Reparameterization and unit-length curve

Reparameterization

A curve can have different parameterizations (even infinite ways).

For parameterized curves \(\gamma: (a, b)\rightarrow\mathbb R^n\) and \(\tilde\gamma: (\tilde a, \tilde b)\rightarrow\mathbb R^n\).
IF \(\exists \phi: (a, b)\rightarrow (\tilde a, \tilde b)\) and its inverse \(\phi^{-1}\) (since bijective) and \(\phi, \phi^{-1}\) are both smooth and

\[\forall s\in (\tilde a, \tilde b). \tilde\gamma(s) = \gamma(\phi(s))\]

\[\forall t\in (a, b). \gamma(t) = \tilde\gamma(\phi^{-1}(t)) = \tilde\gamma(s)\]

THEN \(\tilde\gamma\) and \(\gamma\) are reparameterizations of each other.

Example \(\gamma(t) = (\cos t, \sin t), \tilde \gamma(s)= (\sin s, \cos s)\) are both reparameterizations of each other, parameterizing the same circle \(x^2 + y^2 = 1\).
Take \(\phi(s) = \pi/2 - s, \phi^{-1}(t) = \pi/2 - t\), \(\gamma(t) = \gamma(\phi(s)) = \tilde\gamma(\phi^{-1}(t))\).

Claim 1

If \(\gamma_1\) is a reparameterization of \(\gamma_0\), \(\gamma_2\) is a reparameterization of \(\gamma_1\), then \(\gamma_2\) is a reparameterization of \(\gamma_0\).

proof. Take map \(s_0\) s.t. \(\gamma_1(s_0(t)) = \gamma_0(t)\), take map \(s_1\) s.t. \(\gamma_2(s_1(t)) = \gamma_1(t)\). Then, let \(s_2 = s_1\circ s_0\) so that \(\gamma_2(s_2(t)) = \gamma_2(s_1(s_0(t))) = \gamma_1(s_0(t)) = \gamma_0(t)\). Note that \(s^{-1}_2 = s_0^{-1}\circ s_1^{-1}\) as a composition of invertible functions, is also invertible; and composition of smooth functions is also smooth.

Example: Cissoid of Diocles

The cissoid of Diocles (see below) is the curve

\[\{(r, \theta)\in \mathbb R \times (-\frac{\pi}{2},\frac{pi}{2}) : r = \sin\theta\tan\theta\}\]

Claim the curve can be parameterized by

\[\gamma: (-1, 1)\rightarrow\mathbb R^2, \gamma(t) = (t^2, \frac{t^3}{\sqrt{1-t^2} })\]

proof. Convert the polar coordinates to Cartesian coordinates

\[\begin{align*} (x, y) &= (r\cos\theta, r \sin\theta) \\ &= (\sin^2\theta, \sin^2\theta\tan\theta) \\ &= (\sin^2\theta, \sin^2 \theta \frac{\sin \theta}{\cos\theta})\\ &= (\sin^2\theta, \frac{\sin^3 \theta}{\sqrt{1-\sin^2 \theta} })\\ \end{align*}\]

Therefore, let \(t = \sin\theta\), note that this function is bijective on \((-1, 1)\rightarrow (-\frac{\pi}{2},\frac{pi}{2})\) and both \(\sin\theta\) and \(arc\sin\theta\) are smooth on the specified domain.

Example: Ordinary cusp

A point \(\mathbf p\) of \(\gamma\), corresponding to a parameter value \(t\), is an ordinary cusp if the curve is singular at the point and \(\gamma''(t)\) and \(\gamma'''(t)\) are linearly indepedent and non-zero.

Claim If \(\gamma\) has an ordinary cusp at a point \(\mathbf p\), so does any reparam of it.

proof. Let \(\gamma\) be a parameterized curve and with reparameterization \(\tilde\gamma\) and map \(s\) s.t. \(\tilde\gamma(s(t)) = \gamma(t)\). Assume that for some \(t_0\) s.t. \(\gamma'(t_0) = 0\), \(\gamma''(t_0), \gamma'''(t_0)\) are linearly independent and non-zero.

\[\frac{d\tilde\gamma}{dt}= \frac{d\tilde\gamma}{ds}\frac{ds}{dt} = \frac{d\gamma}{dt} = 0\]

Since \(s\) is smooth, \(ds/dt \neq 0\implies \frac{d\tilde\gamma}{ds} = 0\)

