\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 120, pp. 1--13.\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 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/120\hfil Multiple positive solutions] {Multiple positive solutions to fourth-order singular boundary-value problems in abstract spaces} \author[Yansheng Liu\hfil EJDE-2004/120\hfilneg] {Yansheng Liu} \address{Yansheng Liu \hfill\break Department of Mathematics, Shandong Normal University, Jinan, 250014, China} \email{ysliu6668@sohu.com} \date{} \thanks{Submitted September 22, 2004. Published October 14, 2004.} \thanks{Supported by the Rewarded Foundation for Outstanding Middle and Young Scientist, \hfill\break\indent Shandong Province, China} \subjclass[2000]{34G20, 34B15} \keywords{Banach space; singularity; positive solution; fourth order equation} \begin{abstract} We prove the existence of multiple positive solutions to singular boundary-value problems for fourth-order equations in abstract spaces. Our results improve and extend that obtained in \cite{o1,r1,w1}, even in the scalar case. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \section {Introduction} In this paper, we consider the following singular boundary-value problem (BVP) for fourth-order differential equations in a Banach space $E$: x^{(4)}(t)=f(t, x(t)), \quad 00$such that$\|u\|\leq N\|v\|$if$\theta\leq u\leq v$, where$\theta$denotes the zero element of$E$. Evidently,$C[I, E]$is a Banach space with norm$\|x\|_c:= \max_{t\in I}\|x(t)\|$. Moreover, $$C[I, P]:= \{x\in C[I, E]:\ x(t)\in P,\ t\in I\}$$ is a normal cone of$C[I, E]$with the same normal constant$N$as$P$in$E$. A function$x$is said to be a solution of \eqref{e1.1} subject to \eqref{e1.2} if$x\in C^2[I, E]\cap C^4[(0, 1), E]$satisfies \eqref{e1.1} and boundary conditions \eqref{e1.2}; in addition,$x$is said to be a positive solution if$x(t)>\theta$for$t\in (0, 1)$and$x$is a solution of \eqref{e1.1} with \eqref{e1.2}. Let$u: (0, 1]\to E$be continuous. The abstract generalized integral$\int_0^1 u(t)dt$is said to be convergent if the limit$\lim_{\epsilon\to 0^+}\int_{\epsilon}^1u(t)dt$exists. The convergency or divergency of other kinds of generalized integrals can be defined similarly. For a bounded subset$V$of Banach space$E$, by$\alpha(V)$we denote the Kuratowskii measure of noncompactness of$V$(for details, see \cite{d1,l1}). In this paper, the Kuratowskii measure of noncompactness of bounded set in$E$and$C[I, E]$are denoted by$\alpha(\cdot)$and$\alpha_c(\cdot)$, respectively. To conclude this section, we list three lemmas which will be used in Section 3. \begin{lemma}[\cite{g4}] \label{lem2.1} If$V\subset C[J,E]$is bounded and equicontinuous, then$\alpha(V(t))$is continuous on$J$and$\alpha_c(V)=\max\{\alpha(V(t))| t\in J\}$, where$V(t)=\{x(t)|x\in V\}$. \end{lemma} \begin{lemma}[\cite{n1}] \label{lem2.2} Let$K$be a cone of a real Banach space$E$and$B$:$K\to K$a completely continuous operator. Assume that$B$is order-preserving and positively homogeneous of degree$1$and that there exist$v\in K\setminus \{\theta\}$,$\lambda>0$such that$Bv\geq \lambda v$. Then$r(B)\geq\lambda$, where$r(B)$denotes the spectral radius of$B$. \end{lemma} \begin{lemma}[Fixed point theorem of cone expansion and compression \cite{g4}] \label{lem2.3} Let$P$be a cone of a real Banach space$E$and$ P_{r,s}=\{x\in P: r\leq\|x\|\leq s\}$with$s>r>0$. Suppose that$A$:$P_{r, s}\to P$is a strict contraction such