\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 68, pp. 1--21.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/68\hfil Time scale terminal value problems]
{Terminal value problems for first and second
  order nonlinear equations on time scales}

\author[R. Hilscher, C. C. Tisdell\hfil EJDE-2008/68\hfilneg]
{Roman Hilscher, Christopher C. Tisdell}  % in alphabetical order

\address{Roman Hilscher \newline
  Department of Mathematics and Statistics,
  Faculty of Science, Masaryk University,
  Jan\'a\v ckovo n\'am. 2a, CZ-60200 Brno, Czech Republic}
\email{hilscher@math.muni.cz}

\address{Christopher C. Tisdell \newline
  School of Mathematics and Statistics,
  The University of New South Wales, Sydney NSW 2052, Australia}
\email{cct@unsw.edu.au}

\thanks{Submitted February 11, 2008. Published May 1, 2008.}
\subjclass[2000]{34C99, 39A10}
\keywords{Time scale; terminal value problem; nonlinear equation;
\hfill\break\indent
Banach fixed point theorem; bounded solution; weighted norm}

\begin{abstract}
  In this paper we examine ``terminal'' value problems for dynamic
  equations on time scales -- that is, a dynamic equation
  whose solutions are asymptotic at infinity.
  We present a number of new theorems that guarantee the existence
  and uniqueness of solutions, as well as some comparison-type results.
  The methods we employ feature dynamic inequalities, weighted
  norms, and fixed-point theory.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction} \label{S:intro}

The theory of ``terminal'' value problems, where the problem
consists of a dynamic equation coupled with asymptotic behavior of
the solution at $\infty$, forms an interesting and more challenging
field of research than the theory of initial value problems. This is
due to even the basic results and methods known for initial value
problems, such as the perturbation technique, are unavailable for
use in the setting of terminal value problems. For example, the
existence of a solution to the terminal value problem $x'=f(t,x)$,
$x(\infty)=x_0$, need not imply that the terminal value problem
$x'=f(t,x)\pm\frac{1}{n}$, $x(\infty)=x_0\pm\frac{1}{n}$ has a
solution, see \cite[pg.~1173]{arA.vL81}.

In this work, we examine terminal value problems for ``dynamic
equations on time scales'', which is a new and versatile area of
mathematics that is more general than the fields of differential
equations and difference equations. The area of time scales
originates in the work of Hilger in \cite{sH90}. Such
investigations reveal the bonds and distinctions between the two
areas and also provide a framework with which to more accurately
model stop-start processes.

Our main interest herein is in the qualitative properties of
solutions to terminal value problems on time scales, including the
existence, uniqueness, and comparison theorems. The methods that
we employ involve dynamic inequalities, weighted norms, and fixed
point theory. The motivation for using the weighted (or Bielecki)
norms originates in \cite{ccT.aZ08} and the references quoted
therein, where this method was used in order to prove the
existence and uniqueness results for nonlinear initial value
problems on bounded time scales. The existence of bounded
solutions to {\em initial\/} value problems for second order
dynamic equations and inequalities on unbounded time scales was
studied in \cite{rpA.mB.dO02}, while in \cite{rpA.mB.dO01} results
of this type are given for certain first order dynamic equations.
In the latter three references, as well as in the present paper,
the fixed point theory is utilized.

Our results extend some of the ideas in \cite{arA.vL81} and,  more
recently, those of \cite{lE.aP.ccT?}. More specifically, we
provide some extensions of the comparison results in
\cite{arA.vL81}, which were formulated for terminal value problems
involving ordinary differential equations, to the time scale
environment. Furthermore, compared to \cite{lE.aP.ccT?} we allow
in Section \ref{S:second.order} the nonlinearity $f(t,x^\sigma)$
or $f(t,x^\sigma,x^{\Delta\sigma})$ to be vector valued and we
pose no restriction on the sign of its entries. Also, we assume in
those results that the leading coefficient $r(t)$ is merely
nonzero as opposed to the assumption of its positivity in
\cite{lE.aP.ccT?}. In addition, for the case of positive $r(t)$
and nonnegative nonlinearity $f$ we extend in Section
\ref{S:second.order.matrix} the ideas in \cite{lE.aP.ccT?} from
the scalar case to the matrix/vector case. Some of the main
results (e.g., Theorems \ref{T:compare1}, \ref{T:compare2},
\ref{T:2nd.order}, \ref{T:2nd.order.xD}, \ref{T:2nd.order.matrix}
and \ref{T:2nd.order.xD.matrix}) appear to be new even for the
special case ${\mathbb T}={\mathbb Z}$, that is, for difference
equations.

For additional papers that contain comparison and existence and
uniqueness  results for first-order terminal value problems
involving ordinary differential equations, we refer the reader to
\cite{tgH70,tgH72,rL.pV83,gV76}. For papers dealing with
second-order terminal value problems, the reader is referred to
\cite{tgH72,weS69,weS70}. The methods used in the range of the
aforementioned papers involve differential inequalities and the
fixed-point theorems of Banach or Schauder.

The setup of the paper is the following. In Section \ref{S:pre} we
introduce necessary notation and terminology as well as some
preparatory results about the time scale exponential function. In
Section \ref{S:first.order} we derive an existence and uniqueness
theorem for the terminal value problem of the first order. Then we
continue in deriving comparison results for solutions of first
order dynamic inequalities. In Section \ref{S:second.order} we
consider terminal value problems for second order dynamic
equations with scalar leading coefficient, while in Section
\ref{S:second.order.matrix} we deal with such equations with
matrix leading coefficient and with nonnegative nonlinearity. In
Section \ref{S:examples} we present examples illustrating the
applicability of the obtained results. Finally, in Section
\ref{S:appl} we discuss further applications and extensions, in
particular to nabla dynamic terminal value problems.

\section{Prerequisites and notation} \label{S:pre}

Let $n\in{\mathbb N}$ be a fixed natural number. For a real
symmetric $n\times n$ matrix $A$ we write $A>0$ or $A\geq0$ for
$A$ being a positive definite or positive semidefinite matrix,
respectively. Moreover, if $B$ is a real symmetric $n\times n$
matrix, then we write $A<B$ or $A\leq B$ if $B-A>0$ or $B-A\geq0$,
respectively.

In this paper we will use the vector norm $|\cdot|_\infty$ on
$\mathbb{R}^n$ denoted for simplicity by
$|x|:=|x|_\infty=\max\{|x_i|,\ i=1,\dots,n\}$. Given a number
$0<q\leq\infty$, we use the notation
$\Omega_q:=\{x\in\mathbb{R}^n,\ |x|<q\}$ for the open $q$-ball in
${\mathbb R}^n$. Then we can identify $\Omega_\infty$ with
$\mathbb{R}^n$.

Let ${\mathbb T}$ be a time scale, i.e., a nonempty closed subset
of $\mathbb{R}$, which is bounded below and unbounded above. Then
$a:=\min{\mathbb T}$ exists and we may identify ${\mathbb T}$ with
the time scale interval $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. We shall use the common time scale
notation and terminology e.g. from the books
\cite{mB.aP01,mB.aP03}. In particular, the forward and backward
jump operators are denoted by $\sigma$ and $\rho$, and the
graininess is $\mu(t):=\sigma(t)-t$. As it is common, the sets of
all continuous, rd-continuous, or rd-continuously
$\Delta$-differentiable functions (on a given interval) will be
denoted by ${\rm C}$, ${\rm C}_{\rm rd}$, and ${\rm C}_{\rm rd}^1$, respectively. The $\sup$
norm in the space of bounded $n$-vector functions $x\in{\rm C}$ on
$[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ will be denoted by $\|x\|_0:=\sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}}|x(t)|$. The
$\sup$ norm in the space of bounded $n$-vector functions
$x\in{\rm C}_{\rm rd}^1$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ such that $x^\Delta$ is also bounded will be
denoted by $\|x\|_1:=\max\{\|x\|_0,\,\|x^\Delta\|_0\}$. The
improper integrals used in this paper are defined in the
traditional way as $\int_a^\infty\!f(t)\,\Delta
t:=\lim_{b\to\infty}\int_a^bf(t)\,\Delta t$, see e.g.
\cite[Section~5.6]{mB.gsG03a} and \cite[Section 4]{mB.gsG03}. In
addition, motivated by \cite[Definition~3]{rH.vZ04a}, we adopt the
following terminology.

\begin{definition} \label{D:CrdCC} \rm
 Let $0<q\leq\infty$ and
 $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\times\mathbb{R}^n\to\mathbb{R}^n$ be a
 function. We write $f\in{\rm C}_{\rm rd}\times{\rm C}\times{\rm C}$ and say that $f$
 is {\em ${\rm C}_{\rm rd}\times{\rm C}\times{\rm C}$-continuous\/} on its domain if
 for any $(t_0,x_0,v_0)\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\times\mathbb{R}^n$ and
 any $\varepsilon>0$ there exists $\delta>0$ such that
 $0<|(t-t_0,x-x_0,v-v_0)|<\delta$ implies
  \begin{equation} \label{E:L.CrdCC}
    \big| F(t,x,v)-F(t_0,x_0,v_0) \big| <\varepsilon.
  \end{equation}
 When the point $t_0$ is left-dense and right-scattered at the same
 time, then replace $t_0$ in \eqref{E:L.CrdCC} by $t_0^-$ (the
 left-hand limit).
\end{definition}

In other words, $f\in{\rm C}_{\rm rd}\times{\rm C}\times{\rm C}$ means that $f$ is
continuous at any point
$(t_0,y_0,v_0)\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\times\mathbb{R}^n$ when $t_0$
is right-dense, and that $f$ is {\em jointly regulated}, that is,
$\lim_{n\to\infty}f(t_n,x_n,v_n)$ exists (finite) whenever $t_n\to
t_0^-$ or $t_n\to t_0^+$, and $(x_n,v_n)\to(x_0,v_0)$.

The following result is a minor modification of \cite[Proposition~1]{rH.vZ04a}. It shows that the above continuity concept is
the right one when considering time scale delta integrals involving a ${\rm C}_{\rm rd}\times{\rm C}\times{\rm C}$-continuous function $f$ in the
composition with a ${\rm C}_{\rm rd}^1$ function $x$.

\begin{proposition} \label{P:CrdCC.Crd}
  Let $x\in{\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ and assume that $f\in{\rm C}_{\rm rd}\times{\rm C}\times{\rm C}$ on
 $[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\times\mathbb{R}^n$ with $0<q\leq\infty$. Then
 $f\big(\cdot,x^\sigma(\cdot),x^{\Delta\sigma}(\cdot)\big)\in{\rm C}_{\rm rd}$.
\end{proposition}

Similarly, when the function $f$ is defined only on
$[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q$ we have the following statement.

\begin{proposition} \label{P:CrdC.Crd}
 Let $x\in{\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ and assume that $f\in{\rm C}_{\rm rd}\times{\rm C}$ on
 $[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q$ with $0<q\leq\infty$. Then
 $f\big(\cdot,x^\sigma(\cdot)\big)\in{\rm C}_{\rm rd}$.
\end{proposition}

When considering terminal value problems, we shall use the abbreviation
\begin{equation*}
  x(\infty):=\lim_{t\to\infty}x(t).
\end{equation*}
The vector space of all real $n$-vector functions defined on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ will
 be denoted throughout the paper by ${\mathcal F}$.

