# Limit cycles

• Dec 2nd 2011, 12:11 PM
Darkprince
Limit cycles
Whenever I say ... dot I mean the derivative of ...

I have to show that the system of Ordinary Diff. Eqns

x (dot) = y + x* f(r) / r
y (dot) = -x + y * f(r)/r

r=sqrt(x^2+y^2)

has limit cycles and the limit cycles have radius which correspond to the zeros of f(r)

Also I have to find the direction of motion on the limit cycles.

My workings:

I first found the nullclines.

C(x): y + x* f(r) / r =0 and C(y): -x + y * f(r)/r =0 implies that x=y*f(r) / r

Substituting into C(x) we get that the system has an equilibrium point at
(y * f(r) / r, 0)

Then we consider polar coordinates. Using r^2=x^2 + y^2

Then differentiate to obtain that r dot = (f(r) / r^2 ) * (2x^2 + 2y^2)

Now if 2x^2 + 2y^2 = 0 => r dot = 0

if 2x^2 + 2y^2 > 0 => r dot >0

Now I have to find C1 and C2 s.t they bound a region R that contains no equilibrium points and trap all trajectories so I can use Poincaré–Bendixson theorem to show that there exists a limit cycle.

Any help would be greatly appreciated!

Thank you!
• Dec 3rd 2011, 07:05 PM
Jester
Re: Limit cycles
What is $f(r)?$
• Dec 4th 2011, 02:51 AM
Darkprince
Re: Limit cycles
Quote:

Originally Posted by Danny
What is $f(r)?$

I am not given f(r) at this point, I am given f(r) at a later point, so I have to deduce generally first.
• Dec 6th 2011, 03:30 PM
Darkprince
Re: Limit cycles
Ok, I figured them all out, thank you though :)
• Dec 6th 2011, 03:37 PM
Jester
Re: Limit cycles
Quote:

Originally Posted by Darkprince
Using polar coordinates: x=rcosf, y=rsinf

You don't want to do this. What you what is

$x = r \cos \theta,\;\;\;y = r \sin \theta$.

$f(r)$ is a separate issue
• Dec 6th 2011, 03:56 PM
Darkprince
Re: Limit cycles
Quote:

Originally Posted by Danny
You don't want to do this. What you what is

$x = r \cos \theta,\;\;\;y = r \sin \theta$.

$f(r)$ is a separate issue

What do you mean? When writing f I was meaning theta.
Are my workings not correct?
• Dec 7th 2011, 05:51 AM
Darkprince
Re: Limit cycles
Ok, I figured them all, thank you!