\documentstyle{amsart}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{corollary}{Corollary}
\newtheorem{proposition}{Proposition}
\newtheorem{remark}{Remark}
\newcommand{\refp}[1]{(\ref{#1})}

\begin{document} 
{\noindent\small {\sc Electronic Journal of Differential Equations}\newline
Vol. 1995(1995), No. 03, pp. 1-8. Published March 2, 1995.\newline 
ISSN 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp (login: ftp) 147.26.103.110 or 129.120.3.113 }
\thanks{\copyright 1995 Southwest Texas State University  and 
University of North Texas.} 
\vspace{1.5cm}
\title[\hfilneg EJDE--1995/03\hfil Positive Solutions]{Positive Solutions
 for Higher Order Ordinary Differential Equations} 

\author[P.W. Eloe \& J. Henderson\hfil EJDE--1995/03\hfilneg]
{Paul W. Eloe \& Johnny Henderson }
\address{Department of Mathematics \\
University of Dayton\\
Dayton, OH  45469-2316  USA}
\email{eloe@@udavxb.oca.udayton.edu}
\address{Discrete and Statistical Sciences\\ 
Auburn University\\ 
Auburn, AL 36849-5307  USA}
\email{hendej2@@mail.auburn.edu}
\date{}
\thanks{Submitted on December 4, 1994.}
\subjclass{34B15}
\keywords{Boundary value problems, positive solutions,
superlinear and sublinear, operators on a cone}

\begin{abstract}
Solutions that are positive with respect to a cone are obtained for the
boundary value problem, $u^{(n)} + a(t)f(u) = 0$, $u^{(i)}(0) = u^{(n-2)}(1)
= 0$, $0 \leq i \leq n-2$, in the cases that $f$ is either superlinear or
sublinear.  The methods involve application of a fixed point theorem for
operators on a cone.
\end{abstract}
\maketitle

\section{Introduction}
We are concerned with the existence of solutions for the two-point boundary
value problem,
\begin{equation}  \label {e1}
u^{(n)} + a(t) f(u) = 0, \quad 0 < t < 1,
\end{equation}
\begin{equation}  \label {e2}
u^{(i)}(0) = u^{(n-2)}(1) = 0, \quad 0 \leq i \leq n-2,
\end{equation}
where
\begin{itemize}
\item [(A)] $f: [0, \infty) \to [0, \infty)$ is continuous, and 
\item [(B)] $a: [0, 1] \to [0, \infty)$ is continuous and does not vanish
identically on any subinterval.
\end{itemize}
We remark that, if $u(t)$ is a nonnegative solution of \refp {e1}, \refp {e2}, then
$u^{(n-2)}(t)$ is concave on $[0, 1]$.

Specifically, our aim is to extend the work of Erbe and Wang \cite {LH} to
obtain solutions of \refp {e1}, \refp {e2}, that are positive with respective
to a cone, in the cases when, either (i) $f$ is superlinear, or (ii) $f$ is
sublinear; that is, in the respective cases when, either (i) $f_0 = 0$ and
$f_{\infty} = \infty$, or (ii) $f_0 = \infty$ and $f_{\infty} = 0$, where
$$
f_0 = \lim _{x \to 0^+} \frac {f(x)}{x} \; \mbox { and } \; f_{\infty} = \lim_{x \to
\infty} \frac {f(x)}{x}.
$$

In the case that $n= 2$, the boundary value problem \refp {e1}, \refp {e2}
arises in applications involving nonlinear elliptic problems in annular
regions; see \cite {CB}, \cite {BC}, \cite {XG}, \cite {HW}.  Applications of
\refp {e1}, \refp {e2} can also be made to singular boundary value problems
as in \cite {fb}, \cite {CJ}, \cite {EP}, \cite {JA}, \cite {CD}, \cite {ST},
as well as to extremal point characterizations for boundary value problems in
\cite {PW}, \cite {DH}, \cite {KS}.  In these applications, frequently, only
solutions that are positive are useful.  The results herein are also somewhat
related to those obtained in \cite {CC} and \cite {AF}.  

Our arguments for establishing the existence of solutions of \refp {e1},
\refp {e2} involve concavity properties of solutions that are used in
defining a cone on which a positive integral operator is defined.  A fixed
point theorem due to Krasnosel'skii \cite {MK} is applied to yield a positive
solution of \refp {e1}, \refp {e2}.  

