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Two-variable functions defined on a one-dimensional subset

  1. Dec 7, 2011 #1

    it is known that by using delta-functions we can somehow assign a meaning to "functions" that take a specific value at one point, as a sort of generalization of the Kronecker delta-function in discrete domain.

    Instead I would like to consider the domain R2, and formally define a "function" that takes only specific values along a one-dimensional path inside R2. This path on R2 is defined by the points [itex]\left( x,p(x) \right)[/itex], with p:R--->R.

    Note: if I had had to define a function taking specific values at some specific points, I would have defined it as a finite sum of delta functions [itex]\delta (x,y)[/itex], but in this case I cannot use discrete sums of delta functions.

    Is it possible to express formally such a (non-)function?
    Is it allowed to do something like the following:

    [tex]\int_{R^2}f(\mathbf{t})\delta(\mathbf{t}-\mathbf{p})|d\mathbf{t}| = f(\mathbf{p}) = f(x,p(x))[/tex]

    where the path p is given by [itex]\mathbf{p}(x)=\left( x,p(x) \right)[/itex] , t is some position vector in [itex]R^2[/itex], and [itex]\delta[/itex] is the Dirac delta-function in 2 dimensions.

    Last edited: Dec 7, 2011
  2. jcsd
  3. Dec 7, 2011 #2


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    Homework Helper

    Your guess looks a bit odd - maybe it's just your notation that's not clear. If I understand correctly, you have some function, F, which is a function of a 2d vector x, and you want it to be equal to some other function, say g(x), but only along some path in the 2d space.

    This is how I would probably write the function: first, I would parametrize the curve in [itex]\mathbb{R}^2[/itex] using a parameter t. i.e., the path is some vector [itex]\mathbf{\gamma}(t)[/itex]. Then, I think one can write the function as

    [tex]F(\mathbf{x}) = g(\mathbf{x})\int_{-\infty}^\infty dt~\delta(\mathbf{x}-\mathbf{\gamma}(t)).[/tex]

    When the vector x lies along the path [itex]\mathbf{\gamma}(t)[/itex], the delta function in the integral "pings" and the integral is one. When x does not lie along the path, the delta function doesn't ping and the integral is zero.

    This looks superficially similar to what you wrote, but note my parametrization and hence the integral is one dimensional, whereas your integral seems to be two dimensional.
  4. Dec 7, 2011 #3
    Hi Mute!

    thanks a lot! You got my question despite it was formulated in a confusing manner.
    Your answer was essentially what I wanted to know.
  5. Dec 7, 2011 #4
    Basically on two dimensional space your delta function will be a product as

    δ(x−(x,p(x))) = δ(x1−x)*δ(x2−p(x))

    and so on. For instance a normal delta function that is zero at the point (p,q) of two dimensional space is δ(x1−p)*δ(x2− q).

    And finally, there are not delta functions :p
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