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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 14, pp. 1--14.\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-2011/14\hfil Positive solutions]
{Positive solutions to  generalized second-order three-point
 integral boundary-value problems}

\author[S. Chasreechai, J. Tariboon\hfil EJDE-2011/14\hfilneg]
{Saowaluk Chasreechai, Jessada Tariboon}  % in alphabetical order

\address{Saowaluk Chasreechai \newline
Department of Mathematics, Faculty of Applied Science \\
King Mongkut's University of Technology North Bangkok \\
Bangkok 10800,  Thailand}
\email{slc@kmutnb.ac.th}

\address{Jessada Tariboon \newline
Department of Mathematics, Faculty of Applied Science \\
King Mongkut's University of Technology North Bangkok \\
Bangkok 10800,  Thailand. \newline
Centre of Excellence in Mathematics, CHE, Sri
Ayutthaya Road, Bangkok 10400, Thailand}
\email{jessadat@kmutnb.ac.th}

\thanks{Submitted October 22, 2010. Published January 26, 2011.}
\subjclass[2000]{34B15, 34K10}
\keywords{Positive solution; three-point
boundary value problem; \hfill\break\indent fixed point theorem; cone}

\begin{abstract}
 In this article, by using Krasnoselskii's fixed point theorem,
 we obtain single and multiple positive solutions to the
 nonlinear second-order three-point integral boundary value
 problem
 \begin{gather*}
 u''(t)+a(t)f(u(t))=0,\quad 0<t<T, \\
 u(0)=\beta\int_0^{\eta}u(s)ds,\quad
 \alpha\int_0^{\eta}u(s)ds=u(T),
 \end{gather*}
 where $0<\eta<T$, $0<\alpha<\frac{2T}{\eta^2}$,
 $0<\beta<\frac{2T-\alpha\eta^2}{\eta(2T-\eta)}$ are given constants.
 As an application, we give some examples that illustrate our  results.
\end{abstract}

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

\section{Introduction}

  We are interested in obtaining positive
solutions of the  second-order three-point integral
boundary-value problem (BVP)
\begin{gather}
u''(t)+a(t)f(u(t))=0,\quad t\in(0,T),\label{I1-1}\\
u(0)=\beta\int_0^{\eta}u(s)ds,\quad
\alpha\int_0^{\eta}u(s)ds=u(T),\label{I1-2}
\end{gather}
where $0<\eta<T$, $0<\alpha<\frac{2T}{\eta^2}$,
$0< \beta<\frac{2T-\alpha\eta^2}{\eta(2T-\eta)}$, $f\in C([0, \infty),
[0, \infty))$, $a\in C([0, \infty), [0, \infty))$ and there exists
a $t_0\in (0, T)$, such that $a(t_0)>0$. Set
\[
f_0=\lim_{u\to 0+}\frac{f(u)}{u},\quad
f_{\infty}=\lim_{u\to\infty}\frac{f(u)}{u}.
\]

The study of the existence of solutions of multi-point
boundary-value problems for linear second-order ordinary differential
equations was initiated by Il'in and Moiseev \cite{i1}. Then Gupta
\cite{g2} studied three-point boundary value problems for nonlinear
second-order ordinary differential equations. Since then, the
existence of positive solutions for nonlinear second order
three-point boundary-value problems has been studied by many
authors by using a nonlinear alternative of the Leray-Schauder
approach, coincidence degree theory, the fixed point theorem for
cones and so on. We refer the reader to
\cite{c1,g1,h1,l1,l2,l3,l4,l5,l6,l7,m1,m2,m3,p1,s1,s2,x1}
and the references therein.
However, all of these papers are concerned
with problems with three-point boundary conditions consisting of
restrictions on the slope of the solutions and the solutions
themselves, for example:
\begin{gather*}
u(0)=0,\quad \alpha u(\eta)=u(1); \\
u(0)=\beta u(\eta),\quad \alpha u(\eta)=u(T); \\
u'(0)=0,\quad \alpha u(\eta)=u(1); \\
u(0)-\beta u'(0)=0,\quad \alpha u(\eta)=u(1); \\
\alpha u(0)-\beta u'(0)=0,\quad u'(\eta)+u'(1)=0;\;\text{etc.}
\end{gather*}

Recently, Tariboon \cite{t1} and the author proved the existence of
positive solutions for the three-point boundary-value problem with
integral condition
\begin{gather}
u''(t)+a(t)f(u(t))=0,\quad t\in(0,1), \label{I1-3}\\
u(0)=0,\quad \alpha\int_0^{\eta}u(s)ds=u(1),\label{I1-4}
\end{gather}
where $0<\eta<1$ and $0<\alpha<2/\eta^2$. We note that the
three-point integral boundary conditions \eqref{I1-2} and
\eqref{I1-4} are related to the area under the curve of
solutions $u(t)$ from $t=0$ to $t=\eta$.

The aim of this article is to establish some simple criteria for the
existence of single positive solution for  \eqref{I1-1},
\eqref{I1-2} under $f_0=0$, $f_{\infty}=\infty$
($f$ is superlinear) or $f_0=\infty$, $f_{\infty}=0$
($f$ is sublinear).
Moreover, we establish the existence conditions of two positive
solutions for  \eqref{I1-1}, \eqref{I1-2} under
$f_0=f_{\infty}=\infty$ or $f_0=f_{\infty}=0$. Finally, we give
some examples to illustrate our results. The key tool in our
approach is the Krasnoselskii's fixed point theorem in a cone.

\begin{theorem}[\cite{k1}]\label{Thm1-1}
Let $E$ be a Banach space, and let $K\subset E$ be a cone. Assume
$\Omega_1$, $\Omega_2$ are open subsets of $E$ with
$0\in \Omega_1$, $\overline{\Omega}_1\subset\Omega_2$, and let
$$
A:K\cap(\overline{\Omega}_1\setminus\Omega_2)\to K
$$
be a completely continuous operator such that
\begin{itemize}
\item[(i)]  $\|Au\|\leqslant\|u\|$,  $u\in K\cap\partial\Omega_1$, and
$\|Au\|\geqslant\|u\|$, $u\in K\cap\partial\Omega_2$; or

