Gradient flow of the Dirichlet energy for the curvature of

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Gradient flow of the Dirichlet energy for the curvature of plane curves Gradient flow of the Dirichlet energy for the curvature of plane curves

Yuhan Wu University of Wollongong

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Recommended Citation Recommended Citation Wu, Yuhan, Gradient flow of the Dirichlet energy for the curvature of plane curves, Doctor of Philosophy thesis, School of Mathematics and Applied Statistics, University of Wollongong, 2021. https://ro.uow.edu.au/theses1/1039

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Gradient flow of the Dirichlet energy for the curvature ofplane curves

Yuhan Wu

This thesis is presented as part of the requirements for the conferral of the degree:

Doctor of Philosophy

Supervisor:Glen Wheeler, David Hartley, James McCoy

The University of WollongongSchool of Mathematics and Applied Statistics

January, 2021

This work c© copyright by Yuhan Wu, 2021. All Rights Reserved.

No part of this work may be reproduced, stored in a retrieval system, transmitted, in any form or by anymeans, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of theauthor or the University of Wollongong.

This research has been conducted with the support of a University of Wollongong Faculty of Engineeringand Information Sciences Postgraduate research scholarship.

Declaration

I, Yuhan Wu, declare that this thesis is submitted in partial fulfilment of the requirementsfor the conferral of the degree Doctor of Philosophy , from the University of Wollongong,is wholly my own work unless otherwise referenced or acknowledged. This documenthas not been submitted for qualifications at any other academic institution.

The results of Chapter 4, 5, 6 appear in papers ‘A sixth order flow of plane curves withboundary conditions’, ‘Higher order curvature flows of plane curves with generalisedNeumann boundary conditions’ and ‘Evolution of closed curves by length-constrainedcurve diffusion’ which are under supervision of Assoc. Prof. James McCoy and Dr GlenWheeler.

Yuhan Wu

May 13, 2021

Abstract

This thesis studies curvature flows of planar curves with Neumann boundary conditionand flows of closed planar curves without boundary. We describe the local existencefor them. For the global existence results for curvature flows of planar curves, we con-sider the sixth and higher order curvature flows of planar curves with suitable associatedgeneralised Neumann boundary condition. The conclusion is that the solution of eachflow problem exists for all time and converges to a unique line segment exponentially.Moreover, we study the curve diffusion flow and ideal curve flow of planar curves withconstrained length, as well as the ideal curve flow of planar curves with preserved area.Closed curve satisfying one of these flows exists for all time and converges to a uniqueround circle exponentially.

iv

Acknowledgments

My sincere thanks go to my advisors Assoc. Prof. James McCoy, Dr Glen Wheeler andDr David Hartley for their continuous guidance, support and knowledge.

I would like to express my gratitude to the University of Wollongong, specifically theSchool of Mathematics and Statistics, for providing a friendly environment to study inover the years.

v

Contents

Abstract iv

1 Introduction 11.1 Main results of thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Suggestions for further research . . . . . . . . . . . . . . . . . . . . . . 6

2 Prerequisites 72.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Short time existence 173.1 A sixth order flow of plane curves with boundary conditions . . . . . . . 17

3.1.1 Scalar quasilinear initial boundary value problem . . . . . . . . . 183.1.2 Short time existence for the quasilinear graph problem . . . . . . 26

3.2 Higher order flows of plane curves with boundary conditions . . . . . . . 323.2.1 An equivalent quasilinear scalar graph problem . . . . . . . . . . 333.2.2 Short time existence for the quasilinear graph problem . . . . . . 35

3.3 The closed length-constrained curve diffusion flow . . . . . . . . . . . . 383.3.1 Scalar quasilinear parabolic graph function . . . . . . . . . . . . 393.3.2 Fixed point argument . . . . . . . . . . . . . . . . . . . . . . . . 47

3.4 The closed constrained ideal curve flow . . . . . . . . . . . . . . . . . . 493.4.1 Scalar quasilinear parabolic graph function . . . . . . . . . . . . 493.4.2 Fixed point argument . . . . . . . . . . . . . . . . . . . . . . . . 51

4 A sixth order flow of plane curves with boundary conditions 544.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.2 Controlling the geometry of the flow . . . . . . . . . . . . . . . . . . . . 624.3 Exponential convergence . . . . . . . . . . . . . . . . . . . . . . . . . . 684.4 The unique limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5 Higher order flows of plane curves with boundary conditions 785.1 The gradient flow for the energy . . . . . . . . . . . . . . . . . . . . . . 79

5.1.1 Exponential convergence . . . . . . . . . . . . . . . . . . . . . . 91

vi

CONTENTS vii

5.2 The polyharmonic curve diffusion flow . . . . . . . . . . . . . . . . . . . 965.2.1 Exponential decay . . . . . . . . . . . . . . . . . . . . . . . . . 100

5.3 The unique limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

6 Length-constrained curve diffusion flow 1076.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.2 Controlling the geometry of the flow . . . . . . . . . . . . . . . . . . . . 1156.3 Global existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1236.4 Exponential Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . 1256.5 The unique limiting image . . . . . . . . . . . . . . . . . . . . . . . . . 1306.6 Self-similar solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1326.7 Embeddedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

7 Constrained ideal curve flow 1367.1 Length-constrained ideal curve flow . . . . . . . . . . . . . . . . . . . . 136

7.1.1 Exponential decay . . . . . . . . . . . . . . . . . . . . . . . . . 1387.1.2 Global Existence . . . . . . . . . . . . . . . . . . . . . . . . . . 140

7.2 Area-preserving ideal curve flow . . . . . . . . . . . . . . . . . . . . . . 1457.2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1467.2.2 Global Existence . . . . . . . . . . . . . . . . . . . . . . . . . . 1507.2.3 Exponential decay . . . . . . . . . . . . . . . . . . . . . . . . . 153

Bibliography 163

A Appendix 168A.1 Proof of Theorem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168A.2 Proof of Proposition 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . 169A.3 Proof of Proposition 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173A.4 Referred Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

Chapter 1

Introduction

In differential geometry, the study of curves and surfaces includes local and global prop-erties. Local properties depend only on the behaviour of the curves or surfaces in theneighbourhood of a point. For instance, the curvature of a curve is a local property. Theglobal properties consider the influence of local properties on the behaviour of the entirecurve or surface. Global differential geometry is concerned with the relations betweenlocal and global properties of curves and surfaces.

The properties of curves and surfaces in differential geometry are investigated by differ-ential and integral calculus. The calculus involves differentiable functions, vector fields,differential forms, mappings and various operations of differentiation and integration. Aparabolic evolution equation is a partial differential equation describes processes whichare evolving in time, for example the heat flow shows the transfer of heat from a hotarea to a cool area if there is a different temperature between materials which are next toeach other. A parabolic geometric evolution equation can be called a geometric heat flowwhich is often seen as a gradient flow of a geometric object.

The second order geometric heat flows are well studied. The mean curvature flow canbe seen as the most famous second order geometric flow, it was proposed to describe theformation of grain boundaries in annealing metals which is material science in 1957 byphysicist Mullins in [65] and later in 1959 [66]. Mullins was the first to write down themean curvature flow equation. The mean curvature flow was studied by Gerhard Huiskenfrom the perspective of partial differential equations in 1984 [36]. It is the negative steep-est descent gradient flow for area functional, the surface area is decreasing monotonicallyand stationary if and only if the surface stays minimal. There are some related flowsthat have the same leading order term as mean curvature flow, for example, the volume-preserving mean curvature flow studied by Huisken [37] and area-preserving mean curva-ture flow considered by McCoy [59]. The curve-shortening flow is the one-dimensionalcase of the mean curvature flow. There has been much research on the long-time be-haviour of smooth curves under this flow problem. In [26] and [27], Gage shows that ifa convex plane curve evolving by curve-shortening shrinks to a point, then its limiting

1

CHAPTER 1. INTRODUCTION 2

shape is circular in a weak sense. Afterwards, Gage and Hamilton in [28], Grayson in[33] and [34] study the convergence in C∞ norm.

Second-order curvature flows with free boundaries have been studied for decades. In1989, Huisken considers mean curvature flow of graphs in cylindrical domains [38]. Af-terwards, in [74] and [75], Stahl considers mean curvature flow with a free boundary onan umbilic hypersurface regarding continuation criteria and singularities. They continueto receive significant research attention, for example [9], where Buckland studies bound-ary monotonicity formula. Freire considers mean curvature flow of graphs with constantangle at a free boundary in [25]. Second-order curvature flows with free boundary are alsostudied for example by Koeller [44], Marquardt [57], Mizuno and Tonegawa [64], Edelen[19], Lambert [50], V. Wheeler [84] and [83], McCoy, Mofarreh and Williams [60]. Alsosome results hold for curves with mixed Neumann-Dirichlet boundary conditions in [54],[12]. Results on classification of singularities and the extension beyond singularities ofthe mean curvature flow are obtained by Huisken and Sinestrari in [39], [40], [41] and[42].

Higher order geometric evolution problems have drawn interest in recent years, espe-cially, fourth order geometric equations. The surface diffusion flow is a famous fourthorder geometric evolution equation. In 1957 [65], Mullins first proposed it to model theformation of tiny thermal grooves in phase interfaces. The surface diffusion flow involvessecond derivative of the mean curvature so is a fourth order flow. It is the gradient flowfor surface area in a Sobolev space and decreases the surface area while the volume ispreserved. In [10], Cahn, Elliott and Novick-Cohen prove that the only limit of the Cahn-Hilliard equation with a concentration dependent mobility is surface diffusion. Later in[21], Elliott and Garcke consider motion laws for surface diffusion flow. Giga and Itoconsider self-intersection and loss of convexity of surface diffusion flow in [31] and [32]respectively. Afterwards, in [43], Ito shows the loss of convexity for surface diffusionflow equation. In [24], Escher, Mayer and Simonett consider the solutions to immersedhypersurfaces under surface diffusion flow exist and are unique. Escher and Ito [23] showthe loss of convexity for intermediate surface diffusion flow. they also prove that surfacediffusion flow and intermediate surface diffusion flow develop singularities in finite time.There is also research on surface diffusion flow with boundary conditions. For example[30], Garcke, Ito and Kohsaka study nonlinear stability of stationary plane curves undersurface diffusion flow with specified boundary conditions. Asai and Giga consider sur-face diffusion flow with contact angle boundary conditions in [2]. Global analysis for thesurface diffusion flow is also studied, the theory of singularities for the flow is consideredfor instance by G. Wheeler [79] [80]. In G. Wheeler’s PhD thesis [79], he considers theconstrained surface diffusion flow which has a function of time. This function is chosento coincide with a natural geometric restriction, for instance, different choices of the func-tion can lead to conservation of mixed volumes or a reduction in mass and increase of free


surface energy. He proves that the two-dimensional surface diffusion flow exists for alltime and converges to a round sphere exponentially. By choosing the suitable function oftime, he shows the enclosed volume is monotonically increasing while the surface area isdecreasing.

The curve diffusion flow is the one-dimensional case for the surface diffusion flow. It isthe steepest descent H−1-gradient flow for length and also a fourth order geometric evolu-tion equation. The proof of local existence for curve diffusion flow refers to the standardprocedure in [75] by Stahl, it solves the flow by converting the curve to a graph over theinitial data. Furthermore, in [22], Elliott and Maier-Paape give that the graphs evolvedas curves by curve diffusion flow become non-graphical in finite time. Eventually, it isshown in [8] by Blatt that a large class of higher order hypersurface flows lose convex-ity and embeddedness. In [85], G. Wheeler and V. Wheeler consider the curve diffusionflow of open plane curve with free boundary between on parallel lines. For closed curvesmoving by curve diffusion flow, G. Wheeler proves the area of closed curves is constantin [82]. In [81], G. Wheeler studies the solution of the flow problem by using the isoperi-metric ratio and small initial oscillation of curvature. He proves the closed curve existsfor all time and converges to a simple circle exponentially. We are interested in the closedcurve diffusion flow with constrained length which is studied in Chapter 6.

Another fourth order flow is the Willmore flow, which is the steepest descent L2-gradient flow for Willmore functional. Normally, Willmore functional is the integral ofmean curvature squared. It was first studied by Sophie Germain in the 19th century.Afterwards it received significant attention from Blaschke who first presented its Euler-Lagrange operator in 1929, see Blaschke and Thomsen [7], Blaschke and Reidemeister[6]. For more references, see Simonett [72], Bauer and Kuwert [4], Bernard and Rivière[5], Chen [11] and Kusner [46], McCoy and G. Wheeler [62]. Furthermore, the studiesof Willmore conjecture with minimal surfaces can be found in, for example, Li and Yau[51], Marques and Neves [58], roughly speaking, Willmore conjecture is that there is alower estimate of the Willmore energy for immersed surface.

Elastic flow is the one-dimensional Willmore flows and the steepest descent L2-gradientflow for the elastic energy which is the integral of curvature squared. Local existence forthe flow is shown in [75] by Stahl. In [85], G. Wheeler and V. Wheeler consider the elasticflow have free boundary, supported on parallel lines in the plane. Let us now define E[γ]

for a smooth closed or open planar curve γ by E[γ] = 12∫

γk2

s ds, where ks is the first arc-length derivative of the curvature. Our interest is the L2-gradient flow for this functionalE[γ] which is the sixth order curvature flow studied in Chapter 4.

Fourth-order curvature flows with different boundary conditions have been studied inrecent years. For example, in [29] and [30], stability results are proved by Garcke, Ito andKohsaka for curves under curve diffusion flow that are graphical with nearby equilibriaevolving in bounded domains with free boundary. Other studies include curves moving by


gradient for elastic energy with constrained length and fixed boundary points with zerocurvature by Dall’Acqua, Lin and Pozzi [13], curves moving by Helfrich energy withnatural boundary conditions by Dall’Acqua and Pozzi [14], Helfrich energy for planarcurves is the sum of the Willmore functional and the length functional times a positiveweight factor. The elastic energy with clamped boundary conditions is studied by Lin [52]and Dall’Acqua, Pozzi and Spener [15]. In [67], Novaga and Okabe consider the steepestdescent flow of open planar curves with clamped boundary conditions and symmetricNavier boundary conditions. The most relevant research for us is the study of curvediffusion and elastic flow of curves between parallel lines in [85] by G. Wheeler and V.Wheeler.

Higher-order curvature flows of closed curves without boundary have received some at-tention. Particularly, some works are done by Dziuk, Kuwert and Schätzle [17], Edwards,Gerhardt-Bourke, McCoy, G. Wheeler and V. Wheeler [20], Giga and Ito [32], Parkinsand G. Wheeler [68]. Moreover, McCoy, G. Wheeler and Williams obtain the maximaltime estimate for the closed immersed hypersurfaces which evolve by constrained surfacediffusion flows in [63]. Sixth order flows of closed curves are studied in [1] by Andrews,McCoy, G. Wheeler and V. Wheeler, authors prove the convergence of the solution to anideal curve flow. An ideal curve is a smooth curve with zero normal speed. The appli-cation of the energy in this paper appear in computer aided design [35] by Harary andTal. In Chapter 7, we consider an ideal curve flow that preserves the length while thesigned enclosed area does not decrease. This flow is obtained by adding a suitable globalfunction of time into the flow speed.

When dealing with higher order flows, many of the tools and techniques applied tostudy second order curvature flow cannot be used, such as the maximum principles. Themaximum principle is the important tool to study the behaviour of solutions to the secondorder geometric heat flows and has been used to prove many theorems related to secondorder flows, for example [36], [18], [38] by Huisken, Ecker, Gage and Hamilton. How-ever, some known techniques can be applied to various higher order curvature flow underdifferent conditions. Some inequalities like interpolation inequalities, Sobolev inequality,Young’s inequality, Cauchy-Schwarz inequality and Hölder’s inequality are often usedduring the studies. In [17], Dziuk, Kuwert and Schätzle give an interpolation inequal-ity for closed curves while Dall’Acqua and Pozzi prove an inequality which can handlenon-closed curves in [14].

One technique for studying fourth-order flows is to use curvature integral estimates.It was first used for the Willmore flow by Kuwert and Schätzle. They set up a generalframework which can be used to study different varieties of fourth-order curvature flow in[47], [48] and [49]. The framework includes evolution equations for curvature quantities,integral estimates, short time existence and the concentration compactness alternative.These turn out to be useful for other fourth order flows, even higher order flows whether


they have or do not have a gradient structure. Therefore, this framework can be appliedto related flows to obtain stability results for parabolic problems. Applications and mod-ifications of this framework have been used for the surface diffusion flow by G. Wheelerin [80] and [81], the geometric triharmonic heat flow [61] by McCoy, Parkins and G.Wheeler, and polyharmonic flows [68] by Parkins and G. Wheeler.

About flows of higher even order than four, little research has been done so far. But mo-tivation for them comes from various areas. Applications of sixth order curvature flowsinvolving a thrice-iterated Laplacian are increasing, for example, Ugail and Wilson con-sider the modelling of ulcers in [78], computer graphics studied by Malcolm, Wilson andHagenin [56] and Ugail in [77], interactive design is considered by Kubiesa, Ugail andWilson in [45], and computer designs introduced by Liu and Xu in [55]. Moreover, in[86], Xu, Pan and Bajaj study several geometric flows, including sixth order flows thathave been used for surface blending, N-side hole filling and free-form surface design.Fluid flows and propeller blade design are considered by Tagliabue in [76] and Dekanskiin [16]. In [61], the geometric triharmonic flow for closed surfaces is considered by Mc-Coy, Parkins and G. Wheeler, this flow involves fourth derivatives of the mean curvatureso is a sixth order flow. Parkins and G. Wheeler concern even order flow of closed pla-nar curves in [68]. These interests and applications of sixth order flows provide strongmotivation for the study in Chapter 4.

1.1 Main results of thesis

We summarise the main results of this thesis as the following. In Chapter 2, we givesome basic definitions and notations related to our problems and some inequalities wewill use frequently. Chapter 3 describes the local existence for flows of planar curveswith Neumann boundary condition and flow of closed planar curves without boundary.Some references about short time existence are mentioned in this chapter. In Chapter 4,we consider the sixth order curvature flow of open planar curves with Neumann boundarycondition. We show that the solution of the flow problem exists for all time and convergesto a unique line segment. We also obtain the bounded region of solution. In Chapter5, we consider the higher order curvature flow of planar curves with suitable associatedgeneralised Neumann boundary condition. We generalise the sixth order case where weconsidered the L2-gradient flow for the energy 1

2∫

γk2

s ds, and consider the L2-gradientflow for the energy 1

2∫

γk2

smds with suitable associated generalised Neumann boundaryconditions, where ksm is the mth derivative of the curvature. In this chapter, we follow thesimilar process in Chapter 4, our conclusion is the curve moving under (2m+4)th-ordercurvature flows with generalized Neumann boundary condition converges to a uniquestraight line segment when time goes to infinity. In Chapter 6, we study the curve diffusionflow of planar curves with constrained length. We assume that initial data close to a round


circle, within the sense of normalised L2 oscillation of curvature, closed curve under thisflow exists for all time and converges exponentially fast to a round circle. We also givethe self-similar solutions for this flow problem. In Chapter 7, we study an ideal curve flowof closed curve that preserves the length while the signed enclosed area does not decreaseand the ideal curve flow of closed curve with preserved area. We prove the long-timeexistence of the curves and exponential convergence under the smallness condition.

1.2 Suggestions for further research

In Chapter 4 and Chapter 5, we study the long-time existence for sixth order curvatureflow and (2m+4)th-order curvature flows of planar curves with Neumann boundary con-dition, and the boundaries are two parallel lines with a distance between them. For furtherresearch, the framework we use in these two chapters can be used to study other even or-der curvature flows with the same boundary condition. The results and framework alsocan be adapted to the same or different even order curvature flows of planar curves withNeumann boundary condition, but with the different boundaries, for example, a cone or asimple circle. Moreover, it can be applied to even order curvature flows of planar curveswith other boundary conditions, such as Dirichlet condition. There are some open ques-tions related to closed planar curves, for example, the global existence for sixth ordercurvature flow and higher order curvature flows of closed planar curves without bound-ary, long-time existence of length-constrained or area-preserved curves satisfying theseflows.

Chapter 2

Prerequisites

We introduce some basic definitions and notations related to our problems.Consider the open or closed curve γ to move by the normal velocity F :

∂tγ = F [k]ν ,

here F [k] denotes the normal speed of the curve, ν is unit normal vector field of the curveand k = 〈γss,ν〉 is the scalar curvature, where 〈γss,ν〉 denotes the inner product of γss andν . τ = γu

|γu| = γs is the unit tangent vector field along γ .Firstly, we give the open curves with free boundary. Let η1,η2 : R→ R2 denote two

parallel vertical lines in R2, with a distance |e| 6= 0 between them. Consider a one-parameter family of smooth immersed curves γ : [−1,1]× [0,T )→R2, the two end pointsof the curve lie on η1,η2 respectively and the curves meet the two parallel lines η1,η2

orthogonally. Here e to be any vector such that e is perpendicular to the parallel lines η1,2.We wish the curve γ to move by the normal velocity F . This is called open curves movingunder curvature flow with Neumann boundary condition.

Here we introduce the ideal curves. We consider the energy functional

E[γ] =12

∫γ

k2s ds.

where ks is the derivative of curvature with respect to arc-length and ds the arclengthelement. We are interested in the L2 gradient flow for curves of small initial energy withNeumann boundary condition. As our energy involves the first derivative of curvature,the gradient flow will be sixth order and so six boundary conditions will be needed. Thecorresponding gradient flow has normal speed given by F . Let γ be a smooth curvesatisfying F = 0, that is a stationary solution to the L2-gradient flow of E. We call suchcurves ideal and they are critical points for E.

Secondly, for closed curves, we let γ :R→R2 be a (suitably) smooth embedded regularcurve. We say γ is periodic with period P if there exists a vector V ∈ R2 and a positive

7

CHAPTER 2. PREREQUISITES 8

number P such that, for all m ∈ N,

γ(u+P) = γ(u)+V

and∂

mu γ(u+P) = ∂

mu γ(u).

here ∂ mu denotes the mth derivative of γ .

If V = 0 then γ is closed and we write γ : S1→ R2. The length of γ is

L[γ] =∫ P

0|γu(u)|du

and the signed enclosed area is

A[γ] =−12

∫ P

0〈γ,ν〉|γu|du

where ν is a unit normal vector field on γ , 〈γ,ν〉 denotes the inner product of γ and ν .Throughout this thesis, we will keep our evolving curves γ parametrised by arc-length

s and k2sn denotes (ksn)2, where ksn is the n-th iterated derivative of k with respect to arc-

length and ds = |γu|du.

Definition 1. There are several definitions for both open and closed curves.

(i) The length of γ is

L[γ] =∫

γ

ds.

Note that for open curves, w is not always an integer.

(ii) The average curvature is

k[γ] =1L

∫γ

kds.

(iii) The oscillation of curvature is defined as

Kosc[γ] = L∫

γ

(k− k)2ds.

The following two additional definitions are for closed curves:

(iv) The signed enclosed area is

A[γ] =−12

∫γ

〈γ,ν〉ds.


(v) The isoperimetric ratio of γ is

I[γ] =L2

4πA.

In this thesis, we denote L(t) = L [γ(·, t)], A(t) = A [γ(·, t)], Kosc(t) = Kosc [γ(·, t)] andI(t) = I [γ(·, t)]. For the convenience, we use L to denote L(t) and γ0 to denote initialcurve.

For the winding number ω , we have

ω =1

2π

∫γ

kds.

For closed curves, w ∈ Z, for example, in figure 2.1,

ω[γ1] =1

2π

∫γ1

kds = 0;

ω[γ2] =1

2π

∫γ2

kds = 1;

ω[γ3] =1

2π

∫γ3

kds = 2.

γ1 γ2

γ3

Figure 2.1

However, for open curves with Neumann boundary condition, the winding numbermust be a multiple of 1

2 . For example, in Figure 2.2,

ω[γ4] =1

2π

∫γ4

kds = 1,

ω[γ5] =1

2π

∫γ5

kds =12.


η1 η2

γ4

γ5

Figure 2.2

We give the commutator relation between Euclidean arc-length and time derivatives ofthe evolving curves. Proofs appear for example in [85] by G. Wheeler and V. Wheeler,we provide them again here for completeness.

Lemma 1. The commutator of arc-length and time derivative is given by

∂t∂s = ∂s∂t + kF∂s

and the measure ds evolves by

∂tds =−kFds

Proof. We do the computation

∂t |γu|2 = 2|γu||γu|t = 2〈γu,γtu〉= 2〈γu,(Fν)u〉

= 2F〈γu,νu〉=−2(kF) |γu|2

then∂t |γu|=−kF |γu|,

so we have

∂t∂s−∂s∂t = ∂t

(1|γu|

)∂u =−

∂t |γu||γu|2

∂u

= −−kF |γu||γu|2

∂u =kF|γu|

∂s

= kF∂s.

The result follows.

This commutator relation can make the calculation of derivative of curvature vectorseasier.


Lemma 2. Some basic evolution equations:

τt =−Fsν , νt = Fsτ.

where τ is a unit tangential vector field on γ .

Proof. We apply the above Lemma 1 to the following calculation,

τt = γst = γts + kFγs = (Fν)s + kFτ

= Fsν +Fνs + kFτ

= Fsν−Fkτ + kFτ

= Fsν .

By the orthonormality of τ,ν, 〈τ,ν〉= 0 and 〈νt ,ν〉= 0, so we find

νt =−〈ν ,Fsν〉τ =−Fsτ.

Lemma 3. We have the following corresponding evolution equations for various geomet-

ric quantities:

(i) ddt L =−

∫γ

kFds,

(ii) ∂

∂ t k = Fss + k2F ;(iii) ∂

∂ t ks = Fs3 + k2Fs +3kksF ;(iv) ∂

∂ t kss = Fs4 + k2Fss +5kksFs +4kkssF +3k2s F ;

For each l = 0,1,2, ...,(v) ∂

∂ t ksl = Fsl+2 +∑lj=0 ∂s j (kksl− jF) .

Proof. For (i), under Lemma 1 and Definition 1 (i), we have

ddt

L =ddt

∫γ

ds =−∫

γ

kFds.

For (ii),

γsst = γsts + kFγss = τts + kFγss

= (Fsν)s + kF · kν

=(Fss + k2F

)ν− kFsτ.


The scalar curvature k = 〈γss,ν〉, then by using Lemma 2

∂

∂ tk = kt = 〈γss,νt〉+ 〈γsst ,ν〉

= 〈γss,−Fsτ〉+⟨(

Fss + k2F)

ν− (kFs)τ,ν⟩

= Fss +Fk2.

Working with the scalar quantities, we obtain the evolution equations (iii), (iv),

∂

∂ tks = ∂s∂tk+ kFks

=(Fss + k2F

)s + kksF

= Fs3 + k2Fs +3kksF,

∂

∂ tkss = ∂s∂tks + kFkss

=(Fs3 + k2Fs +3kksF

)s + kkssF

= Fs4 + k2Fss +5kksFs +4kkssF +3k2s F.

For the proof of (v), we use the induction method to get the expression of ksnt , for alln ∈ N∪0,

for n = 0, we have kt = Fss + k2F ;for n = 1, we have kst = Fs3 +∑

1j=0 ∂s j(kks1− jF);

for n = 2, we have ksst = Fs4 +∑2j=0 ∂s j(kks2− jF);

By assuming when n = l, for all l ∈ N, l ≥ 3

ksnt = kslt = Fsl+2 +l

∑j=0

∂s j(kksl− jF),

we have for n = l +1,

ksnt = ksl+1t = kslts + kFksl+1

=

[Fsl+2 +

l

∑j=0

∂s j(kksl− jF)

]s

+ kksl+1F

= Fsl+3 +l

∑j=0

∂s j+1(kksl− jF)+ kksl+1F

= Fsl+3 +l+1

∑j= j+1=1

∂s j(kksl+1− jF)+ kksl+1F

= Fsl+3 +l+1

∑j=0

∂s j(kksl+1− jF).


Here we finish the proof.

The next lemma shows that the winding numbers of curves we study do not changeover time.

Lemma 4. For the open curves with Neumann boundary condition and closed curves, we

obtain ω(t) = ω(0) for all time.

Proof. First for open curves, as mentioned before, the Neumann condition is equivalentto

〈ν(±1, t),e〉= 0,

differentiating it in time implies

Fs(±1, t)〈τ(±1, t),e〉=±|e|Fs(±1, t) = 0,

as |e| 6= 0, we have that Fs(±1, t) = 0.We compute

ddt

∫γ

kds =−∫

γ

Fss +Fk2−Fk2ds =−∫

γ

Fssds =− Fs|0,L = 0,

which implies

ω(t) =1

2π

∫γ

kds =1

2π

∫γ

kds∣∣∣∣t=0

= ω(0).

Then for closed curves, we have

ddt

∫γ

kds =∫

γ

Fss +Fk2−Fk2ds =∫

γ

Fssds = 0,

soω(t) =

12π

∫γ

kds =1

2π

∫γ

kds∣∣∣∣t=0

= ω(0).

2.1 Preliminaries

Here we introduce some inequalities we will use frequently in the following Chapters.First, we state the Poincaré-Sobolev-Wirtinger [PSW] inequalities for open curves.

Proposition 1. Suppose function f : [0,L] → R,L > 0, is absolutely continuous and∫ L0 f dx = 0. Then

∫ L

0f 2dx≤ L2

π2

∫ L

0f 2x dx. (2.1)


Suppose function f : [0,L]→ R,L > 0, is absolutely continuous and f (0) = f (L) = 0.

Then ∫ L

0f 2dx≤ L2

π2

∫ L

0f 2x dx.

Proposition 2. Suppose f : [0,L]→R,L > 0, is absolutely continuous and f (0) = f (L) =

0. Then

‖ f‖2∞ ≤

Lπ

∫ L

0f 2x dx.

Suppose f : [0,L]→ R,L > 0, is absolutely continuous and∫ L

0 f dx = 0. Then

‖ f‖2∞ ≤

2Lπ

∫ L

0f 2x dx. (2.2)

For the closed curves studied in Chapter 6, we will frequently use the following PSWinequalities. For proofs of these see for example [68, Appendix A] by Parkins and G.Wheeler, we also put the proof in Appendix A, 10.

Proposition 3. Suppose function f : R→ R is absolutely continuous and periodic with

period P. Then if∫ P

0 f dx = 0, we have

(i)

∫ P

0f 2dx≤ P2

4π2

∫ P

0f 2x dx, (2.3)

with equality if and only if f (x) = Acos(2π

P x)+Bsin

(2π

P x)

for arbitrary constants A and

B;

(ii)

‖ f‖2∞ ≤

P2π

∫ P

0f 2x dx. (2.4)

To state the next interpolation inequality we will use, we first need to set up somenotations. For normal tensor S and T, we use S?T to denote any linear combination of Sand T. In our setting, S and T will be simply curvature k or its arc-length derivatives. Weuse Pm

n (k) to denote any linear combination of terms of type ∂i1s k ?∂

i2s k ? ... ?∂

ins k, where

m = i1 + i2 + ...+ in is the total number of derivatives.Here we show the following interpolation inequality for the open curves with free

boundary and closed curves. For open curves, we refer to [14, Theorem 4.3] by Dall’Acquaand Pozzi; for closed curves, it appears in Dziuk, Kuwert and Schätzle’s paper [17].

Proposition 4. Let γ : I→R2 be a smooth open or closed curve. Then for any term Pmn (k)


with n≥ 2 that contains derivatives of k of order at most l−1,∫I|Pm

n (k)|ds≤ cL1−m−n‖k‖n−p0,2 ‖k‖

pl,2

where p = 1l

(m+ 1

2n−1)

and c = c(l,m,n). Moreover, if m+ n2 < 2l +1 then p < 2 and

for any ε > 0,

∫I|Pm

n (k)|ds≤∫

I|∂st k|2ds+ cε

−p2−p

(∫I|k|2ds

) n−p2−p

+ c(∫

I|k|2ds

)m+n−1

.

Note that in above, ‖ · ‖0,2 and ‖ · ‖m,2 denotes scale-invariant norms, for example

‖k‖0,2 = L12

(∫k2ds

) 12

and

‖k‖1,2 = L12

(∫k2ds

) 12

+L32

(∫k2

s ds) 1

2

;

Except for the statement of this proposition, in this thesis we will use the notation ‖ · ‖2

to denote the regular unscaled norms, pointing out explicit scaling factors where relevant.In the estimates throughout this thesis, the constants c can be different from line to linewhere they depend only on absolute quantities.

Lastly, we state the following standard inequalities for reference.

Proposition 5 (Hölder inequality). Suppose functions f ,g : I→ R. Assume 1 < p,q≤ ∞

satisfies 1p +

1q = 1. If f ∈ Lp(I) and g ∈ Lq(I), then f g ∈ L1(I) satisfies

∫I| f (x)g(x)|dx≤

(∫I| f (x)|pds

)1/p(∫I|g(x)|qds

)1/q

.

Proposition 6 (Minkowski inequality). Suppose functions f ,g : I→R. Assume 1< p<∞

and f ,g ∈ Lp(I). Then

(∫I| f (x)+g(x)|pds

)1/p

≤(∫

I| f (x)|pds

)1/p

+

(∫I|g(x)|pds

)1/p

.

We use above two inequalities for open curves when I = [−1,1], and apply these twoinequalities to closed curves under I = S1.

Proposition 7 (Grönwall’s inequality). Let f (t) be a non-negative, absolutely continuous

function on [0,T ], which satisfies for

f ′(t)≤ α(t) f (t)+β (t),


where f ′(t) is the time derivative of f (t), φ(t) and ψ(t) are functions on [0,T ]. Then

f (t)≤[

f (0)+∫ t

0β (δ )dδ

]e∫ t

0 α(δ )dδ .

Proof. Let g(t) := f (t) · e−∫ t

0 α(δ )dδ , we have

ddt

g(t) = e−∫ t

0 α(δ )dδ ·[

f ′(t)−α(t) f (t)],

as f ′(t)≤ α(t) f (t)+β (t), then

ddt

(f (t) · e−

∫ t0 α(δ )dδ

)≤ e−

∫ t0 α(δ )dδ ·β (t).

Now integrate and get

f (t) · e−∫ t

0 α(δ )dδ ≤ f (0)+∫ t

0β (δ )dδ .

The result follows.

Proposition 8 (Young’s inequality). If 1 < p,q < ∞ satisfies 1p +

1q = 1. Then we have

ab≤ εap +(ε p)−qp

bq

q.

where a and b are strictly positive real numbers.

Chapter 3

Short time existence

This chapter describes the local existence for flows of open planar curves with Neumannboundary condition and flow of closed planar curves without boundary. Some referencesabout short time existence are mentioned in this chapter.

3.1 A sixth order flow of plane curves with boundary con-ditions

The sixth order curvature flow of plane curves with Neumann boundary condition is de-fined as follows, see more details in Chapter 4.

Definition 2. Let γ : [−1,1]× [0,T )→ R2 be a family of smooth immersion. γ is said to

move under sixth order curvature flow F with homogeneous Neumann boundary condi-

tion, if

∂

∂ t γ(s, t) =−Fν , f or all (s, t) ∈ [−1,1]× [0,T )γ(·,0) = γ0,

〈ν ,νη1,2〉(±1, t) = ks (±1, t) = ks3 (±1, t) = 0, f or all t ∈ [0,T )

(3.1)

where F = ks4 + kssk2− 12k2

s k, ν and νη1,2 are the unit normal fields to γ and η1,2 respec-

tively.

Here we state the way to prove the short time existence for curvature flows of openplanar curves with Neumann boundary condition. The first step is to convert the weaklyparabolic system (3.1) together with boundary conditions to a corresponding quasilinearscalar parabolic equation. This involves fixing a graphical parametrisation over a refer-ence curve. The reference curve here is a straight line segment. The conversion processusing generalised Gaussian coordinates in the case with boundary conditions is describedfor example in [75, Section 2], the case of higher codimension is discussed in [73]. The

17

CHAPTER 3. SHORT TIME EXISTENCE 18

second step is for the scalar parabolic equation with boundary conditions, we considerthe corresponding linearized equation, for which existence of a unique smooth solutionis well-known. By using the solution existence of the linearized problem together withthe general result on the nonlinear evolutionary boundary value problems (for example, in[69, Theorem 4.4]) to see that the scalar graph equation has a unique solution at least fora short time. We then prove scalar graph equation is equivalent to the flow system (3.1),thus a solution to (3.1) exists for a short time. The solution to (3.1) is necessarily notunique due to the possibility of choosing different parametrisations, however the imagecurve is unique. This method also works for the higher order cases (3.6).

3.1.1 Scalar quasilinear initial boundary value problem

This coordinate system (3.1) will be seen as generalized Gaussian coordinates. Let l([−1,1])be a straight line segment which is perpendicular to boundaries η1,η2. Define the fluxlines Φ = Φ(u, ·) to the curve are perpendicular to l([−1,1]) and tangential to η1,η2,see Figure 3.1. Define a neighbourhood U ⊂ R2 of l([−1,1]), Uε := Φ(u,x) : u ∈[−1,1], |x| < ε. In U, let ρ(p0) denote tangential coordinate of p0 on l([−1,1]), thendefine a smooth normal vector field ξ with properties:

〈ξ ,ρ〉|l([−1,1]) = 0, ξ |η1,2⋂Uε∈ T η1,2, ‖ξ‖= 1,

where η1,2∩Uε = p ∈ R2 : p = Φ(u,x),u ∈ η1,2,x ∈ (−ε,ε). Also

Φσ (u,σ) = ξ (Φ(u,σ)) , Φ(u,0) = γ0(u).

Therefore, for any point p = Φ(u,x), we let x(p) is equal to the length of the flux linethrough p between p and intersection point p0 = Φ(u,0) on l([−1,1]). We have

x(p) =∫ x

0|Φσ (u,σ)|dσ =

∫ x

0|ξ (Φ(u,σ)) |dσ .

We define Mt = p∈R2 : p=Φ(u,w(u, t)) ,u∈ [−1,1], here w(u, t) : [−1,1]×[0,σ ]→R and σ ∈ [0,T ).

Lemma 5. Set γ(u, t) : [−1,1]× [0,σ ]→ R2, γ(u, t) := (u,w(u, t)), then the expression

for the evolution of w is as follows,

wt(u, t) = v−6wu6−18v−7vuwu5−22v−7vuuwu4 +141v−8v2uwu4−13v−7vu3wu3

+232v−8vuvuuwu3−561v−9v3uwu3−3v−7vu4wuu +69v−8vuvu3wuu

+48v−8v2uuwuu−699v−9v2

uvuuwuu +945v−10v4uwuu + v−10w2

uuwu4

−4v−11vuw2uuwu3−

12

v−10wuuw2u3 +

212

v−12v2uw3

uu−3v−11vuuw3uu, (3.2)


η1 η2

γ([−1,1])

l([-1,1])ξ ρ

Figure 3.1

where v(u,w(u)) :=√

1+(wu)2.

Proof. As γ(u, t) = (u,w(u, t)), then we have γu(u, t) = (1,wu)(u, t).Let v(u, t) := |γu|(u, t) =

√1+w2

u, then the tangential vector

τ(u, t) =γu

|γu|= v−1(1,wu).

From above, we can get expression of normal vector ν(u, t) directly as follows,

ν(u, t) = v−1(−wu,1).

Differentiating γ with respect to time, then ∂tγ(u, t) = (0,wt(u, t)) and also

∂tγ(u, t) =[

ks4(u, t)+ k2kss(u, t)−12

kk2s (u, t)

]·ν(u, t),

here s =∫ u

u0|γu|du =

∫ uu0

√1+w2

udu,

∂s =∂u

|γu|=

∂u

v.

From ∂tγ ·ν = F , we obtain

(0,wt(u, t)) ·ν = ks4(u, t)+ k2kss(u, t)−12

kk2s (u, t).

As ν(u, t) = v−1(−wu,1)(u, t), then

v−1 ·wt(u, t) = ks4(u, t)+ k2kss(u, t)−12

kk2s (u, t).

Also, we have


γss =1v

∂uγs =1v

∂u

(γu

v

)=

1v· γuu · v− γu · vu

v2

=1v2 · γuu−

1v3 · γuvu =

1v2 (0,wuu)−

vu

v3 (1,wu),

here γu = (1,wu),γuu = (0,wuu).

We can get the expression for the curvature k(u, t),

k(u, t) = 〈γss,ν〉(u, t) =1v3 ·wuu,

and its first, second, third and fourth derivatives are as follows,

ks(u, t) =1v

∂u

(wuu

v3

)=

1v· wu3v3−wuu ·3v2vu

v6 =1v4 wu3−

3v5 vuwuu,

kss(u, t) =1v·∂u

(wu3

v4 −3v5 · vuwuu

)=

1v·(

wu4v4−4wu3v3vu

v8 −3 · uuuwuuv5 + vuwu3v5−5wuuv2uv4

v10

)=

1v5 wu4−

4v6 vuwu3−

3v6 vuuwuu−

3v6 vuwu3 +

15v7 v2

uwuu

=1v5 wu4−

7v6 vuwu3−

3v6 vuuwuu +

15v7 v2

uwuu,

and

ks3(u, t) =1v·∂u

(1v5 wu4−

7v6 vuwu3−

3v6 vuuwuu +

15v7 v2

uwuu

)=

1v·(

wu5v5−5wu4v4vu

v10 −7 · vuuwu3v6 + vuwu4v6−6v5v2uwu3

v12

−3 · vu3wuuv6 + vuuwu3v6−6v5vuvuuwuu

v12

+15 · 2vuvuuwuuv7 + v2uwu3v7−7v6v3

uwuu

v14

).

