Convolution integrals with Dirac's delta and its derivatives

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  • #1
traianus
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Hello everybody,
I am new in this forum and I have a problem. First of all, I am not a mathematician, I am an engineer that is using the Laplace transform. Considering this, please do not be too formal in the answers...

INTRODUCTION:

The LAPLACE transform of a function [tex]f(t)[/tex] is
[tex]\mathcal{L}\left[f(t)\right][/tex]. Now we let the first derivative of [tex]f(t)[/tex] be called with [tex]\dot{f}(t)[/tex] and the second derivative of [tex]f(t)[/tex] be called with [tex]\ddot{f}(t)[/tex]. It is well known that the LAPLACE transforms (the unilateral) for the first and second derivatives are:

First derivative ---> [tex]\mathcal{L}\left[\dot{f}(t)\right] = s\mathcal{L}\left[{f}(t)\right] - f(0)\qquad (1)[/tex]

Second derivative ----> [tex]\mathcal{L}\left[\ddot{f}(t)\right] = s^2\mathcal{L}\left[{f}(t)\right] -sf(0) - \dot{f}(0)\qquad (2)[/tex]
Inverse laplace transform: I indicate it by using the symbol [tex]\mathcal{L}^{-1}\left[f(t)\right][/tex].
Now, from equation [tex](1)[/tex], the inverse Laplace transform of the term
[tex]s\mathcal{L}\left[{f}(t)\right] [/tex] is:

[tex]\mathcal{L}^{-1}\left[s\mathcal{L}\left[{f}(t)\right]\right] = \mathcal{L}^{-1}\left[ \mathcal{L}\left[\dot{f}(t)\right] + f(0)\right] = \dot{f}(t) + \mathcal{L}^{-1}\left[f(0)\right]\qquad (3)[/tex]

Remembering that the inverse Laplace transform of 1 is Dirac's delta function (indicated with [tex]\delta(t)[/tex]), equation [tex](3)[/tex] becomes:

[tex]\fbox{$\mathcal{L}^{-1}\left[s\mathcal{L}\left[{f}(t)\right]\right] = \dot{f}(t) + f(0)\delta(t)$}\qquad (4)[/tex]

Similarly, from equation [tex](2)[/tex], I can obtain the inverse LAPLACE transform of the term [tex]s^2\mathcal{L}\left[{f}(t)\right][/tex]:

[tex]\mathcal{L}^{-1}\left[s^2\mathcal{L}\left[{f}(t)\right]\right] = \mathcal{L}^{-1}\left[ \mathcal{L}\left[\ddot{f}(t)\right] +s f(0) + \dot{f}(0)\right] \qquad (5)[/tex]

or

[tex]\fbox{$\mathcal{L}^{-1}\left[s^2\mathcal{L}\left[{f}(t)\right]\right] = \ddot{f}(t) +\dot{\delta}(t) f(0) + \delta(t)\dot{f}(0)$} \qquad (6)[/tex]

where [tex]\dot{\delta}(t))[/tex] is the derivative of the Dirac distribution.

QUESTION.

Suppose that I like to take a "stupid" but equivalent approach using the CONVOLUTION theorem. Suppose that the goal is to obtain equations (4) and (6) using that theorem. Let's start with equation (4).
The inverse Laplace transform of the Laplace variable s is:

[tex]\mathcal{L}^{-1}\left[s\right] = \dot{\delta}(t)\qquad (7)[/tex]

The inverse of [tex]\mathcal{L}\left[f(t)\right][/tex] is just [tex]f(t)[/tex]. Using the convolution theorem I can find the inverse of the product [tex]s\mathcal{L}[f(t)][/tex] :

[tex]\mathcal{L}^{-1}\left[s\mathcal{L}\left[f(t)\right]\right] = \int\limits_0^t\dot{\delta}(t-u)f(u)\mathrm{d}u\qquad (8)[/tex]

How can I obtain from equation [tex](8)[/tex] equation [tex](4)[/tex]? Should I integrate by parts? I tried it, but I have some theoretical problems with the DIRAC function. Please, help me!

Similar problem if I use the convolution theorem in order to obtain equation [tex](6)[/tex]:

[tex]\mathcal{L}^{-1}\left[s^2\mathcal{L}\left[f(t)\right]\right] = \int\limits_0^t\ddot{\delta}(t-u)f(u)\mathrm{d}u\qquad (9)[/tex]

how do I obtain equation [tex]
(6)[/tex]
?

