# Transform Polar

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• February 3rd 2011, 11:16 AM
dwsmith
Transform Polar
Transform the equation

$u_t=k(u_{xx}+u_{yy})$

to polar coordinates $r=\sqrt{x^2+y^2}$, $\theta=\tan^{-1}\left(\frac{y}{x}\right)$ and specialize the resulting equation to the case when u does not not depend on the angular variable theta.

$\displaystyle u_{xx}=\left[\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial x}\frac{\partial x}{\partial\theta}\right]^2$

$\displaystyle u_{yy}=\left[\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial\theta}\right]^2$

Changing to polar.

$\displaystyle u_{xx}=\left[u_r\frac{x}{r}-u_{\theta}\frac{y}{r}\right]^2=u_{rr}\frac{x^2}{r^2}-u_{r\theta}\frac{2xy}{r^3}+u_{\theta\theta}\frac{y ^2}{r^4}$

$\displaystyle u_{yy}=\left[u_r\frac{y}{r}+u_{\theta}\frac{x}{r}\right]^2=u_{rr}\frac{y^2}{r^2}+u_{r\theta}\frac{2xy}{r^3 }+u_{\theta\theta}\frac{x^2}{r^4}$

$\displaystyle u_{t}=k\left[u_{rr}\left(\frac{x^2+y^2}{r^2}\right)+u_{\theta\t heta}\left(\frac{x^2+y^2}{r^4}\right)\right]=k\left[u_{rr}+\frac{1}{r^2}u_{\theta\theta}\right]$

How do I make u not depend on theta?
• February 3rd 2011, 11:51 AM
Ackbeet
How about setting $u_{\theta\theta}=0?$
• February 3rd 2011, 11:52 AM
dwsmith
Quote:

Originally Posted by Ackbeet
How about setting $u_{\theta\theta}=0?$

So it is ok to just write the equation as $u_t=ku_{rr}$ then?
• February 3rd 2011, 11:53 AM
Ackbeet
Sure. If $u$ does not depend on $\theta,$ it follows that $u_{\theta}=0,$ and then of course, $u_{\theta\theta}=0.$
• February 4th 2011, 01:15 PM
Jester
I hate to be the bearer of bad news but

Quote:

Originally Posted by dwsmith
$\displaystyle u_{xx}\ne\left[\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial x}\frac{\partial x}{\partial\theta}\right]^2$

• February 4th 2011, 01:22 PM
Ackbeet
I think you would have

$\displaystyle u_{xx}+u_{yy}=\frac{1}{r}\,\frac{\partial}{\partia l r}\left(r\,\frac{\partial u}{\partial r}\right)+\frac{1}{r^{2}}\,\frac{\partial^{2} u}{\partial^{2}\theta},$

because you essentially have the Laplacian in two dimensions. If you ignore the z dimension, you can just copy down the version for cylindrical coordinates, which is what I did.
• February 4th 2011, 02:28 PM
dwsmith
Quote:

Originally Posted by Ackbeet

$\displaystyle u_{xx}=\frac{1}{r}\,\frac{\partial}{\partial r}\left(r\,\frac{\partial u}{\partial r}\right)$

How is piece derived from

$\displaystyle u_x=\frac{x}{r}u_x-\frac{y}{r}u_{\theta}$
• February 4th 2011, 02:31 PM
Ackbeet
Let me challenge you right back: how do you get

$\displaystyle u_x=\frac{x}{r}u_x-\frac{y}{r}u_{\theta}?$

The units don't work for that expression. I don't buy it.
• February 4th 2011, 02:34 PM
Ackbeet
See here for a derivation. You'll notice that they've multiplied out the product rule that I have not multiplied out.
• February 4th 2011, 02:55 PM
dwsmith
Quote:

Originally Posted by Ackbeet
See here for a derivation. You'll notice that they've multiplied out the product rule that I have not multiplied out.

On the site, they have

$\displaystyle u_{xx}=\left(\cos{\phi}u_{r}-\frac{\sin{\phi}}{r}u_{\phi}\right)^2$

where $\displaystyle\cos\phi=\frac{x}{r}$

and $\displaystyle\sin\phi=\frac{y}{r}$

$\displaystyle u_{xx}=\left(\frac{x}{r}u_r-\frac{y}{r^2}u_{\phi}\right)^2$

but

$\displaystyle u_{x}=\left(\frac{x}{r}u_r-\frac{y}{r}u_{\phi}\right)$

Why are these different and why didn't the $(u_x)^2=u_{xx}\text{?}$
• February 4th 2011, 03:10 PM
Ackbeet
Please provide a reference for the equation

$\displaystyle u_{x}=\left(\frac{x}{r}u_r-\frac{y}{r}u_{\phi}\right).$

Like I said before, I think this equation is incorrect. The units don't work out. For the first term, the $x$ and $r$ units cancel out, because they're both length. That leaves you with units of $u$ divided by units of length. On the second term, the $y$ and $r$ units cancel out, as before, but the units of $u_{\theta}$ are NOT units of $u$ divided by units of length. Therefore, you can't add the terms because they don't have the same units.

