Euler Euler Everywhere Using the EulerLagrange Equation to Solve Calculus of Variation Problems


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1 Euler Euler Everywhere Using the EulerLgrnge Eqution to Solve Clculus of Vrition Problems Jenine Smllwood Principles of Anlysis Professor Flschk My 12,
2 1. Introduction Clculus of vritions is brnch of the more generl theory of clculus of functionls which dels specificlly with optimizing functionls. In the lte 1600s, John Bernoulli posed the brchistochrone problem, which mrks the beginnings of clculus of vritions. Although such types of problems hd been considered by the ncient Greeks, the problems hd only been solved geometriclly, rther thn nlyticlly. The question tht John Bernoulli introduced in 1696 ws s follows: Given two points A nd B in verticl plne, to find the pth AMB down which movble point M, by virtue of its weight, proceed from A to B in the shortest possible time. 1 The problem ws solved by John Bernoulli, Jmes Bernoulli, Newton, L Hospitl nd Leibniz. Besides the brchistochrone, other stndrd problems include geodesics, minimizing surfces of revolution, nd isoperimetric problems. All of these problems, including the brchistochrone, employ common technique for finding the optiml solution. Ech cn be modelled by J[y] = f(x, y, y ) dx where J : C 1 R nd y C 1 is sought to optimize J. The problem cn then be trnslted into differentil eqution, clled the EulerLgrnge eqution, nd the problem reduces to one which cn be solved using techniques of ordinry differentil equtions. 2. Theoreticl Bckground In the presenttion of some introductory problems of clculus of vritions, we will be considering functionls from C 1 to R of the form J[y] = f(x, y, y ) dx. In order for these functionls to be continuous, we need to use the Sobolev norm on the spce, which is defined to be: y = mx y(x) + mx x b x b y (x). We use this norm so tht two functions will be close if both the functions nd their derivtives re close in the stndrd supnorm. Using the norm y = mx x b y(x) does not stisfy the requirement for continuity for functionls of this form. 1 Goldstine p. 30 2
3 To begin tlking bout necessry conditions for when functionl hs mximum or minimum, we need to hve definition of derivtive for functionl. Definition 1: F is differentible t u if there exists bounded liner mp d u F : C 1 R such tht F (u + h) F (u) d u F (h) = ( h ). 2 Definition 2: Let J[y] be functionl defined on normed liner spce nd let J[h] = J[y + h] J[y] be its increment. Suppose J[h] = φ[h] + ε h where φ[h] is liner functionl nd ε 0 s h 0. Then the principl liner prt of the increment J[h] is clled the vrition of J[h] nd is denoted by δj[h]. 3 Theorem 1: A necessry condition for the differentible functionl J[y] to hve n extremum t y = ŷ is tht its vrition vnish for y = ŷ, i.e., tht δj[h] = 0 for y = ŷ nd ll h stisfying the constrints of the vritionl problem. Proof: By contrdiction. Suppose J[y] hs minimum t ŷ, but there exists h 0 such tht δj[h 0 ] 0. We hve tht J[h] = δj[h] + ε h where ε 0 s h 0. Since ε > 0, nd h > 0, for sufficiently smll h, the sign of J[h] will be the sme s δj[h]. Since ŷ is minimum, this implies tht J[h] 0 for ll h sufficiently smll. So suppose without loss of generlity tht δj[h 0 ] > 0. Then for ny α > 0, since δj is liner, δj[ αh 0 ] = αδj[h 0 ] < 0 which implies tht J[ αh] < 0 which is contrdiction. Since ŷ is minimum, J[h] 0 for ll sufficiently smll h. 4 2 Leg 3 Gelfnd p Gelfnd p. 13 3
4 This leds us to the question, given tht there exists continuous, twicedifferentible function which minimizes the functionl J[h] = f(x, y, y )dx, wht is the differentil eqution which this function must stisfy? This differentil eqution is known s the EulerLgrnge eqution. Theorem 2: Let J[y] be functionl of the form F (x, y, y ) dx, defined on the set of functions y(x) which hve continuous first derivtives in [,b] nd stisfy the boundry conditions y() = A nd y(b) = B. Then necessry condition for J[y] to hve n extremum for given function y = y(x) is tht y stisfy the EulerLgrnge eqution: F y d dx F y = 0 Proof: Let y(x) be the function for which J[y] = F (x, y, y ) dx hs wek extremum. We increment y(x) by h(x), where h() = h(b) = 0 in order for y(x) + h(x) to stisfy the given boundry conditions. Then define J J[y + h] J[y] = = F (x, y + h, y + h ) dx F (x, y, y ) dx [F (x, y + h, y + h ) F (x, y, y )] dx. Using Tylor s theorem we hve: F (x, y + h, y + h ) = F (x, y, y ) + hf y (x, y, y ) + h F y (x, y, y ) + h.o.t from which it follows tht J = = [F (x, y, y ) + hf y (x, y, y ) + h F y (x, y, y ) F (x, y, y )] dx [hf y (x, y, y ) + h F y (x, y, y )] dx. 4
5 We now define Φ[h] [hf y (x, y, y ) + h F y (x, y, y )] dx nd the higher order terms s ε h, so J = Φ[h] + ε h. Since ε h 0 s h 0, this will give us Φ[h] = [hf y(x, y, y ) + h F y (x, y, y )] dx s the vrition of J. Denote the vrition s δj = [hf y (x, y, y ) + h F y (x, y, y )] dx. By Theorem 1, necessry condition for J[y] to hve n extremum for y = y(x) is tht δj = Integrting by prts, we hve (hf y + h F y ) dx = 0 δj = = But h() = h(b) = 0 (hf y + h F y ) dx = 0 hf y dx + F y h b d dx (F y ) dx = 0 = δj = h(f y d dx (f y )) dx = 0 Since this is true for ll h F y d (F dx y ) = 0, which is the EulerLgrnge eqution. 5 The solution of the EulerLgrnge eqution pplied to ny of the bovementioned introductory problems is the twicedifferentible function tht is minimum of the integrl in question. The EulerLgrnge eqution cn be generlized to functionls depending on higherorder derivtives s well s to functionls tht hve more thn one independent vrible. 5 Gelfnd p.14 5
6 3. Bckground on some Introductory Problems There re four clssicl types of problems tht re solved using the clculus of vritions. While the brchistochrone mrked the beginning of interest in the subject, the other three clssicl problems were vritions on the originl brchistochrone problem. The study of geodesics rises from the problem of finding the rc with the shortest possible length lying on the surfce of sphere tht connects two given points on the surfce. In generl, the rc of minimum length connecting two points on given sufce is clled the geodesic for the surfce. 6 On the surfce of sphere, the result is the intersection of the sphere with the plne contining the center of the sphere nd the two given points the gret circle rc. In finding the geodesic of circulr cylinder r = ( cos φ, sin φ, z), the result is z = c 1 φ + c 2, which is twoprmeter fmily of helicl cones lying on the cylinder. 7 Minimum surfce of revolution problems cn generlly be stted s follows: Given two fixed points (x 1, y 1 ), (x 2, y 2 ), we would like to pss n rc through them whose rottion bout the xxis genertes surfce of revolution whose re included in x 1 x x 2 is minimum. The result in this cse is usully ctenry. In solving this problem, we find fmily of ctenries tht pss through one of the points nd stisfies the equtions: (x ) y = b cosh b y 1 = b cosh (x 1 ). b If the point (x 2, y 2 ) is outside the envelope of the fmily, no ctenry will stisfy the endpoint conditions nd the only solution will be the Goldschmidt discontinuous solution, which is defined by 8 x = x 1 0 y y 1 y = 0 x 1 x x 2 x = x 2 0 y y 2. Isoperimetric problems re clss of problems tht hve dded restrictions on the functions besides continuity nd endpoint conditions. A problem would be stted in form such s: 6 Weinstock p Gelfnd p Weinstock p. 30 6