In Section 2, we present some properties of a Green's function which will be
used in defining the positive operator.  We also state the fixed point
theorem from \cite {MK}.  In Section 3, we provide an appropriate Banach
space and cone in order to apply the fixed point theorem yielding solutions
of \refp {e1}, \refp {e2} in both the superlinear and sublinear cases.

\section{Some preliminaries}

In this section, we state a theorem due to Krasnosel'skii, an application of
which will yield in the next section a positive solution of \refp {e1}, \refp
{e2}.  The mapping to which we apply this fixed point theorem will include an
integral whose kernel, $G(t, s)$, is the Green's function for
\begin{equation} \label {e3}
\begin{gathered}
-y^{(n)} = 0,\\
y^{(i)}(0) = y^{(n-2)}(1) = 0, \quad 0 \leq i \leq n -2.
\end{gathered} \end{equation}
Eloe \cite {PE} has shown that, for $0 \leq i \leq n-2$,

\begin{equation}  \label {e4}
\frac {\partial ^i}{\partial t^i} G(t, s) > 0 \mbox { on } (0, 1) \times (0,
1),
\end{equation}
as well as the fact that the function

\begin{equation}  \label {e5}
K(t, s) = \frac {\partial ^{n-2}}{\partial t^{n-2}} G(t, s)
\end{equation}
is the Green's function for

\begin{equation}  \label {e6} \begin{gathered}
-y'' = 0,\\
y(0) = y(1) = 0.
\end{gathered}\end{equation}
We note that 

\begin{equation}  \label {e7}
K(t, s) =  \left \{ \begin{array}{ll}
t(1-s), & 0 \leq t < s \leq 1,\\
s(1-t), & 0 \leq s < t \leq 1,
\end{array} \right. 
\end{equation}
from which it is straightforward that

\begin{equation}  \label {e8}
K(t, s) \leq K(s, s), 0 \leq t, s \leq 1,
\end{equation}
and a nice argument in \cite {LH} shows that

\begin{equation}  \label {e9}
K(t, s) \geq \frac {1}{4} K(s, s), \; \frac {1}{4} \leq t \leq \frac {3}{4},
\; 0 \leq s \leq 1.
\end{equation}

The existence of solutions of \refp {e1}, \refp {e2} is based on an
application of the following fixed point theorem \cite {MK}.  

\begin{theorem}  \label {t21}
Let ${\cal B}$ be a Banach space, and let ${\cal P} \subset {\cal B}$ be a
cone in ${\cal B}$.  Assume $\Omega _1, \Omega _2$ are open subsets of ${\cal
B}$ with $0 \in \Omega _1 \subset \bar {\Omega}_1 \subset \Omega _2$, and let
$$
T: {\cal P} \cap (\bar {\Omega}_2 \backslash \Omega _1) \to {\cal P}
$$
be a completely continuous operator such that, either
\begin{itemize}
\item [(i)] $\|Tu\| \leq \|u\|, u \in {\cal P} \cap \partial \Omega _1$, and
$\|Tu\| \geq \|u\|$, $u \in {\cal P} \cap  \partial \Omega _2$,  or
\item [(ii)] $\|Tu\| \geq \|u\|$, $u \in {\cal P} \cap \partial \Omega _1$,
and $\|Tu\| \leq \|u\|$, $u \in {\cal P} \cap \partial \Omega _2$.
\end{itemize}
Then $T$ has a fixed point in ${\cal P} \cap (\bar {\Omega}_2 \backslash
\Omega _1)$.
\end{theorem}

\section{Existence of solutions}

We are now ready to apply Theorem \ref {t21}.  We remark that $u(t)$ is a
solution of \refp {e1}, \refp {e2} if, and only if,
$$
u(t) = \int_0^1 G(t, s) a(s) f(u(s))ds, \quad 0 \leq t \leq 1.
$$
For our construction, we let 
$$
{\cal B} = \{x \in C^{(n-2)}[0, 1] \mid x^{(i)}(0) = 0, \quad 0 \leq i \leq
n-3\},
$$
with norm, $\|x\| = |x^{(n-2)}|_{\infty}$, where $|\cdot|_{\infty}$ denotes the
supremum norm on $[0, 1]$.  Then $(\cal {B}, \| \cdot \|)$ is a Banach space.

