\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 129, 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} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/129\hfil Existence of positive solutions] {Existence of positive solutions for a multi-point four-order boundary-value problem} \author[L. X. Truong, P. D. Phung\hfil EJDE-2011/129\hfilneg] {Le Xuan Truong, Phan Dinh Phung} % in alphabetical order \address{Le Xuan Truong \newline Department of Mathematics and Statistics, University of Economics, HoChiMinh city, 59C, Nguyen Dinh Chieu Str, District 3, HoChiMinh city, Vietnam} \email{lxuantruong@gmail.com} \address{Phan Dinh Phung \newline Nguyen Tat Thanh University, 300A, Nguyen Tat Thanh Str, District 4, HoChiMinh city, Vietnam} \email{pdphung@ntt.edu.vn} \thanks{Submitted April 4, 2011. Published October 11, 2011.} \subjclass[2000]{34B07, 34B10, 34B18, 34B27} \keywords{Multi point; boundary value problem; Green function; \hfill\break\indent positive solution; Guo-Krasnoselskii fixed point theorem} \begin{abstract} The article shows sufficient conditions for the existence of positive solutions to a multi-point boundary-value problem for a fourth-order differential equation. Our main tools are the Guo-Krasnoselskii fixed point theorem and the monotone iterative technique. We also show that the set of positive solutions is compact. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} Multi-point boundary-value problems for ordinary differential equations arise in a variety of areas in applied mathematics and physics. For this reason the have been investigated by several authors; see for example \cite{HK}-\cite{IM2,HK,IM1,LJG,MA,TNL,SUN}. In this article, we study the existence of positive solutions for the problem \begin{gather}\label{eq1.01} x^{(4)}(t) = \lambda f(t, x(t)), \quad 0 < t < 1, \\ \label{eq1.02} x^{(2k+1)}(0) = 0, \quad x^{(2k)}(1) = \sum_{i=1}^{m-2}\alpha_{ki}x^{(2k)}(\eta_{ki}), \quad k= 0, 1, \end{gather} where $\lambda > 0$, $0 < \eta_{k1} < \eta_{k2} < \dots < \eta_{k,m-2} < 1$, $(k = 0, 1)$ and $\alpha_{ki}$, with $k = 0, 1$; $i = 1, 2, \dots, m-2$, are given positive constants satisfy the conditions \begin{gather} \sum_{i=1}^{m-2}\alpha_{1i}\eta_{1i} \leq 1 < \sum_{i=1}^{m-2}\alpha_{1i}, \label{condition1} \\ \sum_{i=1}^{m-2}\alpha_{0i}\eta_{0i}^{2} < 1 < \sum_{i=1}^{m-2}\alpha_{0i}. \label{condition2} \end{gather} When $m=3$; $\eta_{01} =\eta_0$, $\eta_{11} = \eta_{1}$; $\alpha_{01} = \alpha_0$, $\alpha_{11} = \alpha_1$; and the inhomogeneous term is $f(u(t))$, the problem \eqref{eq1.01}-\eqref{eq1.02} is studied in \cite{EK}. The authors in \cite{EK} obtained several existence results of positive solutions basing the computations of the fixed point index of open subsets of a Banach space relative to a cone and follow from a well-known theorem of Krasnosel'skii. One of the assumptions playing an important role in obtaining positive solution is that $1 < \alpha_i < \frac{1}{\eta_i}$, $i = 0, 1$. The rest of this paper is organized as follows. In section 2, we provide some results which are motivation for obtaining our main results. In section 3 we state and prove several existence results for at least one positive solution. Our main tools are the Guo-Krasnoselskii's fixed point theorem or the monotone iterative technique. Finally, section 4 devoted to the compactness of positive solutions set. \section{Preliminaries} In this article, $C([0, 1])$ denotes the space of all continuous functions $x$ from $[0, 1]$ into $\mathbb{R}$ endowed with the supremum norm $$\|x\| = \sup_{t\in [0, 1]}|x(t)|, \quad x \in C([0, 1]).$$ First we consider the auxiliary linear differential equation $$\label{eq2.01} -x''(t) = g(t), \quad 0 < t < 1,$$ with the boundary conditions $$\label{eq2.02} x'(0) = 0, \quad x(1) = \sum_{i=1}^{m-2}\alpha_ix(\eta_i),$$ where $0 < \eta_1 < \eta_2 < \dots < \eta_{m-2} < 1$ and $\alpha_i$ $(i = 1, 2, \dots, m-2)$ are given positive constants. \begin{lemma}\label{bode2.01} Let $g \in C[0,1]$ be non-negative (non-positive) and $\sum_{i=1}^{m-2}\alpha_i\eta_i \leq 1 < \sum_{i=1}^{m-2}\alpha_i$. Then $$\label{eq2.03} \begin{split} x(t) &= -\int_0^{t}(t-s)g(s)ds + \frac{1}{1-\sum_{i=1}^{m-2}\alpha_i} \Big[ \int_0^{1}(1-s)g(s)ds\\ &\quad - \sum_{i=1}^{m-2}\alpha_i\int_0^{\eta_i}(\eta_i - s)g(s)ds\Big] \end{split}$$ is a unique non-positive (non-negative) solution of \eqref{eq2.01}--\eqref{eq2.02}. \end{lemma} \begin{proof} It is easy to see that \eqref{eq2.03} is a unique solution of \eqref{eq2.01}--\eqref{eq2.02}. If $g(t) \geq 0$ on $[0, 1]$ then \begin{equation*} x'(t) = -\int_0^{t}g(s)ds \leq 0 \end{equation*} and $$x(t) \leq x(0) = \frac{1}{1-\sum_{i=1}^{m-2}\alpha_i} \Big[ \int_0^{1}( 1-s) g( s)ds-\sum_{i=1}^{m-2}\alpha _i\int_0^{\eta _i}( \eta _i-s) g( s) ds\Big]. \label{prooflemma2.1-01}$$ Let $F( \eta ) =\frac{1}{\eta }\int_0^{\eta }(\eta -s) g( s) ds$. We have \begin{equation*} F'( \eta ) =\frac{\eta \int_0^{\eta }g( s)ds -\int_0^{\eta }( \eta -s) g( s) ds}{\eta ^{2}} =\frac{\int_0^{\eta }sg( s) ds}{\eta ^{2}}\geq 0. \end{equation*} This implies $F(\eta_i) \leq F(1)$, for $i = 1, 2, \dots, m-2$; that is, \begin{equation*} \int_0^{\eta _i}( \eta _i-s) g( s) ds \leq \eta_i\int_0^{1}( 1-s) g( s) ds, \quad\text{for }i=1,2,\dots,m-2. \end{equation*} Hence $$\sum_{i=1}^{m-2}\alpha _i\int_0^{\eta _i}( \eta _i-s)g( s) ds \leq \sum_{i=1}^{m-2}\alpha _i\eta_i\int_0^{1}( 1-s) g( s) ds \leq \int_0^{1}(1-s) g( s) ds. \label{prooflemma2.1-02}$$ From \eqref{prooflemma2.1-01} and \eqref{prooflemma2.1-02}, we conclude that $x(t) \leq 0$, for all $t \in [0, 1]$. In the case $g(t) \leq 0$, by similar arguments, we obtain $x(t) \geq 0$, for all $t \in [0, 1]$. This completes the proof. \end{proof} \begin{lemma}\label{bode2.02} Let $g$ be non-positive and non-increasing function in $C[0,1]$ and let $\sum_{i=1}^{m-2}\alpha_i\eta_i^{2} < 1 < \sum_{i=1}^{m-2}\alpha_i$. Then the unique solution \eqref{eq2.03} of \eqref{eq2.01}--\eqref{eq2.02} is nonnegative. Further we have $$\min_{0 \leq t \leq 1}x(t) \geq \gamma \|x\|, \label{eq2.06}$$ where $$\gamma = \frac{1-\sum_{i=1}^{m-2}\alpha_i\eta_i^2}{\sum_{i=1}^{m-2} \alpha_i(1 -\eta_i^2)}. \label{prooflemma2.2-06}$$ \end{lemma} \begin{proof} Because $g(t) \leq 0$ for all $t \in [0,1]$, the unique solution \eqref{eq2.03} of \eqref{eq2.01}--\eqref{eq2.02} is non-decreasing and $$x( t) \geq x( 0) =\frac{1}{1-\sum_{i=1}^{m-2}\alpha_i} \Big[ \int_0^{1}( 1-s) g( s)ds -\sum_{i=1}^{m-2}\alpha _i\int_0^{\eta _i}( \eta _i-s)g( s) ds\Big]. \label{prooflemma2.2-01}$$ Let $F_0(\eta) = \frac{1}{\eta^2}\int_0^{\eta}(\eta - s)g(s)ds$. Then we have \begin{equation*} F_0'( \eta ) =\frac{\eta \int_0^{\eta }g( s)ds -2\int_0^{\eta }( \eta -s) g( s) ds}{\eta ^3} =\frac{\int_0^{\eta }( 2s-\eta ) g( s) ds}{\eta ^3} \end{equation*} It is easy to check that the function $\eta \mapsto \int_0^{\eta }( 2s-\eta ) g( s) ds$ is non-increasing. Thus \begin{equation*} \int_0^{\eta }( 2s-\eta ) g( s) ds \leq 0, \quad \forall \eta \geq 0. \end{equation*} This implies that $F_0'( \eta ) \leq 0$, for all $\eta \geq 0$. Thus $$\begin{split} \sum_{i=1}^{m-2}\alpha _i\int_0^{\eta _i}( \eta _i-s)g( s) ds &=\sum_{i=1}^{m-2}\alpha _i\eta _i^{2}F_0( \eta_i) \geq F_0( 1) \sum_{i=1}^{m-2}\alpha _i\eta_i^{2}\\ &\geq \int_0^{1}( 1-s) g( s) ds. \end{split} \label{prooflemma2.2-02}$$ Combining \eqref{prooflemma2.2-01} and \eqref{prooflemma2.2-02}, we deduce that $x(t) \geq 0$ for all $t \in [0, 1]$. Finally, we need to check inequality \eqref{eq2.06}, or equivalently, $$x(0) \geq \gamma x(1). \label{prooflemma2.2-03}$$ Indeed, it follows from \eqref{eq2.03} that \eqref{prooflemma2.2-03} is equivalent to $$\sum_{i=1}^{m-2}\alpha_i\int_0^{\eta_i}(\eta_i - s)g(s)ds \geq \frac{1-\gamma\sum_{i=1}^{m-2}\alpha_i}{1-\gamma} \int_0^{1}(1-s)g(s)ds. \label{prooflemma2.2-04}$$ By the monotonicity of $F_0$, we have $$\sum_{i=1}^{m-2}\alpha _i\int_0^{\eta _i}( \eta _i-s)g( s) ds =\sum_{i=1}^{m-2}\alpha _i\eta _i^{2}F_0( \eta_i) \geq \sum_{i=1}^{m-2}\alpha _i\eta_i^{2}\int_0^{1}( 1-s) g( s) ds. \label{prooflemma2.2-05}$$ So, it is not difficult to obtain \eqref{prooflemma2.2-04} from \eqref{prooflemma2.2-05} and \eqref{prooflemma2.2-06}. The proof is completed. \end{proof} \begin{remark} \label{rmk2.3} \rm For $t, s \in [0, 1]$, we put $$\begin{split} G(t, s, \alpha_i, \eta_i) &=\begin{cases} s - t,& 0\leq s \leq t \leq 1, \\ 0, & 0 \leq t \leq s \leq 1, \end{cases} \\ &\quad + \overline{\alpha} \begin{cases} 1 - \sum_{i=1}^{m-2}\alpha_i\eta_i + (\sum_{i=1}^{m-2}\alpha_i -1)s, & 0 \leq s \leq \eta_1, \\[4pt] 1 - \sum_{i=2}^{m-2}\alpha_i\eta_i + (\sum_{i=2}^{m-2}\alpha_i -1)s, & \eta_1 \leq s \leq \eta_2, \\ \dots \\ 1 - \sum_{i=k}^{m-2}\alpha_i\eta_i + (\sum_{i=k}^{m-2}\alpha_i -1)s, & \eta_{k-1} \leq s \leq \eta_{k}, \\ \dots \\ 1 - s, & \eta_{m-2} \leq s \leq 1, \end{cases} \label{green.g} \end{split}$$ where $\overline{\alpha} = (1-\sum_{i=1}^{m-2}\alpha_i)^{-1}$. Then \eqref{eq2.03} can be rewrite as $$u(t) = \int_0^{1}G(t, s, \alpha_i, \eta_i)\,g(s)\,ds.$$ \end{remark} Now we consider the linearized equation $$\label{lin.ord.04} x^{(4)}(t) = g(t), \quad 0 < t < 1,$$ subject to the boundary conditions \eqref{eq1.02}. We have the following lemma. \begin{lemma}\label{bode2.03} Let $g \in C[0,1]$ be non-negative and $$\sum_{i=1}^{m-2}\alpha_{1i}\eta_{1i} \leq 1 < \sum_{i=1}^{m-2}\alpha_{1i}, \quad \sum_{i=1}^{m-2}\alpha_{0i}\eta_{0i}^{2} < 1 < \sum_{i=1}^{m-2}\alpha_{0i}.$$ Then \eqref{lin.ord.04}, \eqref{eq1.02} has a unique non-negative solution $$\label{lin.ord.sol} x(t) = \int_0^{1}\Phi(t, s)g(s)ds := Ag(t),$$ where $\Phi(t, s)$ is the Green function $$\label{green_phi} \Phi(t, s) = \int_0^{1}G(t, \tau, \alpha_{0i}, \eta_{0i})G(\tau, s, \alpha_{1i}, \eta_{1i})\,d\tau, \quad \text{for } t, s \in [0, 1].$$ Moreover, we have $\min_{t\in [0, 1]}x(t) \geq \gamma_0 \|x\|$, where $$\gamma_0 = \frac{1-\sum_{i=1}^{m-2}\alpha_{0i} \eta_{0i}^2}{\sum_{i=1}^{m-2}\alpha_{0i}(1 - \eta_{0i}^2)}.$$ \end{lemma} \begin{proof} It follows from Lemma \ref{bode2.01} that \begin{equation*} -x''(t) = \int_0^{1}G(t, s, \alpha_{1i}, \eta_{1i})g(s)\,ds \leq 0 \end{equation*} is non-positive non-increasing for all $t\in [0, 1]$. Thus, by Lemma \ref{bode2.02}, \begin{align*} x(t) &= \int_0^{1}G(t, s, \alpha_{0i}, \eta_{0i})\int_0^{1} G(s, \tau, \alpha_{1i}, \eta_{1i})g(\tau)\,d\tau\,ds\\ &= \int_0^{1}\Big(\int_0^{1}G(t, \tau, \alpha_{0i}, \eta_{0i}) G(\tau, s, \alpha_{1i}, \eta_{1i})\,d\tau\Big)g(s)ds\\ &= \int_0^{1}\Phi(t, s)g(s)ds \geq 0, \quad t \in [0, 1], \end{align*} and $\min_{t\in [0, 1]}x(t) \geq \gamma_0 \|x\|$. The proof is complete. \end{proof} The following result is straightforward and we will omit its proof. \begin{lemma}\label{bode2.04} The operator $A: C([0, 1]) \to C([0, 1])$, defined by \eqref{lin.ord.sol}, be a completely continuous linear operator. If $g$ is a nonnegative function in $C([0, 1])$ then $Ag$ is also nonnegative. \end{lemma} Next we give some properties of the Green function $\Phi(t,s)$ which is used in the sequel. \begin{lemma}\label{bode2.05} Let $$\sum_{i=1}^{m-2}\alpha_{1i}\eta_{1i} \leq 1 < \sum_{i=1}^{m-2}\alpha_{1i}, \quad \sum_{i=1}^{m-2}\alpha_{0i}\eta_{0i}^{2} < 1 < \sum_{i=1}^{m-2}\alpha_{0i}.$$ Then we have \begin{itemize} \item[(1)] $\Phi(t, s) \geq 0$, for all $s, t \in [0, 1]$; \item[(2)] there exists a continuous function $\chi : [0, 1] \to [0, +\infty)$ such that \begin{equation*} \gamma_0\chi(s) \leq \Phi(t, s) \leq \chi(s),\quad \forall s, t \in [0,1]. \end{equation*} \end{itemize} \end{lemma} \begin{proof} From \eqref{green.g} and the assumptions $\sum_{i=1}^{m-2}\alpha_{1i}\eta_{1i} \leq 1 < \sum_{i=1}^{m-2}\alpha_{1i},$ it is easy to check that, for each $s\in [0,1]$, $\tau \mapsto G(\tau, s, \alpha_{1i}, \eta_{1i})$ is a non-positive, non-increasing and continuous function. So by using \eqref{green_phi} and the Lemma \ref{bode2.02}, the function $\Phi(t, s) \geq 0$ for all $s, t \in [0,1]$ and $$\min_{t\in [0,1]}\Phi(t, s) \geq \gamma_0 \|\Phi(\cdot, s)\| = \gamma_0\Phi(1, s).$$ Let $\chi(s) = \Phi(1, s)$. Obviously we have $\gamma_0\chi(s) \leq \Phi(t, s) \leq \chi(s)$. The proof is complete. \end{proof} To study \eqref{eq1.01}-\eqref{eq1.02}, we use the assumption \begin{itemize} \item[(A1)] $f : [0, 1] \times \mathbb{R}^{+} \to \mathbb{R}^{+}$ is continuous \end{itemize} Let $K$ be the cone in $C([0,1])$, consisting of all nonnegative functions and $$P = \{x \in K : \min_{t\in [0, 1]} x(t) \geq \gamma_0 \|x\|\}$$ It is clear that $P$ is also a cone in $C([0,1])$. For each $x\in P$, denote $F(x)(t) = \lambda f(t, x(t))$, $t\in [0,1]$. By the assumption (A1), the operator $F: P \to K$ is continuous. Therefore the operator $T := A\circ F : P \to K$ is completely continuous. On the other hand it is not difficult to check that for $x \in P$ we have $$\min_{0\leq t \leq 1}Tx(t) \geq \gamma_0\|Tx\|$$ using the Lemma \ref{bode2.05}, that is $TP \subset P$. We note that the nonzero fixed points of the operator $T$ are positive solutions of \eqref{eq1.01}-\eqref{eq1.02}. To finish this section we state here the Guo-Krasnoselskii's fixed point theorem (see \cite{KRA}) \begin{theorem}\label{theo.Kras} Let $X$ be a Banach space and $P \subset X$ be a cone in $X$. Assume $\Omega_1, \Omega_2$ are two open