For a two dimensional dynamical system, when does the phase space diagram give an ellipse?

I know about the examples for damped and undamped harmonic oscillators, but our instructor said that the diagram is an ellipse for all systems whose Jacobian gives purely imaginary values in the linear approximation from the fixed points.

I am not able to imagine for what other systems will this be possible, if at all.

  • $\begingroup$ An ellipse is just a circe. You get an exact circle whenever the system is conservative. When the phase space is two-dimensional (which means that the system has just one degree of freedom), it is enough to exhibit a first integral, which is readily provided by the Hamiltonian. A more general result, which yields the $n$-torus $\mathbb T^n$ as a generalisation of the circle, follows from the angle-action variables for integrable systems. $\endgroup$
    – Phoenix87
    Jul 30, 2015 at 8:55

1 Answer 1


Consider a variant of the FitzHugh–Nagumo system:

$$ \begin{align}\dot{x} &= x (a - x) (x - 1) - y \\ \dot{y} &= bx - cy \end{align}$$

with parameters like $a=-0.02; b=0.01; c=0.02$. The eigenvalues of the Jacobian at $(0,0)$ (which is a fix point) are $±\frac{\sqrt{6}}{25}i$.

The attractor for this system looks like this:

Attractor of oscillatory FitzHugh–Nagumo system

Obviously, this is not an ellipsis. But, and that’s what your instructor is probably aiming at: The attractor is topologically equivalent to an ellipsis and that’s what of foremost interest when generally classifying the dynamics.

Note that it’s also possible to obtain a phase diagram that is not an ellipsis for a driven and damped harmonic oscillator, if you select the variables in an unorthodox way.

  • $\begingroup$ Yes, today he talked about integrability of Hamiltonian systems and the fact that this figure needs to be topologically equivalent to a circle came up. $\endgroup$
    – Cheeku
    Aug 1, 2015 at 4:30

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