\begin{remark}  \label {r1}
We note that, for each $x \in {\cal B}$,

\begin{equation}  \label {e10}
|x^{(i)}|_{\infty} \leq \|x\|, \quad 0 \leq i \leq n-2.
\end{equation}
We will seek solutions of \refp {e1}, \refp {e2} which lie in a cone, ${\cal
P}$, defined by
$$
{\cal P} = \{x \in {\cal B} \mid x^{(n-2)}(t) \geq 0 \mbox { on } [0, 1],
\mbox { and } \min _{\frac {1}{4} \leq t \leq \frac {3}{4}} x^{(n-2)}(t) \geq
\frac {1}{4} \|x\|\}.
$$
\end{remark}

\begin{remark}  \label {r2}
We note here that, if $x \in {\cal P}$, then $x^{(i)}(t) \geq 0$ on $[0, 1]$
and $$x^{(i)}(t) \geq \frac {1}{4} \|x\| \frac {(t- \frac
{1}{4})^{n-i-2}}{(n-i-2)!}$$ on $[\frac {1}{4}, \frac {3}{4}]$,
$0 \leq i \leq n-2$.  As a consequence $$x^{(i)}(t) \geq \frac {1}{(n-i-2)!
4^{n-i-1}} \|x\|$$ on $[\frac {1}{2}, \frac {3}{4}]$, $0 \leq i \leq n-2$.
\end{remark}

\begin{theorem}  \label {t31}
Assume that conditions (A) and (B) are satisfied.  If, either
\begin{itemize}
\item [(i)] $f_0 = 0$ and $f_{\infty} = \infty$ (i.e., $f$ is superlinear), or
\item [(ii)] $f_0 = \infty$ and $f_{\infty} = 0$ (i.e., $f$ is sublinear),
\end{itemize}
then \refp {e1}, \refp {e2} has at least one solution in ${\cal P}$.
\end{theorem}
\medskip

\noindent {\bf Proof.}
We begin by defining an integral operator $T: {\cal P} \to {\cal B}$ by

\begin{equation}  \label {e11}
Tu(t) = \int_0^1G(t, s) a(s) f(u(s))\, ds, \quad u \in {\cal P},
\end{equation}
and we seek a fixed point of $T$ in the cone ${\cal P}$ for the respective
cases of $f$ superlinear and $f$ sublinear.

Before dealing with these cases, we make a few observations.  First, if $u
\in {\cal P}$, it follows from \refp {e8} that

\begin{eqnarray*}
(Tu)^{(n-2)}(t) & = & \int_0^1 \frac {\partial ^{n-2}}{\partial t^{n-2}} G(t,
s) a(s) f(u(s))\, ds\\
& = & \int_0^1 K(t, s)a(s)f(u(s))\, ds\\
& \leq & \int_0^1 K(s, s)a(s)f(u(s))\, ds,
\end{eqnarray*}
so that
$$
\|Tu\| = |(Tu)^{(n-2)}|_{\infty} \leq \int_0^1 K(s,s) a(s)f(u(s))\, ds.
$$
In fact, 

\begin{equation}  \label {E12}
\|Tu\| = \int_0^1 K(s,s)a(s) f(u(s))\, ds.
\end{equation}
Next, if $u \in {\cal P}$, it follows from \refp {e9} and \refp {E12} that

\begin{eqnarray*}
\min _{\frac {1}{4} \leq t \leq \frac {3}{4}} (Tu)^{(n-2)}(t) & = & \min
_{\frac {1}{4} \leq t \leq \frac {3}{4}} \int_0^1 K(t, s) a(s) f(u(s)) \, ds\\
& \geq & \frac {1}{4} \int_0^1 K(s, s) a(s) f(u(s))\, ds\\
& \geq & \frac {1}{4} \|Tu\|.
\end{eqnarray*}
Moreover, properties of $G(t, s)$ give that $(Tu)^{(n-2)}(t) \geq 0$, so that
$Tu \in {\cal P}$, and in particular $T: {\cal P} \to {\cal P}$.  Also, the
standard arguments yield that $T$ is completely continuous.

We now turn to the cases of the theorem.
\begin{itemize}
\item [(i)] Assume $f_0 = 0$ and $f_{\infty} = \infty$.  First, dealing with $f_0 =
0$, there exist $\eta > 0$ and $H_1 > 0$ such that $f(x) \leq \eta x$, for $0
< x \leq H_1$, and
$$
\eta \int_0^1 K(s, s) a(s)\, ds \leq 1.
$$

So, if we choose $u \in {\cal P}$ with $\|u\| = H_1$, and if we recall from
Remark \ref {r1} that $|u|_{\infty} \leq  \|u\|$, we have from \refp {e8},

\begin{eqnarray*}
(Tu)^{(n-2)}(t) & = & \int_0^1 K(t, s) a(s) f(u(s))\, ds\\
& \leq & \int_0^1 K(s,s) a(s) f(u(s))\, ds\\
& \leq & \int_0^1 K(s, s) a(s) \eta u(s)\, ds\\
& \leq & \eta \int_0^1 K(s, s) a(s)\, ds \|u\|\\
& \leq & \|u\|, \quad 0 \leq t \leq 1.
\end{eqnarray*}
As a consequence $\|Tu\| = |(Tu)^{(n-2)}(t)|_{\infty} \leq \|u\|$.  Thus, if we
set
$$
\Omega _1 = \{x \in {\cal B} \mid \|x\| < H_1\},
$$
then 
\begin{equation}  \label {e13}
\|Tu\| \leq \|u\|, \mbox { for } u \in {\cal P} \cap \partial \Omega _1.
\end{equation}

Next, dealing with $f_{\infty} = \infty$, there exist $\lambda > 0$ and $\bar {H}_2 >
0$ such that $f(x) \geq \lambda x$, for $x \geq \bar {H}_2$, and
$$
\frac {\lambda}{(n-2)! 4^{n-1}} \int_{\frac {1}{2}}^{\frac  {3}{4}} K(\frac {1}{2}, s)
a(s)\, ds \geq 1.
$$
Now, let $H_2 = \max \{2H_1, (n-2)! 4^{n-1} \bar {H}_2\}$ and set
$$
\Omega _2 = \{x \in {\cal B} \mid \|x\| < H_2\}.
$$
So, if $u \in {\cal P}$ with $\|u\| = H_2$, and if we recall from Remark \ref
{r2} that $u(t) \geq \frac {1}{(n-2)! 4^{n-1}} \|u\| \geq \bar {H}_2$ on $[\frac
{1}{2}, \frac {3}{4}]$, we have 

\begin{eqnarray*}
(Tu)^{(n-2)} (\frac {1}{2}) & = & \int_0^1 K(\frac {1}{2}, s) a(s) f(u(s))\, ds\\
& \geq & \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac {1}{2}, s) a(s) \lambda
u(s)\, ds\\
& \geq & \lambda \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac {1}{2}, s) a(s)
\frac {1}{(n-2)! 4^{n-1}} \|u\|\, ds\\
& = & \frac {\lambda}{(n-2)! 4^{n-1}} \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac
{1}{2}, s) a(s)\, ds\|u\|\\
& \geq & \|u\|,
\end{eqnarray*}
so that $\|Tu\| \geq \|u\|$.  Consequently,

\begin{equation}  \label {e14}
\|Tu\| \geq \|u\|, \mbox { for } u \in {\cal P} \cap \partial \Omega _2.
\end{equation}

Therefore, by part (i) of Theorem \ref {t21} applied to \refp {e13} and \refp
{e14}, $T$ has a fixed point $u(t) \in {\cal P} \cap (\bar {\Omega}_2
\backslash \Omega _1)$ such that $H_1 \leq \|u \| \leq H_2$, and as such,
$u(t)$ is a desired solution of \refp {e1}, \refp {e2}.
(We remark that the arguments carry through, if we had set $H_2 = \max \{H_1,
(n-2)!4^{n-1} \bar {H}_2\}$ and if $H_2 = H_1$, then there is a solution $u
\in {\cal P}$ with $\|u\| = H_1$.)
This completes the case when $f$ is superlinear.

