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How to prove this derivative problem

  1. Dec 7, 2011 #1
    Given that

    [itex]\xi = x + y[/itex] and [itex]\eta = x - y[/itex]



    How do I show that:

    [itex]\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} = \frac{\partial^2}{\partial \xi^2}+\frac{\partial^2}{\partial \eta^2}[/itex]


    I know that

    [itex]\xi^2 + \eta^2 = 2(x^2 + y^2)[/itex] and [itex]\xi^2 - \eta^2 = 4xy[/itex]

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

    Sorry about this newbie question.
     
  2. jcsd
  3. Dec 7, 2011 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

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

    RGV
     
  4. Dec 7, 2011 #3
    There's a problem.
    Suppose you have [itex]f(\xi, \eta) = \xi ^2 + \eta^2[/itex]
     
  5. Dec 7, 2011 #4
    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.
     
  6. Dec 7, 2011 #5
    Sorry, I had to finish, but it seems it's not possible to edit your own messages.
    Anyway:

    [itex]f(\xi,\eta)=\xi^2+\eta^2[/itex]

    This is equal to
    [itex]f(x,y)=2x^2+2y^2[/itex]

    Notice that [itex]\frac{\partial^2 f}{\partial\xi^2}+\frac{\partial^2 f}{\partial\eta^2}=4[/itex]
    while [itex]\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=8[/itex]

    so the thesis doesn't hold, there's something to fix, if I'm not wrong.
     
  7. Dec 7, 2011 #6
    [itex]\frac{\partial}{\partial x} = \frac{\partial}{\partial \zeta}\frac{\partial \zeta}{\partial x} + \frac{\partial}{\partial \eta}\frac{\partial \eta}{\partial x}[/itex]

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

    [itex]\frac{\partial \zeta}{\partial x} = 1[/itex]
    [itex]\frac{\partial \eta}{\partial x} = 1[/itex]

    therefore

    [itex]\frac{\partial}{\partial x} = \frac{\partial}{\partial \zeta} + \frac{\partial}{\partial \eta}[/itex]

    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
     
  8. Dec 7, 2011 #7
    Thanks for the replies...
     
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