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\markboth{\hfil Positive and monotone solutions  \hfil EJDE--2002/??}
{EJDE--2002/??\hfil Panos K. Palamides \hfil}

\begin{document}

\title{\vspace{-1in}
\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2002}(2002), No. ??, pp. 1--16. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)} \vspace{\bigskipamount} \\
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 Positive and monotone solutions of an m-point boundary-value problem 
%
\thanks{
\emph{Mathematics Subject Classifications:} 34B10, 34B18, 34B15. 
\hfil\break \indent
{\em Key words:} multipoint boundary value problems, positive monotone
solution, vector field, \hfill\break\indent
sublinear, superlinear, Kneser's property, solution's funel. 
\hfil\break \indent
\copyright 2002 Southwest Texas State University. \hfil\break \indent
Submitted January 10, 2002. Published February 18, 2002.} }

\date{}
\author{Panos K. Palamides}
\maketitle

\begin{abstract}
  We study the second-order ordinary differential equation 
  $$  y''(t)=-f(t,y(t),y'(t)),\quad 0\leq t\leq 1,  $$
  subject to the multi-point boundary conditions 
  $$
  \alpha y(0)\pm \beta y'(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)\,.
  $$
  We prove the existence of a positive solution (and monotone in some cases)
  under superlinear and/or sublinear growth rate in $f$. Our approach is based
  on an analysis of the corresponding vector field on the $(y,y')$
  face-plane and on Kneser's property for the solution's funnel.
\end{abstract}

\newtheorem{theorem}{Theorem}[section] 
\newtheorem{remark}[theorem]{Remark}

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\section{Introduction}

Recently an increasing interest has been observed in investigating the
existence of positive solutions of boundary-value problems. This interest
comes from situations involving nonlinear elliptic problems in annular
regions. Erbe and Tang \cite{ET} noted that, if the boundary-value problem 
\begin{equation*}
-\Delta u=F(|x|,u)\quad\text{in }R<|x|<\hat{R}
\end{equation*}
with 
\begin{gather*}
u=0\quad\mbox{for }|x| =R,\quad u=0 \quad\mbox{for }|x| =\hat {R}; \quad
\mbox{or} \\
u=0 \quad\mbox{for }|x| =R,\quad \frac{\partial u}{\partial|x|} =0 \quad
\mbox{for } |x| =\hat{R};\quad\mbox{or} \\
\frac{\partial u}{\partial|x|}=0\quad\mbox{for } |x|=R, \quad u=0 \quad
\mbox{for }|x|=\hat{R}
\end{gather*}
is radially symmetric, then the boundary-value problem can be transformed
into the scalar Sturm-Liouville problem 
\begin{gather}
x''(t)=-f(t,x(t)),\quad 0\leq t\leq1,  \label{E0} \\
\alpha x(0)-\beta x'(0)=0,\quad \gamma x(1)+\delta x'(1)=0.
\label{C0}
\end{gather}
where $\alpha$, $\beta$, $\gamma$, $\delta$ are positive constants.

By a positive solution of (\ref{E0})-(\ref{C0}), we mean a function $x(t)$
which is positive for $0<t<1$ and satisfies the differential equation (\ref
{E0}) with the boundary conditions (\ref{C0}).

Erbe and Wang \cite{EW} using Green's functions and the Krasnoselskii's
fixed point theorem on cones proved the existence of a positive solution of (
\ref{E0})-(\ref{C0}), under the following assumptions:

\begin{enumerate}
\item[(B1)]  The function $f$ is continuous and positive on $[0,1]\times
[ 0,\infty )$ and 
\begin{equation}
f_{0}:=\lim_{y\rightarrow 0+}\max_{0\leq t\leq 1}\frac{f(t,y)}{y}=0,\quad
f_{\infty }:=\lim_{y\rightarrow +\infty }\min_{0\leq t\leq 1}\frac{f(t,y)}{y}
=+\infty  \label{SL}
\end{equation}
i.e. $f$ is \emph{superlinear} at both ends points $x=0$ and $x=\infty $; or 
\begin{equation}
f_{0}:=\lim_{y\rightarrow 0+}\min_{0\leq t\leq 1}\frac{f(t,y)}{y}=+\infty
,\quad f_{\infty }:=\lim_{y\rightarrow +\infty }\max_{0\leq t\leq 1}\frac{
f(t,y)}{y}=0.  \label{sl}
\end{equation}
i.e. $f$ is \emph{sublinear} at both $x=0$ and $x=\infty $.

\item[(B2)]  $\rho :=\beta \gamma +\alpha \gamma +\alpha \delta >0$.
\end{enumerate}

Also nonlinear boundary constraints have been studied, among others by
Thompson \cite{TO} and by the author of this paper and Jackson \cite{JP}.
There are common ingredients in these papers: an (assumed) Nagumo-type
growth condition on the nonlinearity $f$ or/and the presence of upper and
lower solutions.

The multi-point boundary-value problem for second-order ordinary
differential equations was initiated by Ilin and Moiseev \cite{IM-1,IM-2}.
Gupta \cite{GU} studied the three-point boundary-value problems for
nonlinear ordinary differential equations. Since then, more general
nonlinear multi-point boundary-value problems have been studied by several
authors. Most of them used the Leray-Schauder continuation theorem,
nonlinear alternatives of Leray-Schauder, coincidence degree theory or a
fixed-point theorem on cones. We refer the reader to \cite{CM, FE, GNT,Ma1}
for some recent results of nonlinear multipoint boundary-value problems.