For completeness we recall the statement of the Banach fixed point
theorem adjusted to the setting of this paper.

\begin{proposition} \label{P:Banach}
  Let $X$ be a Banach space (i.e., a complete normed space) with norm $\|\cdot\|_X$ and let $U\subseteq X$ be its nonempty
  and closed subset. If a mapping $F:U\to U$ is a contraction, i.e., if there exists $L\in(0,1)$ such that
  $\|Fx-Fy\|_X\leq L\,\|x-y\|_X$ for all $x,y\in U$, then $F$ has a unique fixed point, i.e., there exists a unique element
  $x\in U$ such that $x=Fx$. Furthermore, if $x_0\in U$ is arbitrary, and if we set $x_{i+1}:=Fx_i$ for all $i\in{\mathbb N}$,
  then the sequence $\{x_i\}_{i=0}^\infty$ converges in $X$ to the fixed point $x$, and the error between the $i$-th iteration
  $x_i$ and the fixed point $x$ satisfies the estimate
  \begin{equation*}
    \|x_i-x\|_X\leq\frac{L^i}{1-L}\,\|x_1-x_0\|_X, \quad i\in{\mathbb N}.
  \end{equation*}
\end{proposition}

Next we present some important properties of the time scale
exponential function. By definition, see
\cite[Definition~2.30]{mB.aP01}, for an rd-continuous and
regressive function $p:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}$, the time scale
exponential function $e_{p(\cdot)}(t,a)$ is defined to be the
unique solution of the initial value problem $u^\Delta=p(t)\,u$,
$u(a)=1$. In this paper we will utilize the time scale exponential
functions corresponding to the initial value problem
\begin{equation} \label{E:exp.def.u}
  u^\Delta=-p(t)\,u^\sigma, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}, \quad u(a)=1.
\end{equation}
By expanding $u^\sigma$ in \eqref{E:exp.def.u} with the formula
$u^\sigma=u+\mu(t)\,u^\Delta$ and using the regressivity of
$p(\cdot)$, the dynamic equation in \eqref{E:exp.def.u} is
equivalent to the equation $u^\Delta=(\ominus p)(t)\,u$, where
$\ominus p(t):=[-p(t)]/[1+\mu(t)\,p(t)]$. Thus, the time scale
exponential function $e_{\ominus p(\cdot)}(t,a)$ is the unique
solution of the initial value problem \eqref{E:exp.def.u}.
Motivated by \cite[Lemma~3.1]{lE.aP.ccT?}, we have the following
result.

\begin{lemma} \label{L:exp.def}
  Assume that $p:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $p\in{\rm C}_{\rm rd}$, and
  \begin{equation} \label{E:p.int.finite}
    \int_a^\infty p(s)\,\Delta s<\infty.
  \end{equation}
  Let $u(t):=e_{\ominus p(\cdot)}(t,a)$ be the time scale
  exponential function corresponding to the initial value problem
  \eqref{E:exp.def.u}. Then $u(\cdot)$ is positive and decreasing on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, and $\lim_{t\to\infty}u(t)=:u_0\in(0,1)$. Furthermore,
  \begin{equation} \label{E:u.sup.int}
    \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{u(t)} \int_t^\infty [-u^\Delta(s)]\,\Delta s=1-u_0.
  \end{equation}
\end{lemma}

\begin{proof}
Since $p(\cdot)$ is positive, $\ominus p(\cdot)$ is a negative
function. Hence, we have $1+\mu(t)\,(\ominus
p)(t)=1/[1+\mu(t)\,p(t)]>0$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, i.e., $\ominus p(\cdot)$ is
an rd-continuous and positively regressive function. By
\cite[Theorem~2.44]{mB.aP01}, we get that $u(t)>0$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$.
Consequently, $u^\Delta(t)<0$, the function $u(\cdot)$ is
decreasing on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, the limit $u_0$ exists, and $u_0\in[0,1)$.
The fact that actually $u_0>0$ follows from the assumption
\eqref{E:p.int.finite}. We refer to the proof of
\cite[Lemma~3.1]{lE.aP.ccT?} for the details. For the proof of
\eqref{E:u.sup.int}, we have
  \begin{equation*}
    \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{u(t)} \int_t^\infty [-u^\Delta(s)]\,\Delta s=\sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \Big( 1-\frac{u_0}{u(t)} \Big)=1-u_0,
  \end{equation*}
because the function $u$ attains its maximum value $u(a)=1$.
\end{proof}

\section{First order equations} \label{S:first.order}

Consider the first order time scale dynamic equation
\begin{equation} \label{E:1st.order}
  x^\Delta+f(t,x^\sigma)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
\end{equation}
For a given positive number $N$ we define the set
\begin{equation} \label{E:XN.def.1st.order}
  X_N:=\big\{\,x\in{\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}},\ \|x\|_0\leq N\big\},
\end{equation}
Then $X_N$ is a closed subset of the Banach space $({\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_0)$.

\begin{remark} \label{R:psi.Banach} \rm
 Given a function $\psi:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to[c,d]$, $0<c\leq d<\infty$, we
 introduce on the space ${\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ another norm
  \begin{equation*} \label{E:psi.norm.def}
    \|x\|_\psi:=\|x/\psi\|_0=\sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}}\frac{|x(t)|}{\psi(t)}.
  \end{equation*}
 The norm $\|\cdot\|_\psi$ is on ${\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ clearly equivalent to the
 norm $\|\cdot\|_0$, so that $({\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_\psi)$ is also a
 Banach space, compare with \cite[Lemma~3.3]{ccT.aZ08}.
\end{remark}

The following theorem is then our first result.

\begin{theorem} \label{T:1st.order}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\to\mathbb{R}^n$ with $0<q\leq\infty$, $f\in{\rm C}_{\rm rd}\times{\rm C}$, is a function satisfying the
  Lipschitz condition
  \begin{equation} \label{E:Lipschitz.1st.order}
    \big| f(t,x)-f(t,y) \big| \leq k(t)\,|x-y|, \quad\text{for all } t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},\ x,y\in\Omega_q,
  \end{equation}
  where $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm rd}$, and
  \begin{equation} \label{E:k.integral.1st.order}
    \int_a^\infty k(s)\,\Delta s<\infty.
  \end{equation}
  Let $A\in\mathbb{R}^n$ be a given vector. If there exists a number $N\in\mathbb{R}$, $|A|\leq N<q$, such that
  \begin{equation} \label{E:f.integral.1st.order}
    \int_a^\infty \big|f\big(s,x^\sigma(s)\big)\big|\,\Delta s\leq N-|A|, \quad\text{for all } x\in X_N,
  \end{equation}
  where $X_N$ is defined by \eqref{E:XN.def.1st.order}, then the problem \eqref{E:1st.order} has a unique solution $x(t)$
  on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ satisfying $x(\infty)=A$.
\end{theorem}

\begin{proof}
  We will apply the Banach fixed point theorem in the space $({\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_\psi)$ for a suitably chosen function $\psi$.
  Define the operator $F:X_N\to{\mathcal F}$ (the space of $n$-vector functions) by
  \begin{equation*} \label{E:F.def.1st.order}
    [Fx](t):=A+\int_t^\infty f\big(s,x^\sigma(s)\big)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation*}
  It follows from Proposition \ref{P:CrdC.Crd} and assumption \eqref{E:f.integral.1st.order} that $[Fx](t)$ is well-defined for
  all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Furthermore,
  \begin{align*}
    \big|[Fx](t)\big| &\leq |A|+\int_t^\infty \big|f\big(s,x^\sigma(s)\big)\big|\,\Delta s \leq
      |A|+\int_a^\infty \big|f\big(s,x^\sigma(s)\big)\big|\,\Delta s \\
    &\leq |A|+N-|A|=N, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{align*}
  Hence, $\|Fx\|_0\leq N$. Since $Fx$ is $\Delta$-differentiable, hence continuous, it follows that $Fx\in X_N$, and
  \begin{equation} \label{E:Fx.Delta.1st.order}
    [Fx]^\Delta(t)=-f\big(t,x^\sigma(t)\big), \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation}
  Next, motivated by its introduction and use in \cite{ccT.aZ08}, choose the function $\psi(t)$ to be the time scale
  exponential function $e_{\ominus k(\cdot)}(t,a)$. Then,
  by Lemma \ref{L:exp.def}, we have $0<\psi_0\leq\psi(t)\leq1$ for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ with $\psi_0\in(0,1)$, where
  $\psi_0:=\lim_{t\to\infty}\psi(t)$. Thus, by Remark \ref{R:psi.Banach}, $({\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_\psi)$ is a Banach space. By
  using \eqref{E:Lipschitz.1st.order} and \eqref{E:u.sup.int} with $u:=\psi$ and $u_0:=\psi_0$, we have for $x,y\in X_N$
  \begin{align*}
    \|Fx-Fy\|_\psi
    &\leq \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\psi(t)} \int_t^\infty \big| f\big(s,x^\sigma(s)\big)
      -f\big(s,y^\sigma(s)\big) \big|\,\Delta s \\
    &\leq \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\psi(t)} \int_t^\infty k(s)\,\big|x^\sigma(s)-y^\sigma(s)\big|\,\Delta s \\
    &\leq \|x-y\|_\psi \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\psi(t)} \int_t^\infty [-\psi^\Delta(s)]\,\Delta s=(1-\psi_0)\,\|x-y\|_\psi.
  \end{align*}
  Hence, the mapping $F$ is a contraction in $X_N$. By Proposition \ref{P:Banach}, there is a unique function
  $x\in X_N$ such that $x=Fx$, i.e.,
  \begin{equation} \label{E:x.int.eq.1st.order}
    x(t)=A+\int_t^\infty f\big(s,x^\sigma(s)\big)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation}
  By \eqref{E:Fx.Delta.1st.order} and Proposition \ref{P:CrdC.Crd}, $Fx\in{\rm C}_{\rm rd}^1$, and $x$ satisfies equation
  \eqref{E:1st.order}. Finally, from assumption \eqref{E:f.integral.1st.order} it follows that
  \begin{equation*}
    \lim_{t\to\infty} \int_t^\infty f\big(s,x^\sigma(s)\big)\,\Delta s =0,
  \end{equation*}
  and hence, identity \eqref{E:x.int.eq.1st.order} yields that $x(\infty)=A$.
\end{proof}

In Theorem \ref{T:1st.order}  the solution $x(t)$ approaches the
{\em a priori} given limit $A$, and the number $N$ defining the
set $X_N$ then depends on $|A|$. It may be hard to find such
number $N$. On the other hand, in the following result the
solution $x(t)$ approaches a limit $M$, and at the same time the
vector $M$ determines the set in which the contraction mapping $F$
is defined. This type of result is then of the same fashion as
e.g. the results in \cite{lE.aP.ccT?}.

\begin{corollary} \label{C:1st.order.M}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\to\mathbb{R}^n$ with $0<q\leq\infty$, $f\in{\rm C}_{\rm rd}\times{\rm C}$, is a function satisfying the
  Lipschitz condition \eqref{E:Lipschitz.1st.order}, where $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm rd}$, and
  \eqref{E:k.integral.1st.order} holds. If there exists a vector $M\in\mathbb{R}^n$, $|M|<q$, such that
  \begin{equation*} \label{E:f.integral.1st.order.M}
    \int_a^\infty \big|f\big(s,x^\sigma(s)\big)\big|\,\Delta s\leq |M|, \quad\text{for all } x\in X_{2\,|M|},
  \end{equation*}
  then the problem \eqref{E:1st.order} has a unique solution $x(t)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ satisfying $x(\infty)=M$.
\end{corollary}

\begin{proof}
  We let $A:=M$ and $N:=2\,|M|$ in Theorem \ref{T:1st.order}.
\end{proof}

