Minimum time required by a curve to reenter a closed ball with radius equal to the reciprocal of its maximum curvature

Question

I have a continuously differentiable curve $r : \mathbb{R} \to \mathbb{R}^{n}$, $n \ge 2$, with the properties \begin{align} \lvert r'(t) \rvert = 1 \text{ for all } t \in \mathbb{R}, \\ \lvert r'(t) - r'(s) \rvert \le \kappa \lvert t - s \rvert \text{ for all } t,s \in \mathbb{R},\\ \lvert r(0) \rvert = 1 / \kappa, \\ \langle r(0), r'(0) \rangle > 0, \end{align} where $\kappa > 0$.

For all curves with the above properties, I'm looking for a lower bound $t_{0} > 0$ such that $t > 0$ and $\lvert r(t) \rvert = 1 / \kappa$ implies $t > t_{0}$.

Intuitively, the curve that achieves $\lvert r(t) \rvert = 1/\kappa$ in minimum time for given $r(0)$ and $r'(0)$ is a circular arc with radius $1/\kappa$ contained in the plane spanned by $r(0)$ and $r'(0)$ (or any plane containing $r(0)$ if they are parallel). Over all such circular arcs, I believe we would then have $t_{0} = \pi / \kappa$.

I would be grateful for any reference I could cite for such a lower bound on $t$.

Answer 1

Yes, the answer is $\pi/\kappa$.

You can get it as a corollary of the bow lemma (following the same argument as in the hints to 3.25 here)

Connect the origin to $r(0)$ by a smooth convex plane curve $\gamma$ so that the product $\gamma*r$ is piecewise smooth and $C^1$-smooth. By the bow lemma, you get that $|r(t)|\ge |\tilde r(t)|$, where $\tilde r$ is a plane curve with constant curvature $\kappa$ that bends toward the origin and such that $|r(0)|=|\tilde r(0)|$ and $|r(0)|'=|\tilde r(0)|'$. It remains to check that $|\tilde r(\pi/\kappa)|>\tfrac1\kappa$.

Alternatively, you can apply Reshetnayk's majorization theorem, but you will see more technical problems on the way.