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

Dirac distribution: identity

  1. Sep 29, 2012 #1

    I should prove:

    [tex] \delta'(\lambda x) = \dfrac{1}{\lambda \vert \lambda \vert} \delta(x), [/tex]
    where lambda is just a constant.

    If we make use of the scaling property and the definition of the distributional derivative, we find:

    [tex] \left( \delta'(\lambda x), f \right) = \dfrac{1}{\vert \lambda \vert} \left( \delta', f(x/\lambda) \right) = \dfrac{(-1)}{\lambda \vert \lambda \vert} \left( \delta, f'(x/\lambda) \right) = \dfrac{1}{\lambda \vert \lambda \vert} \left( \delta', f(x/\lambda) \right) [/tex]
    Because this is true for all testfunctions f, we have shown the identity.

    Now I think that my last step is wrong.

    Because of
    [tex] f'(x/\lambda) = \left( \dfrac{\partial f}{\partial x} \right) (x/\lambda) \neq \dfrac{\partial}{\partial x} \left( f(x/\lambda) \right) = \dfrac{1}{\lambda} f'(x/\lambda) [/tex]
    there should be an additional factor of lambda if I 'shift' the derivative back to the delta distribution.
    You can also see this by (formally) writing the last step:

    [tex] \dfrac{(-1)}{\lambda \vert \lambda \vert} \left( \delta, f'(x/\lambda) \right) = \dfrac{(-1)}{\lambda \vert \lambda \vert} \int_{-\infty}^{\infty} \delta(x) f'(x/\lambda) \, \mathrm{d}x = \dfrac{1}{\vert \lambda \vert} \int_{-\infty}^{\infty} \delta'(x) f(x/\lambda) \, \mathrm{d} x = \dfrac{1}{\vert \lambda \vert} \left( \delta', f(x/\lambda) \right), [/tex]
    where I used integration by parts.

    So something is really wrong here.
    Maybe someone could help me.
  2. jcsd
  3. Sep 29, 2012 #2


    User Avatar
    Gold Member

    I believe you are right.

    You can also look at this physically the dimensions of this distribution has [1/x].

    So you cannot have on one side dimensions of [1/(x\lambda)] and on the other side dimensions of [1/x\lambda^2].
  4. Sep 29, 2012 #3


    User Avatar
    Homework Helper

    What does the prime denote differentiation with respect to? Is it ##d\delta(\lambda x)/dx## or ##d \delta(\lambda x)/d(\lambda x)##?

    Also, I don't think the identity as you have written it can be correct. Are you trying to prove

    $$\delta'(\lambda x) = \dfrac{1}{\lambda \vert \lambda \vert} \delta'(x)?$$

    i.e., did you make a typo and mean to write the derivative of the dirac function on the RHS?

    You either need both sides to be derivatives of the Dirac function or you need the test function on the LHS and its derivative on the RHS to appear explicitly. Otherwise, for ##\lambda = 1##, your identity reduces to ##\delta'(x) = \delta(x)##, which is not true of course.
  5. Sep 29, 2012 #4
    I don't know. But, I think if you write
    [tex] \delta'(\lambda x)[/tex] you usually mean [tex] \left[ \dfrac{\partial \delta}{\partial x} \right] (\lambda x), [/tex]
    so you first build the derivative and then you evaluate it with lambda*x.

    I also tried ##\partial \delta(\lambda x)/ \partial x##, but I nevertheless cannot show the identity.

    Yes, your are right, it is a typo, thanks. It should be
    [tex] \delta'(\lambda x) = \dfrac{1}{\lambda \vert \lambda \vert} \delta'(x) [/tex]

    But still, I am not able to show the identity. Maybe the identity itself is wrong? What do you think?
  6. Sep 29, 2012 #5
    Ok, I should have somehting.

    [tex] (\delta'(\lambda x), f) = \dfrac{1}{\vert \lambda \vert} \left( \delta', f(x/\lambda) \right) = \dfrac{(-1)}{\vert \lambda \vert} \left( \delta, \partial [ f(x/\lambda) ]/ \partial x \right) = \dfrac{(-1)}{\lambda \vert \lambda \vert} \left( \delta, f'(x/\lambda) \right) = \dfrac{(-1)}{\lambda \vert \lambda \vert} f'(0) = \dfrac{1}{\lambda \vert \lambda \vert} \left( \delta', f \right) [/tex]

    So I used the chain rule in the 3rd step and
    in the last one just the definition of the distributional derivative, i.e., ##(\delta', f) = - f'(0)##
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook