\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 115, pp. 1--7.\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 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/115\hfil Double solutions of three-point boundary-value problems]
{Double solutions of three-point boundary-value problems for
second-order differential equations}

\author[Johnny Henderson\hfil EJDE-2004/115\hfilneg]
{Johnny Henderson} 

\address{Johnny Henderson\hfill\break
Department of Mathematics, Baylor University\\
Waco, TX 76798-7328, USA}
\email{Johnny\_Henderson@baylor.edu}

\date{}
\thanks{Submitted September 16, 2003. Published October 5, 2004.}
\subjclass[2000]{34B15, 34B10, 34B18}
\keywords{Fixed point theorem; three-point; boundary-value problem}

\begin{abstract}
 A double fixed point theorem is applied to yield the
 existence of at least two nonnegative solutions for the three-point
 boundary-value problem for a second-order differential equation,
 $$\displaylines{
 y'' + f(y)=0,\quad 0 \leq t \leq 1,\cr
 y(0) =0,\quad  y(p) - y(1) = 0,
 }$$
 where $0 < p < 1$ is fixed, and $f:\mathbb{R} \to [0, \infty)$ is
 continuous.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}

\section{Introduction}

This paper fits in the rapidly growing literature devoted to applications of
multiple fixed point theorems for boundary value problems for each of ordinary
differential equations, finite difference equations, and dynamic equations on
time scales.  Some of these applications can be found in, to mention a few,
the papers \cite{AnAv} - \cite{AvPe}, \cite{HeTh1} - \cite{Ka} and  \cite{Wo}.
These applications involve in some cases multiple uses of a Guo-Krasnosel'skii
\cite{Ze} fixed point theorem or uses of the Leggett-Williams \cite{LeWi} triple
fixed point theorem.  Other applications have used functional-type cone
expansion-compression fixed point theorems such as found in the above
cited papers \cite{AnAv} - \cite{AvPe}.
In this paper, we apply the Avery-Henderson \cite{AvHe} double fixed point
theorem to obtain at least two positive solutions of the three-point boundary
value problem for the second order differential equation,
\begin{gather} \label{e1}
y'' + f(y) = 0, \quad  0 \leq t \leq 1,\\
y(0)=0, \quad  y(p) - y(1) = 0, \label{e2}
\end{gather}
where $0 < p < 1$ is fixed throughout, and $f:\mathbb{R} \to [0, \infty)$ is continuous.
Multipoint problems such as (\ref{e1}), (\ref{e2}) have received considerable
attention,  often with (\ref{e2}) replaced by $u(1) - \sum_{i=1}^n \alpha_i
u(t_i) = 0$, $a < t_1 < \ldots < t_n < 1$, and $0 < \sum_{i=1}^n \alpha_i < 1$.
For a few such papers, see \cite{An}, \cite{BaFa} - \cite{Gu}, \cite{Ma1} and
\cite{Ma2}.
In Section 2, we provide some background results and we state the double
fixed point theorem.  Then, in Section 3, we impose growth conditions on
$f$ which allow us to apply the fixed point theorem in obtaining double positive
solutions of (\ref{e1}), (\ref{e2}).
We remark that Liu and Ge \cite{LiGe} recently obtained a double fixed
point theorem which would be considered as a dual theorem to the
Avery-Henderson double fixed point theorem.  At the conclusion of this
paper, we state a theorem establishing double solutions of (\ref{e1}), (\ref{e2}) which arise from an application of the Liu-Ge double
fixed point theorem.
In addition, we mention that the term ``nonnegative'' may better describe than ``positive''
one of the solutions of (\ref{e1}), (\ref{e2}).  Yet, if conditions such as
$f(0) > 0$ are satisfied, then our double solutions are indeed positive.

