Master1022
- 590
- 116
- 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)
[tex]\frac{dr}{dt} = r (a - r)[/tex]
[tex]\frac{d\theta}{dt} = -1[/tex]
and the first equation was solved with initial condition: ## r(t = 0) = r_0 ## and ## \theta(t = 0 ) = 0 ## to give:
[tex]r(t) = \frac{a r_0}{r_0 + (a - r_0) e^{-at}}[/tex]
[tex]\theta(t) = -t[/tex]
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:
[tex]P(r) = \frac{a r}{r + (a - r) e^{-2 a \pi}}[/tex]
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:
[tex]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}}[/tex]
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
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)
[tex]\frac{dr}{dt} = r (a - r)[/tex]
[tex]\frac{d\theta}{dt} = -1[/tex]
and the first equation was solved with initial condition: ## r(t = 0) = r_0 ## and ## \theta(t = 0 ) = 0 ## to give:
[tex]r(t) = \frac{a r_0}{r_0 + (a - r_0) e^{-at}}[/tex]
[tex]\theta(t) = -t[/tex]
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:
[tex]P(r) = \frac{a r}{r + (a - r) e^{-2 a \pi}}[/tex]
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:
[tex]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}}[/tex]
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