Proving Periodic Orbit of x' & y' System

[squenshl]

Homework Statement

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)

Homework Equations

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.

 

[dynamicsolo]

...but in terms of theta' I get theta' = 1. If theta' = 1 = constant imply a periodic orbit?

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.

 

[squenshl]

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

 

[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' .

 

[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).

 

[dynamicsolo]

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).

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.

 

[squenshl]

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

 

[dynamicsolo]

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

(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.