Proof needed of link between 2 equations using Laplace Transforms

[chrisAUS]

Homework Statement

Hello all. First of all I should mention that this is not a homework problem. Rather, during workplace research I have come across the following two equations in a journal paper. Here, the authors state that one equation can be shown to be the Laplace Transform of the other, but do not provide a proof. Although I am quite aware of how to use LTs, I am not an expert; in my work I only use mathematics in an applied sense. Therefore I hope that one of the experts on this forum may instead be able to provide a clear step-by-step proof.

Homework Equations

[tex] \mbox{let} f(x,a) = \frac{x}{\sqrt{4 \pi D a^3}}\exp \left(-\frac{(x-v a)^2}{4 D a}\right)[/tex]

According to the paper's authors, taking the Laplace Transform of this equation with respect to [itex]a[/itex] leads to :

[tex] c(x,\lambda) = c_0 \ \exp \left(\frac{x v}{2 D}\left(1-\sqrt{1+\frac{4 D \lambda}{v^2}}\right)\right)[/tex]

The Attempt at a Solution

The authors state that equations (14) and (74) from Zwillinger's 2003 book of mathematical tables and formulae can be used as part of the Laplace Transformation. They are :

[tex]\mbox{if }f(t) = \frac{1}{a-b}\left(a \ \exp^{at} - b \ \exp^{bt}\right) \mbox{ then } \ell \{ f(t) \} = F(s) = \frac{s}{(s-a)(s-b)} \mbox{where }a \mbox{ is not equal to } b \mbox{ (eq. 14)}[/tex]

[tex]\mbox{if }f(t) = \frac{a}{2 \sqrt{\pi t^3}} \exp^{-a^2 / 4t} \mbox{ then } \ell \{ f(t) \} = F(s) = e^{-a \sqrt{s}} \mbox{ (eq. 74)}[/tex]

Thanks very much in advance - being able to prove explicitly the link (via LT) between these two equations will be very useful in my work.

 

[mighty2000]

Thank you for bringing this to our attention. I am happy to provide a proof for the link between these two equations using Laplace Transforms.

First, let's rewrite the given equation as follows:

f(x,a) = \frac{x}{\sqrt{4 \pi D a^3}} \exp \left(-\frac{(x-v a)^2}{4 D a}\right) = \frac{x}{\sqrt{4 \pi D a^3}} \exp \left(-\frac{x^2}{4 D a} + \frac{v^2 a^2}{4 D a}\right)

Now, let's take the Laplace Transform of this equation with respect to a:

\mathcal{L}\{f(x,a)\} = \int_0^{\infty} f(x,a)e^{-sa}da = \frac{x}{\sqrt{4 \pi D}} \int_0^{\infty} \frac{1}{a^{3/2}} \exp \left(-\frac{x^2}{4 D a} + \frac{v^2 a^2}{4 D}\right) e^{-sa}da

Next, we can use equation (74) from Zwillinger's book to simplify the integral:

\int_0^{\infty} \frac{1}{a^{3/2}} \exp \left(-\frac{x^2}{4 D a} + \frac{v^2 a^2}{4 D}\right) e^{-sa}da = \int_0^{\infty} \frac{a}{2 \sqrt{\pi t^3}} \exp \left(-\frac{a^2}{4t} + at - as\right)da = e^{-a\sqrt{s}}

Substituting this back into the original equation, we get:

\mathcal{L}\{f(x,a)\} = \frac{x}{\sqrt{4 \pi D}} \int_0^{\infty} e^{-a\sqrt{s}}da = \frac{x}{\sqrt{4 \pi D}} e^{-a\sqrt{s}} \Big|_0^{\infty} = \frac{x}{\sqrt{4 \pi D}} e^{-a\sqrt{s}}

Finally, using equation (14) from Zwillinger