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Derivative of ##\delta(t)##.

  1. Sep 25, 2015 #1

    Zondrina

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    Hello,

    I am wondering about the derivative of the Dirac delta distribution ##\delta(t)##. I know:

    $$\frac{d}{dt} u(t) = \delta(t)$$

    So what is ##\frac{d}{dt} \delta(t)##?

    How do we take the derivative of a distribution? I've heard about distributional derivatives, but I don't think any of those theorems apply here.
     
  2. jcsd
  3. Sep 25, 2015 #2

    WWGD

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    Last edited: Sep 25, 2015
  4. Sep 25, 2015 #3

    pwsnafu

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    The Dirac delta is the functional which takes a test function ##f(t)## and returns ##f(0)##. In order for this operation to be defined you need a test function which is defined at ##x=0## (ideally you also want continuity there as well).

    The derivative of the Dirac delta is the functional which takes a test function ##f(t)## and returns ##-f'(0)##. This requires the test function to be differentiable at zero.

    The derivative of a distribution is the distributional derivative.
     
  5. Sep 25, 2015 #4

    WWGD

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    Aren't test functions ## C^{\infty} ## ? If 0 is outside of the support, then isn't ##f'=0 ## ?
     
  6. Sep 25, 2015 #5

    pwsnafu

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    You can define test functions however you want. For example you can define discontinuous test function spaces. In the case of Dirac the minimal requirement is existence at zero. It effectively means that the Dirac distribution reduces to the Dirac measure in this case (note that you can't take distributional derivative anymore).
     
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