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Multiple delta function

  1. Sep 29, 2011 #1
    Hi everyone,
    I have trouble understanding the multiple delta function. For one dimensional delta function, we have
    δ([itex]\varphi[/itex](x))=[itex]\sum_{i=1}^{N}[/itex]δ(x−xi)|[itex]\varphi[/itex]′(xi)|
    where xi's (for i = 1 to N) are simple zeros of f(x) and it is known that f(x) has no zeros of multiplicitiy > 1

    but what is the case of multiple delta function
    δ(f(x,y))=?

    PS:This my first time to this forum, I'm not familiar with Latex. Sorry for the caused inconvenience.
     
    Last edited: Sep 29, 2011
  2. jcsd
  3. Sep 29, 2011 #2

    Bill_K

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    Well one thing you could do is to treat y as a parameter and f(x,y) as a function of one variable. Then the zeroes of f(x,y) occur at x = ξi(y), and you have δ(f(x,y)) = ∑ δ(x - ξi(y)) |[∂f(x,y)/∂x]x = ξi(y)|.
     
  4. Sep 30, 2011 #3

    Thank a lot for your help, Bill!!!

    But I wonder whether we can treat y as a parameter and have x = ξi(y), as in f(x,y) the two arguments x and y are independent to each other.
     
  5. Sep 30, 2011 #4

    Bill_K

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    Sure, that's perfectly legal. You will often see this approach used when dealing with the delta function on the light cone δ(x2 - c2t2). Instead of calling it a function of two variables f(x, t) = x2 - c2t2, we define a parameter a = ct and work with a function of one variable, δ(x2 - a2). By the one-dimensional rule this is equal to |1/2a| (δ(x - a) + δ(x + a)), which can then be written |1/2ct| (δ(x - ct) + δ(x + ct)).
     
  6. Oct 1, 2011 #5
    Yes, you are right!
    Thanks again for your help, good luck with you!
     
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