Our attention now turns to the following dynamic equation
\begin{equation} \label{E:1st.order.v2}
  x^\Delta+f(t,x)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
\end{equation}
where $f$ is scalar-valued. Our interest is in obtaining comparison-type theorems for solutions $x$ to \eqref{E:1st.order.v2}
subject to
\begin{equation} \label{E:IC}
  x(\infty) = A.
\end{equation}

\begin{lemma} \label{complem}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\mathbb{R}\to\mathbb{R}$, $f\in{\rm C}_{\rm rd}\times{\rm C}$, and there exist functions $u,v:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}$
  such that: $u(\infty),\; v(\infty)$ both exist,
  \begin{gather}
    u^\Delta(t) + f\big(t,u(t)\big) \ge 0,\quad \mbox{for all } t \in [a,\infty)_{\scriptscriptstyle{\mathbb T}},
     \label{lowineq} \\
    v^\Delta(t) + f\big(t,v(t)\big) \le 0, \quad\mbox{for all } t \in [a,\infty)_{\scriptscriptstyle{\mathbb T}},
     \label{uppineq} \\
    f(t,p) \le f(t,q),\quad \mbox{for all} \quad q \le p. \label{monineq}
  \end{gather}
  If $u(\infty) < v(\infty)$, then $u(t) < v(t)$ for all $t \in [a,\infty)_{\scriptscriptstyle{\mathbb T}}$.
\end{lemma}

\begin{proof}
  Argue by contradiction by assuming that there exists a point
$t_1 \in [a,\infty)_{\scriptscriptstyle{\mathbb T}}$ such that
  \begin{gather}
    u(t_1) \ge v(t_1), \quad \mbox{and}  \label{t_1ineq}\\
    u(t) < v(t), \quad \mbox{for all } t \in (t_1,\infty)_{\scriptscriptstyle{\mathbb T}}.   \label{strineq}
  \end{gather}
  There are two cases to discuss: (a) the point $t_1$ is
  right-scattered; (b) the point $t_1$ is right-dense.
  \par(a) If $t_1$ is right-scattered, then $u^\Delta(t_1) < v^\Delta(t_1)$ and so from \eqref{lowineq} and \eqref{uppineq}
  we obtain
  \begin{equation*}
    -f(t_1,u(t_1)) \le u^\Delta(t_1) < v^\Delta(t_1) \le -f(t_1,v(t_1)),
  \end{equation*}
  which is a contradiction to \eqref{monineq}.
  \par(b) If $t_1$ is right-dense, then \eqref{t_1ineq} is forced to become $u(t_1) = v(t_1)$ by \eqref{strineq} and
  the interme\-dia\-te value theorem. As per part (a) we see
  \begin{equation*}
    -f(t_1,u(t_1)) < -f(t_1,v(t_1)),
  \end{equation*}
  which is impossible since $u(t_1) = v(t_1)$.
  \par Thus, in either case we have  $u(t) < v(t)$ for all $t \in [a,\infty)_{\scriptscriptstyle{\mathbb T}}$.
\end{proof}

We will now apply Lemma \ref{complem} to obtain comparison results
for solutions to \eqref{E:1st.order.v2}, \eqref{E:IC}.

\begin{theorem} \label{T:compare1}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\mathbb{R}\to\mathbb{R}$, $f\in{\rm C}_{\rm rd}\times{\rm C}$, satisfying \eqref{monineq} and there exists a function
  $u:[a,\infty)_{\scriptscriptstyle{\mathbb T}} \to \mathbb{R}$ such that $u(\infty)$ exists and \eqref{lowineq} holds. If $x$ is a solution to \eqref{E:1st.order.v2},
  \eqref{E:IC} and $u(\infty) < A$, then $x(t) > u(t)$ for all $t \in [a,\infty)_{\scriptscriptstyle{\mathbb T}}$.
\end{theorem}

\begin{proof}
  Take $v:=x$ in Lemma \ref{complem} and the result follows.
\end{proof}

Similarly, we have the following result by taking $u:=x$ in Lemma \ref{complem}.

\begin{theorem} \label{T:compare2}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\mathbb{R}\to\mathbb{R}$, $f\in{\rm C}_{\rm rd}\times{\rm C}$,
  satisfying \eqref{monineq} and there exists a function
  $v:[a,\infty)_{\scriptscriptstyle{\mathbb T}} \to \mathbb{R}$ such that $v(\infty)$ exists and \eqref{uppineq}
  holds. If $x$ is a solution to \eqref{E:1st.order.v2},
  \eqref{E:IC} and $v(\infty) > A$, then $x(t) < v(t)$ for all $t \in [a,\infty)_{\scriptscriptstyle{\mathbb T}}$.
\end{theorem}


\section{Second order equations with scalar leading coefficient} \label{S:second.order}

The methods used in Section \ref{S:first.order} to derive the
existence and uniqueness results for the first order equations can
be naturally used in order to derive similar results for the
second order dynamic equations. For the second order setting there
are two cases depending on whether the nonlinearity $f$ involves
the $\Delta$-derivative of $x$ or does not. As we shall see, these
two cases differ in the assumption on the leading coefficient
$r(t)$. Note that as in the previous section the function $f$ can
take both positive and negative values. Consider first the
equation
\begin{equation} \label{E:2nd.order}
  \big(r(t)\,x^\Delta\big)^\Delta+f(t,x^\sigma)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
\end{equation}
We note that while the functions $f$ and $x$ in
\eqref{E:2nd.order} are  $n$-vector valued, the function $r$ will
be (in this section) assumed to be scalar valued. Furthermore,
compared with some recent oscillation and asymptotic results for
second order dynamic equations
\cite{rpA.mB.dO02,rpA.mB.pjyW99,eA.mB.lE.aP02,mB.shS04,mB.shS04a,lE.sH93,lE.aP.pR02,lE.aP.ccT?,pR06}
in which $r(t)>0$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, in this paper we assume (if not
otherwise stated) that $r(t)\neq0$ only. This type of assumption
is common in the oscillation theory of difference equations, see
e.g. \cite{zD.sP07,oD.rH99}, and have also been adopted in some
papers in the time scale setting \cite{oD.sH02,oD.dM07,rH99,rH00}.

The results in this section directly generalize
\cite[Theorems 4.2 and 4.5]{lE.aP.ccT?} to vector valued nonlinearity
$f$ which can take negative values and the leading coefficient $r(t)$
is assumed to be nonzero only.

\begin{remark} \label{R:r.1/r} \rm
  Given a function $r:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}$, $r\in{\rm C}_{\rm rd}$, such that
  \begin{equation} \label{E:r.assume.2nd.order}
    \inf_{t\in[a,b]_{\scriptscriptstyle{\mathbb T}}}\big|r(t)\big|>0, \quad\text{for all } b\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation}
  it follows that $r(t)\neq0$ for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, and the function $\frac{1}{r}$ also belongs to ${\rm C}_{\rm rd}$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Hence, the
  integrals
  \begin{equation*} \label{E:R.bR.def}
    R(t,s):=\int_s^t\frac{1}{r(\tau)}\,\Delta\tau, \quad \bar R(t,s):=\int_s^t\frac{1}{|r(\tau)|}\,\Delta\tau, \quad t,s\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation*}
  are well-defined. Obviously, for a fixed $s\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ both $R(\cdot,s)$ and $\bar R(\cdot,s)$ belong to ${\rm C}_{\rm rd}^1$ with
  ($\Delta$-differentiating with respect to the first argument) $R^\Delta(t,s)=\frac{1}{r(t)}$ and
  $\bar R^\Delta(t,s)=\frac{1}{|r(t)|}>0$ for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Consequently, the function $\bar R(t,s)$ is increasing as $t$
  increases or, for the same reason, as $s$ decreases. Moreover, we have
  \begin{equation} \label{E:R.bR.ineq}
    \big|R(t,s)\big|\leq\bar R(t,s)\leq\bar R(t,a), \quad t,s\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},\ t\geq s.
  \end{equation}
\end{remark}

In connection with these functions we shall frequently use the identities
\begin{equation} \label{E:R.bR.sata}
  R\big(\sigma(s),t\big)=R\big(\sigma(s),a\big)-R(t,a), \quad \bar R\big(\sigma(s),t\big)=\bar R\big(\sigma(s),a\big)-\bar R(t,a),
\end{equation}
for $t,s\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Next we present the first main result of this section.

\begin{theorem} \label{T:2nd.order}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\to\mathbb{R}^n$ with $0<q\leq\infty$, $f\in{\rm C}_{\rm rd}\times{\rm C}$, and $r:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}$,
  $r\in{\rm C}_{\rm rd}$, are given functions satisfying condition \eqref{E:r.assume.2nd.order} and the Lipschitz condition
  \eqref{E:Lipschitz.1st.order}, in which $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm rd}$, and
  \begin{equation} \label{E:k.integral.2nd.order}
    \int_a^\infty \bar R\big(\sigma(s),a\big)\,k(s)\,\Delta s<\infty.
  \end{equation}
  Let $A\in\mathbb{R}^n$ be a given vector. If there exists a number $N\in\mathbb{R}$, $|A|\leq N<q$, such that
  \begin{equation} \label{E:f.integral.2nd.order}
    \int_a^\infty \bar R\big(\sigma(s),a\big)\,\big|f\big(s,x^\sigma(s)\big)\big|\,\Delta s\leq N-|A|,
    \quad\text{for all } x\in X_N,
  \end{equation}
  where $X_N$ is defined by \eqref{E:XN.def.1st.order}, then the problem \eqref{E:2nd.order} has a unique solution $x(t)$
  on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ satisfying $x(\infty)=A$ and $(rx^\Delta)(\infty)=0$.
\end{theorem}

Before proving Theorem \ref{T:2nd.order} we need to establish an
auxiliary lemma.

