How to find fixed points for an equation with $\ln$ term

Question

Find fixed points

$$x'= M + x - \ln(1+x)$$

First step:1

$$0 = M + x - \ln(1+x)$$

I know when $M > 0$, there are no fixed points.

           when $M < 0$, there are two fixed points 

but I do not know how to solve for $x$ and find the two fixed points.

Answer 1

The equation $0=M+x-\ln(1+x)$ is equivalent to $$1=\frac1{1+x}e^{M+x},$$ which is equivalent to $$1+x=e^{M+x},$$ which is equivalent to $$(1+x)e^{-(1+x)}=e^{M-1},$$ which is equivalent to $$-(1+x)e^{-(1+x)}=-e^{M-1}.$$ Let $y=-(1+x),$ hence $$ye^y=-\frac1{e}e^M.$$ The minimum of $ye^y$ over all real $y$ is $-\frac1{e},$ so it is clear to see that for $M\lt0,$ there are exactly two fixed points, while for $M=0,$ there is exactly one fixed point, and for $M\gt0,$ there are no fixed points. The equation $$ye^y=-\frac1{e}e^M$$ is equivalent to $$y=W_{-1}\left(-\frac1{e}e^M\right)$$ or $$y=W_0\left(-\frac1{e}e^M\right).$$ For $M=0,$ we have that $$W_{-1}\left(-\frac1{e}\right)=W_0\left(-\frac1{e}\right)=-1.$$ You may be wondering what exact are these $W_{-1}$ and $W_0$ functions I have been using. Consider $f:\mathbb{R}\to\left[-\frac1{e},\infty\right)$ defined by $f(u)=ue^u.$ As defined, $f$ is surjective, but not injective, so it is not invertible. Nonetheless, the restriction $f|_{(-\infty,-1]}$ is surjective and injective, and similarly, the restriction $f|_{[-1,\infty)}$ is surjective and injective. $W_{-1}$ is defined as the inverse of the former, and $W_0$ as the inverse of the latter.