Dirac delta function proof help

In summary, Dirac delta function is a function that is undefined at x=0, and it has a limit as the series of functions that get more and more peaked near zero.
  • #1
Raze2dust
63
0
[SOLVED] Dirac delta function

Homework Statement


Prove that [tex]\delta(cx)=\frac{1}{|c|}\delta(x)[/tex]


Homework Equations





The Attempt at a Solution



For any function f(x), [tex]
\int_{-\infty}^{\infty}f(x)\delta(cx) dx = \frac{1}{c}\int_{-\infty}^{\infty}f(t/c)\delta(t) dt
[/tex]

where I have used t=cx.
[tex]
=\frac{1}{c}f(0)
[/tex]
This is fine and matches RHS for c>0. But how do we get the mod sign for c<0. Why isn't the above procedure valid for c<0 as well?
 
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  • #2
You have implicitly assumed that c>0 when you changed variables. Either that, or you forgot what else needs to be changed...
 
  • #3
Raze2dust said:

Homework Statement


Prove that [tex]\delta(cx)=\frac{1}{|c|}\delta(x)[/tex]


Homework Equations





The Attempt at a Solution



For any function f(x), [tex]
\int_{-\infty}^{\infty}f(x)\delta(cx) dx = \frac{1}{c}\int_{-\infty}^{\infty}f(t/c)\delta(t) dt
[/tex]

where I have used t=cx.
[tex]
=\frac{1}{c}f(0)
[/tex]
This is fine and matches RHS for c>0. But how do we get the mod sign for c<0. Why isn't the above procedure valid for c<0 as well?

What happens to the limits of integration when you make the change of variable and c is negative?
 
  • #4
nrqed said:
What happens to the limits of integration when you make the change of variable and c is negative?

thanks nrged!
wonder how it didn't strike me after pulling my hair off for one hour
 
  • #5


Consider a spring-mass system with a time-dependent force f(t) applied to the mass.
The situation is modeled by the second-order differential equation
mx"(t) + cx0'(t) + kx(t) = f(t)
where t is time and x(t) is the displacement of the mass from equilibrium. Now suppose
that for t greater than or equal to 0 the mass is at rest in its equilibrium position, so x'(0) = x(0) = 0. At
t = 0 the mass is struck by an “instantaneous” hammer blow. This situation actually occurs
frequently in practice—a system sustains an forceful, almost-instantaneous input. Our goal
is to model the situation mathematically and determine how the system will respond.
In the above situation we might describe f(t) as a large constant force applied on a very
small time interval. Such a model leads to the forcing function
 
  • #6


The Dirac delta function is a special kind of object - it is not a
function in the usual sense, although we sometimes think of it that
way. Basically, it can be thought of as the "function" d(x) such that:

d(x) = 0 for x <> 0
= Infinity for x = 0

and the "size" of the infinity is such that the integral of d(x) with
respect to x from a to b is zero if a and b are of the same sign, and
the integral is exactly unity if a < 0 < b. More precisely, it can be
thought of as the limit of a series of functions that get more and
more peaked near zero, and whose area is always unity.

It is always to be implicitly used inside an integral, even though we
carry out operations such as:

f(x) = d(x) + d(x-2) (when integrated, this has area 2)
 
Last edited:
  • #7


Consider a spring-mass system with a time-dependent force f(t) applied to the mass.
The situation is modeled by the second-order differential equation
mx"(t) + cx0'(t) + kx(t) = f(t)
where t is time and x(t) is the displacement of the mass from equilibrium. Now suppose
that for t greater than or equal to 0 the mass is at rest in its equilibrium position, so x'(0) = x(0) = 0. At
t = 0 the mass is struck by an “instantaneous” hammer blow. This situation actually occurs
frequently in practice—a system sustains an forceful, almost-instantaneous input. Our goal
is to model the situation mathematically and determine how the system will respond.
In the above situation we might describe f(t) as a large constant force applied on a very
small time interval. Such a model leads to the forcing function
f (t) = [tex]\left\{\stackrel{^{\frac{A}{\epsilon},}}{0,else}[/tex] 0 [tex]\leq[/tex] t [tex]\leq[/tex] [tex]\epsilon[/tex]

Here A is some constant and [tex]\epsilon[/tex] is a “small” positive real number. When [tex]\epsilon[/tex] is close to zero the applied force is very large during the time interval 0 [tex]\leq[/tex] t [tex]\leq[/tex] [tex]\epsilon[/tex]
In this case it’s easy to see that for any choice of [tex]\epsilon[/tex] we have
[tex]\int[/tex][tex]\stackrel{\infty^}{f (t) dt = A}[/tex]
[tex]-\infty[/tex]
 
Last edited:
  • #8


k..thanx for the info Horbacz !
:)
 
  • #9


your welcome raze 2 dust. and by the way, I am only 13...
 

1. What is the Dirac delta function?

The Dirac delta function, also known as the unit impulse function, is a mathematical function that is used to represent a point mass or a point charge in physics and engineering. It is defined as zero everywhere except at the origin, where its value is infinite, and its integral over the entire real line is equal to one.

2. How is the Dirac delta function used in proofs?

The Dirac delta function is often used in proofs to simplify mathematical expressions and to model physical phenomena such as impulse forces and point charges. It is a powerful tool in mathematical analysis and is commonly used in the fields of physics, engineering, and signal processing.

3. What is the intuition behind the proof of the Dirac delta function?

The proof of the Dirac delta function involves taking the limit of a sequence of functions that approximate a delta function. These functions have a very narrow peak at the origin and are zero everywhere else. As the peak becomes narrower and taller, the area under the curve remains constant at one, representing the "impulse" or "point mass" at the origin.

4. Why is the Dirac delta function considered a distribution?

The Dirac delta function is considered a distribution because it does not have a well-defined value at any point, but rather it is defined by its properties when integrated with other functions. It is a generalized function that is used to describe point-like objects in mathematical models.

5. Are there any real-life applications of the Dirac delta function?

Yes, the Dirac delta function has many real-life applications. It is commonly used in engineering to model impulse forces and in signal processing to analyze signals. It is also used in physics to model point charges and point masses. Additionally, the Dirac delta function is used in probability theory to represent the probability of a continuous random variable taking a specific value.

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