TITLE: Getting out a system of linear ODEs by knowing the Magnus expansion
QUESTION [5 upvotes]: Assume we are given for a transition between two time points $t_0 = 0$ and $t_1$ a matrix relationship, eventually describing the solution of a system of linear with non-constant coefficients,
$$Y(t_1) = \exp(\Omega(t_1,0))Y(0),$$
or, in a more general setting,
$$Y(t_1) = \exp(\Omega(t_1,0))Y(0)+c(t_1,0),$$
where the $\Omega(t_1,0)$ is the Magnus expansion just as in https://en.wikipedia.org/wiki/Magnus_expansion and http://arxiv.org/pdf/0810.5488.pdf .
If I assume that the system of linear ODEs is given by,
$$\frac{dY}{dt} = A(t)Y(t)$$
or, more generally,
$$\frac{dY}{dt} = A(t)Y(t)+b(t),$$
is there any way to get out some information about $A(t)$ or $\int_0^{t_1} A(t) dt$ or $b(t)$ by just knowing $\Omega(t_1,0))$ (and $c(t_1,0)$).

REPLY [4 votes]: You can directly get the trace of $A$ from the identity
$${\rm det}\,\left[\exp\bigl(\Omega(t,0)\bigr)\right]=\exp\left[\int_0^t {\rm tr}\,A(s)ds\right]$$
The full matrix $A$ is determined by $\Omega$ via the inverse Magnus expansion [see equation 4.2 from this thesis]:
$$A=\sum_{k=0}^{\infty}\frac{(-1)^k}{(k+1)!}\left[\sum_{q=0}^{k}(-1)^q{k\choose q}\Omega^{q}\frac{d\Omega}{dt}\Omega^{k-q}\right]$$
This is useful if $\Omega$ is proportional to some small parameter $\epsilon$, so that the expansion gives you $A$ as a power series in $\epsilon$.