Dirac-Delta Functions and Double Integrals

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In summary, the homework problem is to show that 2C*\delta(x-ct)\delta(x+ct) = \delta(x)\delta(t) using calculus and Dirac-delta properties. This can be done by taking the derivative of the double integral \iint f(x-ct, x+ct) 2c \delta(x-ct) \delta(x+ct) dxdt and then using a parameter transformation to solve the resulting integral.
  • #1
zheng89120
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Homework Statement



show that [itex]\delta[/itex](x-ct)[itex]\delta[/itex](x+ct) = [itex]\delta[/itex](x)[itex]\delta[/itex](t)

P.S. sorry I mean't:

show that 2C*[itex]\delta[/itex](x-ct)[itex]\delta[/itex](x+ct) = [itex]\delta[/itex](x)[itex]\delta[/itex](t)

Homework Equations



calculus and Dirac-delta properties

The Attempt at a Solution



[tex]d/dx \int_{-\infty}^x\delta(x-ct)\delta(x+ct) = \delta(x)\delta(t) dx[/tex]

P.S. sorry I mean't:

[tex]2C*d/dx \int_{-\infty}^x\delta(x-ct)\delta(x+ct) = ...[/tex]

there are a couple of really weird steps that somebody else used after the above
 
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  • #2
zheng89120 said:

Homework Statement



show that [itex]\delta[/itex](x-ct)[itex]\delta[/itex](x+ct) = [itex]\delta[/itex](x)[itex]\delta[/itex](t)

Homework Equations



calculus and Dirac-delta properties

The Attempt at a Solution



[tex]d/dx \int_{-\infty}^x\delta(x-ct)\delta(x+ct) = \delta(x)\delta(t) dx[/tex]

there are a couple of really weird steps that somebody else used after the above

Why are you taking the derivative? How is the Dirac delta function defined?
 
  • #3
Hi zheng89120! :smile:

Do you know how to calculate for any generic function f(u,v):
[tex]\iint f(x-ct, x+ct) 2c \delta(x-ct) \delta(x+ct) dxdt[/tex]

And do you also know how to calculate this double integral after a parameter transformation to (u, v), where u=x-ct and v=x+ct?
 

1. What is a Dirac-Delta function?

A Dirac-Delta function, also known as the Dirac delta distribution or impulse function, is a mathematical function that is defined to be zero everywhere except at one point, where it is infinite. It is commonly used to represent a point mass or impulse in physics and engineering.

2. How is the Dirac-Delta function different from a regular function?

The Dirac-Delta function is different from a regular function in that it is not defined in the traditional sense. It is a distribution, rather than a function, and cannot be evaluated at specific points. It also has properties that are not shared by regular functions, such as the property of being infinite at one point.

3. What are the applications of Dirac-Delta functions?

Dirac-Delta functions have various applications in physics and engineering. They are commonly used to represent point sources, such as a point charge or a point mass, in fields like electromagnetism and fluid mechanics. They are also used in signal processing to model impulses in a system.

4. Can the Dirac-Delta function be integrated?

Yes, the Dirac-Delta function can be integrated over a suitable interval. Its integral over an interval that includes its point of singularity is equal to 1. However, its integral over an interval that does not contain its point of singularity is equal to 0, which is why it is often described as a distribution rather than a function.

5. Are there any limitations to using Dirac-Delta functions?

While the Dirac-Delta function is a useful mathematical tool, it is important to note that it is an idealized function and does not exist in the physical world. It is often used as an approximation in real-world applications. Additionally, its properties can lead to mathematical inconsistencies and must be used with caution.

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