What is the Delta Function Identity?

Click For Summary
SUMMARY

The Delta Function Identity is expressed as f(x) d/dx δ(x) = f(0) d/dx δ(x) - f'(0) δ(x). This identity illustrates the relationship between a function and its derivative when integrated against the Dirac delta function, δ(x). The discussion emphasizes the use of integration by parts to derive the identity, specifically applying the properties of distributions to map functions to their derivatives. The key takeaway is the manipulation of δ(x) and its derivative in the context of functional analysis.

PREREQUISITES
  • Understanding of Dirac delta function properties
  • Familiarity with distribution theory
  • Knowledge of integration by parts technique
  • Basic calculus and differentiation concepts
NEXT STEPS
  • Study the properties of the Dirac delta function in detail
  • Learn about distribution theory and its applications
  • Explore advanced integration techniques, particularly integration by parts
  • Investigate functional analysis and its relevance to differential equations
USEFUL FOR

Mathematicians, physicists, and students studying advanced calculus or functional analysis who seek to understand the applications of the Delta Function Identity in theoretical contexts.

ArcherVillage
Messages
4
Reaction score
0
I know I haven't entered the formulae with the proper syntax, but I'm extremely exhausted at the time of posting, so please just read it and give advice, forgiving me this once for not using proper form (it's basically in latex code format).

Homework Statement



Show [itex]f(x)\frac{d}{dx}\delta(x) = f(0)\frac{d}{dx}\delta(x)-f'(0)\delta(x)[/itex]

Homework Equations


[tex]\delta(x)\intf(x)dx = \int f(x)\delta(x)dx = f(0)[/tex]
[tex](f',\phi):=(f,-\phi ')[/tex]


The Attempt at a Solution


[tex]f(0)\delta'(x) - f'(0)\delta(x)[/tex]
[tex]=\delta'(x)\intf(x)\delta(x)dx-\delta(x)\intf'(x)\delta(x)dx[/tex]
[tex]=...[/tex]
 
Last edited by a moderator:
Physics news on Phys.org
I would use "integration by parts". To integrate
[tex]\int (f(x) d(\delta(x))/dx) dx[/tex]
let u= f(x) and [itex]dv= d(\delta(x))/dx dx[/itex] so that du= f'(x) dx and [itex]v= \delta(x)[/itex].=

Then [tex]\int (f(x) d(\delta(x))/dx) dx= f'(x)\delta(x)- \int f'(x)\delta(x) dx[/tex]
[tex]= f'(x)\delta(x)- f'(0)[/itex].<br /> <br /> So [itex]d(\delta(x))/dx[/itex] is the distribution that maps f(x) to [itex]f'(x)\delta(x)- f'(0)[/itex].[/tex]
 

Similar threads

  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
1K
Replies
7
Views
2K
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K