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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 $t\not\in [a,b]$?


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.

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3 Answers 3

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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.

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    $\begingroup$ Note in the Wikipedia link that for smooth function $f(x)$, to have a flat section, then $f(x)$ must be non-analytical function, since no non-piecewise power series can become a constant value in a non-zero measure interval while not been itself constant in the whole domain, which is due the Identity Theorem $\endgroup$
    – Joako
    Commented Feb 28 at 19:42
  • $\begingroup$ 2D curves Sine curve, Cornu's spiral .. $\endgroup$
    – Narasimham
    Commented Feb 29 at 1:28
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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.

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  • $\begingroup$ With cut-off functions, you are meaning about something like this?: $$f(x) = \begin{cases} 0,\ x\leq -1 \\ 1,\ x \geq 1 \\ \dfrac{1}{4\left(1+\exp\left(\frac{4x}{x^2-1}\right)\right)},\ \text{otherwise}\end{cases}$$ $\endgroup$
    – Joako
    Commented Feb 28 at 20:04
  • $\begingroup$ Not entirely. The graph of this function could very well be seen as the cut-off on the left side of an interval $[1,b]$ for any $b>1$. To cut-off the entire interval you should "flip" it on the right side. It is a bit convoluted, but in this way you could obtain an explicit cut-off function. Beware however that such a cut-off does not approximate the characteristic function of the interval. For that you should add some $\varepsilon$ somewhere to make it precise. I myself prefer the convolution approach as I find more neat, although less explicit. $\endgroup$ Commented Feb 29 at 9:34
  • $\begingroup$ Let me see If I got you: then, by Cut-Off function you talk about smooth bump functions? as example $$g(x) = \begin{cases} 1,\ x=0\\ 0,\ |x|\geq 1\\ \dfrac{1}{1+\exp\left( \frac{1-2|x|}{x^2-|x|} \right)} ,\ \text{otherwise}\end{cases}$$ $\endgroup$
    – Joako
    Commented Feb 29 at 14:50
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    $\begingroup$ Not just a smooth bump function, but the convolution with a smooth bump function. Take for example the interval $I=[-1,1]$, and let $\chi_I$ be its characteristic function. If we call $g$ the smooth function you wrote, then the convolution $$f:=\chi_I*g$$ is what I call a cut-off function, as it is smooth and it "resembles" the characteristic function. $\endgroup$ Commented Feb 29 at 15:30
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    $\begingroup$ What you obtain in the limit $n\to\infty$ is indeed a cut-off for $[-\frac{1}{2},\frac{1}{2}]$. $\endgroup$ Commented Mar 1 at 9:11
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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,

circular arc, tangent line, circular arc

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).

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    $\begingroup$ "Smooth" normally means it must be infinitely differentiable; so having continuous first derivatives is not sufficient. $\endgroup$
    – kaya3
    Commented Feb 29 at 17:17
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    $\begingroup$ But OP's "formal question" only specified $f''$, so it's possible this meets their expectations. $\endgroup$
    – Teepeemm
    Commented Feb 29 at 21:34
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    $\begingroup$ @kaya3 : "Smooth" is one of those floppy words whose meaning is context dependent. If one means infinitely differentiable, one says so. If one's context only requires $C^1$ to be "smooth", one says so. See algebrademystified.wordpress.com/2017/08/20/… for a short discussion of the varied usages of "smooth". OP's formal question only requires that $f''$ is $0$ on a closed interval and is nonzero somewhere else, which my example demonstrates. (It's even $C^2$.) $\endgroup$ Commented Mar 1 at 3:21
  • $\begingroup$ @Teepeemm OP's question specified that the function be smooth and have zero curvature on a finite interval. The question uses the second derivative to specify the latter condition, not the former. $\endgroup$
    – kaya3
    Commented Mar 1 at 11:56
  • $\begingroup$ My "It's even $C^2$." is a typo. I had intended "It's even $C^1$." but didn't notice until too late to edit the comment. (If this is the worst thing I did that day, it was probably a pretty good day.) $\endgroup$ Commented Mar 1 at 17:50

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