Quote:

On the site, they have

$\displaystyle u_{xx}=\left(\cos{\phi}u_{r}-\frac{\sin{\phi}}{r}u_{\phi}\right)^2$...
Actually, they don't have that. What they actually have is this:

$\displaystyle\frac{\partial^{2}}{\partial x^{2}}=\underbrace{\left(\cos(\phi)\,\frac{\partia l}{\partial\rho}-\frac{\sin(\phi)}{\rho}\,\frac{\partial}{\partial\ phi}\right)}_{\text{This is an operator, which {\it acts on}}}\underbrace{\left(\cos(\phi)\,\frac{\partial} {\partial\rho}-\frac{\sin(\phi)}{\rho}\,\frac{\partial}{\partial\ phi}\right)}_{\text{this}.}$

Operators are very slippery creatures. There's a HUGE difference between

$\dfrac{\partial}{\partial\rho}$ and $\dfrac{\partial u}{\partial\rho}.$

Think of the first as an operator: it's going to look to its right and try to find something to operate on (the operation in question is partial differentiation with respect to phi). On the other hand, the second is not an operator, because the differentiation has already taken place. It's just a function.

Does that make sense?
• February 4th 2011, 03:15 PM
dwsmith
Quote:

Originally Posted by Ackbeet
Please provide a reference for the equation

$\displaystyle u_{x}=\left(\frac{x}{r}u_r-\frac{y}{r}u_{\phi}\right).$

$\displaystyle r=\sqrt{x^2+y^2}, \ \ \theta=\tan^{-1}\left(\frac{y}{x}\right)$

$\displaystyle u_x=\frac{x}{r}u_r-\frac{y}{r}u_{\theta}$
• February 4th 2011, 04:17 PM
Ackbeet
Hmm. If you look carefully at the link I gave in post # 9, you will see that

$\displaystyle\frac{\partial}{\partial x}=\frac{\partial\rho}{\partial x}\,\frac{\partial}{\partial\rho}+\frac{\partial\p hi}{\partial x}\,\frac{\partial}{\partial\phi}.$

Note that

$\dfrac{\partial\rho}{\partial x}=\dfrac{x}{\rho},$ and

$\dfrac{\partial\phi}{\partial x}=-\dfrac{\sin(\phi)}{\rho}=-\dfrac{y}{\rho^{2}},$ implying that

$\displaystyle\frac{\partial}{\partial x}=\dfrac{x}{\rho}\,\frac{\partial}{\partial\rho}-\dfrac{y}{\rho^{2}}\,\frac{\partial}{\partial\phi} .$

You can see that the units work out correctly in this expression.
• February 4th 2011, 09:08 PM
dwsmith
Starting from scratch we have:

$\displaystyle r=\sqrt{x^2+y^2}, \ \ \ \theta=\tan^{-1}\left(\frac{y}{x}\right), \ \ \ x=r\cos\theta, \ \ \ y=r\sin\theta$

$\displaystyle\frac{\partial r}{\partial x}=\frac{x}{r}, \ \ \ \frac{\partial r}{\partial y}=\frac{y}{r}$

$\displaystyle\frac{\partial\theta}{\partial x}=-\frac{y}{r^2}, \ \ \ \frac{\partial\theta}{\partial y}=\frac{x}{r^2}$

$\displaystyle u_{x}=u_rr_x+u_{\theta}\theta_x=\frac{x}{r}u_r-\frac{y}{r^2}u_{\theta}=\cos(\theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}$

I found my error.

I forgot to type r^2 for $\theta_{x} \ \ \ \text{and} \ \ \ \theta_{y}$

Geez, guys where was the help in finding that mistake :p
Just kidding. Thanks.
• February 4th 2011, 09:09 PM
dwsmith
Quote:

Originally Posted by Ackbeet
Let me challenge you right back: how do you get

$\displaystyle u_x=\frac{x}{r}u_x-\frac{y}{r}u_{\theta}?$

The units don't work for that expression. I don't buy it.

Typo here u_x should have been u_r
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