Part III. Calculus of variations

Transcription

1 Prt III. Clculus of vritions Lecture notes for MA342H P. Krgeorgis 1/43

2 Introduction There re severl pplictions tht involve expressions of the form Jy) = Lx,yx),y x))dx. For instnce, Jy) could represent re, length, energy nd so on. We re interested in finding the minimum/mximum vlues of Jy) over ll functions yx) tht stisfy some given constrints. A function yx) for which the minimum/mximum vlue is ttined is clled n extreml of Jy). We shll ssume tht both yx) nd the Lgrngin function L re twice continuously differentible. One my impose vrious constrints on yx), but the most common one mounts to specifying its vlues t the endpoints x =,b. This requires the grph of yx) to pss through two given points. 2/43

3 Directionl derivtive Consider functionl of the form Jy) = Lx,yx),y x))dx. Its directionl derivtive in the direction of function ϕ is defined s J Jy +εϕ) Jy) y)ϕ = lim. ε 0 ε This is lso known s the first vrition of Jy). Explicitly, one hs J y)ϕ = Ly ϕ+l y ϕ ) dx. The function ϕ is clled test function, if it is twice continuously differentible nd lso vnishing t the endpoints x =,b. 3/43

4 Directionl derivtive: Proof The directionl derivtive J y)ϕ is defined in terms of Jy +εϕ) Jy) = ) Lx,y +εϕ,y +εϕ ) Lx,y,y ) dx. Denote the integrnd by Fε). Then Tylor series expnsion gives Fε) = L y x,y,y ) εϕ+l y x,y,y ) εϕ +..., where the dots indicte terms which contin higher powers of ε. Once we now combine the lst two equtions, we find tht J Jy +εϕ) Jy) y)ϕ = lim ε 0 ε Fε) b = lim dx = Ly ϕ+l y ϕ ) dx. ε 0 ε 4/43

5 Euler-Lgrnge eqution Theorem 1. Euler-Lgrnge eqution Suppose tht yx) is n extreml of Jy) = Lx,yx),y x))dx. Then yx) must stisfy the Euler-Lgrnge eqution d dx L y = L y. A solution of the Euler-Lgrnge eqution is lso known s criticl point of Jy). Criticl points of Jy) re not necessrily extremls. When the Lgrngin L depends on higher-order derivtives of y, the Euler-Lgrnge eqution involves higher-order derivtives of L. 5/43

6 Euler-Lgrnge eqution: Proof Since yx) is n extreml of Jy), it must stisfy J y)ϕ = 0 = Ly ϕ+l y ϕ ) dx = 0 for every test function ϕ. Integrting by prts, we now get L y d ) [ ] b dx L y ϕdx+ L y ϕ = 0. Since the test function ϕ vnishes t the endpoints, this gives L y d ) dx L y ϕdx = 0 for every test function ϕ. According to the fundmentl lemm of vritionl clculus, we must thus hve L y = d dx L y, s needed. 6/43

7 Fundmentl lemm of vritionl clculus Suppose tht Hx) is continuously differentible with Hx)ϕx)dx = 0 for every test function ϕ. Then Hx) must be identiclly zero. To prove this, consider n rbitrry subintervl [x 1,x 2 ] nd let { x x1 ) ϕx) = 3 x 2 x) 3 } if x 1 x x 2. 0 otherwise Then ϕ is test function which is positive on x 1,x 2 ) nd we hve 0 = Hx)ϕx) dx = x2 x 1 Hx)ϕx)dx. If Hx) is positive/negtive t point, it must hve the sme sign on some intervl [x 1,x 2 ] nd this is not the cse by the lst eqution. 7/43

8 Beltrmi identity Suppose tht yx) is criticl point of the functionl Jy) = Lyx),y x))dx whose Lgrngin function L does not depend on x directly. Then y L y L = constnt. To prove this, one uses the Euler-Lgrnge eqution to find tht d y L y ) = y L y +y d dx dx L y = y L y +y L y. Since Ly,y ) does not depend on x directly, the chin rule gives d dx y L y ) = y L y +y L y = d dx L. 8/43

9 Exmple 1. Shortest pth Consider function yx) whose grph psses through two given points. We wish to minimise the length of its grph Jy) = 1+y x) 2 dx. Since L = 1+y x) 2, the Euler-Lgrnge eqution gives d dx L y = L y = 0 = L y = y x) 1+y x) 2 = c 1. We squre both sides nd then simplify to conclude tht y x) 2 = c 2 1 +c 2 1y x) 2 = y x) = c 2. This shows tht yx) must be liner function. In other words, the shortest pth between ny two points is given by line. 9/43

10 Exmple 2. Miniml surfce, pge 1 Consider function yx) whose grph psses through two given points. We wish to minimise the surfce re of the solid which is obtined by rotting the grph of yx) round the x-xis, nmely Jy) = 2πyx) 1+y x) 2 dx. Since the Lgrngin does not depend on x directly, one hs c = y L y L = 2πyx)y x) 2 1+y x) 2 2πyx) 1+y x) 2 by the Beltrmi identity. It now esily follows tht c 2π = yx) 1+y x) 2 = 1+y x) 2 = yx) /43

11 Exmple 2. Miniml surfce, pge 2 The lst eqution is ctully seprble eqution which gives dy y dx = 2 dy dx 1 = 2 y 2 = 2. Letting y = cosht, we now get y 2 2 = 2 sinh 2 t, hence lso dx sinhtdt = = dt. sinht Since y = cosht by bove, we my finlly conclude tht ) x x 0 x x0 = t = y = cosht = cosh. This is n eqution tht describes the shpe of hnging chin. 11/43

12 Exmple 3. Brchistochrone, pge 1 Consider prticle tht slides long the grph of function yx) from one point to nother under the influence of grvity. We ssume tht there is no friction nd wish to minimise the overll trvel time 1+y Jy) = x) 2 dx. vyx)) To find the speed v, we note tht conservtion of energy gives 1 2 mv2 +mgy = 1 2 mv2 0 +mgy 0. Assuming tht v 0 = 0 for simplicity, the overll trvel time is then 1+y Jy) = x) 2 dx 2gy0 y). 12/43

13 Exmple 3. Brchistochrone, pge 2 Since the Lgrngin does not depend on x directly, one hs c = y L y L = y x) 2 2gy0 y) 1+y x) 2 1+y x) 2 2gy0 y) by the Beltrmi identity. We cler denomintors to get c 2gy 0 y) 1+y x) 2 = 1 nd then we squre both sides to find tht 1+y x) 2 = 2 y 0 yx) = y x) 2 = 2 y 0 +yx). y 0 yx) This is seprble eqution tht cn lso be written s y0 ydy 2 y 0 +y = dx. 13/43

14 Exmple 3. Brchistochrone, pge 3 Letting y 0 y = 2 sin 2 θ gives 2 y 0 +y = 2 cos 2 θ, hence y0 ydy dx = 2 y 0 +y sinθ 2 2 sinθcosθdθ = = 2 1 cos2θ))dθ. cosθ Once we now simplify the integrls, we my finlly conclude tht x 0 x = 2 θ 1 ) 2 sin2θ) = 2 2 ϕ sinϕ), y 0 y = 2 sin 2 θ = cos2θ)) = cosϕ), where ϕ = 2θ. These re the prmetric equtions of cycloid. 14/43

15 Cse 1. Fixed boundry conditions Suppose tht we wish to find the extremls of Jy) = Lx,yx),y x))dx subject to given boundry conditions y) = y 0 nd yb) = y 1. This is the stndrd vritionl problem which implies tht L y d ) [ ] b dx L y ϕdx+ L y ϕ = 0 for every test function ϕ. Since the boundry terms vnish, one cn find the possible extremls by solving the Euler-Lgrnge eqution d dx L y = L y. Thus, we get second-order ODE subject to two boundry conditions. 15/43

