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.

## Help with Converting to Polar Coordinates

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- Proginoskes
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### Re: Help with Converting to Polar Coordinates

brain_ofj wrote: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[r

^{2}-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

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

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

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

### Re: Help with Converting to Polar Coordinates

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.

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.

### Re: Help with Converting to Polar Coordinates

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.

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

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
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### Re: Help with Converting to Polar Coordinates

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

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### Re: Help with Converting to Polar Coordinates

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

why is xdot not ->

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

?

Why is it necessary that I differentiate like mentioned above?

why is xdot not ->

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

?

Why is it necessary that I differentiate like mentioned above?

### Re: Help with Converting to Polar Coordinates

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

why is xdot not ->

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

?

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.

### Re: Help with Converting to Polar Coordinates

Ok. I think that makes sense...

So I have this:

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

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

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

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

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

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

[math]dy/dx = tan(\theta)[/math]

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:

[math]rdot = r(r^2 -1)sin((1/ (r^2 -1)^2)[/math]

[math]\theta dot = 1[/math]

So I have this:

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

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

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

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

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

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

[math]dy/dx = tan(\theta)[/math]

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:

[math]rdot = r(r^2 -1)sin((1/ (r^2 -1)^2)[/math]

[math]\theta dot = 1[/math]

Last edited by brain_ofj on Tue Apr 17, 2012 9:59 pm UTC, edited 1 time in total.

### Re: Help with Converting to Polar Coordinates

brain_ofj wrote:Ok. I think that makes sense...

So I have this:

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

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

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

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

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

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

[math]dy/dx = tan(\theta)[/math]

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. [imath]tan(\theta) = y/x[/imath], and in general, dy/dx is not equal to y/x.

brain_ofj wrote: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.

brain_ofj wrote:edit:

I eventually got it to here:

[math]rdot = r(r^2 -1)sin((1/ (r^2 -1)^2)[/math]

[math]\theta dot = 1[/math]

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

### Re: Help with Converting to Polar Coordinates

Thanks for the help.

My Prof confirmed that [math]\theta dot[/math] is correct for this problem.

My Prof confirmed that [math]\theta dot[/math] is correct for this problem.

- jestingrabbit
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### Re: Help with Converting to Polar Coordinates

[imath]\theta dot[/imath] is never right: try [imath]\dot{\theta}.[/imath]

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