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This problem has been bothering me for a few weeks. Any ideas how to attack? Has this been solved/proved unsolvable/etc? I couldn't really find much online.

Given: Closed curve $C$ in $\mathbb{R}^n$, parametrized as $x = \{x_1(t), x_2(t) \dots x_n(t)\}$. Its curvature then is the derivative of the tangent vector with respect to the arc length element: $k = |\frac{\mathrm{d}{\bf T}}{\mathrm{d}s}|$. $S$ is the length of $C$ (found by $S = \int_C |\mathrm{d}s|$).

Question: Find the curve $C$ that minimizes the integral: $$ \frac{1}{S}\int_Ck\mathrm{d}s $$

Speculation: Essentially this means finding the curve with the lowest average curvature (because integrating curvature along length, then dividing by length). To me this sounds like a calculus of variations problem. But there is a norm of a derivative under the integral sign. And $S$ is also an integral.

Okay, let's write it out (denote average curvature $\hat{k}$): $$ \hat{k} = \frac{\int_C\left|\frac{\mathrm{d}{\bf T}}{\mathrm{d}s}\right|\mathrm{d}s}{\int_C|\mathrm{d}s| } $$

So I'm trying to minimize $\hat{k}$. But I'm not sure how to attack this fraction of integrals.

I'm not necessarily looking for a solution, just ideas how to approach this problem.

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  • $\begingroup$ This looks like it would admit some sort of Gauss–Bonnet formulation... $\endgroup$ – Chappers Jun 22 '15 at 16:22
  • $\begingroup$ Hang on, do you mean the signed or the unsigned curvature? $\endgroup$ – Chappers Jun 22 '15 at 16:50
  • $\begingroup$ In the title you ask for "maximum average curvature" and in the body of your post for the minimum. How so? Perhaps editing your title would clarify. Cheers! $\endgroup$ – Robert Lewis Jun 22 '15 at 16:52
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    $\begingroup$ Unsigned curvature, hence the norm. @RobertLewis Cheers, edited. $\endgroup$ – kbau Jun 23 '15 at 7:37
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    $\begingroup$ For any $\epsilon>0$ there is a curve with average curvature $\epsilon$ and finite area. So the infimum is $0$ and there is no solution. $\endgroup$ – Yves Daoust Jun 23 '15 at 7:59
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Consider a circle of radius $r$. Its average curvature will clearly be $1/r$ (up to a scalar multiple, which I believe is "1" in your formulation).

Now instead, consider

$$ c(t) = \frac{1}{k}(\cos k(2\pi t) , \sin k(2 \pi t ) ) $$ which traverses, $k$ times, a circle of radius $1/k$. Its average curvature will be $k$. Since $k$ can be arbitrary, there is no curve with max average curvature.

You might complain that this curve is selfintersecting (in a big way), but making an arbitrarily small random perturbation will resolve this; as an alternative, stretch it so that it becomes helical, and make the last half-turn be a little wider and join the top of the helix to the bottom. This'll only slightly reduce the average curvature, so the argument still works.

The question of MINIMUM average curvature is more interesting (to me). Surely the answer is "a circle" if there's any answer at all.

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  • $\begingroup$ Ok, fair enough on the max/min argument. I'll rephrase my question appropriately. If looking for min average curvature, then why is a circle "surely" the answer? I don't have a counter example, but I'm not entirely convinced either. Just wondering where you're coming from. $\endgroup$ – kbau Jun 22 '15 at 16:18
  • $\begingroup$ On the subject of minimum average energy curves, see elastic a curves, which are the shapes which minimize stored energy in a bent material. $\endgroup$ – Alex S Jun 22 '15 at 16:18
  • $\begingroup$ Sorry...I was just being whimsical. I suppose that it's possible that a trefoil knot beats it...but it's hard for me to imagine that. I guess one kind of argument (for 3-space) is that if there's a line around which the curve has winding number 2, then you can take this as the $z$-axis of a coordinate system, and in the projection to the xy-plane, there must be a crossing. Now take the "square root" of the curve, i.e., take the longitude coord and divide by 2 in a continuous way, and use the "crossing" to glue up the pieces. That should reduce curvature...I hope. :) $\endgroup$ – John Hughes Jun 22 '15 at 17:01
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It suffices to restrict to plane curves, but see the last section for a concrete example.

I'm assuming that you are talking about the signed curvature in this answer.

In fact, for any closed curve $C$, we have $$ \int_C k \, ds = 2\pi N $$ for some integer $N$, this is called the total curvature or the turning number of $C$. I offer two ways of proving this: one is a simple application of the Gauss–Bonnet theorem, $$ \int_C k \, ds = \int_{\partial M} k_g \, ds = 2\pi \chi(M)-\int_{M} K \, dA = 2\pi \chi(M), $$ where $M$ is the set of which $C$ is the boundary, and $K$, the curvature of $M$, is obviously zero for a plane. $\chi(M)$, of course, is discrete, taking integer values, and since the integral is continuous for curves with nonsingular curvature (and this means that you can't introduce another loop, for example, as it would have to grow from a cusp), it is constant on a class of such curves.


Now, a second, more constructive proof is to consider the explicit form of the curvature: we have (with $'=d/dt$) $$ k=\frac{x'y''-y'x''}{(x'^2+y'^2)^{3/2}}. $$ Therefore $$ k \, ds = \frac{x'y''-y'x''}{x'^2+y'^2} \, dt. $$ Now, there are two ways to go about noticing something about this: Green's theorem or complex analysis. I'll do it by complex analysis because the theorems are probably more familiar. Notice that (and you can work back to this: I did): $$ \frac{1}{i} \frac{x''+iy''}{x'+iy'} = \frac{(y''-ix'')(x'-iy')}{(x'+iy')(x'-iy')} = \frac{(x'y''-y'x'')-i(x'x''+y'y'')}{x'^2+y'^2} $$ Therefore we can write $$ \int_C k \, ds = \Re\left( \frac{1}{i} \int_C\frac{x''+iy''}{x'+iy'} \, dt \right) $$ Ah, but this is the real part of the winding number of $x'+iy'$ about zero: i.e., how many rotations the tangent vector goes through. Well, that has to be a whole number, or the curve isn't smooth. (One criticism that might be levelled here is that we don't know that $x'+iy'$ extends to an analytic function on the interior. Fair enough, but I expect the Green's theorem argument will work anyway. The result is certainly true (as the Gauss–Bonnet argument shows): I'm just trying to make it believable with the minimum framework. See also Total curvature and Turning number on Wikipedia.)


Right, so now that we know the total curvature $\int_C k \, ds$ is an integer, we can answer the question. The homotopy given by $\gamma:[0,1) \to \{ \text{curves} \}$, $\gamma( \lambda) = (1-\lambda)C$, where $ (1-\lambda)C = [ t \mapsto ((1-\lambda)x(t),(1-\lambda)y(t)) ]$, preserves smoothness of the curve, and is continuous, so it must preserve the total curvature (continuous map to a discrete space is locally constant). But it is intuitively obvious (and easy to verify by calculation) that $S(C)$, the length of $C$, undergoes a linear scaling under the action of $\gamma$: $$ S((1-\lambda)C) = (1-\lambda)C. $$ (This should not be surprising: think of perimeters of circles.) It follows that $A(C) := S(C)^{-1} \int_C k \, ds $ has neither maximum nor minimum, even among curves of identical shape (consider $\gamma$, and the similar homotopy that grows $C$ by mapping $\lambda \mapsto (1-\lambda)^{-1}C$).


Now, if you want to talk about the absolute value of the curvature, the result that there is no minimum will still hold, because the circle of radius $r$ has constant curvature $1/r$, so the total curvature is $2\pi$, and the length is $2\pi r$, so $A(C(r)) = 1/r$, which can be made as large or as small as you like.

(Oh, and by the way, you can check that the result about turning number doesn't hold for non-planar curves, just by considering something like $(\cos{t},\sin{t},f(t))$ for some twice-differentiable $f$ with $f(x+2\pi)=f(x)$. This tells you that there's something topologically specific to 2D going on.)

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  • $\begingroup$ Thanks. After a quick read that does look sensible, albeit it's signed curvature (only difference between signed and unsigned is whether we take the norm of the tangent vector derivative, right?). Anyway, I'll have a more thorough read in the next few days just to convince myself completeley. And also maybe try to work out the Green's theorem version of your proof. $\endgroup$ – kbau Jun 23 '15 at 7:44
  • $\begingroup$ This all makes sense...when you have a plane curve. But the original question was about curves in $\mathbb R^n$, for which signed curvature isn't so well-defined. $\endgroup$ – John Hughes Jun 23 '15 at 16:58
  • $\begingroup$ @JohnHughes It shouldn't matter for the result, since $\mathbb{R}^2$ is a submanifold of $\mathbb{R}^n$. But I agree that I should say I'm restricting to plane curves; I'll edit the post. $\endgroup$ – Chappers Jun 23 '15 at 17:03
  • $\begingroup$ "Signed curvature" (for your arguments to work) needs to be relative to the boundary of some surface. But the equator of the sphere is the (oriented) boundary of both the top hemisphere and the (reverse-orientation) lower hemisphere; relative to each of these, its signed curvature is zero, but relative to the equatorial disk, it's clearly nonzero. When you look at a latitude line away from the equation, things get even worse. $\endgroup$ – John Hughes Jun 23 '15 at 17:08

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