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The Dirac Delta function

by arildno
Tags: delta, dirac, function
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arildno
#1
Apr28-05, 06:41 AM
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1. INTRODUCTION
Many students become frustrated when they first meet the Dirac Delta function, typically in a course involving electrostatics, or Laplace transforms.
As it is commonly presented, the Dirac function seems totally meaningless:
Either, it is "defined" as:
[tex]\int_{-\infty}^{\infty}\delta(x)dx=1,\delta(x)=0,x\neq0[/tex]
or, in conjunction with an arbitrary function f:
[tex]\int_{I}\delta(x)f(x)dx=f(0),0\in{I},\int_{I}\delta(x)f(x)dx=0,0\notin{ I}[/tex]
where I is any interval on the real axis.
What sort of function can do these sort of things??
The student's confusion is not markedly lessened when:
the lecturer praises the Dirac function's ability to sample a function value, then the confusion increases when a mathematician scoffs at the whole "definition" and says "it is just a formalism, you've got to think of it as a distribution", and when the lecturer is confronted by this, he backs off a bit and says "yeah, the mathematician is right, but we won't get into distributions in this course, and, anyways, the theory of distributions is unnecessary for PRACTICAL purposes, and that's what this course is all about.."
What shall the poor student think of all of this?
My aim in creating this thread is to make an ultra-short intro to the mathematical way of looking at it.
It won't be very technical, but fairly rigorous, yet, I hope easily understood.
In the next post, I'll look into functionals, and the last post will provide the link between the functional perspective and the common way of "defining" the Dirac function.
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arildno
#2
Apr28-05, 07:40 AM
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A functional can be thought of as a function whose domain is a function space and some range, for example the real numbers.
For example, when we compute "the definite integral" of some (integrable) function on an interval [a,b], we may think of that computation operation as an action done upon f by a functional J(f):
[tex]J(f)=\int_{a}^{b}f(x)dx[/tex]
That is, for any eligible f, the functional J associates f the real number given by the definite integral (which corresponds to the "function value/output" by comparing the action of J with the action of a normal function on a member of its domain)

If we let the space of functions be those which are integrable on any interval I on the real axis, we may form a functional with two "arguments":
[tex]J(f,I)=\int_{I}f(x)dx[/tex]

Now, this particular functional is what is known as a LINEAR functional, because for any constants [tex]\alpha,\beta[/tex] and functions f,g,h satisfying [tex]h=\alpha{f}+\beta{g}[/tex] (i.e, h is a linear combination of f and g), we have:
[tex]J(h,I)=J(\alpha{f}+\beta{g},I)=\int_{I}(\alpha{f}+\beta{g})dx=\alpha\in t_{I}fdx+\beta\int_{I}gdx=\alpha{J}(f,I)+\beta{J}(g,I)[/tex]
by well-known properties of the definite integral.


Must all functionals be defined by the aid of integrals, or must at least all LINEAR functionals be defined with integrals?
By no means.
Here's a counter- example:
Let J(f) be defined by:
[tex]J(f)=f(3)[/tex]
That is, every function in our domain is mapped onto its own value at x=3 by the action of J. (Clearly then, functions in the domain of J must be defined at x=3)
Now, by letting h be as before, we have:
[tex]J(h)=h(3)=\alpha{f}(3)+\beta{g}(3)=\alpha{J}(f)+\beta{J}(g)[/tex]
Thus, J is linear, but there's not the slightest mention of integrals here!!


We are now ready to introduce the Dirac sampling functional:
We want a functional D which to any f assigns it the value f(0), provided 0 is within a specified interval I (that's the sampling feature), but which assigns the value 0 to each f if 0 is not in the specified interval.
Thus, we define in symbols:
[tex]D(f,I)=f(0),0\in{I},D(f,I)=0,0\notin{I}[/tex]
That's it! Period. This is what the Dirac functional "really is".
There's nothing weird about it; it is just a functional among many others, it is trivial to compute its output, and it seems totally unrelated to any features of integrals.
It does, however, have in common with the "integral functional " J the property of llinearity, which is easily proven.

So, what's all this talk about integrals and how the Dirac functional naturally occurs in conjunction with them?
That will be the topic in my next post.
arildno
#3
Apr29-05, 06:33 AM
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While we have seen that there's nothing really strange about the Dirac functional when we consider it simply as a functional, we have yet to see how we may link this object to integrals.

In order to do this, we intoduce a set S of weight functions or comparisons functions with respect to some interval [-a,a] on the real axis.
Let every member [tex]\phi[/tex]of S be a non-negative, integrable function with the following basic specification:
[tex]\phi(x)=0,|x|\geq{a},\int_{-a}^{a}\phi{dx}=1, a>0[/tex]
Surely, there are many functions fulfilling this requirement.
For later simplicity, we'll add the condition that S only includes those functions [tex]\phi[/tex] which are infinitely differentiable.
That behaviour might seem a bit difficult to achieve, so here's an actual example:
Let [tex]\phi(x)[/tex] be defined as:
[tex]\phi(x)=K(a)e^{-\frac{x^{2}}{a^{2}-x^{2}}},|x|<a,\phi(x)=0,|x|\geq{a}, K(a)=\frac{1}{\int_{-a}^{a}e^{-\frac{y^{2}}{a^{2}-y^{2}}}dy}[/tex]
This function is non-negative everywhere,infinitely differentiable, and its definite integral over [-a,a] is 1.

Now, obviously, the integral of such a [tex]\phi[/tex] over any interval I wich is disjoint with [-a,a], (that is, I has no common points with [-a,a],) is obviously zero, since [tex]\phi=0[/tex] on the whole of I.
Also, the integral of [tex]\phi[/tex] over any interval I' which includes [-a,a] as a sub-interval is clearly 1.


Furthermore, S as defined with respect to [-a,a] will include functions [tex]\hat{\phi}[/tex]
which look like the type I mentioned on a strictly SMALLER interval [-b,b], b<a (that is, whose integral over [-b,b] is 1, and [tex]\hat{\phi}=0,|x|\geq{b}[/tex]

Thus, we can form a sequence of [tex]\phi[/tex]-functions, all lying in S, say [tex]\phi_{n}[/tex] where a particular member [tex]\phi_{n}[/tex] fulfills:
[tex]\int_{-a_{n}}^{a_{n}}\phi_{n}dx=1,\phi_{n}(x)=0,|x|\geq{a_{1}}[/tex]
and where [tex]a_{n}[/tex] is a decreasing sequence of positive numbers fulfilling [tex]\lim_{n\to\infty}a_{n}=0[/tex]

That is, the origin lies in the interior of all intervals [tex][-a_{n},a_{n}][/tex].


Now, let F be a set of functions f which are considered continuous and sufficiently differentiable.
We may then form the sequence of well-defined FUNCTIONALS
[tex]D_{n}(f)[/tex]:
[tex]D_{n}(f,I)=\int_{I}f\phi_{n}dx[/tex]


Now, for some finite n, the interval [tex][-a_{n},a_{n}][/tex] will lie either wholly inside of I, or wholly outside of I, depending upon whether (a small neighbour hood of) zero is in I or not.

Due to LATEX problems, this post continues later.

arildno
#4
Apr29-05, 09:05 AM
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The Dirac Delta function

If [tex][-a_{n},a_{n}][/tex] is outside I for n greater than some N, then we evidently have:
[tex]D_{n\geq{N}}(f,I)=0\to\lim_{n\to\infty}D_{n}(f,I)=0[/tex]


If the interval is strictly inside of I, we may write:
[tex]D_{n}(f,I)=\int_{I}f\phi_{n}dx=\int_{-a_{n}}^{a_{n}}f\phi_{n}dx[/tex]

Now, since f was assumed to be continuous, there exist [tex]a_{m}[/tex], so that for any given [tex]\epsilon_{m}>0[/tex] we have:
[tex]|f(x)-f(0)|\leq\epsilon_{m},|x|<a_{m}[/tex]
Thus, we have:
[tex]|D_{m}(f,I)-f(0)|=|\int_{-a_{m}}^{a_{m}}f\phi_{m}dx-f(0)\int_{-a_{m}}^{a_{m}}\phi_{m}dx|=|\int_{-a_{m}}^{a_{m}}(f(x)-f(0))\phi_{m}dx|\leq|=\int_{-a_{m}}^{a_{m}}|f(x)-f(0)|\phi_{m}dx\leq\epsilon_{m}[/tex]
that is, we have [tex]\lim_{n\to\infty}D_{n}=f(0)[/tex], by letting the sequence [tex]\epsilon_{m}[/tex] go to zero as m goes to infinity.

But this means that we have managed to form a proper INTEGRAL representation of our sampling functional D(f,I)!!!

And, furthermore, since we have [tex]D_{n}\to{D}[/tex], we often say instead [tex]\phi_{n}\to\delta[/tex] where [tex]\delta[/tex] is the commonly "defined" Dirac Delta "function".
It must be emphasized, however, that it is the convergence of FUNCTIONALS which is mathematically meaningful.

In my next post, I'll broach the topic of how to define derivatives of the Dirac Delta function(al), and subtleties concerning choices of the set of comparison functions.
arildno
#5
Apr29-05, 11:42 AM
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Now, the example of a comparison function I posted has the additional nice property that all its derivatives vanish at [tex]x=\pm{a}[/tex]

Let us now consider a sequence of such functions which has this property:
[tex]\phi^{(m)}(\pm{a})=0, m=0,1,2,3..[/tex]
We will find a useful definition of the first derivative of the functional D, by investigating the behaviour of a particular sequence of functionals, called [tex]D_{n}^{(1)}(f,I)[/tex] (We focus our attention upon the behaviour of the sequence on the non-trivial intervals [tex][-a_{n},a_{n}][/tex]):
[tex]D_{n}^{(1)}(f,I)=\int_{-a_{n}}^{a_{n}}f\phi^{(1)}dx=f\phi_{n}\mid_{x=a_{n}}-f\phi_{n}\mid_{x=-a_{n}}+\int_{-a_{n}}^{a_{n}}(-f^{(1)})\phi_{n}dx=\int_{-a_{n}}^{a_{n}}(-f^{(1)})\phi_{n}dx=D_{n}(-f^{(1)},I)[/tex]
where we have assumed that f has a continuous first derivative.

Thus, we DEFINE the first derivative of the D to be the functional: [tex]D^{(1)}(f,I)\equiv{D}(-f^{(1)},I)[/tex]

By extension, we may then define the n'th derivative of D as the functional [tex]D^{(n)}(f,I))\equiv{D}((-1)^{n}f^{(n)},I)[/tex]


Note that as long as the [tex]\phi's[/tex] are AT LEAST n times differentiable, we are therefore able to define in a consistent manner the n'th derivative, provided that the f is n times differentiable.
But:
Surely, there are lots of weight functions which would, for example give us the functional D, that is, could be regarded as a sequence "converging" to the delta function [tex]\delta[/tex], but which wasn't n times differentiable?

For example:
If we let [tex]\phi_{n}(x)=\frac{1}{2a_{n}},|x|<a_{n},\phi_{n}(x)=0,|x|\geq{a}[/tex] then we would still have that the sequence of functionals formed using this particular choice of comparison sequence converged to the Dirac sampling functional.
Yet none of the members of that sequence are continuous at [tex]x\pm{a}[/tex], let alone differentiable.

Thus, we must conclude, that depending upon what we want to get out of our limiting procedures, some sets of comparison functions cannot be used at all.
I'll post a few further examples upon this later on.
arildno
#6
Apr29-05, 01:37 PM
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The importance of rate of decay:

We have up till now used comparison functions which are strictly zero outside a finite interval. These therefore kill off any "dangerous" behaviour f exhibits outside that interval.
However, there are lots of function sequences which might be used to approximate the delta function, which are non-zero on the whole axis, for example, the function sequence
[tex]\phi_{n}(x)=\frac{e^{-(\frac{x}{a_{n}})^{2}}}{a_{n}\sqrt{\pi}}[/tex]
will also do, as long as f does not diverge too fast towards infinity.
For example, if [tex]f=Ae^{x^{3}}[/tex], then
[tex]\int_{-\infty}^{\infty}f\phi_{n}dx[/tex]
is divergent for each choice of n, that is, this particular choice of comparison sequence is wholly unsuitable to be used to give an integral representation of D(f,I) (the sampling functional).

