I have a very puzzling question. I have a system of coupled non-linear equations: $$ \begin{aligned} z_1 = e^{az_1+bz_2} \\ z_2 = e^{cz_1+dz_2} \end{aligned} $$ Let's focus on the first equation but everything seems to be the same for the second equation. One solution to this problem comes from the LambertW function: $ye^y = x \rightarrow y = W(x)$. Which this problem would result in: $$ z_1 = -\frac{W(a e^{b z_2})}{a} $$ Now I'm rewriting the equation like this: $$ W(ae^{bz_2}) = -\frac{z_1}{a} $$ And arguing that having an equation like $W(x) = y \rightarrow ye^y = x $.
So I can write $$ -\frac{z_1}{a} e^{-\frac{z_1}{a}} = ae^{bz_2} $$ And now I'm rewriting the equation:
$$ z_1 = - a^2 \exp(\frac{z_1}{a} + b z_2) \\ \ln(z_1) = 2\ln(-a)+\frac{z_1}{a} + b z_2 $$
Coming back to the first equation I can take a log and write the equation as follow: $$ \ln(z_1) = az_1+bz_2 $$ Now I'm using this equality from both equations:
$$ 2\ln(-a)+\frac{z_1}{a} + bz_2 = az_1 + bz_2 $$ So eventually I can simplify the equation: $$ z_1 = \frac{2a}{a^2-1}\ln(-a) $$ We can do the same thing for the other variable: $$ z_2 = \frac{2c}{c^2-1}\ln(-c) $$ The surprising result doesn't depend on the coupling coefficient $b$ and $d$ which is something I don't understand. My first guess is that I'm doing something wrong here. Maybe this step $W(x) = y \rightarrow ye^y = x $. Can you help me where things go wrong?
As @Crostul and @jjagmath mentioned the mistake comes from $$ -\frac{z_1}{a} = W(-a e^{bz_2}) $$ Which should be $$-az_1 = W(-ae^{bz_2})$$ And having that we only get back to the main equation after all the calculations. $$ -az_1 e^{-az_1} = -ae^{bz_2} \rightarrow z_1 = e^{az_1 + bz_2} $$