MathGroup Archive 2008

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Using "Limit" when the limit is a delta function

  • To: mathgroup at
  • Subject: [mg85203] Re: Using "Limit" when the limit is a delta function
  • From: "D. Grady" <D.C.Grady at>
  • Date: Sat, 2 Feb 2008 03:22:36 -0500 (EST)
  • References: <fnuhi4$9te$>

Mathematica tries to make as few assumptions as possible about
undetermined parameters in an expression like this.  So, while in your
head you're probably thinking that D should be a positive real number,
Mathematica is thinking that D could be any random complex number.
The same goes for x.  You can specify assumptions like this using

Assuming[{x \[Element] Reals, d \[Element] Reals, d > 0},
 Limit[1/Sqrt[4 \[Pi] d t] Exp[-x^2/(4 d t)], t -> 0]]


Now the limit will evaluate, and it will give you zero, which again
isn't what you wanted.  Note also that if you assume x is zero

Assuming[{x == 0, d \[Element] Reals, d > 0},
 Limit[1/Sqrt[4 \[Pi] d t] Exp[-x^2/(4 d t)], t -> 0]]


the limit fails to converge, so in a sense, Limit is capturing the
behavior you want, but not evaluating to the form you want.  In fact,
if you look in the documentation for DiracDelta under the Possible
Issues section, they make a point of saying that the Limit of a smooth
function will never be DiracDelta.  The next two paragraphs are an
attempt to explain why I think this is the case, but this is math that
I don't understand really well, so take it with grain of salt.

Although it's easy to see heuristically that the limit of the quantity
you're looking at is a delta 'function,' it's only a function in a
distributional sense.  In the sense of normal, smooth functions that
are Riemann integrable, the delta distribution is not a proper
function at all, and proving rigorously that the limit of your
quantity is a delta distribution is not possible in the setting of
smooth functions.=17

Probably the reason that Limit does not evaluate to DiracDelta for
functions that 'look like' the delta distribution in some limit is
because a function may very well exhibit delta-like behavior as a
parameter becomes small, and yet converge (in a distributional sense)
to a distribution other than the delta distribution, or possibly not
converge at all.


On Feb 1, 1:25=A0am, Jim Rockford <jim.rockfo... at> wrote:
> I'm just wondering whether I can get Mathematica to understand the
> following limit (I'm using version 6.01).
> The standard solution for u(x,t) of the diffusion equation =A0u_t = D
> u_xx =A0 over an infinite domain with initial condition =A0 u(x,0) =
> delta(x) =A0 =A0(Dirac delta) is
> u[x_,t_] = (1/Sqrt[4 Pi D t]) Exp[-x^2/(4 D t)]
> This solution, in fact, defines a family of functions whose limit as
> t-->0 =A0yields the Dirac delta, as the initial condition dictates. =A0In
> Mathematica I tried
> Limit[u[x, t], t -> 0]
> and got nothing from it. =A0Can I get Mathematica to handle this sort of
> limit? =A0I realize that it involves an essential singularity at t=0,
> but I would've guessed that Mathematica is equipped to deal with it.
> An extra option needs to be specified or something?
> Thanks,
> Jim

  • Prev by Date: Re: Drawing on top of the image
  • Next by Date: Possible bug in ListAnimate[] or Manipulate[] v6.0
  • Previous by thread: Using "Limit" when the limit is a delta function
  • Next by thread: Re: Using "Limit" when the limit is a delta function