Introduction to variational calculus: Lecture notes 1

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1 October 10, 2006 Introuction to variational calculus: Lecture notes 1 Ewin Langmann Mathematical Physics, KTH Physics, AlbaNova, SE Stockholm, Sween Abstract I give an informal summary of variational calculus (complementary to the iscussion in the course book. Aims (what I hope you will get out of these notes: (i know a few important examples of variational problem an why it is useful to know how to solve then, (ii how to solve certain stanar types of variational problems, (iii reformulate ODE an PDE problem as variational problems, an, in particular, know Hamilton s principle, (iv the chapter on variational calculus in the course book is rather brief, an thus the present notes shoul provie a more etaile presentation of the theory neee to solve the variational problems in the exercise classes. WARNING: Beware of typos: I type this in quickly. If you fin mistakes please let me know by Examples of variational problems We now formulate a few stanar such problems. We first will give an intuitive escription an then a mathematical formulation. Further below we will iscuss methos how to solve such problems. Problem 1: Fin the shortest path in the plane connecting two ifferent points A an B. Of course we intuitively know the answer to this problem: the straight line connecting these points. However, is instructive to give a precise mathematical formulation of this: Let (, y 0 an (x 1, y 1, < x 1, be the Cartesian coorinates of the points A 1 I thank Christian Aåker for pointing our several typos. 1

2 an B, an y(x a C 2 function, 2 x x 1, such that y(x i = y i for i = 0, 1. Then y(x escribes a path connecting A an B, an the length of this path is Using y = y (xx we get L = B A L[y] = s = x2 + y 2. We now can rephrase Problem 1 in a mathematical language: 1 + y (x 2 x. (1 Problem 1 : Determine the real value C 2 function y on [, x 1 ] (i.e. y is given by y(x with x in [x 1, ], fixe at the en points: y(x i = y i for i = 0, 1, an such that the integral in (1 is minimal. The integral in (1 is an example of a functional: We have a set of functions F, i.e., the set of all C 2 function y on the interval [, x 1 ] such that y(x i = y i for i = 0, 1, an L[y] is assigns to each y in F a real number. Moreover, Problem 1 is a variational problem: we look for the function in F minimizing the functional L[y]. In general, a functional is a mapping that assigns to each element in some function space a real number, an a variational problem amounts to searching for functions which are an extremum (minimum, maximum, or sale points of a given functional. In these notes we restrict ourselves to methos allowing to fin extrema, an I will not iscuss how one can etermine if a known extremum is a minimum, a maximum, or a sale points. One reason for that is that there is no general, simple metho available for that, another that in many variational problems it is not so interesting to know this. Fortunately, in some important variational problems it is obvious (by some special reasons what kin of extremum one has: for example, in Problem 1 any extremum must be a local minimum (since one always can always increase the length of a given the path by a further eformation, an a similar remark applies to Problem 2 below. While in the case above it is possible to guess the solution, it is easy to formulate similar problems where the solution no longer is obvious: 2 A C 2 function is one that is can be ifferentiate twice. We will not be very careful about such technicalities an assume that all functions are sufficiently nice that all erivatives we write are are well-efine: We only write C 2 at some places to remin us that there is some conition on the functions concerning ifferentiability. A more precise mathematical iscussion woul iscuss this in more epth. 2

3 Problem 2: Fin the shortest/extremal path on a given surface embee in three imensions which connects two given points A an B on this surface. To formulate this problem in a mathematical language I assume the surface is given by a C 2 function z = z(x, y in Cartesian coorinates, an that x i, y i, z i = z(x i, y i for i = 0 an i = 1 are the Cartesian coorinates of the two points A an B, respectively. Then the path connecting A an B can be parametrize by a vector value function x(t = (x(t, y(t, 0 t 1, an such that x(0 = = (, y 0 an x(1 = x 1 = (x 1, y 1. The length of such a path is L = B A s = B A x2 + y 2 + z 2, an using x = x (tt, y = y (tt, an z = [z x (x(t, y(tx (t+z y (x(t, y(tx (t]t we get 1 L[x] = x (t 2 + y (t 2 + [z x (x(t, y(tx (t + z y (x(t, y(tx (t] 2 t. (2 0 L[x] again is a functional (now on a ifferent space of functions, an mathematically we again have a variational problem, but now of a function with values in R 2 : Problem 2 : Fin the C 2 function x on the interval [0, 1] with values in (a subset of R 2 (i.e. x is given by x(t = (x(t, y(t with t in [0, 1], fixe at the en points: x(0 = an x(1 = x 1, an which minimizes the functional in (2. The following is an example of a variational problem for a function in two variables: Problem 3: A homogeneous membrane is fixe in a given frame (which is not even. Fin the shape of the membrane. A possible shape of the membrane can be escribe by a function u(x where x = (x, y is a point in a two imensional omain, 3 where the bounary parametrizes the frame where u(x is fixe: u(x on equals some given functions α(x. [To have a specific example: assume that the membrane is a circular isc of raius R. The u(x is a function efine for r < R such that u(r r=r = α(θ where (x, y = r(cos(θ, sin(θ are polar coorinates.] The physical principle etermining the shape of the membrane is: the potential energy is a minimum. The potential energy of a homogeneous membrane can be assume to the proportional to is area, A[u] = S = 1 + u x (x, y 2 + u y (x, y 2 xy (3 3 To be mathematically precise we shoul specify further what kins of we allow: the bounary shoul be a nice curve (which is ifferentiable etc., but for simplicity we will ignore such technicalities here an in the following. 3

4 (since, to increase the area, one nees to stretch the membrane, which increases its potential energy; see our course book for a more etaile motivation; here an in the following, subscripe variables inicate ifferentiation, e.g. u x = u/ x etc. For membrane shapes which o not eviate much from the flat membrane u = 0 one can approximate this functional by expaning the square root in a power series an only taking the leaing non-trivial terms, A [ (u2 x+u 2 y]xy. Dropping the irrelevant constant term xy we get the following (approximate energy functional, J[u] = 1 2 u(x2 2 x = We thus have the following mathematical problem: 1 2 [u x(x, y 2 + u y (x, y 2 ]xy. (4 Problem 3 : Fin the real value functions u on the omain in R 2 which equals to the function α on an which minimizes the functional in (4. The following is a famous variational problem with a constraint given by a functional: Problem 4: Fin shape of a homogeneous chain fixe in the two en points in the earth gravitational fiel. We can moel the chain by a function y(x, x x 1, where (x i, y i = y(x i for i = 0, 1 the the two fixe en points. The physical principle etermining the shape is again: the potential energy is a minimum. We thus nee to fin a mathematical expression for the potential energy: we can think the the chain as a collection of small parts labele by x: each part has the length s = 1 + y (x 2 x an mass m = ρs where ρ is the mass ensity (in kg/m of the chain. The potential energy of such part is gy(xm = gρy(xs, an thus the total energy of the chain is E pot = ρg y(xs. It is important to note that the length of the chain is L = an when searching for the function minimizing the potential energy we only can allow those where L equals the given length L 0 of the chain: Problem 4 : Fin the real-value function y on [, x 1 ] fixe at the en points, y(x i = y i for i = 0, 1, an minimizing the functional J[y] = s, y(x 1 + y (x 2 x (5 4

