Results 1 to 7 of 7

Thread: Find the inverse laplace of 1

  1. #1
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    564

    Find the inverse laplace of 1

    $\displaystyle l^{-1}(1)$ = ??!
    Last edited by mr fantastic; Jan 14th 2010 at 02:38 AM. Reason: Edited post title
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2009
    Posts
    220
    Thanks
    1
    Quote Originally Posted by General View Post
    $\displaystyle l^{-1}(1)$ = ??!
    It's the Dirac delta function.

    $\displaystyle \mathcal{L} [\delta] = \int_0^{\infty} e^{-sx} \delta (x) dx = e^{-s . 0} = 1$ ,

    since

    $\displaystyle \int_{- \infty}^{\infty} \phi (x) \delta (x) dx = \phi(0)$
    Last edited by pomp; Jan 13th 2010 at 02:10 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    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! ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by chisigma View Post
    ... so that it would be...

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

    ... just a little surprising! ...
    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:
    ... and for every $\displaystyle f(*)$ ...

    $\displaystyle \mathcal{L} \{\frac{d}{dt} f(t)\}= s\cdot \mathcal{L} \{f(t)\} - f(0)$ (4)
    this is not true for distributions, where the formula doesn't have the $\displaystyle -f(0)$ term: $\displaystyle \mathcal{L}(T')(s)=s \mathcal{L}(T)$. Indeed, this border term is included in the differentiation: steps in the function $\displaystyle f$ give Dirac terms in the differential of $\displaystyle f$ in the sense of distributions. If you use distributions in the result, you must use a derivative in the sense of distributions for everything to make sense.

    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...)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by Laurent View Post

    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...)
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by CaptainBlack View Post
    what one thinks:

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

    is short hand for is now not at all clear.
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jun 2009
    Posts
    220
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find the Inverse Laplace Transform
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Nov 25th 2011, 10:44 PM
  2. find the inverse of the laplace transform
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Apr 21st 2011, 10:50 PM
  3. How to find these inverse Laplace transforms?
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: Mar 29th 2011, 03:30 AM
  4. Find the inverse laplace transform
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Sep 5th 2010, 11:24 PM
  5. Laplace/Inverse Laplace Questions
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Aug 14th 2010, 11:29 AM

Search Tags


/mathhelpforum @mathhelpforum