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Multivariate Higher Order Derivatives

  1. Dec 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex]h(u,v) = f(u+v, u-v)[/itex]. Show that [itex]f_{xx} - f_{yy} = h_{uv}[/itex] and [itex]f_{xx} + f_{yy} = \frac12(h_{uu}+h_{vv}) [/itex].


    2. Relevant equations



    3. The attempt at a solution
    I'm always confused on how to tackle these types of questions because there isn't an actual function to differentiate.
    So I am assuming here that x = u+v and y = u-v and going from there. So I need to find the first partial with respect to x, which might be something like [itex]f_x = \frac{\partial f}{\partial(u+v)}[/itex] since I need to get it in terms of u and v, but this doesn't seem right. What sort of approach do I need to use on these questions?

    Edit: What I've done now is write [itex]h_{uv} = \frac{\partial}{\partial v}\frac{\partial h}{\partial u} = \frac{\partial}{\partial v} ( \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}) =\frac{\partial}{\partial v} (\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}) [/itex]. I'm having trouble converting from u's and v's to x's and y's.
     
    Last edited: Dec 3, 2013
  2. jcsd
  3. Dec 3, 2013 #2

    ShayanJ

    User Avatar
    Gold Member

    [itex]u=\frac{x+y}{2} [/itex] and [itex] v=\frac{x-y}{2} [/itex] so [itex] f(x,y)=h(\frac{x+y}{2},\frac{x-y}{2}) [/itex]

    So we have:

    [itex]
    f_x=h_u u_x+h_v v_x \Rightarrow f_x=\frac{1}{2}h_u+\frac{1}{2}h_v [/itex]

    [itex]
    f_{xx}=\frac{1}{2}(h_{uu}u_x+h_{uv}v_x+h_{vu}u_x+h_{vv}v_x)
    [/itex]

    I think you can continue yourself.
     
  4. Dec 3, 2013 #3

    Dick

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    Science Advisor
    Homework Helper

    For thinking about these things it's sometimes good to think about the operators without worrying about the functions. Define, for example, ##\partial_u(F)=\frac{\partial F}{\partial u}##. Then wouldn't it be true that ##\partial_u=\partial_x+\partial_y## and ##\partial_v=\partial_x-\partial_y## from the chain rule? It can save you a lot of texing and even spare some confusion. It's kind of the same as your underscore notation for partial derivatives.
     
    Last edited: Dec 3, 2013
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