\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,
\begin{equation}
y(t_1) = y_1, \quad y(t_2) = y_2, \quad
y^{\Delta^j}(t_2)=m  \label{eq3}  % eq 4j
\end{equation}
for $j = 1,2$, and
\begin{equation}
y(t_2) = y_2, \quad y^{\Delta^j}(t_2)=m, \quad
y(t_3)=y_3, \label{eq4} % eq 5j
\end{equation}
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}

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