bounded subsets of $X$ with $0 \in \Omega_1, \overline{\Omega}_1 \subset \Omega_2$ and $T : P \cap (\overline{\Omega}_2 \setminus \Omega_1 ) \to P$ be a completely continuous operator such that \begin{itemize} \item[(i)] $\|Tu\| \leq \|u\|$, $u \in P\cap \partial \Omega_1$ and $\|Tu\| \geq \|u\|$, $u \in P\cap \partial \Omega_2$, or \item[(ii)] $\|Tu\| \geq \|u\|$, $u \in P\cap \partial \Omega_1$ and $\|Tu\| \leq \|u\|$, $u \in P\cap \partial \Omega_2$. \end{itemize} Then $T$ has a fixed point in $P \cap (\overline{\Omega}_2 \setminus \Omega_1 )$. \end{theorem} \section{Existence of positive solutions} We introduce the notation \begin{gather*} f_0:= \liminf_{z\to 0^+}\min_{t\in [0,1]}\frac{f(t, z)}{z}, \quad f^{\infty}:= \limsup_{z\to +\infty}\max_{t\in [0,1]}\frac{f(t, z)}{z}, \\ f^{0}:= \limsup_{z\to 0^+}\max_{t\in [0,1]}\frac{f(t, z)}{z}, \quad f_{\infty}:= \liminf_{z\to +\infty}\min_{t\in [0,1]}\frac{f(t, z)}{z}, \\ A = \Big(\int_0^1\Phi(1, s)ds\Big)^{-1}, \quad B = \frac{A}{\gamma_0}. \end{gather*} \begin{theorem}\label{kras} Assume that {\rm (A1)} holds. Then \eqref{eq1.01}-\eqref{eq1.02} has at least one positive solution for every $\lambda \in \big(\frac{B}{f_0}, \frac{A}{f^\infty}\big)$ if $f_0, f^{\infty} \in (0, \infty)$ satisfy $f_0\gamma_0 > f^\infty$; or $\lambda \in \big(\frac{B}{f_\infty}, \frac{A}{f^0}\big)$ if $f^0, f_{\infty} \in (0, \infty)$ satisfy $f_\infty\gamma_0 > f^0$. \end{theorem} \begin{proof} Set $$\Omega_i = \{x \in C([0, 1]) : \|x\| < R_i\}, \quad i = 1, 2.$$ Then $\Omega_1, \Omega_2$ are two open bounded of $C([0, 1])$ and $0 \in \Omega_1$, $\overline{\Omega}_1 \subset \Omega_2$. \noindent\textbf{Case 1:} $f_0, f^{\infty} \in (0, \infty)$ and $f_0\gamma_0 > f^\infty$. Let $\lambda \in (\frac{B}{f_0}, \frac{A}{f^\infty})$. Then there exists $\varepsilon > 0$ such that $$\frac{B}{f_0 - \varepsilon} < \lambda < \frac{A}{f^\infty + \varepsilon}.$$ Since $f_0 \in (0, \infty)$ there exists $R_1 > 0$ such that $f(t, z) \geq (f_0 - \varepsilon)z$ for all $t \in [0, 1], z \in [0, R_1]$. So if $x \in P$ such that $\|x\| = R_1$, we have $$f(t, x(t)) \geq (f_0 - \varepsilon)x(t) \geq \gamma_0(f_0 - \varepsilon)\|x\|, \quad \forall t \in [0, 1].$$ This implies $$Tx(t) = \lambda\int_0^1\Phi(t, s)f(s, x(s))ds \geq \lambda\gamma_0(f_0 - \varepsilon)\|x\|\int_0^1\Phi(t, s)ds, \quad \forall t\in [0, 1].$$ Hence, for all $x \in P\cap\partial\Omega_1$, $$\|Tx\| \geq \lambda\gamma_0(f_0 - \varepsilon)\max_{0\leq t \leq 1} \Big(\int_0^1\Phi(t, s)ds\Big)\|x\| \geq \|x\|.$$ On the other hand, since $f^\infty \in (0, \infty)$, there exists $R > 0$ such that $f(t, z) \leq (f^\infty + \varepsilon)z$ for all $t \in [0, 1], z \in [R, +\infty]$. Set $R_2 = \max\{R_1 + 1, R\gamma_0^{-1}\}$. Let us $x \in P \cap \partial\Omega_2$. We have $$x(t) \geq \gamma_0 \|x\| = \gamma_0 R_2, \quad \forall t \in [0, 1].$$ So $$Tx(t) = \lambda\int_0^1\Phi(t, s)f(s, x(s))ds \leq \lambda(f^\infty + \varepsilon)\|x\|\int_0^1\Phi(t, s)ds.$$ Consequently, $\|Tx\| \leq \|x\|$ for all $x\in P\cap \partial\Omega_2$. Therefore, using the second part of Theorem \ref{theo.Kras}, we conclude that $T$ has a fixed point in $P \cap \overline{\Omega}_2\setminus \Omega_1$. \noindent\textbf{Case 2:} $f^0, f_{\infty} \in (0, \infty)$ and $f_\infty\gamma_0 > f^0$. Let $\lambda \in (\frac{B}{f_\infty}, \frac{A}{f^0})$. Then there exists $\varepsilon > 0$ such that $$\frac{B}{f_\infty - \varepsilon} < \lambda < \frac{A}{f^0 + \varepsilon}.