\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 86, pp. 1--9.\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 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/86\hfil Positive solutions and eigenvalues] {Positive solutions and eigenvalues of nonlocal boundary-value problems} \author[J. Chu, Z. Zhou\hfil EJDE-2005/86\hfilneg] {Jifeng Chu, Zhongcheng Zhou} % in alphabetical order \address{Jifeng Chu \hfill\break Department of Applied Mathematics, College of Sciences, Hohai University\\ Nanjing 210098, China} \email{jifengchu@hhu.edu.cn} \address{Zhongcheng Zhou \hfill\break School of Mathematics and Finance, Southwest Normal University\\ Chongqing 400715, China} \email{zhouzc@swnu.edu.cn} \date{} \thanks{Submitted April 18, 2005. Published July 27, 2005} \subjclass[2000]{34B15} \keywords{Nonlocal boundary-value problems; positive solutions, eigenvalues; \hfill\break\indent fixed point theorem in cones} \begin{abstract} We study the ordinary differential equation $x''+\lambda a(t)f(x)=0$ with the boundary conditions $x(0)=0$ and $x'(1)=\int_{\eta}^{1}x'(s)dg(s)$. We characterize values of $\lambda$ for which boundary-value problem has a positive solution. Also we find appropriate intervals for $\lambda$ so that there are two positive solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} This paper concerns the ordinary differential equation $$\label{e1.1} x''+\lambda a(t)f(x)=0,\quad \text{a.e. }t\in[0,1]$$ with the boundary conditions \begin{gather} x(0)=0 \label{e1.2}\\ x'(1)=\int_{\eta}^{1}x'(s)dg(s),\label{e1.3} \end{gather} where $\lambda>0$, $\eta\in(0,1)$ and the integral in \eqref{e1.3} is meant in the sense of Riemann-Stieljes. In this paper it is assumed that \begin{itemize} \item[(H1)] The function $f:[0,\infty )\to[0,\infty )$ is continuous. \item[(H2)] The function $a:[0,1]\to [0,\infty )$ is continuous and does not vanish identically on any subinterval. \item[(H3)] The function $g:[0,1]\to\mathbb{R}$ is increasing and such that $g(\eta)=00$ such that $\frac{1}{\eta(f_{\infty }-\varepsilon)B}\leq\lambda \leq\frac{1}{(f_{0}+\varepsilon)A}.$ First, there exists $r>0$ such that $f(x)\leq (f_{0}+\varepsilon)x,\quad 00 such that \[ \frac{1}{\eta(f_{0}-\varepsilon)B}\leq\lambda\leq\frac{1}{(f_{\infty } +\varepsilon)A}.$ First, there exists $r>0$ such that f(x)\geq (f_{0}-\varepsilon)x, \quad 00 such that f(x)\leq M, ~~x\in(0,\infty ). Let R=\max\{2r, \lambda MA\} and set \[ \Omega_{2}=\{x\in K:\|x\|\max\{2r, R_{1}\} such that \[ f(x)\leq f(R),~~~01, \beta>1. \begin{proof}It is easy to see that f_{0}=\infty, f_{\infty }=0 if  0<\alpha<1, 0<\beta <1 and f_{0}=0, f_{\infty }=\infty  if \alpha>1, \beta>1. Then the results can be easily obtained by using Theorem \ref{thm3.1} or Theorem \ref{thm3.2} directly. \end{proof} \section{Twin positive solutions} In this section, we establish the existence of two positive solutions to problem \eqref{e1.1}-\eqref{e1.3}. \begin{theorem} \label{thm4.1} Suppose that {\rm (H1)-(H3)} hold. In addition, assume there exist two constants R>r>0 such that $$\label{e4.1} \max_{0\leq x\leq r}f(x)\leq r/\lambda A, \quad \min_{\eta R\leq x\leq R}f(x)\geq R/\lambda B.