Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Chain rule with functional derivative

  1. Feb 25, 2008 #1
    This is supposedly the chain rule with functional derivative:

    [tex]
    \frac{\delta F}{\delta\psi(x)} = \int dy\; \frac{\delta F}{\delta\phi(y)}\frac{\delta\phi(y)}{\delta\psi(x)}
    [/tex]

    I have difficulty understanding what everything in this identity means. The functional derivative is usually a derivative of a functional with respect to some function, like in the term

    [tex]
    \frac{\delta F}{\delta\phi(y)} := \lim_{\epsilon\to 0} \frac{1}{\epsilon}\big( F(\phi + \epsilon \delta_y) - F(\phi)\big),
    [/tex]

    but isn't the term

    [tex]
    \frac{\delta\phi(y)}{\delta\psi(x)} := ?
    [/tex]

    now a derivative of a function with respect to another function? :confused:
     
  2. jcsd
  3. Feb 26, 2008 #2
    It could be I understood this. If [itex]\phi[/itex] depends on [itex]\psi[/itex] somehow, that means that every [itex]\psi[/itex] can be mapped into a set of functions [itex]\{\phi\}[/itex], [itex]\psi\mapsto\phi_{\psi}[/itex], then with a fixed y there's the natural functional G, [itex]G(\psi) = \phi_{\psi}(y)[/itex].
     
    Last edited: Feb 26, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Chain rule with functional derivative
Loading...