Maybe someone can help me with a problem.
I need a very special parametrization of $f: [0, 1] \rightarrow [0, 1]$ that should be similar to this one. The picture shows
$$f(x) = 2 \frac{g(x)^{q}}{g(x)^{q}+g(2-x)^{q}} \quad \text{with} \quad g(x) = \left(\frac{1}{q} - \frac{1}{2} \right)(1-x)^3-\frac{1}{q}(1-x)+\frac{1}{2} \tag{1}$$
For the picture i have chosen $q = 10$.
What I would like is something more similar to this curve. Around $x=0$ it should fall like $x^q$ (red part) and around $x=1$ it should be close to linear (black part). As $f'(x)\approx 1$ is desired for $x=1$, the function should maintain a slope close to $1$ for $x \in [\varepsilon, 2-\varepsilon]$. The important aspect here is the symmetry of the function with respect to $(1, 1)$, that the slope of the function quickly goes to $0$ as $x$ approaches $0~$ ($f'(x) \rightarrow 0~$ as $~x\rightarrow 0$) and that the function is smooth.
I am having a hard time finding a function that fulfills this, but I'm sure there must be a better solution than equation $(1)$. Can anyone help me?
To kind of explain why I need this: I need to (numerically) calculate the integral of a function that has a (logarithmic) singularity at $x=0$ (explicitly a linear combination of Bessel functions, following the solution of Schrödingers equation for a free particle). In order to do this, the singularity can be approximated by a finite Fourier expansion. If I then integrate numerically, the number of discretization points (and thus computation time) can be drastically reduced by not choosing discretization points on an equidistant grid, but on a graded one. Therefore I need a parametrization that takes my discretization points from an equidistant grid and distributes them as drawn in the example above.
Here is a picture of the solution of Schrodingers equation for a free particle in the given boundary, that motivates my question.
