\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2003(2003), No. 40, pp. 1--7.\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--2003/40\hfil Discrete three-point problem] {Existence of solutions to higher-order discrete three-point problems} \author[Douglas R. Anderson\hfil EJDE--2003/40\hfilneg] {Douglas R. Anderson} \address{Douglas R. Anderson \hfill\break Department of Mathematics and Computer Science\\ Concordia College \\ Moorhead, MN 56562 USA} \email{andersod@cord.edu} \date{} \thanks{Submitted August 19, 2002. Published April 15, 2003.} \subjclass[2000]{39A10} \keywords{Difference equations, boundary-value problem, Green's function, \hfill\break\indent fixed points, cone} \begin{abstract} We are concerned with the higher-order discrete three-point boundary-value problem \begin{gather*} (\Delta^n x)(t)=f(t,x(t+\theta)), \quad t_1\le t\le t_3-1, \quad -\tau\le \theta\le 1\\ (\Delta^i x)(t_1)=0, \quad 0\le i\le n-4, \quad n\ge 4 \\ \alpha (\Delta^{n-3}x)(t)-\beta (\Delta^{n-2}x)(t)=\eta(t), \quad t_1-\tau-1\le t\le t_1 \\ (\Delta^{n-2}x)(t_2)=(\Delta^{n-1}x)(t_3)=0. \end{gather*} By placing certain restrictions on the nonlinearity and the distance between boundary points, we prove the existence of at least one solution of the boundary value problem by applying the Krasnoselskii fixed point theorem. \end{abstract} \maketitle \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} \section{Introduction} We are concerned with the existence of solutions to the higher-order discrete three-point problem \begin{gather} (\Delta^n x)(t)=f(t,x(t+\theta)), \quad t_1\le t\le t_3-1, \quad -\tau\le \theta\le 1 \label{b0} \\ (\Delta^i x)(t_1)=0, \quad 0\le i\le n-4, \quad n\ge 4 \nonumber \\ \alpha (\Delta^{n-3}x)(t)-\beta (\Delta^{n-2}x)(t)=\eta(t), \quad t_1-\tau-1\le t\le t_1 \nonumber \\ (\Delta^{n-2}x)(t_2)=(\Delta^{n-1}x)(t_3)=0. \label{b2} \end{gather} Here we assume \begin{enumerate} \item[$(i)$] any interval $[a,b]$ is the set of integers $\{a,a+1,\cdots,b-1,b\}$; \item[$(ii)$] $t_{i+1}>t_i+n-1$ to avoid overlap in boundary conditions, $i\in\{1,2\}$; \item[$(iii)$] $f:[t_1,t_3-1]\times[0,\infty)\rightarrow[0,\infty)$; \item[$(iv)$] $\alpha, \beta > 0$, \; $t_3-t_1\ge \tau\ge -1$, and $\theta\in[-\tau,1]$ is constant; \item[$(v)$] $\eta:[t_1-\tau-1,t_1]\rightarrow\mathbb{R}$ with $\eta(t_1)=0$; \item[$(vi)$] $x$ is defined on $[t_1-\tau-1,t_3+n-1]$. \end{enumerate} For the rest of this paper we also have the hypotheses \begin{enumerate} \item[(H1)] $G(t,s)$ on $[t_1,t_3+n-1]\times[t_1,t_3-1]$ is the Green's function for the difference equation $$(\Delta^n u)(t)=0, \;\; t\in[t_1,t_3-1]$$ subject to the boundary conditions \eqref{b2} with $\tau=-1$. \item[(H2)] $g(t,s)$ on $[t_1,t_3+2]\times[t_1,t_3-1]$ is the Green's function for the difference equation $$(\Delta^3 u)(t)=0, \;\; t\in[t_1,t_3-1]$$ subject to the boundary conditions \begin{eqnarray} \alpha u(t_1) - \beta(\Delta u)(t_1) = 0 \nonumber\\ (\Delta u)(t_2) = (\Delta^2 u)(t_3) = 0 \label{gbc} \end{eqnarray} for $\alpha, \beta$ as in $(iv)$. \item[(H3)] $\|x\|_{[t_1-\tau-1,t_3+2]}:=\displaystyle\sup_{t_1-\tau-1\le t\le t_3+2}| (\Delta^{n-3}x)(t)|$. \item[(H4)] For $\Xi:=\{t\in[t_1,t_3+n-1]:t_1\le t+\theta\le t_3-1\}$, $$\Xi_h:=\{t\in\Xi: t_2-h\le t+\theta\le t_2+h\}$$ is nonempty for some $h\in(0,t_3-t_2-2)$, which is nonempty by $(ii)$. \end{enumerate} The corresponding Green's function for the discrete homogeneous problem \newline $(\Delta^3 u)(t)=0$ satisfying the boundary conditions \eqref{gbc}, a slight generalization of that in \cite{An1,An2,An3,An4}, is given via $$\label{greensfunction} g(t,s)= \begin{cases} s\in [t_1,t_2-1] &: \begin{cases} u_1(t,s) &:t\le s+1 \\ v_1(t,s) &:t\ge s+1 \end{cases}\\ & \\ s\in [t_2-1,t_3-1] &: \begin{cases} u_2(t,s) &:t\le s+1 \\ v_2(t,s) &:t\ge s+1 \end{cases} \\ \end{cases}$$ for $t\in[t_1,t_3+2]$ and $s\in[t_1,t_3-1]$, where \begin{gather*} u_1(t,s):=\frac{1}{2}(t-t_1)(2s-t-t_1+3)+\frac{\beta}{\alpha}(s-t_1+1), \\ v_1(t,s):=\frac{1}{2}(s-t_1+2)(s-t_1+1)+\frac{\beta}{\alpha}(s-t_1+1), \\ u_2(t,s):=\frac{1}{2}(t-t_1)(2t_2-t-t_1+1)+\frac{\beta}{\alpha}(t_2-t_1), \\ v_2(t,s):=\frac{1}{2}(t\!-\!t_1)(2t_2-t-t_1+1)+\frac{\beta}{\alpha}(t_2\!-\!t_1) +\frac{1}{2}(t\!-\!s\!-\!1)(t\!-\!s\!-\!2). \end{gather*} \begin{remark} \label{rmk1} \rm As in \cite{An2}, it can be shown that if $$\frac{\beta}{\alpha}(t_2-t_1)+1 > \frac{1}{2}(t_3-t_1+2)(t_3+t_1-2t_2+1),$$ then $$g(t,s)>0$$ for all $t\in[t_1,t_3+2]$, $s\in[t_1,t_3-1]$. Note that if the boundary points satisfy $$\label{boundarydistance} t_3-t_2 \le t_2-t_1-1,$$ then the above inequality holds for any choice of $\alpha,\beta > 0$. Thus throughout this paper we assume that \eqref{boundarydistance} holds. Moreover, as in \cite{An3}, we have the following boundedness result. \end{remark} \begin{lemma} \label{lm1} For all $t\in[t_1,t_3+2]$ and $s\in[t_1,t_3-1]$, $$\label{greenbounds} \ell(t)g(t_2,s)\leq g(t,s)\leq g(t_2,s)$$ where $$\label{g} \ell(t):=\min\left\{\frac{t-t_1}{t_2-t_1},\frac{t_3-t+2}{t_3-t_2+2}\right\}.$$ \end{lemma} \begin{remark}\label{discussion} \rm The following discussion is similar to that found in \cite{hong} for a continuous two-point problem on the unit interval. If $x$ is a solution of \eqref{b0}, \eqref{b2}, it can be written as $$x(t)=\begin{cases} x(-\tau;t) & : t_1-\tau-1\leq t\leq t_1 \\ \displaystyle\sum_{s=t_1}^{t_3-1} G(t,s)f(s,x(s+\theta)) & : t_1\le t\le t_3+n-1 \end{cases}$$ where, using standard first-order linear difference equation methods \cite{kp}, $x(-\tau;t)$ satisfies $$(\Delta^{n-3}x)(-\tau;t) = \left(1+\frac{\alpha}{\beta}\right)^{t-t_1}(\Delta^{n-3}x)(t_1) + \frac{1}{\beta}\sum_{s=t}^{t_1-1}\left(1+\frac{\alpha}{\beta}\right)^{t-s-1}\eta(s)$$ for $t\in[t_1-\tau-1,t_1]$. If $u_0$ is the solution of \eqref{b0}, \eqref{b2} with $f\equiv 0$, then $u_0$ satisfies $$\label{unot} (\Delta^{n-3}u_0)(t)=\begin{cases} \displaystyle\frac{1}{\beta}\sum_{s=t}^{t_1-1}\left(1+\frac{\alpha}{\beta}\right)^{t-s-1}\eta(s) & : t_1-\tau-1 \le t \le t_1 \\ 0 & : t_1\leq t\leq t_3+2; \end{cases}$$ note that actually, using the Green's function, $u_0\equiv 0$ on $[t_1,t_3+n-1]$. If $x$ is any solution of \eqref{b0}, \eqref{b2} set $u(t):=x(t)-u_0(t)$. Then $u(t)\equiv x(t)$ on $[t_1,t_3+n-1]$, and $u$ satisfies $$(\Delta^{n-3}u)(t) =\begin{cases} \left(1+\frac{\alpha}{\beta}\right)^{t-t_1}(\Delta^{n-3}u)(t_1) & : t_1-\tau-1 \leq t\leq t_1 \\ \displaystyle\sum_{s=t_1}^{t_3-1} g(t,s)f(s,u(s+\theta)+u_0(s+\theta)) & : t_1\leq t\leq t_3+2. \end{cases}$$ But this implies $$u(t) =\begin{cases} \left(\frac{\beta}{\alpha}\right)^{n-3} \left(1+\frac{\alpha}{\beta}\right)^{t-t_1}(\Delta^{n-3}u)(t_1) & : t_1-\tau-1 \leq t\leq t_1 \\ \displaystyle\sum_{s=t_1}^{t_3-1} G(t,s)f(s,u(s+\theta)+u_0(s+\theta)) & : t_1\leq t\leq t_3+n-1. \end{cases}$$ \end{remark} \section{Existence of at Least One Solution} We are concerned with proving the existence of solutions of the higher-order discrete nonlinear boundary value problem \eqref{b0}, \eqref{b2}. In light of the above discussion in Remark \ref{discussion}, consider the fixed points of the operator $\mathcal{A}$ defined by $$\mathcal{A} u(t) =\begin{cases} \left(\frac{\beta}{\alpha}\right)^{n-3} \left(1+\frac{\alpha}{\beta}\right)^{t-t_1}(\Delta^{n-3}u)(t_1) & : t_1-\tau-1 \leq t\leq t_1 \\ \displaystyle\sum_{s=t_1}^{t_3-1} G(t,s)f(s,u(s+\theta)+u_0(s+\theta)) & : t_1\leq t\leq t_3+n-1, \end{cases}$$ with domain $\{u:[t_1-\tau-1,t_3+n-1]\rightarrow\mathbb{R}\}$. If $\mathcal{A} u=u$, then a solution $x$ of \eqref{b0}, \eqref{b2} would be given by $$x(t) =\begin{cases} \left(\frac{\beta}{\alpha}\right)^{n-3} \left(1+\frac{\alpha}{\beta}\right)^{t-t_1}(\Delta^{n-3}u)(t_1)+u_0(t) & : t_1-\tau-1 \leq t\leq t_1 \\ u(t) & : t_1\leq t\leq t_3+n-1, \end{cases}$$ where $u_0$ satisfies \eqref{unot}. \begin{remark} \rm In the following discussion we will need an $h\in(0,t_3-t_2-2)$; note that for all $t\in[t_2-h,t_2+h]$, we then have $$\label{uofh} \ell(t)\geq \ell(t_2+h+1)=1-\frac{h+1}{t_3-t_2+2}$$ for all $h\in(0,t_3-t_2-2)$, where $\ell$ is given in \eqref{g}. Moreover, let $k,m>0$ such that \begin{align} k^{-1}:&=\sum_{s=t_1}^{t_3-1}g(t_2,s) \label{k} \\ & =\frac{1}{6}(t_2-t_1+1)(t_2-t_1)(3t_3-2t_2-t_1+2) \nonumber \\ & \;\;\;\;\;\; +\frac{\beta}{2\alpha}(t_2-t_1)(2t_3-t_2-t_1+1) \nonumber \end{align} and \begin{align} m^{-1}:&=\ell(t_2+h+1)\sum_{s=t_2-h}^{t_2+h}g(t_2,s) \label{m} \\ &=\frac{1}{6}\left(1-\frac{h+1}{t_3-t_2+2}\right) \left[(t_2-t_1+1)^{\underline{2}}\;(t_2-t_1+3h+5) \right. \nonumber \\ & \;\;\;\;\;\; \left. -(t_2-t_1-h+2)^{\underline{3}} +\frac{3\beta}{\alpha}(4ht_2+2t_2-4ht_1-2t_1-h^2+h)\right], \nonumber \end{align} where we have used the so-called falling factorial power \cite{kp} $$b^{\underline{r}}:=b(b-1)(b-2)\cdots(b-r+1).$$ Finally, set $$\label{Mnot} M_0:= \|u_0\|_{[t_1-\tau-1,t_3+2]}$$ for $u_0$ as in \eqref{unot}. \end{remark} We will employ the following fixed point theorem due to Krasnoselskii \cite{Kr}. \begin{theorem}\label{fixedpt} Let $E$ be a Banach space, $P\subseteq E$ be a cone, and suppose that $\Omega_1$, $\Omega_2$ are bounded open balls of $E$ centered at the origin with $\overline{\Omega}_1\subset\Omega_2$. Suppose further that $\mathcal{A}:P\cap(\overline{\Omega}_2\setminus\Omega_1)\to P$ is a completely continuous operator such that either \begin{enumerate} \item[$(i)$] $\|\mathcal{A} u\| \leq \|u\|$, $u\in P\cap\partial\Omega_1$ and $\|\mathcal{A} u\| \geq \|u\|$, $u\in P\cap\partial\Omega_2$, or \item[$(ii)$] $\|\mathcal{A} u\| \geq \|u\|$, $u\in P\cap\partial\Omega_1$ and $\|\mathcal{A} u\| \leq \|u\|$, $u\in P\cap\partial\Omega_2$ \end{enumerate} holds. Then $\mathcal{A}$ has a fixed point in $P\cap(\overline{\Omega}_2\setminus\Omega_1)$. \end{theorem} \begin{theorem}\label{theorem4} Let $k, m, M_0$ be as in \eqref{k}, \eqref{m}, \eqref{Mnot}, respectively, and suppose the following conditions are satisfied. \begin{itemize} \item[$(C_1)$] There exists $p > 0$ such that $f(t,w)\leq kp$ for $t \in [t_1,t_3-1]$ and $0\leq \|w\| \leq p+M_0$. \item[$(C_2)$] There exists $q > 0$ such that $f(t,w)\geq mq$ for $t \in \Xi_h$ and $q\ell(t_2+h+1)\leq \|w\| \le q$, for $h\in(0,t_3-t_2-2)$ and $\Xi_h$ as in $(H4)$. \end{itemize} Then \eqref{b0}, \eqref{b2} has a solution $x=u+u_0$ such that $\|x\|_{[t_1-\tau-1,t_3+2]}$ lies between $\max\{0,p-M_0\}$ and $q+M_0$. \end{theorem} \begin{proof} Many of the techniques employed here are as in \cite{He, hong}. Let $\mathbb{B}$ denote the Banach space $\{u:[t_1-\tau-1,t_3+n-1]\rightarrow\mathbb{R}\}$ with the norm $$\|u\|_{[t_1-\tau-1,t_3+2]}=\sup_{t\in[t_1-\tau-1,t_3+2]}|(\Delta^{n-3}u)(t)|.$$ Define the cone $\mathbb{P}\subset\mathbb{B}$ by $$\mathbb{P}=\{u\in\mathbb{B}:\displaystyle\min_{t\in[t_2-h,t_2+h]}(\Delta^{n-3}u)(t) \geq \ell(t_2+h+1)\|u\|_{[t_1-\tau-1,t_3+2]}\}.