Polar Coordinates

**dwsmith** · December 24th 2010, 06:22 PM

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.

**Danny** · December 25th 2010, 06:04 AM

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)

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

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

**dwsmith** · December 25th 2010, 11:56 AM

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.