\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 47, pp. 1--12.\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 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/47\hfil Triple-point problems on time scales] {Nonlinear triple-point problems on time scales} \author[Douglas R. Anderson\hfil EJDE-2004/47\hfilneg] {Douglas R. Anderson} \address{Douglas R. Anderson \hfill\break Department of Mathematics and Computer Science, Concordia College, Moorhead, Minnesota 56562 USA} \email{andersod@cord.edu} \date{} \thanks{Submitted March 7, 2004. Published April 6, 2004.} \subjclass[2000]{34B10, 34B15, 39A10} \keywords{Fixed-point theorems, time scales, dynamic equations, cone} \begin{abstract} We establish the existence of multiple positive solutions to the nonlinear second-order triple-point boundary-value problem on time scales, \begin{gather*} u^{\Delta\nabla}(t)+h(t)f(t,u(t))=0, \\ u(a)=\alpha u(b)+\delta u^\Delta(a),\quad \beta u(c)+\gamma u^\Delta(c)=0 \end{gather*} for $t\in[a,c]\subset\mathbb{T}$, where $\mathbb{T}$ is a time scale, $\beta, \gamma, \delta\ge 0$ with $\beta+\gamma>0$, $0<\alpha<\frac{c-a}{c-b}$ and $b\in(a,c)\subset\mathbb{T}$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{cor}[theorem]{Corollary} \newtheorem{defi}[theorem]{Definition} \section{Introduction to the boundary-value problem} We are concerned with proving the existence of multiple positive solutions to the second-order triple-point nonlinear boundary-value problem on a time scale $\mathbb{T}$ given by the time-scale dynamic equation $$\label{bvp} u^{\Delta\nabla}(t)+h(t)f(t,u(t))=0, \quad t\in(a,c)\subset\mathbb{T}$$ with boundary conditions $$\label{bc} u(a)=\alpha u(b)+\delta u^\Delta(a),\quad \beta u(c)+\gamma u^\Delta(c)=0,$$ where $\beta, \gamma, \delta\ge 0$ with $\beta+\gamma>0$, $0<\alpha<\frac{c-a}{c-b}$ and $b\in(a,c)\subset\mathbb{T}$ for $a\in\mathbb{T}_\kappa$, $c\in\mathbb{T}^\kappa$. The function $h\in C_{ld}[a,c]$ is nonnegative with $h(t_0)>0$ for at least one $t_0\in(a,b]$, and the nonlinearity $f:[a,c]\times[0,\infty)\to[0,\infty)$ is continuous such that $f(t,\cdot)>0$ on any subset of $\mathbb{T}$ containing $t_0$. This problem is related to that first studied in the case $\mathbb{T}=\mathbb{R}$ on the unit interval by He and Ge \cite{he} and Ma \cite{ma1,ma2,ma3}, $$\label{maprob} u''+f(t,u)=0,\quad t\in (0,1) \quad u(0)=0,\quad\alpha u(\eta)=u(1)$$ where $0<\eta<1$ and $0<\alpha<1/\eta$. The boundary-value problem \eqref{maprob} has since been extended to general time scales in Anderson \cite{and} and Kaufmann \cite{kauf} as $$u^{\Delta\nabla}(t)+f(t,u(t))=0,\quad u(0)=0,\; \alpha u(\eta)=u(T),$$ and in a slightly different way by Sun and Li \cite{li} $$u^{\Delta\nabla}(t)+a(t)f(t,u(t))=0, \quad u^\Delta(0)=0,\; \alpha u(\eta)=u(T).