Calculus of Variations

Transcription

1 Clculus of Vritions The biggest step from derivtives with one vrible to derivtives with mny vribles is from one to two. After tht, going from two to three ws just more lgebr nd more complicted pictures. Now the step will be from finite number of vribles to n infinite number. Tht will require new set of tools, yet in mny wys the techniques re not very different from those you know. If you ve never red chpter 19 of volume II of the Feynmn Lectures in Physics, now would be good time. It s clssic introduction to the re. For deeper look t the subject, pick up McCluer s book referred to in the Bibliogrphy t the beginning of this book Exmples Wht line provides the shortest distnce between two points? A stright line of course, no surprise there. But not so fst, with few twists on the question the result won t be nerly s obvious. How do I mesure the length of curved (or even stright) line? Typiclly with ruler. For the curved line I hve to do successive pproximtions, breking the curve into smll pieces nd dding the finite number of lengths, eventully tking limit to express the nswer s n integrl. Even with stright line I will do the sme thing if my ruler isn t long enough. Put this in terms of how you do the mesurement: Go to locl store nd purchse ruler. It s mde out of some rel mteril, sy brss. The curve you re mesuring hs been lid out on the ground, nd you move long it, counting the number of times tht you use the ruler to go from one point on the curve to nother. If the ruler mesures in decimeters nd you ly it down 1 times long the curve, you hve your first estimte for the length, 1. meters. Do it gin, but use centimeter length nd you need 18 such lengths: 1.8 meters. Tht s tedious, but simple. Now do it gin for nother curve nd compre their lengths. Here comes the twist: The ground is not t uniform temperture. Perhps you re mking these mesurements over not-fully-cooled lv flow in Hwii. Brss will expnd when you het it, so if the curve whose length you re mesuring psses over hot spot, then the ruler will expnd when you plce it down, nd you will need to plce it down fewer times to get to the end of the curve. You will mesure the curve s shorter. Now it is not so cler which curve will hve the shortest (mesured) length. If you tke the stright line nd push it over so tht it psses through hotter region, then you my get smller result. Let the coefficient of expnsion of the ruler be α, ssumed constnt. For modest temperture chnges, the length of the ruler is l = (1 + α T )l. The length of curve s mesured with this ruler is dl (16.1) 1 + αt Here I m tking T = s the bse temperture for the ruler nd dl is the length you would use if everything styed t this temperture. With this mesure for length, it becomes n interesting problem to discover which pth hs the shortest length. The forml term for the pth of shortest length is geodesic. In section 13.1 you sw integrls tht looked very much like this, though pplied to different problem. There I looked t the time it tkes prticle to slide down curve under grvity. Tht time is the integrl of dt = dl/v, where v is the prticle s speed, function of position long the pth. Using conservtion of energy, the expression for the time to slide down curve ws Eq. (13.6). y x dt = dl (E/m) + gy (16.) Jmes Nering, University of Mimi 1

2 16 Clculus of Vritions In tht chpter I didn t ttempt to nswer the question bout which curve provides the quickest route to the end, but in this chpter I will. Even qulittively you cn see prllel between these two problems. You get shorter length by pushing the curve into region of higher temperture. You get shorter time by pushing the curve lower, (lrger y). In the ltter cse, this mens tht you drop fst to pick up speed quickly. In both cses the denomintor in the integrl is lrger. You cn overdo it of course. Push the curve too fr nd the vlue of dl itself cn become too big. It s blnce. In problems.35 nd.39 you looked t the mount of time it tkes light to trvel from one point to nother long vrious pths. Is the time minimum, mximum, or neither? In these specil cses, you sw tht this is relted to the focus of the lens or of the mirror. This is very generl property of opticl systems, nd is n extension of some of the ides in the first two exmples bove. These questions re sometimes pretty nd elegnt, but re they relted to nything else? Yes. Newton s clssicl mechnics cn be reformulted in this lnguge nd it leds to powerful methods to set up the equtions of motion in complicted problems. The sme ides led to useful pproximtion techniques in electromgnetism, llowing you to obtin high-ccurcy solutions to problems for which there is no solution by other mens. 16. Functionl Derivtives It is time to get specific nd to implement* these concepts. All the preceding exmples cn be expressed in the sme generl form. In stndrd x-y rectngulr coordinte system, Then Eq. (16.1) is dl = dx + dy = dx b y dx 1 + αt (x, y) ( ) dy = dx 1 + y dx (16.3) This mesured length depends on the pth, nd I ve ssumed tht I cn express the pth with y s function of x. No loops. You cn llow more generl pths by using nother prmetriztion: x(t) nd y(t). Then the sme integrl becomes t t 1 ẋ + ẏ dt 1 + αt ( x(t), y(t) ) (16.4) The eqution (16.) hs the sme form b 1 + y dx (E/m) + gy And the trvel time for light through n opticl system is dt = dl v = b 1 + y dx v(x, y) where the speed of light is some known function of the position. * If you find the methods used in this section confusing, you my prefer to look t n lternte pproch to the subject s described in section Then return here.

3 16 Clculus of Vritions 3 In ll of these cses the output of the integrl depends on the pth tken. It is functionl of the pth, sclr-vlued function of function vrible. Denote the rgument by squre brckets. b I[y] = dx F ( x, y(x), y (x) ) (16.5) The specific F vries from problem to problem, but the preceding exmples ll hve this generl form, even when expressed in the prmetrized vribles of Eq. (16.4). The ide of differentil clculus is tht you cn get informtion bout function if you try chnging the independent vrible by smll mount. Do the sme thing here. Now however the independent vrible is the whole pth, so I ll chnge tht pth by some smll mount nd see wht hppens to the vlue of the integrl I. This pproch to the subject is due to Lgrnge. The development in section 16.6 comes from Euler. y δy y + δy I = I[y + δy] I[y] b+ b = dx F ( x, y(x) + δy(x), y (x) + δy (x) ) b dx F ( x, y(x), y (x) ) (16.6) + The (smll) function δy(x) is the verticl displcement of the pth in this coordinte system. To keep life simple for the first ttck on this problem, I ll tke the specil cse for which the endpoints of the pth re fixed. Tht is, y =, b =, δy() =, δy(b) = δy y + δy To compute the vlue of Eq. (16.6) use the power series expnsion of F, s in section.5. F (x + x, y + y, z + z) = F (x, y, z) + F F F x + y + x y z z + F ( x) x + F x y + x y For now look t just the lowest order terms, liner in the chnges, so ignore the second order terms. In this ppliction, there is no x. plus terms of higher order in δy nd δy. Put this into Eq. (16.6), nd F (x, y + δy, y + δy ) = F (x, y, y ) + F F δy + y y δy δi = b dx [ ] F F δy + y y δy (16.7)

