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MathGroup Archive 2005

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Re: Re: (x-y) DiracDelta[x-y] does not simplify to 0

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56338] Re: [mg56297] Re: (x-y) DiracDelta[x-y] does not simplify to 0
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 22 Apr 2005 06:23:24 -0400 (EDT)
  • References: <d42kg5$39t$1@smc.vnet.net> <d45agf$ieu$1@smc.vnet.net> <200504210936.FAA05048@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 21 Apr 2005, at 18:36, Alain Cochard wrote:

> *This message was transferred with a trial version of CommuniGate(tm) 
> Pro*
> yehuda ben-shimol writes:
>
>> As I remember, DiracDelta is singular and has a meaning only under
>> integration.  Anyway the properties of the DiracDelta are kept by
>> Mathematica i.e., Integrate[(x - y)DiracDelta[x - y], {x, -1, 1},
>> {y, -1, 1}] returns 0 as expected
>
> Andrzej Kozlowski writes pretty much the same:
>
>> On the one hand I think the Mathematica implementation of DiracDelta
>> (and KroneckerDelta) leaves a lot to be desired... and that is putting
>> it mildly. (That means I have plenty of much worse examples...).
>>
>> On the other hand, I am not convinced that Mathematica ought to 
>> perform
>> this sort of simplification at all.  DiracDelta is a generalised
>> function.  The statement x DiracDelta[x] == 0 needs a lot of
>> interpreting to make sense of (I prefer to think of it as nonsense).
>> However
>>
>>
>> Integrate[(x-y) DiracDelta[x-y], {x,-Infinity,Infinity}]
>>
>> 0
>>
>> is correct.
>
> I don't understand these reservations. I learned the statement x delta
> = 0 in my lectures on distributions at university, and I checked today
> that it also appears in one of Laurent Schwartz's ("father" of
> distribution theory) books.  Plus I find it very intuitive and it's
> straightforward to demonstrate.
>

As I wrote: it takes a lot of interpreting. You not only have to 
interpret both x and 0 as distributions, but you also have to interpret 
multiplication in a special way, quite different form the normal 
Mathematica interpretation. The statement x DiracDelta[x] ==0 is merely 
an informal short hand for the staatement:

<x DiracDelta[x], f[x]>= <0, f[x]> for any  "test function" f[x], where 
<f,g> stands for Integrate[f[x]*g[x],{-Infinity,Infinity}]

  (One can also approach distributions from the point of view of 
non-standard analysis but this is not implemented in Mathematica and  
the same remarks as below also apply).
One can perform certain further informal manipulations on this 
'identity". But such informal manipulations should only be performed 
carefully by people who know what they are doing and Mathematica 
certainly does not. Because the algebra of distributions is not like 
usual "algebra" (for a start multiplication is not defined for 
arbitrary distributions) so it is very easy to obtain nonsensical 
answers if you manipulate informal expressions containing DiracDelta 
blindly, e.g. the following is complete nonsense:


Solve[x*DiracDelta[x] == 0, x]

{{x -> 0}, {x -> InverseFunction[DiracDelta, 1, 1][0]}}

One can manufacture plenty of such examples  but I think they point I 
am making should be clear.
Since very easy to obtain nonsensical conclusions by manipulating 
expressions involving the Dirac function blindly  and since on the 
other hand there is actually nothing useful that can be done with the 
above "relationship" , I don't think Mathematica should perform any 
such "simplifications" except in the proper formal context, which means 
within Integrate.

Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/andrzej/index.html
http://www.mimuw.edu.pl/~akoz/


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