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Dirac function

  1. Jan 7, 2013 #1
    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.
  2. jcsd
  3. Jan 7, 2013 #2

    Simon Bridge

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    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)
  4. Jan 7, 2013 #3
    Hi Simon,
    Thanks for your response. Unfortunately, I'm still not totally clear. Can you please be more explicity.
    Once again, thank you.
  5. Jan 7, 2013 #4

    Simon Bridge

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    Did you read the link?
  6. Jan 8, 2013 #5
    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]
    [itex]\frac{\partial q(r,z,t)}{\partial t} = -\frac{Q(r,z)}{t} δ(t)[/itex]
    Thank you for your help
  7. Jan 8, 2013 #6


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