\begin{lemma} \label{L:Gg.int}
  Let $g:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}^n$, $g\in{\rm C}_{\rm rd}$, and $r:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}$, $r\in{\rm C}_{\rm rd}$, be given functions such that condition
  \eqref{E:r.assume.2nd.order} holds and
  \begin{equation} \label{E:g.int.assume}
    \int_a^\infty\bar R\big(\sigma(s),a\big)\,|g(s)|\,\Delta s<\infty.
  \end{equation}
  Define the function
  \begin{equation*} \label{E:G.int.def}
    G(t):=\int_t^\infty\!\!\!R\big(\sigma(s),t\big)\,g(s)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation*}
  Then $G(t)$ is well-defined, $G\in{\rm C}_{\rm rd}^1$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, and
  \begin{equation} \label{E:GD.int}
    G^\Delta(t)=-\frac{1}{r(t)}\,\int_t^\infty\!\!\!g(s)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation}
  Moreover,
  \begin{equation} \label{E:G.int.lim}
    \lim_{t\to\infty}G(t)=0, \quad \lim_{t\to\infty}r(t)\,G^\Delta(t)=0.
  \end{equation}
\end{lemma}

\begin{proof}
  First note that, since $\bar R\big(\sigma(s),t\big)\leq\bar R\big(\sigma(s),a\big)$ for $t\geq a$, we have for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$
  \begin{equation*}
    \big|G(t)\big|\leq\int_t^\infty\!\!\!\bar R\big(\sigma(s),t\big)\,\big|g(s)\big|\,\Delta s\leq
    \int_t^\infty\!\!\!\bar R\big(\sigma(s),a\big)\,\big|g(s)\big|\,\Delta s
    =:\bar G(t)\leq\bar G(a).
  \end{equation*}
  Hence, by assumption \eqref{E:g.int.assume}, $G(t)$ is well defined for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Next, since
  $\bar R(t,a)\leq\bar R\big(\sigma(s),a\big)$ for $\sigma(s)\geq t$, the estimate
  \begin{equation*}
    \Big|R(t,a)\int_t^\infty\!\!\!g(s)\,\Delta s\Big|=\bar R(t,a)\,\Big|\int_t^\infty\!\!\!g(s)\,\Delta s\Big|
    \leq\bar G(t), \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation*}
  and the fact that $\bar R(t,a)>0$ for $t>a$ show that $\big|\int_t^\infty\!g(s)\,\Delta s\big|$ and hence
  $\int_t^\infty\!g(s)\,\Delta s$ are finite for any $t\in(a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Fix any $t_0\in(a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Since $g\in{\rm C}_{\rm rd}$,
  the integral $\int_a^{t_0}g(s)\,\Delta s$ exists, and then with respect to the previous conclusion we get that
  $\int_a^\infty\!g(s)\,\Delta s=\big\{\int_a^{t_0}+\int_{t_0}^\infty\big\}g(s)\,\Delta s$ exists finite. The latter then
  implies that
  \begin{equation} \label{E:g.int.lim}
    \lim_{t\to\infty}\int_t^\infty\!\!\!g(s)\,\Delta s=0.
  \end{equation}
  Thus, by using the first expression in \eqref{E:R.bR.sata}, we may write
  \begin{equation} \label{E:G.int.def.expanded}
    G(t)=\int_t^\infty\!\!\!R\big(\sigma(s),a\big)\,g(s)\,\Delta s-R(t,a)\int_t^\infty\!\!\!g(s)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation}
  in which both improper integrals exist finite. This shows that $G$ is a ${\rm C}_{\rm rd}^1$ function on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Using the time scale product
  rule when $\Delta$-differentiating the second term in \eqref{E:G.int.def.expanded} we obtain formula \eqref{E:GD.int}. Finally,
  the first limit in \eqref{E:G.int.lim} follows from the fact that (for example) $G(a)$ is finite, while the second limit in
  \eqref{E:G.int.lim} is a consequence of  formula \eqref{E:GD.int} in combination with the limit \eqref{E:g.int.lim}.
\end{proof}

We are now ready to derive Theorem \ref{T:2nd.order}.

\begin{proof}[Proof of Theorem \ref{T:2nd.order}]
  We will apply the Banach fixed point theorem in the space $({\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_\psi)$ for a suitably chosen function $\psi$.
  Define the operator $F:X_N\to{\mathcal F}$ (the space of $n$-vector functions) by
  \begin{equation*} \label{E:F.def.2nd.order}
    [Fx](t):=A-\int_t^\infty\!\!\!R\big(\sigma(s),t\big)\,f\big(s,x^\sigma(s)\big)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation*}
  Set $g(t):=f\big(t,x^\sigma(t)\big)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Then Proposition \ref{P:CrdC.Crd} yields that $g\in{\rm C}_{\rm rd}$. By assumption
  \eqref{E:f.integral.2nd.order} and Lemma \ref{L:Gg.int}, we have that $[Fx](t)$ is well-defined for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, $Fx\in{\rm C}_{\rm rd}^1$,
  and
  \begin{equation} \label{E:Fx.Delta.2nd.order}
    [Fx]^\Delta(t)=\frac{1}{r(t)}\,\int_t^\infty\!\!\!g(s)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation}
  with the limits
  \begin{equation} \label{E:Fx.lim}
    \lim_{t\to\infty}[Fx](t)=A, \quad \lim_{t\to\infty}r(t)\,[Fx]^\Delta(t)=0.
  \end{equation}
  Furthermore, inequality \eqref{E:R.bR.ineq} and assumption \eqref{E:f.integral.2nd.order} yield that $\big|[Fx](t)\big|\leq N$
  for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, so that $\|Fx\|_0\leq N$ and $Fx\in X_N$.
  \par Next, similarly to the proof of Theorem \ref{T:1st.order}, we choose the function $\psi(t)$ to be the time scale
  exponential function $e_{\ominus p(\cdot)}(t,a)$, where $p(t):=\bar R\big(\sigma(t),a\big)\,k(t)$ for $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. That is,
  $\psi^\Delta(t)=-p(t)\,\psi^\sigma(t)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$.
  Then, by Lemma \ref{L:exp.def}, we have $0<\psi_0\leq\psi(t)\leq1$ for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ with $\psi_0\in(0,1)$, where
  $\psi_0:=\lim_{t\to\infty}\psi(t)$. Thus, by Remark \ref{R:psi.Banach}, $({\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_\psi)$ is a Banach space. By
  using the Lipschitz condition \eqref{E:Lipschitz.1st.order}, we get for any $x,y\in X_N$
  \begin{align}
    \|Fx-Fy\|_\psi &\leq \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\psi(t)} \int_t^\infty\!\!\!\bar R\big(\sigma(s),t\big)\,
      \big| f\big(s,x^\sigma(s)\big)-f\big(s,y^\sigma(s)\big) \big|\,\Delta s \notag \\
    &\leq \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\psi(t)} \int_t^\infty\!\!\!
      \bar R\big(\sigma(s),t\big)\,k(s)\,\big|x^\sigma(s)-y^\sigma(s)\big|\,\Delta s. \label{E:FxFy.psi.hlp1}
  \end{align}
  Now if $t=a$, then for $s\geq t=a$ we have $\bar R\big(\sigma(s),t\big)\,k(s)=\frac{-\psi^\Delta(s)}{\psi^\sigma(s)}$. If
  $t>a$, then for $s\geq t$ the quantity $\bar R\big(\sigma(s),a\big)>0$ and
  $0\leq\bar R\big(\sigma(s),t\big)<\bar R\big(\sigma(s),a\big)$. In this case we have
  \begin{equation} \label{E:bRss.ta}
    \bar R\big(\sigma(s),t\big)\,k(s)=
    \frac{\bar R\big(\sigma(s),t\big)}{\bar R\big(\sigma(s),a\big)}\,\bar R\big(\sigma(s),a\big)\,k(s)\leq
    \frac{-\psi^\Delta(s)}{\psi^\sigma(s)},
  \end{equation}
  and in combination with the previous case we see that inequality \eqref{E:bRss.ta} holds for any $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Thus, we get
  from \eqref{E:FxFy.psi.hlp1} by using \eqref{E:bRss.ta}, the definition of $\|x-y\|_\psi$, and condition \eqref{E:u.sup.int}
  with $u:=\psi$ and $u_0:=\psi_0$ that
  \begin{equation*}
    \|Fx-Fy\|_\psi\leq\|x-y\|_\psi \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\psi(t)} \int_t^\infty\!\!\![-\psi^\Delta(s)]\,\Delta s
    =(1-\psi_0)\,\|x-y\|_\psi.
  \end{equation*}
  Hence, the mapping $F$ is a contraction in $X_N$. By Proposition \ref{P:Banach}, there is a unique function
  $x\in X_N$ such that $x=Fx$, i.e.,
  \begin{equation} \label{E:x.int.eq.2nd.order}
    x(t)=A-\int_t^\infty\!\!\!R\big(\sigma(s),t\big)\,g(s)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation}
  From the limits in \eqref{E:Fx.lim} we get that
  \begin{equation*}
    x(\infty)=[Fx](\infty)=A \quad\text{and}\quad (rx^\Delta)(\infty)=\lim_{t\to\infty}r(t)\,[Fx]^\Delta(t)=0.
  \end{equation*}
  Moreover, equations \eqref{E:Fx.Delta.2nd.order} and \eqref{E:x.int.eq.2nd.order} show that the function $x$ satisfies
  \begin{equation} \label{E:xD.int.f}
    r(t)\,x^\Delta(t)=\int_t^\infty\!\!\!g(s)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation}
  While the right-hand side of \eqref{E:xD.int.f} is $\Delta$-differentiable, it follows that
  \begin{equation*}
    \big(r(t)\,x^\Delta(t)\big)^\Delta=-g(t), \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation*}
  i.e., the function $x$ satisfies the dynamic equation \eqref{E:2nd.order}. The proof is complete.
\end{proof}

Next we turn our attention to the more general dynamic equation
\begin{equation} \label{E:2nd.order.xD}
  \big(r(t)\,x^\Delta\big)^\Delta+f(t,x^\sigma,x^{\Delta\sigma})=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
\end{equation}
As we saw in Theorem \ref{T:2nd.order}, the problem
\eqref{E:2nd.order} which does not involve $x^\Delta$ in $f$ can
be treated within the set $X_N$ consisting of certain continuous
functions $x$. On the other hand, the problem
\eqref{E:2nd.order.xD} must be considered in a narrower space,
because it is implicitly assumed in the form of this equation that
$x^\Delta$ exists throughout the interval $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Therefore, we
introduce the set
\begin{equation} \label{E:XN.def.2nd.order.xD}
  X_N^1:=\big\{\,x\in{\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}},\ \|x\|_1<\infty,\ \|x\|_0\leq N\big\}.
\end{equation}
Then $X_N^1$ is a closed subset of the Banach space $({\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_1)$.

\begin{remark} \label{R:phi.Banach} \rm
  Given a function $\varphi:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to[c,d]$, $0<c\leq d<\infty$, we introduce on the space ${\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ another norm
  \begin{equation*} \label{E:phi.norm.def}
    \|x\|_\varphi:=\max\big\{\|x/\varphi\|_0,\ \|x^\Delta/\varphi\|_0\big\}.
  \end{equation*}
  Since $c\,\|x\|_\varphi\leq\|x\|_1\leq d\,\|x\|_\varphi$, the norm $\|\cdot\|_\varphi$ is on ${\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ equivalent to the
  norm $\|\cdot\|_1$. Hence, $({\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_\varphi)$ is also a Banach space.
\end{remark}

Our main result regarding equation \eqref{E:2nd.order.xD} is the following.

\begin{theorem} \label{T:2nd.order.xD}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\times\mathbb{R}^n\to\mathbb{R}^n$ with $0<q\leq\infty$, $f\in{\rm C}_{\rm rd}\times{\rm C}\times{\rm C}$, and
  $r:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}$, $r\in{\rm C}_{\rm rd}$, are functions satisfying
  \begin{equation} \label{E:r.assume.2nd.order.xD}
    \inf_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}}\big|r(t)\big|\geq r_0>0
  \end{equation}
  for some number $r_0>0$, the Lipschitz condition
  \begin{equation} \label{E:Lipschitz.2nd.order.xD}
    \big| f(t,x,u)-f(t,y,v) \big| \leq k(t)\,\big[|x-y|+|u-v|\big]
  \end{equation}
  for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},\ x,y\in\Omega,\ u,v\in\mathbb{R}^n$, where $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm rd}$, and condition
  \begin{equation} \label{E:k.integral.2nd.order.xD}
    \int_a^\infty \big[\bar R\big(\sigma(s),a\big)+1\big]\,k(s)\,\Delta s<\infty.
  \end{equation}
  Let $A\in\mathbb{R}^n$ be a given vector. If there exists a number $N\in\mathbb{R}$, $|A|\leq N<q$, such that
  \begin{gather}
    \int_a^\infty \bar R\big(\sigma(s),a\big)\,\big|f\big(s,x^\sigma(s),x^{\Delta\sigma}(s)\big)\big|\,\Delta s\leq N-|A|,
      \quad\text{for all } x\in X_N^1, \label{E:f.integral.2nd.order.xD} \\
    \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}}\Big|\int_t^\infty\!\!\!f\big(s,x^\sigma(s),x^{\Delta\sigma}(s)\big)\,\Delta s\Big|<\infty,
      \quad\text{for all } x\in X_N^1, \label{E:int.f.finite}
  \end{gather}
  where $X_N^1$ is defined by \eqref{E:XN.def.2nd.order.xD}, then the problem \eqref{E:2nd.order.xD} has a unique solution $x(t)$
  on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ satisfying $x(\infty)=A$ and $(rx^\Delta)(\infty)=0$.
\end{theorem}

Let us briefly comment on the main differences between the above
Theorems \ref{T:2nd.order.xD} and \ref{T:2nd.order}.

\begin{remark} \label{R:2nd.order.xD} \rm
  (i) In Theorem \ref{T:2nd.order.xD}, the assumption \eqref{E:r.assume.2nd.order.xD} on the function $r(\cdot)$ is stronger
  than the assumption \eqref{E:r.assume.2nd.order} in Theorem \ref{T:2nd.order}. Therefore, functions $r(\cdot)$ decaying to
  zero at infinity are not allowed in Theorem \ref{T:2nd.order.xD} while they are still admissible for
  Theorem \ref{T:2nd.order}.
  \par(ii) It is a part of the proof of Theorem \ref{T:2nd.order.xD} that assumption \eqref{E:f.integral.2nd.order.xD} implies
  the finiteness of the improper integral $\int_t^\infty\!f\big(s,x^\sigma(s),x^{\Delta\sigma}(s)\big)\,\Delta s$ in condition
  \eqref{E:int.f.finite} for every $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ and every
  $x\in X_N^1$.
  \par(iii) Another difference between Theorems \ref{T:2nd.order.xD} and \ref{T:2nd.order} is the presence of the additional
  condition \eqref{E:int.f.finite} in Theorem \ref{T:2nd.order.xD}. Note that this condition is satisfied e.g. when the
  improper integral
  \begin{equation} \label{E:int.f.finite.a}
    \int_a^\infty\!\big|f\big(s,x^\sigma(s),x^{\Delta\sigma}(s)\big)\big|\,\Delta s<\infty
  \end{equation}
  for any $x\in X_N^1$. The latter condition is satisfied, in particular, when $n=1$ and $f(\cdot,\cdot,\cdot)\geq0$ as
  in \cite{lE.aP.ccT?}. Assuming \eqref{E:r.assume.2nd.order.xD}, condition \eqref{E:int.f.finite} in fact means that
  $\big\|[Fx]^\Delta\big\|_0$ is finite, which is needed in order to show that $Fx\in X_N^1$. On the other hand, the
  finiteness of $\big\|[Fx]^\Delta\big\|_0$ is not required in the proof of Theorem \ref{T:2nd.order}, since there we work
  in the set $X_N$ only.
\end{remark}

\begin{proof}[Proof of Theorem \ref{T:2nd.order.xD}]
  The proof is similar to the proof of Theorem \ref{T:2nd.order} and we shall include only the main differences.
  We will apply the Banach fixed point theorem in the space $({\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_\varphi)$ for a suitably chosen function
  $\varphi$. Define the operator $F:X_N^1\to{\mathcal F}$ by
  \begin{equation*} \label{E:F.def.2nd.order.xD}
    [Fx](t):=A-\int_t^\infty\!\!\!R\big(\sigma(s),t\big)\,f\big(s,x^\sigma(s),x^{\Delta\sigma}(s)\big)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation*}
  and set $g(t):=f\big(t,x^\sigma(t),x^{\Delta\sigma}(t)\big)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Then Proposition \ref{P:CrdCC.Crd} yields that $g\in{\rm C}_{\rm rd}$.
  As in the proof of Theorem \ref{T:2nd.order} we get that $[Fx](t)$ is well defined for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, and formulas
  \eqref{E:Fx.Delta.2nd.order} and \eqref{E:Fx.lim} hold. In particular, $\int_t^\infty\!g(s)\,\Delta s$ exists finite for all
  $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Assumption \eqref{E:f.integral.2nd.order.xD} now yields that $\big|[Fx](t)\big|\leq N$ for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, so that
  $\|Fx\|_0\leq N$. Furthermore, from \eqref{E:Fx.Delta.2nd.order} and assumption \eqref{E:int.f.finite} we get
  \begin{equation*}
    \big\|[Fx]^\Delta\big\|_0=\sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}}\frac{1}{|r(t)|}\,\Big|\int_t^\infty\!\!\!g(s)\,\Delta s\Big|
    \leq\frac{1}{r_0}\,\sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}}\Big|\int_t^\infty\!\!\!g(s)\,\Delta s\Big|<\infty.
  \end{equation*}
  This yields that $\|Fx\|_1=\max\big\{\|Fx\|_0,\ \big\|[Fx]^\Delta\big\|_0 \big\}$ is finite, and thus
  $Fx\in X_N^1$.

   Choose the function $\varphi(t)$ to be the time scale exponential function $e_{\ominus p(\cdot)}(t,a)$, where
  $p(t):=\frac{2}{r_0}\,\big[\bar R\big(\sigma(t),a\big)+1\big]\,k(t)$ for $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Then, by Lemma \ref{L:exp.def}, we
  have $0<\varphi_0\leq\varphi(t)\leq1$ for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ with $\varphi_0\in(0,1)$, where
  $\varphi_0:=\lim_{t\to\infty}\varphi(t)$. Thus, by Remark \ref{R:phi.Banach}, $({\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_\varphi)$ is a Banach
  space. Similarly to the proof of Theorem \ref{T:2nd.order} we now deduce that for any $x,y\in X_N^1$
  \begin{equation} \label{E:FxFy.estimate}
    \big\|(Fx-Fy)/\varphi\big\|_0\leq(1-\varphi_0)\,\|x-y\|_\varphi.
  \end{equation}
  Furthermore, by assumption \eqref{E:r.assume.2nd.order.xD} and the Lipschitz condition \eqref{E:Lipschitz.2nd.order.xD},
  \begin{align*}
    &\big\|\big([Fx]^\Delta - [Fy]^\Delta\big)/\varphi\big\|_0 \\
    &= \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\varphi(t)}\,\Big|\frac{-1}{r(t)}\,\int_t^\infty\!\!\!
      \big[g(s)-f\big(s,y^\sigma(s),y^{\Delta\sigma}(s)\big)\big]\,\Delta s\Big| \\
    &\leq \frac{1}{r_0}\,\big[\big\|(x-y)/\varphi\big\|_0+\big\|(x^\Delta-y^\Delta)/\varphi\big\|_0\big]\,
      \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\varphi(t)}\,\int_t^\infty\!\!\!k(s)\,\varphi^\sigma(s)\,\Delta s \\
    &\leq \frac{2}{r_0}\,\|x-y\|_\varphi\,\sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\varphi(t)}
      \,\int_t^\infty\!\!\!k(s)\,\varphi^\sigma(s)\,\Delta s.
  \end{align*}
  Now the choice of $\varphi$, the fact that $\bar R\big(\sigma(s),a\big)+1\geq1$ for any $s\geq t$, and condition
  \eqref{E:u.sup.int} with $u:=\varphi$ and $u_0:=\varphi_0$ yield that
  \begin{align}
    \big\|\big([Fx]^\Delta-[Fy]^\Delta\big)/\varphi\big\|_0 &= \|x-y\|_\varphi\,\sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\varphi(t)}
      \,\int_t^\infty\!\!\!\frac{1}{\bar R\big(\sigma(s),a\big)+1}\,[-\varphi^\Delta(s)]\,\Delta s \notag \\
    &\leq \|x-y\|_\varphi\,\sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}} \frac{1}{\varphi(t)}\,\int_t^\infty\!\!\![-\varphi^\Delta(s)]\,\Delta s \notag \\
    &= (1-\varphi_0)\,\|x-y\|_\varphi. \label{E:FxDFyD.estimate}
  \end{align}
  The estimates \eqref{E:FxFy.estimate} and \eqref{E:FxDFyD.estimate} now
  yield that
  $\|Fx-Fy\|_\varphi\leq(1-\varphi_0)\,\|x-y\|_\varphi$, that is, the mapping $F$ is a contraction in $X_N^1$.
  Hence, by Proposition \ref{P:Banach}, there is a unique function $x\in X_N^1$ such that $x=Fx$. The rest of the proof is the
  same as in Theorem \ref{T:2nd.order}, namely we conclude that the function $x$ satisfies the equation \eqref{E:2nd.order.xD},
  and the limits $x(\infty)=A$ and $(rx^\Delta)(\infty)=0$.
\end{proof}

Similarly as in Corollary \ref{C:1st.order.M}, upon the choice
$A:=M$ and $N:=2\,|M|$ in Theorems \ref{T:2nd.order} and
\ref{T:2nd.order.xD} we can now derive results on the solvability
of the terminal value problems for the equations
\eqref{E:2nd.order} and \eqref{E:2nd.order.xD} with the limits
$x(\infty)=M$ and $(rx^\Delta)(\infty)=0$.

\begin{corollary} \label{C:2nd.order}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\to\mathbb{R}^n$ with $0<q\leq\infty$, $f\in{\rm C}_{\rm rd}\times{\rm C}$, and $r:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}$,
  $r\in{\rm C}_{\rm rd}$, are given functions satisfying condition \eqref{E:r.assume.2nd.order} and the Lipschitz condition
  \eqref{E:Lipschitz.1st.order}, in which $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm rd}$, and \eqref{E:k.integral.2nd.order} holds. If
  there exists a vector $M\in\mathbb{R}^n$, $|M|<q$, such that
  \begin{equation*} \label{E:f.integral.2nd.order.M}
    \int_a^\infty \bar R\big(\sigma(s),a\big)\,\big|f\big(s,x^\sigma(s)\big)\big|\,\Delta s\leq|M|, \quad\text{for all }
    x\in X_{2\,|M|},
  \end{equation*}
  then the problem \eqref{E:2nd.order} has a unique solution $x(t)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ which satisfies $x(\infty)=M$ and
  $(rx^\Delta)(\infty)=0$.
\end{corollary}

\begin{corollary} \label{C:2nd.order.xD}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\times\mathbb{R}^n\to\mathbb{R}^n$ with $0<q\leq\infty$, $f\in{\rm C}_{\rm rd}\times{\rm C}\times{\rm C}$, and
  $r:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}$, $r\in{\rm C}_{\rm rd}$, are functions satisfying \eqref{E:r.assume.2nd.order.xD} for some number $r_0>0$,
  the Lipschitz condition \eqref{E:Lipschitz.2nd.order.xD} in which $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm rd}$, and condition
  \eqref{E:k.integral.2nd.order.xD} holds. If there exists a vector $M\in\mathbb{R}^n$, $|M|<q$, such that
  \begin{gather*}
    \int_a^\infty \bar R\big(\sigma(s),a\big)\,\big|f\big(s,x^\sigma(s),x^{\Delta\sigma}(s)\big)\big|\,\Delta s\leq|M|,
    \quad\text{for all } x\in X_{2\,|M|}^1, \label{E:f.integral.2nd.order.xD.M} \\
    \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}}\Big|\int_t^\infty\!\!\!f\big(s,x^\sigma(s),x^{\Delta\sigma}(s)\big)\,\Delta s\Big|<\infty,
      \quad\text{for all } x\in X_{2\,|M|}^1, \label{E:int.f.finite.M}
  \end{gather*}
  then the problem \eqref{E:2nd.order.xD} has a unique solution $x(t)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ which satisfies $x(\infty)=M$ and
  $(rx^\Delta)(\infty)=0$.
\end{corollary}

