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Re: problem with InverseLaplaceTransform

In article <d8ov0i$39$1 at>,
 "Barry Shaw" < at> wrote:

> I'm pretty new to Mathematica and am having difficulty trying to derive 
> an inverse laplace transform using Mathematica.  When I run the code, 
> Mathematica starts to run (says "running..."), but then just continues 
> to remain like that and doesn't give an if it has crashed 
> or something.  

Do you really, really, really, want to compute this expression in closed 
form? It is possible, but messy. And what do you plan to do with this 
closed-form expression once you have it?

> The code I`m trying to run is below (layout messed up, 
> but if you copy and paste it back into Mathemtica it`s fine):

Some comments:

[1] Note that the first argument to InverseLaplaceTransform is enclosed 
in {{}}. These brackets are not required, but will not cause a problem.  
Perhaps you obtained this expression from a matrix?

[2] It is possible to use subscripted greek variables to make your 
expression easier to read. Alternatively, I've used m1 for muone and la3 
for lambdathree, etc. 

[3] Using Together and Simplify on the expression to be inverted I obtain

  (m1 (la2 + m2 + s) (la3 + m3 + s) (m4 + s) +  
   la1 (m2 (la3 + m3 + s) (m4 + s) + 
   la2 (la3 m4 + m3 (m4 + s))))/  
   (s^2 ((la2 + m2 + s) (la3 + m3 + s) (m4 + s) + la1 ((la3 + m3 + s) 
    (m4 + s) + la2 (la3 + m4 + s))))

By collecting together powers of s in the Denominator and Numerator 
separately, one sees that this expression is of the form

  lp = (a s^3 + b s^2 + c s + d)/(s^2 (e s^3 + f s^2 + g s + h))

where a - h are polynomials in your lambda and mu variables, independent 
of s. Now, Mathematica can quite quickly and easily compute the 
InverseLaplaceTransform of this expression: 

  InverseLaplaceTransform[lp, s, t]

The result is a horrendous collection of Root objects -- for a 
discussion of what these are, and why Mathematica uses this 
representation, see a number of earlier postings by Dan Lichtblau on 
this topic. Each Root object is an indexed root of the cubic in the 
denominator of lp. If you determine the a through h a polynomials in 
your lambda and mu variables and substitute this into the computed 
InverseLaplaceTransform, you have the result you requested -- but I 
doubt that it will be of any practical use to you. The only useful cases 
are likely to be where the cubics in the numerator and denominator 
factor. For example, writing

   lp = (s-a)(s-b)(s-c)/(s^2 (s-d)(s-e)(s-f))

(where a - f are now _different_ constant factors to those in lp above), 
then the InverseLaplaceTransform has a simple closed form.


Paul Abbott                                      Phone: +61 8 6488 2734
School of Physics, M013                            Fax: +61 8 6488 1014
The University of Western Australia         (CRICOS Provider No 00126G)    

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