Show an identity - distribution

[fluidistic]

Homework Statement

Show the following identity (in the sense of distribution): g(\bold x)\delta (\bold x)=g(\bold 0) \delta (\bold x) for a function g.

Homework Equations

No idea.

The Attempt at a Solution

I don't have a concrete idea about what a distribution is (It's an assignment in my electromagnetism course and we've not been introduced the required math so I guess I should search in books, but I don't know which ones and the article http://en.wikipedia.org/wiki/Distribution_(mathematics) in wikipedia doesn't seem easy to understand for me).
From the identity, I get the idea that \delta (\bold x) is 0 everywhere except in \bold x=0 but I'm not sure and even if it were so, I don't really understand why the equality holds.
Any help is greatly appreciated.

 

[Dick]

If d1(x) and d2(x) are equal as distributions, that means the integrals of d1(x)*f(x)dx and d2(x)*f(x)dx are equal for any test function f(x). Can you show that?

 

[gabbagabbahey]

Two distributions are considered equal if you get the same result integrating either over any region. What do you get when you integrate g(\textbf{x})\delta(\textbf{x}) over any volume not enclosing the origin? How about if you integrate it over any volume that does enclose the origin? Compare that to what you get when integrating g(0)\delta(\textbf{x}) in both cases.

 

[Dick]

Two distributions are considered equal if you get the same result integrating either over any region. What do you get when you integrate g(\textbf{x})\delta(\textbf{x}) over any volume not enclosing the origin? How about if you integrate it over any volume that does enclose the origin? Compare that to what you get when integrating g(0)\delta(\textbf{x}) in both cases.

Not quite. I think you want to integrate against an arbitrary test function. Not just f(x)=1.

 

[fluidistic]

Thanks to both,

If d1(x) and d2(x) are equal as distributions, that means the integrals of d1(x)*f(x)dx and d2(x)*f(x)dx are equal for any test function f(x). Can you show that?

Ah ok, I wasn't aware of this. I do not even know what a distribution is. I have in mind the Dirac's delta function. The integral (what kind of integral is required?!) not containing the origin of d(x)f(x)dx would be equal to 0 because d(x)=0 for all x different from x=0, right?
And the integral of d(x)f(x)dx would be equal to a constant if the origin is contained in the domain of integration... right?
I never dealt with this mathematically before. So if you know some books (for physicists better) explaining what a distribution is and how to manipulate them mathematically, I'd be glad.

 

[gabbagabbahey]

Not quite. I think you want to integrate against an arbitrary test function. Not just f(x)=1.

I don't see why. If it is true for f(x)=1, it will be true for any other test function as well. Won't it?

In terms of probability distributions, my statements amounts to saying that if you calculate the probability of measuring \textbf{x} within any region, you will get the same value for both distributions. How could this be true if the two distributions weren't equal?

 

[gabbagabbahey]

I have in mind the Dirac's delta function. The integral (what kind of integral is required?!) not containing the origin of d(x)f(x)dx would be equal to 0 because d(x)=0 for all x different from x=0, right?

When you write \delta(\textbf{x}), can I assume that you are referring to the 3-dimensional Dirac delta which is zero everywhere except the origin? If so, then you need to perform a volume integral, and yes, you will get zero for any volume that does not enclose the origin.

And the integral of d(x)f(x)dx would be equal to a constant if the origin is contained in the domain of integration... right?

Right, but you can do better than just "a constant";

\int_{\mathcal{V}}g(\textbf{x})\delta(\textbf{x})d^3\textbf{x}=g(\mathbf{0})

for any volume \mathcal{V} that encloses the origin.

I never dealt with this mathematically before. So if you know some books (for physicists better) explaining what a distribution is and how to manipulate them mathematically, I'd be glad.

For this type of problem, you may like to think of probability distributions. Integrating a probability distribution p(\textbf{x}) over a given region gives the probability of measuring \textbf{x} to be within that region. If two such distributions give the same probability for any region, they can be considered equal.

 

[fluidistic]

Ok thank you both for all.
No, the exercise doesn't specify the Dirac's delta function in 3 dimensions but only a "distribution function" I believe. I posted the exercise as it appears in the problem we've been given.

