\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 39, pp. 1--10.\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-2006/39\hfil Positive solutions] {Positive solutions of singular fourth-order boundary-value problems} \author[Y. Cui, Y. Zou\hfil EJDE-2006/39\hfilneg] {Yujun Cui, Yumei Zou} % in alphabetical order \address{Department of Applied Mathematics, Shandong University of Science and technology, Qingdao, 266510, China} \email[Y. Cui]{cyj720201@163.com} \date{} \thanks{Submitted September 6, 2005. Published March 21, 2006} \thanks{Supported by grant 10371066 from the National Science Foundation of China} \subjclass[2000]{34A34, 34B15, 45G15} \keywords{Singular boundary value problem; fixed point theorem; \hfill\break\indent positive solution} \begin{abstract} In this paper, we present necessary and sufficient conditions for the existence of positive $C^3[0,1]\cap C^4(0,1)$ solutions for the singular boundary-value problem \begin{gather*} x''''(t)=p(t)f(x(t)),\quad t\in(0,1);\\ x(0)=x(1)=x'(0)=x'(1)=0, \end {gather*} where $f(x)$ is either superlinear or sublinear, $p:(0,1)\to [0,+\infty)$ may be singular at both ends $t=0$ and $t=1$. For this goal, we use fixed-point index results. \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} \allowdisplaybreaks \section{Introduction} In this paper, we consider the fourth order differential equation \begin{gather} x''''(t)=p(t)f(x(t)),\quad t\in(0,1);\label{e1.1}\\ x(0)=x(1)=x'(0)=x'(1)=0.\label{e1.2} \end{gather} where $f(x)$ is either superlinear or sublinear, $p:(0,1)\to [0,+\infty)$ may be singular at both ends $t=0$ and $t=1$. Recently, the existence and multiplicity of positive solutions of \eqref{e1.1}-\eqref{e1.2} in the non-singular case has been extensively studied in the literature; see \cite{m2,k1,y1} and references therein. However for singular fourth order boundary-value problems, the research has proceeded very slowly. Ma and Tisdell \cite{m1} studied the singular sublinear fourth order boundary value problems \begin{gather} x''''(t)=p(t)x^{\lambda}(t),\quad t\in(0,1);\label{e1.3} \\ x(0)=x(1)=x'(0)=x'(1)=0.\label{e1.4} \end{gather} where $\lambda\in (0,1)$ is given, and $p:(0,1)\to [0,\infty)$ may be singular at both ends $t=0$ and $t=1$. Base upon the method of lower and upper solutions, Ma and Tisdell showed that \eqref{e1.3}-\eqref{e1.4} has a positive solution in $C^2[0,1]\cap C^4(0,1)$ if and only if $$0<\int_0^1t^{1+2\lambda}(1-t)^{1+2\lambda}p(t)dt<+\infty.$$ Moreover, this positive solution is in $C^3[0,1]\cap C^4(0,1)$ if and only if $$0<\int_0^1t^{2\lambda}(1-t)^{2\lambda}p(t)dt<+\infty.