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Rigorousness of this Dirac delta formula

  1. Jun 7, 2015 #1

    ShayanJ

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    Is the following equation mathematically rigorous? How can you tell?

    ## \int_{-\infty}^\infty \delta(x-a) \delta(x-b) dx=\delta(a-b)##

    Thanks
     
  2. jcsd
  3. Jun 8, 2015 #2

    Svein

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    That's way outside my specialty, but:
    1. [itex]\delta (x-u) = 0 [/itex] for x ≠ u
    2. [itex]\int_{u-\epsilon}^{u+\epsilon}\delta(x-u)dx=1 [/itex] for any ε>0
    3. Thus, if a≠b, [itex]\int_{a-\epsilon}^{a+\epsilon}\delta(x-a)\delta(x-b)dx=0\cdot \int_{a-\epsilon}^{a+\epsilon}\delta(x-a)dx [/itex] for ε<|b-a|
    4. Symmetrical argument for (x-b)
    5. Thus, if a≠b, the postulate is correct.
    6. Assume a=b. Now the expression becomes [itex]\int_{-\infty}^{\infty}(\delta(x-a))^{2}dx [/itex].
    7. Since [itex]\delta (x-a) = 0 [/itex] for x ≠ a, obviously [itex](\delta (x-a))^{2} = 0 [/itex] for x ≠ a.
    8. Now for the hard part: What about [itex] \int_{a-\epsilon}^{a+\epsilon}(\delta(x-a))^{2}dx[/itex]? Is it equal to [itex] \int_{a-\epsilon}^{a+\epsilon}\delta(x-a)dx[/itex]?
    Anybody want to complete this?
     
  4. Jun 8, 2015 #3

    ShayanJ

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    Oh...I fount a proof myself. Its actually very straightforward. Just substitute the equation ## \delta(x-a)=\frac{1}{2\pi} \int_{-\infty}^\infty dk \ e^{ik(x-a)} ## in the integral and rearrange the resulting triple integral. Seems rigorous to me.
    Actually thinking about the Svein's step 8 led me to this. I don't know why I didn't think about it before that!

    EDIT:
    It seems to me step 8 should be proving that ## \int_{a-\epsilon}^{a+\epsilon} [\delta(x-a)]^2 dx=\delta(0)=\infty##!
     
    Last edited: Jun 8, 2015
  5. Jun 8, 2015 #4
    It is not rigorous. You cannot multiply distributions like that. Things like ##\int_{-\infty}^{+\infty} \delta(x)^2 dx## have no meaning in the classical sense of distributions. While it is possible to extend distribution theory to include this case, this is by no means trivial (and these extensions are typically not very well-behaved).

    Furthermore, something like ##\delta(a-b)## makes no sense outside an integral. And ##\delta(0) = +\infty## is most certainly incorrect.
     
  6. Jun 8, 2015 #5

    ShayanJ

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    What about the integral below?
    ## \int_{-\infty}^\infty f(a) \left[ \int_{-\infty}^\infty \ \left( \int_{-\infty}^\infty e^{ix(k-a)} \ dx \right) \left( \int_{-\infty}^\infty e^{iy(k-b)} \ dy \right) \ dk \right] \ da ##

    A rearrangement of the inner integrals gives ##\int_{-\infty}^\infty e^{ik(b-a)} \ dk##. I'm sure you're right about the product of distributions so it seems to me that either such a rearrangement or the equation ## \delta(x-a)=\int_{-\infty}^\infty e^{ik(x-a)} \ dk ## shouldn't be applicable here for some reason.
     
  7. Jun 8, 2015 #6

    Svein

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    I have thought about my last step, and I have come up with this:

    Since [itex]\int_{a-\epsilon}^{a+\epsilon}\delta(x-a) \delta(x-b)dx=0 [/itex] for |a-b|>ε, ε>0, we certainly have [itex]\int_{a-\frac{1}{n}}^{a+\frac{1}{n}}\delta(x-a) \delta(x-a-\frac{1}{n})dx=0 [/itex] for all n (just choose ε<1/2n). Thus [itex]\lim_{n\rightarrow\infty} \int_{a-\frac{1}{n}}^{a+\frac{1}{n}}\delta(x-a) \delta(x-a-\frac{1}{n})dx=0 [/itex] (just a tiny nagging doubt: Riemann integrals and limits are not necessary compatible).
     
  8. Jun 8, 2015 #7
    The more nagging doubt should be that ##\delta(x-a)\delta(x-b)## does not exist in the usual theory of distributions.
     
  9. Jun 8, 2015 #8

    Svein

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    But - I know that it is possible to construct a function ƒ∈C, ƒ(x)∈[0, 1] with the following specification: Given real numbers a and b with a<b, ƒ(x)=0 for x≤a and ƒ(x)=1 for x≥b. And, of course, ƒ'(x)=0 for x<a and for x>b. Now, obviously ƒ'(x) will approach δ(x-a) as b→a.
    Since ƒ is a function, multiplying two instances of ƒ' is allowable and my argument in post #6 holds with ƒa (=0 for x≤a, =1 for x≥a+(b-a)/3) and ƒb (=0 for x≤b-(b-a)/3, =1 for x≥b).
     
