# Transform Polar

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• February 4th 2011, 09:26 PM
dwsmith
$\displaystyle u_{xx}=\left(\cos(\theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)\left(\cos( \theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)$

Why can't this be computed like $u_{yy}\text{?}$

If I take

$\displaystyle u_{yy}=\left(\frac{y}{r}u_r+\frac{x}{r^2}u_{\theta }\right)^2$

and solve, the solution is correct but it doesn't work with $u_{xx}$. How come?

Also, when I approach

$\displaystyle u_{xx}=\left(\cos(\theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)\left(\cos( \theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)$

$\displaystyle\Rightarrow \cos(\theta)u_r\cos(\theta)u_r-2u_{r\theta}\frac{\sin(\theta)\cos(\theta)}{r^2}+\ frac{\sin(\theta)}{r}u_{\theta}\frac{\sin(\theta)} {r}u_{\theta}$

and

$\displaystyle u_{yy}=\sin(\theta)u_r\sin(\theta)u_r+2u_{r\theta} \frac{\sin(\theta)\cos(\theta)}{r^2}+\frac{\cos(\t heta)}{r}u_{\theta}\frac{\cos(\theta)}{r}u_{\theta }$

Now, I am wary of saying I have cosine and sine squared since I need

$\displaystyle\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)$.

How can I put those two equations together in order to yield the intended results?
• February 4th 2011, 10:10 PM
dwsmith
Quote:

Originally Posted by dwsmith
$\displaystyle u_{xx}=\left(\cos(\theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)\left(\cos( \theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)$

Why can't this be computed like $u_{yy}\text{?}$

If I take

$\displaystyle u_{yy}=\left(\frac{y}{r}u_r+\frac{x}{r^2}u_{\theta }\right)^2$

and solve, the solution is correct but it doesn't work with $u_{xx}$. How come?

Also, when I approach

$\displaystyle u_{xx}=\left(\cos(\theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)\left(\cos( \theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)$

$\displaystyle\Rightarrow \cos(\theta)u_r\cos(\theta)u_r-2u_{r\theta}\frac{\sin(\theta)\cos(\theta)}{r^2}+\ frac{\sin(\theta)}{r}u_{\theta}\frac{\sin(\theta)} {r}u_{\theta}$

and

$\displaystyle u_{yy}=\sin(\theta)u_r\sin(\theta)u_r+2u_{r\theta} \frac{\sin(\theta)\cos(\theta)}{r^2}+\frac{\cos(\t heta)}{r}u_{\theta}\frac{\cos(\theta)}{r}u_{\theta }$

Now, I am wary of saying I have cosine and sine squared since I need

$\displaystyle\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)$.

How can I put those two equations together in order to yield the intended results?

I have it now.

We have

$\displaystyle u_{xx}=\left(\frac{x}{r}u_r-\frac{y}{r^2}u_{\theta}\right)^2=\frac{x}{r}u_r\le ft(\frac{x}{r}u_r\right)-\frac{2}{r}u_{r}\left(u_{\theta}\frac{xy}{r^2}\rig ht)+\frac{y^2}{r^4}u_{\theta\theta}=\frac{x^2}{r}u _r\left(\frac{1}{r}u_r\right)-\frac{2}{r}u_{r}\left(u_{\theta}\frac{xy}{r^2}\rig ht)+\frac{y^2}{r^4}u_{\theta\theta}$

$\displaystyle u_{yy}=\left(\frac{y}{r}u_r+\frac{x}{r^2}u_{\theta }\right)^2=\frac{y}{r}u_r\left(\frac{y}{r}u_r\righ t)+\frac{2}{r}u_{r}\left(u_{\theta}\frac{xy}{r^2}\ right)+\frac{x^2}{r^4}u_{\theta\theta}=\frac{y^2}{ r}u_r\left(\frac{1}{r}u_r\right)+\frac{2}{r}u_{r}\ left(u_{\theta}\frac{xy}{r^2}\right)+\frac{x^2}{r^ 4}u_{\theta\theta}$

$\displaystyle u_r\left(\frac{1}{r}u_r\right)\left(\frac{x^2+y^2} {r}\right)+\left(\frac{x^2+y^2}{r^4}\right)u_{\the ta\theta}=ru_r\left(\frac{1}{r}u_r\right)+\frac{1} {r^2}u_{\theta\theta}$

$\displaystyle\Rightarrow u_t=k\left(ru_r\left(\frac{1}{r}u_r\right)+\frac{1 }{r^2}u_{\theta\theta}\right)$

My r and 1/r are swapped, hmmm....
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