Second Vrition. One-vrible problem Februry 11, 213 Contents 1 Locl vritions 2 1.1 Legendre Tests............................ 2 1.2 Weierstrss Tests........................... 4 1.3 Null-Lgrngins nd convexity................... 8 2 Nonlocl conditions 1 2.1 Sufficient condition for the wek locl minimum.......... 1 2.2 Jcobi vrition............................ 12 2.2.1 Exmples........................... 13 2.3 Distnce on sphere: Columbus problem............. 14 2.4 Nture does not minimize ction.................. 15 1
Sttionry conditions point to possibly optiml trjectory but they do not stte tht the trjectory corresponds to the minimum of the functionl. A sttionry solution cn correspond to minimum, locl minimum, mximum, locl mximum, of sddle point of the functionl. In this chpter, we estblish methods iming to distinguish locl minimum from locl mximum or sddle. In ddition to being solution to the Euler eqution, the true minimizer stisfies necessry conditions in the form of inequlities. We introduce vritionl tests, Weierstrss nd Jcobi conditions, tht supplement ech other. The conclusion of optimlity of the tested sttionry curve u(x) is bsed on comprison of the problem costs I(u) nd I(u + δu) computed t u nd ny close-by dmissible curve u + δu. The closeness of dmissible curve is required to simplify the clcultion nd obtin convenient optimlity conditions. The question whether or not two curves re close to ech other or whether υ(x) is smll depends on wht curves we consider to be close. Below, we work out three tests of optimlity using different definitions of closeness. 1 Locl vritions 1.1 Legendre Tests Consider gin the simplest problem of the clculus of vritions min I(u), I(u) = u(x),x [,b] F (x, u, u )dx, u() = u, u(b) = u b (1) nd function u(x) tht stisfies the Euler eqution nd boundry conditions, F u d dx F u =, u() = u, u(b) = u b, (2) so tht the first vrition δi is zero. Let us compute the increment δ 2 I of the objective cused by the vrition { ɛ δu(x, x ) = 2 φ ( x x ) ɛ if x x < ɛ if x x ɛ (3) where φ(x) is function with the following properties: φ( 1) = φ(1) =, mx φ(x) 1, mx x [ 1,1] x [ 1,1] φ (x) 1 The mgnitude of this Legendre-type vrition tends to zero when ɛ, nd the mgnitude of its derivtive { ( ) δu ɛ φ x x (x, x ) = ɛ if x x < ɛ if x x ɛ tends to zero s well. Additionlly, the vrition is locl: it is zero outside of the intervl of the length 2ɛ. We use these fetures of the vrition in the clcultion of the increment of the cost. 2
= where Expnding F into Tylor series nd keeping the qudrtic terms, we obtin δi = I(u + δu) I(u) = ([ F u d dx F u (F (x, u + δu, u + δu ) F (x, u, u ))dx ] ) δu + Aδu 2 + 2Bδu δu + C(δu ) 2 dx + F u A = 2 F u 2, B = 2 F u u, C = 2 F (u ) 2 x=b x=, (4) nd ll derivtives re computed t the point x t the optiml trjectory u(x). The term in the brckets in the integrnd in the right-hnd side of (4) is zero becuse the Euler eqution is stisfied. Let us estimte the remining terms A(x)(δu) 2 dx = x+ε x ε A(x)(δu) 2 dx x+ε ε 4 A(x)dx = A(x ) ε 5 + o(ε δu 5 ) x ε Indeed, the vrition δu is zero outside of the intervl [x ε, x+ε], hs mgnitude of the order of ε 2 in this intervl, nd A(x) is ssumed to be continuous t the trjectory. Similrly, we estimte x+ε B(x)δu δu dx ε 3 B(x)dx = B(x ) ε 4 + o(ε δu 4 ) x ε x+ε C(x)(δu ) 2 dx ε 2 C(x)dx = C(x ) ε 3 + o(ε δu 3 ) x ε Its derivtive s mgnitude δu is of the order of ε, therefore δu δu s ε ; we conclude tht the lst term in the integrnd in the right-hnd side of (4) domintes. The inequlity δi > implies inequlity which is clled Legendre condition or Legendre test. (u ) 2 (5) Exmple 1.1 (Two-well Lgrngin. I) Consider the Lgrngin F (u, u ) = [(u ) 2 u 2 ] 2 The Legendre test gives the inequlity u 2 = 4(3u 2 u 2 ). 3