$$ But necessary and sufficient conditions for the existence of positive solution of superlinear BVPs \eqref{e1.3}-\eqref{e1.4} still remain unknown. In this paper, by using the fixed point index, we give some necessary and sufficient conditions for the existence of $C^3[0,1]\cap C^4(0,1)$ positive solutions to the singular boundary value problem \eqref{e1.1}-\eqref{e1.2}. In our discussion, by a $C^k[0,1]$ solution $(k=2,3)$ of \eqref{e1.1}-\eqref{e1.2} we mean a function $y(t)\in C^k[0,1]\cap C^4(0,1)$ which satisfies \eqref{e1.2} and \eqref{e1.1} on (0,1). We call a solution $y(t)$ is a positive solution if $y(t)>0$ for $t\in (0,1)$. This paper is organized as follows. Section 2 gives some preliminary lemmas corresponding to \eqref{e1.1}-\eqref{e1.2}. Section 3 is devoted to the the existence of $C^3[0,1]\cap C^4(0,1)$ positive solutions for \eqref{e1.1}-\eqref{e1.2}. At the end of this section we state some lemmas of the fixed point theory, which will be used in Section 3. Let $E$ be a Banach space, $P$ a cone in $E$, $\Omega$ a bounded open set in $E$. \begin{lemma}[\cite{g1}] \label{lem1.1} Let $\theta\in \Omega$, $A:\overline{\Omega}\cap P\to P$ be completely continuous. Suppose that there exists $u_{0}\in P\backslash\{\theta\}$ such that $$u-Au\neq\mu u_{0},\ \forall\ u\in\partial\Omega\cap P,\ \mu\geq0,$$ then the fixed point index $i(A,\ \Omega\cap P,\ P)=0$. \end{lemma} \begin{lemma}[\cite{g1}] \label{lem1.2} Let $\theta\in \Omega$, $A:\overline{\Omega}\cap P\to P$ be completely continuous. Suppose that $$Au\neq\mu u,\quad \forall\ u\in\partial\Omega\cap P,\quad \mu\geq1,$$ then the fixed point index $i(A,\ \Omega\cap P,\ P)$ is equal to 1. \end{lemma} \section{Preliminaries} We give some notations, which will be used below. Let $C[0,1]$, $C^k[0,1]$ and $L^1[0,1]$ be the classical Banach spaces with their usual norms $\|\cdot\|$, $\|\cdot\|_{C^k}$ and $\|\cdot\|_{L^1}$, respectively. Let $AC[0,1]$ be the space of all absolutely continuous functions on [0,1]. Let $$AC^k[0,1]=\{u\in C^k[0,1]: u^{(k)}\in AC[0,1]\}.$$ Clearly $AC^0[0,1]=AC[0,1]$. Let $I$ be an interval of $R$. We denote by $L^1_{\rm loc}I$ the spaces of functions defined by $$L^1_{\rm loc}I=\{u:I\to R: u|_{[c,d]}\in L^1[c,d]\ \ \text{for\ every\ compact\ interval}\ [c,d]\subset I\}.$$ For $n,m\in N$, we denote by $X[n,m]$ the Banach space $$X[n,m]=\{\varphi\in L^1_{\rm loc}(0,1)\Big|\ \int_0^1t^n(1-t)^m|\varphi(t)|dt<+\infty\},$$ equipped with the norm $$\|\varphi\|_{X[n,m]}=\int_0^1t^n(1-t)^m|\varphi(t)|dt.