# How can I show that the trajectory is a circle?

I have this system: $$\frac{dx}{dt} = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}x+\begin{bmatrix}1 \\0\end{bmatrix}i$$

1st: Is this system stable? I think it is stable, but not asymptotically stable.
$$s_{1,2} = \pm j$$ 2nd: How can I tell, that this stands for a circle in the $u_C - i_L$ plane?

• This looks more suitable for physics.SE. Are these not circuits? But, may be we have some control theorists here??? – user45099 Mar 3 '13 at 9:45
• Well the question may have originated from a physical problem but it seems like a legitimate math question belonging here. – Fixed Point Mar 3 '13 at 10:02
• The system is stable because the eigenvalues of $\mathbf{A}$ are $\lambda \in \{\mathrm{i},-\mathrm{i}\}$ (imaginary). The real parts are zero. en.wikibooks.org/wiki/Control_Systems/…. The second question I do not understand. – WG- Mar 3 '13 at 20:25
• Because of the imaginary eigenvalues, the trajectories are circles. And I think I should show something like a circle equation. – madmax Mar 3 '13 at 21:21
• Hmmmm in that case I think they proberbly mean to draw the phase portrait of the systems states $u_c$, $i_L$. So you simulate the system for $L = C$ and put the input $i(t) = 0$ then you plot $x_1$ and $x_2$ versus each other plot(x1,x2) in Matlab. en.wikipedia.org/wiki/Phase_portrait. – WG- Mar 5 '13 at 21:08

Consider the squared euclidean distance between a trajectory at a given time $t$ and the origin:

$V(x(t))=x_1^2(t)+x_2^2(t)$.

Examine it's time-derivative:

$\frac{d}{dt}V(x(t))=\frac{d}{dt}(x_1^2(t)+x_2^2(t))=2(\dot{x}_1x_1+\dot{x}_2x_2)=2i(t)$.

So, if $i(t)\equiv 0$, then for all $t\geq0$

$\frac{d}{dt}V(x(t))=0\Rightarrow V(x(t)) = V(x(0))$.

In other words, the distance between a trajectory at any time $t$ and the origin is the same as it was at time $0$. In turn, this implies that the trajectories are subsets of the circle centred at the origin of radius $\sqrt{x_1^2(0)+x_2^2(0)}$. In addition, the system has a unique equilibrium, the origin, which is not inside the circle (unless $x(0)=0$, in which case $x(t)=0$ for all $t\geq0$) so by uniqueness of solutions (see http://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem), the trajectory must be going round the circle. Lastly, examining the vector field (i.e., the direction in which $\frac{dx}{dt}$ points for different $x$) one can see that the trajectories must be rotating counterclockwise.

By the way, note that the above implies that the system system is stable in the sense of Lyapunov (see http://en.wikipedia.org/wiki/Lyapunov_stability).