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Understanding the Dirac delta

  1. Apr 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Having trouble understanding dirac deltas, I understand what they look like and how you can express one (i.e. from the limiting case of a gaussian) but for the life of me I can't figure out why the results of some integrals featuring dirac deltas equate to what they do.

    2. Relevant equations

    N/A

    3. The attempt at a solution
    An example of one particularly baffling to me is the following (given [itex]u=x+a[/itex], and [itex]du = dx[/itex]):

    1) [itex]\int^{\infty}_{-\infty}\delta(x-u)f(u-a)du = f(x-a)[/itex],

    especially when trying to compare to the 'standard' definition of how the dirac delta behaves under an integral:

    2) [itex]\int^{\infty}_{-\infty}\delta(x-x_{0})f(x)dx = f(x_{0})[/itex],

    in my mind it looks like 1) should be equal to [itex]f(u) = f(x+a)[/itex] because I look at it as [itex]\int^{\infty}_{-\infty}\delta(x-u)f(x)dx[/itex] and compare to 2).
     
  2. jcsd
  3. Apr 12, 2015 #2

    Orodruin

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    Try letting ##g(u) = f(u-a)## and then apply (2) with ##x = u##.
     
  4. Apr 12, 2015 #3
    I don't follow? Why do I do this and also I don't understand, so I have ##\int^{\infty}_{\infty}\delta(x-u)g(u)du## but ##x=u## means I have ##\delta(0)##?
     
  5. Apr 12, 2015 #4

    BruceW

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    I think Orodruin just meant equation (2), but replace x with u, i.e.
    [tex]\int_\infty^\infty \delta (u-u_0) g(u) du = g(u_0)[/tex]
    now just use ##g(u)=f(u-a)## and continue from there.
     
  6. Apr 13, 2015 #5
    ##\int_\infty^\infty \delta(u-u_0)f(u-a)du = f(u_0 - a)## I still don't understand? So ##\int_\infty^\infty \delta(x-u)f(x-a)dx = f(-u-a) = f(-x-2a)##

    I think I'm seriously missing something :S
     
  7. Apr 13, 2015 #6

    Orodruin

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    Yes, you are doing the wrong replacements in the functions. When you have a ##\delta(x-u)## and integrate with respect to ##x##, you need to replace all occurrences of ##x## in the other functions by ##u##, not by ##-u## as you did here.

    Also, when you integrate with respect to ##x##, the result cannot possibly depend on ##x##. Keep track of what is your integration variables and what are free variables!
     
  8. Apr 13, 2015 #7

    BruceW

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    This is the answer, but with ##u_0## replaced with ##x##.

    Uh wait... I'm re-reading your first post. You say given ##u=x+a##, you want to find
    [tex]\int^{\infty}_{-\infty}\delta(x-u)f(u-a)du[/tex]
    But ##\delta(x-u)= \delta(-a)## So this will always be zero (assuming ##a## is some nonzero constant). So equation 1) will just be equal to zero if you want to impose the constraint ##u=x+a##. So I'm guessing you don't want to impose this constraint really, or the answer is not very interesting...
     
  9. Apr 13, 2015 #8
    Is this because, ##\delta(x-u)## is a delta function centered on u? Is it sort of like an orthonormal function where if we turn the integral into a sum along ##x## then the only place where ##\delta(x-u)dx## is non-zero is at ##x=u##? So if we're essentially summing over ##\delta(x-u)g(u)## then you have ##0\times g(u)## for all ##u\neq x##? Then you get ##1\times g(x)##? So the integral equals ##g(x)##?

    I understand it in this context - graphically, is this the only 'motivation' behind replacing blah with blahblah? I thought I was missing something analytic.

    No I'm sorry I shouldn't have included ##u=x+a## that was used in the solutions to get to integral 1) and means nothing in the context of this question I suppose. Please see my above query to Orodruin regarding replacing things :)
     
  10. Apr 13, 2015 #9

    Orodruin

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    Yes, this is essentially the way to understand it. In a more formal setting (distribution theory), the ##\delta## is simply defined such that
    $$
    \int_{-\infty}^{\infty} \delta(x) \varphi(x) dx \equiv \varphi(0),
    $$
    where ##\varphi## is a "nice" function (infinitely differentiable with compact support).
    You can obtain the results for a delta offset from zero by a change of variables.

    Edit: Just to mention, you do not get ##1 \times g##, you informally get ##\infty \times g##, but the infinity is only in one point and integrates to one. It therefore plays the same role as the Kronecker delta in countable sums.
     
  11. Apr 13, 2015 #10
    Apologies that's what I meant, that the 'area' is then just 1 scaled by the 'area' of ##g(u)##. Thanks for all your help guys!
     
  12. Apr 13, 2015 #11

    BruceW

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    good work! :) It's good to really think this stuff through.
     
  13. Apr 13, 2015 #12
    Seriously, thanks for the help! I've got another question I've just posted over in advanced physics on the taylor expansion if you fancy it! :)
     
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