Abel equation

Remarks

Example: for $f(x) = x^2$ on $X = (1, \infty)$, $\alpha(x) = \log_2 \ln x$ is a solution, since $\log_2 \ln(x^2) = \log_2(2 \ln x) = \log_2 \ln x + 1$. Non-example: for $f(x) = 2x$ on $(0, \infty)$, $\alpha(x) = x$ is not a solution, since $\alpha(f(2)) = 4$ while $\alpha(2) + 1 = 3$. If $\alpha$ is a real-valued solution and $p \colon \R \to \R$ satisfies $p(t + 1) = p(t)$, then $\alpha + p \circ \alpha$ is again a solution, so solutions are far from unique without extra conditions. If $f$ has a fixed point $x_0$, no solution is defined at $x_0$, since $\alpha(x_0) = \alpha(x_0) + 1$ is impossible; so the equation is usually posed on a domain without fixed points.