1]$, the function $Ah$
is also positive.
\end{itemize}
\end{lemma}

 The proof of Lemmas \ref{lem2.1}, \ref{lem2.2} are straightforward and
we will omit them.
Now, we shall establish some estimations for the Green function $G(t,s)$.

\begin{lemma} \label{lem2.3}
    Let $\beta \in (0, \frac{\pi}{2})$ and (H1) hold. Then
\begin{itemize}
\item [(i)] $0\leq G(t,s) \leq M $ for all $(t, s) \in [0,1] \times [0,1]$.
\item[(ii)] $G(t, s) \geq M_{0}, \quad \text{for all} \quad (t, s) \in [0,1]
\times [0,\eta_{m-2}]$.
\item [(iii)] There exist a constant $c\in (0,1)$ and a continuous function
$\Phi: [0,1] \to [0, +\infty)$ such that
\begin{equation*}
        c\Phi(s) \leq G(t,s) \leq \Phi(s), \quad \text{for all }
 t, s\in [0,1].
    \end{equation*}
\end{itemize}
\end{lemma}

\begin{proof} \textbf{Part (i).}
 From the Green function $G(t,s)$, we deduce that
\begin{equation}\label{2.08}
    0\leq G(t,s) \leq \frac{\sin{\beta}}{\beta}(1+K_{m}) \equiv M, \quad
 \forall (t, s) \in [0,1] \times [0,1].
\end{equation}
\textbf{Part (ii).}   Put $\eta_{0} = 0$, $\eta_{m-1} = 1$.
For all $t\in [0,1]$ and $s\in [\eta_{k-1}, \eta_{k}]$, we have
\begin{equation} \label{2.09}
\begin{aligned}
    G(t, s) &\geq \frac{K_{m}\cos{\beta t}}{\beta}
\Big[\sin{\beta(1-s)} - \sum_{i=k}^{m-2}\alpha_{i}\sin{\beta(\eta_{i}-s)}\Big]   \\
&\geq  \frac{K_{m}\cos{\beta t}}{\beta}\Big[\sin{\beta(1-s)} -
\sum_{i=k}^{m-2}\alpha_{i}\sin{\beta(1-s)}\Big] \\
&\geq  \frac{K_{m}\cos{\beta t}}{\beta}\Big(1-\sum_{i=k}^{m-2}\alpha_{i}\Big)
\sin{\beta(1-s)}  \\
&\geq  \frac{K_{m}\cos{\beta}}{\beta}\Big(1-\sum_{i=1}^{m-2}\alpha_{i}\Big)
\sin{\beta(1-\eta_{m-2})}\equiv M_{0}  .
\end{aligned}
\end{equation}
Since the above inequality  holds  for  $k = 1, 2, \dots, m-2$,
the proof part (ii) is complete.

\noindent\textbf{Part (iii).} Let
\begin{equation*}
    H(t, s) = \mu(1-s) - G(t, s).
\end{equation*}
We shall show that when $\mu>0$ sufficiently large,
\begin{equation}\label{2.10}
    H(t, s) \geq 0, \quad \forall (t, s) \in [0,1]\times [0,1],
\end{equation}
and that when $\mu>0$ sufficiently small,
\begin{equation}\label{2.11}
    H(t, s) \leq 0, \quad \forall (t, s) \in [0,1]\times [0,1].
\end{equation}

To prove  \eqref{2.10}, we use that from \eqref{2.07},  for all
$t, s \in [0,1]$,
\begin{equation}\label{2.12}
    G(t, s) \leq \frac{1}{\beta}\sin{\beta(1-s)}
+ \frac{K_{m}}{\beta}\sin{\beta(1-s)}\leq \left(K_{m}+1\right)(1-s);
\end{equation}
therefore
\begin{equation}\label{2.13}
    H(t, s) \geq \mu(1-s) - \left(K_{m}+1\right)(1-s) = \left(\mu - K_{m} - 1\right)(1-s).
\end{equation}
So, if we choose $\mu \equiv \mu_{1} \geq K_{m}+1$ then $H(t,s) \geq 0$,
 for all $t, s \in [0,1]$.