Consequently, the solution u of Euler eqution or d dx [(u ) 3 u 2 u ] + u(u ) 2 u 3 =, u() =, u(1) = 1 (3(u ) 2 u 2 )u u((u ) 2 + u 2 ) = u() =, u(1) = 1 corresponds to locl minimum of the functionl if, in ddition, the inequlity 3u 2 u 2 is stisfied in ll points x (, 1). 1.2 Weierstrss Tests The Weierstrss test detects stbility of solution to vritionl problem ginst different kind of vritions the strong locl perturbtions. It is lso locl: it compres trjectories tht coincide everywhere except smll intervl where their derivtives significntly differ. Suppose tht u is the minimizer of the vritionl problem (1) tht stisfies the Euler eqution (2). Additionlly, u should stisfy nother test tht uses different from (3) type of vrition. The vrition is n infinitesiml tringle supported on the intervl [x, x + ε] in neighborhood of point x (, 1) (see??): if x [x, x + ε], u(x) = v 1 (x x ) if x [x, x + αε], v 2 (x x ) αε(v 1 v 2 ) if x [x + αε, x + ε] where v 1 nd v 2 re two rel numbers, α (, 1), nd prmeters α ( < α < 1), v 1 nd v 2 re relted αv 1 + (1 α)v 2 = (6) to provide the continuity of u + u t the point x + ε, or equlity Condition (6) cn be rewritten s u(x + ε ) =. v 1 = (1 α)v, v 2 = αv =, (7) v is n rbitrry rel number. The considered vrition (the Weierstrss vrition) is loclized nd hs n infinitesiml bsolute vlue (if ε ), but, unlike the Legendre vrition, its derivtive ( u) is finite: if x [x, x + ε], ( u) = v 1 if x [x, x + αε], (8) if x [x + αε, x + ε]. v 2 4
The increment is δi = I(u + δu) I(u) = x+αɛ + x x ɛ x +αɛ (F (x, u + δu, u + u ) F (x, u, u ))dx (F (x, u + δu, u + v 1 ) F (x, u, u ))dx (F (x, u + δu, u + v 2 ) F (x, u, u ))dx (9) δi = I(u + δu) I(u) = x+αɛ x F (x, u + δu, u + v 1 )dx + (F (x, u + δu, u + u ) F (x, u, u ))dx x ɛ x +αɛ F (x, u + δu, u + v 2 )dx x+ɛ F (x, u, u )dx (1) x Computing increment δi nd rounding up to ε, we estimte integrnds s F (x, u(x) + u, u(x) + v) = F (x, u(x ), u (x ) + v) + O(1). The smllness of the vrition u is due to smllness of the intervl of the vrition. Also, we ssume tht F (x, u(x), u (x) F (x, u(x ), u (x ) = O(ɛ) x [x, x + ɛ]. Under these ssumptions, we compute the min term of the increment s δi(u, x ) = ε[αf (x, u, u + v 1 ) + (1 α)f (x, u, u + v 2 ) F (x, u, u )] + o(ε) (11) The smllness of the vrition is due to smllness of the intervl of the vrition. Repeting the vritionl rguments nd using the rbitrriness of x, we find tht n inequlity holds δ(u, x) x [, b] (12) for minimizer u. The lst expression yields to the Weierstrss necessry condition. Any minimizer u(x) of (1) stisfies the inequlity αf (x, u, u + (1 α)v) + (1 α)f (x, u, u αv) F (x, u, u ) (13) v, α [, 1] (14) 5
