\documentclass[twoside]{article} \usepackage{amssymb} % used for R in Real numbers \pagestyle{myheadings} \setcounter{page}{129} \markboth{\hfil Multiple Solutions to a Difference Equation \hfil}% {\hfil Susan D. Lauer \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Differential Equations and Computational Simulations III}\newline J. Graef, R. Shivaji, B. Soni, \& J. Zhu (Editors)\newline Electronic Journal of Differential Equations, Conference~01, 1997, pp 129--136. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp 147.26.103.110 or 129.120.3.113 (login: ftp)} \vspace{\bigskipamount} \\ Multiple Solutions to a Boundary Value Problem for an n-th Order Nonlinear Difference Equation \thanks{ {\em 1991 Mathematics Subject Classifications:} 39A10, 34B15. \hfil\break\indent {\em Key words and phrases:} n-th order difference equation, boundary value problem, \hfil\break\indent superlinear, sublinear, fixed point theorem, Green's function, discrete, nonlinear. \hfil\break\indent \copyright 1998 Southwest Texas State University and University of North Texas. \hfil\break\indent Published November 12, 1998.} } \date{} \author{Susan D. Lauer} \maketitle \begin{abstract} We seek multiple solutions to the n-th order nonlinear difference equation $$\Delta^n x(t)= (-1)^{n-k} f(t,x(t)),\quad t \in [0,T]$$ satisfying the boundary conditions $$x(0) = x(1) = \cdots = x(k - 1) = x(T + k + 1) = \cdots = x(T+ n) = 0\,.$$ Guo's fixed point theorem is applied multiple times to an operator defined on annular regions in a cone. In addition, the hypotheses invoked to obtain multiple solutions to this problem involves the condition (A) $f:[0,T] \times {\mathbb R}^+ \to {\mathbb R}^+$ is continuous in $x$, as well as one of the following: (B) $f$ is sublinear at $0$ and superlinear at $\infty$, or (C) $f$ is superlinear at $0$ and sublinear at $\infty$. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \section{Introduction} Define the operator $\Delta$ to be the forward difference $$\Delta u(t) = u(t+1)-u(t),$$ and then for $i \geq 1$ define $$\Delta ^i u(t)=\Delta(\Delta^{i-1} u(t)).$$ For $a \leq b$ integers define $[a,b]=\lbrace a,a+1,\ldots,b-1,b \rbrace $. Let the integers $n, T \geq 2 $ be given, and choose $k \in \lbrace 1,2,\ldots,n-1 \rbrace$. Consider the nth order nonlinear difference equation \begin{equation} \label {e1} \Delta ^n x(t)=(-1)^{n-k} f(t,x(t)), \quad t \in [0,T], \end{equation} satisfying the boundary conditions \begin{equation} \label {e2} x(0)=x(1)= \cdots =x(k-1)=x(T+k+1)= \cdots =x(T+n)=0. \end{equation} To simplify the discussion of the desired properties of the function $f$ define the following four functions: $$\matrix{f_{0,m}={\lim\limits_{u\rightarrow 0^+}} \ {\min\limits_{t \in [k,T+k]}} \ { f(t,u) \over u}, \quad \quad & f_{\infty,m}=\lim\limits_{u\rightarrow +\infty} \ \min\limits_{t \in [k,T+k]} \ {\frac {f(t,u)} {u}}, \cr \cr f_{0,M}=\lim\limits_{u\rightarrow 0^+} \ \max\limits_{t \in [k,T+k]} \ {\frac {f(t,u)} {u}}, {\rm\ \ and} & f_{\infty,M}= \lim\limits_{u\rightarrow +\infty} \ \max\limits_{t \in [k,T+k]} \ {\frac {f(t,u)} {u}.