\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 91, pp. 1--7.\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-2004/91\hfil Solution matching on a time scale] {Solution matching for a three-point boundary-value problem on a time scale} \author[M. Eggensperger, E. R. Kaufmann, N. Kosmatov\hfil EJDE-2004/91\hfilneg] {Martin Eggensperger, Eric R. Kaufmann, Nickolai Kosmatov} % in alphabetical order \address{Martin Eggensperger\hfill\break General Studies\\ Southeast Arkansas College\\ Pine Bluff, Arkansas, USA} \email{meggensperger@seark.edu} \address{Eric R. Kaufmann\hfill\break Department of Mathematics and Statistics\\ University of Arkansas at Little Rock\\ Little Rock, Arkansas 72204-1099, USA} \email{erkaufmann@ualr.edu} \address{Nickolai Kosmatov\hfill\break Department of Mathematics and Statistics\\ University of Arkansas at Little Rock\\ Little Rock, Arkansas 72204-1099, USA} \email{nxkosmatov@ualr.edu} \date{} \thanks{Submitted May 14, 2004. Published July 8, 2004.} \subjclass[2000]{34B10, 34B15, 34G20} \keywords{Time scale; boundary-value problem; solution matching} \begin{abstract} Let $\mathbb{T}$ be a time scale such that $t_1, t_2, t_3 \in \mathbb{T}$. We show the existence of a unique solution for the three-point boundary value problem \begin{gather*} y^{\Delta\Delta\Delta}(t) = f(t, y(t), y^\Delta(t), y^{\Delta\Delta}(t)), \quad t \in [t_1, t_3] \cap \mathbb{T},\\ y(t_1) = y_1, \quad y(t_2) = y_2, \quad y(t_3) = y_3\,. \end{gather*} We do this by matching a solution to the first equation satisfying a two-point boundary conditions on $[t_1, t_2] \cap \mathbb{T}$ with a solution satisfying a two-point boundary conditions on $[t_2, t_3] \cap \mathbb{T}$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \section{Introduction} Bailey, Shampine and Waltman \cite{BSW} were the first to use solution matching techniques to obtain solutions of two-point boundary value problems for the second order equation $y'' = f(x, y, y')$ by matching solutions of initial value problems. Since then, many authors have used this technique on three-point boundary value problems on an interval $[a, c]$ for an $n^{th}$ order differential equation by piecing together solutions of two-point boundary value problems on $[a, b]$, where $b \in (a, c)$ is fixed, with solutions of two-point boundary value problems on $[b, c]$; see for example, Barr and Sherman \cite{BS}, Das and Lalli \cite{DL}, Henderson \cite{JH1, JH2}, Henderson and Taunton \cite{HT}, Lakshmikantham and Murty \cite{LM}, Moorti and Garner \cite{MG}, and Rao, Murty and Rao \cite{RMR}. All the above cited works considered boundary value problems for differential equations. In this work, we will use the solution matching technique to obtain a solution to a three-point boundary value problem for a $\Delta$-differential equation on a time scale. The theory of time scales was introduced by Stephan Hilger, \cite{sh}, as a means of unifying theories of differential equations and difference equations. Three excellent sources about dynamic systems on time scales are the books by Bohner and Peterson \cite{mbap}, Bohner and Peterson \cite{mbap2}, and Kaymakcalan {\em et.\ al.\/}, \cite{kls}. The definitions below can be found in \cite{mbap}. A {\em time scale} $\mathbb{T}$ is a closed nonempty subset of $\mathbb{R}$. For $t < \sup \mathbb{T}$ and $r > \inf \mathbb{T}$, 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*} 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\/}. 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\/}. If $\mathbb{T}$ has a left-scattered maximum at $M$, then we define $\mathbb{T}^\kappa = \mathbb{T} \setminus \{M\}$. Otherwise we define $\mathbb{T}^\kappa = \mathbb{T}$. If $\mathbb{T}$ has a right-scattered minimum at $m$, then we define $\mathbb{T}_\kappa = \mathbb{T} \setminus \{m\}$. Otherwise we define $\mathbb{T}_\kappa = \mathbb{T}$. We say that the function $x$ has a {\em generalized zero (g.z.)}\ at $t$ if $x(t) = 0$ or if $x(\sigma(t)) \cdot x(t) < 0$. In the latter case, we would say the generalized zero is in the real interval $(t, \sigma(t))$. 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 $s \in U$. \noindent{\bf Remarks:} If $\mathbb{T} = \mathbb{R}$, then $x^\Delta(t) = x^\nabla(t) = x'(t)$. If $\mathbb{T} = \mathbb{Z}$, then $x^\Delta(t) = x(t+1) - x(t)$ is the forward difference operator while $x^\nabla(t) = x(t) -x(t-1)$ is the backward difference operator. Let $\mathbb{T}$ be a time scale such that $t_1, t_2, t_3 \in \mathbb{T}$. We consider the existence of solutions of the three-point boundary value problem \begin{gather} y^{\Delta\Delta\Delta}(t) = f(t, y(t), y^\Delta(t), y^{\Delta\Delta}(t)), \, t \in (t_1, t_3) \cap \mathbb{T},\label{eq1}\\ y(t_1) = y_1, \, \, y(t_2) = y_2, \, \, y(t_3) = y_3.\label{eq2} \end{gather} We obtain solutions by matching a solution of \eqref{eq1} satisfying two-point boundary conditions on $[t_1, t_2] \cap \mathbb{T}$ to a solution of \eqref{eq1} satisfying two-point boundary conditions on $[t_2, t_3] \cap \mathbb{T}$. In particular, we will give sufficient conditions such that if $y_1(t)$ is the solution of \eqref{eq1} satisfying the boundary conditions $y(t_1) = y_1, y(t_2) = y_2, y^{\Delta^j}(t_2) = m$, ($j = 1$ or $2$) and $y_2(t)$ is $y(t_2) = y_2, y^{\Delta^j}(t_2) = m, \, y(t_3) = y_3$, (using the same $j$), then the solution of \eqref{eq1}, \eqref{eq2} is \begin{displaymath} y(t) = \begin{cases} y_1(t), & t \in [t_1, t_2] \cap \mathbb{T},\\ y_2(t), & t \in [t_2, t_3] \cap \mathbb{T}\,. \end{cases} \end{displaymath} We will assume that $f:\mathbb{T} \times \mathbb{R}^3 \rightarrow \mathbb{R}$ is continuous and that solutions of initial value problems for \eqref{eq1} exist and are unique on $[t_1,t_3] \cap \mathbb{T}$. Moreover, we require that $t_2 \in \mathbb{T}$ is dense and fixed throughout. In addition to these hypotheses, we suppose that there exists a function $g: \mathbb{T} \times \mathbb{R}^3 \rightarrow \mathbb{R}$ such that: \begin{itemize} \item[(A)] For each $v_3,u_3 \in \mathbb{R}$ the function $f$ satisfies \begin{displaymath} f(t, v_1, v_2, v_3) - f(t, u_1, u_2, u_3) > g(t, v_1 - u_1, v_2 - u_2, v_3 - u_3) \end{displaymath} when $t \in (t_1,t_2] \cap \mathbb{T}, \, u_1 - v_1 \geq 0$, and $u_2 - v_2 < 0$, or when $t \in [t_2, t_3) \cap \mathbb{T}, \, u_1 - v_1 \leq 0$, and $u_2 -v_2 < 0$ \item[(B)] There exists $\varepsilon_1 > 0$ such that, for each $0 < \varepsilon < \varepsilon_1$, \ the initial value problem \begin{gather*} y^{\Delta \Delta \Delta}(t) = g(t, y(t), y^\Delta(t), y^{\Delta \Delta}(t)), \quad t \in [t_1, t_3] \cap \mathbb{T},\\ y(t_2) = 0, \quad y^{\Delta \Delta}(t_2) = 0, \quad y^\Delta(t_2) = \varepsilon, \end{gather*} has a solution $z$ such that $z^\Delta$ does not change sign on $[t_1,t_3] \cap \mathbb{T}$ \item[(C)] There exists $\varepsilon_2 > 0\,$such that, for each $0 < \varepsilon < \varepsilon_2$, the initial value problem \begin{gather*} y^{\Delta \Delta \Delta}(t) = g(t, y(t), y^\Delta(t), y^{\Delta \Delta}(t)), \quad t \in [t_1, t_3] \cap \mathbb{T}\,,\\ y(t_2) = 0, \quad y^\Delta(t_2) = 0, \quad y^{\Delta\Delta}(t_2) = \varepsilon (-\varepsilon) \end{gather*} has a solution $z$ on $[t_2, t_3] \cap \mathbb{T}, ([t_1, t_2] \cap \mathbb{T})$, such that $z^{\Delta \Delta}$ does not change sign on $[t_2, t_3] \cap \mathbb{T}, \, ([t_1, t_2] \cap \mathbb{T})$ \item[(D)] For each $w \in \mathbb{R}$, the function $g$ satisfies $g(t, v_1, v_2, w) \geq g(t, u_1, u_2, w)$ when $t \in (t_1, t_2] \cap \mathbb{T}, \, u_1 - v_1 \geq 0$ and $v_2 > u_2 \geq 0$, or when $t \in [t_2, t_3) \cap \mathbb{T}, \, u_1 - v_1 \leq 0$ and $v_2 > u_2 \geq 0$ \end{itemize} We will need also the following two theorems due to Atici and Guseinov, (Theorems 2.5 and 2.6 in \cite[pg. 79]{FAGG}). \begin{theorem}\label{AG:thm1} If $f :\mathbb{T} \to \mathbb{C}$ is $\Delta$-differentiable on $\mathbb{T}^\kappa$ and if $f^\Delta$ is continuous on $\mathbb{T}^\kappa$, then $f$ is $\nabla$-differentiable on $\mathbb{T}_\kappa$ and \begin{displaymath} f^\nabla(t) = f^\Delta(\rho(t)) \end{displaymath} for all $t \in \mathbb{T}_\kappa$. \end{theorem} \begin{theorem}\label{AG:thm2} If $f \!:\mathbb{T} \to \mathbb{C}$ is $\nabla$-differentiable on $\mathbb{T}_\kappa$ and if $f^\nabla$ is continuous on $\mathbb{T}_\kappa$, then $f$ is $\Delta$-differentiable on $\mathbb{T}^\kappa$ and \begin{displaymath} f^\Delta(t) = f^\nabla(\sigma(t)) \end{displaymath} for all $t \in \mathbb{T}^\kappa$. \end{theorem} \section{Existence and Uniqueness of Solutions} Consider the boundary conditions, $$y(t_1) = y_1, \quad y(t_2) = y_2, \quad y^{\Delta^j}(t_2)=m \label{eq3} % eq 4j$$ for $j = 1,2$, and $$y(t_2) = y_2, \quad y^{\Delta^j}(t_2)=m, \quad y(t_3)=y_3, \label{eq4} % eq 5j$$ for $j = 1,2$, where $y_1, y_2, y_3, m \in \mathbb{R}$. In this section, the solution of \eqref{eq1}, \eqref{eq3}, ($j = 1, 2$) is matched with the solution of \eqref{eq1}, \eqref{eq4}, ($j = 1, 2$) to obtain a unique solution of \eqref{eq1}, \eqref{eq2}. Our first theorem states that solutions of \eqref{eq1}, \eqref{eq3}, $j = 1,2$, and \eqref{eq1}, \eqref{eq4}, $j = 1,2$, are unique. \begin{theorem}\label{thm1} Let $y_1, y_2, y_3 \in \mathbb{R}$, and assume that conditions (A) through (D) are satisfied. Then, given $m \in \mathbb{R}$, each of the boundary value problems \eqref{eq1},\eqref{eq3}, $j = 1, 2$, and \eqref{eq1}\eqref{eq4}, $j = 1, 2$, has at most one solution. \end{theorem} \begin{proof} We will consider only the proof for \eqref{eq1}, \eqref{eq3} with with $j=1$; the arguments for the other cases is similar. Let us assume that there are distinct solutions $\alpha$ and $\beta$ of \eqref{eq1}, \eqref{eq3} (with $j=1$). Define $w \equiv \alpha - \beta$. Then $w(t_1) = w(t_2) = w^\Delta(t_2) = 0$. By uniqueness of solutions of initial value problems for \eqref{eq1} we know that $w^{\Delta\Delta}(t_2) \neq 0$. Without loss of generality, we let $w^{\Delta\Delta}(t_2) < 0$. Since $w(t_1) = 0$ and since $t_2$ is dense, there exists an $r_1 \in (t_1, t_2) \cap \mathbb{T}$ such that $w^{\Delta\Delta}(t)$ has a g.z. at $r_1$, $w^\Delta(t) > 0$ on $[r_1, t_2) \cap \mathbb{T}$, $w(t) < 0$ on $(r_1, t_2] \cap \mathbb{T}$, and $w^{\Delta\Delta}(t) < 0$ on $[r_1, t_2) \cap \mathbb{T}$. From the definition of a generalized zero, we have either $w^{\Delta\Delta}(r_1) = 0$ or $w^{\Delta\Delta}(r_1) \cdot w^{\Delta\Delta}(\sigma(r_1)) < 0$. If $r_1$ is right dense, then $w^{\Delta\Delta}(r_1) = 0$. If $r_1$ is right scattered and $w^{\Delta\Delta}(r_1) \neq 0$, then $w^{\Delta\Delta}(r_1) \cdot w^{\Delta\Delta}(\sigma(r_1)) < 0$. Since $w^{\Delta\Delta}(t) < 0$ on $(r_1, t_2] \cap \mathbb{T}$, $w^{\Delta\Delta}(r_1) > 0$. Thus $w^{\Delta\Delta}(r_1) \geq 0$. Now let $0 < \varepsilon <\frac{1}{2} \min \{ \varepsilon_2, -w^{\Delta\Delta}(t_2) \}$ and let $z_\varepsilon$ satisfy the criteria of hypothesis (C) relative to the interval $[t_1, t_2] \cap \mathbb{T}$; that is \begin{gather*} z_\varepsilon^{\Delta\Delta\Delta}(t) = g(t, z_\varepsilon(t), z_\varepsilon^\Delta(t), z_\varepsilon^{\Delta\Delta}(t)), \quad