Master Thesis. Truong Nguyen Trung Duong. A new geometric proof of Grayson s theorem

Transcription

1 Master Thesis Truong Nguyen Trung Duong A new geometric proof of Grayson s theorem according to B. Andrews & P. Bryan Advisors: M. Soret Defense: Wednesday June 27 th, 2012

2 Acknowledgments The author would like to thank Professor Marc Soret for so many discussions and guidance. Thanks are also to my undergraduate advisor Dr. Huynh Quang Vu for invaluable help throughout my college years.

3 Contents 1 Introduction 2 2 Short time existence Short time existence for parabolic partial differential equation Short time existence for curve shortening flow Long time existence 11 4 Convergence Proof of Theorem Step Step Step Proof of Theorem Bibliography 29 1

4 Chapter 1 Introduction In 1980s, Gage and Hamilton [GH86] showed that the curve shortening flow shrinks a convex embedded curve to a round point. Then by using the result of Gage and Hamilton, Grayson [Gra87] showed that an embedded curve evolves under the curve shortening flow to a round point. In [Gra87], complicated techniques are used to show that an embedded curve shrinks to become circular. After that there are many efforts trying to simplify the proof. For example, Huisken [Hui98] introduced an isoperimetric estimate that controls the ratio of chord length and arc length, and Hamilton [Ham95] gave an estimate controlling the isoperimetric profile to that of a circle of the same area. Then they use the results about the classification of singularities for the curve shortening flow to deduce that an embedded curve must become circular in shape and shrinks to a point. Recently by studying the paper of Huisken [Hui98], Andrews and Bryan [AB11] proved an isoperimetric estimate that gives a proof of Grayson s theorem which does not require the classification of singularities. The goal of this thesis is to explain every detail of the Andrews and Bryan s paper [AB11]. This thesis also gives a simple proof of short time 2

5 existence for the curve shortening flow, a part that is usually proved for general case (mean curvature flow) in literatures. The main theorem in this thesis is Theorem 1.1 (Grayson [Gra87]). Consider the curve shortening flow γ t = kn, (1.1) where k(, t) and N(, t) are respectively the curvature and the unit normal of the curve γ(, t). Let γ 0 : S 1 R 2 be a smooth embedded curve. Then there exists a time T > 0 and a smooth map γ : S 1 [0, T ) R 2 such that γ satisfies (1.1), γ(, 0) = γ 0 ( ), and for each t [0, T ), γ(, t) is a smooth embedded curve. Moreover, when t goes to T, γ(, t) becomes circular in shape, shrinks to a point and the curvature k(, t) of γ(, t) is unbounded. 3

6 Chapter 2 Short time existence In this chapter we recall some facts about parabolic partial differential equation. Then we apply them to prove the local existence of solution of the curve shortening flow. 2.1 Short time existence for parabolic partial differential equation Let I be an interval in R. We consider the nonlinear parabolic partial differential equation u t (x, t) = F (x, t, u, u x, u xx ), (x, t) I (0, T ), (2.1) where F : (x, t, z, p, q) I (0, T ) R 3 F (x, t, z, p, q) R is smooth. Equation (2.1) is called parabolic if F (x, t, z, p, q) is positive in I (0, T ) q R 3. It is called uniformly parabolic if there exist positive numbers λ, Λ such that λ F q Λ in I (0, T ) R3. A function u : (x, t) I (0, T ) u(x, t) R is said to be a classical solution of (2.1) if u is twice continuously differentiable with respect to x, 4

7 continuously differentiable with respect to t and satisfies (2.1). Next we define the parabolic Hölder space C k,α and Hölder space C k,α (see, for instance [LSU68], [Lie96]). Let I be an interval in R. We set Q = I (0, T ) with T > 0. functions u : Q R and v : I R as follows. We define the norms and semi-norms for For α (0, 1] we denote { } u(x, t) u(y, s) [u] α = sup : (x, t), (y, s) Q, ( x y 2 + t s ) α/2 u C(Q) = u Ck,α (Q) = sup (x,t) Q i+2j k [v] α = sup x,y I u(x, t), i+j u x i t j + C(Q) v(x) v(y), x y v C(I) = sup v(x), x I v C k,α (I) = [ ] i v k v x i + i k C(I) x k i+2j=k [ ] i+j u x i t j The parabolic Hölder space C k,α (Q) is the normed vector space consisting of all functions u : Q R such that they are kth continuously differentiable with respect to x and t, and all their partial derivatives of order k are Hölder continuous. The same is for Hölder space C k,α (I). Ck,α (Q) and C k,α (I) are the Banach spaces with the norms Ck,α (Q) and C k,α (I), respectively. To prove the local existence of solution of the nonlinear parabolic partial differential equation (2.1), we use results about the local solvability of linear parabolic partial differential equation (see, for instance [LSU68], [Lie96]). We note here that the Hölder space is necessary. For example, there is a continuous function f such that u = f has no C 2 solution.. α, α 5

