[math-fun] Connecting simple closed curves of the same length
Let C be a rectifiable simple closed curve* (RSCC) in the plane having length L(C) = 2π. Question: --------- Does there always exist a continuous family {C_t} of RSCC such that C_0 = C and C_1 is the unit circle in R^2, such that L(C_t) = 2π for all t in [0,1] ??? —Dan _____ * If C ⊂ R^2 is a RSCC, then C = A([0,1]) for some continuous function A : [0,1] —> R^2 where A(s) = A(t) for s ≠ t if and only if {s,t} = {0,1}, and such that L(C) = sup{∑ ‖A(x_(j+1)) - A(x_j)‖} < oo where the supremum is taken over all 0 = x_0 < x_1 < ... < x_n = 1. The length of C is then defined as L(C).
If it helps, you can dispense with the condition L(C_t) = 2 pi in the hypothesis. Specifically, if you have a continuous family of curves C_t not necessarily satisfying this condition, then you can obtain a continuous family of curves satisfying this condition by scaling them up/down appropriately: C'_t(x) := C_t(x) * (2 pi / L(C_t)) So your question is equivalent to 'is there a homotopy between any two RSCCs such that all of the intermediate steps are also RSCCs?'. It looks like this is a solved problem: https://en.wikipedia.org/wiki/Curve-shortening_flow#Non-smooth_curves Best wishes, Adam P. Goucher
Sent: Thursday, December 10, 2020 at 6:26 PM From: "Dan Asimov" <asimov@msri.org> To: "math-fun" <math-fun@mailman.xmission.com> Subject: [math-fun] Connecting simple closed curves of the same length
Let C be a rectifiable simple closed curve* (RSCC) in the plane having length
L(C) = 2π.
Question: --------- Does there always exist a continuous family {C_t} of RSCC such that
C_0 = C
and
C_1 is the unit circle in R^2, such that
L(C_t) = 2π
for all t in [0,1] ???
—Dan
_____ * If C ⊂ R^2 is a RSCC, then
C = A([0,1])
for some continuous function
A : [0,1] —> R^2
where
A(s) = A(t) for s ≠ t if and only if {s,t} = {0,1},
and such that
L(C) = sup{∑ ‖A(x_(j+1)) - A(x_j)‖} < oo
where the supremum is taken over all 0 = x_0 < x_1 < ... < x_n = 1.
The length of C is then defined as L(C).
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I'm not sure I read that Wikipedia article as saying this is solved for all non-smooth curves. (Or am I missing something?) —Dan
On Thursday/10December/2020, at 12:22 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
If it helps, you can dispense with the condition L(C_t) = 2 pi in the hypothesis. Specifically, if you have a continuous family of curves C_t not necessarily satisfying this condition, then you can obtain a continuous family of curves satisfying this condition by scaling them up/down appropriately:
C'_t(x) := C_t(x) * (2 pi / L(C_t))
So your question is equivalent to 'is there a homotopy between any two RSCCs such that all of the intermediate steps are also RSCCs?'.
It looks like this is a solved problem:
https://en.wikipedia.org/wiki/Curve-shortening_flow#Non-smooth_curves
Best wishes,
Adam P. Goucher
Sent: Thursday, December 10, 2020 at 6:26 PM From: "Dan Asimov" <asimov@msri.org> To: "math-fun" <math-fun@mailman.xmission.com> Subject: [math-fun] Connecting simple closed curves of the same length
Let C be a rectifiable simple closed curve* (RSCC) in the plane having length
L(C) = 2π.
Question: --------- Does there always exist a continuous family {C_t} of RSCC such that
C_0 = C
and
C_1 is the unit circle in R^2, such that
L(C_t) = 2π
for all t in [0,1] ???
—Dan
_____ * If C ⊂ R^2 is a RSCC, then
C = A([0,1])
for some continuous function
A : [0,1] —> R^2
where
A(s) = A(t) for s ≠ t if and only if {s,t} = {0,1},
and such that
L(C) = sup{∑ ‖A(x_(j+1)) - A(x_j)‖} < oo
where the supremum is taken over all 0 = x_0 < x_1 < ... < x_n = 1.
The length of C is then defined as L(C).
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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It says: "Curves with zero Lebesgue measure (including all polygons and piecewise-smooth curves) instantly evolve into smooth curves, after which they evolve as any smooth curve would." and every rectifiable curve has zero Lebesgue measure. Best wishes, Adam P. Goucher
Sent: Thursday, December 10, 2020 at 8:44 PM From: "Dan Asimov" <dasimov@earthlink.net> To: "math-fun" <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Connecting simple closed curves of the same length
I'm not sure I read that Wikipedia article as saying this is solved for all non-smooth curves.
(Or am I missing something?)
—Dan
On Thursday/10December/2020, at 12:22 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
If it helps, you can dispense with the condition L(C_t) = 2 pi in the hypothesis. Specifically, if you have a continuous family of curves C_t not necessarily satisfying this condition, then you can obtain a continuous family of curves satisfying this condition by scaling them up/down appropriately:
C'_t(x) := C_t(x) * (2 pi / L(C_t))
So your question is equivalent to 'is there a homotopy between any two RSCCs such that all of the intermediate steps are also RSCCs?'.
It looks like this is a solved problem:
https://en.wikipedia.org/wiki/Curve-shortening_flow#Non-smooth_curves
Best wishes,
Adam P. Goucher
Sent: Thursday, December 10, 2020 at 6:26 PM From: "Dan Asimov" <asimov@msri.org> To: "math-fun" <math-fun@mailman.xmission.com> Subject: [math-fun] Connecting simple closed curves of the same length
Let C be a rectifiable simple closed curve* (RSCC) in the plane having length
L(C) = 2π.
Question: --------- Does there always exist a continuous family {C_t} of RSCC such that
C_0 = C
and
C_1 is the unit circle in R^2, such that
L(C_t) = 2π
for all t in [0,1] ???
—Dan
_____ * If C ⊂ R^2 is a RSCC, then
C = A([0,1])
for some continuous function
A : [0,1] —> R^2
where
A(s) = A(t) for s ≠ t if and only if {s,t} = {0,1},
and such that
L(C) = sup{∑ ‖A(x_(j+1)) - A(x_j)‖} < oo
where the supremum is taken over all 0 = x_0 < x_1 < ... < x_n = 1.
The length of C is then defined as L(C).
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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