6
$\begingroup$

Let $\gamma: (a,b) \rightarrow \mathbb{R}^2$ be a smooth regular curve such that $\forall s,t \in (a,b)$ , $||\gamma(s)-\gamma(t)||$ is a non-negative real valued function which depends only on $|t-s|$.

Show that $\gamma(t)$ has speed and curvature both constant.


Here is my attempt: I somehow have to use the constraint given of the function $||\gamma(t)- \gamma(s)||$. Since this acts on $(a,b)^2$, I want to handle a nicer funcion defined as follows: $$f(h):= ||\gamma(s+h)-\gamma(s)||^2$$ for $s\in(a,b)$ fixed. Taking the derivative with respect to $h$, denoting with $\langle \cdot , \cdot \rangle$ the Euclidean inner product on $\mathbb{R}^2$

$f'(h)= \frac{d}{dh}\langle \gamma(s+h) - \gamma(s) , \gamma(s+h) - \gamma(s) \rangle = \frac{d}{dh} [\langle \gamma(s+h), \gamma(s+h) \rangle + \langle \gamma(s), \gamma(s) \rangle -2 \langle \gamma(s+h), \gamma(s) \rangle ] = 2[\langle \frac{d}{dh}\gamma(s+h), \gamma(s+h) \rangle - \langle \frac{d}{dh}\gamma(s+h), \gamma(s) \rangle] = 2[\langle \frac{d}{dh}\gamma(s+h), \gamma(s+h) - \gamma(s) \rangle ]$

So $f'(0)=0 \quad \forall s$

I think this is somehow useful, but I don't see how to continue from here.

$\endgroup$
3
$\begingroup$

For the speed it is sufficient to remark that $$|\gamma'(s)|=\lim_{h\to0+}{|\gamma(s+h)-\gamma(s)|\over h}\ .$$ Here the RHS is independent of $s$, by assumption on $\gamma$.

For the curvature we have to extract the value of $\kappa(s)$ from distance measurements. We thereby may assume that $|\gamma'(s)|\equiv1$.

Consider the following model situation: $s\mapsto\gamma(s)=\bigl(x(s),y(s)\bigr)$ is a smooth arclength parametrized curve with $$x(0)=y(0)=0,\quad \dot x(0)=1,\quad \dot y(0)=0,\quad \theta(0)=0\ ,$$ where $\theta(s):={\rm arg}\bigl(\dot x(s),\dot y(s)\bigr)$ is the argument of the tangent direction. Then one has $$\eqalign{\dot x&=\cos\theta,\quad\ddot x=-\sin\theta\cdot\dot\theta,\quad x^{...}=-\cos\theta\cdot\dot\theta^2-\sin\theta\cdot\ddot\theta,\cr\dot y&=\sin\theta,\quad \ddot y=\cos\theta\cdot\dot\theta\cr}$$ identically in $s$, and therefore $$\ddot x(0)=0,\quad x^{...}(0)=-\kappa^2,\qquad \ddot y(0)=\kappa\ ,$$ where $\kappa:=\dot\theta(0)$ is the curvature of $\gamma$ at $(0,0)$. Taylor's theorem then gives $$x(s)=s-{\kappa^2\over6}s^3+O(s^4),\quad y(s)={\kappa\over2}s^2+O(s^3)\qquad(s\to0)\ .$$ It follows that $$|\gamma(s)|^2=\left(s^2-{\kappa^2\over3}s^4+O(s^5)\right)+\left({\kappa^2\over4}s^4+O(s^5)\right)=s^2-{\kappa^2\over12}s^4+O(s^5)\qquad(s\to0)\ ,$$ so that $$\lim_{s\to0}{|\gamma(s)|^2-s^2\over s^4}=-{\kappa^2\over12}\ .$$ Back to our given special curve $\gamma$ this means that $$-{\kappa^2(s)\over12}=\lim_{h\to0}{\bigl|\gamma(s+h)-\gamma(s)\bigr|^2-h^2\over h^4}\ .$$ Here the RHS is independent of $s$, hence $s\mapsto\kappa^2(s)$ is constant. By continuity this implies that $s\mapsto\kappa(s)$ is constant.

$\endgroup$
  • $\begingroup$ Thank you, this helped a lot. What if $\gamma$ is not parametrized by arc-length? Does this line of reasoning still work with more tedious calculations, or do we have to try something else? $\endgroup$ – Siupa Nov 1 '18 at 17:54
  • 1
    $\begingroup$ Part (i) shows that $\gamma$ is parametrized by arc length, up to a constant factor. Assuming $|\gamma'(s)|\equiv1$ instead of $|\gamma'(s)|\equiv c$ is no essential restriction in what follows. $\endgroup$ – Christian Blatter Nov 1 '18 at 19:26
0
$\begingroup$

Your derivative is right but this is not what you should look at. Fro the problem you want $\parallel \gamma^\prime(s) \parallel = cst$. So you should calculate $$\frac{\langle \gamma(s+h)-\gamma(s), \gamma(s+h)-\gamma(s) \rangle}{h^2}$$ as $h \to 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.