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\def\rightheadline{EJDE--2000/73\hfil
Positive solutions for a nonlocal boundary-value problem \hfil\folio}
\def\leftheadline{\folio\hfil G. L. Karakostas \& P. Ch. Tsamatos
\hfil EJDE--2000/73}
\def\pretitle{\vbox{\eightrm\noindent\baselineskip 9pt %
Electronic Journal of Differential Equations,
Vol.~{\eightbf 2000}(2000), No.~73, pp.~1--8.\hfil\break
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\hfill\break
ftp ejde.math.swt.edu (login: ftp)\bigskip} }
\topmatter
\title
Positive solutions for a nonlocal boundary-value problem with
increasing response
\endtitle
\thanks
{\it 2000 Mathematics Subject Classifications:} 34B18. \hfil\break\indent
{\it Key words:} Nonlocal boundary-value problems, positive solutions.
\hfil\break\indent
\copyright 2000 Southwest Texas State University. \hfil\break\indent
Submitted October 26, 2000. Published December 12, 2000.
\endthanks
\author G. L. Karakostas \& P. Ch. Tsamatos \endauthor
\address G. L. Karakostas \hfill\break\indent
Department of Mathematics, University of Ioannina,
451 10 Ioannina, Greece
\endaddress
\email gkarako\@cc.uoi.gr
\endemail
\address P. Ch. Tsamatos \hfill\break\indent
Department of Mathematics, University of Ioannina,
451 10 Ioannina, Greece
\endaddress
\email ptsamato\@cc.uoi.gr \endemail
\abstract
We study a nonlocal boundary-value problem for a second
order ordinary differential equation.
Under a monotonicity condition on the response function, we
prove the existence of positive solutions.
\endabstract
\endtopmatter
\document
\head 1. Introduction \endhead
When looking for positive solutions of the equation
$$u''(t)+a(t)f(u(t))=0,\enskip t\in [0,1],
$$
associated with various boundary conditions the main assumption on the response
function
$f$ is the existence of the limits of $f(u)/u,$ as $u$ approaches $0$
and $+\infty$. Existence of solutions under these conditions has been
shown, for instance, in [1, 4, 5, 6, 7, 11, 18]. Such conditions distinguish
two cases: The sublinear case when the limits are $+\infty$ and $0$,
and the superlinear case when the limits are $0$ and $+\infty$, respectively.
In [16] the authors present a detailed investigation of a twwo-point
boundary-value problem under similar limiting conditions and they
introduce the meaning of the index of convergence.
In this paper, we discuss a general problem with
non-local boundary conditions. We avoid
the limits above, and therefore weaken the restriction of the
function $f$. Instead, we assume that there exist real positive
numbers $u,v$ such that $f(u)\ge \rho u$ and $f(v)< \theta v$,
where $\rho , \theta$ are prescribed positive numbers.
This is a rather weak condition, but we have to pay for it. Indeed,
we assume that the function
$f$ is increasing (not necessarily strictly increasing).
More precisely, we consider the ordinary differential equation
$$
(p(t)x')'+q(t)f(x)=0,\enskip \hbox{a.e.}\enskip t\in [0,1]\tag 1.1 %e
$$
with the initial condition
$$x(0)=0\tag 1.2
$$
and the non-local boundary condition
$$x'(1)=\int_{\eta}^{1}x'(s)dg(s).\tag 1.3
$$
Here $f: \Bbb R\to \Bbb R$ is an increasing
function, the real valued functions $p,q,g$ are defined
at least on the interval $[0,1]$
and $\eta $ is a real number
in the open interval $(0,1)$. Also the integral
in (1.3) is meant in the sense of Riemann-Stieljes.
When (1.1) is an equation of Sturm-Liouville
type, Il'in and Moiseev [12], motivated by a
work of Bitsadze [2] and Bitsadze and Samarskii
[3], investigated the existence of solutions
of the problem
(1.1), (1.2) with the multi-point condition
$$x'(1)=\sum_{i=1}^m\alpha _i x'(\xi _i), \tag 1.4
$$
where the real numbers $\alpha_1, \alpha_2, \dots ,
\alpha_m$ have the same sign. The
formed boundary-value problem (1.1), (1.2), (1.4)
was the subject of some recent papers
(see, e.g. [9, 10]). Condition (1.3) is the continuous version of
(1.4) which happens when $g$ is a piece-wise constant
function that is increasing and has a finitely many jumps.
The question of existence of positive solutions of the boundary-value problem
(1.1)-(1.3) is justified by the large number of papers.
For example one can consult the papers [1, 4, 5, 6, 7, 11, 18] which were
motivated by
Krasnoselskii [17], who presented a complete theory for
positive solutions of operator equations. One of the
more powerful tools exhibited in [17] is
the following general fixed point theorem. This theorem is
an extension of the classical Bolzano-Weierstrass sign theorem for
continuous real valued functions to Banach spaces, when
the usual order is replaced by the order generated by a cone.
\proclaim{Theorem 1.1}
Let ${\Cal B}$ be a Banach space and let $\Bbb K$
be a cone in ${\Cal B}$. Assume that $\Omega_1$ and $\Omega _2 $
are open subsets of ${\Cal B}$, with
$0\in\Omega _1 \subset \overline{\Omega _1 }\subset \Omega _2$,
and let
$$A: \Bbb K\cap (\Omega _2\setminus \overline
{\Omega _1 } )\to \Bbb K$$
be a completely continuous operator such that either
$$\|Au\|\le \|u\|,\enskip u\in \Bbb K \cap \partial
\Omega _1 ,\enskip \|Au\|\ge \|u\|,\enskip
u\in \Bbb K \cap \partial \Omega _2 $$
or
$$\|Au\|\ge \|u\|,\enskip u\in \Bbb K \cap \partial
\Omega _1 ,\enskip \|Au\|\le \|u\|,\enskip
u\in \Bbb K \cap \partial \Omega _2 .$$
Then $A$ has a fixed point in $\Bbb K\cap
(\Omega _2\setminus \overline {\Omega _1 } ).$
\endproclaim
In the literature, boundary-value problems of the form (1.1)-(1.3)
are often solved by using the well
known Leray-Schauder Continuation
Theorem (see, e.g. [9, 10, 13, 19]), or the Nonlinear Alternative (see, e.g.
[8, 15] and the references therein. For another approach see, also, [14]).
On the other hand Krasnoselskii's fixed point theorem, when it is
applied, it provides some additional properties of the solutions,
for instance, positivity (see, e.g. [1, 4, 5, 6, 7, 11, 14]).
However, the more information on the solutions the more restrictions
on the coefficients are needed.
\head 2. Preliminaries and assumptions\endhead
In the sequel we shall denote by $\Bbb R$ the real
line and by $I$ the interval $[0,1]$.
Then $C(I)$ will denote the space of all continuous
functions $x:I\to
\Bbb R$. Let $C_{0}^{1}(I)$ be the
space of all functions $x:I\to
\Bbb R$, whose the first derivative
$x'$ is absolutely continuous on $I$ and $x(0)=0$.
This is a Banach space
when it is furnished with the
norm defined by
$$\| x\|:=\sup\{|x'(t)|: t\in I\},\enskip x\in C_{0}^{1}(I).
$$
We denote by $L^+_1(I)$ the space of
functions $x:I\to \Bbb R^+:=[0, +\infty)$ which are Lebesgue
integrable on I.
Consider the system (1.1), (1.2) and the nonlocal-value condition (1.3).
By a solution of the problem (1.1)-(1.3) we mean a function
$x\in C_{0}^{1}(I)$ satisfying
equation (1.1) for almost all $t\in I$ and condition (1.3).
\par \vskip 0.2cm
Before presenting our results we give our basic assumptions:
\roster
\item"(H1)" $f:\Bbb R\to \Bbb R$ is an increasing continuous function,
with $f(x)\ge 0$, when $x>0$
\item"(H2)" The functions $p,q$ belong to $C(I)$
and they are such that $p>0$, $q\ge 0$ and
$\sup\{q(s): \eta \le {s}\le1\}>0$. Without
loss of generality we can assume that
$p(1)=1$.
\item"(H3)" The function $g: I\to \Bbb R$ is increasing
and such that $g(\eta )=00).$$
The following lemma
is the basic tool in the proof of our main result.
\proclaim{ Lemma 2.1}
If $y\in C(I)$ is a nonnegative and increasing
function, then it holds
$$\int_{\eta}^{1}\Phi (y)(s)dg(s)\ge
\lambda b\int_{0}^{1}q(s)y(s)ds,\enskip b\in [0, b_0 ],$$
where
$$\lambda :=\frac {\int_{\eta}^{1}q(s)ds}
{\int_{0}^{1}q(s)ds}\left ( \sup_{s\in
I}p(s)\right ) ^{-1}.$$
\endproclaim
\demo{Proof}
Since the function $g$ is increasing, for every
$b\in (0, b_0] $ we have
$$g(s)\ge b, \enskip s\in(\eta, 1]. \tag 2.3
$$
Hence it follows that
$$
\aligned
\int_{0}^{1}q(s)y(s)ds
&=\int_{0}^{\eta}q(s)y(s)ds+\int_{\eta}^{1}q(s)y(s)ds\\
&\le y(\eta )\int_{0}^{\eta}q(s)ds+\int_{\eta}^{1}q(s)y(s)ds\\
&\le\frac{\int_{0}^{\eta}q(s)ds}{\int_{\eta}^{1}q(s)ds}
\int_{\eta}^{1}q(s)y(s)ds+
\int_{\eta}^{1}q(s)y(s)ds\\
&= \frac{\int_{0}^{1}q(s)ds}{\int_{\eta}^{1}q(s)ds}
\int_{\eta}^{1}q(s)y(s)ds. \endaligned$$
Now we use assumption $(H_{3})$ and
relation $(2.3)$ to obtain that
$$
\aligned
\int_{0}^{1}q(s)y(s)ds
&\le b^{-1}\frac{\int_{0}^{1}q(s)ds}{\int_{\eta}^{1}
q(s)ds}\int_{\eta}^{1}q(s)y(s)g(s)ds\\
&=-b^{-1}\frac{\int_{0}^{1}q(s)ds}{\int_{\eta}^{1}
q(s)ds}\int_{\eta}^{1}d\left (
\int_{s}^{1}q(\theta )y(\theta )d\theta \right )g(s)\\
&= b^{-1}
\frac{\int_{0}^{1}q(s)ds}{\int_{\eta}^{1}q(s)ds}
\int_{\eta}^{1}\int_{s}^{1}q(\theta )y(\theta
)d\theta dg(s)\\
&\le (\lambda b)^{-1}\int_{\eta}^{1}\frac{1}{p(s)}
\int_{s}^{1}q(\theta )y(\theta )d\theta
dg(s).
\endaligned
$$
The proof is complete.
\qed\enddemo
For convenience we set
$$D:=\int_{\eta}^{1}\Phi (P)(s)dg(s),\quad
H:=\int_{\eta}^{1}\Phi (1) (s)dg(s)$$
and we observe the following:
\proclaim{ Lemma 2.2}
Let $b$ be a fixed real number such that
$$0**0$, by a simple calculation we have the result. Also, if
$D\eta p(0)-H<0$, then $$\sigma \eta =\frac {\alpha \lambda bp(0)\eta} {\alpha
\lambda b+1}D<\frac {\alpha \lambda bH} {\alpha
\lambda b+1}\le H\,.
$$ \enddemo
\head 3. Main results\endhead
Before presenting our main theorem we set $\rho :=\frac{1}{\alpha \sigma
\eta}$ and let $\theta:=\frac{p(0)}{\alpha H+\int_{0}^{1}q(s)ds}$ where
$\sigma$ and
$H$ are the constants defined in Lemma 2.2.
\vskip 0.3cm
\proclaim {Theorem 3.1}
Assume that $f,p,q$ and $g$ satisfy (H1)-(H4). If
\roster
\item"(H5)" There exist $u>0$ and $v>0$ such that
$f(u)\ge \rho u$ and $f(v)< \theta v$,
\endroster
then the boundary-value problem (1.1)-(1.3) admits
at least one positive solution.
\endproclaim
\demo{ Proof}
Our main purpose is to make the appropriate arrangements
so that Theorem 1.1 to be applicable.
Define the set
$$\Bbb K:=\left\{x\in C^1_0 (I): x\ge 0, \enskip x'\ge 0,
\enskip x \enskip \hbox {is} \enskip\hbox {concave} \enskip \hbox{and}\enskip
\int_{\eta}^{1}\Phi (x) (s)dg(s)\ge \sigma \|x\|\right\},$$
which is a cone in $C^1_0 (I)$.
\par
First we claim that the operator $A$ maps $\Bbb K$
into $\Bbb K$. To this end take a point
$x\in\Bbb K$. Then observe that it
holds $Ax\ge 0, (Ax)'\ge 0$ and $(Ax)''\le 0$.
Moreover, we observe that
$$
\aligned
\int_{\eta}^{1}\Phi (Ax)(s)dg(s)
\ge &\alpha \int_{\eta}^{1}\Phi (P)(s)dg(s)
\int_{\eta}^{1}\Phi (f(x))(s)dg(s)\\
=&\alpha D
\int_{\eta}^{1}\frac{1}{p(s)}\int_{s}^{1}
q(\theta)f(x(\theta ))d\theta dg(s)\\
=&\frac{\sigma (\alpha \lambda b
+1)}{\lambda bp(0)}
\int_{\eta}^{1}\frac{1}{p(s)}\int_{s}^{1}
q(\theta )f(x(\theta ))d\theta dg(s)\\
=&\frac{\sigma}{p(0)}\left (\alpha +\frac{1}
{\lambda b}\right )
\int_{\eta}^{1}\frac{1}{p(s)}\int_{s}^{1}
q(\theta )f(x(\theta ))d\theta dg(s)\\
=&\sigma [\frac{\alpha}{p(0)}\int_{\eta}^{1}
\frac{1}{p(s)}\int_{s}^{1}q(\theta)f(x(\theta ))d\theta dg(s)\\
&+\frac{1}{p(0)}\frac{1}{\lambda b}
\int_{\eta}^{1}\frac{1}{p(s)}\int_{s}^{1}q(\theta
)f(x(\theta ))d\theta dg(s)].\endaligned$$
Now we use Lemma 2.1 and get
$$\aligned
\int_{\eta}^{1}\Phi (Ax)(s)dg(s)
\ge&
\sigma [ \frac{\alpha}{p(0)}\int_{\eta}^{1}
\frac{1}{p(s)}\int_{s}^{1}q(\theta
)f(x(\theta ))d\theta dg(s)\\
&+\frac{1}{p(0)}\int_{0}^{1}q(\theta )f(x(\theta))d\theta ]\\
=&\sigma (Ax)'(0)\\
=&\sigma \|(Ax)\| .
\endaligned
$$
This proves our first claim. \par
Now consider an arbitrary $x\in\Bbb K$. The
fact that the function $x$ is
concave implies that
$$\eta x(1)\le x(\eta )\le x(r)\le x(1)\le \|x\|,
\enskip \hbox{for every}\enskip r\in [\eta , 1].$$
So,
$$
\aligned
\sigma\|x\| &\le \int_{\eta}^{1}\Phi (x)(s)dg(s)\\
&=\int_{\eta}^{1}\frac{1}{p(s)}\int_{s}^{1}q(\theta
)x(\theta )d\theta dg(s)\\
&\le x(1)\int_{\eta}^{1}\frac{1}{p(s)}
\int_{s}^{1}q(\theta )d\theta dg(s)\\
&=x(1)\int_{\eta}^{1}\Phi (1)(s)dg(s)\\
&=x(1)H.
\endaligned
$$
Thus we have $x(1)\ge \frac{\sigma\|x\|}{H}$,
which implies that
$$x(r)\ge \frac{\eta \sigma}{H}\|x\|, \enskip r\in [\eta,1].$$
Hence, for every $r\in [\eta, 1]$ we have
$$\frac{\eta \sigma}{H}\|x\|\le x(r)\le \|x\| ,$$
where, notice that, by Lemma 2.2, $\frac{\eta \sigma}{H}\le 1$.
Then, by assumption (H5), there exists
$u>0$ such that $f(u)\ge \rho u$.
Set
$$M:=\frac{H}{\eta\sigma}u$$
and fix a function $x\in\Bbb K$ with $\|x\|=M$. Then
$$\frac{\eta\sigma}{H}M\le x(r) \le M,\enskip
\hbox{for every}\enskip r\in [\eta ,
1]$$ and therefore
$$
\aligned
(Ax)'(1)&\ge \alpha\int_{\eta}^{1}\frac{1}
{p(s)}\int_{s}^{1}q(\theta)f(x(\theta ))d\theta dg(s)\\
&\ge \alpha f(x(\eta ))\int_{\eta}^{1}
\Phi (1)(s)dg(s)=\alpha H f(x(\eta ))\\
&\ge \alpha H f(\frac{\eta\sigma M}{H})=
\alpha H f(u)\ge \alpha H\rho u\\
&=\alpha \rho \eta\sigma M\ge M=\|x\| .
\endaligned
$$
Thus we proved that, if $\|x\| =M$, then $\|Ax\|\ge \|x\|$.\par
Now, again, from
assumption (H5), it follows that there
exists $v>0$ such that $0\le f(v)<
\theta v$. Fix any function $x\in\Bbb K$ with $\|x\| =v$.
Then $0\le x(r)\le v$, $r\in I$. Therefore
$$
\aligned
\|Ax\|=(Ax)'(0) &=\frac{\alpha}{p(0)}\int_{\eta}^{1}\Phi
(f(x))(s)dg(s)+\frac{1}{p(0)}\int_{0}^{1}q(s)f(x(s))ds\\
&=\frac{\alpha}{p(0)}\int_{\eta}^{1}\frac{1}
{p(s)}\int_{0}^{1}q(r)f(x(r))drdg(s)+\frac{1}
{p(0)}\int_{0}^{1}q(s)f(x(s))ds\\
&\le f(v)\left [\frac{\alpha H}{p(0)}+
\frac{1}{p(0)}\int_{0}^{1}q(s)ds\right ]\\
&\le \theta v\left [\frac{\alpha H}{p(0)}+
\frac{1}{p(0)}\int_{0}^{1}q(s)ds\right ]\\
&= v=\|x\| .
\endaligned
$$
So we proved that, if $\|x\| =v$, then $\|Ax\|\le \|x\| .$\par
\par
Finally, we set $\Omega _1 :=\{x\in C_{0}^{1}(I):
\|x\|**