1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Derivative of dirac delta function
Loading...