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\usepackage{mathrsfs}
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\usepackage{booktabs} % Allows the use of \toprule, \midrule and \bottomrule in tables
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%----------------------------------------------------------------------------------------
%	TITLE PAGE
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\title[Laplace Transforms on $\mathbb{T}$]{Laplace Transforms on Time Scales} % The short title appears at the bottom of every slide, the full title is only on the title page

\author{Tom Cuchta} % Your name

\date{\today} % Date, can be changed to a custom date

\begin{document}

\begin{frame}
\titlepage % Print the title page as the first slide
\end{frame}

\begin{frame}
\frametitle{Overview} % Table of contents slide, comment this block out to remove it
\tableofcontents % Throughout your presentation, if you choose to use \section{} and \subsection{} commands, these will automatically be printed on this slide as an overview of your presentation
\end{frame}

%----------------------------------------------------------------------------------------
%	PRESENTATION SLIDES
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%------------------------------------------------
\section{Introduction} % Sections can be created in order to organize your presentation into discrete blocks, all sections and subsections are automatically printed in the table of contents as an overview of the talk
%------------------------------------------------
\begin{frame}
A time scale is a set $\mathbb{T} \subset \mathbb{R}$ closed under the standard topology on $\mathbb{R}$. Define the forward jump operator $\sigma \colon \mathbb{T} \rightarrow \mathbb{T}$ by
$$\sigma(t) = \inf \{ z \in \mathbb{T} \colon z > t \}$$
the graininess $\mu \colon \mathbb{T} \rightarrow \mathbb{R}$ by
$$\mu(t) = \sigma(t)-t$$
and the minimal graininess function $\mu_* \colon \mathbb{T} \rightarrow \mathbb{R}$ by
$$\mu_*(s) = \displaystyle\inf_{\tau \in [s,\infty)\cap \mathbb{T}} \mu(\tau).$$
A function $f \colon \mathbb{T} \rightarrow \mathbb{C}$ is called $\Delta$-differentiable at $t \in \mathbb{T}$ if for every $\epsilon > 0$ there exists $\delta > 0$ such that for all $s \in (t - \delta, t + \delta) \cap \mathbb{T}$ and there exists a number $f^{\Delta}(t)$ such that
$$|[f(\sigma(s))-f(s)]-f^{\Delta}(s)[\sigma(t)-s]| \leq \epsilon |\sigma(t)-s|.$$
Integration is defined so that $\displaystyle\int_s^t f^{\Delta}(\tau) \Delta \tau = f(t)-f(s)$. 
\end{frame}
%------------------------------------------------
\begin{frame}
If $\mathbb{T}$ consists of just isolated points, then
$$f^{\Delta}(t) = \dfrac{f(\sigma(t))-f(t)}{\mu(t)}$$
and
$$\displaystyle\int_a^b f(t) \Delta t = \left\{ \begin{array}{ll}
\displaystyle\sum_{t \in [a,b)\cap\mathbb{T}} \mu(t) f(t) &; a < b \\
0 &; a=b \\
-\displaystyle\sum_{t \in [b,a)\cap\mathbb{T}} \mu(t) f(t) &; a>b.
\end{array} \right.$$
\end{frame}
%------------------------------------------------
\begin{frame}
Note: $\tilde{q}:=q-1$, $\mathbb{Z}^2 := \{n^2 \colon n \in \mathbb{Z}\}$
\hspace*{-16pt}\begin{tabular}{l|l|l|l|l}
Time Scale & $\sigma(t)=$ & $\mu(t) =$ & $f^{\Delta}(t)=$ & $\displaystyle\int_s^t f(\tau) \Delta \tau=$ \\ \hline
$\mathbb{R}$ & $t$ & $0$ & $\displaystyle\lim_{h \rightarrow 0} \dfrac{f(t+h)-f(t)}{h}$ & $\displaystyle\int_s^t f(\tau) d\tau$ \\
$h\mathbb{Z}$ & $t+h$ & $h$ & $\dfrac{f(t+h)-f(t)}{h}$ & $\displaystyle\sum_{k=\frac{s}{h}}^{\frac{t}{h}} hf(hk)$ \\
$q^{\mathbb{N}}; q>1$ & $qt$ & $\tilde{q}t$ & $\dfrac{f(qt)-f(t)}{t\tilde{q}}$ & $\tilde{q}\displaystyle\sum_{k=\log_q(s)}^{\log_q(t)-1} q^kf(q^k)$ \\
$\mathbb{Z}^2$ & $(1+\sqrt{t})^2$ & $2\sqrt{t}+1$ & $\dfrac{f((1+\sqrt{t})^2)-f(t)}{2\sqrt{t}-1}$ & $\displaystyle\sum_{k=\sqrt{s}}^{\sqrt{t}-1} (2k+1)f(k^2)$
\end{tabular}
\end{frame}
%------------------------------------------------
\begin{frame}
We will need some modified sets related to $\mathbb{C}$. First the Hilger complex plane is the set
$$\mathbb{C}_h := \left\{ z \in \mathbb{C} \colon z \neq -\dfrac{1}{h} \right\}$$
where we take $\mathbb{C}_0 := \mathbb{C}$.
The Hilger real part of a complex number $z$ is defined to be
$$\mathrm{Re}_h(z) := \dfrac{1}{h}(|1+hz|-1);$$
it is known that if $h\geq 0$ then $\mathrm{Re}_h(z)$ is an increasing function of $h$.
Define 
$$\mathbb{C}_h(\lambda) := \{ z \in \mathbb{C}_h \colon \mathrm{Re}_h(z) > \lambda \}.$$
\end{frame}
%------------------------------------------------
\section{Exponential Functions}
%------------------------------------------------
\begin{frame}
A function $p \colon \mathbb{T} \rightarrow \mathbb{C}$ is called regressive if for all $t \in \mathbb{T}$,
$$1+\mu(t)p(t) \neq 0.$$
We denote the set of regressive functions with domain $X$ and codomain $Y$ as $\mathscr{R}(X,Y)$. If $1+\mu(t)p(t) > 0$ for all $t$, then we say that $p$ is positively regressive and that $p \in \mathscr{R}^+(\mathbb{T},X)$. We use the notation $\mathscr{R}_c$ to denote constant regressive functions.

Let $p$ be regressive. The exponential function $e_p \colon \mathbb{T} \times \mathbb{T} \rightarrow \mathbb{C}$ is defined as
$$e_p(t,s) = \exp \left\{ \displaystyle\int_s^t \xi_{\mu(\tau)}(p(\tau)) \Delta \tau \right\}$$
where
$$\xi_h(z):=\dfrac{1}{h}Log(1+zh).$$
The solution of the IVP 
$$y^{\Delta}=py; y(s)=1$$
is $e_p(\cdot,s)$.\end{frame}
%------------------------------------------------
\begin{frame}
\begin{tabular}{l|l}
Time Scale & Exponential Function $e_p(t,s)=$ \\ \hline
$\mathbb{R}$ & $\exp\left\{\displaystyle\int_s^t p(\tau) d\tau \right\}$ \\ 
$h\mathbb{Z}$ & $\displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} 1 + hp(hk)$ \\
$q^{\mathbb{N}}; q>1$ & $\displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^k(q-1)$ \\
$\mathbb{Z}^2$ & $\displaystyle\prod_{k=\sqrt{s}}^{\sqrt{t}-1} 1+p(k^2)(2k+1)$
\end{tabular}
\end{frame}
%------------------------------------------------
\begin{frame}
Let $f, g \in \mathscr{R}(\mathbb{T},\mathbb{C})$. Define the operations $\oplus, \ominus$ by the formulas
$$(f \oplus g)(t) := f(t)+g(t) + \mu(t)f(t)g(t)$$
and
$$(f \ominus g)(t) := \dfrac{f(t)-g(t)}{1+\mu(t)g(t)}.$$
Then
\begin{itemize}
\item $e_p(s,s)=1$
\item $e_p(t,s)e_q(t,s)=e_{p\oplus q}(t,s)$
\item $\dfrac{e_p(t,s)}{e_q(t,s)} = e_{p \ominus q}(t,s)$
\item $e_p(\sigma(t),s) = (1+\mu(t)p(t))e_p(t,s)$
\end{itemize}
\end{frame}
%------------------------------------------------
\section{The Laplace Transform}
%------------------------------------------------
\begin{frame}
Let $\lambda \in \mathscr{R}_c^+([s,\infty)\cap\mathbb{T},\mathbb{R})$. Define
$$m_{\lambda}(t,s) := \displaystyle\int_s^t \dfrac{1}{1+ \mu(\tau)\lambda} \Delta \tau.$$
\begin{theorem}(Decay of Exponential Function)
Let $\mathbb{T}$ be a time scale such that $\sup \mathbb{T} = \infty$ and let $s \in \mathbb{T}$ and $\lambda \in \mathscr{R}^+_c([s,\infty)\cap\mathbb{T},\mathbb{R})$. Then for any $z \in \mathbb{C}_{\mu_*(s)}(\lambda)$,
\begin{enumerate}
\item $|e_{\lambda \ominus z}(t,s)| \leq e_{\lambda \ominus \mathrm{Re}_{\mu_*(s)}(z)}(t,s)$
\item $\displaystyle\lim_{t \rightarrow \infty} e_{\lambda \ominus \mathrm{Re}_{\mu_*(s)}(z)}(t,s)=0$
\item $\displaystyle\lim_{t \rightarrow \infty} e_{\lambda \ominus z}(t,s)=0.$
\end{enumerate}
\end{theorem}
\begin{corollary}
For $z \in \mathbb{C}_{\mu_*(s)}(\lambda)$,

$$e_{\lambda \ominus \mathrm{Re}_{\mu_*(s)}(z)}(t,s) \leq \exp \left\{ - \left[\mathrm{Re}_{\mu_*(s)}(z)-\lambda \right] m_{\mathrm{Re}_{\mu_*(s)}(z)}(t,s) \right\}$$
\end{corollary}
\end{frame}
%------------------------------------------------
\begin{frame}
A function $f \in C_{rd}(\mathbb{T},\mathbb{C})$ has exponential order $\alpha$ on $[s,\infty) \cap \mathbb{T}$ if $\alpha \in \mathscr{R}_c^+([s,\infty) \cap \mathbb{T},\mathbb{R})$ and there exists $K>0$ such that $|f(t)| \leq Ke_{\alpha}(t,s)$ for all $t \in [s,\infty) \cap \mathbb{T}$. \\
\hspace{1pt} \\
For some $f \in C_{rd}(\mathbb{T},\mathbb{C})$ of exponential order $\alpha$ define the Laplace transform of $f$ about $s$ by the formula
$$\mathscr{L}\{f\}(z;s) := \displaystyle\int_s^{\infty} f(\tau) e_{\ominus z}(\sigma(\tau),s) \Delta \tau.$$
\end{frame}
%------------------------------------------------
\begin{frame}
\begin{theorem}
Let $f \in C_{rd}([s,\infty)\cap\mathbb{T}, \mathbb{C})$ be of exponential order $\alpha$. Then $\mathscr{L}\{f\}(\cdot;s)$ exists on $\mathbb{C}_{\mu_*(s)}(\alpha)$ and converges absolutely.
\end{theorem}
\begin{proof}
Recall that $\mathrm{Re}_h(z)$ is an nondecreasing function of $h\geq0$. Therefore
$$\mathrm{Re}_{\mu(t)}(z) \geq \mathrm{Re}_{\mu_*(s)}(z)$$
for all $t \in [s, \infty) \cap \mathbb{T}$. Thus
$$|1+ \mu(t)z| \geq 1 + \mu(t) \mathrm{Re}_{\mu_*(s)}(z).$$

Let $t \in [s,\infty)\cap\mathbb{T}$ and $z \in \mathbb{C}_{\mu_*(s)}$. Compute
$$e_{\ominus z}(\sigma(t),s) = \left( 1 - \dfrac{ z\mu(t) }{1+\mu(t)z} \right) e_{\ominus z}(t,s) = \dfrac{e_{\ominus z}(t,s)}{1+ \mu(t)z}.$$
\end{proof}
\end{frame}
%------------------------------------------------