16 Cse 1. Exmple As simple exmple, consider the functionl Jy) = 1 0 y x) 2 dx subject to the boundry conditions y0) = 1 nd y1) = 3. In this cse, the Euler-Lgrnge eqution gives d dx L y = L y = 2y x) = 0 = yx) = x+b. To ensure tht the boundry conditions hold, we must hve 1 = y0) = b, 3 = y1) = +b. Thus, the only possible extreml is given by yx) = 2x+1. 16/43

17 Cse 2. Vrible boundry conditions Suppose tht we wish to find the extremls of Jy) = Lx,yx),y x))dx when no boundry conditions re specified for y) nd yb). Using the sme pproch s before, one finds tht L y d ) [ ] b dx L y ϕdx+ L y ϕ = 0 for every function ϕ. This obviously includes ll test functions, so the Euler-Lgrnge eqution remins vlid, but we must lso hve L y = 0 when x =,b. These conditions re known s the nturl boundry conditions. 17/43

18 Cse 2. Exmple As typicl exmple, consider the functionl Jy) = 1 0 ) y x) 2 +yx)y x) 4yx) dx. Using the Euler-Lgrnge eqution, one finds tht d dx L y = L y = 2y +y = y 4 = yx) = x 2 +x+b. In view of the nturl boundry conditions, we must lso hve 0 = L y = 2y +y = 2 2x) x 2 +x+b when x = 0,1. This gives 2+b = 0 = 3+b 5, so we esily find tht = 5 nd b = 10. In other words, yx) = x 2 +5x /43

19 Cse 3. Severl unknown functions Suppose tht we wish to find the extremls of Jy,z) = subject to given boundry conditions, sy Lx,yx),y x),zx),z x))dx y) = y 0, yb) = y 1, z) = z 0, zb) = z 1. Viewing Jy,z) s function of one vrible, we must then hve d dx L y = L y, d dx L z = L z. In prticulr, one cn find the extremls by solving system of two second-order equtions subject to four boundry conditions. 19/43

20 Cse 3. Exmple As typicl exmple, consider the functionl Jy,z) = 1 0 y x)z x)+yx) 2) dx subject to the conditions y0) = z0) = 0 nd y1) = z1) = 1. The corresponding Euler-Lgrnge equtions re then d dx L y = L y = z = 2y, d dx L z = L z = y = 0. Solving the ltter gives y = x+b, hence lso y = x. Solving the former, we now get z = 2x, so z = 1 3 x3 +cx+d = 1 3 x3 +2x). 20/43

21 Cse 4. Isoperimetric constrints Suppose tht we wish to find the extremls of Jy) = Lx,yx),y x))dx subject to the boundry nd integrl constrints y) = y 0, yb) = y 1, Mx,yx),y x))dx = c. Let us denote by Iy) the integrl tht ppers in the lst eqution. Then the extremls of Jy) re either criticl points of Iy) or else criticl points of Jy) λiy) for some Lgrnge multiplier λ R. In prticulr, one hs to solve the Euler-Lgrnge eqution for two different Lgrngins, nmely M nd lso L λm. 21/43

22 Cse 4. Exmple, pge 1 We determine the possible extremls of Jy) = π 0 y x) 2 dx subject to the boundry nd integrl constrints y0) = yπ) = 0, π 0 yx)sinxdx = 1. Let us denote by Iy) the integrl tht ppers in the lst eqution. Its criticl points must stisfy the Euler-Lgrnge eqution L = yx)sinx = d dx L y = L y nd this mens tht Iy) hs no criticl points t ll. = 0 = sinx 22/43

23 Cse 4. Exmple, pge 2 To find the criticl points of Jy) λiy), we note tht L = y x) 2 λyx)sinx = d dx L y = L y = 2y = λsinx. Integrting this eqution twice, we conclude tht y x) = λ 2 cosx+ = yx) = λ 2 sinx+x+b. Since y0) = yπ) = 0, it is esy to check tht = b = 0. Thus, it remins to determine λ. Using the integrl constrint, we get 1 = π 0 yx)sinxdx = λ 2 π 0 sin 2 xdx = λπ 4. This gives λ = 4 π, so the only possible extreml is yx) = 2 π sinx. 23/43

