\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 70, pp. 1--12.\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-2005/70\hfil Multiple positive solutions] {Multiple positive solutions for singular semi-positone delay differential equation} \author[X. Xu \hfil EJDE-2005/70\hfilneg] {Xian Xu} \address{Xu, Xian \hfill\break Department of Mathematics, Xuzhou Normal University\\ Xuzhou, Jiangsu, 221116, China} \email{xuxian68@163.com} \date{} \thanks{Submitted May 7, 2005. Published June 30, 2005.} \thanks{Supported by grant 04KJB110138 from the Natural Science Foundation of the \hfill\break\indent Jiangsu Education Committee, China} \subjclass[2000]{34B15, 34B25} \keywords{Multiple positive solutions; fixed point index; \hfill\break\indent semi-positone delay differential equation} \begin{abstract} In this paper, we obtain new existence results for multiple positive solutions of a delay singular differential boundary-value problem. Our main tool is the fixed point index method. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \section{Introduction} Singular differential boundary-value problems arise from many branches of basic mathematics and applied mathematics. Many techniques have been developed to establish the existence of positive solutions of various classes of singular differential boundary-value problems; \cite{a1,a2,c1,d1,e1,j1,l1,o1,o2,x1,x2,w1} and the references therein. In particular, the authors of \cite{a1,e1} obtained some existence results for positive solutions of some singular functional differential equations. Motivated by \cite{a1,e1}, in this paper we will consider the following singular delay differential equation $$\begin{gathered} y''+f(t, y(t-a))=0,\quad t\in (0, 1]\backslash\{a\},\\ y(t)=\mu(t),\quad t\in [-a, 0],\\ y(1)=0, \end{gathered} \label{e1.1}$$ where $00,\quad \forall t\in [-a, 0), \label{e1.2} and the nonlinear term$f(t, y)$satisfies $$\phi_0(t)h_0(y)-p(t)\leq f(t, y)\leq\phi(t)(g(y)+h(y)),\label{e1.3}$$$\phi, \phi_0, p$are in$C((0,1], R^+)$,$g$is in$C((0, +\infty),R^+)$,$h_0$,$h$are in$C(R^+, R^+)$, and$R^+=[0,+\infty)$. Problem \eqref{e1.1} is a singular semi-positone boundary-value problem because$p$is allowed to be nonnegative on interval$(0,1)$and$f$may have singularity at$t=0$and$y=0$. Recently, there have been many papers considered the singular semi-positone boundary-value problems; see \cite{a2,l1,w1,x1,j1} and the references therein. But most of these papers paid attention to the existence of positive solutions of the singular semi-positone boundary-value problems and there were fewer papers that discuss the existence of multiple positive solutions of the singular semi-positone problems. To cover up this gap, in this paper we will establish some existence results for multiple positive solutions of singular