To prove of  \eqref {2.11}, we consider two cases:


\textbf{Case 1: $s\in [0, \eta_{m-2}]$.}
 From \eqref{2.09} we can deduce that, for all $t \in [0,1]$,
\begin{equation} \label{2.14}
    H(t, s) =\mu(1-s) - G(t, s) \leq \mu (1-s) - M_{0} \leq \mu -M_{0}.
\end{equation}
So, for $\mu \leq M_{0}$, we have $H(t, s)\leq 0$, for all $t \in [0,1]$,
$s\in [0, \eta_{m-2}]$.

\textbf{Case 2: $s \in [\eta_{m-2}, 1]$.}
 The function $z \mapsto \frac{\sin z}{z}$ is decreasing on $(0, \pi]$,
so we obtain
\[
    \frac{\sin \beta(1-s)}{\beta(1-s)} \geq
\frac{\sin \beta(1-\eta_{m-2})}{\beta(1-\eta_{m-2})}.
\]
Therefore,
\begin{equation} \label{2.15}
\begin{aligned}
    H(t,s)&=\mu(1-s) - G(t,s)  \\
    &\leq \mu(1-s) -\frac{K_{m}\cos \beta}{\beta}\sin\beta(1-s)  \\
    & \leq  \mu(1-s) - K_{m}\cos \beta \frac{\sin\beta(1-s)}{\beta(1-s)}(1-s)  \\
    & \leq  \Big[\mu - K_{m}\cos \beta \frac{\sin\beta(1
 -\eta_{m-2})}{\beta(1-\eta_{m-2})}\Big](1-s).
\end{aligned}
\end{equation}
If we choose
$\mu \leq K_{m}\cos \beta \frac{\sin\beta(1-\eta_{m-2})}{\beta(1-\eta_{m-2})}
\equiv M_{1}$, then $H(t,s)\leq 0$, for all $t\in [0,1]$,
$s\in [\eta_{m-2},1]$.
\smallskip

Hence, for $\mu \equiv \mu_{2} \leq \min\{M_0, M_1\}= M_0$, we have
$H(t,s)\leq 0$, for all $t, s\in [0,1]$.
Finally, by letting $\Phi(s) = \mu_{1}(1-s)$ and $c = \frac{\mu_{2}}{\mu_{1}}$,
 the part $(iii)$ of this lemma is proved.
\end{proof}

Let $K$ be the cone in $C[0,1]$, consisting of all nonnegative functions and
\[
    P=\{x\in K: x(t) \geq c\|x\|, \; \forall t \in [0,1] \}.
\]

It is clear that $P$ is also a cone in $C[0,1]$. For each $x\in P$, denote
$F(x)(t) = g(t, x(t))$, $t\in [0,1]$. From the assumption (H2) we deduce
that the operator $F: P \to K$ is continuous. Therefore, the operator
$T \equiv A\circ F : P \to K$ is a completely continuous.
On the other hand, for $x \in P$, by Lemma \ref{lem2.3} we have
\begin{gather}\label{2.16}
    Tx(t) = \int_{0}^{1}G(t,s)F(x)(s)ds \geq c\int_{0}^{1}\Phi(s)F(x)(s)ds,\\
\label{2.17}
    \|Tx\| =\max_{0\leq t\leq 1}\int_{0}^{1}G(t,s)F(x)(s)ds \leq \int\Phi(s)F(x)(s)ds,
\end{gather}
which implies
\begin{equation}\label{2.18}
    Tx(t) \geq c\|Tx\|.
\end{equation}
Hence, we have the following result.

\begin{lemma} \label{lem2.4}
        The operator $T \equiv A\circ F : P \to P$ is a completely
continuous operator.
\end{lemma}

It is easy to verify the nonzero fixed points of the operator $T$ are
the positive solutions of the problem \eqref{2.01}-\eqref{2.03}.