that one of the following two conditions is satisfied: \begin{itemize} \item[(i)]$Ax\not\leq x$for$x\in P$,$\|x\|=r$and$Ax\not\geq x$for$x\in P$,$\|x\|=s$. \item[(ii)]$Ax\not\geq x$for$x\in P$,$\|x\|=r$and$Ax\not\leq x$for$x\in P$,$\|x\|=s$. \end{itemize} Then, the operator$A$has a fixed point$x\in P$such that$r<\|x\|0, $$where for t\in (0, 1), z(t):= \min\{t, 1-t\} and$$ f_{r, R}(t):= \sup\{\|f(t, x)\|:\ \frac{z(t)}{N}r\leq \|x\|\leq R,\ x\in P\}. $$\item[(H1)] For every [c, d]\subset (0, 1), and positive numbers R_2> R_1> 0, f(t, x) is uniformly continuous on [c, d]\times\overline{P_{R_2}}\setminus P_{R_1} with respect to t. Here P_r=: \{x\in P:\|x\|< r\} for each r>0. \item[(H2)] For every  t\in (0, 1) and every bounded subset D\subset \overline{P_{R_2}}\setminus P_{R_1} (R_2>R_1>0), we have \alpha(f(t, D))\leq l\alpha(D), where l is a constant with l<15. \item[(H3)] There exist \varphi\in L[I, R^+], [c, d]\subset [0, 1], and \varphi^*\in P^* (here P^* denotes the dual cone of P) with \|\varphi^*\|=1 such that$$ \liminf_{x\to\theta,\; x\in P}\varphi^*(f(t, x)) \geq \varphi (t) $$uniformly with respect to t\in [c, d] and \int_c^ds(1-s)\varphi(s)ds > 0. \item[(H4)] There exist functions a \in C[I, R^+] and b\in C[I, P] with a(t)\not\equiv 0 on every subinterval of I such that$$ f(t, x)\geq a(t)x- b(t)\quad {for }\ t\in (0, 1)\ { and }\ x\in P\setminus\{\theta\}. $$\item[(H5)] There exists a positive number R such that$$ N\int_0^1s(1-s)f_{R,R}(s)ds < 8R, $$where f_{R,R}(s) is the same as in (H0).\end{itemize} Note that assumption (H3) is reasonable since f(t, x) is singular at x=\theta. We assume that (H0) holds throughout the remainder of the paper. To overcome the difficulties arising from singularities, we define $$Q:=\{x\in C[I, P]:\ x(t)\geq z(t)x(s)\geq \theta,\ \forall t, s\in I\}.\label{e3.1}$$ It is easy to see that Q is a nonempty (notice t(1-t)\in Q), convex, and closed subset of C[I,E]. Furthermore, Q is a cone of the Banach space C[I, E] and for every x\in Q\setminus\{\theta\}, by \eqref{e3.1} and the normality of the cone P, we have $$\|x(t)\|\geq \frac{z(t)}{N}\|x\|_c > 0\quad \quad\mbox{for }t\in (0, 1).\label{e3.2}$$ Therefore, x is a positive solution of \eqref{e1.1}-\eqref{e1.2} provided that x\in Q\setminus\{\theta\} is a solution of \eqref{e1.1}-\eqref{e1.2}. Define the operator A on Q\setminus\{\theta\} by $$(Ax)(t):= \int_0^1 J(t, \tau)f(\tau, x(\tau))d\tau, \quad\forall x\in Q\setminus\{\theta\},\label{e3.3}$$ where J(t, \tau)=\int_0^1 G(t,s)G(s, \tau)ds\quad and $$G(t, s)=\begin{cases} t(1-s),& 0\leq t\leq s\leq 1;\\ s(1-t),& 0\leq s\leq t\leq 1. \end{cases} \label{e3.4}$$ Now we show the operator A is well defined on Q\setminus\{\theta\}. First we claim that for each x\in Q\setminus\{\theta\}, \int_0^1 G(s, \tau)f(\tau, x(\tau))d\tau  is convergent. In fact, since x\in Q\setminus\{\theta\}, we can see by \eqref{e3.2} that \|x\|_c\neq 0 and$$ \frac{z(t)}{N}\|x\|_c\leq \|x(\tau)\| \leq \|x\|_c\quad \mbox{for each } \tau\in (0, 1). $$This together with G(s, \tau)\leq \tau(1-\tau) for all  s, \tau\in I and (H0) implies that \int_0^1 G(s, \tau)f(\tau, x(\tau))d\tau  is convergent and$$ \int_0^1 \tau(1-\tau)\|f(\tau, x(\tau))\|d\tau < +\infty\quad \mbox{for each } x\in Q\setminus\{\theta\}. $$The Lebesgue dominated convergence theorem yields that for every x\in Q\setminus\{\theta\}, \int_0^1 G(s, \tau)f(\tau, x(\tau))d\tau  is continuous in s on I. Therefore, by \eqref{e3.3} we obtain that Ax\in C^2[I, P] and \begin{gathered} (Ax)^{(4)}(t)=f(t, x(t)), \quad 0\theta for t\in (0, 1) and satisfies \eqref{e1.1}-\eqref{e1.2}. Now we show x\in Q\setminus\{\theta\}. To see this notice that $$x(t)= \int_0^1 G(t, s)\int_0^1 G(s, \tau) f(\tau, x(\tau))d\tau ds\quad \mbox{for } t\in I.