# Moving between Abel's and Schroeder's Functional Equations [closed]

I see a reoccurring tendency in attempts to extend tetration where the author moves easily between Abel's and Schroeder's Functional Equations. The premise with them being topologically conjugate is they have different topological structures that do not map into each other. Can someone provide non-trivial examples of moving between Abel's and Schroeder's Equations that doesn't violate topological conjugacy and an example that does violate topological conjugacy?

Note: The Julia set of the exponential map has "spines" going to positive infinity. I just read that in general, parabolicly neutral fixed points where $$f(0)=0$$ and $$f'(0)=1$$, have tips of the spines that form a straight line out to positive infinity. The rays of the fractal included begins to display a straight line for the exponential map $$a^z$$ where $$f'(0)=1$$ at a fixed point and $$a \approx 1.444$$. The other Julia and Fatou set's spines either arc in or outward. How could mappings like $$\Phi=\log\phi/\log\,\lambda$$, from Alexandre Eremenko's answer, move between spine tips that form a straight line and those which don't?

For some examples, see the work at the Tetration Forum and the following articles.

Trappmann, H., & Kouznetsov, D. (2010). Uniqueness of holomorphic Abel functions at a complex fixed point pair. Aequationes Math., accepted.

Kouznetsov, D., & Trappmann, H. (2010). Portrait of the four regular super-exponentials to base sqrt(2). Math. Comp., 79, 1727–1756.

Kouznetsov, D., & Trappmann, H. (2010). Superfunctions and square root of factorial. Moscow University Physics Bulletin, 65(1), 6–12. –

• Can you cite a specific instance where the author "moves easily" between? The equations are semi-conjugate: if Schroder is solvable then Abel is as well, but not vice versa, as it is well-known. Jan 7, 2020 at 14:08
• If given cause I will invite appropriate authors. Basically all published papers on fractional iteration of the exponential map that I know of.
– user37691
Jan 7, 2020 at 14:21

Let $$f$$ be a function, and $$f(0)=0$$. Schroeder's equation is $$\phi\circ f=\lambda\phi.$$ Applying $$\log$$ to both sides, and dividing on $$\log\lambda$$, we obtain $$\Phi\circ f=\Phi+1,$$ which is Abel's equation with $$\Phi=\log\phi/\log\,\lambda$$. But this is a formal manipulation whose meaning depends on context. In general, these equations are not equivalent.
For example, when $$f$$ is analytic, $$f(0)=0$$, $$\lambda=f'(0)$$, then Schroder equation has an analytic solution with $$\phi(0)=0$$ if $$|\lambda|\neq 1$$. But if $$f(z)$$ is not linear, and $$\lambda=f'(0)=1$$, Schroder's equation does not have even a formal solution, while Abel's equation has a solution.