Geometry and Calculus of Variations Lecture notes

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1 Sum /3/12 23:06 pge 1 #1 Geometry n Clculus of Vritions Lecture notes Mrch 12, 2012 Abstrct This is the type version of Dr A.Gilbert s lecture notes. c 1 Lecture Spce Curves Definition: A spce curve is smooth mp x : I R 3, where I is given intervl in R. We llow I = R. Definition: The following is efine nlogously to plnr curves: velocity, spee, regulr curve, orienttion, reprmetriztion, rc-length, unit tngent, frme. Exmple: Let x(t) = (t, t 2, t 3 ) T (cubic curve). It is esy to see tht x (t) = (1, 2t, 3t 2 ) T, v(t) = 1 + 4t2 + 9t 4 n for the rclength we hve s(t) = t u2 + 9u 4 u Exmple: x(t) = (cos t cos 4t, cos t sin 4t, sin t). Notice tht x(t) = 1 i.e. lies on the unit sphere. x(t) encircles the x 3 xis while climbing from equtor to north pole. By irect computtion v(t) = cos 2 t x = ( sin t cos 4t, sin t sin 4t, cos t) T + cos t( 4 sin t, 4 cos t, 0) T. 1.2 Curvture Unit tngent T = x /v is efine s before. Defn: Let x be regulr spce curve. Curvture k is efine by k(t) = 1 v T. Notice tht for plne curves k = ± 1 v T wheres for spce curves k 0. This is preliminry version of the notes. It will be upte weekly. 1

2 Sum /3/12 23:06 pge 2 #2 () The cubic curve (b) The spheric curve Figure 1: Exmples of spce curves 1.3 Biregulr curves We wnt to efine Frenet frme long x, i.e. group of three mutully orthogonl unit vectors in R 3. We cn o so if x is biregulr. Definition: Regulr curve, i.e. the unit tngent T exists, is biregulr if T 0. Next we efine the Unit norml: unlike plne curves there is whole plne, clle norml plne contining vectors perpeniculr to T. Among the norml vectors we give specil nmes to two prticulr ones: norml n binorml. For biregulr x the unit norml is the vector in the irection T N = T T. Clerly N exists, for T 0. From the efinition it follows tht N n T re orthogonl (to see this ifferentite T (t) 2 = 1 by t). At ech point x(t) we now hve 2 orthogonl unit vectors T n N. To complete the Frenet frme we introuce the binorml B. B = T N it is the vector prouct of T n N. The Frenet Frme is the orthogonl set (3 3 orthogonl mtrix) ( T, N, B) of tngent vectors to R 3 t x(t). Recll tht tngent vectors to R 3 t x(t) nee not point long curve, but unit tngent vector T oes. 2

3 Sum /3/12 23:06 pge 3 #3 Figure 2: Helix with k = = Formule for T n N Let x(t), t I be prmetristion of the curve x then ifferentiting x n using efinitions of T, N n k we conclue x = v T x = v T + v T = v T + v 2 k N. In prticulr if s = t, i.e. the curve is rc length prmetrise, then x = T x = T = k N, So N is unit vector orthogonl T in plne forme by x n x = v T. Exmples: 1) Helix: x(t) = ( cos t, sin t, kt). Clerly x (t) = ( cos t, sin t, 0) points long N. 2) Cubic curve: x(t) = (t, t 2, t 3 ) T we hve t t = 0, x (0) = (1, 0, 0), x (0) = (0, 1, 0). 2 Lecture Structurl equtions We hve efine the Frenet frme for biregulr curve x s orthogonl 3 3 mtrix ( T, N, B) where 3

4 Sum /3/12 23:06 pge 4 #4 T = x v N = T T B = T N We wnt to erive the structurl equtions tht is to relte ( T, N, B ) to orthogonl 3 3 mtrix ( T, N, B). Step 1: Eqution for T From the efinition of curvture n the unit norml we hve combining these two equtions we get tht T = T N k = T v T = kv N Step 2: Eqution for B. It is quite esy to see tht for (t) n b(t) t ( b) = b + b reclling the efinition of B n utilizing this formul we get B = t ( T N) = T N + T N = T N since T n N re prllel. Thus T N is prllel to N which yiels B = τvn for some sclr function τ. Step 3: Eqution for N Notice tht N = B T. Thus N = B T + B T = τvn T + B (kvn) = kvt + τvb Summrizing we get the structurl equtions ( T, N, B ) = ( T, N, B)v 0 k 0 k 0 τ 0 τ 0 (1) Defn: τ : I R is the torsion. Defn: The plne etermine by the unit tngent vector T n the unit norml vector N is clle the osculting plne. Theorem: Curvture n torsion re unchnge by reprmetriztion. 4

5 Sum /3/12 23:06 pge 5 #5 Proof: Step 1: Velocity n spee Let us tke t = g(u), u is the new prmeter n g 0 then x 1 (u) = x(g(u)) n x 1 = x v t g = 1 v 1 g Step 2: Tngent By efinition thus T 1 = x 1 v 1 = T 1 = x 1 1 x T = v 1 v t x t g v 1 = T vg v 1 = T vg v g = ε T where ε = g g = ±1: +1 if g is orienttion preserving or 1 orienttion reversing. Step 3: Curvture The the u-erivtive of the tngents equlity n utilize the structurl equtions v 1 k 1N1 = T 1 u = ε T t g = εvkg N Tknig the mgnitue of both sies n noting tht both N 1 n N re unit vectors we conclue v 1 k 1 = ε vk g k 1 = v v 1 k g = k Moreover N 1 = εvkg v 1 k 1 N = ε v v 1 g N = ε g g = ε2 N = N. Step 4: Torsion Finlly we wnt to show tht torsion is unchnge. By efinition B 1 = T 1 N 1 = ε T N = ε B. Differentiting with respect u then B 1 = ε B t g n using the structurl equtions tking the mgnitue n the proof of the Theorem is now complete. B 1 = v 1 τ 1 N1 = ε( vτ N)g = εb g τ 1 = ε v v 1 g τ = ε g g τ = ε2 τ = τ 2.2 Interprettion of curvture n Torsion The trjectory of prticle moving in spce prouces curve. The osculting plne is forme by unit vectors T n N. For generl spce curves the prticle tries to escpe the osculting plne, for otherwise it is plne curve. When k is constntly zero, T never chnges n the curve is stright line. As the nme curvture suggests, k mesures the rte t which ny nonstright curve tens to eprt from its tngent. When τ is constntly zero, the osculting plne never chnges, n we hve plne curve, with constnt binorml B. Thus the torsion mesures the rte t which twiste curve tens to eprt from its osculting plne. Without loss of generlity we my tke the rc length prmetristion of x n ssume tht T, N, B t x(s 0 ) coincie with the irections of x, y n z-xis. Let X be the projection of the curve onto the 5

6 Sum /3/12 23:06 pge 6 #6 osculting plne t x(s 0 ). Then Then X(s) = x(s) x(s 0 ) B[(x(s) x(s 0 )) B]. X = x B( x B), X = x B( x B) In prticulr t s = s 0, using the fct tht B is orthogonl to T n N, we get X = x = T, X = x = k N. Thus the curvture of X t s = s 0 is k = et[ X, X ]. Thus the projection of the curve onto the osculting plne t x(t 0 ) hs curvture k. Now consier the projection onto the norml plne pssing through x(t 0 ), i.e. forme by N(t 0 ) n B(t 0 ). There the projecte curve hs Frenet frme (t, n), with t = N(t 0 ), n = B(t 0 ) t t = t 0 stisfies t = τn, n = τt. thus τ mesures the rottion in norml plne bout T -xis s we pove long the curve. 2.3 Sphericl imge n Drboux vector Let us efine the vector then ω = v k 2 + τ 2 n ω = vτ T + vk B, T = ω T N = ω N B = ω B which cn be reily verifie. ω is clle Drboux vector. It is the ngulr velocity of the prticle when the point is moving long the curve. 2.4 Formule for k n r Proposition 1 Let x be biregulr curve then k = 1 v 3 x x τ = ( x x ) x v 6 k 2 Proof: Let s ifferentite x three times n use structurl equtions x = v T x = v T + v T = v T + v 2 k N, x = v T + v T + 2vv k N + v 2 k N + v 2 k N, = v T + v vk N + 2vv k N + v 2 k N + v 2 k( kv T + τv B). Hence x x = v T (v T + v 2 k N) = v 3 k T N = v 3 k B. Now tking the mgnitue of this vector we conclue k = 1 v 3 x x. Finlly using the fct tht B is orthogonl to N n T n tking the sclr prouct of x x with x we conclue [ x x ] x = v 3 kv 3 kτ = v 6 k 2 τ n the proof follows. 6

7 Sum /3/12 23:06 pge 7 #7 3 Lecture Locl behviour Theorem 2 If we choose the Frenet frme (the triheron) such tht T (0) = e 1, N(0) = e 2, B(0) = e 3 n tke the rc-length prmetriztion then in this coorintes x(s) s e 1 + k 2 s2 e 2 + kr 6 s3 e 3 here e 1, e 2, e 3 is the stnr orthogonl bsis in R 3. To see this we use Tylor s expnsion x(s) = x(0) + x (0)s + x (0) s 2 + x (0) s ! 3! n use the structurl equtions (remember v = 1!) x = T x = T = kn x = k N + kn = k N + k( kt + rb). Thus ssuming tht x(0) = 0 we get x(s) = T s + k N 2! s2 + k N + k( kt + rb) s ! = T (s k2 3! s3 ) + N( k 2! s2 + k 3! s3 ) + rk 3! s Assuming tht s is very smll we cn ignore the higher powers of s within the coefficients of T, N n B thus x(s) = s e 1 + k 2 s2 e 2 + kr 6 s3 e 3 + = An the result follows. Thus loclly ny biregulr curve is cubic! 3.2 Implicitly efine curves ss k 2 s2 kτ 6 s3 Let f 1, f 2 be functions of three vribles, f 1 : (x, y, z) R, f 2 : (x, y, z) R. Consier C the intersection of the zero level sets of f 1 n f 2, then it is n implicitly efine spce curve, provie tht it is not empty. The tngent plne to the grph of f 1 hs norml f 1 thus the tngent plne, Π 1, t (x 0, y 0, z 0 ) is f 1 (x 0, y 0, z 0 ) (x x 0, y y 0, z z 0 ) = 0. Similrly Π 2, given by f 2 (x 0, y 0, z 0 ) (x x 0, y y 0, z z 0 ) = 0 is the tngent plne to the grph of f 2 t (x 0, y 0, z 0 ). If (x 0, y 0, z 0 ) C then the both Π 1 n Π 2 re tngent tp C t (x 0, y 0, z 0 ). Clerly the tngent vector is colliner to the vector proucts of the normls of Π 1 n Π 2, thus T = f 1(x 0, y 0, z 0 ) f 2 (x 0, y 0, z 0 ) f 1 (x 0, y 0, z 0 ) f 2 (x 0, y 0, z 0 ). Defn: An implicitly efine spce curve is si to be regulr if f 1 f 2 0. Clerly the tngent vector is efine for regulr implicitly efine spce curves. 7

