UNIVERSITA' DEGLI STUDI DI FERRARA

Transcription

1 UNIVERSITA' DEGLI STUDI DI FERRARA DIPARTIMENTO D'INGEGNERIA Corso di lure in Ingegneri Civile ed Ambientle Tesi di Lure "Some exmple of one-dimensionl vritionl problem" Lurendo: AMER SHAABAN Reltore: MICHELE MIRANDA Anno Accdemico

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3 Prefzione In quest tesi bbimo considerto problemi clssici di clcolo di vrizioni consistenti nell ricerc dei minimi funzionli del tipo: J(Y ) = F (x, Y (x), Y (x)) In tli tipi di problemi rientrno d esempio il problem dell geodetic, brchistocron, e problemi isoperimetrici. Per studire tli problemi bbimo richimto lcuni concetti di Anlisi I,II ed bbimo ccennto come tli concetti si estendono nel clcolo delle vrizioni. Prefce In this thesis we considered some clssicl problems of clculus of vritions consistent in reserch of the minimum of functionls of the type: J(Y ) = F (x, Y (x), Y (x)) In these types of problems we re going to see problem of geodesic, Brchistochrone, nd Isoperimetric problems. To study these problems we hve reviewed some concepts from Clculus I,II nd we mentioned how these concepts extend in the clculus of vritions.

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5 Tble of contents 1. Introduction 1.1 Exmple (Shortest Pth Problem) Exmple (Miniml Surfce Revolution) Exmple (Brchistochrone) Exmple (Isoperimetric Problems) Exmple (Chord nd Arc Problem) The Simplest / Fundmentl Problem Mxim nd Minim 3.1 The first neccessry conditions Clculus of Vritions Sttionry Condition Exmple (Shortest Pth Problem) Exmple (Minimum Surfce Problem) Specil Cses of the Euler-Lgrngr Eqution for the Simplest Problem 6. Chnge of vribles Severl Independent Functions of One Vrible Double Integrls Cnonicl Euler-Equtions (Euler-Hmilton) 9.1 The Hmiltonin The Euler-Hmilton (Cnonicl vrition principle) An Extension First Integrls of the Cnonicl Equtions...34

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7 Chpter 1 1. Introduction We re going to introduce some exmple t the beginning in order to present some ppliction of problems in Clculus of Vritions. Exmple 1.1 (Shortest Pth Problem). Let A nd B be two fixed points in spce. Then we wnt to find the shortest curve between these two points. We cn construct the problem digrmmticlly s below. A Y = Y (x) B ds dy b x Figure 1. A simple curve. From bsic geometry (i.e. Pythgors Theorem) we know tht (1.1) ds 2 = 2 + dy 2 = { 1 + (Y ) 2} 2. The second line of this is chieved by noting Y = dy. Now to find the pth between the points A nd B we integrte ds between A nd B, i.e. B ds. We however replce ds using A eqution (1.1) bove nd hence get the expression of the length of our curve J(Y ) = 1 + (Y ) 2. To find the shortest pth, i.e. to minimise J, we need to find the extreml function. Exmple 1.2 (Miniml Surfce of Revolution - Euler). This problem is very similr to the bove but insted of trying to find the shortest distnce, we re trying to find the smllest surfce to cover n re. We gin disply the problem digrmmticlly. 7

8 Y = Y (x) B A ds Y b x Figure 2. A volume of revolution, of the curve Y (x), round the line y = 0. To find the surfce of our shpe we need to integrte 2πY ds between the points A nd B, i.e. B 2πY ds. Substituting ds s bove with eqution (1.1) we obtin our expression of A the size of the surfce re J(Y ) = 2πY 1 + (Y ) 2. To find the miniml surfce we need to find the extreml function. Exmple 1.3 (Brchistochrone). This problem is derived from Physics. If I relese bed from O nd let it slip down frictionless curve to point B, ccelerted only by grvity, wht shpe of curve will llow the bed to complete the journey in the shortest possible time. We cn construct this digrmmticlly below. g O v ds b Y = Y (x) B x Figure 3. A bed slipping down frictionless curve from point O to B. In the bove problem, we wnt to minimise the vrible of time. So, we construct our integrl ccordingly nd consider the totl time tken T s function of the curve Y. 8

