21
$\begingroup$

The curve-shortening flow transforms curves in the plane by moving each point perpendicularly to the curve at a speed proportional to the curvature at that point. It is usually defined for smooth curves (so you can determine the curvature) but according to a survey by White (Math. Int. 1989) it can be extended to rectifiable varifolds. In particular, according to this extension, it should be well-defined for triangles, which should instantly turn into smooth approximations of themselves and then evolve as normal, shrinking to a point. (This would also follow immediately from a positive answer to a previous question about continuity of the curve-shortening flow, assuming that the continuity is in a sufficiently general space of curves to include triangles and not just smooth curves.) Since the flow isn't affected by Euclidean transformations, this point must be a triangle center. Is it a known triangle center (one of the ones listed at the Encyclopedia of Triangle Centers)? If so, which one? If we could just calculate the flow accurately enough we could look it up in ETC but I don't know how to do the calculation.

If this turns out to be known, a reference would be helpful, so that it could be added to the Wikipedia article.

$\endgroup$
  • 1
    $\begingroup$ I think it is known (maybe by Bryant or someone else knowledgeable about exterior differential systems) that it is impossible to determine from a general initial convex curve where the singularity occurs without running the flow (i.e. there is no conservation law that tells you this). It's not clear to me what restricting to triangles gains you so this might not have a nice answer. $\endgroup$ – foliations Nov 2 '15 at 12:18
  • $\begingroup$ Incidentally, there's an example of a triangle undergoing curve-shortening in Cao, Frédéric (2003), Geometric Curve Evolution and Image Processing, Fig. 6.11 (upper right), p. 141. $\endgroup$ – David Eppstein Nov 10 '15 at 6:55
  • 1
    $\begingroup$ According to Theorem 11.1 in Joseph Lauer's paper, A new length estimate for curve shortening flow and low regularity initial data, the curve shortening flow is well defined and unique for all rectifiable curves in the plane. In particular, triangles do indeed flow to a unique point in their interior. $\endgroup$ – Mohammad Ghomi May 22 '17 at 4:29
6
$\begingroup$

I think there can be no expression for this "center of mass" that does not involve running the curve shortening flow. This was proved (along with a lot of other things) for general initial curves by Bryant and Griffiths in this long article (see the continuation of Example 1 in section 4.4 specifically the top of page 53) and I don't see how restricting to triangles helps.

The only thing you can know is the extinction time which is given by $T=A_0/2\pi$ where $A_0$ is the area of the triangle.

You can estimate where the extinction point is by using Huisken's monotonicity formula, but that is probably the best you can do in general.

$\endgroup$
  • 1
    $\begingroup$ Interesting, but I guess I need a little more explanation for why that would be an answer. It says that there is no function F of the curve points and their derivatives that would allow the extinction center of a general curve to be represented as an integral of F over the curve. But in the case of a triangle, the whole triangle can be represented by its three vertices, and the extinction center clearly is a function of those three points. How does not being expressible as an integral in the general curve case prevent a formula in terms of the vertices existing for triangles? $\endgroup$ – David Eppstein Nov 3 '15 at 16:58
  • $\begingroup$ My reasoning is that the curve shortening flow instantaneously turns the triangle into smooth closed curve and it seems highly unlikely that there are any computable criteria that distinguish closed curves that come from a triangle from those that do not. As such, the triangle case is as hard as the general case. This is certainly a heuristic at best and it would be fascinating if there really was a formula, but I am doubtful. $\endgroup$ – foliations Nov 3 '15 at 22:27
  • 1
    $\begingroup$ I guess the point of the Bryant–Griffiths result is that there is no formula in the form of an integral of purely local quantities. But typical formulas for triangle centers are not local; they usually express the coefficients for barycentric or trilinear coordinates as functions that combine information from all three vertices rather than as functions only of a single vertex. $\endgroup$ – David Eppstein Nov 3 '15 at 22:58
4
$\begingroup$

This seems to be addressed by Ben Chow and Dave Glickenstein in:

MR2281928 (2008b:58023) Reviewed 
Chow, Bennett(1-UCSD); Glickenstein, David(1-AZ)
Semidiscrete geometric flows of polygons. 
Amer. Math. Monthly 114 (2007), no. 4, 316–328. 

They show that that the point is the center of gravity of the triangle.

$\endgroup$
  • 5
    $\begingroup$ Glancing at that article, it appears they consider a type of discretized curve shortening flow (that, for instance, preserves polygons) and not the actual curve shortening flow. $\endgroup$ – foliations Nov 2 '15 at 12:10
  • 2
    $\begingroup$ There's also a smooth affine-invariant variation of the curve-shortening flow described by Angenent, Sapiro, and Tannenbaum (JAMS 1998), by making the normal speed be the cube root of curvature instead of the curvature itself. Affine invariance would cause it to converge to the centroid of a triangle as well. But it's the original curve-shortening flow that I'm interested in here. $\endgroup$ – David Eppstein Nov 3 '15 at 1:02
  • 1
    $\begingroup$ Incidentally, I implemented the Chow-Glickenstein discrete flow. You can see an example in this earlier MO question. $\endgroup$ – Joseph O'Rourke Nov 3 '15 at 14:22
2
$\begingroup$

From my understanding, the Brakke flow is not generally unique and so I'm not sure if you can define a unique flow for a polygon like a triangle, which may result in not being able to define a center in this way. Angenent has some work when the curve is in W^2,p, but polygons are not in this space. I'd be interested to know more but I have not been successful at finding a good reference for such small amount of smoothness (help would be appreciated). One could consider the level set formulation, but that may suffer from fattening (which is similar to the nonuniqueness issue). However, most of the literature focuses on nonuniqueness coming from singularities when one starts at a smooth curve, and these would probably not develop the same kind of singularity present in a polygon.

$\endgroup$
  • 1
    $\begingroup$ I was imagining that one obtained a unique solution that, essentially, instantaneously transitioned from a polygon to an almost-polygon with each corner rounded off to the (unique) smooth curve asymptotic to two lines that evolves homothetically, and with the relative scales of the three rounded-off corners determined by the property that the area loss in each corner should be proportional to the exterior angle. Does that not work? $\endgroup$ – David Eppstein Nov 5 '15 at 5:13
  • 3
    $\begingroup$ All triangles are convex so they cannot fatten (and hence have a unique development). $\endgroup$ – foliations Nov 5 '15 at 7:28
  • 1
    $\begingroup$ This paper by Ilmanen, Neves and Schulze is the state of the art arxiv.org/abs/1407.4756 . However, in the specific case of convex initial data you can just use the level set flow -- see for instance section 7.5 of Evans-Sprucks "Motion of level sets by mean curvature I," though you won't get the information about the expanders... $\endgroup$ – foliations Nov 5 '15 at 20:02
  • 1
    $\begingroup$ If you look at their first theorem, Ilmanen et al allow any sort of graph-like singularity (they call these non-regular planar network). However, it seems the flow immediately smooths out so that the only remaining singularities (if any) are triple junctions (they call these regular planar networks). $\endgroup$ – foliations Nov 7 '15 at 18:20
  • 2
    $\begingroup$ I see. I also just found a paper by Lauer on general Jordan curves link.springer.com/article/10.1007%2Fs00039-013-0248-1 $\endgroup$ – David Glickenstein Nov 9 '15 at 21:58
2
$\begingroup$

I formerly had a handwavy argument here showing that the vanishing point is always near (but not equal) the centroid, but have now replaced it by a more rigorous argument. This doesn't entirely answer the question (which center is the vanishing point), but should at least help eliminate many centers as definitely not being it, because most centers do not have the same property of always being near the centroid.

First, to eliminate the centroid itself: intuitively, the vanishing point shouldn't be the centroid by the idea presented in the answer by foliations: the centroid has a local formula and the vanishing point shouldn't have such a formula. (This should also be true of the Spieker center, the centroid of the triangle's perimeter.) But a little more directly: in a highly-obtuse isosceles triangle, all triangle centers lie on the altitude of the triangle, but the centroid is always exactly 1/3 of the way along the altitude from the base (and the Spieker center is almost on the base), while the vanishing point should be closer to the incenter, near the midpoint of the altitude.

Next, let's show that the vanishing point is near the centroid. The previous handwavy argument used a vague concept of "aspect ratio" (see comments); to make this more concrete let's use the isoperimetric ratio $\rho=L^2/A$ which is known to decreases monotonically through the evolution of a convex curve (Gage 1984).

