First vrition (one-vrible problem) Jnury 14, 2013 Contents 1 Sttionrity of n integrl functionl 2 1.1 Euler eqution (Optimlity conditions)............... 2 1.2 First integrls: Three specil cses................. 5 1.3 Vritionl problem s limit of finite-dimensionl problem.. 8 2 Sttionrity of boundry terms 10 2.1 Vrition of boundry conditions.................. 10 2.2 Broken extreml nd the Weierstrss-Erdmn condition..... 11 3 Functionl dependent on higher derivtives 12 4 Non-fixed intervl 14 4.1 Trnsverslity condition....................... 14 4.2 Extreml broken t n unknown point............... 16 5 Severl minimizers 17 5.1 Euler equtions nd first integrls................. 17 5.2 Vritionl boundry conditions.................. 20 1
Since, however, the rules for isoperimetric curves or, in modern terms, extreml problems were not sufficiently generl, the fmous Euler undertook the tsk of reducing ll such investigtions to generl method which he gve in the work Essy on new method of determining the mxim nd minim of indefinite integrl formuls ; n originl work in which the profound science of the clculus shines through. Even so, while the method is ingenious nd rich, one must dmit tht it is not s simple s one might hope in work of pure nlysis. In Essy on new method of determining the mxim nd minim of indefinite integrl formuls, by Lgrnge, 1760 1 Sttionrity of n integrl functionl The technique ws developed by Euler, who lso introduced the nme Clculus of vritions in 1766. The method is bsed on n nlysis of infinitesiml vritions of minimizing curve. The min scheme of the vritionl method is s follows: Assume tht the optiml curve u(x) exists mong smooth (twice-differentible curves). Compre the optiml curve with close-by trjectories u(x) + δu(x), where δu(x) is smll in some sense. Using the smllness of δu, we simplify the comprison, deriving necessry conditions for the optiml trjectory u(x) Vritionl methods yield to only necessry conditions of optimlity becuse it is ssumed tht the compred trjectories re close to ech other; on the other hnd, they re pplicble to gret vriety of extreml problems clled vritionl problems. 1.1 Euler eqution (Optimlity conditions) Consider the problem clled the simplest problem of the clculus of vritions min u I(u), I(u) = F (x, u, u )dx, u() = u, u(b) = u b, (1) Here integrnt F clled the Lgrngin, is twice differentible function of its three rguments, I(u) is clled the cost functionl. It is not known priori whether the minimizer u 0 (x) is smooth, but let us ssume tht it is twice differentible function of x. For exmple, consider the re of the surfce of revolution. According to the clculus, the re J of the surfce is A(r) = π r(x) 1 + r (x) 2 dx, where r(x) is the vrible distnce from the xes OX of rottion. The problem of miniml re of such surfce I = min r(x) A(u), r() = R, r(b) = R b 2
is vritionl problem. To void trivil solution, vlues r() nd r(b) re fixed. To derive necessry condition of optimlity of minimizer u 0 we use the ides of clculus, computing n nlog of the derivtive of I with respect to u (clled the functionl derivtive) nd setting it to zero. We suppose tht function u 0 = u 0 (x) is minimizer nd replce u 0 with test function u 0 + δu, ssuming tht the norm δu of the vrition δu is infinitesiml. The test function u 0 +δu stisfies the sme boundry conditions s u 0. If indeed u 0 is minimizer, the increment of the cost δi(u 0 ) = I(u 0 + δu) I(u 0 ) is nonnegtive: δi(u 0 ) = 1 0 (F (x, u 0 + δu, (u 0 + δu) ) F (x, u 0, u 0))dx 0. (2) If δu is not specified, the eqution (2) is not too informtive. However, prticulr form of the vrition δu simplifies (2) nd llows for finding n eqution for the minimizer. Clculus of vritions suggests set of tests tht differ by vrious ssumed form of vritions δu. Euler Lgrnge Equtions The simplest vritionl condition (the Euler Lgrnge eqution) is derived ssuming tht the vrition δu is infinitesimlly smll nd loclized: δu = { ρ(x) if x [x0, x 0 + ε], 0 if x is outside of [x 0, x 0 + ε]. (3) Here ρ(x) is continuous function tht vnishes t points x 0 nd x 0 + ε nd is constrined s follows: ρ(x) < ε, ρ (x) < ε x. (4) The integrnd t the perturbed trjectory cn be expnded into Tylor series, F (x, u 0 + δu, (u 0 + δu) ) = F (x, u 0, u 0) + (x, u 0, u 0) δu + (x, u 0, u 0) δu + o(δu, δu ) Here, δu is derivtive of the vrition δu, δu = (δu), o(δu, δu ) denotes higher order terms which norms re smller thn δu nd δu when ε 0. Substituting this expression into (2) nd collecting liner (with respect to ε) terms, we rewrite (2) s δi(u 0 ) = ( ) (δu) + (δu) dx + o(ε) 0. (5) where The F is clculted t the exmined trjectory u 0. To simplify nottions, we omit index ( 0 ) below. 3
The vritions δu nd (δu) re mutully dependent nd (δu) cn be expressed in terms of δu. Integrtion by prts of the underlined term in (5) gives nd we obtin (δu) dx = 0 δi(u 0 ) = ( d dx ) δu dx + S F (x, u, u )δu dx + δu where S F denotes the functionl derivtive, δu x=b x= x=b x= + o(ε), (6) S F (x, u, u ) = d dx +. (7) The nonintegrl term in the right-hnd side of (6) is zero, becuse the boundry vlues of u re prescribed s u() = u nd u(b) = u b ; therefore their vritions δu x= nd δu x=b equl zero, δu x= = 0, δu x=b = 0 Due to the rbitrriness of δu, we rrive t the following Theorem 1.1 (Sttionrity condition) Any differentible nd bounded minimizer u 0 of the vritionl problem (1) is solution to the boundry vlue problem S F (x, u, u ) = d dx = 0 x (, b); u() = u, u(b) = u b, (8) clled the Euler Lgrnge eqution. The Euler Lgrnge eqution is lso clled the sttionry condition of optimlity becuse it expresses sttionrity of the vrition. Remrk 1.1 The sttionrity test lone does not llow to conclude whether u is true minimizer or even to conclude tht solution to (8) exists. For exmple, the function u tht mximizes I(u) stisfies the sme Euler Lgrnge eqution. The tests tht distinguish miniml trjectory from other sttionry trjectories re discussed in Chpter??. In this derivtion, it is ssumed tht the extreml u(t) is twice differentible function of x. Using the chin rule, the left-hnd side of eqution (8) cn be rewritten s S F (x, u, u ) = 2 F 2 u + 2 F u + 2 F x (9) 4
Exmple 1.1 Compute the Euler eqution for the problem I = min u(x) 1 We compute = u, The minimizer u 0 (x) is 0 F (x, u, u )dx u(0) = 1, u(1) =, F = 1 2 (u ) 2 + 1 2 u2 = u nd the Euler eqution becomes u u = 0 in (0, 1), u(0) = 1, u(1) =. u 0 (x) = cosh(x) + cosh(1) sinh(1) sinh(x). The definition of the wek solution nturlly rises from the vritionl formultion tht does not check the behvior of the minimizer in ech point but in ech infinitesiml intervl. The minimizer cn chnge its vlues t severl points, or even t set of zero mesure without effecting the objective functionl. In mbiguous cses, one should specify in wht sense (Riemnn, Lebesgue) the integrl is defined nd chnge the definition of vrition ccordingly. 1.2 First integrls: Three specil cses In severl cses, the Euler eqution (8) cn be integrted t lest once. These re the cses when Lgrngin F (x, u, u ) does not depend on one of the rguments. Below, we investigte these cses. Assume tht F = F (x, u), nd the mini- Lgrngin is independent of u miztion problem is I(u) = min u(x) 1 0 F (x, u)dx (10) In this cse, the vrition does not involve integrtion by prts, nd the minimizer does not need to be continuous. Euler eqution (8) becomes n lgebric reltion for u = 0 (11) Curve u(x) is determined in ech point independently of neighboring points. The boundry conditions in (8) re stisfied by jumps of the extreml u(x) in the end points; these conditions do not ffect the objective functionl t ll. Exmple 1.2 Consider the problem I(u) = min u(x) 1 0 (u sin x) 2 dx, u(0) = 1; u(1) = 0. The miniml vlue J(u 0 ) = 0 corresponds to the discontinuous minimizer sin x if 0 x 1 u 0 (x) = 1 if x = 0 0 if x = 1 5