9 now using v = ds dt T(Y ) = nd rerrnging we chieve = dt ds v. Finlly using the formul v 2 = 2gY we obtin b 1 + (Y = ) 2. 2gY Thus to find the smllest possible time tken we need to find the extreml function. Exmple 1.4 (Isoperimetric problems). These re problems with constrints. A simple exmple of this is trying to find the shpe tht mximises the re enclosed by rectngle of fixed perimeter p. y A x Figure 4. A rectngle with sides of length x nd y. We cn see clerly tht the constrint equtions re (1.2) A = xy (1.3) p = 2x + 2y. By rerrnging eqution (1.3) in terms of y nd substituting into (1.2) we obtin tht we require tht da = 0 nd thus p A = x 2 p 2x da = p 2 2x 2 2x = 0 x = p 4 9

10 nd finlly substituting bck into eqution (1.3) gives us y = 1 (p 12 ) 2 p = p 4. Thus squre is the shpe tht mximises the re. Exmple 1.5 (Chord nd Arc Problem). Here we re seeking the curve of given length tht encloses the lrgest re bove the x-xis. So, we seek the curve Y = y (y is reserved for the solution nd Y is used for the generl cse). We describe this digrmmticlly below. Y = Y (x) AREA b x Figure 5. A curve Y (x) bove the x-xis. We hve the re of the curve J(Y ) to be J(Y ) = 0 Y where J(Y ) is mximised subject to the length of the curve Y 0 where c is given constnt. K(Y ) = (Y ) 2 = c, 10

11 To solve the descriped problems we first prove tht the problem is well posed; tht is the minimum exists nd then find the stisfied neccessry conditions tht llows us to chrcterize the solutions. Existence of minimum Suppose tht the lgrngin F (x, Y, p) stisfies the following conditions: i) F (x, Y, p) nd F p (x, Y, p) re continuous in (x, Y, p); ii) F (x, Y, p) is convex in p; iii) F (x, Y, p) hs superliner growth. then there exists minimizer of J(Y ) := F (x, Y, Y ) I in the clss C(α, β) := {Y H 1,1 ((, b), R) : Y () = α, Y (b) = β} Where α, β re fixed rel numbers. Here H 1,1 {(, b), R} is Sobolev spce of functions Y = (, b) R such tht Y (x) 2, Y '(x) 2 re integrble in (, b). i.e Y (x) 2 < + Y (x) 2 < +. It is possible to prove tht if Y (x, Y, p) end p F p (x, Y, p) re C 1 functions nd then the minim of J re C 2 functions nd then Euler-Lgrnge equtions tht we re going to write re well defined. 11

12 2. The Simplest / Fundmentl Problem Exmples 1.1, 1.2 nd 1.3 re ll specil cses of the simplest/fundmentl problem. y b B y A Y b x Figure 6. The Simplest/Fundmentl Problem. Suppose A(, y ) nd B(b, y b ) re two fixed points nd consider set of curves (2.1) Y = Y (x) joining A nd B. Then we seek member Y = y(x) of this set which minimises the integrl (2.2) J(Y ) = F(x, Y, Y ) where Y () = y, Y (b) = y b. We note tht exmples 1.1 to 1.3 correspond to specifiction of the bove generl cse with the integrnd An extr exmple is J(Y ) = F = 1 + (Y ) 2, F = 2πY 1 + (Y ) 2, 1 + (Y F = ) 2. 2gY { 1 2 p(x)(y ) } 2 q(x)y 2 + f(x)y Now, the curves Y in (2.2) my be continuous, differentible or neither nd this ffects the problem for J(Y ). We shll suppose tht the functions Y = Y (x) re continuous nd hve continuous derivtives suitble number of times. Thus the functions (2.1) belong to set Ω of dmissible functions. We define Ω precisely s. 12

13 { (2.3) Ω = Y } Y continuous nd dy. continuous, k = 1 So the problem is to minimise J(Y ) in (2.2) over the functions Y in Ω where Y () = y nd Y (b) = y b. This bsic problem cn be extended to much more complicted problems. Exmple 2.1. We cn extend the problem by considering more derivtives of Y. So the integrnd becomes F = F(x, Y, Y, Y ), i.e. F depends on Y s well s x, Y, Y nd here Ω = C 2 ([, b]) Exmple 2.2. We cn consider more thn one function of x. Then the integrnd becomes F = F(x, Y 1, Y 2, Y 1, Y 2 ), so F depends on two (or more) functions Y k of x. Exmple 2.3. Finlly we cn consider functions of more thn one independent vrible. So, the integrnd would become F = F(x, y, Φ, Φ x, Φ y ), where subscripts denote prtil derivtives. So, F depends on functions Φ(x, y) of two independent vribles x, y. This would men tht to clculte J(Y ) we would hve to integrte more thn once, for exmple J(Y ) = F dy. Note. The integrl J(Y ) is numericl-vlued function of Y, which is n exmple of functionl. Definition: Let R be the rel numbers nd Ω set of functions. Then the function J : Ω R is clled functionl. Then we cn sy tht the clculus of vritions is concerned with mxim nd minim (extremum) of functionls. 13

14 Chpter 2 3. Mxim nd Minim 3.1. The First Necessry Condition (i) We use ides from elementry clculus of functions f(u). f(u) u Figure 7. Plot of function f(u) with minimum t u =. If f(u) f() for ll u ner on both sides of u = this mens tht there is minimum t u =. The consequences of this re often seen in n expnsion. Let us ssume tht there is minimum t f() nd Tylor expnsion exists bout u = such tht (3.1) f( + h) = f() + hf () + h2 2 f () + O 3 (h 0). Note tht we define df(, h) := hf () to be the first differentil. As there exists minimum t u = we must hve (3.2) f( + h) f() for h ( δ, δ) by the bove comment. Now, if f () 0, sy it is positive nd h is sufficiently smll, then (3.3) sign{ f = f( + h) f()} = sign{df = hf ()} ( 0). In eqution (3.3) the L.H.S. 0 becuse f hs minimum t nd hence eqution (3.2) holds nd lso the R.H.S. > 0 if h > 0. However this is contrdiction, hence df = 0 which f () = 0. (ii) For functions f(u, v) of two vribles; similr ides hold. Thus if (, b) is minimum then f(u, v) f(, b) for ll u ner nd v ner b. Then for some intervls ( δ 1, δ 1 ) nd ( δ 2, δ 2 ) we hve tht (3.4) { δ1 u + δ 1 b δ 2 v b + δ 2 14

15 gives minimum / mximum t (, b) f(, b). The corresponding Tylor expnsion is (3.5) f( + h, b + k) = f(, b) + hf u (, b) + kf v (, b) + O 2. We note tht in this cse the first derivtive is df(, b, h, k) := hf u (, b)+kf v (, b). For minimum (or mximum) t (, b) it follows, s in the previous cse tht necessry condition is (3.6) df = 0 f u = f v = 0 t (, b). (iii) Now considering functions of multiple (sy n) vribles, i.e. f = f(u 1, u 2,..., u n ) = f(u) we hve the Tylor expnsion to be (3.7) f( + h) = f() + h f() + O 2. Thus the sttement from the previous cse (3.6) becomes (3.8) df = 0 f() = Clculus of Vritions Now we consider the integrl (3.9) J(Y ) = F(x, Y, Y ). Suppose J(Y ) hs minimum for the curve Y = y. Then (3.10) J(Y ) J(y) for Y Ω = {Y Y C 2, Y () = y, Y (b) = y b }. To obtin informtion from (3.10) we expnd J(Y ) bout the curve Y = y by tking the so-clled vried curves (3.11) Y = y + εξ like u = + h in (3.2). We cn represent the consequences of this expnsion digrmmticlly. 15

