Poincare Bendixson's Theorem

1. Apr 9, 2012

Rubik

1. The problem statement, all variables and given/known data
System in polar coordinates
$\dot{r}$ = 2r - r3(2 + sin($\theta$)),
$\dot{\theta}$ = 3 - r2

Use a trapping region to show there is at least one periodic orbit?

2. Relevant equations

By using Poincare Bendixson's Theorem

3. The attempt at a solution

I am struggling to know where to start.. Do I start by considering
g($\theta$) = 2 + sin($\theta$). Any help would be greatly appreciated?

2. Apr 9, 2012

Dick

Your g(θ) is always between 3 and 1. Try and find a small value r1 such that r' on the circle r=r1 is always positive. Now find a larger value of r2 such that r' on the circle r=r2 is always negative. Doesn't that make the region between the two circles a trapping region? Now think about fixed points.

3. Apr 10, 2012

Rubik

Okay so does this mean

1≤g(θ)≤3 So you get

$\dot{r}$ = 2r - r3g(θ)
$\dot{r}$ > 2r - 3r3 > 0 and so 2/3 > r2

$\dot{r}$ < 2r - r3 < 0 and so r2 > 2

and √(2/3) < r < √2

4. Apr 10, 2012

Dick

No, you don't really have to solve for anything. Pick a small value of r like r=1/2. Can you show r'>0 if r=1/2?

5. Apr 10, 2012

Rubik

So if I have to give bounds for the periodic orbit I can pick any value for r(small) provided r'>0 and then choose another value (this time larger) and show r'<0? How would I then determine its stability characteristics?

6. Apr 10, 2012

Dick

If you can find such values then you have a trapped region between the two circles. Can you have a fixed point in between? Try and figure out why not. Where can a fixed point be? Then use Poincare Bendixson clearly, yes?

Last edited: Apr 10, 2012
7. Apr 10, 2012

Rubik

There can't be any fixed points in trapping region. There can be a fixed point at the origin? So to determing the stability I determine the fixed point at the origin?

8. Apr 10, 2012

Dick

Sure there is a fixed point at the origin. But that doesn't matter since it's not in your trapping region. Then what does your theorem tell you?

9. Apr 10, 2012

Rubik

So the flow enters the trapping region about the origin which has no critical points which means it is a stable periodic orbit in the trapping region..

10. Apr 10, 2012

Dick

Not quite. It tells you there IS a periodic orbit in the trapping region. What's the exact statement of the theorem you are using?

11. Apr 10, 2012

Rubik

Suppose $\alpha$(x0) enters and does not leave some closed and bounded domain D that contains no critical points. This means that $\phi$(x0, t) $\in$ D for all t≥$\tau$, for some $\tau$≥0. Then there is at least one periodic orbit in D and this orbit is in the $\omega$-limit set of x0.

What does this mean in terms of stability? I have no idea how to determine the stability?

12. Apr 10, 2012

Dick

I'm maybe overextending myself here, but didn't the original problem ask you to show that there is a periodic orbit? I don't think it said anything about stability?

13. Apr 10, 2012

Rubik

Yes it did, but I didn't realise there was a second part that asks to determine its stability characteristics..

14. Apr 10, 2012

Dick

Ok, dynamical systems aren't really something I know well, so I might have to recuse myself here. Sorry.

15. Apr 10, 2012

Rubik

No problem thanks so much for your help!