Can a smooth curve have a segment of straight line?

Question

Setting: we are given a smooth curve $\gamma: \mathbb{R} \rightarrow \mathbb{R}^n$

Informal Question: Is it possible that $\gamma$ is a straight line on $[a,b]$, but not a straight line on $[a,b]^c$?

Formal Question: It is possible that $\gamma''(t)=0$ for all $t\in [a,b]$, while $\gamma''(t)\neq 0$ for some $t\not\in [a,b]$?

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The motivation for me to ask this question is that the textbook we use in our geometry class discusses only smooth curves on bounded open intervals. While I know that the curvature $\gamma''$ can be zero on a point (for instance: $\gamma(t)=(t,\sin t)$ has zero curvature on $\{n\pi:n\in\mathbb{Z}\}$), I can not come up with an example of a smooth curve $\gamma$ such that $\gamma''$ is $0$ on some interval $(a,b)$.

I think such an example if exists will be interesting to see in GGB, but I failed to come up with one due to my inexperience. Thanks for any help.

Answer 1

Yes, such curves exist. A "famous" one is the graph of the function $$f(x)=\begin{cases} \exp\left(\frac{-1}{1-x^2}\right) & -1 \lt x \lt 1\\0&\text{else}\end{cases}. $$ This example is important in differential geometry, in the context of partitions of unity. See https://en.wikipedia.org/wiki/Bump_function.

Answer 2

Using Friedrichs mollifiers you are able to construct smooth functions which are identically one, hence with second derivative identically $0$, in a neighborhood of any compact set. One can obtain such smooth functions, also called "cut-off functions", via convolution with a bump function. I suggest the wikipedia page https://en.wikipedia.org/wiki/Mollifier for an overview. Roughly, let $\varphi:\mathbb{R}\to\mathbb{R}$ be defined as $$\varphi(x):=\begin{cases}\frac{e^{\frac{-1}{1-x^2}}}{C}\quad\ |x|\le 1\\ 0\quad \qquad |x|>1\end{cases},$$ where $C$ is a suitable normalizing factor. For each $\varepsilon>0$ define $\varphi_\varepsilon(x):=\frac{1}{\varepsilon}\varphi(\frac{x}{\varepsilon})$.

Now let $I=[a,b]$. Denote with $\chi_I$ the characteristic function of $I$ and with $"*"$ the convolution product between functions. The function $$\chi_{\varepsilon}(x):=\chi_I*\varphi_\varepsilon(x)$$ is identically "1" on $[a+\varepsilon,b-\varepsilon]$. The procedure can be done similarly for compact sets in $\mathbb{R}^n$, provided one defines $\varphi(x)=\varphi(|x|)$, i.e. as a function of the norm of the argument, making it radial (one obtains a bell). I hope this answers the question.

Answer 3

You should be able to come up with examples that are circular arcs and line segments (that are tangent to the arcs where they are met). For instance,

This is the part of the circle centered at $(-1/2,1/10)$ with radius $1$ between angles $\pi$ and $3\pi/2$, the horizontal line segment from $(-1/2,1/10)$ to $(1/2,1/10)$, then the part of the circle centered at $(1/2,1/5)$ of radius $1/10$ between the angles $3\pi/2$ and $11\pi/6$.

Note: There is nothing special about the line segment being horizontal -- that's just easy for me to find an example off the top of my head. As long as the line segment(s) are tangent to the circles at the points where the lines meet their arcs, the result is continuous and has continuous first derivatives, i.e., is $C^1$. The second (and higher) derivatives are not continuous at the points where these arcs and segments meet. Also, a detail of the example is that the graphed function is continuous on $[-3/2,1/2+\sqrt{3}/20]$, but the derivative is only continuous on $(-3/2,1/2+\sqrt{3}/20]$ (being undefined where the tangent is vertical).