16 y b y Y = y(x) + εξ(x) A εξ B Y = y(x) x b Figure 8. Plot of y(x) nd the expnsion y(x) + εξ(x). Since ll curves Y, including y, go through A nd B, it follows tht (3.12) ξ() = 0 nd ξ(b) = 0. Now substituting eqution (3.11) into (3.10) gives us (3.13) J(Y ) = J(y + εξ) J(y) for ll y + εξ Ω nd substituting (3.11) into (3.9) gives us (3.14) J(y + εξ) = F(x, y + εξ, y + εξ ). Now to del with this expnsion we tke fixed x in (, b) nd tret y nd y s independent vribles. Recll Y nd Y re independent nd the Tylor expnsion of two vribles (eqution (3.5)) from bove. Then we hve (3.15) f(u + h, v + k) = f(u, v) + h f u + k f v + O 2. Now we tke u = y, h = εξ, v = y, k = εξ, f = F. Then (3.14) implies O 2 = O ( (h, k) ) J(Y ) = J(y + εξ) = { F(x, y, y ) + εξ } y + εξ y + O(ε2 ) = J (y) + ε δj + O 2. We note tht δj is clculus of vritions nottion for dj where we hve 16

17 (3.16) nd we lso hve tht (3.17) { ξ } δj = y + ξ y = liner terms in ε = first vrition of J y = { } (x, Y, Y ) Y Y =y,y =y. Now δj is nlogous to the liner terms in (3.1), (3.5), (3.7). Then if J(Y )hs minimum t Y = y, then (3.18) δj = 0 the first necessry condition for minimum. Proof. Suppose δj 0 then J(Y ) J(y) = δj + O 2. For smll enough εξ then sign{j(y ) J(y)} = sign{δj}. We hve tht δj > 0 or δj < 0 for some vried curves corresponding to εξ. However there is minimum of J t Y = y L.H.S. 0. This is contrdiction. For our J, we hve by (3.17) (3.19) δj = If we integrte 2nd term by prts we obtin ( ξ ) y + ξ y. (3.20) δj = nd so we hve ξ { y d } y + ] b [ ξ y }{{} =0 (3.21) δj = Note. For nottionl purposes we write ξ { y d } y. (3.22) f, g = f(x)g(x) which is n inner (or sclr) product. Also, for our J, write 17

18 (3.23) J (y) = y d s the derivtive of J. Then we cn express δj s (3.24) δj = ξ, J (y). y This gives us the Tylor expnsion (3.25) J(y + εξ) = J(y) +ε ξ, J (y) +O 2. }{{} δj We compre this with the previous cses of Tylor expnsion (3.1), (3.5) nd (3.7). Now collecting our results together we get Theorem 3.1. A necessry condition for J(Y ) to hve n extremum (mximum or minimum) t Y = y is (3.26) δj = ξ, J (y) = 0 for ll dmissible ξ. i.e. To estblish n extremum, we need to exmine the sign of 4. Sttionry Condition (δj = 0) Our next step is to see wht cn be deduced from the condition (4.1) δj = 0. In detil this is (4.2) J(Y ) hs n extremum t Y = y δj(y, ξ) = 0. Definition: y is criticl curve (or n extreml), i.e. y is solution of δj = 0. J(y) is sttionry vlue of J nd (3.26) is sttionry condition. For our cse J (Y ) = J = J(y + ξ) J(y) = totl vrition of J. We consider this lter in prt two of this thesis. ξ, J (y) = ξj (y) = 0. F(x, Y, Y ), we hve 18

19 (4.3) J (y) = y d y. To del with equtions (4.2) nd (4.3) we use the Euler-Legrnge Lemm. Using this Lemm we cn stte Theorem 4.1. A necessry condition for (4.4) J(Y ) = F(x, Y, Y ), with Y () = y nd Y (b) = y b, to hve n extremum t Y = y is tht y is solution of (4.5) J (y) = y d = 0 y with < x < b nd F = F(x, y, y ). This is known s the Euler-Lgrnge eqution. The bove theorem is the Euler-Lgrnge vritionl principle nd it is sttionry principle. Exmple 4.1 (Shortest Pth Problem). We now revisit Exmple 1.1 with the mechnisms tht we hve just set up. So we hd tht F(x, Y, Y ) = 1 + (Y ) 2 previously but insted we write F(x, y, y ) = 1 + (y ) 2. Now y = 0, y = The Euler-Lgrnge eqution for this exmple is 2y (y ) 2 = y 1 + (y ) 2. ( ) 0 d y = 0 < x < b 1 + (y ) 2 y 1 + (y ) 2 = const. y = α y = αx + β where α, β re rbitrry constnts. So, y = αx + β defines the criticl curves. We require more informtion to estblish minimum. Exmple 4.2 (Minimum Surfce Problem). Now revisiting Exmple 1.2, we hd F(x, Y, Y ) = 2πY 1 + (Y ) 2 but gin we write F(x, y, y ) = 2πy 1 + (y ) 2. We drop the 2π to give us 19

20 y = 1 + (y ) 2, So, we re left with the Euler-Lgrnge eqution y = yy 1 + (y ) 2. ( ) 1 + (y ) 2 d yy = (y ) 2 Now solving the differentil eqution leves us with for finite y in (, b). We therefore hve ( 1 + (y ) 2) 3 2 { 1 + (y ) 2 yy } = 0, (4.6) 1 + (y ) 2 = yy. Strt by rewriting y in terms of y nd y. Then substituting gives us (4.7) y = dy = dy dy dy = y dy dy = 1 2 d(y ) 2 dy. So, substituting (4.7) into (4.6), the Euler-Lgrnge eqution implies tht 1 + (y ) 2 = 1 ) 2 2 yd(y dy dy y = 1 d(y ) (y ) = 1 dz z ln y = 1 2 ln ( 1 + (y ) 2) + C which is first integrl. y = C 1 + (y ) 2, 20

21 Now integrting gin gives us y = 1+(y 2 ) y the utonomous differentil eqution suppose tht y = v(y) then we hve y = v (y).y = v (y)v(y) so we get vv = 1+v2 y = v v 1+v 2 = 1 y Integrte both sides we get 1 2 log(1+v2 ) = log y +c 1 log(1+v 2 ) = logy 2 +c 1 nd 1 + v 2 = c 1 y 2, v= c 1 y 2 1, so y = c 1 y 2 1 divide by c 1 y 2 1 we get y c1 y 2 1 = 1 thus c 1 y 2 = cosh 2 α, y = 1 c1 coshα, dy = 1 c1 senhα integrting 1 senhα c1 senhα dα = x + c 2 then 1 c1 α = x + c 2 we hve α = c 1 x + c 2, α = rccosh( c 1 y) = y=cosh( c 1 x + c 2 ) where C 1 nd C re rbitrry constnts. We note tht this is ctenry curve. 21

22 Chpter 3 5. Specil Cses of the Euler-Lgrnge Eqution for the Simplest Problem For F(x, y, y ) the Euler-Lgrnge eqution is 2nd order differentil eqution in generl. There re specil cses worth noting. (1) y = 0 F = F(x, y ) nd hence y is missing. The Euler-Lgrnge eqution is y =0 d = 0 y d = 0 y = c y on orbits (extremls or criticl curves), first integrl. If this cn be for y thus then y(x) = f(x, c) + c 1. y = f(x, c) Exmple 5.1. F = x 2 + x 2 (y ) 2. Hence y first integrl y solution. = 0. So Euler-Lgrnge eqution hs = c 2x 2 y = c y = c 2x 2 y = c 2x + c 1, which is the (2) x = 0 nd so F = F(y, y ) nd hence x is missing. For ny differentible F(x, y, y ) we hve by the chin rule on orbits. Now, we hve on orbits. When x df = x + y = ( d x + = x + d y + y ) y y + y ( y ) y d ( F y ) = y x = 0, this gives y y 22

23 on orbits. First integrl. ( d F y ) = 0 on orbits y F y F y = c Note. G := F y F y = G(x, y, y ) is the Jcobi function. Exmple 5.2. F = 2πy 1 + (y ) 2. An exmple of cse 2. So we hve then the first integrl is F y = 2πyy 1 + (y ) 2. F = y F y { } (y ) 2 = 2πy 1 + (y ) (y )2 2πy = 1 + (y ) 2 = const. = 2πc y = c 1 + (y ) 2 - First Integrl. This rose in Exmple 4.2 s result of integrting the Euler Lgrnge eqution once. (3) = 0 nd hence F = F(x, y), i.e. y is missing. The Euler-Lgrnge eqution is y 0 = y d = y y = 0 but this isn t differentil eqution in y. Exmple 5.3. F(x, y, y ) = y lny + xy nd then the Euler-Lgrnge eqution gives y d = 0 ln y 1 + x = 0 y ln y = 1 + x y = e 1+x 23

24 6. Chnge of Vribles In the shortest pth problem bove, we note tht J(Y ) does not depend on the coordinte system chosen. Suppose we require the extreml for (6.1) J(r) = θ1 θ 0 r2 + (r ) 2 dθ where r = r(θ) nd r = dr. The Euler-Lgrnge eqution is dθ (6.2) To simplify this we cn r d = 0 dθ r () chnge vribles in (6.2), or (b) chnge vribles in (6.1). For exmple, in (6.1) tke ( r r2 + (r ) d 2 dθ r r2 + (r ) 2 ) = 0. So, we obtin x = r cosθ y = r sin θ = dr cosθ r sin θ dθ dy = dr sin θ + r cosθ dθ. 2 + dy 2 = dr 2 + r 2 dθ 2 { ( 1 + (y ) 2) ( ) } 2 dr 2 = + r 2 dθ dθ (y ) 2 = r 2 + (r ) 2 dθ. This is just the smllest pth problem, hidden in polr coordintes. We know the solution to this is J(r) J(y) = 1 + (y ) 2. Now the Euler-Lgrnge eqution is y = 0 y = αx + β. Now, r sin θ = αr cosθ + β r = β sin θ α cosθ 24

25 7. Severl Independent functions of One Vrible Consider generl solution of the simplest problem. (7.1) J(Y 1, Y 2,...,Y n ) = Here ech curve Y k goes through given end points F(x, Y 1,...,Y n, Y 1,...,Y n ). (7.2) Y k () = y k, Y k (b) = y kb for k = 1,..., n. Find the criticl curves y k, k = 1,..., n. Tke (7.3) Y k = y k (x) + εξ k (x) where ξ k () = 0 = ξ k (b). By tking Tylor expnsion round the point (x, y 1,...,y n, y 1,...,y n) we find tht the first vrition of J is (7.4) δj = with n ξ k, J (y 1,...,y n ). k=1 (7.5) J k (y 1,...,y n ) = d y k for k = 1,...,n. The sttionry condition δj = 0 (7.6) n ξ k, J k = 0 k=1 (7.7) ξ k, J k = 0 for ll k = 1,..., n. Thus, by the Euler-Lgrnge Lemm, this implies y k (7.8) J k = 0 for k = 1,...,n. i.e. (7.9) d y k for k = 1,...,n. (7.9) is system of n Euler-Lgrnge equtions. These re solved subject to (7.3). y k = 0 25

26 Exmple 7.1. F = y 1 y y 2 1y 2 + y 1y 2 nd so we obtin equtions Consider { J 1 = 0 y y 1y 2 d (y 2 ) = 0 J 2 = 0 2y 1 y 2 + y1 2 d (y 1) = 0 df = n ( x + y k y + ) y k y k k k=1 the generl chin rule. = n ( x + k=1 y k d y k + ) y y k k on orbits. = n ( x + d k=1 nd this is gin on orbits. Now we obtin on orbits. { d F n ) y k y k y k y k=1 k } = x { F } y k We define G = y. Then If x dg = x. = 0, i.e. F does not depend implicitly on x, we hve { dg = d F } y k = 0 y 26

27 on orbits. i.e. F n k=1 y k = c y k on orbits, where c is constnt. This is first integrl. i.e. G is constnt on orbits. This is the Jcobi integrl. 27

28 8. Double Integrls Here we look t functions of 2 vribles, i.e. surfces, (8.1) Φ = Φ(x, y). We then tke the integrl (8.2) J(Φ) = R F(x, y, Φ, Φ x, Φ y ) dy with Φ ϕ B = given on the boundry R of R. Suppose J(Φ) hs minimum for Φ = ϕ. Hence R is some closed regioon in the xy-plne nd Φ x = Φ, Φ x y = Φ. Assume F hs y continuous first nd second derivtives with respect to x, y, Φ, Φ x, Φ y. Consider J(ϕ + εξ) = R F(x, y, ϕ + εξ, ϕ x + εξ x, ϕ y + εξ y ) dy nd expnd in Tylor series { = F(x, y, ϕ, ϕ x, ϕ y ) + εξ } R ϕ + εξ x + εξ y + O 2 dy ϕ x ϕ y (8.3) = J(ϕ) + δj + O 2 where we hve { (8.4) δj = ε ξ } R ϕ + ξ x + ξ y dy ϕ x ϕ y is the first vrition. For n extremum t ϕ, it is necessry tht (8.5) δj = 0. To use this we rewrite (8.4) (8.6) δj = ε R { ξ ϕ + ( ξ ) + ( ξ ) ξ ( ) ξ ( )} x ϕ x y ϕ y x ϕ x y ϕ y Now by Green s Theorem we hve (8.7) So, R R P x dy = P dy R Q y dy = Q R P = ξ ϕ x Q = ξ ϕ y dy. 28

29 { (8.8) δj = ε ξ R ϕ x ϕ x y } ϕ y If we choose ll functions Φ including ϕ to sttisfy ( dy + ε ξ dy ξ ). R ϕ x ϕ y ϕ = ϕ B then ϕ + εξ = ϕ B ξ = 0 on R on R on R. Hence (8.8) simplifies to δj = R ξ, J (ϕ). { ξ(x, y) ϕ x ϕ x y } ϕ y dy By simple extension of the Euler-Lgrnge lemm, since ξ is rbitrry in R, we hve δj = 0 ϕ(x, y) is solution of (8.9) J (ϕ) = ϕ x ϕ x y ϕ y = 0. This is the Euler-Lgrnge Eqution - prtil differentil eqution. We seek the solution ϕ which tkes the given vlues ϕ B on R. Exmple 8.1. F = F(x, y, ϕ, ϕ x, ϕ y ) = 1 2 ϕ2 x ϕ2 y + fϕ, where f = f(x, y) is given. The Euler-Lgrnge eqution is So, ϕ = f ϕ x = ϕ x ϕ y = ϕ y. ϕ x So, i.e. we re left with ϕ x y ϕ y = 0 f x ϕ x y ϕ y = 0. which is Poisson s Eqution. 2 ϕ x ϕ y 2 = f, 29

30 Exmple 8.2. F = F(x, t, ϕ, ϕ x, ϕ t ) = 1 2 ϕ2 x 1 2c 2 ϕ 2 t. The Euler-Lgrnge eqution is This is the clssicl wve eqution. x x 0 x ϕ x t = 0 ϕ x t ϕ ( t ) 1 c 2ϕ t = 0 2 ϕ x 2 = 1 c 2 2 ϕ t 2 30

31 Chpter 4 9. Cnonicl Euler-Equtions (Euler-Hmilton) In this chpter we sketch different pproch to some problems in clculus of vritions, the Hmiltonin pproch The Hmiltonin We hve the Euler-Lgrnge equtions (9.1) d y k which give the criticl curves y 1,...,y n of y k = 0 (9.2) J(Y 1,...,Y n ) = F(x, Y 1,..., Y n, Y 1,...,Y n) Equtions (9.1) form system of n 2nd order differentil equtions. We shll now rewrite this s system of 2n first order differentil equtions. First we introduce new vrible (9.3) p i = y i i = 1,...,n. p i is sid to be the vrible conjugte to y i. We suppose tht equtions (9.3) cn be solved to give y s function ψ i of x, y j, p j (j = 1,..., n). Then it is possible to define new function H by the eqution (9.4) H(x, y 1,...,y n, p 1,...,p n ) = n p i y i F(x, y 1,...,y n, y 1,..., y n ) i=1 where y i = ψ i(x, y j, p j ). The function H is clled the Hmiltonin corresponding to (9.2). Now look t the differentil of H which by (9.4) is 31