8 Theorem 2.1 (existence, uniqueness and a priori estimate, Theorem 5.1 in [LSU68] page 320). Consider u t (x, t) = a(x, t)u xx (x, t) + b(x, t)u x (x, t) + c(x, t)u(x, t) f(x, t) (2.2) on Q = S 1 (0, T ) where it is uniformly parabolic, and the coefficients are 2π periodic and in C k,α (Q) for some (k, α). Let u 0 be in C k+2,α (S 1 ). Then there exists a unique solution u in C k+2,α (Q) satisfying u(, 0) = u 0 and a constant C which depends on λ, Λ, k, α such that ) u Ck+2,α (Q) ( f C Ck,α (Q) + u 0 C k+2,α (S 1 ). By using Theorem 2.1 we can prove the local existence of solution of the nonlinear partial differential equation (2.1). Theorem 2.2 (local existence). Consider (2.1) on Q = S 1 (0, T ) where F is smooth and uniformly parabolic. Let u 0 be in C k+2,α (S 1 ) for some (k, α). Suppose further that all kth partial derivatives of F are Hölder continuous. Then there exists a positive time t 0 T such that (2.1) has a solution u in C k+2,α (Q(t 0 )) with Q(t 0 ) = S 1 (0, t 0 ) satisfying u(, 0) = u 0. Proof. We use the inverse function theorem (see [Lan02]) to prove this theorem. We first prove the theorem for the case u 0 = 0. We set Q(t) = S 1 (0, t) and introduce the functional F : X = { u C k+2,α (Q(t)) u(, 0) = 0 } C k,α (Q(t)) defined by F(u) = u t F (x, t, u, u x, u xx ). Then F is well-defined and the derivative of F at u 1 is given by ( F DF(u 1 )(u) = u t z u + F p u x + F ) q u xx, for u C k+2,α (Q(t)), where the derivatives are taken at u 1. 6

9 The uniform parabolicity of F and Theorem 2.1 imply that DF(u 1 )(u) = 0 has unique solution u = 0. This means that the derivative DF(u 1 ) is one-to-one. In other words, F is invertible. So by the inverse function theorem, there is a δ > 0 such that for f F(u 1 ) Ck,α (Q(t)) < δ there exists a unique u C k+2,α (Q(t)) satisfying u u 1 Ck+2,α (Q(t)) < δ and F(u) = f. Furthermore by choosing u 1 = F (x, t, 0, 0, 0)t, we get F(u 1 ) goes to 0 when t goes to 0. In fact, there is a t 0 > 0 such that F(u 1 ) Ck,α (Q(t 0 )) < δ. This implies that equation F(u) = 0 is solvable. For the case u 0 0 we proceed as follows: If u is a solution of u t = F (x, t, u, u x, u xx ) with u t=0 = u 0 then v = u u 0 is a solution of v t = F (x, t, v u 0, v x u 0x, v xx u 0xx ) = G(x, t, v, v x, v xx ) with v t=0 = 0. On the other hand, if v is a solution of v t = F (x, t, v, v x, v xx ) with v t=0 = 0 then u = v + u 0 is a solution of u t = F (x, t, u u 0, u x u 0x, u xx u 0xx ) = G(x, t, u, u x, u xx ) with u t=0 = u 0. This means we have shown that (2.1) has a solution. The proof is complete. 2.2 Short time existence for curve shortening flow Let k 1 be a positive integer and let α be in (0, 1]. We consider the curve shortening flow γ t = kn, γ t=0 = γ 0, (2.3) where γ 0 is an embedded curve in C k 1,α (S 1 ), and k(, t) and N(, t) are respectively the curvature and the unit normal of the curve γ(, t). Proposition 2.3. Consider the initial value problem γ t = kn + G(γ, k)t, (2.4) 7

10 where G is a bounded smooth function. Let γ be a solution of (2.4). Then there exists a smooth function ϕ : S 1 [0, T ) S 1 such that γ 1 (, t) = γ(ϕ(, t), t) is a solution of (2.3). Proof. We have γ 1 t (p, t) = γ p(ϕ(p, t), t)ϕ t (p, t) + γ t (ϕ(p, t), t) = γ p (ϕ(p, t), t)ϕ t (p, t) + kn + G(γ, k)t. So if γ 1 satisfies (2.3), then ϕ must solve the equation ϕ t (p, t) = G(γ(ϕ(p, t), t), k(ϕ(p, t), t)). (2.5) γ p (ϕ(p, t), t) This is a first order differential equation with parameter p. So by the standard theory of ordinary differential equation, (2.9) has a unique smooth solution ϕ : S 1 [0, T ) S 1. Proposition 2.3 implies the following corollary. It says that to find a solution of the curve shortening flow (2.3) we just need to solve the equation γ, N = k. t Corollary 2.4. If γ satisfies γ, N = k, then it can be reparametrized to t a solution of the curve shortening flow (2.3). Thanks to Corollary 2.4, we now consider the following initial value problem γ t, N = k, γ t=0 = γ 0. (2.6) Suppose that we have a solution γ : S 1 [0, T ) R 2 which satisfies (2.6). If for small t we can express γ as following. γ(p, t) = γ 0 (p) + u(p, t)n 0 (p), 8

