Re: DiracDelta multiplication question ...
- To: mathgroup at smc.vnet.net
- Subject: [mg38998] Re: DiracDelta multiplication question ...
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Fri, 24 Jan 2003 05:04:19 -0500 (EST)
- Organization: Universitaet Leipzig
- References: <b0oq3d$bjf$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
because Mathematica has no build-in rule to handle
product of delta-Functions
but now it has
Unprotect[DiracDelta]
DiracDelta /: DiracDelta[a_]^n_Integer := DiracDelta[a]
DiracDelta /: DiracDelta[a_]* DiracDelta[b_] /; a =!= b := 0
Protect[DiracDelta]
But you should keep in mind that only expression
Integrate[a_*DiracDelta[t],{t,-Infinity,Infinity}]/;
FreeQ[a,_DiracDelta]
are covered by the theory of distributions.
Regards
Jens
Matt Flax wrote:
>
> Why doesn't this multiplication equal zero ?
>
> In[1]:= DiracDelta[2 + t] DiracDelta[t]
> Out[1]= DiracDelta[t] DiracDelta[2 + t]
>
> A delta multiplied by a delayed delta = 0 ... right ?
>
> Also like this :
>
> In[2]:= DiracDelta[2 + t] DiracDelta[t] // SimplifyDiracDelta
> Out[2]= SimplifyDiracDelta[DiracDelta[t] DiracDelta[2 + t]]
>
> thanks
> Matt