Logistic map type function with controlled steepness on either side

Question

I am looking for a function that must have the following requirements:

1. $f(1) = f(-1) = 0$
2. $f(x) > 0, \forall x \in (-1,1)$
3. $f$ is differentiable.

Additionally, I would like it to be parametrizable to control the steepness with which it rises and falls around $x = -1$ and $x = 1$

The usual logistic map satisfies the main requirements but is not parametrizable:

$f(x) = (1-x)(1+x)$

logistic map

Another candidate:

$f(x) = \frac{\sin{\frac{\pi(1+x)}{2}}}{\sqrt{\epsilon^2+\sin^2{\frac{\pi(1+x)}{2}}}} \sqrt{1+\epsilon^2}$

symmetric

This is great and has a nice property of $max(f(x)) = 1$ but is symmetric.

Is there a similar function in which the steepness of either side can be controlled? Extra points if it is normalised as well.

Answer 1

Do you need to enforce the condition $\max f(x)=1$? Otherwise, any function of the form

$$f(x)=\sum_{j=1}^{n}c_j (1-x)^{a_j}(1+x)^{b_j}\tag{1}$$

with $a_j, b_j, c_j > 0$ will have the desired property.

Moreover, by adjusting the values of $n$, $a_j$'s, $b_j$'s, $c_j$'s, functions of the form $\text{(1)}$ can approximate any continuous function $g:[-1,1]\to[0,\infty)$ with $g(-1)=g(1)=0$ uniformly.

Note, however, that $f$ in $\text{(1)}$ or $\text{(2)}$ need not be unimodal, meaning that $f$ may have several peaks.

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A special case of $\text{(1)}$,

$$f(x) = c_1\left(\frac{1-x}{2}\right)^a(1+x)+c_2(1-x)\left(\frac{1+x}{2}\right)^b \tag{2} $$

with $c_1,c_2>0$ and $a,b > 1$, satisfies

$$f'(-1) = c_1 \qquad\text{and}\qquad f'(1)=-c_2.$$

Moreover, if $a = b = 2$, then $f$ is unimodal.