\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 147, pp. 1--29.\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 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/147\hfil Multiple positive solutions] {Multiple positive solutions for singular $m$-point boundary-value problems with nonlinearities depending on the derivative} \author[Y. Ma, B. Yan\hfil EJDE-2008/147\hfilneg] {Ya Ma, Baoqiang Yan} % in alphabetical order \address{Department of Mathematics, Shandong Normal University, Jinan, 250014, China} \email[Ya Ma]{maya-0907@163.com} \email[Baoqiang Yan]{yanbqcn@yahoo.com.cn} \thanks{Submitted August 13, 2008. Published October 24, 2008.} \thanks{Supported by grants 10871120 from the National Natural Science, and Y2005A07 \hfill\break\indent from the Natural Science of Shandong Province, China} \subjclass[2000]{34B10, 34B15} \keywords{$m$-point boundary-value problem; singularity; positive solutions; \hfill\break\indent fixed point theorem} \begin{abstract} Using the fixed point theorem in cones, this paper shows the existence of multiple positive solutions for the singular $m$-point boundary-value problem \begin{gather*} x''(t)+a(t)f(t,x(t),x'(t))=0,\quad 00,\; t\in (0,1);\label{e2.3}\\ f\in C([0,1]\times \mathbb{R}_+ \times \mathbb{R}_-,[0,+\infty)); \label{e2.4} \end{gather} There exists $g\in C([0,+\infty)\times (-\infty,0],[0,+\infty))$ such that $$f(t,x,y)\leq g(x,y),\forall (t,x,y)\in [0,1]\times \mathbb{R}_+ \times \mathbb{R}_-. \label{e2.5}$$ For $x\in P$ and $t\in [0,1]$, define operator \begin{aligned} (Ax)(t)&=-\int_{0}^{t}(t-s)a(s)f(s,x(s)x'(s))ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s) a(s)f(s,x(s),x'(s))\Big)ds\,. \end{aligned} \label{e2.6} \begin{lemma}[\cite{m3}] \label{lem2.3} Assume \eqref{e2.1}. Then for $y\in C[0,1]$ the problem $$\begin{gathered} x''+y(t)=0,t\in(0,1) \\ x'(0)=\sum_{i=1}^{m-2}b_ix'(\xi_i),\quad x(1)=\sum_{i=1}^{m-2}a_ix(\xi_i) \end{gathered}\label{e2.7}$$ has a unique solution $$x(t)=-\int_{0}^{t}(t-s)y(s)ds+Mt+N,\label{e2.8}$$ where, \begin{gather*} M=\frac{\sum_{i=1}^{m-2}b_i\int_{0}^{\xi_i}a(s)y(s)ds} {\sum_{i=1}^{m-2}b_i-1},\\ \begin{aligned} N&=\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)y(s)ds-\sum_{i=1}^{m-2}a_i \int_{0}^{\xi_i}(\xi_i-s)a(s)y(s)ds\\ &\quad -\frac{\sum_{i=1}^{m-2}b_i\int_{0}^{\xi_i}a(s)y(s)ds}{\sum_{i=1}^{m-2}b_i-1} (1-\sum_{i=1}^{m-2}a_i\xi_i)\Big). \end{aligned} \end{gather*} Further, if $y\geq 0$, for all $t\in [0,1]$, $x$ satisfies $$\inf_{t\in[0,1]}x(t)\geq \gamma\|x\|_1,\label{e2.9}$$ where $\gamma=\big(\sum_{i=1}^{m-2}a_i(1-\xi_i)\big)/ \big(1-\sum_{i=1}^{m-2}a_i\xi_i\big)$. \end{lemma} \begin{lemma} \label{lem2.4} Suppose \eqref{e2.3}--\eqref{e2.5}hold. Then $A:P\to P$ is a completely continuous operator. \end{lemma} \begin{proof} For $x\in P$, from \eqref{e2.6}, one has \begin{aligned} (Ax)(t)&\geq -\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x(s),x'(s))ds \Big)\\ &\geq \frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\ &\quad -\sum_{i=1}^{m-2}a_i \int_{0}^{1}(\xi_i-s)a(s)f(s,x(s),x'(s))ds\\ &\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)f(s,x(s),x'(s))ds \geq 0, \quad t\in[0,1], \end{aligned} \label{e2.10} \begin{aligned} |(Ax)(t)| &=-\int_{0}^{1}(t-s)a(s)f(s,x(s)x'(s))ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s) a(s)f(s,x(s),x'(s))ds \Big)\\ &\leq \frac{1}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds \\ &\leq \frac{1}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)f(s,x(s),x'(s))ds\\ &\leq \frac{1}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)ds \max_{0\leq c\leq \|x\|,-\|x\|\leq c'\leq 0}g(c,c') <+\infty, \quad t\in[0,1], \end{aligned}\label{e2.11} \begin{aligned} |(Ax)'(t)| &=\big|-\int_{0}^{t}a(s)f(s,x(s)x'(s))ds\big|\\ &=\int_{0}^{t}a(s)f(s,x(s)x'(s))ds\\ &\leq \int_{0}^{1}a(s)f(s,x(s)x'(s))ds\\ &\leq \int_{0}^{1}a(s)ds\max_{0\leq c\leq \|x\|, -\|x\|\leq c'\leq 0}g(c,c') <+\infty, \end{aligned}\label{e2.12} which implies that $A$ is well defined. \begin{align*} (Ax)(0) &=\frac{1}{1-\sum_{i=1}^{m-2}a_i} (\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i} (\xi_i-s)a(s)f(s,x(s),x'(s))ds)\\ &\geq \frac{1}{1-\sum_{i=1}^{m-2}a_i}(\sum_{i=1}^{m-2}a_i \int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\ &\quad - \sum_{i=1}^{m-2}a_i\int_{0}^{1}(\xi_i-s)a(s)f(s,x(s),x'(s))ds)\\ &= \sum_{i=1}^{m-2}a_i(1-\xi_i)\int_{0}^{1}a(s)f(s,x(s),x'(s))ds. \end{align*} On the other hand, \begin{align*} \|Ax\|_{2} &=\max_{t\in[0,1]}|(Ax)'(t)|\\ &=\max_{t\in[0,1]}|-\int_{0}^{t}a(s)f(s,x(s)x'(s))ds|\\ &=\int_{0}^{1}a(s)f(s,x(s)x'(s))ds . \end{align*} Then $$(Ax)(0)\geq \delta\|Ax\|_{2}.