# Dirac function

1. Jan 7, 2013

Hi All,
I have a problem in understanding the concept of dirac delta function. Let say I have a function, q(r,z,t) and its defined as q(r,z,t)= δ(t)Q(r,z), where δ(t) is dirac delta function and Q(r,z) is just the spacial distribution.
My question are:
1. How can I find the time derivative of this function, that is, $\frac{\partial q(r,z,t)}{\partial t}$?
2. will hankel transformation of $\frac{\partial q(r,z,t)}{\partial t}$ be equal to zero (even when Q(r,z) $\neq$ 0)?

FM

2. Jan 7, 2013

### Simon Bridge

The delta function is actually a distribution, and is not differentiable in the classical sense. In order to consider such differentiation, we have to revert to generalized derivatives. This is done by assuming a certain level of differentiability on f and some vanishing conditions.
-- Kreizhn (post #2)

3. Jan 7, 2013

Hi Simon,
Thanks for your response. Unfortunately, I'm still not totally clear. Can you please be more explicity.
Once again, thank you.
FM

4. Jan 7, 2013

### Simon Bridge

5. Jan 8, 2013

Yes, I did, but I didn't fully grasp it. Anyway, this is what I can come up with, please take a look and let me know if it makes (physical) sense.
Definition: q(r,z,t)=δ(t)Q(r,z)
$\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) \frac{d}{dt}[δ(t)]$
$\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) δ^{'}(t)$
$\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) \frac{t}{t} δ^{'}(t)$
since: $x δ^{'}(x) = -δ(x)$
Hence,
$\frac{\partial q(r,z,t)}{\partial t} = -\frac{Q(r,z)}{t} δ(t)$