8 Sum /3/12 23:06 pge 8 #8 () (b) (c) Figure 3: Exmples of implicit spce curves 3.3 Exmples ) Let f 1 = x x 2 2 x 1 n f 2 = x 3 x 1 b) Let f 1 = x x x n f 2 = x 1 + x 2 + x 3 1. Thus f 1 = 0 is the unit sphere n f 2 = 0 is plne so intersection is circle in R 3. c) Another exmple is the intersection of S 1 : x 2 = x 2 1 n S 2 : x 3 = x Torsion Exmples Exmple 1: Biregulr plne curve hs τ = 0. This is simple exercise. The converse sttement is lso true. A biregulr curve with τ = 0 is plnr! (Hint: Use structurl equtions). Exmple 2: A helix hs constnt curvture n torsion. It is stightforwr to verify tht for helix x(t) = ( cos t, sin t, bt) T k = 2 + b 2, τ = b 2 + b 2. Conversely let us show tht if k n τ re constnts then x is helix. If τ = 0 then x is plne curve with constnt curvture thus it is circle, for k = θ s, θ is n rgument function (Lecture 6) n hence θ = ks + C where C is n rbitrry constnt. Hence x = T = (cos(kt + C), sin(kt + C)) T implying tht x = 1 k (sin(kt + C) + A, cos(kt + C)) + B)T which is circle of rius R = 1/k centere t (A, B). Thus we ssume tht τ 0. Let s tke the rc length prmetristion of the curve. Then Drboux vector ω = τ T + k B is constnt, for ω = τ T + k B = τ(k N) + k( τ N) = 0. Without loss of generlity we my ssume tht ω is colliner to e 3 so ω = e 3 k2 + τ 2. Next ω T = (τt + kb) T = τ, thus the unit tngent vector mkes constnt ngle α with the e 3 τ xis n cos α =. k 2 +τ 2 Furthermore x 3 ( x ω) = = x ω = T s s ω = cos α thus x 3 (s) = s cos α + C, where C is n rbitrry constnt. Next let us consier the projection of the curve onto the plne orthogonl to ω = e 3, i.e. X = x ( x ω)ω. For this plne curve we hve X = x ( x ω)ω, X = x 8

9 Sum /3/12 23:06 pge 9 #9 In prticulr the spee of X is V = ( x ) 2 2( x ω) 2 + ( x ω) 2 = 1 ( x ω) 2 = 1 τ 2 thus V is constnt. If (t, n) is the Frenet frme of X the projecte curve, then from Frenet equtions we hve k N = x = X = s ( X ) = s (V t) = V t = V k X n, where k X is the curvture of the projecte curve X. Equting the right n left hn sies n tking the mgnitues gives k X = k/v = const. Thus X is circle. 4 Lecture Surfces Defn: A locl surfce in R 3 is smooth, injective mp x : D R 3 with continuous inverse x 1 : x(d) D where D is omin (open, connecte set) in R 2. x : (u, v) x(u, v) Remrks: 1) The ssumption of smoothness implies tht x is somewht istorte version of D 2) x is injective points of x(d) re lbelle (coorinte) by corresponing points of D 3) the inverse mpping x 1 is continuous prevents ner self-intersection Uner these conitions x is homeomorphism, i.e. bijection with continuous inverse x Surfces in R 3 If locl surfce x lies in set Σ R 3 then x is locl surfce in Σ. A surfce in R n is subset Σ R 3 such tht for ech point p of Σ there exists locl surfce in Σ whose imge contins neighborhoo N of p in Σ. 4.3 Exmples Sphere: Let us consier x(u, v) = sin u cos v sin u sin v cos u, (u, v) D = (0, π) (0, 2π). The trce of x is unit sphere excluing semi-circle connecting North n South poles. Line u = const is line of ltitue n line v = const is line of longitue. Ellipsoi: Let us consier x(u, v) = sin u cos v b sin u sin v c cos u, (u, v) D = (0, π) (0, 2π). Here, b, c re non-zero constnts. The trce is n ellipsoi with n rc omitte. 9

