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A parametrization of a 1-dimensional curve is called regular if its velocity is always positive. For example, the following parametrization:

$$x(t)=t^3, y(t)=t^6$$

is not regular because its velocity is 0 in $t=0$.

But, this same curve can be re-parametrized as:

$$x(t)=t, y(t)=t^2$$

and this second parametrization is regular because its velocity anywhere is at least 1.

So, my question is: given a non-regular parametrization of a curve, is there an algorithm to tell whether the curve has a regular parametrization, and find it if it exists?

Erel Segal-Halevi
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    What about natural parametrization? – Michael Galuza Jul 20 '15 at 06:03
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    If we perform a reparametrization of any curve with parameter $t$ to parameter $\tau = \tau(t)$ then $v_\tau = v_t \left|\frac{dt}{d\tau}\right|$. In your case $v_t = 3t^2\sqrt{1 + 4t^{6}}$ so we can for example cancel the first factor by picking $3t^2 = \frac{d\tau}{dt}\to \tau = t^3$. If $\tau = \int v_t dt$ is a valid parametriation then $v_\tau = 1$ everywhere. – Winther Jul 20 '15 at 06:46
  • @Winther what do you mean by "if $\tau$ is a valid parametrization"? How can I tell whether it is a valid parametrization? – Erel Segal-Halevi Jul 20 '15 at 08:26
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    I just added that since I don't know too much about this. I was thinking the integral might have singularities, it might have branches so that it can only describe parts of the curve etc. I don't know if this is a show-stopper. But $\tau(t)$ as described above seems to be closely related to what Michael mentioned above namely the so-called natural parametrisation (as mentioned here). – Winther Jul 20 '15 at 08:42

1 Answers1

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$\newcommand{\Reals}{\mathbf{R}}$Let $I$ be a non-empty open interval of real numbers, and $\gamma:I \to \Reals^{n}$ a continuously-differentiable path. (This hypothesis is arguably substantial, but appears not out of line with the OP's intent.)

At each point where $\gamma$ is regular, define the unit tangent field $$ T(t) = \frac{\gamma'(t)}{\|\gamma'(t)\|}. $$ By continuity, the zero set of $\gamma'$ is a closed subset of $I$; that is, the domain of $T$ is an open subset of $I$. Without loss of generality, we may assume the domain of $T$ is the complement of a discrete set. (Loosely, if there are intervals on which $\gamma' = 0$, "excise them and push their endpoints together").

Claim: The following are equivalent:

(i) $\gamma$ has a regular $C^{1}$ parametrization.

(ii) $T$ has a continuous extension to $I$.

(i) implies (ii): The unit tangent field is independent of monotone reparametrization in the sense that if $\Gamma(t) = \gamma(\tau(t))$ for some differentiable function $\tau$ with positive derivative, then the unit tangent field of $\Gamma$ at $t$ is the unit tangent field of $\gamma$ at $\tau(t)$. If $\gamma$ has a regular $C^{1}$ reparametrization $\Gamma$, then the unit tangent field of $\Gamma$ continuously extends the unit tangent field of $T$.

(ii) implies (i). If $T$ has a continuous extension to $I$ (which we continue to denote $T$), then $$ \Gamma(s) = \int_{0}^{s} T(t)\, dt $$ is a regular parametrization of $\gamma$.

For the curve in question, we have $$ \gamma(t) = (t^{3}, t^{6}),\qquad \gamma'(t) = (3t^{2}, 6t^{5}),\qquad T(t) = \frac{(3t^{2}, 6t^{5})}{\sqrt{9t^{4} + 36t^{10}}} = \frac{(1, 2t^{3})}{\sqrt{1 + 4t^{6}}}, $$ which has a continuous extension at $t = 0$ (given by the same formula).

For a cusp, say, we have $$ \gamma(t) = (t^{2}, t^{3}),\qquad \gamma'(t) = (2t, 3t^{2}),\qquad T(t) = \frac{(2t, 3t^{2})}{\sqrt{4t^{2} + 9t^{6}}} = \frac{t(2, 3t)}{|t|\sqrt{4 + 9t^{2}}}, $$ which does not extend continuously to $0$.

Andrew D. Hwang
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  • @andrew-d-hwang I don't think the demostration of (ii) implies (i) is correct, because that integral is not a reparametrization of $\gamma$. – P. W. Maunt Aug 15 '20 at 12:03
  • @P.W.Maunt Just came across your comment. The formula for $\Gamma(s)$ may differ from standard convention, but it looks to me that $s$ is an arc length parameter for $\gamma$, i.e., $\Gamma'(s) = T(s)$. (Alternatively, the chain rule gives $\frac{d}{dt} \Gamma(s) = T(s) s'(t) = \gamma'(t)$.) Does that not seem right? – Andrew D. Hwang Jan 09 '22 at 14:15