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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 82, 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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2005/82\hfil Two positive solutions]
{Two positive solutions for second-order functional
and ordinary boundary-value problems}

\author[K. G. Mavridis, P. Ch. Tsamatos\hfil EJDE-2005/82\hfilneg]
{Kyriakos G. Mavridis, Panagiotis Ch. Tsamatos}  % in alphabetical order

\address{Kyriakos G. Mavridis \hfill\break
Department of Mathematics, University of Ioannina, P. O. Box 1186,
451 10 Ioannina, Greece}
\email{kmavride@otenet.gr}

\address{Panagiotis Ch. Tsamatos \hfill\break
Department of Mathematics, University of Ioannina, P. O. Box 1186,
451 10 Ioannina, Greece}
\email{ptsamato@cc.uoi.gr}

\date{}
\thanks{Submitted February 22, 2005. Published July 15, 2005.}
\subjclass[2000]{34K10, 34B18}
\keywords{Boundary value problems; Positive solutions}


\begin{abstract}
 In this paper we use a fixed point theorem due to Avery and Henderson
 to prove, under appropriate conditions, the existence of at least
 two positive solutions for a second-order functional and ordinary
 boundary-value problem.
\end{abstract}

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

\section{Introduction}

Throughout the recent years, an increasing interest has been observed in
finding conditions that guarantee the existence of positive
solutions for boundary-value problems. The well known
Guo-Krasnoselskii fixed point theorem \cite{g1,k5} has been proved
to be a useful tool to achieve such conditions, while this same
theorem can be applied repeatedly to prove the existence of
multiple positive solutions (see \cite{e1,e2,k1,k3,m1,w1,w2} and the
references therein). Besides this theorem, there are a number of
others, referring to Banach spaces ordered by proper cones, that
provide conditions such that certain boundary-value problems have
multiple positive solutions, for example the Leggett-Williams
fixed point theorem \cite{k2,l1} and the Avery-Henderson fixed
point theorem \cite{a2}.

At this point we would like to stress the recent increase in the
number of papers dealing with functional boundary-value problems,
usually specifically with the existence of positive solutions for
these problems (see \cite{k1,k4,l2,l3,m2} and the references
therein). Here we will first study a functional boundary-value
problem and then a separate section will be devoted to briefly
outlining the analogues of our results for the ordinary case,
since even these analogues are novel. We will use the
Avery-Henderson fixed point theorem (\cite{a2}, see also
\cite{l3,l5}); for other papers using this theorem we refer to
\cite{a3,h1,l3,l4,l5}. It is remarkable that under certain
conditions the Avery-Henderson fixed point theorem can give
results similar to those of the Krasnoselskii fixed point theorem,
i.e. existence of solutions whose norm is upper and lower bounded
by specific constants. Corollaries $3.5$ and $4.3$ provide such
results.

Let $\mathbb{R}$ be the set of real numbers,
$\mathbb{R}^{+}:=\{x\in\mathbb{R}: x\geq0\}$ and $I:=[0,1]$. Also,
let $0\leq q<1$ and $J:=[-q,0]$. For every closed interval $B\subseteq J\cup I$
we denote by $C(B)$ the Banach space of all continuous real functions
$\psi:B\to\mathbb{R}$ endowed with the usual sup-norm
$$
\|\psi\|_{B}:=\sup\{|\psi(s)|:s\in B\}.
$$
Also, we define the set
$$
C^{+}(B):=\{\psi\in C(B): \psi\geq0\}.
$$
If $x\in C(J\cup I)$ and $t\in I$, then we denote by $x_{t}$ the element of
$C(J)$ defined by
$$
x_{t}(s)=x(t+s), \quad s\in J.
$$
Now, consider the equation
\begin{equation}
\left(p(t)x'(t)\right)'+f(t,x_{t})=0, \quad t\in I,\label{e1.1}
\end{equation}
along with the boundary conditions
\begin{gather}
x_{0}=\phi, \label{e1.2} \\
 ax(1)+bp(1)x'(1)=0,
\label{e1.3}
\end{gather}
where $f:\mathbb{R}^{+}\times C^{+}(J)
\to\mathbb{R}^{+}$, $p:I\to(0,+\infty)$ and
$\phi: J\to\mathbb{R}^{+}$ are continuous functions, $p$ is also
nondecreasing and such that
$0<\int_{0}^{1}\frac{1}{p(s)}ds<+\infty$, and $a,b\in\mathbb{R}$.
Also assume that
$\phi(0)=0$.

This paper is motivated by and extends the results of \cite{a1,k1}
and is organized as follows. In section $2$ we present the definitions and the lemmas we
are going to use, including the fixed point theorem, due to  Avery and
 Henderson \cite{a2}.
Section $3$ contains the results for the functional case and section $4$ the results for the
ordinary case. Finally, in section $5$ we give some applications of our results.

\section{Preliminaries and some basic lemmas}

\subsection*{Definition}
A function $x\in C(J\cup I)$ is a solution of the boundary-value problem
\eqref{e1.1}--\eqref{e1.3} if $x$ satisfies equation \eqref{e1.1},
the boundary condition \eqref{e1.3} and,
moreover $x|J=\phi$.

Define $P:I\to\mathbb{R}^{+}$ and $A:I\to\mathbb{R}$, as
$$
P(t):=\int_{0}^{t}\frac{1}{p(s)}ds,\quad t\in I,
\quad \hbox{and}\quad
A(t):=a\int_{t}^{1}\frac{1}{p(s)}ds+b,\quad t\in I.
$$
At this point we make our first assumption:

\begin{itemize}
\item[(H1)] For the constants $a,b$ and the function $p$,
$$
A(0)=aP(1)+b\neq0\quad\hbox{and}\quad aA(0)=a(aP(1)+b)\leq0.
$$
\end{itemize}

We remark that assumption (H1) is equivalent to
$$
a\neq-\frac{b}{P(1)}\quad\hbox{and}\quad
\min\{0,-\frac{b}{P(1)}\}\leq a\leq\max\{0,-\frac{b}{P(1)}\}.
$$

\begin{lemma} \label{lem2.1}
A function $x\in C(J\cup I)$ is a solution of the boundary-value problem
\eqref{e1.1}--\eqref{e1.3} if and only if $x$ is a fixed point of the operator
$T:C(J\cup I)\to C(J\cup I)$, with
$$
Tx(t)= \begin{cases}
  \phi(t), &t\in J\\
  \int_{0}^{1}G(t,s)f(s,x_{s})ds,& t\in I,
\end{cases}
$$
where
\begin{equation}
G(t,s)=\frac{1}{A(0)} \begin{cases}
  P(t)A(s), & 0\leq t\leq s\leq1\\
  P(s)A(t), & 0\leq s\leq t\leq1.
\end{cases}\label{e2.1}
\end{equation}
\end{lemma}

\begin{proof}
It is well known (see \cite{a1}) that the Green's function for the
homogenous boundary-value problem consisting of the equation
$(p(t)x'(t))'=0$ and the boundary conditions $x(0)=0$ and
\eqref{e1.3} is given by the formula \eqref{e2.1}. Therefore,
since $\phi(0)=0$, the proof is obvious.
\end{proof}

In the sequel, we need the following definitions:

 Let $\mathbb{E}$ be a real Banach
space. A cone in $\mathbb{E}$ is a nonempty, closed set
$\mathbb{P}\subset\mathbb{E}$ such that
\begin{itemize}
\item[(i)] $\kappa u+\lambda v\in\mathbb{P}$ for all $u$, $v\in\mathbb{P}$
and all $\kappa$, $\lambda\geq 0$

\item[(ii)] $u$, $-u\in\mathbb{P}$ implies $u=0$.
\end{itemize}


Let $\mathbb{P}$ be a cone in a real Banach space $\mathbb{E}$. A
functional $\psi :\mathbb{P}\to\mathbb{E}$ is said to be
increasing on $\mathbb{P}$ if $\psi(x)\leq\psi(y)$, for any $x$,
$y\in\mathbb{P}$ with $x\leq y$, where $\leq$ is the partial
ordering induced to the Banach space by the cone $\mathbb{P}$,
i.e.
$$
x\leq y \quad\hbox{if and only if}\quad y-x\in\mathbb{P}.
$$

Let $\psi$ be a nonnegative functional on a cone $\mathbb{P}$.
For each $d>0$ we denote by $\mathbb{P}(\psi,d)$ the set
$$
\mathbb{P}(\psi,d):=\{x\in\mathbb{P} : \psi(x)<d\}.
$$
\smallskip

At this point we can present  Theorem \ref{thm2.2},
due to Avery and  Henderson (\cite{a2}, see also \cite{l3,l5}).
This theorem will be used to show that our boundary-value problem
 \eqref{e1.1}--\eqref{e1.3} has at least
two distinct positive solutions and, moreover, for each of these
solutions, we have an upper bound at some specific point of its
domain and a lower bound at some other specific point of its
domain. Also, both solutions are concave and nondecreasing on $I$.

\begin{theorem} \label{thm2.2}
Let $\mathbb{P}$ be a cone in a real Banach
space $\mathbb{E}$. Let $\alpha$ and $\gamma$ be increasing,
nonnegative, continuous functionals on $\mathbb{P}$, and let $\theta$
be a nonnegative, continuous functional on $\mathbb{P}$ with $\theta(0)=0$ such
that, for some $c>0$ and $\Theta>0$,
$\gamma(x)\leq\theta(x)\leq\alpha(x)$ and $\|x\|\leq \Theta\gamma(x)$,
for all $x\in\overline{\mathbb{P}(\gamma,c)}$. Moreover, suppose there
exists a completely continuous operator
$T:\overline{\mathbb{P}(\gamma,c)}\to\mathbb{P}$ and $0<a<b<c$ such that
$$
\theta(\lambda x)\leq\lambda\theta(x), \quad\hbox{for}\quad
0\leq\lambda\leq1\quad\hbox{and}\quad x\in \partial
\mathbb{P}(\theta,b),
$$
and
\begin{itemize}
\item[(i)] $\gamma(Tx)>c$, for all $x\in \partial
\mathbb{P}(\gamma,c)$, \item[(ii)] $\theta(Tx)<b$, for all
$x\partial \in \mathbb{P}(\theta,b)$, \item[(iii)]
$\mathbb{P}(\alpha,a)\neq\emptyset$, and $\alpha(Tx)>a$, for all
$x\in\partial \mathbb{P}(\alpha,a)$,
\end{itemize}
or
\begin{itemize}
\item[(i')] $\gamma(Tx)<c$, for all $x\in\partial
\mathbb{P}(\gamma,c)$, \item[(ii')] $\theta(Tx)>b$, for all
$x\in\partial \mathbb{P}(\theta,b)$, \item[(iii')]
$\mathbb{P}(\alpha,a)\neq\emptyset$, and $\alpha(Tx)<a$, for all
$x\in\partial \mathbb{P}(\alpha,a)$.
\end{itemize}
Then $T$ has at least two fixed points $x_{1}$ and $x_{2}$ belonging to
$\overline{\mathbb{P}(\gamma,c)}$ such that
$$
a<\alpha(x_{1})\quad\hbox{and}\quad\theta(x_{1})<b,
$$
and
$$
b<\theta(x_{2})\quad\hbox{and}\quad\gamma(x_{2})<c.
$$
\end{theorem}

\section{Main Results}

Define the set
$$
\mathbb{K}:=\{x\in C(J\cup I):x(t)\geq 0,\quad t\in J\cup I,\;
x|I\hbox{ is nondecreasing and concave}\},
$$
which is a cone in $C(J\cup I)$. The following lemma (see
\cite{k4}) will be needed later.

\begin{lemma} \label{lem3.1}
Let $x:I\to\mathbb{R}$ be a nonnegative, nondecreasing and concave function.
Then, $x(t)\geq t\|x\|_{I}$, $t\in I$.
\end{lemma}

\begin{proof}
For any $t\in I$, since $x$ is nonnegative, nondecreasing and concave,
 we have
$$
x(t)=x((1-t)0+t)\geq(1-t)x(0)+tx(1)\geq tx(1)=t\|x\|_{I}.
$$
\end{proof}

Now let
$0< r_{1}\leq r_{2}\leq r_{3}\leq 1$
and consider the following functionals:
\begin{gather*}
\gamma(x)=x(r_{1}),\quad x\in\mathbb{K}, \\
\theta(x)=x(r_{2}),\quad x\in\mathbb{K}, \\
\alpha(x)=x(r_{3}),\quad x\in\mathbb{K}.
\end{gather*}
It is easy to see that $\alpha, \gamma$ are nonnegative, increasing
and continuous functionals on $\mathbb{K}$,
$\theta$ is nonnegative and continuous on $\mathbb{K}$ and $\theta(0)=0$.
Also, it is straightforward to see that
\begin{equation}
\gamma(x)\leq\theta(x)\leq\alpha(x),\quad x\in\mathbb{K},
\label{e3.1}
\end{equation}
since $x\in \mathbb{K}$ is nondecreasing on $I$. Furthermore,
for any $x\in\mathbb{K}$, by Lemma \ref{lem3.1} we have
$\gamma(x)=x(r_{1})\geq r_{1}\|x\|_{I}$.
So
\begin{equation}
\|x\|_{I}\leq\frac{1}{r_{1}}\gamma(x),\quad x\in\mathbb{K}. \label{e3.2}
\end{equation}
Additionally, by the definition of $\theta$ we obtain
$$
\theta(\lambda x)=\lambda\theta(x),\quad 0\leq\lambda\leq1,\quad x\in\mathbb{K}.
$$
At this point, we state the following assumptions:
\begin{itemize}
\item[(H2)]
There exist $M>0$, continuous function $u:I\to\mathbb{R}^{+}$
and a function $L:\mathbb{R}^{+}\to\mathbb{R}^{+}$, which is nondecreasing
on $[0,M]$, such that
\begin{gather*}
f(t,y)\leq u(t)L(\|y\|_{J}),\quad t\in I, \; y\in C^{+}(J), \\
L(M)\int_{0}^{1}G(r_{2},s)u(s)ds< Mr_{2}.
\end{gather*}

\item[(H3)] There exist constants $\delta\in(0,1)$, $\eta_{1},\eta_{3}>0$
and functions $\tau:I\to[0,q]$, continuous $v:I\to\mathbb{R}^{+}$ and
nondecreasing $w:\mathbb{R}^{+}\to\mathbb{R}^{+}$ such that
$$
f(t,y)\geq v(t)w(y(-\tau(t))),\quad t\in X, \; y\in C^{+}(J),
$$
where $X:=\{t\in I:\delta\leq t-\tau(t)\leq1\}$, $\sup\{v(t):t\in X\}>0$,
$$
w(\eta_{i})\int_{X}G(r_{i},s)v(s)ds>\frac{\eta_{i}}{\delta},\quad i\in\{1,3\},
$$
$0<\eta_{3}<M\delta r_{2}<\eta_{1}$,
and $M$ is defined in (H2).
\end{itemize}

The following lemma can be found in \cite{k4}. The proof is
provided for the sake of completeness.

\begin{lemma} \label{lem3.2}
If $x:I\to\mathbb{R}$ is a differentiable function with
$x'(t)\geq0$, $t\in I$, and $p:I\to\mathbb{R}$ is a positive and
nondecreasing function such that $(p(t)x'(t))'\leq0$, $t\in I$,
then $x$ is concave.
\end{lemma}

\begin{proof}
Let $t_{1},t_{2}\in I$, with $t_{1}\leq t_{2}$. Since $(p(t)x'(t))'\leq0$, $t\in I$, $px'$ is a nonincreasing
function on $I$, so $p(t_{1})x'(t_{1})\geq p(t_{2})x'(t_{2})$. Therefore, we have
\begin{align*}
p(t_{1})(x'(t_{2})-x'(t_{1}))
&=p(t_{1})x'(t_{2})-p(t_{1})x'(t_{1})\\
&\leq p(t_{1})x'(t_{2})-p(t_{2})x'(t_{2})\\
&=x'(t_{2})(p(t_{1})-p(t_{2})).
\end{align*}
Since $p$ is nondecreasing on $I$, we have $p(t_{1})-p(t_{2})\leq0$, so
$$
p(t_{1})(x'(t_{2})-x'(t_{1}))\leq
x'(t_{2})(p(t_{1})-p(t_{2}))\leq0.
$$
However $p(t_{1})\geq0$, therefore we get that $x'(t_{1})\geq
x'(t_{2})$, which implies that $x'$ is nonincreasing on $I$.
Consequently, $x$ is concave.
\end{proof}

The following lemma will be needed for the proof of Theorem \ref{thm3.4}.

\begin{lemma} \label{lem3.3}
Suppose that {\rm (H1)} holds. Then
\begin{itemize}
\item[(i)] $\frac{A(t)}{A(0)}>0$, $t\in I$. \item[(ii)]
$\frac{A'(t)}{A(0)}\geq0$, $t\in I$. \item[(iii)] $Tx(t)\geq0$,
$t\in I$, $x\in \mathbb{K}$. \item[(iv)] $(Tx)'(t)\geq0$, $t\in
I$, $x\in \mathbb{K}$. \item[(v)] $T(\mathbb{K})\subseteq
\mathbb{K}$.
\end{itemize}
\end{lemma}

\begin{proof}
(i) For any $t\in I$, keeping in mind that $aA(0)\leq0$, we get
\begin{align*}
\frac{A(t)}{A(0)}
&=\frac{a\int_{t}^{1}\frac{1}{p(s)}ds+b}{A(0)}\\
&=\frac{a}{A(0)}\int_{t}^{1}\frac{1}{p(s)}ds+\frac{b}{A(0)}\\
&\geq\frac{a}{A(0)}\int_{0}^{1}\frac{1}{p(s)}ds+\frac{b}{A(0)}\\
&=\frac{a\int_{0}^{1}\frac{1}{p(s)}ds+b}{A(0)}=1.
\end{align*}
Therefore, $\frac{A(t)}{A(0)}\geq0$, $t\in I$.

\noindent(ii) Since $p(t)>0$, $t\in I$, and, by (H1),
$\frac{a}{A(0)}\leq0$, for any $t\in I$ we have
$$
\frac{A'(t)}{A(0)}=\frac{-\frac{a}{p(t)}}{A(0)}
=-\frac{a}{A(0)}\frac{1}{p(t)}\geq0.
$$
(iii) By the definition of $T$, we get
$$
Tx(t)=\frac{A(t)}{A(0)}\int_{0}^{t}P(s)f(s,x_{s})ds+P(t)\int_{t}^{1}
\frac{A(s)}{A(0)}f(s,x_{s})ds, \enskip t\in I.
$$
Moreover, if $x\in\mathbb{K}$ then $x_{t}\geq0$, $t\in I$. By the
definition of $f$ we have $f(t,x_{t})\geq0$, $t\in I$.
So using (i) we conclude that $Tx(t)\geq0$, $t\in I$.

\noindent (iv) By the definition of $T$, for every $t\in I$ we get
\begin{align*}
(Tx)'(t)
&=A'(t)\int_{0}^{t}\frac{1}{A(0)}P(s)f(s,x_{s})ds
+A(t)\frac{1}{A(0)}P(t)f(t,x_{t})\\
&\quad+P'(t)\int_{t}^{1}\frac{1}{A(0)}A(s)f(s,x_{s})ds
-P(t)\frac{1}{A(0)}A(t)f(t,x_{t})\\
&=\frac{A'(t)}{A(0)}\int_{0}^{t}P(s)f(s,x_{s})ds
+P'(t)\int_{t}^{1}\frac{A(s)}{A(0)}f(s,x_{s})ds.
\end{align*}
Therefore, using (ii) and the easily provable facts that $P'(t)\geq0$,
$t\in I$, and $f(t,x_{t})\geq0$, $t\in I$, $x\in\mathbb{K}$,
we conclude that $(Tx)'(t)\geq0$, $t\in I$.

\noindent (v) From \eqref{e1.1}, for every $t\in I$ and $x\in\mathbb{K}$
we have
$$
(p(t)x'(t))'=-f(t,x_{t})\leq0.
$$
So, since for $x\in\mathbb{K}$ we have $x_{t}\geq0$, $t\in I$, by
the definition of $f$ and, according to Lemma \ref{lem3.2}, we
have that $Tx$ is concave. This, along with (iii) and (iv),
completes the proof.
\end{proof}

\begin{theorem} \label{thm3.4}
Suppose that assumptions {\rm (H1)--(H3)} hold and
$\|\phi\|_{J}<M$. Then the boundary-value problem
\eqref{e1.1}--\eqref{e1.3} has at least two solutions $x_{1}$,
$x_{2}$, which are concave and nondecreasing on $I$, positive on
$J\cup I$ and such that $x_{1}(r_{3})>\frac{\eta_{3}}{\delta}$,
$x_{1}(r_{2})<Mr_{2}$, $x_{2}(r_{2})>Mr_{2}$ and
$x_{2}(r_{1})<\frac{\eta_{1}}{\delta}$.
\end{theorem}

\begin{proof} First of all, we observe that, because of (H1), $f(t,\cdot)$ maps
bounded sets into bounded sets. Therefore $T$ is a completely continuous operator.
Additionally, according to Lemma \ref{lem3.3} we have
$T:\overline{\mathbb{K}(\gamma,c)}\to\mathbb{K}$.

Now we set $\beta_{1}=\frac{\eta_{1}}{\delta}$, $\beta_{2}=Mr_{2}$
and $\beta_{3}=\frac{\eta_{3}}{\delta}$. Let $x\in \partial
\mathbb{K}(\gamma,\beta_{1})$.
 Then $\gamma(x)=x(r_{1})=\beta_{1}$ and so $\|x\|_{I}\geq\beta_{1}$.
Having in mind assumption $(H_{3})$, we get
\begin{align*}
\gamma(Tx)
&=(Tx)(r_{1})\\
&=\int_{0}^{1}G(r_{1},s)f(s,x_{s})ds\\
&\geq\int_{X}G(r_{1},s)f(s,x_{s})ds\\
&\geq\int_{X}G(r_{1},s)v(s)w(x_{s}(-\tau(s)))ds\\
&=\int_{X}G(r_{1},s)v(s)w(x(s-\tau(s)))ds\\
&\geq\int_{X}G(r_{1},s)v(s)w(x(\delta))ds.
\end{align*}
Additionally, by assumption (H3), the definition of $\mathbb{K}$ and
Lemma \ref{lem3.1}, we have
\begin{align*}
\gamma(Tx)
&\geq\int_{X}G(r_{1},s)v(s)w(\delta\|x\|_{I})ds\\
&\geq w(\delta\beta_{1})\int_{X}G(r_{1},s)v(s)ds\\
&=w(\eta_{1})\int_{X}G(r_{1},s)v(s)ds\\
&>\frac{\eta_{1}}{\delta}=\beta_{1}.
\end{align*}
This means that condition (i) of Theorem \ref{thm2.2} is satisfied.

Now let $x\in \partial \mathbb{K}(\theta,\beta_{2})$. Then
$\theta(x)=x(r_{2})=\beta_{2}$ and so
$$
\|x\|_{I}\leq\frac{1}{r_{2}}x(r_{2})=\frac{1}{r_{2}}\beta_{2}=M.
$$
Also we assumed that $\|\phi\|_{J}\leq M$, so $\|x\|_{J\cup I}\leq M$.
Now, by $(H_{2})$, we have
\begin{align*}
\theta(Tx)
&=Tx(r_{2})\\
&=\int_{0}^{1}G(r_{2},s)f(s,x_{s})ds\\
&\leq\int_{0}^{1}G(r_{2},s)u(s)L(\|x_{s}\|_{J})ds\\
&\leq\int_{0}^{1}G(r_{2},s)u(s)L(M)ds\\
&=L(M)\int_{0}^{1}G(r_{2},s)u(s)ds
<Mr_{2}=\beta_{2}.
\end{align*}
So condition (ii) of Theorem \ref{thm2.2} is also satisfied.

Now, define the function $y:J\cup I\to\mathbb{R}$ with
$y(t)=\frac{\beta_{3}}{2}$. Then it is obvious that
$\alpha(y)=\frac{\beta_{3}}{2}<\beta_{3}$, so
$\mathbb{K}(\alpha,\beta_{3})\neq\emptyset$. Also, for any $x\in
\partial \mathbb{K}(\alpha,\beta_{3})$ we have
$\alpha(x)=x(r_{3})=\beta_{3}$. Therefore, $\|x\|_{I}\geq
\beta_{3}$. Now, having in mind assumption $(H_{3})$ and as in the
case of the functional $\gamma$ above, we get
$$
\alpha(Tx)=Tx(r_{3})\geq\int_{X}G(r_{3},s)v(s)w(x(\delta))ds\,.
$$
Taking into account assumption (H3), the definition of $\mathbb{K}$ and
Lemma \ref{lem3.1}, we have
$$
\alpha(Tx)=w(\eta_{3})\int_{X}G(r_{3},s)v(s)ds>\frac{\eta_{3}}{\delta}=\beta_{3}.
$$
Consequently, assumption (iii) of Theorem \ref{thm2.2} is satisfied.
The result can now be obtained by applying Theorem \ref{thm2.2}.
\end{proof}

The solutions $x_{1}$, $x_{2}$ obtained  in Theorem \ref{thm3.4}
are both nondecreasing. Thus, in the special case when
$r_{1}=r_{2}=r_{3}=1$, we have that
$x_{i}(r_{j})=x_{i}(1)=\|x_{i}\|$, $i=1,2$, $j=1,2,3$. Therefore,
we have the following corollary of Theorem \ref{thm3.4}.

\begin{corollary} \label{coro3.5}
Suppose that assumptions {\rm (H1)--(H3)} hold for
$r_{1}=r_{2}=r_{3}=1$ and furthermore $\|\phi\|_{J}\leq M$. Then
the boundary-value problem \eqref{e1.1}--\eqref{e1.3} has at least
two solutions $x_{1}$, $x_{2}$, which are concave and
nondecreasing on $I$, positive on $J\cup I$ and such that
$$
\frac{\eta_{3}}{\delta}<\|x_{1}\|<M<\|x_{2}\|<\frac{\eta_{1}}{\delta}.
$$
\end{corollary}

\section{The Ordinary Case}

Suppose that $q=0$. Then $J=\{0\}$, so the boundary-value problem
\eqref{e1.1}--\eqref{e1.3} is reformulated as follows
\begin{gather}
(p(t)x'(t))'+f(t,x(t))=0, \quad t\in I, \label{e4.1}\\
x(0)=0, \label{e4.2} \\
ax(1)+bp(1)x'(1)=0, \label{e4.3}
\end{gather}
where $f:\mathbb{R}^{+}\times\mathbb{R}^{+}\to\mathbb{R}^{+}$,
 $p:I\to(0,+\infty)$ are continuous
functions, $p$ is also nondecreasing and lower bounded by a strictly
positive number and $a,b\in\mathbb{R}$. Note that equation \eqref{e4.1}
is equivalent to the  form
$$
(p(t)x'(t))'+f(t,x_{t}(0))=0, \quad t\in I
$$
and $C^{+}(\{0\})\equiv \mathbb{R}^{+}$, so
$f:\mathbb{R}^{+}\times C^{+}(\{0\})\to\mathbb{R}^{+}$.

Now, the analogue of Lemma \ref{lem2.1} for this case is as follows.

\begin{lemma} \label{lem4.1}
A function $x\in C(I)$ is a solution of the
boundary-value problem \eqref{e4.1}--\eqref{e4.3} if and only if $x$
is a fixed point of the operator $\widehat{T}:C(I)\to C(I)$, with
$$
\widehat{T}x(t)=\int_{0}^{1}G(t,s)f(s,x(s))ds, \quad t\in I,
$$
where $G$ is defined in Lemma $\ref{lem2.1}$.
\end{lemma}

Assumptions (H2), (H3), for the special case $q=0$, are stated as follows:
\begin{itemize}
\item[(H2')] There exist $M>0$, continuous function
$u:I\to\mathbb{R}^{+}$ and a function
$L:\mathbb{R}^{+}\to\mathbb{R}^{+}$ which is nondecreasing on
$[0,M]$, such that
\[
f(t,y)\leq u(t)L(y)\quad\text{ for all} \quad t\in I\quad\text{
and}\quad y\in\mathbb{R}^{+} \] and
$$
L(M)\int_{0}^{1}G(r_{2},s)u(s)ds< Mr_{2}.
$$

\item[(H3')] There exist constants $\delta\in(0,1)$, $\eta_{1},\eta_{3}>0$
 and functions $v:I\to\mathbb{R}^{+}$ continuous and
$w:\mathbb{R}^{+}\to\mathbb{R}^{+}$ nondecreasing, such that
\[
f(t,y)\geq v(t)w(y)\quad\text{ for all} \quad t\in
Z:=[\delta,1]\quad\text{ and}\quad y\in \mathbb{R}^{+}
\]
and
$$
w(\eta_{i})\int_{Z}G(r_{i},s)v(s)ds>\frac{\eta_{i}}{\delta},\quad i\in\{1,3\},
$$
where $0<\eta_{3}<M\delta r_{2}<\eta_{1}$, and $M$ is defined in
(H2').
\end{itemize}

Therefore, we have the following results, which are the analogue
of Theorem \ref{thm3.4} and Corollary \ref{coro3.5} respectively.

\begin{theorem} \label{thm4.2}
Suppose that assumptions {\rm (H1), (H2)', (H3')} hold. Then the
boundary-value problem \eqref{e4.1}--\eqref{e4.3} has at least two
solutions $x_{1}$, $x_{2}$, which are concave, nondecreasing and
positive on $I$, such that $x_{1}(r_{3})>\frac{\eta_{3}}{\delta}$,
$x_{1}(r_{2})<Mr_{2}$, $x_{2}(r_{2})>Mr_{2}$ and
$x_{2}(r_{1})<\frac{\eta_{1}}{\delta}$.
\end{theorem}

\begin{corollary} \label{coro4.3}
Suppose that {\rm (H1), (H2'), (H3')} hold for
$r_{1}=r_{2}=r_{3}=1$. Then the boundary-value problem
\eqref{e4.1}--\eqref{e4.3} has at least two solutions $x_{1},
x_{2}$, which are concave, nondecreasing and positive on $I$, such
that
$$
\frac{\eta_{3}}{\delta}<\|x_{1}\|<M<\|x_{2}\|<\frac{\eta_{1}}{\delta}.
$$
\end{corollary}


\section{Applications}


\noindent{\bf 1.} Consider the boundary-value problem
\begin{gather}
(e^{t}x'(t))'+\exp\big(\frac{t+5}{20}x(t-\frac{1}{2})\big)=0,
\quad t\in I:=[0,1] \label{e5.1}\\
x_{0}(t)=\phi(t):=|t|,\quad t\in J:=[-\frac{1}{2},0], \label{e5.2}\\
x(1)-2ex'(1)=0. \label{e5.3}
\end{gather}
Obviously, $f(t,y):=\exp((t+5)y/20)$ is positive on
$\mathbb{R}^{+}\times C^{+}(J)$, $\phi$ is positive on $J$ and $p(t):=e^{t}$
is positive and nondecreasing on $I$. Also we have $a=1$ and $b=-2$,
so $P(t)=1-e^{-t}$, $t\in I$, $A(t)=e^{-t}-e^{-1}-2$, $t\in I$,
$A(0)=-(1+e^{-1})\neq0$, $aA(0)=-(1+e^{-1})\leq0$ and
$$
G(t,s)=\begin{cases}
  -\frac{(1-e^{-t})(e^{-s}-e^{-1}-2)}{1+e^{-1}}, & 0\leq t\leq s\leq1\\[3pt]
  -\frac{(1-e^{-s})(e^{-t}-e^{-1}-2)}{1+e^{-1}}, & 0\leq s\leq t\leq1.
\end{cases}
$$
Set $r_{1}=1/3$, $r_{2}=1/2$ and $r_{3}=2/3$.
Define $L(t)=e^{0.3t}$, $t\in\mathbb{R}^{+}$, and $u(t)=1$, $t\in I$.
Since
$$
e^{0.3M}+\frac{1+e^{-1}}{2e^{-1}+e^{-0.5}-e^{-1.5}}M<0
$$
for $M=2$, assumption (H2) is satisfied.

Additionally, set $\delta=1/5$, $\tau(t)=1/2$,
$t\in I$, $v(t)=1$, $t\in I$ and $w(t)=e^{0.25t}$, $t\in\mathbb{R}^{+}$.
Then, $X=[\frac{7}{10},1]$ and the inequalities in assumption (H3) take
the forms
\begin{gather*}
e^{0.25\eta_{1}}+\frac{5(1+e^{-1})}{(1-e^{-1/3})(e^{-0.7}-1.3e^{-1})
-0.6}\eta_{1}>0, \\
e^{0.25\eta_{3}}+\frac{5(1+e^{-1})}{(1-e^{-2/3})(e^{-0.7}-1.3e^{-1})-0.6}
\eta_{3}>0,
\end{gather*}
which are satisfied for $\eta_{1}=29$ and $\eta_{3}=0.004$.

Finally, it is obvious that $0<0.004<0.2<29$ and $\|\phi\|_{J}\leq
2$, so we can apply Theorem \ref{thm3.4} to get that the
boundary-value problem \eqref{e5.1}--\eqref{e5.3} has at least two
concave and nondecreasing on $[0,1]$ and positive on
$[-\frac{1}{2},1]$ solutions $x_{1}$, $x_{2}$, such that
$$
x_{1}(\frac{2}{3})>0.02,\quad
x_{1}(\frac{1}{2})<1,\quad
x_{2}(\frac{1}{2})>1,\quad
x_{2}(\frac{1}{3})<145.
$$

\noindent{\bf 2.} Consider the boundary-value problem
\begin{gather}
x''(t)+\big(x(t)-\frac{4}{5}\big)^{5}+1=0, \quad t\in I:=[0,1], \label{e6.1}\\
x(0)=0, \label{e6.2}\\
2x'(1)-x(1)=0. \label{e6.3}
\end{gather}
Obviously, $f(t,y):=(y-\frac{4}{5})^{5}+1$ is positive on
$\mathbb{R}^{+}\times \mathbb{R}^{+}$ and
$p(t):=1$ is positive and nondecreasing on $I$. Also we have
$a=-1$ and $b=2$, so $P(t)=t$, $t\in I$, $A(t)=t+1$, $t\in I$, $A(0)=1\neq0$,
$aA(0)=-1\leq0$ and
$$
G(t,s)= \begin{cases}
  t(s+1), & 0\leq t\leq s\leq 1\\
  s(t+1), & 0\leq s\leq t\leq 1.
\end{cases}
$$
Set $r_{1}=2/5$, $r_{2}=3/5$ and $r_{3}=4/5$.
Define $L(t):=(t-\frac{4}{5})^{5}+1$, $t\in\mathbb{R}^{+}$, and
$u(t)=1$, $t\in I$. Since
$$
\big(M-\frac{4}{5}\big)^{5}+1-\frac{5}{6}M<0
$$
for $M=1.5$, assumption (H2') is satisfied.

Additionally, set $\delta=9/10$, $v(t)=1$, $t\in I$ and
$w(t)=(t-\frac{4}{5})^{5}+1$, $t\in\mathbb{R}^{+}$. Then,
$Z=[\frac{9}{10},1]$ and the inequalities in assumption (H3') take
the forms
$$
\big(\eta_{1}-\frac{4}{5}\big)^{5}+1-\frac{2500}{351}\eta_{1}>0,
\quad
\big(\eta_{3}-\frac{4}{5}\big)^{5}+1-\frac{2500}{351}\eta_{3}>0,
$$
which are satisfied for $\eta_{1}=3$ and $\eta_{3}=0.1$.
Finally, it is obvious that
$$
0<\eta_{3}<\frac{27}{50}M<\eta_{1},
$$
so we can apply Theorem \ref{thm4.2} to conclude that the boundary-value problem
\eqref{e6.1}--\eqref{e6.3}
has at least two solutions $x_{1}$, $x_{2}$, which are concave,
 nondecreasing and positive on $[0,1]$,
such that
$$
x_{1}(\frac{3}{4})>\frac{1}{9},\quad
x_{1}(\frac{1}{2})<\frac{9}{10},\quad
x_{2}(\frac{1}{2})>\frac{9}{10},\quad
x_{2}(\frac{1}{4})<\frac{10}{3}.
$$

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\end{document}