11 where N 0 is the unit normal of γ 0. For simplicity we assume that γ 0 is parametrized by arc length. Then γ u k = t, N = t N 0, N = u N0, N. t So u t = k N 0, N. (2.7) If u is a solution of equation (2.7) then γ(, t) = γ 0 ( )+u(, t)n 0 ( ) satisfies (2.6). In fact, γ u t, N = t N 0, N = u t N 0, N = k N 0, N N 0, N = k. We now compute the unit tangent T (, t), the unit normal N(, t), the curvature k(, t) of the curve γ(, t), and the inner product N 0, N, namely T = (1 k 0u)T 0 + u p N 0 [ (1 k0 u) 2 + u 2 p] 1/2, N = u pt 0 + (1 k 0 u)n 0 [ (1 k0 u) 2 + u 2 p] 1/2, N 0, N = k = (1 k 0u)u pp + 2k 0 u 2 p + k 0 u p u + k 3 0u 2 2k 2 0u + k 0 [ (1 k0 u) 2 + u 2 p] 3/2, 1 k 0 u [ (1 k0 u) 2 + u 2 p] 1/2, where k 0 is the curvature of γ 0 and T 0 is the unit tangent of γ 0. By injecting k and γ into equation (2.7), we obtain u t = (1 k 0u)u pp + 2k 0 u 2 p + k 0 u p u + k0u 3 2 2k0u 2 + k 0 (1 k 0 u) [ ] (1 k 0 u) 2 + u 2 p 1 = [ ]u (1 k0 u) 2 + u 2 pp + 2k 0u 2 p + k 0 u p u + k0u 3 2 2k0u 2 + k 0 p (1 k 0 u) [ ]. (1 k 0 u) 2 + u 2 p (2.8) If t is small, u and u p is bounded. In other words, there exists t 0 > 0 such that u 1/(2k 0max ) and u p 1/(2k 0max ) for t [0, t 0 ]. Then (1 k 0 u) 2 + u 2 p u k 0 + k 2 0u 2 + u 2 p 4 + 1/(2k 0max ) 2. 9

12 1 So [ ] 1 (1 k 0 u) 2 +u 2 4+1/(2k 0max > 0. This means that (2.8) is uniformly ) 2 p parabolic partial differential equation. Then it is straightforward to prove the local existence of solution by applying Theorem 2.2. Thus we have proved the following theorem. Theorem 2.5 (local existence for curve shortening flow). Let γ 0 be an embedded curve in C k 1+2,α (S 1 ). Then the curve shortening flow γ t = kn (2.9) has a solution γ in C k 1+2,α (S 1 (0, t 0 )) such that for each t [0, t 0 ), γ(, t) is an embedded curve, and γ(, 0) = γ 0 ( ). 10

13 Chapter 3 Long time existence Let T be the maximal time such that the solution of (2.9) exists. Then we call γ : S 1 [0, T ) R 2 the maximal solution of the curve shortening flow. From now on for simplicity a solution always means a maximal solution. The curvature k(, t) of γ(, t) becomes unbounded when t goes to T. In fact, if the curvature is uniformly bounded then the curve γ(, T ) is smooth and we can extend the solution to the time interval [0, T + ɛ) by using Theorem 2.5. This contradicts the fact that γ is a maximal solution. So the curvature must be unbounded when t goes to T. Furthermore, we have some properties of γ. Proposition 3.1 (uniqueness). Let γ 1 and γ 2 be two solutions of (2.9) in C k+2,α (S 1 [0, T )). If γ 1 (, 0) = γ 2 (, 0), then γ 1 (, t) = γ 2 (, t) for all t > 0. Proof. With the same technique as above, for t 0 small, we can express the curves γ 1 and γ 2 as follows. γ 1 (p, t) = γ 1 (p, 0) + u 1 (p, t)n 1 (p), γ 2 (p, t) = γ 2 (p, 0) + u 2 (p, t)n 2 (p), for t [0, t 0 ), where γ 1 (, 0) = γ 2 (, 0), and N 1 and N 2 are respectively the 11

