What is the limit distance to the base function if offset curve is a function too?

Question

I asked a question about parallel functions in here . I understood that offset curves that are the parallels of a function may not be functions after J.M.'s answer. I got new questions after that answer.

Q1) What is the limit distance to the base function if offset curve is a function too?

Q2) It can be shown as geometrically that all parallel curves of line and half circle are also functions. What is the whole function family defination for such functions?

Thanks for answers

Please see parallel curve examples below. (Thanks to J.M. for the graphs)

Answer 1

I will try to answer question 1 about "limit distance".

For a parametric curve $x=x(t)$, $y=y(t)$ to have an equation of the form $y=g(x)$, we need $x$ to be a strictly increasing function of $t$. Suppose we have a smooth function $y=f(x)$ and consider its parallel curve at distance $d$ (measured upward; $d$ could be positive or negative). Then $$x(t)=t-d\frac{f'(t)}{\sqrt{1+(f'(t)^2}}$$ If $x'>0$ for all $t$, then the parallel curve is also the graph of a function. Computation shows (after a simplification) that $$x'(t)=1-d\frac{f''(t)}{(1+f'(t)^2)^{3/2}}$$ So $x(t)$ is strictly increasing when $$d\frac{f''(t)}{(1+f'(t)^2)^{3/2}}<1$$ and fails to be strictly increasing if the reverse inequality holds. You will find the critical value of $d$ by considering the values of $f''(t)/{(1+f'(t)^2)^{3/2}}$. Not incidentally, the latter quantity is the curvature of the graph $y=f(x)$.

Answer 2

A different and somewhat more abstract viewpoint is given by considering the "squared distance function" defined by $\rho(x,y)=d((x,y),G)^2$ for all $(x,y)\in \Bbb R^2$. Here $G=\{(u,v):v=f(u)\}$ is the graph of your function, considered as a subset of $\Bbb R^2$, and $d((x,y),G)$ is the distance from $(x,y)$ to $G$, which is the same as the distance from $(x,y)$ to the closest point in $G$. (Assume that your function $f$ is continuous on $\Bbb R$ so that $G$ is a closed subset of $\Bbb R^2$.)

If your function $f$ is smooth then $\rho(x,y)$ will be smooth on a neighborhood of $G$. More precisely, if $f$ is smooth of class $C^k$ with $k\geq 2$ then $\rho(x,y)$ will be smooth of class $C^k$ near the graph $G$. (See this.) From now on, we always assume $k\geq 2$. We have used the squared distance function to get smoothness on the graph $G$, in the same way that the function $x^2$ is smooth at $x=0$ wheras the function $|x|$ is not.

How far away from $G$ will $\rho(x,y)$ be smooth? Let $(x,y)$ be some point in $\Bbb R^2$ and let $(u,v)$ be the point on $G$ which is closest to $(x,y)$ (assume that there is only one such point). If the distance between the two points is less than the radius of curvature of $G$ at $(u,v)$ then we are guaranteed that the squared distance function will be smooth at $(x,y)$.

Now compute the radius of curvature $r(u,v)$ of $G$ at a general point $(u,v)$ on the graph $G$. If the radius of curvature is bounded from below on $G$, so that we have $r(u,v)\geq c$ for some $c>0$ and all $(u,v)$ on the graph, then the squared distance will be smooth on the set of points $\{(x,y):d((x,y),G)<c\}$. You can then define parallel graphs on this set as level curves for the distance function $d(\cdot,G)$.

Can something go wrong here? Yes, if there are more than one point on $G$ which is nearest to $(x,y)$. For general curves this can be a problem, but since your curve is the graph of a function this problem cannot occur when $d(x,y)<c$, where $c$ is the uniform lower bound from the last paragraph for the radius of curvature.

I called this approach more abstract, since it is not so easy to get explicit formulas for the distance function $d(\cdot, G)$. Nevertheless, this function (or the function $\rho$) is often a useful tool for studying curves and higher dimensional surfaces.