\documentclass[reqno]{amsart} \usepackage{amssymb} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2003(2003), No. 82, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \thanks{\copyright 2003 Southwest Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE--2003/82\hfil Positive solutions to three-point BVPs] {Positive solutions of a three-point boundary-value problem on a time scale} \author[Eric R. Kaufmann \hfil EJDE--2003/82\hfilneg] {Eric R. Kaufmann} \address{Eric R. Kaufmann \hfill\break\indent Department of Mathematics and Statistics\hfill\break\indent University of Arkansas at Little Rock\hfill\break\indent Little Rock, Arkansas 72204-1099, USA} \email{erkaufmann@ualr.edu} \date{} \thanks{Submitted May 9, 2003. Published August 9, 2003.} \subjclass[2000]{34B10, 34B15, 34G20} \keywords{Time scale, cone, boundary-value problem, positive solutions} \begin{abstract} Let $\mathbb{T}$ be a time scale such that $0, T \in \mathbb{T}$. We consider the second order dynamic equation on a time scale \begin{gather*} u^{\Delta\nabla}(t) + a(t)f(u(t)) = 0, \quad t \in (0,T) \cap \mathbb{T},\\ u(0) = 0, \quad \alpha u(\eta) = u(T), \end{gather*} where $\eta \in (0, \rho(T)) \cap \mathbb{T}$, and $0 < \alpha \inf \mathbb{T}$, we define the {\em forward jump operator}, $\sigma$, and the {\em 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 $t \in \mathbb{T}$. If $\sigma(t) > t$, $t$ is said to be {\em right scattered}, and if $\sigma(t) = t$, $t$ is said to be {\em right dense\ }(rd). If $\rho(t) < t$, $t$ is said to be {\em left scattered}, and if $\rho(t) = t$, $t$ is said to be {\em left dense\ }(ld). A function $f$ is left-dense continuous, ld-continuous, $f$ is continuous at each left dense point in $\mathbb{T}$ and its right-sided limits exist at each right dense points in $\mathbb{T}$. For $x:\mathbb{T} \to \mathbb{R}$ and $t \in \mathbb{T}$, (assume $t$ is not left scattered if $t = \sup \mathbb{T}$), we define the {\em delta derivative} of $x(t)$, $x^\Delta (t)$, to be the number (when it exists), with the property that, for each $\varepsilon > 0,$ there is a neighborhood, $U$, of $t$ such that \begin{displaymath} \big\vert x(\sigma(t)) - x(s) - x^\Delta (t) (\sigma(t) - s)\big \vert \leq \varepsilon \vert \sigma(t) - s \vert, \end{displaymath} for all $s \in U$. For $x:\mathbb{T} \to \mathbb{R}$ and $t \in \mathbb{T}$, (assume $t$ is not right scattered if $t = \inf \mathbb{T}$), we define the {\em nabla derivative} of $x(t)$, $x^\nabla (t)$, to be the number (when it exists), with the property that, for each $\varepsilon > 0,$ there is a neighborhood, $U$, of $t$ such that \begin{displaymath} \big \vert x(\rho(t)) - x(s) - x^\nabla (t) (\rho(t) - s) \big \vert \leq \varepsilon \vert \rho(t) - s \vert, \end{displaymath} for all $s \in U$. If $\mathbb{T} = \mathbb{R}$ then $f^\Delta(t) = f^\nabla(t) = f'(t)$. If $\mathbb{T} = \mathbb{Z}$ then $f^\Delta(t) = f(t+1) - f(t)$ is the forward difference operator while $f^\nabla(t) = f(t) - f(t-1)$ is the backward difference operator. In 1998, Ma \cite{ma1} showed the existence of a positive solution to the second order three-point boundary-value problem \begin{gather*} u'' + a(t) f(u) = 0, \quad t \in (0, 1),\\ u(0) = 0, \quad \alpha u(\eta) = u(1), \end{gather*} where $0 < \eta < 1, 0 < \alpha < 1/\eta$ and $f$ was either superlinear or sublinear. Later Cao and Ma \cite{caoma} extended these results to the $m$-point eigenvalue problem $u'' + \lambda a(t) f(u, u') = 0, u(0) = 0, \sum_{i=1}^{m-2} a_i u(\xi_i) = u(1)$. Ma and Raffoul \cite{mayr} showed the existence of positive solutions for a three-point boundary-value problems for difference equation. Recently, Anderson \cite{da} showed the existence of at least one positive solution (using the Krasnosel'ski\u{\i} fixed point theorem) and the existence of at least three positive solutions (using the Leggett-Williams fixed point theorem) for the three-point boundary-value problem on a time scale. For other references on multi-point boundary-value problems we refer the reader to the papers \cite{ma3, ma2, ynr} and references therein. Many authors have studied the existence of multiple positive solutions for boundary value problems for differential and difference equations, see \cite{ehkauf, ehkos, erk, kk, kos} and references therein. The book \cite{ARW} is an excellent source for information on the theory of positive solutions for differential, difference and integral equations. One of the first papers to consider countably many positive solutions for boundary-value problems on a time scale is \cite{kk2}. In this paper, we show the existence of multiple positive solutions for a second order three-point boundary-value problem on a time scale. Let $\mathbb{T}$ be a time scale such that $0, T \in \mathbb{T}$ and denote the set of all left-dense continuous functions from $\mathbb{T}$ to $E \subseteq \mathbb{R}$ by $C_{ld}(\mathbb{T}, E)$. Consider the second order dynamic equation \begin{gather} u^{\Delta\nabla}(t) + a(t)f(u(t)) = 0, \quad t \in (0, T) \cap \mathbb{T}, \label{eq1}\\ u(0) = 0, \quad \alpha u(\eta) = u(T), \label{eq2} \end{gather} where $\eta \in (0, \rho(T)) \cap \mathbb{T}$, and $0 < \alpha < T/\eta$. We will assume throughout that $f :\mathbb{T} \to [0, +\infty)$ is continuous. We will also assume that $a \in C_{ld}(\mathbb{T}, [0, +\infty))$ and there exists at least one $t_0 \in [\eta, T) \cap \mathbb{T}$ such that $a(t_0) > 0$. Define \begin{displaymath} f_0 = \lim_{u \to 0^+} \frac{f(u)}{u} \quad \mbox{and}\quad f_{\infty} = \lim_{u \to \infty} \frac{f(u)}{u}. \end{displaymath} Note that $f_0 = 0$ and $f_\infty = \infty$ correspond to $f$ being superlinear and that $f_0 = \infty$ and $f_\infty = 0$ correspond to $f$ being sublinear. We show the existence of two positive solutions for the boundary-value problem (\ref{eq1}), (\ref{eq2}) when $f_0 = 0$ and $f_\infty = 0$ and when $f_0 = \infty$ and $f_\infty = \infty$. In section 2 we state some lemmas that will be needed in order to prove our main theorems. We also define an operator whose fixed points are solutions to (\ref{eq1}), (\ref{eq2}) and state a fixed point theorem due to Krasnosel'ski\u{\i}, see \cite{Kra}. In section 3 we state and prove two theorems for the existence of two positive solutions of (\ref{eq1}), (\ref{eq2}). We begin section 4 with a modification of Lemma \ref{lemma4} in section 2. This new lemma will allow us to define a sequence of cones in which we will find fixed points of our operator. % ---------------------------------------------------------------- \section{Preliminaries} We will need the following lemmas, whose proofs can be found in Anderson \cite{da}, in order to prove our main theorems. Consider the linear boundary-value problem \begin{gather} u^{\Delta\nabla}(t) + y(t) = 0, \quad t \in (0, T) \cap \mathbb{T}, \label{eq3}\\ u(0) = 0, \quad \alpha u(\eta) = u(T). \label{eq4} \end{gather} \begin{lemma}\label{lemma1} If $\alpha \eta \neq T$ then for $y \in C_{ld}(\mathbb{T}, \mathbb{R})$ the boundary-value problem (\ref{eq3}), (\ref{eq4}) has the unique solution \begin{displaymath} u(t) = -\int_0^t \! (t-s)y(s) \,\nabla s - \frac{\alpha t}{T - \alpha \eta} \int_0^{\eta} \! (\eta - s) y(s) \, \nabla s + \frac{t}{T - \alpha \eta} \int_0^T \!