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Delta problems

  1. May 19, 2010 #1
    Hi alll,

    I have an integral which includes a Kronecker delta:

    I = \int_{u=0}^{a} \int_{v=0}^{a} F(u) G(v) \delta_{u,v} \, \mathrm{d}u \, \mathrm{d}v

    I know that for a 1D integral there exists the special property: \int F(u) DiracDelta(u-a) = F(a)

    However, is there something equivalent for the problem I have stated in the first equation? I was thinking perhaps that:

    I = \int F(u) G(u) du dv or something similar....


  2. jcsd
  3. May 19, 2010 #2


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    Fixing your latex to make it readable:

    I = \int_{u=0}^{a} \int_{v=0}^{a} F(u) G(v) \delta_{u,v} du dv
  4. May 19, 2010 #3


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    Are you sure that it is a Kronecker Delta and not a Dirac Delta?
  5. May 19, 2010 #4
    Yes I believe so, but I did derive the equation myself. I thought there would be a way to convert between the two types of delta functions...
  6. May 19, 2010 #5
    I have just obtained the following simplification, which may make the next step easier:

    I = \int_{u=0}^{a} \int_{v=0}^{a} F(u) F(v) \delta_{u,v} \, \mathrm{d}u \, \mathrm{d}v

    i.e. the two functions are the same, just with a different variable.

  7. May 19, 2010 #6
    I think it should be zero... the only thing that makes the integral with a Dirac Delta produce something different of zero is that it assumes infinite value at a certain point. For any finite value, which is the case for any function by definition (the Dirac Delta is NOT a function by definition), the Riemann integral should be zero, as it's not altered by a single point.
  8. May 19, 2010 #7


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    You're really sure it's a Kronecker delta? Were you always dealing with integrals, or did you have sums that you approximated as integrals?

    How did you arrive at this equation? Kronecker deltas do not take continuous variables as their arguments, so I would suspect that you either converted sums to integrals but didn't convert the Kronecker delta or you made some other error. Please tell us how you got the Kronecker so that we can try to figure out what the problem is.
  9. May 20, 2010 #8
    Yes it is definitely Kronecker.

    I also think that this integral is zero. When I plot the function, it looks like a 2D function in a 3D environment and thus the "area" under the curve is infinitesimal.

    Thanks for your advise on this!

  10. May 20, 2010 #9


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    If this is, in fact, a continuous Kronecker [itex]\delta[/itex], [itex]\delta_{xy}= 1[/itex] if x= y, 0 other wise, then [itex]F(u)G(u)\delta_{uv}[/itex] is 0 except on the line u= v. Since that is one dimensional it has two dimensional measure 0 and the double integral of any function on that set is 0.
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