Notes on the Calculus of Variations and Optimization. Preliminary Lecture Notes

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1 Notes on the Clculus of Vritions nd Optimiztion Preliminry Lecture Notes Adolfo J. Rumbos c Drft dte November 14, 17

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3 Contents 1 Prefce 5 Vritionl Problems 7.1 Miniml Surfces Linerized Miniml Surfce Eqution Vibrting String Indirect Methods Geodesics in the plne Fundmentl Lemms The Euler Lgrnge Equtions Convex Minimiztion Gâteux Differentibility A Minimiztion Problem Convex Functionls Convex Minimiztion Theorem Optimiztion with Constrints Queen Dido s Problem Euler Lgrnge Multiplier Theorem An Isoperimetric Problem A Some Inequlities 87 A.1 The Cuchy Schwrz Inequlity B Theorems About Integrtion 89 B.1 Differentiting Under the Integrl Sign B. The Divergence Theorem C Continuity of Functionls 93 C.1 Definition of Continuity

4 4 CONTENTS

5 Chpter 1 Prefce This course is n introduction to the Clculus of Vritions nd its pplictions to the theory of differentil equtions, in prticulr, boundry vlue problems. The clculus of vritions is subject s old s the Clculus of Newton nd Leibniz. It rose out of the necessity of looking t physicl problems in which n optiml solution is sought; e.g., which configurtions of molecules, or pths of prticles, will minimize physicl quntity like the energy or the ction? Problems like these re known s vritionl problems. Since its beginnings, the clculus of vritions hs been intimtely connected with the theory of differentil equtions; in prticulr, the theory of boundry vlue problems. Sometimes vritionl problem leds to differentil eqution tht cn be solved, nd this gives the desired optiml solution. On the other hnd, vritionl methods cn be successfully used to find solutions of otherwise intrctble problems in nonliner prtil differentil equtions. This interply between the theory of boundry vlue problems for differentil equtions nd the clculus of vritions will be one of the mjor themes in the course. We begin the course with n exmple involving surfces tht spn wire loop in spce. Out of ll such surfces, we would like to find, if possible, the one tht hs the smllest possible surfce re. If such surfce exists, we cll it mimiml surfce. This exmple will serve to motivte lrge portion of wht we will be doing in this course. The miniml surfce problem is n exmple of vritionl problem. In vritionl problem, out of clss of functions (e.g., functions whose grphs in three dimensionl spce yield surfce spnning given loop) we seek to find one tht optimizes (minimizes or mximizes) certinty quntity (e.g., the surfce re of the surfce). There re two pproches to solving this kind of problems: the direct pproch nd the indirect pproch. In the direct pproch, we try to find minimizer or mximizer of the quntity, in some cses, by considering sequences of functions for which the quntity under study pproches mximum or minimum, nd then extrcting subsequence of the functions tht converge in some sense to the sought fter optiml solution. In the indirect method of the Clculus of Vritions, which ws developed first 5

6 6 CHAPTER 1. PREFACE historiclly, we first find necessry conditions for given function to be n optimizer for the quntity. In cses in which we ssume tht functions in the clss under study re differentible, these conditions, sometimes, come in the form of differentil equtions, or system of differentil equtions, tht the functions must stisfy, in conjunction with some boundry conditions. This process leds to boundry vlue problem. If the boundry vlue problem cn be solved, we cn obtin cndidte for n optimizer of the quntity ( criticl point ). The next step in the process is to show tht the given cndidte is n optimizer. This cn be done, in some cses, by estblishing some sufficient conditions for function to be n optimizer. The indirect method in the Clculus of Vritions is reminiscent of the optimiztion procedure tht we first lern in first single vrible Clculus course. Conversely, some clsses of boundry vlue problems hve prticulr structure in which solutions re optimizers (minimizers, mximizers, or, in generl, criticl points ) of certin quntity over clss of functions. Thus, these differentil equtions problems cn, in theory, be solved by finding optimizers of certin quntity. In some cses, the existence of optimizers cn be chieved by direct method in the Clculus of Vritions. This provides n pproch, known s the vritionl pproch in the theory of differentil equtions.

7 Chpter Exmples of Vritionl Problems.1 Miniml Surfces Imgine you tke twisted wire loop, s tht pictured in Figure.1.1, nd dip it into sop solution. When you pull it out of the solution, sop film spnning the wire loop develops. We re interested in understnding the mthemticl properties of the film, which cn be modeled by smooth surfce in three z y x Ω Figure.1.1: Wire Loop 7

8 8 CHAPTER. VARIATIONAL PROBLEMS dimensionl spce. Specificlly, the shpe of the sop film spnning the wire loop, cn be modeled by the grph of smooth function, u: Ω R, defined on the closure of bounded region, Ω, in the xy plne with smooth boundry Ω. The physicl explntion for the shpe of the sop film relies on the vritionl principle tht sttes tht, t equilibrium, the configurtion of the film must be such tht the energy ssocited with the surfce tension in the film must be the lowest possible. Since the energy ssocited with surfce tension in the film is proportionl to the re of the surfce, it follows from the lest energy principle tht sop film must minimize the re; in other words, the sop film spnning the wire loop must hve the shpe of smooth surfce in spce contining the wire loop with the property tht it hs the smllest possible re mong ll smooth surfces tht spn the wire loop. In this section we will develop mthemticl formultion of this vritionl problem. The wire loop cn be modeled by the curve determined by the set of points: (x, y, g(x, y)), for (x, y) Ω, where Ω is the smooth boundry of bounded open region Ω in the xy plne (see Figure.1.1), nd g is given function defined in neighborhood of Ω, which is ssumed to be continuous. A surfce, S, spnning the wire loop cn be modeled by the imge of C 1 mp given by Φ: Ω R 3 Φ(x, y) = (x, y, u(x, u)), for ll x Ω, (.1) where Ω = Ω R is the closure of Ω, nd u: Ω R is function tht is ssumed to be C in Ω nd continuous on Ω; we write u C (Ω) C(Ω). Let A g denote the collection of functions u C (Ω) C(Ω) stisfying u(x, y) = g(x, y), for ll (x, y) Ω; tht is, A g = {u C (Ω) C(Ω) u = g on Ω}. (.) Next, we see how to compute the re of the surfce S u = Φ(Ω), where Φ is the mp given in (.1) for u A g, where A g is the clss of functions defined in (.). The grid lines x = c nd y = d, for rbitrry constnts c nd d, re mpped by the prmetriztion Φ into curves in the surfce S u given by y Φ(c, y)

9 .1. MINIMAL SURFACES 9 nd x Φ(x, d), respectively. The tngent vectors to these pths re given by ( Φ y =, 1, u ) y nd respectively. The quntity Φ x = (.3) ( 1,, u ), (.4) x Φ x Φ y x y (.5) gives n pproximtion to the re of portion of the surfce S u tht results from mpping the rectngle [x, x + x] [y, y + y] in the region Ω to the surfce S u by mens of the prmetriztion Φ given in (.1). Adding up ll the contributions in (.5), while refining the grid, yields the following formul for the re S u : re(s u ) = Φ x Φ y dxdy. (.6) Ω Using the definitions of the tngent vectors Φ x nd Φ y in (.3) nd (.4), respectively, we obtin tht ( Φ x Φ y = u ) x, u y, 1, so tht or Φ x Φ y = 1 + ( ) u + x Φ x Φ y = 1 + u, ( ) u, y where u denotes the Eucliden norm of u. We cn therefore write (.6) s re(s u ) = 1 + u dxdy. (.7) The formul in (.7) llows us to define mp by A(u) = Ω Ω A: A g R 1 + u dxdy, for ll u A g, (.8) which gives the re of the surfce prmetrized by the mp Φ: Ω R 3 given in (.1) for u A g. We will refer to the mp A: A g R defined in (.8) s the re functionl. With the new nottion we cn restte the vritionl problem of this section s follows:

10 1 CHAPTER. VARIATIONAL PROBLEMS Problem.1.1 (Vritionl Problem 1). Out of ll functions in A g, find one such tht A(u) A(v), for ll v A g. (.9) Tht is, find function in A g tht minimizes the re functionl in the clss A g. Problem.1.1 is n instnce of wht hs been known s Plteu s problem in the Clculus of Vritions. The mthemticl question surrounding Pteu s problem ws first formulted by Euler nd Lgrnge round 176. In the middle of the 19 th century, the Belgin physicist Joseph Plteu conducted experiments with sop films tht led him to the conjecture tht sop films tht form round wire loops re of miniml surfce re. It ws not until 1931 tht the Americn mthemticin Jesse Dougls nd the Hungrin mthemticin Tibor Rdó, independently, cme up with the first mthemticl proofs for the existence of miniml surfces. In this section we will derive necessry condition for the existence of solution to Problem.1.1, which is expressed in terms of prtil differentil eqution (PDE) tht u A g must stisfy, the miniml surfce eqution. Suppose we hve found solution, u A g, of Problem.1.1 in u A g. Let ϕ: Ω R by C function with compct support in Ω; we write ϕ Cc (Ω) (we show construction of such function in the Appendix). It then follows tht u + tϕ A g, for ll t R, (.1) since ϕ vnishes in neighborhood of Ω nd therefore u + tϕ = g on Ω. It follows from (.1) nd (.9) tht A(u) A(u + tϕ), for ll t R. (.11) Consequently, the function f : R R defined by f(t) = A(u + tϕ), for ll t R, (.1) hs minimum t, by virtue of (.1) nd (.1). It follows from this observtion tht, if f is differentible t, then f () =. (.13) We will see next tht, since we re ssuming tht u C (R) C(Ω) nd ϕ Cc (Ω), f is indeed differentible. To see why this is the cse, use (.1) nd (.8) to compute f(t) = 1 + (u + tϕ) dxdy, for ll t R, (.14) Ω where (u + tϕ) = u + t ϕ, for ll t R,

11 .1. MINIMAL SURFACES 11 by the linerity of the differentil opertor. It then follows tht (u + tϕ) = ( u + t ϕ) ( u + t ϕ) = u u + t u ϕ + t ϕ u + t ϕ ϕ = u + t u ϕ + t ϕ, so tht, substituting into (.14), f(t) = 1 + u + t u ϕ + t ϕ dxdy, for ll t R. (.15) Ω Since the integrnd in (.15) is C 1, we cn differentite under the integrl sign (see Appendix) to get f u ϕ + t ϕ (t) = dxdy, (.16) 1 + u + t u ϕ + t ϕ Ω for ll t R. Thus, f is differentible nd, substituting for t in (.16), f u ϕ () = dxdy. (.17) 1 + u Ω Hence, if u is minimizer of the re functionl in A g, it follows from (.1) nd (.17) tht u ϕ dxdy =, for ll ϕ 1 + u C c (Ω). (.18) Ω The sttement in (.18) provides necessry condition for the existence of minimizer of the re functionl in A g. We will next see how (.18) gives rise to PDE tht u C (Ω) C(Ω) must stisfy in order for it to be minimizer of the re functionl in A g. First, we integrte by prts (see Appendix) in (.18) to get Ω ( ) u u n ϕ dxdy + ϕ ds =, (.19) 1 + u Ω 1 + u for ll ϕ Cc (Ω), where the second integrl in (.19) is pth integrl round the boundry of Ω. Since ϕ Cc (Ω) vnishes in neighborhood of the boundry of R, it follows from (.19) tht ( ) u ϕ dxdy =, for ll ϕ C 1 + u c (Ω). (.) Ω By virtue of the ssumption tht u is C functions, it follows tht the divergence term of the integrnd (.) is continuous on Ω, it follows from the

12 1 CHAPTER. VARIATIONAL PROBLEMS sttement in (.) tht ( ) u =, in Ω. (.1) 1 + u The eqution in (.1) is second order nonliner PDE known s the miniml surfce eqution. It provides necessry condition for function u C (Ω) C(Ω) to be minimizer of the re functionl in A g. Since, we re lso ssuming tht u A g, we get tht u must solve the boundry vlue problem (BVP): ( ) u 1 + u = in Ω; u = g on Ω. (.) The BVP in (.) is clled the Dirichlet problem for the miniml surfce eqution. The PDE in (.1) cn lso be written s (1 + u y)u xx u x u y u xy + (1 + u x)u yy =, in Ω, (.3) where the subscripted symbols red s follows: u x = u x, u y = u y, nd u xx = u x, u yy = u y, u xy = u y x = u x y = u yx. (.4) The fct tht the mixed second prtil derivtives in (.4) re equl follows from the ssumption tht u is C function. The eqution in (.3) is nonliner, second order, elliptic PDE.. The Linerized Miniml Surfce Eqution For the cse in which the wire loop in the previous section is very close to horizontl plne (see Figure..), it is resonble to ssume tht, if u A g, u is very smll throughout R. We cn therefore use the liner pproximtion t 1 + t, for smll t, (.5)

13 .. LINEARIZED MINIMAL SURFACE EQUATION 13 z y x R Figure..: Almost Plnr Wire Loop to pproximte the re function in (.8) by A(u) [1 + 1 ] u dxdy, for ll u A g, so tht Ω A(u) re(ω) + 1 Ω u dxdy, for ll u A g. (.6) The integrl on the right hnd side of the expression in (.6) is known s the Dirichlet Integrl. We will use it in these notes to define the Dirichlet functionl, D : A g R, D(u) = 1 u dxdy, for ll u A g. (.7) Thus, in view of (.6) nd (.7), Ω A(u) re(ω) + D(u), for ll u A g. (.8) Thus, ccording to (.8), for wire loops close to horizontl plne, miniml surfces spnning the wire loop cn be pproximted by solutions to the following vritionl problem, Problem..1 (Vritionl Problem ). Out of ll functions in A g, find one such tht D(u) D(v), for ll v A g. (.9)