delay differential semi-positone boundary value problem \eqref{e1.1}. It is difficult to show the existence and multiplicity results for positive solutions of semi-positone problems. For our purpose, a special cone will be needed to establish the multiplicity results for positive solutions of semi-positone problems. Let$P=\{x\in C[-a,1]:x(t)\geq 0 \mbox{ for all } t\in [-a,1]\}$. It is well known that$C[-a,1]$is a real Banach space with the maximum norm$\|x\|=\max_{t\in [-a,1]}|x(t)|$, and$P$a normal cone of$C[-a,1]$. By a positive solution of \eqref{e1.1} we mean a function$x\in P$satisfying \eqref{e1.1} and$x(t)\not\equiv 0$. Throughout this paper, we will assume that \eqref{e1.2} and \eqref{e1.3} hold. \section {Preliminary Lemmas} Let us list some assumptions to be used. \begin{itemize} \item[(H1)]$g:(0,+\infty)\to R^+$is continuous and decreasing,$h, \ h_0:R^+\to R^+$are continuous and increasing. \item[(H2)] For any constant$k_0>0$, $$\int^a_0[\phi(s)g(\mu(s-a))+g(k_0s)+p(s)]ds<+\infty.$$ \item[(H3)] There exists$R_0\geq 2\int^1_0p(s)ds$such that $$R_0-2\sqrt{B(R_0)}>A, \label{e*}$$ where \begin{gather*} A=\int^a_0[\phi(s)(g(\mu(s-a))+h(\mu(s-a)+1))+p(s)]ds + \int^1_a\phi(s)g(\frac{1}{2}R_0a(s-a))ds, \\ B(R_0)=2\sup_{t\in [a,1]}[\phi(t)+p(t)]\int^{R_0}_0[g(\frac{1}{2}s) +h(s+1)+1]ds. \end{gather*} \item[(H4)] There exist$u_1>R_0$and$[\alpha, \beta]\subset (a,1)$such that $$\alpha(1-\beta-a)h_0(\frac{1}{2}u_1)\int^{\beta+a}_{\alpha+a}\phi_0(s)ds>u_1.$$ \end{itemize} \begin{remark} \label{rmk1}\rm The nonlinear term$f$of the form \eqref{e1.3} in the case$\phi_0(t)=p(t)=0$for all$t\in [0,1]$has been studied by many authors \cite{a1,o2,e1,x1}. \end{remark} Let$Q=\{x\in P:x(t)\geq\|x\|t(1-t)$for$t\in [0,1]\}$. It is easy to see that$Q$is a cone of$C[-a,1]$. For each$x\in P$, let $$[x(t-a)]^*=\max\{x(t-a)+x_0(t-a)-w(t-a), \widetilde{x}_0(t-a)\},\quad \forall t\in [0,1],$$ where \begin{gather*} x_0(t)= \begin{cases} \mu(t), &t\in [-a,0],\\ 0,&t\in (0,1], \end{cases} \\ \widetilde{x}_0(t)= \begin{cases} 0, &t\in [-a,0],\\ \frac{1}{2}R_0t(1-t),&t\in (0,1], \end{cases} \\ w(t)= \begin{cases} 0, &t\in [-a,0],\\ \int^1_0G(t,s)p(s)ds,&t\in (0,1], \end{cases} \\ G(t,s)= \begin{cases} s(1-t), &s\leq t,\\ t(1-s),&s>t. \end{cases} \end{gather*} For each positive integer$n$, let us define an operator$T_n:P\to P$by $$\label{e2.1} (T_nx)(t)=\begin{cases} 0,&t\in [-a,0];\\ \int^1_0 G(t,s)[f(s,[x(s-a)]^*+n^{-1})+p(s)]ds,&t\in (0,1].\\ \end{cases}$$ \begin{lemma}[\cite{g1}] \label{lem1} Let$X$be a retract of the real Banach space$E$and$X_1$be a bounded convex retract of$X$. Let$U$be a nonempty open set of$X$and$U\subset X_1$. suppose that$A:X_1\to X$is completely continuous,$A(X_1)\subset X_1$and$A$has no fixed points on$X_1\backslash U$. Then$i(A, U, X)=1$. \end{lemma} \begin{lemma} \label{lem2} Suppose that (H1) and (H2) hold. Then$T_n:P\to Q$is a completely continuous operator for each positive integer$n$. \end{lemma} \begin{proof} Let$n$be a fixed positive integer, and$y=T_nx$for some$x\in P$. Suppose that$\|y\|_{[0,1]}=y(t_0)$for some$t_0\in (0,1)$, where$\|y\|_{[0,1]}=\max_{t\in [0,1]}|y(t)|$. Since $$y(t)\geq y(s)=0,\quad t\in [-a,1],\; s\in [-a,0],$$ it follows that$\|y\|=\|y\|_{[0,1]}$. It is easy to see that $$y''(t)=-f(t, [x(t-a)]^*+n^{-1})-p(t)\leq 0,\ \forall t\in (0, 1].$$ Therefore, the graph of$y(t)$is concave down on$(0,1)$. For any$0\leq t\leq t_0$, we have $$y(t)=y((1-\frac{t}{t_0})\cdot 0+\frac{t}{t_0}t_0) \geq \|y\|t(1-t).$$ Similarly, $$y(t)=y(\frac{1-t}{1-t_0}t_0+(1-\frac{1-t}{1-t_0})\cdot 1) \geq \|y\|t(1-t),\quad \forall t_0\leq t\leq 1.$$ Hence,$T_n:P\to Q$. Now, we show that$T_n$is a completely continuous operator for every positive integer$n$. It is easy to see that$T_n$is a continuous and bounded operator for every positive integer$n$. Let$B\subset P$be a bounded set such that$\|x\|\leq L$for all$x\in B$and some$L>0$. Then, we can easily see that $$x_0(t-a)+\widetilde{x}_0(t-a)\leq [x(t-a)]^*\leq \|x\|+\|w\| +\|\mu\|+R_0, \quad \forall x\in B, \; t\in [0,1]. \label{e2.2}$$ Thus, for any$t_1,\ t_2\in [0,1], we have \begin{aligned} &|(T_nx)(t_1)-(T_nx)(t_2)|\\ &\leq \int^1_0|G(t_1,s)-G(t_2,s)|[\phi(s)(g(x_0(s-a) +\widetilde{x}_0(s-a))\\ &\quad +h(L+\|w\|+\|\mu\|+R_0+1))+p(s)]ds. \end{aligned} \label{e2.3} Then the uniform continuity ofG(t,s)$on$[0,1]\times [0,1]$, \eqref{e2.3} and the assumption (H2) imply that$T_n(B)$is an equicontinuous set on [0,1]. Obviously,$T_n(B)$is equicontinuous on$[-a,0]$. Thus,$T_n:P\to Q$is a completely continuous operator. The proof is complete. \end{proof} \begin{lemma} \label{lem3} Let$\Omega_0=\{x\in Q:\|x\|a.} By \eqref{e2.7}, \eqref{e2.8} and \eqref{e2.9}, we have \begin{align*} -z_0''(t)&\leq \phi(t)(g(z_0(t-a)-w(t-a))+h(z_0(t-a)+1))+p(t)\\ &\leq \phi(t)(g(\frac{1}{2}z_0(t-a))+h(z_0(t-a)+1))+p(t),\quad \forall t\in [a,t_0]. \end{align*} Sincez_0'(t-a)\geq z_0'(t)$for$t\in [a,t_0]$, we have $$-z_0''(t)z_0'(t)\leq [\phi(t)(g(\frac{1}{2}z_0(t-a))+h(z_0(t-a)+1))+p(t)]z_0'(t-a),$$ for all$t\in [a,t_0]$. Integrating from$t(t\in [a, t_0])$to$t_0, we have \begin{align*} &[z_0'(t)]^2\\ &\leq 2\sup_{t\in [a,1]}[\phi(t)+p(t)]\int^{t_0}_t[g(\frac{1}{2}z_0(s-a)) +h(z_0(s-a)+1)+1] z_0'(s-a)ds\\ &\leq 2\sup_{t\in [a,1]}[\phi(t)+p(t)]\int^{z_0(t_0-a)}_{z_0(t-a)} [g(\frac{1}{2}s)+h(s+1)+1]ds\\ &\leq 2\sup_{t\in [a,1]}[\phi(t)+p(t)]\int_0^{z_0(t_0)} [g(\frac{1}{2}s)+h(s+1)+1]ds\\ &=B(R_0)\\ \end{align*} and so $$z_0'(t)\leq \sqrt{ B(R_0)},\quad \forall t\in [a,t_0].\label{e2.11}$$ Then integrating froma$to$t_0$, we have $$z_0(t_0)\leq z_0(a)+\sqrt{B(R_0)}.