Curve Flows with a Global Forcing Term

Abstract

We consider embedded, smooth curves in the plane which are either closed or asymptotic to two lines. We study their behaviour under curve shortening flow with a global forcing term. We prove an analogue to Huisken’s distance comparison principle for curve shortening flow for initial curves whose local total curvature does not lie below $-\pi$ and show that this condition is sharp. With that, we can exclude singularities in finite time for bounded forcing terms. For immortal flows of closed curves whose forcing terms provide non-vanishing enclosed area and bounded length, we show convexity in finite time and smooth and exponential convergence to a circle. In particular, all of the above holds for the area preserving curve shortening flow.

1 Introduction

Let $\Sigma _0\subset \mathbb {R}^2$ be an embedded, smooth curve, parametrised by the embedding $X_0:{\mathcal {I}}\rightarrow \mathbb {R}^2$, where ${\mathcal {I}}\in \{\mathbb {S}^1,\mathbb {R}\}$. We seek a one-parameter family of maps $X:{\mathcal {I}}\times [0,T)\rightarrow \mathbb {R}^2$ with $X(\,\cdot \,,0)=X_0$ satisfying the evolution equation

$$\begin{aligned} {\dfrac{\partial X}{\partial t}}(p,t)=\big (h(t)-\kappa (p,t)\big )\varvec{\nu }(p,t) \end{aligned}$$

(1.1)

for $(p,t)\in {\mathcal {I}}\times (0,T)$, where the vector $\varvec{\nu }$ is the outward pointing unit normal to the curve $\Sigma _t:=X({\mathcal {I}},t)$, $\kappa$ is the curvature function and T is th maximal time of existence. The global term h is smooth and smoothly bounded whenever the curvature is bounded. For the curve shortening flow (CSF), $h\equiv 0$. For closed curves, the enclosed area preserving curve shortening flow (APCSF) has the global term

$$\begin{aligned} h(t)=\frac{2\pi }{L_t}\,, \end{aligned}$$

(1.2)

where $L_t=L(\Sigma _t)$ is the length of the curve. The length preserving curve flow (LPCF) has the global term

$$\begin{aligned} h(t)=\frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t\,. \end{aligned}$$

(1.3)

The total curvature of a curve $\Sigma _t=X({\mathcal {I}},t)$ is given by

$$\begin{aligned} \alpha (t):=\int _{\Sigma _t}\kappa \,ds_t\,, \end{aligned}$$

(1.4)

where $\alpha =2\pi$ if the curve $\Sigma =X(\mathbb {S}^1)$ is embedded, closed and positively oriented. For ${\mathcal {I}}=\mathbb {R}$, we assume that $\Sigma _t=X_0(\mathbb {R},t)$ is, up to translation, smoothly asymptotic to two distinct time-independent lines for $p\rightarrow -\infty$ and $p\rightarrow \infty$, where we also assume that

$$\begin{aligned} \alpha (t)\equiv \alpha _0\in (-\pi ,\pi ) \end{aligned}$$

(1.5)

as well as

$$\begin{aligned} \int _{\Sigma _0}|\kappa |\,ds<\infty \,. \end{aligned}$$

(1.6)

Note that (1.5) follows from the evolution equation of $\alpha (t)$, see Lemma 3.1, and the asymptotic behaviour of the curve.

The APCSF was first studied by Gage [8]. He proved that initially embedded, closed, convex curves stay embedded, smooth and convex, and converge smoothly to a circle of radius $\sqrt{A_0/\pi }$, where $A_0=A(\Sigma _0)$ is the enclosed area of the initial curve. In [21], Maeder-Baumdicker studied APCSF for convex curves with Neumann boundary on a convex support curve and showed smooth convergence to an arc for sufficiently short, convex, embedded initial curves. She proved a monotonicity formula and excluded type-I singularities for embedded, convex curves under the APCSF. For the LPCF, Pihan [25] showed that initially embedded, closed, convex curves stay embedded, smooth and convex, and converge smoothly and exponentially to a circle of radius $L_0/2\pi$.

In this paper, we will adapt theory from CSF. For CSF in the plane, Gage–Hamilton and Grayson [9, 11] showed that all embedded, closed initial curves stay embedded until they smoothly and exponentially shrink to a round point. In [18], Huisken gave a different proof for this result by bounding the ratio of the extrinsic distance

$$\begin{aligned} d(p,q,t):=\Vert X(q,t)-X(p,t)\Vert _{\mathbb {R}^2} \end{aligned}$$

and the intrinsic distance

$$\begin{aligned} l(p,q,t):=\int _p^qds_t \end{aligned}$$

for curves $\Sigma _t=X(\mathbb {R},t)$ with asymptotic ends, respectively, the extrinsic distance and the function

$$\begin{aligned} \psi (p,q,t):=\frac{L_t}{\pi }\sin \!\left( \frac{\pi \,l(p,q,t)}{L_t}\right) \end{aligned}$$

for curves $\Sigma _t=X(\mathbb {S}^1,t)$, below away from zero, and by applying singularity theory for CSF. In [3], Andrews and Bryan found an explicit function to proof curvature bounds via the distance comparison principle. To analyse curvature blow-ups, one distinguishes between type-I and type-II singularities and rescales the curve near a point of highest curvature. Using his famous monotonicity formula in [17], Huisken showed that if an immersed curve develops a type-I singularity under CSF, the curves $\Sigma _t$ have to be asymptotic to a homothetically shrinking solution around the singular point. Abresch and Langer [1] had previously classified all embedded, homothetically shrinking solutions of CSF as circles. One concludes, in case of a type-I singularity, that the curves shrink to a round point. For the type-II singularities, Hamilton [12] and Altschuler [2] showed that each rescaling sequence converges to a translating solution. For curves in the plane, the only solution of this kind is the so-called grim reaper which is, for all $\tau \in \mathbb {R}$, given by the graph of the function $u(\sigma ,\tau )=\tau -\log \cos (\sigma )$, where $\sigma \in (-\pi /2,\pi /2)$. On the grim reaper $\inf (d/l)=0$, so that type-II singularities can be excluded. Since $T<\infty$ and a singularity has to form, it has to be of type I.

This paper is structured as follows. In Sect. 2, we state evolution equations for the geometric quantities under (1.1) and draw first conclusions. In Sect. 3, we consider angles of tangent vectors and derive a strong maximum principle for the local total curvature

$$\begin{aligned} \theta (p,q,t):=\int _p^q\kappa \,ds_t\,. \end{aligned}$$

(1.7)

In the subsequent sections, we study the flow (1.1) for embedded, positively oriented, smooth initial curves $\Sigma _0=X_0({\mathcal {I}})$ with

$$\begin{aligned} \theta _0(p,q)=\int _p^q\kappa \,ds\ge -\pi \end{aligned}$$

(1.8)

for all $p,q\in {\mathcal {I}}$. Note that for convex curves $\theta _0\ge 0$. Figure 1 is an example for condition (1.8), where all the angles lay between $-\pi$ and $3\pi$, e. g. $\theta (p,q)=-\pi$, $\theta (q,p)=3\pi$, $\theta (q,r)=2\pi$, $\theta (r,q)=0$, $\theta (r,p)=\pi$.

Fig. 1

Example for condition (1.8)

In Sects. 4 and 5, we modify the distance comparison principles from [18] and prove that, for

$$\begin{aligned} h(t)\in {\left\{ \begin{array}{ll} [0,\infty )&{}\qquad \text { for }{\mathcal {I}}=\mathbb {R}\\ \left[ 0,\frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t+\frac{2\pi }{L_t}\right] &{} \qquad \text { for }{\mathcal {I}}=\mathbb {S}^1\end{array}\right. } \end{aligned}$$

(1.9)

and if the initial embedding $\Sigma _0$ satisfies (1.8), the ratio d/l for ${\mathcal {I}}=\mathbb {R}$ and $d/\psi$ for ${\mathcal {I}}=\mathbb {S}^1$ is bounded from below away from zero uniformly in time. We conclude that the curves $\Sigma _t$ stay embedded for all $t\in [0,T)$. We also show that the condition (1.8) is sharp, that is, one can construct initial curves which violate (1.8) arbitrarily mildly and for which the resulting flow self-intersects in finite time. An example is the initial curve in Fig. 2 with length sufficiently large compared to the $C^{3,\alpha }$-norm of its embedding and for which $\min _{{\mathcal {I}}\times {\mathcal {I}}}\theta _0<-\pi$, e. g. $\theta (p_1,p_2)<-\pi$.

Fig. 2

Counterexample 5.5

In Sect. 6, we assume that $T<\infty$ and there exist constants $0<c,C<\infty$ so that h satisfies (1.9) and additionally

$$\begin{aligned} L_t\ge c \qquad \text { and }\qquad 0\le h(t)\le C \end{aligned}$$

(1.10)

for $t\in [0,T)$ and study curvature blow-ups via parabolic rescaling. We use the distance comparison principles from Sects. 4 and 5 in the same fashion as for CSF in [18] to exclude type-II singularities and conclude that the flow exists for all positive times.

In Sect. 7, we assume that a solution is immortal, that is, it exists for all positive times and the global term satisfies the following. Let $\delta \in (0,\infty )$ be given so that $\delta A_0$ is the desired limit area and

$$\begin{aligned} h(t)=(1-\gamma )\frac{2\pi }{L_t}+\gamma \frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t\,, \end{aligned}$$

(1.11)

where $\gamma =(\delta -1)A_0/\big (L_0^2/4\pi -A_0\big )$. We prove that the curves become convex in finite time. The global term above ensures that the enclosed area is bounded away from zero and the length is bounded away from infinity throughout the flow. In Sect. 8, we assume that an immortal solution of (1.1) with h satisfying (1.11) is convex. We expand Gage’s and Pihan’s results and show smooth and exponential convergence to a round circle.

Note that the global term (1.2) of the APCSF satisfies conditions (1.9), (1.10) and (1.11). The global term (1.3) of the LPCF satisfies a priori (1.9) and (1.11).

This paper extends results from the author’s PhD thesis [5].

2 Evolution Equations and First Consequences

Let ${\mathcal {I}}\in \{\mathbb {S}^1\!,\mathbb {R}\}$ and $X:{\mathcal {I}}\rightarrow \mathbb {R}^2$ be a smooth, embedded curve with length element $v:{\mathcal {I}}\rightarrow \mathbb {R}$ by $v(p):=\big \Vert {\dfrac{d }{d p}}X(p)\big \Vert$. For a fixed point $p_0\in {\mathcal {I}}$, the arc length parameter $s:{\mathcal {I}}\rightarrow [0,L]$ is given by $s(p):=\int _{p_0}^pv(r)\,dr$, so that $ds=vdp$ and ${\dfrac{d }{d s}}=\frac{1}{v}{\dfrac{d }{d p}}$. For $\Sigma =X(\mathbb {S}^1)$, the arc length parameter is given by $s:\mathbb {S}^1\rightarrow \mathbb {S}^1_{L/2\pi }$ and ${\tilde{X}}:=X\circ s^{-1}:\mathbb {S}^1_{L/2\pi }\rightarrow \mathbb {R}^2$ parametrises $\Sigma$ by arc length. For $\Sigma =X(\mathbb {R})$, the arc length parameter is given by $s:\mathbb {R}\rightarrow \mathbb {R}$. The unit tangent vector field $\varvec{\tau }$ to $\Sigma$ in direction of the arc length parametrisation is given by $\varvec{\tau }:={\dfrac{d }{d s}}{\tilde{X}}$. The outward unit normal is given by $\varvec{\nu }:=(\varvec{\tau }_2,-\varvec{\tau }_1)$. We define the curvature by

$$\begin{aligned} \kappa :=-\left\langle {\dfrac{d \varvec{\tau }}{d s}},\varvec{\nu }\right\rangle =\left\langle \varvec{\tau },{\dfrac{d \varvec{\nu }}{d s}}\right\rangle \end{aligned}$$

and the curvature vector by $\varvec{\kappa }:=-\kappa \varvec{\nu }$. The Frenet–Serret equations read as

$$\begin{aligned} {\dfrac{d \varvec{\tau }}{d s}}=-\kappa \varvec{\nu }\qquad \text { and }\qquad {\dfrac{d \varvec{\nu }}{d s}}=\kappa \varvec{\tau }\,. \end{aligned}$$

Let $X:{\mathcal {I}}\!\times [0,T)\rightarrow \mathbb {R}^2$ be a one-parameter family of maps. For fixed $t\in [0,T)$, we can parametrise $\Sigma _t=X({\mathcal {I}},t)$ by arc length via the arc length parameter $s(\,\cdot \,,t)$, where $s({\mathcal {I}},t)\in \{\mathbb {S}^1_{L_t/2\pi },\mathbb {R}\}$ and the arc length parametrisation is given by ${\tilde{X}}(\,\cdot \,,t)=X(\,\cdot \,,t)\circ s^{-1}(\,\cdot \,,t):s({\mathcal {I}},t)\rightarrow \mathbb {R}^2$. The evolution equation (1.1) applied to the arc length parametrisation reads

$$\begin{aligned} {\dfrac{\partial {\tilde{X}}}{\partial t}}(s,t)={\dfrac{\partial ^2{\tilde{X}}}{\partial s^2}}(s,t)+h(t){\tilde{\varvec{\nu }}}(s,t) \end{aligned}$$

for $s\in s({\mathcal {I}},t)$, where ${\tilde{\varvec{\nu }}}(s,t)={\tilde{\varvec{\nu }}}(s(p,t),t)=\varvec{\nu }(p,t)$ and we used the identity $\Delta _{\Sigma }{\tilde{X}}={\dfrac{d 2}{d s^2}}{\tilde{X}}=\varvec{\kappa }$ for the curvature vector. Whenever we will calculate via the arc length parametrisation, we will do so at a fixed time. Since the images $X({\mathcal {I}},t)={\tilde{X}}(s({\mathcal {I}},t),t)$ are the same and X and ${\tilde{X}}$ only differ by a tangential diffeomorphism, we will omit the “$\sim$” in the following above geometric quantities related to ${\tilde{X}}$ if these depend on s rather than p.

Lemma 2.1

(Gage  [8]) Let $X:{\mathcal {I}}\times (0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1). Then, for $t\in (0,T)$,

$$\begin{aligned} \begin{array}{llll} &{}{\dfrac{\partial v}{\partial t}}=\kappa (h-\kappa )v\,,\qquad &{}{\dfrac{\partial }{\partial t}}{\dfrac{\partial }{\partial s}}={\dfrac{\partial }{\partial s}}{\dfrac{\partial }{\partial t}}-\kappa (h-\kappa ){\dfrac{\partial }{\partial s}}\,, \\ &{}{\dfrac{\partial \varvec{\tau }}{\partial t}}=-{\dfrac{\partial \kappa }{\partial s}}\varvec{\nu }\,,\qquad \qquad &{}{\dfrac{\partial \varvec{\nu }}{\partial t}}={\dfrac{\partial \kappa }{\partial s}}\varvec{\tau }\,,\qquad {\dfrac{\partial \kappa }{\partial t}}={\dfrac{\partial ^2\kappa }{\partial s^2}}-(h-\kappa )\kappa ^2\,. \end{array} \end{aligned}$$

Corollary 2.2

(Huisken  [16, Thm. 1.3]) Let $X:\mathbb {S}^1\!\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) and let $\kappa \ge 0$ on $\Sigma _0$. Then $\kappa >0$ on $\Sigma _t$ for all $t\in (0,T)$.

Lemma 2.3

(Gage  [8]) Let $X:\mathbb {S}^1\!\times (0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1). Then, for $t\in (0,T)$,

$$\begin{aligned} {\dfrac{d A}{d t}}=hL-2\pi \qquad \text { and }\qquad {\dfrac{d L}{d t}}=2\pi h-\int _{\Sigma _t}\kappa ^2\,ds_t\,. \end{aligned}$$

Proposition 2.4

Let $X:{\mathcal {I}}\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) with initial curve $\Sigma _0$. If $T<\infty$, then $\max _{p\in {\mathcal {I}}}|\kappa (p,t)|\rightarrow \infty$ for $t\rightarrow T$.

Proof

Like in [25, Sect. 6.3] (see also [5, Chap. 4]), we can bound the derivatives of the curvature in terms of the curvature as long as the curvature is bounded. The proposition then follows like in [15, Thm. 8.1]. $\square$

3 Angles and Local Total Curvature

We want to exploit the relationship between angles of tangent vectors and local total curvatures and prove a strong maximum principle for the latter.

Define $\vartheta :{\mathcal {I}}\times [0,T)\rightarrow \mathbb {S}^1$ to be the angle between the $x_1$-axis and the tangent vector, so that

$$\begin{aligned} \vartheta (p,t)={\left\{ \begin{array}{ll} \arccos (\langle {\mathbf {e}}_1,\varvec{\tau }(p,t)\rangle )&{}\qquad \text { if }\langle {\mathbf {e}}_2,\varvec{\tau }(p,t)\rangle \ge 0 \\ 2\pi -\arccos (\langle {\mathbf {e}}_1,\varvec{\tau }(p,t)\rangle )&{}\qquad \text { if }\langle {\mathbf {e}}_2,\varvec{\tau }(p,t)\rangle <0\,. \end{array}\right. } \end{aligned}$$

Since $\varvec{\nu }=(\varvec{\tau }_2,-\varvec{\tau }_1)$,

$$\begin{aligned} \cos (\vartheta )=\langle {\mathbf {e}}_1,\varvec{\tau }\rangle =-\langle {\mathbf {e}}_2,\varvec{\nu }\rangle \quad \text { and }\quad \sin (\vartheta )=\langle {\mathbf {e}}_2,\varvec{\tau }\rangle =\langle {\mathbf {e}}_1,\varvec{\nu }\rangle \,. \end{aligned}$$

(3.1)

For a fixed time $t\in [0,T)$, we can define the angle ${\tilde{\vartheta }}$ via the arc length parameter by ${\tilde{\vartheta }}:s({\mathcal {I}},t)\rightarrow [0,2\pi )$. As explained earlier, we can omit the “$\sim$” for simplicity.

Lemma 3.1

(see Gage–Hamilton  [9, Lem. 3.1.5]) Let $X:{\mathcal {I}}\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) with initial curve $\Sigma _0$. Then

$$\begin{aligned} {\dfrac{\partial \vartheta }{\partial s}}=\kappa \qquad \text { and }\qquad {\dfrac{\partial \vartheta }{\partial t}}={\dfrac{\partial \kappa }{\partial s}} \quad \text { for }\; t\in (0,T). \end{aligned}$$

Like in (1.7), we define the total local curvature $\theta :{\mathcal {I}}\times {\mathcal {I}}\times [0,T)\rightarrow \mathbb {R}$ by

$$\begin{aligned} \theta (p,q.t):=\int _p^q\kappa (r,t)\,ds_t\,, \end{aligned}$$

(3.2)

where we integrate in direction of the parametrisation. The total curvature $\alpha (t)$ is given by the full integral over the curvature as stated in (1.4). For ${\mathcal {I}}=\mathbb {S}^1$ and $p,q\in [0,2\pi )$, we set

$$\begin{aligned} \theta (p,q.t)= {\left\{ \begin{array}{ll} \int _p^q\kappa \,ds_t&{}\qquad \text { if }p\le q\\ \int _p^{2\pi }\kappa \,ds_t+\int _0^q\kappa \,ds_t&{}\qquad \text { if }q<p\,. \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} 2\pi =\theta (p,q,t)+\theta (q,p,t) \end{aligned}$$

(3.3)

for all $p,q\in \mathbb {S}^1$. For ${\mathcal {I}}=\mathbb {R}$ and $p<q$, we set $\theta (q,p,t)=-\theta (p,q,t)$. By Lemma 3.1,

$$\begin{aligned} \theta (p,q) =\int _p^q\kappa \,ds =\int _p^q\frac{1}{v}{\dfrac{\partial \vartheta }{\partial r}}\,vdr =\vartheta (q)-\vartheta (p)+2\pi \omega (p,q)\,, \end{aligned}$$

(3.4)

where $\omega \in \mathbb {Z}$ is the local winding number. Hence, $\theta$ is the angle between the tangent vectors at two points on the curve modulo the local winding number. If a curve $\Sigma =X({\mathcal {I}})$ is embedded and convex, then $0\le \theta (p,q)<\alpha$ for all $p,q\in {\mathcal {I}}$. For ${\mathcal {I}}=\mathbb {S}^1$ and fixed $p\in \mathbb {S}^1$,

$$\begin{aligned} \lim _{q\searrow p}\theta (p,q) =\lim _{q\searrow p}\int _p^q\kappa \,vdr =0 \end{aligned}$$

(3.5)

and

$$\begin{aligned} \lim _{q\nearrow p}\theta (p,q) =\lim _{q\nearrow p}\int _0^q\kappa \,vdr+\int _p^{2\pi }\kappa \,vdr =\int _{\mathbb {S}^1}\kappa \,vdr =2\pi \,. \end{aligned}$$

(3.6)

Hence, $\theta$ is discontinuous along the diagonal $\{p=q\}\subset \mathbb {S}^1\!\times \mathbb {S}^1$.

Lemma 3.2

Let $\Sigma =X(\mathbb {S}^1)$ be an embedded, closed curve. Then

$$\begin{aligned} \sup _{\mathbb {S}^1\!\times \mathbb {S}^1}\theta =2\pi -\min _{\mathbb {S}^1\!\times \mathbb {S}^1}\theta \,. \end{aligned}$$

Proof

Let the maximum of $\theta$ be attained at $p_0,q_0\in \mathbb {S}^1$, that is, by (3.3),

$$\begin{aligned} \max _{\mathbb {S}^1\!\times \mathbb {S}^1}\theta =\theta (p_0,q_0)=2\pi -\theta (q_0,p_0)\,. \end{aligned}$$

(3.7)

Then, for all $p,q\in \mathbb {S}^1$, $p\ne q$, by (3.3) and (3.7),

$$\begin{aligned} 2\pi -\theta (q_0,p_0) =\theta (p_0,q_0) \ge \theta (q,p) =2\pi -\theta (p,q)\,. \end{aligned}$$

Consequently, $\theta (q_0,p_0)\le \theta (p,q)$ for all $p,q\in \mathbb {S}^1$, $p\ne q$, which implies with (3.7),

$$\begin{aligned} \qquad \qquad \qquad \qquad \qquad \min _{\mathbb {S}^1\!\times \mathbb {S}^1}\theta =\theta (q_0,p_0) =2\pi -\max _{\mathbb {S}^1\!\times \mathbb {S}^1}\theta \,.\qquad \qquad \qquad \qquad \qquad \qquad \end{aligned}$$

$\square$

Lemma 3.3

Let $\Sigma =X(\mathbb {R})$. Then

$$\begin{aligned} \sup _{\mathbb {R}\times \mathbb {R}}\theta \le \alpha -2\inf _{\mathbb {R}\times \mathbb {R}}\theta \,. \end{aligned}$$

Proof

For $p,q\in \mathbb {R}$, $p<q$,

$$\begin{aligned} \qquad \qquad \alpha =\int _{-\infty }^p\kappa \,ds+\int _p^q\kappa \,ds+\int _q^\infty \kappa \,ds \ge 2\inf _{\mathbb {R}\times \mathbb {R}}\theta +\theta (p,q)\,.\qquad \qquad \qquad \end{aligned}$$

$\square$

For $t\in [0,T)$, we define

$$\begin{aligned} \theta _{\min }(t):=\min _{(p,q)\in \mathbb {S}^1\!\times \mathbb {S}^1}\theta (p,q,t)\le 0 \quad \text { and }\quad \theta _{\inf }(t):=\inf _{(p,q)\in \mathbb {R}\times \mathbb {R}}\theta (p,q,t)\le \alpha \,. \end{aligned}$$

Theorem 3.4

Let $X:{\mathcal {I}}\times (0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1). Then

$$\begin{aligned} \left( {\dfrac{\partial }{\partial t}}-\Delta _{\Sigma _t}\right) \theta (p,q,t)=0 \end{aligned}$$

for all $p,q\in {\mathcal {I}}$ ($p\ne q$ for ${\mathcal {I}}=\mathbb {S}^1$) and $t\in (0,T)$. Moreover, let $t_0\in (0,T)$.

• (i) For ${\mathcal {I}}=\mathbb {S}^1$, suppose $\theta _{\min }(t_0)<0$, then $\theta _{\min }(t_0)<\theta _{\min }(t)$ for all $t\in (t_0,T)$.

• (ii) For ${\mathcal {I}}=\mathbb {R}$, let (1.5) be satisfied. Suppose $\theta _{\inf }(t_0)<\min \{0,\alpha \}$, then $\theta _{\inf }(t_0)<\theta _{\inf }(t)$ for all $t\in (t_0,T)$.

Proof

We differentiate $\theta$ at $p,q\in {\mathcal {I}}$ ($p\ne q$ for ${\mathcal {I}}=\mathbb {S}^1$) in direction of $\varvec{\tau }$ with $\varvec{\tau }(\theta ) = {\dfrac{\partial }{\partial s}}\theta$, and use Lemma 3.1 to obtain $\varvec{\tau }_p(\theta ) =-\kappa _p$ and $\varvec{\tau }_q(\theta ) =\kappa _q$, as well as

$$\begin{aligned} \Delta _{\Sigma _t}\theta =\varvec{\tau }_p^2(\theta )+\varvec{\tau }_q^2(\theta ) =\varvec{\tau }_q(\kappa _q)-\varvec{\tau }_p(\kappa _p) ={\dfrac{\partial \vartheta _q}{\partial t}}-{\dfrac{\partial \vartheta _p}{\partial t}} ={\dfrac{\partial \theta }{\partial t}}\,. \end{aligned}$$

(3.8)

Since X is smooth, $\theta$ is smooth in ${\mathcal {I}}\times {\mathcal {I}}\times [0,T)$.

(i) Let ${\mathcal {I}}=\mathbb {S}^1$. The set $S:=\mathbb {S}^1\!\times \mathbb {S}^1\setminus \{p=q\}$ is an oriented cylinder. By (3.5) and (3.6), the closure ${\bar{S}}$ has two boundaries

$$\begin{aligned} (\partial S)_-=\left\{ (p,p)\,\left| \, p\in \mathbb {S}^1\right\} \quad \text { and }\quad (\partial S)_+\right. =\left\{ \left( \lim _{r\nearrow p}r,p\right) \,\bigg |\, p\in \mathbb {S}^1\right\} \,, \end{aligned}$$

where $\theta \equiv 0$ on $(\partial S)_-\times [0,T)$ and $\theta \equiv 2\pi$ on $(\partial S)_+\times [0,T)$. The claim now follows from the strong maximum principle with boundary conditions.

(ii) Let ${\mathcal {I}}=\mathbb {R}$. By (1.5), $\Sigma _t$ is up to translation smoothly asymptotic to two time-independent lines, that is, $|\varvec{\tau }_p\kappa (p,t)|\rightarrow 0$ uniformly for $p\rightarrow \pm \infty$, by (3.8), $\lim _{p\rightarrow \pm \infty }\vartheta _p$ is constant in time. For $p\in \mathbb {R}$, define

$$\begin{aligned} \theta ^-(p,t):=\lim _{q\rightarrow -\infty }\theta (q,p,t) \quad \text { and }\quad \theta ^+(p,t):=\lim _{q\rightarrow \infty }\theta (p,q,t)\,. \end{aligned}$$

If $\theta _{\inf }<\min \{0,\alpha \}$, $\theta _{\inf }$ either equals $\theta _{\inf }^+$ or $\theta _{\inf }^-$, or is attained at a local minimum. Suppose $\theta _{\inf }<0$ is attained at a local minimum. We have that $\theta \equiv 0$ on $\{p=q\}\times [0,T)=\partial \{p\ne q\}\times [0,T)$. The strong maximum principle yields that $\theta$ is strictly decreasing. Suppose $\theta _{\inf }(t)<\min \{0,\alpha \}$ equals $\theta _{\inf }^+$ or $\theta _{\inf }^-$. By (3.8),

$$\begin{aligned} \left( {\dfrac{\partial }{\partial t}}-\Delta _{\Sigma _t}\right) \theta ^{\pm }=0\,. \end{aligned}$$

If $\theta _{\inf }^{\pm }<\min \{0,\alpha \}$, then $\theta _{\inf }^{\pm }$ is attained at a point $p\in \mathbb {R}$. The strong maximum principle yields that $\theta _{\inf }^{\pm }$ is strictly increasing as long as $\theta _{\inf }^{\pm }<\min \{0,\alpha \}$. $\square$

We define the extrinsic distance $d:{\mathcal {I}}\times {\mathcal {I}}\times [0,T)\rightarrow \mathbb {R}$ by

$$\begin{aligned} d(p,q,t):=\Vert X(q,t)-X(p,t)\Vert _{\mathbb {R}^2} \end{aligned}$$

and the vector ${\mathbf {w}}:\big ({\mathcal {I}}\times {\mathcal {I}}\times [0,T)\big )\setminus \{d=0\}\rightarrow \mathbb {R}^2$ by

$$\begin{aligned} {\mathbf {w}}(p,q,t):=\frac{X(q,t)-X(p,t)}{d(p,q,t)}\,. \end{aligned}$$

Lemma 3.5

Let $\Sigma =X({\mathcal {I}})$ be an embedded curve and $p,q\in {\mathcal {I}}$ with $d(p,q)\ne 0$. Let $\langle {\mathbf {w}},\varvec{\tau }_p\rangle =\langle {\mathbf {w}},\varvec{\tau }_q\rangle =\cos (\beta /2)$ for $\beta \in [0,\pi ]$. Then either

• (i) $\langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =-\sin (\beta /2)$ and $\theta (p,q)=2\pi k+\beta$,

• (ii) $\langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =\sin (\beta /2)$ and $\theta (p,q)=2\pi k-\beta$, or

• (iii) $\langle {\mathbf {w}},\varvec{\nu }_p\rangle =\langle {\mathbf {w}},\varvec{\nu }_q\rangle =\pm \sin (\beta /2)$ and $\theta (p,q)=2\pi k$

for $k\in \mathbb {Z}$.

Proof

The angles are invariant under rotations in the plane, thus we may assume ${\mathbf {w}}={\mathbf {e}}_1$. Since $\beta \in [0,\pi ]$, the definition (3.1) of $\vartheta \in [0,2\pi )$ yields

$$\begin{aligned} \cos (\vartheta _p)=\langle {\mathbf {e}}_1,\varvec{\tau }_p\rangle =\cos \!\left( \frac{\beta }{2}\right) \ge 0 \quad \text { and }\quad \sin (\vartheta _p)=\langle {\mathbf {e}}_1,\varvec{\nu }_p\rangle =\pm \sin \!\left( \frac{\beta }{2}\right) \end{aligned}$$

as well as

$$\begin{aligned} \cos (\vartheta _q)=\langle {\mathbf {e}}_1,\varvec{\tau }_q\rangle =\cos \!\left( \frac{\beta }{2}\right) \ge 0 \quad \text { and }\quad \sin (\vartheta _q)=\langle {\mathbf {e}}_1,\varvec{\nu }_q\rangle =\pm \sin \!\left( \frac{\beta }{2}\right) \,. \end{aligned}$$

Hence, by (3.4),

$$\begin{aligned} \vartheta _p,\vartheta _q\in \left\{ \frac{\beta }{2},2\pi -\frac{\beta }{2}\right\} \quad \text { and }\quad \theta =2\pi \omega +\vartheta _q-\vartheta _p\,, \end{aligned}$$

(3.9)

where $\omega \in \mathbb {Z}$.

(i) Assume that $\sin (\vartheta _p)=-\sin (\beta /2)\le 0$ and $\sin (\vartheta _q)=\sin (\beta /2)\ge 0$. From (3.9), it follows that $\vartheta _p=2\pi -\beta /2$, $\vartheta _q=\beta /2$ and $\theta =2\pi (\omega +1)+\beta$.

(ii) Assume that $\sin (\vartheta _p)=\sin (\beta /2)\ge 0$ and $\sin (\vartheta _q)=-\sin (\beta /2)\le 0$. From (3.9), it follows that $\vartheta _p=\beta /2$, $\vartheta _q=2\pi -\beta /2$ and $\theta =2\pi (\omega +1)-\beta$.

(iii) Assume that $\sin (\vartheta _p)=\pm \sin (\beta /2)$ and $\sin (\vartheta _q)=\pm \sin (\beta /2)$. From (3.9), it follows that either $\vartheta _p=\vartheta _q=\beta /2$ or $\vartheta _p=\vartheta _q=2\pi -\beta /2$ and thus $\theta =2\pi \omega$. $\square$

4 Distance Comparison Principle for Noncompact Curves

We adapt the methods from Huisken [18] to obtain estimates that imply a certain noncollapsing behaviour of the evolving curves.

The intrinsic distance $l:{\mathcal {I}}\times {\mathcal {I}}\times [0,T)\rightarrow \mathbb {R}$ is given by

$$\begin{aligned} l(p,q,t):=\int _p^qv(r,t)\,dr\,. \end{aligned}$$

We set $d/l\equiv 1$ on $\{p=q\}\times [0,T)$, then $d/l\in C^0(\mathbb {R}\times \mathbb {R}\times [0,T))$. Embedded curves satisfy $(d/l)(p,q)>0$ for all $p,q\in \mathbb {R}$. If a curve is not a line, then there exist $p,q\in \mathbb {R}$ so that $d(p,q)<l(p,q)$ and thus $\inf _{\mathbb {R}\times \mathbb {R}}(d/l)<1$.

Lemma 4.1

Let $\Sigma =X(\mathbb {R})$ be an embedded curve. Let $p,q\in \mathbb {R}$, $p\ne q$, such that $\Sigma$ crosses the connecting line between X(p) and X(q) at X(r) with $r\notin [p,q]$. Then (d/l)(p, q) cannot be the infimum.

Proof

Let $X(\mathbb {R})$ cross the connecting line between X(p) and X(q) at X(r). Then $X(r)=X(p)+{\mathbf {w}}(p,q)\Vert X(r)-X(p)\Vert$. Set $d:=d(p,q)$, $d_1:=d(p,r)$ and $d_2:=d(r,q)$. Then $d=d_1+d_2$. Furthermore, set $l:=l(p,q)$, $l_1:=l(p,r)$ and $l_2:=l(r,q)$. If $p<q<r$, then $l<l_1$ and $d/l>d_1/l_1$. If $r<p<q$, then $l<l_2$ and $d/l>d_2/l_2$. Thus, d/l cannot be a global minimum. $\square$

Now we can prove a similar result to [18, Thm. 2.1].

Theorem 4.2

Let $\Sigma _0=X_0(\mathbb {R})$ be a smooth, embedded curve satisfying (1.5) and (1.8). Let $X:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) satisfying (1.9), and with initial curve $\Sigma _0$. Then there exists a constant $c(\Sigma _0)>0$ such that

$$\begin{aligned} \inf _{(p,q,t)\in \mathbb {R}\times \mathbb {R}\times [0,T)}\frac{d}{l}(p,q,t)\ge c\,. \end{aligned}$$

Proof

Let $\Sigma _0=X_0(\mathbb {R})$ be an embedded curve satisfying (1.5) and (1.8). Then

$$\begin{aligned} \inf _{t\in [0,T)}\lim _{p\rightarrow \infty }\frac{d}{l}(p,-p,t)=:c_1(\Sigma _0)\in (0,1] \end{aligned}$$

(4.1)

for all $t\in (0,T)$. Lemma 3.3 implies that $\theta _0\in [-\pi ,\alpha +2\pi ]$. From the maximum principle for $\theta$, Theorem 3.4, it follows that

$$\begin{aligned} \theta (p,q,t)\in (-\pi ,\alpha +2\pi ) \end{aligned}$$

(4.2)

for all $p,q\in \mathbb {S}^1$ and $t\in (0,T)$. Since d/l is continuous and initially positive, there exists a time $T'\in (0,T]$ so that $d/l>0$ on $[0,T')$. Fix $t_0\in (0,T')$. If $\Sigma _{t_0}$ is a line, then $d/l\equiv 1$ on $\mathbb {R}\times \mathbb {R}$. Assume that $\Sigma _{t_0}$ is not a line so that $\inf _{\mathbb {R}\times \mathbb {R}}(d/l)<1$ at $t_0$. Let $p,q\in \mathbb {R}$, $p\ne q$, be points where a local spatial minimum of d/l at $t_0$ is attained and assume w.l.o.g. that $s(p,t_0)<s(q,t_0)$. We have for all $\xi \in T_{X(p,t_0)}\Sigma _{t_0}\bigoplus T_{X(q,t_0)}\Sigma _{t_0}$,

$$\begin{aligned} 0<\frac{d}{l}(p,q,t_0)<1\,,\quad \xi \!\left( \frac{d}{l}\right) \!(p,q,t_0)=0 \quad \text { and }\quad \xi ^2\!\left( \frac{d}{l}\right) \!(p,q,t_0)\ge 0\,. \end{aligned}$$

In the following, we always calculate at the point $(p,q,t_0)$. The spatial derivatives of d and l are all given in [18] (for detailed calculations, see [5, Lems. 6.2 and 7.4]). The first spatial derivative of d/l at $(p,t_0)$ in direction of the vector $\xi =\varvec{\tau }_p\oplus 0$ is given by

$$\begin{aligned} 0=(\varvec{\tau }_p\oplus 0)\!\left( \frac{d}{l}\right) =-\frac{1}{l}\langle {\mathbf {w}},\varvec{\tau }_p\rangle +\frac{d}{l^2}\,. \end{aligned}$$

(4.3)

At $(q,t_0)$ and for the vector $\xi =0\oplus \varvec{\tau }_q$, we have

$$\begin{aligned} 0=(0\oplus \varvec{\tau }_q)\!\left( \frac{d}{l}\right) =\frac{1}{l}\langle {\mathbf {w}},\varvec{\tau }_q\rangle -\frac{d}{l^2}\,. \end{aligned}$$

(4.4)

Since $d/l\in (0,1)$, and by (4.3) and (4.4), there exists $\beta \in (0,\pi )$ with

$$\begin{aligned} \langle {\mathbf {w}},\varvec{\tau }_p\rangle =\langle {\mathbf {w}},\varvec{\tau }_q\rangle =\frac{d}{l} =\cos \!\left( \frac{\beta }{2}\right) \in (0,1)\,. \end{aligned}$$

(4.5)

By Lemma 3.5, (4.2) and (4.5), either

$$\begin{aligned} \langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =-\sin \!\left( \frac{\beta }{2}\right)&\quad \text { and }\quad 2\pi k+\beta =\theta \in (0,\pi )\cup (2\pi ,\alpha +2\pi )\,, \\ \langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =\sin \!\left( \frac{\beta }{2}\right)&\quad \text { and }\quad 2\pi k-\beta =\theta \in (-\pi ,0)\cup (\pi ,2\pi )\,,\quad \text { or } \\ \langle {\mathbf {w}},\varvec{\nu }_p\rangle =\langle {\mathbf {w}},\varvec{\nu }_q\rangle =\pm \sin \!\left( \frac{\beta }{2}\right)&\quad \text { and }\quad \theta \in \{0,2\pi \} \end{aligned}$$

for $k\in \mathbb {Z}$. We use the evolution equation (1.1) and Lemma 2.1 to differentiate the ratio in time,

$$\begin{aligned} {\dfrac{\partial }{\partial t}}\left( \frac{d}{l}\right)&=\frac{1}{l}\big (h\langle {\mathbf {w}},\varvec{\nu }_q-\varvec{\nu }_p \rangle +\langle {\mathbf {w}},\varvec{\kappa }_q-\varvec{\kappa }_p\rangle \big ) \nonumber \\&\quad +\frac{d}{l^2}\left( \int _p^q\kappa ^2\,ds_t-h\int _p^q\kappa \,ds_t\right) \,. \end{aligned}$$

(4.6)

We are now considering four different cases.

(i) Assume that

$$\begin{aligned} \langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =-\sin \!\left( \frac{\beta }{2}\right) \quad \text { and }\quad \beta =\theta \in (0,\pi )\,. \end{aligned}$$

(4.7)

Adding (4.3) to (4.4) yields $\langle {\mathbf {w}},\varvec{\tau }_q-\varvec{\tau }_p\rangle =0$. For unit tangent vectors, we have

$$\begin{aligned} \langle \varvec{\tau }_p+\varvec{\tau }_q,\varvec{\tau }_q-\varvec{\tau }_p\rangle =\Vert \varvec{\tau }_p\Vert ^2-\Vert \varvec{\tau }_q\Vert ^2=0\,. \end{aligned}$$

Thus, ${\mathbf {w}}$ and $\varvec{\tau }_p+\varvec{\tau }_q$ are both perpendicular to $\varvec{\tau }_q-\varvec{\tau }_p$ and are therefore parallel, that is, $\measuredangle ({\mathbf {w}},\varvec{\tau }_q+\varvec{\tau }_p)=0$. Using $\Vert {\mathbf {w}}\Vert =1$, we calculate

$$\begin{aligned} 0\le 2\frac{d}{l} =\langle {\mathbf {w}},\varvec{\tau }_q+\varvec{\tau }_p\rangle =\Vert \varvec{\tau }_q+\varvec{\tau }_p\Vert \,. \end{aligned}$$

(4.8)

By (4.7),

$$\begin{aligned} \langle {\mathbf {w}},\varvec{\kappa }_q-\varvec{\kappa }_p\rangle =-\kappa _q\langle {\mathbf {w}},\varvec{\nu }_q\rangle +\kappa _p\langle {\mathbf {w}},\varvec{\nu }_p\rangle =-(\kappa _p+\kappa _q)\sin \!\left( \frac{\theta }{2}\right) \,. \end{aligned}$$

(4.9)

We differentiate d/l at $(p,q,t_0)$ twice with respect to the vector $\xi =\varvec{\tau }_p\ominus \varvec{\tau }_q$ and calculate with (4.8) and (4.9),

$$\begin{aligned} 0&\le (\varvec{\tau }_p\ominus \varvec{\tau }_q)^2\left( \frac{d}{l}\right) =\frac{1}{l}\langle {\mathbf {w}},\varvec{\kappa }_q-\varvec{\kappa }_p\rangle =-\frac{1}{l}(\kappa _p+\kappa _q)\sin \!\left( \frac{\theta }{2}\right) \,. \end{aligned}$$

We abbreviate $\kappa :=(\kappa _p+\kappa _q)/2$ and obtain

$$\begin{aligned} 2\kappa \sin \!\left( \frac{\theta }{2}\right) \le 0\,. \end{aligned}$$

Since $\sin (\theta /2)>0$ for $\theta \in (0,\pi )$, we conclude $\kappa \le 0$ and

$$\begin{aligned} h-\kappa \ge h>h-\frac{\theta }{l}\,. \end{aligned}$$

Furthermore, the inequality

$$\begin{aligned} \sin \!\left( \frac{\theta }{2}\right)>\frac{\theta }{2}\cos \!\left( \frac{\theta }{2}\right) >0 \end{aligned}$$

holds for $\theta \in (0,\pi )$. Hence,

$$\begin{aligned} (h-\kappa )\sin \!\left( \frac{\theta }{2}\right) >\frac{\theta }{2}\cos \!\left( \frac{\theta }{2}\right) \left( h-\frac{\theta }{l}\right) \end{aligned}$$

(4.10)

for $\theta \in (0,\pi )$. Cauchy–Schwarz and the definition (3.2) of $\theta$ imply

$$\begin{aligned} \int _p^q\kappa ^2\,ds_t \ge \frac{1}{l}\left( \int _p^q\kappa \,ds_t\right) ^{2} =\frac{\theta ^2}{l}\,. \end{aligned}$$

(4.11)

Then (4.6), (4.5), (4.7), (4.9), (4.10) and (4.11) yield

$$\begin{aligned} {\dfrac{\partial }{\partial t}}\left( \frac{d}{l}\right)&\ge \frac{2}{l}\left( (h-\kappa )\sin \!\left( \frac{\theta }{2}\right) -\frac{\theta }{2}\cos \!\left( \frac{\theta }{2}\right) \left( h-\frac{\theta }{l}\right) \right) >0 \end{aligned}$$

at $(p,q,t_0)$.

(ii) Assume that $\langle {\mathbf {w}},\varvec{\nu }_p\rangle =\langle {\mathbf {w}},\varvec{\nu }_q\rangle =\pm \sin (\beta /2)$ and $\theta =0$. By (4.5),

$$\begin{aligned} \varvec{\tau }_p=\varvec{\tau }_q\qquad \text { and }\qquad \varvec{\nu }_p=\varvec{\nu }_q\,. \end{aligned}$$

(4.12)

We differentiate d/l at $(p,q,t_0)$ twice with respect to the vector $\xi =\varvec{\tau }_p\oplus \varvec{\tau }_q$ and calculate with (4.12),

$$\begin{aligned} 0&\le (\varvec{\tau }_p\oplus \varvec{\tau }_q)^2\left( \frac{d}{l}\right) =\frac{1}{l}\langle {\mathbf {w}},\varvec{\kappa }_q-\varvec{\kappa }_p\rangle \,. \end{aligned}$$

(4.13)

We conclude with (3.2), (4.6), (4.12), (4.13),

$$\begin{aligned} {\dfrac{\partial }{\partial t}}\left( \frac{d}{l}\right) \ge \frac{d}{l^2}\int _p^q\kappa ^2\,ds_t >0 \end{aligned}$$

at $(p,q,t_0)$.

(iii) Assume that $\langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =\sin (\beta /2)$ and $-\beta =\theta \in (-\pi ,0)$. Again by continuity of d/l, there exists $t_1(\Sigma _0)\in (0,T')$ so that

$$\begin{aligned} \inf _{\mathbb {R}\times \mathbb {R}}\frac{d}{l}(\,\cdot \,,\,\cdot \,,t)\ge \frac{1}{2}\inf _{\mathbb {R}\times \mathbb {R}}\frac{d}{l}(\,\cdot \,,\,\cdot \,,0) \end{aligned}$$

(4.14)

for all $t\in [0,t_1]$. If $t_0\in (t_1,T)$, Theorem 3.4 and (4.2) yield $-\pi<\theta _{\inf }(t_1)\le \theta _{\inf }(t_0)<0$ so that (4.5) and the monotone behaviour of the cosine on $(-\pi ,0)$ imply

$$\begin{aligned} \frac{d}{l}=\cos \!\left( \frac{\theta }{2}\right) \ge \cos \!\left( \frac{\theta _{\inf }(t_0)}{2}\right) \ge \cos \!\left( \frac{\theta _{\inf }(t_1)}{2}\right) >0\,. \end{aligned}$$

(4.15)

We deduce with (4.14) and (4.15) that

$$\begin{aligned} \frac{d}{l}\ge \min \left\{ \frac{1}{2}\inf _{\mathbb {R}\times \mathbb {R}}\frac{d}{l}(\,\cdot \,,\,\cdot \,,0),\cos \!\left( \frac{\theta _{\inf }(t_1)}{2}\right) \right\} =:c_2(\Sigma _0)>0 \end{aligned}$$

(4.16)

at $(p,q,t_0)$.

(iv) Assume that $\theta \in (\pi ,\alpha +2\pi )$. By (4.5), $\langle {\mathbf {w}},\varvec{\tau }_q\rangle =\langle {\mathbf {w}},\varvec{\tau }_p\rangle =d/l\in (0,1)$. Since $\Sigma _{t_0}$ is embedded with ends going to $\infty$ and $X(\,\cdot \,,t_0)$ is continuous, the curve has to cross the line segment between $X(p,t_0)$ and $X(q,t_0)$ at least once at $X(r,t_0)$ with $r\notin [p,q]$. Lemma 4.1 implies that d/l cannot attain the infimum at $(p,q,t_0)$ (it could still, however, attain a local minimum at this point). Hence,

$$\begin{aligned} \frac{d}{l}>\inf _{\mathbb {R}\times \mathbb {R}}\frac{d}{l}(\,\cdot \,,\,\cdot \,,t_0)\,, \end{aligned}$$

where $\inf _{\mathbb {R}\times \mathbb {R}}d/l(\,\cdot \,,\,\cdot \,,t_0)$ is either the infimum from (4.1) or a local minimum as discussed in cases (i), (ii) and (iii).

Assume that d/l falls below $c:=\min \{c_1,c_2\}$, where $c_1$ and $c_2$ are given in (4.1) and (4.16), and attains $\Lambda \in (0,c)$ for the first time at time $t_2\in (0,T)$ and points $p,q\in \mathbb {S}^1$, $p\ne q$, so that

$$\begin{aligned} c>\Lambda =\frac{d}{l}(p,q,t_2)=\inf _{\mathbb {R}\times \mathbb {R}}\frac{d}{l}(\,\cdot \,,\,\cdot \,,t_2) \end{aligned}$$

(4.17)

is the infimum and

$$\begin{aligned} {\dfrac{\partial }{\partial t}}_{|_{t=t_2}}\!\left( \frac{d}{l}\right) (p,q,t)\le 0\,. \end{aligned}$$

(4.18)

Cases (i) and (ii) contradict (4.18), and cases (iii) and (iv) contradict (4.17). $\square$

Corollary 4.3

Let $\Sigma _0=X_0(\mathbb {R})$ be a smooth, embedded curve satisfying (1.5) and (1.8). Let $X:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) satisfying (1.9), and with initial curve $\Sigma _0$. Then $\Sigma _t=X(\mathbb {R},t)$ is embedded for all $t\in (0,T)$.

Remark 4.4

Counterexample 5.5 shows that in order for embeddedness to be preserved, it is crucial to assume that the initial local total curvature lies above $-\pi$.

5 Distance Comparison Principle for Closed Curves

We continue to adapt the methods from Huisken [18]. Let $X(\mathbb {S}^1)$ be a circle of radius R. Then

$$\begin{aligned} \frac{d(p,q)}{2} =\frac{L}{2\pi }\sin \!\left( \frac{\pi l(p,q)}{L}\right) \end{aligned}$$

for all $p,q\in \mathbb {S}^1$. This motivates the definition of the function $\psi :\mathbb {S}^1\!\times \mathbb {S}^1\!\times [0,T)\rightarrow \mathbb {R}$ with

$$\begin{aligned} \psi (p,q,t):=\frac{L_t}{\pi }\sin \!\left( \frac{\pi l(p,q,t)}{L_t}\right) \,, \end{aligned}$$

(5.1)

where $L_t<\infty$. We set $d/\psi \equiv 1$ on $\{p=q\}\times [0,T)$, then $d/\psi \in C^0(\mathbb {S}^1\!\times \mathbb {S}^1\!\times [0,T))$.

Remark 5.1

Since $\sin (\pi -\alpha )=\sin (\alpha )$, we have $\psi (p,q,t)=\psi (q,p,t)$. Hence, we will later assume that $l\le L/2$. Embedded curves satisfy $d/\psi >0$. If a closed curve $\Sigma _t$ is not a circle, then there exist $p,q\in \mathbb {S}^1$ so that $d(p,q,t)<\psi (p,q,t)$ and thus $\min _{\mathbb {S}^1\!\times \mathbb {S}^1}(d/\psi )<1$.

Lemma 5.2

Let $\Sigma =X(\mathbb {S}^1)$ be an embedded, closed curve. Let $p,q\in \mathbb {S}^1$, $p\ne q$, such that $\Sigma$ crosses the connecting line between X(p) and X(q). Then $(d/\psi )(p,q)$ cannot be a global minimum.

Proof

Let $\Sigma =X(\mathbb {S}^1)$ be an embedded, closed curve that crosses the connecting line between X(p) and X(q). That is, there exists an $r\in \mathbb {S}^1$, $r\ne p,q$, with $X(r)=X(p)+{\mathbf {w}}(p,q)\Vert X(r)-X(p)\Vert$. Set $d:=d(p,q)$, $d_1:=d(p,r)$ and $d_2:=d(r,q)$. Then

$$\begin{aligned} d=d_1+d_2\,. \end{aligned}$$

(5.2)

Furthermore, set $l:=l(p,q)$, $l_1:=l(p,r)$ and $l_2:=l(r,q)$ and $\psi :=\psi (p,q)$, $\psi _1:=\psi (p,r)$ and $\psi _2:=\psi (r,q)$ and assume that $d/\psi$ attains its global minimum at (p, q). We parametrise $\Sigma$ by arc length, so that $s(p)=0$. Then we have either $0=s(p)<s(r)<s(q)$ with $l=l_1+l_2$ and

$$\begin{aligned} \frac{\pi }{L}\psi =\sin \!\left( \frac{\pi l}{L}\right) <\sin \!\left( \frac{\pi l_1}{L}\right) +\sin \!\left( \frac{\pi l_2}{L}\right) =\frac{\pi }{L}\psi _1+\frac{\pi }{L}\psi _2\,, \end{aligned}$$

or we have $0=s(p)<s(q)<s(r)$ with $l=L-(l_1+l_2)$ and likewise

$$\begin{aligned} \frac{\pi }{L}\psi&=\sin \!\left( \frac{\pi l}{L}\right) =\sin \!\left( \frac{\pi L}{L}-\frac{\pi (l_1+l_2)}{L}\right) <\sin \!\left( \frac{\pi l_1}{L}\right) +\sin \!\left( \frac{\pi l_2}{L}\right) \\&=\frac{\pi }{L}\psi _1+\frac{\pi }{L}\psi _2\,. \end{aligned}$$

Since $d/\psi$ is a global minimum, we can estimate with (5.2) and the above,

$$\begin{aligned} \frac{d_1}{\psi _1}\ge \frac{d}{\psi }>\frac{d_1+d_2}{\psi _1+\psi _2} \qquad \text { and }\qquad \frac{d_2}{\psi _2}\ge \frac{d}{\psi }>\frac{d_1+d_2}{\psi _1+\psi _2} \end{aligned}$$

so that

$$\begin{aligned} d_1(\psi _1+\psi _2)>(d_1+d_2)\psi _1 \qquad \text { and }\qquad d_2(\psi _1+\psi _2)>(d_1+d_2)\psi _2\,. \end{aligned}$$

Adding both inequalities yields a contradiction. Thus, $d/\psi$ cannot be a global minimum. $\square$

Now we can prove a similar result to [18, Thm. 2.3].

Theorem 5.3

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded curve satisfying (1.8). Let $X:\mathbb {S}^1\!\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) satisfying (1.9) and with initial curve $\Sigma _0$. Then there exists a constant $c(\Sigma _0)>0$ such that

$$\begin{aligned} \inf _{(p,q,t)\in \mathbb {S}^1\!\times \mathbb {S}^1\!\times [0,T)}\frac{d}{\psi }(p,q,t)\ge c\,. \end{aligned}$$

Proof

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be an embedded closed curve satisfying (1.8). Lemma 3.2 implies that $\theta _0\in [-\pi ,3\pi ]$. From the maximum principle for $\theta$, Theorem 3.4, it follows that

$$\begin{aligned} \theta (p,q,t)\in (-\pi ,3\pi ) \end{aligned}$$

(5.3)

for all $p,q\in \mathbb {S}^1$ and $t\in (0,T)$. Since $d/\psi$ is continuous and initially positive, there exists a time $T'\in (0,T]$ so that $d/\psi >0$ on $[0,T')$. Fix $t_0\in (0,T')$. If $\Sigma _{t_0}$ is a circle, then Remark 5.1 yields that $d/\psi \equiv 1$ on $\mathbb {S}^1\!\times \mathbb {S}^1$. Assume that $\Sigma _{t_0}$ is not a circle so that $\min _{\mathbb {S}^1\!\times \mathbb {S}^1}(d/\psi )<1$ at $t_0$. Let $p,q\in \mathbb {S}^1$, $p\ne q$, be points where a local spatial minimum of $d/\psi$ at $t_0$ is attained and assume w.l.o.g. that $s(p,t_0)<s(q,t_0)$. Again by Remark 5.1, we can assume that $l(p,q,t_0)\le L_{t_0}/2$. We have for all $\xi \in T_{X(p,t_0)}\Sigma _{t_0}\bigoplus T_{X(q,t_0)}\Sigma _{t_0}$,

$$\begin{aligned} 0<\frac{d}{\psi }(p,q,t_0)<1\,,\quad \xi \!\left( \frac{d}{\psi }\right) \!(p,q,t_0)=0 \quad \text { and }\quad \xi ^2\!\left( \frac{d}{\psi }\right) \!(p,q,t_0)\ge 0\,. \end{aligned}$$

In the following, we always calculate at the point $(p,q,t_0)$. The spatial derivatives of d and $\psi$ are all given in [18] (for detailed calculations, see [5, Cor. 7.12 and Thm. 7.21]). The first spatial derivative of $d/\psi$ at $(p,t_0)$ in direction of the vector $\xi =\varvec{\tau }_p\oplus 0$ is given by

$$\begin{aligned} 0=(\varvec{\tau }_p\oplus 0)\!\left( \frac{d}{\psi }\right) =-\frac{1}{\psi }\langle {\mathbf {w}},\varvec{\tau }_p\rangle +\frac{d}{\psi ^2}\cos \!\left( \frac{\pi l}{L}\right) \,. \end{aligned}$$

(5.4)

At $(q,t_0)$ and for the vector $\xi =0\oplus \varvec{\tau }_q$, we have

$$\begin{aligned} 0=(0\oplus \varvec{\tau }_q)\!\left( \frac{d}{\psi }\right) =\frac{1}{\psi }\langle {\mathbf {w}},\varvec{\tau }_q\rangle -\frac{d}{\psi ^2}\cos \!\left( \frac{\pi l}{L}\right) \,. \end{aligned}$$

(5.5)

Since $l\in (0,L/2]$ and $d/\psi \in (0,1)$, and by (5.4) and (5.5), there exists $\beta \in (0,\pi ]$ with

$$\begin{aligned} \langle {\mathbf {w}},\varvec{\tau }_p\rangle =\langle {\mathbf {w}},\varvec{\tau }_q\rangle =\frac{d}{\psi }\cos \!\left( \frac{\pi l}{L}\right) =\cos \!\left( \frac{\beta }{2}\right) \in [0,1)\,. \end{aligned}$$

(5.6)

By Lemma 3.5 and (5.3), either

$$\begin{aligned} \langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =-\sin \!\left( \frac{\beta }{2}\right)&\quad \text { and }\quad 2\pi k+\beta =\theta \in (0,\pi ]\cup (2\pi ,3\pi )\,, \\ \langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =\sin \!\left( \frac{\beta }{2}\right)&\quad \text { and }\quad 2\pi k-\beta =\theta \in (-\pi ,0)\cup [\pi ,2\pi )\,,\quad \text { or } \\ \langle {\mathbf {w}},\varvec{\nu }_p\rangle =\langle {\mathbf {w}},\varvec{\nu }_q\rangle =\pm \sin \!\left( \frac{\beta }{2}\right)&\quad \text { and }\quad \theta \in \{0,2\pi \} \end{aligned}$$

for $k\in \mathbb {Z}$. We are now considering three different cases.

(i) Assume that

$$\begin{aligned} \langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =-\sin \!\left( \frac{\beta }{2}\right) \quad \text { and }\quad \beta =\theta \in (0,\pi ]\,. \end{aligned}$$

(5.7)

By (5.6), also

$$\begin{aligned} \langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =-\frac{d}{\psi }\sin \!\left( \frac{\pi l}{L}\right) \end{aligned}$$

so that

$$\begin{aligned} 2\sin \!\left( \frac{\theta }{2}\right) =\langle {\mathbf {w}},\varvec{\nu }_q-\varvec{\nu }_p\rangle =2\frac{d}{\psi }\sin \!\left( \frac{\pi l}{L}\right) \,. \end{aligned}$$

(5.8)

Like in the proof of Theorem 4.2, $\measuredangle ({\mathbf {w}},\varvec{\tau }_q+\varvec{\tau }_p)=0$. Using $\Vert {\mathbf {w}}\Vert =1$ and (5.6), we calculate

$$\begin{aligned} 0\le 2\frac{d}{\psi }\cos \!\left( \frac{\pi l}{L}\right) =\langle {\mathbf {w}},\varvec{\tau }_q+\varvec{\tau }_p\rangle =\Vert \varvec{\tau }_q+\varvec{\tau }_p\Vert \,. \end{aligned}$$

(5.9)

Since $d/\psi <1$, and again by (5.6),

$$\begin{aligned} \cos \!\left( \frac{\theta }{2}\right) =\frac{d}{\psi }\cos \!\left( \frac{\pi l}{L}\right) <\cos \!\left( \frac{\pi l}{L}\right) \,. \end{aligned}$$

(5.10)

As $\pi l/\!L\in (0,\pi /2]$ and the cosine function is axially symmetric and monotonically decreasing on $(0,\pi /2]$, (5.10) implies

$$\begin{aligned} \theta >\frac{2\pi l}{L}\,. \end{aligned}$$

(5.11)

By (5.7),

$$\begin{aligned} \langle {\mathbf {w}},\varvec{\kappa }_q-\varvec{\kappa }_p\rangle =-\kappa _q\langle {\mathbf {w}},\varvec{\nu }_q\rangle +\kappa _p\langle {\mathbf {w}},\varvec{\nu }_p\rangle =-(\kappa _p+\kappa _q)\sin \!\left( \frac{\theta }{2}\right) \,. \end{aligned}$$

(5.12)

We differentiate $d/\psi$ at $(p,q,t_0)$ twice with respect to the vector $\xi =\varvec{\tau }_p\ominus \varvec{\tau }_q$ and calculate, using the definition (5.1) of $\psi$, (5.9) and (5.12),

$$\begin{aligned} 0&\le (\varvec{\tau }_p\ominus \varvec{\tau }_q)^2\!\left( \frac{d}{\psi }\right) =-\frac{1}{\psi }(\kappa _p+\kappa _q)\sin \!\left( \frac{\theta }{2}\right) +\frac{4\pi ^2d}{L^2\psi }\,. \end{aligned}$$

We abbreviate $\kappa :=(\kappa _p+\kappa _q)/2$ and obtain

$$\begin{aligned} 2\kappa \sin \!\left( \frac{\theta }{2}\right) \le \frac{4\pi ^2}{L^2}d\,. \end{aligned}$$

(5.13)

Since the sine function is positive and monotonically increasing on $(0,\pi /2]$, we conclude with $d/\psi <1$, (5.1), (5.11) and (5.13) that

$$\begin{aligned} 2\kappa \sin \!\left( \frac{\theta }{2}\right) \le \frac{4\pi ^2}{L^2}d<\frac{4\pi ^2}{L^2}\psi =\frac{4\pi ^2}{L^2}\frac{L}{\pi }\sin \!\left( \frac{\pi l}{L}\right) <\frac{4\pi }{L}\sin \!\left( \frac{\theta }{2}\right) \,. \end{aligned}$$

Since $\sin (\theta /2)>0$ for $\theta \in (0,\pi ]$, we can divide by it to obtain with Cauchy–Schwarz,

$$\begin{aligned} \kappa <\frac{2\pi }{L}\le \frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t\,. \end{aligned}$$

(5.14)

The definition (3.2) of $\theta$ and (5.11) imply

$$\begin{aligned} -h\int _p^q\kappa \,ds_t+\int _p^q\kappa ^2\,ds_t \ge -h\theta +\frac{\theta ^2}{l} >-\theta \left( h-\frac{2\pi }{L}\right) . \end{aligned}$$

(5.15)

We use the evolution equation (1.1) and Lemma 2.1 to differentiate the ratio in time and obtain by (5.8), (5.10), (5.12) and (5.15),

$$\begin{aligned}&{\dfrac{\partial }{\partial t}}\!\left( \frac{d}{\psi }\right) \nonumber \\&\;=\frac{1}{\psi }\big (h\langle {\mathbf {w}},\varvec{\nu }_q-\varvec{\nu }_p\rangle +\langle {\mathbf {w}},\varvec{\kappa }_q-\varvec{\kappa }_p\rangle \big ) -\frac{d}{\psi ^2}\cos \!\left( \frac{\pi l}{L}\right) \left( h\int _p^q\kappa \,ds_t-\int _p^q\kappa ^2\,ds_t\right) \nonumber \\&\;\quad +\frac{d}{\pi \psi ^2}\left( \int _{\Sigma _t}\kappa ^2\,ds_t-2\pi h\right) \left( \sin \!\left( \frac{\pi l}{L}\right) -\frac{\pi l}{L}\cos \!\left( \frac{\pi l}{L}\right) \right) \nonumber \\&\;\ge \frac{2}{\psi }(h-\kappa )\sin \!\left( \frac{\theta }{2}\right) -\frac{\theta }{\psi }\cos \!\left( \frac{\theta }{2}\right) \left( h-\frac{\theta }{l}\right) \nonumber \\&\;\quad +\frac{1}{\psi }\left( \frac{1}{\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2h\right) \left( \sin \!\left( \frac{\theta }{2}\right) -\frac{\pi l}{L}\cos \!\left( \frac{\theta }{2}\right) \right) \nonumber \\&\;=\frac{2}{\psi }\left( \frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-\kappa \right) \left( \sin \!\left( \frac{\theta }{2}\right) -\frac{\theta }{2}\cos \!\left( \frac{\theta }{2}\right) \right) \nonumber \\&\;\quad +\frac{\theta }{\psi }\cos \!\left( \frac{\theta }{2}\right) \left( \frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-\kappa -h+\frac{\theta }{l}-\frac{l}{L\theta }\int _{\Sigma _t}\kappa ^2\,ds_t+\frac{h2\pi l}{L\theta }\right) \,, \end{aligned}$$

(5.16)

where we just added a zero in the last step. Since h satisfies (1.9), that is, $h\le \int _{\Sigma _t}\kappa ^2\,ds_t/2\pi +2\pi /L$, we estimate with (5.11) and (5.14),

$$\begin{aligned}&\frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-\kappa -h+\frac{\theta }{l}-\frac{l}{L\theta }\int _{\Sigma _t}\kappa ^2\,ds_t+\frac{h2\pi l}{L\theta } \nonumber \\&\quad>\left( \frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-h\right) \left( 1-\frac{2\pi l}{L\theta }\right) +\frac{\theta }{l}\left( 1-\frac{2\pi l}{L\theta }\right) \nonumber \\&\quad >\left( \frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t+\frac{2\pi }{L}-h\right) \left( 1-\frac{2\pi l}{L\theta }\right) \ge 0\,. \end{aligned}$$

(5.17)

Furthermore, the inequality

$$\begin{aligned} \sin \!\left( \frac{\theta }{2}\right) >\frac{\theta }{2}\cos \!\left( \frac{\theta }{2}\right) \end{aligned}$$

(5.18)

holds for all $\theta \in (0,\pi ]$. Thus, we conclude with, (5.14), (5.16), (5.17) and (5.18) that

$$\begin{aligned} {\dfrac{\partial }{\partial t}}\!\left( \frac{d}{\psi }\right) >0 \end{aligned}$$

at $(p,q,t_0)$.

(ii) Assume that $\langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =\sin (\beta /2)$ and $-\beta =\theta \in (-\pi ,0)$. Again by continuity of $d/\psi$, there exists $t_1(\Sigma _0)\in (0,T')$ so that

$$\begin{aligned} \min _{\mathbb {S}^1\!\times \mathbb {S}^1}\frac{d}{\psi }(\,\cdot \,,\,\cdot \,,t)\ge \frac{1}{2}\min _{\mathbb {S}^1\!\times \mathbb {S}^1}\frac{d}{\psi }(\,\cdot \,,\,\cdot \,,0) \end{aligned}$$

(5.19)

for all $t\in [0,t_1]$. For $t_0\in (t_1,T)$, Theorem 3.4 applied to the initial time $t_1$ and (5.3) yield $-\pi<\theta _{\min }(t_1)<\theta _{\min }(t_0)<0$ so that the monotone behaviour of the cosine on $(-\pi ,0)$ implies

$$\begin{aligned} \cos \!\left( \frac{\theta _{\min }(t_0)}{2}\right)>\cos \!\left( \frac{\theta _{\min }(t_1)}{2}\right) >0\,. \end{aligned}$$

(5.20)

From $0<l\le L/2$ it follows that $1>\cos (\pi l/\!L)\ge 0$ so that, by (5.6), (5.20) and again the monotone behaviour of the cosine on $(-\pi ,0)$,

$$\begin{aligned} \frac{d}{\psi }>\frac{d}{\psi }\cos \!\left( \frac{\pi l}{L}\right) =\cos \!\left( \frac{\theta }{2}\right) \ge \cos \!\left( \frac{\theta _{\min }(t_0)}{2}\right) >\cos \!\left( \frac{\theta _{\min }(t_1)}{2}\right) \,. \end{aligned}$$

(5.21)

With (5.19) and (5.21), we deduce that

$$\begin{aligned} \frac{d}{\psi }\ge \min \left\{ \frac{1}{2}\min _{\mathbb {S}^1\!\times \mathbb {S}^1}\frac{d}{\psi }(\,\cdot \,,\,\cdot \,,0),\cos \!\left( \frac{\theta _{\min }(t_1)}{2}\right) \right\} =:c(\Sigma _0)>0 \end{aligned}$$

at $(p,q,t_0)$.

(iii) Assume that $\theta \in \{0\}\cup (\pi ,3\pi )$ or $\theta =\pi$ and $\langle {\mathbf {w}},\varvec{\nu }_p\rangle =-\langle {\mathbf {w}},\varvec{\nu }_q\rangle =1$. By (5.6), $\langle {\mathbf {w}},\varvec{\tau }_q\rangle =\langle {\mathbf {w}},\varvec{\tau }_p\rangle \in [0,1)$. Since $\Sigma _{t_0}$ is closed and $X(\,\cdot \,,t_0)$ is continuous, $\Sigma _{t_0}$ has to cross the straight line segment between $X(p,t_0)$ and $X(q,t_0)$ at least once. Lemma 5.2 implies that the ratio $d/\psi$ cannot have a global minimum at $(p,q,t_0)$ (it could still, however, attain a local minimum at this point). Hence,

$$\begin{aligned} \frac{d}{\psi }>\min _{\mathbb {S}^1\!\times \mathbb {S}^1}\frac{d}{\psi }(\,\cdot \,,\,\cdot \,,t_0)\,, \end{aligned}$$

where $\min _{\mathbb {S}^1\!\times \mathbb {S}^1}(d/\psi )(\,\cdot \,,\,\cdot \,,t_0)$ is attained at a point which was treated in cases (i) and (ii).

Assume that $d/\psi$ falls below c and attains $\Lambda \in (0,c)$ for the first time at time $t_2\in (0,T)$ and points $p,q\in \mathbb {S}^1$, $p\ne q$, so that

$$\begin{aligned} c>\Lambda =\frac{d}{\psi }(p,q,t_2)=\min _{\mathbb {S}^1\!\times \mathbb {S}^1}\frac{d}{\psi }(\,\cdot \,,\,\cdot \,,t_2) \end{aligned}$$

(5.22)

is a global minimum and

$$\begin{aligned} {\dfrac{\partial }{\partial t}}_{|_{t=t_2}}\!\left( \frac{d}{\psi }\right) (p,q,t)\le 0\,. \end{aligned}$$

(5.23)

Case (i) contradicts (5.23), and cases (ii) and (iii) contradict (5.22). $\square$

Corollary 5.4

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded curve satisfying (1.8). Let $X:\mathbb {S}^1\!\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) satisfying (1.9) and with initial curve $\Sigma _0$. Then $\Sigma _t=X(\mathbb {S}^1\!,t)$ is embedded for all $t\in (0,T)$.

The next example shows, why the condition $\min \theta _0\ge -\pi$ is sharp.

Counterexample 5.5

Gage [8, p. 53] suggested the following counterexample. Pihan [25, Sect. 5.4] gave an incomplete proof for its validity which we will fix here. If we allow local total curvature smaller than $-\pi$, then there exist counterexamples for any given minimum $\min \theta _0<-\pi$. For the curve in Fig. 2, $\theta _{\min }=\theta (p_1,p_2)<-\pi$. We will construct a solution of (1.1) with embedded initial curve $\Sigma _0$ that intersects itself in finite time. Fix $K_0>0$. Let ${\mathcal {S}}$ be the set of all smooth, embedded curves in $\mathbb {R}^2$ that satisfy

$$\begin{aligned} \min \theta _0<-\pi \,,\quad \Vert X_0\Vert _{C^{3,\alpha }(\mathbb {S}^1)}\le K_0 \quad \text { and }\quad L(\Sigma )=L_0\ge 8\pi K_0\,, \end{aligned}$$

(5.24)

where $L_0$ is chosen big enough so that curves like in Fig. 2 are in ${\mathcal {S}}$. By the short time existence, see  [16, p. 36] or  [25, Thms. 4.3 and Corollary 4.4], there exists a time $T=T(K_0)$ so that

$$\begin{aligned} \Vert X\Vert _{C^{3,\alpha ;1\lfloor \alpha /2\rfloor }(\mathbb {S}^1\!\times [0,T/2])}\le K_1(K_0)\,. \end{aligned}$$

In particular,

$$\begin{aligned} \left| {\dfrac{\partial X^1}{\partial t}}(p,t)-{\dfrac{\partial X^1}{\partial t}}(p,0)\right| \le K_1t^{\alpha /2}\,, \end{aligned}$$

(5.25)

where $X^1:=\langle X,{\mathbf {e}}_1\rangle$, and, by (1.1) and (5.25),

$$\begin{aligned} -K_1t^{\alpha /2} \le (h(t)-\kappa (p,t))\varvec{\nu }^1(p,t)-(h(0)-\kappa (p,0))\varvec{\nu }^1(p,0) \le K_1t^{\alpha /2} \end{aligned}$$

(5.26)

for all $p\in \mathbb {S}^1$ and for all $t\in [0,T/2]$, where $\varvec{\nu }^1:=\langle \varvec{\nu },{\mathbf {e}}_1\rangle$. Assume $h(0)>0$ and set

$$\begin{aligned} t_1=t_1(K_0):=\min \left\{ \frac{T}{2},\left( \frac{h(0)}{2K_1}\right) ^{-\alpha /2}\right\} \,. \end{aligned}$$

(5.27)

Then (5.26) holds for $t\in [0,t_1]$. Let $\Sigma \in {\mathcal {S}}$ be a curve like in Fig. 2, which is symmetric about the $x_2$-axis. Let $p,q\in \mathbb {S}^1$ be located as in the picture so that

$$\begin{aligned} \varvec{\nu }(p,0)=-\varvec{\nu }(q,0)=-{\mathbf {e}}_1\,\qquad \text { and }\qquad \kappa (p,0)=\kappa (q,0)=0\,. \end{aligned}$$

(5.28)

We estimate with (1.1), (5.26), (5.27) and (5.28),

$$\begin{aligned} {\dfrac{\partial X^1}{\partial t}}(p,t)&=(h(t)-\kappa (p,t))\varvec{\nu }^1(p,t) \le (h(0)-\kappa (p,0))\varvec{\nu }^1(p,0)+K_1t_1^{\alpha /2} \nonumber \\&\le -h(0)+\frac{h(0)}{2} =-\frac{h(0)}{2} \end{aligned}$$

(5.29)

and likewise

$$\begin{aligned} {\dfrac{\partial X^1}{\partial t}}(q,t)&=(h(t)-\kappa (q,t))\varvec{\nu }^1(q,t) \ge (h(0)-\kappa (q,0))\varvec{\nu }^1(q,0)-K_1t_1^{\alpha /2} \nonumber \\&\ge h(0)-\frac{h(0)}{2} =\frac{h(0)}{2} \end{aligned}$$

(5.30)

for $t\in [0,t_1]$. Since $\min \theta _0<-\pi$, we can smoothly deform a curve like in Fig. 2 to achieve arbitrarily small distance between X(p, 0) and X(q, 0) without exceeding the upper bound $K_0$ in (5.24) or changing the length or enclosed area. Hence, we can choose an embedded initial curve $\Sigma _0$ with

$$\begin{aligned} X^1(p,0)=-X^1(q,0)\le \frac{h(0)t_1}{4}\,. \end{aligned}$$

(5.31)

Then, by (5.29) and (5.31)

$$\begin{aligned} X^1(p,t_1)=X^1(p,0)+\int _0^{t_1}{\dfrac{\partial X^1}{\partial t}}(p,t)\,dt \le \frac{h(0)t_1}{4}-\frac{h(0)t_1}{2}<0 \end{aligned}$$

and by (5.30) and (5.31)

$$\begin{aligned} X^1(q,t_1)=X^1(q,0)+\int _0^{t_1}{\dfrac{\partial X^1}{\partial t}}(q,t)\,dt \ge -\frac{h(0)t_1}{4}+\frac{h(0)t_1}{2}>0 \end{aligned}$$

so that the curve has crossed itself by the time $t_1$.

6 Singularity Analysis

Proposition 2.4 states that the curvature blows up if $T<\infty$. In this section, we assume $T<\infty$ and investigate curvature blow-ups for embedded flows (1.1) that satisfy (1.8), (1.9) and (1.10). We adapt techniques from the theory of CSF to show that the curvature does not blow up in finite time and conclude $T=\infty$.

Proposition 2.4 motivates the following definition. We say that a solution $X:{\mathcal {I}}\times [0,T)\rightarrow \mathbb {R}^2$ of (1.1) develops a singularity at $T\le \infty$ if $\max _{p\in \mathbb {S}^1}|\kappa (p,t)|\rightarrow \infty$ for $t\nearrow T$.

Lemma 6.1

Let $X:{\mathcal {I}}\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) satisfying (1.10) and with maximal time $T<\infty$. Then, for all $t\in (0,T)$,

$$\begin{aligned} \max _{p\in \mathbb {S}^1}|\kappa (p,t)|\ge \frac{1}{2\sqrt{T-t}}\,. \end{aligned}$$

Proof

The proof is as in [17, Lem. 1.2], see also [21, Prop. 4.1] or [5, Lem. 9.5]. $\square$

Like for CSF, we distinguish between two kinds of singularities according to the blow-up rate from Lemma 6.1. Let $X:{\mathcal {I}}\times (0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) with $T<\infty$. We say that a singularity is of type-I, if there exists a constant $C_0>0$ so that

$$\begin{aligned} \max _{p\in \mathbb {S}^1}|\kappa (p,t)|\le \frac{C_0}{\sqrt{T-t}} \end{aligned}$$

for all $t\in (0,T)$. A singularity is said to be of type-II, if such a constant does not exist, that is,

$$\begin{aligned} \limsup _{t\rightarrow T}\max _{p\in \mathbb {S}^1}|\kappa (p,t)|\sqrt{T-t}=\infty \,. \end{aligned}$$

Type-I singularities have already been exploited in [21, Sect. 4]. We refer also to [26, Sect. 11] for a characterisation of singularities for almost Brakke flows with bounded global terms, using a monotonicity formula and a result of [20].

Theorem 6.2

Mäder-Baumdicker  [21, Prop. 4.12] Let $X:{\mathcal {I}}\times (0,T)\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) satisfying (1.10) and with $T<\infty$. Then a type-I singularity cannot form at T.

Proof

In [21, Prop. 4.12], the theorem is only stated for convex curves. But the proof does not use the convexity, see also [5, Sect. 9.4]. By Corollaries 4.3 and 5.4, initially embedded curves stay embedded. Since the global term is bounded, it will vanish in any limit flow of a type-I rescaling where we rescale by the maximal curvature. Also, since the lengths of the curves are bounded away from zero, the curves of any limit flow will be of infinite length. Like in the analysis in [17] of type-I singularities of mean curvature flow, a monotonicity formula, see  [21, Proposition 4.9] or [5, Theorem 8.5], yields that any limit flow of a type-I rescaling is an embedded homothetically shrinking solution of CSF with non-vanishing curvature. By [1], this is an embedded shrinking circle. This contradicts the unbounded length. $\square$

To investigate type-II singularities, we want to rescale the curves $\Sigma _t$ near a singular point as $t\rightarrow T<\infty$. The following rescaling technique for type-II singularities was introduced in [13, Proof of Thm. 16.4] for Ricci flow and applied to type-II singularities of MCF in [19, p. 11]. Let $(p_k,t_k)_{k\in \mathbb {N}}$ be a sequence in ${\mathcal {I}}\times [0,T-1/k]$ with

$$\begin{aligned} T_k:=\kappa ^2(p_k,t_k)\left( T-\frac{1}{k}-t_k\right) =\max _{(p,t)\in \mathbb {S}^1\!\times [0,T-1/k]}\left( \kappa ^2(p,t)\left( T-\frac{1}{k}-t\right) \right) \end{aligned}$$

for each $k\in \mathbb {N}$. We set $\lambda _k^2:=\kappa ^2(p_k,t_k)$, $\alpha _k:=-\lambda _k^2t_k$ and define the rescaled embeddings $X_k:{\mathcal {I}}\times [\alpha _k,T_k]\rightarrow \mathbb {R}^2$ by

$$\begin{aligned} X_k(p,\tau ):=\lambda _k\left( X\!\left( p,t_k+\frac{\tau }{\lambda _k^2}\right) -X(p_k,t_k)\right) \,. \end{aligned}$$

(6.1)

Theorem 6.3

Let $X:{\mathcal {I}}\times (0,T)\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) with $T<\infty$ and satisfying (1.6) for ${\mathcal {I}}=\mathbb {R}$ and (1.10) for ${\mathcal {I}}\in \{\mathbb {S}^1,\mathbb {R}\}$. Then there exists a sequence of intervals $0\in I_k\subset \mathbb {R}$ and rescaled embeddings

$$\begin{aligned} \left( {\bar{X}}_{k}:I_k\times [\alpha _k,T_k]\rightarrow \mathbb {R}^2\right) _{k\in \mathbb {N}} \end{aligned}$$

that converges for $k\rightarrow \infty$ along a subsequence, uniformly and smoothly on compact subsets $I\times J\subset \mathbb {R}\times \mathbb {R}$ with $0\in I$ and compact subsets in $\mathbb {R}^2$ to a maximal, smooth, strictly convex or strictly concave limit solution $X_{\infty }:\mathbb {R}\times \mathbb {R}\rightarrow \mathbb {R}^2$ which satisfies

$$\begin{aligned} {\dfrac{\partial X_{\infty }}{\partial \tau }}(p,\tau )=-\kappa _{\infty }(p,\tau )\varvec{\nu }_{\infty }(p,\tau )\,. \end{aligned}$$

Moreover, $L(\Sigma _\tau ^{\infty })=\infty$ for all $\tau \in \mathbb {R}$, $X_{\infty }(0,0)=0$, $\sup _{\mathbb {R}\times \mathbb {R}}|\kappa _{\infty }|=|\kappa _{\infty }(0,0)|\,{=}\,1$.

Proof

The convergence follows similar lines to those of [6, Rem. 4.22(2)] and [21, Prop. 4.7]. For details, see also [5, Thm. 9.13]. The strict convexity/concavity is proofed like in [2, Thms. 5.14 and 7.7], where we use that, away from T, the coefficients in the evolution equation for the curvature are bounded and that $\int _{\Sigma _t}|\kappa |\,ds_t<\infty$ (see property (1.6) for ${\mathcal {I}}=\mathbb {R}$). A more detailed proof can be found in [22, Prop. 4.3.2] or in [5, Prop. 9.16]. $\square$

We now can proceed as in [18, Thm. 2.4].

Theorem 6.4

Let $\Sigma _0=X_0({\mathcal {I}})$ be a smooth, embedded curve satisfying (1.5) and (1.6) for ${\mathcal {I}}=\mathbb {R}$ as well as (1.8), (1.9) and (1.10) for ${\mathcal {I}}\in \{\mathbb {S}^1,\mathbb {R}\}$. Let $X:{\mathcal {I}}\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) with $T<\infty$ and initial curve $\Sigma _0$. Then a type-II singularity cannot form at T.

Proof

Theorem 6.3 yields that the limit flow consists of strictly convex or concave curves $\Sigma ^{\infty }_\tau$ for $\tau \in \mathbb {R}$ satisfying $\sup _{\mathbb {R}\times \mathbb {R}}|\kappa _{\infty }|=|\kappa _{\infty }(0,0)|=1$. If $\kappa _{\infty }<0$, we change the direction of parametrisation so that $\kappa _{\infty }>0$. Since the curvature attains its maximum at the point $(0,0)\in \mathbb {R}\times \mathbb {R}$, [14, Main Theorem B] yields that $X_{\infty }$ is a translating solution of CSF. [2, Thm. 8.16] implies that $\Sigma ^{\infty }_\tau$ is the grim reaper for every $\tau \in \mathbb {R}$. The grim reaper is asymptotic to two parallel lines of distance $\pi$ from inside. Let $\tau \in \mathbb {R}$. We can find a sequence of points $(p_j,q_j)_{j\in \mathbb {N}}$ in $\mathbb {R}\times \mathbb {R}$ with $d_{\infty }(p_j,q_j,\tau )\le \pi$ for all $j\in \mathbb {N}$ and $l_{\infty }(p_j,q_j,\tau )\rightarrow \infty$ for $j\rightarrow \infty$. Hence,

$$\begin{aligned} \inf _{\mathbb {R}\times \mathbb {R}}\frac{d_{\infty }}{l_{\infty }}(\,\cdot \,,\,\cdot \,,\tau )=0\,. \end{aligned}$$

However, like in [18, Thms. 2.4 and 2.5] (for details, see [5, Thm. 9.21]), the lower bound $\inf _{\mathbb {R}\times \mathbb {R}\times [0,T)}(d/l)\ge c$ from Theorem 4.2 and the lower bound $\inf _{\mathbb {S}^1\!\times \mathbb {S}^1\!\times [0,T)}(d/\psi )\ge c$ from Theorem 5.3 imply that

$$\begin{aligned} \inf _{\mathbb {R}\times \mathbb {R}}\frac{d_{\infty }}{l_{\infty }}(\,\cdot \,,\,\cdot \,,\tau )\ge c \end{aligned}$$

for every limit flow of rescalings according to (6.1). $\square$

Corollary 6.5

Let $\Sigma _0=X_0({\mathcal {I}})$ be a smooth, embedded curve satisfying (1.5) and (1.6) for ${\mathcal {I}}=\mathbb {R}$ as well as (1.8), (1.9) and (1.10) for ${\mathcal {I}}\in \{\mathbb {S}^1,\mathbb {R}\}$. Let $X:{\mathcal {I}}\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) with initial curve $\Sigma _0$. Then $T=\infty$.

Proof

By Theorems 6.2 and 6.4, neither a type-I nor a type-II singularity can form at T so that curvature stays bounded on [0, T] by a constant $C(\Sigma _0,T)$. We can extend the flow beyond T and repeat the above argument. Hence, for every time $T'<\infty$, there exists a constant $C(\Sigma _0,T')<\infty$ so that $\max _{p\in \mathbb {S}^1}|\kappa (p,t)|\le C$ for all $t\in [0,T')$. Applying Proposition 2.4 yields that the short time solution can be extended to a smooth solution on $(0,\infty )$. $\square$

7 Convexity in Finite Time

In this section, we show that a smooth, embedded solution $X:\mathbb {S}^1\!\times (0,\infty )\rightarrow \mathbb {R}^2$ of (1.1) with a global term h satisfying (1.11) becomes convex in finite time.

Remark 7.1

We observe that, by Lemma 2.3,

$$\begin{aligned} h=\frac{1}{L}\left( 2\pi +{\dfrac{d A}{d t}}\right) =\frac{1}{2\pi }\left( \int _{\Sigma _t}\kappa ^2\,ds_t+{\dfrac{d L}{d t}}\right) \end{aligned}$$

(7.1)

and

$$\begin{aligned} -{\dfrac{d }{d t}}\left( \frac{L^2}{4\pi }-A\right) =\frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \,. \end{aligned}$$

(7.2)

In respect of (7.2), choose

$$\begin{aligned} {\dfrac{d A}{d t}}=\gamma \left( \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right) \end{aligned}$$

(7.3)

and

$$\begin{aligned} \frac{L}{2\pi }{\dfrac{d L}{d t}} =\frac{1}{4\pi }{\dfrac{d L^2}{d t}} =-(1-\gamma )\left( \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right) \,, \end{aligned}$$

(7.4)

where $\gamma \in \mathbb {R}$. Then, (7.1) yields

$$\begin{aligned} h=(1-\gamma )\frac{2\pi }{L}+\frac{\gamma }{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t\,. \end{aligned}$$

For arbitrary $\gamma <0$, however, the positivity of h is not guaranteed.

Lemma 7.2

Let $X:\mathbb {S}^1\!\times [0,T)\rightarrow \mathbb {R}^2$ be a solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Then, A and L are monotone and there exist constants $0<c<C<\infty$ such that $c\le A,L\le C$ on [0, T) and

$$\begin{aligned} \frac{L}{2\pi }\int _{\Sigma _\tau }\kappa ^2\,ds_\tau -2\pi \in L^1([0,T))\,. \end{aligned}$$

Proof

By (7.3) and (7.4),

$$\begin{aligned} (1-\gamma ){\dfrac{d A}{d t}}=-\frac{\gamma }{4\pi }{\dfrac{d L^2}{d t}} \end{aligned}$$

(7.5)

so that, with $\delta \in (0,\infty )$ and

$$\begin{aligned} \gamma =\frac{(\delta -1)A_0}{L_0^2/4\pi -A_0}\in \left( -\frac{A_0}{L_0^2/4\pi -A_0},\infty \right) \,, \end{aligned}$$

integrating (7.5) yields

$$\begin{aligned} (1-\gamma )A_t+\gamma \frac{L^2_t}{4\pi } =(1-\gamma )A_0+\gamma \frac{L^2_0}{4\pi } =\delta A_0 \end{aligned}$$

(7.6)

for all $t\in (0,T)$. For $\delta \in (0,1)$, we have $\gamma <0$ and $-(1-\gamma )<0$, so that by (7.3) and (7.4),

$$\begin{aligned} {\dfrac{d A}{d t}}<0 \qquad \text { and }\qquad {\dfrac{d L}{d t}}<0\,. \end{aligned}$$

Hence, A and L are uniformly bounded away from infinity. By (7.6),

$$\begin{aligned} (1-\gamma )A_t >(1-\gamma )A_t+\gamma \frac{L^2_t}{4\pi } =\delta A_0 \end{aligned}$$

and so that, by the isoperimetric inequality, A and L are uniformly bounded away from zero. For $\delta \in [1,L_0^2/4\pi A_0]$, we have $\gamma \in [0,1]$ and by (7.3) and (7.4),

$$\begin{aligned} {\dfrac{d A}{d t}}\ge 0 \qquad \text { and }\qquad {\dfrac{d L}{d t}}\le 0\,. \end{aligned}$$

Hence, A and L are uniformly bounded away from zero and infinity. For $\delta >L_0^2/4\pi A_0$, we have $\gamma >1$ and by (7.3) and (7.4),

$$\begin{aligned} {\dfrac{d A}{d t}}>0 \qquad \text { and }\qquad {\dfrac{d L}{d t}}>0\,. \end{aligned}$$

Hence, A and L are uniformly bounded away from zero. By (7.6),

$$\begin{aligned} \gamma \frac{L^2_t}{4\pi } <(1-\gamma )A_t+\gamma \frac{L^2_t}{4\pi } =\delta A_0 \end{aligned}$$

and so that A and L are uniformly bounded away from infinity. The uniform bounds on the area and length from above and (7.2) yield

$$\begin{aligned} \frac{L}{2\pi }\int _{\Sigma _\tau }\kappa ^2\,ds_\tau -2\pi \,d\tau \in L^1([0,T))\,. \end{aligned}$$

                                                                   $\square$

Like in [21, Sect. 7], we use the following Gagliardo–Nirenberg interpolation inequality.

Theorem 7.3

(Gagliardo–Nirenberg interpolation inequality,  [23, p. 125], see also  [4, Thm. 3.70]) Let $f\in C^\infty (\mathbb {S}^1)$. Let $p>2$ and $\sigma \in [0,1)$ with $\sigma =1/2-1/p$. Then there exist constants $C_1=c_1(p,\sigma )$ and $C_2=c_2(p,\sigma )$ such that

$$\begin{aligned} \left( \int _{\mathbb {S}^1}|f|^p\,dx\right) ^{1/p}&\le C_1\left( \int _{\mathbb {S}^1}\left( {\dfrac{d f}{d x}}\right) ^{2}dx\right) ^{\sigma /2} \left( \int _{\mathbb {S}^1}f^2\,dx\right) ^{(1-\sigma )/2} \nonumber \\&\quad +C_2\left( \int _{\mathbb {S}^1}f^2\,dx\right) ^{1/2}\,. \end{aligned}$$

Lemma 7.4

(see proof of  [21, Cor. 7.5]) Let $f\in C^1((0,\infty ))\cap L^1((0,\infty ))$ with $f\ge 0$ and ${\dfrac{d }{d t}}f\le C(C+f)^3$ for $C\ge 0$. Then $f(t)\rightarrow 0$ for $t\rightarrow \infty$.

Lemma 7.5

Let $X:\mathbb {S}^1\!\times (0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) and h satisfying (1.11). Then there exists a constant $C>0$ such that, for all $t\in (0,\infty )$,

$$\begin{aligned} {\dfrac{d }{d t}}\left( \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right) \le C\left( C+\frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right) ^3\,. \end{aligned}$$

Proof

By Lemma 7.2, we can estimate

$$\begin{aligned} h\le C\left( 1+\int _{\Sigma _t}\kappa ^2\,ds_t\right) \,, \end{aligned}$$

(7.7)

where C depends on $\gamma$ and the lower bound on L. We deduce with Theorem 7.3 for $p=4$ and $\sigma =1/4$, the estimate $(a+b)^4\le C(a^4+b^4)$, and Young’s inequality for $p=q=2$,

$$\begin{aligned} \int _{\Sigma _t}\kappa ^4\,ds_t&\le \left( C\left( \int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t\right) ^{1/8} \left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^{3/8} +C\left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^{1/2}\right) ^4 \nonumber \\&\le \delta \int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t +C(\delta )\left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^3 +C\left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^{2} \end{aligned}$$

(7.8)

for a constants $C>0$. Again, by Theorem 7.3 for $p=3$ and $\sigma =1/6$, the estimate $(a+b)^3\le C(a^3+b^3)$, and Young’s inequality for $p=4$ and $q=4/3$,

$$\begin{aligned}&\int _{\Sigma _t}\kappa ^3\,ds_t \nonumber \\&\quad \le \left( C\left( \int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t\right) ^{1/12} \left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^{5/12} +C\left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^{1/2}\right) ^3 \end{aligned}$$

(7.9)

$$\begin{aligned}&\quad \le \delta \int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t +C(\delta )\left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^{5/3} +C\left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^{3/2}\,. \end{aligned}$$

(7.10)

Multiplying $\int _{\Sigma _t}\kappa ^2\,ds_t$ to (7.9) yields with Young’s inequality for $p=4$ and $q=4/3$ that

$$\begin{aligned}&\int _{\Sigma _t}\kappa ^2\,ds_t\int _{\Sigma _t}\kappa ^3\,ds_t \nonumber \\&\quad \le \left( C\left( \int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t\right) ^{1/12} \left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^{9/12} +C\left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^{5/6}\right) ^3\nonumber \\&\quad \le \delta \int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t +C(\delta )\left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^3 +C\left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^{5/2}. \end{aligned}$$

(7.11)

We use Lemma 2.1, integration by parts, (7.7), and (7.8), (7.10), (7.11) with $\delta =1/3$ to calculate,

$$\begin{aligned} {\dfrac{d }{d t}}\int _{\Sigma _t}\kappa ^2\,ds_t&=2\int _{\Sigma _t}\left( \kappa {\dfrac{\partial ^2\kappa }{\partial s^2}}-(h-\kappa )\kappa ^3\right) ds_t +\int _{\Sigma _t}\kappa ^3(h-\kappa )\,ds_t \nonumber \\&=-2\int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}ds_t -h\int _{\Sigma _t}\kappa ^3\,ds_t +\int _{\Sigma _t}\kappa ^4\,ds_t \nonumber \\&\le C\sum _{p=2,\frac{3}{2},\frac{5}{3},\frac{5}{2},3}\left( \int _{\Sigma _t}\kappa ^2\,ds_t\right) ^p \le C\left( 1+\int _{\Sigma _t}\kappa ^2\,ds_t\right) ^3 \end{aligned}$$

(7.12)

for all $t\in (0,\infty )$. By Lemma 2.3, the bounds on L from Lemma 7.2, (7.7) and (7.12),

$$\begin{aligned}&{\dfrac{d }{d t}}\left( \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right) =\frac{1}{2\pi }{\dfrac{d L}{d t}}\int _{\Sigma _t}\kappa ^2\,ds_t +\frac{L}{2\pi }{\dfrac{d }{d t}}\int _{\Sigma _t}\kappa ^2\,ds_t \nonumber \\&\quad \le \frac{1}{2\pi }\left( 2\pi h-\int _{\Sigma _t}\kappa ^2\,ds_t\right) \int _{\Sigma _t}\kappa ^2\,ds_t +C\left( 1+\int _{\Sigma _t}\kappa ^2\,ds_t\right) ^3 \nonumber \\&\quad \le C\left( 1+\int _{\Sigma _t}\kappa ^2\,ds_t\right) ^2 +C\left( 1+\int _{\Sigma _t}\kappa ^2\,ds_t\right) ^3 \nonumber \\&\quad \le C\left( C+\frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right) ^3\,. \end{aligned}$$

$\square$

Lemma 7.6

Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Then,

$$\begin{aligned} Lh\rightarrow 2\pi \qquad \text { and }\qquad \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t\rightarrow 2\pi \end{aligned}$$

for $t\rightarrow \infty$, and there exist a time $t_0\ge 0$ and constants $0<c<C<\infty$ such that $\inf _{[t_0,\infty )}h\ge c$ and

$$\begin{aligned} \sup _{[0,\infty )}h +\sup _{[0,\infty )}\left| {\dfrac{d h}{d t}}\right| +\sup _{[0,\infty )}\int _{\Sigma _t}\kappa ^2\,ds_t \le C\,. \end{aligned}$$

Proof

Lemmata 7.2, 7.4 and 7.5 yield that

$$\begin{aligned} \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t\rightarrow 2\pi \end{aligned}$$

(7.13)

for $t\rightarrow \infty$. By Lemma 7.2, L is bounded away from zero and infinity. By (7.3) and (7.13),

$$\begin{aligned} {\dfrac{d A}{d t}}=\int _{\Sigma _t}(h-\kappa )\,ds_t=Lh-2\pi \rightarrow 0 \end{aligned}$$

for $t\rightarrow \infty$. Hence, there exist a time $t_0\in [0,\infty )$ and constants $0<c<C<\infty$ so that $c\le h\le C$ on $[t_0,\infty )$. By (7.4) and (7.13),

$$\begin{aligned} {\dfrac{d L}{d t}}=\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi h\rightarrow 0 \end{aligned}$$

for $t\rightarrow \infty$ so that there exists $0<C<\infty$ with $\big |{\dfrac{d }{d t}}L\big |+\int _{\Sigma _t}\kappa ^2\,ds_t\le C$ on $[0,\infty )$. This yields $\big |{\dfrac{d h}{d t}}\big |\le C$. $\square$

Lemma 7.7

Let $X:\mathbb {S}^1\!\times (0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1). Then there exists a constant $C>0$ such that

$$\begin{aligned} {\dfrac{d }{d t}}\int _{\Sigma _t}(h-\kappa )^2\,ds_t&\le -\int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t +C{\dfrac{d h}{d t}}\int _{\Sigma _t}(h-\kappa )\,ds_t \\&\quad +C\sum _{i=1}^5\left( \int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^{p_i}h^{q_i} \end{aligned}$$

for all $t\in (0,\infty )$, where $p_i\in [1,3]$ and $q_i\in [0,2]$.

Proof

We follow the lines of [21, Lems. 7.3 and 7.4]. Write $\kappa =h-(h-\kappa )$. Then

$$\begin{aligned} (h-\kappa )^3\kappa =h(h-\kappa )^3-(h-\kappa )^4 \end{aligned}$$

and

$$\begin{aligned} (h-\kappa )^2\kappa ^2 =h^2(h-\kappa )^2-2h(h-\kappa )^3+(h-\kappa )^4\,. \end{aligned}$$

Lemma 2.1 and integration by parts yields

$$\begin{aligned}&{\dfrac{d }{d t}}\int _{\Sigma _t}(h-\kappa )^2\,ds_t \nonumber \\&\,=\int _{\Sigma _t}(h-\kappa )^3\kappa \,ds_t +2{\dfrac{d h}{d t}}\int _{\Sigma _t}(h-\kappa )\,ds_t +2\int _{\Sigma _t}(h-\kappa ) \left( -{\dfrac{\partial ^2\kappa }{\partial s^2}}+(h-\kappa )\kappa ^2\right) \,ds_t \nonumber \\&\,=-2\int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}ds_t +2{\dfrac{d h}{d t}}\int _{\Sigma _t}(h-\kappa )\,ds_t +\int _{\Sigma _t}(h-\kappa )^4\,ds_t \nonumber \\&\qquad -3h\int _{\Sigma _t}(h-\kappa )^3\,ds_t +2h^2\int _{\Sigma _t}(h-\kappa )^2\,ds_t\,. \end{aligned}$$

(7.14)

Like in [21, Cor. 7.4], we use Theorem 7.3 with $p=4$ and $\sigma =1/4$ and Young’s inequality with $p=4/3$ and $q=4$ as well as for $p=q=2$, to estimate

$$\begin{aligned}&\int _{\Sigma _t}(h-\kappa )^4\,ds_t \nonumber \\&\quad \le \left( C\left( \int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t\right) ^{1/8} \left( \int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^{3/8} +C\left( \int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^{1/2}\right) ^4 \nonumber \\&\quad \le \frac{1}{2}\int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t +C\left( \int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^3 +C\left( \int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^{2}. \end{aligned}$$

(7.15)

Again by Theorem 7.3 with $p=3$ and $\sigma =1/6$ and Young’s inequality for $p=3/2$ and $q=3$ as well as for $p=4$ and $q=4/3$ we obtain

$$\begin{aligned} 3h\int _{\Sigma _t}(h-\kappa )^3\,ds_t&\le 3h\left( C\left( \int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t\right) ^{1/12} \left( \int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^{5/12}\right. \nonumber \\&\quad +\left. C\left( \int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^{1/2}\right) ^3 \nonumber \\&\le \frac{1}{2}\int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t +Ch^{4/3}\left( \int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^{5/3} \nonumber \\&\quad +Ch\left( \int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^{3/2}. \end{aligned}$$

(7.16)

Altogether, (7.14), (7.15), (7.16) yield the claim. $\square$

Lemma 7.8

Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Then

$$\begin{aligned} \int _{\Sigma _t}(h-\kappa )^2\,ds_t\rightarrow 0 \end{aligned}$$

for $t\rightarrow \infty$ and

$$\begin{aligned} \int _0^\infty \int _{\Sigma _t}(h-\kappa )^2\,ds_t\,dt+\int _0^\infty \int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t\,dt<\infty \,. \end{aligned}$$

Proof

Similar to [16, p. 47], Lemma 2.3 yields

$$\begin{aligned} \int _{\Sigma _t}(h-\kappa )^2\,ds_t =Lh^2-4\pi h+\int _{\Sigma _t}\kappa ^2\,ds_t =h{\dfrac{d A}{d t}}-{\dfrac{d L}{d t}}\,. \end{aligned}$$

Lemmata 7.2 and 7.6 imply for $0<\varepsilon<\tau <\infty$,

$$\begin{aligned} \int _\varepsilon ^\tau \int _{\Sigma _t}(h-\kappa )^2\,ds_t\,dt =\sup _{t\in [\varepsilon ,\tau ]}h(t)(A_\tau -A_\varepsilon )+(L_\varepsilon -L_\tau )\le C\,. \end{aligned}$$

We let $\varepsilon \rightarrow 0$ and $\tau \rightarrow \infty$ to obtain

$$\begin{aligned} \int _0^\infty \int _{\Sigma _t}(h-\kappa )^2\,ds_t\,dt<\infty \,. \end{aligned}$$

(7.17)

By Lemma 7.6,

$$\begin{aligned} {\dfrac{d h}{d t}}\int _{\Sigma _t}(h-\kappa )\,ds_t\,dt \le \sup _{[0,\infty )}\left| {\dfrac{d h}{d t}}\right| |Lh-2\pi | \le C\,, \end{aligned}$$

so that Lemma 7.7 implies

$$\begin{aligned} {\dfrac{d }{d t}}\int _{\Sigma _t}(h-\kappa )^2\,ds_t \le C\left( 1+\int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^3\,. \end{aligned}$$

Like in [21, Cor. 7.5], Lemma 7.4 yields

$$\begin{aligned} \int _{\Sigma _t}(h-\kappa )^2\,ds_t\rightarrow 0 \end{aligned}$$

for $t\rightarrow \infty$. Consequently, there exists a time $t_0\ge 0$ so that

$$\begin{aligned} \int _{\Sigma _t}(h-\kappa )^2\,ds_t<1 \end{aligned}$$

for all $t>t_0$, and thus

$$\begin{aligned} \left( \int _{\Sigma _t}(h-\kappa )^2\,ds_t\right) ^{p}\le \int _{\Sigma _t}(h-\kappa )^2\,ds_t \end{aligned}$$

(7.18)

for all $p\ge 1$ and $t>t_0$. By Lemma 7.2, ${\dfrac{d }{d t}}A$ has a sign so that

$$\begin{aligned} \int _\varepsilon ^\tau \left| \int _{\Sigma _t}(h-\kappa )\,ds_t\right| \,dt =|A_\tau -A_\varepsilon | \le C\,, \end{aligned}$$

where $C>0$ is independent of time. Sending $\varepsilon \rightarrow 0$ and $\tau \rightarrow \infty$ yields with Lemma 7.6,

$$\begin{aligned} \int _0^\infty {\dfrac{d h}{d t}}\int _{\Sigma _t}(h-\kappa )\,ds_t\,dt \le \sup _{[0,\infty )}\left| {\dfrac{d h}{d t}}\right| \int _0^\infty \left| \int _{\Sigma _t}(h-\kappa )\,ds_t\right| \,dt \le C. \end{aligned}$$

Thus, with Lemma 7.7, (7.17) and (7.18) we obtain

$$\begin{aligned} \int _{t_0}^\infty \int _{\Sigma _t}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_t\,dt \le \int _{\Sigma _{t_0}}(h-\kappa )^2\,ds_t\,dt +C+C\int _{t_0}^\infty \int _{\Sigma _t}(h-\kappa )^2\,ds_t\,dt<\infty \,. \end{aligned}$$

Since $\Sigma _t$ is smooth for $t\in [0,t_0]$, the claim follows. $\square$

Theorem 7.9

Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Then there exists a time $T_0\ge 0$ such that $\Sigma _t$ is strictly convex for $t>T_0$.

Proof

By Lemma 7.6,

$$\begin{aligned} h\ge c_h>0 \end{aligned}$$

(7.19)

on $[t_0,\infty )$ for $t_0\ge 0$ and $c_h>0$. Lemma 7.8 implies that there exists a sequence $(t_k)_{k\in \mathbb {N}}$ with $t_k\rightarrow \infty$ for $k\rightarrow \infty$ so that

$$\begin{aligned} \int _{\Sigma _{t_k}}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}ds_{t_k}\rightarrow 0 \end{aligned}$$

(7.20)

for $k\rightarrow \infty$. Hence, there exists $k_0\in \mathbb {N}$ so that for all $k\ge k_0$

$$\begin{aligned} \int _{\Sigma _{t_k}}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}ds_{t_k}<1\,. \end{aligned}$$

(7.21)

We employ [10, Thm. 7.26(ii)] to obtain that $W^{1,2}(\mathbb {S}^1)$ is compactly embedded in $C^0(\mathbb {S}^1)$. Furthermore, $C^0(\mathbb {S}^1)\subset L^2(\mathbb {S}^1)$, and $\Vert f\Vert _{L^2(\mathbb {S}^1)} \le \sqrt{2\pi }\Vert f\Vert _{C^0(\mathbb {S}^1)}$ for every $f\in C^0(\mathbb {S}^1)$. Hence, $C^0(\mathbb {S}^1)$ is continuously embedded in $L^2(\mathbb {S}^1)$. Let $f\in W^{1,2}(\mathbb {S}^1)$. By Ehrling’s lemma, for all $\varepsilon >0$, there exists a constant $C(\varepsilon )>0$ so that

$$\begin{aligned} \Vert f\Vert _{C^0(\mathbb {S}^1)}\le \varepsilon \Vert f\Vert _{W^{1,2}(\mathbb {S}^1)}+C(\varepsilon )\Vert f\Vert _{L^2(\mathbb {S}^1)}\,. \end{aligned}$$

(7.22)

Lemma 7.8 and (7.20) yield $h(t_k)-\kappa (\,\cdot \,,t_k)\in W^{1,2}(\mathbb {S}^1)$ for each $k\in \mathbb {N}$. Hence, we can use (7.21) and (7.22) to estimate

$$\begin{aligned} \max _{p\in \mathbb {S}^1}|h(t_k)-\kappa (p,t_k)|&\le \varepsilon \left( \int _{\Sigma _{t_k}}\left( {\dfrac{\partial \kappa }{\partial s}}\right) ^{2}\,ds_{t_k}\right) ^{1/2} +\varepsilon \left( \int _{\Sigma _{t_k}}(h-\kappa )^2\,ds_{t_k}\right) ^{1/2} \nonumber \\&\quad +C(\varepsilon )\left( \int _{\Sigma _{t_k}}(h-\kappa )^2\,ds_{t_k}\right) ^{1/2} \nonumber \\&\le \varepsilon +C(\varepsilon )\left( \int _{\Sigma _{t_k}}(h-\kappa )^2\,ds_{t_k}\right) ^{1/2} \end{aligned}$$

(7.23)

for all $k\ge k_0$. Choose $\varepsilon =c_h/4$ to deduce with (7.23)

$$\begin{aligned} \max _{p\in \mathbb {S}^1}|h(t_k)-\kappa (p,t_k)| \le \frac{c_h}{4}+C\left( \int _{\Sigma _{t_k}}(h-\kappa )^2\,ds_{t_k}\right) ^{1/2}\,. \end{aligned}$$

(7.24)

Lemma 7.8 implies that there exists $k_1\ge k_0$ so that for all $k\ge k_1$

$$\begin{aligned} \int _{\Sigma _{t_k}}(h-\kappa )^2\,ds_{t_k}<\left( \frac{c_h}{4C}\right) ^2\,. \end{aligned}$$

By (7.24),

$$\begin{aligned} \max _{p\in \mathbb {S}^1}|h(t_{k_1})-\kappa (p,t_{k_1})|\le \frac{c_h}{2}\,. \end{aligned}$$

With (7.19), we conclude that $\kappa >0$ at $t_{k_1}$. From Corollary 2.2 it follows that $\kappa >0$ for all $t>t_{k_1}$. Hence, the claim holds for $T_0=t_{k_1}$. $\square$

8 Longtime Behaviour

In this section, we show that convex solutions of (1.1) that exist for all positive times converge exponentially and smoothly to a round circle. This was already shown in [8] for the APCSF and in [25] for the LPCF. We repeat and extend the arguments here for h satisfying (1.11) for the sake of completeness. We mostly follow the lines of [9, Sect. 5] for rescaled convex CSF, [8] for convex APCSF, and [25, Chap. 7] for convex LPCF. For further details, see [5, Chap. 11].

