Neccesary condition for a family of functions to have a period-doubling bifurcation

Question

I have the following definition of period-doubling bifurcation for a one-parameter family of functions $\left\{{ F_{\lambda} }\right\}$ :

Definition. A family of functions $\left\{{ F_{\lambda} }\right\}$ undergoes a period-doubling bifurcation at the parameter value $\lambda=\lambda_{0}$ if there exist an open interval $I$ and an $\epsilon >0$ such that:

1. For each $\lambda \in [\lambda_{0}-\epsilon,\lambda_{0}+\epsilon]$, there exist a unique fixed point $p_{\lambda}$ for $F_{\lambda}$ in $I$.

2. For $\lambda \in (\lambda_{0}-\epsilon, \lambda_{0})$ , $F_{\lambda}$ has not cycles of period $2$ in $I$ and $p_{\lambda}$ is attracting(resp. repelling).

3. For $\lambda \in (\lambda_{0},\lambda_{0}+\epsilon)$, there exist a unique 2-cycle $q_{\lambda}^1,q_{\lambda}^2$ in $I$ with $F_{\lambda}(q_{\lambda}^1)=q_{\lambda}^2$. This 2-cycle is attracting (resp., repelling). Meanwhile, the fixed point $p_{\lambda}$ is repelling(resp., attracting).

4. As $\lambda \rightarrow{} \lambda_{0}$, $q_{\lambda}^i \rightarrow p_{\lambda_{0}}$.

If the above conditions are valid with the symbol > instead of <, we also say that the family have a period-doubling bifurcation. Althought this is certainly a complicated definition, there exist simple examples of this kind of bifurcation. For instance, the family $F_{c}(x)=x^2+c$ has a period-doubling bifurcation at $c=-3/4$, in which case, we can take $\epsilon=1$ and $I=(\infty,-1/2)$.

My question is: why this definition implies that $F_{\lambda_{0}}'(p_{\lambda_{0}})=-1$?. I mean, it's clear for me that there is only two possibilities: $1$ or $-1$. But I still don't see why we can discard the first case. Can anyone explain me please?.

Thank you in advance.

Answer 1

First off, it's pretty easy to see (in the context of the quadratic family) why $F_{\lambda}'(p_{\lambda})=1$ doesn't guarantee period doubling. If you graph $$F_{\lambda}(x) = x^2 + \lambda$$ for $\lambda$ in a neighborhood of $1/4$, there simply is no period-doubling bifurcation since, for $\lambda>1/4$, there is no fixed point to double:

To be clear, though, there is a bifurcation - just one of a different type. As $\lambda$ increases through $1/4$, two fixed points merge and annihilate each other. This type of bifurcation is called a saddle-node bifurcation. I guess this highlights the true importance of the condition that $|F_{\lambda}'(p)|=1$; that's where some type of bifurcation can occur but there's no immediate guarantee as to which type.

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To answer your question as to why period doubling occurs only when the derivative is $-1$, assume that $x_1$ and $x_2$ form a 2-cycle for $F_{\lambda}$ and assume (WLOG) that $x_1<x_2$. Then, over the interval $I=[x_1,x_2]$, the average rate of change of $F_{\lambda}$ is $$\frac{F_{\lambda}(x_2)-F_{\lambda}(x_1)}{x_2-x_1} = \frac{x_1-x_2}{x_2-x_1} = -1.$$ By the mean value theorem, there is a point $c\in(x_1,x_2)$ such that $F_{\lambda}'(c)=-1$. But, in the context of your definition, you have an $\varepsilon$ decreasing to zero which forces the $x_1$ and $x_2$ to collapse on the fixed point. Thus, $F_{\lambda}(p_\lambda)=-1$.