Flow by Curvature

Question

In the case of plane curves the flow by curvature is, $$\frac{\partial F}{\partial u}=kN,$$ where $F:S^1\times [0, T)\rightarrow \mathbb R^2$ is a family of plane curves, $k$ is the curvature and $N$ is the inner unit normal..How to show the above curvature equation is equivalent to $$\frac{\partial x}{\partial t}=\frac{\partial^2 x}{\partial s^2}\ \textrm{e}\ \frac{\partial y}{\partial t}=\frac{\partial^2 y}{\partial s^2},$$ where $s$ is the arc length parameter and $F=(x, y)$?

Answer 1

The curvature vector of $F(\cdot, u)$ is $\vec{\kappa} = \frac{\partial^2 F}{\partial s^2}(\cdot,u)$, the unit normal is $N = \frac{\vec{\kappa}}{\|\vec{\kappa}\|}$ and the curvature is $k = \|\vec{\kappa}\|$. The right-hand side of the flow equation is just the curvature vector of the time $u$ curve, so by writing the equation out in Cartesian coordinates you get the result.