5 uner the constraint for a given L 0. L[y] = There are many more examples, of course y (x 2 x = L 0 (6 2. Metho of solution As iscusse, a variational problem amounts to extremizing some functional. The stanar metho to solve such a problem is to convert it into a ODE or PDE problem. It also can be useful to reformulate a given ODE or PDE problem as variational problem: this can provie a starting point for an approximate solution. 2.1 Functionals epening on one function of one variable. In this section I iscuss in etail the simples kin of variational problems. All essential ieas behin variational calculus appear alreay here, an the generalization of these ieas to more complicate cases in the following sections is quite straightforwar (I will be rather brief there. Problem 1 above is a (rather trivial example of the following kin of problem: Problem A: Given a functional J[u] = F(u(x, u (x, xx (7 where F = F(u, u, x is a C 2 function of three real variables. Fin the function(s u on the interval [, x 1 ] fixe at the en points, u(x i = y i for i = 0, 1, an extremizing this functional. To avoi confusion I shoul make clear here that I slightly abuse notation an use the symbols u an u in two ifferent meanings: firstly, in F(u, u, x as a symbols for the first an secon variable of the function F, an seconly, as symbol for the function u on [, x 1 ] an its erivative. This is not the same, an a peant woul have written F = F(r, p, x an use ifferent symbols. However, once one has thought this through carefully there is no more nee to make such istinctions: as we will see, it is quite useful to be somewhat vague about what one means by u an u since this avois lengthy notation. However, I know that this ientifying u with u(x an u with u (x which is use in many physics texts (an here can be quite confusing for beginners, an I hope my remarks here an below will help to avoi this confusion. (Similar remarks apply to my iscussions of other variational problems below: in these cases I will also abuse notation in a similar manner without further comments. Below I erive an (try to explain the following important 5

6 Fact A: All solutions of the variational Problem A above satisfy the following so-calle Euler-Lagrange equations, u(x x with the bounary conitions y(x i = y i for i = 0, 1. u (x = 0 (8 Remarks on notation: A few important remarks are in orer here: as iscusse, the function F epens on three variables: F = F(u, u, x, an in the integran in (7 the first variable u is replace by u(x an the secon variable u by u (x. 4 Then means F u(x u(u(x, u (x, x (partial erivative of F(u, u, x with respect to the first variable, an replacing u by u(x an u by u (x, an similarly means u (x F u (u(x, u (x, x. Thus if, e.g., then F = F(u, u, x = (u 2 + xu 3 u(x = 3xu(x2, u (x = 2u (x. It is also important to istinguish partial an total erivative here: we have x u (x = x (u(x, u (x, x u (x = 2 F u(x u (x u (x + u (x 2u (x + 2 F x u (x etc. If you fin this confusing I recommen that you try to use the more careful notation when trying to unerstan the erivation below I trust that you will soon be convince that the (common abuse of notation I use is much more convenient. Derivation of the Fact above. Step I: The iea is to consier a small variation the the function, change y(x to a functions y(x+εη(x where ε is a small parameter an η(x is a small variation. Since the en points are fixe we only consier functions η(x such that η(x i = 0 for i = 0, 1. (9 In general, the change of the functional by this variation: δj = J[y + εη] J[y], will be proportional to ε. However, if y is an extremum of J[y] then the variation will vanish faster. Thus, the function y on [, x 1 ] with fixe en points is an extremum of the functional in (7 if J[y + εη] ε = 0 (10 4 A peant woul have written:...f = F(r, p, x... r by u(x...p by u (x. 6

7 for all η on [, x 1 ] such that η(x i = 0 for i = 0, 1. Analogy proviing aitional motivation: It might be useful at this point to recall a similar problem which you know very well: Fin the extremal points of a C 2 function f(x in N variables, x = (x 1, x 2,...,x N. To check if some point x is an extremum we can go a little away from this point an check how the function changes: x is extremum if, for an arbitrary vector η, [f(x + εη f(x]/ε vanishes as ε 0 or, equivalently, if f(x + εη ε = 0 for all η R N. Note that η specifies the irection in which we go away from the point x, an to know if we have an extremum we nee to check all possible irections. By the chain rule this is equivalent to the following well-known necessary conition for extremal points, of course: f(x = 0 (i.e., f(x/ x j = 0 for j = 1, 2,..., N. Derivation of the Fact above. Step II: We now show how to erive (8 from (10: We compute (to come from the first to the secon line below we interchange ifferentiation with respect to ε an integration, an we use the chain rule of ifferentiation J[y + εη] ε = ε = t1 t 0 t1 = u (x η(x F(u(x + εη(x, u (x + εη (x, xx t 0 [ u(x η(x + ] u (x η (x x }{{} x 1 + x= = x ( u (x η (x η(x x η(x ( u(x x u (x u (x x (11 which must hol true for every C 2 function η such that η(x i = 0 for i = 0, 1. The key step here was a partial integration so that the integran is of the form η(x (, an by that we get a bounary term (the first term. Since η(x i = 0 for i = 0, 1 this bounary term vanishes for all allowe variational functions η, an we can conclue that (10 hols true if an only if (8 hols true. The completes our erivation of the Fact above. Remark: In the last step we use the following fact: The integral η(xf(xx vanishes for all C 2 functions η on [, x 1 ] if an only if F(x = 0. This is very plausible: take a sequence of functions η(x converging to δ(x, an we get F( 0 for arbitrary (strictly speaking this fact requires a more careful proof which is beyon the scope of my notes. 7

