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General solution to PDE: f(x-y) = g(y-x) for arbitrary functions f and g?

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Quote:
    " PDE: ∂u/∂x + ∂u/∂y = 0
    The general solution is u(x,y) = f(x-y) where f is an arbitrary function.
    Alternatively, we can also say that the general solution is u(x,y) = g(y-x) where g is an arbitrary function. The two answers are equivalent since u(x,y) = g(y-x) = f[-(x-y)] "


    I don't see why the two different representations above [u(x,y) = f(x-y) and u(x,y) = g(y-x)] would describe exactly the SAME general solution. Why can we freely switch the order of x and y?
    Also, I don't understand why u(x,y) = g(y-x) = f[-(x-y)].

    2. Relevant equations
    N/A

    3. The attempt at a solution
    I was thinking of odd and even functions? But I don't think the quote is meant to restrict only to these special functions...


    Thanks for any help!:smile:
     
  2. jcsd
  3. Sep 15, 2009 #2

    LCKurtz

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    Maybe looking at an example will help you. Say you have the general solution f(x-y) and I claim the general solution is g(y-x). So you say, look, I can get exp(x-y) by taking f(x-y)=exp(x-y).

    I say, so can I. I will take g(y-x) = exp(-(y-x))

    It's because f(x-y) = f(-(y-x)) which I can call g(y-x). Clear as mud?
     
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