Proving the Delta Function Identity Using the Local Behavior of Functions

[jumbogala]

Homework Statement

See http://mathworld.wolfram.com/DeltaFunction.html

I want to show (6) on that page. I can show it using (7), but we aren't supposed to do that. I already proved (5), and my prof says to use the fact that (5) is true to get the answer.

Homework Equations

The Attempt at a Solution

Here's what I tried:
δ(x² - a²) = δ((x-a)(x+a))

I'm not sure how to use (5), because here a is not multiplying x. I'm not sure where to go from here.

 

[sunjin09]

Imagine what the delta look like in the neighborhood of a and -a, i.e., when one factor goes to zero, the other factor is pretty much constant over that entire neighborhood.

 

[jumbogala]

I don't really know what it looks like. I know that δ(x) is zero everywhere except at x = 0. At x = 0, it's infinity.

I know that δ(x-a) is the same as above except that now it's infinity at x = a.

But I don't know what δ(x²) looks like.

 

[Dick]

I don't really know what it looks like. I know that δ(x) is zero everywhere except at x = 0. At x = 0, it's infinity.

I know that δ(x-a) is the same as above except that now it's infinity at x = a.

But I don't know what δ(x²) looks like.

Near x=a, δ((x-a)(x+a)) pretty much looks like δ((x-a)*2a). That's sunjin09's point.

 

[jumbogala]

I don't understand why it looks like that though. I am having problems visualizing it.

I don't get how you know what it looks like unless it's just δ(x) or δ(x-a) by itself.

 

[Dick]

I don't understand why it looks like that though. I am having problems visualizing it.

I don't get how you know what it looks like unless it's just δ(x) or δ(x-a) by itself.

Near x=a, (x+a) is nearly 2a. You can't visualize that?

 

[jumbogala]

Ohh okay, I see that near x = a, (x+a) is about 2a. So we're just making an approximation and plugging it into the delta function, is that right?

I wasn't sure what the delta function itself looked like, not what x+a looks like.

 

[Dick]

Ohh okay, I see that near x = a, (x+a) is about 2a. So we're just making an approximation and plugging it into the delta function, is that right?

I wasn't sure what the delta function itself looked like, not what x+a looks like.

Yes, I think you are ok with hand waving through this. Near x=(-a) the value of (x-a) is nearly -2a. So split it into two delta functions at the two values where x^2-a^2 vanishes.

 

[jumbogala]

Alright, that makes a lot more sense now. So basically, we're saying:

δ((x-a)(x+a)) = δ((x-a)*2a) + δ((x+a)*(-2a))

Is it okay to do that because it's zero elsewhere (within the delta function)?

 

[Dick]

Alright, that makes a lot more sense now. So basically, we're saying:

δ((x-a)(x+a)) = δ((x-a)*2a) + δ((x+a)*(-2a))

Is it okay to do that because it's zero elsewhere? It seems a little odd to split a multiplication up like that.

Yes, I think it's ok to do that because it's zero elsewhere. It's not a formal proof, but the answer is correct.

 

[sunjin09]

The idea is δ(f(x)) is zero except at f(x0)=0, so all that matters is the local behavior of f(x) near x0, so you can approximate f(x) around x0 by f(x)≈f'(x0)(x-x0). Since all the zeros of f(x) must be accounted for, you easily derive the general formula (7) mentioned in your original post. This is certainly not a formal proof, as Dick pointed out, but I think you can have a formal but still not rigorous proof by using a test function, i.e., try evaluate ∫ δ(f(x))*g(x) dx and see what you get.