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Currently, I am facing this problem:

Given two real functions $A( \vec x )$ and $B( \vec x ):\Bbb R^N\to \Bbb R$, I want to find a third real, monotonic function $f(x):\Bbb R\to\Bbb R$ such that:

$$A\big(f( \vec x )\big)=f\big(B( \vec x )\big)$$

where the simplified the notation writing $f( \vec x )$ means $$ f(x_1,x_2,\ldots,x_n) = \big(f(x_1), f(x_2),\ldots, f(x_n)\big). $$

I am interested in either having a formula/method for finding $f$, or even just having a proof that $f$ exists (or doesn't) under some specific conditions. Eventually, I am interested also in the solution in the case $N=1$.

Also, does this type of problem have a specific name?

Thank you very much!

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  • $\begingroup$ Fixed! I hope now it's fun enough $\endgroup$
    – JustaRegularSchroedinger
    Commented Feb 10, 2020 at 9:37
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    $\begingroup$ you might also want to consider changing the name of your profile; the adjectives you use to characterise Schrödinger are inappropriate, IMO. $\endgroup$
    – Carlo Beenakker
    Commented Feb 10, 2020 at 9:49
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    $\begingroup$ Ok, I also changed the nick to a more appropriate one. I guess the next answer I'll receive is not gonna be on the question I asked, but on the way, I use the comma. Maybe I should ask this same question on a grammar forum... $\endgroup$
    – JustaRegularSchroedinger
    Commented Feb 10, 2020 at 9:58
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    $\begingroup$ It seems too easy: if $A(x_1,x_2)=1+x_1$ and $B(x_1,x_2)=x_1+x_2$ then you are looking for $f$ such that $1+f(x_1)=f(x_1+x_2)$ which cannot be, for example if $x_2=0$. Maybe you missed some conditions? (Or did I?) $\endgroup$
    – Yaakov Baruch
    Commented Feb 10, 2020 at 10:09
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    $\begingroup$ Also, does this type of problem have a specific name? Given two maps $A$ and $B$, trying to find an $f$ such that $A \circ f =f \circ B$ is a problem of semi-conjugacy. $\endgroup$
    – Laurent Berger
    Commented Feb 10, 2020 at 13:12

1 Answer 1

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Existence of such $f$ is a very strong condition on $A$ and $B$ which is called semi-conjugacy. Of course, for generic $A$ and $B$ function $f$ does not exist. Suppose for simplicity that $N=1$. Then it is clear that the image of any fixed point of $B$ under $f$ is a fixed point of $A$. Furthermore, when $N=1$, your equation implies that $$A^n\circ f=f\circ B^n$$ where $A^n$ means the $n$-th iterate. This implies that periodic points of $B$ are mapped to periodic points of $A$ of the same period. So we have some very complicated relation between $A$ and $B$.

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  • $\begingroup$ Alexandre, thank you very much! I think I need to study a little more the theory of topological conjugacy, as there are some points that I don't fully understand. By adding slightly more stringent conditions (i.e. A() and B() continuous and f derivable) is it possible to find a solution to the problem? Otherwise, what are the typical hypotheses to solve it? (if they exist) Eventually, could you suggest me a book or article that could help me in understanding better how to solve the problem? Again, thanks a lot! $\endgroup$
    – JustaRegularSchroedinger
    Commented Feb 10, 2020 at 14:27

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