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Let $\omega$ be a positive real number and let $A,B\in\mathbb{R}^{n\times n}$ be $n\times n$ real matrices. Consider the following linear time-varying dynamical system $$ \dot{x}(t) = (A + \cos(\omega t)B)x(t),\quad x(0)=x_0\in\mathbb{R}^{n}. $$

Let $\Phi(t,0)$ denote the state-transition matrix of the above system.

My question. Is it true that $$ \lim_{\omega \to \infty} \Phi(t,0) = e^{At} \ \ \ ? $$

Some remarks. If matrices $A$ and $B$ commute, this is true. Indeed, in this case it holds $$ \Phi(t,0) = e^{\int_0^t A + \cos(\omega \tau)B\, \mathrm{d}\tau}. $$ In the non-commuting case, I didn't manage to prove this. The main issue here is that a closed-form expression of $\Phi(t,0)$ does not exist, apparently. The only (perhaps useful) idea that I had so far is to exploit the Peano-Baker expansion of $\Phi(t,0)$. However, even with this tool, I couldn't provide an answer to my question.

Thanks for your help!

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    $\begingroup$ We can write the solution as $x=e^{At}y$, so we get that $y$ satisfies $y'=\cos(\omega t)C(t)y$, where $C(t)=e^{-At}Be^{At}$, and for the state-transition matrix $\Lambda(t) = e^{\int_0^t \cos(\omega \tau)C(\tau)\,d\tau}$. We see that $\Phi(t)=e^{At}\Lambda(t)$ and we must prove thus $\Lambda(t)\to I$ when $\omega\to\infty$, which seems true using integration by parts. $\endgroup$ – user90189 Jul 11 '18 at 3:58
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    $\begingroup$ @user90189: Thank you. When you have time, could you please turn your comment into an answer (perhaps elaborating a little more the different steps)? So that I will be able to accept it. $\endgroup$ – Ludwig Jul 11 '18 at 4:39
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    $\begingroup$ @user90189 I'm looking forward to your answer. I'm interested, in particular, in the proof that $y'=\cos(\omega t)C(t)y$ has state-transition matrix $\exp\bigl(\int_0^t\cos(\omega \tau)C(\tau)\,d\tau\bigr)$. $\endgroup$ – user539887 Jul 11 '18 at 8:19
  • $\begingroup$ $C(t)$ and $\int_0^t \cos(\omega s) C(s)ds$ do not necessarily commute... $\endgroup$ – copper.hat Jul 11 '18 at 14:48
  • $\begingroup$ @Ludwig, I was fooled by the form of the equation, I cannot prove that $\Lambda(t)=\exp(\int_0^t\cos(\omega \tau)C(\tau)\,d\tau)$. Sorry for misleading you. $\endgroup$ – user90189 Jul 11 '18 at 16:19
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Let me turn my comment into an answer: if we write $x(t)=e^{At}y(t)$, then $y$ satisfies the differential equation $\dot{y}=\cos(\omega t)C(t)y$, where $C(t)=e^{-At}Be^{At}$. We see thus that $\Phi(t,0)=e^{At}\Lambda(t,0)$, where $\Lambda(t,0)$ is the state-transition matrix associated to $\cos(\omega t)C(t)$. Hence, we must prove that $\Lambda(t,0)y_0\to y_0$ when $\omega\to\infty$.

To show this, we integrate the differential equation to get $y(t)=y_0+\int_0^t\cos(\omega s)C(s)y(s)\,ds$, so we must prove that the integral term on the right goes to zero, but $y$ depends on $\omega$, then we write also $y_\omega$. We integrate by parts to get

$$ \frac{1}{\omega}\big[-\int_0^t\sin(\omega s)(C(s)y_\omega(s))'\,ds+\sin(\omega t)C(t)y_\omega(t)\big] \\ = \frac{1}{\omega}\big[-\int_0^t\sin(\omega s)(C'(s)y_\omega(s)+\cos(\omega s)C(s)^2y_\omega(s))\,ds+\sin(\omega t)C(t)y_\omega(t)\big]\hspace{0.5cm}(*) $$

We want to make $\omega$ goes to infinity, but we must first get a uniform upper bound of $y_\omega$. Take inner product at both sides of the differential equation to reach $\frac{1}{2}\dot{(|y|^2)}=\cos(\omega t)\langle C(t)y,y\rangle\le |C(t)||y|^2$. By Gronwall's inequality we get $|y(t)|\le R(t)|y_0|$, where $R$ is a function independent from $\omega$. Use this in equation (*), to conclude that $y(t)\to y_0$ as $\omega\to\infty$ for fixed $t$, so $\Lambda(t,0)\to I$ as $\omega\to \infty$.

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  • $\begingroup$ +1: Very cute. ${}$ $\endgroup$ – copper.hat Jul 11 '18 at 23:22
  • $\begingroup$ Ingenious approach, thanks for sharing this! Out of curiosity, I was wondering if your approach can be somehow exploited to prove that for the (more general) system $\dot{x}=(A+\sum_{i=1}^m \cos(\omega_i t) B_i)x$, $\omega_i>0$, it holds $\lim_{\omega_i\to \infty}\Phi(t,0)=e^{At}$. $\endgroup$ – Ludwig Jul 12 '18 at 5:04
  • $\begingroup$ @Ludwig, I think it works. $\endgroup$ – user90189 Jul 12 '18 at 14:47

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