$$ Now let $G(t,s)$ be the Green's function of the linear problem \begin{gather*} x''''(t)=0,\quad t\in(0,1);\\ x(0)=x(1)=x'(0)=x'(1)=0, \end{gather*} which can be explicitly given by $$G(t,s)=\frac16\begin{cases} t^2(1-s)^2[(s-t)+2(1-t)s],& 0\leq t\leq s\leq 1,\\ s^2(1-t)^2[(t-s)+2(1-s)t], & 0\leq s\leq t\leq 1. \end{cases}$$ It is clear that for all $t,s\in [0,1]$, $$\frac13t^2(1-t)^2s^2(1-s)^2\leq G(t,s)\leq \frac12t^2(1-t)^2,\ G(t,s)\leq \frac12s^2(1-s)^2.\label{e2.1}$$ Suppose that $\varphi\in X[2,2]$. We denote $$T(\varphi)(t)=\int^1_0 G(t,s)\varphi(s)ds,$$ i.e. \begin{align*} T(\varphi)(t)&= \frac16\int_0^t s^2(1-t)^2[(t-s)+2(1-s)t]\varphi(s)ds\\ &\quad +\frac16\int_t^1 t^2(1-s)^2[(s-t)+2(1-t)s]\varphi(s)ds. \end{align*} \begin{lemma}[\cite{m1}] \label{lem2.1} Let $\varphi\in X[2,2]$. Then $T(\varphi)(t)$, $[T(\varphi)]'(t)$, $[T(\varphi)]''(t)$, $[T(\varphi)]'''(t)$ are $AC_{\rm loc}(0,1)\cap C^1(0,1)$, and $$[T(\varphi)]''''(t)=\varphi(t),\quad \text{a.e. } t\in (0,1).$$ \end{lemma} \begin{lemma}[\cite{m1}] \label{lem2.2} Let $\varphi\in X[2,2]$. Then $$T(\varphi)(0)=T(\varphi)(1)=T(\varphi)'(0)=T(\varphi)'(1)=0.$$ \end{lemma} \begin{lemma}[\cite{m1}] \label{lem2.3} Let $\varphi\in L^1(0,1)$. Then $[T(\varphi)](t)\in AC^3[0,1]$. \end{lemma} \section{Main Result} We shall assume the following conditions: \begin{itemize} \item[(H1)] $f:[0,\infty)\to [0,\infty)$ is continuous and nondecreasing in $x$, $f(x)>0$ on $(0,\infty)$, and there exists $\lambda>1$ such that $$f(cx)\leq c^{\lambda}f(x),\ \ \forall\ c\geq 1,\ x\in [0,+\infty).\label{e3.1}$$ \item[(H2)] $p:(0,1)\to[0,\infty)$ is continuous, $\int_0^1s^2(1-s)^2p(s)ds<+\infty$, and there exists $\theta\in (0,1/2)$ such that $$0<\int_\theta^{1-\theta}s^2(1-s)^2p(s)ds.$$ \item[(H3)] $0\leq \limsup_{x\to 0+}\frac{f(x)}{x}0$. \end{itemize} \end{remark} \begin{proof}[Proof of Theorem \ref{thm3.1}] Necessity. Let $x\in C^2[0,1]\cap C^4(0,1)$ be a positive solution of \eqref{e1.1} and \eqref{e1.2}. Then by the fact \begin{align*} x''(t)&= \frac16\int_0^t\{2s^2[(t-s)+2(1-s)t]-4s^2(1-t)[1+2(1-s)]\} p(s)f(x(s))ds\\ &\quad +\frac16\int_t^1\{2(1-s)^2[(s-t)+2(1-t)s]+4t(1-s)^2[-1-2s]\} p(s)f(x(s))ds. \end{align*} we have that \begin{gather*} x''(0)=\int_0^1(1-s)^2sp(s)f(x(s))ds>0, \\ x''(1)=\int_0^1s^2(1-s)p(s)f(x(s))ds>0. \end{gather*} and accordingly, there exist $I_1,I_2\in(0,+\infty)$ such that $$I_1t^2(1-t)^2\leq x(t)\leq I_2t^2(1-t)^2,\quad t\in [0,1].$$ Let $c_1\geq \max\{1,1/I_1\}$, then $$t^2(1-t)^2\leq c_1x(t),\quad t\in [0,1].