\label{e2.12}$$ On the other hand, by \eqref{e2.8} and \eqref{e2.9}, we have $$-z_0''(t)\leq \phi(t)[g(\mu(t-a))+h(\mu(t-a)+1)]+p(t), \quad \forall t\in (0,a].$$ Integrating from$t(t\in (0,a))$to$a$, by \eqref{e2.11}, we have $$z'_0(t)\leq z_0'(a)+\int^a_0[\phi(s)(g(\mu(s-a))+h(\mu(s-a)+1)+p(s)]ds\\ \leq \sqrt{B(R_0)}+A.$$ Then integrating from 0 to$a$, we have $$z_0(a)\leq \sqrt{B(R_0)}+A.\label{e2.13}$$ It follows from \eqref{e2.12} and \eqref{e2.13} that $$z_0(t_0)\leq 2\sqrt{B(R_0)}+A\label{e2.14}$$ Since$z_0(t_0)=R_0$, from \eqref{e2.10} and \eqref{e2.14}, we have $$R_0\leq 2\sqrt{B(R_0)}+A,$$ which is a contradiction to (H3). Thus \eqref{e2.4} holds. By the properties of fixed point index, we have $$i(T_n,\Omega_0, Q)=i(\theta, \Omega_0, Q)=1.$$ The proof is completed. \end{proof} \begin{remark} \label{rmk2} \rm The inequality \eqref{e*} played an important role in the proof of Lemma \ref{lem3}. This type of inequality has been employed extensively in the literature \cite{a1,e1,o1}. The main idea of our proof of Lemma \ref{lem3} is derived from \cite{a1}. \end{remark} \section{Main Results} \begin{theorem} \label{thm1} Suppose that (H1)-(H4) hold. Assume that $$\lim_{x\to +\infty}\frac{h(x)}{x}=0.\label{e3.1}$$ Then \eqref{e1.1} has at least two positive solutions. \end{theorem} \begin{proof} For each positive integer$n$, let us define the operator$T_n$by \eqref{e2.1}. It follows from Lemma \ref{lem2} that$T_n:Q\to Q$is a completely continuous operator for every$n$. Let$\delta$be a positive number such that $$0<\delta<\min\big\{1,(\int^1_0\phi(s)ds)^{-1}\big\}.$$ It follows from \eqref{e3.1} that there exists$R>u_1$such that$h(x)\leq \delta x$for all$x\geq R$. Since$h:R^+\to R^+$is increasing, then $$h(x)\leq \delta x+h(R), \quad \forall x\in R^+.$$ By (H4), there exists$\widetilde{u}_1>u_1$such that $$\alpha(1-\beta-a)h(\frac{1}{2}\widetilde{u}_1) \int^{\beta+a}_{\alpha+a}\phi_0(s)ds>\widetilde{u}_1. \label{e3.2}$$ Put $$R_1=\max\big\{2\widetilde{u}_1,\frac{2[A+\|w\|+ (\|w\|+\|\mu\|+R_0+1+h(R))\int^1_0\phi(s)ds]}{1-\delta\int^1_0\phi(s)ds} \big\},\label{e3.3}$$ \begin{gather*} \Omega_0=\{x\in Q:\|x\|u_1\},\\ U_{01}=\{x\in Q:\|x\|\widetilde{u}_1\}. \end{gather*} It is easy to see that$\Omega_0$,$\Omega_1$,$\Omega_{01}$and$U_{01}$are bounded open convex sets of$Q$, and that $$\Omega_0\subset \Omega_1,\quad \Omega_{01}\subset \Omega_1,\quad U_{01}\subset \Omega_1,\quad \Omega_0\cap \Omega_{01}=\emptyset, \quad U_{01}\subset \Omega_{01}.$$ For each positive integer$n$and$x\in \overline{\Omega}_1, by \eqref{e2.2} and \eqref{e3.3}, we have \begin{aligned} &(T_nx)(t)\\ &\leq\int^1_0G(t,s)[\phi(s)(g([x(s-a)]^*+n^{-1})+h([x(s-a)]^*+n^{-1}))+p(s)]ds\\ &\leq \int^a_0\phi(s)(g(\mu(s-a))ds +\int_a^1\phi(s)(g(\frac{1}{2}R_0a(s-a))ds\\ &+h(\|x\|+\|w\|+\|\mu\|+R_0+1))\int^1_0\phi(s)ds+\|w\|\\ &\leq A+\|w\|+[\delta(\|x\|+\|w\|+\|\mu\|+R_0+1)+h(R)]\int^1_0\phi(s)ds\\ &\leq A+\|w\|+[\delta R_1+\|w\|+\|\mu\|+R_0+1+h(R)]\int^1_0\phi(s)ds\\ &u_1,\quad \forall t\in [\alpha, \beta], \end{align*} and so,T_n(\overline{\Omega}_{01})\subset\Omega_{01}$for every positive integer$n$. Also by Lemma \ref{lem1}, we have $$i(T_n, \Omega_{01}, Q)=1\label{e3.6}$$ for every positive integer$n$. Similarly, by \eqref{e3.2} we can show that $$i(T_n, U_{01},Q)=1\label{e3.7}$$ for every positive integer$n$. It follows from \eqref{e3.7} that for every positive integer$n$,$T_n$has at least one fixed point$\widetilde{x}_n\in \overline{U}_{01}. It is easy to see that \begin{align*} [\widetilde{x}_n(t-a)]^* &=\begin{cases} \mu(t-a),&t\in [0,a] \\ \widetilde{x}_n(t-a)-w(t-a),&t\in (a,1] \end{cases} \\ &\geq\begin{cases} \mu(t-a),& t\in (0,a]\\ \frac{1}{2}\widetilde{x}_n(t-a),& t\in (a,1] \end{cases} \\ & \geq x_0(t-a)+\widetilde{x}_0(t-a),\quad t\in [0,1], \end{align*} and so $$g([\widetilde{x}_n(t-a)]^*+n^{-1})\leq g(x_0(t-a)+\widetilde{x}_0(t-a)), t\in (0,1]\backslash\{a\}.\label{e3.8}$$ Using essentially the same argument as in Lemma \ref{lem3}, we see that there existst_n\in(0,1)$such that$\widetilde{x}'_n(t_n)=0$, and $$-\widetilde{x}''_n(t)\leq \phi(t) [g(x_0(t-a)+\widetilde{x}_0(t-a)+n^{-1})+h(R_1+1)]+p(t),$$ for all$t\in (0,1]$. Integration yields $$|\widetilde{x}_n'(t)|\leq \int^1_0[\phi(s)(g(x_0(s-a)+ \widetilde{x}_0(s-a))+h(R_1+1))+p(s)]ds,$$ for all$t\in(0,1]$. This means that$\{ \widetilde{x}_n\}$is equicontinuous on$[0,1]$. Since$ \widetilde{x}_n(t)=0$for$t\in [-a,0]$,$\{\widetilde{x}_n\}$is equicontinuous on$[-a, 0]$. Therefore, the Arzela-Ascoli Theorem guarantees the existence of a subsequence$\{ \widetilde{x}_{n_i}\}$of$\{ \widetilde{x}_n\}$and a function$x_{01}\in \overline{U}_{01}$with$\widetilde{x}_{n_i}$converging uniformly on$[-a,1]$to$x_{01}$as$i\to \infty$. From$ \widetilde{x}_n =T_n \widetilde{x}_n, by \eqref{e3.8} and using the Lebesgue dominated convergence Theorem, we have \begin{align*} x_{01}(t)&=\begin{cases} 0,&t\in [-a,0];\\ \int^1_0G(t,s)[f(s,[x_{01}(s-a)]^*)+p(s)]ds,&t\in (0,1]. \end{cases} \\ &=\begin{cases} 0,&t\in [-a,0];\\ \int^1_0G(t,s)[f(s,x_{01}(s-a)+x_0(s-a)\\ -w(s-a))+p(s)]ds,& t\in (0,1]. \end{cases} \end{align*} Lety_{01}(t)=x_{01}(t)+x_0(t)-w(t)$for$t\in [-a,1]$. Then, we have $$y_{01}(t)=\begin{cases} \mu(t), &t\in [-a,0];\\ \int^1_0G(t,s)f(s, y_{01}(s-a))ds,&t\in (0,1]. \end{cases}$$ It is easily verfied that$y_{01}$is a positive solution of \eqref{e1.1}. It follows from \eqref{e3.5}, \eqref{e3.6} and Lemma \ref{lem3} that for every positive integer$n$$$i(T_n,\Omega_1\backslash( \bar{\Omega}_{01}\cup\bar{\Omega}_0), Q)=i(T_n, \Omega_1,Q)-i(T_n,\Omega_{01}, Q)-i(T_n, \Omega_0, Q)=-1.