For my project, I have to model a system with a step-like function (like the Heaviside Step Function), but with different limiting behavior. However, I was unable to come up with one, so I came here.
In particular, I need a function $f: [0,\infty)\to \mathbb [0,1]$, with the properties -
- $f(x)$ is increasing monotically.
- $f(0) = 0$
- $\lim_{x\to \infty} f(x) = 1$
- $f(x)$ is analytic (if this is not possible, a function which is differentiable enough for linear stability analysis to be done without worries of non-differentiability)
- $f'(0) = 0$
- $\lim_{x\to \infty}f'(x) = 0$
Thanks in advance!
Let $h:\mathbb R\to [0,\infty]$ be strictly increasing with $h(-\infty)=0$ and $h(\infty)=\infty$. Let $g:\mathbb R \to \mathbb R$ be a function like your $f$, except that $g(-\infty)=0$ instead of property 3. in your question. Then $f:=g\circ h$ is a good candidate (you need to check that $f'(0)=0$ though). For example $$ f(x) = \begin{cases}\frac{\tanh(\log x)+1}{2} & x>0 \\ 0 & x=0 \end{cases} $$ should work, if I understand your criteria right.
EDIT: Actually, my function is equal to $\frac{1}{1+x^{-2}}$. You can also introduce parameters: $$ f(x) = \frac{1}{1+\beta x^{-\alpha}}, \quad \alpha>1, \beta>0 $$ Or you can use $$ f(x) = 1-e^{-\beta x^\alpha}, \quad \alpha>1, \beta>0 $$ These are based on using $1-{}$(a bell-shaped curve).