\section{Proofs and  corollaries of  main Theorems}

At first, by using the same method as in \cite{h1} and the monotone iterative
technique, combining with Lemmas \ref{lem2.1}--\ref{lem2.4}, we prove
Theorems \ref{thm1.1} and \ref{thm1.2}.
For the convenience of the reader, let us state  the following
Guo-Krasnoselskii's fixed point theorem \cite{g1}.

\begin{theorem}[Guo-Krasnoselskii] \label{thmGK}
  Let $X$  be a Banach space, and let  $P \subset X$  be a cone.
Assume  $\Omega_{1}$, $\Omega_{2}$  are two open bounded subsets of
 $X$  with  $0\in \Omega_{1}$, $\overline{\Omega}_{1} \subset \Omega_{2}$
 and let $T: P\cap (\overline{\Omega}_{2} \backslash \Omega_{1}) \to P$
be a completely continuous operator such that
\begin{itemize}
\item [(i)] $\|Tu\| \leq \|u\|$, $u \in P \cap \partial\Omega_{1}$, and
 $\|Tu\| \geq \|u\|$, $u \in P \cap \partial\Omega_{2}$, or
\item [(ii)] $\|Tu\| \geq \|u\|$, $u \in P \cap \partial\Omega_{1}$, and
 $\|Tu\| \leq \|u\|$, $u \in P \cap \partial\Omega_{2}$.
\end{itemize}
 Then $T$  has a fixed point in
$P\cap (\overline{\Omega}_{2} \backslash \Omega_{1})$.
\end{theorem}

\begin{proof}[Proof of  Theorem \ref{thm1.1}]
Let
$$
\Omega_{1}=\{ x\in C[0,1]| : \|x\|<R_1 \}, \quad
\Omega_{2}=\{ x\in C[0,1]|: \|x\|<R_2\} \,.
$$
Then $\Omega _{1},\Omega _{2}$ are open bounded subsets of
$C[0,1]$ with $ 0 \in \Omega _{1},\overline{\Omega _{1}}\subset \Omega _{2}$.

\textbf{Case \eqref{a1}.}
For $ x\in P $ with $ \|x\|=R_1, $ we have
$$
g(s, x(s)) = f(s, x(s)) +\beta
^{2} x(s) \leq \frac {R_1} {M} = \frac {\|x\|} {M}.
$$
So
\[
\|Tx\| = \max_{t\in \lbrack 0,1]} \int_{0}^{1}G(t,s)g(s, x(s))ds
\leq \frac {\|x\|} {M} \,ax_{t\in \lbrack 0,1]} \int_{0}^{1}G(t, s)ds
\leq \|x\|\,.
\]
This implies
\begin{equation} \label{eq3.01a}
\qquad \|Tx\|\leq \|x\|,\quad \forall x\in P\cap \partial \Omega _{1}.
\end{equation}
On the other hand, for $ x\in P $ with $ \|x\|=R_2$, we have
\begin{align*}
T x(t) &=  \int_{0}^{1}G(t,s) \Big ( f(s, x(s)) +\beta
^{2} x(s) \Big) ds \\
&\geq \frac {R_2} {M_0 \eta_{m-2}} \int_{0}^{\eta_{m-2}}G(t,s) ds
\geq R_2 = \|x\|,
\end{align*}
accordingly
\begin {equation} \label{eq3.01b}
\qquad \|Tx\| \geq \|x\|, \quad \forall x\in P\cap \partial \Omega _{2}.
\end {equation}
By \eqref{eq3.01a}-\eqref{eq3.01b} and the first part of
Theorem \ref{thmGK}, it follows that $T$ has a fixed point $x_0$ in
$P\cap (\overline{\Omega}_{2} \backslash \Omega_{1})$,
such $ x_0 $ is a positive solution of \eqref{2.01}-\eqref{2.03}.

\textbf{Case \eqref{a2}.}  In this case, using the same method, by the second
part of Theorem \ref{thmGK}, we obtain the same result as above.

The proof is complete.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1.2}]
We define
\[
    P_{[R_{1}, R_{2}]}=\{x\in P: R_{1}\leq \|x\| \leq R_{2}\}.
\]
Let $x\in P_{[R_{1}, R_{2}]}$, then
$cR_{1}\leq c\|x\| \leq x(t) \leq \|x\| \leq R_{2}$, for all
$t\in [0,1]$. So, we have
\begin{gather*}
    Tx(t) = \int_{0}^{1}G(t,s)g(s,x(s))ds
\leq \frac{R_{2}}{M}\int_{0}^{1}G(t,s)ds \leq R_{2},
\\
    Tx(t) = \int_{0}^{1}G(t,s)g(s,x(s))ds \geq
\int_{0}^{1}G(t,s)g(s,cR_{1})ds \geq R_{1},
\end{gather*}
which implies  $TP_{[R_{1}, R_{2}]} \subset P_{[R_{1}, R_{2}]}$.