(T - s) y(s) \, \nabla s. \end{displaymath} \end{lemma} \begin{lemma}\label{lemma2} If $u(0) = 0$ and $u^{\Delta \nabla} \leq 0$, then $\frac{u(s)}{s} \leq \frac{u(t)}{t}$ for all $s, t \in (0,T] \cap \mathbb{T}$ with $t \leq s$. \end{lemma} \begin{lemma} Let $0 < \alpha < T/\eta$. If $y \in C_{ld}(\mathbb{T}, \mathbb{R})$ and $y \geq 0$ then the solution $u$ of boundary-value problem (\ref{eq3}), (\ref{eq4}) satisfies $u(t) \geq 0$ for all $t \in [0,T] \cap \mathbb{T}$. \end{lemma} \begin{lemma}\label{lemma3} Let $\alpha \eta > T$. If $y \in C_{ld}(\mathbb{T}, \mathbb{R})$ and $y \geq 0$ then the boundary-value problem (\ref{eq3}), (\ref{eq4}) has no nonnegative solution. \end{lemma} In view of Lemma \ref{lemma3}, we will assume that $\alpha \eta < T$ for the rest of the paper. Our Banach space is $\mathcal{B} = C_{ld}(\mathbb{T}, \mathbb{R})$ with norm $\|u\| = \sup_{t \in [0,T]\cap \mathbb{T}} |u(t)|$. Define the operator $A\!: \mathcal{B} \to \mathcal{B}$ by \begin{align*} Au(t)&= -\int_0^t \!(t-s)a(s)f(u(s))\, \nabla s - \frac{\alpha t}{T - \alpha \eta} \int_0^\eta \! (\eta - s) a(s) f(u(s)) \, \nabla s\\ & \quad + \frac{t}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(u(s)) \, \nabla s. \end{align*} The function $u$ is a solution of the boundary-value problem (\ref{eq1}), (\ref{eq2}) if and only if $u$ is a fixed point of the operator $A$. \begin{lemma}\label{lemma4} Let $0 < \alpha \eta < T$. If $y \in C_{ld}(\mathbb{T}, [0, \infty))$, then the unique solution $u$ of (\ref{eq3}), (\ref{eq4}) satisfies \begin{equation}\label{ineq1} \min_{t \in [\eta, T]\cap \mathbb{T}} u(t) \geq \gamma \|u \| \end{equation} where \begin{displaymath} \gamma = \min \Big \{ \frac{\alpha \eta}{T}, \frac{\alpha(T - \eta)}{T - \alpha \eta}, \frac{\eta}{T} \Big \}. \end{displaymath} \end{lemma} \noindent {\bf Remark: } Since $\alpha \eta < T$ and since $\eta < T$, it follows that $0 < \gamma < 1$. \begin{definition} \rm Let $\mathcal B$ be a Banach space and let $\mathcal P \subset \mathcal B$ be closed and nonempty. Then $\mathcal P$ is said to be a cone if \begin{enumerate} \item $\alpha u + \beta v \in \mathcal P$ for all $u, v \in \mathcal P$ and for all $\alpha, \beta \geq 0$, and \item $u, -u \in \mathcal P$ implies $u = 0$. \end{enumerate} \end{definition} As in \cite{da} we define the cone \begin{displaymath} \mathcal{P} = \{u \in \mathcal{B} \! : u(t) \geq 0, t \in \mathbb{T} \mbox{ and } \min_{t \in [\eta, T]\cap \mathbb{T}} u(t) \geq \gamma \|u\| \}. \end{displaymath} From Lemma \ref{lemma4} we have $A \! : \mathcal{P} \to \mathcal{P}$. Standard arguments show that the operator $A$ is completely continuous. Before we state the fixed point theorem, we establish some inequalities. Since both $a$ and $f$ are nonnegative then for all $u \in \mathcal{B}$ \begin{equation}\label{A-ineq1} Au(t) \leq \frac{t}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(u(s)) \, \nabla s. \end{equation} Furthermore, \begin{align*} Au(\eta) & = -\int_0^\eta \!(\eta-s)a(s)f(u(s))\, \nabla s - \frac{\alpha \eta} {T - \alpha \eta} \int_0^\eta \! (\eta - s) a(s) f(u(s)) \, \nabla s \\ &\quad + \frac{\eta}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & = -\frac{T}{T - \alpha \eta}\int_0^\eta \! (\eta - s) a(s) f(u(s)) \, \nabla s + \frac{\eta}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & = \frac{\eta T}{T - \alpha \eta} \int_\eta^T \! a(s) f(u(s)) \, \nabla s + \frac{T}{T - \alpha \eta} \int_0^\eta \! s a(s) f(u(s)) \, \nabla s \\ &\quad - \frac{\eta}{T - \alpha \eta} \int_0^T \! s a(s) f(u(s)) \, \nabla s\\ & \geq \frac{\eta}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) f(u(s)) \, \nabla s. \end{align*} That is, \begin{equation}\label{A-ineq2} Au(\eta) \geq \frac{\eta}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) f(u(s)) \, \nabla