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The Gage-Grayson-Hamilton curve-shortening flows along the normal to the curve:


           CurveShortening
           (Image from Andrejs Treibergs presentation PDF.)
The result is remarkable avoidance of self-intersection, regardless of how convoluted is the original curve, and ultimate convergence to a "round point." See, e.g., this YouTube Video.

My question is:

Q. Has this flow been studied along vectors at some fixed angle to the normal? E.g., at $90^\circ$ counterclockwise of the normal, tangent to the curve?

Presumably, for any angle $\alpha > 0$ counterclockwise of the normal, the curve might self-intersect during its evolution. But perhaps it still converges to a limit shape?

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  • $\begingroup$ In this generalization it seems that a circle would just "rotate". Would you consider this case also as having converged to a limit shape? $\endgroup$ – Menachem Nov 13 '16 at 1:05
  • $\begingroup$ @Menachem: Yes, exactly, the circle remains a circle; let's call that its limit shape. The question is: What happens to other shapes, both for $\alpha=90^\circ$, and for arbitrary fixed $\alpha$? $\endgroup$ – Joseph O'Rourke Nov 13 '16 at 1:34
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    $\begingroup$ FYI: A curve shortening flow toy by Anthony here. Also, the flow is along the curvature vector $\kappa n$ instead of the normal $n$. $\endgroup$ – Arctic Char Nov 13 '16 at 1:55
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    $\begingroup$ It has always been my intuition that movement in the direction of the tangent only affect the parametrization not the embedding, so projecting your speed vector to its normal part should not change the shape of the evolution. If I find the time I'll try to make it rigorous and post it as an answer. $\endgroup$ – Thomas Richard Nov 13 '16 at 12:39
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    $\begingroup$ If $\partial_t \gamma = -k n + v t$ with $\gamma : \mathbb{S}^1 \times (0, T) \to \mathbb{R}^2$, then letting $\phi : \mathbb{S}^1 \times (0, T) \to \mathbb{S}^1$ satisfy $\partial_t \phi = - v \circ \phi$ you find that $\bar{\gamma}(x, t) = \gamma(\phi(x, t), t)$ satisfies $\partial_t \bar{\gamma} = -\bar{k} \bar{n}$. $\endgroup$ – Paul Bryan Nov 13 '16 at 18:47

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