List Archives >  Maple User Group List Archive >  Archive by date >  This Month By Date >  This Month By Topic

[MUG] Dirac delta

Search email archive for  

[MUG] Dirac delta
Author: Herman Jaramillo    Posted: Mon, 05 Aug 2002 12:00:11 -0500
>> From: Herman Jaramillo "HJaramillo" Dear Maple Users: When I compute the integral of the Dirac delta function between 0 and infinity using Maple 7, I get "1". When I do it using Mathematica I get "1/2". When I try to do it using distribution theory by choosing a test function which has bounded support on the positive real line and using the limit from the left and the right of the interval of integration (epsilon, infinity) as epsilon goes to 0 I get that the integral should be "0". Does anyone understand this?. Can somebody tell me where I can I find a formal proof using distribution theory whether this integral is 1,1/2 or zero?. Best Regards. Herman. Herman Jaramillo phone: 713 689 6503 Senior Research Scientist fax: 713 689 6100 WesternGeco, email: "herman.jaramillo" P.O. Box 2469 Houston, Texas 77252-2469

[MUG] Re: Dirac delta
Author: Maple User Group    Posted: Wed, 7 Aug 2002 16:23:54 -0400 (
>> From: Maple User Group "maple_gr" | >> From: Herman Jaramillo "HJaramillo" |Dear Maple Users: | |When I compute the integral of the Dirac delta function between 0 and |infinity using Maple 7, I get "1". When I do it using Mathematica I |get "1/2". | |When I try to do it using distribution theory by choosing a test |function which has bounded support on the positive real line and using |the limit from the left and the right of the interval of integration |(epsilon, infinity) as epsilon goes to 0 I get that the integral |should be "0". -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Tue, 6 Aug 2002 11:30:51 -0400 To: "maple-list" From: "Gerald A. Edgar" "edgar" Subject: Dirac delta The delta "function" is defined so that int(f(x)*delta(x),x=-infinity .. infinity) = f(0) for any continuous function f. For discontinuous functions, it is strictly speaking not defined, so getting various answers in various programs is not unexpected. An alternative to think of this: the antiderivative of delta(x) is the Heaviside function, which jumps at 0. What is the value of Heaviside(x) exactly at the jump point? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Tue, 6 Aug 2002 11:45:36 -0400 To: "maple-list" From: Steve Campbell "slc" Subject: Dirac delta Your taking the left and right limit is only correct if the thing being integrated is a nice function and one is using something like the usual Riemann integral. The dirac delta is NOT A FUNCTION. It is a distribution. delta dt is a point measure. There are examples where the limit to infinity of a function does not work right either. Ie., you get a Reimann integral but not a Lebesque. The 1/2 sounds like there is some sort of Fourier series lurking around. You have to define exactly what space you are working on but I would go with a value of 1 for the usual differential equations type definition. Steve Campbell -- +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Department of Mathematics Phone: 1-919-515-3300 Box 8205 FAX: 1-919-515-3798 North Carolina State University email: "slc" Raleigh, NC 27695-8205 USA http://www4.ncsu.edu/eos/users/s/slc/www/index.html {Personal) http://www.math.ncsu.edu/ (Department) +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Tue, 6 Aug 2002 11:58:47 -0400 (EDT) From: John Robert Kitchin "jkitchin" To: "maple-list" Subject: Dirac delta The integral of a dirac delta function over any interval that contains it is by definition 1 (it is zero in any interval that does not contain it). I don't know how you got 1/2 out of mathematica, my version correctly outputs 1. Check out this website, it is one of many I found by googling dirac delta function: http://www.chm.uri.edu/urichm/chm532/delta/node4.html it also gives an example of deriving the delta function from the limit of a distribution. j -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Tue, 6 Aug 2002 12:25:54 -0400 (EDT) From: Andrzej Hanyga "hanyga" Subject: Dirac delta To: "maple-list" You are applying the delta function with a singularity at 0 to a distribution (the Heaviside function) with a singularity at 0 too, or multiplying two distributions. Such products are not defined. In general products of distributions are defined only in Colombeau's theory. In both tests it was just garbage in garbage out. AH =============================================================================== Andrzej Hanyga tel. 1-(617)-384 9624 fax 1-(617)-495 9837 Visiting Scholar Dept of Engineering and Applied Sciences, Harvard University Pierce Hall, 29 Oxford Street Cambridge MA

[MUG] Re: Dirac delta
Author: Maple User Group    Posted: Wed, 7 Aug 2002 16:24:10 -0400 (
>> From: Maple User Group "maple_gr" -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Tue, 6 Aug 2002 15:13:43 -0400 (EDT) From: Fred Chapman "fwchapma" To: "maple-list" Subject: Dirac delta By definition, Dirac's delta distribution satisfies (in Maple notation) infinity / | | f(x) Dirac(x) dx = f(0) | / -infinity In your case, f(x) is implicitly defined as the Heaviside function H(x), because infinity infinity / / | | | Dirac(x) dx = | H(x) Dirac(x) dx | | / / 0 -infinity where { 0 x < 0 H(x) = { { 1 0 < x The value of your integral is therefore H(0). The problem is that the definition of H(0) is itself ambiguous! Some people define H(0) = 1 so that H(x) is right-continuous at x = 0; some people define H(0) = 0 so that H(x) is left-continuous at x = 0; some people define H(0) = 1/2, which is the average (arithmetic mean) of the right-sided and left-sided limits 1 and 0. I recommend that you make an arbitrary choice for the value of H(0), defining it to be always 0 or always 1/2 or always 1 according to what makes the most sense in the context of your larger problem. Whatever value you assign to H(0) is also the value of your original integral. You can't avoid making an arbitrary choice, but you can apply your choice consistently throughout your work, which is the important thing here. For a practical introduction to distribution theory, I recommend Bernard Friedman's books: "Principles and Techniques of Applied Mathematics" (Dover, 1990) and "Lectures on Applications-Oriented Mathematics" (Wiley, 1991). For a more in-depth theoretical treatment, I recommend A.H. Zemanian's "Distribution Theory and Transform Analysis: An Introduction to Generalized Functions, with Applications" (Dover, 1987). ---------------- http://www.scg.uwaterloo.ca/~fwchapma/ ---------------- Frederick W. Chapman, Ph.D. Student UW Office: Math & Computer 5162 Department of Applied Mathematics UW Phone: (519) 888-4567 x6672 University of Waterloo E-Mail: "fwchapman" Waterloo, Ontario, N2L 3G1, Canada Curriculum Vitae: see home page -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Tue, 6 Aug 2002 17:05:11 -0700 (PDT) From: Robert Israel "israel" To: "maple-list" Subject: Dirac delta The answer is that the question is not well-defined. The Dirac delta is a Borel probability measure on the real line with a unit mass at the origin. So you can integrate this measure over any Borel set, in particular an interval from 0 to infinity. But when you said "between 0 and infinity" you didn't specify whether the interval is [0,infinity) or (0,infinity), i.e. whether or not 0 is included in the set. The integral over [0,infinity) is 1, the integral over (0,infinity) is 0. Mathematica's answer of 1/2 is not the integral over any set (which doesn't mean it's necessarily wrong - you could interpret it as integrating a function H(x) that is 0 for x < 0, 1/2 for x = 0 and 1 for x > 0). Robert Israel "israel" Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 7 Aug 2002 15:41:27 +0200 (CEST) From: "Richard B. Kreckel" "kreckel" To: "maple-list" Subject: Dirac delta It is not defined. Both Maple and Mathematica define it in some way they consider meaningful or elegant. Mathematica obviously defines it to be symmetric. You may "represent" the Dirac delta function as any ordinary sharply peaked function (such that their integral is normalized to unity) and take the limit where the peak gets infinitely sharp. Take step functions for instance, you may chose a step sitting just right of zero, or just left of zero or centered around zero. From the point of view of distribution theory it does not matter, that representation is at best a visualization anyway. > Does anyone understand this?. Can somebody tell me where I can I find > a formal proof using distribution theory whether this integral is > 1,1/2 or zero?. There is no such thing. Regards -richy. -- Richard B. Kreckel "Richard.Kreckel" <http://wwwthep.physik.uni-mainz.de/~kreckel/> -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: "Daniel Willard" "willardd" To: "maple-list" Subject: Dirac delta Date: Wed, 7 Aug 2002 09:51:59 -0400 If your delta function is "delta(x)" then Mathematica is considered correct. If the limit as epsilon goes to zero gets all the way to zero, then 0 is an acceptable answer. Question of open versus closed interval. | >> From: Herman Jaramillo "HJaramillo" |Dear Maple Users: | |When I compute the integral of the Dirac delta function between 0 and |infinity using Maple 7, I get "1". When I do it using Mathematica I |get "1/2". | |When I try to do it using distribution theory by choosing a test |function which has bounded support on the positive real line and using |the limit from the left and the right of the interval of integration |(epsilon, infinity) as epsilon goes to 0 I get that the integral |should be "0".

[MUG] Re: Dirac delta
Author: John Harper    Posted: Thu, 8 Aug 2002 14:51:46 +1200 (
>> From: John Harper "John.Harper" I once had the problem of using a Dirac delta function at an end-point in a research paper (Geophys.J.Roy.Astron.Soc. 55, 87-110, 1978; see p.103). I said "delta denotes an asymmetric Dirac delta function such that int(f(x)*delta(x),x=0..1)=f(0)". I had to give a definition because the difficulty arises so seldom that there is no standard way out of it. Nobody complained. I needed only x in [0,1], otherwise I might have said int(f(x)*delta(x),x=-infinity..infinity) = int(f(x)*delta(x),x=0..infinity) = f(0). John Harper, School of Mathematical and Computing Sciences, Victoria University, PO Box 600, Wellington, New Zealand e-mail "john.harper" phone (+64)(4)463 5341 fax (+64)(4)463 5045

Previous by date: [MUG] Re: How to force a common factor from a simbolic matrix, Robert Israel
Next by date: [MUG] Re: A bug?, Maple User Group
Previous thread: [MUG] Maple 7: `process[forck]' in batch mode, Jerome BENOIT
Next thread: [MUG] A bug?,  Carl Eberhart