\item[(ii)] $\|Au\|\geqslant\|u\|$,  $u\in K\cap\partial\Omega_1$,
and $\|Au\|\geqslant\|u\|$, $u\in K\cap\partial\Omega_2$.
\end{itemize}
Then $A$ has a fixed point in
$K\cap(\overline{\Omega}_2\setminus\Omega_1)$.
\end{theorem}

\section{Preliminaries}

We now state and prove several lemmas before stating our main
results.

\begin{lemma}\label{Lem-1}
Let $\beta\neq\frac{2T-\alpha\eta^2}{\eta(2T-\eta)}$. Then for
$y\in C[0,T]$, the problem
\begin{gather}\label{P2-1}
u''+y(t)=0,\quad t\in(0,T), \\
\label{P2-2}
u(0)=\beta\int_0^{\eta}u(s)ds,\quad
\alpha\int_0^{\eta}u(s)ds=u(T),
\end{gather}
has a unique solution
\begin{align*}
u(t) &= \frac{(\beta-\alpha)t-\beta
T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{\eta}(\eta-s)^2y(s)ds \\
&\quad +\frac{2(1-\beta\eta)t+\beta\eta^2}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^T
(T-s)y(s)ds-\int_0^t(t-s)y(s)ds.
\end{align*}
\end{lemma}

\begin{proof}
 From \eqref{P2-1}, we have $u''(t) =-y(t)$.
For $t\in[0,T)$, integrating from $0$ to $t$, we obtain
\[
u'(t)=u'(0)-\int_0^ty(s)ds.
\]
For $t\in[0,T]$, integrating from $0$ to $t$, we obtain
\[
u(t)=u(0)+u'(0)t-\int_0^t\Big(\int_0^xy(s)ds\Big)dx;
\]
i.e.,
\begin{equation}\label{P2-3}
u(t)=u(0)+u'(0)t-\int_0^t(t-s)y(s)ds:=A+Bt-\int_0^t(t-s)y(s)ds.
\end{equation}
Integrating \eqref{P2-3} from $0$ to $\eta$, where
$\eta\in (0,T)$, we have
\begin{align*}
\int_0^{\eta}u(s)ds &= \eta A+\frac{\eta^2}{2}B-\int_0^{\eta}
\Big(\int_0^x(x-s)y(s)ds\Big)dx \\
&= \eta A+\frac{\eta^2}{2}B-\frac{1}{2}\int_0^{\eta}(\eta-s)^2y(s)ds.
\end{align*}
Since $u(0)=A$,
\[
u(T)= A+BT-\int_0^T(T-s)y(s)ds.
\]
By \eqref{P2-2}, from $u(0)=\beta\int_0^{\eta}u(s)ds$ we have
\[
(1-\beta\eta)A-\frac{\beta\eta^2}{2}B=-\frac{\beta}{2}
\int_0^{\eta}(\eta-s)^2y(s)ds,
\]
and from $u(T)=\alpha\int_0^{\eta}u(s)ds$ we have
\[
(1-\alpha\eta)A+\big(T-\frac{\alpha\eta^2}{2}\big)B
=\int_0^T(T-s)y(s)ds -\frac{\alpha}{2}\int_0^{\eta}(\eta-s)^2y(s)ds.
\]
Therefore,
\begin{align*}
A&=\frac{\beta\eta^2}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^T(T-s)
y(s)ds \\
&\quad -\frac{\beta T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{\eta}(\eta-s)^2y(s)ds
\end{align*}
\begin{align*}
B&=\frac{2(1-\beta\eta)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
 \int_0^T(T-s)y(s)ds \\
&\quad +\frac{(\beta-\alpha)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{\eta}(\eta-s)^2y(s)ds.
\end{align*}
Hence, \eqref{P2-1}-\eqref{P2-2} has a unique solution
\begin{align*}
u(t) &= \frac{(\beta-\alpha)t-\beta
T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{\eta}(\eta-s)^2y(s)ds \\
&\quad +\frac{2(1-\beta\eta)t+\beta\eta^2}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^T
(T-s)y(s)ds-\int_0^t(t-s)y(s)ds.
\end{align*}
\end{proof}

\begin{lemma}\label{Lem-2}
Let $0<\alpha<\frac{2T}{\eta^2}$,
$0<\beta<\frac{2T-\alpha\eta^2}{\eta(2T-\eta)}$.
If $y\in C(0,T)$ and
$y(t)\geqslant 0$ on $(0,T)$, then the unique solution $u$ of
\eqref{P2-1}-\eqref{P2-2} satisfies $u(t)\geqslant 0$ for
$t\in[0,T]$.
\end{lemma}

\begin{proof}
 It is known that the graph of $u$ is concave
down on $[0, T]$ from $u''(t)=-y(t)\leqslant 0$, we obtain
\begin{equation}\label{P2-4}
\int_0^{\eta}u(s)ds\geqslant \frac{1}{2}\eta
\big(u(0)+u(\eta)\big),
\end{equation}
where $\frac{1}{2}\eta \big(u(0)+u(\eta)\big)$ is the area of
the trapezoid under the curve $u(t)$ from $t=0$ to $t=\eta$ for
$\eta\in (0, T)$. Combining \eqref{P2-4} with \eqref{P2-2}, we
can get
\begin{gather}\label{P2-5}
u(0)\geqslant \frac{\beta\eta}{2-\beta\eta}u(\eta), \\
\label{P2-6}
u(T)\geqslant \frac{\alpha\eta}{2-\beta\eta}u(\eta),
\end{gather}
such that
\begin{equation}\label{P2-7}
 2-\beta\eta>2-\frac{2T-\alpha\eta^2}{2T-\eta}
=\frac{2(T-\eta)+2\eta^2}{2T-\eta}>0.
\end{equation}

