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Homework Help: Simple Functional Derivatives

  1. Dec 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Hey, can I just check these functional derivatives?:

    1) [tex] \frac{\delta F[g]}{\delta g(y)}[/tex] where [tex] F[g] = \int dx \left[ \frac{1}{\sqrt{1+(g'(x))^2}} - 2g(x) + 5 \right]\>. [/tex]

    2) [tex] \frac{\delta F[a,b,g]}{\delta g(y)}[/tex] where [tex] F[a,b,g] = \int d^4x \left[ A(\partial_{\mu} g(x))a(x)b(x) + Bg^3(x) \right]\>. [/tex]

    2. Relevant equations

    [tex] \frac{\delta F[g]}{\delta g(y)} = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ F[g(x) +\epsilon\delta(x-y)] - F[g(x)] \right]\>. [/tex]

    3. The attempt at a solution

    The answers I get are 1) -2 and 2) 3Bg2(y).

    Thanks in advance!
    Last edited: Dec 18, 2012
  2. jcsd
  3. Dec 18, 2012 #2


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    Homework Helper

    How did you get your answers? I don't think they are correct. It looks like you have assumed that ##\delta g'(x)/\delta g(y) = 0##. This is not true; you need to use ##\delta g'(x)/\delta g(y) = \delta'(x-y)##, the derivative of the dirac delta function.
  4. Dec 18, 2012 #3
    Oh. Yes, I did assume that. I'll have another go.
    Last edited: Dec 18, 2012
  5. Dec 18, 2012 #4
    So, for (2),

    [tex] F[a,b,g] = \int d^4x \left[ A(\partial_{\mu} g(x))a(x)b(x) + Bg^3(x) \right]\>. [/tex]

    [tex] \implies \frac{\delta F[a,b,g]}{\delta g(y)} = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \int d^4x\>[A(\partial_{\mu}(g(x)+\epsilon \delta(x-y)))a(x)b(x) - A(\partial_{\mu}g(x))a(x)b(x) + B{(g(x)+\epsilon \delta(x-y))}^3-Bg^3(x)] \right][/tex]

    [tex] = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[\int d^4x\>A\epsilon\>\partial_{\mu}(\delta(x-y))\>a(x)b(x) + B(3g^2(x)\epsilon\delta(x-y)+3g(x)\epsilon^2\delta^2(x-y)+\epsilon^3\delta^3(x-y)) \right] [/tex]

    [tex] = \left[\int d^4x\>A\>\partial_{\mu}(\delta(x-y))\>a(x)b(x) + B(3g^2(x)\delta(x-y)) \right]\>. [/tex]

    Then apparently [tex] \int \delta'(x-y)f(x) = - \int \delta(x-y)f'(x) [/tex] so,

    [tex] \frac{\delta F[a,b,g]}{\delta g(y)} = 3Bg^2(y) - A\partial_{\mu}[a(y)b(y)]\>. [/tex]

    And similarly, (1) gives -2g''(y)-2?
    Last edited: Dec 18, 2012
  6. Dec 20, 2012 #5
    Does this seem correct? =/
  7. Dec 23, 2012 #6


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    Sorry for late reply. I've been travelling. Your (2) looks superficially good - the problem I see is that you have a ##\partial_\mu##, but there is no indication that the index is being summed over, so either you have a dangling index or you're missing another term with a ##\mu## index (which could change the result - e.g., if the derivative were really ##\partial^\mu\partial_\mu##, then this is a second derivative, so when you apply the dirac derivative rule twice the minus signs cancel out).

    Your (1) still looks incorrect - the first term should be ##-\frac{d}{dx}\left[g'(x)/(1+(g'(x))^2)^{3/2}\right]##, evaluated at x = y, which I don't think reduces to ##-2g''(y)##.
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