Supremum size of room contours had outdoors device round having constantly much less than unity curvature

I remain in the procedure of confirming that if a room contour (in $R^3$) has boundless size and also the curvature often tends in the direction of $0$ as the all-natural parameter $s$ often tends to infinity, the contour has to be boundless - i.e. not had in any kind of round of limited distance. This appears proper with ease, yet I have no warranty it is proper, unless I am missing out on something noticeable. One means to confirm my suspicion, I have actually reasoned, is to make use of a lemma that any kind of contour had outdoors device round with curvature constantly much less than one have to have a limited upper bound on its size (perhaps $2π$, yet maybe better for all I recognize).

Just how might one deal with confirming such an upper bound exists, or if it exists? It could additionally behave to recognize what the bound especially is, also. I've assumed it could be feasible to posture this as a variational trouble - making best use of size - and afterwards lowering it right into a less complex trouble, yet that seems hellishly made complex. Ideas?

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2019-12-02 02:47:26
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Here is something along these lines, originating from price quotes making use of primary calculus, yet it is much weak than the lemma you intend to confirm.

Allow $\gamma$ be a normal $C^2$ contour in $\mathbb{R}^3$ parametrized relative to arclength $s$ (I will certainly think for simpleness that $s$ begins at $0$). If the curvature $\|\gamma''(s)\|$ is constantly much less than $K$ and also $\gamma$ is had in a round of distance $R\leq\frac{1}{4K}$, after that the size of $\gamma$ is much less than $\frac{1}{K}(1-\sqrt{1-4KR})$.

To see this, write $\gamma(s)=\gamma(0)+s\gamma'(0)+\int_0^s(s-t)\gamma''(t)dt$ (acquired as in this Wikipedia article). Relocating $s\gamma'(0)$ to the left and also $\gamma(s)$ to the right and also using the triangular inequality, $$ \begin{align} s &\leq \|\gamma(s)-\gamma(0)\|+\|\int_0^s(s-t)\gamma''(t)dt\| \\ &\leq 2R +\int_0^s\|(s-t)\gamma''(t)\|dt \\ &\lt 2R +K\int_0^s(s-t)dt \\ &=2R+\frac{K}{2}s^2. \end{align} $$

For this inequality to constantly hold, $s$ has to continue to be smaller sized than the tiniest origin of $\frac{K}{2}x^2-x+2R$. Hence, $s\lt\frac{1}{K}(1-\sqrt{1-4KR})$ as asserted.

You can still use this to confirm the initial outcome as mentioned, yet with any luck a person comes with a more powerful outcome.

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2019-12-03 04:24:52
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