Solving a non-linear integro-differential equation I am trying to solve the following equation
$$ f^2(x) - g^2(x) = \alpha\int_0^x f(u) (x-u)du $$
For $\alpha=0$ we get $f=g$. I would like to see how the solution moves away from $g$ when I increase the value of $\alpha$.
In order to derive a closed-form solution I first tried Laplace transforms since
$$ \mathcal{L} \left\{ \int_0^\cdot f(u) (\cdot-u)du \right\}(s) = \frac{1}{s^2} \mathcal{L}\left\{ f\right\}(s) =: \frac{F(s)}{s^2} $$
The problem is that I did not found any general formula for the Laplace transform of $f^2$.
Is there any ways to relate the Laplace transform of $f^2$ to $F$ (or a polynomial/series in $F$) ? If not is there any other method to solve this problem analytically ?

With the function $g(x)$ supposed to be given, on can express the unknown function $f(x)$ as a series of the parameter $\alpha$. This allows to see how $f(x)$ moves away from $g(x)$  when the value of $\alpha$ incresses :
$f(x) = g(x)+\alpha g_1(x) + \alpha^2 g_2(x)+...$  
The analytical form of $g_1(x)$ and $g_2(x)$ are shown below. One could continue with the same method to compute the next terms, but this will be more and more arduous.

On a purely formal way, one can derive the recurrence formula giving $g_k(x)$ from the preceeding terms $g(x), g_1(x), g_2(x), ..., g_{k-1}(x)$ But this involves multiple integrals which will be arduous to compute, depending on the given $g(x)$. This is also without considering the question of the convergence of so complicated series.

