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Starting with the definition of the Dirac delta function,

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Starting with the definition of the Dirac delta function, show that [itex] \delta( \sqrt{x}) [/itex]


    um... i have looked in my book and looked online for a problem like this and i really have no clue where to start. the only time i have used the dirac delta function is in an integral with another function and never with it in this form, only like delta(x-a).


    looking for guidance.
     
  2. jcsd
  3. Mar 4, 2009 #2

    Mark44

    Staff: Mentor

    Show that [itex]\delta( \sqrt{x}) [/itex] is what?
     
  4. Mar 4, 2009 #3
    sorry about that.

    equal to 0
     
  5. Mar 4, 2009 #4

    Mark44

    Staff: Mentor

    How is the Dirac delta function defined? The problem asks you to start with this definition.
     
  6. Mar 4, 2009 #5
    from what i get out of the text. it is what replaces an inner product that vanishes if x doesn't equal x'.

    I also know that it's integral is unity.
     
  7. Mar 4, 2009 #6

    Avodyne

    User Avatar
    Science Advisor

    [tex]\int_{-\infty}^{+\infty}dx\,f(x)\delta(x)=f(0)[/tex]

    Now, if only the argument of the delta function was [itex]\sqrt{z}[/itex] instead of [itex]x[/itex] ... hmmm, how could we make this happen?
     
  8. Mar 4, 2009 #7
    let x= [itex] \sqrt{z} [/itex] then dx=(1/2)z^(-1/2)dz

    and we get [itex] \int_{-\infty}^{\infty}dz z^{-1/2} f(z^{-1/2})\delta(z^{-1/2}) [/itex]
     
  9. Mar 4, 2009 #8
    is this a valid assumption

    treat [itex] \delta(\sqrt{x}) [/itex] as [itex] \delta(\sqrt{x}-0) [/itex]
     
  10. Mar 4, 2009 #9

    Hurkyl

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    Staff Emeritus
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    Gold Member

    Aside: I know the hand-wavy argument is clear, but does anyone know of a source that actually defines the result of composing a distribution with a function of some sort?
     
  11. Mar 5, 2009 #10
    so assuming i made the substitution right and that was what i was supposed to do, i still dont see how i proved that it equals zero.

    is that supposed to say if i plug in zero for x i get zero?

    and even if i treat [itex]
    \delta(\sqrt{x})
    [/itex] as [itex]
    \delta(\sqrt{x}-0)
    [/itex]
    i still don't have a function to plug zero into.

    i am still quite confused about what i am trying to do.
     
  12. Mar 5, 2009 #11

    George Jones

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    Section 7.4.d Composition of [itex]\delta[/itex] with a function, from Mathematics for Physics and Physicists by Walter Appel.

    The idea, as usual, is to to use the distribution obtained by integrating a locally integrable function against test functions to motivate a more general definition.

    1. Use locally integrable [itex]g[/itex] and integration to generate a distribution [itex]G[/itex].

    2. Use locally integrable [itex]g \circ f[/itex] and integration to generate a distribution denoted by [itex]G \circ f[/itex].

    3. If [itex]x[/itex] is the integration variable in 2., make the substitution [itex]y = f(x)[/itex].

    4. Relate [itex]G \circ f[/itex] to [itex]G[/itex].

    5. Use 4. to motivate the definition of [itex]T \circ f[/itex] in tems of an arbitrary distribution [itex]T[/itex] and differentiable and bijective function [itex]f[/itex].

    If I get time tomorrow, I might type in the details.
     
    Last edited: Mar 5, 2009
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