14 unit normals of γ 1 (, 0) and γ 2 (, 0). Then u 1 and u 2 are two solutions of the nonlinear parabolic partial differential equation u t = u t=0 = 0, 1 [ ]u (1 k0 u) 2 + u 2 pp + 2k 0u 2 p + k 0 u p u + k0u 3 2 2k0u 2 + k 0 p (1 k 0 u) [ ], (1 k 0 u) 2 + u 2 p where k 0 is the curvature of γ 1 (, 0). So by the maximum principle, we get u 1 = u 2 for t [0, t 0 ). By continuing the procedure with the initial curve γ 1 (, t 0 ), we thus proved that γ 1 (, t) = γ 2 (, t) for t > 0. Proposition 3.2 (preserving embeddedness). Let γ be a solution of (2.9) in C k+2,α (S 1 [0, T )). Then γ(, t) is embedded if γ(, 0) is embedded. Proof. We prove this proposition by contradiction. Suppose that there is a time t 0 > 0 such that γ(, t 0 ) has self-intersection but not for t < t 0 (note that γ(, t) is embedded for t small). We can find p, q S 1 with p q such that γ(p, t 0 ) = γ(q, t 0 ). We set R = ( a, a) ( b, b). Up to rotation and translation and because of the uniform boundedness of the curvature of γ on small time interval ( ɛ + t 0, t 0 + ɛ), we can assume that γ(p, t 0 ) is the origin and R γ are graphs of two functions u 1 and u 2 for t ( ɛ + t 0, t 0 + ɛ) with u 2 u 1, u 2 (0, t 0 ) = u 1 (0, t 0 ) and u 2 > u 1 for t < t 0. Then u 1 and u 2 are two solutions of u t =. 1 + u 2 x (This equation can be obtained when viewing curves γ(, t) locally as graphs of functions). The maximum principle implies that u 2 = u 1 for t < t 0. It contradicts the fact that u 2 > u 1 for t < t 0. This completes the proof. Proposition 3.3 (containment principle). Let γ 1 and γ 2 be two solutions of (2.9) in C k+2,α (S 1 [0, T )). Let D(t) be a bounded domain enclosed by 12 u xx

15 γ 2 (, t). Suppose that γ 1 (, 0) is in D(0). Then γ 1 (, t) is also in D(t) for all t > 0. Proof. If there exists time t > 0 such that γ 1 (, t) D(t). Then there exists t 0 < t such that γ 1 (, t 0 ) touches γ 2 (, t 0 ) at the point γ 2 (p 0, t 0 ) (boundary of D(t 0 )). Up to rotation and translation, we can assume that γ 2 (p 0, t 0 ) is the origin. Because the curvature of γ 1 (, t) and γ 2 (, t) are uniformly bounded on small time interval ( ɛ + t 0, t 0 + ɛ), there exists an interval I = ( a, a) such that γ 1 (, t) and γ 2 (, t) are respectively the graphs of functions u 1 (, t) and u 2 (, t) over I for t ( ɛ+t 0, t 0 +ɛ). We can also assume that u 1 (, t 0 ) u 2 (, t 0 ). Furthermore, the hypothesis asserts that for t > t 0 there exists a point p such that u 1 (p, t) < u 2 (p, t). By a simple computation, we get u 1 and u 2 are two solutions of the parabolic partial differential equation u t = u xx, 1 + u 2 x with u 2 u 1 at time t 0. So the maximum principle implies that u 2 u 1 for t [t 0, t 0 + ɛ). This obtains a contradiction. The proof is complete. 13

16 Chapter 4 Convergence In this chapter we shall prove Theorem 1.1. We follow the paper of Andrews and Bryan [AB11]. Let γ 1 : (p, t 1 ) S 1 [0, T 1 ) γ 1 (p, t 1 ) R 2 be a solution of (2.9). We now normalize the flow. We would like to work with the flow which exists for all time. We set γ(p, t) = 2π L[γ 1 (, t 1 )] γ 1(p, t 1 ), (4.1) where t = T = t1 0 T1 0 ( ) 2 2π dt 2, L[γ 1 (, t 2 )] ( ) 2 2π dt 2. L[γ 1 (, t 2 )] It is straightforward to check that the length of γ(, t) is 2π. We also have the following proposition. Proposition 4.1. γ(p, t), as in (4.1), is a solution of the normalized curve shortening flow problem γ t = k2 γ + kn, (4.2) 14

17 where k is the curvature of the normalized curve γ(, t) at time t and k 2 (t) = k(p, t) 2 ds(p) is the average curvature. 1 2π Proof. We just compute the derivative of γ with respect to t. We get t γ(p, t) = ( ) 2π t L[γ 1 (, t 1 )] γ 1(p, t 1 ) 2π d ( = L[γ1 (, t L[γ 1 (, t 1 )] 2 1 )] ) 2π γ 1 (p, t 1 ) + dt L[γ 1 (, t 1 )] t γ(p, t 1) 2π dt 1 d ( = L[γ1 (, t L[γ 1 (, t 1 )] 2 1 )] ) 2π dt 1 γ 1 (p, t 1 ) + γ 1 (p, t 1 ) dt dt 1 L[γ 1 (, t 1 )] dt t 1 ( ) 2 2π L[γ1 (, t 1 )] d ( = L[γ1 (, t L[γ 1 (, t 1 )] 2 1 )] ) γ 1 (p, t 1 ) 2π dt 1 ( ) 2 2π L[γ1 (, t 1 )] + γ 1 (p, t 1 ) L[γ 1 (, t 1 )] 2π t 1 = 1 k 1 (p, t 1 ) 2 γ 1p (p, t 1 ) dpγ 1 (p, t 1 ) + L[γ 1(, t 1 )] k 1 (p, t 1 )N(p, t 1 ) 2π S 2π 1 = 1 ( k 1 (p, t 1 ) L[γ ) 2 1(, t 1 )] γ p (p, t) dpγ(p, t) + k(p, t)n(p, t) 2π S 2π 1 = 1 k(p, t) 2 γ p (p, t) dpγ(p, t) + k(p, t)n(p, t) 2π S 1 = k 2 (t)γ(p, t) + k(p, t)n(p, t). Before giving the proof of Theorem 1.1, we need a result about the comparison between the chord length d and the arc length l of two points on the curve γ(, t). The idea is to find a function f which behaves like function 2 sin(x/2) such that d f(l) because we know that d = 2 sin(l/2) for the flow of unit circles. Theorem 4.2 (Andrews and Bryan [AB11]). Let γ : S 1 [0, T ) R 2 be a smooth embedded solution of the normalized curve shortening flow (4.2) with fixed total length 2π. Then there exists t 0 R such that for p, q S 1 and 15

