# Product of opposite Heaviside Steps

1. Sep 25, 2011

### ivl

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. Sep 25, 2011

### Studiot

What is your definition of the heaviside step function?

Mine is

H(t) = 0 for t$\leq$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

3. Sep 25, 2011

### Dickfore

Strictly speaking, $h(x)$ does not have a limit at $x = 0$. 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 $h(x) ( 1 - h(x))$?

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

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 $x = 0$ is irrelevant.

However, if you differentiate this product and use $h'(x) = \delta(x)$, 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 $Y(x)$, you will get:
$$h(x) \left( h(x) - 2 h(0) \right) = 0$$
For $x < 0$, this is satisfied because of the first factor. For $x > 0$ it goes to $1 - 2 h(0) = 0$. This gives another choice of $h(0)$.

Last edited: Sep 25, 2011
4. Sep 25, 2011

### ivl

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!

5. Sep 25, 2011

### Studiot

Difficult to comment further without more input from ivl.

Are you studying the Heaviside expansion theorem, Laplace transforms, Z transforms or what?

Last edited: Sep 25, 2011
6. Sep 25, 2011

### Dickfore

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 $\eta$ and the transform is:
$$\tilde{h}(\eta) = \frac{1}{i \eta}, \mathrm{Im}(\eta) < 0$$
Alternatively, you can consider $\eta$ as a purely real number, but add a small infinitesimal imaginary part $\eta \rightarrow \eta - i \epsilon, \; \epsilon > 0$.

Then, expanding the product in the definition of $y(x)$, 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 $s_{1} = i \epsilon$ and $s_{2} = \eta - i \epsilon$. You can analyze all the possible cases and evaluate it.

7. Sep 26, 2011

### ivl

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!

8. Sep 26, 2011

### Studiot

I agree different authors use different definitions for H(t)

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

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