  10. Jun 8, 2015 #9

    jasonRF

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    Of course I agree that the product of distributions is not defined int he usual theory of distributions, but the convolution of two distributions is (with some requirements on the supports of the distributions-compact support makes things work out nice I think). However, I thinkn that the intent of what the OP wrote is correct - if you convolve a delta function with a delta function you get another delta function:
    ## \delta_{x_1} \ast \delta_{x_2} = \delta_{x_1+x_2)} ## (or in usual engineering notation, ##\delta(x-x_1)\ast\delta(x-x_2)=\delta(x-x_1-x_2)##).

    The integral representation of the convolution of two deltas is of course inspired by the definition for nice functions: ## \left( f \ast g \right) (x) = \int_{-\infty}^{\infty} dy \, f(y) g(x-y) ##. Engineers will usually denote it ## f(x) \ast g(x) ##.

    So while the original equation in the post may not be rigorous, if interpreted as a symbolic representation of the convolution it will work out fine in applications.

    jason
     
  11. Jun 14, 2015 #10
    δ(x) = limε→0 1/√ε exp(-x2/ε)

    ∫δ(x-a)δ(x-b)dx = limε→0 1/ε ∫exp(-(x-a)2/ε-(x-b)2/ε) dx

    = limε→0 1/ε ∫exp(-2x2/ε+2(a+b)x/ε-a2/ε-b2/ε)dx

    = limε→0 1/√ε exp(-(a-b)2/ε + O(ε) )
    = δ(a-b)
     
  12. Jun 14, 2015 #11

    pwsnafu

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    Dirac is an equivalence class of said functions (specifically delta sequences), so you need to demonstrate your result is independent of the choice of the function. So no that's not rigorous either.
     
    Last edited: Jun 14, 2015
  13. Jun 15, 2015 #12
    Why does that matter? You might say sin(x) is the equivalent class of all series definitions of sin(x). As long as one series converges you don't need to use the rest of them.
     
  14. Jun 15, 2015 #13

    pwsnafu

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    There are results that hold for only specific choices of delta sequences. The most important example is ##\delta^2##. Certain choices of delta sequences converge, other choices do not.

    A more general theory is the sequential intrinstic product of Schwartz distributions. The idea is simple, if ##\phi## and ##\psi## are distributions then define ##\phi \cdot \psi := \lim_{n \to \infty} \phi (\psi * \delta_n)##. It turns out that shrinking the set of allowable delta sequences allows you to multiply a larger set of distributions.
     
    Last edited: Jun 15, 2015
  15. Jun 29, 2015 #14

    Ben Niehoff

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    Remember that an integral is a limit, and you can't go interchanging the order of limits willy-nilly. Especially when, as in your case, they do not converge.
     
  16. Jun 29, 2015 #15
    Hello! I'm interested in learning something about distributions and how to give rigorous meaning to the Dirac function. Does anyone know about any good textbook?
     
  17. Jun 29, 2015 #16
    What math do you know?
     
  18. Jun 29, 2015 #17
    I have some knowledge in single and multivariable analysis, and a little bit of Lebesgue measure theory in [itex]\mathbb{R}^n[\itex]. For next year I'll be taking some probability and statistics theory subjects and maybe functional analysis.
     
  19. Jun 29, 2015 #18
    The notions of distributions depend quite a bit on functional analysis. So you should probably study that first. You don't want the books which don't assume functional analysis (such as Strichartz).
     
  20. Jul 11, 2015 #19
    Read Lighthill's Introduction to Fourier Analysis and Generalized Functions
    Make sure to review Taylor Series and the Remainder theorem
     
  21. Jul 11, 2015 #20

    jasonRF

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    Does Lighthill discuss convolution at all? I don't remember reading about it there, and I can't find it in the index or a quick perusal, but perhaps it is in there somewhere.
     
  22. Jul 12, 2015 #21
    δ∫δ(x-a)*δ(x-b) dx = δ(a-b), this is only true if for a≠ b, the integral is zero (and a blow for a = b but focus on that later), define δ(x-a) to be defined from a-τ to a+τ and have a value 1/2τ (for simplicity), now consider δ(x-a)δ(x-b), for a very small τ ]a-τ,a+τ[ ∩ ]b-τ,b+τ[ = ∅ (because we can pick a τ where a+τ<=b-τ for a<b which yield to τ<(b-a)/2 ), so for a ≠ b and taking lim τ -> 0, it's relevant that δ(x-a)*δ(x-b) vanish and so does the integral, if we pick now a = b, the problem reduces to ∫δ(x-a)δ(x-a) dx, which the previous definition this is a function defined on ]a-τ,a+τ[ which a value of 1/4τ2, so the integral is lim τ-> 0 2*τ/4τ2 which surely blows to infinity, whith these 2 behaviour one might say Oh, ∫δ(x-a)δ(x-a) dx = δ(a-a) = δ(0), because it blows to infinity and ∫δ(x-a)*δ(x-b) dx = 0 for a ≠ b, so ∫δ(x-a)*δ(x-b)*dx = δ(a-b)
     
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