The reder my recognize in this inequlity the definition of convexity, or the condition tht the grph of the function F (.,., z) (considered s function of the third rgument u lies below the the chord in between the points there the chord meet the grph. The Weierstrss condition requires convexity of the Lgrngin F (x, y, z) with respect to its third rgument z = u. The first two rguments x, y = u here re the coordintes x, u(x) of the testing minimizer u(x). Recll tht the tested minimizer u(x) is solution to the Euler eqution. The Weierstrss test is stronger thn the Legendre test becuse convexity implies nonnegtivity of the second derivtive. It compres the optiml trjectory with lrger set of dmissible trjectories. Exmple 1.2 (Two-well Lgrngin. II) Consider gin the Lgrngin It is convex s function of u if F (u, u ) = ( (u ) 2 u 2) 2 u u The solution u of Euler eqution corresponds to locl minimum of the functionl if, the inequlity u (x) u(x) is stisfied in ll points x (, 1). The Weierstrss test is stronger thn Legendre test, u 2 u 2 > 1 3 u2 Remrk 1.1 Convexity of the Lgrngin does not gurntee the existence of solution to vritionl problem. It sttes only tht differentible minimizer (if it exists) is stble ginst fine-scle perturbtions. However, the minimum my not exist t ll or be unstble to other vritions. If the solution of vritionl problem fils the Weierstrss test, then its cost cn be decresed by dding infinitesiml centered wiggles to the solution. The wiggles re the Weierstrss tril functions, which decrese the cost. In this cse, we cll the vritionl problem ill-posed, nd we sy tht the solution is unstble ginst fine-scle perturbtions. Remrk 1.2 Weierstrss condition is lwys stisfied in the geometric optics. The Lgrngin depends on the derivtive s L = 1+y 2 v(y) 2 L y 2 = 1 v(y)(1 + y 2 ) 3 2 nd its second derivtive is lwys nonnegtive if v >. It is physiclly obvious tht the fstest pth is stble to short-term perturbtions. Weierstrss condition is lso lwys stisfied in the Lgrngin mechnics. The Lgrngin depends on the derivtives of the generlized coordintes through the 6
Figure 1: The construction of Weierstrss E-function. The grph of convex function nd its tngent plne. kinetic energy T = 1 2 qr(q) q nd its Hessin equls generlized inerti R which is lwys positive definite. Physiclly speking, inerti does not llow for infinitesiml oscilltions becuse they lwys increse the kinetic energy while potentil energy is insensitive to them. Weierstrss E-function Weierstrss suggested convenient test for convexity of Lgrngin, the so-clled E-function equl to the difference between the vlue of Lgrngin L(x, u, ẑ) in tril point u, z = z nd the tngent hyperplne L(x, u, u ) (ẑ u ) T L(x,u,u ) u to the optiml trjectory t the point u, u : E(L(x, u, u, ẑ) = L(x, u, ẑ) L(x, u, u ) (ẑ u ) T L(x, u, u ) u (15) Function E(L(x, u, u, ẑ) vnishes together with the derivtive E(L) ẑ when ẑ = u : E(L(x, u, u, ẑ) ẑ=u =, ẑ E(L(x, u, u, ẑ) ẑ=u =. According to the bsic definition of convexity, the grph of convex function is greter thn or equl to tngent hyperplne. Therefter, the Weierstrss condition of minimum of the objective functionl cn be written s the condition of positivity of the Weierstrss E-function for the Lgrngin, E(L(x, u, u, ẑ) ẑ, x, u(x) where u(x) tested trjectory. Exmple 1.3 Check the optimlity of Lgrngin L = u 4 φ(u, x)u 2 + ψ(u, x) where φ nd ψ re some functions of u nd x using Weierstrss E-function. The Weierstrss E-function for this Lgrngin is E(L(x, u, u, ẑ) = [ ẑ 4 φ(u, x)ẑ 2 + ψ(u, x) ] or [ u 4 φ(u, x)u 2 + ψ(u, x) ] (ẑ u )(4u 3 2φ(u, x)u). E(L(x, u, u, ẑ) = (ẑ u ) 2 ( ẑ 2 + 2ẑu φ + 3u 2). As expected, E(L(x, u, u, ẑ) is independent of n dditive term ψ nd contins qudrtic coefficient (ẑ u ) 2. It is positive for ny tril function ẑ if the qudrtic π(ẑ) = ẑ 2 2ẑu + φ 3u 2 7
