Asymptotic behavior of the isoperimetric deficit for expanding convex plane curves

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 ${G \left(z \right) :\left(0, \infty \right) \rightarrow \left( 0, \infty \right)}$ is a strictly increasing function. We show that if ${{\rm lim}_{z \rightarrow \infty} G \left(z \right) = \infty}$, then the isoperimetric deficit ${D \left(t \right) : = L^{2}\left(t \right) -4 \pi A \left(t \right)}$ of the flow converges to zero. On the other hand, if ${{\rm lim}_{z \rightarrow \infty}G \left(z \right) = \lambda \in (0,\infty)}$, then for any number d ≥ 0 and ${\varepsilon > 0}$, one can choose an initial curve C ₀ so that its isoperimetric deficit ${D \left(t \right)}$ satisfies ${\left \vert D \left(t \right) -d \right \vert < \varepsilon}$ for all ${t \in [0, \infty)}$. 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 ${\lambda}$.