$$ In this paper there is a nexus between the boundary conditions at $a$ and $b$ instead of at $b=\eta$ and $c=T$, and we add the $\delta\ge 0$ term as well. Consequently these results are new for differential and difference equations as well as for dynamic equations on general time scales. For more on time scales and positive solutions, see the books by Bohner and Peterson \cite{bp,bp2} and the following articles: \cite{akin,and2,ah,atici,aa,chyan,erbepete1,erbepete2,hend,hilger}. \section{preliminary results} To prove the main existence results we will employ several straightforward lemmas. These lemmas are based on the linear boundary-value problem \begin{gather}\label{line3} u^{\Delta\nabla}(t)+y(t)=0 \quad t\in(a,c)\subset\mathbb{T} \\ \label{line4} u(a)=\alpha u(b)+\delta u^\Delta(a),\quad \beta u(c)+\gamma u^\Delta(c)=0. \end{gather} \begin{lemma}\label{lemma1} Let $$\label{dee} d:=\gamma(1-\alpha)+\beta\left[(c-a)-\alpha(c-b)+\delta\right].$$ If $d\neq 0$, then for $y\in C_{ld}[a,c]$ the boundary-value problem \eqref{line3}, \eqref{line4} has the unique solution \label{unique} \begin{aligned} u(t)=& \frac{1}{d}\left(\gamma+\beta(c-t)\right) \Big[ \int_a^c (s-a+\delta)y(s)\nabla s \\ & - \alpha\int_b^c (s-b)y(s)\nabla s \Big] - \int_t^c (s-t)y(s)\nabla s. \end{aligned} \end{lemma} \begin{proof} Let $u$ be as in \eqref{unique}. Then the delta derivative of $u$ is given by $$u^{\Delta}(t) = -\frac{\beta}{d} \left[\int_a^c (s-a+\delta)y(s)\nabla s - \alpha\int_b^c (s-b)y(s)\nabla s \right] + \int_t^c y(s)\nabla s$$ and $$u^{\Delta\nabla}(t)=-y(t),$$ so that $u$ given in \eqref{unique} is a solution of \eqref{line3}. It is routine to check that the boundary conditions \eqref{line4} are met by $u$ in \eqref{unique} as well. \end{proof} \begin{lemma}\label{lemma2} If $d>0$ and $y\in C_{ld}[a,c]$ with $y\geq 0$, the unique solution $u$ of \eqref{line3}, \eqref{line4} given in \eqref{unique} satisfies $$u(t)\geq 0, \quad t\in[a,c]\subset\mathbb{T}.$$ \end{lemma} \begin{proof} From the fact that $u^{\Delta\nabla}(t)=-y(t)\le 0$, we know that if $u(a)\ge 0$ and $u(c)\ge 0$, then $u(t)\ge 0$ for $t\in[a,c]$. For $0 < \alpha \le 1$, \begin{align*} u(c) & = \frac{\gamma}{d}\left[\int_a^c (s-a+\delta)y(s)\nabla s - \alpha\int_b^c (s-b)y(s)\nabla s\right] \\ &\ge \frac{\gamma}{d}\left[(1-\alpha)\int_b^c (s-b)y(s)\nabla s + \delta\int_a^c y(s)\nabla s\right] \\ &\ge 0. \end{align*} For $1 < \alpha < \frac{c-a}{c-b}$, \begin{align*} u(c) & = \frac{\gamma}{d}\left[\int_a^b (s-a+\delta)y(s)\nabla s+\int_b^c (s-a-\alpha(s-b)+\delta)y(s)\nabla s\right]\\ &\ge \frac{\gamma}{d}\left[\int_a^b (s-a+\delta)y(s)\nabla s+\int_b^c (c-a-\alpha(c-b)+\delta)y(s)\nabla s\right]\\ &\ge 0 \end{align*} since $\alpha<(c-a)/(c-b)$. Finally, \begin{align*} u(a) & = \frac{1}{d}\left(\gamma + \beta(c-a)\right) \Big[\int_a^c(s-a+\delta)y(s)\nabla s-\alpha\int_b^c(s-b)y(s)\nabla s\Big] -\frac{d}{d}\int_a^c(s-a)y(s)\nabla s\\ & = \frac{\alpha}{d}(\gamma+\beta(c-b))\int_a^b(s-a)y(s)\nabla s+\frac{\alpha}{d}(b-a)\int_b^c(\gamma+\beta(c-s))y(s)\nabla s \\ & \quad +\frac{\delta}{d}\int_a^c (\gamma+\beta(c-s))y(s)\nabla s \\ &\ge 0. \end{align*} \end{proof} \begin{lemma}\label{lemma3} Let $d>0$. If $y\in C_{ld}[a,c]$ with $y$ nonnegative but nontrivial, then the unique solution $u$ as in \eqref{unique} of \eqref{line3}, \eqref{line4} satisfies $$\inf_{t\in[a,b]}u(t)\ge k\|u\|, \quad \|u\|:=\sup_{t\in[a,c]}|u(t)|,$$ where $$\label{kdef} k:=\min\Big\{\frac{\alpha(c-b)}{c-a}, \frac{c-b}{c-a}, \frac{\alpha(b-a)}{c-a-\alpha(c-b)} \Big\}\in(0,1).$$ \end{lemma} \begin{proof} Note that $u^{\Delta\nabla}(t)=-y(t)\le 0$ for all $t\in(a,c)$, so that $$\min_{t\in[a,b]}u(t)=\min\{u(a), u(b)\}.$$ Then for any $\tau\in[a,c)$, $$\eta(t):=u(t)-\big(\frac{c-t}{c-\tau}\big)u(\tau)$$ satisfies $\eta(\tau)=0$, $\eta(c)=u(c)\ge 0$, and $\eta^{\Delta\nabla}(t)=u^{\Delta\nabla}(t)\le 0$ on $[\tau,c)$. In particular, $$\frac{u(t)}{c-t} \ge \frac{u(\tau)}{c-\tau}$$ for all $t\in[\tau,c)$. Now fix $\tau\in[a,c)$ such that $u(\tau)=\|u\|$. If $a\le \tau \le b$ and $u(a)\le u(b)$, then $$\frac{\alpha u(b)}{c-b}\ge \frac{\alpha u(\tau)}{c-\tau} \ge \frac{\alpha u(\tau)}{c-a},$$ rewritten with the boundary condition at $b$ as $$u(a)-\delta u^\Delta(a)\ge\big(\frac{c-b}{c-a}\big)\alpha \|u\|.$$ Therefore, $$\min_{t\in[a,b]}u(t)=u(a)\ge \big(\frac{c-b}{c-a}\big)\alpha\|u\|,$$ since $u^\Delta(a)\ge 0$ in this case. If $a\le \tau \le b$ and $u(b)\le u(a)$, then $$u(b)\ge \big(\frac{c-b}{c-\tau}\big)u(\tau) \ge \big(\frac{c-b}{c-a}\big)u(\tau),$$ so that $$\min_{t\in[a,b]}u(t)=u(b)\ge \big(\frac{c-b}{c-a}\big)\|u\|.$$ If $b < \tau < c$, then $u(a)=\min_{t\in[a,b]}u(t)$, and by the concavity of $u$ and a secant line we have \begin{align*} u(\tau) &\le u(a)+\Big(\frac{u(b)-u(a)}{b-a}\Big)(c-a) \\ & = \frac{\left[c-a-\alpha(c-b)\right]u(a)-\delta (c-a)u^\Delta(a)}{\alpha(b-a)} \\ &\le \frac{\left[c-a-\alpha(c-b)\right]u(a)}{\alpha(b-a)}, \end{align*} since again $u^\Delta(a)\ge 0$. Consequently, $$\min_{t\in[a,b]}u(t)=u(a)\ge\frac{\alpha(b-a)}{c-a-\alpha(c-b)}\|u\|.$$ \end{proof} \section{Existence of at Least Two Positive Solutions} We apply the Avery-Henderson Fixed Point Theorem \cite{AvHe} to prove the existence of at least two positive solutions to the nonlinear boundary-value problem \eqref{bvp}, \eqref{bc}, where $h\in C_{ld}[a,c]$ is nonnegative with $h(t_0)>0$ for at least one $t_0\in(a,b]$,and the nonlinearity $f:[a,c]\times[0,\infty)\to[0,\infty)$ is continuous such that $f(t,\cdot)>0$ on any subset of $\mathbb{T}$ containing $t_0$. The solutions are the fixed points of the operator $A$ defined by \begin{align*} A u(t) &= \frac{1}{d}\left(\gamma+\beta(c-t)\right) \Big[ \int_a^c (s-a+\delta)h(s)f(s,u(s))\nabla s \\ &\quad - \alpha\int_b^c (s-b)h(s)f(s,u(s))\nabla s \Big] - \int_t^c (s-t)h(s)f(s,u(s))\nabla s \end{align*} by Lemma \ref{lemma1}. Notationally, the cone $P$ has subsets of the form $P(\phi,r):=\{u\in P:\phi(u)r$ for all $u\in \partial P(\phi,r)$, \item[(ii)] $\theta(A u)p$ for all $u\in\partial P(\eta,p)$, \end{itemize} then $A$ has at least two fixed points $u_1$ and $u_2$ such that $$p < \eta(u_1) \text{ with } \theta(u_1) < q \quad\text{ and }\quad q < \theta(u_2) \text{ with }\phi(u_2) < r.$$ \end{theorem} Let $\mathcal{S}$ denote the Banach space $C[\rho(a),\sigma(c)]$ with the supremum norm. Define the cone $P\subset\mathcal{S}$ by $$\label{P} P=\{u\in\mathcal{S}: u(t)\ge 0, u \mbox{ is concave, and } \min_{t\in[a,b]}u(t) \ge k\|u\|\},$$ where $k$ is given in \eqref{kdef}. Finally, let the nonnegative, increasing, continuous functionals $\phi$, $\theta$, and $\eta$ be defined on the cone $P$ by $$\phi(u):=\min_{t\in[a,b]} u(t), \quad \theta(u):=\max_{t\in[b,c]} u(t), \quad \eta(u):=\max_{t\in[a,c]} u(t).$$ Observe that, for each $u\in P$, the concavity of $u$ implies \begin{gather}\label{gpd} \phi(u)\le\theta(u)\le\eta(u), \\ \label{normgam} \|u\| = \le \frac{1}{k}\min_{t\in[a,b]}u(t) = \frac{1}{k}\phi(u)\le\frac{1}{k}\theta(u)\le\frac{1}{k}\eta(u). \end{gather} \begin{theorem}\label{averyhen2} Let $d>0$ and $k$ be as in \eqref{kdef}. Suppose there exist positive numbers $0 < p < q < r$ such that the function $f$ satisfies the following conditions: \begin{enumerate} \item[$(i)$] $f(s,u) > pM$ for $s\in[a,b]$ and $u\in[kp,p]$, \item[$(ii)$] $f(s,u) < qm$ for $s\in[a,c]$ and $u\in[0,q/k]$, \item[$(iii)$] $f(s,u) > rM$ for $s\in[a,b]$ and $u\in[r,r/k]$ \end{enumerate} for some positive constants $m$ and $M$. Then the second-order boundary-value problem \eqref{bvp}, \eqref{bc}, has at least two positive solutions $u_1$ and $u_2$ such that \begin{gather*} p < \max_{t\in[a,c]}u_1(t) \quad \text{with } \max_{t\in[b,c]}u_1(t) < q, \\ q < \max_{t\in[b,c]}u_2(t) \quad \text{ with } \min_{t\in[a,b]}u_2(t) < r. \end{gather*} \end{theorem} \begin{proof} For $u\in P$, $u(t)\ge k\|u\|$ for all $t\in[a,b]$. By Lemma \ref{lemma3}, $A(P)\subset P$. Standard arguments show that $A:P\to P$ is completely continuous. For any $u\in P$, \eqref{gpd} and \eqref{normgam} imply $$\phi(u)\le\theta(u)\le\eta(u), \quad \|u\|\le\frac{1}{k}\phi(u).