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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 53, 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 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/53\hfil Fixed point theorems]
{Two modifications of the Leggett-Williams fixed point theorem
and their applications}

\author[K. G. Mavridis\hfil EJDE-2010/53\hfilneg]
{Kyriakos G. Mavridis} 

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

\thanks{Submitted February 15, 2010. Published April 14, 2010.}
\subjclass[2000]{47H10, 34B40, 34K10, 34B18}
\keywords{Leggett-Williams fixed point theorem;
positive solutions of \hfill\break\indent
boundary value problems; functional second
order differential equations}

\begin{abstract}
 This article presents two modifications of the Leggett-Williams
 fixed point theorem, and two applications of these
 results to a terminal and to a boundary value problem for
 ordinary differential equations.
\end{abstract}

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

\section{Introduction}\label{sect:001}

This article presents two modifications of the fixed point theorem
named after  Leggett and  Williams \cite{lg1979}, published in 1979,
as well as two applications of these results to a terminal and
to a boundary value problem for ordinary differential equations.
The widely used version of the Leggett-Williams fixed point
theorem provides conditions which ensure the existence of at
least three fixed points. However this version of the theorem is
only an extension of the original result, presented in the same
paper by the authors, which is in turn a modification, but not a true
extension, of the well-known Krasnoselskii fixed point theorem.
The original result presents conditions which guarantee the existence
of at least one fixed point, just like the Krasnoselskii fixed point
theorem does. One of the differences between the two theorems,
lies in the sets chosen to replace the order intervals, which are
present in the class of fixed point theorems based on the classical
Schauder theorem. Namely, at the Krasnoselskii fixed point theorem
that set is of the form $\{x: a\leq\|x\|\leq b\}$, where
$a,b\in(0,+\infty)$, whilst at the Leggett-Williams fixed point
theorem that set is of the form $\{x: a\leq\alpha(x)\text{ and }
 \|x\|\leq b\}$, where $a,b\in(0,+\infty)$ and $\alpha$ is a
properly chosen functional. Although the two approaches are not
easily comparable, using the functional $\alpha$, which, by
its definition, cannot coincide with the norm, allows for easier
calculations and more versatile results. In this context,
although the essence of the two theorems, and indeed lots of others
based on them, is more or less the same, it is preferable to use
the Leggett-Williams approach.

Here, following the ideas demonstrated in
\cite{a1998, aho2008a, aho2008b, aho2008c}, see also
\cite{a1976, k2008, z1986}, the set $\{x: a\leq\alpha(x) \text{ and }
 \|x\|\leq b\}$ is replaced by the set $\{x: a\leq\alpha(x) \text{ and }
 \beta(x)\leq b\}$, where $\beta$ is another properly chosen functional.
This modification is an extension of the original Leggett-Williams fixed
point theorem, since the functional $\beta$ can coincide with the norm.
A closely related result can be found in \cite{aha2010}.
Additionally, going a step further, the set $\{x: a\leq\alpha(x)
\text{ and } \|x\|\leq b\}$ is replaced by the set
$\{x: u\preceq A(x)\text{ and }B(x)\preceq v\}$, where $A,B$
are operators and $u,v$ are functions. Since we are not aware
of any known proofs for these results, we use the fixed point index
to prove them.

A specific example regarding a terminal value problem and another
one regarding a boundary value problem, both for second order
differential equations, are provided to demonstrate the applicability
of the results and to pinpoint the advantages of their use.
These problems are well-known in the literature, for example
the terminal value problem is studied in \cite{mm1999, w2004, z1994}
and the boundary value problem is studied in \cite{kt2000}.
It is worth mentioning that, to the best of our knowledge, the results
we obtain here are new.

\section{Existence Theorems}\label{sect:002}

Let $\mathbb{R}$ be the set of real numbers. For any interval
$I\subseteq\mathbb{R}$ and any set $S\subseteq\mathbb{R}$,
by $C(I,S)$ we denote the set of all continuous functions defined
on $I$, which have values in $S$.

\begin{lemma}[Fixed Point Index]\label{lemm:001}
Let $Q$ be a retract of a Banach space $E$. For every open subset
$U$ of $Q$ and every completely continuous map
$A:\overline{U}\to Q$ which has no fixed points on
$\partial U$ (i.e. the boundary of $U$), there exists an integer
$i(A,U,Q)$ satisfying the following
\begin{itemize}
\item[(i)]
if $A:\overline{U}\to U$ is a constant map, then $i(A,U,Q)=1$.
\item[(ii)]
if $U_{1}$ and $U_{2}$ are disjoint open subsets of $U$ such that
$A$ has no fixed points on $\overline{U}\backslash(U_{1}\cup U_{2})$,
then $i(A,U,Q)=i(A,U_{1},Q)+i(A,U_{2},Q)$, where
$i(A,U_{k},Q)=i(A|\overline{U_{k}},U_{k},Q)$, $k=1,2$.
\item[(iii)]
if $I$ is a compact interval in $\mathbb{R}$ and
$h:I\times\overline{U}\to Q$ is a continuous map with relatively
compact range such that $h(\lambda,x)\ne x$ for
$(\lambda,x)\in I\times\partial U$, then $i(h(\lambda,\cdot),U,Q)$
is well-defined and independent of $\lambda$.
\item[(iv)]
if $i(A,U,Q)\ne0$, then $A$ has at least one fixed point in $U$.
\item[(v)]
if $Q_{1}$ is a retract of $Q$ and $A(\overline{U})\subset Q_{1}$,
then $i(A,U,Q)=i(A,U\cap Q_{1},Q_{1})$, where
$i(A,U\cap Q_{1},Q_{1})=i(A|\overline{U\cap Q_{1}},U\cap Q_{1},Q_{1})$.
\item[(vi)]
if $V$ is open in $U$ and $A$ has no fixed points in
$\overline{U}\backslash V$, then $i(A,U,Q)=i(A,V,Q)$.
\end{itemize}
\end{lemma}

\begin{definition}\label{defi:004} \rm
Let $E$ be a real Banach space. A nonempty closed convex set
$K\subseteq E$ is called a cone if it satisfies the following
two conditions
\begin{itemize}
\item[(i)]
for every $x\in K$ and $\lambda\geq0$ it holds that $\lambda x\in K$,
\item[(ii)]
if $x\in K$ and $-x\in K$ then $x=0$.
\end{itemize}
Every cone induces an ordering in $E$ given by
\[
x\leq y\quad\text{if and only if}\quad  y-x\in K.
\]
\end{definition}

\begin{definition}\label{defi:005} \rm
A map $\alpha$ is said to be a concave positive functional on a cone
$K$ of a real Banach space $E$ if $\alpha:K\to[0,+\infty)$
is continuous and
\[
\alpha(\lambda x+(1-\lambda)y)\geq \lambda\alpha(x)+(1-\lambda)\alpha(y)
\]
for all $x,y\in K$ and $\lambda\in[0,1]$. Similarly, we say that
the map $\beta$ is a convex positive functional on a cone $K$
of a real Banach space $E$ if $\beta:P\to[0,+\infty)$ is continuous and
\[
\beta(\lambda x+(1-\lambda)y)\leq \lambda\alpha(x)+(1-\lambda)\alpha(y)
\]
for all $x,y\in K$ and $\lambda\in[0,1]$.
\end{definition}

\begin{theorem}[Leggett-Williams \cite{lg1979}]\label{theo:005}
Let $K$ be a cone in a Banach space $E$ and define the sets
\[
K_{\epsilon_{1}}:=\{x\in K: \|x\|\leq\epsilon_{1}\},
\text{ for } \epsilon_{1}>0
\]
and
\[
S(\beta,\epsilon_{2},\epsilon_{3}):=\{x\in K:
\epsilon_{2}\leq\beta(x)\text{ and }\|x\|\leq\epsilon_{3}\},
\]
for $\epsilon_{3}>\epsilon_{2}>0$ and any concave positive functional
$\beta$ defined on the cone $K$, with $\beta(x)\leq\|x\|$.

Suppose that $c\geq b>a>0$, $\alpha$ is a concave positive functional
with $\alpha(x)\leq\|x\|$ and $A:K_{c}\to K$ is a completely continuous
operator, such that
\begin{itemize}
\item[(i)]
$\{x\in S(\alpha,a,b): \alpha(x)>a\}\ne\emptyset$, and $\alpha(Ax)>a$ if $x\in S(\alpha,a,b)$,
\item[(ii)]
$Ax\in K_{c}$ if $x\in S(\alpha,a,c)$,
\item[(iii)]
$\alpha(Ax)>a$ for all $x\in S(\alpha,a,c)$ with $\|Ax\|>b$.
\end{itemize}
Then $A$ has a fixed point in $S(\alpha,a,c)$.
\end{theorem}

