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MathGroup Archive 2004

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Factor 2 error in Inverse Laplace Transform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg51232] Factor 2 error in Inverse Laplace Transform
  • From: p-valko at tamu.edu (Peter Valko)
  • Date: Sat, 9 Oct 2004 04:18:58 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

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[1][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[1][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


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