Valeriy Slastikov Spring, f(x, u(x), u (x)) dx, (1)

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1 Clculus of Vritions Vleriy Slstikov Spring, D Clculus of Vritions. We re going to study the following generl problem: Minimize functionl I(u) = subject to boundry conditions u() = α, u(b) = β. f(x, u(x), u (x)) dx In order to correctly set up this problem we hve to ssume certin properties of f(x, u, ξ) nd function u(x). Since we know how to integrte continuous functions it is resonble to ssume tht function f : [, b] R R R is continuous. In fct in this section we will need stronger ssumption tht f C 2 ([, b] R R), mening tht f, f nd f re continuous functions. Concerning function u : [, b] R we my ssume tht u C 1 ([, b]) nd u() = α, u(b) = β. It seems resonble since minimiztion problem involves u (x) (tht needs to be continuous since we know how to integrte continuous functions) nd boundry conditions. In fct in this section we will use bit stronger regulrity ssumption u C 2 ([, b]). Note: we cn chnge boundry conditions by removing constrints t one or both ends. In generl, one cn not prescribe u t the end points (this will be explined lter). We re redy to set up the problem with mthemticl precision: Define I(u) where f C 2 ([, b] R R), u C 1 ([, b]). Problem: where X = {u C 1 ([, b]) : u() = α, u(b) = β}. f(x, u(x), u (x)) dx, (1) inf I(u), (2) u X We wnt to find function u (x) such tht u X nd u delivers minimum to the functionl I(u) (we cll such function minimizer). In order to do this we need to understnd wht it mens to minimize the functionl I(u). Definition 1.1 Function u X is minimizer (or globl minimizer) of the functionl I(u) on set X if I(u ) I(u) for ll u X. In generl it is very difficult to solve (2), however one cn try to chrcterize minimizers. Let s consider simple exmple from Clculus. 1

2 Exmple 1.1 Let g C 1 ([, b]) nd we wnt to find minimizer of g. By well known result of Anlysis we know tht ny continuous function on compct intervl (closed nd bounded) chieves its minimum (nd mximum). Obviously the minimizer is just some point x [, b]. In order to find it we hve to use the definition, i.e. g(x ) g(x) for ll x [, b]. Assume tht x (, b) then for ny x (, b) we hve nd f(x) f(x ) x x, if x > x f(x) f(x ) x x, if x < x. Tking limit x x we obtin f (x ) nd f (x ). Therefore we hve f (x ) =. Note: we must ssume x (, b) in order to obtin f (x ) =, explin it. In order to find minimizer we must find ll points where f (x) = nd compre vlues t these points with vlues t the ends of the intervl (f(), f(b)). In order to chrcterize minimizers of I(u) we proceed in the similr wy. We know tht if u is minimizer then I(u ) I(u) for ll u X. Tke ny function h C 1([, b]), obviously u + th X nd therefore I(u ) I(u + th) for ll h C([, 1 b]). We cn define function F (t) = I(u + th) nd notice tht since u is minimizer of I(u) then t = is minimizer of F (t) on R. Therefore we must hve F () = (see exmple). Note: we priori ssumed existence of minimizer u X for I(u) nd just trying to chrcterize it if it exists. In the exmple bove for 1D function existence ws gurnteed by simple result of Anlysis (look it up). We chrcterize minimizers in the following theorem Theorem 1.2 Let f C 2 ([, b] R R) (f f(x, u, ξ)) nd { inf I(u) u X where X = {u C 1 ([, b]) : u() = α, u(b) = β}. Then 1. If (3) dmits minimizer u X C 2 ([, b]) then } f(x, u(x), u (x)) dx = m, (3) d ( fξ (x, u (x), u dx (x)) ) = f u (x, u (x), u (x)) x (, b) (4) u () = α, u (b) = β. 2. If some function u X C 2 ([, b]) stisfies (4) nd if (u, ξ) f(x, u, ξ) is convex (s function of two vribles) for every x [, b] then u is minimizer of (3). 2

3 Proof. 1) As we discussed before, if u X is minimizer of I(u) then F () = (see the definition of F bove). Let s find F (): by definition of derivtive It s not difficult to see tht I(u + th) I(u ) t F F (t) F () I(u + th) I(u ) () = lim = lim. t t t t = f(x, u (x) + th(x), u (x) + th (x)) f(x, u (x), u (x)) t dx. (5) Pssing to the limit nd using stndrd Anlysis results to interchnge integrl nd limit (for instnce Dominted Convergence Theorem) we obtin F () = Using stndrd rules of differentition we hve d dt f(x, u (x) + th(x), u (x) + th (x)) t= dx. d dt f(x, u (x) + th(x), u (x) + th (x)) t= = f u (x, u (x), u (x))h(x) + f ξ (x, u (x), u (x))h (x). (6) Therefore if u X is minimizer of I(u) we must hve f u (x, u (x), u (x))h(x) + f ξ (x, u (x), u (x))h (x) dx = for ll h C 1 ([, b]). This is wek form of Euler-Lgrnge eqution. It is not cler how to del with this integrl identity nd therefore we wnt to obtin pointwise identity. We cn use integrtion by prts to obtin f ξ (x, u (x), u (x))h (x) dx d ( = fξ (x, u (x), u dx (x)) ) h(x) dx + f ξ (x, u (x), u (x))h(x) b. (7) Reclling tht h() = h(b) = we hve [ f u (x, u (x), u (x)) d ( fξ (x, u (x), u dx (x)) )] h(x) dx = for ll h C 1 ([, b]). Using Fundmentl Lemm of Clculus of Vritions (proved fter this theorem) we obtin f u (x, u (x), u (x)) d ( fξ (x, u (x), u dx (x)) ) = x (, b). Conditions u () = α, u (b) = β follow from the fct u X. 2) Let u X C 2 ([, b]) is solution of (4). Since f(x, u, ξ) is convex in lst two vribles we hve f(x, u(x), u (x)) f(x, u (x), u (x)) + f u (x, u (x), u (x))(u(x) u (x)) + f ξ (x, u (x), u (x))(u (x) u (x)) (8) 3

4 for every u X (this inequlity will be shown lter for convex function of one vrible, extend it). We cn integrte both prts of this inequlity over (, b) to obtin f(x, u(x), u (x)) dx f(x, u (x), u (x)) dx + + f u (x, u (x), u (x))(u(x) u (x)) dx f ξ (x, u (x), u (x))(u (x) u (x)) dx. (9) Using integrtion by prts nd noting tht u u C 1 ([, b]) we obtin [ I(u) I(u ) + f u (x, u (x), u (x)) d ( fξ (x, u (x), u dx (x)) )] (u(x) u (x)) dx. (1) Since u stisfies (4) we hve I(u) I(u ) for ll u X. Theorem is proved. Remrk. Note tht we did not prove existence of minimizer in this theorem. In the first sttement we showed tht if u is minimizer then it solves Euler-Lgrnge eqution. In the second sttement we showed tht if u solves Euler-Lgrnge eqution nd the function f is convex in lst two vribles then u is minimizer. But we did not prove existence of solution for Euler-Lgrnge eqution nd therefore even the cse of convex f we don t know if there is minimizer. Now we prove Fundmentl Lemm of Clculus of Vritions (FLCV) Lemm 1.1 Let g C([, b]) nd then g(x) = on [, b]. g(x)η(x) dx = for ll η C ([, b]), Proof. Suppose there exists point c (, b) such tht g(c). We my ssume without loss of generlity tht g(c) >. By continuity of function g(x) there exists n open intervl (s, t) (, b) contining point c such tht g(x) > on (s, t). Tking η(x) to be positive on (s, t) with η(s) = η(t) = we obtin the contrdiction with g(x)η(x) dx =. Therefore g(x) = for ll x (, b). By continuity of g we obtin g(x) = on [, b]. Lemm is proved. For convenience we lso prove simple fct tht for g C 1 ([, b]) convexity is equivlent to g(x) g(y) + g (y)(x y) for ll x, y (, b). Proof. Let g be convex then by definition g(αx + (1 α)y) αg(x) + (1 α)g(y) for ll α (, 1) nd x, y in (, b). After simple rerrngement we hve g(y + α(x y)) g(y) α g(x) g(y). 4

5 Tking limit s α we obtin We proved it in one direction. Now we ssume g (y)(x y) g(x) g(y). g(x) g(y) + g (y)(x y) for ll x, y (, b). Using this we hve nd g(x) g(αx + (1 α)y) + (1 α)g (αx + (1 α)y)(x y) g(y) g(αx + (1 α)y) αg (αx + (1 α)y)(x y). Multiplying first inequlity by α, second inequlity by 1 α nd dding them we obtin Result is proved. αg(x) + (1 α)g(y) g(αx + (1 α)y). Note tht this fct is true lso for functions defined on R n. Let s consider severl exmples. Exmple 1.2 (Brchistochrone problem) In this exmple we hve 1 + ξ 2 f(u, ξ) = u, nd X = {u C 1 ([, 1]) : u() =, u(1) = β, u(x) > for x (, 1) }. solutions of Euler-Lgrnge eqution. Using theorem 1.2 we cn esily find Euler-Lgrnge equtions ( u 1 + (u = u[1 + (u ) ]) ) 2 2 4u 3. We would like to find u We cn multiply both prts by u[1+(u ) 2 ] nd obtin ( ) u u u[1 + (u ) 2 ] u[1 + (u ) 2 ] = u 2u 2. It is cler tht this is equivlent to Now we obtin ( (u ) 2 ) = u[1 + (u ) 2 ] ( ) 1. u 1 u = (u ) 2 u[1 + (u ) 2 ] + C 5

6 nd simplifying this expression we obtin u[1 + (u ) 2 ] = 2µ, where µ > is some constnt tht we cn find from boundry conditions. Seprting vribles we obtin u dx = 2µ u du. Now we integrte both prts x = u y 2µ y dy. We cn mke nturl substitution y = µ(1 cos θ) to obtin x = µ(θ sin θ). Therefore solution to this problem cn be represented in the following prmetric form: u(θ) = µ(1 cos θ), x(θ) = µ(θ sin θ). It is cler tht u() = so tht first boundry condition is stisfied nd we cn find µ by pplying second boundry condition. Note tht we don t know if this solution is minimizer of the problem. For this we hve to prove either existence of minimizer or convexity of f(u, ξ). However from physicl model we cn see tht there must be solution here (therefore minimizer should exist) nd since solution to the problem is unique it must be minimizer (I guess tht ws the resoning in 17-th century). Exmple 1.3 (Miniml surfce of revolution) In this exmple we hve f(u, ξ) = 2πu 1 + ξ 2, nd X = {u C 1 ([, 1]) : u() = α, u(1) = β, u(x) > }. We would like to find solutions of Euler-Lgrnge eqution. Agin we cn use the theorem 1.2 to obtin ( ) uu = 1 + (u ) (u ) 2 uu Multiplying both prts by 1+(u ) 2 nd integrting we obtin u (u ) 2 = C, or (u ) 2 = u2 2 1 for some constnt >. We cn seprte vribles to obtin the solution here but there is better wy: we serch for solution in the following form u(x) = cosh f(x). Plugging this into eqution we obtin [f (x)] 2 = 1 nd therefore either f(x) = x + µ or f(x) = x + µ. Since cosh x is even function nd µ is ny constnt we hve u(x) = cosh x + µ. Continue to find the solutions of this problem ssuming u() = u(1) = α >. The number of solutions will depend on α. 6

