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The curve-shortening flow is $$ \frac{\partial C}{\partial t} = \kappa n $$ where $\kappa$ is the curvature, and $n$ is the unit normal vector. For a smooth Jordan curve $C\subset\mathbb R^2$ (closed curve without crossings), this flow has many nice properties, e.g. the only singularity happens when the curve collapses to a point.

I am curious to learn whether the following "one-sided version": $$ \frac{\partial C}{\partial t} = f(\kappa) n $$ has similar nice properties. Here, $f:\mathbb R\to \mathbb R_{\ge 0}$ is a $C^\infty$ function satisfying $f(x) =0, \forall x\le 0$ and $f(x)=x, \forall x\ge 1$ and $f'(x)>0, \forall x> 0$.

Precise question: Let $C\subset \mathbb R^2$ be a non-compact simple curve which is periodic in the sense that it is invariant under the translation $(x,y)\mapsto (x+1,y)$. Is the above "one-sided curve-shortening flow" applied to $C$ defined for all times? and does it converge to a horizontal line in $\mathbb R^2$?

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    $\begingroup$ Nice question. My guess is that simplicity (no crossings)---a feature of curve shortening---is not preserved. Consider a U-shape. The inside of the U has negative $\kappa$ and will remain fixed. Meanwhile the bottom of the U has positive $\kappa$ and will collide with the fixed portion. $\endgroup$
    – Joseph O'Rourke
    Commented Jan 25, 2023 at 23:54
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    $\begingroup$ Your precise question is clearly not true if you consider the curve given by the graph of $y=sin(\pi x)$ as, for instance, the curvature with respect to the upward normal is negative on $[0,1]$ so the flow won't move this part. (It's a bit unclear what conventions you are using for the sign of the curvature -- for instance with the usual meanings of curvature and normal, the flow you write down is backwards parabolic when it is not degenerate). $\endgroup$
    – RBega2
    Commented Jan 26, 2023 at 1:06
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    $\begingroup$ Another pathological aspect of this flow is that in the example I wrote down the tangent at $(0,0)$ is constant in time and lies one the line $x=y$, so even if the positively curved part does converge to the line segment it can't do so in $C^1$. $\endgroup$
    – RBega2
    Commented Jan 26, 2023 at 1:44
  • $\begingroup$ @JosephO'Rourke. Thank you for your comment. You have answered my question (in the negative). If you want to post it as a (short) answer, I will accept it. $\endgroup$
    – André Henriques
    Commented Jan 26, 2023 at 13:28
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    $\begingroup$ The curves shortening flow is (in an appropriate gauge -- e.g. written as a normal graph) a strictly parabolic equation -- you flow is extremely degenerate when the curve is not convex. On a more concrete level the evolution of curvature in CSF is a semilinear heat equation while the evolution of curvautre in your example looks likes some sort of non-linear heat flow where the the thermal conductivity is supported on the part where the curvature is positive (i.e. the negative curved regions seem to act as perfect insulators) and so that part is static. $\endgroup$
    – RBega2
    Commented Jan 26, 2023 at 14:40

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Repeating my comment as requested, it seems to me that simplicity (no crossings)---a feature of curve shortening---is not preserved. Consider a $U$-shape. The inside of the $U$ (red in figure) has negative $\kappa$ and will remain fixed, because

$f(\kappa) =0, \forall \kappa \le 0 \;,$

i.e., negative curvature points remains fixed.

Meanwhile the bottom of the $U$ (blue) has positive $f'(\kappa) > 0$ and will evolve to collide with the fixed portion.

enter image description here

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