Quick question about Dirac delta functions

black_hole

What does the square of a Dirac delta function look like? Is the approximate graph the same as that of the delta function?

tiny-tim

yes

(only more so! )

George Jones

The square of a Dirac delta function is not defined.

micromass

George is right. There is no way that we can make sense of squares of the Dirac Delta function. Distributions are very strange things and their properties are subtle. Nice operations like squaring don't always make sense.

So the square doesn't exist and so the approximate graph also doesn't exist.

rbj

often this question comes up in an electrical engineering class when one is faced with convolving an impulse [itex] \delta(t) [/itex] with another impulse. i.e. what would happen if you had a linear, time-invariant system with impulse response [itex] h(t) = \delta(t) [/itex] and you input to that system [itex] x(t) = \delta(t) [/itex]. obviously, the output should be [itex] y(t) = \delta(t) [/itex], but how do you get that from the convolution integral?

Mute

It's important for students to realize that the convolution of two delta functions can be well-defined, as the integral contains both a dummy variable and a free variable: ##\int d\tau~\delta(\tau)\delta(t-\tau) = \delta(t)##. One can interpret the integral as being done over one of the delta functions to give the result (there's probably a much more formal and rigorous way to do this). The problem with something like ##\int d\tau~\delta(\tau)\delta(\tau)## is that the result is naively ##\delta(0)##, which doesn't have a well-defined interpretation (though in some contexts, like quantum field theory, for example, you might see it taken to mean the volume a system which is to be taken to grow infinitely large at the end of the calculation).

Although it looks like the convolution integral will generate the squared delta function integral when t = 0, one has to remember that 'functions' like the delta function are meant to exist under integrals, so t has to remain a free variable. Fixing its value doesn't really make much sense.

Of course, this is a physicist's way of looking at the issue, so there are some gaps in the formality and rigor, and mathematicians should feel free to shore it up (or tear it down, as the case may be) with the appropriate rigor.

pwsnafu

Well, first you need to know the actual definition of convolution of distributions

First, let ##f(x)## and ##g(y)## be distributions on ##\Omega \subset \mathbb{R}##. We define the two variable distribution ##f \oplus g## as
##(f \oplus g, \, \phi(x,y)) := (f(x), \, (g(y), \, \phi(x,y)))##.

We now define ##(f * g, \phi) := (f(x) \oplus g(y), \phi(x+y))##.

Now this is well-defined in any of the following cases:

1. f or g has compact support,
2. both have their support bounded from below,
3. both have their support bounded from above.

Dirac has compact support therefore the convolution exists.

Edit: Now that I think about it, the simples method would be a approximation to identity argument. You would still need to show that the identity holds irrespective of your choice of representative though.