Let $a_i\geq 0$ for $i=1,\dots ,m-2$ and let $\xi_i$ satisfy $0<\xi
_{1}<\xi_{2}<\dots <\xi_{m-2}<1$. Ma \cite{Ma2} applied a fixed-point theorem
on cones to prove the existence of a positive solution of 
\begin{equation*}
\begin{gathered} u''+a(t)f(u)=0\\ u(0)=0,\quad
u(1)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i) \end{gathered}
\end{equation*}
under superlinearity or sublinearity assumptions on $f$. He also assumed the
following

\begin{enumerate}
\item[($\Gamma 1$)]  $a\in C([0,1],[0,\infty ))$, $f\in C([0,\infty
),[0,\infty ))$, and there exists $t_{0}\in $ $[\xi_{m-2},1]$ such that $
a(t_{0})>0$

\item[($\Gamma 2$)]  For $i=1,\dots ,m-2$, $a_i\geq 0$ and $
\sum_{i=1}^{m-2}a_i\xi_i<1$.
\end{enumerate}
Recently, Gupta \cite{GU1} obtained existence results for the boundary-value
problem 
\begin{gather*}
y''(t)=f(t,y(t),y'(t))+e(t),\quad 0\leq t\leq 1 \\
y(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i),
\end{gather*}
by using the Leray-Schauder continuation theorem, under smallness
assumptions of the form 
\begin{equation*}
|f(t,y,y')| \leq p(t)|y| +q(t)| y'| +r(t)\quad\text{and}
\quad C_{1}\| p(t)\| +C_{2}\|q(t)\| \leq1,
\end{equation*}
with $p(t)$, $q(t)$, $r(t)$ and $e(t)$ in $L^{1}(0,1)$ and $C_{1}$ and $
C_{2} $ constants.

In this paper, we consider the problem of existence of positive solutions
for the m-point boundary-value problem 
\begin{gather}
y''(t)=-f(t,y(t),y'(t)),\quad 0\leq t\leq1,  \label{E}
\\
\alpha y(0)-\beta y'(0)=0,\quad
y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i).  \label{C}
\end{gather}
We assume $\alpha>0$, $\beta>0$, the function $f$ is continuous, and 
\begin{equation}
f(t,y,y')\geq0,\quad \text{for all }t\in[0,1],\;y\geq 0\,\;
y'\in\mathbb{R}.  \label{A1}
\end{equation}
The presence of the third variable $y'$ in the function $
f(t,y,y')$ causes some considerable difficulties, especially, in the
case where an approach relies on a fixed point theorem on cones and the
growth rate of $f(t,y,y')$ is sublinear or superlinear. We overcome
this predicament, by extending below the concept-assumptions (\ref{SL}) and (
\ref{sl}) as follows:

Suppose that for any $M>0$, 
\begin{equation}
\begin{gathered} f_{0,0}:=\lim_{(y,y')\to (0,0)}\max_{0\leq t\leq1}
\frac{f(t,y,y')}{y}=0 \\ f_{+\infty}:=\lim_{y\to +\infty}\min_{0\leq
t\leq1}\frac {f(t,y,y')}{y}=+\infty, \quad\mbox{for }|y'|\leq M
\end{gathered}  \label{A2S}
\end{equation}
i.e. $f$ is \emph{jointly superlinear} at the end point $(0,0)$ and \emph{
uniformly superlinear }at\emph{\ }$+\infty$. Similarly 
\begin{equation}
\begin{gathered} f_{0}:=\lim_{y\to 0+}\min_{0\leq t\leq1}\frac{f(t,y,y')}
{y}=+\infty,\quad\mbox{for } |y'|\leq M.\\
f_{+\infty,+\infty}:=\lim_{(y,y')\to (+\infty,+\infty)} \max_{0\leq
t\leq1}\frac{f(t,y,y')}{y}=0, \end{gathered}  \label{A2s}
\end{equation}
i.e. $f$ is \emph{jointly sublinear} at $(+\infty,+\infty)$ and \emph{
uniformly sublinear }at $0$.

Furthermore there exist $\bar{l}\in(0,\infty]$, such that for every $\bar{M}
>0$ 
\begin{equation}
\lim_{y'\to -\infty}\max_{0\leq t\leq1}\frac{f(t,y,y')}{
y'}=-\bar{l},\quad\text{for }y\in[0,\bar{M}]  \label{A3}
\end{equation}
i.e. $f(t,y,.)$ is \emph{linear or superlinear }at $-\infty$ \ and for every 
$\bar{\eta}>0$ 
\begin{equation}
\lim_{y'\to 0}\min_{0\leq t\leq1}\frac{f(t,y,y')}{y'}
=0, \quad\text{for }y\in(0,\bar{\eta}).  \label{A4}
\end{equation}
i.e. $f(t,y,.)$ is \emph{superlinear }at $0$.

\begin{remark}\label{R1} \rm
Note that the differential equation (\ref{E}) defines a
vector field whose  properties will be crucial for our study. More
specifically, we look at the $(y,y')$ face semi-plane $(y>0)$. From
the sign condition on $f$ (see assumption (\ref{A1})), we immediately see
that $y''<0$. Thus any trajectory $(y(t),y'(t))$, $t\geq0$,
emanating from the semi-line
\[
E_{0}:=\{(y,y'):\alpha y-\beta y'=0,\;y>0\}
\]
``trends'' in a natural way, (when $y'(t)>0$) toward the positive
$y$-semi-axis and then (when $y'(t)<0$) trends toward the negative $y'$-semi-axis.
Lastly, by setting a certain growth rate on $f$ (say superlinearity) we can
control the vector field, so that some trajectory satisfies the given
boundary condition
\[
y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)
\]
at the time $t=1$. These properties will be referred as ``\emph{The nature
of the vector field''} throughout the rest of paper.
\end{remark}

So the technique presented here is different to that given in the above
mentioned papers \cite{GU1, EW,DEH, GNT,ET}, but it is closely related with
those in \cite{JP,Ma2}. Actually, we rely on the above ''nature of the
vector field'' and on the simple shooting method. Finally, for completeness
we refer to the well-known Kneser's theorem (see for example Copel's
text-book \cite{Co}).

\begin{theorem} \label{Th1}
Consider the system
\begin{equation}
\;x''=f(t,x,x'),\quad (t,x,x')\in\Omega
:=[\alpha,\beta]\times\mathbb{R}^{2n},\label{*}
\end{equation}
with the function $f$ continuous. Let $\hat{E}_{0}$ be a continuum
(compact and connected) set in $\Omega_{0}:=\{(t,x,x')\in
\Omega:t=\alpha\}$ and let $\mathcal{X}(\hat{E}_{0})$ be the family of \ all
solutions of (\ref{*}) emanating from $\hat{E}_{0}$. If any solution
$x\in\mathcal{X}(\hat{E}_{0})$ is defined on the interval $[\alpha,\tau]$,
then the set (\emph{cross-section }at the point $\tau$)
\[
\mathcal{X}(\tau;\hat{E}_{0}):=\left\{  (x(\tau),x'(\tau
)):x\in\mathcal{X}(\hat{E}_{0})\right\}
\]
is a continuum in $\mathbb{R}^{2n}$. \end{theorem}

Now consider (\ref{E})-(\ref{C}) with the following notation. 
\begin{gather*}
\sigma:=\sum_{i=1}^{m-2}\alpha_i\xi_i<1, \quad
\sigma^{\ast}:=\sum_{i=1}^{m-2}\alpha_i\xi_i+\frac{\beta}{\alpha} \Big\{
\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1\Big\} <1, \\
K_{0}:=\max\Big\{ \frac{2\alpha}{\beta},\;2\big[ \frac{\alpha+\beta}{\beta}-
\frac{\sigma}{\xi_{m-2}}\big] \Big\} , \\
\mu_{0}:=\min\Big\{ (1-m^{\ast})\frac{\varepsilon\alpha}{\beta},\;2\big[ 
\frac{\varepsilon(\alpha+\beta)}{\beta}-1\big] \text{ }\Big\}
\end{gather*}
where $\beta/(\alpha+\beta)<\varepsilon<1$ and $\sigma^{\ast}<m^{\ast}<1$.

So by (\ref{A3}), for any $\bar{K}\in(0,\bar{l})$ there exists $H>0$ such
that 
\begin{equation}
\min_{0\leq t\leq1}f(t,y,y')>-\bar{K}y',\;\;0\leq y\leq H
\big( 1+\frac{\alpha}{\beta}\big)\quad\text{and}\quad y'<-H.
\label{99}
\end{equation}
By the superlinearity of $f(t,y,y')$ at $y=+\infty\;$(see condition (
\ref{A2S})), for any $K^{\ast}>K_{0}$ there exists $H^{\ast}>H$ such that 
\begin{equation}
\min_{0\leq t\leq1}f(t,y,y')>K^{\ast}y,\ y\geq H^{\ast}\quad\text{and
} \quad -2H\leq y'\leq\frac{\alpha}{\beta}H.  \label{100}
\end{equation}
Similarly by the superlinearity of $f(t,y,y')$ at $(0,0)$, for any $
0<\mu^{\ast}<\mu_{0}$ there is an $\eta^{\ast}>0$ such that 
\begin{equation}
0<y\leq\eta^{\ast}\text{ and }0<y'\leq\frac{\alpha}{\alpha+\beta }
\eta^{\ast}\Rightarrow\max_{0\leq t\leq1}f(t,y,y')\leq\mu^{\ast}y.
\label{101}
\end{equation}
Also consider the rectangle 
\begin{equation*}
D:=\big[ 0,\;\big( 1+\frac{\alpha}{\beta}\big) H\big] \times\big[-2H,\;\frac{
\alpha}{\beta}H\big]
\end{equation*}
and define a bounded continuous modification $F$ of $f$ such that 
\begin{equation*}
F(t,y,y')=f(t,y,y'),\;(t,y,y')\in\left[ 0,1\right]
\times D.
\end{equation*}

\section{An m-point boundary-value Problem}

We consider now the boundary-value problem 
\begin{equation}
\begin{gathered} y''+F(t,y,y')=0. \\ \alpha y(0)-\beta y'(0)=0,\quad
y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i) \end{gathered}  \label{E*}
\end{equation}

\begin{theorem}
\label{Th3} Assume that (\ref{A1}) holds and
\begin{equation}
\sigma^{\ast}=\sum_{i=1}^{m-2}\alpha_i\xi_i+\frac{\beta}{\alpha}\Big\{
\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1\Big\}  <1. \label{MP}
\end{equation}
Then the boundary-value problem (\ref{E})-(\ref{C}) has a positive solution
provided that:
\begin{itemize}
\item  The function $f$ is superlinear (see (\ref{A2S})) along with
(\ref{A3}), or
\item  The function $f$ is sublinear (see (\ref{A2s})), (\ref{A4}) holds
and in addition,
\begin{equation}
\Big(  \sum_{i=1}^{m-2}\alpha_i\xi_i\Big)
\Big[  \frac{1}{2\xi_{m-2}}+\frac{\alpha}{2\beta}\Big]>1.\label{MP*}
\end{equation}
\end{itemize}
Furthermore, there exists a positive number $H$ such that
\[
0<y(t)\leq H\quad\text{and}\quad -2H\leq y'(t)
\leq\frac{\alpha}{\beta}H,\quad 0\leq t\leq1,
\]
for any such solution.
\end{theorem}

\paragraph{Proof}

\textbf{1}) \emph{\ Superlinear case}. Since $f_{\infty }=+\infty $ and in
view of (\ref{100}), for any $K^{\ast }>K>K_{0}$ there exists $H^{\ast }\geq
H>0$ such that 
\begin{equation}
\min_{0\leq t\leq 1}f(t,y,y')>Ky,\;\;\ y\geq H\quad \text{and}\quad 
\frac{\alpha }{\beta }H\geq y'\geq -2H.  \label{1}
\end{equation}
Consider the function 
\begin{equation*}
W(P):=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)-y(1),
\end{equation*}
where $y\in \mathcal{X}(P_{1})$ is any solution of differential equation (
\ref{E*}) starting at the point $P_{1}:=(y_{1},y_{1}')\in E_{0}$
with $y_{1}=H$.

By the assumption (\ref{A1}) (i.e. the nature of the vector field, see
Remark \ref{R1}) it is obvious that $y(t)\geq y_{1}=H$ and $
y'(t)\leq y_{1}'=\frac{\alpha}{\beta}y_{1}=\frac{\alpha}{
\beta}H,$\ for all $t$ in a sufficiently small neighborhood of $t=0$.

Let's suppose that there is $t^{\ast}\in(0,1]$ such that 
\begin{equation*}
y(t)\geq H,\;-2H\leq y'(t)\leq\frac{\alpha}{\beta}H,\;0\leq
t<t^{\ast }\text{ and }y(t^{\ast})=H
\end{equation*}
or 
\begin{equation*}
y(t)\geq H,\;-2H\leq y'(t)\leq\frac{\alpha}{\beta}H,\;0\leq
t<t^{\ast}\text{ and }y'(t^{\ast})=-2H.
\end{equation*}

Consider first the case: $y(t^{\ast})=H$. Then since $P_{1}\in E_{0}$, by
the Taylor's formula we get $t\in[0,t^{\ast}]$ \ such that 
\begin{equation}
H=y(t^{\ast})\leq H\big[ 1+\frac{\alpha}{\beta}\big] -\frac{1}{2}
f(t,y(t),y'(t))  \label{2}
\end{equation}
and thus 
\begin{equation*}
H\frac{2\alpha}{\beta}\geq f(t,y(t),y'(t)).
\end{equation*}
But since $y(t)\geq H$ and $-2H\leq y'(t)\leq\frac{\alpha}{\beta }
H,\ 0\leq t<t^{\ast}$ by (\ref{1}), we have 
\begin{equation*}
f(t,y(t),y'(t))\geq\min_{0\leq t\leq1}f(t,y(t),y'(t))\geq
Ky(t))\geq KH
\end{equation*}
and so we obtain $H2\alpha/\beta\geq KH$ contrary to the choice $K>\frac{
2\alpha}{\beta}$. Furthermore, by (\ref{2}), 
\begin{equation}
H\leq y(t)<H\big[ 1+\frac{\alpha}{\beta}\big],\quad 0\leq t\leq1.  \label{21}
\end{equation}

We recall also (see (\ref{99})) that for any $\varepsilon^{\ast}\in\left(
1,\;\min\left\{ 2,1+\bar{l})\right\} \right) $ there exists $\bar{K}
\in\left( \varepsilon^{\ast}-1,\;\bar{l}\right) $ such that 
\begin{equation}
\min_{0\leq t\leq1}f(t,y,y')>-\bar{K}y',\quad 0\leq y\leq H
\big[ 1+\frac{\alpha}{\beta}\big] \quad\text{and}\quad y'<-H.
\label{210}
\end{equation}
We shall prove that 
\begin{equation}
\frac{\alpha}{\beta}H\geq y'(t)\geq-\varepsilon^{\ast}H>-2H,\quad
0\leq t\leq1.  \label{22}
\end{equation}
Indeed, since $y'(t)$ is decreasing on $\left[ 0,1\right]$, let's
assume that there exist $t_{0},t_{1}\in(0,1)$ such that 
\begin{equation*}
y'(t_{0})=-H\;,\;-\varepsilon^{\ast}H<y'(t)<-H,\;t_{0}\leq
t\leq t_{1}\quad\text{and}\quad y'(t_{1})=-\varepsilon^{\ast}H
\end{equation*}
Then by (\ref{21})-(\ref{22}), for some $\bar{t}\in(t_{0},t_{1})$, we get 
\begin{align*}
-\varepsilon^{\ast}H=&y'(t_{1}) =y'(t_{0})-f(\bar{t},y(\bar{t
}),y'(\bar{t})) \\
\leq&-H+\bar{K}y'(\bar{t})\leq-H+\bar{K}y'(t_{0}) \\
=&-H-\bar{K}H,
\end{align*}
Thus we get another contradiction $\bar{K}\leq\varepsilon^{\ast}-1$. On the
other hand by the concavity of the solution $y\in\mathcal{X}(P_{1})$ (due to
the assumption (\ref{A1})), we know that the function $y(\xi)/\xi$, 
$0<\xi\leq1$ is decreasing and so 
\begin{equation}
\frac{y(\xi_i)}{\xi_i}\geq\frac{y(\xi_{m-2})}{\xi_{m-2}} ,\quad
i=1,2,\dots ,m-2.  \label{4}
\end{equation}
Thus in view of (\ref{21}) 
\begin{align*}
W(P_{1})=&\sum_{i=1}^{m-2}\alpha_iy(\xi_i)-y(1)=\sum_{i=1}
^{m-2}\alpha_i\xi_i\frac{y(\xi_i)}{\xi_i}-y(1) \\
\geq&\Big[ \sum_{i=1}^{m-2}\alpha_i\xi_i\Big] \frac{H}{\xi_{m-2}}
-y(1)=\sigma\frac{H}{\xi_{m-2}}-y(1).
\end{align*}
where we recall that $\sigma=\sum_{i=1}^{m-2}\alpha_i\xi_i<1$.
Consequently by Taylor's formula, 
\begin{equation*}
W(P_{1})\geq\frac{\sigma}{\xi_{m-2}}H-\big( y_{1}+\frac{\alpha}{\beta }y_{1}-
\frac{1}{2}f(t^{\ast},y(t^{\ast}),y'(t^{\ast}))\big)
\end{equation*}
Thus by (\ref{1}), (\ref{21}) and (\ref{22}), we get 
\begin{align*}
W(P_{1})\geq&\frac{\sigma}{\xi_{m-2}}H-\big[ 1+\frac{\alpha}{\beta }\big] H+
\frac{1}{2}Ky(t^{\ast}) \\
\geq&\frac{\sigma}{\xi_{m-2}}H-\big[ 1+\frac{\alpha}{\beta}\big] H+\frac{1}{2
}KH.
\end{align*}
In this way we get 
\begin{equation}
W(P_{1})\geq 0,  \label{5}
\end{equation}
since by the choice of $K$ at (\ref{1}), we have 
\begin{equation*}
K>2\big[1+\frac{\alpha}{\beta}-\frac{\sigma}{\xi_{m-2}}\big].
\end{equation*}
Similarly by the superlinearity of $f(t,y,y')$ at $(0,0)$, for any $\mu>0$ 
there is an $\eta>0$ such that 
\begin{equation}
0<y\leq\eta\text{ and }0\leq y'\leq\frac{2\alpha\varepsilon}{\beta }
\eta\quad\text{imply}\quad\max_{0\leq t\leq1}f(t,y,y')<\mu y,
\label{6}
\end{equation}
where $\frac{\beta}{\alpha+\beta}<\varepsilon<1$. We choose now (see (\ref
{101})) 
\begin{equation}
\mu^{\ast}\leq\mu<\mu_{0}=\;\min\big\{ (1-m^{\ast})\frac{\varepsilon\alpha }{
\beta},\;2\big[ \frac{\varepsilon(\alpha+\beta)}{\beta}-1\big] \big\}
\label{7}
\end{equation}
and then clearly $\eta\geq\eta^{\ast}.\;$