t \in [t_1, t_3] \cap \mathbb{T},\\ z_\varepsilon(t_2) = z_\varepsilon^\Delta(t_2) = 0, \quad z_\varepsilon^{\Delta\Delta}(t_2) = -\varepsilon \end{gather*} and $z_\varepsilon^{\Delta\Delta}$ does not change sign in $[t_1, t_2] \cap \mathbb{T}$. Set $Z \equiv w - z_\varepsilon$. Then $Z(t_2) = Z^\Delta(t_2) = 0$, and $Z^{\Delta\Delta}(t_2) < 0$. Moreover, $Z^{\Delta\Delta}(r_1) = w^{\Delta\Delta}(r_1) - z^{\Delta\Delta}_\varepsilon(r_1) > 0$, and $Z^{\Delta\Delta}(t_2)<0$ imply that there exists an $r_2 \in [r_1, t_2) \cap \mathbb{T}$ such that $Z^{\Delta\Delta}$ has a g.z.\ at $r_2$ and $Z^{\Delta\Delta}(t) < 0$ on $(r_2, t_2] \cap \mathbb{T}$. As above, since $Z^{\Delta\Delta}$ has a g.z.\ at $r_2$, $Z^{\Delta\Delta}(r_2) \geq 0$. Also, $Z^\Delta(t) > 0$ and $Z(t) < 0$ on $[r_2, t_2) \cap \mathbb{T}$. When $\sigma(r_2) > r_2$, \begin{displaymath} Z^{\Delta\Delta\Delta}(r_2) = \frac{Z^{\Delta\Delta}(\sigma(r_2)) - Z^{\Delta\Delta}(r_2)}{\sigma(r_2) - r_2} < 0\,. \end{displaymath} When $\sigma(r_2) = r_2$, \begin{displaymath} Z^{\Delta\Delta\Delta}(r_2) = \lim_{t \to r_2^+} \frac{Z^{\Delta\Delta}(t)}{t-r_2} < 0\,. \end{displaymath} Regardless of wether $r_2$ is right dense or right scattered we have, from the definition of the delta derivative, that $Z^{\Delta\Delta\Delta}(r_2) < 0$. From conditions (A) and (D) we have \begin{align*} Z^{\Delta\Delta\Delta}(r_2) & = w^{\Delta\Delta\Delta}(r_2) - z_\varepsilon^{\Delta\Delta\Delta}(r_2)\\ & > g(r_2, w(r_2), w^\Delta(r_2), w^{\Delta\Delta}(r_2)) - g(r_2, z_\varepsilon(r_2), z_\varepsilon^\Delta(r_2), z_\varepsilon^{\Delta\Delta}(r_2))\\ & \geq 0. \end{align*} That is, $Z^{\Delta\Delta\Delta}(r_2) > 0$, which is a contradiction. Our assumption must be wrong and consequently \eqref{eq1} \eqref{eq3} has at most one solution. \end{proof} \begin{theorem}\label{thm2} Assume that hypotheses (A) through (D) are satisfied. Then \eqref{eq1}, \eqref{eq2} has at most one solution. \end{theorem} \begin{proof} Assume that there exist two distinct solutions $\alpha$ and $\beta$ of \eqref{eq1}, \eqref{eq2}. Define $w = \alpha - \beta$. Then $w(t_1) = w(t_2) = w(t_3)=0$. From Theorem \ref{thm1}, $w^\Delta(t_2) \neq 0$ and $w^{\Delta\Delta}(t_2) \neq 0$. Without loss of generality let $w^\Delta(t_2) = \alpha^\Delta(t_2) - \beta^\Delta(t_2) > 0$. By Theorem \ref{AG:thm2} we have $w^\nabla(t_2) = w^\Delta(t_2) > 0$. Then there exist points $r_1 \in (t_1, t_2) \cap \mathbb{T}$ and $r_2 \in (t_2, t_3) \cap \mathbb{T}$ such that $w^\Delta$ has a g.z. at $r_1$ and $r_2$ and $w^\Delta(t) > 0$ on ($r_1,r_2) \cap \mathbb{T}$. Let $\varepsilon = \frac{1}{2}\min \{ \varepsilon_1, w^\Delta(t_2)\}$ and let $z_\varepsilon$ be the solution of the initial value problem $z^{\Delta\Delta\Delta}_\varepsilon(t) = g(t, z_\varepsilon(t), z_\varepsilon^\Delta(t), z_\varepsilon^{\Delta\Delta}), t \in [t_1, t_3] \cap \mathbb{T}, z_\varepsilon(t_2) = 0, z^\Delta(t_2) = \varepsilon, z_\varepsilon(t_2) = 0$. By condition (B), $z^\Delta_\varepsilon$ does not change sign on $[t_1, t_3] \cap \mathbb{T}$. Define $Z \equiv w - z_\varepsilon$. Then $Z(t_2) = 0, Z^\Delta(t_2) > 0$, and $Z^{\Delta\Delta}(t_2) = w^{\Delta\Delta}(t_2) \neq 0$. There are two cases to consider. \noindent\textbf{Case 1: $Z^{\Delta\Delta}(t_2) < 0$.} Recall that $w^\Delta$ has a g.z.\ at $r_1$. If $r_1$ is right dense, then $w^\Delta(r_1) = 0$. If $r_1$ is right scattered, then either $w^\Delta(r_1) = 0$ or $w^\Delta(\sigma(r_1)) \cdot w^\Delta(r_1) < 0$. In the latter case since $w^\Delta(t) > 0$ on $(r_1, r_2) \cap \mathbb{T}$, we have $w^\Delta(r_1) < 0$. Regardless of wether $r_1$ is right dense or right scattered we have $Z^\Delta(r_1) = w^\Delta(r_1) - z^\Delta_\varepsilon(r_1) \leq 0$. Since $Z^\Delta(r_1) \leq 0$ and $Z^{\Delta\Delta}(t_2) < 0$, there exists an $r_3 \in (r_1, t_2] \cap \mathbb{T}$ such that $Z^{\Delta\Delta}$ has a g.z. at $r_3$ and $Z^{\Delta\Delta}(t) < 0$ on $(r_3, t_2] \cap \mathbb{T}$. On the one hand, if $\sigma(r_3) > r_3$, then \begin{displaymath} Z^{\Delta\Delta\Delta}(r_3) = \frac{Z^{\Delta\Delta}(\sigma(r_3)) - Z^{\Delta\Delta}(r_3)}{\sigma(r_3) - r_3} < 0\,. \end{displaymath} If $\sigma(r_3) = r_3$, then \begin{displaymath} Z^{\Delta\Delta\Delta}(r_3) = \lim_{t \to r_3^+} \frac{Z^{\Delta\Delta}(t)}{t - r_3} < 0\,. \end{displaymath} Regardless of wether $r_3$ is right dense or right scattered we have, from the definition of the delta derivative, that $Z^{\Delta\Delta\Delta}(r_3) < 0$. On the other hand, from conditions (A) and (D) we have \begin{align*} Z^{\Delta\Delta\Delta}(r_3) & = w^{\Delta\Delta\Delta}(r_3) - z_\varepsilon^{\Delta\Delta\Delta}(r_3)\\ & > g(r_3, w(r_3), w^\Delta(r_3), w^{\Delta\Delta}(r_3)) - g(r_3, z_\varepsilon(r_3), z_\varepsilon^\Delta(r_3), z_\varepsilon^{\Delta\Delta}(r_3))\\ & \geq 0\,. \end{align*} That is, conditions (A) and (D) imply that $Z^{\Delta\Delta\Delta}(r_3) > 0$ which is a contradiction. Consequently, $Z^{\Delta\Delta}(t_2) \not < 0$. \noindent\textbf{Case 2: $Z^{\Delta\Delta}(t_2) > 0$.} Again, we know that $w^\Delta$ has a g.z.\ at $r_2$. If $\sigma(r_2) = r_2$, then $w^\Delta(r_2) = 0$. If $\sigma(r_2) > r_2$, then either $w^\Delta(r_2) = 0$ or $w^\Delta(r_2) > 0$ and $w^\Delta(\sigma(r_2)) < 0$ or $w^\Delta(r_2) < 0$ and $w^\Delta(\rho(r_2)) > 0$. Consequently, either $Z^\Delta(r_2) < 0$ or $Z^\Delta(\sigma(r_2)) < 0$. Since $Z^\Delta(r^*) < 0$, (where $r^* = r_2$ or $r^* = \sigma(r_2)$), and since $Z^{\Delta\Delta}(t_2) > 0$, there exists $r_4 \in (t_2, r^*)$ such that $Z^{\Delta\Delta}$ has a g.z.\ at $r_4$, $Z^{\Delta\Delta}(t)> 0$ on $[t_2, r_4) \cap \mathbb{T}$, and $Z^{\Delta\Delta}$ does not have a g.z.\ in $[t_2, r_4) \cap \mathbb{T}$. We now obtain a contradiction. On the one hand, we can use the definition of the $\Delta$-derivative to calculate $Z^{\Delta\Delta\Delta}(r_4)$. If $\rho(r_4) = r_4$, then by Theorem \ref{AG:thm1} we have $Z^{\Delta\Delta\Delta}(r_4) = Z^{\Delta\Delta\nabla}(r_4) = \lim_{t \to r_4^-} \frac{Z^{\Delta\Delta}(t) - 0}{t - r_4} < 0.$ If $\rho(r_4) < r_4$, then either $\sigma(r_4) = r_4$ or $\sigma(r_4) > r_4$. If $\sigma(r_4) = r_4$, then \begin{displaymath} Z^{\Delta\Delta\Delta}(r_4) = \lim_{t \to r_4^+} \frac{Z^{\Delta\Delta}(t)}{t - r_4} < 0. \end{displaymath} If $\sigma(r_4) > r_4$, then \begin{displaymath} Z^{\Delta\Delta\Delta}(r_4) = \frac{Z^{\Delta\Delta}(\sigma(r_4)) - Z^{\Delta\Delta}(r_4)}{\sigma(r_4) - r_4} < 0. \end{displaymath} In any case, we have, by definition of the $\Delta$-derivative, that $Z^{\Delta\Delta\Delta}(r_4) < 0$. On the other hand, we have from conditions (A) and (D), \begin{align*} Z^{\Delta\Delta\Delta}(r_4) & = w^{\Delta\Delta\Delta}(r_4) - z_\varepsilon^{\Delta\Delta\Delta}(r_4)\\ & > g(r_4, w(r_4), w^\Delta(r_4), w^{\Delta\Delta}(r_4)) - g(r_4, z_\varepsilon(r_4), z_\varepsilon^\Delta(r_4), z_\varepsilon^{\Delta\Delta}(r_4))\\ & \geq 0. \end{align*} Conditions (A) and (D) imply $Z^{\Delta\Delta\Delta}(r_4) > 0$ which is a contradiction. Thus $Z^{\Delta\Delta}(t_2) \not > 0$. Since $Z^{\Delta\Delta}(t_2) \neq 0$ and $Z^{\Delta\Delta}(t_2) < 0$ and $Z^{\Delta\Delta}(t_2) > 0$ lead to contradictions, our original assumption must be false. As such, the boundary value problem \eqref{eq1}, \eqref{eq2} has at most one solution and the theorem is proved. \end{proof} Now given $m \in \mathbb{R}$, let $\alpha(x,m), \beta(x,m), u(x,m)$ and $v(x,m)$ denote the solutions, when they exist, of the boundary value problems for \eqref{eq1},\eqref{eq3} and \eqref{eq1},\eqref{eq4}, $j = 1, 2$, respectively. \begin{theorem}\label{thm3} Suppose that (A) through (D) are satisfied and that, for each $m \in \mathbb{R}$, there exist solutions of \eqref{eq1}, \eqref{eq3} and \eqref{eq1}, \eqref{eq4}, $j = 1, 2$. Then $u^\Delta(t_2, m)$ and $\alpha^{\Delta\Delta}(t_2, m)$ are strictly increasing functions of $m$ whose range is $\mathbb{R}$, and $v^\Delta(t_2, m)$ and $\beta^{\Delta\Delta}(t_2, m)$ are strictly decreasing functions of $m$ with ranges all of $\mathbb{R}$. \end{theorem} \begin{proof} The strictness'' of the conclusion arises from Theorem \ref{thm1}. We will prove the theorem with respect to the solution $\alpha(t, m)$. Let $m_1 > m_2$ and let $w(t) \equiv \alpha(t, m_1) - \alpha(t, m_2)$. Then when $w(t_1) = w(t_2) = 0, w^\Delta(t_2) > 0$, and $w^{\Delta\Delta}(t_2) \neq 0$. Assume that $w^{\Delta\Delta}(t_2) < 0$. Then there exists an $r_1 \in (t_1, t_2) \cap \mathbb{T}$ such that $w^{\Delta}$ has a g.z.\/ at $r_1$ and $w^\Delta(t) > 0$ on $(r_1, t_2] \cap \mathbb{T}$. By continuity, there exists an $r_2 \in (r_1, t_2) \cap \mathbb{T}$ such that $w^{\Delta\Delta}$ has a g.z.\ at $r_2$ and $w^{\Delta\Delta}(t) < 0$ on $(r_2, t_2] \cap \mathbb{T}$. Note that $w(t) < 0$ on $[r_2, t_2) \cap \mathbb{T}$. Let $0 < \varepsilon < \min \{ \varepsilon_2, -w^{\Delta\Delta}(t_2) \}$ and let $z_\varepsilon$ be the solution of the initial value problem satisfying conditions of (C), and set $Z \equiv w - z_\varepsilon$. Then $Z(t_2) = 0, Z^\Delta(t_2) = w^\Delta(t_2) > 0$, and $Z^{\Delta\Delta}(t_2) < 0$. Furthermore $Z^{\Delta\Delta}(r_2) \geq 0$. Thus there exist $r_3 \in (r_2, t_2) \cap \mathbb{T}$ such that $Z^{\Delta\Delta}(r_3) = 0$ and $Z^{\Delta\Delta}(t) < 0$ on $(r_3, t_2]$. Then $Z^\Delta(t) > 0$ and $Z(T) < 0$ on $[r_3, t_2)$. As in the proofs of Theorems \ref{thm1} and \ref{thm2}, we can then argue that $Z^{\Delta\Delta\Delta}(r_3) < 0$ and $Z^{\Delta\Delta\Delta}(r_3) > 0$, which is again a contradiction. Thus $w^{\Delta\Delta}(t_2) > 0$ and consequently, $\alpha^{\Delta\Delta}(t_2, m)$ is strictly increasing as a function of $m$. We now show that $\{\alpha^{\Delta\Delta}(t_2, m) \big | m \in \mathbb{R}\} = \mathbb{R}$. Let $k \in \mathbb{R}$ and consider the solution $u(x,k)$ of the \eqref{eq1}, \eqref{eq3} (with $j=2$) with $u$ as specified above. Consider also the solution $\alpha(x, u^\Delta(t_2,k))$, of \eqref{eq1}, \eqref{eq3} (with $j=1)$. Then $\alpha(x, u^\Delta(t_2, k))$ and $u(x,k)$ are solutions of \eqref{eq1}, \eqref{eq3}. Hence, by Theorem \ref{thm1}, $\alpha(x, u^\Delta(t_2, k)) \equiv u(x,k)$. Therefore, $\alpha^{\Delta\Delta}(t_2, u^\Delta(t_2,k)) = k$ and so $\{\alpha^{\Delta\Delta}(t_2, m) : m \in \mathbb{R}\} = \mathbb{R}$. The other three parts are established in a similar manner and the proof is complete. \end{proof} \begin{theorem} Assume the hypothesis of Theorem \ref{thm3}. Then \eqref{eq1}, \eqref{eq2} has a unique solution. \end{theorem} \begin{proof} By Theorem \ref{thm3}, there exists a unique $m_0$ such that $u^\Delta(t_2, m_0) = v^\Delta(t_2, m_0)$. Also $u^{\Delta\Delta}(t_2, m_0) = m_0 = v^{\Delta\Delta}(t_2, m_0)$. Then, \begin{displaymath} y(t) = \begin{cases} u(t,m_0) = y_1(t), & t_1 \leq t \leq t_2,\\ v(t,m_0) = y_2(t), & t_2 \leq t \leq t_3, \end{cases} \end{displaymath} is a solution of \eqref{eq1}, \eqref{eq2}. By Theorem \ref{thm2}, $y(t)$ is the unique solution. \end{proof} \begin{thebibliography}{00} \bibitem{FAGG} F. M. Atici and G. Sh. 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