Formlly, the discontinuous minimizer contrdicts the ssumption posed when the Euler eqution ws derived. To be consistent, we need to repet the derivtion of the necessry condition for the problem (10) without ny ssumption on the continuity of the minimizer. This derivtion is quite obvious. Lgrngin is independent of u If Lgrngin does not depend on u, F = F (x, u ), Euler eqution (8) cn be integrted once: = constnt (12) The first order differentil eqution (12) for u is the first integrl of the problem; it defines quntity tht stys constnt everywhere long the optiml trjectory. To find the optiml trjectory, it remins to integrte the first order eqution (12) nd determine the constnts of integrtion from the boundry conditions. Exmple 1.3 Consider the problem I(u) = min u(x) 1 0 (u cos x) 2 dx, u(0) = 1; u(1) = 0. The first integrl is = u (x) cos x = C Integrting, we find the minimizer, u(x) = sin x + Cx + C 1. The constnts C nd C 1 re found from nd the boundry conditions: C 1 = 1, C = 1 sin 1, minimizer u 0 nd the cost of the problem become, respectively u 0 (x) = sin x (sin 1 + 1)x + 1 I(u 0 ) = (sin 1 + 1) 2. Notice tht the Lgrngin in the exmple (1.2) is the squre of difference between the minimizer u nd function sin x, nd the Lgrngin in the exmple (1.3) is the squre of difference of their derivtives. In the problem (1.2), the minimizer coincides with sin x, nd jumps to stisfy the prescribed boundry vlues. The minimizer u in the exmple (1.3) cnnot jump. Consider continuous pproximtion of derivtive u of discontinuous function; it increses in the proximity of the point of discontinuity nd is unbounded, such growth would increse the objective functionl, nd therefore it would be nonoptiml. We del with such problems below in Chpter??. 6
Lgrngin is independent of x If F = F (u, u ), eqution (8) hs the first integrl: W (u, u ) = constnt (13) where W (u, u ) = u F Indeed, compute the x-derivtive of W (u, u ) which must be equl to zero by virtue of (13): d dx W (u, u ) = [ ( 2 u + F u u + 2 F u 2 )] u u = 0 where the expression in squre brckets is the derivtive of the first term of W (u, u ). Cncelling the equl terms, we bring this eqution to the form ( u 2 F 2 u + 2 F u ) = 0 (14) The expression in prenthesis coincide with the left-hnd-side term S(x, u, u ) of the Euler eqution in the form (9), simplified for the considered cse (F is independent of x, F = F (u, u )). W is constnt t ny solution u(x) of Euler eqution. Insted of solving the Euler eqution, we my solve the first-order eqution W = 0 obtining the sme solution. Exmple 1.4 Consider the Lgrngin F = 1 2 [ (u ) 2 ω 2 u 2] The Euler eqution is The first integrl is u + ω 2 u = 0 W = 1 2 ( ω 2 u 2 + (u ) 2) = C 2 = constnt Let us check the constncy of the first integrl. The solution u of the Euler eqution is equl u = A cos(ωx) + B sin(ωx) where A nd B re constnts. Substituting the solution into the expression for the first integrl, we compute W = (u ) 2 + ω 2 u 2 = [ Aω sin(cx) + Bω cos(ωx)] 2 +ω 2 [A cos(ωx) + B sin(ωx)] 2 = ω 2 (A 2 + B 2 ) We hve shown tht W is constnt t the optiml trjectory. In mechnicl ppliction, W is the totl energy of the oscilltor. 7
1.3 Vritionl problem s limit of finite-dimensionl problem Here, we derive Euler eqution for finite-dimensionl problem tht pproximte the simplest vritionl problem min I(u), I(u) = u(x) F (x, u, u )dx Consider clss of piece-wise constnt discontinuous functions U N : ū(x) U N, if ū(x) = u i x [ + in ] (b ) A function ū in U N is defined by n N-dimensionl vector {u 1,... u N }. Rewriting the vritionl problem for this clss of minimizers, we replce the derivtive u (x) with finite difference Diff (u i ) Diff (u i ) = 1 (u i u i 1 ), = b N ; (15) when N, this opertor tends to the derivtive. The vritionl problem is replced by finite-dimensionl optimiztion problem: min u 1,...,u N 1 I N N I N = F i (u i, z i ), z i = Diff (z i ) = 1 (z i z i 1 ) (16) i=1 Compute the sttionry conditions for the minimum of I N (u) I N i = 0, i = 1...., N. Only two terms, F i nd F i+1, in the bove sum depend on u i : the first depends on u i directly nd lso through the opertor z i = Diff (u i ), nd the second only through z i+1 = Diff (u i+1 ): df i = i + i 1 du i i z i, df i+1 = i+1 1 du i z i+1. df k = 0, k i, k i + 1 du i Collecting the terms, we write the sttionry condition with respect to u i : I N = i + 1 ( i i i z ) i+1 = 0 z or, reclling the definition (15) of Diff -opertor, the form I N = ( ) i i+1 Diff = 0. i i z 8
The initil nd the finl point u 0 nd u N enter the difference scheme only once, therefore the optimlity conditions re different. They re, respectively, N+1 Diff (u N+1 ) = 0; o Diff (u 0 ) = 0. Formlly pssing to the limit N, Diff d dx, z u replcing the index ( i ) with continuous vrible x, vector of vlues {u k } of the piece-wise constnt function with the continuous function u(x), difference opertor Diff with the derivtive d dx ; then nd N F i (u i, Diff u i ) i=1 i i Diff ( i+1 z ) F (x, u, u )dx. d d x The conditions for the end points become the nturl vritionl conditions: (0) = 0, (T ) = 0, Remrk 1.2 So fr, we followed the forml scheme of necessry conditions, thereby tcitly ssuming tht ll derivtives of the Lgrngin exist, the increment of the functionl is correctly represented by the first term of its power expnsion, nd the limit of the sequence of finite-dimensionl problems exist nd does not depend on the prtition {x 1,... x N } if only x k x k 1 0 for ll k. We lso indirectly ssume tht the Euler eqution hs t lest one solution consistent with boundry conditions. If ll the mde ssumptions re correct, we obtin curve tht might be minimizer becuse it cnnot be disproved by the sttionry test. In other terms, we find tht is there is no other close-by clssicl curve tht correspond to smller vlue of the functionl. This sttement bout the optimlity seems to be rther wek but this is exctly wht the clculus of vrition cn give us. On the other hnd, the vritionl conditions re universl nd, being ppropritely used nd supplemented by other conditions, led to very detiled description of the extreml s we show lter in the course. Remrk 1.3 In the bove procedure, we ssume tht the limits of the components of the vector {u k } represent vlues of smooth function in the close-by points x 1,..., x N. At the other hnd, u k re solutions of optimiztion problems with the coefficients tht slowly vry with the number k. We need to nswer the question whether the solution of minimiztion problem tends to is differentible function of x; tht is whether the limit u k u k 1 lim k x k x k 1 exists nd this is not lwys the cse. We ddress this question lter in Chpter?? 9
2 Sttionrity of boundry terms 2.1 Vrition of boundry conditions Vritionl conditions nd nturl conditions The vlue of minimizer my not be specified on one or both ends of the intervl [, b]. In this cse, these vlues re clculted by the minimiztion of the gol functionl together with the minimizer. Consider vritionl problem where the boundry vlue t the right end b of the intervl is not defined nd the functionl directly depends on this vlue, min I(u), I(u) = u(x):u()=u F (x, u, u )dx + f(u(b)) (17) The Euler eqution for the problem remin the sme, S(x, u, u ) = 0, but this time it must be supplemented by vritionl boundry condition tht comes from the requirement of the sttionrity of the minimizer with respect to vrition of the boundry term. This term is ( + f ) δu(b) x=b The first term comes from the integrtion by prt in the derivtion of Euler eqution, see (6), nd the second is the vrition of the lst term in the objective functionl (17). Becuse the sign of the vrition δu(b) is rbitrry, the sttionrity condition hs the form + f x=b = 0 (18) x=b This equlity provides the missing boundry condition t the endpoint x = b for the second order Euler eqution. Similr condition cn be derived for the point x = if the vlue t this point is not prescribed. Exmple 2.1 Minimize the functionl 1 1 I(u) = min u 0 2 (u ) 2 dx + Au(1), u(0) = 0 Here, we wnt to minimize the endpoint vlue nd we do not wnt the trjectory be too steep. The Euler eqution u = 0 must be integrted with boundry conditions u(0) = 0 nd (see (18)) u (1) + A = 0 The extreml is stright line, u = Ax. The cost of the problem is I = 1 2 A2. If f = 0, the condition (18) becomes = 0 (19) x=b nd it is clled the nturl boundry condition. Exmple 2.2 Consider the problem F = (x)(u ) 2 +φ(x, u) where (x) 0 The nturl boundry condition is u x=b = 0. 10