If you have some answers (but not too technical because I am not a mathematician) please let me know:

traianus@gmail.com
 
Last edited:

Answers and Replies

  • #2
traianus
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Nobody knows?
Basically, even if you do not know the Laplace transforms, the problem I have is to demonstrate that

[tex]\int\limits_0^t\ddot{\delta}(t-u)f(u)\mathrm{d}u = \ddot{f}(t) +\dot{\delta}(t) f(0) + \delta(t)\dot{f}(0)[/tex]

and I have to demonstrate that

[tex]\int\limits_0^t\dot{\delta}(t-u)f(u)\mathrm{d}u = \dot{f}(t) + f(0)\delta(t)[/tex]
 
  • #3
Hurkyl
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I would imagine this is a straightforward application of integration by parts.
 
  • #4
traianus
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Yes, I think so too, but I have some problems with the Dirac and its derivatives. So, can you please show me how to do it?
 
  • #5
Hurkyl
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Where do you have problems?
 
  • #6
traianus
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Can you please post how you perform the integration by parts?
 
  • #7
traianus
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Enybody knows the solution of this problem? I do not think it is so difficul if you know well the properties of the delta functions. I obviously do not. So please post here the solution!
Thank !
 
  • #8
lurflurf
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[tex]\int_{0^-}^{t^+} \delta'(t-u)f(u) du=-\delta(t-u)f(u)|_{0^-}^{t^+} +\int_{0^-}^{t^+} \delta(t-u)f'(u) du=f(0)\delta(t)+f'(0)[/tex]
the one sided limits are needed to avoid problems that arise when the singular point of the dirac delta is an endpoint of an integration interval.
 
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  • #9
lurflurf
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traianus said:
Can you please post how you perform the integration by parts?
[tex]\int_a^b u(x)v'(x) dx=u(x)v(x)|_a^b-\int_a^b u'(x)v(x) dx[/tex]
 
  • #10
traianus
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Thank you lurflurf for your answer. I obtained your equation integrating by parts like you did, but I had/have a conceptual problem in the integration by parts of the delta function. In particular, can you explain with all details how you calculate the term
[tex]-\delta(t -u)f(u) [/tex] between the limits [tex]0^- [/tex]
and [tex]t^+[/tex] ?
thank you for your help.
Traianus
 
  • #11
traianus
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Should I give up? No body knows?
 
  • #12
Galileo
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The first endpoint at t is a bit of a problem, because of the singularity from the delta function. We do have: [itex]\delta(t-u) \to 0[/itex] as [itex]u\to t[/itex].
At the other endpoint, it is simply [itex]\delta(t)[/itex]. So:

[tex]-\delta(t-u)f(u)|^{u=t}_{u=0}=-(0)f(t)+\delta(t)f(0)=f(0)\delta(t)[/tex]
 
  • #13
traianus
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Thank you Galileo. Finally we reached the point in which I have the problem.
I figured out that [tex]\delta(t-u)\rightarrow 0[/tex] as [tex]u\rightarrow t[/tex] because I knew the final result of the integration by parts. But how do you explain that [tex]\delta(t-u)\rightarrow 0[/tex] as [tex]u\rightarrow t[/tex]? This is my REAL problem. Please explain with ALL the details why this is true. Thank you!
 
  • #14
Galileo
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Because the spike is at t. At all other values it is zero, so naturally the limit as u approaches t is zero.
Suppose you have a function that is constant everywhere except at one point (f(x)=C, except at x=a.) Then you still have lim(x->a)f(x)=C. What matters is the value the function approaches as x->a, not the value of the function at a.
 
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  • #15
traianus
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Thank you for your answer. The function I have is [tex]\delta (t-u)[/tex], which is similar (in concept) to the function [tex]\delta(t)[/tex]. The function [tex]\delta(t)[/tex] is infinite when [tex]t = 0[/tex]. So I deduce that [tex]\delta (t-u)[/tex] is infinite when [tex]t-u=0[/tex]. Right? If so something does not work. Please convince me!
 
  • #16
lurflurf
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traianus said:
Thank you for your answer. The function I have is delta (t-u), which is similar (in concept) to the function delta(t). The function delta(t) is infinite when t = 0. So I deduce that delta (t-u) is infinite when t-u=0. Right? If so something does not work. Please convince me!
t is never zero in delta(t-t+) the t+ means we always chose a number larger than t. Thus we always get zero. It is like delta(0+) we chose a number larger than zero (but in a limit process ever closer to zero) thus we always get zero.
delta(0+)=0
delta(t-t+)=0
this is much like the nonsingular case define
f(x)=0 x!=0(x not zero)
f(x)=1 x=0
f(0+)=0
Limits care about the journey not the destination. The whole point of using limits is to find out what happen as we approch a value when we do not like or do not care about what happens at the value.
 

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