Consider the following matrix function $$ f(t) = \cos(\omega_1t) A_1 + \cos(\omega_2t) A_2, \quad t\ge 0, $$ where $A_1$, $A_2$ are real square matrices and $\omega_1$, $\omega_2$ positive numbers.

Now, consider the following series of nested integrals $$ S(t)=\sum_{n=1}^\infty \int_0^{t} f(t_1)\left(\int_0^{t_1} f(t_2)\cdots \left(\int_0^{t_{n-1}} f(t_n)\, \mathrm{d}t_n\right)\cdots\mathrm{d}t_2\right) \mathrm{d}t_1. $$

If $\omega_1=\omega_2=\bar \omega$ then the above series converges to $$ \exp\left({(A_1+A_2) \int_0^t \cos(\bar \omega \tau )\, \mathrm{d}\tau \,}\right)-I, $$ so that, in this case, a (rather loose) upper bound to the 2-norm of $S(t)$ is $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1, $$ where $\alpha:=\max\{\|A_1\|,\|A_2\|\}$.

My question. With reference to the case $\omega_1\ne \omega_2$ does the following upper bound hold true $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1 $$ where $\bar \omega:=\min\{\omega_1,\omega_2\}$? [If this is not true, is it possible to find an upper bound on $\|S(t)\|$ depending on $\bar \omega$?]

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A (perhaps) simplifying assumption. From the comments of some of you, I've realized that finding such a bound is a rather difficult task for general matrices $A_1$ and $A_2$. So let us suppose that $A_1$ and $A_2$ are such that $[A_1,A_2]=V$ where $V$ is a skew-symmetric matrix and $[\cdot,\cdot]$ denotes the matrix commutator. Does this assumption lead to a simplification of the problem?

Any comment is very welcome.

  • $\begingroup$ If you differentiate the expression, you obtain the differential equation $\dot X(t)=f(t)X(t)$, where $X(t)=I+S(t)$. If $A_1$ and $A_2$ don't commute, there is no reason to expect solutions of this differential equation to remain bounded. $\endgroup$ May 22 '18 at 5:53
  • 1
    $\begingroup$ @AnthonyQuas why not? I mean, that's the question no? $\endgroup$
    – lcv
    May 23 '18 at 2:03
  • $\begingroup$ Are your matrices real or complex? Skew symmetric and real or just skew symmetric? $\endgroup$
    – lcv
    May 25 '18 at 19:49
  • $\begingroup$ @lcv: matrices $A_1$ and $A_2$ are real and the commutator $[A_1,A_2]$ is (real and) skew-symmetric. $\endgroup$
    – Ludwig
    May 25 '18 at 19:59

I know that what I write here is rather a comment, but it is quite long, so I formulate it as an answer.

Perhaps one could look at the problem in the light of the Magnus expansion. It is a way of writing the transition matrix for the IVP $$ X'(t) = A(t) X(t), \quad X(0) = I $$ as $$ \exp{\Omega(t)}, $$ where $\Omega(t)$ is the sum of a series whose terms are nested integrals of nested commutators. See Expansions that grow on trees by A. Iserles, or The Magnus expansion and some of its applications by S. Blanes et al. In your case, the matrix function $A(\cdot)$ is periodic (when $\omega_1/\omega_2$ is rational), or quasi-periodic. Regarding the former, see Floquet theory: exponential perturbative treatment by F. Casas et al.

Magnus expansions are notoriously difficult to deal with, but perhaps for your choice of matrices $A_1$ and $A_2$ the formulas for $\Omega(t)$ simplify (let's hope so!). Then you can have some estimate.

  • $\begingroup$ Thanks for your input! Suppose that $[A_1,A_2]=S$, where $S$ is skew-symmetric (here square brackets denote commutator). Do you think this assumption could be helpful? $\endgroup$
    – Ludwig
    May 24 '18 at 16:57
  • $\begingroup$ Yes, for any $x \in \mathbb{R}^n$ there holds $\langle e^{tS}x, e^{tS}x \rangle' = \langle Se^{tS}x, e^{tS}x \rangle + \langle e^{tS}x, Se^{tS}x \rangle = \langle e^{tS}x, (S - S) e^{tS}x \rangle = 0$, that is, $\lVert e^{tS} \rVert = 1$ for all $t \in \mathbb{R}$. $\endgroup$
    – user539887
    May 24 '18 at 20:22
  • $\begingroup$ So, do you mean that, from the fact $\|e^{tS}\|=1$, a bound on (the norm of) the Magnus expansion can be found? (I don't see how...) $\endgroup$
    – Ludwig
    May 24 '18 at 20:32
  • 1
    $\begingroup$ No, I don't know exactly: just thinking aloud. $\endgroup$
    – user539887
    May 24 '18 at 20:41
  • $\begingroup$ Is $f(t)$ anti-hermitian? That would help $\endgroup$
    – lcv
    May 25 '18 at 3:38

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