\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 123, pp. 1--13.\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-2008/123 \hfil Multi-point problems on time scales] {Existence of countably many positive solutions for $n$th-order $m$-point boundary-value problems on time scales} \author[S. Liang, J. Zhang\hfil EJDE-2008/123\hfilneg] {Sihua Liang, Jihui Zhang, Zhiyong Wang} \address{Sihua Liang \newline Institute of Mathematics, School of Mathematics and Computer Sciences, Nanjing Normal University, 210097, Jiangsu, China. \hfill\break College of Mathematics, Changchun Normal University, Changchun 130032, Jilin, China} \email{liangsihua@163.com} \address{Jihui Zhang \newline Institute of Mathematics, School of Mathematics and Computer Sciences, Nanjing Normal University, 210097, Jiangsu, China} \email{jihuiz@jlonline.com} \address{Zhiyong Wang \newline Department of Mathematics, Nanjing University of Information Science and Technology, Nanjing 210044, Jiangsu, China} \email{mathswzhy@126.com} \thanks{Submitted February 20, 2008. Published September 4, 2008.} \subjclass[2000]{34B18} \keywords{Time scales; positive solutions; singular boundary-value; \hfill\break\indent fixed-point index theory} \begin{abstract} In this paper, we study the existence of positive solutions for the nonlinear $n$-th order with $m$-point singular boundary-value problem. By using the fixed point index theory and a new fixed point theorem in cones, the existence of countably many positive solutions for a nonlinear singular boundary value problem are obtained. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{example}[theorem]{Example} \newtheorem{remmark}[theorem]{Remmark} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} In this paper, by introducing a new operator, improving and generating a $p$-Laplace operator for some $p > 1$, we study the existence of countably many positive solutions for $n$-th order with $m$-point nonlinear boundary-value problems \begin{equation}\label{e1.1} (\varphi(u^{\Delta^{n-1}})(t))^\nabla + a(t)f(u(t), u^{\Delta}(t), \dots, u^{\Delta^{n-2}}(t)) = 0, \quad 0 < t < T, \end{equation} subject to the boundary conditions \begin{equation} \begin{gathered} u^{\Delta^{i}}(0) = 0, \quad i = 0, 1, \dots, n -3, \\ u^{\Delta^{n-2}}(0) = \sum_{i=1}^{m-2}\alpha_iu^{\Delta^{n-2}}(\xi_i),\quad u^{\Delta^{n-1}}(T) = 0, \end{gathered} \label{e1.2} \end{equation} where $\varphi: R \to R$ is the increasing homeomorphism and positive homomorphism and $\varphi(0) = 0$. $ \xi_i \in [0, T]_{\mathbf{T}}$ with $ 0 < \xi_1 < \xi_2 < \dots < \xi_{m-2} < T$ and $\alpha_i$ satisfy $\alpha_i \in [0, T]_{\mathbf{T}}$, $0 < \sum_{i=1}^{m-2} \alpha_i < 1$. $a(t): [0, T]_{\mathbf{T}} \to [0, +\infty)$ and has countably many singularities in $[0, T]_{\mathbf{T}}$. A projection $\varphi: R \to R$ is called an increasing homeomorphism and positive homomorphism, if the following conditions are satisfied: \begin{itemize} \item[(1)] if $x \leq y$, then $\varphi(x) \leq \varphi(y)$, for all $x, y \in R$; \item[(2)] $\varphi$ is a continuous bijection and its inverse mapping is also continuous; \item[(3)]$\varphi(xy) = \varphi(x)\varphi(y)$, for all $x,y \in [0, +\infty)$. \end{itemize} In the above definition, we can replace condition (3) by the following stronger condition: \begin{itemize} \item[(4)] $\varphi(xy) = \varphi(x)\varphi(y)$, for all $x, y \in \mathbb{R}$, where $\mathbb{R} = (-\infty, +\infty)$. \end{itemize} \begin{remmark}\label{rem1.1} \rm If conditions (1), (2) and (4) hold, then $\varphi$ is homogenous generating a $p$-Laplace operator; i.e., $\varphi(x) = |x|^{p - 2}x$, for some $p > 1$. \end{remmark} Moreover, throughout this paper the following conditions hold: \begin{itemize} \item[(C1)] $f: [0, +\infty) \to [0, +\infty)$ is continuous; \item[(C2)] $a: [0, T]_{\mathbf{T}} \to [0, +\infty)$ and has countably many singularities in $[0, T]_{\mathbf{T}}$, i.e., there exists a sequence $\{t_i\}_{i = 1}^\infty$ such that $0 < t_{i+1} < t_{i} < \frac{T}{2}$, $\lim_{i \to \infty} t_i = t_0 < \frac{T}{2}$, and $t_0 \in [0, T]_{\mathbf{T}}$. $\lim_{t \to t_i} a(t) = \infty,\ i = 1, 2,\dots$, and $a(t)$ does not vanish identically on any subinterval of $[0, T]_{\mathbf{T}}$. Moreover \[ 0 < \int_0^T a(s)\nabla s < +\infty. \] \end{itemize} Recently, there is much attention paid to the existence of positive solutions for three-point boundary-value problems on time scales, see \cite{a2,b1,b2,h1,k1,l4,s1,z1} and references therein. However, there are not many results concerning the increasing homeomorphism and positive homomorphism operator on time scales. A time scale $\mathbb{T}$ is a nonempty closed subset of $\mathbb{R}$. We make the blanket assumption that $0,T$ are points in $\mathbb{T}$. By an interval $(0,T)$, we always mean the intersection of the real interval $(0,T)$ with the given time scale; that is $(0,T)\cap\mathbb{T}$. Anderson \cite{a2} discussed the dynamic equation on time scales: \begin{gather} u^{\Delta\nabla}(t)+a(t)f(u(t))=0, \quad t\in(0, T), \label{e1.3}\\ u(0)=0, \quad \alpha u(\eta)=u(T).\label{e1.4} \end{gather} He obtained some results for the existence of one positive solution of the problem \eqref{e1.3} and \eqref{e1.4} based on the limits $f_{0}=\lim_{u\to 0^{+}}\frac{f(u)}{u}$ and $f_{\infty}=\lim_{u\to \infty}\frac{f(u)}{u}$. He also obtained the existence of at least three positive solutions. Kaufmann \cite{k1} studied the problem \eqref{e1.3} and \eqref{e1.4} and obtained existence results of finitely many positive solutions and countably many positive solutions. Zhou and Su \cite{z1} studied the quasi-linear equation with $p$-Laplacian operator: \begin{gather} (\phi_{p}(u^{(n-1)}))'+g(t)f(u(t),u'(t), \dots, u^{(n-2)}(t))=0,\quad 0\inf\mathbb{T}$, define the forward jump operator $\sigma$ and backward jump operator $\rho$, respectively, by \begin{gather*} \sigma(t)=\inf\{\tau\in\mathbb{T}:\tau> t\}\in\mathbb{T}, \\ \rho(r)=\sup\{\tau\in\mathbb{T}:\tau< r\}\in\mathbb{T}. \end{gather*} for all $t, r\in\mathbb{T}$. If $\sigma(t)>t$, $t$ is said to be right scattered, and if $\rho(r)0$, there is a neighborhood $U$ of $t$ such that $$ |f(\sigma(t))-f(s)-f^{\Delta}(t)(\sigma(t)-s)|\leq\epsilon|\sigma(t)-s|, $$ for all $s\in U$. \end{definition} For $f:\mathbb{T}\to \mathbb{R}$ and $t\in\mathbb{T}_{\kappa}$, the nabla derivative of $f$ at $t$ is the number $f^{\nabla}(t)$, (provided it exists), with the property that for each $\epsilon>0$, there is a neighborhood $U$ of $t$ such that $$ |f(\rho(t))-f(s)-f^{\nabla}(t)(\rho(t)-s)|\leq\epsilon|\rho(t)-s|, $$ for all $s\in U$. \begin{definition} \label{def2.3}\rm A function $f$ is left-dense continuous (i.e. ld-continuous), if $f$ is continuous at each left-dense point in $\mathbb{T}$ and its right-sided limit exists at each right-dense point in $\mathbb{T}$. It is well-known that if $f$ is ld-continuous, then there is a function $F(t)$ such that $F^{\nabla}(t)=f(t)$. In this case, it is defined that $$ \int_a^b f(t)\nabla t=F(b)-F(a). $$ \end{definition} If $u^{\Delta\nabla}(t)\leq0$ on $[0, T]$, then we say $u$ is concave on $[0, T]$. \begin{definition} \label{def2.4}\rm Let $(E, \ \|.\|)$ be a real Banach space. A nonempty, closed, convex set $P \subset E$ is said to be a cone provided the following are satisfied: \begin{itemize} \item[($a$)] if $y \in P$ and $\lambda\ \geq 0$, then $\lambda y \in P$; \item[($b$)] if $y \in P$ and $-y \in P$, then $y = 0$. \end{itemize} If $P \subset E$ is a cone, we denote the order induced by $P$ on $E$ by $\leq$, that is, $x \leq y$ if and only if $y - x \in P$. \end{definition} \begin{definition} \label{def2.5}\rm Given a nonnegative continuous functional $\gamma$ on a cone $P$ of $E$, for each $d > 0$ we define the set $$ P(\gamma, d) = \{x \in P : \gamma(x) < d\}. $$ \end{definition} The following fixed point theorems are fundamental and important for the proofs of our main results. \begin{theorem}[\cite{g1}] \label{thm2.1} Let $E$ be a Banach space and $P \subset E$ be a cone in $E$. Let $r> 0$ define $\Omega_r = \{ x \in P : \|x\| < r \}$. Assume that $A : P \bigcap \overline{\Omega}_r \to P$ is completely continuous operator such that $Ax \neq x$ for $x \in \partial \Omega_r $. \begin{itemize} \item[(i)] If $\|Ax\| < \|x\|$ for $x \in \partial \Omega_r$, then $i(A, \Omega_r,\ P)=1$. \item[(ii)] If $\|Ax\| > \|x\|$ for $x \in \partial \Omega_r$, then $i(A, \Omega_r,\ P)=0$. \end{itemize} \end{theorem} \begin{theorem}[\cite{r1}] \label{thm2.2} Let $P$ be a cone in a Banach space $E$. Let $\alpha$, $\beta$ and $\gamma$ be three increasing, nonnegative and continuous functionals on $P$, satisfying for some $c > 0$ and $M > 0$ such that $$ \gamma(x) \leq \beta(x) \leq \alpha(x),\indent \|x\| \leq M\gamma(x) $$ for all $x \in \overline{P(\gamma, c)}$. Suppose there exists a completely continuous operator $A : \overline{P(\gamma, c)} \to P$ and $0 < a < b < c$ such that \begin{itemize} \item[(i)] $\gamma(Ax) < c$, for all $x \in \partial P(\gamma, c)$; \item[(ii)] $\beta(Ax) > b $, for all $x \in \partial P(\beta, b)$; \item[(iii)] $P(\alpha, a) \neq \emptyset$, and $\alpha(Ax) < a$, for all $x \in \partial P(\alpha, a)$. Then $A$ has at least three fixed points $x_1$, $x_2$, $x_3 \in \overline{P(\gamma, c)}$ such that \[ 0 \leq \alpha(x_1) < a < \alpha(x_2),\hskip 0.5cm \beta(x_2) < b < \beta(x_3), \hskip 0.5cm \gamma(x_3) < c. \] \end{itemize} \end{theorem} \section{Preliminaries and Lemmas} In the rest of this article, $\mathbf{T}$ is closed subset of $\mathbb{R}$ with $0 \in {\mathbf{T}}_\kappa$, $T \in {\mathbf{T}}^\kappa$. And $$ E =\big\{u\in C_{ld}^{{n-2}}[0,T]: u^{\Delta^{i}}(0)=0,\ 0\leq i\leq n-3\big\}. $$ Then $E$ is a Banach space with the norm $\|u\|=\sup_{t\in[0, T]}|u^{\Delta^{n-2}}(t)|$. And let $$ P=\big\{u\in E: u^{\Delta^{n-2}}(t)\geq0,\, u^{\Delta^{n-2}}(t) \text{is concave nondecreasing on $[0,T]$}\big\}. $$ Obviously, $P$ is a cone in $E$. Set $P_r=\{u\in P:\|u\|\leq r\}$. We can easily get the following Lemmas. \begin{lemma}\label{lem3.1} Suppose condition {\rm (C2)} holds. Then there exists a constant $\theta \in \max\{t \in {T}|\ 0 < t < \frac{T}{2}\}$ that satisfies $$ 0 < \int_{\theta}^{T - \theta}a(s)\nabla s < +\infty. $$ Furthermore, the function $$ H(t) = \int_t^{T-t_1}\varphi^{-1}\Big(\int_s^{T - t_1}a(\tau)\nabla\tau \Big)\Delta s + \frac{\sum_{i=1}^{m-2}\alpha_i\int_{t_1}^t\varphi^{-1}\big(\int_s^t a(\tau)\nabla\tau \big)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_i} $$ is continuous and positive on $[t_1, T - t_1]$. Furthermore there exists a constant $L > 0$ such that \[ L = \min_{t \in [t_1, T - t_1]}H(t) > 0. \] \end{lemma} \begin{proof} At first, it is easily seen that $H(t)$ is continuous on $[t_1, T - t_1]$. Let \begin{gather*} H_1(t) =\int_t^{T-t_1}\varphi^{-1}\Big(\int_s^{T - t_1}a(\tau)\nabla\tau \Big)\Delta s, \\ H_2(t) = \frac{\sum_{i=1}^{m-2}\alpha_i\int_{t_1}^t\varphi^{-1}\left(\int_s^t a(\tau)\nabla\tau \right)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_i}. \end{gather*} Then from condition {\rm (C2)}, we know that $H_1(t)$ strictly monotone decreasing on $[t_1, T - t_1]$ and $H_1(T - t_1) = 0$. Similarly function $H_2(t)$ is strictly monotone increasing on $[t_1, T - t_1]$ and $H_2(t_1) = 0$. Since $H_1(t)$ and $H_2(t)$ are not equal to zero at the same time. So the function $H(t) = H_1(t) + H_2(t)$ is positive on $[t_1, T - t_1]$, which implies $L = \min_{t \in [t_1, T - t_1]} H(t) > 0$. \end{proof} \begin{lemma}\label{lem3.2} If $u \in P$. Then $$ u^{\Delta^{n-2}}(t) \geq \frac{\theta}{T}\|u\|, \quad t \in[\theta, T - \theta], $$ \end{lemma} The proof of the above lemma is similar to the proof of in \cite{h1}, so we omit it. Now, we define a mapping $F: P\to C_{ld}^{n-1}[0,T]$ by \begin{equation}\label{e3.1} (Fu)(t)= \int_0^t \int_0^{\zeta_1}\dots\int_0^{\zeta_{n-3}}w(\zeta_{n-2})\Delta\zeta_{n-2}\Delta\zeta_{n-3}\dots \Delta\zeta_1, \end{equation} where \begin{align*} w(\zeta_{n-2}) &= \int_0^{\zeta_{n-2}} \varphi^{-1}\Big(\int_s^T