Services & Resources / Wolfram Forums / MathGroup Archive
-----

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 smc.vnet.net
  • Subject: [mg85246] Re: Using "Limit" when the limit is a delta function
  • From: danl at wolfram.com
  • Date: Mon, 4 Feb 2008 03:10:47 -0500 (EST)
  • References: <fnuhi4$9te$1@smc.vnet.net>

On Feb 1, 1:25=A0am, Jim Rockford <jim.rockfo... at gmail.com> 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

I don't know if the approach below will exactly do what you want, but
it might give some ideas for alternatives you could try. As you
observe, you cannot get directly, from Limit, a limiting form of a
function that behaves as a delta distribution. But you can use simple
forms, in conjunction with Integrate, to emulate the behavior of that
distribution.

Here is one such. Since it plays nice with Integrate, I use the
rectangular pulse form that, in the limit, behaves like a delta
functional. The function I integrate against is (x-x0)^n with n
assumed to be nonnegative. The idea being, you can form real analytic
functions in a neighborhood of x0 using sums of these.

In[42]:= Limit[
 Integrate[
  Boole[Abs[x] <= a]*1/(2*a)*(x + x0)^n, {x, -Infinity, Infinity},
  Assumptions -> {0 < a < 1/1000, n >= 0, Element[x0, Reals]}],
 a -> 0]

Out[42]= x0^n

Compare to:

In[43]:= Integrate[DiracDelta[x]*(x + x0)^n, {x, -Infinity, Infinity},
  Assumptions -> {Element[x0, Reals]}]

Out[43]= x0^n


Daniel Lichtblau
Wolfram Research



  • Prev by Date: Re: How edit a saved palette?
  • Next by Date: Re: Possible bug in ListAnimate[] or Manipulate[] v6.0
  • Previous by thread: Re: Using "Limit" when the limit is a delta function
  • Next by thread: Re: question about the function "go to"