7 Find the curve y = y(x) for which the functionl J[y] = F (x, y, y ) dx hs n extremum, where the dmissible curves stisfy the boundry condtions y() = A, y(b) = B nd re such tht nother functionl K[y] = G(x, y, y ) dx tkes fixed vlue l. 9 The best known problem of this clss is tht of finding the closed curve of given perimeter for which the re is mximum, the solution being circle. 4. The Brchistochrone We begin some exmples of the use of the EulerLgrnge eqution with clssicl clculus of vritions problem, the brchistrochrone. Severl solutions to the brchistochrone problem were put forth, most notbly by Newton, Leibniz, L Hospitl, nd the Bernoulli brothers. John Bernoulli, who initilly posed the problem, gve two solutions, one which used Fermt s principle of lest time. The two problems re similr in tht Fermt showed tht ry of light moves through medium on pth to rech boundry in the shortest time possible. Bernoulli used this ide by dividing up the spce between the two fixed endpoints nd ssigning n index of refrction inversely proportionl to the prticle s velocity in tht lyer. Then s the prticle moves through ech slice, he used Snell s Lw to determine how the prticle bent s it crossed the interfce. 10 The method used below ws not vilble to Bernoulli t the time when he introduced the brchistochrone problem. Restting the originl problem, we hve: Given two points (x 1, y 1 ), (x 2, y 2 ), find the curve tht tkes the lest time to go between these points under the influence of grvity. We ssume tht prticle is descending the curve y(x) with n initil velocity v i. If we let s be the rclength of the curve y(x), then the velocity 9 Gelfnd p Goldstine p. 39 7
8 v of the prticle is ds. This gives us: dt So we hve ds dt = ds dx = 1 + (y ) 2 = ds = 1 + (y ) 2 dx v = ds dt = t = = x2 x 1 x2 x 1 ds v 1 + (y ) 2 dx v Since the prticle is moving down frictionless surfce, energy is conserved, implying tht 1 2 mv2 1 2 mv2 i = mg(y y 1 ) where m is the mss of the prticle nd g is the ccelertion due to grvity. This gives us Letting k = v2 i, we hve 2g v 2 = 2g(y y 1 + v2 i 2g ) v = = t = 2g(y y 1 + k) 1 2g x2 x (y ) 2 y y1 + k dx To solve this, we cn now use the EulerLgrnge eqution. We hve functionl of the form J[y] = F (y, y ) dx. Since our integrnd does not depend on x, we cn derive specil cse of the EulerLgrnge eqution. Since F is only function of y nd y, = F y d dx F y = F y F y yy F y y y = 0. Multiplying through by y, nd dding nd subtrcting F y y, we hve y F y + y F y (y ) 2 F y y (y )y F y y y F y = 0 8
9 = d dx (F y F y ) = 0 = F y F y = C In our problem, the integrnd is F (y, y 1+(y ) = ) y y 2, nd so the Euler 1 +k Lgrnge eqution will be F y F y = If we let C = 1/ 2, then we hve 1 + (y ) 2 (y y 1 + k) (y ) 2 = C (y y 1 + k) 1 + (y ) 2 1 = (y y 1 + k) 1 + (y ) = (y ) 2 = 2 (y y 1 + k) y y 1 + k y y1 + k = x = 2 (y y 1 + k) dy. Letting y y 1 + k = 2 sin 2 θ 2 dy = 2 sin θ 2 cos θ 2 dθ gives us 2 sin 2 θ 2 x = 2 2 cos 2 θ sin θ 2 cos θ 2 dθ 2 = 2 sin 2 θ 2 dθ = x = (1 cosθ) dθ = x = (θ sin θ) + C 2 So we hve prmeterized eqution for the curve given by y = (y 1 k) + (1 cos θ) x = C 2 + (θ sin θ) 9
10 which is the prmetriztion of cycloid, generted by the motion of fixed point on the circumference of circle of rdius Isoperimetric Problems In this section, we will look t nother clss of problems, clled isoperimetric problems, which involve optimizing functionl with n dded constrint. Recll tht the originl isoperimetric problem seeks to find curve which encloses the lrgest re given curves of prescribed length. The problem tht we will discuss now is tht of the hnging rope. In this problem, the gol is to find the shpe of flexible rope of uniform density which hngs t rest with its endpoints fixed. Solutions to this, nd other isoperimetric problems, cn be obtined using the method of Lgrnge