$\displaystyle l^{-1}(1)$ = ??!

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- Jan 13th 2010, 12:45 PMGeneralFind the inverse laplace of 1
$\displaystyle l^{-1}(1)$ = ??!

- Jan 13th 2010, 01:01 PMpomp
- Jan 14th 2010, 01:18 AMchisigma
The question is a source of some little controversial... according to the definition of $\displaystyle \delta (*)$ given in Wolfram MathWorld , probably the most 'reliable' mathematical site in the Web, is...

http://mathworld.wolfram.com/DeltaFunction.html

$\displaystyle \delta (t)= \frac {d}{dt} \mathcal {H} (t)$ (1)

... where $\displaystyle \mathcal {H} (*)$ is the 'Heaviside Step Function' defined as...

http://mathworld.wolfram.com/HeavisideStepFunction.html

$\displaystyle \mathcal{H} (t)= \left\{\begin{array}{cc} 0,&\mbox { if } t<0\\ \frac{1}{2},&\mbox{ if } t=0\\1,&\mbox{ if } t>0\end{array}\right.$ (2)

Now, according to the basic properties of the Laplace Transform...

http://mathworld.wolfram.com/LaplaceTransform.html

..., we have...

$\displaystyle \mathcal{L} \{\mathcal{H} (t)\}= \frac {1}{s}$ (3)

... and for every $\displaystyle f(*)$ ...

$\displaystyle \mathcal{L} \{\frac{d}{dt} f(t)\}= s\cdot \mathcal{L} \{f(t)\} - f(0)$ (4)

Now if we combine (1), (2), (3) and (4) we obtain...

$\displaystyle \mathcal {L} \{\delta(t)\}= s\cdot \frac{1}{s} -\frac{1}{2} = \frac{1}{2}$ (5)

... so that it would be...

$\displaystyle \mathcal {L}^{-1} \{1\}= 2\cdot \delta(t)$ (6)

... just a little surprising! (Itwasntme) ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jan 14th 2010, 01:58 AMLaurent
All you're proving here, is that delta can not be considered as a function. The inverse laplace transform of 1 does not exist as a function. However, the laplace inverse of 1

__is__$\displaystyle \delta$, which is the Dirac**distribution**, in the sense of distributions.

When you say:

Quote:

... and for every $\displaystyle f(*)$ ...

$\displaystyle \mathcal{L} \{\frac{d}{dt} f(t)\}= s\cdot \mathcal{L} \{f(t)\} - f(0)$ (4)

By the way, as a distribution, the Heaviside function doesn't have a specific value at 0. It is customary to choose it to be 0 for symmetry reasons, but it could be anything when we consider $\displaystyle \mathcal{H}$ as a distribution (for instance, when we differentiate it).

And Pomp's justification was also correct (except for the use of the word "function" and a slight abuse of notation in writing delta in an integral like physicists do...) - Jan 14th 2010, 02:57 AMCaptainBlack
Actually I don't think it is, he ought to more careful since $\displaystyle 0$ is one of the limits of integration and what one thinks:

$\displaystyle \int_0^{\infty} e^{-st}\delta(t) \; dt$

is short hand for is now not at all clear. If we take it as the limit of integrals where $\displaystyle \delta$ is replaced by functions whose support becomes more and more concentrated near the origin and with integrals all $\displaystyle 1$ what we get depends on the form of these functions and in this case could give any value between $\displaystyle 0$ and $\displaystyle 1$.

The sort of intuitive use that physicists and engineers make of $\displaystyle \delta$ fails here and one has to be more careful.

CB - Jan 14th 2010, 03:27 AMLaurent
Sure, that's the kind of thing I meant talking about "abuses of notation"... The integral should run from $\displaystyle -\infty$ (or any $\displaystyle -\varepsilon$) to $\displaystyle +\infty$ to be "a little" less controversial. Laplace transform is defined for distributions that are supported by $\displaystyle [0,\infty)$ (but not for any of these) hence the values on the negative axis don't matter. In fact, it is really not easy to define the Laplace transform of distributions properly; formally, it should be $\displaystyle \mathcal{L}(T)(z)=\langle T, t\mapsto e^{-tz}\rangle$ when one is able to make this make sense.

- Jan 14th 2010, 05:06 AMpomp
Seems I inadvertently opened a can of worms!

The notation I used was for convenience. Thanks for the pointers, I'll be more careful in future.