Conjugacy in the Logistic Family

The case of $\mu \in (1,2)$.

Pick two parameters in the interval $(1,2)$. We will find an explicit topological conjugacy between the two maps restricted to $[0,1]$. (It is not hard to extend the conjugacy outside the interval, but this will be enough to understand how to build such conjugacies.)

Below we plot $F$ with the diagonal over the interval $[0,1]$.

Here is the plot of $G$:

Recall that these maps have fixed points at zero and at $p_\mu=\frac{\mu-1}{\mu}$.

Check that $p$ is fixed by $F$:

We'll use $q$ for the corresponding fixed point of $G$:

Notice that $F$ restricts to a map $$F|_{(0,p)} : (0,p) \to (0,p),$$ and this map is a homeomorphism.

Furthermore, we have $F(x)>x$ for $x \in (0,p)$. Therefore, all orbits in $(0,p)$ converge monotonically to $p$.

We choose any point $a \in (0,p)$ to construct a fundamental domain $\big[a,b\big)$ for $F|_{(0,p)}$. This means that for every point $x \in (0,p)$ there is a unique $n \in {\mathbb Z}$ such that $$F|_{(0,p)}^n(x) \in [a,b).$$ To do this we must take $b=F(a)$.

Now we find a fundamental domain $[c,d)$ for the restriction $G|_{(0,q)}$.

The first stage of the topological conjugacy

We can now produce a topological conjugacy $h_1:[0,p] \to [0,q]$, which should satisfy $G \circ h_1 = h_1 \circ F$.

To do this, we first select a homeomorphism $h_0:[a,b) \to [c,d)$ between the fundamental domains. For simplicity, we'll just use an affine linear map (though any orientation-preserving homeomorphism will work):

Let's check that this does what we want:

The topological conjugacy $h_1$ must satisfy $0 \mapsto 0$ and $p \mapsto q$.

For $x \in (0,p)$, we will be defining $$h_1(x) = G^{-n} \circ h_0 \circ F^n(x)$$ where $n$ is the integer such that $F^n(x) \in [a,b)$. Computing this $n$ and $F^n(x)$ is not too difficult. Basically, if $x<a$, then we continue applying $F$ until the point lies in $[a,b)$, and if $x>b$ we continue applying $F^{-1}$ until the point lies in $[a,b)$.

For this second possibility, we need to an inverse of $F$. There really are two inverses, and sage can actually compute them:

This will be our left and right inverses:

Here is a sanity check:

Let's do the same for $G$:

Now lets turn our attention to computing $n$ and $F^n(x)$. Let's suppose that $x<a$. Let's pick a particular example and try it.

The idea is to repeat applying $F$ until our point becomes bigger than or equal to $a$. At that moment it will lie in $[a,b)$.

Now we need to check if $x$ is still less than $a$.

We have to repeat it again:

Now we check again:

To be sure, lets see if $x \in [a,b)$:

To perform these actions automatically, we can use a while loop:

We remark that we are doing exact arithmetic. If our input is floating point, we will get more readable answers:

Here is a function implementing the full topological conjugacy $h_1:[0,p] \to [0,q]$:

Let's see what $h_1$ looks like:

Let's check the conjugacy equation:

Note that $h_1$ is continuous but not differentiable (despite the apparent smoothness!):

The second stage:

The point $1/2$ is special because it is the critical point. Concretely, $F(1/2)$ is the only point with only one preimage. The same holds for $G$. So, $1/2$ must be sent to $1/2$ by a conjugacy.

Observe that $(p,\frac{1}{2}] \subset W^s(p)$. This is because for $x \in (p, \frac{1}{2}]$, we have $F(x) \in (p, \frac{1}{2}]$ and $F(x) < x$.

We set $a_1=F(\frac{1}{2})$ and $b_1=G(\frac{1}{2})$:

Now we define a homeomorphism between the fundamental domains: $h_2:(a_1, \frac{1}{2}] \to (b_1, \frac{1}{2}]$:

We now extend the previous conjugacy $h_1:[0,p] \to [0,q]$ to a conjugacy $h_3:[0,\frac{1}{2}] \to [0, \frac{1}{2}]$:

Here we plot h3:

Here we graphically check the conjugacy equation:

We look at the derivative for fun:

An interesting thing to look at is what is happening for $h$ near the fixed point $p$. Recall that we showed the conjugacy can not be a $C^1$ homeomorphism in a neighborhood of $p$. From the above graph, it looks like $h'(p)=0$, which means that $h$ cannot have a differentiable inverse. Here is a graph of $h$ in a small neighborhood of $p$:

The third stage:

Now we extend to the full interval $[0,1]$.

Suppose that $x \in (\frac{1}{2},1]$. We'll want out conjugacy $h$ to send $x$ to some point $y = h(x) \in (\frac{1}{2},1]$. Observe that $F(x)=F(1-x)$ and $G(y)=G(1-y)$. Our map $h_3$ defined above satisfies $$h_3 \circ F(1-x) = G \circ h_3(1-x).$$ We claim that if we define $$h_4(x) = \begin{cases} h_3(x) & \text{if }x \in [0, \frac{1}{2}] \\ 1-h_3(1-x) & \text{if } x \in (\frac{1}{2},1] \end{cases}$$ then we'll get a conjugacy. To see this fix an $x > \frac{1}{2}$. Then $h_4(x)>\frac{1}{2}$ and so $$G \circ h_4(x) = G\big(1-h_3(1-x)\big)=G\big(h_3(1-x)\big)$$ and since $F(x)<\frac{1}{2}$ we have $$h_4 \big(F(x)\big)=h_3 \circ F(x)= h_3 \circ F(1-x) = G \circ h_3(1-x).$$ Thus $G \circ h_4(x) = h_4 \circ F(x)$ on $[0,1]$.

Graphically checking the conjugacy: