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There's a question in my book that asks us to show how the Hopf bifurcation theorem fails for the system $$\dot{r}=\mu r$$ $$ \dot{\theta}=1$$.

I don't see how it does though. From what I understand, you need to examine the eigenvalues of the jacobian at the origin. Here, I'm getting $\lambda=\mu\pm i2$.

So $\Re (\lambda(0)) = 0$,

$\Im (\lambda(0)) \neq 0$

and $\frac{d\Re}{d\mu}\neq 0$.

Does that not mean that there is a hopf bifurcation at $\mu=0$ and that the theorem works? I don't really understand the question so I'd appreciate if someone could explain.

The notes suggested using Matlab to plot the system but I don' see how that helps either.

The Hopf Bifurcation Theorem according to my notes is:

Let $x = f (x,\mu)$, $x \in \mathcal{R}^n$, $\mu \in \mathcal{R}$, have a fixed point $x =0$ for all μ. Suppose the eigenvalues of $Df(0,\mu)$ have negative real part, except for a pair $\lambda(\mu) = \alpha(\mu)+i\omega(\mu)$ and its complex conjugate $\bar{\lambda}(\mu)$, for which $\lambda(0) =i\omega$, and suppose (transversality) $\alpha'(0) \neq 0$ (i.e. the eigenvalues cross the imaginary axis at a non zero rate). Then for small $\mu$, there exists a one-parameter family of periodic orbits, $x_\mu(t)$, in either $\mu ≥ 0$ or $\mu ≤ 0$, whose stability is opposite to that of the coexisting fixed point.

Pat
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saxonryan
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  • This system has no equilibrium. Also I do not see how you get that Jacobian, as far as I see it, the Jacobian has eigenvalues $\lambda = 0, \mu$ everywhere. – smallStackBigFlow Mar 05 '21 at 10:12
  • Sorry, I've only started studying this module and I'm still making sense of it. I converted it to cartesian coordinates so $\dot{x} = \mu x -y$ and $\dot{y} = \mu y +x$ and got the following quadratic for my eigenvalues $\lambda^2-2\mu+\mu^2+1=0$. Is there a more obvious way of doing this? – saxonryan Mar 05 '21 at 11:05
  • Is an equilibrium the same as a fixed point? There are other examples in the book whereby I have to show that the Hopf bifurcation theorem fails by plotting them. Would you have an idea of how you would show the theorem fails? Would I plot the solution on the cartesian plane and should that show me something or have I the wrong idea? – saxonryan Mar 05 '21 at 11:09
  • Could you perhaps tell us which book you're talking about? That would be much more useful information than “my book”. – Hans Lundmark Mar 06 '21 at 15:43
  • Also, check whether the theorem in question says something about the creation of a limit cycle at the bifurcation. If so, there's no way that the theorem could apply here, since in this example there are no limit cycles for any $\mu$. – Hans Lundmark Mar 06 '21 at 15:55
  • Sorry, I've added the definition to my question. I understand that a linear system has no limit cycles but this version of the theorem doesn't mention limit cycles. I'm also unclear on what a 'one parameter family of periodic orbits is'. – saxonryan Mar 10 '21 at 17:26

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