Multiple delta function

  • Thread starter fuwuchen
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  • #1
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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.
 
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Answers and Replies

  • #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)|.
 
  • #3
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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)|.


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.
 
  • #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)).
 
  • #5
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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)).

Yes, you are right!
Thanks again for your help, good luck with you!
 

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