How to solve $1-x = \frac{ax}{e^{ax}-1}$

Question

I am looking for an explicit solution to the equation

$$1-x = \frac{\alpha x}{e^{\alpha x} -1}$$

I tried this with the Lambert-W function but can't get a sensible solution. How would you approach this? And does a closed form solution exist for $\alpha \neq 1$?

For Lambert-W, I can only find a solution for $\alpha = 1$ where $$x = W\left(-\frac{1}{e} \right)+1$$ (I set $\alpha = 1$ for $(1-x)e^{\alpha x}-1+x = \alpha x$ which gives $(x-1)e^{x-1} = -\frac{1}{e}$ - then I apply Lambert W and get $$x = W\left(-\frac{1}{e}\right)+1$$

Answer 1

$$1-x=\frac{\alpha x}{e^{\alpha x}-1}$$ $$(1-x)(e^{\alpha x}-1)=\alpha x$$ $$(1-x)e^{\alpha x}-1+x=\alpha x$$

Your equation is an equation of elementary functions. For rational $\alpha$, the equation is related to an algebraic equation in dependence of $x$ and $e^x$. Because the terms $x,e^x$ are algebraically independent and the equation is irreducible, we don't know how to rearrange the equation for $x$ by only elementary operations (means elementary functions). A theorem of Lin (1983) proves, if Schanuel's conjecture is true, that irreducible algebraic equations involving both $x$ and $e^x$ don't have solutions in the elementary numbers.

$$(1-x)e^{\alpha x}=\alpha x+1-x$$ $$(1-x)e^{\alpha x}=(\alpha-1)x+1$$ $$\frac{1-x}{(\alpha-1)x+1}e^{\alpha x}=1$$ $$\frac{1-x}{x+\frac{1}{\alpha-1}}e^{\alpha x}=\alpha-1$$

We see, your equation cannot be solved in terms of Lambert W but in terms of Generalized Lambert W:

$$\frac{x-1}{x+\frac{1}{\alpha-1}}e^{\alpha x}=1-\alpha$$ $$\frac{x-1}{x-\frac{1}{1-\alpha}}e^{\alpha x}=1-\alpha$$ $$x=W\left(^{\ \ \ 1}_{\frac{1}{1-a}};1-a\right)$$

The inverse relation of your kind of equations is what Mezö et al. call $r$-Lambert function. They write: "Depending on the parameter $r$, the $r$-Lambert function has one, two or three real branches and so the above equations can have one, two or three solutions"

So we have a closed form for $x$, and the representations of Generalized Lambert W give some hints for calculating $x$.

[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mező/Baricz 2017] Mező, I.; Baricz, Á.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

Answer 2

The most I can get to is by setting $X=ax-a$:

$\begin{align}(1-x)(e^{ax}-1)=ax &\iff -\frac 1a(ax-a)(e^ae^{ax-a}-1)=(ax-a)+a\\\\ &\iff-\frac 1aX(e^ae^X-1)=X+a\\\\ &\iff Xe^X=(1-a)e^{-a}X-a^2e^{-a}\end{align}$

But I don't think there is a general closed form for $Xe^X=\alpha X+\beta$.

Answer 3

Let $ax=t$ and rewrite the equation as $$e^{-t}=\frac 1 {1-a}\,\, \frac {t-a} {t-\frac a{1-a}}$$ which has a solution in terms of the generalized Lambert function (have a look at equation $(4)$).

Back to the original equation means that you are looking for the inverse of $$a=1-\frac{1+W\left(e^{x-1} (x-1)\right)}{x}$$