\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 13, pp. 1--6.\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-2005/13\hfil Positive solutions] {Positive solutions for a class of singular boundary-value problems} \author[D. D. Hai\hfil EJDE-2005/13\hfilneg] {Dang Dinh Hai} \address{Dang Dinh Hai \hfill\break Department of Mathematics, Mississippi State University, Mississippi State, MS 39762, USA} \email{dang@math.msstate.edu} \date{} \thanks{Submitted August 25, 2005. Published January 26, 2006.} \subjclass[2000]{34B16, 34B18} \keywords{Positive solutions; singular; sub-super solutions} \begin{abstract} This paper concerns the existence and multiplicity of positive solutions for Sturm-Liouville boundary-value problems. We use fixed point theorems and the sub-super solutions method to two solutions to the problem studied. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \textbf{Introduction} Consider the boundary-value problem \label{e1} \begin{gathered} Lu=\lambda f(t,u),\quad 00$,$q\geq 0$on$[0,1]$,$\alpha ,\beta ,\gamma ,\delta \geq 0$with$\alpha \delta +\alpha \gamma +\beta \gamma >0$,$f:(0,1)\times [ 0,\infty)\to [ 0,\infty )$, and$\lambda $is a positive parameter. The existence and nonexistence of positive solutions of problem \eqref{e1} with$f$possibly singular have been established by Choi \cite{Ch}, Dalmasso \cite{Dal}, Wong \cite{W}, and recently by Erbe and Mathsen \cite{Erb}. In this paper, we shall obtain positive solutions to \eqref{e1} under assumptions less stringent than in \cite{Erb}. In particular, we do not need the condition that$f(t,u)$be nondecreasing in$u$, which is essential in \cite{Ch,Dal,Erb,W}. Our approach depends on fixed point theorems and sub-super solutions method. \section{Main results} Let$G(t,s)$be the Green's function for \eqref{e1}. Then$u$is a solution of \eqref{e1} if and only if $u(t)=\lambda \int_{0}^{1}G(t,s)f(s,u(s))ds.$ Recall that $G(t,s)=\begin{cases} c^{-1}\phi (t)\psi (s) &\text{if }t\leq s \\ c^{-1}\phi (s)\psi (t) &\text{if }t>s, \end{cases}$ where$\phi $and$\psi $satisfy% $$\begin{gathered} L\phi =0,\quad \phi (0)=\beta ,\quad \phi '(0)=\alpha \\ L\psi =0,\quad \psi (1)=\delta ,\quad \psi '(1)=-\gamma \end{gathered} \label{e2.1}$$ and$c=-r(t)(\phi (t)\psi '(t)-\phi '(t)\psi (t))>0$. Note that$\phi '>0$on$(0,1]$,$\psi '<0$on$[0,1)$. We shall make the following assumptions: \begin{itemize} \item[(H1)]$f:(0,1)\times [ 0,\infty )\to [ 0,\infty )$is continuous \item[(H2)] For each$M>0$, there exists a continuous function$g_{M}$on$(0,1)$such that$f(t,u)\leq g_{M}(t)$for$t\in (0,1)$,$0\leq u\leq M$, and $\int_{0}^{1}G(s,s)g_{M}(s)ds<\infty .$ \item[(H3)] There exist an interval$I\subset (0,1)$and a function$m\in L^{1}(I)$with$m\geq 0$,$m\not\equiv 0$such that for every$a>0$, there exists$r_{a}>0$such that $f(t,u)\geq am(t)u\quad \text{for }t\in I,\;u\in (0,r_{a})$ \item[(H4)] There exist an interval$J\subset (0,1)$and a positive number$d$such that $f(t,u)\geq du\quad \text{for }t\in J,\;u\geq 0.$ \item[(H5)] There exist an interval$I_{1}\subset (0,1)$and a function$m_{1}\in L^{1}(I_{1})$with$m_{1}\geq 0,m_{1}\not\equiv 0$such that for every$b>0$, there exists$R_{b}>0$such that $f(t,u)\geq bm_{1}(t)u\quad \text{for }t\in I_{1},\;u\geq R_{b}.$ \end{itemize} Our main results are stated as follows. \begin{theorem} \label{thm1} Let \textrm{(H1)--(H3)} hold. Then there exists$\lambda _{0}>0$such that \eqref{e1} has a positive solution for$0<\lambda <\lambda _{0}$. If, in addition, \textrm{(H5)} holds, then \eqref{e1} has at least two positive solutions for$0<\lambda <\lambda _{0}$. \end{theorem} \begin{theorem} \label{thm2} Let \textrm{(H1)--(H4)} hold. Then there exists$\lambda ^{\ast }>0$such that \eqref{e1} has a positive solution for$0<\lambda <\lambda ^{\ast }$and no positive solution for$\lambda >\lambda ^{\ast }$. \end{theorem} \begin{remark} \label{rmk3} \rm Let$f(t,u)=m(t)g(u)$, where$g:[0,\infty )\to [ 0,\infty )$be continuous with$\lim_{u\to 0^{+}}\frac{g(u)}{u}=\infty $,$\lim_{u\to \infty }\frac{g(u)}{u}=\infty $, and$m\in L^{1}(0,1)$with$m\geq 0$,$m\not\equiv 0$. Then$f$satisfies (H1)--(H3) and (H5) and therefore Theorem \ref{thm1} applies. If we take$m(t)=1/\sqrt{t}$,$g(u)=u^{p}+u^{q}+h(u)$, where$p<1\leq q$and$h$is a nonnegative continuous function, then it is easily seen that$f(t,u)$satisfies (H1)--(H5) and Theorem \ref{thm2} applies. However, the results in \cite{Ch,Dal,Erb,W} may not apply since$g$may not be nondecreasing. \end{remark} To prove our main results, we first establish the following results. \begin{lemma}\label{lem4} Let$h\in L^{1}(0,1)$be such that$h\geq 0$and let$u$satisfy \begin{gather*} Lu=h\quad \text{in }(0,1) \\ \alpha u(0)-\beta u'(0)=0,\quad \gamma u(1)+\delta u'(1)=0. \end{gather*} Then $u(t)\geq |u|_{0}p(t),$ where$p(t)=\min \big( \frac{\phi (t)}{|\phi |_{0}}, \frac{\psi (t)}{|\psi|_{0}}\big) $, and$\|\cdot\|_{0}$denotes the supremum norm. \end{lemma} \begin{proof} We proceed as in \cite{Dan}. It is easy to see that $u(t)=\int_{0}^{1}G(t,s)h(s)ds.$ Let$|u|_{0}=u(t_{0})$for some$t_{0}\in (0,1)$. We verify that $\frac{G(t,s)}{G(t_{0},s)}\geq p(t).$ If$t,t_{0}\leq s$then $\frac{G(t,s)}{G(t_{0},s)}=\frac{\phi (t)}{\phi (t_{0})} \geq \frac{\phi (t)}{|\phi |_{0}},$ and if$t_{0}\leq s\leq t$then $\frac{G(t,s)}{G(t_{0},s)}=\frac{\phi (s)\psi (t)}{\phi (t_{0})\psi (s)}\geq \frac{\psi (t)}{\psi (s)}\geq \frac{\psi (t)}{|\psi |_{0}}$ since$\phi (s)\geq \phi (t_{0})$. The other two cases are treated in a similar manner. Hence $u(t)\geq p(t)u(t_{0})=|u|_{0}p(t).$ \end{proof} \begin{lemma} \label{lem5} Let \textrm{(H1)--(H3)} hold. Then for each$\lambda >0$, there exists$c_{\lambda }>0$such that if$u$is a nonzero solution of \eqref{e1} then$|u|_{0}\geq c_{\lambda }$. Furthermore,$(c_{\lambda })$is nondecreasing in$\lambda $. \end{lemma} \begin{proof} Let$p_{0}=\min_{t\in I}p(t)$, where$p$is defined in Lemma \ref{lem4}, and $$K= \int_{I}G(\frac{1}{2},s)m(s)ds .