\begin{theorem}\label{theo:001}
Let $I\subseteq\mathbb{R}$ and $E$ be the Banach space of all bounded
functions $x\in C(I,\mathbb{R})$ endowed with the norm
\[
\|x\|:=\sup\{|x(t)|: t\in I\},\; x\in C(I,\mathbb{R}).
\]
Suppose that
\begin{itemize}
\item
$K$ is a cone in $E$ and for any $\epsilon>0$
\[
K_{\epsilon}:=\{x\in K:\|x\|\leq\epsilon\}
\]
\item
$0<a<b<c<d$ are real numbers
\item
$T:K_{d}\to K$ is completely continuous
\item
$\alpha$ is a concave positive functional and $\beta$ is a
convex positive functional such that $\alpha(x)\leq\beta(x)$, $x\in K$
\end{itemize}
and set
\[
K_{\alpha,\beta}(a,b):=\{x\in K: \alpha(x)\geq a\text{ and }
\beta(x)\leq b\}.
\]
If
\begin{itemize}
\item[(i)]
${\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{c})\neq\emptyset$
(i.e. the internal of $K_{\alpha,\beta}(a,b)\cap K_{c}$ with respect
to $K_{d}$ is non-empty) and for $x\in K_{\alpha,\beta}(a,b)\cap K_{c}$
it holds that
\[
\beta(Tx)<b,\quad \alpha(Tx)>a
\]
\item[(ii)]
$Tx\in K_{d}$ for $x\in\overline{K_{\alpha,\beta}(a,b)\cap K_{d}}$
\item[(iii)]
$\beta(Tx)<b$ and $\alpha(Tx)>a$ for $x\in K_{\alpha,\beta}(a,b)\cap K_{d}$ with $\|Tx\|>c$
\end{itemize}
then the operator $T$ has at least one fixed point
\[
y\in{\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{d})
\]
i.e.
$\alpha(y)>a$, $\beta(y)<b$, $\|y\|\leq d$.
\end{theorem}

\begin{proof}
Suppose that $x\in\partial({\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)
\cap K_{d}))$ is a fixed point of operator $T$. Then
\begin{equation}\label{equa:003}
\alpha(x)=a\ \text{or}\ \beta(x)=b.
\end{equation}
Also, obviously either
\[
x\in K_{\alpha,\beta}(a,b)\cap K_{c}
\]
or
$ \|x\|>c$.
\begin{itemize}
\item
If $x\in K_{\alpha,\beta}(a,b)\cap K_{c}$ then according to assumption $(i)$, we have
\[
\beta(x)=\beta(Tx)<b\text{ and }\alpha(x)=\alpha(Tx)>a,
\]
which contradicts \eqref{equa:003}.
\item
If $\|x\|>c$ then
\[
\|Tx\|=\|x\|>c,
\]
so according to assumption $(iii)$, we have
\[
\beta(Tx)<b\text{ and }\alpha(Tx)>a,
\]
which contradicts \eqref{equa:003}.
\end{itemize}
So $T$ has no fixed points on
$\partial\big({\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{d})\big)$.

Since ${\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{c})\neq\emptyset$,
we choose $x_{0}\in{\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{c})$
 and define the map
\[
h:[0,1]\times\overline{{\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)
\cap K_{d})}\to K_{d}
\]
by
$h(t,x)=(1-t)Tx+tx_{0}$.
It is easy to see that $h$ is continuous and
\[
h\left([0,1]\times\overline{{\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{d})}\right)
\]
is relatively compact.

Suppose there exists
\[
(t,x)\in[0,1]\times\partial\big({\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)
\cap K_{d})\big)
\]
such that $h(t,x)=x$. Then
\[
\alpha(x)=a\quad\text{or}\quad \beta(x)=b.
\]
\begin{itemize}
\item
If $\|Tx\|>c$ then by assumption $(iii)$ we have
\[
\beta(Tx)<b\ \ \text{ and }\ \ \alpha(Tx)>a,
\]
so
\begin{itemize}
\item
if $\alpha(x)=a$ then
\begin{align*}
\alpha(x)
&=\alpha(h(t,x))
=\alpha((1-t)Tx+tx_{0})\\
&\geq(1-t)\alpha(Tx)+t\alpha(x_{0})
>(1-t)a+ta\\
&=a
\end{align*}
which contradicts $\alpha(x)=a$.
\item
if $\beta(x)=b$ then
\begin{align*}
\beta(x)
&=\beta(h(t,x))
=\beta((1-t)Tx+tx_{0})\\
&\leq(1-t)\beta(Tx)+t\beta(x_{0})
<(1-t)b+tb\\
&=b
\end{align*}
which contradicts $\beta(x)=b$.
\end{itemize}
\item
If $\|Tx\|\leq c$ then
\begin{align*}
\|x\|
&=\|h(t,x)\|
=\|(1-t)Tx+tx_{0}\|\\
&\leq(1-t)\|Tx\|+t\|x_{0}\|
<(1-t)c+tc\\
&=c,
\end{align*}
therefore by assumption $(i)$ we have
\[
\beta(Tx)<b\ \ \text{ and }\ \ \alpha(Tx)>a,
\]
so
\begin{itemize}
\item
if $\alpha(x)=a$ then
\begin{align*}
\alpha(x)
&=\alpha(h(t,x))
=\alpha((1-t)Tx+tx_{0})\\
&\geq(1-t)\alpha(Tx)+t\alpha(x_{0})
>(1-t)a+ta\\
&=a
\end{align*}
which contradicts $\alpha(x)=a$.
\item
if $\beta(x)=b$ then
\begin{align*}
\beta(x)
&=\beta(h(t,x))
=\beta((1-t)Tx+tx_{0})\\
&\leq(1-t)\beta(Tx)+t\beta(x_{0})
<(1-t)b+tb\\
&=b
\end{align*}
which contradicts $\beta(x)=b$.
\end{itemize}
\end{itemize}
Consequently, for each
\[
(t,x)\in[0,1]\times\partial\left({\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{d})\right)
\]
it holds that
$h(t,x)\ne x$.
So, by Lemma \ref{lemm:001}, we have
\[
i\left(T,{\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{d}),K_{d}\right)
=i\left(x_{0},{\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{d}),K_{d}\right)
=1.
\]
Hence, operator $T$ has at least one fixed point
$y\in{\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{d})$, i.e.
\[
\alpha(y)>a,\ \beta(y)<b, \quad \|y\|\leq d.
\]
\end{proof}