$$ Therefore,$T_n$has at least one fixed point$\bar{x}_n\in \Omega_1\backslash(\bar{\Omega}_{01}\cup \bar{\Omega}_0)$for every positive integer$n$. For every positive integer$n$, there is at least one point$t_n\in [\alpha,\beta]$such that$\bar{x}_n(t_n)\leq u_1$. In a similar way as above, we can show that there exist a subsequence$\{\bar{x}_{n_i}\}$of$\{\bar{x}_n\}$,$x_1\in \overline{\Omega_1\backslash( \bar{\Omega}_{01}\cup \bar{\Omega}_0)}$and a point$t_0\in [\alpha,\beta]$such that$\bar{x}_{n_i}$convergence uniformly on [-a,1] to$x_1$as$i\to \infty$, and$x_1(t_0)\leq u_1$. Let$y_1(t)=x_1(t)+x_0(t)-w(t)$for$t\in [-a,1]$. Then$y_1$is a positive solution of \eqref{e1.1}. Since $$x_1(t_0)\leq u_1<\widetilde{u}_1\leq x_{01}(t_0),$$$y_{01}$and$y_1$are two distinct positive solutions of \eqref{e1.1}. \end{proof} \begin{theorem} \label{thm2} Suppose that (H1)-(H4) hold, and that there exists$R_1>u_1$such that \begin{gather*} A+\|w\|+h(R_1+\|w\|+\|\mu\|+1)\int^1_0\phi(s)dsR_1$ such that $$A+\|w\|+h(\bar{R}_1+\|w\|+\|\mu\|+1)\int^1_0\phi(s)ds<\bar{R}_1. \label{e3.11}$$ Let us define the open sets $\Omega_0$, $\Omega_{01}$, $\Omega_1$ and $U_{01}$ as in Theorem \ref{thm1}. Let $U_1 =\{x\in Q:\|x\|<\bar{R}_1\}$. It is easy to see that for any $x\in \bar{\Omega}_1$ and $t\in [0,1]$ $$R_1+\|w\|+\|\mu\|\geq[x(t-a)]^*\geq \begin{cases} \mu(t-a),&t\in [0, a];\\ \frac{1}{2}R_0a(t-a),&t\in[a,1]. \end{cases}$$ Since $h:R^+\to R^+$ is increasing, we have $$h([x(t-a)]^*+n^{-1})\leq h(R_1+\|w\|+\|\mu\|+1),\forall t\in [0,1].$$ Therefore, by \eqref{e3.9}, we have \begin{align*} |(T_nx)(t)| &\leq A+\|w\|+h(R_1+\|w\|+\|\mu\|+1)\int^1_0\phi(s)ds\\ & 2(\alpha(1-\beta)\sup_{t\in [0,1]}\int^{\beta+a}_{\alpha+a}G(t,s) \phi_0(s)ds)^{-1}. \label{e3.13} By \eqref{e3.10}, there exists $\widetilde{R}>\bar{R}_1$ such that $h_0(y)\geq My$ for any $y\geq \widetilde{R}$. Set $R_2=2\widetilde{R }\alpha^{-1}(1-\beta)^{-1}$, \Omega_2=\{x\in Q:\|x\|\widetilde{R},\quad t\in [\alpha+a,\beta+a]. \end{align*} Therefore, \begin{align*} R_2\geq y_0(t)&\geq \int^1_0G(t,s)\phi_0(s)h_0([y_0(s-a)]^*)ds\\ &\geq \int^{\beta+a}_{\alpha+a}G(t,s)\phi_0(s)h_0(y_0(s-a)-w(s-a))ds\\ &\geq \frac{1}{2}MR_2\alpha(1-\beta)\int^{\beta+a}_{\alpha+a}G(t,s)\phi_0(s)ds, \quad t\in [0,1]. \end{align*} Hence $$M\leq 2(\alpha(1-\beta)\sup_{t\in [0,1]}\int^{\beta+a}_{\alpha+a}G(t,s) \phi_0(s)ds)^{-1},$$ which is a contradiction to \eqref{e3.13}. Hence, \eqref{e3.14} holds. From the properties of the fixed point index, we have $$i(T_n, \Omega_2, Q)=0.