$$ By (H3), there exists$r_{\lambda }\in (0,1)$such that $\frac{f(t,u)}{u}\geq \frac{2m(t)}{\lambda p_{0}K}\quad \text{for }t\in I,\;0\bar{u}(t) \\ f(t,v) & \text{if } \underline{u}(t)\leq v\leq \bar{u}(t) \\ f(t,\underline{u}(t))+\frac{\underline{u}(t)-v}{1+v^{2}} & \text{if }v\leq \underline{u}(t). \end{cases}$ For each$v\in C[0,1]$, let$u=Tv$be the solution of \begin{gather*} Lu=\bar{f}(t,v),\quad 0\bar{u}(t_{0})$. Let $w=u-\bar{u}$ and $t_{1}\in [ 0,1]$ be such that $w(t_{1})=\max_{0\leq t\leq 1}w(t)>0$. If $t_{1}\in (0,1)$ then $w'(t_{1})=0$ and $(rw')'(t_{1})\leq 0$, which implies that $Lw(t_{1})\geq 0$. On the other hand, $Lw(t_{1})=Lu(t_{1})-L\bar{u}(t_{1})\leq -\frac{w(t_{1})}{1+u^{2}(t_{1})}<0,$ a contradiction. Suppose that $t_{1}=0$. Then $w'(0)\leq 0$, and since $\alpha w(0)-\beta w'(0)\leq 0$, we have a contradiction if $\alpha >0$. If $\alpha =0$ then $\beta >0$ and therefore $w'(0)=0$. Since $-(rw')'(t)+qw(t)\equiv Lw(t)<0$ for small $t>0$, it follows by integrating that $(rw')(t)>0$ and so $w'(t)>0$ for small $t>0$, a contradiction. Similarly, we reach a contradiction if $t_{1}=1$. Hence $u\leq \bar{u}$ on $(0,1)$. The lower inequality can be derived in a similar manner. \end{proof} In view of Lemmas \ref{lem4} and \ref{lem5}, we see that $u$ is a positive solution of \eqref{e1} if and only if $u$ is a solution of \begin{gathered} Lu=\lambda \tilde{f}(t,u),\quad 0p_{1}$such that $\tilde{f}(s,u)\geq bm_{1}(s)u\quad \text{for }s\in I_{1},\;u\geq R_{b}.$ We claim that $u\in \mathbb{K}\text{ and }u\geq A_{\lambda }u\;\Rightarrow\; |u|_{0}\neq R_{b}p_{1}^{-1}$ Suppose that$u\in \mathbb{K}$and$u\geq A_{\lambda }u$. If$|u|_{0}=R_{b}p_{1}^{-1}then it follows from Lemma \ref{lem4} that $u(s)\geq R_{b}p_{1}^{-1}p(s)\geq R_{b}\quad \text{for }s\in I_{1}.$ Hence \begin{align*} R_{b}p_{1}^{-1} &=|u|_{0}\geq u\big( \frac{1}{2}\big) \\ &\geq \lambda \int_{0}^{1}G\big( \frac{1}{2},s\big) \tilde{f}(s,u(s))ds \\ &\geq bR_{b}\lambda \Big( \int_{I_{1}}G\big( \frac{1}{2},s)\big) m_{1}(s)ds\Big) =2R_{b}p_{1}^{-1}, \end{align*} a contradiction, and the claim is proved. By Krasnoselskii's fixed point Theorem, \cite{K},A_{\lambda }$has a fixed point$\tilde{u}$in$\mathbb{K}$with$1<|\tilde{u}|_{0}0$such that \eqref{e1} has a positive solution and let$\lambda ^{\ast}=\sup \Lambda $. By Theorem \ref{thm1} and Lemma \ref{lem6},$0<\lambda ^{\ast }<\infty $. Let$0<\lambda <\lambda ^{\ast }$. Then there exists$\lambda _{0}>0$such that$\lambda <\lambda _{0}$and$\eqref{e1}_{\lambda _{0}}$has a positive solution$u_{\lambda _{0}}$. Then$u_{\lambda _{0}}satisfies $u_{\lambda _{0}}(t)\geq c_{\lambda _{0}}p(t)\geq c_{\lambda }p(t),$ and therefore \begin{align*} Lu_{\lambda _{0}}(t) &=\lambda _{0}f(t,u_{\lambda _{0}}(t))\\ &=\lambda_{0}f(t,\max (u_{\lambda _{0}}(t),c_{\lambda }p(t)) \\ &\geq \lambda f(t,\max (u_{\lambda _{0}}(t),c_{\lambda }p(t))\\ &=\lambda \tilde{f}(t,u_{\lambda _{0}}(t)), \end{align*} i.e.,u_{\lambda _{0}}$is a supersolution of \eqref{e1'}. Since$0$is a subsolution of \eqref{e1'}, it follows from Lemma \ref{lem7} that \eqref{e1'} has a solution$u_{\lambda }$with$0\leq u_{\lambda }\leq u_{\lambda _{0}}$. Thus$u_\lambda\$ is a positive solution of \eqref{e1}, completing the proof of Theorem \ref{thm2}. \end{proof} \begin{thebibliography}{9} \bibitem{Ch} Y. S. Choi, \emph{A singular boundary value problem arising from near-ignition analysis of flame structures}, Diff. Integral Eqns. \textbf{4} (1991), 891-895. \bibitem{Dal} R. 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