Showing that two functions are topologically conjugate

Question

Let $Q_{c}(x) = x^{2} + c$. Show that if $c < \frac{1}{4}$, then there is a unique $\mu > 1$ such that $Q_{c}$ is topologically conjugate to $F_{\mu}(x) = \mu x(1 - x)$ via a map of the form $h(x) = ax + b$.

The definition for topologically conjugate is that maps $f : A \rightarrow A$ and $g : B \rightarrow B$ are topologically conjugate iff. there is a homeomorphsim $\phi : A \rightarrow B$ such that $f \circ h = g \circ h$.

What I tried doing was using $h(x) = ax + b$ and then setting $F_{\mu} \circ h = Q_{c} \circ h$ and then solving. But eventually this led to nowhere.

Any ideas and hints are appreciated.

Answer 1

We have to find under which condition we can write, for any $x$:

$$a(\mu x(1-x))+b=(ax+b)^2+c\tag{1}$$

(in the order indicated by @TokenToucan)

Let us give to both sides of (1) the classical factorization ("completing the square"):

$$-\mu a\left(\left(x-\tfrac12\right)^2-\tfrac14\right)+b=a^2\left(x+\tfrac{b}{a}\right)^2+c$$

$$-\mu a\left(x-\tfrac12\right)^2 + \left(b + \tfrac14 \mu a\right) =a^2\left(x+\tfrac{b}{a}\right)^2+c$$

By unicity of this decomposition, we can identify corresponding terms :

$$\begin{cases}-\mu a &=&a^2\\ \frac{b}{a}&=&-\frac12\\b + \frac14 \mu a&=&c\end{cases}\tag{2}$$

Eliminating $b$ and $\mu$, we get the equation :

$$a^2+2a+4c=0 \ \ \iff \ \ (a+1)^2=1-4c \tag{3}$$

which is possible if and only if

$$c<\dfrac14 \tag{4}$$

Remark : a generalisation with Chebyshev polynomials :https://math.stackexchange.com/q/1114988 .

Answer 2

I see that the functions you are given are those or the Mandelbrot-Julia fractal and the logistic iteration. I assume that the task is to show that both discrete dynamical systems are essentially the same. In that regard I find it more intuitive in thinking in sequences than in conjugate maps, while the computations and results remain the same.

You have the iteration $x_{n+1}=x_n^2+c$ and you want to eliminate the constant term. Try $y_n=x_n+a$ $$ y_{n+1}=x_{n+1}+a=(y_n-a)^2+c+a=y_n^2-2ay_n+c+a^2+a $$ Now you have got to solve $0=a^2+a+c$ which has a real solution if $c$ is real and $c\le\frac14$.

After that rescale $z_n=by_n$ so that the coefficient structure on the right side matches the desired structure of the logistic map.