Simplify above equation, we have


ks3(u, t) =1v6 ·wu5−

5v7 vuwu4−

7v7 vuuwu3−

7v7 vuwu4 +

42v8 v2

uwu3−3v7 vu3wuu

− 3v7 vuuwu3 +

18v8 vuvuuwuu +

30v8 vuvuuwuu +

15v8 v2

uwu3−105v9 v3

uwuu

=1v6 wu5−

12v7 vuwu4−

10v7 vuuwu3 +

57v8 v2

uwu3−3v7 vu3wuu

+48v8 vuvuuwuu−

105v9 v3

uwuu,

ks4 =1v·∂u

(1v6 wu5−

12v7 vuwu4−

10v7 vuuwu3 +

57v8 v2

uwu3−3v7 vu3wuu +

48v8 vuvuuwuu

−105v9 v3

uwuu

)=

1v·(

wu6v6−6v5vuwu5

v12 −12vuuwu4v7 + vuwu5v7−7v6v2

uwu4

v14

−10vu3wu3v7 + vuuwu4v7−7v6vuvuuwu3

v14

+572vuvuuwu3v8 + v2

uwu4v8−8v7v3uwu3

v16

−3vu4wuuv7 + vu3wu3v7−7v6vuvu3wuu

v14

−1053v2

uvuuwuuv9 + v3uwu3v9−9v8v4

uwuu

v18

+48v2

uuwuuv8 + vuvu3wuuv8 + vuvuuwu3v8−8v7v2uvuuwuu

v16

)=

1v7 wu6−

6v8 vuwu5−

12v8 vuuwu4−

12v8 vuwu5 +

84v9 v2

uwu4−10v8 vu3wu3−

10v8 vuuwu4

+70v9 vuvuuwu3 +

114v9 vuvuuwu3 +

57v9 v2

uwu4−456v10 v3

uwu3−3v8 vu4wuu−

3v8 vu3wu3

+21v9 vuvu3wuu +

48v9 v2

uuwuu +48v9 vuvu3wuu +

48v9 vuvuuwu3−

384v10 v2

uvuuwuu

−315v10 v2

uvuuwuu−105v10 v3

uwu3 +945v11 v4

uwuu

=1v7 wu6−

18v8 vuwu5−

22v8 vuuwu4 +

141v9 v2

uwu4−13v8 vu3wu3 +

232v9 vuvuuwu3

−561v10 v3

uwu3−3v8 vu4wuu +

69v9 vuvu3wuu +

48v9 v2

uuwuu−699v10 v2

uvuuwuu

+945v11 v4

uwuu.

Then here we have F(u, t) as the following equation:


F(u, t) = ks4(u, t)+ k2kss(u, t)−12

kk2s (u, t)

=1v7 wu6−

18v8 vuwu5−

22v8 vuuwu4 +

141v9 v2


232v9 vuvuuwu3

−561v10 v3

uwu3−3v8 vu4wuu +

69v9 vuvu3wuu +

48v9 v2

uuwuu−699v10 v2

uvuuwuu

+945v11 v4

uwuu +1v6 w2

uu ·(

1v5 wu4−

7v6 vuwu3−

3v6 vuuwuu +

15v7 v2

uwuu

)−1

2· 1

v3 wuu ·(

1v8 w2

u3 +9

v10 v2uw2

uu−6v9 vuwuuwu3

)=

1v7 wu6−

18v8 vuwu5−

22v8 vuuwu4 +

141v9 v2


232v9 vuvuuwu3

−561v10 v3

uwu3−3v8 vu4wuu +

69v9 vuvu3wuu +

48v9 v2

uuwuu−699v10 v2

uvuuwuu

+945v11 v4

uwuu +1

v11 w2uuwu4−

7v12 vuw2

uuwu3−3

v12 vuuw3uu +

15v13 v2

uw3uu

−12· 1

v11 wuuw2u3−

92· 1

v13 v2uw3

uu +3

v12 vuw2uuwu3

=1v7 wu6−

18v8 vuwu5−

22v8 vuuwu4 +

141v9 v2


232v9 vuvuuwu3

−561v10 v3

uwu3−3v8 vu4wuu +

69v9 vuvu3wuu +

48v9 v2

uuwuu−699v10 v2

uvuuwuu

+945v11 v4

uwuu +1

v11 w2uuwu4−

4v12 vuw2

uuwu3−3

v12 vuuw3uu +

212v13 v2

uw3uu

−12· 1

v11 wuuw2u3,

also

wt(u, t) = v ·F(u, t)

= v−6wu6−18v−7vuwu5−22v−7vuuwu4 +141v−8v2uwu4−13v−7vu3wu3

+232v−8vuvuuwu3−561v−9v3uwu3−3v−7vu4wuu +69v−8vuvu3wuu

+48v−8v2uuwuu−699v−9v2

uvuuwuu +945v−10v4uwuu + v−10w2

uuwu4

−4v−11vuw2uuwu3−

12

v−10wuuw2u3 +

212

v−12v2uw3

uu−3v−11vuuw3uu.


Now we calculate the boundary condition for scalar initial-boundary-value graph prob-lem. At the boundary, the Neumann boundary condition and νη1,2 = (1,0) imply

0 = 〈ν ,νη1,2〉(u, t) =−v−1wu,

which can give us wu(±1, t) = 0.


Also at the boundary we can have v(±1, t) = 1, vu = 0, then

ks(±1, t) =1v4 wu3(±1, t)− 3

v5 vuwuu(±1, t) = 0

implies wu3(±1, t) = 0.The third boundary condition in (3.1) shows

ks3(±1, t) =1v6 wu5−

12v7 vuwu4−

10v7 vuuwu3 +

57v8 v2

uwu3

− 3v7 vu3wuu +

48v8 vuvuuwuu−

105v9 v3

uwuu

= 0

which implies wu5(±1, t) = 0.Thus, the boundary conditions in (3.1) can be written as

wu(±1, t) = wu3(±1, t) = wu5(±1, t) = 0.

We transform the problem (3.1) to a scalar initial-boundary-value graph problem asfollows,

∂w∂ t (u, t) = f (u, t) = (vF)(u, t), f or all (u, t) ∈ [−1,1]× [0,σ ]

w(·,0) = w0, f or all u ∈ [−1,1]wu = wu3 = wu5 = 0, f or all (u, t) ∈ ±1× [0,σ ]

(3.3)

here v(u, t) =√

1+(wu)2 in Lemma 5, F(u, t) denotes the curvature flow of γ(u, t), andf (u, t) is shown in (3.2).

In the evolution of the graph function w(u, t), the coefficient of the principal part isv−6 only contains the first derivative of w, the rest of the evolution is purely nonlinear,which consists the first to fifth derivative of w. Therefore, the evolution of the graphfunction is quasilinear. Then we check the parabolicity, as the graph function is a sixthorder equation, it is strictly parabolic if the coefficient of the principle part is definite

positive. Again the coefficient of the principal part is v−6 =(√

1+(wu)2)−6

> 0. Thus,the evolution of the graph function is parabolic and quasilinear.

Next, we prove that problems (3.1) and (3.3) are equivalent. For any solution γ (u, t) to(3.1), there is only one solution w(u, t) to (3.3) and vice versa.

We let w(u, t) = w(φ(u, t), t),

γ(φ(u, t), t) = (φ(u, t), w(φ(u, t), t)) := γ(u, t),

also ν(u, t) = ν (φ(u, t), t), τ(u, t) = τ (φ(u, t), t) and F(u, t) = F (φ(u, t), t),


Definition 3. Define φ : [−1,1]× [0,σ ]→ [−1,1] by the following system of ordinary

differential equations:

ddt φ(u, t) =−(Dγ)−1 ·

(∂

∂ t γ

)T(φ(u, t), t)

φ(u,0) = u,(3.4)

where αT :=α−〈α, ν〉· ν denotes the tangential component of a vector α and ν(φ(u, t), t)=

v−1 ·(−wφ ,1

)and τ(φ(u, t), t) = v−1 ·(1,wφ ) denote normal vector and tangential vector

field respectively, wφ (φ(u, t), t) = ∂w∂φ

(φ(u, t), t).

At least for a short time, φ is a diffeomorphism on [−1,1], this is equivalent to that ∂φ

∂ t

is tangential to the boundaries η1,2, i.e.

u ∈ η1,2 =⇒ φ(u, t) ∈ η1,2, f or all t ∈ [0,σ ].

At the boundary, we have τ = (1,0) which yields

〈 ∂

∂ tγ, τ〉(φ(±1, t), t) =

⟨(0,

∂w∂ t

),(1,0)

⟩= 0,

we also have

〈ν , τ〉(φ(±1, t), t) = 0.

Combining these two relations and using equation αT := α −〈α, ν〉 · ν , we obtain atthe boundary

⟨Dγ · dφ

dt, τ

⟩(φ(±1, t), t)

=

⟨−(

∂

∂ tγ

)T

, τ

⟩(φ(±1, t), t)

=

⟨− ∂

∂ tγ, τ

⟩(φ(±1, t), t)+

⟨∂

∂ tγ, ν

⟩· 〈ν , τ〉(φ(±1, t), t)

= 0.

This means that Dγ · dφ

dt is tangential to η1,2, thus dφ

dt must be tangential to η1,2.

Lemma 6. If w(u, t) is a solution of (3.3). Then there exists a unique solution γ(u, t) :[−1,1]× [0,σ ]→ R2 of (3.1).


Proof. We calculate

ddt

γ(u, t) =ddt

γ (φ(u, t), t)

=∂

∂φγ (φ(u, t), t) · dφ

dt(u, t)+

∂

∂ tγ (φ(u, t), t)

= −(

∂

∂ tγ

)T

(φ(u, t), t)+∂

∂ tγ (φ(u, t), t)

= 〈 ∂

∂ tγ, ν〉 · ν (φ(u, t), t)

=

⟨(0,

∂ w∂ t

), ν

⟩· ν(φ(u, t), t)

=

⟨(0, Fv) ,v−1

(−wφ

1

)⟩·ν(φ(u, t), t)

= F ν(φ(u, t), t)

= Fν(u, t).

Also at the boundary, 〈ν ,τ〉(±1, t) = 〈ν , τ〉(φ(±1, t), t) = 0 and ks(±1, t) = ks3(±1, t) =0.

Thus, there is a unique γ(u, t) satisfies (3.1) when w(u, t) is a solution of (3.3).

Lemma 7. If γ(u, t) satisfies (3.1), then for a short time, there is a unique w(u, t) :[−1,1]× [0,σ ]→ R satisfies (3.3).

Proof. We do the time derivative of γ(u, t),

ddt

γ(u, t) =(

ddt

φ(u, t),ddt

w(φ(u, t), t)),

also we haveφ(u, t) = γ(u, t) · (1,0),

w(φ(u, t), t) = γ(u, t) · (0,1),

then we get the following two equations,

dφ

dt(u, t) =

ddt

γ(u, t) · (1,0) = Fν(u, t) · (1,0) = F ν(φ(u, t), t) · (1,0)

= F · v−1 · (−wφ ,1)

(10

)= −Fv−1wφ (φ(u, t), t) ,


dwdt

(φ(u, t), t) =ddt

γ(u, t) · (0,1) = Fν(u, t) · (0,1) = F ν (φ(u, t), t) · (0,1)

= −F · v−1 · (−wφ ,1)

(01

)= Fv−1 (φ(u, t), t) .

As φ is a diffeomorphism on [−1,1], then

dwdt

(φ(u, t), t) =∂ w∂φ

(φ(u, t), t)dφ

dt(u, t)+

∂ w∂ t

(φ(u, t), t) .

Set y := φ(u, t), for all (y, t) ∈ [−1,1]× [0,σ ], γ(y, t) := (y,w(y, t)). Thus, we have

∂w∂ t

(y, t)∣∣∣∣y=φ(u,t)

=dwdt

(y, t)− ∂w∂y

(y, t)dydt

(u, t)

= Fv−1 (y, t)+wyFv−1wy (y, t)

= Fv−1 ·(1+w2

y)= Fv−1v2

= (Fv)(y, t)|y=φ(u,t) .


Lemma 6 and Lemma 7 show that our original problem (3.1) and the scalar graphproblem (3.3) are equivalent.

3.1.2 Short time existence for the quasilinear graph problem

Here we prove that (3.3) has a unique solution for a short time. Here we refer to [69]. Thegeneral Theorem 4.4 in [69] can be used to prove nonlinear evolutionary problem withboundary conditions is well posed. See this theorem and some notations in Appendix A,Theorem 16.

In order to use this result in [69], we need to check three conditions, we give the fol-lowing lemma:

Lemma 8. The boundary conditions in (3.3) satisfy the compatibility condition that is,

for all (u, t) ∈ η1,2× [0,σ ], we have

∂j

t wu

∣∣∣t=0

= ∂j

t wu3

∣∣∣t=0

= ∂j

t wu5

∣∣∣t=0

= 0, j = 0,1,2, ...,n.

Proof. We have v(±1,0) =√

1+(wu)2 = 1, vu|t=0 = wuwuu√1+(wu)2

= 0, vun|t=0 = 0,n =

0,1,2,3, ....

vt |t=0 =[1+(wu)

2]−1/2 ·wu ·wut

∣∣∣t=0

=[1+(wu)

2]−1/2 ·wu ·wtu

∣∣∣t=0

= 0


vtn|t=0 = 0,n = 0,1,2,3, ....

also∂u∂t = ∂s∂t = ∂t∂s− kF∂s = ∂t∂u.

B(t,w) = (B1(t,w),B2(t,w),B3(t,w)) = (wu(±1, t),wu3(±1, t),wu5(±1, t)) = 0

First, we check compatibility condition for B1(t,w) = wu(±1, t), we do time derivativesof it.

j = 0: ∂j

t B1(t,w)∣∣∣t=0

=B1(0,w) = wu(±1,0) = 0;j = 1:

∂j

t B1(t,w)∣∣∣t=0

= ∂tB1(t,w)|t=0 = wut(±1, t)|t=0 = wtu(±1, t)|t=0

= (−vF)u|t=0 = −vuF− vFu|t=0

= 0

j = 2: ∂j

t B1(t,w)∣∣∣t=0

= ∂ 2t B1(t,w)

∣∣t=0 = ∂uwt2(±1, t)|t=0 = ∂uwt2(±1,0) = 0,

j = 3: ∂j

t B1(t,w)∣∣∣t=0

= ∂ 3t B1(t,w)

∣∣t=0 = ∂uwt3(±1, t)|t=0 = ∂uwt3(±1,0) = 0,

......j = n: ∂

jt B1(t,w)

∣∣∣t=0

= ∂ nt B1(t,w)|t=0 = ∂uwtn(±1, t)|t=0 = ∂uwtn(±1,0) = 0. ly

Secondly, we check compatibility condition for B2(t,w) = wu3(±1, t), we do timederivatives of it.

j = 0: B2(0,w) = wu3(±1,0) = 0;j = 1: ∂

jt B2(t,w)

∣∣∣t=0

= ∂twu3(t,w)|t=0 = ∂ 3u wt(±1, t)

∣∣t=0 = ∂ 3

u wt(±1,0) = 0,

j = 2: ∂j

t B2(t,w)∣∣∣t=0

= ∂ 2t wu3(t,w)

∣∣t=0 = ∂ 3

u wt2(±1, t)∣∣t=0 = ∂ 3

u wt2(±1,0) = 0,

j = 3: ∂j

t B2(t,w)∣∣∣t=0

= ∂ 3t wu3(t,w)

∣∣t=0 = ∂ 3

u wt3(±1, t)∣∣t=0 = ∂ 3

u wt3(±1,0) = 0,......j = n: ∂

jt B2(t,w)

∣∣∣t=0

= ∂ nt wu3(t,w)|t=0 = ∂ 3

u wtn(±1, t)∣∣t=0 = ∂ 3

u wtn(±1,0) = 0.Thirdly, we check compatibility condition for B3(t,w) =ws5(±1, t), we do time deriva-

tives of it.j = 0: B3(0,w) = ws5(±1) = 0;j = 1: ∂tB3(t,w)|t=0 = ∂twu5(t,w)|t=0 = ∂ 5

u wt(±1, t)∣∣t=0 = ∂ 5

u wt(±1,0) = 0,

j = 2: ∂j

t B3(t,w)∣∣∣t=0

= ∂ 2t wu5(t,w)

∣∣t=0 = ∂ 5

u wt2(±1, t)∣∣t=0 = ∂ 5

u wt2(±1,0) = 0,

j = 3: ∂ 3t B3(t,w)

∣∣t=0 = ∂ 3

t wu5(t,w)∣∣t=0 = ∂ 5

u wt3(±1, t)∣∣t=0 = ∂ 5

u wt3(±1,0) = 0,......j = n: ∂

jt B3(t,w)

∣∣∣t=0

= ∂ nt wu5(t,w)|t=0 = ∂ 5

u wtn(±1, t)∣∣t=0 = ∂ 5

u wtn(±1,0) = 0.Thus we prove that our boundary condition

B(t,w) = (B1(t,w),B2(t,w),B3(t,w)) = (wu(±1, t),wu3(±1, t),wu5(±1, t))


satisfies compatibility condition.

Lemma 9. The boundary conditions in (3.3) satisfy the normal boundary conditions.

Proof. We check our boundary condition satisfy the normal boundary condition. ForB1(t,w) = wu(±1, t), we have

B1 = ∂u,Bp1 = ∂uγ(u, t) = vτ(u, t), u ∈ η1,2,

BP1 (u,ν) = vτ ·νη1,2 =−1 6= 0,

here BP1 is the principal part of B1, νη1,2 = νη1,2(u) denotes the inward normal vector to

η1,2 at u. Then we prove that B1 is normal.For B2(t,w) = wu3(±1, t), we have

B2 = ∂3u ,B

p2 = ∂

3u γ(u, t) =−vτ(u, t), u ∈ η1,2,

BP2 (u,ν) =−vτ(u, t) ·νη1,2 = 1 6= 0,

here BP2 is the principal part of B2. Then we prove that B2 is normal.

For B3(t,w) = wu5(±1, t), we have

B3 = ∂5u ,B

p3 = ∂

5u γ(u, t) = vτ(u, t), u ∈ η1,2,

BP3 (u,ν) = vτ(u, t) ·νη1,2 =−1 6= 0,

here BP3 is the principal part of B3. Then we prove that B3 is normal. Thus, our boundary

condition satisfies the normal boundary condition.

Now we do the linearization at any a ∈ w : [−1,1]× [0,σ ]→ R for the factors in thequasilinear problem is (3.3). Here z ∈C∞

0 ([0,T ), [−1,1]).

v(w) =√

1+(wu)2, v(a+ εz) =√

1+(a+ εz)2u

For general constant n, we have

ddε

v−n(a+ εz)∣∣∣∣ε=0

=d

dε

[1+[(a+ εz)u]

2]−n/2

∣∣∣∣ε=0

=d

dε

[1+a2

u +2εauzu + ε2z2

u]−n/2

∣∣∣∣ε=0

= −n2[1+a2

u +2εauzu + ε2z2

u]−n/2−1 ·

(2auzu +2z2

uε)∣∣∣

ε=0

= −n(1+a2u)−(n+2)/2au · zu.


Asddu

v(w) =wuwuu√1+w2

u,

then we havevu(a+ εz) =

(a+ εz)u · (a+ εz)uu√1+[(a+ εz)u]

2,

here we linearize vu at a,

ddε

vu(a+ εz)∣∣∣∣ε=0

=d

dε

(a+ εz)u · (a+ εz)uu√1+[(a+ εz)u]

2

∣∣∣∣∣∣ε=0

=d

dε

auauu + εauzuu + εauuzu + ε2zuzuu√1+a2

u +2εauzu + ε2z2u

∣∣∣∣∣ε=0

=(auzuu +auuzu +2εzuzuu) ·

√1+a2

u +2εauzu + ε2z2u

1+a2u +2εauzu + ε2z2

u

∣∣∣∣∣ε=0

−

(auauu + εauzuu + εauuzu + ε2zuzuu)·(1+a2

u +2εauzu + ε2z2u)− 1

2


u

·(2auzu +2εz2

u)

2

]∣∣∣∣∣ε=0

=(auzuu +auuzu) ·

√1+a2

u−a2uauu

(1+a2

u)− 1

2 zu

1+a2u

= au(1+a2

u)− 1

2 · zuu +

[auu(1+a2

u)− 1

2 −a2uauu

(1+a2

u)− 3

2

]· zu.

Computing the second derivatives of v(u, t) with respect to u,

vuu(a+ εz) = ∂uvu(a+ εz)

= ∂uauauu + εauzuu + εauuzu + ε2zuzuu√


u

=

[(a2

uu +auau3 +2εauuzuu + εauzu3 + εau3wu + ε2z2uu + ε2zuzu3

)1+a2

u +2εauzu + ε2z2u

·√


u

]

−

(auauu + εauzuu + εauuzu + ε2zuzuu)(


u)− 1

2


u(auauu + εauuzu + εauzuu + ε

2zuzuu)],


then do the linearization at any a, we have

ddε

vuu(a+ εz)∣∣∣∣ε=0

=(2auuzuu +auzu3 +au3zu) ·

√1+a2

u ·(1+a2

u)

(1+a2u)

2

+

(a2

uu +auau3)· 1

2

(1+a2

u)− 1

2 (2auzu) ·(1+a2

u)

(1+a2u)

2

−(a2

uu +auau3)·√

1+a2u ·2auzu

(1+a2u)

2 −(auzuu +auuzu) ·

(1+a2

u)− 1

2 auauu(1+a2

u)

(1+a2u)

2

−auauu ·

(−1

2

)·(1+a2

u)− 3

2 ·2auzu ·auauu ·(1+a2

u)

(1+a2u)

2

−auauu ·

(1+a2

u)− 1

2 · (auuzu +auzuu) ·(1+a2

u)

(1+a2u)

2

+auauu ·

(1+a2

u)− 1

2 ·auauu ·2auzu

(1+a2u)

2

= au(1+a2

u)− 1

2 · zu3 +

[2auu

(1+a2

u)− 1

2 −2a2uauu ·

(1+a2

u)− 3

2

]· zuu

+

[au3(1+a2

u)− 1

2 −a2uau3 ·

(1+a2

u)− 3

2 −3aua2uu ·(1+a2

u)− 3

2

+3a3ua2

uu ·(1+a2

u)− 5

2

]· zu.

Moreover,

ddε

vu3(a+ εz)∣∣∣∣ε=0

= g(au)zu4 +g(au,auu)zu3 +g(au,auu,au3)zuu

+g(au,auu,au3,au4)zu,

ddε

vu4(a+ εz)∣∣∣∣ε=0

= g(au)zu5 +g(au,auu)zu4 +g(au,auu,au3)zu3

+g(au,auu,au3,au4)zuu

+g(au,auu,au3,au4,au5)zu,

where g are different functions depending only on derivatives of a(u, t).Then the linearization for f at any a ∈ w : [−1,1]× [0,σ ]→ R is


ddε

f (a+ εz)∣∣∣∣ε=0

= fa(a)z(u, t)

= − ddε

v−6(a+ εz)∣∣∣∣ε=0

au6− v−6(a)zu6 −18d

dεv−7(a+ εz)

∣∣∣∣ε=0

vu(a)au5

−18v−7(a)d

dεvu(a+ εz)

∣∣∣∣ε=0

au5−18v−7(a)vu(a)zu5 + ......

= −v−6(a) · zu6 +g5 (a,au,auu) · zu5 +g4 (a,au,auu,au3) · zu4

+g3 (a,au,auu, ...,au4) · zu3 +g2 (a,au,auu, ...,au5) · zuu

+g1 (a,au,auu, ...,au6) · zu,

here g are different functions only depending on derivatives of a(u, t) and all smooth inspace and time.

As the boundary condition

B(t,z) = (zu(±1, t),zu3(±1, t),zu5(±1, t)) ,

We setB(t,a(u, t)) = (au(±1, t),au3(±1, t),au5(±1, t)) ,

Ba(t,a(u, t))z(u, t) = (zu(±1, t),zu3(±1, t),zu5(±1, t)) .

Therefore, we get the linear problem

∂ z∂ t (u, t) = fa(a)z(u, t)+g(t), f or all (u, t) ∈ [−1,1]× [0,σ ]

Ba(t,a(t))z(t) = (zu,zu3,zu5)(±1, t) = 0, f or all t ∈ [0,σ ]

z(·,0) = 0, f or all (u, t) ∈ [−1,1]×t = 0(3.5)

We can obtain that there is a unique solution z ∈ C∞ ([−1,1], [0,T )) of the linearizedproblem by using the classical results on linear parabolic boundary value problem ([53],Ch IV, 6.4). This result is shown in Appendix A, Theorem 17. Here we state the classicalresult as the following theorem for our problem:

Proposition 9. Let z0 : [−1,1]→ R be a smooth immersion. There exists a maximal

T ∈ (0,∞] such that the linear sixth order problem (3.5) admits a unique solution in the

space C∞ ([−1,1], [0,T )).

For the unique solution of (3.5) z ∈ C∞ ([−1,1], [0,T )), we can see that for suitableconstants ck > 0 and integers b,k ≥ 0, it satisfies ‖z‖k ≤ ck [a,g]b,k , the definition of[a,g]b,k can be found in Appendix A. From Lemma 8, Lemma 9 and Proposition 9, weprove (3.3) satisfies the conditions in Theorem 16 in Appendix A. Then (3.3) has a unique


smooth solution w for some T0 > 0.As we proved that (3.3) is equivalent to (3.1), thus (3.1) has a unique solution for

finite time. Short time existence for sixth order curvature flow of curves with Neumannboundary condition is given here.

Theorem 1. There exists a smooth solution γ : [−1,1]×[0,T )→R2, unique up to parametri-

sation, of the flow (3.1) with speed F, satisfying Neumann boundary conditions and

ks = ks3 = 0 at the boundary and with initial curve γ (·,0) = γ0 compatible with the bound-

ary conditions.

3.2 Higher order flows of plane curves with boundaryconditions

The definition for (2m+4)th order curvature flow of open curves with generalized Neu-mann boundary condition is as follows, for more details in Chapter 5.

Definition 4. Let γ : [−1,1]× [0,T ]→ R2 be a family of smooth immersion. γ is said

to move under (2m+ 4)th order curvature flow F with generalized Neumann boundary

condition, if∂

∂ t γ(s, t) =−Fν , f or all (s, t) ∈ [−1,1]× [0,T )γ(·,0) = γ0, f or all s ∈ [−1,1]< ν ,νη1,2 > (±1, t) = 0, f or all t ∈ [0,T )ks = ...= ks2m−1 = ks2m+1 = 0, f or all (s, t) ∈ ±1× [0,T )

(3.6)

where F = (−1)m+1ks2m+2 +∑mj=1(−1) j+1kksm+ jksm− j − 1

2kk2sm , m ∈ N∪0, ν and νη1,2

are the unit normal fields to γ and η1,2 respectively.

We are also interested in one-parameter families of curves γ (·, t) satisfying the poly-harmonic curvature flow

∂

∂ tγ(s, t) = (−1)m+1 ks2m+2ν , (3.7)

here general m ∈ N∪0. Above ν is the smooth choice of unit normal such that theabove flow is parabolic in the generalised sense.

Lemma 10. While a solution to the flow (3.7) with generalised Neumann boundary con-

ditions exists, we haveddt

L =−∫

γ

k2sm+1ds,

where L denotes the length of the curve.


In view of this lemma and the separation of the supporting parallel lines η1,2, the lengthL of the evolving curve γ (·, t) remains bounded above and below under the flow (3.7).

3.2.1 An equivalent quasilinear scalar graph problem

Firstly, we transform the given problem into an equivalent quasilinear scalar graph prob-lem. See section 3.1.1 for the definitions of Gaussian coordinates and w.

Lemma 11. : Set γ : [−1,1]× [0,σ ]→ R2, γ(u, t) := (u,w(u, t)),

(i) Tangential vector field is τ(u) = v−1(1,wu), here v(u,w(u)) :=√

1+(wu)2,

(ii) Normal vector field is ν(u) = v−1 (−wu,1),(iii) The evolution for w(u, t) is

wt(u, t)

= (−1)m+1v2m+4

∑q=2

g2m+4−qwuq− 12

v−2wuu

(m+2

∑p=2

gm+2−pwup

)2

+m

∑j=1

(−1) j+1v−2wuu

m+ j

∑l=2

gm+ j+2−lwul

m− j

∑n=2

gm− j+2−nwun

here g2m+4−q = g2m+4−q(v,vu, ...,vu2m+4−q) is a function only depending on v,vu, ...,

vu2m+4−q , similarly, gm+ j+2−l = gm+ j+2−l(v,vu, ...,vum+ j+2−l),

gm− j+2−n = gm− j+2−n(v,vu, ...,vum− j+2−n), gm+2−p = gm+2−p(v,vu, ...,vum+2−p).

(iv) The boundary conditions are

wu(±1, t) = wu3(±1, t) = wu5(±1, t) = ..= wu2m+3(±) = 0.

Proof. We give the proof of (iii) directly,

∂tγ(u, t) =

[(−1)m+1ks2m+2 +

m

∑j=1

(−1) j+1kksm+ jksm− j −12

kk2sm

]·ν(u, t).

So

(0,wt(u, t)) ·ν = (−1)m+1ks2m+2 +m

∑j=1


kk2sm ,

1v·wt(u, t) = (−1)m+1ks2m+2 +

m

∑j=1


kk2sm.

For the calculations for k,ks, ...,ks4 , see Lemma 5. By looking for the pattern, we can


have that for m ∈ N,

ksm = g0(v)wum+2 +g1(v,vu)wum+1 +g2(v,vu,vuu)wum +gs(v,vu,vuu,vu3)wum−1 + ...

+gm(v,vu, ...,vum)wuu

=m+2

∑p=2

gm+2−p(v,vu, ...,vum+2−p)wup ,

here g0 = v−m−3, gm+2−p = gm+2−p(v,vu, ...,vum+2−p) is a function only depending onv,vu, ...,vum+2−p .

Furthermore, we can write ks2m+2,ksm+ j ,ksm− j as follows:

ks2m+2 =2m+4

∑q=2

g2m+4−q(v,vu, ...,vu2m+4−q)wuq,

ksm+ j =m+ j

∑l=2

gm+ j+2−l(v,vu, ...,vum+ j+2−l)wul ,

ksm− j =m− j

∑n=2

gm− j+2−n(v,vu, ...,vum− j+2−n)wun.

where g, g, g are functions only depending on derivatives of v.

1v·wt(u, t) = (−1)m+1ks2m+2 +

m

∑j=1


kk2sm

= (−1)m+12m+4

∑q=2

g2m+4−qwuq− 12

v−3wuu

(m+2

∑p=2

gm+2−pwup

)2

+m

∑j=1

(−1) j+1v−3wuu

m+ j

∑l=2

gm+ j+2−lwul

m− j

∑n=2

gm− j+2−nwun

= −F(u, t).

Therefore, we have

wt(u, t) = (−1)m+1v2m+4

∑q=2

g2m+4−qwuq− 12

v−2wuu

(m+2

∑p=2

gm+2−pwup

)2

+m

∑j=1

(−1) j+1v−2wuu

m+ j

∑l=2

gm+ j+2−lwul

m− j

∑n=2

gm− j+2−nwun. (3.8)

We define a scalar initial-boundary-value problem: Denote w : [−1,1]× [0,σ ]→ R,


γ(u, t) := (u,w(u, t)) and γ : [−1,1]× [0,σ ]→ R2.

∂w∂ t (u, t) = (−vF)(u, t), f or all (u, t) ∈ [−1,1]× [0,σ ]

w(·,0) = w0, f or all u ∈ [−1,1]wu = wu3 = wu5 = ..= wu2m+3 = 0, f or all (u, t) ∈ η1,2× [0,σ ]

(3.9)

here v(u,w(u, t)) :=√

1+(wu)2 and F(u, t) denotes the curvature flow of γ(u, t).The proof that equations (3.6) and (3.9) are equivalent is same as in Lemma 6 and

Lemma 7.

3.2.2 Short time existence for the quasilinear graph problem

Here we prove that scalar quasilinear initial-boundary-value problem (3.9) has a uniquesolution for a short time. We refer to [69]. See Theorem 16 in Appendix A. In order touse this theorem, we need to prove the following two lemmas and one proposition.

Lemma 12. The boundary conditions in (3.9) satisfy the compatibility condition, for all

(u, t) ∈ η1,2× [0,σ ], we have

∂j

t wu

∣∣∣t=0

= ∂j

t wu3

∣∣∣t=0

= ...= ∂j

t wu2m+3

∣∣∣t=0

= 0, j = 0,1,2, ...,n.

Proof. In scalar graph equation (3.9), we can see that when t = 0,

v(±1,0) =√

1+(wu)2 = 1, vu|t=0 =wuwuu√1+(wu)2

= 0, vun |t=0 = 0,n = 0,1,2,3, ....

vt |t=0 =[1+(wu)

2]−1/2 ·wu ·wut

∣∣∣t=0

=[1+(wu)

2]−1/2 ·wu ·wtu

∣∣∣t=0

= 0.

The boundary conditions are

B(t,w) = (B1(t,w),B2(t,w),B3(t,w), ...,Bm+2(t,w))

= (wu(±1, t),wu3(±1, t),wu5(±1, t), ...,wu2m+3(±1, t))

= 0.

First, we check compatibility condition for B1(t,w)=wu(±1, t), B2(t,w)=wu3(±1, t),B3(t,w) = wu5(±1, t), similarly see Lemma 6.

Generally, we check compatibility condition for Bm+2(t,w) = wu2m+3(±1, t), we dotime derivatives of it.


j = 0: Bm+2(0,w) = wu2m+3(±1) = 0;j = 1: ∂

jt Bm+2(t,w)

∣∣∣t=0

= ∂twu2m+3(t,w)|t=0 = ∂ 2m+3u wt(±1, t)

∣∣t=0 = 0;

j = 2: ∂j

t Bm+2(t,w)∣∣∣t=0

= ∂ 2t wu2m+3(t,w)

∣∣t=0 = ∂ 2m+3

u wt2(±1, t)∣∣t=0 = 0;

j = 3: ∂j

t Bm+2(t,w)∣∣∣t=0

= ∂ 3t wu2m+3(t,w)

∣∣t=0 = ∂ 2m+3

u wt3(±1, t)∣∣t=0 = 0;

......j = n: ∂

jt Bm+2(t,w)

∣∣∣t=0

= ∂ nt wu2m+3(t,w)|t=0 = ∂ 2m+3

u wtn(±1, t)∣∣t=0 = 0.

Thus, we have proved that our boundary condition

B(t,w) = (B1(t,w),B2(t,w),B3(t,w), ...,Bm+2(t,w))

= (ws(±1, t),ws3(±1, t),ws5(±1, t), ...,ws2m+3(±1, t))

satisfies compatibility condition.

Lemma 13. The boundary conditions in (3.9) satisfy the normal boundary conditions.

Proof. For B1(t,w)=wu(±1, t), we have the operator B1 = ∂u, BP1 = ∂uγ = vτ , BP

1 (u,ν)=

vτ ·νη1,2 6= 0,u ∈ η1,2, here BP1 is the principal part of B1, νη1,2 = νη1,2(u) denotes the in-

ward normal vector to η1,2 at u. Thus, we prove that B1 is normal.For B2(t,w) = wu3(±1, t), we have B2 = ∂ 3

u , Bp2 = ∂ 3

u γ =−vτ , BP2 (u,ν) =−vτ ·νη1,2 6=

0,u ∈ η1,2. Thus, we get that B2 is normal.For B3(t,w) = wu5(±1, t), we have B3 = ∂ 5

u ,BP3 = ∂ 5

u γ = vτ , BP3 (u,ν) = vτ · νη1,2 6=

0,u ∈ η1,2. Thus, we prove that B3 is normal.......For Bm+2(t,w)=wu2m+3(±1, t), we have Bm+2 = ∂ 2m+3

u , Bpm+2 = ∂ 2m+3

u γ =(−1)m+1vτ ,BP

m+2(u,ν) = (−1)m+1vτ ·νη1,2 6= 0,u ∈ η1,2, here BPm+2 is the principal part of Bm+2.

Thus we prove that Bm+2 is normal.Then our boundary condition satisfies the normal boundary condition.

In problem (3.9), we let f (w) := wt(u, t) in (3.8). When m = 0,

f (w) = wt =−v−4wu4 +7v−5vuwu3 +3v−5vuuwuu−15v−6v2uwuu.

Now we do the linearization of f (w) at a when m = 0,

v(w) =√

1+(wu)2, v(a+ εz) =√

1+(a+ εz)2u.

When m = 0, from above calculations, we can get


ddε

f (a+ εz)∣∣∣∣ε=0

= − ddε

v−4(a+ εz)∣∣∣∣ε=0·au4− v−4(a) · zu4

+7d

dεv−5(a+ εz)

∣∣∣∣ε=0· vu(a)au3 +7v−5(a)

ddε

vu(a+ εz)∣∣∣∣ε=0

au3

+3d

dεv−5(a+ εz)

∣∣∣∣ε=0· vuu(a)auu +3v−5(a)

ddε

vuu(a+ εz)∣∣∣∣ε=0

auu

−15d

dεv−6(a+ εz)

∣∣∣∣ε=0· v2

u(a)auu−15v−6(a)d

dεv2

u(a+ εz)∣∣∣∣ε=0

auu

+7v−5(a)vu(a) · zu3 +3v−5(a)vuu(a) · zuu−15v−6(a)v2u(a) · zuu

= −v−4(a) · zu4 +7v−5vu(a) · zu3 +3v−5vuu(a) · zuu−15v−6v2u(a) · zuu

+4au4(1+a2u)−3au · zu−35vu(a)au3(1+a2

u)− 7

2 au · zu

−15vuu(a)auu(1+a2u)− 7

2 au · zu +90v2u(a)auu(1+a2

u)−4au · zu

+7v−5(a)au3au(1+a2

u)− 1

2 · zuu +3v−5(a)auu ·au(1+a2

u)− 1

2 · zu3

+7v−5(a)au3

[auu(1+a2

u)− 1

2 −a2uauu

(1+a2

u)− 3

2

]· zu

+3v−5(a)auu ·[

2auu(1+a2

u)− 1

2 −2a2uauu ·

(1+a2

u)− 3

2

]· zuu

+3v−5(a)auu ·[

au3(1+a2

u)− 1

2 −a2uau3 ·

(1+a2

u)− 3

2 −3aua2uu ·(1+a2

u)− 3

2

+3a3ua2

uu ·(1+a2

u)− 5

2

]· zu

−30v−6(a)vu(a)auuau(1+a2

u)− 1

2 · zuu

−30v−6(a)vu(a)auu

[auu(1+a2

u)− 1

2 −a2uauu

(1+a2

u)− 3

2

]· zu

= −v−4(a) · zu4 +g [a,au,auu] · zu3 +g [a,au,auu,au3 ] · zuu +g [a,au,auu,au3,au4] · zu.

Then it is natural to get the linearization for general m is

ddε

f (a+ εz)∣∣∣∣ε=0

= −v−2m+4(a) · zu2m+4 +g2m+3 [a,au,auu] · zu2m+3 +g2m+2 [a,au,auu,au3 ] · zu2m+2

+......+g3 [a,au,auu, ...,au2m+2] · zu3 +g2 [a,au,auu, ...,au2m+3] · zuu

+g1 [a,au,auu, ...,au2m+4] · zu

= fa(a)z(u, t).


The boundary condition is

B(t,w) = (wu(±1, t),wu3(±1, t),wu5(±1, t), ...,wu2m+3(±1, t)) .

We setB(t,a(t)) = (au(±1, t),au3(±1, t),au5(±1, t), ...,au2m+3(±1, t)) ,

Ba(t,a(t))z(t) = (zu(±1, t),zu3(±1, t),zu5(±1, t), ...,zu2m+3(±1, t)) ,

Thus, our linearized scalar problem is∂ z∂ t (u, t) = fa(a)z(u, t)+g(t), f or all (u, t) ∈ [−1,1]× [0,σ ]

(zu,zu3, ...,zu2m+3)(±1, t) = 0, f or all t ∈ [0,σ ]

z(·,0) = 0.

(3.10)

Referring to the classical results on linear parabolic boundary value problem ([53], ChIV, 6.4) which is shown in Appendix A, Theorem 17, we obtain the solution of the linearproblem (3.10) is unique and exists.

Proposition 10. There is always a unique solution for the linear problem (3.10) in the

space C∞ ([−1,1], [0,T )).

For the unique solution of (3.10) z ∈ C∞ ([−1,1], [0,T )), we can see that for suitableconstants ck > 0 and integers b,k ≥ 0, it satisfies ‖z‖k ≤ ck [a,g]b,k . From Theorem 16 inAppendix A, the problem (3.9) has a unique solution for finite time. As we proved thatgraph scalar problem (3.9) is equivalent to our original problem (3.6) in Lemma 6 andLemma 7, thus (3.6) has a unique solution for finite time.

Short time existence for (2m+ 4)th order curvature flow with Generalized Neumannboundary condition is proved.

Theorem 2. There exists a smooth solution γ : [−1,1]×[0,T )→R2, unique up to parametri-

sation, of the system (3.6) with speed F and smooth initial curve γ (·,0) = γ0 compatible

with the boundary conditions.

Directly, we can get the short time existence for flow (3.7) satisfying Neumann bound-ary condition and ks = ... = ks2m−1 = ks2m+1 = 0 at the boundary and with smooth initialcurve γ (·,0) = γ0 compatible with the boundary conditions, the solution is also unique upto parametrisation and smooth for 0≤ t < ∞.

3.3 The closed length-constrained curve diffusion flow

The framework of short time existence for flow of closed planar curves without boundaryis that we first write the length-constrained curve diffusion flow as a graph over the initial


curve for unknown function of time, we have the scalar quasilinear parabolic problem.Secondly, we prove there is a unique solution for the graph problem and the length-constrained curve diffusion flow is invariant under tangential diffeomorphisms. Thenthese is a unique solution for length-constrained flow with the unknown time function.Thirdly, we use the Schauder fixed point theorem to prove the unique solution exists forour original problem with specific h(t).

We consider one-parameter families of immersed closed curves γ : S1× [0,T )→ R2.The energy functional

L(γ) =∫

γ

|γu|du.

The curve diffusion flow is the steepest descent gradient flow for length in H−1. We de-fine the constrained curve diffusion flow here, see more details about this flow in Chapter6.

Definition 5. Let γ : S1× [0,T )→ R2 be a C4,α -regular immersed curve. The length

constrained curve diffusion flow∂tγ =−(kss−h(t))ν , f or all (s, t) ∈ S1× [0,T )

γ|t=0 = γ0, f or all s ∈ S1 (3.11)

where ν denotes a unit normal vector field on γ .

To preserve length of the evolving curve γ(·, t), we take

h(t) =−∫

γk2

s ds

2πω,

where w denotes the winding number of γ(·, t).

3.3.1 Scalar quasilinear parabolic graph function

Firstly we write γ : S1× [0,T )→ R2 as a graph for unknown function of time h(t) overthe initial curve γ0, using ν(u, t), τ(u, t) to denote the normal and tangential vector fieldsof the curve γ(u, t) respectively, then 〈ν ,τ〉(u, t) = 0, ν(u, t) = rotπ/2τ(u, t), see in Figure3.2. Let f : R× [0,T )→ R, ν0(u) = ν(u,0) write

γ(u, t) = γ0(u)+ f (u, t)ν0(u).

then take the derivative on both sides with respect to t, we have

(∂tγ)(u, t) = (∂t f )(u, t) ·ν0(u),

then(∂t f )(u, t)ν0(u) = (h(t)− kss(u, t))ν(u, t),


γ0(u)

ν0 τ0

Figure 3.2

calculating the inner product with ν0 on both sides,

(∂t f )(u, t) = (h(t)− kss(u, t))〈ν(u, t),ν0(u)〉.

Let V := |γu|, then ∂s =∂u|γu| =

∂uV , τ(u, t) = γs(u, t) =

γu(u,t)V and V0 = |∂uγ0|, τ0 =

∂uγ0|∂uγ0| =

∂uγ0V0

, (V0)u = 〈τ0,(γ0)uu〉. We also have

νu = |γu| ·νs =−kτ · |γu|, ∂uν0 =−k0τ0|∂uγ0|.

As γ(u, t) = γ0(u)+ f (u, t)νo(u), differentiating with respect with u,

γu = ∂uγ0 +(∂u f )ν0 +(∂uν0) f = ∂uγ0 +(∂u f )ν0 +(−k0τ0|∂uγ0|) f

= ∂uγ0 + fuν0− k0τ0V0 f = τ0V0 + fuν0− k0τ0V0 f

= τ0V0(1− k0 f )+ fuν0,

we get V 2 = |γu|2 =V 20 (1− k0 f )2 + f 2

u , then

V = |γu|=√

V 20 (1− k0 f )2 + f 2

u

and

τ(u, t) =γu(u, t)|γu(u, t)|

=τ0V0(1− k0 f )+ fuν0

V.

Differentiating V 2 with respect with u,

2VVu = 2V0(V0)u(1− k0 f )2 +2V 20 (1− k0 f )[−(k0)u f − k0 fu]+2 fu fuu,


Vu =V0(V0)u(1− k0 f )2 +V 2

0 (1− k0 f )[−(k0)u f − k0 fu]+ fu fuu

V.

Here we give several calculations τs = kν ,νs =−kτ , (τ0)u =V0k0ν0,(ν0)u =−V0k0τ0,then calculating the derivatives of γ of u,

γuu = (τ0)uV0(1− k0 f )+ τ0(V0)u(1− k0 f )+ τ0V0[−(k0)u f − k0 fu]+ fuuν0 + fu(ν0)u

= V 20 k0ν0(1− k0 f )+ τ0(V0)u(1− k0 f )+ τ0V0[−(k0)u f − k0 fu]+ fuuν0

+ fuV0(−k0τ0)

= V 20 k0ν0(1− k0 f )+ τ0(V0)u(1− k0 f )+ τ0V0[−(k0)u f −2k0 fu]+ fuuν0,

γuuu = 2V0(V0)u · k0 ·ν0(1− k0 f )+V 20 (k0)u ·ν0(1− k0 f )+V 2

0 k0(ν0)u(1− k0 f )

+V 20 k0ν0[−(k0)u f − k0 fu]+ (τ0)u(V0)u(1− k0 f )+ τ0(V0)uu(1− k0 f )

+τ0(V0)u[−(k0)u f − k0 fu]+ (τ0)uV0[−(k0)u f −2k0 fu]

+τ0(V0)u · [−(k0)u f −2k0 fu]+ τ0V0[−(k0)u f −2k0 fu]u + fuuuν0 + fuu(ν0)u

= 2V0(V0)uk0ν0(1− k0 f )+V 20 (k0)u ·ν0(1− k0 f )−V 3

0 k20τ0(1− k0 f )

+V 20 k0ν0[−(k0)u f − k0 fu]+V0k0(V0)uν0(1− k0 f )+ τ0(V0)uu(1− k0 f )

+τ0(V0)u[−(k0)u f − k0 fu]+V 20 k0ν0[−(k0)u f −2k0 fu]

+τ0(V0)u[−(k0)u f −2k0 fu]+ τ0V0[−(k0)uu f −3(k0)u fu−2k0 fuu]

+ fuuuν0− fuuV0k0τ0,

γuuuu = 2(ν0)u ·V0 · (V0)uk0(1− k0 f )+2ν0[V0(V0)uk0(1− k0 f )]u

+(ν0)u ·V 20 (k0)u(1− k0 f )+ν0[V 2

0 (k0)u(1− k0 f )]u− (τ0)u ·V 30 k2

0(1− k0 f )

−τ0 · [V 30 · k2

0(1− k0 f )]u− (ν0)u ·V 20 k0[(k0)u f + k0 fu]

−ν0 · [V 20 k0((k0)u f + k0 fu)]u +(ν0)uV0k0(V0)u(1− k0 f )

+ν0 · [V0k0(V0)u(1− k0 f )]u +(τ0)u(V0)uu(1− k0 f )+ τ0 · [(V0)uu · (1− k0 f )]u

−(τ0)u · (V0)u[(k0)u f + k0 fu]− τ0 · [(V0)u · ((k0)u f + k0 fu)]u

−(ν0)u ·V 20 k0[(k0)u f +2k0 fu]−ν0 · [V 2

0 k0((k0)u f +2k0 fu)]u

−(τ0)u · (V0)u[(k0)u f +2k0 fu]− τ0 · [(V0)u((k0)u f +2k0 fu)]u

−(τ0)u ·V0[(k0)uu f +3(k0)u fu +2k0 fuu]

−τ0 · [V0((k0)uu f +3(k0)u fu +2k0 fuu)]u +(ν0)u fu3 +ν0 fu4− (τ0)u fuuV0k0

−τ0( fuuV0k0)u.



γuuuu = −τ0 ·2 ·V 20 (V0)uk2

0(1− k0 f )+2ν0[V0(V0)uk0(1− k0 f )]u

−τ0 ·V 30 k0(k0)u(1− k0 f )+ν0 · [V 2

0 · (k0)u · (1− k0 f )]u−ν0 ·V 40 k3

0(1− k0 f )

−τ0 · [V 30 · k2

0 · (1− k0 f )]u + τ0 ·V 30 · k2

0[(k0)u f + k0 fu]

−ν0 · [V 20 k0((k0)u f + k0 fu)]u− τ0 ·V 2

0 k20(V0)u(1− k0 f )

+ν0 · [V0k0(V0)u(1− k0 f )]u +ν0V0k0(V0)uu(1− k0 f )+ τ0 · [(V0)uu(1− k0 f )]u

−ν0V0k0(V0)u((k0)u f + k0 fu)− τ0 · [(V0)u · ((k0)u f + k0 fu)]u

+τ0V 30 k2

0[(k0)u f +2k0 fu]−ν0 · [V 20 k0((k0)u f +2k0 fu)]u

−ν0V0k0(V0)u · [(k0)u f +2k0 fu]− τ0 · [(V0)u · ((k0)u f +2k0 fu)]u

−ν0V 20 k0[(k0)uu f +3(k0)u fu +2k0 fuu]

−τ0 · [V0 · ((k0)uu f +3(k0)u fu +2k0 fuu)]u− τ0V0k0 fu3 +ν0 fu4−ν0V 20 k2

0 fuu

−τ0 · ( fuuV0k0)u.

Now we show the expressions of the curvature k(u, t) and its derivatives, as

γss =1V

∂u (γs) =1V

∂u

(γu(u, t)

V

)=

1V

γuu ·V − γu ·Vu

V 2 =γuuV − γuVu

V 3 ,

first we calculate the curvature,

k = 〈γss,ν〉= 〈γss,rotπ/2(γs)〉

= 〈γuuV − γuVu

V 3 ,rotπ/2

(γu

V

)〉= 〈γuu

V 2 −γu ·Vu

V 3 ,1V· rotπ/2(γu)〉

=1

V 3 〈γuu,rotπ/2(γu)〉,

here are the first and second derivatives of the curvature,

ks =1V·∂uk

=1V· 1V 6 ·

[〈γuuu,rotπ/2(γu)〉 ·V 3 + 〈γuu,∂urotπ/2(γu) ·V 3

−〈γuu,rotπ/2(γu)〉 ·3V 2Vu]

=〈γu3,rotπ/2(γu)〉

V 4 +〈γuu,∂urotπ/2(γu)〉

V 4 −3〈γuu,rotπ/2(γu)〉 ·Vu

V 5 ,


kss =1V·∂uks

=1V· 1V 8

[〈γu4,rotπ/2(γu)〉 ·V 4 + 〈γuuu,∂urotπ/2(γu) ·V 4

−〈γuuu,rotπ/2(γu)〉 ·4 ·V 3 ·Vu]

+1V· 1V 8

[〈γu3,∂urotπ/2(γu)〉 ·V 4 + 〈γuu,∂

2u rotπ/2(γu) ·V 4

−〈γuu,∂urotπ/2(γu)〉 ·4 ·V 3 ·Vu]

− 1V· 3V 10

[〈γu3,rotπ/2(γu)〉 ·Vu ·V 5 + 〈γuu,∂urotπ/2(γu)〉 ·Vu ·V 5

+〈γuu,rotπ/2(γu)〉 ·Vuu ·V 5−〈γuu,rotπ/2(γu) ·Vu ·5 ·V 4 ·Vu

]=〈γu4,rotπ/2(γu)〉

V 5 +2〈γu3,∂urotπ/2(γu)〉

V 5 −7〈γu3,rotπ/2(γu)〉 ·Vu

V 6

+〈γuu,∂

2u rotπ/2(γu)〉

V 5 −7〈γuu,∂urotπ/2(γu) ·Vu

V 6 −3〈γuu,rotπ/2(γu)〉 ·Vuu

V 6

+15〈γuu,rotπ/2(γu) ·V 2

u

V 7 .

Finally, we can get the time derivative of the function f (u, t),

∂t f (u, t)

= (h(t)− kss(u, t))〈ν(u, t),ν0(u)〉

= (h(t)− kss(u, t)) ·1V· 〈rotπ/2(γu),ν0(u)〉

=1V·(

h(t)−[〈γu4,rotπ/2(γu)〉

V 5 +2〈γu3 ,∂urotπ/2(γu)〉


V 6

+〈γuu,∂

2u rotπ/2(γu)〉

V 5 −7〈γuu,∂urotπ/2(γu) ·Vu


V 6


u

V 7

])· 〈rotπ/2(γu),ν0(u)〉

=1V·

[h(t)−

(2〈γu3,∂urotπ/2(γu)〉


V 6 +〈γuu,∂

2u rotπ/2(γu)〉

V 5

−7〈γuu,∂urotπ/2(γu) ·Vu


V 6


u

V 7

)]· 〈rotπ/2(γu),ν0(u)〉

− 1V 6 〈γu4,rotπ/2(γu)〉 · 〈rotπ/2(γu),ν0(u)〉

= Q( f ). (3.12)


In Q( f ), the highest order term about f is in the last term

− 1V 6 〈γu4,rotπ/2(γu)〉 · 〈rotπ/2(γu),ν0(u)〉,

as γu = τ0V0(1− k0 f )+ fuν0, then rotπ/2γu = ν0V0(1− k0 f )− fuτ0, calculating the innerproduct we get

〈rotπ/2 (γu) ,ν0〉=V0(1− k0 f ),

and

〈γu4,rotπ/2(γu)〉

= 〈γu4,ν0V0(1− k0 f )− fuτ0〉

= 2[V0(V0)uk0(1− k0 f )]u ·V0 · (1− k0 f )+ [V 20 (k0)u · (1− k0 f )]u ·V0 · (1− k0 f )

−V 50 k3

0(1− k0 f )2− [V 20 k0((k0)u f + k0 fu)]u ·V0 · (1− k0 f )

+[V0k0(V0)u(1− k0 f )]u ·V0(1− k0 f )+V 20 k0(V0)uu(1− k0 f )2

−V 20 k0(V0)u · [(k0)u f + k0 fu](1− k0 f )− [V 2

0 k0 · ((k0)u f +2k0 fu)]u ·V0 · (1− k0 f )

−V 20 k0(V0)u · [(k0)u f +2k0 fu] · (1− k0 f )+ fu4V0(1− k0 f )

−V 30 k0[(k0)uu f +3(k0)u fu +2k0 fuu](1− k0 f )−V 3

0 k20 fuu(1− k0 f )

+2V 20 (V0)uk2

0(1− k0 f ) fu +V 30 k0(k0)u(1− k0 f ) fu +[V 3

0 k20(1− k0 f )]u fu

−V 30 k2

0[(k0)u f + k0 fu] fu +V 20 k2

0(V0)u(1− k0 f ) fu− [(V0)uu(1− k0 f )]u fu

+[(V0)u · ((k0)u f + k0 fu]u fu−V 30 k2

0[(k0)u f +2k0 fu] fu

+[(V0)u · ((k0)u f +2k0 fu)]u fu +[V0((k0)uu f +3(k0)u fu +2k0 fuu)]u fu

+V0k0 fu3 fu +( fuuV0k0)u fu.

It is easy to see that the highest order term in above is fu4V0(1− k0 f ). Then the highestorder term in Q( f ) in (3.12) is − 1

V 6 · fu4V0(1− k0 f ) ·V0(1− k0 f ) =−V 20 (1−k0 f )2

V 6 · fu4 .Thus, we can write (3.12) as follows

∂t f (u, t) = Q( f )

= −V 2

0 (1− k0 f )2

V 6 · fu4 +g(h, f , fu, fuu, fu3),

here V = |γu|=√

V 20 (1− k0 f )2 + f 2

u and g depends only on h(t), f , fu, fuu, fu3 .Our quasilinear parabolic problem is as follows,(∂t f )(u, t) = Q( f ) =−V 2

0 (1−k0 f )2

V 6 · fu4 +g(h, f , fu, fuu, fu3),(u, t) ∈ S1× [0,T )f (·,0) = 0,

(3.13)


where g depends only on h(t), f , fu, fuu, fu3 . We write g as

g = g3(h, f , fu, fuu, fu3

)fu3 +g2

(h, f , fu, fuu

)fuu +g1

(h, f , fu

)fu +g0(h, f ) f ,

here gl are functions depending on h(t), f , ..., ful , l = 0,1,2,3. Also

V = |γu|=√

V 20 (1− k0 f )2 + f 2

u .

To prove that there is a unique solution for quasilinear parabolic problem (3.13), werefer to [3, Main Theorem 5], see Appendix A, Theorem 18.

In the evolution of the graph function f (u, t), the coefficient of the principal part (seeDefinition 10) is −V 2

0 (1−k0 f )2

V 6 only contains the first derivative of f , the rest of the evolu-tion is purely nonlinear, which consists the first, second, and third derivative of f . There-fore, the evolution of the graph function is quasilinear. Then we check the parabolicity,as the graph function is a fourth order equation, it is strictly parabolic if the coefficientof the principle part is definite negative. Again the coefficient of the principal part is−V 2

0 (1−k0 f )2

V 6 < 0 when f 6= 1k0

, then the fourth order equation is strictly parabolic. Thus,the evolution of the graph function is parabolic and quasilinear.

Now we linearized Q( f ) at f0 = 0, here f0 = f (u,0).

V ( f0 + ε f ) =√

V 20 (1− k0ε f )2 + ε f 2

u ,

ddε

Q( f0 + ε f )∣∣∣∣ε=0

=d

dεQ(ε f )

∣∣∣∣ε=0

= − ddε

(−

V 20 (1− k0ε f )2

V 6(ε f )

)∣∣∣∣ε=0

(ε f )u4 |ε=0−V 2

0 (1− k0ε f )2

V 6(ε f )

∣∣∣∣ε=0· d

dε(ε f )u4

∣∣∣∣ε=0

+3

∑l=0

ddε

gl(ε f )

∣∣∣∣∣ε=0

(ε f )ul |ε=0 +3

∑l=0

gl(ε f )

∣∣∣∣∣ε=0

· ddε

(ε f )ul

∣∣∣∣ε=0

= −V−40 · fu4 +

3

∑l=0

gl(ε f )

∣∣∣∣∣ε=0

· ful .

Then our linearized scalar graph problem at f0 = 0 is(∂t f )(u, t) =−V−4

0 · fu4 +gl · ful +g(t), f or all (u, t) ∈ S1× [0,T )f (·,0) = 0.

As f ∈ C4,1,α [S1× [0,T )], thus the leading coefficient −V−4

0 and gl(ε f )|ε=0, l =

0,1,2,3 and g(t) are continuous at u, t and uniformly bounded. We also can see thatthe leading coefficient satisfies Legendre-Hadamard condition in Definition 11. There-fore, we can refer to Theorem 18 in Appendix A and proof that there is a unique solution


for the quasilinear scalar graph problem (3.13) when h(t) is an unknown function of time.We need to prove this coincides with the problem:

∂tγ =−(kss− h(t))ν , f or all (s, t) ∈ S1× [0,T )γ|t=0 = γ0, f or all s ∈ S1 (3.14)

Here we use the method of applying a tangential diffeomorphism in Definition 3 inChapter 3 at each time step to our flow to ensure that the domain of our graph function isindependent of time.

Lemma 14. Length-constrained curve diffusion flow is invariant under tangential diffeo-

morphisms.

Proof. The immersion γ(u, t) = γ0(u)+ f (u, t)ν0(u) can be written as

γ(u, t) = (φ(u, t), f (φ(u, t), t)) .

We define φ : S1× [0,σ ]→ S1 by the system of ordinary differential equation as (3.4).We have already shown that the graph f satisfies a quasilinear fourth order evolution.

However, in computing the evolution of γ(u, t), we only used that the normal part of thespeed is equal to −kss + h. That is, this formulation of constrained curve diffusion flow is(

∂

∂ tγ

)⊥=(−kss + h

)·ν ,

where (·)⊥ denotes normal projection. This differs from our desired evolution by a tan-gential diffeomorphisms φ satisfying

Dγ ·(

∂φ

∂ t

)=−

(∂γ

∂ t

)>,

where (·)> denotes tangential projection, see (3.4).

Consider now an evolution(

∂

∂ t γ

)⊥=(−kss + h

)·ν . As φ(s, t) : S1× [0,σ ]→ S1 sat-

isfies

Dγ (φ(u, t), t) ·(

∂φ

∂ t(u, t)

)=−

(∂γ

∂ t(φ(u, t), t)

)>.

We set γ(u, t) = γ (φ(u, t), t) then

∂

∂ tγ =

ddt

γ (φ(s, t), t) = Dγ · ∂φ

∂ t+

∂γ

∂ t=−

(∂γ

∂ t

)>+

∂γ

∂ t=

(∂

∂ tγ

)⊥=

(−kss + h

)·ν .



Therefore, for a short time, there is unique solution for the constrained curve diffusionflow (3.14).

3.3.2 Fixed point argument

We still need to prove that there is a fixed point argument for h(t) in our problem. Beforegiving the fixed point argument, we calculate d2

dt2 h(0)≤ c(γ0) first.In Definition 5, we have that

h(t) =−∫

k2s ds

2πω.

We do the first derivative of h(t) with respect to time, here ω doesn’t change under theflow. We use kslt to denote ∂

∂ t ksl , l = 0,1,2, ....

ddt

h(t) =−1

2πω· d

dt

∫γ

k2s ds

=−1

2πω·(∫

γ

k2s

∂

∂ tds+

∫γ

∂

∂ tk2

s ds)

=−1

2πω·(

k2s kFds+

∫γ

2kskstds)

=−1

2πω·[

k2s k(−kss +h(t))ds+2

∫γ

ks(−Fs3−Fsk2−3Fksk)ds]

=−1

2πω·[−∫

γ

kk2s kssds+h(t)

∫γ

k2s kds+2

∫γ

ksks5ds+2∫

γ

ksk2ks3ds

+6∫

γ

k2s kkssds−6

∫γ

k2s kh(t)ds

]=

−12πω

·[

5∫

γ

kssk2s kds−5h(t)

∫γ

k2s kds+2

∫γ

k2s3ds+2

∫γ

ks3ksk2ds]

=−1

2πω·[

5∫

γ

kssk2s kds+2

∫γ

k2s3ds+

52

h(t)∫

γ

kssk2ds−2∫

γ

k2ssk

2ds

−4∫

γ

kssk2s kds

]=

−12πω

·[∫

γ

kssk2s kds+2

∫γ

k2s3ds+

52

h(t)∫

γ

kssk2ds−2∫

γ

k2ssk

2ds]

=−1

2πω·[∫

γ

−13

k4s ds+2

∫γ

k2s3ds+

52

h(t)∫

γ

kssk2ds−2∫

γ

k2ssk

2ds]

=1

2πω·[−2∫

γ

k2s3ds+

13

∫γ

k4s ds− 5

2h(t)

∫γ

kssk2ds+2∫

γ

k2ssk

2ds].

As we need to have the second time derivative of h(0) is bounded, so we only careabout the highest order of d2

dt2 h(t), here we check the first term in above,

ddt

∫k2

s3ds =∫

γ

k2s3

∂

∂ tds+2

∫γ

ks3ks3tds,


where

ks3t = ∂t∂sks2 = ∂sks3t− kFks3 = ∂s(∂t∂s)ks− kFks3

= ∂s(∂skst− kFks2)− kFks3 = ∂2s kst−∂s(kFks2)− kFks3

= ∂2s (−Fs3−Fsk2−3Fksk)−∂s(kFks2)− kFks3

= ks7 +g1 (t,k,ks,kss,ks3,ks4,ks5,ks6) ,

where g1 only depends on k,ks, ...,ks6 and t from h(t). then

ddt

∫γ

k2s3ds =

∫γ

k2s3

∂

∂ tds+2

∫γ

ks3ks3tds

= −2∫

γ

k2s5ds+g2 (t,k,ks,kss,ks3,ks4,ks5,ks6) ,

where g2 only depends on k,ks, ...,ks6 and t from h(t). Thus, the second derivative of h(t)

with respect to time is

d2

dt2 h =1

2πω· d

dt

[−2∫

γ

k2s3ds+

13

∫γ

k4s ds− 5

2h(t)

∫γ

kssk2ds+2∫

γ

k2ssk

2ds]

=1

2πω·[

4∫

γ

k2s5ds+g(t,k,ks,kss,ks3 ,ks4)

],

where g is a function only depending on k,ks,kss,ks3,ks4 and t from h(t).The highest order term in above is easily seen to be

∫k2

s5ds. We can see that d2

dt2 h(0)is bounded if γ0 ∈C7,α(S1). Before we show that at least one of the functions h coincidewith our given constrained function h, we need the following fixed point theorem, theproof of Theorem 3 can refer to the proof of Theorem 2.7 in [79].

Theorem 3. (Schauder fixed point theorem) Let I be a compact, convex subset of a Banach

space B and let J be a continuous map of I into itself. Then J has a fixed point.

Then, by referring to [79, Theorem 2.7], we get the short time existence for the length-constrained curve diffusion flow.

Theorem 4. Let γ0 : S1 → R2 be a C7,α -regular immersed curve. Then there exists a

maximal T ∈ (0,∞] such that the constrained curve diffusion flow γ : S× [0,T )→ R2

∂tγ =−(kss−h(t))ν , f or all (s, t) ∈ S1× [0,T )

γ|t=0 = γ0, f or all s ∈ S1

is uniquely solvable with γ of degree C4,1,α (S1× [0,T )).

For the proof of this theorem, see Appendix A.


3.4 The closed constrained ideal curve flow

The framework of short time existence for closed constrained ideal curve flow is similarto the previous section (the short time existence for the length-constrained curve diffusionflow). We consider one-parameter families of immersed closed curves γ : S1× [0,T )→R2. The energy functional is

E(γ) =∫

γ

k2s ds.

We define the constrained ideal curve flow here, see more details about this flow inChapter 7.

Definition 6. Let γ : S1× [0,T )→R2 be a C6,α -regular immersed curve. The constrained

ideal curve flow∂tγ = (ks4 + k2kss− 1

2kk2s +h(t))ν , f or all (s, t) ∈ S1× [0,T )

γ|t=0 = γ0, f or all s ∈ S1 (3.15)

where ν denotes a unit normal vector field on γ .

To preserve length and area of the evolving curve γ(·, t), we take h(t) as1

2πω

(−∫

γk2

ssds+ 72∫

γk2

s k2ds)

and 52L∫

γkk2

s ds respectively.

3.4.1 Scalar quasilinear parabolic graph function

Firstly we write γ : S1× [0,T )→ R2 as a graph for unknown function of time h(t) overthe initial curve γ0, using ν(u, t), τ(u, t) to denote the normal and tangential vector fieldsof the curve γ(u, t) respectively, then take the derivative on both sides with respect to t,we have

(∂t f )(u, t) = (ks4 + k2kss−12

kk2s + h(t))〈ν(u, t),ν0(u)〉.

Referring to Section 3.3.1, we have the expressions for γu,γuu,γu3 and γu4 . Also wecalculate k,ks and kss. It is easy to see that the highest order term in γu6 is ν0 fu6 .

As rotπ/2γu = ν0V0(1− k0 f )− fuτ0 and 〈rotπ/2 (γu) ,ν0〉 = V0(1− k0 f ). Then, we canget the time derivative of the fuction f (u, t),

∂t f (u, t) = (ks4 + k2kss−12

kk2s + h(t))〈ν(u, t),ν0(u)〉

=1V

(1

V 7 〈γu4,rotπ/2(γu)〉+g(γ,γu, ...γu5)

)· 〈rotπ/2(γu),ν0(u)〉

=V 2

0 (1− k0 f )2

V 8 fu6 +g(h, f , fu, ..., fu5)

= Q( f ). (3.16)


Here V = |γu|=√

V 20 (1− k0 f )2 + f 2

u and g depends only on h(t), f , fu, ..., fu5 .To prove that there is a unique solution for quasilinear parabolic problem, we refer to

[3, Main Theorem 5], see Appendix A, Theorem 18. Our quasilinear parabolic problemsatisfies the conditions in Theorem 18.

(∂t f )(u, t) = Q( f ) = V 2

0 (1−k0 f )2

V 8 fu6 +g(h, f , fu, ..., fu5),(u, t) ∈ S1× [0,T )f (·,0) = 0,

(3.17)

where g depends only on h(t), f , fu, ..., fu5 . We write g as

g = g5(h, f , fu, fuu, fu5

)fu5 + ...+g1

(h, f , fu

)fu +g0(h, f ) f ,

gl are functions depending on h(t), f , ..., ful , l = 0,1,2,3.In the evolution of the graph function f (u, t), the coefficient of the principal part (see

Definition 10) is V 20 (1−k0 f )2

V 8 only contains the first derivative of f , the rest of the evolutionis purely nonlinear, which consists the first, second, and third derivative of f . Therefore,the evolution of the graph function is quasilinear. Then we check the parabolicity, asthe graph function is a fourth order equation, it is strictly parabolic if the coefficient of

the principle part is positive. Again the coefficient of the principal part is V 20 (1−k0 f )2

V 8 > 0when f 6= 1

k0, then the fourth order equation is strictly parabolic. Thus, the evolution of

the graph function is parabolic and quasilinear.Now we linearized Q( f ) at f0 = 0, here f0 = f (u,0). Then our linearized scalar graph

problem at f0 = 0 is(∂t f )(u, t) =V−6

0 · fu6 +gl · ful +g(t), f or all (u, t) ∈ S1× [0,T )f (·,0) = 0.

As f ∈C6,1,α [S1× [0,T )], thus the leading coefficient V−6

0 and gl(ε f )|ε=0, l = 0,1, ...,5

and g(t) are continuous at u, t and uniformly bounded. We also can see that the leadingcoefficient satisfies Legendre-Hadamard condition in Definition 11. Therefore, we canrefer to Theorem 18 and proof that there is a unique solution for the quasilinear scalargraph problem (3.17) when h(t) is an unknown function of time.

We need to prove this coincides with the problem:∂tγ = (ks4 + k2kss− 1

2kk2s + h(t))ν , f or all (s, t) ∈ S1× [0,T )

γ|t=0 = γ0, f or all s ∈ S1 (3.18)

By applying a tangential diffeomorphism in Definition 3 in Chapter 3 at each time stepto our flow, we can have that the domain of our graph function is independent of time.


Lemma 15. Constrained ideal curve flow is invariant under tangential diffeomorphisms.

Therefore, for a short time, there is unique solution to the constrained ideal curve flow(3.18).

3.4.2 Fixed point argument

We still need to prove that there is a fixed point argument for h(t) in our problems. Beforegiving the fixed point argument, we calculate d2

dt2 h(0)≤ c(γ0).

The length-constrained ideal curve flow

In Definition 6, for length-constrained ideal curve flow, we have that

h(t) =1

2πω

(−∫

γ

k2ssds+

72

∫γ

k2s k2ds

),

see more details about this flow in Section 7.1.We do the first derivative of h(t) with respect to time, here ω doesn’t change under the

flow. See Lemma 4 in Chapter 2.

ddt

h(t) =−1

2πω· d

dt

(∫γ

k2ssds− 7

2

∫γ

k2s k2ds

)=

−12πω

·[−2∫

γ

k2s5ds+g(t,k,ks, ...,ks4)

]

where g only depends on t,k,ks, ...,ks4 .As we need to have the second time derivative of h(0) is bounded, so we only care

about the highest order of d2

dt2 h(t). Thus, the second derivative of h(t) with respect to timeis

d2

dt2 h =−1

2πω·[

4∫

γ

k2s8ds+g(t,k,ks, ...,ks7)

],

where g is a function only depending on k,ks, ...,ks7 and t from h(t).The highest order term in above is easily seen to be −2

πω

∫k2

s8ds. We can see that d2

dt2 h(0)is bounded if γ0 ∈C10,α(S1). Before we show that at least one of the functions h coincidewith our given constrained function h, we need the fixed point theorem, Theorem 3.

Then, by referring to [79, Theorem 2.7], we get the short time existence for the length-constrained ideal curve flow.


Theorem 5. Let γ0 : S1 → R2 be a C10,α -regular immersed curve. Then there exists a

maximal T ∈ (0,∞] such that the length-constrained curve diffusion flow γ : S× [0,T )→R2

∂tγ = (ks4 + k2kss− 1

2kk2s +h(t))ν , f or all (s, t) ∈ S1× [0,T )


where h(t) = 12πω

(−∫

γk2

ssds+ 72∫

γk2

s k2ds)

, is uniquely solvable with γ of degree

C6,1,α (S1× [0,T )).

The proof of this theorem is similar to the proof of Theorem 4.

The area-preserving ideal curve flow

In Definition 6, for area-preserving ideal curve flow, we let that

h(t) =5

2L

∫γ

kk2s ds,

see more details about this flow in Section 7.2.We do the first derivative of h(t) with respect to time, here w doesn’t change under the

flow.

ddt

h(t) =ddt

(5

2L·∫

γ

kk2s ds)

=52

[−L−2

(ddt

L)∫

γ

kk2s ds+L−1 d

dt

∫γ

kk2s ds]

=52

[L−2

∫γ

kFds∫

kk2s ds+L−1

∫γ

(Fss +Fk2)k2

s ds

+2L−1∫

γ

kks(Fs3 + k2Fs +3kksF

)ds−L−1

∫γ

kk2s · kFds

]=

52·[−5L−1

∫γ

ks4ks3ks−L−1∫

γ

k2s4kds+g(t,k,ks,kss,ks3)

]

where g only depends on t,k,ks,kss,ks3 .As we need to have the second time derivative of h(0) is bounded, so we only care

about the highest order of d2

dt2 h(t). Thus, the second derivative of h(t) with respect to timeis

d2

dt2 h = 5 ·[

3L−1∫

γ

ks7ks6ksds+L−1∫

γ

k2s7kds+g(t,k,ks, ...,ks6)

],


where g is a function only depending on k,ks, ...,ks6 and t from h(t).The highest order factor in above is easily seen to be ks7 . We can see that d2

dt2 h(0) isbounded if γ0 ∈ C9,α(S1). Before we show that at least one of the functions h coincidewith our given constrained function h, we need the fixed point theorem, Theorem 3.

Then, by referring to [79, Theorem 2.7], we get the short time existence for the area-preserving ideal curve flow.

Theorem 6. Let γ0 : S→ R2 be a C9,α -regular immersed curve. Then there exists a

maximal T ∈ (0,∞] such that the area-preserving ideal curve flow γ : S× [0,T )→ R2

∂tγ = (ks4 + k2kss− 1

2kk2s +h2(t))ν ,

γ|t=0 = γ0

where h(t) = 52L∫

γkk2

s ds, is uniquely solvable with γ of degree C6,1,α (S1× [0,T )).

Chapter 4

A sixth order flow of plane curves withboundary conditions

4.1 Introduction

Let η1,η2 :R→R2 denote two parallel vertical lines in R2, with distance |e| 6= 0 betweenthem. We call η1,η2 supporting lines for the flow. Here e to be any vector such that e

is perpendicular to the parallel lines η1,2, see Figure 4.1. We consider one-parameterfamilies of smooth immersed curves γ : [−1,1]× [0,T )→ R2 meeting two parallel lineswith Neumann boundary condition together with other boundary conditions and

γ(−1, ·) ∈ η1(R),γ(1, ·) ∈ η2(R).

The energy functional is

E(γ) =12

∫γ

k2s ds,

where k2s = (ks)

2.

η1 η2

γ

eντν

Figure 4.1

54

CHAPTER 4. SIXTH ORDER FLOW OF CURVES 55

We are interested in the L2-gradient flow for curves of the energy functional E withNeumann boundary conditions. As our energy involves the first derivative of curvature,the gradient flow will be sixth order and so three pairs boundary conditions will be needed.

This chapter is organised as follows. Firstly, we calculate the normal variation of theenergy and obtain the L2-gradient flow under Neumann and other chosen boundary con-ditions. We show that the length of the curve does not increase under the small energy as-sumption. Moreover, under the small energy condition and Neumann boundary condition,the winding number of curves remains zero. By examining the oscillation of curvatureunder the flow we are able to show that small energy curves with zero winding numberremain embedded. We then show that curvature and all curvature derivatives in L2 arebounded under the flow in section 4.2. We can see that these bounds are independent oftime which means the solutions exist for all time. Furthermore, in section 4.3 we give asmaller energy condition and obtain that the L2-norm of the second derivative of curvaturedecays exponentially under this condition. Then naturally, we show that the curvature andall curvature derivatives decay to zero exponentially. A stability argument implies that thesolution converges to a straight line segment which is unique in section 4.4. By the ex-ponential convergence of the flow speed, we show that the limiting line segment is in abounded region of the initial curve.

Throughout this chapter, we use c to denote constants which can be absolute constantsor depend on some parameters under different circumstances.

The corresponding gradient flow has normal speed given by F , that is

∂tγ = Fν , (4.1)

where ν denotes a unit normal vector field on γ and has to be chosen so that the flow isparabolic in the generalised sense.

Here we give the evolution of ks,

kst = kts +Fksk = (Fss +Fk2)s +Fksk = Fs3 +Fsk2 +3Fksk,

ddt

∫γ

k2s ds = 2

∫γ

kskstds−∫

γ

k2s (kF)ds

= 2∫

γ

ksFs3ds+2∫

γ

ksk2Fsds+5∫

γ

k2s kFds

= −2∫

γ

F(

ks4 + kssk2− 12

k2s k)

ds+ 2ksFss|∂γ− 2kssFs|∂γ

+ 2ks3F |∂γ

+ 2ksk2F∣∣∂γ,


where ∂γ denotes the two end points of γ on η1,η2.Then we state the evolution equation of the energy E,

ddt

12

∫γ

k2s ds = −

∫γ

F(

ks4 + kssk2− 12

k2s k)

ds+ ksFss|∂γ

− kssFs|∂γ+ ks3F |

∂γ+ ksk2F

∣∣∂γ. (4.2)

Thus, the normal variation γ = γ + εFν yields

ddε

E[γ]∣∣∣∣ε=0

=−∫

γ

(ks4 + k2kss−

12

kk2s

)Fds+

[ksFss− kssFs + ks3F + k2ksF

]∂γ.

In view of the integral in the above, we wish to take

F = ks4 + kssk2− 12

k2s k

and we prove that the above boundary term is equal to zero in Lemma 16, i.e.

[ksFss− kssFs + ks3F + k2ksF

]∂γ

= 0. (4.3)

Lemma 16. Under boundary condition 〈ν ,νη1,2 · γ〉(±1, t) = ks(±1, t) = ks3(±1, t) = 0,

(4.3) holds.

Proof. Differentiating the Neumann boundary condition < ν ,νη1,2 > (±1, t) = 0, i.e. <ν(±1, t),e >= 0 in time yields Fs(±1, t) < τ(±1, t),e >= ±|e|Fs(±1, t) = 0. As thedistance between η1 and η2 is |e|, |e| 6= 0 so we must have that

Fs(±1, t) = 0.

In addition to the Neumann boundary condition, we also assume the no curvature fluxcondition ks(±1, t) = 0. and ksss(±1, t) = 0 at the boundary. We obtain

ksFss|∂γ− kssFs|∂γ

+ ks3F |∂γ− ksk2F

∣∣∂γ

= 0.

Under the condition in Lemma 16, (4.2) becomes

ddt

12

∫γ

k2s ds =−

∫γ

F(

ks4 + kssk2− 12

k2s k)

ds =−∫

γ

F2ds (4.4)

where F = ks4 + kssk2− 12k2

s k.


We haveddt

E =−‖∂tγ‖22

which yields the flow F is the steepest descent gradient flow of E in L2.Any multiple of F is parallel to the gradient flow. For convenience we normalise the

coefficient of the highest order term, then we study the corresponding gradient flow

∂tγ = Fν =

(ks4 + k2kss−

12

kk2s

)ν .

Definition 7. Let γ be a smooth curve satisfying F = ks4 + kssk2− 12k2

s k = 0 that is a

stationary solution to the L2-gradient flow of E. We call such curves ideal.

Thus, the free boundary value problem that we wish to consider for the flow is thefollowing:

(∂tγ)(u, t) = (Fν)(u, t), f or all (u, t) ∈ [−1,1]× [0,T )γ(−1, t) ∈ η1(R),γ(1, t) ∈ η2(R), f or all t ∈ [0,T )〈ν ,νη1〉(−1, t) = 〈ν ,νη2〉(1, t) = 0, f or all t ∈ [0,T )ks(±1, t) = ks3(±1, t) = 0. f or all t ∈ [0,T )

(4.5)

The following is the main theorem in this chapter:

Theorem 7. Let γ0 be a smooth embedded regular curve, γ : [−1,1]× [0,T )→ Rn be a

solution to (4.1). If the initial curve γ0 satisfies ω = 0 and

L3(0)‖ks(0)‖22 ≤√

74−810

·π3,

where L(0) is the length of γ0 and k(0) = k(·,0) is the curvature, then the flow exists for

all time T = ∞ and γ(·, t) converges exponentially to a horizontal line segment γ∞ in the

C∞ topology.

In order to prove Theorem 7, we need to have the short time existence proved in Theo-rem 1 in Chapter 3.

Lemma 17. The hypothesis of Theorem 7 implies that ω(t) = ω(0) = 0.

It follows immediately that the average curvature k satisfies

k :=1L

∫γ

kds≡ 0.

Proof. From Lemma 4, we obtain

ω(t) = ω(0) = 0.


η1 η2

γ∞

Figure 4.2

Then we have ∫γ

kds∣∣∣∣t=0

= 2ωπ = 0.


k =1L

∫γ

kds≡ 0.

Here we show that for small energy the length of the evolving curve does not increase.

Lemma 18. Let γ : [−1,1]× [0,T )→ R2 be a solution to (4.1). Under the assumptions

of Theorem 7, then

L(t)≤ L(0),

for all t ∈ [0,T ).

Proof. Using Lemma 3, ddt L = d

dt∫

γds =−

∫γ

kFds.

Now using integration by parts and the boundary conditions, we have

ddt

L = −∫

γ

k2ssds+

72

∫γ

k2s k2ds

≤ −∫

γ

k2ssds+

72‖k‖2

∞

∫γ

k2s ds

≤ −∫

γ

k2ssds+

7L3

π3 ‖kss‖22

∫γ

k2s ds

≤ −(

1− 7L3

π3 ‖ks‖22

)·∫

γ

k2ssds,


where we have also used the PSW inequalities (2.1) and (2.2) in Proposition 1 and Propo-sition 2. We use ‖ksl‖2

2 to denote∫

γk2

sl ds, l = 0,1,2, .... As∫

γkds =

∫γ

ksds = 0, then

‖k‖2∞ ≤ 2L

π‖ks‖2

2 ≤2L2

π3 ‖kss‖22.

The assumption L3(0)‖ks(0)‖22 ≤

√74−810 ·π3 ≤ π3

7 is that the energy is small at t=0. Weknow that for all t is that the energy is decreasing in (4.4) since the flow is the L2 gradientflow. We obtain that L3 ∫

γk2

s ds≤ π3

7 for all time. The claim follows.

The following lemma shows the expression for ∂tksl , where l ∈ N∪0.

Lemma 19. Let γ : [−1,1]× [0,T )→ R2 be a solution to (4.1). The evolution of the `-th

derivative of curvature

∂tks` = ks`+6 + ∑q+r+u=`

(c1ksq+4ksrksu + c2ksq+3ksr+1ksu + c3ksq+2ksr+2ksu

+c4ksq+2ksr+1ksu+1)+ ∑a+b+c+d+e=`

c5ksaksbkscksd kse

for constants c1,c2,c3,c4,c5 ∈ R and a,b,c,d,e,q,r,u≥ 0.

Proof. First under Lemma 3, we calculate the evolution of the curvature

∂tk = ks6 +

(kssk2− 1

2k2

s k)

ss+

(ks4 + kssk2− 1

2k2

s k)

k2

= ks6 +

(ks3k2 +2kssksk− kssksk−

12

k3s

)s+ ks4k2 + kssk4− 1

2k2

s k3

= ks6 + ks4k2 +2ks3ksk+ ks3ksk+ k2ssk+ kssk2

s −32

kssk2s + ks4k2 + kssk4− 1

2k2

s k3

= ks6 +2ks4k2 +3ks3ksk+ k2ssk−

12

kssk2s + kssk4− 1

2k2

s k3 (4.6)

and the derivative of ks with respect to t,

∂tks = ∂t∂sk = ∂s∂tk+ kFks

= ∂t∂sk = ∂s∂tk+ k(

ks4 + kssk2− 12

k2s k)

ks

= ∂s

(ks6 +2ks4k2 +3ks3ksk+ k2

ssk−12

kssk2s + kssk4− 1

2k2

s k3)+ ks4ksk

+kssksk3− 12

k3s k2

= ks7 +2ks5k2 +8ks4ksk+5ks3ks2k+52

ks3k2s + ks3k4 +4kssksk3−2k3

s k2, (4.7)


then we calculate the derivative of kss with respect of t,

∂tkss = ∂t∂sks = ∂s∂tks + k(

ks4 + kssk2− 12

k2s k)

kss

= ∂s(ks7 +2ks5k2 +8ks4ksk+5ks3ks2k+52

ks3k2s + ks3k4 +4kssksk3−2k3

s k2)

+ks4kssk+ k2ssk

3− 12

kssk2s k2

= ks8 +2ks6k2 +4ks5ksk+8ks5ksk+8ks4kssk+8ks4k2s +5ks4kssk+5k2

s3k

+5ks3kssks +52

ks4k2s +5ks3kssks + ks4k4 +4ks3ksk3 +4ks3ksk3 +4k2

ssk3

+12kssk2s k2−6kssk2

s k2−4k4s k+ ks4kssk+ k2

ssk3− 1

2kssk2

s k2

= ks8 +2ks6k2 +12ks5ksk+14ks4kssk+212

ks4k2s +5ks4kssk+10k2

s3k+ ks4k4

+8ks3ksk3 +5k2ssk

3 +112

kssk2s k2−4k4

s k. (4.8)

It is clear to see that when l ≥ 2 we will have

∂tksl−1 = ∂t∂sksl−2 = ∂s∂tksl−2 + k(

ks4 + kssk2−12

k2s k)

ksl−1

= ksl+5 + ∑q+r+u=l−1


+c4ksq+2ksr+1ksu+1)+ ∑a+b+c+d+e=l+1

c5ksaksbkscksd kse .

for constants c1, c2, c3, c4, c5 ∈ R and a,b,c,d,e,q,r,u≥ 0.Thus, we have

∂tksl = ksl+6 + ∑q+r+u=l


+c4ksq+2ksr+1ksu+1)+ ∑a+b+c+d+e=l+2

c5ksaksbkscksd kse (4.9)

for constants c1,c2,c3,c4,c5 ∈ R and a,b,c,d,e,q,r,u≥ 0.This is the conclusion of this Lemma.

