Let $X(s) = (x(s),y(s))$ be a regular and smooth curve. Define $\tau(s) = \langle X(s), T(s)\rangle$ and let $X(s)$ be such that $v = \langle X(s), N(s) \rangle = \kappa(s)$, where $\kappa(s)$ is the curvature. Consider the system of differential equations: $$\tau'(s) = 1+(v(s))^2 \\ v'(s) = -\tau(s)v(s)$$

Now that I have provided some context (and if more is needed I'll happily provide it), can anyone make sense of the outlined part? I'm not really getting what that funcion really is (I guess I'm too used to seing functions in the usual $y = f(x)$ way).

Proof. $\,$Here $k=\nu$ so the system for $\tau$ and $\nu$ becomes \begin{cases} \tau'=1+\nu^2,\\ \nu'=-\tau\nu. \end{cases} The phase portrait appears in Figure 2-4(a). There are no fixed points, $\tau$ is increasing and $\nu$ never changes sign. The trivial solution $\tau=s$, $\nu=0$, corresponds to $X$ being a line through the origin, making the expansion vacuous. To find the other trajectories, ${{\color{red}{\boldsymbol{[}}}\kern{-3pt}{\color{red}{\boldsymbol{[}}}}\kern{-3pt}{\color{red}{\boldsymbol{[}}}\!$ note that the function $\nu^2e^{\tau^2+\nu^2}=k^2e^{r^2}$ is constant $\!{{\color{red}{\boldsymbol{]}}}\kern{-3pt}{\color{red}{\boldsymbol{]}}}}\kern{-3pt}{\color{red}{\boldsymbol{]}}}$. By symmetry, it suffices to look at trajectories in the upper half-plane $(\nu\gt0)$. Since $\tau'\ge1$, it follows that $\lim_{s\to\pm\infty}\tau=\pm\infty$ and $\lim_{s\to\pm\infty}\nu=0$. Therefore, the corresponding curve $X$ is convex with $k\to0$ on each end.
$\hspace{0.4cm}$ To get a better description of the curves, we rewrite $k=\nu$ as an ODE for $y$ as a function of $x$: $$y''(x)=(1+y'(x)^2)(y(x)-xy'(x)).$$

For anyone wondering, this is where I got it from,


You need to think of it as $$ s\mapsto v(s)^2e^{\tau(s)^2+v(s)^2}. $$ Since $$ \begin{align}(v^2e^{\tau^2+v^2})' &=2vv'e^{\tau^2+v^2}+v^2e^{\tau^2+v^2}(2\tau\tau'+2vv')\\ &=2v(-\tau v)e^{\tau^2+v^2}+v^2e^{\tau^2+v^2}(2\tau(1+v^2)+2v(-\tau v))\\ &=e^{\tau^2+v^2}(-2\tau v^2+2\tau v^2+2\tau v^4-2\tau v^4)=0, \end{align} $$ the function $v^2e^{\tau^2+v^2}$ (of time) is constant.

  • $\begingroup$ Perfect. Thanks! $\endgroup$ – Matheus Andrade Apr 25 '18 at 22:58

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