\section{Background preliminaries and a double fixed point theorem}

In this section, we provide some background from the theory of cones in
Banach spaces, and we then state a double fixed point theorem for a cone
preserving operator.
\begin{definition} \rm
Let $(\mathcal{B}, \| \cdot \|)$ be a real Banach space.  A
nonempty, closed, convex set $\mathcal{P \subset \mathcal{B}}$ is said to be a {\em cone}
provided the following are satisfied:
\begin{itemize}
\item[(a)] If $y \in \mathcal{P}$ and $\lambda \geq 0$ , then $\lambda y \in \mathcal{P}$;
\item[(b)] If $y \in \mathcal{P}$ and $-y \in \mathcal{P}$ , then $ y = 0$.
\end{itemize}
\end{definition}

Every cone $\mathcal{P} \subset \mathcal{B}$ induces a partial ordering, $\leq$, on $\mathcal{B}$
defined by
$$
x \leq y \quad \text{if and only if} \quad  y-x \in \mathcal{P}.
$$
\begin{definition} \rm
Given a cone $\mathcal{P}$ in a real Banach space $\mathcal{B}$, a functional $\psi: \mathcal{P}
\to R$ is said to be {\em increasing} on $\mathcal{P}$, provided $\psi(x) \leq \psi (y)$,
for all $x,y \in \mathcal{P}$ with $x \leq y$.
\end{definition}
\begin{definition} \rm
Given a nonnegative continuous functional $\gamma$ on a cone $\mathcal{P}$ of a real
Banach space $\mathcal{B}$, ({\em i.e.}, $\gamma:\mathcal{P} \to [0, \infty)$ continuous),
we define, for each $d > 0$, the convex set
$$
\mathcal{P}(\gamma,d) = \{x \in \mathcal{P}: \gamma(x) < d \}.
$$
\end{definition}
Our main results concerning multiple positive solutions of (\ref {e1}), (\ref {e2})
will arise as applications of the following fixed point theorem due to Avery
and Henderson \cite {AvHe}.

\begin{theorem} \label{t2.1}
Let $\mathcal{P}$ be a cone in a real Banach space $\mathcal{B}$.  Let $\alpha$ and
$\gamma$ be increasing, nonnegative, continuous functionals on $\mathcal{P}$, and
let $\theta$ be a nonnegative continuous functional on $\mathcal{P}$ with $\theta (0) = 0$
such that, for some $c > 0$ and $M > 0$,
$$
\gamma(x) \leq \theta (x) \leq \alpha(x) \quad \text{and} \quad
 \|x\| \leq M \gamma(x),
$$
for all $ x \in \overline{\mathcal{P}(\gamma,c)}$. Suppose there exist a completely
continuous operator $A: \overline{\mathcal{P}(\gamma,c)} \to \mathcal{P}$ and $0 < a < b < c$
such that
$$
\theta(\lambda x) \leq \lambda \theta(x), \quad \text{for} \ 0 \leq \lambda \leq 1
\text{ and }  x \in \partial \mathcal{P}(\theta,b),
$$
and
\begin{itemize}
\item[(i)] $\gamma (Ax) > c$, for all $x \in \partial \mathcal{P}(\gamma,c)$;
\item[(ii)] $\theta(Ax) < b$, for all $x \in \partial \mathcal{P}(\theta,b)$;
\item[(iii)] $\mathcal{P}(\alpha,a) \neq \emptyset$, and $\alpha(Ax) > a$,
for all $x \in \partial \mathcal{P}(\alpha,a)$.
\end{itemize}
Then $A$ has at least two fixed points, $x_1$ and $x_2$ belonging to
$\overline{\mathcal{P}(\gamma,c)}$ such that
$$
a < \alpha(x_1), \quad\text{with }  \theta(x_1) < b,
$$
and
$$
b < \theta(x_2), \quad\text{with } \gamma(x_2) < c.
$$
\end{theorem}

