I am wondering if there is a solution for (and a method to solve) the following functional equation
$$\frac{d}{dx}f(x) = 1 - \beta f^{-1}(x)$$
in $[0,1]$, with $0<\beta<1$, and initial condition $f(0)=0$. Here $f^{-1}(x)$ refers to the inverse function of $f(x)$. To avoid problems with the definition of $f^{-1}$, we can think of the problem as $f'(x) = 1 - \beta H\left(f^{-1}(x)\right)$ with $H(\cdot)$ being equal to one if $f^{-1}(x)$ is not defined in $[0,1]$.
I haven't been able to find an analytical or numerical solution. The problem with a numerical solution is that if you start from $x=0$ and build to the right, after two steps you can't compute the derivative.
We have the following known:
$$\require{cancel}f(0)=0$$
Take the $f^{-1}$ of both sides:
$$\xcancel{f^{-1}(f}(0))=0=f^{-1}(0)$$
$$f^{-1}(0)=0$$
Thus, look at the original functional equation and consider $x=0$.
$$f(0)=1-\beta f^{-1}(0)$$
$$0=1$$
Contradiction. Therefore, there cannot exist an analytic solution to this problem, and the solution cannot exist at $x=0$ with the given values.
If you neglect your initial condition, though, see that we have:
$$f(x)=mx+b$$
Putting in the values, we get
$$f(x)=i\sqrt\beta x+\frac{\beta-i\beta\sqrt\beta}{\beta^2+\beta}$$
where $i=\sqrt{-1}$