Help with Converting to Polar Coordinates

[brain_ofj]

I need a little help with converting a nonlinear system to polar coordinates. I'm a bit stuck and could use a point in the right direction (this is for homework).

the system is
xdot = y + x[(x^2) + (y^2) -1]*sin(1/(((x^2) + (y^2) -1)^2))
ydot = -x + y[(x^2) + (y^2) -1]*sin(1/(((x^2) + (y^2) -1)^2))

Now, if it were just
((x^2) + (y^2) -1)^2), I'd set r = (x^2 + y^2)^1/2 subbing in partially...

xdot = y + x[r -1]*sin(1/(((r^2 -1)^2)) ...

but, uh, this is where I'm stuck.

How do I handle that leading 'y' and '-x' and the sin() ?

I would appreciate any help you can give. Thank you.

[Proginoskes]

I need a little help with converting a nonlinear system to polar coordinates. I'm a bit stuck and could use a point in the right direction (this is for homework).

the system is
xdot = y + x[(x^2) + (y^2) -1]*sin(1/(((x^2) + (y^2) -1)^2))
ydot = -x + y[(x^2) + (y^2) -1]*sin(1/(((x^2) + (y^2) -1)^2))

Now, if it were just
((x^2) + (y^2) -1)^2), I'd set r = (x^2 + y^2)^1/2 subbing in partially...

xdot = y + x[r -1]*sin(1/(((r^2 -1)^2)) ...

but, uh, this is where I'm stuck.

How do I handle that leading 'y' and '-x' and the sin() ?

I would appreciate any help you can give. Thank you.

First of all ...

xdot = y + x[r² -1]*sin(1/(((r^2 -1)^2)) ...

Second of all, x = r*cos(theta) and y = r*sin(theta). If I'm doing this correctly (and I'm not 100% sure that I am), that would make

{dx \over dt} = r\cdot-\sin(\theta)\cdot{d\theta\over dt} + {dr\over dt}\cdot \cos (\theta)

and you get a similar expression for y which you can substitute.

As for the sin\left(1 \over (r^2-1)^2\right), I'm stymied.

[PM 2Ring]

Are you just trying to convert to polar, or do you also want to integrate the resulting differential equations?

To convert, differentiate r² = x² + y² and tan(theta) = y / x with respect to t then plug in the given expressions for xdot and ydot.

If you want to integrate the polar differential equations, the one for theta_dot simplifies considerably (if I haven't screwed up the algebra ) and is easy to integrate, the expression for r_dot is a bit more complicated, and I don't see an easy way to integrate it.

[brain_ofj]

No, I'm just trying to convert to polar coordinates. I'm then to show that there are an infinite number of limit cycles...
That last bit should be doable but I'm stuck on the conversion to polar coordinates.

To convert, differentiate r² = x² + y² and tan(theta) = y / x with respect to t then plug in the given expressions for xdot and ydot.

Uh, ok, now I'm confused. Why do I need to differentiate?

[doogly]

You just need to reach back to the simple implicit diff and chain rule topics from calc one, and all will become clear.

[brain_ofj]

No, I'm sorry, it's not clear why...

why is xdot not ->

xdot = rsin(\theta) + rcos(\theta)(r^2 -1)sin(1/((r^2 -1)^2)

?

Why is it necessary that I differentiate like mentioned above?

[PM 2Ring]

No, I'm sorry, it's not clear why...

why is xdot not ->

\dot{x} = r sin\theta + r cos\theta(r^2 -1)sin(1/(r^2 -1)^2)

?

Why is it necessary that I differentiate like mentioned above?

That expression for xdot is correct, but we want to get rid of x, y and their derivatives and express everything in terms of r and theta. So you need to come up with expressions for rdot and thetadot in terms of r and theta, and ideally to separate the variables, i.e., express rdot in terms of r and thetadot in terms of theta. By taking the time derivatives I suggested earlier you'll be able to form expressions for rdot and thetadot in terms of x, xdot, y and ydot. You then plug in the given equations for xdot and ydot into those expressions for rdot and thetadot and with the help of r² = x² + y² you'll be able to eliminate all the x and y stuff.

[brain_ofj]

Ok. I think that makes sense...

So I have this:

xdot = y + x(x^2 + y^2 -1)sin(1/ (x^2 + y^2 -1)^2)

ydot = -x + y(x^2 + y^2 -1)sin(1/ (x^2 + y^2 -1)^2)

x = rcos(\theta), y = rsin(\theta), r = (x^2 + y^2)^1/2

dx/dt = r*-sin(\theta)d(\theta)/dt + dr/dtcos(\theta)

dy/dt = rcos(\theta)d(\theta/dt + dr/dtsin(\theta)

dr/dt = 1/2*sqrt(x^2 + y^2) * (2x*xdot + 2y*ydot)

dy/dx = tan(\theta)

And then sub in dx/dt and dy/dt in to dr/dt and dy/dx ? Is that right?
That will still case weirdness with the sin(1/( )^2) though, right?

edit:
I eventually got it to here:

rdot = r(r^2 -1)sin((1/ (r^2 -1)^2)

\theta dot = 1

[PM 2Ring]

Ok. I think that makes sense...

So I have this:

xdot = y + x(x^2 + y^2 -1)sin(1/ (x^2 + y^2 -1)^2)

ydot = -x + y(x^2 + y^2 -1)sin(1/ (x^2 + y^2 -1)^2)

x = rcos(\theta), y = rsin(\theta), r = (x^2 + y^2)^1/2

dx/dt = r*-sin(\theta)d(\theta)/dt + dr/dtcos(\theta)

dy/dt = rcos(\theta)d(\theta/dt + dr/dtsin(\theta)

dr/dt = 1/2*sqrt(x^2 + y^2) * (2x*xdot + 2y*ydot)

dy/dx = tan(\theta)

And then sub in dx/dt and dy/dt in to dr/dt and dy/dx ? Is that right?

Yes, you can do that, but it's more tedious than doing it the way that I suggested earlier. BTW, that last line is wrong. tan(\theta) = y/x, and in general, dy/dx is not equal to y/x.

That will still cause weirdness with the sin(1/( )^2) though, right?

Yes, but you don't need to worry too much about that since you're not integrating.

edit:
I eventually got it to here:

rdot = r(r^2 -1)sin((1/ (r^2 -1)^2)

\theta dot = 1

Yay! Your expression for \dot{r} is correct, but I think you've got the sign wrong for \dot\theta.

[brain_ofj]

Thanks for the help.
My Prof confirmed that

\theta dot

is correct for this problem.

[jestingrabbit]

\theta dot is never right: try \dot{\theta}.