Help on Fredholm Eqn of the First Kind

Steve Zissou

Hello humans,
Can you offer advice on the following situation?
[itex] g(s)=\int_{a}^{b}K(t,s)f(t)dt [/itex]
However I understand that if K can be expressed as
[itex] K(t,s)=K(t-s) [/itex]
then we can say
[itex]f(t)=\mathcal{F}^{-1}\left [ \frac{\mathcal{F}[g(t)]}{\mathcal{F}[K(t)]} \right ] [/itex]
Where the fancy F is Fourier, natch. Although I am fuzzy on what happened to a and b. Anyway, in my case, my function looks like this:
[itex] g(s)=\int_{0}^{\infty}t f(t)dt [/itex]
Can you offer any tips, advice, et cetera?
Thanks

Steve Zissou

By the way the idea here is to solve for f, knowing g.
Thanks

Mute

K(t,s) = K(t-s) is not a good enough condition to be able to use the fourier transform method to solve the equation. Your integration limits also have to be [itex]\pm \infty[/itex].

You could solve the equation numerically using quadrature methods, since then it's basically a linear algebra problem [itex]\mathbf{g} = K\mathbf{f}[/itex].

Steve Zissou

Thanks man. I appreciate it.

Steve Zissou

followup:
What if we say
[itex] t=e^{\omega u/2} [/itex], so [itex] dt=\frac{\omega}{2}e^{\omega u/2}du [/itex]
and then
[itex] g=\int_{0}^{\infty}\frac{\omega}{2}e^{\omega u}f(u)du [/itex]
which means just taking the inverse Laplace transform of g in order to get f?
Just an idea?

Mute

I overlooked the part of your original post where you said your equation was

[tex]g(s) = \int_0^\infty dt~t f(t).[/tex]

Unless g(s) happens to be a constant, that is an ill-formed equation. The integration kernel does not depend on s, so the integral is a constant. If g is constant, you can have infinitely many solutions for f.

Steve Zissou

Good point, you're right, in the interest of expedience I haven't written it right. The real problem looks like this:
[itex] g(s)=\int_{0}^{\infty}tf(t,s)dt [/itex]
I know g, I am trying to find f.

Mute

Hm, well, general solutions to integral equations are hard to come by. You can't apply the Laplace transform as is, for instance, since it will only decouple the integrand if the upper limit is s. You could try, for example, assuming that [itex]f(t,s) = \tilde{f}(s-t)\Theta(s-t)[/itex], where Theta is a Heaviside side function. Then your equation would be of the form

[tex]g(s) = \int_0^s dt~t \tilde{f}(s-t),[/tex]

and the Laplace transform would factor the integral for you:

[tex]\mathcal L g(u) = \frac{\mathcal L f(u)}{u},[/tex]
which you could then inverse Laplace transform. However, since you have to assume a particular form of the solution here, it might not be general, or perhaps the inverse transform won't exist, etc.

Another possibility is to consider the Mellin transform. The Mellin transform of a function f(x) is

[tex]\varphi(y) = \int_0^\infty dx x^{y-1} f(x).[/tex]
(it's actually related to the Laplace transform by a change of variables). This is almost the form of your equation, if your g were g(s,y=2). If you can choose a suitable generalization of g(s) to some g(s,y), then the solution to

[tex]g(s,y) = \int_0^\infty dt~t^{y-1} f(t,s)[/tex]
would simply be the inverse Mellin transform of g:

[tex]f(t,s) = \frac{1}{2\pi i}\mathcal M^{-1}[g(s,y)](t) = \int_{c-i\infty}^{c+i\infty} dy t^{-y} g(s,y).[/tex]

(see the wikipedia page for the Mellin transform definition and what the inverse equation means)

But I'm not sure if you can really get the f(t,s) that you actually want out of that solution.

Another cause for concern that I have with this approach is that there are effectively infinitely many ways you might generalize g(s) to g(s,y) such that g(s,2) is the g(s) you're starting with. I don't know if that means there are infinitely many solutions, or if some solutions just won't work (e.g., the inverse Mellin transform doesn't exist because your g doesn't satisfy the Mellin inversion theorem requirements), or something else.

haruspex

My immediate reaction is to get rid of the t factor by change of variable, u = t², f(t) = 2h(u).
If g(s) can be written as Ʃa_ns^-n, n > 0, then there's a solution h(u, s) = e^-usƩa_nu^n-1/(n-1)! (the details may be wrong).
Given any solution h = h(u, s), h + k(u,s) is also a solution iff ∫₀^∞k(u,s)du is identically 0.