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Variation of Dirac delta function

  1. Jun 4, 2012 #1
    Is it possible to take the variation of the Dirac delta function, by that I mean take the functional derivative of the Dirac delta function?
  2. jcsd
  3. Jun 4, 2012 #2
    yes you can (its laplace transform is s) and you can even take the derivative of this one although in practice i m not really sure how you can use it
  4. Oct 2, 2012 #3
    Could someone describe this procedure in math symbols? Or maybe point me to a link? Thanks.
  5. Oct 2, 2012 #4


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    Homework Helper

    I suppose in principle you could do it. I doubt it would be useful.

    $$\frac{\delta~\delta(t)}{\delta~\delta(t')} = \delta(t-t').$$

    $$\frac{\delta~\delta(t)}{\delta \eta(t')} = 0,$$
    where ##\eta(t')## is an arbitrary function not related to the dirac delta function.
  6. Oct 3, 2012 #5
    You ask how it would be useful. Let me give some context. We know
    [tex]\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1[/tex]
    Then suppose that the dirac delta is composed with some other function, f(x). We get
    [tex]\int_{ - \infty }^{ + \infty } {\delta (f(x))f'(x)dx} = 1[/tex]
    Now, we also know that the variation of a constant is zero,
    [tex]{\rm{\delta }}[{\rm{constant]}} = 0[/tex]
    So we should have,
    [tex]{\rm{\delta }}[\int_{ - \infty }^{ + \infty } {\delta (f(x))f'(x)dx} {\rm{]}} = {\rm{\delta }}[1{\rm{]}} = 0[/tex]
    But the variation of a definite integral is the definite integral of the variation. A variation is like differentiation which commutes with integration. So we have,
    [tex]{\rm{\delta }}[\int_{ - \infty }^{ + \infty } {\delta (f(x))f'(x)dx} {\rm{]}} = \int_{ - \infty }^{ + \infty } {{\rm{\delta }}[\delta (f(x))f'(x)]dx} [/tex]
    And in order for this to be identically zero, then the integrand must be zero, or
    [tex]{\rm{\delta }}[\delta (f(x))f'(x)] = 0[/tex]
    Thus the question about variations of dirac delta functions.

    I suppose you could use the chain rule to break this down to
    [tex]{\rm{\delta }}[\delta (f(x))]f'(x) + \delta (f(x)){\rm{\delta }}[f'(x)] = 0[/tex]
    So what's the next step? Is it true that
    [tex]{\rm{\delta }}[\delta (f(x))] = \frac{d}{{dx}}\{ \delta (f(x))\} \,{\rm{\delta }}x[/tex]
  7. Oct 9, 2012 #6
    As I understand it, if
    [tex]F\left[ {y\left( x \right),z\left( x \right)} \right][/tex]
    [tex]\delta F = \frac{{\partial F}}{{\partial y}}\delta y + \frac{{\partial F}}{{\partial z}}\delta z[/tex]
    So the variation of the Dirac delta of a function, S(x), would be
    [tex]\delta \delta (S(x)) = \frac{{\partial \delta (S(x))}}{{\partial S}}\delta S(x)[/tex]
    And from the wikipedia.com site

    [itex]x\frac{{d\delta (x)}}{{dx\,\,\,\,\,\,\,\,\,\,\,}} = - \delta (x).[/itex] OR [itex]\frac{{d\delta (x)}}{{dx\,\,\,\,\,\,\,\,\,\,\,}} = - \frac{{\delta (x)}}{x}[/itex]

    [tex]\frac{{\partial \delta (S(x))}}{{\partial S}} = - \frac{{\delta (S(x))}}{{S(x)}}[/tex]
    And since
    [tex]\delta S(x) = \frac{{dS(x)}}{{dx}}dx[/tex]
    We have,
    [tex]\delta \delta (S(x)) = \frac{{\partial \delta (S(x))}}{{\partial S}}\delta S(x) = - \frac{{\delta (S(x))}}{{S(x)}}\delta S(x) = - \frac{{\delta (S(x))}}{{S(x)}}\frac{{dS(x)}}{{dx\,\,\,\,\,\,\,\,\,}}dx[/tex]
    Does this all seem right so far? Or is there something special about the Dirac delta that these techniques don't apply? I don't work with the math of variation very often. So I'd appreciate some help. Thank you.
    Last edited: Oct 9, 2012
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