Consider a smooth convex curve $C$ with diameter $D$, and let $w$ be the width perpendicular to the diameter; then $D=\Theta(L)$ and $A=\Theta(wD)$ so $w=\Theta(D/\rho)$. Let $xy$ be a chord of $C$ perpendicular to $D$ through the centroid. Then $xy$ must cross $D$ somewhere within its middle third, and the tangents to $C$ at $x$ and $y$ must form an angle with each other (and with the diameter segment) that is $O(1/\rho)$, else $xy$ would be longer than $w$. We consider all of the symbols $D$, $w$, $x$, and $y$ to vary continuously as the curve evolves, and use $w_0$ (etc.) to refer to their initial values. Note that, because $C$ remains confined to a rectangular box with width $w_0$, it will always be the case that $w=O(w_0)$.

At any point in the evolution of the curve, the same calculation shows that the angle between the two tangents is small, $O(w/D)$. It follows that the rates of area loss on the two sides of $xy$ differ from each other by at most a factor of $1+O(w/D)$, and we know that the total rate of area loss is constant, so the rate at which the area difference between the two sides changes is $O(w/D)$. It follows that the speed at which segment $xy$ can move is $O(1/D)$, because a faster movement would create too much area difference from one side to the other.

Now consider any time period within which the diameter decreases from $D_t$ to $D_t/2$. Surrounding the curve by pair of width-$w$ grim reaper curves (one in each direction), and using the avoidance principle for curve shortening (if a curve is surrounded by another curve, the two curves cannot cross) shows that this happens in time $O(w_tD_t)$. Within this time period, the centroid can only move a distance of $O(w_t)$: the speed of $xy$ controls its motion parallel to the diameter and the length of $xy$ controls its perpendicular motion. After repeating this argument $\Theta(\log\rho)$ times we will have $D\le w_0$, after which the vanishing point is bounded within the $O(w_0)$ remaining diameter of the curve. Therefore, the vanishing point of any smooth curve is within distance $O(D\frac{\log\rho}{\rho})$ of the centroid.

Triangles aren't smooth but immediately become smooth as soon as you start evolving them, so the same argument applies. I'm not sure whether the logarithmic factor in the bound above is necessary, or whether the vanishing point is always within the closer distance $O(D/\rho)$ of the centroid.

$\endgroup$
  • $\begingroup$ Could you please clarify: What is the aspect ratio of a triangle? $\endgroup$ – Joseph O'Rourke Nov 7 '15 at 0:46
  • 1
    $\begingroup$ You can define aspect ratio as the ratio between the lengths of the sides of the minimum enclosing rectangle, as the ratio between longest edge and the altitude to that edge, as the ratio between inradius and minimum enclosing radius, or as the ratio between inradius and circumradius. These numbers are different from each other but it doesn't make much difference: it's something that's $O(1)$ when all angles are bounded away from zero and $\omega(1)$ otherwise. $\endgroup$ – David Eppstein Nov 7 '15 at 3:38
  • 1
    $\begingroup$ Another way to estimate where the vanishing point is to use Huisken's monotonicity formula. Specifically, scale the triangle so the area is $2\pi$ (so flow disappears at time $T=1$). Now compute the integral of $\Phi_{\mathbf{x_0}}(\mathbf{x})=(4\pi)^{-1/2} e^{-|\mathbf{x}-\mathbf{x}_0|^2/4}$ over the triangle. Here $\mathbf{x}_0=(x_0,y_0)$. By Huisken monotonicity, if this number is less than $\sqrt{\frac{2\pi}{e}}$ then the flow cannot disappear at $\mathbf{x}_0$. $\endgroup$ – foliations Nov 7 '15 at 18:16
  • $\begingroup$ @foliations — the Huisken monotonicity estimate looks like it should be particularly accurate for near-circular curves (e.g. when the curve is exactly a circle then the estimate says only one point, the circle center, will work), but not very informative for convex curves with high isoperimetric ratio. The same is true for another simpler (known) estimate, that the vanishing point is in a circle concentric with the minimum containing circle with area = containing circle area – curve area. Do you have any thoughts on how those two estimates might compare to each other in their strength? $\endgroup$ – David Eppstein Nov 11 '15 at 7:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.