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A simple application of dirac delta shift theorem help

  1. Feb 13, 2012 #1
    A "simple" application of dirac delta "shift theorem"...help

    1. The problem statement, all variables and given/known data

    show that for a, b, c, d positive:

    δ(a/b-c/d) = bdδ(ad-bc)

    2. Relevant equations

    ∫f(x)δ(x-a)dx = f(a)

    3. The attempt at a solution

    Ok so I start with

    ∫δ(a/b-c/d)f(x)dx

    But I am not sure how to apply the shift theorem. It seems I need to somehow relate a/b and x so that I can get it in the form of the shift theorem. But trying integration by substitution I always get tangled up. If I let u=a/b, then I can't relate dx to du to intergrate.

    If I just say let's call a/b as "x". then dx = what?

    ugh, this is a simple problem too. Seems like its an easy canditate for one of the first proofs shown after learning about the shift theorem so I feel pretty dumb that I'm not sure even where to start...
     
  2. jcsd
  3. Feb 13, 2012 #2

    lanedance

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    Re: A "simple" application of dirac delta "shift theorem"...help

    thats not quite the shift theorem but a integral prepty of the delta function, though maybe it is...

    I'm thinking you may want to consider a substitution, now first can you solve
    [tex]\int f(x)\delta(mx-c)dx [/tex]
     
    Last edited: Feb 13, 2012
  4. Feb 13, 2012 #3

    lanedance

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    Re: A "simple" application of dirac delta "shift theorem"...help

    then how about associating x with a, and then the something around the substitution ybd = xd-cd. I haven't totally nailed it but it feels close
     
  5. Feb 13, 2012 #4
    Re: A "simple" application of dirac delta "shift theorem"...help

    Let u= mx ; du = mdx

    (1/m)∫f(u/m)δ(u-c)du

    Zero everywhere except where u=c therefore ..

    sorry, I'm stuck
     
  6. Feb 13, 2012 #5

    lanedance

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    Re: A "simple" application of dirac delta "shift theorem"...help

    [tex] \int f(x)\delta(mx-c)dx [/tex]

    you pretty much has it, but I'd do as follows, let u = mx-c, then x=(u+c)/m, dx=du/m
    [tex] \int f(\frac{u+c}{m})\delta(u)\frac{1}{m}du [/tex]

    which gives
    [tex] \int f(x)\delta(mx-c)dx =\frac{1}{m} \int f(\frac{x+c}{m})\delta(u) du= \frac{1}{m} f(\frac{c}{m})[/tex]

    so you can see there are a few similarities in transformation to the original problem, it just working up the substitution correctly
     
  7. Feb 13, 2012 #6
    Re: A "simple" application of dirac delta "shift theorem"...help

    But my equation doesn't even have an x. So that's where I get tangled up.

    I'm missing something obvious.

    What I mean is, no matter what I substitute, I'm not doing to be able to determine what to replace for dx.
     
    Last edited: Feb 13, 2012
  8. Feb 13, 2012 #7

    lanedance

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    Re: A "simple" application of dirac delta "shift theorem"...help

    or how about letting x(a) = a/b-c/d, then y(a) = bdx(a) = ad-bc, and dy = bd.dx

    then yo uhave sometihing like
    [tex] \int \delta(bdy) dy= \int \delta(x)bd.dx
    [/tex]
     
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