Lemma 20. All odd derivatives of the curvature are equal to zero at the boundary.

Proof. Our boundary conditions are 〈ν(±1, t),e〉= ks(±1, t) = ks3(±1, t) = 0.Differentiating the Neumann boundary condition in time implies

Fs(±1, t)〈τ(±1, t),e〉=±|e|Fs(±1, t) = 0


0 = Fs(±1, t) = −ks5− ks3k2−2kssksk+ kssksk+12

k3s

= −ks5− ks3k2− kssksk+12

k3s .

As ks(±1, t) = 0,ks3(±1, t) = 0, we have ks5(±1, t) = 0. Also

Fs3(±1, t) = −ks7−(ks3k2)

ss− (kksks2)ss +12(k3

s)

ss

= −ks7−(ks4k2 +2ks3kks

)s−(k2

s ks2 + kk2s2 + kksks3

)s +

(32

k2s ks2

)s

= −ks7− ks5k2−2ks4ksk−2ks4ksk−2ks3k2s −2ks3kssk−2k2

ssks− ks3k2s

−k2ssks−2ks3kssk− ks3k2

s − ks3kssk− ks4ksk+3k2ssks +

32

ks3k2s

= ks7(±1, t),

we get ks7(±1, t) = 0.Let us give the induction argument, we assume that for all p ∈ 0,1, ...,n,

ks2p−1(±1, t) = 0.

The evolution of the l−th derivative of curvature is given in Lemma 19 by



+c4ksq+2ksr+1ksu+1)+ ∑a+b+c+d+e=l

c5ksaksbkscksd kse

for constants c1,c2,c3,c4,c5 ∈ R and a,b,c,d,e,q,r,u≥ 0.The inductive hypothesis implies that, for l odd and less than or equal to 2n−5, for all

n≥ 3, the derivative ksl vanished at the boundary. Take l = 2n−5. Then we have

ks(2n+1) = ∂tks2n−5− ∑2q1+2r1+2u1=2n−5

(ks(2q1+4) ∗ ks(2r1) ∗ ks(2u1) + ks(2q1+3) ∗ ks(2r1+1) ∗ k(2su

1)

+ks(2q1+2) ∗ ks(2r1+2) ∗ ks(2u1) + ks(2q1+2) ∗ ks(2r1+1) ∗ ks(2u1+1)

)− ∑

2a1+2b1+2c1+2d1+2e1=2n−5ks(2a1) ∗ ks(2b1) ∗ ks(2c1) ∗ ks(2d1) ∗ ks(2e1),

here ∗ denotes a linear combination of terms with absolute coefficient. We assume that


ks2p−1(±1, t) = 0, we remove all terms with an odd number of derivatives of k. We con-clude at the boundary

ks(2n+1) = 0,

as required.

4.2 Controlling the geometry of the flow

Under Lemma 19, we can now show that∫

γk2

sl ds are all bounded, l = 0,1,2, ....For l = 1, from the energy E = 1

2∫

γk2

s ds is decreasing, it is clear that

∫γ

k2s ds≤

∫γ

k2s ds∣∣∣∣t=0

. (4.10)

For l = 0, using the PSW inequality (2.1) in Proposition 1,

∫γ

k2ds≤ L2

π2

∫γ

k2s ds≤ L2

π2

∫γ

k2s ds∣∣∣∣t=0

. (4.11)

Then, under the conditions in Theorem 7, we obtain that∫

γk2ds and

∫γ

k2s ds are bounded.

For l = 2, we give the following lemma.

Lemma 21. Let γ : [−1,1]× [0,T )→ R2 be a solution to (4.1). Under the assumptions

of Theorem 7, then there exists a universal constant c ∈ (0,∞) such that for all t ∈ [0,T )

‖kss‖22 ≤

∥∥kss‖22∣∣t=0 + c.

Proof. Applying (4.8) in Lemma 19, we have

ddt

∫γ

k2ssds = −2

∫γ

k2s5ds+4

∫γ

k2s4k2ds−8

∫γ

ks4ks3kskds+11∫

γ

ks4k2sskds

+5∫

γ

ks4kssk2s ds+10

∫γ

k2s3ksskds+20

∫γ

ks3k2ssksds+2

∫γ

ks4kssk4ds

+16∫

γ

ks3kssksk3ds+9∫

γ

k3ssk

3ds+232

∫γ

k2ssk

2s k2ds

−8∫

γ

kssk4s kds. (4.12)

We estimate above terms separately as the following,

9∫

γ

k3ssk

3ds =−9∫

γ

(k2

ssk3)

s ksds =−18∫

γ

ks3kssksk3ds−27∫

γ

k2ssk

2s k2ds.


As ∫γ

ks3k2ssksds =−

∫γ

kss(k2

ssks)

s ds =−2∫

γ

ks3k2ssksds−

∫γ

k4ssds,

we can see that

20∫

γ

ks3k2ssksds =−20

3

∫γ

k4ssds.

As∫γ

k2s3ksskds =−

∫γ

(ks3kssk)s kssds =−∫

γ

ks4k2sskds−

∫γ

k2s3ksskds−

∫γ

ks3k2ssksds,

after previous equation, we have

10∫

γ

k2s3ksskds =−5

∫γ

ks4k2sskds−5

∫γ

ks3k2ssksds,

−8∫

γ

kssk4s kds =

85

∫γ

k6s ds = 4

∫γ

ks3k3s k2ds+12

∫γ

k2ssk

2s k2ds.

Moreover,

−8∫

γ

ks4ks3kskds = 4∫

γ

k2s3ksskds+4

∫γ

k2s3k2

s ds

= −2∫

γ

ks4k2sskds+

23

∫γ

k4ssds+4

∫γ

k2s3k2

s ds.

Substituting above five calculations into (4.12), we get

ddt

∫γ

k2ssds = −2

∫γ

k2s5ds+4

∫γ

k2s4k2ds−2

∫γ

ks4k2sskds+

23

∫γ

k4ssds+4

∫γ

k2s3k2

s ds

+11∫

γ

ks4k2sskds+5

∫γ

ks4kssk2s ds−5

∫γ

ks4k2sskds−5

∫γ

ks3k2ssksds

−203

∫γ

k4ssds+2

∫γ

ks4kssk4ds+16∫

γ

ks3kssksk3ds−18∫

γ

ks3kssksk3ds

−27∫

γ

k2ssk

2s k2ds+

232

∫γ

k2ssk

2s k2ds+4

∫γ

ks3k3s k2ds+12

∫γ

k2ssk

2s k2ds

= −2∫

γ

k2s5ds+4

∫γ

k2s4k2ds+4

∫γ

ks4k2sskds− 13

3

∫γ

k4ssds+4

∫γ

k2s3k2

s ds

+4∫

γ

ks4k2sskds+2

∫γ

ks4kssk4ds−2∫

γ

ks3kssksk3ds− 72

∫γ

k2ssk

2s k2ds

+4∫

γ

ks3k3s k2ds,


here we wish to absorb −2∫

γks3kssksk3ds into other terms, we calculate

−2∫

γ

ks3kssksk3ds =12

∫γ

ks4kssk4ds+12

∫γ

k2s3k4ds,

then

ddt

∫γ

k2ssds = −2

∫γ

k2s5ds+4

∫γ

k2s4k2ds+4

∫γ

ks4k2sskds− 13

3

∫γ

k4ssds+4

∫γ

k2s3k2

s ds

+5∫

γ

ks4kssk2s ds+

52

∫γ

ks4kssk4ds+12

∫γ

k2s3k4ds− 7

2

∫γ

k2ssk

2s k2ds

+4∫

γ

ks3k3s k2ds. (4.13)

We estimate the seven positive terms 4∫

γk2

s4k2ds,4∫

γks4k2

sskds, 5∫

γks4kssk2

s ds,52∫

γks4kssk4ds, 1

2∫

γk2

s3k4ds, 4∫

γks3k3

s k2ds above by using interpolation inequality inProposition 4, we obtain

4∫

γ

k2s4k2ds+4

∫γ

ks4k2sskds+5

∫γ

ks4kssk2s ds =

∫γ

P84 (k)ds≤ ε

∫γ

k2s5ds+ c‖k‖22

2 ,

and

52

∫γ

ks4kssk4ds =∫

γ

P66 (k)ds≤ ε

∫γ

k2s5ds+ c‖k‖22

2 .

Here we use interpolation inequality,∫γ

P64 (k)ds≤ ε‖ks4‖2

2 + c‖k‖182 ,

together with PSW inequality (2.2) in Proposition 2,∫

γkds = 0,

‖k‖2∞ ≤

2Lπ‖ks‖2

2,

we have as the following two estimates,

12

∫γ

k2s3k4ds ≤ 1

2‖k‖2

∞

∫γ

k2s3k2ds≤ 1

2‖k‖2

∞

∫γ

P64 (k)ds

≤ ‖k‖2∞

(ε‖ks4‖2

2 + c‖k‖182)

≤ 2Lπ‖ks‖2

2

(ε

L2

π2‖ks5‖22 + c‖k‖18

2

)= ε

2L3

π3 ‖ks‖22‖ks5‖2

2 + c2Lπ‖ks‖2

2‖k‖182 ,


and

4∫

γ

ks3k3s k2ds ≤ 4‖k‖2

∞

∫γ

|ks3k3s |ds≤ 4‖k‖2

∞

∫γ

P64 (k)ds

≤ ‖k‖2∞(ε‖ks4‖2

2 + c‖k‖182 )

≤ 2Lπ‖ks‖2

2

(ε

L2

π2‖ks5‖22 + c‖k‖18

2

)= ε

2L3

π3 ‖ks‖22‖ks5‖2

2 + c2Lπ‖ks‖2

2‖k‖182 .

Applying PSW inequalities (2.1) and (2.2) in Proposition 1 and Proposition 2, as∫

γks3ds=∫

γks4ds = 0, we have

‖ks3‖2∞ ≤

2Lπ‖ks4‖2

2 ≤2Lπ

L2

π2‖ks5‖22,

we estimate the fifth term in (4.13),

4∫

γ

k2s3k2

s ds≤ ‖ks3‖2∞

∫γ

k2s ds≤ 8L3

π3 ‖ks5‖22‖ks‖2

2,

then using the Hölder inequality, we estimate the fourth term in (4.13),

−133

∫γ

k4ssds≤−13

3L

(∫γ

k2ssds)2

.

Substituting the above calculations into (4.13), we obtain

ddt

∫γ

k2ssds ≤

[−2+4ε +(8+4ε)

L3

π3‖ks‖22

]‖ks5‖2

2

+c‖k‖222 + c

Lπ‖ks‖2

2‖k‖182 −

133L

(∫γ

k2ss

)2

Assume that

−2+4ε +(8+4ε)L3

π3‖ks‖22 ≤ 0,

then

(8+4ε)L3

π3 ‖ks‖22 ≤ 2+4ε,


we get that under condition

‖ks‖22 ≤

(2+4ε)π3

(8+4ε)L3 ,

the following inequality holds

ddt

∫γ

k2ssds≤−c‖kss‖2

2 + c(‖k‖2

2),

i.e.∫

γk2

ssds is bounded, which is the result of this lemma.

We then show that all curvature derivatives in L2 are bounded under the flow.

Lemma 22. Let γ : [−1,1]× [0,T )→ R2 be a solution to (4.1). Under the assumption of

Theorem 7, we have T = ∞ and there exist absolute constants c such that

‖ksl‖22 ≤

∥∥ksl‖22∣∣t=0 + c.

Proof. Applying (4.9) in Lemma 19, we have

ddt

∫γ

k2sl =

∫γ

2ksl ∂tksl ds+∫

γ

k2sl ∂tds

=∫

γ

2ksl ksl+6ds+2∫

γ∑

q+r+u=l(c1ksl ksq+4ksrksu + c2ksl ksq+3ksr+1ksu

+c3ksl ksq+2ksr+2ksu + c4ksl ksq+2ksr+1ksu+1)ds

+2∫

γ∑

a+b+c+d+e=l+2c5ksl ksaksbkscksd kseds−

∫γ

k2sl ks4kds

−∫

γ

k2sl kssk3ds+

12

∫γ

k2sl k2

s k3ds

=∫

γ

2ksl ksl+6ds+2∫

γ∑

q+r+u=l(c1ksl ksq+4ksrksu + c2ksl ksq+3ksr+1ksu


+2∫

γ∑

a+b+c+d+e=l+2c5ksl ksaksbkscksd kseds.

We have already proved that when l = 0,1,2, the results hold in (4.11), (4.10) andLemma 21. So here we let l ≥ 3.

It is easy to see that when l ≥ 3, in the term

2∫

γ∑

a+b+c+d+e=l+2c5ksl ksaksbkscksd kseds

always has a≥ 1.


By integration by parts, when a≥ 1, we have

ddt

∫γ

k2sl = −

∫γ

2ksl+3ds+2∫

γ∑

q+r+u=l(c1ksl+2ksq+2ksrksu + c1ksl+1ksq+2ksr+1ksu

+c2ksl+1ksq+2ksr+1ksu + c3ksl ksq+2ksr+2ksu + c4ksl ksq+2ksr+1ksu+1)ds

+2∫

γ∑

a+b+c+d+e=l+2(c5ksl+1ksa−1ksbkscksd kse

+c5ksl ksa−1ksb+1kscksd kse)ds.

By using interpolation inequality, we have∫γ

∑q+r+u=l

(c1ksl+2ksq+2ksrksu + c1ksl+1ksq+2ksr+1ksu + c2ksl+1ksq+2ksr+1ksu


=∫

γ

P2l+44 (k)ds≤ ε

∫γ

k2sl+3ds+ c‖k‖4l+14

2 .

Again, we can get

2∫

γ∑

a+b+c+d+e=l+2(c5ksl+1ksa−1ksbkscksd kse + c5ksl ksa−1ksb+1kscksd kse)ds

=∫

γ

P2l6 (k)ds≤ ε

∫γ


2 .

Above estimates give us

ddt

∫γ

k2sl = −

∫γ

[2ksl+3ds+P2l+4

4 (k)+P2l6 (k)+P2l

6 (k)]

ds

≤ −∫

γ

2ksl+3ds+ ε

∫γ


2 + ε

∫γ


2

+ε

∫γ


2

≤(−1+ ε +

L2

π2 ε

)∫γ


2 + c‖k‖2l+102

≤ −c∫

γ

k2sl ds+ c

(‖k‖2

2).

Then‖ksl‖2

2 ≤∥∥ksl‖2

2∣∣t=0 + c.

as required.

Pointwise bounds on all derivatives of curvature now follow from PSW inequality (2.1)in Proposition 1. From PSW inequality (2.2) in Proposition 2 and

∫γ

ksl ds = 0, we have‖ksl‖2

∞ ≤ 2Lπ‖ksl+1‖2

2, then∫

γk2

sl ds bounds in L∞. It follows that the solution of the flowremains smooth up to and including the final time T from which we apply local existence


if T < ∞. This shows that the flow exists for all time, that is, T = ∞.

4.3 Exponential convergence

We show that under a smaller energy assumption, the second derivative of curvature de-cays exponentially in L2.

Lemma 23. Let γ : [−1,1]× [0,∞)→ R2 be a solution to (4.1). Assume ω(0) = 0 and if

at some time t0 the energy satisfies

‖ks‖22 <

√74−810

· π3

L3(0). (4.14)

Then there exists a universal constant δ ∈ (0,∞) such that for all t > t0

‖kss‖22 ≤ e−δ (t−t0) ‖kss‖2

2∣∣t=t0

.

Proof. The condition (4.14) remains true for all t > t0 by following the monotonicity ofthe energy. Applying (4.13) in Lemma 21,

ddt

∫γ

k2ssds = −2

∫γ

k2s5ds+4

∫γ

k2s4k2ds+4

∫γ

ks4k2sskds− 13

3

∫γ

k4ssds

+4∫

γ

k2s3k2

s ds+5∫

γ

ks4kssk2s ds+

52

∫γ

ks4kssk4ds+12

∫γ

k2s3k4ds

−72

∫γ

k2ssk

2s k2ds+4

∫γ

ks3k3s k2ds.

Here we handle term 4∫

γks4k2

sskds by doing integration by parts,

4∫

γ

ks4k2sskds =−4

∫γ

ks3(k2ss)sds =−8

∫γ

k2s3ksskds−4

∫γ

ks3k2ssks,

here we notice that

−4∫

γ

ks3k2ssksds = 4

∫γ

kss(k2ssks)sds = 8

∫γ

ks3k2ssksds−4

∫γ

k4ssds,

−4∫

γ

ks3k2ssksds =

43

∫γ

k4ssds,

then

4∫

γ

ks4k2sskds = −8

∫γ

(k2s3k)sksds−4

∫γ

ks3k2ssks

= 16∫

γ

ks4ks3kskds+8∫

γ

k2s3k2

s ds+43

∫γ

k4ssds.


So, we can get

ddt

∫γ

k2ssds = −2

∫γ

k2s5ds+4

∫γ

k2s4k2ds+4

∫γ

k2s3k2

s ds+4∫

γ

ks4k2sskds

+5∫

γ

ks4kssk2s ds−3

∫γ

k4ssds+2

∫γ

ks4kssk4ds− 72

∫γ

k2ssk

2s k2ds

+4∫

γ

ks3k3s k2ds−2

∫γ

ks3kssksk3ds

= −2∫

γ

k2s5ds+4

∫γ

k2s4k2ds+12

∫γ

k2s3k2

s ds+16∫

γ

ks4ks3kskds

+5∫

γ

ks4kssk2s ds−3

∫γ

k4ssds+

52

∫γ

ks4kssk4ds− 72

∫γ

k2ssk

2s k2ds

+4∫

γ

ks3k3s k2ds+

12

∫γ

k2s3k4ds. (4.15)

Again, by applying integration by parts to four terms in above (4.15), for the fourthterm,

16∫

γ

ks4ks3kskds =−16∫

γ

ks4ks3kskds−16∫

γ

k2s3ksskds−16

∫γ

k2s3k2

s ds,

then

16∫

γ

ks4ks3kskds =−8∫

γ

k2s3ksskds−8

∫γ

k2s3k2

s ds.

Also we have

5∫

γ

ks4kssk2s ds = −5

∫γ

k2s3k2

s ds−10∫

γ

ks3k2ssksds

= −5∫

γ

k2s3k2

s ds+103

∫γ

k4ssds,

52

∫γ

ks4kssk4ds = −52

∫γ

k2s3k4ds−10

∫γ

ks3kssksk3ds

= −52

∫γ

k2s3k4ds+5

∫γ

k3ssk

3ds+15∫

γ

k2ssk

2s k2ds,

and

4∫

γ

ks3k3s k2ds = −12

∫γ

k2ssk

2s k2ds−8

∫γ

kssk4s kds

= −12∫

γ

k2ssk

2s k2ds+

85

∫γ

k6s ds.


Substituting above equations into (4.15), we get

ddt

∫γ

k2ssds = −2

∫γ

k2s5ds+4

∫γ

k2s4k2ds−

∫γ

k2s3k2

s ds−8∫

γ

k2s3ksskds−2

∫γ

k2s3k4ds

+13

∫γ

k4ssds− 1

2

∫γ

k2ssk

2s k2ds+5

∫γ

k3ssk

3ds+85

∫γ

k6s ds. (4.16)

Again, we calculate several terms in above by using integration by parts, for the fourthterm in (4.16),

−8∫

γ

k2s3ksskds = 8

∫γ

k2s3k2

s ds−8∫

γ

k2s4k2ds−8

∫γ

ks5ks3k2ds.

For the sixth term,

13

∫γ

k4ssds =−

∫γ

ksks3k2ssds =

∫γ

k2ssks3ksds+

∫γ

ks4kssk2s ds+

∫γ

k2s3k2

s ds,

then

13

∫γ

k4ssds =−

∫γ

ksks3k2ssds =

12

∫γ

ks4kssk2s ds+

12

∫γ

k2s3k2

s ds.

The last two terms are rewritten as follows,

∫γ

k3ssk

3ds =−3∫

γ

k2ssk

2s k2ds+

12

∫γ

k2s3k4ds+

12

∫γ

ks4kssk4ds,

85

∫γ

k6s ds = 4

∫γ

ks3k3s k2ds+12

∫γ

k2ssk

2s k2ds.

Hence, we obtain

ddt

∫γ

k2ssds = −2

∫γ

k2s5ds−4

∫γ

k2s4k2ds+

152

∫γ

k2s3k2

s ds+12

∫γ

ks4kssk2s ds

−8∫

γ

k2s4k2ds−8

∫γ

ks5ks3k2ds+12

∫γ

k2s3k4ds− 7

2

∫γ

k2ssk

2s k2ds

+4∫

γ

ks3k3s k2ds+

52

∫γ

ks4kssk4ds.

By using PSW inequalities (2.1) in Proposition 1 and (2.2) in Proposition 2, as∫

γksl ds=

0, l = 0,1,2,3,4, we have ‖ksl‖2∞ ≤ 2L

π‖ksl+1‖2

2 and ‖ksl‖22 ≤

L2

π2‖ksl+1‖22.


We can estimate the terms in above equation separately as follows,

152

∫γ

k2s3k2

s ds≤ 152‖ks3‖2

∞‖ks‖22 ≤

15L3

π3 · ‖ks‖22‖ks5‖2

2,

12

∫γ

ks4kssk2s ds≤ 1

2‖ks4‖∞‖kss‖∞‖ks‖2

2 ≤L3

π3‖ks‖22‖ks5‖2

2,

and

−8∫

γ

ks5ks3k2ds ≤ 8‖k‖2∞‖ks3‖2‖ks5‖2 ≤

16L3

π3 ‖ks‖22‖ks5‖2

2,

12

∫γ

k2s3k4ds≤ 1

2‖k‖4

∞‖kss‖2‖ks4‖2 ≤2L6

π6 ‖ks‖42‖ks5‖2

2,

also

4∫

γ

ks3k3s k2ds≤ 4‖ks3‖∞‖ks‖∞‖k‖2

∞‖ks‖22 ≤

8L6

π6 ‖ks‖4‖ks5‖22,

52

∫γ

ks4kssk4ds≤ 52‖k‖4

∞‖kss‖2‖ks4‖2 ≤10L6

π6 ‖ks‖42‖ks5‖2

2.

Finally, by combining above calculations we obtain

ddt

∫γ

k2ssds≤

(−2+

32L3

π3 ‖ks‖22 +

20L6

π6 ‖ks‖42

)‖ks5‖2

2.

If the coefficient of ‖ks5‖22 satisfies

−2+32L3

π3 ‖ks‖22 +

20L6

π6 ‖ks‖42 < 0

i.e.

‖ks‖22 <

√74−810

· π3

L3 ,

Estimating L from above by L(0), if at time t0, we have

‖ks‖22 <

√74−810

· π3

L3(0),


then

ddt

∫γ

k2ssds≤−c‖ks5‖2

2 ≤ c(

L6

π6

)‖kss‖2

2 ≤−δ‖kss‖22,

Therefore, we obtain

‖kss‖22 ≤ ‖kss‖2

2∣∣t=t0

e−δ (t−t0).

This is the end of the proof.

Secondly, under Lemma 23, we obtain exponential decay of all curvature derivativesto zero by a standard induction argument involving integration by parts and the curvaturebounds.

Lemma 24. Let γ : [−1,1]× [0,∞)→ R2 be a solution to (4.1). Under the assumption of

Lemma 23. So ‖ksl‖22 and ‖ksl‖2

∞ exponentially decay for t > t0.

Proof. As we have proved that when t→ ∞,∫

γk2

ssds exponentially decays in Lemma 23,we can obtain

‖k‖22 ≤

L2

π2‖ks‖22 ≤

L4(0)π4 ‖kss‖2

2 ≤L4(0)

π4 e−δ t ‖kss‖22∣∣t=0 ,

then∫

γk2ds exponentially decays. Similarly

‖ks‖22 ≤

L2

π2‖kss‖22 ≤

L2(0)π2 ‖kss‖2

2 ≤L2(0)

π2 e−δ t ‖kss‖22∣∣t=0 ,

then∫

γk2

s ds exponentially decays.

∫γ

k2s3ds =−

∫γ

kssks4ds≤(∫

γ

k2ssds) 1

2(∫

γ

k2s4ds

) 12

as(∫

γk2

ssds) 1

2 ≤(

e−δ t ‖kss‖22

∣∣t=0

) 12 , and

(∫γ

k2s4ds

) 12 is bounded in L2, then

∫γ

k2s3ds

exponentially decays.

∫γ

k2s4ds =−

∫γ

ks3ks5ds≤(∫

γ

k2s3ds

) 12(∫

γ

k2s5ds

) 12

as(∫

γk2

s5ds) 1

2 is bounded in L2, then∫

γk2

s4ds exponentially decays.

By induction argument, we assume that∫

γk2

sl ds exponentially decays,

∫γ

k2sl+1ds =−

∫γ

ksl ksl+2ds≤(∫

γ

k2sl ds) 1

2(∫

γ

k2sl+2ds

) 12


as(∫

γk2

sl+1ds) 1

2 is bounded in L2, then∫

γk2

sl+1ds exponentially decays.

Thus, we have∫

γk2

sl ds exponentially decays in L2 for all t > t0. As ‖ksl‖2∞ ≤ L

π‖ksl+1‖2

2,we have

∫γ

k2sl ds exponentially decays in L∞ for all t > t0.

Note that under the assumption of Theorem 7, we have T = ∞. From above lemma, wecan obtain uniform bounds for all derivatives of the evolving curve γ .

Proposition 11. Suppose γ0 : S1→R2 solves (4.1) and satisfies the conditions of Theorem

7. Then for all l ∈ N0,

‖∂ul γ‖∞ ≤ c(l)+ cl

∑p=0‖∂sl γ0‖∞,

where c(l) is a constant only depending on l, E(0), L(0).

Proof. We claim that for l, p ∈ N0,

∂t∂sl γ = ν

l

∑p=0

(P4+p

1+l−p(k)+P2+p3+l−p(k)

)+ τ

l

∑p=0

(P4+p

1+l−p(k)+P2+p3+l−p(k)

). (4.17)

We prove this by induction. First, F = P41 (k)+P2

3 (k) so the equation holds for l = 0. Forq ∈ N0, we do the differentiation

∂t∂sl+1γ = kF ·∂sl+1γ +∂s(∂t∂sl γ)

= ∂s

[ν

l

∑p=0

(P4+p

1+l−p(k)+P2+p3+l−p(k)

)+ τ

l

∑p=0

(P4+p

1+l−p(k)+P2+p3+l−p(k)

)]+kF ·∂sl τ

= kFν ∑p+q=l

Ppq (k)+ kFτ ∑

p+q=lPp

q (k)+ν

l

∑p=0

(P4+p

2+l−p(k)+P2+p4+l−p(k)

)+τ

l

∑p=0

(P4+p

2+l−p(k)+P2+p4+l−p(k)

)+ν

l

∑p=0

(P5+p

1+l−p(k)+P3+p3+l−p(k)

)+τ

l

∑p=0

(P5+p

1+l−p(k)+P3+p3+l−p(k)

)= ν ∑

p+q=l

(Pp+4

q+2 (k)+Pp+2q+4 (k)

)+ν

l

∑p=0

(P4+p

2+l−p(k)+P2+p4+l−p(k)

)+ν

l+1

∑p=1

(P4+p

2+l−p(k)+P2+p4+l−p(k)

)+ τ ∑

p+q=l

(Pp+4

q+2 (k)+Pp+2q+4 (k)

)+τ

l

∑p=0

(P4+p

2+l−p(k)+P2+p4+l−p(k)

)+ τ

l+1

∑p=1

(P4+p

2+l−p(k)+P2+p4+l−p(k)

)= ν

l+1

∑p=0

(P4+p

2+l−p(k)+P2+p4+l−p(k)

)+ τ

l+1

∑p=0

(P4+p

2+l−p(k)+P2+p4+l−p(k)

)


as required. Integrating (4.17) and using Lemma 24, we find

‖∂sl γ‖∞ ≤ ‖∂sl γ0‖∞ + c∫ t

0e−ct ′dt ′ ≤ ‖∂sl γ0‖∞ + c(l).

As u is the initial space parameter before reparametrisation by arc-length, set v = |∂uγ|.Referring to the proof of Theorem 3.1 in [17], then for any function Φ : S→ R, we have

∂ul Φ = v∂sl Φ+Pl(|∂uΦ|, ...,∂ul−1 |∂uΦ|,Φ, ...∂sl−1Φ)

where Pl is a polynomial. We can get

‖∂ul γ‖∞ ≤ c‖∂uγ‖‖∂sl γ‖∞ +‖Pl‖∞.

Then we obtain‖∂ul γ‖∞ ≤ c‖∂sl γ0‖∞ + c(l).

as required.

The next lemma states that along a subsequence of times, there is convergence to astraight line segment.

Lemma 25. Let γ : [−1,1]× [0,T )→ R2 be a solution to (4.1). Under the assumption of

Theorem 7, there exist a subsequence of time t j such that

γ(·, t j)→ γ∞

uniformly with γ∞ a straight line segment.

Proof. Global existence implies

limt→∞

∫ t

0

∫γ

F2dsdt = E(0)− limt→∞

E(t)

= ‖ks‖22

t=0− limt→∞‖ks‖2

2

t ≤ ‖ks‖22|t=0.

So there exists a subsequence, t j→ ∞ such that∫γ

|F(γ(·, t j))|2ds→ 0, as j→ ∞.

For any ε > 0,∫

γF2ds is eventually smaller than ε . Let T0,N be such that for j > N,

we have t j > T0 and ∫γ

F2(γ(·, t j))ds < ε.


Then ‖ks‖22 decays and becomes eventually arbitrarily small on the full sequence. This

implies there exist a limit curve γ∞ that the flow converges to which satisfies k = 0, thatis, γ∞ is a straight line segment.

4.4 The unique limit

As we have proven that the derivative of curvature decays to zero in L2 along the flow,standard theory implies that the flow converges to a solution of k(γ∞) = 0. The only thingleft to prove is that the limit is unique.

Lemma 26. The image of limiting straight line segment is unique.

Proof. In order to prove that γ∞ is the unique limit, we refer to [1, Theorem A.1], Supposethere exists a sequence s j ⊂ [0,∞), s j→ ∞, such that γ(·,s j)→ γ∞ 6= γ∞, in C∞.

Consider the functionalG[γ] =

∫γ

|γ− γ∞|2ds.

Since γ∞ and γ∞ are smooth, it follows that

lims j→∞

G[ f (·,s j)] 6= 0. (4.18)

We estimate by using Lemma 1,∣∣∣∣ ddt

G∣∣∣∣ =

ddt

∫γ

|γ− γ∞|2 ds

= 2∫

γ

|γ− γ∞| ·ddt|γ− γ∞|ds+

∫γ

|γ− γ∞|2 ·ddt

ds

= 2∫

γ

F · |γ− γ∞|ds+∫

γ

kF · |γ− γ∞|2 ds

then we have∣∣∣∣ ddt

G∣∣∣∣ =

∫γ

F · |γ− γ∞| · (2+ k |γ− γ∞|)ds

≤(∫

γ

F2ds) 1

2

·[∫

γ

|γ− γ∞|2 (2+ k |γ− γ∞|)2 ds] 1

2

≤(∫

γ

F2ds) 1

2

·[∫

γ

|γ− γ∞|2(

4+ k2 |γ− γ∞|2)

ds] 1

2

≤ c(∫

γ

F2ds) 1

2

·(

4∫

γ

|γ− γ∞|2 + k2 |γ− γ∞|4 ds) 1

2

≤ c(∫

γ

F2ds) 1

2

·(∫

γ

|γ|2ds+∫

γ

|γ∞|2ds+∫

γ

|k|2 · |γ|4ds+∫

γ

|k|2 · |γ∞|4ds) 1

2

.


Now we show that∫

γ|γ|2ds,

∫γ|γ∞|2ds,

∫γ|k|2 · |γ|4ds and

∫γ|k|2 · |γ∞|4ds are uniformly

bounded.First, we can see that under the exponential decay of curvature and its derivatives in

C∞, we obtain that

‖γ(·, t)‖2∞ =

(∫∞

t|F |dt

)2

≤ c(∫

∞

te−δ tdt

)2

≤ cδ

e−δ t ≤ c1,

then ∫γ

|γ|2ds≤ ‖γ‖2∞

∫γ

ds≤ c1L(0)≤ c2,∫γ

|k|2 · |γ|4ds≤ ‖γ‖4∞

∫γ

|k|2ds≤ c3.

From above two estimates, it is clear that∫

γ|γ∞|2ds and

∫γ|k|2 · |γ∞|4ds are bounded as

well.Thus, ∣∣∣∣ d

dtG∣∣∣∣≤ c‖F‖2,

By the exponential decay of curvature and its derivatives in L2, we have that

G[ f (·,s j)]≤ c∫

∞

s j

‖F‖2dt ≤ c∫

∞

s j

e−δ tdt = ce−δ s j

it follows thatlim

s j→∞G[ f (·,s j)] = 0,

which is in contradiction with (4.18).This proves that there does not exist a sequence s j, so the flow must converge in L2

to a unique line segment. We can obtain the exponential convergence in C∞ to a uniqueline segment as the curvature and all its derivatives exponentially decay.

This finishes the proof.

Then we have the following lemma,

Lemma 27. The flow exists globally (T = ∞) and converges smoothly to a straight line

segment.

Remark 1. While we don’t know the precise height of the limiting straight horizontal line

segment for the curve. However, ‖ksl‖∞ decays exponentially in Lemma 24 shows that the

solution curve remains within a bounded distance of the initial curve: for any x,


|γ(x, t)− γ(x,0)| ≤∫ t

0

∣∣∣∣∂γ

∂ t(x, t)

∣∣∣∣dt

=∫ t

0

∣∣∣∣ks4 + kssk2− 12

k2s k∣∣∣∣dt

≤ c∫ t

0e−δ tdt =

cδ

(1− e−δ t

)≤ c

δ.

Chapter 5

Higher order flows of plane curves withboundary conditions

Let γ : [−1,1]× [0,T )→ R2 be a smooth immersed regular curve, η1,η2 denote twoparallel vertical lines inR2, with a distance |e| 6= 0 between them. Two end points γ(±1, t)of the curve meet two parallel lines η1,2 orthogonally,

γ(−1, ·) ∈ η1(R),γ(1, ·) ∈ η2(R).

This chapter considers one-parameter families of smooth immersed curves γ : [−1,1]×[0,T )→ R2 moving by either the L2-gradient flow for any m ∈ N∪0

∂γ

∂ t=

[(−1)m+1ks2m+2 +

m

∑j=1


kk2sm

]ν , (5.1)

or polyharmonic curve diffusion flow

∂γ

∂ t= (−1)m+1ks2m+2ν (5.2)

for the energy

12

∫γ

k2smds, (5.3)

with γ(·,0) := γ0 and generalised Neumann boundary conditions. Here ksm denotes them-th derivative of curvature with respect to the arc length parameter s and ν is the unitnormal vector of curve γ .

78

CHAPTER 5. HIGHER ORDER FLOWS OF CURVES 79

5.1 The gradient flow for the energy

In order to get the gradient flow, first we calculate the evolution of∫

γk2

smds,

ddt

∫γ

k2smds = 2

∫γ

ksmksmtds−∫

γ

k2sm(kF1)ds. (5.4)

By using (ii) in Lemma 3, the first term in (5.4) is

2∫

γ

ksmksmtds = 2∫

γ

ksm ·

[(F1)sm+2 +

m

∑j=0

∂s j(kksm− jF1)

]ds

= 2∫

γ

ksm · (F1)sm+2ds+2∫

γ

ksm ·m

∑j=0

∂s j(kksm− jF1)ds

= 2∫

γ

F ·

[(−1)m+2ks2m+2 +

m

∑j=0

(−1) jkksm+ jksm− j

]ds. (5.5)

The first term in (5.5)

2∫

γ

ksm · (F1)sm+2ds

= −2∫

γ

ksm+1(F1)sm+1ds+2 ksm(F1)sm+1|∂γ

= 2∫

γ

ksm+2(F1)smds−2 ksm+1(F1)sm |∂γ

+2 ksm(F1)sm+1|∂γ

= −2∫

γ

ksm+3(F1)sm−1ds+2 ksm+2(F1)sm−1|∂γ−2 ksm+1(F1)sm|

∂γ+2 ksm(F1)sm+1|∂γ

= ......

= 2∫

γ

(−1)m+2ks2m+2F1ds+2m+1

∑j=0

(−1) j ksm+ j(F1)sm+1− j |∂γ. (5.6)

The second term in (5.5)

2∫

γ

ksm ·m

∑j=0

∂j

s (kksm− jF1)ds

= 2∫

γ

kk2smF1ds−2

∫γ

ksm+1

m

∑j=1

∂j−1

s (kksm− jF1)ds+2 ksm

m

∑j=1

∂s j−1(kksm− jF1)

∣∣∣∣∣∂γ

= 2∫

γ

kk2smF1ds−2

∫γ

ksm+1kksm−1F1ds+2∫

γ

ksm+2

m

∑j=2

∂s j−2(kksm− jF1)ds

−2 ksm+1

m

∑j=2


∣∣∣∣∣∂γ

+2 ksm

m

∑j=1


∣∣∣∣∣∂γ

= ......

= 2∫

γ

m

∑j=0

(−1) jkksm+ jksm− jF1ds+2m

∑l=1

(−1)l−1ksm+l−1

m

∑j=l

∂s j−l(kksm− jF1)

∣∣∣∣∣∂γ

. (5.7)


Substituting (5.6) and (5.7) into (5.5) we have

2∫

γ

ksmksmtd = 2∫

γ

F1 ·

[(−1)m+2ks2m+2 +

m

∑j=0

(−1) jkksm+ jksm− j

]ds

+ 2m+1

∑j=0

(−1) jksm+ j(F1)sm+1− j

∣∣∣∣∣∂γ

+2m

∑l=1

(−1)l−1ksm+l−1

m

∑j=l


∣∣∣∣∣∂γ

, (5.8)

here in order to have the L2 gradient flow, we make one choice of boundary conditionssuch that the following equation is satisfied,

m+1

∑j=0

(−1) jksm+ j(F1)sm+1− j

∣∣∣∣∣∂γ

+m

∑l=1

(−1)l−1ksm+l−1

m

∑j=l


∣∣∣∣∣∂γ

= 0. (5.9)

Substituting (5.8) and (5.9) into (5.4),

ddt

12

∫γ

k2smds =

∫γ

F1 ·

[(−1)m+2ks2m+2 +

m

∑j=0

(−1) jkksm+ jksm− j −12

kk2sm

]ds

= −∫

γ

F1 ·

[(−1)m+1ks2m+2 +

m

∑j=0

(−1) j+1kksm+ jksm− j +12

kk2sm

]ds

= −∫

γ

F1 ·

[(−1)m+1ks2m+2 +

m

∑j=1


kk2sm

]ds.

Under the condition that (5.9) holds, for the corresponding L2 gradient flows of theassociated curvature-dependent energy

12

∫γ

k2smds,

we would take the normal flow speed

F1 = (−1)m+1ks2m+2 +m

∑j=1


kk2sm. (5.10)

We can see that the gradient flow is of order 2m+4 from above equation. It should have2m+4 boundary conditions for a unique solution. Our choice of boundary conditions arethe classical Neumann condition and all odd curvature derivatives up to order 2m+1 equalto zero at both boundaries. Note that there are other possible boundary conditions. Here


we see each pair of boundary conditions at ±1 as one condition, then the total number ofconditions is m+2. Therefore, we give the following lemma.

Lemma 28. The boundary condition (5.9) holds under (5.28), the number of the condi-

tions is 2m+4 on both boundaries which is the same as the highest order of the derivatives

of the curvature.

Proof. Differentiating the Neumann boundary condition < ν ,νη1,2 > (±1, t) = 0, i.e. <ν(±1, t),e >= 0 in time yields (F1)s(±1, t)< τ(±1, t),e >=±|e|(F1)s(±1, t) = 0. Now|e| 6= 0 so we must have that

(F1)s(±1, t) = 0.

The evolution of ks(±1, t) = 0 in time gives

∂tks = kts + kF1ks =[(F1)ss +Fk2]

s + kF1ks = (F1)s3 +Fsk2 +3F1ksk,

as ks(±1, t) = (F1)s(±1, t) = 0, we must have that

(F1)s3(±1, t) = 0.

Again, differentiating ks3(±1, t) = 0 in time yields

∂tksss = kssts + kF1ksss

=[(F1)s4 +(k2F1)ss +(kksF1)s + kkssF1

]s + kF1ksss

= (F1)s5 +(k2F1)s3 +(kksF1)ss +(kkssF1)s + kks3F1

= (F1)s5 +6kskssF1 +7k2s (F1)s +4kskssF1 +5kks3F1

+9kkss(F1)s +7kks(F1)ss + k2(F1)s3 + k2s F1,

as ks(±1, t) = ks3(±1, t) = (F1)s(±1, t) = (F1)s3(±1, t) = 0, we obtain that

(F1)s5(±1, t) = 0.

Generally, for each odd derivative of k equal to zero on the boundary, we get that thenext odd derivative of F1 is also equal to zero on the boundary by the similar argument.We assume that all odd derivative of k up to order 2m+1 are equal to zero on the bound-ary, then keep differentiating ks5(±1, t) = ks7(±1, t) = ...= ks2m+1(±1, t) = 0 in time, weobtain

(F1)s7(±1, t) = (F1)s9(±1, t) = ...= (F1)s2m+3 = 0.


Thus

(F1)s(±1, t) = (F1)s3(±1, t) = (F1)s5(±1, t) = ...= (F1)s2m+3 = 0. (5.11)

When m+ j is odd, from (5.28), we have ksm+ j(±1, t) = 0; when m+ j is even, from(5.11), we have (F1)sm+1− j(±1, t) = 0. Then for any m, j ∈N∪0, the first term in (5.9),

m+1

∑j=0

(−1) jksm+ j(F1)sm+1− j

∣∣∣∣∣∂γ

= 0.