4 16 Clculus of Vritions 4 For exmple, Let F = x + y + y on the intervl x 1. Tke bse pth to be stright line from (, ) to (1, 1). Choose for the chnge in the pth δy(x) = ɛx(1 x). This is simple nd it stisfies the boundry conditions. I[y] = I[y + δy] = dx [ x + y + y ] = dx [ x + x + 1 ] = 5 3 [x + ( x + ɛx(1 x) ) ( ) ] ɛ(1 x) = ɛ ɛ (16.8) The vlue of Eq. (16.7) is δi = 1 dx [ yδy + y δy ] = 1 dx [xɛx(1 x) + ɛ(1 x)] = 1 6 ɛ Return to the generl cse of Eq. (16.7) nd you will see tht I ve explicitly used only prt of the ssumption tht the endpoint of the pth hsn t moved, = b =. There s nothing in the body of the integrl itself tht constrins the chnge in the y-direction, nd I hd to choose the function δy by hnd so tht this constrint held. In order to use the equtions δy() = δy(b) = more generlly, there is stndrd trick: integrte by prts. You ll lwys integrte by prts in these clcultions. b dx F y δy = b dx F d δy y dx = F b b y δy dx d ( ) F dx y δy(x) This expression llows you to use the informtion tht the pth hsn t moved t its endpoints in the y direction either. The boundry term from this prtil integrtion is F b y δy = F y ( b, y(b) ) δy(b) F y (, y() ) δy() = Put the lst two equtions bck into the expression for δi, Eq. (16.7) nd the result is δi = b dx [ F y d ( )] F δy dx y (16.9) Use this expression for the sme exmple F = x + y + y with y(x) = x nd you hve δi = 1 dx This is sort of like Eq. (8.16), [ y d ] dx y δy = 1 dx [x ] ɛx(1 x) = 1 6 ɛ df = G. d r = grd f. d r = f. d r = f x k dx k = f x 1 dx 1 + f x dx + The differentil chnge in the function depends linerly on the chnge d r in the coordintes. It is sum over the terms with dx 1, dx,.... This is precise prllel to Eq. (16.9), except tht the sum over discrete index k is now n integrl over the continuous index x. The chnge in I is liner functionl of

5 16 Clculus of Vritions 5 the chnge δy in the independent vrible y; this δy corresponds to the chnge d r in the independent vrible r in the other cse. The coefficient of the chnge, insted of being clled the grdient, is clled the functionl derivtive though it s essentilly the sme thing. δi δy = F y d ( ) F, δi[y, dx y δy] = dx δi ( x, y(x), y (x) ) δy(x) (16.1) δy nd for chnge, I ve indicted explicitly the dependence of δi on the two functions y nd δy. This prllels the eqution (8.13). The sttement tht this functionl derivtive vnishes is clled the Euler- Lgrnge eqution. Return to the exmple F = x + y + y, then δi δy = δ δy 1 dx [ x + y + y ] = y d dx y = y y Wht is the minimum vlue of I? Set this derivtive to zero. y y = = y(x) = A cosh x + B sinh x The boundry conditions y() = nd y(1) = 1 imply y = B sinh x where B = 1/ sinh 1. The vlue of I t this point is I[B sinh x] = 1 dx [ x + B sinh x + B cosh x ] = 1 + coth 1 (16.11) 3 Is it minimum? Yes, but just s with the ordinry derivtive, you hve to look t the next order terms to determine tht. Compre this vlue of I[y] = to the vlue 5/3 found for the nerby function y(x) = x, evluted in Eq. (16.8). Return to one of the exmples in the introduction. Wht is the shortest distnce between two points, but for now ssume tht there s no temperture vrition. Write the length of pth for function y between fixed endpoints, tke the derivtive, nd set tht equl to zero. b L[y] = dx 1 + y, so δl δy = d y dx = y + y y 1 + y 1 + y ( ) = y 1 + y 3/ ( ) = 1 + y 3/ For minimum length then, y =, nd tht s stright line. Surprise! Do you relly hve to work through this mildly messy mnipultion? Sometimes, but not here. Just notice tht the derivtive is in the form δl δy = d dx f(y ) = (16.1) so it doesn t mtter wht the prticulr f is nd you get stright line. f(y ) is constnt so y must be constnt too. Not so fst! See section 16.9 for nother opinion.

6 16 Clculus of Vritions Brchistochrone Now for tougher exmple, gin from the introduction. In Eq. (16.), which of ll the pths between fixed initil nd finl points provides the pth of lest time for prticle sliding long it under grvity. Such pth is clled brchistochrone. This problem ws first proposed by Bernoulli (one of them), nd ws solved by severl people including Newton, though it s unlikely tht he used the methods developed here, s the generl structure we now cll the clculus of vritions ws decdes in the future. Assume tht the prticle strts from rest so tht E =, then T [y] = x For the minimum time, compute the derivtive nd set it to zero. dx δt 1 + y g δy = d y 3/ dx 1 + y gy (16.13) y y 1 + y = This is strting to look intimidting, leding to n impossibly* messy differentil eqution. Is there nother wy? Yes. Why must x be the independent vrible? Wht bout using y? In the generl setup leding to Eq. (16.1) nothing chnges except the symbols, nd you hve Eqution (16.13) becomes I[x] = dy F (y, x, x ) δi δx = F x d ( ) F dy x T [x] = y dy (16.14) 1 + x gy (16.15) The function x does not pper explicitly in this integrl, just its derivtive x = dx/dy. This simplifies the functionl derivtive, nd the minimum time now comes from the eqution δi δx = d ( ) F dy x = (16.16) This is much esier. d()/dy = mens tht the object in prentheses is constnt. F x = C = 1 gy x 1 + x Solve this for x nd you get (let K = C g) x = dx dy = K y, 1 K so x(y) = y K dy y 1 K y This is n elementry integrl. Let = 1/K x(y) = dy y = y y dy y + y y = dy (y ) + (y ) * Only improbbly. See problem 16.1.

7 16 Clculus of Vritions 7 Mke the substitution (y ) = z in the first hlf of the integrl nd (y ) = sin θ in the second hlf. x(y) = 1 1 dz z + cos θdθ sin θ = z + θ = ( ) y (y ) + sin 1 + C The boundry condition tht x() = determines C = π/, nd the other end of the curve determines : x(y ) = x. You cn rewrite this s x(y) = ( ) y y y + cos 1 This is cycloid. Wht s cycloid nd why does this eqution describe one? See problem 16.. (16.17) x-independent In Eqs. (16.15) nd (16.16) there ws specil cse for which the dependent vrible ws missing from F. Tht mde the eqution much simpler. Wht if the independent vrible is missing? Does tht provide comprble simplifiction? Yes, but it s trickier to find. I[y] = Use the chin rule to differentite F with respect to x. Multiply the Lgrnge eqution (16.18) by y to get dx F (x, y, y ) δi δy = F y d ( ) F dx y = (16.18) df dx = F x + dy F dx y + dy F dx y (16.19) y F y d F y dx y = Now substitute the term y ( F/ y) from the preceding eqution (16.19). The lst two terms re the derivtive of product. df dx F x dy F dx y d F y dx y = df dx F x d [ ] y F dx y = (16.) If the function F hs no explicit x in it, the second term is zero, nd the eqution is now derivtive [ ] d F y F dx y = nd y F y F = C (16.1) This is lredy down to first order differentil eqution. The combintion y F y F tht ppers on the left of the second eqution is importnt. It s clled the Hmiltonin.

8 16 Clculus of Vritions Fermt s Principle Fermt s principle of lest time provides formultion of geometricl optics. When you don t know bout the wve nture of light, or if you ignore tht spect of the subject, it seems tht light trvels in stright lines t lest until it hits something. Of course this isn t fully correct, becuse when light hits piece of glss it refrcts nd its pth bends nd follows Snell s eqution. All of these properties of light cn be described by Fermt s sttement tht the pth light follows will be tht pth tht tkes the lest* time. dl T = dt = v = 1 n dl (16.) c The totl trvel time is the integrl of the distnce dl over the speed (itself function of position). The index of refrction is n = c/v, where c is the speed of light in vcuum, so I cn rewrite the trvel time in the bove form using n. The integrl n dl is clled the opticl pth. From this ide it is very esy to derive the rule for reflection t surfce: ngle of incidence equls ngle of reflection. It is eqully esy to derive Snell s lw. (See problem 16.5.) I ll look t n exmple tht would be difficult to do by ny mens other thn Fermt s principle: Do you remember wht n sphlt rod looks like on very hot dy? If you re driving cr on blck colored rod you my see the rod in the distnce pper to be covered by wht looks like wter. It hs sort of mirror-like sheen tht is lwys receding from you the hot rod mirge. You cn never ctch up to it. This hppens becuse the rod is very hot nd it hets the ir next to it, cusing strong temperture grdient ner the surfce. The density of the ir decreses with rising temperture becuse the pressure is constnt. Tht in turn mens tht the index of refrction will decrese with the rising temperture ner the surfce. The index will then be n incresing function of the distnce bove ground level, n = f(y), nd the trvel time of light will depend on the pth tken. n dl = f(y) dl = f(y) 1 + x dy = f(y) 1 + y dx (16.3) Wht is f(y)? I ll leve tht for moment nd then fter crrying the clcultion through for while I cn pick n f tht is both plusible nd esy to mnipulte. y rod x Should x be the independent vrible, or y? Either should work, nd I chose y becuse it seemed likely to be esier. (See problem 16.6 however.) The integrnd then does not contin the dependent vrible x. minimize n dl = f(y) 1 + x dy = x f(y) = C 1 + x d [ ] f(y) 1 + x dy x = Solve for x to get f(y) x = C ( 1 + x ) so x = dx dy = C f(y) C (16.4) * Not lwys lest. This just requires the first derivtive to be zero; the second derivtive is ddressed in section Fermt s principle of sttionry time my be more ccurte, but Fermt s principle of lest time is entrenched in the literture.

9 16 Clculus of Vritions 9 At this point pick form for the index of refrction tht will mke the integrl esy nd will still plusibly represent relity. The index increses grdully bove the rod surfce, nd the simplest function works: f(y) = n (1 + αy). The index increses linerly bove the surfce. x(y) = C n (1 + αy) C dy = C αn 1 dy (y + 1/α) C /α n This is n elementry integrl. Let u = y + 1/α, then u = (C/αn ) cosh θ. x = C dθ θ = αn αn C (x x ) y = 1 α + C cosh ( (αn αn /C)(x x ) ) C nd x re rbitrry constnts, nd x is obviously the center of symmetry of the pth. You cn relte the other constnt to the y-coordinte t tht sme point: C = n (αy + 1). Becuse the vlue of α is smll for ny rel rodwy, look t the series expnsion of this hyperbolic function to the lowest order in α. y y + α(x x ) / (16.5) When you look down t the rod you cn be looking t n imge of the sky. The light comes from the sky ner the horizon down towrd the rod t n ngle of only degree or two. It then curves up so tht it cn enter your eye s you look long the rod. The shimmering surfce is reflection of the distnt sky or in this cse n utomobile mirge Electric Fields The energy density in n electric field is ɛ E /. For the sttic cse, this electric field is the grdient of potentil, E = φ. Its totl energy in volume is then W = ɛ * dv ( φ) (16.6) Wht is the minimum vlue of this energy? Zero of course, if φ is constnt. Tht question is too loosely stted to be much use, but keep with it for while nd it will be possible to turn it into something more precise nd more useful. As with ny other derivtive tken to find minimum, chnge the independent vrible by smll mount. This time the vrible is the function φ, so relly this quntity W cn more fully be written s functionl W [φ] to indicte its dependence on the potentil function. W [φ + δφ] W [φ] = ɛ = ɛ * Donld Collins, Wrren Wilson College dv ( (φ + δφ) ) ɛ dv ( φ. δφ + ( δφ) ) dv ( φ)

10 Now pull out vector identity from problem 9.36, 16 Clculus of Vritions 1. ( f g ) = f. g + f. g nd pply it to the previous line with f = δφ nd g = φ. W [φ + δφ] W [φ] = ɛ dv [.(δφ φ) δφ φ ] + ɛ dv ( δφ) The divergence term is set up to use Guss s theorem; this is the vector version of integrtion by prts. W [φ + δφ] W [φ] = ɛ d A.( φ)δφ ɛ dv δφ φ + ɛ dv ( δφ) (16.7) If the vlue of the potentil φ is specified everywhere on the boundry, then I m not llowed to chnge it in the process of finding the chnge in W. Tht mens tht δφ vnishes on the boundry. Tht mkes the boundry term, the d A integrl, vnish. Its integrnd is zero everywhere on the surfce of integrtion. In looking for minimum energy I wnt to set the first derivtive to zero, nd tht s the coefficient of the term liner in δφ. δw δφ = ɛ φ = The function tht produces the minimum vlue of the totl energy (with these fixed boundry conditions) is the one tht stisfies Lplce s eqution. Is it relly minimum? Yes. In this instnce it s very esy to see tht. The extr term in the chnge of W is dv ( δφ). Tht is positive no mtter wht δφ is. Tht the correct potentil function is the one hving the minimum energy llows for n efficient pproximte method to solve electrosttic problems. I ll illustrte this using n exmple borrowed from the Feynmn Lectures in Physics nd tht you cn lso solve exctly: Wht is the cpcitnce of length of coxil cble? (Neglect the edge effects t the ends of the cble of course.) Let the inner nd outer rdii be nd b, nd the length L. A chrge density λ is on the inner conductor (nd therefore λ on the inside of the outer conductor). It cretes rdil electric field of size λ/πɛ r. The potentil difference between the conductors is V = b dr λ πɛ r = λ ln(b/) πɛ (16.8) The chrge on the inner conductor is λl, so C = Q/ V = πɛ L / ln(b/), where V = V b V. The totl energy stisfies W = C V /, so for the given potentil difference, knowing the energy is the sme s knowing the cpcitnce. This exct solution provides lbortory to test the efficcy of vritionl pproximtion for the sme problem. The ide behind this method is tht you ssume form for the solution φ(r). This ssumed form must stisfy the boundry conditions, but it need not stisfy Lplce s eqution. It should lso hve one or more free prmeters, the more the better up to the limits of your ptience. Now compute the totl energy tht this tril function implies nd then use the free prmeters to minimize it. This function with minimum energy is the best pproximtion to the correct potentil mong ll those with your ssumed tril function.

11 16 Clculus of Vritions 11 Let the potentil t r = be V nd t r = b it is V b. An exmple function tht stisfies these conditions is φ(r) = V + (V b V ) r b (16.9) The electric field implied by this is E = φ = ˆr (V V b )/(b ), constnt rdil component. From (16.6), the energy is ɛ b L πr dr ( ) dφ = ɛ dr b ( ) Vb V L πr dr = 1 b πlɛ b + b V Set this to C V / to get C nd you hve C pprox = πlɛ b + b How does this compre to the exct nswer, πɛ L/ ln(b/)? Let x = b/. C pprox C = 1 b + b ln(b/) = 1 x + 1 x 1 ln x x: rtio: Assuming constnt mgnitude electric field in the region between the two cylinders is clerly not correct, but this estimte of the cpcitnce gives remrkble good result even when the rtio of the rdii is two or three. This is true even though I didn t even put in prmeter with which to minimize the energy. How much better will the result be if I do? Insted of liner pproximtion for the potentil, use qudrtic. φ(r) = V + α(r ) + β(r ), with φ(b) = V b Solve for α in terms of β nd you hve, fter little mnipultion, Compute the energy from this. φ(r) = V + V r + β(r )(r b) (16.3) b W = ɛ b [ ] V L πr dr + β(r b) b Rerrnge this for esier mnipultion by defining β = γ V/(b ) nd c = ( + b)/ then W = ɛ ( ) V ((r Lπ ) [ ] c) + c dr 1 + γ(r c) b = ɛ ( ) V Lπ [ c(b ) + γ(b ) 3 /6 + cγ (b ) 3 /1 ] b

12 16 Clculus of Vritions 1 γ is free prmeter in this clcultion, replcing the originl β. To minimize this energy, set the derivtive dw/dγ =, resulting in the vlue γ = 1/c. At this vlue of γ the energy is W = ɛ Lπ ( V b ) [ 1 ( b ) ] (b )3 6(b + ) (16.31) Except for the fctor of V / this is the new estimte of the cpcitnce, nd to see how good it is, gin tke the rtio of this estimte to the exct vlue nd let x = b/. C = ln x 1 ] x + 1 (x 1) [1 x 1 3(x + 1) C pprox x: rtio: (16.3) For only one prmeter djustment, this provides very high ccurcy. This sort of technique is the bsis for mny similr procedures in this nd other contexts, especilly in quntum mechnics Discrete Version There is nother wy to find the functionl derivtive, one tht more closely prllels the ordinry prtil derivtive. It is due to Euler, nd he found it first, before Lgrnge s discovery of the tretment tht I ve spent ll this time on. Euler s method is perhps more intuitive thn Lgrnge s, but it is not s esy to extend it to more thn one dimension nd it doesn t esily lend itself to the powerful mnipultive tools tht Lgrnge s method does. This lternte method strts by noting tht n integrl is the limit of sum. Go bck to the sum nd perform the derivtive on it, finlly tking limit to get n integrl. This turns the problem into finite-dimensionl one with ordinry prtil derivtives. You hve to decide which form of numericl integrtion to use, nd I ll pick the trpezoidl rule, Eq. (11.15), with constnt intervl. Other choices work too, but I think this entils less fuss. You don t see this pproch s often s Lgrnge s becuse it is hrder to mnipulte the equtions with Euler s method, nd the nottion cn become quite cumbersome. The trpezoidl rule for n integrl is just the following picture, nd ll tht you hve to hndle with ny cre re the endpoints of the integrl. y y k k+1 b x k = + k, k N where N = [ ] b N 1 1 dx f(x) = lim f() + f(x k ) + 1 f(b) 1 x k x k+1 The integrl Eq. (16.5) involves y, so in the sme spirit, pproximte this by the centered difference. y k = y (x k ) ( y(x k+1 ) y(x k 1 ) ) / This evlutes the derivtive t ech of the coordintes {x k } insted of between them. The discrete form of (16.5) is now I discrete = F (, y(), y () ) F ( x k, y(x k ), (y(x k+1 ) y(x k 1 ))/ ) + F ( b, y(b), y (b) ) N 1 + 1

13 16 Clculus of Vritions 13 Not quite yet. Wht bout y () nd y (b)? The endpoints y() nd y(b) ren t chnging, but tht doesn t men tht the slope there is fixed. At these two points, I cn t use the centered difference scheme for the derivtive, I ll hve to use n symmetric form to give I discrete = F (, y(), (y(x 1 ) y(x ))/ ) + F ( b, y(b), (y(x N ) y(x N 1 ))/ ) F ( x k, y(x k ), (y(x k+1 ) y(x k 1 ))/ ) N (16.33) When you keep the endpoints fixed, this is function of N 1 vribles, {y k = y(x k )} for 1 k N 1, nd to find the minimum or mximum you simply tke the prtil derivtive with respect to ech of them. It is not function of ny of the {x k } becuse those re defined nd fixed by the prtition x k = + k. The clumsy prt is keeping trck of the nottion. When you differentite with respect to prticulr y l, most of the terms in the sum (16.33) don t contin it. There re only three terms in the sum tht contribute: l nd l ± 1. In the figure N = 5, nd the l = coordinte (y ) is being chnged. For ll the indices l except the two next to the endpoints (1 nd N 1), this is y l I discrete = [ F ( x y l 1, y l 1, (y l y l )/ ) + l F ( x l, y l, (y l+1 y l 1 )/ ) + F ( x l+1, y l+1, (y l+ y l )/ )] An lternte stndrd nottion for prtil derivtives will help to keep trck of the mnipultions: D 1 F is the derivtive with respect to the first rgument The bove derivtive is then [D F ( x l, y l, (y l+1 y l 1 )/ ) + 1 [ D3 F ( x l 1, y l 1, (y l y l )/ ) D 3 F ( x l+1, y l+1, (y l+ y l )/ )]] (16.34) There is no D 1 F becuse the x l is essentilly n index. If you now tke the limit, the third rgument in ech function returns to the derivtive y evluted t vrious x k s: [ D F ( x l, y l, y l ) 1 + D3 F [ ( x l 1, y l 1, y l 1) D3 F ( x l+1, y l+1, y )] ] l+1 = [ D F ( x l, y(x l ), y (x l ) ) (16.35) + 1 [ D3 F ( x l 1, y(x l 1 ), y (x l 1 ) ) D 3 F ( x l+1, y(x l+1 ), y (x l+1 ) )]] Now tke the limit tht, nd the lst prt is precisely the definition of (minus) the derivtive with respect to x. This then becomes 1 I y disc D F ( x l, y l, y ) d l l dx D 3F ( x l, y l, y l ) (16.36)

14 16 Clculus of Vritions 14 Trnslte this into the nottion tht I ve been using nd you hve Eq. (16.1). Why did I divide by in the finl step? Tht s the equivlent of looking for the coefficient of both the dx nd the δy in Eq. (16.1). It cn be useful to retin the discrete pproximtion of Eq. (16.34) or (16.35) to the end of the clcultion. This llows you to do numericl clcultions in cses where the nlytic equtions re too hrd to mnipulte. Agin, not quite yet. The two cses l = 1 nd l = N 1 hve to be hndled seprtely. You need to go bck to Eq. (16.33) to see how they work out. The fctors show up in different plces, but the finl nswer is the sme. See problem It is curious tht when formulting the problem this wy, you don t seem to need prtil integrtion. The result cme out utomticlly. Would tht be true with some integrtion method other other thn the trpezoidl rule? See problem Clssicl Mechnics The clculus of vritions provides wy to reformulte Newton s equtions of mechnics. The results produce efficient techniques to set up complex problems nd they give insights into the symmetries of the systems. They lso provide lternte directions in which to generlize the originl equtions. Strt with one prticle subject to force, nd the force hs potentil energy function U. Following the trditions of the subject, denote the prticle s kinetic energy by T. Picture this first in rectngulr coordintes, where T = m v / nd the potentil energy is U(x 1, x, x 3 ). The functionl S depends on the pth [x 1 (t), x (t), x 3 (t)] from the initil to the finl point. The integrnd is the Lgrngin, L = T U. S[ r ] = t t 1 L ( r, r ) dt, where L = T U = m The sttement tht the functionl derivtive is zero is Set this to zero nd you hve δs = L d ( ) L = U d ( ) mẋk δx k x k dt ẋ k x k dt (ẋ 1 + ẋ + ẋ ) 3 U(x1, x, x 3 ) (16.37) mẍ k = U x k, or m d r dt = F (16.38) Tht this integrl of L dt hs zero derivtive is F = m. Now wht? This my be elegnt, but does it ccomplish nything? The first observtion is tht when you stte the problem in terms of this integrl it is independent of the coordinte system. If you specify pth in spce, giving the velocity t ech point long the pth, the kinetic energy nd the potentil energy re well-defined t ech point on the pth nd the integrl S is too. You cn now pick whtever bizrre coordinte system tht you wnt in order to do the computtion of the functionl derivtive. Cn t you do this with F = m? Yes, but computing n ccelertion in n odd coordinte system is lot more work thn computing velocity. A second dvntge will be tht it s esier to hndle constrints by this method. The technique of Lgrnge multipliers from section 8.1 will pply here too. Do the simplest exmple: plne polr coordintes. The kinetic energy is T = m (ṙ + r φ )

15 16 Clculus of Vritions 15 The potentil energy is function U of r nd φ. With the of the Lgrngin defined s T U, the vritionl derivtive determines the equtions of motion to be S[r, φ] = t t 1 L ( r(t), φ(t) ) dt δs δr = L r d L dt ṙ = mr φ U r m r = δs δφ = L φ d L = U dt φ φ m d ( r dt φ ) = These re the components of F = m in polr coordintes. If the potentil energy is independent of φ, the second eqution sys tht ngulr momentum is conserved: mr φ. Wht do the discrete pproximte equtions (16.34) or (16.35) look like in this context? Look t the cse of one-dimensionl motion to understnd n exmple. The Lgrngin is L = m ẋ U(x) Tke the expression in Eq. (16.34) nd set it to zero. or du dx (x l) + 1 [ m(xl x l )/ m(x l+ x l )/ ] = m x l+ x l + x l ( ) = du dx (x l) (16.39) This is the discrete pproximtion to the second derivtive, s in Eq. (11.1) Endpoint Vrition Following Eq. (16.6) I restricted the vrition of the pth so tht the endpoints re kept fixed. Wht if you don t? As before, keep terms to the first order, so tht for exmple t b δy is out. Becuse the most common ppliction of this method involves integrls with respect to time, I ll use tht integrtion vrible tb + t b S = dt L ( t, y(t) + δy(t), ẏ(t) + δẏ(t) ) tb t + t tb + t b [ = dt L(t, y, ẏ) + L t + t y [ tb + t b ] t+ t = dt L(t, y, ẏ) + t b t δy + L ẏ δẏ ] tb t tb t dt L ( t, y(t), ẏ(t) ) dt L ( t, y(t), ẏ(x) ) t [ ] L L dt δy + y ẏ δẏ Drop qudrtic terms in the second line: nything involving (δy) or δyδẏ or (δẏ). Similrly, drop terms such s t δy in going to the third line. Do prtil integrtion on the lst term tb t dt L d δy ẏ dt = L tb tb ẏ δy dt d t t dt ( ) L δy(t) (16.4) ẏ The first two terms, with the t nd t b, re to first order [ tb + t b ] t+ t dt L(t, y, ẏ) = L ( (t b, y(t b ), ẏ(t b ) ) t b L ( (t, y(t ), ẏ(t ) ) t t b t

16 16 Clculus of Vritions 16 This produces n expression for S S =L ( (t b, y(t b ), ẏ(t b ) ) t b L ( (t, y(t ), ẏ(t ) ) t + L ẏ (t b)δy(t b ) L tb ẏ (t )δy(t ) + dt t [ L y d dt ( )] L δy ẏ (16.41) Up to this point the mnipultion ws stright-forwrd, though you hd to keep ll the lgebr in good order. Now there re some rerrngements needed tht re not ll tht esy to nticipte, dding nd subtrcting some terms. Strt by looking t the two terms involving t b nd δy(t b ) the first nd third terms. The chnge in the position of this endpoint is not simply δy(t b ). Even if δy is identiclly zero the endpoint will chnge in both the t-direction nd in the y-direction becuse of the slope of the curve (ẏ) nd the chnge in the vlue of t t the endpoint ( t b ). The totl movement of the endpoint t t b is horizontl by the mount t b, nd it is verticl by the mount ( δy + ẏ t b ). To incorporte this, dd nd subtrct this second term, ẏ t, in order to produce this combintion s coefficient. y y + δy t b δy ẏ t b y(t) t b L ( (t b, y(t b ), ẏ(t b ) ) t b + L ẏ (t b)δy(t b ) [ = L(t b ) t b L ẏ (t b)ẏ(t b ) t b [ = L L ] ẏ ẏ t b + L ẏ ] [ L + ] [ ẏ t b + δy ẏ (t b)ẏ(t b ) t b + L Do the sme thing t t, keeping the pproprite signs. Then denote p = L, H = pẏ L, y = δy + ẏ t ẏ H is the Hmiltonin nd Eq. (16.41) becomes Noether s theorem. [ ] tb tb S = p y H t dt t + t [ L y d dt ] ẏ (t b)δy(t b ) (16.4) ( )] L δy (16.43) ẏ If the equtions of motion re stisfied, the rgument of the lst integrl is zero. The chnge in S then comes only from the trnsltion of the endpoint in either the time or spce direction. If t is zero, nd y is the sme t the two ends, you re trnslting the curve in spce verticlly in the grph. Then t b S = p y = [ p(t b ) p(t ) ] y = t

17 16 Clculus of Vritions 17 If the physicl phenomenon described by this eqution is invrint under spcil trnsltion, then momentum is conserved. If you do trnsltion in time insted of spce nd S is invrint, then t is the sme t the strt nd finish, nd S = [ H(t b ) + H(t ) ] t = This is conservtion of energy. Write out wht H is for the cse of Eq. (16.37). If you write this theorem in three dimensions nd require tht the system is invrint under rottions, you get conservtion of ngulr momentum. In more complicted system, especilly in field theory, there re other symmetries, nd they in turn led to conservtion lws. For exmple conservtion of chrge is ssocited with symmetry clled guge symmetry in quntum mechnicl systems. The eqution (16.1), in which the vrition δy hd the endpoints fixed, is much like directionl derivtive in multivrible clculus. For directionl derivtive you find how function chnges s the independent vrible moves long some specified direction, nd in the vritionl cse the direction ws specified to be with functions tht were tied down t the endpoints. The development of the present section is in the spirit of finding the derivtive in ll possible directions, not just specil set Kinks In ll the preceding nlysis of minimizing solutions to vritionl problems, I ssumed tht everything is differentible nd tht ll the derivtives re continuous. Tht s not lwys so, nd it is quite possible for solution to one of these problems to hve discontinuous derivtive somewhere in the middle. These re more complicted to hndle, but just becuse of some extr lgebr. An integrl such s Eq. (16.5) is perfectly well defined if the integrnd hs few discontinuities, but the prtil integrtions leding to the Euler-Lgrnge equtions re not. You cn pply the Euler-Lgrnge equtions only in the intervls between ny kinks. If you re deling with problem of the stndrd form I[x] = b dt L(t, x, ẋ) nd you wnt to determine whether there is kink long the pth, there re some internl boundry conditions tht hve to hold. Roughly speking they re conservtion of momentum nd conservtion of energy, Eq. (16.44), nd you cn show this using the results of the preceding section on endpoint vritions. S = tb t dt L(t, x, ẋ) = tm t dt L + tb t m dt L t t m t b Assume there is discontinuity in the derivtive ẋ t point in the middle, t m. The eqution to solve is still δs/δx =, nd for vritions of the pth tht leve the endpoints nd the middle point lone you hve to stisfy the stndrd Euler-Lgrnge differentil equtions on the two segments. Now however you lso hve to set the vrition to zero for pths tht leve the endpoints lone but move the middle point. t x t t m t b t t m t b Apply Eq. (16.43) to ech of the two segments, nd ssume tht the differentil equtions re lredy stisfied in the two hlves. For the sort of vrition described in the lst two figures, look t the endpoint vritions of the two segments. They produce [ ] tm [ ] tb δs = p x H t + p x H t = [ p x H t ] (t m) [ p x H t ] (t + m) = t t m

18 16 Clculus of Vritions 18 These represent the contributions to the vrition just bove t m nd just below it. This hs to vnish for rbitrry t nd x, so it sys p(t m) = p(t + m) nd H(t m) = H(t + m) (16.44) These equtions, clled the Weierstrss-Erdmnn conditions, re two equtions for the vlues of the derivtive, ẋ, on the two side of t m. The two equtions for the two unknowns my tell you tht there is no discontinuity in the derivtive, or if there is then it will dictte the lgebric equtions tht the two vlues of ẋ must stisfy. More dimensions mens more equtions of course. There is clss of problems in geometry coming under the generl heding of Plteu s Problem. Wht is the miniml surfce tht spns given curve? Here the functionl is da, giving the re s function of the function describing the surfce. If the curve is circle in the plne, then the minimum surfce is the spnning disk. Wht if you twist the circle so tht it does not quite lie in plne? Now it s tough problem. Wht if you hve two prllel circles? Is the pproprite surfce cylinder? (No.) This subject is nothing more thn the mthemticl question of trying to describe sop bubbles. They re not ll spheres. Do kinks hppen often? They re rre in problems tht usully come up in physics, nd it seems to be of more concern to those who pply these techniques in engineering. For n exmple tht you cn verify for yourself however, construct wire frme in the shpe of cube. You cn bend wire or you cn mke it out of solder, which is much esier to mnipulte. Attch hndle to one corner so tht you cn hold it. Now mke sop solution tht you cn use to blow bubbles. (A trick to mke it work better is to dd some glycerine.) Now dip the wire cube into the sop nd see wht sort of sop film will result, filling in the surfces mong the wires. It is not wht you expect, nd hs severl fces tht meet t surprising ngles. There is squre in the middle. This surfce hs minimum re mong surfces tht spn the cube. Exmple In Eq. (16.1), looking t the shortest distnce between two points in plne, I jumped to conclusion. To minimize the integrl b f(y )dx, use the Euler-Lgrnge differentil eqution: f y d dx f y = f (y )y = This seems to sy tht f(y ) is constnt or tht y =, implying either wy tht y = Ax + B, stright line. Now tht you know tht solutions cn hve kinks, you hve to look more closely. Tke the prticulr exmple f(y ) = αy 4 βy, with y() =, nd y() = b (16.45) One solution corresponds to y = nd y(x) = bx/. Cn there be others? Apply the conditions of Eq. (16.44) t some point between nd. Cll it x m, nd ssume tht the derivtive is not continuous. Cll the derivtives on the left nd right (y ) nd (y + ). The first eqution is p = L y = 4αy 3 βy, nd p(x m) = p(x + m) 4α(y ) 3 β(y ) = 4α(y + ) 3 β(y + ) [ (y ) (y + ) ][ (y + ) + (y + )(y ) + (y ) β/α ] =