Now, let $x_{0}(t) = R_{2}, t\in [0,1]$ then $x_{0}\in P_{[R_{1}, R_{2}]}$.
We put
\begin{equation}\label{eq3.01}
    x_{n+1}= Tx_{n} = T^{n+1}x_{0}, \quad n = 1, 2, \dots
\end{equation}
Since $TP_{[R_{1}, R_{2}]} \subset P_{[R_{1}, R_{2}]}$ we have
$x_{n} \in P_{[R_{1}, R_{2}]}$, for all $n \in \mathbb{Z}_{+}$.
By the Lemma \ref{lem2.4}, we can deduce that there exists a subsequence
$\{x_{n_k}\}$ of $\{x_{n}\}$ such that
\begin{equation}\label{eq3.02}
    \lim_{k\to +\infty}x_{n_k} = x_{1}^{*}\in P_{[R_{1}, R_{2}]}.
\end{equation}
On the other hand, from the assumption (H2'), it is clear that
$T: P_{[R_{1}, R_{2}]} \to P_{[R_{1}, R_{2}]}$ is nondecreasing.
Therefore, since
\[
    0\leq x_{1}(t) \leq \|x_1\| \leq R_{2} = x_{0}(t), t\in [0,1],
\]
we have
$Tx_{1}\leq Tx_{0}, \quad i.e., x_{2} \leq x_{1}$.
By induction, then
\begin{equation}\label{eq3.03}
    x_{n+1} \leq x_{n}, \quad \text{for all } n=1, 2, \dots.
\end{equation}
Combining \eqref{eq3.02} and \eqref{eq3.03}, we obtain
    \begin{equation}\label{eq3.04}
    \lim_{n\to +\infty}x_{n} = x_{1}^{*}.
\end{equation}
Letting $n \to +\infty$ in \eqref{eq3.01} yields $Tx_{1}^{*} =x_{1}^{*}$.

Let $\widehat{x}_{0}(t) = R_{1}$,$ t\in [0,1]$ and
$\widehat{x}_{n+1}=T\widehat{x}_{n}$, $n=1, 2, \dots$. We have
$\widehat{x}_{0}\in P_{[R_{1}, R_{2}]}$ which implies that
$\widehat{x}_{n} \in P_{[R_{1}, R_{2}]}$, for all $n \in \mathbb{Z}_{+}$.
Moreover, from the assumptions of Theorem \ref{thm1.2} and from the definition
 of the operator $T$,
\[
    \widehat{x}_{1}(t)=T\widehat{x}_{0}(t)\geq \int_{0}^{1}
G(t,s)g(s,cR_{1})ds\geq R_{1}=\widehat{x}_{0}(t), \quad t\in [0,1].
\]
Therefore, by using the arguments as above, we deduce that
$\widehat{x}_{n} \to x^{*}_{2}\in P_{[R_{1}, R_{2}]}$
and $Tx^{*}_{2}=x^{*}_{2}$. The proof is complete.
\end{proof}

Next, in order to present the first corollary, we will use the following
notation:
\begin{gather*}
f^{0}=\limsup_{x\to 0^{+}}\max_{t\in[0,1]}\frac{f(t,x)}{x}, \quad
f^{\infty}=\limsup_{x\to +\infty}\max_{t\in[0,1]}\frac{f(t,x)}{x}, \\
f_{0}=\liminf_{x\to 0^{+}}\min_{t\in[0,1]}\frac{f(t,x)}{x}, \quad
f_{\infty}=\liminf_{x\to +\infty}\min_{t\in[0,1]}\frac{f(t,x)}{x}.
\end{gather*}

\begin{corollary}
    Let {\rm (H1)--(H2)} hold. Then the boundary-value problem
\eqref{2.01}-\eqref{2.03} has at least one positive solution in the case
\begin{itemize}
\item[(i)]  $f^{0} \leq - \beta^2+ \frac{1}{M} $  and
 $ f_{\infty}\geq  \frac{1}{M_{0}\eta_{m-2}}$,
(in particular $f^{0} = - \beta^{2}$, $f_{\infty} = \infty$); or
\item[(ii)] $f_{0} \geq  \frac{1}{M_{0}\eta_{m-2}} $  and
$ f_{\infty}\leq - \beta^2+ \frac{1}{M}$,
(in particular $f_{0} = \infty$, $f^{\infty} = -\beta^{2}$).
\end{itemize}
\end{corollary}

\begin{proof}
    It is easy to verify that conditions of Theorem \ref{thm1.1} can be obtained
from conditions (i) or (ii) of this corollary. We omit the proof.
\end{proof}

We close this section with the following result.

\begin{corollary} \label{coro3.2}
    Let {\rm (H1), (H2')} hold.  Further assume
\begin{equation}\label{eq3.05}
    \liminf_{x\to +\infty}\sup_{t\in[0,1]}\frac{f(t,x)}{x}
\leq -\beta^2+\frac{1}{M}, \quad
\Big(\text{in particular } \liminf_{x\to +\infty}\sup_{t\in[0,1]}
\frac{f(t,x)}{x} =-\beta^2\Big)
\end{equation}
and
\begin{equation}\label{eq3.06}
    \limsup_{x\to 0^{+}}\inf_{t\in[0,1]}\frac{f(t,x)}{x}
\geq \frac{1}{M_{0}\eta_{m-2}}, \quad
\Big(\text{in particular } \limsup_{x\to 0^{+}}
\inf_{t\in[0,1]}\frac{f(t,x)}{x} = +\infty\Big).
\end{equation}
Then there exist two constants $0<R_{1}<R_2$ such that the problem
\eqref{2.01}-\eqref{2.03} has positive solutions $x_{1}^{*}$,
$x_{2}^{*}$ ($x_{1}^{*}$ and $x_{2}^{*}$ may coincide)  with
\begin{equation}
  R_{1}\leq \|x_{1}^{*}\| \leq R_{2} \quad \text{and}\quad
\lim_{n\to +\infty}T^{n}x_{0} = x_{1}^{*}, \quad \text{where }
x_{0}(t) = R_{2},\; t\in [0,1],
\end{equation}
\begin{equation}
  R_{1}\leq \|x_{2}^{*}\| \leq R_{2} \quad \text{and}\quad
\lim_{n\to +\infty}T^{n}\widehat{x}_{0} = x_{2}^{*}, \quad \text{where }
 \widehat{x}_{0}(t) = R_{1},\; t\in [0,1].
\end{equation}
\end{corollary}

Clearly, from the assumptions of this corollary, the conditions of
Theorem \ref{thm1.2} hold. So we omit the proof.

\section{Existence of infinitely many positive solutions}

 In this section we  give sufficient conditions for existence of
infinitely many positive solutions. For this purpose, we assume that there
exists a sequence $\{R_{n}\}_{n=1}^{\infty} \subset \mathbb{R}$ such that
$0<R_{n}<cR_{n+1}$ and for all $n\in \mathbb{N}$,
\begin{itemize}
\item [(H3)] $f(t,x) + \beta^{2}x \leq \frac{R_{2n+1}}{M}$,
 for all $(t,x) \in [0,1]\times [cR_{2n-1}, R_{2n-1}]$,
\item [(H4)] $f(t,x) + \beta^{2}x \geq \frac{R_{2n}}{M_{0}\eta_{m-2}}$,
\quad for all $(t,x) \in [0,1]\times [cR_{2n}, R_{2n}]$.
\end{itemize}


\begin{theorem} \label{thm4.1}
 Assume {\rm (H1)--(H4)} hold. Then the boundary-value problem
\eqref{2.01}-\eqref{2.03} has infinitely many positive solutions
 $\{x_{n}\}_{n\in \mathbb{N}}$ satisfying
$ R_{2n-1} \leq \|x_{n}\| \leq R_{2n}$, for $n\in \mathbb{N}$.
\end{theorem}

