The question is for a 2D system, but for the sake of simplicity, let's consider a 1D system $\dot{x} = \mu + x^2$. Then for $\mu < 0$ the fixed point $x = \pm\sqrt{\mu}$ varies along the $x$-axis as $\mu$ varies. On the other hand for the system $\dot{x} = x (\mu + x^2)$, we have three fixed points for a given $\mu$, but one of them $x = 0$ does not vary with respect to $\mu$.

For a 2D system $(\dot{x}, \dot{y}) = f(x, y; \mu)$, a Hopf bifurcation is usually introduced as a bifurcation of a fixed point that changes from stable to unstable (as a stable spiral changes to an unstable spiral or the other way around).

My question is, can this fixed point vary in the $(x, y)$ plane, or does it always remain at a fixed point e.g. $(x_0, y_0) = (0, 0)$? So far, for the examples I have seen, the fixed point does indeed seem to stay fixed in the $(x, y)$ plane, but I don't how to show that this is always the case from the definition of a Hopf bifurcation, which is vague at the level of Strogatz's book.

Related question: As we vary $\mu$ just past the bifurcation point $\mu_c$, a limit cycle appears. Is this limit cycle always elliptical?

  • 1
    $\begingroup$ For the sake of simplicity when people study bifurcation problems they make a coordinates transformation that keeps an equilibrium or fixed point at the same point at the origin. That's a feature of many normal forms. And no, limit cycle is not always elliptical, though for small $\mu - \mu_c$ it is very close to elliptical. However, if you are dealing with planar system of form $\dot{r} = f(r), \dot{\theta} = g(\theta, r)$, then limit cycles are always circles. $\endgroup$
    – Evgeny
    Commented Jul 6, 2018 at 11:46
  • 1
    $\begingroup$ Hopf bifurcation is just the appearance or disappearance of a limit cycle in a dynamical system by continuously varying parameters. If you check out the wikipedia article for Hopf bifurcation, you can see an animated example of a system with a moving "center" equilibrium as it goes through two Hopf bifurcations. $\endgroup$
    – Alex Jones
    Commented Jul 6, 2018 at 11:58
  • 1
    $\begingroup$ @AlexanderJ93 Thanks! That's an excellent animation $\endgroup$
    – LMZ
    Commented Jul 6, 2018 at 12:41

1 Answer 1


Yes, the coordinates of the equilibrium can vary. Here is an example: $$ \dot N=rN\left(1-\frac{N}{K}\right)-\frac{CNP}{A+N},\\ \dot P=P\left(-d+\frac{BN}{A+N}\right), $$ where $r,K,C,A,B,d$ are some positive parameters.

I will leave all the computations to you, but probably the easiest way to see what happens in this system is to plot the null-clines and analyze their behavior depending on the parameters.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .