Uniformly Compressing Mean Curvature Flow

Abstract

Michor and Mumford showed that the mean curvature flow is a gradient flow on a Riemannian structure with a degenerate geodesic distance. It is also known to destroy the uniform density of gridpoints on the evolving surfaces. We introduce a related geometric flow which is free of these drawbacks. Our flow can be viewed as a formal gradient flow on a certain submanifold of the Wasserstein space of probability measures endowed with Otto’s Riemannian structure. We obtain a number of analytic results concerning well-posedness and long-time stability which are, however, restricted to the 1D case of evolution of loops.

1 Introduction

The mean curvature flow [16, 47] is a geometric flow which describes the behavior of a k-dimensional submanifold $M_t\subset \mathbb {R}^d$, $1\le k<d$, which evolves over time t according to the law:

(1)

where x is an arbitrary point of $M_t$, and $\vec H(x)$ is the mean curvature vector of the submanifold in x. It has huge variety of applications ranging from formation of grain boundaries in metals to image processing. The mean curvature flow (MCF) is the formal negative gradient flow of the volume functional ${{\mathrm{vol}}}\simeq {\mathcal {H}}^{k}$,

$$\begin{aligned} \partial _t M_t = -{{\mathrm{grad}}}{{\mathrm{vol}}}(M_t), \end{aligned}$$

(2)

where the “manifold of k-dimensional submanifolds” is equipped with the $L^2$ Riemannian structure (see [9, 30] or our formula (76)), and the gradient is understood in the sense of this structure. Hence, as usual in the context of gradient flows [5, 49], the volume functional, being the driving entropy of the gradient flow, decays with time in the following way:

$$\begin{aligned} \frac{d}{\mathrm{{d}}t} {{\mathrm{vol}}}(M_t) = -\langle {{\mathrm{grad}}}{{\mathrm{vol}}}(M_t),{{\mathrm{grad}}}{{\mathrm{vol}}}(M_t)\rangle _{M_t} =-\int _{M_t}|\vec H|^2\, \mathrm{{d}}{\mathcal {H}}^{k}. \end{aligned}$$

(3)

There is an avalanche of works about the theoretical and numerical aspects of the mean curvature flow. One can introduce minimizing movement schemes à la de Giorgi, which exploit the gradient flow structure [2, 29]. The solutions should a priori collapse in finite time, but singular behavior before the breakdown does not permit existence of smooth solutions up to the final moment. There exist various strategies to go beyond the first singularity. The pioneering work [12] relaxed the notion of solution so that the evolving objects were barely varifolds. The level-set approach [39] led to solutions in the Crandall–Lions viscosity sense [6, 15, 19]. For the curve-shortening flow, i.e., when $k=1$, one can define and construct a weak solution as a limit of certain curves which live in an ambient space of higher dimension and which are called ramps [3, 41]. Other types of weak solutions for the curve-shortening flow were introduced and studied in [7, 8, 13, 17, 22].

There are two issues which mar the overall harmony. The first one is that the underlying Riemannian distance is degenerate (i.e., one can connect any two surfaces with a path of arbitrarily small Riemannian length), cf. [9, 10, 30,31,32], which is unpleasant since a nondegenerate metric space structure is an important precondition in treatment of gradient flows, see [5, 44]. The second issue is that the Hausdorff measure ${\mathcal {H}}^{k}$ is not uniformly contracted by the flow, that is, if $T_t:x(0)\mapsto x(t)$ is the flow operator describing the trajectories of material particles forming $M_t$ in the ambient space $\mathbb {R}^d$, then the property

(4)

is violated except for some very special scenarios as a shrinking sphere. From the numerical perspective, this means that the flow destroys the uniform density of gridpoints on $M_t$, which is unwelcome and may cause computational instabilities [33, 48]. For the curve-shortening flow in the plane, this can be fixed [1, 33] by adding a certain tangential motion to the right-hand side of (1) in order to conserve the uniform density of the moving particles without affecting the evolution of the shapes.

In this paper, we suggest a different approach which simultaneously eliminates the two above-mentioned drawbacks of the mean curvature flow, and which is applicable for any dimensions k and d. The idea is to consider the flow which is the closest possible to the original MCF (1) in a certain least-squares sense among the flows which uniformly contract the k-Hausdorff measure (in other words, which respect the uniform density of gridpoints). We employ the infinite-dimensional manifold $\mathcal {A}_k$ (consisting of objects of the form $\eta =\lambda \xi$, where $\lambda >0$ is a scalar which quantifies the volume of M, and $\xi$ is a volume-preserving immersion in the sense of [21]) endowed with the parametrization-invariant $L^2$ metric. Our flow is driven by the orthogonal projection of the mean curvature onto $T\mathcal {A}_k$. We dub the resulting object the uniformly compressing mean curvature flow (UCMCF) because the evolving surfaces can be thought of as being constituted by fluid particles density of which depends merely on time (the surfaces in question up to a time-dependent constant are incompressible membranes [18, 20, 34]). The UCMCF is by construction the negative gradient flow of the volume functional on $\mathcal {A}_k$. It is a genuinely geometric flow in the sense that the evolution of submanifolds $M_t=\eta (t)(M)$ does not depend on their parametrization.

Unlike the tangentially corrected MCF [1, 33], our flow differs from the classical MCF in the normal direction and thus the geometrical evolution of the submanifolds along the two flows do not coincide in general (although they do coincide for a shrinking sphere). Nevertheless, we show that the qualitative behavior of UCMCF is quite similar to the one of the usual MCF, and thus it may be used as a substitute for the MCF in applications.

We will observe that the our flow collapses in finite time, and in order to study the evolution and stability of the shapes before the breakdown, we need to renormalize the problem both in time and in space. Surprisingly enough, our normalized flow is also a gradient flow: namely, the positive gradient flow of the $L^2$-mass on the space of volume-preserving immersions. Our recent work [46] studies the gradient flow of a different functional (potential energy) on a similar Riemannian structure, which turns out to be a model for an overdamped fall of an inextensible string in a gravitational field. A similar mechanical interpretation for our normalized UCMCF is an overdamped motion of an inextensible loop ($k=1$) or an incompressible membrane ($k>1$) repelled from the origin with the force field identically equal to the radius vector.

Other gradient flows of inextensible strings were considered in [24, 35,36,37,38]. In those papers, additional fourth-order terms coming from the bending energy appear, which help to secure nondegenerate parabolicity of the equations and decrease the difficulties created by the Lagrange multiplier, cf. (18), (19) below.

Both the original and normalized flows can be viewed as formal gradient flows on certain submanifolds of the Wasserstein space [40, 49, 50] of probability measures endowed with Otto’s Riemannian structure, cf. Sect. B.2. in the Appendix.

We will mostly work with the immersed curves $1=k<d$. The Appendix is devoted to the general case $1\le k<d$. It turns out to be more convenient to analyze the normalized flow, which allows us to descry the asymptotic behavior of the shapes near the breakdown. We show local strong well-posedness of the problem. We characterize the steady states, and prove global existence of strong solutions for the initial data which are close to the steady states which maximize the driving $L^2$-mass, i.e., to simple circles. We establish the exponential decay of such a global solution to a steady state. We address the global solvability for any Lipschitz initial curve which does not need to be close to the equilibria, and we prove existence of suitably defined weak solutions. Our approach is based on approximation of the gradient flow on the manifold of volume-preserving immersions by Hilbertian gradient flows; in particular, we do not use the ramps. Unlike [3, 7, 8, 17, 22, 41], our weak solutions are $H^1$-regular in time.

In this paper, we make a technical and geometrically nonrestrictive assumption that the center of mass is fixed at the origin; otherwise the center of mass would fly away to infinity.

2 The Flow

2.1 Uniformly Compressing Curve-Shortening Flow

Let $\mathbb {S}^1\simeq \mathbb {R}/\mathbb {Z}$ denote the circle of length 1. For $d\ge 2$ let $\mathcal {K}$ be the space of closed curves $\mathcal {K}:=\{\eta : \eta \in H^2(\mathbb {S}^1;\mathbb {R}^d), \int _0^1\eta (s) \mathrm{{d}}s =0\}$. Let $L:\mathcal {K}\rightarrow \mathbb {R}$, $L(\eta ):=\int _0^1|\partial _s\eta |\ \mathrm{{d}}s$ be the length functional. We consider the space of immersed curves with the constant speed parametrization, i.e.,

$$\begin{aligned} \mathcal {A}:=\{\eta \in \mathcal {K}: \&|\partial _s\eta (s)|=L(\eta )>0 \text { for all } s\in \mathbb {S}^1\}. \end{aligned}$$

Arguing as Theorem A.1 in [43] we see that $\mathcal {A}$ is a smooth Hilbert submanifold of $\mathcal {K}$, with the tangent space

$$\begin{aligned} T_\eta \mathcal {A}=&\{w\in H^2(\mathbb {S}^1;\mathbb {R}^d): \frac{\mathrm{{d}}}{\mathrm{{d}}s}\left( \partial _s w(s)\cdot \partial _s \eta (s)\right) =0 \text { for a.e. } s\in \mathbb {S}^1\\&\text { and }\int _0^1w(s)\ \mathrm{{d}}s=0\},\ \eta \in \mathcal {A}. \end{aligned}$$

We endow $\mathcal {K}$ with the Riemannian metric

$$\begin{aligned} \langle v,w\rangle _{T_\eta \mathcal {K}}:=\int _0^1 v(s)\cdot w(s)|\partial _s\eta (s)|\ \mathrm{{d}}s, \end{aligned}$$

which is invariant under reparametrization, cf. [31]. It induces a metric on $\mathcal {A}$:

$$\begin{aligned} \langle v,w\rangle _{T_\eta \mathcal {A}}=\int _0^1 v(s)\cdot w(s) L(\eta )\ \mathrm{{d}}s. \end{aligned}$$

(5)

Proposition 2.1

The Riemannian distance $d_{\mathcal {A}}$ is nondegenerate.

Proof

Take any two closed curves $\eta _0,\eta _1\in \mathcal {A}$. Renormalizing if needed, we can suppose that $\Vert \eta _0-\eta _1\Vert _{L^2({\mathcal {S}}^1)}=1$. We claim that

$$\begin{aligned} \mathrm{{d}}_{\mathcal {A}}(\eta _0,\eta _1)\ge m:=\frac{1}{2} \min (L(\eta _0),\Vert \eta _0\Vert ^2_{L^2({\mathcal {S}}^1)},2). \end{aligned}$$

(6)

If not, there exists a $C^1$ path $\eta :[0,1]\rightarrow \mathcal {A}$, $\eta (0)=\eta _0$, $\eta (1)=\eta _1$, with Riemannian length

$$\begin{aligned} \begin{aligned} {\mathcal {L}}(\eta )&:=\int _0^1\sqrt{\langle \dot{\eta }(t),\dot{\eta }(t)\rangle _{T_{\eta (t)}\mathcal {A}}} \,\mathrm{{d}}t\\&=\int _0^1 \Vert \dot{\eta }(t)\Vert _{L^2({\mathcal {S}}^1)} \sqrt{L(\eta (t))}\,\mathrm{{d}}t< m. \end{aligned} \end{aligned}$$

(7)

Since by the Minkowski inequality and integration by parts

$$\begin{aligned} \int _0^1\Vert \dot{\eta }(t)\Vert _{L^2({\mathcal {S}}^1)}\, \mathrm{{d}}t \ge \Vert \eta _1-\eta _0\Vert _{L^2(\mathbb {S}^1)}{=} 1, \end{aligned}$$

we have $L(\eta (t))<m^2$ for some t. Due to continuity of $L(\eta (t))$ (and recall that $L(\eta _0)\ge 2m$ from (6)), there exists $t_*\in (0,1)$ such that $L(\eta (t_*))=m$ and $L(\eta (t))>m$ for $t<t_*$. Then

$$\begin{aligned} {\mathcal {L}} (\eta )\ge {\mathcal {L}} (\eta |_{(0,t_*)})\ge \sqrt{m}\Vert \eta _0-\eta (t_*)\Vert _{L^2({\mathcal {S}}^1)}. \end{aligned}$$

(8)

By Wirtinger’s inequality,

$$\begin{aligned} \Vert \eta (t_*)\Vert _{L^2({\mathcal {S}}^1)}\le \frac{1}{2\pi } \Vert \partial _s\eta (t_*)\Vert _{L^2({\mathcal {S}}^1)}=\frac{m}{2\pi }. \end{aligned}$$

(9)

Combining (7)–(9), we infer

$$\begin{aligned} \sqrt{2m}\le \Vert \eta _0\Vert _{L^2({\mathcal {S}}^1)}\le \Vert \eta _0-\eta (t_*)\Vert _{L^2({\mathcal {S}}^1)}+\Vert \eta (t_*)\Vert _{L^2({\mathcal {S}}^1)}\le \sqrt{m}\left( 1 +\frac{1}{2\pi }\right) , \end{aligned}$$

which is a contradiction. $\square$

We are interested in the formal gradient flow of the length functional $L(\eta )=\int _0^1|\partial _s\eta | \ \mathrm{{d}}s$, $\eta \in \mathcal {A}$, under the metric (5):

$$\begin{aligned} \partial _t\eta =-{{\mathrm{grad}}}_{\mathcal {A}} L(\eta (t)). \end{aligned}$$

(10)

In order to derive the PDE formulation of (10), we compute (formally) the orthogonal projection from $T_\eta \mathcal {K}$ (which can be identified with $\mathcal {K}$) onto $T_\eta \mathcal {A}$ with respect to the metric (5) (cf. Proposition 3.2 in [43]).

Lemma 2.2

Let $\eta \in \mathcal {A}\cap H^4(\mathbb {S}^1;\mathbb {R}^d)$. The orthogonal projection $P_\eta : T_\eta \mathcal {K}\rightarrow T_\eta \mathcal {A}$ is given by

$$\begin{aligned} P_\eta (z)&=z-\partial _s(\sigma \partial _s\eta ),\text { where } \sigma : {\mathbb {S}}^1\rightarrow \mathbb {R}\text { solves }\nonumber \\ L^2\partial _{ss}\sigma -|\partial _{ss} \eta |^2\sigma&= \partial _sz\cdot \partial _s\eta +const, \end{aligned}$$

(11)

$$\begin{aligned} \int _0^1\sigma (s) \mathrm{{d}}s&=0. \end{aligned}$$

(12)

Proof

• (1) We first show that for any $\sigma$ satisfying (12), the vector field $\partial _s(\sigma \partial _s\eta )$ is orthogonal to $T_\eta \mathcal {A}$. Indeed, given any $w\in T_\eta \mathcal {A}$,

  $$\begin{aligned} \langle \partial _s(\sigma \partial _s\eta ), w\rangle _{T_\eta \mathcal {K}}&=L(\eta )\int _0^1 \partial _s(\sigma \partial _s\eta ) \cdot w \ \mathrm{{d}}s\\&=L(\eta )\sigma \partial _s\eta \cdot w \big |_{s=0}^{s=1}-L(\eta )\int _0^1\sigma \partial _s\eta \cdot \partial _s w\ \mathrm{{d}}s\\&=0. \end{aligned}$$

  The first term is zero since we work on $\mathbb {S}^1$. The second term is zero because $\partial _s\eta \cdot \partial _sw=const$ and $\sigma$ has mean zero.

• (2) Next we show $w=z-\partial _s(\sigma \partial _s\eta )\in T_\eta \mathcal {A}$. We will mainly check the condition $\partial _sw\cdot \partial _s\eta =const$. Indeed,

  $$\begin{aligned} \partial _sw\cdot \partial _s\eta =\partial _sz\cdot \partial _s\eta -\partial _{ss}\sigma |\partial _s\eta |^2-2\partial _s\sigma \partial _{ss}\eta \cdot \partial _s\eta -\sigma \partial _{sss}\eta \cdot \partial _s\eta . \end{aligned}$$

  The constant speed parametrization $|\partial _s\eta |=L=const$ yields $\partial _{ss}\eta \cdot \partial _s\eta =0$ and $\partial _{sss}\eta \cdot \partial _s\eta =-|\partial _{ss}\eta |^2$. Thus

  $$\begin{aligned} \partial _sw\cdot \partial _s\eta =\partial _s z\cdot \partial _s\eta -\partial _{ss}\sigma L^2 +\sigma |\partial _{ss}\eta |^2, \end{aligned}$$

  which is constant by (11).

$\square$

Since $\mathcal {A}$ is a smooth submanifold of the space ${\mathcal {K}}$, the general definition of the gradient [27] implies that

$$\begin{aligned} {{\mathrm{grad}}}_{\mathcal {A}}L(\eta )=P_\eta ({{\mathrm{grad}}}_{{\mathcal {K}}}L(\eta )). \end{aligned}$$

Standard calculus of variations shows that the first $L^2$-variation of $L(\eta )$ is

$$\begin{aligned} \delta L(\eta )=-\partial _s\left( \frac{\partial _s\eta }{|\partial _s\eta |}\right) . \end{aligned}$$

But

$$\begin{aligned} \langle {{\mathrm{grad}}}_{{\mathcal {K}}}L(\eta ),w\rangle _{T_\eta \mathcal {K}}=\int _0^1 \delta L(\eta )(s)\cdot w(s)\ \mathrm{{d}}s, \end{aligned}$$

for every $w\in T_\eta \mathcal {K}\simeq {\mathcal {K}}$. We conclude that

$$\begin{aligned} {{\mathrm{grad}}}_{{\mathcal {K}}}L(\eta )=-\frac{1}{|\partial _s\eta |}\partial _s\left( \frac{\partial _s\eta }{|\partial _s\eta |}\right) . \end{aligned}$$

By Lemma 2.2, the orthogonal projection of the negative gradient in ${\mathcal {K}}$ to the tangent space $T_\eta \mathcal {A}$ is

$$\begin{aligned} P_\eta (-{{\mathrm{grad}}}_{{\mathcal {K}}} L(\eta ))=\frac{\partial _{ss}\eta -\partial _s(\sigma \partial _s\eta )}{L^2}, \end{aligned}$$

where $\sigma :\mathbb {S}^1\rightarrow \mathbb {R}$ satisfies

$$\begin{aligned} L^2\partial _{ss}\sigma -\sigma |\partial _{ss}\eta |^2=\partial _{sss}\eta \cdot \partial _s\eta +const=-|\partial _{ss}\eta |^2+const, \quad \int _0^1\sigma \mathrm{{d}}s =0. \end{aligned}$$

To determine the constant, we integrate in s and thus obtain $const=\int _0^1(1-\sigma )|\partial _{ss}\eta |^2ds$. Letting $\tilde{\sigma }:=1-\sigma$, we get the expression for the gradient flow (10):

$$\begin{aligned} \partial _t\eta (t,s)=L(t)^{-2}\partial _s(\tilde{\sigma }(t,s)\partial _s\eta (t,s)), \end{aligned}$$

(13)

where $L(t):=L(\eta (t))$, and the Lagrange multiplier $\tilde{\sigma }(t,s)$ satisfies

$$\begin{aligned} \begin{aligned}&\partial _{ss}\tilde{\sigma }(t,s)-{L(t)^{-2}}\tilde{\sigma }(t,s)|\partial _{ss}\eta (t,s)|^2=-L(t)^{-2}\int _0^1\tilde{\sigma }(t,s)|\partial _{ss}\eta (t,s)|^2 \mathrm{{d}} s \\&\quad \text { for all }(t,s) \text {and } \int _0^1\tilde{\sigma }(t,s)\ \mathrm{{d}}s =1\text { for all } t. \end{aligned} \end{aligned}$$

Remark 2.3

(Sign of $\tilde{\sigma }$) From $\Vert \partial _{ss}\eta -P_\eta (\partial _{ss}\eta )\Vert _{L^2(\mathbb {S}^1)}\le \Vert \partial _{ss}\eta \Vert _{L^2(\mathbb {S}^1)}$ as well as the orthogonality $\partial _s\eta \cdot \partial _{ss}\eta =0$ everywhere, it follows that

$$\begin{aligned} \Vert \partial _{ss}\eta \Vert _{L^2(\mathbb {S}^1)}^2\ge \Vert \partial _s\sigma \Vert _{L^2(\mathbb {S}^1)}^2+\Vert \sigma \partial _{ss}\eta \Vert _{L^2(\mathbb {S}^1)}^2. \end{aligned}$$

Using $\tilde{\sigma }=1-\sigma$ we can rewrite the above inequality as

$$\begin{aligned} 2\int _{\mathbb {S}^1}\tilde{\sigma }|\partial _{ss}\eta |^2 \mathrm{{d}}s\ge \int _{\mathbb {S}^1}\tilde{\sigma }^2|\partial _{ss}\eta |^2 \mathrm{{d}}s + \int _{\mathbb {S}^1}(\partial _s\sigma )^2 \mathrm{{d}}s. \end{aligned}$$

Thus, $\int _{\mathbb {S}^1}\tilde{\sigma }|\partial _{ss}\eta |^2 \mathrm{{d}}s \ge 0$ and the equality holds if and only if $\tilde{\sigma }\equiv 0$ on $\mathbb {S}^1$, which cannot happen since the mean of $\tilde{\sigma }$ is 1. Then by (13) $\tilde{\sigma }$ satisfies an inhomogeneous elliptic equation with a negative inhomogeneity, and provided $\eta (t)\in C^2(\mathbb {S}^1)$ by the strong maximum principle we infer that $\tilde{\sigma }>0$.

Remark 2.4

Assume $\eta (t,s)$ is a classical solution to Eq. (13) emanating from $\eta _0$ which satisfies the constraints $|\partial _s\eta _0(s)|\equiv L(\eta _0)$ and $\int _0^1\eta _0(s)\mathrm{{d}}s=0$. Then the constraints are preserved along the flow, i.e.,

$$\begin{aligned} |\partial _s\eta (t,s)|\equiv L(t)\text { and }\int _0^1\eta (t,s)\mathrm{{d}}s=0 \text { for each }t. \end{aligned}$$

To show the first equality, we let $Z(t,s):=|\partial _s\eta (t,s)|^2-L(t)^2$. Then a direct computation shows that Z satisfies

$$\begin{aligned} \partial _tZ&=L^{-2}\tilde{\sigma }\partial _{ss}Z+2L^{-2}\partial _s\tilde{\sigma }\partial _sZ+2\left( \partial _{ss}\tilde{\sigma }-L^{-2}\tilde{\sigma }|\partial _{ss}\eta |^2-L\partial _tL\right) \\&=L^{-2}\tilde{\sigma }\partial _{ss}Z+2L^{-2}\partial _s\tilde{\sigma }\partial _sZ\quad \text { with }Z(0,s)=0. \end{aligned}$$

Here we have used the equation of $\tilde{\sigma }$ (note that by (i) of Proposition 2.5 below, the right-hand side of the equation of $\tilde{\sigma }$ is equal to $L\partial _tL$). By Remark 2.3 above $\tilde{\sigma }(t,s)>0$ for the classical solutions $\eta$. Thus, $Z(t,s)=0$ by the strong maximum principle for parabolic equations. To show that $\int _0^1\eta =0$ is preserved along the flow, we integrate the equation along $\mathbb {S}^1$ to obtain $\partial _t\int _0^1\eta (t,s)\mathrm{{d}}s =0$.

2.2 Evolution of the Length and the $L^2$-Mass

We now exploit the gradient flow structure to derive some evolution properties of the length and $L^2$-mass of the curves. Throughout this subsection, we assume the existence of the $H^2$ solutions to the gradient flow (13).

The first proposition is about the evolution of the length functional.

Proposition 2.5

Let $\eta$ be a solution to (13). Then

• (i) $\partial _tL(\eta )=-L^{-3}\int _0^1\tilde{\sigma }|\partial _{ss}\eta |^2 ds$.

• (ii) $\partial _{tt}L(\eta )^2 \ge 0$.

Proof

1. Since $|\partial _s\eta (s,t)|=L(t)$, then

$$\begin{aligned} \partial _t L&=\partial _t |\partial _s\eta (t,s)|\\&= \frac{\partial _s\eta (t,s)}{|\partial _s\eta (t,s)|}\cdot \partial _{ts}\eta (t,s) \\&=\frac{\partial _s\eta (t,s)}{L}\cdot \partial _s\left( \frac{\tilde{\sigma }\partial _{ss}\eta }{L^2}+ \frac{\partial _s\tilde{\sigma }\partial _s\eta }{L^2}\right) . \end{aligned}$$

Note that the LHS does not depend on s. Thus, an integration in s from 0 to 1 and an integration by parts yield the desired equality.

2. Let $\lambda (t):=-L(t)\partial _tL(t)=L^{-2}\int _0^1\tilde{\sigma }|\partial _{ss}\eta |^2ds$. It suffices to show that $\partial _t\lambda (t)\le 0$. Note that one can rewrite the equation of $\tilde{\sigma }$ as

$$\begin{aligned} \partial _{ss}\tilde{\sigma }-L^{-2}\tilde{\sigma }|\partial _{ss}\eta |^2 = -\lambda . \end{aligned}$$

Let N be the unit inner normal to the curve $\eta$, and let k be such that $\partial _{ss}\eta =kN$. Differentiation of the equation of $\tilde{\sigma }$ in time yields

$$\begin{aligned} \partial _{ss}\dot{\tilde{\sigma }}-\dot{\tilde{\sigma }}k^2 L^{-2}-2\tilde{\sigma } k\dot{k} L^{-2} + 2L^{-3}\partial _tL \tilde{\sigma }k^2 =-\dot{\lambda }. \end{aligned}$$

We multiply the above equation by $\tilde{\sigma }$ and integrate in s. Integrating by parts, we get

$$\begin{aligned} \dot{\lambda }&=-\int _0^1 \dot{\tilde{\sigma }}\partial _{ss}\tilde{\sigma } \mathrm{{d}}s +L^{-2}\int _0^1 \dot{\tilde{\sigma }}\tilde{\sigma } k^2 \mathrm{{d}}s +2L^{-2}\int _0^1 \tilde{\sigma }^2k\dot{k} \mathrm{{d}}s-2L^{-3}\partial _tL\int _0^1\tilde{\sigma }^2k^2 \mathrm{{d}}s\\&=\lambda \int _0^1\dot{\tilde{\sigma }} \mathrm{{d}}s +2L^{-2}\int _0^1 \tilde{\sigma }^2k\dot{k} \mathrm{{d}}s-2L^{-3}\partial _tL\int _0^1\tilde{\sigma }^2k^2 \mathrm{{d}}s\\&=2L^{-2}\int _0^1 \tilde{\sigma }^2k\dot{k} \mathrm{{d}}s-2L^{-3}\partial _tL\int _0^1\tilde{\sigma }^2k^2 \mathrm{{d}}s. \end{aligned}$$

Here in the second equality, we have used the equation of $\tilde{\sigma }$. The last equality is due to $\int _0^1\tilde{\sigma }(t,s) \mathrm{{d}}s\equiv 1$. Now we compute $k\dot{k}$. Differentiating the equation of $\eta$ in s, we infer

$$\begin{aligned} \partial _{ts}\eta = L^{-2}\left( \partial _{ss}\tilde{\sigma } \partial _s\eta + 2\partial _{s}\tilde{\sigma }\partial _{ss}\eta +\tilde{\sigma } \partial _{sss}\eta \right) . \end{aligned}$$

Using that $\partial _{sss}\eta \cdot \partial _s\eta =-k^2$ and $\partial _{sss}\eta \cdot N=\partial _sk$, we get

$$\begin{aligned} L^2\partial _{t s}\eta&=\left( \partial _{ss}\tilde{\sigma } -L^{-2}k^2 \tilde{\sigma }\right) \partial _s\eta +\left( 2\partial _s\tilde{\sigma }k+\tilde{\sigma }\partial _sk\right) N+\tilde{\sigma } R\\&=-\lambda \partial _s\eta +\left( 2\partial _s\tilde{\sigma }k+\tilde{\sigma }\partial _sk\right) N+\tilde{\sigma }R \end{aligned}$$

where $R:=\partial _{sss}\eta -(L^{-2}\partial _{sss}\eta \cdot \partial _s\eta )\partial _s\eta -(\partial _{sss}\eta \cdot N)N$ and we have used the equation of $\tilde{\sigma }$. Since $\partial _sN\cdot N=0$ and $N\cdot R=0$, we have

$$\begin{aligned} L^2\partial _{t ss}\eta \cdot \partial _{ss}\eta =-\lambda k^2+k\partial _s\left( 2\partial _s\tilde{\sigma }k+\tilde{\sigma }\partial _sk\right) +\tilde{\sigma }(\partial _sR\cdot \partial _{ss}\eta ). \end{aligned}$$

But $R\cdot \partial _{ss}\eta =0$, hence $\partial _sR\cdot \partial _{ss}\eta = -R\cdot \partial _{sss}\eta =-|R|^2$. Thus,

$$\begin{aligned} L^2\partial _{t ss}\eta \cdot \partial _{ss}\eta =L^2 k\dot{k}=-\lambda k^2+k\partial _s\left( 2\partial _s\tilde{\sigma }k+\tilde{\sigma }\partial _sk\right) -\tilde{\sigma }|R|^2. \end{aligned}$$

(14)

Until this moment we implicitly assumed that $\partial _{ss} \eta \ne 0$. However, (14) is still valid in the points with $\partial _{ss} \eta = 0$ since we can make an agreement that $k=| R|=0$ in those points. Plugging the expression (14) into the equation for $\dot{\lambda }$, we derive

$$\begin{aligned} \dot{\lambda }&=-2L^{-4}(\lambda +L\partial _tL)\int _0^1\tilde{\sigma }^2k^2 \mathrm{{d}}s \\&\quad + 2L^{-4}\int _0^1\tilde{\sigma }^2k\partial _s(2\partial _s\tilde{\sigma }k+\tilde{\sigma }\partial _sk)\mathrm{{d}}s - 2L^{-4}\int _0^1\tilde{\sigma }^3|R|^2 \mathrm{{d}}s. \end{aligned}$$

The first term on the right-hand side vanishes due to the definition of $\lambda$. Then an integration by parts yields

$$\begin{aligned} \begin{aligned} L^4\dot{\lambda }&=-2\int _0^1 \partial _s(\tilde{\sigma }^2 k )(\tilde{\sigma }\partial _s k)\mathrm{{d}}s - 4\int _0^1\partial _s(\tilde{\sigma }^2k)\partial _s\tilde{\sigma } k \mathrm{{d}}s -2 \int _0^1\tilde{\sigma }^3|R|^2 \mathrm{{d}}s\\&=-2\int _0^1\tilde{\sigma }\left( \tilde{\sigma }\partial _s k + 2\partial _s\tilde{\sigma } k\right) ^2 \mathrm{{d}}s-2 \int _0^1\tilde{\sigma }^3|R|^2 \mathrm{{d}}s. \end{aligned} \end{aligned}$$

(15)

Note that $\tilde{\sigma }\ge 0$ by Remark 2.3. Thus, $\dot{\lambda }\le 0$ and we complete the proof for (ii). $\square$

In the next proposition, we show that the $L^2$-mass of the solution decays with constant speed.

Proposition 2.6

Let $\eta$ be a solution to the gradient flow (13). Let $M(t):=\frac{1}{2}\int _0^1|\eta (t,s)|^2 \mathrm{{d}}s$ be the $L^2$-mass. Then $\partial _tM(t)=-1$.

Proof

We multiply the equation of $\eta$ in (13) by $\eta$ and integrate in s. An integration by parts implies

$$\begin{aligned} \partial _tM(t)=-L^{-2}\int _0^1 \tilde{\sigma }|\partial _s\eta |^2 \mathrm{{d}}s. \end{aligned}$$

Using that $|\partial _s\eta |=L$ and $\int _0^1\tilde{\sigma } \mathrm{{d}}s=1$, we obtain the desired equality. $\square$

An immediate consequence of Proposition 2.6 is that the flow becomes extinct in finite time.

Corollary 2.7

$M(t)\rightarrow 0$ as $t\rightarrow t^*$, where $t^*=M(0)=\frac{1}{2}\int _0^1|\eta _0(s)|^2 \mathrm{{d}}s$.

It is also possible to obtain the decay rate of L(t) near the extinction time $t^*$.

Corollary 2.8

Let $L_0:=L(\eta _0)>0$ and let $t^*$ be the extinction time as in Corollary 2.7. Then for all $t\in [0,t^*)$,

• (i) $2\sqrt{2}\pi \sqrt{t^*-t}\le L(t)\le L_0\sqrt{\frac{t^*-t}{t^*}}$,

• (ii) $4\pi ^2\le -L(t)\partial _tL(t)\le -(L\partial _tL)(t)\big |_{t=0}$.

Proof

1. The lower bound follows directly from the Wirtinger’s inequality. Indeed, by Proposition 2.6

$$\begin{aligned} M(t)=t^*-t\text { for } t\in [0,t^*). \end{aligned}$$

On the other hand, by Wirtinger’s inequality

$$\begin{aligned} M(t)\le \frac{1}{2}\frac{1}{4\pi ^2}\int _0^1|\partial _s\eta |^2ds=\frac{L(t)^2}{8\pi ^2}. \end{aligned}$$

Here we have used the assumption $\int _0^1\eta =0$. Combining the above two inequalities together, we obtain the lower bound $L(t)\ge 2\sqrt{2}\pi \sqrt{t^*-t}$.

Next we show the upper bound. Using $\partial _tM(t)=\int _0^1\eta \cdot \partial _t\eta =-1$ and the Hölder’s inequality, we deduce

$$\begin{aligned} \int _0^1|\partial _t\eta |^2 \mathrm{{d}}s \ge \frac{1}{\int _0^1|\eta |^2}=\frac{1}{2M(t)}=\frac{1}{2(t^*-t)}. \end{aligned}$$

(16)

On the other hand, by the gradient flow structure

$$\begin{aligned} \partial _tL= -L\int _0^1|\partial _t\eta |^2 \mathrm{{d}}s. \end{aligned}$$

(17)

Thus, $-\partial _t\ln L\ge \frac{1}{2(t^*-t)}$. Then an integration in t from 0 to t yields

$$\begin{aligned} L(t)\le L_0\sqrt{\frac{t^*-t}{t^*}}, \quad t\in [0,t^*). \end{aligned}$$

2. By (ii) of Proposition 2.5, $t\mapsto -L\partial _t L$ is monotone decreasing. Thus, $-L\partial _tL \le -L\partial _tL \big |_{t=0}$. To see the lower bound, we note that by (17), (16) and Wirtinger’s inequality

$$\begin{aligned} -L(t)\partial _tL(t)=L(t)^2\int _0^1|\partial _t\eta |^2 \ge \frac{L(t)^2}{2M(t)}\ge 4\pi ^2 . \end{aligned}$$

$\square$

3 Normalized Flow

3.1 Renormalization

One can ask as the $L^2$-mass M(t) goes to zero, whether the curve becomes circular. To study this problem, we plan to show that the isoperimetric ratio $\frac{2M(t)}{L^2(t)}$ goes to the optimal constant in the Wirtinger’s inequality $\frac{1}{4\pi ^2}$ as $t\rightarrow t^*$. For that and for many other purposes, it is convenient to renormalize the flow.

We first introduce a slow time variable. More precisely, for $t\in [0,t^*)$ let

$$\begin{aligned} \tau (t):=-\ln L(t). \end{aligned}$$

Note that by Proposition 2.5 (i) and Corollary 2.8, $\tau (t)$ is monotone increasing in t and $\tau \rightarrow +\infty$ iff $t\rightarrow t^*$; this is also clear in view of (88). Next we consider the normalization

$$\begin{aligned} \xi (\tau , s ):=\frac{\eta (t(\tau ),s)}{L(t(\tau ))} \end{aligned}$$

One advantage of using such renormalization is that the curve $\xi (\tau ,s)$ has the unit speed parametrization, i.e.,

$$\begin{aligned} L(\tau )&=|\partial _s\xi (\tau ,s)|=1 \text { for all } (\tau ,s)\in [0,\infty )\times \mathbb {S}^1. \end{aligned}$$

(18)

A direct computation shows that $\xi$ satisfies the equation

$$\begin{aligned} \partial _\tau \xi =\partial _s(\sigma \partial _s\xi )+\xi . \end{aligned}$$

(19)

Here $\sigma (\tau , s)$ can be viewed as a Lagrange multiplier coming from the constraint (18). It satisfies

$$\begin{aligned} \partial _{ss}\sigma -\sigma |\partial _{ss}\xi |^2=-1, \quad \sigma (\tau , 0)=\sigma (\tau ,1). \end{aligned}$$

(20)

Indeed, from the change of variable $\frac{\mathrm{{d}}\tau }{\mathrm{{d}}t}=-\frac{\partial _tL(t)}{L(t)}$. Thus,

$$\begin{aligned} \partial _\tau \xi =\frac{\partial _t\eta }{L}\frac{\mathrm{{d}}t}{\mathrm{{d}}\tau }-\eta \frac{\partial _tL}{L^2}\frac{\mathrm{{d}}t}{\mathrm{{d}}\tau }=-\frac{\partial _t\eta }{\partial _tL}+\frac{\eta }{L}. \end{aligned}$$

Using the equation of $\eta$, we have

$$\begin{aligned} \partial _\tau \xi =-\frac{\partial _s(\tilde{\sigma }\partial _s\eta )}{L^2\partial _tL}+\frac{\eta }{L}. \end{aligned}$$

Letting $\sigma :=\frac{\tilde{\sigma }}{-L\partial _tL}$ and writing the above equation in terms of $\xi$, we arrive at (19). To derive (20), we can either use the equation of $\tilde{\sigma }$, or use the above equation of $\xi$ together with the constraint $|\partial _s\xi |\equiv 1$.