Let now $y\in\mathcal{X}(P_{0})$ be a solution of differential equation (\ref
{E*}) starting at the point $P_{0}:=(y_{0},y_{0}')\in E_{0}$ with $
y_{0}=\varepsilon\eta$. We shall show that 
\begin{equation}
\varepsilon\eta\leq y(t)\leq\eta\quad\text{and}\quad m^{\ast}\frac{
\alpha\varepsilon }{\beta}\eta\leq y'(t)\leq\frac{\alpha\varepsilon}{
\beta} \eta,\quad 0\leq t\leq1,  \label{8}
\end{equation}
where we recall that 
\begin{equation*}
\sigma^{\ast}=\sum_{i=1}^{m-2}\alpha_i\xi_i+\frac{\beta}{\alpha} \Big\{
\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1\Big\}<m^{\ast}<1.
\end{equation*}
Indeed, if there is a least $t^{\ast}\in(0,1]$ such that $m^{\ast}\frac
{\alpha\varepsilon}{\beta}\eta=y'(t^{\ast})$, and 
\begin{equation*}
\varepsilon\eta\leq y(t)\leq\eta\quad\text{and}\quad m^{\ast}\frac{
\alpha\varepsilon}{\beta}\eta\leq y'(t) \leq\frac{\alpha\varepsilon}{
\beta}\eta,\quad 0\leq t<t^{\ast},
\end{equation*}
then again by Taylor's formula, 
\begin{equation*}
m^{\ast}\frac{\alpha\varepsilon}{\beta}\eta=y'(t^{\ast})=y_{0}\frac{
\alpha}{\beta}-f(t,y(t),y'(t))\geq y_{0}\frac{\alpha }{\beta}-\mu
y(t)\geq\varepsilon\eta\frac{\alpha}{\beta}-\mu\eta,
\end{equation*}
and hence we obtain the contradiction $\mu\geq\left( 1-m^{\ast}\right) \frac{
\varepsilon\alpha}{\beta}$, due to the choice of $\mu$ at (\ref{7}).
Similarly we may prove the first inequality of (\ref{8}).

Consider once again the function 
\begin{equation*}
W(P)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)-y(1)
\end{equation*}
and then by (\ref{4}), 
\begin{equation}
\begin{aligned}
W(P_{0})=&\sum_{i=1}^{m-2}\alpha_iy(\xi_i)-y(1)=\sum_{i=1}
^{m-2}\alpha_i\xi_i\frac{y(\xi_i)}{\xi_i}-y(1)\\ \leq&\Big[
\sum_{i=1}^{m-2}\alpha_i\xi_i\Big] \frac{y(\xi_{1})}{\xi_{1}}-y(1).
\end{aligned}  \label{9}
\end{equation}
Now in view of (\ref{8}), 
\begin{equation*}
\frac{y(\xi_{1})}{\xi_{1}}=\frac{1}{\xi_{1}}\Big\{y(0)+\int_{0}^{\xi
_{1}}y'(s)ds\Big\}\leq \frac{\varepsilon \eta }{\xi_{1}}+\frac{
\alpha \varepsilon }{\beta }\eta \;
\end{equation*}
and 
\begin{equation*}
y(1)=y(0)+\int_{0}^{1}y'(s)ds\geq \varepsilon \eta +m^{\ast }\frac{
\alpha \varepsilon }{\beta }\eta .
\end{equation*}
Consequently by (\ref{9}), 
\begin{equation}
\begin{aligned} W(P_{0})\leq&\Big[ \sum_{i=1}^{m-2}\alpha_i\xi_i\Big]
\big(\frac{\alpha\varepsilon}{\beta}\eta+\frac{\varepsilon\eta}{\xi_{1}}
\big) -m^{\ast}\frac{\alpha\varepsilon}{\beta}\eta-\varepsilon\eta\\
=&\Big(\sum_{i=1}^{m-2}\alpha_i\xi_i-m^{\ast}\Big)
\frac{\alpha\varepsilon}{\beta}\eta+\Big\{ \sum_{i=1}^{m-2}\alpha_i\frac
{\xi_i}{\xi_{1}}-1\Big\} \eta\varepsilon\leq 0 \end{aligned}  \label{10}
\end{equation}
due to the choice of $m^{\ast }>\sum_{i=1}^{m-2}\alpha_i\xi_i+\frac{
\beta }{\alpha }\big\{\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1
\big\}$.

It is now clear that the function $W=W(P)$, $P\in[P_{0},P_{1}]$ is
continuous and thus by the Kneser's property (see Theorem \ref{Th1}), (\ref
{5}) and (\ref{10}), we get a point $P\in[P_{0},P_{1}]$ (we chose the last
one to the ``left'' of $P_{1}$) such that $W(P)=0$. This fact clearly means
that there is a solution $y\in\mathcal{X}(P)$ of equation (\ref{E*}), such
that 
\begin{equation*}
W(P)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)-y(1)=0.
\end{equation*}