}}$$ We seek to prove the existence of multiple positive solutions to (\ref{e1}) and (\ref{e2}) where \begin{itemize} \item [(A)] $f:[0,T] \times {\mathbb R} ^+ \rightarrow {\mathbb R} ^+ $ is continuous in $ x $ , where ${\mathbb R} ^+$ denotes the nonnegative reals. \end{itemize} We also require that one of the following sublinearity and superlinearity conditions on the function $f$ holds: \begin{itemize} \item [(B)] $f_{0,m}=+\infty$ \quad and \quad $f_{\infty,m}=+\infty$, or \item [(C)]$f_{0,M}=0$ \quad \quad and \quad $f_{\infty,M}=0$. \end{itemize} We apply Guo's Fixed point theorem, Guo and Lakshmikantham \cite {GL}, using cone methods to accomplish this. This technique was first applied to differential equations in the landmark paper by Erbe and Wang \cite {EW} using Krasnosel'ski{\u i}'s fixed point theorem, Krasnosel'ski{\u i} \cite {Kr}. A key to applying this fixed point theorem involves discrete concavity of solutions of the boundary value problem in conjunction with a lower bound on an appropriate Green's function. This work constitutes a complete generalization of the paper by Eloe, Henderson and Kaufmann \cite {EHK} which we use extensively. We also utilize techniques from Hartman \cite {Ha}, Merdivenci \cite {Me2}, and Peterson \cite {Pe}. Extensive use of the results by Eloe \cite {El} concerning a lower bound for the Green's function will be made. \section{Preliminaries} Let $G(t,s)$ be the Green's function for the disconjugate boundary value problem \begin{equation} \label {e3} Lx(t) \equiv \Delta ^n x(t) = 0, \quad t \in [0,T] \end{equation} and satisfying (\ref{e2}). The characterization of the Green's function can be found in Kelley and Peterson \cite{KP}. We will use $G(t,s)$ as the kernel of an integral operator preserving a cone in a Banach space, the setting for our fixed point theorem. A closed, non-empty subset $\cal P$ of a Banach space $\cal B$ is said to be a $cone$ provided (i) $au+bv \in \cal P$ for all $u,v \in \cal P$ and for all $a,b \geq 0$, and (ii) $u, -u \in \cal P$ implies $u=0$. Repeated application of the following fixed point theorem from Guo, Guo and Lakshmikantham \cite {GL}, will yield two solutions to (\ref{e1}) and (\ref{e2}). \begin{theorem} \label {t1} Let $\cal B$ be a Banach space and ${\cal P}\subset {\cal B}$ be a cone. Let $\Omega_1$ and $\Omega_2$ be two bounded open sets in $\cal B$ such that $0 \in \Omega _1 \subset \overline{\Omega}_1 \subset \Omega _2$. Let $${\cal H} : \quad {\cal P} \cap ( \overline {\Omega} _2 \setminus \Omega _1) \rightarrow {\cal P} $$ be a completely continuous operator satisfying either \begin{itemize} \item [(i)] $\| {Hx} \| \leq \| x \|$, $x \in {\cal P} \cap {\partial} \Omega _1$, and $ \| Hx \| \geq \| x \|$, $x \in {\cal P} \cap \partial \Omega _2 $, or \item [(ii)] $\| {Hx} \| \geq \| x \|$, $x \in {\cal P} \cap \partial \Omega _1$, and $ \| Hx \| \leq \| x \|$, $x \in {\cal P} \cap \partial \Omega _2 $. \end{itemize} Then $\cal H$ has a fixed point in ${\cal P} \cap \left( {\overline {\Omega}} _2 \setminus \Omega _1 \right )$. \end{theorem} Two applications of \ref {t1} to the problem (\ref{e1}) and (\ref{e2}) following along the lines of methods incorporated by Eloe, Henderson and Kaufmann \cite {EHK} will be performed. Note that $x(t)$ is a solution of (\ref{e1}) and (\ref{e2}) if and only if $$x(t) = (-1)^{n-k} \sum\limits_{s=0} ^{T} G(t,s)f(s,x(s)), \qquad t \in [0,T+n]\,.$$ Hartman \cite {Ha} extensively studied the boundary value problem (\ref{e1}) and (\ref{e2}) with $(-1)^{n-k} f(t,u) \geq 0$. Eloe \cite {El} employed lemmas from Hartman to arrive at the following theorem that gives a lower bound for the solution to the class of boundary value problems studied by Hartman. \begin{theorem} \label {t2} Assume that $u$ satisfies the difference inequality $(-1)^{n-k} \Delta ^n u(t) \geq 0, \enspace t \in [0,T]$, and the homogeneous boundary conditions, (\ref{e2}). Then for $t \in [k,T+k]$, $$(-1)^{n-k} \Delta ^n u(t) \geq \frac {T! \enspace \nu !} {(T+ \nu )!} \| u \|,$$ where $\|u\|=\max\limits _{t \in [k,T+k]} \vert u(t) \vert$ and $\nu = \max \{k,n-k\}$. \end{theorem} We remark that Agarwal and Wong \cite {AW3} have recently sharpened the inequality of Theorem \ref {t2}. This sharper inequality is of little consequence for this work. Eloe also contributed the following corollary. \begin{corollary} \label {c1} Let $G(t,s)$ denote the Green's function for the boundary value problem, (\ref{e3}) and (\ref{e2}). Then for all $s \in [0,T], \enspace t \in [k,T+k]$, $$(-1)^{n-k} \Delta ^n G(t,s) \geq \frac {T! \enspace \nu !} {(T+ \nu )!} \| G(\cdot,s) \|,$$ where $\|G(\cdot , s)\|=\max\limits _{t \in [k,T+k]} \vert G(t,s) \vert$ and $\nu = \max \{k,n-k\}$. \end{corollary} To fulfill the hypotheses of Theorem \ref {t1} let {$ { {\cal B}=\lbrace u:[0,T+n]\rightarrow {\mathbb R} \vert}\\ u(0)=u(1)= \cdots =u(k-1)=u(T+k+1)= \cdots =u(T+n)=0 \rbrace$ with $\| u \| =\max \limits _{t \in [k,T+k]} \vert u(t) \vert$. Now $( {\cal B}, \| \cdot \|)$ is a Banach space. Let \begin{equation} \label {e4} \sigma = \frac {T! \enspace \nu !} {(T+ \nu )!} \end{equation} with $\nu = \max \lbrace k,n-k \rbrace$ and define a cone $${\cal P} = \lbrace u \in {\cal B} \enspace \vert \enspace u(t) \geq 0 \enspace {\tt on } \enspace [0,T+n] \enspace {\tt and } \enspace \min\limits _{t \in [k,T+k]} u(t) \geq \sigma \|u\| \rbrace . $$ \section{Main Results} We first seek two solutions to the case when $f$ is sublinear at 0 and superlinear at $\infty$. Define \begin{equation} \label {e5} \eta = \left ( \sum \limits _{s=0} ^{T} \|G( \cdot , s \| \right ) ^{-1}. \end{equation} \begin{theorem} \label {t3} Assume $f(t,x)$ satisfies conditions (A) and (B). Suppose there exists $p > 0$ such that if $0 \leq u(t) \leq p$, $t \in [0, T]$, then $f(t,u) \leq \eta p$. Then the