7 Exmple 1.4 (Convexity) In this exmple we ssume tht f(ξ) is convex on R nd X = {u C 1 ([, b]) : u() = α, u(b) = β, }. We will prove existence of minimizer for I(u) = f(u (x)) dx in this cse nd find it explicitly. By convexity of f we know tht f(x) f(y) + f (y)(x y). We cn define u (x) = β α b (x ) + α. Using bove inequlity we see tht for ny u X It is cler tht u β α (x) b f(u (x)) f(u (x)) + f (u (x))(u (x) u (x)). nd therefore integrting both prts over (, b) we obtin Therefore u is minimizer of I(u). f(u (x)) dx Now we give severl exmples of nonexistence. f(u (x)) dx. Exmple 1.5 (Non-convexity implies no minimizer) We consider f(ξ) = e ξ2 C 1 ([, 1]) : u() =, u(1) =, } nd wnt to show tht for the problem nd X = {u there is no solution. We see tht Euler-Lgrnge eqution is inf X 1 f(u (x)) dx d dx f (u ) =. And therefore u = const tht implies in this cse (using boundry conditions) tht u(x). It is cler tht this is mximizer of the problem, not minimizer. Moreover, if we tke u n (x) = n(x 1 2 )2 n 4 we hve u n X for ny n N. Clculting I(u n ) we obtin 1 I(u n ) = e 4n2 (x 1 2) 2 dx = 1 n e x2 dx, 2n n s n. Therefore inf X I(u) = but obviously no function cn deliver this infimum nd so minimizer does not exist. Exmple 1.6 (Not smooth minimizers) Consider f(ξ) = (ξ 2 1) 2, X = {u C 1 ([, 1]) : u() =, u(1) =, }. Show tht there re no minimizers for this problem. However there re infinitely mny minimizers tht re piecewise smooth. 7

8 1.1 Problems with constrints - Lgrnge multipliers In mny problems rising in pplictions there exist some integrl constrints on the set of functions where we look for minimizer. One bsic exmple is Ctenry problem where we hve to find the shpe of the hnging chin of fixed length (see exmple below). In order to chrcterize the minimizer of the problem (2) with integrl constrint G(u) g(x, u(x), u (x)) dx = (11) we cn not simply tke vrition of the functionl I(u) with respect to ny function h C 1 ([, b]) since we cn violte this integrl constrint. Therefore we hve to invent method tht llows us to derive Euler-Lgrnge eqution in this sitution. This method is clled Method of Lgrnge Multipliers. Let s stte the problem tht we re solving here with mthemticl precision. We investigte minimizers of the following problem where We cn prove the following theorem inf u Y Y = X {u C 1 ([, b]) : G(u) f(x, u(x), u (x)) dx, (12) g(x, u(x), u (x)) dx = }. Theorem 1.3 Let f, g C 2 ([, b] R R) nd u be minimizer of the problem (12) bove such tht g ξ (x, u (x), u (x))w (x) + g u (x, u (x), u (x))w(x) dx (13) for some w C 1([, b]). Then u stisfies Euler-Lgrnge equtions corresponding to the following problem inf I(u) λg(u). (14) (u,λ) X R Proof. Let u be minimizer of (12). We would like to tke vrition o I(u) preserving constrint G(u) =. By (13) (rescling if needed) we know tht there exists function w C 1 ([, b]) such tht g ξ (x, u (x), u (x))w (x) + g u (x, u (x), u (x))w(x) dx = 1. (15) Let v C 1 ([, b]) be ny nd w be s bove, we define eeewsddq4 nd F (ε, h) = I(u + εv + hw) = G(ε, h) = G(u + εv + hw) = f(x, u (x) + εv(x) + hw(x), u (x) + εv (x) + hw (x)) dx (16) g(x, u (x) + εv(x) + hw(x), u (x) + εv (x) + hw (x)) dx. (17) 8

9 We know tht G(, ) = G(u ) = (since u stisfies constrint (11)) nd G h (, ) = 1 (since u stisfies (15)). Using Implicit Function Theorem we cn deduce tht there exists ε nd continuously differentible function t(ε) such tht t() = nd G(ε, t(ε)) = for ll ε [ ε, ε ]. Therefore we know tht for ll ε [ ε, ε ] we hve u + εv + t(ε)w Y. Now we cn tke vrition of I(u). We consider F (ε, t(ε)) = I(u + εv + t(ε)w), by the sme rguments s before since u is minimizer of I(u) nd since F (, ) = I(u ) we know tht is minimizer of function F (ε, t(ε)) nd therefore Computing this derivtive we obtin It s not difficult to see tht F h (, ) = d dε F (ε, t(ε)) ε= =. F ε (, ) + F h (, )t () =. f ξ (x, u (x), u (x))w (x) + f u (x, u (x), u (x))w(x) dx, nd since u nd w re fixed functions F h (, ) is just constnt tht we denote s λ F h (, ). We hve to understnd wht is t () nd in order to do this we use the fct tht G(ε, t(ε)) = for ll ε [ ε, ε ] nd compute = d dε G(ε, t(ε)) ε= = G ε (, ) + G h (, )t (). Since G h (, ) = 1 then t () = G ε (, ). Therefore we hve tht if u is minimizer of I(u) then We my rewrite this s F ε (, ) λg ε (, ) =. f u (x, u (x), u (x))v(x) + f ξ (x, u (x), u (x))v (x) dx λ g ξ (x, u (x), u (x))v (x) + g u (x, u (x), u (x))v(x) dx = (18) for ll v C 1 ([, b]). Using integrtion by prts, FLCV nd reclling tht G(u ) = we obtin d ( fξ (x, u (x), u dx (x)) ) f u (x, u (x), u (x)) ( d = λ dx ( gξ (x, u (x), u (x)) ) ) g u (x, u (x), u (x)) g(x, u (x), u (x)) dx = This is exctly Euler-Lgrnge eqution for the problem (14). Theorem is proved. x (, b) (19) 9

10 Remrk. Think bout why we need condition with w. Exmple 1.7 (Ctenry problem) Solve this problem. Exmple 1.8 (Isotropic-nemtic phse trnsition in liquid crystls) In this exmple we discuss 2D Onsger s model tht describes isotropic-nemtic phse trnsition in liquid crystlline system. The liquid crystl molecule cn be viewed s rod. Since we re considering 2D system the orienttion prmeter of the rod is just one ngle φ [, 2π). The corresponding free energy functionl reds, 2π [ F[ρ] := τρ(φ) ln ρ(φ) 1 2π ] 2 ρ(φ) cos 2(φ ψ) ρ(ψ) dψ dφ. (2) Here ρ(φ) is probbility density of rod orienttions (we ssume ρ > with constrint 2π ρ(φ) dφ = 1). Vrying the temperture τ one cn put the system in isotropic stte ρ = 1 2π (for high tempertures) nd nemtic stte ρ δ(φ φ ) (for low tempertures). We hve problem with constrint nd hve to use Lgrnge multipliers method to find Euler- Lgrnge eqution. The Euler-Lgrnge eqution for criticl points of the free energy (2) reds 2π cos 2(φ ψ) ρ(ψ) dψ + τ ln ρ(φ) = µ, (21) where µ is the Lgrnge multiplier which ssures tht integrl of ρ is unity. thermodynmic potentil Φ by reltion ρ(φ) =: Z 1 exp{φ(φ)}, Z = 2π Let us introduce the exp{φ(φ)} dφ. (22) Note tht both ρ nd Φ implicitly depend on the temperture τ. The Euler-Lgrnge eqution (21) cn be written s τ Φ(φ) = 2π Differentiting this eqution twice with respect to φ, we obtin cos 2(φ ψ) ρ(ψ) dψ. (23) Φ (φ) + 4Φ(φ) =. (24) Now let us show tht ny solution of this eqution produces legitimte solution of the Euler-Lgrnge eqution (21) for some the temperture τ. An rbitrry solution of (24) is given by Φ(φ) = β cos 2(φ φ ) for some β nd φ (shifting φ by π/2 chnges sign of β). Define cndidte solution of (21), ϱ(φ), by ϱ(φ) := Z 1 e β cos 2(φ φ ), Z = Observe tht β = genertes the uniform density ϱ = ρ, 2π e β cos 2ψ dψ. (25) ρ(φ) = 1 2π, (26) which solves (21) for ll tempertures τ. From now on we will ssume β >. Let us verify tht ny ϱ given by (25) stisfies the Euler-Lgrnge eqution (21) for some τ. Substituting ϱ nd Φ into (23) we obtin tht 2π βτ cos 2(φ φ ) = Z 1 cos 2(φ ψ) e β cos 2(ψ φ) dψ (27) 1

11 should hold for ll φ nd some constnt (independent on φ) τ. Using trigonometric identity cos 2(φ ψ) = cos 2(φ φ ) cos 2(ψ φ ) + sin 2(φ φ ) sin 2(ψ φ ) nd integrting (the term with sin 2(φ φ ) disppers) we obtin 2π βτ cos 2(φ φ ) = Z 1 cos 2(φ φ ) cos 2ψ e β cos 2ψ dψ. (28) The fct tht dependence on φ ppers on both sides only through cos 2(φ φ ) gurntees tht τ (if exists for given β) does not depend on φ, s required. Using the definition of Z, (25), we cn rewrite (28) s, Noticing tht τ = 1 β d dβ ln Z = 1 β d 2π dβ ln e β cos 2ψ dψ. (29) I (β) = 1 2π e β cos 2ψ dψ, 2π where I (β) is the modified Belles function, we obtin τ = 1 β d dβ ln I (β) = 1 I 1 (β) β I (β). (3) This eqution gives us ll vlues of β nd therefore ll criticl points depending on the temperture τ (except Φ(φ) ). 11

