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I would sincerely appreciate if anyone can tell me how to solve g(x) defined by the following functional equation:

$h(t) = \int_0^t f(2t-x)g(x)dx$ for $0\leq t\leq \infty$?

where: f(x) is a known function (actually a probability density functions defined on $[0^+,\infty$) with finite first and second order moments. h(t) is a known function and g(x), the unknown function, is a pdf defined on $[0^+,\infty$ with finite first and second order moments.

Assume $f(x)=g(x)=h(x)=0$ when $x<0$ and $f(x) \geq 0, g(x)\geq 0, and h(x)\geq 0$ when $x \geq 0$.

I understand basic convolution stuff(Laplace transform etc) but have minimum explosures to functional analysis. The formulation looks somewhat similar to convolution but not exactly the same. Also it seems related to Wiener-Hopf integral equation but I am unfamiliar with it.

Numerical solution is OK. But any kind of analytic insight or closed-form solution under specific class of functions will be very beneficial. In particular I am interested in the case when f(t) follows a (truncated) Gaussian distribution.

Any textbook/web link/paper recommendation is highly appreciated.

Many thanks!

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  • $\begingroup$ The problem I posed belongs to Volterra equations of the first kind. So numerical solutions seem to be the answer. $\endgroup$
    – Haining Yu
    Commented Oct 12, 2010 at 13:34
  • $\begingroup$ There will need to be some compatibility condition on $h$ and $f$ by the way you defined it. Doing a change of variable on $x$, your functional equation is the same as $h(t) = \int_0^t f(t+x) g(t-x) dx$. So if $f$ were a truncated Gaussian, such that $f(x) = 0$ for all $x > x_0$, then necessarily $h(x) = 0$ for all $x > x_0$ and your functional equation gives no information on $g$ for $x > x_0$. $\endgroup$
    – Willie Wong
    Commented Oct 13, 2010 at 23:12
  • $\begingroup$ Thank you for your comment. When I mean truncated Gaussian, what I really mean is $f(x)\leq 0$ when $f(x)\leq 0$. $\endgroup$
    – Haining Yu
    Commented Oct 14, 2010 at 17:40

1 Answer 1

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It sounds like the functions you're dealing with are pretty nice and admit Laplace transforms. So we know that for two functions $f$ and $g$, the Laplace transform of their "convolution" (as written below) is:

$\mathcal{L}(f\star g)=\mathcal{L}\int_0^t f(t-x)g(x)dx=F(s)G(s)$

where $F$ and $G$ are the Laplace transforms of $f,g$ respectively. Look at the proof in this link here. For $f(2t-x)$, following the notation in the link and working bottom up, the only difference that occurs is that instead of $t=\sigma+\tau$, you replace with $2t=\sigma+\tau$. This gives

$\mathcal{L}\int_0^t f(2t-x)g(x)dx=\frac{1}{2}\int_0^\infty \int_0^\infty f(\sigma)e^{-s(\sigma+\tau)/2}d\sigma g(\tau)d\tau=\frac{1}{2}F(s/2) G(s/2)$

So for your equation, you get:

$H(s)=\frac{1}{2}F(s/2)G(s/2)$

and you can now solve for $g$ by using the Laplace transform inverse.

As far as references go, try "Introduction to integral equations with applications" By Abdul J. Jerri.

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  • $\begingroup$ There is a problem with you change of variables. The interior integral of the double integral cannot be just $\int_0^\infty$, it should be $\int_\tau^\infty$. So I think the solution you wrote down using Laplace transform is not correct. $\endgroup$
    – Willie Wong
    Commented Oct 13, 2010 at 23:46
  • $\begingroup$ Thank you both for commenting. I tried to following the link provided and work my way bottom up. It turns out to be difficult (if still possible). The difficulty arises when the change of variable takes place. When you doa variable change, two things have to be met simutaneously: 1) the term for $f(\cdot)$ has to have contain one variable, but not two ($t$ and $x$), and 2) the integral range after the change of variable has to be something like $[\tau,\infty)$. It turns out that such change of variable is hard to find -- because of coefficient of 2 in the term $2t-\tau$. $\endgroup$
    – Haining Yu
    Commented Oct 14, 2010 at 17:38
  • $\begingroup$ Looks like I screwed up. Perhaps it's still salvageable? $\endgroup$
    – Alex R.
    Commented Oct 14, 2010 at 18:54
  • $\begingroup$ Volterra equations seems a viable option. $\endgroup$
    – Haining Yu
    Commented Oct 15, 2010 at 1:51

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