# Closed-form solution of a linear time-varying system

Consider the following three-dimensional linear time-varying system $$\label{eq1}\tag{\star} \dot{x}(t)=(A \cos(\omega_1 t) + B \cos(\omega_2 t) )x(t), \ \ x(0)\in\mathbb{R}^{3},$$ where $\omega_1$ and $\omega_2$ are positive real numbers such that $\omega_1\ne \omega_2$ and $$A=\begin{bmatrix}a & -a & 0 \\ -a & a & 0 \\ 0 & 0 & 0\end{bmatrix}, \quad B=\begin{bmatrix}0 & 0 & 0 \\ 0 & -b & b \\ 0 & b & -b\end{bmatrix}$$ with $a,b$ positive real numbers. Notice that $A$, $B$ do not commute. However, the commutator $$[A,B]=AB-BA = \begin{bmatrix}0 & ab & -ab \\ -ab & 0 & ab \\ ab & -ab & 0\end{bmatrix}$$ is a skew-symmetric matrix (I don't know if this can be useful though).

My questions are:

1. Does there exist an explicit closed-form expression for the solution of \eqref{eq1}?
2. If not, is it possible to find a bound on $\|x(t)\|$ which explicitly depends on $\omega_1$ and $\omega_2$?

I've been stuck on this problem for a while now. So I would greatly appreciate any kind of comment or help. Thanks.

• You may find this useful math.stackexchange.com/questions/770679/… – caverac May 27 '18 at 1:37
• Is $\omega_1/\omega_2$ rational? – Kwin van der Veen May 29 '18 at 8:24
• Why do you think the bound should depend on $\omega_1$ and $\omega_2$. It is more likely that it will depend on $x_0$. Have you simulated the system to get some insight into the behaviour? – MrYouMath May 29 '18 at 14:36
• @KwinvanderVeen: If assuming $\omega_1/\omega_2$ rational simplifies the problem, then yes, we can make this assumption. – Ludwig May 29 '18 at 15:38
• @MrYouMath: Yes, I've run some simulations. The behavior is not very clear though. It seems that as $\omega_1$ and $\omega_2$ grow, the trajectories are bounded and closer to zero. – Ludwig May 29 '18 at 15:40