\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 159, 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 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/159\hfil Positive solutions of fourth-order four-point BVP] {Multiple positive solutions of fourth-order four-point boundary-value problems with changing sign coefficient} \author[Z. Fang, C. Li, C. Bai \hfil EJDE-2008/159\hfilneg] {Zheng Fang, Chunhong Li, Chuanzhi Bai} \address{Zheng Fang \newline School of Science\\ Jiangnan University \\ Wuxi, Jiangsu 214122, China} \email{fangzhengm@yahoo.com.cn} \address{Chunhong Li \newline Department of Mathematics \\ Huaiyin Teachers College\\ Huaian, Jiangsi 223300, China} \email{lichshy2006@126.com} \address{Chuanzhi Bai \newline Department of Mathematics \\ Huaiyin Teachers College\\ Huaian, Jiangsi 223300, China} \email{czbai8@sohu.com} \thanks{Submitted September 3, 2008. Published December 3, 2008.} \subjclass[2000]{34B10, 34B15} \keywords{Four-point boundary-value problem; positive solution; \hfill\break\indent fixed point on a cone} \begin{abstract} In this paper, we investigate the existence of multiple positive solutions of the fourth-order four-point boundary-value problems \begin{gather*} y^{(4)}(t) = h(t) g(y(t), y''(t)), \quad 0 < t < 1, \\ y(0) = y(1) = 0, \\ a y''(\xi_1) - b y'''(\xi_1) = 0, \quad c y''(\xi_2) + d y'''(\xi_2) = 0, \end{gather*} where $0 < \xi_1 < \xi_2 < 1$. We show the existence of three positive solutions by applying the Avery and Peterson fixed point theorem in a cone, here $h(t)$ may change sign on $[0, 1]$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Recently, several authors have studied the existence of positive solutions to boundary-value problems for fourth-order differential equations. For details; see, for example, \cite{b2,h1,l1,l2,y1,z1,z2,z3}. Zhong, Chen and Wang \cite{z3} investigated the fourth-order nonlinear ordinary differential equation $$y^{(4)}(t) - f(t, y(t), y''(t)) = 0, \quad 0 \leq t \leq 1, \label{e1.1}$$ with the four-point boundary conditions $$\begin{gathered} y(0) = y(1) = 0, \\ a y''(\xi_1) - b y'''(\xi_1) = 0, \quad c y''(\xi_2) + d y'''(\xi_2) = 0, \end{gathered} \label{e1.2}$$ where $f \in C([0, 1] \times [0, \infty) \times (- \infty, 0], [0, \infty))$, $a, b, c, d$ are nonnegative constants, and $0 \leq \xi_1 < \xi_2 \leq 1$. Some results on the existence of at least one positive solution to BVP \eqref{e1.1}-\eqref{e1.2} are obtained by using the Krasnoselskii fixed point theorem. Their key result reads as follows. \begin{lemma}[{\cite[Lemma 2.2]{z3}}] \label{lem1.1} If $\alpha = ad + bc + ac(\xi_2 - \xi_1) \not= 0$ and $h(t) \in C[\xi_1, \xi_2]$, then the boundary-value problem \begin{equation*} \begin{gathered} u^{(4)}(t) = h(t), \quad 0 < t < 1, \\ u(0) = u(1) = 0, \\ a u''(\xi_1) - b u'''(\xi_1) =0, \quad c u''(\xi_2) + d u'''(\xi_2) = 0 \end{gathered} \end{equation*} has a unique solution $$u(t) = \int_0^1 G_1(t, s) \int_{\xi_1}^{\xi_2} G_2(s, \tau) h(\tau) \,d\tau\, ds, \label{e1.3}$$ where $$\label{e1.4} \begin{gathered} G_1(t, s) = \begin{cases} s(1 - t), & 0 \leq s \leq t \leq 1, \\ t(1 - s), & 0 \leq t < s \leq 1, \end{cases}\\ G_2(t, s) = \frac{1}{\alpha} \begin{cases} (a(s - \xi_1) + b)(d + c(\xi_2 - t)), & s < t \leq 1, \ \xi_1 \leq s \leq \xi_2,\\ (a(t - \xi_1) + b)(d + c(\xi_2 - s)), & 0 \leq t \leq s, \ \xi_1 \leq s \leq \xi_2. \end{cases} \end{gathered}$$ \end{lemma} Unfortunately this lemma is wrong. Indeed, by \cite[Lemma 2.1]{b1}, expression \eqref{e1.3} should be replaced by \begin{aligned} u(t) & = \int_0^1 G_1(t, s) \int_t^{\xi_1} (\tau - s) h(\tau) \,d\tau\,ds \\ & \quad + \frac{1}{\delta} \int_0^1 G_1(t, s) \int_{\xi_1}^{\xi_2} (a(\xi_1 - s) - b)(c(\xi_2 - \tau) + d) h(\tau) \,d\tau\,ds, \end{aligned} \label{e1.5} where $\delta = a d + b c + a c (\xi_2 - \xi_1) >0$. So the conclusions in \cite{z3} should be reconsidered. If $f(t, y(t), y''(t))$ in \eqref{e1.1} are replaced by $h(t) g(y(t), y''(t))$, then \eqref{e1.1} reduces to $$y^{(4)}(t) - h(t) g(y(t), y''(t)) = 0, \quad 0 \leq t \leq 1, \label{e1.6}$$ where $h \in C[0, 1]$ and $g \in C([0, \infty) \times (- \infty, 0], [0, \infty))$. To the authors' knowledge, no one has studied the existence of positive solutions for problem \eqref{e1.6}, \eqref{e1.2} using the assumption that $h(t)$ changes sign. Hence, the aim of this paper is to investigate the existence of positive solutions of the BVP \eqref{e1.6} and \eqref{e1.2} by using a triple positive fixed-point theorem of Avery and Peterson in \cite{a1}. \section{Preliminaries} Let $E = \{y \in C^2[0, 1]:y(0) = y(1) = 0\}$. Then we have the following lemma. \begin{lemma}[\cite{z3}] \label{lem2.1} For $y \in E$, we get $$\|y\|_{\infty} \leq \|y^{\prime}\|_{\infty} \leq \|y''\|_{\infty},$$ where $\|y\|_{\infty} = \sup_{t \in [0, 1]}|y(t)|$. \end{lemma} By Lemma \ref{lem2.1}, $E$ is a Banach space with the norm $\|y\|= \|y''\|_{\infty}$. We define the operator $T : E \to E$ by $$T y(t) = \int_0^1 G_1(t, s) (Q y)(s) ds, \label{e2.1}$$ where $G_1(t, s)$ as in \eqref{e1.4}, and \begin{aligned} (Q y)(s) & = \int_{\xi_1}^s (\tau - s) h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau \\ & \quad + \frac{1}{\delta} \int_{\xi_1}^{\xi_2} (b - a(\xi_1 - s))(c(\xi_2 - \tau) + d) h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau. \end{aligned} \label{e2.2} Here, $\delta = a d + b c + a c (\xi_2 - \xi_1) > 0$. From \cite[Lemma 2.1]{b1}, we easily know that $u(t)$ is a solution of the four-point boundary-value problem \eqref{e1.6}, \eqref{e1.2} if and only if $u(t)$ is a fixed point of the operator $T$. It is rather straightforward to show that $$0 \leq G_1(t, s) \leq G_1(s, s), \quad 0 \leq t, s \leq 1, \label{e2.3}$$ and $$G_1(t, s) \geq \omega G_1(s, s), \quad t \in [\omega, 1 - \omega], \quad s \in [0, 1], \label{e2.4}$$ where $$0 < \omega < \min \{\xi_1, 1-\xi_2\} < \frac{1}{2}. \label{e2.5}$$ For the convenience of the reader, we present some definitions from the cone theory in Banach spaces. \subsection*{Definition} The map $\alpha$ is said to be a nonnegative continuous concave functional on a cone $P$ of a real Banach space $E$ provided that $\alpha: P \to [0, \infty)$ is continuous and $$\alpha(tx +(1-t)y) \ge t \alpha(x) + (1-t)\alpha(y), \quad \forall x, y \in P, \quad 0 \le t \le 1.$$ Similarly, we say the map $\beta$ is a nonnegative continuous convex functional on a cone $P$ of a real Banach space $E$ provided that $\beta: P \to [0, \infty)$ is continuous and $$\beta(tx +(1-t)y) \le t \beta(x) + (1-t)\beta(y), \quad \forall x, y \in P, \quad 0 \le t \le 1.$$ Let $\gamma$ and $\theta$ be nonnegative continuous convex functionals on $P$, $\alpha$ be a nonnegative continuous concave functional on $P$, and $\psi$ be a nonnegative continuous functional on $P$. Then for positive real numbers $a, b, c$, and $d$, we define the following convex sets: \begin{gather*} P(\gamma, d) = \{x \in P : \gamma(x) < d\}, \\ P(\gamma, \alpha, b, d) = \{x \in P : b \le \alpha(x), \gamma(x) \le d \}, \\ P(\gamma, \theta, \alpha, b, c, d) = \{x \in P : b \le \alpha(x), \theta(x) \le c, \gamma(x) \le d \}, \\ R(\gamma, \psi, a, d) = \{x \in P : a \le \psi(x), \gamma(x) \le d\}. \end{gather*} The following fixed-point theorem due to Avery and Peterson is fundamental in the proof of our main result. \begin{lemma}[\cite{a1}] \label{lem2.2} Let $P$ be a cone in a real Banach space $E$. Let $\gamma$ and $\theta$ be nonnegative continuous convex functionals on $P$, $\alpha$ be a nonnegative continuous concave functional on $P$, and $\psi$ be a nonnegative continuous functional on $P$ satisfying $\psi(\lambda x) \le \lambda \psi(x)$ for $0 \le \lambda \le 1$, such that for some positive numbers $M$ and $d$, $$\alpha(x) \le \psi(x) \quad \text{and}\quad \|x\| \le M \gamma(x),$$ for all $x \in \overline{P(\gamma, d)}$. Suppose $T: \overline{P(\gamma, d)} \to \overline{P(\gamma, d)}$ is completely continuous and there exist positive numbers $a, b$, and $c$ with $a < b$ such that \begin{itemize} \item[(i)] $\{x \in P(\gamma, \theta, \alpha, b, c, d) : \alpha(x) >b\} \neq \empty$ and $\alpha(Tx) >b$ for $x \in P(\gamma, \theta, \alpha, b, c, d)$; \item[(ii)] $\alpha(Tx) >b$ for $x \in P(\gamma, \alpha, b, d)$ with $\theta(Tx) >c$; \item[(iii)] $0 \not\in R(\gamma, \psi, a, d)$ and $\psi(Tx) 0$, $b \geq a \xi_1$ and $d \geq c(1 - \xi_2)$, we have $b > 0$ and $d > 0$. \end{remark} For convenience of notation, we set \begin{gather} M = \int_0^{\xi_1} - \tau h(\tau) d \tau + \int_{\xi_2}^1 -(1 - \tau) h(\tau) d\tau + \big(\xi_2 - \xi_1 + \frac{b d}{\delta}\big) \int_{\xi_1}^{\xi_2} h(\tau) d\tau, \label{e3.6}\\ m = \min \{m_1, m_2\}, \label{e3.7} \end{gather} where $$\begin{gathered} m_1 = \frac{b d}{\delta} \int_{\xi_1}^{\xi_2} h(\tau) d \tau \int_{\xi_1}^{\xi_2} G_1 (\omega, s)ds, \\ m_2 = \frac{b d}{\delta} \int_{\xi_1}^{\xi_2} h(\tau) d \tau \int_{\xi_1}^{\xi_2} G_1 (1 - \omega, s)ds. \end{gathered} \label{e3.8}$$ We are now in a position to present and prove our main results. \begin{theorem} \label{thm3.3} Let $b \geq a \xi_1$ and $d \geq c(1 - \xi_2)$. Assume {\rm (H1)--(H2)} hold. Suppose there exist constants $0 < p < q < \min \{\omega, \frac{1}{8}\} r$ such that \begin{itemize} \item[(H3)] $g(u, v) \le r/M$, for $(u, v) \in [0, r] \times [- r, 0]$, \item[(H4)] $g(u, v) > q/m$, for $(u, v) \in [q,q/\omega] \times [- r, 0]$, \item[(H5)] $g(u, v) < 8 p/M$, for $(u, v) \in [0, p] \times [- r, 0]$, \end{itemize} where $M, m$ are as in \eqref{e3.6}-\eqref{e3.7}, then \eqref{e1.6}, \eqref{e1.2} has at least three positive solutions $y_1$, $y_2$, and $y_3$ such that \begin{gather*} \max _{0 \leq t \leq 1} |y_i''(t)| \le r, \quad \text{for } i=1, 2, 3; \\ \min _{\omega \le t \le 1 - \omega}|y_1(t)| > q; \quad p < \max _{0 \leq t \leq 1}|y_2(t)|; \\ \min_{\omega \le t \le 1-\omega}|y_2(t)| < q; \quad \max _{0 \leq t \leq 1} |y_3(t)| < p. \end{gather*} \end{theorem} \begin{proof} From Lemma \ref{lem3.1}, $T : P \to P$ is completely continuous. We now show that all the conditions of Lemma \ref{lem2.2} are satisfied. If $y \in \overline {P(\gamma, r)}$, then $\gamma(y) = \max_{0 \leq t \leq 1}|y''(t)| \le r$. By Lemma \ref{lem2.1}, we have $\max_{0 \leq t \leq 1}|y(t)| \le r$, then assumption (H3) implies $g(y(t), y''(t)) \le r/M$. On the other hand, from \eqref{e3.1}-\eqref{e3.3}, we have \begin{aligned} \max _{0 \leq t \leq \xi_1} (Q y)(t) & \leq \int_0^{\xi_1} - \tau h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau \\ & \quad + \frac{1}{\delta} \int_{\xi_1}^{\xi_2} b (c(\xi_2 - \tau) + d) h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau \\ & \leq \int_0^{\xi_1} - \tau h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau \\ & \quad + \frac{1}{\delta} b(c(\xi_2 - \xi_1) + d) \int_{\xi_1}^{\xi_2} h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau, \end{aligned} \label{e3.9} \begin{aligned} \max _{\xi_1 \leq t \leq \xi_2} (Q y)(t) & \leq \frac{1}{\delta} \int_{\xi_1}^t (b + a(t - \xi_1))(c (\xi_2 - \tau) + d) h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau \\ & \quad + \frac{1}{\delta} \int_t^{\xi_2} (b + a(t - \xi_1))(c(\xi_2 - \tau) + d) h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau \\ & = \frac{1}{\delta} \int_{\xi_1}^{\xi_2} (b + a(t - \xi_1))(c(\xi_2 - \tau) + d) h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau \\ & \leq \frac{1}{\delta} (b + a(\xi_2 - \xi_1))(c(\xi_2 - \xi_1) + d) \int_{\xi_1}^{\xi_2} h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau, \end{aligned} \label{e3.10} and \begin{aligned} \max _{\xi_2 \leq t \leq 1} (Q y)(t) & \leq \frac{1}{\delta} \int_{\xi_1}^{\xi_2} d (b + a(\tau - \xi_1)) h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau \\ & \quad + \int_{\xi_2}^1 - (1 - \tau) h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau \\ & \leq \frac{1}{\delta} d (b + a(\xi_2 - \xi_1)) \int_{\xi_1}^{\xi_2} h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau \\ & \quad + \int_{\xi_2}^1 - (1 - \tau) h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau. \end{aligned} \label{e3.11} By \eqref{e3.9}-\eqref{e3.11}, we get \begin{align} \gamma(T y) & = \max _{t \in [0, 1]}|(T y)''(t)| = \max _{t \in [0, 1]}|(Q y)(t)| \notag \\ & = \max \Big\{\max _{0 \leq t \leq \xi_1}|(Q y)(t)|, \max _{\xi_1 \leq t \leq \xi_2}|(Q y)(t)|, \max _{\xi_2 \leq t \leq 1}|(Q y)(t)|\Big\} \notag \\ & \leq \int_0^{\xi_1} - \tau h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau + \int_{\xi_2}^1 - (1 - \tau) h(\tau)g\big(y(\tau), y''(\tau)\big) d \tau \notag \\ & \quad + \frac{1}{\delta} (b + a(\xi_2 - \xi_1))(c(\xi_2 - \xi_1) + d) \int_{\xi_1}^{\xi_2} h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau \label{e3.12} \\ & \leq \frac{r}{M} \Big(\int_0^{\xi_1} - \tau h(\tau) d \tau + \int_{\xi_2}^1 - (1 - \tau) h(\tau)d\tau + \big(\xi_2 - \xi_1 + \frac{b d}{\sigma}\big) \int_{\xi_1}^{\xi_2} h(\tau) d\tau\Big) \notag \\ & = \frac{r}{M} M = r. \notag \end{align} Hence, $T: \overline {P(\gamma, r)} \to \overline {P(\gamma, r)}$. To check condition (i) of Lemma \ref{lem2.2}, we choose $y(t) = q/\omega$, $0 \le t \le 1$. It is easy to see that $y(t)= q/\omega \in P(\gamma, \theta, \alpha, q,q/\omega, r)$ and $\alpha(y)=q/\omega> q$, and so $\{y \in P(\gamma, \theta, \alpha, q, q/\omega, r) : \alpha(y) > q\} \neq \emptyset$. Hence, if $y \in P(\gamma, \theta, \alpha, q, q/\omega,r)$, then $q \le y(t) \le q/\omega$, $- r \leq y''(t) \le 0$ for $\omega \le t \le 1 - \omega$. From assumption (H4), we have $g(y(t), y''(t))> b/m$ for $\omega \le t \le 1 - \omega$, and by the definitions of $\alpha$ and the cone $P$, we distinguish two cases as follows: \noindent Case (1): $\alpha(T y)= (T y)(\omega)$. By \eqref{e3.4} and \eqref{e3.2}, we have \begin{align*} & \alpha(T y) \\ &=(Ty)(\omega) = \int_0^1 G_1 (\omega, s) (Q y)(s) ds \\ & > \int_{\xi_1}^{\xi_2} G_1 (\omega, s) (Q y)(s) ds\\ & \geq \frac{1}{\delta} \int_{\xi_1}^{\xi_2} G_1 (\omega, s)ds \Big[\int_{\xi_1}^s b d h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau + \int_s^{\xi_2} b d h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau \Big] \\ & = \frac{b d}{\delta} \int_{\xi_1}^{\xi_2} G_1 (\omega, s)ds \int_{\xi_1}^{\xi_2} h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau \\ & \geq \frac{b d}{\delta} \frac{q}{m} \int_{\xi_1}^{\xi_2} G_1(\omega, s)ds \int_{\xi_1}^{\xi_2} h(\tau) d \tau \\ &= \frac{q}{m} \cdot m_1 \geq q. \end{align*} \noindent Case (2): $\alpha(T y)= (T y)(1 - \omega)$. Similarly, we obtain \begin{align*} \alpha(T y)&=(T y)(1 - \omega) > \int_{\xi_1}^{\xi_2} G_1 (1 - \omega, s) (Q y)(s) ds \\ & \geq \frac{b d}{\delta} \frac{q}{m} \int_{\xi_1}^{\xi_2} G_1 (1 - \omega, s)ds \int_{\xi_1}^{\xi_2} h(\tau) d \tau \\ & = \frac{q}{m} \cdot m_2 \geq q. \end{align*} i.e., $$\alpha(T y) > q, \quad \forall y \in P(\gamma, \theta, \alpha, q, \frac{q}{\omega}, r).$$ This show that condition (i) of Lemma \ref{e2.2} is satisfied. Secondly, we have $$\alpha(T y) = \min _{\omega \leq t \leq 1-\omega} |(Ty)(t)| \geq \omega \|Ty\|_{\infty} = \omega \theta(T y) > \omega \frac{q}{\omega} = q,$$ for all $y \in P(\gamma, \alpha, q, r)$ with $\theta(T y) > q/\omega$. Thus, condition (ii) of Lemma \ref{lem2.2} is satisfied. We finally show that (iii) of Lemma \ref{lem2.2} also holds. Clearly, as $\psi(0)=0 < p$, there holds that $0 \not\in R(\gamma, \psi, p, r)$. Suppose that $y \in R(\gamma, \psi, p, r)$ with $\psi(y) = p$. Then, by (H5) and \eqref{e3.12}, we get \begin{align*} \psi(T y) &= \max_{0 \leq t \leq 1}|(T y(t)| = \max_{0 \leq t \leq 1} \int_0^1 G_1(t, s) (Q y)(s)ds \\ &= \max _{0 \leq t \leq 1} \Big|\int_0^{\xi_1} G_1(t, s) (Q y)(s) ds + \int_{\xi_1}^{\xi_2} G_1(t, s) (Q y)(s) ds \\ &\quad + \int_{\xi_2}^1 G_1(t, s) (Q y)(s) ds \Big| \\ & \leq \max _{0 \leq t \leq 1} \Big[\max _{0 \leq s \leq \xi_1}(Q y)(s) \int_0^{\xi_1} G_1(t, s) ds + \max _{\xi_1 \leq s \leq \xi_2}(Q y)(s) \int_{\xi_1}^{\xi_2} G_1(t, s) ds \\ & \quad + \max _{\xi_2 \leq s \leq 1}(Q y)(s) \int_{\xi_2}^1 G_1(t, s) ds \Big] \\ & \leq \max \big\{\max _{0 \leq s \leq \xi_1}(Q y)(s), \ \max _{\xi_1 \leq s \leq \xi_2}(Q y)(s), \ \max _{\xi_2 \leq s \leq 1} (Q y)(s) \big\} \max _{0 \leq t \leq 1} \int_0^1 \! G_1(t, s) ds \\ & \leq \max _{0 \leq t \leq 1} \int_0^1 G_1(t, s) ds \Big[\int_0^{\xi_1} - \tau h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau \\ &\quad + \int_{\xi_2}^1 - (1 - \tau) h(\tau) g\big(y(\tau), y''(\tau)\big) d \tau \\ & \quad + \frac{1}{\delta} (b + a(\xi_2 - \xi_1))(c(\xi_2 - \xi_1) + d) \int_{\xi_1}^{\xi_2} h(\tau) g\big(y(\tau), y''(\tau)\big) d\tau \Big] \\ & \leq \max _{0 \leq t \leq 1} \int_0^1 G_1(t, s) ds \cdot \frac{8 p}{M} \Big[\int_0^{\xi_1} - \tau h(\tau) d \tau + \int_{\xi_2}^1 - (1 - \tau) h(\tau)d\tau \\ & \quad + \big(\xi_2 - \xi_1 + \frac{b d}{\sigma}\big) \int_{\xi_1}^{\xi_2} h(\tau) d\tau\Big] \\ & = \frac{1}{8} \cdot \frac{8 p}{M}\cdot M = p. \end{align*} So, condition (iii) of Lemma \ref{lem2.2} is satisfied. Therefore, an application of Lemma \ref{lem2.2} imply the boundary-value problem \eqref{e1.6}, \eqref{e1.2} has at least three positive solutions $y_1, y_2$, and $y_3$ such that \begin{gather*} \max _{0 \leq t \leq 1}|y_i''(t)| \le r, \quad \text{for } i=1, 2, 3; \quad \min _{\omega \le t \le 1 - \omega}|y_1(t)| > q; \\ p < \max _{0 \leq t \leq 1}|y_2(t)|, \quad \min_{\omega \le t \le 1-\omega}|y_2(t)| < q; \quad \max _{0 \leq t \leq 1} |y_3(t)| < p. \end{gather*} The proof is complete. \end{proof} Now, we give an example to demonstrate our result. Consider the fourth-order four-point boundary-value problem \begin{gather} y^{(4)}(t) = h(t) g(y(t), y''(t)), \quad 0 < t < 1, \label{e3.13} \\ \begin{gathered} y(0) = y(1) = 0, \\ y''(\frac{1}{3}) - y'''(\frac{1}{3}) = 0, \quad y''(\frac{2}{3}) + y'''(\frac{2}{3}) = 0, \end{gathered} \label{e3.14} \end{gather} where $\xi_1 = \frac{1}{3}$, $\xi_2 = \frac{2}{3}$, $h(t) = 9 \pi \sin (3t-1)\pi$, and $g(u, v) = \begin{cases} \frac{u^2}{2} - (\frac{v}{150})^3, & 0 \leq u \leq 1, \ v \leq 0, \\ 11 \sqrt[4]{u-1} - (\frac{v}{150})^3 + \frac{1}{2}, & 1 < u \leq 9, \ v \leq 0, \\ 11 \sqrt[4]{8} + \frac{1}{2} - (\frac{v}{150})^3, & u > 9, \ v \leq 0. \end{cases}$ It is easy to check that the functions $h$ and $g$ satisfy (H1) and (H2). Set $\omega = 1/3$. It follows from a direct calculation that \begin{align*} M &= 9 \pi \Big[\int_0^{1/3} - \tau \sin (3 \tau - 1)\pi d \tau + \int_{2/3}^1 -(1 - \tau)\sin (3 \tau - 1)\pi d\tau \\ & \quad + \frac{16}{21} \int_{1/3}^{2/3} \sin (3 \tau - 1)\pi d\tau\Big] \\ & = \frac{46}{7}, \end{align*} and $$m = 9 \pi \cdot \frac{3}{7} \int_{1/3}^{2/3} \sin (3 \tau - 1)\pi d \tau \cdot \min \Big\{\int_{1/3}^{2/3} G (\frac{1}{3}, s)ds, \int_{1/3}^{2/3} G (\frac{2}{3}, s)ds\Big\} = \frac{2}{7}.$$ Choose $p = 1$, $q = 3$ and $r = 130$, then we have \begin{gather*} g(u, v) \leq 1.151 < 1.21 = \frac{8 p}{M}, \quad \text{for } 0 \leq u \leq 1, \ - 130 \leq v \leq 0; \\ g(u, v) \geq 14.232 > 10.5 = \frac{q}{m}, \quad \text{for } 3 \leq u \leq 9, \ - 130 \leq v \leq 0; \\ g(u, v) \leq 19.651 < 19.78 = \frac{r}{M}, \quad \text{for } 0 \leq u \leq 130, \ - 130 \leq v \leq 0. \end{gather*} Noticing that $b > \xi_1 a$ and $d > (1 - \xi_2)c$ hold, then all conditions of Theorem \ref{thm3.3} hold. Hence, by Theorem \ref{thm3.3}, BVP \eqref{e3.13}, \eqref{e3.14} has at least three positive solutions $y_1$, $y_2$ and $y_3$ such that \begin{gather*} \max _{0 \leq t \leq 1} |y_i''(t)| \le 130, \quad \text{for } i=1, 2, 3; \quad \min _{\frac{1}{3} \le t \le \frac{2}{3}}|y_1(t)| > 3; \\ 1 < \max _{0 \leq t \leq 1}|y_2(t)|, \quad \min_{\frac{1}{3} \le t \le \frac{2}{3}}|y_2(t)| < 3 \quad \max _{0 \leq t \leq 1} |y_3(t)| < 1. \end{gather*} \subsection*{Acknowledgements} The authors are grateful to the anonymous referee for his or her useful suggestions. This research was supported by grant 10771212 from the National Natural Science Foundation of China, and by grant 06KJB110010 from the Natural Science Foundation of Jiangsu Education Office. \begin{thebibliography}{00} \bibitem{a1} R. Avery, A. Peterson; \emph{Three positive fixed points of nonlinear operators on order Banach spaces}, Comput. Math. Appl. 42 (2001) 313-322. \bibitem{b1} C. Bai, D. Yang, H. 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