inordinatum

Physics and Mathematics of Disordered Systems

Tricks for inverting a Laplace Transform, part II: Products and Convolutions

with 4 comments

EDIT: In the meanwhile, I have continued the series of posts on Laplace Transform inversion. You can find the subsequent articles here: part I (guesses based on series expansions), part III, part IV (substitutions), part V (pole decomposition). Enjoy!

Following the previous post on inverting Laplace transforms, here is another trick up the same alley. This one actually considers a generalization of the previous case

Find P(x) such that \int_0^\infty P(x)e^{\lambda x}dx=\left(1-q\lambda\right)^a \left(1-\lambda\right)^b.

As usual, the built-in InverseLaplaceTransform function from Mathematica 8 fails to give a result. To obtain a closed formula manually, note that each of the factors can be easily inverted:
\int_0^\infty e^{\lambda x}R_{a,q}(x)\,dx=(1-q\lambda)^a
has the solution
R_{a,q}(x)=\frac{e^{-\frac{x}{q}}\left(\frac{x}{q}\right)^{-1-a}}{q \Gamma(-a)}.

Hence, using the fact that Laplace transforms of convolutions give products, the solution for P(x) can be written as a convolution:
P(x) = \int_0^x dx' R_{a,q}(x')R_{b,1}(x-x').

Computing the integral gives the following expression for P(x) in terms of the hypergeometric function \,_1 F_1:

P(x) = \frac{e^{-x}q^a x^{-1-a-b}}{\Gamma(-a-b)} \,_1 F_1\left(-a,-a-b,\frac{q-1}{q}x\right)

Enjoy!!

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Written by inordinatum

September 3, 2011 at 10:34 am

4 Responses

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  1. […] This is a continuation of the series of articles on Laplace transforms… You can find part I here and part II here. […]

  2. […] on tricks for the inversion of Laplace transforms. You can find the previous parts here: part I, part II, part […]

  3. […] the series of posts on Laplace Transform inversion. You can find the subsequent articles here: part II, part III, part IV. […]

  4. […] transforms. You can find the previous parts here: part I (guesses based on series expansions), part II (products and convolutions), part III, part IV […]


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