Exploring PDE Solutions in Polar Coordinates

[Dustinsfl]

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.

 

[HallsofIvy]

Use the chain rule.

$$\frac{\partial u}{\partial x}= \frac{\partial r}{\partial x}\frac{\partial u}{\partial r}+ \frac{\partial \theta}{\partial x}\frac{\partial u}{\partial \theta}$$
and
$$\frac{\partial u}{\partial y}= \frac{\partial r}{\partial y}\frac{\partial u}{\partial r}+ \frac{\partial \theta}{\partial y}\frac{\partial u}{\partial \theta}$$


Since $r= (x^2+ y^2)^{1/2}$,
$$\frac{\partial r}{\partial x}= \frac{1}{2}(x^2+ y^2)^{-1/2}(2x)$$
[tex]= \frac{x}{(x^2+ y^2)^{1/2}}= \frac{r cos(\theta)}{r}= cos(\theta)[/itex]
and
$$\frac{\partial r}{\partial y}= \frac{1}{3}(x^2+ y^2)^}{-1/2}(2x)$$
[tex]= \frac{y}{(x^2+ y^2)^{1/2}}= \frac{r sin(\theta)}{r}= sin(\theta)[/itex]

Since $\theta= tan^{-1}(y/x)$
$$\frac{\partial \theta}{\partial x}= \frac{1}{1+\frac{y^2}{x^2}}\left(-\frac{y}{x^2}\right)$$
$$= -\frac{y}{x^2+ y^2}= -\frac{r sin(\theta)}{r^2}= -\frac{1}{r}sin(\theta)$$
and
$$\frac{\partial \theta}{\partial y}= \frac{1}{1+ \frac{y^2}{x^2}}\left(\frac{1}{x}\right)$$
[tex]= \frac{x}{x^2+ y^2}= \frac{r cos(\theta)}{r^2}= \frac{1}{r}cos(\theta)[/itex]

 

[Dustinsfl]

Thanks.