18 t 0, d(p, q, t) f(l(p, q, t), t t 0 ), (4.3) where d(p, q, t) = γ(q, t) γ(p, t) and l(p, q, t) = min { q γ p p dp, 2π q γ p p dp }, and f is defined by f(x, t) = 2e t arctan ( e t sin( x)) for t R 2 and x [0, 2π]. Furthermore, T = and sup p S 1 k(p, t) 2 1+2e2(t 0 t) 4 for t [0, T ). We remark that inequality (4.3) could be used to measure the embeddedness of the flow γ. In fact, if there exist p, q S 1 and t 0 such that d(p, q, t) = 0 then f(l(p, q, t), t t 0 ) = 0. So l(p, q, t) = 0. This implies p = q. 4.1 Proof of Theorem 4.2 We divide the proof into three steps Step 1 In this step we prove the inequality (4.3) for t = 0, that is the inequality (4.3) holds for the initial curve γ 0 ( ) = γ(, 0). By taking the derivative of f with respect to t, we get t f(x, t) = 2et arctan (e t sin( x2 ) ) + 2e t e t sin( x) e 2t sin 2 ( x) 2 = 2e t g (e t sin( x2 ) ), where g(x) = arctan(x) x. We have 1+x 2 g(0) = 0, g (x) = y 1 + y2 2y 2 2 (1 + y 2 ) 2 = 2y 2 (1 + y 2 ) 2 > 0. 16

19 So f is a strictly increasing function with respect to t. We denote F (p, q, t) = d(p, q, 0) f(l(p, q, 0), t) for p, q S 1, p q and t R. Then F t theorem, = f(l(p, q, 0), t) > 0. Thanks to the implicit function t a(p, q) = inf { e t F (p, q, t) 0 } is a smooth function and the following equality holds at 0 < d < 2 sin(l/2), d(p, q, 0) = f(l(p, q, 0), log(a(p, q))). (4.4) We can extend the function a to S 1 S 1 such that a is continuous on S 1 S 1. We have the following proposition. Proposition 4.3. The function a is defined by max { 4k 0 (p) 2 1, 0 } a(p, p) = 2 is continuous, where k 0 is the curvature of curve γ 0. Proof. We just need to prove that a is continuous at points (p, p). Let x 1, x 2 be in S 1 and suppose that γ(, 0) is parametrized by arc length. Because of the Taylor expansion at x 1, we get γ 0 (x 2 ) = γ 0 (x 1 ) + (x 2 x 1 )γ 0,p (x 1 ) + (x 2 x 1 ) 2 γ 0,pp (x 1 ) 2 + (x 2 x 1 ) 3 γ 0,ppp (x 1 ) + o( x 2 x 1 4 ) 6 = γ 0 (x 1 ) + (x 2 x 1 )T (x 1 ) + (x 2 x 1 ) 2 k 0 (x 1 )N(x 1 ) 2 + (x 2 x 1 ) 3 ( k0,p (x 1 )N(x 1 ) k 0 (x 1 ) 2 T (x 1 ) ) + o( x 2 x 1 4 ). 6 17

