Consider the matrix differential equation $$X'(t) = \beta M(t)X(t)$$

where $M$ is a non-inversible matrix and $M(t)$ does not commute with $$\exp \left(\int_a^t M(s)ds \right)$$ and, hence, there is no simple solution. I now $X_0(t)$ the solution of $M(t)X(t)=0$ that can be considered as the approximate solution when $\beta \gg 1$.

How can I re-use $X_0(t)$ to get a second more precise approximation of $X(t)$ for big $\beta$ ?


1 Answer 1


The formal solution to your problem is $$ X(t) = T \exp\left(\beta \int_0^t M(s) ds\right) X(0) $$ where $T$ is the time-ordering operator.

(I do not see any simple way how to construct a series expansion in $1/\beta$ as already the first order would contain an equation as this. )

  • $\begingroup$ thanks, what is the time-ordering operator ? $\endgroup$
    – Nichola
    Jul 27, 2020 at 20:16
  • $\begingroup$ It is an operator which sorts the products (here of exponents of increments of the integral) according to time: earlier to the right, later to the left. Please see en.wikipedia.org/wiki/Path-ordering#Time_ordering $\endgroup$
    – Nikodem
    Jul 27, 2020 at 21:37

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