When m+ l is even, from (5.28), we have ksm+l−1(±1, t) = 0; when m+ l is odd, thenumber of derivatives in ∂s j−l(kksm− jF1) is m− j+( j− l) = m− l which is odd, then from(5.28) and (5.11),

m

∑l=1

m

∑j=l


∣∣∣∣∣∂γ

= 0,

then for any m, j ∈ N∪0, the second term in (5.9),

m

∑l=1

(−1)l−1ksm+l−1

m

∑j=l


∣∣∣∣∣∂γ

= 0.

Thus, we prove that under (5.28), (5.9) holds. We choose (5.28) to be the ‘generalisedNeumann boundary conditions’ for our curves.

Here we check our boundary condition satisfies special cases:When m = 0, the boundary condition (5.28) is < ν ,νη1,2 > (±1, t) = ks(±1, t) = 0 in

[85];When m = 1, we need < ν ,νη1,2 > (±1, t) = ks(±1, t) = ks3(±1, t) = 0 in Chapter

3.

The definition for 2m+4th order curvature flow with natural boundary condition is asfollows,

Definition 8. Let γ : [−1,1]× [0,T )→ R2 be a family of smooth immersion, which meet-

ing two parallel lines η1,2 with m+2 generalised Neumann boundary conditions (5.28).γ is said to move under (2m+4)th order curvature flow (5.1), if

∂

∂ t γ(s, t) =−F1ν , f or all (s, t) ∈ [−1,1]× [0,T )γ(·,0) = γ0,

< ν ,νη1,2 >= ks = ...= ks2m−1 = ks2m+1 = 0, f or all (s, t) ∈ η1,2× [0,T )

where F1 = (−1)m+1ks2m+2 +∑mj=1(−1) j+1kksm+ jksm− j− 1

2kk2sm denotes the 2m+4th order


curvature vector of the immersions, m ∈N∪0, ν and νη1,2 are the unit normal fields to

γ and η1,2 respectively.

The main result in this section is:

Theorem 8. Let γ : [−1,1]× [0,T )→ R2 be a family of smooth embedded or immersed

curves as in Definition 8. If the initial curve γ0 satisfies ω = 0 and has sufficiently small

energy, that is

L2m+1(0)∫

γ

k2sm(0)ds≤ ε, (5.12)

where k(0) = k(·,0), for some positive ε depending only on m, then there exists a smooth

solution γ : [−1,1]× [0,∞)→R2 to the L2-gradient flow for the energy (5.3) with γ(·,0) =γ0. The solution γ converges to a straight line segment exponentially and is unique. The

distance between the limiting curve and γ0 is finite.

The ε in above theorem can be calculated from to (5.33).



k :=1L

∫γ

kds≡ 0.

For the proof, see Lemma 17.

Lemma 30. With the natural Neumann boundary condition (5.28), a solution to the flow

(5.1) satisfies ks2l+1 = 0 and (F1)s2l+1 = 0 on the boundary for all l ∈ N∪0.

Proof. Calculate the first derivative of the flow equation (5.10) with respect to the arc-length s,

(F1)s =

[(−1)m+1ks2m+2 +

m

∑j=1


kk2sm

]s

= (−1)m+1ks2m+3 +m

∑j=1

(−1) j+1ksksm+ jksm− j +m

∑j=1

(−1) j+1kksm+ j+1ksm− j

+m

∑j=1

(−1) j+1kksm+ jksm− j+1−12

ksk2sm− kksmksm+1 .

From (5.28), we must haveks2m+3(±1, t) = 0,


(F1)s3 = (−1)m+1ks2m+5 +m

∑j=1

(−1) j+1∂

2s (ksksm+ jksm− j)+

m

∑j=1

(−1) j+1∂

2s (kksm+ j+1ksm− j)

+m

∑j=1

(−1) j+1∂

2s (kksm+ jksm− j+1)−

12

ks3k2sm−3kssksmksm+1−3ksk2

sm+1

−3kksm+1ksm+2− ksksmksm+2− kksmksm+3.

Again from (5.28), we obtain

ks2m+5(±1, t) = 0.

Let us give the induction argument, assuming that for n = 2,3, ...

ks2n+2m−1(±1, t) = 0,

∂tksl = (F1)sl+2 +l

∑h=0

∂hs (kksl−hF1)

= (−1)m+1ks2m+l+4 +m

∑j=1

(−1) j+1∂

l+2s (kksm+ jksm− j)−

12

∂l+2s (kk2

sm)

+l

∑h=0

∂hs

[kksl−h

((−1)m+1ks2m+2 +

m

∑j=1


kk2sm

)]

= (−1)m+1ks2m+l+4 +m

∑j=0

(−1) j+1∑

q+r+u=l+2cqruksqksm+ j+rksm− j+u

+(−1)m+1l

∑h=0

∑q+r+u=h

¯cqruksqksl−h+rks2m+2+u

+l

∑h=0

m

∑j=0

(−1) j+1∑

q+r+u+v+w=hcqruvwksqksrksl−h+uksm+ j+vksm− j+w ,

for constants cqru, ¯cqru,cqruvw ∈ R with q,r,u,v,w≥ 0.The inductive hypothesis implies that for l odd and less than or equal to 2n+ 2m− 1,

the derivatives ksl vanishes on the boundary. Here we take l = 2n− 3, then under thehypothesis ks2n+2m−1(±1, t) = 0, we removed all terms with an odd number of derivativesof k, we conclude

k2n+2m+1 = 0,n = 2,3, ...

Above result together with boundary conditions (5.28) yields ks2l+1(±1, t) = 0, p =

0,1,2, ....We have

(F1)s(±1, t) = (F1)s3(±1, t) = (F1)s5(±1, t) = ...= (F1)s2m+3 = 0


and later prove that all odd derivatives of the curvature are equal to zero at the boundary.

(F1)sn = (−1)m+1ks2m+2+n +m

∑j=1

(−1) j+1 (kksm+ jksm− j)sn−12(kk2

sm)

sn

= (−1)m+1ks2m+2+n +m

∑j=1

(−1) j+1∑

l1+l2+l3=nksl1 ksm+ j+l2 ksm− j+l3

−12 ∑

l1+l2+l3=nksl1 ksm+l2 ksm+l3 , (5.13)

here n = 2l +1, l ∈ N∪0.If m is even, j is even, then m+ j,m− j are even. l1, l2, l3 have at least one is odd. So we

know 2m+2+n is odd, l1,m+ j+l2,m− j+l3 has at least one is odd, also l1,m+l2,m+l3has at least one is odd. Thus, all three terms in (5.13) are zero at the boundary: ks2m+2+n =

0, ∑mj=1(−1) j+1

∑l1+l2+l3=n ksl1 ksm+ j+l2 ksm− j+l3 = 0 and 12 ∑l1+l2+l3=n ksl1 ksm+l2 ksm+l3 = 0.

If m is even, j is odd, then m + j,m− j are odd. l1, l2, l3 have one or three areodd. So l1,m+ j + l2,m− j + l3 has at least one is odd, also l1,m+ l2,m+ l3 has atleast one is odd. Thus, all three terms in (5.13) are zero at the boundary: ks2m+2+n = 0,

∑mj=1(−1) j+1

∑l1+l2+l3=n ksl1 ksm+ j+l2 ksm− j+l3 = 0 and 12 ∑l1+l2+l3=n ksl1 ksm+l2 ksm+l3 = 0

If m and j are odd, the result is the same as when m, j are even. If m is odd, j is even,the result is the same as when m is even, j is odd.

Thus (F1)sn = 0 holds for n = 2l +1, any l,m, j ∈ N∪0 and j ≤ m.

Next lemma for the flow (5.1) shows that the length of the evolving curve does notincrease under the smallness assumption of initial energy (5.12).

Lemma 31. Suppose γ0 satisfies the conditions of Theorem 8. Then, under the flow (5.1)

with normal speed (5.10), the length of γ does not increase.

Proof. We use Lemma 3 (i), by using integration by parts and the boundary conditions,we get

ddt

L = −∫

γ

kF1ds

= −∫

γ

k

[(−1)m+1ks2m+2 +

m

∑j=1


kk2sm

]ds

= −∫

γ

[(−1)2m+2k2

sm+1 +m

∑j=1

(−1) j+1k2ksm+ jksm− j −12

k2k2sm

]ds

= −∫

γ

k2sm+1ds+

m

∑j=1

(−1) j∫

γ

k2ksm+ jksm− jds+12

∫γ

k2k2smds. (5.14)

Estimating the second and third terms in (5.14), here P2m4 (k) is defined in Chapter 2,


m

∑j=1

(−1) j∫

γ


∫γ

k2k2smds

=m

∑j=1

(−1)2 j∫

γ

ksm(k2ksm− j

)s j ds+

12

∫γ

k2k2smds

=∫

γ

P2m4 (k)ds≤

∫γ

∣∣P2m4 (k)

∣∣ds,

In the above, integration by parts gives us that highest order derivative of k in all termsis ksm , we can write these terms as the form

∫γ

P2m4 (k)ds. Thus, we estimate using inter-

polation inequality in Proposition 4,

∫γ

∣∣P2m4 (k)

∣∣ds ≤ cL1−2m−4‖k‖2m+3m+1

2 · ‖k‖2m+1m+1

m+1,2

≤ cL1−2m−4(‖k‖

2m+3m+1

2 ‖ksm+1‖2m+1m+1

2 +‖k‖42

)≤ c‖k‖

2m+3m+1

2 ‖ksm+1‖2m+1m+1

2 + cL1−2m−4/2‖k‖42

= c‖k‖22 · ‖k‖

1m+12 · ‖ksm+1‖

2m+1m+1

2 + cL−2m−1‖k‖22‖k‖2

2

≤ c‖k‖22 ·(

L2

π2

) m+12(m+1)

· ‖ksm+1‖1

m+12 · ‖ksm+1‖

2m+1m+1

2

+cL−2m−1‖k‖22 ·(

L2

π2

)m+1

· ‖ksm+1‖22

= cL‖k‖22 · ‖ksm+1‖2

2

≤ cL ·(

L2

π2

)m

‖ksm‖22 · ‖ksm+1‖2

2

= c1(m)L2m+1‖ksm‖22 · ‖ksm+1‖2

2. (5.15)

Substituting (5.15) into (5.14), we have the length of the curve does not increase.

ddt

L = −∫

γ

k2sm+1ds+

m

∑j=1

(−1) j∫

γ


∫γ

k2k2smds

≤ −[

1− c1(m)L2m+1∫

γ

k2smds

]·∫

γ

k2sm+1ds

≤ 0, (5.16)

where c1(m)L2m+1‖ksm‖22 < 1 which is the small energy condition.

Since∫

γk2

smds does not increase under the flow, then if the term L2m+1 ∫γ

k2smds is suffi-

ciently small for the initial curve, then L is nonincreasing under the flow (5.1).


Here we show that L2 norm of all curvature derivatives are bounded.

Proposition 12. Suppose γ0 satisfies the conditions of Theorem 8. Then, under the flow

(5.1), we have

∫γ

k2sl ds≤ c(l,m),

for all l ∈ N∪0.

Proof. As when l ≤ m,∫

γk2

sl ds is obviously bounded vis PSW inequality (2.1) in Propo-

sition 1 and Lemma 31. As∫

γksl ds = 0, we have ‖ksl‖2

2 ≤L2

π2‖ksl+1‖22.

For any l, we have via Lemma 3,

ddt

∫γ

k2sl ds

= 2∫

γ

ksl ksltds−∫

γ

k2sl · kF1ds

= −2∫

γ

F ·

[(−1)l+1ks2l+2 +

l

∑d=1

(−1)d+1kksl+d ksl−d −12

kk2sl

]ds

= −2∫

γ

[(−1)m+1ks2m+2 +

m

∑j=1


kk2sm

]·[

(−1)l+1ks2l+2 +l

∑d=1

(−1)d+1kksl+d ksl−d −12

kk2sl

]ds.


ddt

∫γ

k2sl ds

= (−1)m+l+12∫

γ

ks2m+2ks2l+2ds+2m

∑j=1

(−1) j+l+2∫

γ

kksm+ jksm− jks2l+2ds

+2(−1)l+1∫

γ

kk2smks2l+2ds+2

l

∑d=1

(−1)d+m+1∫

γ

kksl+d ksl−d ks2m+2ds

+2m

∑j=1

l

∑d=1

(−1)d+ j+1∫

γ

k2ksl+d ksl−d ksm+ jksm− jds

+l

∑d=1

(−1)d+1∫

γ

k2ksl+d ksl−d k2smds+(−1)m+1

∫γ

kks2m+2k2sl ds

+m

∑j=1

(−1) j+1k2ksm+ jksm− jk2sl −

12

∫γ

k2k2smk2

sl ds. (5.17)


For each l > m, we examine each of the nine terms on the right-hand side of (5.17) inturn.

(−1)m+l+12∫

γ

ks2m+2ks2l+2ds = (−1)2l+12∫

γ

k2sm+l+2ds,

here as 2m+2 and 2l +2 are both even, so these boundary terms all have odd derivativesof curvature, i.e. they are all zero.

2m

∑j=1

(−1) j+l+2∫

γ

kksm+ jksm− jks2l+2ds =m

∑j=1

(−1)2l+ j−m+3∫

γ

(kksm+ jksm− j)sl−m+1ksl+m+1ds,

here 2l +2 is even, the order in kksm+ jksm− j is 2m which is even as well, so we can makesure that the boundary terms all have odd derivatives of the curvature, they are all zero.

2(−1)l+1∫

γ

kk2smks2l+2ds = 2(−1)2l−m+2

∫γ

(kk2sm)sl−m+1ksl+m+1ds,

here 2l + 2 is even, the order in kk2sm is 2m which is even as well, so we can make sure

that the boundary terms all have odd derivatives of the curvature, they are all zero.

2l

∑d=1

(−1)d+m+1∫

γ


= 2m

∑d=1

(−1)d+m+1∫

γ

kksl+d ksl−d ks2m+2ds+2l

∑d=m+1

(−1)d+m+1∫

γ


= 2m

∑d=1

(−1)2m+2∫

γ

ksm+d+1(kksl+d ksl−d)sm−d+1ds

+2l

∑d=m+1

(−1)2d∫

γ

ksl+m+1(kksl−d ks2m+2)sd−m−1ds

= 2m

∑d=1

∫γ

ksm+d+1(kksl+d ksl−d)sm−d+1ds+2l

∑d=m+1

∫γ

ksl+m+1(kksl−d ks2m+2)sd−m−1ds,

here in the second line in above, for the first term, 2m+2 is even, the order in kksl+d ksl−d is2l which is even as well, so the boundary terms all have odd derivatives of the curvature,they are all zero;

For the second term, when l +d and l−d are even, then the reason is the same, whenl +d and l−d are odd, as the total order in

∫γ

kkl+dksl−d ks2m+2ds is 2m+2l +2, the orderin boundary terms are always 1 less than 2m+ 2l + 2: 2m+ 2l + 1 which is odd, then atleast one term in boundary terms has odd order, i.e. all the boundary terms are zero.


2m

∑j=1

l

∑d=1

(−1)d+ j+1∫

γ


= 2m

∑j=1

m

∑d=1

(−1)d+ j+1∫

γ


+2m

∑j=1

l

∑d=m+1

(−1)d+ j+1∫

γ


= 2m

∑j=1

m

∑d=1

(−1)m+ j+2∫

γ

(k2ksl+d ksl−d ksm− j)sm−d+1ks j+d−1ds

+2m

∑j=1

l

∑d=m+1

(−1)2d+ j−m∫

γ

ksl+m+1(k2ksl−d ksm+ jksm− j)sd−m−1ds,

here as the total order in∫

γk2ksl+d ksl−d ksm+ jksm− jds is 2m+2l +2, the order in boundary

terms are always 1 less than 2m+2l+2: 2m+2l+1 which is odd, then at least one factorin each boundary term has odd order, i.e. all the boundary terms are zero.

For the same reason as above, the boundary terms in the following four equalities areall zero as well.

l

∑d=1

(−1)d+1∫

γ

k2ksl+d ksl−d k2smds

=m

∑d=1

(−1)d+1∫

γ

k2ksl+d ksl−d k2smds+

l

∑d=m+1

(−1)d+1∫

γ

k2ksl+d ksl−d k2smds

=m

∑d=1

(−1)m+2∫

γ

ksd−1(k2ksl+d ksl−d ksm

)sm+1−d ds

+l

∑d=m+1

(−1)2d−m∫

γ

ksl+m+1(k2ksl−d k2

sm)

sd−m−1 ds,

(−1)m+1∫

γ

kks2m+2k2sl ds = (−1)2m+2

∫γ

ksm+1(kk2

sl

)sm+1 ds,

and

m

∑j=1

(−1) j+1k2ksm+ jksm− jk2sl =

m

∑j=1

(−1) j+m+2ks j−1(k2ksm− jk2

sl

)sm+1 ,

−12

∫γ

k2k2smk2

sl ds =−12(−1)m+1

∫γ

ksl−m−1(k2k2

smksl)

sm+1 ds.


Substituting above nine equations into (5.17), we have for each l > m,

ddt

∫γ

k2sl ds

= (−1)2l+12∫

γ

k2sm+l+2ds+

m

∑j=1

(−1)2l+ j−m+3∫

γ

(kksm+ jksm− j)sl−m+1ksl+m+1ds

+2(−1)2l−m+2∫

γ

(kk2sm)sl−m+1ksl+m+1ds+2

m

∑d=1

∫γ

ksm+d+1 (kksl+d ksl−d)sm−d+1 ds

+2l

∑d=m+1

∫γ

ksl+m+1 (kksl−d ks2m+2)sd−m−1 ds

+2m

∑j=1

m

∑d=1

(−1)m+ j+2∫

γ

(k2ksl+d ksl−d ksm− j

)sm−d+1 ks j+d−1ds

+2m

∑j=1

l

∑d=m+1

(−1)2d+ j−m∫

γ

ksl+m+1(k2ksl−d ksm+ jksm− j

)sd−m−1 ds

+m

∑d=1

(−1)m+2∫

γ

ksd−1(k2ksl+d ksl−d ksm

)sm+1−d ds

+l

∑d=m+1

(−1)2d−m∫

γ

ksl+m+1(k2ksl−d k2

sm)

sd−m−1 ds

+(−1)2m+2∫

γ

ksm+1(kk2

sl

)sm+1 ds+

m

∑j=1

(−1) j+m+2ks j−1(k2ksm− jk2

sl

)sm+1

−12(−1)m+1

∫γ

ksl−m−1(k2k2

smksl)

sm+1 ds.


ddt

∫γ

k2sl ds

= (−1)2l+12∫

γ

k2sm+l+2ds+

∫γ

P2m+2l+24 (k)ds+

∫γ

P2m+2l6 (k)ds

≤ (−1)2l+12∫

γ

k2sm+l+2ds+

∫γ

∣∣∣P2m+2l+24 (k)

∣∣∣ds+∫

γ

∣∣∣P2m+2l6 (k)

∣∣∣ds

= −2∫

γ

k2sm+l+2ds+

∫γ

∣∣∣P2m+2l+24 (k)

∣∣∣ds+∫

γ

∣∣∣P2m+2l6 (k)

∣∣∣ds, (5.18)

where no higher than (m+ l +1)-th order derivative appears in the∫

γP2m+2l+2

4 (k)ds termand no higher than (m+ l)-th derivative appears in

∫γ

P2m+2l6 ds. Note that l +m+ 1 ≥

2m+2 because of l > m. Then we estimate then using interpolation inequality in Propo-sition 4,

For the second term∫

γ

∣∣∣P2m+2l+24 (k)

∣∣∣ds in (5.18), we have


∫γ

∣∣∣P2m+2l+24 (k)

∣∣∣ds ≤ ε

∫γ

k2sm+l+2ds+ cε

−(2m+2l+3) ·(∫

γ

k2ds)2m+2l+5

+c(∫

γ

k2ds)2m+2l+5

. (5.19)

For the third term∫

γ

∣∣∣P2m+2l6 (k)

∣∣∣ds in (5.18), we obtain

∫γ

∣∣∣P2m+2l6 (k)

∣∣∣ds ≤ ε

∫γ

k2sm+l+2ds+ cε

−(m+l+1) ·(∫

γ

k2ds)2m+2l+5

+c(∫

γ

k2ds)2m+2l+5

. (5.20)

Substituting (5.19) and (5.20) into (5.18), therefore we obtain

ddt

∫γ

k2sl ds

= −2∫

γ

k2sm+l+2ds+

∫γ

∣∣∣P2m+2l+24 (k)

∣∣∣ds+∫

γ

∣∣∣P2m+2l6 (k)

∣∣∣ds

≤ −(2−2ε) ·∫

γ

k2sm+l+2ds+

[cε−(2m+2l+3)+ cε

−(m+l+1)+ c]·(∫

γ

k2ds)2m+2l+5

≤ −(2−2ε) ·(

π2

L2

)m+2 ∫γ

k2sl ds+ c

(‖k‖2

2)

= −c∫

γ

k2sl ds+ c

(‖k‖2

2), (5.21)

from which the result follows.

5.1.1 Exponential convergence

We give that all derivatives of the curvature decay exponentially for all time.

Remark 2. In view of above Proposition 12, in fact the solution to (5.1) exist for all time,

T = ∞.

Proof. From (5.18) in Proposition 12, we have

ddt

∫γ

k2sl ds

= −2∫

γ

k2sm+l+2ds+

∫γ

∣∣∣P2m+2l+24 (k)

∣∣∣ds+∫

γ

∣∣∣P2m+2l6 (k)

∣∣∣ds

≤ −(2−2ε) ·∫

γ

k2sm+l+2ds+

[cε−(2m+2l+3)+ cε

−(m+l+1)+ c]·(∫

γ

k2ds)2m+2l+5

.


Let l = 0, then we obtain

ddt

∫γ

k2ds≤ c(∫

γ

k2ds)2m+5

,

from which it follows that ∫γ

k2ds≥ c(T − t)−1

2m+4 .

We show that as in [17, Theorem 3.1] that if the maximal existence time T of a solution(5.1) is finite, then the curvature must blow up in L2. We get a contradiction with theexponential decay of

∫γ

k2ds in Proposition 13, Thus, we prove that the solution to (5.1)exists for all time, that is, T = ∞.


(5.1),∫

γk2ds decays exponentially for all time.

Proof. We use Lemma 3, the boundary condition and (5.10)

ddt

∫γ

k2ds = 2∫

γ

kktds−∫

γ

k2 · kF1ds

= 2∫

γ

k[(F1)ss + k2F1

]ds−

∫γ

k3F1ds

= 2∫

γ

kssF1ds+∫

γ

k3F1ds

= 2∫

γ

kss

[(−1)m+1ks2m+2 +

m

∑j=1


kk2sm

]ds

+∫

γ

k3

[(−1)m+1ks2m+2 +

m

∑j=1


kk2sm

]ds

= 2(−1)2m+1∫

γ

k2sm+2ds+2

m

∑j=1

(−1) j+1+ j−1ksm+1 (kkssksm− j)s j−1

+∫

γ

(k2smk)s · ksds+(−1)2m+2

∫γ

(k3)sm+1ksm+1ds

+m

∑j=1

(−1) j+1+ j−1ksm+1(k4ksm− j

)j−1 +

12

∫γ

(ksmk4)sksm−1ds

= −2∫

γ

k2sm+2ds+2

m

∑j=1

ksm+1 (kkssksm− j)s j−1 +∫

γ

(k2smk)s · ksds

+∫

γ

(k3)

sm+1 ksm+1ds+m

∑j=1

ksm+1(k4ksm− j

)j−1

+12

∫γ

(ksmk4)

s ksm−1ds. (5.22)

For the first term in (5.22), we get it by m integrations by parts and using Lemma 30.Next, we will deal with each of the above terms separately. The second, third and fourth


terms in (5.22)

2m

∑j=1


γ

(k2

smk)

s · ksds+∫

γ

(k3)

sm+1 ksm+1ds

=∫

γ

P2m+24 (k)ds≤

∫γ

∣∣P2m+24 (k)

∣∣ds.

Estimating it by the interpolation inequality in Proposition 4,∫γ

∣∣P2m+24 (k)

∣∣ds

≤ cL1−2m−2−4‖k‖2m+5m+2

2 · ‖k‖2m+3m+2

m+2,2

≤ cL1−2m−2−4(‖k‖

2m+5m+2

2 ‖ksm+2‖2m+3m+2

2 +‖k‖42

)≤ c‖k‖

2m+5m+2

2 ‖ksm+2‖2m+3m+2

2 + cL1−2m−2−4/2‖k‖42

= c‖k‖22‖k‖

1m+22 · ‖ksm+2‖

2m+3m+2

2 + cL−2m−3‖k‖22 · ‖k‖2

2

≤ c‖k‖22 ·(

L2

π2

) m+22(m+2)

· ‖ksm+2‖1

m+2 · ‖ksm+1‖2m+3m+2

2 + cL−2m−3‖k‖22 ·(

L2

π2

)m+2

‖ksm+2‖22

= cL‖k‖22 · ‖ksm+2‖2

2 ≤ cL(

L2

π2

)m

‖ksm‖22 · ‖ksm+2‖2

2

= c2(m)L2m+1‖ksm‖22‖ksm+2‖2

2. (5.23)

For the fifth and sixth terms in (5.22), we have

m

∑j=1

ksm+1(k4ksm− j

)j−1 +

12

∫γ

(ksmk4)

s ksm−1ds =∫

γ

P2m6 (k)ds≤

∫γ

∣∣P2m6 (k)

∣∣ds,

using the interpolation inequalities, we also have∫γ

∣∣P2m6 (k)

∣∣ds ≤ cL1−2m−6‖k‖4m+10m+2

2 · ‖k‖2m+2m+2

m+2,2

≤ cL1−2m−6(‖k‖

4m+10m+2

2 ‖ksm+2‖2m+2m+2

2 +‖k‖62

)≤ c‖k‖

4m+10m+2

2 ‖ksm+2‖2m+2m+2

2 + cL1−2m−6/3‖k‖62

= c‖k‖42 · ‖k‖

2m+22 ‖ksm+2‖

2m+2m+2

2 + cL−2m−1‖k‖42 · ‖k‖2

2

≤ c‖k‖42 ·(

L2

π2

) 2(m+2)2(m+2)

‖ksm+2‖2

m+22 + cL−2m−1‖k‖4

2 ·(

L2

π2

)m+2

‖ksm+2‖22

= cL2‖k‖42 · ‖ksm+2‖2

2

≤ cL2 ·(

L2

π2

)2m

‖ksm‖42 · ‖ksm+2‖2

2

= c3(m)L4m+2‖ksm‖42 · ‖ksm+2‖2

2. (5.24)


Substituting (5.23) and (5.24) into (5.22), we obtain

ddt

∫γ

k2ds = −2∫

γ

k2sm+2ds+2

m

∑j=1


γ

(k2smk)s · ksds

+∫

γ

(k3)sm+1ksm+1ds+m

∑j=1

ksm+1(k4ksm− j

)j−1 +

12

∫γ

(ksmk4)sksm−1ds

≤ −[2− c2(m)L2m+1‖ksm‖2

2− c3(m)L4m+2‖ksm‖42]·∫

γ

k2sm+2ds. (5.25)

From Lemma 31, we know that L does not increase when the initial curve has suffi-ciently small energy. Also, the energy does not increase, so the coefficient on the righthand side in (5.25) is smaller than −δ if it holds initially, for some δ > 0. Then we have

ddt

∫γ

k2ds ≤ −δ

∫γ

k2sm+2ds

≤ −δ (L(0),m)∫

γ

k2ds

where

c2(m)L2m+1‖ksm‖22 + c3(m)L4m+2‖ksm‖4

2 < 2,

then we get ∫γ

k2ds≤ ce−δ t ,

hence the result.

We have that∫

γk2ds decays exponentially. By induction argument, all derivatives of

the curvature decay exponentially.


(5.1),∫

γk2

sl ds and ‖ksl‖2∞ exponentially decay for all time and any l ∈ N∪0.

Proof. We calculate the curvature derivative decay in L2 by using integration by parts,∫γ

k2s ds =−

∫γ

kkssds≤ ‖k‖2‖kss‖2 ≤ ce−δ1t ,

where we have eliminated the boundary term as it contains an odd derivative of curvatureks which is equal to zero. The

∫γ

k2ssds is bounded by Proposition 12, the exponential

convergence follows from Proposition 16.We next compute ∫

γ

k2ssds =−

∫γ

ksks3ds≤ ‖ks‖2‖ks3‖2 ≤ ce−δ2t ,


where similarly∫

γk2

s3ds is bounded by Proposition 12, so then the exponential conver-gence follows from the previous step.

Moreover for any l ∈ N∪0, as from Proposition 12, we know that all derivatives ofthe curvature are bounded. We have∫

γ

k2sl ds =−

∫γ

kksl+1ds≤ ‖k‖2‖ksl+1‖2 ≤ ce−δlt .

Above c are all different constants. Now we obtain that all derivatives of the curvaturehave exponential convergence. Moreover, exponential convergence in L∞ of γ to γ∞ fol-lows, as ‖ksl‖2

∞ ≤ Lπ‖ksl+1‖2

2 in Proposition 2, we have∫

γk2

sl ds exponentially decays inL∞.

From above proposition, we can also obtain uniform bounds for all derivatives of theevolving curve γ .



‖∂ul γ‖∞ ≤ c(l)+l

∑p=0‖∂sl γ0‖∞,


Proof. We claim that for l, p ∈ N0,

∂t∂sl γ = ν

l

∑p=0

(P2m+2+p

1+l−p (k)+P2m+p3+l−p(k)

)+ τ

l

∑p=0

(P2m+2+p

1+l−p (k)+P2m+p3+l−p(k)

). (5.26)

We prove this by induction. First, F = P2m+21 (k)+P2m

3 (k) so the equation holds for l = 0.For q ∈ N0, we do the differentiation

∂t∂sl+1γ

= kF ·∂sl+1γ +∂s(∂t∂sl γ)

= ∂s

[ν

l

∑p=0

(P2m+2+p

1+l−p (k)+P2m+p3+l−p(k)

)+ τ

l

∑p=0

(P2m+2+p

1+l−p (k)+P2m+p3+l−p(k)

)]+kF ·∂sl τ

= kFν ∑p+q=l

Ppq (k)+ kFτ ∑

p+q=lPp

q (k)+ν

l

∑p=0

(P2m+2+p

2+l−p (k)+P2m+p4+l−p(k)

)+τ

l

∑p=0

(P2m+2+p

2+l−p (k)+P2m+p4+l−p(k)

)+ν

l

∑p=0

(P2m+3+p

1+l−p (k)+P2m+1+p3+l−p (k)

)+τ

l

∑p=0

(P2m+3+p

1+l−p (k)+P2m+1+p3+l−p (k)

)


= ν ∑p+q=l

(Pp+2m+2

q+2 (k)+Pp+2mq+4 (k)

)+ν

l

∑p=0

(P2m+2+p

2+l−p (k)+P2m+p4+l−p(k)

)+ν

l+1

∑p=1

(P2m+2+p

2+l−p (k)+P2m+p4+l−p(k)

)+ τ ∑

p+q=l

(Pp+2m+2

q+2 (k)+Pp+2mq+4 (k)

)+τ

l

∑p=0

(P2m+2+p

2+l−p (k)+P2m+p4+l−p(k)

)+ τ

l+1

∑p=1

(P2m+2+p

2+l−p (k)+P2m+p4+l−p(k)

)= ν

l+1

∑p=0

(P2m+2+p

2+l−p (k)+P2m+p4+l−p(k)

)+ τ

l+1

∑p=0

(P2m+2+p

2+l−p (k)+P2m+p4+l−p(k)

)

as required. Integrating (5.26) and using Proposition 14, we find

‖∂sl γ‖∞ ≤ ‖∂sl γ0‖∞ + c∫ t

0e−ct ′dt ′ ≤ ‖∂sl γ0‖∞ + c(l).

As u is the initial space parameter before reparameterization by arc-length, set v = |∂uγ|.Referring to the proof of Theorem 3.1 in [17], then for any function Φ : S→ R, we have

∂luΦ = v∂

ls Φ+Pl(v, ...,∂ l−1

u v,Φ, ...∂ l−1s Φ)

where Pl is a polynomial. Then we obtain

‖∂ul γ‖∞ ≤ ‖∂sl γ0‖∞ + c(l).

as required.

To finish the proof of Theorem 8, we need two lemmas in Section 5.3. Lemma 36 givesthat there is a subsequence ti→ ∞ such that γ(·, ti)→ γ∞ in C∞

([−1,1],R2). Lemma 38

proves that the exponential convergence of γ to a unique horizontal straight segment. Thiscompletes the proof of Theorem 8.

5.2 The polyharmonic curve diffusion flow

This section considers the dual spaces to the Sobolev spaces Hm+1, which are denoted byH−(m+1) and consist of the bounded linear functionals L : Hm+1→ R. We establish ourresult for the polyharmonic curve diffusion flow (5.2) for each fixed m ∈ N∪0 whichis the gradient flow for length in the Sobolev spaces H−(m+1).

Since curvature depends on second derivatives of γ with respect to s, the flow (5.2) hasorder 2m+4. For a solution to the flow (5.2), we have F2 =(−1)m+1ks2m+2 is the H−(m+1)-gradient flow of the length

∫γ|γu|du, then using Lemma 3 (i) and m+ 1 integrations by


parts, we obtain

ddt

L(γ) = −∫

γ

kF2ds

= (−1)m+2∫

γ

kks2m+2ds

= (−1)m+1∫

γ

ksks2m+1ds+(−1)m kks2m+1|∂γ

= (−1)m∫

γ

kssks2mds+(−1)m−1 ksks2m|∂γ+(−1)m kks2m+1|∂γ

= ......

= (−1)2∫

γ

ksmksm+2ds+(−1)1 ksm−1ksm+2|∂γ+ ...+(−1)m−1 ksks2m|∂γ

+(−1)m kks2m+1|∂γ

= −∫

γ

k2sm+1ds+(−1)0 ksmksm+1|∂γ

+(−1)1 ksm−1ksm+2|∂γ+ ...

+(−1)m−1 ksks2m|∂γ+(−1)m kks2m+1|∂γ

= −∫

γ

k2sm+1ds+

m

∑j=0

(−1)m− jks jks2m+1− j

∣∣∣∣∣∂γ

. (5.27)

We can see from above, each of the boundary terms contains an odd derivative of k withthe order up to 2m+1.

Lemma 32. The boundary terms ∑mj=0 ks jks2m+1− j

∣∣∣∂γ

= 0 in above holds under the follow-

ing:

< ν ,νη1,2 > (±1, t) = ks(±1, t) = ...= ks2m−1(±1, t) = ks2m+1(±1, t) = 0 (5.28)

the number of the conditions is 2m+4 on both boundaries which is the same as the highest

order of the derivatives of the curvature.

Lemma 33. While a solution to the flow (5.2) with boundary condition (5.28), we have

ddt

L =−∫

γ

k2sm+1ds. (5.29)

Proof. The result follows from (5.27),

ddt

L =−∫

γ

k2sm+1ds+

m

∑j=0

(−1)m− jks jks2m+1− j

∣∣∣∣∣∂γ

.

From above lemma, the boundary terms ∑mj=0(−1)m− jks jks2m+1− j

∣∣∣∂γ

= 0. Then we have

the conclusion.


For special cases, when m = 0, the flow (5.2) is the classical curve diffusion flow:

∂

∂ tγ = F2ν =−kssν .

The case m = 1 can be seen as the geometric triharmonic heat flow of curves γ:

∂

∂ tγ = F2ν = ks4ν .

Remark 3. In fact, under the flow (5.2), from (5.29) the length L is strictly decreasing

unless γ is a straight line segment.

Because in (5.29), it is easy to see that

ddt

L =−∫

γ

k2sm+1ds≤ 0,

the length L is strictly decreasing except ddt L = 0.

From ddt L = 0, we get ksm+1 ≡ 0, therefore the curvature of the curve is zero. And the

two end points of the curve meet two parallel vertical lines η1,2 orthogonally from theNeumann boundary condition. Thus, the only smooth solutions of d

dt L = 0 are horizontalline segments.

In view of Lemma 33 and the distance |e| of two parallel lines η1,2, the length L of theevolving curve γ(·, t) remains bounded above and below under the flow (5.2).

Definition 9. Let γ : [−1,1]× [0,T )→ R2 be a family of smooth immersion whose ends

meet the parallel lines η1,2 with m+2 generalised Neumann boundary conditions (5.28).γ is said to move under (2m+4)th order curvature flow (5.2), if

∂

∂ t γ(s, t) =−F2ν , f or all (s, t) ∈ [−1,1]× [0,T )γ(·,0) = γ0,

< ν ,νη1,2 >= ks = ...= ks2m−1 = ks2m+1 = 0, f or all (s, t) ∈ η1,2× [0,T )

where F2 = (−1)m+1ks2m+2 , m ∈ N∪0, ν and νη1,2 are the unit normal fields to γ(±1)and η1,2 respectively.

Theorem 9. Let γ : [−1,1]→R2 be a smooth embedded or immersed curve in Definition

9. If the initial curve γ0 satisfies ω = 0 and the curvature k(0) of γ0 is sufficiently small in

L2, that is

L(0)∫

γ

k2(0)ds≤ ε (5.30)

where ε is positive and depends only on m, then there exists a smooth solution γ : [0,∞)→R2 to (5.2) with γ(·,0) = γ0. The solution γ converges to a horizontal line segment expo-


nentially and is unique up to parametrization. The distance between the limit curve and

the initial curve is finite.

The ε in above theorem can be calculated from (5.34). The proof of this theoremfollows by the same argument as in the previous Section 5.1.

Lemma 34. The hypothesis of Theorem 9 implies that ω(t) = ω(0) = 0. The average

curvature k satisfies k := 1L∫

γkds≡ 0.

For the proof, see Lemma 17.

Lemma 35. All odd derivatives of the curvature equal to zero at the boundary,

ks2p+1(±1, t) = 0, p = 0,1,2, ...

Proof. Here we do the first spatial derivative of the normal speed

F2 = (−1)m+1k2m+2,

then (F2)s = (−1)m+1ks2m+3 = 0, yielding

ks2m+3(±1, t) = 0.

Calculate the third derivatives (F2)s3 = (−1)m+1ks2m+5 = 0, we obtain

ks2m+5(±1, t) = 0.

Let us give the induction argument, we assume that for n = 2,3, ...

ks2n+2m−1(±1, t) = 0,

F2 = (−1)m+1ks2m+2,

∂tksl = Fsl+2 +l

∑h=0

∂sh (kksl−hF2)

= (−1)m+1ks2m+l+4 +l

∑h=0

∂hs[kksl−h(−1)m+1ks2m+2

]= (−1)m+1ks2m+l+4 +(−1)m+1

l

∑h=0

∑q+r+u=h

cqruksqksl−h+rks2m+2+u,

for constants cqru ∈ R with q,r,u,v,w≥ 0.


The inductive hypothesis implies that for l odd and less than or equal to 2n+ 2m− 1,the derivatives ksl vanishes on the boundary. Here we take l = 2n− 3, then under thehypothesis ks2n+2m−1(±1, t) = 0, we removed all terms with an odd number of derivativesof k, we conclude

k2n+2m+1 = 0,n = 2,3, ...

Together with boundary conditions (5.28), we get ks2p+1(±1, t) = 0, p = 0,1,2, ...

5.2.1 Exponential decay

Next we show that∫

γk2ds decays exponentially under the smallness assumption in Theo-

rem 9.

Proposition 16. If γ0 satisfies the conditions of Theorem 9, then under the flow (5.2), we

have ∫γ

k2ds≤∫

γ

k2(0)ds · e−δ t .

where δ > 0 depends on ε and L(0).

Proof. Under the flow (5.2), by using Lemma 33, the boundary conditions and integra-tion, we obtain

ddt

∫γ

k2ds = 2∫

γ

kktds−∫

γ

k2 · kF2ds

= 2∫

γ

k[(F2)ss + k2F2

]ds−

∫γ

k3F2ds

= 2∫

γ

kssF2ds+∫

γ

k3F2ds

= 2∫

γ

kss(−1)m+1ks2m+2ds+∫

γ

k3(−1)m+1ks2m+2ds

= 2(−1)2m+1∫

γ

k2sm+2ds+(−1)2m+2

∫γ

(k3)

sm+1 ksm+1ds

= −2∫

γ

k2sm+2ds+

∫γ

(k3)

sm+1 ksm+1ds. (5.31)

The second term∫γ

(k3)

sm+1 ksm+1ds =∫

γ

P2m+24 (k)ds≤

∫γ

∣∣P2m+24 (k)

∣∣ds.

Since the highest order derivative in P2m+24 (k) is ksm+1 , we now estimate using the in-

terpolation inequality in Proposition 4 in the first step and then PSW inequality (2.1) in


Proposition 1, as ‖k‖22 ≤

L2(m+2)

π2(m+2)‖ksm+2‖22, we have

∫γ

∣∣P2m+24 (k)

∣∣ds ≤ cL1−2m−2−4‖k‖2m+5m+2

2 · ‖k‖2m+3m+2

m+2,2

≤ c‖k‖2m+5m+2

2 ‖ksm+2‖2m+3m+2

2 + cL1−2m−2−4/2‖k‖42

= c‖k‖22‖k‖

1m+22 ‖ksm+2‖

2m+3m+2

2 + cL−2m−3‖k‖22‖k‖2

2

≤ c‖k‖22 ·(

L2

π2

) m+22(m+2)

‖ksm+2‖1

m+22 · ‖ksm+2‖

2m+3m+2

2

+cL−2m−3‖k‖22 ·(

L2

π2

)m+2

‖ksm+2‖22

= c(m)L‖k‖22 · ‖ksm+2‖2

2.

Combining above inequality with (5.31), we have

ddt

∫γ

k2ds = −2∫

γ

k2sm+2ds+

∫γ

(k3)

sm+1 ksm+1ds

≤ −[2− c(m)L‖k‖2

2]·∫

γ

k2sm+2ds. (5.32)

Suppose initially cL∫

γk2ds ≤ 2− 2δ , for some δ > 0. Then, at least for a short time,

cL∫

γk2ds≤ 2− δ . We obtain

ddt

∫γ

k2ds≤−δ

∫γ

k2sm+2ds≤−δ

(π2

L2

)m+2 ∫γ

k2ds≤−δ

(π2

L(0)2

)m+2 ∫γ

k2ds

where we have used again Lemma 33. The result follows∫γ

k2ds≤∫

γ

k2(0)ds · e−δ t .

Next we show that all curvature derivatives remain bounded under the flow in L2. Thisproof here is considerably more direct than the process for the flow (5.1).


(5.2), we have for all l ∈ N∪0, ∫γ

k2sl ds≤ cl,

for constants cl .

