\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 147, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2013 Texas State University - San Marcos.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2013/147\hfil Positive solutions] {Positive solutions for a third-order three-point boundary-value problem} \author[F. J. Torres \hfil EJDE-2013/147\hfilneg] {Francisco J. Torres} % in alphabetical order \address{Francisco J. Torres \newline Departmento de Matem\'atica, Universidad de Atacama, Avenida Copayapu 485, Copiap\'o, Chile} \email{ftorres@mat.uda.cl} \thanks{Submitted March 29, 2013. Published June 27, 2013.} \thanks{Partially supported by DIUDA grant 221181, Universidad de Atacama.} \subjclass[2000]{34B18, 37C25} \keywords{Cone; fixed point index; fixed point theorem; positive solution} \begin{abstract} In this article, we study the existence of positive solutions to a nonlinear third-order three point boundary value problem. The main tools are Krasnosel'skii fixed point theorem on cones, and the fixed point index theory. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} In this article, we are interested in the existence of single and multiple positive solutions to nonlinear third-order three-point boundary-value problem \begin{gather}\label{ecua} u'''(t)+a(t)f(t,u(t))=0 ,\quad 00$, then$\lambda x \in K$\item it$x\in K$and$-x\in K$, then$x=0$. \end{enumerate} \end{definition} \begin{definition} \label{def2.2} \rm An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets. \end{definition} \begin{remark}\rm \label{rmk1} By the positive solution of \eqref{ecua}, \eqref{bord} we understand a function$u(t)$wich is positive on$[0,1]$and satisfies the differential equation \eqref{ecua} and the boundary conditions \eqref{bord}. \end{remark} We shall consider the Banach space$E=C[0,1]$equipped with standard norm $\|u\|=\max_{0 \leq t\leq 1} |u(t)|\,.$ The proof of existence of solution is based on an applications of the following theorems. \begin{theorem}[\cite{Deimling,Guo}] \label{teor1} Let$E$be a Banach space and let$K\subseteq E$be a cone. Assume$\Omega _{1}$and$\Omega_{2}$are open subsets of$E$with$0\in \Omega_{1}\subseteq \overline{\Omega _{1}}\subseteq\Omega_{2}$and let $T:K\cap (\overline{\Omega_{2}}\backslash\Omega_{1})\to K$ be completely continuous such that \begin{itemize} \item[(i)]$\|Tu\|\leq \|u\|$if$u\in K\cap \partial \Omega_{1}$and$\|Tu\|\geq \|u\|$if$u\in K\cap \partial \Omega_{2}$; or \item[(ii)]$\|Tu\|\geq \|u\|$if$u\in K\cap \partial \Omega_{1}$and$\|Tu\|\leq\|u\|$if$u\in K\cap \partial \Omega_{2}$\,. \end{itemize} Then$T$has a fixed point in$K\cap (\overline{\Omega_{2}}\backslash\Omega_{1})$\end{theorem} \begin{theorem}[\cite{Deimling,Guo}]\label{teor2} Let$E$be a Banach space and$K$be a cone of$E$. For$r>0$, define$K_{r}=\{u\in K :\|u\|\leq r\}$and assume that$T:K_{r}\to K$is a completely continuous operator such that$Tu\neq u$for$u\in \partial K_{r}$\begin{enumerate} \item If$\|Tu\|\leq \|u\|$for all$u\in\partial K_{r}$, then$i(T,K_{r},K)=1$\item If$\|Tu\|\geq \|u\|$for all$u\in\partial K_{r}$, then$i(T,K_{r},K)=0$. \end{enumerate} \end{theorem} Consider