s. \end{equation} The inequalities (\ref{A-ineq1}) and (\ref{A-ineq2}), which are also found in \cite{da} and \cite{ynr}, will play critical roles in the proofs of our main theorems. We will also need the following fixed point theorem found in \cite{Kra} \begin{theorem}[Krasnosel'ski\u{\i}] \label{thm1} Let $\mathcal B$ be a Banach space and let $\mathcal P \subset \mathcal B$ be a cone. Assume $\Omega_1, \, \Omega_2$ are bounded open balls of $\mathcal B$ such that $0 \in \Omega_1 \subset \overline{\Omega}_1 \subset \Omega_2$. Suppose that \begin{displaymath} A\!: \mathcal P \cap (\overline{\Omega}_2 \setminus \Omega_1) \to \mathcal P \end{displaymath} is a completely continuous operator such that, either \begin{enumerate} \item $\|Au\| \leq \|u\|, \; u \in \mathcal P \cap \partial \Omega_1$ and $\|Au\| \geq \|u\|, \; \; u \in \mathcal P \cap \partial \Omega_2$, or \item $\|Au\| \geq \|u\|, \; u \in \mathcal P \cap \partial \Omega_1$ and $\|Au\| \leq \|u\|, \; \; u \in \mathcal P \cap \partial \Omega_2$. \end{enumerate} Then $A$ has a fixed point in $\mathcal P \cap (\overline{\Omega}_2 \setminus \Omega_1)$. \end{theorem} % ---------------------------------------------------------------- \section{Two Positive Solutions} In this section we use Theorem \ref{thm1} to establish the existence of two positive solutions of the boundary-value problem (\ref{eq1}), (\ref{eq2}). In Theorems \ref{mainthm1} and \ref{mainthm2}, the inequalities we derive that are based on $f_0$ and $f_\infty$ are similar to those found in \cite{da}, \cite{ma1}, and \cite{ynr} and are included for completeness. \begin{theorem}\label{mainthm1} Assume that $f$ satisfies conditions \begin{itemize} \item[(A1)] $f_0 = +\infty, \, f_{\infty} = +\infty$; and \item[(B1)] there exists a $p > 0$ such that if $0 \leq x \leq p$ then $ f(x) \leq \mu p $ where $\mu = \Big ( \frac{T}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s \Big )^{-1}$. \end{itemize} Then the boundary-value problem (\ref{eq1}), (\ref{eq2}) has at least two positive solutions $u_1, u_2 \in \mathcal{P}$ such that \begin{displaymath} 0 < \|u_1\| \leq p \leq \|u_2\|. \end{displaymath} \end{theorem} \begin{proof} Choose $m > 0$ such that \begin{equation}\label{ineq2} \frac{m\eta \gamma}{T - \alpha \eta} \int_\eta^T \!(T - s) a(s) \, \nabla s \geq 1. \end{equation} By condition (A1), ($f_0 = +\infty$), there exists an $0 < r < p$ such that \begin{equation}\label{f0-larger-m} f(u) \geq m u \end{equation} for all $0 \leq u \leq r$. Let $u \in \mathcal{P}$ with $\|u\| = r$. From (\ref{A-ineq2}), (\ref{ineq1}), (\ref{ineq2}) and (\ref{f0-larger-m}) we have \begin{align*} Au(\eta) & \geq \frac{\eta}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & \geq \frac{m\eta}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) u(s) \, \nabla s\\ & \geq \Big ( \frac{m\eta\gamma}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) \, \nabla s \Big ) \|u\|\\ & \geq \|u\|. \end{align*} Define $\Omega_1 = \{ u \in \mathcal{B} \! : \|u \| < r\}$. From the above string of inequalities we have \begin{equation}\label{kros-cond-1} \|Au\| \geq \|u\|, \quad u \in \mathcal{P} \cap \partial \Omega_1. \end{equation} Now consider $u \in \mathcal{P}$ with $\|u\| \leq p$. From (\ref{A-ineq1}) and condition (B1) we have \begin{align*} Au(t) & \leq \frac{t}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & \leq \frac{T}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & \leq \Big ( \frac{T}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s \Big ) \mu p = p. \end{align*} Define $\Omega_2 = \{ u \in \mathcal{B} \! : \|u \| < p\}$. Then \begin{equation}\label{kros-cond-2} \|Au\| \leq \|u\|, \quad u \in \mathcal{P} \cap \partial \Omega_2. \end{equation} Using the inequalities (\ref{kros-cond-1}) and (\ref{kros-cond-2}) there exists, by Theorem \ref{thm1}, a fixed point $u_1$ of $A$ in $\mathcal{P} \cap (\bar{\Omega}_2 \setminus \Omega_1)$. This fixed point satisfies $r \leq \|u_1\| \leq p$. Using condition (A1) again, ($f_{\infty} = \infty$), we know there exists an $R_1 > p$ such that \begin{equation}\label{f8-larger-M} f(u) \geq Mu \end{equation} for all $u \geq R_1$ where $M$ is chosen so that \begin{equation}\label{ineq3} \frac{M\eta \gamma} {T - \alpha \eta} \int_\eta^T \!(T - s) a(s) \, \nabla s \geq 1. \end{equation} Set $R = R_1/\gamma$ and pick $u \in \mathcal{P}$ so that $\|u\| = R$. Since $0 < \gamma < 1$ then $R > R_1 > p$. Furthermore, $\min_{t \in [\eta, T]\cap \mathbb{T}} u(t) \geq \gamma R \geq R_1$. From (\ref{A-ineq2}), (\ref{ineq1}), (\ref{f8-larger-M}), and (\ref{ineq3}) we have \begin{align*} Au(\eta) & \geq \frac{\eta}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & \geq \frac{M\eta}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) u(s) \, \nabla s\\ & \geq \Big ( \frac{M\eta \gamma}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) \, \nabla s \Big ) \|u\|\\ & \geq \|u\|. \end{align*} Define $\Omega_3 = \{ u \in \mathcal{B} \! : \|u \| < R\}$. Then \begin{equation}\label{kros-cond-3} \|Au\| \geq \|u\|, \quad u \in \mathcal{P} \cap \partial \Omega_3. \end{equation} Theorem \ref{thm1} together with (\ref{kros-cond-2}) and (\ref{kros-cond-3}) implies that there exists a fixed point $u_2$ of $A$ that satisfies $p \leq \|u_2\| \leq R$ and the proof is complete. \end{proof} \begin{theorem}\label{mainthm2} Assume that $f$ satisfies conditions \begin{itemize} \item[(A2)] $f_0 = 0, \, f_{\infty} = 0$; and \item[(B2)] there exists a $q > 0$ such that if $\gamma q \leq x \leq q$ then $f(x) \geq \nu q $ where $\nu = \Big( \frac{\eta}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) \, \nabla s \Big)^{-1}$. \end{itemize} Then the boundary-value problem (\ref{eq1}), (\ref{eq2}) has at least two positive solutions $u_1, u_2 \in \mathcal{P}$ such that \begin{displaymath} 0 < \|u_1\| \leq q \leq \|u_2\|. \end{displaymath} \end{theorem} \begin{proof} Choose $m > 0$ such that \begin{equation} \frac{T m}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s \leq 1. \end{equation} By condition (A2), ($f_0 = 0$), there exists an $0 < r < q$ such that \begin{equation} f(u) \leq mu \end{equation} for all $0 \leq u \leq r$. Define $\Omega_1 = \{ u \in \mathcal{B} : \|u\| < r\}$ and let $u \in \mathcal{P} \cap \partial \Omega_1$. Then, \begin{align*} Au(t) & \leq \frac{t}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & \leq \frac{T m}{T - \alpha \eta} \int_0^T \! (T - s) a(s) u(s) \, \nabla s\\ & \leq \Big( \frac{T m}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s \Big) \|u\|\\ & \leq \|u\|. \end{align*} And so, \begin{equation}\label{thm2kros-cond-1} \|Au\| \leq \|u\|, \quad u \in \mathcal{P} \cap \partial \Omega_1. \end{equation} Now define $\Omega_2 = \{u \in \mathcal{B} : \|u\| < q\}$. Notice that if $u \in \mathcal{P} \cap \partial \Omega_2$ then \begin{displaymath} \min_{t \in [\eta, T]\cap \mathbb{T}} u(t) \geq \gamma \|u \| \geq \gamma q. \end{displaymath} By condition (B2) we have \begin{align*} Au(t) & \geq \frac{\eta}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & \geq \Big ( \frac{\eta}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) \, \nabla s \Big ) \nu q\\ & = q \, = \, \|u\|. \end{align*} Hence, \begin{equation}\label{thm2kros-cond-2} \|Au\| \geq \|u\|, \quad u \in \mathcal{P} \cap \partial \Omega_2. \end{equation} Consider the