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. $\lim_{x\to \infty}f'(x) = 0$

Thanks in advance!

  • 1
    $\begingroup$ Condition 3 is incompatible with $f$ being a map $[0,\infty)\to [0,{\color{red}1}]$ $\endgroup$ Commented Jan 19, 2020 at 17:06
  • $\begingroup$ $f$ maps to $[0,1]$ and $\lim_{x\rightarrow \infty}f(x)=\infty$? $\endgroup$
    – cangrejo
    Commented Jan 19, 2020 at 17:09
  • $\begingroup$ Sorry, stupid mistake. Please have a look at it now. $\endgroup$
    – Ishan Deo
    Commented Jan 19, 2020 at 18:32

1 Answer 1


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).


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