10 Sum /3/12 23:06 pge 10 #10 Grph of function: Now let f : D R be smooth function. Consier x(u, v) = u v f(u, v). Note tht not ll surfces cn be represente s grphs. See the exmple of unit sphere. p(t) Surfce of revolution: Let t q(t) 0 be curve with trce in the (x 1, x 2 ) plne n rotte it bout the x 1 xis to get p(u) x(u, v) = q(u) cos v. q(u) sin v 4.4 Regulrity For surfce x, the prtil velocities t x(u, v) re tngent vectors to R 3 t x(u, v) given by x u = x 1 u x 2 u x 3 u, x v = x 1 v x 2 v x 3 v. Interprettion: The spce curves (lines) u x(u, v 0 ) n v x(u 0, v), for fixe u 0 n v 0 hve velocities x u n x v. Defn: The locl surfce is regulr if x u n x v re linerly inepenent t ech point of D. 4.5 Displcement n re [ ] δu Consier smll isplcement t ech the point p δv 0 = (u 0, v 0 ) D. Then it is mppe by the [ ] δu mtrix [x u, x v ] to [x u, x v ]. The regulrity implies tht x δv u n x v hve istinct irections t ech p 0 D. Hence smll rectngle (u 0 + δu) (v 0 + δv) in (u, v) spce is pproximtely mppe, by x to the prllelogrm with sies x u δu, x v δv. If x u, x v re linerly inepenent then the mtrix [x u, x v ], which pproximtes x, efines n invertible liner trnsformtion between bove rectngle n prllelogrm. The Inverse Function Theorem sys tht the sme hols for the exct mp x: Aroun ech point p 0 D, there exists neighborhoo u on which x is smooth bijection between N n x(n), where N is neighborhoo of x(p 0 ). 4.6 Regulrity of previous exmples The sphere is regulr locl surfce since x u n x v re linerly inepenent. Similrly it follows tht ellipsoi is regulr too. 10

11 Sum /3/12 23:06 pge 11 #11 5 Lecture Regulr implicitly efine surfces Defn: Let f : R 3 R be smooth function n c R be such tht the level set Σ = {x R 3 : f(x) = c} is non-empty. If f 0, x Σ then Σ is regulr implicitly efine surfce. Theorem: A regulr implicitly efine curve surfce is surfce in R 3. Geometric inpterprettion: f is the norml to Σ, if x3 x 3 surfce contins no verticl irections surfce loclly grph of x 3 (x 1, x 2 ). Exmples: f(x) = x 2 1 f(x) = x 2 3 x 2 1 x 2 2 Quric surfce k 1 x k 2 x k 3 x 2 3 = 1 6 Lecture Curve in surfce The curves lying in surfce provie importnt informtion bout the surfce. Defn: Let Σ be surfce. The curve α : I R 3 is curve in Σ if α(i) Σ. When Σ = x(d) is locl surfce we cn pull bck α to D, since it is much esier to work out in D thn in x(d) Lemm Let α be curve in Σ, Then there exists unique smooth curve w : I D so tht α(t) = x(w(t)). Proof: Define w = x 1 α : I D, so α = x (x 1 α). Exmple: D = (0, 2π) (0, ), x(u, v) = (cos u, sin u, v) T. Tke w 1 (t) = (2t, π/t) T, x(w 1 (t)) = (cos 2t, sin 2t, π/4) T w 2 (t) = (π/2, t) T, x(w 2 (t)) = (0, 1, t) T w 3 (t) = (2t, t) T, x(w 3 (t)) = (cos 2t, sin 2t, t) T 6.2 Tngent spce to surfce Σ Recpitulting: T p R 3, the tngent spce to R 3 t p, is spce of ll tngent vectors (p, v) to R 3 t p. Defn: Let p Σ. A tngent vector v to R 3 t p is tngent to Σ if v is the velocity t p of some curve in Σ. The set of ll such tngent vectors to Σ t p is the tngent spce (plne) to Σ t p, written T p Σ. Next we chrcterise of tngent spce: Lemm Let x(d) be locl surfce in Σ n let p = x(w 0 ) x(d). Then T p Σ is the subspce of T p R 3 spnne by x u (w 0 ) n x v (w 0 ). Proof: Step1: T p Σ Spn(x u, x v ). Let t w(t) be curve in D, such tht w(0) = w 0. Then t x(w(t)) is curve in x(d), pssing through p when t = 0, nt its velocity t p is By efinition the velocity is in T p Σ. t [x(w(t))] t=0 = x u (w 0 )u (0) + x v (w 0 )v (0). 11

12 Sum /3/12 23:06 pge 12 #12 Step2: Spn(x u, x v ) T p Σ. For ny λ, µ R : is velocity of curve in x(d). Summriging Spn(x u, x v ) = T p Σ. λx u (w 0 ) + µx v (w 0 ) = t [x(w 0 + t(λ, µ) T )] t=0 7 Lecture Vector fiels on surfce Defn: A vector fiel Z on surfce Σ is n ssignment to ech p Σ of tngent vector Z(p) to R 3 t p. A unit norml vector fiel N is vector fiel on Σ such tht t ech p Σ : N(p) = 1, N(p) T p Σ On regulr locl surfce x, smooth unit norml fiel is given by N = x u x v x u x v where s for regulr implicitly efine surfce N = f f. Exmple: The norml vector fiel of the sphere x(u, v) = N n for the grph of function z = g(x, y) it is N = f. f = ( gx, gy,1) 1+g 2 x +gy First Funmentl form of Surfce Definition: Let p : x( w 0 ) be point in regulr surfce Σ. The first funmentl form is symmetric biliner form on T p Σ efine by I p (X, Y) = X Y, X, Y T p Σ where The mtrix tht efines this biliner form is [ E F F G 7.3 Exmples Sphere Cyliner Grph Surfce of revolution E = x u x u, F = x u x v, G = x v x v. ] 12

13 Sum /3/12 23:06 pge 13 #13 8 Lecture Arclength Let z(t) = x( w), w(t) = [ u(t) v(t) Hence the rclength of z from z() to z(b) is ] be regulr curve lying in Σ. Its velocity is [ ] z u (t) = (x u x v ) (t) v (t) z (t) t = [ ] E F Exmple 1: For the sphere = F G w (t) = ( 1, 2) hence s = π 0 [1 + 4 [ sin2 t] 1/2 t. ] E F Exmple 2: For the cyliner = F G (1, 1) T hence s = 2π t = 2 2π. 8.2 Reprmetristion (E(u ) 2 + 2F u v + G(v ) 2 ) 1/2 t. [ sin 2 u [ ]. Tke w(t) = (π t, 2t), t (o, π). ]. For w(t) = (t, t) T, t (0, 2π) w (t) = We cn reprmetrise s usul: if s z(s), lying in Σ, is rclength prmetrise then its unit spee n I(z, z ) = 1 Exmple: Cone hs prmetristion x(u, v) = u cos v [ ] [ ] u sin v E F 2 0 n = F G 0 u u 2. Let w(t) = (t, 2 ln t), t (0, ). Then w = (1, 2 t ) n hence s(t) = t 2 = 2t so rc length prmetristion is z 1 (s) = x(w 1 (s)), where w 1 (s) = ( s 2, 2 ln( s 2 ))T Isometry Defn: If S 1 n S 2 re surfces, smooth mp φ : S 1 S 2 is clle locl isometry if it tkes ny curve in S 1 to curve of the sme length in S 2. If locl isometry φ exists we sy S 1 n S 2 re loclly isomorphic. Theorem: A smooth φ : S 1 S 2 is locl isometry if n only if the mtrices of first funmentl forms re equl. [ ] E F Exmple: Plne (u, v, 0) n cyliner (cos u, sin u, v) re isomorphic sonce = F G 8.4 Geoesics [ Motivtion Look for nlogue in surfce Σ of stright lines in plne. Stright lines re chrcterise by: zero ccelertion n istnce minimising between two points. Definition: A curve z in Σ R 3 is geoesic of Σ if its ccelertion z is lwys norml to Σ. ]. Since z is norml to Σ we get z z (s z is tngent to Σ) thus z 2 z = const. Hence we conclue tht geoesics hve constnt spee. t = 2 z z = 0 implying tht 13