the three-point boundary-value problem \begin{gather}\label{ecua2} u'''+h(t)=0 ,\quad 00$ : $f(t,u)0$ : $f(t,u)>r \rho$, $\rhoR>0$. \end{remark} \section{Existence of Positive Solutions}\label{sec.2} \begin{theorem}\label{teor7} Assume that the conditions on $a$, $f$ and {\rm(a)} hold. Then \eqref{ecua}, \eqref{bord} has at least one positive solution. \end{theorem} \begin{proof} Since $f^{0}=0$, $\exists \ H_{1}>0$ such that $f(t,u)\leq \varepsilon u$, for all $t\in[0,1]$, $00$. Then for $u\in K\cap \partial \Omega_{1}$, with $\Omega_{1}=\{u\in X:\|u\|0$ such that $f(t,u)\geq\delta u$, for all $t\in [\sigma,1]$ with $\bar{H_{2}}\leq u$ and $\delta>0$. Then for $u\in K\cap\partial\Omega_{2}$, where $\Omega_{2}=\{u\in X:\|u\| \bar{H_{2}}$. So, by \eqref{not1}, we obtain \begin{align*} (Tu)(1) &= \int_0^{1}g(1,s)a(s)f(s,u)ds+\frac{\alpha }{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds \\ &\geq \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)f(s,u)ds+\frac{\alpha }{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)f(s,u)ds \\ &\geq \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)\delta u(s)ds+\frac{\alpha }{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)\delta u(s)ds \\ &\geq \delta\Big[\int_{\sigma}^{1}\frac{\gamma}{2}s(1-s)a(s) ds+\frac{\alpha\gamma}{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)ds\Big]\|u\|\,. \end{align*} If $\delta N \geq 1$, then $$Tu(1)\geq\|u\|$$ which implies that $$\|Tu\|\geq \|u\|\,.$$ Therefore, by Theorem \ref{teor1}, the operator $T$ has at least one fixed point, which is a positive solution of \eqref{ecua}, \eqref{bord}. \end{proof} \begin{theorem}\label{teor8} Assume that the conditions on $a$, $f$ and {\rm (b)} hold. Then \eqref{ecua}, \eqref{bord} has at least one positive solution. \end{theorem} \begin{proof} Since $f_0=\infty$, there exists $H_{1}>0$ such that $f(t,u)\geq \xi u$, for all $t\in [\sigma,1]$, $00$; thus, for $u\in K\cap \partial \Omega_{1}$, with $\Omega_{1}=\{u\in X:\|u\|0$ such that $f(t,u)\leq\lambda u$, for all $t\in [0,1]$ with $\bar{H_{2}}\leq u$ and $\lambda>0$. We consider two cases: \noindent\textbf{Case 1.} Suppose $f$ is bounded. Let $L$ such that $f(t,u)\leq L$ and $\Omega_{2}=\{u\in X:\|u\|\max\{2H_{1},\bar{H_{2}}\}$ such that $f(t,u)\leq f(t,H_{2})$ with $00$ and $0<\varepsilon_{1}0$ and $\varepsilon_{2}> 0$ such that $f(t,u)\geq(r+\varepsilon_{2})u$ for $u\geq \bar{H_{2}}$ and $\sigma\leq t \leq 1$. Let $H_{2}=\max\{2H_{1},\frac{\bar{H_{2}}}{\gamma}\}$ and $\Omega_{2}=\{u\in X:\|u\| \bar{H_{2}}$. So, by \eqref{not1} we obtain \begin{align*} Tu(1) &= \int_0^{1}g(1,s)a(s)f(s,u)ds+\frac{\alpha }{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds \\ &\geq \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)(r+\varepsilon_{2})u ds+\frac{\alpha }{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)(r+\varepsilon_{2})uds \\ &\geq (r+\varepsilon_{2})\Big[\int_{\sigma}^{1}\frac{\gamma}{2}s(1-s)a(s)ds+\frac{\alpha\gamma }{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)ds\Big]\|u\| \\ &= (r+\varepsilon_{2})N\|u\| > \|u\|\,. \end{align*} Thus, $\|Tu\| > \|u\|$. Therefore, by Theorem \ref{teor1}, the operator $T$ has at least one fixed point, which is a positive solution of \eqref{ecua}, \eqref{bord}. \end{proof} \begin{theorem}\label{teor10} Assume that the conditions on $a$, $f$ and {\rm (f)} hold. Then \eqref{ecua}, \eqref{bord} has at least one positive solution. \end{theorem} \section{Multiplicity results} \begin{theorem}\label{teor11} Assume that the conditions on $a$, $f$, {\rm (c)} and {\rm (g)} hold. Then \eqref{ecua}, \eqref{bord} has at least two positive solutions. \end{theorem} \begin{proof} Since $f_0=\infty$, $\exists H_{1}>0$ where $0ru$ with 0 \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)ru ds+\frac{\alpha }{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)ru ds \\ &> r\left[\int_{\sigma}^{1}\frac{\gamma}{2}s(1-s)a(s)ds+\frac{\gamma\alpha }{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)ds\right]\|u\| \\ &= r N\|u\|=\|u\| \end{align*} Thus,\|Tu\|>\|u\|$. Therefore, by Theorem \ref{teor2} $i(T,K_{H_{1}},K)=0\,.$ Since$f_{\infty}=\infty$, there exists$\bar{H_{2}}>\rho$such that$f(t,u)>ru$with$u\geq \bar{H_{2}}$,$t\in [\sigma,1]$. Let$H_{2}=\frac{\bar{H_{2}}}{\gamma}$and$\Omega_{2}=\{u\in X:\|u\| \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)ru ds+\frac{\alpha }{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)ru ds \\ &> r\Big[\int_{\sigma}^{1}\frac{\gamma}{2}s(1-s)a(s)ds+\frac{\gamma\alpha }{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)ds\Big]\|u\| \\ &= r N\|u\|=\|u\|\,. \end{align*} Thus, $\|Tu\|>\|u\|$. Therefore, by Theorem \ref{teor2} $i(T,K_{H_{2}},K)=0\,.$ On the other hand, let $\Omega_{3}=\{u\in X:\|u\|<\rho\}$. For any $u\in K\cap \partial\Omega_{3}$, we get from (e) that $f(t,u)1$, the conclusions of Theorem \ref{teor7}, hold. \item[(b)] If $f(t,u)=1+u^{\alpha}$, $\alpha \in(0,1)$ the conclusions of Theorem \ref{teor8}, hold. \end{itemize} \end{example} \begin{example} \rm Let $f(t,u)=\lambda t \ln{(1+u)}+u^2$, fix $\lambda >0$, sufficiently small. Clearly $f^{0}=\lambda$ and $f_{\infty}=\infty$. By Theorem \ref{teor9}, \eqref{ecua} and \eqref{bord} have at least one positive solution. \end{example} \begin{example} \rm Let $f(t,u)=u^2e^{-u}+ \mu \sin{u}$, fix $\mu > 0$ sufficiently large. Then $f_0=\mu$ and $f^{\infty}=0$. By Theorem \ref{teor10}, \eqref{ecua} and \eqref{bord} have at least one positive solution. \end{example} \begin{example} \rm Consider the boundary-value problem \begin{gather} \label{ecuaex} u'''(t)+u^{b}+u^{c}=0 ,\quad 01$. Then$f_0=\infty$and$f_{\infty}=\infty$. By a simple calculation,$M=2/21$then$R=21/2$. On the other hand, we could choose$ \rho =1$, then$f(t,u)\leq 2< \frac{21}{2}1=R \rho$for$(t,u)\in[0,1]\times[0,\rho]\$. By Theorem \ref{teor11}, \eqref{ecuaex} and \eqref{bordex} have at least two positive solutions. \end{example} \begin{thebibliography}{99} \bibitem{Bai} Z. Bai; Existence of Solutions for Some Third-order Boundary Value Problems, \emph{Electron. J. Differential Equations,} 25 (2008), 1--6. \bibitem{Deimling} K. 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