Genralized functions

1. Sep 8, 2007

aphrodasic

hi
can anyone tell me how could one go about to prove
x* (delta)' ~ -delta

where delta is the dirac delta funtion of x.
~ approximately equal

delta' = first derivative of delta

i know this can be done by using the concept of GENERALISED FUNTIONS.
WHICH INVOLVES MUTLIPLYING THE LHS OF THE EQUATION WITH A GOOD FUNCTION and taking limits to infinity to get the RHS..
I tried but there is something wrong.
can anyone help me out?

2. Sep 8, 2007

Hurkyl

Staff Emeritus
What is your definition of the derivative of a generalized function?

In the one I'm most familiar with, the definition of the derivative of a generalized function $\varphi$, its that it's the generalized function satisfying
$$\int_{-\infty}^{+\infty} \varphi'(x) f(x) \, dx = -\int_{-\infty}^{+\infty} \varphi(x) f'(x) \, dx$$
for all test functions f.

3. Sep 8, 2007

aphrodasic

yes, that is the correct defination.

4. Sep 8, 2007

Hurkyl

Staff Emeritus
$$x \delta'(x) = -\delta(x)$$
is true if and only if
$$\int_{-\infty}^{+\infty} x \delta'(x) f(x) \, dx = \int_{-\infty}^{+\infty} (-\delta(x)) f(x) \, dx = -f(0)$$
for all test functions f, right? So what did you get when you used the definition of derivative?

5. Sep 8, 2007

aphrodasic

i expanded the LHS of your equation using PArts.
where u = x(delta') and v = f
and i got stuck.
i am still not able to trace the steps which you did to land on to the expression which you posted in the previous step.

6. Sep 8, 2007

matt grime

Isn't one of the important points in doing integration by parts that you choose something to be du/dx, and the other part to be v? Setting u= somthing and v=something seems lke you're not sure how to integrate by parts.

There is clearly only one choice to make for du/dx....

7. Sep 8, 2007

Hurkyl

Staff Emeritus
I was simply restating what you needed to prove, not actually providing a proof.

I'm not sure what you're trying to do with IBP; if you were setting $u = x \delta'(x)$ and $v = f(x)$, that would allow you to do something with
$$\int u \, dv = \int f'(x) \delta'(x) x \, dx$$
but it wouldn't help at all when the integrand is $\delta'(x) x f(x) \, dx$.

Did you try applying the definition of the derivative of a generalized function?
(This is equivalent to formally doing integration by parts with a certain choice, but I find it easier to think about by treating it as a rule in of itself)

Last edited: Sep 8, 2007
8. Sep 8, 2007

aphrodasic

yes i used the defination of the GF,
taking u = delta'*f'; v = x
and solving the integral gives me zero. something is going wrong..

9. Sep 8, 2007

aphrodasic

i was wondering can i use the following defination for delta fucntion
delta = n/(pi)^.5 *exp (-(n*x)^2)
having done this i can get my answer.
but i do not know if this is correct?

10. Sep 8, 2007

Hurkyl

Staff Emeritus
Huh? There are no u's and v's in the definition I quoted...

I agree with matt's assessment, though; you seem to have forgotten integration by parts. If you want to use integration by parts here, I really think it would be worthwhile to reopen your calculus textbook and do some integration by parts exercises before proceeding with this problem.

11. Sep 8, 2007

aphrodasic

i am not sure what is going wrong here..
integral u(x)v(x) dx = u(x)integral v(x) - integral( (derivative u(x)) (integral v(x)) )
where u(x) = something and v(x) = somethign else.

but anyways, i solved my way through the problem, if only i can replace delta by the defination of gaussian functions delta= n/(pi)^.5 *exp (-(n*x)^2) where n--> infinity.
i get my result

12. Sep 8, 2007

Hurkyl

Staff Emeritus
You use a different notation for IBP than I've ever seen. I'm used to:
$$\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx$$
So your v is my v'.

Anyways, it's clear why your attempts at integration by parts isn't working: you have no idea what delta''(x) is, so it doesn't help to differentiate delta'(x).

Last edited: Sep 8, 2007