$$ So by (H1), \begin{align*} \int_0^1p(s)f(s^2(1-s)^2)ds &\leq \int_0^1p(s)f(c_1x(s))ds\\ &\leq c_1^{\lambda}\int_0^1p(s)f(x(s))ds\\ &= c_1^{\lambda}\int_0^1x''''(s)ds \\ &\leq c_1^{\lambda}[x'''(1)-x'''(0)]<\infty. \end{align*} On the other hand, if $c_2\leq \min\{1/2,1/I_2\}$, then $$t^2(1-t)^2\geq c_2x(t),\ \ \ t\in [0,1].$$ So by (H1) and \eqref{e3.3}, $$\int_0^1p(s)f(s^2(1-s)^2)ds\geq \int_0^1p(s)f(c_2x(s))ds\geq c_2^{\lambda}\int_0^1p(s)f(x(s))ds\geq 0$$ Notice that $\int_0^1p(s)f(x(s))ds>0$, for otherwise $p(s)f(x(s))\equiv0$ on (0,1). In this case \eqref{e1.1}-\eqref{e1.2} has only trivial solution $x\equiv 0$. This contradicts the assumption that $x$ is a positive solution. Thus \eqref{e3.2} holds. \smallskip Sufficiency. Suppose that \eqref{e3.2} holds. we define a set $P\subset C[0,1]$ by \begin{align*} P=\big\{&x\in C[0,1]: \exists c_x>0,\ 0\leq x(t)\leq c_xt^2(1-t)^2,\\ & x(t)\geq \frac23t^2(1-t)^2\|x\|,\ t\in [0,1]\big\}. \end{align*} By its definition, it is easy to verify that $P$ is a cone. We define $T:P\to C[0,1]$ by $$T(x)(t)=\int^1_0 G(t,s)p(s)f(x(s))ds,\quad t\in[0,1],\ x\in P.$$ In the following, we prove that $T:P\to P$ is completely continuous. \noindent 1. We first show that $T:P\to P$ is well defined. For $x\in P$, there exist $c_x\geq 1$ such that $0\leq x(t)\leq c_xt^2(1-t)^2$ and for $t\in [0,1]$, by \eqref{e2.1}, we get $$(Tx)(t)=\int_0^1G(t,s)p(s)f(x(s))ds \leq \frac12 c_x^{\lambda}t^2(1-t)^2\int_0^1p(s)f(s^2(1-s)^2)ds.$$ This implies that $p(t)f(x(t))\in L^1[0,1]$, by Lemma \ref{lem2.3}, we have $Tx\in C[0,1]$. Let $c_{Tx}=\frac12 c_x^{\lambda}\int_0^1p(s)f(s^2(1-s)^2)ds$. By (3.2), we know $c_{Tx}>0$, so $$(Tx)(t)\leq c_{Tx}t^2(1-t)^2,\ \ t\in [0,1].$$ In addition, for $t\in [0,1]$, by \eqref{e2.1}, we get $$(Tx)(t)=\int_0^1G(t,s)p(s)f(x(s))ds\geq \frac13t^2(1-t)^2\int_0^1s^2(1-s)^2p(s)f(x(s))ds,\label{e3.4}$$ and $$(Tx)(t)=\int_0^1G(t,s)p(s)f(x(s))ds\leq \frac12\int_0^1s^2(1-s)^2p(s)f(x(s))ds.$$ Hence $$\|Tx\|\leq \frac12\int_0^1s^2(1-s)^2p(s)f(x(s))ds.$$ Combining the above with \eqref{e3.4}, we have $$(Tx)(t)\geq \frac13t^2(1-t)^2\int_0^1s^2(1-s)^2p(s)f(x(s))ds\geq \frac23t^2(1-t)^2\|Tx\|,$$ i.e., $T(P)\subset P$. \smallskip \noindent 2. We show that $T:P\to P$ is compact. Let $D\subset P$ be bounded, i.e., $\|x\|\leq M$ for all $x\in D$ and some $M>0$. It is clear that if $x\in P$ satisfies $x\in D$, by (H2) we have $$|(Tx)(t)|\leq \frac12\int_0^1s^2(1-s)^2p(s)f(x(s))ds\leq \frac12\int_0^1s^2(1-s)^2p(s)f(M)ds.