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Proof involving Dirac Delta function

  1. Sep 14, 2012 #1
    Prove that
    [itex] x \frac{d}{dx} [\delta (x)] = -\delta (x) [/itex]
    this is problem 1.45 out of griffiths book by the way.
    2. Relevant equations
    I attempted to use integration by parts as suggest by griffiths using [itex] f = x , g' = \frac{d}{dx}[/itex]
    This yields [itex] x [\delta (x)] - \int \delta (x)dx[/itex]

    next I tried taking the derivative of both sides so that I would get my original
    [itex] x \frac{d}{dx} [\delta (x)] = x \frac{d}{dx}[\delta (x)] [/itex]

    and so I'm back where I started. I have also tried using a definite integral so that

    [itex]\int _{-\infty }^{\infty }x \frac{d}{dx}[\delta (x)] \text{dx} = \int_{-\infty }^{\infty } x[\delta (x)] \, dx - \int_{-\infty }^{\infty } [\delta (x)] \, dx [/itex]

    but we know that [itex]\int_{-\infty }^{\infty } x[\delta (x)] \, dx = 0 [/itex] so our equation simplifies.
    However this didn't get me any closer to solving the problem either.

    I also have the second part of the problem regarding the step function which is defined as
    [itex] \theta (x)
    \begin{array}{ll}
    \{ &
    \begin{array}{ll}
    1 & x>0 \\
    0 & x\leq 0 \\
    \end{array}
    \\
    \end{array}
    [/itex] Show that [itex] \frac{d}{dx}\theta (x) =\delta (x)[/itex]

    This something I can grasp by stating that

    [itex]
    \frac{\Delta \theta }{\text{$\Delta $x}}(x=0 )\longrightarrow 1/0, \text{ as } \text{$\Delta $x}\longrightarrow 0.[/itex]
    which is clearly an undefined(or infinite) slope, and at every other point on this graph the derivative is going to be 0. However I want to know if this is sufficiently rigorous, and if it isn't what a step in the right direction might be.
     
  2. jcsd
  3. Sep 14, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    For the first part you should integrate against a test function f(x). Show that integrating f(x)*x*δ'(x) is the same as integrating f(x)*(-δ(x)).
     
  4. Sep 16, 2012 #3
    Alright thanks I will try that. Can you give some sort of reasoning for why this method works to get the answer? It just seems strange to me to introduce a dummy function.
     
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