It remains to be proved that the so obtaining solution $y=y(t)$ is actually
a bounded function. Indeed, by the choice of $P$, the continuity of $y(t)$
with respect initial values, (\ref{5}) and (\ref{10}), it follows that 
\begin{equation*}
y(t)>0,\quad 0\leq t\leq1,
\end{equation*}
because if 
\begin{equation*}
y(t)>0,\;0\leq t<1\quad\text{and}\quad y(1)=0,
\end{equation*}
then $W(P)>0$. Moreover by the nature of the vector field (see Remark \ref
{R1}), there is $t_{P}\in\left( 0,1\right) $ such that the so obtaining
solution $y\in\mathcal{X}(P)$ is strictly increasing on $\left[ 0,t_{p}
\right] $, strictly decreasing on $\left[ t_{p},1\right] $ and further is
strictly positive on $\left[ 0,1\right] $. Also it holds $y(t)\leq H$, 
$0\leq t\leq1$, i.e. 
\begin{equation}
0<y(t)\leq H,\quad 0\leq t\leq 1.  \label{111}
\end{equation}
Indeed, let's assume that there exist $t_{0},t_{1}\in[0,1]$ such that 
\begin{equation*}
y(t)\leq H,\quad 0\leq t<t_{0},\quad y(t_{0})=H,\quad y(t)\geq H,
\;\text{and}\; y'(t)\geq 0, \quad t_{0}\leq t\leq t_{1}.
\end{equation*}
Then we have $0<y'(t_{0})<\frac{\alpha}{\beta}y(t_{0})\leq\frac
{\alpha}{\beta}H$ and further by (\ref{1}), for some $\bar{t}
\in(t_{0},t_{1}) $ 
\begin{align*}
H\leq y(t_{1})=&y(t_{0})+(t_{1}-t_{0})y'(t_{0})-\frac{1}{2} f(\bar{t}
,y(\bar{t}),y'(\bar{t})) \\
\leq& H\big[ 1+\frac{\alpha}{\beta}\big] -\frac{K}{2}y(\bar{t}) \\
\leq& H\big[ 1+\frac{\alpha}{\beta}\big] -\frac{K}{2}H.
\end{align*}
Thus we get the contradiction $K<2\alpha/\beta$. Also by assumption (\ref{A3}
), we may show (exactly as at (\ref{22})) that the above solution $y\in
\mathcal{X}(P)$ implies further the inequalities 
\begin{equation}
\frac{\alpha}{\beta}H\geq y'(t)\geq-\varepsilon^{\ast}H\geq2H,\quad
0\leq t\leq 1.  \label{13}
\end{equation}
and hence by (\ref{111}) and the definition of the modification $F$, the
obtaining solution of (\ref{E*}) is actually a solution of the original
equation (\ref{E}). \smallskip

\noindent \textbf{2)} \emph{Sublinear case}. We choose $\varepsilon_{0}>
\frac{\alpha +\beta }{\beta }$ and recall that 
\begin{equation*}
\sum_{i=1}^{m-2}\alpha_iy(\xi_i)+\frac{\beta }{\alpha }\Big\{
\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1\Big\}<m^{\ast }<1.
\end{equation*}
Since $f_{+\infty ,+\infty }=0$, for $\mu <\min \big\{(1-m^{\ast })\frac{
\alpha }{\varepsilon_{0}\beta },\;\frac{2}{\varepsilon_{0}}[\varepsilon
_{0}-\frac{\alpha +\beta }{\beta }]\big\}$, there exists $H>0$ such that 
\begin{equation}
\max_{0\leq t\leq 1}f(t,y,y')<\mu y,\quad y\geq H,\quad \text{and}
\quad \frac{\alpha }{\beta }H\geq y'\geq m^{\ast }\frac{\alpha }{
\beta }H.  \label{71}
\end{equation}
Let's consider a point $P_{0}:=(y_{0},y_{0}')\in E_{0}$ with $
y_{0}=H$. We will prove first that for any solution $y\in \mathcal{X}(P_{0})$
, 
\begin{equation}
H\leq y(t)\leq \varepsilon_{0}H\quad \text{and}\quad \frac{m^{\ast }\alpha 
}{\beta }H\leq y'(t)\leq \frac{\alpha }{\beta }H,\quad 0\leq t\leq
1.  \label{72}
\end{equation}
Let us suppose that this is not the case. Then by the assumption (\ref{A1}),
there is $t^{\ast }\in [ 0,1]$ such that 
\begin{equation}
\begin{gathered} H\leq y(t)\leq\varepsilon_{0}H,\quad
\frac{m^{\ast}\alpha}{\beta }H\leq y'(t)\leq\frac{\alpha}{\beta}H,\quad
0<t<t^{\ast},\\ \text{and}\quad y(t^{\ast})=\varepsilon_{0}H\quad
\text{or}\quad y'(t)=\frac{m^{\ast}\alpha}{\beta}H. \end{gathered}
\label{720}
\end{equation}
Assume that $y(t^{\ast })=\varepsilon_{0}H.\;$Then by the Taylor's formula,
(\ref{71}) and (\ref{720}) we obtain $t\in [ 0,t^{\ast }]$ such that 
\begin{align*}
\varepsilon_{0}H=y(t^{\ast })=& y_{0}[1+\frac{\alpha }{\beta }]-\frac{1}{2}
f(t,y(t),y'(t)) \\
<& H[1+\frac{\alpha }{\beta }]+\frac{1}{2}\mu y(\bar{t})\leq H[1+\frac{
\alpha }{\beta }]+\frac{1}{2}\mu \varepsilon_{0}H
\end{align*}
and hence it contradicts 
\begin{equation*}
\mu <\frac{2}{\varepsilon_{0}}\big[\varepsilon_{0}-\frac{\alpha +\beta }{
\beta }\big].
\end{equation*}
Let's suppose now that $y'(t^{\ast })=m^{\ast }\frac{\alpha }{\beta 
}H$. Then again by (\ref{71}) and (\ref{720}), we obtain 
\begin{equation*}
m^{\ast }\frac{\alpha }{\beta }H=y'(t^{\ast })=y_{0}^{\prime
}-f(t,y(t),y'(t))\geq \frac{\alpha }{\beta }H-\mu y(t)\geq \frac{
\alpha }{\beta }H-\mu \varepsilon_{0}H,
\end{equation*}
which contradicts $\mu <(1-m^{\ast })\alpha /(\varepsilon_{0}\beta )$.

Consider the function $W(P)$. Then 
\begin{align*}
W(P_{0})=&\sum_{i=1}^{m-2}\alpha_iy(\xi_i)-y(1)=\sum_{i=1}
^{m-2}\alpha_i\xi_i\frac{y(\xi_i)}{\xi_i}-y(1) \\
\leq&\Big[\sum_{i=1}^{m-2}\alpha_i\xi_i\Big] \frac{y(\xi_{1})} {\xi_{1}}
-y(1)
\end{align*}
and so by the second inequality of (\ref{72}) (see also (\ref{10})), we get 
\begin{equation}
\begin{aligned} W(P_{0}) \leq&\Big[\sum_{i=1}^{m-2}\alpha_i\xi_i\Big]
\big(\frac{\alpha}{\beta}H+\frac{H}{\xi_{1}}\big)
-m^{\ast}\frac{\alpha}{\beta}H-H\\
=&\big(\sum_{i=1}^{m-2}\alpha_i\xi_i-m^{\ast}\big)
\frac{\alpha}{\beta}H+\big(\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}
-1\big) H\leq 0 \end{aligned}  \label{74}
\end{equation}
due to the fact that $m^{\ast}>\sum_{i=1}^{m-2}\alpha_i\xi_i+\frac{\beta 
}{\alpha}\big\{  \sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1\big\}$.

