Continuous-time Dynamical System - Periodic Trajectory

Another good problem. Consider the matrix A, which is of the form:

$A = \left[\begin{matrix} 0&-a_1&a_2\\a_1&0&-a_3\\-a_2&a_3&0\end{matrix}\right]$

The characteristic equation of this matrix is:

$\lambda^3 + (a_1^2 + a_2^2+a_3^2)\lambda =0$

It can be seen that:

$a_1^2 + a_2^2+a_3^2=1$

So the eigen values of $A$ are $(0,\pm i)$ .

The eigenvalue decomposition of $A$ reads:

$A = V D V^{-1}$

Where $D$ is:

$D = \left[\begin{matrix} i&0&0\\0& -i&0\\0&0&0\end{matrix}\right]$

The general solution of the given system is:

$\mathrm{v}(t) = V \mathrm{e}^{Dt} V^{-1} \mathrm{v}(0)$ $\mathrm{v}(t) = V \left[\begin{matrix} \mathrm{e}^{it}&0&0\\0& \mathrm{e}^{-it}&0\\0&0&1\end{matrix}\right] V^{-1} \mathrm{v}(0)$

The solution of this system is periodic when:

$\mathrm{v}(t +T) =\mathrm{v}(t) = V \left[\begin{matrix} \mathrm{e}^{i(t+T)}&0&0\\0& \mathrm{e}^{-i(t+T)}&0\\0&0&1\end{matrix}\right] V^{-1} \mathrm{v}(0)$

The above is only possible when $T = 2\pi$ . This means that a point on the trajectory which would be diametrically opposite to the initial condition $\mathrm{v}(0)$ would be:

$\mathrm{v}(\pi) = V \left[\begin{matrix} \mathrm{e}^{i \pi}&0&0\\0& \mathrm{e}^{-i \pi}&0\\0&0&1\end{matrix}\right] V^{-1} \mathrm{v}(0)$

Therefore, the radius is:

$R = \frac{\lvert \mathrm{v}(\pi) - \mathrm{v}(0) \rvert}{2} = 4.779$