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MathGroup Archive 2006

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Re: InverseLaplaceTransform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg72103] Re: InverseLaplaceTransform
  • From: bghiggins at ucdavis.edu
  • Date: Mon, 11 Dec 2006 04:55:49 -0500 (EST)
  • References: <ele6c7$rkt$1@smc.vnet.net>

Raja,

Make use of the Convolution theorem by noting that you can finf the
InverseLaplace Transform of the following functions:

In[7]:=
InverseLaplaceTransform[
  E^(((V - Sqrt[4*A*L + 4*A*R*s + V^2])*x)/(2*A)), s, t]

Out[7]=
(E^(-((t*(4*A*L + V^2))/(4*A*R)) + (V*x)/(2*A) -
     (R*x^2)/(4*A*t))*x)/(2*A^2*Sqrt[Pi]*R*
   Sqrt[t^3/(A^3*R^3)])

In[8]:=
InverseLaplaceTransform[c1/s, s, t]

Out[8]=
c1

You are then left with doing an integral. Incidently, this is what
Mathematica gives you when you try to invert the complete expression.
However the integral does not converge on the interval because of a
singularity of the form 1/Sqrt[s^3]. You may be able to do the integral
by substracting out the singularity- I will leave this or an exercise

Cheers,

Brian


Raja wrote:
> Hi!
> Can you help me deriving this InverseLaplaceTransform with mathematica?
>
> InverseLaplaceTransform[c1*Exp[(V - (4 A L + V^2 + 4 A R
> s)^(1/2))x/2/A]/s,s,t]
>
> I know that the solution can be found in Carslaw and Jager but I want
> to try to solve with Mathematica but didn't succeded.
> Thanks,
> Raja


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