boundary value problem (\ref{e1}) and (\ref{e2}) has at least two positive solutions $u_1 , u _ 2 \in {\cal P}$ satisfying $0 \leq \|u_1 \| \leq p \leq \|u _2 \|$. \end{theorem} \paragraph{proof} Define a summation operator ${\cal H}: \enspace {\cal P} \rightarrow {\cal B}$ by \begin{equation} \label {e6} {\cal H}x(t)=(-1)^{n-k} \sum\limits_{s=0} ^{T} G(t,s)f(s,x(s)), \qquad x \in {\cal P} \end{equation} Now ${\cal H}: \enspace {\cal P} \rightarrow {\cal P}$ and is completely continuous. Choose $\alpha > 0$ such that \begin{equation} \label {e7} \alpha \sigma ^2 \sum\limits _{s=k} ^{T} \|G( \cdot , s ) \| \geq 1. \end{equation} By the sublinearity of $f$ at 0 there exists $00$ such that \begin{equation} \label {e10} \omega \sigma ^2 \sum\limits _{s=k} ^{T} \|G( \cdot , s ) \| \geq 1. \end{equation} By the superlinearity of $f$ at infinity there exists $R_1 >0$ such that $f(t,u) \geq \omega u $ for all $u \geq R _1$, $t \in [0,T+n]$. Let $R=\max \lbrace 2p, R_1 \rbrace$. Now for $x \in {\cal P}$ with $\|x\|=R$ \begin{eqnarray*} {\cal H}x(t) & = & (-1)^{n-k} \sum\limits_{s=0} ^{T} G(t,s)f(s,x(s)) \\ & \geq & \sigma \sum\limits_{s=0} ^{T} \| G( \cdot,s) \| f(s,x(s)) \\ & \geq & \omega \sigma \sum\limits_{s=0} ^{T} \| G(\cdot ,s) \| x(s) \\ & \geq & \omega \sigma ^2 \sum\limits_{s=k} ^{T} \| G(\cdot ,s)\| ^{ } \|x\| \\ & \geq & \|x\|, \qquad t \in [k,T+k]. \end{eqnarray*} Therefore $\|{\cal H} x \| \geq \|x\|$. Hence if we set $$\Omega _3 = \lbrace u \in {\cal B} \enspace\vert\enspace \| u\| 0$ such that if $\sigma q \leq u(t) \leq q$, $t \in [k,T+k]$, then $f(t,u) \geq \tau q$, where \begin{equation} \label {e12} \tau = \left ( \sigma \sum \limits _{s=k} ^{T} \|G( \cdot , s ) \| \right ) ^{-1} . \end{equation} Then the boundary value problem (\ref{e1}) and (\ref{e2}) has at least two positive solutions $u_1 , u _ 2 \in {\cal P}$ such that $0 \leq \|u_1 \| \leq q \leq \|u _2 \|$. \end{theorem} \paragraph{proof} Define the summation operator as in (\ref{e6}) and define $\eta$ as in (\ref{e5}). By the superlinearity of $f$ at 0 there exists $00$, there exists a $\xi>0$ such that for all $u \geq 0$, $t \in [0,T+k]$, $f(t,u) \leq \xi + \varepsilon u$. Let $\varepsilon = {\eta \over 2}$, where $\eta$ is defined by (\ref{e5}) and select a corresponding $\xi$. Let $R= \max \lbrace 2q,2{\xi \over \eta} \rbrace $. Then for $x \in {\cal P}$ with $\|x\|=R$ \begin{eqnarray*} {\cal H}x(t) & = & (-1)^{n-k} \sum\limits_{s=0} ^{T} G(t,s)f(s,x(s)) \\ & \leq & \sum\limits_{s=0} ^{T} \| G( \cdot,s) \| [ \xi + \varepsilon x(s)] \\ & \leq & \xi \sum\limits_{s=0} ^{T} \| G(\cdot ,s)\| \enspace + \enspace \varepsilon \sum\limits_{s=0} ^{T} \| G(\cdot ,s)\| x(s) \\ & \leq & {\xi \over \eta} + \varepsilon \sum\limits_{s=0} ^{T} \| G(\cdot ,s)\| ^ { } \|x\| \\ & \leq & {R \over 2} + {\|x\| \over 2} = \|x\|, \qquad t \in [0,T+k]. \end{eqnarray*} Therefore $\|{\cal H} x \| \leq \|x\|$. Hence if we set $$\Omega _3 = \lbrace u \in {\cal B} \enspace\vert\enspace \| u\|