\begin{definition}\label{defi:002} \rm
Let $I\subseteq\mathbb{R}$ be bounded and $f,g\in C(I,\mathbb{R})$.
We define the relation $\preceq$ by
\[
f\preceq g\quad \text{if and only if}\quad
 f(t)\leq g(t),\; \forall t\in I
\]
and the relation $\prec$ by
\[
f\prec g\quad \text{if and only if}\quad  f(t)<g(t),\; \forall t\in I.
\]
\end{definition}

\begin{definition}\label{defi:003} \rm
Let $I\subseteq\mathbb{R}$ be bounded and
$A\in C(C(I,\mathbb{R}),C(I,\mathbb{R}))$. We say that operator
$A$ satisfies
\begin{itemize}
\item
property P1 if and only if
\[
(1-t)A(x)+tA(y)\preceq A((1-t)x+ty),\quad \forall x,y\in C(I,\mathbb{R}),\ \forall t\in[0,1].
\]
\item
property P2 if and only if
\[
A((1-t)x+ty)\preceq(1-t)A(x)+tA(y),\quad \forall x,y\in C(I,\mathbb{R}),\ \forall t\in[0,1].
\]
\item
property P3 if and only if
\[
Ax(t)\geq0,\ \forall x\in C(I,\mathbb{R}),\quad \forall t\in[0,1].
\]
\end{itemize}
\end{definition}

\begin{theorem}\label{theo:003}
Let $I\subseteq\mathbb{R}$ be bounded and $E$ be the Banach space of
all bounded functions $x\in C(I,\mathbb{R})$ endowed with the norm
\[
\|x\|:=\sup\{|x(t)|: t\in I\},\quad x\in C(I,\mathbb{R}).
\]
Suppose that
\begin{itemize}
\item
$K$ is a cone in $E$ and for any $\epsilon>0$
\[
K_{\epsilon}:=\{x\in K:\|x\|\leq\epsilon\}
\]
\item
$0<c<d$ are real numbers
\item
$T:K_{d}\to K$ is completely continuous
\item
$A$ is an operator satisfying properties P1 and P3, and $B$ is an operator satisfying properties P2 and P3, such that $A(x)\preceq B(x)$, $x\in K$
\item
$u,v\in C(I,[0,+\infty))$ with $u\prec v$
\end{itemize}
and set
\[
K_{A,B}(u,v):=\{x\in K: u\preceq A(x)\text{ and }B(x)\preceq v\}.
\]
If
\begin{itemize}
\item[(i)]
${\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{c})\neq\emptyset$
(i.e. the internal of $K_{A,B}(u,v)\cap K_{c}$ with respect to $K_{d}$
is non-empty) and for $x\in K_{A,B}(u,v)\cap K_{c}$ it holds that
\[
B(Tx)\prec v\quad \text{and}\quad u\prec A(Tx)
\]
\item[(i)]
$Tx\in K_{d}$ for $x\in\overline{K_{A,B}(u,v)\cap K_{d}}$
\item[(i)]
$B(Tx)\prec v$ and $u\prec A(Tx)$ for $x\in K_{A,B}(u,v)\cap K_{d}$
with $\|Tx\|>c$
\end{itemize}
then the operator $T$ has at least one fixed point
\[
y\in{\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{d})
\]
i.e.
\[
u\prec A(y),\quad B(y)\prec v,\quad \|y\|\leq d.
\]
\end{theorem}

\begin{proof}
Suppose that
$x\in\partial\left({\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{d})\right)$
is a fixed point of operator $T$. Then
\begin{equation}\label{equa:006}
Ax(t_{0})=u(t_{0})\ \text{or}\ Bx(t_{0})=v(t_{0}),\ \text{for some}\ t_{0}\in I.
\end{equation}
Also, obviously either
\[
x\in K_{A,B}(u,v)\cap K_{c}
\]
or
$\|x\|>c$.

\begin{itemize}
\item
If $x\in K_{A,B}(u,v)\cap K_{c}$ then according to assumption $(i)$,
we have
\[
B(x)=B(Tx)\prec v\text{ and }u\prec A(Tx)=A(x),
\]
which contradicts \eqref{equa:006}.
\item
If $\|x\|>c$ then
\[
\|Tx\|=\|x\|>c,
\]
so according to assumption $(iii)$, we have
\[
B(Tx)\prec v\text{ and }u\prec A(Tx),
\]
which contradicts \eqref{equa:006}.
\end{itemize}
So $T$ has no fixed points on
$\partial\left({\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{d})\right)$.