2.2 Broken extreml nd the Weierstrss-Erdmn condition The clssicl derivtion of the Euler eqution requires the existence of ll second prtils of F, nd the solution u of the second-order differentil eqution is required to be twice-differentible. In some problems, F is only piece-wise twice differentible; in this cse, the extreml consists of severl curves solutions of the Euler eqution tht re computed t the intervls of smoothness. We consider the question: How to join these pieces together? The first continuity condition is continuity of the (differentible) minimizer u(x) [u] + = 0 long the optiml trjectory u(x) (20) Here [z] + = z + z denotes the jump of the vrible z. The extreml u is differentible, the first derivtive u exists t ll points of the trjectory. This derivtive does not need to be continuous. Insted, Euler eqution requests the differentibility of to ensure the existence of the term d dx in the Euler eqution. Integrting the sttionrity condition (8), we obtin sttionrity in the integrl form or If x S F (x, u, u )dx = x (x, u, u ) x = ( d (x, u, u ) dx (x, u, ) u ) dx = 0 (x, u, u ) x 0 dx + (x, u, u ) (21) x= is bounded t the optiml trjectory, the right-hnd side is continuous function of x, nd so is the left-hnd side. This requirement of continuity of n optiml trjectory is clled the Weierstrss-Erdmn condition on broken extreml. Theorem 2.1 At ny point of the optiml trjectory, the Weierstrss-Erdmn condition must be stisfied: [ ] + = 0 long the optiml trjectory u(x). (22) Exmple 2.3 (Broken extreml) Consider the Lgrngin F = 1 2 c(x)(u ) 2 + 1 { c1 if x [, x 2 u2, c(x) = ), c 2 if x (x, b] where x is point in (, b). The Euler eqution is held everywhere in (, b) except of the point x, d dx [c 1u ] u = 0 if x [, x ) d dx [c 2u ] u = 0 if x (x, b], 11
At x = x, the continuity conditions hold, u(x 0) = u(x + 0), c 1 u (x 0) = c 2 u (x + 0). The derivtive u (x) itself is discontinuous; its jump is determined by the jump in the coefficients: u (x + 0) u (x 0) = c 1 c 2 These conditions together with the Euler eqution nd boundry conditions determine the optiml trjectory. 3 Functionl dependent on higher derivtives Consider more generl type vritionl problem with the Lgrngin tht depends on the minimizer nd its first nd second derivtive, J = F (x, u, u, u )dx The Euler eqution is derived similrly to the simplest cse: The vrition of the gol functionl is ( δj = δu + δu + ) δu dx Integrting by prts the second term nd twice the third term, we obtin ( δj = d ) dx + d2 dx 2 δu dx [ + δu + δu d ] x=b δu (23) dx The sttionrity condition becomes the fourth-order differentil eqution d 2 dx 2 d dx x= + = 0 (24) supplemented by two nturl boundry conditions on ech end, δu [ = 0, δu d ] dx = 0 t x = nd x = b (25) or by the correspondent min conditions posed on the minimizer u nd its derivtive u t the end points. Exmple 3.1 The equilibrium of n elstic bending bem correspond to the solution of the vritionl problem L min w(x) 0 ( 1 2 (E(x)w ) 2 q(x)w)dx (26) 12
where w(x) is the deflection of the point x of the bem, E(x) is the elstic stiffness of the mteril tht cn vry with x, q(x) is the lod tht bends the bem. Any of the following kinemtic boundry conditions cn be considered t ech end of the bem. (1) A clmped end: w() = 0, w () = 0 (2) simply supported end w() = 0. (3) free end (no kinemtic conditions). Let us find eqution for equilibrium nd the missing boundry conditions in the second nd third cse. The Euler eqution (24) becomes The equtions (25) become (Ew ) q = 0 (, b) δu (Eu ) = 0, δu ((Ew ) ) = 0 In the cse (2) (simply supported end), the complementry vritionl boundry condition is Eu = 0, it expresses vnishing of the bending momentum t the simply supported end. In the cse (3), the vritionl conditions re Eu = 0 nd (Ew ) = 0; the lst expresses vnishing of the bending force t the free end (the bending momentum vnishes here s well). Generliztion The Lgrngin F (x, u, u,..., u (n)) dependent on first k derivtives of udependent on higher derivtives of u is considered similrly. The sttionry condition is the 2k-order differentil eqution d dx dk +... + ( 1)k dx k (k) = 0 supplemented t ech end x = nd x = b of the trjectory by k boundry conditions [ ] δu (k 1) (k) x=,b = 0 [ d ] δu (k 2) (k 1) dx (k) x=,b = 0... [ d dx d(k 1) +... + ( 1)k dx (k 1) ] δu (k) x=,b = 0 If u is vector minimizer, u cn be replced by vector but the structure of the necessry conditions sty the sme. 13