does not hve rel roots, or if φ(u, x) 2u 2 If this condition is violted t point of n optiml trjectory u(x), the trjectory is nonoptiml. Vector-Vlued Minimizer The Legendre nd Weierstrss conditions nd cn be nturlly generlized to the problem with the vector-vlued minimizer. If the Lgrngin is twice differentible function of the vector u = z, the Legendre condition becomes He(F, z) (16) (see Section??) where He(F, z) is the Hessin... 2 F z1 He(F, z) = 2 z 1 z n......... z 1 z n... 2 F zn 2 nd inequlity in (16) mens tht the mtrix is nonnegtive definite (ll its eigenvlues re nonnegtive). The Weierstrss test requires convexity of F (x, y, z) with respect to the third vector rgument. 1.3 Null-Lgrngins nd convexity Find the Lgrngin cnnot be uniquely reconstructed from its Euler eqution. Similrly to ntiderivtive, it is defined up to some term clled null-lgrngin. Definition 1.1 The Lgrngins φ(x, u, u ) for which the opertor S(φ, u) of the Euler eqution (??) identiclly vnishes S(φ, u) = u re clled Null-Lgrngins. Null-Lgrngins in vritionl problems with one independent vrible re liner functions of u. Indeed, the Euler eqution is second-order differentil eqution with respect to u: d dx ( u φ ) u φ = 2 φ (u ) 2 u + 2 φ u u u + 2 φ u x φ. (17) u The coefficient of u is equl to 2 φ (u ). If the Euler eqution holds identiclly, 2 this coefficient is zero, nd therefore φ u does not depend on u. Hence, φ linerly depends on u : φ(x, u, u ) = u A(u, x) + B(u, x); A = 2 φ u u, 8 B = 2 φ u x φ u. (18)
Additionlly, if the following equlity holds A x = B u, (19) then the Euler eqution vnishes identiclly. In this cse, φ is null-lgrngin. We notice tht the Null-Lgrngin (18) is simply full differentil of function Φ(x, u): φ(x, u, u ) = d Φ Φ(x, u) = dx x + Φ u u ; equtions (19) re the integrbility conditions (equlity of mixed derivtives) for Φ. The vnishing of the Euler eqution corresponds to the Fundmentl theorem of clculus: The equlity dφ(x, u) dx = Φ(b, u(b)) Φ(, u()). dx tht does not depend on u(x) only on its end-points vlues. Exmple 1.4 Function φ = u u is the null-lgrngin. Indeed,we check ( ) d dx u φ u φ = u u. Null-Lgrngins nd Convexity The convexity requirements of the Lgrngin F tht follow from the Weierstrss test re in greement with the concept of null-lgrngins (see, for exmple [?]). Consider vritionl problem with the Lgrngin F, min u 1 F (x, u, u )dx. Adding null-lgrngin φ to the given Lgrngin F does not ffect the Euler eqution of the problem. The fmily of problems min u 1 (F (x, u, u ) + tφ(x, u, u )) dx, where t is n rbitrry number, corresponds to the sme Euler eqution. Therefore, ech solution to the Euler eqution corresponds to fmily of Lgrngins F (x, u, z) + tφ(x, u, z), where t is n rbitrry rel number. In prticulr, Lgrngin cnnot be uniquely defined by the solution to the Euler eqution. The stbility of the minimizer ginst the Weierstrss vritions should be property of the Lgrngin tht is independent of the vlue of the prmeter t. It should be common property of the fmily of equivlent Lgrngins. On the other hnd, if F (x, u, z) is convex with respect to z, then F (x, u, z) + tφ(x, u, z) is lso convex. Indeed, φ(x, u, z) is liner s function of z, nd dding the term tφ(x, u, z) does not ffect the convexity of the sum. In other words, convexity is chrcteristic property of the fmily. Accordingly, it serves s test for the stbility of n optiml solution. 9
