# Construction of a Step-Like Function

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 -

1. $$f(x)$$ is increasing monotically.
2. $$f(0) = 0$$
3. $$\lim_{x\to \infty} f(x) = 1$$
4. $$f(x)$$ 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)
5. $$f'(0) = 0$$
6. $$\lim_{x\to \infty}f'(x) = 0$$

• Condition 3 is incompatible with $f$ being a map $[0,\infty)\to [0,{\color{red}1}]$ Commented Jan 19, 2020 at 17:06
• $f$ maps to $[0,1]$ and $\lim_{x\rightarrow \infty}f(x)=\infty$? Commented Jan 19, 2020 at 17:09
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).