On the Chord Length to Arc Length Ratio for Open Curves Undergoing a Length-Rescaled Curvature Flow

Transcription

1 On he Chord engh o Arc engh Raio for Oen Curves Undergoing a engh-rescaed Curvaure Fow Erik Forseh Juy 3, 2007 The engh-rescaed Curvaure Fow We begin by defining a few roeries of he new fow. e φ be he curve arameer and be a ime arameer, so ha curves undergoing he fow can be described by aramerizaions X(φ, ). e be he oa engh of he curve. Then he evouion equaion is given by: ( ) X = κn + κ 2 ds X () Under his fow, we now have ha = 0 (2) which is a condiion ha arises from inegraion and does no hod oinwise. We ve aso go ha: ( ) A = 2π κ 2 ds X, N ds (3) where N is he uni norma vecor, and: κ = κ + κ 3 κ κ 2 ds (4) 2 Theorem e d and denoe he chord engh and arc engh resecivey beween any wo oins and q on an oen curve, so ha d = X(, ) X(q, ) (5)

2 We now rove he foowing: = q dφ (6) Theorem: e X : Γ [0, T ] R 2 be an embedded souion of he engh-rescaed curvaure fow, where Γ S so ha is smoohy defined on Γ Γ. Then he minimum of d/ on Γ is nondecreasing; i is sricy increasing uness d/ and Γ is a sraigh ine segmen. Proof. d and are smooh funcions off he diagona, so i suffices o show ha whenever heir raio aains a saia minimum for some air of oins (, q) Γ Γ a some ime 0 [0, T ], we have ha ( ) d d (, q, 0 ) 0 (7) d Assume wihou oss of generaiy ha q and ha s() s(q) a 0. Then by assumion we have ha, foowing Huisken s noaion, he firs and second variaions obey: δ(ξ) = 0 (8) δ 2 (ξ) 0 (9) for variaions ξ T Γ 0 T q Γ 0. I wi be hefu o rearamerize he curve ocay around and q using arcengh arameers u and v resecivey, so ha he curve is described near hese oins by X(u, 0 ) and X(v, 0 ). Then we need o define severa vecors before coninuing. Have e and e 2 denoe he uni angen vecors aong he curve a and q resecivey: e = (u, 0) u e 2 = (v, 0) v (0) () e ω denoe he uni vecor in he direcion from o q : ω = X(v, ) X(u, ) d (2) Then noe ha our firs variaion obeys a eibniz rue: 2

3 δ = δ(d) d δ() (3) 2 so ha we need ony comue he variaions of d and individuay. In order o comue he firs variaion of d, we need o firs comue d u, d v, and d. We ause o do ha now (e he curvaure vecor κ denoe κn in he cacuaions o foow): d = d d u = X(u, ) X(v, ), e d d v = X(u, ) X(v, ), e 2 d (X(u, ) X(v, )), κ(u, ) + ( = ω, e (4) = ω, e 2 (5) κ 2 ds ) X(u, ) κ(v, ) ( κ 2 ds ) X(v, ) = ω, κ(u, ) κ(v, ) ( κ 2 ds ) ω, X(u, ) X(v, ) = ω, κ(v, ) κ(u, ) + d κ 2 ds (6) We now consider he vanishing of he firs variaion aong e and e 2. We firs comue he firs variaions of d and in hese direcions and hen ug hem ino he roduc rue for he raio d/ : δ(e 0)(d) = D e d = e, d = d u = ω, e (7) δ(e 0)() = (8) δ(0 e 2 )(d) = D e2 d = e 2, d = d v = ω, e 2 (9) δ(0 e 2 )() = (20) so, ugging hese ino equaion (3) we ve go δ(e 0) = d 2 ω, e = 0 (2) δ(0 e 2 ) = ω, e 2 d 2 = 0 (22) from which i foows ha 3

4 ω, e = ω, e 2 = d (23) which we need o kee in mind for fuure cacuaions. Now we urn o he second variaion, for which we can wrie: δ 2 = δ2 (d) 2 δ(d)δ() d(δ())2 3 d 2 δ2 () 0 (24) We wi consider wo cases here: Case : e = e 2 A variaions of now vanish so we need ony consider he firs erm in equaion (24). Thus, we need o comue he second variaion of d : ( ) ( ) δ 2 (e e 2 )(d) = H(d), = d uu + 2d uv + d vv (25) Now, using he reaions we derived in equaion (23), we can comue hese second arias, for which i shoud urn ou ha: d uu = d d ω, κ(u, ) (26) 2 d vv = d d + ω, κ(v, ) (27) 2 d uv = d 2 d (28) Pugging hese ino (25), we we ge ha δ 2 (e e 2 ) = ω, κ(v, ) κ(u, ) 0 (29) Case 2: e e 2 We now choose ξ = e e 2 so ha and δ(e e 2 )() = 2 (30) δ(e e 2 )(d) = ω, e + e 2 (3) 4

5 We erform cacuaions exacy anaogous o hose in Case, comuing he new arias and he new second variaion of d. We end u being abe o concude he same inequaiy ha we go for Case in equaion (29). See he searae wrie-u Case 2 for Theorem o see hese cacuaions in deai. We are now ready o urn our aenion o a quaniy of greaer ineres: he ime derivaive of he raio d/. Wih wha we aready know, we can go ahead and wrie: ( d ) = d d 2 = ω, κ(v, ) κ(u, ) + d κ 2 ds d 2 (32) In order o say more abou his exression we need o come u wih an exression for. Though i may seem ha since his is a engh-rescaed fow ha aricuar erm shoud vanish, we need o kee in mind ha whie he oa engh of he curve canno change, we don know wha is haening ocay, and refers ony o he arc engh beween wo aricuar oins and q on he curve. To begin wih, reca ha we can exress using equaion (6), so ha we can hen wrie q = dφ (33) Then we mus find an exression for he inegrand here: = D E, 2 X = D E, 2 X ( ( = T, κn + κ 2 ds ) X ) = T, κ N κ 2 T + ( ) κ 2 ds T = κ 2 + κ 2 ds (34) So, inegraing (34) wih resec o φ from o q, we ge ha κ 2 ds q = κ 2 ds (35) Wih his exression in hand, we reurn o equaion (32) and wrie: ( d ) = ω, κ(v, ) κ(u, ) + d R κ 2 ds + d 2 q κ2 ds d = ω, κ(v, ) κ(u, ) + d 2 q κ2 ds d 2 q κ2 ds R κ 2 ds 5

6 This as erm is obviousy greaer han zero exce on he diagona of Γ Γ, where i is equa o zero. Thus: d q 2 κ 2 ds 0 (36) and he heorem has been roven. 6