\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 41, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/41\hfil Positive solutions for delay three-point BVP] {Positive solutions for a functional delay second-order three-point boundary-value problem} \author[C. Bai, X. Xu \hfil EJDE-2006/41\hfilneg] {Chuanzhi Bai, Xinya Xu} % in alphabetical order \address{Chuanzhi Bai \hfill\break 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{Xinya Xu \hfill\break Department of Mathematics \\ Huaiyin Teachers College\\ Huaian, Jiangsi 223001, China} \email{xinyaxu@hytc.edu.cn} \date{} \thanks{Submitted August 10, 2005. Published March 26, 2006.} \thanks{Supported by grant 03KJD110056 from the Natural Science Foundation of Jiangsu \hfill\break\indent Education Office and Jiangsu Planned Projects for Postdoctoral Research Funds} \subjclass[2000]{34K10, 34B10} \keywords{Functional delay differential equation; boundary value problem; \hfill\break\indent positive solutions; fixed point index} \begin{abstract} We establish criteria for the existence of positive solutions to the three-point boundary-value problems expressed by second-order functional delay differential equations of the form \begin{gather*} - x''(t) = f(t, x(t), x(t - \tau), x_t), \quad 0 0$, let $C(J)$ be the Banach space of all continuous functions $\psi : [- \tau, 0] =: J \to \mathbb{R}$ endowed with the sup-norm $$ \|\psi\|_J := \sup \{|\psi(s)|: s \in J\}. $$ For any continuous function $x$ defined on the interval $[-\tau, 1]$ and any $t \in I =:[0, 1]$, the symbol $x_t$ is used to denote the element of $C(J)$ defined by $$ x_t(s) = x(t + s), \quad s \in J. $$ Set $$ C^+(J) =: \{\psi \in C(J) : \psi(s) \geq 0, s \in J\}. $$ In this paper, motivated and inspired by \cite{b1, h1, j1, k1}, we apply a fixed point theorem in cones to investigate the existence of positive solutions for three point boundary-value problems of second-order functional delay differential equation \begin{equation} \begin{gathered} - x''(t) = f(t, x(t), x(t - \tau), x_t), \quad 0 < t < 1, \\ x_0 = \phi, \quad x(1) = x(\eta), \end{gathered} \label{e1.2} \end{equation} where $0 < \tau < 1/4$, $\tau < \eta < 1$, $f : I \times \mathbb{R}^+ \times \mathbb{R}^+ \times C^+(J) \to \mathbb{R}^+$ is a continuous function, and $\phi$ is an element of the space $$ C_0^+(J) =: \{\psi \in C^+(J) : \psi(0) = 0\}. $$ We need the following well-known lemma. See \cite{d1} for a proof and further discussion of the fixed-point index $i(A, K_r, K)$. \begin{lemma}\label{lem1.1} Assume that $E$ is a Banach space, and $K \subset E$ is a cone in $E$. Let $K_r = \{x \in K: \|u\| < r \}$. Furthermore, assume that $A : K \to K$ is a compact map, and $A x \not= x$ for $x \in \partial K_r = \{x \in K; \|x\| = r\}$. Then, one has the following conclusions. \begin{itemize} \item[(1)] If $\|x\| \leq \|A x\|$ for $x \in \partial K_r$, then $i(A, K_r, K) = 0$. \item[(2)] If $\|x\| \geq \|A x\|$ for $x \in \partial K_r$, then $i(A, K_r, K) = 1$. \end{itemize} \end{lemma} \section{Preliminaries and some lemmas} In the sequel we shall denote by $C_0(I)$ the space all continuous functions $x : I \to \mathbb{R}$ with $x(0)= 0$. This is a Banach space when it is furnished with usual sup-norm $$ \|x\|_I := \sup \{|x(s)| : s \in I\}. $$ We set $$ C_0^+(I) := \{x \in C_0(I) : x(t) \geq 0, t \in I\}. $$ For each $\phi \in C_0^+(J)$ and $x \in C_0^+(I)$ we define \[ x_t(s; \phi) := \begin{cases} \phi(t + s),& t + s \leq 0, \; t \in I, \; s \in J,\\ x(t + s), & 0 \leq t + s \leq 1, \; t \in I, \; s \in J, \end{cases} \] and observe that $x_t(\cdot; \phi) \in C^+(J)$. It is easy to check that $\varphi_1(t) = \sin \frac{\pi}{\eta + 1} t$ is the eigenfunction related to the smallest eigenvalue $\lambda_1 = \frac{\pi^2}{(\eta + 1)^2}$ of the eigenproblem $$ - x'' = \lambda x, \quad x(0)= 0, \quad x(1) = x(\eta). $$ By \cite{h1}, the Green's function for the three-point boundary-value problem $$ - x'' = 0, \quad x(0) = 0, \quad x(1) = x(\eta), $$ is given by \[ G(t,s)= \begin{cases} t, &t\leq s\leq \eta,\\ s, &s\leq t \text{ and } s \leq \eta,\\ \frac{1 - s}{1 - \eta} t, &t \leq s \text{ and } s \geq \eta,\\ s + \frac{\eta - s}{1 - \eta} t, & \eta \leq s \leq t. \end{cases} \] \begin{lemma}\label{lem2.1} Suppose that $G(t, s)$ is defined as above. Then we have the following results: \begin{itemize} \item[(1)] $0 \leq G(t,s) \leq G(s,s),\quad 0 \leq t,s \leq 1,$ \item[(2)] $G(t, s) \geq \eta t G(s, s), \quad 0 \leq t, s \leq 1$. \end{itemize} \end{lemma} \begin{proof} It is easy to see that (1) holds. To show that (2) holds, we distinguish four cases:\vspace{2mm} \noindent $\bullet$ If $t \leq s \leq \eta$, then $$G(t, s) = t \geq \eta t s = \eta t G(s, s).$$ $\bullet$ If $s \leq t$ and $s \leq \eta$, then $$G(t, s) = s \geq \eta t s = \eta t G(s, s).$$ $\bullet$ If $t \leq s $ and $s \geq \eta$, then $$G(t, s) = \frac{1 - s}{1 - \eta} t \geq \eta s t \frac{1 - s}{1 - \eta} = \eta t G(s, s).$$ $\bullet$ Finally, if $\eta \leq s \leq t$, then \begin{equation*} \begin{aligned} G(t, s) &= s - \frac{s - \eta}{1 - \eta} t \geq s - \frac{s - \eta}{1 - \eta} = \frac{\eta (1 - s)}{1 - \eta}\\ &\geq t s \frac{\eta (1 - s)}{1 - \eta} = \eta t \frac{s (1 - s)}{1 - \eta} = \eta t G(s, s). \end{aligned} \end{equation*} \end{proof} \begin{remark} {\rm If $s \leq \eta$ and $s \geq \eta$, then $G(s, s) = s$ and $G(s, s) = \frac{s(1 - s)}{1 - \eta}$, respectively. } \end{remark} For convenience, let \[ x(t; \phi) := \begin{cases} \phi(t),& - \tau \leq t \leq 0, \\ x(t), & 0 \leq t \leq 1. \end{cases} \] Suppose that $x(t)$ is a solution of BVP (\ref{e1.2}), then it can be written as $$ x(t) = \int_0^1 G(t, s) f(s,x(s), x(s - \tau; \phi), x_s(\cdot; \phi))ds, \quad t \in I. $$ Let $K \subset C_0(I)$ be a cone defined by $$ K = \{x \in C_0^+(I) : x(t) \geq \eta t \|x\|_I, \ \forall t \in I\}. $$ For each $x \in K$ and $t \in I$, we