Again, it is important to realize that the FUNCTIONAL D(f,I) is perfectly well behaved for that choice of f; we have merely chosen a comparison sequence which won't yield an integral representation of that functional!!!

In short then, my choice of S (infinitely differentiable vanishing outside a finite interval) can be regarded as the CORE SET for construction of valid integral representations of the Dirac sampling functional for continuous f's.
Other choices might well work for some particular continuous f's, but not for other continuous f's, or those choices are unsuitable for defining derivatives of the functionals involved.

Discontinuous f's:
Let's look at the following ugly f:
[tex]f(x)=0,x\neq{0},f(0)=1[/tex]
Now, we have as before, equally unproblematically:
[tex]D(f,I)=f(0)=1,0\in{I},D(f,I)=0,x\notin{I}[/tex]

But how should we manage an INTEGRAL representation of D(f,I)???
If we try any of the comparison functions we've used so far, all we'd ever get out is 0.
The simplest way of managing this, is to introduce a sequence of [tex]f_{n}[/tex]-functions converging to f (in some sense of the word):
[tex]f_{n}(x)=1, |x|<a_{n},f_{n}(x)=0,|x|\geq{a}_{n}[/tex]

Thus, we have on the non-trivial interval:
[tex]D_{n}(f_{n},I)=\int_{-a_{n}}^{a_{n}}f_{n}\phi_{n}dx=1=f_{n}(0)[/tex]
and we are entitled to say [tex]\lim_{n\to\infty}D_{n}=D[/tex] in this case as well..
arildno
#7
Apr30-05, 06:44 AM
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The last example with a discontinuous f shows us how we may form an integral representation of Dirac's sampling function for an extremely wide class of functions.

Let's simply assume that f is defined on the whole real axis, and in particular, at the origin.
Then the following sequence of functions converges pointwise to f:
[tex]f_{n}(x)=f_{n}(0),|x|<a_{n},f_{n}(x)=f(x),|x|\geq{a}_{n}[/tex]
Thus, the product function [tex]f_{n}\phi_{n}[/tex] is integrable everywhere irrespective of the integrability of [tex]f_{n}[/tex], assuming that the [tex]\phi_{n}'s[/tex] are members of the "core set" of comparison functions mentioned earlier.

Furthermore, it is readily seen that we have the convergence of functionals: [tex]D_{n}(f_{n},I)\to{D}(f,I)[/tex] where I is a set on the real axis upon which we are allowed to define "integration"
(that is, a sufficiently nice region; a finite interval, for example)

That is, as long as the set I is nice, or that f has a sufficiently nice domain, then we see that D(f,I) can be given some sort of integral representation.

But, since this holds for just about any f's and sets we're interested in, it means that we won't ever (or, almost never) face any trouble with using the conventional approach either.
That is, we are justified in using the [tex]\delta[/tex]-formalism without any further thinking, being assured that there exists some rigorous manner of doing it as well..
saltydog
#8
Apr30-05, 09:39 AM
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Hello Arildno. I'm a few parts back and have the following question:

The following relation (from Part 2) seems to be at the heart of the theory and I wish to understand it fully. You say:

If for any [itex]\epsilon_m> 0[/itex], we can find an [itex]a_m[/itex] such that:

[tex]|f(x)-f(0)|\leq\epsilon_{m},|x|<a_m[/tex]

Then:

[tex]|D_{m}(f,I)-f(0)|=|\int_{-a_{m}}^{a_{m}}f\phi_{m}dx-f(0)\int_{-a_{m}}^{a_{m}}\phi_{m}dx|=|\int_{-a_{m}}^{a_{m}}(f(x)-f(0))\phi_{m}dx|\leq|=\int_{-a_{m}}^{a_{m}}|f(x)-f(0)|\phi_{m}dx\leq\epsilon_{m}[/tex]