With the same argument as in Sect. 2 it is not hard to see that the normalized flow (19) and (20) can be viewed as the positive gradient flow of the $L^2$-mass $M(\xi ):=\frac{1}{2}\int _0^1|\xi (s)|^2$ on the manifold of immersed curves with arc-length parametrization, cf. (78),

$$\begin{aligned} \tilde{\mathcal {A}}:=\left\{ \xi \in H^2(\mathbb {S}^1;\mathbb {R}^d): \ |\partial _s\xi (s)|=1 \text { for all } s\in \mathbb {S}^1, \ \int _0^1\xi (s)\mathrm{{d}}s =0\right\} \end{aligned}$$

with respect to the $L^2(\mathbb {S}^1;\mathbb {R}^d)$-induced metric (see also Appendix B and our recent work [46] which analyzes the gradient flow of the potential energy on a space very similar to $\tilde{\mathcal {A}}$).

The normalized flow (19)–(20) can be interpreted in the spirit of [46] as an overdamped motion of an inextensible loop particles of which are repelled from the origin with the force equal to the radius vector.

Using either the gradient flow structure for the normalized flow or tracing the change of variables and normalization, one obtains the monotonic quantities along the flow.

Proposition 3.1

Let $\xi$ be a solution to the normalized flow (19)–(20). Then

• (i) $\tau \mapsto \int _0^1|\xi (\tau ,s)|^2\mathrm{{d}}s$ is monotone increasing. Moreover, $\int _0^1|\xi (\tau ,s)|^2\mathrm{{d}}s \le \frac{1}{4\pi ^2}$ for all $\tau$.

• (ii) $\tau \mapsto \int _0^1\sigma (\tau ,s)\mathrm{{d}}s$ is monotone increasing with $\int _0^1\sigma (\tau ,s)\mathrm{{d}}s\le \int _0^1|\xi (\tau ,s)|^2\mathrm{{d}}s$ for all $\tau$.

• (iii) As $\tau \rightarrow \infty$, we have $\int _0^1\sigma (\infty ,s)\mathrm{{d}}s=\int _0^1|\xi (\infty ,s)|^2\mathrm{{d}}s$.

Proof

1. We multiply (19) by $\partial _\tau \xi$ and integrate in s from 0 to 1. After an integration by part and using $\partial _s\xi \cdot \partial _{s\tau }\xi =0$ (which follows from $|\partial _s\xi |=1$), we obtain that

$$\begin{aligned} \int _0^1 |\partial _\tau \xi |^2 \mathrm{{d}}s = \partial _\tau \left( \frac{1}{2}\int _0^1|\xi |^2 \mathrm{{d}}s\right) . \end{aligned}$$

(21)

Thus, $\tau \rightarrow \int _0^1|\xi |^2$ is monotone increasing. The upper bound follows from the Wirtinger’s inequality

$$\begin{aligned} \int _{0}^1|\xi |^2 \ \mathrm{{d}}s \le \frac{1}{4\pi ^2} \int _0^1|\partial _s\xi |^2 \mathrm{{d}}s = \frac{1}{4\pi ^2}. \end{aligned}$$

Again we have used that $\int _0^1\xi (\tau ,s)\mathrm{{d}}s=0$.

2. From the definition of $\sigma$, we have

$$\begin{aligned} \int _0^1\sigma (\tau ,s)\mathrm{{d}}s=\frac{\int _0^1\tilde{\sigma }\mathrm{{d}}s}{-L\partial _tL}=\frac{1}{-L\partial _tL(t(\tau ))}. \end{aligned}$$

By (ii) of Proposition 2.5, $-L\partial _tL$ is monotone decreasing in t thus in $\tau$. Thus, $\int _0^1\sigma (\tau ,s)\mathrm{{d}}s$ is monotone increasing in $\tau$. To prove the upper bound, we multiply (19) by $\xi$ and integrate in s. Integrating by parts and using that $|\partial _s\xi |=1$ we have

$$\begin{aligned} \frac{1}{2}\partial _t\int _0^1|\xi |^2 \mathrm{{d}}s =\int _0^1|\xi |^2 \mathrm{{d}}s-\int _0^1\sigma \mathrm{{d}}s. \end{aligned}$$

(22)

By (i), the left-hand side is larger than or equal to zero. Thus, the upper bound follows.

3. By (i) and (ii), $\lim _{\tau \rightarrow \infty }\int _0^1|\xi (\tau ,s)|^2\mathrm{{d}}s$ and $\lim _{\tau \rightarrow \infty }\int _0^1\sigma (\tau ,s)\mathrm{{d}}s$ exist and in the limit $\int _0^1|\xi (\infty ,s)|^2\mathrm{{d}}s-\int _0^1\sigma (\infty ,s)\mathrm{{d}}s\ge 0$. To see the limit is actually zero, we argue by contradiction. Suppose not, then there exist $\epsilon >0$ and $M_\epsilon >0$ such that $\int _0^1|\xi (\tau ,s)|^2\mathrm{{d}}s-\int _0^1\sigma (\tau ,s)\mathrm{{d}}s\ge \epsilon$ for all $\tau \ge M_\epsilon$. By (22) for any $\tau _1>\tau _2\ge M_\epsilon$,

$$\begin{aligned} \int _0^1|\xi (\tau _1,s)|^2\mathrm{{d}}s-\int _0^1|\xi (\tau _2,s)|^2\mathrm{{d}}s\ge 2\epsilon (\tau _1-\tau _2). \end{aligned}$$

By (i) the left-hand side is bounded from above by $\frac{1}{4\pi ^2}$. However, the right-hand side procedes to infinity as $\tau _1\rightarrow \infty$, which is a contradiction. Thus, we have shown that the limit is zero. $\square$

In the rest of the paper, we will work with the normalized flow (19)–(20). In Sect. 3.2, we show local well-posedness of the problem in some Hölder class. In Sect. 3.3, we explore the stationary solutions. In Sect. 3.4, we address global well-posedness and long time asymptotics of the solution for the initial data which are close to steady states, and show exponential decay of the solutions to a steady state. In Sect. 3.5, we address the global solvability without restrictions on the initial data but in a generalized sense. We stress that there is a one-to-one correspondence between the normalized flow and the original gradient flow (Sects. 2 and 3, resp.). After the backward change of variable and renormalization, we can infer the well-posedness of the original flow and the asymptotics of the flow near the extinction time.

3.2 Local Well-Posedness

In this section, we show the local well-posedness of the normalized flow (19)–(20). First, we introduce the function spaces we will work with. Given $\alpha ,\beta \in [0,1)$, $T\in (0,\infty )$ and $k\in \mathbb {N}\cup \{0\}$, let

$$\begin{aligned} C^{k+\alpha , \beta }([0,T]\times \mathbb {S}^1):=\{\xi :[0,T]\times \mathbb {S}^1\rightarrow \mathbb {R}^d: \Vert \xi \Vert _{k+\alpha ,\beta }<\infty \}, \end{aligned}$$

where

$$\begin{aligned} \Vert \xi \Vert _{k+\alpha ,\beta }:=\sup _{t}\Vert \xi (t,\cdot )\Vert _{C^{k+\alpha }(\mathbb {S}^1)}+\sup _s\sum _{j=0}^k\Vert \partial _s^j\xi (\cdot ,s)\Vert _{C^{\beta }([0,T])}. \end{aligned}$$

Here, we use $C^{k+\alpha }(\mathbb {S}^1)$ ($C^{\beta }([0,T])$) to denote the usual Hölder spaces for functions only depending on one variable. Similarly, let

$$\begin{aligned} C^{k+\alpha ,1+\beta }([0,T]\times \mathbb {S}^1):=\{\xi :[0,T]\times \mathbb {S}^1\rightarrow \mathbb {R}^d: \Vert \xi \Vert _{k+\alpha ,\beta }+\Vert \partial _t\xi \Vert _{k+\alpha ,\beta }<\infty \}. \end{aligned}$$

The local well-posedness result we want to prove in this section is as follows:

Theorem 1

Given any initial datum $\xi _0\in C^{2+\alpha }(\mathbb {S}^1)$ with $|\partial _s\xi _0(s)|=1$, $\int _0^1\xi _0(s)\mathrm{{d}}s=0$ , there exists $T>0$, which depends on $\Vert \xi _0\Vert _{C^{2+\alpha }(\mathbb {S}^1)}$, such that the Cauchy problem

$$\begin{aligned} \begin{aligned} \partial _t\xi =\partial _s(\sigma \partial _s\xi )+\xi ,\\ \partial _{ss}\sigma -|\partial _{ss}\xi |^2\sigma =-1,\\ \xi (0,s)=\xi _0(s), \end{aligned} \end{aligned}$$

(23)

has a unique solution $\xi \in C^{2+\alpha ,\alpha /2}([0,T]\times \mathbb {S}^1)$, $\partial _t\xi \in C^{\alpha , \alpha /2}([0,T]\times \mathbb {S}^1)$.

Remark 3.2

As in Remark 2.4, if $\xi$ is the solution emanating from $\xi _0$ provided by Theorem 1, then $\int _0^1\xi (t,s)\mathrm{{d}}s=0$ and $|\partial _s\xi (t,s)|\equiv 1$ for all $t\in [0,T]$.

The proof of Theorem 1 is based on the Banach fix point theorem, where we show the solution map $\xi \mapsto \sigma _\xi \mapsto \tilde{\xi }$ is a contraction in the Banach space $C^{2+\alpha ,\alpha /2}([0,T]\times \mathbb {S}^1)$. The proof is divided into several lemmas.

Lemma 3.3

For any $\xi \in C^{2+\alpha , \alpha /2}([0,T]\times \mathbb {S}^1)$ such that $\Vert \partial _{ss}\xi (t,\cdot )\Vert _{L^2(\mathbb {S}^1)}\ne 0$ for all $t\in [0,T]$, there exists a unique solution $\sigma \in C^{2+\alpha /2,\alpha /4}([0,T]\times \mathbb {S}^1)$ to the ODE

$$\begin{aligned} \begin{aligned}&\partial _{ss}\sigma (t,s)-\sigma (t,s)|\partial _{ss}\xi (t,s)|^2=-1. \end{aligned} \end{aligned}$$

(24)

The solution satisfies the estimate

$$\begin{aligned} \Vert \sigma \Vert _{2+\alpha /2,\alpha /4}\le C, \end{aligned}$$

for some constant C depending on d and $\Vert \partial _{ss}\xi \Vert _{\alpha ,\alpha /2}$, which is uniformly bounded if $\Vert \partial _{ss}\xi \Vert _{\alpha ,\alpha /2}$ is uniformly bounded.

Proof

Let $k(t,s):=|\partial _{ss}\xi (t,s)|$. Since $\Vert k(t,\cdot )\Vert _{L^2}\ne 0$, the inhomogeneous equation has a unique solution $\sigma (t,s)$ for each t. Furthermore, by the regularity theory for the elliptic equations $\sigma (t,\cdot )\in C^{2+\alpha }(\mathbb {S}^1)$ for each $t\in [0,T]$ and one has the estimate

$$\begin{aligned} \Vert \sigma \Vert _{2+\alpha ,0}\le C\left( \Vert k\Vert _{\alpha ,0}+1\right) \end{aligned}$$

(25)

for some universal $C>0$ (cf. Lemma 3.4 below for the estimate of $\Vert \sigma \Vert _{L^\infty }$).

To derive the regularity in t, we consider the equation for $\sigma (t_1,\cdot )-\sigma (t_2,\cdot )$ for any $0\le t_1<t_2\le T$:

$$\begin{aligned}&\partial _{ss}\left( \sigma (t_1,s)-\sigma (t_2,s)\right) -k(t_1,s)^2\left( \sigma (t_1,s)-\sigma (t_2,s)\right) \\&\quad =\sigma (t_2,s)\left( k^2(t_1,s)-k^2(t_2,s)\right) . \end{aligned}$$

By the Schauder estimate,

$$\begin{aligned} \left\| \sigma (t_1,s)-\sigma (t_2,s)\right\| _{2+\alpha /2,0}\le \tilde{C}\Vert k(t_1,s)-k(t_2,s)\Vert _{\alpha /2,0}, \end{aligned}$$

where $\tilde{C}$ is a constant depending on $\Vert \sigma \Vert _{\alpha /2,0}$ and $\Vert k\Vert _{\alpha /2,0}$, and it is uniformly bounded if $\Vert k\Vert _{\alpha /2,0}$ is uniformly bounded (recall the estimate (25) for $\sigma$). Here, we have used

$$\begin{aligned} \Vert \sigma (t_2,s)\left( k^2(t_1,s)-k^2(t_2,s)\right) \Vert _{\alpha /2,0}\le 6\Vert \sigma \Vert _{\alpha /2,0}\Vert k\Vert _{\alpha /2,0}\Vert k(t_1,\cdot )-k(t_2,\cdot )\Vert _{\alpha /2,0} \end{aligned}$$

to estimate the right-hand side. Since $k(t,s)\in C^{\alpha ,\alpha /2}([0,T]\times \mathbb {S}^1)$, it is not hard to see that

$$\begin{aligned} \Vert k(t_1,s)-k(t_2,s)\Vert _{\alpha /2,0}\le C\Vert k\Vert _{\alpha ,\alpha /2}|t_1-t_2|^{\alpha /2-\alpha /4} \end{aligned}$$

for some universal $C>0$. Combining the last two inequalities, we obtain the desired estimate of $\sigma (t,s)$. $\square$

Next we state the pointwise upper and lower bound on $\sigma (t,\cdot )$ in terms of $\Vert k(t,\cdot )\Vert _{L^2(\mathbb {S}^1)}$. This is a direct consequence of the upper and lower bound of the Green’s function of the Schrödinger operator $\partial _{ss}-k^2$ with the periodic boundary conditions (cf. Proposition A.3 in [43]).

Lemma 3.4

Let $\xi$ and $\sigma$ be the same as in Lemma 3.3. Then

$$\begin{aligned} \frac{e^{-\rho /2}}{\rho }\le \sigma (t,s)\le 1+\frac{1}{\rho },\text { where }\rho =\rho (t)=\int _0^1|\partial _{ss}\xi (t,s)|^2 \mathrm{{d}}s, \end{aligned}$$

for all $s\in \mathbb {S}^1$.

Fix an initial datum $\xi _0(s)$ as in Theorem 1. Firstly, we note that $\int _0^1|\partial _{ss}\xi _0(s)|^2 \mathrm{{d}}s \ge 4\pi ^2$. To see this let $Z:=\partial _s\xi _0$. From the assumptions on $\xi _0$ and using periodicity, we have $|Z(s)|\equiv 1$ and $\int _0^1Z=0$. Then the claimed inequality follows from the Wirtinger’s inequality $\int _0^1|\partial _sZ|^2 \mathrm{{d}}s\ge 4\pi ^2\int _0^1|Z|^2\mathrm{{d}}s =4\pi ^2$. Next we let $\delta _0:=\frac{1}{2}\Vert \partial _{ss}\xi _0\Vert _{L^2(\mathbb {S}^1)}\ge \pi$ and let

$$\begin{aligned} M_{\xi _0}:=\{\xi \in C^{2+\alpha ,\alpha /2}([0,T]\times \mathbb {S}^1):&\ \Vert \xi (t,s)\Vert _{2+\alpha ,\alpha /2}\le \Vert \xi _0\Vert _{2+\alpha }+\delta _0,\\&\text { and }\Vert \partial _{ss}\xi (t,\cdot )-\partial _{ss}\xi _0(\cdot )\Vert _{L^2(\mathbb {S}^1)}\le \delta _0\}. \end{aligned}$$

It is not hard to see that $M_{\xi _0}$ is a closed convex subset in $C^{2+\alpha ,\alpha /2}([0,T]\times \mathbb {S}^1)$.

Lemma 3.5

Given any $\xi \in M_{\xi _0}$, let $\sigma =\sigma _\xi$ be as in Lemma 3.3. Then there is $T_0=T_0(\Vert \xi _0\Vert _{2+\alpha })$ sufficiently small, such that for any $T\in (0,T_0]$ there exists a unique solution $\tilde{\xi }(t,s)\in M_{\xi _0}$ to the initial value problem

$$\begin{aligned} \begin{aligned} \partial _t\tilde{\xi }=\partial _s(\sigma \partial _s\tilde{\xi })+\tilde{\xi } \text { in } (0,T)\times \mathbb {S}^1,\quad \tilde{\xi }\big |_{t=0}=\xi _0. \end{aligned} \end{aligned}$$

(26)

Proof

Given any $\xi \in M_{\xi _0}$, by the triangle inequality $\delta _0\le \Vert \partial _{ss}\xi (t,\cdot )\Vert _{L^2(\mathbb {S}^1)}\le 3\delta _0$ for any $t\in [0,T]$. Thus, by Lemma 3.3, there exists a unique solution $\sigma =\sigma _\xi$ to (24) in the class $C^{2+\alpha /2, \alpha /4}([0,T]\times \mathbb {S}^1)$. Moreover, it satisfies $\Vert \sigma \Vert _{2+\alpha /2,\alpha /4}\le C$, where C depends on $\Vert \xi _0\Vert _{2+\alpha }$ since $\xi \in M_{\xi _0}$. From Lemma 3.4, we have that $\sigma \ge c_0>0$ for some $c_0$ depending on $\Vert \xi _0\Vert _{2+\alpha }$. Thus, the Eq. (26) is parabolic. By the classical well-posedness results for the parabolic equations, there is a unique solution $\tilde{\xi }\in C^{2+\alpha /2,1+\alpha /4}([0,T]\times \mathbb {S}^1)$ to the Eq. (26), and $\Vert \tilde{\xi }\Vert _{2+\alpha /2,1+\alpha /4}\le \tilde{C}$, where $\tilde{C}$ depends on $\Vert \xi _0\Vert _{2+\alpha }$ (cf. Sect. 9.2 in [26]).

Next we claim there is a small enough $T_0>0$ depending on $\Vert \xi _0\Vert _{2+\alpha }$, such that $\tilde{\xi }\in M_{\xi _0}$ for any $T\in (0,T_0]$. Indeed, the definition of the Hölder class and an interpolation yield that

$$\begin{aligned} \Vert \tilde{\xi }\Vert _{2+\alpha ,\alpha /2}\le \Vert \xi _0\Vert _{2+\alpha }+ 2T^{1-\alpha /4}\Vert \tilde{\xi }\Vert _{2+\alpha /2,1+\alpha /4}\le \Vert \xi _0\Vert _{2+\alpha }+ 2T^{1-\alpha /4}\tilde{C}. \end{aligned}$$

By the Hölder regularity of the solution, $\Vert \partial _{ss}\tilde{\xi }(t,\cdot )-\partial _{ss}\xi _0(\cdot )\Vert _{L^2(\mathbb {S}^1)}\le 2\tilde{C}t^{\alpha /2}$ for any $t\in [0,T]$. The claim then follows by taking $T_0$ sufficiently small depending on $\Vert \xi _0\Vert _{2+\alpha }$. $\square$

Lemma 3.6

Let $\xi , \tilde{\xi }\in M_{\xi _0}$ be as in Lemma 3.5. Then there is $T>0$ sufficiently small depending on $\Vert \xi _0\Vert _{2+\alpha }$, such that the mapping $\xi \mapsto \tilde{\xi }$ is a contraction.

Proof

Given $\xi _1, \xi _2\in M_{\xi _0}$, let $\sigma _1$ and $\sigma _2$ be the solutions to (24) with respect to $\xi _1$ and $\xi _2$ correspondingly. Let $k_i:=|\partial _{ss}\xi _i|$, $i=1,2$. Then $\sigma _1-\sigma _2$ satisfies the equation

$$\begin{aligned} \partial _{ss}(\sigma _1-\sigma _2)-k_1^2(\sigma _1-\sigma _2)=\sigma _2(k_1^2-k_2^2). \end{aligned}$$

By the similar arguments as in Lemma 3.3 (with slightly more involved estimates when dealing with the regularity in time due to the more complicated right-hand side), we have

$$\begin{aligned} \Vert \sigma _1-\sigma _2\Vert _{2+\alpha /2,\alpha /4}\le C\Vert \xi _1-\xi _2\Vert _{2+\alpha ,\alpha /2} \end{aligned}$$

(27)

for some $C=C(\Vert \xi _0\Vert _{2+\alpha })$.

Let $\tilde{\xi }_1, \tilde{\xi }_2$ be the solutions to (26) with respect to $\sigma _1, \sigma _2$ correspondingly. Then $\zeta :=\tilde{\xi }_1-\tilde{\xi }_2$ satisfies the equation

$$\begin{aligned} \partial _t\zeta =\partial _s(\sigma _1\partial _s\zeta )+\zeta +\partial _s\left( (\sigma _1-\sigma _2)\partial _s\tilde{\xi }_2\right) ,\quad \zeta (0,s)=0. \end{aligned}$$

By the parabolic Schauder estimate, we have

$$\begin{aligned} \Vert \zeta \Vert _{3+\alpha /2,1+\alpha /4}\le C\Vert \sigma _1-\sigma _2\Vert _{2+\alpha /2,\alpha /4} \end{aligned}$$

(28)

for some $C=C(\Vert \xi _0\Vert _{2+\alpha })$. An interpolation together with (28) and (27) yields

$$\begin{aligned} \Vert \zeta \Vert _{2+\alpha ,\alpha /2}&\le CT^{1-\alpha /4}\Vert \zeta \Vert _{3+\alpha /2,1+\alpha /4}\\&\le CT^{1-\alpha /4}\Vert \sigma _1-\sigma _2\Vert _{2+\alpha /2,\alpha /4}\\&\le CT^{1-\alpha /4}\Vert \xi _1-\xi _2\Vert _{2+\alpha ,\alpha /2}. \end{aligned}$$

Here C might be different from line to line but all depend on $\Vert \xi _0\Vert _{2+\alpha }$. Choosing T to be sufficiently small depending on $\Vert \xi _0\Vert _{2+\alpha }$, we obtain

$$\begin{aligned} \Vert \tilde{\xi }_1-\tilde{\xi }_2\Vert _{2+\alpha ,\alpha /2}=\Vert \zeta \Vert _{2+\alpha ,\alpha /2}\le \frac{1}{2}\Vert \xi _1-\xi _2\Vert _{2+\alpha ,\alpha /2}, \end{aligned}$$

which completes the proof. $\square$

In the end, we provide a proof for the local-posedness of our problem (23).

Proof of Theorem 1

By Lemma 3.6, the mapping $\xi \mapsto \tilde{\xi }$ is a contraction on $M_{\xi _0}$ provided T is sufficiently small depending on $\xi _0$. By the Schauder fixed point theorem, there is a unique function $\xi \in M_{\xi _0}$ with $\xi =\tilde{\xi }$. It is not hard to see that such fixed point $\xi$ is a solution to the initial value problem (23). The solution is Hölder $\alpha /2$ continuous in t up to $t=0$. The regularity of $\partial _t\xi$ for $t>0$ follows immediately from the interior regularity of the parabolic equation. $\square$

3.3 Stationary Solutions to the Normalized Flow

Our goal is to study the global well-posedness of the normalized Eqs. (19)–(20), and the long-time asymptotics of the solution. Before that, we investigate the stationary solutions to the normalized equation, and show in this case the Lagrange multiplier $\sigma$ satisfies an ODE which has a first integral. Just like as for the conventional curve-shortening flow [1], the stationary solutions are not necessarily circles in our case. However, we will show that the circle (with $\sigma \equiv \frac{1}{4\pi ^2}$) is the only solution to the ODE if one assumes that the curve is simple and $\int _0^1\sigma \mathrm{{d}}s$ satisfies a lower bound $\int _0^1\sigma \mathrm{{d}}s \ge \frac{27}{32}\frac{1}{4\pi ^2}$.

We start by recalling the stationary equation $\xi :{\mathcal {S}}^1\rightarrow \mathbb {R}^d$

$$\begin{aligned} \begin{aligned} \partial _s(\sigma \partial _s\xi )+ \xi&=0,\ |\partial _s\xi |=1. \end{aligned} \end{aligned}$$

(29)

Proposition 3.7

Let $\xi \in C^2(\mathbb {S}^1;\mathbb {R}^d)$ be a solution to (29). Then $\xi$ is a plane curve curvature of which satisfies $k(s)>0$ for all s.

Proof

Differentiating the equation in s and using the constraint $|\partial _s\xi |=1$ it is not hard to see that $\sigma$ satisfies

$$\begin{aligned} \partial _{ss}\sigma -\sigma |\partial _{ss}\xi |^2 = -1. \end{aligned}$$

We have $\sigma \in C^2(\mathbb {S}^1)$ by the elliptic estimate. Furthermore, by the strong maximum principle $\sigma >0$. This together with the equality $-\xi =\partial _s\sigma \partial _s\xi +\sigma \partial _{ss}\xi$ implies that $\xi$ is a plane curve, since $\xi , \partial _s\xi$ and $\partial _{ss}\xi$ are in the same plane.

For the arc-length parametrized curve, we have $\partial _{ss}\xi =k N$, where N is the unit inner normal along the curve and k is the curvature. Differentiating the equation of $\xi$ and using $\partial _sN=-k\partial _s\xi$, we get

$$\begin{aligned} \left( \partial _{ss}\sigma -\sigma k^2+1\right) \partial _s\xi + \left( 2\partial _s\sigma k +\sigma \partial _sk\right) N=0. \end{aligned}$$

Thus, $\sigma \partial _s k+2\partial _s \sigma k =0$, which implies

$$\begin{aligned} \sigma ^2 k=const. \end{aligned}$$

(30)

Since $\int _0^1k(s)\mathrm{{d}}s=2\pi$ for a regular closed plane curve, and $\sigma >0$, one has $k(s)>0$ for all $s\in \mathbb {S}^1$. $\square$

In the next proposition, we derive a first-order ODE of $\sigma$.

Proposition 3.8

Let $\xi \in C^{2}(\mathbb {S}^1;\mathbb {R}^d)$ be a solution to (29). Let $\tau :=\sigma ^2$. Then $\tau$ satisfies the first integral

$$\begin{aligned} \frac{1}{2}(\partial _s\tau )^2+V(\tau )=\lambda ,\quad V(\tau ):=4\tau ^{3/2}-6\bar{\tau } \tau . \end{aligned}$$

Here $\bar{\tau }=\int _0^1\tau ^{1/2}ds$ and $\lambda =V(\tau _e)$ is a fixed constant, where $\tau _e$ is any extreme value of $\tau$. Moreover, we have $\lambda \in [-2\bar{\tau }^3,0)$.

Proof

First multiplying (29) by $\xi$ and an integration yield

$$\begin{aligned} \int _0^1\sigma \mathrm{{d}}s =\int _0^1|\xi |^2 \mathrm{{d}}s. \end{aligned}$$

(31)

Then, we multiply (29) by $\partial _s\xi$ and use $|\partial _s\xi |=1$ to obtain

$$\begin{aligned} \partial _s\sigma +\frac{1}{2} \partial _s|\xi |^2=0 \text { for all } s\in \mathbb {S}^1, \end{aligned}$$

which together with (31) gives

$$\begin{aligned} \sigma +\frac{1}{2} |\xi |^2=\frac{3}{2}\bar{\sigma },\quad \bar{\sigma }:=\int _0^1\sigma . \end{aligned}$$

(32)

On the other hand, (29) together with the orthogonality $\partial _s\xi \cdot \partial _{ss}\xi =0$ yields

$$\begin{aligned} |\partial _s\sigma |^2+\sigma ^2k^2 =|\xi |^2. \end{aligned}$$

(33)

Since $\sigma k^2=\partial _{ss}\sigma +1$ by the equation of $\sigma$, we obtain from (33)

$$\begin{aligned} |\partial _s\sigma |^2+\sigma (\partial _{ss}\sigma + 1)=\frac{1}{2}\partial _{ss}(\sigma ^2) + \sigma =|\xi |^2, \end{aligned}$$

which together with (32) gives the ODE of $\sigma$

$$\begin{aligned} \frac{1}{2}\partial _{ss}(\sigma ^2)=3\bar{\sigma }- 3 \sigma . \end{aligned}$$

(34)

Remember that we have set $\tau :=\sigma ^2$. Let us rewrite the above equation in terms of $\tau$ as $\partial _{ss}\tau =6\bar{\tau } -6\tau ^{1/2}$, where $\bar{\tau }=\bar{\sigma }=\int _0^1\tau ^{1/2} \mathrm{{d}}s$. Multiplying both sides by $\partial _s\tau$ and integrating, we obtain

$$\begin{aligned} \frac{1}{2}(\partial _s\tau )^2+V(\tau )=\lambda ,\quad V(\tau ):=4\tau ^{3/2}-6\bar{\tau } \tau . \end{aligned}$$

(35)

At s such that $\tau (s)=\tau _e$, we have $\partial _s\tau (s)=0$, thus $\lambda =V(\tau _e)$ from (35). The potential V satisfies $V''(\tau )=3\tau ^{-1/2}$, hence it is convex on $(0,\infty )$. We note from (32) and the definition $\tau =\sigma ^2$ that $\tau \in (0,3\bar{\tau }/2)$. This implies $V(\tau )\in [-2\bar{\tau }^3, 0)$ with $\min V(\tau )=V(\bar{\tau }^2)=-2\bar{\tau }^3$. Thus, $\lambda =V(\tau _e)\in [-2\bar{\tau }^3, 0)$. $\square$

If $\xi$ is an m-covered circle, $m\in \mathbb {N}$, then it is easy to see from (32)–(33) that $\sigma =\frac{1}{4\pi ^2m^2}$. In general, it is possible to apply the method for the proof of Theorem A in [1] to classify solutions $\tau$ (hence $\sigma$) to the ODE (35). In the next proposition, we show that if $\xi$ is simple and $\sigma$ is close to $\frac{1}{4\pi ^2}$ ($m=1$), then $\xi$ is a circle and $\sigma =\frac{1}{4\pi ^2}$.

Proposition 3.9

Let $\xi \in C^{2}(\mathbb {S}^1;\mathbb {R}^d)$ be a solution to (29). Assume $\xi$ is simple, i.e., $\xi (s_1)\ne \xi (s_2)$ for $s_1\ne s_2$. Assume $\int _0^1\sigma \ \mathrm{{d}}s\ge \frac{27}{32}\frac{1}{4\pi ^2}$. Then $\sigma \equiv \frac{1}{4 \pi ^2}$, and $\xi$ is a circle centered at 0 with radius $\frac{1}{2\pi }$.

Proof

1. By Proposition 3.8, $\xi$ is a plane curve with curvature $k>0$. Moreover, $\xi$ is simple by assumption. Thus, by the four-vertex theorem, k(s) has at least four critical points. Let $0\le s_1<s_2<\cdots {<} s_J\le 1$ with index $J\ge 4$ be the critical points. Since $\sigma ^2 k = const$ and $\sigma >0$, $\sigma$ has the same critical points at $s_i$, i.e., $\partial _s\sigma (s_i)=0$ for $i\in J$. We claim that

$$\begin{aligned} \frac{1}{|I_i|}\int _{I_i}\sigma (s)\mathrm{{d}}s =\bar{\sigma },\quad I_i=(s_i,s_{i+1}),\ |I_i|=s_{i+1}-s_i, \ i\in J. \end{aligned}$$

(36)

Indeed, an integration of (33) in s over $I_i$ together with (32) gives

$$\begin{aligned} \int _{I_i}(\partial _s\sigma )^2 + \int _{I_i}\sigma ^2k^2 =\int _{I_i} |\xi |^2 = \int _{I_i}(3\bar{\sigma }-2\sigma ). \end{aligned}$$

On the other hand, multiplying the equation of $\sigma$ by $\sigma$ and an integration by parts yield,

$$\begin{aligned} \int _{I_i}(\partial _s\sigma )^2 + \int _{I_i}\sigma ^2k^2= \int _{I_i}\sigma . \end{aligned}$$

Here, we have used $\partial _s\sigma (s_i)=0$; hence, the boundary term in the integration by parts vanishes. From the above two equalities, we conclude

$$\begin{aligned} \int _{I_i}(3\bar{\sigma }-2\sigma )= \int _{I_i}\sigma . \end{aligned}$$

Thus, (36) follows.

2. We show that if $\bar{\sigma } \ge \frac{27}{32}\frac{1}{4\pi ^2}$, then $\sigma (s)\equiv \bar{\sigma }$ in the intervals $I_i$ with $|I_i|\le \frac{1}{4}$. In particular, since ${J}\ge 4$ there is always an open interval where $\sigma \equiv \bar{\sigma }$ there.

Take $I_i$ such that $|I_i|\le \frac{1}{4}$. For simplicity, we write I instead of $I_i$ in the sequel. We multiply both sides of (34) by $\sigma$ and integrate from $s_i$ to $s_{i+1}$. An integration by parts gives

$$\begin{aligned} \int _{I}\sigma (\partial _s\sigma )^2 \mathrm{{d}}s = 3\int _{I} (\sigma ^2-\bar{\sigma }\sigma ) \mathrm{{d}}s. \end{aligned}$$

(37)

On the other side, by a generalized Beckner-type inequality (see [14, Lemma 4] with $q=\frac{4}{3}$, $p=\frac{3}{2}$ and $f=\sigma ^{3/2}$; see also [25] for a link with “unbalanced optimal transport”), we have

$$\begin{aligned} |I|^{-1/2}\Vert \sigma \Vert _{L^2(I)}\left( \Vert \sigma \Vert _{L^2(I)}^2 -|I|^{-1} \Vert \sigma \Vert _{L^1(I)}^2\right) \le \frac{9}{2}\frac{|I|^2}{4\pi ^2}\int _I \sigma (\partial _s\sigma )^2 \mathrm{{d}}s. \end{aligned}$$

(38)

Here, we have used that $C_P=\frac{|I|^2}{4\pi ^2}$ is the optimal Poincaré constant with respect to the interval I. Combining (37) and (38), we arrive at

$$\begin{aligned} |I|^{-1/2}\Vert \sigma \Vert _{L^2(I)}\left( \Vert \sigma \Vert _{L^2(I)}^2 -|I|^{-1} \Vert \sigma \Vert _{L^1(I)}^2\right) \le \frac{27}{2}\frac{|I|^2}{4\pi ^2} \left( \Vert \sigma \Vert _{L^2(I)}^2 - |I|^{-1} \Vert \sigma \Vert _{L^1(I)}^2\right) . \end{aligned}$$

Note that if $|I|\le \frac{1}{4}$ and $\bar{\sigma }> \frac{27}{32}\frac{1}{4\pi ^2}$, then

$$\begin{aligned} |I|^{-1/2}\Vert \sigma \Vert _{L^2(I)}> \frac{27}{2}\frac{|I|^2}{4\pi ^2}. \end{aligned}$$

Indeed, this immediately follows from Hölder’s inequality and (36):

$$\begin{aligned} |I|^{-1/2}\Vert \sigma \Vert _{L^2(I)}\ge |I|^{-1}\Vert \sigma \Vert _{L^1}=\bar{\sigma }>\frac{27}{2}\frac{|I|^2}{4\pi ^2}. \end{aligned}$$

Thus, $\Vert \sigma \Vert _{L^2(I)}^2=|I|^{-1}\Vert \sigma \Vert _{L^1(I)}^2$. From the equality case of the Hölder’s inequality $|\sigma |=const$ a.e. in I. This together with the continuity of $\sigma$ yields that $\sigma (s)\equiv \bar{\sigma }$ in I.

3. In the last step, we show that $\sigma (s)=1/(4\pi ^2)$ for all $s\in \mathbb {S}^1$. Indeed, from step 2 above there exists an interval, say, $(0,s_0)$ for some small $s_0>0$ such that $\sigma \equiv \bar{\sigma }$ there. Let $\tau :=\sigma ^2$. By the Picard theorem the initial value problem ($\tau =\sigma ^2$)

$$\begin{aligned} \partial _{ss}\tau = 6 (\bar{\sigma }-\sqrt{\tau }),\quad \tau (s_0)=\bar{\sigma }^2, \quad \partial _s\tau (s_0)=0, \end{aligned}$$

has a unique solution in $(s_0,s_0+\delta )$ for some $\delta >0$. Since the constant function $\tau \equiv \bar{\sigma }^2$ is a solution, thus necessarily $\tau =\bar{\sigma }^2$ in $(s_0, s_0+\delta )$. This shows that $\tau$, thus $\sigma$, is identically $\bar{\sigma }$ in the whole circle $\mathbb {S}^1$.

With this at hand, (32) yields that $|\xi |^2=const=\bar{\sigma }$ in $\mathbb {S}^1$. Thus, $\xi$ is a circle centered at the origin. Since the length of the curve is equal to 1, then $|\xi |=\frac{1}{2\pi k}$ for $k\in \{1,2,\cdots \}$. By our assumption $\bar{\sigma } \ge \frac{27}{32}\frac{1}{4\pi ^2}$ (or because the curve is simple), we should have $|\xi |=\frac{1}{2\pi }$ and $\bar{\sigma }=\frac{1}{4\pi ^2}$. $\square$

Remark 3.10

From the proof of Proposition 3.9, one can see that we only need the existence of two critical points $s_1$, $s_2$ of the curvature function k(s) with $|s_1-s_2|\le 1/4$. This can also be achieved by assuming the symmetry property $\xi (s)=-\xi (s+1/2)$ instead of assuming that the curve is simple.

3.4 Global Well-Posedness and Exponential Stability

In this section, we study the global well-posedness of the normalized flow (19)–(20) under the assumption that initially the curve is $C^2$ close to the circle

$$\begin{aligned} w_0(s):=\frac{1}{2\pi }(\cos (2\pi s), \sin (2\pi s), 0,\cdots , 0). \end{aligned}$$

(39)

We will mainly show the uniform (in time) boundedness of curvature $|\partial _{ss}\xi |^2$, which yields that the time T in the local well-posedness result has a uniform lower bound. The main idea of the proof is to show that under some smallness assumption at the initial time, the parabolicity is preserved along the flow, i.e., $\sigma >c_0$ pointwise for any $t<T$, where $c_0>0$ is some absolute constant.

The proof of the result is based on a dynamical system approach. We let

$$\begin{aligned} \mathcal {C}:=\{c w_0(s+\theta ): c,\theta \in \mathbb {R}\} \end{aligned}$$

denote the manifold generated by $w_0$ which is invariant under the dilation and rotation. For each $t\in (0,T)$, we decompose

$$\begin{aligned} \xi (t,s)=\tilde{\xi }(t,s)+c(t)w_0(s+\theta (t)), \end{aligned}$$

(40)

where $c(t)w_0(\cdot +\theta (t))$ is the $L^2$ projection of $\xi (t)$ onto $\mathcal {C}$, i.e., for each t fixed

$$\begin{aligned} c(t)w_0(\cdot +\theta (t))\in {\text {argmin}}_{w\in \mathcal {C}} \left\{ \int _{\mathbb {S}^1}|\xi (t,\cdot )-w(\cdot )|^2d\mathcal {H}^1\right\} . \end{aligned}$$

We remark that minimum are achieved by considering the minimization problem over the finite dimensional parameter space

$$\begin{aligned} \inf _{c,\theta \in \mathbb {R}}\mathcal {F}_{\xi }(c,\theta ),\quad \mathcal {F}_\xi (c,\theta ):= \int _{{\mathcal {S}}^1}|\xi (\cdot )-cw_0(\cdot +\theta )|^2 d\mathcal {H}^1. \end{aligned}$$

The first derivatives $\partial \mathcal {F}_\xi (c,\theta )/\partial c$ and $\partial \mathcal {F}_{\xi }(c,\theta )/\partial \theta$ vanish at the minimizers $(c(t),\theta (t))$, yielding the following orthogonality conditions

$$\begin{aligned} \int _0^1 \tilde{\xi }(t,s) \cdot w_0 (s+\theta (t)) \mathrm{{d}}s = \int _0^1 \tilde{\xi }(t,s)\cdot c(t)\partial _sw_0(s+\theta (t)) \mathrm{{d}}s =0. \end{aligned}$$

(41)

Note that since $c\mapsto \mathcal {F}_{\xi }(c,\theta )$ is strictly convex, there is indeed a unique c(t) associated with $\xi (t,\cdot )$ for each $t\in [0,T]$.

Now we derive the evolution of the parameters c(t) and $\theta (t)$ (assume for now they are differentiable. For more detailed discussion, we refer to Lemma 3.11). By using the equation of $\xi$, we obtain the equation of $\tilde{\xi }$

$$\begin{aligned} \begin{aligned} \partial _t\tilde{\xi }(t,s)&=\tilde{\xi }(t,s)+\partial _s(\sigma (t,s)\partial _s\tilde{\xi }(t,s))+\\&\quad +c(t)\partial _s\left( (\sigma (t,s)-\frac{1}{4\pi ^2})\partial _s w_0(s+\theta (t))\right) -\dot{c}(t)w_0(s+\theta (t))\\&\quad -c(t)\dot{\theta }(t)\partial _sw_0(s+\theta (t)), \end{aligned} \end{aligned}$$

(42)

where we have used the relation $w_0+\frac{1}{4\pi ^2}\partial _{ss}w_0=0$. We multiply (42) by $\tilde{\xi }(t,\cdot )$, $w_0(\cdot +\theta (t))$ and $\partial _sw_0(\cdot +\theta (t))$ and integrate over $\mathbb {S}^1$. Using the orthogonality condition (41) we quantify the evolution of $\Vert \xi (t,\cdot )\Vert _{L^2}^2$ as well as of the parameters c(t) and $\theta (t)$ for $t<T$:

$$\begin{aligned} \begin{aligned} \frac{1}{2}\partial _t\int _0^1|\tilde{\xi }|^2 \mathrm{{d}}s&= \int _0^1|\tilde{\xi }|^2 -\int _0^1\sigma |\partial _s\tilde{\xi }|^2 \mathrm{{d}}s -c(t)\int _0^1\sigma \partial _sw_0(\cdot +\theta (t))\cdot \partial _s\tilde{\xi } \mathrm{{d}}s, \\ \dot{c}(t)&=\left( 1-4\pi ^2\int _0^1\sigma \mathrm{{d}}s\right) c(t)-4\pi ^2\int _0^1\sigma \partial _sw_0(\cdot +\theta (t))\cdot \partial _s\tilde{\xi } \mathrm{{d}}s,\\ c(t)\dot{\theta }(t)&=4\pi ^2\int _0^1\sigma w_0(\cdot +\theta (t))\cdot \partial _s\tilde{\xi } \mathrm{{d}}s. \end{aligned} \end{aligned}$$

(43)

From the unit speed constraint $|\partial _s\xi (t,s)|=|\partial _sw_0(s+\theta (t))|= 1$ for all $(t,s)\in [0,T]\times \mathbb {S}^1$, we obtain

$$\begin{aligned} |\partial _s\tilde{\xi }(t,s)|^2+2c(t)\partial _s\tilde{\xi }(t,s)\cdot \partial _sw_0(s+\theta (t))+c(t)^2=1. \end{aligned}$$

(44)

Integrating over $\mathbb {S}^1$ and using (41) (note that $\int _0^1\partial _s\tilde{\xi }\cdot \partial _sw_0 \mathrm{{d}}s=-\int _0^1\tilde{\xi }\cdot \partial _{ss}w_0=0$ since $\partial _{ss}w_0=-4\pi ^2w_0$) yields

$$\begin{aligned} \int _0^1|\partial _s\tilde{\xi }|^2 \mathrm{{d}}s =1-c(t)^2. \end{aligned}$$

(45)

Here and in the sequel for brevity, we write $w_0$ and $\partial _sw_0$ instead of $w_0(s+\theta (t))$ and $\partial _sw_0(s+\theta (t))$, respectively.

Lemma 3.11

Suppose $\xi (t,s)\in C^{2+\alpha ,\alpha /2}([0,T]\times \mathbb {S}^1)$, $\partial _t\xi (t,s)\in C^{\alpha ,\alpha /2}((0,T]\times \mathbb {S}^1)$ is a classical solution to the normalized flow (19)–(20). Suppose the $L^2$ projection of $\xi (0,\cdot )$ onto $\mathcal {C}$ is not 0, i.e. $c(0)\ne 0$. Then there exist $t_0\in (0,T]$ depending on c(0), $\Vert \xi (0,\cdot )\Vert _{C^2(\mathbb {S}^1)}$, and parameters $(c(t),\theta (t))\in C^0([0,t_0])\cap C^1((0,t_0])$, such that (40)–(43) are satisfied for all $t\in (0,t_0)$.

Proof

Given a classical solution $\xi$, there exist $(c(0),\theta (0))\in \mathbb {R}\times [0,2\pi )$ be such that

$$\begin{aligned} \mathcal {F}_{\xi (0)}(c(0),\theta (0))=\min _{(c,\theta )\in \mathbb {R}\times [0,2\pi )} \mathcal {F}_{\xi (0)}(c,\theta ). \end{aligned}$$

Indeed, since $\mathcal {F}_{\xi }$ is continuous, strictly convex in c and $2\pi$-periodic in $\theta$, the minimum is realized. We consider the ODE system

$$\begin{aligned} \begin{aligned} {\dot{c}}(t)&= c(t)-4\pi ^2 \int _0^1 \sigma \partial _s w_0(s+\theta (t)) \cdot \partial _s\xi (t,s) \mathrm{{d}}s,\\ c(t)\dot{\theta }(t)&=4\pi ^2 \int _0^1\sigma w_0(s+\theta (t))\cdot \partial _s\xi (t,s)\mathrm{{d}}s,\\ (c(t),\theta (t))\big |_{t=0}&=(c(0),\theta (0)). \end{aligned} \end{aligned}$$

(46)

Using $|\partial _s\xi |=1$ and the bound of $\sigma$ in Lemma 3.4, we have that the right-hand side of the system is bounded by constants depending on $\Vert \xi (0,\cdot )\Vert _{C^2(\mathbb {S}^1)}$. Thus, by the Picard–Lindelöf theorem, there exists a unique solution $(c(t),\theta (t))$ defined in $[0,t_0]$ for some $t_0\in (0,T]$, value of which lies in a neighborhood of $(c(0),\theta (0))$. Here $t_0$ and the size of the neighborhood depend on c(0) and $\Vert \xi (0,\cdot )\Vert _{C^2(\mathbb {S}^1)}$. Furthermore, from the t-regularity of $\xi$ and $\sigma$, we can conclude that $(c(t),\theta (t))\in C^0([0,t_0])\cap C^1((0,t_0])$.

Let $\tilde{\xi }(t,s):=\xi (t,s)-c(t)w_0(s+\theta (t))$, $t\in [0,t_0]$. From the equation of $\xi$, one easily derives the equation of $\tilde{\xi }$ in (42). Moreover, using $|\partial _sw_0|=1$ and $\partial _sw_0\cdot w_0=0$, the ODE system (46) can be rewritten in terms of $\tilde{\xi }$ as in (43). We claim that (42) together with the ODE for $(c(t),\theta (t))$ in (43) implies the orthogonality conditions (41). Indeed, let

$$\begin{aligned} A(t):=\int _0^1 \tilde{\xi }(t,s)\cdot w_0(s+\theta (t)) \mathrm{{d}}s,\ B(t):=\int _0^1 \tilde{\xi }(t,s)\cdot \partial _s w_0(s+\theta (t)) \mathrm{{d}}s,\ t\in [0,t_0]. \end{aligned}$$

We have $A(0)=B(0)=0$. Direct differentiation yields

$$\begin{aligned} \begin{aligned} {\dot{A}}(t)&= \int _0^1\partial _t\tilde{\xi }(t,s)\cdot w_0(s+\theta (t)) \mathrm{{d}}s+ \dot{\theta }(t) B(t),\\ {\dot{B}}(t)&= \int _0^1 \partial _t\tilde{\xi }(t,s) \cdot \partial _s w_0(s+\theta (t))\mathrm{{d}}s - \frac{\dot{\theta }(t)}{4\pi ^2} A(t). \end{aligned} \end{aligned}$$

(47)

Multiplying (42) by $w_0(s+\theta (t))$ and $\partial _s w_0(s+\theta (t))$, and using the ODE of $(c(t),\theta (t))$ in (43), we obtain

$$\begin{aligned} \begin{aligned} \int _{0}^1 \partial _t\tilde{\xi }(t,s)\cdot w_0(s+\theta (t)) \mathrm{{d}}s = A(t),\ \int _0^1 \partial _t\tilde{\xi }(t,s) \cdot \partial _s w_0(s+\theta (t))\mathrm{{d}}s = B(t). \end{aligned} \end{aligned}$$

(48)

Applying (48) to (47) yields

$$\begin{aligned} \left( \begin{array}{ll} {\dot{A}}(t)\\ {\dot{B}}(t) \end{array}\right) = \left( \begin{array}{ll} 1 &{} \dot{\theta }(t)\\ -\frac{\dot{\theta }(t)}{4\pi ^2} &{} 1 \end{array} \right) \left( \begin{array}{ll} A(t)\\ B(t). \end{array}\right) \end{aligned}$$

Since the coefficient matrix is nonsingular and since $A(0)=B(0)=0$ we have that $A(t)=B(t)=0$ for all $t\in [0,t_0]$, which completes the proof for the claim. $\square$

The rest of the argument goes as follows:

• (i) in Lemmas 3.12 and 3.13 we prove the decay estimate for $t\mapsto \Vert \xi (t,\cdot )\Vert _{L^2}$ and $t\mapsto \Vert \partial _s\xi (t,\cdot )\Vert _{L^2}$, $t\in (0,t_0]$, under the initial assumptions that $\int _0^1\sigma (0,s)\mathrm{{d}}s \ge \frac{2}{3}\frac{1}{4\pi ^2}$ and $c(0)^2\ge \frac{1}{2}$. As a corollary of the decay estimate and Proposition 3.1 (ii), these initial assumptions are preserved along the flow. We remark that the initial assumptions are always satisfied if initially $\xi$ is sufficiently close to the stationary solution $w_0$ (cf. Theorem 2).

• (ii) in Lemma 3.16, we derive pointwise oscillation estimate for the Lagrange multiplier $\sigma$. This is a crucial step, since $\sigma$ appears as the parabolicity coefficient in the equation of $\xi$.

• (iii) Lemma 3.16 together with the decay estimate in (i) would yield the uniform (in $t_0$) boundedness of $\Vert \xi (t,\cdot )\Vert _{C^{2+\alpha }(\mathbb {S}^1)}$. Thus, we have a lower bound on the time step in the iteration procedure. These are proved in Theorem 2, where we show the global well-posedness results for initial datum sufficiently close to $w_0$.

In the sequel $t_0$, $\tilde{\xi }$ and $(c(t), \theta (t))$ are those given by Lemma 3.11 associated with a classical solution $\xi (t,s)\in C^{2+\alpha , \alpha /2}([0,T]\times \mathbb {S}^1)$, $\partial _t\xi (t,s)\in C^{\alpha ,\alpha /2}((0,T]\times \mathbb {S}^1)$ with $c(0)\ne 0$. We also let

$$\begin{aligned} \bar{\sigma }(t):=\int _0^1\sigma (t,s)\mathrm{{d}}s. \end{aligned}$$

Lemma 3.12

Assume that

$$\begin{aligned} \bar{\sigma }(0) \ge \frac{2}{3}\frac{1}{4\pi ^2}, \end{aligned}$$

(49)

Then $t\mapsto \Vert \tilde{\xi }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)}$ is monotone decreasing in $[0,t_0]$. Moreover,

$$\begin{aligned} \Vert \tilde{\xi }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)}\le e^{-\frac{1}{3}t}\Vert \tilde{\xi }(0,\cdot )\Vert _{L^2(\mathbb {S}^1)}, \quad t\in [0,t_0]. \end{aligned}$$

(50)

Proof

We will derive a differential inequality on $\Vert \tilde{\xi }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)}$ using its evolution equation from (43). First, we multiply $\sigma$ to the both sides of (44) and integrate over $\mathbb {S}^1$

$$\begin{aligned} (1-c^2)\bar{\sigma }=\int _0^1\sigma |\partial _s\tilde{\xi }|^2 \mathrm{{d}}s +2c\int _0^1\sigma \partial _s\tilde{\xi }\cdot \partial _sw_0 \mathrm{{d}}s. \end{aligned}$$

(51)

Thus, (43) can be rewritten as

$$\begin{aligned} \frac{1}{2}\partial _t\int _0^1|\tilde{\xi }|^2&= \int _0^1 |\tilde{\xi }|^2 - \int _0^1\sigma |\partial _s\tilde{\xi }|^2 -\frac{1-c^2}{2}\bar{\sigma } +\frac{1}{2}\int _0^1\sigma |\partial _s\tilde{\xi }|^2 \mathrm{{d}}s\\&=\int _0^1 |\tilde{\xi }|^2 - \frac{1}{2}\int _0^1\sigma |\partial _s\tilde{\xi }|^2-\frac{1-c^2}{2}\bar{\sigma }. \end{aligned}$$

Applying (45) to the last term of the above equation, we obtain

$$\begin{aligned} \begin{aligned} \frac{1}{2}\partial _t\int _0^1|\tilde{\xi }|^2&=\int _0^1 |\tilde{\xi }|^2 - \frac{1}{2}\int _0^1\sigma |\partial _s\tilde{\xi }|^2-\frac{1}{2}\bar{\sigma }\int _0^1|\partial _s\tilde{\xi }|^2 \mathrm{{d}}s\\&\le \int _0^1 |\tilde{\xi }|^2-\frac{1}{2}\bar{\sigma }\int _0^1|\partial _s\tilde{\xi }|^2 ds. \end{aligned} \end{aligned}$$

(52)

We claim that (40) implies

$$\begin{aligned} \int _0^1|\tilde{\xi }(t,\cdot )|^2 \mathrm{{d}}s\le \frac{1}{16\pi ^2}\int _0^1|\partial _s\tilde{\xi }(t,\cdot )|^2 \mathrm{{d}}s. \end{aligned}$$

(53)

To see this, one can employ Fourier expansion of $\tilde{\xi }(t,\cdot )$. By $\int _0^1\tilde{\xi }=0$ as well as the orthogonality condition (41), one finds that the zero and first-order Fourier coefficients of $\tilde{\xi }$ are zero.

Applying (53) to (52), we obtain

$$\begin{aligned} \frac{1}{2}\partial _t\int _0^1|\tilde{\xi }|^2\le -\left( 8\pi ^2\bar{\sigma }(t)-1\right) \int _0^1 |\tilde{\xi }|^2. \end{aligned}$$

Since $t\mapsto \bar{\sigma }(t)$ is monotone increasing, cf. Proposition 3.1 (ii), from (49) we have $8\pi ^2\bar{\sigma }(t)-1\ge \frac{1}{3}$ for all $t\in [0,t_0]$. Thus,

$$\begin{aligned} \partial _t\int _0^1|\tilde{\xi }|^2\le -\frac{2}{3}\int _0^1 |\tilde{\xi }|^2, \quad t\in (0,t_0]. \end{aligned}$$

This implies that $t\mapsto \Vert \tilde{\xi }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)}$ is monotone decreasing. An integration in t results in (50), which completes the proof. $\square$

The next lemma concerns the decay estimate of $t\mapsto \Vert \partial _s\tilde{\xi }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)}^2$.

Lemma 3.13

Assume that $\bar{\sigma }(0)$ satisfy the same assumptions as in Lemma 3.12. Assume furthermore that

$$\begin{aligned} c(0)^2\ge \frac{1}{2}. \end{aligned}$$

(54)

Then $t\mapsto \Vert \partial _s\tilde{\xi }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)}$ is monotone decreasing. Moreover,

$$\begin{aligned} \Vert \partial _s\tilde{\xi }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)}\le e^{-\frac{1}{16}t}\Vert \partial _s\tilde{\xi }(0,\cdot )\Vert _{L^2(\mathbb {S}^1)} ,\quad t\in [0,t_0]. \end{aligned}$$

Proof

By (45) and the expression of $\dot{c}$ in (43),

$$\begin{aligned} \frac{1}{2}\partial _t\int _0^1|\partial _s\tilde{\xi }|^2 =-c(t)\dot{c}(t)=-\left( 1-4\pi ^2\bar{\sigma }\right) c^2 + 4\pi ^2 c\int _0^1\sigma \partial _sw_0\cdot \partial _s\tilde{\xi }. \end{aligned}$$

Applying (51) to the last term in the above equation, we obtain

$$\begin{aligned} \begin{aligned} \frac{1}{2}\partial _t\int _0^1|\partial _s\tilde{\xi }|^2&=-\left( 1-4\pi ^2\bar{\sigma }\right) c^2+4\pi ^2\left( \frac{1-c^2}{2}\bar{\sigma } -\frac{1}{2}\int _0^1\sigma |\partial _s\tilde{\xi }|^2\right) \\&\le -c^2+4\pi ^2\bar{\sigma } c^2+4\pi ^2\bar{\sigma }\frac{1-c^2}{2}. \end{aligned} \end{aligned}$$

(55)

Next we note that

$$\begin{aligned} 4\pi ^2\bar{\sigma }\le 1-\frac{3}{4}\int _0^1|\partial _s\tilde{\xi }|^2. \end{aligned}$$

(56)

Indeed, by (ii) of Proposition 3.1$\bar{\sigma }\le \int _0^1|\xi |^2$, which in terms of $\tilde{\xi }$ and c reads $\bar{\sigma } \le \int _0^1|\tilde{\xi }|^2+\frac{c^2}{4\pi ^2}$. By (53), $\bar{\sigma }\le \frac{1}{16\pi ^2}\int _0^1|\partial _s\tilde{\xi }|^2 + \frac{c^2}{4\pi ^2}$, which combined with (45) gives (56).

With (56) at-hand, one can bound $\partial _t\int _0^1|\partial _s\tilde{\xi }|^2$ in terms of $\int _0^1|\partial _s\tilde{\xi }|^2 \mathrm{{d}}s$ as

$$\begin{aligned} \frac{1}{2}\partial _t\int _0^1|\partial _s\tilde{\xi }|^2 \mathrm{{d}}s\le \int _0^1|\partial _s\tilde{\xi }|^2 \mathrm{{d}}s\left( -\frac{1}{4}+\frac{3}{8}\int _0^1|\partial _s\tilde{\xi }|^2 \mathrm{{d}}s\right) . \end{aligned}$$

(57)

Indeed, since $c^2$ and $1-c^2$ are nonnegative (cf. (45)), applying (56) to (55) we have that

$$\begin{aligned} \frac{1}{2}\partial _t\int _0^1|\partial _s\tilde{\xi }|^2\le -c^2 + \left( 1-\frac{3}{4}\int _0^1|\partial _s\tilde{\xi }|^2\right) c^2 + \left( 1-\frac{3}{4}\int _0^1|\partial _s\tilde{\xi }|^2\right) \frac{1-c^2}{2}. \end{aligned}$$

Substituting $c^2$ by $1-\int _0^1|\partial _s\tilde{\xi }|^2 \mathrm{{d}}s$ (cf. (45)) and after some algebraic manipulations, we arrive at (57).

Thus, if $\int _0^1|\partial _s\tilde{\xi }(0,s)|\mathrm{{d}}s=1-c(0)^2\le \frac{1}{2}$ by (54), then by (57) $\int _0^1|\partial _s\tilde{\xi }(t,s)|\mathrm{{d}}s\le \frac{1}{2}$ for all $t\in [0,t_0]$. Moreover,

$$\begin{aligned} \frac{1}{2}\partial _t\int _0^1|\partial _s\tilde{\xi }|^2 \mathrm{{d}}s\le -\frac{1}{16}\int _0^1|\partial _s\tilde{\xi }|^2 \mathrm{{d}}s, \end{aligned}$$

for all $t\in (0,t_0]$. Solving the above differential inequality, we complete the proof. $\square$

The exponential decay of $t\mapsto \Vert \partial _s\tilde{\xi }(t,\cdot )\Vert _{L^2}$ immediately yields the convergence of the multiplicative factor c(t) due to (45).

Corollary 3.14

Under the same assumptions as in Lemma 3.13, we have

$$\begin{aligned} |c(t)^2-1|\le e^{-\frac{1}{8}t}|c(0)^2-1| \text { for all } t\in [0,t_0]. \end{aligned}$$

Remark 3.15

The decay estimate on c(t) implies that $t_0=T$, where $[0,t_0]$ is the interval of the definition of the solution to the ODE (46) in Lemma 3.11. Indeed, since c(t), $t\in [0,t_0]$, is bounded away from zero and has uniformly bounded modulus (cf. Corollary 3.14), the solution to (46) can be extended for longer time as long as $\Vert \xi (t,\cdot )\Vert _{C^{2+\alpha }(\mathbb {S}^1)}$ remains bounded.

In order to show the global existence, we need to control the norm $\Vert \xi (t,s)\Vert _{2+\alpha ,0}$ along the flow. For this purpose, it suffices to find a pointwise upper and lower bound on the ellipticity coefficient $\sigma (t,s)$. The next lemma says that if $\frac{1}{4\pi ^2}-\bar{\sigma }$, $\Vert \tilde{\xi }\Vert _{L^2}$ and $\Vert \partial _s\tilde{\xi }\Vert _{L^2}$ are sufficiently small at $t=0$, then the oscillation $\sigma (t,s)-\bar{\sigma }(t)$ and $|\xi |^2(t,s)-\int _0^1|\xi |^2 (t,s)\mathrm{{d}}s$ are under control along the flow.

Lemma 3.16

Given $\epsilon \in (0,\frac{1}{(32\pi ^2)^2}]$, if

$$\begin{aligned} \left( \frac{1}{4\pi ^2}-\bar{\sigma }(0)\right) + \left\| \tilde{\xi }(0,\cdot )\right\| ^2_{L^2(\mathbb {S}^1)}+\left\| \partial _s\tilde{\xi }(0,\cdot )\right\| ^2_{L^2(\mathbb {S}^1)}\le \epsilon , \end{aligned}$$

(58)

then for all $(t,s)\in [0,T]\times \mathbb {S}^1$, we have

$$\begin{aligned} \left| |\xi (t,s)|^2-\int _0^1|\xi (t,s)|^2ds \right| \le 2\sqrt{\epsilon }, \quad \left| \sigma (t,s)-\bar{\sigma }(t)\right| \le 3\sqrt{\epsilon }. \end{aligned}$$

In particular, $\sigma$ satisfies

$$\begin{aligned} \frac{1}{4\pi ^2}-4\sqrt{\epsilon }\le \sigma (t,s)\le \frac{1}{4\pi ^2}+3\sqrt{\epsilon }\text { for all }(t,s)\in [0,T]\times \mathbb {S}^1. \end{aligned}$$

Proof

1. We observe that by the monotonicity properties from Proposition 3.1, Lemmas 3.12 and 3.13 with $t_0=T$ (cf. Remark 3.15), the smallness assumption (58) indeed holds for all $t\in [0,T]$. Here, we have used that (58) implies (54).

2. We show that

$$\begin{aligned} \int _0^1|\partial _t\xi (t,s)|^2 \mathrm{{d}}s =\int _0^1|\xi (t,s)|^2 \mathrm{{d}}s -\bar{\sigma }(t) \le \epsilon \text { for all } t\in (0,T]. \end{aligned}$$

(59)

Indeed, the first equality is due to (21) and (22). To see the second inequality, we note that by the assumption (58) and the observation above,

$$\begin{aligned} \int _0^1|\xi |^2 \mathrm{{d}}s -\bar{\sigma }&= \left( \int _0^1|\xi |^2-\frac{1}{4\pi ^2}\right) +\left( \frac{1}{4\pi ^2}-\bar{\sigma }\right) \\&=\int _0^1|\tilde{\xi }|^2 \mathrm{{d}}s +(c^2-1)\frac{1}{4\pi ^2}+\left( \frac{1}{4\pi ^2}-\bar{\sigma }\right) \\&=\int _0^1|\tilde{\xi }|^2 \mathrm{{d}}s -\frac{1}{4\pi ^2}\int _0^1|\partial _s\tilde{\xi }|^2\mathrm{{d}}s +\left( \frac{1}{4\pi ^2}-\bar{\sigma }\right) \le \epsilon , \end{aligned}$$

where in the second last inequality, we have used (45).

3. We estimate the oscillation of $\sigma$ and $|\xi |^2$. First, we show that the oscillation of $\sigma$ is bounded by the oscillation of $|\xi |^2$: for all $(t,s)\in [0,T]\times \mathbb {S}^1$

$$\begin{aligned} \left| \sigma (t,s)-\bar{\sigma }(t)\right| \le \sqrt{\epsilon }+\left| \frac{1}{2}|\xi (t,s)|^2-\frac{1}{2}\int _0^1|\xi (t,s)|^2\mathrm{{d}}s\right| . \end{aligned}$$

(60)

For this, we multiply the equation of $\xi$ by $\partial _s\xi$ and get $\partial _t\xi \cdot \partial _s\xi = \partial _s\sigma + \xi \cdot \partial _s\xi = \partial _s\left( \sigma +\frac{1}{2}|\xi |^2\right)$. An integration in s yields

$$\begin{aligned} \left| \sigma (t,s)+\frac{1}{2}|\xi (t,s)|^2 - \int _0^1\left( \sigma (t,s)+\frac{1}{2}|\xi (t,s)|^2\right) \mathrm{{d}}s \right| \le \int _0^1\left| \partial _t\xi (t,s)\cdot \partial _s\xi (t,s)\right| \mathrm{{d}}s. \end{aligned}$$

By Hölder’s inequality and (59),

$$\begin{aligned} \int _0^1\left| \partial _t\xi (t,s)\cdot \partial _s\xi (t,s)\right| \mathrm{{d}}s \le \left( \int _0^1|\partial _t\xi (t,s)|^2ds\right) ^{1/2}\le \sqrt{\epsilon }. \end{aligned}$$

Combining the above two inequalities together, we obtain (60).

Next we rewrite the oscillation of $|\xi |^2$ by using $|\tilde{\xi }|^2$:

$$\begin{aligned} \left| \frac{1}{2}|\xi (t,s)|^2-\frac{1}{2}\int _0^1|\xi (t,s)|^2\mathrm{{d}}s\right| =\left| \frac{1}{2}|\tilde{\xi }(t,s)|^2-\frac{1}{2}\int _0^1|\tilde{\xi }(t,s)|^2\mathrm{{d}}s+2c\tilde{\xi }\cdot w_0\right| . \end{aligned}$$

Using the fundamental theorem of calculus as well as the step 1, one can bound the oscillation of $|\tilde{\xi }|^2$ as

$$\begin{aligned} \left| |\tilde{\xi }(t,s)|^2-\int _0^1|\tilde{\xi }(t,s)|^2\mathrm{{d}}s \right|&\le 2\int _0^1|\partial _s\tilde{\xi }\cdot \tilde{\xi }| \mathrm{{d}}s\le 2\Vert \partial _s\tilde{\xi }(t,\cdot )\Vert _{L^2}\Vert \tilde{\xi }(t,\cdot )\Vert _{L^2}\le 2\epsilon . \end{aligned}$$

For the term $c\tilde{\xi }\cdot w_0$, we use $|c|\le 1$, $|w_0|=\frac{1}{2\pi }$ and the above oscillation estimate to get

$$\begin{aligned} |c\tilde{\xi }\cdot w_0|\le \frac{1}{2\pi }|\tilde{\xi }(t,s)| \le \frac{1}{2\pi }\left( \int _0^1|\tilde{\xi }(t,s)|^2ds+2\epsilon \right) ^{1/2}\le \frac{1}{2\pi }\sqrt{3\epsilon }. \end{aligned}$$

Combining together, we have for all $(t,s)\in [0,T]\times \mathbb {S}^1$

$$\begin{aligned} \left| \frac{1}{2}|\xi (t,s)|^2-\frac{1}{2}\int _0^1|\xi (t,s)|^2ds\right|&\le \epsilon +2\frac{1}{2\pi }\sqrt{3\epsilon }\le 2\sqrt{\epsilon }. \end{aligned}$$

Combining this with (60), we obtain the desired oscillation estimate for $\sigma$. Taking $\epsilon \le \frac{1}{(32\pi ^2)^2}$ and using the step 1, we obtain the pointwise lower bound for $\sigma (t,s)$. The upper bound follows from (60) and Proposition 3.1. $\square$

At the end of this section, we show the global well-posedness of the normalized flow in the Hölder class $C^{2+\alpha ,\alpha /2}$, under the assumption that initially the curve is sufficiently close to the stationary solution $w_0$.

Theorem 2

Let $w_0$ be the stationary solution defined in (39). Given an initial datum $\xi _0\in \tilde{\mathcal {A}}\cap C^{2+\alpha }(\mathbb {S}^1)$, which satisfies

$$\begin{aligned} \Vert \xi _0-w_0\Vert _{H^2(\mathbb {S}^1)}\le \epsilon _0 \end{aligned}$$

(61)

for some small universal constant $\epsilon _0>0$, the Cauchy problem (23) has a solution

$$\begin{aligned} \xi \in C^{2+\alpha , \alpha /2}([0,\infty )\times \mathbb {S}^1), \partial _t\xi \in C^{\alpha ,\alpha /2}((0,\infty )\times \mathbb {S}^1). \end{aligned}$$

Proof

Let $\xi \in C^{2+\alpha ,\alpha /2}([0,T]\times \mathbb {S}^1)$, $T\in (0,1)$ depending on $\Vert \xi _0\Vert _{C^{2+\alpha }(\mathbb {S}^1)}$, be the classical solution to the Cauchy problem (23) with the initial datum $\xi _0$ (cf. Theorem 1). Let $\sigma$ be the Lagrange multiplier such that $\xi _0+\partial _s(\sigma \partial _s\xi _0)\in T_{\xi _0}\tilde{\mathcal {A}}$. Similar arguments as in Lemma 2.2 yield

$$\begin{aligned} \partial _{ss}\sigma -|\partial _{ss}\xi _0|^2\sigma =-1. \end{aligned}$$

(62)

Note that from (62), we have

$$\begin{aligned} \int _0^1\sigma =\Vert \partial _s(\sigma \partial _s\xi _0)\Vert _{L^2}^2\le \Vert \xi _0\Vert _{L^2}^2, \end{aligned}$$

which is no larger than $\frac{1}{4\pi ^2}$ by Wirtinger’s inequality. We want to show if $\epsilon _0$ is sufficiently small, then the smallness assumption (58) from Lemma 3.16 is satisfied.

To see this, we first consider $\tilde{\xi }_0:=\xi _0-cw_0(\cdot +\theta )$, where $cw_0(\cdot +\theta )\in {\text {argmin}}_{w\in \mathcal {C}} \Vert \xi _0-w\Vert _{L^2}$. Then (61) implies

$$\begin{aligned} \Vert \tilde{\xi }_0\Vert _{L^2(\mathbb {S}^1)}\le \epsilon _0\text { and }\Vert \partial _s\tilde{\xi }_0\Vert _{L^2(\mathbb {S}^1)}\le 2\pi \sqrt{\epsilon _0(1+\pi ^{-1})}. \end{aligned}$$

Indeed, the first inequality is immediate since by the definition of the projection $\Vert \tilde{\xi }_0\Vert _{L^2(\mathbb {S}^1)}\le \Vert \xi -w_0\Vert _{L^2(\mathbb {S}^1)}$. The second inequality follows from (45) having observed that

$$\begin{aligned} \frac{c^2}{4\pi ^2}= & {} \Vert \xi _0\Vert _{L^2(\mathbb {S}^1)}^2-\Vert \tilde{\xi }_0\Vert _{L^2(\mathbb {S}^1)}^2\ge \Vert w_0\Vert _{L^2(\mathbb {S}^1)}^2-2\Vert w_0\Vert _{L^2(\mathbb {S}^1)}\Vert \xi -w_0\Vert _{L^2(\mathbb {S}^1)}\\&-\Vert \tilde{\xi }_0\Vert _{L^2(\mathbb {S}^1)}^2 \ge \frac{1}{4\pi ^2}- \epsilon _0(1+\pi ^{-1}). \end{aligned}$$

Next, by viewing (62) as a perturbation of

$$\begin{aligned} \partial _{ss}\sigma -|\partial _{ss}w_0|^2\sigma =\partial _{ss}\sigma -4\pi ^2\sigma =-1, \end{aligned}$$

with periodic boundary conditions, solution of which is the constant function $\frac{1}{4\pi ^2}$, we have that the solution to (62) satisfies

$$\begin{aligned} \left| \frac{1}{4\pi ^2}-\int _0^1\sigma \mathrm{{d}}s\right| \le C \epsilon _0, \end{aligned}$$

for some universal constant $C>0$. Hence if $\epsilon _0$ is sufficiently small but universal, the smallness assumption (58) holds.

Now we apply Lemma 3.16 to $\xi$ and get

$$\begin{aligned} \frac{1}{8\pi ^2}\le \sigma (t,s)\le \frac{1}{2\pi ^2}, \end{aligned}$$

(63)

in $[0,T]\times \mathbb {S}^1$. Furthermore, for each fixed t, $\sigma (t,\cdot )$ is Lipschitz continuous with uniformly bounded Lipschitz constant. Indeed, an integration of the equation of $\sigma$ (cf. (24)) yields $\int _{\mathbb {S}^1} \sigma |\partial _{ss}\eta |^2 =1$, which implies that $\int _{\mathbb {S}^1}|\partial _{ss}\sigma (t,s)|ds\le 2$. By the regularity theory for the parabolic equations, cf. [28],

$$\begin{aligned} \sup _{t\in [T/2,T]} \Vert \xi (t,\cdot )\Vert _{C^{2+\alpha }(\mathbb {S}^1)}\le C (T^{-(2+\alpha )}+1)\Vert \xi \Vert _{L^\infty ([T/4,T]\times \mathbb {S}^1)}, \end{aligned}$$

(64)

for some $C>0$ only depending on d and $\alpha$ (this is the interior Schauder estimate for parabolic equations. Since our solutions satisfy periodic bound conditions, the estimate holds globally in $\mathbb {S}^1$). In the meanwhile, the smallness assumption (58) of Lemma 3.16 is satisfied for all $t\in [0,T]$ due to the monotonicity properties of $\int _0^1\sigma (t,s)\mathrm{{d}}s$, $\Vert \tilde{\xi }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)}$ and $\Vert \partial _s\tilde{\xi }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)}$ (Proposition 3.1 (ii), Lemmas 3.12 and 3.13). In particular,

$$\begin{aligned} \Vert \xi \Vert _{L^\infty ([T/4,T]\times \mathbb {S}^1)}\le \Vert cw_0\Vert _{L^\infty ([T/4,T]\times \mathbb {S}^1)}+ \Vert \tilde{\xi }\Vert _{L^\infty ([T/4,T]\times \mathbb {S}^1)}\le \bar{C}, \end{aligned}$$

(65)

where $\bar{C}$ is an absolute constant. We apply Theorem 1 starting from $t=T$. The solution $\xi$ extends in the Hölder class $C^{2+\alpha ,\alpha /2}$ up to $t=T+T_0$ for some $T_0>0$. Lemma 3.16 applies and yields (63) and (65) in $[0,T+T_0]\times \mathbb {S}^1$. Again by the interior Schauder estimate

$$\begin{aligned} \sup _{t\in [\frac{T+T_0}{2},T+T_0]} \Vert \xi (t,\cdot )\Vert _{C^{2+\alpha }(\mathbb {S}^1)}\le C ((T+T_0)^{-(2+\alpha )}+1)\Vert \xi \Vert _{L^\infty ([0,T+T_0]\times \mathbb {S}^1)}. \end{aligned}$$

Since $\Vert \xi (t,\cdot )\Vert _{C^{2+\alpha }(\mathbb {S}^1)}$ cannot blow up as time getting large, the time steps have a uniform lower bound $\tilde{T}_0>0$. Repeating the above arguments from $T+2\tilde{T}_0$, $T+3\tilde{T}_0$ and so on, we obtain the global existence of the Cauchy problem (23) in the Hölder class $C^{2+\alpha ,\alpha /2}$. $\square$

As a by-product, we also obtain the exponential decay of our solution to the stationary solution $w_0$ under the initial smallness assumption.

