\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 17, pp. 1--14.\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 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/17\hfil Three-point problems on time scales] {Positive solutions to a generalized second-order three-point boundary-value problem on time scales} \author[H. Luo, Q. Ma\hfil EJDE-2005/17\hfilneg] {Hua Luo, Qiaozhen Ma} % in alphabetical order \address{Hua Luo\hfill\break College of Mathematics and Information Science, Northwest Normal University, \hfill\break Lanzhou 730070, Gansu, China} \email{luohua@nwnu.edu.cn} \address{Qiaozhen Ma \hfill\break College of Mathematics and Information Science, Northwest Normal University, \hfill\break Lanzhou 730070, Gansu, China} \email{maqzh@nwnu.edu.cn} \date{} \thanks{Submitted September 1, 2004. Published February 1, 2005.} \subjclass[2000]{34B18, 39A10} \keywords{Time scales; three-point boundary value problems; cone; fixed points;\hfill\break\indent positive solutions } \begin{abstract} Let $\mathbb{T}$ be a time scale with $0,T \in \mathbb{T}$. We investigate the existence and multiplicity of positive solutions to the nonlinear second-order three-point boundary-value problem \begin{gather*} u^{\Delta\nabla}(t)+a(t)f(u(t))=0,\quad t\in[0, T]\subset \mathbb{T},\\ u(0)=\beta u(\eta),\quad u(T)=\alpha u(\eta) \end{gather*} on time scales $\mathbb{T}$, where $0<\eta0$,\ $\beta\ge0$,\ $\eta\in (0, T)\subset\mathbb{T}$ are given constants. Clearly if $\beta=0$, then \eqref{e1.4} reduces to \eqref{e1.2}. We also point out that when $\mathbb{T}=\mathbb{R},\ \beta=0$, \eqref{e1.3}-\eqref{e1.4} becomes a boundary-value problem of differential equations and just is the problem considered in \cite{MaRy1}; when $\mathbb{T}=\mathbb{Z},\ \beta=0$, \eqref{e1.3}-\eqref{e1.4} becomes a boundary-value problem of difference equations and just is the problem considered in \cite{MaRy2}. We will use Guo-Krasnoselskii's fixed-point theorem and Leggett-Williams fixed-point theorem to investigate the existence and multiplicity of positive solutions for the problem \eqref{e1.3}-\eqref{e1.4}. Our main results extend the main results of Ma\cite{MaRy1}, Anderson\cite{Ande3}, Ma and Raffoul\cite{MaRy2}. \vskip 2mm 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 a positive solution of \eqref{e1.3}-\eqref{e1.4}, the main tool being the Guo-Krasnoselskii's fixed-point theorem. Next in Section 4, we give a multiplicity result by using the Leggett-Williams fixed-point theorem. Finally we give two examples to illustrate our results in Section 5. \section{Preliminaries} For convenience, we list here the following definitions which are needed later. A time scale $\mathbb{T}$ is an arbitrary nonempty closed subset of real numbers $\mathbb{R}$. The operators $\sigma$ and $\rho$ from $\mathbb{T}$ to $\mathbb{T}$, defined by \cite{Hilge}, \begin{gather*} \sigma(t)=\inf\{\tau\in\mathbb{T}:\tau>t\}\in\mathbb{T}, \\ \rho(t)=\sup\{\tau\in\mathbb{T}:\taut$, respectively. Let$f:\mathbb{T}\to \mathbb{R}$and$t\in \mathbb{T}$(assume$t$is not left-scattered if$t=\sup\mathbb{T}$), then the {\it delta derivative of$f$at the point$t$} is defined to be the number$f^{\Delta}(t)$(provided it exists) with the property that for each$\epsilon>0$there is a neighborhood$U$of$t$such that $$|f(\sigma(t))-f(s)-f^{\Delta}(t)(\sigma(t)-s) |\le | \sigma(t)-s|, \quad \text{for all } s\in U.