From the graph of $u$ being concave down on $[0, T]$ again, we obtain
\begin{equation}\label{P2-8}
\frac{u(\eta)-u(0)}{\eta}\geqslant \frac{u(T)-u(0)}{T}.
\end{equation}
Using \eqref{P2-5}, \eqref{P2-6} and \eqref{P2-8}, we
obtain
\[
\frac{2-2\beta\eta}{\eta}u(\eta)\geqslant
\frac{(\alpha-\beta)\eta}{T}u(\eta).
\]
If $u(0)< 0$, then $u(\eta)<0$. It implies
$\frac{2T-\alpha\eta^2}{\eta(2T-\eta)}\leqslant\beta$, a
contradiction to $\beta<\frac{2T-\alpha\eta^2}{\eta(2T-\eta)}$. If
$u(T)<0$, then $u(\eta)<0$, and the same contradiction emerges.
Thus, it is true that $u(0)\geqslant 0$, $u(T)\geqslant 0$,
together with the concavity of $u$, we have
$u(t)\geqslant 0$ for $t\in[0, T]$.
This proof is complete.
\end{proof}

\begin{lemma}\label{Lem-3}
Let $\alpha\eta^2\neq 2T$,
$\beta>\max\big\{\frac{2T-\alpha\eta^2}{\eta(2T-\eta)}, 0\big\}$.
If $y\in C(0,T)$ and $y(t)\geqslant 0$ for $t\in[0,T]$, then
problem \eqref{P2-1}-\eqref{P2-2} has no positive solutions.
\end{lemma}

\begin{proof}
 Suppose that \eqref{P2-1}-\eqref{P2-2} has a positive solution $u$
satisfying $u(t)\geqslant 0$, $t\in[0,T]$ and there is a
$\tau_0\in(0,T)$ such that $u(\tau_0)>0$.

If $u(T)>0$, then $\int_{0}^{\eta}u(s)ds>0$. It implies
\begin{equation}
u(0)=\beta\int_0^{\eta}u(s)ds>\frac{2T-\alpha\eta^2}
{\eta(2T-\eta)}\int_0^{\eta}u(s)ds\geqslant\frac{\eta T
(u(0)+u(\eta))-\eta^2u(T)} {\eta(2T-\eta)};
\end{equation}
that is
\begin{equation}
\frac{u(T)-u(0)}{T}>\frac{u(\eta)-u(0)}{\eta},
\end{equation}
which is a contradiction to the concavity of $u$.

If $u(T)=0$, then $\int_0^{\eta}u(s)ds=0$. When
$\tau_0\in(0,\eta)$, we obtain $u(\tau_0)>u(T)=0>u(\eta)$, which
contradicts the concavity of $u$. When $\tau_0\in(\eta, T)$,
we obtain $u(\eta)\leqslant 0=u(0)< u(\tau_0)$, which contradicts the
concavity of $u$ again. Therefore, no positive solutions
exist.
\end{proof}

Let $E=C[0, T]$, then $E$ is a Banach space with respect to the norm
\[
\|u\|=\sup_{t\in[0, T]}|u(t)|.
\]

\begin{lemma}\label{Lem-4}
Let $0<\alpha<\frac{2T}{\eta^2}$, $0<
\beta<\frac{2T-\alpha\eta^2}{\eta(2T-\eta)}$. If $y\in C(0,T)$ and
$y(t)\geqslant 0$ for $t\in[0,T]$, then the unique solution to
problem \eqref{P2-1}-\eqref{P2-2} satisfies
\begin{equation}\label{P2-9}
\min_{t\in[0, T]}u(t)\geqslant\gamma\|u\|,
\end{equation}
where
\begin{equation}\label{P2-10}
\gamma:=\min\Big\{\frac{\alpha\eta(T-\eta)}
{T(2-\beta\eta)-\alpha\eta^2},\frac{\alpha\eta^2}
{(2-\beta\eta)T},\frac{\beta\eta(T-\eta)}{(2-\beta\eta)T},
\frac{\beta\eta^2}{(2-\beta\eta)T}\Big\}.
\end{equation}
\end{lemma}

\begin{proof}
 From the fact that $u''(t)=-y(t)\leqslant 0$, we know that the
graph of $u(t)$ is concave down on $[0,T]$.
If $u(t)$ is maximum at $t=\tau_1$, then $\|u\|=u(\tau_1)$. We
divide the proof into two cases.

Case (i) If $u(0)\geqslant u(T)$ and $\min_{t\in[0,T]}u(t)=u(T)$, then
either $0\leqslant \tau_1\leqslant \eta<T$, or
$0<\eta<\tau_1<T$.
If $0\leqslant \tau_1\leqslant \eta<T$, then
\begin{align*}
u(\tau_1) &\leqslant u(T)+\frac{u(T)-u(\eta)}{T-\eta}(\tau_1-T) \\
&\leqslant u(T)+\frac{u(T)-u(\eta)}{T-\eta}(0-T) \\
&=\frac{Tu(\eta)-\eta u(T)}{T-\eta} \\
&\leqslant \frac{T\Big[\frac{2-\beta\eta}{\alpha\eta}\Big]-\eta}{T-\eta}u(T)
\quad \text{(by \eqref{P2-6})} \\
&= \frac{T(2-\beta\eta)-\alpha\eta^2}{\alpha\eta(T-\eta)}u(T).
\end{align*}
This implies
\[
\min_{t\in[0,T]}u(t)\geqslant
\frac{\alpha\eta(T-\eta)}{T(2-\beta\eta)-\alpha\eta^2}\|u\|.
\]
If $0<\eta<\tau_1<T$, from
\[
\frac{u(\eta)}{\eta}\geqslant \frac{u(\tau_1)}{\tau_1}\geqslant
\frac{u(\tau_1)}{T},
\]
together with \eqref{P2-6}, we have
\[
u(T)\geqslant \frac{\alpha\eta^2}{(2-\beta\eta)T}\,u(\tau_1).
\]
This implies
\[
\min_{t\in[0,T]}u(t)\geqslant
\frac{\alpha\eta^2}{(2-\beta\eta)T}\,\|u\|.
\]

