\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 38, 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} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/38\hfil Positive solutions] {Positive solutions for second-order singular three-point boundary-value problems with sign-changing nonlinearities} \author[C. Ji, B. Yan\hfil EJDE-2010/38\hfilneg] {Caisheng Ji, Baoqiang Yan} % in alphabetical order \address{Caisheng Ji \newline Department of Mathematics, Shandong Normal University, Jinan 250014, China} \email{jicaisheng@163.com} \address{Baoqiang Yan \newline Department of Mathematics, Shandong Normal University, Jinan 250014, China} \email{yanbqcn@yahoo.com} \thanks{Submitted July 30, 2009. Published March 14, 2010.} \thanks{Supported by grants 10871120 from the fund of National Natural Science, J07WH08 from \hfill\break\indent the Shandong Education Committee, and Y2008A06 from the Shandong Natural Science} \subjclass[2000]{34B10, 34B15} \keywords{Singular three-point boundary-value problem; sign-changing; \hfill\break\indent nonlinearity; positive solution; fixed point} \begin{abstract} In this article, we study the existence and uniqueness of the positive solution for a second-order singular three-point boundary-value problem with sign-changing nonlinearities. Our main tool is a fixed-point theorem. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \section{Introduction} In this article, we consider the second-order boundary-value problem \begin{gather} x''(t)+f(t,x(t))=0,\quad 00$. Moreover, when$g(t)$is a sign-changing function in$[0,1]$and$f$is nondecreasing and without any singular points, using the fixed point theorem of strict-set-contractions, Bing Liu \cite{l1} established the existence of at least two positive solutions for \eqref{e1.3}. When$g(t)>0$and$f$is a given sign-changing function without any singular points and any monotonicity, using the increasing operator theory and approximation process, Xian Xu \cite{x1} showed at least three solutions for the three-point boundary-value problem \eqref{e1.3}. In addition, the existence of solutions of nonlinear multi-point boundary-value problems have been studied by many other authors; the readers are referred to \cite{l1,m1,x2,x3} and the references therein. Motivated by \cite{h1,y1}, the purpose of this article is to examine the existence and the uniqueness of the positive solution of \eqref{e1.1}-\eqref{e1.2} under the assumption that$f$may be singular at$t=0$,$x=0$and be superlinear at$x=\infty$and change sign. There are only a few papers considering \eqref{e1.1}-\eqref{e1.2} under this assumptions. We try to fill this gap in the literature with this paper. In this article, we use the following assumptions: \begin{itemize} \item[(H1)]$f(t,x)\in C((0,1]\times(0,+\infty),(-\infty,+\infty))$, \item[(H2)]$k(t), a(t), b(t)\in C((0,1],(0,+\infty))$,$tk(t)\in L(0,1]$, \item[(H3)] there exist$F(x)\in C((0,+\infty), (0,+\infty))$,$G(x)\in C([0,+\infty),[0,+\infty))$such that$f(t,x)\leq k(t)(F(x)+G(x))$. \item[(S1)]$f(t,x)\geq a(t)$hold for$01$, such that$\int_1^R\frac{dy}{F(y)}\cdot (1+\frac{\bar{G}(R)}{F(R)})^{-1} >\int_0^1 sk(s)ds$, where$\bar{G}(R)=\max_{s\in [0,R]} G(s)$. \end{itemize} This paper is organized as follows. In Section 2, we give some preliminaries. In Section 3, we obtain the existence of at least one positive solution for \eqref{e1.1}-\eqref{e1.2}, and show an application of our