Dynamical Systems: how to find equation for Poincare map?

[Master1022]

Homework Statement

Using the solution given in the previous part, show that for $a > 0$ one can define the Poincare map: $P: \mathbb{R}^{+} \righarrow \mathbb{R}^{+}$ by: (expression below)

Relevant Equations

Differential equations

Hi,

I was attempting a question on the dynamical systems topic of Poincare maps, and was struggling to understand a certain part of it.

Knowledge from prior parts of the questions:
There was a system which we converted to polar coordinates to get: ($a$ is an arbitrary real constant)
\frac{dr}{dt} = r (a - r)
\frac{d\theta}{dt} = -1

and the first equation was solved with initial condition: $r(t = 0) = r_0$ and $\theta(t = 0 ) = 0$ to give:
r(t) = \frac{a r_0}{r_0 + (a - r_0) e^{-at}}
\theta(t) = -t
We can see that all solutions tend to a limit cycle ($r = a$) as $t \rightarrow \infty$ for $a > 0$.

Question I am stuck on: Using the solution given in the previous part, show that for $a > 0$ one can define the Poincare map: $P: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ by $r_n = P(r_{n - 1})$ with:
P(r) = \frac{a r}{r + (a - r) e^{-2 a \pi}}

Attempt:
This should be quite a simple thing to do, but I don't quite understand each step in the logic to get there. So I know the Poincare map can be thought of as a plane where we mark the intersections of the trajectory with the plane. This means that a fixed point on a Poincare map corresponds to a limit cycle for the full trajectory.

This plane will always be at the same angle $\pm 2 n \pi$ (the x-axis). Are we interested in all the times $t$ when $= - 2 n \pi$? These times $t$ can be found by doing: $-t = -2 n \pi \rightarrow t = 2 n \pi$.

Substituting this into the expression for $r(t)$ yields:
r(t) = \frac{a r_0}{r_0 + (a - r_0) e^{-2a n \pi}} \rightarrow r_n = \frac{a r_0}{r_0 + (a - r_0) e^{-2a n \pi}}

However, I am not sure how to show that the expression is equal to $P(r)$. What is the reasoning behind that?

Thanks in advance

 

[pasmith]

Don't solve for r_n = r(2n\pi) subject to r(0) = r_0. Instead solve for r(2\pi) subject to r(0) = r_{n-1}.