\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 74, pp. 1--25.\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 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/74\hfil Multiple positive solutions] {Multiple positive solutions for singular boundary-value problems with derivative dependence on finite and infinite intervals} \author[B. Yan\hfil EJDE-2006/74\hfilneg] {Baoqiang Yan} \address{Baoqiang Yan \newline Department of Mathematics, Shandong Normal University, Ji-nan, 250014 China\newline and School of Mathematics, Trinity College, Dublin University, Dublin 2, Ireland} \email{bqyan819@beelink.com} \date{} \thanks{Submitted March 16, 2006. Published July 10, 2006.} \thanks{Supported by grants 10571111 from the National Natural Science and Y2005A07 from the \hfill\break\indent natural science of Shandong Province} \subjclass[2000]{34B16, 34B18, 34B40} \keywords{Krasnoselskii's theorem; singular boundary value problems; \hfill\break\indent fixed point theorem of cone expansion and compression} \begin{abstract} In this paper, Krasnoselskii's theorem and the fixed point theorem of cone expansion and compression are improved. Using the results obtained, we establish the existence of multiple positive solutions for the singular second-order boundary-value problems with derivative dependance on finite and infinite intervals. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In \cite{a1}, by an alternative method to Leray-Schauder and sequential technique, Agarwal and O'Regan considered the singular boundary-value problem $$\begin{gathered} \frac{1}{p}(py')'+\Phi(t)f(t,y,py')=0, \quad t\in(0,1)\\ \alpha y(0)-\beta\lim_{t\to 0^+}p(t)y'(t)=0,\quad y(1)=0 \end{gathered} \label{e1.1}$$ and obtained the existence of one solution to equation \eqref{e1.1} when $\alpha=0$ or $\beta=0$. In \cite{p1}, by a generalization of the Kneser's property (continuum) of the cross-sections of the solutions funnel, Palamides and Galanis considered the following problems $$\begin{gathered} \frac{1}{p}(py')'+\Phi(t)f(t,y,py')=0, \quad t\in(0,+\infty)\\ y(0)=0,\quad \lim_{t\to +\infty}p(t)y'(t)=0 \end{gathered} \label{e1.2}$$ and also obtained the existence of one positive and monotone unbounded solution. There are some other results on the existence of at least one solution for equation \eqref{e1.1}, \eqref{e1.2}, and we refer the reader also to \cite{a2,a3,b1,b2,b3,b4,c1,g1,o1,o2,o3,o4}. Moreover, under the condition that $p\equiv 1$, $\beta=0$ and $f$ has no singularity at $x=0$ and $px'=0$, in \cite{h1}, using pairs of lower and upper solutions, Henderson and Thompson considered the existence of three solutions for equation \eqref{e1.1} and in \cite{y1}, by reducing the equation \eqref{e1.1} to a quasi-linear one, I.Yermachenko and F.Sadyrbaev obtained the existence of multiple solutions for equation \eqref{e1.1} also. Up to now, there are fewer results on the existence of multiple positive solutions to equation \eqref{e1.1}, \eqref{e1.2} if $f(t,x,px')$ is singular at $x$ and is related to $px'$. Motivated by this, in this paper, we discuss the existence of multiple positive solutions to equation \eqref{e1.1}, \eqref{e1.2} when $f(t,x,px')$ is singular at $x=0$. There are three sections in our paper. In section 2, in order to overcome the difficulty from $px'$, we improve the Krasnoselskii's theorem and fixed point theorem of cone expansion and compression on unbounded set in a Banach space with a special norm. In section 3, we establish special cones, and using obtained theorems, present the existence of multiple positive solutions to equation \eqref{e1.1}. In section 4, we consider the existence of multiple positive solutions to equation \eqref{e1.2}. \section{The improvement of the Krasnoselskii's theorem and fixed point theorem of cone expansion and compression} In this section, we improve the Krasnoselskii's theorem and fixed point theorem of cone expansion and compression in a Banach space with a special norm. In \cite{g2}, Granas and Dugudji presented the theory of fixed point index on unbounded open sets which has same basic properties as those in the theory of fixed point index on bounded open sets \cite{g4}. The degree theory on bounded open sets and unbounded open sets can be found in \cite{d1,g2,g3,k1,l1,z1}. According to the the theory of fixed point index on unbounded open sets in Chapter 4 of \cite{g2}, it is easy to obtain following result. Let $E$ be a real Banach space containing a cone $P$. \begin{lemma} \label{lem2.1} Assume $\Omega\subseteq E$, $\theta\in\Omega$, $\Omega\cap P$ is a relatively open set in $P$. Let $A:P\cap\overline{\Omega}\to P$ be continuous with relatively compact $A(P\cap\overline{\Omega})$. Suppose that $$Ax\neq \mu x,\quad \forall x\in P\cap\partial\Omega,\; \mu\geq 1.\label{e2.1}$$ Then $i(A,P\cap\Omega,P)=1$. \end{lemma} \begin{proof} Let $H(t,x)=tAx$, $t\in[0,1]$ and $x\in P\cap\partial\Omega$. Then $H:[0,1]\times(P\cap\overline{\Omega})\to P$ is continuous, and the continuity of $H(t,x)$ in $t$ is uniform with respect to $x\in P\cap\overline{\Omega}$. Moreover, $H(t,P\cap\overline{\Omega})$ is relatively compact for every $t\in[0,1]$. Evidently, $H(t,x)\neq x$ for $x\in P\cap\partial\Omega$ and $0\leq t\leq 1$. Hence, by the homotopy invariance and normality of fixed point index, we have $$i(A,P\cap\Omega,P)=i(\theta,P\cap\Omega,P)=1.$$ The proof is complete. \end{proof} Now we consider a real Banach space in a special case. Assume that $E$ is a linear space and it satisfies three conditions: \begin{enumerate} \item There is a norm $x\to \|x\|_1$ on $x\in E$ and under $\|\cdot\|_1$, $E$ is a normed linear space (not complete) \item There is another semi-norm $\|\cdot\|_2$ \item Under $\|x\|=\max\{\|x\|_1,\|x\|_2\}$, $E$ is a Banach space. \end{enumerate} For example, for $x\in C^1([0,1],R)$, under $\|x\|_1=\max_{t\in[0,1]}|x(t)|$, $C^1([0,1],R)$ is an incomplete normed linear space. Let $\|x\|_2=\max_{t\in[0,1]}|x'(t)|$. Obviously, $\|\cdot\|_2$ is a semi-norm of $C^1([0,1],R)$. If we define $\|x\|=\max\{\|x\|_1,\|x\|_2\}$, $C^1([0,1],R)$ is a Banach space. Assume that $P$ is a cone of $E$ and $\Omega\subset E$ is a open set with $\sup_{x\in\overline{\Omega}}\|x\|_1<+\infty$. Since $\sup_{x\in\overline{\Omega}}\|x\|_1\leq \sup_{x\in\overline{\Omega}}\|x\|$, it is possible that $\Omega$ is unbounded in $E$. We have the following lemma(the ideas coming from \cite{g4}). \begin{lemma} \label{lem2.2} With $E$ and $P$ as above, assume that $\Omega\subseteq E$ is an open set with $\sup_{x\in\overline{\Omega}}\|x\|_1<+\infty$. Let $A:P\cap \overline{\Omega}\to P$ be continuous with relatively compact $A(P\cap\overline{\Omega})$ and $B:P\cap\partial\Omega\to P$ be continuous with relatively compact $B(P\cap\partial {\Omega})$. Suppose that \begin{itemize} \item[(a)] $\inf_{x\in P\cap\partial\Omega}\|Bx\|_1>0$; \item[(b)] $x-Ax\neq tBx$, for all $x\in P\cap\partial\Omega$, $t\geq 0$. \end{itemize} Then, we have $$i(A,P\cap \Omega,P)=0.