\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 68, pp. 1--11.\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-2006/68\hfil Existence of positive solutions] {Existence of positive solutions for boundary-value problems for singular higher-order functional differential equations} \author[C. Bai, Q. Yang, J. Ge\hfil EJDE-2006/68\hfilneg] {Chuanzhi Bai, Qing Yang, Jing Ge} \address{Chuanzhi Bai \newline Department of Mathematics \\ Huaiyin Teachers College\\ Huaian, Jiangsi 223001, China; \newline and Department of Mathematics\\ Nanjing University \\ Nanjing 210093, China} \email{czbai8@sohu.com} \address{Qing Yang \hfill\break Department of Mathematics \\ Huaiyin Teachers College\\ Huaian, Jiangsi 223001, China} \email{yangqing3511115@163.com} \address{Jing Ge \hfill\break Department of Mathematics \\ Huaiyin Teachers College\\ Huaian, Jiangsi 223001, China} \email{gejing0512@163.com} \date{} \thanks{Submitted April 21, 2006. Published July 6, 2006.} \thanks{Supported by the Natural Science Foundation of Jiangsu Education Office and by \hfill\break\indent Jiangsu Planned Projects for Postdoctoral Research Funds} \subjclass[2000]{34K10, 34B16} \keywords{Boundary value problem; higher-order; positive solution; \hfill\break\indent functional differential equation; fixed point} \begin{abstract} We study the existence of positive solutions for the boundary-value problem of the singular higher-order functional differential equation \begin{gather*} (L y^{(n-2)})(t) + h(t) f(t, y_t) = 0, \quad \text{for } t \in [0, 1], \\ y^{(i)}(0) = 0, \quad 0 \leq i \leq n - 3, \\ \alpha y^{(n-2)}(t) - \beta y^{(n-1)} (t) = \eta (t), \quad \text{for } t \in [- \tau, 0],\\ \gamma y^{(n-2)}(t) + \delta y^{(n-1)}(t) = \xi (t), \quad \text{for } t \in [1, 1 + a], \end{gather*} where $ Ly := -(p y')' + q y$, $p \in C([0, 1],(0, + \infty))$, and $q \in C([0, 1], [0, + \infty))$. Our main tool is the fixed point theorem on a cone. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \section{Introduction} As pointed out in \cite{e1}, boundary-value problems associated with functional differential equations arise from problems in physics, from variational problems in control theory, and from applied mathematics; see for example \cite{g1,g3}. Many authors have investigated the existence of solutions for boundary-value problems of functional differential equations; see \cite{d1, h1, n1, t2}. Recently an increasing interest in studying the existence of positive solutions for such problems has been observed. Among others publication, we refer to \cite{b1, b2, h2, h3, k1, w1}. In this paper, we investigate the existence of positive solutions for singular boundary-value problems (BVP) of an $n$-th order ($n \geq 3$) functional differential equation (FDE) of the form \begin{gather} (L y^{(n-2)})(t) + h(t) f(t, y_t) = 0, \quad \text{for } t \in [0, 1], \label{e1.1}\\ y^{(i)}(0) = 0, \quad 0 \leq i \leq n - 3, \label{e1.2} \\ \alpha y^{(n-2)}(t) - \beta y^{(n-1)} (t) = \eta (t), \quad \text{for } t \in [- \tau, 0], \label{e1.3}\\ \gamma y^{(n-2)}(t) + \delta y^{(n-1)}(t) = \xi (t), \quad \text{for } t \in [1, 1 + a], \label{e1.4} \end{gather} where $ Ly := -(p y')' + q y$, $p \in C([0, 1],(0, + \infty))$, and $q \in C([0, 1], [0, + \infty))$; $\alpha,\beta, \gamma, \delta \geq 0$, and $\alpha \delta + \alpha \gamma + \beta \gamma > 0$; $ \eta \in C([- \tau, 0], \mathbb{R})$, $\xi \in C([1, b], \mathbb{R})$ ($b = 1 + a$), and $\eta (0) = \xi (1) = 0$; $h \in C((0, 1), \mathbb{R})$ ($h(t)$ is allowed to have singularity at $t = 0$ or $1$); $f \in C([0, 1] \times D, \mathbb{R})$, $D = C([- \tau, a], \mathbb{R})$, for every $t \in [0, 1]$, $y_t \in D$ is defined by $y_t(\theta) = y(t + \theta), \theta \in [- \tau, a]$. The study of higher-order functional differential equation has received also some attention; see for example \cite{d1, h2, t1}. Recently, Hong et al. \cite{h4} imposed conditions on $f(t, y^t)$ to yield at least one positive solution to \eqref{e1.1}-\eqref{e1.4} for the special case $ h(t) \equiv 1$, $p(t) \equiv 1$, and $ q(t) \equiv 0$. They applied the Krasnosel'skii fixed-point theorem. The purpose of this paper is to establish the existence of positive solutions of the singular higher-order functional differential equation \eqref{e1.1} with boundary conditions \eqref{e1.2}-\eqref{e1.4} under suitable conditions on $f$. \section{Preliminaries} To abbreviate our discussion, we assume the following hypotheses: \begin{itemize} \item[{\rm (H1)}] $G(t, s)$ is the Green's function of the differential equation $$ (Ly^{(n - 2)})(t) = 0, \quad 0 < t < 1 $$ subject to the boundary condition \eqref{e1.2}-\eqref{e1.4} with $\tau = a = 0$. \item[{\rm (H2)}] $g(t, s)$ is the Green's function of the differential equation $$ Ly(t) = 0 \quad t \in (0, 1) $$ subject to the boundary conditions \[ \alpha y(0) - \beta y' (0) = 0,\quad \gamma y(1) + \delta y'(1) = 0, \] where $\alpha, \beta, \gamma$ and $ \delta$ are as in \eqref{e1.3} and \eqref{e1.4}. \item[{\rm (H3)}] $h \in C((0, 1), [0, + \infty))$ and satisfies $$ 0 < \int_0^1 g(s, s) h(s) ds < + \infty.$$ \item[{\rm (H4)}] $f \in C([0, 1] \times D^+, \, [0, \infty))$, where $D^+ = C([- \tau, a], [0, + \infty))$. \item[{\rm (H5)}] $ \eta \in C([- \tau, 0], [0, +\infty))$, $ \xi \in C([1, 1 + a], [0, + \infty))$, and $\eta (0) = \xi (1) =0$. \end{itemize} It is easy to see that $$ \frac{\partial^{n-2}}{\partial t^{n-2}}G(t, s) = g(t, s), \quad t, s \in [0, 1]. $$ It is also well known that the Green's function $g(t, s)$ is \[ g(t, s) = \frac{1}{c} \begin{cases} \phi (s) \psi(t), & \text{if} \ 0 \leq s \leq t \leq 1, \\ \phi (t) \psi(s), & \text{if} \ 0 \leq t \leq s \leq 1, \end{cases} \] where $\phi$ and $\psi$ are solutions, respectively, of \begin{gather} L \phi = 0, \quad \phi(0) = \beta, \quad \phi' (0) = \alpha, \label{e2.1} \\ L \psi = 0, \quad \psi(1) = \delta, \quad \psi' (1) = - \gamma. \label{e2.2} \end{gather} One can show that $c = - p(t) (\phi (t) \psi' (t) - \phi' (t) \psi (t)) > 0$ and $\phi' (t) > 0$ on $(0, 1]$ and $\psi' (t) < 0$ on $[0, 1)$. Clearly \begin{equation} g(t, s) \leq g(s, s), \quad 0 \leq t, s \leq 1. \label{e2.3} \end{equation} By {\rm (H3)}, there exists $t_0 \in (0, 1)$ such that $h(t_0) > 0$. We may choose $\varepsilon \in (0, 1/2)$ such that $t_0 \in (\varepsilon, 1 - \varepsilon)$. Then for $\varepsilon \leq t \leq 1 - \varepsilon$ we have $\phi (\varepsilon) \leq \phi (t) \leq \phi (1 - \varepsilon)$ and $\psi (1 - \varepsilon) \leq \psi (t) \leq \psi(\varepsilon)$. Also for $(t, s) \in [\varepsilon, 1 - \varepsilon] \times (0, 1)$ \begin{equation} \frac{g(t, s)}{g(s, s)} \geq \min \big\{ \frac{\psi (1 - \varepsilon)}{ \psi (s)}, \frac{\phi (\varepsilon)}{\phi (s)} \big\} \geq \min \big\{ \frac{\psi (1 - \varepsilon)}{ \psi (0)}, \frac{\phi (\varepsilon)}{\phi (1)} \big\} := \sigma. \label{e2.4} \end{equation} Let $E = C^{(n-2)}([- \tau, b]; \mathbb{R})$ with a norm $\|u\|_{[-\tau, b]} = \sup_{-\tau \leq t \leq b}|u^{(n-2)}(t)|$ for $u \in E$. Obviously, $E$ is a Banach space. And let $C = C^{(n-2)}([- \tau, a], \mathbb{R})$ be a space with norm $\|\psi \|_{[- \tau, a]} = \sup_{- \tau \leq t \leq a} |\psi^{(n-2)}(x)|$ for $\psi \in C$. Let $$ C^+ = \{\psi \in C: \psi(x) \geq 0, x \in [- \tau, a]\}. $$ It is easy to see that $C^+$ is a subspace of $C$. Define a cone $K \subset E$ as follows: \begin{equation} K = \{ y \in E : y(t) \geq 0, \; \min _{t \in [\varepsilon, 1 - \varepsilon]} y^{(n-2)}(t) \geq \overline{\sigma} \|y\|_{[- \tau, b]} \},\label{e2.5} \end{equation} where $\overline{\sigma} = \frac{1}{b} \min \{\varepsilon, \sigma\}$, $\sigma$ is as in \eqref{e2.4}. For each $\rho > 0$, we define $K_{\rho} = \{y \in K : \|y\|_{[-\tau, b]} < \rho \}$. Furthermore, we define a set $\Omega_{\rho} $ as follows: $$ \Omega_{\rho} = \big\{y \in K : \min_{\varepsilon \leq t \leq 1 - \varepsilon} y^{(n - 2)}(t) < \overline{\sigma} \rho \big\}. $$ Similar to the \cite[Lemma 2.5]{l1}, we have \begin{lemma}\label{lem2.1} $\Omega_{\rho}$ defined above has the following properties: \begin{itemize} \item[{\rm (a)}] $\Omega_{\rho}$ is open relative to $K$. \item[{\rm (b)}] $K_{\overline{\sigma} \rho } \subset \Omega_{\rho} \subset K_{\rho}$. \item[{\rm (c)}] $y \in \partial \Omega_{\rho}$ if and only if $\min _{\varepsilon \leq t \leq 1 - \varepsilon} y^{(n-2)}(t) = \overline{\sigma} \rho. $ \item[{\rm (d)}] If $y \in \partial \Omega_{\rho}$, then $\overline{\sigma} \rho \leq y^{(n-2)}(t) \leq \rho$ for $t \in [\varepsilon, 1 - \varepsilon]$. \end{itemize} \end{lemma} To obtain the positive solutions of \eqref{e1.1}-\eqref{e1.4}, the following fixed point theorem in cones will be fundamental. \begin{lemma}\label{lem2.2} Let $K$ be a cone in a Banach space $E$. Let $D$ be an open bounded subset of $E$ with $D_K = D \cap K \neq \emptyset$ and $\overline{D}_K \neq K$. Assume that $A : \overline{D}_K \to K$ is a compact map such that $x \neq Ax$ for $x \in \partial D_K$. Then the following results hold. \begin{itemize} \item[{\rm (1)}] $ \ \|Ax\| \leq \|x\|, \quad x \in \partial D_K,$ then $i_K (A, D_K) = 1$. \item[{\rm (2)}] If there exists $ e \in K \backslash \{0\}$ such that $ x \neq Ax + \lambda e$ for all $ x \in \partial D_K$ and all $ \lambda > 0$, then $i_K (A, D_K) = 0$. \item[{\rm (3)}] Let $U$ be an open set in $E$ such that $\overline{U} \subset D_K$. If $i_K(A, D_K) = 1$ and $i_K(A, U_K) = 0$, then $A$ has a fixed point in $D_K \backslash \overline{U}_K$. The same results holds if $i_K(A, D_K) = 0$ and $i_K(A, U_K) = 1$. \end{itemize} \end{lemma} Suppose that $y(t)$ is a solution of \eqref{e1.1}-\eqref{e1.4}, then it can be written as \[ y(t) = \begin{cases} y(- \tau; t), & - \tau \leq t \leq 0,\\ \int_0^1 G(t, s) h(s) f(s, y_s)ds, & 0 \leq t \leq 1,\\ y(b; t), & 1 \leq t \leq b, \end{cases} \] where $y(- \tau; t)$ and $y(b; t)$ satisfy \[ y^{(n-2)}(- \tau; t) = \begin{cases} e^{\frac{\alpha}{\beta}t} \left(\frac{1}{\beta} \int_t^0 e^{- \frac{\alpha}{\beta}s}\eta(s)ds + y^{(n-2)}(0)\right), & t \in [- \tau, 0], \; \beta \neq 0, \\ \frac{1}{\alpha}\eta(t), & t \in [- \tau, 0], \;\beta = 0, \end{cases} \] and \[ y^{(n-2)}(b; t) = \begin{cases} e^{- \frac{\gamma}{\delta}t} \left(\frac{1}{\delta} \int_1^t e^{\frac{\gamma}{\delta}s}\xi(s)ds + e^{\frac{\gamma}{\delta}}y^{(n-2)}(1)\right), & t \in [1, b], \; \delta \neq 0, \\ \frac{1}{\gamma}\xi(t), & t \in [1, b], \;\delta = 0. \end{cases} \] Throughout this paper, we assume that $u_0(t)$ is the solution of \eqref{e1.1}-\eqref{e1.4} with $f \equiv 0$, and $\|u_0\|_{[- \tau, b]} =: M_0$. Clearly, $u_0^{(n-2)}(t)$ can be expressed as follows: \[ u_0^{(n-2)}(t) = \begin{cases} u_0^{(n-2)}(- \tau; t), & - \tau \leq t \leq 0, \\ 0, & 0 \leq t \leq 1, \\ u_0^{(n-2)}(b; t), & 1 \leq t \leq b. \end{cases} \] where \[ u_0^{(n-2)}(- \tau; t) = \begin{cases} \frac{1}{\beta} e^{\frac{\alpha}{\beta}t} \int_t^0 e^{- \frac{\alpha}{\beta}s}\eta(s)ds, & t \in [- \tau, 0], \; \beta \neq 0, \\ \frac{1}{\alpha} \eta(t), & t \in [- \tau, 0], \; \beta = 0, \end{cases} \] and \[ u_0^{(n-2)}(b; t) = \begin{cases} \frac{1}{\delta} e^{- \frac{\gamma}{\delta}t} \int_1^t e^{\frac{\gamma}{\delta}s} \xi(s)ds, & t \in [1, b], \; \delta \neq 0,\\ \frac{1}{\gamma} \xi(t), & t \in [1, b], \; \delta = 0. \end{cases} \] Let $y(t)$ be a solution of BVP \eqref{e1.1}-\eqref{e1.4} and $u(t) = y(t) - u_0(t)$. Noting that $u(t) \equiv y(t)$ for $0 \leq t \leq 1$, we have \[ u^{(n-2)}(t) = \begin{cases} u^{(n-2)}(- \tau; t),& - \tau \leq t \leq 0,\\ \int_0^1 g(t, s) h(s) f(s, u_s + (u_0)_s)ds, & 0 \leq t \leq 1,\\ u^{(n-2)}(b; t),& 1 \leq t \leq b, \end{cases} \] where \[ u^{(n-2)}(- \tau; t) = \begin{cases} e^{\frac{\alpha}{\beta}t} \int_0^1 g(0, s) h(s) f(s, u_s + (u_0)_s)ds,& t \in [- \tau, 0],\; \beta \neq 0,\\ 0, & t \in [- \tau, 0], \; \beta = 0, \end{cases} \] and \[ u^{(n-2)}(b; t) = \begin{cases} e^{- \frac{\gamma}{\delta}(t - 1)} \int_0^1 g(1, s) h(s) f(s, u_s + (u_0)_s)ds,& t \in [1, b],\; \delta \neq 0,\\ 0, & t \in [1, b], \; \delta = 0. \end{cases} \] It is easy to see that $y(t)$ is a solution of BVP \eqref{e1.1}-\eqref{e1.4} if and only if $u(t) = y(t) - u_0(t)$ is a solution of the operator equation \begin{equation} u(t) = Au(t) \quad \text{for } t \in [- \tau, b]. \label{e2.6} \end{equation} Here, operator $A : E \to E$ is defined by \[ Au(t) := \begin{cases} B_1 u(t), & - \tau \leq t \leq 0, \\ \int_0^1 G(t, s) h(s) f(s, u_s + (u_0)_s)ds, & 0 \leq t \leq 1,\\ B_2 u(t), & 1 \leq t \leq b, \end{cases} \] where \[ B_1 u(t) := \begin{cases} \big(\frac{\beta}{\alpha}\big)^{n-2} e^{\frac{\alpha}{\beta}t} \int_0^1 g(0, s) h(s) f(s, u_s + (u_0)_s)ds, & \beta \neq 0,\; \alpha \neq 0,\\ \frac{t^{n-2}}{(n-2)!} \int_0^1 g(0, s) h(s) f(s, u_s + (u_0)_s)ds, & \beta \neq 0,\; \alpha = 0,\\ 0, & \beta = 0 \end{cases} \] for each $t \in [- \tau, 0]$, and \[ B_2 u(t) := \begin{cases} \big(-\frac{\delta}{\gamma}\big)^{n-2}e^{-\frac{\gamma}{\delta}(t - 1)} \int_0^1 g(1, s) h(s)f(s, u_s + (u_0)_s)ds, & \delta \neq 0,\; \gamma \neq 0, \\ \frac{t^{n-2}}{(n-2)!} \int_0^1 g(1, s) h(s)f(s, u_s + (u_0)_s)ds, & \delta \neq 0,\; \gamma = 0, \\ 0, & \delta = 0 \end{cases} \] for any $t \in [1, b]$. Obviously, \[ (Au)^{(n-2)}(t) := \begin{cases} (B_1 u)^{(n-2)}(t), & - \tau \leq t \leq 0,\\ \int_0^1 g(t, s) h(s) f(s, u_s + (u_0)_s)ds, & 0 \leq t \leq 1,\\ (B_2 u)^{(n-2)}(t), & 1 \leq t \leq b, \end{cases} \] where \[ (B_1 u)^{(n-2)}(t) := \begin{cases} e^{\frac{\alpha}{\beta}t} \int_0^1 g(0, s) h(s) f(s, u_s + (u_0)_s)ds, & t \in [- \tau, 0], \ \beta \neq 0,\\ 0, & t \in [- \tau, 0],\; \beta = 0, \end{cases} \] and \[ (B_2 u)^{(n-2)}(t) := \begin{cases} e^{-\frac{\gamma}{\delta}(t - 1)} \int_0^1 g(1, s) h(s)f(s, u_s + (u_0)_s)ds, & t \in [1, b], \; \delta \neq 0, \\ 0, & t \in [1, b], \; \delta = 0. \end{cases} \] \begin{lemma}\label{lem2.3} With the above notation, $A(K) \subset K$. \end{lemma} \begin{proof} By the assumptions of {\rm (H1)-(H5)}, it is easy to know that $Au \in E$ and $Au \geq 0$ for any $u \in K$. Moreover, it follows from \begin{gather*} 0 \leq (Au)^{(n-2)}(t) \leq (Au)^{(n-2)}(0) \quad \text{for } - \tau \leq t \leq 0 \\ 0 \leq (Au)^{(n-2)}(t) \leq (Au)^{(n-2)}(1) \quad \text{for } 1 \leq t \leq b \end{gather*} that $\|Au\|_{[- \tau, b]} = \|Au\|_{[0, 1]}$. By \eqref{e2.3} we have, for any $u \in K$ and $t \in [0, 1]$ that \begin{equation} \|Au\|_{[- \tau, b]} = \|Au\|_{[0, 1]} \leq \int_0^1 g(s, s) h(s)f(s, u_s + (u_0)_s) ds. \label{e2.7} \end{equation} From \eqref{e2.4}, we get \begin{equation} \begin{aligned} \min _{\varepsilon \leq t \leq 1 - \varepsilon} (Au)^{(n-2)}(t) &= \min _{\varepsilon \leq t \leq 1 - \varepsilon } \int_0^1 g(t, s) h(s) f(s, u_s + (u_0)_s) ds\\ &\geq \sigma \int_0^1 g(s, s) h(s) f(s, u_s + (u_0)_s) ds\\ &\geq \overline{\sigma} \int_0^1 g(s, s) h(s) f(s, u_s + (u_0)_s) ds. \end{aligned} \label{e2.8} \end{equation} In view of \eqref{e2.7} and \eqref{e2.8}, we obtain $$ \min _{\varepsilon \leq t \leq 1 - \varepsilon} (Au)^{(n-2)}(t) \geq \overline{\sigma} \|Au\|_{[- \tau, b]}, \quad u \in K, $$ which implies $A(K) \subset K$. \end{proof} Let \begin{gather*} C_{[k, r]}^+ = \{\varphi \in C^+ : \ k \leq \|\varphi\|_{ [- \tau, a]} \leq r \}, \\ C_{[k, \infty)}^+ = \{\varphi \in C^+ : \ k \leq \|\varphi\|_{[- \tau, a]} < \infty \}, \end{gather*} where $ 0 \leq k < r$. \begin{lemma}\label{lem2.4} $A : K \to K$ is completely continuous. \end{lemma} \begin{proof} We apply a truncation technique (cf. \cite{s1}). We define the function $h_m$ for $m \geq 2$, by \[ h_m(t) = \begin{cases} \min \big\{h(t), h(\frac{1}{m})\big\}, & 0 < t \leq \frac{1}{m},\\ h(t), & \frac{1}{m} < t < 1 - \frac{1}{m}, \\ \min \big\{h(t), h(\frac{m - 1}{m})\big\}, & \frac{m - 1}{m} \leq t < 1. \end{cases} \] It is clear that $h_m(t)$ is nonnegative and continuous on $[0, 1]$. We define the operator $A_m$ by \[ A_m u(t) := \begin{cases} B_{1 m} u(t), & - \tau \leq t \leq 0, \\ \int_0^1 G(t, s) h_m(s) f(s, u_s + (u_0)_s)ds, & 0 \leq t \leq 1,\\ B_{2 m} u(t), & 1 \leq t \leq b, \end{cases} \] where \[ B_{1 m} u(t) := \begin{cases} \left(\frac{\beta}{\alpha}\right)^{n-2} e^{\frac{\alpha}{\beta}t} \int_0^1 g(0, s) h_m (s) f(s, u_s + (u_0)_s)ds, & \beta \neq 0,\; \alpha \neq 0,\\ \frac{t^{n-2}}{(n-2)!} \int_0^1 g(0, s) h_m(s) f(s, u_s + (u_0)_s)ds, & \beta \neq 0,\; \alpha = 0,\\ 0, & \beta = 0 \end{cases} \] for each $t \in [- \tau, 0]$, and \[ B_{2 m} u(t) := \begin{cases} \big(-\frac{\delta}{\gamma}\big)^{n-2}e^{-\frac{\gamma}{\delta}(t - 1)} \int_0^1 g(1, s) h_m(s)f(s, u_s + (u_0)_s)ds, & \delta \neq 0,\; \gamma \neq 0, \\ \frac{t^{n-2}}{(n-2)!} \int_0^1 g(1, s) h_m(s)f(s, u_s + (u_0)_s)ds, & \delta \neq 0,\; \gamma = 0, \\ 0, & \delta = 0 \end{cases} \] for any $t \in [1, b]$. By Lemma \ref{lem2.3}, it is easy to check that $A_m : K \to K$. And, $A_m$ is continuous, the proof is similar to that of \cite[Theorem 2.1]{h3}. Next let $B \subset K$ be a bounded subset of $K$, and $M_1 > 0$ be a constant such that $\|u\|_{[- \tau, b]} \leq M_1$ for $u \in B$. Noting that if $x_t \in C = C^{n-2}([- \tau, a], \mathbb{R})$, then $x_t^{(n-2)} \in C([- \tau, a], \mathbb{R})$, and $x_t^{(n-2)} (\theta) = x^{(n-2)}(t + \theta)$, $\theta \in [- \theta, a]$. Thus \begin{equation} \begin{aligned} \|u_s + (u_0)_s\|_{[- \tau, a]} &= \sup_{- \tau \leq \theta \leq a} |(u^s + u_0^s)^{(n-2)}(\theta)|\\ &\leq \sup_{- \tau \leq \theta \leq a} |(u^{(n-2)}(s + \theta)| + \sup_{- \tau \leq \theta \leq a}|u_0^{(n-2)}(s + \theta)| \\ &\leq \sup _{- \tau \leq t \leq b} |u^{(n-2)}(t)| + \sup _{- \tau \leq t \leq b} |u_0^{(n-2)}(t)| = \|u\|_{[- \tau, b]} + \|u_0\|_{[- \tau, b]}\\ &\leq M_1 + M_0 := M_2 \end{aligned} \label{e2.9} \end{equation} for $ u \in B$ and $s \in [0, 1]$. Hence, there exists a constant $M_3 > 0$ such that \begin{equation} |f(s, u_s + (u_0)_s)| \leq M_3, \quad {\rm on} \ [0, 1] \times C_{[0, M_3]}^+, \label{e2.10} \end{equation} since $f$ is continuous on $[0, 1] \times C^+$. For $u \in B$ we have \[ (A_m u)^{(n-2)}(t) := \begin{cases} (B_{1 m} u)^{(n-2)}(t), & - \tau \leq t \leq 0,\\ \int_0^1 g(t, s) h_m(s) f(s, u_s + (u_0)_s)ds, & 0 \leq t \leq 1,\\ (B_{2 m} u)^{(n-2)}(t), & 1 \leq t \leq b, \end{cases} \] where \[ (B_{1 m} u)^{(n-2)}(t) := \begin{cases} e^{\frac{\alpha}{\beta}t} \int_0^1 g(0, s) h_m(s) f(s, u_s + (u_0)_s)ds, & t \in [- \tau, 0], \; \beta \neq 0,\\ 0, & t \in [- \tau, 0],\; \beta = 0, \end{cases} \] and \[ (B_{2 m} u)^{(n-2)}(t) := \begin{cases} e^{-\frac{\gamma}{\delta}(t - 1)} \int_0^1 g(1, s) h_m(s)f(s, u_s + (u_0)_s)ds, & t \in [1, b], \; \delta \neq 0, \\ 0, & t \in [1, b], \; \delta = 0. \end{cases} \] These and \eqref{e2.10} imply $(A_m u)^{(n-2)} (t)$ is continuous and uniformly bounded for $u \in B$. So $(A_m u)' (t)$ is continuous and uniformly bounded for $u \in B$ also. The Ascoli-Arzela Theorem implies that $A_m$ is a completely continuous operator on $K$ for any $m \geq 2$. Moreover, $A_m$ converges uniformly to $A$ as $m \to \infty$ on any bounded subset of $K$. To see this, note that if $u \in K$ with $\|u\|_{[- \tau, b]} \leq M$, then from {\rm (H3)} and $0 \leq h_n(s) \leq h(s)$, \begin{align*} |(A_m u)^{(n-2)}(t) - (A u)^{(n-2)}(t)| & = \Big|\int_0^{\frac{1}{m}} g(t, s) [h(s) - h_m(s)] f(s, u_s + (u_0)_s)ds \ \\ & \quad + \int_{\frac{m - 1}{m}}^1 g(t, s) [h(s) - h_m(s)] f(s, u_s + (u_0)_s)ds \Big|\\ & \leq \int_0^{\frac{1}{m}} g(s, s)|h(s) - h_m(s)| f(s, u_s + (u_0)_s)ds \\ & \quad + \int_{\frac{m - 1}{m}}^1 g(s, s) |h(s) - h_m(s)| f(s, u_s + (u_0)_s)ds \\ & \leq 2 \overline{M} \Big[\int_0^{\frac{1}{m}} g(s, s) h(s) ds + \int_{\frac{m - 1}{m}}^1 g(s, s) h(s) ds\Big]\\ & \to 0 \quad \text{as } m \to \infty, \end{align*} where $\overline{M} := \max_{t \in [0, 1], \varphi \in C_{[0, M + M_0]}^+} f(t, \varphi)$. Thus, we have $$ \|A_m u - A u\|_{[- \tau, b]} = \|A_m u - A u\|_{[0, 1]} \to 0, \quad n \to \infty, $$ for each $u \in K$ with $\|u\|_{[- \tau, b]} \leq M$. Hence, $A_m$ converges uniformly to $A$ as $m \to \infty$ and therefore $A$ is completely continuous also. This completes the proof of Lemma \ref{lem2.4}. \end{proof} \section{Main results} For convenience, we introduce the following notation. Let \begin{gather*} \omega = \Big(\int_0^1 g(s, s) h(s) ds\Big)^{-1}; \quad N = \Big(\min _{\varepsilon \leq t \leq 1 - \varepsilon} \int_{\varepsilon}^{1 - \varepsilon} g(t, s) h(s) ds \Big)^{- 1}; \\ f_{\overline{\sigma} \rho }^{\rho} = \inf \big\{ \min _{t \in [\varepsilon, 1 - \varepsilon]} \frac{f(t, \varphi)}{\rho} : \varphi \in C_{[\overline{\sigma} \rho, \rho + M_0]}^{+} \big\}; \\ f_0^{\rho} = \sup \big\{ \max _{t \in [0, 1]} \frac{f(t, \varphi)}{\rho} : \varphi \in C_{[0, \rho + M_0]}^{+} \big\}; \\ f^{\mu} = \lim _{\| \varphi \|_{[- \tau, a]} \to \mu} \sup \max _{t \in [0, 1]} \frac{f(t, \varphi)}{\| \varphi \|_{[- \tau, a]}}; \\ f_{\mu} = \lim _{\| \varphi \|_{[- \tau, a]} \to \mu} \inf \min _{t \in [\varepsilon, 1 - \varepsilon]} \frac{f(t, \varphi)}{\| \varphi \|_{[- \tau, a]}}, \quad (\mu := \infty \ {\rm or} \ 0^+). \end{gather*} Now, we impose conditions on $f$ which we assure that $i_K(A,K_{\rho}) = 1$. \begin{lemma}\label{lem3.1} Assume that \begin{equation} f_0^{\rho} \leq \omega \quad \text{and} \quad u \neq Au \quad \text{for } u \in \partial K_{\rho}. \label{e3.1} \end{equation} Then $i_K(A, K_{\rho}) = 1$. \end{lemma} \begin{proof} For $u \in \partial K_{\rho}$, we have $\|u_s + (u_0)_s \|_{[- \tau, a]} \leq \rho + M_0,$ for all $s \in [0, 1]$, i.e., $u_s + (u_0)_s \in C_{[0, \rho + M_0]}$ for any $s \in [0, 1]$. It follows from \eqref{e3.1} that for $t \in [0, 1]$, \begin{align*} (Au)^{(n-2)}(t) & = \int_0^1 g(t, s) h(s) f(s, u_s + (u_0)_s)ds\\ & \leq \int_0^1 g(s, s)h(s) f(s, u_s + (u_0)_s)ds\\ & < \rho \omega \int_0^1 g(s, s) h(s)ds \\ &= \rho = \|u\|_{[- \tau,b]}. \end{align*} This implies that $\|Au\|_{[- \tau, b]} < \|u\|_{[- \tau, b]}$ for $u \in \partial K_{\rho}$. By Lemma \ref{e2.2} (1), we have $i_K(A, K_{\rho}) = 1$. \end{proof} Let $u \in \partial \Omega_{\rho}$, then for any $s \in [\varepsilon, 1 - \varepsilon]$, we have by Lemma \ref{lem2.1}(c) that \begin{align*} \|u_s + (u_0)_s\|_{[- \tau, a]}& = \sup _{\theta \in [- \tau, a]} (u^{(n-2)}(s + \theta) + (u_0)^{(n-2)}(s + \theta))\\ & \geq \sup _{\theta \in [- \tau, a]} u^{(n-2)}(s + \theta) \ \text{(since $(u_0)^{(n-2)}(t) \geq 0$ for $t\in [- \tau, b]$)}\\ & \geq u^{(n-2)}(s) \\ &\geq \min _{t \in [\varepsilon, 1 - \varepsilon]} u^{(n-2)}(t) = \overline{\sigma} \rho. \end{align*} By Lemma \ref{lem2.1}(b), we have $\overline{\Omega}_{\rho} \subset \overline{K}_{\rho}$, that is $ \|u\|_{[- \tau, b]} \leq \rho.$ Thus, from \eqref{e2.9} we get $$ \|u_s + (u_0)_s\|_{[- \tau, a]} \leq \|u\|_{[- \tau, b]} + \|u_0\|_{[- \tau, b]} \leq \rho + M_0, \quad \forall s \in [0, 1]. $$ Hence \begin{equation} u_s + (u_0)_s \in C_{[\overline{\sigma} \rho, \ \rho + M_0]}^+, \quad \text{for } u \in \partial \Omega_{\rho} ,\; s \in [\varepsilon, 1 - \varepsilon]. \label{e3.2} \end{equation} Next, we impose conditions on $f$ which assure that $i_K(A,\Omega_{\rho}) = 0$. \begin{lemma}\label{lem3.2} If $f$ satisfies the condition \begin{equation} f_{\overline{\sigma} \rho}^{\rho} \geq N \overline{\sigma} \quad \text{ and} \quad u \neq Au \quad \text{for } u \in \partial \Omega_{\rho}. \label{e3.3} \end{equation} Then $i_K(A, \Omega_{\rho}) = 0$. \end{lemma} \begin{proof} Let \[ e(t) = \begin{cases} - \frac{t^{n-1}}{(1 + \tau) (n-1)!},& n \text{ is odd},\; - \tau \leq t \leq 0,\\ \frac{t^n}{(1 + \tau)^2 n!},& n \text{ is even },\; - \tau \leq t \leq 0,\\ \frac{t^{n-1}}{b (n-1)!}, & 0 \leq t \leq b. \end{cases} \] It is easy to verify that $e \in C^{(n-2)}([- \tau, b], \mathbb{R})$, $e(t) \geq 0$ for $t \in [- \tau, b]$, and \[ e^{(n-2)}(t) = \begin{cases} - \frac{t}{1 + \tau},& n \text{ is odd},\; - \tau \leq t \leq 0,\\ \frac{t^2}{2 (1 + \tau)^2},& n \text{ is even}, \; - \tau \leq t \leq 0,\\ \frac{t}{b}, & 0 \leq t \leq b, \end{cases} \] which implies that $e \in K$ and $\|e\|_{[- \tau, b]} =1$, that is $e \in \partial K_1$. We claim that $$ u \neq Au + \lambda e, \quad u \in \partial \Omega_{\rho}, \quad \lambda > 0. $$ In fact, if not, there exist $u \in \partial \Omega_{\rho}$ and $\lambda_0 > 0$ such that $ u = Au + \lambda_0 e$. Then by \eqref{e3.2} for $t \in [\varepsilon, 1 - \varepsilon]$, we get \begin{align*} u^{(n-2)} (t) & = (Au)^{(n-2)} (t) + \lambda_0 e^{(n-2)}(t) \\ & = \int_0^1 g(t, s) h(s) f(s, u_s + (u_0)_s ) ds + \lambda_0 e^{(n-2)}(t) \\ & \geq \int_{\varepsilon}^{1 - \varepsilon} g(t, s) h(s) f(s, u_s + (u_0)_s ) ds + \lambda_0 \frac{\varepsilon}{b} \\ & \geq \min _{\varepsilon \leq t \leq 1 - \varepsilon} \int_{\varepsilon}^{1 - \varepsilon} g(t, s) h(s) f(s, u_s + (u_0)^s ) ds + \lambda_0 \overline{\sigma}\\ & \geq \rho N \overline{\sigma} \min _{\varepsilon \leq t \leq 1 - \varepsilon} \int_{\varepsilon}^{1 - \varepsilon} g(t, s) h(s) ds + \lambda_0 \overline{\sigma}\\ & \geq \overline{\sigma} \rho + \lambda_0 \overline{\sigma}. \end{align*} This implies that $\overline{\sigma} \rho \geq \overline{\sigma} \rho + \lambda_0 \overline{\sigma}$, a contradiction. Moreover, it is easy to check that $u \neq Au$ for $u \in \partial \Omega_{\rho}$ from \eqref{e3.3}. Hence, by Lemma \ref{lem2.2} (2), it follows that $i_K(A, \Omega_{\rho}) =0$. \end{proof} \begin{theorem}\label{thm3.3} If one of the following conditions holds: \begin{itemize} \item[{\rm (H7)}] There exist $\rho_1, \rho_2, \rho_3 \in (0, \infty)$ with $\rho_1 < \overline{\sigma} \rho_2$ and $\rho_2 < \rho_3$ such that $$ f_0^{\rho_1} \leq \omega, \quad f_{\overline{\sigma} \rho_2}^{\rho_2} \geq N \overline{\sigma}, \quad u \neq Au \quad \text{for} \ u \in \partial \Omega_{\rho_2} \quad \text{and} \quad f_0^{\rho_3} \leq \omega. $$ \item[{\rm (H8)}] There exist $\rho_1, \ \rho_2, \ \rho_3 \in (0, \infty)$ with $\rho_1 < \rho_2 < \overline{\sigma} \rho_3$ such that $$f_{\overline{\sigma} \rho_1}^{\rho_1} \geq N \overline{\sigma}, \quad f_0^{\rho_2} \leq \omega, \quad u \neq Au \quad \text{for } u \in \partial K_{\rho_2} \quad \text{and} \quad f_{\overline{\sigma} \rho_3}^{\rho_3} \geq N \overline{\sigma}. $$ \end{itemize} Then BVP \eqref{e1.1}-\eqref{e1.4} has two positive solutions. Moreover, if in {\rm (H7)}, $f_0^{\rho_1} \leq \omega$ is replaced by $f_0^{\rho_1} < \omega$, then \eqref{e1.1}-\eqref{e1.4} has a third positive solution $u_3 \in K_{\rho_1}$. \end{theorem} \begin{proof} Suppose that {\rm (H7)} holds. We show that either $A$ has a fixed point $u_1$ in $\partial K_{\rho_1}$ or in $\Omega_{\rho_2} \setminus \overline{K_{\rho_1}}$. If $u \neq Au$ for $u \in \partial K_{\rho_1} \cup \partial K_{\rho_3}$, by Lemmas \ref{lem3.1}-\ref{lem3.2}, we have $i_K(A, K_{\rho_1}) = 1$, $i_K(A, \Omega_{\rho_2}) = 0$ and $i_K(A, K_{\rho_3}) = 1$. Since $\rho_1 < \overline{\sigma} \rho_2$, we have $\overline{K_{\rho_1}} \subset K_{\overline{\sigma} \rho_2} \subset \Omega_{\rho_2}$ by Lemma \ref{lem2.1} (b). It follows from Lemma \ref{lem2.2} that $A$ has a fixed point $u_1 \in \Omega_{\rho_2} \setminus \overline{K_{\rho_1}}$. Similarly, $A$ has a fixed point $u_2 \in K_{\rho_3} \setminus \overline{\Omega_{\rho_2}}$. The proof is similar when {\rm (H8)} holds. \end{proof} As a special case of Theorem \ref{thm3.3} we obtain the following result. \begin{corollary} \label{coro3.4} Let $\xi(t) \equiv 0$, $\eta(t) \equiv 0$. If there exists $\rho > 0 $ such that one of the following conditions holds: \begin{itemize} \item[{\rm (H9)}] $0 \leq f^0 < \omega$, $f_{\overline{\sigma} \rho}^{\rho} \geq N \overline{\sigma}$, $ u \neq Au$ for $u \in \partial \Omega_{\rho}$ and $0 \leq f^{\infty} < \omega$. \item[{\rm (H10)}] $ N < f_0 \leq \infty$, $f_0^{\rho} \leq \omega$, $u \neq Au$ for $u \in \partial K_{\rho}$ and $N < f_{\infty} \leq \infty$. \end{itemize} Then BVP \eqref{e1.1}-\eqref{e1.4} has two positive solutions. \end{corollary} \begin{proof} From $\xi(t) \equiv 0, \eta(t) \equiv 0$, it is clear that $u_0(t) \equiv 0$ for $t \in [- \tau, b]$, thus $M_0 = 0$. We now show that ${\rm (H_9)}$ implies ${\rm (H_7)}$. It is easy to verify that $0 \leq f^0 < \omega$ implies that there exists $\rho_1 \in (0, \overline{\sigma} \rho)$ such that $f_0^{\rho_1} < \omega$. Let $k \in (f^{\infty}, \omega)$. Then there exists $r > \rho$ such that $\max_{t \in [0, 1]} f(t, \varphi) \leq k \|\varphi\|_{[- \tau, a]}$ for $\varphi \in C_{[r, \infty)}^+$ since $0 \leq f^{\infty} < \omega$. Let $$ l = \max \{\max_{t \in [0, 1]} f(t, \varphi) : \varphi \in C_{[0, r]}^+ \}, \quad \text{and} \quad \rho_3 > \max \big\{\frac{l}{\omega - k}, \ \rho \big\}. $$ Then we have $$ \max _{t \in [0, 1]} f(t, \varphi) \leq k \|\varphi\|_{[- \tau, a]} + l \leq k \rho_3 + l < \omega \rho_3 \quad \text{for } \varphi \in C_{[0, \rho_3]}^+. $$ This implies that $f_0^{\rho_3} < \omega$ and {\rm (H7)} holds. Similarly, {\rm (H10)} implies {\rm (H8)}. \end{proof} By a similar argument to that of Theorem \ref{thm3.3}, we obtain the following results on existence of at least one positive solution of \eqref{e1.1}-\eqref{e1.4}. \begin{theorem}\label{thm3.5} If one of the following conditions holds: \begin{itemize} \item[{\rm (H11)}] There exist $\rho_1, \rho_2 \in (0, \infty)$ with $\rho_1 < \overline{\sigma} \rho_2$ such that $$ f_0^{\rho_1} \leq \omega \quad \text{and}\quad f_{\overline{\sigma} \rho_2}^{\rho_2} \geq N \overline{\sigma}. $$ \item[{\rm (H12)}] There exist $\rho_1, \rho_2 \in (0,\infty)$ with $\rho_1 < \rho_2$ such that $$ f_{\overline{\sigma} \rho_1}^{\rho_1} \geq N \overline{\sigma} \quad \text{and} \quad f_0^{\rho_2} \leq \omega. $$ \end{itemize} Then BVP \eqref{e1.1}-\eqref{e1.4} has a positive solution. \end{theorem} As a special case of Theorem \ref{thm3.5}, we obtain the following result. \begin{corollary} \label{coro3.6} Let $\xi(t) \equiv 0, \eta(t) \equiv 0$. 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