20 So d(x 1, x 2, 0) 2 = γ 0 (x 2 ) γ 0 (x 1 ) 2 = x 2 x 1 2 x 2 x 1 4 k 0 (x 1 ) 2 + x 2 x 1 5 k 0 (x 1 )k 0,p (x 1 ) + O( x 2 x 1 6 ) 3 6 = x 2 x 1 2 x 2 x 1 4 k 0 (x 1 ) 2 + O( x 2 x 1 5 ) ( 3 = x 2 x x ) 2 x 1 2 k 0 (x 1 ) 2 + O( x 2 x 1 3 ) 3 ( = x 2 x x 2 x 1 2 k 0 (p) 2 + O ( ( x 1 p + x 2 p ) x 2 x 1 2)) 3 ( = l(x 1, x 2, 0) 2 1 l(x ( )) 1, x 2, 0) 2 k 0 (p) 2 + O( x 1 p + x 2 p ) 3 (4.5) By taking the square root both sides of the equality (4.5), we get d(x 1, x 2, 0) = l(x 1, x 2, 0) (1 l(x ( )) 1, x 2, 0) 2 k 0 (p) 2 + O( x 1 p + x 2 p ) 6 (4.6) By using Taylor expansion, we get f(x, t) = x 1 + 2e 2t x 3 + O(x 4 ), 24 f ( l(x 1, x 2, 0), log(a(x 1, x 2 )) ) = l(x 1, x 2, 0) 1 + 2a(x 1, x 2 ) 2 l(x 1, x 2, 0) O ( l(x 1, x 2, 0) 4). Thanks to the equalities (4.4), (4.6) and (4.7), we obtain l l3 6 We finally get ( k0 (p) 2 + O( x 1 p + x 2 p ) ) = l 1 + 2a2 l 3 + O ( l 4). 24 lim a(x 1, x 2 ) = x 1 p,x 1 p max{4k0 (p) 1, 0}. 2 (4.7) 18

21 Because of the equality (4.4) and Proposition 4.3, we obtain d(p, q, 0) f(l(p, q, 0), log(a(p, q))) f(l(p, q, 0), t 0 ), where t 0 = log(sup p,q S 1 a(p, q) ). This means that the equality (4.3) holds for the initial time t = 0. This finishes the proof of step Step 2 We now show that the equality (4.3) holds for positive time t > 0. We set Z(p, q, t) = d(p, q, t) f(l(p, q, t), t t 0 ) and set Z ɛ (p, q, t) = Z(p, q, t) + ɛe Ct (C will be chosen later) for ɛ > 0. We see that function Z is continuous on S 1 S 1 [0, T ) and smooth at points p q. We prove that Z ɛ 0 for every ɛ > 0. Suppose that there exists time t 2 such that Z ɛ (p, q, t) < 0 for p, q S 1. Then there are p 0, q 0 and t 1 < t 2 such that Z ɛ (p 0, q 0, t 1 ) = So at the point (p 0, q 0, t 1 ), Z(p 0, q 0, t 1 ) = ɛe Ct 1. Z t ɛ(p 0, q 0, t 1 ) = Z(p t 0, q 0, t 1 ) + ɛce Ct 1 0. inf Z ɛ (p, q, t) = 0. (4.8) p,q S 1,t [0,t 1 ] Furthermore, we have Now let ξ, η R. We set σ(u) = (p 0 + ξu, q 0 + ηu, t 1 ). We suppose that the smooth embedded curve γ(, t 1 ) is parametrized by arc length. We 19

22 compute the derivative u Z ɛ(σ(u)), namely u Z ɛ(σ(u)) = u d(σ(u)) f u l(σ(u)) = = γ(q0 + ηu, t 1 ) γ(p 0 + ξu, t 1 ) f u u η p γ(q 0 + ηu, t 1 ) ξ p γ(p 0 + ξu, t 1 ), f (η ξ) ( T = η (q0 + ηu, t 1 ), W ) f ( q0 +ηu p 0 +ξu ) ds γ(q 0 + ηu, t 1 ) γ(p 0 + ξu, t 1 ) γ(q 0 + ηu, t 1 ) γ(p 0 + ξu, t 1 ) ( + ξ ) T (p 0 + ξu, t 1 ), W + f, where f is the derivative with respect to first argument, T (p 0 + ξu, t 1 ) = γ(p p 0 + ξu, t 1 ) and T (q 0 + ηu, t 1 ) = γ(q p 0 + ηu, t 1 ) are the tangent vectors and W = γ(q 0 + ηu, t 1 ) γ(p 0 + ξu, t 1 ) γ(q 0 + ηu, t 1 ) γ(p 0 + ξu, t 1 ). By virtue of (4.8), we get u Z ɛ(σ(u)) u=0 =0. So by choosing ξ = η = 1, we obtain We have two cases: 1. T (q 0 ) = T (p 0 ) W, 2. T (q 0 ) and T (p 0 ) bisect. f = T (q 0 ), W = T (p 0 ), W. 1. We first rule out the first case. If T (p 0 ) = T (q 0 ) and the vector γ(q 0, t 1 ) γ(p 0, t 1 ) is in the region Ω bounded by the smooth embedded curve F (, t 1 ) then then one of the two normal vectors at points p 0, q 0 must point out Ω. Thus the vector γ(q 0, t 1 ) γ(p 0, t 1 ) must intersect the curve F (, t 1 ) at a point r 0 is such that p 0 < r 0 < q 0. Then we get d(p 0, q 0, t 1 ) = d(p 0, r 0, t 1 ) + d(r 0, q 0, t 1 ), l(p 0, q 0, t 1 ) = min { l(p 0, r 0, t 1 ) + l(r 0, q 0, t 1 ), 2π l(p 0, r 0, t 1 ) l(r 0, q 0, t 1 ) }. 20