\item [(ii)] Now, assume $f_0 = \infty$ and $f_{\infty} = 0$.  Dealing with $f_0 =
\infty$, there exist $\bar {\eta} > 0$ and $J_1 > 0$ such that $f(x) \geq \bar
{\eta} x$, for $0 < x \leq J_1$, and 
$$
\frac {\bar {\eta}}{(n-2)! 4^{n-1}} \int_{\frac {1}{2}}^{\frac {3}{4}}
K(\frac {1}{2}, s) a(s) \, ds \geq 1.
$$
This time, we choose $u \in {\cal P}$ with $\|u\| = J_1$.  Since $|u|_{\infty}
\leq \|u\| = J_1$, we have $f(u(s)) \geq \bar {\eta} u(s)$, $0 \leq s \leq
1$.  Also, we know $u(s) \geq \frac {1}{(n-2!) 4^{n-1}} \|u\|$, $\frac {1}{2}
\leq  s \leq \frac {3}{4}$.  Thus, 

\begin{eqnarray*}
(Tu)^{(n-2)} (\frac {1}{2}) & = & \int_0^1 K(\frac {1}{2}, s) a(s) f(u(s))\, ds\\
& \geq & \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac {1}{2}, s) a(s) \bar {\eta}
u(s)\, ds\\
& \geq & \frac {\bar {\eta}}{(n-2)! 4^{n-1}} \int_{\frac {1}{2}}^{\frac
{3}{4}} K(\frac {1}{2}, s)a(s)\, ds \|u\|\\
& \geq & \| u\|,
\end{eqnarray*}
and in particular, $\|Tu\| \geq \|u\|$.  Setting
$$
\Omega _1 = \{x \in {\cal B} \mid \|x\| < J_1\},
$$
we conclude

\begin{equation}  \label {e15}
\|Tu\| \geq \|u\|, \mbox { for } u \in {\cal P} \cap \partial \Omega _1.
\end{equation}
\end{itemize}

For the  final part of this case, we deal with $f_{\infty} = 0$.  There exist
$\bar {\lambda} > 0$ and $\bar {J}_2 > 0$ such that, $f(x) \leq \bar
{\lambda} x$, for $x \geq \bar {J}_2$, and
$$
\bar {\lambda} \int_0^1 K(s, s) a(s) \, ds \leq 1.
$$
There are two further sub-cases to be considered:

(I) We suppose first that $f$ is bounded.  Then, there exists $N > 0$  such
that $f(x) \leq N$, for all $0 < x < \infty$.  Let $J_2 = \max \{2 J_1, N \int_0^1
K(s, s) a(s)\, ds\}$.  Then, for $u \in {\cal P}$ with $\|u\| = J_2$, since
$|u|_{\infty} \leq \|u\|$ and $K(t, s) \leq K(s, s)$, $0 \leq s, t \leq 1$, we
have

\begin{eqnarray*}
(Tu)^{(n-2)} (t) & = & \int_0^1 K(t, s) a(s) f(u(s))\, ds\\
& \leq & N \int_0^1 K(s, s) a(s) \, ds\\
& \leq & J_2\\
& = & \|u\|, \quad 0 \leq t \leq 1.
\end{eqnarray*}
Consequently, $\|Tu\| \leq \|u\|$.

(II)  For the second sub-case, suppose that $f$ is unbounded.  Then, there
exists $J_2 > \max \{2J_1, \bar {J}_2\}$ such that $f(x) \leq f(J_2)$, for $0
< x \leq J_2$.  We now choose $u \in {\cal P}$ with $\|u\| = J_2$.  Again,
recalling $|u|_{\infty} \leq \|u\|$ and $K(t, s) \leq K(s, s)$ leads to

\begin{eqnarray*}
(Tu)^{(n-2)}(t) & = & \int_0^1 K(t, s) a(s) f(u(s))\, ds\\
& \leq & \int_0^1 K(s, s) a(s) f(J_2) \, ds\\
& \leq & \bar {\lambda} \int_0^1 K(s, s) a(s) \, ds J_2\\
& \leq & \|u\|, \quad 0 \leq t \leq 1.
\end{eqnarray*}
Thus, $\|Tu\| \leq \|u\|$.

We conclude from each sub-case, (I) and (II), if we set
$$
\Omega _2 = \{x \in {\cal B} \mid \|x\| < J_2\},
$$
then

\begin{equation}  \label {e16}
\|Tu\| \leq \|u\|, \mbox { for } u \in {\cal P} \cap \partial \Omega _2.
\end{equation}

Therefore, by part (ii) of Theorem \ref {t21} applied to \refp {e15} and
\refp {e16}, $T$ has a fixed point $u(t) \in {\cal P} \cap (\Omega _2
\backslash \Omega _1)$ such that $J_1 \leq \|u\| \leq J_2$, and $u(t)$ is a
sought solution of \refp {e1}, \refp {e2}.  This completes the argument for
the case of $f$ sublinear.

The proof is complete. \quad$\Box$


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\end{document}