I am studying bifurcation and had a system like this: $$dx/dt=ux-y-x(x^2+y^2),$$ $$dy/dt=x+uy-y(x^2+y^2).$$ I want to determine whether a Hopf bifurcation would occur. I wrote the system into polar coordinates: $$dr/dt=ur-r^3,$$ $$d\theta/dt=1.$$ So I have a unstable limit cycle at $$r=\sqrt{u},$$ when u is positive. Can I then conclude that a Hopf bifurcation do occur? Since the spiral inside and outside the limit cycle towards different direction? But then I am confused by the question "at what value of $u$ a Hopf bifurcation occurs"? What does that mean? Thanks!

  • 3
    $\begingroup$ Everything you've done is right. The question "at what value of $u$ does a Hopf bifurcation occur?" is asking for the value of $u$ for which the behavior of the system changes. When $u$ is less than this value there is no limit cycle, and when $u$ is greater than this value a limit cycle exists. $\endgroup$ Commented Oct 29, 2014 at 1:44
  • $\begingroup$ Thanks a lot! That make sense!:D $\endgroup$
    – Sissi Sue
    Commented Oct 29, 2014 at 16:38

1 Answer 1


I'm writing this to add some details for anyone who may come across this in the future.

First I'd like to clarify that you have a stable limit cycle (when the limit cycle appears). To see this, set $r = \sqrt{u} + \epsilon$ and compute $\dot{r}$: I find that $\dot{r} < 0$. Similarly, setting $r = \sqrt{u} - \epsilon$ gives $\dot{r} > 0$ . Then the limit cycle is attractive. This will help us identify the bifurcation that takes place here.

To see why a Hopf bifurcation occurs, consider values of $u < 0$. Here we see that $\dot{r} < 0$ for every $r$ and hence $r \to 0$ as $t \to \infty$ so that the origin is a stable equilibrium; because $\dot{\theta} = 1$, trajectories spiral into the origin. In addition, for $u > 0$, the origin is an unstable equilibrium and trajectories spiral outward from it. As you've noted above, a limit cycle appears at $r = \sqrt{\mu}$, and we just decided that this limit cycle is stable.

To "see" the Hopf bifurcation take place, consider the Jacobian of the system (in Cartesian coordinates) at the origin, \begin{equation} A = \left(\begin{array}{cr} u & -1 \\ 1 & u \end{array}\right). \end{equation} The eigenvalues of $A$ are $\lambda = u \pm i$. We see that as $u$ goes from negative to positive, both eigenvalues cross the imaginary axis (which is the boundary of stability) from left to right, which is the hallmark of a supercritical Hopf bifurcation. A supercritical Hopf bifurcation occurs when a stable fixed point becomes unstable and sheds a stable limit cycle. The supercriticality coincides with what we identified above: a stable fixed point sheds a stable limit cycle and the fixed point changes its stability.


You must log in to answer this question.

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