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The chain rule for 2nd+ order partial differential equations

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    w= f(x,y)
    x = u + v Verify that Wxx - Wyy = Wuv
    y = u - v


    2. Relevant equations



    3. The attempt at a solution

    I know how to find Wu or Wv but I have no idea on how to proceed to find the 2nd order derivative (or 3rd,4rth etc.. obviously). I assume this somehow uses the chain rule but I had no idea on how to apply it in this situation. Thank you!
     
  2. jcsd
  3. Nov 1, 2011 #2

    HallsofIvy

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    I presume that your "w" and "W" are the same.

    Since W(x,y)= f(x,y), x=u+ v, and y= u- v,
    [tex]W_u= \frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial v}= f_x(1)+ f_y(1)= f_x+ f_y[/tex]

    [tex]W_v= \frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial v}= f_x(1)+ f_y(-1)= f_x- f_y[/tex]

    You already have that, right?

    Then [itex]W_{uu}= (W_u)_u= (f_x+ f_y)_u[/itex]
    [tex]= \left(\frac{\partial f_x}{\partial x}\frac{\partial x}{\partial u}+ \frac{\partial f_x}{\partial y}\frac{\partial y}{\partial u}\right)+ \left(\frac{\partial f_y}{\partial x}\frac{\partial x}{\partial u}+ \frac{\partial f_y}{\partial y}\frac{\partial y}{\partial u}\right)[/tex]
    [tex]= \left(f_{xx}(1)+ f_{xy}(1)\right)+\left(f_{xy}(1)+ f_{yy}(1)\right)= f_{xx}+ 2f_{xy}+ f_{yy}[/tex]

    Can you try the others, [itex]W_{uv}[/itex] and [itex]W_{vv}[/itex] yourself?
     
  4. Nov 1, 2011 #3
    Wow! Thanks for that helpful explanation! I get it now !
     
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