Understanding Multiple Delta Function in 1D and Multidimensional Spaces

[fuwuchen]

Hi everyone,
I have trouble understanding the multiple delta function. For one dimensional delta function, we have
δ($\varphi$(x))=$\sum_{i=1}^{N}$δ(x−xi)|$\varphi$′(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.

 

[Bill_K]

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 = ξ_i(y), and you have δ(f(x,y)) = ∑ δ(x - ξ_i(y)) |[∂f(x,y)/∂x]_{x = ξi(y)}|.

 

[fuwuchen]

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 = ξ_i(y), and you have δ(f(x,y)) = ∑ δ(x - ξ_i(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.

 

[Bill_K]

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)).

 

[fuwuchen]

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!