second condition in (A2), $f_{\infty} = 0$. There exists an $R_1 > q$ such that $f(u) \leq Mu$ for all $u \geq R_1$ where $M$ is chosen so that \begin{displaymath} \frac{T M}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s \leq 1. \end{displaymath} There are two cases to consider: $f$ is bounded or $f$ is unbounded. Suppose that $f$ is bounded. Let $K$ be such that $f(u) \leq K$ for all $u$ and choose \begin{displaymath} R = \max \Big \{ 2q, \frac{T K}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s \Big \}. \end{displaymath} Let $u \in \mathcal{P}$ be such that $\|u\| = R$. Then \begin{align*} Au(t) & \leq \frac{t}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & \leq \frac{T K}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s\\ & \leq R. \end{align*} Hence $\|Au\| \leq \|u\|$. Now suppose that $f$ is unbounded. From condition (A2) there exists an $R \geq \frac{R_1}{\gamma}$ such that $f(u) \leq f(R)$ for all $0 < u \leq R$. Since $\gamma < 1$ then $q < R_1 < \frac{R_1}{\gamma} = R$. Let $u \in \mathcal{P}$ be such that $\|u\| \leq R$. Then \begin{align*} Au(t) & \leq \frac{t}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & \leq \frac{T}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(R) \, \nabla s\\ & \leq \frac{T M R}{T - \alpha \eta} \int_0^T \! (T - s) a(s) u \, \nabla s\\ & \leq R. \end{align*} Hence $\|Au\| \leq \|u\|$. In either case, if we define $\Omega_3 = \{ \mathcal{B} : \|u\| < R\}$ then $\Omega_2 \subsetneq \Omega_3$ and \begin{equation}\label{thm2kros-cond-3} \|Au\| \geq \|u\|, \quad u \in \mathcal{P} \cap \partial \Omega_3. \end{equation} The conclusion of the theorem follows by applying Theorem \ref{thm1} to the inequalities (\ref{thm2kros-cond-1}), (\ref{thm2kros-cond-2}), (\ref{thm2kros-cond-3}). \end{proof} % ---------------------------------------------------------------- \section{Countably Many Positive Solutions} In this section we show the existence of countably many positive solutions when $f$ satisfies an oscillatory like growth about a wedge. We begin with a modification of Lemma \ref{lemma4}. \begin{lemma}\label{lemma5} Let $0 < \alpha \eta < T$. Let $\tau_k \in \mathbb{T} \cap (\eta, \rho(T))$. If $y \in C_{ld}(\mathbb{T}, [0, \infty))$, then the unique solution $u$ of (\ref{eq3}), (\ref{eq4}) satisfies \begin{displaymath} \min_{t \in [\tau_k, T]\cap \mathbb{T}} u(t) \geq \gamma_k \|u \| \end{displaymath} where \begin{displaymath} \gamma_k = \min \Big \{ \frac{\alpha \eta \tau_k}{T^2}, \frac{\alpha \eta (T - \tau_k)}{\tau_k(T - \alpha \eta)}, \frac{\eta \tau_k}{T^2} \Big \} \end{displaymath} \end{lemma} \begin{proof} Suppose that $0 \leq \alpha < 1$. Let $t_0 \in (0, T) \cap \mathbb{T}$ be such that $u(t_0) = \|u\|$. By the second boundary condition in (\ref{eq2}) we have $u(\eta) \geq u(T)$. There are two cases to consider. Suppose $t_0 \leq \eta < \rho(T)$ then $\min_{t\in [\tau_k, T]\cap \mathbb{T}} u(t) = u(T)$ and \begin{align*} u(t_0) & \leq u(T) + \frac{u(T) - u(\tau_k)}{T - \tau_k} ( 0 - T)\\ & = \frac{-\alpha \tau_k u(\eta) + T u(\tau_k)}{T - \tau_k}. \end{align*} By Lemma \ref{lemma2}, we know that $\frac{u(\tau_k)}{\tau_k} \leq \frac{u(\eta)}{\eta}$. Hence \begin{align*} u(t_0) & \leq \frac{\frac{\tau_k}{\eta} T u(\eta) - \alpha \tau_k u(\eta)}{T - \tau_k}\\ & \leq \frac{\tau_k(T - \alpha \eta)}{\eta(T - \tau_k)} u(\eta)\\ & \leq \frac{\tau_k(T - \alpha \eta)}{\alpha \eta(T - \tau_k)} u(T). \end{align*} Consequently, $\min_{t \in [\tau_k, T]\cap \mathbb{T}} u(t) = u(T) \geq \frac{\alpha \eta (T - \tau_k)}{\tau_k(T - \alpha \eta)} \|u\|$. Now suppose that $\eta \leq t_0 \leq T$. Again we have $\min_{t \in [\tau_k, T]\cap \mathbb{T}} u(t) = u(T) $. By Lemma \ref{lemma2} we know $\frac{u(\eta)}{\eta} \geq \frac{u(t_0)}{t_0}$. Hence, $u(\eta) \geq \eta \frac{u(t_0)}{t_0}$ and so, $u(T) = \alpha u(\eta) \geq \alpha \eta \frac{u(t_0)}{t_0} \geq \frac{\alpha \eta}{T} u(t_0)$. Since $\tau_k < T$, then $\min_{t \in [\tau_k, T]\cap \mathbb{T}} u(t) = u(T) \geq \frac{\alpha \eta}{T^2} \|u\|$. If $1 \leq \alpha < T/\eta$. Then $u(\eta) \leq u(T)$. Let $t_0$ be such that $u(t_0) = \|u\|$. In this case $t_0 \in [\eta, T] \cap \mathbb{T}$ and $\min_{t \in [\eta, T]\cap \mathbb{T}} u(t) = u(\eta)$. From Lemma \ref{lemma2} we have $u(\eta) \geq \eta \frac{u(t_0)}{t_0}$. Consequently, \begin{displaymath} \min_{t \in [\eta, T]\cap \mathbb{T}} u(t) = u(\eta) \geq \frac{\eta}{t_0} \|u\| \geq \frac{\eta}{T} \|u\| \geq \frac{\eta \tau_k}{T^2} \|u\|, \end{displaymath} Since $\tau_k \geq \eta$, then $\min_{t \in [\tau_k, T]\cap \mathbb{T}} u(t) \geq \frac{\eta \tau_k}{T^2} \|u\|$ and the proof is complete. \end{proof} In Theorem \ref{thm11} we show the existence of a countably infinite number of solutions. We will require that there exists at least one right dense point $\tau^* \in \mathbb{T} \cap (\eta, \rho(T))$. In addition, we will need a countable collection of cones. For each $k \in \mathbb{N}$ define the cone \begin{displaymath} \mathcal{P}_k = \{ u \in \mathcal{B}: u(t) \geq 0, t \in \mathbb{T} \mbox{ and } \min_{t \in [\tau_k, T] \cap \mathbb{T}} u(t) \geq \gamma_k \|u\| \}. \end{displaymath} \begin{theorem}\label{thm11} Let $\tau^* \in \mathbb{T}$ be r.d. and suppose that $\tau^* > \eta$. Let $\{\tau_k\} \subset \mathbb{T}$ be such that $\eta < \tau_1 < \rho(T)$ and $\tau_k \downarrow \tau^*$. Let $t_0$ be such that $t_0 \in [\tau_1 , T) \cap \mathbb{T}$ and $a(t_0) > 0$. Let $\{A_k\}_{k=1}^\infty$ and $\{B_k\}_{k=1}^\infty$ be such that \begin{displaymath} A_{k+1} < \gamma_k B_k < B_k < C B_k < A_k, \quad k \in \mathbb{N} \end{displaymath} where \begin{displaymath} C = \max \Big \{ \Big ( \frac{\eta}{T - \alpha \eta} \int_{\tau_1}^T \! (T - s) a(s) \, \nabla s \Big ) ^{-1}, 1 \Big \}. \end{displaymath} Assume \begin{itemize} \item[(A3)] $f(x) \leq M A_k$ for all $x \in [0, A_k], k \in \mathbb{N}$ where \begin{displaymath} M < \Big ( \frac{T}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s \Big )^{-1}; \end{displaymath} and \item[(B3)] $f(x) \geq C B_k$ for all $x \in [\gamma_k B_k, B_k]$. \end{itemize} Then the boundary-value problem (\ref{eq1}), (\ref{eq2}) has infinitely many positive solutions $\{u_k\}_{k=1}^\infty$ such that $B_k \leq \|u_k\| \leq A_k$ for all $k \in \mathbb{N}$. \end{theorem} \begin{proof} First note that since $\frac{T}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s \geq \frac{\eta}{T - \alpha \eta} \int_{\tau_1}^T \! (T - s) a(s) \, \nabla s$, it follows that $M < C$ (otherwise the theorem is vacuously true). Fix $k \in \mathbb{N}$. Define $\Omega_{1k} = \{u \in \mathcal{B} \! : \|u\| < A_k\}$. Let $u \in \mathcal{P}_k \cap \partial \Omega_{1k}$. Then $u(t) \leq A_k = \|u\|$ for all $t \in [0, T] \cap \mathbb{T}$. So, \begin{align*} Au(t) & \leq \frac{t}{T - \alpha \eta} \int_0^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & \leq \frac{T M}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s A_k\\ & \leq A_k. \end{align*} Thus $\|Au\| \leq A_k = \|u\|$ for $u \in \mathcal{P}_k \cap \partial \Omega_{1k}$. Now define $\Omega_{2k} = \{u \in \mathcal{B} \! : \|u\| < B_k\}$ and let $u \in \mathcal{P}_k \cap \partial \Omega_{2k}$. Let $t \in [\tau_k, T] \cap \mathbb{T}$. Then \begin{displaymath} B_k = \|u\| \geq u(t) \geq \min_{t \in [{\tau_k}, T]\cap \mathbb{T}} u(t) = \geq \gamma_k \|u\| = \gamma_k B_k. \end{displaymath} So, \begin{align*} Au(\eta) & \geq \frac{\eta}{T - \alpha \eta} \int_\eta^T \! (T - s) a(s) f(u(s)) \, \nabla s\\ & \geq \Big ( \frac{\eta}{T - \alpha \eta} \int_{\tau_1}^T \! (T - s) a(s) \, \nabla s \Big ) C B_k\\ & \geq B_k. \end{align*} Thus $\|Au\| \geq B_k = \|u\|$ for all $u \in \mathcal{P}_k \cap \partial \Omega_{2k}$. By Theorem \ref{thm1} there exists a fixed point $u_k$ of $A$ such that $B_k \leq \|u_k\| \leq A_k$. Since $k$ was arbitrary the result follows and the proof is complete. \end{proof} The proofs for our last two theorems require slight modifications of the proofs for Theorems \ref{mainthm1}, \ref{mainthm2} and \ref{thm11} and as such will be omitted. Theorem \ref{thm12} shows the existence of an odd number of solutions and Theorem \ref{thm13} shows the existence of an even number of solutions. It is not difficult to establish other theorems of these forms stating the existence of multiple positive solutions. \begin{theorem}\label{thm12} Let $m \geq 1$ be a fixed integer. Let $\{\tau_k\}_{k=1}^m \subset \mathbb{T}$ be such that $\eta < \tau_{k+1} < \tau_k < \rho(T)$. Let $t_0$ be such that $t_0 \in [\tau_1 , T) \cap \mathbb{T}$ and $a(t_0) > 0$. Let $\{A_k\}_{k=1}^{m}$ and $\{B_k\}_{k=1}^{m}$ be such that \begin{displaymath} A_{k+1} < \gamma_k B_k < B_k < C B_k < A_k, \quad k = 1, 2, \dots m - 1 \end{displaymath} and $0 < B_m < C B_m < A_m$ where \begin{displaymath} C = \max \Big \{ \Big ( \frac{\eta}{T - \alpha \eta} \int_{\tau_1}^T \! (T - s) a(s) \, \nabla s \Big ) ^{-1}, 1 \Big \}. \end{displaymath} Assume \begin{itemize} \item[(A4)] $f_0 = 0$ and $f_{\infty} = +\infty$; \item[(B4)] $f(x) \leq M A_k$ for all $x \in [0, A_k], k = 1, 2, \dots m$ where \begin{displaymath} M < \Big ( \frac{T}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s \Big )^{-1}; \end{displaymath} and \item[(C4)] $f(x) \geq C B_k$ for all $x \in [\gamma_k B_k, B_k], k = 1, 2, \dots m$. \end{itemize} Then the boundary-value problem (\ref{eq1}), (\ref{eq2}) has at least $2m+1$ positive solutions $\{u_k\}_{k=1}^{2m+1}$ such that $0 < \|u_{2m+1}\| \leq B_{m} \leq \|u_{2m}\| \leq A_{m} \leq \dots \leq B_1 \leq \|u_2\| \leq A_1 \leq \|u_1\| < \infty$. \end{theorem} \begin{theorem}\label{thm13} Let $m \geq 1$ be a fixed integer. Let $\{\tau_k\}_{k=1}^m \subset \mathbb{T}$ be such that $\eta < \tau_{k+1} < \tau_k < \rho(T)$. Let $t_0$ be such that $t_0 \in [\tau_1 , T) \cap \mathbb{T}$ and $a(t_0) > 0$. Let $\{A_k\}_{k=1}^m$ and $\{B_k\}_{k=1}^{m-1}$ be such that \begin{displaymath} A_{k+1} < \gamma_k B_k < B_k < C B_k < A_k, \quad k = 1, 2, \dots m-1, \end{displaymath} and $A_m > 0$, where \begin{displaymath} C = \max \Big \{ \Big ( \frac{\eta}{T - \alpha \eta} \int_{\tau_1}^T \! (T - s) a(s) \, \nabla s \Big ) ^{-1}, 1 \Big \}. \end{displaymath} Assume \begin{itemize} \item[(A5)] $f_0 = +\infty$ and $f_{\infty} = +\infty$; \item[(B5)] $f(x) \leq M A_k$ for all $x \in [0, A_k], k = 1, 2, \dots m$ where \begin{displaymath} M < \Big ( \frac{T}{T - \alpha \eta} \int_0^T \! (T - s) a(s) \, \nabla s \Big )^{-1}; \end{displaymath} and \item[(C5)] $f(x) \geq C B_k$ for all $x \in [\gamma_k B_k, B_k]$, $k = 1, 2, \dots m-1$. \end{itemize} Then the boundary-value problem (\ref{eq1}), (\ref{eq2}) has at least $2m$ positive solutions $\{u_k\}_{k=1}^{2m}$ such that $0 < \|u_{2m}\| \leq A_{m} \leq \|u_{2m}\| \leq B_{m-1} \leq \dots \leq B_1 \leq \|u_2\| \leq A_1 \leq \|u_1\| < \infty$. \end{theorem} \begin{thebibliography}{00} \bibitem{ARW} R. P. Agarwal, D. O'Regan and P. J. Y. Wong, ``Positive Solutions of Differential, Difference and Integral Equations,'' Kluwer, Dordrecht, 1999. \bibitem{da} D. 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