I have some issues to understand the Hopf bifurcation. To keep it simple I would like to discuss the 2 dimensional case.

In that case the eigenvalues of the Jacobian at a critical point look like

\begin{equation} \lambda_{1/2} = \frac{1}{2}\left[\mu \pm \sqrt{\mu^2 - 4\beta} \right] \end{equation}

Assuming that $4\beta > \mu^2$ we have \begin{equation} \begin{matrix} \mu < 0: &\text{ stable spiral}\\ \mu = 0: &\text{ limit cycle}\\ \mu > 0: &\text{ non-stable spiral} \end{matrix} \end{equation}

I can test if the limit cycle is stable by adding a small value $\epsilon$ to the critical point and calculate the time-derivative.

A supercritical Hopf-Bifurcation occurs when a stable spiral ($\mu < 0 $) changes into a unstable spiral ($\mu > 0 $) surrounded by a elliptical limit cycle. [Strogatz1994]

In this image I understand how the hopf-bifurcation works:

negative real part: stable spiral

no real part: limit cycle

positive real part: unstable spiral

enter image description here

But the next image confuses me. Here exists a limit cycle for non-zero real parts. I cannot explain the existance of the limit cycle. Can someone help me? enter image description here

  • $\begingroup$ Sorry for my bad english. I unstable spiral has eigenvalues with non-zero real and imaginary parts. The real part of the eigenvalue of a limit cycle is zero. So why can spirals and limit cycles coexist? $\endgroup$ – Alexander Tille May 18 '17 at 7:32
  • $\begingroup$ The type of limit cycle is determined by eigenvalues of Jacobi matrix that is computed at fixed point of Poincaré mapping. In your case this mapping is one-dimensional interval mapping and Jacobi matrix computed at fixed point has only one real eigenvalue. That's not in accordance with your undestanding of limit cycle. $\endgroup$ – Evgeny May 18 '17 at 7:55
  • 1
    $\begingroup$ Oh, I get it now. You are confusing terminology and linear case with non-linear. Limit cycle is a closed trajectory which has no closed trajectories in some of its neighbourhoods. When you have linear system with purely imaginary eigenvalues, you don't have limit cycles: all trajectories are closed and in any neighbourhood of any trajectory there is another closed trajectory. Andronov-Hopf bifurcation shows what really happens in non-linear system: your intution and habits from linear case don't work here, in typical case only one closed trajectory appears after the bifurcation. $\endgroup$ – Evgeny May 19 '17 at 21:04
  • $\begingroup$ Ok, thanks. I think I got your point $\endgroup$ – Alexander Tille May 22 '17 at 12:50

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