# Homework Help: Periodic orbit

1. Aug 20, 2011

### squenshl

1. The problem statement, all variables and given/known data
Prove that the following system has at least one periodic orbit.
x' = x - y - 2x3 - 2xy2 + x2sqrt(x2 + y2)
y' = x + y - 2x2y - 2y3 + xysqrt(x2 + y2)

2. Relevant equations

3. The attempt at a solution
I converted to polar coordinates to get r' = (xx'+yy')/r and theta' = (xy'-yx')/r2
For r' I got something ugly but in terms of theta' I get theta' = 1. If theta' = 1 = constant imply a periodic orbit? Any help please.

2. Aug 20, 2011

### dynamicsolo

That will only imply that the angular speed along the trajectory is a constant; that doesn't guarantee by itself that the trajectory even forms a closed curve.

Try using polar coordinates in the other direction. Transform x , y , x' , and y' on both sides of the equation: you find something cancels that will simply r' considerably. (Note that an r2 can be factored out of the third and fourth terms in both differential equations.) Both the x' and the y' equations will lead to the same result.

3. Aug 21, 2011

### squenshl

Cheers, not sure what you mean by polar coordinates in the other direction though.

Last edited: Aug 21, 2011
4. Aug 21, 2011

### dynamicsolo

Use $x = r \cos \theta$ and $y = r \sin \theta$ , and also differentiate these to get expressions for x' and y' to use on the left-hand side, rather than trying to express r' and $\theta \prime$ in terms of x' and y' .

5. Aug 21, 2011

### squenshl

Thanks.
Doing it the first way I get r' = r3cos(theta)-2r3+r, theta' = 1. How do I interpret these to show that this system has a periodic orbit (if I can indeed interpret these).

6. Aug 21, 2011

### dynamicsolo

Starting from $x\prime = x - y - 2x r^{2} + \frac{x^{2}}{r}$ ,

we get $(r \cos\theta)\prime = r ( 1 - 2r^{2} ) \cos\theta - r\sin\theta + r \cos^{2}\theta .$

Since $(r \cos\theta)\prime = r\prime \cos\theta - r\sin\theta$,

we get to cancel the r sin(theta) term and divide through by cos(theta).

I believe you should get $r \prime = r ( 1 - 2r^{2} ) + r \cos\theta .$
Handling the y' equation leaves us with exactly the same result.

Now plainly we won't have r' = 0 everywhere, but is there a value of r for which r' will be zero on average over one or more cycles of theta? If so, then we've found a periodic orbit.

7. Aug 21, 2011

### squenshl

Awesome. Cheers mate, but I got r' = r(1-2r2) + r3cos(theta).

Last edited: Aug 21, 2011
8. Aug 21, 2011

### dynamicsolo

(expletive) I'm sorry, I've been reading that last term wrong all this time (I should not try to do anything on PF after 1 AM).

This should be $x\prime = x - y - 2x r^{2} + r x^{2}$ , which leads to

we get $(r \cos\theta)\prime = r ( 1 - 2r^{2} ) \cos\theta - r\sin\theta + r^{3} \cos^{2}\theta .$

So your equation $r \prime = r ( 1 - 2r^{2} ) + r^{3} \cos\theta$ was correct.

This does not invalidate the rest of what I suggested (fortunately); it doesn't even change the value of r required (you want to "zero out" the first term on the right-hand side).

If you plot your system of equations on a grapher that can handle vector fields, you'll find that the "orbit" constitutes what is called a "limit cycle". The "flow" outside this orbit is attracted to it, and so is the flow within the orbit.

Last edited: Aug 21, 2011