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Genralized functions

  1. Sep 8, 2007 #1
    hi
    can anyone tell me how could one go about to prove
    x* (delta)' ~ -delta

    where delta is the dirac delta funtion of x.
    ~ approximately equal

    delta' = first derivative of delta

    i know this can be done by using the concept of GENERALISED FUNTIONS.
    WHICH INVOLVES MUTLIPLYING THE LHS OF THE EQUATION WITH A GOOD FUNCTION and taking limits to infinity to get the RHS..
    I tried but there is something wrong.
    can anyone help me out?
     
  2. jcsd
  3. Sep 8, 2007 #2

    Hurkyl

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    What is your definition of the derivative of a generalized function?

    In the one I'm most familiar with, the definition of the derivative of a generalized function [itex]\varphi[/itex], its that it's the generalized function satisfying
    [tex]
    \int_{-\infty}^{+\infty} \varphi'(x) f(x) \, dx
    =
    -\int_{-\infty}^{+\infty} \varphi(x) f'(x) \, dx
    [/tex]
    for all test functions f.
     
  4. Sep 8, 2007 #3
    yes, that is the correct defination.
     
  5. Sep 8, 2007 #4

    Hurkyl

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    So, your identity
    [tex]x \delta'(x) = -\delta(x)[/tex]
    is true if and only if
    [tex]\int_{-\infty}^{+\infty} x \delta'(x) f(x) \, dx =
    \int_{-\infty}^{+\infty} (-\delta(x)) f(x) \, dx = -f(0)
    [/tex]
    for all test functions f, right? So what did you get when you used the definition of derivative?
     
  6. Sep 8, 2007 #5
    i expanded the LHS of your equation using PArts.
    where u = x(delta') and v = f
    and i got stuck.
    i am still not able to trace the steps which you did to land on to the expression which you posted in the previous step.
     
  7. Sep 8, 2007 #6

    matt grime

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    Isn't one of the important points in doing integration by parts that you choose something to be du/dx, and the other part to be v? Setting u= somthing and v=something seems lke you're not sure how to integrate by parts.

    There is clearly only one choice to make for du/dx....
     
  8. Sep 8, 2007 #7

    Hurkyl

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    I was simply restating what you needed to prove, not actually providing a proof.



    I'm not sure what you're trying to do with IBP; if you were setting [itex]u = x \delta'(x)[/itex] and [itex]v = f(x)[/itex], that would allow you to do something with
    [tex]\int u \, dv = \int f'(x) \delta'(x) x \, dx[/tex]
    but it wouldn't help at all when the integrand is [itex]\delta'(x) x f(x) \, dx[/itex].



    Did you try applying the definition of the derivative of a generalized function?
    (This is equivalent to formally doing integration by parts with a certain choice, but I find it easier to think about by treating it as a rule in of itself)
     
    Last edited: Sep 8, 2007
  9. Sep 8, 2007 #8
    yes i used the defination of the GF,
    taking u = delta'*f'; v = x
    and solving the integral gives me zero. something is going wrong..
     
  10. Sep 8, 2007 #9
    i was wondering can i use the following defination for delta fucntion
    delta = n/(pi)^.5 *exp (-(n*x)^2)
    having done this i can get my answer.
    but i do not know if this is correct?
     
  11. Sep 8, 2007 #10

    Hurkyl

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    Huh? There are no u's and v's in the definition I quoted...

    I agree with matt's assessment, though; you seem to have forgotten integration by parts. If you want to use integration by parts here, I really think it would be worthwhile to reopen your calculus textbook and do some integration by parts exercises before proceeding with this problem.
     
  12. Sep 8, 2007 #11
    i am not sure what is going wrong here..
    integral u(x)v(x) dx = u(x)integral v(x) - integral( (derivative u(x)) (integral v(x)) )
    where u(x) = something and v(x) = somethign else.

    but anyways, i solved my way through the problem, if only i can replace delta by the defination of gaussian functions delta= n/(pi)^.5 *exp (-(n*x)^2) where n--> infinity.
    i get my result
     
  13. Sep 8, 2007 #12

    Hurkyl

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    You use a different notation for IBP than I've ever seen. I'm used to:
    [tex]\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx[/tex]
    So your v is my v'.


    Anyways, it's clear why your attempts at integration by parts isn't working: you have no idea what delta''(x) is, so it doesn't help to differentiate delta'(x).
     
    Last edited: Sep 8, 2007
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