1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Product of a delta function and functions of its arguments

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data
    I am trying to determine whether
    $$f(x)g(x')\delta (x-x') = f(x)g(x)\delta (x-x') = f(x')g(x')\delta(x-x')$$
    where [itex]\delta(x-x')[/itex] is the Dirac delta function and [itex] f,g[/itex] are some arbitrary (reasonably nice?) functions.
    2. Relevant equations
    The defining equation of a delta function:
    $$\int_{-\infty}^{\infty} \delta(x-x')f(x')dx' = f(x)$$
    (I'm supposing [itex]x'[/itex] is the variable of integration, but it shouldn't matter I think)

    3. The attempt at a solution
    It appears that from the integral definition of the delta function,
    $$\int_{-\infty}^{\infty}f(x)g(x')\delta (x-x') h(x')dx' = f(x)g(x)h(x) = \int_{-\infty}^{\infty}f(x)g(x)\delta (x-x') h(x')dx' = \int_{-\infty}^{\infty}f(x')g(x')\delta (x-x') h(x')dx'$$
    for all [itex]h(x)[/itex]. Thus, the above identity appears to be correct. However, when I ask WolframAlpha a special case of this question, the answer is that the identity is false. How can I determine which it is?

    Any comments/suggestions would be highly appreciated!
     
  2. jcsd
  3. Apr 7, 2016 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I don't have an answer, but if you replace ##y## with any concrete number, then the equality becomes correct. So wolfram is interpreting ##y## as a variable, while in your proof you treat it as a constant.
     
  4. Apr 7, 2016 #3

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Can someone give a counterexample ?
     
  5. Apr 7, 2016 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    For which special case did WolframAlpha say it is false? AFIK what you wrote is true, at least if ##f## and ##g## are "nice" functions.

    If we understand that ##x'## is the integration variable, then the four operators
    [tex] L_1 = f(x') g(x) \delta(x'-x)\\
    L_2 = f(x) g(x') \delta(x'-x)\\
    L_3 = f(x')g(x') \delta(x'-x)\\
    L_4= f(x) g(x) \delta(x'-x)
    [/tex]
    produce the same result when applied to any test function ##h(x')##, so in the generalized-function sense we have ##L_1 = L_2 = L_3=L_4##.

    I think it is important that ##x'## and ##x## are "independent"; if they somehow vary together (for example, if ##x = \phi(x')##) then, of course, everything can change.
     
  6. Apr 8, 2016 #5
    Thanks! I indeed meant that ##x## and ##x'## are independent. The special case I entered into WolframAlpha is ##xy\delta (x-y) \overset{?}{=} x^2\delta (x-y)## to which it returned false. Following micromass' comment, I entered ##y=5## and it returned true as an "alternate form" while still answering false as the main answer. Apparently something's wrong with WolframAlpha itself then.

    Thanks again!

    EDIT: As a follow up: To prove some equality of generalized functions, is it sufficient to simply show as above that they give the same result for all possible test functions? In other words, can such functions be distinguished in any way without considering their action on a test function?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Product of a delta function and functions of its arguments
  1. Delta Function (Replies: 4)

  2. Delta function (Replies: 3)

  3. DElta function (Replies: 4)

  4. The delta function (Replies: 0)

  5. Delta function (Replies: 1)

Loading...