$$ Using the arguments as in Case 1, we can find $R_2 > R_1 > 0$ such that $\|Tx\| \leq \|x\|$, for all $x \in P\cap \partial \Omega_1$ and $\|Tx\| \geq \|x\|$, for all $x \in P\cap \partial \Omega_2$. So $T$ has a fixed point in $P \cap \overline{\Omega}_2\setminus \Omega_1$ which is a positive solution of \eqref{eq1.01}-\eqref{eq1.02}, using the Theorem \ref{theo.Kras}. \end{proof} Next, we add the following assumption \begin{itemize} \item[(A2)] The function $f(t, x)$ is nondecreasing about $x$. \end{itemize} Using the monotone iterative technique, we get the following result. \begin{theorem}\label{theo.iterative} Let {\rm (A1)} and {\rm (A2)} hold. Assume that there exist two positive numbers $R_1 < R_2$ such that $$0 < R_1\sup_{t\in [0, 1]}f(t, R_2) < \gamma_0 R_2\inf_{t\in [0, 1]}f(t, \gamma_0 R_1).$$ Then if $$\lambda \in \big[\frac{BR_1}{\inf_{t\in [0, 1]}f(t, \gamma_0 R_1)}, \, \frac{AR_2}{\sup_{t\in [0, 1]}f(t, R_2)}\big]$$ then \eqref{eq1.01}-\eqref{eq1.02} has positive solutions $x^*_1, x^*_2$ ($x^*_1$ and $x^*_2$ may coincide) with $$R_1 \leq \|x_1^*\| \leq R_2 \quad \text{and} \quad \lim_{n\to \infty}T^nx_0 = x_1^*,\quad\text{where } x_0(t) = R_2, \quad\forall t \in [0, 1];$$ and $$R_1 \leq \|x_2^*\| \leq R_2 \quad \text{and} \quad \lim_{n\to \infty}T^n \overline{x}_0 = x_2^*, \quad \text{where } \overline{x}_0(t) = R_1, \quad \forall t \in [0, 1].$$ \end{theorem} \begin{proof} Set $$P_{[R_1, R_2]} = \{x \in P : R_1 \leq \|x\| \leq R_2\}.$$ Let $x\in P_{[R_1, R_2]}$. It's clear that $\gamma_0 R_1 \leq \gamma_0\|x\| \leq x(t) \leq \|x\| \leq R_2$, for all $t\in [0, 1]$. So $$Tx(t) =\lambda\int_0^1\Phi(t, s)f(s, x(s))ds \leq \lambda\int_0^1\Phi(t, s)f(s, R_2)ds \leq R_2,$$ and $$Tx(t) \geq \lambda\int_0^1\Phi(t, s)f(s, \gamma_0 R_1)ds \geq \frac{AR_1}{\gamma_0}\int_0^1\Phi(t, s)ds \geq AR_1\int_0^1\Phi(1, s)ds = R_1.$$ This implies that $TP_{[R_1, R_2]} \subset P_{[R_1, R_2]}$. Let $x_0(t) = R_2$ for all $t\in [0, 1]$. It is evident that $x_0 \in P_{[R_1, R_2]}$. We consider the sequence in $P_{[R_1, R_2]}$, $\{x_n\}_{n \in \mathbb{N}}$, defined by $$\label{ite.01} x_{n} = Tx_{n-1} = T^{n}x_0, \quad n =1, 2, \dots.$$ Because $T$ is the completely continuous operator, there exists a subseqence $\{x_{n_k}\}$ of $\{x_n\}$ which uniformly converges to $x^{*}_1 \in C([0, 1])$. On the other hand we can see that $T : P_{[R_1, R_2]} \to P_{[R_1, R_2]}$ is a nondecreasing operator using the assumption (A2). Therefore, since $$0 \leq x_1(t) \leq \|x_1\| \leq R_2 = x_0(t), \quad \forall t \in [0, 1],$$ we have $Tx_1 \leq Tx_0$, that is $x_2 \leq x_1$. Similarly by induction we deduce that $x_{n+1} \leq x_n$ for all $n \in \mathbb{N}$. Therefore, we can conclude that the sequence $\{x_n\}$ uniformly converges to $x^*$. Letting $n \to +\infty$ in \eqref{ite.01} yields $Tx^*_1 = x^*_1$. Let $\overline{x}_0(t) = R_1$ for all $t \in [0, 1]$ and $\overline{x}_n = T\overline{x}_{n-1}$ for $n = 1, 2, \dots$. It is clear that $x_n \in P_{[R_1, R_2]}$ for all $n \in \mathbb{N}$. Moreover, by definition of the operator $T$, we have \begin{align*} \overline{x}_1(t) &= T\overline{x}_0(t) = \lambda\int_0^1\Phi(t, s)f(s, \overline{x}_0(s))ds\\ &\geq \lambda\int_0^1\Phi(t, s)f(s, \gamma_0 R_1)ds \geq R_1 \equiv \overline{x}_0(t), \end{align*} for $t\in [0, 1]$. Therefore, by using the arguments as above, we deduce that $\{\overline{x}_n\}$ converges uniformly to $x^*_2 \in P_{[R_1, R_2]}$ and $Tx^*_2 = x^*_2$. The proof is complete. \end{proof} \begin{example} \label{examp3.3} \rm Let $a, b, c, d$ be positive numbers such that $5bc > 42ad$. We consider the boundary-value problem \begin{gather*} x^{(4)}(t) = (t^2 + 1)\frac{ax^2(t) + bx(t)}{cx(t) + d}, \quad 0 < t < 1,\\ x'(0) = x^{(3)}(0) = 0, \\ x(1) = \frac{3}{2}x(\frac{3}{4}), \,\, x''(1) = \frac{4}{3}x''(\frac{1}{2}). \end{gather*} We have $\gamma_0 = \frac{5}{21}$, $$G( t,\tau, \alpha_{01}, \eta_{01} ) =\begin{cases} \tau -t & \text{if } 0\leq \tau \leq t\leq 1 \\ 0 & \text{if } 0\leq t\leq \tau \leq 1 \end{cases} +\begin{cases} \frac{1}{4}-\tau & \text{if } 0\leq \tau \leq \frac{3}{4} \\ 2\tau -2 & \text{if } \frac{3}{4}\leq \tau \leq 1 \end{cases}$$ and $$G_1( \tau ,s, \alpha_{11}, \eta_{11}) =\begin{cases} s-\tau & \text{if } 0\leq s\leq \tau \leq 1 \\ 0 & \text{if } 0\leq \tau \leq s\leq 1 \end{cases} -\begin{cases} 1+s & \text{if } 0\leq s\leq \frac{1}{2} \\ 3( 1-s) & \text{if } \frac{1}{2}\leq s\leq 1\,. \end{cases}$$ By doing some calculating, $\Phi(t, s)$ is defined as follows: For $s \leq t$, \begin{align*} \Phi ( t,s) &= -\frac{1}{6}( s-t) ^3 \\ &\quad +\begin{cases} -\frac{5}{32}s+( \frac{1}{2}t^{2}+\frac{5}{32}) ( s+1) -\frac{1}{8}s^{2}+\frac{1}{6}s^3+\frac{47}{384} & \text{if } 0\leq s\wedge s\leq \frac{1}{2} \\ -\frac{5}{32}s-( 3s-3) ( \frac{1}{2}t^{2}+\frac{5}{32}) -\frac{1}{8}s^{2}+\frac{1}{6}s^3+\frac{47}{384} & \text{if } \frac{1}{2}\leq s\wedge s\leq \frac{3}{4} \\ -(3s-3) ( \frac{1}{2}t^{2}+\frac{5}{32}) -\frac{1}{3}( s-1) ^3 & \text{if } s\leq 1\wedge \frac{3}{4}\leq s; \end{cases} \end{align*} and for $t \leq s$, $$\Phi ( t,s) = +\begin{cases} -\frac{5}{32}s+( \frac{1}{2}t^{2}+\frac{5}{32}) ( s+1) -\frac{1}{8}s^{2}+\frac{1}{6}s^3+\frac{47}{384} & \text{if } 0\leq s\wedge s\leq \frac{1}{2} \\ -\frac{5}{32}s-( 3s-3) ( \frac{1}{2}t^{2}+\frac{5}{32} ) -\frac{1}{8}s^{2}+\frac{1}{6}s^3+\frac{47}{384} & \text{if } \frac{1}{2}\leq s\wedge s\leq \frac{3}{4} \\ -( 3s-3) ( \frac{1}{2}t^{2}+\frac{5}{32}) -\frac{1}{3}( s-1) ^3 & \text{if } s\leq 1\wedge \frac{3}{4}\leq s \end{cases}$$ So $A = \big(\int_0^1\Phi(1, s)ds\big)^{-1} = 103/128$. Now we set $$f(t, x) = (t^2 + 1)\frac{ax^2 + bx}{cx + d}.$$ Then $f: [0, 1] \times \mathbb{R}^{+} \to \mathbb{R}^+$ is continuous and \begin{gather*} f_0 = \lim_{x\to 0^+}\min_{0\leq t \leq 1}\frac{f(t, x)}{x} =\lim_{x\to 0^+}\frac{ax^2 + bx}{cx^2 + dx} = \frac{b}{d}, \\ f^{\infty} = \lim_{x\to \infty}\max_{0\leq t \leq 1} \frac{f(t, x)}{x} =2\lim_{x\to \infty}\frac{ax^2 + bx}{cx^2 + dx} = \frac{2a}{c}; \end{gather*} that is, $\gamma_0f_0 > f^{\infty}$. Thus, by Theorem \ref{kras}, we conclude that for each $\lambda \in (\frac{2163d}{640b}, \frac{103c}{256a})$ our problem has at least one positive solution. \end{example} \section{Compactness of the set of positive solutions} \begin{theorem}\label{compactness} Let {\rm (A1)} hold. Assume that we have $$\label{compact} f_0, f^{\infty} \in (0, \infty), \quad f_0\gamma_0 > f^\infty \quad \text{and} \quad \lambda \in \big(\frac{B}{f_0}, \frac{A}{f^\infty}\big).