$$ Similarly, for$t\in \mathbb{T}$(assume$t$is not right-scattered if$t=\inf\mathbb{T}$), the {\it nabla derivative of$f$at the point$t$} is defined in \cite{atici} to be 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 $$|f(\rho(t))-f(s)-f^{\nabla}(t)(\rho(t)-s) |\le | \rho(t)-s |, \quad \text{for all } s\in U.$$ A function$f$is {\it 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\cite{Bohne} 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 $$\int^b_a f(t)\nabla t=F(b)-F(a).$$ For the rest of this article,$\mathbb{T}$denotes a time scale with$0, T\in\mathbb{T}$. Also we denote the set of left-dense continuous functions from$[0, T]\subset\mathbb{T}$to$E\subset\mathbb{R}$by$C_{ld}([0, T],\ E)$, which is a Banach space with the maximum norm$\|u\|=\max_{t\in[0, T]}|u(t)|$. We now state and prove several lemmas before stating our main results. \begin{lemma}\label{lemm1} Let$\beta\neq\frac{T-\alpha\eta}{T-\eta}$. Then for$y\in C_{ld}([0, T],\ \mathbb{R}), the problem \begin{gather} u^{\Delta\nabla}(t)+y(t)=0, \quad t\in[0, T]\subset\mathbb{T}, \label{e2.1} \\ u(0)=\beta u(\eta),\quad u(T)=\alpha u(\eta) \label{e2.2} \end{gather} has a unique solution \begin{aligned} u(t)&=-\int_0^t (t-s)y(s)\nabla s+\frac{(\beta-\alpha)t-\beta T} {(T-\alpha\eta)-\beta(T-\eta)} \int_0^\eta (\eta-s)y(s)\nabla s \\ &\quad +\frac{(1-\beta)t+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)y(s)\nabla s. \end{aligned} \label{e2.3} \end{lemma} \begin{proof} From \eqref{e2.1}, we have $$u(t)=u(0)+u^\Delta (0)t-\int_0^t (t-s)y(s)\nabla s\ :=A+Bt- \int_0^t (t-s)y(s)\nabla s.$$ Since \begin{align*} u(0)=A;\\ u(\eta)=A+B\eta-\int_0^\eta (\eta-s)y(s)\nabla s;\\ u(T)=A+BT-\int_0^T (T-s)y(s)\nabla s, \end{align*} by \eqref{e2.2} fromu(0)=\beta u(\eta)$, we have $$(1-\beta)A-B\beta\eta=-\beta\int_0^\eta (\eta-s)y(s)\nabla s;$$ from$u(T)=\alpha u(\eta), we have $$(1-\alpha)A+B(T-\alpha\eta)=\int_0^T (T-s)y(s)\nabla s-\alpha \int_0^\eta (\eta-s)y(s)\nabla s.$$ Therefore, \begin{align*} A&=\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)y(s)\nabla s \\ &\quad -\frac{\beta T}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^\eta (\eta-s)y(s)\nabla s;\\ B&=\frac{1-\beta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)y(s)\nabla s\\ &\quad -\frac{\alpha-\beta} {(T-\alpha\eta)- \beta(T-\eta)} \int_0^\eta (\eta-s)y(s)\nabla s\,, \end{align*} from which it follows that \begin{align*} u(t)=&\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)y(s)\nabla s\\ &\quad -\frac{\beta T}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^\eta (\eta-s)y(s)\nabla s\\ &\quad +\frac{(1-\beta)t}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)y(s)\nabla s\\ &\quad -\frac{(\alpha-\beta)t} {(T-\alpha\eta)- \beta(T-\eta)} \int_0^\eta (\eta-s)y(s)\nabla s -\int_0^t (t-s)y(s)\nabla s\\ &=-\int_0^t (t-s)y(s)\nabla