Can mathematica solve this analysis difficult problem ? "Functional" equation

Given two continuous functions $G,H: mathbb R^+ rightarrow mathbb R$ such that $frac{G}{H}$ is positive and strictly increasing, I am looking for two continuous functions $f_0, f_1 :[0,1] rightarrow mathbb R$ such that $f_0(0)=1,f_0(1)=0,f_1(0)=0,f_1(1)=1$ and for all $s<t$ in $[0,1]$,

(f[1, t] (f[1, s] G[1] - G[s]) - f[1, s] G[t]) H[1] + 
 f[0, t] G[0] (f[1, s] H[1] - H[s]) + 
 f[0, s] G[0] (f[0, t] H[0] + f[1, t] H[1] - H[t]) + G[s] H[t]

has separable variable, i.e it can be rewritten as $p(s)u(t)$. I showed that this is equivalent to

-(f[1, s] f[1, t] G[1] H[1] - f[1, t] G[s] H[1] - f[1, s] G[t] H[1] + 
     f[0, t] G[0] (f[1, s] H[1] - H[s]) + 
     f[0, s] G[0] (f[0, t] H[0] + f[1, t] H[1] - H[t]) + 
     G[s] H[t]) (f[1, s] f[1, u] G[1] H[1] - f[1, u] G[s] H[1] - 
    f[1, s] G[u] H[1] + f[0, u] G[0] (f[1, s] H[1] - H[s]) + 
    G[u] H[s] + 
    f[0, s] G[0] (f[0, u] H[0] + f[1, u] H[1] - H[u])) + (f[0, s]^2 G[
      0] H[0] + f[1, s]^2 G[1] H[1] - 2 f[1, s] G[s] H[1] + 
    2 f[0, s] G[0] (f[1, s] H[1] - H[s]) + 
    G[s] H[s]) (f[1, t] f[1, u] G[1] H[1] - f[1, u] G[t] H[1] - 
    f[1, t] G[u] H[1] + f[0, u] G[0] (f[1, t] H[1] - H[t]) + 
    G[u] H[t] + f[0, t] G[0] (f[0, u] H[0] + f[1, u] H[1] - H[u]))

being equal to $0$ for all $u<s<t$ in $[0,1]$.

I managed to find some particular solutions: if $H=1$, then a solution is

f[1, t_] := (G[t] - G[0])/(G[1] - G[0])
f[0, t_] := (G[1] - G[t])/(G[1] - G[0])

Another one: If $G(0)=0$, then $f_0$ disappears from the equations and a solution for $f_1$ is again

f[1, t_] := (G[t] - G[0])/(G[1] - G[0])