Solution to the functional equation $(x(t))^p(1-t^2)=x\big(\frac{2t}{1-t^2}\big)$ for $x(t).$

Question

Let $p\in\mathbb{R}$ be a fixed real number. I am trying to find any and all functions $x(t)$ such that $$(1-t^2)(x(t))^p=x\bigg(\frac{2t}{1-t^2}\bigg)$$ Is there a solution to this functional equation? If so, what is it, and for which values of $p?$ I am hoping to find a closed form solution as a function of $t$ and $p.$

I have solutions for $p=0$ and $p=2.$ When $p=0,$ I was able to find two functions that solve this equation, and they are $$x(t)=-\frac{2}{t^2}\pm\frac{1}{t}\sqrt{1+\frac{1}{t^2}}.$$ When $p=2,$ I obtain the piecewise function below as a solution: $$x(t)=\left\{ \begin{array}{ll} \frac{1}{\sqrt{1+t^2}} & t\in[-1,1] \\ -\frac{1}{\sqrt{1+t^2}} & |t|>1 \\\end{array} \right..$$ I don't know if these are the only solutions. Please help me. Anything helps. Thank you very much.

Answer 1

Solving functional equations is not necessarily impossible, but in general quite bit of work. It would be therefore good idea to give people a more compelling reason to want to invest some time in helping you solve such a problem. Specially since you actually ask for a generic solution with arbitrary $p$. So why do you actually want to solve this particular problem? If it is of the type, let's see whether I/we can you underestimate the difficulties of such a question.

Anyway, for the case $p=1$ there is the general solution $f(x) = c \frac{ \arctan x}{x}$ for any constant $c$.

Answer 2

I solved my original problems! The general solution to $u(t)\big(x(t)\big)^p=(x\circ f)(t)$ is $$x(t):=\prod_{n=1}^{\infty}\big(g_n(t)\big)^{p^{n-1}}$$ where $$g_n:=u\circ f^{-n}.$$ This is, of course, assuming $f$ is invertible and this product of functions converges (point-wise?). I am currently trying to find conditions for this by looking at the the corresponding log-sum. Thank you for your insight and techniques which ultimately led to my solution. I know user90369 is anonymous, but, if you'd like, I am willing to cite you in a paper I plan on publishing which uses this result. I can't thank you enough. Also, thanks to Ronald Blaak whose solution made me realize that if this equation has one solution and $p=1$, then any constant multiple must also be a solution.