s+\frac{(\beta-\alpha)t-\beta T} {(T-\alpha\eta)-\beta(T-\eta)} \int_0^\eta (\eta-s)y(s)\nabla s \\ &\quad +\frac{(1-\beta)t+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)y(s)\nabla s. \end{align*} The functionu$presented above is a solution to the problem \eqref{e2.1}-\eqref{e2.2}, and the uniqueness of$u$is obvious. \end{proof} \begin{lemma}\label{lemm2} Let$0<\alpha<\frac{T}{\eta}$,\$0\le \beta<\frac{T-\alpha\eta}{T-\eta}$. If$y\in C_{ld}([0, T],\ [0, \infty))$, then the unique solution$u$of the problem \eqref{e2.1}-\eqref{e2.2} satisfies $$u(t)\ge 0,\quad t\in[0, T]\subset\mathbb{T}.$$ \end{lemma} \begin{proof} It is known that the graph of$u$is concave down on$[0, T]$from$u^{\Delta\nabla}(t)=-y(t)\le 0$, so $$\frac{u(\eta)-u(0)}{\eta}\ge\frac{u(T)-u(0)}{T}.$$ Combining this with \eqref{e2.2}, we have $$\frac{1-\beta}{\eta} u(\eta)\ge\frac{\alpha-\beta}{T} u(\eta).$$ If$u(0)<0$, then$u(\eta)<0$. It implies that$\beta\ge\frac{T-\alpha\eta}{T-\eta}$, a contradiction to$\beta<\frac{T-\alpha\eta}{T-\eta}$. If$u(T)<0$, then$u(\eta)<0$, and the same contradiction emerges. Thus, it is true that$u(0)\ge 0$,\$u(T)\ge 0$, together with the concavity of$u$, we have $$u(t)\ge 0,\quad t\in[0, T]\subset\mathbb{T}.$$ as required. \end{proof} \begin{lemma}\label{lemm3} Let$\alpha\eta\neq T$,$\beta>\max\{\frac{T-\alpha\eta}{T-\eta},0\}$. If$y\in C_{ld}([0, T], [0, \infty))$, then problem \eqref{e2.1}-\eqref{e2.2} has no nonnegative solutions. \end{lemma} \begin{proof} Suppose that problem \eqref{e2.1}-\eqref{e2.2} has a nonnegative solution$u$satisfying$u(t)\ge 0, t\in[0, T]$and there is a$t_0\in(0, T)$such that$u(t_0)>0$. If$u(T)>0$, then$u(\eta)>0$. It implies $$u(0)=\beta u(\eta)>\frac{T-\alpha\eta}{T-\eta} u(\eta)=\frac {Tu(\eta)-\eta u(T)}{T-\eta},$$ that is $$\frac{u(T)-u(0)}{T}>\frac{u(\eta)-u(0)}{\eta},$$ which is a contradiction to the concavity of$u$. If$u(T)=0$, then$u(\eta)=0$. When$t_0\in (0, \eta)$, we get$u(t_0)>u(\eta)=u(T)$, a violation of the concavity of$u$. When$t_0\in (\eta, T)$, we get$u(0)=\beta u(\eta)=0=u(\eta)T$. \end{remark} \begin{lemma}\label{lemm4} Let$0<\alpha<\frac{T}{\eta}$,$0<\beta<\frac{T-\alpha\eta}{T-\eta}$. If$y\in C_{ld}([0, T],[0, \infty))$, then the unique solution to the problem \eqref{e2.1}-\eqref{e2.2} satisfies $$\min_{t\in [0, T]} u(t)\ge \gamma \|u\|, \label{e2.4}$$ where $$\gamma:=\min\Big\{\frac{\alpha(T-\eta)}{T-\alpha\eta},\; \frac{\alpha\eta}{T},\; \frac{\beta(T-\eta)}{T},\; \frac{\beta\eta}{T}\Big\}. \label{e2.5}$$ \end{lemma} \begin{proof} It is known that the graph of$u$is concave down on$[0, T]$from$u^{\Delta\nabla}(t)=-y(t)\le 0$. We divide the proof into two cases. \noindent Case 1.\;$0<\alpha<1$, then$\frac{T-\alpha\eta}{T-\eta}>\alpha$. For$u(0)=\beta u(\eta)=\frac \beta\alpha u(T)$, it may develop in the following two possible directions. \\ (i)\;$0<\alpha\le\beta$. It implies that$u(0)\ge u(T)$, so $$\min_{t\in[0, T]} u(t)=u(T).