How to prove this derivative problem

[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 :(

Sorry about this newbie question.

 

[Ray Vickson]

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 :(

Sorry about this newbie question.

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

RGV

 

[Quinzio]

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

 

[Spartakhus]

Thanks for the reply. Could you write me an example or tell me where I can read about this?

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

 

[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.

 

[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

 

[Spartakhus]

Thanks for the replies...