On the solution of the functional equation $(x(t))^{p}u(t)=(x\circ f)(t)$

Question

I am given a real number $p$ and two real valued functions $f(t),u(t)$ on $\mathbb{R}$ You can assume $u(t)$ and $f(t)$ are as nice as they need to be, including that they're invertible.

I am trying to find a function $x(t)$ that solves the equation $$(x(t))^{p}u(t)=(x\circ f)(t).$$ Ideally, the solution for $x(t)$ is some expression involving $u(t)$ and $f(t).$

Also, observe if $t^{*}$ is a fixed point of $f$ and $p\neq1,$ then $$x(t^{*})=(u(t^{*}))^{\frac{1}{1-p}}.$$

Answer 1

I'm not even sure that this is possible to solve without being given $f$. However, since you said that I could assume whatever I liked about $f$, I'm going to assume that $$f(f(t))=t$$

By the original functional equation, $$x(t)^pu(t)=x(f(t))$$ and by my assumption, $$x(f(t))^pu(f(t))=x(f^2(t))=x(t)$$ and so I now have the system of equations $$x(f(t))=x(t)^pu(t)$$ $$x(f(t))^pu(f(t))=x(t)$$ and by substitution $$x(t)^{p^2}u(t)^{p}u(f(t))=x(t)$$ $$x(t)^{p^2-1}=\frac{1}{u(t)^pu(f(t))}$$ $$x(t)=\sqrt[p^2-1]{\frac{1}{u(t)^pu(f(t))}}$$

So that is the solution for some cases. A few of these cases include $$f(x)=c-x, c\in \mathbb R$$ $$f(x)=\pm\frac{1}{x}$$

Answer 2

We can relax the condition on invertibility and get the general solution $$x(t):=\prod_{n=1}^{\infty} \big(h_{n}(t)\big)^{-1/p^{n}}$$ where $h_{n}:=u\circ f^{n-1}.$