$$ Assume$\|u\|=u(t_1)$,\$t_1\in [0, T)$, then either$0\le t_1\le\eta<\rho(T)$, or$0<\eta\frac{\alpha\eta}{T} u(t_1), $$so that, \min_{t\in[0, T]} u(t)\ge\frac{\alpha\eta}{T} \|u\|. \noindent(ii)\; 0<\beta<\alpha. It implies that u(0)\le u(T), so$$ \min_{t\in[0, T]} u(t)=u(0). $$Assume \|u\|=u(t_2), t_2\in (0, T], then either 0\alpha\ge 1, then \frac{T-\alpha\eta}{T-\eta}\le\alpha. In this case, \beta<\alpha is true. It implies that u(0)\le u(T). So,$$ \min_{t\in [0, T]} u(t)=u(0). $$Assume \|u\|=u(t_2),\ t_2\in(0, T] again. Since \alpha\ge 1, it is known that u(\eta)\le u(T), together with the concavity of u, we have 0<\eta\le t_2\le T. Similar to the above discussion,$$ \min_{t\in [0, T]} u(t)\ge\frac{\beta\eta}{T} \|u\|. $$Summing up, we have$$ \min_{t\in [0, T]} u(t)\ge\gamma \|u\|, $$where$$ 0<\gamma=\min\Big\{\frac{\alpha(T-\eta)}{T-\alpha\eta},\; \frac{\alpha\eta}{T},\; \frac{\beta(T-\eta)}{T},\; \frac{\beta\eta}{T}\Big\}<1. $$This completes the proof. \end{proof} \begin{remark} \label{rmk2.6} \rm If \beta=0, Anderson obtained the inequality in \cite[Lemma 7]{Ande3} that is$$ \min_{t\in [\eta, T]} u(t)\ge r\|u\|, $$where$$r\,:=\min\Big\{\frac{\alpha(T-\eta)}{T-\alpha\eta},\ \frac{\alpha\eta}{T},\ \frac{\eta}{T}\Big\}.$$\end{remark} The following two theorems, Theorem \ref{thm1} (Guo-Krasnoselskii's fixed-point theorem)and Theorem \ref{thm2} (Leggett-Williams fixed-point theorem), will play an important role in the proof of our main results. \begin{theorem}[\cite{GuoDj}]\label{thm1} Let E be a Banach space, and let K\subset E be a cone. Assume \Omega_1,\Omega_2 are open bounded subsets of E with 0\in \Omega_1, \ \overline\Omega_1\subset\Omega_2, and let$$ A:K\cap(\overline \Omega_2 \setminus \Omega_1)\longrightarrow K $$be a completely continuous operator such that either \begin{enumerate} \item[(i)] \|Au\| \le \|u\|, \ \ u\in K\cap \partial \Omega_1, and \|Au\| \ge\|u\|, \ \ u\in K\cap \partial \Omega_2; or \item[(ii)] \|Au\| \ge \|u\|, \ \ u\in K\cap \partial \Omega_1, and \|Au\| \le \|u\|, \ \ u\in K\cap \partial \Omega_2 \end{enumerate} hold. Then A has a fixed point in K\cap (\overline \Omega_2 \setminus \Omega_1). \end{theorem} \begin{theorem}[\cite{Legge}] \label{thm2} Let P be a cone in the real Banach space E. Set \begin{gather} P_c:=\{x\in P:\|x\|b\}\neq\emptyset and \psi(Ax)>b for all x\in P(\psi, b, d); \item[(ii)]\|Ax\|b for x\in P(\psi, b, c) with \|Ax\|>d. \end{enumerate} Then A has at least three fixed points x_1, x_2 and x_3 in \overline P_c  satisfying$$ \|x_1\|b,\quad a<\|x_3\|\quad\text{with } \psi(x_3)0$. \end{enumerate} Define $$f_0=\lim_{u\to 0^+}\frac{f(u)}{u},\quad f_{\infty}=\lim _{u\to\infty}\frac{f(u)}{u}.$$ For the boundary-value problem \eqref{e1.3}-\eqref{e1.4}, we establish the following existence theorem by using Theorem \ref{thm1} (Guo-Krasnoselskii's fixed-point theorem). \begin{theorem}\label{onesolution} Assume (A1), (A2) hold, and$0<\alpha<\frac T\eta$,$0<\beta<\frac{T-\alpha\eta}{T-\eta}$. If either \begin{enumerate} \item[(C1)]$f_0=0$and$f_\infty=\infty$({\it$f$is superlinear}), or \item[(C2)]$f_0=\infty$and$f_\infty=0$({\it$f$is sublinear}), \end{enumerate} \noindent then problem \eqref{e1.3}-\eqref{e1.4} has at least one positive solution. \end{theorem} \begin{proof} It is known that$0<\alpha<\frac