\label{e2.2}$$ \end{lemma} \begin{proof} Suppose that the $E_1$ is a Banach space completion of $E$ under norm $\|x\|_1$. By the extension theorem of Dugundji \cite{d2}, we can extend $B$ to a continuous operator from $P\cap\overline{\Omega}$ into $P$ such that $$B(P\cap\overline{\Omega})\subseteq \overline{{\rm co}} B(P\cap\partial(\Omega)) \subseteq (\overline{{\rm co}}B(P\cap \partial\Omega))_1,\label{e2.3}$$ where $(\overline{{\rm co}}B(P\cap \partial\Omega))_1$ is the closure of $B(P\cap \partial\Omega)$ under the norm $\|\cdot\|_1$ and the followings are similar. Let $F=B(P\cap\partial\Omega)$, then $(\overline{{\rm co}}B(P\cap\partial\Omega))_1=(\overline{{\rm co}}F)_1 =(\overline{M})_1$, where $$M=\{y=\sum_{i=1}^n\lambda_iy_i : y_i\in F,\lambda_i\geq 0,\sum_{i=1}^n\lambda_i=1; n=1,2,\dots\}.$$ We first prove $$\inf_{y\in(\overline{M})_1}\|y\|_1>0.\label{e2.4}$$ Denote by $E_0$ the subspace of $E$ spanned by $F$ under norm $\|\cdot\|_1$. Since $B(P\cap\partial\Omega)$ is relatively compact in $E$ under norm $\|\cdot\|$, we know that $B(P\cap\partial\Omega)$ is relatively compact in $E_1$ under norm $\|\cdot\|_1$. Therefore, $E_0$ is separable. Evidently, $P_0=P\cap E_0$ is a cone of $E_0$ and $F\subseteq P_0$. By property of the cone \cite[Theorem 1.4.1]{g4}, there exists $f_0\in E_0^*$ such that $f_0(y)>0$ for any $y\in P_0$ with $y\neq \theta$. We claim that $$\inf_{y\in F}f_0(y)=\sigma>0.\label{e2.5}$$ In fact, if $\sigma=0$, then there exists $\{y_k\}\subseteq F$ such that $f_0(y_k)\to 0$. By the relative compactness of $F$ in $E$, there is a subsequence $\{y_{k_i}\}$ of $\{y_k\}$ such that $y_{k_i}\to y_0\in P$ and $y_0\in E_0$. Then $y_0\in P_0$, and so $f_0(y_{k_i})\to f_0(y_0)=0$. Hence, $y_0=\theta$ and $\|y_{k_i}\|_1\to 0$, which contradicts hypothesis (a). Thus, \eqref{e2.5} holds. For any $y=\sum_{i=1}^{n}\lambda_iy_i\in M$, where $y_i\in F$, $\lambda_i\geq 0$ and $\sum_{i=1}^n\lambda_i=1$, we have $$f_0(y)=\sum_{i=1}^n\lambda_if_0(y_i)\geq \sum_{i=1}^n\lambda_i\sigma=\sigma,$$ and therefore $$f_0(y)\geq\sigma, \forall y\in(\overline{M})_1.\label{e2.6}$$ Since $(\overline{M})_1=(\overline{{\rm co}}F)_1$ is compact, there exists a $z_0\in(\overline{M})_1$ such that $$\inf_{y\in(\overline{M})_1}\|y\|_1=\|z_0\|_1.\label{e2.7}$$ By \eqref{e2.6}, $f_0(z_0)\geq\sigma$, and this implies that $z_0\neq \theta$. It follows therefore from \eqref{e2.7} that \eqref{e2.4} holds. By \eqref{e2.3} and \eqref{e2.4}, we get $$\inf_{x\in P\cap{\overline{\Omega}}}\|Bx\|_1=\sigma>0.\label{e2.8}$$ Now, it is easy to show that \eqref{e2.2} holds. In fact, if $i(A,P\cap\Omega,P)\neq 0$, then by the hypothesis (b) and the homotopy invariance property of fixed point index, we have $$i(A+tB,P\cap\Omega,P)=i(A,P\cap\Omega,P)\neq 0,\forall t>0.$$ In particular, choosing $t_0>\frac{a+c}{\sigma}$, where $a=\sup_{x\in\overline{\Omega}}\|x\|_1$ and $c=\sup_{x\in P\cap\overline{\Omega}}\|Ax\|_1$, we have $$i(A+t_0B, P\cap\Omega,P)\neq 0,$$ and so, by the solution property of fixed point index, there exists an $x_0\in P\cap\Omega$ such that $Ax_0+t_0Bx_0=x_0$. Hence $$t_0=\frac{\|x_0-Ax_0\|_1}{\|Bx_0\|_1}\leq\frac{a+c}{\sigma},$$ which is a contradiction. The proof is complete. \end{proof} \begin{corollary} \label{coro2.1} Assume that $\Omega$ is an open set with $\sup_{x\in\overline{\Omega}}\|x\|_1<+\infty$. Let $A:P\cap\overline{\Omega}\to P$ be continuous with relatively compact $A(P\cap\overline{\Omega})$. If there exists $u_0>\theta$ such that $$x-Ax\neq t u_0, \forall x\in P\cap\partial\Omega, t\geq 0,\label{e2.9}$$ then \eqref{e2.2} holds. \end{corollary} \begin{proof} Since $\|\cdot\|_1$ is a norm, $u_0>\theta$ implies that $\|u_0\|_1>0$. Hence, the corollary follows directly from Lemma \ref{lem2.2} by putting $Bx=u_0$ for any $x\in P\cap\partial\Omega$. \end{proof} \begin{corollary} \label{coro2.2} Assume that $\Omega$ is an open set with $\sup_{x\in\overline{\Omega}}\|x\|_1<+\infty$. Let $A:P\cap\overline{\Omega}\to P$ be continuous with relatively compact $A(P\cap\overline{\Omega})$. If $$Ax\not\leq x, \quad \forall x\in P\cap\partial\Omega ,\label{e2.10}$$ then \eqref{e2.2} holds. \end{corollary} \begin{proof} Choose an $u_0>\theta$. Then, $$x-Ax\neq t u_0, \forall x\in P\cap\partial\Omega, \ t\geq 0.$$ By Corollary \ref{coro2.1}, \eqref{e2.2} holds. \end{proof} \begin{lemma} \label{lem2.3} Assume that $\Omega$ is an open set with $\sup_{x\in\overline{\Omega}}\|x\|_1<+\infty$. Let $A:P\cap\overline{\Omega}\to P$ be continuous with relatively compact $A(P\cap\overline{\Omega})$. Suppose that \begin{itemize} \item[(i)] $\inf_{x\in P\cap\partial\Omega}\|Ax\|_1>0$; and \item[(ii)] $Ax\neq \mu x,\forall x\in P\cap\partial\Omega,0<\mu<1$. \end{itemize} Then \eqref{e2.2} holds. \end{lemma} \begin{proof} Taking $B=A$ in Lemma \ref{lem2.2}, we see that condition (a) of Lemma \ref{lem2.2} is the same as condition $(i)$ of Lemma \ref{lem2.3}. Also, condition (b) of Lemma \ref{lem2.2} is true. In fact, if there exist $x_0\in P\cap\partial\Omega$ and $t_0\geq 0$ such that $x_0-Ax_0=t_0Ax_0$, then $Ax_0=\mu x_0$, where $\mu_0=(1+t_0)^{-1}$. Evidently $0<\mu_0\leq 1$, which contradicts the condition $(ii)$. Thus, \eqref{e2.2} follows from Lemma \ref{lem2.2}. \end{proof} \begin{lemma} \label{lem2.4} Assume that $\Omega$ is an open set with $\sup_{x\in\overline{\Omega}}\|x\|_1<+\infty$. Let $A:P\cap\overline{\Omega}\to P$ be continuous with relatively compact $A(P\cap\overline{\Omega})$. Suppose that \begin{itemize} \item[(i')] $Ax\neq \mu x, \forall x\in P\cap\partial\Omega, 0\leq\mu\leq 1$, and \item[(ii')] the set $\{\|Ax\|_1^{-1}Ax|x\in P\cap\partial\Omega\}$ is relatively compact. \end{itemize} Then \eqref{e2.2} holds. \end{lemma} \begin{proof} Let $A_1x=\alpha(\|Ax\|_1)^{-1}Ax$ for $x\in P\cap\partial\Omega$, where $\alpha=\sup_{x\in P\cap\partial\Omega}\|Ax\|_1>0$. Then, by hypotheses, $A_1:P\cap\partial\Omega\to P$ is continuous with relatively compact $A_1(P\cap\partial{\Omega})$. By the extension theorem, $A_1$ can be extended to a continuous operator from $P\cap\overline{\Omega}$ into $P$ with relatively compact $A_1(P\cap{\Omega})$. We now prove that $A_1$ satisfies the condition (i) and (ii) of Lemma \ref{lem2.3}. In fact, first we have $$\inf_{x\in P\cap\partial\Omega}\|A_1x\|_1=\sigma>0.$$ Secondly, if there exists $x_0\in P\cap\partial\Omega$ and $0<\mu_0\leq 1$ such that $A_1x_0=\mu_0 x_0$, then $Ax_0=\lambda_0x_0$, where $\lambda_0=\mu_0\alpha^{-1}\|Ax_0\|_1$. Evidently, $0<\lambda_0\leq\mu_0\leq 1$, which contradicts hypothesis (i'). Hence, by Lemma \ref{lem2.3}, we have $$i(A_1,P\cap\Omega,P)=0.\label{e2.11}$$ Now, we prove $$(1-t)Ax+tA_1x\neq x, \forall x\in P\cap\partial\Omega, 0\leq t\leq 1.\label{e2.12}$$ If there is an $x_1\in P\cap\partial\Omega$ and a $0\leq t_1\leq 1$ such that $(1-t_1)Ax_1+t_1A_1x_1=x_1$, then $Ax_1=\mu_1x_1$, where $\mu_1=[1+t_1(\alpha/\|Ax_1\|_1-1)]^{-1}$, $0\leq \mu_1\leq 1$, in contradiction with hypothesis (i'). Hence, by \eqref{e2.11}, \eqref{e2.12} and the homotopy invariance of fixed point index, we get $$i(A,P\cap \Omega,P)=i(A_1,P\cap\Omega,P)=0.$$ The proof is complete. \end{proof} \begin{theorem} \label{thm2.5} Let $\Omega_1$ and $\Omega_2$ be two open in $E$ such that $\theta\in\Omega_1$ and $\overline{\Omega}_1\subseteq\Omega_2$ with $\sup_{t\in\overline{\Omega}_2}\|x\|_1<+\infty$. Let $A:P\cap(\overline{\Omega}_2-\Omega_1)\to P$ be continuous with relatively compact $A(P\cap(\overline{\Omega}_2-\Omega_1))$. Suppose that one of the two conditions \begin{itemize} \item[(H1)] $Ax\not\geq x$, $\forall x\in P\cap\partial\Omega_1$ and $Ax\not\leq x$, $\forall x\in P\cap\partial\Omega_2$, \item[(H2)] $Ax\not\leq x$, $\forall x\in P\cap\partial\Omega_1$ and $Ax\not\geq x$, $\forall x\in P\cap\partial\Omega_2$ \end{itemize} is satisfied. Then $A$ has at least one fixed point in $P\cap(\Omega_2-\overline{\Omega}_1)$. \end{theorem} \begin{proof} By the extension theorem (Dugundji \cite{d2}), $A$ has a completely continuous extension (also noted by $A$) from $P\cap\overline{\Omega}_2$ from $P\cap\overline{\Omega}_2$ to $P$. First we assume that (H1) is satisfied, i.e., it is the case of cone expansion. It is easy to see that $$Ax\neq \mu x,\quad \forall x\in P\cap\partial\Omega_1,\; \mu\geq1,\label{e2.13}$$ since, otherwise, there exists $x_0\in P\cap\partial\Omega_1$ and $\mu_0\geq 1$ such that $Ax_0=\mu_0x_0\geq x_0$, in contradiction with (H1). Now, from \eqref{e2.1} and Lemma \ref{lem2.1}, we obtain $$i(A,P\cap \Omega_1,P)=1.\label{e2.14}$$ On the other hand, by Corollary \ref{coro2.2}, we have $$i(A,P\cap \Omega_2,P)=0.\label{e2.15}$$ It follows therefore from \eqref{e2.14} and \eqref{e2.15} and additivity property of fixed point index that $$i(A,P\cap(\Omega_2-\overline{\Omega}_1),P) =i(A,P\cap\Omega_2,P)-i(A,P\cap\Omega_1,P)=-1\neq 0.\label{e2.16}$$ Hence, by the solution property of fixed point index, $A$ has at least one fixed point in $P\cap(\Omega_2-\overline{\Omega}_1)$. Similarly, when (H2) is satisfied, instead of \eqref{e2.14}, \eqref{e2.15}, we have $i(A,P\cap\Omega_2,P)=1$, and $i(A,P\cap(\Omega_2-\overline{\Omega}_1),P)=1$. As a result we also can assert that $A$ has at least one fixed point in $P\cap(\Omega_2-\overline{\Omega}_1)$. The proof is complete. \end{proof} We remark that this theorem improves \cite[theorem 2.3.3]{g4} because the condition that $\Omega_1$ and $\Omega_2$ are bounded is not necessary. \begin{theorem} \label{thm2.6} Let $\Omega_1=\{x\in E|\|x\|_11$ such that $Ax_0=\mu_0x_0$ and hence $\|Ax_0\|=\mu_0\|x_0\|>\|x_0\|$, in contradiction with (H3). Thus, by \eqref{e2.13}, Lemma \ref{lem2.1}, \eqref{e2.14} holds. On the other hand, it is also easy to verify $$Ax\neq \mu x,\forall x\in P\cap\partial\Omega_2,0<\mu<1.\label{e2.17}$$ In fact, if there are $x_1 \in P\cap\partial\Omega_2$ and $0<\mu_1<1$ such that $Ax_1=\mu_1x_1$, then $$\|Ax_1\|_1=\mu_1\|x_1\|_1<\|x_1\|_1,$$ in contradiction with (H3). In addition, by (H3) we have $$\inf_{x\in P\cap\partial\Omega_2}\|Ax\|_1\geq \inf_{x\in\partial\Omega_2}\|x\|_1>0.\label{e2.18}$$ It follows from \eqref{e2.17}, \eqref{e2.18} and Lemma \ref{lem2.3} that \eqref{e2.15} holds. As before, \eqref{e2.14} and \eqref{e2.15} imply \eqref{e2.16}, and therefore $A$ has at least one fixed point in $P\cap(\Omega_2-\overline{\Omega}_1)$. \end{proof} We remark that this theorem improves the the Krasnoselskii's theorem in \cite{g4} because the condition that $\Omega_1$ and $\Omega_2$ are bounded is not necessary. \section{The existence of multiple positive solutions to equation \eqref{e1.1}} In this section, we consider \eqref{e1.1} and suppose that $f\in C([0,1]\times R^+_0\times R,R^+)$, $p\in C([0,1],R)\cap C((0,1),R^+_0)\cap C^1((0,1),R)$ with $\int_0^1\frac{1}{p(r)}dr<+\infty$, $\Phi\in C((0,1),R^+_0)$ and $\alpha\geq 0$, $\beta\geq 0$ (not equal to $0$ at the same time); here $R^+=[0,+\infty)$, $R^+_0=(0,+\infty)$, $R=(-\infty,+\infty)$. Let \begin{gather*} \rho^2=\beta+\alpha\int_0^1\frac{1}{p(r)}dr \quad(\rho>0),\\ u_1(t)=\frac{1}{\rho}\int_t^1\frac{1}{p(r)}dr,\quad v_1(t)=\frac{1}{\rho}(\beta+\alpha\int_0^t\frac{1}{p(r)}dr),\\ G_1(t,s)=\begin{cases} u_1(t)v_1(s)p(s),& 0\leq s\leq t\leq 1\\ v_1(t)u_1(s)p(s),& 0\leq t\leq s\leq 1. \end{cases} \end{gather*} Assume that $C^1_p[0,1]=\{x:[0,1]\to R|$ $x(t)$ is continuous on $[0,1]$ and $p(t){x'}(t)$ is continuous on $[0,1]$ also with $\max_{t\in[0,1]}p(t)|{x'}(t)|<+\infty\}$ (see \cite{o3}). For $x\in C^1_p$, let $\|x\|_1=\max_{t\in[0,1]}|x(t)|$, $\|x\|_2=\max_{t\in[0,1]}p(t)|x'(t)|$ and $\|x\|=\max\{\|x\|_1,\|x\|_2\}$. It is easy to see that $C^1_p$ satisfies the conditions (1), (2) and (3) of the Banach space $E$ in section 2. Obviously, $x(t)\in C^1_p$ is a solution to equation \eqref{e1.1} if and only if $x(t)$ is a solution of the following integral equation $$x(t)=\int_0^1G_1(t,s)\Phi(s)f(s,x(s),p(s)x'(s))ds,\quad t\in[0,1].$$ Let $P=\{x\in C^1_p|x(t)\geq\gamma_1(t)\|x\|_1\}$, where $\gamma_1(t)= u_1(t)v_1(t)\frac{1}{u_1(0)v_1(1)}$ for all $t\in[0,1]$. \begin{lemma} \label{lem3.1} Assume that $l\in L^1[0,1]$ with $l(t)>0$ for all $t\in(0,1)$ and $q(t)=\int_0^1G_1(t,s)l(s)ds$, $t\in[0,1]$. Then $$q(t)\geq\gamma_1(t)\max_{s\in[0,1]}q(s).$$ \end{lemma} \begin{proof} Suppose $q(t_0)=\max_{s\in[0,1]}q(s)$. Then \begin{align*} \frac{G_1(t,s)}{G_1(t_0,s)} &=\begin{cases} \frac{v_1(t)u_1(s)p(s)}{u_1(t_0)v_1(s)p(s)},&0\leq t\leq s\leq t_0\leq 1\\ \frac{u_1(t)v_1(s)p(s)}{v_1(t_0)u_1(s)p(s)},&0\leq t_0\leq s\leq t\leq 1\\ \frac{v_1(t)u_1(s)p(s)}{v_1(t_0)u_1(s)p(s)},&0\leq t,t_0\leq s\leq 1\\ \frac{u_1(t)v_1(s)p(s)}{u_1(t_0)v_1(s)p(s)},&0\leq s\leq t,t_0\leq 1\\ \end{cases}\\ &=\begin{cases} u_1(t)v_1(t)\frac{u_1(s)}{u_1(t_0)}\frac{1}{v_1(s)u_1(t)},&0\leq t\leq s\leq t_0\leq 1\\ u_1(t)v_1(t)\frac{v_1(s)}{v_1(t_0)}\frac{1}{u_1(s)v_1(t)},&0\leq t_0\leq s\leq t\leq 1\\ u_1(t)v_1(t)\frac{1}{u_1(t)v_1(t_0)},&0\leq t,t_0\leq s\leq 1\\ u_1(t)v_1(t)\frac{1}{u_1(t_0)v_1(t)},&0\leq s\leq t,t_0\leq 1\\ \end{cases} \\ & \geq \begin{cases} u_1(t)v_1(t)\frac{1}{v_1(1)u_1(0)},&0\leq t\leq s\leq t_0\leq 1\\ u_1(t)v_1(t)\frac{1}{u_1(0)v_1(1)},&0\leq t_0\leq s\leq t\leq 1\\ u_1(t)v_1(t)\frac{1}{u_1(0)v_1(1)},&0\leq t,t_0\leq s\leq 1\\ u_1(t)v_1(t)\frac{1}{u_1(0)v_1(1)},&0\leq s\leq t,t_0\leq 1\\ \end{cases}\\ &=\gamma_1(t). \end{align*} As a consequence, \begin{align*} q(t)&=\int_0^1G_1(t,s)l(s)ds =\int_0^1\frac{G_1(t,s)}{G_1(t_0,s)}G_1(t_0,s)l(s)ds \\ &\geq\gamma_1(t)\int_0^1G_1(t_0,s)l(s)ds=\gamma_1(t)\max_{s\in[0,1]}q(s). \end{align*} The proof is complete. \end{proof} Now we will list some conditions for convenience: \begin{itemize} \item[(H1)] There exists a $k\in C([0,1],R^+_0)$, a $g\in C(R^+_0,R^+_0)$ and a decreasing continuous function $h\in C(R^+_0, R^+_0)$ such that $$f(t,x,z)\leq k(t)g(x),\quad \forall x\in R^+_0, z\in R, t\in[0,1],$$ where $\frac{g(x)}{h(x)}$ is an increasing function and $\int_0^{1}p(s)\Phi(s)k(s)h(c{\gamma}_1(s))ds<+\infty$ for each $c>0$; \item[(H2)] $$\sup_{c\in R^+_0}\frac{ch(c)}{u_1(0)v_1(1)\int_0^{1}p(s) \Phi(s)k(s)h(c{\gamma}_1(s))ds g(c)}>1;$$ \item[(H3)] There exists a $k_1\in C([0,1],R^+_0)$ and a $g_1\in C(R^+_0,R^+_0)$ with $f(t,x,z)\geq k_1(t)g_1(x)$, for all $(t,x,z)\in[0,1]\times R^+_0\times (-\infty,+\infty)$ such that $$\lim_{x\to+\infty}\frac{g_1(x)}{x}=+\infty,$$ where $\int_0^{1}p(s)\Phi(s)k_1(s)ds<+\infty$; \item[(H4)] For any $c>0$, there exists a $\psi_c\in C([0,1],R^+_0)$ such that $f(t,x,z)\geq \psi_{c}(t)$ for all $(t,x,z)\in [0,1]\times (0,c]\times (-\infty,+\infty)$ with $\int_0^{1}p(s)\Phi(s)\psi_c(s)ds<+\infty$. \end{itemize} For given $n\in\{1,2,\dots\}$, let $f_n(t,x,z)=f(t,\max\{\frac{1}{n},x\},z)$ and for $x\in P$, define $$(A_nx)(t)=\int_0^1G_1(t,s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds,\quad n\in\{1,2,\dots\},\; t\in[0,1].