\section{Second order equations with matrix leading coefficient}
\label{S:second.order.matrix}

In this section we present existence and uniqueness results for
second order $n$-vector dynamic equations of the form
\eqref{E:2nd.order} and \eqref{E:2nd.order.xD}, but in which the
leading coefficient is a nonnegative $n\times n$ matrix function
and $f$ is a nonnegative $n$-vector function. The method for
proving such results is a combination of the approach from
\cite{lE.aP.ccT?}, where scalar valued dynamic equations with
nonnegative $f$ were considered, with the methods presented in
Section \ref{S:second.order}.

Since in this section the nonlinearity $f$ and solutions $x$ will
have nonnegative entries, we modify the notation accordingly. If a
vector $x\in\mathbb{R}^n$ has nonnegative entries, we shall denote
it by $x\geq0$. Similarly, for two vectors $x,y\in\mathbb{R}^n$ we
write $x\leq y$ provided $y-x\geq0$, i.e., their entries are
compared componentwise.

For a real $n\times n$ matrix $A$ we shall use any norm
$\|\cdot\|$ compatible with (e.g., induced by) the vector norm $|\cdot|$,
for example, the maximum row sum matrix norm
$\|A\|:=\|A\|_\infty=\max\{\sum_{j=1}^n|a_{ij}|,\ i=1,\dots,n\}$.
We also require, that the matrix norm is monotone; that is,
\begin{equation} \label{E:matrix.norm.monotone}
  \begin{aligned}
   &\text{if $A$ and $B$ are symmetric with nonnegative} \\
   &\text{entries and $0<A\leq B$, then $\|A\|\leq\|B\|$}.
  \end{aligned}
\end{equation}
We refer to \cite{dsB05} for more properties of matrix norms.
Throughout this section we will denote by
$\mathbb{R}_+:=[0,\infty)$ the set of all nonnegative real
numbers. Similarly, the set of all such $n$-tuples will be denoted
by $\mathbb{R}_+^n:=[0,\infty)^n$. Accordingly with the notation
introduced in Section \ref{S:pre}, we denote by $\Omega_q^+$ the
intersection of the open $q$-ball $\Omega_q$ with
$\mathbb{R}_+^n$, that is $\Omega_q^+:=\{x\in\mathbb{R}_+^n,\
|x|<q\}$.

For any vector $M\in\mathbb{R}_+^n$ we define the set
\begin{equation} \label{E:YM.def.2nd.order}
  Y_M:=\big\{\,x\in{\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}},\ 0\leq x(t)\leq M, \text{ for all } t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}\big\}.
\end{equation}
Then $Y_M$ is a closed subset of the Banach space $({\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_0)$. Consider the second order dynamic equation
\begin{equation} \label{E:2nd.order.matrix}
  \big(P(t)\,x^\Delta\big)^\Delta+f(t,x^\sigma)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
\end{equation}
In this section we denote by $\lambda_0(t)$ the smallest
eigenvalue of the symmetric matrix $P(t)$. Similarly to Remark
\ref{R:r.1/r}, for $t,s\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ we define the symmetric $n\times n$
matrix
\begin{equation*} \label{E:Qts.def}
  Q(t,s):=\int_s^tP^{-1}(\tau)\,\Delta\tau.
\end{equation*}
We emphasize that in this section the matrix $P(t)$ and the $n$-vectors $f(t,x)$ or $f(t,x,y)$ have only nonnegative
entries.

The following two results generalize \cite[Theorems~4.2 and~4.5]{lE.aP.ccT?} to vector valued nonlinearity $f$ and matrix
valued leading coefficient $r(t)$.

\begin{theorem} \label{T:2nd.order.matrix}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q^+\to\mathbb{R}_+^n$ with $0<q\leq\infty$, $f\in{\rm C}_{\rm rd}\times{\rm C}$, and
  $P:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}_+^{n\times n}$, $P\in{\rm C}_{\rm rd}$, $P(t)$ positive definite for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, are given functions satisfying
  \begin{equation*} \label{E:P.assume.2nd.order.matrix}
    \inf_{t\in[a,b]_{\scriptscriptstyle{\mathbb T}}}\lambda_0(t)>0, \quad\text{for all } b\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation*}
  and the Lipschitz condition \eqref{E:Lipschitz.1st.order}, in which
  $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm rd}$, and
 \begin{equation*} \label{E:k.integral.2nd.order.matrix}
    \int_a^\infty \big\|Q\big(\sigma(s),a\big)\big\|\,k(s)\,\Delta s<\infty,
  \end{equation*}
  where the matrix norm $\|\cdot\|$ is compatible with the vector norm $|\cdot|$ and monotone in the sense of condition
  \eqref{E:matrix.norm.monotone}. If there exists a vector $M\in\mathbb{R}_+^n$, $|M|<q$, such that
  \begin{equation} \label{E:f.integral.2nd.order.matrix}
    \int_a^\infty Q\big(\sigma(s),a\big)\,f\big(s,x^\sigma(s)\big)\,\Delta s\leq M, \quad\text{for all } x\in Y_M,
  \end{equation}
  where $Y_M$ is defined by \eqref{E:YM.def.2nd.order}, then the problem \eqref{E:2nd.order.matrix} has a unique solution
  $x(t)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ satisfying $x(\infty)=M$ and $(rx^\Delta)(\infty)=0$. Furthermore,
  \begin{equation} \label{E:xxD.PQ}
    x(t)\geq Q(t,a)\,P(t)\,x^\Delta(t), \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation}
\end{theorem}

\begin{proof}
  The proof is a combination of the proofs of Theorem \ref{T:2nd.order} and \cite[Theorem~4.2]{lE.aP.ccT?}. With the notation
  $g(s):=f\big(s,x^\sigma(s)\big)$, the operator
  \begin{equation*} \label{E:T.def.2nd.order}
    [Tx](t):=M-\int_t^\infty\!\!\!Q\big(\sigma(s),t\big)\,g(s)\,\Delta s, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation*}
  maps the set $Y_M$ into itself. Indeed, $Tx\in{\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}}\subseteq{\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ and since the functions $P$ and $f$ have nonnegative
  entries, we have
  \begin{equation} \label{E:Tx.Delta.2nd.order}
    [Tx]^\Delta(t)=P^{-1}(t)\,\int_t^\infty\!\!\!g(s)\,\Delta s\geq0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation}
  and $[Tx](t)\leq M$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ and $[Tx](a)\geq0$. This yields that $[Tx](\cdot)$ is nondecreasing so that $[Tx](t)\geq0$ on
  $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. Furthermore,
  \begin{equation} \label{E:Q.lim.hlp1}
    0\leq\int_t^\infty\!\!\!Q\big(\sigma(s),t\big)\,g(s)\,\Delta s=\int_t^\infty\!\!\!Q\big(\sigma(s),a\big)\,g(s)\,\Delta s
    -Q(t,a) \int_t^\infty\!\!\!g(s)\,\Delta s
  \end{equation}
  implies that
  \begin{equation*}
    0\leq Q(t,a) \int_t^\infty\!\!\!g(s)\,\Delta s\leq\int_t^\infty\!\!\!Q\big(\sigma(s),a\big)\,g(s)\,\Delta s,
  \end{equation*}
  so that, by assumption \eqref{E:f.integral.2nd.order.matrix},
  \begin{equation} \label{E:Q.lim.hlp2}
    \lim_{t\to\infty} \int_t^\infty\!\!\!Q\big(\sigma(s),a\big)\,g(s)\,\Delta s =0= \lim_{t\to\infty}
    Q(t,a) \int_t^\infty\!\!\!g(s)\,\Delta s.
  \end{equation}
  Equations \eqref{E:Q.lim.hlp1} and \eqref{E:Q.lim.hlp2} now yield that
  \begin{equation*} \label{E:Q.lim.hlp3}
    \lim_{t\to\infty} \int_t^\infty\!\!\!Q\big(\sigma(s),t\big)\,g(s)\,\Delta s =0,
  \end{equation*}
  Thus, the definition of $Tx$ implies that $[Tx](\infty)=M$. The contraction property of $T$ is proven similarly as in
  Theorem \ref{T:2nd.order} but with $p(t):=\big[\big\|Q\big(\sigma(t),a\big)\big\|+1\big]\,k(t)$. Here $\|\cdot\|$ is
  the previously discussed matrix norm, for which we have
  $\big\|Q\big(\sigma(s),t\big)\big\|\leq\big\|Q\big(\sigma(s),a\big)\big\|$ for $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ and $s\geq t$.

