Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Delta function of two variable function

  1. Jun 20, 2012 #1
    Hi
    Iknow that if we have delta function of one variables function, then we can write it as:

    [itex]\delta (f(x)) = \sum \frac{\delta(x-x0)}{f'(x0)} [/itex]

    but how we can write a function of two variables:

    [itex]\delta (f(x,y))[/itex]
     
  2. jcsd
  3. Jun 20, 2012 #2
    What exactly do you mean by delta function? Are you referring to dirac delta function or something else?
     
  4. Jun 20, 2012 #3

    Dirac delta function
     
  5. Jun 21, 2012 #4
    I am not aware of such an identity. What is the sum over?

    The dirac delta function for a single variable is simply ##\delta [f(x)] = f(0) ##. In two variables it would be ##\delta [f(x,y)] = f(0,0)##. Of course you can shift the delta function to any point off the origin as well.
     
  6. Jun 22, 2012 #5
    This is a well-known and easily provable identity. The sum is over the zeroes x0 of f.
    I think there's some confusion here. A distribution can indeed be represented as a linear functional which takes (test) functions as inputs. But here, when we say ##\delta (f(x)) ##, we don't mean the functional ##\delta ## evaluated at the function f. Rather, informally we are thinking of the Dirac delta as a function of numbers, and we are forming the composition of the Dirac delta function and the function f. So we mean the "function" ##\delta (u)## evaluated at the number u=f(x).
     
  7. Jun 22, 2012 #6

    pwsnafu

    User Avatar
    Science Advisor

    One way to go about this is by the observation that every real number has an associated constant function [itex]g_c (t) = c[/itex]. Trivally every constant function is a distribution. Then [itex]\delta(f(x)) = \delta \circ f \circ g_x[/itex] and if the resulting expression is also a constant function you can use the duality backwards to "claim" [itex]\delta(f(x)) \in \mathbb{R}[/itex].

    I remember hearing about this in a lecture some time in my undergraduate years, but I haven't seen it mentioned in the literature.
     
  8. Jun 22, 2012 #7
    Would you mind elaborating on this? I've worked out the case where ##f## is injective below, but I don't see how one would define composition where ##f## vanishes at more than one point.


    If ##d, g, f## are test functions and ##f## is injective and ##f(x_0)=0## then
    $$\langle d\circ f, g \rangle = \int d\circ f(x)\cdot g(x) dx $$. A u-substitution yields
    $$ = \int\frac{d(u)\cdot g\circ f^{-1}(u)}{f'(f^{-1}(u))}du = \left\langle d(u), \frac{ g\circ f^{-1}(u)}{f'(f^{-1}(u))} \right\rangle$$

    Thus this should be taken as the definition for composition of a general distribution with such an ##f##. Inserting the dirac delta then, I see that
    $$\langle \delta\circ f,g\rangle = \frac{ g\circ f^{-1}(0)}{f'(f^{-1}(0))} = \frac{ g(x_0)}{f'(x_0)} = \left\langle\frac{\delta_{x_0}}{f'(x_0)} ,g\right\rangle $$
    which recovers the formula (excusing my sloppy u-substitution).

    To reiterate my question, if ##f## vanishes at several points, what motivation is there to define the OP's formula?
     
  9. Jun 22, 2012 #8
  10. Jun 22, 2012 #9
    One approach you might try is to calculate in a similar fashion ##\delta (f(x)g(x))##, where f(x) has a single zero at x1 and g(x) has a single zero at x2.
     
  11. Jun 22, 2012 #10
    The delta function is non-zero only when it's input is zero. The OP's formula picks out the points where the argument of the delta function is non-zero and expands the expression into delta functions centred at these points.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Delta function of two variable function
Loading...