\label{e3.1}$$ \begin{lemma} \label{lem3.2} Assume the condition (H1) holds. Then, for every $n\in\{1,2,\dots\}$, $A_n:P\to P$ is continuous and for any $r>0$ and $B_r=\{x\in C^1_p|\|x\|_1\leq r\}$, $A_n(P\cap B_r)$ is relatively compact. \end{lemma} \begin{proof} First, for a given $n\in\{1,2,\dots\}$, we show that $A_nP\subseteq P$. For any $x\in P$, we have \begin{align*} |(A_nx)(t)|&=\big|\int_0^1G_1(t,s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\big|\\ &\leq\int_0^1G_1(t,s)\Phi(s)f(s,\max\{\frac{1}{n},x(s)\},p(s){x'}(s))ds\\ &\leq\int_0^1G_1(t,s)\Phi(s)k(s)g(\max\{\frac{1}{n},x(s)\})ds\\ &\leq\int_0^1G_1(t,s)\Phi(s)k(s)h(\max\{\frac{1}{n},x(s)\})\frac{g(\max\{\frac{1}{n},x(s)\})}{h(\max\{\frac{1}{n},x(s)\})}ds\\ &\leq\int_0^1G_1(t,s)\Phi(s)k(s)ds h(\frac{1}{n})\frac{g(\max\{\frac{1}{n},\|x\|_1\})}{h(\max\{\frac{1}{n},\|x\|_1\})}\\ &<+\infty \end{align*} and \begin{align*} |p(t)(A_nx)'(t)| &=|-\frac{1}{\rho}\int_0^tv_1(s)p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\quad +\frac{\alpha}{\rho}\int_t^1u_1(s)p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds|\\ &\leq\frac{K}{\rho}\int_0^1p(s)\Phi(s)f(s,\max\{\frac{1}{n},x(s)\},p(s){x'}(s))ds\\ &\leq\frac{K}{\rho}\int_0^1p(s)\Phi(s)k(s)g(\max\{\frac{1}{n},x(s)\})ds\\ &\leq\frac{K}{\rho}\int_0^1p(s)\Phi(s)k(s)h(\max\{\frac{1}{n},x(s)\})\frac{g(\max\{\frac{1}{n},x(s)\})}{h(\max\{\frac{1}{n},x(s)\})}ds\\ &\leq\frac{K}{\rho}\int_0^1p(s)\Phi(s)k(s)ds h(\frac{1}{n})\frac{g(\max\{\frac{1}{n},\|x\|_1\})}{h(\max\{\frac{1}{n}, \|x\|_1\})}\\ &<+\infty, \end{align*} where $K=\max\{\alpha u_1(0), v_1(1)\}$ and the following is same as before. Then, $A_n$ is well defined. Moreover, from Lemma \ref{lem3.1}, for any $x\in P$, we have $$(A_nx)(t)\geq \gamma_1(t)\max_{s\in[0,1]}|(A_nx)(s)| =\gamma_1(t)\|A_nx\|_1, \quad \forall t\in [0,1].$$ Consequently, $A_nP\subseteq P$. Second, we show $A_n:P\to P$ is continuous. Assume that $\lim_{m\to+\infty}x_m=x_0$, which means there exists an $M>1/n$ such that $\|x_m\|\leq M$, for all $m\in\{0,1,2,\dots\}$. Then \begin{aligned} f_n(t,x_m(t),p(t){x'}_m(t)) &\leq k(t)g(\max\{\frac{1}{n},x_m(t)\})\\ &=k(t)h(\max\{\frac{1}{n},x_m(t)\}) \frac{g(\max\{\frac{1}{n},x_m(t)\})}{h(\max\{\frac{1}{n},x_m(t)\})}\\ &\leq k(t)h(\frac{1}{n})\frac{g(M)}{h(M)}, \quad m\in \{1,2,\dots\}.\end{aligned} \label{e3.2} Since $$f_n(t,x_m(t),p(t)x'_m(t))\to f_n(t,x_0(t),p(t)x'_0(t)),\quad \mbox{as } m\to+\infty,$$ from \eqref{e3.2}, the Lesbegue Dominated Convergence Theorem guarantees \begin{align*} &\lim_{m\to+\infty}\max_{t\in[0,1]}|(A_nx_m)(t)-(A_nx_0)(t)|\\ &=\lim_{m\to+\infty}\max_{t\in[0,1]} |\int_0^1G_1(t,s)\Phi(s)f_n(s,x_m(s),p(s)x_m'(s))ds\\ &\quad -\int_0^1G_1(t,s)\Phi(s)f_n(s,x_0(s),p(s)x_0'(s))ds|\\ &\leq\lim_{m\to+\infty}\max_{t\in[0,1]} \int_0^1G_1(t,s)\Phi(s)\Big|f_n(s,x_m(s),p(s)x_m'(s)) \\ &\quad -f_n(s,x_0(s),p(s)x_0'(s))\Big|ds\\ &\leq\lim_{m\to+\infty} u_1(0)v_1(1)\int_0^1p(s)\Phi(s) \Big|f_n(s,x_m(s),p(s)x_m'(s))\\ &\quad -f_n(s,x_0(s),p(s)x_0'(s))\Big|ds =0 \end{align*} and \begin{align*} &\lim_{m\to+\infty}\max_{t\in[0,1]}|p(t)(A_nx_m)'(t)-p(t)(A_nx_0)'(t)|\\ &=\lim_{m\to+\infty}\max_{t\in[0,1]} \Big|-\frac{1}{\rho}\int_0^tv_1(s)p(s)\Phi(s)[f_n(s,x_m(s),p(s){x'_m}(s))\\ &\quad -f_n(s,x_0(s),p(s){x'_0}(s))]ds +\frac{\alpha}{\rho}\int_t^1u_1(s)p(s)\Phi(s)[f_n(s,x_m(s),p(s)x_m'(s))\\ &\quad -f_n(s,x_0(s),p(s)x_0'(s))]ds\Big|\\ &\leq \lim_{m\to+\infty}\frac{K}{\rho}\int_0^1p(s)\Phi(s) \Big|f_n(s,x_m(s),p(s)x_m'(s))-f_n(s,x_0(s),p(s)x_0'(s))\Big|ds\\ &=0, \end{align*} which mean that $$\lim_{m\to+\infty}\|A_nx_m-A_nx_0\|=0.$$ Finally, we show $A_n(B_r\cap P)$ is relatively compact. Obviously, $B_r$ is an unbounded set in $C^1_p$. Without loss of generality, we suppose $r>1/n$. Then, for any $x\in B_r\cap P$, we have \begin{align*} \max_{t\in[0,1]}|(A_nx)(t)| &=\max_{t\in[0,1]}|\int_0^1G_1(t,s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds|\\ &=\max_{t\in[0,1]}\int_0^1G_1(t,s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\leq\max_{t\in[0,1]}\int_0^1G_1(t,s)\Phi(s)k(s)h(\max\{\frac{1}{n},x(s)\})\frac{g(\max\{\frac{1}{n},x(s)\})}{h(\max\{\frac{1}{n},x(s)\})}ds\\ &\leq\max_{t\in[0,1]}\int_0^1G_1(t,s)\Phi(s)k(s)ds h(\frac{1}{n})\frac{g(r)}{h(r)}\\ &\leq u_1(0)v_1(1)\int_0^1p(s)\Phi(s)k(s)ds h(\frac{1}{n})\frac{g(r)}{h(r)} \end{align*} and \begin{align*} \max_{t\in[0,1]}|p(t)(A_nx)'(t)| &=\max_{t\in[0,1]}|-\frac{1}{\rho}\int_0^tv_1(s)p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\quad +\frac{\alpha}{\rho}\int_t^1u_1(s)p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds|\\ &\leq\frac{K}{\rho}\int_0^1p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\leq\frac{K}{\rho}\int_0^1p(s)\Phi(s)k(s)h(\max\{\frac{1}{n},x(s)\})\frac{g(\max\{\frac{1}{n},x(s)\})}{h(\max\{\frac{1}{n},x(s)\})}ds\\ &\leq\frac{K}{\rho}\int_0^1p(s)\Phi(s)k(s)ds h(\frac{1}{n})\frac{g(r)}{h(r)}, \end{align*} which means that $A_n(B_r\cap P)$ is bounded. Assume that $t, t'\in[0,1]$. Then, for $x\in B_r\cap P$, we have \begin{align*} |(A_nx)(t)-(A_nx)(t')| &=|\int_0^1G_1(t,s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\quad -\int_0^1G_1(t',s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds|\\ &\leq\int_0^1|G_1(t,s)-G_1(t',s)|\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\leq\int_0^1|G_1(t,s)-G_1(t',s)|\Phi(s)k(s)dsh(\frac{1}{n})\frac{g(r)}{h(r)} \end{align*} and \begin{align*} &|p(t)(A_nx)'(t)-p(t')(A_nx)'(t')|\\ &=|-\frac{1}{\rho}\int_0^tv_1(s)p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\quad +\frac{1}{\rho}\int_0^{t'}v_1(s)p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\quad +\frac{\alpha}{\rho}\int_t^1u_1(s)p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\quad -\frac{\alpha}{\rho} \int_{t'}^1u_1(s)p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds|\\ &\leq 2\frac{K}{\rho}|\int_t^{t'}p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))|ds\\ &\leq 2\frac{K}{\rho}|\int_t^{t'}p(s)\Phi(s)k(s)ds|h(\frac{1}{n}) \frac{g(r)}{h(r)}. \end{align*} Then, for any $\varepsilon>0$, we can choose $\delta>0$ small enough such that $$|(A_nx)(t)-(A_nx)(t')|<\varepsilon,\quad |p(t)(A_nx)'(t)-p(t')(A_nx)'(t')|<\varepsilon,$$ for all $x\in B_r\cap P$, $|t-t'|<\delta$, $t,t'\in[0,1]$. Consequently, $\{(A_n(B_r\cap P))(t)\}$ and $\{p(t)(A_n(B_r\cap P))'(t)\}$ is equicontinuous on $[0,1]$. Consequently, from Arzela-Ascoli theorem, $A_n(B_r\cap P)$ is relatively compact. The proof is complete. \end{proof} \begin{theorem} \label{thm3.3} Assume that (H1)--(H4) hold. Then \eqref{e1.1} has at least two positive solutions. \end{theorem} \begin{proof} From Lemma \ref{lem3.2}, for each $n\in\{1,2,\dots\}$, $A_n:P\to P$ is continuous operator and for any $r>0$, $A_n(B_r\cap P)$ is relatively compact. From (H2), choose $R_0>0$ such that $$\frac{R_0h(R_0)}{u_1(0)v_1(1)\int_0^{1}p(s)\Phi(s)k(s)h(R_0{\gamma}_1(s))ds g(R_0)}>1.