Consider the dynamical system:


Find all fixed points and limit cycles for:

a) $~~a=2$




For all three values/ranges that we are considering there will be a fixed point at the origin. To make the behavior clearer, I can rewrite the first equation as follows:


a) $~~a=2 \implies \dot{r}=(2r^3-r^5-r)(1-r)$

The radius will either grow, shrink or stay constant depending on the value of r:

If$~~r=1\implies \dot{r}=0$ If$~~r<1\implies \dot{r}<0$ If$~~r>1\implies \dot{r}>0$

So there is an unstable limit cycle at $r=1$

b) It appears that there is no change for this case, only that paths will be repelled from the limit cycle faster (what is a better way to word this?). Is this correct?

c) There seems to be a second limit cycle spawning from r=1 who's radius increases as a increases. Although I'm not sure why my lecturer has picked the value $~2\sqrt{2}$, since it occurs beyond this value. I was only able to find this by using mathematica and plotting the paths for different initial conditions and different values of a. How would one go about showing this? And why do you think the value $~2\sqrt{2}$ was chosen?



$$\dot{r}=r(r-1)(r^4-ar^2+1)$$ For $a\in[0,2)$ it holds true that $$r^4-ar^2+1>0\qquad \forall r\in\mathbb{R}$$ When $a>2$ $$\dot{r}=r(r-1)\left(r^2-\frac{a+\sqrt{a^2-4}}{2}\right)\left(r^2-\frac{a-\sqrt{a^2-4}}{2}\right)$$ and equivalently $$\dot{r}=r(r-1)\left(r-r_1\right)\left(r-r_2\right)\left(r+r_1\right)\left(r+r_2\right)$$ where $$r_1:=\sqrt{\frac{a-\sqrt{a^2-4}}{2}}\\ r_2:=\sqrt{\frac{a+\sqrt{a^2-4}}{2}}.$$ Thus

1) if $a\in (0,2)$ then $r=0$ is a stable equilibrium, $r=1$ is an unstable limit cycle

2) if $a>2$ then $r=0$ is a stable equilibrium, $r=r_1$ is an unstable limit cycle, $r=1$ is a stable limit cycle and $r=r_2$ is an unstable limit cycle.


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