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Functional derivative

  1. Jun 25, 2010 #1
    For my current research, I need to prove the following:

    [tex]\int_0^1 \frac{dC(q(x) + k'(q'(x) - q(x)))}{dk'}\,dk' = \int_0^1 \int_L^U p(q(x) + k(q'(x) - q(x)))(q'(x)-q(x)) dx dk[/tex]

    where [tex]C(q(x)) = \int_0^1 \int_L^U p(kq(x)) q(x)\,dx\,dk [/tex]



    Here's what I've tried using the definition of functional derivative:

    [tex]
    \frac{\partial C(q(x))}{\partial q(x)}
    [/tex]

    [tex]
    = \lim_{\delta q(x) \rightarrow 0} \frac{C[q(x) + \delta q(x)] - C[q(x)]}{\delta q(x)}
    [/tex]

    [tex]
    = \int_L^U \int_0^1 \frac{\partial p(kq(x))}{\partial q(x)}q(x) + p(kq(x)) dk dx
    [/tex]

    My guess is that

    [tex]
    \frac{dC(q(x) + k'(q'(x) - q(x)))}{dk'} = \frac{\partial C(q(x) + k'(q'(x) - q(x)))}{\partial (q(x) + k'(q'(x) - q(x)))} \frac{d(q(x) + k'(q'(x) - q(x)))}{dk'}
    [/tex]

    but I'm not sure what to do next. Any help will be greatly appreciated!

    Thank you very much!
    Alice
     
    Last edited: Jun 25, 2010
  2. jcsd
  3. Aug 17, 2010 #2
    Hello, alicexigao. I can't really help you prove the identity. But I think it maybe able to evaluate to something more simple.

    [tex]\int _0^1\frac{d C(q(x)+k'(q'(x)-q(x)))}{d k'}d k'[/tex]

    [tex]=\int _0^1d C(q(x)+k'(q'(x)-q(x)))[/tex]

    [tex]=[ C(q(x)+k'(q'(x)-q(x)))]_0^1[/tex]

    [tex]=C(q(x)+(q'(x)-q(x)))-C(q(x))[/tex]

    [tex]=C(q'(x))-C(q(x))[/tex]

    Now, you can do the substitution and continue.
     
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