\label{e3.3}$$ Without loss of the generality, suppose $R_0\geq 1/n$, $n\in\{1,2,\dots\}$. Set $$\Omega_1=\{x\in C^1_p|\|x\|_10, we can choose a \delta>0 such that$$ |x_{n}^{(1)}(t')-x_{n}^{(1)}(t'')|<\varepsilon,\quad |p(t')(x_{n}^{(1)})'(t')-p(t'')(x_{n}^{(1)})'(t'')|<\varepsilon, $$for all n\in\{1,2,\dots\}, |t'-t''|<\delta, t',t''\in[0,1], which implies that \{x_n^{(1)}(t)\} and \{p(t)(x_n^{(1)})'(t)\} are equi-continuous on [0,1]. The Arzela-Ascoli theorem guarantees that there is a subsequence \{x_{n_j}^{(1)}\} of \{x_n^{(1)}\} with \lim_{j\to+\infty}x_{n_j}^{(1)}=x_0^{(1)}. From \eqref{e3.6}, we have$$ x_0(t)\geq k^*\gamma_1(t),\quad \forall t\in[0,1]. Then, for t\in(0,1), if j is big enough, we have \begin{align*} &|f_{n_j}(t,x_{n_j}^{(1)}(t),p(t)(x_{n_j}^{(1)})'(t)) -f(t,x_0^{(1)}(t),p(t)(x_0^{(1)})'(t))| \\ &=|f(t,x_{n_j}^{(1)}(t),p(t)(x_{n_j}^{(1)})'(t)) -f(t,x_0^{(1)}(t),p(t)(x_0^{(1)})'(t))|\to 0,\quad\mbox{as } j\to+\infty \end{align*} and \begin{aligned} f_{n_j}(t,x_{n_j}^{(1)}(t),p(t)(x_{n_j}^{(1)})'(t)) &=f(t,\max\{\frac{1}{n_j},x_{n_j}^{(1)}(t)\},p(t)(x_{n_j}^{(1)})'(t))\\ &\leq k(t)h(k^*\gamma_1(t))\frac{g(R_0)}{h(R_0)}. \end{aligned} \label{e3.9} Then the Lesbegue Dominated Convergence Theorem guarantees that \begin{aligned} x_0^{(1)}(t) &=\lim_{j\to+\infty}x_{n_j}^{(1)}(t)\\ &=\lim_{j\to+\infty}\int_0^{1}G_1(t,s)\Phi(s)f(s, \max\{\frac{1}{n_j},x_{n_j}^{(1)}(s)\}, p(s)(x_{n_j}^{(1)})'(s))ds\\ &=\int_0^{1}G_1(t,s)\Phi(s)f(s,x_0^{(1)}(s),p(s)(x_0^{(1)})'(s))ds. \end{aligned} \label{e3.10} Obviously \|x_0^{(1)}\|_1\leq R_0. Thus \eqref{e3.3} can guarantee \|x_0^{(1)}\|_1R such that $$g_1(x)>N^*x, \forall x\geq R'.\label{e3.11}$$ Now we define $$\Omega_2=\{x\in \big|\|x\|_1<\frac{R'}{c^*}\}.\label{e3.12}$$ We might as well suppose that \frac{R'}{c^*}>1, R'>1. Then we have $$A_nx\not\leq x\quad \text{for all } x\in\partial{\Omega_2}\cap P, \; n\in\{1,2,\dots\}.\label{e3.13}$$ Otherwise, suppose there exists x_0\in \partial{\Omega_2}\cap P with A_nx_0\leq x_0. Since x_0\in \partial(\Omega_2)\cap P, $$\min_{t\in[a^*,b^*]}x_0(t)\geq\min_{t\in[a^*,b^*]}\gamma_1(t)\|x\|_1 >c^*\frac{R'}{c^*}=R'>1.\label{e3.14}$$ Then, for t\in[a^*, b^*], from \eqref{e3.11} and \eqref{e3.14}, we have \begin{align*} x_0(t) &\geq(A_nx_0)(t)\\ &=\int_0^{1}G_1(t,s)\Phi(s)f_n(s,x_0(s),p(s)x_0'(s))ds\\ &\geq\int_{a^*}^{b^*}G_1(t,s)\Phi(s)f(s,\max\{\frac{1}{n},x_0(s)\},p(s)x_0'(s))ds\\ &\geq\int_{a^*}^{b^*}G_1(t,s)\Phi(s)k_1(s)g_1(\max\{\frac{1}{n},x_0(s)\})ds\\ &>\int_{a^*}^{b^*}G_1(t,s)\Phi(s)k_1(s)N^*x_0(s)ds\\ &>\int_{a^*}^{b^*}G_1(t,s)\Phi(s)k_1(s)dsN^*c^*\frac{R'}{c^*}\\ &>\frac{R'}{c^*}, \quad \forall n\in\{1,2,\dots\} \end{align*} which implies \|x_0\|_1>R'/c^*. This contradicts to x_0\in P\cap\partial\Omega_2. From \eqref{e3.4} and \eqref{e3.13}, the Theorem \ref{thm2.5} guarantees that A_n has a fixed point x_{n}^{(2)}\in(\Omega_2-\overline{\Omega}_1)\cap P, n\in\{1,2,\dots\}. It is easy to see that $$x_n^{(2)}(t)\geq\gamma_1(t)\|x_n^{(2)}\|_1\geq\gamma_1(t)R_0, \quad n\in\{1,2,\dots\}.\label{e3.15}$$ By proof similar to \eqref{e3.6}, \eqref{e3.7}, \eqref{e3.8}, we know that \{x_n^{(2)}\} is relatively compact in C^1_p. Then there exists a subsequence \{x_{n_i}^{(2)}\} of \{x_{n}^{(2)}\} with \lim_{i\to+\infty}x_{n_i}^{(2)}=x_0^{(2)}. And moreover, by similar proof as \eqref{e3.10}, we get x_0^{(2)}(t) is a positive solution to equation \eqref{e1.1} with \frac{R'}{c^*}>\|x_0^{(2)}\|_1>R_0. Consequently, equation \eqref{e1.1} has at least two different positive solution x_0^{(1)}(t) and x_0^{(2)}(t). The proof is complete. \end{proof} \begin{example} \label{exa3.1} \rm Now we consider $$\begin{gathered} x''+\frac{1}{16}t^{-1/2}(1-t)^{-1/4}(x^{-1/4} +x^{2})(1+\cos^2(x'))=0, \quad t\in(0,1)\\ \lim_{t\to 0+}x'(t)=0=x(1). \end{gathered} \label{e3.16}$$ Then \eqref{e3.16} has at least two positive solutions. \end{example} To prove that \eqref{e3.16} has at least two positive solutions, we apply Theorem \ref{thm3.3} with \Phi(t)=\frac{1}{16}t^{-1/2}(1-t)^{-1/4}, p(t)\equiv 1, f(t,x,z)=(x^{-1/4}+x^{2})(1+\cos^2(z)), k(t)\equiv 1, g(x)=2(x^{-1/4}+x^{2}), h(x)=x^{-1/4}, \gamma_1(t)=1-t, k_1(t)\equiv 1, g_1(x)=x^{-1/4}+x^{2}, \Psi_c(t)=c^{-1/4}. It is easy to verify that (H1)--(H4) hold. Hence, \eqref{e3.16} has at leat two positive solutions. \begin{example} \label{exa3.2} \rm Now we consider $$\begin{gathered} x''+\frac{1}{12\pi}t^{-1/4}(1-t)^{-1/4}(x^{-1/4}+x^{2})(\pi+\arctan {x'})=0, \quad t\in(0,1)\\ x(0)=0=x(1). \end{gathered} \label{e3.17}$$ Then \eqref{e3.17} has at least two positive solutions. \end{example} To prove that \eqref{e3.17} has at least two positive solutions, we apply Theorem \ref{thm3.3} with \Phi(t)=\frac{1}{12\pi}t^{-1/4}(1-t)^{-1/4}, p(t)\equiv 1, f(t,x,z)=(x^{-1/4}+x^{2})(\pi+\arctan {x'}), k(t)\equiv 1, g(x)=\frac{3\pi}{2}(x^{-1/4}+x^{2}), h(x)=x^{-1/4}, \gamma_1(t)=t(1-t), k_1(t)\equiv 1, g_1(x)=x^{-1/4}+x^{2}, \Psi_c(t)=c^{-1/4}. It is easy to verify that (H1)--(H4) hold. Hence, \eqref{e3.17} has at leat two positive solutions. \section{The existence of multiple positive solutions to equation \eqref{e1.2}} In this section, we consider \eqref{e1.2} and suppose that f\in C(R^+\times R^+_0\times R^+,R^+), p\in C(R^+,R)\cap C(R^+_0,R^+_0)\cap C^1(R^+_0,R) with \int_0^{+\infty}\frac{1}{p(r)}dr=+\infty, \Phi\in C(R^+_0,R^+); here R^+=[0,+\infty), R^+_0=(0,+\infty), R=(-\infty,+\infty). Let G_2(t,s)=\begin{cases} u_2(t)v_2(s)p(s),& a\leq s\leq t<+\infty\\ v_2(t)u_2(s)p(s),& 0\leq t\leq s<+\infty, \end{cases} $$where u_2(t)=1 and  v_2(t)=\int_0^t\frac{1}{p(r)}dr for all t\in R^+. Let  C^1_{\infty}=\big\{x:[0,+\infty)\to R| x(t) is continuous on R^+ and p(t){x'}(t) is continuous on R^+ also with \lim_{t\to+\infty}\frac{x(t)}{1+v_2(t)} exists and \sup_{t\in[0,+\infty)}p(t)|{x'}(t)|<+\infty\big\}. For x\in C^1_{\infty}, let$$ \|x\|_1=\sup_{t\in[0,+\infty)}\frac{|x(t)|}{1+v_2(t)}\quad\text{and}\quad \|x\|_2=\sup_{t\in[0,+\infty)}p(t)|x'(t)|. $$It is easy to see that \|\cdot\|_1 is a norm of C^1_{\infty} and \|\cdot\|_2 is a semi-norm of C^1_{\infty}. Now Let \|x\|=\max\{\|x\|_1,\|x\|_2\}. Obviously, C^1_{\infty} satisfies (1), (2) and (3) of the Banach space E in section 2. It is easy to prove that if x(t)\in C^1_{\infty} is a solution to integral equation$$ x(t)=\int_0^{\infty}G_2(t,s)\Phi(s)f(s,x(s),p(s)x'(s))ds, \quad t\in R^+, $$then x(t) is a solution to \eqref{e1.2}. Let  P=\{x\in C^1_{\infty}|x(t)\geq\gamma_2(t)\|x\|_1,\; \forall