\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 105, pp. 1--9.\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-2006/105\hfil Existence of positive solutions] {Existence of positive solutions for singular eigenvalue problems} \author[M. Feng, W. Ge\hfil EJDE-2006/105\hfilneg] {Meiqiang Feng, Weigao Ge} % in alphabetical order \address{Meiqiang Feng\newline Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081, China\newline Department of Fundamental Sciences, Beijing Information Technology Institute, Beijing 100101, China} \email{meiqiangfeng@sina.com} \address{Weigao Ge \newline Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081, China} \email{gew@bit.edu.cn} \date{} \thanks{Submitted May 10, 2006. Published September 8, 2006.} \thanks{Supported by grants 10371006 from the National Nature Science Foundation of China, and \hfill\break\indent 20050007011 from the Doctoral Program Foundation of Education Ministry of China} \subjclass[2000]{34B15} \keywords{Positive solution; nonexistence; existence; complete continuity; \hfill\break\indent singularity} \begin{abstract} In this paper, we discuss the existence, nonexistence, and multiplicity of positive solutions for a class of singular eigenvalue problems. Some of our theorems are new, while others extend earlier results obtained by Zhang and Kong \cite{z1}. The interesting point is that the authors obtain the relation between the existence of solutions and the parameter $\lambda$. The arguments are based on the fixed point index theory and the upper and lower solutions method. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \section{Introduction} The deformations of an elastic beam are described by a fourth-order differential equation $$u^{(4)}=f(t,u,u'').$$ Most of the available literature on fourth-order boundary value problems, for example \cite{a1,a2,g1,g3,j1,o1,w1}, discusses the case when $f$ is either continuous or a Caratheodory function and is concerned with the existence and uniqueness of positive solutions for boundary value problems for the above differential equation. However, only a small number of papers have discussed fourth-order singular eigenvalue problems; see for example \cite{y1,z3}. In this paper, we study the fourth-order singular differential equation $$u^{(4)}(t)=\lambda g(t)f(u(t)),\quad 0< t <1 , \label{e1.1}$$ subject to one of the following boundary conditions: \begin{gather} u(0)=u(1)=u''(0)=u''(1)=0, \label{e1.2}\\ u(0)=u'(1)=u''(0)=u'''(1)=0, \label{e1.3} \end{gather} where $\lambda > 0$. The following assumptions will stand throughout this paper: \begin{itemize} \item[(H1)] $f\in C([0,+ \infty ), (0,+ \infty ))$ and is nondecreasing on $[0,+\infty)$. Furthermore, there exist $\bar \delta>0,m\geq 2$ such that $f(u)>\bar \delta u^{m},u\in[0,+\infty)$; \item[(H2)] $g \in C((0,1), (0,+ \infty ))$ and $0<\int_{0}^{1}s(1-s)g(s)ds<+\infty$. \end{itemize} It is the purpose of this paper to obtain the existence and the nonexistence of positive solutions, and multiplicity results for the eigenvalue problems (EP) \eqref{e1.1}-\eqref{e1.2} and \eqref{e1.1}-\eqref{e1.3} by employing new technique (different from the one used in \cite{z1}). Very few papers discuss the connection between the existence of solutions and the parameter $\lambda$. The work done by others \cite{z3} does not cover the general case given in \eqref{e1.1}-\eqref{e1.2} and \eqref{e1.1}-\eqref{e1.3}. In this paper, we use mainly the following fixed point index theory to obtain multiplicity results for \eqref{e1.1}-\eqref{e1.2} and \eqref{e1.1}-\eqref{e1.3}. \begin{lemma}[\cite{g2}] \label{lem1.1} Let $P$ be a cone in a real Banach space $E$ and $\Omega$ be a bounded open subset of $E$ with $\theta \in \Omega$. Suppose $A:P \cap \bar \Omega \to P$ is a completely continuous operator, that satisfies $$Ax=\mu x,\; x \in P \cap \partial\Omega \Longrightarrow \mu<1.$$ Then $i(A,P \cap \Omega ,P)=1$. \end{lemma} \begin{lemma}[\cite{g2}] \label{lem1.2} Suppose $A:P \cap \bar \Omega \to P$ is a completely continuous operator, and satisfies: \begin{enumerate} \item $\inf_{x \in P \cap \partial\Omega}\|Ax\|>0$; \item $Ax=\mu x,x \in P \cap \partial\Omega \Longrightarrow \mu\not\in(0,1].$\\ Then $i(A,P \cap \Omega ,P)=0$. \end{enumerate} \end{lemma} In Section 2, we provide some necessary background. In particular, we state some properties of the Green's function associated with \eqref{e1.1}-\eqref{e1.2} and some Lemmas. In Section 3, we present our main result and discuss an example. \section{Preliminaries} For the convenience of the reader, we present here the necessary definitions and Lemmas. Let $E=C[0,1]$ be a real Banach space with the norm $\|u\|=\max_{0 \leq t\leq 1}|u(t)|$. Let $S=\{\lambda >0\text{ such that \eqref{e1.1} has at least one solution}\}$ and $P=\{u\in E : u(t)\geq 0,t\in [0,1]\}$. It is clear that $P$ is a cone of $E$. We deal first with \eqref{e1.1}-\eqref{e1.2}. Define $$G_{1}(t,\xi)=\begin{cases} t(1-\xi), & 0\leq t \leq \xi \leq 1,\\ \xi(1-t), & 0\leq \xi \leq t \leq 1. \end{cases}$$ \begin{align*} G(t,s)&=\int_{0}^{1}G_{1}(t,\xi)G_{1}(\xi,s)d \xi \\ &=\begin{cases} t(1-s)\frac{2s-s^2-t^2}{6}, \quad 0\leq t \leq s \leq 1,\\ s(1-t)\frac{2t-t^2-s^2}{6}, \quad 0\leq s \leq t \leq 1. \end{cases} \end{align*} It is easy to prove that $G_{1}(t,s)$ and $G(t,s)$ have the following properties. \begin{proposition} \label{prop2.1} For all $t,s\in[0,1]$, we have $$\begin{gathered} G_{1}(t,s)>0,\quad\text{for }(t,s)\in (0,1)\times(0,1); \\ G_{1}(t,s)\leq G_{1}(s,s)=s(1-s),\quad \text{for } 0\leq t, s \leq 1;\\ 0\leq G_{1}(t,s)\leq \frac{1}{4},\quad\text{for } 0\leq t, s \leq 1;\\ G(t,s)\leq \frac{1}{6}G_{1}(s,s)=\frac{1}{6}s(1-s),\quad \text{for } 0\leq t, s \leq 