Solve the following non-linear system of equations $x = \alpha \log(y/(1-y)), y = \alpha \log(x/(1-x)) - \beta$ in terms of $\alpha, \beta$

Question

I have the following non-linear equations,

$$x = \alpha \log(y/(1-y))$$ $$y = \alpha \log(x/(1-x)) - \beta $$

where $\alpha, \beta$ are constants, $\log$ is the natural logarithm.

I wish to solve for $x$.

Is it possible? It seems that by plugging in $x$ into the $y$ equation, I get something like $$y = \alpha \log(\alpha \log(y/(1-y))/(1-\alpha \log(y/(1-y)))) - \beta$$ and I have no idea if it is possible to recover an analytical solution of $y$ in terms of $\alpha$ and $\beta$

Answer 1

I do not think that you could obtain any analytical solution and then, hoping that I am correct, you will need to consider some numerical method.

Consdering your equations $$x=\alpha \log \left(\frac{y}{1-y}\right) \tag 1$$ $$y=\alpha \log \left(\frac{x}{1-x}\right)-\beta \tag 2$$ as you did, the second equation becomes, $$y=\alpha \log \left(\frac{\alpha \log \left(\frac{y}{1-y}\right)}{1-\alpha \log \left(\frac{y}{1-y}\right)}\right)-\beta\tag 3 $$ which does not look very pleasant.

Still thinking about numerical method, I should define $$z=\alpha \log \left(\frac{y}{1-y}\right)$$ to write the final equation as $$\frac{e^{\frac{y+\beta }{\alpha }}}{1+e^{\frac{y+\beta }{\alpha }}}=\alpha \log \left(\frac{y}{1-y}\right)\tag 4$$ which is probably better conditioned.