\label{e3.15}$$ It follows from \eqref{e3.12} and \eqref{e3.15} that $$i(T_n, \Omega_2\backslash\bar{U}_1, Q)=i(T_n, \Omega_2, Q)-i(T_n, U_1, Q)=-1$$ for every positive integern$. Hence,$T_n$has at least one fixed point$\widetilde{x}_n\in \Omega_2\backslash\bar{U}_1$. Using essentially the same argument as in Theorem \ref{thm1}, we can show that there exist a subsequence$\{ \widetilde{x}_{n_i}\}$of$\{\widetilde{x}_n\}$, and$x_3\in \overline{\Omega_2\backslash\bar{U}_1}$such that$\widetilde{x}_{n_i}\to x_3(i\to +\infty)$, and$y_3=x_3+x_0-w$is a positive solution of \eqref{e1.1}. The proof is completed. \end{proof} \begin{corollary} \label{coro1} Suppose that (H1)-(H3) hold. Moreover, there exist$R_i,\ u_i(i=1,2,\dots, n)$with$R_0u_i, \quad i=1,2,\dots, n, \\ \lim_{x\to+\infty}\frac{h_0(x)}{x}=+\infty. \end{gather*} Then \eqref{e1.1} has at least $2n+1$ positive solutions. \end{corollary} \begin{corollary} \label{coro2} Suppose that (H1)-(H3) hold, and $\lim_{x\to +\infty}\frac{h_0(x)}{x}=+\infty$. Then \eqref{e1.1} has at least one positive solution. \end{corollary} We remark that the multiplicity results for positive solutions of singular semi-positone delay differential equations are new. Obviously, we can use the ideas of this paper to establish multiplicity results for positive solutions of the more general delay equation. \subsection*{Example} Consider the delay differential boundary-value problem $$\begin{gathered} y''+40\big[\frac{1}{\sqrt{y(t-\frac{1}{4})}}+h(y(t-\frac{1}{4}))\big] =\frac{1}{t^{1/4}}, \quad t\in (0,1]\backslash\{\frac{1}{4}\},\\ y(t)=-t, \quad t\in [-\frac{1}{4}, 0],\\ y(1)=0, \end{gathered} \label{e3.16}$$ where $h:R^+\to R^+$ is increasing, $h(y)=y^{1/4}$ for $y\in [0, 2\times 10^4]$, and $\lim_{x\to+\infty}\frac{h(x)}{x}=0\,.$ \noindent\textbf{Claim:} If there exits $u_1>2\times 10^4$ such that $h(\frac{1}{2}u_1)>2u_1$, then the boundary-value problem \eqref{e3.16} has at least two positive solutions. \begin{proof} Let $\phi(t)=\phi_0(t)=40$ for $t\in [0,1]$, $a=1/4$, $g(y)=1/\sqrt{y},$ $p(t)=1/t^{1/4}$, $\mu(t)=-t$ for $t\in [-\frac{1}{4}, 0]$. It is easy to see (H1) and (H2) hold. Set $R_0=10^4$. Then, we have \begin{align*} A&\leq \int^{1/4}_0\big[40(\frac{1}{\sqrt{\frac{1}{4}-s}}+h(\frac{1}{4}-s+1)) +\frac{1}{s^{1/4}}\big]ds+\int^1_{1/4} \frac{40}{\sqrt{\frac{R_0}{8}(s-\frac{1}{4})}}\\ &\leq 80\sqrt{\frac{1}{4}-s}\Big|^0_{1/4}+10h(\frac{5}{4}) +\frac{4}{3}+\frac{320}{\sqrt{R_0}}\\ &=40+10h(\frac{5}{4})+\frac{4}{3}+\frac{320}{\sqrt{R_0}} \leq 80, \end{align*} \begin{align*} B(R_0)&=2\times (40+\sqrt{2})\int^{R_0}_0 \big(\frac{1}{\sqrt{s/2}}+(s+1)^{\frac{1}{4}}+1\big)ds\\ &< 84(2\sqrt{2R_0}+\frac{4}{5}(R_0+1)^{5/4}+R_0)\\ &<340\sqrt{R_0}+80(R_0+1)^{5/4}+84R_0<8875040, \end{align*} and $$R_0-2\sqrt{B(R_0)}>R_0-2\sqrt{8875040}>80,$$ which implies (H3). Put $[\alpha,\beta]=[\frac{1}{3}, \frac{1}{2}]$, $h(y)=h_0(y)$ for $y\in R^+$. It is easy to check that (H4) holds. Thus, by Theorem \ref{thm1}, the conclusion holds. The proof is completed. \end{proof} \begin{thebibliography}{00} \bibitem{a1} R. P. Agarwal and D. 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