have \begin{equation} \begin{aligned} \|x_t(\cdot; \phi)\|_J &= \sup_{s \in [- \tau, 0]}|x_t(s; \phi)|\\ &= \max \left \{ \begin{array}{ll} \sup_{s \in [- \tau, 0]} |x(t + s)|,& {\rm if} \ t + s \in I,\\ \sup_{s \in [- \tau, 0]} |\phi(t + s)|,& {\rm if} \ t + s \leq 0 \end{array} \right\} \\ &\leq \max \{\|x\|_I, \|\phi\|_J\}, \label{e2.1} \end{aligned} \end{equation} and \begin{equation} \|x_t(\cdot; \phi)\|_J \geq \sup_{s \in [- \tau, 0]}\{x(t + s): t + s \in I\} \geq x(t) \geq \eta t \|x\|_I. \label{e2.2} \end{equation} Define an operator $A_{\phi} : K \to C_0(I)$ as follows: $$ (A_{\phi} x)(t) := \int_0^1 G(t, s) f(s, x(s), x(s - \tau; \phi), x_s(\cdot; \phi))ds, \quad t \in I.$$ Firstly, we have the following result. \begin{lemma}\label{lem2.3} $ A_{\phi} (K) \subset K$. \end{lemma} \begin{proof} For any $x \in K$, we observe that $(A_{\phi} x)(0) = 0$. By Lemma \ref{lem2.1} (1), we have $(A_{\phi} x)(t) \geq 0$, $t \in I$. It follows from Lemma \ref{lem2.1} (1) and (2) that \begin{align*} (A_{\phi} x)(t) &\geq \eta t \int_0^1 G(s, s)f(s, x(s), x(s - \tau; \phi), x_s(\cdot; \phi))ds \\ &\geq \eta t \|A_{\phi} x\|_I, \quad t \in I. \end{align*} Thus, $A_{\phi} (K) \subset K$. \end{proof} Secondly, similar to the proof of Theorem 2.1 in \cite{h3}, we get that \begin{lemma}\label{lem2.4} $A_{\phi} : K \to K$ is completely continuous. \end{lemma} We formulate some conditions for $f(t, u, v, \psi)$ as follows which will play roles in this paper. \begin{itemize} \item[(H1)] $ \displaystyle{\lim \inf \limits_{u + v + \|\psi\|_J \to + \infty} \min \limits_{t \in [0, 1]} \frac{f(t, u, v, \psi)}{u + v + \|\psi\|_J} = \infty.}$ \item[(H2)] $ \displaystyle{\lim \sup \limits_{u + v + \|\psi\|_J \to + \infty} \max \limits_{t \in [0, 1]} \frac{f(t, u, v, \psi)}{u + v + \|\psi\|_J} = 0.}$ \item[(H3)] $\displaystyle{\lim \inf \limits_{u + v + \|\psi\|_J \to + \infty} \min \limits_{t \in [0, 1]} \frac{f(t, u, v, \psi)}{u + v + \|\psi\|_J} > \frac{1}{3} \frac{\pi^2}{(\eta + 1)^2} ( 1 + M), }$ \noindent where $$ M = \frac{\pi^2 \tau (\eta - \tau) + 3 \pi(1 - \eta^2) + \pi \tau (\eta + 1) + 3 (\eta + 1)^2}{\pi \eta (\eta + 1) \left(\int_0^{\eta - \tau} t \sin \frac{\pi}{\eta + 1}(t + \tau) dt + 2 \int_0^{\eta} t \sin \frac{\pi}{\eta + 1} t dt\right)}.$$ \item[(H4)] There is a $h_1 > 0$ such that $0 \leq u \leq h_1$, $0 \leq v \leq \max\{h_1, \|\phi\|_J\}$, $0 \leq \|\psi\|_J \leq \max\{h_1, \|\phi\|_J\}$, and $0 \leq t \leq 1$ implies $$ f(t, u, v, \psi) < \mu h_1, \quad {\rm where} \ \mu = \left(\int_0^1 G(s, s)ds\right)^{- 1} = \frac{6}{1 + \eta + \eta^2}.$$ \item[(H5)] There is a $h_2 > 0$ such that $ \frac{1}{4}\eta h_2 \leq u \leq h_2$, $(\frac{1}{4} - \tau)\eta h_2 \leq v \leq h_2$, $\frac{1}{4}\eta h_2 \leq \|\psi\|_J \leq h_2$, and $ 0 \leq t \leq 1$ implies $$ f(t, u, v, \psi) > b h_2, \quad {\rm where} \ b = \left(\int_{1/4}^{3/4} G\left(\frac{1}{2}, t\right)dt\right)^{- 1}.$$ \end{itemize} In the following, we give some lemmas which will be used in this paper. \begin{lemma}\label{lem2.5} If {\rm (H1)} is satisfied, then there exist $0 < r_0 < \infty$ such that $$ i(A_{\phi}, K_r, K) = 0, \hspace{8mm} r \geq r_0.$$ \end{lemma} \begin{proof} Choose $L > 0$ such that $$ \eta \big(\frac{3}{4} - \tau\big) L \int_{1/4}^{3/4} G\big(\frac{1}{2}, s\big)ds > 1. $$ For $u, v \geq 0$ and $\psi \in C^+(J)$, {\rm (H1)} implies that there is $r_1 > 0$ such that \begin{equation} f(t, u, v, \psi) \geq L (u + v + \|\psi\|_J), \quad u + v + \|\psi\|_J \geq r_1, \quad 0 \leq t \leq 1. \label{e2.3} \end{equation} Choose $r_0 > \frac{4 r_1}{3(1 - 4 \tau) \eta}$. For $x \in \partial K_r$, $r \geq r_0$, we have by the definition of $K$ and (\ref{e2.2}) that \begin{gather*} x(t - \tau) \geq \eta (t - \tau) \|x\|_I \geq \eta \big(\frac{1}{4} - \tau\big) r > \frac{1}{3} r_1, \quad \frac{1}{4} \leq t \leq \frac{3}{4}, \\ \|x_t(\cdot; \phi)\|_J \geq x(t) \geq \eta t \|x\|_I \geq \frac{1}{4} \eta r > \frac{1}{3} r_1, \quad \frac{1}{4} \leq t \leq \frac{3}{4}. \end{gather*} So by (\ref{e2.3}), we have for such $x$, \begin{align*} (A_{\phi} x)\big(\frac{1}{2}\big) &\geq \int_{1/4}^{3/4} G\big(\frac{1}{2}, t\big) f(t, x(t), x(t - \tau; \phi), x_t(\cdot; \phi))dt \\ &= \int_{1/4}^{3/4} G\big(\frac{1}{2}, t\big) f(t, x(t), x(t - \tau), x_t(\cdot; \phi))dt \\ &\geq L \int_{1/4}^{3/4} G\big(\frac{1}{2}, t\big) [x(t) + x(t - \tau) + \|x_t(\cdot; \phi)\|_J] dt \\ &\geq \eta \big(\frac{3}{4} - \tau\big) L r \int_{1/4}^{3/4} G\big(\frac{1}{2}, t\big) dt > r = \|x\|_I. \end{align*} This shows that $$ \|A_{\phi} x\|_I > \|x\|_I, \quad \forall x \in \partial K_r. $$ It is obvious that $A_{\phi} x \not= x$ for $x \in \partial K_r$. Therefore, by Lemma \ref{lem1.1}, we conclude that $i(A_{\phi}, K_r, K) = 0$. \end{proof} \begin{lemma}\label{lem2.6} If {\rm (H2)} is satisfied, then there exists $0 < R_0 < \infty$ such that $$ i(A_{\phi}, K_R, K) = 1 \quad {\rm for} \ R \geq R_0. $$ \end{lemma} \begin{proof} By (H2), for any $ 0 < \varepsilon < \frac{1}{3} \big(\int_0^1 G(s, s) ds\big)^{- 1}$, $u, v \geq 0$ and $\psi \in C^+(J)$, there exists $R' > 0$ such that $$ f(t, u, v, \psi) \leq \varepsilon (u + v + \|\psi\|_J), \quad u + v + \|\psi\|_J \geq R', \quad 0 \leq t \leq 1.$$ Putting $$ C := \max \limits_{0 \leq t \leq 1} \max \limits_{ 0 \leq u, v, \ u + v + \|\psi\|_J \leq R'} |f(t, u, v, \psi) - \varepsilon (u + v + \|\psi\|_J)| + 1,$$ then \begin{equation} f(t, u, v, \psi) \leq \varepsilon (u + v + \|\psi\|_J) + C, \quad {for } u, v \geq 0, \ \psi \in C^+(J), \ t \in I. \label{e2.4} \end{equation} Choose $$ R_0 > (C + 2 \varepsilon \|\phi\|_J) \int_0^1 G(s, s) ds \Big/ \Big(1 - 3 \varepsilon \int_0^1 G(s, s) ds\Big). $$ Let $R \geq R_0$ and consider a point $x \in \partial K_R$. By the definition of $x(t; \phi)$, we get \begin{equation} x(s - \tau; \phi) \leq \max \{\|x\|_I, \|\phi\|_J\}, \quad \forall s \in I. \label{e2.5} \end{equation} By (\ref{e2.1}), (\ref{e2.4}) and (\ref{e2.5}), for $x \in \partial K_R$, $R \geq R_0$, and $t \in I$, \begin{align*} (A_{\phi} x)(t) &= \int_0^1 G(t, s) f(s, x(s), x(s - \tau; \phi), x_s(\cdot; \phi))ds \\ &\leq \int_0^1 G(s, s) f(s, x(s), x(s - \tau; \phi), x_s(\cdot; \phi))ds \\ &\leq \int_0^1 G(s, s) [\varepsilon (x(s) + x(s - \tau; \phi) + \|x_s(\cdot; \phi)\|_J) + C] ds \\ &\leq \int_0^1 G(s, s) [\varepsilon (\|x\|_I + 2 \max \{\|x\|_I, \|\phi\|_J\}) + C] ds \\ &\leq \int_0^1 G(s, s) [\varepsilon(3 \|x\|_I + 2 \|\phi\|_J) + C] ds \\ &= 3 \varepsilon R \int_0^1 G(s, s) ds + (C + 2 \varepsilon \|\phi\|_J) \int_0^1 G(s, s) ds\\ &< R = \|x\|_I. \end{align*} Thus, $\|A_{\phi} x\|_I < \|x\|_I$ for $x \in \partial K_R$. Hence, by Lemma \ref{lem1.1}, $i(A_{\phi}, K_R, K) = 1$. \end{proof} \begin{lemma}\label{lem2.7} If {\rm (H4)} is satisfied, then $i(A_{\phi}, K_{h_1}, K) = 1$. \end{lemma} \begin{proof} Let $ x \in \partial K_{h_1}$, then we have by (\ref{e2.1}) and (\ref{e2.5}) that $$ 0 \leq x(t - \tau; \phi) \leq \max\{h_1, \|\phi\|_J\}, \quad 0 \leq t \leq 1, $$ and $$ 0 \leq \|x_t(\cdot; \phi)\|_J \leq \max\{h_1, \|\phi\|_J\}, \quad 0 \leq t \leq 1. $$ Thus, from (H4) we obtain \begin{align*} (A_{\phi} x)(t) &\leq \int_0^1 G(s, s) f(s, x(s), x(s - \tau; \phi), x_s(\cdot; \phi))ds \\ & < \mu h_1 \int_0^1 G(s, s) ds = h_1 = \|x\|_I, \quad 0 \leq t \leq 1. \end{align*} This shows that $$ \|A_{\phi} x\|_I < \|x\|_I, \quad \forall x \in \partial K_{h_1}.$$ Hence, Lemma \ref{lem1.1} implies $i(A_{\phi}, K_{h_1}, K) = 1$. \end{proof} \begin{lemma}\label{lem2.8} If {\rm (H5)} is satisfied, then $i(A_{\phi}, K_{h_2}, K) = 0$. \end{lemma} \begin{proof} For $x \in \partial K_{h_2}$ , we have $$ h_2 = \|x\|_I \geq x(t - \tau) \geq \eta (t - \tau) \|x\|_I \geq \eta \big(\frac{1}{4} - \tau\big) h_2, \quad \frac{1}{4} \leq t \leq \frac{3}{4}, $$ and $$h_2 = \|x\|_I \geq \sup \limits_{s \in [- \tau, 0]} x(t + s) = \|x_t(\cdot; \phi)\|_J \geq x(t) \geq \eta t \|x\|_I \geq \frac{1}{4} \eta h_2, $$ for $ \frac{1}{4} \leq t \leq\frac{3}{4}$. It follows from (H5) that \begin{align*} (A_{\phi} x)\big(\frac{1}{2}\big) &\geq \int_{1/4}^{3/4} G\big(\frac{1}{2}, t\big) f(t, x(t), x(t - \tau), x_t(\cdot; \phi))dt \\ & > b h_2 \int_{1/4}^{3/4} G\big(\frac{1}{2}, t\big)dt = h_2 = \|x\|_I. \end{align*} This shows that $$ \|A_{\phi} x\|_I > \|x\|_I, \quad \forall x \in \partial K_{h_2}.$$ Therefore, by Lemma \ref{lem1.1}, we conclude that $i(A_{\phi}, K_{h_2}, K) = 0$. \end{proof} \section{Main results} \begin{theorem}\label{thm3.1} Assume that {\rm (H3)} and {\rm (H4)} are satisfied, then BVP \eqref{e1.2} has at least one positive solution. \end{theorem} \begin{proof} According to Lemma \ref{lem2.7}, we have that \begin{equation} i(A_{\phi}, K_{h_1}, K) = 1. \label{e3.1} \end{equation} Fix $ m > 1$, and let $g(t, u, v, \psi) = (u + v + \|\psi\|_J)^m$ for $u, v \geq 0$ and $\psi \in C^+(J)$. Then $g(t, u, v, \psi)$ satisfy (H1). Define $B_{\phi} : K \to K$ by $$ (B_{\phi} x)(t) := \int_0^1 G(t, s) g(s, x(s), x(s - \tau; \phi), x_s(\cdot; \phi))ds, \quad t \in I. $$ Then $B_{\phi}$ is a completely continuous operator. One has from Lemma \ref{lem2.5} that there exists $0 < h_1 < r_0 < \infty$, such that $r \geq r_0$ implies \begin{equation} i(B_{\phi}, K_r, K) = 0. \label{e3.2} \end{equation} Define $H_{\phi} : [0, 1] \times K \to K$ by $H_{\phi}(s, x) = (1 - s) A_{\phi} x + s B_{\phi} x$, then $H_{\phi}$ is a completely continuous operator. By the condition (H3) and the definition of $g$, for $u, v \geq 0$, $\psi \in C^+(J)$, and $t \in I$, there are $\varepsilon > 0$ and $r' > r_0$ such that \begin{gather*} f(t, u, v, \psi) \geq \frac{1}{3} (\lambda_1 (1 + M) + \varepsilon) (u + v + \|\psi\|_J), \quad u + v + \|\psi\|_J > r',\\ g(t, u, v, \psi) \geq \frac{1}{3} (\lambda_1 (1 + M) + \varepsilon) (u + v + \|\psi\|_J), \quad u + v + \|\psi\|_J > r', \end{gather*} where $\lambda_1 = \frac{\pi^2}{(\eta + 1)^2}$. We define \begin{equation*} \begin{aligned} C&:= \max \limits_{0 \leq t \leq 1} \max \limits_{0 \leq u, v, u + v + \|\psi\|_J \leq r'} |f(t, u, v, \psi) - \frac{1}{3}[\lambda_1 (1 + M) + \varepsilon] (u + v + \|\psi\|_J)|\\ & \ + \max \limits_{0 \leq t \leq 1} \max \limits_{0 \leq u, v, u + v + \|\psi\|_J \leq r'} |g(t, u, v, \psi) - \frac{1}{3}[\lambda_1 (1 + M) + \varepsilon] (u + v + \|\psi\|_J)| + 1. \end{aligned} \end{equation*} It follows that \begin{equation} f(t, u, v, \psi) \geq \frac{1}{3}[\lambda_1(1 + M) + \varepsilon](u + v + \|\psi\|_J) - C, \ u, v \geq 0, \ \psi \in C^+(J), \ t \in I,\label{e3.3} \end{equation} \begin{equation} g(t, u, v, \psi) \geq \frac{1}{3}[\lambda_1(1 + M) + \varepsilon](u + v + \|\psi\|_J) - C, \ u, v \geq 0, \ \psi \in C^+(J), \ t \in I. \label{e3.4} \end{equation} We claim that there exists $r_1 \geq r'$ such that \begin{equation} H_{\phi}(s, x) \not= x, \quad \forall s \in [0, 1], \quad x \in K, \quad \|x\| \geq r_1. \label{e3.5} \end{equation} In fact, if $H_{\phi}(s_1, z) = z$ for some $z \in K$ and $ 0 \leq s_1 \leq 1$, then $z(t)$ satisfies the equation \begin{equation} \begin{aligned} - z''(t) &= (1 - s_1) f(t, z(t), z(t - \tau; \phi), z_t(\cdot; \phi))\\ & \quad + s_1 g(t, z(t), z(t - \tau; \phi), z_t(\cdot; \phi)), \quad 0 < t < 1, \label{e3.6} \end{aligned} \end{equation} and the boundary condition \begin{equation} z(0) = 0, \quad z(1) = z(\eta). \label{e3.7} \end{equation} From the above condition, there exists $\xi \in (\eta, 1)$ such that $z'(\xi) = 0$. Multiplying left side of (\ref{e3.6}) by $\varphi_1(t) = \sin \frac{\pi}{\eta + 1} t$ and then integrating from $0$ to $\xi$, after integrating two times by parts, we get from $z'(\xi) = 0$ that \begin{equation} \int_0^{\xi} - z''(t) \varphi_1(t) dt = \varphi_1'(\xi) z(\xi) + \lambda_1 \int_0^{\xi} z(t) \varphi_1(t) dt. \label{e3.8} \end{equation} By (\ref{e3.6}) and (\ref{e3.7}), we have that $- z''(t) \geq 0 $ for each $t \in I$. Thus we obtain from (\ref{e3.6}), (\ref{e3.8}) and (H3) that \begin{equation} \begin{aligned} \lambda_1 \int_0^1 z(t) \varphi_1(t) dt & \geq \lambda_1 \int_0^{\xi} z(t) \varphi_1(t) dt\\ & = \int_0^{\xi} - z''(t) \varphi_1(t) dt - \varphi_1'(\xi) z(\xi)\\ & \geq \int_0^{\eta} - z''(t) \varphi_1(t) dt - \|\varphi_1'\|_I \|z\|_I\\ & = (1 - s_1) \int_0^{\eta} f(t, z(t), z(t - \tau; \phi), z_t(\cdot; \phi)) \varphi_1(t) dt\\ & \quad + s_1 \int_0^{\eta} g(t, z(t), z(t - \tau; \phi), z_t(\cdot; \phi)) \varphi_1(t) dt - \frac{\pi}{\eta + 1} \|z\|_I. \label{e3.9} \end{aligned} \end{equation} Combining (\ref{e2.2}), (\ref{e3.3}), (\ref{e3.4}) and (\ref{e3.9}), we get \begin{align*} &\lambda_1 \int_0^1 z(t) \varphi_1 (t) dt \\ &\geq \frac{1}{3} (1 - s_1)(\lambda_1 (1 + M) + \varepsilon) \int_0^{\eta} [z(t) + z(t - \tau; \phi) + \|z_t(\cdot; \phi)\|_J]\varphi_1(t) dt \\ &\quad - (1 - s_1) C \int_0^{\eta} \varphi_1 (t) dt \\ &\quad + \frac{1}{3} s_1 (\lambda_1 (1 + M) + \varepsilon)\int_0^{\eta} [z(t) + z(t - \tau; \phi) + \|z_t(\cdot; \phi)\|_J]\varphi_1(t) dt \\ &\quad - s_1 C \int_0^{\eta} \varphi_1 (t) dt - \frac{\pi}{\eta + 1}\|z\|_I \\ & = \frac{1}{3}(\lambda_1 (1 + M) + \varepsilon) \int_0^{\eta} [z(t) + z(t - \tau; \phi) + \|z_t(\cdot; \phi)\|_J]\varphi_1(t) dt \\ &\quad - C \int_0^{\eta} \varphi_1 (t) dt - \frac{\pi}{\eta + 1}\|z\|_I \\ &\geq \frac{1}{3} (\lambda_1 (1 + M) + \varepsilon) \int_0^{\eta} [2 z(t) + z(t - \tau; \phi)] \varphi_1(t) dt \\ &\quad - C \int_0^{\eta} \varphi_1 (t) dt - \frac{\pi}{\eta + 1}\|z\|_I \\ &\geq \frac{1}{3} (\lambda_1 (1 + M) + \varepsilon)\Big(2 \int_0^{\eta} z(t) \varphi_1(t) dt + \int_{\tau}^{\eta} z(t - \tau; \phi)\varphi_1(t) dt \Big) \\ &\quad - C \int_0^{\eta} \varphi_1 (t) dt - \frac{\pi}{\eta + 1}\|z\|_I \\ & = \frac{1}{3} (\lambda_1 (1 + M) + \varepsilon)\Big(2 \int_0^{\eta} z(t) \varphi_1(t) dt + \int_{\tau}^{\eta} z(t - \tau)\varphi_1(t) dt \Big) \\ &\quad - C \int_0^{\eta} \varphi_1 (t) dt - \frac{\pi}{\eta + 1}\|z\|_I \\ &= \frac{1}{3} (\lambda_1 (1 + M) + \varepsilon)\Big(2 \int_0^{\eta} z(t) \varphi_1(t) dt + \int_0^{\eta - \tau} z(t)\varphi_1(t + \tau) dt \Big) \\ &\quad - C \int_0^{\eta} \varphi_1 (t) dt - \frac{\pi}{\eta + 1}\|z\|_I, \end{align*} then we have \begin{equation} \begin{aligned} &(\lambda_1 M + \varepsilon) \Big(\int_0^{\eta - \tau} z(t) \varphi_1(t + \tau) dt + 2 \int_0^{\eta} z(t) \varphi_1(t) dt \Big)\\ &\leq \lambda_1 \int_0^{\eta - \tau} z(t) [\varphi_1(t) - \varphi_1(t + \tau)] dt + \lambda_1 \int_{\eta - \tau}^1 z(t) \varphi_1(t) dt\\ & \quad + 2 \lambda_1 \int_{\eta}^1 z(t) \varphi_1(t) dt + 3 C \int_0^{\eta} \varphi_1(t) dt + \frac{3\pi}{\eta + 1}\|z\|_I\\ & \leq \lambda_1 \tau (\eta - \tau) \|\varphi_1'\|_I \|z\|_I + \lambda_1 (1 - \eta + \tau) \|\varphi_1\|_I \|z\|_I\\ &\quad + 2 \lambda_1 (1 - \eta) \|\varphi_1\|_I \|z\|_I + 3 C \eta \|\varphi_1\|_I + \frac{3\pi}{\eta + 1}\|z\|_I \\ &= \lambda_1 \big[\frac{\pi}{\eta + 1} \tau (\eta - \tau) + 3 (1 - \eta) + \tau + \frac{3(\eta + 1)}{\pi} \big] \|z\|_I + 3 C \eta. \label{e3.10} \end{aligned} \end{equation} We also have \begin{equation} \begin{aligned} &\int_0^{\eta - \tau} z(t) \varphi_1(t + \tau) dt + 2 \int_0^{\eta} z(t) \varphi_1(t) dt\\ &\geq \eta \|z\|_I \int_0^{\eta - \tau} t \varphi_1(t + \tau) dt + 2 \eta \|z\|_I \int_0^{\eta} t \varphi_1(t) dt, \label{e3.11} \end{aligned} \end{equation} which together with (\ref{e3.10}) leads to \begin{align*} &(\lambda_1 M + \varepsilon) \eta \Big( \int_0^{\eta - \tau} t \varphi_1(t + \tau) dt + 2 \int_0^{\eta} t \varphi_1(t) dt \Big) \|z\|_I\\ &\leq \lambda_1 \big[\frac{\pi}{\eta + 1} \tau (\eta - \tau) + 3 (1 - \eta) + \tau + \frac{3(\eta + 1)}{\pi} \big] \|z\|_I + 3 C \eta\\ &= \lambda_1 \frac{1}{\pi (\eta + 1)} \big[\pi^2 \tau (\eta - \tau) + 3\pi(1 - \eta^2) + \pi \tau (\eta + 1) + 3(\eta + 1)^2 \big] \|z\|_I + 3 C \eta; \end{align*} i.e., \begin{align*} \|z\|_I &\leq \frac{3 C}{\varepsilon \big(\int_0^{\eta - \tau} t \varphi_1(t + \tau) dt + 2 \int_0^{\eta} t \varphi_1(t) dt\big)} \\ &= \frac{3 C}{\varepsilon \big(\int_0^{\eta - \tau} t \sin \frac{\pi}{\eta + 1}(t + \tau) dt + 2 \int_0^{\eta} t \sin \frac{\pi}{\eta + 1} t dt\big)}: = \overline{r}. \end{align*} Let $r_1 = 1 + \max \{r', \overline{r}\}$. We obtain (\ref{e3.5}) and consequently, by (\ref{e3.2}) and homotopy invariance of the fixed-point index, we have \begin{equation} \begin{aligned} i(A_{\phi}, K_{r_1}, K) &= i(H_{\phi}(0, \cdot), K_{r_1}, K)\\ & = i(H_{\phi}(1, \cdot), K_{r_1}, K) = i(B_{\phi}, K_{r_1}, K) = 0. \label{e3.12} \end{aligned} \end{equation} Use (\ref{e3.1}) and (\ref{e3.12}) to conclude that $$ i(A_{\phi}, K_{r_1}\setminus \overline{K}_{h_1}, K)= - 1. $$ Hence, $A_{\phi}$ has fixed points $x_*$ in $K_{r_1}\setminus \overline{K}_{h_1}$, which means that $x_*(t)$ is a positive solution of BVP (\ref{e1.2}) and $\|x_*\|_I > h_1$. Thus, the proof is complete. \end{proof} By Lemmas \ref{lem2.6} and \ref{lem2.8}, we have the following result. \begin{theorem}\label{thm3.2} Assume that {\rm (H2)} and {\rm (H5)} are satisfied, then BVP \eqref{e1.2} has at least one positive solution. \end{theorem} Finally, we obtain from Lemmas \ref{lem2.7} and \ref{lem2.8} the following result. \begin{theorem}\label{thm3.3} If {\rm (H4)} and {\rm (H5)} are satisfied, then BVP (\ref{e1.2}) has at least one positive solution. \end{theorem} \begin{thebibliography}{00} \bibitem{b1} C. Z. Bai, J.Fang; \emph{Existence of multiple positive solutions for functional differential equations}, Comput. Math. Appl. 45 (2003) 1797-1806. \bibitem{d1} K. Deimling; \emph{Nonlinear Functional Analysis}, Springer-Verlag, New York, 1985. \bibitem{e1} L.H. Erbe, Q.K. Kong; \emph{Boundary value problems for singular second order functional differential equations}, J. Comput. Appl. Math. 53 (1994) 377-388. \bibitem{h1} J. 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