T\eta$,$0<\beta<\frac{T-\alpha\eta}{T-\eta}$. From Lemma \ref{lemm1},$u$is a solution to the boundary-value problem \eqref{e1.3}-\eqref{e1.4} if and only if$u$is a fixed point of operator$A$, where$Ais defined by \begin{aligned} &Au(t)\\ &=-\int^t_0 (t-s)a(s)f(u(s))\nabla s +\frac{(\beta-\alpha)t-\beta T}{(T-\alpha\eta)-\beta(T-\eta)} \int^{\eta}_0 (\eta-s)a(s)f(u(s))\nabla s\\ &\quad+\frac{(1-\beta)t+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int^T_0 (T-s)a(s)f(u(s))\nabla s. \end{aligned} \label{e3.1} Denote $$K=\{u\in C_{ld}([0,T], \mathbb{R}) : u \ge 0, \min_{t\in [0, T]} u(t) \ge\gamma \|u\|\},$$ where\gamma$is defined in \eqref{e2.5}. It is obvious that$K$is a cone in$C_{ld}([0,T], \mathbb{R})$. Moreover, from (A1), (A2), Lemma \ref{lemm2} and Lemma \ref{lemm4},$AK\subset K$. It is also easy to check that$A: K\to K$is completely continuous. \subsection*{Superlinear case}$f_0=0$and$f_\infty =\infty$. Since$f_0=0$, we may choose$H_1>0$so that$f(u) \le \epsilon u$, for$0 0$satisfies $$\epsilon\frac {T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)} \int^T_0(T-s)a(s)\nabla s\le 1.$$ Thus, if we let $$\Omega_1=\{u\in C_{ld}([0,T],\ \mathbb{R})\ : \ \|u\| 0 such that f(u)\ge \rho u,\ \text{for}\ u \ge \hat H_2, where \rho >0 is chosen so that$$ \rho\gamma\frac{T-\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int^T_0 sa(s) \nabla s \ge 1. $$Let H_2 =\max \{2H_1,\; \frac{\hat H_2}\gamma \} and$$ \Omega_2=\{u\in C_{ld}([0, T], \mathbb{R}) : \|u\|0$ such that $f(u)\ge M u$ for $00$ satisfies $$M\gamma\frac{T-\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T sa(s)\nabla s\ge 1.$$ Let $$\Omega_3=\{u\in C_{ld}([0,T], \mathbb{R}) : \|u\| 0 so that f(u) \le \lambda u for u\ge \hat H_4, where \lambda>0 satisfies$$ \lambda\frac{T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)a(s)\nabla s\le 1. $$Choose H_4 =\max \{2H_3,\ \frac {\hat H_4}\gamma \}. Let$$ \Omega_4=\{u\in C_{ld}([0, T],\ \mathbb{R}) : \|u\|0$from$0<\eta<\rho(T)$,$0<\alpha<\frac T\eta$,$0<\beta<\frac{T-\alpha\eta}{T-\eta}$. Using Theorem \ref{thm2}(the Leggett-Williams fixed-point theorem), we established the following existence theorem for the boundary-value problem \eqref{e4.1}-\eqref{e4.2}. \begin{theorem}\label{threesolution} Assume (A3) holds, and$0<\alpha<\frac T\eta$,$0<\beta<\frac{T-\alpha\eta}{T-\eta}$. Suppose there exists constants$0b,\quad a<\|u_3\|\quad\text{with } \min_{t\in[0, T]}(u_3)(t)b$, let$d=b/\gamma$, then$\{u\in P(\psi, b, d):\psi(u)>b\}\neq\emptyset$. For$u\in P(\psi, b, d)$, we have$b\le u(t)\le b/\gamma,\ t\in[0, T]$. Combining with (D2), we get $$f(t, u)\ge\frac b\delta,\quad t\in[\eta, T].