\section{Double positive solutions of (\ref{e1}), (\ref{e2})}

In this section, we impose growth conditions of $f$ and then apply
Theorem \ref{t2.1} to establish the existence of double positive
solutions of (\ref {e1}), (\ref {e2}).  We note that from the
nonnegativity of $f$, a solution $y$ of (\ref {e1}), (\ref {e2})
is both nonnegative and concave on $[0,1]$, and in addition, assumes
its maximum in the interval $(p,1)$. We will apply Theorem \ref{t2.1} to
a completely continuous operator whose kernel, $G(t,s)$, is the Green's function for
\begin{equation} \label{e3}
-y'' = 0,
\end{equation}
satisfying (\ref {e2}).  In this instance,
\begin{equation} \label{e4}
    G(t,s)=\begin{cases}
    t,&t\leq s\leq p,\\
    s,&s\leq t \text{ and } s \leq p,\\
    \frac{1-s}{1-p} t,&t \leq s \text{ and } s \geq p,\\
    s + \frac{p-s}{1-p} t,&p\leq s\leq t.\end{cases}
\end{equation}
Properties of $G(t,s)$ for which we will make use include
\begin{gather} \label{e5}
G(t,s) \leq G(s,s),\quad  0 \leq t,s \leq 1, \\  \label{e6}
G(t,s) \geq G(p,s), \quad  p \leq t \leq 1, \ 0 \leq s \leq 1.
\end{gather}
Let the Banach space $\mathcal{B} = C[0,1]$ be equipped with the
norm $\|y\| = \max_{0 \leq t \leq 1} |y(t)|$, and choose the cone
$\mathcal{P} \subset \mathcal{B}$ defined by
$$
\mathcal{P} = \{y \in \mathcal{B}: y \text{ is concave and nonnegative-valued on }
[0,1], \text{ and } y(p) = y(1)\}.
$$
For the remainder of the paper, fix $r \in (p,1)$, and define the nonnegative,
increasing functionals, $\gamma, \theta$ and $\alpha$, on $\mathcal{P}$ by
\begin{gather*}
   \gamma (y) = \min_{p \leq t \leq r} y(t) = y(p) =  y(1),\\
   \theta (y) = \max_{0 \leq t \leq p}y(t) =  y(p), \\
   \alpha (y) = \max_{0 \leq t \leq r}y(t).
\end{gather*}
We observe that, for each $y \in \mathcal{P}$,
\begin{equation} \label{e7}
\gamma (y) = \theta (y) \leq \alpha (y).
\end{equation}
In addition, for each $y \in \mathcal{P}$,
\begin{equation} \label{e8}
\|y \| \leq \frac{1}{p} y(p) \leq \frac{1}{p} \gamma (y).
\end{equation}
Finally, we note that
\begin{equation} \label {e9}
\theta (\lambda y) = \lambda \theta (y), \ \ 0 \leq \lambda \leq 1,\ \
\text{and}\ \ y \in \partial \mathcal{P} (\theta, b).
\end{equation}
We now state growth conditions on $f$ so that (\ref{e1}), (\ref{e2}) has at
least two positive solutions.

\begin{theorem} \label {t3.1}
Let
\[
0 < a < \frac{r[r(1-r) + p(r-p)]}{p(1-p)} b < \frac{r[r(1-r) +
p(r-p)]}{(1-p)} c\,,
\]
and suppose that $f$ satisfies the following conditions:
\begin{itemize}
  \item[(A)]  $f(w) > \frac{2c}{p(1-p)}$, if $c \leq w \leq \frac{c}{p}$,
  \item[(B)]  $f(w) < \frac{2b}{p}$, if $0 \leq w \leq \frac{b}{p}$,
  \item[(C)]  $f(w) > \frac{2(1-p)a}{r[r(1-r) + p(r-p)]}$, if $0 \leq w \leq a$.
\end{itemize}
Then, the three point boundary value problem (\ref{e1}), (\ref{e2}) has at
least two positive solutions, $x_1$ and $x_2$, such that
$$
a < \max_{0 \leq t \leq r} x_1(t), \quad\text{ with } \max_{0 \leq t \leq p}
x_1(t) < b,
$$
and
$$
b < \max_{0 \leq t \leq p} x_2(t), \quad \text{ with } \min_{p \leq t \leq r}
x_2(t) < c.
$$
\end{theorem}