$$ Then the boundary-value problem \eqref{e1.1}-\eqref{e1.3} has at least one positive solution x\in K with r\leq\|x\|\leq R. \end{theorem} \begin{proof} For x\in\partial K_{r}=\{x\in K:\|x\|=r\}, we have f(x(t))\leq r/\lambda A for t\in[0,1]. Then we have \begin{align*} (T_{\lambda}x)(t) &\leq \frac{\lambda}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))dr\,dg(s)+ \lambda\int_{0}^{1}\int_{s}^{1}a(r)f(x(r))\,dr\,ds \\ &\leq \frac{\lambda }{1-g(1)}\frac{r}{\lambda A }\int_{\eta}^{1}\int_{s}^{1}a(r)dr\,dg(s)+ \lambda\frac{r}{\lambda A}\int_{0}^{1}\int_{s}^{1}a(r)\,dr\,ds = r\,. \end{align*} As a result, \|T_{\lambda}x\|\leq\|x\|, \forall x\in\partial K_{r}. For x\in\partial K_{R}, we have f(x(t))\geq R/\lambda B for t\in[\eta,1]. Then we have \begin{align*} (T_{\lambda}x)(1) &\geq \frac{\lambda }{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))drdg(s)+ \lambda\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))\,dr\,ds \\ &\geq \frac{\lambda }{1-g(1)}\frac{R}{\lambda B }\int_{\eta}^{1}\int_{s}^{1}a(r)drdg(s)+ \lambda\frac{R}{\lambda B}\int_{\eta}^{1}\int_{s}^{1}a(r)\,dr\,ds = R. \end{align*} As a result, \|T_{\lambda}x\|\geq\|x\|, for all x\in\partial K_{R}. Then we can obtain the result by using Theorem \ref{thm2.1}. \end{proof} \begin{remark} \label{rmk4.2}\rm In Theorem \ref{thm4.1}, if condition (4.1) is replaced by \[ \max_{0\leq x\leq R}f(x)\leq R/\lambda A, \quad \min_{\eta r\leq x\leq r}f(x)\geq r/\lambda B. Then \eqref{e1.1} has also a solution $x\in K$ with $r\leq\|x\|\leq R$. \end{remark} For the remainder of this section, we need the following condition: \begin{itemize} \item[(H4)] $\sup_{r>0}\min_{\eta r\leq x\leq r}f(x) >0$. \end{itemize} Let $\lambda^{*}=\sup_{r>0}\frac{r}{A\max_{0\leq x\leq r }f(x)}, \quad \lambda^{**}=\inf_{r>0}\frac{r}{B\min_{\eta r \leq x\leq r }f(x)}.$ We can easily obtain that $0<\lambda^{*}\leq\infty$ and $0\leq \lambda^{**}<\infty$ by using (H1) and (H4). \begin{theorem} \label{thm4.3} Suppose that {\rm (H1)-(H4)} hold. In addition, assume that $f_{0}=\infty$ and $f_{\infty }=\infty$. Then the boundary-value problem \eqref{e1.1}-\eqref{e1.3} has at least two positive solutions for any $\lambda\in(0,\lambda^{*})$. \end{theorem} \begin{proof} Define $$h(r)=\frac{r}{A\max_{0\leq x\leq r}f(x)}.$$ Using the condition (H1), $f_{0}=\infty$ and $f_{\infty }=\infty$, we can easily obtain that $h:(0,\infty)\to (0,\infty)$ is continuous and $$\lim_{r\to0}h(r)=\lim_{r\to\infty }h(r)=0.$$ So there exists $r_{0}\in(0,\infty)$ such that $h(r_{0})=\sup_{r>0}h(r)=\lambda^{*}$. For $\lambda\in(0,\lambda^{*})$, there exist two constants $r_{1}, r_{2} (00}g(r)=\lambda^{**}$. For $\lambda\in(\lambda^{**},\infty)$, there exist two constants \$r_{1},r_{2} (0