# Question about limit cycles and linear systems

it says:

A limit cycle is however an intrinsically nonlinear concept: a linear system cannot have a limit cycle.

Is there a proof somewhere to show this result? What is the fundamental insight here?

• Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Feb 13 '16 at 8:25

The answer to your question is not so difficult. In your notes, the author defines a limit cycle as an isolated periodic trajectory, i.e. nearby there can not be any periodic trajectories. Of course, it is easy to construct periodic trajectories for linear systems, just look at

$$x'(t) = y(t), \quad y'(t) = -x(t), \quad x(0) = x_0, \quad y(0) = 0.$$

However, if the system is linear, by slightly changing the initial conditions you can build many new periodic trajectories. In this case, such a nearby trajectory would be generated by putting

$$x(0) = (1 \pm \epsilon)x_0$$

with $\epsilon \ll 1$.

I think you can see that the a linear dynamical system does not have limit cycle in this way:

For a linear dynamical system, the Fourier modes are the eigenvectors of the linear operator, and the solution of the system is some system specific amplitudes (constant in time) multiply the Fourier modes. So the phase portrait ($\dot{x}$ vs $x$) must be cyclic.

In the case of a nonlinear system, projecting the system to Fourier modes will no longer give constant amplitude. Instead the amplitude will consist of Fourier modes that have other wave numbers. In some cases, these amplitudes will result in a phase portrait that spirals in a limit cycle.