Since ${\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{c})\neq\emptyset$, we choose $x_{0}\in{\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{c})$ and define the map
\[
h:[0,1]\times\overline{{\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{d})}\to K_{d}
\]
by
\[
h(t,x)=(1-t)Tx+tx_{0}.
\]
It is easy to see that $h$ is continuous and
\[
h\left([0,1]\times\overline{{\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{d})}\right)
\]
is relatively compact.

Suppose there exists
\[
(t,x)\in[0,1]\times\partial\left({\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{d})\right)
\]
such that $h(t,x)=x$. Then
\[
Ax(t_{0})=u(t_{0})\quad \text{or}\quad
Bx(t_{0})=v(t_{0}), \text{ for some } t_{0}\in I.
\]
\begin{itemize}
\item
If $\|Tx\|>c$ then by assumption $(iii)$ we have
\[
B(Tx)\prec v\ \ \text{ and }\ \ u\prec A(Tx),
\]
so
\begin{itemize}
\item
if $Ax(t_{0})=u(t_{0})$ then
\begin{align*}
A(x)
&=A(h(t,x))
=A((1-t)Tx+tx_{0})\\
&\succeq(1-t)A(Tx)+tA(x_{0})
\succ(1-t)u+tu\\
&=u
\end{align*}
which contradicts $Ax(t_{0})=u(t_{0})$.
\item
if $Bx(t_{0})=v(t_{0})$ then
\begin{align*}
B(x)
&=B(h(t,x))
=B((1-t)Tx+tx_{0})\\
&\preceq(1-t)B(Tx)+tB(x_{0})
\prec(1-t)v+tv\\
&=v
\end{align*}
which contradicts $Bx(t_{0})=v(t_{0})$.
\end{itemize}
\item
If $\|Tx\|\leq c$ then
\begin{align*}
\|x\|
&=\|h(t,x)\|
=\|(1-t)Tx+tx_{0}\|\\
&\leq(1-t)\|Tx\|+t\|x_{0}\|
<(1-t)c+tc\\
&=c,
\end{align*}
therefore by assumption $(i)$ we have
\[
B(Tx)\prec v\quad\text{and}\quad  u\prec A(Tx),
\]
so
\begin{itemize}
\item
if $Ax(t_{0})=u(t_{0})$ then
\begin{align*}
A(x)
&=A(h(t,x))
=A((1-t)Tx+tx_{0})\\
&\succeq(1-t)A(Tx)+tA(x_{0})
\succ(1-t)u+tu\\
&=u
\end{align*}
which contradicts $Ax(t_{0})=u(t_{0})$.
\item
if $Bx(t_{0})=v(t_{0})$ then
\begin{align*}
B(x)
&=B(h(t,x))
=B((1-t)Tx+tx_{0})\\
&\preceq(1-t)B(Tx)+tB(x_{0})
\prec(1-t)v+tv\\
&=v
\end{align*}
which contradicts $Bx(t_{0})=v(t_{0})$.
\end{itemize}
\end{itemize}
Consequently, for each
\[
(t,x)\in[0,1]\times\partial\left({\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{d})\right)
\]
it holds that
$h(t,x)\ne x$.
So, by Lemma \ref{lemm:001}, we have
\[
i\left(T,{\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{d}),K_{d}\right)
=i\left(x_{0},{\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{d}),K_{d}\right)
=1.
\]
Hence, operator $T$ has at least one fixed point
$y\in{\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{d})$, i.e.
\[
u\prec A(y),\quad B(y)\prec v,\quad \|y\|\leq d.
\]
\end{proof}

\section{An Application of Theorem \ref{theo:001} to a Terminal
 Value Problem}\label{sect:003}