$$ Consider the mapping $\mathcal{A}:\mathbb{P}\rightarrow\mathbb{B}$ via $$\mathcal{A} u(t) =\begin{cases} \left(\frac{\beta}{\alpha}\right)^{n-3} \left(1+\frac{\alpha}{\beta}\right)^{t-t_1}(\Delta^{n-3}u)(t_1) & : t_1-\tau-1 \leq t\leq t_1 \\ \displaystyle\sum_{s=t_1}^{t_3-1} G(t,s)f(s,u(s+\theta)+u_0(s+\theta)) & : t_1\leq t\leq t_3+n-1. \end{cases}$$ Then $$\Delta^{n-3}(\mathcal{A} u)(t) =\begin{cases} \left(1+\frac{\alpha}{\beta}\right)^{t-t_1}\displaystyle\sum_{s=t_1}^{t_3-1} g(t_1,s)f(s,u(s+\theta)+u_0(s+\theta)) & \\ \displaystyle\sum_{s=t_1}^{t_3-1} g(t,s)f(s,u(s+\theta)+u_0(s+\theta)) & \end{cases}$$ so that $\Delta^{n-3}(\mathcal{A} u)(t)\leq \Delta^{n-3}(\mathcal{A} u)(t_1)$ for $t_1-\tau-1\le t\le t_1$. In other words, $\|\mathcal{A} u\|_{[t_1-\tau-1,t_3+2]} = \|\mathcal{A} u\|_{[t_1,t_3+2]}$. It follows for $h\in(0,t_3-t_2-2)$ and $t\in[t_2-h,t_2+h]$ that \begin{align*} \Delta^{n-3}(\mathcal{A} u)(t) & = \sum_{s=t_1}^{t_3-1} g(t,s)f(s,u(s+\theta)+u_0(s+\theta)) \\ & \ge \ell(t)\sum_{s=t_1}^{t_3-1} g(t_2,s)f(s,u(s+\theta)+u_0(s+\theta)) \\ & \ge \ell(t_2+h+1)\|\mathcal{A} u\|_{[t_1-\tau-1,t_3+2]} \end{align*} by properties of the Green's function \eqref{greenbounds}, so that $\mathcal{A}:\mathbb{P}\rightarrow\mathbb{P}$. Without loss of generality, we may assume $0 < p < q$. Define the bounded open balls $$\Omega_p = \{u \in \mathbb{B}: \|u\|_{[t_1-\tau-1,t_3+2]} < p \},$$ and $$\Omega_q = \{u \in \mathbb{B}: \|u\|_{[t_1-\tau-1,t_3+2]} < q \};$$ then $0 \in \Omega_p \subset \Omega_q$. If $u\in\mathbb{P}\cap\partial\Omega_p$, then $\|u\| = p$ and $$|(\Delta^{n-3}u)(t)+(\Delta^{n-3}u_0)(t)|\le p+M_0$$ for all $t\in[t_1,t_3+2]$. As a result, \begin{align*} \|\mathcal{A} u\|&= \sum_{s=t_1}^{t_3-1}g(t_2,s)f(s,u(s+\theta)+u_0(s+\theta)) \\ &\leq kp\sum_{s=t_1}^{t_3-1}g(t_2,s) \\ &=p \\ &=\|u\| \end{align*} using $(C_1)$ and \eqref{k}. Thus, $\|\mathcal{A} u\| \leq \|u\|$ for $u \in \mathbb{P} \cap \partial \Omega_p$. Similarly, let $u\in\mathbb{P}\cap\partial\Omega_q,$ so that $\|u\| = q$. Then for $s\in\Xi_h$, $$(\Delta^{n-3}u)(s+\theta)\ge \min_{t\in[t_2-h,t_2+h]}(\Delta^{n-3}u)(t) \geq \|u\|\ell(t_2+h+1)$$ for all $h\in(0,t_3-t_2-2)$ and $\ell(\cdot)$ as in \eqref{uofh}. As a result, $$q\ell(t_2+h+1) \leq (\Delta^{n-3}u)(s+\theta)+(\Delta^{n-3}u_0)(s+\theta) \leq q$$ for $s\in\Xi_h$, since $\Delta^{n-3}u_0\equiv 0$ on $[t_1,t_3+2]$ by Remark \ref{discussion}. It follows that \begin{align*} \|\mathcal{A} u\|&=\sum_{s=t_1}^{t_3-1}g(t_2,s)f(s,u(s+\theta)+u_0(s+\theta)) \\ &\geq \sum_{s\in\Xi_h}g(t_2,s)f(s,u(s+\theta)+u_0(s+\theta)) \\ &\geq mq\ell(t_2+h+1)\sum_{s=t_2-h}^{t_2+h}g(t_2,s) \\ &=q \\ &= \|u\| \end{align*} by $(C_2)$ and \eqref{m}. Consequently, $\|\mathcal{A} u\| \geq \|u\|$ for $u \in \mathbb{P} \cap \partial \Omega_q$. By Theorem~\ref{fixedpt}, $\mathcal{A}$ has a fixed point $u \in \mathbb{P} \cap (\overline{\Omega}_q \setminus \Omega_p)$; i.e., $p\le \|u\|\le q$. Therefore the discrete problem \eqref{b0}, \eqref{b2} has a solution $x=u+u_0$ such that $p-M_0 \leq \|x\| \leq q+M_0$, if \$M_0