14 Sum /3/12 23:06 pge 14 #14 9 Lecture 19 Let us fin the geoesics of the sphere n cyliner. Sphere: First consier the sphere S of rius r, x x = r 2. If z is geoesic then z (t) is colliner to the norml of S t p = z(t) i.e. to z(t) since z is colliner to the norml. Using this observtion we compute t ( z z) = z z + z z. Clerly z z = 0. But z is colliner to the norml, by the efinition of geoesic, n z is colliner to the norml hence z z =. Combining we get t ( z z) = 0. This ientity, in prticulr, implies tht the plne Π pssing through 0 n forme by the vectors z n z hs constnt norml N (iepenent of t). Since z Π it implies tht Thus the geoesics re the gret circles. N z(t) = 0. Cyliner: It is simple exercise. See lso Lecture Distnce minimistion Theorem: Let Σ = x(d) be regulr, locl surfce n z = x w regulr curve connecting A = z(), B = z(b). Then the rclength of z J[ z] = is sttionry implies tht z is geoesic. ( z (s) z (s)) 1/2 s We wnt to show tht the ccelertion z is colliner to n. Let z be curve for which the minimum is relise. Without loss of generlity we my ssume tht z is of spee one, i.e. z z = 1 or equivlently E(u ) 2 + 2F u v + G(v ) 2 = 1 where z(t) = x w(t) n w(t) = (u(t), v(t)), t (, b) is curve in omin D. Introuce z(t, ε) = x(w + εh) where ε > 0 is smll n h() = h(b) = 0 thus z(t, ε) leves the points A, B on the surfce unchnge. Using Tylor s expnsion in ε t ε = 0, it is esy to see tht z(t, ε) = z(t, 0) + ε z ε (t, 0) + ε2 (... ) By efinition z ε (t, ε) = x x(w + εh) = ε u h 1 + x v h 2 where h = (h 1, h 2 ). Hence t ε = 0 we hve z ε (t, 0) = x u(w(t))h 1 (t) + x v (w(t))h 2 (t)- liner combintion of prtil velocities hence z ε (t, 0) T pσ. Denote this tngent vector by T = z ε (t, 0) = x u(w(t))h 1 (t) + x v (w(t))h 2 (t). Thus z(t, ε) = z + εt + ε 2 (... ) n z (t, ε) = z (t) + εt + ε 2 (... ). It follows tht z (t, ε) z (t, ε) = [ z (t) + εt + ε 2 (... )] [ z (t) + εt + ε 2 (... )] = z (t) z (t) + 2 z(t) T + ε 2 ( ) therefore 14

15 Sum /3/12 23:06 pge 15 #15 ε z (τ, ε) z (τ, ε)τ = z (τ, ε) z ε 2T z (τ, ε) + ε(... ) (τ, ε)τ = 2 z (τ, ε) z (τ, ε) t ε = 0 we get T z z (τ) z (τ) = 0 Since z is of spee one, i.e. z (τ) z (τ) = 1, we conclue fter integrtion by prts 0 = T z τ = [T(b) z (b) T() z()] T z τ = T z τ for T(b) = T() = 0 (recll tht T = x u (w(t))h 1 (t) + x v (w(t))h 2 (t) n h() = h(b) = 0). Decompose vector z = z n + z T where z n is colliner to norml n n z T T pσ-the tngentil component. Then 0 = T z = T [ z n + z T ] = T z T Since T T p Σ is n rbitrry tngent vector, for h is rbitrry, we my choose T = z [ z T ] 2 τ = 0 thus the tngentil component of z vnishes n thus z is colliner to the norml n. 10 Lecture Energy The energy functionl of curves in Σ is E[ z] = where we set g( z (τ)) = z (τ) z (τ). g( z (τ))τ = z (τ) z (τ)τ T so tht The erivtion of this sttement is similr to the one for the rc length function J (see lecture 19). As bove we tke z(t, ε) = x(w(t) + εh(t)). Then ε thus t ε = 0 we hve tht g( z (τ, ε))τ = z g( z (t, ε)) z g( z (τ)) T τ = 0. z(τ, ε)τ ε Noting tht z g( z ) = z we conclue, fter integrtion by prts tht 0 = z T = z T hence the tngentil component of the ccelertion z vnishes n z is colliner to the norml n. In other wors if E[ z] is sttionry on curves then τ ( g) is norml to Σ. Now g = 2 z hence E sttionry lso implies tht z is norml to Σ i.e. extremls of E re lso geoesics of Σ. Remrk: For the rclength we hve to impose the constrint z = 1 otherwise we get the trces of geoesics. Plus the wkwr term. For E the constnt spee conition comes utomticlly n no wkwr terms re present. 15

