Characterizing conditions for $\tanh{(kx-b)}=x$ to have 1/2/3 fixed points.

Question

I am trying to understand what are the conditions for $\tanh{(kx-b)}$ to have 1 or 2 or 3 fixed points. That is I am trying to characterize conditions on $k$ and $b$ for which equation $\tanh{(kx-b)}=x$ will have exactly 1, or exactly 2, or exactly 3 solutions.

Is there an easy way to do that?

Thank you in advance.

Answer 1

For ease of interpretation, let us rewrite the equation as the intersection of an arbitray straight line with the $\tanh$ curve:

$$px+q=\tanh x.$$

By looking at a plot, you easily see that there is a single intersection for slopes higher than the maximum slope of the $\tanh$, i.e. when $p\ge1$, and for $p\le0$.

Now for fixed $0<p<1$, you get $1$, $2$ or $3$ solutions depending on the value of $q$. More precisely, the tangency condition ensures $2$ roots and separates the other cases:

$$(px_0+q)'=p=(\tanh x_0)'=1-\tanh^2x_0\iff x_0=\pm\tanh^{-1}\sqrt{1-p},$$ and $$q=\pm(\tanh x_0-px_0).$$