14 14 CHAPTER. VARIATIONAL PROBLEMS It cn be shown tht necessry condition for u A g to be solution to the Vritionl Problem..1 is tht u solves the boundry vlue problem u = in Ω; (.3) u = g on Ω, where u = u xx + u yy, the two dimensionl Lplcin. The BVP in (.3) is clled the Dirichlet Problem for Lplce s eqution..3 Vibrting String Consider string of length L (imgine guitr string or violin string) whose ends re locted t x = nd x = L long the x xis (see Figure.3.3). We ssume tht the string is mde of some mteril of (liner) density ρ(x) (in units of mss per unit length). Assume tht the string is fixed t the end points nd L x Figure.3.3: String of Length L t Equilibrium is tightly stretched so tht there is constnt tension, τ, cting tngentilly long the string t ll times. We would like to model wht hppens to the string fter it is plucked to configurtion like tht pictured in Figure.3.4 nd then relesed. We ssume tht the shpe of the plucked string is described L x Figure.3.4: Plucked String of Length L by continuous function, f, of x, for x [, L]. At ny time t, the shpe of the string is described by function, u, of x nd t; so tht u(x, t) gives the verticl displcement of point in the string locted t x when the string is in the equilibrium position pictured in Figure.3.3, nd t time t. We then hve tht u(x, ) = f(x), for ll x [, L]. (.31) In ddition to the initil condition in (.31), we will lso prescribe the initil speed of the string, u (x, ) = g(x), for ll x [, L], (.3) t

15 .3. VIBRATING STRING 15 where g is continuous function of x; for instnce, if the plucked string is relesed from rest, then g(x) = for ll x [, L]. We lso hve the boundry conditions, u(, t) = u(l, t) =, for ll t, (.33) which model the ssumption tht the ends of the string do not move. The question we would like to nswer is: Given the initil conditions in (.31) nd (.3), nd the boundry conditions in (.33), cn we determine the shpe of the string, u(x, t), for ll x [, L] nd ll times t >? We will nswer this questions in subsequent chpter in these notes. In this section, though, we will derive necessry condition in the form of PDE tht u must stisfy in order for it to describe the motion of the vibrting string. In order to find the PDE governing the motion of the string, we will formulte the problem s vritionl problem. We will use Hmilton s Principle in Mechnics, or the Principle of Lest Action. This principle sttes tht the the pth tht configurtions of mechnicl system tke from time t = to t = T is such tht quntity clled the ction is minimized (or optimized) long the pth. The ction is defined by A = T [K(t) V (t)] dt, (.34) where K(t) denotes the kinetic energy of the system t time t, nd V (t) its potentil energy t time t. For the cse of string whose motion is described by smll verticl displcements u(x, t), for ll x [, L] nd ll times t, the kinetic energy is given by K(t) = 1 L ( ) u ρ(x) (x, t) dx. (.35) t To see how (.35) comes bout, note tht the kinetic energy of prticle of mss m is K = 1 mv, where v is the speed of the prticle. Thus, for smll element of the string whose projection on the x xis is the intervl [x, x+ x], so tht its pproximte length is x, the kinetic energy is, pproximtely, K 1 ( ) u ρ(x) (x, t). (.36) t Thus, dding up the kinetic energies in (.36) over ll elements of the string dding in length to L, nd letting x, yields the expression in (.35), which we rewrite s K(t) = 1 L ρu t dx, for ll t, (.37) where u t denotes the prtil derivtive of u with respect to t.

16 16 CHAPTER. VARIATIONAL PROBLEMS In order compute the potentil energy of the string, we compute the work done by the tension, τ, long the string in stretching the string from its equilibrium length of L, to the length t time t given by L 1 + u x dx; (.38) so tht [ ] L V (t) = τ 1 + u x dx L, for ll t. (.39) Since we re considering smll verticl displcements of the string, we cn linerize the expression in in (.38) by mens of the liner pproximtion in (.5) to get L L 1 + u x dx [1 + 1 u x] dx = L + 1 L 1 u x dx, so tht, substituting into (.39), V (t) 1 L τu x dx, for ll t. (.4) Thus, in view of, we consider the problem of optimizing the quntity A(u) = T L [ 1 ρu t 1 ] τu x dxdt, (.41) where we hve substitute the expressions for K(t) nd V (t) in (.37) nd (.4), respectively, into the expression for the ction in (.34). We will use the expression for the ction in (.41) to the define functionl in the clss of functions A defines s follows: Let R = (, L) (, T ), the crtesin product of the open intervls (, L) nd (, T ). Then, R is n open rectngle in the xt plne. We sy tht u A if u C (R) C(R), nd u stisfies the initil conditions in (.31) nd (.3), nd the boundry conditions in (.33). Then, the ction functionl, A: A R, is defined by the expression in (.41), so tht A(u) = 1 R [ ρu t τu ] x dxdt, for u A. (.4) Next, for ϕ Cc (R), note tht u + sϕ A, since ϕ hs compct support in R, nd therefore ϕ nd ll its derivtives re on R. We cn then define rel vlued function h: R R by h(s) = A(u + sϕ), for s R, (.43)

17 .3. VIBRATING STRING 17 Using the definition of the functionl A in (.4), we cn rewrite h(s) in (.43) s h(s) = 1 [ ρ[(u + sϕ)t ] τ[(u + sϕ) x ] ] dxdt so tht = 1 R R [ ρ[ut + sϕ t ] τ[u x + sϕ x ] ] dxdt, h(s) = A(u) + s [ρu t ϕ t τu x ϕ x ] dxdt + s A(ϕ), (.44) R for s R, where we hve used the definition of the ction functionl in (.4). It follows from (.44) tht h is differentible nd h (s) = [ρu t ϕ t τu x ϕ x ] dxdt + sa(ϕ), for s R. (.45) R The principle of lest ction implies tht, if u describes the shpe of the string, then s = must be c criticl point of h. Hence, h () = nd (.45) implies tht [ρu t ϕ t τu x ϕ x ] dxdt =, for ϕ Cc (R), (.46) R is necessry condition for u(x, t) to describe the shpe of vibrting string for ll times t. Next, we use the integrtion by prts formuls R R ψ ϕ x dxdt = R ψϕn 1 ds R ψ x ϕ dxdt, for C 1 functions ψ nd ϕ, where n 1 is the first component of the outwrd unit norml, n, on R (wherever this vector is defined), nd ψ ϕ t dxdt = ψ ψϕn ds t ϕ dxdt, R where n is the second component of the outwrd unit norml, n, (see Problem 1 in Assignment #8), to obtin ρu t ϕ t dxdt = ρu t ϕn ds R R R t [ρu t]ϕ dxdt, so tht since ϕ hs compct support in R. Similrly, R R R ρu t ϕ t dxdt = R t [ρu t]ϕ dxdt, (.47) τu x ϕ x dxdt = R x [τu x]ϕ dxdt. (.48)

18 18 CHAPTER. VARIATIONAL PROBLEMS Next, substitute the results in (.47) nd (.48) into (.46) to get [ t [ρu t] ] x [τu x] ϕ dxdt =, for ϕ Cc (R). (.49) R Thus, pplying the Fundmentl Lemm of the Clculus of Vritions (see the next chpter in these notes), we obtin from (.49) tht ρ u t τ u =, in R, (.5) x since we re ssuming tht u is C, ρ is continuous function of x, nd τ is constnt. The PDE in (.5) is clled the one dimensionl wve eqution. It is sometimes written s u t = τ u ρ x, or where u t = c u x, (.51) c = τ ρ, where the cse in which ρ is ssumed to be constnt. The wve eqution in (.5) or (.51) is second order, liner, hyperbolic PDE.

