Assume I have a system of three equations
$$\begin{cases} \frac{-x_1}{1+x_1-b_1} + \ln \left( 1+x_1-b_1 \right) = \frac{y}{a_1}\\ \frac{-x_2}{1+x_2-b_2} + \ln \left( 1+x_2-b_2 \right) = \frac{y}{a_2}\\ \frac{a_1}{1+x_1-b_1} + \frac{a_2}{1+x_2-b_2} = y \end{cases}$$
We know that $a_1,a_2,b_1,b_2$ are all positive real numbers which are our know constants ($a_1,a_2$ are arbitrary and can be adjusted, more explanation at the bottom of the page). Also $x_1,x_2,y \in \mathbb{R}^+$ are our variables. We know $x_1-b_1 \ge 0$, $x_2-b_2 \ge 0$, $1-b_1 \le 0$ and $1-b_2 \le 0$ always holds true.
It can be shown that $x_1,x_2$ can be solved like this
$$\begin{cases} x_1 = \exp \left( W_0\left( -(1-b_1)e^{-(1+\frac{y}{a_1})} \right) +1 + \frac{y}{a_1} \right) - (1-b_1) \\ x_2 = \exp \left( W_0\left( -(1-b_2)e^{-(1+\frac{y}{a_2})} \right) +1 + \frac{y}{a_2} \right) - (1-b_2) \end{cases}$$
Where $W_0$ is the zeroth vranch of Lambert_W function. Then we substitute $x_1,x_2$ in the third equation to get this natural recursion
$$f(y) = \\ \frac{a_1}{\exp \left( W_0\left( -(1-b_1)e^{-(1+\frac{y}{a_1})} \right) +1 + \frac{y}{a_1} \right)} + \frac{a_2}{\exp \left( W_0\left( -(1-b_2)e^{-(1+\frac{y}{a_2})} \right) +1 + \frac{y}{a_2} \right)} \\ = y$$
I want to solve this last equation using the fixed point iteration and first guess $y^0 \ge 0$ (I omitted the proof that root is unique and always exist, just take derivative and check function at zero and infinity, since the function is continuous). For a fixed point iteration $f(y)=y$ to converge to the root, the sufficient condition is that $|f'(y)|\le 1$. Now we calculate the derivative (for Lambert-W, $\frac{dW(z)}{dz} = \frac{W(z)}{z(1+W(z))}$)
$$f'(y) = \frac{-1}{\exp \left( W_0\left( -(1-b_1)e^{-(1+\frac{y}{a_1})} \right) +1 + \frac{y}{a_1} \right)\left( 1+W_0\left( -(1-b_1)e^{-(1+\frac{y}{a_1})} \right) \right)} + \\ \frac{-1}{\exp \left( W_0\left( -(1-b_2)e^{-(1+\frac{y}{a_2})} \right) +1 + \frac{y}{a_2} \right)\left( 1+W_0\left( -(1-b_2)e^{-(1+\frac{y}{a_2})} \right) \right)}$$
As it can be seen, we always have $f'(y)\le 0$. Also $\lim_{y\to\infty}f(y) = 0$. Therefore if $|f'(0)|\le 1$ then the fixed point iteration always converges but unfortunately
$$|f'(0)| = \frac{1}{\exp \left( W_0\left( -(1-b_1)e^{-1} \right) +1\right)\left( 1+W_0\left( -(1-b_1)e^{-1} \right) \right)} + \\ \frac{1}{\exp \left( W_0\left( -(1-b_2)e^{-1} \right) +1 \right)\left( 1+W_0\left( -(1-b_2)e^{-1} \right) \right)}$$
Which means $|f'(0)|\le 1$ is not always true necesssary.
============> My question is <============
-How can I guarantee the convergence of fixed point iteration?
-The parameters $a_1,a_2$ are arbitrary, in the sense that I can scale them with a positive factor, for example $\beta \ge 0$, $(a_1,a_2) \Rightarrow (\beta a_1,\beta a_2)$ can be substituted. I also calculated $\frac{\partial f'(y)}{\partial a_i}, i \in {1,2}$ and found out that $f'(y)$ is a decreasing function of $a_1,a_2$. Can I scale $a_1,a_2$ to guarantee convergence? Please notice that at origin, i.e. $f(0)$, dependency on $a_1,a_2$ vanishes.
Any help, hint, clue would be appreciated!