\input amstex \documentstyle{amsppt} \loadmsbm \magnification=\magstephalf \hcorrection{1cm} \vcorrection{-6mm} \nologo \TagsOnRight \NoBlackBoxes \headline={\ifnum\pageno=1 \hfill\else% {\tenrm\ifodd\pageno\rightheadline \else \leftheadline\fi}\fi} \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)$$ and we observe the following: \proclaim{ Lemma 2.2} Let $b$ be a fixed real number such that $$00$, 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\|