$$ So $T(D)$ is uniformly bounded. Next we prove that $\|(Tx)'\|\leq N$ for all $x\in D$ and some $N>0$. In fact, for $x\in D$. By Lemma \ref{lem2.3}, we know $Tx\in C^2[0,1]$ and \begin{align*} & |(Tx)'(t)|\\ &= \Big| \frac16\int_0^t\{-2s^2(1-t)[(t-s)+2(1-s)t]+s^2(1-t)^2[1+2(1-s)]\}p(s)f(x(s))ds\\ &\quad +\frac16\int_t^1\{2t(1-s)^2[(s-t)+2(1-t)s]+t^2(1-s)^2[-1-2s]\}p(s)f(x(s))ds\Big|\\ & \leq \frac16\int_0^t\{2s^2(1-s)[(1-s)+2(1-s)]+s^2(1-s)^2[1+2(1-s)]\}p(s)f(M)ds\\ &\quad +\frac16\int_t^1\{2t(1-s)^2[s+2s]+s^2(1-s)^2[1+2s]\}p(s)f(M)ds\\ & \leq \frac96\int_0^ts^2(1-s)^2p(s)f(M)ds \quad +\frac96\int_t^1s^2(1-s)^2p(s)f(M)ds\\ &= \frac32\int_0^1s^2(1-s)^2p(s)f(M)ds=N. \end{align*} This means that $T(D)$ is equicontinuous. From the Ascoli-Arzela theorem, $T(D)$ is relatively compact. This completes the proof that $T$ is compact. \smallskip \noindent 3. We prove $T:P\to P$ is continuous. Assume that $x_n,x\in P$ and $x_n\to x$. Then there exists $M>0$ such that $\|x\|\leq M,\ \|x_n\|\leq M$ for every $n>0$. Since $f(x)$ is continuous, we have $$|f(x_n(s))-f(x(s))|\to 0,\quad \text{as } n\to \infty,\quad \forall\ s\in [0,1],$$ and $$|f(x_n(s))-f(x(s))|\leq 2f(M),\quad \forall\ t\in [0,1],\ (n=1,2,3\dots).$$ Consequently, for all $t\in [0,1]$, $$\|(Tx_n)(t)-(Tx)(t)\|\leq \int_0^1s^2(1-s)^2p(s)|f(x_n(s))-f(x(s))|ds\to 0. \label{e3.5}$$ We now show $$\|Tx_n-Tx\|\to 0\quad \text{as } n\to \infty).\label{e3.6}$$ If \eqref{e3.6} is not true, then there exist a positive number $\varepsilon>0$ and a sequence $\{x_{n_i}\}\subset\{x_n\}$ such that $$\|Tx_{n_i}-Tx\|\geq\varepsilon,\quad (i=1,2,3\ldots).\label{e3.7}$$ Since $\{x_n\}$ is bounded, $\{Tx_n\}$ is relatively compact and there is a subsequence of $\{Tx_{n_i}\}$ which converges in $C[0,1]$ to some $y\in C[0,1]$. Without loss of generality, we may assume that $\{Tx_{n_i}\}$ itself converges to $y$: $$\|Tx_{n_i}-y\|\to 0,\quad \text{as } i\to \infty.\label{e3.8}$$ By virtue of \eqref{e3.5} and \eqref{e3.8}, we have $y=Tx$, and so, \eqref{e3.8} contradicts \eqref{e3.7}. Hence, \eqref{e3.6} holds, and the continuity of $T$ is proved. To sum up, we have proved $T:P\to P$ is completely continuous. For all $x\in P$, from the above proof, we know $Tx\in P$, By Lemma \ref{lem2.1} and Lemma \ref{lem2.2}, the fixed point of the equation $$Tx=x,\quad x\in P.$$ is the solution of \eqref{e1.1}-\eqref{e1.2}. Next we will look for the fixed point. By the first part of (H3), there exist $1>r>0$, $\varepsilon>0$ such that $0\max\{\theta r,1\}$, $\varepsilon_1>0$ such that $$f(x)\geq (m_1+\varepsilon_1)x,\quad x\geq R_1.$$ Let $R_2>\frac{3R_1}{2\theta^2(1-\theta)^2}$, and $B_{R_2}=\{x\in C[0,1]: \|x\|0$, otherwise there is a fixed point in $\partial B_{R_2}\cap P$ and this would complete the proof. Let $\eta=\min_{t\in[\theta,1-\theta]}x_1(t)$. Then if $t\in[\theta,1-\theta]$, we have \begin{align*} x_1(t)&= \int_0^1G(t,s)p(s)f(x_1(s))ds+\mu_1t^2(1-t)^2\\ &\geq \int_\theta^{1-\theta}G(t,s)p(s)f(x_1(s))ds+\mu_1t^2(1-t)^2\\ &\geq (m_1+\varepsilon_1)\int_\theta^{1-\theta}G(t,s)p(s)x_1(s)ds +\mu_1t^2(1-t)^2\\ &\geq \eta(m_1+\varepsilon_1)\int_\theta^{1-\theta}G(t,s)p(s)ds+\mu_1t^2(1-t)^2\\ &\geq \eta+\eta\varepsilon_1\int_\theta^{1-\theta}G(t,s)p(s)ds+\mu_1t^2(1-t)^2. \end{align*} Therefore, $$x_1(t)>\eta,\quad t\in [\theta,1-\theta],$$ which is a contradiction. According to Lemma \ref{lem1.1}, we get $$i(T, B_{R_2}\cap P,\ P)=0.\label{e3.10}$$ By \eqref{e3.9} and \eqref{e3.10}, we have $$i(T,\ (B_{R_2}\overline{\backslash{B_r}})\cap P,\ P) =i(T,\ B_{R_2}\cap P,\ P)-i(T,\ B_r\cap P,\ P)=-1.$$ Then $T$ has at least a fixed point $x^*$ in $(B_{R_2}\overline{\backslash{B_r}})\cap P$ satisfying $01$ such that $x^*\leq r_{x^*}t^2(1-t)^2$, then \begin{align*} \int_0^1p(s)f(x^*(s))ds &\leq \int_0^1p(s)f(r_{x^*}s^2(1-s)^2)ds\\ &\leq r_{x^*}^{\lambda}\int_0^1p(s)f(s^2(1-s)^2)ds<+\infty, \end{align*} that is $p(t)f(x^*(t))\in L^1(0,1)$, then by Lemma \ref{lem2.3}, we have $x^*\in AC^3[0,1]$, so $x^*$ is a $C^3[0,1]\cap C^4(0,1)$ positive solution of \eqref{e1.1}-\eqref{e1.2}. This completes the proof of sufficiency. \end{proof} \begin{corollary} \label{coro1} Let $p$ be as above, $0<\int_0^1s^2(1-s)^2p(s)ds<+\infty$, and $\lambda>1$. Then BVP \eqref{e1.3}-\eqref{e1.4} has at least a positive solution in $C^3[0,1]\cap C^4(0,1)$ \end{corollary} \begin{proof} The hypotheses on the function $p(s)$ implies $0<\int_0^1(s(1-s))^{2\lambda}p(s)ds<+\infty$ for $\lambda>1$. The result now follows from Theorem \ref{thm3.1}. \end{proof} \begin{theorem} \label{thm3.2} Assume that (H1) and (H2) are satisfied. If $$\lim_{x\to 0+}\frac{f(x)}{x}=0, \quad \lim_{x\to +\infty}\frac{f(x)}{x}+\infty,$$ Then a necessary and sufficient condition for \eqref{e1.1}-\eqref{e1.2} to have a positive solution in $C^3[0,1]\cap C^4(0,1)$ is that $$\int_0^1p(s)f(s^2(1-s)^2)ds<+\infty.$$ \end{theorem} \begin{proof} Clearly (H1)-(H3) hold, and result follows from Theorem \ref{thm3.1}. We omit the detail. \end{proof} Next, we shall study \eqref{e1.1}-\eqref{e1.2} in the sublinear case. We assume: \begin{itemize} \item[(H1')] $f:[0,\infty)\to [0,\infty)$ is continuous and nondecreasing in $x$, $f(x)>0$ on $(0,\infty)$, and there exists $0<\lambda_1<1$ such that $$f(cx)\geq c^{\lambda_1}f(x),\quad \forall\ c\in(0,1),\ x\in [0,+\infty).