dirac function

femiadeyemi

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, [itex]\frac{\partial q(r,z,t)}{\partial t}[/itex]?
2. will hankel transformation of [itex]\frac{\partial q(r,z,t)}{\partial t}[/itex] be equal to zero (even when Q(r,z) [itex] \neq [/itex] 0)?

Thank you in advance.
FM

Simon Bridge

http://www.physicsforums.com/showthread.php?t=372548
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)

femiadeyemi

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

Simon Bridge

Did you read the link?

femiadeyemi

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)
[itex]\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) \frac{d}{dt}[δ(t)][/itex]
[itex]\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) δ^{'}(t)[/itex]
[itex]\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) \frac{t}{t} δ^{'}(t)[/itex]
since: [itex]x δ^{'}(x) = -δ(x) [/itex]
Hence,
[itex]\frac{\partial q(r,z,t)}{\partial t} = -\frac{Q(r,z)}{t} δ(t)[/itex]
Thank you for your help
FM

HallsofIvy

Did you understand it well enough to grasp what a "distribution" or "generalized function" is?