On the other hand, since $f_{0}=+\infty $, for any $K>\max \{\frac{2(\alpha
-\beta )}{\beta },\frac{2\alpha }{\beta }\}$ there exist $\eta \in (0,H)$
such that 
\begin{equation}
\min_{0\leq t\leq 1}f(t,y,y')>Ky,\;0<y\leq \eta \quad \text{and}
\quad -\eta \leq y'\leq \frac{\alpha }{\beta }\frac{\eta }{2}.
\label{75}
\end{equation}
Consider a point $P_{1}:=(y_{1},y_{1}')\in E_{0}$ with $y_{1}=\frac{
\eta }{2}$ and any $y\in \mathcal{X}(P_{1})$. As above, by Taylor's formula,
(\ref{75}) and the choice $K>\max \{\frac{2(\alpha -\beta )}{\beta },\frac{
2\alpha }{\beta }\}$ we can easily prove that 
\begin{equation}
\frac{\eta }{2}\leq y(t)\leq \eta ,\quad 0\leq t\leq 1.  \label{76}
\end{equation}
We choose now $\varepsilon_{0}^{\ast }\in \left( 1,2\right) $ and then\ by
Assumption (\ref{A4}), there exist $\bar{\eta}_{0}\in (0,\eta )$ and 
\begin{equation}
0<K^{\ast }<\min \Big\{\frac{\varepsilon_{0}^{\ast }-1}{\varepsilon
_{0}^{\ast }},\;\min \{1,\frac{2\beta }{\alpha }\}[\frac{\sigma \xi_{m-2}}{2
}]^{-1}[\frac{\sigma }{2\xi_{m-2}}+\frac{\sigma \alpha }{2\beta }-1]\Big\}
\label{760}
\end{equation}
such that 
\begin{equation}
\max_{0\leq t\leq 1}f(t,y,y')<K^{\ast }|y'|,\quad \frac{
\eta }{2}\leq y\leq \eta ,\quad \text{and}\quad -2\bar{\eta}_{0}\leq
y'<\frac{\alpha \eta }{2\beta }.  \label{761}
\end{equation}
Besides (\ref{76}) we shall prove that 
\begin{equation}
\frac{\alpha }{\beta }\frac{\eta }{2}\geq y'(t)\geq -\bar{\eta}
_{0}>-\eta ,\quad 0\leq t\leq 1.  \label{78}
\end{equation}
Indeed since $y'(t)$ is decreasing on $\left[ 0,1\right] $ and $
\varepsilon_{0}^{\ast }\in \left( 1,2\right) $ is arbitrary, let's assume
that there exist $t_{0},t_{1}\in [ 0,1]$ such that $y^{\prime
}(t_{0})=-\bar{\eta}_{0}$, 
\begin{equation*}
-2\bar{\eta}_{0}<-\varepsilon_{0}^{\ast }\bar{\eta}_{0}\leq y^{\prime
}(t)\leq -\bar{\eta}_{0},\quad t_{0}\leq t<t_{1},\quad \text{and}
\;\;y'(t_{1})=-\varepsilon_{0}^{\ast }\bar{\eta}_{0}.
\end{equation*}
Thus by (\ref{76})-(\ref{761}), we have for some $\bar{t}\in (t_{0},t_{1})$ 
\begin{equation*}
-\varepsilon_{0}^{\ast }\bar{\eta}_{0}=y'(t_{1})=y^{\prime
}(t_{0})-f(\bar{t},y(\bar{t}),y'(\bar{t}))\geq -\bar{\eta}
_{0}+K^{\ast }y'(\bar{t})\geq -\bar{\eta}_{0}-K^{\ast }\varepsilon
_{0}^{\ast }\bar{\eta}_{0},
\end{equation*}
and so, we get another contradiction $K^{\ast }\geq (\varepsilon_{0}^{\ast
}-1)/\varepsilon_{0}^{\ast }$, due to (\ref{760}).

Now as above (see (\ref{4}) and (\ref{74})), we have 
\begin{equation*}
W(P_{1})\geq\Big[ \sum_{i=1}^{m-2}\alpha_i\xi_i\Big] \frac
{y(\xi_{m-2})}{\xi_{m-2}}-y(1)=\sigma\frac{y(\xi_{m-2})}{\xi_{m-2}}-y(1).
\end{equation*}
Consequently by (\ref{76}) and the Taylor's formula, 
\begin{align*}
W(P_{1})\geq&\frac{\sigma}{\xi_{m-2}}\Big( y_{1}+\frac{\alpha}{\beta }
y_{1}\xi_{m-2}-\frac{\xi_{m-2}^{2}}{2}f(\bar{t},y(\bar{t}),y'(\bar
{t}))\medskip\Big) -\eta \\
=&\frac{\sigma}{\xi_{m-2}}\frac{\eta}{2}+\sigma\frac{\alpha}{\beta} \frac{
\eta}{2}-\frac{1}{2}\sigma\xi_{m-2}f(\bar{t},y(\bar{t}), y'(\bar{t}
))-\eta
\end{align*}
Thus by (\ref{76}) and (\ref{78}), we get 
\begin{align*}
W(P_{1})\geq& \frac{\sigma}{\xi_{m-2}}\frac{\eta}{2}+\sigma\frac{\alpha }{
\beta}\frac{\eta}{2}-\frac{\sigma\xi_{m-2}}{2}K^{\ast}| y'(\bar{t}
)| -\eta \\
\geq& \frac{\sigma}{\xi_{m-2}}\frac{\eta}{2}+\sigma\frac{\alpha}{\beta} 
\frac{\eta}{2}-\frac{\sigma\xi_{m-2}}{2}K^{\ast}\hat{\eta}-\eta,
\end{align*}
where $\hat{\eta}:=\max\{\eta,\frac{\alpha\eta}{2\beta}\}$. In this way, by
the assumption (\ref{MP*}) and the choice of $K^{\ast}$ at (\ref{760}), we
get 
\begin{equation*}
W(P_{1})\geq0.
\end{equation*}
Thus as at the superlinear case, we obtain a point $P\in[ P_{0},P_{1}]$ such
that $W(P)=0$ and this clearly completes the proof.

