solution uniqueness of non-linear Fredholm equations the equation is 
$F(x)=G\left(\int k(x,y)f(y)dy\right)$  $(*)$
where $f(x)=\frac{dF(x)}{dx}$ is unknown and it's required to be non-negative. With integration by parts we'll have the form of a non-linear Fredholm equation. 
In addition, k(x,y)>0 is square integrable, $k_x=\frac{dk(x,y)}{dx}<0$, $k_y=\frac{dk(x,y)}{dy}<0$; And $G$ is a differentiable, weakly increasing function ranging in $[0,1]$. 
I can prove uniqueness for the case when $k_{xy}>0$ and $G$ is convex (or conversely $k_{xy}<0$ while $G$ is concave). Yet I believe the result should hold more generally. 
 A: Here's my partial proof:
Suppose we have two solutions to $(*)$, $F^1(x)$ and $F^2(x)$, let $f^1(x)$, $f^2(x)$ denote their derivatives. 
Take derivatives of $(*)$ wrp $x$:
$f^i(x)=g(\int k(x,y)f(y)dy))\int k_x(x,y)f^i(y)dy)$,  $i=1,2$
where $g$ is the derivative of G. 
Now let $A$ be the region where $F^1(x)\geq F^2(x)$ and B=$\mathbb{R}\backslash A$.
$f^i(x)=g(\int k(x,y)f(y)dy))(\int_A k_x(x,y)f^i(y)dy+\int_B k_x(x,y)f^i(y)dy)$
Consider $F^i$ in B as a truncated probability distribution function, then if $k_{xy}>0$, $\int_B k_x(x,y)f^2(y)dy))<\int_B k_x(x,y)f^1(y)dy))$ as $F^1(x)<F^2(x)$ in $B$.
Let $m=\int _A f^1(x)dx=\int _A f^2(x)dx$.
Now consider the operator $T^1h(x)=\int_A g(\int k(x,u)f(u)du))(k_x(x,y)+\int_B k_x(x,u)f^1(u)du/m)h(y)dy$, it has a positive  eigenfunction $f^1(x)$, thus has spectral radius 1 (Krein-Rutman theorem).
Similarly $T^2h(x)=\int_A g(\int k(x,u)f(u)du))(k_x(x,y)+\int_B k_x(x,u)f^2(u)du/m)h(y)dy$ also has spectral radius 1. 
However in $A$, $F^1(x)>F^2(x)$, $\int k(x,y)f^1(y)dy>\int k(x,y)f^2(y)dy$ as $k_y(x,y)<0$. When $G$ is convex, $g(\int k(x,y)f^1(y)dy))>g(\int k(x,y)f^2(y)dy))$.
The integral kernel of $T^1$ is greater than that of $T^2$, it should have greater spectral radius. Thus we reach a contradiction. 