\label{e2.13}$$ By Lemma \ref{lem2.3}, we have $(Ax)(t)\geq \gamma \|Ax\|_{1}.$ As a result $Ax\in P$, which implies $AP\subseteq P$. By a standard argument, we know that $A:P\to P$ is continuous and completely continuous. \end{proof} \section{Singularities at $x'=0$ but not at $x=0$} In this section the nonlinearity $f$ may be singular at $x'=0$ but not at $x=0$. We will assume that the following conditions hold. \begin{itemize} \item[(H1)] $a(t)\in C(0,1)\cap L^1[0,1]$, $a(t)>0$, $t\in(0,1)$ \item[(H2)] $f(t,u,z)\leq h(u)[g(z)+r(z)]$, where $f\in C([0,1]\times \mathbb{R}_+\times \mathbb{R}_-,\mathbb{R}_+)$, $g(z)>0$ continuous and nondecreasing on $\mathbb{R}_-$, $h(u)\geq 0$ continuous and nondecreasing on $\mathbb{R}_+$, $r(z)>0$ continuous and non-increasing on $(-\infty,0]$; \item[(H3)] $$\sup_{c\in \mathbb{R}_+}{\frac{c }{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1} {1-\sum_{i=1}^{m-2}a_i}}I^{-1}(h(c)\int_0^1 a(s)ds)}}>1,$$ where $I(z)={\int_z^0{\frac{du}{g(u)+r(u)}}}$, $z\in \mathbb{R}_-$; \item[(H4)] There exists a function $g_1\in C([0,+\infty)\times(-\infty,0],[0,+\infty))$, such that $f(t,u,z)\geq g_1(u,z),\forall (t,u,z)\in [0,1]\times \mathbb{R}_+\times \mathbb{R}_-$, and $\lim_{u\to +\infty}\frac{g_1(u,z)}{u}=+\infty$, uniformly for $z\in \mathbb{R}_-$. \item[(H5)] There exists a function $\Psi_H\in C([0,1],\mathbb{R}_+)$ and a constant $0\leq \delta<1$ such that $f(t,u,z)\geq \Psi_H(t)u^\delta$, for all $(t,u,z)\in [0,1]\times[0,H]\times \mathbb{R}_-$. \end{itemize} For $n\in\{1,2,\dots\}$ and $x\in P$, define operator \begin{aligned} (A_{n}x)(t) &=-\int_{0}^{t}(t-s)a(s)f(s,x(s),-|x'(s)|-\frac{1}{n})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x(s),-|x'(s)|-\frac{1}{n})ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x(s),-|x'(s)|-\frac{1}{n})ds \Big),\quad t\in [0,1]. \end{aligned} \label{e3.1} \begin{theorem} \label{thm3.1} Suppose {\rm (H1)--(H5)} hold. Then \eqref{e1.1} has at least two positive solutions $x_{0,1},x_{0,2}\in C^1[0,1]\cap C^2(0,1)$ with $x_{0,1}(t),x_{0,2}(t)>0$, $t \in [0,1]$. \end{theorem} \begin{proof} Choose $R_1>0$ such that $${\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}} I^{-1}(h(R_1)\int_0^1 a(s)ds)}}>1.\label{e3.2}$$ From the continuity of $I^{-1}$ and $h$, we can choose $\varepsilon>0$ and $\varepsilon1, \label{e3.3}$n_0\in\{1,2,\dots\}$with${\frac{1}{n_0}} < \min\{\varepsilon,{\delta/2} \}$and let$N_0 =\{n_0,n_0+1, \dots\}$. Lemma \ref{lem2.4} guarantees that for$n\in N_0$,$A_n:P\to Pis a completely continuous operator. Let $$\Omega_1=\{x\in C^1[0,1]:\| x \| 0, x_0(t)>0,t\in[0,1]. Differentiating \eqref{e3.5}, we have \begin{gathered} x''_0(t)+\mu_0 a(t)f(t,x_0(t),x_0'(t)-{\frac{1}{n}})=0,\quad 0R_1 such that $$g_1(x,y)\geq N^{*}x ,\quad \forall x\geq R_2,\; y\in \mathbb{R}_-. \label{e3.11}$$ Let \Omega_2=\{x\in C^1[0,1]:\|x\|<\frac{R_2}{a^*}\}. Then$$ Ax \not\leq x,\quad \forall x\in \partial \Omega_2 \cap P. $$In fact, if there exists x_0\in \partial \Omega_2 \cap P with x_0\geq A_{n}x_0. By the definition of the cone and Lemma \ref{lem2.3}, one has$$ x_0(t)\geq \gamma\|x_0\|_1\geq \gamma x(0)\geq \gamma\delta\|x_0\|_2,\quad x_0(t)\geq \frac{R_2}{a^*}>R_2,\quad \forall t\in[0,1], from \eqref{e3.11}, \begin{align*} \gamma x_0(t)&\geq \gamma A_{n}x_0(t)\\ &=\gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds \Big)\Big)\\ &\geq \gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\ &\quad-\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds \Big)\Big)\\ &\geq \gamma\Big(\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{t}(t-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\ &\quad -\frac{\sum_{i=1}^{m-2}a_i\int_{0}^{1}(\xi_i-s)a(s)f(s,x_0(s), -|x_0'(s)|-\frac{1}{n})ds}{1-\sum_{i=1}^{m-2}a_i}\Big)\\ &=\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\ &\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)g_1(x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\ &\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)dsN^{*} x_0(s)\\ &\geq a^*\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)dsN^{*}\frac{R_2}{a^*} >\frac{R_2}{a^*}. \end{align*} Then \|x_0\|\geq \gamma\|x_0\|_1>\frac{R_2}{a^*}, which is a contradiction to x_0\in \partial \Omega_2 \cap P. Then \eqref{e3.10} holds. From Lemma \ref{lem2.2}, \begin{aligned} i(A_n,\Omega_2\cap P,P)=0. \end{aligned} \label{e3.12} which with \eqref{e3.9} guarantee that \begin{aligned} i(A_n,(\Omega_2-\bar{\Omega}_1)\cap P,P)=-1. \end{aligned} \label{e3.13} From this equality and \eqref{e3.9}, A_n has two fixed points with x_{n,1}\in \Omega_1\cap P,x_{n,2}\in (\Omega_2-\bar{\Omega}_1)\cap P. For each n\in N_0, there exists x_{n,1}\in\Omega_1\cap P such that x_{n,1}=A_nx_{n,1}; that is, \begin{aligned} x_{n,1}(t) &=-\int_{0}^{t}(t-s)a(s)f(s,x_{n,1}(s),-|x'_{n,1}(s)|-\frac{1}{n})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n,1}(s),-|x'_{n,1}(s)|-\frac{1}{n})ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s) f(s,x_{n,1}(s),-|x'_{n,1}(s)|-\frac{1}{n})ds \Big). \end{aligned} \label{e3.14} As in the proof of \eqref{e3.5}, we have x'_{n,1}(t)\leq 0, t \in (0,1) and x'_{n,1}(t)=-\int_0^t a(s)f(s,x_{n,1}(s),x_{n,1}'(s) -{\frac{1}{n}})ds,\quad n\in N_0,\; t\in(0,1). Now we consider \{x_{n,1}(t)\}_{ n\in N_0} and \{x' _{n,1}(t)\}_{n\in N_0}. Since \| x_{n,1}\|\leq R_1, it follows that \begin{gather} \text{\{x _{n,1}(t)\} is uniformly bounded on [0,1],} \label{e3.15}\\ \text{\{x' _{n,1}(t)\} is uniformly bounded on [0,1].} \label{e3.16} \end{gather} Then $$\text{\{x _{n,1}(t)\} is equicontinuous on [0,1].} \label{e3.17}$$ As in the proof as \eqref{e3.6}, \begin{gathered} x''_{n,1}(t)+a(t)f(t,x_{n,1}(t),x_{n,1}'(t)-{\frac{1}{n}})=0,\quad 00, there exists \epsilon'>0 such that $$|I^{-1}(s_1)-I^{-1}(s_2)|<\bar\epsilon,\quad \forall |s_1-s_2|<\epsilon',\; s_1, s_2\in[0,I(-R_1-\epsilon)].\label{e3.22}$$ And \eqref{e3.19} guarantees that for \epsilon'>0, there exists \delta'>0 such that $$|I( x'_{n,1}(t_2)-{\frac{1}{n}})-I( x'_{n,1}(t_1)-{\frac{1}{n}})|<\epsilon', \quad \forall\ |t_1-t_2|<\delta',\; t_1,t_2\in[0,1].\label{e3.23}$$ From this inequality and \eqref{e3.22}, \begin{aligned} | x'_{n,1}(t_2)- x'_{n,1}(t_1)| &=| x'_{n,1}(t_2)-{\frac{1}{n}}-( x'_{n,1}(t_1)-{\frac{1}{n}})|\\ &=|I^{-1}(I( x'_{n,1}(t_2)-{\frac{1}{n}}))-I^{-1}(I( x'_{n,1}(t_1) -{\frac{1}{n}}))|\\ &< \bar\epsilon, \quad \forall |t_1-t_2|<\delta',\; t_1,t_2\in[0,1]; \end{aligned} \label{e3.24} that is, $$\text{\{x' _{n,1}(t)\} is equicontinuous on [0,1].} \label{e3.25}$$ From \eqref{e3.15}--\eqref{e3.17}, \eqref{e3.25} and the Arzela-Ascoli Theorem, \{x _{n,1}(t)\} and \{x'_{n,1}(t)\} are relatively compact on C[0,1], which implies there exists a subsequence \{x_{n_j,1}\} of \{x_{n,1}\} and function x_{0,1}(t)\in C[0,1] such that \lim_{j\to +\infty}\max_{t\in[0,1]}|x_{n_j,1}(t)-x_{0,1}(t)|=0, \quad \lim_{j\to +\infty}\max_{t\in[0,1]}|x'_{n_j,1}(t)-x'_{0,1}(t)|=0. $$Since x'_{n_j,1}(0)=0, x_{n_j,1}(1)=\sum_{i=1}^{m-2}a_i x_{n_j,1}(\xi_i), x'_{n_j,1}(t)<0, x_{n_j,1}(t)>0, t\in(0,1), j\in\{1,2,\dots\}, $$x'_{0,1}(0)=0, \quad x_{0,1}(1)=\sum_{i=1}^{m-2}a_i x_{0,1}(\xi_i), \quad x'_{0,1}(t)\leq 0,\quad x_{0,1}(t)\geq 0, \quad t\in(0,1).\label{e3.26}$$ For (t,x_{n_j,1}(t),x_{n_j,1}'(t)-{\frac{1}{n}})\in [0,1]\times [0,R_1+\epsilon]\times (-\infty,0), from (H5) there exists a function \Psi_{R_1}\in C([0,1],\mathbb{R}_+) such that$$ f(t,x_{n_j,1}(t),x_{n_j,1}'(t)-{\frac{1}{n_j}})ds \geq \Psi_{R_1}(t)(x_{n_j,1}(t))^\delta,\quad 0\leq \delta<1. Then, for n\in N_0, \begin{align*} x_{n_j,1}(t) &=-\int_{0}^{t}(t-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n_j,1}(s), x'_{n_j,1}(s)-\frac{1}{n_j})ds \Big)\\ &\geq -\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s) f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds \Big)\\ &\geq \frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\quad -\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}(\xi_i-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &=\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)\Psi_{R_1}(s)(x_{n_j,1}(s))^\delta ds\\ &\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)\Psi_{R_1}(s)\gamma^{\delta}ds\|x_{n_j,1}\|_1^\delta, \end{align*} which implies \|x_{n_j,1}\|_1\geq \Big(\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)} {1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)\Psi_{R_1}(s) \gamma^{\delta}ds\Big)^{\frac{1}{1-\delta}}, $$and$$ x_{n_j,1}(t)\geq \Big(\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)} {1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)\Psi_{R_1}(s)\gamma^{\delta}ds\Big) ^{\frac{1}{1-\delta}} =a_0>0 . Thus \begin{align*} x_{n_j,1}'(t) &=-\int_{0}^{t}a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\leq -\int_{0}^{t}a(s)\Psi_{R_1}(s)(x_{n_j,1}(s))^\delta ds\\ &\leq -\int_{0}^{t}a(s)\Psi_{R_1}(s) ds a_0^\delta ,\quad t\in[0,1], n\in N_0. \end{align*} Consequently, \begin{gather*} \inf_{j\geq 1}\min_{s\in[\frac{1}{2},t]}|x'_{n_j,1}(s)| >0,\quad t\in[\frac{1}{2},1),\\ \inf_{j\geq 1}\min_{s\in[t,\frac{1}{2}]}|x'_{n_j,1}(s)| >0, \quad t\in(0,\frac{1}{2}]. \end{gather*} Sincex_{n_j,1}'(t)-x_{n_j,1}'({\frac{1}{2}}) =- \int_{1/2}^{t} a(s)f(s,x_{n_j,1}(s),x_{n_j,1}'(s)-{\frac{1}{n_j}})ds,\quad t\in(0,1), and \begin{align*} f(t,x_{n_j,1}(t),x_{n_j,1}'(t)-{\frac{1}{n_j}}) &\leq h(x_{n_j,1}(t))[g(x_{n_j,1}'(t)-{\frac{1}{n_j}})+r(x_{n_j,1}'(t)-{\frac{1}{n_j}})]\\ &\leq h(\frac{R_1}{\gamma})[g(-\int_{0}^{t}a(s)\Psi_{R_1}(s) ds a_0^\delta)+r(-\frac{R_1}{\gamma\delta}-\epsilon)], \end{align*} letting j\to +\infty, the Lebesgue Dominated Convergence Theorem guarantees that $$x_{0,1}'(t)-x_{0,1}'({\frac{1}{2}})= - \int_{1/2}^{t} a(s)f(s,x_{0,1}(s),x_{0,1}'(s))ds,\quad t\in(0,1). \label{e3.27}$$ Differentiating, we have x''_{0,1}(t)+a(t)f(t,x_{0,1}(t),x'_{0,1}(t))=0,\quad 01,00. If $$\mu<\sup_{c\in \mathbb{R}_+}\frac{I(-\frac{c(1-\sum_{i=1}^{m-2}a_i)} {1+\sum_{i=1}^{m-2}a_i\xi_i})}{1+c^b+c^d}. \label{e3.28}$$ Then \eqref{e1.1} has at least two positive solutions $x_{0,1},x_{0,2}\in C^1[0,1]\cap C^2(0,1)$. We apply Theorem \ref{thm3.1} with $g(z)=(-z)^{-a},r(z)=1,h(u)=\mu(1+u^b+u^d),\Psi(t)=\mu,g_1(u,z)=\mu u^b$. (H1), (H2), (H4), (H5) hold. Also \begin{align*} &\sup_{c\in \mathbb{R}_+}{\frac{c}{-{\frac{\sum_{i=1}^{m-2} a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}I^{-1}(h(c)\int_0^1 a(s)ds)}}\\ &=\sup_{c\in \mathbb{R}_+}{\frac{c}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1} {1-\sum_{i=1}^{m-2}a_i}}I^{-1}(\mu(1+c^b+c^d))}}, \end{align*} and \eqref{e3.28} guarantees that (H3) holds. \section{Singularities at $x'=0$ and $x=0$} In this section the nonlinearity $f$ may be singular at $x'=0$ and $x=0$. We assume that the following conditions hold. \begin{itemize} \item[(P1)] $a(t)\in C[0,1]$, $a(t)>0$, $t\in(0,1)$; \item[(P2)] $f(t,u,z)\leq [h(u)+\omega(u)][g(z)+r(z)]$, where $f\in C([0,1]\times \mathbb{R}_+\times \mathbb{R}_-,\mathbb{R}_+)$, $g(z)>0$ continuous and non-increasing on $(-\infty,0]$, $\omega(u)>0$ continuous and non-increasing on $[0,+\infty)$, $h(u)\geq 0$ continuous and nondecreasing on $\mathbb{R}_+$, $r(z)>0$ continuous and nondecreasing on $\mathbb{R}_-$; \item[(P3)] $$\sup_{c\in \mathbb{R}_+}{\frac{c}{-{\frac{\sum_{i=1}^{m-2} a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[-\max_{t\in [0,1]}a(t)(ch(c)+\int_0^c \omega(s)ds)])}}>1,$$ where $I(z)={\int_z^0{\frac{udu}{g(u)+r(u)}}}$, $z\in \mathbb{R}_-$, $\int_{0}^{a}\omega(s)ds < +\infty$; \item[(P4)] There exists a function $g_1\in C( [0,+\infty)\times (-\infty,0],[0,+\infty))$, such that $f(t,u,z)\geq g_1(u,z), \forall (t,u,z)\in [0,1]\times \mathbb{R}_+\times \mathbb{R}_-$, and $\lim_{u\to +\infty}\frac{g_1(u,z)}{u}=+\infty$, uniformly for $z\in \mathbb{R}_-$. \item[(P5)] There exists a function $\Psi_H\in C([0,1],\mathbb{R}_+)$ with $f(t,u,z)\geq \Psi_H(t)$, for all $(t,u,z)\in [0,1]\times[0,H]\times[-H,0)$. \end{itemize} For $n\in\{1,2,\dots\}$, $x\in P$, $t\in [0,1]$, define operator \begin{aligned} (A_{n}x)(t) &=-\int_{0}^{t}(t-s)a(s)f(s,x(s)+\frac{1}{n},-|x'(s)|-\frac{1}{n})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x(s)+\frac{1}{n},-|x'(s)|-\frac{1}{n})ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x(s) +\frac{1}{n},-|x'(s)|-\frac{1}{n})ds\Big). \end{aligned} \label{e4.1} \begin{theorem} \label{thm4.1} Suppose {\rm (P1)--(P5)} hold. Then \eqref{e1.1} has at least two positive solutions $x_{0,1},x_{0,2}\in C^1[0,1]\cap C^2(0,1)$ and $x_{0,1}(t),x_{0,2}(t)>0$, $t \in [0,1]$. \end{theorem} \begin{proof} Choose $R_1>0$ such that $${\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1} {1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[-\max_{t\in[0,1]}a(t)(R_1 h(R_1)+\int_0^{R_1} \omega(s)ds)])}}>1.