2 Nonlocl conditions 2.1 Sufficient condition for the wek locl minimum We ssume tht trjectory u(x) stisfies the sttionry conditions nd Legendre condition. We investigte the increment cused by nonlocl vrition δu of n infinitesiml mgnitude: υ < ε, υ < ε, vrition intervl is rbitrry. To compute the increment, we expnd the Lgrngin into Tylor series keeping terms up to O(ɛ 2 ). Recll tht the liner of ɛ terms re zero becuse the Euler eqution S(u, u ) = for u(x) holds. We hve where δi = r S(u, u )δu dx + r δ 2 F dx + o(ɛ 2 ) (2) δ 2 F = 2 F u 2 (δu)2 + 2 2 F u u (δu)(δu ) + 2 F (u ) 2 (δu ) 2 (21) Becuse the vrition is nonlocl, we cnnot neglect υ in comprison with υ. No vrition of this kind cn improve the sttionry solution if the qudrtic form ) is positively defined, Q(u, u ) = ( u 2 u u u u (u ) 2 Q(u, u ) > on the sttionry trjectory u(x) (22) This condition is clled the sufficient condition for the wek minimum. It neglects the reltion between δu nd δu nd trets them s two independent tril functions. If the sufficient condition is stisfied, no trjectory tht is smooth nd sufficiently close to the sttionry trjectory cn increse the objective functionl of the problem compred with the objective t tht tested sttionry trjectory. Notice tht the first term 2 F u is nonnegtive becuse of the Legendre condition. 2 Problem 2.1 Show tht the sufficient condition is stisfied for the Lgrngins F 1 = 1 2 u2 + 1 2 (u ) 2 nd F 2 = 1 u (u ) 2 1
Exmple: Sttionry solution is not minimizer. The Weierstrss test does not gurntee tht the sttionry solution is minimizer. One cn find nonlocl vrition tht decreses the cost below its sttionry vlue. Consider the problem: r ( ) 1 I = min u 2 (u ) 2 c2 2 u2 dx u() = ; u(r) = A (23) where c is constnt. The first vrition δi i δi = r s zero if u(x) stisfies the Euler eqution The sttionry solution u(x) is ( u + c 2 u ) δu dx u + c 2 u =, u() =, u(r) = A. (24) u(x) = ( ) A sin(cx) sin(cr) The Weierstrss test is stisfied, becuse the dependence of the Lgrngin on the derivtive u is convex, 2 L u = 1. The sufficient condition of locl minimum 2 is not stisfied becuse 2 L u = c 2. 2 Let us show tht the sttionry condition does not correspond to minimum of I is the intervl s length r is lrge enough. We simply demonstrte vrition tht improves the sttionry trjectory by decresing cost of the problem. Compute the second vrition (21): r ( ) 1 δ 2 I = 2 (δu ) 2 c2 2 (δu)2 dx (25) Since the boundry conditions t the ends of the trjectory re fixed, the vrition δu stisfies homogeneous conditions δu() = δu(r) =. Let us choose the vrition s follow: { ɛx( x), x δu = x > where the intervl of vrition [, ] is not greter tht [, r], r. Computing the second vrition of the gol functionl from (25), we obtin The increment δ 2 I is positive only if δ 2 I() = ɛ2 6 3 (1 c 2 2 ), r < r crit, r crit = 11 1 c.
