How to find a solution for the functional equation $\phi(f(x))=f'(x)\phi(x)$?

Question

Given $f$, how to find some $\phi$ so that $\phi(f(x))=f'(x)\phi(x)$?

If you noticed, the given functional equation is of the form of Julia's Equation. However, I cannot seem to find a good article on this equation, so I've come here to ask for methods that are used to solve this functional equation. It would also help me if you could provide an example.

Answer 1

Not quite an answer, but a good start.

If $\Psi(x)$ is the Schroder function of $f$, then $\Psi(f(x)) = \lambda\Psi(x)$. Therefore $\Psi'(f(x))f'(x) = \lambda\Psi'(x)$ and if $\phi = \frac{1}{\Psi'(x)}$ then $\phi(f(x)) = \frac{f'(x)}{\lambda}\phi(x)$. The domains of holomorphy depend on where $\Psi$ is defined. Usually this is locally about a fixed point, which means $\phi$ is holomorphic in a neighborhood of this fixed point. If $\Psi$ (or $f$) have no critical points in the immediate basin, then $\phi$ is holomorphic there. I'm not certain, but I think this is the best you can do using the Schroder function.

To find $\Psi(x)$ we generally use Schroder's limit about a fixed point. WLOG assume $f(0) = 0$. Let $\lambda = f'(0)$. If $0<|\lambda| < 1$ then

$$\Psi(x) = \lim_{n\to\infty}\frac{f^{\circ n}(x)}{\lambda^n}$$

if $|\lambda| > 1$ then letting $g(x) = f^{-1}(x)$

$$\Psi(x) =\lim_{n\to \infty}g^{\circ n}(x)\lambda^n$$

This gives a solution in a neighborhood of $0$, generally it can be expanded. The degenerate cases $\lambda = 0$ or $|\lambda| = 1$ are much more sporadic and typically no solution exists.

In the general case, I'm not sure how you would remove $\lambda$ from the functional equation. Usually Julia's equation is used when $\lambda = 1$; sadly this solution doesn't follow from the above methods. I might have to crack open a book or two to recall how you actually construct the equation you are after.

Edit:

So as I seemed to remember, the solution is only when $\lambda = 1$, there is no solution when $\lambda$ is anything else. Therefore the best you get in the general case is what I just wrote above. However, for an exhausting construction when $\lambda = 1$ observe the references of

https://www.math.ucla.edu/~matthias/pdf/itlog-final.pdf

This was really clearly written.The papers of interest, as I gather, are by Ecalle. He's the wiz when $\lambda = 1$.