4 Non-fixed intervl 4.1 Trnsverslity condition Free boundry Consider now the cse when the intervl [, b] is not fixed, but the end point is to be chosen so tht it minimizes the functionl. Let us compute the difference between the two functionls over two different intervls = δi = +δx (F (x, u + δu, u + δu ) F (x, u, u )) dx + The second integrl is estimted s +δx b F (x, u + δu, u + δu )dx +δx b F (x, u, u )dx F (x, u + δu, u + δu )dx F (x, u + δu, u + δu )dx = F (x, u, u ) x=b δx + o( δu, δx ) nd the first integrl is computed s before with integrtion by prts: S F (x, u, u )δu dx + δu(b) = 0 x=b 1. Suppose tht no boundry conditions re imposed t the minimizer t the point x = b. Becuse of rbitrriness of δx nd δu, we rrive t the conditions: S F (x, u, u ) = 0 x (, b), = 0, x=b nd F (x, u, u ) x=b = 0. (27) Euler eqution for the extreml stisfies n extr boundry condition (27), but hs lso n dditionl degree of freedom: unknown coordinte b. Exmple 4.1 Consider the problem s ( ) 1 min u(x),s 2 u 2 u + x dx u(0) = 0. 0 The Euler eqution u + 1 = 0 nd the condition t u(0) = 0 corresponds to the extreml u = 1 2 x2 + Ax, u = x + A where A is prmeter. The condition = u = 0 t the unknown right end x = s gives s = A. The trnsverslity condition F = 0 or ( u + x) x=a=s = 1 2 s2 s 2 + s = s (1 12 ) s = 0 We find s = 2, u = 1 2 x2 + 2x. 14
2. Next, consider the problem in which the boundry dt t b is prescribed, u = β, but the vlue of b is not known. In the perturbed trjectory, the boundry condition is u(b + δx) = β. The vlue of u(b + δx) is n extrpoltion of u(x) s follows u(b + δx) = u(b) + u (b)δx + o( δu, δx ) Therefore, the vlue (u + δu) x=b depends on δx, u(b) = β u (b)δx or δu(b) = u (b)δx. Combining the depending on δx terms, we obtin the condition ( F (b)) x=b u δx Becuse δx is rbitrry, the boundry conditions re: u = β nd ( F (x, u, u ) u ) x=b = 0. (28) Notice tht the condition (28) t the unknown end is identicl to the first integrl (13) of the problem in the cse when F (u, u ) is independent of x. This integrl is constnt long the trjectory. therefore condition (28) cnnot be stisfied t n isolted point. 3. Finlly, consider the problem when the rjectory ends t curve. If the boundry vlue depends on b, u(b) = φ(b), then δu = φ δx u δx. The sttionrity conditions become ( u(b) = φ(b), F (u φ ) ) x=b = 0. (29) The next exmple dels with constrint t the unknown length of the intervl nd the boundry dt. Exmple 4.2 Find the shortest pth between the origin nd curve φ(x). The pth length is given by s I = min 1 + y 2 dx, u(0) = 0 y(x),s 0 At the end point x the pth meets the curve, therefore y(s) = φ(s) or The Euler eqution δy = φ (s)δs (30) y = y = C 1 + y 2 shows tht y = constnt, therefore the pth is stright line, y = Ax s expected. At the point s, the vrition is ( u ) y F δx + y y δy = 1 δx + y δu 1 + y 2 1 + y 2 The sttionrity gives the reltion. δx + y δu = 0. Compring it with the constrint (30), we conclude tht y (s)φ (s) = 1, or tht the shortest pth is line orthogonl to the curve φ(x), s it is expected. 15
4.2 Extreml broken t n unknown point Combining the techniques, we my ddress the problem of en extreml broken in n unknown point. The position of this point is determined from the minimiztion requirement. Assume tht Lgrngin hs the form { F (x, u, u F (x, u, u ) = ) if x (, ξ) F + (x, u, u ) if x (ξ, b) where ξ is n unknown point in the intervl (, b) of the integrtion. The Euler eqution is { SF (u) if x (, ξ) S F (u) = S F+ (u) if x (ξ, b) The sttionrity conditions t the unknown point ξ consist of sttionrity of the trjectory + = 0 (31) nd sttionrity of the position of the trnsit point F + (u) u + + = F (u) u. (32) They re derived by the sme procedure s the conditions t the end point. The vrition δx of the trnsit point δx = δx + = δx increses the first prt of the trjectory nd decreses the second prt, or vise vers, which explins the structure of the sttionrity conditions. In prticulr, if the Lgrngin is independent of x, the condition (32) expresses the constncy of the first integrl (13) t the point ξ. Exmple 4.3 Consider the problem with Lgrngin { F (x, u, u + u ) = 2 + b + u 2 if x (, ξ) u 2 if x (ξ, b) nd boundry conditions The Euler eqution is u() = 0, u(b) = 1 S F (u) = { + u b + u = 0 if x (, ξ) u = 0 if x (ξ, b) The solution to this eqution tht stisfies the boundry conditions is ( ) u + (x) = C 1 sinh b+ + (x ) if x (, ξ) u (x) = C 2 (x b) + 1 if x (ξ, b) ; it depends on three constnts ξ, C 1, nd C 2 (Notice tht the coefficient does not enter the Euler equtions). These constnts re determined from three conditions 16