1. \end{gathered} \label{e2.1}$$ \end{proposition} \begin{proposition} \label{prop2.2} For all $t \in [\theta,1-\theta]$, we have $$G_{1}(t,s)\geq \theta G_{1}(s,s),\quad \theta \in (0,\frac{1}{2}),\; s\in [0,1]. \label{e2.2}$$ In fact $$\frac{G_{1}(t,s)}{G_{1}(s,s)}= \begin{cases} \frac{t}{s}, & 0\leq t \leq s \leq 1,\\ \frac{1-t}{1-s},& 1\geq t \geq s \geq 0. \end{cases} \geq \begin{cases} t\geq \theta, & t\leq s,\\ 1-t\geq \theta, &t\geq s. \end{cases}$$ Therefore, for all $t \in [\theta,1-\theta]$, we have $$G_{1}(t,s)\geq \theta G_{1}(s,s),\quad \theta \in (0,\frac{1}{2}),\; s \in [0,1].$$ \end{proposition} \begin{definition} \label{def2.1} \rm Let $\alpha(t)\in C^2[0,1] \cap C^4(0,1)$. We say that $\alpha$ is a lower solution of \eqref{e1.1}-\eqref{e1.2} if it satisfies \begin{gather*} \alpha^{(4)}(t)\leq \lambda g(t)f(u(t)), \quad 0< t <1 , \\ \alpha(0)\leq 0, \alpha(1)\leq 0, \alpha''(0)\geq 0 ,\quad \alpha''(1)\geq 0. \end{gather*} \end{definition} \begin{definition} \label{def2.2}\rm Let $\beta(t)\in C^2[0,1] \cap C^4(0,1)$. We say that $\beta$ is an upper solution of \eqref{e1.1}-\eqref{e1.2} if it satisfies \begin{gather*} \beta^{(4)}(t)\geq \lambda g(t)f(u(t)), \quad 0< t <1 , \\ \beta(0)\geq 0, \beta(1)\geq 0, \beta''(0)\leq 0 ,\quad \beta''(1)\leq 0. \end{gather*} \end{definition} First, we consider the following eigenvalue problem $$\begin{gathered} u^{(4)}(t)=\lambda g(t)f(u(t)), \quad 0< t <1 , \\ u(0)=u(1)=u''(0)=u''(1)=h\geq 0. \end{gathered} \label{e2.3}$$ Define $T_{\lambda}^h: E \to E$ by $$T_{\lambda}^h u(t) =h+\int_{0}^{1}G(t,s)\lambda g(s)f(u(s))ds-\int_{0}^{1}G_{1}(t,s)hds. \label{e2.4}$$ From \eqref{e2.4}, it is easy to obtain the following lemma, which is proved by a direct computation. \begin{lemma} \label{lem2.1} Suppose that (H1) and (H2) are satisfied. Then \eqref{e1.1}-\eqref{e1.2} has a solution $u$ if and only if $u$ is a fixed point of $T_{\lambda}^0$. \end{lemma} To prove the following results we define the cone $$Q=\{u\in C[0,1]|u(t)\geq 0,\min_{\theta\leq t \leq 1- \theta}u(t) \geq M_{\theta}\|u\|\} \label{e2.5}$$ where $\|u\|=\max_{t\in [0,1]}|u(t)|$, $M_{\theta}=\theta^2(1-6\theta^2+4\theta^3)$, $\theta \in (0,\frac{1}{2})$. It is clear that $Q\subset P$. \begin{lemma} \label{lem2.2} Suppose that (H1) and (H2) are satisfied. Then $T_{\lambda}^{0}Q\subset Q$ is completely continuous and nondecreasing. \end{lemma} \begin{proof} For any $u\in P$, by \eqref{e2.1} and \eqref{e2.4}, we have \begin{align*} T_{\lambda}^0 u(t)&=\int_{0}^{1}G(t,s)\lambda g(s)f(u(s))ds \\ &\leq\frac{1}{6}\int_{0}^{1}\lambda s(1-s)g(s)f(u(s))ds. \end{align*} Therefore, $\|T_{\lambda}^0 u\|\leq\frac{1}{6}\int_{0}^{1}\lambda s(1-s)g(s)f(u(s))ds.$ On the other hand, by \eqref{e2.2}, for any $\theta\leq t\leq 1-\theta$, we have $$G(t,s)=\int_{0}^{1}G_{1}(t,\xi)G_{1}(\xi,s)d \xi \geq M_{\theta}\frac{1}{6}s(1-s).\label{e2.6}$$ Therefore, \begin{align*} \min_{\theta\leq t\leq 1-\theta}T_{\lambda}^0 u(t) &=\min_{\theta\leq t\leq 1-\theta}\int_{0}^{1}G(t,s)\lambda g(s)f(u(s))ds\\ & \geq M_{\theta}\frac{1}{6}\int_{0}^{1}\lambda s(1-s)g(s)f(u(s))ds \\ & \geq M_{\theta}\|T_{\lambda}^0 u\|. \end{align*} Hence $T_{\lambda}^0 P\subset Q$ and then $T_{\lambda}^0 Q\subset Q$ by $Q\subset P$. By similar arguments in \cite{a2,o1,z1,z3}, $T_{\lambda}^0: Q\to Q$ is completely continuous. Since f is increasing on $[0,+ \infty)$, it is easy to obtain that $T_{\lambda}^{0}$ is nondecreasing on $[0,+ \infty)$. \end{proof} \begin{remark} \label{rmk2.1}\rm Reasoning as in the proofs of Lemmas \ref{lem2.1} and \ref{lem2.2}, we conclude that $T_{\lambda}^{h}:Q\to Q$ is completely continuous and that $u(t)$ is a solution of \eqref{e2.3} if and only if $u(t)$ is a fixed point of $T_{\lambda}^{h}$. \end{remark} \begin{lemma} \label{lem2.3} Suppose that $\lambda \in S,S_{1}=(\lambda,+\infty)\cap S\not\equiv \emptyset$. Then there exists $R(\lambda )>0$, such that $\|u_{\lambda'}\|\leq R(\lambda )$, where $\lambda'\in S_{1}$, and $u_{\lambda'}\in Q$ is a solution of \eqref{e1.1}-\eqref{e1.2} with $\lambda'$ instead of $\lambda$. \end{lemma} \begin{proof} For any $\lambda'\in S$, let $u_{\lambda'}$ be a solution of \eqref{e1.1}-\eqref{e1.2} with $\lambda'$ instead of $\lambda$. Then $$u_{\lambda'}(t)= T_{\lambda'}^{0}u_{\lambda'}(t) = \int_{0}^{1}G(t,s)\lambda'g(s)f(u_{\lambda'}(s))ds.$$ Let $R(\lambda) =\max\{[\frac{1}{6}\lambda'M_{\theta}^{m+1}\bar\delta \int_{\theta}^{1-\theta}G_{1}(s,s) g(s)ds]^{-1},1\}$. Next we shall prove that $\|u_{\lambda'}\|\leq R(\lambda )$. Indeed, if $\|u_{\lambda'}\|< 1$, the result is easily obtained. On the other hand, if $\|u_{\lambda'}\|\geq 1$, then we have by (H1) and \eqref{e2.6}, \begin{align*} \frac{1}{\|u_{\lambda'}\|} &\geq \frac{\min_{\theta\leq t\leq 1-\theta}u_{\lambda'}(t)} {\|u_{\lambda'}\|^{2}}\\ & = \frac{1}{\|u_{\lambda'}\|^{2}}\min_{\theta\leq t\leq 1-\theta} \int_{0}^{1}G(t,s)\lambda' g(s)f(u_{\lambda'}(s))ds\\ & \geq \frac{1}{\|u_{\lambda'}\|^{2}}M_{\theta} \int_{\theta}^{1-\theta}\frac{1}{6}G_{1}(s,s)\lambda' g(s) \bar\delta u_{\lambda'}(s)^{m}ds\\ & \geq \frac{1}{\|u_{\lambda'}\|^{2}}M_{\theta}^{m+1}\int_{\theta}^{1-\theta} \frac{1}{6}G_{1}(s,s)\lambda' g(s)\bar\delta \|u_{\lambda'}\|^{m}ds\\ & \geq \frac{1}{6}\lambda' M_{\theta}^{m+1}\bar\delta\int_{\theta}^{1-\theta}G_{1}(s,s)g(s)ds. \end{align*} Therefore, $\|u_{\lambda'}\|\leq R(\lambda)$ and the conclusion of Lemma \ref{lem2.3} follows. \end{proof} \begin{lemma}[\cite{d1}] \label{lem2.4} Suppose that $f:[0,+\infty) \to (0,+\infty)$ is continuous and increasing. If $s,s_{0}$ and $M$ are such that $00$, then there exist $\bar s\in(s,s_{0}),h_{0}\in (0,1)$ such that $$sf(u+h)<\bar s f(u),u\in[0,M],h\in(0,h_{0}).$$ \end{lemma} \section{Main results} In this section, we apply Lemmas \ref{lem1.1} and \ref{lem1.2} to establish nonexistence and existence of positive solutions, as well as multiplicity results for \eqref{e1.1}-\eqref{e1.2} and \eqref{e1.1}-\eqref{e1.3}. Our approach depends on the upper and lower solutions method and the fixed point index theory. We deal with \eqref{e1.1}-\eqref{e1.2} first. \begin{theorem} \label{thm3.1} Let (H1) and (H2) be satisfied. Then there exists $0<\lambda^{*}<+\infty$ such that \begin{enumerate} \item EP \eqref{e1.1}-\eqref{e1.2} has no solution for $\lambda >\lambda^{*}$; \item EP \eqref{e1.1}-\eqref{e1.2} has at least one positive solution for $\lambda =\lambda^{*}$; \item EP \eqref{e1.1}-\eqref{e1.2} has at least two positive solutions for $0<\lambda <\lambda^{*}$. \end{enumerate} \end{theorem} \begin{proof} First, we prove that the conclusion (1) of Theorem \ref{thm3.1} holds. If $\beta(t)$ is a solution of the boundary-value problem $$\begin{gathered} u^{(4)}(t)=g(t) \quad 0< t <1 , \\ u(0)=u(1)=u''(0)=u''(1)=0, \end{gathered} \label{e3.1}$$ then, by Lemma \ref{lem2.1}, we have $\beta(t)=\int_{0}^{1}G(t,s)g(s)ds$. Let $\beta_{0}=\max_{t\in [0,1]}\beta(t)$; therefore, by (H1) and \eqref{e2.4}, $$T_{\lambda}^{0}\beta(t)\leq T_{\lambda}^{0}\beta_{0} =\int_{0}^{1}G(t,s)\lambda g(s)f(\beta_{0})ds<\beta(t),\quad \forall 0<\lambda<\frac{1}{f(\beta_{0})}.$$ This implies that $\beta(t)$ is an upper solution of $T_{\lambda}^{0}$. On the other hand, let $\alpha(t)\equiv 0,t\in [0,1]$, then $\alpha(t)$ is a lower solution of $T_{\lambda}^{0}$, and $\alpha(t)<\beta(t),t\in[0,1]$. Clearly $T_{\lambda}^{0}$ is completely continuous on $[\alpha,\beta]$. Therefore, $T_{\lambda}^{0}$ has a fixed point $u_{\lambda}\in [\alpha,\beta]$, and therefore $u_{\lambda}$ is a solution of \eqref{e1.1}-\eqref{e1.2}. Hence, for any $0<\lambda<\frac{1}{f(\beta_{0})}$, we have $\lambda \in S$, which implies that $S\not=\emptyset$. On the other hand, if $\lambda_{1}\in S$, then we must have $(0,\lambda_{1})\subset S$. In fact, let $u_{\lambda_{1}}$ be a solution of \eqref{e1.1}-\eqref{e1.2}. Then, by Lemma \ref{lem2.1}, we have $$u_{\lambda_{1}}(t)=T_{\lambda_{1}}^{0}u_{\lambda_{1}}(t),t\in [0,1].$$ Therefore, for any $\lambda\in (0,\lambda_{1})$, by \eqref{e2.4}, we have \begin{align*} T_{\lambda}^{0}u_{\lambda_{1}}(t) &=\int_{0}^{1}G(t,s)\lambda g(s) f(u_{\lambda_{1}}(s)ds \\ & \leq \int_{0}^{1}G(t,s)\lambda_{1} g(s) f(u_{\lambda_{1}}(s)ds \\ & =T_{\lambda_{1}}^{0}u_{\lambda_{1}}(t)\\ & =u_{\lambda_{1}}(t), \end{align*} which implies that $u_{\lambda_{1}}$ is an upper solution of $T_{\lambda}^{0}$. Combining this with the fact that $\alpha(t)\equiv 0$ ($t\in [0,1]$) is a lower solution of $T_{\lambda}^{0}$, then, by Lemma \ref{lem2.1}, EP \eqref{e1.1}-\eqref{e1.2} has a solution, therefore $\lambda \in S$. Thus we have $(0,\lambda_{1})\subset S$. Let $\lambda^{*}=\sup S$, now we prove that $\lambda^{*}<+\infty$. If this is not true, then we must have $N\subset S$, where $N$ denotes natural number numbers. Therefore, for any $n\in N$, by Lemma \ref{lem2.1}, there exists $u_{n}\in Q$ satisfying $$u_{n}=T_{n}^{0}u_{n} =\int_{0}^{1}G(t,s)n g(s) f(u_{n}(s)ds.