Proof. Under the flow (5.2), by using integration by parts and all odd derivatives of the


curvature equal to zero, then for l = 0,1,2, ..., we have

ddt

∫γ

k2sl ds

= 2∫

γ

ksl ksltds−∫

γ

k2sl · kF2ds

= 2∫

γ

ksl · (−1)m+1ks2m+l+4 +2∫

γ

ksl · (−1)m+1l

∑h=0

∑q+r+u=h

cqruksqksl−h+rks2m+2+u

−∫

γ

k2sl · k · (−1)m+1ks2m+2ds

= −2∫

γ

k2sm+l+2ds+2 · (−1)2m+2

∫γ

ksl+m+1 ·l

∑h=0

∑q+r+u=h

cqruksqksl−h+rksm+1+u

+2 · (−1)2m+2∫

γ

ksl ·l

∑h=0

∑q+r+u=h

cqru (ksqksl−h+r)sm+1 ksm+1+u

−(−1)2m+2∫

γ

(k2

sl · k)

sm+1 · ksm+1ds

= −2∫

γ

k2sm+l+2ds+

∫γ

P2m+2l+24 (k)ds

where the highest order of derivatives of k of k in the second term∫

γP2m+2l+2

4 (k)ds ism+ l +1. Using interpolation inequality in Proposition 4 we have for any ε > 0,

∫γ

P2m+2l+24 (k)ds≤ ε

∫γ

k2sm+l+2ds+ c(m, l,s)

(∫γ

k2ds)2m+2l+5

.

Combining above, we obtain by taking ε < 2

ddt

∫γ

k2sl ds = −2

∫γ

k2sm+l+2ds+

∫γ

P2m+2l+24 (k)ds

≤ −(2− ε)∫

γ

k2sm+l+2ds+ c

(∫γ

k2ds)2m+2l+5

.

As∫

γk2ds decays exponentially in Proposition 16, it is bounded by a constant. Using

now PSW inequality (2.1) in Proposition 1, −∫

γk2

sm+l+2ds≤−(

π2

L2

)m+2 ∫γ

k2sl ds, then we

obtain

ddt

∫γ

k2sl ds ≤ −(2− ε)

∫γ

k2sm+l+2ds+ c

(∫γ

k2ds)2m+2l+5

≤ −(2− ε)

(π2

L2(0)

)m+2 ∫γ

k2sl ds+ c1

= −c2

∫γ

k2sl ds+ c1,


the result follows.

Remark 4. In view of above Proposition 17, the solution to (5.2) exist for all time, T = ∞.

The proof of above remark follows the proof in Remark 2.Proposition 16 and 17 imply via interpolation that all curvature derivatives decay expo-

nentially in L2 and in L∞ via PSW inequalities (2.1) in Proposition 1 and (2.2) in Propo-sition 2. We have that

∫γ

k2ds decays exponentially. By induction, all derivatives of thecurvature decay exponentially.

Proposition 18. Suppose γ0 satisfies the conditions of Theorem 9. Then, while a solution

to the flow (5.2) there exists constants c > 0 and δl > 0, depending only on ε and L(0)such that, for all l ∈ N∪0, ∫

γ

k2sl ds≤ ce−δlt .

The quantities ‖ksl‖∞ also decay exponentially for all l.

Proof. The proof here is similar to Proposition 14. From Proposition 17, we know thatall derivatives of the curvature are bounded. Thus, the curvature derivative decay in L2

follows by standard integration by parts, we show the first two calculations:∫γ

k2s ds =−

∫γ

kkssds≤ ‖k‖2‖kss‖2 ≤ ce−δ1t .

From Proposition 17, we have∫

γk2

ssds is bounded, then Proposition 16 gives the exponen-tial convergence of

∫γ

k2s ds.

We next compute ∫γ

k2ssds =−

∫γ

kks3ds≤ ‖k‖2‖ks3‖2 ≤ ce−δ2t .

Generally, ∫γ

k2sl ds =−

∫γ

kksl+1ds≤ ‖k‖2‖ksl+1‖2 ≤ ce−δlt .

Above c are all different constants. Continue doing this, we obtain the exponentialdecay of all curvature derivatives in L2. Exponential convergence in L∞ of γ to γ∞ nowfollows, as ‖ksl‖2

∞ ≤ Lπ‖ksl+1‖2

2 in Proposition 2, we have∫

γk2

sl ds exponentially decays inL∞.




‖∂ul γ‖∞ ≤ c(l)+l

∑p=0‖∂sl γ0‖∞,



The proof of above proposition can refer to the proof of Proposition 15.In Section 5.3, Lemma 37 implies there exists an immersion γ∞ : [−1,1]→ R2 sat-

isfying the boundary conditions and a subsequence ti → ∞ such that γ(·, ti) → γ∞ inC∞([−1,1],R2). Since ‖k‖∞ → 0, Lemma 38 implies the curve γ∞ is a unique straight

line segment. This completes the proof of Theorem 9.

5.3 The unique limit

Lemma 36. Assume γ : [−1,1]× [0,T )→ R2 satisfies flow (5.1) and conditions :

c1(m)L2m+1‖ksm‖22 < 1 and c2(m)L2m+1‖ksm‖2

2 + c3(m)L4m+2‖ksm‖42 < 2. (5.33)

So there exists a subsequence ti ⊂ [0,∞)such that γ(·, ti)→ γ∞ with γ∞ a unique straight

line.

Lemma 37. Assume γ : [−1,1]× [0,T )→ R2 satisfies flow (5.2) and

c(m)L‖k‖22 < 2. (5.34)

So there exists a subsequence ti ⊂ [0,∞)such that γ(·, ti)→ γ∞ with γ∞ a unique straight

line.

For curves satisfying both curvature flows (5.1) and (5.2), we have the same proof:

Proof. Under different conditions, as we have proven that the curvature decays to zero inL2 along the flow, ‖k‖2

2 ≤ e−ct ‖k‖22

∣∣t=0 , So there exists a subsequence ti ⊂ [0,∞) such

that ∫γ

k2ds→ 0,as ti→ ∞.

This implies there exist a limit curve γ∞ that

γ(·, ti)→ γ∞,

with γ∞ satisfying the boundary conditions. Then the flow converges to which satisfiescurvature k [γ∞] = 0, that is, γ∞ is a straight line.

The only issue for Theorem 9 and Theorem 8 is that this does not uniquely determinethe limit. The following lemma completes the proof under the decay in the curvature.


Lemma 38. Suppose the initial curves γ0 satisfies the conditions of Theorem 9 and The-

orem 8 respectively. Then, under the flows (5.2), (5.1), the limits of the curves are both

unique.



∫γ

|γ− γ∞|2ds.


lims j→∞

G[ f (·,s j)] 6= 0. (5.35)


G∣∣∣∣ =

ddt

∫γ

|γ− γ∞|2 ds

= 2∫

γ


∫γ

|γ− γ∞|2 ·ddt

ds

= 2∫

γ

F · |γ− γ∞|ds+∫

γ

kF · |γ− γ∞|2 ds

=∫

γ

F · |γ− γ∞| · (2+ k |γ− γ∞|)ds

≤(∫

γ

F2ds) 1

2

·[∫

γ

|γ− γ∞|2 (2+ k |γ− γ∞|)2 ds] 1

2

≤(∫

γ

F2ds) 1

2

·[∫

γ

|γ− γ∞|2(

4+ k2 |γ− γ∞|2)

ds] 1

2

≤ c(∫

γ

F2ds) 1

2

·(

4∫

γ

|γ− γ∞|2 + k2 |γ− γ∞|4 ds) 1

2

≤ c(∫

γ

F2ds) 1

2

·(∫

γ

|γ|2ds+∫

γ

|γ∞|2ds+∫

γ

|k|2 · |γ|4ds+∫

γ

|k|2 · |γ∞|4ds) 1

2

.

Now we show that∫

γ|γ|2ds,

∫γ|γ∞|2ds,





‖γ(·, t)‖2∞ =

(∫∞

t|F |dt

)2

≤ c(∫

∞

te−δ tdt

)2

≤ cδ

e−δ t ≤ c1,


then ∫γ

|γ|2ds≤ ‖γ‖2∞

∫γ

ds≤ c1L(0)≤ c2,∫γ

|k|2 · |γ|4ds≤ ‖γ‖4∞

∫γ

|k|2ds≤ c3.


γ|γ∞|2ds and



dtG∣∣∣∣≤ c‖F‖2,


G[ f (·,s j)]≤ c∫

∞

s j

‖F‖2dt ≤ c∫

∞

s j


it follows thatlim

s j→∞G[ f (·,s j)] = 0,

which is in contradiction with (5.35).This proves that there does not exist a sequence s j, the convergence of flow in L2 to

a straight line segment is unique. We can obtain the exponential convergence in C∞ to aunique line segment as the curvature and all its derivatives exponentially decay.


Remark 5. Although we cannot find the precise height of the limiting straight horizontal

line segment for both curves, that the flow speed decays exponentially shows that the

solution curve remains in a bounded region of the initial curve: for any x,

|γ(x, t)− γ(x,0)| ≤∫ t

0

∣∣∣∣∂γ

∂ t(x, t)

∣∣∣∣dt ≤ c∫ t

0e−δ tdt =

cδ

(1− e−δ t

).

Chapter 6

Length-constrained curve diffusion flow

6.1 Introduction

This chapter considers one-parameter families of immersed closed curves γ : S1× [0,T )→R2. See Figure 6.1. We consider the energy functional

L[γ] =∫

γ

|γu|du.

The curve diffusion flow has normal speed given by F , that is

∂tγ = Fν ,

Under the evolution of the functional L which is also the first variation of the energy, astraightforward calculation yields

γ

ν τ

Figure 6.1

107

CHAPTER 6. LENGTH-CONSTRAINED CURVE DIFFUSION FLOW 108

ddt

∫γ

|γu|du = −∫

γ

Fk |γu|du.

For the flow to be the steepest descent gradient flow of length functional in H−1, werequire

F =−kss.

In this chapter, we study the length-constrained curve diffusion flow. To preserve lengthof the evolving curve γ(s, t), we take

h(t) =−∫

γk2

s ds

2πω,

where ω denotes the winding number of γ , which is defined in Chapter 2. Then we obtain

F =−kss(s, t)+h(t).

Define the length-constrained curve diffusion flow γ : S1× [0,T )→ R2 as follows,∂tγ =−(kss−h(t))ν ,

γ(·,0) = γ0.(6.1)

A curve moving under length-constrained curve diffusion flow fixes length and in-creases area. The area is enclosed by the closed plane curve. However, a curve satisfyingregular curve diffusion flow fixes area and reduces length. We can say that the length-constrained curve diffusion flow is ‘dual’ to curve diffusion flow.

In this chapter, our goal is to show there is an immortal solution of (6.1) convergingexponentially fast to a simple round circle. It is necessary to assume that initial data closeto a round circle.

The short time existence for flow problem (6.1) is given in Theorem 4 in Chapter 3.Our main theorem in this chapter is:

Theorem 10. Suppose γ0: S1× [0,T )→ R2 is a regular smooth immersed closed curve

with A(0)> 0 and ω(0) = 1. Then there exists a constant K∗ > 0 such that if

Kosc(0)< K∗, I(0)<4π2

4π2−K∗,

then the length-constrained curve diffusion flow (6.1) exists for all time and converges

exponentially in C∞ to a round circle with radius L02π

.

From our calculation (6.7) in Proposition 21, we know that

K∗ ' 0.05.


γ∞

L(0)2π

Figure 6.2

The limiting curve is showed in Figure 6.2.The structure of the proof of Theorem 10 is as follows. Firstly, we show that the length

of curve satisfying (6.1) is constrained, and area is increasing. Under the condition thatthe oscillation and isoperimetric ratios of the curve are bounded in Theorem 10, we havein section 6.2 that Kosc remains under control under (6.1) and all curvature derivativesin L2 are bounded under the flow. Secondly, we move on to the analysis of the globalbehaviour of the flow, we prove the maximal time is infinite in section 6.3, and when timegoes to infinity, the derivatives of curvature in L∞ decays exponentially in section 6.4. Theoscillation goes to zero when time goes to infinity, thus the curve subconverges to a roundcircle. Finally, we show in section 6.5 this round circle is unique which is our conclusion.We also give the self-similar solutions for this length-constrained curve diffusion flow.

The short time existence is shown in Chapter 3, Theorem 4. From Lemma 4, we obtainthe following Lemma.


Lemma 40. Suppose γ : S1× [0,T )→ R2 solves (6.1) and satisfies the assumptions of

Theorem 10, we have

L = L(0),


Proof. Using integration by parts, we obtain

0 =ddt

L = −∫

γ

k (kss−h)ds

= −∫

γ

k2s ds−h

∫γ

kds

= −∫

γ

k2s ds−2πωh(t),


In Lemma 39, we have proofed that ω(t) = 1, as h(t) =−∫

γk2

s ds2π

≤ 0, thus under thesmall energy assumption we have the length is a constant.

Lemma 41. Suppose γ : S1× [0,T )→R2 solves (6.1). Under the assumption of Theorem

10,

A(t)≥ A(0),


Proof. Using integration by parts, we compute the evolution of the area functional,

ddt

A =∫

γ

(kss−h)ds =−h(t)L(0) =

∫γ

k2s dµ

2πwL(0)≥ 0.

Thus, we have the area is increasing under the small energy condition.

Lemma 42. Suppose γ : S1× [0,T )→R2 solves (6.1). The evolution of the `-th derivative

of curvature


cqruksq+2ksrksu−h(t) ∑a+b=l

cabksaksb. (6.2)

for constants cqru,cab ∈ R with a,b,q,r,u≥ 0.

Proof. First, we calculate the evolution of the curvature

∂tk = Fss +Fk2

= (−kss +h(t))ss +(−kss +h(t))k2

= −ks4− kssk2 +h(t)k2

and the derivative of ks with respect to t,

∂tks = ∂t∂sk = ∂s∂tk+ kFks

= ∂s(−ks4− kssk2 +h(t)k2)+ kks (−kss +h(t))

= −ks5− ks3k2−2kssksk+2h(t)ksk− kssksk+h(t)ksk

= −ks5− ks3k2−3kssksk+3h(t)ksk,

then we calculate the derivative of kss with respect of t,

∂tkss = ∂t∂sks = ∂s∂tks + kFkss

= ∂s(−ks5− ks3k2−3kssksk+3h(t)ksk

)+ kkss (−kss +h(t))

= −ks6− ks4k2−2ks3ksk−3ks3ksk−3k2ssk−3kssk2

s +3h(t)kssk

+3h(t)k2s − k2

ssk+h(t)kssk

= −ks6− ks4k2−5ks3ksk−4k2ssk−3kssk2

s +4h(t)kssk+3h(t)k2s . (6.3)


It is straightforward to assume that when l ≥ 2,

∂tksl−1 = ∂t∂sksl−2 = ∂s∂tksl−2 + kFksl−1

= −ksl+3 + ∑q+r+u=l−1

˜cqruksq+2ksrksu +h(t) ∑a+b=l−1

˜cabksaksb,

for constants ˜cqru, ˜cab ∈ R with a,b,q,r,u≥ 0.Thus, we have

∂tksl = ∂t∂sksl−1 = ∂s∂tksl−1 + kFksl

= ∂s

(−ksl+3 + ∑

q+r+u=l−1˜cqruksq+2ksrksu +h(t) ∑

a+b=l−1˜cabksaksb

)+kksl (−kss +h(t))

= −ksl+4 + ∑q+r+u=l

cqruksq+2ksrksu +h(t) ∑a+b=l

cabksaksb.

for constants cqru,cab ∈Rwith a,b,q,r,u≥ 0, which is the conclusion of this Lemma.

The following evolution equation for various geometric quantities under the flow willbe used in our analysis.

Lemma 43. Under the flow (6.1),(i) d

dt L = 0;(ii) d

dt A =−h(t)L(0);(iii) d

dt∫

γk2ds =−2

∫γ

k2ssds+3

∫γ

k2k2s ds+h(t)

∫γ

k3ds;

(iv) ddt Kosc =−2L(0)

∫γ

k2ssds+3L(0)

∫γ

(k− k

)2 k2s ds+6L(0)k

∫γ

(k− k

)k2

s ds

+2k2L(0)∫

γk2

s ds+L(0)h(t)[∫

γ(k− k)3ds+3k

∫γ(k− k)2ds

];

(v) ddt∫

γk2

s ds =−2∫

γk2

s3ds+2∫

γk2k2

ssds+ 13∫

γk4

s ds+5h(t)∫

γkk2

s ds;(vi) d

dt∫

γk2

ssds =−2∫

γk2

s4ds+2∫

γk2

s3k2ds−4∫

γk3

sskds−3∫

γk2

ssk2s ds+7h

∫γ

k2sskds;

Moreover, for l ∈ N∪0,(vii) d

dt∫

γk2

sl ds =−2∫

γk2

sl+2ds+∫

γksl Pl+2

3 (k)ds+h(k)∫

γksl Pl

2(k)ds.

Here L = L(0) is the constant length of the evolving curve γt .

Proof. (i) (ii) are proven in Lemma 40 and Lemma 41 respectively.(iii) Calculate the evolution of

∫γ

k2ds,

ddt

∫γ

k2ds =∫

γ

2k∂k∂ t

ds−∫

γ

k2 ∂

∂ tds = 2

∫γ

k(Fss + k2F

)ds−

∫γ

k2kFds

= −2∫

γ

ksFsds+2∫

γ

k3Fds−∫

γ

k3Fds

= 2∫

γ

kssFds+∫

γ

k3Fds = 2∫

γ

(kss +

12

k3)

Fds.


For constrained curve diffusion flow, we can have

ddt

∫γ

k2ds = 2∫

γ

(kss +

12

k3)(−kss +h(t))ds

= −2∫

γ

k2ssds+3

∫γ

k2k2s ds+2h(t)

∫γ

(kss +

12

k3)

ds

= −2∫

γ

k2ssds+3

∫γ

k2k2s ds+h(t)

∫γ

k3ds.

(iv) We compute the evolution of the oscillation,

ddt

Kosc =ddt

L(0)∫

γ

(k− k)2ds = L(0)ddt

∫γ

(k− k)2ds

=∫

γ

2(k− k)(Fss + k2F

)ds−

∫γ

(k− k)2kFds

= −2L(0)∫

γ

k2ssds+3L(0)

∫γ

(k− k)2k2s ds+6kL(0)

∫γ

(k− k)2k2s ds

+2k2L(0)∫

γ

k2s ds+2L(0)h(t)

∫γ

(k− k)k2ds−h(t)L(0)∫

γ

(k− k)2kds

= −2L(0)∫

γ

k2ssds+3L0

∫γ


∫γ

(k− k)2k2s ds

+2k2L(0)∫

γ

k2s ds+2L(0)h(t)

∫γ

(k− k)3ds+4kL(0)h(t)∫

γ

(k− k)kds

−2k2L0h∫

γ

(k− k)ds−hL(0)∫

γ

(k− k)3− kL(0)h∫

γ

(k− k)2ds

= −2L(0)∫

γ

k2ssds+3L(0)

∫γ


∫γ

(k− k)k2s ds

+2k2L(0)∫

γ

k2s ds+3kL(0)h

∫γ

(k− k)2ds+L(0)h(t)∫

γ

(k− k)3ds

= −2L(0)∫

γ

k2ssds+3kh(t)Kosc +3L(0)

∫γ

(k− k)2k2s ds

+6kL(0)∫

γ

(k− k)k2s ds+2k2L(0)

∫γ

k2s ds+L(0)h(t)

∫γ

(k− k)3ds.

(v) Using integration by parts,

ddt

∫γ

k2s ds = 2

∫γ

ks ·∂ks

∂ tds−

∫γ

k2s

∂

∂ tds

=∫

γ

ks ·(−2Fs3−2Fsk2−6Fksk+ kskF

)ds

=∫

γ

−2ks3Fs−2ksk2Fs−5k2s kFds

=∫

γ

(−2k2

s3−2ks3ksk2−5kssk2s k)

ds+5h∫

γ

k2s kds

= −2∫

γ

k2s3ds+2

∫γ

k2k2ssds+

13

∫γ

k4s ds+5h

∫γ

k2s kds.


(vi) By using (6.3) in Lemma 42, and

−10∫

γ

ks3ksskskds = 5∫

γ

k3sskds+5

∫γ

k2ssk

2s ds,

−2∫

γ

ks4kssk2ds = 2∫

γ

k2s3k2ds−2

∫γ

k3sskds−2

∫γ

k2ssk

2s ds.

We obtain

ddt

∫γ

k2ssds

= 2∫

γ

kss∂

∂tkssds−

∫γ

k2sskFds

= 2∫

γ

kss(−ks6− ks4k2−5ks3ksk−4k2

ssk−3kssk2s +4h(t)kssk+3h(t)k2

s)

ds

−∫

γ

k2ssk (−kss +h(t))ds

= −2∫

γ

k2s4ds−2

∫γ

ks4kssk2ds−10∫

γ

ks3ksskskds−7∫

γ

k3sskds

−6∫

γ

k2ssk

2s ds+6h(t)

∫γ

kssk2s ds+7h(t)

∫γ

k2sskds

= −2∫

γ

ks4ds−2∫

γ

ks4kssk2−∫

γ

k2ssk

2s ds−2

∫γ

k3sskds+6h(t)

∫γ

kssk2s ds

+7h(t)∫

γ

k2sskds

= −2∫

γ

k2s4ds+2

∫γ

k2s3k2ds−4

∫γ

k3sskds−3

∫γ

k2ssk

2s ds+6h

∫γ

kssk2s ds

+7h(t)∫

γ

k2sskds

= −2∫

γ

k2s4ds+2

∫γ

k2s3k2ds−4

∫γ

k3sskds−3

∫γ

k2ssk

2s ds+7h(t)

∫γ

k2sskds.

(vii) Here we use (6.2) in Lemma 42, for l ∈ N,

ddt

∫γ

k2sl ds = 2

∫γ

ksl ∂tksl ds−∫

γ

kFk2sl ds

= 2∫

γ

ksl

(−ksl+4 + ∑

q+r+u=lcqruksq+2ksrksu +h(t) ∑

a+b=lcabksaksb

)ds

−∫

γ

kk2sl (kss−h(t))ds

= −2∫

γ

k2sl+2ds+

∫γ

ksl Pl+23 (k)ds+h(t)

∫γ

ksl Pl2(k)ds.

We finish the proof here.


Lemma 44. For n ∈ N,

∫γ

k2sn−1ds≤

(∫γ

k2ds) 1

n(∫

γ

k2snds

) n−1n

.

Proof. The result is obvious for n = 1. By induction, we assume that

∫γ

k2si−1ds≤

(∫γ

k2ds) 1

i(∫

γ

k2sids) i−1

i

(6.4)

and use this to show

∫γ

k2sids≤

(∫γ

k2ds) 1

i+1(∫

γ

k2si+1ds

) ii+1

.

By integration by parts and the Hölder inequality, we have

∫γ

k2sids =−

∫γ

ksi−1ksi+1ds≤(∫

γ

k2si−1ds

) 12(∫

γ

k2sl+1ds

) 12

.

Inserting above on the right-hand side of (6.4), we get

∫γ

k2sids≤

(∫γ

k2ds) 1

2i(∫

γ

k2sids) i−1

2i(∫

γ

k2si+1ds

) 12

.

In other words (∫γ

k2sl ds) i+1

2i

≤(∫

γ

k2ds) 1

2i(∫

γ

k2si+1ds

) 12

,

which implies ∫γ

k2sids≤

(∫γ

k2ds) 1

i+1(∫

γ

k2si+1ds

) 1i+1

as required.

Lemma 45. For each n ∈ N, the global term h(t) may be estimated as

|h(t)| ≤ 12π

(∫γ

k2ds)1− 1

n(∫

γ

k2snds

) 1n

.

Proof. Here we again use induction argument. As h(t) =−∫

γk2

s ds2πw , it is obvious for n = 1.


Assume that

|h(t)| ≤ 12π

(∫γ

k2ds)1− 1

i(∫

γ

k2sids) 1

i

(6.5)

and it shows

|h(t)| ≤ 12π

(∫γ

k2ds)1− 1

i+1(∫

γ

k2si+1ds

) 1i+1

.

From Lemma 44, we have

∫γ

k2sids≤

(∫γ

k2ds) 1

i+1(∫

γ

k2si+1ds

) 1i+1

.

Substituting this into (6.5) we obtain

|h(t)| ≤ 12π

(∫γ

k2ds)1− 1

i[(∫

γ

k2ds) 1

i+1(∫

γ

k2sids) i

i+1] 1

i

which simplifies to the required expression.

From Lemma 4 in Chapter 2, we can have the following lemma.

Lemma 46. Suppose γ : S1× [0,T )→ R2 solves (6.1) and

ω(0) =1

2π

∫γ

kds = 1,

then

ω(t) = ω(0) = 1.

By previous lemma, we get that the average curvature k satisfies

k :=1L

∫γ

kds =2π

L(0).

6.2 Controlling the geometry of the flow

Here we show that Kosc is a L1 function in time:

Lemma 47. Suppose γ: S1× [0,T )→ R2 solves our flow (6.1). Then,

‖Kosc‖1 ≤L2(0)

2π

[L(0)4π−A(0)

]where A(0) denotes the signed enclosed area of the initial curve.


Proof. From Lemma 43 (ii), we have

Kosc = L(0)∫

γ

(k− k)2ds≤ L2(0)4π2 L(0)‖ks‖2

2 ≤L2(0)

2π

ddt

A,

then

‖Kosc‖1 ≤L2(0)

2π[A(t)−A(0)]≤ L2(0)

2π

[L2(0)

4π−A(0)

].

Next, we show that Kosc remains bounded under (6.1) if initially Kosc is sufficientlysmall and the isoperimetric ratio I(0) = L2(0)

4πA(0) is sufficiently close to 1.

Proposition 20. Suppose γ: S1× [0,T )→ R2 solves our flow (6.1) and satisfies∫

γkds =

2π . Then

Kosc(0)< K∗, I(0)<4π2

4π2−K∗,

implies Kosc ≤ 2K∗, for all t ∈ [0,T ).

Proof. Suppose we establish a contradiction these exists a first time T ∗ < T for whichKosc = 2K∗. In view of Proposition 21,

Kosc ≤ Kosc(0)+16π3ω3

L2(0)[A(t)−A(0)]

Since

A(t)−A(0)≤ L2(0)4π−A(0) =

L2(0)4π

[1− 1

I(0)

],

we haveKosc ≤ Kosc(0)+4π

2[

1− 1I(0)

]<

32

K∗

which is a contradiction. We obtain the conclusion that Kosc < 2K∗for all t ∈ [0,T ).As I(t) = L2(0)

4πA(t) , we know that A(t) is increasing, A(t) ≥ A(0), then I(t) is decreasingwhen t→ ∞.

Proposition 21. Suppose γ : S1× [0,T )→ R2 solves (6.1). If there exists a T ∗ such that

for t ∈ [0,T ∗), we have

Kosc ≤ 2K∗,

then during this time we have

Kosc(t)≤ Kosc(0)+16π3ω3

L2(0)[A(t)−A(0)]


Proof. We start with the evolution equation of Lemma 43 (iv),

ddt

Kosc = −2L(0)∫

γ

k2ssds+3L(0)

∫γ

(k− k

)2 k2s ds+6L(0)k

∫γ

(k− k

)k2

s ds

+2k2L(0)∫

γ

k2s ds+L(0)h(t)

[∫γ

(k− k)3ds+3k∫

γ

(k− k)2ds].

We estimate using PSW inequality (2.4) in Proposition 4, ‖ks‖2∞ ≤

L(0)2π‖kss‖2

2,

3L(0)∫

γ

(k− k)2k2s ds≤ 3L(0)

2πKosc‖kss‖2

2

and

6kL(0)∫

γ

(k− k)k2s ds≤ 6ωL(0)

√Kosc‖kss‖2

2.

Although we can neglect the negative term h(t)∫

γ

(k− k

)2 ds, we need to estimate theother h(t) term as follows:

∫γ

k2s ds =−

∫γ

kkssds =−∫

γ

(k− k

)kssds≤

(∫γ

(k− k

)2 ds) 1

2(∫

γ

k2ssds) 1

2

and ∫γ

(k− k

)3 ds≤ ‖k− k‖∞

∫γ

(k− k

)2 ds,

thus

L(0)h(t)∫

γ

(k− k

)3 ds ≤ L(0)2πω

[∫γ

(k− k

)3 ds] 3

2

‖k− k‖∞‖kss‖2

≤ L(0)4π2√

2πωK

32osc‖kss‖2

2.

Using Lemma 43 (ii), we have

ddt

Kosc +L(0)(

2− 14π2√

2πωK

32osc−

32π

Kosc−6ωK12osc

)∫γ

k2ssds≤ 16π3ω3

L2(0)dAdt

. (6.6)

Here we take 2K∗ as the smallest positive solution of

2− 14π2√

2πωK

32osc−

32π

Kosc−6ωK12osc = 0, (6.7)

the coefficient of∫

γk2

ssds remains positive on the interval [0,T ∗). Then by integration intime, the result follows.


When ω = 1, we can estimate (6.7) to get 2K∗ ≈ 0.1.Moving on to proof the curvature derivatives in L2 are bounded under the flow. We start

with the L2 norm of first curvature derivative.

Proposition 22. Suppose γ : S1× [0,T )→ R2 solves (6.1) and satisfies the assumptions

of Theorem 10, then there exist a constant c depending only on γ0 such that

‖ks‖22 ≤ c.

Proof. Using Lemma 43 (v),

ddt

∫γ

k2s ds = −2

∫γ

k2s3ds+2

∫γ

k2k2ssds+

13

∫γ

k4s ds+5h

∫γ

k2s kds

= −2∫

γ

k2s3ds+

53

∫γ

k4s ds+2

∫γ

k2k2ssds+4

∫γ

kk2s kssds

+5h∫

γ

k2s kds. (6.8)

For the h(t) term we estimate using integration by parts

5h∫

γ

k2s kds =

52(−h)

∫γ

kssk2ds

and ∫γ

kssk2ds =∫

γ

kss(k− k

)2 ds+2k∫

γ

ksskds

=∫

γ

kss(k− k

)2 ds+2k∫

γ

kss(k− k

)ds

≤ 14ε

∥∥k− k∥∥2

2 + ε

∫γ

(k− k

)2 k2ssds− 4π

L(0)

∫γ

k2s ds

≤ 14εL(0)

Kosc +εKosc

2π‖ks3‖2

2−4π

L(0)‖ks‖2

2

=1

4εL(0)Kosc +

εKosc

2π‖ks3‖2

2.

It follows that

5h∫

γ

k2s kds =

52(−h)

∫γ

kssk2ds

≤ 52(−h)

14L(0)

Kosc +52(−h)

Kosc

2π‖ks3‖2

2

=5Kosc(−εh)

4π‖ks3‖2

2 +5(−h)8εL(0)

Kosc.

Since∫

γk4

s ds =−3∫

γkk2

s kssds≤ 12∫

γk4

s ds+ 92∫

γk2k2

ssds, we have


53

∫γ

k4s ds+4

∫γ

kk2s kssds≤ 27

∫γ

k2k2ssds.

where ∫γ

k2k2ssds =

∫γ

(k− k)2k2ssds+2k

∫γ

kk2ssds− k2

∫γ

k2ssds

=∫

γ

(k− k)2k2ssds+2k

∫γ

(k− k)k2ssds+ k2

∫γ

k2ssds

= 2∫

γ

(k− k)2k2ssds+2k2

∫γ

k2ssds

≤ Kosc

π‖ks3‖2

2 +2k2‖ks‖2‖ks3‖2

≤ Kosc

π‖ks3‖2

2 +1

108‖ks3‖2

2 +108k4‖ks‖22

≤ Kosc

π‖ks3‖2

2 +1

108‖ks3‖2

2 +54k4

√L(0)Kosc

π‖ks3‖2

≤ Kosc

π‖ks3‖2

2 +1

108‖ks3‖2

2 +1

108‖k3

s‖2 +543k8L(0)Kosc

2π2

≤(

Kosc

π+

154

)‖ks3‖2

2 +543k8L(0)Kosc

2π2 .

Substituting these estimates into (6.8), we obtain

ddt

∫γ

k2s ds = −2‖ks3‖2

2 +53

∫γ

k4s ds+2

∫γ

k2k2ssds+4

∫γ

kk2s kssds+5h

∫γ

k2s kds

= −2‖ks3‖22 +27

∫γ

k2k2ssds+5h

∫γ

k2s kds

≤ −2‖ks3‖22 +27

[(Kosc

π+

154

)‖ks3‖2

2 +543k8L(0)Kosc

2π2

]+

5Kosc(−h)4π

‖ks3‖22 +

5(−h)8L(0)

Kosc

=

[−2+27

(Kosc

π+

154

)+

5Kosc(−εh)4π

]‖ks3‖2

2

+

[542272k8L(0)

π2 +5(−h)8εL(0)

]Kosc.

Because h =−∫

γk2

s dµ

2πw ≤ 0, then −h =∫

γk2

s dµ

2πw = 2πwL(0)

dAdt

12πw = 1

L(0)dAdt and

−L(0)h =dAdt

,

−L(0)[h(t)−h(0)] = A(t)−A(0)≤ L2(0)4π−A(0) =

L2(0)4π

[1− 1

I(0)

]≤ L2(0)K∗

32π3 ,


−h(t)≤−h(0)+L(0)K∗

32π3 . (6.9)

In order to get

−2+27(

Kosc

π+

154

)+

5Kosc(−εh)4π

≤ 0,

we need to have

Kosc ≤6π

108−5εh,

here there is a small ε satisfies 6π

108−5εh > K∗(in Theorem 10)), ie ε(−h)< 6π/K∗−1085 .

Then we only needKosc ≤ K∗.

If so, we find

ddt

∫γ

k2s ds+‖ks3‖2

2 ≤[

542272k8L(0)π2 +

5(−h)8L(0)

]Kosc,

andddt

∫γ

k2s ds≤− 16π4

L4(0)‖ks‖2

2 +

[542272k8L(0)

π2 +5(−h)8εL(0)

]Kosc.

Assume there exists δ0 ≥ 0, such that∫

γk2

s ds ≥ c for t ∈ [δ0,δ1], (here δ0 can be 0),here c is a very small constant.

Since −h is bounded by (6.9) and Kosc ∈ L1, so

∫γ

[542272k8L(0)

π2 +5(−h)8εL(0)

]Kosc

‖ks‖22≤

∫γ

[542272k8L(0)

π2 +5(−h)8εL(0)

]· Kosc

c≤ c‖Kosc‖1 ≤ c,

and ∫γ

k2s ds≤ c · e−

16π4

L4(0)(t−δ0)

= c1

Then when t /∈ [δ0,δ1], we take c1 = c, then we still have∫

γk2

s ds < c = c1.

Also we know that after enough time, ‖k‖2∞ ≤ 2

(2ωπ

L(0)

)2≤ c .

Now we have that all curvature derivatives in L2 are bounded under the flow. Thesebounds are independent of time implies solutions exist for all time.

First, we have the following lemma,

Lemma 48. Suppose F =−kss +h(t), then for m≥ 0 the derivatives of the curvature ksm

satisfy:∂

∂ tksm + ksm+4 = Pm+2

3 (k)+h(t)Pm2 (k).


Proof. For m = 0 this follows from Proposition (3) (ii).

kt = Fss + k2F =−ks4− kssk2 +h(t)k2.

For m > 1, we obtain using Proposition (3) (v),

∂

∂ tksm = Fsm+2 +

m

∑j=0

∂sm(kkl− jF

)= −ksm+4 +Pm+2

3 (k)+h(t)Pm2 (k).

The result follows.

Lemma 49. Suppose γ : S1× [0,T )→ R2 solves (6.1) and satisfies the assumptions of

Theorem 10, there exists absolute constants cm > 0 such that for any m ∈ N∪0,

‖ksm‖22 ≤ cm.

It follows that ‖ksm‖2∞ is bounded as well.

Proof. Lemma 48 above yields for m≥ 0

ddt

∫γ

k2smds = 2

∫γ

ksm∂

∂ tksmds+

∫γ

k2smkFds

= 2∫

γ

ksm ·[−ksm+4 +Pm+2

3 (k)+h(t)Pm2 (k)

]ds+

∫γ

k2sm · k · [−kss +h(t)]ds

= −2∫

γ

k2sm+2ds+

∫γ

ksmPm+23 (k)ds+h(t)

∫γ

ksmPm2 (k)ds.

From interpolation inequality and by doing integration by parts, we achieve the esti-mates of last two terms on the right-hand side,

∫γ

ksmPm+23 (k)ds≤ ε1

∫γ

k2sm+2ds+ c1

(∫γ

k2ds)2m+5

,

where c1 is a constant depending only on ε1, m.

h(t)∫

γ

ksmPm2 (k)ds ≤ −h(t)ε2

∫γ

k2sm+1ds−h(t)c2

(∫γ

k2ds)2m+2

≤ −h(t)ε2 ·L2(0)4π2

∫γ

k2sm+2ds−h(t)c2

(∫γ

k2ds)2m+2

where constant c2 depends only on ε2, m.


Then we have

ddt

∫γ

k2smds = −2

∫γ

k2sm+2ds+

(ε1− ε2 ·

h(t)L2(0)4π2

)∫γ

k2sm+2ds

+c1

(∫γ

k2ds)2m+5

−h(t)c2

(∫γ

k2ds)2m+2

which yields if we let ε1− ε2 · h(t)L2(0)4π2 = 1,

ddt

∫γ

k2smds+

∫γ

k2sm+2ds≤ c1

(∫γ

k2ds)2m+5

−h(t)c2

(∫γ

k2ds)2m+2

. (6.10)

Here ∫γ

k2ds =∫

γ

(k− k

)ds+2k

∫γ

kds− k2∫

γ

ds

=Kosc

L(0)+2

2π

L(0)·2π−

(2π

L(0)

)2

·L(0)

=Kosc

L(0)+

4π2

L(0),

as Kosc ≤ 2K∗ in Proposition 21,∫

γk2ds is bounded. Also −h(t) is bounded by (6.9),

therefore on the right-hand side of (6.10), c1

(∫γ

k2ds)2m+5

− h(t)c2

(∫γ

k2ds)2m+2

isbounded.

So we have

ddt

∫γ

k2smds≤−

∫γ

k2sm+2ds+ c.

it follows

‖ksm‖22 ≤ cm.

and the L∞ estimates follow immediately by Proposition 2.

The following proposition shows that there is an upper bound of time for which thecurvature of a solution of (6.1) is not strictly positive.

Proposition 23. Suppose γ : S1× [0,T )→ R2 solves (6.1) and satisfies the assumptions

of Theorem 10, Then

Lt ∈ [0,∞) : k(·, t)≯ 0 ≤ L2(0)4π3

(L2(0)

4π−A(0)

).

In the above, k(·, t)≯ 0 means that there exists a p such that k(p, t)≤ 0. This estimate

is optimal in the sense that the right-hand side is zero for a simple circle.


Proof. Rearranging γ in time if necessary, we may assume that for t0 >L2(0)4π3

(L2(0)

4π−A(0)

)k(·, t)≯ 0, f or all t ∈ [0, t0);

k(·, t)> 0, f or all t ∈ [t0,∞).

We have that k always has a zero when t ∈ [0, t0).

As 2π =∫

γkds≤ L1/2

(∫γ

k2ds)1/2

, then

∫γ

k2ds≥ 4π2

L(0)

and

ddt

A =

∫γ

k2s ds

2πwL(0)≥ L(0)

2π· 4π2

L2(0)

∫γ

k2ds

=L(0)2π· 4π2

L2(0)· 4π2

L(0)=

8π3

L2(0).

For any t ∈ [0, t0),

A(t)−A(0)≥ 8π3

L2(0)t,

As t0 >L4(0)32π4 , then

A(t) ≥ A(0)+8π3

L2(0)· L

2(0)4π3

(L2(0)

4π−A(0)

)= A(0)+

L2(0)2π−2A(0)

≥ L2(0)2π−A(0).

As A(0)< L2(0)4π

, then

A(t)>L2(0)

4π.

We establish a contradiction to the isoperimetric inequality. Because from the isoperi-metric inequality, we know that A(t)≤ L2(0)

4π. Thus, the result follows.

6.3 Global existence

In this section, we prove that the flow exists globally. In order to prove that the solutionof (6.1) exists for all time (T = ∞), we refer to [81, Corollary 4.1], see also [17, Theorem3.1]. It is shown in the following Theorem 11.


Theorem 11. Let γ : S1× [0,T )→ R2 be a maximal solution of (6.1). If T < ∞, then∫γ

k2ds≥ c(T − t)−1/4.

Proof. Let m = 0 in (6.10) in Lemma 48 we have

ddt

∫γ

k2ds+∫

γ

k2ssds≤ c1

(∫γ

k2ds)5

−h(t) · c2

(∫γ

k2ds)2

, (6.11)

where h(t) =−∫

γk2

s ds2π

.Then we estimate two terms evolving h(t) above as follows,

(−h) ·(∫

γ

k2ds)2

=1

2π

∫γ

k2s ds(∫

γ

k2ds)−1

2(∫

γ

k2ds) 5

2

≤ ε

[∫γ

k2s ds ·

(∫γ

k2ds)−1

2]2

+1

16π2ε

(∫γ

k2ds)5

= ε

[−∫

γ

kkssds ·(∫

γ

k2ds)−1

2]2

+1

16π2ε

(∫γ

k2ds)5

≤ ε

[(∫γ

k2ds) 1

2

·(∫

γ

k2ssds) 1

2

·(∫

γ

k2ds)−1

2]2

+1

16π2ε

(∫γ

k2ds)5

= ε

∫γ

k2ssds+

116π2ε

(∫γ

k2ds)5

.

Substituting above estimate into (6.11), we obtain

ddt

∫γ

k2ds+[1− ε · c2]∫

γ

k2ssds≤

(c1 +

c2

16π2ε

)(∫γ

k2ds)5

.

Let ε small enough to have 1− ε · c2 ≥ 0, then

ddt

∫γ

k2ds≤ c(∫

γ

k2ds)5

,

here c = c1 +c2

16π2ε.

By integrating above inequality on [t, t], where t→ T ,


Thus, ∫γ

k2ds≥ limt→T

c(t− t)−1/4,

∫γ

k2ds≥ c(T − t)−1/4.

where c can be different from line to line. hence the result.

Corollary 1. Suppose γ : S1× [0,T )→ R2 solves (6.1) and satisfies the assumptions of

Theorem 10. Then, T = ∞.

Proof. Suppose on the contrary that γ satisfies the conditions of Proposition 20 and T <

∞. Then by the Theorem 11, ‖k‖22→ ∞ as t→ T . However,

Kosc = L(0)∫

γ

(k− k

)2 ds = L(0)‖k‖22−2k2π + k2L2(0) = L(0)‖k‖2

2−2π2,

then Kosc→ ∞ as t → T . This is in contradiction with Proposition 20. We conclude thatit must be the case that T = ∞.