\label{e4.2}$$ From the continuity of $I^{-1}$ and $h$, we can choose $\epsilon>0$ and $\epsilon 0$, $x_0(t)>0,t\in[0,1]$. Differentiating \eqref{e4.5}, we have \begin{gathered} x''_0(t)+\mu_0 a(t)f(t,x_0(t)+\frac{1}{n},x_0'(t) -{\frac{1}{n}})=0,\quad 0R_2$, for all$t\in [0,1]$,$x_0(t)+\frac{1}{n}>R_2, from \eqref{e3.11}, \begin{align*} \gamma x_0(t) &\geq \gamma A_{n}x_0(t)\\ &=\gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)| -\frac{1}{n})ds\\ &\quad+\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\ &-\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s) +\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds \Big)\Big)\\ &\geq \gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)| -\frac{1}{n})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s) +\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds \Big)\Big) \\ &\geq \gamma\Big(\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{t}(t-s) a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\ &\quad -\frac{\sum_{i=1}^{m-2} a_i\int_{0}^{1} (\xi_i-s)a(s)f\big(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n}\big)ds} {1-\sum_{i=1}^{m-2}a_i}\Big)\\ &=\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\ &\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)g_1(x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\ &\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)dsN^{*} x_0(s)\\ &\geq a^*\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)dsN^{*}\frac{R_2}{a^*} >\frac{R_2}{a^*}. \end{align*} Then\|x_0\|\geq \gamma\|x_0\|_1>\frac{R_2}{a^*}$, which is a contradiction to$x_0\in \partial \Omega_2 \cap P$. Then \eqref{e4.11} holds. From Lemma \ref{lem2.2}, $$i(A_n,\Omega_2\cap P,P)=0. \label{e4.12}$$ This equality and \eqref{e4.10} guarantee, $$i(A_n,(\Omega_2-\bar{\Omega}_1)\cap P,P)=-1. \label{e4.13}$$ From this equality and \eqref{e4.10},$A_n$has two fixed points with$x_{n,1}\in \Omega_1\cap P,x_{n,2}\in (\Omega_2-\overline{\Omega}_1)\cap P$. For each$n\in N_0$, there exists$x_{n,1}\in\Omega_1\cap P$with$x_{n,1}=A_nx_{n,1}; that is, \begin{aligned} x_{n,1}(t) &=-\int_{0}^{t}(t-s)a(s)f(s,x_{n,1}(s)+\frac{1}{n},-|x'_{n,1}(s)|-\frac{1}{n})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n,1}(s)+\frac{1}{n},-|x'_{n,1}(s)| -\frac{1}{n})ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n,1}(s) +\frac{1}{n},-|x'_{n,1}(s)|-\frac{1}{n})ds \Big). \end{aligned}\label{e4.14} As in the same proof of \eqref{e3.5}, x'_n(t)\leq 0$,$t \in (0,1)$and $$x'_{n,1}(t)=-\int_0^t a(s)f(s,x_{n,1}(s)+\frac{1}{n},x_{n,1}'(s) -{\frac{1}{n}})ds,\quad n\in N_0,\; t\in(0,1).$$ Now we consider$\{x_{n,1}(t)\}_{ n\in N_0}$and$\{x' _{n,1}(t)\}_{n\in N_0}$, since$\| x_{n,1}\|\leq R_1$, it follows that \begin{gather} \text{$\{x _{n,1}(t)\}$is uniformly bounded on$[0,1]$,} \label{e4.15}\\ \text{$\{x' _{n,1}(t)\}$is uniformly bounded on$[0,1]$.} \label{e4.16} \end{gather} Then $$\text{\{x _{n,1}(t)\} is equicontinuous on [0,1].} \label{e4.17}$$ As in the same proof of \eqref{e3.6}, \begin{gathered} x''_{n,1}(t)+a(t)f(t,x_{n,1}(t)+\frac{1}{n},x_{n,1}'(t)-{\frac{1}{n}})=0, \quad 00$, there exists $\epsilon'>0$ such that $$|I^{-1}(s_1)-I^{-1}(s_2)|<\bar\epsilon, \forall\ |s_1-s_2|<\epsilon', s_1, s_2\in[0,I(-R_1-\epsilon)].\label{e4.22}$$ Then \eqref{e4.19} guarantees that for $\epsilon'>0$, there exists $\delta'>0$ such that $$|I( x'_{n,1}(t_2)-{\frac{1}{n}})-I( x'_{n,1}(t_1)-{\frac{1}{n}})|<\epsilon', \quad \forall |t_1-t_2|<\delta',t_1,t_2\in[0,1]. \label{e4.23}$$ From this inequality and \eqref{e4.22}, \begin{aligned} | x'_{n,1}(t_2)- x'_{n,1}(t_1)| &=| x'_{n,1}(t_2)-{\frac{1}{n}}-( x'_{n,1}(t_1)-{\frac{1}{n}})|\\ &=|I^{-1}(I( x'_{n,1}(t_2)-{\frac{1}{n}}))-I^{-1}(I( x'_{n,1}(t_1)-{\frac{1}{n}}))|\\ &< \bar\epsilon, \quad \forall |t_1-t_2|<\delta',t_1,t_2\in[0,1]; \end{aligned}\label{e4.24} that is, $$\text{\{x' _{n,1}(t)\} is equicontinuous on [0,1].}\label{e4.25}$$ From \eqref{e4.15}--\eqref{e4.17}, \eqref{e4.25} and the Arzela-Ascoli Theorem, $\{x_{n,1}(t)\}$ and $\{x'_{n,1}(t)\}$ are relatively compact on $C[0,1]$, which implies, there exists a subsequence $\{x_{n_j,1}\}$ of $\{x_{n,1}\}$ and function $x_{0,1}(t)\in C[0,1]$ such that $$\lim_{j\to +\infty}\max_{t\in[0,1]}|x_{n_j,1}(t)-x_{0,1}(t)|=0, \quad \lim_{j\to +\infty}\max_{t\in[0,1]}|x'_{n_j,1}(t)-x'_{0,1}(t)|=0.$$ Since $x'_{n_j,1}(0)=0$, $x_{n_j,1}(1)=\sum_{i=1}^{m-2}a_i x_{n_j,1}(\xi_i)$, $x'_{n_j,1}(t)<0$, $x_{n_j,1}(t)>0$, $t\in(0,1)$, $j\in\{1,2,\dots\}$, $$x'_{0,1}(0)=0, x_{0,1}(1)=\sum_{i=1}^{m-2}a_i x_{0,1}(\xi_i), x'_{0,1}(t)\leq 0, x_{0,1}(t)\geq 0,\quad t\in(0,1).\label{e4.26}$$ For $(t,x_{n_j,1}(t)+{\frac{1}{n_j}},x_{n_j,1}'(t)-{\frac{1}{n_j}})\in [0,1]\times [0,R_1+\epsilon]\times (-\infty,0)$, from (P5) there exists a function $\Psi_{R_1}\in C([0,1],\mathbb{R}_+)$ such that $$f(t,x_{n_j,1}(t)+{\frac{1}{n_j}},x_{n_j,1}'(t)-{\frac{1}{n_j}})ds \geq \Psi_{R_1}(t).$$ Then, for $n\in N_0$, \begin{align*} x_{n_j,1}(t) &=-\int_{0}^{t}(t-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s) -\frac{1}{n_j})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds \Big)\\ &\geq -\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s) -\frac{1}{n_j})ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} (\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds )\\ &\geq \frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\quad -\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}(\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &=\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds\\ &\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)\Psi_{R_1}(s) ds =a_0, \end{align*} and \begin{align*} x_{n_j,1}'(t) &=-\int_{0}^{t}a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s) -\frac{1}{n_j})ds\\ &\leq -\int_{0}^{t}a(s)\Psi_{R_1+\epsilon}(s) ds,\quad t\in[0,1],\;n\in N_0. \end{align*} Thus, \begin{gather*} \inf_{j\geq 1}\min\{\min_{s\in[\frac{1}{2},t]}x_{n_j,1}(s), \min_{s\in[\frac{1}{2},t]}|x'_{n_j,1}(s)|\}>0,\quad t\in[\frac{1}{2},1),\\ \inf_{j\geq 1}\min\{\min_{s\in[t,\frac{1}{2}]}x_{n_j,1}(s), \min_{s\in[t,\frac{1}{2}]}|x'_{n_j,1}(s)|\} >0, \quad t\in(0,\frac{1}{2}]. \end{gather*} From (P2), we have \begin{align*} &f\big(t,x_{n_j,1}(t)+{\frac{1}{n_j}},x_{n_j,1}'(t)-{\frac{1}{n_j}}\big)\\ &\leq [h(x_{n_j,1}(t)+{\frac{1}{n_j}})+\omega(x_{n_j,1}(t) +{\frac{1}{n_j}})][g(x_{n_j,1}'(t)-{\frac{1}{n_j}})+r(x_{n_j,1}'(t) -{\frac{1}{n_j}})]\\ &\leq [h(\frac{R_1}{\gamma}+\epsilon)+\omega(a_0)] [r(-\int_{0}^{t}a(s)\Psi_{R_1+\epsilon}(s)ds ) +g(-\frac{R_1}{\gamma\delta}-\epsilon)]. \end{align*} and since $$x_{n_j,1}'(t)-x_{n_j,1}'({\frac{1}{2}}) =- \int_{1/2}^{t}a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x_{n_j,1}'(s) -{\frac{1}{n_j}})ds,\quad t\in(0,1),$$ letting $j\to +\infty$, the Lebesgue Dominated Convergence Theorem guarantees that $$x_{0,1}'(t)-x_{0,1}'({\frac{1}{2}})= - \int_{1/2}^{t} a(s)f(s,x_{0,1}(s),x_{0,1}'(s))ds,t\in(0,1).\label{e4.27}$$ Differentiating, we have x''_{0,1}(t)+a(t)f(t,x_{0,1}(t),x'_{0,1}(t))=0,\quad 01, 00. If $$\mu<\sup_{c\in \mathbb{R}_+}\frac{I(-\frac{c(1-\sum_{i=1}^{m-2}a_i)} {\sum_{i=1}^{m-2}a_i\xi_i+1})} {\max_{t\in [0,1]}a(t)(c+c^{1-d}+ \frac{c^{1+b}}{1+b})}. \label{e4.28}$$ Then equation \eqref{e1.1} has at least two positive solutions x_{0,1},x_{0,2}\in C^1[0,1]\cap C^2(0,1). We apply Theorem \ref{thm4.1} with g(z)=1, r(z)=(-z)^{-a}, h(u)=\mu u^{-d}, \omega(u)=\mu(1+u^b), \Psi(t)=\mu, g_1(u,z)=\mu u^b. Note that (P1), (P2), (P4) and (P5) hold, and that \begin{align*} &\sup_{c\in \mathbb{R}_+}{\frac{c }{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}I^{-1}[\max_{t\in [0,1]}a(t)(ch(c)+\int_0^c \omega(s)ds)]}}\\ &=\sup_{c\in \mathbb{R}_+}{\frac{c }{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}I^{-1} [\mu\max_{t\in [0,1]}a(t)(c+c^{1-d}+ \frac{c^{1+b}}{1+b})]}} \end{align*} Then \eqref{e4.28} guarantees that (P3) holds. \section{Singularities at x=0 but not x'=0} In this section the nonlinearity f may be singular at x=0, but not at x'=0. We assume that the following conditions hold. \begin{itemize} \item[(S1)] a(t)\in C[0,1], a(t)>0, t\in(0,1); \item[(S2)] f(t,u,z)\leq [h(u)+\omega(u)]r(z), where f\in C([0,1]\times \mathbb{R}_+\times \mathbb{R}_-,\mathbb{R}_+), \omega(u)>0 continuous and non-increasing on [0,+\infty), h(u)\geq 0 continuous and nondecreasing on \mathbb{R}_+, r(z)>0 continuous and nondecreasing on \mathbb{R}_-; \item[(S3)] \sup_{c\in \mathbb{R}_+}{\frac{c }{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[-\max_{t\in [0,1]}a(t)(ch(c)+\int_0^c \omega(s)ds)])}}>1, where I(z)={\int_z^0{\frac{udu}{r(u)}}}, z\in \mathbb{R}_-, \int_{0}^{a}\omega(s)ds<+\infty, a\in \mathbb{R}_+; \item[(S4)] There exists a function g_1\in C([0,+\infty)\times (-\infty,0],[0,+\infty)), such that f(t,u,z)\geq g_1(u,z), \forall (t,u,z)\in [0,1] \times \mathbb{R}_+\times \mathbb{R}_-, and \lim_{u\to +\infty}\frac{g_1(u,z)}{u}=+\infty, uniformly for z\in \mathbb{R}_-. \item[(S5)] There exists a function \Psi_H\in C([0,1],\mathbb{R}_+) and a constant 0\leq \delta such that f(t,u,z)\geq \Psi_H(t)(-z)^\delta, for all (t,u,z)\in [0,1]\times[0,H]\times[-H,0). \end{itemize} For n\in\{1,2,\dots\}, x\in P, define operator \begin{aligned} (A_{n}x)(t) &=-\int_{0}^{t}(t-s)a(s)f(s,x(s)+\frac{1}{n},-|x'(s)|)ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x(s)+\frac{1}{n},-|x'(s)|)ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x(s) +\frac{1}{n},-|x'(s)|)ds\Big),\quad t\in [0,1]. \end{aligned} \label{e5.1} \begin{theorem} \label{thm5.1} Suppose {\rm (S1)--(S5)} hold. Then \eqref{e1.1} has at least two positive solutions  x_{0,1}, x_{0,2} in C^1[0,1]\cap C^2(0,1) with x_{0,1}(t),x_{0,2}(t)>0, t \in [0,1]. \end{theorem} \begin{proof} Choose R_1>0 such that $${\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1} {1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[-\max_{t\in[0,1]}a(t)(R_1 h(R_1) +\int_0^{R_1} \omega(s)ds)])}}>1.\label{e5.2}$$ From the continuity of  I^{-1} and h, we can choose \epsilon>0 and \epsilon1. \label{e5.3} Let n_0\in\{1,2,\dots\} with {\frac{1}{n_0}} < \min\{\epsilon,{\delta/2} \} and let N_0 =\{n_0,n_0+1, \dots\}. Then Lemma \ref{lem2.4} guarantees that for n\in N_0, A_n:P\to P is a completely continuous operator. Let \Omega_1=\{x\in C^1[0,1]:\| x \| 0$,$x_0(t)>0,t\in[0,1]$. Differentiating \eqref{e5.5}, we have \begin{gathered} x''_0(t)+\mu_0 a(t)f(t,x_0(t)+\frac{1}{n},\quad x_0'(t))=0,\;0R_2$ for all $t\in[0,1]$. Then $x_0(t)+\frac{1}{n}>R_2$. From \eqref{e3.11}, \begin{align*} \gamma x_0(t) &\geq \gamma A_{n}x_0(t)\\ &=\gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\ &-\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds \Big)\Big)\\ &\geq \gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\ &-\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds \Big)\Big)\\ &\geq \gamma\Big(\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\ &\quad -\frac{\sum_{i=1}^{m-2}a_i\int_{0}^{1}(\xi_i-s)a(s)f(s,x_0(s) +\frac{1}{n},x_0'(s)}{1-\sum_{i=1}^{m-2}a_i})\Big)\\ &=\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\ &\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)g_1(x_0(s)+\frac{1}{n},x_0'(s))ds\\ &\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)dsN^{*} x_0(s)\\ &\geq a^*\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2} a_i}\int_{0}^{1}a(s)dsN^{*}\frac{R_2}{a^*} >\frac{R_2}{a^*}; \end{align*} that is, $\|x_0\|\geq \gamma\|x_0\|_1>\frac{R_2}{a^*}$, which is a contradiction to $x_0\in \partial \Omega_2 \cap P$. Then \eqref{e5.11} holds. From Lemma \ref{lem2.2}, $$i(A_n,\Omega_2\cap P,P)=0. \label{e5.12}$$ This equality and (5.10) guarantee $$i(A_n,(\Omega_2-\bar{\Omega}_1\cap P,P)=-1. \label{e5.13}$$ From this equality and (5.10), $A_n$ has two fixed points with $x_{n,1}\in \Omega_1\cap P,x_{n,2}\in (\Omega_2-\bar{\Omega}_1)\cap P$. For each $n\in N_0$, there exists $x_{n,1}\in\Omega_1\cap P$ such that $x_{n,1}=A_nx_{n,1}$; that is, for $t\in [0,1]$, \begin{aligned} x_{n,1}(t) &=-\int_{0}^{t}(t-s)a(s)f(s,x_{n,1}(s)+\frac{1}{n},-|x'_{n,1}(s)|)ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n,1}(s)+\frac{1}{n},-|x'_{n,1}(s)|)ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n,1}(s) +\frac{1}{n},-|x'_{n,1}(s)|)ds\Big). \end{aligned}\label{e5.14} As in the proof of \eqref{e3.5}, $x'_n(t)\leq 0$, $t \in (0,1)$ and $$x'_{n,1}(t)=-\int_0^t a(s)f(s,x_{n,1}(s)+\frac{1}{n},x_{n,1}'(s))ds,\quad n\in N_0,\; t\in(0,1).