19 If the slope is not continuous, the second fctor must vnish. 16 Clculus of Vritions 19 (y + ) + (y + )(y ) + (y ) β/α = This is one eqution for the two unknown slopes. For the second eqution, use the second condition, the one on H. H = y f y f, nd H(x m) = H(x + m) H = y [ 4αy 3 βy ] [ αy 4 βy ] = 3αy 4 βy [ (y ) (y + ) ][ (y + ) 3 + (y + ) (y ) + (y + )(y ) + (y ) 3 β ( (y + ) + (y ) ) /3α ] = Agin, if the slope is not continuous this is (y + ) 3 + (y + ) (y ) + (y + )(y ) + (y ) 3 β ( (y + ) + (y ) ) /3α = These re two equtions in the two unknown slopes. It looks messy, but look bck t H itself first. It s even. Tht mens tht its continuity will lwys hold if the slope simply chnges sign. Cn this work in the other (momentum) eqution? (y + ) = (y ) (y + ) + (y + )(y ) + (y ) β/α = is now (y + ) = β/α As long s α nd β hve the sme sign, this hs the solution (y + ) = ± β/α, (y ) = β/α (16.46) The boundry conditions on the originl problem were y() = nd y() = b. Denote γ = ± β/α, nd x 1 = / + b/γ, then y = { γx ( < x < x1 ) b γ(x ) (x 1 < x < b) b y 1 x (16.47) The pths lbeled, 1, nd re three solutions tht mke the vritionl functionl derivtive vnish. Which is smllest? Does tht nswer depend on the choice of the prmeters? See problem Are there ny other solutions? After ll, once you ve found three, you should wonder if it stops there. Yes, there re mny infinitely mny in this exmple. They re chrcterized by the sme slopes, ±γ, but they switch bck nd forth severl times before coming to the endpoint. The sme internl boundry conditions (p nd H) pply t ech corner, nd there s nothing in their solutions sying tht there is only one such kink. Do you encounter such weird behvior often, with n infinite number of solutions? No, but you see from this exmple tht it doesn t tke very complicted integrnd to produce such pthology.

20 16 Clculus of Vritions 16.1 Second Order Except for couple of problems in optics in chpter two,.35 nd.39, I ve mostly ignored the question bout minimum versus mximum. Does it mtter in clssicl mechnics whether the integrl, L dt is minimum or not in determining the equtions of motion? No. In geometric optics, does it mtter whether Fermt s principle of lest time for the pth of the light ry is relly minimum? Yes, in this cse it does, becuse it provides informtion bout the focus. In the clcultion of minimum energy electric potentils in cpcitor does it mtter? No, but only becuse it s lwys minimum. In problems in quntum mechnics similr to the electric potentil problem, the fct tht you re deling sometimes with minimum nd sometimes not leds to some serious technicl difficulties. How do you ddress this question? The sme wy you do in ordinry clculus: See wht hppens to the second order terms in your expnsions. Tke the sme generl form s before nd keep terms through second order. Assume tht the endpoints re fixed. b I[y] = dx F ( x, y(x), y (x) ) I = I[y + δy] I[y] b = dx F ( x, y(x) + δy(x), y (x) + δy (x) ) b dx F ( x, y(x), y (x) ) b [ F F = dx δy + y y δy + F ( ) F δy + y y y δyδy + F ( δy ) ] y (16.48) If the first two terms combine to zero, this sys the first derivtive is zero. Now for the next terms. Recll the similr question tht rose in section How cn you tell if function of two vribles hs minimum, mximum, or neither? The nswer required looking t the mtrix of ll the second derivtives of the function the Hessin. Now, insted of mtrix s in Eq. (8.31) you hve n integrl. ( ) ( ) fxx f d r, H d r = ( dx dy ) xy dx b f yx dx ( δy δy ) f yy dy ( Fyy F yy F y y F y y ) ( ) δy δy In the two dimensionl cse f = defines minimum if the product d r, H d r is positive for ll possible directions d r. For this new cse the directions re the possible functions δy nd its derivtive δy. The direction to look first is where δy is big. The reson is tht I cn hve very smll δy tht hs very big δy : 1 3 sin ( 1 6 x ). If I is to be positive in every direction, it hs to be positive in this one. Tht requires F y y >. y + δy δy y

21 16 Clculus of Vritions 1 Is it relly tht simple? No. First the δy terms cn be importnt too, nd second y cn itself hve severl components. Look t the ltter first. The finl term in Eq. (16.48) should be b F dx δy y mδy m y n n This set of prtil derivtives of F is t ech point long the curve Hessin. At ech point it hs set of eigenvlues nd eigenvectors, nd if ll long the pth ll the eigenvlues re lwys positive, it meets the first, necessry conditions for the originl functionl to be minimum. If you look t n exmple from mechnics, for which the independent vrible is time, these y n terms re then ẋ n insted. Terms such s these typiclly represent kinetic energy nd you expect tht to be positive. An exmple: T T S[ r ] = dt L(x, y, ẋ, ẏ, t) = dt 1[ẋ + ẏ + γtẋẏ x y ] This is the ction for prticle moving in two dimensions (x, y) with the specified Lgrngin. The eqution of motion re δs = x ẍ γ(tÿ + ẏ) = δx δs = y ÿ γ(tẍ + ẋ) = δy If γ = you hve two independent hrmonic oscilltors. The mtrix of derivtives of L with respect to ẋ = ẏ 1 nd ẏ = ẏ is ( ) L 1 γt = ẏ m ẏ n γt 1 ( ) ( ) 1 1 The eigenvlues of this mtrix re 1 ± γt, with corresponding eigenvectors nd. This 1 1 Hessin then sys tht S should be minimum up to the time t = 1/γ, but not fter tht. This is lso singulr point of the differentil equtions for x nd y. Focus When the Hessin mde from the δy terms hs only positive eigenvlues everywhere, the preceding nlysis might led you to believe tht the functionl is lwys minimum. Not so. Tht condition is necessry; it is not sufficient. It sys tht the functionl is minimum with respect to rpidly oscillting δy. It does not sy wht hppens if δy chnges grdully over the course of the integrl. If this hppens, nd if the length of the intervl of integrtion is long enough, the δy terms my be the smll ones nd the (δy) my then dominte over the whole length of the integrl. This is exctly wht hppens in the problems.35,.39, nd When this hppens in n opticl system, where the functionl T = dl/v is the trvel time long the pth, it signls something importnt. You hve focus. An ordinry light ry obeys Fermt s principle tht T is sttionry with respect to smll chnges in the pth. It is minimum if the pth is short enough. A focus occurs when light cn go from one point to nother by mny different pths, nd for still longer pths the pth is neither minimum nor mximum.

22 16 Clculus of Vritions In the integrl for T, where the strting point nd the ending point re the source nd imge points, the second order vrition will be zero for these long, grdul chnges in the pth. The stright-line pth through the center of the lens tkes lest time if its strting point nd ending point re closer thn this source nd imge. The sme pth will be sddle (neither mximum nor minimum) if the points re frther prt thn this. This sort of observtion led to the development of the mthemticl subject clled Morse Theory, topic tht hs hd pplictions in studying such diverse subjects s nucler fission nd the grvittionl lensing of light from qusrs. Thin Lens This provides simple wy to understnd the bsic eqution for thin lens. Let its thickness be t nd its rdius r. p q n r t Light tht psses through this lens long the stright line through the center moves more slowly s it psses through the thickness of the lens, nd tkes time T 1 = 1 c (p + q t) + n c t Light tht tke longer pth through the edge of the lens encounters no glss long the wy, nd it tkes time T = 1 [ p c + r + ] q + r If p nd q represent the positions of source nd the position of its imge t focus, these two times should be equl. At lest they should be equl in the pproximtion tht the lens is thin nd when you keep terms only to the second order in the vrition of the pth. T = 1 [ ] p 1 + r c /p + q 1 + r /q = 1 [p ( 1 + r /p ) + q ( 1 + r /q )] c Equte T 1 nd T. (p + q t) + nt = [p ( 1 + r /p ) + q ( 1 + r /q )] (n 1)t = r p + r q 1 p + 1 (n 1)t = q r = 1 (16.49) f This is the stndrd eqution describing the focusing properties of thin lenses s described in every elementry physics text tht even mentions lenses. The focl length of the lens is then f = r /(n 1)t. Tht is not the expression you usully see, but it is the sme. See problem Notice tht this eqution for the focus pplies whether the lens is double convex or plno-convex or meniscus:. If you llow the thickness t to be negtive (equivlent to sying tht there s n extr time dely t the edges insted of in the center), then this result still works for diverging lens, though the nlysis leding up to it requires more thought.

23 16 Clculus of Vritions 3 Exercises 1 For the functionl F [x] = x() + π dt ( x(t) + ẋ(t) ) nd the function x(t) = 1 + t, evlute F [x]. For the functionl F [x] = 1 dt x(t) with the boundry conditions x() = nd x(1) = 1, wht is the minimum vlue of F nd wht function x gives it? Strt by drwing grphs of vrious x tht stisfy these boundry conditions. Is there ny reson to require tht x be continuous function of t? 3 With the functionl F of the preceding exercise, wht is the functionl derivtive δf/δx?

24 16 Clculus of Vritions 4 Problems 16.1 You re ner the edge of lke nd see someone in the wter needing help. Wht pth do you tke to get there in the shortest time? You cn run t speed v 1 on the shore nd swim t probbly slower speed v in the wter. Assume tht the edge of the wter forms stright line, nd express your result in wy tht s esy to interpret, not s the solution to some qurtic eqution. Ans: Snell s Lw. 16. The cycloid is the locus of point tht is on the edge of circle tht is itself rolling long stright line pebble cught in the tred of tire. Use the ngle of rottion s prmeter nd find the prmetric equtions for x(θ) nd y(θ) describing this curve. Show tht it is Eq. (16.17) In Eq. (16.17), describing the shortest-time slide of prticle, wht is the behvior of the function for y? In figuring out the series expnsion of w = cos 1 (1 t), you my find it useful to tke the cosine of both sides. Then you should be ble to find tht the two lowest order terms in this expnsion re w = t t 3/ /1. You will need both terms. Ans: x = y 3 / / The dimensions of n ordinry derivtive such s dx/dt is the quotient of the dimensions of the numertor nd the denomintor (here L/T). Does the sme sttement pply to the functionl derivtive? 16.5 Use Fermt s principle to derive both Snell s lw nd the lw of reflection t plne surfce. Assume two stright line segments from the strting point to the ending point nd minimize the totl trvel time of the light. The drwing pplies to Snell s lw, nd you cn compute the trvel time of the light s function of the coordinte x t which the light hits the surfce nd enters the higher index medium. h 1 x L n h 16.6 Anlyze the pth of light over rodwy strting from Eq. (16.3) but using x s the independent vrible insted of y () Fill in the steps leding to Eq. (16.31). And do you understnd the point of the rerrngements tht I did just preceding it? Also, cn you explin why the form of the function Eq. (16.3) should hve been obvious without solving ny extr boundry conditions? (b) When you cn explin tht in few words, then wht generl cubic polynomil cn you use to get still better result? 16.8 For the function F (x, y, y ) = x + y + y, explicitly crry through the mnipultions leding to Eq. (16.41) Use the explicit vrition in Eq. (16.8) nd find the minimum of tht function of ɛ. Compre tht minimum to the vlue found in Eq. (16.11). Ans: Do either of the functions, Eqs. (16.9) or (16.3), stisfy Lplce s eqution? For the function F (x, y, y ) = x + y + y, repet the clcultion of δi only now keep ll the higher order terms in δy nd δy. Show tht the solution Eq. (16.11) is minimum Use the techniques leding to Eq. (16.1) in order to solve the brchistochrone problem Eq. (16.13) gin. This time use x s the independent vrible insted of y.