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Consider the equation: \begin{equation} \dot{x} = Mx, \end{equation} where \begin{equation} M = \begin{pmatrix} i\omega_1 & 0 & \cdots & 0 \\ 0 & i\omega_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & i\omega_n \end{pmatrix}. \end{equation}

I have seen it stated that if the fraction $\frac{\omega_i}{\omega_j}$ is rational for all $i$ and $j$ then the trajectory is periodic. If any of these quotients are irrational then the trajectory is quasi-periodic. (Quasi-periodic motion wiki: http://en.wikipedia.org/wiki/Quasiperiodic_motion)

What is the proof of this?

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The solution to your equation is:

$$ {\bf x}(t) = \left(\begin{matrix} A_0 e ^{i\omega_0t} \\ A_1 e ^{i\omega_1t} \\ A_2 e ^{i\omega_2t} \\ ... \end{matrix} \right) $$

So it's basically a whole lot of simple harmonic oscillators. Periodic motion means there must be some time $\tau$ where all the different oscillators have completed an integral number of cycles i.e.

$$\begin{align} \omega_0\tau = n_0 2\pi \\ \omega_1\tau = n_1 2\pi \\ \omega_2\tau = n_2 2\pi \\ ... \end{align}$$

for some integers $n_0$, $n_1$, etc. The ratio $\omega_i/\omega_j$ is just $n_i/n_j$, and since we've decided all the $n$s are integers the ratio must be a rational number.

Conversely, if there were any $\omega_i/\omega_j$ that was irrational there would never be a time when the $i$th and $j$th oscillators had bioth completed an integral number of cycles so there can't be any periodic motion.

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