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.

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    $\begingroup$ How is this a differential equation? $\endgroup$ – MathematicsStudent1122 Sep 28 '16 at 0:16
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    $\begingroup$ $$f(0)=0\implies f^{-1}(0)=0\\\implies f(0)\ne1-\beta f^{-1}(0)$$ $\endgroup$ – Simply Beautiful Art Sep 28 '16 at 0:24
  • $\begingroup$ I solved it! $\color{white}{;)}$ $\endgroup$ – Simply Beautiful Art Sep 28 '16 at 1:13

We have the following known:


Take the $f^{-1}$ of both sides:



Thus, look at the original functional equation and consider $x=0$.

$$f(0)=1-\beta f^{-1}(0)$$


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:


Putting in the values, we get

$$f(x)=i\sqrt\beta x+\frac{\beta-i\beta\sqrt\beta}{\beta^2+\beta}$$

where $i=\sqrt{-1}$

  • $\begingroup$ If you want to see what happens when you assume the solution is real valued, check the revisions. $\endgroup$ – Simply Beautiful Art Sep 28 '16 at 1:12

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