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Product of opposite Heaviside Steps

  1. Sep 25, 2011 #1


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    Dear all,

    I would like to give you the physics context in which this question emerged, but that would be a very long explanation (sorry!). So, since the question is almost self contained, I am just going to tell you what it is.

    Consider the product Y(x)=H(x)(1-H(x)), where H(x) is the Heaviside step function. Of course, Y(x) is zero everywhere. Except, at the origin there is a big problem: the value of Y(0)=H(0)(1-H(0)) depends on the choice of H(0).

    To make things worse, in the context where this question emerged, there is no particular reason to choose a specific value for H(0).

    To make things even worse, apparently one cannot use the theory of distributions, since the product of distributions is ill-defined!

    Does anyone have any suggestions for a rigorous way to deal with the problem?

    Thanks a lot!
  2. jcsd
  3. Sep 25, 2011 #2
    What is your definition of the heaviside step function?

    Mine is

    H(t) = 0 for t[itex]\leq[/itex]0
    H(t) = 1 for t>0

    This H(t) and 1-H(t) are well defined at t=0

    H(t)(1-H(t)) = (0)x(1) = 0
  4. Sep 25, 2011 #3
    Strictly speaking, [itex]h(x)[/itex] does not have a limit at [itex]x = 0[/itex]. For all other points, the following identity holds:
    h^2(x) = h(x), \ x \neq 0

    So, for those points, what will you get if you expand out the product [itex]h(x) ( 1 - h(x))[/itex]?

    Mind you, however, that the limit of this function is still not defined at [itex]x = 0[/itex]. You may want to make it continuous at [itex]x = 0[/itex]. What are the choices for defining [itex]h(0)[/itex]?

    Nevertheless, for integration purposes, this set where the product does not have a limit is of measure zero, and the function does not behave like a Dirac delta-function, so the definition of the value at [itex]x = 0[/itex] is irrelevant.

    However, if you differentiate this product and use [itex]h'(x) = \delta(x)[/itex], then you will get, by the product rule:
    \frac{d}{d x} h(x) (1 - h(x)) = h'(x)(1 - h(x)) + h(x) (-h'(x)) = (1 - 2 h(x)) \, \delta(x) = (1 - 2 h(0)) \, \delta(x)

    So, if you find the antiderivative back again, you will get:
    Y(x) = (1 - 2 h(0)) h(x)
    Equating the two definitions of [itex]Y(x)[/itex], you will get:
    h(x) \left( h(x) - 2 h(0) \right) = 0
    For [itex]x < 0[/itex], this is satisfied because of the first factor. For [itex]x > 0[/itex] it goes to [itex]1 - 2 h(0) = 0[/itex]. This gives another choice of [itex]h(0)[/itex].
    Last edited: Sep 25, 2011
  5. Sep 25, 2011 #4


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    Thanks for the replies!

    @studiot: the problem is precisely the fact that different definitions give different Y(x).

    @dickfore: your final argument, in favour of h(0)=0.5, is interesting. I am not competent enough to say if it conclusive...

    Thanks again!
  6. Sep 25, 2011 #5
    Difficult to comment further without more input from ivl.

    What is your application?

    Are you studying the Heaviside expansion theorem, Laplace transforms, Z transforms or what?
    Last edited: Sep 25, 2011
  7. Sep 25, 2011 #6
    Ah, yes, you can also consider the Fourier transform of a heaviside step function:

    \tilde{h}(\eta) = \int_{-\infty}^{\infty}{h(x) e^{-i \eta x} \, dx} = \int_{0}^{\infty}{e^{-i \eta x} \, dx} = \left. \frac{e^{-i \eta x}}{-i \eta} \right|^{\infty}_{0}
    The integral converges on the upper bound iff:
    |e^{-i \eta x}| = e^{\mathrm{Re}(-i \eta x)} e^{x \mathrm{Im}(\eta)} \rightarrow 0 \Leftrightarrow \mathrm{Im}(\eta) < 0
    This means that the Fourier transform only exists in the lower half-plane of the complex number [itex]\eta[/itex] and the transform is:
    \tilde{h}(\eta) = \frac{1}{i \eta}, \mathrm{Im}(\eta) < 0
    Alternatively, you can consider [itex]\eta[/itex] as a purely real number, but add a small infinitesimal imaginary part [itex]\eta \rightarrow \eta - i \epsilon, \; \epsilon > 0[/itex].

    Then, expanding the product in the definition of [itex]y(x)[/itex], and applying the convolution theorem for the product of the two Heaviside step functions, we get:
    \tilde{y}(\eta) = \tilde{h}(\eta) - \int_{-\infty}^{\infty}{\frac{ds}{2 \pi} \, \tilde{h}(s) \tilde{h}(\eta - s)}
    The poles of the integrand are at [itex]s_{1} = i \epsilon[/itex] and [itex]s_{2} = \eta - i \epsilon[/itex]. You can analyze all the possible cases and evaluate it.
  8. Sep 26, 2011 #7


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    Hi all!

    @studiot: Sorry, I refrained from giving further input, as I now realise that the question has multiple answers. Consider for example the last equation put forward by Dickfore, expressing the product Y(x) of opposite step functions as a convolution. The convolution contains two poles, whence it is possible to choose different paths. Different paths will give different results for Y(0).

    @Dickfore: see my reply to studiot above. Is this right?

    Thanks both!
  9. Sep 26, 2011 #8
    I agree different authors use different definitions for H(t)

    Attached is the reason I like mine - it also happens to be the original.

    Attached Files:

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