24 Lgrnge multipliers Theorem 2. Lgrnge multipliers Suppose tht yx) is n extreml of Jy) = Lx,yx),y x))dx subject to the integrl constrint Iy) = c, where Iy) = Mx,yx),y x))dx. Then the extreml yx) must be either criticl point of Iy) or else criticl point of Jy) λiy) for some Lgrnge multiplier λ R. This theorem is closely relted to the corresponding theorem for the extrem of function fx,y) subject to constrint gx,y) = c. 24/43

25 Lgrnge multipliers: Proof, pge 1 Let ϕ,ψ be some given test functions nd define f,g: R 2 R by fε,δ) = Jy +εϕ+δψ), gε,δ) = Iy +εϕ+δψ). We note tht g0,0) = Iy) = c by ssumption, while gε,0) g0,0) Iy +εϕ) Iy) g ε 0,0) = lim = lim = I y)ϕ. ε 0 ε ε 0 ε Suppose yx) is not criticl point of Iy). We cn then find test function ϕ such tht I y)ϕ 0. According to the implicit function theorem, we cn thus find function ε = εδ) with ε0) = 0 nd c = gε,δ) = Iy +εϕ+δψ) in neighbourhood of δ = 0, nmely for smll enough δ. 25/43

26 Lgrnge multipliers: Proof, pge 2 Since yx) is n extreml of Jy) subject to the given constrint, fε,δ) = Jy +εϕ+δψ) ttins n extremum t 0,0) subject to the constrint gε,δ) = c. Using stndrd clculus result, we conclude tht either g0,0) = 0 or else f0,0) = λ g0,0) for some λ R. One my esily exclude the former cse since g ε 0,0) = I y)ϕ 0 by bove. We thus hve f0,0) = λ g0,0) = f δ 0,0) = λg δ 0,0) = J y)ψ = λi y)ψ for ll test functions ψ, so yx) is criticl point of Jy) λiy). 26/43

27 Second vrition The first vrition of Jy) is defined s the limit J y)ϕ = lim ε 0 Jy +εϕ) Jy) ε nd one cn use Tylor series expnsion to derive the formul J y)ϕ = Ly ϕ+l y ϕ ) dx. The second vrition of Jy) is defined s the limit J Jy +εϕ) Jy) εj y)ϕ y)ϕ = lim ε ε2 nd one cn use Tylor series expnsion to derive the formul J y)ϕ = L yy ϕ 2 +2L yy ϕϕ +L y y ϕ ) 2) dx. 27/43

28 Second vrition: Sketch of proof According to the definition of Jy), one hs Jy +εϕ) Jy) = ) Lx,y +εϕ,y +εϕ ) Lx,y,y ) dx. Denote the integrnd by Fε). Then Tylor series expnsion gives Fε) = ε L y ϕ+l y ϕ ) + ε2 L yy ϕ 2 +2L yy ϕϕ +L y 2 y ϕ ) 2) +..., where the dots indicte terms which contin higher powers of ε. Since the liner terms correspond to the first vrition J y)ϕ, it esily follows tht the qudrtic terms correspond to J y)ϕ. 28/43

29 Necessry condition Theorem 3. Necessry condition If the functionl Jy) ttins locl minimum t the function yx), then we must ctully hve J y)ϕ 0 for ll functions ϕ. This condition resembles the second derivtive test from dvnced clculus. It is closely relted to the expnsion Jy +εϕ) = Jy)+εJ y)ϕ+ ε2 2 J y)ϕ+... nd the fct tht J y)ϕ = 0 for ll criticl points of Jy). It my hppen tht J y)ϕ > 0 for ll functions ϕ, even though Jy) hs no locl minimum. In prticulr, the given condition is necessry but not sufficient for the existence of locl minimum. 29/43

