Consider the ordered time derivative operator: $$ D_t := \overset{\longleftarrow}{\frac{\partial }{\partial t}} $$
Here the arrow indicates differentiation of operators appearing to the left of the symbol $D_t$. It is easy to see that for any two time-dependent functions $F(t)$ and $G(t)$, we have \begin{equation} F(t)e^{D_t \delta t}G(t) = F(t + \delta t)G(t). \end{equation}
I would like to consider the following expression:
$$ \exp(D_t + A) $$
where $A$ is a time-indepdent matrix. I am looking to expand the above exponential in the form
$$ \exp(D_t + A) = \exp(A) + \cdots. $$
For instance, if $A$ and $B$ are time-indepdnent matrices, the operator version of variation of paramters implies:
$$ e^{(A+B)} = e^{A} + \sum_{l=1}^{p-1} \int_0^1 \int_0^{s_{1}}\cdots \int_0^{s_{l-1}} e^{A(1-s_1)} B e^{A (s_1-s_2)}B \cdots e^{A (s_{l-1}-s_l)}B e^{As_l} ds_l \cdots ds_2 ds_1 + R_p(A,B). $$
Ideally, I am looking to derive a similar expression in the case where one of the arguments in the exponential is an unbounded operator. It looks like this problem should be well studied.
Could someone point out relevant references or provide hints on how to derive an expression like above.
Since $A$ is assumed as time-independent, it commutes with $D_t$. Indeed, one has : $$ f[D_t,A]g = f\left(D_tA - AD_t\right)g = f_tAg - (fA)_tg = f_tAg - f_tAg = 0, $$ where the $t$-subscript denotes partial differentiation with respect to time. In consequence, the exponential can be separated simply as $e^{D_t+A} = e^{D_t}e^A$, hence $$ f(t)\,e^{D_t+A}\,g(t) = f(t+1)\,e^A\,g(t). $$