By Lemma 47, Kosc ∈ L1 ([0,∞)) we conclude Kosc→ 0 as t → ∞ and therefore limitcurves are circles of circumference length L(0) under the conditions of Proposition 20.

6.4 Exponential Convergence

To complete the proof of Theorem 10, it remains to show that the limit circle is unique andconvergence is exponential. Since we have convergence to circle of radius L(0)

2π, we can be

sure that k(s, t) ∈[

π

L(0) ,3π

L(0)

]say for all t ≥ t1. For such times we also have ‖k‖∞ ≤ 3π

L(0) .

Lemma 50. Suppose γ : S1× [0,T )→R2 solves (6.1) and satisfies conditions of Theorem

10, we have T = ∞ and there exists absolute constants c,δ > 0 such that

‖kss‖22 ≤ ce−δ t .

Proof. Using Lemma 43 (vi),

ddt

∫γ

k2ssds = −2

∫γ

k2s4ds+2

∫γ

k2s3k2ds−4

∫γ

k3sskds

−3∫

γ

k2ssk

2s ds+7h

∫γ

k2s kds

= −2∫

γ

k2s4ds+2

∫γ

k2s3k2ds−4

∫γ

k3sskds

−3∫

γ

k2ssk

2s ds+7(−h)

∫γ

ks3kskds, (6.12)


then we calculate the second term in above,

∫γ

k2s3k2ds =

∫γ

k2s3

(k− k

)2 ds− k2∫

γ

k2s3ds+2k

∫γ

k2s3kds

≤ 12

∫γ

k2s3k2ds+ k

∫γ

k2s3ds+

∫γ

k2s3

(k− k

)2 ds,

and

∫γ

k2s3k2ds ≤ 2k2

∫γ

k2s3ds+2

∫γ

k2s3

(k− k

)2 ds

= −2k2∫

γ

ks4kssds+2∫

γ

k2s3

(k− k

)2 ds

≤ 2k2(

σ1

∫γ

k2s4ds+

14σ1

∫γ

k2ssds)+2‖ks3‖2

2Kosc

L(0)

= 2k2[

σ1

∫γ

k2s4ds+

14σ1

∫γ

ks4(k− k

)ds]+2‖ks3‖2

2Kosc

L(0)

≤ 2k2[

σ1

∫γ

k2s4ds+

14σ1

(σ2

∫γ

k2s4ds+

14σ2

Kosc

L(0)

)]+

2π

Kosc‖ks4‖22

= 2k2(

σ1 +σ2

4σ1

)∫γ

k2s4ds+2k2 1

16σ1σ2

Kosc

L(0)+

2π

Kosc‖ks4‖22.

Choosing σ1 =ε1

2k2 and σ2 =ε2

1k4 , we have

∫γ

k2s3k2ds ≤ 2

(ε1 +

Kosc

π

)∫γ

k2s4ds+

k8

4ε31

Kosc

L(0),

now we compute the third term in (6.12),

−4∫

γ

k3sskds = 4

∫γ

k2ssk

2s ds+8

∫γ

ks3ksskskds

≤ 8∫

γ

k2ssk

2s ds+4

∫γ

k2s3k2ds,

As we see several similar terms appear in above calculation, we put the second, thirdand fourth terms in (6.12) together,

2∫

γ

k2s3k2ds−4

∫γ

k3sskds−3

∫γ

k2ssk

2s ds

= 6∫

γ

k2s3k2ds+5

∫γ

k2ssk

2s ds.


The following estimation

5∫

γ

k2ssk

2s ds ≤ 5‖ks‖2

2

∫γ

k2ssds

≤ 5L(0)2π

(∫γ

k2ssds)2

=5L(0)

2π

[∫γ

ks4(k− k

)ds]2

≤ 5Kosc

2π

∫γ

k2s4ds.

yields

2∫

γ

k2s3k2ds−4

∫γ

k3sskds−3

∫γ

k2ssk

2s ds

≤ 6[

2(

ε1 +Kosc

π

)∫γ

k2s4ds+

k8

4ε31

Kosc

L(0)

]+

5Kosc

2π

∫γ

k2s4ds

=

(12ε1 +

292

Kosc

π

)∫γ

k2s4ds+

3k8

2ε31

Kosc

L(0).

Moving on to the last term in (6.12),

7(−h)∫

γ

ks3kskds ≤ 7(−h)L(0)2π

√L(0)2π‖k‖∞

∫γ

kssds ·∫

γ

ks4ds

≤ 14ε2

∫γ

k2s4ds+ ε2 ·49h2

(L(0)2π

√L(0)2π

)2

‖k‖2∞

∫γ

k2ssds

≤ 14ε2

∫γ

k2s4ds+

ε2

4

∫γ

k2s4ds+ ε2

(49h2L3(0)

8π3

)2

‖k‖4∞Kosc.

Thus, we get

ddt

∫γ

k2ssds = −2

∫γ

k2s4ds+2

∫γ

k2s3k2ds−4

∫γ

k3sskds−3

∫γ

k2ssk

2s ds

+7(−h)∫

γ

ks3kskds

≤(−2+12ε1 +20

Kosc

π

)∫γ

k2s4ds+

3k8

2ε31 L(0)

Kosc +1

4ε2

∫γ

k2s4ds

+ε2

4

∫γ

k2s4ds+ ε2

(49h2L3(0)

8π3

)2

‖k‖4∞Kosc,

ddt

∫γ

k2ssds ≤

(−2+12ε1 +

292

Kosc

π+

14ε2

+ε2

4

)∫γ

k2s4ds

+

[ε2

(49h2L3(0)

8π3

)2

‖k‖4∞ +

384ε3

1 L9(0)

]Kosc.


When Kosc ≤ K∗, we can have −2+12ε1 +292

Koscπ

+ 14ε2

+ ε24 =−c(L(0))≤ 0.

After enough time, we have ‖k‖2∞ ≤ 2

(2wπ

L(0)

)2≤C and from Lemma (45), we know h2

is bounded, thus,

ddt

∫γ

k2ssds≤−c(L(0))

π4

L4(0)‖kss‖2

2 + cKosc =−δ‖kss‖22 + cKosc,

as Kosc ∈ L1([0,∞)), we apply Grönwall’s inequality in Proposition 7 to obtain∫γ

k2ssds≤ ce−δ t

Exponential decay of the higher curvature derivatives follows by interpolation usingthe uniform bounds on

∫γ

k2smds in Lemma 49.


10, So ‖ksl‖22 and ‖ksl‖2

∞ exponentially decays for any l ∈ N.

Proof. As we have proved that when t→ ∞,∫

γk2

ssds exponentially decays in Lemma 50,we can obtain

∫γ

k2s3ds =−

∫γ

kssks4ds≤(∫

γ

k2ssds) 1

2(∫

γ

k2s4ds

) 12

as(∫

γk2

ssds) 1

2 ≤(

e−δ t ‖kss‖22

∣∣t=0

) 12 , and

(∫γ

k2s4ds

) 12 is bounded in L2 in Lemma 49,

then∫

γk2

s3ds exponentially decays.By induction argument, we assume that

∫γ

k2sl ds exponentially decays,

∫γ

k2sl+1ds =−

∫γ

ksl ksl+2ds≤(∫

γ

k2sl ds) 1

2(∫

γ

k2sl+2ds

) 12

as(∫

γk2

sl+1ds) 1

2 is bounded in Lemma 49, then∫

γk2

sl+1ds exponentially decays.

Thus, we have∫

γk2

sl ds exponentially decays in L2. As ‖ksl‖2∞ ≤ L

2π‖ksl+1‖2

2, we have inL∞ norm decays exponentially.




‖∂ul γ‖∞ ≤ c(l)+l

∑p=0‖∂sl γ0‖∞,



Proof. We claim that for l, p ∈ N0, l ≥ 1,

∂t∂sl γ = ν

l

∑p=0

(P2+p

1+l−p(k)+Ppl−p(k)

∫γ

P22 (k)ds

)

+τ

l

∑p=0

(P2+p


∫γ

P22 (k)ds

). (6.13)

We prove this by induction. First, we have ∂tγ = Fν =[P2

1 (k)+∫

γP2

2 (k)ds]

ν , we do thedifferentiation

∂t∂sl+1γ

= kF ·∂sl+1γ +∂s(∂t∂sl γ)

= kF ·∂sl τ +∂s

[ν

l

∑p=0

(P2+p


∫γ

P22 (k)ds

)]

+∂s

[τ

l

∑p=0

(P2+p


∫γ

P22 (k)ds

)]

= kFν ∑p+q=l

Ppq (k)+ν

l

∑p=0

(P2+p

2+l−p(k)+Pp1+l−p(k)

∫γ

P22 (k)ds

)

+kFτ ∑p+q=l

Ppq (k)+ τ

l

∑p=0

(P2+p


∫γ

P22 (k)ds

)

+ν

l

∑p=0

(P3+p

1+l−p(k)+P1+pl−p (k)

∫γ

P22 ds)+ τ

l

∑p=0

(P3+p

1+l−p(k)+P1+pl−p (k)

∫γ

P22 ds)

= ν ∑p+q=l

(Pp+2

q+2 (k)+Ppq+1(k)

∫γ

P22 ds)+ν

l

∑p=0

(P2+p


∫γ

P22 ds)

+ν

l+1

∑p=1

(P2+p


∫γ

P22 ds)+ τ ∑

p+q=l

(Pp+2

q+2 (k)+Ppq+1(k)

∫γ

P22 ds)

+τ

l

∑p=0

(P2+p


∫γ

P22 ds)+ τ

l+1

∑p=1

(P2+p


∫γ

P22 ds)

= ν

l+1

∑p=0

(P2+p


∫γ

P22 ds)+ τ

l+1

∑p=0

(P2+p


∫γ

P22 ds)

as required. Integrating (6.13) and using Lemma 51, we find

‖∂sl γ‖∞ ≤ ‖∂sl γ0‖∞ + c∫ t

0e−ct ′dt ′ ≤ ‖∂sl γ0‖∞ + c(l).

As u is the initial space parameter before reparameterization by arc-length, set v = |∂uγ|.


Referring to the proof of Theorem 3.1 in [17], then for any function Φ : S→ R, we have

∂luΦ = v∂

ls Φ+Pl(v, ...,∂ l−1

u v,Φ, ...∂ l−1s Φ)


‖∂ul γ‖∞ ≤ ‖∂sl γ0‖∞ + c(l).

as required.

Using Lemma 50 with Lemma 6.1 gives in turn exponential decay of∫

γk2

s ds, applyingthe PSW inequality (2.3) in Proposition 3 in the first tep in the following estimate,

‖ks‖22 ≤

L2

4π2‖kss‖22 ≤

L2(0)4π2 ‖kss‖2

2 ≤L2(0)4π2 e−δ t ‖kss‖2

2∣∣t=0 ≤ ce−δ t ,

then as h =−∫

γk2

s ds2π

, hence we get the decay for h(t).Also

∫γ

(k− k

)2 ds decays exponentially as follows,

∫γ

(k− k

)2 ds≤ L2

4π2

∫γ

k2s ds≤ ce−δ t ,

and the corresponding L∞ norms,

‖k− k‖2∞ ≤

L(0)2π‖k− k‖2

2 ≤ ce−δ t .

This implies subconvergence of the flow to circles with perimeter length L(0).


10, there exist a subsequence of time t j such that

γ(·, t j)→ γ∞

uniformly with γ∞ a circle.

Corollary 2. Suppose γ : S1× [0,T )→ R2 solves our flow and satisfies the assumptions

of Theorem 10. then the curve diffusion flow exists globally (T = ∞) and converges expo-

nentially fast to a round circle with radius L(0)2π

.

6.5 The unique limiting image

Lemma 53. The image of limiting round circle is unique.




∫γ

|γ− γ∞|2ds.


lims j→∞

G[ f (·,s j)] 6= 0. (6.14)


G∣∣∣∣ =

ddt

∫γ

|γ− γ∞|2 ds

= 2∫

γ


∫γ

|γ− γ∞|2 ·ddt

ds

= 2∫

γ

F · |γ− γ∞|ds+∫

γ

kF · |γ− γ∞|2 ds

=∫

γ

F · |γ− γ∞| · (2+ k |γ− γ∞|)ds

≤(∫

γ

F2ds) 1

2

·[∫

γ

|γ− γ∞|2 (2+ k |γ− γ∞|)2 ds] 1

2

≤(∫

γ

F2ds) 1

2

·[∫

γ

|γ− γ∞|2(

4+ k2 |γ− γ∞|2)

ds] 1

2

≤ c(∫

γ

F2ds) 1

2

·(

4∫

γ

|γ− γ∞|2 + k2 |γ− γ∞|4 ds) 1

2

≤ c(∫

γ

F2ds) 1

2

·(∫

γ

|γ|2ds+∫

γ

|γ∞|2ds+∫

γ

|k|2 · |γ|4ds+∫

γ

|k|2 · |γ∞|4ds) 1

2

Now we show that∫

γ|γ|2ds,

∫γ|γ∞|2ds,





‖γ(·, t)‖2∞ =

(∫∞

t|F |dt

)2

≤ c(∫

∞

te−δ tdt

)2

≤ cδ

e−δ t ≤ c1,

then ∫γ

|γ|2ds≤ ‖γ‖2∞

∫γ

ds≤ c1L(0)≤ c2,∫γ

|k|2 · |γ|4ds≤ ‖γ‖4∞

∫γ

|k|2ds≤ c3.


γ|γ∞|2ds and




dtG∣∣∣∣≤ c‖F‖2,


G[ f (·,s j)]≤ c∫

∞

s j

‖F‖2dt ≤ c∫

∞

s j


it follows thatlim

s j→∞G[ f (·,s j)] = 0,

which is in contradiction with (6.14).This proves that there does not exist a sequence s j, the convergence of flow in L2 to

a straight line segment is unique. We can obtain the exponential convergence in C∞ to aunique line segment as the curvature and all its derivatives exponentially decay.


Therefore, we obtain the limiting curve image is unique.Exponential decay of the speed allows us to compute the bounded distance between the

limit solution and initial curve γ0. For any x ∈ [−1,1], we have

|γ(x, t)− γ(x,0)|=∣∣∣∣∫ t

0

∂γ

∂ t(x,τ)

∣∣∣∣≤ ∫ t

0|h(τ)−κss|dτ ≤ c

δ

(1− e−δ t

)≤ c

δ.

It follows that solution converge exponentially to a translate of the circle, with no cor-rection for translations.

6.6 Self-similar solutions

A self-similar solution to a curvature flow equation such as (6.1) is a solution whoseimage maintains the same shape as it evolves. Its image changes in time only by trans-lation, scaling and rotation. In our setting, the constrained length rules out expandingand contracting self-similar solutions, so we focus on stationary solutions, translators androtators. Here we refer to [20].

Lemma 54. The only smooth, closed stationary solutions to (6.1) are multiply-covered

circles.

Proof. Such solutions satisfyh(t)− kss ≡ 0,


which is 12πw

∫γ

k2s ds− kss ≡ 0.

Integrating this equation over γ , we obtain

L(0)2π

∫γ

k2s ds≡ 0,

so these closed curves have ks ≡ 0 and are circles, where the length is controlled by theinitial length L(0).

A family of curves γ : S1× [0,T )→ R2 evolving purely by translation satisfies

γ(u, t) = γ0(u)+Vt + γs(u, t)φ(u, t)

for some constant vectors V and smooth diffeomorphism φ . As φ(u, t) is the tangentialdiffeomorphism, then ⟨

γsddt

φ ,ν

⟩= 0.

In this case, if ∂γ

∂ t (u, t) = Fν = [−kss +h(t)] ·ν(u, t),⟨∂γ

∂ t,ν

⟩= 〈V,ν〉,

then γ must satisfies−kss +h(t)≡ 〈V,ν〉.

We call the solution γ a translator. If V = 0, then the solution γ is stationary, a trivialtranslator.

Lemma 55. Let γ : S1→ R2 be a smooth, closed, translating solution to (6.1). Then γ is

trivial; that is, γ(S1) is a standard round circle.

Proof. Integrating −kss +h(t)≡ 〈V,ν〉 by k gives

∫γ

[−kss +h(t)]ds =∫

γ

〈V,ν〉ds.

As V is a constant vector and the curve is closed, we have∫γ

〈V,ν〉ds = 0,

and ∫γ

−kss +h(t)ds =∫

γ

h(t)ds = h(t)∫

γ

ds = h(t)L(0),

then we haveh(t)L(0) =

∫γ

〈V,ν〉ds = 0.


As L(0) 6= 0, then h(t) = 0 and we have

−kss ≡ 〈V,ν〉.

Integrating by parts, as γss = kv, we obtain∫γ

k2s ds = −

∫γ

kkssds =∫

γ

〈V,kν〉ds

=∫

γ

〈V,γss〉ds =∫

γ

∂s〈V,γs〉ds

= 0

yields ks ≡ 0 on γ which gives the curvature is constant and γ must be a round circle.

Lemma 56. Let γ : S1→ R2 be a smooth, closed, rotating solution to (6.1). Then γ(S1)

is a standard round circle.

Proof. A family of curves γ : S× [0,T )→ R2 evolving purely by rotation satisfies

−kss +h(t)≡ 2S(t)〈γs,γ〉.

Since γ is closed, integrating −kss +h(t)≡ 2S(t)〈γs,γ〉 by k gives

∫γ

−kss +h(t)ds =∫

γ

2S(t)〈γs,γ〉ds = 2S(t)∫

γ

〈γs,γ〉ds

As∫

γ〈γs,γ〉ds= 1

2∫

γ∂s|γ|2ds= 0 and

∫γ−kss+h(t)ds=

∫γ

h(t)ds= h(t)∫

γds= h(t)L(0),

we haveh(t)L(0) = 2S(t)

∫γ

〈γs,γ〉ds = 0.

As L(0) 6= 0, then h(t) = 0 and we have

−kss ≡ 2S(t)〈γs,γ〉.

Integrating by parts, as 〈ν ,γs〉= 0, we obtain∫γ

k2s ds = −

∫γ

kkssds = 2S(t)∫

γ

〈kγs,γ〉ds

= −2S(t)∫

γ

〈νs,γ〉ds = 2S(t)∫

γ

〈ν ,γs〉ds

= 0

yields ks ≡ 0 on γ , which gives us that the curvature is constant. Thus, we conclude thatγ is a round circle.


6.7 Embeddedness

The result that solutions with sufficiently small oscillation of curvature remain embeddedfor all time applies by referring to [82, Theorem 1.6] (see also [51, Theorem 6]). Thistheorem is showed as follows.

Theorem 12. Suppose γ : S1→ R2 is a smooth immersed curve with winding number w

and let m denote the maximum number of times γ intersects itself in any one point; that is∫γ

kds = 2ωπ and m(γ) = supx∈R2

∣∣γ−1(x)∣∣ .

Then,

Kosc(γ)≥ 16m2−4ω2π

2.

Proposition 25. Any solution of (6.1) with initial embedded curve γ0 satisfying the as-

sumption of Theorem 10 remains embedded for all time.

Proof. First we find the point x0 where the curve γ intersects itself the most. Let m denotethe maximum number of times γ intersects itself in any one point, m(γ)= supx∈R2 |γ−1(x)|.We know that γ−1(x0) = m. When m = 2 the curve γ has one intersection.

It is proven in Proposition 20 that Kosc ≤ 2K∗. Let m = 2, by assumption Kosc(0)< K∗

in Theorem 10, we have 2K∗ < 16 ·22−4π2 = 64−4π2, which means that the curve doesnot have any intersection.

As the initial curve is embedded, thus the curve γ is an embedded curve for each t.

Chapter 7

Constrained ideal curve flow

Let us define the energy for the smooth closed planar curve γ : S1× [0,T )→ R2 by

E[γ] =12

∫γ

k2s ds.

We define the constrained ideal curve flow as follows,∂tγ =

(ks4 + k2kss− 1

2kk2s +h(t)

)ν , f or all (s, t) ∈ S1× [0,T )


Denote that ν is the normal vector field of γ . We let F := G+ h(t) = ks4 + k2kss−12kk2

s + h(t). If let the curve move under the suitable six-order curvature flow, then wecan control the length and the area of this closed curve. In this chapter, we will study twoconstrained ideal curve flow: the length-constrained ideal curve flow and area-preservingideal curve flow.

7.1 Length-constrained ideal curve flow

Suppose the curve evolves by the sixth-order curvature flow∂tγ =

(ks4 + k2kss− 1

2kk2s +h1(t)

)ν , f or all (s, t) ∈ S1× [0,T )

γ|t=0 = γ0, f or all s ∈ S1 (7.1)

To preserve length of the evolving curve γ we take

h1(t) =1

2πω

(−∫

γ

k2ssds+

72

∫γ

k2s k2ds

)(7.2)

where w denotes the winding number of γ .Our main theorem is as follows:

136

CHAPTER 7. CONSTRAINED IDEAL CURVE FLOW 137

Theorem 13. Suppose γ0 :R→R2 be a regular smooth immersed closed curve with fixed

length L(0) and ω(0) = 1, for a small absolute constant ε , satisfies condition

E(0)≤ ε

the length-constrained ideal curve flow (7.1) with initial data γ0 exists for all time and

converges exponentially to a round circle with radius L(0)2π

.

The short time existence is shown in Chapter 3, Theorem 5. From Lemma 4, we canget that ω(t) = ω(0) = 1.

We show the curve remains embedded with the smallness condition in Theorem 13.


sumptions of Theorem 13 remains embedded for all time.

Under the assumption, we have the ε is small enough to satisfy the condition in [82,Theorem 1.6]. This theorem is showed as Theorem 12 in Chapter 6.

The structure of the proof of Theorem 13 is as follows. Firstly, in section 7.1.1 we focuson estimating the energy which appears to be decaying exponentially. With this result inhand, in section 7.1.2 we move on to the analysis of the global behaviour of the flow. Weprove that the curve exists for all time and converges exponentially fast to a round circleunder the stated conditions. Finally, we show this round circle is unique, then completethe proof.

Lemma 57. Under the flow (7.1), while a solution exists, it satisfies

L(t) = L(0).

Proof. We compute

ddt

L(t) =∫

γ

kFds

=∫

γ

kks4ds+∫

γ

k3kssds− 12

∫γ

kk2s ds+h1(t)

∫γ

kds

=∫

γ

k2ssds− 7

2

∫γ

k2s k2ds+2πh1(t)

As h1(t) = 12π

(−∫

γk2

ssds+ 72∫

γk2

s k2ds)

in (7.2), the result follows.

Here, we establish a suitable bound on h in terms of the L2 norms of k.

Lemma 58. For a constant c > 0, the global term h1(t) may be estimated as

h1(t)≤ c(∫

γ

k2ds)5

.


Proof. By using interpolation inequality in Lemma 4, we estimate (7.2)

h1(t) =1

2π

(−∫

γ

k2ssds+

72

∫γ

k2s k2ds

)≤ 1

2π

[−∫

γ

k2ssds+ ε

∫γ

k2ssds+ c

(∫γ

k2ds)5]

≤ c(∫

γ

k2ds)5


Now we show a useful proposition.

Proposition 27. Let γ0 : S→ R2 be a smooth immersed curve. Then there exist universal

constants c and ε1 = ε1(L(0))> 0 depending only on w such that

E(0)< ε1 =⇒∫

γ

G2ds≥ cL−6E.

For the proof of above proposition see [1, Proposition 7.2]. We also put the proof inAppendix A.2.

Next we prove that the energy decays exponentially by using proposition 27.

Proposition 28. There exists an ε2 = ε2(L(0)) such that

E(0)< ε2 =⇒ ddt

E(t)≤−c2

∫γ

k2s4ds.

Proof. From Lemma 57, we have h1 =− 12π

∫γ

kGds, then

ddt

E =−∫

γ

G2ds+1

2π

∫γ

kGds∫

γ

Gds,

Let us first note the estimate∣∣∣∣∫γ

Gds∣∣∣∣≤ 5

2‖k‖∞

∫k2

s ds

≤ 5E(‖k− k‖∞ +

2π

L

)≤ 5E

[(L2(0)E

2π2

)1/2

+2π

L(0)

]:= f1(E,L(0)) .

Using this we find


12π

∫γ

Gds∫

γ

kGds≤ f1(E,L(0))2π

∫γ

kGds

≤ 12

∫γ

G2 ds+[ f1(E,L(0))]

2

8π2

∫γ

k2 ds

≤ 12

∫γ

G2 ds+[ f1(E,L(0))]

2

8π2

(L2(0)2π2 E +

4π2

L(0)

).

Now observe that the second expression satisfies

[ f1(E,L(0))]2

8π2

(L2(0)2π2 E +

4π2

L(0)

)≤ c0E2(1+E2) .

Recall the estimate ∫γ

G2 ds≥ c∫

γ

k2s4 ds− c1E2(1+E3)

from the estimate (A.3) in the proof of Proposition 27, ε2 ≤ ε1. Combining these, we find

ddt

E ≤−c∫

γ

k2s4 ds+ c2E2(1+E3) .

Now observe that

E ≤ 12

(L(0)2π

)6 ∫γ

k2s4 ds

which impliesddt

E ≤∫

γ

k2s4 ds

(c3E(1+E3)− c

).

Therefore, for E(0) small enough depending only on L(0) and ω , we have

ddt

E ≤−c2

∫γ

k2s4 ds ,

as required.

Proposition 28 implies both uniform L1-control on ‖ks4‖22 as well as exponential decay

of the ideal curve energy.

Corollary 3. Under the conditions of Proposition 28, we have

E(t)≤ E(0)e−δ t

where δ = δ (L(0)) and ∫ T

0‖ks4‖2

2(t)dt ≤ 2c

E(0) .


7.1.2 Global Existence

Let us now give two useful lemmas before showing uniform bounds for all derivatives ofcurvature.

Lemma 59. For any immersed curve γ : S→ R2 we have the estimate

‖k‖∞ ≤ 2L(0)12 E(0)+

2π

L(0):=C0.

where C0 only depends on L(0) and E(0).

Proof. We calculate k = k− k + k ≤∫

γ|ks|ds+ 2π

L . Taking a supremum and using theHölder inequality, we find

‖k‖∞ ≤ 1L

(√L3‖ks‖2

2 +2π

)≤ 2L(0)

12 E(0)+

2π

L(0).

Lemma 60. Under the flow (7.1), for E(0) sufficiently small, there is a corresponding

constant such that

|h(t)| ≤ ch,

where ch only depends on L(0) and E(0).

Proof. We have from Lemma 3 (v) and Proposition 4

ddt

∫γ

k2s`ds ≤ −2

∫γ

k2sl+3ds+2

∫γ

ksl

[Pl+4

3 (k)+Pl+25 (k)

]−h(t)

∫γ

k2sl kds

≤ (−2+ ε)∫

γ

k2s`+3ds+ c(ε)

(∫γ

k2ds)2`+7

−h(t)∫

k2s`k ds, (7.3)

For l = 2, we have

ddt

∫γ

k2ssds ≤ −(2− ε)

∫γ

k2s5ds+ c

(∫γ

k2ds)11

+1

2π

∫γ

k2ssds

∫γ

kk2ssds

− 74π

∫γ

k2k2s ds

∫γ

kk2ssds

Using Lemma 59 and Hölder inequality, we estimate


12π

∫γ

k2ssds

∫γ

kk2ssds ≤ C0

2π

(∫γ

k2ssds)2

=C0

2π

(−∫

γ

ksks3

)2

≤ C0

πE(t)

∫γ

k2s3ds≤ C0E(0)L4(0)

16π5

∫γ

k2s5ds

− 74π

∫γ

k2k2s ds

∫γ

kk2ssds≤

7C30

2πE(t)

∫γ

k2ssds≤

7C30E(0)L6(0)

128π7

∫γ

k2s5ds

Hence

ddt

∫γ

k2ssds ≤ −

(2− ε−C0E(0)L4(0)

16π5 −7C3

0E(0)L6(0)128π7

)∫γ

k2s5ds+ c

(∫γ

k2ds)11

Assuming E(0) is small enough, we can get that∫

γk2

ssds is bounded,∫

γk2

ssds≤ c2, herec2 only depends on L(0) and E(0).

We estimate from (7.2)

|h(t)| ≤ 12π

∫γ

k2ssds+

74π

∫γ

k2k2s ds

≤ 12π

∫γ

k2ssds+

74π‖k‖2

∞

∫γ

k2s ds

≤ 12π

c2 +7

4πC0E(0) = ch

where ch only depends on L(0) and E(0).

Proposition 29. Under the flow (7.1), for E(0) sufficiently small, there are corresponding

constants such that ∫γ

k2s`ds≤ cl

Proof. For ` = 0 we have the uniform estimate via the inequality in Proposition 3 andCorollary 3: ∫

γ

k2 ds =∫

γ

(k− k)2 ds+4π2

L(0)≤ L2(0)

2π2 E(t)+4π2

L(0).

For ` = 1, it is the first estimate is in Corollary 3. For each ` ≥ 2 we have from (7.3)and using Proposition 4, Proposition 3 and Lemma 60,


ddt

∫γ

k2s`ds ≤ (−2+ ε)

∫γ

k2s`+3ds+ c(ε)

(∫γ

k2ds)2`+7

−h(t)∫

γ

k2s`kds

≤ (−2+ ε)∫

γ

k2s`+3ds+ c(ε)

(∫γ

k2ds)2`+7

+ |h(t) |∣∣∣∣∫

γ

k2s`kds

∣∣∣∣≤ (−2+2ε)

∫γ

k2s`+3ds+ c(ε)

(∫γ

k2ds)2`+7

+ c(ε)(∫

γ

k2ds)2`+2

.

Thus we get the conclusion∫

k2sl ds≤ cl .

Proposition 30. Under the flow (7.1), with E(0) < ε , we have that there exist constants

δ` depending only on L0 and ω such that

∫γ

k2s`ds≤ e−δ`t

∫γ

k2s`ds∣∣∣∣t=0

,

for `≥ 1, t ∈ [0,T ]. We can also have ksl exponentially decays in L∞.

Proof. Note that ε satisfies the requirements in Proposition 28 and Lemma 60.When t→ ∞,

∫γ

k2s ds exponentially decays in Corollary 3

Kosc ≤L3(0)4π2

∫γ

k2s ds≤ ce−δ t ,

then Kosc exponentially decays.

∫γ

k2ssds =

∫γ

(k− k)ks4ds≤[∫

γ

(k− k)2ds]1/2(∫

γ

k2s4ds

)1/2

,

as∫

γk2

s4ds is bounded in L2 in Proposition 29, then∫

γk2

ssds exponentially decays.We let l ≥ 1, ∫

γ

k2sl+1ds≤

(∫γ

k2s ds)1/2(∫

γ

k2s2l+1ds

)1/2

,

as∫

γk2

s2l+1ds is bounded in L2, then∫

γk2

sl+1ds exponentially decays in L2. Also ‖ksl‖2∞ ≤

L2π‖ksl+1‖2

2 So we have ksl exponentially decays in L∞.

We have that all curvature derivatives remain bounded in L2 and L∞ from above propo-sition. That these bounds are independent of T implies T = ∞, then we get the followingcorollary.

Corollary 4. Suppose γ : S1× [0,T )→ R2 solves (7.1) and satisfies the conditions of

Theorem 13. Then T = ∞.


From proposition 30, we can also obtain uniform bounds for all derivatives of the evolv-ing curve γ .



‖∂ul γ‖∞ ≤ c(l)+l

∑p=0‖∂sl γ0‖∞,


Proof. We claim that for l, p ∈ N0, l ≥ 1,

∂t∂sl γ = ν

l

∑p=0

[P4+p

1+l−p(k)+P2+p3+l−p(k)+Pp

l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

+τ

l

∑p=0

[P4+p


l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]. (7.4)

We prove this by induction. First, ∂tγ =Fν =

P41 (k)+P2

3 (k)+∫

γ

[P4

2 (k)+P24 (k)

]ds

ν ,for q ∈ N0, we do the differentiation

∂t∂sl+1γ = kF ·∂sl+1γ +∂s(∂t∂sl γ)

= kF ·∂sl τ

+∂s

[ν

l

∑p=0

(P4+p


l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds)]

+∂s

[τ

l

∑p=0

(P4+p


l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds)]

= kFν ∑p+q=l

Ppq (k)+ kFτ ∑

p+q=lPp

q (k)

+ν

l

∑p=0

[P4+p


1+l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

+ν

l

∑p=0

[P5+p

1+l−p(k)+P3+p3+l−p(k)+P1+p

l−p (k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

+τ

l

∑p=0

[P4+p


1+l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

+τ

l

∑p=0

[P5+p

1+l−p(k)+P3+p3+l−p(k)+P1+p

l−p (k)∫

γ

(P4

2 (k)+P24 (k)

)ds]


= ν ∑p+q=l

[Pp+4

q+2 (k)+Pp+2q+4 (k)+Pp

q+1(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

+ν

l

∑p=0

[P4+p


1+l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

+ν

l+1

∑p=1

[P4+p


1+l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

+τ ∑p+q=l

[Pp+4

q+2 (k)+Pp+2q+4 (k)+Pp

q+1(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

+τ

l

∑p=0

[P4+p


1+l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

+τ

l+1

∑p=1

[P4+p


1+l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

= ν

l+1

∑p=0

[P4+p


l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

+τ

l+1

∑p=0

[P4+p


l−p(k)∫

γ

(P4

2 (k)+P24 (k)

)ds]

as required. Integrating (7.4) and using Proposition 30, we find

‖∂sl γ‖∞ ≤ ‖∂sl γ0‖∞ + c∫ t

0e−ct ′dt ′ ≤ ‖∂sl γ0‖∞ + c(l).

As u is the initial space parameter before reparameterization by arc-length, set v = |∂uγ|.Referring to the proof of Theorem 3.1 in [17], then for any function Φ : S→ R, we have

∂luΦ = v∂

ls Φ+Pl(v, ...,∂ l−1

u v,Φ, ...∂ l−1s Φ)


‖∂ul γ‖∞ ≤ ‖∂sl γ0‖∞ + c(l).

as required.

In order to complete the proof of Theorem 13 it remains to show that the limit circle isunique. Here we refer to [1, Theorem A.1] to conclude full convergence of the flow. Thistheorem is stated in Appendix A.4, Theorem 19.

Uniform boundedness of γ and all its derivatives in Proposition 31 imply the first hy-pothesis of [1, Theorem A.1] is satisfied. For the second hypothesis, using Proposition 30we note that


∫γ

F2ds≤ c1e−c2t .

This implies ∫ T

0

(∫γ

F2ds) 1

2

dt ≤ c1

∫ T

0e−

c22 tdt ≤ c

where c is a constant depending only on w, E(0) and L. For the third hypothesis, uniformboundedness of all derivatives of γ in Proposition 31 gives that for any sequence t j→ ∞,the C∞-norm of γ(·, t j) is uniformly bounded. We have the exponential decay of theenergy, so E[γ(·, t j)]→ 0, which implies that a subsequence γ(·, t j) converges to a smoothcircle in C∞-topology. Therefore, we conclude full convergence of the flow.

Exponential decay of the speed allows us to bound the region of the plane in which thesolution lies relative to γ0 via standard argument. We may bound the distance travelled byany point on the initial curve γ0 as follows

|γ(x, t)−γ(x,0)|=∣∣∣∣∫ t

0

∂γ

∂ t(x,v)

∣∣∣∣≤ ∫ t

0

∣∣∣∣ks4 + k2kss−12

kk2s +h1(t)

∣∣∣∣dv≤ cσ

(1− e−σt)≤ c

σ.

7.2 Area-preserving ideal curve flow

For the smooth closed planar curve γ : S1× [0,T )→ R2 by energy E[γ] = 12∫

γk2

s ds. Wesuppose the curve evolves by the sixth-order curvature flow

∂tγ = (ks4 + k2kss− 1

2kk2s +h2(t))ν , f or all (s, t) ∈ S1× [0,T )

γ|t=0 = γ0, f or all s ∈ S1 (7.5)

To preserve area of the evolving curve γ we take

h2(t) =5

2L

∫γ

kk2s ds (7.6)

where L denotes the length of curve γ ,The main theorem is as follows:

Theorem 14. Suppose γ0 : S1 → R2 be a regular smooth immersed closed curve with

constant area A(0)> 0 and ω(0) = 1,

E(0)≤ ε,

the area-preserving ideal curve flow (7.5) with initial data γ0 exists for all time and con-


verges exponentially to a round circle with radius√

A/π .

In above theorem, ω denotes the winding number of curve (7.5). From Lemma 4, wecan get the winding number ω(t) = ω(0) = 1. In order to prove Theorem 14, we need tohave short time existence of a solution to (7.5), see Theorem 6 in Chapter 3. We show thecurve remains embedded with the smallness condition in Theorem 14.



This section is organised as follows. In section 7.2.1, we show some preliminaries.In finite time, the energy decreases, the length is bounded. We also give the estimatefor h2(t) term. With these results in hand, in section 7.2.3 we prove that the energydecays exponentially. Then, in section 7.2.2 we show that the curve exists for all time andconverges exponentially fast to a round circle under the stated conditions, completing theproof of Theorem 14.

Lemma 61. Under the flow (7.5), while a solution exists, it satisfies

A(t) = A(0).

Proof. We compute

ddt

A =−∫

γ

Fds =52

∫γ

kk2s ds−h2(t)L = 0

As h2(t) == 52L∫

γkk2

s ds in (7.6), the result follows.

7.2.1 Preliminaries

Lemma 62. Under the flow (7.5), E is monotone decreasing.

Proof. We have F = G+h2(t) = ks4 +k2kss− 12kk2

s +h2(t). As h2(t) = 52L∫

γkk2

s ds, which

is also h2(t) =− 1L∫

γGds. Because

∫γ

Gds≤(∫

γG2ds

) 12 ·L 1

2 , we have

ddt‖ks‖2

2 = −∫

γ

(G+h)Gds =−∫

γ

G2ds−h∫

γ

Gds

= −∫

γ

G2ds+1L

∫γ

Gds∫

γ

Gds

≤ −∫

γ

G2ds+1L

L∫

γ

G2ds = 0


The∫

γk2

s ds is monotone decreasing.

∫γ

k2ds =∫

γ

(k− k)2ds+2k∫

γ

kds− k2ds

=∫

γ

(k− k)2ds+8π2

L− 4π2

L=∫

γ

(k− k)2ds+4π2

L

≤ L2

4π2

∫γ

k2s ds+

4π2

L≤ L2

4π2 E0 +4π2

L

We know from above that∫

γk2ds is bounded for finite time. Also

Kosc ≤L3

4π2‖ks‖22 ≤

L3

4π2 E0.

Lemma 63. Under the flow (7.5), L is bounded in finite time.

Proof. As

k = k− k+ k ≤∫

γ

|ks|ds+2π

L,

by Hölder inequality, we have

‖k‖∞ ≤√

2LE +2π

L.

Then, we estimate the evolution of L by using above estimate,

ddt

L = −∫

γ

kFds =−∫

γ

k2ssds+

72

∫γ

k2k2s ds− 5π

L

∫γ

kk2s ds

≤ −4π2

L2

∫γ

k2s ds+

72‖k‖2

∞

∫γ

k2s ds+

(5π

L

)2 ∫γ

k2k2s ds+

14

∫γ

k2s ds

≤ −4π2

L2

∫γ

k2s ds+

72‖k‖2

∞

∫γ

k2s ds+

25π2

L2 ‖k‖2∞

∫γ

k2s ds+

14

∫γ

k2s ds

=

[−4π2

L2 +72‖k‖2

∞ +25π2

L2 ‖k‖2∞ +

14

]∫γ

k2s ds

≤

[−4π2

L2 +

(72+

25π2

L2

)(√L‖ks‖2 +

2π

L

)2

+14

]∫γ

k2s ds

≤[−4π2

L2 +

(72+

25π2

L2

)(L‖ks‖2

2 +4π2

L2 +4π

L1/2‖ks‖2

)+

14

]∫γ

k2s ds

= −4π2‖ks‖2

2L−2 +72‖ks‖4

2 ·L+14π2‖ks‖2

2L−2 +14π‖ks‖2‖ks‖22L−

12 +25π

2‖ks‖42L−1

+100π4‖ks‖2

2L−4 +100π3‖ks‖2 · ‖ks‖2

2L−52 +

14‖ks‖2

2


≤ 72‖ks‖4

2 ·L+10π2‖ks‖2

2L−2 +14π‖ks‖32L−

12 +25π

2‖ks‖42L−1 +100π

4‖ks‖22L−4

+100π3‖ks‖3

2L−52 +

14‖ks‖2

2

=

(72‖ks‖4

2 +10π2‖ks‖2

2L−3 +14π‖ks‖32L−

32 +25π

2‖ks‖42L−2 +100π

4‖ks‖22L−5

+100π3‖ks‖3

2L−72 +

14‖ks‖2

2L−1)·L (7.7)

On the right-hand side of above estimate, we can see that the powers of L are all nega-tive. As the evolving closed curve has preserved area A, then the length is bounded belowby L := 2

√πA. Also the energy is monotone decreasing, we obtain We obtain

ddt

L≤ c ·L (7.8)

where c only depending on E(0), L.From (7.8), we estimate the evolution equation of lnL,

ddt

lnL = L−1 ddt

L≤ c.

Integrating on both sides,

lnL(t)≤ lnL(0)+ ct

which isL(t)≤ L(0)ect .

Thus, we obtain that length is bounded in finite time.

In the following lemma, we estimate the global term h2(t).

Lemma 64. For any constants c1,c2 > 0, the global term h2(t) may be estimated as

|h2(t)| ≤ c1L−358 ‖k‖3

2 + c2L−158 ‖k‖

742 ‖kss‖

542

Proof. By using interpolation inequality,

|h2(t)| =5

2L·∣∣∣∣∫

γ

kk2s ds∣∣∣∣= 5

2L

∫γ

|P23 (k)|ds≤ 5

2LcL1−3−2 · ‖k‖

742 · ‖k‖

542,2

=5

2LcL−4‖k‖

742

(L

58‖k‖

542 +L

258 ‖kss‖

542

)=

52

cL−358 ‖k‖3

2 +52

cL−158 ‖k‖

742 ‖kss‖

542


Here we prove that ‖kss‖22 is bounded.

Fs4 =−ks8−10kskssks3−152

k2s ks4−5kk2

s3−10kkssks4−3kksks5 + k2ks6,

from lemma 3 (v), we have

ddt

∫γ

k2ssds = 2

∫γ

kss∂tkssds+∫

γ

k2sskFds

= −2∫

γ

kssFs4ds−2∫

γ

kssk2Fssds−6∫

γ

ksskskFs−2∫

γ

kssk2s F−3

∫γ

k2sskFds

= −2∫

γ

k2s5ds−4

∫γ

k4ssds+3

∫γ

k2s kssks4ds+3

∫γ

kk2ssks4ds+6

∫γ

k2s k2

s3

+32

∫γ

k4kssks4ds− 132

∫γ

k2ssk

2s k2ds+

∫γ

ks3k3s k2ds+2

∫γ

k2s kssks4ds

+5∫

γ

k2k2s ks4ds+2

∫γ

k3kskssds+2h∫

γ

kssk2s ds+3h

∫γ

k2sskds

= −2∫

γ

k2s5ds+3

∫γ

k2s kssks4ds+3

∫γ

kk2ssks4ds+6

∫γ

k2s k2

s3 +32

∫γ

k4kssks4ds

+∫

γ

ks3k3s k2ds+2

∫γ

k2s kssks4ds+5

∫γ

k2k2s ks4ds+2

∫γ

k3kskssds+2h∫

γ

kssk2s ds

+3h∫

γ

k2sskds

≤ (−2+8ε)∫

γ

k2s5ds+ c+2h2

∫γ

kssk2s ds+3h2

∫γ

k2sskds

= (−2+8ε)∫

γ

k2s5ds+ c+3h2

∫γ

k2sskds

As h2 =5

2L∫

γkk2

s ds, we can get

3h2

∫γ

k2sskds =

152L

∫γ

kk2s ds ·

∫γ

k2sskds≤ 15

2L‖k‖2

∞

∫γ

k2s ds

∫γ

k2ssds

≤ 152L

(√LE +

2π

L

)2

·E ·(

L2

4π2

)3

·∫

γ

k2s5ds

From Lemma 62, we know that E is decreasing. So when t > t, we can have E ≤ ε .Thus

3h2

∫γ

k2sskds≤ εc(L)

∫γ

k2s5ds = ε

∫γ

k2s5ds


ddt

∫γ

k2ssds ≤ (−2+8ε)

∫γ

k2s5ds+ c+3h2

∫γ

k2sskds

≤ (−2+9ε)∫

γ

k2s5ds+ c

(‖k‖2

2)

≤ −c1‖kss‖22 + c

(‖k‖2

2)

From above we know that ‖kss‖22 is bounded, then we have h2(t) is bounded as the

following.

|h2(t)| ≤ c1L−358 ‖k‖3

2 + c2L−158 ‖k‖

742 ‖kss‖

542 .

7.2.2 Global Existence

In above, we show that the length is bounded for finite time intervals, and the h(t) term isbounded as well. In this section, firstly, we give the boundness of curvature derivatives inL2 and L∞. This proposition will be used to prove Theorem 15 and Proposition 40 later.Then we show that T = ∞.

Proposition 33. Suppose γ0 : S1→ R2 solves (7.5). Then there exist constants cl, cl > 0depending only on γ0 such that for all l ∈ N0,

‖ksl‖22 ≤ cl and ‖ksl‖2

∞ ≤ cl.

Proof. From Lemma 3 (v), we have

ddt

∫γ

k2sl ds

= −∫

γ

2k2sl+3ds+

∫γ

P2l+44 (k)ds+

∫γ

P2l6 (k)ds+

∫γ

P2l+35 (k)ds+h2(t)

∫γ

ksl Pl2(k)ds

≤ −∫

γ

2k2sl+3ds+ ε

∫γ


2 + ε

∫γ


2 + ε

∫γ

k2sl+2ds

+c‖k‖2l+102 +h2(t)

∫γ

ksl Pl2(k)ds

≤ (−2+3ε)∫

γ


2 + c‖k‖2l+102 + c‖k‖4l+14

2 +h2(t)∫

γ

ksl Pl2(k)ds

From Lemma 64 we know that


|h2(t)| ≤ c(L)‖k‖32 + c(L)‖k‖

742

and

∫γ

ksl Pl2(k)ds ≤ c(L)

(∫γ

k2ds) l+ 5

22l+2

(∫γ

k2ds) 2l+ 1

22l+2

+

(∫γ

k2sl+1ds

) 2l+ 12

2l+2

= c(L)

(∫γ

k2ds) 3

2

+

(∫γ

k2ds) l+ 5

22l+2(∫

γ

k2sl+1ds

) 2l+ 12

2l+2

,

‖k‖522 =

(L

12

(∫γ

k2ds) 1

2) 5

2

= L54

(∫γ

k2ds) 5

4

,

‖k‖322,2 = L

34

(∫γ

k2ds) 3

4

+L54

(∫γ

k2s2ds

) 34

.

By using Young’s inequality, we have

h2(t)∫

γ

ksl Pl2(k)ds

≤ c‖k‖62 + c‖k‖

194

2 + c‖k‖8l+112l+2

2 · ‖ksl+1‖4l+12l+22 + c‖k‖

11l+174l+4

2 · ‖ksl+1‖4l+12l+22

= c(∫

γ

k2ds)3

+ c(∫

γ

k2ds) 19

8

+ c ·(∫

γ

k2ds) 8l+11

4l+4

·(∫

γ

k2sl+1ds

) 4l+14l+4

+c(∫

γ

k2ds) 11l+17

8l+8

·(∫

γ

k2sl+1ds

) 4l+14l+4

≤ ε

2·

(∫γ

k2sl+1ds

) 4l+14l+4 ·

4l+44l+1

4l+44l+1

+2ε·

(∫γ

k2ds) 8l+11

4l+4 ·4l+4

3

4l+43

+ε

2·

(∫γ

k2sl+1ds

) 4l+14l+4 ·

4l+44l+1

4l+44l+1

+2ε·

(∫γ

k2ds) 11l+17

8l+8 ·4l+4

3

4l+43

+ c(∫

γ

k2ds)3

+ c(∫

γ

k2ds) 19

8

= ε

∫γ

k2sl+1ds+ c

(∫γ

k2ds) 8l+11

3

+

(∫γ

k2ds) 11l+17

6

+ c(∫

γ

k2ds)3

+ c(∫

γ

k2ds) 19

8

= ε

∫γ

k2sl+1ds+ c‖k‖

16l+113

2 + c‖k‖11l+17

32 + c‖k‖6

2 + c‖k‖194

2 .


Then we have

ddt

∫γ

k2sl ds

≤ (−1+3ε)∫

γ


2 + c‖k‖2l+102 + c‖k‖4l+14

2 +h2(t)∫

γ

ksl Pl2ds

≤ (−1+3ε)∫

γ


2 + c‖k‖2l+102 + c‖k‖4l+14

2 + ε

∫γ

k2sl+1ds

+c‖k‖16l+11

32 + c‖k‖

11l+173

2 + c‖k‖62 + c‖k‖

194

2

= (−1+4ε)∫

γ


2 + c‖k‖2l+102 + c‖k‖4l+14

2 + c‖k‖16l+11

32

+c‖k‖11l+17

32 + c‖k‖6

2 + c‖k‖194

2 .

We get that all derivatives of the curvature are bounded in L2. As ‖ksl‖2∞ ≤ L

2π‖ksl+1‖2

2,all derivatives of the curvature are bounded in L∞ follows.

We show that if the maximal existence time is finite, then the curvature must blow upin L2. For the similar proof, see Chapter 6.

Theorem 15. Suppose γ0 : S1→ R2 be a maximal solution of (7.5). If T < ∞ then

∫γ

k2ds∣∣∣∣t≥ c(T − t)−

16

Proof. Now we refer to [17, Theorem 3.1], assume ‖k‖2 is bounded for all t < T , then wecan get that γ can extend smoothly to S1× [0,T ] by short time existence, which contradictsthe maximality of T and therefore that k cannot be uniformly bounded in L2 on [0,T ).

For l = 0 in Proposition 33 we have

ddt

∫γ

k2ds ≤ c‖k‖142 + c‖k‖10

2 + c‖k‖113

2 + c‖k‖173

2 + c‖k‖62 + c‖k‖

194

2

Using Young’s inequality yields

ddt

∫γ

k2ds ≤ c(∫

γ

k2ds)7

+ c(∫

γ

k2ds)5

+ c(∫

γ

k2ds) 11

6

+ c(∫

γ

k2ds) 17

6

+c(∫

γ

k2ds)3

+ c(∫

γ

k2ds) 19

8

Since∫

γk2ds blows up as t → T , all the others term can be absorbed by the power 7


term, i.e. ddt∫

γk2ds≤ c

(∫γ

k2ds)7

. Thus we obtain

∫γ

k2ds∣∣∣∣t≥ c(T − t)−

16 .

Then, we get the T = ∞ by a contradiction.

Corollary 5. Suppose γ0 : S1→ R2 solves (7.5), then T = ∞.

Proof. Suppose on the contrary that γ satisfies the conditions of Theorem 14 and T < ∞.We know from Theorem 15 above that ‖k‖2

2→ ∞ as t→ T .

Kosc = L∫

γ

(k− k)2ds = L‖k‖22−2π

2→ ∞

this contradicts that Kosc is bounded in Lemma 62, then T = ∞ and ‖k‖22 does not blow

up.


Before proving the energy decays exponentially, we show several useful propositions anda lemma which shows that the length is bounded for all time.

Proposition 34. For all time, if(L3E

)(t)≤ ε0, we obtain∫

γ

G2ds+h2

∫γ

Gds≥ cL−6E

Proof.

F− F =(k2− k2)kss−

12

kk2s +h2(t),

∫γ

G2ds+h2

∫γ

Gds =∫

γ

F2ds−h2

∫γ

Gds−h22L (7.9)

here h2(t) = 52L∫

kk2s ds.

∫γ

F2ds =∫

γ

F2ds+2∫

γ

F(F− F)ds+∫

γ

(F− F)2ds

≥ 12

∫γ

F2ds−∫

γ

(F− F)2ds

≥ c1

2

∫γ

k2s4ds− cL−3E2−

∫γ

(F− F)2ds (7.10)


Now F− F =(k2− k2)kss− 1

2kk2s +h2(t), so

∫γ

(F− F)2ds ≤ 2∫

γ

(k2− k2)2

k2ssds+

12

∫γ

k2k4s ds

+2h∫

γ

[(k2− k2)kss−

12

kk2s

]ds+h2L (7.11)

Substituting (7.10) into (7.9), we have∫γ

G2ds+h2

∫γ

Gds

=∫

γ

F2ds−h2

∫γ

Gds−h22L

≥ c1

2

∫γ

k2s4ds− cL−3E2−

[∫γ

(F− F)2ds+h2

∫γ

Gds+h22L]

(7.12)

From (7.11), we have∫γ

(F− F)2ds+h2

∫γ

Gds+h22L

≤ 2∫

γ

(k2− k2)2

k2ssds+

12

∫γ

k2k4s ds+2h2

∫γ

[(k2− k2)kss−

12

kk2s

]ds+

h22L+h2

∫γ

(ks4 + k2kss−

12

kk2s

)ds+h2

2L

≤ 2∫

γ

(k2− k2)2

k2ssds+

12

∫γ

k2k4s ds+3h2

∫γ

(k2kss−

12

kk2s

)ds+2h2

2L

(7.13)

For the first two terms in (7.13), we have

2∫

γ

(k2− k2)2

k2ssds+

12

∫γ

k2k4s ds

≤ 2‖k+ k‖2∞

(∫γ

|ks|ds)2 ∫

γ

k2ssds+

12‖k‖2

∞

∫γ

k4s ds

≤ 4‖k+ k‖2∞ ·LE ·

∫γ

k2ssds+

12‖k‖2

∞

[3L2π·E ·

∫γ

k2ssds+

3L2π·E ·

∫γ

|ksks3|ds]

=

(4L‖k+ k‖2

∞ +3L4π‖k‖2

∞

)·E∫

γ

k2ssds+

3L4π‖k‖2

∞ ·E∫

γ

|ksks3|ds (7.14)

ask = k− k+ k ≤

∫γ

|ks|ds+2π

L,

by Hölder inequality, we have

‖k‖∞ ≤√

2LE +2π

L,


‖k+ k‖∞ =√

2LE +4π

L.

here h2(t) = 52L∫

kk2s ds,

So the third and fourth terms can be estimated as the following,

3h2

∫γ

(k2kss−

12

kk2s

)ds+2h2

2L

=152L

∫γ

kk2s ds ·

∫γ

k2kssds− 154L

∫γ

kk2s ds ·

∫γ

kk2s ds+2L

(5

2L

∫γ

kk2s ds)2

= −15L

(∫γ

kk2s ds)2

− 154L

(∫γ

kk2s ds)2

+252L

(∫γ

kk2s ds)2

= −254L

(∫γ

kk2s ds)2

≤ 0 (7.15)

Substituting (7.14) and (7.15) into (7.13), we have

∫γ

(F− F)2ds+h2

∫γ

Gds+h22L

≤(

4L‖k+ k‖2∞ +

3L4π‖k‖2

∞

)·E∫

γ

k2ssds+

3L4π‖k‖2

∞ ·E∫

γ

|ksks3|ds

= c3L−1E∫

γ

k2ssds+ c4L−1E

∫γ

|ksks3|ds (7.16)

where c3 = 4L2‖k+ k‖2∞ + 3L2

4π‖k‖2

∞, c4 =3L2

4π‖k‖2

∞, c3 and c4 are bounded by functiononly involve ε and L as follows.

c3 ≤ 4L−1 ·L−2(√

2L3E +4π

)2+

34πL

L−2(√

2L3E +2π

)2

≤ 4L−3(√

2ε +4π

)2+

34π

L−3(√

2ε +2π

)2

and

c4 ≤3

4πL−3

(√2ε +2π

)2

We have the following three inequalities

c3L−1E∫

γ

k2ssds ≤ cc3L−1E

(∫γ

k2s ds) 2

3(∫

γ

k2s4ds

) 13

≤ cc3L−1E(2E)23

(∫γ

k2s4ds

) 13

≤ c1

8

∫γ

k2s4ds+ cc

323 L−

32 E

52 (7.17)


c4L−1E∫

γ

|ksks3|ds≤ c · c61

2πE(∫

γ

k2s ds) 1

2(∫

γ

k2s4ds

) 12

E ≤ c1

8

∫γ

k2s4ds+ c · c2

6E3

Substituting above two inequalities into (7.16), we have

∫γ

(F− F)2ds+h2

∫γ

Gds+h22L ≤ c1

4

∫γ

k2s4ds+ cc

323 L−

32 E

52 + cc2

6E3 (7.18)

We substitute (7.18) into (7.12), we have

∫γ

G2ds+h2

∫γ

Gds

=∫

γ

F2ds−h2

∫γ

Gds−h22L

≥ c1

2

∫γ

k2s4ds−29

ω8π

8L−3E2−[∫

γ

(F− F)2ds+h2

∫γ

Gds+h22L]

≥ c1

2

∫γ

k2s4ds−29

ω8π

8L−3E2− c1

4

∫γ

k2s4ds− cc

323 L−

32 E

52 − cc2

6E3

≥ c1

4

∫γ

k2s4ds−29

ω8π

8L−3E2− cc323 L−

32 E

52 − cc2

6E3

≥ c1

4

(4ω2π2

L2

)3

E−29ω

8π

8L−3E2− cc323 L−

32 E

52 − cc2

6E3

= c1L6 E−29

ω8π

8L−3E2− cc323 L−

32 E

52 − cc2

6E3 (7.19)

Then, rearrange (7.19), we have

L9(∫

γ

G2ds+h2

∫γ

Gds)≥ cL3E−29

ω8π

8L6E2− cc323 L

152 E

52 − cc2

6L9E3

≥ cL3E−29ω

8π

8 (L3E)2− cc

323(L3E

) 52 − cc2

6(L3E

)3

As(L3E

)(t)≤ ε0 for all time, thus∫

γ

G2ds+h2

∫γ

Gds≥ cL−6E

holds for all time.

The following proposition can relax the condition(L3E

)(t)≤ ε0 in Proposition 34.


Proposition 35. Suppose γ0 : S1→ R2 solves (7.5) and under the assumption

E(0)≤ ε,

we have for all t in [0,∞], (L3E

)(t)≤

(L3E

)(0).

Proof. From (7.7), we have

ddt

L = −∫

γ

kFds =−∫

γ

k2ssds+

72

∫γ

k2k2s ds− 5π

L

∫γ

kk2s ds

≤ −4π2

L2

∫γ

k2s ds+

72‖k‖2

∞

∫γ

k2s ds+

∫γ

k2k2s ds+

(5π

L

)2 14

∫γ

k2s ds

≤ −4π2

L2

∫γ

k2s ds+

72‖k‖2

∞

∫γ

k2s ds+‖k‖2

∞

∫γ

k2s ds+

25π2

4L2

∫γ

k2s ds

As‖k‖∞ ≤

1L

(√2L3E +2π

),

also in Proposition 34, for all time, we have(L3E

)(t)≤ ε0, then

ddt

L ≤[

92

(√2L3E +2π

)2+

9π2

4

]·L−2

∫γ

k2s ds

≤[

92

(√2ε0 +2π

)2+

9π2

4

]·L−2

∫γ

k2s ds

= c(ε0)L−2∫

γ

k2s ds (7.20)

From Proposition 34,∫

γG2ds + h

∫γ

Gds ≥ cL−6E, and −∫

γG2ds− h

∫γ

Gds ≤ 0 inLemma 62, we do the time derivative of L3E as follows,

ddt

L3E = L3 ddt

E +2L2Eddt

L

≤ −L3(∫

γ

G2ds+h∫

γ

Gds)+2L2E · c(ε0)L−2

∫k2

s ds

≤ L3(∫

γ

G2ds+h∫

γ

Gds)·[−1+ ¯cε0L3E

]In above, if

(L3E

)(0)≤min

ε0,

1¯cε0− ε2

= ε ·L3(0) = ε1, i.e.

E(0)≤ ε,

then ddt L3E ≤−cL3

(∫γ

G2ds+h∫

γGds

)≤ 0, thus

(L3E

)(t)≤ ε1 ≤ ε0 holds for all time.


By using Proposition 35, we can get the following proposition which is better thanProposition 34.


E(0)≤ ε,

we have ∫γ

G2ds+h∫

γ

Gds≥ cL−6E

holds for all time.

Before proving the energy decays exponentially, we need to show that the length isbounded for all time.

Lemma 65. Suppose γ0 : S1→ R2 solves (7.5) and under the assumption

E(0)≤ ε,

the length is bounded for infinite time.

Proof. From Proposition 36 and (7.20) in Proposition 35, we can get

ddt

lnL≤ cL−3E ≤ cL3(∫

γ

G2ds+h∫

γ

Gds)

In the end of the proof for Proposition 35, we have ddt L3E ≤−cL3

(∫γ

G2ds+h∫

γGds

)≤

0. By applying this inequality, we get

ddt

lnL≤−cddt

L3E,

thuslnL(t)≤ lnL(0)+ c

(L3E

)(0).

which gives an upper bound for length.

Above lemma shows the length is bounded under the smallness energy assumption,however, we can have a better result shown in the following proposition, where the small-ness energy condition is not needed.

Proposition 37. Define E = A32 E. (Note that E ≤ E0 for any flow.) For any smooth

embedded curve where k has a zero:

L≤√

A4Eπ

.


If k does not have a zero (i.e. the curve is convex) then

L≤√

A(

2√

π +Eπ+2

√E2

π2 +2√π

E).

Proof. If the curve is embedded and star-shaped, then we claim the inequality

L≤ 2A‖k‖∞

holds. This is easy to see:

L =∫|γs|2ds =

∫k(−〈γ,ν〉)ds≤ 2A‖k‖∞.

In order to upgrade this inequality to hold for all embedded curves, we flow γ by area-preserving curve shortening flow. Suppose that ‖k‖∞ < L

2A . This curvature bound ispreserved along the flow. However, the flow converges to a circle, which has of courseL

2A = 1r = k; this is a contradiction.

In order to obtain the two estimates, observe that if k has a zero then as it is periodic itmust contain two zeros, then an integral splitting argument (see [82]) implies that

‖k‖2∞ ≤

L2π

∫γ

k2s ds≤ L

πE .

Then

L≤ A2√

L√π

√E .

Rearranging gives the first estimate. Then for the second estimate, k does not have a zero.We have

‖k− k‖2∞ ≤

Lπ

E,

as L2 ≥ 4πA, we can get

L ≤ 2A‖k‖∞ ≤ 2A‖k− k+ k‖∞

≤ 2A(‖k− k‖∞ +‖k‖∞

)≤ 2A ·

√Lπ

E +2A ·2π

L

≤ 2A√

E√π·√

L+2√

Aπ.

Rearranging this estimate gives the following

(√L)2− 2A

√E√

π·√

L−2√

Aπ ≤ 0.


By solving this question, we get

√L≤ A

√E√

π+

√A2E

π+2√

Aπ,

then

L ≤√

A

2√

π +A

32 Eπ

+2

√A3E2

π2 +2A

32 E√π

≤√

A

2√

π +Eπ+2

√E2

π2 +2√π

E

which is the second estimate.

Note that in either case, the bound on the right-hand side is uniformly bounded underthe flow. This means that so long as the curve remains embedded, length is well-controlledunder the flow.

We show the curve remains embedded with the smallness condition in Theorem 14.



Under the assumption, we have the ε is small enough to satisfy the condition in [82,Theorem 1.6]. This theorem is showed as Theorem 12 in Chapter 6.

Finally, we can get the exponential decaying of the energy by using above Lemma 65and Proposition 36.


E(0)≤ ε,

there exist constants c,δ , we have ∫γ

k2s ds≤ ce−δ t

for all time.

Proof. From Lemma 65 and Proposition 36, we get

ddt

∫γ

k2s ds =−

∫γ

G2ds−h∫

γ

Gds≤−c∫

γ

k2s ds,

which means∫

γk2

s ds exponentially decays, where c depends on ε and L.


The following proposition shows that the curvature derivatives decay exponentially inL2 and L∞.


14. Then there exists constants cl > 0 depending only on γ0 such that for all l ∈ N0,

‖ksl‖22 ≤ cle−δ t

and ksl exponentially decays in L∞.

Proof. When t→ ∞,

∫γ

k2s ds =−

∫γ

(k− k)kssds≤(∫

γ

(k− k)2ds) 1

2(∫

γ

k2ssds) 1

2

,

as∫

γk2

ssds is bounded in L2 in Proposition 33 and∫

γk2

s ds≤ ce−δ t in Proposition 39, then∫γ

k2s ds exponentially decays.

∫γ

k2ssds =−

∫γ

ksks3ds≤(∫

γ

k2s ds) 1

2(∫

γ

k2s3ds

) 12

,

as∫

γk2

s3ds is bounded in L2 in Proposition 33, then∫

γk2

ssds exponentially decays.We let l ≥ 1,

∫γ

k2sl+1ds≤

(∫γ

k2s ds) 1

2(∫

γ

k2s2l+1ds

) 12

,

as∫

γk2

s2l+1ds is bounded in L2, then∫

γk2

sl+1ds exponentially decays in L2. Also ‖ksl‖2∞ ≤

L2π‖ksl+1‖2

2 So we have ksl exponentially decays in L∞.

We can obtain uniform bounds for all derivatives of the evolving curve γ by using aboveProposition 40.



‖∂ul γ‖∞ ≤ c(l)+l

∑p=0‖∂sl γ0‖∞,

where c(l) is a constant only depending on l, E(0), L.

The proof of above proposition is similar to Proposition 31.For the proof of Theorem 14, it remains to show that the full convergence of the flow.

Here again we refer to [1, Theorem A.1]. There are three hypotheses in this theorem weneed to check. Uniform boundedness of γ and all its derivatives in Proposition 41 implythe first hypothesis is satisfied.


For the second hypothesis, using Proposition 40 we note that∫γ

F2ds≤ c1e−c2t .

This implies ∫ T

0

(∫γ

F2ds) 1

2

dt ≤ c1

∫ T

0e−

c22 tdt ≤ c

where c is a constant depending only on w, E(0) and L.

For the third hypothesis, Proposition 41 gives that for any sequence t j → ∞, the C∞-norm of γ(·, t j) is uniformly bounded. We have the exponential decay of the energy, soE[γ(·, t j)]→ 0, which implies that a subsequence γ(·, t j) converges to an smooth w-circlein C∞-topology. Therefore we apply [1, Theorem A.1] to conclude full convergence ofthe flow.

Exponential decay of the speed allows us to bound the region of the plane in which thesolution lies relative to γ0 via standard argument. We may bound the distance travelled byany point on the initial curve γ0 as follows

|γ(x, t)−γ(x,0)|=∣∣∣∣∫ t

0

∂γ

∂ t(x,v)

∣∣∣∣≤ ∫ t

0

∣∣∣∣ks4 + k2kss−12

kk2s +h2(t)

∣∣∣∣dv≤ cσ

(1− e−σt)≤ c

σ.

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Appendix A

Appendix

A.1 Proof of Theorem 4

For the proof of Theorem 4 in Chapter 3, we refer to [79, Theorem 2.7] and give the proofas the following.

Proof. We consider the family of initial value problem

∂

∂ tγh =−(kss + h)ν ,

where h ∈C1([0,T )) is a known function of time and γ0 = γ(·,0) ∈C4(S1). The process-ing arguments give short time existence for each γh. We will now show that at least oneof the functions h coincide with our given constrained function h, which is normal a ratioof integrals of curvature and not a priori known function of time. We prove that h satisfiesthe initial condition d2

dt2 h(t)≤ c(γ0), which we note forces some measure of regularity onthe immersion γ0, γ0 = γ(·,0) ∈C7(S1).

Let S =C1([0,T )) for some σ > 0 which will be chosen. The theorem will be provedif we can apply Theorem 3 with the mapping P : S→ S defined by

Ph = h.

Noting that C1([0,σ ]) is a compact, convex subset of the Banach space C1([0,T ]), weneed to demonstrate that P maps S into itself and is continuous. Both of these followfrom the assumption d2

dt2 h ≤ c(γ0). In particular, we have that d2

dt2 h ≤ c(γ0) and so ddt h is

continuous on [0,σ ] for some σ > 0 , and so

ddt

h =ddt

Ph ∈C1([0,σ ]).

This also knows that P′ is bounded in the operator norm on C1([0,σ ]) and so P is contin-uous. Therefore we apply Theorem 3 and deduce at least one of the functions h coincides

168

APPENDIX A. APPENDIX 169

with the given constraint function h on an interval [0,σ ] ⊂ [0,T ). The initial conditiond2

dt2 h(t) ≤ c(γ0) will give a sequence σi. The maximal time for length-constrained curvediffusion flow is T = limi→∞ σi.

A.2 Proof of Proposition 27

Lemma 66. Let

F = ks4 + k2kss.

Then there exists constants c1, c2 such that∫F2ds≥ c1

∫γ

k2s4ds− c2L−3E2.

For the proof of this Lemma see [1, Lemma 7.1], we also show the proof as follows.

Proof. Consider the Fourier series for k:

k = ∑p

ap exp(

i2π

Lps).

Then

κ0 = ∑p

ap

[(4π2

L2 p2)2

− 4π2

L2 ω2 4π2

L2 p2

]exp(

i2π

Lps)

= ∑p

ap

(4π2

L2

)2

p2 (p2−ω2)exp

(i2π

Lps).

This implies

∫γ

κ20 ds = ∑

p|ap|2

(4π2

L2

)4

p4 (p2−ω2)2

L.

We calculate

a±w =1L

∫γ

k exp(±i

2π

Lωs)

ds =1L

∫γ

(k− 2πω

L

)exp(±i

2π

Lωs)

ds.

This implies

|a±ω| ≤ 1L

∫γ

∣∣∣∣k− 2πω

L

∣∣∣∣ · ∣∣∣∣exp(±i

2π

Lωs)− exp(±iθ)

∣∣∣∣ds

+1L

∣∣∣∣∫γ

(k− 2πω

L

)exp(±iθ)ds

∣∣∣∣ .In the above we have again used θ to denote the tangential angle.


Noting that

∫γ

(k− 2πω

L

)exp(±iθ)ds =

∫γ

θs exp(±iθ)ds− 2πω

L

∫γ

τds = 0

and ∣∣∣∣2π

Lωs−θ

∣∣∣∣ ≤ ∫ s

0

∣∣∣∣ dds

(2π

Lωs−θ

)∣∣∣∣ds

≤∫ s

0

∣∣∣∣2π

Lω− k

∣∣∣∣ds

≤ L∫

γ

|ks|ds

≤√

2L32 E

12

we obtain

|a±ω| ≤ 1L

∫γ

∣∣∣∣k− 2πω

L

∣∣∣∣ · ∣∣∣∣exp(±i

2π

Lωs)− exp(±iθ)

∣∣∣∣ds

≤ 1L

∫γ

|ks|ds∫

γ

∣∣∣∣(±i2π

Lωs)− (±iθ)

∣∣∣∣ds

≤ L(∫

γ

|ks|ds)2

≤ 2L2E.

Now ∫γ

k2s4ds = ∑

p

(4π2

L2

)4

p8|ap|2L,

so we have

∫γ

κ20 ds = ∑

p|ap|2

(4π2

L2

)4

p4 (p2−ω2)2

L

= ∑|p|6=ω

|ap|2(

4π2

L2

)4

p4 (p2−ω2)2

L

≥ ∑|p|6=ω,0

|ap|2(

4π2

L2

)4

p8(

1− ω2

p2

)2

L

We define Cω by

(1− ω2

p2

)2

≥min

(1− ω2

(ω−1)2

)2

,

(1− ω2

(ω +1)2

)2

:=Cω .


Then

∫γ

κ20 ds ≥ Cω

∫γ

k2s4ds−

(4π2

L2

)4

ω8 (|a+ω |2 + |a−ω |2

)L

≥ Cω

∫γ

k2s4ds−4ω

8(

4π2

L2

)4

L5E2

≥ Cω

∫γ

k2s4ds−45

ω8π

8L−3E2,

As for all time, ω = 1 and L = L(0), then there exist constants c1, c2 such that∫F2ds≥ c1

∫k2

s4ds− c2L−3(0)E2.

as required.

Secondly, we show the curvature bound here.

Lemma 67. For any immersed curve γ : S→ R2 we have the estimate

L‖k‖∞ ≤√

L3‖ks‖22 +2ωπ.

Here ω is the winding number of γ .

Proof. We calculate k = k− k+ k ≤∫

γ|ks|ds+ 2ωπ

L . Taking a supremum and using theHölder inequality, we find

‖k‖∞ ≤1L

(√L3‖ks‖2

2 +2ωπ

).

Now we show the proof of Proposition 27 by using above two lemmas.

Proof. Using above lemmas and ω = 1, we have∫γ

G2ds =∫

γ

F2ds+2∫

γ

F(G− F)ds+∫

γ

(G− F)2ds

≥ 12

∫γ

F2ds−∫

γ

(G− F)2ds

≥ c1

2

∫k2

s4ds−29π

8L−3(0)E2−∫

γ

(G− F)2ds (A.1)

Now G− F =[k2−

(2π

L

)2]

kss− 12kk2

s , so

∫γ

(G− F)2ds≤ 2∫

γ

[k2−

(2π

L

)2]2

k2ssds+

12

∫γ

k2k4s ds.


Now the curvature bound yields∫γ

(G− F)2ds

≤ 2[(√

2LE +2π

L

)+

(2π

L

)]2

·(∫

γ

|ks|ds)2 ∫

γ

k2ssds

+12

(√2LE +

2π

L

)2 ∫k4

s ds

≤ 4√

2(√

2L3E +4π

)2 EL

∫γ

k2ssds+

12

(√2L3E +2π

)2L−2

∫k4

s ds (A.2)

Now the Gagliardo-Nirenberg Sobolev inequality yields universal constants c3,c4 suchthat inequalities

c1

∫γ

k2ssds ≤ c1c3

(∫γ

k2s ds) 2

3(∫

γ

k2s4ds

) 13

≤ c1c3(2E)23

(∫γ

k2s4ds

) 13

≤ Lc1

8E

∫γ

k2s4ds+ cc

321 c

323 L−

12 E

32

and

c2

∫γ

k4s ds ≤ c2c4L2

(∫γ

k2s ds) 3

2(∫

γ

k2s4ds

) 12

≤ c2c4L2(2E)32

(∫γ

k2s4ds

) 12

≤ L2c1

8

∫γ

k2s4ds+ cc2

2c24E3L2

hold. Using these with c1 = 4√

2(√

2L3E +4π

)2and c2 =

12

(√2L3E +2π

)2in combi-

nation with (A.2) above yields∫γ

(G− F)2ds≤ c1

4

∫γ

k2s4ds+ c

[((L3E)

32 +1

)(L−

32 E

52

)+((L3E)2 +1

)E3]

Plugging this into (A.1) we find∫γ

G2ds

≥ c1

2

∫γ

k2s4ds−29

π8L−3E2−

∫γ

(G− F)2ds

≥ c1

4

∫γ

k2s4ds− c

[L−3E2 +

((L3E)

32 +1

)(L−

32 E

52

)+((L3E)2 +1

)E3]

≥ c1

4

(4π2

L2

)3

E− c[L−3E2 +

((L3E)

32 +1

)(L−

32 E

52

)+((L3E)2 +1

)E3].(A.3)


This implies

L9∫

γ

F2ds≥ a(L3E)−b(L2E)2−g(L3E)52 −d(L3E)3− e(L3E)4− f (L3E)5

where a,b,g,d,e, f are universal constants. Therefore, for L3E small enough, we have

L9∫

γ

G2ds≥ a2

L3E,

as required.

A.3 Proof of Proposition 3

Proof. For the constrained problem, the associated Euler-Lagrange equation is

L = f 2 +λ f 2x ,

with extremal functions satisfying

∂L∂ f− d

dx

(∂F∂ fx

)= 2 f −2λ fxx = 0.

That is to say

fxx−1λ

f = 0. (A.4)

This means that

0≤∫ P

0f 2x dx =−

∫ P

0f fxxdx =− 1

λ

∫ P

0f 2dx,

which forces λ < 0. By standard arguments, we conclude from (A.4) that our extremalfunction is

f (x) = Acos

(x√|λ |

)+Bsin

(x√|λ |

). (A.5)

Here A,B are constants. The periodicity of f forces f (0) = f (P), so

A = Acos

(P√|λ |

)+Bsin

(P√|λ |

). (A.6)


Also, the requirement that∫ P

0 f dx = 0 forces

Asin

(P√|λ |

)−Bcos

(P√|λ |

)=−B. (A.7)

Combining (A.6) and (A.7),

A2 = A2 cos

(P√|λ |

)+ABsin

(P√|λ |

)

and

B2 = B2 cos

(P√|λ |

)−ABsin

(P√|λ |

)

meaning that

A2 +B2 =(A2 +B2)cos

(P√|λ |

).

We concludeP√|λ |

= 2nπ

for some n ∈ Z/0 to be determined. Hence

f (x) = Acos(

2nπxP

)+Bsin

(2nπx

P

). (A.8)

A quick calculation yields ∫ P

0f 2dx =

(A2 +B2

2

)P,

and ∫ P

0f 2x ds =

(2nπ

P

)2(A2 +B2

2

)P.

Hence for any of our extremal functions f ,

∫ P0 f 2ds∫ P0 f 2

x dx=

(P

2nπ

)2

≤ P2

4π2 ,

with equality if and only if n = 1. Thus our constrained function f that maximises theratio

∫ P0 f 2ds∫ P0 f 2

x dxis given by

f (x) = Acos(

2π

Px)+Bsin

(2π

Px),


with ∫ P

0f 2dx≤ P2

4π2

∫ P

0f 2x ds

amongst all continuous and P−periodic function with∫ P

0 f dx = 0.Since

∫ P0 f dx = 0 and f is P− periodic we conclude that there exists distinct 0 ≤ p <

q < P such thatf (p) = f (q) = 0.

Next, the fundamental theorem of calculus tells us that for any x ∈ (0,P),

12[ f (x)]2 =

∫ x

pf fxdx =

∫ x

qf fxdx.

Hence

[ f (x)]2 =∫ x

pf fxdx−

∫ q

xf fxdx≤

∫ q

p| f fx|dx≤

∫ P

0| f fx|dx

≤(∫ P

0f 2dx ·

∫ P

0f 2x dx) 1

2

≤ P2π

∫ P

0f 2x dx.

A.4 Referred Theorems

The general Theorem 4.4 in [69] is shown as below.

Theorem 16. Let T1 > 0, Ω ∈ Rn, φ ∈ H∞(Ω) and h ∈C∞ ([0,T1]) , H∞ (∂Ω) satisfy the

compatibility condition

h( j)(0) = ∂j

t B(t,u(t))∣∣∣t=0

on ∂Ω, j = 0,1,2, ....

and normal boundary conditions. Assume that there are b ≥ 0 and ck > 0 and an open

neighbourhood U of φ in H∞ (Ω) so that any 0 < T ≤ T1 and

u ∈W = w ∈C∞ ([0,T ],H∞(Ω)) : w(t) ∈U, t ∈ [0,T ]

such that for any f (t) ∈C∞ ([0,T ],H∞(Ω)) the linear problemwt(t) = Fu(t,u(t))w(t)+ f (t), in Ω, t ∈ [0,T ]Bu(t,u(t))w(t) = 0, on ∂Ω, t ∈ [0,T ]w(0) = 0.

admits for any f ∈C∞0 ([0,T ],H∞(Ω)) a unique solution w∈C∞

0 ([0,T ],H∞(Ω)) satisfying


the estimates

‖w‖k ≤ ck [u; f ]b,k , k = 0,1,2, ...

where ck > 0 are suitable constants and b≥ 0 is an integer.

Then the nonlinear problemut = F(t,u), in Ω, t ∈ [0,T ]B(t,u) = h(t), on ∂Ω, t ∈ [0,T ]u(0) = φ .

has a unique solution u ∈C∞0 ([0,T0],H∞(Ω)).

For any integer k ≥ 0, the Sobolev space Hk(Ω) is equipped with its natural norms

‖u‖k =

(∑|α|≤k

∫Ω

|∂ αu(x)|2ds

)1/2

, u ∈ Hk(Ω).

[u; f ]b,k is defined by the corresponding Sobolev norms of u, f .[u; f ]b,k = sup

‖u‖b+i1 · ... · ‖u‖b+ir · ‖ f‖b+ j

, where the "sup" is running over all in-

tegers i1, ..., ir, j ≥ 0, i1 + ...+ ir + j ≤ k, where 0 ≤ r ≤ k. More properties of the terms[ ]b,k can be found in [70].

The normal boundary condition is defined as follows:Let B jp

j=1 be a set of differential operators B j = B j(u,∂ ) of order m j given by

B j = B j(x,∂ ) = ∑|β |≤m j

b j,β (u)∂β , j = 1, ..., p

with b j,β ∈C∞(η1,2). The set B jpj=1 is called normal if m j 6= mi for j 6= i and if for any

x ∈ η1,2 we have BPj (u,ν) 6= 0, j = 1, ..., p where ν = νη1,2 denotes the inward normal

vector to η1,2 at u and BPj denotes the principal part of B j.

Here we give a simple definition for the principal part of an operator.

Definition 10. The principal part of a differential operator is the part which contains the

highest order partial derivatives.

The classical results on linear parabolic boundary value problem ([53], Ch IV, 6.4)there is a unique solution w ∈C∞

0 ([0,T ], [−1,1]) of the linearized problem. We state thistheorem as follows.

Theorem 17. Let Ω be an open set in Rn and Γ be the infinitely differentiable boundary

of Ω. ∂w∂ t (u, t) = Aw+ f (t), f or all (u, t) ∈Ω× [0,T ]B jw = g j, f or all (u, t) ∈ Γ× [0,T ]w(·,0) = w0, f or all (u, t) ∈Ω×t = 0

(A.9)


where A are given by Aφ = ∑|p|,|q|≤m(−1)|p|Dpx(apqDq

xφ), the coefficients apq satisfy

apq ∈ D(Ω× [0,T ]). The boundary operators B j, 0 ≤ j ≤ m− 1, defined by B jφ =

∑|h|≤m j b jhDhxφ . where the functions b jh = b jh(x, t) satisfy b jh =D(Ω×t = 0).

Let r ≥ 0, g j,w0, f be given with

g j ∈ H2(r+1)m−m j−1/2,(r+1)−(m j+1/2)/2m (Γ× [0,T ]) , 0≤ j ≤ m−1,

w0 ∈ H2(r+1/2)m(Ω),

f ∈ H2rm,r(Ω× [0,T ])

and with the compatibility relation. Then problem (A.9) admits a unique solution in the

space H2(r+1)m,r+1(Ω× [0,T ]).

The following theorem in [3, Main Theorem 5] give us that there is a unique solutionfor nonlinear parabolic problem.

Theorem 18. Let E × (0,w) be a vector bundle over M× (0,w), where M a smooth

closed manifold, and let U be a section Γ(E × (0,w)). Consider the following initial

value problem:

P(U) := ∂tU−F(x, t,U,∂xU, ...,∂ 2mu U) = 0 in E× (0,w),

U(M,0) =U0,

with U0 ∈C2m,1,α(Ew). The linearized operator of P at U0 in the direction V is then given

by

∂P[U0]V = ∂tU +(−1)m∑|I|≤2m

AI(x, t,U0,∂xU0, ...,∂2mu U0))∂IV.

Suppose that the following conditions are satisfied:

1) The leading coefficient Aai1 j1...im jmb satisfies the symmetry condition Aai1 j1...im jm

b =

Ab j1i1... jmima

2)The leading coefficient satisfies the Legendre-Hadamard condition with constant λ

3) There exists a uniform constant Λ < ∞ such that ∑|I|≤2m |AI|α,Ew ≤ Λ

4) F is a continuous function of all its arguments

here | · |α,Ew denotes the Holder norm, |u|α,Ew = [u]α,Ew := supX 6=Y∈Ew|u(X)−u(Y )|

d(X ,Y )α , also

u ∈C2m,1(P) : [u]2m,1,α,P<∞. F is the principal symbol of the linearized operator at the

initial time.

Then there exists a unique solution U ∈C2m,1,β (Ew), where β < α , fir some short time

tε > 0 to the above initial value problem. Furthermore, if U0 and all the coefficients of the

linearized operator are smooth, this solution is smooth.

Definition 11. If there exists a positive constant λ ∈ R such that the function f satisfies

| f | ≥ λ . We say that function f satisfies Legendre-Hadamard condition.


We state the Theorem A.1 in [1] as follows.

Theorem 19. Let (Nn,h) be an n-dimensional Riemannian manifold and Mm be an m-

dimensional manifold with n > m. Suppose γ : Mm× [0,∞)→ Nn is a one-parameter

family of smooth isometric immersions satisfying

∂tγ = F.

Suppose furthermore that

1) (uniform bounds) We have the estimates∫M|γ|2dµ ≤ c1 and

∫M|k|2|γ|4dµ ≤ c2

for time-independent constants c1 and c2.

2) (L1−L2 Velocity) The L2-norm of the velocity is uniformly L1 in time, that is,

∫ T

0

(∫M|F |2dµ

)1/2

dt ≤ c3

for a constant c3 that does not depend on T .

3) (Subconvergence) There exists a smooth immersion γ∞ : Mm→ Nn and a sequence

t j ⊂ [0,∞)→ ∞, such that γ(·, t j)→C∞

γ∞,

Then γ converges to γ∞.

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