\label{e3.6}$$ This and \frac{G(t, s)}{G(\xi, s)}= \begin{cases} \frac{t(1-s)}{\xi (1-s)}\geq t\geq z(t), & t, \xi\leq s;\\[3pt] \frac{s(1-t)}{s(1-\xi)}\geq 1-t\geq z(t),& t, \xi\geq s;\\[3pt] \frac{t(1-s)}{s(1-\xi)}\geq t\geq z(t), & ts>\xi \end{cases} \label{e3.7} yield$$ x(t)\geq z(t)\int_0^1 G(\xi,s)\int_0^1 G(s, \tau)f(\tau, x(\tau))d\tau\quad \mbox{for } t, \xi\in I, $$which implies x\in Q\setminus\{\theta\}. On the other hand, it is easy to see by \eqref{e3.6} that Ax=x. This completes the proof. \end{proof} Consequently, the existence of positive solution for \eqref{e1.2} is equivalent to that of fixed point of A in Q\setminus\{\theta\}. By \eqref{e3.5} and the process similar to the proof of Lemma \ref{lem3.1}, we also obtain the following Lemma. \begin{lemma} \label{lem3.2} A(Q\setminus\{\theta\})\subset Q. \end{lemma} \begin{lemma} \label{lem3.3} For every pair of positive numbers R_2 and R_1 with R_2>R_1>0, A: \overline{Q_{R_2}}\setminus Q_{R_1}\to Q is a strict set contraction, where Q_r:= \{x\in Q:\ \|x\|_c0). \end{lemma} \begin{proof} First, under the assumptions for R_2 and R_1, (H0) guarantees that for each x\in\overline{Q_{R_2}}\setminus Q_{R_1}, $$\int_0^1 \tau(1-\tau)\|f(\tau, x(\tau))\|d\tau \leq \int_0^1\tau(1-\tau)f_{R_1, R_2}(\tau)d\tau < +\infty \label{e3.8}$$ which implies A: \overline{Q_{R_2}}\setminus Q_{R_1}\to Q is bounded. Next we show A: \overline{Q_{R_2}}\setminus Q_{R_1}\to Q is continuous. To see this from \eqref{e3.3} it follows that for x\in\overline{Q_{R_2}}\setminus Q_{R_1} and t_1, t_2\in I, $$\|(Ax)(t_1)-(Ax)(t_2)\|\leq \int_0^1 |G(t_1, s)-G(t_2, s)|ds\int_0^1 G(s, \tau)\|f(\tau, x(\tau))\|d\tau. \label{e3.9}$$ This and \eqref{e3.8} yield that for every subset V\subset \overline{Q_{R_2}}\setminus Q_{R_1}, (AV)(t) is equicontinuous on I, where (AV)(t)= \{(Ax)(t): x\in V\}, t\in I. Let x_n, x\in \overline{Q_{R_2}}\setminus Q_{R_1} with \|x_n-x\|_c\to 0  as n\to +\infty . This implies$$ \|x_n(t)-x(t)\|\to 0\quad \mbox{as } n\to +\infty \quad\mbox{for } t\in I. $$From Lebesgue dominated convergence theorem and \eqref{e3.8}, it follows that$$ \|(Ax_n)(t)-(Ax)(t)\|\to 0\quad \mbox{as } n\to +\infty. $$Thus, \{(Ax_n)(t)\} is relatively compact for every t\in I. From this and the equicontinuity of \{Ax_n(t)\} by the Ascoli-Arzela theorem, we obtain that \{Ax_n\} is a relatively compact subset of Q. Now it remains to show \|Ax_n-Ax\|_c\to 0 as n\to +\infty. In fact, if this is not true, then there is a constant \epsilon_0>0 and a subsequence \{x_{n_i}\} of \{x_n\} such that \|Ax_{n_i}-Ax\|_c\geq \epsilon_0 (i=1,2,\dots). However, the relative compactness of \{Ax_n\} implies that \{Ax_{n_i}\} contains a subsequence which converges in C[I, P]. Without loss of generality, we may assume that \{Ax_{n_i}\} itself converges to y, that is, \|Ax_{n_i}-y\|_c\to 0  as i\to +\infty . So we have y=Ax. This is a contradiction. Therefore, A is continuous. Finally, we show A: \overline{Q_{R_2}}\setminus Q_{R_1}\to Q is a strict set contraction, that is, there exists k\in (0, 1) such that \alpha_c(AV)\leq k\alpha_c(V)\quad for each V\subset\overline{Q_{R_2}}\setminus Q_{R_1}. Fix V\subset\overline{Q_{R_2}}\setminus Q_{R_1}. Let $$(A_nx)(t):= \int_{1/n}^{1-(1/n)} J(t, s)f(s, x(s))ds\quad \mbox{for } x\in V.