12 2 Second Vrition nd Stbility In the previous section we introduced minimiztion problem nd lerned method to find criticl points of the functionl I(u) using Euler-Lgrnge equtions. However we still don t know nything bout how to solve the originl minimiztion problem. In prticulr we don t understnd if the criticl points tht we find by solving Euler-Lgrnge equtions re stble or unstble (in some sense). This section is devoted to investigting stbility of the criticl point of I(u) using second vrition. 2.1 Vrious Derivtives In this section we introduce two derivtives tht might be useful when deling with the first nd second vritions. Definition 2.1 Let u, v C 1 ([, b]) nd I : C 1 ([, b]) R be functionl (see for instnce (1)). We define Gteux derivtive of I t the point u in the direction v s I I(u + tv) I(v) (u, v) = lim. t t If the bove limit exists for ll v C 1 ([, b]) we sy tht I is Gteux differentible. We lredy implicitly used Gteux derivtive in the previous section when we were tking first vrition of the functionl I(u). Definition 2.2 Let u, v C 1 ([, b]) nd I : C 1 ([, b]) R be functionl (see for instnce (1)). We sy tht I is Frechet differentible t u if there exists bounded liner opertor A u : C 1 ([, b]) R such tht I(u + v) I(v) A u (v) lim C 1 =. t v C 1 If the bove limit exists for ll u C 1 ([, b]) we sy tht I is Frechet differentible nd cll denote its derivtive s DI(u). Note tht in the bove definitions you cn chnge C 1 ([, b]) nd R to ny Bnch spces X, Y. 2.2 Computing Second Vrition Now we re redy to compute the second vrition of the functionl I(u) = defined on X = {u C 1 ([, b]), u() = α, u(b) = β}. f(x, u(x), u (x)) dx Definition 2.3 Assuming f C 2 we define the second vrition of I t the point u X in the direction φ C 1 ([, b]) s follows Here derivtives re tken in Gteux sense. δ 2 I(u)(φ) = d2 dt 2 t= I(u + tφ). 12

13 It is not difficult to compute this derivtive for the bove functionl I(u) explicitly. δ 2 I(u)(φ) = ( φ 2 f uu (x, u, u ) + 2φφ f uξ (x, u, u ) + φ 2 f ξξ ) dx. We cn integrte this expression by prts tking into ccount φ C 1 ([, b]) to obtin [ δ 2 I(u)(φ) = (φ 2 f uu (x, u, u ) d ] ) dx f uξ(x, u, u ) + φ 2 f ξξ dx. The second vrition is extremely importnt in the investigtion of the stbility nd locl minimlity of the criticl point of functionl I(u). 2.3 Second Vrition nd Stbility We know tht if f R R is twice differentible function nd x is criticl point of f then f (x ) =. If we wnt to understnd more bout nture of x, i.e. if it is locl minimum, locl mximum or sddle point we know from bsic nlysis tht we hve to investigte the sign of f (x ). In prticulr if f (x ) > then x is locl minimum, i.e. there exists δ > such tht f(x ) f(x) for ll x (x δ, x + δ) ; if f (x ) < then x is locl mximum, i.e. there exists δ > such tht f(x ) f(x) for ll x (x δ, x + δ) ; if f (x ) = then nothing definite cn be sid bout properties of x nd in order to conclude you need to consider higher order derivtives. Remrk. Recll tht we used nlogy between 1D problems nd infinite dimensionl problem in the previous section when finding necessry conditions on the criticlity of u. Think bout how you cn prove bove sttements using convexity (concvity) of f or Tylor expnsion of f ner x. Cn you trnsfer your ides to the infinite dimensionl cse? We re interested in minimistion problem nd therefore we wnt to single out locl minimisers (if we cnnot find globl minimizer). We lredy understnd tht locl minimisers of I(u) hve to stisfy Euler-Lgrnge equtions. Now we wnt to show tht locl minimisers hve to stisfy more restrictive condition. Proposition 2.4 Let u X = {u C 1 ([, b]), u() = α, u(b) = β} be locl minimizer of the functionl I(u) defined in (1), i.e. there exists δ > such tht I(u ) I(u) for ll u X such tht u u C 1 < δ. Then δ 2 I(u )(φ), for ll φ C([, 1 b]). Proof Assume tht there is φ C 1 ([, b]) such tht δ2 I(u )(φ ) <. Define function F (t) = I(u + tφ ). It s cler tht t = is criticl point of F (t) nd F () =, F () <. Now you cn use 1D rguments to show tht there exists δ > such tht F () > F (t) for ll t (t δ, t + δ), t nd this is in contrdiction with locl minimlity of u. Bsed on this result we introduce severl definitions 13