\begin{proof}
We begin by defining a completely continuous integral operator $A:\mathcal{B}
\to \mathcal{B}$ by
$$
Ax(t) = \int_0^1 G(t,s) f(x(s)) ds, \ x \in \mathcal{B}, \ \ 0 \leq t \leq 1.
$$
Solutions of (\ref{e1}), (\ref{e2}) are fixed points of $A$ and conversely.  Our
proof consists of showing the conditions of Theorem \ref{t2.1} are satisfied.
First, we choose $x \in \overline{\mathcal{P} (\gamma,c)}$. By the nonnegativity
of $f$ and $G$, for $0 \leq t \leq 1$,
$$
Ax(t) = \int_0^1 G(t,s) f(x(s)) ds \geq 0.
$$
Moreover, $(Ax)''(t) = - f(x(t)) \leq 0$, and so $ (Ax)(t)$ is concave on
$[0,1]$. Since $G(t,s)$ satisfies the boundary conditions (\ref{e2}) as a
function of $t$, we have $(Ax)(p) = (Ax)(1)$. Thus, $Ax \in \mathcal{P}$ and
$A: \overline{\mathcal{P} (\gamma, c)} \to \mathcal{P}$.
We now consider property (i) of Theorem \ref{t2.1}.  If we choose $x \in
\partial \mathcal{P}(\gamma,c)$, then $\gamma (x) = \min_{p \leq t \leq r}
x(t)$ $= x(p) = c$. Since $x \in \mathcal{P}$, $x(t) \geq c$, $p \leq t \leq 1$.
By recalling $\|x\| \leq \frac{1}{p} \gamma (x) = \frac{1}{p} x(p) =
\frac{c}{p}$, we have
$$
c \leq x(t) \leq \frac{c}{p}, \ p \leq t \leq 1.
$$
As a consequence of (A),
$$
f(x(s)) > \frac{2c}{p(1-p)}, \ p \leq s \leq 1.
$$
Also, $Ax \in \mathcal{P}$, and so
\begin{align*}
  \gamma (Ax)  & =  (Ax)(p) \\
  & =  \int_0^1 G(p,s)f(x(s))ds  \\
  & \geq  \int_p^1 G(p,s)f(x(s))ds \\
  & =  \int_p^1 \big( \frac{1-s}{1-p} \big)pf(x(s))ds \\
  & >  \frac{2c}{p(1-p)} \int_p^1\big( \frac{1-s}{1-p} \big)pds\\
  & =  c.
\end{align*}
We conclude that (i) of Theorem \ref{t2.1} is satisfied.
We next address (ii) of Theorem \ref{t2.1}.  We choose $x \in
\partial \mathcal{P}(\theta, b)$. Then $\theta (x)$ $= \max_{0 \leq t
\leq p} x(t) =$ $x(p) = b$. Then $0 \leq x(t) \leq b$, $0 \leq t \leq p$, and
since $x \in \mathcal{P}$, we also have $b \leq x(t) \leq \|x\|$, $p \leq t
\leq 1$. Moreover, $\|x\| \leq \frac{1}{p} \gamma (x) \leq \frac{1}{p}
\theta (x) = \frac{b}{p}$. So,
$$
0 \leq x(t) \leq \frac{b}{p}, \quad 0 \leq t \leq 1.
$$
From (B),
$$
f(x(s)) < \frac{2b}{p}, \quad 0 \leq s \leq 1.
$$
$Ax \in \mathcal{P}$, and so
\begin{align*}
  \theta (Ax)  & =  (Ax)(p) \\
  & =  \int_0^1 G(p,s)f(x(s))ds  \\
  & <  \frac{2b}{p}\int_0^1 G(p,s)ds \\
  & =  \frac{2b}{p} \big[\int_0^p G(p,s)ds +\int_p^1 G(p,s)ds \big] \\
  & =  \frac{2b}{p} \big[\int_0^p sds +\int_p^1 \big(\frac{1-s}{1-p}\big)
  p ds \big] \\
  & =  b.
\end{align*}
In particular, (ii) of Theorem \ref{t2.1} holds.
For the final part, we turn to (iii) of Theorem \ref{t2.1}.  If we first
define $y(t) = \frac{a}{2}, 0 \leq t \leq 1$, then $\alpha(y) = \frac{a}{2}
< a$, and $\mathcal{P} (\alpha, a) \neq \emptyset$.
Now, let us choose $x \in \partial \mathcal{P} (\alpha, a)$. Then, for some
$r_0 \in (p,1)$, $\alpha (x) = \max_{0 \leq t \leq r} x(t)$ $= x(r_0) =a$. So,
in particular
$$
0 \leq x(t) \leq a, \ \ 0 \leq t \leq r.
$$
From assumption (C),
$$
f(x(s)) > \frac{2(1-p)a}{r[r(1-r)+p(r-p)]},\quad 0 \leq s \leq r.
$$
As before, $Ax \in \mathcal{P}$, and so for some $\rho_0 \in (p,1)$,
\begin{align*}
  \alpha (Ax)  & =  (Ax)(\rho_0) \\
  & \geq  (Ax)(r)\\
  & =  \int_0^1 G(r,s)f(x(s))ds  \\
  & \geq  \int_0^r G(r,s)f(x(s))ds \\
  & >  \frac{2(1-p)a}{r[r(1-r)+p(r-p)]}
  \Big[\int_0^p sds +\int_p^r s + \big(\frac{p-s}{1-p}\big)rds \Big] \\
  & =  a.
\end{align*}
Thus, (iii) of Theorem \ref{t2.1} is also satisfied.  Hence, there exist
at least two fixed points of $A$ which are positive solutions $x_1$ and
$x_2$, belonging to $\overline{\mathcal{P} (\gamma, c)}$, of the boundary value
problem (\ref{e1}), (\ref{e2}) such that
$$
a < \alpha (x_1), \quad\text{with } \theta (x_1) < b,
$$
and
$$
b < \theta(x_2), \quad \text{with } \gamma (x_2) < c.
$$
The proof is complete.
\end{proof}

\subsection*{Example}
For $0 < p < r < 1$ fixed and
\[
0 < a < \frac{r[r(1-r)+p(r-p)]}{p(1-p)} b < \frac{r[r(1-r) +
p(r-p)]}{(1-p)} c,
\]
 if $f:\mathbb{R} \to [0, \infty)$ is defined by
$$
f(w) = \begin{cases}
\frac{b}{p} + \frac{(1-p)a}{r[r(1-r)+p(r-p)]}, & w \leq \frac{b}{p},\\
\ell(w), & \frac{b}{p} \leq w \leq c,\\
\frac {2c}{p(1-p)} + 1, & c \leq w,
\end{cases}
$$
where $\ell(w)$ satisfies $\ell'' = 0$, $\ell(\frac{b}{p}) = \frac{b}{p} +
\frac{(1-p)a}{r[r(1-r)+p(r-p)]}$ and $\ell(c) = \frac {2c}{p(1-p)}+1$,
then by Theorem \ref{t3.1}, the boundary value problem (\ref {e1}), (\ref {e2})
has at least two positive solutions.

\subsection*{Remark}
 Liu and Ge recently obtained a double fixed point
theorem \cite[Lemma 2, p. 553]{LiGe} which could be considered as a type of
dual to Theorem \ref{t2.1} in that, conditions are given for the existence
of double fixed points when inequalities (i), (ii) and (iii) of
Theorem \ref{t2.1} are reversed.  We provide in this remark, as an application
of the Liu-Ge double fixed point, a dual result to Theorem \ref{t3.1} for
double positive solutions of (\ref{e1}), (\ref{e2}).  Because of close
similarity of its proof to that of Theorem \ref{t3.1}, we will omit the proof.
For convenience of notation, we will define
$$
\lambda = \max_{0 \leq t \leq r} \int_0^1 G(t,s)ds.
$$

\begin{theorem} \label {t3.2}
Let $0 < a < b < c$ be such that $
0 < a < \min\{pb, \frac{2\lambda b}{p(1-p)}\}< \frac{2\lambda c}{p}$,
and suppose that $f$ satisfies the following conditions:
\begin{itemize}
  \item[(A)]  $f(w) < \frac{2c}{p}$, if $0 \leq w \leq \frac{c}{p}$,
  \item[(B)]  $f(w) > \frac{2b}{p(1-p)}$, if $b \leq w \leq \frac{b}{p}$,
  \item[(C)]  $f(w) < \frac{a}{\lambda}$, if $0 \leq w \leq \frac{a}{p}$.
\end{itemize}
Then, the three point boundary value problem (\ref{e1}), (\ref{e2}) has at
least two positive solutions, $x_1$ and $x_2$, such that
$$
a < \max_{0 \leq t \leq r} x_1(t), \quad\text{ with } \max_{0 \leq t \leq p}
x_1(t) < b,
$$
and
$$
b < \max_{0 \leq t \leq p} x_2(t), \quad \text{ with } \min_{p \leq t \leq r}
x_2(t) < c.
$$
\end{theorem}

\subsection*{Acknowledgments}
The author expresses his gratitude to the
referee for pointing out that a result such as Theorem \ref{t3.2} could
be obtained from the Liu-Ge double fixed point theorem.


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\end{document}