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Chain rule with functional derivative

  1. Jun 24, 2008 #1
    Given that

    [tex]F = \int{f[h(s),s]ds}[/tex]

    does

    [tex]\frac{\partial}{\partial h}ln(F)=\frac{1}{F}\frac{\delta F}{\delta h}=\frac{1}{F}\frac{\partial f}{\partial h}[/tex]

    ?
     
  2. jcsd
  3. Jun 24, 2008 #2

    Ben Niehoff

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  4. Jun 25, 2008 #3
    I don't think that means anything now. Shouldn't you be calculating

    [tex]
    \frac{\delta}{\delta h(s')} \textrm{ln}(F)
    [/tex]

    with some fixed [tex]s'[/tex]?
     
  5. Jun 25, 2008 #4

    If so, then is

    [tex]
    \frac{\delta}{\delta h(s')} \textrm{ln}(F) = \frac{1}{F[h(s')]}\frac{\delta F}{\delta h(s')}
    [/tex]

    ?


    Basically I'm not sure if the same kind of chain rule applies to functional derivatives as to regular ones. I haven't been able to find an example of a derivative of a function containing a functional like this natural log one.
     
  6. Jun 25, 2008 #5
    [tex]
    \frac{\delta}{\delta h(s')} \textrm{ln}(F[h]) = \frac{1}{F[h]}\frac{\delta F[h]}{\delta h(s')}
    [/tex]

    F depends only on the mapping h, not on variable s. It seems to be okey to leave the parameter [tex]h[/tex] (which is a mapping itself) out, and write [tex]F=F[h][/tex], but [tex]F[h(s')][/tex] would not make sense.

    It's like here. If you have [tex]f:\mathbb{R}^3\to\mathbb{R}[/tex], and [tex]x\in\mathbb{R}^3[/tex], then you can write [tex]f(x)[/tex], but [tex]f(x_3)[/tex] (from [tex]x=(x_1,x_2,x_3)[/tex]) would not make sense.

    I have the same problem. I have never encountered a good source on functional differentiation. I can merely calculate, which is often enough, though. If you keep your head clear about simple things like what are parameters for each function, there are not many alternatives left for calculation rules.
     
  7. Jun 25, 2008 #6
    Thanks
     
  8. Jun 25, 2008 #7

    Ben Niehoff

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    Couldn't one take the mathematical definition on the Wikipedia page I linked, and then prove whether the chain rule is valid or not? I think it would be analogous to proving the chain rule for ordinary derivatives via the limit definition of the derivative:

    [tex]\frac{df}{dx} = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}[/tex]

    I would have done this yesterday for you, but I have other stuff I need to do right now.
     
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