 Thus, the Banach Theorem (Proposition \ref{P:Banach}) yields a unique function $x\in Y_M$ with $Tx=x$, i.e.,
  $x\in{\rm C}[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ and $0\leq x(t)\leq M$ (componentwise) for all $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$. From \eqref{E:Tx.Delta.2nd.order} we get
  $x^\Delta=[Tx]^\Delta\in{\rm C}_{\rm rd}[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ which implies that $x\in{\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}}$.
  Moreover, since $Q(\cdot,a)$ is positive definite and
  increasing, it follows that either some of its eigenvalues tend monotonically to $\infty$ or all its eigenvalues are bounded
  and in this case $Q(t,a)$ converges to some constant matrix $Q_0>0$ as $t\to\infty$. But in any of these two cases the
  second limit in \eqref{E:Q.lim.hlp2} implies that
  \begin{equation} \label{E:Q.lim.hlp4}
    \lim_{t\to\infty} \int_t^\infty\!\!\!g(s)\,\Delta s =0.
  \end{equation}
  Hence, by \eqref{E:Tx.Delta.2nd.order} and \eqref{E:Q.lim.hlp4},
  \begin{equation*}
    (Px^\Delta)(\infty)=\lim_{t\to\infty} P(t)\,[Tx]^\Delta(t)=\int_t^\infty\!\!\!g(s)\,\Delta s =0.
  \end{equation*}
  Furthermore, from $x=Tx$ we get
  \begin{align*}
    x(t) &\geq M-\int_a^\infty\!\!\!Q\big(\sigma(s),a\big)\,g(s)\,\Delta s+
      Q(t,a)\,P(t)\,P^{-1}(t)\int_t^\infty\!\!\!g(s)\,\Delta s,
  \end{align*}
  so that by assumption \eqref{E:f.integral.2nd.order.matrix} and by using formula \eqref{E:Tx.Delta.2nd.order} in
  $x^\Delta(t)=[Tx]^\Delta(t)$ we have inequality \eqref{E:xxD.PQ}. The proof is now complete.
\end{proof}

Next we define the set
\begin{equation}
\begin{aligned}
  Y_M^1:=\big\{&x\in{\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}}, &\ 0\leq x(t)\leq M,\ x^\Delta(t)\geq
   0  \\
    &\text{for all } t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},\ \|x^\Delta\|_0<\infty\big\}.
\end{aligned}    \label{E:YM1.def.2nd.order}
\end{equation}
Then $Y_M^1$ is a closed subset of the Banach space $({\rm C}_{\rm rd}^1[a,\infty)_{\scriptscriptstyle{\mathbb T}},\|\cdot\|_1)$.
 Consider the second order dynamic equation
\begin{equation} \label{E:2nd.order.xD.matrix}
  \big(P(t)\,x^\Delta\big)^\Delta+f(t,x^\sigma,x^{\Delta\sigma})=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
\end{equation}

\begin{theorem} \label{T:2nd.order.xD.matrix}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q^+\times\mathbb{R}_+^n\to\mathbb{R}_+^n$ with $0<q\leq\infty$,
  $f\in{\rm C}_{\rm rd}\times{\rm C}\times{\rm C}$, and $P:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}_+^{n\times n}$, $P\in{\rm C}_{\rm rd}$, $P(t)$ positive definite for all
  $t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}$, are given functions satisfying
  \begin{equation*} \label{E:P.assume.2nd.order.xD.matrix}
    \inf_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}}\lambda_0(t)\geq r_0>0
  \end{equation*}
  for some number $r_0>0$, and the Lipschitz condition \eqref{E:Lipschitz.2nd.order.xD}, in which
  $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm rd}$, and
 \begin{equation*} \label{E:k.integral.2nd.order.xD.matrix}
    \int_a^\infty \big[\big\|Q\big(\sigma(s),a\big)\big\|+1\big]\,k(s)\,\Delta s<\infty,
  \end{equation*}
  where the matrix norm $\|\cdot\|$ is compatible with the vector norm $|\cdot|$ and monotone in the sense of condition
  \eqref{E:matrix.norm.monotone}. If there exists a vector $M\in\mathbb{R}_+^n$, $|M|<q$, such that
  \begin{gather*}
    \int_a^\infty Q\big(\sigma(s),a\big)\,f\big(s,x^\sigma(s),x^{\Delta\sigma}(s)\big)\,\Delta s\leq M, \quad\text{for all }
      x\in Y_M^1, \label{E:f.integral.2nd.order.xD.matrix} \\
    \sup_{t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}}\Big|\int_t^\infty\!\!\!f\big(s,x^\sigma(s),x^{\Delta\sigma}(s)\big)\,\Delta s\Big|<\infty,
      \quad\text{for all } x\in Y_M^1, \label{E:int.f.finite.matrix}
  \end{gather*}
  where $Y_M^1$ is defined by \eqref{E:YM1.def.2nd.order}, then the problem \eqref{E:2nd.order.xD.matrix} has a unique solution
  $x(t)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ satisfying $x(\infty)=M$ and $(rx^\Delta)(\infty)=0$. Furthermore, inequality \eqref{E:xxD.PQ} holds.
\end{theorem}

\begin{proof}
  The proof is a combination of the proofs of Theorem \ref{T:2nd.order.xD} and \cite[Theorem~4.5]{lE.aP.ccT?}. The proof of
  the contraction now uses the function
  \begin{equation*}
    p(t):=\frac{2}{r_0}\,\big[\big\|Q\big(\sigma(t),a\big)\big\|+1\big]\,k(t).
  \end{equation*}
  The details are omitted.
\end{proof}

\section{Examples} \label{S:examples}

We now present some examples to illustrate how to apply the main ideas
 of this paper.

\begin{example} \label{Ex:first.order} \rm
  Consider the scalar-valued terminal value problem
  \begin{equation*}
    x^\Delta = \dfrac{(x^\sigma)^2}{9\,e^\sigma(t,a)}, \quad t \in [a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation*}
  and $x(\infty) =1$. We claim that this problem has a unique
  solution $x$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$.
\end{example}

\begin{proof}
    Our objective is to show that the conditions of Theorem \ref{T:1st.order} are satisfied. We choose $q:=3$ to form $\Omega_q$.
    Next, we see for all $t \in [a,\infty)_{\scriptscriptstyle{\mathbb T}}$ and all $u,v \in \Omega_3$ we have
    \begin{equation*}
      \dfrac{|u^2 - v^2|}{9\,e^\sigma(t,a)} \le \dfrac{6\,|u-v|}{9\,e^\sigma(t,a)}
    \end{equation*}
    and so \eqref{E:Lipschitz.1st.order} holds for
    \begin{equation*}
      k(t):= \dfrac{2}{3\,e^\sigma(t,a)}, \quad t \in [a,\infty)_{\scriptscriptstyle{\mathbb T}}.
    \end{equation*}
    Furthermore, it is not difficult to show that the left-hand side of \eqref{E:k.integral.1st.order} is $\frac{2}{3}$ and so
    inequality \eqref{E:k.integral.1st.order} holds for the above defined function $k(\cdot)$.
    \par
    Now choose $N:=2$ to form $X_N$.  See that for all $x \in X_2$ we then have
    \begin{equation*}
      \int_a^\infty \dfrac{[x^\sigma(s)]^2}{9\,e^\sigma(s,a)}\, \Delta s \le \int_a^\infty \dfrac{4}{9\,e^\sigma(s,a)}\, \Delta s
      = \dfrac{4}{9}
    \end{equation*}
    and so \eqref{E:f.integral.1st.order} holds. Thus, the claim follows from Theorem \ref{T:1st.order}.
\end{proof}


\begin{example} \label{Ex:second.order.scalar} \rm
  In this example we illustrate the applicability of Theorem \ref{T:2nd.order}. Let $n=1$, $q=\infty$,
  ${\mathbb T}={\mathbb Z}$, $a=0$, and
  \begin{equation*}
    r(t):=\frac{(-1)^t}{t+1}, \quad f(t,x):=\frac{1}{(t+1)^{3+\beta}}\,\frac{x}{1+x^2}, \quad\text{with } \beta>0.
  \end{equation*}
  Then we claim that the assumptions of Theorem \ref{T:2nd.order} are
  satisfied.
\end{example}

\begin{proof}
    First note that $r(t)$ changes its sign but
    \begin{equation*}
      \inf_{t\in[0,b]_{{\scriptscriptstyle{\mathbb Z}}}}\big|r(t)\big|=
      \inf_{t\in[0,b]_{{\scriptscriptstyle{\mathbb Z}}}}\frac{1}{t+1}=\frac{1}{b+1}>0 \quad\text{for every }
      b\in[0,\infty)_{{\scriptscriptstyle{\mathbb Z}}},
    \end{equation*}
    hence condition \eqref{E:r.assume.2nd.order} holds. It follows that
    \begin{equation*}
      R(t,s)=\sum_{i=s}^{t-1}\frac{i+1}{(-1)^i}=\sum_{i=s+1}^t(-1)^{i-1}\,i, \quad \bar R(t,s)=\sum_{i=s+1}^ti=
      \frac{t(t+1)}{2}-\frac{s(s+1)}{2}.
    \end{equation*}
    Moreover, $f(t,x)$ is Lipschitz with $k(t):=1/(t+1)^{3+\beta}$, i.e., condition \eqref{E:Lipschitz.1st.order} holds.
    With the estimate $(i+1)(i+2)\leq2(i+1)^2$, a simple calculation shows that
    \begin{equation*}
      \int_0^\infty\!\! \bar R(s+1,0)\,k(s)\,\Delta s=\sum_{i=0}^\infty
      \frac{(i+1)\,(i+2)}{2\,(i+1)^{3+\beta}} \leq \sum_{i=0}^\infty \frac{1}{(i+1)^{1+\beta}}<\infty,
    \end{equation*}
    by the integral criterion for infinite series. Thus, condition \eqref{E:k.integral.2nd.order} holds. Finally,
    since the function $x/(1+x^2)$ is bounded on $\mathbb{R}$ (by $\frac{1}{2}$), we get for any sequence $x\in X_N$
    that
    \begin{align*}
      \int_0^\infty\!\! \bar R(s+1,0)\,\big|f\big(s,x(s+1)\big)\big|\,\Delta s & \leq \frac{1}{2} \sum_{i=0}^\infty
        \frac{1}{(i+1)^{\beta+1}} = \frac{1}{2} + \frac{1}{2} \int_1^\infty \!\!\!\!\frac{1}{\tau^{\beta+1}}\,{\mathrm d}\tau \\
      &\leq \frac{1}{2}+\frac{1}{2\beta}=\frac{\beta+1}{2\beta}.
    \end{align*}
    Since the above estimate is independent of $x(\cdot)$, we may choose $N:=|A|+\frac{\beta+1}{2\beta}$, and then
    inequality \eqref{E:f.integral.2nd.order} holds. Hence, by Theorem \ref{T:2nd.order}, for any $A\in\mathbb{R}$
    the terminal value problem
    \begin{gather*}
      \Delta\Big( \frac{(-1)^t}{(t+1)^{3+\beta}}\,\Delta x(t) \Big)
      + \frac{1}{(t+1)^{3+\beta}}\,\frac{x(t+1)}{1+x^2(t+1)}=0,
        \quad t\in[0,\infty)_{{\scriptscriptstyle{\mathbb Z}}}, \\
      \lim_{t\to\infty} x(t)=A, \quad \lim_{t\to\infty} \frac{(-1)^t}{(t+1)^{3+\beta}}\,\Delta x(t)=0,
    \end{gather*}
    has a unique solution $x(\cdot)$ on $[0,\infty)_{{\scriptscriptstyle{\mathbb Z}}}$.
  \end{proof}


\begin{example} \label{Ex:second.order.scalar.r} \rm
  Note that the leading coefficient $r(t)$ from Example \ref{Ex:second.order.scalar} is not allowed in
  Theorem \ref{T:2nd.order.xD}, since
  \begin{equation*}
    \inf_{t\in[0,\infty)_{{\scriptscriptstyle{\mathbb Z}}}}\big|r(t)\big|=
    \inf_{t\in[0,\infty)_{{\scriptscriptstyle{\mathbb Z}}}}\frac{1}{t+1}=0,
  \end{equation*}
  contradicting condition \eqref{E:r.assume.2nd.order.xD}. However, one can consider a leading coefficient such as
  $r(t)=(-1)^t$ on $[0,\infty)_{{\scriptscriptstyle{\mathbb Z}}}$, which is admissible in Theorem \ref{T:2nd.order.xD}.
\end{example}

\begin{example} \label{Ex:second.order.scalar.xD} \rm
  In this example we illustrate the applicability of
Theorem \ref{T:2nd.order.xD}. Let $n=1$, $q=\infty$,
  ${\mathbb T}=\mathbb{R}$, $a=0$, and
  \begin{equation*}
    r(t):=1, \quad f(t,x,y):=\frac{1}{(t+1)^{2+\beta}}\,
    \Big( \cos x+\frac{\sin y}{1+\sin^2y}\Big),
    \quad\text{with } \beta>0.
  \end{equation*}
  Then we claim that the assumptions of Theorem \ref{T:2nd.order.xD}
  are satisfied.

\begin{proof}
    We have $R(t,s)=\bar R(t,s)=\int_s^t1\,{\mathrm d}\tau=t-s$, and
    the Lipschitz condition \eqref{E:Lipschitz.2nd.order.xD}
    is satisfied with the function $k(t):=1/(t+1)^{2+\beta}$. Conditions \eqref{E:k.integral.2nd.order.xD} and
    \eqref{E:f.integral.2nd.order.xD} are verified similarly as in Example \ref{Ex:second.order.scalar}. Condition
    \eqref{E:int.f.finite} follows from \eqref{E:int.f.finite.a} in Remark \ref{R:2nd.order.xD}(iii) and from the estimate
    \begin{equation*}
      \int_0^\infty \big|f\big(s,x(s),x'(x)\big)\big|\,{\mathrm d}s \leq \frac{3}{2} \int_0^\infty \!\!\!\!
      \frac{1}{(s+1)^{2+\beta}}\,{\mathrm d}s=\frac{3}{2\,(\beta+1)}<\infty
    \end{equation*}
    for every $x\in X_N^1$, since the function $\cos x+\frac{\sin y}{1+\sin^2y}$ is bounded on $\mathbb{R}$ (by
    $\frac{3}{2}$). Hence, by Theorem \ref{T:2nd.order.xD}, for any $A\in\mathbb{R}$ the terminal value problem
    \begin{gather*}
      x'' + \frac{1}{(t+1)^{2+\beta}}\,\Big( \!\cos x+\frac{\sin x'}{1+\sin^2x'}\Big)=0, \quad t\in[0,\infty), \\
      \lim_{t\to\infty} x(t)=A, \quad \lim_{t\to\infty} x'(t)=0,
    \end{gather*}
    has a unique solution $x(\cdot)$ on $[0,\infty)$.
  \end{proof}
\end{example}

The  two examples above motivate the following corollaries of
Theorems \ref{T:2nd.order} and \ref{T:2nd.order.xD}, in which the
existence of the number $N$ is guaranteed from the assumed
estimates on the data.

\begin{corollary} \label{C:second.order.scalar}
  Assume that $g:\Omega_q\to\mathbb{R}^n$ with $0<q\leq\infty$, $g\in{\rm C}^1$, and $r:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}$, $r\in{\rm C}_{\rm rd}$, and
  $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm rd}$, are given functions satisfying condition \eqref{E:r.assume.2nd.order},
  \begin{equation} \label{E:second.order.scalar.k1}
    \int_a^\infty \! \bar R\big(\sigma(s),a\big)\,k(s)\,\Delta s\leq k_1<\infty,
  \end{equation}
  $\big|g(x)\big|\leq M_1$ on $\Omega_q$, and $g'(x)$ is bounded on $\Omega_q$. Then for any $A\in\mathbb{R}^n$ such that
  $|A|<q-M_1k_1$ (in particular, for any $A\in\mathbb{R}^n$ if $q=\infty$) the problem
  \begin{equation*}
    \big(r(t)\,x^\Delta\big)^\Delta+k(t)\,g(x^\sigma)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation*}
  has a unique solution $x(\cdot)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ satisfying $x(\infty)=A$ and $(rx^\Delta)(\infty)=0$.
\end{corollary}
\begin{proof}
  Upon taking $f(t,x):=k(t)\,g(x)$ and $N:=|A|+M_1\,k_1$ we show that these data satisfy the assumptions of
  Theorem \ref{T:2nd.order}.
\end{proof}

\begin{corollary} \label{C:second.order.scalar.xD}
  Assume that $g:\Omega_q\times\mathbb{R}^n\to\mathbb{R}^n$ with $0<q\leq\infty$, $g\in{\rm C}^1$, and $r:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to\mathbb{R}$,
  $r\in{\rm C}_{\rm rd}$, and $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm rd}$, are given functions satisfying conditions \eqref{E:r.assume.2nd.order.xD},
  \eqref{E:second.order.scalar.k1}, \eqref{E:k.integral.1st.order}, and $\big|g(x,y)\big|\leq M_1$ on
  $\Omega_q\times\mathbb{R}^n$, and $g_x(x,y)$ and $g_y(x,y)$ are bounded on $\Omega_q\times\mathbb{R}^n$. Then for any
  $A\in\mathbb{R}^n$ such that $|A|<q-M_1k_1$ (in particular, for any $A\in\mathbb{R}^n$ if $q=\infty$) the problem
  \begin{equation*}
    \big(r(t)\,x^\Delta\big)^\Delta+k(t)\,g(x^\sigma,x^{\Delta\sigma})=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
  \end{equation*}
  has a unique solution $x(\cdot)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ satisfying $x(\infty)=A$ and $(rx^\Delta)(\infty)=0$.
\end{corollary}

\begin{proof}
  Upon taking $f(t,x,y):=k(t)\,g(x,y)$ and $N:=|A|+M_1\,k_1$ we show that these data satisfy the assumptions of
  Theorem \ref{T:2nd.order.xD}. Note that condition \eqref{E:int.f.finite.a} from Remark \ref{R:2nd.order.xD}(iii) is used
  in order to verify condition \eqref{E:int.f.finite}.
\end{proof}

In Example \ref{Ex:second.order.scalar} we had
$k(t)=1/(t+1)^{3+\beta}$, $k_1=\frac{\beta+1}{\beta}$,
$g(x)=x/(1+x^2)$, and $M_1=\frac{1}{2}$, while in Example
\ref{Ex:second.order.scalar.xD} we had $k(t)=1/(t+1)^{2+\beta}$,
$k_1=\frac{1}{\beta}$, $g(x,y)=\cos x+\frac{\sin y}{1+\sin^2y}$,
and $M_1=\frac{3}{2}$.

\section{Further applications and extensions} \label{S:appl}

Nabla dynamic equations, see e.g. \cite{drA.jB.lE.aP.hnT03}, can
be considered as a dual version of $\Delta$-differential
equations. The backward graininess function is denoted by
$\nu(t):=t-\rho(t)$. The spaces of ld-continuous and
ld-continuously $\nabla$-differentiable functions are accordingly
denoted by ${\rm C}_{\rm ld}$ and ${\rm C}_{\rm ld}^1$. Similarly to Definition
\ref{D:CrdCC} we have the notion of $f\in{\rm C}_{\rm ld}\times{\rm C}\times{\rm C}$,
and results corresponding to Propositions \ref{P:CrdCC.Crd} and
\ref{P:CrdC.Crd} now hold true for the $\nabla$-setting. The
$\nabla$-exponential function having the properties concluded in
Lemma \ref{L:exp.def} corresponds to the dynamic equation
$u^\nabla=-p(t)\,u$, see \cite[Section~3.2]{drA.jB.lE.aP.hnT03}.
Consequently, all the results of this paper extend directly to the
corresponding results for the $\nabla$-differential equations. As
examples of such results we have the following.

\begin{theorem} \label{T:1st.order.nabla}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\to\mathbb{R}^n$ with $0<q\leq\infty$ is a function satisfying $f\in{\rm C}_{\rm ld}\times{\rm C}$ and
  the Lipschitz condition \eqref{E:Lipschitz.1st.order}, where $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm ld}$, and
  \begin{equation} \label{E:k.integral.1st.order.nabla}
    \int_a^\infty k(s)\,\nabla s<\infty.
  \end{equation}
  Let $A\in\mathbb{R}^n$ be a given vector. If there exists a number $N\in\mathbb{R}$, $|A|\leq N<q$, such that
  \begin{equation*} \label{E:f.integral.1st.order.nabla}
    \int_a^\infty \big|f\big(s,x(s)\big)\big|\,\nabla s\leq N-|A|, \quad\text{for all } x\in X_N,
  \end{equation*}
  where $X_N$ is defined by \eqref{E:XN.def.1st.order}, then the problem
  \begin{equation} \label{E:1st.order.nabla}
    x^\nabla+f(t,x)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}.
  \end{equation}
  has a unique solution $x(t)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ satisfying $x(\infty)=A$.
\end{theorem}

\begin{corollary} \label{C:1st.order.M.nabla}
  Assume that $f:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\times\Omega_q\to\mathbb{R}^n$ with $0<q\leq\infty$ is a function satisfying $f\in{\rm C}_{\rm ld}\times{\rm C}$ and
  the Lipschitz condition \eqref{E:Lipschitz.1st.order}, where $k:[a,\infty)_{\scriptscriptstyle{\mathbb T}}\to(0,\infty)$, $k\in{\rm C}_{\rm ld}$, and
  \eqref{E:k.integral.1st.order.nabla} holds. If there exists a vector $M\in\mathbb{R}^n$, $|M|<q$, such that
  \begin{equation*} \label{E:f.integral.1st.order.M.nabla}
    \int_a^\infty \big|f\big(s,x(s)\big)\big|\,\nabla s\leq |M|, \quad\text{for all } x\in X_{2\,|M|},
  \end{equation*}
  then the problem \eqref{E:1st.order.nabla} has a unique solution $x(t)$ on $[a,\infty)_{\scriptscriptstyle{\mathbb T}}$ satisfying $x(\infty)=M$.
\end{corollary}

The corresponding results to Theorems \ref{T:2nd.order},
\ref{T:2nd.order.xD}, \ref{T:2nd.order.matrix},
\ref{T:2nd.order.xD.matrix} and Corollaries
\ref{C:second.order.scalar}, \ref{C:second.order.scalar.xD} now
hold under appropriate assumptions for the second order
$\nabla$-dynamic equations
\begin{gather*}
  \big(r(t)\,x^\nabla\big)^\nabla+f(t,x)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}, \\
  \big(r(t)\,x^\nabla\big)^\nabla+f(t,x,x^\nabla)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}, \\
  \big(P(t)\,x^\nabla\big)^\nabla+f(t,x)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}, \\
  \big(P(t)\,x^\nabla\big)^\nabla+f(t,x,x^\nabla)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}, \\
  \big(r(t)\,x^\nabla\big)^\nabla+k(t)\,g(x)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}}, \\
  \big(r(t)\,x^\nabla\big)^\nabla+k(t)\,g(x,x^\nabla)=0, \quad t\in[a,\infty)_{\scriptscriptstyle{\mathbb T}},
\end{gather*}
respectively.

\subsection*{Acknowledgements}
The first author was supported by the Grant Agency of the Academy of Sciences of the Czech Republic under grant KJB100190701,
by the Czech Grant Agency under grant 201/07/0145, and by the research project MSM 0021622409 of the Ministry of Education,
Youth, and Sports of the Czech Republic. He wishes to thank the University of New South Wales for the hospitality provided
while conducting this research project.

The authors are grateful to the anonymous referee for detailed
reading and constructive comments which helped improve the
presentation of the results. In particular, his/her request to
provide examples illustrating the applicability of the second
order problems which led us to establishing Corollaries
\ref{C:second.order.scalar} and \ref{C:second.order.scalar.xD} are
kindly appreciated.

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\end{document}