Re: problem with InverseLaplaceTransform

*To*: mathgroup at smc.vnet.net*Subject*: [mg58035] Re: problem with InverseLaplaceTransform*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Thu, 16 Jun 2005 06:43:53 -0400 (EDT)*Organization*: The University of Western Australia*References*: <d8ov0i$39$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

In article <d8ov0i$39$1 at smc.vnet.net>, "Barry Shaw" <barry.shaw at qub.ac.uk> 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 answer....as 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. Cheers, Paul -- 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) AUSTRALIA http://physics.uwa.edu.au/~paul http://InternationalMathematicaSymposium.org/IMS2005/