$$ \boldsymbol{u}' - \boldsymbol{A}(t) \boldsymbol{u} = \boldsymbol{f}(t)$$

be solved using an integration factor method analogous to that used in the non-system case? (i.e. $u'-a(t)u=f(t)$). Giving the following solution:

$$ \boldsymbol{u} = e^{-\int^t A(k)dk}\left (\int^t e^{\int^s A(k)dk}\boldsymbol{f}(s) ds + \boldsymbol{c}\right )$$

where the matrix exponential defined in the usual way. If you rearrange the solution and differentiate you find $$ \frac {\text{d}e^{-\int^t A(k)dk}} {\text{d}t} \boldsymbol{u} + e^{-\int^t A(k)dk} \boldsymbol{u}' = e^{-\int^t A(k)dk} \boldsymbol{f} $$

If this is indeed the solution, it would then imply that $$ \frac {\text{d}e^{-\int^t A(k)dk}} {\text{d}t} = - e^{-\int^t A(k)dk} \boldsymbol{A} $$

However this doesn't seem to be the case from Eq. 2.1 of http://aip.scitation.org/doi/pdf/10.1063/1.1705306 referenced on the Wiki page for matrix exponentials.

EDIT: Disproved on page 6 here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.