t\in R^+\}, where \begin{gather*} \gamma_2(t)=\begin{cases} \int_0^{t}\frac{1}{p(r)}dr, & t\in[0,\tau]\\ 1, & t\in(\tau,+\infty) \end{cases} \\ \widetilde{\gamma}_2(t)=\frac{\gamma_2(t)}{ 1+v_2(t)},\quad t\in R^+; \end{gather*} here \int_0^{\tau}\frac{1}{p(r)}dr=1. Suppose that  x=(1+v_2(t))y, t\in R^+ and F(t,y,z)=f(t,(1+v_2(t))y,z)=f(t,x,z). Now we will list some conditions for convenience: \begin{itemize} \item[(H1)] There exists a k\in C(R^+,R^+_0), a g\in C(R^+_0,R^+_0) and a decreasing continuous function h\in C(R^+_0, R^+_0) such that$$ F(t,y,z)\leq k(t)g(y),\quad \forall (y,z)\in R^+_0\times R^+_0, t\in R^+, $$where  \frac{g(y)}{h(y)} is an increasing function and \int_0^{\infty}p(s)\Phi(s)k(s)h(c\tilde{\gamma}_2(s))ds<+\infty for each c>0 \item[(H2)]$$\sup_{c\in R^+_0}\frac{ch(c)}{\int_0^{\infty}p(s) \Phi(s)k(s)h(c\widetilde{\gamma}_2(s))ds g(c)}>1 $$\item[(H3)] There exists a k_1\in C(R^+,R^+_0) and a g_1\in C(R^+_0,R^+_0) with F(t,y,z)\geq k_1(t)g_1(y), for all (t,y,z)\in[0,+\infty)\times R^+_0\times R^+ such that$$ \lim_{y\to+\infty}\frac{g_1(y)}{y}=+\infty, $$where  \int_0^{\infty}p(s)\Phi(s)k_1(s)ds<+\infty \item[(H4)] for any c>0, there exists a \psi_c\in C(R^+,R^+_0) such that F(t,y,z)\geq \psi_{c}(t) for all (t,y,z)\in R^+\times (0,c]\times R^+ with \int_0^{\infty}p(s)\Phi(s)\psi_c(s)ds<+\infty. \end{itemize} Let  C_{l}=\big\{x:R^+\to R| x(t) is continuous on R^+ and \lim_{t\to+\infty}x(t) exists \big\} with norm \|x\|_l=\sup_{t\in[0,+\infty)}|x(t)|. From \cite{m1}, we know that C_{l} is a Banach space and following theorem is true. \begin{theorem}[\cite{m1}] \label{thm4.1} Let M\subseteq C_l(R^+,R). Then M is relatively compact in the space C_l(R^+,R) if the following conditions hold: \begin{itemize} \item[(a)] M is bounded in C_l \item[(b)] the functions belonging to M are locally equi-continuous on R^+; \item[(c)] the functions from M are equiconvergent, that is, given \varepsilon>0, there corresponds T(\varepsilon)>0 such that |x(t)-x(+\infty)|<\varepsilon for any t\geq T(\varepsilon) and x\in M. \end{itemize} \end{theorem} \begin{theorem}[\cite{o4}] \label{thm4.2} Let M\subseteq C_{\infty}^1(R^+,R). Then M is relatively compact in C_{\infty}^1(R^+,R) if the following conditions hold: \begin{itemize} \item[(a)] M is bounded in C_{\infty}^1; \item[(b)] the functions belonging to \{y\big|y(t)=\frac{x(t)}{1+v_2(t)}, x\in M\} and the functions belonging to \{y\big|y(t)=p(t)x'(t), x\in M\} are locally equi-continuous on R^+; \item[(c)] the functions from \{y\big| y(t)=\frac{x(t)}{1+v_2(t)}, x\in M\} and the functions from \{y\big| y(t)=p(t)x'(t), x\in M\} are equi-convergent at +\infty. \end{itemize} \end{theorem} \begin{lemma}[\cite{o4}] \label{lem4.3} Assume that \overline{\Phi}(t)\in C(R^+_0,R^+) with \int_0^{+\infty}p(s)\overline{\Phi}(t)dt<+\infty and F(t)=\int_0^{\infty}G_2(t,s)\overline{\Phi}(s)ds. Then$$ F(t)\geq\gamma_2(t)\frac{F(\tau)}{1+v_2(\tau)},\quad \forall t\in R^+, \; \tau\in R^+, $$and$$ F(t)\geq\gamma_2(t)\|F\|_1,\quad \forall t\in R^+,\; \lim_{t\to+\infty}\frac{F(t)}{1+v_2(t)}=0. \end{lemma} Let f_n(t,x,z)=f(t,\max\{\frac{1}{n}(1+v_2(t)),x\},z), \; n\in \{1,2,\dots\} and for x\in P, n\in\{1,2,\dots\}, t\in R^+, define $$(A_nx)(t)=\int_0^{\infty}G_2(t,s)\Phi(s)f_n(s,x(s),p(s)x'(s))ds\,. \label{e4.1}$$ \begin{lemma} \label{lem4.4} Assume that the condition (H1) holds. Then for each n\in\{1,2,\dots\}, A_n:P\to P is continuous and for any r>0 and  B_r=\{x\in C^1_{\infty}|\|x\|_1\leq r\}, A_n(P\cap B_r) is relatively compact for each n\geq 1. \end{lemma} \begin{proof} First, we show that A_nP\subseteq P. For any x\in P, we have \begin{align*} |(A_nx)(t)| &=\big|\int_0^{\infty}G_2(t,s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\big|\\ &\leq\int_0^{\infty}G_2(t,s)\Phi(s)f(s,\max\{\frac{1}{n}(1+v_2(s)),x(s)\}, p(s){x'}(s))ds\\ &=\int_0^{\infty}G_2(t,s)\Phi(s)F(s,\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\}, p(s){x'}(s))ds\\ &\leq\int_0^{\infty}G_2(t,s)\Phi(s)k(s)g(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\})ds\\ &\leq\int_0^{\infty}G_2(t,s)\Phi(s)k(s)h(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\}) \frac{g(\max\{\frac{1}{n},\frac{x(s)}{1+v_2(s)}\})}{h(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\})}ds\\ &\leq\int_0^{\infty}G_2(t,s)\Phi(s)k(s)ds h(\frac{1}{n})\frac{g(\max\{\frac{1}{n},\|x\|_1\})}{h(\max\{\frac{1}{n}, \|x\|_1\})}<+\infty \end{align*} and \begin{align*} |p(t)(A_nx)'(t)| &=\big|\int_t^{\infty}p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\big|\\ &\leq\int_t^{\infty}p(s)\Phi(s)f(s,\max\{ \frac{1}{n}(1+v_2(s)),x(s)\},p(s){x'}(s))ds\\ &=\int_t^{\infty}p(s)\Phi(s)F(s,\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\},p(s){x'}(s))ds\\ &\leq\int_t^{\infty}p(s)\Phi(s)k(s)g(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\})ds\\ &\leq\int_t^{\infty}p(s)\Phi(s)k(s)h(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\}) \frac{g(\max\{\frac{1}{n},\frac{x(s)}{1+v_2(s)}\})}{h(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\})}ds\\ &\leq\int_0^{\infty}p(s)\Phi(s)k(s)ds h(\frac{1}{n}) \frac{g(\max\{\frac{1}{n},\|x\|_1\})}{h(\max\{\frac{1}{n},\|x\|_1\})}\\ &<+\infty. \end{align*} Since (A_nx)(t)\geq 0, it is easy to see that A_n is well defined. Moreover, from Lemma \ref{lem4.3}, for any x\in P, we have (A_nx)(t)\geq \gamma_2(t)\|A_nx\|_1, \quad \forall t\in R^+. Consequently, A_nP\subseteq P. Second, we show A_n:P\to P is continuous. Assume that \lim_{m\to+\infty}x_m=x_0, which means there exists an M>1/n such that \|x_m\|\leq M, for all m\in\{0,1,2,\dots\}. Then \begin{aligned} f_n(t,x_m(t),p(t){x'}_m(t)) &=F(t,\max\{\frac{1}{n},\frac{x_m(t)}{1+v_2(t)}\},p(t)x_m'(t))\\ &\leq k(t)g(\max\{\frac{1}{n},\frac{x_m(t)}{1+v_2(t)}\})\\ &\leq k(t)h(\max\{\frac{1}{n},\frac{x_m(t)}{1+v_2(t)}\}) \frac{g(\max\{\frac{1}{n},\frac{x_m(t)}{1+v_2(t)}\})} {h(\max\{\frac{1}{n},\frac{x_m(t)}{1+v_2(t)}\})} \\ &\leq k(t)h(\frac{1}{n})\frac{g(M)}{h(M)},\quad m\in\{1,2,\dots\}. \end{aligned} \label{e4.2} Since f_n(t,x_m(t),p(t)x'_m(t))\to f_n(t,x_0(t),p(t)x'_0(t)),\quad \text{as }m\to+\infty, from \eqref{e4.2}, the Lesbegue Dominated Convergence Theorem guarantees that \begin{align*} &\lim_{m\to+\infty}\sup_{t\in[0,+\infty)} \frac{|(A_nx_m)(t)-(A_nx_0)(t)|}{1+v_2(t)}\\ &=\lim_{m\to+\infty}\sup_{t\in[0,+\infty)} \big|\int_0^{\infty}\frac{G_2(t,s)}{1+v_2(t)}\Phi(s)(f_n(s,x_m(s),p(s)x_m'(s))\\ &\quad -f_n(s,x_0(s),p(s)x_0'(s)))ds\Big|\\ &\leq \lim_{m\to+\infty}\sup_{t\in[0,+\infty)} \int_0^{\infty}\frac{G_2(t,s)}{1+v_2(t)}\Phi(s)|f_n(s,x_m(s),p(s)x_m'(s))\\ &\quad -f_n(s,x_0(s),p(s)x_0'(s))|ds\\ &=\lim_{m\to+\infty}\int_0^{\infty}p(s)\Phi(s)|f_n(s,x_m(s),p(s)x_m'(s))\\ &\quad -f_n(s,x_0(s),p(s)x_0'(s))|ds =0 \end{align*} and \begin{align*} &\lim_{m\to+\infty}\sup_{t\in[0,+\infty)}|p(t)(A_nx_m)'(t)-p(t)(A_nx_0)'(t)|\\ &=\lim_{m\to+\infty}\sup_{t\in[0,+\infty)} \Big|\int_t^{\infty}p(s)\Phi(s)f_n(s,x_m(s),p(s)x_m'(s))ds \\ &\quad -\int_t^{\infty}p(s)\Phi(s)f_n(s,x_0(s),p(s)x_0'(s))ds\Big|\\ &\leq\lim_{m\to+\infty}\sup_{t\in[0,+\infty)} \int_t^{\infty}p(s)\Phi(s)|f_n(s,x_m(s),p(s)x_m'(s))\\ &\quad -f_n(s,x_0(s),p(s)x_0'(s))|ds\\ &=\lim_{m\to+\infty}\int_0^{\infty}p(s)\Phi(s) \big|f_n(s,x_m(s),p(s)x_m'(s))-f_n(s,x_0(s),p(s)x_0'(s))\big|ds =0, \end{align*} which implies \lim_{m\to+\infty}\|A_nx_m-A_nx_0\|=0. Finally, we show A_n(B_r\cap P) is relatively compact. Obviously, B_r is a unbounded set in C^1_{\infty}. Without loss of generality, we suppose r>1/n. Then, for any x\in B_r\cap P, we have \begin{align*} &\sup_{t\in[0,+\infty)}\frac{|(A_nx)(t)|}{1+v_2(t)}\\ &=\sup_{t\in[0,+\infty)}|\int_0^{\infty} \frac{G_2(t,s)}{1+v_2(t)}\Phi(s)f_n(s,x(s),p(s){x'}(s))ds|\\ &=\sup_{t\in[0,+\infty)}\int_0^{\infty} \frac{G_2(t,s)}{1+v_2(t)}\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\leq\sup_{t\in[0,+\infty)}\int_0^{\infty} \frac{G_2(t,s)}{1+v_2(t)}\Phi(s)k(s)h(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\})\frac{g(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\})}{h(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\})}ds\\ &\leq\sup_{t\in[0,+\infty)}\int_0^{\infty} \frac{G_2(t,s)}{1+v_2(t)}\Phi(s)k(s)ds h(\frac{1}{n})\frac{g(r)}{h(r)}\\ &\leq\int_0^{\infty}p(s)\Phi(s)k(s)ds h(\frac{1}{n})\frac{g(r)}{h(r)} \end{align*} and \begin{align*} &\sup_{t\in[0,+\infty)}|p(t)(A_nx)'(t)|\\ &=\sup_{t\in[0,+\infty)} |\int_t^{\infty}p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds|\\ &=\sup_{t\in[0,+\infty)}\int_t^{\infty}p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\leq\sup_{t\in[0,+\infty)}\int_t^{\infty}p(s)\Phi(s)k(s)h(\max \{\frac{1}{n},\frac{x(s)}{1+v_2(s)}\}) \frac{g(\max\{\frac{1}{n},\frac{x(s)}{1+v_2(s)}\})}{h(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\})}ds\\ &\leq\sup_{t\in[0,+\infty)}\int_t^{\infty}p(s)\Phi(s)k(s)ds h(\frac{1}{n}) \frac{g(r)}{h(r)}\\ &=\int_0^{\infty}p(s)\Phi(s)k(s)ds h(\frac{1}{n})\frac{g(r)}{h(r)}, \end{align*} which implies A_n(P\cap B_r) is bounded. Assume that t, t'\in R^+. Then, for x\in B_r\cap P, we have \begin{align*} &|\frac{(A_nx)(t)}{1+v_2(t)}-\frac{(A_nx)(t')}{1+v_2(t')}|\\ &=|\int_0^{\infty}\frac{G_2(t,s)}{1+v_2(t)} \Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\quad -\int_0^{\infty}\frac{G_2(t',s)}{1+v_2(t')}\Phi(s)f_n(s,x(s), p(s){x'}(s))ds|\\ &\leq \int_0^{\infty}|\frac{G_2(t,s)}{1+v_2(t)}-\frac{G_2(t',s)}{1+v_2(t')} |\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\\ &\leq \int_0^{\infty}|\frac{G_2(t,s)}{1+v_2(t)}-\frac{G_2(t',s)}{1+v_2(t')}| \Phi(s)k(s)dsh(\frac{1}{n})\frac{g(r)}{h(r)} \end{align*} and \begin{align*} &|p(t)(A_nx)'(t)-p(t')(A_nx)'(t')|\\ &=\big|\int_t^1p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds -\int_{t'}^1p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\big|\\ &=\big|\int_t^{t'}p(s)\Phi(s)f_n(s,x(s),p(s){x'}(s))ds\big|\\ &\leq\big|\int_t^{t'}p(s)\Phi(s)k(s)ds\big|h(\frac{1}{n})\frac{g(r)}{h(r)}. \end{align*} For any \varepsilon>0, T>0, we can choose \delta>0 small enough such that |\frac{(A_nx)(t)}{1+v_2(t)}-\frac{(A_nx)(t')}{1+v_2(t')}|<\varepsilon,\quad |p(t)(A_nx)'(t)-p(t')(A_nx)'(t')|<\varepsilon, for all x\in B_r\cap P, |t-t'|<\delta, t,t'\in[0,T]. Consequently, \{\frac{(A_n(B_r\cap P))(t)}{1+v_2(t)}\} and \{p(t)(A_n(B_r\cap P))'(t)\} is locally equi-continuous on [0,+\infty). Moreover, Lemma \ref{lem4.3} guarantees that \begin{align*} &\lim_{t\to+\infty}\sup_{x\in P\cap B_r}\big|\frac{(Ax)(t)}{1+v_2(t)}\big|\\ &\leq\lim_{t\to+\infty}\int_0^{\infty}\frac{G_2(t,s)}{1+v_2(t)} \Phi(s)f_n(s,x(s),p(s)x'(s))ds\\ &\leq\lim_{t\to+\infty}\sup_{x\in P\cap B_r}\int_0^{\infty}\frac{G_2(t,s)}{1+v_2(t)}\Phi(s)k(s)g(\max\{\frac{1}{n}, \frac{x(s)}{1+v_2(s)}\})ds\\ &\leq\lim_{t\to+\infty}\frac{\int_0^{\infty}G_2(t,s)\Phi(s)k(s)ds} {1+v_2(t)}h(\frac{1}{n})\frac{g(r)}{h(r)} =0\end{align*} and \begin{align*} \lim_{t\to+\infty}\sup_{x\in P\cap B_r}|p(t)(Ax)'(t)| &\leq\lim_{t\to+\infty}\int_t^{\infty}p(s)\Phi(s)f_n(s,x(s),p(s)x'(s))ds\\ &\leq\lim_{t\to+\infty}\sup_{x\in P\cap B_r}\int_t^{\infty}p(s)\Phi(s)k(s)g(\max\{\frac{1}{n},x(s)\})ds\\ &\leq\lim_{t\to+\infty}\int_t^{\infty}p(s)\Phi(s)k(s)dsh(\frac{1}{n}) \frac{g(r)}{h(r)} =0, \end{align*} which imply that the functions belonging to \{\frac{(A(P\cap B_r))(t)}{1+v_2(t)}\} and the functions belonging to \{p(t)(A(P\cap B_r))'(t)\} are equi-convergent. As a result, from Lemma \ref{lem4.3}, A_n(B_r\cap P) is relatively compact. The proof is complete. \end{proof} \begin{theorem} \label{thm4.5} Assume that (H1)--(H4) hold. Then \eqref{e1.2} has at least two positive solutions. \end{theorem} \begin{proof} From Lemma \ref{lem4.4}, for each n\in\{1,2,\dots\}, A_n:P\to P is continuous operator and for any r>0, A_n(B_r\cap P) is relatively compact. From (H2), choose R_0>0 such that $$\frac{R_0h(R_0)}{\int_0^{\infty}p(s)\Phi(s)k(s)h(R_0\tilde{\gamma}_2(s))ds g(R_0)}>1.\label{e4.3}$$ Without loss of the generality, suppose R_0\geq 1/n. Set \Omega_1=\{x\in C^1_{\infty}|\|x\|_10, T>0, we can choose a \delta>0 such that if |t'-t''|<\delta and t', t''\in [0,T], then |\frac{x_{n}^{(1)}(t')}{1+v_2(t')}-\frac{x_{n}^{(1)}(t'')}{1+v_2(t'')}| <\varepsilon,\quad |p(t')(x_{n}^{(1)})'(t')-p(t'')(x_{n}^{(1)})'(t'')|<\varepsilon, $$for all n\in\{1,2,\dots\}, which implies that \{\frac{x_n^{(1)}(t)}{1+v_2(t)}\} and \{p(t)(x_n^{(1)})'(t)\} are locally equi-continuous on [0,+\infty). Thus, Theorem \ref{thm4.2} guarantees that there is a convergent subsequence \{x_{n_j}^{(1)}\} of \{x_n^{(1)}\} with \lim_{j\to+\infty}x_{n_j}^{(1)}=x_0^{(1)}. From \eqref{e4.7}, we have$$ x_0^{(1)}(t)\geq k^*\gamma_2(t),\quad \forall t\in[0,+\infty). Then, for t\in(0,+\infty), if j is big enough, we have \begin{align*} &|f_{n_j}(t,x_{n_j}^{(1)}(t),p(t)(x_{n_j}^{(1)})'(t)) -f(t,x_0^{(1)}(t),p(t)(x_0^{(1)})'(t))|\\ &=|f(t,x_{n_j}^{(1)}(t),p(t)(x_{n_j}^{(1)})'(t)) -f(t,x_0^{(1)}(t),p(t)(x_0^{(1)})'(t))| \to 0, \end{align*} as j\to+\infty, and \begin{align*} f_{n_j}(t,x_{n_j}^{(1)}(t),p(t)(x_{n_j}^{(1)})'(t)) &=f(t,\max\{\frac{1}{n_j}(1+v_2(t)),x_{n_j}^{(1)}(t)\}, p(t)(x_{n_j}^{(1)})'(t))\\ &\leq k(t)h(k^*\tilde{\gamma}_2(t))\frac{g(R_0)}{h(R_0)}. \end{align*} Then the Lesbegue Dominated Convergence Theorem guarantees that \begin{aligned} & x_0^{(1)}(t)\\ &=\lim_{j\to+\infty}x_{n_j}^{(1)}(t)\\ &=\lim_{j\to+\infty}\int_0^{\infty}G_2(t,s)\Phi(s)f(s, \max\{\frac{1}{n_j}(1+v_2(s)),x_{n_j}^{(1)}(s)\}, p(s)(x_{n_j}^{(1)})'(s))ds\\ &=\int_0^{\infty}G_2(t,s)\Phi(s)f(s,x_0^{(1)}(s),p(s)(x_0^{(1)})'(s))ds, \end{aligned} \label{e4.12} which implies x_0^{(1)}(t) is a positive solution to \eqref{e1.2}. Obviously, \|x_0^{(1)}\|_1\leq R_0. Then \eqref{e4.3} can guarantee \|x_0^{(1)}\|_1R such that $$g_1(y)>N^*y, \quad \forall y\geq R'.\label{e4.13}$$ Now we define $$\Omega_2=\{x\in C^1_{\infty}\big|\|x\|_1<\frac{R'}{c^*}\}.\label{e4.14}$$ We might as well suppose that \frac{R'}{c^*}>1, R'>1. Then we have $$A_nx\not\leq x, \quad n\in\{1,2,\dots\}\label{e4.15}$$ for all x\in\partial{\Omega_2}\cap P. Otherwise, suppose there exists x_0\in\partial{\Omega_2}\cap P with A_nx_0\leq x_0. Since x_0\in \partial(\Omega_2)\cap P, \min_{t\in[a^*,b^*]}\frac{x_0(t)}{1+v_2(t)} \geq\min_{t\in[a^*,b^*]}\tilde{\gamma}_2(t) \sup_{t\in[0,+\infty)}\frac{|x(t)|}{1+v_2(t)}>c^*\frac{R'}{c^*}=R'>1.  