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\begin{document}
\begin{frame}{}
\begin{center}{\LARGE Dynamic Matrix Exponential via Matrix Cylinder Transformation}\end{center}
\begin{center}{\underline{Tom Cuchta} (Fairmont State University), David Grow (Missouri S\&T), Nick Wintz (Lindenwood University)} \\
\vspace{0.2in}
{\small International Conference on Differential \& Difference Equations and Applications \\
Lisbon, Portugal, 1-5 July 2019}\end{center} 
\end{frame}

\begin{frame}{Preliminaries}
If $A \in \mathbb{C}^{n \times n}$ is diagonalizable, then Jordan canoncial form is
\[ A = J^{-1}\mathrm{diag}(\lambda_1,\ldots,\lambda_n)J,\]
where the $\lambda_i$ are eigenvalues of $A$. If $f \colon U\subset\mathbb{C} \rightarrow \mathbb{C}$ is a function so that $f(\lambda_i)$ exists for $i=1,2,\ldots,n$, then we define the matrix analogue of $f$ by
\[ f(A)=J^{-1}\mathrm{diag}(f(\lambda_1),\ldots,f(\lambda_n))J.\]
\textit{note: the formula exists but is more complicated for nondiagonalizable matrices}
\end{frame}

\begin{frame}{Preliminaries}
We easily obtain $\Log(A)$, $\exp(A)$, and $|A|$. 

We also extend the function $\mathrm{sign} \colon \mathbb{C} \setminus \{ z \in \mathbb{C} \colon \mathrm{Re}(z)=0 \} \rightarrow \{1,-1\}$ defined by
\[ \mathrm{sign}(z) = \left\{ \begin{array}{ll}
1, & \quad \mathrm{Re}(z)>0 \\
-1, & \quad \mathrm{Re}(z)<1.
\end{array} \right.\]
To a matrix $\text{sign}$ function.
\end{frame}

\begin{frame}{Preliminaries}
\begin{theorem}[\cite{zbMATH05267840},Theorem 10.2]
For $A,B \in \mathbb{C}^{n \times n}$, 
\[ \exp((A+B)t)=\exp(At)\exp(Bt)\]
for all $t \in \mathbb{C}$ iff $AB=BA$.
\end{theorem}

\begin{theorem}[\cite{zbMATH05267840},Theorem 11.3]
Suppose $B, C \in \mathbb{C}^{n \times n}$ have no negative eigenvalues and $BC=CB$. If for every eigenvalue $\lambda_j$ of $B$ and corresponding eigenvalue $\eta_j$ of $\mathbb{C}$,
\[ \mathrm{arg}(\lambda_j)+\mathrm{arg}(\eta_j)| < \pi,\]
then $\Log(BC)=\Log(B)+\Log(C)$.
\end{theorem}
\end{frame}

\begin{frame}{Preliminaries}
Using standard branch cut $(-\infty,0]$, we obtain logarithms at $x<0$ as
\[ \log(x) = \log(|x|)+i\pi.\]
From this, if $A$ is Hermitian (hence diagonalizable), then 
\begin{eqnarray*}
\lefteqn{\Log(A) = J^{-1} \Bigg( \mathrm{diag}(\Log(|\lambda_1|),\ldots,\Log(|\lambda_n|)} \\
& &  + \mathrm{diag} \left(\dfrac{1-\mathrm{sign}(\lambda_1)}{2}, \ldots, \dfrac{1-\mathrm{sign}(\lambda_n)}{2} \right) i\pi \Bigg)J, 
\end{eqnarray*}
or more compactly,
\[ \Log(A) = \Log(|A|)+i\pi \dfrac{I-\mathrm{sign}(A)}{2}.\]
\end{frame}


\begin{frame}{Preliminaries}
Let $\mathbb{T}$ be a time scale. \\
\vspace{0.3in}
A function $f \colon \mathbb{T} \rightarrow \mathbb{C}$ is called regressive if $\forall t \in \mathbb{T}$, $1+\mu(t)f(t) \neq 0$. \\
\vspace{0.3in}
More generally, a function $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is called regressive if $\forall t \in \mathbb{T}^{\kappa}$, $1+\mu(t)A(t)$ is invertible. \\
\vspace{0.3in}
Sometimes we will use the symbol $a_{\ell}$ to denote right-scattered points.
\end{frame}

\begin{frame}{Scalar cylinder transformation $\xi$} 
Hilger complex plane (``regressive constants" with respect to $h$) is 
\[\mathbb{C}_h=\left\{ \begin{array}{ll}
\mathbb{C} \setminus \left\{ -\dfrac{1}{h} \right\}, & \quad h>0, \\
\mathbb{C}, & \quad h=0. 
\end{array} \right.\]
and the strip 
\[ \mathbb{Z}_h=\left\{z \in \mathbb{C} \colon -\dfrac{\pi}{h} < \mathrm{Im}(z) \leq \dfrac{\pi}{h} \right\}.\]

Cylinder transformation $\xi_h \colon \mathbb{C}_h \rightarrow \mathbb{Z}_h$ is
\[ \xi_h(z)=\left\{ \begin{array}{ll}
\dfrac{1}{h} \log(1+hz), & \quad h>0, \\
z, & \quad h=0.
\end{array} \right.\]
\end{frame}

\begin{frame}{Preliminaries}
Let $\mathcal{F}=\{[a,b) \cap \mathbb{T} \colon a,b \in \mathbb{R}\}$ (ring of subsets of $\mathbb{T}$) and define $m_1([a,b))=b-a$.  \\
\vspace{0.5in}
The unique Carath\'{e}odory extension of $m_1$ to the $\sigma$-algebra generated by $\mathcal{F}$ is called the $\mu_{\Delta}$ measure. \\
\vspace{0.5in}
This measure expresses the $\Delta$-integral as a ``Lebesgue $\Delta$-integral" allowing us to define the function space $L_{\mathrm{loc}}^1(\mathbb{T},\mu_{\Delta})$, the locally $\Delta$-integrable functions.
\end{frame} 

\begin{frame}{Scalar dynamic exponential $e_p$}
For certain $p \colon \mathbb{T} \rightarrow \mathbb{C}$, exponential function $e_p \colon \mathbb{T} \times \mathbb{T} \rightarrow \mathbb{C}$ is defined by
\[ e_p(t,s) = \exp \left( \displaystyle\int_s^t \xi_{\mu(\tau)}(p(\tau)) \Delta \tau \right).\]
Coincidentally, if $y(t)=e_p(t,s)$, then $y$ satisfies the initial value problem
\[ y^{\Delta}=py, \quad y(s)=1.\]
Typically $p$ is chosen to be regressive and rd-continuous, but in \cite{zbMATH06527142}, the exponential was studied for $p \in L^1_{\mathrm{loc}}(\mathbb{T},\mu_{\Delta})$. 
\end{frame}

\begin{frame}{Matrix dynamic exponential $e_A$}
For certain $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$, $e_A \colon \mathbb{T} \times \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is \textbf{defined} to be the solution of the matrix dynamic equation 
\[ y^{\Delta}=Ay, \quad y(s)=I.\]
\textbf{No matrix cylinder transformation is typically mentioned.}
\end{frame}

\begin{frame}{Matrix cylinder transformation $\Xi$}
$\mathbb{C}_h^{n \times n}$: set of matrices $X$ in $\mathbb{C}^{n \times n}$ for which $1+hX$ is invertible 

$\mathbb{Z}_h^{n \times n}$: set of $\mathbb{C}^{n \times n}$ matrices whose eigenvalues lie in $\mathbb{Z}_h$ 

\textbf{Matrix cylinder transformation} $\Xi \colon \mathbb{C}_h^{n \times n} \rightarrow \mathbb{Z}_h^{n \times n}$ is
\[ \Xi_h(A)= \left\{ \begin{array}{ll}
\dfrac{1}{h} \Log\left(1+hA \right), & \quad h>0, \\
A, & \quad h=0.
\end{array} \right. \]
\textbf{Immediate Consequence}: $\Xi$ applies the scalar cylinder transformation $\xi$ to the eigenvalues of $A$
\end{frame}

\begin{frame}{Matrix cylinder transformation}
\begin{lemma}
If $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is a regressive Hermitian matrix function, then 
\begin{enumerate}[(i)]
    \item $I + \mu(t)A(t)$ is Hermitian for all $t \in \mathbb{T}$,
    \item if $A(t)$ is positive semi-definite for all $t \in \mathbb{T}$, then $\Xi_{\mu(t)}(A(t))$ is Hermitian for all $t \in \mathbb{T}$, and
    \item if $A(t)$ is positive definite for all $t \in \mathbb{T}$, then $\Xi_{\mu(t)}(A(t))$ is positive definite for all $t \in \mathbb{T}$.
\end{enumerate}
\label{positivedefimpiesiplusmuaposdefandxihermitian}
\end{lemma}
Proof is routine calculation.
\end{frame}