$$ Since$u\in P(\psi, b, d)$, then there are two cases that either$\psi(Au)(t)=Au(0)$, or$\psi(Au)(t)=Au(T). As the former holds, we have \begin{align*} \psi(Au)(t) &=\frac{-\beta T}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^\eta (\eta-s)f(s, u(s))\nabla s\\ &\quad +\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)f(s, u(s))\nabla s\\ &=\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_\eta^T T f(s, u(s))\nabla s\\ &\quad+\frac{\beta T}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^\eta s f(s, u(s))\nabla s\\ &\quad-\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T s f(s, u(s))\nabla s\\ &>\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_\eta^T T f(s, u(s))\nabla s\\ &\quad-\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_\eta^T s f(s, u(s))\nabla s\\ &\ge \frac{b\beta\eta}{\delta[(T-\alpha\eta)-\beta(T-\eta)]} \int_\eta^T (T-s)\nabla s \ge b. \end{align*} As the later holds, we have \begin{align*} &\psi(Au)(t)\\ &=-\int_0^T (T-s)f(s, u(s))\nabla s+\frac{(\beta-\alpha)T-\beta T}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^\eta (\eta-s)f(s, u(s))\nabla s\\ &\quad +\frac{(1-\beta)T+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)f(s, u(s))\nabla s\\ &=\frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)}\int_0^T (T-s)f(s, u(s))\nabla s\\ &\quad -\frac{\alpha T}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^\eta (\eta-s)f(s, u(s))\nabla s\\ &=\frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)}\int_\eta^T T f(s, u(s))\nabla s\\ &\quad -\frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)}\int_0^T s f(s, u(s))\nabla s\\ &\quad +\frac{\alpha T}{(T-\alpha\eta)-\beta(T-\eta)} \int^\eta_0 s f(s, u(s))\nabla s\\ &> \frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_\eta^T T f(s, u(s))\nabla s\\ &\quad -\frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_\eta^T s f(s, u(s))\nabla s\\ &\ge \frac{b\alpha\eta}{\delta[(T-\alpha\eta)-\beta(T-\eta)]} \int_\eta^T (T-s)\nabla s \ge b. \end{align*} So,\psi(Au)>b,\ u\in P(\psi, b, b/\gamma)$, as required. For the condition (iii) of the Theorem \ref{thm2}, we can verify it easily under our assumptions using Lemma \ref{lemm4}. Here $$\psi(Au)=\min_{t\in [0, T]} Au(t) \ge \gamma\|Au\| > \gamma \frac b\gamma = b$$ as long as$u\in P(\psi, b, c)$with$\|Au\|>b/\gamma$. Since all conditions of Theorem \ref{thm2} are satisfied. We say the problem \eqref{e4.1}-\eqref{e4.2} has at least three positive solutions$u_1$,\$u_2$,\$u_3$with $$\|u_1\|b,\quad a<\|u_3\|\quad\text{with }\psi(u_3)1. In this case,$$ \lim_{u\to 0^+} \frac{f(u)}{u}=\lim_{u\to 0^+}u^{p-1}=0, \quad \lim_{u\to\infty} \frac{f(u)}{u}=\lim_{u\to \infty}u^{p-1}=\infty, $$and (C1) of Theorem \ref{onesolution} holds. So the problem \eqref{e5.1}-\eqref{e5.2} has at least one positive solution by Theorem \ref{onesolution}. \noindent Case 2. p<1. In this case,$$ \lim_{u\to 0^+} \frac{f(u)}{u}=\lim_{u\to 0^+}\frac 1{u^{1-p}}=\infty, \quad \lim_{u\to\infty} \frac{f(u)}{u}=\lim_{u\to \infty}\frac 1{u^{1-p}}=0, $$and (C2) of Theorem \ref{onesolution} holds. So the problem \eqref{e5.1}-\eqref{e5.2} has at least one positive solution by Theorem \ref{onesolution}. Therefore, the boundary-value problem \eqref{e5.1}-\eqref{e5.2} has at least one positive solution when p\neq 1. \end{example} \begin{example} \label{exa5.2} Let \mathbb{T}=\{0\}\cup\{1/2^n:n\in\mathbb{N}_0\}. Considering the boundary-value problem on \mathbb{T} \begin{gather} u^{\Delta\nabla}(t)+\frac{2005u^3}{u^3+5000}=0,\quad t\in[0, 1]\subset \mathbb{T}, \label{e5.3}\\ u(0)=\frac13 u(\frac1{16}),\quad