Proving the scaling property of the Delta function

In summary, the given equation can be proven by substituting u = at and transforming the integration boundaries, and taking the absolute value of a into account. This helps with the proof of the scaling function by allowing for a more simplified calculation.
  • #1
caesius
24
0

Homework Statement



Prove that [tex]\delta(at)=\frac{1}{abs(a)}\delta(t)[/tex]

Hint: Show that [tex]\int\phi(t)\delta(at)dt=\frac{1}{abs(a)}\phi(0)[/tex]

(the limits of integration are from -inf to +inf btw, I couldn't find how to put them in..)

Homework Equations





The Attempt at a Solution



Ok. I understand that the integral is only defined for at = 0, i.e., t = 0. But if I follow this logical then I evaluate the integral to be phi(0), not phi(0)/a. Where does the a come from?

And I'm not entirely sure how proving that relationship helps with the proof of the scaling function...
 
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  • #2
That trick only works when the argument to the delta is the bare integration variable.
So try substituting u = a t, so you get something like
[tex]\int_{-\infty}^{\infty} f(u) \delta(u) \, du = f(0)[/tex]
(and click the formula to see how I did the boundaries)
 
  • #3
Ok I've substituted x = at, now I get

[tex]
\int_{-\infty}^{\infty}\phi(\frac{x}{a})\delta(x)d\frac{x}{a} = \frac{1}{a}\int_{-\infty}^{\infty}\phi(\frac{x}{a})\delta(x)dx = \frac{1}{a}\phi(\frac{0}{a}) = \frac{1}{a}\phi(0)
[/tex]

I don't quite see how I'm allowed to take the 1/a out of the d/dx operater and put it at the front though. And why should it be the abs(a) not a?

And I still can't see how this helps prove anything.

Cheers, Benjamin
 
  • #4
You are allowed to take the a outside, because it is just a constant: du = a dt, so dt = (1/a) du, and you can put constants outside the integration as in
[tex] \int \tfrac12 x^2 \, dx = \tfrac12 \int x^2 \, dx[/tex]

The absolute value of a comes from transforming the integration boundaries. If a < 0, then -infinity goes to +infinity in the transformed integral, and vice versa. However, in the final formula you simply want -infinity to be the lower boundary, and +infinity as the upper boundary. So if a < 0 you have to swap them, at the cost of a minus sign, whereas for a > 0 you don't. Do you have to multiply the whole thing by sgn(a), and then you can shorten sgn(a) / a to 1/abs(a).
 

1. What is the scaling property of the Delta function?

The scaling property of the Delta function, also known as the Dirac delta function, states that when the argument of the function is multiplied by a constant, the value of the function at that point is also multiplied by the same constant.

2. How is the scaling property of the Delta function used in mathematics?

The scaling property of the Delta function is commonly used in mathematical equations and models to represent a point mass or impulse at a specific location. It is also used in the theory of distributions to define a generalized function that can be integrated against other functions in a meaningful way.

3. What is the mathematical expression for the scaling property of the Delta function?

The mathematical expression for the scaling property of the Delta function is:

δ(ax) = |a|⁻¹δ(x)

Where a is the scaling factor and δ(x) is the Delta function.

4. Can the scaling property of the Delta function be proven?

Yes, the scaling property of the Delta function can be proven using mathematical techniques such as substitutions and limit proofs. It can also be proven using the definition of the Delta function as a limit of a sequence of functions.

5. What are some applications of the scaling property of the Delta function?

The scaling property of the Delta function has various applications in different fields of science and engineering. Some examples include signal processing, quantum mechanics, and fluid dynamics. It is also used in solving differential equations and in the representation of point sources in physical systems.

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