%------------------------------------------------
\begin{frame}
\begin{proof}
Now using the Decay Theorem and that $f$ is of exponential order $\alpha$ compute
$$\begin{array}{ll}
\left| \displaystyle\int_s^t f(\tau) e_{\ominus z}(\sigma(\tau),s) \Delta \tau \right| \leq \displaystyle\int_s^t |f(\tau) e_{\ominus z}(\sigma(\tau),s)| \Delta \tau \\
\leq K \displaystyle\int_s^t |e_{\alpha}(t,s)e_{\ominus z}(\sigma(t),s)| \Delta \tau \\
= K \displaystyle\int_s^t \dfrac{\left| e_{\alpha \ominus z}(\tau,s) \right|}{\left| 1 + \mu(\tau) z \right|} \Delta \tau \\
\leq K \displaystyle\int_s^t \dfrac{e_{\alpha \ominus \mathrm{Re}_{\mu_*(s)}(z)}(\tau,s)}{1+\mu(\tau) \mathrm{Re}_{\mu_*(s)}(z)} \Delta \tau \\
= \dfrac{K}{\alpha - \mathrm{Re}_{\mu_*(s)}(z)} \displaystyle\int_s^t \dfrac{\alpha - \mathrm{Re}_{\mu_*(s)}(z)}{1+\mu(\tau)\mathrm{Re}_{\mu_*(s)}(z)}e_{\alpha \ominus \mathrm{Re}_{\mu*(s)}(z)}(\tau,s) \Delta \tau \\

\end{array}$$
\end{proof}
\end{frame}
%------------------------------------------------
\begin{frame}
\begin{proof}
$$\begin{array}{ll}
\left| \displaystyle\int_s^t f(\tau) e_{\ominus z}(\sigma(\tau),s) \Delta \tau \right| \leq \dfrac{K}{\alpha - \mathrm{Re}_{\mu_*(s)}(z)} \displaystyle\int_s^t \dfrac{\alpha - \mathrm{Re}_{\mu_*(s)}(z)}{1+\mu(\tau)\mathrm{Re}_{\mu_*(s)}(z)}e_{\alpha \ominus \mathrm{Re}_{\mu*(s)}(z)}(\tau,s) \Delta \tau \\
= \dfrac{K}{\alpha - \mathrm{Re}_{\mu_*(s)}(z)} \displaystyle\int_s^t (\alpha \ominus \mathrm{Re}_{\mu_*(s)}(z))e_{\alpha \ominus \mathrm{Re}_{\mu_*(s)}(z)}(\tau,s) \Delta \tau \\
= \dfrac{K}{\alpha - \mathrm{Re}_{\mu_*(s)}(z)} \displaystyle\int_s^t e_{\alpha \ominus \mathrm{Re}_{\mu_*(s)}(z)}^{\Delta}(\tau,s) \Delta \tau \\
= \dfrac{K}{\mathrm{Re}_{\mu_*(s)}(z)-\alpha} \left( 1 - e_{\alpha \ominus \mathrm{Re}_{\mu_*(s)}(z)}(t,s) \right)
\end{array}$$
\end{proof}
\end{frame}
%------------------------------------------------
\begin{frame}
\begin{proof}
We have shown
$$\left| \displaystyle\int_s^t f(\tau) e_{\ominus z}(\sigma(\tau),s) \Delta \tau \right| \leq \dfrac{K}{\mathrm{Re}_{\mu_*(s)}(z)-\alpha} \left( 1 - e_{\alpha \ominus \mathrm{Re}_{\mu_*(s)}(z)}(t,s) \right)$$
so now let $t \rightarrow \infty$ and the Decay Theorem yields
$$\displaystyle\int_s^{\infty} \left| f(\tau) e_{\ominus z}(\sigma(\tau),s) \right| \Delta \tau \leq \dfrac{K}{\mathrm{Re}_{\mu_*(s)}(z)-\alpha}$$
for all $z \in \mathbb{C}_{\mu_*(s)}(\alpha)$. QED
\end{proof}
\end{frame}
%------------------------------------------------
\begin{frame}
\begin{theorem}Let $f \in C_{rd}([s,\infty)\cap\mathbb{T},\mathbb{C})$ be of exponential order $\alpha$. Then the Laplace transform $\mathscr{L}\{f\}$ converges uniformly in $\mathbb{C}_{\mu_*(s)}(\beta)$ where $\beta > \alpha$. 
\end{theorem}
\begin{proof}
Let $t \in [s,\infty) \cap \mathbb{T}$ and $z \in \mathbb{C}_{\mu_*(s)}(\alpha)$. Using an inequality from our previous proof and the corollary to the Decay Theorem, we see
$$\begin{array}{ll}
\displaystyle\int_t^{\infty} |f(\tau)e_{\ominus z}(\sigma(\tau),s)|\Delta \tau &\leq \displaystyle\int_s^{\infty} |f(\tau)e_{\ominus z}(\sigma(\tau),s)|\Delta \tau \\
&\leq \dfrac{K}{\mathrm{Re}_{\mu_*(s)}(z) - \alpha} e_{\alpha \ominus \mathrm{Re}_{\mu_*(s)}(z)}(t,s) \\
&\leq \dfrac{K}{\mathrm{Re}_{\mu_*(s)}(z) - \alpha} \exp \left\{ -[\mathrm{Re}_{\mu_*(s)}(z)-\alpha] m_{\mathrm{Re}_{\mu_*(s)}(z)}(t,s) \right\}.
\end{array}$$
\end{proof}
\end{frame}
%------------------------------------------------
\begin{frame}
\begin{proof}
If $z \in \mathbb{C}_{\mu_*(s)}(\alpha)$ then $\mathrm{Re}_{\mu_*(s)}(z) > \alpha$ and thus for all $\tau \in [s, t) \cap \mathbb{T}$,
$$\dfrac{1}{1 + \mu(\tau) \mathrm{Re}_{\mu*(s)}(z)} < \dfrac{1}{1+ \mu(\tau) \alpha}.$$
Therefore
$$\begin{array}{ll}
m_{\mathrm{Re}_{\mu_*(s)}(z)}(t,s) &= \displaystyle\int_s^{t} \dfrac{1}{1+\mu(\tau) \mathrm{Re}_{\mu_*(s)}(z)} \Delta \tau  \\
&\leq \displaystyle\int_s^t \dfrac{1}{1+ \mu(\tau) \alpha} \Delta \tau \\
&= m_{\alpha}(t,s). 
\end{array}$$

If $z \in \mathbb{C}_{\mu_*(s)}(\beta)$ for some $\beta > \alpha$ then $\mathrm{Re}_{\mu_*(s)}(z) > \beta$ and so $\alpha - \mathrm{Re}_{\mu_*(s)}(z) < \alpha - \beta$. Thus
$$\dfrac{1}{\mathrm{Re}_{\mu_*(s)}(z) - \alpha} \leq \dfrac{1}{\beta - \alpha} \mathrm{\hspace{3pt} and \hspace{3pt}} \alpha - \mathrm{Re}_{\mu_*(s)}(z) \leq \alpha - \beta.$$
\end{proof}
\end{frame}
%------------------------------------------------
\begin{frame}
\begin{proof}
Combining these inequalities yields
$$\displaystyle\int_t^{\infty} |f(\tau)e_{\ominus z}(\sigma(\tau,s)| \Delta \tau \leq \dfrac{K}{\beta - \alpha} \exp \left\{ -(\beta - \alpha) m_{\alpha}(t,s) \right\}.$$
Since we know that $\displaystyle\lim_{t \rightarrow \infty} m_{\alpha}(t,s) = \infty$, we see that a $t \in [s,\infty) \cap \mathbb{T}$ can be chosen to make the right-hand-side arbitrarily small. Therefore we have shown that for every $\epsilon > 0$ there exists $r \in [s,\infty) \cap \mathbb{T}$ such that
$$\left| \displaystyle\int_t^{\infty} f(\tau) e_{\ominus z}(\sigma(t),s) \Delta \tau \right| \leq \epsilon$$
for all $t \in [r,\infty) \cap \mathbb{T}$ and all $z \in \mathbb{C}_{\mu_*(s)}(\beta)$. QED
\end{proof}
\end{frame}
%------------------------------------------------
\begin{frame}
Some further properties:
\begin{itemize}
\item (Convolution) Define $\hat{f}$ to be the solution of the shifting problem: for $t,t_0,s \in \mathbb{T}$ with $t \geq s \geq t_0$
$$u^{\Delta_t}(t,\sigma(s))=-u^{\Delta_s}(t,s); u(t,t_0)=f(t)$$
Define convolution by 
$$(f*g)(t) = \displaystyle\int_{t_0}^t \hat{f}(t,\sigma(s))g(s) \Delta s$$
then
$$\mathscr{L}\{f*g\}(z) = \mathscr{L}\{f\}(z)\mathscr{L}\{g\}(z)$$
\item (Inverse transform) Suppose $0 < \mu_{\min} \leq \mu(t) \leq \mu_{\max} < \infty$. If $\displaystyle\int_{c-i\infty}^{c+i\infty} \left| \mathscr{L}\{f\}(z) \right| |dz| < \infty$ and the poles of $\mathscr{L}\{f\}(z)$ are regressive constants $\{z_1,\ldots,z_n\}$ of finite order then
$$f(t) = \displaystyle\sum_{i=1}^n Res_{z=z_i} e_z(t,0) \mathscr{L}\{f\}(z)$$
\end{itemize}
\end{frame}
%------------------------------------------------
\begin{frame}
\begin{itemize}
\item (Transform of derivative) $\mathscr{L}\{y^{\Delta^n}\}(z)=z^n\mathscr{L}\{y\}(z) - \displaystyle\sum_{j=0}^{n-1} z^j y^{\Delta^{n-j-1}}(s)$
\item (Differentiation of transform) For $f \in C_{rd}([s,\infty)\cap\mathbb{T},\mathbb{C})$,
$$-\dfrac{d}{dz}\mathscr{L}\{f\}(z;s) = \mathscr{L}\{m_z(\sigma(\cdot),s)f\}(z;s)$$
\item (Integration) Define $v(x;t,s) := \displaystyle\int_x^{\infty} e_{x \ominus y}(t,s)dy$. If $\displaystyle\int_s^{\infty} e_{\alpha}(\tau,s) v(x;\sigma(t),s)e_{\ominus x}(\sigma(\tau),s)\Delta \tau < \infty$ for each $x \in \mathbb{R}_{\mu_*(s)}(\alpha)$ then for all $x \in \mathbb{R}_{\mu_{*}(s)}(\alpha)$
$$\displaystyle\int_x^{\infty} \mathscr{L}\{f\}(y;s) dy = \mathscr{L}\{v(x;\sigma(\cdot),s)f\}(x;s)$$
\end{itemize}
\end{frame}
%------------------------------------------------
\begin{frame}
Laplace transform table