Case (ii) If $u(0)\leqslant u(T)$ and $\min_{t\in[0,T]}u(t)=u(0)$, then
either $0<\tau_1<\eta<T$, or $0<\eta\leqslant \tau_1\leqslant T$.
If $0<\tau_1<\eta<T$, from
\[
\frac{u(\eta)}{T-\eta}\geqslant
\frac{u(\tau_1)}{T-\tau_1}\geqslant \frac{u(\tau_1)}{T},
\]
together with \eqref{P2-5}, we have
\[
u(0)\geqslant \frac{\beta\eta(T-\eta)}{(2-\beta\eta)T}\,u(\tau_1).
\]
Hence
\[
\min_{t\in[0,T]}u(t)\geqslant
\frac{\beta\eta(T-\eta)}{(2-\beta\eta)T}\,\|u\|.
\]
If $0<\eta\leqslant \tau_1\leqslant T$, from
\[
\frac{u(\tau_1)}{T}\leqslant \frac{u(\tau_1)}{\tau_1}\leqslant
\frac{u(\eta)}{\eta},
\]
together with  \eqref{P2-5}, we have
\[
u(0)\geqslant \frac{\beta\eta^2}{(2-\beta\eta)T}\,u(\tau_1).
\]
This implies
\[
\min_{t\in[0,T]}u(t)\geqslant
\frac{\beta\eta^2}{(2-\beta\eta)T}\,\|u\|.
\]
This completes the proof.
\end{proof}

In the rest of this article,  we assume that
$0<\alpha<2T/\eta^2$, $0<\beta<\frac{2T-\alpha\eta^2}{\eta(2T-\eta)}$.
It is easy to see that  \eqref{I1-1}-\eqref{I1-2} has a solution
$u=u(t)$ if and only if $u$ is a solution of the operator equation
\begin{align*}
u(t) &= \frac{(\beta-\alpha)t-\beta
T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{\eta}(\eta-s)^2a(s)f(u(s))ds \\
&\quad+\frac{2(1-\beta\eta)t+\beta\eta^2}{(2T-\alpha\eta^2)
 -\beta\eta(2T-\eta)}\int_0^T
(T-s)a(s)f(u(s))ds \\
&\quad -\int_0^t(t-s)a(s)f(u(s))ds
\triangleq  A u(t).
\end{align*}
Denote
\begin{equation}
K=\big\{u\in E:u\geqslant
0,\min_{t\in[0,T]}u(t)\geqslant\gamma\|u\|\big\},
\end{equation}
where $\gamma$ is defined in \eqref{P2-10}.

It is obvious that $K$ is a cone in $E$. Moreover from Lemma
\ref{Lem-2} and Lemma \ref{Lem-4}, $A(K)\subset K$. It is also
easy to check that $A:K\to K$ is completely continuous. In
the following, for the sake of convenience, set
\begin{gather*}
\Lambda_1=\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)
-\beta\eta(2T-\eta)} \int_0^{T}T(T-s)a(s)ds, \\
\Lambda_2=\frac{\gamma(2-\beta\eta)(T-\eta)}{(2T-\alpha\eta^2)
-\beta\eta(2T-\eta)}\int_0^{T} sa(s)ds.
\end{gather*}

\section{Main results}

Now we are in the position to establish the main result.

\begin{theorem} \label{Thm3-1}
Problem \eqref{I1-1}-\eqref{I1-2} has at least one positive
solution under the assumptions:
\begin{itemize}
\item[(H1)]  $f_0=0$ and $f_{\infty}=\infty$ (superlinear); or
\item[(H2)]  $f_0=\infty$ and $f_{\infty}=0$ (sublinear).
\end{itemize}
\end{theorem}

\begin{proof}
 At first, let (H1) hold. Since
$f_0=\lim_{u\to 0^+}(f(u)/u)=0$ for any
$\varepsilon\in(0,\Lambda_1^{-1}]$, there exists $\rho_*$ such that
\begin{equation}\label{M3-1}
f(u)\leqslant \varepsilon u\quad \text{for }u\in[0,\rho_*].
\end{equation}
Let $\Omega_{\rho_*}=\{u\in E:\|u\|<\rho_*\}$ for any
$u\in K\cap\partial\Omega_{\rho_*}$. From \eqref{M3-1}, we obtain
\begin{align*}
Au(t) &\leqslant \frac{(\beta-\alpha)t-\beta
T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{\eta}(\eta-s)^2a(s)f(u(s))ds \\
&\quad +\frac{2(1-\beta\eta)t+\beta\eta^2}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^T
(T-s)a(s)f(u(s))ds \\
&\leqslant  \frac{\beta t}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{\eta}(\eta-s)^2a(s)f(u(s))ds \\
&\quad+\frac{2t+\beta\eta^2}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^T
(T-s)a(s)f(u(s))ds \\
&\leqslant  \frac{\beta T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{\eta}(\eta-s)^2a(s)f(u(s))ds \\
&\quad +\frac{2T+\beta\eta^2}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^T
(T-s)a(s)f(u(s))ds \\
&\leqslant
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}(T^2-sT)a(s)f(u(s))ds \\
&\leqslant
\varepsilon\rho_*\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)ds\\
&=\varepsilon\Lambda_1\rho_*\leqslant \rho_*=\|u\|,
\end{align*}
which yields
\begin{equation}\label{M3-2}
\|Au\|\leqslant\|u\|\quad \text{for }
u\in K\cap\partial\Omega_{\rho_*}.
\end{equation}
Further, since
$f_{\infty}=\lim_{u\to\infty}(f(u)/u)=\infty$,  for
any $M^*\in[\Lambda_2^{-1},\infty)$, there exists $\rho^*>\rho_*$
such that
\begin{equation}\label{M3-3}
f(u)\geqslant M^*u\quad \text{for }u\geqslant\gamma\rho^*.
\end{equation}
Set $\Omega_{\rho^*}=\{u\in E:\|u\|<\rho^*\}$ for
 $u\in K\cap\partial\Omega_{\rho^*}$. Since $u\in K$,
$\min_{t\in[0,T]}u(t)\geqslant\gamma\|u\|=\gamma\rho^*$. Hence,
for any $u\in K\cap\Omega_{\rho^*}$, from \eqref{M3-3} and
\eqref{P2-7}, we obtain
\begin{align*}
Au(\eta)
&= \frac{(\beta-\alpha)\eta-\beta
T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{\eta}(\eta-s)^2a(s)f(u(s))ds \\
&\quad +\frac{(2-\beta\eta)\eta}{(2T-\alpha\eta^2)
 -\beta\eta(2T-\eta)}\int_0^T
(T-s)a(s)f(u(s))ds\\
&\quad -\int_0^{\eta}(\eta-s)a(s)f(u(s))ds \\
&=\frac{(2-\beta\eta)\eta}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^T
(T-s)a(s)f(u(s))ds \\
&\quad +\frac{1}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\\
&\quad\times \int_0^{\eta}
(\eta-s)\Big[-(2-\beta\eta)T+\big(\beta(T-\eta)+\alpha\eta\big)s\Big]a(s)f(u(s))ds \\
&\geqslant \frac{(2-\beta\eta)\eta}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^T
(T-s)a(s)f(u(s))ds \\
&\quad +\frac{-T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{\eta}
(\eta-s)(2-\beta\eta)a(s)f(u(s))ds \\
&\geqslant \frac{(2-\beta\eta)\eta}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^T
(T-s)a(s)f(u(s))ds \\
&\quad +\frac{-T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{T}
(\eta-s)(2-\beta\eta)a(s)f(u(s))ds \\
&=\frac{(2-\beta\eta)(T-\eta)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
 \int_0^{T} sa(s)f(u(s))ds \\
&\geqslant \gamma\rho^*M^*\frac{(2-\beta\eta)(T-\eta)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{T}
sa(s)ds\\
&=M^*\Lambda_2\rho^*\geqslant\rho^*=\|u\|,
\end{align*}
which implies
\begin{equation}\label{M3-4}
\|Au\|\geqslant\|u\|\quad \text{for }u\in
K\cap\partial\Omega_{\rho^*}.
\end{equation}