results. \section{Preliminaries} \begin{lemma}[\cite{d1}] \label{lem2.1} Let$E$be a Banach space,$R>0$,$B_R=\{x\in E:\|x\|\leq R\}$,$F:B_R\to E$be a completely continuous operator. If$x\neq \lambda F(x)$for any$x\in E$with$\|x\|=R$and$0<\lambda<1$, then$F$has a fixed point in$B_R$. \end{lemma} Let$n>[\frac{1}{\eta}+1]$be a natural number,$d_n=\min\{b(t):t\in[\frac{1}{n},1]\}$,$b_n=\min\{d_n,\frac{1}{n}\}$,$C_n=\{x: x\in C[\frac{1}{n},1]\}$with norm$ \|x\|=\max\{|x(t)|, \frac{1}{n}\leq t\leq 1\}$. It is easy to see that ($C_n, \| \cdot \|$) is a Banach space. Inspired by \cite{y1}, we define$T_n$as $$(T_nx)(t)=b_n+\int_{\frac{1}{n}}^1 G_{\frac{1}{n},1}(t,s) f(s,\max\{b_n,x(s)\})ds,\quad x\in C_n,\; t\in[\frac{1}{n},1],$$ where $G_{\frac{1}{n},1}(t,s)=\begin{cases} G_1(t,s), &\frac{1}{n}<\eta \leq s,\\ G_2(t,s), &\frac{1}{n}\leq s\leq \eta, \end{cases}$ \begin{gather*} G_1(t,s)=\begin{cases} \frac{1}{1- \alpha \eta-(1- \alpha) \frac{1}{n}}(1-s) (t- \frac{1}{n}),& \frac{1}{n}\leq t\leq s\leq 1, \\ \frac{1}{1- \alpha \eta-(1- \alpha) \frac{1}{n}}[ \alpha (t-s)(\eta - \frac{1}{n})-(t-1)(s- \frac{1}{n})], &\eta \leq s\leq t \leq 1, \end{cases} \\ G_2(t,s)=\begin{cases} \frac{(1-\alpha \eta)(t-\frac{1}{n})-s(1-\alpha) (t-\frac{1}{n})}{1- \alpha \eta-(1- \alpha) \frac{1}{n}}, &\frac{1}{n}\leq t\leq s\leq 1,\\ \frac{(1-\alpha \eta)(s-\frac{1}{n})-t(1-\alpha) (s-\frac{1}{n})}{1- \alpha \eta-(1- \alpha) \frac{1}{n}}, &\frac{1}{n}\leq s\leq t\leq 1, \end{cases} \end{gather*} and$G_{\frac{1}{n},1}(t,s)$is Green's function to the boundary-value problem \begin{gather*} x''(t)=0,\quad \frac{1}{n}x(1)$, we can get a point $t_1\in (\frac{1}{n},\eta)$ such that $x(t_1)=x(1)$. Let $\gamma=\sup \{t_1:\ t_1\in (\frac{1}{n},\eta), x(t_1)=x(1)\}$. It follows that $x(\gamma)=x(1)$ and $x(t)(1+\frac{\bar{G}(R)}{F(R)})\int_0^1sk(s)ds$. Hence \eqref{e2.2} holds. It follows from Lemma \ref{lem2.1} and \eqref{e2.2} that $T_n$ has a fixed point $x_n$ in $C_n$. Using $x_n$ and 1 in the place of $x$ and $\lambda$ in \eqref{e2.2}, we obtain easily $b_n\leq x_n(t)\leq R, t\in [\frac{1}{n},1]$. The proof is complete. \end{proof} \begin{lemma} \label{lem2.4} For a fixed $h\in (0,\min\{\frac{1}{2},\eta\})$, suppose $m_{n,h}=\min\{x_n(t), t\in[h,1]\}$. Then $m_h=\inf\{m_{n,h}\}>0$. \end{lemma} \begin{proof} Since $x_n(t)\geq b_n>0$, we get $m_h\geq 0$. For any fixed natural numbers $n$ ($n>[\frac{1}{\eta}]+1$), let $t_n\in[h,1]$ such that $x_n(t_n)=\min\{x_n(t),t\in [h,1]\}$. If $m_h=0$, there exist a countable set $\{n_i\}$ such that $$\lim_{n_i\to+\infty}x_{n_i}(t_{n_i})=0 .\label{e2.14}$$ So there exist $N$ such that $x_{n_i}(t_{n_i})\leq \min\{b(t), t\in [\frac{h}{2},1]\}$, $n_i>N$. Then we have two cases. Case 1. There exist $n_k\in \{n_i\}, n_k>N$ and $t^*_{n_k}\in [\frac{h}{2},h]$ such that $x_{n_k}(t^*_{n_k})\geq x_{n_k}(t_{n_k})$. By the same argument in Lemma \ref{lem2.3}, we can get $t'_{n_k}, t''_{n_k}\in [\frac{h}{2},1], t'_{n_k}N$. And so we have $$\lim_{n_i\to +\infty }x_{n_i}(t)=0, \quad t\in[\frac{h}{2},h]. \label{e2.17}$$ On the other hand for any $t\in[\frac{h}{2},h]$, \begin{aligned} x_{n_i}(t)&=\frac{2}{h}\int_{\frac{h}{2}}^t(t-\frac{h}{2})(h-s) f(s,x_{n_i}(s))ds\\ &\quad +\frac{2}{h}\int_t^h(s-\frac{h}{2})(h-t) f(s,x_{n_i}(s))ds+x_{n_i}(\frac{h}{2})+x_{n_i}(h)\\ &\geq \frac{2}{h}[\int_{\frac{h}{2}}^t(t-\frac{h}{2})(h-s)a(s)ds +\int_t^h(s-\frac{h}{2})(h-t)a(s)ds]>0, \end{aligned} \label{e2.18} which contradicts \eqref{e2.17}. The proof is complete. \end{proof} \section{Main Result} \begin{theorem} \label{thm3.1} If {\rm (S1)--(S3)} hold, the three-point boundary-value problem \eqref{e1.1}-\eqref{e1.2} has at least one positive solution. \end{theorem} \begin{proof} For any natural numbers $n\geq [\frac{1}{\eta}+1]$, it follows from Lemma \ref{lem2.3} that there exist $x_n\in C_n, b_n\leq x_n\leq R$ satisfying \eqref{e2.1}. Now we divide the proof into three steps. Step 1. There exist a convergent subsequence of $\{x_n\}$ in (0,1]. For a natural number $k\geq \max\{3,[\frac{1}{\eta}]+1\}$, it follows from Lemma \ref{lem2.4} that $00, t\in(0,1]$ by Lemma \ref{lem2.4}. Step 2. $x(t)$ satisfies \eqref{e1.1}. Fixed $t\in (0,1]$, we may choose $h\in (0,\min\{\frac{1}{2},\eta\})$ such that $t\in (h,1]$ and \begin{aligned} x_n(t)&=-\int_h^t(t-s)f(s,x_n(s))ds\\ &\quad +\frac{t-h}{1-\alpha \eta-h(1-\alpha)} [\int_h^1(1-s)f(s,x_n(s))ds-\alpha \int_h^{\eta}(\eta -s) f(s,x_n(s))ds]\\ &\quad +x_n(h)+\frac{(t-h)(1-\alpha)}{1-\alpha \eta-h(1-\alpha)} (b_n-x_n(h)), \quad t\in (h,1].\label{e3.2} \end{aligned} Letting $n\to +\infty$ in \eqref{e3.2}, we have \begin{aligned} x(t)&=-\int_h^t(t-s)f(s,x(s))ds +\frac{t-h}{1-\alpha \eta-h(1-\alpha)}\\ &\quad\times \big[\int_h^1(1-s)f(s,x(s))ds -\alpha \int_h^{\eta}(\eta -s)f(s,x(s))ds\big]\\ &\quad +x(h)+\frac{(t-h)(1-\alpha)}{1-\alpha \eta-h(1-\alpha)}(-x(h)), \quad t\in (h,1].\label{e3.3} \end{aligned} Differentiating \eqref{e3.3}, we get the desired result. Step 3. $x(t)$ satisfies \eqref{e1.2}. Let $$t_n=\inf \{t:x_n(t)=\|x_n\|, x'_n(t)=0, t\in[\frac{1}{n},1]\},$$ where $\|x_n\|=\max _{\frac{1}{n}\leq t\leq 1}x_n(t)\leq R$. Then $$t_n\in[\frac{1}{n},1], \quad x_n(t_n)=\|x_n\|, \quad x'_n(t_n)=0.$$ Using $x_n(t), 1$ and $t_n$ in place of $x(t), \lambda$ and $t'$ in Lemma \ref{lem2.3}, we obtain easily by \eqref{e2.13} $$\int_{b_n}^{\|x_n\|}\frac{dx}{F(x)} \leq (1+\frac{\bar{G}(R)}{F(R)})\int_0^{t_n}\int_t^1k(s)\,ds\,dt. \label{e3.4}$$ It follows from \eqref{e3.4} and Lemma \ref{lem2.4} that $0x_2(\eta)$. That is to say $z(\eta)>0, 0z(1)>0, \quad t\in(t_1,t_2). \] Letting$s(t)=z(t)-z(1)$, we have that$s(t_1)=s(t_2)=0$and$s(t)>0, t\in(t_1,t_2)$. It follows from \eqref{e1.1} and the monotonicity of$f(t,\cdot)$that$s''(t)=z''(t)\geq 0, t\in (t_1,t_2)$. An elementary form of the maximum principle implies$s(t)\leq 0$for all$t\in (t_1,t_2)$and hence a contradiction. Then,$x_1(\eta)=x_2(\eta)$, which also yields that$x_1(1)=x_2(1)$. That is to say$z(0)=z(\eta)=z(1)=0$. We next claim that$x_1(t)=x_2(t), t\in(0,\eta)$. In fact, if it is not true, without loss of generality, we can get$x_1(t_0)>x_2(t_0)$for some$t_0\in(0,\eta)$. Let$t_3=\max\{t\in(0,t_0), z(t)=0\}, t_4=\min\{t\in(t_0,\eta), z(t)=0\}$(note$z(\eta)=0)$). Then$z(t_3)=z(t_4)=0$and$z(t)>0, t\in(t_3,t_4)$. Let$s_1(t)=z_1(t)-z_2(t)$,$t\in[t_3,t_4]$. Then$s_1(t)>0$for all$t\in[t_3,t_4]$. On the other hand, the monotonicity of$f(t,\cdot)$implies that$s''_1(t)\geq 0, t\in (t_3,t_4)$. An elementary form of the maximum principle implies$s_1(t)\leq 0$for all$t\in (t_3,t_4)$and hence a contradiction. The same argument yields that$x_1(t)=x_2(t), t\in(\eta,1)$. Hence we get$x_1(t)=x_2(t), t\in[0,1]\$. Thus the result is proved. \end{proof} \subsection*{Example} Consider the second order singular three-point boundary-value problem \begin{gather} x''(t)+\frac{1}{4}(x^2(t)+\frac{1}{x^2(t)}-\frac{x^3(t)}{t^5} -\frac{1}{t^2})=0,\quad 0