23 We observe that for x, y > 0 and x + y < 2π, f(x + y) f(x) + f(y). Moreover, f(x, t) = f(2π x, t). We thus get Z(p 0, q 0, t 1 ) = d(p 0, q 0, t 1 ) f(l(p 0, q 0, t 1 ), t 1 ) = d(p 0, r 0, t 1 ) + d(r 0, q 0, t 1 ) f(l(p 0, r 0, t 1 ) + l(r 0, q 0, t 1 ), t 1 ) > d(p 0, r 0, t 1 ) + d(r 0, q 0, t 1 ) f(l(p 0, r 0, t 1 ), t 1 ) f(l(r 0, q 0, t 1 ), t 1 ) = d(p 0, r 0, t 1 ) f(l(p 0, q 0, t 1 ), t 1 ) + d(r 0, q 0, t 1 ) f(l(r 0, q 0, t 1 ), t 1 ) = Z(p 0, r 0, t 1 ) + Z(r 0, q 0, t 1 ). So either Z(p 0, r 0, t 1 ) < Z(p 0, q 0, t 1 ) or Z(r 0, q 0, t 1 ) < Z(p 0, q 0, t 1 ). contradicts the equality (4.8). This 2. The second case is more complicated than the first case. We firstly calculate the second derivative 2 Z u 2 ɛ (σ(u)), for simplicity we denote p = 21

24 p 0 + ξu and q = q 0 + ηu, namely 2 u Z ɛ(σ(u)) = η ( T (q, t1 ), W ) f + ξ ( ) T (p, t 2 1 ), W + f u u ( = η η ) p T (q, t 1), W + T (q, t 1 ), u W f u l(p, q, t 1) ( ) + ξ ξ p T (p, t 1), W T (p, t 1 ), u W + f u l(p, q, t 1) = η 2 p T (q, t 1), W + η T (q, t 1 ), u W ηf u l(p, q, t 1) ξ 2 p T (p, t 1), W ξ T (p, t 1 ), u W + ξf u l(p, q, t 1) = η 2 κ(p, t 1 )N(p, t 1 ), W + η2 d ηξ Tp, T q d η2 Tq, W 2 ξη + Tp, W T q, W d d η 2 f + ηξf η 2 κ(p, t 1 )N(p, t 1 ), W ηξ ξ 2 Tp, T q + d + ηξ Tp, W T q, W ξ2 Tp, W 2 + ηξf ξ 2 f ( d d 1 ( = ξ ) ) 2 1 T (p), W 2 k(p, t 1 )N(p, t 1 ), W f d ( ) 1 ( ) + 2ηξ T (p), W T (q), W T (p), T (q) + f d + η 2 ( 1 d ( 1 T (q), W 2 ) + k(q, t 1 )N(q, t 1 ) f ). By the fact that T (p 0 ) and T (q 0 ) bisect, we can set T (p 0 ), W = T (q 0 ), W = cos θ and T (p 0 ), T (q 0 ) = cos 2θ = 2 cos 2 θ 1. By taking ξ = 1 and η = 1, d 22

25 we get 2 u Z ɛ(σ(u)) 2 = 1 ) (1 T (p 0 ), W 2 k(p 0, t 1 )N(p 0, t 1 ), W f u=0 d 2 ( ) T (p 0 ), W T (q 0 ), W T (p 0 ), T (q 0 ) 2f d + 1 ) (1 T (q 0 ), W 2 + k(q 0, t 1 )N(q 0, t 1 ) f d = 1 ( 1 cos 2 θ ) κ(p 0, t 1 )N(p 0, t 1 ), W f d 2 ( cos 2 θ 2 cos 2 ϕ + 1 ) 2f d + 1 ( 1 cos 2 θ ) + κ(q 0, t 1 )N(q 0, t 1 ), W f d = κ(q 0, t 1 )N(q 0, t 1 ) κ(p 0, t 1 )N(p 0, t 1 ), W 4f 0. Next we compute the evolution of the arc length l(p, q, t 1 ) and the chord length d(p, q, t 1 ), namely γ γ d(p, q, t) = (q, t) (p, t), W t t t = κ 2 (t)γ(q, t) + κ(q, t)n(q, t) κ 2 (t)γ(p, t) κ(p, t)n(p, t), W = κ(q, t)n(q, t) κ(p, t)n(p, t), W + κ 2 (t)d(p, q, t), (4.9) 23

26 t l(p, q, t) = t = = = = q p q p q p q p q p γ p (p, t) dp t γ p(p, t) dp t γ γ p (p, t) p(p, t), dp γ p (p, t) ( ) k p 2 (t)γ(p, t) + k(p, t)n(p, t), k 2 (t) γ p (p, t) dp = κ 2 (t)l(p, q, t) q p q p k(p, t) 2 γ p (p, t) dp k(p, t) 2 γ p (p, t) dp. γ p (p, t) dp γ p (p, t) (4.10) 24