Therefore, from \eqref{M3-2}$, \eqref{M3-4}$ and Theorem
\ref{Thm1-1}, it follows that $A$ has a fixed point in
$K\cap(\overline{\Omega}_{\rho^*}\setminus\Omega_{\rho_*})$ such
that $\rho_*\leqslant\|u\|\leqslant\rho^*$.

Next, let (H2) hold. In view of $f_0=\lim_{u\to
0^+}(f(u)/u)=\infty$ for any $M_*\in[\Lambda_2^{-1}, \infty)$,
there exists $r_*>0$ such that
\begin{equation}\label{M3-5}
f(u)\geqslant M_*u\quad \text{for }0\leqslant u\leqslant r_*.
\end{equation}
Set $\Omega_{r_*}=\{u\in E:\|u\|<r_*\}$ for $u\in
K\cap\partial\Omega_{r_*}$. Since $u\in K$, it follows that
$\min_{t\in[0,T]}u(t)\geqslant\gamma\|u\|=\gamma r_*$. Thus from
\eqref{M3-5} for any $u\in K\cap\partial\Omega_{r_*}$, we have
\begin{align*}
Au(\eta) &= \frac{(\beta-\alpha)\eta-\beta
T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{\eta}(\eta-s)^2a(s)f(u(s))ds \\
&\quad +\frac{(2-\beta\eta)\eta}{(2T-\alpha\eta^2)
 -\beta\eta(2T-\eta)}\int_0^T (T-s)a(s)f(u(s))ds\\
&\quad -\int_0^{\eta}(\eta-s)a(s)f(u(s))ds \\
&\geqslant \gamma r_*M_*\frac{(2-\beta\eta)(T-\eta)}{(2T-\alpha\eta^2)
-\beta\eta(2T-\eta)}\int_0^{T}
sa(s)ds\\
&=M_*\Lambda_2r_*\geqslant r_*=\|u\|,
\end{align*}
which yields
\begin{equation}\label{M3-6}
\|Au\|\geqslant\|u\|\quad \text{for } u\in K\cap\partial\Omega_{r_*}.
\end{equation}

Since $f_{\infty}=\lim_{u\to \infty}(f(u)/u)=0$,  for
any $\varepsilon_1\in(0,\Lambda_1^{-1}]$, there exists $r_0>r_*$
such that
\begin{equation}\label{M3-7}
f(u)\leqslant \varepsilon_1u\quad \text{for } u\in[r_0,\infty).
\end{equation}
We have the next two cases:

Case (i): Suppose that $f(u)$ is unbounded, then from $f\in
C([0,\infty),[0,\infty))$, we know that there is $r^*>r_0$ such
that
\begin{equation}\label{M3-8}
f(u)\leqslant f(r^*)\quad \text{for } u\in[0,r^*].
\end{equation}
Since $r^*>r_0$,  from \eqref{M3-7} and \eqref{M3-8}, one
has
\begin{equation}\label{M3-9}
f(u)\leqslant f(r^*)\leqslant
\varepsilon_1r^*\quad \text{for }u\in[0,r^*].
\end{equation}
For $u\in K$, $\|u\|=r^*$, from \eqref{M3-9}, we obtain
\begin{align*}
Au(t)
&\leqslant \frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)
-\beta\eta(2T-\eta)} \int_0^{T}T(T-s)a(s)f(u(s))ds \\
&\leqslant \varepsilon_1r^*
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)ds \\
&= \varepsilon_1\Lambda_1r^*\leqslant r^*=\|u\|.
\end{align*}

