From where do we get $\varphi(w) = w + \frac 1 w$ for $w^2$ and $z^2 - 2$?

Question

In Iteration of Quadratic Polynomials, Julia Sets,

we must find some (invertible?) $\varphi$ s.t. $g = \varphi^{-1}(f(\varphi(z)))$ in order to study $\{g^{\circ n}(z)\}_{n=1}^{\infty}$ the iterates of $g$ assuming we have studied $\{f^{\circ n}(z)\}_{n=1}^{\infty}$ the iterates of $f$

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Case 1.

For $f(z) = z^2 + c$ and $g(z) = az^2+bz+d$

we can study $\{g^{\circ n}(z)\}_{n=1}^{\infty}$ by studying $\{f^{\circ n}(z)\}_{n=1}^{\infty}$, the iterates of $f$ because $$g^{\circ n}(z) = \varphi_1^{-1}(f^{\circ n}(\varphi_1(z)))$$

where $$\varphi_1(z) = az + \frac b 2$$ for some appropriate domain and range

Case 2.

For $g(w) = w^2$ and $f(z) = z^2 - 2$

we can study $\{f^{\circ n}(z)\}_{n=1}^{\infty}$ by studying $\{g^{\circ n}(z)\}_{n=1}^{\infty}$ because $$f^{\circ n}(z) = \varphi^{-1}_2(g^{\circ n}(\varphi_2(z)))$$

where $$\varphi_2(w) = w + \frac 1 w$$ for some appropriate domain and range

From where did $\varphi_2(w)$ come? It doesn't seem to be in the form $\varphi_1(z) = az + \frac b 2$, though I'm thinking there's some substitution to be done (hence the use of $w$ and not $z$)

Answer 1

I think you just have a mix up in the notations for $\varphi_2(w) = w + \frac{1}{w}$. Do not use the $w$ variable, stick to just $z$. So assume the conformal change of variables is
$$\varphi_2(z) = z + \frac{1}{z}$$

Assume $g(z) = z^2$ and $f(z) = z^2 - 2$. Then you can show that $$\varphi_2 \circ g = f \circ \varphi_2$$ This is equivalent to $$f = \varphi_2 \circ g \circ \varphi^{-1}_2$$ Let's check $\varphi_2 \circ g = f \circ \varphi_2$. It is enough to do that and everything else follows from it. Thus we have to verify the identity $$ \varphi_2 \big( g(z)\big) = \varphi_2( z^2 ) = f\big(\varphi_2(z)\big) = \big(\varphi_2(z)\big)^2 - 2$$ Indeed $$ \varphi_2 \big( g(z)\big) = \varphi_2( z^2 ) = z^2 + \frac{1}{z^2} $$ $$f\big(\varphi_2(z)\big) = \big(\varphi_2(z)\big)^2 - 2 = \left( z + \frac{1}{z}\right)^2 - 2 = z^2 + 2 \, z \, \frac{1}{z} + \frac{1}{z^2} - 2 =$$ $$= z^2 + 2 + \frac{1}{z^2} - 2 = z^2 + \frac{1}{z^2}$$

Answer 2

You ask, where did $\varphi_2$ come from? I can't tell you exactly what the original author was thinking but the point is that it does what you want it to do. That is, if $E$ denotes the exterior of the unit disk and $F$ denotes $$F = C \setminus [-2,2],$$ then $\varphi_2:E\to F$ in one-to-one fashion and (as the first reply correctly shows) it conjugates $f$ to $g$.

Now, you're absolutely correct that $\varphi_2$ is not of the form $az+b/2$ but that's not required. Generally, the nicer the conjugacy function the nicer the relationship between the functions you're comparing. If the conjugacy had the form $az+b/2$, then an orbit $g$ would be geometrically similar to some orbit of $f$. That's just not the case here, though. In fact, the chaotic orbits of $g$ lie on the unit circle and the chaotic orbits of $f$ lie on the interval $[-2,2]$. The function $\varphi_2$ is just a semi-conjugacy there because it's not one-to-one. The overall effect of $\varphi_2$ on the region of interest looks like so: