# Polar Coordinates

• Dec 24th 2010, 07:22 PM
dwsmith
Polar Coordinates
Solving a PDE with Polar coordinates

$yu_x-xu_y=0$

$x=r\cos{\theta} \ \mbox{and} \ y=r\sin{\theta}$

$u(r,\theta)$

Does $u_x\Rightarrow u_r \ \mbox{or} \ u_{\theta} \ \mbox{and why?}$

Thanks.
• Dec 25th 2010, 07:04 AM
Jester
You need to use the chain rule

$u_x = u_r r_x + u_{\theta} \theta_x$

$u_y = u_r r_y + u_{\theta} \theta_y$

or if you use Jacobians (the way I'd do it)

$u_x =\dfrac{\partial(u,y) }{\partial(r,\theta)}/\dfrac{\partial(x,y)}{\partial(u,\theta)}$

$u_y =\dfrac{\partial(x,u) }{\partial(r,\theta)}/\dfrac{\partial(x,y)}{\partial(u,\theta)}$.
• Dec 25th 2010, 12:56 PM
dwsmith
Danny,

I haven't done anything with the Jacobian in a long time. Could you show me how it is done?

Thanks.

$\displaystyle u_x=\cos{\theta}u_r-\frac{\sin{\theta}}{r}u_{\theta}$

$\displaystyle u_y=\sin{\theta}u_r+\frac{\cos{\theta}}{r}u_{\thet a}$

$\displaystyle yu_x-xu_y=r\sin{\theta}\left[\cos{\theta}u_r-\frac{\sin{\theta}}{r}u_{\theta}\right]-r\cos{\theta}\left[\sin{\theta}u_r+\frac{\cos{\theta}}{r}u_{\theta}\r ight]$

$\displaystyle r\sin{\theta}\cos{\theta}u_r-\sin^2{\theta}u_{\theta}-r\cos{\theta}\sin{\theta}-\cos^2{\theta}u_{\theta}=0$

$\displaystyle -u_{\theta}(\sin^2{\theta}+\cos^2{\theta})=0\Righta rrow -u_{\theta}=0\Rightarrow u_{\theta}=0$

$\displaystyle\int u_{\theta}d\theta=\int 0d\theta\Rightarrow u(r,\theta)=f(r)$

The solution I am supposed to conclude is $u(x,y)=f(x^2+y^2)$, f arbitrary.

How do I finish off from $u(r,\theta)=f(r)\mbox{?}$

Thanks.
• Dec 25th 2010, 03:05 PM
zzzoak
From
$
u_{\theta}=0 \; \Rightarrow \; u(r,\theta)=f(r).
$
• Dec 25th 2010, 03:08 PM
dwsmith
Quote:

Originally Posted by zzzoak
From
$
u_{\theta}=0 \; \Rightarrow \; u(r,\theta)=f(r).
$

How does that transform to $u(x,y)=f(x^2+y^2)\mbox{?}$
• Dec 25th 2010, 03:40 PM
zzzoak
$
f(r)=f( \; \sqrt{x^2+y^2} \; )=g( \; x^2+y^2 \; ).
$
• Dec 25th 2010, 03:42 PM
dwsmith
And making the change from u(r, theta) to u(x,y) is fine?
• Dec 25th 2010, 04:01 PM
zzzoak
We convert
$
u(r,\theta)=U(x,y)
$

and
$
f(r)=F(x^2+y^2)
$

and have finally
$
U(x,y)=F(x^2+y^2).
$

We may check inserting it to the initial equation.