30 Necessry condition: Sketch of proof Using the definition of J y)ϕ, one cn write Jy +εϕ) = Jy)+εJ y)ϕ+ ε2 2 J y)ϕ+ ε2 2 Ry,ε,ϕ) for some reminder term R such tht Ry,ε,ϕ) 0 s ε 0. Since yx) is point of locl minimum, this implies tht 0 Jy +εϕ) Jy) = ε2 2 J y)ϕ+ ε2 2 Ry,ε,ϕ) for ll smll enough ε. Letting ε 0, we conclude tht ) J y)ϕ = lim J y)ϕ+ry,ε,ϕ) 0. ε 0 30/43

31 Legendre condition Theorem 4. Legendre condition If the functionl Jy) ttins locl minimum t the function yx), then one hs L y y x,y,y ) 0 throughout the intervl [,b]. As simple exmple, consider the functionl Jy) = 1 1 x 1+y x) 2 dx. In this cse, the Lgrngin L is esily seen to stisfy L y = xy x) 1+y x) 2 = L y y = x 1+y x) 2 ) 3/2. Since L y y chnges sign on the given intervl, one finds tht Jy) hs neither locl minimum nor locl mximum. 31/43

32 Legendre condition: Sketch of proof, pge 1 If it hppens tht L y y < 0 t some point, then L y y < 0 on some intervl [x 0 ε,x 0 +ε] by continuity. Consider the test function { ) } sin ϕx) = 3 πx x0 ) ε if x x 0 ε. 0 otherwise This function is bounded for ll ε > 0, but its derivtive becomes unbounded s ε 0. One thus expects the second vrition J y)ϕ = L yy ϕ 2 +2L yy ϕϕ +L y y ϕ ) 2) dx to become negtive s ε 0. This contrdicts our previous theorem which sserts tht J y)ϕ 0 t point of locl minimum. It remins to show tht J y)ϕ < 0 for ll smll enough ε > 0. 32/43

33 Legendre condition: Sketch of proof, pge 2 Since the Lgrngin is twice continuously differentible, one hs J y)ϕ = x0 +ε x 0 ε x0 +ε x 0 ε L yy ϕ 2 +2L yy ϕϕ +L y y ϕ ) 2) dx C 1 ϕ 2 +C 2 ϕϕ C 3 ϕ ) 2) dx for some constnts C 1,C 2,C 3 > 0. Letting u = π ε x x 0), we get J y)ϕ ε π π π C 1 + 3πC ) 2 9π2 C 3 ε ε 2 sin 4 u cos 2 u du = 2εC 1 +6πC 2 C 4 ε for some constnts C 1,C 2,C 4 > 0 nd the result now follows. 33/43

34 Poincré inequlity Suppose tht ϕ is test function on the intervl [,b]. Then ϕx) = ϕx) ϕ) x ϕ y) dy, so one my use the Cuchy-Schwrz inequlity to find tht ϕx) 2 x dy ϕ y) 2 dy = x ) ϕ x) 2 dx. Integrting over [, b], we thus obtin the Poincré inequlity ϕx) 2 dx b )2 2 ϕ x) 2 dx. 34/43

35 Sufficient condition Theorem 5. Sufficient condition Suppose tht yx) is criticl point of Jy) = Lx,yx),y x))dx subject to the boundry conditions y) = y 0 nd yb) = y 1. Suppose lso tht there exists some constnt δ > 0 such tht J y)ϕ δ ϕ x) 2 dx for ll test functions ϕ. Then Jy) ttins locl minimum t yx). 35/43