27 Thanks to the equalities (4.9) and (4.10), we get ɛce Ct 1 t Z(p 0, q 0, t 1 ) = d t (p 0, q 0, t 1 ) f l t (p 0, q 0, t 1 ) f t = k(q 0, t 1 )N(q 0, t 1 ) k(p 0, t 1 )N(p 0, t 1 ), W + k 2 (t 1 )d(p 0, q 0, t 1 ) q ) f (k 2 (t 1 )l(p 0, q 0, t 1 ) k(p, t 1 ) 2 γ p (p, t 1 ) dp f p t = k(q 0, t 1 )N(q 0, t 1 ) k(p 0, t 1 )N(p 0, t 1 ), W ( ) + k 2 (t) d(p 0, q 0, t 1 ) f l(p 0, q 0, t 1 ) q0 + f k(p, t 1 ) 2 γ p (p, t 1 ) dp f p 0 t = k(q 0, t 1 )N(q 0, t 1 ) k(p 0, t 1 )N(p 0, t 1 ), W + k 2 (t 1 ) ( ɛe Ct 1 + f(l(p 0, q 0, t 1 ), t 1 ) f l(p 0, q 0, t 1 ) ) + f q0 p 0 k(p, t) 2 γ p (p, t) dp f t 4f + κ 2 (t) ( ɛe Ct 1 + f(l(p 0, q 0, t 1 ), t 1 ) f l(p 0, q 0, t 1 ) ) + f q0 p 0 k(p, t) 2 γ p (p, t) dp f t. (4.11) For convenience, we rewrite the inequality (4.11) as follow. ɛce Ct 1 4f + k 2( ɛe Ct 1 + f f l ) + f q0 p 0 k 2 f t. (4.12) Now let us estimate some terms in the inequality (4.12). We see that f is concave, so if l > 0, (f f l) = f l > 0 and f f l > 0. Thanks to Holder s inequality, we get k 2 = 1 2π 1 0 k 2 1 ( k) = 1. 2π 2π 0 25

28 Because of 0 l π, f (l, t) = namely q0 p 0 k 2 1 l cos(l/2) 1+e 2t sin 2 (l/2) > 0. We estimate q 0 p 0 k 2, ( q0 p 0 ) 2 k θ2 l, where θ is the angle between T (p 0 ) and T (q 0 ). Thanks to the equalities f = cos(θ/2) = T (q 0 ), W = T (p 0 ), W, we get θ = 2 arccos(f ). We obtain Lf = 4f + f f l + 4f arccos 2 (f ) l ɛce Ct 1 + ɛk 2 e Ct 1 < 0, by choosing C = 2 max{k 2 (t) 0 t t 1 }. f t We observe that the function h(z) = arccos 2 (z) is a convex function. Thanks to this property, we get We finally obtain h(f ) h(cos(l/2)) + h ( cos(l/2) )( f cos(l/2) ) = l2 4 l ( f cos(l/2) ). sin(l/2) Lf = 4f + f f l + 4f arccos 2 (f ) l = 4f + f f l + 4f h(f ) f l t 4f + f f l + 4f ( l 2 l 4 l = 4f + f 4f =: L 1 f. sin(l/2) f t ( f cos(l/2) )) f sin(l/2) t ( f cos(l/2) ) f t The computation gives us that f is in fact a solution of the equation L 1 f = 0. This contradicts with the fact that L 1 f < 0. The proof of step 2 is complete. 26

29 4.1.3 Step 3 We are now going to prove that the curvature k of the flow is uniformly bounded and the maximal time T =. 1. We denote a(p, q, t) = inf { e t0 s : d(p, q, t) f(l(p, q, t), s t 0 ) }. By doing the same argument as in step 1, we get max{4k(p, t)2 1, 0} a(p, p, t) = 2 sup a(p, q, t) p,q S 1,p q e t 0 t. So sup p S 1 k(p, t) 2 1+2e2(t 0 t) 4 for t [0, T ). 2. If T < then the curvature k of γ is bounded, namely k(p, t) sup 0 t T 1+2e 2(t 0 t) 4 <. The unnormalized curve can be computed from the normalized one by setting λ(t) = L[γ 1(, 0)] e t 0 k2 (t 2 )dt 2, 2π γ 1 (p, t 1 ) = λ(t)γ(p, t), where t 1 = t 0 λ(t 2) 2 dt 2 for t [0, T ), and p S 1. This implies that the curvature k 1 of γ 1 is uniformly bounded. This is impossible because we have shown that the curvature k 1 (, t 1 ) becomes unbounded when t 1 goes to T 1. 27

30 4.2 Proof of Theorem 1.1 We now have enough ingredient to prove Theorem 1.1. We have kds = L = 2π and (k 1) 2 ds k 2 ds 2 = (k 2 1)ds 2e 2(t 0 t). kds + L So when t goes to, k(, t) converges to 1. This means geometrically that the normalized curve γ(, t) converges to unit circle. In other words, the unnormalized curve γ 1 (, t 1 ) becomes circular in shape and shrinks to a point whenever t 1 goes to T 1. This proves Theorem

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