Differential equations rest points Let $0 < d < a < 2d$  and $\eta > 0.$ Consider the following differential equation:
\begin{equation}
\dot{x} = \frac{\exp(\eta^{-1}ax)}{\exp(\eta^{-1}ax)+\exp(\eta^{-1}d)}-x
\end{equation}
Fix $a$ and $d.$ For sufficiently large $\eta,$ I believe it can be shown that the above system has a unique restpoint. I am interested in finding out the lowest value of $\eta$ where the above system will have a unique restpoint. Call such an $\eta$ to be $\eta^{\ast}$ and the corresponding restpoint to be $x_{\eta^{\ast}}.$ Is it possible to write an explicit formula for $\eta^{\ast}$ and $x_{\eta^{\ast}}$ as a function of $a$ and $d$?
Thanks a lot in advance.
 A: The condition for a equilibrium is a equation with linear and exponential factors wich can not be solved with the Lambert's W function. So I'm affraid that the best you can do analytically is to derive a sufficient condition for a unique solution.
Consider the function: $$f(x) = \frac{\exp(\eta^{-1}ax)}{\exp(\eta^{-1}ax)+\exp(\eta^{-1}d)}-x = \frac{1}{1+\exp(-\eta^{-1}ax+\eta^{-1}d)}-x.$$
Observe that $$f(0)=\frac{1}{1+\exp(0+\eta^{-1}d)}-0=\frac{1}{1+\exp(\eta^{-1}d)}>0$$ and $$f\left(\frac{d}{a}\right)=\frac{1}{1+\exp(0)}-\frac{d}{a}=\frac{1}{2}-\frac{d}{a}<0,$$
because of your condition $a<2d$. So, by the intermediate value theorem, there exist one solution of $f=0$ between $0$ and $d/a$.
For a unique solution, is a sufficient condition that $f$ be strictly decreasing for $x>0$. We have that
$$f'(x) = \frac{-1}{(1+\exp(-\eta^{-1}ax+\eta^{-1}d))^2}\exp(-\eta^{-1}ax+\eta^{-1}d)(-\eta^{-1}a)-1,$$
after some manipulations we find
$$f'(x) = \frac{-\exp^2(-\eta^{-1}ax+\eta^{-1}d)+(\eta^{-1}a-2)\exp(-\eta^{-1}ax+\eta^{-1}d)-1}{(1+\exp(-\eta^{-1}ax+\eta^{-1}d))^2},$$
wich, by doing $A=\exp(-\eta^{-1}ax+\eta^{-1}d),$ becomes
$$f'(x) = \frac{-A^2+(\eta^{-1}a-2)A-1}{(1+A)^2}.$$
Note that $A>0$ for all $x$. For $f'(x)$ be negative for all $x$, we need $$F(A)=-A^2+(\eta^{-1}a-2)A-1<0$$ for all $A>0$. This can be satisfied in two ways:
i) if the vertex of $F$ be negative, wich leads to $\frac{1}{2}(\frac{a}{\eta}-2)<0$, that is $\frac{a}{2}<\eta$,
or
ii) the maximum of $F$ be negative, wich leads to $\frac{a}{4}<\eta$.
So $\frac{a}{4}<\eta$ guarantees a unique solution.