$$ Let $K=[\frac{\bar\delta M_{\theta}^{m+1}}{6} \int_{\theta}^{1-\theta}G_{1}(s,s) g(s) ds]^{-1}$. Suppose $\|u_{n}\|\geq 1$. Then we have \begin{align*} 1&\geq\frac{1}{\|u_{n}\|}\\ &\geq\frac{\min_{\theta\leq t\leq 1-\theta}u_{n}(t)}{\|u_{n}\|^{2}}\\ &= \frac{1}{\|u_{n}\|^2} \min_{\theta\leq t\leq 1-\theta}\int_{0}^{1}G(t,s)n g(s) f(u_{n}(s)ds\\ &\geq \frac{1}{\|u_{n}\|^2}M_{\theta} \int_{\theta}^{1-\theta}\frac{1}{6}G_{1}(s,s)n g(s)\bar\delta u_{n}(s)^m ds\\ &\geq \frac{1}{\|u_{n}\|^2}M_{\theta}^{m+1} \int_{\theta}^{1-\theta}\frac{1}{6}G_{1}(s,s)n g(s)\bar\delta \|u_{n}\|^m ds\\ &\geq \frac{1}{6}n M_{\theta}^{m+1}\bar\delta \int_{\theta}^{1-\theta}G_{1}(s,s) g(s) ds. \end{align*} If $\|u_{n}\|\leq 1$, then \begin{align*} 1\geq\|u_{n}\| &\geq\min_{\theta\leq t\leq 1-\theta}\int_{0}^{1}G(t,s)n g(s) f(u_{n}(s)ds \\ &\geq M_{\theta} \int_{\theta}^{1-\theta}\frac{1}{6}G_{1}(s,s)n g(s)f(0)ds. \end{align*} Hence $n\leq \{K,(M_{\theta} \int_{\theta}^{1-\theta}\frac{1}{6} G_{1}(s,s)g(s)f(0)ds)^{-1}\}$, this contradicts the fact that $N$ is unbounded; therefore $\lambda^{*}<+\infty$, and the proof of the conclusion (1) is complete. Secondly, we verify the conclusion (2) of Theorem \ref{thm3.1}. Let $\{\lambda_{n}\}\subset [\frac{\lambda^{*}}{2}, \lambda^{*}),\lambda_{n}\to \lambda^{*}(n\to \infty)$, $\{\lambda_{n}\}$ be an increasing sequence. Suppose $u_{n}$ is solution of \eqref{e1.1} with $\lambda_n$ instead of $\lambda$. By Lemma \ref{lem2.3}, there exists $R(\frac{\lambda^{*}}{2})>0$ such that $\|u_{n}\|\leq R(\frac{\lambda^{*}}{2})$, $n=1,2,\cdots$. Hence $u_{n}$ is a bounded set. It is clear that $\{u_{n}\}$ is an equicontinuous set of $C[0,1]$. Therefore, by the Ascoli-Arzela theorem, it follows that $\{u_{n}\}$ is compact set, and therefore $\{u_{n}\}$ has a convergent subsequence. Without loss of generality, we suppose that $u_{n}$ is convergent: $u_{n}\to u^{*}(n\to +\infty)$. Since $u_{n}=T_{\lambda_{n}}^{0}u_{n}$, by control convergence theorem ($f$ is bounded), we have $u^{*}=T_{\lambda^{*}}^{0}u^{*}$. Therefore, by Lemma \ref{lem2.1}, $u^{*}$ is a solution of \eqref{e1.1}-\eqref{e1.2} with $\lambda^*$ instead of $\lambda$. Hence the conclusion (2) of Theorem \ref{thm3.1} holds. Finally, we prove the conclusion (3) of Theorem \ref{thm3.1}. Let $\alpha(t)\equiv h$ $(t\in[0,1])$. Then for any $\lambda \in (0,\lambda^{*})$, $\alpha(t)$ is a lower solution of \eqref{e2.3}. On the other hand, by Lemma \ref{lem2.3}, there exists $R(\lambda)>0$ such that $\|u_{\lambda'}\|\leq R(\lambda)$, $\lambda'\in [\lambda, \lambda^{*}]$, where $u_{\lambda'}$ is a solution of \eqref{e1.1} with $\lambda'$ instead of $\lambda$. Also by Lemma \ref{lem2.4}, there exist $\bar \lambda \in [\lambda, \lambda^{*}],h_{0}\in(0,1)$ satisfying $$\lambda f(u+h)<\bar \lambda f(u),u\in[0,R(\lambda)],h\in (0,h_{0}).