a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\quad+ \frac{\sum_{i=1}^{m-2}\alpha_{i} \int_{0}^{\xi_{i}}\varphi^{-1}\big(\int_s^T a(\tau)f(u(\tau), u^{\prime}(\tau), \dots, u^{(n-2)}(\tau))\nabla\tau \big) \Delta s}{1 - \sum_{i=1}^{m-2}\alpha_{i}}. \end{align*} Then it is easy to see that \begin{align*} (Fu)^{\Delta^{n-2}}(t) &= \int_0^{t} \varphi^{-1}\Big(\int_s^T a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\quad + \frac{\sum_{i=1}^{m-2}\alpha_{i} \int_{0}^{\xi_{i}}\varphi^{-1}\Big(\int_s^T a(\tau)f(u(\tau), u^{\prime}(\tau), \dots, u^{(n-2)}(\tau))\nabla\tau \Big)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_{i}}\\ &\geq 0, \quad 0 \leq t \leq T. \end{align*} \begin{align*} (Fu)^{\Delta^{n-1}}(t) &= \varphi^{-1}\Big(\int_t^T a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\\ &\geq 0, \quad 0 \leq t \leq T. \end{align*} We also have $$ [\varphi((Fu)^{\Delta^{n-1}})(t)]^{\nabla} = -a(t)f(u(t), u^{\Delta}(t), \dots, u^{\Delta^{n-2}}(t)) \leq 0. $$ Together with $\varphi$ is a increasing operator, we know $(Fu)^{\Delta^{n-2}}$ is a concave function. This shows $F(P) \subset P$. Using the Arzela-Ascoli Theorem, we obtain the following lemma. \begin{lemma}\label{lem3.4} The operator $F:P\to P$ is completely continuous. \end{lemma} \begin{lemma}\label{lem3.5} Suppose that conditions {\rm (C1), (C2)} hold. Then the solution $u(t)\in P$ of \eqref{e1.1}, \eqref{e1.2} satisfies $$ u(t) \leq u^{\Delta}(t) \leq \dots \leq u^{\Delta^{n-3}}(t), \quad 0 \leq t \leq T, $$ and for $\theta\in(0,\frac{T}{2})$ in Lemma \ref{lem3.1}, we have $$ u^{\Delta^{n-3}}(t) \leq \frac{T}{\theta}u^{\Delta^{n-2}}(t), \quad \theta \leq t \leq T - \theta. $$ \end{lemma} The proof of the above lemma is similar to the proof of in \cite [lemma 2.4]{z1}. \section{Main results} For notational convenience, we define $$ \lambda_1 = \frac{1}{L},\quad \lambda_2 = \frac{(1 - \sum_{i = 1}^{m - 2}\alpha_i)}{\int_0^T\varphi^{-1} \Big(\int_s^Ta(\tau)\nabla\tau\Big)\Delta s}. $$ The main results of this paper are the following. \begin{theorem}\label{thm4.1} Suppose that conditions {\rm (C1)-(C2)} hold. Let $\{\theta_k\}_{k = 1}^\infty$ be such that $\theta_k \in (t_{k + 1},\ t_k)\ (k=1, 2,\dots)$. Let $\{r_k\}_{k = 1}^\infty$ and $\{R_k\}_{k = 1}^\infty$ be such that $$ R_{k + 1} < \frac{\theta_k}{T} r_k < r_k < m r_k < R_k, \quad mr_k \leq MR_k,\quad k = 1, 2,\dots. $$ Furthermore for each natural number $k$ we assume that $f$ satisfy: \begin{itemize} \item[(C3)] $f(v_1, v_2,\dots,v_{n-1}) \geq \varphi (m r_k)$ for all $0 \leq v_1, v_2, \dots, v_{n-2} \leq \frac{T}{\theta_k}r_k$, $\frac{\theta_k}{T}r_k \leq v_{n-1} \leq r_k$; \item[(C4)] $f(v_1, v_2,\dots,v_{n-1}) \leq \varphi (M R_k)$ for all $0 \leq v_1, v_2,\dots, v_{n-1} \leq R_k$. \end{itemize} Where $m \in (\lambda_1, \infty)$, $M \in (0, \lambda_2)$. Then the boundary-value problem \eqref{e1.1}, \eqref{e1.2} has infinitely many solutions $\{u_k\}_{k = 1}^\infty$ such that $$ r_k \leq \|u_k\| \leq R_k,\quad k = 1, 2,\dots. $$ \end{theorem} \begin{proof} Since $0 < t_0 < t_{k + 1} < \theta_k < t_k < \frac{T}{2},\ k = 1, 2,\dots$, for any $k \in \mathbb{N}$ and $u \in P$, by the Lemma \ref{lem3.2} we have \begin{equation}\label{e4.1} u^{\Delta^{n-2}}(t) \geq \frac{\theta_k}{T} \|u\|, \quad t \in [\theta_k, T - \theta_k]. \end{equation} Consider the sequences $\{\Omega_{1, k}\}_{k = 1}^\infty$ and $\{\Omega_{2, k}\}_{k = 1}^\infty$ of open subsets of $E$ defined by \begin{gather*} \Omega_{1, k} = \{u \in P :\|u\| < r_k\},\quad k = 1, 2,\dots, \\ \Omega_{2, k} = \{u \in P :\|u\| < R_k\},\quad k = 1, 2,\dots. \end{gather*} For a fixed $k$ and $u \in \partial\Omega_{1, k}$. From (\ref{e4.1}) we have \[ r_k = \|u\| \geq u^{\Delta^{n-2}}(t) \geq \frac{\theta_k}{T} \|u\| = \frac{\theta_k}{T} r_k,\quad t \in [\theta_k, T - \theta_k]. \] Since $(t_1, T - t_1) \subset [\theta_k, T - \theta_k]$, in the following we consider three cases: \noindent (i) If $\xi_1 \in [t_1, T - t_1]$. In this case, from (\ref{e3.1}), condition (C3) and Lemma \ref{lem3.1}, we have \begin{align*} \|Fu\| &= (Fu)^{\Delta^{n-2}}(T)\\ &= \int_0^T \varphi^{-1}\Big(\int_s^T a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\quad + \frac{\sum_{i=1}^{m-2}\alpha_{i} \int_{0}^{\xi_{i}}\varphi^{-1}\Big(\int_s^T a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau \Big)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_{i}}\\ &\geq\int_{\xi_1}^{T-t_1} \varphi^{-1}\Big(\int_s^{T-t_1} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\quad + \frac{\sum_{i=1}^{m-2}\alpha_{i} \int_{t_1}^{\xi_{1}}\varphi^{-1}\Big(\int_s^{\xi_1} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau \Big)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_{i}}\\ &\geq (mr_k)\Big[ \int_{\xi_1}^{T - t_1} \varphi^{-1}\Big(\int_s^{T - t_1} a(\tau)\nabla\tau\Big)\Delta s\Big.\\ &\quad \Big. + \frac{\sum_{i=1}^{m-2}\alpha_i }{1 - \sum_{i=1}^{m-2}\alpha_i}\int_{t_1}^{\xi_1}\varphi^{-1}\Big(\int_s^{\xi_1} a(\tau)\nabla\tau \Big)\Delta s\Big]\\ &= mr_k H(\xi_1) > mr_kL > r_k = \|u\|. \end{align*} \noindent(ii) If $\xi_1 \in [0, t_1]$. In this case, from (\ref{e3.1}), condition $(C_3)$ and Lemma\ref{lem3.1}, we have \begin{align*} \|Fu\| &\geq \int_0^T \varphi^{-1}\Big(\int_s^T a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\geq\int_{t_1}^{T-t_1} \varphi^{-1}\Big(\int_s^{T-t_1} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\geq (mr_k)\Big[ \int_{t_1}^{T - t_1} \varphi^{-1}\Big(\int_s^{T - t_1} a(\tau)\nabla\tau\Big)\Delta s\Big]\\ &= mr_k H(t_1) > mr_kL > r_k = \|u\|. \end{align*} \noindent (iii) If $\xi_1 \in [T - t_1, T]$. In this case, from (\ref{e3.1}), condition $(C_3)$ and Lemma\ref{lem3.1}, we have \begin{align*} \|Fu\| &\geq \frac{\sum_{i=1}^{m-2}\alpha_{i} \int_{t_1}^{T-t_1}\varphi^{-1}\big(\int_s^{T-t_1} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau \big)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_{i}}\\ &\geq (mr_k)\Big[\frac{\sum_{i=1}^{m-2}\alpha_i }{1 - \sum_{i=1}^{m-2}\alpha_i}\int_{T - t_1}^{t_1}\varphi^{-1}\Big(\int_s^{T-t_1} a(\tau)\nabla\tau \Big)\Delta s\Big]\\ &= mr_k H(T-t_1) > mr_kL > r_k = \|u\|. \end{align*} Thus in all cases, an application of Theorem \ref{thm2.1} implies \begin{equation}\label{e4.2} i(F,\ \Omega_{1, k},\ P) = 0. \end{equation} On the another hand, let $u(t) \in \partial\Omega_{2, k}$, we have $u^{\Delta^{n-2}}(t) \leq \|u\| = R_k$, by $(C_4)$ we have \begin{align*} \|Fu\| &= (Fu)^{(n-2)}(T)\\ &= \int_0^T \varphi^{-1}\Big(\int_s^T a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\quad + \frac{\sum_{i=1}^{m-2}\alpha_{i} \int_{0}^{\xi_{i}}\varphi^{-1}\Big(\int_s^T a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau \Big)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_{i}}\\ &\leq\int_{0}^{T} \varphi^{-1}\Big(\int_s^{T} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\quad + \frac{\sum_{i=1}^{m-2}\alpha_{i} \int_{0}^{\xi_{m-2}}\varphi^{-1}\big(\int_s^{T} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau \big)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_{i}}\\ &\leq MR_k\frac{1 }{1 - \sum_{i=1}^{m-2}\alpha_i}\Big[ \int_{0}^{T} \varphi^{-1}\Big(\int_s^{T} a(\tau)\nabla\tau\Big)\Delta s \Big]\\ &= R_k = \|u\|. \end{align*} Thus Theorem \ref{thm2.1} implies \begin{equation}\label{e4.3} i(T,\ \Omega_{2, k},\ P) = 1. \end{equation} Hence since $r_k < R_k$ for $k \in \mathbb{N}$, (\ref{e4.2}) and (\ref{e4.3}), it follows from additivity of the fixed-point index that $$ i(T,\ \Omega_{2, k}\backslash \overline{\Omega}_{1, k},\ P) = 1 \quad \text{for } k \in \mathbb{N}. $$ Thus $F$ has a fixed point in $\Omega_{2, k}\backslash\overline{\Omega}_{1, k}$ such that $r_k \leq \|u_k\| \leq R_k$. Since $k \in \mathbb{N}$ was arbitrary, the proof is complete. \end{proof} To use Theorem \ref{thm2.2}, let $\theta_k < r_k < 1 - \theta_k$ and $\theta_k$ of Theorem \ref{thm4.1}, we define the nonnegative, increasing, continuous functionals \begin{gather*} \gamma_k(u) = \max_{\theta_k \leq t \leq r_k}u^{\Delta^{n-2}}(t) = u^{\Delta^{n-2}}(r_k), \\ \beta_k(u) = \min_{r_k \leq t \leq T - \theta_k}u^{\Delta^{n-2}}(t) = u^{\Delta^{n-2}}(r_k), \\ \alpha_k(u) = \max_{\theta_k \leq t \leq T - \theta_k}u^{\Delta^{n-2}}(t) = u^{\Delta^{n-2}}(T - \theta_k). \end{gather*} It is obvious that for each $u \in P$, $$ \gamma_k(u) \leq \beta_k(u) \leq \alpha_k(u). $$ In addition, by Lemma \ref{lem3.2}, for each $u \in P$, $$ \gamma_k(u) = u^{\Delta^{n-2}}(r_k) \geq \frac{\theta_k}{T}\|u\|. $$ Thus $$ \|u\| \leq \frac{T}{\theta_k}\gamma_k(u) \quad\text{ for all } u \in P. $$ For convenience, we denote \begin{gather*} \lambda = \frac{1}{1 - \sum_{i=1}^{m-2}\alpha_i}\Big[ \int_{0}^{T} \varphi^{-1}\Big(\int_s^{T} a(\tau)\nabla\tau\Big)\Delta s \Big], \\ \eta_k = \int_{\theta_k}^{r_k} \varphi^{-1}\Big(\int_s^{T-\theta_k} a(\tau)\nabla\tau\Big)\Delta s. \end{gather*} \begin{theorem}\label{thm4.2} Suppose {\rm (C1)-(C2)} hold. Let $\{\theta_k\}_{k = 1}^\infty$ be such that $\theta_k \in (t_{k + 1},\ t_k)$ ($k=1, 2,\dots$). Let $\{a_k\}_{k = 1}^\infty$, $\{b_k\}_{k = 1}^\infty$ and $\{c_k\}_{k = 1}^\infty$ be such that $$ c_{k + 1} < a_k < \frac{\theta_k}{T}b_k < b_k < c_k, \quad\text{and } \rho_k b_k < \eta_k c_k,\text{ for } k = 1, 2, \dots. $$ Furthermore for each natural number $k$ we assume that $f$ satisfies: \begin{itemize} \item[(C5)] $f(v_1, v_2,\dots,v_{n-1}) < \varphi (\frac{c_k}{\lambda})$, for all $ 0 \leq v_1, v_2, \dots, v_{n-1} \leq \frac{T}{\theta_k}c_k$; \item[(C6)] $f(v_1, v_2,\dots,v_{n-1}) > \varphi (\frac{b_k}{\eta_k})$, for all $0 \leq v_1, v_2, \dots, v_{n-2} \leq \frac{T}{\theta_k}b_k$, $ b_k \leq v_{n-1}(t) \leq \frac{T}{\theta_k}b_k$; \item[(C7)] $f(v_1, v_2,\dots,v_{n-1}) < \varphi (\frac{a_k}{\lambda})$, for all $ 0 \leq v_1, v_2, \dots, v_{n-1} \leq \frac{T}{\theta_k}a_k$. \end{itemize} Then the boundary-value problem \eqref{e1.1}, \eqref{e1.2} has three infinite families of solutions $\{u_{1k}\}_{k = 1}^\infty$ $\{u_{2k}\}_{k = 1}^\infty$ and $\{u_{3k}\}_{k = 1}^\infty$ satisfying $$ 0 \leq \alpha_k(u_{1k}) < a_k < \alpha_k(u_{2k}), \quad \beta_k(u_{2k}) < b_k < \beta_k(u_{3k}), \quad \gamma(u_{3k}) < c_k, $$ for $n \in \mathbb{N}$. \end{theorem} \begin{proof} We define the completely continuous operator $F$ by \ref{e3.1}. So it is easy to check that $F : \overline{P(\gamma_k, c_k)} \to P$, for $k \in \mathbb{N}$. We now show that all the conditions of Theorem \ref{thm2.2} are satisfied. To make use of property (i) of Theorem \ref{thm2.2}, we choose $u \in \partial P(\gamma_k, c_k)$. Then $\gamma_k(u) = \max_{\theta_k \leq t \leq r_k}u^{\Delta^{n-2}}(t) = u^{\Delta^{n-2}}(r_k) = c_k$, this implies that $0 \leq u^{\Delta^{n-2}}(t) \leq c_k$ for $[0, r_k]$. If we recall that $\|u\| \leq \frac{T}{\theta_k}\gamma_k (u) = \frac{T}{\theta_k}c_k$. So we have $$ 0 \leq u^{\Delta^{i}}(t) \leq \frac{T}{\theta_k}c_k, \quad 0 \leq t \leq T,\; i = 0,1,\dots,n-1. $$ Then assumption (C5) implies $$ f(u(t), u^{\Delta}(t), \dots, u^{\Delta^{n-2}}(t)) < \varphi \big(\frac{c_k}{\lambda}\big), \quad 0 \leq t \leq T. $$ Therefore \begin{align*} \gamma_k(Fu) &= \max_{\theta_k \leq t \leq r_k}(Fu)^{\Delta^{n-2}}(t) = (Fu)^{\Delta^{n-2}}(r_k)\\ &\leq\int_{0}^{T} \varphi^{-1}\Big(\int_s^{T} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\quad + \frac{\sum_{i=1}^{m-2}\alpha_{i} \int_{0}^{\xi_{m-2}}\varphi^{-1}\left(\int_s^{T} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau \right)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_{i}}\\ &\leq \frac{c_k}{\lambda}\frac{1 }{1 - \sum_{i=1}^{m-2}\alpha_i}\Big[ \int_{0}^{T} \varphi^{-1}\Big(\int_s^{T} a(\tau)\nabla\tau\Big)\Delta s \Big]\\ &= c_k. \end{align*} Hence condition (i) is satisfied. Secondly, we show that (ii) of Theorem \ref{thm2.2} is fulfilled. For this we select $u \in \partial P(\beta_k, b_k)$. Then $\beta_k(u) = \min_{r_k \leq t \leq T - \theta_k}u^{\Delta^{n-2}}(t) = u^{\Delta^{n-2}}(r_k) = b_k$, this fact implies that $u^{\Delta^{n-2}}(t) \geq b_k$, for $r_k \leq t \leq T$. Noticing that $\|u\| \leq \frac{T}{\theta_k}\gamma_k(u) \leq \frac{T}{\theta_k}\beta_k(u) = \frac{T}{\theta_k}b_k$, we have $$ b_k \leq u^{\Delta^{n-2}}(t) \leq \frac{T}{\theta_k}b_k, \quad\text{for}\ r_k \leq t \leq T. $$ By (C6), we have $$ f(u(t), u^{\Delta}(t), \dots, u^{\Delta^{n-2}}(t)) > \varphi \big(\frac{b_k}{\eta_k}\big). $$ Therefore, \begin{align*} \beta_k(Fu) &= \min_{r_k \leq t \leq T - \theta_k}(Fu)^{\Delta^{n-2}}(t) = (Fu)^{\Delta^{n-2}}(r_k)\\ &= \int_0^{r_k} \varphi^{-1}\Big(\int_s^T a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\quad + \frac{\sum_{i=1}^{m-2}\alpha_{i} \int_{0}^{\xi_{i}}\varphi^{-1}\Big(\int_s^T a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau \Big)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_{i}}\\ &\geq\int_{\theta_k}^{r_k} \varphi^{-1}\Big(\int_s^{T-\theta_k} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &= \frac{b_k}{\eta_k}\Big[ \int_{\theta_k}^{r_k} \varphi^{-1}\Big(\int_s^{T-\theta_k} a(\tau)\nabla\tau\Big)\Delta s \Big]\\ &= b_k. \end{align*} Hence condition (ii) is satisfied. Finally, we verify that (iii) of Theorem \ref{thm2.2} is satisfied. Noting that $u^{\Delta^{n-2}}(t) \equiv \frac{a_k}{4}$, $0 \leq t \leq T$ is a member of $P(\alpha_k, a_k)$ and $\alpha_k(u) = \frac{a_k}{4} < a_k$. So $P(\alpha_k, a_k) \neq \emptyset$. Now let $u \in \partial P(\alpha_k, a_k)$. Then $\alpha_k(u) = \max_{\theta_k \leq t \leq T - \theta_k}u^{\Delta^{n-2}}(t) = u^{\Delta^{n-2}}(T - \theta_k) = a_k$. This implies that $0 \leq u^{\Delta^{n-2}}(t) \leq a_k, 0 \leq t \leq T - \theta_k$. Noticing that $\|u\| \leq \frac{T}{\theta_k}\gamma_k(u) \leq \frac{T}{\theta_k}\alpha_k(u) = \frac{T}{\theta_k}a_k$. Then we get $$ 0 \leq u^{\Delta^{i}}(t) \leq \frac{a_k}{r_k}, \quad 0 \leq t \leq T,\; i = 0,1,\dots,n-1. $$ Then assumption (C7) implies $$ f(u(t), u^{\Delta}(t), \dots, u^{\Delta^{n-2}}(t)) < \varphi \big(\frac{a_k}{\lambda}\big), \quad 0 \leq t \leq T. $$ As before, we get \begin{align*} \alpha_k(Fu) &= \max_{\theta_k \leq t \leq T - \theta_k} (Fu)(t) = (Fu)^{\Delta^{n-2}}(T - \theta_k) \\ &\leq\int_{0}^{T} \varphi^{-1}\Big(\int_s^{T} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau\Big)\Delta s\\ &\quad + \frac{\sum_{i=1}^{m-2}\alpha_{i} \int_{0}^{\xi_{m-2}}\varphi^{-1}\big(\int_s^{T} a(\tau)f(u(\tau), u^{\Delta}(\tau), \dots, u^{\Delta^{n-2}}(\tau))\nabla\tau \big)\Delta s}{1 - \sum_{i=1}^{m-2}\alpha_{i}}\\ &\leq \frac{a_k}{\lambda}\frac{1 }{1 - \sum_{i=1}^{m-2}\alpha_i}\Big[ \int_{0}^{T} \varphi^{-1}\Big(\int_s^{T} a(\tau)\nabla\tau\Big)\Delta s \Big]\\ &= a_k. \end{align*} Thus (iii) of Theorem \ref{thm2.2} is satisfied. Since all hypotheses of Theorem \ref{thm2.2} are satisfied, the assertion follows. \end{proof} \begin{remmark}\label{rem4.2} \rm If we add the condition of $a(t)f(u(t), u^{\Delta}(t), \dots, u^{\Delta^{n-2}}(t)) \not\equiv 0$, $t \in [0, T]$, to Theorem \ref{thm4.2} we can get three infinite families of positive solutions $\{u_{1k}\}_{k = 1}^\infty$, $\{u_{2k}\}_{k =1}^\infty$, and $\{u_{3k}\}_{k = 1}^\infty$ satisfying $$ 0 < \alpha_k(u_{1k}) < a_k < \alpha_k(u_{2k}), \quad \beta_k(u_{2k}) < b_k < \beta_k(u_{3k}), \quad \gamma(u_{3k}) < c_k, $$ for $n \in \mathbb{N}$. \end{remmark} \begin{remmark}\label{rem4.3} \rm The same conclusions of Theorem \ref{thm4.1} and Theorem \ref{thm4.2} hold when conditions (1), (2) and (4) are satisfied. Especially, for $p$-Laplacian operator $\varphi(x) = |x|^{p - 2}x$, for some $p> 1$, our conclusions are also true and new. \end{remmark} \section{Applications} There exists a function $a(t)$ satisfying condition (C2). \begin{example} \label{exa5.1} \rm Let ${\mathbf{T}} \equiv 1$ and \[ \delta = 2\big(\frac{\pi^2}{3} - \frac{9}{4}\big),\quad t^{\ast} = \frac{15}{32}, \quad t_i = t^{\ast} - \sum_{k = 1}^{i} \frac{1}{2(k + 1)^4}, \quad i = 1, 2,\dots. \] Consider the function $a(t) : [0, 1] \to (0, \infty)$, $a(t) = \sum_{i = 1}^{\infty}a_i(t)$, $t \in [0, 1]$, where \[ a_i(t) = \begin{cases} \frac{1}{(2i-1)(2i+ 1)(t_{i + 1} + t_i)}, &0 \leq t < \frac{t_{i + 1} + t_i}{2},\\ \frac{1}{\delta(t_i - t)^{1/2}}, &\frac{t_{i + 1} + t_i}{2} \leq t < t_i,\\ \frac{1}{\delta(t - t_i)^{1/2}}, & t_i < t \leq \frac{t_i + t_{i - 1}}{2},\\ 0, &\frac{t_i + t_{i - 1}}{2} < t \leq t_1. \\ \frac{1}{2(2i-1)(2i + 1)(1 - t_1 )}, & t_1 \leq t \leq 1. \end{cases} \] It is easy to check that $t_1 = \frac{7}{16} < \frac{1}{2}$, $t_i - t_{i+1} = \frac{1}{2(i + 2)^4}$, $i = 1, 2,\dots$ (denote $\sum_{i =1}^\infty \frac{1}{i^4} = \frac{\pi^4} {90}$), and \[ t_{0} = \lim_{i \to \infty}t_i = \frac{15}{32} - \sum_{k = 1}^{\infty} \frac{1}{2(k + 1)^4} = \frac{31}{32} - \frac{\pi^4}{180} > \frac{1}{5} \] and because $\sum_{i = 1}^\infty 1/i^2 = \pi^2/6$, we have \begin{align*} &\sum_{i = 1}^\infty \int_0^1 a_i(t)\nabla t \\ &= \sum_{i =1}^\infty \frac{1}{(2i-1)(2i + 1)} + \frac{1}{\delta}\sum_{i = 1}^\infty \Big[\int_{\frac{t_{i + 1} + t_i}{2}}^{t_i} \frac{1}{(t_i -t)^{1/2}}\nabla t + \int_{t_i}^{\frac{t_i + t_{i - 1}}{2}}\frac{1}{(t -t_i)^\frac{1}{2}}\nabla t\Big]\\ &= \frac{1}{2} + \frac{\sqrt{2}}{\delta}\sum_{i = 1}^{\infty}\left[(t_i - t_{i + 1})^{1/2} + (t_{i - 1} - t_{i })^{1/2}\right]\\ &= \frac{1}{2} + \frac{1}{\delta}\sum_{i = 1}^{\infty}\Big[\frac{1}{(i + 2)^2} + \frac{1}{(i + 1)^2}\Big]\\ &= \frac{1}{2} + \frac{1}{\delta}\big[\frac{\pi^2}{3} - \frac{9}{4} \big] = 1. \end{align*} Therefore, \[ \int_0^1 a(t)\nabla t = \sum_{i = 1}^\infty \int_0^1 a_i(t)\nabla t =1 < \infty. \] Which implies that Condition (C2). \end{example} \begin{example} \label{exam5.2} \rm As an example we mention the boundary-value problem \begin{gather} \label{e5.1} [\varphi(u^{\Delta^2})]^\nabla + a(t)f(u(t)) = 0, \quad t \in [0, 1]_{\mathbf{T}}, \\ \begin{gathered} u(0)= u^{\Delta}(0) = 0, \\ u^{\Delta^{2}}(0) = \frac{1}{4}u^{\Delta^{2}}(\frac{1}{4}) + \frac{1}{2}u^{\Delta^{2}}(\frac{1}{2}), \quad u^{\Delta^{3}}(1) = 0, \end{gathered} \label{e5.2} \end{gather} where \[ \varphi(u) = \begin{cases} \frac{u^7}{1 + u^2}, & u \leq 0,\\ u^2, &u > 0, \end{cases} \] and \[ f(u(t)) = \begin{cases} M^2R_{1}^2, & u \in (R_1, +\infty),\\ m^2r_{k}^2 + \frac{M^2R_{k}^2 - m^2r_{k}^2}{R_k -r_k}(u - r_k), & u \in [r_k, R_k ],\\ m^2r_{k}^2, & u \in (\frac{\theta_k}{T}r_k, r_k),\\ M^2R_{k+1}^2 + \frac{m^2r_{k}^2 - M^2R_{k+1}^2}{\frac{\theta_k}{T} r_k - R_{k+1}}(u - R_{k+1}), & u \in (R_{k+1}, \frac{\theta_k}{T}r_k],\\ 0 , & u = 0. \end{cases} \] Since $H'(t) \leq 0$, So it is easy to see by calculating that \[ L = \min_{[t_1, 1 - t_1]}H(t) = H(1-t_1) = \frac{\sum_{i=1}^{m-2}\alpha_i}{1-\sum_{i=1}^{m-2}\alpha_i}\frac{1}{1 - t_1}\big[\frac{2}{3}(1-2t_1)^{\frac{3}{2}}\big] = \frac{8}{9}. \] Then \[ \lambda_1 = \frac{1}{L} = \frac{9}{8}, \quad \lambda_2 = \frac{2}{5}. \] Therefore, we take $m = 10 \in (\frac{9}{8}, +\infty)$, $M = \frac{1}{5} \in (0, \frac{2}{5})$ and let \[ \theta_k = t^{\ast} - \frac{1}{2}\Big(\sum_{i=1}^{k+1}\frac{1}{2(i+1)^4} + \sum_{i=1}^{k}\frac{1}{2(i+1)^4}\Big) \in (0, \frac{15}{32}) \] For $R_k = \frac{1}{800^k}$ and $r_k = \frac{1}{300\times 800^k}$, $k= 1, 2,\dots$ we have \[ \frac{1}{800^{k+1}} < \frac{\theta_k}{300\times 800^k } < \frac{1}{300 \times 800^k} < \frac{m}{300\times 800^k} < \frac{1}{800^k}. \] After some simple calculation we have \begin{gather*} f(u) \geq \varphi(mr_k) = m^2r_{k}^2 \quad \text{for } u \in [\mu_kr_k, r_k]; \\ f(u) \leq \varphi(MR_k) = M^2R_{k}^2 \quad \text{for}\ \ u \in [0, R_k]. \end{gather*} Then by Theorem \ref{thm4.1}, the boundary-value problem (\ref{e5.1}) and (\ref{e5.2}) has infinitely many solutions $\{u_k\}_{k = 1}^\infty$ such that \[ \frac{1}{300\times 800^k} \leq \|u_k\| \leq \frac{1}{800^k},\quad k = 1, 2,\dots. \] \end{example} \begin{remmark}\label{rem5.3} \rm From the Example \ref{exam5.2}, we can see that $\varphi$ is not odd, then the boundary value problem with $p$-Laplacian operator \cite{h1,z1} do not apply to Example \ref{exam5.2}. 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