Theorem 3

Under the same assumptions as in Theorem 2, we have for all $t>0$,

$$\begin{aligned} \Vert \xi (t,s)-w_0(s+\theta _\infty )\Vert _{L^\infty (\mathbb {S}^1)}\le C e^{-t/16}\Vert \xi _0-w_0\Vert _{H^2(\mathbb {S}^1)}, \end{aligned}$$

for some universal constant $C>0$ and some constant $\theta _\infty$.

Proof

By Lemmas 3.12, 3.16 and Corollary 3.14, if $\epsilon _0:=\Vert \xi _0-w_0\Vert _{H^2(\mathbb {S}^1)}$ is sufficiently small (say $\epsilon _0\le \frac{1}{(100\pi ^2)^2}$), then for all $t>0$

$$\begin{aligned} \Vert \tilde{\xi }(t,\cdot )\Vert _{L^\infty (\mathbb {S}^1)}+(1-c^2(t))\le Ce^{-\frac{1}{16}t}\epsilon _0, \end{aligned}$$

for some absolute constant $C>0$. We still need to estimate the evolution of $\theta (t)$. Note that from the expression of $\dot{\theta }$ in (43)

$$\begin{aligned} |\dot{\theta }(t)|\le \frac{4\pi ^2}{c(t)}\frac{1}{2\pi }\max _{s\in \mathbb {S}^1}\sigma (t,s)\Vert \partial _s\tilde{\xi }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)},\quad t>0. \end{aligned}$$

By (60) if $\epsilon _0$ in (61) is sufficiently small, then $\sigma (t,s)\le \int _0^1\sigma (t,s) \mathrm{{d}}s + 3\sqrt{\epsilon _0}\le \frac{1}{2\pi ^2}$. Thus, combining Corollary 3.14 and Lemma 3.13, we infer

$$\begin{aligned} |\dot{\theta }(t)|\le e^{-\frac{1}{16}t}\Vert \partial _s\tilde{\xi }(0,\cdot )\Vert _{L^2(\mathbb {S}^1)},\quad t>0. \end{aligned}$$

This implies that $\lim _{t\rightarrow \infty }\theta (t)$ exists. Letting $\theta _\infty$ denote the limit, we obtain

$$\begin{aligned} |\theta (t)-\theta _\infty |\le 16e^{-\frac{1}{16}t}\Vert \partial _s\tilde{\xi }(0,\cdot )\Vert _{L^2}\le Ce^{-\frac{1}{16}t}\epsilon _0, \end{aligned}$$

after an integration from t to $\infty$. Thus, by the triangle inequality,

$$\begin{aligned} \left| \xi (t,s)-w_0(s+\theta _\infty )\right|&\le \left| \xi (t,s)-c(t)w_0(s+\theta (t))\right| +\left| (c(t)-1)w_0(s+\theta (t))\right| \\&\quad +\left| w_0(s+\theta (t))-w_0(s+\theta _\infty )\right| \\&\le |\tilde{\xi }(t,s)|+\frac{1}{2\pi }|c(t)-1|+ |\theta (t)-\theta _\infty |\\&\le Ce^{-\frac{1}{16}t}\epsilon _0. \end{aligned}$$

$\square$

Remark 3.17

The Hessian (92) of the $L^2$-mass is strictly negative-definite on the tangent vectors $\tilde{\xi }\in T_{w_0}\tilde{\mathcal {A}}$ which satisfy (53), i.e., on those which are orthogonal to the pure rotations:

$$\begin{aligned} \langle Hess\; M(w_0) \tilde{\xi }, \tilde{\xi }\rangle _{T_{w_0}\tilde{\mathcal {A}}}&=\int _0^1 |\tilde{\xi }|^2(s)\, \mathrm{{d}}s-\int _0^1 \varsigma (s) \, \mathrm{{d}}s\le -3\int _0^1 |\tilde{\xi }|^2(s)\, \mathrm{{d}}s\nonumber \\&=-3 \langle \tilde{\xi }, \tilde{\xi }\rangle _{T_{w_0}\tilde{\mathcal {A}}}. \end{aligned}$$

(66)

Here $\varsigma$ is the initial tension of a geodesic emanating from $w_0$ at the direction $\tilde{\xi }$, which satisfies

$$\begin{aligned} \partial _{ss}\varsigma - |\partial _{ss} w_0|^2 \varsigma +|\partial _s\tilde{\xi }|^2=0, \end{aligned}$$

(67)

(see [42, 45]). Indeed, since $|\partial _{ss} w_0|\equiv 2\pi$, an integration of (67) together with (53) implies

$$\begin{aligned} \int _0^1 \varsigma (s) \, \mathrm{{d}}s = \frac{1}{4\pi ^2} \int _0^1 |\partial _s\tilde{\xi }|^2(s)\, \mathrm{{d}}s \ge 4 \int _0^1 |\tilde{\xi }|^2(s)\, \mathrm{{d}}s, \end{aligned}$$

which gives (66). Then one can anticipate the exponential decay (Theorem 3) of the gradient flow in a neighborhood of $w_0$ via a Bakry–Émery argument, cf. [49]. However, such argument is not applicable in our situation since the Riemannian connection of $\tilde{\mathcal {A}}$ is not smooth and $(\tilde{\mathcal {A}}, d_{\tilde{A}})$ is not a geodesic metric space, cf. Theorem 4.2 in [43] and [11, 34].

3.5 Global Existence Without Restrictions on the Initial Data

We conclude by showing global solvability of the normalized flow in a generalized sense without any restrictions on the initial data. It is an adaptation of the approach we recently developed in [46] for a different gradient flow.

We begin by rewriting our flow in a form which explicitly involves the arc length parametrization constraint (cf. the beginning of the Sect. 3):

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \partial _t\xi &{}= \partial _s(\sigma \partial _s\xi ) + \xi \\ |\partial _s\xi | &{} =1 \end{array}\right. } \text { for } (t,s)\in Q_\infty :=(0,\infty )\times \mathbb {S}^1. \end{aligned} \end{aligned}$$

(68)

Definition 3.18

Given an initial datum $\xi _0\in W^{1,\infty }(\mathbb {S}^1;\mathbb {R}^d)$ with $|\partial _s\xi _0(s)|\le 1$ for a.e. $s\in {\mathcal {S}}^1$, we call a pair $(\xi ,\sigma )$ a generalized solution to the normalized UCMCF if the following hold

• (i) $\xi \in L^\infty _{loc}([0,\infty ); W^{1,\infty }({\mathcal {S}}^1))^d$, $\partial _t\xi \in L^2_{loc}([0,\infty );L^2({\mathcal {S}}^1))^d$, $\sigma \in L^2_{loc}([0,\infty ); H^1({\mathcal {S}}^1))$ and $\sigma \partial _s\xi \in L^2_{loc}([0,\infty ); H^1({\mathcal {S}}^1))^d$.

• (ii) The pair $(\xi , \sigma )$ satisfies for a.e. $(t,s)\in Q_\infty$

  $$\begin{aligned} \partial _t\xi (t,s) = \partial _s(\sigma (t,s) \partial _s\xi (t,s))&+ \xi , \end{aligned}$$

  (69)
  $$\begin{aligned} \sigma (t,s) \left( |\partial _s\xi (t,s)|^2 -1\right)&=0, \end{aligned}$$

  (70)
  $$\begin{aligned} |\partial _s\xi (t,s)|&\le 1, \end{aligned}$$

  (71)

  and the initial condition

  $$\begin{aligned} \xi (0,s)=\xi _0(s) \text { for a.e. }s. \end{aligned}$$

• (iii) The solution $\xi$ satisfies the energy dissipation inequality

  $$\begin{aligned} \int _{{\mathcal {S}}^1} |\partial _t\xi (t,s)|^2 \mathrm{{d}}s\le \int _{{\mathcal {S}}^1} \xi \cdot \partial _t\xi (t,s) \mathrm{{d}}s \end{aligned}$$

  (72)

  for a.e. $t\in (0,\infty )$.

Remark 3.19

(Strong and weak constraint) The generalized solutions in Definition 3.18 are not required to satisfy the strong constraint $|\partial _s\xi |= 1$ but merely the relaxed one

$$\begin{aligned} \sigma \left( |\partial _s\xi |^2 -1\right) =0,\quad |\partial _s\xi |\le 1. \end{aligned}$$

In the next remark, we will show that under the regularity assumptions $\xi \in C^1(\overline{Q_\infty })\cap C^2(Q_\infty )$ and $|\partial _s\xi _0|=1$, the generalized solutions solve (68) in the classical sense. However, without the regularity assumptions, we do not know whether the constraint $|\partial _s\xi |=1$ is satisfied or not.

Remark 3.20

(Relation with the classical solution) It is not hard to see that if $(\xi ,\sigma )$ is a $C^2$ regular solution to (68), then it is also a generalized solution in the sense of Definition 3.18; in particular, (71) and (72) become strict equalities. On the other hand, we claim that any generalized solution $(\xi ,\sigma )$ with $\xi \in C^1(\overline{Q_\infty })\cap C^2(Q_\infty )$ and $|\partial _s\xi _0|=1$ solves (68).

Formally, this claim is a trivial consequence of (70) since $\sigma$ is expected to be strictly positive by the strong maximum principle, cf. Remark 2.3. However, we cannot guarantee the strict positivity of $\sigma$ for the generalized solutions. Nevertheless, to prove the claim it suffices to show that the open set $U:=\{(t,s)\in Q_\infty : |\partial _s\xi (t,s)|<1\}$ is empty. Suppose not, then $\sigma =0$ a.e. in U due to (70). This implies that $\partial _t\xi =\xi$ in U, whence $\partial _t(|\partial _s\xi |^2)=2 |\partial _s\xi |^2$. For each $(t_0,s_0)\in U$, let $t_1=\inf \{t\ge 0: (t,t_0)\times \{ s_0\}\subset U\}$. If $t_1=0$ then

$$\begin{aligned} |\partial _s\xi (t_1,s_0)|=1, \end{aligned}$$

(73)

due to our assumption about $\xi _0$, and if $t_1>0$ then (73) also holds by the continuity of $\partial _s\xi$. From $\partial _t(|\partial _s\xi |^2)\ge 0$ in U, we immediately deduce that

$$\begin{aligned} |\partial _s\xi (t_0,s_0)|^2\ge |\partial _s\xi (t_1,s_0)|^2= 1, \end{aligned}$$

arriving at a contradiction.

The next theorem concerns the global existence of the generalized solution without any smallness or closeness assumption on the initial datum. We stress that the theorem does not cover Theorem 2 due to the relaxation of the unit speed constraint in (68).

Theorem 4

For every $\xi _0\in W^{1,\infty }(\mathbb {S}^1;\mathbb {R}^d)$ with $|\partial _s\xi _0(s)|\le 1$ for a.e. $s\in \mathbb {S}^1$, there exists a (global in time) generalized solution $(\xi ,\sigma )$ to the normalized UCMCF, and $\sigma (t,s)\ge 0$ for almost every $(t,s)\in Q_\infty$.

The proof mimics the one of [46, Theorem 3] and has the following outline. We rewrite (68) as a first-order system, and approximate it by Hilbertian gradient flows. Let $\kappa :=\sigma \partial _s\xi$, then the problem in the new variables $(\xi , \kappa , \sigma )$ would read

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \partial _t\xi =\partial _s\kappa +\xi \\ \kappa =\sigma \partial _s\xi \\ \sigma =\kappa \cdot \partial _s\xi . \end{array}\right. } \end{aligned} \end{aligned}$$

(74)

For $\epsilon >0$, let

$$\begin{aligned}&F^\epsilon :\mathbb {R}^d\rightarrow \mathbb {R}^d,\quad F^\epsilon (\kappa ):=\epsilon \kappa +\frac{\kappa }{\sqrt{\epsilon +|\kappa |^2}},\\&G^\epsilon (\tau ):=(F^\epsilon )^{-1}(\tau ), \end{aligned}$$

and consider the problem

$$\begin{aligned} \partial _t\xi ^\epsilon = \partial _s(G^\epsilon (\partial _s\xi ^\epsilon ))+\xi ^\epsilon \text { in } Q_\infty . \end{aligned}$$

(75)

Let us introduce the functional

$$\begin{aligned} \mathcal {E}^\epsilon (\xi ):=\int _{\mathbb {S}^1}\epsilon \left( \frac{|G^\epsilon (\partial _s \xi )|^2}{2}-\frac{1}{\sqrt{\epsilon +|G^\epsilon (\partial _s \xi )|^2}}\right) -\frac{1}{2} |\xi |^2. \end{aligned}$$

Then (75) can be interpreted as a negative gradient flow of $\mathcal {E}^\epsilon$ with respect to the flat Hilbertian structure of $L^2(\mathbb {S}^1;\mathbb {R}^d)$. The existence of a unique smooth solution $\xi ^\epsilon : [0,\infty )\times [0,1]\rightarrow \mathbb {R}^d$ to (75) follows from Amann’s theory [4]. The solutions satisfy uniform energy estimates as in [46, Proposition 3.1]. Moreover, $\partial _s \xi ^\epsilon$ has a uniform $L^\infty$ bound as in [46, Proposition 3.3]. We now set

$$\begin{aligned} \begin{aligned} \kappa ^\epsilon := G^\epsilon (\partial _s\xi ^\epsilon ),\quad \sigma ^\epsilon := G^\epsilon (\partial _s\xi ^\epsilon )\cdot \partial _s\xi ^\epsilon \ge 0. \end{aligned} \end{aligned}$$

Arguing as in the proof of [46, Proposition 3.4 and Theorem 3], we can pass to the limit and obtain a solution $(\xi , \kappa , \sigma )$ to (74). The pair $(\xi , \sigma )$ solves the normalized UCMCF in the sense of Definition 3.18. We refer to [46] for the full implementation.

Remark 3.21

One can adapt the approach of [46, Section 6] to construct backward generalized solutions to the normalized UCMCF. It seems, however, that all one can get in this way is the trivial solution $(\xi ,\sigma )(t)=(e^t \xi _0,0)$, $t\le 0$. It satisfies (69)–(72), and is smooth provided $\xi _0$ is smooth, but obviously violates the strong constraint $|\partial _s\xi |=1$. This contrasts with Remark 3.20 and with [46] where smoothness implied the strong constraint. Consequently, the method of [46, Section 6] for constructing two different solutions emanating from an initial datum $\xi _0$ with $|\partial _s\xi _0|=1$ is not applicable. This leads us to conjecture the uniqueness of the generalized solutions to the normalized UCMCF.

Appendices

Appendix A. Higher Dimensional UCMCF

In this Appendix, we describe how our approach can be implemented in the case of evolution of surfaces. For the sake of simplicity of the presentation, we work with embeddings of a compact manifold into the ambient space $\mathbb {R}^d$, but this can be generalized in various directions—in particular, one can consider immersions instead of embeddings.

1.1 A.1. Riemannian Structure

Fix a smooth, compact-connected k-dimensional submanifold ${\mathcal {M}}$ of $\mathbb {R}^d$. Without loss of generality in the sequel, we assume that ${{\mathrm{vol}}}(\mathcal {M})=1$. Let $\mathcal {K}_k$ be the space of $H^m$-regular embeddings $\eta : {\mathcal {M}} \rightarrow \mathbb {R}^d$, $\int _{{\mathcal {M}}} \eta \ \mathrm{{d}}{\mathcal {H}}^k =0$, $m> \frac{n +2}{2}$. Each element $v\in T_\eta \mathcal {K}_k$ can be identified with a vector field $v:{\mathcal {M}} \rightarrow \mathbb {R}^d$. We endow the space ${\mathcal {K}}_k$ with the parametrization-invariant $L^2$ Riemannian metric (cf. [30])

$$\begin{aligned} \langle v,w\rangle _{T_\eta \mathcal {K}_k}:=\int _{\eta (\mathcal {M})} (v\cdot w)\circ (\eta ^{-1})\ \mathrm{{d}}{\mathcal {H}}^{k}=\int _{\mathcal {M}} v\cdot w\ J\eta \ \mathrm{{d}}\mathcal {H}^k, \end{aligned}$$

(76)

which has a degenerate Riemannian distance [10]. Here $J\eta (x):=\sqrt{\det (d\eta _x)^*\circ (d\eta _x)}$ for each $x\in \mathcal {M}$ is the Jacobian of $\eta$. Let

$$\begin{aligned} {{\mathrm{vol}}}:\mathcal {K}_k\rightarrow \mathbb {R}, \ {{\mathrm{vol}}}(\eta )={\mathcal {H}}^k({\eta ({\mathcal {M}})} ), \end{aligned}$$

be the volume functional. By Sobolev embedding $H^{m}\subset C^1$, ${{\mathrm{vol}}}$ is continuous w.r.t. the $H^{m}$-topology of ${\mathcal {K}}_k$. Sometimes, we will also use a flat metric $\langle \cdot , \cdot \rangle ^*$ on $\mathcal {K}_k$:

$$\begin{aligned} \langle v,w\rangle _{T_\eta \mathcal {K}_k}^*:=\int _{{\mathcal {M}}} v\cdot w\, \mathrm{{d}}{\mathcal {H}}^k. \end{aligned}$$

(77)

We consider the submanifold of ${\mathcal {K}}_k$ consisting of uniformly dilating embeddings, i.e.,

Let us also define the submanifold of $\mathcal {A}_k$ consisting of volume-preserving embeddings

$$\begin{aligned} \tilde{\mathcal {A}}_k:=\{\eta \in \mathcal {A}_k, \ {{\mathrm{vol}}}(\eta )=1\}. \end{aligned}$$

(78)

The tangent space at $\eta \in \tilde{\mathcal {A}}_k$ is

$$\begin{aligned} T_\eta \tilde{\mathcal {A}}_k=\{h\in T_\eta \mathcal {K}_k: {{\mathrm{div}}}_{\eta (\mathcal {M})} (h\circ \eta ^{-1})=0\}, \end{aligned}$$

and thus it is not hard to verify that $T_\eta \mathcal {A}_k=\{h\in T_\eta \mathcal {K}_k: {{\mathrm{div}}}_{\eta (\mathcal {M})} (h\circ \eta ^{-1})=const\}$. The metrics (76) and (77) induce metrics on $\mathcal {A}_k$: for $\eta \in \mathcal {A}_k$

$$\begin{aligned}&\langle v,w\rangle _{T_\eta \mathcal {A}_k}=\int _{{\mathcal {M}}} v\cdot w\, {{\mathrm{vol}}}(\eta )\, \mathrm{{d}}{\mathcal {H}}^k, \end{aligned}$$

(79)

$$\begin{aligned}&\langle v,w\rangle _{T_\eta \mathcal {A}_k}^*=\int _{{\mathcal {M}}} v\cdot w\, \mathrm{{d}}{\mathcal {H}}^k. \end{aligned}$$

(80)