$$ Let $u_{\bar \lambda}$ be a solution of \eqref{e1.1}-\eqref{e1.2} with $\bar \lambda$, and $\bar u_{\lambda}(t)=u_{\bar \lambda}+h$, $h\in (0,h_{0})$. Then \begin{align*} \bar u_{\lambda}(t) &=u_{\bar \lambda}+h\\ & =\int_{0}^{1}G(t,s) \bar \lambda g(s) f(u_{\bar \lambda}(s))ds+h\\ & \geq h+\int_{0}^{1}G(t,s) \lambda g(s) f(u_{\bar \lambda}(s)+h)ds-\int_{0}^{1}G_{1}(t,s)hds\\ & = T_{\lambda}^{h}\bar u_{\lambda}(t). \end{align*} Combining this with $\bar u_{\lambda}(0)=\bar u_{\lambda}(1)\geq h$, $\bar u_{\lambda}''(0)=0\leq h$, $\bar u_{\lambda}''(1)=0\leq h$, we have the $\bar u_{\lambda}(t)$ is an upper solution of \eqref{e2.3}. Therefore \eqref{e2.3} has solution. Let $v_{\lambda}(t)$ be a solution of \eqref{e2.3}. Let $\Omega=\{u\in Q |u(t)0$ such that $\bar R^{m-1}\int_{\theta}^{1-\theta}G(\frac{1}{2},s) \lambda g(s)\bar \delta M_{\theta}^mds>1$. Therefore, for any $R>\bar R$ and $B_{R}\subset Q$, by \eqref{e3.3}, $$\|T_{\lambda}^{0}u\|>\|u\| >0,u\in \partial B_{R}, \label{e3.4}$$ where $B_{R}=\{u\in Q| \|u\|0, \quad t,s \in (0,1),\\ \hat{G}(t,s) \leq \hat{G}(s,s) = s, \quad t,s \in [0,1],\\ \hat{G}(t,s) \geq \alpha \hat{G}(s,s) ,\quad t \in[\alpha,1-\alpha],\; s \in [0,1],\\ \tilde G(t,s) \leq \frac{1}{2}s, \quad t,s \in [0,1];\\ \tilde G(t,s) \geq \frac{1}{2}M_{\alpha}s, \quad t \in [\alpha,1-\alpha],\; s \in [0,1] \end{gather*} where$M_{\alpha}= \alpha^{2}(1-2\alpha)$. \end{proposition} Define the cone $$\hat Q=\{u\in C[0,1]|u(t)\geq 0,\min_{\alpha\leq t \leq 1- \alpha}u(t) \geq M_{\alpha} \|u\|\}$$ and let \begin{itemize} \item[(H3)]$g \in C((0,1), (0,+ \infty ))$and$0<\int_{0}^{1}s g(s)ds<+\infty$\end{itemize} \begin{theorem} \label{thm3.2} Let (H1) and (H3) be satisfied. Then there exists$0<\lambda^{*}<+\infty$such that: \begin{enumerate} \item EP \eqref{e1.1}-\eqref{e1.3} has no solution for$\lambda >\lambda^{*}$; \item EP \eqref{e1.1}-\eqref{e1.3} has at least one positive solution for$\lambda =\lambda^{*}$; \item EP \eqref{e1.1}-\eqref{e1.3} has at least two positive solutions for$0<\lambda <\lambda^{*}$. \end{enumerate} \end{theorem} As an example we consider the eigenvalue problem \begin{gathered} u^{(4)}(t)=\lambda \frac{1}{t(1-t)}2^{2u},\quad 00, m=2$, we have $f(u)=2^{2u}=\bar\delta 2^{2u}>u^{2}=u^{m}>0$. So that (H1) is satisfied. \subsection*{Acknowledgments} The author is thankful to the referee for his/her valuable suggestions regarding the original manuscript. \begin{thebibliography}{00} \bibitem{a1} 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{a2} A. R. 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