Derivative of Dirac Delta function

In summary, the conversation is about evaluating the derivative of the dirac delta function when paired with a function g(x) and using the integral notation and integration by parts to solve it. There is also a discussion about the validity of using the normal distribution and taking the limit as the standard deviation approaches 0. The conversation ends with a suggestion to consult a book for a proof and a reference to a similar problem being discussed in a forum.
  • #1
LedPhoton
10
0
Hello I'm trying to figure out how to evaluate(in the distribution sense)
[itex]\delta'(g(x))[/itex]. Where [itex]\delta(x)[/itex] is the dirac delta function. Please notice that what I want to evaluate is not [itex] \frac{d}{dx}(\delta(g(x)))[/itex] but the derivative of the delta function calculated in g(x).
If anyone could post a proof, an idea to find the proof or a link it would be greatly appreciated!
 
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  • #2
[itex] \delta' [/itex] is a linear operator on functions. Are you referring to its value when you pair it with g(x), as in
$$ \langle \delta',\, g\rangle = \int_{\mathbb{R}} \delta'(x)g(x)\, dx \, ?$$
I'll assume you are. In that case, you use the integral notation above and then symbolically do integration by parts. Don't worry if it is not a well-defined operation because the answer you get is literally the definition of what you want.

Look here under distributional derivatives for more info:
http://en.wikipedia.org/wiki/Dirac_delta_function
 
  • #3
No I'm sorry if I wasn't clear. I understand the value of
[itex]\int \delta'(x)g(x)dx[/itex]
I'm asking the value of
[itex]\int \delta'(g(x))\phi(x)dx[/itex]
Where [itex]\phi(x)[/itex] is the test function. Here it is not immediately obvious to me how to integrate by parts. I thought about this(but I am unsure of whether it is correct):
Assume that g(x) is an invertible function with as many derivatives as necessary(to keep things simple for now), so we substitute [itex]y = g(x)[/itex] and get
[itex]\int \delta'(y)\frac{\phi(g^{-1}(y))}{g'(g^{-1}(y))} dy[/itex] Now I could integrate by parts and get
[itex]-\int \delta(y)\frac{d}{dy}(\frac{\phi(g^{-1}(y))}{g'(g^{-1}(y))}) dy[/itex]
Do you think my reasoning is correct up to here?
 
  • #4
The dirac delta is just a normal distribution who's standard deviation approaches 0. Take the derivative of the normal dist. then take the limit as stdev =>0. I'm not sure if that's a valid way to do the problem, but its what I would try.
 
  • #5
Ok thanks, I'll try that
 
  • #6
Your calculation looks right to me, and what Aero said makes sense too. As far as proof goes, I can't remember exactly how general the rules are for changing variables like that. A book like Friedlander would probably have it...

Check out the very last post here for a similar problem:
https://www.physicsforums.com/showthread.php?t=201774&page=2
There is no proof, but there is a citation.

If g is not injective, then in the end, when you evaluate against a test function, you should get a sum of terms, one for each zero of g. If g' and g are simultaneously 0 at any point, then I don't think the distribution is well-defined.
 

FAQ: Derivative of Dirac Delta function

1. What is the Dirac Delta function?

The Dirac Delta function, denoted as δ(x), is a mathematical function used in the field of calculus to represent an infinitely sharp spike at a particular point. It is defined as zero everywhere except at the origin, where it is infinite.

2. What is the purpose of the Dirac Delta function?

The Dirac Delta function is commonly used in mathematics and physics to model point sources or idealized impulses. It is also used as a tool for solving differential equations and evaluating integrals.

3. How is the derivative of the Dirac Delta function defined?

The derivative of δ(x) is defined as a distribution, rather than a traditional function. It is denoted as δ'(x) and is equal to zero everywhere except at the origin, where it is not well-defined. However, when integrated over any interval containing the origin, the result is 1.

4. What is the relationship between the Dirac Delta function and the Heaviside step function?

The Heaviside step function, also known as the unit step function, is defined as 0 for negative x and 1 for positive x. It can be thought of as the integral of the Dirac Delta function. In other words, the Dirac Delta function is the derivative of the Heaviside step function.

5. How is the Dirac Delta function used in practice?

The Dirac Delta function is used in various fields of science and engineering, such as signal processing, quantum mechanics, and fluid mechanics. It is also used in the study of distributions and generalized functions, and plays a crucial role in the development of the theory of distributions.

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