Does there exist $f:(0,\infty)\to(0,\infty)$ such that $f'=f^{-1}$?

Question

Recently the following question was posed: does there exist a differentiable bijection $f:\mathbb R\to\mathbb R$ such that $f'=f^{-1}$? (Here, $f^{-1}$ is the inverse of $f$ with respect to composition of functions.) The answer, as it turns out, is negative, because such a bijection is monotonic, which implies that the derivative cannot be bijective.

There seem to be no such problems if we instead search for a bijection $f:(0,\infty)\to(0,\infty)$ such that $f'=f^{-1}$, so the following variation of the question seems interesting:

Does there exist a differentiable bijection $f:(0,\infty)\to(0,\infty)$ such that $f'=f^{-1}$?

(Edit: as mentioned at the end of this question, the question of existence has been resolved. Is the solution unique?)

Here's the first idea: if there is a function $f:(0,\infty)\to(0,\infty)$ such that $f'(x)=f^{-1}(x)$ holds for all $x\in(0,\infty)$, then $$f'(f(x))=f^{-1}(f(x))=x$$ also must hold for all $x$, since $f$ is bijective. This immediately reminds us of the chain rule, so we multiply by $f'(x)$ to yield $$f'(f(x))f'(x)=xf'(x)$$ which is equivalent to $$f'(f(x))f'(x)=xf'(x)+f(x)-f(x).$$ This implies (integrate from $1$ to $x$, for instance) that there is a primitive function $F$ of $f$ such that $$f(f(x)) = x f(x) - F(x)$$ holds for all $x\in(0,\infty)$, so we may try solving this equation instead. I don't know if this is any easier than the original problem, though. Maybe some kind of fixed point principle might work to show existence?

It would be very nice if it turns out that there is a nice characterization of such functions.

Edit: At the link pointed out by Christian Blatter in the comments there is an explicit solution ($f(x)=\frac{x^\phi}{\phi^{\phi-1}}$, where $\phi=\frac{1+\sqrt5}{2}$ is the golden ratio), so maybe it would also be interesting to know:

Is this solution unique?

Answer 1

I cannot find a general solution for this. But I may find one: The given DE eqn. is equivalent to $$ \frac{dg^{-1}}{dx} = g$$ setting $g = f^{-1}$. And this is in turn equivalent to $$ g^{-1}(x) = \int g(x)dx$$ or $$x = g(\int g(x)dx). $$ Try $g = ax^n$. Then you will find $$ x = a(\frac{a}{n+1}x^{n+1})^{n} = \frac{a^{n+1}}{(n+1)^{n}}x^{n(n+1)}.$$ Then by the root formula you will find two $n$ from the equation $n(n+1)=1$, and for these $n$ you also find suitable $a$.

And try if a linear combination of this solution works, i.e., if the solution space is linear.

If the above works, probably you can extend this idea even further: you man consider any function that is expressible as an infinite sum, i.e., the polynomials of $\infty$ power.

The above two processes, though they may indeed work, will not be that easy I guess.