0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ How is this a differential equation? $\endgroup$ – MathematicsStudent1122 Sep 28 '16 at 0:16
  • 1
    $\begingroup$ $$f(0)=0\implies f^{-1}(0)=0\\\implies f(0)\ne1-\beta f^{-1}(0)$$ $\endgroup$ – 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
1
$\begingroup$

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 $i=\sqrt{-1}$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.