\begin{tabular}{l|l}
$y(t)$ & $\mathscr{L}\{y\}(z)$ \\ \hline
$1$ & $\dfrac{1}{z}$ \\
$t$ & $\dfrac{1}{z^2}$ \\
$h_k(t,s)$ & $\dfrac{1}{z^{k+1}}$ \\
$e_{\alpha}(t,s)$ & $\dfrac{1}{z-\alpha}$ \\
\end{tabular} 
\end{frame}
%------------------------------------------------
\begin{frame}
\frametitle{References}
\footnotesize{
\begin{thebibliography}{99} % Beamer does not support BibTeX so references must be inserted manually as below
\bibitem{} The Convolution on Time Scales. M. Bohner, G. Guseinov. Abstract and Applied Analysis. Volume 2007.
\bibitem{} Properties of the Laplace Transform on Time Scales with Arbitrary Graininess - M. Bohner, G. Guseinov, B. Karpuz. Integral Transforms and Special Functions. Vol.22, No.11, November 2011, 785-800.
\bibitem{} Further properties of the Laplace transform on time scales with arbitrary graininess. M. Bohner, G. Guseinov, B. Karpuz. Integral Transforms and Special Functions, 2013. Vol. 24, No.4, 289-301.
\bibitem{} M. Bohner and A. Peterson, Dynamic Equations on Time Scales: An Introduction with Applications, Birkhäuser,
Boston, 2001.
\bibitem{} J.M. Davis, I.A. Gravagne, B.J. Jackson, R.J. Marks II, and A.A. Ramos, The Laplace transform on time scales
revisited, J. Math. Anal. Appl. 332(2) (2007), pp. 1291–1307.
\end{thebibliography}
}
\end{frame}
\end{document} 