How to prove this derivative problem

1. Dec 7, 2011

Spartakhus

Given that

$\xi = x + y$ and $\eta = x - y$

How do I show that:

$\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} = \frac{\partial^2}{\partial \xi^2}+\frac{\partial^2}{\partial \eta^2}$

I know that

$\xi^2 + \eta^2 = 2(x^2 + y^2)$ and $\xi^2 - \eta^2 = 4xy$

But I do not know how to handle these derivatives :(

2. Dec 7, 2011

Ray Vickson

Use the chain rule to relate $\partial /\partial x,$ etc, to $\partial/ \partial \xi$ and $\partial / \partial \eta .$

RGV

3. Dec 7, 2011

Quinzio

There's a problem.
Suppose you have $f(\xi, \eta) = \xi ^2 + \eta^2$

4. Dec 7, 2011

Spartakhus

I know that dy/dx = dy/du*du/dx
But I don't know how to apply this in case of d/dx.

5. Dec 7, 2011

Quinzio

Sorry, I had to finish, but it seems it's not possible to edit your own messages.
Anyway:

$f(\xi,\eta)=\xi^2+\eta^2$

This is equal to
$f(x,y)=2x^2+2y^2$

Notice that $\frac{\partial^2 f}{\partial\xi^2}+\frac{\partial^2 f}{\partial\eta^2}=4$
while $\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=8$

so the thesis doesn't hold, there's something to fix, if I'm not wrong.

6. Dec 7, 2011

genericusrnme

$\frac{\partial}{\partial x} = \frac{\partial}{\partial \zeta}\frac{\partial \zeta}{\partial x} + \frac{\partial}{\partial \eta}\frac{\partial \eta}{\partial x}$

you following (I'm using zeta because I can't remember what that other squiggle is called)

$\frac{\partial \zeta}{\partial x} = 1$
$\frac{\partial \eta}{\partial x} = 1$

therefore

$\frac{\partial}{\partial x} = \frac{\partial}{\partial \zeta} + \frac{\partial}{\partial \eta}$

do that again, and do the same for y, you should pick up some negatives on the y side that will cancel out everything and leave you with what you're trying to get

7. Dec 7, 2011

Spartakhus

Thanks for the replies...