I'm trying to work out a bifurcation diagram for the function given by
$x'=x+\tanh(rx) $
for all values of $r$.
I know that $x_*=0$ is a fixed point for all values of $r$. I can verify graphically that two more fixed points will appear for values of $r<-1$, however I am unable to explain quantitatively why this is so. Playing around with graphs makes me believe that as $r$ gets smaller and smaller (approaches $-\infty$, the new fixed points $x_*$ approach $\pm1$.
Can someone please give me an explanation as to why this is the case? I don't have much (any) experience with hyperbolic functions, so perhaps something is escaping me here. It does not seem as though there is an easy way to explicitly solve for $x_*$ in terms of $r$.
I am thinking you mean to calculate the following:
$$x_{n+1}=x_n+\tanh(rx_n)$$
And you have clearly seen that $x_*=0$ is a fixed point for all $r$ since
$$0=0+\tanh(0)$$
If you take the limit $r\to-\infty$, the problem changes to the following:
$$x_{n+1}=\begin{cases}x_n+1&;x_n<0\\x_n-1&;x_n>0\\x_n&;x_n=0\end{cases}$$
So, $x_*=\pm1$ are not actually fixed points in this scenario. For example:
$$x_k=0.5$$
$$x_{k+1}=-0.5$$
$$x_{k+2}=0.5$$
And so on, the values will alternate around $0$, and if $a_0$ was an integer, it would converge to $0$.
Lastly, the problem is to find the following solution:
$$\lim_{n\to\infty}x_n=L$$
Assuming the limit exists, we are tasked with solving
$$L=L+\tanh(rL)$$
$$0=\tanh(rL)$$
$$L=\frac{\operatorname{arctanh}(0)}r$$
which has the single real solution
$$L=0$$
Thus, $a_*=0$ is the only fixed point.