8 Bounary conitions. Above we consiere a variational problem where the allowe functions were fixe at the en points. There are other such problems where this is not the case, an then one also gets Euler-Lagrange equations but with ifferent bounary conitions. The erivation of the Euler-Lagrange equations above is important since it also shows how the correct bounary conitions are. To show that I consier the erivation of the following variant of Fact A above: Fact B: All functions u(x extremizing the functional J[u] in (7 (without any constraints at the en points! are solutions of the Euler-Lagrange equations in (8 with the bounary conitions u (x = 0, x=x0 u (x = 0. (12 x=x1 This is easy to unerstan from the computation in (11: the bounary term in the last line now no longer vanishes automatically (since η( an η(x 1 can now be arbitrary, an to have it zero we have to require (12. Of course, one coul also have a variational problem where η( is fixe but η(x 1 is arbitrary - in this case we only get the secon conition in (12, etc. Remark on how to solve such Euler-Lagrange equations: In general, the Euler- Lagrange equation in (8 is a ODE of orer 2 (since / u (x typically contains some power of u (x, which, when ifferentiate, gives a term u (x. The general solution of this equation thus contains two arbitrary constant which nee to be fixe by two conitions: either the conitions in Fact A above, or otherwise the conitions in Fact B. Anyway, solving a 2n orer ODE is not always easy, an there are two important special cases where the problem can be simplifie to a 1st orer ODE: Fact C: (a For a function F = F(u (x, x inepenent of u(x, the Euler-Lagrange equations in (8 are equivalent to for an arbitrary integration constant C. u (x = C (13 (b For a function F = F(u(x, u(x inepenent of x, the Euler-Lagrange equations in (8 are implie by u (x u (x F = C (14 for an arbitrary integration constant C. 8

9 In these two cases we only nee to solve a 1st orer ODE which is a lot simpler. The proof of this is simple: For (a we note that F being inepenent of u(x means that / u(x = 0 an thus (8 becomes x u (x = 0. equivalent to (13. To prove (b we ifferentiate (14 with respect to x: ( u (x x u (x F = 0, an by a simple computation using the chain rule we see that this is equivalent to u (x u (x + u (x x u (x u(x u (x u (x u (x = 0 where we use / x = 0. The terms proportional to u (x cancel, an pulling out the common factor u (x in the two remaining terms we get the Euler-Lagrange equations in (8. I stress that Fact C is important for solving many examples. To illustrate this I consier the following example: Extremize the functional S[x] = t1 t 0 ( m 2 x (t 2 V (x(t, t. (15 As I iscuss below, this is the action functional for a particle with mass m > 0 moving in x-irection an in an external potential V (x, t epening on position an time t. The Euler-Lagrange equations in this example are nothing but Newton s equations, mx V (x(t, t (t = x(t (16 (check that!. Moreover, in case V (x, t = V (x is inepenent of t then Fact C (b above implies that m 2 x (t 2 + V (x(t = C, (17 which is nothing but energy conservation. This is useful since the latter ODE can be solve (in principle by separation: x = t + const. 2 (C V (x m This you knew before: the energy of a system is conserve only for potentials which o not explicitly epen on time. 9

10 For further examples see the course book. 2.2 Functionals epening on several functions of one variable. Problem 2 above is a special case of the following generalizing Problem A above: Fact D: Consier the following functional J[x] = t1 t 0 F(x(t,x (t, tt (18 where x = (x 1, x 2,...,x N is a collection of N functions (or equivalently, a function with values in R N an F = F(x,x, t a C 2 function of 2n + 1 variables x, x, an t. Then the function x on [t 0, t 1 ] extremizes this functional if the following Euler-Lagrange equations are fulfille: x j (t t = 0 for all j = 1, 2,..., N. (19 (t x j The erivation is very similar to the one above an we therefore are rather sketchy: The conition for x to extremize J[x] is J[x + εη] ε = 0 (20 for variational functions η on [t 0, t 1 ] with values in R N, an by a computation similar to the one in (11 we fin J[x + εη] ε = ( t1 ( N N = x j (t η j(t + x j (tη j (t t where t 0 = BT + j=1 t1 t 0 j=1 ( η j (t x j (t t x j (t j BT = j=1 (21 t N 1 x j (tη j(t (22 t=t 0 is a bounary term obtaine by partial integration, as above. If the function x is fixe at the en points then this bounary term vanishes, otherwise it will provie some bounary conitions, as above (we will now spell out these ifferent cases in more 10

11 etail: I hope it will be clear in particular examples, an if not the safe metho is to go back to the computation in (21 an treat the bounary term in more etail. Anyway, we see that for the expression in (21 to be zero for every function η we nee the Euler-Lagrange equations in (19 to hol true. In general this is a system of N ODEs of orer 2 which are ifficult to solve. Let me mention an Important example in physics. I. For a mechanical system, the functional S = t1 t 0 ( kinetic energy potential energy t is calle action of this system, an the Hamilton principle states that the time evolution of a system is such that its action is extremal. In many important examples the action is a functional of the form S[q] = t1 t 1 N ( mj 2 q j(t 2 U(q 1 (t,...,q N (t t (23 j=1 where q is a function on [t 0, t 1 ] with values in R N (e.g. for a system with 2 particles moving in R 3 we have N = 6, m j = M 1 (mass of particle 1 for j = 1, 2, 3, m j = M 2 (mass of particle 3 for j = 4, 5, 6, q = (x 1, y 1, z 1, x 2, y 2, z 2 the collection of all Cartesian coorinates of the particles, an U the sum of two-boy an external potentials. In this case it is easy to see that Hamilton s principle implies the usual Newton s equations by Fact D above: m j q j t (t = U(q(t, q j (t for j = 1, 2,..., N (24 (check that!. One thus can use the action functional to efine a mechanical system (rather than Newton s equations. This has many avantages Functionals epening on functions of several variable. Problem 3 above is a special case of the following: Problem E: Consier the following functional J[u] = F(u(x, y, u x (x, y, u y (x, y, x, yxy (25 where is some omain in R 2, an F = F(u, u x, u y, x, y a C 2 function in 5 variables. Fin the C 2 function(s u on which are fixe on the bounary of an which extremize this functional. 11

12 As above one can make precise this conition on u by J[u + εη] ε = 0 (26 for all functions η = η(x, y on. The following computation similar to the one in (11, J[u + εη] ε = F(u(x, y + εη(x, y, u x (x, y + εη x (x, y, u y (x, y + εη y (x, y, x, yxy ε ( = η(x, y + u(x, y u x (x, y η x(x, y + u }{{} y (x, y η y(x, y xy }{{} x ( ux(x,y η(x,y η(x,y x ux(x,y ( = BT + η(x, y u(x, y x u x (x, y xy (27 y u y (x, y where BT = ( x ( η(x, y + u x (x, y y ( u y (x, y η(x, y xy ( = η(x, yy u x (x, y u y (x, y η(x, yx (28 is a bounary term (we use Green s theorem which vanishes since u is fixe on, an thus the allowe variation functions η vanish on. We thus can conclue, as above: Fact E: All solutions of the Problem E above satisfy the following Euler-Lagrange equations u(x, y x u x (x, y y u y (x, y = 0 (29 where u is fixe. Again, it is imporant to istinguish partial an total erivatives, e.g., = 2 F u x + 2 F u xx + 2 F u xy + 2 F x u x u u x u y u x x u x u 2 x (we suppress arguments x, y of u an its erivatives etc. Similarly as above there are also moifie variational problems where u is allowe to vary on all of or on parts of it. In this case one also gets the Euler-Lagrange 12

