Help needed understanding the steps in this Poincare map example

**ralphp** · June 18th 2010, 03:30 PM

Take a system of equations:
$\dot{r}=r(1-r^2)$
$\dot{\theta}=1$

Let S be the positive x-axis and $r_0$ is an initial condition on S.

Since $\dot{\theta}=1$ the first return to S occurs at $t=2\pi$ .
Then $r_1=P(r_0)$ where $r_1$ satisfies
$\int_{r_0}^{r_1}\frac{dr}{r(1-r^2)} = \int_0^{2\pi}dt = 2\pi \Rightarrow r_1$

Hence $P(r)=[1+e^{-4\pi}(r^{-2}-1)]^{-\frac{1}{2}}$

I don't understand how the "hence" bit is derived. I've spent a fair amount of time googling, and I believe partial fractions may hold part of the solution, but honestly I'm a little lost...

Many thanks!!!

**Ackbeet** · June 18th 2010, 06:03 PM

The way a Poincare map works is by taking pictures, once per "period", of phase space. In this case, what you really have is the following relation:

$\int_{r_{n}}^{r_{n+1}}\frac{dr}{r(1-r^{2})}=2\pi$ .

Moreover, the Poincare map looks like the following: $r_{n+1}=P(r_{n})$ .

So, perform the integral I just mentioned, and solve the equation for $r_{n+1}$ as a function of $r_{n}$ . That, I think, will get you the "hence" part.

**ralphp** · June 19th 2010, 01:59 AM

Thanks! That was very helpful.

I've managed to work through the problem and get the result given - or rather, I worked forward from the integral (I had to "cheat" and use Maple - clearly a bit of calculus revision is in order), and back from the "Hence P(r)=" bit, and met in the middle. But at least I now can see the method even if my integration skills are rusty, and that's better than being stuck on a page of notes thinking "what the hell?!?" I'm not easily able to leave something I don't understand and just continue, and this had me stuck for a day and a half.

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