14 Definition 2.5 The criticl point u X of the functionl I(u) is clled loclly stble if δ 2 I(u )(φ) for ll φ C ([, b]); strictly loclly stble if δ2 I(u )(φ) > for ll φ C ([, b]); nd uniformly loclly stble in spce Y if δ 2 I(u )(φ) > µ φ 2 Y for ll φ C ([, b]). Remrk. Here spce Y does not necessrily coincides with X nd usully is much weker thn Y. The typicl exmple for Y in our frmework would be either W 1,p (, b) or L p (, b) for some p > 1. In generl, there is no hope to tke Y = C 1 ([, b]) nd prove uniform positivity of the second vrition for I. Tht s why you need to study Sobolev nd L p spces if you wnt to do reserch in Clculus of Vritions. There is nother necessry condition (Legendre condition) for locl stbility tht directly follows from definition of locl stbility. Proposition 2.6 Let I(u) be defined s usully (see (1)), u is criticl point of I(u) nd δ 2 I(u )(φ) for ll φ C ([, b]) then f ξξ (x, u (x), u (x)) for ll x [, b]. Proof Proof is simple exercise. Try to use mollifier in the neighbourhood of point x to show tht f ξξ (x, u (x), u (x)). 2.4 Locl Minimlity In the previous section we introduced the notion of locl stbility. However, in order to show locl minimlity it is in generl not enough to hve locl stbility. Definition 2.7 We sy tht u X is locl minimizer of I(u) if there exists δ > such tht I(u ) I(u) for ll u X such tht u u X < δ. Proposition 2.8 Uniform locl stbility implies locl minimlity. Proof Left s n exercise. 3 Importnt Exmples In this section we study severl importnt exmples tht shows how to obtin the informtion on the second vrition nd how to use it to show locl (nd sometimes globl) minimlity. Exmple 1. Let s define I(u) in the following wy ( 1 I(u) = 2 u ) 4 (u2 1) 2 dx. We define set X = {u C 1 ([, R]), u() =, u(r) = 1} nd wnt to solve the following problem inf I(u). u X At this point we know tht in order to find solution to this problem we hve to find criticl points by solving Euler-Lgrnge equtions. There is no gurntee tht wht we find is minimizer but we will try to investigte its stbility lter. 14

15 Euler-Lgrnge equtions It is cler tht Euler-Lgrnge equtions re u = u(u 2 1), u() =, u(r) = 1. If you wnt to find the solution here you cn reduce this eqution to first order eqution tht you might be ble to solve. Multiplying both prts of the bove eqution by u nd integrting from to x we obtin 1 ( u (s) 2) 1 ( ds = (u 2 (s) 1) 2) ds x 2 x 4 This gives us the following reltion 1 2 u (R) u (x) 2 = 1 ( (u 2 (R) 1) 2) 1 ( (u 2 (x) 1) 2). 4 4 Now we cn use boundry dt to obtin u (x) 2 = u (R) ( (u 2 (x) 1) 2). We now hve u (x) = ± u (R) ((u2 (x) 1) 2 ) it s not difficult to show tht u (x) cnnot chnge sign (try to elborte on it) nd therefore u (x) = u (R) ((u2 (x) 1) 2 ). This is seprble eqution nd one cn obtin the solution s x = u du K + 12 (u2 1) 2, where K > should be determined from the boundry condition u(r) = 1. Note, tht you hve obtined unique solution to the Euler-Lgrnge eqution nd therefore if minimizer exists then it hs to be the minimizer. One cn show the existence of minimizer in this problem using Sobolev spces nd Direct Methods. Since we don t know nything bout Sobolev spce nd Direct methods yet, we proceed by investigting stbility of this solution. Stbility In order to do this we hve to study second vrition of the functionl I(u). computtion gives us δ 2 I(u)(φ) = φ 2 + (3u 2 1)φ 2, Strightforwrd where φ C (, R). We wnt to show tht δ2 I(u)(φ) > for ll nontrivil φ C (, R). If you look t the expression for the second vrition you notice tht it is not cler if it is positive or negtive unless you hve extremely good informtion bout behvior of solution u. At this poit we know tht u(x) > for x > (try to understnd why this is the cse). Any other informtion bout 15

16 solution comes with price of solving ODE (tht in this prticulr cse one cn do) nd using obtined informtion to estimte the sign of the integrl. In generl, obtining some informtion bout solution of the nonliner second order boundry vlue problem is extremely difficult problem. Therefore we will use nice trick, clled Hrdy trick. The ide is to control nonlinerity through the Euler-Lgrnge equtions. Since we know tht u > we cn represent ny function φ C (, R) s product φ(x) = u(x)ψ(x) for function ψ C (, R). It s cler tht when ψ runs over the whole C (, R) φ lso runs over the whole C (, R) nd therefore we cn sfely tke this new representtion for φ. Now the second vrition becomes δ 2 I(u)(φ) = (uψ) 2 + (3u 2 1)(uψ) 2 = (uψ) 2 + (u 2 1)(uψ) 2 + 2u 4 ψ 2, An now we try to control the nonlinerity. We know tht u = u(u 2 1) nd therefore we single out this nonlinerity in the eqution nd substitute u insted. The second vrition becomes δ 2 I(u)(φ) = u ψ + ψ u 2 + u uψ 2 + 2u 4 ψ 2. After integrting by prts second term nd bsic lgebr we obtin δ 2 I(u)(φ) = = u 2 ψ 2 + ψ 2 u 2 + 2u uψ ψ u 2 ψ 2 2u uψ ψ + 2u 4 ψ 2 (31) ψ 2 u 2 + 2u 4 ψ 2 >. (32) We hve shown tht second vrition is strictly positive. Therefore we proved tht u is strictly loclly stble. Now we wnt to investigte locl minimlity. In generl, to investigte locl minimlity one hs to show uniform locl stbility in some spce. However in this prticulr cse we will be ble to prove evenglobl minimlity of u by direct clcultion. Globl minimlity Let s tke two energies I(v) nd I(u), where u is our criticl point nd v X is ny function. We wnt to show tht I(v) > I(u). We first show tht I(u + φ) I(u) > for ny φ C (, R). Direct computtion yeilds I(u + φ) I(u) = 1 2 ( φ 2 + (u 2 1)φ 2) + (u 2 φ 2 + uφ φ4 ). It is cler tht the first integrl is positive due to Hrdy trick (supply detils here) nd the seconf integrl is contins full squre. Therefore we obtin I(u + φ) > I(u) for ny nontrivil φ C (, R). Now we cn use the density rguments (supply detils here, the ide is tht C is dense in C 1, i.e. for ny function w in C 1 there exists sequence of functions {w n} in C such tht w n w C 1 ) to show tht I(u + φ) > I(u) for ny nontrivil φ C 1(, R). Since v u C1 (, R) we conclude tht u is globl minimizer of I. Let me stress gin: we could hve shown tht u is globl minimizer by using direct methods nd uniqueness of the criticl point. However, in rel life problems you rrely hve solvble Euler- Lgrnge equtions nd in most cses you don t know nything bout uniqueness of solution for 16

