Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Product of opposite Heaviside Steps

  1. Sep 25, 2011 #1

    ivl

    User Avatar

    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:
    [tex]
    h^2(x) = h(x), \ x \neq 0
    [/tex]

    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:
    [tex]
    \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)
    [/tex]

    So, if you find the antiderivative back again, you will get:
    [tex]
    Y(x) = (1 - 2 h(0)) h(x)
    [/tex]
    Equating the two definitions of [itex]Y(x)[/itex], you will get:
    [tex]
    h(x) \left( h(x) - 2 h(0) \right) = 0
    [/tex]
    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

    ivl

    User Avatar

    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:

    [tex]
    \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}
    [/tex]
    The integral converges on the upper bound iff:
    [tex]
    |e^{-i \eta x}| = e^{\mathrm{Re}(-i \eta x)} e^{x \mathrm{Im}(\eta)} \rightarrow 0 \Leftrightarrow \mathrm{Im}(\eta) < 0
    [/tex]
    This means that the Fourier transform only exists in the lower half-plane of the complex number [itex]\eta[/itex] and the transform is:
    [tex]
    \tilde{h}(\eta) = \frac{1}{i \eta}, \mathrm{Im}(\eta) < 0
    [/tex]
    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:
    [tex]
    \tilde{y}(\eta) = \tilde{h}(\eta) - \int_{-\infty}^{\infty}{\frac{ds}{2 \pi} \, \tilde{h}(s) \tilde{h}(\eta - s)}
    [/tex]
    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

    ivl

    User Avatar

    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:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Product of opposite Heaviside Steps
  1. Heaviside Method (Replies: 3)

Loading...