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Reparameterization and unit-length curve

Reparameterization

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

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

∀s∈(a~,b~).γ~(s)=γ(ϕ(s))\forall s\in (\tilde a, \tilde b). \tilde\gamma(s) = \gamma(\phi(s))
∀t∈(a,b).γ(t)=γ~(ϕ−1(t))=γ~(s)\forall t\in (a, b). \gamma(t) = \tilde\gamma(\phi^{-1}(t)) = \tilde\gamma(s)

THEN γ~\tilde\gamma and γ\gamma are reparameterizations of each other.

Example γ(t)=(cos⁡t,sin⁡t),γ~(s)=(sin⁡s,cos⁡s)\gamma(t) = (\cos t, \sin t), \tilde \gamma(s)= (\sin s, \cos s) are both reparameterizations of each other, parameterizing the same circle x2+y2=1x^2 + y^2 = 1.
Take ϕ(s)=π/2−s,ϕ−1(t)=π/2−t\phi(s) = \pi/2 - s, \phi^{-1}(t) = \pi/2 - t, γ(t)=γ(ϕ(s))=γ~(ϕ−1(t))\gamma(t) = \gamma(\phi(s)) = \tilde\gamma(\phi^{-1}(t)).

Claim 1

If γ1\gamma_1 is a reparameterization of γ0\gamma_0, γ2\gamma_2 is a reparameterization of γ1\gamma_1, then γ2\gamma_2 is a reparameterization of γ0\gamma_0.

proof. Take map s0s_0 s.t. γ1(s0(t))=γ0(t)\gamma_1(s_0(t)) = \gamma_0(t), take map s1s_1 s.t. γ2(s1(t))=γ1(t)\gamma_2(s_1(t)) = \gamma_1(t). Then, let s2=s1∘s0s_2 = s_1\circ s_0 so that γ2(s2(t))=γ2(s1(s0(t)))=γ1(s0(t))=γ0(t)\gamma_2(s_2(t)) = \gamma_2(s_1(s_0(t))) = \gamma_1(s_0(t)) = \gamma_0(t). Note that s2−1=s0−1∘s1−1s^{-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,θ)∈R×(−π2,pi2):r=sin⁡θtan⁡θ}\{(r, \theta)\in \mathbb R \times (-\frac{\pi}{2},\frac{pi}{2}) : r = \sin\theta\tan\theta\}

Claim the curve can be parameterized by

γ:(−1,1)→R2,γ(t)=(t2,t31−t2)\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

(x,y)=(rcos⁡θ,rsin⁡θ)=(sin⁡2θ,sin⁡2θtan⁡θ)=(sin⁡2θ,sin⁡2θsin⁡θcos⁡θ)=(sin⁡2θ,sin⁡3θ1−sin⁡2θ)\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⁡θt = \sin\theta, note that this function is bijective on (−1,1)→(−π2,pi2)(-1, 1)\rightarrow (-\frac{\pi}{2},\frac{pi}{2}) and both sin⁡θ\sin\theta and arcsin⁡θarc\sin\theta are smooth on the specified domain.

Example: Ordinary cusp

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

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

proof. Let γ\gamma be a parameterized curve and with reparameterization γ~\tilde\gamma and map ss s.t. γ~(s(t))=γ(t)\tilde\gamma(s(t)) = \gamma(t). Assume that for some t0t_0 s.t. γ′(t0)=0\gamma'(t_0) = 0, γ′′(t0),γ′′′(t0)\gamma''(t_0), \gamma'''(t_0) are linearly independent and non-zero.

dγ~dt=dγ~dsdsdt=dγdt=0\frac{d\tilde\gamma}{dt}= \frac{d\tilde\gamma}{ds}\frac{ds}{dt} = \frac{d\gamma}{dt} = 0

Since ss is smooth, ds/dt≠0  ⟹  dγ~ds=0ds/dt \neq 0\implies \frac{d\tilde\gamma}{ds} = 0

d2γ~dt2=ddt(dγ~dsdsdt)=d2γ~ds2dsdtdsdt+dγ~dsd2sdt2=d2γ~ds2(dsdt)2dγ~ds=0d3γ~dt3=ddt(d2γ~ds2(dsdt)2+dγ~dsd2sdt2)=d3γ~ds3(dsdt)3+2d2γ~ds2dsdtd2sdt2+d2γ~ds2dsdtd2sdt2+dγ~dsd3sdt3=d3γ~ds3(dsdt)3+3d2γ~ds2dsdtd2sdt2dγ~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.

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Regular point and regular curves

A point γ(t)\gamma(t) is a regular point if γ′(t)≠0\gamma'(t)\neq 0, otherwise γ(t)\gamma(t) is a singular point.
A parameterized curve γ\gamma is a regular curve if ∀t∈(a,b).γ′(t)≠0\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 γ:R→R2,γ(t)=(cos⁡2t,sin⁡2t)\gamma: \mathbb R\rightarrow \mathbb R^2, \gamma(t) = (\cos^2 t, \sin^2 t) is not regular

γ′(t)=(−2cos⁡tsin⁡t,2sin⁡tcos⁡t)=(−sin⁡(2t),sin⁡(2t))\gamma'(t) = (-2\cos t\sin t, 2\sin t\cos t) = (-\sin(2t), \sin(2t))
γ′(k2π)=(0,0),∀k∈Z\gamma'(\frac{k}{2}\pi) = (0, 0), \forall k\in \mathbb Z

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

γ′(t)=(1,sinh⁡t)≠0\gamma'(t) = (1, \sinh t)\neq 0

Properties of regular curves

Claim Any reparameterization of a regular curve is regular.

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

1=dϕdsdϕ−1dt1 = \frac{d\phi}{ds}\frac{d\phi^{-1} }{dt}

Hence dϕds≠1\frac{d\phi}{ds}\neq 1. So that

γ~′(s)=γ~′(ϕ(t))ϕ′(t)≠0,∀s\tilde\gamma'(s) = \tilde\gamma'(\phi(t))\phi'(t) \neq 0, \forall s

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

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

s′(t)=∥γ′(t)∥=x′(t)2+y′(t)2s'(t) = \|\gamma'(t)\| = \sqrt{x'(t)^2 + y'(t)^2}

Known that square root is smooth on (0,∞)(0, \infty), and we have that γ′(t)≠0\gamma'(t) \neq 0 so that x′(t)2+y′(t)2>0x'(t)^2 + y'(t)^2 > 0. Therefore, s′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 γ^\hat\gamma be a unit-speed reparametrization of γ\gamma. Since γ^\hat\gamma is unit-speed, i.e. ∀t.∥γ^′(t)∥=1\forall t. \|\hat\gamma'(t)\| = 1. It is regular, then γ\gamma is a reparametrization of γ^\hat\gamma, hence also regular.

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

γ′(t)=γ~(s(t))s′(t)\gamma'(t) = \tilde\gamma(s(t))s'(t)

Take arc-length of both sides,

∥γ~(s(t))s′(t)∥=∥γ~(s(t))∥∣s′(t)∣=∥γ~(s(t))s′(t)∥s′(t)=∥γ′(t)∥=s′(t)\| \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 ∥γ~(s(t))∥=1\|\tilde\gamma(s(t))\|= 1.

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

proof. note that

∥γ′(t)∥=∥γ~(u(t))u′(t)∥∥γ~′(s(t))∥∣s′(t)∣=∥γ~′(u(t))∥∣u′(t)∣∣s′(t)∣=∣u′(t)∣\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)=±s+cu(t) = \pm s + c