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

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.

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What is $\mathrm{step}(x)$ supposed to be, the unit step function? In which case, formulae 17-27 here might be useful. – J. M. May 7 '12 at 2:26
It might help if you provided some context. Also, echoing @J.M., what is the value of step(0)? – copper.hat May 7 '12 at 4:06
Thanks for your replies! @J.M.: Yes, step(x) is unit step function. I've check the link but it hasn't helped me yet. In most of the formulas there (still checking some), the limit for $t \rightarrow 0$ ($\epsilon \rightarrow 1$ in my notation) is $y=0.5$, and not $y=x$. I've tried to fix this adding some linear term, but I've realized there is an additional property I'd like to be satisfied: $y_\epsilon (0) = 0$ for any $\epsilon$ – stefano May 7 '12 at 4:31
@copper.hat: In several optimal control problems, the optimal control $y$ is function of a costate $x\ge 0$. In bang-bang solutions, $y(x)$ is a step function, with $y(x) = 0$ for $x\in[0,1)$, and $y(x) = 1$ for $x\in [1,\infty)$. In minimum-effort solutions, $y(x) = x$. I am looking for a family of $C^1$ functions that can help my code transition between one case and the other (I need differentiability to use numerical continuation methods). Stefano – stefano May 7 '12 at 4:31
I don't follow, but am curious. When you change the cost function from, say, min. effort to min. time, the solution will change, typically from something smooth to some extreme value control. But surely that should happen by changing the cost, not by how you represent the solution 'vector'? – copper.hat May 7 '12 at 5:37

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