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Asymptotic behavior of the isoperimetric deficit for expanding convex plane curves

Asymptotic behavior of the isoperimetric deficit for expanding convex plane curves We consider the expansion of a convex closed plane curve C 0 along its outward normal direction with speed G (1/ k ), where k is the curvature and $${G \left(z \right) :\left(0, \infty \right) \rightarrow \left( 0, \infty \right)}$$ G z : 0 , ∞ → 0 , ∞ is a strictly increasing function. We show that if $${{\rm lim}_{z \rightarrow \infty} G \left(z \right) = \infty}$$ lim z → ∞ G z = ∞ , then the isoperimetric deficit $${D \left(t \right) : = L^{2}\left(t \right) -4 \pi A \left(t \right)}$$ D t : = L 2 t - 4 π A t of the flow converges to zero. On the other hand, if $${{\rm lim}_{z \rightarrow \infty}G \left(z \right) = \lambda \in (0,\infty)}$$ lim z → ∞ G z = λ ∈ ( 0 , ∞ ) , then for any number d ≥ 0 and $${\varepsilon > 0}$$ ε > 0 , one can choose an initial curve C 0 so that its isoperimetric deficit $${D \left(t \right)}$$ D t satisfies $${\left \vert D \left(t \right) -d \right \vert < \varepsilon}$$ D t - d < ε for all $${t \in (0, \infty)}$$ t ∈ ( 0 , ∞ ) . Hence, without rescaling, the expanding curve C t will not become circular. It is close to some expanding curve P t , where each P t is parallel to P 0 . The asymptotic speed of P t is given by the constant $${\lambda}$$ λ . http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Journal of Evolution Equations Springer Journals

Asymptotic behavior of the isoperimetric deficit for expanding convex plane curves

Journal of Evolution Equations , Volume 14 (4) – Dec 1, 2014

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References (13)

Publisher
Springer Journals
Copyright
Copyright © 2014 by Springer Basel
Subject
Mathematics; Analysis
ISSN
1424-3199
eISSN
1424-3202
DOI
10.1007/s00028-014-0238-2
Publisher site
See Article on Publisher Site

Abstract

We consider the expansion of a convex closed plane curve C 0 along its outward normal direction with speed G (1/ k ), where k is the curvature and $${G \left(z \right) :\left(0, \infty \right) \rightarrow \left( 0, \infty \right)}$$ G z : 0 , ∞ → 0 , ∞ is a strictly increasing function. We show that if $${{\rm lim}_{z \rightarrow \infty} G \left(z \right) = \infty}$$ lim z → ∞ G z = ∞ , then the isoperimetric deficit $${D \left(t \right) : = L^{2}\left(t \right) -4 \pi A \left(t \right)}$$ D t : = L 2 t - 4 π A t of the flow converges to zero. On the other hand, if $${{\rm lim}_{z \rightarrow \infty}G \left(z \right) = \lambda \in (0,\infty)}$$ lim z → ∞ G z = λ ∈ ( 0 , ∞ ) , then for any number d ≥ 0 and $${\varepsilon > 0}$$ ε > 0 , one can choose an initial curve C 0 so that its isoperimetric deficit $${D \left(t \right)}$$ D t satisfies $${\left \vert D \left(t \right) -d \right \vert < \varepsilon}$$ D t - d < ε for all $${t \in (0, \infty)}$$ t ∈ ( 0 , ∞ ) . Hence, without rescaling, the expanding curve C t will not become circular. It is close to some expanding curve P t , where each P t is parallel to P 0 . The asymptotic speed of P t is given by the constant $${\lambda}$$ λ .

Journal

Journal of Evolution EquationsSpringer Journals

Published: Dec 1, 2014

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