I wonder if there is a minor typo in that last relation: a stray absolute value sign and equal sign appearing right after the [itex]\leq[/itex] sign. Should it read:

[tex]|D_{m}(f,I)-f(0)|=|\int_{-a_{m}}^{a_{m}}f\phi_{m}dx-f(0)\int_{-a_{m}}^{a_{m}}\phi_{m}dx|=|\int_{-a_{m}}^{a_{m}}(f(x)-f(0))\phi_{m}dx|\leq\int_{-a_{m}}^{a_{m}}|f(x)-f(0)|\phi_{m}dx\leq\epsilon_{m}[/tex]

To me, that relation really shows why:

[tex]\lim_{n\to\infty}D_{n}[f,I]=f(0)[/tex]

and I want to make sure I have it understood correctly. Can you confirm this suspicion?

Thanks,
Salty
arildno
#9
Apr30-05, 11:06 AM
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You're absolutely right; they shouldn't be there..
arildno
#10
May1-05, 11:50 AM
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While the technique I espoused in part 5 for the formation of an integral representation of the sampling functional is okay enough, I guess, coupling the limiting processes of [tex]\phi_{n},D_{n},f_{n}[/tex] isn't really necessary, and rather inelegant.

Thus, I would prefer the following:
Let [tex]f_{m}(x)=f(0),|x|<c_{m},f_{m}(x)=f(x),|x|\geq{c}_{m}[/tex]
where [tex]c_{m}[/tex] is a sequence of positive numbers converging to 0, and let [tex]a_{n}[/tex] be a similar sequence.

Define therefore on the non-trivial interval [tex][-a_{n},a_{n}][/tex] the functional:
[tex]D_{n}(f_{m},I)=\int_{-a_{n}}^{a_{n}}f_{m}\phi_{n}dx[/tex]
that is, we have:
[tex]\lim_{n\to\infty}D_{n}(f_{m},I)=D(f_{m},I)[/tex]
and we have:
[tex]\lim_{m\to\infty}D(f_{m},I)=\lim_{m\to\infty}f(0)=f(0)=D(f,I)[/tex]
I like this somewhat better..
Cincinnatus
#11
Mar24-06, 09:12 PM
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This thread was very helpful to me Arildno, thanks for taking the time to write it.
HerbM
#12
Jul9-06, 11:20 PM
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On the MIT OCW (Open CourseWare) the course on Differential Equations 18.03 includes Laplace transforms and a very clear* explanation of the Dirac delta function in Lecture #22.

*It's clear to me and I am mathematically naive at the college level (just learning DiffEq and other stuff now.)

Anyone not familiar with Laplace transforms should go back to Lecture #19 and proceed from 19-22.

--
Herb Martin
Lee
#13
Nov9-06, 10:04 AM
P: 56
When solving problems involving the dirac dela function, can you recommend any sort of method or steps to follow to evaluate intergrals involving the dirac delta function.
arildno
#14
Nov9-06, 10:13 AM
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Quote Quote by Lee
When solving problems involving the dirac dela function, can you recommend any sort of method or steps to follow to evaluate intergrals involving the dirac delta function.
Find f(0). That's about it.
Please do not ask questions in this tutorial thread; instead, open a new thread or use the PM option.
Warr
#15
Dec6-06, 12:25 AM
P: 122
thanks, very helpful :)
gammamcc
#16
Feb21-07, 11:21 PM
P: 151
Wouldn't the concept of Dirac delta function be simpler by the use of basic measure theory? In fact, Lebesgue measure in not even needed: You can use Radon measure theory a la Buck's Advanced Calculus. In a naive sense, any partition of the real line (x-axis, say) would give weight 1 of the set containing 0 and weight 0 for any other set. Starting with a continuous function on a finite interval of the x-axis, work out Riemann sums with this modification (instead of multiplying function values by lengths of intervals, you are multiply the function by 0 except for the value you choose on the interval containing x=0). Then, take the limit as the norm of the partition goes to zero like freshman calc and the result obtains.

Those who want an over analysis of it can refer to http://en.wikipedia.org/wiki/Radon_measure
(probably won't help you in your tenure evaluation).


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