\[\begin{align*} \frac{d^2\tilde\gamma}{dt^2} &= \frac{d}{dt}(\frac{d\tilde\gamma}{ds}\frac{ds}{dt})\\ &= \frac{d^2\tilde\gamma}{ds^2}\frac{ds}{dt}\frac{ds}{dt} +\frac{d\tilde\gamma}{ds}\frac{d^2s}{dt^2}\\ &= \frac{d^2\tilde\gamma}{ds^2}(\frac{ds}{dt})^2&\frac{d\tilde\gamma}{ds}=0\\ \frac{d^3\tilde\gamma}{dt^3} &= \frac{d}{dt}(\frac{d^2\tilde\gamma}{ds^2}(\frac{ds}{dt})^2+\frac{d\tilde\gamma}{ds}\frac{d^2s}{dt^2})\\ &= \frac{d^3\tilde\gamma}{ds^3}(\frac{ds}{dt})^3 + 2\frac{d^2\tilde\gamma}{ds^2}\frac{ds}{dt}\frac{d^2s}{dt^2} + \frac{d^2\tilde\gamma}{ds^2}\frac{ds}{dt}\frac{d^2s}{dt^2} + \frac{d\tilde\gamma}{ds}\frac{d^3s}{dt^3}\\ &= \frac{d^3\tilde\gamma}{ds^3}(\frac{ds}{dt})^3 + 3\frac{d^2\tilde\gamma}{ds^2}\frac{ds}{dt}\frac{d^2s}{dt^2}&\frac{d\tilde\gamma}{ds}=0 \end{align*}\]

Obviously, the two vectors aren't multiple of each other.

Source code

import numpy as np
import matplotlib.pyplot as plt

t = np.arange(-0.999, 1., 0.001)
plt.plot(t*t, t*t*t / (1 - t*t)**0.5)
plt.savefig("../assets/reparam.jpg")

Regular point and regular curves

A point \(\gamma(t)\) is a regular point if \(\gamma'(t)\neq 0\), otherwise \(\gamma(t)\) is a singular point.
A parameterized curve \(\gamma\) is a regular curve if \(\forall t \in (a, b). \gamma'(t)\neq 0\).
Note that for the same level curve, it might be parameterized into both regular curve and non-regular curve.

Example of Regular Curves

Claim the circle with parameterization \(\gamma: \mathbb R\rightarrow \mathbb R^2, \gamma(t) = (\cos^2 t, \sin^2 t)\) is not regular

\[\gamma'(t) = (-2\cos t\sin t, 2\sin t\cos t) = (-\sin(2t), \sin(2t))\]

\[\gamma'(\frac{k}{2}\pi) = (0, 0), \forall k\in \mathbb Z\]

Claim \(\gamma(t) = (t, \cosh t)\) is regular

\[\gamma'(t) = (1, \sinh t)\neq 0\]

Properties of regular curves

Claim Any reparameterization of a regular curve is regular.

proof. Let \(\gamma(t)\) be a regular curve and \(\tilde\gamma(s)\) be reparameterization with \(t = \phi(s) = \phi(\phi^{-1}(t))\). Differentiating the equation will give

\[1 = \frac{d\phi}{ds}\frac{d\phi^{-1} }{dt}\]

Hence \(\frac{d\phi}{ds}\neq 1\). So that

\[\tilde\gamma'(s) = \tilde\gamma'(\phi(t))\phi'(t) \neq 0, \forall s\]

Claim If \(\gamma(t)\) is regular, then its arc-length function \(s\) is smooth.

proof. Let \(\gamma(t) = (x(t), y(t))\), note that we already have

\[s'(t) = \|\gamma'(t)\| = \sqrt{x'(t)^2 + y'(t)^2}\]

Known that square root is smooth on \((0, \infty)\), and we have that \(\gamma'(t) \neq 0\) so that \(x'(t)^2 + y'(t)^2 > 0\). Therefore, \(s'\) is also smooth.

Unit-speed Reparameterization

Claim A parametrized curve has a unit-speed reparametrization if and only if it is regular.

proof.
\(\Rightarrow\) Let \(\hat\gamma\) be a unit-speed reparametrization of \(\gamma\). Since \(\hat\gamma\) is unit-speed, i.e. \(\forall t. \|\hat\gamma'(t)\| = 1\). It is regular, then \(\gamma\) is a reparametrization of \(\hat\gamma\), hence also regular.

\(\Leftarrow\) Let \(\gamma\) be a regular curve, then it has a smooth arc-length function \(s\). By inverse function theorem, \(s\) is injective with an open interval image \(s: (a,b)\rightarrow (c, d)\). Then, we take some \(\tilde \gamma\) s.t. \(\gamma(t) = \tilde \gamma(s(t))\). Differentiate both sides and take the arc-length, we have that

\[\gamma'(t) = \tilde\gamma(s(t))s'(t)\]

Take arc-length of both sides,

\[\| \tilde\gamma(s(t))s'(t)\| = \|\tilde\gamma(s(t))\||s'(t)| = \| \tilde\gamma(s(t))s'(t)\|s'(t) = \|\gamma'(t)\| = s'(t)\]

so that we can conclude that \(\|\tilde\gamma(s(t))\|= 1\).

Corollary A parametrized curve \(\gamma\) has a unit-speed reparameterization IFF the reparametrization map \(u\) follows that \(u(t) = \pm s + c\), where \(c\) is a constant.

proof. note that

\[\begin{align*} \|\gamma'(t)\| &= \|\tilde\gamma(u(t))u'(t)\|\\ \|\tilde\gamma'(s(t))\||s'(t)| &= \|\tilde\gamma'(u(t))\||u'(t)|\\ |s'(t)| &= |u'(t)| \end{align*}\]

Thus, it implies that \(u(t) = \pm s + c\)