PDE

meks08999 · #1 $Old$ October 25th, 2009, 07:18 PM

Need help with this problem. Problem is attached.

Thanks

TheEmptySet · #2 $Old$ October 25th, 2009, 07:51 PM

Quote:

Originally Posted by meks08999 $View Post$

Need help with this problem. Problem is attached.

Thanks

Something is wrong......

$u(x,y)=\psi(s)$ where $s=x^2+y^2$

taking some derivatives we get

$u_x=\frac{d \psi}{ds}(2x)$ and

$u_y=\frac{d \psi}{ds}(2y)$

Now

$yu_x+xu_y=2xy\frac{d \psi}{ds}+2xy\frac{d \psi}{ds}=4xy\frac{d \psi}{dx} \ne 0$

meks08999 · #3 $Old$ October 25th, 2009, 08:04 PM

What do you mean by something is wrong?

TheEmptySet · #4 $Old$ October 25th, 2009, 08:16 PM

Quote:

Originally Posted by meks08999 $View Post$

What do you mean by something is wrong?

The preposed solution is not a solution to the PDE.

If you think about the problem geometrically

$yu_x+xu_y=0$ the left hand side is the directional derviative of the function.

This states the function is constant in the direction of the vector $y \vec{i} +x \vec{j}$

This tells us that $\frac{dy}{dx}=\frac{x}{y} \iff y^2=x^2+c \iff y^2-x^2=c$

This is a family of hyperbola's not circles as the problem asks......

Your PDE needs to have one of the terms negative for that to be a solution.

$u(x,y)=\psi(y^2-x^2)$ is a solution to the pde.

I.e if your problem read

$-yu_x+xu_y=0$ or $yu_x-xu_y=0$

$u(x,y)=\psi(y^2+x^2)$ would be a solution.

meks08999 · #5 $Old$ October 26th, 2009, 05:19 PM

There is an error in the problem. Thanks for pointing it out. I just received an email from my teacher just a little bit ago saying there was an incorrect sign.

It was supposed to be yu_x-xu_y=0.

TheEmptySet · #6 $Old$ October 26th, 2009, 06:00 PM

Quote:

Originally Posted by meks08999 $View Post$

There is an error in the problem. Thanks for pointing it out. I just received an email from my teacher just a little bit ago saying there was an incorrect sign.

It was supposed to be yu_x-xu_y=0.

To change variables you need to use the chain rule

$u(x,y)=u(r\cos(\theta),r\sin(\theta))$

Hint: Find $\frac{\partial u }{\partial \theta}$

Danny · #7 $Old$ October 27th, 2009, 07:07 AM

Use these chain rules

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

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