19 Chpter 3 Indirect Methods in the Clculus of Vritions We begin this chpter by discussing very simple problem in the Clculus of Vritions: Given two points in the plne, find smooth pth connecting those two points tht hs the shortest length. A solution of this problem is clled geodesic curve between the two points. This exmple is simple becuse we know the nswer to this question from Eucliden geometry. Nevertheless, the solutions tht we present here serve to illustrte both the direct nd indirect methods in the clculus of vrition. It will certinly be good introduction to the indirect methods. 3.1 Geodesics in the plne Let P nd Q denote two points in the xy plne with coordintes (x o, y o ) nd (x 1, y 1 ), respectively. We consider the clss, A, of smooth pths tht connect P to Q. One of theses pths is shown in Figure We ssume tht pths re given prmetriclly by the pir of functions (x(s), y(s)), for s [, 1], where x: [, 1] R nd y : [, 1] R re differentible functions with continuous derivtives in some oven intervl tht contins [, 1], such tht (x(), y()) = (x o, y o ) nd (x(1), y(1)) = (x 1, y 1 ). We my write this more succinctly s A = {(x, y) C 1 ([, 1], R ) (x(), y()) = P nd (x(1), y(1)) = Q}. (3.1) We define functionl, J : A R, by J(x, y) = 1 (x (s)) + (y (s)) ds, for ll (x, y) A. (3.) 19

20 CHAPTER 3. INDIRECT METHODS y Q P x o x 1 x Figure 3.1.1: Pth connecting P nd Q Thus, J(x, y) is the rc length long the pth from P to Q prmetrized by (x(s), y(s)), for s [, 1]. We would like to solve the following vritionl problem: Problem (Geodesic Problem 1). Out of ll pths in A, find one, (x, y), such tht J(x, y) J(u, v), for ll (u, v) A (3.3) Observe tht the expression for J in (3.) cn be written s J(x, y) = 1 (x (s), y (s)) ds, for ll (x, y) A, (3.4) where (, ) in the integrnd in (3.4) denotes the Eucliden norm in R. We will first use this exmple to illustrte the direct method in the Clculus of Vritions. We begin by showing tht the functionl J defined in (3.4) is bounded below in A by Q P, or (x 1, y 1 ) (x o, y o ) = (x 1 x o ) + (y 1 y o ), the Eucliden distnce from P to Q; tht is Q P J(u, v), for ll (u, v) A. (3.5) Indeed, it follows from the Fundmentl Theorem of Clculus tht for ny (u, v) A, or (u(1), v(1)) (u(), v()) = Q P = 1 1 (u (s), v (s)) ds, (u (s), v (s)) ds, for ny (u, v) A. (3.6)

21 3.1. GEODESICS IN THE PLANE 1 Now, using the fct tht Q P = (Q P ) (Q P ), the dot product (or Eucliden inner product) of Q P with itself, we obtin from (3.6) tht or Q P = (Q P ) Q P = 1 1 (u (s), v (s)) ds, for ny (u, v) A, (Q P ) (u (s), v (s)) ds, for ny (u, v) A. (3.7) Thus, pplying the Cuchy Schwrz inequlity to the integrnd of the integrl on the right hnd side of (3.7), we get tht or or Q P 1 Q P Q P Q P (u (s), v (s)) ds, for ny (u, v) A, 1 (u (s), v (s)) ds, for ny (u, v) A, Q P Q P J(u, v), for ny (u, v) A. (3.8) For the cse in which P Q, we see tht the estimte in (3.5) follows from the inequlity in (3.8). Now, it follows from (3.5) tht the functionl J is bounded from below in A by Q P. Hence, the infimum of J over A exists nd Q P inf J(u, v). (3.9) (u,v) A Next, we see tht the infimum in (3.9) is ttined nd is Q P. Indeed, let (x(s), y(s)) = P + s(q P ), for s [, 1], (3.1) the stright line segment connecting P to Q. Note tht (x(), y()) = P nd (x(1), y(1)) = Q, nd (x, y) is differentible pth with (x (s), y (s)) = Q P, for ll s [, 1]. (3.11) Thus, (x, y) belongs to the clss A defined in (3.1). Using the definition of the functionl J in (3.4) nd the fct in (3.11), compute J(x, y) = 1 Consequently, we get from (3.9) tht Q P ds = Q P. inf J(u, v) = Q P. (3.1) (u,v) A

22 CHAPTER 3. INDIRECT METHODS Furthermore, the infimum in (3.1) is ttined on the pth given in (3.1). This is in ccord with the notion from elementry Eucliden geometry tht the shortest distnce between two points is ttined long stright line segment connecting the points. Next, we illustrte the indirect method in the Clculus of Vritions, which is the min topic of this chpter. We consider the specil prmetriztion (x(s), y(s)) = (s, y(s)), for x o s x 1, where y : [x o, x 1 ] R is continuous function tht is differentible, with continuous derivtive, in n open intervl tht contins [x o, x 1 ]. Here we re ssuming tht x o < x 1, y(x o ) = y o nd y(x 1 ) = y 1. Thus, the pth connecting P nd Q is the grph of differentible function over the intervl [x 1, x ]. This is illustrted in Figure We ll hve to define the clss A nd the functionl J : A R in different wy. Put nd A = {y C 1 ([x o, x 1 ], R) y(x o ) = y o nd y(x 1 ) = y 1 }, (3.13) J(y) = x1 x o 1 + (y (s)) ds, for ll y A. (3.14) We would like to solve the vritionl problem: Problem 3.1. (Geodesic Problem ). Out of ll functions in A, find one, y, such tht J(y) J(v), for ll v A (3.15) In the indirect method, we ssume tht we hve solution of the optimiztion problem, nd then deduce conditions tht this function must stisfy; in other words, we find necessry condition for function in competing clss to be solution. Sometimes, the necessry conditions cn be used to find cndidte for solution of the optimiztion problem ( criticl point ). The next step in the indirect method is to verify tht the cndidte indeed solves the optimiztion problem. Thus, ssume tht there is function y A tht solves Geodesic Problem ; tht is, y stisfy the estimtes in (3.15). Next, let η : [x o, x 1 ] R denote C 1 function such tht η(x o ) = nd η(x 1 ) = (we will see lter in these notes how this function η cn be constructed). It then follows from the definition of A in (3.13) tht y + tη A, for ll t R.

23 3.1. GEODESICS IN THE PLANE 3 It then follows from (3.15) tht Next, define f : R R by J(y) J(y + tη), for ll t R. (3.16) f(t) = J(y + tη), for ll t R. (3.17) It follows from (3.16) nd the definition of f in (3.17) tht f hs minimum t ; tht is f(t) f(), for ll t R. Thus, if it cn be shown tht the function f defined in (3.17) is differentible, we get the necessry condition which cn be written in terms of J s Next, we proceed to show tht the function f(t) = J(y + tη) = f () =, (3.18) d dt [J(y + tη)] t= =. (3.19) x1 x o 1 + (y (s) + tη (s)) ds, for t R, is differentible function of t. Observe tht we cn write f s f(t) = x1 x o 1 + (y (s)) + ty (s)η (s) + t (η (s)) ds, for t R. (3.) To see show tht f is differentible, we need to see tht we cn differentite the expression on the right hnd side of (3.) under the integrl sign. This follows from the fct tht the prtil derivtive of the integrnd on the right hnd side of 3., 1 + (y (s)) t + ty (s)η (s) + t (η (s)), or y (s)η (s) + t(η (s)) 1 + (y (s)) + ty (s)η (s) + t (η (s)), for t, y nd s, for s in some open intervl contining [x o, x 1 ]. It then follows from the results in Appendix B.1 tht f given in (3.) is differentible nd f (t) = x1 y (s)η (s) + t(η (s)) ds, for t R, (3.1) x o 1 + (y (s)) + ty (s)η (s) + t (η (s)) (see Proposition B.1.1). Evluting the expression for f (t) in (3.1) t t =, we obtin tht x1 f y (s)η (s) () = ds. (3.) 1 + (y (s)) x o

24 4 CHAPTER 3. INDIRECT METHODS Thus, in view of (3.18) nd (3.), we see tht necessry condition for y A, where A is given in (3.13), to be minimizer of the functionl J : A R given in (3.14), is tht where x1 x o y (s) 1 + (y (s)) η (s) ds =, for ll η C 1 o ([x o, x 1 ], R), (3.3) C 1 o ([x o, x 1 ], R) = {η C 1 ([x o, x 1 ], R) η(x o ) = nd η(x 1 ) = }, (3.4) the clss of C 1, rel vlued functions in [x o, x 1 ] tht vnish t the end points of the intervl [x o, x 1 ]. We will see in the next section tht if the condition in (3.3) holds true for every η C 1 o ([x o, x 1 ], R), then y (s) 1 + (y (s)) = c 1, for ll s [x o, x 1 ], (3.5) where c 1 is constnt (see the second fundmentl lemm in the Clculus of Vritions, Lemm 3..8 on pge 9 in these notes). Now, squring on both sides of (3.5), (y (s)) 1 + (y (s)) = c 1, for ll s [x o, x 1 ]. (3.6) It follows from (3.6) tht c 1 1 (otherwise we would conclude tht 1 =, which is impossible). Hence, we cn solve (3.6) for (y (s)) to obtin from which we conclude tht (y (s)) = c 1 c, for ll s [x o, x 1 ], (3.7) 1 y (s) = c, for ll s [x o, x 1 ], (3.8) where c is constnt. We cn solve the differentil eqution in (3.8) to obtin the generl solution y(s) = c s + c 3, for ll s [x o, x 1 ], (3.9) where c 1 nd c is constnts. Since we re lso ssuming tht y A, where A is given in (3.13), it follows tht y must stisfy the boundry conditions y(x o ) = y o nd y(x 1 ) = y 1. We therefore get the following system of equtions tht c nd c 3 must solve { c x o + c 3 = y o ; c x 1 + c 3 = y 1,

25 3.. FUNDAMENTAL LEMMAS 5 or { x o c + c 3 = y o ; x 1 c + c 3 = y 1. (3.3) Solving the system in (3.3) for c nd c 3 yields c = y 1 y o x 1 x o nd c 3 = x 1y o x o y 1 x 1 x o. Thus, using the expression for y in (3.9), y(s) = y 1 y o x 1 x o s + x 1y o x o y 1 x 1 x o. (3.31) Note tht the expression in (3.31) is the eqution of stright line tht goes through the points (x o, y o ) nd (x 1, y 1 ). Thus, we hve shown tht cndidte for minimizer of the rc length functionl J defined in (3.14) over the clss given in (3.13) is stright line segment connecting the point P to the point Q. It remins to show tht the function y in (3.31) is in indeed minimizer of J in A, nd tht it is the only minimizer of J in A. This will be done in subsequent section in these notes. 3. Fundmentl Lemms in the Clculus of Vritions In the previous section we found necessry condition for function y A, where A = {y C 1 ([x o, x 1 ], R) y(x o ) = y o nd y(x 1 ) = y 1 }, (3.3) to be minimizer of the rc length functionl J : A R, J(y) = x1 x o 1 + (y (x)) dx, for y C 1 ([x o, x 1 ], R), (3.33) over the clss A. We found tht, if y A is minimizer of J over the clss A, then y must stisfy the condition x1 x o y (x) 1 + (y (x)) η (s) dx =, for ll η C 1 o ([x o, x 1 ], R). (3.34) This is necessry condition for y A to be minimizer of the functionl J : A R defined in (3.33). We then invoked fundmentl lemm in the Clculus of Vritions to deduce tht the condition in (3.34) implies tht y (x) 1 + (y (x)) = c 1, for ll x [x o, x 1 ], (3.35)

26 6 CHAPTER 3. INDIRECT METHODS nd for some constnt c 1. This is lso necessry condition for y A being minimizer of the functionl JA R defined in (3.33), where A given in given in (3.3). We will see in this section tht the differentil eqution in (3.35) follows from the condition in (3.34) provided tht the function y 1 + (y ) is known to be continuous. This is the cse, for instnce, if y is ssumed to come from certin clsses of differentible functions defined on closed nd bounded intervl. We ll strt by defining these clsses of functions. Let, b R, nd ssume tht < b. Definition The clss C([, b], R) consists of rel vlued functions, f : [, b] R, defined on the closed nd bounded intervl [, b], nd which re ssumed to be continuous on [, b]. It cn be shown tht C([, b], R) is liner spce (or vector) spce in which the opertions re point wise ddition (f + g)(x) = f(x) + g(x), for ll x [, b], nd ll f, g C([, b], R), nd sclr multipliction (cf)(x) = cf(x), for ll x [, b], ll c R nd ll f C([, b], R). Definition 3... The clss C o ([, b], R) consists of ll functions in C([, b], R) tht vnish t nd b; in symbols, C o ([, b], R) = {f C([, b], R) f() = nd f(b) = }. We note tht C o ([, b], R) is liner subspce of C([, b], R). Definition The clss C 1 ([, b], R) consists of ll functions f : U R tht re differentible in n open intervl U tht contins [, b] nd such tht f is continuous on [, b]. We note tht C 1 ([, b], R) is liner subspce of C([, b], R). Definition The clss C 1 o ([, b], R) consists of ll functions f C 1 ([, b], R) tht vnish t the end points of the intervl [, b]; thus, C 1 o ([, b], R) = {f C 1 ([, b], R) f() = nd f(b) = }. We begin by stting nd proving the following bsic lemm for the clss C([, b], R).

27 3.. FUNDAMENTAL LEMMAS 7 Lemm 3..5 (Bsic Lemm 1). Let f C([, b], R) nd ssume tht f(x) for ll x [, b]. Suppose tht Then, f(x) = for ll x [, b]. f(x) dx =. Proof: Suppose tht f C([, b], R), f(x) for ll x [, b], nd f(x) dx =. (3.36) Assume, by wy of contrdiction, tht there exists x o (, b) with f(x o ) >. Then, since f is continuous t x o, there exists δ > such tht (x o δ, x o + δ) (, b) nd x (x o δ, x o + δ) f(x) f(x o ) < f(x o). (3.37) Now, using the tringle inequlity, we obtin the estimte f(x o ) f(x) f(x o ) + f(x) ; so tht, in view of (3.37) nd the ssumption tht f is nonnegtive on [, b], f(x o ) < f(x o) from which we get tht It follows from (3.38) tht + f(x), for x o δ < x < x o + δ, f(x) > f(x o), for x o δ < x < x o + δ. (3.38) xo+δ x o δ f(x) dx > xo+δ x o δ f(x o ) Thus, since we re ssuming tht f on [, b], f(x) dx xo+δ x o δ dx = δf(x o ). f(x) dx > δf(x o ) >, which is in direct contrdiction with the ssumption in (3.36). Consequently, f(x) = for ll x (, b). By the continuity of f on [, b], we lso get tht f(x) = for ll x [, b], nd the proof of the lemm is now complete. Lemm 3..6 (Bsic Lemm ). Let f C([, b], R) nd ssume tht d c f(x) dx =, Then, f(x) = for ll x [, b]. for every (c, d) (, b).

28 8 CHAPTER 3. INDIRECT METHODS Proof: Suppose tht f C([, b], R) nd d c f(x) dx =, for every (c, d) (, b). (3.39) Arguing by contrdiction, ssume tht f(x o ) for some x o (, b). Without loss of generlity, we my ssume tht f(x o ) >. Then, by the continuity of f, there exists δ > such tht (x o δ, x o + δ) (, b) nd or x (x o δ, x o + δ) f(x) f(x o ) < f(x o), x (x o δ, x o + δ) f(x o ) f(x o) from which we get tht It follows from (3.4) tht < f(x) < f(x o ) + f(x o), f(x) > f(x o), for x o δ < x < x o + δ. (3.4) xo+δ x o δ f(x) dx > xo+δ x o δ f(x o ) dx = δf(x o ) >, which is in direct contrdiction with (3.39). Hence, it must be the cse tht f(x) = for < x < b. It then follows from the continuity of f tht f(x) = for ll x [, b]. Next, we stte nd prove the first fundmentl lemm of the Clculus of Vritions. A version of this result is presented s Bsic Lemm in Section 3 1 in [Wei74]. Lemm 3..7 (Fundmentl Lemm 1). Let G C([, b], R) nd ssume tht G(x)η(x) dx =, Then, G(x) = for ll x [, b]. Proof: Let G C([, b], R) nd suppose tht for every η C o ([, b], R). G(x)η(x) dx =, for every η C o ([, b], R). (3.41) Arguing by contrdiction, ssume tht there exists x o (, b) such tht G(x o ). Without loss of generlity, we my lso ssume tht G(x o ) >. Then, since G is continuous on [, b], there exists δ > such tht (x o δ, x o + δ) (, b) such tht G(x) > G(x o), for x o δ < x < x o + δ. (3.4)

29 3.. FUNDAMENTAL LEMMAS 9 Put x 1 = x o δ nd x = x o + δ nd define η : [, b] R by, if x x 1 ; η(x) = (x x 1 )(x x), if x 1 < x x ;, if x < x b. (3.43) Note tht η is continuous on [, b] nd tht η() = η(b) = ; so tht, η C o ([, b], R). Observe lso, from the definition of η in (3.43) tht η(x) >, for x 1 < x < x. (3.44) It lso follows from the definition of η in 3.43 tht G(x)η(x) dx = so tht, in view of (3.44) nd (3.4), x x 1 G(x)η(x) dx > G(x o) G(x)η(x) dx; x x 1 η(x) dx >, which is in direct contrdiction with the ssumption in (3.41). Consequently, G(x) = for ll x (, b). The continuity of G on [, b] then implies tht G(x) = for ll x [, b]. The following result is the one tht we used in the exmple resented in the previous section. We shll refer to it s the second fundmentl lemm in the Clculus of Vritions. Lemm 3..8 (Fundmentl Lemm ). Let G C([, b], R) nd ssume tht G(x)η (x) dx =, for every η C 1 o ([, b], R). Then, G(x) = c for ll x [, b], where c is constnt. Proof: Let G C([, b], R) nd ssume tht Put G(x)η (x) dx =, for every η C 1 o ([, b], R). (3.45) c = 1 b the verge vlue of G over [, b]. Define η : [, b] R by η(x) = x G(x) dx, (3.46) (G(t) c) dt, for x [, b]. (3.47)

30 3 CHAPTER 3. INDIRECT METHODS Then, η is differentible function, by virtue of the Fundmentl Theorem of Clculus, with η (x) = G(x) c, for x [, b], (3.48) which defines continuous function on [, b]. Consequently, η C 1 ([, b], R). Observe lso, from (3.47) tht nd η(b) = η() = (G(t) c) dt = (G(t) c) dt =, G(t) dt c dt =, in view of the definition of c in (3.46). It then follows tht η C 1 o ([, b], R). Next, compute (G(x) c) dx = = where we hve used (3.48); so tht, (G(x) c)(g(x) c) dx (G(x) c)η (x) dx, (G(x) c) dx = G(x)η (x) dx c η (x). Thus, using the ssumption in (3.45) nd the Fundmentl Theorem of Clculus, (G(x) c) dx =, (3.49) since η Co 1 ([, b], R). It follows from (3.49) nd the Bsic Lemm 1 (Lemm 3..5) tht G(x) = c for ll x [, b], since (G c) is [, b]. The following lemm combines the results of the first nd the second fundmentl lemms in the Clculus of Vritions. Lemm 3..9 (Fundmentl Lemm 3). Let f : [, b] R nd g : [, b] R be continuous rel vlues functions defined on [, b]. Assume tht [f(x)η(x) + g(x)η (x)] dx =, Then, g is differentible in (, b) nd for every η C 1 o ([, b], R). d [g(x)] = f(x), dx for ll x (, b).

31 3.3. THE EULER LAGRANGE EQUATIONS 31 Proof: Let f C([, b], R), g C([, b], R), nd ssume tht Put [f(x)η(x) + g(x)η (x)] dx =, for every η C 1 o ([, b], R). (3.5) F (x) = x f(t) dt, for x [, b]. (3.51) Then, by the Fundmentl Theorem of Clculus, F is differentible in (, b) nd F (x) = f(x), for x (, b). (3.5) Next, let η C 1 o ([, b], R) nd use integrtion by prts to compute f(x)η(x) dx = f(x)η(x) b = F (x)η (x) dx, F (x)η (x) dx since η() = η(b) = ; consequently, we cn rewrite the condition in (3.5) s [g(x) F (x)]η (x) dx =, for every η C 1 o ([, b], R). (3.53) We cn then pply the second fundmentl lemm (Lemm 3..8) to obtin from (3.53) tht g(x) F (x) = C, for ll x [, b] nd some constnt C, from which we get tht g(x) = F (x) + C, for ll x [, b]. (3.54) It follows from (3.54), (3.51) nd the Fundmentl Theorem of Clculus tht g is differentible with derivtive g (x) = F (x), for ll x (, b); so tht g (x) = f(x) for ll x (, b), in view of (3.5). 3.3 The Euler Lgrnge Equtions In the previous two sections we sw how the second fundmentl lemm in the Clculus of Vritions (Lemm 3..8) cn be used to obtin the differentil eqution y 1 + (y ) = c 1, (3.55)

32 3 CHAPTER 3. INDIRECT METHODS where c 1 is constnt, s necessry condition for function y C 1 ([x o, x 1 ], R) to be minimizer of the rc length functionl, J : C 1 ([x o, x 1 ], R) R: J(y) = x1 over the clss of functions x o 1 + (y (x)) dx, for C 1 ([x o, x 1 ], R), (3.56) A = {y : C 1 ([x o, x 1 ], R) y(x o ) = y o nd y(x 1 ) = y 1 }. (3.57) The differentil eqution in (3.55) nd the boundry conditions y(x o ) = y o nd y(x 1 ) = y 1, defining the clss A in (3.57), constitute boundry vlue problem. In Section 3.1 we were ble to solve this boundry vlue problem to obtin the solution in (3.31), stright line segment from the point (x o, y o ) to the point (x 1, y 1 ). This is cndidte for minimizer of the rc length functionl J given in (3.56) over the clss A in (3.57). In this section we illustrte the procedure employed in the Sections 3.1 nd 3. in the cse of generl functionl J : C 1 ([, b], R) R of the form J(y) = F (x, y(x), y (x)) dx, for y C 1 ([, b], R), (3.58) where F : [, b] R R R is continuous function of three vribles. An exmple of function F in the integrnd in (3.58) with vlue F (x, y, z) is F (x, y, z) = 1 + z, for ll (x, y, z) [x o, x 1 ] R R, which is used in the definition of the rc length functionl J given in (3.56). In the following exmple we provide nother exmple of F (x, y, z) tht comes up in the celebrted brchistochrone problem, or the curve of shortest descent time. A version of this problem is lso discussed on pge 19 of [Wei74] (note tht is tht version of the problem, the positive y xis points downwrds, while in the version discussed here, it points upwrds s shown in Figure 3.3.). Exmple (Brchistochrone Problem). Given points P nd Q in verticl plne, with P higher tht Q nd to the left of Q (see Figure 3.3.), find the curve connecting P to Q long which prticle of mss m descends from P to Q in the shortest possible time, ssuming tht only the force of grvity is cting on the prticle. The sketch in Figure 3.3. shows possible curve of descent from P to Q. Observe lso from the figure tht we hve ssigned coordintes (, y o ) to P nd (x 1, y 1 ) to Q, where x 1 > nd y o > y 1. We ssume tht the pth from P to Q is the grph of C 1 function y : [, x 1 ] R such tht y() = y o nd y(x 1 ) = y 1.

33 3.3. THE EULER LAGRANGE EQUATIONS 33 y y o P y 1 Q (x1, y 1 ) x 1 x Figure 3.3.: Pth descending from P to Q The rc-length long the pth from the point P to ny point on the pth, s function of x, is then given by so tht, s(x) = x 1 + (y (t)) dt, for x x 1 ; s (x) = 1 + (y (x)), for < x < x 1, (3.59) by the Fundmentl Theorem of Clculus. The speed, v, of the prticle long the pth t ny point on the curve is given by v = ds dt. (3.6) We cn use (3.59) nd (3.6) to obtin formul for the descent time, T, of the prticle: x1 s (x) T = dx, v or x1 1 + (y (x)) T = dx. (3.61) v It thus remins to compute v in the denomintor of the integrnd in (3.61). The speed of the prticle will depend on the loction of the prticle long the pth. If we ssume tht the prticle is relesed from rest, the v = t P. To find the speed t other points on the pth, we will use the lw of conservtion of energy, which sys tht the totl mechnicl energy of the system is conserved; tht is, the totl energy remins constnt throughout the motion. The totl energy of this prticulr system is the sum of the kinetic energy nd the potentil energy of the prticle of mss m. Totl Energy = Kinetic Energy + Potentil Energy.

34 34 CHAPTER 3. INDIRECT METHODS At P the prticle is t rest, so while its potentil energy is Kinetic Energy t P =, Potentil Energy t P = mgy o, where g is the grvittionl ccelertion. Thus, Totl Energy t P = mgy o. At ny point (x, y(x)) on the pth the totl energy is Totl Energy t (x, y) = 1 mv + mgy. Thus, the lw of conservtion of energy implies tht or fter cncelling m; so tht, from which we get tht 1 mv + mgy = mgy o, 1 v + gy = gy o, v = g(y o y), v = g(y o y). (3.6) This gives us n expression for v s function of y. Note tht we need to ssume tht ll the pths connecting P to Q under considertion must stisfy y(x) < y o for ll < x < x 1. Substituting the expression for v in (3.6) into the denomintor of the integrnd in (3.61), we obtin the expression for the time of descent x1 1 + (y (x)) T (y) = dx, (3.63) g(yo y(x)) for y C 1 ([, x 1 ], R) in the clss A = {y C 1 ([, x 1 ], R) y() = y o, y(x 1 ) = y 1, nd y(x) < y o for < x < x 1 }. (3.64) We would like to minimize the time of descent functionl in (3.63) over the clss of functions A defined in (3.64). Note tht, if y A is minimizer of the functionl T given in (3.63), then y is lso minimizer of x1 1 + (y (x)) gt (y) = dx, for y A. yo y(x) Thus, we will seek minimizer of the functionl J : A R given by x1 1 + (y (x)) J(y) = dx, for y A. (3.65) yo y(x)

35 3.3. THE EULER LAGRANGE EQUATIONS 35 The functionl J derived in Exmple (the Brchistochrone Problem), corresponds to function F : [, x 1 ] (, y o ) R R defined by F (x, y, z) = 1 + z yo y, for (x, y, z) [, x 1] (, y o ) R, in the generl functionl J given in (3.58). We will see mny more exmples of this generl clss of functionls in these notes nd in the homework ssignments. The generl vritionl problem we would like to consider in this section is the following: Problem 3.3. (Generl Vritionl Problem 1). Given rel numbers nd b such tht < b, let F : [, b] R R R denote continuous function of three vribles, (x, y, z), with x [, b], nd y nd z in the set of rel numbers (in some cses, s in the Brchistochrone problem, we might need to restrict the vlues of y nd z s well). Define the functionl J : C 1 ([, b], R) R by J(y) = F (x, y(x), y (x)) dx, for y C 1 ([, b], R), (3.66) For rel numbers y o nd y 1, consider the clss of functions If possible, find y A such tht A = {y C 1 ([, b], R) y() = y o, y(b) = y 1 }. (3.67) J(y) J(v), for ll v A, (3.68) or J(y) J(v), for ll v A. (3.69) Thus, in Problem 3.3. we seek minimum, in the cse of (3.68), of the functionl J in (3.66) over the clss A in (3.67) ( minimiztion problem), or we seek mximum, in the cse of (3.69), of J over A ( mximiztion problem). In generl, we cll either problem (3.68) or (3.69) n optimiztion problem. We will see in these notes tht, in order to nswer the questions posed in the Generl Vritionl Problem 1, we need to impose dditionl conditions on the function F. We will see wht those conditions re s we ttempt to solve the problem. Suppose tht we know priori tht the functionl J defined in (3.66) is bounded from below in A given in (3.67). Thus, it mkes sense to sk whether there exists y A t which J is minimized; tht is, is there y A for which (3.68) holds true? We begin by ssuming tht this is the cse; tht is, there exists y A for which J(y) J(v), for ll v A. (3.7)

36 36 CHAPTER 3. INDIRECT METHODS As in one of our solutions of the geodesic problem presented in Section 3.1, we next seek to find necessry conditions for y A to be minimizer of J defined in (3.66) over the clss A given in (3.68). Let η : [, b] R denote C 1 function such tht η() = nd η(b) = ; so tht η C 1 o ([, b], R). It then follows from the definition of A in (3.67) tht Thus, we get from (3.7) tht Define g : R R by y + tη A, for ll t R. J(y) J(y + tη), for ll t R. (3.71) g(t) = J(y + tη), for ll t R. (3.7) It follows from (3.7) nd the definition of g in (3.7) tht g hs minimum t ; tht is g(t) g(), for ll t R. Thus, if it cn be shown tht the function g defined in (3.7) is differentible, we get the necessry condition which cn be written in terms of J s g () =, d dt [J(y + tη)] t= =. (3.73) Hence, in order to obtin the necessry condition in (3.73), we need to mke sure tht the mp t J(y + tη), for t R, (3.74) is differentible function of t t t =, where, ccording to (3.66), J(y + tη) = F (x, y(x) + tη(x), y (x) + tη (x)) dx, for t R. (3.75) According to Proposition B.1.1 in Appendix B.1 in these notes, the question of differentibility of the mp in (3.74) reduces to the continuity of the prtil derivtives y [F (x, y, z)] = F y(x, y, z) nd z [F (x, y, z)] = F z(x, y, z). Thus, in ddition to being continuous, we will ssume tht F hs continuous prtil derivtives with respect to y nd with respect to z, F y nd F z, respectively. Mking these dditionl ssumptions, we cn pply Proposition B.1.1 to the expression in (3.75) to obtin, using the Chin Rule s well, d [J(y + tη)] = dt [F y (x, y + tη, y + tη )η + F z (x, y + tη, y + tη )η ] dx, (3.76)

37 3.3. THE EULER LAGRANGE EQUATIONS 37 for ll t R, where we hve written y for y(x), y for y (x), η for η(x), nd η for η (x) in the integrnd of the integrl on the right hnd side of (3.76). Substituting for t in (3.76) we then obtin tht d dt [J(y + tη)] t= = [F y (x, y, y )η + F z (x, y, y )η ] dx. Thus, the necessry condition in (3.73) for y A to be minimizer of J in A, for the cse in which F is continuous with continuous prtil derivtives F y nd F z, is tht [F y (x, y, y )η + F z (x, y, y )η ] dx =, for ll η C 1 o ([, b], R). (3.77) Since we re ssuming tht F y nd F z re continuous, we cn pply the third fundmentl lemm in the Clculus of Vritions (Lemm 3..9) to obtin from (3.77) tht the mp x F z (x, y(x), y (x)), for x [, b], is differentible for ll x (, b) nd d dx [F z(x, y(x), y (x))] = F y (x, y(x), y (x)), for ll x (, b). (3.78) The differentil eqution in (3.78) is clled the Euler Lgrnge eqution ssocited with the functionl J defined in (3.66). It gives necessry condition for function y A to be n optimizer of J over the clss A given in (3.67). We restte this fct, long with the ssumptions on F, in the following proposition. Proposition (Euler Lgrnge Eqution). Let, b R be such tht < b nd let F : [, b] R R be continuous function of three vribles (x, y, x) [, b] R R with continuous prtil derivtives with respect to y nd with respect to z, F y nd F z, respectively. Define J : C 1 ([, b], R) R by J(y) = For rel numbers y o nd y 1, define F (x, y(x), y (x)) dx, for y C 1 ([, b], R). (3.79) A = {y C 1 ([, b], R) y() = y o, y(b) = y 1 }. (3.8) A necessry condition for y A to be minimizer, or mximizer, of J over the clss A is tht y solves the Euler Lgrnge eqution d dx [F z(x, y(x), y (x))] = F y (x, y(x), y (x)), for ll x (, b). (3.81)

38 38 CHAPTER 3. INDIRECT METHODS In the indirect method of the Clculus of Vritions, s it pplies to the Generl Vritionl Problem 1 (Problem 3.3.), we first seek for function y A, where A is given in (3.8), tht solves the Euler Lgrnge eqution in (3.81). This leds to the two point boundry vlue problem d dx [F z(x, y(x), y (x))] = F y (x, y(x), y (x)), y() = y o, y(b) = y 1. for ll x (, b); (3.8) A solution of the boundry vlue problem in (3.8) will be cndidte for minimizer, or mximizer, of the functionl J in (3.79) over the clss A given in (3.8). The second step in the indirect method is to verify tht the cndidte is minimizer or mximizer. In the reminder of this section we give exmples of the boundry vlue problems in (3.8) involving the Euler Lgrnge eqution. In subsequent sections we will see how to verify tht solution of the boundry vlue problem in (3.8) yields minimizer for lrge clss of problems. Exmple (Geodesic Problem, Revisited). In Problem 3.1. (The Geodesic Problem ), we looked t the problem of minimizing the functionl, given by J(y) = over the clss In this cse, x1 J : C 1 ([x o, x 1 ], R) R, x o 1 + (y (x)) dx, for ll y C 1 ([x o, x 1 ], R), (3.83) A = {y C 1 ([x o, x 1 ], R) y(x o ) = y o, y(x 1 ) = y 1 }. (3.84) F (x, y, z) = 1 + z, for (x, y, z) [x o, x 1 ] R R. So tht, F y (x, y, z) = nd F z (x, y, z) = z 1 + z, for (x, y, z) [x o, x 1 ] R R. The Euler Lgrnge eqution ssocited with the functionl in (3.83) is then [ ] d y (x) =, for x (x o, x 1 ). dx 1 + (y (x)) Integrting this eqution yields y (x) 1 + (y (x)) = c 1, for x (x o, x 1 ),

39 3.3. THE EULER LAGRANGE EQUATIONS 39 for some constnt of integrtion c 1. This is the sme eqution in (3.5) tht we obtined in the solution of Geodesic Problem. The solution of this eqution subject to the boundry conditions ws given in (3.31); nmely, y(x o ) = y o nd y(x 1 ) = y 1 y(x) = y 1 y o x 1 x o x + x 1y o x o y 1 x 1 x o, for x o x x 1. (3.85) The grph of the function y given in (3.85) is stright line segment form the point (x o, y o ) to the point (x 1, y 1 ). We will see in subsequent section tht the function in (3.85) is the unique minimizer of the rc length functionl defined in (3.83) over the clss A given in (3.84). Exmple (The Brchistochrone Problem, Revisited). In Exmple we sw tht the time of descent form point P (, y o ) to point Q(x 1, y 1 ), where x 1 > nd y o > y 1, long pth tht is the grph of function y C 1 ([, x 1 ], R), with y() = y o nd y(x 1 ) = y 1, is proportionl to J(y) = x1 1 + (y (x)) yo y(x) dx, for y A, (3.86) where A is the clss A = {y C 1 ([, x 1 ], R) y() = y o, y(x 1 ) = y 1, nd y(x) < y o for < x < x 1 }. (3.87) In this cse, the function F corresponding to the functionl J in (3.86) is given by 1 + z F (x, y, z) = yo y, for x x 1, y < y o, nd z R. We then hve tht 1 + z F y (x, y, z) = (y o y), for x x 1, y < y 3/ o, nd z R, nd F z (x, y, z) = z 1 + z y o y, for x x 1, y < y o, nd z R. Thus, the Euler Lgrnge equtions ssocited with the functionl J given in (3.86) is [ ] d y (x) dx 1 + (y (x)) y o y(x) 1 + (y (x)) = (y o y(x)), for < x < x 1. (3.88) 3/

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