 

[gabbagabbahey]

No, the exercise doesn't specify the Dirac's delta function in 3 dimensions but only a "distribution function" I believe. I posted the exercise as it appears in the problem we've been given.

The identity you are asked to prove isn't true for just any old distribution \delta(\textbf{x}). It is true for the 3d Dirac Delta distribution however, and if this is for an EM course I'd go ahead and assume that it is indeed the Dirac Delta. (Are you by chance studying from Jackson? He uses the same notation.)

 

[fluidistic]

The identity you are asked to prove isn't true for just any old distribution \delta(\textbf{x}). It is true for the 3d Dirac Delta distribution however, and if this is for an EM course I'd go ahead and assume that it is indeed the Dirac Delta. (Are you by chance studying from Jackson? He uses the same notation.)

Ok, I'll take note of this.
I just started the upper undergrad level electromagnetism course and I'm suffering a lot. I should be using a course I'll be taking next semester (mathematical methods used in physics). We're studying the course at Jackson's level and I've heard many of our exercises are taken from Jackson's book but I didn't check it out. So it's likely the same notation as Jackson, yes.

 

[gabbagabbahey]

Section 1.2 in Jackson covers the Dirac Delta fairly succinctly, and he gives references for further reading on it too.

 

[fluidistic]

Section 1.2 in Jackson covers the Dirac Delta fairly succinctly, and he gives references for further reading on it too.

Thank you so much, I'll take a look.

 

[Dick]

I don't see why. If it is true for f(x)=1, it will be true for any other test function as well. Won't it?

In terms of probability distributions, my statements amounts to saying that if you calculate the probability of measuring \textbf{x} within any region, you will get the same value for both distributions. How could this be true if the two distributions weren't equal?

Integrating over regions is sort of the same as saying they are the same for step functions. I'd rather do it for the usual class of test functions, where f(x) is CONTINUOUS with compact support. Otherwise you get into awkwardness like what is the integral of delta(0)*f(x), where f(x)=1 for x in [0,1] and 0 otherwise. For the same reason, I'd also be happier if the statement of the problem had said g(x) is continuous at 0.

 

[gabbagabbahey]

Integrating over regions is sort of the same as saying they are the same for step functions.

I guess, but the regions can be made arbitrarily small.

Both of our methods are easy to apply to this particular distribution, I just thought mine might be more intuitively obvious/easy to picture.

 

[Dick]

I guess, but the regions can be made arbitrarily small.

Both of our methods are easy to apply to this particular distribution, I just thought mine might be more intuitively obvious/easy to picture.

If f(x)=1 at x=0 and f(x)=0 otherwise, are you ok with saying integral f(x)*delta(0)dx=1?

 

[gabbagabbahey]

If f(x)=1 at x=0 and f(x)=0 otherwise, are you ok with saying integral f(x)*delta(0)dx=1?

Do you mean, am I okay with saying \int_{-\infty}^{\infty} f(x)\delta(x)dx=1? If so, yes. Should I not be okay with it?

 

[Dick]

Do you mean, am I okay with saying \int_{-\infty}^{\infty} f(x)\delta(x)dx=1? If so, yes. Should I not be okay with it?

No, I don't think you should. \int_{-\infty}^{\infty} f(x)\delta(x)dx is completely undefined. The only point I'm trying to make here is \int_{-\infty}^{\infty} f(x)\delta(0)dx=f(0) only if f is continuous at 0. It's the continuous bit I'm missing here.

 

[gabbagabbahey]

No, I don't think you should. \int_{-\infty}^{\infty} f(x)\delta(x)dx is completely undefined.

I don't see why. Could you show me a proof of that?

The only point I'm trying to make here is \int_{-\infty}^{\infty} f(x)\delta(0)dx=f(0) only if f is continuous at 0. It's the continuous bit I'm missing here.

Keep in mind that this problem is posed on a physics assignment, so it is naturally lacking a certain amount of mathematical rigor. In context, I would assume that by "function", the questioner is referring to a continuous function.

 

[fluidistic]

No, I don't think you should. \int_{-\infty}^{\infty} f(x)\delta(x)dx is completely undefined. The only point I'm trying to make here is \int_{-\infty}^{\infty} f(x)\delta(0)dx=f(0) only if f is continuous at 0. It's the continuous bit I'm missing here.

I just checked once again the exercise and it doesn't mention the word "continuous". Maybe it's ill posed.
Edit: and about the rigorousness of the mathematics, I believe I should be very rigorous. One of the professors to help us in this course is very rigorous. He proves everything, puts \vec 0 instead of just 0 when we're dealing with the null vector, etc.

 

[Dick]

I don't see why. Could you show me a proof of that?

<br /> \int_{-\infty}^{\infty} f(x)\delta(x)dx<br /> is just plain sloppy. You know that right? I don't need to prove it's undefined. delta(x) is 'infinite' at x. You are integrating an 'infinite' thing times f(x) for all x. You meant to write delta(0).

 

[gabbagabbahey]

<br /> \int_{-\infty}^{\infty} f(x)\delta(x)dx<br /> is just plain sloppy. You know that right? I don't need to prove it's undefined. delta(x) is 'infinite' at x. You are integrating an 'infinite' thing times f(x) for all x. You meant to write delta(0).

We are both talking about the Dirac Delta right? It's zero everywhere except at x=0... \delta(0) is undefined.

 

[Dick]

I just checked once again the exercise and it doesn't mention the word "continuous". Maybe it's ill posed.
Edit: and about the rigorousness of the mathematics, I believe I should be very rigorous. One of the professors to help us in this course is very rigorous. He proves everything, puts \vec 0 instead of just 0 when we're dealing with the null vector, etc.

gabbagabbahey is right, if it's a physics question 'continuous' may be implicit. If you want to make it more rigorous, put in the 'continuous' for f(x) and g(x). But you probably aren't being asked to prove it with epsilons and deltas.

 

[Dick]

We are both talking about the Dirac Delta right? It's zero everywhere except at x=0... \delta(0) is undefined.

You are right. I'm flubbing this one up seriously. Apologies. My head must be elsewhere. If continuity of everything is assumed (except the delta) and it's a physics course, I think your explanation works fine.

 

[gabbagabbahey]

You are right. I'm flubbing this one up seriously. Apologies. My head must be elsewhere.

What's her name?

If continuity of everything is assumed (except the delta) and it's a physics course, I think your explanation works fine.

To be honest, I'm not sure why g(\textbf{x}) would need to be continuous at the origin (although it would definitely have to be defined there). None of the definitions of the Dirac Delta that I can recall seeing, mention this. Jackson says \int_{\mathbb{R}^3}f(\textbf{x})\delta(\textbf{x})d^3x=f(\mathbf{0}) "for any function f". Does a more rigorous treatment require continuity?

 

[Dick]

What's her name?

No her, just being an airhead. But, yes, a rigorous treatment needs continuity. Again, take f(0)=1 and f(x)=0 otherwise. To really define the delta function you take a representation of the delta function, you represent the delta function as the limit of ordinary functions. Like d_n(x) where d_n(x) is n if |x|<1/(2n) and 0 otherwise. Then define the integral of f(x)*delta(x) to be the limit of the integral of f(x)*d_n(x) as n->infinity. Since the integral of f(x)*d_n(x)=0 for all n, you are clearly in trouble if you claim integral f(x)*delta(x)=f(0)=1. Because f(x) isn't continuous at zero. That's all.

 

[gabbagabbahey]

But, yes, a rigorous treatment needs continuity. Again, take f(0)=1 and f(x)=0 otherwise. To really define the delta function you take a representation of the delta function, you represent the delta function as the limit of ordinary functions. Like d_n(x) where d_n(x) is n if |x|<1/(2n) and 0 otherwise. Then define the integral of f(x)*delta(x) to be the limit of the integral of f(x)*d_n(x) as n->infinity. Since the integral of f(x)*d_n(x)=0 for all n, you are clearly in trouble if you claim integral f(x)*delta(x)=f(0)=1. Because f(x) isn't continuous at zero. That's all.

Okay, thanks Dick.

 

[Dick]

Okay, thanks Dick.

Thank YOU. I was being a bit incoherent in the context of the physics problem.