The induced Riemannian metric on $\tilde{\mathcal {A}}_k$ (both from $\langle \cdot , \cdot \rangle$ and $\langle \cdot , \cdot \rangle ^*$) is then

$$\begin{aligned} \langle v,w\rangle _{T_\eta \tilde{\mathcal {A}}_k}=\langle v,w \rangle ^*_{T_\eta \tilde{\mathcal {A}}_k}=\int _{{\mathcal {M}}} v\cdot w\, \mathrm{{d}}{\mathcal {H}}^k. \end{aligned}$$

(81)

1.2 A.2. Orthogonal Projection

By [11, Theorem 7], the orthogonal projection ${\tilde{P}}_\eta :T_\eta {\mathcal {K}}_k\rightarrow T_\eta \tilde{\mathcal {A}}_k$ (with respect to the invariant $L^2$ metric $\langle \cdot , \cdot \rangle _{T_\eta \mathcal {K}_k}$) is

$$\begin{aligned} {\tilde{P}}_\eta (z)=z-\sigma \vec H(\eta )- d \eta _\cdot {{\mathrm{grad}}}^g\sigma , \end{aligned}$$

(82)

where $\vec H(\eta )$ is the vectorial mean curvature (the trace of the second fundamental form) corresponding to the embedding $\eta$, the metric $g=\eta ^*\cdot$ is the pull-back of the Euclidean metric $\cdot$ on $\mathbb {R}^d$, and $\sigma :{\mathcal {M}} \rightarrow \mathbb {R}$ is a Lagrange multiplier, which is a solution to

$$\begin{aligned} \Delta _g \sigma -|\vec H(\eta )|^2 \sigma ={{\mathrm{div}}}_{\eta (\mathcal {M})}(z\circ \eta ^{-1}). \end{aligned}$$

Employing the methods of [11, 34], one can derive from (82) that the map $P_\eta : T_\eta \mathcal {K}_k\rightarrow T_\eta \mathcal {A}_k$,

$$\begin{aligned} \begin{aligned} P_\eta (z)=z-\sigma \vec H(\eta )- d \eta _\cdot {{\mathrm{grad}}}^g\sigma , \text { where }\\ \Delta _g \sigma -|\vec H(\eta )|^2 \sigma ={{\mathrm{div}}}_{\eta (\mathcal {M})}(z\circ \eta ^{-1})+const,\quad \int _{{\mathcal {M}}}\sigma \,\mathrm{{d}}{\mathcal {H}}^k=0, \end{aligned} \end{aligned}$$

(83)

is an orthogonal projection. The key observations in the proof of this claim are that the volume density for $\eta \in \mathcal {A}_k$ is constant (equal to ${{\mathrm{vol}}}(\eta )$) and the identity following from the divergence formula

$$\begin{aligned} \begin{aligned}&\int _{{\mathcal {M}}}w\cdot \left( \sigma \vec H(\eta )+ d \eta _\cdot {{\mathrm{grad}}}^g \sigma \right) {{\mathrm{vol}}}(\eta )\, \mathrm{{d}}{\mathcal {H}}^k \\&=-\int _{\eta ({\mathcal {M}})}(\sigma \circ \eta ^{-1}) {{\mathrm{div}}}_{\eta (\mathcal {M})}(w\circ \eta ^{-1}) \,\mathrm{{d}}{\mathcal {H}}^k\\&=-const\int _{{\mathcal {M}}}\sigma {{\mathrm{vol}}}(\eta )\,\mathrm{{d}}{\mathcal {H}}^k, \end{aligned} \end{aligned}$$

(84)

for any $w\in T_\eta \mathcal {A}_k$. Here in the last equality, we have used $w\in T_\eta \mathcal {A}_k$ and the characterization of the tangent space $T_\eta \mathcal {A}_k$.

1.3 A.3. The Gradient Flow

The UCMCF is the gradient flow

$$\begin{aligned} \partial _t \eta =- {{\mathrm{grad}}}_{\mathcal {A}_k} {{\mathrm{vol}}}(\eta ), \end{aligned}$$

(85)

of the volume functional on the space $\mathcal {A}_k$ under the metric (79). By construction, the flow operator

$$\begin{aligned} T_t:\eta (0,s)\mapsto \eta (t,s), \ s\in {\mathcal {M}}, \end{aligned}$$

complies with (4).

By the first variation of area formula, the negative ${\mathcal {K}}_k$-gradient of the volume functional is simply $\vec H$, and

$$\begin{aligned} \langle \vec H (\eta ), \eta \rangle _{T_\eta \mathcal {K}_k}=-k{{\mathrm{vol}}}(\eta ). \end{aligned}$$

(86)

With the projection (83) at hand, by an argument similar to the one from the Sect. 2.1, we can express the gradient flow (85) in the following form:

$$\begin{aligned} \partial _t\eta =\tilde{\sigma }\vec H (\eta )+ \mathrm{{d}} \eta _\cdot {{\mathrm{grad}}}^g\tilde{\sigma }, \quad \int _{{\mathcal {M}}}\tilde{\sigma }\,\mathrm{{d}}{\mathcal {H}}^k=1,\quad \eta \in \mathcal {A}_k. \end{aligned}$$

(87)

A direct computation yields that if a pair $(\eta (t),\tilde{\sigma }(t))$ solves (87), and $r:{\mathcal {M}} \rightarrow {\mathcal {M}}$ is a volume-preserving diffeomorphism, then $(\eta (t)\circ r,\tilde{\sigma }(t)\circ r)$ also solves (87). Note that the reparametrizations r which do not preserve the Hausdorff measure on ${\mathcal {M}}$ are ruled out automatically by our construction. Thus, UCMCF is a truly geometric flow since it does not depend on possible reparametrizations of the evolving submanifold $\eta (t)({\mathcal {M}})$. This claim will become completely transparent after we recast our flow into a parametrization-free form (97).

Let

$$\begin{aligned} M(\eta ):=\frac{1}{2} \int _{{\mathcal {M}}} |\eta |^2\, \mathrm{{d}} {\mathcal {H}}^k, \end{aligned}$$

be the $L^2$-mass functional. We are going to see that this functional decays with a constant speed along our gradient flow, cf. Proposition 2.6. Indeed, since $\eta \in T_\eta \mathcal {A}_k$ (since ${{\mathrm{div}}}_{\eta (\mathcal {M})}(\eta \circ \eta ^{-1})=k$),

$$\begin{aligned} \begin{aligned} \partial _t M(\eta )&= \int _{{\mathcal {M}}} \eta \cdot \partial _t \eta \, \mathrm{{d}}{\mathcal {H}}^k= \frac{1}{{{\mathrm{vol}}}(\eta )} \langle \eta , \partial _t\eta \rangle _{T_\eta \mathcal {K}_k}= \frac{1}{{{\mathrm{vol}}}(\eta )} \langle \eta , \partial _t\eta \rangle _{T_\eta \mathcal {A}_k}\\&=-\frac{1}{{{\mathrm{vol}}}(\eta )} \langle \eta , {{\mathrm{grad}}}_{\mathcal {A}_k} {{\mathrm{vol}}}(\eta ) \rangle _{T_\eta \mathcal {A}_k}=-\frac{1}{{{\mathrm{vol}}}(\eta )} \langle \eta , {{\mathrm{grad}}}_{{\mathcal {K}}_k} {{\mathrm{vol}}}(\eta ) \rangle _{T_\eta \mathcal {K}_k} \\&=\frac{1}{{{\mathrm{vol}}}(\eta )} \langle \eta , \vec H(\eta ) \rangle _{T_\eta \mathcal {K}_k} =-k, \end{aligned} \end{aligned}$$

by (86). Thus, our flow collapses in finite time $t_*=\frac{1}{k}M(\eta _0)$.

Remark A.1

The Riemannian distances $d_{\mathcal {A}^*_k}$ and $d_{\tilde{\mathcal {A}}_k}$ on submanifolds of $\mathcal {K}_k$ are nondegenerate since they are controlled from below by the Hilbertian distance $d_{\mathcal {K}^*_k}$ (which is induced by the Riemannian metric $\langle \cdot , \cdot \rangle ^*$ in (80)). We do not know whether the Riemannian distance $d_{\mathcal {A}_k}$ is nondegenerate in general, but as we observed in Proposition 2.1, the conjecture is true for $k=1$. Nevertheless, we have got backup ways to render UCMCF as a gradient flow with respect to a nondegenerate distance. Namely, the gradient flow

$$\begin{aligned} \partial _t \eta =- {{\mathrm{grad}}}_{\mathcal {A}^*_k} \left( \ln {{\mathrm{vol}}}(\eta )\right) . \end{aligned}$$

(88)

reproduces (85). This is immediate by observing that

$$\begin{aligned} {{\mathrm{grad}}}_{\mathcal {A}_k} {{\mathrm{vol}}}(\eta )=\frac{1}{{{\mathrm{vol}}}(\eta )}{{\mathrm{grad}}}_{\mathcal {A}^*_k} {{\mathrm{vol}}}(\eta )= {{\mathrm{grad}}}_{\mathcal {A}^*_k} \left( \ln {{\mathrm{vol}}}(\eta )\right) . \end{aligned}$$

Other options are presented in Appendix B.

Appendix B. Normalized Flow

The general higher dimensional UCMCF (87) can be renormalized in the same way as the curve-shortening flow. Namely, the new time is

$$\begin{aligned} \tau (t):=-\ln {{\mathrm{vol}}}(\eta (t)), \end{aligned}$$

and the new unknown functions are

$$\begin{aligned} \xi (\tau , s ):=\frac{\eta (t(\tau ),s)}{{{\mathrm{vol}}}(\eta (t(\tau )))},\ \sigma (\tau , s)=\frac{\tilde{\sigma }(t(\tau ),s)}{-({{\mathrm{vol}}}\partial _t{{\mathrm{vol}}})(\eta (t(\tau )))}. \end{aligned}$$

Then the pair $(\xi ,\sigma )$ solves the equation

$$\begin{aligned} \partial _\tau \xi =\xi +\sigma \vec H (\xi )+ \mathrm{{d}} \xi _\cdot {{\mathrm{grad}}}^{\xi ^*\cdot }\sigma ,\quad \xi \in \tilde{\mathcal {A}}_k. \end{aligned}$$

(89)

Employing the characterization of the orthogonal projection ${\tilde{P}}_\xi :T_\xi {\mathcal {K}}_k\rightarrow T_\xi \tilde{\mathcal {A}}_k$, we immediately rewrite (89) as a positive gradient flow of the $L^2$-mass:

$$\begin{aligned} \partial _\tau \xi ={\tilde{P}}_\xi \xi ={\tilde{P}}_\xi {{\mathrm{grad}}}_{\mathcal {K}_k} M(\xi ) = {{\mathrm{grad}}}_{\tilde{\mathcal {A}}_k} M(\xi ). \end{aligned}$$

(90)

1.1 B.1. Evolution of the Averaged Lagrange Multiplier

In order to illustrate the power of the gradient flow structure (90), we will formally derive a neat formula for the evolution of the mean of $\sigma$ along the UCMCF trajectories, thereby generalizing (ii) in Proposition 3.1. We first observe that the geodesics in $\tilde{\mathcal {A}}_k$ are determined by the condition $\partial _{\tau \tau }\gamma \perp T_{\gamma }\tilde{\mathcal {A}}_k$, which can be expressed as

$$\begin{aligned} \partial _{\tau \tau }\gamma =\varsigma \vec H (\gamma )+ d \gamma _\cdot {{\mathrm{grad}}}^{\gamma ^*\cdot }\varsigma ,\quad \gamma \in \tilde{\mathcal {A}}_k, \end{aligned}$$

(91)

cf. [11, 34]. Then we can calculate the Hessian of the $L^2$-mass, taking into account (84) (with $w=\gamma$ and ${{\mathrm{div}}}_{\gamma (\mathcal {M})}(\gamma \circ \gamma ^{-1})=k$):

$$\begin{aligned} \begin{aligned} \langle Hess\; M(\gamma ) \dot{\gamma }, \dot{\gamma }\rangle _{T_{\gamma }\tilde{\mathcal {A}}_k}=\frac{\mathrm{{d}}^2 M (\gamma (t))}{\mathrm{{d}}^2t}&=\int _{{\mathcal {M}}} \partial _{\tau \tau }\gamma \cdot \gamma +\partial _\tau \gamma \cdot \partial _\tau \gamma \, \mathrm{{d}}{\mathcal {H}}^k\\&=-k\int _{{\mathcal {M}}} \varsigma \, \mathrm{{d}}{\mathcal {H}}^k+\langle \dot{\gamma }, \dot{\gamma }\rangle _{T_{\gamma }\tilde{\mathcal {A}}_k}. \end{aligned} \end{aligned}$$

(92)

Now we compute in two different ways the second time derivative of the $L^2$-mass along a trajectory $\xi (t)$ of the gradient flow (90). On the one hand,

(93)

where $\varsigma (t)$ is the Lagrange multiplier (which may be referred to as the tension) corresponding to the geodesic passing through $\xi (t)$ at the direction $\dot{\xi }(t)$, see (91). On the other hand, employing (84) and orthogonality of the projection ${\tilde{P}}_\xi$, we find that

$$\begin{aligned} \begin{aligned} \langle {{\mathrm{grad}}}_{\tilde{\mathcal {A}}_k} M(\xi ),&{{\mathrm{grad}}}_{\tilde{\mathcal {A}}_k} M(\xi ) \rangle _{T_{\xi }\tilde{\mathcal {A}}_k} =\langle {\tilde{P}}_\xi \xi , {\tilde{P}}_\xi \xi \rangle _{L^2({\mathcal {M}})}\\&=\langle \xi , \xi \rangle _{L^2({\mathcal {M}})}+\langle \xi , {\tilde{P}}_\xi \xi - \xi \rangle _{L^2({\mathcal {M}})} \\&= \langle \xi , \xi \rangle _{L^2({\mathcal {M}})}+\left\langle \xi , \sigma \vec H (\xi )+ \mathrm{{d}} \xi _\cdot {{\mathrm{grad}}}^{\xi ^*\cdot }\sigma \right\rangle _{L^2({\mathcal {M}})}\\&=2M(\xi )-k\int _{{\mathcal {M}}} \sigma \,\mathrm{{d}}{\mathcal {H}}^k. \end{aligned} \end{aligned}$$

(94)

This yields the upper bound

$$\begin{aligned} k\int _{{\mathcal {M}}} \sigma \,\mathrm{{d}}{\mathcal {H}}^k\le 2M(\xi ). \end{aligned}$$

(95)

Differentiating (94) in time and comparing with (93), we deduce

(96)

In the particular case ${\mathcal {M}}={\mathcal {S}}^1$, it is known [42, 45] that the Lagrange multipliers $\varsigma$ related to the geodesics are always nonnegative, so we infer (ii) in Proposition 3.1.

1.2 B.2. Relation with the Optimal Transport

We now establish a link with the optimal transport theory [49, 50] by explaining how our flow (90) may be formally viewed as a gradient flow on a submanifold of the Wasserstein space. We recall [40, 49] that the space ${\mathcal {P}}_2(\mathbb {R}^d)$ of probability measures with finite second moments admits a formal Riemannian structure so that the 2-Wasserstein distance coincides with the geodesic distance. The mapping

is a restriction of Otto’s Riemannian submersion [40]. We refer to [23] for a basic introduction to Riemannian submersions. Actually, $\Pi$ is a bijection up to volume-preserving diffeomorphisms on $\mathcal {M}$. Set $\rho =\Pi (\xi )$, where $\xi =\xi (\tau )$ is a solution to the normalized UCMCF. Since the $L^2$-mass functional M is invariant with respect to volume-preserving changes of variables on ${\mathcal {M}}$, it is consistent to define $M(\rho ):=M(\xi )$. Set ${\mathcal {B}}=\Pi (\tilde{\mathcal {A}}_k)$. Then ${\mathcal {B}}$ may be formally viewed as a submanifold of ${\mathcal {P}}_2(\mathbb {R}^d)$. We claim that the evolution of $\rho$ satisfies

$$\begin{aligned} \partial _\tau \rho = {{\mathrm{grad}}}_{{\mathcal {B}}} M(\rho ). \end{aligned}$$

(97)

Indeed, fix time $\tau _0$, $\xi =\xi (\tau _0)$, and $\rho =\Pi (\xi (\tau _0))$. We need to show that

$$\begin{aligned} \langle \partial _\tau \rho ,v\rangle _{T_\rho {{\mathcal {P}}_2}}= \langle {{\mathrm{grad}}}_{{\mathcal {B}}} M(\rho ),v\rangle _{T_\rho {{\mathcal {P}}_2}}, \end{aligned}$$

(98)

for an arbitrary $v\in T_\rho {{\mathcal {B}}}$. Let $\tilde{\rho }(\tau )$ be a curve in ${\mathcal {B}}$ satisfying $\tilde{\rho }(\tau _0)=\rho$, $\partial _\tau \tilde{\rho }(\tau _0)=v$. Let $\tilde{\xi }$ be the horizontal lift of the curve $\tilde{\rho }$ with respect to the submersion $\Pi$ passing through $\xi$ at $\tau _0$. Denote by $\partial _\tau \xi _h$ the horizontal component of $\partial _\tau \xi$. Then at time $\tau _0$

$$\begin{aligned}&\langle \partial _\tau \rho ,v\rangle _{\rho }=\langle \partial _\tau \Pi (\xi ),\partial _\tau \Pi (\tilde{\xi })\rangle _{\rho }=\langle d\Pi (\xi )\cdot \partial _\tau \xi ,d\Pi (\xi )\cdot \partial _\tau \tilde{\xi }\rangle _{\rho }\\&\quad = \langle \partial _\tau \xi _h,\partial _\tau \tilde{\xi }\rangle _{\xi }=\langle \partial _\tau \xi ,\partial _\tau \tilde{\xi }\rangle _{\xi }\\&\quad =\langle {{\mathrm{grad}}}_{\tilde{\mathcal {A}}_k} M(\xi ),\partial _\tau \tilde{\xi }\rangle _{\xi }=\partial _\tau M(\tilde{\xi })=\partial _\tau M(\tilde{\rho })=\langle {{\mathrm{grad}}}_{{\mathcal {B}}} M(\rho ),\partial _\tau \tilde{\rho }\rangle _{\rho }\\&\quad =\langle {{\mathrm{grad}}}_{{\mathcal {B}}} M(\rho ),v\rangle _{\rho }. \end{aligned}$$

Note that the geodesic distance on ${\mathcal {B}}$ is a priori nondegenerate since it is controlled from below by the 2-Wasserstein distance. The strategy above is applicable to the unnormalized flow. Indeed, the continuation of $\Pi$ defined by

is still a restricted Otto’s submersion. Then ${\mathcal {B}}^*=\Pi (\mathcal {A}_k^*)$ may be formally viewed as a submanifold of ${\mathcal {P}}_2(\mathbb {R}^d)$, and we are allowed to set ${{\mathrm{vol}}}(\rho ):={{\mathrm{vol}}}(\eta )$ for $\rho =\Pi (\eta )\in {\mathcal {B}}^*$. Then (88) can be recast as

$$\begin{aligned} \partial _t \rho =- {{\mathrm{grad}}}_{{\mathcal {B}}^*} \ln ({{\mathrm{vol}}}(\rho )). \end{aligned}$$

(99)