32 (9.5) dh = n ( p i dy i + y i dp i) n x i=1 = x + n i=1 ( y idp i y i using (9.3). For H = (x, y 1,...,y n, p 1,...,p n ) then we hve Comprison with (9.5) gives (9.6) i=1 ) i=1 ( y i dy i + dh = H n ( H dy i + H ) + dp i. x y i p i y i = H p i dy i = H p i = H y i y i dp i = H y i ) y dy i i for i = 1,..., n. Equtions (9.6) re the cnonicl Euler-Lgrnge equtions ssocited with the integrl (9.2). Exmple 9.1. Tke (9.7) J(Y ) = ( α(y ) 2 + βy 2) where α, β re given functions of x. For this F(x, y, y ) = α(y ) 2 + βy 2 if so φ = = 2αy y = 1 y 2α φ. The Hmiltonin H is, by (9.4), H = py F = py α(y ) 2 βy 2 with y = 1 2α φ = p 1 2α p α 1 βy 2 4α 2p2 = 1 4α p2 βy 2 in correct vribles x, y, p. Cnonicl equtions re then (9.8) dy = H p = 1 2α p dp = H y = 2βy. 32

33 The ordinry Euler-Lgrnge eqution for J(Y ) is y d = 0 2βy d y (2αy ) = 0 which is equivlent to (9.8) 9.2. The Euler-Hmilton (Cnonicl) vritionl principle. Let I(P, Y ) = { P dy } H(x, Y, P) be defined for ny dmissible independent functions P nd Y with Y () = y, Y (b) = y b, i.e. { y t x = Y = y B = y b t x = b. Suppose I(P, Y ) is sttionry t Y = y, P = p. Tke vried curves Then we obtin I(p + εη, y + εξ) = where we hve Y = y + εξ P = p + εη. {(p + εη) d (y + εξ) H(x, y + εξ, p + εη) } { p dy = + εηdy + εpdξ + ε2 η dξ = I(p, y) + δi + O 2. δi = ε = ε { η dy { η } + pdξ ξ H y η H p ( dy H ) ( dp ξ p + H )} y } H(x, y, p) εξ H y εη H p O 2 + [ εpξ ] b }{{} =0 If ll curves Y, including y, go through (, y ) nd (b, y b ) then ξ = 0 t x = nd x = b. Then δi = 0 (p, y) re solutions of with dy = H p dp = H y ( < x < b) 33

34 y = y B = { y y b t x = t x = b Note. If Y = y B no boundry conditions re required on p An extension. Modify I(P, Y ) to be I mod (P, Y ) = { P dy } H(x, P, Y ) [ ] b P(Y y B ) Here P nd Y re ny dmissible functions. I mod is sttionry t P = p, Y = y. Y = y+εξ, P = p + εy nd then mke δi mod = 0. You should find tht (y, p) solve with y = y B on [, b]. dy = H p dp = H y 10. First Integrls of the Cnonicl Equtions A first integrl of system of differentil equtions is function, which hs constnt vlue long ech solution of the differentil equtions. We now look for first integrls of the cnonicl system (10.1) dy i = H p i nd hence of the system (10.2) dy i = y i d y i dp i = H y i y i which is equivlent to (10.1). Tke the cse where (i = 1,...,n) = 0 (i = 1,...,n) i.e. F = F(y 1,...,y n, y 1,...,y n ). Then x = 0. is such tht H x = 0 nd hence H = n p i y i F i=1 34

35 On criticl curves (orbits) this gives 0 dh = H n ( H x + dy i y i + H ) dp i. p i = = 0 n i=1 i=1 ( H H + H ( 1) H ) y i p i p i y i which implies H is constnt on orbits. Consider now n rbitrry differentible function W = W()x, y 1,..., y n, p 1,..., p n ). Then on orbits. dw = W x + n i=1 ( W dy i y i + W ) dp i = W n ( W p i x + H W ) H y i=1 i p i p i y i Definition: We define the Poisson brcket of X nd Y to be [X, Y ] = with X = X(y i, p i ) nd Y = Y (y i, p i ). n i=1 ( X Y X ) Y y i p i p i y i Then we hve on orbits. In the cse when W x dw = W x = 0, we hve + [W, H] So, dw = [W, H]. dw = 0 [W, H] = 0. 35