13 equations but with ifferent bounary conitions: if u(x, y is allowe to vary on then one gets the following bounary conitions, ( n x u x + n y u y = 0. (30 where n = (n x, n y is the unit vector normal to the curve. In various important examples one has F = const. u where the ots inicate terms inepenent of u, an in these cases (30 correspons to Neumann bounary contions. To obtain (30, we note that if is parametrize by a curve x(t = (x(t, y(t, 0 t 1, then we can write the bounary term above as BT = 1 where we suppress the arguments an use 0 ( u x n x + u y n y x=x(t,y=y(t η(x(t, y(ts (31 y (tt = n x (ts, x (tt = n y (ts, s = x (t 2 + y (x 2 t. If η(x(t, y(t is allowe to be non-zero on (parts of, we must require the conition in (30 (on these very parts. It is easy to generalize this to functional of functions u in N variables which, obviously, also are important in physics: Fact F: Functions u on the omain in R N extremizes the functional J[u] = F(u(x, u(x,x N x, (32 u = (u x1,...,u xn, F a C 2 function of 2N + 1 variables u, u, x, if the following Euler-Lagrange equation hols true: N u(x j=1 x j u xj (x = 0. (33 The erivation is as above an recommene as exercise to the reaer. The most general case are functionals of functions u on a omain in R N with values in R M : u = (u 1,...,u M is a function of the variable x = (x 1,...,x N. Again, it is straightforwar erive the Euler-Lagrange equations for such cases. I only write them own an leave it to the reaer to fill in the etails: N u k (x j=1 x j (u k xj (x 13 = 0 for k = 1, 2,..., M. (34

14 This is a couple system of PDEs an, in general, ifficult to solve. I only mention that action functionals use to efine important theories in physics like electroynamics or general relativity are of this type. Important example in physics. II. Consier the oscillations of a homogeneous membrane escribe by a function u = u(x, t where x in are spatial coorinates an t > 0 time. As iscusse in more etail in our course book, the kinetic energy of the membrane is E kin = ρ 2 u t(x, t 2 xy with ρ > 0 the mass ensity of the membrane (in kg/m 2, an the potential energy is S E pot = 2 [u x(x, t 2 + u y (x, t 2 ]xy with a constant S > 0 characterizing material properties of the membrane. Thus the action functional for the membrane is S[u] = t1 t 0 ( ρ 2 u2 t S 2 [u2 x + u2 y ] xy, (35 (we suppress the common arguments t, x, y of the erivatives of the function u. Hamilton s principle can also be applie to this membrane, an from Fact F we obtain the following equations of motion for the membrane: ρu tt S[u xx + u yy ] = 0. This is the wave equation which we stuie repeately in this course, of course. However, using variational calculus together with simple physical reasoning we now erive this equation, which otherwise woul have been rather ifficult. In a similar manner one can erive our string moel erive an iscusse extensively in this course in a much simpler way using variational calculus. 2.4 Variational problems with constraints given by functionals. Problem 4 above is a special case of the following important class of problems: Problem G: Given two functionals J[u] = F(u(x, u (x, xx (36 an K[u] = G(u(x, u (x, xx (37 14