$$ Now we consider $\{x_{n,1}(t)\}_{ n\in N_0}$ and $\{x' _{n,1}(t)\}_{n\in N_0}$, since $\| x_{n,1}\|\leq R_1$, \begin{gather} \text{$\{x _{n,1}(t)\}$ is uniformly bounded on $[0,1]$,} \label{e5.15}\\ \text{$\{x' _{n,1}(t)\}$ is uniformly bounded on $[0,1]$.} \label{e5.16} \end{gather} Then $$\text{\{x _{n,1}(t)\} is equicontinuous on [0,1].}\label{e5.17}$$ As in the proof of \eqref{e3.6}, \begin{gathered} x''_{n,1}(t)+a(t)f(t,x_{n,1}(t)+\frac{1}{n},x_{n,1}'(t))=0,00$, there exists$\epsilon'>0$such that $$|I^{-1}(s_1)-I^{-1}(s_2)|<\bar\epsilon, \quad \forall |s_1-s_2|<\epsilon',\; s_1, s_2\in[0,I(-R_1-\epsilon)].\label{e5.22}$$ Also \eqref{e5.19} guarantees that for$\epsilon'>0$, there exists$\delta'>0$such that $$|I( x'_{n,1}(t_2))-I( x'_{n,1}(t_1))|<\epsilon', \quad \forall\ |t_1-t_2|<\delta',\; t_1,t_2\in[0,1].\label{e5.23}$$ From this inequality and \eqref{e5.22}, $$\begin{gathered} | x'_{n,1}(t_2)- x'_{n,1}(t_1)| =|I^{-1}(I( x'_{n,1}(t_2)))-I^{-1}(I( x'_{n,1}(t_1)))|< \bar\epsilon, \\ \forall |t_1-t_2|<\delta',\;t_1,t_2\in[0,1]; \end{gathered} \label{e5.24}$$ that is, $$\text{\{x' _{n,1}(t)\} is equi-continuous on [0,1].}\label{e5.25}$$ From \eqref{e5.15}--\eqref{e5.17}, \eqref{e5.25} and the Arzela-Ascoli Theorem,$\{x _{n,1}(t)\}$and$\{x'_{n,1}(t)\}$are relatively compact on$C^1[0,1]$. This implies, there exists a subsequence$\{x_{n_j,1}\}$of$\{x_{n,1}\}$and function$x_{0,1}(t)\in C^1[0,1]$such that $$\lim_{j\to +\infty}\max_{t\in[0,1]}|x_{n_j,1}(t)-x_{0,1}(t)|=0, \quad \lim_{j\to +\infty}\max_{t\in[0,1]}|x'_{n_j,1}(t)-x'_{0,1}(t)|=0.$$ Since$x'_{n_j,1}(0)=0$,$x_{n_j,1}(1)=\sum_{i=1}^{m-2}a_i x_{n_j,1}(\xi_i)$,$x'_{n_j,1}(t)<0$,$x_{n_j,1}(t)>0$,$t\in(0,1)$,$j\in\{1,2,\dots\}$, $$x'_{0,1}(0)=0, x_{0,1}(1)=\sum_{i=1}^{m-2}a_i x_{0,1}(\xi_i), x'_{0,1}(t)\leq 0, x_{0,1}(t)\geq 0, \;t\in(0,1).\label{e5.26}$$ For$(t,x_{n_j,1}(t)+{\frac{1}{n_j}}$,$x_{n_j,1}'(t))\in [0,1]\times [0,R_1+\epsilon]\times (-\infty,0)$, from (S5) there exists a function$\Psi_{R_1}\in C([0,1],\mathbb{R}_+)$such that $$f(t,x_{n_j,1}(t)+{\frac{1}{n_j}},x_{n_j,1}'(t))ds\geq \Psi_{R_1}(t)(-x_{n_j,1}'(t))^{\delta},\quad 0\leq \delta<1.$$ Then, for$n\in N_0, \begin{align*} x_{n_j,1}'(t) &=-\int_{0}^{t}a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\ &\leq -\int_{0}^{t}a(s)\Psi_{R_1+\epsilon}(s)(-x'_{n_j,1}(s))^\delta ds,\quad t\in[0,1],\; n\in N_0, \end{align*} which implies $$x_{n_j,1}'(t)\leq -(1-\delta)(\int_{0}^{t}a(s) \Psi_{R_1+\epsilon}(s)ds)^{\frac{1}{1-\delta}} ,$$ and \begin{align*} x_{n_j,1}(t) &=-\int_{0}^{t}(t-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} (\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds )\\ &\geq -\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\ &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} (\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\ &\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds )\\ &\geq \frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1} (1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\ &\quad -\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1} (\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\ &=\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\ &\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)\Psi_{R_1+\epsilon}(s)(-x'_{n_j,1}(s))^\delta ds\\ &\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i} \int_{0}^{1}a(s)\Psi_{R_1+\epsilon}(s) ((1-\delta)(\int_{0}^{s}a(\tau)\Psi_{R_1+\epsilon}(\tau)d\tau) ^{\frac{1}{1-\delta}})^\delta ds\\ &=:F,\quad t\in [0,1]. \end{align*} Since \begin{align*} &f(t,x_{n_j,1}(t)+{\frac{1}{n_j}},x_{n_j,1}'(t))\\ &\leq [h(x_{n_j,1}(t)+{\frac{1}{n_j}}) +\omega(x_{n_j,1}(t)+{\frac{1}{n_j}})]r(x_{n_j,1}'(t))\\ &\leq [h(\frac{R_1}{\gamma}+\epsilon)+\omega(F)]r((\delta-1) \Big(\int_{0}^{t}a(s)\Psi_{R_1+\epsilon}(s)ds\Big)^{\frac{1}{1-\delta}})), \end{align*} and $$x_{n_j,1}'(t)-x_{n_j,1}'({\frac{1}{2}}) =-\int_{1/2}^{t} a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x_{n_j,1}'(s))ds,\ \quad t\in(0,1),$$ lettingj\to +\infty$, the Lebesgue Dominated Convergence Theorem guarantees that $$x_{0,1}'(t)-x_{0,1}'({\frac{1}{2}})= - \int_{1/2}^{t} a(s)f(s,x_{0,1}(s),x_{0,1}'(s))ds,\ t\in(0,1).\label{e5.27}$$ Differentiating, we have$$x''_{0,1}(t)+a(t)f(t,x_{0,1}(t),x'_{0,1}(t))=0,\quad 01,00$. 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