1
$\begingroup$

It's well known that the derivative of the matrix exponential in direction $V$ is given by \begin{align} D_V(t,A)&=\int_0^t \exp((t-s)A)V\exp(sA)ds \end{align} In [1] they state that one can also write it as (eqn. 104) \begin{align} D_V(t,A)&=\int_0^t \exp(sA)V\exp((t-s)A)ds \end{align} Why are these equivalent?

[1] Najfeld, Igor, and Timothy F. Havel. "Derivatives of the matrix exponential and their computation." Advances in applied mathematics 16, no. 3 (1995): 321-375.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Begin with the first integral formula, and apply the substitution $\sigma = t - s$. We find that $$ \int_0^t \exp((t-s)A)V\exp(sA)ds = \\ \int_{\sigma(0)}^{\sigma(t)} \exp(\sigma A) V \exp((t - \sigma )A)\cdot (-1)d \sigma =\\ -\int_t^0 \exp(\sigma A)V \exp((t - \sigma A))\,d\sigma =\\ \int_0^t \exp(\sigma A)V \exp((t - \sigma A))\,d\sigma, $$ which is the desired formula.

$\endgroup$
5
  • $\begingroup$ Thanks. I was working through it myself and got to the third line and got confused. What does an integral from 0 to t mean and why can we invert it to get its negative? $\endgroup$ Commented Oct 12, 2021 at 3:34
  • $\begingroup$ @mlstudent I'm not sure how to answer your question about what the integral "means". In general, $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$; switching the limits of an integral multiplies it by $-1$. $\endgroup$ Commented Oct 12, 2021 at 3:36
  • $\begingroup$ Actually I found a discussion here math.stackexchange.com/questions/1577675/…. $\endgroup$ Commented Oct 12, 2021 at 3:39
  • $\begingroup$ @mlstudent This statement holds for real-valued integrals, perhaps it would be helpful to convince yourself of that simpler statement $\endgroup$ Commented Oct 12, 2021 at 3:39
  • $\begingroup$ Sorry when I said from 0 to t I meant from t to 0. $\endgroup$ Commented Oct 12, 2021 at 3:42

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.