u(1)=8u(\frac1{16}), \label{e5.4} \end{gather} When taking T=1, \eta=1/16, \alpha=8, \beta=1/3, and$$ f(t,u)=f(u)=\frac{2005u^3}{u^3+5000},\quad u\ge 0, $$we prove the solvability of the problem \eqref{e5.1}-\eqref{e5.2} by means of Theorem \ref{threesolution}. It is clear that f(\cdot) is continuous and increasing on [0,\infty). We can also seen that$$ 0<\alpha\eta=\frac 12<1=T,\quad 0<\beta(T-\eta)=\frac 5{16}10,\quad \frac 15<\|u_3\|\quad\text{with}\quad \min_{t\in[0, T]}(u_3)(t)<10.$\$ \end{example} \begin{thebibliography}{99} \bibitem{atici} F. M. Atici and G. Sh. Guseinov; On Green's functions and positive solutions for boundary value problems on time scales, \emph{J. Comput. Appl. Math.}, {\bf 141}(2002), 75-99. \bibitem{avery} R. I. Avery and D. R. Anderson; Existence of three positive solutions to a second-order boundary value problem on a measure chain, \emph{J. Comput. Appl. Math.}, {\bf 141}(2002), 65-73. \bibitem{agarw} R. P. Agarwal and D. O'Regan; Nonlinear boundary value problems on time scales, \emph{Nonlinear Anal.}, {\bf 44}(2001), 527-535. \bibitem{erbep} L. H. Erbe and A. C. Peterson; Positive solutions for a nonlinear differential equation on a measure chain, \emph{Mathematical and Computer Modelling}, {\bf 32}(2000), 571-585. \bibitem{maluo} Ruyun Ma and Hua Luo; Existence of solutions for a two-point boundary value problem on time scales, \emph{Appl. Math. Comput.}, {\bf 150}(2004), 139-147. \bibitem{Ande1} D. R. Anderson; Nonlinear triple-point problems on time scales, \emph{Electron. J. Differential Equations}, {\bf 47}(2004), 1-12. \bibitem{Ande2} D. R. Anderson and R. I. Avery; An even-order three-point boundary value problem on time scales, \emph{J. Math. Anal. Appl.}, {\bf 291}(2004), 514-525. \bibitem{Kaufm} E. R. Kaufmann; Positive solutions of a three-point boundary value problem on a time scale, \emph{Electron. J. Differential Equations}, {\bf 82}(2003), 1-11. \bibitem{Ande3} D. R. Anderson; Solutions to second order three-point problems on time scales, \emph{J. Difference Equations and Applications}, {\bf 8}(2002), 673-688. \bibitem{MaRy1} Ruyun Ma; Positive solutions of a nonlinear three-point boundary-value problem, \emph{Electron. J. Differential Equations}, {\bf 34}(1999), 1-8. \bibitem{MaRy2} Ruyun Ma and Y. N. Raffoul; Positive solutions of three-point nonlinear discrete second-order boundary value problem, \emph{J. Difference Equations and Applications}\ {\bf 10}(2004), 129-138. \bibitem{Hilge} S. Hilger; Analysis on measure chains -- a unified approach to continuous and discrete calculus, \emph{Results Math.}, {\bf 18}(1990), 18-56. \bibitem{Bohne} M. Bohner and A. C. Peterson, editors; \emph{Advances in Dynamic Equations on Time Scales}, Birkh\"auser, Boston, 2003. \bibitem{GuoDj} Dajun Guo and V. Lakshmikantham; Nonlinear problems in abstract cones, \emph{Academic Press, San Diego}, 1988. \bibitem{Legge} R. W. Leggett and L. R. Williams; Multiple positive fixed points of nonlinear operators on ordered Banach spaces, \emph{Indiana. University Math. J.}\ {\bf 28}(1979), 673-688. \end{thebibliography} \end{document}