Heaviside function and dirac delta

  • Thread starter KateyLou
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  • #1
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Homework Statement



Hi there, i am trying to do a proof that H'(t)= δ(t)

Homework Equations



We have been given the following:

F is a smooth function such that lim (t-->±∞)F(t)=0
Therefore the integral between ±∞ of [H(t)F(t)]'=[H(t)F(t)]-∞=0

I understand it up until this point; however next it says:
"Integration by parts:
(1) = Integral between ±∞ of [H(f)'F(t)]dt
(2) = -the integral between ±∞ of H(t)F'(t)dt
(3) = -the integral between ∞ and 0 of F'(t)dt
(4) = [-F(t)]0
(5) = F(0)
(6) = Integral between ±∞ of δ(t)F(t)dt

The Attempt at a Solution



I dont know where they have got theequation from in (1) or (2) or (3)! I get 4 though and 5! Although i dont then get 6!

I think if i knew where (1) came from i maybe could get through the rest but i just dont know where it has come from?
 

Answers and Replies

  • #2
gabbagabbahey
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Homework Statement



Hi there, i am trying to do a proof that H'(t)= δ(t)

Homework Equations



We have been given the following:

F is a smooth function such that lim (t-->±∞)F(t)=0
Therefore the integral between ±∞ of [H(t)F(t)]'=[H(t)F(t)]-∞=0

Do you understand that this says [tex]\int_{-\infty}^{\infty} \left(\frac{d}{dt}(H(t)F(t))\right)dt=0[/tex]?

Since this is true for all functions F(t) such that lim (t-->±∞)F(t)=0, what must then be true of [tex]\frac{d}{dt}(H(t)F(t))[/tex]?

Equation (1) follows from using that along with the product rule :[tex]\frac{d}{dt}(H(t)F(t))=H'(t)F(t)+H(t)F'(t)[/tex]
 
  • #3
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Yes i think I get the =0 part (by the way how do you write equations like that?)
And thank you re: the product rule - not sure how i managed to miss that! will go back to it and see how far i get this time!
 
  • #4
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Right i have now completed this up to the last stage - i do not understand how it has gone from F(0) to [tex]\int_{-\infty}^{\infty} \left(\delta (t)F(t)\right)dt[/tex]
 
  • #5
HallsofIvy
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And that is what defines the delta "function" (I have put "function" in quotes because it is not, of course, a "function", it is a "distribution" or "generalized function").
 
  • #6
Hurkyl
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What definition are you using for the derivative here? Note that the heaviside function is discontinuous, so it isn't differentiable using the derivative you used in Calc I. (similarly, H(x)F(x) is not differentiable)

The generalizations I know of the derivative take (1) = (2) to be the definition. (With the requirement that F be a sufficiently good function; here we need at least differentiable and vanishing at infinity)

As for the last question, what do you know about the Dirac delta distribution? Even if you know only one thing about it, it's probably the thing you need to use....
 
  • #7
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i dont know what definition i am using... I have taken that straight from the maths notes we were given... (sorry!)

What do i know about the dirac delta function...well, the intergral of it is one, and that happens at t=0?
 
  • #8
Hurkyl
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What do i know about the dirac delta function...well, the intergral of it is one, and that happens at t=0?
D'oh, you got me. :frown: If that's the only thing you've been taught about the dirac delta distribution, then you are incapable of finishing this problem. (And so shame on your teacher for assigning it!) The step you are stuck upon is often used as the definition of the dirac delta.
 
  • #9
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ha! ok!
So what is this definition then? I think i shall go an google it...
 
  • #10
Hurkyl
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ha! Ok!
So what is this definition then? I think i shall go an google it...

(5) = (6)
 
  • #11
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oh! lol!
Thank you guys :-)
 

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