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Math Help - Mixed derivative with change of variables

  1. #1
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    Mixed derivative with change of variables

    Hello,

    I'm wondering how I can show that for all functions which satisfy:
    \frac{\partial ^2 f}{\partial x^2} - \frac{d^2 f}{dy^2} = 0
    that the following is true:
    \frac{\partial ^2 f}{\partial s \partial t} = 0

    Where f(x,y) = f(x(s,t),y(s,t)) = f(s,t)

    Thanks very much for your help
    Jon
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  2. #2
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    Quote Originally Posted by jonmondalson View Post
    Hello,

    I'm wondering how I can show that for all functions which satisfy:
    \frac{\partial ^2 f}{\partial x^2} - \frac{d^2 f}{dy^2} = 0 I assume that those d's in the second term should be partials?
    that the following is true:
    \frac{\partial ^2 f}{\partial s \partial t} = 0

    Where f(x,y) = f(x(s,t),y(s,t)) = f(s,t)
    That result is not true. For example, if f(x,y) = xy, x(s,t)=s and y(s,t)=t then \frac{\partial ^2 f}{\partial x^2} = \frac{\partial^2 f}{\partial y^2} = 0 but \frac{\partial ^2 f}{\partial s \partial t} = 1.
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  3. #3
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    What is true is that if you let s= x+ y, t= s- y, then the equation becomes
    \frac{\partial^2 f}{\partial s\partial t}= 0
    and, from that, f(s, t)= F(s)+ G(t)= F(x+y)+ G(x- y) where F and G are any differentiable functions of a single variable.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    What is true is that if you let s= x+ y, t= s- y, then the equation becomes
    \frac{\partial^2 f}{\partial s\partial t}= 0
    and, from that, f(s, t)= F(s)+ G(t)= F(x+y)+ G(x- y) where F and G are any differentiable functions of a single variable.
    Thanks bud, but can I ask, how do you calculate \frac{\partial^2 f}{\partial s\partial t}, for example when s= x+ y and t= s- y, as in your example?
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