Calculus of Variations Lagrangian formalism is the main tool of theoretical classical mechanics. Calculus of Variations is a part of Mathematics which Lagrangian formalism is base on. In this section, we iscuss the basics of the Calculus of Variations an, in particular, consier some simple applications. This will provie us with the mathematical language necessary for formulating the Lagrangian Mechanics. The key object of the Calculus of Variations is a functional, a real-value function efine on some class of functions. That is the argument of a functional is a function of one or many variables. The functional assigns a real value to a function of the given class. Typically, one eals with functionals of the form F [f] = g(f, f,, (1 where f( is any function of a given class, f f ( is its erivative, an g is a given function of three variables. Apart from being ifferentiable, the functions of the given class can be subject to some constraint. Say, f( = y a, f( b = y b. (2 The efinition of functional is straightforwarly generalize to the cases of more than one function f, more than one variable, an higher-orer (partial erivatives. Eample 1. Length of a line in a plane. Suppose we have a smooth line in the y-plane passing through the two given points (, y a an ( b, y b. The line is suppose to be specifie by the equation y = f(. (3 The length of the line is a functional of f. Let us make sure that it can be represente in the form (1. We have l[f] = l = ( 2 + (y 2 = ( 2 + (f 2 = 1 + (f 2. (4 We see that l[f] has the form (1 with g(f, f, = 1 + (f 2. (5 Note that in this eample the function g oes not epen eplicitly on f an. Eample 2. Potential energy of a fleible cable suspene from two fie points. The cable is suppose to be at rest. Choose the y ais perpenicular to the surface of the Earth, the cable being represente by a curve (9 the y plane, with the constraint (2. Then its potential energy is given by (we o not care about the units, an thus neglect the imensional proportionality coefficient U[f] = y l = f 1 + (f 2. (6 (The potential energy of the element l is proportional to its hight y an its length l. We see that l[f] has the form (1 with g(f, f, = f 1 + (f 2. (7 Here the function g eplicitly epens on f an f, but not on. 1
Eample 3. Length of a line in a surface. Suppose we have a smooth line L in the surface S, passing through the two given points (, y a, z a an ( b, y b, z b. Let the surface S an the line L be specifie by the equations z = s(, y, (8 y = f(, z = s(, f(, (9 respectively. Let us show that the length of the line is a functional of the form (1. First we introuce convenient notation s(, y s(, y s =, s y =, (10 y an note that Then, where y = f, z = s + s y y = s + s y f. (11 l[f] = l = g(f, f, = ( 2 + (y 2 + (z 2 = g(f, f,, (12 1 + f 2 + (s (, f + s y (, f f 2. (13 A typical problem which arises in connection with a functional F [f] is the problem of fining such a function f that minimizes the functional. In the contet of above eamples, this problem comes from natural questions like: What is the shortest istance between two points in a given surface? What is the shape of the suspene cable (corresponing to the minimum of the potential energy at a given length? Calculus of Variations provies mathematical tools for solving the problem. Suppose the function f is a (local minimum/maimum of the functional F. Then, for any small variation of the function f variation of the functional is suppose to be sign-efinite. Let us see what are the implications of this fact. We introuce the symbol δf( for an infinitesimal variation of the function f an the symbol δf for corresponing variation of the functional F. That is With a simple observation that we write F [f + δf] F [f]. (14 f f + δf f f + (δf (15 g( f + δf, f + (δf, Then we take into account the smallness of δf: g( f + δf, f + (δf, g(f, f, + f f δf + g(f, f,. (16 δf + f (δf. (17 f (δf. (18 Doing by parts the secon integral, with Eq. (2 taken into account [meaning δf( = δf( b = 0 ], we have b (δf = (δf. (19 f f That is A( δf, (20 2
where A( = f f. (21 We see that the variation δf is a linear functional of δf. The linearity means that the sign of δf changes if δf δf. On the other han, the conition of minimum/maimum requires that the sign of variation o not change. This is only possible if A( 0, an we arrive at the celebrate Euler s equation, the funamental equation of the Calculus of Variations: f f = 0. (22 As an easy eample, let us make sure that the shortest istance between two points is a straight line. Applying Eq. (22 to the function g of the Eample 1, see Eq. (5, we reaily get f ( = 0 f( = C 1 + C 2, (23 where C 1 an C 2 are some constants. An this is the equation of a straight line. The constants are fie by the bounary conitions (2. Alternate form of Euler s equation If f 0, then Euler equation is equivalent to ( g f f = 0. (24 Inee, by the chain rule we have Taking into account that we then get g = f f + f + f. (25 f ( g f = f + f f f f This means that Eq. (24 is equivalent to f ( f = ( + f f f, (26. (27 f = 0, (28 f which, in its turn, is equivalent to the original Euler s equation (22 if f 0. To put it ifferent, any solution of Eq. (22 is simultaneously a solution of Eq. (24, but, in contrast to Eq. (22, equation (24 has also a trivial solution f = const. Now if g = g(f, f, that is g oes not epen eplicitly on, Eq. (24 simplifies to an immeiately yiels the first integral ( g f = 0, (29 f g f f = const. (30 3
Soap film. The equilibrium shape of a soap film correspons to a (local minimum of its surface energy. The latter is just proportional to the surface area. Hence, to fin the equilibrium shape of a soap film one has to fin the (local minimum of the surface area uner the appropriate bounary conitions. Consier simplest eample when the film is aially symmetric being stretche between two parallel coaial wire circles. Let the center of the first circle be at = 0 an the center of the secon circle be at = h; the two raii being R a an R b, respectively; being the symmetry ais. The surface is parameterize by the raius r at a given : The surface area, A, is then a functional of f: We see that A[f] = r = f(, f(0 = R a, f(h = R b. (31 2πr l = 2πr ( 2 + (r 2 g = 2π f = 2π h 0 f 1 + f 2. (32 1 + f 2 (33 oes not epen eplicitly on n we can use the Euler equation in the form of its first integral (30. This yiels f 1 + f 2 ff2 = C 1 + f 2 0, (34 where C 0 is a constant. A simple algebra then leas to f = C 0 1 + f 2 f 2 = C0 2 (1 + f 2 f = ± (f/c 0 2 1. (35 We get a simple ifferential equation an easily solve it: f = ± (f/c 0 2 1. (36 f (f/c0 2 1 = ±. (37 C 0 cosh 1 (f/c 0 = ± + const. (38 ( 0 f( = C 0 cosh. (39 Note that the sign is absorbe by the constant 0. The solution (39 has two free constants which are fie by two bounary conitions: { cosh(0 /C 0 = R a /C 0, (40 cosh((h 0 /C 0 = R b /C 0. C 0 4
Functional of many functions The generalization of the theory to the case of more than one function, f f (1, f (2,..., f (m, is quite straightforwar. Now we have F [f (1, f (2,..., f (m ] = g(f (1, f (2,..., f (m, f (1, f (2,..., f (m,, (41 Introucing the variations an evaluating δf, we get ( = y (j a, ( b = y (j b (j = 1, 2,..., m. (42 + δ, (43 A j ( = A j ( δ, (44. (45 Now if we are looking for minimum/maimum of the functional F, then by efinition we have δf 0 for a minimum, or δf 0 for a maimum. But this is only possible if A j ( 0 (j = 1, 2,..., m. (46 Inee, if for some j = j 0 we ha A j0 ( 0, then we coul set δ 0 for all j s but j 0 an write A j0 ( δf (j 0. (47 An then we just repeat the reasoning we use to prove A( 0 in the case of just one function. Hence, for a set of functions {f (1, f (2,..., f (m } to correspon to a minimum/maimum of the functional (41 the following Euler s equations have to be satisfie: = 0 (j = 1, 2,..., m. (48 This system of m ifferential equations efines the functions {f (1, f (2,..., f (m } up to 2m free constants. The constants are then fie by the bounary conitions (42. Important Theorem What is the generalization, if any, of the alternative Euler s equation (24 for the case of many functions? In the case of m functions, we cannot replace the whole system of m equations (48 with alternative ones, equivalent to (24. The generalization of Eq. (24 comes in the form of a theorem that may seem too abstract, but actually is etremely important being irectly relevant to the conservation of energy in Lagrangian mechanics. Consier the quantity (we use the same letter we use for energy, an o it on purpose E(f (1 (,..., f (m (, f (1 (,..., f (m (, = g. (49 5
The theorem says that if the functions {f (1, f (2,..., f (m } minimize/maimise the functional (41, then for the quantity E(f (1 (,..., f (m (, f (1 (,..., f (m (, the following relation takes place E =. (50 In particular, if the function g oes not eplicitly epen on, then the right-han sie is zero an the quantity E is just a constant. The proof is a straightforwar generalization of Eqs. (25-(27: = +. (51 Hence, g = E m (j f + = + m ( +. (52. (53 So far we i just ientical transformations. Now we take into account that the functions {f (1, f (2,..., f (m } minimize/maimise the functional (41 an are suppose thus to satisfy Euler s equations (48. Eqs. (48 state that each term in the brackets in the r.h.s. of (53 is ientically equal to zero, which completes the proof. Lagrange Multipliers Let us recall the problem of the fleible cable (Eample 2. This problem involves an etra feature, a constraint that the total length of the curve be fie. To solve problems like that we have to unerstan how to treat the constraints. Actually, the iea is the same as in the calculus of functions: one can use Lagrange multipliers. Suppose we want to fin an etremum of the functional (1 on the set of functions subject to the bounary conitions (2 an also the constraint that where C is a given constant an Q[f] = Q[f] = C, (54 q(f, f, (55 is a given functional. (In our Eample 2, Q[f] is the length of the curve. Consier a new functional F [f] = F [f] λq[f], (56 where λ is a fie real number. Uner the constraint (54, an etremum of the functional F is simultaneously an etremum of the functional F, an vice versa, since the two iffer by the fie constant λc. Now consier a genuine (i.e. unconstraine etremum of the functional (56. At an arbitrary λ, this genuine etremum is not suppose to have anything to o with the etremum uner the constraint. However, normally their eists a special choice of λ = λ at which the function f (λ, the genuine etremum of the functional (56, just satisfies the conition Q[f (λ ] = C. (57 6
Clearly, the function f (λ is the solution to the original problem, since the genuine etremum automatically implies the etremum uner given (an satisfie constraint. Once we know the value of λ, the problem is solve. An to fin λ we simply solve the problem of the genuine etremum of the functional (56 with an arbitrary (unefine λ, an then a posteriori fi the proper value λ = λ by requiring that the conition (57 be satisfie. Solution to the problem of the fleible cable. We have (see Eample 2 Hence, F [f] = g(f, f, = f q(f, f, = [g(f, f, λq(f, f, ] = 1 + (f 2, (58 1 + (f 2. (59 (f λ 1 + (f 2. (60 With this particular form of the functional F it is very convenient to introuce the new function because then we will have F [ f] = f = f λ, (61 f 1 + ( f 2, (62 which is (up to a numeric factor is the functional which we have minimize alreay in the contet of the problem of the soap film. We just write the answer, Eq. (39: ( 0 f( = C 0 cosh C 0 ( 0 f( = C 0 cosh + λ. (63 The values of the three constants, C 0, 0, λ, shoul be chosen to satisfy two bounary conitions, Eq. (2, an the constraint that the total length of the curve is equal to a given value l. C 0 7