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Derivative of Dirac Delta function

  1. Jun 2, 2012 #1
    Hello I'm trying to figure out how to evaluate(in the distribution sense)
    [itex]\delta'(g(x))[/itex]. Where [itex]\delta(x)[/itex] is the dirac delta function. Please notice that what I want to evaluate is not [itex] \frac{d}{dx}(\delta(g(x)))[/itex] but the derivative of the delta function calculated in g(x).
    If anyone could post a proof, an idea to find the proof or a link it would be greatly appreciated!
     
  2. jcsd
  3. Jun 2, 2012 #2
    [itex] \delta' [/itex] is a linear operator on functions. Are you referring to its value when you pair it with g(x), as in
    $$ \langle \delta',\, g\rangle = \int_{\mathbb{R}} \delta'(x)g(x)\, dx \, ?$$
    I'll assume you are. In that case, you use the integral notation above and then symbolically do integration by parts. Don't worry if it is not a well-defined operation because the answer you get is literally the definition of what you want.

    Look here under distributional derivatives for more info:
    http://en.wikipedia.org/wiki/Dirac_delta_function
     
  4. Jun 2, 2012 #3
    No I'm sorry if I wasn't clear. I understand the value of
    [itex]\int \delta'(x)g(x)dx[/itex]
    I'm asking the value of
    [itex]\int \delta'(g(x))\phi(x)dx[/itex]
    Where [itex]\phi(x)[/itex] is the test function. Here it is not immediately obvious to me how to integrate by parts. I thought about this(but I am unsure of whether it is correct):
    Assume that g(x) is an invertible function with as many derivatives as necessary(to keep things simple for now), so we substitute [itex]y = g(x)[/itex] and get
    [itex]\int \delta'(y)\frac{\phi(g^{-1}(y))}{g'(g^{-1}(y))} dy[/itex] Now I could integrate by parts and get
    [itex]-\int \delta(y)\frac{d}{dy}(\frac{\phi(g^{-1}(y))}{g'(g^{-1}(y))}) dy[/itex]
    Do you think my reasoning is correct up to here?
     
  5. Jun 3, 2012 #4
    The dirac delta is just a normal distribution who's standard deviation approaches 0. Take the derivative of the normal dist. then take the limit as stdev =>0. I'm not sure if that's a valid way to do the problem, but its what I would try.
     
  6. Jun 3, 2012 #5
    Ok thanks, I'll try that
     
  7. Jun 3, 2012 #6
    Your calculation looks right to me, and what Aero said makes sense too. As far as proof goes, I can't remember exactly how general the rules are for changing variables like that. A book like Friedlander would probably have it...

    Check out the very last post here for a similar problem:
    https://www.physicsforums.com/showthread.php?t=201774&page=2
    There is no proof, but there is a citation.

    If g is not injective, then in the end, when you evaluate against a test function, you should get a sum of terms, one for each zero of g. If g' and g are simultaneously 0 at any point, then I don't think the distribution is well-defined.
     
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