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Suppose we have a smooth real-valued curve $\alpha(s):[0,1] \mapsto (x(s),y(s))$ satisfying $x(0)=x(1)$ and $y(0)=y(1)$.

I have a few questions which are probably basic but are confusing me.

First, since $\alpha$ is only defined in a right-neighborhood of $0$, and a left-neighborhood of $1$,

then is $\alpha'(0) = (x'(0), y'(0))$ where $x'(0) = \lim_{h \rightarrow 0^+} \frac{x(h)-x(0)}{h}$ and $y'(0) = \lim_{h \rightarrow 0^+} \frac{y(h)-y(0)}{h}$

$\alpha'(1) = (x'(1), y'(1))$ where $x'(1) = \lim_{h \rightarrow 0^+} \frac{x(1)-x(1-h)}{h}$ and $y'(0) = \lim_{h \rightarrow 0^+} \frac{y(1)-y(1-h)}{h}$ ?

Second question, is it necessarily true that $x'(0)=x'(1)$ and $y'(0)=y'(1)$?

I'm thinking of a simple example, $\alpha(s) = (\cos(2\pi s), \sin(2 \pi s)$), then this is a smooth closed curve with $\alpha'(s) = (- 2 \pi \sin (2 \pi s), 2 \pi \cos (2 \pi s))$ and $x'(0)=x'(1)$ and $y'(0)=y'(1)$, but I'm not sure how this has to be argued in general. Any insights appreciated.

I'm also imagining the curvature at $s=0$ and $s=1$ will be the same up to a sign.

IntegrateThis
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  • Note that you can alternatively write $\displaystyle \alpha'(0)=\lim{h\to 0^+} \frac{\alpha(t)-\alpha(0)}{h}.$ To obtain your expressions, write $$\alpha(t)-\alpha(0)=(x(t)-x(0))(1,0)+(y(t)-y(0))(0,1)$$ and apply the limits to each term separately. – Semiclassical Aug 30 '21 at 04:10
  • @Semiclassical right typo, fixed. – IntegrateThis Aug 30 '21 at 04:12
  • As for the second question, consider a curve which crosses itself (e.g., a Lissajous curve.) Now suppose the common point at $t=0,1$ is at one of the intersections. The points are certainly the same by assumptions, but is there any reason for the derivatives to be the same? – Semiclassical Aug 30 '21 at 04:19
  • @Semiclassical ok so the answer for the second question is no, but my first question, which admittedly was not phrased well, is what exactly is the convention for the definition for the derivatives at the endpoints where you do not have a full neighborhood around the point? – IntegrateThis Aug 30 '21 at 04:41
  • See the first part of https://math.stackexchange.com/a/126192/137524 for some discussion of your concern here. – Semiclassical Aug 30 '21 at 04:44

1 Answers1

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So to recap the discussions in the chat, the definition of a differentiable function on a closed interval is based on the one-sided limits or semi-derivatives as per the question https://math.stackexchange.com/a/126192/137524.

As for the second question, it is not necessarily the case that a closed smooth curve has derivative/velocity vectors at the endpoints that is equal at the endpoints.

IntegrateThis
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  • Looks right. One question which this may suggest: If a closed curve also has matching first derivatives at its endpoints, must the second derivatives also match (i.e., curvature)? I believe the answer is no, but I don't recall a simple example of such. – Semiclassical Aug 30 '21 at 05:03