multipliers. Given functionl I = F (x, y, y ) dx we wish to constrin it by nother functionl J = G(x, y, y ) dx. Define I = I + λjwhichimpliesi = F = F + λg. Using the sme method to find the EulerLgrnge eqution s before, it cn be shown tht for the constrined problem, the EulerLgrnge eqution will be Fy d dx F y = 0. To find our rope of interest, we ssume tht the rope is hnging in the verticl plne. We let y = y(x) be rope with the designted boundry conditions nd prescribed length, nd σ be the constnt mss per unit length of the rope. Then the potentil energy of portion of the rope hving length ds will be gyσds, where g is the ccelertion due to grvity. This implies tht the totl potentil energy of the rope is l x2 σgyds = σg y 1 + (y ) 2 dx, 0 x 1 where (x 1, y 1 ), (x 2, y 2 ) re the fixed endpoints. The constrint in this problem is tht J = x 2 x (y ) 2 dx, i.e. the rclength of the rope hs length l. Using Lgrnge multipliers, we re now interested in optimizing I = σgy 1 + (y ) 2 dx + λ 1 + (y ) 2 dx ( = σg 1 + (y ) 2 y + λ ) dx. σg 11 Weinstock p
11 The EulerLgrnge eqution for this problem will then be σg 1 + (y ) 2 d ( σgy y + λ ) = 0. dx 1 + (y ) 2 σg Since F is independent of x, we cn write the EulerLgrnge eqution s y ( y + λ ) ( σg 1 + (y σg ) 2 y + λ ) σg = C 1 ( = σg y + λ ) (y ) 2 σg 1 + (y ) 1 + (y ) 2 2 = C 1 = (σgy + λ) 1 = C (y ) 2 σgy 1 + (y ) 2 Solving for y, we hve (y ) 2 = 1 (σgy + λ) 2 1 C1 2 dy = x = C 1 (σgy + λ) 2 1 = C 1 σg ln σgy + λ + (σgy + λ) C 2 = C 1 σg cosh 1 (σgy + λ) + C 2 σgy + λ 1 Writing y in terms of x we hve y = 1 ( ) σg(x σg cosh C2 ) λ C 1 σg. We cn see tht the shpe of the hnging rope tht we re interested in is ctenry with verticl xis. By pplying ny given boundry conditions nd knowing the length of the rope we cn determine the constnts λ, C 1, nd C 2. 11
12 6. Conclusion There re mny vritions to the problems which hve been discussed. The brchistochrone problem cn be restted s vrible endpoint problem, where we look for curve where prticle reches verticl line in the lest mount of time. For the isoperimetric problems, number of finite subsidry conditions cn be dded for one specific functionl to be extremized. Although ll of the problems presented here hve only involved optimizing functionls tht result in function with continuous first nd second derivtives, the methods cn esily be extended to problems which involve solutions with higher order derivtives. The problems cn lso be generlized to involve optimizing functionls tht depend on n continuously differentible rguments. In ech of these cses, we cn derive the pproprite EulerLgrnge eqution for which the extreml function must stisfy. All of the problems discussed bove hve smooth solutions s the extremizing function. However, mny vritionl problems hve no smooth solutions, nd for these problems we must expnd the clss of dmissible functions to piecewise smooth curves. The EulerLgrnge eqution must then be stisfied on ech piecewise intervl. The EulerLgrnge eqution is tough cookie nd is centrl to mny of the problems of clculus of vritions. It is remrkble tht so mny problems from so mny different settings ultimtely boil down to solving this sme eqution. But then, the unifying power of the EulerLgrnge eqution is probbly why mthemticins re still using this eqution more thn two hundred yers fter its conception. 12
13 Bibliogrphy [1] I. M. Gelfnd nd S. V. Fomin. Clculus of Vritions. PrenticeHll, Inc., Englewood Cliffs, New Jersey, [2] Hermn H. Goldstine. A History of the Clculus of Vritions from the 17th Through the 19th Century. SpringerVerlg, New York, [3] Joceline Leg. Lecture notes, April [4] Robert Weinstock. Clculus of Vritions. Dover Publictions, Inc., New York,
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