\label{e3.10}$$ By \eqref{e3.8} we know $$(A_nx)(t)\to (Ax)(t)\quad \mbox{as } n\to +\infty\quad \mbox{for each } x\in V \mbox{ and } t\in I.\label{e3.11}$$ This implies$$ d_H((A_nV)(t), (AV)(t))\to 0\quad \mbox{as } n\to +\infty \quad \mbox{for each } t\in I, $$where d_H(\cdot, \cdot) denotes the Hausdorff metric. Thus, by the property of noncompactness measure, $$\alpha((A_nV)(t))\to\alpha((AV)(t))\quad \quad\mbox{for }t\in I.\label{e3.12}$$ In what follows, we estimate \alpha((A_nV)(t)) for each t\in I. Note that x(s)\geq z(s)x(\tau)\geq \theta\quad for s, \tau \in I and x\in V. Thus,$$ \frac{R_1}{nN}\leq \frac{\|x\|_c}{nN}\leq \|x(s)\|\leq R_2\quad \quad\mbox{for } s\in[\frac{1}{n}, 1-\frac{1}{n}]. $$By the definition of integration and \eqref{e3.4}, respectively, we have$$ \int_{1/n}^{1-(1/n)} J(t, s)f(s, x(s))ds\in(1-\frac{2}{n})\overline{co}(\{J(t,s)f(s, x(s)): s\in[\frac{1}{n}, 1-\frac{1}{n}]\} $$and$$ J(t, s)=\int_0^1 G(t, \tau)G(\tau, s)d\tau\leq \int_0^1 \tau^2(1-\tau)^2d\tau =\frac{1}{30}\quad \mbox{for all } t, s\in I. These, (H1), (H2), and Lemma \ref{lem2.1} guarantee that \begin{aligned} \alpha((A_nV)(t)) &=\alpha (\{\int_{1/n}^{1-(1/n)} J(t, s)f(s, x(s))ds |\ x\in V\})\\ &\leq (1-\frac{2}{n})\alpha (\overline{co}\{J(t, s)f(s, x(s)) |\ s\in [\frac{1}{n},\ 1-\frac{1}{n}],\ x\in V\})\\ &\leq \alpha(\{J(t, s)f(s, x(s))|\ s\in [\frac{1}{n},\ 1-\frac{1}{n}],\ x\in V\})\\ &\leq \frac{1}{30}\alpha(\{f(s, x(s))|\ s\in [\frac{1}{n},\ 1-\frac{1}{n}],\ x\in V\})\\ &\leq \frac{1}{30}\alpha(f(I_n\times V(I_n))\leq \frac{1}{30}l\cdot\alpha V(I_n))\leq \frac{1}{15}l\alpha_c(V), \end{aligned}\label{e3.13} where I_n= [\frac{1}{n}, 1-\frac{1}{n}] and V(I_n)= \{x(s): x\in V,\ s\in I_n\}. Combining Lemma \ref{lem2.1} with \eqref{e3.9}, \eqref{e3.12}, and \eqref{e3.13} again, one can obtain\alpha_c(AV)= \max_{t\in I}\alpha ((AV)(t))\leq \frac{1}{15}l\cdot \alpha_c(V). $$This implies A is a strict set contraction with k=\frac{1}{15}l< 1 from \overline{Q_{R_2}}\setminus Q_{R_1} to Q. \end{proof} Using (H4), we define an operator L on C[I, R] by$$ (Lu)(t):= \int_0^1 J(s, t)a(s)u(s)ds=\int_0^1 G(\tau, t)\int_0^1 G(s, \tau)a(s)u(s)dsd\tau $$for u\in C[I, R] where J is given by \eqref{e3.4}, and a is the same as in (H4). It is easy to see L: C[I, R]\to C[I, R] is a completely continuous positive operator. Note that if v(t)=t(1-t) on I, then \|v\|_c=\frac{1}{4}. By G(t, \tau)\geq t\tau (1-t)(1-\tau) for t, \tau\in I, we know$$ (Lv)(t)\geq \int_0^1\tau ^2(1-\tau)^2d\tau\int_0^1 s^2(1-s)^2a(s)ds\cdot v(t)=\delta_0 v(t)\quad\quad\mbox{for }t\in I, where  \delta_0 =\frac{1}{30}\int_0^1 s^2(1-s)^2a(s)ds > 0. From Lemma \ref{lem2.2} it follows that the spectral radius r(L)\geq \delta_0>0. So the well-known Krein-Rutman theorem \cite{n1} guarantees that there exists an p\in C[I, R^+] with p(t)\not\equiv 0 on I such that $$(Lp)(t)=\int_0^1J(s,t)a(s)p(s)ds=r(L)p(t)\quad\quad\mbox{for }t\in I. \label{e3.14}$$ From \eqref{e3.7} one deduces that \begin{align*} p(t) &=\frac{1}{r(L)}\int_0^1(\int_0^1 G(s, \tau)G(\tau, t)d\tau)a(s)p(s)ds\\ &\geq \frac{1}{r(L)}\int_0^1 z(s)\int_0^1 G(\xi, \tau)G(\tau, t)d\tau)a(s)p(s)ds\\ &\geq \frac{1}{r(L)}\int_0^1 z(s)a(s)p(s)ds\cdot J(\xi, t)\quad\mbox{for all } t, \xi\in I. \end{align*} Therefore, $$p(t)\geq \delta J(\xi, t)\quad{\rm for\ all}\ t, \xi\in I, \label{e3.15}$$ where \delta:= \frac{1}{r(L)}\int_0^1 z(s)a(s)p(s)ds. This and \eqref{e3.14} guarantees that $$\int_0^1 p(t)a(t)dt\geq \delta r(L).