       Re: Factor 2 error in Inverse Laplace Transform

• To: mathgroup at smc.vnet.net
• Subject: [mg51286] Re: Factor 2 error in Inverse Laplace Transform
• From: p-valko at tamu.edu (Peter Valko)
• Date: Tue, 12 Oct 2004 01:57:50 -0400 (EDT)
• References: <ck87ta\$9mf\$1@smc.vnet.net> <ckajqc\$m6v\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Maxim,

As far as your Log example is considered, it really shows
inconsistency.
The other examples are also interesting, but you make very general
comments and they bother me a bit. Somehow I feel they are unfair.

For instance, in your second example, first you consider the
input-output pair
In=
Integrate[DiracDelta[x - a], {x, 0, 1}]
Out=
UnitStep[1 - a]*UnitStep[a]
then you are substituting a->1 into the result and Mathematica gives 0.

You compare this to the pair
In=
Integrate[DiracDelta[x], {x, 0, 1}]
Out=
1/2

and draw the conclusion that something is wrong.
I respectfully disaggree with you.

In the Help of Integrate, Mathematica explicitely warns us:
"When an integrand depends on a parameter, the indefinite integral
should be considered valid for "generic" values of the parameter. For
certain values the reported integral may be meaningless,..."

In general, it is extremely easy to ask Mathematica to "derive" a parametric
result, and then to substitute something very special into the

Personally, I am not really interested in that.
My examples try to mimic the situation when a user wants to solve a
specific problem, arrives at a Laplace transform and tries to find the
original using the "InverseLaplaceTransform" service of Mathematica.
In such a situation a factor of 2 error is dangerous and I think with
a little modification Mathematica could avoid it.

For instance here is Problem 3:
InverseLaplaceTransform[s/(1+s)^(3/2),s,t]
It gives an answer that is wrong by a factor of two, while
Problem 3a:
InverseLaplaceTransform[Apart[s/(1+s)^(3/2)],s,t]
gives the right answer because then Mathematica finds other rules to apply and
they are correct.

I suggest to put a little "safeguard" into InverseLaplaceTransform and
force it to avoid this trap. This can be done without redefining
DiracDelta and UnitStep. (Those are used extensively by Mathematica and
therefore any change would impact the whole system in too many ways.)

Regards
Peter

ab_def at prontomail.com (Maxim) wrote in message news:<ckajqc\$m6v\$1 at smc.vnet.net>...
> p-valko at tamu.edu (Peter Valko) wrote in message news:<ck87ta\$9mf\$1 at smc.vnet.net>...
> > Hi,
> >
> > InverseLaplaceTransform is an extremely useful part of Mathematica
> > (since v 4.1).
> > However, in the following simple problem it gives the wrong answer:
> > Problem 1:
> > InverseLaplaceTransform[s/(s+1),s,t]
> > -1/(2*E^t)+DiracDelta[t]
> > where the factor 2 is completely wrong.
> >
> > To see that I slightly rewrite Problem 1 into
> > Problem 1a:
> > InverseLaplaceTransform[Apart[s/(s+1)],s,t]
> > and then I get the correct answer:
> > -E^(-t)+DiracDelta[t]
> >
> > Of course one can "Unprotect" InverseLaplaceTransform and teach it to
> > give the correct answer but that is not the point.
> > (Also one can start a long debate about the meaning of DiracDelta in
> > Mathematica, but that is also not the point here. )
> >
> > There are several similar simple examples when the wrong factor of two
> > shows up, for instance
> > Problem 2:
> > InverseLaplaceTransform[ s ArcTan[1/s],s,t]
> >
> > Using the Trace one can find out that all these "factor 2" errors have
> > a common origin.
> > Solving Problem 1 Mathematica calculates the convolution integral
> >
> >  Integrate[E^(-t+x)*Derivative[DiracDelta][x],{x,0,t}]
> >
> > and because the lower limit is exactly zero,the factor 2 shows up in
> > -1/(2*E^t), that is Mathematica "halves" the Dirac delta and all its
> > derivatives at the origin.
> >
> > I think the InverseLaplaceTransform function could be much improved if
> > the above convolution integral would be evaluated more carefully.
> >
> > For instance, doing it in two steps:
> >  res1=Integrate[E^(-t+x)*Derivative[DiracDelta][x],{x,-eps,t},
> >  Assumptions -> eps>0];
> >  res2=res1/.eps -> 0
> > would give the right result.
> > (This caution is necessary only, if generalized functions are involved
> > in the integration.)
> >
> > I wonder if further examples/suggestions are welcome in this group
> > regarding InverseLaplaceTransform???
> >
> > Peter
>
> I'd say that this is two messes mixed together. One is a rather poor
> implementation of integral transforms, especially when there are
> generalized functions involved. For example:
>
> In:=
> LaplaceTransform[InverseLaplaceTransform[Log[p], p, t], t, p] - Log[p]
> LaplaceTransform[InverseLaplaceTransform[PolyGamma[p], p, t], t, p] -
> PolyGamma[p]
>
> Out=
> EulerGamma
>
> Out=
> EulerGamma
>
> We can see that LaplaceTransform/InverseLaplaceTransform aren't
> consistenly defined for these functions (here Mathematica doesn't
> internally take integrals of distributions). Another issue is the
> question of how the integral of DiracDelta on [0,a] should be
> interpreted:
>
> In:=
> Integrate[DiracDelta[x], {x, 0, 1}]
> Integrate[DiracDelta[x - a], {x, 0, 1}]
> Integrate[DiracDelta[x - 1], {x, 1, 2}]
>
> Out=
> 1/2
>
> Out=
> UnitStep[1 - a]*UnitStep[a]
>
> Out=
> 0
>
> The value of the first integral is by convention taken to be 1/2;
> however, substituting a=0 into Out we obtain 1, and making the
> change of variables x->x-1 (the third integral) we get 0. Similarly
> for the integrals of the DiracDelta derivatives:
>
> In:=
> Integrate[DiracDelta'[x]*phi[x], {x, -eps, Infinity}]
> Integrate[DiracDelta'[x]*phi[x], {x, 0, Infinity}]
> Integrate[DiracDelta'[x]*x, {x, 0, Infinity}]
>
> Out=
> (-DiracDelta[eps])*phi - phi'
>
> Out=
> 0
>
> Out=
> -(1/2)
>
> The value of the first integral for eps<0 is incorrect in any case,
> and Out and Out are not consistent with each other.
>
> Maxim Rytin
> m.r at inbox.ru

```

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