Need help with this problem. Problem is attached.

Thanks

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- Oct 25th 2009, 06:18 PM #1

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- Oct 25th 2009, 06:51 PM #2
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

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

- Oct 25th 2009, 07:04 PM #3

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- Oct 25th 2009, 07:16 PM #4
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.

I.e if your problem read

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

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

- Oct 26th 2009, 04:19 PM #5

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- Oct 26th 2009, 05:00 PM #6

- Oct 27th 2009, 06:07 AM #7