- #1
Cathr
- 67
- 3
I started studying distribution theory and I am struggling with the understanding of some basic concepts. I would hugely appreciate any help, made as simple as possible, because by now I'm only familiar with the formalism, but not all the meaning behind.
The concepts I am struggling with are the following:
1. As a distribution, the Dirac delta function is defined by ∫ δ(x) φ(x) dx = δ(φ) = φ(0)
There are several things I do not understand here:
1.1. What is the meaning of φ(0)? There is a wide range of bump functions, so the value of phi in zero could vary for each test function, so why do we write that it is equal to the generalized delta function?
1.2. Can we write δ(x) outside the integral? Won't it have the same meaning as δ(φ)?
2. We have (f(0) φ(0))' - basically we want to derive the product of two functions.
What we obtain is f(0)'φ(0) + f(0)φ(0)'. The first term equals zero and we are left with f(0) multiplied by the derivative of φ(0)'. But by definition, the defivative of the delta function is MINUS the derivative of φ in zero. So the first result, by the classical derivation is wrong. I don't understand why...
3. This one gives me the most headaches. The goal is to determine (x-a)δa. To do this let's call f=x-a and T=δa (shouldn't we write Tδa ?).
So we have the following properties: fT(φ)=T(f φ)=f(a)φ(a)=0.
I don't understand the notations, and why can we include f in the parantheses along with φ, when φ is a test function, infinitely differentiable and f is not.
Also, suppose we look at the function f as a distribution => we write Tf . What should we do if we have:
gTf (φ) = ?
1° Tgf (φ) - we evaluate g as a distribution as well
2° Tf (g φ) - we multiply g with phi as in the previous example.
May someone please clarify the notations and the meaning begind? That would save my life...
Thank you very much in advance for the answers and for the patience.
EDIT: I think I understood a part of my question 3 - finally, the two variants that I wrote 1° and 2° are the same. However, how do we prove that ∫ (x-a) δ(x-a) φ(x) dx = 0?
The concepts I am struggling with are the following:
1. As a distribution, the Dirac delta function is defined by ∫ δ(x) φ(x) dx = δ(φ) = φ(0)
There are several things I do not understand here:
1.1. What is the meaning of φ(0)? There is a wide range of bump functions, so the value of phi in zero could vary for each test function, so why do we write that it is equal to the generalized delta function?
1.2. Can we write δ(x) outside the integral? Won't it have the same meaning as δ(φ)?
2. We have (f(0) φ(0))' - basically we want to derive the product of two functions.
What we obtain is f(0)'φ(0) + f(0)φ(0)'. The first term equals zero and we are left with f(0) multiplied by the derivative of φ(0)'. But by definition, the defivative of the delta function is MINUS the derivative of φ in zero. So the first result, by the classical derivation is wrong. I don't understand why...
3. This one gives me the most headaches. The goal is to determine (x-a)δa. To do this let's call f=x-a and T=δa (shouldn't we write Tδa ?).
So we have the following properties: fT(φ)=T(f φ)=f(a)φ(a)=0.
I don't understand the notations, and why can we include f in the parantheses along with φ, when φ is a test function, infinitely differentiable and f is not.
Also, suppose we look at the function f as a distribution => we write Tf . What should we do if we have:
gTf (φ) = ?
1° Tgf (φ) - we evaluate g as a distribution as well
2° Tf (g φ) - we multiply g with phi as in the previous example.
May someone please clarify the notations and the meaning begind? That would save my life...
Thank you very much in advance for the answers and for the patience.
EDIT: I think I understood a part of my question 3 - finally, the two variants that I wrote 1° and 2° are the same. However, how do we prove that ∫ (x-a) δ(x-a) φ(x) dx = 0?
Last edited: