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Distributions: Convolution product

  1. May 17, 2009 #1
    So I have some problems and I tried to resolve them, I also have the results so I can check them but I'm curious if I made them good.

    P1: (H*δ)'=?, where H is the heavisede distrobution and δ is diracs distributin. So I tried liek this : <(H*δ)',φ>=-<H*δ,φ'>=-<δ,<H,φ'>>, <H,φ'>=∫φ'dx=φ => -<δ,φ>=<(H*δ)',φ> => (H*δ)'=δ. This is the result given so I think it's OK.

    P2:Here is my real problem, here I have doubts-unclearityes so it looks like this :
    F[δ*δ]=?, Where F[] is the Fourier transform. My solution: <F[δ*δ],φ>=<δ*δ,F[φ]>=
    <δ,<δ,F[φ]>, <δ,F[φ]>=F[φ](0)=∫φe^0dx=∫φdx=∫1φdx=<T1,φ> =>
    <δ,<δ,F[φ]>=<δ,<T1,φ>> So I by definition <δ,φ>=φ(0) so this is<T1,φ>(0) but I don't really know what tis means... does it mean <T1,φ>? Because if it means this then the solution given by the book is correct and is T1, but if it is not or even if it is could anyone tell me why is it like this ? Because in ∫φdx(0) I don't really see the reason.
     
  2. jcsd
  3. May 17, 2009 #2
    P1:

    Something weird is going on when you are removing the convolutions here:
    I'm not sure what you're doing, but on the face of it, it seems false. <H,φ'> is a number, but there it is used as if it is a function. There may be some subtle notation issue here but I don't see it. Could you please elaborate on what you are thinking of when you do this step?

    When I look at -<H*δ,φ'>, the part that immediately jumps out to me is that H*δ can be rewritten as just H (the delta function is the identity under convolution), and if you do that substitution then the result falls out.

    The same issue about convolutions as part a arises, so I don't really see what you're doing.

    The way I would go about attacking this is to use 2 facts. Fact 1: F[u*v] = C FF[v] where the constant C is (2pi)^(d/2) or whatever depending on your fourier transform convention. I think this "fact" is often even use as the definition for the convolution of distributions that have locally integrable fourier transforms. Fact 2: F[δ] = 1/C. Combining these 2 facts you can solve the problem.

    For intuition as to why F[δ] = const: the Fourier transform takes functions bunched up at zero and spreads them out, and takes functions that are spread out and bunches them up at zero, while keeping the norm the same. The happy medium is a gaussian of standard width, whose fourier transform is itself. For skinnier and skinnier functions, their Fourier transform get wider and wider. For the most ultimately skinny "function" - the delta distribution - the Fourier transform is the widest function - a constant. This result can be proven rigorously using the definitions (<F[δ],v>=<δ,F[v]>=F[v](0)= 1/C int v = <1/C,v> for v in the Schwarz space)
     
    Last edited: May 17, 2009
  4. May 17, 2009 #3
    -<H*δ,φ'>=-<δ,<H,φ'>> this is the definiton of the convolution product in a distribution, as I know( www.emis.de/journals/NSJOM/Papers/23_1/NSJOM_23_1_013_027.pdf ). But you are talking abot F[δ] which I did not write anywhere. As you see my problem is that i can't compute <δ,<T1,φ>> , where as you could figure out T1<=>f(x)=1 .
     
  5. May 17, 2009 #4
    Oh I see. They are somewhat abusing the notation in the document. They use the shorthand notation:

    "<(f*g)(x),φ(x)>" = ∫ f(y) ∫ g(x)φ(x+y) dxdy := "<f(y),<g(x),φ(x+y)>>"

    I really don't like the way they are doing it since it is easy to get confused about what is a function and what is a number, and what is the argument of a function vs a parameter. But OK, lets do it their way. Then the step in question is:

    <δ(y),<T1(x),φ(x+y)>> = <δ(y), ∫ T1(x)φ(x+y) dx> = (∫ φ(x+y) dx)(0) = ∫φ(x+0) dx = <T1,φ>
     
  6. May 18, 2009 #5
    Thank you very much.
     
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