Page 20.2: Definition for a Matrix Exponential Function. Properties

Theorem

Let \(A, B\in \mathcal{R}\) and \(t, s, r\in \mathbb{T}\). Then

1. \(e_0(t, s)=I\), \(e_A(t, t)=I\),

2. \(e_A(\sigma(t), s)=(I+\mu(t)A(t))e_A(t, s)\),

3. \(e_A(s, t)=e_A^{-1}(t, s)=e_{\ominus A^*}^*(t, s)\),

4. \(e_A(t, s) e_A(s, r)=e_A(t, r)\),

5.  \(e_A(t, s)e_B(t, s)=e_{A\oplus B}(t, s)\) if \(e_A(t, s)\) and \(B\) commute.

Proof

1. Consider the IVP
\begin{equation*}
Y^{\Delta}=O,\quad Y(s)=I.
\end{equation*}
Then its unique solution is
\begin{equation*}
e_O(t, s)=I.
\end{equation*}
Now we consider the IVP
\begin{equation*}
Y^{\Delta}=A(t)Y,\quad Y(s)=I.
\end{equation*}
By the definition of \(e_A(\cdot, s)\) we obtain
\begin{equation*}
e_A(s, s)=I.
\end{equation*}

2.  By Theorem  the definition of \(e_A(\cdot, s)\), we have
\begin{eqnarray*}
e_A(\sigma(t), s)&=& e_A(t, s)+\mu(t) e_{A}^{\Delta}(t, s)\\ \\
&=& e_A(t, s)+\mu(t) A(t) e_A(t, s)\\ \\
&=&(I+\mu(t)A(t))e_A(t, s).
\end{eqnarray*}

3.  Let
\begin{equation*}
Z(t)=\left(e_A^{-1}(t, s)\right)^*.
\end{equation*}
Then
\begin{eqnarray*}
Z^{\Delta}(t)&=&\left(\left(e_A^{-1}(t, s)\right)^{\Delta}\right)^*\\ \\
&=& -\left(e_A^{-1}(t, s) e_A^{\Delta}(t, s)\left(e_A(\sigma(t), s)\right)^{-1}\right)^*\\ \\
&=& -\left(e_A^{-1}(t, s) A(t) e_A(t, s)\left(e_A(\sigma(t), s)\right)^{-1}\right)^*\\ \\
&=& -\left(A(t) \left(e_A(\sigma(t), s)\right)^{-1}\right)^*\\ \\
&=& -\left(A(t)\left((I+\mu(t) A(t))e_A(t, s)\right)^{-1}\right)^*\\ \\
&=& -\left(A(t) e_A^{-1}(t, s) \left(I+\mu(t)A(t)\right)^{-1}\right)^*\\ \\
&=& \left(e_A^{-1}(t, s) \left(-\left(I+\mu(t) A(t)\right)^{-1}A(t) \right)\right)^*\\ \\
&=& \left(e_A^{-1}(t, s)(\ominus A)(t)\right)^*\\ \\
&=& (\ominus A)^* (t) \left(e_A(t, s)\right)^*\\ \\
&=& \left(\ominus A^*\right) (t) Z(t).
\end{eqnarray*}
Therefore
\begin{equation*}
Z(t)=e_{\ominus A^*}(t, s),
\end{equation*}
i.e.,
\begin{equation*}
e_{\ominus A^*}(t, s)=\left(e_A^{-1}(t, s)\right)^*,
\end{equation*}
or
\begin{equation*}
e_A^{-1}(t, s)= e_{\ominus A^*}^*(t, s).
\end{equation*}

4.  Let
\begin{equation*}
Z(t)= e_A(t, s) e_A(s, r).
\end{equation*}
Then
\begin{eqnarray*}
Z^{\Delta}(t)&=& e_A^{\Delta}(t, s) e_A(s, r)\\ \\
&=& A(t) e_A(t, s) e_A(s, r)\\ \\
&=& A(t) Z(t),\\ \\
Z(r)&=&e_A(r, s) e_A(s, r)\\ \\
&=& e_A^{-1}(s, r) e_A(s, r)\\ \\
&=& I\\ \\
&=& e_A(r, r).
\end{eqnarray*}
Therefore
\begin{equation*}
e_A(t, s) e_A(s, r)=e_A(t, r).
\end{equation*}

5.  Let
\begin{equation*}
Z(t)= e_A(t, s) e_B(t, s).
\end{equation*}
Then
\begin{eqnarray*}
Z^{\Delta}(t)&=& e_A^{\Delta}(t, s) e_B^{\sigma}(t, s) +e_A(t, s) e_B^{\Delta}(t, s)\\ \\
&=& A(t) e_A(t, s) (I+\mu(t)B(t))e_B(t, s)\\ \\
&&+B(t) e_A(t, s) e_B(t, s)\\ \\
&=& A(t)(I+\mu(t)B(t))e_A(t, s) e_B(t, s) +B(t) e_A(t, s) e_B(t, s)\\ \\
&=& (A(t)+B(t) +\mu(t)A(t)B(t))e_A(t, s) e_B(t, s)\\ \\
&=& (A\oplus B)(t) Z(t),\\ \\
Z(s)&=& e_A(s, s) e_B(s, s)\\ \\
&=& I.
\end{eqnarray*}
Consequently
\begin{equation*}
e_{A\oplus B}(t, s)=e_A(t, s) e_B(t, s).
\end{equation*}
This completes the proof.

Theorem

Let \(A\in \mathcal{R}\) and \(t, s\in \mathbb{T}\). Then
\begin{equation*}
\left(e_A(s, t)\right)_t^{\Delta}=-e_A(s, \sigma(t))A(t).\end{equation*}

Proof
We have
\begin{eqnarray*}
\left(e_A(s, t)\right)_t^{\Delta}&=& \left(e_A^{-1}(t, s)\right)_t^{\Delta}\\ \\
&=& \left(e_{\ominus A^*}^*(t, s)\right)_t^{\Delta}\\ \\
&=& \left(\left(e_{\ominus A^*}(t, s)\right)_t^{\Delta}\right)^*\\ \\
&=& \left((\ominus A^*)(t) e_{\ominus A^*}(t, s)\right)^*\\ \\
&=& \left(-\left(I+\mu(t)A^*(t)\right)^{-1}A^*(t) e_{\ominus A^*}(t, s)\right)^*\\ \\
&=& \left(-A^*(t)\left(I+\mu(t) A^*(t)\right)^{-1}e_{\ominus A^*}(t, s)\right)^*\\ \\
&=& \left(-A^*(t) e_{\ominus A^*}(\sigma(t), s)\right)^*\\ \\
&=& -e_{\ominus A^*}^*(\sigma(t), s) A(t)\\ \\
&=& -e_A(s, \sigma(t))A(t).
\end{eqnarray*}
This completes the proof.