\label{e3.16}$$ Now we are ready to give the main result of the present paper. \begin{theorem} \label{thm3.1} Assume that (H0)--(H5) hold and r(L)>1. Then \eqref{e1.1} subject to \eqref{e1.2} has at least two positive solutions. \end{theorem} \begin{proof} Set $$K:= \{x\in Q:\quad \int_0^1p(t)a(t)x(t)dt\geq \delta r(L)x(s),\quad \forall s\in I\},\label{e3.17}$$ where Q is given by \eqref{e3.1}, a(t) is given by (H4), p(t), r(L), and \delta are given by \eqref{e3.14} and \eqref{e3.15}. By \eqref{e3.16} it is easy to see K\setminus\{\theta\}\neq\emptyset and K is also a cone of C[I, E]. We now prove that the operator A defined by \eqref{e3.3} maps Q\setminus\{\theta\} into K. In fact, for x\in Q\setminus\{\theta\}, it follows from \eqref{e3.3}, \eqref{e3.15}, and \eqref{e3.2} that \begin{align*} \int_0^1p(t)a(t)(Ax)(t)dt &=\int_0^1p(t)a(t)\int_0^1 J(t, s)f(s, x(s))dsdt\\ &=\lim_{n\to +\infty} \int_0^1 p(t)a(t)\int_{1/n}^{1-(1/n)} J(t, s)f(s, x(s))dsdt\\ &= \lim_{n\to +\infty} \int_{1/n}^{1-(1/n)} f(s, x(s))ds\int_0^1 J(t, s)a(t)p(t)dt\\ &= r(L)\lim_{n\to +\infty} \int_{1/n}^{1-(1/n)} p(s)f(s, x(s))ds \\ &\geq \delta r(L)\lim_{n\to +\infty} \int_{1/n}^{1-(1/n)} J(\tau, s)f(s, x(s))ds\\ &= \delta r(L)\lim_{n\to +\infty} \int_0^1 J(\tau, s)f(s, x(s))ds \\ &= \delta r(L)(Ax)(\tau)\quad \mbox{for all } \tau\in I, \end{align*} which implies A(Q\setminus\{\theta\})\subset K. Consequently, we obtain by Lemma \ref{lem3.2} and Lemma \ref{lem3.3} that A: \overline{K_{R_2}}\setminus K_{R_1}\to K is a strict set contraction for every pair of positive numbers R_2 and R_1 with R_2> R_1>0, where K_{R_1}= \{x\in K: \|x\|_c< R_1\}. Choose a positive number R_0 with R_0 > R such that R_0> \frac{N\|b\|_c}{30\delta(r(L) - 1)}\int_0^1 a(t)p(t)dt. $$We proceed to prove $$Ax\not\leq x\quad\mbox{for all } x\in\partial K_{R_0}. \label{e3.18}$$ Suppose, on the contrary, there exists an x_0\in\partial K_{R_0} such that Ax_0\leq x_0. Therefore,$$ x_0(t)\geq (Ax_0)(t) = \int_0^1 J(t,s)f(s, x_0(s))ds\quad\mbox{for all } t\in I. Multiply by p(t)a(t) and integrate from 0 to 1 to obtain \begin{align*} & \int_0^1 p(t)a(t)x_0(t)dt \geq \int_0^1 p(t)a(t)\int_0^1 J(t, s)f(s, x_0(s))dsdt\\ &\geq \int_0^1 p(t)a(t)\int_0^1 J(t, s)a(s)x_0(s)dsdt - \int_0^1\int_0^1 J(t, s)a(t)p(t)b(s)dsdt\\ &= \int_0^1(\int_0^1 J(t, s)a(t)p(t)dt)a(s)x_0(s)ds -\int_0^1\int_0^1 J(t, s)a(t)p(t)b(s)dsdt\\ &= r(L)\int_0^1 p(s)a(s)x_0(s)ds - \int_0^1\int_0^1 J(t, s)a(t)p(t)b(s)dsdt. \end{align*} This and \eqref{e3.17} yield \begin{align*} \int_0^1\int_0^1 J(t, s)a(t)p(t)b(s)dsdt &\geq (r(L) -1)\int_0^1 p(s)a(s)x_0(s)ds\\ &\geq (r(L)-1)\delta x_0(\tau)\geq \theta\quad\mbox{for all } \tau\in I. \end{align*} The normality of the cone P and |J(t, s)|\leq \frac{1}{30} for all t, s\in I guarantee that \frac{N\|b\|_c}{30}\int_0^1 a(t)p(t)dt\geq \delta(r(L) - 1)R_0. $$This is a contradiction with the selection of R_0. Consequently, \eqref{e3.18} holds. In what follows we show $$Ax\not\geq x\quad\mbox{for all } x\in\partial K_R. \label{e3.19}$$ If this is false, then there exists an x_1\in\partial K_R such that x_1\leq Ax_1, that is,$$ \theta\leq x_1(t)\leq (Ax_1)(t)\quad\quad\mbox{for }t\in I. $$Since x_1\in K\subset Q, we get$$ x_1(t)\geq z(t)x_1(\tau)\geq \theta\quad\quad\mbox{for }t, \tau\in I. $$As a result,$$\frac{z(t)}{N}R = \frac{z(t)}{N}\|x_1\|_c \leq \|x_1(t)\| \leq R\quad\quad\mbox{for }t\in I. $$This implies$$ \|f(t, x_1(t))\| \leq f_{R, R}(t)\quad\quad\mbox{for }t\in (0, 1). Combining the above with (H5) we know \begin{align*} \|x_1(t)\|&\leq N\|(Ax_1)(t)\|\leq N\|\int_0^1 J(t, s)f(s, x_1(s))ds\|\\ &\leq N \int_0^1 J(t, s)\|f(s, x_1(s))\|ds\\ &\leq N\int_0^1 J(t, s)f_{R, R}(s)ds\\ &\leq N\int_0^1 G(t, \tau)(\int_0^1 s(1-s)f_{R, R}(s)ds)d\tau\\ &\leq \frac{N}{8}\int_0^1 s(1-s)f_{R, R}(s)ds < R\quad\quad\mbox{for }t\in I. \end{align*} This is a contradiction. Then \eqref{e3.19} follows. Finally, we prove that there exists a positive number R' with R' < R such that $$Ax\not\leq x\quad\quad\mbox{for }x\in \partial K_{R'}.\label{e3.20}$$ In fact, by (H3), given \epsilon\in (0, \int_c^d J(\frac{1}{2},s)\phi(s)ds), there exists an R''> 0 such that $$\phi^*(f(t, x(t))\geq \phi(t) -\epsilon\quad\quad\mbox{for }t\in [c, d] \mbox{ and } x\in P_{R''}\setminus\{\theta\}.\label{e3.21}$$ Choose $$R':= \min\big\{\frac{R}{2},\ R'', \ \int_c^d J(\frac{1}{2}, s)\phi(s)ds- \epsilon\big\}. \label{e3.22}$$ Now we are ready to prove \eqref{e3.20} holds. Suppose, on the contrary, there exists an x_2\in \partial K_{R'} such that Ax_2\leq x_2. Then by \eqref{e3.2} we know \frac{z(t)}{N}R'\leq \|x_2(t)\| \leq R'\leq R''\quad\quad\mbox{for }t\in (0, 1). This and \eqref{e3.21} guarantee that \begin{align*} \phi^*(x_2(t))&\geq \int_0^1 J(t, s)\phi^*(f(s, x_2(s))ds\\ &\geq \int_c^d J(t, s)[\phi(s) - \epsilon]ds \\ &> \int_c^d J(t, s)\phi(s)ds - \frac{\epsilon}{2}\quad\quad\mbox{for }t\in (0, 1). \end{align*} Consequently, \|x_2\|_c\geq \phi^*(x_2(\frac{1}{2}))\geq \int_c^d J(\frac{1}{2}, s)\phi(s)ds - \frac{\epsilon}{2} > R'. $$This is a contradiction with x_2\in \partial K_{R'}. Then the conclusion follows from Lemma \ref{lem2.3}. \end{proof} \begin{remark} \label{rmk3.2}\rm If r(L)>1 is replaced with \int_0^1 s^2(1-s)^2a(s)ds > 30 in Theorem \ref{thm3.1}, the conclusion of Theorem \ref{thm3.1} also holds. In fact, by v(t)=t(1-t)\in Q and G(t, s)\geq ts(1-t)(1-s) for t, s\in I, and using the definition of the operator L, one can obtain$$ Lv\geq \frac{1}{30}\int_0^1 s^2(1-s)^2a(s)ds\cdot v. $$Then by Lemma \ref{lem2.2}, r(L)>1 follows. \end{remark} For the next theorem we replace (H4) by \begin{itemize} \item[(H'4)] There exists an \psi^*\in P^* with \|\psi^*\| = 1 and a subinterval [c', d']\subset (0, 1) such that$$ \lim_{\|x\|\to +\infty,\; x\in P}\frac{\psi^*(f(t, x))}{\|x\|} > \mu $$uniformly with respect to t\in [c', d'], where$$ \mu= N\Big(\min\{c', 1-d'\}\cdot\int_{c'}^{d'} J(\frac{1}{2}, s)ds\Big)^{-1}. $$\end{itemize} \begin{theorem} \label{thm3.2} Assume that (H0)-(H3), (H'4), and (H5) hold. Then \eqref{e1.1} subject to \eqref{e1.2} has at least two positive solutions. \end{theorem} \begin{proof} Consider the operator A in Q\setminus\{\theta\}, where Q is defined by \eqref{e3.1}. From the proof of Theorem \ref{thm3.1}, it is not difficult to see that \eqref{e3.19} and \eqref{e3.20} hold for x\in\partial Q_{R'} and x\in\partial Q_{R}, respectively; where R' is given by \eqref{e3.22} and R is given by (H5). It remains to show that there exists a positive number R_0 with R_0> R such that \eqref{e3.18} holds for x\in\partial Q_{R_0}. By (H'4), there exist an \epsilon>0 and an R_1>0 such that $$\psi^*(f(t, x))\geq (\mu+\epsilon)\|x\|\quad\quad\mbox{for }t\in[c', d'],\ x\in P, \mbox{ and }\|x\|\geq R_1.