# Transform Polar

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Feb 3rd 2011, 11:16 AM
dwsmith
Transform Polar
Transform the equation

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

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

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

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

Changing to polar.

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

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

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

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

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

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

Quote:

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

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

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

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.
• Feb 4th 2011, 02:28 PM
dwsmith
Quote:

Originally Posted by Ackbeet

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

How is piece derived from

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

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

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

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

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

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

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

but

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

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

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

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

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

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

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

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

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?
• Feb 4th 2011, 03:15 PM
dwsmith
Quote:

Originally Posted by Ackbeet
Please provide a reference for the equation

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

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

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

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

Note that

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

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

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

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

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

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

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

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

I found my error.

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

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

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

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

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

Typo here u_x should have been u_r
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last