\begin{proof}
 Let $\Omega_{n} = \{x\in C[0,1]: \|x\| < R_{n}\}$.
Then $0 \in \Omega_{n}$ and $\overline{\Omega_{n}} \subset \Omega_{n+1}$,
for $n\in \mathbb{N}$. In the following, we show that for all
$n \in \mathbb{N}$,
\begin{gather}\label{eq3.07}
    \|Tx\| \leq \|x\|, \quad \forall x\in P\cap \partial\Omega_{2n-1},\\
\label{eq3.08}
    \|Tx\| \geq \|x\|, \quad \forall x\in P\cap \partial\Omega_{2n}.
\end{gather}
First, for $x\in P\cap \partial\Omega_{2n-1}$, $s\in [0,1]$, we have
\begin{equation}
    cR_{2n-1}= c\|x\| \leq x(s) \leq \|x\| =R_{2n-1}.
\end{equation}
So, by the assumption (H3),
\begin{equation}
    g(s,x(s))\leq \frac{R_{2n-1}}{M}.
\end{equation}
Consequently,
\begin{equation}
    \|Tx\| = \max_{t\in[0,1]}\int_{0}^{1}G(t,s)g(s,x(s))ds
\leq R_{2n-1} =\|x\|, \quad t\in [0,1],
\end{equation}
which implies \eqref{eq3.07}.

Next, for $x\in P\cap \partial\Omega_{2n}$ and $s\in [0,1]$, we have
$cR_{2n} \leq x(s) \leq R_{2n}$. Then, from (H4) it follows that
for $t\in [0,1]$,
\begin{equation}
    Tx(t) = \int_{0}^{1}G(t,s)g(s,x(s))ds
\geq \int_{0}^{\eta_{m-2}}G(t,s)g(s,x(s))ds \geq R_{2n}
= \|x\|\, .
\end{equation}

So, we obtain \eqref{eq3.08}. The inequalities \eqref{eq3.07} and
\eqref{eq3.08} prove that  $T$ satisfies condition (i) of
 Therorem 1.1 in $P\cap(\overline{\Omega}_{2n}\backslash \Omega_{2n-1})$.
Therefore $T$ has a fixed point
$x_{n}\in P\cap(\overline{\Omega}_{2n}\backslash \Omega_{2n-1})$.
This implies that $ R_{2n-1} \leq \|x_{n}\| \leq R_{2n}$.
This completes the proof.
\end{proof}


\subsection*{Example}
Let $\alpha, \beta \in \mathbb{R}_{+}$ and
$\rho: \mathbb{R}_{+} \to \mathbb{R}$ such that
 $\alpha x \leq \rho(x)\leq \beta x$, for all $x\in \mathbb{R}_{+}$.
 We consider the function $f: [0,1]\times \mathbb{R}_{+} \to \mathbb{R}$,
defined by
\[
    f(t,x) = \begin{cases}
        f_{1}(t,x) & \text{if } R_{2n-2} \leq x \leq cR_{2n-1},\\
        f_{2}(t,x) & \text{if } cR_{2n-1} \leq x \leq R_{2n-1}, \\
        f_{3}(t,x) & \text{if } R_{2n-1} \leq x \leq cR_{2n}, \\
        f_{4}(t,x) & \text{if } cR_{2n} \leq x \leq R_{2n},
    \end{cases}
\]
for all $n\in \mathbb{N}$, where $R_{0} = 0$,
$\{R_{n}\}_{n=1}^{+\infty} \subset \mathbb{R}$ such that
$0<R_{n}<cR_{n+1}$ and
\begin{gather*}
    f_1(t, x) = \frac{cR_{2n-1} -x}{cR_{2n-1} -R_{2n-2}}f_4(t,x)
+  \frac{R_{2n-2} -x}{R_{2n-2} -cR_{2n-1}}f_{2}(t,x), \\
 f_2(t,x) =t\rho\Big(\frac{x}{\beta M}\Big) - \beta^{2}x, \\
 f_3(t,x) = \frac{cR_{2n} -x}{cR_{2n} -R_{2n-1}}f_2(t,x)
+  \frac{R_{2n-1} -x}{R_{2n-1} -cR_{2n}}f_{4}(t,x), \\
 f_{4}(t,x)=(t+1)\rho\Big(\frac{x}{\alpha M_{0}\eta_{m-2}}\Big)
-\beta^{2}\ln(1+x).
\end{gather*}

It is clear that $f$ is a function continuous on
$[0,1]\times [0,+\infty)$. Since the inequality
$x - \ln(1+x) \geq 0, \quad \forall x \geq 0$, so
$f(t,x)+\beta^{2}x \geq 0, \quad \forall t \in [0,1], x\in [0,+\infty)$.
Moreover, by the properties of the function $\rho$, we can deduce that
the assumptions (H3) and (H4) of  Theorem \ref{thm4.1} hold.

\section{Compactness of the set of positive solutions}

\begin{theorem} \label{thm5.1}
Let {\rm (H1)--(H2)} hold. In addition, suppose that there exists
a constant $\alpha\in (0,1)$ such that
\begin{equation}\label{eq4.01}
f_{0}\geq \frac{1}{M\eta_{m-2}} \quad \text{and} \quad
f^{\infty} \leq -\beta^{2} + \frac{\alpha}{M} \quad
\Big(\text{in particular } f_{0}=\infty,\;  f^{\infty}=-\beta^{2} \Big).
\end{equation}
Then the set of positive solutions of the problem
\eqref{2.01}--\eqref{2.03} is nonempty and compact.
\end{theorem}

\begin{proof}
    Put $\Sigma =\{x\in P: x= Tx\}$.
By Theorem \ref{thm1.1}, we have $\Sigma \neq \emptyset$.
We shall show that $\Sigma$ is compact.
 From assumption \eqref{eq4.01}, there exists $R>0$ such that
\begin{equation}
    f(t,x) \leq \left(-\beta^2 + \frac{\alpha}{M}\right)x, \quad
\forall t\in [0,1], x\geq R.
\end{equation}
Therefore,
\begin{equation}
    g\left(t,x(t)\right)=f\left(t,x(t)\right) + \beta^2x(t)
\leq \frac{\alpha}{M}x(t) + \gamma, \quad \forall t\in [0,1],
\end{equation}
where
$\gamma=\max\{g(t,x): (t,x)\in [0,1]\times [0,R]\}$.
For $x\in \Sigma$ and $t\in [0,1]$, we have
\begin{equation}
    x(t) = \int_{0}^{1}G(t,s)g\left(s,x(s)\right)ds\leq M\int_{0}^{1}\left(\frac{\alpha}{M}x(s) + \gamma\right)ds \leq \alpha \|x\| + M\gamma,
\end{equation}
so
\begin{equation}\label{eq4.02}
    \|x\|\leq \frac{M\gamma}{1-\alpha}, \quad \forall x\in \Sigma.
\end{equation}
From the compactness of the operator $T:P \to P$, it follows
from \eqref{eq4.02} that $T(\Sigma)$, and then
$\Sigma\subset T(\Sigma)$ are relatively compact.

To prove $\Sigma$ is closed, let $\{x_{n}\} \subset \Sigma$ be a sequence
and $\lim_{n \to +\infty} \|x_n -\widehat{x}\| =0$.
For  $t\in [0,1]$, we have
\begin{align*}
&\big|\widehat{x}(t) - \int_{0}^{1}G(t,s)g\left(s,\widehat{x}(s)\right)ds\big|  \\
&\leq |\widehat{x}(t) - x_{n}(t)|
 +\big|x_{n}(t) - \int_{0}^{1}G(t,s)g\left(s,x_{n}(s)\right)ds\big|  \\
&\quad +\big|\int_{0}^{1}G(t,s)g\left(s,x_{n}(s)\right)ds
  - \int_{0}^{1}G(t,s)g\left(s,\widehat{x}(s)\right)ds\big|  \\
&\leq |\widehat{x}(t) - x_{n}(t)|
 + M \int_{0}^{1}\left| g\left(s,x_{n}(s)\right)
 - g\left(s,\widehat{x}(s)\right)\right|ds .
\end{align*}
Let $n \to +\infty$, by the continuity of $g$, we deduce that
$\big|\widehat{x}(t) - \int_{0}^{1}G(t,s)g\left(s,\widehat{x}(s)\right)ds\big|
 =0$. So
\begin{equation}
    \widehat{x}(t)=\int_{0}^{1}G(t,s)g\left(s,\widehat{x}(s)\right)ds, \quad
t\in [0,1],
\end{equation}
which implies that $\widehat{x}\in \Sigma$. Therefore, $\Sigma$ is closed.
The proof of Theorem \ref{thm4.1} is complete.
\end{proof}


\subsection*{Open problem}
With the assumptions of Theorem \ref{thm5.1}, is the set of positive
solutions discrete or continuum?


\subsection*{Acknowledgements}
 The authors wish to express their sincere thanks to the anonymous
referee for his/her helpful comments and remarks.

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\end{document}