Honestly, I'm kind of stuck... I was looking at the following recursively defined functions. Given $f_0$ and $f_1$ let
$$ f_n\left(x\right) = \int_0^1 F\left(x,s\right) f_{n-1}\left(s\right) + G\left(x,s\right) f_{n-2}\left(s\right) ds $$
for $n > 1$. Is there some trick to solve such kind of recursions?
Thanks in advance.
On
Hint:
Let us consder the simpler case such that $G\equiv0$.
If I am right, when you iterate the integral, you can use Fubini and swap the order of integration, like
$$\int_{t=0}^1 F(x,t)\int_{s=0}^1F(t,s) f(s)\,ds\,dt=\int_{s=0}^1\left(\int_{t=0}^1 F(x,t)F(t,s)\,dt\right) f(s)\,ds\\ =\int_{s=0}^1F_2(x,s) f(s)\,ds.$$
Then, provided you can obtain the iterated kernels $F_k$ and $G_k$, the solution follows linearly,
$$f_n(x)=\int_{s=0}^1F_n(x,s) f_0(s)\,ds.$$
In the case of two terms, the solution will also take a linear form,
$$f_n(x)=\int_{s=0}^1\left(H_n(x,s) f_0(s)+K_n(x,s) f_1(s)\right)\,ds.$$
This is only a very partial answer, but if ${u}_{n} \left(x\right) = {\left({f}_{n-1} \left(x\right) , {f}_{n} \left(x\right)\right)}^{\top }$, the relation writes
$${u}_{n} = \left[\begin{array}{cc}0&I\\ A&B \end{array}\right] {u}_{n-1} = T {u}_{n-1}$$
and $A , B$ are the integral operators with kernels $G$ and $F$. A trick if any, to study the sequence ${u}_{n+1} = {T}^{n} {u}_{1}$ is to study the spectrum of the operator $T$ in a convenient functional space. It is likely that ${T}^{2}$ is a compact operator and Fredholm's theory may apply.