\label{e3.23}$$ Choose$$ R_0\geq \max\big\{R+1,\ \frac{NR_1}{\min\{c', 1-d'\}}\big\}. $$We proceed to prove \eqref{e3.18} holds for x\in\partial Q_{R_0}. If this is false, then there exists an x_0\in\partial Q_{R_0} such that Ax_0\leq x_0. By \eqref{e3.1} we know$$ x_0(t)\geq z(t)x_0(s)\geq \min\{c',1-d'\}\cdot x_0(s)\geq \theta\quad\quad \mbox{for }t\in[c', d'] \mbox{ and } s\in I. $$>From the normality of the cone P it follows that$$ N\|x_0(t)\| \geq \min\{c',\ 1-d'\}\cdot\|x_0\|_c = \min\{c',\ 1-d'\}\cdot R_0\quad\quad\mbox{for }t\in[c',\ d'], that is, \|x_0(t)\|\geq R_1 for t\in [c,\ d']. This, \eqref{e3.23}, and \eqref{e3.3} guarantee that \begin{align*} R_0 &\geq \psi^*(x_0(\frac{1}{2}))\\ &\geq \int_0^1 J(\frac{1}{2}, s)\psi^*(f(s, x_0(s))ds\\ &\geq \int_{c'}^{d'} J(\frac{1}{2}, s)(\mu+\epsilon)\|x_0(s)\|ds \\ &\geq \frac{\min\{c', 1-d'\}}{N}\int_{c'}^{d'} J(\frac{1}{2}, s)(\mu+\epsilon)R_0ds > R_0, \end{align*} which is a contradiction. Then the statement of Theorem \ref{thm3.2} follows. \end{proof} \begin{corollary} \label{coro0} Suppose (H0), (H1), (H2), and one of the following conditions are satisfied: \begin{itemize} \item[(i)] (H3) and (H5). \item[(ii)] (H4), (H5), and r(L)>1. \item[(iii)] (H'4) and (H5). \end{itemize} Then \eqref{e1.1} subject to \eqref{e1.2} has at least one positive solution. \end{corollary} \begin{remark} \label{rmk3.3} \rm By the same method used above, we can study the existence of multiple positive solutions of second order nonlinear singular boundary-value problems in scalar or in abstract space. \end{remark} \section{Examples} \begin{example} \label{ex1} \rm Consider the boundary-value problem consisting of a finite system of fourth-order scalar nonlinear differential equations. \begin{gathered} x_n^{(4)}(t)=\frac{1}{\sqrt{t(1-t)}}\big(x_n^{\frac{3}{2}} + \frac{1}{\max_{1\leq i\leq m}|x_i|}\big), \quad 00 and $$\int_0^1 t(1-t)f_{1, 1}(t)dt\leq \int_0^1[\sqrt{t(1-t)} + \frac{1}{\sqrt{t(1-t)}}]dt < (\frac{1}{8} + \pi) < 8.$$ Therefore, by Theorem \ref{thm3.1} or Theorem \ref{thm3.2}, our conclusion follows. \end{proof} \begin{example} \label{ex2} \rm Consider the boundary-value problem consisting of an infinite system of fourth order scalar nonlinear differential equations. $$\begin{gathered} x_n^{(4)}(t)=f_n(t, x(t)), \quad t\in (0, 1); \\ x_n(0)=x_n(1)=x_n''(0)=x_n''(1)=0,\quad (n=1, 2,\dots ). \end{gathered} \label{e4.2}$$ where \begin{gather*} f_1(t,x)=\frac{1}{\sqrt{t(1-t)}}\big(x_1 + \frac{x_2}{2} + (\sum_{i\geq 1}|x_i|)^2 + b_1\big),\\ f_2(t,x)=\frac{1}{\sqrt{t(1-t)}}\big( \frac{x_2}{2} + \frac{x_3}{3}+ \frac{1}{\sum_{i\geq 1}|x_i|} + b_2\big),\\ f_n(t, x)= \frac{1}{\sqrt{t(1-t)}}\big(\frac{x_n}{n} + \frac{x_{n+1}}{n+1} + b_n\big),\quad n= 3, 4, \dots;\\ x=(x_1, x_2, \dots),\quad b_i\geq 0 \quad(i=1, 2, \dots),\quad \sum_{i\geq 1}b_i\leq 1. \end{gather*} \textbf{Claim:} System \eqref{e4.2} has at least two positive solutions $x^*(t)= (x_1^*(t), x_2^*(t),\dots )$ and $x^{**}(t)= (x_1^{**}(t), x_2^{**}(t),\dots )$ such that $$0< \sum_{1\leq i,\; t\in[0, 1]}|x_i^*(t)| < 1 < \sum_{1\leq i,\;t\in[0, 1]}|x_i^{**}(t)|.$$ \end{example} \begin{proof} Let $E= l^{1}$ with norm $\|x\|= \sum_{i\geq 1}|x_i|$ and $$P=\{x=(x_1, x_2, \dots ): x_n\geq 0 \quad\mbox{for }n=1, 2, \dots \}.