$$ \item[(H3')] $0\leq \limsup_{x\to +\infty}\frac{f(x)}{x}1$, $\varepsilon_3>0$ such that $x\geq R_3$ implies $f(x) \leq (M_1-\varepsilon_3)x$. Let $M=\max\{f(x): 0\leq x\leq R_3\}$, then $$f(x)\leq (M_1-\varepsilon_3)x+M,\quad x\in [0,+\infty).$$ Choose $R_4>\max\{M\varepsilon_3^{-1},1\}$. Let $B_{R_4}=\{x\in C[0,1]: \|x\|0$ such that $00$, otherwise there is a fixed point in $\partial B_{r_1}\cap P$ and this would complete the proof. Let $\eta=\min_{t\in[\theta,1-\theta]}x_2(t)$. Then if $t\in[\theta,1-\theta]$, we have \begin{align*} x_2(t)&= \int_0^1G(t,s)p(s)f(x_2(s))ds+\mu_2t^2(1-t)^2\\ &\geq \int_\theta^{1-\theta}G(t,s)p(s)f(x_2(s))ds+\mu_2t^2(1-t)^2\\ &\geq (m_1+\varepsilon_5)\int_\theta^{1-\theta}G(t,s)p(s)x_2(s)ds +\mu_2t^2(1-t)^2\\ &\geq \eta(m_1+\varepsilon_5)\int_\theta^{1-\theta}G(t,s)p(s)ds+\mu_2t^2(1-t)^2\\ &\geq \eta+\eta\varepsilon_5\int_\theta^{1-\theta}G(t,s)p(s)ds+\mu_2t^2(1-t)^2. \end{align*} Therefore, $$x_2(t)>\eta,\quad t\in [\theta,1-\theta].$$ which is a contradiction. According to Lemma 1.1, we get $$i(T,\ B_{r_1}\cap P,\ P)=0.\label{e3.13}$$ By \eqref{e3.12} and \eqref{e3.13}, we have $$i(T,\ (B_{R_4}\overline{\backslash{B_{r_1}}})\cap P,\ P) =i(T,\ B_{R_4}\cap P,\ P)-i(T,\ B_{r_1}\cap P,\ P)=1.$$ Then $T$ has at least a fixed point $x^*$ in $(B_{R_4}\overline{\backslash{B_{r_1}}})\cap P$, satisfying $01,\\ x(0)=x(1)=x'(0)=x'(1)=0, \end{gather*} has a solution$x\in C^3[0,1]\cap C^4(0,1)$with$x(t)>0$on$(0,1)$. To see this, we will apply Theorem \ref{thm3.1} with$p(t)=t^{-5/2}(1-t)^{-4/3}$,$f(x)=x^{\lambda}(\lambda>1)$. Clearly (H1) holds. Note that $$\int_0^1p(s)s^2(1-s)^2ds=\int_0^1s^{-1/2}(1-s)^{2/3}ds\leq 2.$$ Consequently (H2) holds (with$\theta=1/4$). Also note that (H3) holds since $$\lim_{x\to 0+}\frac{f(x)}{x}=0, \quad \lim_{x\to +\infty}\frac{f(x)}{x}=+\infty.$$ Finally note that$\int_0^1p(s)f(s^2(1-s)^2)ds=\int_0^1p(s)(s(1-s))^{2\lambda}ds<+\infty\$. The result now follows from Theorem \ref{thm3.1}. \end{example} \begin{thebibliography}{00} \bibitem{a1} H. Asakawa, \emph{Nonresonant singular two-point boundary value problems}, Nonlinear Analysis. TMA., 44(2001):791-809. \bibitem{d1} K. Deimling, \emph{Nonlinear Functional Analysis}, New York: Springer-Verlag, 1985. \bibitem{g1} D. Guo, V. Lakshmikantham and X. Liu, \emph{Nonlinear integral equations in abstract spaces}, Kluwer Academic Publishers, 1996. \bibitem{g2} D. Guo, J. 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