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Homework Help: Derivative of dirac delta function

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data

    show

    [tex]x\frac{d}{dx}\delta(x)=-\delta)(x)[/tex]

    using the gaussian delta sequence ([tex]\delta_n[/tex]) and treating [tex]\delta(x)[/tex] and its derivative as in eq. 1.151.

    2. Relevant equations
    the gaussian delta sequence given in the book is
    [tex]\delta_n=\frac{n}{\sqrt{\pi}}e^{-n^2x^2}[/tex]

    and eq 1.151 is just part of the definition of the delta function:
    [tex]f(0)=\displaystyle\int_{-\infty}^{\infty}f(x)\delta(x)dx[/tex]


    3. The attempt at a solution

    thus far, I have tried substitution the derivative of [tex]\delta_n(x)[/tex] for the derivative of the delta function, and then taking the limit as n goes to infinity, but that got me nowhere. I have also tried integrating both sides to see where it got me, but that was nowhere useful. The problem is I just don't understand how the derivative of the delta function works on its own.
     
  2. jcsd
  3. Jan 25, 2010 #2
    The delta function is actually a distribution, and is not differentiable in the classical sense. In order to consider such differentiation, we have to revert to generalized derivatives. This is done by assuming a certain level of differentiability on f and some vanishing conditions.

    If f is a smooth function with compact support on a set D, the generalized derivative v' of a distribution v is any function w such that
    [tex] \int_D v' f dx = - \int_D w f' dx [/tex]
    Since the delta function is a distribution, it only truly makes sense to characterize its derivative under integration.
     
  4. Jan 25, 2010 #3
  5. Jan 25, 2010 #4
    Okay, I realize that what I gave you may not be entirely helpful since you have to use the Gaussian sequence.

    We know that
    [tex] \displaystyle \lim_{n\to\infty} \delta_n = \delta [/itex]
    and that it only makes sense to consider the delta function under the integral. So try evaluating
    [tex]\lim_{n\to\infty} \int_{-\infty}^\infty x \frac{d}{dx} \delta_n \ dx [/itex]
    then show that the value you get is equivalent to
    [tex] \int_{-\infty}^\infty -\delta(x) \ dx [/tex]
     
  6. Jan 25, 2010 #5
    your first answer, and the fact that it only makes sense under integration, actually got me doing just what you suggested, so thanks. THanks for the links too, I was looking for good resources on the delta function. I think I have it now.
     
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