17 Euler-Lgrnge eqution. Therefore you usully show existence of minimizer by direct methods but then in order to investigte criticl points of the functionl I(u) you hve to study second vrition. Exmple 2. We define I(u) = 1 u 2 nd would like to solve the following problem inf I(u), u X { where X = u C 1(, 1), } 1 u2 = 1. It is cler tht we hve problem with constrint nd we wnt to reformulte it using the frmework of Lgrnge multipliers. We lwys strt with Euler-Lgrnge equtions nd we know tht in this cse we should find Euler-Lgrnge equtions for the following problem A simple clcultion yields inf (u,λ) C 1 (,1) R 1 ( 1 ) u 2 λ u 2 1. u + λu =, u() =, u(1) =, 1 u 2 = 1. Solving differentil eqution we obtin λ > (why this is the cse?) nd the generl form of the solution u(x) = A cos( λx) + B sin( λx) Plugging in the boundry dt we obtin u(x) = A sin( λx), λ = π 2 k 2, k N. In order to find constnt A we use constrint 1 u2 = 1 nd it gives us A = 2. Now we found ll criticl points u(x) = ± 2 sin(πkx), k N. We ctully wnt to minimize 1 u 2 nd therefore we cn plug in the criticl points in this expression to obtin 1 u 2 = π 2 k 2. It s cler tht this expression is minimizer when k = 1 nd therefore we found minimizers u(x) = ± 2 sin(πx). Try to use Euler-Lgrnge equtions to show directly tht Exmple 3. (Geodesics on surfce) λ = 1 u 2. 17

18 We define surfce S(x, y, z) = nd wnt to find geodesics on this surfce. By definition geodesic curve between the points A nd B is curve tht lies on the surfce nd hs the miniml length mong ll curves connecting points A nd B. Assume the curve in R 3 is given prmetriclly s x(t) = (x(t), y(t), z(t)) then the distnce between points A = x(t ) nd B = x(t 1 ) is given by L(x) = t1 t x (t) 2 + y (t) 2 + z (t) 2 dt. We lso know tht this curve hs to lie on the surfce S nd therefore S(x(t), y(t), z(t)) =. We obtin the following minimiztion problem inf L(x), x X where X = {x C 1 ([t, t 1 ]), x(t ) = A, x(t 1 ) = B, S(x(t)) = }. It is cler tht this is minimiztion problem with constrint nd therefore unless we cn express one of the vribles (x, y, z) in terms of the other two using S(x, y, z) = we hve to use method of Lgrnge multipliers. Although here the constrint is pointwise nd therefore you will hve Lgrnge multiplier s function of t, not s constnt when you hve integrl constrint (think bout it s stisfying S(x(t), y(t), z(t)) = t ech point t nd therefore requiring different Lgrnge multiplier t ech point t). Try to find Euler-Lgrnge equtions here nd solve this problem ssuming S is sphere. 4 Direct Methods Direct methods re used to prove existence of minimizers, costruct minimizer or minimizing sequence. The min ide is tht you cn costruct minimizing sequence by minimizing your functionl on the finite dimensionl subspce of your originl infinite dimensionl spce. However, this construction usully does not provide you with explicit minimizing sequence. The best you cn hope for is to show existence of the minimizing sequence nd its convergence to some function tht presumbly will be minimizer. 4.1 Ritz Method 4.2 Frmework on proving Existence There is generl frmework on proving existence of minimizers of the functionls defined on Bnch spces. Let me give you very specific exmple nd explin how this frmework works here. It is not difficult to generlize this exmple to your specific needs but in order to do this you need to know functionl nlysis nd mesure theory. Exmple. (Existence of minimizer) We consider the following functionl I(u) = ( 1 2 u ) 4 (u2 1) 2 dx. 18

19 We ssume u() = nd u(r) = 1 nd we wnt to show existence of minimizer in the spce C 1 ([, R]) stisfying the prescribed boundry conditions. In order to do this we hve to relx the spce in which we look for minimizer nd first show tht it exists in the Sobolev spce H 1 (, R). Define the following spce H 1 (, R) = { u(x) : u(x) is integrble nd You cn think of H 1 (, R) s completion of C (, R) in the H1 norm ( u H 1 = ( u 2 + u 2) } dx is bounded, u() = u(r) =. ( u 2 + u 2) ) 1/2 dx. In fct, this is equivlent definition of H 1. To ccomodte boundry conditions you tke ny nice function u (x) bounded in H 1 tht stisfies these conditions (in this prticulr cse we tke u (x) = x/r) nd represent your unknown u s u(x) = u (x) + v(x), where v H 1 (, R). Now your minimiztion problem is reduced to inf I(u + v). v H 1(,R) Note tht u is fixed nd your unknown function is v H 1. This is stndrd vritionl problem on Sobolev spces. The frmework for solving this problem is the following. Step 1. We first wnt to show tht I(u + v) C for ny v H 1. In this prticulr cse it is trivil, since I(u + v) for ll v H 1. This gurntees the existence of minimizing sequence v n H 1 such tht I(u + v n ) inf I(u + v). v H 1(,R) Step 2. Now we wnt to show tht this minimizing sequence is bounded in H 1 nd therefore wekly compct there. This is exctly why we needed to relx the spce, where you look for minimizer. since it s impossible to show compctness of minimizing sequence in C 1 unless you crry out very specific construction. Since {v n } is minimizing sequence then we know tht I(u + v n ) C, where C is constnt independent of n. It is strightforwrd to show tht v n 2 C, where C is constnt independent of n (we use C to denote generic constnt). Provide detils to show tht bove is true. Using Poincre s inequlity (prove it) v n 2 C v n 2 C 19