The most dngerous vrition corresponds to the mximl vlue = r. This increment is negtive when r is sufficiently lrge, r > r crit. In this cse δ 2 I() is negtive, δ 2 I() < We conclude tht the sttionry solution does not correspond to the minimum of I if the length of the trjectory is lrger thn r crit. If the length is smller thn r crit, the sitution is inconclusive. It could still be possible to choose nother type of vrition different from considered here tht disproves the optimlity of the sttionry solution. 2.2 Jcobi vrition The Jcobi necessry condition chooses the most sensitive long nd shllow vrition nd exmines the increment cused by such vrition. It complements the Weierstrss test tht investigtes stbility of sttionry trjectory to strong loclized vritions. Jcobi condition tries to disprove optimlity by testing stbility ginst optiml nonlocl vritions with smll mgnitude. Assume tht trjectory u(x) stisfies the sttionry nd Legendre conditions but does not stisfy the sufficient conditions for wek minimum, tht is Q(u, u ) in (22) is not positively defined, S(u, u ) =, (u ) 2 >, Q(u, u ) (u ) 2 To derive Jcobi condition, we consider n infinitesiml nonlocl vrition: δu = O(ɛ) 1 nd δu = O(ɛ) 1 nd exmine the expression (21) for the second vrition. When n infinitesiml nonlocl vrition is pplied, the increment increses becuse of ssumed positivity of nd decreses becuse of ssumed nonpositivity of the mtrix Q. Depending on the length r of the intervl of integrtion nd of chosen form of the vrition δu, one of these effects previls. If the second effect is stronger, the extreml fils the test nd is nonoptiml. Jcobi conditions sks for the choice of the best δu of the vrition. The expression (21) itself is vritionl problem for δu which we renme here s v for short; the Lgrngin is qudrtic of v nd v nd the coefficients re functions of x determined t the sttionry trjectory u(x) which is ssumed to be known: where δi = r [ Av 2 + 2B v v + C(v ) 2] dx, v() = v(r) = (26) A = 2 F u 2, B = 2 F u u, C = 2 F (u ) 2 12
The problem (26) is the vritionl problem for the unknown vrition v. Its Euler eqution: d dx (Cv + Bv) Av =, v(r ) = v(r conj ) = [r, r conj ] [, r] (27) is solution to Storm-Liouville problem The point r nd r conj re clled the conjugte points. The problem is homogeneous: If v(x) is solution nd c is rel number, cv(x) is lso solution. Jcobi condition is stisfied if the intervl does not contin conjugte points, tht is if there is no nontrivil solutions to (27) on ny subintervl of [r, r conj ] [, r], tht is if there re no nontrivil solutions of (27) with boundry conditions v(r ) = v(r conj ) =. If this condition is violted, thn there exist fmily of trjectories { u + αv if x [r, r U(x) = conj ] u if x [, r]/[r, r conj ] tht deliver the sme vlue of the cost. Indeed, v is defined up to multiplier: If v is solution, αv is solution too. These trjectories hve discontinuous derivtive t the points r nd r conj. Such discontinuity leds to contrdiction to the Weierstrss-Erdmn condition which does not llow broken extreml t these points. 2.2.1 Exmples Exmple 2.1 (Nonexistence of the minimizer: Blow up) Consider gin problem (23) r ( ) 1 I = min u 2 (u ) 2 c2 2 u2 dx u() = ; u(r) = A The sttionry trjectory nd the second vrition re given by formuls (24) nd (25), respectively. Insted of rbitrry choosing the second vrition (s we did bove), we choose it s solution to the homogeneous problem (27) for v = δu v + c 2 v =, r =, u() =, u(r conj ) =, r conj r (28) This problem hs nontrivil solution v = ɛ sin(cx) if the length of the intervl is lrge enough to stisfy homogeneous condition of the right end. We compute cr conj = π or r( conj ) = π c The second vrition δ 2 I is positive when r is smll enough, δ 2 I = 1 ( ) π 2 r ɛ2 r 2 c2 > if r < π c In the opposite cse r > π c, the increment is negtive which shows tht the sttionry solution is not minimizer. 13
To clrify this, let us compute the sttionry solution (24). We hve ( ) ( ) A u(x) = sin(cx) nd I(u) = A2 π 2 sin(cr) sin 2 (cr) r 2 c2 When r increses pproching the vlue c π, the mgnitude of the sttionry solution indefinitely grows, nd the cost indefinitely decreses: lim I(u) = r c π The cost of the problem cnnot increse if r increses. Indeed, the trjectory tht correspond to the sme cost is esily constructed. Indeed, let u(x), x [, r] be minimizer (recll, tht u() = ) nd the cost functionl is I r. In lrger intervl x [, r + d], the dmissible trjectory { if < x < d û(x) = u(x d) if d x r + d corresponds to the sme cost I r. Obviously, this trjectory of the Euler eqution is not minimizer if r > π c, becuse it corresponds to finite cost I(u) >. Remrk 2.1 Compring this result with the result in Exmple (2.1), we see tht for this exmple the optiml choice of vrition improved the length of the criticl intervl t only.65%. 2.3 Distnce on sphere: Columbus problem This simple exmple illustrtes the use of second vrition without single clcultion. We consider the problem of geodesics (shortest pth) on sphere. Sttionrity Let us prove tht geodesics is prt of the gret circle. Suppose tht geodesics is different curve, or tht it exists n rc C, C tht is prt of the geodesics but does not coincide with the rc of the gret circle. Let us perform vrition: Replce this rc with its mirror imge the reflection cross the plne tht psses through the ends C, C of this rc nd the center of the sphere. The reflected curve hs the sme length of the pth nd it lies on the sphere, therefore the new pth remins geodesics. On the other hnd, the new pth is broken in two points C nd C, nd therefore cnnot be the shortest pth. Indeed, consider prt of the pth in n infinitesiml circle round the point C of brekge nd fix the points A nd B where the pth crosses tht circle. This pth cn be shorten by rc of gret circle tht psses through the points A nd B. To demonstrte this, it is enough to imgine humn-size scle on Erth: The infinitesiml prt of the round surfce becomes flt nd obviously the shortest pth correspond to stright line nd not to zigzg line with n ngle. 14
Jcobi-type vrition The sme considertion shows tht the length of geodesics is no lrger thn π times the rdius of the sphere, or it is shorter thn the gret semicircle. Indeed, if the length of geodesics is lrger thn the gret semicircle one cn fix two opposite points the poles of the sphere on the pth nd turn on n rbitrry ngle the xis the prt of geodesics tht psses through these points. The new pth lies on the sphere, hs the sme length s the originl one, nd is broken t the poles, thereby its length is not miniml. We conclude tht the minimizer does not stisfy Jcobi test if the length of geodesics is lrger thn π times the rdius of the sphere. Therefore, geodesics on sphere is prt of the gret circle tht joins the strt nd end points nd which length is less tht hlf of the equtor s length. Remrk 2.2 The rgument tht the solution to the problem of shortest distnce on sphere bifurctes when its length exceeds hlf of the gret circle ws fmously used by Columbus who rgued tht the shortest wy to Indi psses through the Western route. As we know, Columbus wsn t be ble to prove or disprove the conjecture becuse he bumped into Americn continent discovering New World for better nd for worse. 2.4 Nture does not minimize ction The next exmple dels with system of multiple degrees of freedom. Consider the vritionl problem with the Lgrngin L = n i=1 1 2 m u2 i 1 2 C(u i u i 1 ) 2, u() = u We will see lter in Chpter?? tht this Lgrngin describes the ction of chin of prticles with msses m connected by springs with constnt C. Sttionrity is the solution to the system m i ü i + C( u i+1 + 2u i u i 1 ), u() = u Tht describes dynmics of the chin. The continuous limit of the chin dynmics is the dynmics of n elstic rod. The second vrition (here we lso use the nottion v = δu) δ 2 L = n i=1 1 2 mv i 2 1 2 C(v i v i 1 ) 2, v =, v n = corresponds to the Euler eqution the eigenvlue problem m v = C m Av 15
where v(t) = [v 1 (t),..., v n (t)] is the vector of vritions nd A = The problem hs solution 2 1... 1 2 1... 1 2..................... 2. v(t) = α k v k sin ω k t v() = v(t conj ) =, T conj T where v k re the eigenvectors, α re coefficients found from initil conditions, nd ω k re the squre roots of eigenvlues of the mtrix A. Solving the chrcteristic eqution for eigenvlues det(a ω 2 I) = we find tht these eigenvlues re C ω k = 2 m sin2 ( C m πk n ), k = 1,... n The Jcobi condition is violted if v(t) is consistent with the homogeneous initil nd finl conditions tht is if the time intervl is short enough. Nmely, the condition is violted when the durtion T is lrger thn π m T mx(ω k ) 2π C. The continuous limit of the chin is chieved when the number N of msses indefinitely growth nd ech mss decreses correspondingly s m(n) = m() N. The distnce between msses decreses, the stiffness of one link increses s C(N) = C()N s it become N times shorter. Correspondingly, C(N) m(n) = N C() m() nd the mximl eigenvlue ω N tends to infinity s N. This implies Jcobi condition is violted t ny finite time intervl, or tht ction J of the continuous system is not minimized t ny finite time intervl. Wht is minimized in clssicl mechnics? Lgrngin mechnics sttes tht differentil equtions of Newtonin mechnics correspond to the sttionrity of ction: the integrl of difference between kinetic T nd potentil V energies. Kinetic energy is qudrtic form of velocities q i of prticles, nd potentil energy depends only on positions (generlized coordintes) q i of them T (q, q) = 1 2 n q T R(q) q i V = V (q) 16
We ssume tht T is convex function of q nd q, nd V is convex function of q. As we hve seen t the bove exmples, ction L = T V does not stisfy Jcobi condition becuse kinetic nd potentil energies, which both re convex functions or q nd q, enter the ction with different signs. Generlly, the ction is sddle function of q nd q. The notion tht Newton mechnics is not equivlent to minimiztion of universl quntity, hd significnt philosophicl implictions, it destroyed the hypothesis bout universl optimlity of the world. The miniml ction principle cn be mde miniml principle, in the Minkovski spce. Formlly, we replce time t with the imginry vrible t = iτ nd use the second-order homogeneity of T : T (q, q) = 1 2 qt R(q) q = q τ T R(q)q τ The Lgrngin, considered s function of q nd q τ insted of q nd q, become negtive of convex function if potentil energy V nd inerti R(q) re convex. It become formlly equl to the first integrl (the energy) L(q, q τ ) = q τ T R(q)q τ V (q) tht is conserved in the originl problem. The locl mximum of the vritionl problem, or J = min q(τ) t t ( L(q, q τ ))dτ does exist, since the Lgrngin L(q, q τ ) is convex with respect to q nd q τ. Exmple 2.2 The Lgrngin L for n oscilltor L = 1 ( m u 2 Cu 2) 2 becomes ˆL = 1 2 ( mu 2 + Cu 2 ). The Euler Eqution for ˆL corresponds to the solution m u Cu = u(τ) = A cosh(ωτ) + B sinh(ωτ), ω = C m The sttionry solution stisfies Weierstrss nd Jcobi conditions. Returning to originl nottions t = iτ we obtin A cos(ωt) + B sin(ωt) the correct solution of the originl problem. Remrkble, tht this solution is unstble, but its trnsform to Minkovski spce is stble. These ides hve been developed in the specil theory of reltivity (world lines). 17