\begin{remark} \label{R2} \rm
By the choice of $m^{\ast}\in\big(\sum_{i=1}^{m-2}\alpha_i
\xi_i+\frac{\beta}{\alpha}\big\{  \sum_{i=1}^{m-2}\alpha_i\frac{\xi_i
}{\xi_{1}}-1\big\}  ,\;1\big)$ and following Ma \cite{Ma2}, we may
easily show that for
\[
\sum_{i=1}^{m-2}\alpha_i\xi_i\geq1,
\]
there is not (positive) solution $y\in\mathcal{X}(P)$ of the BVP
(\ref{E})-(\ref{C}). Indeed, if there is one, then
\[
y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)=\sum_{i=1}^{m-2}\alpha_i
\xi_i\frac{y(\xi_i)}{\xi_i}
\geq\sum_{i=1}^{m-2}\alpha_i\xi_i\frac{y(\xi^{\ast})}{\xi^{\ast}}
>\frac{y(\xi^{\ast})}{\xi^{\ast}},
\]
where clearly $\xi^{\ast}=\xi_{m-2}$ and this contradicts the concavity of
the solution $y=y(t)$.
Furthermore we must seek the monotone (obviously increasing) solutions of
(\ref{E})-(\ref{C}), only for the case $\sum_{i=1}^{m-2}\alpha_i\geq1$, since otherwise we get
\[
0=W(P)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)-y(1)<\Big[  \sum_{i=1}
^{m-2}\alpha_i-1\Big]  y(1)<0.
\]
The question of existence of such a monotone solution remains open. However
we can obtain a strictly decreasing solution for the boundary-value problem
\begin{equation}
\begin{gathered}
y''(t)=-f(t,y(t),y'(t)),\quad 0\leq t\leq1, \\
\alpha y(0)+\beta y'(0)=0,\quad
y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i).
\end{gathered} \label{C*}
\end{equation}
where $\alpha\geq0$ $\ $and $\beta>0$. \end{remark}

\begin{remark} \label{R3} \rm
Suppose that the concept of jointly sublinearity is modified to
\begin{equation}
\begin{gathered}
f_{0}:=\lim_{y\to 0+}\min_{0\leq t\leq1}\frac{f(t,y,y')}
{y}=+\infty,\quad \mbox{for } |y'|\leq M.\\
f_{\infty,-\infty}:=\lim_{(y,y')\to (+\infty,-\infty)}
\max_{0\leq t\leq1}\frac{f(t,y,y')}{y}=0.
\end{gathered} \label{A2*}
\end{equation}
Then, following almost the same line as above (under the obvious
modifications) we may prove the next theorem.
\end{remark}

\begin{theorem}
Assume that (\ref{A1}) holds and further
\[
\sigma^{\ast}=\sum_{i=1}^{m-2}\alpha_i\xi_i+\frac{\beta}{\alpha}
\Big\{\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1\Big\}  <1.
\]
Then the boundary-value problem (\ref{C*}) has a positive strictly
decreasing solution provided that:
\begin{itemize}
\item  The function $f$ is superlinear (see (\ref{A2S})) along with
(\ref{A3}), or

\item  The function $f$ is sublinear (see (\ref{A2*})), (\ref{A4}) is true
and in addition,
\[
\sum_{i=1}^{m-2}\alpha_i\xi_i\big[  \frac{1}{\xi_{m-2}}-\frac{\alpha
}{\beta}\big]  >1.
\]
\end{itemize}
Furthermore there exists a positive number $H$ such that
\[
0<y(t)\leq H\quad\text{and}\quad -2H\leq y'(t)\leq-\frac{\alpha}{\beta
}H,\quad 0\leq t\leq1,
\]
for any such solution.
\end{theorem}

\begin{remark} \label{R4} \rm
Again, as in Remark \ref{R2}, we may show that for
\[
\sum_{i=1}^{m-2}\alpha_i\xi_i\geq1,
\]
there is no (positive) solution $y\in\mathcal{X}(P)$ of the BVP
(\ref{C*}). Furthermore we must seek the possible solutions of
(\ref{C*}) only for the case
\[
\sum_{i=1}^{m-2}\alpha_i\leq1,
\]
since otherwise, by the monotonicity of $y(t)$, we get the contradiction
\[
0=W(P)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)\geq\Big[  \sum_{i=1}^{m-2}
\alpha_i-1\Big]  y(1)>0.
\]
\end{remark}

Finally consider the boundary-value problem 
\begin{equation}
\begin{gathered} y''+f(t,y,y')=0, y'(0)=0,\quad
y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i). \end{gathered}  \label{D}
\end{equation}
Then following almost the same lines as above, we may prove the next theorem.

\begin{theorem}
Assume that (\ref{A1}) holds and
\[
\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}<1.
\]
Then the boundary-value problem (\ref{D}) has a positive strictly
decreasing solution provided that
\begin{itemize}
\item  The function $f$ is superlinear (see (\ref{A2S})) along with
(\ref{A3}), or

\item  The function $f$ is sublinear (see (\ref{A2*})), (\ref{A4}) holds
and in addition
\[
\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{m-2}}>1.
\]
\end{itemize}
Furthermore there exists a positive number $H$ such that
\[
0<y(t)\leq H\quad\text{and}\quad -2H\leq y'(t)\leq0,\quad 0\leq t\leq 1,
\]
for any such solution.
\end{theorem}

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\noindent \textsc{Panos K. Palamides}\newline
Naval Academy of Greece, Piraeus 183 03, Greece\newline
and\newline
Department of Mathematics, Univ. of Ioannina, \newline
451 10 Ioannina, Greece \newline
e-mail:ppalam@otenet.gr

\end{document}