20 we obtin v n H 1 C nd this implies wek compctness of the sequence {v n }. Using Bnch- Aloglou theorem we deduce existence of subsequence {v nk } {v n } (we will not relbel it nd will still use v n ) such tht v n v wekly in H 1. Moreover, using compctness inclusion of H 1 (, R) Lp (, R) for ny p 1 we know tht v n v in L p (, R). Step 3. Now we wnt to show wek lowersemicontinuity of the functionl I, i.e. if v n v wekly in H 1 then lim inf I(u + v n ) I(u + v). It is strightforwrd to see tht this is true (use wek convergence in H 1 L p ). Therefore we obtin tht nd strong convergence in inf I(u) = lim inf I(u + v n ) I(u + v), v H 1 nd therefore v is minimizer of this problem tht implies tht u = u + v is minimizer of the originl problem. We hve shown tht minimizer exists in the spce H 1. We understnd tht if we compre the energy for this minimizer u with the energy of ny function φ from C 1, we will hve I(u) I(φ) but we don t know if u C 1. Step 4. In order to show tht this minimizer tht we found is ctully in C 1 we need to derive wek Euler-Lgrnge equtions tht will be true in H 1 nd then show tht ny solution of these equtions will be in C 1. This is usully done using regulrity theory of differentil equtions. In this prticulr cse we lredy know tht Euler-Lgrnge hs unique smooth solution nd therefore we re done. 5 Dulity Principle Let L(t, u, ξ) be convex function in ξ vrible nd I(u) = L(t, u(t), u (t)) dt. Euler-Lgrnge equtions for I(u) re given s We know tht L u = d L dt ξ. We don t worry bout boundry conditions here. We would like to obtin different formultion for minimiztion problem nd s consequence for Euler-Lgrnge equtions. For tht we define dul function of L s follows H(t, u, p) = sup{pξ L(t, u, ξ)}. ξ We wnt to prove the following result on equivlence of Euler-Lgrnge nd some Hmiltonin system. Theorem 5.1 Let u C 2 ([, b]) be solution of the Euler-Lgrnge eqution L u (t, u(t), u (t)) = d L dt ξ (t, u(t), u (t)), 2

21 where L(t, u, ξ) is Lgrngin strictly convex in ξ vrible. We define nd p(t) = L ξ (t, u(t), u (t)) H(t, u, p) = sup{pξ L(t, u, ξ)}. ξ Then (u(t), p(t)) stisfies the following Hmiltonin system Proof It is not difficult to find where ξ is found from eqution u (t) = H (t, u(t), p(t)) p p (t) = H (t, u(t), p(t)) u H(t, u, p) = pξ(t, u, p) L(t, u, ξ(t, u, p)), p = L (t, u, ξ). ξ It is cler tht if L(t, u, ξ) is strictly convex in ξ then this eqution hs unique solution ξ(t, u, p). And therefore if we tke p(t) defined in the conditions of the theorem then ξ = u (t) is unique solution of this eqution. Let s compute H p = ξ + p ξ p L ξ ξ p = ξ = u On the other hnd H u = p ξ u L u L ξ ξ u = L u Using the definition p(t) = L ξ (t, u(t), u (t)) nd tking derivtive we hve Theorem is proved. The opposite result holds, nmely p (t) = d L dt ξ = L u = H u. Theorem 5.2 Let u, p C 2 ([, b]) be solution of the following system u (t) = H (t, u(t), p(t)) p p (t) = H (t, u(t), p(t)), u 21

22 where H(t, u, p) is strictly convex in p vrible. We define L(t, u, ξ) = sup{pξ H(t, u, p)}. p Then u(t) stisfies the following Euler-Lgrnge equtions Proof Proof is left s n exercise. L u (t, u(t), u (t)) = d L dt ξ (t, u(t), u (t)). Remrk. Note tht if L(u, ξ) is independent of t vrible then Hmiltonin H(u(t), p(t)) = const long the trjectories of the Hmiltonin system nd therefore we obtin different formultion of the Euler-Lgrnge equtions u (t) L ξ (t, u(t), u (t)) L(u(t), u (t)) = const In the definition of Hmiltonin H(t, u, p) we used the following trnsformtion tht is clled Legendre- Fenchel trnsform. Definition 5.3 Let f(x) fe continuous function then f (s) = sup{sx f(x)} x is clled its dul function (or Legendre-Fenchel trnsform of f(x)). Lemm 5.1 Let f(x) be continuous function then f (s) is convex function. convex then f (x) = f(x). Moreover, if f(x) is Proof Proof is left s n exercise. 22