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Suppose a curve $S$ in 2-D is parameterized by $$S=\{(u(s),v(s)): s\in\mathbb{R}\},$$ where $u$ and $v$ are $C^{1,\alpha}$ for $\alpha \in (0,1)$ with $$u(s+2\pi)=u(s)+2\pi \qquad \text{and} \qquad v(s+2\pi)=v(s) \qquad \text{for all } s \in \mathbb{R}.$$

They claim that $$u'(s)^2 + v'(s)^2 \ne 0 \qquad \text{for all } s \in \mathbb{R}.$$

I think there must be a simple reason for it, but I don't see it. Can someone please help me on this?

I don't think the periodicity has anything with the claim.

dh16
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1 Answers1

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Here is what I am thinking:

If $u(s+2\pi) = u(s) + 2\pi$, then we know that $u(s)$ is not a constant function. Thus, the derivative of $u$ with respect to $s$ cannot be $0$ for all $s$.

We also do not know for certain what $v(s)$ is. From what we are given, it could be constant, or it could be periodic.

Notice that for $u'(s)^2 + v'(s)^2=0$ to be true, both $u'(s)$ and $v'(s)$ would need to be $0$. But we've already shown above that $u'(s) \ne 0$

WaveX
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