\begin{frame}{Matrix cylinder transformation}
\begin{lemma}
If $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is a regressive Hermitian matrix function with each $a_{ij} \in L^1_{\mathrm{loc}}(\mathbb{T},\mu_{\Delta})$, then the function $t \mapsto \Xi_{\mu(\tau)}(A(t))$ is locally integrable.\label{cylinderisintegrable}
\end{lemma}
\begin{proof}
Since $A$ is locally integrable 
$\displaystyle\sum_{t_0 \leq a_{\ell} < t} \mu(a_{\ell}) \lVert A(a_{\ell}) \rVert_1 \leq \displaystyle\int_{t_0}^t \lVert A(\tau) \rVert_1 \Delta \tau < \infty.$
So $\mu(a_{\ell}) \lVert A(a_{\ell}) \rVert_1 \leq \dfrac{1}{2}$ for all but finitely many indices $\ell$. If $\mu(\tau) = 0$, then $\lVert \Xi_{\mu(\tau)}(A(\tau)) \rVert_1 = \lVert A(\tau) \rVert_1$. Now suppose $\mu(\tau) > 0$. Then $\tau=a_{\ell}$ for some ${\ell}$. By definition, $\left\lVert \Xi_{\mu(\tau)}(A(\tau)) \right\rVert_1 = \dfrac{1}{\mu(a_{\ell})} \left\lVert \Log(I+\mu(a_{\ell})A(a_{\ell}))\right\rVert_1.$
We know
$$\left| \ln(1+y) \right| \leq \left\{ \begin{array}{ll}
y, & \quad 0 \leq y \leq \dfrac{1}{2} \\
|y|\ln(4), & \quad -\dfrac{1}{2} \leq y < 0.
\end{array} \right.$$
\end{proof} 
\end{frame}

\begin{frame}{Matrix cylinder transformation}
\begin{proof}
Since $r\Big(\mu(a_{\ell})A(a_{\ell})\Big) \leq \lVert \mu(a_{\ell})A(a_{\ell}) \rVert_1 \leq \dfrac{1}{2}$, we use power seriees for matrix logarithm to write
$$\begin{array}{ll}
\lVert \Log(I + \mu(a_{\ell}) A(a_{\ell}) \rVert_1 &= \left\lVert \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k (\mu(a_{\ell})A(a_{\ell}))^k}{k} \right\rVert \\
&\leq \displaystyle\sum_{k=1}^{\infty} \dfrac{(\mu(a_{\ell})\lVert A(a_{\ell})\rVert_1)^k}{k} \\
&= -\ln \left( 1 - \mu(a_{\ell}) \lVert A(a_{\ell}) \rVert_1 \right) \\
&\leq \ln(4) \mu(a_{\ell}) \lVert A(a_{\ell}) \rVert_1.
\end{array}$$
Therefore
\[\lVert \Xi_{\mu(a_{\ell})}(A(a_{\ell})) \rVert_1 = \dfrac{1}{\mu(a_{\ell})} \lVert \Log(I+\mu(a_{\ell})A(a_{\ell})) \rVert_1 \leq \ln(4) \lVert A(a_{\ell}) \rVert_1.\]
\end{proof}
\end{frame}

\begin{frame}{Matrix cylinder transformation}
\begin{lemma}
If $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is a regressive Hermitian matrix function with each $a_{ij} \in L^1_{\mathrm{loc}}(\mathbb{T},\mu_{\Delta})$, then the function $t \mapsto \Xi_{\mu(\tau)}(A(t))$ is locally integrable.\label{cylinderisintegrable}
\end{lemma}

\begin{proof}
Hence, 
\[\begin{array}{ll}
\left\lVert \displaystyle\int_{t_0}^t \Xi_{\mu(\tau)}(A(\tau)) \Delta \tau \right\rVert_1 &\leq \displaystyle\int_{t_0}^t \left\lVert \Xi_{\mu(\tau)}(A(\tau)) \right\rVert_1 \Delta \tau \\
&\leq \ln(4) \displaystyle\int_{t_0}^t \lVert A(\tau) \rVert_1 \Delta \tau < \infty,
\end{array}\]
completing the proof.
\end{proof}
\end{frame}