15 where F = F(u, u, x an G = G(u, u, x are two C 2 functions of three real variables. Fin the function(s u on the interval [, x 1 ] fixe at the en points, u(x i = y i for i = 0, 1, an extremizing the functional J[u] uner the constraint K[u] = K 0 where K 0 is a given constant. This kin of problem can be reuce to a Problem A above by Lagrange s multiplicator metho; see Fact G below. I will now give a erivation, but only motivate it by analogy with functions on R N : Lagrange s multiplicator metho for functions on R N : 5 Suppose we want to solve the following problem: Fin the minimum of a function f on R N with the constraint that another function g on R N has a fixe given value g 0. Geometrically, all points x in R N satisfying g(x = g 0 efine a hypersurface of imension N 1 in R N, an our problem is to extremize f on this hypersurface (for N = 2 this hypersurface is a curve, for N = 3 a surface, etc.. To check if a point x fulfills these requirements we check what happens if we go away from this point, remaining on this hypersurface: if x is on this hypersurface an x + εη is on this hypersurface as well then g(x + εη g(x = g 0 g 0 = 0, an taking the erivative of this with respect to ε gives g(x + εη ε = η g(x = 0. (38 For f to be extremal on this hypersurface we nee that f(x + εη ε = η f(x = 0. (39 for all vectors η satisfying the conition in (38 (this means that f only changes faster then linear in ε if we go away from x in the irection η/ η tangent to the hypersurface by a istance ε η, similarly as iscusse after Fact A above. Now g(x (for fixe x is a vector in R N, an thus (38 efines a hyperplane of imension N 1 (this is just the tangent plane to the above-mentione hypersurface, of course. The conition in (39 means that f(x is also orthogonal to this very hyperplane, an this obviously can be true if an only if f(x an g(x are parallel, i.e., there must exist a constant λ such that f(x = λ g(x. (40 The latter is obviously equivalent to the function f λ (x = f(x λg(x (41 5 To unerstan the following I suggest you write out everything for N = 2 first: in this case you can raw pictures an thus nicely visualize the argument geometrically, which I something I recommen. 15

16 being extremal: the extremal points x = x λ will epen on λ, an the value of λ is then to be fixe so that g(x λ = g 0. The paramter λ introuce here is often calle Lagrange multiplier. To summarize: The problem extremize the function f on R N with the constraint that another function g on R N has a fixe given value g 0 is equivalent to extremize the function f λg on R N an etermine the parameter λ so that the extremal point fulfills the constraint. This argument can be generalize to functionals an use to erive the following Fact G: Problem G above is equivalent to extremizing the functional J[u] λk[u] (42 an then etermining the parameter λ so that the constraint K[u] = K 0 is fulfille: To solve Problem G one solves the Euler-Lagrange equations (F λg u(x (F λg = 0 (43 x u (x (which has solutions u epening on the parameter λ, of course an then fixing λ by the constraint. It is easy to generalize this metho to other kins of functions an/or to variational problems with several constraints given by functional: Minimizing any kin of functional J[u] with constraints K J [u] = k J, J = 1, 2,..., L, is equivalent to minimizing J[u] L λ J K J [u] (44 J=1 an then fixing the parameters λ J so that all constraints are fulfille: The solutions of the Euler-Lagrange equations which amount to minimizing the functional in (44 will epen on the parameters λ 1, λ 2,...,λ L, an this is exactly the number of free parameters neee to fulfill the L constraints. Again I leave etails to the intereste reaer. 2.5 Variational problems with higher erivatives. I finally shortly iscuss variational problems involving higher erivatives an leaing to higher orer ifferential equations. For simplicity I only consier such problems for real-value functions u of one real variable the generalization to other cases is straightforwar: We consier a functional J[u] = F(u(x, u (x, u (x, xx (45 16

17 with F = F(u, u, u, x a C 2 function epening now on four variables: the integran of this functional epens not only on the function u an its erivative u but also on the the secon erivative u. The Euler-Lagrange equations for such functional are u(x x u (x + 2 x 2 u (x = 0. (46 It is straightforwar to prove this by generalizing our erivation of Fact A in Section 2.1 above I only note that the last term comes from two partial integrations, an it has a plus sign since ( 2 = +: one minus sign for each partial integration. Note that the equation in (46 is in general a ODE of orer 4. We also note that in such problems it is usually worth going though the erivation of the Euler-Lagrange equations from (10 since this allows one to obtain the correct bounary conitions. For example, one may fix u an u in x = but leave u an u in x = x 1 arbitrary. One then obtains Euler-Lagrange equations with particular bounary conitions which one erives by computing (10 allowing variational functions such that η( = η ( = 0 but with η(x 1 an η (x 1 arbitrary: keeping track of the bounary terms obtaine by the partial integrations one obtains the correct bounary conitions. 17