1. ## PDE

Need help with this problem. Problem is attached.

Thanks

2. Originally Posted by meks08999
Need help with this problem. Problem is attached.

Thanks
Something is wrong......

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

taking some derivatives we get

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

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

Now

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

3. What do you mean by something is wrong?

4. Originally Posted by meks08999
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

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

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

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

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.

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

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

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

5. 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.

6. Originally Posted by meks08999
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

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

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

7. Use these chain rules

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

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