$$ Then $P$ is a normal cone in $E$ and the normal constant is $N=1$. System \eqref{e4.2} can be regarded as a boundary-value problem form \eqref{e1.1} with \eqref{e1.2}, where $x=(x_1, x_2, \dots )$, $$f(t, x)= (f_1(t, x_1, \dots ),\dots, f_n(t, x_1, \dots ),\dots,).$$ Evidently, $f\in C[ (0,1)\times P\setminus\{\theta\}, P]$ and is singular at $t=0$, $t=1$, and $x=\theta =(0,0,\dots, )$. Note that for $t\in (0, 1)$ and $R\geq r>0$, $$f_{r, R}(t)\leq \frac{1}{\sqrt{t(1-t)}}\big( 2R + R^2 + \frac{1}{rt(1-t)} + \|b\|\big)$$ So, (H0) is satisfied. In addition, (H1) is obvious. As in \cite[Example 2.1.2]{g4}, one can see (H2) is satisfied with $l=0$. Choosing $\phi^*=\psi^*=(1, 1, 0,\dots, 0, \dots)$, we know that (H3) and (H'4) holds for \eqref{e4.1}. From $$\int_0^1 s(1-s)f_{1,1}(s)ds\leq \int_0^1 ((3+\|b\|)\sqrt{s(1-s)} + \frac{1}{\sqrt{s(1-s)}})ds \leq (\frac{4}{8} +\pi) < 8,$$ it follows that (H5) is satisfied. By Theorem \ref{thm3.2}, our conclusion follows. \end{proof} \begin{thebibliography}{00} \bibitem{a1} A. R. Aftabizadeh, \emph{Existence and uniqueness theorems for fourth-order boundary value problem}, J. Math. Anal. Appl., 116 (1986), 415-426. \bibitem{a2} R. P. Agarwal, \emph{Some new results on two point boundary value problems for higher order differential equations}, Funkcial. Ekvac., 29 (1986), 197-212. \bibitem{d1} K. Deimling, \emph{Ordinary differential equations in Banach spaces}, LNM 886. New York, Berlin: Springer-Verlag, 1987. \bibitem{e1} P. W. Eloe and J. Henderson, \emph{Nonlinear boundary value problems and a priori bounds on solutions}, SIAM J. Math. Anal., 15 (1984), 642-647. \bibitem{g1} A. Granas, R. B. Guenther, and J. W. Lee, \emph{Nonlinear boundary value problems for some classes of ordinary differential equations}, Rocky Mountain J. Math., 10 (1980),35-58. \bibitem{g3} J. R. Graef and B. Yang, \emph{On a nonlinear boundary value problem for fourth order equations}, Appl. Anal., 72 (1999), 439-448. \bibitem{g4} D. Guo, V. Lakshmikantham and X. Liu. \emph{Nonlinear Integral Equations in Abstract Spaces}, Kluwer Academic Publishers, 1996. \bibitem{g2} C. P. Gupta, \emph{Existence and uniqueness theorems for the bending of an elastic beam equation}, Appl. Anal., 26 (1988), 289-304. \bibitem{j1} L. K. Jackson, \emph{Existence and uniqueness of solutions of boundary value problems for Lipschitz equations}, J. Differential Equations, 32 (1979), 76-90. \bibitem{l1} V. Lakshmikantham and S. Leela, \emph{Nonlinear differential equations in abstract spaces}, Pergamon, Oxford, 1981. \bibitem{m1} R. Ma, J. Jihui and F. Shengmao, \emph{The method of lower and upper solutions for fourth-order two-point boundary value problems}, J. Math. Anal. Appl., 215 (1997),415-422. \bibitem{m2} R. H. Martin, \emph{Nonlinear operators and differential equations in Banach spaces}, John Wiley and Sons, New York, 1976. \bibitem{n1} R. D. Nussbaum, \emph{Eigenvectors of nonlinear positive operators and the linear Krein-Rutman theorem}. In \emph{Fixed Point Theory} LNM 886. New York, Berlin: Springer-Verlag, 1980, 309-330. \bibitem{o1} D. O'Regan, \emph{Solvability of some fourth (and higher) order singular boundary value problems}, J. Math. Anal. Appl., 161 (1991),78-116. \bibitem{r1} D. O'Regan, \emph{Fourth (and higher) order singular boundary value problems}, Nonlinear Anal., 14 (1990), 1001-1038. \bibitem{w1} Z. Wei, \emph{Positive solutions of singular boundary value problems of fourth order differential equations}, Acta Mathematica Sinica, 42 (1997), 715-722. \end{thebibliography} \end{document}