Let $\gamma:[a,b]\times \mathbb R\to\mathbb R$ be a flow of plane curves given by $$\dot\gamma=\frac{\kappa'}{\vert\gamma'\vert}JT+\frac{1}{2}\kappa^2T$$ where $T$ is the unit tangent vector and $\kappa$ the curvature of the curve $\gamma_t:[a,b]\to\mathbb R^2$, $\gamma_t(x):=\gamma(x,t)$. We use the sign convention $T'=\kappa vN$.

How can I show that $\gamma_0$ is an arc-length paramatrization if and only if $\gamma_t$ is an arc-length parametrization?

Not sure how to start, any ideas?

  • $\begingroup$ What is $J$?! Unclear $\endgroup$ – Behnam Esmayli May 21 '16 at 21:27
  • $\begingroup$ @Behnam $\mathbf{J}$ is a rotation matrix which maps the tangent vector $\mathbf{T}=(T_{x},T_{y})$ into $(-T_{y},T_{x})$. $\endgroup$ – Ng Chung Tak May 21 '16 at 22:15

We want to show $|\gamma'| = 1$ is preserved in time; so it suffices to show $$\frac 1 2\partial_t |\gamma'|^2 = \langle \gamma', \dot\gamma' \rangle = 0.$$ Substituting in the given expression for $\dot \gamma$, we get

$$ \langle \gamma ', \frac \partial {\partial s}\left(\frac{\kappa'}{\vert\gamma'\vert}JT+\frac{1}{2}\kappa^2T \right)\rangle. $$

Computing the derivative with the product rule along with $T' = \kappa |\gamma'|N$ (so $(JT)' = J(T') = - \kappa |\gamma'| T$) and taking only the tangential components (since $\langle \gamma', N \rangle = 0$), we get

$$\langle \gamma', \frac{\kappa'}{\vert\gamma'\vert} (- \kappa |\gamma'| T) + \kappa \kappa' T \rangle.$$

These two terms cancel out and thus we have the desired result.


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