$$ It is clear that $\theta(0)=0$, and for all $u\in P$, $\lambda\in[0,1]$ we have $$\theta(\lambda u) = \max_{t\in[b,c]}(\lambda u)(t) = \lambda\max_{t\in[b,c]} u(t) = \lambda\theta(u).$$ Since $0\in P$ and $p>0$, $P(\eta,p)\ne\emptyset$. In the following claims, we verify the remaining conditions of Theorem \ref{averyhen}. \smallskip \noindent {\bf Claim 1:} If $u\in\partial P(\eta,p)$, then $\eta(A u)>p$: Since $u\in\partial P(\eta,p)$, $kp\le u(t)\le\|u\|=p$ for $t\in[a,b]$. Define $$\label{bigm} M:=d\Big(\min\{\alpha,1\}(\gamma+\beta(c-b))\int_a^b(s-a+\delta)h(s)\nabla s \Big)^{-1}.$$ Thus \begin{align*} \eta(A u) & = \max_{t\in[a,c]} A u(t) \\ &\ge A u(b) \\ & = \frac{1}{d}\left(\gamma+\beta(c-b)\right) \Big[ \int_a^c (s-a+\delta)h(s)f(s,u(s))\nabla s\\ &\quad - \alpha\int_b^c (s-b)h(s)f(s,u(s))\nabla s \Big] - \int_b^c (s-b)h(s)f(s,u(s))\nabla s \\ & = \frac{\gamma+\beta(c-b)}{d}\int_a^b (s-a+\delta)h(s)f(s,u(s))\nabla s\\ & \quad+\frac{b-a+\delta}{d}\int_b^c (\gamma+\beta(c-s))h(s)f(s,u(s))\nabla s \\ &\ge \frac{\gamma+\beta(c-b)}{d}\int_a^b (s-a+\delta)h(s)f(s,u(s))\nabla s \\ & > pM\frac{\gamma+\beta(c-b)}{d}\int_a^b (s-a+\delta)h(s)\nabla s\\ &\ge p \end{align*} by hypothesis $(i)$ and \eqref{bigm}. \smallskip \noindent {\bf Claim 2:} If $u\in\partial P(\theta,q)$, then $\theta(A u)r$: Since $u\in\partial P(\phi,r)$, from \eqref{normgam} we have that $\displaystyle\min_{t\in[a,b]}u(t)=r$ and $r\le \|u\|\le \frac{r}{k}$. Since $\delta\ge 0$, as in the proof of Claim 1 we have \begin{align*} A u(b)&=\frac{\gamma+\beta(c-b)}{d}\int_a^b (s-a+\delta)h(s)f(s,u(s))\nabla s\\ &\quad +\frac{b-a+\delta}{d}\int_b^c (\gamma+\beta(c-s))h(s)f(s,u(s))\nabla s. \end{align*} In a similar fashion from the proof of Lemma \ref{lemma2} we have \begin{align*} A u(a) &=\alpha\Big[\frac{\gamma+\beta(c-b)}{d}\int_a^b (s-a)h(s)f(s,u(s))\nabla s\\ &\quad+\frac{b-a}{d}\int_b^c (\gamma+\beta(c-s))h(s)f(s,u(s))\nabla s\Big] \\ & \quad +\frac{\delta}{d}\int_a^c(\gamma+\beta(c-s))h(s)f(s,u(s))\nabla s. \end{align*} By concavity, $$\min_{t\in[a,b]} A u(t) = \min\{A u(a), A u(b)\}.$$ If $\alpha\ge 1$, then $A u(a)\ge A u(b)$. If $0<\alpha<1$, then $A u(a) \ge \alpha A u(b)$. It follows that \begin{align*} \phi (A u) & = \min_{t\in[a,b]} A u(t) \\ &\ge \min\{\alpha,1\}A u(b) \\ &\ge \min\{\alpha,1\}\big(\frac{\gamma+\beta(c-b)}{d}\big) \int_a^b (s-a+\delta)h(s)f(s,u(s))\nabla s \\ & > \min\{\alpha,1\}rM\big(\frac{\gamma+\beta(c-b)}{d}\big) \int_a^b (s-a+\delta)h(s)\nabla s \\ &\ge r \end{align*} by hypothesis $(iii)$ and \eqref{bigm}. Therefore the hypotheses of Theorem \ref{averyhen} are satisfied and there exist at least two positive fixed points $u_1$ and $u_2$ of $A$ in $\overline{P(\phi,r)}$. Thus, the second-order triple-point boundary-value problem \eqref{bvp}, \eqref{bc}, has at least two positive solutions $u_1$ and $u_2$ such that \begin{gather*} p < \eta(u_1) \quad \mbox{with } \theta(u_1) < q, \\ q < \theta(u_2) \quad \mbox{with } \phi(u_2) < r. \end{gather*} \end{proof} \begin{cor}\label{ahcor} Let $d>0$. If there exists $q>0$ such that the function $f$ satisfies the following conditions: \begin{enumerate} \item[$(i)$] $\displaystyle\liminf_{u\to 0^+}\frac{f(s,u)}{u} > M/k$ for $s\in[a,b]$, \item[$(ii)$] $f(s,u) < qm$ for $s\in[a,c]$ and $u\in[0,q/k]$, \item[$(iii)$] $\displaystyle\liminf_{u\to\infty}\frac{f(s,u)}{u} > M$ for $s\in[a,b]$, \end{enumerate} then the second-order boundary-value problem \eqref{bvp}, \eqref{bc}, has at least two positive solutions $u_1$ and $u_2$ such that \begin{gather*} p < \max_{t\in[a,c]}u_1(t) \quad \text{with } \max_{t\in[b,c]}u_1(t) < q, \\ q < \max_{t\in[b,c]}u_2(t) \quad \text{with } \min_{t\in[a,b]}u_2(t) < r. \end{gather*} \end{cor} \begin{proof} From assumption $(i)$ of the corollary we know there exists $p\in(0,q)$ such that $$\frac{f(s,u)}{u} > \frac{M}{k}, \quad u\in(0,p], \; s\in[a,b].$$ In particular, $$f(s,u) > uM/k \ge pk(M/k) = pM, \quad u\in[kp,p], \; s\in[a,b],$$ and $(i)$ of Theorem \ref{averyhen2} holds. Now let $$f_\infty:=\liminf_{u\to\infty}\Big(\min_{s\in[a,b]}\frac{f(s,u)}{u}\Big),$$ and $\eta\in(M,f_\infty)$. Then there exists $r'>q$ such that $\min_{s\in[a,b]}f(s,u)\ge \eta u$, $u\in[r',\infty)$. Set $$\varpi:=\min\big\{\min_{s\in[a,b]}f(s,u):u\in[0,r']\big\}$$ and take $$r>\max\big\{r', \frac{\varpi}{\eta-M}\big\}.$$ Then $$\min_{s\in[a,b]}f(s,u)\ge \eta u-\varpi \ge \eta r-\varpi > rM, \quad u\in[r,\infty),$$ so that $(iii)$ of Theorem \ref{averyhen2} holds. The conclusion follows. \end{proof} Similar to Corollary \ref{ahcor}, one can prove the following statement. \begin{cor} \label{coro7} Let $d>0$. If there exists $q>0$ such that the function $f$ satisfies the following conditions: \begin{enumerate} \item[$(i)$] $\displaystyle\limsup_{u\to 0^+}\frac{f(s,u)}{u} < m$ for $s\in[a,c]$, \item[$(ii)$] $f(s,u) > qM$ for $s\in[a,b]$ and $u\in[kq,q]$, \item[$(iii)$] $\displaystyle\limsup_{u\to\infty}\frac{f(s,u)}{u} < mk$ for $s\in[a,c]$, \end{enumerate} then the second-order boundary-value problem \eqref{bvp}, \eqref{bc}, has at least two positive solutions. \end{cor} \section{Existence of at Least Three Positive Solutions} To prove the existence of at least three positive solutions to \eqref{bvp}, \eqref{bc} we will use the Leggett-Williams fixed point theorem \cite{gu,le}: \begin{theorem}\label{LWTHM} Let $P$ be a cone in the real Banach space $\mathcal{S}$, $A:\overline{P_r} \to \overline{P_r}$ be completely continuous and $\phi$ be a nonnegative continuous concave functional on $P$ with $\phi(u)\le\|u\|$ for all $u\in \overline{P_r}$. Suppose there exists $0 q\}\neq\emptyset$ and $\phi(A u)>q$ for all $u\in P(\phi,q,\ell)$; \item[$(ii)$] $\|A u\| < p$ for $\|u\|\le p$; \item[$(iii)$] $\phi(A u) > q$ for $u\in P(\phi,q,r)$ with $\|A u\| > \ell$. \end{enumerate} Then $A$ has at least three fixed points $u_1$, $u_2$, and $u_3$ in $\overline{P_r}$ satisfying: $$\|u_1\| < p, \quad \phi(u_2) > q, \quad p < \|u_3\|\;\; \mbox{with}\;\; \phi(u_3) < q.$$ \end{theorem} Again define the continuous concave functional $\phi:P\to[0,\infty)$ to be $\phi(u):=\min_{t\in[a,b]} u(t)$, the cone $P$ as in \eqref{P}, $M$ as in \eqref{bigm}, and $$m:=d\Big((\gamma+\beta(c-a))\int_a^c(s-a+\delta)h(s)\nabla s \Big)^{-1}.$$ Moreover, we take $$P_r:= \{u\in P: \|u\| \ < r \}, \quad P(\phi,p,q) := \{u\in P : p\le\phi(u),\; \|u\| \le q \}.$$ \begin{theorem}\label{3sols} Let $d>0$. Suppose that there exist constants 0 p \mbox{ with } \min_{t\in[a,b]}u_3(t) < q. \end{theorem} \begin{proof} Again the solutions are the fixed points of the operator A defined by \begin{align*} A u(t) &= \frac{1}{d}\left(\gamma+\beta(c-t)\right) \Big[ \int_a^c (s-a+\delta)h(s)f(s,u(s))\nabla s\\ &\quad - \alpha\int_b^c (s-b)h(s)f(s,u(s))\nabla s \Big] - \int_t^c (s-t)h(s)f(s,u(s))\nabla s. \end{align*} The conditions of Theorem \ref{LWTHM} will now be shown to be satisfied. For all u\in P we have \phi(u)\le\|u\|. If u\in\overline{P_r}, then \|u\|\le r and assumption (D1) implies f(s,u(s))\le rm for s\in[a,c]. Consequently, \begin{align*} \|A u\| & = \max_{t\in[a,c]} A u(t) \\ &\le \max_{t\in[a,c]} \frac{1}{d}\left(\gamma+\beta(c-t)\right) \Big[ \int_a^c (s-a+\delta)h(s)f(s,u(s))\nabla s\\ &\quad - \alpha\int_b^c (s-b)h(s)f(s,u(s))\nabla s \Big] \\ &\le \frac{1}{d}\left(\gamma+\beta(c-a)\right) \int_a^c (s-a+\delta)h(s)f(s,u(s))\nabla s \\ & < \frac{rm}{d}\left(\gamma+\beta(c-a)\right) \int_a^c (s-a+\delta)h(s)\nabla s \\ & = r. \end{align*} This proves that A: \overline{P_r}\to\overline{P_r}. Similarly, if u\in P_p, then assumption (D3) yields f(s,u(s)) q, so that \{u\in P(\phi,q,q/k): \phi(u)>q\}\neq\emptyset. Consequently, if u\in P(\phi,q,q/k), then q\le u(s)\le q/k when s\in[a,b]. From assumption (D2) we have that f(s,u(s))\ge qM$$for s\in[a,b]. As in Claim 3 of the proof of Theorem \ref{averyhen2},$$ \min_{t\in[a,b]} A u(t) = \min\{A u(a), A u(b)\}. If \alpha\ge 1, then A u(a)\ge A u(b); if 0<\alpha<1, then A u(a) \ge \alpha A u(b). It follows that \begin{align*} \phi (A u) & = \min_{t\in[a,b]} A u(t) \\ &\ge \min\{\alpha,1\}A u(b) \\ &\ge \min\{\alpha,1\}\big(\frac{\gamma+\beta(c-b)}{d}\big) \int_a^b (s-a+\delta)h(s)f(s,u(s))\nabla s \\ & > \min\{\alpha,1\}qM\big(\frac{\gamma+\beta(c-b)}{d}\big) \int_a^b (s-a+\delta)h(s)\nabla s \\ &\ge q \end{align*} by hypothesis (D2) and \eqref{bigm}. Therefore, \phi(A u) > q, \quad u\in P(\phi,q,q/k), $$so that condition (i) of Theorem \ref{LWTHM} holds. To check on Theorem \ref{LWTHM} (iii), we suppose that u\in P(\phi,q,r) with \|A u\|>q/k. Then, Lemma \ref{lemma3} and the definition of \phi yield$$ \phi(A u) = \min_{t\in[a,b]}A u(t) \ge k\|A u\| > kq/k = q. $$\end{proof} Using the ideas in the proof of the above theorem, we can establish the existence of an arbitrary odd number of positive solutions of \eqref{bvp}, \eqref{bc}, assuming the right conditions on the nonlinearity f. \begin{theorem} \label{thm10} Let d>0. Suppose that there exist constants$$ 0 < p_1 < q_1 < q_1/k < p_2 < q_2 < q_2/k < p_3 < \dots < p_n, \quad n\in\{2,3,4,\cdots\}, $$such that \begin{enumerate} \item[(D1)] f(s,u)\ge q_iM for s\in[a,b], u\in[q_i,q_i/k]; \item[(D2)] f(s,u) < p_im for s\in[a,c], u\in[0,p_i], \end{enumerate} where k is given in \eqref{kdef}. Then the boundary-value problem \eqref{bvp}, \eqref{bc} has at least 2n-1 positive solutions. \end{theorem} \section{Example} Let \mathbb{T}= \{1-(1/2)^{\mathbb{N}_0}\}\cup\{1\}. Taking a=0, b=31/32, c=\beta=1, \alpha=20, \gamma=1/19, and \delta=3/4, we have d=1/8 and k=1/32. If we let h(s)\equiv 1, then m=57/680 and M=77824/71145. Suppose$$ f(t,u)=f(u):=\frac{2000e^u}{99e^{700}+e^u}, \quad t\in[0,1], \quad u\ge 0. $$Clearly f is always increasing. If we take p=701, q=705, and r=24000, then$$ 0 < p < q < q/k < r. $$We check that (D1), (D2), and (D3) of Theorem \ref{3sols} are satisfied. Since f(715)\approx 1999.94 and \lim f(u)=2000,$$ f(u)\le 24000m\approx 2011.76, \quad u\in[0,r], $$so that (D1) is met. To verify (D2), note that f(705)\approx 1199.72, so that$$ f(u)\ge 705M\approx 771.18, \quad u\in[q,32q]. $$Lastly, as f(701)\approx 53.45,$$ f(u) < 701m\approx 58.76, \quad u\in[0,p] $$and (D3) holds. Therefore by Theorem \ref{3sols}, the boundary-value problem$$ u^{\Delta\nabla}(t)+f(u(t))=0, \quad u(0)=20u(31/32)+\frac{3}{4}u^\Delta(0), \quad u(1)+\frac{1}{19}u^\Delta(1)=0 $$has at least three positive solutions u_1, u_2, u_3 satisfying$$ \|u_1\| < 701, \quad 705 < \min_{t\in[0,31/32]}u_2(t), \quad \|u_3\| > 701 \quad \mbox{with } \min_{t\in[0,31/32]}u_3(t) < 705.\$ \begin{thebibliography}{99} \bibitem{akin} E. Akin, Boundary value problems for a differential equation on a measure chain, \emph{Panamerican Mathematical Journal}, 10:3 (2000) 17--30. \bibitem{and} D. R. Anderson, Solutions to second-order three-point problems on time scales, \emph{J. Difference Eq. 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