16 Sum /3/12 23:06 pge 16 # Surfces of revolution The question of fining the geoesics of given surfce my be very complicte. However if some itionl properties of surfce is known it cn be reuce to system of ODE s which my be solve explicitly. Such exmple is surfce of revolution. For convenience we tke (θ, z) s inepenent vribles x(θ, z) = R(z) cos θ R(z) sin θ z Here z is the height of the point n R(z) > 0 is given function. The prtil velocities n the first funmentl form re x θ (θ, z) = R(z) sin θ R(z) cos θ 0, x z (θ, z) = R (z) cos θ R (z) sin θ 1 E = R 2, F = 0, G = (R ) hence the energy functionl is E = R 2 (θ ) 2 + (R 2 + 1) z 2 Consier the Euler-Lgrnge equtions (remember tht the unknown functions re θ(t), z(t)) t (2R2 θ ) = 0 t (2(R 2 + 1)z ) z (R2 (θ ) 2 + (R 2 + 1) z 2 ) = 0 Unlike the first eqution the secon one is very complicte. Inste of simplifying it we concentrte on the first one n R 2 (θ ) 2 + (R 2 + 1) z 2 = 1 (since the spee of geoesic is constnt). Thus t (2R2 θ ) = 0 R 2 (θ ) 2 + (R 2 + 1) z 2 = 1 The first eqution mens tht R 2 θ = C, C is constnt. If C = 0 then θ = k-constnt giving the meriins s geoesics. Notice tht z = 0 is not necessrily geoesic. If C 0 then θ = C R 2, then fter iviing the secon eqution by θ 2 we get ( ) z R 2 + (R ) θ = 1 (θ ) 2 = R4 C 2 or equivlently ( ) z 2 [ ] R 4 θ = C 2 1 R2 (R 2 + 1). 16

17 Sum /3/12 23:06 pge 17 #17 Now ssume tht z is function of θ, i.e. in the omin D of (θ, z)-coorintes, z cn be represente s grph over θ-xis, then by chin rule we hve or equivlently z t = z θ θ t θ = z θ fter substituting into the eqution bove results z ( ) 2 [ ] z R 4 = θ C 2 1 R2 (R 2 + 1). This is n ODE with unknown function z = z(θ) n ny solution of this eqution represents geoesic in (θ, z)-plne (the omin D) Cyliner An interesting cse is when R(z) = ρ-constnt which correspons to the cyliner. Then R = 0 n the ODE is ( ) 2 z = ρ2 [ ρ 2 θ C 2 C 2] = A Using the first Euler-Lgrnge eqution we euce θ = C ρ -constnt thus θ(t) = C 2 ρ t + D where D is n 2 rbitrry constnt. Thereby z = Aθ + B = AC ρ t + AD + B where A, B re constnts. Thus 2 ρ cos( C ρ t + D) 2 z(t) = ρ sin( C ρ t + D) 2 t + AD + B AC ρ 2 is the generl form of the geoesic for cyliner. It is helix if C 0, A 0, verticl line if C = 0, A 0 n circle if C 0, A = 0. 17