Then for $t\in[a^*, b^*]$, from \eqref{e4.13}, we have \begin{align*} \frac{x_0(t)}{1+v_2(t)}&\geq\frac{(A_nx_0)(t)}{1+v_2(t)}\\ &=\int_0^{+\infty}\frac{G_2(t,s)}{1+v_2(t)}\Phi(s)f_n(s,x_0(s),p(s)x_0'(s))ds\\ &\geq\int_{a^*}^{b^*}\frac{G_2(t,s)}{1+v_2(t)}\Phi(s)f(s, \max\{\frac{1}{n}(1+v_2(s)),x_0(s)\},p(s)x_0'(s))ds\\ &\geq\int_{a^*}^{b^*}\frac{G_2(t,s)}{1+v_2(t)} \Phi(s)k_1(s)g_1(\max\{\frac{1}{n},\frac{x_0(s)}{1+v_2(s)}\})ds\\ &>\int_{a^*}^{b^*}\frac{G_2(t,s)}{1+v_2(t)}\Phi(s)k_1(s)N^* \frac{x_0(s)}{1+v_2(s)}ds\\ &>\int_{a^*}^{b^*}G_2(t,s)\Phi(s)k_1(s)dsN^*c^*\frac{R'}{c^*}\\ &>\frac{R'}{c^*}, \end{align*} which implies $\|x_0\|_1>R'/c^*$. This contradicts to $x_0\in P\cap\partial\Omega_2$. Then \eqref{e4.15} is true. From \eqref{e4.5} and \eqref{e4.15}, Theorem \ref{thm2.5} guarantees that $A_n$ has a fixed point $x_{n}^{(2)}\in(\Omega_2-\overline{\Omega}_1)\cap P$. For the set $\{x_n^{(2)}\}$, since $\|x_n^{(2)}\|_1=\sup_{t\in[0,+\infty)}\frac{|x_n^{(2)}(t)|}{1+v_2(t)}\leq \frac{R'}{c^*}$, (H4) guarantees that there is a $\psi_{R'/c^*}(t)$ such that $$f(t,x,z)=F(t,y,z)\geq \psi_{\frac{R'}{c^*}}(t), \quad \forall (t,y,z)\in[0,+\infty)\times(0,\frac{R'}{c^*}]\times[0,+\infty). \label{e4.16}$$ By proof as in \eqref{e4.7}, \eqref{e4.8}, \eqref{e4.9}, \eqref{e4.10} and \eqref{e4.11}, we can prove that $\{x_n^{(2)}\}$ is relatively compact in $C^1_{\infty}$, which means that there exists a subsequence $\{x_{n_i}^{(2)}\}$ of $\{x_{n}^{(2)}\}$ with $\lim_{i\to+\infty}x_{n_i}^{(2)}=x_0^{(2)}$. By proof as in \eqref{e4.12} $x_0^{(2)}(t)$ is a positive solution to equation \eqref{e1.2} with $\frac{R'}{c^*}>\|x_0^{(2)}\|_1>R_0$. Consequently, equation \eqref{e1.2} has at least two different positive solutions $x_0^{(1)}(t)$ and $x_0^{(2)}(t)$. The proof is complete. \end{proof} \begin{example} \label{exa4.1} \rm Now we consider $$\begin{gathered} x''+\frac{1}{16}e^{-t}t^{-1/4}((1+t)^{1/2} x^{-1/2}+\frac{1}{(1+t)^3}x^3)(1+\frac{{x'}^2}{1+{x'}^2})=0,\; t\in(0,+\infty)\\ x(0)=0,\quad \lim_{t\to+\infty}x'(t)=0. \end{gathered} \label{e4.17}$$ Then, equation \eqref{e4.17} has at least two positive solutions. \end{example} To prove that \eqref{e4.17} has at least two positive solutions, we apply Theorem \ref{thm4.5} with $\Phi(t)=\frac{1}{16}e^{-t}t^{-1/4}$, $p(t)\equiv 1$, $f(t,x,z)=((1+t)^{1/2} x^{-1/2}+\frac{1}{(1+t)^3}x^3)(1+\frac{{x'}^2}{1+{x'}^2})$, $k(t)\equiv 1$, $g(y)=2(y^{-1/2}+y^3)$, $h(x)=y^{-1/2}$, $\gamma_1(t)=\begin{cases} t,&t\in[0,1]\\ 1,&t\in(0,+\infty) \end{cases}$, $\tilde{\gamma}_2(t)=\frac{\gamma_2(t)}{1+t}$, $k_1(t)\equiv 1$, $g_1(y)=y^{-1/4}+y^{3}$, $\Psi_c(t)=c^{-1/4}$. It is easy to verify that (H1)--(H4) hold. Hence, \eqref{e4.17} has at leat two positive solutions. \subsection*{Acknowledgements} The author wants to thanks Professor Temoney for his guidance, and the anonymous referee for his/her suggestions. \begin{thebibliography}{00} \bibitem{a1} R. P. Agarwal, D. O'Regan; \emph{Nonlinear superlinear singular and nonsingular second order boundary value problems}, Journal of Differential Equations, 143 (1998), 60-95. \bibitem{a2} R. P. Agarwal, D. O'Regan, V. Lakshmikantham, S. Leela; \emph{Existence of positive solutions for singular initial and boundary value problems via the classical upper and lower solution approach}, Nonlinear Analysis, 50 (2002), 215-222. \bibitem{a3} R. P. Agarwal, D. O'Regan; \emph{Singular Problems on the Infinite Interval Modelling Phenomena in Draining Flows}, IMA Journal of Applied Mathematics, 66 (2001), 621-635. \bibitem{b1} J. V. Baxley; \emph{Existence and uniqueness for nonlinear boundary value problems on infinite intervals}, J. Math. Anal. Appl., vol. 147 (1990), 127-133. \bibitem{b2} L. E. Bobisud; \emph{Existence of solutions for nonlinear singular boundary value problems}, Appl. Anal., vol. 35 (1990), 43-57. \bibitem{b3} L. E. Bobisud; \emph{Existence of positive solutions to some nonlinear singular boundary value problems on finite and infinite intervals}, J. Math. Anal. Appl., vol. 173 (1993), 69-83. \bibitem{b4} L. E. Bosbisud, D. O'Regan, W. D. Royalty; \emph{Solvability of some nonlinear boundary value problems}, Nanlinear Anal., vol. 12 (1988), 855-869. \bibitem{c1} Shaozhu Chen, Yong Zhang; \emph{Singular boundary value problems on a half-line}, J. Math. Anal. Appl., vol. 195 (1995), 449-468. \bibitem{d1} K. Deimling; \emph{Nonlinear Functional Analysis}, Springer-Verlag, New York, 1985. \bibitem{d2} Dugundji; \emph{An extension of Tietze's theorem}, Pacific J. Math., vol. 1 (1951), 353-367. \bibitem{g1} A. Granas, R. B. Guenther, J. W. Lee and D. O'Regan; \emph{Boundary value problems on infinite intervals and semiconductor devices}, J. Math. Anal. Appl., vol. 116 (1986), 335-345. \bibitem{g2} A. Granas, J. Dugudji, \emph{Fixed point Theory}, Springer, 2003. \bibitem{g3} Dajun Guo, \emph{Nonlinear Functional Analysis}, Shandong Science and Technique Press, Jinan, 1985. \bibitem{g4} Dajun Guo and V. Lakshmikantham; \emph{Nonlinear problems in abstract cones}, Academic Press. Inc. New York, 1988. \bibitem{h1} J. Henderson; H. B. Thompson \emph{Existence of multiple solutions for second order boundary value problems}, Journal of Differential Equations, vol. 166 (2000), 444-454. \bibitem{k1} A. G. Kartsatos, \emph{On the perturbation theorem of $m$-accretive operators in Banach spaces}, Proceeding of American Math. Society, vol. 124 no. 6 (1996), 1811-1820. \bibitem{l1} N. G. Lioyd, \emph{Degree theory}, Cambridge Univ.Press, Cambridge, 1978. \bibitem{m1} M. Meehan, D. O'Regan; \emph{Existence theory for nonlinear Fredholm and Volterra Integral Equations on Half-Open Intervals}, Nonlinear Anal. , vol. 35 (1999), 355-387. \bibitem{o1} D. O'Regan, \emph{Positive solutions for a class of boundary value problems on infinite intervals}, Nonlinear Diff. Eqns. Appl., vol. 1 (1994), 203-228. \bibitem{o2} D. O'Regan, \emph{Theory of Singular Boundary Value Problems}, World Scientific, Singapore, 1994. \bibitem{o3} D. O'Regan, \emph{Existence theory for ordinary differential equations}, Kluwer, Dordrecht, 1997. \bibitem{o4} D. O'Regan, Baoqiang Yan \& R.P.Agarwal; \emph{Nonlinear boundary value problems on the semi-infinite intervals: an upper and lower solution approach}, (to appear) \bibitem{p1} P. K. Palamides and G. N. Galanis; \emph{Positive, unbounded and monotone solutions of the singular second Painlev equation on the half-line}, Nonlinear Analysis, vol. 57 (2004), 401-419. \bibitem{y1} I. Yermachenko, F.Sadyrbaev; \emph{Types of solutions and multiplicity results for two-point nonlinear boundary-value problems}, Nonlinear Analysis, in press. \bibitem{z1} G. Zhang, C. Wang, Y. Liu; \emph{Fixed point theorems on unbounded sets}, Journal of Northeast University, vol. 22 No. 2 (2001), 226-228. (in Chinese) \end{thebibliography} \end{document}