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Change of Variables Question with chain rule

  1. Apr 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider the function of two variables:
    u(x,y) = f(x-y) + g(x+(1/3)y)
    where f(s) and g(t) are each arbitrary functions of a single variable.

    Using the change of variables:
    s = x-y
    t = x-(1/3)y

    use the chain rule to determine the first and second derivatives of u with respect to x and y in terms of derivatives of f and g.

    Hence, show that the second derivatives satisfy
    u_xx = 2u_xy + 3u_yy

    where u_xx is the second derivative of u with respect to x, etc.

    3. The attempt at a solution
    My attempt, along with the original question paper, is attached as a PDF. It looks very fiddly but I have attempted the question a few times and still can’t satisfy the last equation.
     

    Attached Files:

    Last edited: Apr 10, 2015
  2. jcsd
  3. Apr 10, 2015 #2

    BvU

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    Gold Member

    Dear RY&T,

    It's not all that difficult to use the subscript button on the green menu bar. makes life a lot easier for you and us....

    Anyway, I'm not completely happy with ux = uf fs + ug gt
    uf and ug are trivially 1 and I propose to carry it a little further:
    ux = fs fx + gt gx
    where fx and gx are trivially 1 also, leaving ux = fs + gt
    (which incidentally may well be what you intended).

    Similarly, uy = fs fy + gt gy = - fs + gt/3

    The next step (on scan2 -- by the way: PF frowns on posting pictures when you can also type it out using the buttons -- or better: LaTeX) is a derailment:

    uxx is the derivative of fs + gt and that certainly is not the same as ux ux :H !

    Simply repeat what you did with u to get ux:

    uxx = ( fs + gt )x = ( fs )x + ( gt )x

    analogously with uyy and uxy (which I think is (uy)x -- someone correct me if I am wrong here -- too long ago :rolleyes: )

    - [edit] corrected myself -- see post #4
     
    Last edited: Apr 11, 2015
  4. Apr 10, 2015 #3
    I agree with you that uf and ug are trivially 1; however, with the uxx there is the derivative operator with respect to x acting on ux to give uxx. I wasn't multiplying ux by ux.
     
  5. Apr 11, 2015 #4

    BvU

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    Hi,

    I see what you mean. It just looked a lot like multiplying. Still a derailment to interpret ##\partial\over \partial x## as ##(f_s {\partial\over \partial f }+ g_t {\partial\over \partial g})##.

    The incomplete way to write ##u_x## avenges itself somewhat.

    Instead you want to differentiate ##u_x = u_f f_s s_x + u_g g_t t_x = f_s+g_t ## as ##(f_s)_x + (g_t)_x##. Write it out.

    [edit]Or, if you want to write the whole thing, e.g. for the first term:

    ##{\partial\over \partial x} u_f f_s s_x = u_{fx} f_s s_x+u_f f_{sx} s_x+u_f f_s s_{xx}##​

    And treat ##f_s## in ##(f_s)_x## the same way as you treated ##f## in ##(f)_x##


    :rolleyes: I have now also corrected myself in post #2. Really had to look it up: ##f_{xy} = (f_x)_y = \partial_{yx}f = {\partial^2f\over \partial y \partial x}={\partial\over \partial y}\left(\partial f\over \partial x\right )##

    -
     
    Last edited: Apr 11, 2015
  6. Apr 11, 2015 #5

    Mark44

    Staff: Mentor

    Yes, the notation is counterintuitive, with ##f_{xy}## being read one way and ##\frac{\partial}{\partial y}\left( \frac{\partial f}{\partial x} \right)## being read the other way. IOW, with ##f_{xy}##, you read this left to right, meaning that you differentiate wrt x first, then with respect to y.
     
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