\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 34, 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 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/34\hfil Multi-point problems on time scales] {Positive solutions of nonlinear m-point boundary-value problem for p-Laplacian dynamic equations on time scales} \author[Y. Sang, H. Xi\hfil EJDE-2007/34\hfilneg] {Yanbin Sang, Huiling Xi} \address{Yanbin Sang \newline Department of Mathematics, North University of China, Taiyuan 030051, Shanxi, China} \email{syb6662004@163.com} \address{Huiling Xi \newline Department of Mathematics, North University of China, Taiyuan 030051, Shanxi, China} \email{cxhhhl@126.com} \thanks{Submitted May 15, 2006. Published February 27, 2007.} \subjclass[2000]{34B18} \keywords{Time scale; three-point boundary-value problem; cone; fixed point;\hfill\break\indent positive solution} \begin{abstract} In this paper, we study the existence of positive solutions to nonlinear $m$-point boundary-value problems for a $p$-Laplacian dynamic equation on time scales. We use fixed point theorems in cones and obtain criteria that generalize and improve known results. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \section{Introduction} 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,k1,l2,s1} and references therein. However, there are not many results concerning the $p$-Laplacian problems 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^{\triangle\nabla}(t)+a(t)f(u(t))=0, \quad t\in(0, T), \label{e1.1}\\ u(0)=0, \quad \alpha u(\eta)=u(T).\label{e1.2} \end{gather} He obtained some results for the existence of one positive solution of the problem \eqref{e1.1} and \eqref{e1.2} based on the limits $f_{0}=\lim_{u\to 0^{+}}\frac{f(u)}{u}$ and $f_{\infty}=\lim_{u\to \infty}\frac{f(u)}{u}$ as well as existence of at least three positive solutions. Kaufmann \cite{k1} studied the problem \eqref{e1.1} and \eqref{e1.2} and obtained existence results of finitely many positive solutions and countably many positive solutions. Sun and Li \cite{s1} considered the existence of positive solutions of the following dynamic equations on time scales \begin{gather} u^{\triangle\nabla}(t)+a(t)f(t, u(t))=0, \quad t\in(0, T), \label{e1.3}\\ \beta u(0)-\gamma u^{\triangle}(0)=0,\quad \alpha u(\eta)=u(T).\label{e1.4} \end{gather} They obtained the existence of single and multiple positive solutions of the problem \eqref{e1.3} and \eqref{e1.4} by using a fixed point theorem and Leggett-Williams fixed point theorem, respectively. In this paper concerns the existence of positive solutions of the $p$-Laplacian dynamic equations on time scales \begin{gather} (\phi_{p}(u^{\Delta}))^{\nabla}+a(t)f(t, u(t))=0,\quad t\in(0, T),\label{e1.5}\\ \phi_{p}(u^{\Delta}(0))=\sum_{i=1}^{m-2}a_{i}\phi_{p}(u^{\Delta}(\xi_{i})), \quad u(T)=\sum_{i=1}^{m-2}b_{i}u(\xi_{i})\label{e1.6} \end{gather} where $\phi_{p}(s)$ is $p$-Laplacian operator, i.e., $\phi_{p}(s)=|s|^{p-2}s$, $p>1$, $\phi_{p}^{-1}=\phi_{q}$, $\frac{1}{p}+\frac{1}{q}=1$, $0<\xi_{1}<\dots<\xi_{m-2}<\rho(T)$, and $a_{i}$, $b_{i}$, $a$, $f$ satisfy: \begin{itemize} \item[(H1)] $a_{i},\ b_{i}\in [0,+\infty)$ satisfy $0<\sum_{i=1}^{m-2}a_{i}<1$, and $\sum_{i=1}^{m-2}b_{i}<1$, $T\sum_{i=1}^{m-2}b_{i}\geq \sum_{i=1}^{m-2}b_{i}\xi_{i}$; \item[(H2)] $a(t)\in C_{\rm ld}((0, T), [0, +\infty))$ and there exists $t_{0}\in (\xi_{m-2}, T)$, such that $a(t_{0})>0$; \item[(H3)] $f\in C([0, T]\times[0, +\infty), [0, +\infty))$. \end{itemize} We point out that when $\mathbb{T}=\mathbb{R}$ and $p=2$, \eqref{e1.5}, \eqref{e1.6} becomes a boundary-value problem of differential equations and is the problem considered in \cite{m1}. Our main results extend and include the main results of \cite{m1}. The rest of the paper is arranged as follows. We state some basic time scale definitions and prove several preliminary results in Section 2. Section 3 is devoted to the existence of positive solutions of \eqref{e1.5}, \eqref{e1.6}, the main tool being a fixed point theorem for cone-preserving operators. \section{Preliminaries} For convenience, we list the following definitions which can be found in \cite{a1,a3,b1,b2,h1}. \begin{definition} \label{def2.1}\rm A time scale $\mathbb{T}$ is a nonempty closed subset of real numbers $\mathbb{R}$. For $t<\sup\mathbb{T}$ and $r>\inf\mathbb{T}$, define the forward jump operator $\sigma$ and backward jump operator $\rho$, respectively, by \begin{gather*} \sigma(t)=\inf\{\tau\in\mathbb{T}\mid\tau> t\}\in\mathbb{T}, \\ \rho(r)=\sup\{\tau\in\mathbb{T}\mid\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 $$ |f(\sigma(t))-f(s)-f^{\triangle}(t)(\sigma(t)-s)|\leq\epsilon|\sigma(t)-s|, $$ for all $s\in U$. \end{definition} For $f:\mathbb{T}\to \mathbb{R}$ and $t\in\mathbb{T}_{k}$, 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 $u^{\triangle\nabla}(t)\leq0$ on $[0, T]$, then we say $u$ is concave on $[0, T]$. By a positive solution of \eqref{e1.5}, \eqref{e1.6}, we understand a function $u(t)$ which is positive on $(0, T)$, and satisfies \eqref{e1.5}, \eqref{e1.6}. To prove the main results in this paper, we will employ several lemmas. These lemmas are based on the linear boundary-value problem \begin{gather} (\phi_{p}(u^{\Delta}))^{\nabla}+h(t)=0,\ \ t\in(0, T),\label{e2.1}\\ \phi_{p}(u^{\Delta}(0))=\sum_{i=1}^{m-2}a_{i}\phi_{p}(u^{\Delta}(\xi_{i})), \ \ u(T)=\sum_{i=1}^{m-2}b_{i}u(\xi_{i})\label{e2.2} \end{gather} \begin{lemma} \label{lem2.1} For $h\in C_{\rm ld}[0,T]$ the BVP \eqref{e2.1}--\eqref{e2.2} has the unique solution \begin{equation} u(t)=-\int_{0}^{t}\phi_{q}\Big(\int_{0}^{s}h(\tau)\nabla\tau-A\Big) \Delta s+B, \label{e2.3} \end{equation} where \begin{gather*} A=-\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}h(\tau)\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}},\\ B=\frac{\int_{0}^{T}\phi_{q}\Big(\int_{0}^{s}h(\tau)\nabla \tau-A\Big)\Delta s -\sum_{i=1}^{m-2}b_{i}\int_{0}^{\xi_{i}}\phi_{q} \Big(\int_{0}^{s}h(\tau)\nabla\tau-A\Big)\Delta s}{1-\sum_{i=1}^{m-2}b_{i}} \end{gather*} \end{lemma} \begin{proof} Let $u$ be as in \eqref{e2.3}. By \cite[Theorem 2.10(iii)]{a3}, taking the delta derivative of \eqref{e2.3}, we have $$ u^{\Delta}(t)=-\phi_{q}\Big(\int_{0}^{t}h(\tau)\nabla\tau-A\Big), $$ moreover, we get $$ \phi_{p}(u^{\Delta})=-\Big(\int_{0}^{t}h(\tau)\nabla \tau-A\Big), $$ taking the nabla derivative of this expression yields $(\phi_{p}(u^{\Delta}))^{\nabla}=-h(t)$. And routine calculation verify that $u$ satisfies the boundary value conditions in \eqref{e2.2}, So that $u$ given in \eqref{e2.3} is a solution of \eqref{e2.1} and \eqref{e2.2}. It is easy to see that the BVP $$ (\phi_{p}(u^{\Delta}))^{\nabla}=0,\quad \phi_{p}(u^{\Delta}(0))=\sum_{i=1}^{m-2}a_{i}\phi_{p}(u^{\Delta}(\xi_{i})), \quad u(T)=\sum_{i=1}^{m-2}b_{i}u(\xi_{i}) $$ has only the trivial solution. Thus $u$ in \eqref{e2.3} is the unique solution of \eqref{e2.1},\ \eqref{e2.2}. The proof is complete. \end{proof} \begin{lemma} \label{lem2.2} Assume (H1) holds, For $h\in C_{\rm ld}[0,T]$ and $h\geq 0$, then the unique solution $u$ of \eqref{e2.1}--\eqref{e2.2} satisfies $u(t)\geq 0$, for $t\in [0,T]$. \end{lemma} \begin{proof} Let $$ \varphi_{0}(s)=\phi_{q}\Big(\int_{0}^{s}h(\tau)\nabla\tau-A\Big). $$ Since $$ \int_{0}^{s}h(\tau)\nabla \tau-A =\int_{0}^{s}h(\tau)\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}h(\tau)\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}} \geq 0, $$ it follows that $\varphi_{0}(s)\geq 0$. According to Lemma \ref{lem2.1}, we get \begin{align*} u(0)&=B\\ &=\frac{\int_{0}^{T}\varphi_{0}(s)\Delta s -\sum_{i=1}^{m-2}b_{i}\int_{0}^{\xi_{i}}\varphi_{0}(s)\Delta s}{1-\sum_{i=1}^{m-2}b_{i}}\\ &=\frac{\int_{0}^{T}\varphi_{0}(s)\Delta s -\sum_{i=1}^{m-2}b_{i}\Big(\int_{0}^{T}\varphi_{0}(s)\Delta s-\int_{\xi_{i}}^{T}\varphi_{0}(s)\Delta s\Big)}{1-\sum_{i=1}^{m-2}b_{i}}\\ &= \int_{0}^{T}\varphi_{0}(s)\Delta s+\frac{\sum_{i=1}^{m-2}b_{i}\int_{\xi_{i}}^{T}\varphi_{0}(s)\Delta s}{1-\sum_{i=1}^{m-2}b_{i}} \geq 0. \end{align*} and \begin{align*} u(T)&=-\int_{0}^{T}\varphi_{0}(s)\Delta s+B\\ &= -\int_{0}^{T}\varphi_{0}(s)\Delta s+\frac{\int_{0}^{T}\varphi_{0}(s)\Delta s-\sum_{i=1}^{m-2}b_{i}\int_{0}^{\xi_{i}}\varphi_{0}(s)\Delta s}{1-\sum_{i=1}^{m-2}b_{i}}\\ &=\frac{\sum_{i=1}^{m-2}b_{i}\int_{\xi_{i}}^{T}\varphi_{0}(s)\Delta s}{1-\sum_{i=1}^{m-2}b_{i}}\geq 0. \end{align*} If $t\in (0,T)$, we have \begin{align*} u(t) &=-\int_{0}^{t}\varphi_{0}(s)\Delta s + \frac{1}{1-\sum_{i=1}^{m-2}b_{i}}\Big[ \int_{0}^{T}\varphi_{0}(s)\Delta s -\sum_{i=1}^{m-2}b_{i}\int_{0}^{\xi_{i}}\varphi_{0}(s)\Delta s\Big]\\ &\geq -\int_{0}^{T}\varphi_{0}(s)\Delta s + \frac{1}{1-\sum_{i=1}^{m-2}b_{i}}\Big[ \int_{0}^{T}\varphi_{0}(s)\Delta s -\sum_{i=1}^{m-2}b_{i}\int_{0}^{\xi_{i}}\varphi_{0}(s)\Delta s\Big]\\ &=\frac{1}{1-\sum_{i=1}^{m-2}b_{i}} \Big[ -\big(1-\sum_{i=1}^{m-2}b_{i}\big)\int_{0}^{T}\varphi_{0}(s)\Delta s+\int_{0}^{T}\varphi_{0}(s)\Delta s\\ &\quad -\sum_{i=1}^{m-2}b_{i}\int_{0}^{\xi_{i}}\varphi_{0}(s)\Delta s\Big]\\ &=\frac{1}{1-\sum_{i=1}^{m-2}b_{i}} \sum_{i=1}^{m-2}b_{i}\int_{\xi_{i}}^{T}\varphi_{0}(s)\Delta s\geq 0. \end{align*} So $u(t)\geq 0$, $t\in [0,T]$. The proof is complete. \end{proof} \begin{lemma} \label{lem2.3} Assume (H1) holds, if $h\in C_{\rm ld}[0,T]$ and $h\geq 0$, then the unique solution $u$ of \eqref{e2.1}--\eqref{e2.2} satisfies $$ \inf_{t\in [0,T]}u(t)\geq \gamma \| u\|, $$ where $$ \gamma=\frac{\sum_{i=1}^{m-2}b_{i}(T-\xi_{i})}{T-\sum_{i=1}^{m-2}b_{i}\xi_{i}}, \quad \|u\|=\sup_{t\in [0,T]}|u(t)|. $$ \end{lemma} \begin{proof} It is easy to check that $u^{\Delta}(t)=-\varphi(t)\leq 0$, this implies $$ \| u \|=u(0),\quad \min_{t\in [0,T]}u(t)=u(T). $$ It is easy to see that $u^{\Delta}(t_{2})\leq u^{\Delta}(t_{1})$ for any $t_{1}, t_{2}\in [0,T]$ with $t_{1}\leq t_{2}$. Hence $u^{\Delta}(t)$ is a decreasing function on $[0,T]$. This means that the graph of $u^{\Delta}(t)$ is concave down on $(0,T)$. For each $i\in \{1, 2, \dots, m-2\}$, we have $$ \frac{u(T)-u(0)}{T-0}\geq \frac{u(T)-u(\xi_{i})}{T-\xi_{i}}, $$ i.e., $Tu(\xi_{i})-\xi_{i}u(T)\geq (T-\xi_{i})u(0)$, so that $$ T\sum_{i=1}^{m-2}b_{i}u(\xi_{i})- \sum_{i=1}^{m-2}b_{i}\xi_{i}u(T) \geq \sum_{i=1}^{m-2}b_{i}(T-\xi_{i})u(0). $$ With the boundary condition $u(T)=\sum_{i=1}^{m-2}b_{i}u(\xi_{i})$, we have $$ u(T)\geq \frac{\sum_{i=1}^{m-2}b_{i}(T-\xi_{i})}{T-\sum_{i=1}^{m-2}b_{i} \xi_{i}}u(0). $$ This completes the proof. \end{proof} Let the norm on $C_{\rm ld}[0,T]$ be the sup norm. Then $C_{\rm ld}[0,T]$ is a Banach space. It is easy to see that \eqref{e1.5}--\eqref{e1.6} has a solution $u=u(t)$ if and only if $u$ is a fixed point of the operator \begin{equation} (Au)(t)=-\int_{0}^{t}\phi_{q} \Big(\int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla\tau-\tilde{A}\Big)\Delta s+\tilde{B},\label{e2.4} \end{equation} where \begin{gather*} \tilde{A}=-\frac{\sum_{i=1}^{m-2}a_{i} \int_{0}^{\xi_{i}}a(\tau)f(\tau,u(\tau))\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}, \\ \begin{aligned} \tilde{B}&=\Big[\int_{0}^{T}\phi_{q} \Big(\int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla \tau -\tilde{A}\Big)\Delta s \\ &\quad -\sum_{i=1}^{m-2}b_{i} \int_{0}^{\xi_{i}}\phi_{q} \Big(\int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla\tau-\tilde{A}\Big)\Delta s \Big]\frac{1}{1-\sum_{i=1}^{m-2}b_{i}}. \end{aligned} \end{gather*} Denote $$ K=\big\{u: u\in C_{\rm ld}[0, T], u(t)\geq 0, \inf_{t\in [0,T]} u(t) \geq\gamma \| u\| \big\}, $$ where $\gamma$ is the same as in Lemma \ref{lem2.3}. It is obvious that $K$ is a cone in $C_{\rm ld}[0, T]$. By Lemma \ref{lem2.3}, $A(K)\subset K$. It is easy to see that $A:K\to K$ is completely continuous. \begin{lemma} \label{lem2.4} Let $$\varphi(s)=\phi_{q}\Big(\int_{0}^{s} a(\tau)f(\tau,u(\tau))\nabla \tau-\tilde{A}\Big). $$ For $\xi_{i},(i=1,\dots, m-2)$, then $$ \int_{0}^{\xi_{i}}\varphi(s)\Delta s \leq \frac{\xi_{i}}{T}\int_{0}^{T}\varphi(s)\Delta s . $$ \end{lemma} \begin{proof} Since $$ \int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla\tau-\tilde{A} =\int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)f(\tau,u(\tau))\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}} $$ which greater than or equal to zero, we have $\varphi(s)\geq 0$. Now, for all $t\in (0,T]$, we have $$ \Big(\frac{\int_{0}^{t}\varphi(s)\Delta s}{t}\Big)^{\Delta} =\frac{t\varphi(t)-\int_{0}^{t}\varphi(s)\Delta s}{t\sigma(t)} \geq 0. $$ In fact, Let $\psi(t)=t\varphi(t)-\int_{0}^{t}\varphi(s)\Delta s$, taking the delta derivative of the above expression, we have $$ \psi^{\Delta}(t)=t\varphi^{\Delta}(t)\geq 0. $$ Hence, $\psi(t)$ is a nondecreasing function on $[0,T]$. i.e. $\psi(t)\geq 0$. For all $t\in (0,T]$, \begin{equation} \frac{\int_{0}^{t}\varphi(s)\Delta s}{t} \leq\frac{\int_{0}^{T}\varphi(s)\Delta s}{T}. \label{e2.5} \end{equation} By \eqref{e2.4}, for $\xi_{i},(i=1,\dots, m-2)$, we have $$ \int_{0}^{\xi_{i}}\varphi(s)\Delta s \leq \frac{\xi_{i}}{T}\int_{0}^{T}\varphi(s)\Delta s . $$ The proof is complete. \end{proof} The following well-known result of the fixed point theorems is needed in our arguments. \begin{lemma}[\cite{g1}] \label{lem2.5} Let $K$ be a cone in a Banach space $X$. Let $D$ be an open bounded subset of $X$ with $D_{K}=D\cap K\neq \phi$ and $\overline{D_{K}}\neq K$. Assume that $A:\overline{D_{K}}\to K$ is a compact map such that $x\neq AK$ for $x\in \partial D_{K}$. Then the following results hold: \begin{enumerate} \item If $\|Ax\|\leq\|x\|$ for $x\in \partial D_{K}$, then $i(A, D_{K}, K)=1$; \item If there exists $x_{0}\in K\backslash\{\theta\}$ such that $x\neq Ax+\lambda x_{0}$, for all $x\in \partial D_{K}$ and all $x>0$, then $i(A, D_{K}, K)=0$; \item Let $U$ be an open set in $X$ such that $\overline{U}\subset D_{K}$. If $i(A, U, K)=1$ and $i(A, D_{K}, K)=0$, then $A$ has a fixed point in $D_{K}\backslash\overline{U}_{K}$. The same results holds, if $i(A, U, K)=0$ and $i(A, D_{K}, K)=1$. \end{enumerate} \end{lemma} We define $$ K_{\rho}=\{u(t)\in K:\|u\|<\rho\},\quad \Omega_{\rho}=\{u(t)\in K:\min_{\xi_{m-2}\leq t\leq T}u(t)<\gamma\rho\}. $$ \begin{lemma}[\cite{l1}] \label{lem2.6} The set $\Omega_{\rho}$ defined above has the following properties: \begin{itemize} \item[(a)] $K_{\gamma\rho}\subset \Omega_{\rho}\subset K_{\rho}$; \item[(b)] $\Omega_{\rho}$ is open relative to K; \item[(c)] $X\in\partial \Omega_{\rho}$ if and only if $\min_{\xi_{m-2}\leq t\leq T}x(t)=\gamma\rho$; \item[(d)] If $x\in\partial \Omega_{\rho}$, then $\gamma\rho\leq x(t)\leq \rho$ for $t\in [\xi_{m-2}, T]$. \end{itemize} \end{lemma} For our convenience, we introduce the following notation: \begin{gather} f_{\gamma\rho}^{\rho}=\min\big\{\min_{\xi_{m-2}\leq t\leq T}\frac{f(t, u)}{\phi_{p}(\rho)}:u\in[\gamma\rho, \rho]\big\}, \nonumber \\ f_{0}^{\rho}=\max\big\{\max_{0\leq t\leq T}\frac{f(t, u)}{\phi_{p}(\rho)}:u\in[0, \rho]\big\}, \nonumber \\ f^{\alpha}=\lim_{u\to \alpha}\sup\max_{0\leq t\leq T}\frac{f(t, u)}{\phi_{p}(u)}, \quad f_{\alpha}=\lim_{u\to \alpha}\inf\max_{\xi_{m-2}\leq t\leq T}\frac{f(t, u)}{\phi_{p}(u)},\quad (\alpha:=\infty\text{ or } 0^{+}), \nonumber \\ m=\Big\{\frac{1}{1-\sum_{i=1}^{m-2}b_{i}} \int_{0}^{T}\phi_{q}\Big[\int_{0}^{s}a(\tau)\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}\Big]\Delta s\Big\}^{-1}, \label{e2.6} \\ M=\Big\{\frac{T\sum_{i=1}^{m-2}b_{i}- \sum_{i=1}^{m-2}b_{i}\xi_{i}}{T(1-\sum_{i=1}^{m-2}b_{i})} \int_{0}^{T}\!