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 $c(t_1,0)$).


1 Answer 1


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 $\epsilon$.

  • $\begingroup$ Many thanks for your answer. Regarding the second, more useful equation, I guess I do not know $\frac{d\Omega}{dt}$, which might be a problem. $\endgroup$
    – tobias
    Commented Oct 5, 2015 at 10:51
  • $\begingroup$ if you don't know $d\Omega/dt$, then you will want to stop at the lowest order term $\Omega=\epsilon\int A(s)ds+{\rm order}(\epsilon^2)$ $\endgroup$ Commented Oct 5, 2015 at 11:16
  • $\begingroup$ Hi Carlo, I am insecure how to interpret your answer. Could you perhaps explain in a little more detail? Many thanks for your help in advance! $\endgroup$
    – tobias
    Commented Oct 5, 2015 at 12:41
  • $\begingroup$ it all depends on whether or not there is a small parameter in your problem; if you know that $\Omega$ is proportional to some parameter $\epsilon\ll 1$, then you try to relate $A$ to $\Omega$ in a power series in $\epsilon$; so you might truncate the sum over $k$ in the inverse Magnus expansion at the first few terms; the very first term relates $\Omega$ to the integral of $A$. $\endgroup$ Commented Oct 5, 2015 at 12:45
  • $\begingroup$ Do you have some examples for such scenarios? That would be great as I could then possibly relate it to my problems. $\endgroup$
    – tobias
    Commented Oct 5, 2015 at 12:53

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