Periodic Orbits on Cylinder: Bendixson's Theorem

[funcosed]

Homework Statement

Show that the following vector field on the cylinder has a periodic orbit,

v' = -v
Θ' = 1

Homework Equations

Bendixson's theorem: Suppose D is a simply connected open subset of R^2. If the expression div(f,g) = ∂f/∂x + ∂g/∂y is not identically zero and does not change the sign in D, the there are no periodic orbits in the autonomous system in D.

The Attempt at a Solution

f = v' =-v
g = Θ' = 1
div(f,g) = -1 which would mean there are no periodic orbits or am I missing something??

 

[HallsofIvy]

Explain your terms please. I can guess that \theta is the angle measured at the axis to the cylinder but what is "v"? Is it the other cylindrical dimension, the height on the cylinder or is the rate of change of \theta?

In any case, "div(v, Θ)= v'+ Θ'= -v+ 1, not -1.

 

[funcosed]

Sorry that was unclear, the theorem has been restated above and I don't know what v is, all the information I have on the system is in the question.

 

[dirk_mec1]

Your answer is correct. Note that your initial problem was incorrect formulated (and thus confused Halls).

 

[funcosed]

Yes but the question asks to show it does have a periodic solution, whereas the theorem shows it doesnt. I think it might have something to do with an orbit overlapping the region but not sure how to show this.

 

[Joseph7]

Are you sure that the Bendixson's criteria is what determines if it has periodic orbit?

You are correct- the Bendixson's criteria does not hold. But this has periodic orbit. Any thoughts.
thanks

 

[HallsofIvy]

If the system is really v'= -v, \theta'= 1 then they equations are "uncoupled" and the simplest thing is just to solve the equations. v'= -v gives, immediately, v= ce^{-t} and \theta'= 1 gives \theta= t+ d. No, there are no periodic solutions to that!

 

[Joseph7]

I am sorry, are you suggesting that there is no periodic orbit. But the question is to show that it has periodic orbit. Shouldnt one make the conversion: x=vcos(theta) and y=vsing(theta) and solve of x' and y'
DO you think there is an error in the statement of the problem?
thank you very much.