\phi_{q}\Big[\int_{0}^{s}a(\tau)\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}\Big]\Delta s\Big\}^{-1} \label{e2.7} \end{gather} \begin{lemma} \label{lem2.7} If $f$ satisfies the conditions \begin{equation} f^{\rho}_{0}\leq \phi_{p}(m)\quad\text{and}\quad u\neq Au \label{e2.8} \end{equation} for $u\in\partial K_{\rho}$, then $i(A, K_{\rho},K)=1$. \end{lemma} \begin{proof} By \eqref{e2.6} and \eqref{e2.8}, for all $u\in\partial K_{\rho}$, we have \begin{align*} &\int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla\tau-\tilde{A}\\ &=\int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)f(\tau,u(\tau))\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}\\ &\leq \Phi_{p}(\rho)\phi_{p}(m)\Big[\int_{0}^{s}a(\tau)\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}\Big],\\ \end{align*} so that \begin{align*} \varphi(s) &=\Phi_{q}\Big(\int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla\tau-\tilde{A}\Big) \\ &\leq \rho m\Phi_{q}\Big[\int_{0}^{s}a(\tau)\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}\Big]. \end{align*} Therefore, by \eqref{e2.4}, we have \begin{align*} \| Au\| &\leq\tilde{B}=\frac{1}{1-\sum_{i=1}^{m-2}b_{i}} \Big(\int_{0}^{T}\varphi(s)\Delta s -\sum_{i=1}^{m-2}b_{i} \int_{0}^{\xi_{i}}\varphi(s)\Delta s\Big)\\ &\leq \frac{1}{1-\sum_{i=1}^{m-2}b_{i}} \int_{0}^{T}\varphi(s)\Delta s \\ &\leq \rho m \frac{1}{1-\sum_{i=1}^{m-2}b_{i}} \int_{0}^{T}\phi_{q}\Big[\int_{0}^{s}a(\tau)\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}\Big]\Delta s\\ &=\rho=\|u\|. \end{align*} This implies $\|Au\|\leq\|u\|$ for $u\in\partial\ K_{\rho}$. By Lemma \ref{lem2.5}(1), we have $i(A, K_{\rho}, K)=1$. \end{proof} \begin{lemma} \label{lem2.8} If $f$ satisfies the conditions \begin{equation} f^{\rho}_{\gamma \rho}\geq \Phi_{p}(M \gamma)\quad\text{and}\quad u\neq Au \label{e2.9} \end{equation} for $u\in\partial \Omega_{\rho}$, then $i(A, \Omega_{\rho}, K)=0$. \end{lemma} \begin{proof} Let $e(t)\equiv 1$, for $t \in [0, T]$; then $e\in\partial K_{1}$. We claim that $u\neq Au+\lambda e$ for $u\in\partial\ \Omega_{\rho}$, and $\lambda >0$. In fact, if not, there exist $u_{0}\in \partial\Omega$, and $\lambda_{0}>0$ such that $u_{0} =Au_{0}+\lambda_{0}e$. By \eqref{e2.7} and \eqref{e2.9}, for $t\in [0,T]$, we have \begin{align*} &\int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla\tau-\tilde{A}\\ &=\int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)f(\tau,u(\tau))\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}\\ &\geq \Phi_{p}(\rho)\phi_{p}(M\gamma)\Big[\int_{0}^{s}a(\tau)\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}\Big],\\ \end{align*} so that \begin{align*} \varphi(s) &=\Phi_{q}\Big(\int_{0}^{s}a(\tau)f(\tau,u(\tau))\nabla\tau-\tilde{A}\Big)\\ &\geq \rho M\gamma\Phi_{q}\Big[\int_{0}^{s}a(\tau)\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}\Big]. \end{align*} Applying \eqref{e2.4} and Lemma \ref{lem2.4}, it follows that \begin{align*} u_{0}(t) &=Au_{0}(t)+\lambda_{0}e(t)\\ &\geq -\int_{0}^{T}\varphi(s)\Delta s + \frac{1}{1-\sum_{i=1}^{m-2}b_{i}}\Big(\int_{0}^{T}\varphi(s)\Delta s-\sum_{i=1}^{m-2}b_{i}\int_{0}^{\xi_{i}}\varphi(s)\Delta s\Big)+\lambda_{0}\\ &= \frac{\sum_{i=1}^{m-2}b_{i}}{1-\sum_{i=1}^{m-2}b_{i}} \int_{0}^{T}\varphi(s)\Delta s -\frac{\sum_{i=1}^{m-2}b_{i}\int_{0}^{\xi_{i}}\varphi(s)\Delta s}{1-\sum_{i=1}^{m-2}b_{i}}+\lambda_{0} \\ &\geq \frac{\sum_{i=1}^{m-2}b_{i}}{1-\sum_{i=1}^{m-2}b_{i}} \int_{0}^{T}\varphi(s)\Delta s- \frac{\sum_{i=1}^{m-2}b_{i}\xi_{i}}{T(1-\sum_{i=1}^{m-2}b_{i})} \int_{0}^{T}\varphi(s)\Delta s +\lambda_{0}\\ &= \frac{T\sum_{i=1}^{m-2}b_{i}-\sum_{i=1}^{m-2}b_{i}\xi_{i}} {T(1-\sum_{i=1}^{m-2}b_{i})}\int_{0}^{T}\varphi(s)\Delta s+\lambda_{0} \\ &\geq \gamma \rho M \frac{T\sum_{i=1}^{m-2}b_{i}- \sum_{i=1}^{m-2}b_{i}\xi_{i}}{T(1-\sum_{i=1}^{m-2}b_{i})}\\ &\quad\times \int_{0}^{T}\phi_{q}\Big[\int_{0}^{s}a(\tau)\nabla \tau +\frac{\sum_{i=1}^{m-2}a_{i}\int_{0}^{\xi_{i}}a(\tau)\nabla \tau}{1-\sum_{i=1}^{m-2}a_{i}}\Big]\Delta s +\lambda_{0} \\ &=\gamma \rho+\lambda_{0} \end{align*} This implies $\gamma \rho\geq \gamma \rho+\lambda_{0}$, a contradiction. Hence, by Lemma \ref{lem2.5} (2), it follows that $i(A, \Omega_{\rho}, K)=0$. \end{proof} \section{Existence of Positive Solutions} We now present our results on the existence of positive solutions for \eqref{e1.5}--\eqref{e1.6} under the assumptions: \begin{itemize} \item[(H4)] There exist $\rho_{1}, \rho_{2} \in (0, +\infty)$ with $\rho_{1}<\gamma \rho_{2}$ such that $$ f^{\rho_{1}}_{0}\leq \phi_{p}(m), f^{\rho_{2}}_{\gamma \rho_{2}}\geq \phi_{p}(M \gamma); $$ \item[(H5)] There exist $\rho_{1}, \rho_{2} \in (0, +\infty)$ with $\rho_{1}<\rho_{2}$ such that $$ f^{\rho_{2}}_{0}\leq \phi_{p}(m), f^{\rho_{1}}_{\gamma \rho_{1}}\geq \phi_{p}(M \gamma). $$ \end{itemize} \begin{theorem} \label{thm3.1} Assume that (H1)--(H3) and either (H4) or (H5) hold. Then \eqref{e1.5}--\eqref{e1.6} has a positive solution. \end{theorem} \begin{proof} Assume that (H4) holds. We show that $A$ has a fixed point $u_{1}$ in $\Omega_{\rho_{2}}\backslash \overline{K}_{\rho_{1}}$. By Lemma \ref{lem2.7}, we have $$ i(A, K_{\rho_{1}}, K)=1. $$ By Lemma \ref{lem2.8}, we have $$i(A, K_{\rho_{2}}, K)=0.$$ By Lemma \ref{lem2.6} (a) and $\rho_{1}<\gamma \rho_{2}$, we have $\overline{K}_{\rho_{1}}\subset K_{\gamma \rho_{2}}\subset \Omega_{\rho_{2}}$. It follows from Lemma \ref{lem2.5}(3) that $A$ has a fixed point $u_{1}$ in $\Omega_{\rho_{2}}\backslash \overline{K}_{\rho_{1}}$, The proof is similar when $H_{5}$ holds, and we omit it here. The proof is complete. \end{proof} As a special case of Theorem \ref{thm3.1}, we obtain the following result, under assumptions \begin{itemize} \item[(H6)] $0\leq f^{0}<\phi_{p}(m)$ and $\phi_{p}(M)< f_{\infty}\leq\infty$; \item[(H7)] $0\leq f^{\infty}<\phi_{p}(m)$ and $\phi_{p}(M)< f_{0}\leq\infty$. \end{itemize} \begin{corollary} \label{coro3.2} Assume that (H1)--(H3) and either (H6) or (H7) hold. Then \eqref{e1.5}--\eqref{e1.6} has a positive solution. \end{corollary} For the next result we use the following assumptions: \begin{itemize} \item[(H8)] There exist $\rho_{1}, \rho_{2}, \rho_{3} \in (0, +\infty)$ with $\rho_{1}< \gamma \rho_{2}$ and $\rho_{2}< \rho_{3}$ such that $$ f^{\rho_{1}}_{0}\leq \phi_{p}(m), f^{\rho_{2}}_{\gamma \rho_{2}}\geq \phi_{p}(M \gamma), u\neq Au, \forall\ u\in \partial \Omega_{\rho_{2}}\quad\text{and}\quad f^{\rho_{3}}_{0}\leq \phi_{p}(m); $$ \item[(H9)] There exist $\rho_{1}, \rho_{2}, \rho_{3} \in (0, +\infty)$ with $\rho_{1}< \rho_{2}< \gamma \rho_{3}$ such that $$ f^{\rho_{2}}_{0}\leq \phi_{p}(m), f^{\rho_{1}}_{\gamma \rho_{1}}\geq \phi_{p}(M \gamma), u\neq Au, \forall\ u\in \partial K_{\rho_{2}}, \quad\text{and}\quad f^{\rho_{3}}_{\gamma \rho_{3}}\geq \phi_{p}(M \gamma). $$ \end{itemize} \begin{theorem} \label{thm3.2} Assume that (H1)--(H3) and either (H8) or (H9) hold. Then \eqref{e1.5}--\eqref{e1.6} has two positive solutions. Moreover, if in (H8), $f_{0}^{\rho_{1}}\leq\phi_{p}(m)$ is replaced by $f^{\rho_{1}}_{0}<\phi_{p}(m)$, then \eqref{e1.5}--\eqref{e1.6} has a third positive solution $u_{3}\in K_{\rho_{1}}$. \end{theorem} \begin{proof} Assume that (H8) holds. We show that either $A$ has a fixed point $u_{1}$ in $\partial K_{\rho_{1}}$ or in $\Omega_{\rho_{2}}\backslash \overline{K}_{\rho_{1}}$. If $u\neq Au$ for $u\in\partial K_{\rho_{1}}\cup\partial K_{\rho_{3}}$, then by Lemmas \ref{lem2.7} and \ref{lem2.8}, we have $$ i(A, K_{\rho_{1}}, K)=1,\quad i(A, K_{\rho_{3}}, K)=1, \quad i(A, K_{\rho_{2}}, K)=0. $$ By Lemma \ref{lem2.6} (a) and $\rho_{1}<\gamma\rho_{2}$, we have $\overline{K}_{\rho_{1}}\subset K_{\gamma \rho_{2}}\subset \Omega_{\rho_{2}}$. It follows from Lemma \ref{lem2.5} (3) that $A$ has a fixed point $u_{1}$ in $\Omega_{\rho_{2}}\backslash \overline{K}_{\rho_{1}}$. Similarly, $A$ has a fixed point in $K_{\rho_{3}}\backslash \overline{\Omega}_{\rho_{2}}$. The proof is similar when (H9) holds and we omit it here. The proof is complete. \end{proof} As a special case of Theorem \ref{thm3.2}, we obtain the following result, using the assumptions: \begin{itemize} \item[(H10)] $0 \leq f^{0}< \phi_{p}(m)$, $f^{\rho}_{\gamma \rho}\geq \phi_{p}(M \gamma)$, $u\neq Au$, for all $u\in \partial \Omega_{\rho}$ and $0 \leq f^{\infty}< \phi_{p}(m)$; \item[(H11)] $\phi_{p}(m)0$ such that either (H10) or (H11) hold, then \eqref{e1.5}--\eqref{e1.6} has two positive solutions. \end{corollary} Note that when $\mathbb{T}=\mathbb{R}$, $(0, T)=(0, 1)$, and $p=2$, Theorems \ref{thm3.1} and \ref{thm3.2} here improve \cite[Theorem 3.1]{m1}. \begin{thebibliography}{0} \bibitem{a1} R. P. Agarwal, D. O'Regan, \emph{Nonlinear boundary value problems on time scales}, Nonl. Anal. 44 (2001), 527-535. \bibitem{a2} D. R. Anderson, \emph{Solutions to second-order three-point problems on time scales}, J. Differ. Equ. Appl. 8 (2002), 673-688. \bibitem{a3} F. M. Atici, G. 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