t the unknown point ξ which express (1) continuity of the extreml (2) Weierstrss-Erdmn condition (3) trnsverslity condition u + (ξ) = u (ξ), + u +(ξ) = u (ξ), + (u +(ξ)) 2 + b + u(ξ) 2 = (u (ξ)) 2. The trnsverslity condition sttes the equlity of two first integrl. It is simplified to C 2 1b + = C 2 2 From the Weierstrss-Erdmn condition, we find C 1 + b + cosh q = C 2, where q = b + + (ξ ) The first condition nd the definition of q llows for determintion of ξ: cosh q = +, ξ = + + b + cosh 1 + Finlly, we define constnts C 1 nd C 2 from the continuity C 1 sinh q = 1 + C 2 (ξ b) nd trnsverslity conditions: C 1 = sinh q b + (ξ b), C b+ 2 = sinh q b + (ξ b), 5 Severl minimizers 5.1 Euler equtions nd first integrls The Euler eqution cn be nturlly generlized to the problem with the vectorvlued minimizer I(u) = min u F (x, u, u )dx, (33) where x is point in the intervl [, b] nd u = (u 1 (x),..., u n (x)) is vector function. We suppose tht F is twice differentible function of its rguments. 17
Let us compute the vrition δi(u) equl to I(u + δu) I(u), ssuming tht the vrition of the extreml nd its derivtive is smll nd loclized. To compute the Lgrngin t the perturbed trjectory u+δu, we use the expnsion F (x, u + δu, u + δu ) = F (x, u, u ) + n i=1 i δu i + n δu i i=1 i We cn perform n independent vritions of ech component of vector u pplying vritions δ i u = (0,..., δu i..., 0). The increment of the objective functionl should be zero for ech of these vrition, otherwise the functionl cn be decresed by one of them. The sttionrity condition for ny of considered vritions coincides with the one-minimizer cse. δ i I(u) = ( δu i + δu i i i ) dx 0 i = 1,..., n. Proceeding s before, we obtin the system of n second-order differentil equtions, d = 0, i = 1,... n (34) dx i nd the boundry term i n x=b δu i = 0 (35) i=1 i If the vlue of u i () or u i (b) is not prescribed, the nturl boundry conditions x= or i x=b, respectively, must be stisfied. i The vector form of the system (34), S F (u) = d dx x=b = 0, δut = 0 (36) is identicl to the sclr Euler eqution. This system corresponds to n definition of differentition with respect to vector rgument u. x= Exmple 5.1 Consider the problem with the integrnd x= F = 1 2 u 2 1 + 1 2 u 2 2 u 1 u 2 + 1 2 u2 1 (37) The system of sttionrity conditions is computed to be d dx d dx 1 2 1 = u 1 + u 2 u 1 = 0 2 = (u 2 u 1 ) = 0. If consists of two differentil equtions of second order for two unknowns u 1 (x) nd u 2 (x). 18
First integrls The first integrls tht re estblished for the specil cses of the sclr Euler eqution, cn lso be derived for the vector eqution. 1. If F is independent of u k, then one of the Euler equtions degenertes into lgebric reltion: = 0 k nd the one of differentil eqution in (34) becomes n lgebric one. The vrible u k (x) cn be discontinuous function of x in n optiml solution. Since the Lgrngin is independent of u k, the discontinuities of u k(x) my occur long the optiml trjectory. 2. If F is independent of u k, the first integrl exists: k = constnt For instnce, the second eqution in Exmple 5.1 cn be integrted nd replced by u 2 u 1 = constnt 3. If F is independent of x, F = F (u, u ) then first integrl exist Here W = u T T u = F = constnt (38) n i=1 u i i For the Exmple 5.1, this first integrl is computed to be ( 1 W = u 2 1 + u 2 (u 2 u 1) 2 u 2 1 + 1 2 u 2 2 u 1 u 2 + 1 ) 2 u2 1 = 1 ( u 2 2 1 + u 2 2 u 2 1) = constnt These three cses do not exhust ll possible first integrls for vector cse. For exmple, if the functionl depends only on, sy (u 1 + u 2 ), one cn hope to find new invrints by chnging the vribles. We discuss this mtter below in Sections?? nd??. Trnsverslity nd Weierstrss-Erdmn conditions These conditions re quite nlogous to the sclr cse nd their derivtion is strightforwrd. We simply list them here. The expressions, i = 1..., n remin continuous t every point of n i optiml trjectory, including the points where u i is discontinuous. If the end point of the trjectory is unknown, the condition t the end point is stisfied. u T F = 0 19