36 Sufficient condition: Sketch of proof, pge 1 We use Tylor s theorem with reminder to write Jy +εϕ) = Jy)+ ε2 2 L yy ϕ 2 +2L yy ϕϕ +L y y ϕ ) 2) dx with the second derivtives of L evluted t point of the form x,y +tεϕ,y +tεϕ ), 0 t 1. Since L is twice continuously differentible, this implies tht Jy +εϕ) = Jy)+ ε2 2 J y)ϕ + ε2 2 R 1 ϕ 2 +2R 2 ϕϕ +R 3 ϕ ) 2) dx for some functions R 1,R 2,R 3 which pproch zero s ε 0. We now estimte the integrl tht ppers on the right hnd side. 36/43

37 Sufficient condition: Sketch of proof, pge 2 Let us denote the lst integrl by I. We then hve I ) R 1 + R 2 ϕx) 2 dx+ ) R 2 + R 3 ϕ x) 2 dx nd we cn use the Poincré inequlity to conclude tht I Rε,x) ϕ x) 2 dx for some positive function R which pproches zero s ε 0. In view of our ssumption on J y)ϕ, this implies tht Jy +εϕ) Jy) ε2 2 δ Rε,x)) ϕ x) 2 dx 0 for ll smll enough ε, so Jy) ttins locl minimum t yx). 37/43

38 Sufficient condition: Exmple Consider the shortest pth exmple tht corresponds to the cse Jy) = 1+y x) 2 dx. In this cse, the criticl points re lines, nmely functions yx) whose derivtive is constnt, sy y x) = c. One cn esily check tht L y y = 1 1+y x) 2 ) 3/2 = 1 1+c 2 ) 3/2 = δ for some constnt δ > 0, while L yy = L yy = 0. This implies tht J y)ϕ = δϕ x) 2 dx, so the sufficient condition is stisfied nd Jy) hs locl minimum. 38/43

39 Invrince If there is trnsformtion x,y) x,y ) such tht Lx,y,y )dx = Lx,y,y )dx for ll < b, we sy tht Jy) is invrint under the given trnsformtion. A very common exmple is time invrince x = x+ε, y = y. This cse rises whenever L is independent of x, for instnce. Another common exmple is trnsltion invrince x = x, y = y +ε. This cse rises whenever L is independent of y, for instnce. 39/43

40 Noether s theorem Theorem 6. Noether s theorem Suppose Jy) is invrint under fmily of trnsformtions x,y) x,y ) = fx,y,ε),gx,y,ε)) such tht x = x nd y = y when ε = 0. Then the quntity Q = α L y L y ) +βly is independent of x whenever yx) is criticl point of Jy) nd α = x ε, β = y ε=0 ε. ε=0 The Beltrmi identity is very specil cse of this theorem. 40/43

41 Noether s theorem: Sketch of proof, pge 1 We simplify the invrince condition using Tylor series expnsion nd keeping the liner terms only. This gives the identity Lx,y,y ) = Lx,y,y )+x x) d dx Lx,y,y ) +y y)l y x,y,y )+y y )L y x,y,y ). Let us express this identity in the more compct form L = L+ x d dx L+ y L y + y L y. Keeping liner terms s before, we get x = x x = αε nd y = y x) yx) = y x) y x )+y x ) yx) = y x) x+βε = β αy )ε. 41/43

42 Noether s theorem: Sketch of proof, pge 2 We now integrte the identity bove. Since x = x+αε, we hve Ldx = = = L dx L+ x L+αε ) d dx L+ y L y + y L y dx d dx L+ y L y + y L y +α ε L Rerrnging terms nd integrting by prts, we conclude tht 0 = [ ] b αε L) dx+ y L y + ) dx. y L y d ) dx L y dx. Here, the rightmost integrl is zero by the Euler-Lgrnge eqution. 42/43

43 Noether s theorem: Sketch of proof, pge 3 In view of our computtion bove, we must thus hve [ ] b 0 = αε L+ y L y [ = αε L+β αy )ε L y ] b. Since the endpoints, b re rbitrry, this ctully mens tht Q = αl+β αy )L y is independent of x. Rerrnging terms, we conclude tht Q = αl y L y )+βl y is independent of x. This finlly completes the proof. 43/43