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I've got the following problem I'm working on which is related to some of my research:

Solve:

$f(x) = \int_{-\infty}^x G(x,y)f(y)f(x-y)dy$

for f, given $G$ which has whatever smoothness properties you might need.

There are two possible simplifications that can be made regarding G: either

1: $G(x,y) \equiv 0$ when $y>x$ and thus we continue the integral up to $\infty$

or

2: $G$ is symmetric, $G(x,y) = G(y,x)$ (but the integral must extend to x, not $\infty$)

These are the only bounds I have so far to put on G.

Hoping you guys might have some ideas. This question comes about in a few problems related to the study of the structure of solids.

-edit

I'm not really looking for a general solution, since its highly unlikely it will be possible. I'm really interested in finding particular solutions, assumptions, or techniques that might be helpful in understanding the problem.

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I'm confused: it sounds like you are saying $G(x,y) = G(y,x)$ and $G(x,y) = 0$ for $y>x$. That would mean $G(x,y)=0$ unless $x=y$ and therefore $F(x) = 0$ for all $x$. –  Noah Stein May 29 '12 at 21:00
    
Written in this way is very difficult to help. What is $G$ and what is unknown, $f$ or $F$? –  Jon May 29 '12 at 21:05
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I can't imagine a reason why there could be non-zero solutions... The map $f\mapsto f - Q(f)$ with a quadratic Q is locally invertible at 0, so that the solution f=0 should be isolated. On the other hand, it is difficult to imagine an invariant set which is not a nbd of 0. –  Pietro Majer May 30 '12 at 5:14
    
Pietro, there do exist non-zero solutions, for example the Weibull distribution. If f(x) = 0 w/ x<0, $G(x,y) = C*x^{\gamma}$ we can pull it out. –  Chris May 30 '12 at 17:01
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1 Answer

The simplest approach one can think is an iterative one. Let us consider the given equation $$ f(x)=\int_{-\infty}^xG(x,y)f(y)f(x-y)dy. $$ Now, we assume as a first iterate $f^{(0)}(x)=1$ and so $$ f^{(1)}(x)=\int_{-\infty}^xdyG(x,y) $$

$$ f^{(2)}(x)=\int_{-\infty}^xdyG(x,y)\int_{-\infty}^ydwG(y,w)\int_{-\infty}^{x-y}dzG(x-y,z) $$

and so on. One is granted the existence of the n-th iterate provided the integral of $G$ exists and is properly bounded.

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We also need the integrand to be Lipschitz, right? –  Chris May 30 '12 at 17:10
    
Chris, I think so. –  Jon May 30 '12 at 19:46
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