inordinatum

Physics and Mathematics of Disordered Systems

Posts Tagged ‘trick

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

Tricks for inverting a Laplace Transform, part I: Guesses based on Series

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 II (products and convolutions), part III, part IV (substitutions), part V (pole decomposition). Enjoy!

Working with probability distributions one frequently meets the problem of inverting a Laplace transform, which is notoriously difficult. An example that I met recently is the following:

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

The built-in InverseLaplaceTransform function from Mathematica 8 fails to give a result. However, a closed formula for P(x) can be obtained using a trick: Consider integer v, for which the right-hand side becomes a rational function. Take its Laplace inverse, and then extend the result analytically for non-integer v>0.

Since this may be useful for other cases, too, here a detailed account of how one proceeds:

For small integer values of v, one obtains (e.g. using Mathematica):

v=0: P(x)=\delta(x)
v=1: P(x)=q \delta(x)+(1-q)e^{-x}
v=2: P(x)=q^2 \delta(x)+2q(1-q)e^{-x}+q(1-q)^2 x e^{-x}
v=3: P(x)=q^3 \delta(x)+3q^2(1-q)e^{-x}+3q(1-q)^2 x e^{-x}+\frac{1}{2}(1-q)^3 x^2 e^{-x}
v=4: P(x)=q^4 \delta(x)+4q^3(1-q)e^{-x}+6q^2(1-q)^2 x e^{-x}+2q(1-q)^3 x^2 e^{-x}+\frac{1}{6}(1-q)^4 x^3 e^{-x}
v=5: P(x)=q^5 \delta(x)+5q^4(1-q)e^{-x}+10q^3(1-q)^2 x e^{-x}+5q^2(1-q)^3 x^2 e^{-x}+\frac{5}{6}q(1-q)^4 x^3 e^{-x}+\frac{1}{24}(1-q)^5 x^4 e^{-x}

By trying out, one can now guess a general formula for integer v:

P(x)=q^v \delta(x)+e^{-x}\sum_{k=1}^v \frac{1}{(k-1)!}\binom{v}{k}q^{v-k} (1-q)^k x^{k-1}

For example using Mathematica’s Sum command, one can re-write this in terms of a hypergeometric function:

P(x)=q^v \delta(x) + e^{-x}(1-q)q^{-1+v}v \,_1 F_1(1-v,2,-\frac{1-q}{q}x)

This expression now has a well-defined meaning for any v>0, and is the result we looked for. Actually, Mathematica even knows how to take its Laplace transform, which can be used to check its validity.

More tricks for similar problems to come later!

Written by inordinatum

July 28, 2011 at 6:44 pm