\begin{frame}{New matrix exponential $E_A$}
For $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ with locally integrable components, we define $E_A \colon \mathbb{T} \times \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ by 
\[ E_A(t,s) = \exp \left( \displaystyle\int_s^t \Xi_{\mu(\tau)}(A(\tau)) \Delta \tau \right).\]
\begin{proposition}
If $A$ is regressive, then $E_A(t,t)=I$ and $E_{\mathbf{0}}(t,s)=I$.
\end{proposition}
\begin{proposition}
If $A$ is regressive, then $E_A^{-1}(t,s)=E_A(s,t)$.
\end{proposition}
\end{frame}

\begin{frame}{Example}
In $\mathbb{T}=\mathbb{N}_0$ taking $A(t)=\begin{bmatrix} t+2 & (t+2)^2 \\ (t+2)^3 & (t+2)^4 \end{bmatrix}$, we obtain
\[e_A(1,0)=I+A(0)=\begin{bmatrix} 3&4 \\ 8&17 \end{bmatrix}=E_A(1,0), \]
and 
\[e_A(2,0)=(I+A(1))(I+A(0))=\begin{bmatrix} 84 & 169 \\ 737 & 1502 \end{bmatrix},\]
while
\[\begin{array}{ll}
E_A(2,0)&=\exp\left( \Log(I+A(0))+ \Log(I+A(1)) \right) \\
&= \exp \left( \begin{bmatrix} \dfrac{\ln(19)}{9} + \dfrac{\ln(85)}{28} & \dfrac{2\ln(19)}{9} + \dfrac{3\ln(85)}{28} \\ \dfrac{4\ln(19)}{9} + \dfrac{9\ln(85)}{28} & \dfrac{8\ln(19)}{9} + \dfrac{27\ln(85)}{28} \end{bmatrix} \right) \\
&\approx \begin{bmatrix} 
96.9684 & 240.686 \\ 
582.728 & 1463.05		
\end{bmatrix}
\end{array}\]
\end{frame}

\begin{frame}{Properties}
\begin{center}\begin{tabular}{|l|l|}
\multicolumn{2}{c}{Properties of $e_A$ and $E_A$} \\
\hline
$e_A(t,t_0)$ & $E_A(t,t_0)$ \\
\hline
$e_A(t,t)=I$ & $E_A(t,t)=I$ \\
$e_0(t,t_0)=I$ & $E_0(t,t_0)=I$ \\
$e_A(t,t_0)=e_A(t_0,t)^{-1}$ & $E_A(t,t_0)=E_A(t_0,t)^{-1}$ \\
$e_A^{\sigma}=(I+\mu A)e_A$ & $E_A^{\sigma}=\exp\Big( \mu\Log(I+\mu A) + \Log(E_A) \Big)$ \\
\hline
\end{tabular}\end{center}

\end{frame}

\begin{frame}{Continuous and discrete part}
\begin{theorem}
If $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is regressive, Hermitian matrix function with locally integrable components, then for all $t \in \mathbb{T}$ with $t \geq t_0$,
\begin{eqnarray*}
\lefteqn{E_A(t,t_0)=\exp \left(\displaystyle\int_{[t_0,t]\cap\mathbb{T}}A(\tau) \mathrm{d}\tau \right.} \\
&&\left.+ \displaystyle\sum_{t_0 \leq a_{\ell} < t} \left[ \Log(|I+\mu(a_{\ell})A(a_{\ell})|)+i\pi \dfrac{I-\mathrm{sign}(I+\mu(a_{\ell})A(a_{\ell})}{2} \right] \right) 
\end{eqnarray*}
\end{theorem}
The scalar case then has, for $m$ dependent on $t$,
\begin{corollary}
\[e_p(t,t_0)=(-1)^m \exp \left( \displaystyle\int_{[t_0,t]\cap\mathbb{T}} p(\tau) \mathrm{d}\tau \right) \displaystyle\prod_{t_0 \leq a_{\ell} < t} (1+\mu(a_{\ell})p(a_{\ell}) )\]
\end{corollary}
\end{frame}

\begin{frame}{Trace Inequality}
\begin{theorem}[\cite{zbMATH00647292}, (8)]
If $A$ and $B$ are positive semi-definite, then 
\[ \mathrm{Tr}\Big(\exp(A+B)\Big) \leq \mathrm{Tr}\Big(\exp(A)\exp(B))\Big).\]
\end{theorem}
\begin{corollary}
\begin{eqnarray*}
\lefteqn{\mathrm{Tr}(E_A(t,t_0))} \\
&&\leq \mathrm{Tr} \left( \exp \left( \displaystyle\int_{[t_0,t]\cap\mathbb{T}} A(\tau) \mathrm{d}\tau \right) \right. \left. \exp \left( \displaystyle\sum_{t_0 \leq a_{\ell} <t} \Log(1+\mu(a_{\ell})A(a_{\ell})) \right) \right)  \end{eqnarray*}
\end{corollary}
\end{frame}

\begin{frame}{Distributing $\exp$ through $E_A$}
\begin{theorem}
Let $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ be a regressive Hermitian matrix function where $a_{ij}$ are locally integrable. If $A$ satisfies 
\begin{enumerate}[(i)]
\item for all $t_1, t_2 \in [t_0,t) \cap \mathbb{T}$, $A(t_1)$ commutes with $A(t_2)$, and
\item for all $t_1 \in [t_0,t) \cap \mathbb{T}$, $\displaystyle\int_{[t_0,t)\cap\mathbb{T}} A(\tau) \mathrm{d}\tau$ commutes with $A(t_1)$,
\end{enumerate}
then 
\[E_A(t,t_0) = \exp \left( \displaystyle\int_{[t_0,t]\cap\mathbb{T}} A(\tau) \mathrm{d}\tau \right) \displaystyle\prod_{t_0 \leq a_{\ell} < t} (I+\mu(a_{\ell})A(a_{\ell}).\]
\end{theorem}
\end{frame}

\begin{frame}{Measurable eigenvectors}
\begin{theorem} If $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is a matrix function with each $a_{ij} \in L^1_{\mathrm{loc}}(\mathbb{T},\mu_{\Delta})$ and $A(t)$ is Hermitian for every $t \in \mathbb{T}$, then the distinct eigenvalues $\lambda_n(t) > \lambda_{n-1}(t) > \ldots > \lambda_{n-m}(t)$ of $A(t)$, where $m \leq n$, are measurable and have corresponding measurable eigenvector functions which form an orthonormal basis of $\mathbb{C}^{n \times n}$.\label{eigenvectorsmeasurable}
\end{theorem}

\end{frame} 

\begin{frame}{Diagonalizer}
The previous theorem is needed to prove the existence of $U$ in the following theorem.
\begin{theorem}
If $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is a matrix function with each $a_{ij}$ locally integrable and $A(t)$ is positive definite for all $t \in \mathbb{T}$, then there is a $\mu_{\Delta}$-measurable unitary matrix function $U(t)$ satisfying
\[ E_{U*AU}(t,t_0) = \mathrm{diag}\left(e_{\lambda_1}(t,t_0), \ldots, e_{\lambda_n}(t,t_0) \right),\]
where the $\lambda_i$ are the (not necessarily distinct) eigenvalues functions of $A$.
\end{theorem}
\end{frame} 

\begin{frame}{Dynamic integral equation}
Earlier:
\begin{lemma}
If $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is a regressive Hermitian matrix function with each $a_{ij} \in L^1_{\mathrm{loc}}(\mathbb{T},\mu_{\Delta})$, then the function $t \mapsto \Xi_{\mu(\tau)}(A(t))$ is locally integrable.\label{cylinderisintegrable}
\end{lemma}
Thus the integral of $\Xi$ is locally absolutely continuous allowing us to prove
\begin{lemma}
If $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is regressive Hermitian with each $a_{ij} \in L^1_{\mathrm{loc}}(\mathbb{T},\mu_{\Delta})$, then the function $t \mapsto A(t)E_A(t,t_0)$ is locally integrable. 
\end{lemma}

\end{frame} 

\begin{frame}{Dynamic integral equation}
\begin{theorem} If $A \colon \mathbb{T} \rightarrow \mathbb{C}^{n \times n}$ is a regressive Hermitian matrix function with each $a_{ij}$ locally integrable, then the function $Y(t)=E_A(t,t_0)$ satisfies
\[ Y(t) = I + \displaystyle\int_{t_0}^t F(\tau,Y(\tau))\Delta \tau,\]
where
\begin{eqnarray*}
\lefteqn{F(t,Y(t))} \\
&&= \left\{ \begin{array}{ll}
A(t)Y(t), & \quad \mu(t)=0, \\
\dfrac{\exp\Big( \Log(I+\mu(t)A(t))+\Log(Y(t)) \Big)-Y(t)}{\mu(t)}, &\quad \mu(t)>0.
\end{array} \right.
\end{eqnarray*}
\end{theorem}
\end{frame} 

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\end{frame}

\begin{frame}{Thanks}
Obrigado!
\end{frame}

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%\bibliographystyle{plain}
%\bibliography{citations}
%\end{frame}

\end{document}
