I have a system of equations with two equations containing the Lambert $W$ function as follows,

$$\begin{cases} x = 1 - W_0(\frac{C_1 e^{y + 1}}{y + 1}) \\ y = 1 - W_0(\frac{C_2 e^{x + 1}}{x + 1}) \end{cases}$$

I have already solved the system numerically using a simple iterative method. A gentleman told me that the answer to the system is $y = \frac{k_1 - k_2 x}{k_1 x - k_2}$ where $k_1 = C_2 - C_1$ and $k_2 = C_2 - C_1$ but I have no clue how can one compute the answer.

  1. How can I get the answer to the system (i.e the steps)?
  2. While we have two curves in the system of equation, the intersection (i.e the answer of the system) must be a point not a curve. But $y = \frac{k_1 - k_2 x}{k_1 x - k_2}$ is a hyperbola. How can I describe it?
  • $\begingroup$ Where do the $k_1$ and $k_2$ suddenly come from? $\endgroup$ – Raskolnikov Mar 14 '11 at 21:37
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    $\begingroup$ OK, for the solution to be right $k_1=C_1-C_2$ and $k_2=-(C_1+C_2)$. (Although some multiple would work as well.) My suggestion: invert the Lambert W function. $\endgroup$ – Raskolnikov Mar 14 '11 at 21:44
  • $\begingroup$ Sorry, I fixed the question. $\endgroup$ – Helium Mar 14 '11 at 21:47
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    $\begingroup$ Thanks a lot. I solved the first part then. Since $W_0(f(y)) = 1 - x$, we can say $(1 - x)e^{1 - x} = f(y)$. We do the same thing with the other equation and the rest is straightforward. But I am still confused. Why I got a hyperbola as an answer instead of a point? Did we lost something? I mean a part of the information from the equations? $\endgroup$ – Helium Mar 14 '11 at 23:13
  • $\begingroup$ Yes, you combine the two equations into one, thereby you still have one independent equation to use. So you could for example pick one of the original equations and keep it. $\endgroup$ – Raskolnikov Mar 14 '11 at 23:16

$$\begin{cases} 1-x = W_0(\frac{C_1 e^{y + 1}}{y + 1}) \\ 1-y = W_0(\frac{C_2 e^{x + 1}}{x + 1}) \end{cases}$$ Using the poperty of the Lambert function $Y=W(X)$ invertion : $Ye^Y=X$ $$\begin{cases} (1-x)e^{1-x}=\frac{C_1 e^{y + 1}}{y + 1} \\ (1-y)e^{1-y}=\frac{C_2 e^{x + 1}}{x + 1} \end{cases}$$ $$C_1C_2e^{x+y}=C_2(1-x)(y+1)=C_1(1-y)(x+1)$$ $$C_2(1-x)y+C_1(x+1)y= C_1(x+1)-C_2(1-x)$$ $$y=\frac{C_1(x+1)-C_2(1-x)}{C_2(1-x)+C_1(x+1)}$$ This leads to the expected relationship : $$y=\frac{(C_1+C_2)x+C_1-C2}{(C_1-C_2)x+C_1+C_2}$$ Then we have to separate the equation with unknown $x$ from the equation with unknown $y$. $$y+1= \frac{2C_1(x+1)x}{ (C_1-C_2)x+C_1+C_2}$$ $$(1-x)e^{1-x}= \frac{C_1 e^{y + 1}}{y + 1} =\frac{C_1 e^{\frac{2C_1(x+1)x}{ (C_1-C_2)x+C_1+C_2}}}{\frac{2C_1(x+1)x}{ (C_1-C_2)x+C_1+C_2}}$$ $$2x(x+1)(1-x)e^{1-x}=((C_1-C_2)x+C_1+C_2) e^{\frac{2C_1(x+1)x}{ (C_1-C_2)x+C_1+C_2}}$$ This equation contains the unknown $x$ only. A similar one can be found containing $y$ only. But, the more likely, those equations cannot be solved for the unknown expressed on closed form.


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