There exists a unique stable limit cycle around $(0,0)$

Question

Consider the system $$ x'=\mu x-y-x\sqrt{x^2+y^2}, \, y'=x+\mu y-y\sqrt{x^2+y^2} $$ I am working on that show that (1) as $\mu>0$, there exists a unique stable limit cycle around $(0,0)$ (2) as $\mu\le 0$, there is no limit cycle around $(0,0)$. And draw the phase portraits for these two cases. Also, draw the bifurcation diagram.

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My proof: note that the only critical point is $(0,0)$ and $$Df(0,\mu) =\begin{pmatrix} \mu & -1 \\ 1 & \mu \end{pmatrix}$$

Then $(0,0)$ is stable if $\mu<0$ and $(0,0)$ is unstable if $\mu>0$.

For $\mu=0$, consider the polar coordinate: $$ r'=r(\mu-r), \theta'=1 $$

So $\mu=0$, $(0,0)$ is a stable focus.

For $\mu>0$, there is a limit cycle $(\mu \cos t, \mu \sin t)$. [** Question: But how to say that is unique?**]

Since as $\mu<0$, $(0,0)$ is unstable, then there is no limit cycle.

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The phase portraits:

The bifurcation diagram: The red line is $r=\mu$.

Answer 1

I guess you can study everything is polar coordinate, as the system in that case is reduced to 1-d. You have

$$ \dot{r} = r(m-r) =: f_\mu(r) $$

with fixed points $r=0$ and $r= \mu$. The stability is controlled by by

$$ f'_\mu(r):=\frac{d}{dr}f_\mu(r) =\mu - 2r $$. Going by cases:

• $\mu < 0$ : The only acceptable fixed point is $r=0$ (since $r$ must be positive). We see it is stable cause

$$ f'_\mu(0) = \mu <0 $$

So there is no limit cycle, and the origin attracts all trajectories.

• $\mu > 0$: Now both fixed points are acceptable and we see

$$ f'_\mu(r=0) = \mu >0 $$ Meaning that $r=0$ is repelling. And

$$ f'_\mu(r=\mu) = -\mu < 0 $$ Which means that $r=\mu$ is repelling. This proves the uniqueness of the limit cycle (and actually, also its existence), as all the trajectories on the plane are attracted to it (both from the 'inside' and from the 'outside' the closed orbit). A stable periodic orbit does not necessarily have this property. Thus, proving that 0 is unstable is not enough.

• $\mu = 0$: This is tricky as if you use the formulas from before you would get $f'(\mu)=0$ and $f'(0)=0$. But it is wrong to proceed in this way. First we notice that the equation for $\mu=0$ reads: $$ f_\mu(r) = -r^2 $$ Making the only equilibrium ($r=0$) attracting ($f'_\mu(0)=-r <0$ for all $r>0$).

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It might be nice to know that $f(r) = x(\mu-x)$ is the famous logistic equation, that can be solved analytically.