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Dirac delta function confusion

  1. Aug 19, 2007 #1

    radou

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    OK, I'm currently reading Hughes' Finite Element Method book, and I'm stuck on a chapter the goal of which is to prove that the Galerkin solution to a boundary value problem is exact at the nodes.

    So, the author first speaks about the Dirac delta function: "Let [itex]\delta_{y}(x) = \delta(x-y)[/itex] denote the Dirac delta function." Now, what exactle does this mean? Is it simply an operator, where [itex]\delta[/itex] can be any function, and y any real number?

    Further on, the author points out that, for a continuous function w on [0, 1], we write: "[itex](w, \delta_{y}) = \int_{0}^1 w(x)\delta(x-y)dx = w(y)[/itex]", so "[itex]\delta_{y}[/itex] sifts out the value of w at y". I don't understand where this result comes from.

    Any help is highly appreciated.
     
  2. jcsd
  3. Aug 19, 2007 #2
    I think that's the universal definition and exposition of the dirac-delta distribution in physics textbooks...I never seem to understand them. :frown: This is why physicists should never be allowed to create mathematics. :tongue2:. Sorry that I can't offer any help.
     
  4. Aug 19, 2007 #3

    arildno

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    Dearly Missed

  5. Aug 19, 2007 #4

    radou

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  6. Aug 19, 2007 #5

    matt grime

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    The definition of d(x) is that it is the generalized distribution with the propetry that d(x)=0 for x not equal to 0, and the integral of f(x)d(x)dx (apologies for the d's) equals f(0).

    It can be shown that formally, d(x) is 'the derivative of the Heaviside step function' and the limit in the distribution theoretic sense (whatever that means) of various functions.

    Anyway, the important thing is how it behaves, and that is when you integrate it it picks out the value of a function at 0 (and you can of course generalize this to make it pick out the value at y, as your d_y(x) does).
     
  7. Aug 19, 2007 #6
    On the other mathematicians can be critized too. A rigorous definition of a distribution as a mapping that maps functions to numbers is often unnecessarily abstract.

    My advice radou is that at least don't underestimate representations of delta function, that work like

    [tex]
    \lim_{n\to\infty} \int\limits_{-\infty}^{\infty} dx\; f(x)\delta_n(x) = f(0).
    [/tex]

    Where [itex]\delta_n[/itex] are some well defined functions. There are several cases where this is precisely what is needed, and the actual delta function is not necessary for the rigorous calculation.

    Another example. Assume that [itex]f(x)[/itex] approaches zero in the infinity. You can calculate quite easily

    [tex]
    \int\limits_{-\infty}^{\infty} dx\; \theta(x-x')\; Df(x) = -f(x'),
    [/tex]

    without ever using integration by parts and [itex]D\theta(x-x')=\delta(x-x')[/itex]. In fact, if rigorous meaning is wanted for the derivative of the step function, this calculation should be carried out first.

    (btw. I have never taken any course on distributions, and have not read specifically about them either. This is only the way I have learned to see them.)
     
  8. Aug 19, 2007 #7

    Hurkyl

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    I disagree; abstraction or not, it's what physicists already do. IMHO, it's a disservice to all to pretend otherwise.

    It's fine1 if you want to retain the integral notation, and focus on the fact that evey distribution is a limit of representable ones -- but please acknowledge the fact that you're doing functional calculus, not the stuff you learned in Calc I.




    1: although, IMHO, you mentally encumber yourself by doing so.
     
  9. Aug 22, 2007 #8
    I agree that there are cases where distributions come handy, but still defend my original point. It is an error to think, that always when a calculation can be carried out heuristically with delta function, the rigor way would go through distributions.

    For example, assume that f:R->R is continuous and vanishes at infinities, and you want to calculate

    [tex]
    D_x \int\limits_{-\infty}^{\infty} dy\; \theta(x-y) f(y) = f(x),
    [/tex]

    or that f:R^3->R vanishes at infinities and is smooth and you want to calculate

    [tex]
    \int d^3x\; \frac{x-x'}{|x-x'|^3} \cdot\nabla f(x) = -4\pi f(x'),
    [/tex]

    or that f:R->R is L_1 integrable and continuous and you want to calculate

    [tex]
    \lim_{L\to\infty} \int\limits_{-L}^{L}\frac{dk}{2\pi} \Big(\lim_{L'\to\infty}\int\limits_{-L'}^{L'} dx'\; f(x') e^{-ikx'} \Big) e^{ikx} = \cdots =\lim_{L\to\infty} \int\limits_{-\infty}^{\infty} dx'\; f(x') \int\limits_{-L}^{L} \frac{dk}{2\pi} e^{ik(x-x')} = f(x).
    [/tex]

    A mere rigorous definition of a distribution does not tell how to perform these calculations rigorously, and when they are calculated rigorously, distributions are useless.
     
    Last edited: Aug 22, 2007
  10. Aug 22, 2007 #9

    Hurkyl

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    If you take care to only write expressions that denote valid Lesbegue integrations of real functions, then you do not need to use distributions. (Note that that does not imply they are useless. One of the reasons to use distributions is that they make elementary calculus easier) But that is not the situation about which I speak. I am referring to situations where one writes an integrand with a delta in it, or one writes an equation like

    [tex]\frac{d}{dx} \left[ \int _{-\infty}^{+\infty} \theta(x - y) f(y) \, dy \right] = f(x)[/tex]

    where f is only known to be continuous and vanish at infinity -- note that, if you interpret this as Riemann integration, this identity is invalid. For example, it fails for [itex]f(t) = 2t / (t^2 + 1)[/itex] because the integral does not exist.

    OTOH, that identity is perfectly valid, given your hypotheses, if the integral and the derivative is meant in the distributional sense, as you can check directly by verifying that

    [tex]- \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \theta(x - y) f(y) g'(x) \, dx \, dy
    = \int_{-\infty}^{+\infty} f(x) g(x) \, dx[/tex]

    for any test function g. (Of course, the identity follows in a much easier way by using the calculus of distributions)


    (I put my dy on the right because I understand that grammar. I've never seen the lecture on how to properly parse the other convention)
     
    Last edited: Aug 22, 2007
  11. Aug 24, 2007 #10

    Haelfix

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    The only times physicists really ever get worried about distribution theory (outside of constructive field theory, where they are heavily utilized) is when they encounter a multiplication of two delta functions, which is clearly ill defined, even by our standards.

    It means you have to stop, go back 8 steps, pick up your old dusty tome on the subject and go through things rigorously and usually the mistake or the problem dissappears.
     
  12. Aug 25, 2007 #11

    arildno

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    Hmm..from what I know, multiplication of distributions IS ill-defined, and hence, distribution theory will be of no help whatsoever for the physicist wanting to multiply them.

    Instead, he'll need to step out of his world of idealized sampling functions, and to multiply sharply decaying functions instead, i.e, use a more precise modelling of the phenomena than he usually have to bother with.
     
  13. Aug 25, 2007 #12
    I guess the only resolution would be to have the product of two distributions depending on two different variables, a double integral kind of thing. But physicist manage to suffocate every clue about the real meaning of what they're doing using their fancy integral notation anyway.
     
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