Case (ii) Suppose that $f(u)$ is bounded, say $f(u)\leqslant N$
for all $u\in [0, \infty)$. Taking
$r^*\geqslant\max\{N/\varepsilon_1, r_*\}$, for $u\in K$,
$\|u\|=r^*$, we have
\begin{align*}
Au(t)
&\leqslant \frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)
 -\beta\eta(2T-\eta)} \int_0^{T}T(T-s)a(s)f(u(s))ds \\
&\leqslant N
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)ds \\
&\leqslant \varepsilon_1r^*
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)ds \\
&=\varepsilon_1\Lambda_1r^*\leqslant r^*=\|u\|.
\end{align*}
Hence, in either case, we always may set
$\Omega_{r^*}=\{u\in E:\|u\|<r^*\}$ such that
\begin{equation}\label{M3-10}
\|Au\|\leqslant\|u\|\quad \text{for }u\in K\cap\partial\Omega_{r^*}.
\end{equation}

Hence, from \eqref{M3-6}, \eqref{M3-10} and Theorem
\ref{Thm1-1}, it follows that $A$ has a fixed point in
$K\cap(\overline{\Omega}_{\rho^*}\setminus\Omega_{\rho_*})$ such
that $r_*\leqslant\|u\|\leqslant r^*$. The proof is complete.
\end{proof}

\begin{theorem} \label{Thm3-2}
Suppose that the following assumptions are satisfied:
\begin{itemize}
\item[(H3)] $f_0=f_{\infty}=\infty$,
\item[(H4)] There exists a constant $\rho_1>0$, such that
$f(u)\leqslant\Lambda_1^{-1}\rho_1$ for $u\in[0,\rho_1]$.
\end{itemize}
Then  \eqref{I1-1}, \eqref{I1-2} has at least two positive
solutions $u_1$ and $u_2$ such that
\[
0<\|u_1\|<\rho_1<\|u_2\|.
\]
\end{theorem}

\begin{proof}
At first, in view of $f_0=\lim_{u\to0^+}(f(u)/u)=\infty$, for any
$M_*\in[\Lambda_2^{-1},\infty)$, there exists
$\rho_*\in(0,\rho_1)$ such that
\begin{equation}\label{M3-11}
f(u)\geqslant M_*u,\quad \text{for }0\leqslant u\leqslant\rho_*.
\end{equation}
Set $\Omega_{\rho_*}=\{u\in E:\|u\|<\rho_*\}$ for $u\in
K\cap\partial\Omega_{\rho_*}$. Since $u\in K$, then
$\min_{t\in[0,T]}u(t)\geqslant\gamma\|u\|=\gamma\rho_*$. Thus from
\eqref{M3-11}, for any $u\in K\cap\partial\Omega_{\rho_*}$, we obtain
\begin{align*}
Au(\eta) &= \frac{(\beta-\alpha)\eta-\beta
T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{\eta}(\eta-s)^2a(s)f(u(s))ds \\
&\quad +\frac{(2-\beta\eta)\eta}{(2T-\alpha\eta^2)
 -\beta\eta(2T-\eta)}\int_0^T (T-s)a(s)f(u(s))ds\\
&\quad -\int_0^{\eta}(\eta-s)a(s)f(u(s))ds \\
&\geqslant \gamma
\rho_*M_*\frac{(2-\beta\eta)(T-\eta)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{T}
sa(s)ds\\
&=M_*\Lambda_2\rho_*\geqslant \rho_*=\|u\|,
\end{align*}
which implies
\begin{equation}\label{M3-12}
\|Au\|\geqslant\|u\|\quad \text{for }u\in
K\cap\partial\Omega_{\rho_*}.
\end{equation}
Next, since $f_{\infty}=\lim_{u\to\infty}(f(u)/u)=\infty$,
then for any $M^*\in[\Lambda_2^{-1},\infty)$, there exists
$\rho^*>\rho_1$ such that
\begin{equation}\label{M3-13}
f(u)\geqslant M^*u,\quad \text{for }u\geqslant\gamma\rho^*.
\end{equation}
Set $\Omega_{\rho^*}=\{u\in E:\|u\|<\rho^*\}$ for
 $u\in K\cap\partial\Omega_{\rho^*}$. Since $u\in K$, then
$\min_{t\in[0,T]}u(t)\geqslant\gamma\|u\|=\gamma \rho^*$. Thus
from \eqref{M3-13} for any $u\in K\cap\partial\Omega_{\rho^*}$,
we have
\begin{align*}
Au(\eta) &= \frac{(\beta-\alpha)\eta-\beta
T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
 \int_0^{\eta}(\eta-s)^2a(s)f(u(s))ds \\
&\quad +\frac{(2-\beta\eta)\eta}{(2T-\alpha\eta^2)
 -\beta\eta(2T-\eta)}\int_0^T
(T-s)a(s)f(u(s))ds\\
&\quad -\int_0^{\eta}(\eta-s)a(s)f(u(s))ds \\
&\geqslant \gamma
\rho^*M^*\frac{(2-\beta\eta)(T-\eta)}{(2T-\alpha\eta^2)
-\beta\eta(2T-\eta)}\int_0^{T} sa(s)ds\\
&=M^*\Lambda_2\rho^*\geqslant \rho^*=\|u\|,
\end{align*}
which implies
\begin{equation}\label{M3-14}
\|Au\|\geqslant\|u\|\quad \text{for }u\in
K\cap\partial\Omega_{\rho^*}.
\end{equation}
Finally, let $\Omega_{\rho_1}=\{u\in E:\|u\|<\rho_1\}$ for any
$u\in K\cap\partial\Omega_{\rho_1}$. Then from (H4) we obtain
\begin{align*}
Au(t)&\leqslant
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)f(u(s))ds \\
&\leqslant \Lambda_1^{-1}\rho_1
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)ds \\
&\leqslant \rho_1=\|u\|,
\end{align*}
which yields
\begin{equation}\label{M3-15}
\|Au\|\leqslant\|u\|\quad \text{for }u\in
K\cap\partial\Omega_{\rho_*}.
\end{equation}
Thus, from \eqref{M3-12}, \eqref{M3-14} and \eqref{M3-15},
it follows from Theorem \ref{Thm1-1} that $A$ has a fixed point
$u_1$ in $K\cap(\overline{\Omega}_{\rho_1}\setminus\Omega_{\rho_*})$,
and a fixed point $u_2$ in
$K\cap(\overline{\Omega}_{\rho^*}\setminus\Omega_{\rho_1})$. Both
are positive solutions of  \eqref{I1-1}, \eqref{I1-2}
and
$0<\|u_1\|<\rho_1<\|u_2\|$.
The proof is complete.
\end{proof}

\begin{theorem} \label{Thm3-3}
Suppose that the following assumptions are satisfied:
\begin{itemize}
\item[(H5)]
 $f_0=f_{\infty}=0$,