$$ Then the set of positive solutions of \eqref{eq1.01}-\eqref{eq1.02} is nonempty and compact. \end{theorem} \begin{proof} Put $S = \{x \in P : x = Tx\}$. By Theorem \ref{kras} $S$ is nonempty. We shall show that $S$ is compact in $C([0, 1])$. First we claim that $S$ is a closed subset of $C([0, 1])$. Indeed, assume that $\{x_n\}_{n \in \mathbb{N}}$ be a sequence in $S$ and $\lim_{n \to \infty}\|x_n - x\| = 0$. Then for each $t\in [0, 1]$, we have \begin{align*} &\big|x(t) - \lambda\int_0^1\Phi(t, s)f(s, x(s))ds\big|\\ &\leq |x(t) - x_n(t)| +\big|x_n(t) - \lambda\int_0^1\Phi(t, s)f(s, x_n(s))ds\big| \\ &\quad + \lambda\big|\int_0^1\Phi(t, s)f(s, x(s))ds - \int_0^1\Phi(t, s)f(s, x_n(s))ds\big|. \end{align*} This implies \begin{align*} &\big|x(t) - \lambda\int_0^1\Phi(t, s)f(s, x(s))ds\big|\\ & \leq |x(t) - x_n(t)| + \lambda\int_0^1\Phi(t, s)|f(s, x(s))-f(s, x_n(s))|ds, \end{align*} because $x_n = Tx_n$ for all $n \in \mathbb{N}$. Let $n \to \infty$ in the last inequality we can deduce that $$x(t) = \lambda\int_0^1\Phi(t, s)f(s, x(s))ds, \quad \forall t \in [0, 1],$$ using the continuity of the function $f$ and the dominated convergence theorem. So $x \in S$ and $S$ is closed in $C([0, 1])$. It remains to check that $S$ is relatively compact in $C([0, 1])$. Let \eqref{compact} holds. Choosing $\varepsilon^* > 0$ such that $$\frac{B}{f_0 - \varepsilon^*}< \lambda < \frac{A}{f^\infty + \varepsilon^*}.$$ Clearly there exists a constant $R > 0$ such that $f(t, z) \leq (f^\infty + \varepsilon^*)z$, for all $t \in [0, 1]$ and $z \in [R, \infty)$. Hence $$f(t, x(t)) \leq (f^\infty + \varepsilon^*)x(t) + \beta, \, t\in [0, 1],$$ where $\beta = \max\{f(t, z) : (t, z) \in [0, 1] \times [0, R]\}$. So, for $x \in S$ and for every $t \in [0, 1]$, we have \begin{align*} x(t) &= \lambda\int_0^1\Phi(t, s)f(s, x(s))ds \\ &\leq \lambda\int_0^1\Phi(t, s)[(f^\infty + \varepsilon^*)x(s) + \beta]ds\\ &\leq \frac{\lambda}{A}(f^\infty + \varepsilon^*)\|x\| + \frac{\lambda\beta}{A}. \end{align*} We can deduce from this inequality that $\|x\| \leq \frac{\lambda\beta}{A - \lambda(f^\infty + \varepsilon^*)}$; that is, $S$ is bounded in $C([0, 1])$. By the compactness of the operator $T : P \to P$ we conclude that $S = T(S)$ is relatively compact. The proof is complete. \end{proof} \begin{remark} \label{rmk4.2} \rm Assume that $f^0, f_{\infty} \in (0, \infty)$, $f_\infty\gamma_0 > f^0$, $f^\infty \leq f^0$ and $$\lambda \in \big(\frac{B}{f_\infty}, \frac{A}{f^0}\big).$$ Thanks to Theorem \ref{theo.Kras}, the set of positive solutions $S$ of the problem \eqref{eq1.01} \eqref{eq1.02} is nonempty. Moreover by the similar arguments we can show that $S$ is compact in $C([0, 1])$. \end{remark} \subsection*{Acknowledgements} The authors wish to express their gratitude to the anonymous referee for his/her helpful comments and remarks. \begin{thebibliography}{00} \bibitem{EK} M. Eggensperger, N. Kosmatov; \emph{Positive solutions of a fourth-order multi-point boundary value problem}, Comm. Math. Anal. \textbf{6} (2009), pp. 22-30. \bibitem{HK} J. Henderson, N. Kosmatov; \emph{The existence and muliplicity of constant sign solution to a three-point boundary value problem}, Comm. Appl. Nonlinear Anal. 14 (2007), pp 63-78. \bibitem{IM1} V. 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