Homotopy between $y=x$ and $y = {\rm step}(x-1)$?

Question

I am looking for a homotopy between $y=x$ to $y = {\rm step}(x-1)$ (to the limit; a very steep sigmoid is also ok). More formally, I look for a family $$y_\epsilon \in C^1(A,A)$$ where $$A = [0,\infty)$$ $$\epsilon \in [0, 1)$$ such that $$y_0 = y$$ $$\lim_{\epsilon \rightarrow 1} y_\epsilon = {\rm step}(x-1)$$ Any hint is appreciated.

Answer 1

How about: $$ y_\epsilon(x) = (1-\epsilon) x + \epsilon\left( \frac{1}{2} + \frac{1}{\pi} \arctan\left(\frac{x-1}{1-\epsilon}\right)\right) $$ Clearly $y_0(x) = x$ and $$ \lim_{\epsilon \uparrow 1} y_\epsilon(x) = \lim_{\epsilon \uparrow 1} \left( \frac{1}{2} + \frac{1}{\pi}\arctan\left(\frac{x-1}{1-\epsilon}\right)\right) = \cases{1 & x > 1 \\ 1/2 & x=1 \\ 0 & x<1} $$