Is there a systematic procedure to Solve Abel's, Böttcher's, or Schröder's Equation

Question

I've been interested greatly in the study of functional equations for some time now, I've learnt many different techniques for their solution. Currently I have been studying superfunctions and fractional iterations of functions. In all these subjects I have been led to three main equations all tied together. The first one I stumbled across in an effort to extend tetration to the reals (the forth operation above exponentiation) is called Abel's Equation, $$ f(g(x))=f(x)+1. $$ I studied it and searched through many works on methods for its solution, but in all of them I could neither find a general method nor a simple step by step systematic solution procedure. With the other two, i.e. Böttcher's equation, $$ f(g(x))=f(x)^n, $$ and Schröder's equation, $$ f(g(x))=sf(x), $$ I ran into the same problems. So my question is, is there a easy to follow neat systematic procedure in which I can use to solve any one of these equations to get the family or families of functions that satisfy the relations, and if so where I can find it?

Answer 1

A rough heuristic which I use for these is below:

Abel's Equation:

$$ f(g(x)) = f(x) + 1 \rightarrow f(g(x))-f(x)=1 $$

I haven't found an easy way to give a closed form / series representation of this always. But if $g$ is sufficiently "nice" you can find $f$ by doing the following: consider some point $x=p$ then the sequence of points $x_1 = g(p), x_2= g(g(p)), x_3= g(g(g(p)))\ \dots \ x_k = g^{k}(p), ... $ let $f(x_k) = k$ then you want to interpolate this set of points as a curve

Another way to view the same procedure is to let $f$ be the function such that counts how many times do you need to apply $g^{-1}$ (the function inverse! not the reciprocal) to a value $x$ to get a standard value $p$. So if $g = x+1$ then $g^{-1} = x-1$ and if we let the standard value $p=0$ then $f$ asks how many times do you have to apply $x-1$ to a value $x$ to get it 0. If $x$ is an integer then this is just $x$ times $f = x$ over the integers (we then just say $f = x$ for all inputs to simplify).

We can let $g = 2x$ then let $p=1$ and if $x$ is a power of $x=2^n$ we can have that $f = \log_2(x)$. So we have defined $x$ for all powers of $2$ and we can now just use that same definition over all real/complex numbers.

IF we let $g=x^2$ then let $p=2$ and if $x$ is a power of a power of $2$ so $x=2^{2^{n}}$ then $f=\log_2(\log_2(x))$. Etc...

Schröder's Equation:

$$ f(g(x)) - sf(x) = 0 $$

This is a little trickier to solve. First we need a function $H$ such that $H(g^{k}(x)) \rightarrow 0 $ as $k \rightarrow \infty$ and $k \rightarrow -\infty$ for some set of $x$. If $g^{-1}$ isn't well defined this can be a little subtle but there is usually a way to make this work. Also $H$ needs to approach $0$ "fast" depending on the choice of $s$.

Then we consider the function

$$ f(x) = ... s^2H(g^{-2}(x)) + sH(g^{-1}(x)) + H(x) + s^{-1}H(g(x)) + s^{-2}H(g^2(x)) + ... \sum_{k=-\infty}^{\infty} s^{-k} H(g^{k}(x)) $$

Note that $s^k$ is just a normal exponential and $g^k$ is function composition.

Clearly this obeys the desired functional equation and WILL converge if $H$ goes to 0 sufficiently fast for repeated applications of $g$ and $g^{-1}$.

We given an example now. Suppose we wish to solve

$$ F(x^2) - 2F(x) = 0 $$

In this case $g(x) = x^2$ and $g^{-1}(x)=x^{\frac{1}{2}} = e^{\frac{\ln(x)}{2}}$ (we can use the complex exponential with a branch cut to define this in the complex plane).

Now depending on the absolute value of $x$ we have differing behavior

$$ g^{+\infty}(x) = \begin{cases} \infty & \text{if $|x|>1$} \\ 0 & \text{if $|x|< 1$} \\ 1 & \text{if $x$ is a $2^n$ root of unity} \\ \text{undefined} & \text{otherwise} \end{cases} $$ $$ g^{-\infty}(x) = \begin{cases} 0 & \text{if $x$ = 0} \\ 1 & \text{otherwise} \end{cases} $$

So a good $H(x) \rightarrow 0$ quickly if $x \rightarrow 0,1,\infty$. We can basically make whatever we want as long as we meet that condition. We consider for example $H(x) = x^2(1-x)^2e^{-x^2}$

Then a solution can be written quite simply as

$$ f(x) =... x^{\frac{x}{4}}(1-x^{\frac{1}{8}})e^{-\frac{x}{4}} + x^{\frac{x}{2}}(1-x^{\frac{1}{4}})e^{-\frac{x}{2}} + x(1-x^{\frac{1}{2}})e^{-x} + x^2(1-x)^2e^{-x^2} + x^4(1-x^2)^2e^{-x^4} + x^8(1-x^4)^2e^{-x^8} + ... = \sum_{k=-\infty}^{\infty} x^{2^{1+k}}(1 - x^{2^k})^2 e^{-x^{2^k}} $$

This is some curve and it obeys our target equation. Of course $f(x) = \ln(x)$ is a much simpler solution! BUT, the point is that there are really uncountably many other exotic solutions that can be built up by changing $H$ (a fun exploration might be to ask which $H$ produces the natural log?)

Bötchers Equation:

Botchers Equation (this is my first time seeing it) is basicaly just the multiplicative version of Schroder's Equation...

If we have an $H$ that goes to 1 sufficiently quickly for $g^k$ and $g^{-k}$ as $k\rightarrow \infty$ then the following should work:

$$f(x) = ... H(g^{-2}(x)^{n^2} H(g^{-1}(x))^n H(x) H(g(x))^{\frac{1}{n}} H(g^2(x))^{\frac{1}{n^2}} ... \prod_{k=-\infty}^{\infty} H(g^{k}(x))^{n^{-k}} $$

Further Considerations:

These same tricks apply for ANY associative functional equation. So if $T$ is a binary associative operator $U,V$ are operators that distribute over $T$ then $T(U[f],V[f]) = \text{Id}_{T}$ supports an $H$ construction like we did for Schroder and Botcher.

Now it might be tempting to consider functional equations with multiple functional shifts such as

$$ f(g_1(x)) + f(g_2(x)) + f(x) = 0 $$

These are substantially more difficult to solve. And basically instead of looking at the chain $... g^{-k} ... g^{-2}, g^{-1}, g, g^1, g^2, ... g^{k} ... $ you need to look at the Cayley Diagram formed by $g_1, g_2$ (and this might shock you but this Diagram is almost NEVER a free group diagram)

Once you have nailed down the diagram then you need to find $H$ that decay properly for all the limit directions of that diagram and then you can sum.