For any interval $I\subseteq\mathbb{R}$ and any set $S\subseteq\mathbb{R}$, by $C^{2}(I,S)$ we denote the set of all twice continuously differentiable functions defined on $I$, which have values in $S$. Let $J$ be an unbounded interval in $\mathbb{R}^{+}$. It is easy to see that the set
\[
BC^{2}(J,\mathbb{R}^{+}):=\{x\in C^{2}(J,\mathbb{R}^{+}): x
 \text{ is bounded }\}
\]
endowed with the norm
\[
\|x\|:=\sup_{t\in J}|x(t)|,\quad x\in BC^{2}(J,\mathbb{R}^{+}),
\]
is a Banach space. We are looking for functions
$x\in BC^{2}(J,\mathbb{R}^{+})$ which satisfy the second order
differential equation
\begin{equation}\label{equa:001}
x''(t)+f(t,x(t))=0,\quad  t\in J
\end{equation}
as well as the terminal condition
\begin{equation}\label{equa:002}
\lim_{t\to+\infty}x(t)=\xi,
\end{equation}
where $\xi\in\mathbb{R}^{+}$, $f:J\times\mathbb{R}^{+}\to\mathbb{R}^{+}$
is continuous and
\[
\int_{t}^{+\infty}\int_{s}^{+\infty}f(\sigma,y(\sigma))\,d\sigma \,ds\leq\xi,
\quad \text{for every }t\in J\text{ and every }
y\in BC^{2}(J,\mathbb{R}^{+}).
\]
Define the following set $K$, which is a cone in
$BC^{2}(J,\mathbb{R}^{+})$
\[
K:=\{x\in BC^{2}(J,\mathbb{R}^{+}): x(t)\geq0,\ x'(t)\geq0
\text{ and }x''(t)\leq0, \text{ for all } t\in J\}.
\]

\begin{lemma}\label{lemm:002}
Let $\epsilon>0$. A function $x\in K_{\epsilon}$ is a solution of the
terminal value problem \eqref{equa:001}--\eqref{equa:002} if and
only if $x$ is a fixed point of the operator
$T:K_{d}\to C(J,\mathbb{R}^{+})$ defined by the formula
\[
Ty(t):=\xi-\int_{t}^{+\infty}\int_{s}^{+\infty}f(\sigma,y(\sigma))\,
d\sigma\, ds,\quad t\in J.
\]
\end{lemma}

\begin{definition}\label{defi:001} \rm
A set $U$ of real valued functions defined on the interval $J$
is called equiconvergent at $\infty$ if all functions in $U$ are
convergent in $\mathbb{R}$ at the point $\infty$ and, in addition,
for each $\epsilon>0$, there exists $T\equiv T(\epsilon)>0$ such that,
for all functions $u\in U$, it holds
\[
|u(t)-\lim_{s\to\infty}u(s)|<\epsilon,\quad\text{for every } t\geq T.
\]
\end{definition}

\begin{lemma}\label{lemm:003}
Let $U$ be an equicontinuous and uniformly bounded subset of the
Banach space $BC^{2}(J,\mathbb{R})$. If $U$ is equiconvergent
at $\infty$, it is also relatively compact.
\end{lemma}

\begin{lemma}\label{lemm:004}
Let $\epsilon>0$. Operator $T$ is completely continuous and maps
$K_{\epsilon}$ into $K$.
\end{lemma}

\begin{theorem}\label{theo:002}
Let $r_{1},r_{2}\in J$, with $r_{1}<r_{2}$, and $0<a<b<c<d$,
with $d\geq\xi$.
Also, define the functionals $\alpha(x)=x'(r_{2})$, $x\in K$,
 and $\beta(x)=x'(r_{1})$, $x\in K$. Suppose that for any
$x\in K_{\alpha,\beta}(a,b)\cap K_{c}$ as well as for any
$x\in K_{\alpha,\beta}(a,b)\cap K_{d}$ with $\|Tx\|>c$, it holds that
\[
\int_{r_{1}}^{+\infty}f(s,x(s))ds<b\quad{and}\quad
\int_{r_{2}}^{+\infty}f(s,x(s))ds>a.
\]
Then the terminal value problem \eqref{equa:001}--\eqref{equa:002} has
at least one solution $y$ such that
\[
y'(r_{2})>a,\quad y'(r_{1})<b,\quad \|y\|\leq d.
\]
\end{theorem}

\begin{proof}
Obviously $\alpha$ is a concave positive functional and $\beta$ is
a convex positive functional such that
\[
\alpha(x)\leq\beta(x),\quad x\in K.
\]
It is easy to see that any function $x\in K$ with
\[
x(t)=\lambda t,\quad t\in[r_{1},r_{2}],
\]
where $\lambda\in(a,b)$, and such that $\|x\|<c$, belongs to
${\rm int}_{K_{d}}(K_{\alpha,\beta}(a,b)\cap K_{c})$.
Also, since $d\geq\xi$, it is obvious that
\[
Tx\in K_{d},\quad \forall x\in\overline{K_{\alpha,\beta}(a,b)\cap K_{d}}.
\]
The rest of the proof is easy.
\end{proof}

\begin{corollary}\label{coro:001}
The terminal value problem
\begin{gather}\label{equa:004}
x''(t)+\frac{1}{t^{3}+x(t)}=0,\quad t\in[1,+\infty), \\
\label{equa:005}
\lim_{t\to+\infty}x(t)=2.
\end{gather}
has at least one non-negative solution $y$ such that
\[
y'(3)>\frac{1}{20},\quad y'(2)<\frac{1}{6},\quad
\sup_{t\in[1,+\infty)}|y(t)|\leq2.
\]
\end{corollary}

\begin{proof}
The result follows from Theorem \ref{theo:002}, for $r_{1}=2$,
$r_{2}=3$, $a=\frac{1}{20}$, $b=\frac{1}{6}$, $c=1$, $d=2$ and $\xi=2$.
We notice that, for any $t\in[1,+\infty)$ and $x\in K_{d}$, it holds
\[
\int_{t}^{+\infty}\int_{s}^{+\infty}\frac{1}{\sigma^{3}
+x(\sigma)}\,d\sigma \,ds
\leq\int_{t}^{+\infty}\int_{s}^{+\infty}\frac{1}{\sigma^{3}}\,d\sigma \,ds
=\frac{1}{2t}
\leq2=\xi,
\]
and, for any $x\in K_{d}$, we have
\[
\int_{2}^{+\infty}\frac{1}{s^{3}+x(s)}ds
\leq\int_{2}^{+\infty}\frac{1}{s^{3}}ds<\frac{1}{6}
\]
and
\[
\int_{3}^{+\infty}\frac{1}{s^{3}+x(s)}ds
\geq\int_{3}^{+\infty}\frac{1}{s^{3}+2}ds>\frac{1}{20}.
\]
\end{proof}

\section{An Application of Theorem \ref{theo:003} to a Boundary
Value Problem}\label{sect:004}

Consider the second order boundary value problem
\begin{gather}\label{equa:007}
x''(t)-f(t,x(t))=0,\quad  t\in[0,1],\\
\label{equa:008}
x(0)=0,\quad x'(1)=ax'(0)
\end{gather}
where $f:[0,1]\times\mathbb{R}^{+}\to\mathbb{R}^{+}$ is continuous
and $a>1$. Define the following set $K$, which is a cone in
$C([0,1],\mathbb{R}^{+})$,
\[
K:=\{x\in C([0,1],\mathbb{R}^{+}): x(t)\geq0,\; \forall t\in[0,1]
\text{ and }x'(t)\geq0,\; \forall t\in[0,1]\}.
\]

\begin{lemma}\label{lemm:005}
Let $\epsilon>0$. A function $x\in K_{\epsilon}$ is a solution
of the boundary value problem \eqref{equa:007}--\eqref{equa:008}
if and only if $x$ is a fixed point of the operator
$T:K_{\epsilon}\to C([0,1],\mathbb{R}^{+})$ defined by the formula
\[
Ty(t):=\frac{t}{a-1}\int_{0}^{1}f(s,y(s))ds
+\int_{0}^{t}\int_{0}^{s}f(\sigma,y(\sigma))\,d\sigma \,ds,\quad
 t\in[0,1].
\]
\end{lemma}

\begin{lemma}\label{lemm:006}
Let $\epsilon>0$. Operator $T$ is completely continuous and maps
$K_{\epsilon}$ into $K$.
\end{lemma}

\begin{theorem}\label{theo:004}
Let $u,v\in C([0,1],\mathbb{R}^{+})$ with $u\prec v$, $u'(t)\geq0$,
$\forall t\in[0,1]$, and $v'(t)\geq0$, $\forall t\in[0,1]$.
Also, define the operators $A,B$ by
\[
A(x)=B(x)=x',\quad x\in K.
\]
and let $0<c<d$, with
\[
\sup_{t\in[0,1]}Tx(t)<d,\quad
\forall x\in\overline{K_{A,B}(u,v)\cap K_{d}}.
\]
Suppose that for any $x\in K_{A,B}(u,v)\cap K_{c}$ as well as
for any $x\in K_{A,B}(u,v)\cap K_{d}$ with $\|Tx\|>c$, it holds that
\[
u(t)<\frac{1}{a-1}\int_{0}^{1}f(s,x(s))ds+\int_{0}^{t}f(s,x(s))ds<v(t),\quad
t\in[0,1].
\]
Then the boundary value problem \eqref{equa:007}--\eqref{equa:008}
has at least one solution $y$ such that
\[
u\prec y',\ y'\prec v\text{ and }\|y\|\leq d.
\]
\end{theorem}

\begin{proof}
It is easy to see that
${\rm int}_{K_{d}}(K_{A,B}(u,v)\cap K_{c})\ne\emptyset$
and
\[
Tx\in K_{d},\quad \forall x\in\overline{K_{A,B}(u,v)\cap K_{d}}.
\]
The rest of the proof is easy.
\end{proof}

\begin{corollary}\label{coro:002} \rm
The boundary value problem
\begin{gather}\label{equa:009}
x''(t)-(1+\sin^{2}(x(t)))=0,\quad t\in[0,1], \\
\label{equa:010}
x(0)=0,\quad x'(1)=2x'(0).
\end{gather}
has at least one non-negative solution $y$ such that
\[
t<y(t)<t^{2}+2t,\ \forall t\in[0,1],\quad\text{and}\quad
\sup_{t\in[0,1]}|y(t)|\leq3.
\]
\end{corollary}

\begin{proof}
The result follows from Theorem \ref{theo:004}, for $c=1$, $d=3$,
$u(t)=1$, $t\in[0,1]$, and $v(t)=2(t+1)$, $t\in[0,1]$.
We notice that, for any $t\in[0,1]$, it holds
\[
1<\int_{0}^{1}(1+\sin^{2}(x(s)))ds+\int_{0}^{t}(1+\sin^{2}(x(s)))ds
\]
and
\[
\int_{0}^{1}(1+\sin^{2}(x(s)))ds+\int_{0}^{t}(1+\sin^{2}(x(s)))ds
<2(t+1).
\]
Also, for any $x\in K_{d}$, we have
\[
\int_{0}^{t}\int_{0}^{s}(1+\sin^{2}(x(\sigma)))\,d\sigma \,ds
\leq2\int_{0}^{t}\int_{0}^{s}\,d\sigma \,ds
=t^{2}\leq1,\quad \forall t\in[0,1],
\]
and
\[
\frac{t}{2}\int_{0}^{1}\left(1+\sin^{2}(x(s))\right)ds
\leq t\int_{0}^{1}ds
=t \leq1,\quad \forall t\in[0,1],
\]
so
\[
\sup_{t\in[0,1]}Tx(t)\leq2<d.
\]
\end{proof}

\subsection*{Acknowledgments}
The author wishes to thank Professors Panagiotis Ch. Tsamatos
and Theodor Vidalis for their valuable help during the preparation
of this article.

\begin{thebibliography}{99}
\bibitem{a1976} H. Amann;
Fixed point equations and nonlinear eigenvalue problems in
ordered Banach spaces, \textit{SIAM Rev.} 18 (1976), no. 4, 620--709.

\bibitem{a1998} R. I. Avery;
 A generalization of the Leggett-Williams fixed point theorem, \textit{MSR Hot-Line} 2 (1998), 9--14.

\bibitem{aha2010} R. I. Avery, J. Henderson and D. R. Anderson;
 Existence of a positive solution to a right focal boundary value problem, \textit{Electron. J. Qual. Theory Differ. Equ.} 2010, No. 5, 6 pp.

\bibitem{aho2008a} R. Avery, J. Henderson and D. O'Regan;
 Four functionals fixed point theorem, \textit{Math. Comput. Modelling} 48 (2008), no. 7-8, 1081--1089.

\bibitem{aho2008b} R. Avery, J. Henderson and D. O'Regan;
Functional compression-expansion fixed point theorem,
\textit{Electron. J. Differential Equations} 2008, No. 22, 12 pp.

\bibitem{aho2008c} R. Avery, J. Henderson and D. O'Regan;
 Six functionals fixed point theorem, \textit{Commun. Appl. Anal.} 12 (2008), no. 1, 69--81.

\bibitem{kt2000} G. L. Karakostas and P. Ch. Tsamatos;
 Positive solutions of a boundary-value problem for second order
ordinary differential equations,
\textit{Electron. J. Differential Equations} 2000, No. 49, 9 pp.

\bibitem{k2008} M. K. Kwong;
 On Krasnoselskii's cone fixed point theorem,
\textit{Fixed Point Theory Appl.} 2008, Art. ID 164537, 18 pp.

\bibitem{lg1979} R. W. Leggett and L. R. Williams;
 Multiple positive fixed points of nonlinear operators on ordered
Banach spaces, \textit{Indiana Univ. Math. J.} 28 (1979), no. 4, 673--688.

\bibitem{mm1999} H. Maagli and S. Masmoudi;
 Existence theorem of nonlinear singular boundary value problem,
\textit{Nonlinear Anal.} 46 (2001), 465--473.

\bibitem{w2004} E. Wahlen;
 Positive solutions of second--order differential equations,
\textit{Nonlinear Anal.} 58 (2004), 359--366.

\bibitem{z1986} E. Zeidler;
 \textit{Nonlinear functional analysis and its applications. I.
 Fixed-point theorems. Translated from the German by Peter R. Wadsack},
Springer-Verlag, New York, 1986.

\bibitem{z1994} Z. Zhao;
 Positive solutions of nonlinear second order ordinary differential
equations, \textit{Proc. Amer. Math. Soc.} 121 (1994), no. 2, 465--469.

\end{thebibliography}

\end{document}