5.2 Vritionl boundry conditions Consider the vrition of the boundry term (35) which we rewrite here for convenience δu 1 +... + x=b 1 δu n = 0 (39) n x= If ll vritions δu i (), δu i (b) re free, it produces 2n boundry conditions i = 0, x = nd x = b for Euler equtions (34). In the opposite cse, when the vlues of ll minimizers re prescribed t the end points, u i () = u i, u i (b) = u b i, i = 1,..., n then the eqution (39) is stisfied, becuse ll vritions re zero. δu i () = 0, δu i (b) = 0, i = 1,..., n If the vlues of severl components of u() or u(b) re not given, the vritions of these components re free nd the corresponding nturl boundry condition supplements the boundry conditions: For ech i = 1,..., n one of the two conditions holds Either = 0 or δu i x=,b = 0. (40) x=,b i The totl number of the conditions t ech endpoint is n. The missing min boundry conditions re supplemented by the nturl conditions tht express the optimlity of the trjectory. This number grees with the number of boundry conditions needed to solve the boundry vlue problem for Euler eqution. In generl cse, p reltions (p < 2n) between boundry vlues of u re prescribed, β k (u 1 (),..., u n (), u 1 (b),..., u n (b), ) = 0 (41) t the end points x = nd x = b. In this cse, we need to find 2n p supplementry vritionl constrints t these points tht together with (41) give 2n boundry conditions for the Euler eqution (35) of the order 2n. The conditions (41) re stisfied t ll perturbed trjectories, β k (w + δw) = 0 where 2n dimensionl vector w is the direct sum of u() nd u(b) defined s: w k = u k () if k = 1,..., n w k = u k n (b) if k = n + 1,..., 2n. 20
The vrition δw i is constrint by liner system T β k δw = 0, w k = 1,..., p which shows tht the constrints (41) re stisfied t the vried trjectory. A mtrix form of these conditions is P δw = 0, where β 1 β 1... 1 n P =......... β p β 1... p n nd we my lso ssume tht these conditions re linerly independent. Then the solution to the system is vector δw d of the form δw d = Qv where v is n rbitrry (2n p)-dimensionl rbitrry vector nd (2n p) n mtrix Q is supplement (orthogonl mtrix) to P tht is defined s solution to the mtrix eqution P Q = 0 which leds to P δw d = P Qδv = 0 for ny v. Any dmissible vrition δw d mkes the fist vrition (36) of the objective functionl vnish; correspondingly, we hve ( ) T δw d = 0 Using the representtion of δw d nd the rbitrriness of the potentils v, we conclude tht the fist vrition vnishes is the coefficient by ech of these potentils is zero or ( ) T Q = 0 (42) This representtion provides the 2n p missing boundry conditions. Exmple 5.2 Consider gin the vritionl problem with the Lgrngin (37) ssuming tht the following boundry conditions re prescribed u 1 () = 1, β(u 1 (b), u 2 (b)) = u 2 1(b) + u 2 2(b) = 1 Find the complementry vritionl boundry conditions. At the point x =, the vrition δu 1 is zero, nd δu 2 is rbitrry. The vritionl condition is = u 2() u 1 () = 0 x= 2 Since the conditionu 1 () = 1 is prescribed, it becomes u 2() = 1 21
At the point x = b, the vritions δu 1 nd δu 2 re connected by the reltion β 1 δu 1 + β 2 δu 2 = 2u 1 δu 1 + 2u 2 δu 2 = 0 which implies the representtion (δu = Qδv) δu 1 = u 2 δv, δu 2 = u 1 δv where δv is n rbitrry sclr. The vritionl condition t x = b becomes ( u 2 + ) 1 u 1 δv = ( u 1u 2 + (u 2 u 1 )u 1 ) x=b δv = 0 δv 2 x=b or u 1u 2 + u 1 u 2 u 2 1 x=b = 0. We end up with four boundry conditions: u 1 () = 1, u 2() = 1, u 2 1(b) + u 2 2(b) = 1, u 1 (b)u 2(b) u 1 (b) u 2 (b) u 1 (b) 2 = 0. The second rw conditions re vritionl, they re obtined from the minimiztion requirement. Periodic boundry conditions Consider vritionl problem with periodic boundry conditions u() = u(b). The vritionl boundry conditions re = x= x=b They re obtined from the expression (39) nd equlities δu() = δu(b). 22