Lemma 8.1

Isoperimetric inequality, Gage [7] For a closed, convex $C^2$-curve in the plane,

$$\begin{aligned} \int _{\Sigma }\kappa ^2\,ds\ge \frac{\pi L}{A} \end{aligned}$$

with equality if and only if the curve is a circle.

Lemma 8.2

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded, convex curve. Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a solution of (1.1) with initial curve $\Sigma _0$. Then there exists a constant $C=C(\Sigma _0)>0$, such that, for all $t>0$,

$$\begin{aligned} \left( \frac{L^2}{A}-4\pi \right) \le C\exp \!\left( -\int _0^t\frac{2\pi }{A}\,d\tau -\log \frac{A_t}{A_0}\right) \,. \end{aligned}$$

Proof

We follow the lines of  [8, Cor. 2.4] and  [25, Lem. 7.7] and use Lemma 8.1 to estimate for $t>0$

$$\begin{aligned} {\dfrac{d }{d t}}\!\left( \frac{L^2}{A}-4\pi \right)&=-\frac{2L}{A}\left( \int _{\Sigma _t}\kappa ^2\,ds_t-2\pi h\right) -\frac{L^2}{A^2}\left( Lh-2\pi \right) \\&\le -\frac{L}{A}\left( \frac{2\pi L}{A}-4\pi h+\frac{L^2}{A}h-\frac{2\pi L}{A}\right) =-\frac{hL}{A}\left( \frac{L^2}{A}-4\pi \right) \,. \end{aligned}$$

By (7.1),

$$\begin{aligned} \qquad \qquad \int _0^t\frac{hL}{A}\,d\tau =\int _0^t\frac{2\pi }{A}+{\dfrac{d }{d t}}\log A\,d\tau =\int _0^t\frac{2\pi }{A}\,d\tau +\log \frac{A_t}{A_0}\,.\qquad \qquad \qquad \qquad \qquad \end{aligned}$$

$\square$

Proposition 8.3

(Bonnesen isoperimetric inequality,  [24, Thm. 4 (21)]) For an embedded, closed curve $\Sigma$ in the plane,

$$\begin{aligned} \frac{L^2}{A}-4\pi \ge \frac{\pi ^2}{A}(r_{{{\,\mathrm{circ}\,}}}-r_{{{\,\mathrm{in}\,}}})^2\ge 0\,, \end{aligned}$$

where $r_{{{\,\mathrm{circ}\,}}}$ and $r_{{{\,\mathrm{in}\,}}}$ are the circumscribed and inscribed radius of $\Sigma$.

Proposition 8.4

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded, convex curve. Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Then

$$\begin{aligned} r_{{{\,\mathrm{circ}\,}}}(t)-r_{{{\,\mathrm{in}\,}}}(t)\rightarrow 0 \end{aligned}$$

for $t\rightarrow \infty$ and $\Sigma _t=X(\mathbb {S}^1\!,t)$ converges in $C^0$ to a circle of radius

$$\begin{aligned} R:=\lim _{t\rightarrow \infty }\frac{L_t}{2\pi }=\lim _{t\rightarrow \infty }\sqrt{\frac{A_t}{\pi }}\in (0,\infty )\,. \end{aligned}$$

Moreover, for all $\beta \in (0,1)$ there exist a time $t_0>0$ and a constant $C>0$ such that, for all $t\ge t_0$,

$$\begin{aligned} \left( \frac{L^2}{4\pi }-A\right) \le C\exp \!\left( -\frac{2\beta t}{R^2}\right) \quad \text { and }\quad \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \le C\exp \!\left( -\frac{\beta t}{R^2}\right) \,. \end{aligned}$$

Proof

By Lemma 7.2,

$$\begin{aligned} \int _0^t\frac{2\pi }{A}\,d\tau +\log \frac{A_t}{A_0} \ge \frac{2\pi t}{C}+\log \frac{c}{A_0} \rightarrow \infty \end{aligned}$$

for $t\rightarrow \infty$. Lemma 8.2 and the bounds from Lemma 7.2 imply

$$\begin{aligned} \frac{L}{2\pi }-\sqrt{\frac{A}{\pi }}\rightarrow 0 \end{aligned}$$

for $t\rightarrow \infty$. Also, $L/2\pi =\sqrt{A/\pi }$ only holds on a circle. Proposition 8.3 yields $r_{{{\,\mathrm{circ}\,}}}(t)-r_{{{\,\mathrm{in}\,}}}(t)\rightarrow 0$ for $t\rightarrow \infty$. Let $\beta \in (0,1)$ and $\varepsilon (\beta ,R)>0$ so that

$$\begin{aligned} \left( 1-\varepsilon R^2\right) \ge \beta \,. \end{aligned}$$

We can choose $t_0(\beta )>0$ so that for all $t\ge t_0$,

$$\begin{aligned} \left( \frac{1}{R^2}-\varepsilon \right) \le \frac{\pi }{A}\,. \end{aligned}$$

Hence,

$$\begin{aligned} -\int _0^t\frac{2\pi }{A}\,d\tau \le -2\left( 1-\varepsilon R^2\right) \frac{t}{R^2} \le -\frac{2\beta t}{R^2} \end{aligned}$$

and again by the bounds on A from Lemma 7.2,

$$\begin{aligned} \left( \frac{L^2}{4\pi }-A\right) =\frac{A}{4\pi }\left( \frac{L^2}{A}-4\pi \right) \le C\exp \!\left( -\frac{2\beta t}{R^2}\right) \end{aligned}$$

(8.1)

for all $t\ge t_0$. Let $f\in C^2([0,\infty ))$. Since $C^2([0,\infty ))$ is compactly embedded in $C^1([0,\infty ))$ and $C^1([0,\infty ))$ is continuously embedded in $C^0([0,\infty ))$, Ehrling’s Lemma yields that for every $\delta >0$, there exists $C(\delta )>0$ so that

$$\begin{aligned} \Vert f\Vert _{C^1([0,\infty ))}\le \delta \Vert f\Vert _{C^2([0,\infty ))}+C(\delta )\Vert f\Vert _{C^0([0,\infty ))}\,. \end{aligned}$$

We set $\delta =1/2$ and conclude

$$\begin{aligned} \sup _{[0,\infty )}\left| {\dfrac{d f}{d t}}\right| \le \sup _{[0,\infty )}\left| {\dfrac{d ^2f}{d t^2}}\right| +C\sup _{[0,\infty )}|f|\,. \end{aligned}$$

(8.2)

Let $\eta >0$ and define $f_\eta :[0,\infty )\rightarrow \mathbb {R}$ by $f_\eta (t):=f(\eta t)$. Then

$$\begin{aligned} {\dfrac{d f_\eta }{d t}}(t)=\eta {\dfrac{d f}{d t}}(\eta t) \qquad \text { and }\qquad {\dfrac{d ^2f_\eta }{d t^2}}(t)=\eta ^2{\dfrac{d ^2f}{d t^2}}(\eta t) \end{aligned}$$

as well as with (8.2),

$$\begin{aligned} \sup _{[0,\infty )}\left| {\dfrac{d f}{d t}}\right|&=\frac{1}{\eta }\sup _{[0,\infty )}\left| {\dfrac{d f_\eta }{d t}}\right| \le \frac{1}{\eta }\sup _{[0,\infty )}\left| {\dfrac{d ^2f_\eta }{d t^2}}\right| +\frac{C}{\eta }\sup _{[0,\infty )}|f_\eta | \nonumber \\&=\eta \sup _{[0,\infty )}\left| {\dfrac{d ^2f}{d t^2}}\right| +\frac{C}{\eta }\sup _{[0,\infty )}|f|\,. \end{aligned}$$

(8.3)

By Lemmata 7.5 and 7.6, there exists a time $t_1\ge t_0$ so that for all $t\ge t_1$,

$$\begin{aligned} {\dfrac{d }{d t}}\left( \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right) \le C\left( C+\frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right) ^3 \le C(C+1)^3\,. \end{aligned}$$

(8.4)

We choose

$$\begin{aligned} \eta =\exp \!\left( -\frac{\beta t}{R^2}\right) \end{aligned}$$

to obtain by (7.2), (8.1), (8.3) and (8.4), for all $t\ge t_1$,

$$\begin{aligned} \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi&\le \sup _{[t,\infty )}\left( \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right) \\&\le \eta c\sup _{[t,\infty )}\left( 1+\frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right) ^3 +\frac{C}{\eta }\sup _{[t,\infty )}\left| \frac{L^2}{4\pi }-A\right| \\&\le C\exp \!\left( -\frac{\beta t}{R^2}\right) \,. \end{aligned}$$

$\square$

By Corollary 2.2, $\Sigma _t$ is strictly convex for all $t>0$. Like introduced in Sect. 3, let $\vartheta :\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}$ be the angle between the $x_1$-axis and the tangent vector at the point X(p, t). Since $\Sigma _t$ is strictly convex on $(0,\infty )$, $\vartheta (\,\cdot \,,t)$ is injective for each $t\in (0,\infty )$. We want to use $\vartheta$ as spatial coordinate and define $\tau$ to be a new time variable so that $\tau =t$ as well as

$$\begin{aligned} {\dfrac{d \tau }{d t}}=1\qquad \text { and }\qquad {\dfrac{\partial \vartheta }{\partial \tau }}=0\,. \end{aligned}$$

(8.5)

The spatial derivative transforms according to $\frac{1}{v}{\dfrac{\partial }{\partial p}}={\dfrac{\partial }{\partial s}}=\kappa {\dfrac{\partial }{\partial \vartheta }}$. In the following, we use the coordinates $(\vartheta ,\tau )$ on $\mathbb {S}^1\!\times (0,\infty )$.

Lemma 8.5

(Gage–Hamilton  [9, Lem. 4.1.3] and Pihan  [25, Lem. 6.12]) Let $X:\mathbb {S}^1\!\times (0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, strictly convex solution of (1.1). Then, for $\tau \in (0,\infty )$,

$$\begin{aligned} {\dfrac{\partial \kappa }{\partial \tau }}=\kappa ^2{\dfrac{\partial ^2\kappa }{\partial \vartheta ^2}}-(h-\kappa )\kappa ^2\,. \end{aligned}$$

For $\tau >0$, we define

$$\begin{aligned} m(\tau ):=\max _{{\bar{\tau }}\in [0,\tau ]}\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,{\bar{\tau }})\,. \end{aligned}$$

(8.6)

Lemma 8.6

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded, convex curve. Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Then there exists a constant $C(\Sigma _0)>0$ such that, for all $\tau >0$,

$$\begin{aligned} \int _{\mathbb {S}^1}\left( {\dfrac{\partial \kappa }{\partial \vartheta }}\right) ^{2}d\vartheta \le \int _{\mathbb {S}^1}\kappa ^2\,d\vartheta +C(m(\tau )+1)\,. \end{aligned}$$

Proof

We follow similar lines to [8, Lem. 3.4 and Cor. 3.5] and [25, Lem. 6.9]. We observe that

$$\begin{aligned} 0<\int _{\Sigma _t}\kappa ^2\,ds_t =\int _{\mathbb {S}^1}\kappa \,d\vartheta \le 2\pi \max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau ) \end{aligned}$$

(8.7)

and use the time independency (8.5) of $\vartheta$, Lemma 8.5, integration by parts to estimate

$$\begin{aligned} {\dfrac{d }{d \tau }}\int _{\mathbb {S}^1}\left( \kappa ^2-\left( {\dfrac{\partial \kappa }{\partial \vartheta }}\right) ^{2}-2h\kappa \right) d\vartheta&=\int _{\mathbb {S}^1}2\left( \kappa +{\dfrac{\partial ^2\kappa }{\partial ^2\vartheta }}-h\right) {\dfrac{\partial \kappa }{\partial \tau }}\,d\vartheta -2{\dfrac{d h}{d \tau }}\int _{\mathbb {S}^1}\kappa \,d\vartheta \nonumber \\&=2\int _{\mathbb {S}^1}\kappa ^2\left( \kappa +{\dfrac{\partial ^2\kappa }{\partial ^2\vartheta }}-h\right) ^{2}d\vartheta -2{\dfrac{d h}{d \tau }}\int _{\mathbb {S}^1}\kappa \,d\vartheta \nonumber \\&\ge -2{\dfrac{d h}{d \tau }}\int _{\mathbb {S}^1}\kappa \,d\vartheta \end{aligned}$$

(8.8)

for all $\tau >0$. By (1.11),

$$\begin{aligned} {\dfrac{d h}{d \tau }}\int _{\mathbb {S}^1}\kappa \,d\vartheta =-(1-\gamma )\frac{2\pi }{L^2}{\dfrac{d L}{d \tau }}\int _{\mathbb {S}^1}\kappa \,d\vartheta +\frac{\gamma }{4\pi }{\dfrac{d }{d \tau }}\left( \int _{\mathbb {S}^1}\kappa \,d\vartheta \right) ^2\,. \end{aligned}$$

(8.9)

By Lemma 7.2, ${\dfrac{d }{d \tau }}L$ has a sign so that

$$\begin{aligned} \int _\varepsilon ^\tau \left| {\dfrac{d L}{d {\bar{\tau }}}}\right| \,d{\bar{\tau }}\le |L_\tau -L_0|\le C \end{aligned}$$

for all $0<\varepsilon<\tau <\infty$. We integrate (8.9) from $\varepsilon$ to $\tau$ and conclude with $\varepsilon \rightarrow 0$, the upper bound from Lemma 7.6, the definition (8.6) of m and (8.7),

$$\begin{aligned} \int _0^\tau {\dfrac{d h}{d {\bar{\tau }}}}\int _{\mathbb {S}^1}\kappa \,d\vartheta \,d{\bar{\tau }} \le C\max _{{\bar{\tau }}\in [0,\tau ]}\int _{\mathbb {S}^1}\kappa \,d\vartheta +C\left( \int _{\mathbb {S}^1}\kappa (\vartheta ,\tau )\,d\vartheta \right) ^2 \le C(m(\tau )+1) \end{aligned}$$

for all $\tau \in (0,\infty )$. Hence, integrating (8.8) and the bounds from Lemma 7.6 yield the claim. $\square$

For $\tau >0$, define

$$\begin{aligned} m^*(\tau ):=1+\frac{\sqrt{m(\tau )+1}}{\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )}\,. \end{aligned}$$

(8.10)

Lemma 8.7

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded, convex curve. Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Let $\tau >0$, $\vartheta _1,\vartheta _2\in \mathbb {S}^1$ and $\delta \in (0,\pi /2]$. If $|\vartheta _1-\vartheta _2|<\delta$, then there exists $C\ge \sqrt{2\pi }$ with

$$\begin{aligned} |\kappa (\vartheta _1,\tau )-\kappa (\vartheta _2,\tau )|<Cm^*(\tau )\sqrt{\delta }\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )\,. \end{aligned}$$

Proof

We follow similar lines to [9, Paragraph 4.3.6] and [25, Lem. 7.1]. Lemma 7.2 provides

$$\begin{aligned} \max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )\ge \frac{L}{2\pi }\ge c>0\,. \end{aligned}$$

Let $\delta \in (0,\pi /2]$. For $|\vartheta _1-\vartheta _2|<\delta$, Cauchy–Schwarz and Lemma 8.6 imply

$$\begin{aligned} |\kappa (\vartheta _1,\tau )-\kappa (\vartheta _2,\tau )|&\le |\vartheta _1-\vartheta _2|^{1/2}\left( \int _{\vartheta _1}^{\vartheta _2} \left( {\dfrac{\partial \kappa }{\partial \vartheta }}(\vartheta ,\tau )\right) ^{2}d\vartheta \right) ^{1/2} \nonumber \\&\le \sqrt{\delta }\left( \int _{\mathbb {S}^1}\kappa ^2(\vartheta ,\tau )\,d\vartheta +C(m(\tau )+1)\right) ^{1/2} \nonumber \\&\le \sqrt{\delta }\left( \sqrt{2\pi }\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )+\sqrt{C(m(\tau )+1)}\right) \nonumber \\&\le \sqrt{\delta }C\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )\left( 1+\frac{\sqrt{m(\tau )+1}}{\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )}\right) \,, \end{aligned}$$

where we used $\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )>0$ for $\tau >0$. $\square$

Lemma 8.8

(Gage–Hamilton  [9, Cor. 5.2]) Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded, convex curve. Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Let $\varepsilon \in (0,1)$ and $\tau >0$. Then

$$\begin{aligned} \max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )r_{{{\,\mathrm{in}\,}}}(\tau ) \le \left\{ (1-\varepsilon )\left[ 1-K\!\left( \left( \frac{\varepsilon }{Cm^*(\tau )}\right) ^2\right) \left( \frac{r_{{{\,\mathrm{circ}\,}}}(\tau )}{r_{{{\,\mathrm{in}\,}}}(\tau )}-1\right) \right] \right\} ^{-1}\,, \end{aligned}$$

where $K:(0,\pi ]\rightarrow [0,\infty )$ is a positive decreasing function with $K(\omega )\rightarrow \infty$ for $\omega \searrow 0$ and $K(\pi )=0$.

Proof

The proof follows with the help of Lemma 8.7 and can be found in [9, Cor. 5.2] and [25, Lem. 7.11]. For details, see also [5, Lem. 11.11]. $\square$

Corollary 8.9

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded, convex curve. Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). For every $\varepsilon \in (0,1)$, there exists a time $\tau _0>0$ such that, for all $\tau \ge \tau _0$,

$$\begin{aligned} \max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )r_{{{\,\mathrm{in}\,}}}(\tau )\le \frac{1}{(1-\varepsilon )^2}\,. \end{aligned}$$

Proof

We extend the proof of [9, Prop. 5.3] and [25, Cor. 7.12]. Proposition 8.4 implies that, for every $\delta >0$, there exists a time $\tau _0(\delta )>0$ so that $r_{{{\,\mathrm{circ}\,}}}(\tau )-r_{{{\,\mathrm{in}\,}}}(\tau )\le \delta$ for all $\tau \ge \tau _0$, and thus

$$\begin{aligned} \frac{r_{{{\,\mathrm{circ}\,}}}(\tau )}{r_{{{\,\mathrm{in}\,}}}(\tau )}-1\le \frac{\delta }{r_{{{\,\mathrm{in}\,}}}(\tau )}\,. \end{aligned}$$

(8.11)

Recall the definitions (8.6) and (8.10) of m and $m^*$. We define

$$\begin{aligned} I_1:=\{\tau \ge \tau _0\;|\;m(\tau )=\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )\} \end{aligned}$$

and

$$\begin{aligned} I_2:=\{\tau \ge \tau _0\;|\;m(\tau )>\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )\}\,. \end{aligned}$$

Then, m is monotonically increasing on $I_1$ and constant on every connected subinterval of $I_2$. By Lemma 7.2, L is uniformly bounded from above. Hence, there exists a constant $c>0$ so that $\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )\ge c$ for all $\tau \in [\tau _0,\infty )$ and

$$\begin{aligned} m^*(\tau )\le 2+\frac{1}{c} \end{aligned}$$

for $\tau \in I_1$. We distinguish between three cases.

• (i) Assume that $\sup _{[\tau _0,\infty )}m<\infty$. Then $\sup _{[\tau _0,\infty )}m^*<\infty$.

• (ii) Assume that $\sup _{[\tau _0,\infty )}m=\infty$ and $\sup \{\tau \in I_2\}=:\tau _1<\infty$. Then $[\tau _1,\infty )\subset I_1$ and

  $$\begin{aligned} \sup _{[\tau _0,\infty )}m^*=\sup _{I_1}m^*<2+\frac{1}{c}\,. \end{aligned}$$

• (iii) Assume that $\sup _{[\tau _0,\infty )}m=\infty$ and $\sup \{\tau \in I_2\}=\infty$. Assume there exists $\tau _2\in [\tau _0,\infty )$ so that $(\tau _2,\infty )\subset I_2$, then $m(\tau )=m(\tau _2)<\infty$ for all $\tau \in (\tau _2,\infty )$. This contradicts $\sup _{[\tau _0,\infty )}m=\infty$. Hence, $I_2$ consists of infinitely many disjoint open intervals $I_{2,k}$, $k\in \mathbb {N}$ and $\sup _{I_1}m^*\le 2+1/c$. Define the sequence

  $$\begin{aligned} \big (\tau _k:=\sup \{\tau \in I_{2,k}\}\in I_1\big )_{k\in \mathbb {N}}\,. \end{aligned}$$

  Then $\tau _k\rightarrow \infty$ for $k\rightarrow \infty$ and for all $k\in \mathbb {N}$, and since $\tau _k\in I_1$,

  $$\begin{aligned} m(\tau )=m(\tau _k)=\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau _k) \end{aligned}$$

  as well as

  $$\begin{aligned} m^*(\tau )\le 1+\frac{m(\tau )+1}{\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )} =1+\frac{m(\tau _k)+1}{\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau _k)} \le 2+\frac{1}{c} \end{aligned}$$

  for all $\tau \in I_{2,k}$. Hence,

  $$\begin{aligned} \sup _{\tau \in [\tau _0,\infty )}m(\tau ) =\sup _{\tau \in I_1\cup I_2}m(\tau ) \le 2+\frac{1}{c}\,. \end{aligned}$$

Thus, for any $\tau \ge \tau _0$, $m^*$ is independent of time. Recall that K, as defined in Lemma 8.8, is a positive decreasing function that satisfies $K(\omega )\rightarrow \infty$ for $\omega \searrow 0$ and $K(\pi )=0$. By Proposition 8.4, $r_{{{\,\mathrm{in}\,}}}(\tau )\ge c>0$ for all $\tau \ge 0$. Hence, for given $\varepsilon \in (0,1)$, we can choose $\delta >0$ and $\tau _0(\delta )>0$ so that

$$\begin{aligned} \frac{\delta }{r_{{{\,\mathrm{in}\,}}}(\tau )}\le \frac{\varepsilon }{K\big ((\varepsilon /Cm^*)^2\big )} \end{aligned}$$

(8.12)

for all $\tau \ge \tau _0$. Combining (8.11) and (8.12) yields

$$\begin{aligned} \frac{r_{{{\,\mathrm{circ}\,}}}(\tau )}{r_{{{\,\mathrm{in}\,}}}(\tau )}-1\le \frac{\varepsilon }{K\big ((\varepsilon /Cm^*)^2\big )} \end{aligned}$$

so that

$$\begin{aligned} 1-\varepsilon \le 1-K\!\left( \left( \frac{\varepsilon }{Cm^*}\right) ^2\right) \left( \frac{r_{{{\,\mathrm{circ}\,}}}(\tau )}{r_{{{\,\mathrm{in}\,}}}(\tau )}-1\right) \end{aligned}$$

for all $\tau \ge \tau _0$. This and Lemma 8.8 imply

$$\begin{aligned} \max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )r_{{{\,\mathrm{in}\,}}}(\tau ) \le \frac{1}{(1-\varepsilon )^2} \end{aligned}$$

for any $\tau \ge \tau _0$. $\square$

Corollary 8.10

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded, convex curve. Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Then

$$\begin{aligned} \frac{\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )}{\min _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )}\rightarrow 1\,,\qquad \kappa (\vartheta ,\tau )\rightarrow \frac{1}{R}\qquad \text { and }\qquad h(\tau )\rightarrow \frac{1}{R} \end{aligned}$$

for every $\vartheta \in \mathbb {S}^1$ and for $\tau \rightarrow \infty$, where R is given in Proposition 8.4.

Proof

We follow the lines of  [25, Cor. 7.14]. By Proposition 8.4, $\Sigma _\tau$ is strictly convex for $\tau \in (0,\infty )$. Like in [9, Thm. 5.4], [25, Prop. 7.13] or [5, Prop. 11.13], we first conclude with the help of Corollary 8.9 that $\kappa (\vartheta ,\tau )r_{{{\,\mathrm{in}\,}}}(\tau )\rightarrow 1$ for all $\vartheta \in \mathbb {S}^1$ and for $\tau \rightarrow \infty$. Hence, it also holds that $\max _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )r_{{{\,\mathrm{in}\,}}}(\tau )\rightarrow 1$ and $\min _{\vartheta \in \mathbb {S}^1}\kappa (\vartheta ,\tau )r_{{{\,\mathrm{in}\,}}}(\tau )\rightarrow 1$ for $\tau \rightarrow \infty$ and the first claim follows. By Proposition 8.4, the curve converges to a circle of radius R. This yields the second claim. The third claim follows from Lemma 7.6 and $L\rightarrow 2\pi R$. $\square$

Theorem 8.11

(Pihan  [25, Prop. 7.17]) Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded, convex curve. Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Then, for all $n\in \mathbb {N}$, ${\dfrac{\partial ^n}{\partial \vartheta ^n}}\kappa \rightarrow 0$ uniformly for $\tau \rightarrow \infty$. Hence, the curves converge uniformly in $C^\infty$ to a circle of radius R.

Proof

The proof uses Corollary 8.10 and can be found in [25, Prop. 7.17] or [5, Thm. 11.17]. $\square$

We summarise our results in the following and two theorems.

Theorem 8.12

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded curve. Let $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ be a smooth, embedded solution of (1.1) with initial curve $\Sigma _0$ and h satisfying (1.11). Then the evolving surfaces $\Sigma _t=X(\mathbb {S}^1\!,t)$ are contained in a uniformly bounded region of the plane for all times. And, for all $\beta \in (0,1)$, there exists a time-independent constant $C>0$ such that, for all $t\ge 0$,

• (i) $|\max _{p\in \mathbb {S}^1}\kappa (p,t)-\min _{p\in \mathbb {S}^1}\kappa (p,t)|\le C\exp \!\left( -\frac{\beta }{R^2}t\right)$,

• (ii) $|\kappa (p,t)-1/R|\le C\exp \!\left( -\frac{\beta }{R^2}t\right)$ for all $p\in \mathbb {S}^1$,

• (iii) $|h(t)-1/R|\le C\exp \!\left( -\frac{\beta }{R^2}t\right)$, and

• (iv) $\left| {\dfrac{\partial ^n}{\partial t^m}}{\dfrac{\partial ^n}{\partial p^n}}\kappa (p,t)\right| \le C\exp \!\left( -\frac{\beta }{(n+2m+1)R^2}t\right)$ for all $p\in \mathbb {S}^1$ and all $n,m\in \mathbb {N}$.

Hence, the solution converges smoothly and exponentially to a circle of radius R.

Proof

By Theorem 7.9, there exists a time $T_0>0$ so that the curves are strictly convex on $(T_0,\infty )$. Like in [9, 25, Sect. 7.5] and [5, Sect. 11.4], we can show for convex curves with the help of Wirtinger’s inequality and the smooth convergence of Theorem 8.11 exponential decay of the $L^2$-norm of the derivative of the curvature. The proof is independent of the particular form of h, which is why we do not repeat it here. Interpolation inequalities then yield that for $\beta \in (0,1)$ and $m,n\in \mathbb {N}\cup \{0\}$, $m+n>0$, there exist constants $C_{n,m}>0$ such that

$$\begin{aligned} \max _{\vartheta \in \mathbb {S}^1}\left| {\dfrac{\partial ^m}{\partial \tau ^m}}{\dfrac{\partial ^n\kappa }{\partial \vartheta ^n}}(\vartheta ,\tau )\right| \le C_{n,m}\exp \!\left( -\frac{\beta \tau }{(n+2m+1)R^2}\right) \end{aligned}$$

(8.13)

for $\tau$ large enough. To prove (i), we follow the lines of [25, Prop. 7.27]. For $t\ge 0$, let $p_1,p_2\in \mathbb {S}^1$ be the points where the curvature attains its maximum and minimum. By Lemma 7.2 and (8.13), there exists a time-independent constant $C>0$ so that

$$\begin{aligned} \left| \max _{p\in \mathbb {S}^1}\kappa (p,t){-}\min _{p\in \mathbb {S}^1}\kappa (p,t)\right| {=}|\kappa (p_2,t)-\kappa (p_1,t)| {\le }\int _{\Sigma _t}\left| {\dfrac{\partial \kappa }{\partial s}}\right| ds_t {\le } C\exp \!\left( -\frac{\beta t}{R^2}\right) \,. \end{aligned}$$

for t large enough. To show claim (ii), we observe that for embedded, closed, convex curves,

$$\begin{aligned} \min _{p\in \mathbb {S}^1}\kappa (p) \le \frac{1}{r_{{{\,\mathrm{circ}\,}}}} =\sqrt{\frac{\pi }{A_{{{\,\mathrm{circ}\,}}}}} \le \sqrt{\frac{\pi }{A}} \le \frac{2\pi }{L} \le \frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t \le \max _{p\in \mathbb {S}^1}\kappa (p)\,. \end{aligned}$$

(8.14)

By the intermediate value theorem and (8.14), there exist points $p_0,p_1,p_2\in \mathbb {S}^1$ with $\kappa (p_0,t)=\sqrt{\pi /A}$, $\kappa (p_1,t)=2\pi /L$ and $\kappa (p_2,t)=\int _{\Sigma _t}\kappa ^2\,ds_t/2\pi$, so that for $p\in \mathbb {S}^1$ with (8.13),

$$\begin{aligned} \left| \kappa (p,t)-\sqrt{\frac{\pi }{A}}\right| =|\kappa (p,t)-\kappa (p_0,t)| \le \int _{\Sigma _t}\left| {\dfrac{\partial \kappa }{\partial s}}\right| ds_t \le C\exp \!\left( -\frac{\beta t}{R^2}\right) \end{aligned}$$

(8.15)

and likewise

$$\begin{aligned} \left| \kappa (p,t)-\frac{2\pi }{L}\right| +\left| \kappa (p,t)-\frac{1}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t\right| \le C\exp \!\left( -\frac{\beta t}{R^2}\right) \end{aligned}$$

(8.16)

for t large enough. Furthermore, Proposition 8.4 and yields

$$\begin{aligned} \left| \sqrt{\frac{\pi }{A}}-\frac{1}{R}\right| \le \frac{\sqrt{|A-\pi R^2|}}{R\sqrt{A}} \le C\sqrt{\int _t^\infty \left| {\dfrac{d A}{d \tau }}\right| \,d\tau } \le C\int _t^\infty \left| {\dfrac{d A}{d \tau }}\right| \,d\tau \end{aligned}$$

(8.17)

and

$$\begin{aligned} \left| \frac{2\pi }{L}-\frac{1}{R}\right| =\frac{|2\pi R-L|}{LR} \le C\int _t^\infty \left| {\dfrac{d L}{d \tau }}\right| \,d\tau \,. \end{aligned}$$

(8.18)

By (7.2), (7.3) and Proposition 8.4, there exists a constant $C>0$ so that

$$\begin{aligned} \sqrt{\int _t^\infty \left| {\dfrac{d A}{d \tau }}\right| \,d\tau }&=\sqrt{\gamma \left( \frac{L^2}{4\pi }-A\right) } \le C\exp \!\left( -\frac{\beta t}{R^2}\right) \end{aligned}$$

(8.19)

for t large enough. By (8.15), (8.16), (8.17), (8.18) and (8.19), for $p\in \mathbb {S}^1$,

$$\begin{aligned} \left| \kappa (p,t)-\frac{1}{R}\right| \le \left| \kappa (p,t)-\sqrt{\frac{\pi }{A}}\right| +\left| \sqrt{\frac{\pi }{A}}-\frac{1}{R}\right| \le C\exp \!\left( -\frac{\beta t}{R^2}\right) \,. \end{aligned}$$

(8.20)

Likewise with Proposition 8.4, (8.15) and (8.16),

$$\begin{aligned} |\kappa (p,t)-h(t)|&\le \left| \kappa (p,t)-\frac{2\pi }{L}\right| +\frac{|\gamma |}{L}\left| \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \right| \nonumber \\&\le C\exp \!\left( -\frac{\beta t}{R^2}\right) \end{aligned}$$

(8.21)

for t large enough. The boundedness of the curvature on $[0,T_0]$ yields the claim for all $t\ge 0$. For claim (iii), we estimate with (8.20) and (8.21),

$$\begin{aligned} \left| h-\frac{1}{R}\right| \le C\exp \!\left( -\frac{\beta t}{R^2}\right) \end{aligned}$$

for all $t\ge 0$. For claim (iv), we use Lemma 2.1 and (8.21) to estimate

$$\begin{aligned} {\dfrac{\partial v}{\partial t}} =\kappa (h-\kappa )v \le C\left( \exp \!\left( -\frac{\beta t}{R^2}\right) \right) v \end{aligned}$$

for all $t\ge 0$. Hence, $v\ge C$ on $\mathbb {S}^1\!\times [0,\infty )$ and the claim follows with (8.13), for every $m,n\in \mathbb {N}\cup \{0\}$, $m+n>0$. To show that the curves stay in a bounded region, we observe that with (8.21),

$$\begin{aligned} \Vert X(p,t)-X(p,0)\Vert _{\mathbb {R}^2} \le \int _0^t|\kappa (p,\tau )-h(\tau )|\,d\tau \le C\int _0^t\exp \!\left( -\frac{\beta \tau }{R^2_0}\right) \,d\tau \le C \end{aligned}$$

for all $p\in \mathbb {S}^1$ and $t\in (0,\infty )$, where C is independent of time. $\square$

Remark 8.13

All the proofs leading up to Theorem 8.12 also work, if we prescribe the derivative of the area or the length by a function $g\in C^\infty ([0,\infty ))\cap L^1([0,\infty ))$. If we prescribe the derivative of the area, Lemma 2.3 and (7.1) yield

$$\begin{aligned} {\dfrac{d A}{d t}}=g\,,\quad h=\frac{2\pi +g}{L} \quad \text { and }\quad {\dfrac{d L}{d t}} =-\frac{2\pi }{L}\left( \frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi -g\right) \,, \end{aligned}$$

(8.22)

where either

$$\begin{aligned} -2\pi<g\le 0\,,&\quad {\dfrac{d g}{d t}}\ge 0\quad \text { and }\quad \int _0^\infty g\,dt>-A_0\,,\qquad \text { or } \\ 0\le g<\frac{L_t}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi \,,&\quad {\dfrac{d g}{d t}}\le 0\quad \text { and }\quad \int _0^\infty g\,dt\le \frac{L_0^2}{4\pi }-A_0\,, \end{aligned}$$

since need A and L to be monotone and bounded and we will need h to be positive in Remark 8.15. If we prescribe the derivative of the length, Lemma 2.3 and (7.1) yield

$$\begin{aligned} {\dfrac{d L}{d t}}=g\,,\quad h=\frac{1}{2\pi }\left( \int _{\Sigma _t}\kappa ^2\,ds_t+g\right) \quad \text { and }\quad {\dfrac{d A}{d t}} =\frac{L}{2\pi }\int _{\Sigma _t}\kappa ^2\,ds_t-2\pi +\frac{Lg}{2\pi }\,, \end{aligned}$$

where either

$$\begin{aligned} -\int _{\Sigma _t}\kappa ^2\,ds_t+\frac{4\pi }{L_t}<g\le 0\,,&\quad {\dfrac{d g}{d t}}\ge 0\quad \text { and }\;\int _0^\infty g\,dt>-L_0\,,\quad \text { or } \\ 0\le g\,,&\quad {\dfrac{d g}{d t}}\le 0\quad \text { and }\quad \int _0^\infty g\,dt<\infty \,, \end{aligned}$$

since again need A and L to be monotone and bounded. Then Theorem 8.12 holds with the addition in the cases

• (ii) $|\kappa (p,t)-1/R|\le C\exp \!\left( -\frac{\beta }{R^2}t\right) +C\int _t^\infty g\,d\tau$ for all $p\in \mathbb {S}^1$, and

• (iii) $|h(t)-1/R|\le C\exp \!\left( -\frac{\beta }{R^2}t\right) +C\int _t^\infty g\,d\tau +Cg(t)$

for all $\beta \in (0,1)$ and $t\ge 0$, where $C>0$ is time-independent.

Theorem 8.14

Let $\Sigma _0=X_0(\mathbb {S}^1)$ be a smooth, embedded curve satisfying (1.8). Then there exists a unique, smooth, embedded solution $X:\mathbb {S}^1\!\times [0,\infty )\rightarrow \mathbb {R}^2$ to (1.1) with initial curve $\Sigma _0$ and h satisfying (1.2). The evolving curves $\Sigma _t=X(\mathbb {S}^1\!,t)$ are contained in a uniformly bounded region and converge smoothly and exponentially to a circle of radius R.

Proof

By the short time existence, there exists a unique solution $X\in C^\infty (\mathbb {S}^1\!\times [0,T))$ By Lemma 7.2, $c\le L\le C$ so that h is uniformly bounded from above and below away from zero. By Corollary 5.4, the curves remain embedded on (0, T). Corollary 6.5 yields that $T=\infty$. Hence, we can apply Theorem 8.12. $\square$

Remark 8.15

Theorem 8.14 also holds for h satisfying (8.22).