\item[(H6)] There exists a constant $\rho_2>0$, such that
\[
f(u)\geqslant\Lambda_2^{-1}\rho_2\quad
\text{for }u\in[\gamma\rho_2,\rho_2].
\]
\end{itemize}
Then \eqref{I1-1}, \eqref{I1-2} has at least two positive
solutions $u_1$ and $u_2$ such that
\[
0<\|u_1\|<\rho_2<\|u_2\|.
\]
\end{theorem}


\begin{proof}
 Firstly, since $f_0=\lim_{u\to
0^+}(f(u)/u)=0$, for any $\varepsilon\in (0,\Lambda_1^{-1}]$,
there exists $\rho_*\in(0,\rho_2)$ such that
\begin{equation}\label{M3-16}
f(u)\leqslant\varepsilon u,\quad \text{for }u\in[0,\rho_*].
\end{equation}
Let $\Omega_{\rho_*}=\{u\in E:\|u\|<\rho_*\}$ for any
$u\in K\cap\partial\Omega_{\rho_*}$. Then from \eqref{M3-16},
we obtain
\begin{align*}
Au(t)&\leqslant
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)f(u(s))ds \\
&\leqslant \varepsilon\rho_*
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)ds \\
&\leqslant \varepsilon\Lambda_1\rho_*\leqslant\rho_*=\|u\|,
\end{align*}
which implies
\begin{equation}\label{M3-17}
\|Au\|\leqslant\|u\|\quad \text{for }u\in
K\cap\partial\Omega_{\rho_*}.
\end{equation}
Secondly, in view of $f_{\infty}=\lim_{u\to
\infty}(f(u)/u)=0$, for any $\varepsilon_1\in(0, \Lambda_1^{-1}]$
there exists $\rho_0>\rho_2$, such that
\begin{equation}\label{M3-18}
f(u)\leqslant \varepsilon_1u,\quad \text{for }u\in[\rho_0,\infty).
\end{equation}
We consider the next two cases.

 Case (i): Suppose that $f(u)$ is unbounded. Then from $f\in
C([0,\infty),[0,\infty))$, there exists $\rho^*>\rho_0$ such that
\begin{equation}\label{M3-19}
f(u)\leqslant f(\rho^*),\quad \text{for }u\in[0, \rho^*].
\end{equation}
Since $\rho^*>\rho_0$, from \eqref{M3-18} and
\eqref{M3-18} one has
\begin{equation}\label{M3-20}
f(u)\leqslant f(\rho^*)\leqslant
\varepsilon_1\rho^*,\quad \text{for }u\in [0,\rho^*].
\end{equation}
For $u\in K$, and $\|u\|=\rho^*$, from \eqref{M3-20}, we obtain
\begin{align*}
Au(t)&\leqslant
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)f(u(s))ds \\
&\leqslant \varepsilon_1\rho^*
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)ds \\
&\leqslant \varepsilon_1\Lambda_1\rho^*\leqslant
\rho^*=\|u\|.
\end{align*}

 Case (ii): Suppose that $f(u)$ is bounded, say $f(u)\leqslant L$
for all $u\in[0,\infty)$. Taking
$\rho^*\geqslant\max\{L/\varepsilon_1, \rho_0\}$, for $u\in K$
with $\|u\|=\rho^*$, we have
\begin{align*}
Au(t)&\leqslant
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)f(u(s))ds \\
&\leqslant L
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)ds \\
&\leqslant \varepsilon_1\rho^*
\frac{2T+\beta(T+\eta^2)}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}
\int_0^{T}T(T-s)a(s)ds \\
&\leqslant \varepsilon_1\Lambda_1\rho^*\leqslant
\rho^*=\|u\|.
\end{align*}
Hence, in either case, we always may set $\Omega_{\rho^*}=\{u\in
E:\|u\|<\rho^*\}$ such that
\begin{equation}\label{M3-21}
\|Au\|\leqslant\|u\|\quad \text{for }u\in
K\cap\partial\Omega_{\rho^*}.
\end{equation}
Finally, set $\Omega_{\rho_2}=\{u\in E:\|u\|<\rho_2\}$ for $u\in
K\cap\partial\Omega_{\rho_2}$. Since $u\in K$,
$\min_{t\in[0, T]}u(t)\geqslant\gamma\|u\|=\gamma\rho_2$.
Hence, for any $u\in K\cap\partial\Omega_{\rho_2}$, and
(H6), we have
\begin{align*}
Au(\eta)
&= \frac{(\beta-\alpha)\eta-\beta
T}{(2T-\alpha\eta^2)-\beta\eta(2T-\eta)}\int_0^{\eta}(\eta-s)^2a(s)f(u(s))ds \\
&\quad +\frac{(2-\beta\eta)\eta}{(2T-\alpha\eta^2)
 -\beta\eta(2T-\eta)}\int_0^T
(T-s)a(s)f(u(s))ds\\
&\quad -\int_0^{\eta}(\eta-s)a(s)f(u(s))ds \\
&\geqslant \gamma \rho_2\Lambda_2^{-1}
\frac{(2-\beta\eta)(T-\eta)}{(2T-\alpha\eta^2)
-\beta\eta(2T-\eta)}\int_0^{T} sa(s)ds\geqslant\rho_2=\|u\|,
\end{align*}
which yields
\begin{equation}\label{M3-22}
\|Au\|\geqslant\|u\|\quad \text{for }u\in
K\cap\partial\Omega_{\rho_2}.
\end{equation}
Thus, since $\rho_*<\rho<\rho^*$ and from \eqref{M3-17},
\eqref{M3-21} and \eqref{M3-22}, it follows from Theorem
\ref{Thm1-1} that $A$ has a fixed point $u_1$ in
$K\cap(\overline{\Omega}_{\rho_2}\setminus\Omega_{\rho_*})$, and a
fixed point $u_2$ in
$K\cap(\overline{\Omega}_{\rho^*}\setminus\Omega_{\rho_2})$. Both
are positive solutions of  \eqref{I1-1}, \eqref{I1-2}
and
$0<\|u_1\|<\rho_2<\|u_2\|$.
The proof is complete.
\end{proof}

\section{Some examples}
In this section,  to illustrate our results, we consider
some examples.

\begin{example}\rm
Consider the boundary-value problem
\begin{gather}
u''(t)+t^2u^p=0,\quad 0<t<e^2,\label{E4-1}\\
u(0)=\frac{2}{9}\int_0^eu(s)ds,\quad u(e^2)=
\frac{2}{3}\int_0^eu(s)ds.\label{E4-2}
\end{gather}
Set $\alpha=2/3$, $\beta=2/9$, $\eta=e$, $T=e^2$,
$a(t)=t^2$, $f(u)=u^p$. We can show that
\[
0<\alpha=\frac{2}{3}<2=\frac{2T}{\eta^2},\quad
0<\beta=\frac{2}{9}<\frac{4}{3(2e-1)}
=\frac{2T-\alpha\eta^2}{\eta(2T-\eta)}.
\]

 Case I: $p\in(1,\infty)$. In this case, $f_0=0$,
$f_{\infty}=\infty$ and (H1) holds. Then
\eqref{E4-1}, \eqref{E4-2} has at least one positive solution.

Case II: $p\in(0,1)$ In this case, $f_0=\infty$, $f_{\infty}=0$
and (H2) holds. Then
\eqref{E4-1}, \eqref{E4-2} has at least one positive solution.
\end{example}

\begin{example}\rm Consider the boundary-value problem
\begin{gather}
u''(t)+\frac{1}{8^3}(4-t)^{1/2}(u^{1/2}+u^2)=0,\quad
 0<t<4,\label{E4-3}\\
u(0)=\frac{1}{10}\int_0^1u(s)ds,\quad u(4)=
2\int_0^1u(s)ds.\label{E4-4}
\end{gather}
Set $\alpha=2$, $\beta=1/10$, $\eta=1$, $T=4$,
$a(t)=\frac{1}{8^3}(4-t)^{1/2}$,
$f(u)=u^{1/2}+u^2$. We can show that
$0<\alpha=2<8=2T/\eta^2$,
$0<\beta=1/10<6/7=(2T-\alpha\eta^2)/(\eta(2T-\eta))$. Since
$f_0=f_{\infty}=\infty$, then $(H_3)$ holds. Again
$\Lambda_1^{-1}=((2T-\alpha\eta^2)-\beta\eta(2T-\eta))/
(2T+\beta(T+\eta^2))(\int_0^{T}T(T-s)a(s)ds)^{-1}=530/85$,
because $f(u)$ is monotone increasing function for $u\geqslant 0$,
taking $\rho_1=4$, then when $u\in[0,\rho_1]$, we obtain
\[
f(u)\leqslant f(4)=18<\frac{530}{85}\rho_1=\Lambda_1^{-1}\rho_1,
\]
which implies (H4) holds. Hence, by Theorem \ref{Thm3-2},
BVP \eqref{E4-3}, \eqref{E4-4} has at least two positive
solutions $u_1$ and $u_2$ such that
$0<\|u_1\|<4<\|u_2\|$.
\end{example}

\begin{example}\rm
Consider the boundary-value problem
\begin{gather}
u''(t)+e^{32}u^2e^{-u}=0,\quad 0<t<\frac{4}{5},\label{E4-5}\\
u(0)=2\int_0^{1/5}u(s)ds,\quad u\Big(\frac{4}{5}\Big)=
20\int_0^{1/5}u(s)ds.\label{E4-6}
\end{gather}
Set $\alpha=20$, $\beta=2$, $\eta=1/5$, $T=4/5$,
$a(t)\equiv e^{32}$, $f(u)=u^2e^{-u}$. We can show that
$0<\alpha=20<40=2T/\eta^2$, $0<\beta=2<20/7
=(2T-\alpha\eta^2)/(\eta(2T-\eta))$,
$\gamma=\min\{\alpha\eta(T-\eta)/
(T(2-\beta\eta)-\alpha\eta^2),\alpha\eta^2/ ((2-\beta\eta)T),
\beta\eta(T-\eta)/((2-\beta\eta)T),\beta\eta^2/((2-\beta\eta)T)\}=\min\{
5,5/8,3/16,1/16\}=1/16$. Since $f_0=f_{\infty}=0$, then (H5)
holds. Again
$\Lambda_2^{-1}=((2T-\alpha\eta^2)-\beta\eta(2T-\eta))/
(\gamma(2-\beta\eta)(T-\eta))(\int_0^{T}sa(s)ds)^{-1}=\frac{25}{2}e^{-32}$,
since $f(u)$ is monotone decreasing function for $u\geqslant 2$,
taking $\rho_2=32$,when $u\in[\gamma\rho_2,\rho_2]$, we obtain
\[
f(u)\geqslant f(32)=1024e^{-32}>400e^{-32}=\Lambda_2^{-1}\rho_2,
\]
which implies (H6) holds. Hence, by Theorem \ref{Thm3-3},
BVP \eqref{E4-5}, \eqref{E4-6} has at least two positive
solutions $u_1$ and $u_2$ such that
$0<\|u_1\|<32<\|u_2\|$.
\end{example}

\subsection*{Acknowledgements}
The authors would like to thank Dr. Elvin James Moore for his
valuable advice. This research is supported by the Centre of
Excellence in Mathematics, Thailand.

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